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The  Tellurian  Globe  is  made  in  [two  sizes.  Eight 
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Bt  dr.  MALCOLM  MACVICAR,  Ph.D.,  LL.D., 
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Handbook  of  the  Mac  Vicar  Tellurian  Jllobe, 

for  the  use  of  Teachers,  Schools  and  Families ;  oon- 
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School  Booics  Published  by  Daicson  Bros, 


By  professor  ANDREW, 

0/  the  University  of  McGill  College, 

The  New  Dramatic  Beader  ;  Comprising  a  Selec- 
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ductory hints  to  Readers.     Price,  75  cents. 

By  J.  D.  MORELL,  LL.D., 
n.  M.  Inspector  of  Schools,  England, 

A  Complete  Manual  of  Spelling  on  the  Prin- 
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Canadian  Elementary  School  Atlas;  For  the 

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Nos«  If  3*  3*  4  dc  6, Common  School  Sorles. 

No«.  6  &  7* Bunlneso  Series. 

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COMPLETE  ARITHMETIC, 


ORAL  AND  WRITTEN. 


Designed  for  the  Use  of  Common  and  High  Schools 
AND  Collegiate  Institutes. 


REV.    D.    H.    MACVICAR,    LL.D., 

PRLVCIPAL,  PrbSBTTBRIAH  COLLBOl,  MOHTRIAL. 


■  «»  ■ 


MONTREAL: 
DAWSON   BROTHERS,  PUBLISHERS 

1879. 


OA  /03 

1^31 


Entered  according  to  Act  of  Parliament  of  Canada,  in   the  year  1879,  byl 
Dawson  Brothers,  in  the  Office  of  the  Minister  of  Agriculture. 


PREFACE. 


THIS  work  is  based  upon  the  Complete  Arith- 
metic prepared  two  years  ago  by  M.  JMacVicar, 
Mi.D.,  LL.D.,  Principal  of  the  State  Normal  School, 
'otsdam,  N.  Y.,  and  the  undersigned.  In  the  present 
vohime  the  order  is  changed  throughout,  and  the 
work  is  specially  adapted  to  the  wants  of  the  Com- 
mon and  High  Schools,  and  Collegiate  Institutes  of 
'unada. 

Attention  is  invited  to  the  Properties  of  Numbers, 
rreatest  Common  Divisor,  Fractions,  Decimals,  Com- 
[)ound  Numbers,  The  Metric  System,  Business  Arith- 
metic, Ratio  and  Proportion,  Alligation,  and  Square 
md  Cube  Root,  with  the  belief  that  the  treatment 
will  be  found  new  and  an  improvement  upon  former 
uethods. 

In  every  subject  the  pupil  is  first  required  to 
baster  elements  or  preparatory  steps  and  propositions 
In  order  to  fit  him  for  the  more  advanced  and  com- 
)lex  operations. 
A  systematic  drill  is  provided  on  Oral  and  Written 
Exercises  and  Review  and  Test  Questions,  by  which 
[he  principles  and  processes  of  numbers,  with  their 


iv 


F  liEFAC  E. 


applications  to  j)racticul  busiiiesri,  will  be  permanently 
fixed  in  the  mind. 

In  all  cases  in  which  it  is  possible,  each  process  is 
presented  objectively,  so  that  the  truth  is  exhibited  to 
the  eye  and  thus  clearly  defined  in  the  mind. 

The  entire  drill  and  discussions  are  believed  to  be 
so  arranged,  and  so  thorough  and  complete,  that  by 
passing  through  them  the  pupil  cannot  fail  to  acijuire 
such  a  knowledge  of  principles  and  facts,  and  to 
receive  such  mental  discipline,  as  will  prepare  him 
properly  for  the  study  of  the  higher  mathematics. 

Invaluable  aid  in  methods  of  presentation  and  the 

management  of   class  work  will  be  found   in   the 

Teacher's  Edition  of  Dr.  M.  MacVicar's  Complete 

Arithmetic,  published  by  Taintor  Brothers,  Merrill 

&Co.,  N.  Y. 

D.  H.  MacVICAE. 

Montreal,  July,  1879. 


CONTENTS. 


Page 

Notation  and  Numeua- 

TION 1 

Roman  Notation 9 

Revikw  and  Test  Ques- 
tions   10 

Addition 11 

Canadian  Money 18 

Rkview  and  Test  Ques- 
tions   20 

Subtraction 21 

Review  and  Test  Ques- 
tions   27 

Multiplication 28 

Review  and  Test  Ques- 
tions   40 

Division 42 

Division  by  Factors  ...  55 
Review  and  Test  Ques- 
tions   60 

Properties  op  Numbers  66 

Exact  Division 67 

Prime  Numbers  70 

Factoring 72 

Cajjcellation 74 

Greatest  Com.  Divisor.  77 

Least  Com.  Multiple.  .  85 


Pagk 
Review  and  Test  Ques- 
tions     89 

Fractions  90 

Reduction 93 

Addition 102 

Subtraction 105 

Multiplication 107 

Division 114 

Complex  Fractions 119 

Review  and  Test  Ques- 
tions   126 

Decimal  Fractions 128 

Notation  and  Numera- 
tion   129 

Reduction 134 

Addition 142 

Subtraction 143 

mui.tiplication 144 

Division 140 

Review  and  Test  Ques- 
tions   152 

Denominate  Numbers..  154 

Units  op  Weight 155 

Comparative  Units  op 

Weight 156 

Units  of  Length 168 


VI 


CONTENTS, 


Page 

Units  op  Surfaoe 171 

Units  of  Volume.  .  . .  174 
Units  of  Capacity.  . . .  180 
Comparative   Units  of 

Capacity 182 

Units  of  Time 183 

Units  of  Money 186 

Metuic  System 188 

Decimal  Related  Units  188 
Comparative  Table  of 

Decimal  Units 191 

Duodecimals 192 

Longitude  and  Time  . .  193 
Review  and  Test  Ques- 
tions   195 

Business  Arithmetic...  196 

Aliquot  Parts 199 

Business  Problems 202 

Applications 217 

Profit  and  Loss 218 

Commission 220 

Insurance 222 

Stocks  224 

Duties  or  Customs 228 

Review  and  Test  Ques- 
tions  5)29 

Interest 230 

Method     by     Aliquot 

Parts 232 

Method    by    Six    Per 

Cent 234 

Method  by  Decimals  . .  236 


Page 

Exact  Interest. 237 

Compound  Interest 242 

Interest  Tables 244 

Annual  Interest 246 

Partial  Payments 247 

Discount 251 


253 
256 
257 
261 


,  Bank  Discount 

']  Exchange 

Inland  Exchange 

Foreign  Exchange.  . . 
Equation  of  Payments.  266 
Review  and  Test  Ques- 
tions      275 

Ratio 276 

Proportion 281 

SiMPiJE  Proportion 285 

Compound  Proportion.  289 

Partnership 292 

Alligation  Medial. 296 

i  Alligation  Alternate.  297 
Involution 302 

j  Evolution 304 

Progressions 320 

Arithmetical  Progres- 
sion   321 

Geometrical    Progres- 
sion   323 

Annuities 325 

Mensuration 328 

Review  and  Test  Ex- 
amples   342 

Answers 358 


ARITHMETIC. 


NOTATION    AND    NUMERATION. 


NUMBERS    PROM    1    TO    1000. 
Art,  1.     Numbers  are  expressed  by  means  of  ten  figiirei, 

r.».o«,.  0    1  2  3  J,.  5  6   7  8  (j 

Snmea,  Naught,  Que,  Two,  Three,  Four,  Five,  Six,  Seven,  Eight,  Nine. 


Observe  regarding  the  ten  figures  : 

1.  The  uaugld  is  also  called  cipher  or  zero,  and  when  written 
alone  stands  for  no  number. 

2.  The  other  nine   figures  are  called  digits  or  significant 
figures,  and  each  stands  for  the  number  written  under  it. 

:3.  Any  number  of  objects  not  greater  than  nine  is  expressed 
[by  one  figure. 

Thus,  4  boys,  2  girls,  6  pens,  9  desks,  9  windows. 

One  or  luiitg  is  the  foundation  of  all  numbers,  and  they  are 
j<lerived  from  it  by  the  process  oi  addition.  Thus,  two  is  formed 
by  adding  one  and  one  ;  three  by  adding  one  and  two  ;  five  by 
idding  one  and  four,  etc. 


2 


NOTATION    AND     NUMER  ATION, 


2,  Th£  names  of  numbers  ore  formed  by  combining  the 
wtmes  of  the  figures  used  to  express  the  numbers  with  the  names 
of  the  orders  of  groups  represented. 

1.  11  is  ten  and  one,  read  Eleven,  12  ia  ten  and  two, 
read  Twelve,  These  two  numbers  are  the  only  exceptions  to 
the  proposition. 

2.  The  namen  of  numbers  from  twelve  to  two  tens  are  formed 
by  changing  teii  into  teen,  and  prefixing  the  name  of  the  digit 
which  expresses  how  many  the  number  is  greater  than  ten. 
The  name  of  the  digit,  when  necessary  to  combine  properly 
with  teen,  is  changed,  thus  : 

13  is  three  and  ten,  or  Thir-teeii  ;  three  changed  to  thir. 

14  is  four  and  ten,  or  Four-teen. 

15  is  five  and  ten,  or  Fif-teen  ;  fvce  changed  to^. 
Hi  is  six  and  ten,  or  Six-teen, 

17  is  seven  and  ten,  or  Seven-teen, 

IS  ia  eight  and  ten,  or  Elgh-teen  ;  eight  changed  to  eigh. 

10  is  nine  &ud  ten,  OT  Nine-teen,  ' 

3.  The  name  of  any  number  of  tens  from  o?t€  ten  to  ten  tens 
is  formed  by  changing  the  word  tens  to  ty  and  prefixing  the 
name  of  the  digit  which  expresses  the  required  number  of  tens, 
making  the  necessary  changes  in  the  name  of  the  digits  to 
combine  properly  with  ty,  thus : 

20  is  two  tens,  or  Twen-ty  ;  two  changed  to  twen. 

30  is  three  tens,  or  Thir-ty  ;  three  changed  to  thir, 

40  is  four  tens,  or  Ffn*-ty  ;  four  changed  to  for. 

50  is  five  tens,  or  Fif-ty  ;  fwe  changed  to  ff. 

00  is  six  tens,  or  Six-ty, 

70  is  seven  tens,  or  Seven -ty, 

80  is  eight  tens,  or  Eiyh-ty  ;  eight  changed  to  eigh. 

DO  is  nine  tens,  or  Nine-ty, 


EXAMPLES,  3 

4.  Tens,  and  ones  or  units,  when  written  together,  are  read 
by  uniting  the  two  names  in  one,  thus  : 

5^  is  five  tens  and  two,  read  Fifty-two» 

74  is  seven  tens  and  four,  read  Seventy -four, 

5.  Hundreds  are  read  by  naming  the  digit  that  expresses 
them.    Thus,  200  is  read  Two  hundred. 

6.  Hundreds,  tens,  and  units,  when  written  together,  are 
read  by  uniting  the  three  names.  Thus,  683  is  read  Six  hun- 
dred eighty-three. 


EXAMPLES    FOR    PRACTICE. 

15,  Read  the  following  numbers  : 

1.  14.     13.    10.     10.    18.  14.  11.     15.    12.    17. 

2.  35.    74.     40.    80.     60.  90.  20.    50.    84.    93.     99. 

3.  504.    210.     700.    412.  820.  395.    602.    903.    509. 

4.  783.     625.    934.    478.  899.  369.     707.    900. 

5.  808.     112.     273.     90fc.  990.  999.     777.     224. 

Name  the  orders  in  the  following  numbers,  commencing  at 
the  ri^ht.     Thus,  839  is  9  units,  3  tens,  and  8  hundreds. 

6.  494.    760.    897.     475.     906.     780.    934.    983.     316. 

7.  572.    409.    603.     942.     307.     85£.    903.    870. 

Read  and  analyze  the  following,  thus  : 

8.  40  horses.  Analtsis. — Forty-six  horfies  may  be  regarded  as 
4  groups  of  (en  horses  and  6  single  horses,  or  as  46  single  horees. 

9.  542  trees,  analysis.— Five  hundred  forty-two  trees  may  be  re- 
garded ax  5  groups  of  one  hundred  trees  each,  4  groups  of  ten  trees  each, 
and  2  nnffle  trees  ;  or  it  may  be  regarded  as  54  groups  of  ten  trees  each  and 
2  single  trees ;  or  as  542  single  trees. 

10.  27  tops.    76  rings.    34  boys.    17  men.    95  sheep. 

11.  107  tables.     128  beds.    204  chairs.     512  stoves. 

12.  696  beetles.     478  bees.     930  flies. 

13.  424  robins.     500  hawks.     775  blackbirds. 

14.  196  lamb.''.     430  goats.     555  cows.     009  foxes. 

15.  203.    940.    608.     490.    836.     593.    909. 


NOT  ATI  0  X     A  XD     X  U.V  V  T?  A  TTON. 


BEADING    IiAHGE    NUMBERS. 

4.  T?ie  names  of  the  ordevx  in  large  numbers  are  formed  hy 
giving  a  new  name  to  the  order  in  every  third  j'iacti  counting 

FROM  the  UNITS. 

1.  We  indicate  the  orders  to  which  new  names  are  ai)plied 
hy  inserting  a  comma  at  the  left  of  every  third  figure, 
counting  from  the  right, 

The  commas  are  inserted  and  the  names  applied  ;  thus. 


Ml 


i 


i 

^         n         ^         ^ 

8«,073,490,4«0,;5  79 


6 


5thPerioa.    4th  Period.   3d  Period.    2d  Period.    Ist  Period. 

2.  The  commas  separate  the  number  into  sets  of  three 
figures.    Each  set  is  called  a  Period, 

3.  The  rigkt-hand  order  in  each  period  has  a  new  name,  as 
shown  in  the  illustration.  The  figure  in  this  place  expresses 
Oiix^s  of  the  given  name. 

4.  The  second  figure  in  (;ach  period  expresses  tenSf  and  the 
third  hundretlH  of  whatever  the  first  order  is  called. 

For  example,  the  figures  in  the  third  period  of  the  ab(,ve 
number  are  490,  and  the  right-hand  order  is  called  millions  ; 
h<^nce  the  period  is  read,  four  hundred  ninety  millions. 

5.  The  figures  in  each  period  are  read  in  the  same  manner 
as  thoy  would  be  if  there  were  but  one  period  in  the  numl)er. 

Thus,  in  the  above  number,  the  fifth  period  is  read,  cifjhty- 
six  trillions,  the  fourth  is  read,  seventy-three  billions.  There 
bein<r  no  hundreds  expressed  in  the  fourth  period  nothing  is 
said  ab  >ut  liundreds.  Each  succeeding  period  is  read  in  the 
same  manner. 


EXAMPLES.  5 

6.  A  large  number  can  be  read  aa  easilj^  as  a  number  of 
three  places,  when  the  followiii<^  names  of  the  fii-st  order  on 
the  right  of  the  successive  periods  are  fixed  in  the  memory ; 


PEKIODS. 

NAMES. 

PERIODS. 

NAMES. 

1st. 

Units. 

7th. 

Quintiliions 

2d. 

Thousands. 

8th. 

Sextillions. 

3d. 

Millions. 

9th. 

Septillions. 

4th. 

Billions. 

10th. 

Octillions. 

5th. 

Trillions. 

nth. 

Nonil  lions. 

Gth. 

Quadrillions. 

12th. 

Decillions. 

From  these  illustrations,  we  obtain  for  reading  numbers 
the  follotving 

RULE. 

I.  Begin  at  the  right  and  separate  the  7iumber,  by  inserting 
commasy  into  periods  of  three  figures  each. 

II.  Begin  at  the  left  and  read  the  hundredx,  tens,  and  ones 
of  each  period,  giving  the  name  of  the  ones  in  each  case  except  in 
the  last  peHod. 


EXAMPLES     FOR     PRACTICE. 

5.  Point  off  and  read  the  following  numbers :       ■    '■ 

1.  400.  704.  393.  3348.  5592.  9347.  6043. 

2.  74085.  93061.  452034.  290620.  48207604. 

3.  1401.  4033.  6306.  8300.  3080.  5906.  310S. 

4.  65003.  64004.  99040.  30307.  406205.  340042: 
5.507009.  85004.  230060.  903560.  100001. 

6.  2060.  50040.  3040000.  2406007.  50:]0063. 

7.  3000000.  40006003.  30304090.  400006000.  300000804. 

8.  900800800800.  4005008004.  307000060080. 

9.  804042.  85064.  9002005.  100100100101. 

10.  3000050030.  8300400706005.  9000100130004. 

11.  97304306590724059034.  3000600000596034006670. 


6 


NOTATION     AND     NUMERATION. 


WHITING    LABGE    NUMBEBS. 

G.  Nurnhers  are  written  one  'period  at  a  time  and  in  the  order 
in  which  the  periods  are  read. 

Observe  regarding  this  proposition  . 

1.  Each  period  in  a  number  except  the  one  at  the  left  must 
contain  three  figures.  Hence  the  places  for  which  significant 
figures  are  not  given  must  be  filled  with  ciphers. 

Thus,  three  hundred  seven  million,  four  thousand,  eighty- 
two,  is  written  307,004,082.  Observe  in  this  number  a 
significant  figure  is  given  only  for  the  hundreds  and  ones  in 
the  million's  period,  hence  the  ten's  place  is  filled  with  a  cipJier. 
For  a  like  reason  the  ten's  and  hundred's  place  in  the 
thousand's  period  and  the  hundred's  place  in  the  unit's  period 
are  filled  with  ciphers. 

2.  When  a  number  is  read,  a  period  in  which  all  the  orders 
are  wanting  is  not  named.  Care  must  therefore  be  taken  to 
notice  such  periods  and  fill  their  places  in  each  case  with 
three  ciphers. 

For  example,  in  the  number  seven  million  three  hundred 
four,  the  thousands  period  is  not  named,  but  when  the  num- 
ber is  expressed  in  figures  its  place  is  tilled  with  three  ciphers ; 
thus,  7,000,304. 

RULE. 

Begin  at  tJie  left  and  write  the  figures  expressing  the  hun- 
dreds, tens,  and  ones  of  each  period  in  their  proper  order,  filling 
with  cipJiers  all  x)eriods  or  places  where  no  signifixiant  figures  are 
given. 

EXAMPLES    FOR    PRACTICE. 
7.  Express  in  figures  the  following  numbers  : 

1.  Three  hundred  four.  Five  hundred  sixty.  Eight  hun- 
dred ninety.    Three  hundred  seventy-seven. 

2.  Tliree  hundred  five.     Elight  thousand  thirty. 


DEFINITIONS,  7 

3.  Twelve  hundred.    Twenty-seven  hundred. 

4.  Ten  tens.     One  hundred  tens.     Ten  tens  and  five.    Two 
hundr(;d  tens  and  sixteen.     Eight  hundred  six  tens. 

5.  Ten  thousand.     Ten  thousand  four.     Twenty  thousand. 
Twenty  tliousand  fifty-nine. 

6.  Eleven  thousand  eleven.    One  million  five. 

7.  Eighty  million  seventy  thousand  ninety. 

8.  Three  thousand  three  hundred  six  million. 

9.  Eight  hundred  thirty-two  tens.    Six  hundred  seven  tens. 
Two  thousand  sixty-five  tens. 

10.  Fifty-nine  thousand  million.     Seven  thousand  and  six 
million.     Forty-four  billion  seven. 

11.  77  million  1  thousand  5.    409  trillion  4  million  6. 

12.  12  billion  205  thousand  49.    1  trillion  1  million  1. 


0    DEFINITIONS. 

8.  A  Unit  is  a  single  thing,  or  group  of  single  things, 
regarded  as  one ;  as,  one  ox,  one  yard,  one  ten,  one  hundred. 

9.  Units  are  of  two  kinds — Mathematical  and  Com- 
mon. A  matTiematical  unit  is  a  single  thing  which  has  a  fixed 
value  ;  as,  one  yard,  one  quart,  one  Tumr,  one  ten.  A  common 
unit  is  a  single  thing  which  has  no  fixed  value ;  as,  one  house, 
one  tree,  one  garden,  one  farm. 

10.  A  Number  is  a  unit,  or  collection  of  units;  a.B,  one 
man,  three  houses,  four,  six  hundred. 

Observe,  the  number  is  "  the  how  many"  and  is  represented  by 
whatever  answers  the  question.  How  many?    Thus,  in  the 
^expression  seven  yards,  seven  represents  the  number. 

1 1.  The  Unit  of  a  Number  is  one  of  the  things  num- 
bered.  Thus,  the  unit  of  eight  bushels  is  one  bushel,  of  five 
boys  is  one  boy,  of  nine  is  one. 

1 2.  A  Concrete  Number  is  a  number  which  is  applied 
!  to  objects  that  are  named ;  as,  four  chairs,  ten  beUs, 


8 


NOTATION    AND     NUMERATION. 


ill, 


1J$.  An  Abstract  Number  is  a  number  which  is  not 
applied  to  any  named  objects ;  as,  nine^  Jlue,  thirteen. 

14:.  LiUce  Numbers  are  such  as  have  the  same  unit. 
Thus,  four  windows  and  eleven  windows  are  like  numbers, 
eight  aud  ten,  three  hundred  and  seven  hundred, 

15.  Unlike  Numbers  are  such  as  have  different  units.  I 
Thus,  twelve  yards  and  five  days  are  unlike  numbers,  also  six  j 
cents  and  nine  minutes. 

16.  Fi ff tires  are  characters  used  to  express  numbers. 

17.  The  Value  of  a  figure  is  the  number  which  itj 
represents. 

18.  The  Simitle  or  Absolute  Value  of  a  figure  is  the  | 
number  it  represents  when  standing  alone,  as  8. 

19.  The  Local  or  Representative  Value  of  a  figure  isj 
the  number  it  represents  in  consequence  of  the  place  it] 
occupies.  Thus,  in  66  the  6  in  the  second  place  from  the  rigln  | 
represents  a  number  ten  times  as  great  as  the  6  in  the  first  | 
place. 

20.  Notation  is  the  method  of  writing  numbers  by] 
means  of  figures  or  letters- 

21.  Numeration  is  the  method  of  reading  numbers] 
which  are  expressed  by  figures  or  letters. 

22.  A  Seafe  in  Arithmetic  is  a  succession  of  mathematical! 
units  which  increase  or  decrease  in  value  according  to  a  fixed] 
order. 

23t  A  Deeiinal  Scale  is  one  in  which  the  fixed  order  of| 
increase  or  decrease  is  uniformly  ten. 

This  is  the  scale  used  in  expressing  numbers  by  figures. 

24.  Arithmetic  is  the  Science  of  Numbers  and  the  Artf 
of  Computation. 


B  0  M  A  ^'    N  OTATIO  X, 


9 


ROMAN    NOTATION. 

25.  Cluiracters  Used, — The  liomau  Notation  expresses 
numbers  by  seven  letters  and  a  dasL. 


Letters,— I,       V,      X,       L,         C,  D.  M. 

r«//ie«.-One,   Five,    Ten,    Fifty,   u^«^,^  „  Fiv«^^  ^^^O,.  ^^^^ 


2C5.  Laws  of  llonian  Notation ,— The  above  sevoii 
letters  and  the  dash  are  used  in  accordance  with  the  following 
laws: 

1.  Repeating  a  letter  repeats  its  value. 

Thus,  I  denotes  one;  II,  two;  III,  three;  X,  ten;  XX,  two 
tens,  or  twenty. 

2.  ^Mle>l  a  letter  ia  placed  at  the  left  of  one  of  greater  value, 
the  difference  (f  their  values  is  tJie  7iumber  expreaseil. 

Thus,  IV  denotes  four  ;  IX,  nine:  XL,  forty. 

3.  W/ien  a  letter  is  placed  at  the  right  of  one  of  greater  value, 
the  sum  of  their  values  is  the  number  expressed. 

Thus,  VI  denotes  six  ;  XI,  eleven  ;  LX,  sixty. 

4.  A  dash  placed  over  a  letter  multiplies  its  value  hy  one 
thousand. 

Thus,  XI  denotes  eleven  thousand  ;  V,  five  thousand ;  VI,  six 
thousand. 

EXERCISE     FOR    PRACTICE. 

27.  Express  the  following  numbers  by  Roman  Notation : 


1.  Six. 

2.  Four. 

3.  Three. 

4.  Two. 


5.  Nine. 

6.  Sixteen. 

7.  Thirteen. 

8.  Seventeen. 


13.  Thirty -eight. 

14.  Thirty- nine. 

15.  Forty-six. 

16.  Forty-seven. 


9.  Nineteen. 

10.  Fourteen, 

11.  Twenty. 
13.  Seventy-five 

17.  One  hundred  twenty-seven.     Seven  hundred  four. 

18.  Nine  hundred  forty-nine.     Ninety-five. 

19.  One  thousand.     Nine  thousand.    Fifty  thousand. 

2 


10 


NOTATION    AND     NUMERATION, 


20.  Four  thousand.    One  hundred  thousand.    Eight  hundred 
thi)usand.     Ninety  thousand. 

21.  2800.     1875.    0053.     7939.    4854.    10365.    ^5042. 


22.  Read   the   following:    MIXj    MDLXIV ;   X;   D;  MM;| 
MD ;  DVII ;  MDCCCLXXVI ;  ML  ;  DLX. 


REVIEW    AND    TEST    QUESTIONS. 

ii8.    Study  carefully  and  answer  each  of  the  following! 
questions : 

1.  Define  a  scale.     A  decimal  scale. 

2.  How  many  figures  are  required  to  express  numbers  in  tliej 
decimal  scale,  and  why  ? 

3.  Explain  the  use  of  the  cipher,  and  illustrate  by  exampKsJ 

4.  State  reasons  why  a  scale  is  necessary  in   expressing! 
numbers. 

5.  Explain  the  use  of  each  of  the  three  elements— ^gr?/r^^' 
place,  and  comma — in  expressing  numbers. 

6.  Wliat  is  meant  by  the  simple  or  absolute  value  of  figures  !| 
What  by  the  local  or  representative  value  ? 

7.  How  is  the  local  value  of  a  figure  affected  by  changing  itj 
from  the  first  to  the  third  place  in  a  number  ? 

8.  How  by  changing  a  figure  from  the  second  to  the  fourth ! 
From  the  fourth  to  the  ninth  ? 

9.  Explain  how  the  names  of  numbers  from  twelve  to  twentj 
ar(i  formed.     From  twenty  to  nine  hundred  ninety. 

10.  What  is  meant  by  a  period  of  figures  ? 

11.  Explain  how  the  name  for  each  order  in  any  period  isj 
foi-med. 

12.  State  the  name  of  the  right-hand  order  in  each  of  tli^ 
first  six  periods,  commencing  with  units. 

13.  State  the  two  things  mentioned  in  (6)  which  must 
observed  when  writing  large  numbers. 

14.  Give  a  rule  for  reading   numbers ;   also   for  writing 
numbers. 


ADDITION 


20.  Tlie  Addition  Table  consists  of  the  sums  of  the 
lumbers  from  1  to  0  inclusive,  taken  two  at  a  time.     These 

11118  must  at  first  be  found  by  counting ;  but  when  found, 
|i«y  Hhould  l>e  fixed  in  the  memory  so  that  they  can  be  given 
Hight  of  the  figures. 

30.  To  find  the  mim  of  two  or  more  numbers^  each  expressed 
one  figure,  1/y  using  the  Addition  Table, 

1.  Find  the  sum  of  7,  9,  and  8. 

SOLUTION,— (1)  We  know  at  once  from  the  memorized  remits  of  the 
Idiiion  table,  that  the  pum  of  7  and  9  is  16  or  1  ^en  and  6  units. 
\{'l)  We  add  the  8  unitti  to  the  6  uuitB  of  the  last  result  and  know  in 
kc  sami!  manner  that  the  sum  of  the  8  and  6  le  14,  or  1  (en  and  4  unit*. 
jniting  tliia  ten  with  the  ten  found  by  adding  the  7  and  9,  we  have  2  tens 
|i(l  4  units.,  or  %\.    Hence  the  sum  of  7,  9,  and  8  is  84. 

2.  The  process  in  finding  the  sum  of  any  column  of  figures 
)iisists  in  noting  the  tens  which  the  column  makes. 

Thus,  suppose  the  figures  in  a  column  to  be  9,  6,  8,  5,  and  7. 
)mmencing  with  9  we  note  that  9  and  6  make  1  ten  and  5. 
^'  add  the  8  to  the  5  and  we  have  another  ten  and  3,  making 
tons  and  3.  We  add  the  5  to  the  3,  making  2  tens  and  8. 
^e  now  add  the  7  to  the  8  and  we  have  another  ten  and  5, 
iking  in  all  3  tens  and  5  units,  or  35. 

|3.  Be  careful  to  observe  that  in  practice  each  new  number 
added  to  the  excess  of  the  tens   mentally^  and   nothing 
Lined  but  results 

jFor  example,  in  finding   the  sum  of  a   column  consisting 
the   figures  9,  2,  8,  5,  7  and  4,  commencing  with    9    the 
Hilts  should  be  named,  thus,  ni?ie,  eleven,  nineteen,  twenty- 
nir,  thirty-one,  thirty  five. 


IS 


ADDITION. 


4.  The  numbers  to  bo  added  are  callt-d  Addends.  The  result 
found  is  called  the  tyuni  or  Amount,  and  the  process  <'J  riniHii;r 
the  sum  is  called  Addition.  The  tii'fjn  +,  read  i)Ius,  placed 
between  numbers  ;  thus,  G  +  3  +  10,  hIjovvs  that  these  numbers 
are  to  be  added.  The  sign  — ,  read  ('(juals,  denotes  that  wluit 
is  writt(?n  l)ef()re  it  is  equal  to  what  is  written  after  it  ;  thus, 
8  +  ()  =^  14,  is  read  8  plus  0  eijuals  14, 

5.  To  become  expert  and  accurate  in  adding  you  muht 
practice  on  columns  first  of  three  figures,  then  four,  then  fui , 
until  you  can  giv(;  the  sums  of  such  columns  at  sight,  iou 
rmst  al.<o  at  the  same  time  apply  tliis  practice  on  long 
columns,  so  as  to  aciiuirc  the  habit  of  hohiing  the  tcnn  in  your 
mind  while  you  pcrfonn  the  addition. 

Examples  for  this  practice  can  be  copied  from  the  following: 
table : 


ARITHMETICAL    DRILL    TABLE    NO.    1. 


A. 

B. 

c. 

D. 

E. 

F. 

G. 

H. 

I. 

J. 

^^H    4}..  . 

1. 

rJ 

-^ 

7 

1 

O 

8 

5 

^ 

2 

^mfice  nu 

2. 

1 

1^ 
O 

8 

5 

2 

5 

o 

8 

3 

5 

Band  ace 

:$. 

s 

0 

5 

6 

■# 

8 

6 

^ 

8 

G 

1 

4. 

5 

5 

7 

G 

G 

8 

7 

9 

5 

9 

I 

4 
6 

8 
8 

7 
G 

5 
9 

8 
1 

G 
9 

9 

5 

G 

8 

G 

7 
9 

Hcach  CO 
Border. 

7. 

8 

4- 

8 

^ 

O 

8 

9 

5 

O 

I    Fiudi 

8. 

7 

9 

/v 

8 

5 

9 

8 

G 

8 

o 

I    (1.) 

O. 

0 

5 

9 

7 

G 

8 

9 

7 

^ 

9 

I      ^ 

lO. 

0 

8 

5 

7 

8 

9 

G 

7 

8 

S 

wk 

11. 

7 

5 

9 

^ 

7 

8 

5 

G 

0 

Wf^m          ^ 

12. 

9 

8 

6 

8 

9 

J^ 

7 

9 

7 

8 

^Pi'^t  exam 

A  n  I  r  U  M  KTIC  A  L      TAD  L  I\ 


13 


l\\,  ^'opy  from  this  table  oxamph's  aa  follows  : 

1.  Coininonce  with  ^'olunin  A  oppoyito  I  und  copy  three 
numlxTs  for  t1u>  first  example,  tlion  oppoHite  li  and  copy  three 
nuinluMrt  for  the  Hccond  cxanipl".  and  no  on  to  the  bottom  of 
tlie  column.  The  fir.st  hIx  examplch  copied  from  column  A  in 
this  way  are  , 


(1) 
} 

1 
S 


(8.) 


3 

5 


(8.) 

3 

5 


(4.) 

(5.) 

(0.) 

5 

4 

6 

A 

6 

8 

9 

8 

7 

8.  Copy  examples  with  three  numbers  from  each  column  in 
tlie  name  way,  and  i)ractice  on  finding  the  sums  as  directed  for 
I  memorizing  the  addition  table. 

3.  Copy  in  the  same  manner  examples  with  four  numbers, 
fice  numbers,  and  so  on  up  to  ten  numbers. 
Continue  to  practice  in  this  way  until  you  can  add  rapidly 
land  accurately. 


ILLUSTRATION    OF    PROCESS. 

»*{2.  Prob.  1. — To  find  the  sum  of  two  or  more  numbers, 
)ach  containing  only  one  order  of  units,  and  all  the  same 
)rder. 


Find  the  sum  of 

(1.)       (2.)  (3.) 

60  600 

90  900 

^  JOO 

230  2300 


Explanation.— 1.  The  Biim  of  8,  9,  and 
6  is  found  by  forming  groups  of  ten.  Thut», 
8  and  9  make  1  ten  and  7 ;  7  and  6  make  1 
ten  and  3 ;  hence,  8,  9,  and  6  make  2  tens 
and  3,  or  33. 

2.  The  snm  of  8,  9,  and  6  is  the  same 
whether  these  flt^ures  express  units,  tens, 
or  hundreds,  etc.  Hence,  when  their  sum 
is  found,  if  they  express  units,  as  in  the 
lifit  example,  the  sum  is  units  ;  if  they  express  tens,  as  in  the  second  ex- 
ample, the  sum  is  tens  ;  [{hundreds,  hundreds,  etc. 


6 
9 

_8 

23 


14 


ADDITION, 


SIGHT 

EXERCISES. 

* 

• 

Find  the 

sum  of 

1.    70  +  9  +  5. 

5.    3500  +  80  +  2. 

9. 

500  +  300. 

3.    900  + 

50  +  8. 

6.    7 

006  +  800. 

10. 

8000  +  6000 

3.     900  + 

60  +  7. 

7.    90  +  80. 

11. 

6000  +  5000 

4.    3000 

+  50  +  3. 

8.    70  +  80. 

12. 

200  +  400. 

(13.) 

(14.) 

(15.) 

(16.) 

(17.) 

(18.) 

30 

200 

9000 

40000 

800 

9000 

60 

400 

3000 

60000 

600 

5000 

80 

700 

7000 

70000 

700 

3000 

(19.) 

(30.) 

(31.) 

(22.) 

(33.) 

(34.) 

40 

900 

5000 

.50000 

700 

8000 

50 

300 

3000 

80000 

300 

3000 

70 

800 

7000 

60000 

400 

6000 

33.    Prob.  2.- 

-To  find 

the  sum  of 

any  two  or  more 

numbers. 

Find  the  sum  of  985,  854,  and  698. 


(1.) 

ANALYSIS. 

(3.) 

985 

=      900+80+5 

985) 

854 

-^      800  +50+4 

854 

■  Addends 

698 

=      600+90+8 

698  1 

17^ 

2537 

Sum. 

320  [ 

=    2300  +  220  +  17 

2300) 

2537 

Explanation.— 1.  The  orders  of  units  in  the  ntunbers  to  be  added  i 
indeiwndent  of  each  other,  and  may  ])e  separated  as  ehown  in  the  analymX 

2.  The  Hum  of  each  order  u  found  by  finding  the  eum  of  the  figures  e^ 
pressing  that  order  (32). 


EXERCISES     FOR     PRACTICE, 


15 


00+300. 
000  +  6000. 
.OOO  +  5O00J 
100  +  400. 

(18.) 
9000 
5000 
3000 


(24.) 
8000 
3000 
6000 


0  or  mord 


Addends. 


Sam. 


3.  The  Bums  of  the  separate  orders  may  be  united  into  one  sum,  a9 
phown  in  the  aualyf  is  ;  or, 

4.  By  commencing  with  the  units'  order,  the  number  of  tens  found  can 
at  once  be  added  to  the  tens'  order;  so  with  the  hundreds  found  by  ailding 
the  tens'  order,  etc.,  and  thus  the  sum  may  be  found  in  one  operation,  as 
shown  in  (2). 

From  these  illustrations  we  obtain  the  following 


RULE.      ' 

iJ4:.    /  Write  t?vc  numbers  to  he  added  in  such  a  manvfir  that 
units  of  the  same  order  will  stand  in  the  same  column. 
IF.  Add  each  column  separately,  commencing  uith  the  units. 

III.  When  the  sum  of  any  column  U  expressed  by  two  or  more 
figure*,  place  the  right-hnnd  figure  under  the  column,  and  add 
the  nuinber  expressed  by  the  remaining  figures  to  the  next 
column. 

IV.  Write  under  the  last  column  its  entire  sum. 

Proof. — Add  the  numbers  by  commencing  at  the  top  of  the 
columns.    If  tlce  results  agree,  the  work  is  probably  correct. 

EXERCISES     FOR     PRACTICE. 

IJo.  For  practice  with  abstract  numbers,  copy  from  Table 
No.  1,  page  12,  examples  as  follows  : 


Three  Numbers  of  Ttiree  Plaeest 
1.  Use  any  three  consecutive  columns,  as  A,  B,  C. 


Com- 


mence opposite  1  and  copy  three  numbers  for  ilie  first  example, 
then  opposite  2  and  copy  three  more  for  the  second  example, 
and  so  on  to  the  bottom  of  the  table. 
The  first  six  examples  copied  in  this  way  are  as  follows  : 


1 

■  (1.) 

(2.) 

(3.) 

(4.) 

(5.) 

(6.) 

be  added  oil 

1234 

138 

395 

557 

487 

68G 

the  awa/yfl^l 

|l38 

395 

557 

487 

686 

848 

he  figures  CM 

|395 

557 

487 

686 

848 

797 

16 


ADDITION, 


3.  Copy  in  the  same  manner  examples  with  three  numbers 
from  columns  b,  c,  d  ;  c,  D,  e  ;  D,  E,  F ;  E,  F,  G ;  P,  G,  H  ; 
G,  H,  I ;  and  H,  i,  J. 

Four  Numbers  of  Four  Flacea* 

36.  1.  Copy  as  before  the  numbers  from  any  four  consec- 
utive columns,  as  c,  d,  e,  f.  Commence  in  each  case  oppo- 
site 1  for  the  first  number  of  the  first  example,  opposite  2  for 
the  first  number  of  the  second  example,  and  so  on  to  the 
bottom  of  the  table. 

2.  Copy  in  the  same  manner  examples  from  A,  B,  c,  D ; 
B,  c,  D,  E  ;  D,  E,  F,  G  ;  E,  P,  G,  H  ;  F,  G,  H,  I ;  and  G,  H,  I,  J. 

■i 

Numbers  of  Five  Fiaees. 

37.  Continue  the  practice  by  copying  numbers  of  five 
places,  as  already  directed.  Commence  with  examples  of  five 
numbers,  then  six,  then  seven,  and  so  on. 


ORAL     EXAMPLES. 

38.     1.  A  farmer  sold  60  bushels  of  wheat  to  one  man,  40 
to  another,  and  20  to  another  ;  how  many  bushels  did  he  sell  ? 

Solution.— He  BOld  as  many  bushels  as  the  Bum  of  60,  40,  and  20,  which 
is  120.    Hence  he  sold  120  bushels. 

2.  Mr.  Amaron  owns  30  acres  of  land,  Mr.  Cruchet  owns  50, 
and  Mr.  Easty  70  ;  how  many  acres  do  they  all  own  ? 

3.  R.  \Vliillans  sold  a  cow  for  $40  and  fifteen  sheep  for  $65 ; 
how  much  did  he  receive  for  the  cow  and  sheep  ? 

4.  A  lady  paid  ,f  34  for  a  shawl,  $45  for  a  dress,  and  $7  for  a 
scarf ;  how  much  did  she  pay  for  all  ? 

5.  A  boy  bought  4  bails  and  paid  40  cents  for  each  ball. 
How  much  did  he  pay  for  the  three  ? 

6.  W.  D.  Russel  sold  a  tub  of  butter  for  $27,  a  cheese  for 
$24,  and  some  beans  for  $16 ;  how  much  money  did  he  receive  1 


EXAMPLES, 


17 


7.  A  tailor  sold  a  coat  for  $25,  a  vest  for  $6,  and  a  hat  for 
$5  ;  how  many  dollars  did  he  get  for  all  ? 

8.  A  lady  gave  $72  for  a  watch,  ^32  for  a  chain,  $2  for  a 
key,  and  ij^8  for  a  case  ;  what  did  she  give  for  all  ? 


WRITTEN     EXAMPLES. 

39.  1.  A  newsboy  sold  244  papers  in  January,  301  in  Feb- 
ruary, 278  in  March,  and  390  in  April  ;  how  many  papers  did  he 
sell  in  the  four  months?  Ans.  1213. 

2.  A  grocer  paid  $375  for  coffee,  $280  for  tea,  $564  for 
|suf]:ar,  !^108  for  dried  apples,  and  $198  for  spices;  what  was 
I  the  amount  of  the  purchases  ?  Ans.  $1525. 

3.  In  a  city  containing  4  wards,  there  are  340  voters  in  the 
[first  ward,  533  in  the  second,  311  in  the  third,  and  425  in  the 
I  fourth  ;  how  many  voters  in  the  city? 

4  Norman  D.  Warren  has  a  house  worth  $850,  and  five  more 
3ach  worth  $975 ;  what  is  the  value  of  the  six  "'' 

5.  In  1870  the  population  of  Albany  was  G9452,  Utiea 
>879S,   Syracuse    43081,    Rochester    63424,    Buffalo    117778; 

,'hat  was  the  united  population  of  these  cities  ? 

6.  What  is  the  distance  from  the  Gulf  of  St.  Lawrence  to 
jake  Michigan,  passing  up  the  River  St.  Lawrence  750  miles, 
jake  Ontario  180  miles,  Niagara  River  34  miles.  Lake  Erie 
ioO  miles,  Detroit  River  23  miles,  Lake  and  River  St.  Clair 
15  miles,  and  Lake  Huron  260  miles  ?  Ans.  1542  miles. 

7.  A  man  bought  a  house  for  $3420  ;  he  paid  $320  to  have 
jt  painted,  and    $40  to  have  it  shingled  ;   for  what  amount 

mst  he  sell  it  in  order  to  gain  $250?  u4/i«.  $4030. 

8.  Bought  a  horse  for  $275  and  a  carriage  for  $342  ;  sold 
|he  horse  at  an  advance  of  $113  and  the  carriage  at  an  advance 
^f  !{;65 ;  how  much  did  I  get  for  both  ?  Ans.  $795. 

9.  A  grain  dealer  paid  $1420  for  a  lot  of  flour,  and  $680  for 
lot  of  meal :   he  gained  $342  on  the  flour  and  $175  on  the 

leal  ;  how  much  did  he  receive  for  both  lots  ? 


18 


ADDITION, 


10.  Bought  3  house-lots  ;  the  first  cost  $325,  the  second  $15 
more  than  the  first,  and  the  third  as  much  as  both  the  others  ; 
what  was  the  cost  of  the  whole  ?  An%.  $1330. 


CANADIAN    MONEY. 

40.  Tlie  sign  $  stands  for  the  word  dollars*  Thus,  $13 
is  read  13  doUara. 

41.  The  letters  ct.  stand  for  cents.  Thus,  57  ct.  is  read 
fifty-seven  cents. 

43.  When  dollars  and  cents  are  written  together,  the  cents 
are  separated  from  the  dollars  by  a  ( , ).  Thus,  $42  and  58  ct. 
are  written  $42.58. 

43.  When  the  number  of  cents  is  less  than  10,  i  cipher 
must  occupy  the  first  place  at  the  right  of  the  period.  Thus, 
$8  and  4  ct.  are  written  $8.04. 

44.  In  arranging  the  numbers  for  adding,  do.  lars  must  be 
placed  under  dollars  and  cents  under  cents,  in  such  a  manner 
that  the  periods  in  the  numbers  stand  over  each  other,  thus : 

(1.)  (3.)  (3.) 

$370.84  $3497.03  $53.70 

43.09  69.50  786. 

706.40  240.84  9.08 


^A/■RITTEN     EXAMPLES. 

45.  Read,  arrange,  and  add  the  following  : 

1.  $4.75  +  $3083.09  +  $72.50  +  $9.32  +  $384. 

2.  $93.48  +  $406.30  +  $8.07  +  $5709.80. 

3.  $500  +  $93.05  +  $364.80  +  $47.09. 

Express  in  figures  the  following : 

4.  Nine  hundred  six  dollars  and  seventy-five  cents. 

5.  Seventy-five  dollars  and  thirty -eight  cents. 

6.  Three  hundred  twelve  dollars  and  nine  cents. 

7.  Eighty -four  cents  ;  seven  cents  ;  three  cents. 


DEFINITIONS, 


1& 


8.  Find  the  sum  of  $206.08,  $5.54,  and  $396.03. 

9.  A  farmer  sold  a  quantity  of  wheat  for  $97.75,  of  barley  for 
.$42.06,  of  oats  for  $39.50.  How  much  did  he  receive  for  the 
whole?  '^^^'  $179.31. 

10.  A  man  bought  a  horse  for  $345.50,  a  carriage  for  $182.90, 
and  sold  them  so  as  to  gain  on  both  $85.50.  How  much  were 
they  sold  for?  Arts.  $613.90. 

11.  Bought  a  house  for  $4268.90,  furniture  for  $790.07,  car- 
peting  $380.60,  and  made  repairs  on  the  house  amounting  to 
$307.05.     How  much  did  the  whole  cost?         Ans.  $5746.62. 

12.  A  man  is  in  debt  to  one  man  $773.60,  to  another  $600.50, 
to  another  $73.08,  to  another  $305.04 ;  how  much  does  he  owe 
in  all  ?  Ans.  $1751.22. 

13.  A  furniture  dealer  sold  a  bedroom  set  for  $125.86,  a 
bookcase  for  $85.09,  and  3  rocking-chairs  for  $5.75  each.  How 
much  did  he  receive  for  the  whole  ?  Ans.  $228.20. 

14.  D.  N.  Mac  Vicar  bought  a  saw  mill  for  $8394.75,  and  sold 
it  so  as  to  gain  $590.85  ;  for  how  much  did  he  sell  it  ? 

15.  A  lady  after  paying  $23.85  for  a  shawl,  $25.50  for  a  dress, 
$2.40  for  gloves,  and  $4.08  for  ribbon,  finds  she  has  $14.28 
left ;  how  much  had  she  at  first  ?  Am.  $70.11. 


DEFINITIONS. 


46.  Ailditiofi  is  the  process  of  uniting  two  or  more  num- 
bers into  one  number. 

47.  Adilenils  are  the  numbers  added. 

48.  The  Sam  or  Amount  is  the  number  found  by  addi- 
tion. 

40.  The  Process  of  Afldition  consists  in  forming  units 
of  the  same  order  into  groups  of  ten,  so  as  to  express  their 
amount  in  terms  of  a  higher  order. 

60.  The  Sign  of  Addition  is  4-,  and  is  read  plus. 
When  placed  between  two  or  more  numbers,  thus,  8  +  3  +  6  +  2 
+  9,  it  means  that  they  are  to  be  added. 


20 


ADDITION, 


■ii 


51.  The  Sign  of  Equdlity  is  =,  and  is  read  equals,  or 
equal  to  ;  thus,  9  +  4  =  13  is  read,  nine  plus  four  equals  thirteen. 

r>2.  Principles. — /.  Only  numbers  of  the  same  denomind' 
Hon  and  units  of  the  same  order  can  be  added. 

IT.  TJie  sum  is  of  the  same  denomination  as  the  addends. 

HI.  The  whole  is  equal  to  the  sum  of  all  the  parts. 


REVIEW    AND    TEST    QUESTIONS, 

53.     1.  Define  Addition,  Addends,  and  Sum  or  Amount, 
3.  Name  each  step  in  the  process  of  Addition. 

3.  Why  place  the  numbers,   preparatory  to  adding,  units 
under  units,  tens  under  tens,  etc.  ? 

4.  Why  commence  adding  with  the  units'  column  ? 

5.  What  objections  to  adding  the  columns  in  an  irregular 
order?     Illustrate  by  an  example. 

6.  Construct,  and  explain  the  use  of  the  addition  table. 

7.  How  many  combinations  in  the  table,  and  how  found  ? 

8.  Explain  carrying  in  addition.     What  objection  to  the  use 
of  the  word? 

9.  Define  counting,  and  illustrate  by  an  example. 

10.  Write  five  examples  illustrating  the  general  problem  of 
addition,  "Given  all  the  parts  to  find  the  whols." 

11.  State  the  difference  between  the  addition  of  objects  and 
the  addition  of  numbers. 

12.  Show  how  addition  is  performed  by  using  the  addition 
table. 

13.  What  is  meant  by  the  denomination  of    a    number? 
Wliat  by  units  of  the  same  order  ? 

14.  Show  by  analysis  that  in  adding  numbers  of  two  or  more 
places,  the  orders  are  treated  as  independent  of  each  other. 


SUBTRACTION. 


54,  The  difference  between  two  numbers  is  the  amount  that 
one  number  i.s  greater  than  the  other.  ThuH,  7  is  2  greater 
than  5  ;  hence  2  is  the  difference  between  7  and  5. 


ILLUSTRATION     OF     PROCESS. 

*>5.  Prob.  I. — To  find  the  difference  between  two 
numbers,  each  containing  only  one  order  of  units  and 
both  the  same  order. 


Find  the  difference  between 


(1.) 

8 
3 


5 


m 


(3.) 
800 
300 

500 


Explanation.— 1.  The  differ- 
ence between  8  and  3  is  found  by 
making  8  into  two  parts,  one  of 
which  is  3,  the  other  5,  the  differ- 
ence. 

2.  The  difference  between  8 
and  3  is  the  Bamo,  whether  these 
fi<?ure8  express  units,  tens,  or 
hundreds,  etc.  Hence,  when  their  difference  is  found,  if  they  exi)res8 
unit?,  as  in  the  first  example,  the  difference  is  units;  if  they  express  tens, 
aji  in  the  second  example,  tlie  difference  is  tens;  if  hundreds,  huuu.eds, 
etc. 

SIGHT    EXERCISES. 
Find  the  difference  between  the  following  numbers  : 

(!■)  (3.)         .  (3.)  (4.) 


70 
20 

(0.) 
13 
J 

(11.) 

170 

GO 


800 

m 

(7.) 
130 
JO 

(12.) 
13000 
4000 


600 

80 

200 

30 

(8.) 

(0.) 

1300 

150 

400 

70 

(13.) 

(14.) 

loOOO 

12000 

0000 

3000 

(5.) 

9000 

5000 

(10) 

1500 

700 

(15.) 
18000 
7000 


22 


suit  TRA  CTIO  X. 


56.    Prob.  II.-— To  find  the  difference  between  any  two 
numbers. 

Find  the  difference  between  853  and  495. 


ANALYSIS. 

Minuend, 

853 

=       700 

4- 

140 

+ 

13 

Subtrahend, 

495 

=     400 

+ 

90 

+ 

5 

Difference, 

358 

=     300 

+ 

50 

+ 

8 

BxPLANATioN.— 1.  The  5  units  cannot  be  taken  from  the  3  units ; 
hence  1  of  the  5  tens  is  added  to  the  3  units,  makin;,'  13,  as  shown  in  the 
analysis,  and  the  5  units  are  then  taken  from  13,  leaving  8  units. 

2.  One  ten  has  been  taken  from  the  6  tens  in  tlie  minuend,  leaving 
4  tens.  The  9  tens  of  the  nubtrahend  cannot  be  taken  from  the  4  tens  that 
are  left.  Hence  1  of  the  8  hundreds  is  added  to  the  4  tens,  making  14  tens, 
or  140,  as  shown  in  the  analysis.  The  9  tens  are  taken  from  the  14  tcus, 
leaving  5  tens. 

3.  One  hundred  has  been  taken  from  the  8  hundreds,  leaving  7  hundreds. 
Hence  the  difference  between  8.53  and  495  is  358. 

From  these  illustrations  we  obtain  the  following 


RULE. 

57,  I.  Write  the  subtrahend  under  the  minuend,  placing  wiits 
oft/ie  same  order  in  the  same  column. 

II.  Begin  at  the  right,  and  subtract  the  number  of  units 
of  ^ach  order  of  the  subtrahend  from  the  number  of  units 
of  the  corresponding  order  of  the  minuend,  and  write  the  result 
beneath. 

III.  If  the  number  of  units  of  any  order  of  the  subtrahend  is 
greater  than  the  number  of  units  of  the  corresponding  order  of 
the  minuend,  increase  the  latter  by  10  and  subtract ;  then  dimin- 
ish by  I  the  units  of  the  next  higher  order  of  the  minuend  and 
proceed  as  before. 

Proof. — Add  the  remainder  to  the  subtrahend  ;  if  the  sum  is 
equal  to  the  minuend,  the  work  is  probably  correct. 


ILLUSTRATION     OF    PROCESS, 


33 


EXAMPLES     FOR    PRACTICE. 

58.  Copy  examples  for  practice  with  abstract  numbers  from 
Arithmetical  Table  No.  1,  on  page  12,  as  follows : 

Examples  with  Ttiree  JtHgures. 

1.  Take  the  numbers  from  columns  a,  b,  c.  For  the  first 
example  use  the  numbers  opposite  1  and  12  ;  for  the  second 
those  opposite  ti  and  J5 ;  then  3  and  4,  4  and  5,  and  so  on 
to  the  bottom  of  the  columns.  The  first  six  examples  are  as 
follows 


(1.) 

(2.) 

(3.) 

(4.) 

(5.) 

(6.) 

234 

395 

557 

557 

686 

848 

138 

138 

395 

487 

487 

686 

2.  Copy  examples  in  the  same  manner  from  columns  B,  c,  D  ; 
then  c,  D,  E ;  D,  B,  P ;  E,  F,  G ;  P,  G,  n ;  G,  H,  i ;  and 
n,  I,  J. 

Examples  with  Four  Figures* 

50.  For  examples  with  four  figures,  copy  the  numbers  for 

[the  first  set  from  columns  A,  B,  c,  D  ;   for  the  second,  from 

B,  c,  D,  E  ;   the  third,  c,  D,  e,  P  ;   the  fourth,  D,  E,  F,  G  ;    the 

lit'th,   E,  F,  G,  H  ;    the  sixth,    F,  G,  H,  i ;    and  tho  seventh, 

|g,  h,  I,  j. 

Examples  with  Six  ligures. 

00.  For  examples  with  six  figures  copy  the  numbers  from 
the  columns  as  follows :  first  set,  a,  b,  c,  d,  e,  f  ;  second  set, 
I.  c.  D,  E,  F,  G  ;  third  set,  c,  D,  e,  f,  g,  ii  ;  fourth  set,  D,  e,  f, 
},  n,  I ;  fifth  set,  E,  p,  G,  H,  I,  J. 

Let  all  these  examples  be  worked  out  of  school  and  between 
recitations,  and  brought  to  class  on  paper  for  the  correction  of 
me  answers. 


m 


24 


S  UliTIi  A  CTION, 


WRITTEN     EXAMPLES. 

61.  1.  The  independence  of  the  United  States  was  declared 
in  1770;  how  long  after  tliat  event  is  the  year  1876? 

3.  A  man  deposited  $1050  in  a  bank  and  afterwards  drew  out 
$105  ;  how  much  was  left?  Aiis.  $1485. 

3.  The  population  of  a  city  in  1860  was  22529  and  in  lb  10 
it  was  28798 ;  what  was  the  increase  ?  Ans.  6209. 

4.  The  height  of  Mt.  Etna  is  10840  feet  and  that  of  Mt. 
Vesuvius  3948  feet ;  how  many  feet  higher  is  Etna  than 
Vesuvius?  ^/<«.  0892  feet. 

5.  The  number  of  pupils  attending  school  in  Boston  in  1870 
was  38944,  and  of  these  35442  attended  the  public  schools  :  how 
many  in  all  other  schools?  Ans.  3502  pujuls. 

C.  The  sum  of  two  numbers  is  7427,  and  the  smaller  number 
is  1487 ;  what  is  the  greater  ?  Ans.  5940. 

■  7.  A  man  bought  four  houses,  for  which  he  paid  .jJl.jOOO;  for 
the  first  he  paid  $3180,  for  the  second  $2783,  and  for  the  third 
$4789 ;  how  much  did  ho  pay  for  the  fourth  ? 

Solution.— If  the  man  paid  $159<)0  for  the  four  houses  and  the  sum 
of  |;J186  +  $-2783  +  |;478!>,  which  is  $10758  for  three  of  them,  lie  iiui>-t 
have  paid  for  the  fourtli  the  diflcrencc  between  $15960  aud  $10758 
which  is  $5202.  .     _  t- 


8.  A  man's  salary  is  $1300  a  year,  and  he  has  money  at 
interest  which  brings  him  $125  more ;  if  his  expenses  art.'  (875, 
how  much  can  he  save  ?  Ans.  $5j0. 

9.  A  has  $6185,  B  has  $15181,  C  has  $858  less  than  A 
and  B  together,  and  D  has  as  much  as  all  the  rest ;  how  much 
has  D?  Ans.  $41874. 

10.  Warren  Xewhall  deposited  $302  in  the  Montreal  Bank 
on  Monday,  $760  on  Tuesday,  and  $882  on  Thursday  ;  on 
Wednesday  he  drew  out  $380,  on  Friday  $350,  and  on  Satur- 
day $200 ;  how  much  remained  on  deposit  at  the  end  of  the 
week?  Ans.  $1074. 

11.  My  property  is  valued  at  $7090,  and  I  owe  a  debt  of 


EXAMPLES' 


25 


|600,  another  of  $1247,  and  another  of  |420 ;  wlint  am  I 
worth  V  Ans.  !*^4820. 

12.  A  merchant  paid  $4570  for  goods ;  he  sold  a  part  of  them 
for  .t;3480,  and  the  rust  for  $2724;  how  much  did  he  gain  by 
the  transaction ?  Ana.  J?  1034. 

i:j.  A  man  deposits  $1110  in  the  bank  at  one  time,  and 
$1004  at  another  ;  he  then  draws  out  $786  at  one  tinu*,  $Go4  at 
another,  $489  at  another;  how  much  still  remained  in  the 
birnk?  Am.  $245. 

14.  J.  Locke  bought  a  farm  for  $4750,  and  built  a  house  and 
barn  upon  it  at  a  cost  of  $4475,  and  then  sold  tlie  whole  for 
$8090 ;  how  much  did  he  lose  ?  Am.  $1135. 

15.  A  grain  dealer  bought  9710  bushels  of  grain  ;  he  then 
sold  3348  bushels  at  one  time  and  5303  bushels  at  another ; 
how  many  bushels  had  he  left  ?  An8.  1005  bushels. 

10.  Find  the  difference  between  $527.03  and  $204.39. 

Explanation.— Write  the  Bnbtrabend  under  the  min- 
uend, bo  that  dollars  are  under  dollars  and  centi^  under 
cents.  Snbtract  as  if  the  numbers  were  abstract,  and 
place  a  period  in  the  result  between  the  second  and  third 
figures  from  the  right.    The  figures  on  the  left  of  the 


$527.03 
204.39 


$202.04 
period  express  dollars  and  tho^e  on  the  ri-^'ht  cents 


17.  I  received  $352.07,  and  paid  out  of  this  sum  to  one  man 
$73.12,  to  another  $112.57 ;  how  much  have  I  left  of  the  money  ? 

Ai,f>.  $100.38. 

18.  A  lady  had  $23.37,  and  paid  out  of  this  $7.19  for  flour, 
$3.07  for  sugar,  $2.05  for  butter;  how  much  had  she  left? 

19.  A  farmer  sold  $153  worth  of  wheat,  $54.75  of  barley,  and 
$29.0")  of  oats.  He  paid  out  of  the  money  receivetl  to  one  man 
$:i2.13,  to  another  $109.55;  how  much  had  he  left? 

20.  Three  men  are  to  pay  a  debt  of  $6809.  The  first  man 
pays  $3905.38,  the  second  $2001.70  ;  how  much  has  the  third  to 
pay?  An».  $901.92. 

21.  A  merchant  sold  in  one  day  $782.17  of  goods.  He  re- 
ceived in  cash  $459.58;  how  much  did  he  sell  on  credit? 

22.  A  man  owns  five  farms  containing  in  all  3256  acres,  and 
jsellstwoof  them  containing  together  876  acres.  How  many 
I  acres  has  he  left?  Ana.  2380. 

3 


96 


aUBTRA  CTION. 


DEFINITIONS. 

62.  Subtraction  is  tho  process  of  finding  the  difference 
between  two  numbers. 

an.  The  3luiU€lid  is  the  greater  of  two  numbers  whose 
difference  is  to  be  found. 

<14:.  Tho  Siibtrahenil  is  the  smaller  of  two  numbers 
whose  difference  is  to  be  found. 

05.  The  Difference  or  Remainder  is  tlie  result  ob- 
tained by  subtraction. 

GO.  Tho  Process  of  Subtraction  consists  in  comparing 
two  numbers,  and  resolving?  the  greater  into  two  parts,  one  of 
which  is  equal  to  tho  less  and  the  other  to  the  difference  of  the 
numbers. 

67.  The  Sign  of  Subtraction  is  — ,  and  is  called  minus. 
When  placed  between  'two  numbers,  it  indicates  that  their  dif- 
ference is  to  be  found  ;  thus,  14  —  6  is  read,  14  minus  0,  and 
means  that  the<liffereuce  between  14  and  G  is  to  be  found. 

68.  Parentheses  (  )  denote  that  the  numbers  enclosed 
between  them  are  to  be  considered  as  one  number. 


69.  A  Vinculum 


affects  numbers  in  the  same 


manner  as  parentheses.  Tbus,  19  +  (13 — 5),  or  19  +  13  —  5 
signifies  that  the  difference  between  13  and  5  is  to  be  added 
to  19. 

70.  Principles.— 7.  Only  like  numbers  and  units  of  th\ 
same  order  can  he  subtracted. 

II.  The  minuend  is  the  sum  of  the  subtrahend  and  difference, 
or  th^  minuend  is  the  whole  of  which  the  subtrahend  and  differ- 
ence are  the  parts, 

in.  An  equal  increase  or  decrease  of  the  m,inuend  and  siibtra-\ 
hend  does  not  change  the  difference. 


RE  VIE  W, 


27 


REVIEW    AND    TEST    QUESTIONS. 

71.  1.  Define  the  process  of  subtraction,  lllustrato  each 
ett'p  by  an  example. 

3.  Explain  how  subtraction  bhould  bo  performed  when  an 
order  in  the  subtrahend  is  greater  than  the  corresiionding  order 
in  the  minuend.     Illustrate  by  an  example. 

8.  Indicate  the  difference  between  the  subtraction  of  numbers 
and  the  subtraction  of  objects. 

4.  When  is  the  result  in  subtraction  a  remainder,  and  when 
n  difference? 

5.  Show  that  so  far  as  the  process  with  numbers  is  concerned, 
the  result  is  always  a  difference. 

6.  Prepare  four  original  examples  under  each  of  the  following 
problems,  and  explain  the  method  of  solution  : 

Prob.  I. — Oiten  the  whole  and  one  of  the  parts  to  find  the 
[  other  part. 

Prob.  II. — Gicen  tTie  sum  of  four  numbers  and  three  of  them 
\to  find  the  fourth. 

7.  Construct  a  Subtraction  Table. 

8.  Define  counting  by  subtraction. 
0.  Show  that  counting  by  addition,  when  we  add  a  number 

larger  than  one,  necessarily  involves  counting  by  subtraction. 

10.  What  is  the  difference  between  the  meaning  of  denomi- 
ition  and  orders  of  units  ? 

11.  State  Principle  III  and  illustrate  its  meaning  by  an 
Bxaini)le. 

13.  Show  that  the  difference  between  63  and  9  is  the  same  as 
fhe  difference  between  (63  + 10)  and  (9  + 10). 

13.  Show  that  28  can  be  subtracted  from  93,  without  aualyz- 
[ng  the  minuend  as  in  (56).  by  adding  10  to  each  number. 

14.  What  must  be  added  to  each  number,  to  subtract  275 
rom  820  without  analyzing  the  minuend  as  in  (56)  ? 

15.  What  is  meant  by  borrovring  and  carrying  in  subtrac- 
ion? 


w/? 


MULTIPLICATION. 


72.  The  MnlUpUcation  Table  consists  of  the  products 
of  numbers  from  2  to  12  inclusive.  These  products  are  found 
by  addition,  and  then  memorized  so  that  they  can  be  given  at 
sight  of  their  factors. 

73.  To  memorize  the  Multiplication  Table. 
Pursue  the  folloveing  course  : 

1.  Write  on  your  slate  in  two  sets  and  in  irregular  order 
2  times  2  are,  3  times  fJ  are,  and  so  on,  up  to  12  times  2  are, 
thus : 

(1.)  (2.) 

2  times  2  are  7  timr>s  2  are 
5  times  2  are                       11  times  2  are 

3  times  2  are  9  times  2  are 
8  times  2  are                       12  timers  2  are 

4  times  2  are  G  times  2  are 
G  times  2  are                       10  times  2  are 

2.  Find,  by  adding,  the  product  of  each  example  and  write  it 
after  the  w^rd  "  are." 

3-  Read  very  carefully  the  two  sets  several  times,  then  erase 
the  products  and  writ(^  them  again  from  memory  ns  you  read 
the  example.  Continue  to  erase  and  write  the  products  in 
this  way  until  they  are  firmly  fixed  in  your  memory. 

4.  Write  on  your  slate  a  series  of  ticoa,  and  write  under  them 
in  irregular  order  the  numbers  from  2  to  12  inclusive ;  thus, 

222    2    22    22     2    2    2 

i  ^  ^  '1  Q.  'Ill  I  U  I M 

Write  the  product  under  each  example  as  you  repeat  men- 
tally the  number  of  twos.    Continue  to  erase  and  write  again  and 


EXAMPLES. 


29 


erase 

read 

zts  in 

I  them 

18, 


a^n,  until  each  product  ia  called  up  to  your  mind  just  as 
soon  as  you  look  at  the  two  numbers. 

5.  Pursue  the  same  course   in    memorizing    the  products 
of  3's,  4'8,  5's,  6's,  7'8,  8'8,  and  9's. 

MULTIPLIER    ONE    FIGURE. 

PREPARATORY     STEPS. 

74.  Step  I. — Find  by  using  the  Multiplication  Table  the 
product  of  eiich  of  tlie  following : 

Thus,  5  X  7  =  35,  5  tens  x  7  =  35  tens,  500  x  7  =  3500. 

Find  the  product  of 

1.  8  X  6 ;  8  tens  x  6 ;  8  hundred  x  6 ;  8000  x  6. 

2.  9  X  7  ;  90  X  7  ;  900  X  7  ;  9000  x  7. 

3.  3  X  5 ;  30  X  5 ;  300  X  5  ;  3000  x  5. 

4.  7000  X  3  ;  500  X  6 ;  8000  x  4  ;  4000  x  4. 

5.  60000  X  9 ;  900000  x  7;  5000000  x  5. 

75.  Step  XL — The  orders  in  a  number  are  independent  of 
each  other ;  hence,  to  find  any  number  of  times  a  given  nuniber, 
we  muUiply  each  order  separately,  thus : 

To  find  6  times  748,  we  regard  the  748  =  700  +  40  +  8. 
We  know  from  memonzed  results  that  G  times  8  are  48,  that 
6  times  40  are  240,  and  that  6  times  700  are  4200.  Having 
taken  each  of  the  three  parts  of  748  G  times,  the  sum  of  these 
products  must  be  6  times  748.  Hence,  48  +  240  +  4200  = 
4488 1=  6  times  748. 

EXERCISE    FOR    PRACTICE. 
Multiply  and  explain,  as  shown  in  Step  H,  each  of  the 


2     1 

foUowing : 

*      ' 

^ 

w   1 

1.  242  X  4. 

6. 

735  X  8. 

9. 

637  X  4. 

1 

2.  432  X  3. 

6. 

507  X  6. 

10. 

482  X  8. 

men-     I 

3.  321  X  2. 

7. 

389  X  5. 

11. 

795  X  9. 

nand  ■ 

4.  612  X  7. 

8. 

837  X  6. 

12. 

359  X  7. 

1 


*l 


j'V^r. 


\'  m 


M 


30 


MULTIPLICATION. 


76,  The  method  of  finding  the  mm  of  two  or  more  times  a 
given  number  by  using  memorized  results  is  called  Multipli- 
cation. The  number  taken  is  called  the  Multiplicand,  and 
the  number  wliich  denotes  how  many  times  the  multiplicand 
is  taken  is  called  the  Multiplier. 

ILLUSTRATION    OF    PROCESS. 

77.  Prob.  I. — To  multiply  any  number  by  numbers 
less  than  lo. 


How  many  are  4  times  369  ? 

(1.)     ANALYSIS. 

i     9x4  = 

369  X  4  =  ]   GO  X  4  = 

(  300  X  4  = 


36 

240 

1200 

1476 


(2.) 
869 

4 

1476 


Explanation.— 1.  The  869  is  equal  to  the  three  parts,  9,  60,  and  ^K). 

2.  By  taking  each  of  these  parts  four  timeB,  the  369  is  taken  four  times. 
Hence,  to  fiud  4  times  ;i69,  the  9  is  taken  4  times ;  then  the  60  ;  then  the 
300,  as  shown  in  the  analysis. 

3.  Uniting  the  36,  the  240,  and  the  1200  in  one  number,  we  have  4  times 
969.    Ilence,  1476  is  4  times  309. 

4.  In  practice,  no  analysis  is  made  of  the  number.  We  commence  with 
the  units  and  multiply  thus : 

(1.)  4  times  9  units  are  36  units  or  3  tens  and  6  units.  We  write  the 
6  units  in  the  units'  place  and  reserve  the  4  tens  to  add  to  the  product  of 
the  tens. 

(2.)  4  times  6  tens  are  24  tens,  and  the  3  tens  reserved  are  27  tens  or 
2  hundred  and  7  tens.  We  write  the  7  tens  in  the  tens'  place,  aud  reserve 
the  2  hundred  to  add  to  the  product  of  the  hundreds. 

(3.)  We  proceed  in  the  same  manner  with  hundreds,  thousands,  etc. 

From  these  illustrations,  we  obtain  the  following 


RULE. 

78.  Begin  at  the  right  hand  and  multiply  each  order  of 
the  multiplicand  by  the  multiplier.  Write  in  the  product, 
in  each  case,  the  units  of  the  result,  and  add  the  tens  to 
the  next  higher  result. 


EXAMPLES. 


31 


EXAMPLES     FOR 

PRACTICE. 

Perform  the 

multiplication  in  the 

following : 

1.      837x3. 

7.      986  x  2. 

18.     579x9. 

2.    5709x5. 

8.    7093x5. 

14.  90703  X  7. 

3.  83095x6. 

9.  50739x8. 

15.  29073x8. 

4.  39706x5. 

10.  79060  X  6. 

16.  40309  X  7. 

5.  95083  K  4. 

11.  79350x3. 

17.  73290x8. 

6.  70G39>8. 

12.  60790x5. 

18.  30940x6. 

i 


7U.  Continue  the  practice  with  abstract  numbers  by  taking 
examples  from  Arithmetical  Table  No.  1,  page  12,  in  the  fol- 
lowing order: 

Thi*ee  Figures  in  the  Multiplicand. 

1.  Use  three  columns  and  copy  for  multiplicands  each  num- 
ber in  Ihe  colunms,  commencing  at  the  top  of  the  Table. 

2.  Take  as  multii)lier  the  figure  immediately  under  the  right- 
hand  figure  of  the  multiplicand. 

T]ie  first  six  examples  taken  in  this  way  from  columns  A, 
B,  C,  are 


(1.) 

(2.) 

(3.) 

(4.) 

(5.) 

(6.) 

234 

138 

395 

557 

487 

686 

8 

5 

7 

7 

6 

8 

3.  Let  examples  be  copied  in  this  way  from  columns  A,  b,  c  ; 
B,  C,  D  ;  C,  D,  E ;  D,  E,  F  ;  E,  F,  G  ;  F,  G,  n  ;  G,  H,  I  ;  and  n,  I,  J. 


Four  Figures  in  the  Multijylicand, 

1.  Use  four  columns,  and  copy  the  multiplicands  and  multi- 
pliers in  the  same  way  as  \nth  three  figures,  taking  the  multi- 
pliers from  the  first  column  on  the  right. 

2.  Copy  from  columns  A,  b,  c,  D ;  then  B,  C,  D,  E ;  C,  D,  E,  F ; 

D,  E,  F,  G  -,  E,  F,  G,  H  ;  F,  G,  H,  I. 


32 


MULTIPLICA  TION, 


Six  Figures  in  the  Multiplicand. 

1.  Copy,  as  already  directed,  examples  from  columns  A,  B,  C, 
D,  E,  F  ;  then  B,  C,  D,  E,  F,  G  ;  C,  D,  E,  F,  G,  H  ;  D,  E,  F,  G,  H,  I ; 
and  E,  F,  G,  u,  I,  J.  Take  the  multipliers  from  the  right-hand 
column  used. 

.  2.  Let  the  examples  from  each  of  these  sets  be  worked  at 
your  seat  between  recitations  or  out  of  school. 


ORAL    EXAMPLES. 

80.  1.  Bought  4  barrels  of  tlour,  at  ^12  a  barrel,  and  a 
barrel  of  crackers  for  $G  ;  how  much  did  the  whole  cost  ? 

Solution.— The  whole  cost  four  times  $12,  plus  $G,  whidi  is  $54, 

2.  If  it  requires  5  yards  of  cloth  to  make  a  coat,  and  1  yard 
to  make  a  vest,  how  many  yards  will  make  9  of  each  ?  12  of 
each  ?    7  of  each  ?  ' 

3.  Bought  12  chairs  at  |3  each,  a  sofa  at  |47,  and  8  taHes  at 
$9  each  ;  how  much  did  the  whole  cost  ? 

4.  Gave  $7  each  to  5  men,  paid  for  10  yards  of  cloth  at  ^  a 
yard,  and  for  a  coat  .$17  ;  how  much  money  have  I  spent  ? 

5.  At  7  dollars  a  cord,  what  will  6  cords  of  v>^ood  cost? 
8  cords?    11  cords?    9  cords?    13  cords?  ,,  ,, 


WRITTEN     EXAMPLES. 

81.    6.  How  much  will   7  acres  of  land   cost,  at  $285  an 
acre?  Ans.  $1995. 

Solution.— 7  acres  will  copt  7  timcp  $285.    7  time?  $285=7  times 
$5  +  7  times  $80  +  7  times  $200  =  $19!>5.    Hence,  7  acres  cost  $1095. 

7.  What  will  be  the  cost  of  building  213  yards  of  iron  fence, 
at  3  dollars  a  yard?  Aits.  Go9  dollars. 

8.  What  will  647  cords  of  wood  cost  at  $G  a  cord  ? 

9.  There   are   5280  f"5et  in  a  mile;    how   many   foot  in   12 
miles?  Ans.  63360  feet. 


PREPARATOnr    STEPS, 


33 


10.  I  sold  852  yds.  of  cloth  at  3  dollars  a  yard ;  bow  much 
money  did  I  receive  ?  Ana.  $2556. 

11.  There  are  4  fartliings  in  one  penny  ;  how  many  farthings 
in  379  pennies  ?  Ans.  1516  farthings. 

12.  William  Robb  went  to  market  with  $485  ;  he  paid  for  20 
barrels  of  flour  at  $8  a  barrel ;  16  boxes  of  soap  at  $3  a  box  ; 
and  3  tubs  of  butter  at  $12  a  tub  ;  how  much  money  did  he 
have  left  ?  Ans.  $241. 

13.  Sold  89  bushels  of  beans  at  $2  a  bushel,  and  7  loads  of 
hay  at  $19  a  load ;  how  much  did  I  receive  for  both  ? 

14.  A  merchant  bought  12  hogsheads  of  molasses  at  $50  a 
hogshead,  and  sold  the  whole  for  $524  ;  how  much  did  he  gain 
by  the  transaction  ?  Ans.  $76. 


MULTIPLIERS    10    AND    ABOVE. 

PREPARATORY    STEPS. 

83.  Step  I. — To  multiply  any  number  by  10,  100,  1000, 
and  so  on. 

1.  A  figure  is  multiplied  by  10  by  moving  it  one  place  to 
the  left,  by  100  by  moving  it  two  places,  etc.  Thus,  4 
expresses  four,  40  expresses  10  fours,  400  expresses  100 
fours,  etc. 

2.  A  cipher  placed  at  the  right  of  a  number  moves  each 
significant  figure  in  it  one  place  to  the  left  ;  hence,  multiplies 
it  by  10. 

Thus,  in  372  the  2  is  in  the  first  place,  the  7  in  the  second, 
and  the  3  in  the  third  ;  but  in  3720  the  2  is  in  the  second, 
the  7  in  the  third,  and  the  3  in  the  fourth  place;  hence, 
annexing  the  cipher  has  removed  each  figure  one  place  to  the 
left,  and  consequently  multiplied  each  order  in  the  nimiber 
by  10. 

3.  In  like  manner  annexing  two  ciphers,  three  ciphers,  etc., 
multiplies  a  number  by  100,  1000.  etc.,  respectively. 


34 


MULTIPLICA  TION 


83.  Step  ll.^rTo  multiply  by  using  the  parts  of  the 
multiplier. 

1.  The  multiplier  may  be  made  into  any  desired  parts,  and 
the  multiplicand  taken  separately  the  number  of  times  ex- 
pressed by  each  part.  The  sum  of  the  products  thus  found  is 
the  required  product. 

Thus,  to  find  9  times  12  we  may  take  4  times  12  which  are 
48,  then  5  times  12  which  are  60.  4  times  12  plus  5  times  12 
are  9  times  12  ;  hence,  48  plus  60,  or  108,  are  9  times  12. 

2.  WV  1  we  multiply  by  one  of  the  equal  parts  of  the 
muiiiplier,  we  find  one  of  the  equal  parts  of  the  required 
product.  Hence,  by  multiplying  the  part  thus  found  by  the 
number  of  such  parts,  we  find  the  required  product. 

For  eram^^l  ,  to  find  12  times  64  we  may  proceed  thus : 


'  '   ^    AVA1,T8I8. 

(2.) 

■:4  X    4  -    ■:'^r,[  = 

04  y      i  :..  ...    "  ' 

64 

=  3  times  256. 

4 

256 

64  X  12  ■-=  ;Go 

3 

768 

(1.)  Observe,  that  12  =  4  +  4  +  4 ;  hence,  4  is  one  of  the  3 
equal  parts  of  12.  - 

(2.)  That  64  is  taken  12  times  by  taking  it  4  times  +  4  times 
+  4  times,  as  shown  in  the  analysis. 

(3.)  That  4  times  64,  or  256,  is  one  of  the  3  equal  parts  of 
12  times  64.  Hence,  multiplying  256  by  3  gives  12  times  64, 
or  768. 

3.  In  multiplying  by  20,  30,  and  so  on  up  to  90,  we  invari- 
ably multiply  by  10  one  of  the  equal  parts  of  these  numbers, 
and  then  by  the  number  of  such  parts. 

For  example,  to  multiply  43  by  30,  we  take  10  times  43,  or 
430,  and  multiply  this  product  by  3 ;  430  x  3  =  1290,  which 
is  30  times  43. 


ILLUSTRATION    OF    PROCESS, 


35 


We  multiply  in  the  same  manner  by  200,  300,  etc.,  2000, 
:3000,  etc. ;  multiplying  first  by  100,  1000,  etc.,  then  the 
product  thus  found  by  the  number  of  lOO's,  lOOO's,  etc. 

ILLUSTRATION    OF    PROCESS. 

84.  Prob.  II. — To  multiply  by  a  number  containing 
only  one  order  of  units. 

1.  Multiply  347  by  500. 


(1.)    ANALTSIB. 

(2.) 

First  step, 

347  X  100  =     34700 

347 

Secoud  step, 

34700  X  5  =  173500 

500 

173500 

Explanation.— 500  is  equal  to  5  times  100 ;  hence,  by  taking  347, 
as  \xx  first  step^  100  times,  5  times  this  result,  or  5  times  34700,  as  shown 
in  second  step,  will  make  500  times  347.  Hence  173500  is  500 
times  347. 

2.  In  practice  we  multiply  first  bv  the  significant  figure,  and 
annex  to  the  product  as  many  ciphers  as  there  are  ciphers  in  the 
multiplier,  as  shown  in  (2) ;  hence  the  following 

RULE. 

85.  Multiply  by  the  significant  figure  and  annex  as  many 
ciphers  to  the  result  as  there  are  ciphers  in  the  multi2)lier. 


86. 


EXAMPLES     FOR    PRACTICE. 


or 
ch 


(1.) 

(2.) 

(3.) 

(4.) 

Multiply 

34 

256 

573 

968 

By 

50 

70 

90 

60 

(5.) 

(6.) 

(7.) 

(8.) 

Multiply 

3465 

8437 

2769 

4763 

By 

600 

300 

800 

200 

36 


MULTIPLICATION, 


(9.) 

(10.) 

(11.) 

(12.) 

Multiply 

70 

850 

7300 

8300 

By 

40 

80 

600 

900 

(13.) 

(14.) 

(15.) 

(16.) 

Multiply 

326 

184 

972 

262 

By 

80 

700 

500 

20 

87.    Prob.  III.— To  multiply  by  a  number  containing^ 
two  or  more  orders  of  units. 

1.  Multiply  539  by  374. 

(1.)   ANALYSIS. 


r539  X 

539  X  374  =  ^539  X 

(539  X 


374 

4  =      2156 

70  =     37730 

300  =  161700 

201586 


(3.) 
539     Multiplicand. 
Multiplier. 

1st  partial  product. 
2d  partial  product. 
3d  partial  product. 

Whole  product. 


Explanation.— 1.  The  multiplier,  374,  is  analyzed  into  the  parts  4,  70, 
and  300,  according  to  (83). 

2.  The  multiplicand,  53{>,  is  taken  flrt<t  4  times  =  2156  (77);  then 
70  times  =  37730  (84) ;  then  300  times  =  161700  (84). 

3.  4  times  +  70  times  +  300  times  are  equal  to  374  times ;  hence  the 
sum  of  the  partial  products,  2156,  37730,  and  1G1700,  is  equal  to  374  times 
689  =  201.J86. 

4.  Observe,  that  in  practice  we  arrange  the  partial  products  as  shown 
in  (2),  omitting  the  ciphers  at  the  right,  and  placing  the  first  significant 
figure  of  each  product  under  the  order  to  which  it  belongs.  Hence  the 
following 

BULE. 

88.  /.  Write  the  multiplier  under  the  multiplicand,  so  that 
units  of  the  same  order  stand  in  t?ie  same  column. 

11.  Multiply  tlie  multiplicand  by  each  significant  figure  in 
the  multiplier,  successively,  beginning  at  the  right,  and  plaice  the 
right-hand  figure  of  each  partial  product  under  the  order  of  the 
multiplier  used.  Add  the  partial  prodttcts,  tchich  wiM  give  the 
product  required. 


EXAMPLES, 


87 


Proof. — /.  Repeat  the  work.  II.  Use  the  multiplicand  at 
multiplier;  if  the  remdts  are  the  same  the  work  is  probably 
correct. 

EXAMPLES     FOR    PRACTICE. 
80,  Copy  examples  from  Arithmetical  Table  No.  1,  page  12. 

Multiplicand  five  figures,  MtiltlpHer  three, 

1.  Take  the  multiplicands  in  order,  commencing  opposite 
1,  from  columns  a,  b,  c,  d,  e  ;    b,  c,  d,  e,  f  ;    c,  d,  e,  f,  g  ; 

D,  e,  f,  g,  If ;  and  e,  f,  g,  h,  i. 

2.  Take  the  multipliers  in  each  set  from  the  three  right-hand 
columns  used  for  multiplicands,  the  number  immediately  under 
the  multiplicand. 

Multiplicand  six  figures,  Multiplier  five, 

1.  Take  the  multiplicands  in  order  from  columns  a,  b,  c,  D, 

E,  F  ;    B,  C,  D,  E,  F,  G  ;  C,  D,  E,  F,  G,  II  ,  and   D,  E,  F,  G,  H,  I. 

2.  Take  the  multipliers  in  each  set  from  the  five  right-hand 
columns  used  for  multiplicand. 


WRITTEN     EXAMPLES. 

90.  1.  If  you  should  buy  2682  barrels  of  flour,  at  $9  a 
barrel,  and  pay  $15838  down,  how  much  would  you  still  owe 
for  the  flour?  Ans.  $8:i00. 

2.  A  man  left  $2400  to  his  wife,  $3254  to  each  of  his  five 
daughters,  and  the  remainder  of  his  property,  amounting  to 
$4960  to  his  only  son  ;  what  was  the  value  of  his  estate  ? 

3.  Sold  5  oxen  at  $75  each,  3  horses  at  $256  each,  a  carriage 
at  $325,  and  a  plow  for  $25 ;  how  much  did  I  receive  for  the 
whole?  Ans.  $1493. 

4.  I  bought  8  barrels  of  sugar,  at  $54  a  barrel  ;  3  barrels  of 
it  were  spoiled  by  exposure,  but  the  rest  was  sold  at  $72  a 
barrel ;  how  much  did  I  lose  on  the  sugar  ?  Ans.  $72. 


38 


MULTIPLICATION, 


\,1 


It 


5.  There  are  63  ^lons  in  a  liogshcad ;  how  many  gallons 
in  8290  hogsheads  ?  Am.  522:^70. 

6.  If  an  acre  yields  38  bushels  of  wheat,  how  many  bushels 
may  be  raised  on  372  acres  V  Ans,  14i:jG  bushels. 

7.  If  27G  men  cau  do  a  jwece  of  work  in  517  da^s,  in  what 
time  cotdd  one  man  do  the  same  work  ?      A)i8.  l'i2UU2  days. 

8.  A  man  owns  2  orchards,  in  each  of  which  tlici-e  are  21  rows 
of  trees,  with  213  trees  in  each  row  ;  how  many  trees  do  both 
orchards  contain  ?  Ans.  8940  trees. 

9.  I  bought  14  cows  at  39  dollars  each,  and  29  oxen  at  G3  dol- 
lars each  ;  how  much  did  I  pay  for  all?  Ans.  $2373. 

10.  If  2  tons  of  hay,  worth  $13  a  ton,  winter  one  cow,  what 
will  be  the  cost  of  wintering  348  cows  ?  Ans.  $9048. 

11.  Franco  contains  20373G  square  miles,  and  the  popula- 
tion averages  17G  per  square  mile  ;  what  is  the  entire  popula- 
tion? Ans.  35857536. 

13.  A  square  mile  contains  640  acres ;  find  the  cost  of  36 
square  miles  at  $45  an  acre.  Ans.  $1036800. 

13.  What  is  the  cost  of  5  yards  of  cloth  at  $2.25  a  yard. 

Solution.— Since  1  yard  costs  $2.25,  5  yards  will  coi-i  5  times  $2.25, 
which  is  -til. 25. 

Observe,  tliat  when  the  multiplicand  contains  cent?,  wo  multiply  with- 
out regard  to  the  period,  and  insert,  a  period  between  the  second  and  third 
figures  of  the  result.  The  two  figures  at  the  right  express  the  cents  in 
the  answer. 


14.  A  fruit  merchant  bought  295  baskets  of  peaches  at  $1.25 
a  basket ;  finding  that  43  baskets  were  worthless,  he  sold  the 
rest  at  $1.75  ;  how  much  did  he  make  on  the  transaction  ? 

15.  A  farmer  sold  57  bushels  of  beans  at  $2.36  per  bushel, 
and  285  bushels  of  wheat  at  $1.75.  How  much  did  he  receive 
for  both?  Ans.  $633.27. 

16.  A  drover  bought  94  head  of  cattle  at  $39  a  head  and  236 
sheep  at  $3.89  a  head.  He  sold  the  cattle  at  a  gain  of  $9  a 
head  and  the  sheep  at  a  loss  of  $.  75  a  head  ;  what  was  the  total 
amount  of  the  sale,  and  the  gain  on  the  transaction  ? 

Ans.  Amount  of  sale,  $5313.04 ;  Gain,  $729. 


DEFINITIONS, 


3i> 


17.  A  merchant  bought  473  yards  of  cloth  at  $1.25  a  yard  ; 
147  were  damaged  and  had  to  be  sold  at  $.07  a  yard.  He  8f)ld 
the  remainder  at  $1.58  a  yard ;  did  he  gain  or  lose  on  the 
transaction,  and  how  much?  Ana.  $21.99  gain. 

18.  A  mechanic  employed  on  a  building  78  days  received 
$2.75  a  day.  His  family  expenses  during  the  same  time  were 
$1.80  a  day  ;  how  much  did  he  save  ?  Ans.  $09.42. 

19.  A  merchant  purchased  10  pieces  of  cloth,  each  containing 
48  yards,  at  $2.75  a  yard.  He  sold  the  entire  lot  at  an  advance 
of  $.45  per  yard.  How  much  did  he  pay  for  the  cloth,  and 
what  was  his  entire  gain  ? 

20.  Bought  107  bushels  of  wheat  at  $1.65  a  bushel,  and 
287  bushels  of  oats  at  $.37  a  bushel.  I  sold  the  wheat  at  a 
loss  of  4  cents  on  a  bushel,  and  34  bushels  of  oats  at  a  gain  of 
18  cents  a  bushel,  the  remainder  at  a-  gain  of  13  cents.  What 
did  I  gain  on  the  transaction  ? 


a 


DEFINITIONS. 

91.  Multiplication  is  the  process  of  ttiking  one  number 
as  many  times  as  there  are  units  in  another. 

92.  The  Multiplicand  is  the  number  taken,  or  multi- 
plied.      .  • 

93.  The  Multiplier  is  the  number  which  denotes  how 
many  times  the  multiplicand  is  taken. 

94.  The  Product  is  the  result  obtained  by  multipli- 
cation. 

95.  A  Partial  Product  is  the  result  obtained  by 
multiplying  by  one  order  of  units  in  the  multiplier,  or  by  any 
part  of  the  multiplier. 

96.  The  Total  or  Tlliole  Product  is  the  sum  of  all  the 
partial  products. 

97.  The  Process  of  Multiplication  consists,  first. 


40 


MUL  TIP  Lie  A  TION. 


in  finding  partial  products  by  usinf?  the  memorized  results  of 
the  Multiplication  Table ;  second,  iu  uniting  these  partial 
products  by  addition  into  a  total  product. 

08.  A  Ffictor  is  one  of  the  cqnnl  pttrta  of  a  number. 
Thus,  12  is  composed  of  six  2*8,  four  ^'s,  three  4's,  or  two  O's ; 
hence,  2,  3,  4,  and  0  are  factors  of  12. 

The  multiplicand  and  mnltiplior  are  factorH  of  the  product.  Thus, 
87  X  2.5  =  925.  The  product  Wi5  \h  conlpo^'ed  of  (weniy-Jlve  37'h,  or  t/iirly- 
seven  25'f .    Ilcuce,  both  37  and  2.5  are  equal  parts  or  factors  of  926. 

90.  The  Sign  of  Mult nd't cation  is  x ,  and  is  read 
times,  or  multiplied  by. 

When  placed  between  two  number?,  it  denotes  that  either  is  to  be  mul- 
tiplied by  the  other.  Thus,  8x6  shows  that  8  is  to  be  talien  6  times,  or 
that  6  in  to  be  taken  8  times ;  hence  it  may  be  read  either  8  times  ti  or 
6  times  8. 

lOO.  Principles. — /  The  midtiplieand  may  he  either  an 
abstract  or  concrete  number. 

11.  The  multiplier  is  alirays  an  abstract  number. 

in.  The  2)roduct  is  of  the  same  denomination  as  the  mxdtipli 
cand. 


KEVIEW    AND    TEST    QUESTIONS. 

101.     1.  Define   Multiplication,   Multiplicand,  Multiplier, 
and  Product. 

2.  What  is  meant  by  Partial  Product?    Illustrate  by  an 
example. 

3.  Define  Factor,  and  illustrate  by  examples. 

4.  What  are  the  factors  of  G  ?    14?    15?    9?    20?   24?   25? 
27?    32?    10?    30?    50?    and    70? 

5.  Show  that  the  multiplicand  and  multiplier  are  factors  of 
the  product. 

6.  What  must  the  denomination  of  the  product  always  be, 
and  whv  ? 


REVIEW, 


41 


an 
of 


7.  Explain  tlio  procoaa  in  each  of  the  following  cases,  and 
illustrate  by  examples  : 

I.  To  multiply  hy  numbers  less  than  10. 
II.  To  multiply  by  10,  100,  1000,  and  so  on. 

III.  To  multiply  l)y  one  order  of  units. 

IV.  To  inulti])ly  by  two  or  more  order  of  units  (Hti). 
V.  To  multij)ly  l)y  the  factors  of  a  number  (83 — 2), 

8.  Give  a  rule  for  the  third,  fourth,  and  fifth  cases. 

9.  (Jive  a  rule  for  the  shortest  method  of  working  examples 
where  both  the  multiplicand  and  multiplier  have  one  or  more 
ciphers  on  the  right  ? 

10.  Show  how  multiplication  may  be  performed  by  addition. 

11.  Explain  the  construction  of  the  Multiplication  Table, 
and  illustrate  its  use  in  multiplying. 

12.  Why  may  the  cijdiers  be  omitted  at  the  right  of  partial 
products? 

13.  Why  commence  multiplying  the  units'  order  in  the 
multiplicand  first,  then  the  tens',  and  p<^  on  ?  Illustrate  your 
answer  by  an  example. 

14.  Multiply  8795  by  G29,  multiplying  first  by  the  tens,  then 
by  the  hundreds,  and  last  V)y  the  units. 

15.  Multiply  3572  by  483,  commencing  with  the  thousands 
of  the  multiplicand  and  hundreds    f  the  multiplier. 

10.  Show  that  hnndrcds  multiplied  by  hundreds  will  give 
ten  thoymnds  in  the  product. 

17.  Multiplying  thousands  by  thousands,  what  order  will 
the  product  be? 

18.  Name  at  sight  the  loirest  order  which  each  of  the  follow- 
ing examples  will  give  in  the  product  : 

(1.)  8000  X  3000  ;  2000000  x  3000  ;  5000000000  x  7000. 
(2.)  40000  X  20000 ;  7000000  x  4000000. 

19.  What  orders  in  3928  can  be  multiplied  by  each  order  in 
473,  and  not  have  any  order  in  the  product  less  than  thousands? 


^  ■ 


DIVISION. 


<    i 


102.  To  apply  the  Multiplication  Table  in  finding  at  (rigJit 
how  many  times  a  nuriiber  expressed  fry  one  figure  is  contained  in 
any  nutnber  not  greater  than  9  times  the  given  number. 

Pursue  the  following  course  : 

I.  Write  on  your  islute  in  irregular  order  the  products  of  the 
Multiplieaiion  Tablo,  commencing  with  the  products  of  2. 
Write  immediately  before,  the  number  whose  products  you 
La>  e  taken  ;  thus, 

^JJO   .i)l   2)11,,   2)6   2)12  2)10 


/^ 


2.  Write  under  the  line  from  memory  the  number  of  2's  in 
10,  in  4,  in  14,  etc.  When  this  is  done,  erase  each  of  these 
results,  and  rewrite  and  erase  again  and  again,  until  you  can 
give  the  quotients  at  sight  of  the  other  two  numbers. 

3.  Look  at  the  numbers  and  question  yourself.  Thus,  you 
say  mentally,  t^cos  in  ten,  and  you  follow  with  the  answer, 
five ;  tiros  in  four,  tico  ;  twos  m  fourteen,  seven. 

4.  Omit  the  questions  entirely,  and  pass  your  eye  along  the 
exani])les  and  name  the  results  ;  tlius,^cf,  two,  seven,  etc. 

lOJJ,  Practice  as  above  directed  on  each  of  the  following  : 


1. 

3)0 

3)12 

3)15 

3)9 

3)24 

3)18 

3)21 

2. 

4)12 

4)24 

4)32 

^L^ 

4)16 

4)28 

4)20 

3. 

5)15 

5)25 

5)10 

6)30 

5)20 

5)45 

5)35 

4. 

0)  12 

0)24 

6)30 

6)18 

6)^42 

6)54 

6)30 

5. 

7)14 

7  )  35 

7)49 

7)21 

7)42 

7  )  56 

7)28 

6. 

8)  16 

8)40 

8)24 

8)56 

8)32 

8)64 

8)48 

7. 

9)27 

9)45 

9)18 

9)5-1 

9)72 

9)36 

9)03 

PREPARATORY    STEPS. 


43 


DIVISORS    PROM    2    TO    12. 

PREPARATORY    STEPS. 

104.  Step  I. — To  dicide  when  the  quotient  is  expressed  by 
two  or  mare  peaces,  but  contains  only  one  order  of  iinitn. 

1.  Regard  the  dividend  as  made  iuto  equal  parts,  divide  one 
of  these  equal  parts  by  the  given  divisor  aud  multiply  the 
quotient  by  the  number  of  eqm\]  parts  ;  thus, 

Take  for  exampU'  GO  divided  by  3.  VVe  know  6  is  one  of  the 
10  equal  parts  of  GO.  We  know  also  that  there  are  2  threes  in 
G,  and  that  each  G  in  the  GO  must  contain  2  threes.  Then  as  60 
contains  10  times  G.  it  must  contain  10  times  2  threes,  or  20 
threes.     Hence  the  quotient  of  GO  divided  by  3  is  20. 

2.  The  equal  parts  of  the  dividend  which  wo  divide  may  be 
expresst^l  by  two  or  more  fii^ures. 

Take,  for  example,  3500  divided  by  7.  Here  we  divide  first 
the  35  by  7.  We  know  that  35  is  one  of  the  100  e<iual  parts 
of  3500.  We  know  also  that  then;  are  5  sevens  in  35,  and  that 
each  35  in  3500  must  contain  5  sevens.  Wc  know,  therefore, 
that  as  3500  contains  100  times  35,  it  must  contain  100  times 
5  sevens,  or  500  sevens.  Hence  the  quotient  of  3500  divided  by 
7  is  500. 

3.  When  there  is  only  one  order  in  the  quotient,  it  can  be 
given  at  signt  of  the  dividend  and  divisor. 

Thus,  in  dividing  2700  by  9.  you  know  at  once  that  there  are 
3  nines  in  27,  and  hence  that  there  are  300  nines  in  2700. 


n 

Ll.».           EXAMPLES     FOR     ] 

PRACTICE. 

1. 

80  ^2. 

C. 

350  ^  5. 

11.     3G00  -*-  12. 

2. 

90  -f-3. 

7. 

320  -f-  8. 

12.     51()0  -f-    6. 

3. 

60 -f- 2. 

8. 

4200  -^  7. 

13.     .5G(H)  -T-    8. 

4. 

120-^4 

9. 

1^500  H-  5. 

14.     4400  H-  11. 

5. 

180-^9. 

10. 

7200  ^  0. 

15.     9G()0  -T-  12. 

106.     Step  II. — To  dicide  ichen  the  quotient  contains  two 
or  more  orders  of  units. 


44 


DIVISION. 


I  if 


I  4 


n 


Observe  carefully  the  following : 

1.  Each  order  of  the  dividend  may  contain  the  divisor  an 
exact  number  of  times.  In  this  case  the  division  of  each  order 
is  perfonned  independently  of  the  others. 

For  example,  to  divide  888  by  3,  we  may  separate  the  orders 
thus 

800  -^  2  =  400  ) 
888-^8=^    80 -^2=     40  (=444 
8-h2  =      4) 


-=l 


2.  When  each  order  does  not  contain  the  divisor  an  exact 
number  of  times,  we  take  the  largest  part  of  the  dividend 
which  we  know  does  contain  it. 

Thus,  in  dividing  92  by  4,  we  observe  at  once  that  80  is 
the  largest  part  of  the  dividend  which  we  know  contains  4  an 
exact  number  of  times.  We  divide  80,  and  obtain  20  as  the 
quotient.  We  have  now  left  of  the  dividend  undivided  1  ten 
and  2  units,  which  make  12  units.  We  know  that  12  contains 
3  times  4,  and  we  have  already  foimd  that  80  contains 
20  times  4.  Hence  80  +  12,  or  93,  must  contain  20  +  3  or 
23  times  4. 

EXAMPI-ES    FOR    PRACTICE. 

107.  Perform  the  division  in  the  following  examples,  and 
explain  each  step  in  the  process,  as  above : 


1. 

180  -J-  6. 

13. 

85  -«-  5. 

33. 

87  -4-3. 

2. 

192  -=-  6. 

13. 

940  -4-  2. 

24. 

870  -i-  3. 

3. 

272  -J-  8. 

14. 

93-4-4 

25. 

8700  -4-  3. 

4. 

405  -i-  5. 

15. 

240  -f-  8. 

26. 

9800  ^  7. 

5. 

245-5-7. 

16. 

272  -f-  8. 

27. 

8000  -^  5. 

6. 

8888^4. 

17. 

360  -^  9. 

28. 

9000  -T-  6. 

7. 

9693  -4-  3. 

18. 

387  -4-  9. 

29. 

4200-7. 

8. 

684 --2. 

19. 

200  ^  5. 

30. 

4620  -5-  7. 

9. 

90  4-4. 

20. 

4826  -4-  3. 

31. 

3600  -J-  8. 

10. 

84-4-6. 

21. 

0396  -4-  3. 

32. 

4050  H-  9. 

11. 

780  ^  3. 

22. 

8480  -J-  4. 

33. 

2680  -1-  4 

1 


ILLUSTRATION    OF    PROCESS. 


45 


an 
ler 

;r8 


LCt 

ad 

is 
an 
be 
en 
ns 
ns 
or 


id 


ILL.USTRA.TION    OF    PROCESS. 

108.    Prob.  I. — To  divide  any  number  by  any  divisor 
not  greater  than  Z2. 

1.  Divide  986  by  4. 


ANALYSIS. 

4  )  9««  (  300 


4  X  200  = 


4  X  40    = 


4x6 


800_ 

180 
100 

26 
24 

% 


40 


0 

246| 


Explanation.  —  Follow  the  analysis 
and  uoticc  each  step  in  the  procesn ; 
thiiH, 

1.  We  commence  by  dividing  the 
higher  order  of  units.  We  icnow  that  9, 
the  figure  expreiiisiDg  hundreds,  contains 
twice  the  divisor  4,  and  1  remaining. 
Hence  900  contains,  according  to  (106— 
2),  200  times  the  divisor  4,  and  100  re- 
maining. We  multiply  the  divisor  4  by 
200,  and  subtract  the  product  800  from 
986,  leaving  186  of  the  dividend  yet  to  be  divided. 

2.  We  know  that  18,  the  number  expressed  by  the  two  left-hand  flgnres 
of  the  undivided  dividend,  contains  4  times  4,  and  2  remaining.  Hence 
18  tens,  or  180,  conf^ins,  according  to  (106—2),  40  times  4,  and  20  remain- 
ing. We  multiply  the  divisor  4  by  40,  and  subtract  the  product  160  from 
186,  leaving  26  yet  to  be  divided. 

3.  We  know  that  26  contains  G  times  4,  and  2  remaining,  which  is  less 
than  the  divisor,  hence  the  division  is  completed. 

4.  We  have  now  found  that  there  are  200  fours  in  800,  40  fours  in  160, 
and  6  fours  in  26,  and  2  remaining  ;  and  we  know  that  800  +  160  +  26  =  986. 
Hence  986  contains  (200  +  40  +  6),  or  246  fours,  and  2  remaining.  The 
remainder  is  placed  over  the  divisor  and  written  after  the  quotient; 
thus,  246|. 


EXAMPLES    FOR    PRACTICE. 
109.  Solve  and  explain  as  above  each  of  the  following : 


1. 

51  -^3. 

2. 

72-1-2. 

3. 

96-*- 6. 

4. 

a")  -f-  5. 

6. 

98-*- 4. 

6. 

89-*- 7. 

7. 

45-^3. 

8. 

54-1-6. 

9. 

395  -f-  4. 

10. 

367  ^  8. 

11. 

935  -5-  5. 

12. 

895  -*-  7. 

13. 

352  -*-  2. 

14. 

794  -H  5. 

15. 

865-4-9. 

16. 

593  -*-  6. 

17. 

48506  -f-  8. 

18. 

73040  -*-  4. 

19. 

50438  -f-  3. 

20. 

49050  -*-  7. 

21. 

20607  -f-  7. 

22. 

72352  -t-  8. 

23. 

46846  -*-  7. 

24. 

50430  -*-  9. 

^ 


46 


DIVISTON. 


SHORT    AND    LONG    DIVISION    COMPARED. 

no.  Compare  carefully  the  following  forms  of  writing  the 
work  in  division  : 


(1.) 

FORM  USED  FOR  EXPLANATION. 

Two  steps  in  the  process  written. 

4  )  986  (  200 
4x200=      800      40 


4x    40: 


4x     6= 


180    _6 

160    246 

26 
24 


(2.) 

(3.) 

LONG  DITI8I0N. 

SHORT  DIVISION. 

One  njtep  written. 

Entirely  mental. 

4  )  986  (  246 

4)986 

8 

240f 

18 

16 

»■ 

26 

24 

i:  • 

^tl 


Observe  carefully  the  following  : 

1.  The  division  is  performed  by  a  successive  division  of 
parts  of  the  dividend. 

2.  There  are  three  steps  in  the  process :  First,  finding  the 
quotient  figures ;  Second,  multiplying  the  divisor  by  the  quo- 
tient figures ;  IViird,  subtracting  from  the  undivided  dividend 
the  part  that  has  been  divided,  to  find  what  remains  yet  to  be 
divided. 

3.  In  (1),  the  form  for  explanation,  the  numbers  used  in  the 
second  and  third  steps  of  the  process  are  written.  This  is  done 
to  avoid  taxing  the  memory  with  them,  and  thus  concentrate 
the  whole  attention  on  the  explanation. 

4.  In  (2),  the  form  called  Long  Dwision,  the  numbers  used 
in  the  second  step  in  the  process  are  held  in  the  memory,  and 
those  used  in  the  third  step  are  only  partially  written,  the 
ciphers  on  the  right  being  omitted.  This  method  is  always 
used  when  the  divisor  is  greater  than  12. 

5.  In  (3),  the  form  called  Short  Division,  all  the  numbers 
used  in  the  process  are  held  in  the  memory,  the  quotient  only 
being  oxpressed.  This  method  should  invariably  be  used  in 
practice  when  the  divisor  is  not  greater  than  13. 


EX  A?r  PLES. 


47 


AEITHMETICAL    DRILL    TABLE    NO.    2. 


A. 

B. 

c. 

D. 

E. 

F. 

o. 

H. 

I. 

J. 

1. 

'9 

•^ 

J 

7 

0 

8 

G 

9 

-^ 

8 

ti» 

i 

6 

8 

6 

9 

^ 

9 

7 

3 

G 

3. 

^ 

9 

/V 

^ 

G 

3 

7 

5 

4 

9 

4. 

.9 

^ 

2 

^ 

G 

3 

8 

G 

9 

5. 

.5 

7 

6 

8 

^ 

9 

G 

2 

5 

2 

O. 

7 

6' 

8 

3 

7 

5 

) 
O 

8 

4- 

8 

7. 

■# 

■9 

■^ 

9 

0 

G 

7 

3 

s 

8. 

tf 

J 

G 

5 

8 

4 

9 

5 

G 

9 

». 

<9 

5 

2 

7 

3 

8 

^ 

9 

7 

10. 

3 

9 

^ 

2 

9 

3 

8 

7 

9 

3 

11. 

9 

^ 

8 

Jj- 

7 

5 

3 

-^ 

G 

G 

12. 

5 

9 

.9 

8 

5 

9 

G 

8 

I 

9 

EXAMPL'dlS     FOR    PRACTICE. 

111.  Copy,  as  follows,  examples  with  one  figure  in  the 
divisor  from  the  above  Table,  and  perform  the  work  in  each 
case  by  Short  Division.   * 


Tfirce  Figures  in  the  Dividetul. 

1.  Commence  opposite  2,  and  take  the  numbers  for  the 
dividends  from  the  columns  in  the  same  manner  and  order  as 
was  done  in  multiplication. 

2.  Take  as  divisor  the  figure  immediately  above  the  right- 
band  figure  of  the  dividend. 

The  first  six  examples  from  columns  A,  B,  C,  are : 

(1.)  (2.)  (3.)  (4.)  (5.)  (6.) 

3)168        8)424        4)392         2)576       6)768        8)424 


48 


DIVISION, 


Five  Figures  in  the  Dividend. 

1.  Commencing  opposite  2,  take  the  dividends  from  columns 

A,  B,  C,  D,  E  ;     B,  C,  D,  E,  F  ;    C,  D,  E,  F,  O  ;    D,  E,  F,  G,  H  ;    E,  F, 

G,  u,  I ;  and  f,  g,  h,  i,  J. 

2.  Take  as  divisor  the  figure  immediately  above  the  right- 
hand  figure  of  the  dividend. 


ORAL    EXAMPLES. 


k 


li-t 
ii 


112.  1.  A  party  of  ten  boys  went  fishing  ;  they  had  a  boat 
for  every  two  boys ;  how  many  boats  had  they  ? 

Solution.— They  had  as  many  boats  as  2  boys  are  contained  times  in 
10  boys,  which  is  5.    Hence  they  liad  5  boats. 

2.  George  earns  9  cents  a  day ;  how  many  days  must  he  work 
to  earn  27  cents? 

3.  A  man  buys  63  pounds  of  sugar  ;  how  many  weeks  will  it 
last,  if  his  family  use  9  pounds  a  week  ? 

4.  There  are  35  windows  on  one  side  of  a  building,  arranged 
in  5  rows  ;  how  many  windows  in  each  row  ? 

5.  How  many  ranks  of  6  soldiers  each  will  48  soldiers  make  ? 
42  soldiers  ?    60  soldiers  ?    72  soldiers  ? 

6.  At  8  dollars  apiece,  how  many  trunks  can  be  bought  for 
48  dollars  ?    For  56  dollars  ?    For  96  dollars  ? 

7.  When  5  ploughs  cost  $40,  what  is  the  cost  of  3  ploughs  ? 

Solution.— If  5  ploughs  cost  $40,  one  plough  will  cost  as  many  dollars 
as  5  is  contained  times  in  40,  which  is  8.  Hence,  one  plough  costs  $8. 
Three  ploughs  will  cost  3  times  $8,  which  is  $24.    Hence,  etc. 

8.  Daniel  paid  28  cents  for  4  oranges,  and  Luke  bought  7  at 
the  same  price  ;  how  much  did  Luke  pay  for  his? 

9.  If  you  can  earn  54  dollars  in  G  weeks,  how  much  can  you 
earn  in  8  weeks  ? 

10.  When  5  yards  of  cloth  can  be  bought  for  30  dollars,  how 
many  yards  of  the  same  cloth  can  be  bought  for  32  dollars  ? 

11.  When  88  dollars  will  pay  for  11  barrels  of  flour,  how 
many  barrels  can  be  bought  for  74  dollars  ? 


i: 


C: 

f( 


a 

P 

t] 


a 
b 


EXAMPLES. 


49 


"WRITTEN    EXAMPLES. 

1 13.  1.  At  $3  a  cord,  how  many  cords  of  wood  could  be 
bought  for  $093  ?    For  $900  ?  Ansicera.  231 ;  303. 

2.  A  father  left  $9850,  which  he  wished  to  be  divided  equally 
%iuong  his  seven  sons  and  three  daughters ;  how  luuch  did 
each  one  receive  ?  Aus.  $985. 

3.  If  a  man  walk  at  the  rate  of  4  miles  an  hour,  in  how 
many  hours  can  he  walk  840  miles?  Ans.  210  hours. 

4.  How  many  barrels  of  Hour  can  be  made  of  588  bushels  of 
wheat,  if  it  takes  4  bushels  to  make  a  barrel  ?  Ans.  147. 

5.  A  certain  laborer  saves  $13  a  month  ;  how  many  months 
will  it  take  him  to  save  $1453?  Ans.  121  months. 

6.  How  long  will  a  man  be  employed  in  cutting  175  cords  of 
wood,  if  he  cut  7  cords  each  week  ?  Ans,  25  weeks. 

7.  There  are  004800  seconds  in  a  week  ;  how  many  seconds 
in  one  day?  .4 /i«.  80400  seconds. 

8.  How  many  baskets,  each  of  which  holds  6  pecks,  would 
be  needed  to  hold  804  pecks  of  apples  ?  Ans.  134. 

9.  How  many  revolutions  will  be  made  by  a  wheel  11  feet  in 
circumference,  in  running  one  mile,  which  is  equal  to  5280 
feet?  Ans.  480. 

10.  A  man  distributed  $282  among  poor  persons,  giving 
each  $0  ;  how  many  persons  received  the  money?     Ans.  47. 

11.  A  furniture  dealer  expended  $413  in  purchasing  chairs 
at  $7  a  dozen  ;  how  many  dozen  did  he  buy  ? 

12.  A  merchant  expended  $534  in  purchasing  boots  at  $0  a 
pair,  which  he  sold  at  $8  a  pair ;  how  much  did  he  gain  on  the 
transaction  ? 

13.  A  farmer  sold  184  bushels  of  wheat  at  $1.50  per  bushel, 
and  expended  the  amount  received  in  buying  sheep  at  $4  a 
head ;  how  many  sheep  did  he  buy  ? 

14.  A  grain  dealer  sold  912  bushels  of  com  at  $.75  a  bushel, 
and  expended  the  amount  received  in  buying  flour  at  $9  a 
barrel ;  how  many  barrels  of  flour  did  he  purchase? 


60 


DIVISION. 


DIVISORS    GREATER    THAN    12. 

PREPARATORY     STEPS. 

114.  Step  I. — Examples  with  one  order  of  units  in  the 
quotient,  where  the  quotient  figure  can  be  found  at  once  by  divid- 
ing by  the  left-hand  figure  of  the  divisor  ;  thus, 


Divide  13000  by  34. 

34  )  13G00  (  400 
13000 


Here  observe  that  3,  the  Icn-hand  figure 
of  the  divisor,  is  contained  4  times  in  13, 
the  two  left-hand  (i^'ures  of  the  dividend, 
and  that  M  multiplied  by  4  equals  136. 
Hence  34  is  contained  4  times  in  136,  and, 


according  to  (104),  400  times  in  13600. 


EXAMPLES     FOR    PRACTICE. 

115.  Divide  and  explain  each  of  the  following  examples: 


1.  1680-^74. 

2.  2790-4-93. 

3.  3280-4-83. 

4.  3780-4-87. 

5.  6480-5-02. 


6.  47100-7-53. 

7.  51500-J-03. 

8.  5900-4-29. 

9.  33000-4-07. 
10.  59500-5-74. 


11.  765000-4-95. 

12.  107000-5-59. 

13.  280000^46. 

14.  436000^36. 

15.  6230004-89. 


11<>.    Step  II. — Examples  tcith  one  order  in  the  quotient, 
where  the  quotient  figure  must  be  found  by  trial. 

In  examples  of  this  kind,  we  proceed  thus : 


Divide  1709  by  287. 

PTRST  TRIAL. 

287  )  1709  (  8 
2296 


SKCONS  TRIAL. 

287  )  1769  (  7 
2009 


1.  We  divide  as  before  by  2,  the  left-hand 
figure  of  the  divisor,  and  find  the  quotient  8. 
This  course  will  always  give  the  largest  pos- 
ftiUe  quotient  figure.  Multiplying  the  divisor 
287  by  8,  we  observe  at  once  that  the  product 
2296  is  greater  than  the  dividend  1769.  Hence 
287  is  not  contained  8  times  in  1769. 

2.  We  erase  the  8  and  2296  and  try  7  as  the 
quotient  figure.  Multiplying  287  by  7,  we 
observe  agjiin  that  the  product  2009  is  greater 
than  the  dividend  1769.  Hence  287  is  not 
contained  7  times  in  1769. 


EXAMPLES, 


61 


THIRD  TRIAL. 

287  )  1769  (  6 
1782 

47 


3.  We  erase  the  7  and  2009,  and  try  6  as  the  quo- 
tient  figure.  Multiplying  287  by  6,  we  obpcrve  that 
the  product  1722  in  less  than  the  dividend  1769. 
Subtracting  1722  from  1769,  we  have  47  remaining, 
a  number  lef  b  than  the  diviKor  287.  Hence  287  is 
contained  6  timcn  in  1769  and  47  remains. 


EXAMPLES     FOR     PRACTICE. 

117.  Find  the  quoticixts  r.nd  rcmaiuders  in  each  of  the  fol- 
lowing : 


1. 

1194-27. 

8. 

4275  ^4')8. 

15. 

215400-5-356. 

2. 

236-^40. 

9. 

o93(>-5-C43. 

10. 

430900 -T- 588. 

3. 

lOO-f-39. 

10. 

9758 -^r.82. 

17. 

028400^898. 

4. 

410-J-G8. 

11. 

3657^739. 

18. 

80(5700  ^903. 

6. 

248-^38. 

12. 

7890-1-490 

19. 

190000^379. 

6. 

845-^07. 

13. 

4705 -^  (58. 

20. 

587500^-825. 

7. 

605^84. 

14. 

9850-^  39. 

21. 

477400-V-493. 

( , 

22.    784- 

n32. 

043(5-4-27. 

7357 

■^857 

:  .i 

23.    02G- 

1-82. 

8708^40. 

327G8 

■4-760. 

118. 

divisor. 


ILLUSTRATION     OF    PROCESS. 

Prob.  II. — To  divide  any  number  by  any  given 


1.  Divide  21524  by  59. 

59  )  21524  (  364 
177 

882 
354 

284 
236 

48 


Explanation.— 1.  We  find  how  many 
times  the  (livif'or  is  contained  in  the  few- 
est of  the  left-hand  flgurcH  of  the  dividend 
which  will  contain  it. 

59  is  contained  3  times  in  215,  with  a 
remainder  38;  hence,  according  to 
(104-1),  it  ii*  contained  ^iOO  limes  in 
81500,  with  a  remainder  :i800. 

2.  We  annex  the    figure  in  the  next 
lower  order  of  the  dividend  to  the  remain- 
der of  the  previous  division,  and  divide 
the  number  thus  found  by  the  divisor. 
2  tens  annexed  to  380  tens  make  .382  tens.    59  is  contained  6  times  in 
383,  with  a  remainder  28;  hence,  according  to  (104-1),  It  is  contained 
60  times  in  3820,  with  a  remainder  280. 

3.  We  annex  the  next  lower  figure  and  proceed  as  before. 


52 


DIVISION, 


juuci 

each  fitcp. 

step. 

69 

X     300 

^ 

17700 

69 

X        GO 

= 

3540 

59 

X          4 

= 

236 

59 

X      3G4 

= 

21476 

4  units  annexed  to  380  nnlts  make  284  nnits.  59  is  contained  4  times  in 
384,  Willi  a  remainder  of  48,  a  number  smaller  tlmn  tlic  divinor,  hence  tlie 
division  is  completed,  and  we  have  found  that  50  is  contained  364  times  in 
31624,  with  a  remainder  48. 

Observe  carefully  the  following  analysis  of  the  process  in  the 
preceding  example ; 

Multiplying  the  divisor  by     Part  of  dividend     Part  of  divided  dividend 
the  part  of  the  quotient  divided  each  subtracted  from  the 

part  undivided. 
21524 
17700 

3834 
3540 

284 
236 

•f  It 

From  these  illustrations  we  obtain  the  following 


RULE. 

no,  /.  Find  lioic  many  times  t7ie  divisor  is  contained  in 
the  least  number  of  orders  at  the  left  of  the  dividend  that  tnll 
contain  it,  and  vyrite  the  result  for  the  first  figure  of  the 
quotient. 

IT.  Multiply  the  divisor  by  this  quotient  figure,  and  subtra4it 
the  result  from  the  part  of  the  dividend  that  was  used  ;  to  the 
remainder  annex  tJie  next  lower  order  of  the  dividend  for  a  new 
partial  dividend  and  divide  as  before.  Proceed  in  this  manner 
loith  each  order  of  the  dividend. 

III.  If  there  be  at  last  a  remainder,  place  it  after  tJie  quotient, 
vAth  tJie  divisor  underneath. 

Proof. — Multiply  the  divisor  by  tJie  quotient  and  add  the 
remainder,  if  any,  to  the  product.  This  result  will  be 
equal  to  the  dividend,  when  the  division  has  been  performed 
correctly. 


EXAMPLES, 


63 


120. 


EXAMPLES    FOR    PRACTICE. 


1. 

9225  -r-    45. 

2. 

18450  -f-  90. 

3. 

20840  4-  61. 

4. 

280135  -?-  89. 

5. 

17472  -5-  21. 

6. 

255708  -T-  81. 

7. 

72144  -i-    72. 

8. 

9590  -5-  70. 

9. 

137505  -T-  309. 

10. 

59644  -{-  62. 

11. 

467775  -^  105. 

13. 

264375  -^  705. 

13. 

1292928  -^  312. 

14. 

289520  -f-  517. 

15. 

2750283  -i-  603. 

16. 

1143723  -4-  509. 

17. 
1& 
19. 

20. 
21. 
22. 
23. 
24. 
25. 
26. 
27. 
28. 
29. 
30. 
31. 
32. 


13824 

35904 

142692 

1678306 

31809868 

04109742 

5332114 

19014604 

10205721 

7977489 

203812983 

31907835 

61142488 

119836687 

406070736 

330445150 


-i-  128. 
-J-  204. 
H-  517. 
H-  313. 
H-4004. 
-^  706. 
-*-  4321. 
-f-  406. 
-f-  3243. 
-f-  923. 
-^5049. 
-^4005. 
-f-  4136. 
-^  3041. 
-5-  8056. 
-H  3145. 


121.  Additional  examples  for  practice  should  be  taken  from 
Arithmetical  Table  No.  2,  page  47,  as  follows :    . 

Dividend  four  ftgureSf  Biviaor  two. 

1.  Take  the  dividends  in  order  from  columns  A,  B,  C,  D ; 

B,  C,  D,   B ;     C,  D,   E,   P ;     D,   E,  F,  G ;     E,   F,  G,   H  ;     F,   G,  H,   I  ; 
G,  H,  I,  J. 

2.  Take  as  divisors  in  each  set  the  figures  immediately  above 
the  dividend,  in  the  two  right-hand  columns  of  those  used. 


Dividend  six  flgnres.  Divisor  three, 

1.  Take  the  dividends  in  order  from  columns  A,  B,  c,  D,  E, 
F  ;  B,  C,  D,  E,  F,  G  ;  C,  D,  E,  P,  G,  H  ;  D,  E,  P,  G,  H,  I  ;  E,  F,  O, 
H,  I,  J. 

2.  Take  the  divisors  as  before  from  the  three  right-hand 
columns  of  those  used  for  dividend. 


54 


Diviaioir, 


WRITTEN     EXAMPLES. 

122t  1.  A  hogshead  of  molausoH  containB  G3  ^llons  ;  how 
many  hogsheads  in  1GU02  gallons  ? 

Solution.— As  ouo  hogehcad  contains  63  gallons,  10008  gallons  will 
make  a»  many  bogHhuadu  an  (Vi  i»  cuutalncd  timcH  in  10002.  16002  +  G:i  = 
2&1.    Ilcnce  thuru  arc  2&4  bogsticadb  in  HMyif.  gallons. 

2.  An  anny  contractor  furnished  horses,  at  $72  each,  to  the 
amount  of  !j;llol204;  how  many  did  he  furnish  ?  Aji8.  15712. 

3.  A  man  winlies  to  carry  to  market  2C2JJ  bushels  of  jM)ta- 
toes ;  if  he  carries  61  bushels  at  a  load,  how  many  loads  will 
they  makeV  Ans.  43  loads. 

4.  A  man  paid  |1548  for  a  farm  at  the  rate  of  $43  an  acre  ; 
how  manv  acres  did  the  farm  contain  ?  Ans.  30  acres. 

5.  A  certain  township  contains  192000  acres  ;  how  many 
square  miles  in  the  township,  there  being  640  acres  in  a  square 
mile?  Ans.  300  miles. 

6.  How  many  acres  of  land  at  $200  an  acre,  can  be  bought 
for  $53400  ?  Ans.  267  acres. 

7.  A  certain  product  is  43964  and  one  of  the  factors  is  58  ; 
what  is  the  other  factor  ?  Ans.  758. 

8.  At  what  yearly  salary  will  a  man  earn  40800  dollars  in 
34  years?  Ans.  $1200. 

9.  If  light  travels  192000  miles  in  a  second,  in  how  many 
seconds  will  it  travel  691200000  miles?  Ans.  3000. 

10.  Henry  Morgan  divided  $47400  into  3  equal  parts,  one 
of  wliich  he  gave  to  his  wife  ;  the  rest,  after  paying  a  debt  of 
$3280,  he  divided  equally  among  4  children ;  what  did  each 
child  receive  ?  Ans.  ^70S0. 

11.  A  piano  maker  expended  in  one  year  for  material  $20041, 
and  for  labor  $4925,  paying  each  week  the  same  amount ;  whnt 
was  his  weekly  expense  ? 

12.  A  farmer  in  Ontario  raised  in  one  year  13475  '  -wl**^' 
wheat ;  the  average  yield  was  49  bushels  per  acre  ;        »  mui 
acres  did  he  have  sown  ? 


I 


P  h'  /;  /• .  I  A'  .  1  r  O  A'  1'    S  T  KP  s. 


65 


in 


nc 
of 
cli 


DIVISION    BY    FACTORS. 

PREPARATORY     STEPS. 

l!2fl.  Step  I. — Any  munher  may  he  e.rpreased  in  terms  of 
one  of  its  fdctors  by  takimj  anuthev  facto v  as  the  Unit,    (11.) 

Thus,  12  =  4  +  4  +  4  ;  honco,  12  raay  be  expreeetxl  as 
2  fours,  the  four  being  tho  tcnit  of  the  number  3. 

Write  the  following  numbers : 

1.  Express  12  as  2*8  ;  as  'S'h  ;  as  4's  ;  as  C's. 

2.  Express  80  as  3'8  ;  as  Oh  ;  as  IBs  ;  as  12's  ;  as  6'b. 
8.  Ex])reHS  4.")  as  5*8  ;  as  3*8  ;  as  9*8  ;  as  IS's. 

4.  Express  42  and  24  each  as  O'h. 

5.  Express  45  and  225  each  as  U's  ;  as  o's  ;  as  3'8. 

1124,  Stkp  II. —  When  a  number  is  made  into  three  or  more 
factors,  any  tico  or  7nore  of  them  may  be  regarded  as  the  unit 
of  the  71  Umber  expressed  by  the  remaining  factors. 

For  example,  24  =  3  x  4  x  2.  This  may  be  expressed 
thus,  24  =  3  (4  twos).  Hero  the  3  expresses  the  number  of 
4  twos ;  hence,  (4  twos)  is  regarded  as  the  unit  of  the 
number  3. 

Write  the  following : 

1.  Express  12  as  (3  twos) ;  as  (2  twos) ;  as  (2  threes). 

2.  Express  30  as  (3  twos) ;  as  {2  Jives) ;  as  (3  fees).  . 

3.  Express  42  and  126  each  as  (2  snrns) ;  as  (7  threes) 

4.  Exi)res8  75,  225,  and  375  each  as  (5  threes). 

5.  Express  6t),  198,  and  264  each  as  (11  threes)  and  as 
(2  elevens).  ' 

125.  Step  III. —  When  the  same  factor  is  made  the  unit 
of  both  the  dividend  and  divisor,  the  division  is  performed  as  if 
the  numbers  were  concrete. 

Thus,  60  -4-  12  may  be  expressed,  20  threes  h-  4  threes,  and 
the  division  performed  in  the  same  manner  "-  in  6  feet  -4- 
3  feet.  4  threes  are  contained  5  times  in  20  threes  ;  hence, 
12  is  contained  5  times  in  GO. 


56 


DIVISION, 


The  division  may  be  perfonned  in  this  way  when  the  factors 
are  connected  by  the  sign  of  multiplication  ;  thns,  CO  -5-  12  = 
(20  X  3)  -5-  (4  X  3).  We  can  regard  as  before  the  3  rs  the  unit 
of  both  dividend  and  divisor,  and  hence  say,  4  threes  are  con- 
tained  5  times  in  20  threes. 

Perform  the  division  in  eacli  of  the  following  eicamples, 
without  performing  the  multiplication  indicated : 


1. 

25  threes  -*-    5  threes  =  ? 

5.  (64x   9)-*.(  8x   9)  =  ? 

2. 

42  eights  -^    C  eights  =  ? 

6.  (49xl3)-i-(  7x13)  =  ? 

3. 

88  twos    -J-  11  twos    -  ? 

7.  (96  x   7)-T.(12x   7)  =  ? 

4. 

108  fives     ^    9  fives     =  ? 

8.  (78  X  11) -f- (20x11)  =  ? 

ILLUSTRATION    OF    PROCESS. 

126.    Pros.  III.— To  divide  by  using  the  factors  of 
the  divisor. 


Ex.  1.  Divide  375  by  35 

5  )J15 
7  jives  )  03  fives 


Explanation.—*.  The  divisor  .35  =  1  Jives, 
2.  Dividing  the  315  by  5.  we  fiud  that  315  := 

"  3.  The  OS  Jives  contain  9  timee  TJlres ;  hence 

815  contains  9  times  1  Jives  or  9  times  35. 


Ex.2.  Divide  350  by  24. 

2  [359 

3  twos  I  179  twos       and  1  remaining  =     1 

4  (3  twos)    I  59  (3  twos)  and  2  twos  remaining  =    4 

Quotient,     14  and  3  (3  twos)  remaining  =  18 

True  rcmaiuder,  23 

Explanation.— 1.  The divieor 34  =  4x3^2  —  4f3 fttot). 

8.  Dividing  359  by  2,  wo  find  that  3.59  =  179  ftco"  and  1  uuit  rpmaining. 

3.  Dividing  179  twos  by  3  twos,  we  find  that  179  twos  -  r>9  {}i  twos)  and 

3  ttvos  rcma!.iing. 

4.  Dividing  59  (3  twos)  by  4  (3  twos),  we  find  that  50  !3  tivos)  contain 

4  (3  ttvos)  14  times  :tnd  3  (3  twos)  remaining. 

Hence  859,  which  is  equal  to  59  (3  tivos)  and  2  ttr<M>  + 1,  contains  4  (3  twos)^ 
or  94,  }4  times,  and  3  (3  twos)  +  S  twos  +  1,  or  S«,  remaining. 


t€ 

ni 

Wi 

in 


PREPARATORY    STEPS. 


57 


From  these  illustrations  we  obtain  the  following 

RULE. 

127.  /.  Resolve  the  divisor  into  convenient  factors  ;  divide 
the  dividend  by  one  of  these  factors,  the  quotient  thus  obtained 
by  another,  and  so  on  until  all  the  factors  have  been  used.  The 
last  quotient  will  be  the  true  quotient. 

II.  The  true  remainder  isfovnd  by  multiplying  each  remain- 
der, after  the  first,  by  all  the  divisors  preceding  its  own,  and 
finding  the  sum  of  these  products  and  the  first  remainder. 


EXERCISE     FOR    PRACTICE. 

128.    Examples  for  practice  in  dividinpf  by  the  factors  of 
the  divisor: 


t 

.376-^ 

100. 

10. 

19437  -^  40. 

iL 

8975  -4- 

100. 

11. 

13658  -f-  42. 

3. 

76423  -5- 

1000. 

13. 

27780  -r-    60. 

4. 

92708 -j- 

1000. 

18. 

7169  -J-  90. 

5. 

774-^ 

18. 

14. 

4947  -T-  108. 

6. 

873  H- 

24. 

15. 

S0683  -5-  400. 

7. 

4829  ^ 

28. 

16. 

75947  -J-  900. 

8. 

15836  -^ 

30. 

17. 

8460  -H  180. 

0. 

7859-5- 

84 

18. 

14025  H-  165. 

land 
8).     • 


ONE    ORDER    IN    DIVISOR. 

PREPARATORY     STEPS. 

129.    Step  l.—To  divide  by  10,  100,  1000,  etc. 

1.  Observe,  the  figure  in  the  second  place  in  a  number  denotes 
tens,  and  this  figure,  with  those  to  the  left  of  it,  express  the 
number  of  tens.  Hence,  to  find  how  many  tens  in  a  numl)er, 
we  cut  off  the  right-hand  figure. 

Thus,  in  7309  tiie  6  denotes  tens,  and  736  the  number  of  tens 
in  7369 ;  hence  7369  -5-  10  =  786,  and  9  remaining. 


4 


58 


DIVISION, 


2.  In  like  maimer  the  figure  in  the  third  place  denotes 
hundreds,  the  figure  in  the  f mirth  place  thousands,  otc.  Hence 
by  cutting  off  tiDO  figures  at  the  right,  we  divide  by  100 ;  by 
cutting  off  three,  we  divide  by  1000,  etc.  The  figures  cut  off 
are  the  remainder. 

Qlve  the  quotient  and  remainder  of  the  following  at  sight : 


587  -^    10. 
463  -5-  100. 


8973  -5-  100. 
50380  -*-  100. 


73265  ^    1000. 
58307  -^  10000. 


130.  Step  II. — A  number  c-onsisting  of  only  one  order  of 
nnits,  contains  two  factors  which  can  be  given  at  sight. 

Thus,  20  =  2  X  10.    400  =  4  X  100.     7000  =  7  x  1000. 

Observe  that  the  sJi-fuificant  flf^urs  of  the  number,  in  each  case,  is  one 
factor  and  that  the  other  factor  i;*  1  with  as  many  ciphers  annexed  as  there 
are  ciphers  at  the  ri^ht  of  the  eigniflcaut  fl<{ure. 


ILLUSTRATION    OF    PROCESS. 

liil,    Prob.  IV.— To  divide  when  the  divisor  consists 
of  only  one  order  of  units.  '  * 

1.  Divide  8T36  by  500. 
5)87|36 


17  and  230  remaining. 


Explanation.— 1.   We  divide  first 

by  the   factor   100.    This  is  done  by 

cutting:  off  36,  the  nnlts  and  tens  at  the 

right  of  thPdividon(?. 

2.  We  divide  the  quotient,  87  hundreds,  by  the  factor  5.  which  sivos  a 

quotient  of  17  and  2  hundred  remaining,  which  added  to  36  gives  236,  the 

true  remainder. 

EXAMPLES    FOR    PRACTICE. 

132.  Divide  and  explain  each  of  the  following  examples : 

13.  03nf*0  -4-  800. 
1 1.  79:565  --  3000. 
15.  57842  -^  5000. 
10.  90000  ^    900. 

17.  40034  ^    600. 

18.  20306  -h    700. 


1. 

752  -*-  200. 

7. 

8365  -J-  1000. 

o 

/W. 

593  -f-  30. 

8. 

5973  H-  400. 

3. 

80>  -f-  50. 

9. 

62850  -J-  4000. 

4. 

938  -h  600. 

10. 

06462  -i-  6000. 

5. 

452  -i-  300. 

11. 

86352  -i-    900. 

6. 

983  H-  700. 

12. 

49730-*-  800. 

DE  FIN  IT  IONS* 


09 


lenotes 

Hence 

00 ;  by 

cut  off 

orht : 

.    1000. 
-  10000. 

^rdcr  of 

m. 

>e,  is  one 
1  as  Ihcre 


consists 


vide  firpt 

donf  by 

I'lis  at  the 

rh  clvoH  a 
js  236,  the 


nplos : 

^  800. 
-4-  :^000. 
-H  r)000. 

^  000. 
-J-  000. 
-*-    700. 


DEFINITIONS. 

133.  Division  is  tho  iirocess  of  finding  how  maDj  times 
one  number  is  contained  in  another. 

134.  The  Dividend  is  the  number  divided. 

135.  The  Divisor  is  the  number  by  which  the  dividend 
is  divided. 

1 30.  The  Quotient  is  the  result  obtained  by  division. 

137.  The  Remainder  is  the  part  of  the  dividend  left 
after  the  division  is  performed. 

138.  A  l*artial  Dividend  is  any  part  of  the  dividend 
which  is  divided  in  one  operation. 

139.  A  Partial  Quotient  is  any  part  of  the  quotient 
which  expresses  the  number  of  times  the  divisor  is  contained 
in  a  partial  dividend. 

140.  The  I*roeess  of  Division  consists,  J^r.<(^  in  finding 
the  partial  quotients  by  means  of  memorized  results;  second, 

lin  nniltiplying  the  di\i8or  by  the  partial  quotients  to  find  the 
partial  dividends  ;  third,  in  subtracting  the  partial  dividends 

jfrom  the  part  of  the  di\idend  that  remains  undivided,  to  find 
tlu'  i)art  yet  to  Ije  divided. 

141.  Short  Division  is  that  form  of  division  in  which 
lo  step  of  the  process  is  written. 

142.  TjOnff  Division  is  tlint  form  of  division  in  which 
[he  third  step  of  the  i)roce88  is  written. 

143.  The  Sign  of  Division  is  ■^-.  ami  is  read  divided  by. 
\W\ou  placed  between  two  numbers,  it  dfuoti-s  that  the  number 

•f  »n'  it  is  to  be  divide<l  by  the  number  atter  it  ;  thus,  2b  -«-  7 
read  28  divided  by  7. 

I^lvlcion  Ip  aluo  cxprcpsod  hy  plnclni?  the  dividend  above  the  dhi>»or, 
[ith  a  uhort  horizontal  line  between  them ;  Ihue,  Y  i"^  read, ;» ilividod  by  5. 


60 


Dl  VISION, 


144.    Principles.—/.  The  dividend  and  divia&r  muttt  be 
numbers  of  the  same  denomination. 

II.  The  denomination  of  t?ie  quotient  is  determined  by  the 
nnture  of  the  probk in  soloed. 

III.  The  remainder  is  of  the  same  denomination  as   the 
dividend. 


KEVIEW    AND    TEST    QUESTIONS. 

14f'>.     1.  Dofiue  Division,  and  illustrate  each  step  in  the 
process  by  exaraploa 

2.  Explain  and  illustnito  by  examples  Partial  Dividend, 
Partial  Quotient,  and  Uemainder. 

8.  Pre|)arc  two  examples  illustratinjir  each  of  the  following 
problems : 

I.  Given  all  the  parts,  to  find  the  whole. 
II.  Given  t\\o,  whole  and  one  of  the  parts,  to  find  the 
other  part. 

III.  Given  one  of  the  equal  parts  and  the  number  of 

parts,  to  find  the  whole. 

IV.  Given  the  whole  and  the  size  of  one  of  the  parts, 

to  find  the  number  of  parts. 
V.  Given  the?  whole  and  the  number  of  equal  parts,  to 
find  tlu>  Bi'/e  of  one  of  the  parts. 

4.  Show  that  45  can  be  f^xpressed  as  nines,  as  fives,  as  threes. 

5.  What  is  meant  by  true  remainder,  and  how  found? 

6.  Explain  division  by  factors.     Illustrate  by  an  example. 

7.  ^Vhy  cut  off  as  many  figures  at  the  light  of  the  dividend 
as  there  are  ciphers  at  the  right  of  the  divisor  ?  Illustrate  by 
an  example. 

8.  Give  a  rule  for  dividing  by  a  number  with  one  or  more 
cii)hers  at  the  right.  Illustrate  the  steps  in  the  process  by  an 
exam])le. 

9.  Explain  the  difference  between  Long  and  Short  Division, 
and  show  that  the  process  in  both  cases  is  performed 
mentally. 


AP  P  LI  C  A  TlOyS. 


Gl 


nu8t  be 
by  the 
aa   the 


in  the 
ivideiul, 
jUowing 

find  the 
mber  of 
le  parts, 
parts,  to 

IS  threes. 

iple. 
Idivideud 
itratu  by 

or  more 
ss  by  an 

>i  vision, 
jrtbrmed 


10.  Illustrate  each  of  the  following  problems  by  three  ex- 
amples : 

VI.  Uiven  tho  final  quotient  of  a  continued  division, 
the  true  remainder,  and  the  several  divisors,  to 
find  the  dividend. 
VII.  (jJiven  the  i)r()duct  of  a  continued  multiplication 
and  the;  several  multipliers,  to  find  the  multi- 
plicand. 
VIII.  Oiven  tlw  sum  and  the  difference  of  two  numl)er8, 
to  find  the  numbers. 


APPLICATIONS. 

140.  Prob.  I. — To  find  the  cost  when  the  number  of 
units  and  the  price  of  one  unit  is  given. 

Ex.  1.  What  is  the  cost  of  42  yards  of  silk,  at  |2.36  a  yard  ? 

Solution.— If  one  yard  cost  $2M,  H  yards  must  coet  42  times  12.36. 
Ileuce,  *-J.3«  x  42  =  $mA'i,  is  the  cost  of  42  yards. 

Find  the  cost  of  the  following : 

2.  118  stoves,  at  jflS  for  each  stove.  Aus.  |;3384. 

3.  25J)  yards  of  broadcloth,  at  ^2.84  per  yard. 

4.  436  bushels  wh.at,  at  $1.7()  a  bushel.      Atts.  J|;767.3(>. 

5.  2  farms,  each  containing  139  acres,  at  $73.75  jxt  acre. 
«;.  84  tons  of  coal,  at  17.84  per  ton.  Ans.  ^058.56. 

7.  218  barrels  of  apples,  at  $2.90  per  barrel. 

8.  432  yards  cloth,  at  !f;1.75  i)er  yard. 

0.  34C  bushels  of  wheat,  at  !?1.73  a  bushel. 
10.  897  pounds  butter,  at  $.37  per  pound. 

147.  Prob.  II.— To  find  the  price  per  unit  when  the 
cost  and  number  of  units  are  given. 

Ex.  1.  Bought  25  cows  for  $1175  ;  how  much  did  each  cost? 

SoLmoN.— since  85  cows  coPt  $1175,  cnch  cow  cont  an  many  dollars  as 
26  is  contained  times  iu  1173.  Hence,  1175+2.5  =  47,  the  number  of  dollara 
each  cow  cost. 


€2 


AP PLICA  TlOyS, 


Find  the  price  of  the  following  : 

2.  If  42  tons  of  hay  cost  $546,  what  is  the  price  per  ton  ? 
'S.  Bought  268  yards  cloth  for  $804  ;  how  much  did  I  pay  per 
yard?  Ann.  $y. 

4.  Paid  $1029  for  147  barrels  flour;  what  did  I  pay  per 
barrel?  A  us.  $7. 

5.  Sold  190  acres  laud  for  $10102;  how  much  did  I  receive 
an  acre  V  Am.  $52. 

0.  The  total  cost  for  conducting  a  certain  school  for  14  years 
was  $252000  ;  what  was  the  yearly  exi)ense  ?     Ans.  $18000. 

7.  Received  $980  for  28  weeks'  work ;  what  was  my  wages 
per  week?  Ann.  $35. 

8.  A  merchant  pays  his  clerks  for  half  a  year  $1872,  How 
much  is  this  per  week  ?  Ans.  $72. 

148.  PiiOB.  III.— To  find  the  cost  when  the  number 
of  units  and  the  price  of  two  or  more  units  are  giyen. 

1.  At  .$15  for  8  cords  of  wood,  what  is  the  cost  of  39  cords  ? 

Solution  1.— Since  3  cords  C08t  $15,  3f.i  cords  must  cost  as  mauy  tiiuet 
$15  SB  .3  Is  contained  times  in  39.  Hence,  Jirtit  step,  39+3  =  13 ;  second  step, 
$15  X  13  =  ♦1!>.5,  tlic  cost  of  39  cords. 

Solution  2.— Since  3  cords  cost  $15,  eacli  cord  coat  as  many  dollars  as  3 
is  contained  times  in  15 ;  bcncc  cacli  cord  cost  |5,  and  39  cords  cost  39  times 
$5,  or  1195.  Wcncc, first  step,  15+3  =  5;  second  step,  $5x29  =  $195,  the 
cost  of  :39  cords. 

Observe  careftilly  the  difference  between  these  two  solutions.  Let  both 
be  used  in  practice.  Tulve  for  each  example  the  one  by  which  the  divisioa 
can  be  most  readiiy  performed. 


Solve  and  explain  the  following : 

2.  If  0  stoves  cost  $135,  what  is  the  cost,  at  the  same  rate,  of 
84  stoves  ? 

8.  Paid  $28  for  4  barrels  of  flour ;  how  much,  at  the  same 
rate,  will  I  pay  for  164  barrels  ?  Ana.  $1148. 

4.  A  farmer  paid  for  64  sheep  $256  ;  what  is  the  cost,  at  the 
same  rate,  of  793?  An9.  $3173. 


A  PP  LIGATIONS, 


oa 


5.  A  man  travelled  by  railroad  5184  miles  in  0  days  ;  how 
many  miles,  at  the  uame  rate,  will  he  travel  in  54  days  ? 

n.  A  book-keeper  receives  for  his  service  at  the  rate  of  $024 
for  13  weeks  ;  what  is  his  yearly  salary?  Ans.  $2496. 

7.  A  merchant  bought  150  yards  of  cotton  for  $18;  how 
much  will  he  pay,  at  the  same  rate,  for  1350  yards? 

14t>.  PiiOB.  IV.— '."o  find  the  number  of  units  when 
the  cost  and  the  price  per  unit  are  given. 

1.  At  $7  a  ton,  how  many  tons  of  coal  can  be  bought  for 

$6r)8? 

Solution.— since  1  ton  can  be  bought  for  17,  there  can  be  as  many  tons 
boui^'lit  fur  $038  a.s  ^1  is  contuined  titncti  in  |ti58.  Hence,  $6G8-«-|7  c  M, 
the  uumber  of  tona  that  can  be  bought  for  |668. 

Solve  and  explain  the  following : 

2.  For  $935,  how  many  barrels  of  pears  can  be  bought  at  $11 
a  barrel  ?  Ans.  85. 

3.  How  many  horses  can  be  bought  for  $9928,  at  $130  per 


hor 


sc 


Ans.  73. 


4.  A  mechanic  received  at  one  time  from  his  enii)loyer  $357. 
He  was  paid  at  the  rate  of  $21  a  week  :  how  many  weeks  had 
\w  worked?  Ann.  17. 

5.  A  umn  paid  for  a  farm  $6134,  at  $38  per  fvcre;  how  many 
acres  does  the  farm  contain  ? 

0.  Wm.  Henry  paid  for  walnut  lumber  $27795,  at  $85  a 
thousand  feet ;  how  many  thousand  feet  did  he  buy  ? 


150.  Prob.  V. — To  find  the  number  of  units  that  can 
be  purchased  for  a  given  sum  when  the  cost  of  two  or 
more  units  is  given. 

1.  Wlien  8  bushels  of  wheat  can  be  bought  for  $12,  how 
many  bushels  can  be  lM)ught  for  $(5348  ? 

PoLUTioN.— Since  8  bashcl!)  can  be  bouirht  for  $12,  there  can  be  an  many 
times  8  buHhelt*  bought  for  ♦(i:M8  an  $12  is  contained  time»  in  $;G:}-18.  Ileuce, 
flrM  step,  $«»48-»-|t2  =  52{);  nermd  step,  529x8  =  4232,  the  uumber  of 
buHhele  that  can  be  bought  for  $6348. 


:| 


64 


A  PI' LI  ('A  rroNS, 


Solve  and  explain  the  following : 

2.  If  30  pounds  of  sugar  cost  $4,  how  many  pounds  can  be 
bouglit  fur  $375  ?  Ati^.  3375  |)ouuds. 

3.  The  cost  of  4  boxes  of  ortinges  is  $12.  How  many  boxes, 
at  the  same  rate,  can  be  bought  for  $552  ? 

4.  A  farmer  sold  butter  at  $35  a  hundred  jwunds,  and 
received  $1715  ;  how  many  pounds  did  he  sell  ? 

5.  Wlien  peaches  are  sold  at  $0  for  8  baskets,  how  many 
baskets  must  a  man  sell  to  receive  !3!582  ?     Aus.  770  baskets. 

0.  A  carpenter  was  paid  at  the  rate  of  $42  for  12  days,  and 
received  $588 ;  how  many  days  was  lie  employed  ? 

7.  At  $09  for  12  cords  of  wood,  how  many  cords  can  be 
bought  for  $900  ?  Ans.  108  cords. 

REVIEW     EXAMPLES. 


1/51,  1.  I  sold  75  pounds  of  butter  at  20  cents  a  pound, 
and  laid  out  the  proceeds  in  coffee  at  00  cents  a  pound  ;  how 
many  jwunds  of  coffee  did  I  buy  ?  A)is.  25  pounds. 

2.  Bought  a  quantity  of  wood  for  S3959,  and  sold  it  for 
$0095,  thus  gaining  $3  on  each  cord  sold  ;  how  much  wood 
did  I  buy?  An8.  712  cords. 

3.  I  sold  a  farm  of  850  acres  at  $45  an  acre,  and  another  farm 
of  175  acres  at  $75  an  acre  ;  how  much  more  did  the  first  farm 
bring  than  the  second  ?  *4/<«.  $2025. 

4.  I  paid  $8900  for  8  city  lots,  and  sold  them  at  a  loss  of  $12 
on  each  lot ;  how  much  did  I  receive  for  3  of  them? 

5.  The  expenses  of  a  young  lady  at  school  were  $75  for 
tuition,  $20  for  books,  $08  for  clothes,  $17  for  railroad  fare, 
$5  a  week  for  board  for  42  weeks,  and  $30  for  other  exi^enses ; 
what  was  the  total  expense  ?  Ans.  $420. 

0.  I  bought  27  acres  of  land  at  $41  an  acre,  and  20  acres  at 
$27  an  acre,  and  sold  the  whole  at  $43  an  acre ;  how  much  did 
I  gain  or  lose  ?  Ans.  $470  gain. 

7.  What  is  the  total  cost  of  45  acres  of  land  at   $17  an 


APPLICATIONS, 


G5 


acre,  two  horses  at  $132  each,  a  yoke  of  oxen  for  $130,  a  horee- 
rake  for  .t«5,  and  a  plough  for  117  ?  Aiis.  $1241. 

8.  A  grocer  bought  28  barrels  of  apples  for  $84:  how  much 
will  he  pay  at  the  same  rate  for  168  barrels  ?         Ans.  $504. 

9.  The  sum  of  two  numbers  is  73,  and  their  difference  47 ; 
what  are  the  numbers  ? 

Solution.— The  pum  73  is  equal  to  the  greater  unmber  plun  the  lesB, 
and  the  Ichs  number  plus  the  diflcrunce  47,  are  equal  to  the  greater ;  hence, 
if  47  be  added  to  the  Hum  73,  wc  have  twice  the  greater  number.  Ilence, 
jtrst  Htep,  73  +  47  =  130 ;  stoond  t<tej},  120  +  2  =  00,  the  greater  number ; 
(/lird  slepy  GO  — 47  :=  13,  the  lets  number. 

10.  Two  men  owed  together  $3057,  one  of  them  owed  $235 
more  than  the  other  ;  what  was  each  man's  debt  ? 

11.  The  sum  of  two  numbers  is  8076,  and  the  difference  452  ; 
what  are  the  numbers?  Ana.  4714  and  4262. 

12.  A  house  and  lot  are  worth  $7394.  The  house  is  valued 
at  .$2462  more  than  the  h)t ;  what  is  each  worth  ? 

13.  If  15  cows  can  subsist  on  a  certain  quantity  of  hay  for 
10  days,  how  long  will  the  same  suffice  for  3  cinvs  V 

14.  Bought  576  barrels  of  apples  at  $36  for  every  8  barrels, 
and  sold  them  at  $33  for  every  6  barrels  ;  how  much  did  I  gain 
on  the  transaction?  Ans.  $576. 

15.  Bought  at  Guelph  cattle  at  $47  a  head  to  the  amount  of 
•i'2061,  and  sold  them  at  Toronto  for  $3528  ;  what  was  the  sell- 
ing i^rice  per  head  ?  Ans.  $56. 

16.  .\  fanner  i>aid  $22541  for  two  farms,  and  the  difference  in 
the  cost  of  the  farms  was  $3471.  The  ])rice  of  the  farm  for 
which  he  paid  the  smaller  sum  was  $64  an  acre,  and  of  the 
other  $87  an  nrre.     How  many  acres  in  each  farm  ? 

17.  A  fami-r  sold  62  bushels  of  wheat  for  $50,  also  14  cords 
of  wood  at  §5  a  cord.  4  tons  of  hay  at  <^15  a  ton,  and  2  co\r9 
at  !530  apiece  ;  he  took  in  payment  *145  in  money,  a  coat 
worth  !j:50,  a  horse-rake  worth  $21,  and  the  balance  in  clover- 
seed  at  $4  a  bushel ;  how  many  bushels  of  seed  did  he 
receive  ?  Ans.  6  bushels. 


PROPERTIES    OF    NUMBERS. 


DEFINITIONS. 

1512.  An  Integer  is  a  uumber  that  expresses  bow  many 
there  are  in  a  collection  of  whole  things. 

Thus,  8  yards,  12  houses,  32  dollars,  10  tables,  18  windows, 
25  horses,  etc 

153.  An  Exact  Divisor  is  a  number  that  will  divide 
another  number  without  a  remainder. 

Thus,  3  or  5  is  an  exact  divisor  of  15  ;  4  or  6  is  an  exact  divi- 
sor of  24  and  36,  etc. 

All  numbers  with  reference  to  exact  divisors  are  either  prime 
or  composite. 

154.  A  Prime  Number  is  a  number  that  has  no  exact 
divisor  besides  1  and  itself. 

Thus,  1,  3.  5,  7,  11.  13,  17,  19,  23,  29,  31,  37,  etc.,  are  prime 
numbers. 

155.  A   Composite  Number  is  a  number  that   has 
other  exact  divisors  l)e8ides  1  and  itself. 

Thus,  6  is  divisible  by  either  2  or  3  ;  12  is  divisible  by  either 
8  or  4 ;  hence  0  and  12  are  composite  numbers. 

150.  A  Prime  Divisor  is  a  prime  number  used  as  a 
divisor. 

Thus,  in  35  h-  7,  7  is  a  prime  divisor. 

157.  A  Composite  Divisor  is  a  composite  number  used 
as  a  divisor. 

Thus,  in  18  +  6,  0  is  a  comix)site  divison 


EXACT    DIVISION. 


«T 


EXACT    DIVISION. 


158.  The  following  teats  of  exact  division  should  be  care- 
fully studied  and  fixed  in  the  memory  for  future  uae. 

Puup.  I. — A  didsor  of  any  number  is  a  dimorofany  number 
oftimea  that  number. 

Thus,  12  =  3  fours.  Hence,  13  x  6  =  3  fours  x  6  =  18  fours. 
Hut  18  fours  are  divisible  by  4  Hence,  12  x  G,  or  72,  is  divisi- 
ble by  4. 

Prop.  II. — A  divisor  of  each  of  tico  or  more  numbers  is  a 
divisor  of  their  sum. 

Thus,  5  is  a  divisor  of  10  and  30  ;  that  is,  10  =  2  fives,  and 
30  =  «  fives.  Hence,  10  +  30  =  2  fives  +  0  fives  =  8  fives.  But 
8  fives  are  divisible  by  5.  Hence,  5  is  a  divisor  of  the  sum  of 
10  and  30. 

Prop.  III. — A  divisor  of  each  of  two  nmnbcrs  is  a  divisor  of 
thiir  difference. 

Thus,  3  is  a  divisor  of  27  and  15  ;  that  is,  27  =  9  threes  and 
15  =  5  throes.  Hence,  27  —  15  =  1)  threes  —  5  threes  = 
4  times.  But  4  threes  are  divisible  by  3.  Hence  3  is  a  divisor 
of  the  difference  between  27  and  15. 

Prop.  IV.—Ani/  number  ending  with  a  cipJier  is  divisible  by 
the  dicisors  of  10,  viz.,  2  and  5. 

Thus,  370  =  37  times  10.  Hence  is  divisible  by  2  and  5,  the 
divisors  of  10,  according  to  Prop.  I. 

Prop.  V. — Any  nnmber  is  divisible  by  either  of  the  divisoi's  of 
10,  ichcn  its  right-hand  figure  is  divisible  by  the  same. 

Thus,  498  =  490  +  8.    Each  of  these  parts  is  divisible  by  2, 
Hence  the  number  498  is  divisible  by  2,  according  to  Prop.  II. 
In  the  same  way  it  may  be  shown  that  495  is  divisible  by  5. 


68 


FROrERTlEU     OF    M  U  M  U  E  It  S . 


Puop.  VI. — Any  nnmber  ending  with  two  ciphers  is  divisible 
by  the  dmnuTH  of  lUO,  via.,  2,  4,  5,  10,  20,  25,  arul  50. 

'I'huH,  8!HK)  =  bU  times  100.  Hence  is  divisible  by  any  of  the 
divi.soiH  ol"  100,  according  to  Prop.  1. 

Piior.  V'll. — Any  number  is  divisible  by  any  one  of  the  dicisors 
of  100,  irhen  the  number  npnsacd  by  its  two  r'ujht-hund  Jiy tires 
it!  diinnible  by  the  same. 

Thus,  4075  =  4000  +  75.  Any  divisor  of  100  is  a  divisor  of 
4000  (Prop.  VI).  Hence,  any  divisor  of  100  which  will  divide 
75  is  a  divisor  of  4075  (Prop.  II). 

Pkop.  VIII. — Any  number  ending  irith  three  cipfiers  is  divin- 
ble  by  the  divisors  of  1000,  viz.,  2,  4,  5,  8,  10,  20,  25,  40,  50,  100, 
125,  200,  250,  and  500. 

Thus,  83000  =  83  times  1000.  Hence  is  divisible  by  any  of 
the  divisors  of  1000,  according  to  Proj).  I. 

Prop.  IX. — Any  nnmber  is  divitiible  by  any  one  of  the  dirisors 
of  1000,  when  the  number  expressed  by  its  three  right-hind 
figures  is  dirisihle  by  the  same. 

Thus,  02025  =  02000  +  025.  Any  divisor  of  1000  is  a  divisor 
of  02000  (Prop.  VIII).  Hence,  any  divisor  of  1000  which  will 
divide  025  is  a  divisor  of  02025  (Prop.  II). 

Prop.  X. — Any  number  is  divisible  by  0,  if  the  sum  of  its  digits 
is  divisible  by  0. 

This  proposition  may  be  shown  thus : 

(1.)  486  =  400  +  80  +  6. 

(2.)  100  =  00  +  1  •-  11  nines  +  1.  Hence,  400  =  44  nines 
+  4,  and  is  divisible  by  0  with  a  remainder  4. 

(3.)  10  =  0  +  1  =  1  nine  +  1.  Hence,  80  =  8  nines  +  8,  and 
is  divisible  by  0  with  a  remainder  8. 

(4.)  From  the  foregoing  it  follows  that  400  +  80  +  6,  or  486,  is 
divisible  by  0  with  a  remainder  4  +  8  +  6,  the  sum  of  the  digits. 
Hence,  if  the  sum  of  the  digits  is  divisible  by  0,  the  number 
486  is  divisible  by  0  (Prop.  II). 


^XACT    Divisloy, 


G9 


divUible 
y  of  the 

:  divisors 

I  Jiyurea 

ivLsor  of 

II  divide 


is  divisi- 
,  50,  100, 

by  any  of 


ic  di'risors 
ight-Juind 

n  divisor 
hicli  will 


'  its  digits 


Vnov.Xl.^Any  number  is  divisible  by  3,  if  the  sum  of  its 
digiti*  ut  dicisible  by  3. 

Thirt  proposition  i.s  shown  In  the  same  manner  as  Prop.  X  ; 
as  3  divides  10,  100,  1000,  etc.,  with  a  remainder  1  in  each  cu»e. 

Puoi*.  XII.— .l/<iy  number  is  dicisible  by  \ I,  if  the  differmre 
of  tin  sums  oft/ie  digits  in  the  odd  and  eceu  places  is  ziro  or  is 
divisible  by  \l. 

This  may  be  shown  thus  : 

(1.)  4028  =  4000  +  5)00  +  20  +  8. 

(2.)  1000  =  91  elevens  -  1.     Hence,  4000  =  364  elevens  -  4. 

(3.)  100  =  9  elevens  +  1.     Ilene-,  900  =  81  eleven.^  +  9. 

(4.)  10  =  1  eleven  —  1.     Hence,  20  =  2  elevens  —  2. 

(5.)  From  the  fore^oin^  it  follows  that  4928  =  304  eh-vens 
+  HI  elevens  +  2  elevens  —  4  +  9—2  +  8. 

But  -4  +  9-2  +  8  =  11.  Hence,  4928  -  364  elevens 
+  81  elevens  +  2  elevens  +  1  eleven  =  448  elevens,  und  is 
therefore  divisible  by  11. 

The  same  course  of  reasoning  applies  where  the  difference  is 
minus  or  zero.     Hence,  etc. 

EXAMPLES    FOR    PRACTICE. 

1*»5>.  Find  exact  divisors  of  each  of  the  following  numbers 
by  jijiplylng  the  foregoing  tests : 


1. 

470. 

12. 

9375. 

23. 

5478. 

o 

975. 

13. 

15264. 

24. 

3825. 

44  nines    M 

3. 
4. 

23ai. 
4500. 

14. 
15. 

37128. 
28475. 

25. 
26. 

8094. 
3270. 

5. 

8712. 

16. 

47000. 

27. 

3003. 

+  8,  and   M 

6. 

9736. 

17. 

69392. 

28. 

8004. 

1. 

5725. 

18. 

34605. 

29. 

7007. 

or  486, is   M 

8. 

8375. 

19. 

38745. 

80. 

1005. 

le  digits.   ■ 

9. 

6000. 

20. 

53658. 

31. 

9009. 

1  number   ■ 

10. 

a500. 

21. 

25839. 

32. 

3072. 

11. 

3025. 

22. 

21762 

33. 

8008. 

n 


70 


PROPERTIES     OF    NUMBERS, 


PRIME    :^:UMBERS. 

PREPARATORY     PFxOPOSITIONS. 

lOO.  Prop.  I.— All  even  numbert  are  ditmble  by  2  and 
consequently  all  even  numbers,  ejc^pt  .',  are  composite. 

Ilenco,  in  finding  the  prime  numbere,  we  cancel  as  composite 
all  even  numbers  rxcept  2. 

Thus.    3,  4y  5,  0,  7,  $f  0,  10,  11,  a,    and  so  on. 

Prop.  II. — AV/cA  unniber  in  the  stries  of  odd  numbers  is  2 
greater  than  the  munber  immediately  precediny  it. 

Thus,  the  numljers  left  after  cancelling  the  even  numbers 
are. 
8  5  T  •  11  13.  and  so  on. 

8       8  +  3       5  +  3        7  +  2       9  +  2        li  .  2 

Prop.  III. — Tn  the  series  of  odd  numbers,  erery  third  num- 
ber from  i{  is  dir/nihle  by  H,  erery  FiFTn  it  umber  from  5  is 
dirinihlc  by  5,  and  so  on  irith  each  numbtr  in  the  series. 

This  proposition  may  l)e  shown  thus  : 

According  to  Proj;.  II,  the  serit-B  of  odd  numbers  incnmso 
by  2's.  Hence  the  third  number  from  U  is  found  by  adding 
2  three  times,  thus: 

t  5  7  0 

I  8  +  2  3  +  2  +  3  8  +  3-^2  +  2 


From  this  it  will  be  seen  that  9,  the  third  numlM»r  from  3, 
is  composed  of  Jl,  p'us  5{  twos,  and  in  divisible  by  U  (Prop.  II) ; 
and  HO  with  the  third  numlM^r  from  9.  and  so  on. 

By  the  same  courst^  of  reasoning,  onch  fifth  number  in  the 
series,  counting  from  5,  may  U;  shown  to  be  divisible  l)y  5  ; 
and  80  with  any  other  number  in  the  wries  ;  hence  the  follow- 
ing method  of  finding  the  prime  nambcrs. 


PRIME    NUMBERS, 


71 


ILLUSTRATION     OF    PROCESS. 

1CI1.    Pnon— To  find  all  the  Prime  Numbers  from  x  to 
any  given  number. 

Find  all  the  prime  numbers  from  1  to  03. 


1 

S 

5 

7 

9 

11 

13 

15 

a  6 

17 

19 

21 

8  7 

23 

25 

t 

27 

3  » 

29 

ai 

33 

3  11 

35 

6  7 

37 

39 

3  13 

41 

48 

45 

3  S  !•  15 

47 

49 

7 

51 

3  17 

53 

55 

S  11 

01 
9  ]• 

50 

ei 

63 

3  7  »  SI 

Expi-akatios.— 1.  Arrange  the  pcrlos  of  odd  nnmb<>n>  «n  line?,  at  con- 
vcnii-ut  dii'tuiicc)'  from  each  other,  an  Bbown  iu  illustration. 

•i.  Write  3  under  every  third  number  from  3,  5  under  wcryjlffh  nnrabcr 
from  o.  7  under  every  stventh  number  from  7.  nnd  ho  on  with  eaeh  of  tho 
Ml  her  i.umbe.'r. 

3  The  terms  under  which  the  nnmbere  nre  written  are  compof>ite,  and 
ibc  numberi"  written  umlerare  their  fuotorc,  accordini;  to  Prop.  III.  All 
tlic  remniniuf;  number>*  arc  prime. 

Henre  all  tlu^  prinin  nunilK»r8  from  1  to  ftii  rel,2,  3,  5,  7, 
11,  l:],  17.  10,  23,  20,  31,  37,  41,  43,  47,  53,  59,  01. 


EXAMPLES     FOR     PRACTICE. 

\iVZ,     1.  Find  all  tin'  prime  niini1)er8  from  1  to  87. 
2.  Find  all  tin*  prinv  nunilM'rs  froDi  32  to  100. 
•1   Find  nil  tin*  primr  ntiiiilxrs  from  84  to  157. 
4.  Find  all  the  prime  num1)ers  from  1  to  200. 
•°>.  Find  all  tin-  prim<'  n:  ^hers  from  200  to  400. 
0.  Show  by  nn  examine  thnt  eN'ery  seventh  number  from 
sevi-n,  hi  the  eorics  of  odd  numbers,  is  divimibk'  by  seven. 


72 


PROPERTIES     OF    X  r  JI  Ji  E  R  S  , 


i| 


FACTORI^^G. 


PREPARATORY    STEPS. 


10S5.  Step  I. — Find  by  irtupection  all  the  exact  diciaors  of 
euch  of  the  following  numbers,  and  xcrite  them  in  order  on  your 
date,  thus    6  =  3x2,    10  =  5  x  2. 


1. 

G 

10 

14 

15 

21 

22 

26 

2. 

33 

34 

35 

38 

39 

46 

51 

3. 

55 

57 

58 

62 

65 

69 

74 

4. 

77 

79 

82 

85 

86 

87 

01 

5. 

3 

11 

115 

119 

123 

129 

141 

Refer  to  the  results  on  your  slate  and  obsi'rve 

(1.)  Euch  prime  exact  divisor  is  culled  n  prime  factor  of  the 

number  of  whicli  it  is  a  divisor. 
(2.)  Each  number   is   equal   to  the  prtxluct  of    its    pnwjt' 

factors. 

Step  II. — The  same  prime  factor  may  enter  into  a  mtmn  . 
two  or  more  times.  Thus,  18  =  2  x  3  x  3.  Hence  the  prime 
factor,  3,  enters  twice  into  18. 

Resolve  the  following  numbers  into  their  prime  factors,  and 
name  how  many  times  each  factor  enters  into  a  number. 


1. 

4 

8 

16 

32 

64 

9 

27 

2. 

18 

20 

28 

40 

44 

45 

50 

3. 

54 

50 

75 

80 

98 

100 

108 

4. 

71 

25 

49 

125 

121 

213 

343 

one 


are 


DEFINITIONS. 

164.  A  Factor  U  one  of  the  equal  parts  of  a  number,  or 
e  of  its  exact  divisors. 

Thus,  15  is  romposed  of  djices,  or  5  threes;  hence,  5  and  3 
e  factors  o»'  15.       . 


FA  CTORiy  G. 


73 


165.  A  Pritne  Factor  is  a  prime  number  which  is  a 
factor  of  a  given  number. 
Thus,  5  is  a  prime  factor  of  30. 

IGG.  A  Composite  factor  is  a  composite  number  which 
lb  a  factor  of  a  givi  n  nunil>er. 
Thus,  (J  is  a  com|x)site  factor  of  24. 

107.  Pactorhtff  is  the  process  of  resolving  a  coini»o8ite 
number  into  its  factors. 

H58.  An  K' poHcnt  is  a  small  figure  placed  at  the  ri^ht 
of  a  number  and  a  littlo  above,  to  show  how  many  times  tlu; 
nuuib»T  is  used  as  a  factor. 

Tims,  8'  =  3  X  a  X  3  X  3  X  3.  The  5  at  the  right  of  3  denotes 
that  the  3  is  used  5  times  as  a  factor. 

KM),  A  Common  Factor  is  a  number  that  is  a  factor  of 
each  of  two  or  more  numbers. 

Thus,  3  is  a  factor  of  0,  J),  13,  and  15;  hence  is  a  common 
fiictor. 

170.  The  Greatest  Common  Factor  \h  the  greatest 
number  that  is  a  factor  of  each  of  two  or  mon*  numbers. 

Thus,  4  is  the  greatest  number  that  is  a  factor  of  8  and  also 
of  13.     lleu<.e  4  is  the  greatest  common  factor  of  b  and  13. 


ILLUSTRATION     OF    PROCESS. 


171.  Find  the  prime  factors  of  402. 


Explanation— 1.  We  observe  that  the  number  4*i3  is 
dlvicibU'  by  i,  the  Hinallei^t  prlino  nuinber.  Ilonci?  \v««  di- 
vide by  a. 

4.  We  obeerve  that  th«  flrft  quotient,  SWt.  in  divinlble  by 
3,  wliich  lt«  a  prime  number.     Ilfut-e  we  divide  by  .3. 

3.  We  observe  tliaf  the  neennd  (juoilent,  77.  In  dlvicible 
by  7.  whieh  \x  a  prime  number.    Ilenee  we  diviiU*  )>y  7. 
4.  The  third  (luotient,  11,ir*a  prime  uumlM-r,    IIcuco  the  prime  tacturs 
)f  4ti2  are  8,  3,  7,  and  11 ;  that  b,  -«»  =  ^I  x  8  <  7  x  11. 


3  )  4({3 
3)23) 

11 


74 


PROPERTIES     OF    NUMBERS, 


Any  composite  number  may  be  factored  in  the  same  manner 
Hence  the  following 

RULE. 

1712.  Dioide  tJi^  giren  nvinher  hf  any  prime  number  that  is 
an  exact  divisoi',  and  the  reuniting  quotient  by  amtther,  and  so 
continve  the  dimion  uvtil  the  quotient  is  a  prime  number.  The 
several  didsors  and  the  bust  quotient  are  the  rt quirt  d  prims 
factms. 

EXAMPI-ES     FOR     PRACTICE. 


17:J.  Find  the 

prime  factors  of  the  folio v 

k  ing  1 

lumbers: 

1.  «30. 

12. 

19175. 

23. 

9100. 

2.  210. 

13. 

10028. 

24. 

5184. 

3.  1380. 

14. 

1250. 

25. 

8030. 

4.  402. 

15. 

10323. 

20. 

410.5. 

5.  8130. 

10. 

2240. 

27. 

02500. 

0.  1470. 

17. 

0400. 

28. 

81000. 

7.  4301. 

18. 

4515. 

29. 

04000. 

8.  3234. 

19. 

1000. 

80. 

45500. 

9.  11025. 

20. 

2310. 

31. 

10875 

10.  30030. 

21. 

7854. 

32. 

18590. 

11.  14000. 

22. 

54.50. 

33. 

10380. 

CANCELLATION. 


PREPARATORY     PROPOSITIONS. 

174.  Study  carefully  the  following  propositions  : 

Prop.  I. — Rejecting  a  factor  from  a  number  divides  the  num- 
ber by  that  factor. 

Thus,  72  =  24  X  3.  Hence,  njrcting  the  factor  3  fron.  72,  we 
have  34,  the  (piotient  of  72  diviiled  by  3. 

P«OP.  II. — Dividing  both  dividend  and  divisor  by  the  same 
number  docs  not  change  (hv  quotient. 

Thus,  CC  -^  12  --  20  thm-.i  +  4  threes  =  5. 


CA  y  CELL  A  Tioy, 


75 


Obporve  that  the  unit  three,  in  20  threes  -4-  4  threes,  does  not 
in  nny  way  nft'ect  tho  pIzo  of  the  (luotient ;  therefore,  it  may  l)e 
rejected  RntI  the  quotient  will  not  be  clianped. 

Hence,  dividing  both  the  dividend  (>0  and  the  divisor  12  by 
3  does  not  change  tho  quotient. 


ILLUSTRATION     OF    PROCESS. 


1 75.     Ex.  1.  Divide  402  by  42. 

Explanation.— We  divide  I'oth  the  dlvipor 
niid  dividend   l>y  ♦>.    Accoidiii;,'  to  Prop.  1!, 
the  quotient  Ic  uot  changed. 
Hence,  T7+7  =  4<a+42  =11. 


fi  )  402  _  T7  _ 
(5  M2  ~  7  ~ 


Ex.  2.  Divide  05  x  24  x  55  by  39  x  15  x  35. 
vu         8         11 

0^  X  ;^v(  X  0^ 

$0  X  10  X  3^^ 
3  IL  1 


8^11  _88_ 
3  X  7  "21  "^^' 


Explanation.—!.  Wo  divide  nny  factor  In  the  dividend  byanynnmbcr 
that  will  divide  n  factor  in  the  divisor. 

TliiiH,  65  in  the  dividend  and  16  in  the  divisor  are  divided  each  by  5.  lu 
the  >anie  manner,  M  and  .'iS,  i:}  and  39,  S-l  and  3  are  divided. 

The  remainlnf,'  factors,  H  inid  11.  in  the  dividend  are  prime  to  each  of  tho 
rcinalninjj  factors  in  tlie  divis^or.  Hence,  no  further  dlvlHlon  can  l)e  jjcr- 
formed. 

2.  We  divide  the  product  of  8  and  11,  the  remaining  faotort*  in  tho  divi- 
dend, by  the  product  of  .S  and  7,  the  remaininu  factors  in  the  divii««)r,  and 
find  as  a  (jiiotlent  IV.  1  ^^bich,  accordin;?  to  ^114— II),  b  equal  lu  the  quo- 
tient of  65  -  21  X  .15  divided  by  .39  ^  15  x  .35. 


u 


.Ml  similar  cases  may  be  treated  in  the  same  manner  ;  hence, 
the  following 

RULE. 

17(5.  /.  Caned  nil  the  fdetorn  that  arc  common  to  the  (lin- 
den d  dud  dirisor. 

IF.  Din'dr  the  prod  net  of  the  remninlnrj  fartorK  of  (he  dtiidend 
hjf  the  product  of  the  remaining  factors  of  the  ditiitor.  77ie  re- 
sult irill  hi  the  qui>ti>'nt  rcf  fired. 


76 


P  R  0  P  E  RTI E  S     0  F    X  LMUERS, 


\A^RITTEN     EXAMPLES. 


Ans.  16f. 

Am.  \\. 

Ann.  20. 
Ans.  15  J 1. 
Ans.  \m\. 


Ann.  5/, 


4t.' 


177.     1.  Divide  847.')  by  52"). 

2.  Divido  OOOD  by  GOOO. 

3.  Divido  8  X  15  X  40  by  10  X  24. 

4.  Divide  ;];528  by  210. 

5.  Divide  12.">00  by  75, 
0.  Divide  4!)  x  25  x  12  by  10  x  30  x  5. 

7.  Divide  12  >  lU  x  27  by  42  x  14.  Ann.  27. 

8.  Divido  04  x  81  X  25  by  24  x  27.  Ana.  200. 

9.  Multiply  8  times  00  by  5  times  18  and  divide  the  prcMluct 
by  33  times  72.  Ana.  20. 

10.  What  is  the  (luotient  of  10  times  5  times  4  divided  by  8 
times  20?  Ann.  2. 

11.  How  many  barrels  of  Hour,  at  12  dollars  a  barrel,  are 
worth  ns  much  a»s  \{\  rords  of  wood,  at  3  dollars  a  cord  '.' 

12.  If  10,  12,  HI,  and  42  are  the  factors  of  the  dividend,  and 
12,  5,  24,  and  7  aro  the  factors  of  the  diviwjr,  what  is  the 
quotient?  Ann.  42. 

13.  When  a  laborer  can  buy  30  bu.shels  of  potatoes,  at  4 
shillings  a  buHhi^l,  with  the  earnings  of  24  days,  how  many 
shillinps  does  he  earn  a  <lay  ?  Ans.  0  Hhillinp:s. 

14.  At  18  dolhirn  a  week,  liow  many  weeks  must  a  man  work 
to  pny  3  debts  of  180  dollars  eacli  ?  Ans.  30  weeks. 

15.  How  many  loads  of  potatoes,  each  containing  15  bushels, 
at  42  cents  a  bushel,  will  pay  for  12  rolls  of  cariK'ting,  each 
contiiininfjf  .50  yardn,  at  75  cints  n  yard  ?  Ans.  80  loads. 

'«»  How  inan>  pounds  of  tea,  at  72  cents  a  pound,  would  i>ay 
for  3  hogsheads  t>f  sugar,  each  weighing  1404  pounds,  at  15 
cents  a  pound  ?  Ana.  015  jwunds. 

17.  A  man  exchanged  75  bu.shels  of  onions,  at  00  cents  a 
bushel,  for  a  number  of  boxes  of  tea,  containing  25  pounds 
each,  at  54  cents  a  ix)uud  \  how  many  boxes  did  he  receive  ? 


GREATEST    COMMON    DIVISOR,       77 


GREATEST    COMMO::^    DIVISOR. 


PREPARATORY     STEPS. 

1 7H.    Step  I. — Find  hy  inspection  an  exact  ditisor  for  each 
of  the  fulioiting  setn  of  n  u inherit : 

1.  3,  U,  ir»,  and  12.       4.  18,  45,  27.  and  72. 

2.  7.  14,  21,  and  :m.       5.  80,  84,  108,  and  60. 
X    8,  12,  :{«.  and  28.       6.  42,  70,  28,  and  112. 

Stki'  U.—Find  hy  inspection  the  grcdtrat  nunilnr  that  is  an 
41  art  dirisor  of  each  of  the  f  dinting  jmirn  of  niimberti : 

1.     r,.  25.  3.    «,  120.  5.    25,  750. 


7     O 


4.   la,  laoo. 


0.     45,  9000. 


Find  in  the  nam**  manner  the  greatest  exact  divisor  of  the 
following : 
7.     14,  :J5.  0.     3(5,  96.  11.    84.  1^2. 

H.     25,  45.  10.     72,  108.  12.    8H.  121. 

Stkp  III. — ErprcftM  the  vfitnbcrit  in  eacJi  of  the  foregoing 
cvamjilen  in  terms  of  thtir  grintcnt  exact  dirimr. 

ThiiH,  the  jifreatest  exact  divisor  of  16  and  40  is  8,  hence  16 
may  l>e  expresaed  as  2  eights,  and  40  as  5  eights. 

DEFINITIONS. 

1  79.  A  Common  Divisor  is  a  number  that  is  an  exact 
(livisor  of  each  of  two  or  more  niiml)ers. 
Thus,  5  is  a  divisor  of  10,  15,  and  20. 

18<K  Thf  Greatest  Common  Divisor  is  the  gn'atest 
huiuImt  tliat  is  an  exact  divisor  of  each  of  two  or  more 
mnulters. 

Thus,  3  is  the  greatest  exact  divisor  of  each  of  the  cumlHtrs 
0  and  15.     Hence  A  in  tlieir  yfrtatest  common  divisor. 

IHl.  Numl»er8  are  primv  to  each  other  when  they 
httve  no  couMUuu  divisor  U'sitles  1 ;  thus,  8,  9,  25. 


m 


19 


78 


PROPERTIES     OF    A  UMBERS, 


METHOD    BY    FACTORING. 

PREPARATORY    PROPOSITIONS. 
1H*2,  Illustrate  thu  following  propoeitiou  by  examples. 

The  greatest  common  divisor  is  t/ie  product  of  the  prime  fac- 
tors that  a^e  common  to  all  the  given  numbers  ;  thus, 
42=     7x2x3=     7  sixes ; 
60  =  11  X  2  X  3  =  11  sixea 
7  and  11  being  prime  to  each  other,  6  must  be  the  greatest 
common  divisor  of  7  sixes  and  11  sixes.     But  G  is  the  product 
of  2  and  3,  the  common  prime  factors  ;   hence  the  greatest 
common  divisor  of  42  and  04  is  the  product  of  their  common 
prime  factors. 


m 


ILLUSTRATION    OF    PROCESS. 

1  Hli,    PuoB.  I. — To  find  the  Greatest  Common  Divisor 
of  two  or  more  numbers  by  factoring. 

Find  the  greatest  common  divi8<*r  of  08.  70,  and  154. 

m  .  (2.) 

2)70        2)154  2)98 70    J54 

7_)_35         7 )  77         Or,        7  )  4i)      35        77 
5 


2)98 
7)49 


11  7        5 

2x7  =  greatest  common  divisor. 


11 


Exi'LANATioN.— 1,  We  resolve  each  of  the  numberti  Into  their  prime 
factor!*,  as  ohowii  In  (1)  or  (8). 

2  We  obnerve  that  3  uiid  7  are  the  only  prime  toctorn  common  to  all 
the  niimberH,  Hence  the  product  of  2  and  7,  or  14,  according  to  (182;, !» 
thf  greuteft  common  dlvlaor  ot*.KS,  70,  and  154, 

The  greatt'st  common  divisor  of  any  two  or  more  numbers  is 
found  in  the  same  manner  ;  hence  the  following 

RULE. 

184.  Resolve  each  number  into  its  prime  factorx,  and  find 
the  product  of  the  prime  factors  that  are  common  to  all  the 
nnmhi  rs. 


GREATEST     COMMON    DIVISOR, 


79 


EXAMPLES     FOR     PRACTICE. 


IS."*.  Fiiul  the  gn-ntest  common  diviHor  of 


1. 

70. 

15. 

210. 

2. 

ao. 

105. 

3. 

(5;]. 

lo.">. 

117. 

4. 

78. 

i<r,. 

117. 

r.. 

Vi\, 

',>:51, 

:50(J. 

(J. 

11-2. 

19(J. 

272. 

7. 

187, 

221. 

;}23. 

H. 

40."), 

r>(iT. 

324. 

0. 

225. 

525, 

300. 

10. 

«H. 

102. 

238 

11. 

(i(). 

l:i2, 

231 

12. 

105. 

245, 

315 

13. 

138. 

181. 

322 

14. 

105, 

2j?0. 

::4r) 

15. 

147. 

483 

10. 

228. 

2Ti;, 

348 

17. 

8-10. 

312. 

408 

18. 

300. 

315. 

4U5. 

bJTHOD    BY    DIVISION. 


PREPARATORY     PROPOSITIONS. 

18(>.  Lot  tlio  two  followin*^  piojMwitions  be  carefully  studied 
niul  illustrated  by  other  examples,  bofon*  attempting  to  find 
tlu'  greatest  common  divisor  by  this  ineth(xl. 

Phop  I. —  77(/'  greatest  rommon  din'mr  of  two  mtmher^  U  the 
(jrentcd  common  divisor  of  the  smaller  number  and  their  differ- 

enee. 

Thus,  3  is  the  greatest  common  divisor  of  15  and  27. 

Hence,     15  =  5  threes        and        27  =  9  tlirees ; 
and    0  threes  —  5  threes  —  4  threes. 

But  9  and  5  are  prime  to  each  otlier ;  hence,  4  and  5  must  be 
prime  to  each  other,  for  if  not,  their  common  divi.sor  will 
divide  their  sum,  according  to  (158— II),  an<l  Im-  a  common 
divisor  of  9  and  5. 

Therefore,  *{  is  the  greatest  common  divisor  of  5  threes  and 
4  threes,  or  of  15  and  12.  Hence,  tlw>  greatest  common  divis4)r 
of  two  numb«'rs  is  tho  greatest  common  divisor  of  the  smaller 
numlx^r  and  their  ditlVrence. 


.«■■ 
It 


I- 

m 


•3t 


80 


PROPERTIES     OF    NUMBERS, 


22-0=  10 
10  -  0  =  10 
10-0=    4 


Pnor.  \\.  —  T?w  f/rratest  common  divisor  of  two  numbers  U  the 
grentvHt  ronimon  dirimrr  of  (he  nmiiUcr  number  and  the  remainder 
after  the  division  of  the  greater  by  the  Uhh. 

TliJH  proposition  may  Ix-  illuHtruted  tlius: 

1.  Sul)tnut  (I  from  22.  thru  from  the  dif- 
fcr(;iK*<>,  1(1,  etc.,  until  u  remainder  le»s  thau 
0  in  o1)taincd. 

3.  Obscrvt;  tlial  the  number  of  tinu'H  0  lias 
))e«'n  Hubtracted  is  tin- (juotientof  22  divided 
by  0,  and  honre  that  the  remainder,  4,  is  the  remainder  after 
the  divirtion  of  22  by  (I. 

W.  According  to  Prop.  I,  the  jjreatest  common  dlvJKor  of  22 
and  0  is  the  grcatcHt  ('onimon  divisor  of  their  difTerence,  10, 
and  0.  It  is  also,  a<'<'()rdin/j;  to  the  same  Proposition,  the  great- 
ent  connnon  divisor  of  10  and  0.  and  of  4  and  0.  But  4  is  the 
n-mainder  after  division  and  0  the  snuiUer  number.  Hence 
tlie  <i^reatest  ronim<m  divisor  of  2?  and  0  is  tlw  greatest  common 
divisor  of  the  smaller  nuniljer  and  the  remainder  after  division. 


ILLUSTRATION     OF     PROCESS. 

187.    Puon.  II.— To  find  the  Greatest  Common  DiTisor 
of  two  or  more  numbers  by  continued  division. 

Find  the  greatest  annmon  divisor  of  23  and  170. 


28  )  170  (  0 
108 

8  )  2S  (  3 
24 

4)8(2 
• 
0 


K,\rLANATioN.  1.  We  divide  176  by  88, 
find  And  8  for  a  irmnlnder ;  then  we  divide 
2H  hy  R,  iiiid  tlnd  I  Tur  a  remainder;  then  we 
divide  a  t)y  i.  and  find  0  for  a  rcnmiiidcr. 

2.  Afrordinfj  to  Prop.  II.  the  >,'reatert 
common  divisor  of  28  and  17(1  h  the  name  M 
tlic  prratoHf  common  divisor  of  28  and  8,  ftl«o 
of  8  and  4.  Bnt  4  In  the  eroatoHt  common 
divii^or  of  8  and  t.  Flence  4  1h  the  greatont 
common  divisor  of  28  and  17C. 


Note. — If  the  giv«>n  numbers  have  not  a  eommon  factor,  they 
cannot  have  a  common  divisor  ^'n*ater  tlian  unity,  and  are 
either  prime  numbers  or  prime  to  each  other. 


GREATEST    COMMON    DIVISOR, 


81 


The  greatoRt  common  divisor  of  any  two  numlKjre  is  found 
in  the  same  manner  ;  hence  the  following 

188.  Iti'LE. — Divide  the  greater  number  by  the  lens,  tJun  the 
ItHS  nuinbirht/  the  veiriditKhr,  then  thi  fast  dirisor  hi/ the  UtH 
remniinh  r,  (tud  ho  on  until  nothing  remains.  The  taut  dicinor 
i*  the  grin  tent  common  dicutur  sought. 

To  find  tlip  pn'fttoHt  common  divisor  of  throe  or  more  num- 
Im-Mh  l)y  tills  method  we  liuve  tlio  following 

180.  RiiiE. — Find  the  greatest  common  divisor  of  tiro  of  the 
iiunilwri*,  then  of  the  common  dirimr  thus  found  and  a  third 
number,  and  so  on  icith  a  fourth,  ffth,  etc.,  number. 


ARITHMETICAL    DRILL    TABLE    NO.    3. 

ItM).  Taljle  forOral  Exercises  iiiCJreatest  Common  Divisor, 
jilid  for  Oral  and  Written  Exercises  in  Least  Common  Multiple. 


1. 
4. 

8. 

!>. 
M>. 
11. 
12. 


A. 

18 
21 
20 
20 
18 
21 

■'iO 
IS 

')0 


B. 

10 
9 

27 
32 
35 
^2 
42 
10 
^5 

70 

55 


72    48 


c. 

u 

15 

S 

36 

40 

54. 
03 
32 

72 
00 

no 

84 


i>. 

G 

6 

16 

15 

45 


12 
18 


F.   O. 

16      8 


"0 


fl 


40 
56 

27 
60 

77 
60 


2 


12 

48 
28 
72 
63 
80 
22 
96 


24 
12 
10 

24 

14 
56 

48 

SI 

44 

88 
36 


12 
28 
30 
36 
35 
24 
64 
36 
66 
24 
108 


1 


■  \'L 


•^, 


■-h- 


■*m 


^ 


A/. 


IMAGE  EVALUATION 
TEST  TARGET  (MT-3) 


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II 


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2.2 

1.8 


1.25      1.4 

1.6 

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82 


PROPERTIES     OF    NUMBERS, 


ARITHMETICAL    DRILL    TABLE    NO.    4. 

101.  Table  for  Written  Exercises  in  Greatest  Common 
Divisor  and  Least  Common  Multiple. 


A. 

B. 

c. 

D. 

E. 

F. 

1. 

30 

36 

154 

176 

88 

198 

a. 

48 

210 

72 

54 

84 

126 

n. 

252 

396 

264 

480 

220 

792 

4. 

60 

120 

40 

420 

175 

195 

6. 

132 

26j^ 

396 

462 

594 

528 

6. 

UO 

105 

420 

156 

315 

585 

7. 

96 

280 

112 

192 

336 

840 

8. 

198 

315 

297 

693 

567 

594 

9. 

210 

350 

240 

300 

720 

630 

10. 

132 

220 

264 

308 

660 

528 

11. 

168 

480 

504 

420 

252 

540 

13. 

156 

312 

130 

364 

273 

351 

EXAMPLES     FOR    PRACTICE. 

10i2.  Find  the  greatest  common  divisor  of  the  following: 


1. 

357  and  483. 

6. 

385  and  1085. 

3. 

195  and  465. 

7. 

356  and  808. 

3. 

418  and  330. 

8. 

195  and  483. 

4. 

803,  546,-  and  124. 

9. 

546,  4641,  and  364. 

5. 

455  and  1085. 

10. 

465, 1365,  and  215. 

19«$.  Continue  the  practice  in  finding  the  greatest  common 
divisor  of  abstract  numbers  by  taking  examples  from  the 
above  Arithmetical    Tables.     Let    all    the  examples    taken 


e  BEATS  ST     VOilMOy    D  I  VI  SOB.       83 


I.    4. 
Common 

F. 

198 
12G 

792 
195 
52S 
58') 
8JfO 
50^ 
I  GclO 
528 
5J^0 
351 


owing : 


common 
I  from  the 
los    token 


from  Table  No.  3  be  worked  orally  and  in  sets  in  the  same 
manner  as  directed  for  written  exercises. 

JExutnples  with  Two  Auinbcrs. 

194.    FiEST  Set. — Take  examples  for  written  exercises 
from  the  first  line  of  Table  No.  4,  thus  : 


(1.)  30  36 
(2.)  36  154 
(3.)    154    170 


(4.)    176      88 
(^.)      88    196 


Observe  that  in  each  new  example,  the  first  number  taken 
in  the  last  example  is  omitted  and  a  new  number  added. 

Take  in  the  same  manner  examples  from  each  line  in  the 
table. 

19,1.  Second  Set. — Take  examples  from  the  first  and 
second  lines  thus : 


(1.)  30   48 

(3.)  154 

72 

(5.)   88   84 

(2.)  30  210 

(4.)  176 

54 

(6.)  198  126 

To  present  other  examples,  omit  the  first  line  and  use  the 
second  and  thii'd,  then  the  third  and  fourth,  and  so  on  to  the 
bottom  of  the  table. 

190.  TiiiKD  Set. — Take  the  numbers  from  the  first  and 
thh'd  line,  then  from  the  second  andfoyrth,  then  from  the  tJiird 
^nd  fifth,  etc.,  to  the  bottom  of  the  table. 

Examples  with  Uiree  Nntnbers. 

197.    First  Set. — Take  examples  from  each  line,  thus: 


(1.)    30      30    154 
(2.)    36    154    176 


(3.)    154    170      88 
(4.)    170      88    198 


198.  Second  Set. — Take  the  numbers  from  the  first, 
second,  and  third  lines,  then  from  the  second,  third,  and  fourth, 
and  so  on  to  the  bottom  of  the  table. 


>?''>■§ 


m 


■*'  ■  -  t; 


<M 


'M. 


1  •  >!l 


84 


PROPERTIES     OF    NUMBERS, 


WRITTEN     EXAMPLES. 

199.  1.  I  have  rooms  12  feet,  15  feet,  and  24  feet  wide; 
what  is  the  width  of  the  widest  carpeting  that  will  fit  any  room 
in  ray  house?  A  us.  8  feet. 

2.  Divide  the  greatest  common  divisor  of  48,  72,  UO,  and  120 
by  the  greatest  common  divisor  of  21,  30,  39,  and  84, 

3.  In  the  city  of  Montreal,  some  of  the  sidewalks  are 
48  inches  wide,  some  CO  inches,  and  others  72  inches ;  what 
is  the  widest  flagging  that  can  be  used  in  each  of  these  side- 
walks without  cutting  ?  Anfi.  12  inches. 

4.  D.  White  owns  in  Hamilton  3  lots  of  equal  depth,  the  first 
having  a  front  of  72  feet,  the  second  144  feet,  and  the  third 
108  feet,  which  he  wishes  to  divide  into  as  many  lots  as  pos- 
sible having  equal  fronts ;  how  many  feet  will  each  front  con- 
tain ?  Ans.  30  feet. 

5.  A  teamster  agrees  to  cart  132  barrels  of  flour  for  a  mer- 
chant on  Monday,  84  barrels  on  Wednesday,  and  108  barrels 
on  Friday;  what  is  the  largest  number  he  can  carry  at  a  load, 
and  yet  have  the  same  number  in  each  ?        Afis.  12  barrels. 

6.  I  have  a  lot  whose  sides  measure,  respectively,  42  feet, 
84  feet,  112  feet,  and  126  feet ;  I  wish  to  enclose  it  with  boards 
having  the  greatest  possible  unifonn  length  ;  Avhat  will  be  the 
length  of  each  board  ?  Ano.  14  feet. 

7.  A  merchant  has  three  pieces  of  cloth  containing  respec- 
tively 42,  98,  and  84  yards,  which  he  proposes  to  sell  in  dress 
patterns  of  uniform  size.  What  is  the  largest  number  of  yards 
the  dress  i>atterns  can  contain  so  that  there  may  be  nothing 
left  of  either  piece  ? 

8.  If  two  farms  containing  each  nn  exact  number  of  acres 
were  purchased  for  $8132  and  $0270  respectively,  what  is  the 
highest  uniform  price  per  acre  th.at  could  have  been  paid,  and 
in  this  case  how  many  acres  in  each  farm  ? 


LEAST     COJfJlOX    MULTIPLE, 


85 


LEAST    OOMMOjS"    MULTIPLE. 

PREPARATORY     PROPOSITIONS. 

200.  Study  carefully  each  of  the  following  propositions : 

Prop.  I. — A  multiple  of  a  number  contaiTis  as  a  factar  each 

prime  factor  of  the  number  as  muiiy  times  as  it  enters  into  the 

number. 

Thusi,  60,  which  Is  a  multiple  of  12,  contains  5  times  12,  or  5  times 
2x2x3,  the  prime  factors  of  12.  Hence,  each  of  the  prime  factors  of  12 
enters  as  a  factor  into  60  as  many  times  as  it  enters  into  12. 

Prop.  ll.—  T7ie  least  common  multiple  of  two  or  more  given 

numbers  must  contain,  as  a  factor,  each  prime  factor  in  those 

numbers  the  greatest  number  of  times  that  it  enters  into  any  one 

of  tliem. 

Thus,  12  =  2  X  2  X  3,  and  9  =  3x3.  The  prime  factors  in  12  and  9  are  2 
and  3.  A  multiple  of  12,  according  to  Prop.  I,  must  contain  2  as  a  factor 
twice  and  3  once.  A  multiple  of  9,  according  to  the  same  propopition,  must 
contain  3  as  a  factor  twice.  Hence  a  number  which  is  a  multiple  of  both 
12  and  9  must  contain  2  as  a  factor  twice  and  3  twice,  which  is  equal  to 
2  X  2  X  3  X  3  =  36.    Hence  36  is  the  least  common  multiple  of  12  and  9. 


'•,  ?! 


'4    ll 
11'" 


'^i>\ 


DEFINITIONS. 

201.  A  Multiple  of  a  number  is  a  number  that  is  exactly 

divisible  by  the  given  number. 

Thus,  24  is  divisible  by  8 ;  hence,  24  is  a  multiple  of  8. 

202.  A  Conunon  Multifile  of  two  or  more  numbers  is 

a  number  that  is  exactly  divisible  by  each  of  them. 

Thus,  36  Is  divisible  by  each  of  the  numbers  4,  9,  and  12 ;  hence,  36  is  a 
common  multiple  of  4,  9,  and  12. 

2015.  The  Least  Common  Multiple  of  two  or  more 

numbers  is  the  least  number  that  is  exactly  divisible  by  each 

of  them. 

Thus,  24  is  the  least  number  that  is  divisible  by  each  of  the  numbers  6 
and  8 ;  hence,  24  is  the  least  common  multiple  of  6  and  8. 


m 


86  PROPERTIES     OF    NUMBERS, 


METHOD    BY    FACTORING. 

ILLUSTRATION  OF  PROCESS. 

204.  Prob.  I. — To  find,  by  factoring,  the  least  com- 
mon multiple  of  two  or  more  numbers. 

Find  the  least  common  multiple  of  18,  24, 15,  and  35. 

Explanation.— 1.  We  observe  that 
3  is  a  factor  of  18,  a4,  and  13.  Dividing 
these  numbers  by  3,  we  write  the  quo- 
tients with  35,  in  the  second  line. 

2.  Observing  that  2  is  a  factorof  6  and 

8,  we  divide  as  before,  and  find  the  third 

line  of  numbers.    Dividing  by  5,  we  find 

the  fourth  line  of  numbers,  which  are  priiLe  to  each  other;  hence  cannot 

be  further  divided. 

3.  Observe  tlie  divisors  8,  2,  and  5  are  all  the  factors  that  are  common  to 
any  two  or  more  of  the  given  numbers,  and  th  quotients  3. 4,  and  7  are  the 
factors  that  belong  each  only  to  one  number.  Therefore  the  divisors  and 
quotients  together  contain  each  of  the  prime  factors  of  18,  24, 15,  and  35  as 
many  times  as  it  enters  into  any  one  of  these  numbers.  Thus,  the  divisors 
Sand  2,  with  the  quotient  3,  are  the  prime  factors  of  18;  and  so  with  the 
other  numbers. 

Hence,  according  to  (200—11),  the  continued  product  of  the  divisors  3, 
2,  and  5,  and  the  quotients  3,  4,  and  7,  which  is  equal  to  2520,  is  the  least 
common  multiple  of  18,  24, 15,  and  85. 

From  this  illustration  we  have  the  following 


3 

18 

24 

15 

35 

2 

6 

8 

5 

35 

5 

3 

4 

5 

35 

3 

4 

% 

RULE. 

205.  L  Write  the  numbers  in  a  line,  and  divide  by  any 
prime  fci^tor  that  is  contained  in  any  two  or  more  of  them, 
placing  the  quotients  and  the  undivided  numbers  in  the  line 
below. 

II.  Oj)erate  upon  the  second  line  ofnumbem  in  the  mme  man- 
ner, and  so  on  until  a  line  of  numbers  that  are  prime  to  each 
other  is  found. 

III.  Find  the  continued  product  of  the  divisors  used  and  the 
numbers  in  the  last  line  ;  this  icill  give  tlie  least  common  multiple 
of  th^  given  numbers. 


LEAST     rOJfJfO.Y    MULTIPLE, 


87 


Ml 


WRITTEN     EXAMPLES. 

200.  1.  What  is  the  least  number  of  cents  that  can  be 
exactly  expended  in  oranges,  whether  they  cost  4,  o,  or  6  cents 
apiece  ? 

2.  What  is  the  least  common  multiple  of  the  nine  digits  ? 

3.  What  is  the  smallest  quantity  of  milk  that  will  exactly 
fill  either  six-quart,  nine-quart,  or  twelve-quart  cans? 

4.  ^Vhat  is  the  smallest  sum  of  money  that  I  can  exactly  lay 
out  in  calves  at  14  dollars  each,  cowa  at  38  dollars  each,  or 
oxen  at  57  dollars  each  ?  Ans.  798  dollars. 

5.  A  can  lay  42  rows  of  shingles  on  my  house  in  a  day,  and 
B  can  lay  50  rows  ;  what  is  the  least  number  of  rows  that  will 
give  a  number  of  full  days'  work  to  either  A  or  B  ? 

6.  What  is  the  width  of  the  narrowest  street  across  which 
stepping-stones  either  3,  4,  or  9  feet  long  will  exactly  reach  ? 

7.  Three  separate  parties  are  measuring  the  distance  from 
the  city  hall,  Kingston,  to  the  University,  Toronto ;  one  party 
uses  a  chain  33  feet  long,  another  a  chain  66  feet  long,  and  the 
third  a  chain  50  feet  long,  marking  each  chain's  length  with  a 
stake  ;  at  what  intervals  of  space  will  three  stakes  he  driven 
at  the  same  place  ?  Ans.  Every  1650  feet. 


M 


m^ 


METHOD  BY  GREATEST  COMMON  DIVISOR. 

ILLUSTRATION    OF    PROCESS. 

207.  Prob.  II.— To  find,  by  using  the  greatest  com- 
mon divisor,  the  least  common  multiple  of  two  or  more 
numbers. 

Find  the  least  common  multi])le  of  195  and  255. 

Explanation.—!.  We  find  the  greatest  common  divitor  of  195  and  255, 
which  is  15. 

2.  The  greatest  common  divipor,  15,  according  to  (182),  contains  all 
the  prime  factors  that  are  common  to  1'15  and  255.  Dividing  each  of  these 
numbers  by  15,  we  find  the  factors  that  are  not  common,  namely,  13  and  17. 


l-f 


-m 


Mm 


1..^. 


88 


PROPERTIES     OF    N  U  Jif  B  E  R  S . 


3.  The  common  divisor  15  and  the  qnoticnt  13  contain  all  the  prime 
factors  of  195,  and  the  common  divisor  15  and  the  quotient  17  contain  all 
the  prime  factors  of  255. 

Hence,  accor(!'ng  to  (^200—11),  the  continued  product  of  the  common 
divisor  15  and  the  quotients  13  and  17,  which  is  3315,  is  the  least  common 
multiple  of  195  and  255. 

The  least  common  .  uiltiple  of  any  two  numbers  is  found  in 
the  same  manner ;  hence  the  following 

RULE. 

208.  /.  Find  the  grefftent  common  dicisar  of  the  two  given 
numbers,  and  divide  each  of  the  numbers  by  this  divisor. 

II,  Find  the  continued  product  of  the  greatest  common  divisor 
and  the  qvotients ;  this  will  give  the  least  common  midtiple  of 
the  two  given  numbers. 

To  find  the  least  common  multiple  of  three  or  more  numbers 
by  this  method,  we  have  the  foUowinpf 

RULE. 

200*  Find  the  least  common  mtdtiple  of  two  of  them;  then 
find  the  least  common  multiple  of  the  multiple  thus  found  and 
the  third  number ,  and  so  on  with  four  or  more  numbers.   , 


i:XAMPLES     FOR     PRACTICE. 


Ill 


m 


[!* 


210.  Find  the  least  common  multiple  of 


1.  110  and  165. 

2.  91  and  182. 

3.  78  and  195. 

4.  143  and  165. 


5.  385  and  455. 

6.  154  and  231. 

7.  462  and  546. 

8.  364  and  637. 


211.  For  further  practice  take  examples  with  two  numbers 
from  Table  No.  4,  page  82,  as  directed  in  (194),  (195),  and 
(196) ;  and  examples  with  three  numbers  from  Table  No.  3, 
as  directed  in  (197)  and  (198). 

Continue  to  practice  with  abstract  numbers  until  you  can 
find  the  least  common  multiple  of  two  or  more  numbers 
accurately  and  rapidly. 


LEAST     COMMOX    MULTIPLE, 


89 


BEVIEW    AND    TEST    QUESTIONS. 


212.  1.  Define  Prime  Number,  Composite  Number,  and 
Exact  Divisor,  and  illustrate  each  by  an  exami)]e. 

2.  What  is  meant  by  an  Odd  Number?     An  Even  Number? 

3.  Show  that  if  an  even  number  is  divisible  by  an  odd  num- 
ber, the  quotient  must  be  even. 

4.  Name  the  prime  numbers  from  1  to  40. 

5.  Why  are  all  even  numbers  except  2  composite  ? 

6.  State  how  you  would  show,  in  the  series  of  odd  numbers, 
that  every  fifth  number  from  5  is  divisible  by  5. 

7.  What  is  a  Factor  ?    A  Prime  Factor  ? 

8.  What  are  the  prime  factors  of  81  ?    Of  64  ?    Of  125  ? 

9.  Show  that  rejecting  the  same  factor  from  the  divisor  and 
dividend  does  not  change  the  quotient. 

10.  Explain  Cancellation,  and  illustrate  by  an  example. 

11.  Give  reasons  for  calling  an  exact  divisor  a  measure. 

12.  What  is  a  Common  Measure  ?  The  Greatest  Common 
Measure  ?    Illustrate  each  answer  by  an  example. 

13.  Show  that  the  greatest  common  divisor  of  42  and  114  is 
the  greatest  common  divisor  of  42  and  the  remainder  after  the 
division  of  114  by  42. 

14.  Explain  the  rule  for  finding  the  greatest  common  divisor 
by  factoring  ;  by  division. 

15.  Why  must  we  finally  get  a  common  divisor  if  the  greater 
of  two  numbers  be  divided  by  the  less,  and  the  divisor  by  the 
remainder,  and  so  on  ? 

16.  What  is  a  Multiple  ?    The  Least  Common  Multiple  ? 

17.  Explain  how  the  Least  Common  Multiple  of  two  or  more 
numbers  is  found  by  using  their  greatest  common  divisor. 

18.  Prove  that  a  number  is  divisible  by  9  when  the  sum  of 
its  digits  is  divisible  by  9. 

19.  Prove  that  a  number  is  divisible  by  11  when  the  differ- 
ence of  the  sums  of  the  digits  in  the  odd  and  even  places 
is  zero. 


5'  '^  -»  1 


FRACTIONS. 


PREPARATORY    PROPOSITIONS. 

213.    Prop.  I. — Any  thing  regarded  as  a  whole  can  be 
divided  Into  unequal  or  equal  parts ;  thus, 


WHOLE. 


PARTS. 


Hi' 
■iH' 


(1.) 

(2.) 


1.  Equal  parts  of  a  whole  are  called  Fractions. 

2.  Into  what  kind  of  parts  can  a  pear  be  divided  ?     A  bushel 
of  wheat?    A  slate?    A  garden  ?    Anything? 

3.  Make  $12  into  unequal  parts  in  six  ways,  and  into  equal 
parts  in  five  ways  ? 

4.  In  how  many  ways  can  15  be  made  into  equal  parts? 
Into  unequal  parts  ? 

Prop.  II. — The  same  whole  can  be  divided  into  equal  parts  of 
different  s-izea ;  thus,  .        •    • 


WHOLE. 


EQUAL  PARTS. 


Halves. 
Thirds. 
Fourths. 
Fifths. 


1.  Observe,  the  equal  parts  are  named  by  using  the  ordinal 
corresponding  with  the  number  of  parts.  Thus,  when  the 
whole  is  made  into  three  parts,  one  part  is  called  a  third,  when 


PREPARATORY    PROPOSITIONS,       91 


into  four  parts,  one  part  is  called  a  fourth,  and  so  on  to  any 
mimber  of  parts. 

2.  When  the  whole  is  made  into  ten  equal  parts,  what  is 
one  part  called  ?  Into  sixteen  equal  parts  ?  Into  twenty-four  ? 
Into  forty-three  ? 

3.  What  are  the  largest  equal  parts  that  can  be  made  of  a 
whole  ?    The  next  largest  ?    The  next  largest  ? 

4.  What  ib  meant  by  one-half  of  an  apple?  One-third? 
One-fifth  ? 

5.  What  is  meant  by  two-thirds  of  a  line?  Of  an  hour? 
Of  a  day? 

0.  How  would  you  find  the  fourth  of  anything?  The 
seventh?    The  tenth ? 

7.  Find  the  third  of  6.    Of  12.    Of  15.    Of  24.     Of  48. 

8.  If  a  whole  is  made  into  twelve  equal  parts,  how  would 
you  name  three  parts?  Seven  parts?  Five  parts?  Nine 
parts  ? 

9.  How  many  halves  make  a  whole?  How  many  thirds f 
How  many  sevenths?  How  many  tenths?  How  many  fif- 
teenths ? 

Prop.  III. — Equal  parts  of  a  whole,  or  Fractions,  are 

expressed  by  two  numbers  written  one  over  the  other,  with  a  line 
between  them  ;  thus, 

Numerator,        lL    Shows  the  number  of  equal  parts  In  the  fraction. 
Dividing  Line,    ^    Shows  that  4  and  5  express  a  fraction. 
Denominator,     ^    Shows  the  number  of  equal  parts  in  the  whole. 

Read,  Four -fifths. 


,'  t 


1. 

Read  the  following : 

3      7      6       9 

"5>  F»  ^»  TTF? 

I  If. 

* 

2. 

What  does  f  signify 

?      f?      If? 

23  9 

:Vi)  • 

3. 

Express  in  numbers  three-fifths  • 

nine- 

■thirteenths ; 

eleven- 

titteenths. 

4. 

Read  the  following : 

fi  M->  \l 

Uh 

ih  ni 

5. 

Write  in  numbers 

eight-twentieths 

;  twelve-sixteenths;                        j 

fifteen -seventieths ;  nine- 

■fortieths. 

■ 

92 


FHA  t'TIONS. 


C.  What  does  Numerator  mean  ?  Denominator  f  Dividing 
lino  ?     Tirniit  of  a  fraction  ? 

7.  How  is  a  fraction  oxpreflsod  by  numbers? 

8.  Name  tlie  tenns  of  I  Viy.  ifl.  n-  Vh  ill 

9.  Express  in  numbers  seven-nintlis  ?    Nineteen  forty-fifths. 

Prop.  IV. — The  me  or  value  of  the  same  kind  of  equal  parts 
depends  upon  the  size  or  value  of  the  whale  of  which  they  are 
parts;  tlius, 


f 


WHOLE. 


EQUAL  PARTS. 


Hnlvpft, 
IlalvvH, 


1.  Tlio  equal  parts  in  the  illustration,  although  halves  in 
both  cases,  are  unequal  in  size,  because  the  wholes  are  unequal 
in  size. 

2.  Which  is  the  larger,  the  half  of  $4  or  the  half  of  |6  ? 

8.  Which  is  the  smaller,  the  fourth  of  13  inches,  or  the 
fourth  of  20  inches,  and  why? 

4.  If  oft'ored  the  half  of  either  of  two  farms,  which  would 
you  take,  and  why  ? 

Prop.  V. — The  size  or  value  of  the  equal  parts  of  a  whole 
diminish  as  the  number  of  parts  increase^  or  increase  as  the 
number  of  parts  diminish  ;  thus. 


WHOLE. 


EQUAL  PABT8. 


Tttirda. 

Fourths* 

Fifths. 


1.  Which  is  the  greater,  one-half  or  one-third?  One-fourth 
or  one-fifth  ?    One-sixth  or  one-ninth,  and  why  ? 

2.  How  much  is  J  of  $48  smaller  than  J  of  it? 

3.  Upon  what  two  things  does  the  value  of  one-half,  one- 
third,  one-fourth,  one-fifth,  etc.,  depend?  Illustrate  your 
answer  by  two  examples. 


PHE1*AKAT0RY    PROPOSITIONS,       93 


DEPIinTIONS. 


214«  A  Ft*actional  Unit  is  one  of  the  equal  parts  of 
fiuything  regarded  as  a  whole. 

iil5.  A  Fraction  is  one  or  more  of  the  equal  parts  of 
anything  regarded  as  a  whole. 

21($.  The  Unit  of  n  Fraction  is  the  unit  or  whole 
which  is  considered  as  divided  into  equal  parts. 

217.  The  Nnincrator  is  the  number  above  the  dividing 
line  in  the  expression  of  a  fraction,  and  indicates  how  many 
equal  parts  are  in  the  fraction. 

218.  The  Denominator  is  the  number  below  the 
dividing  line  in  the  expression  of  a  fraction,  and  indicates  how 
many  equal  jmrts  are  in  the  whole. 

210,  The  Terms  of  a  fraction  are  the  numerator  and 
denominator. 

220.  Taken  together,  the  tei^ms  of  a  fraction  are  called  a 
Fraction,  or  Fractional  Namher, 

221.  Hence,  the  word  Fraction,  means  one  or  more  of  the 
equal  parts  of  anything,  or  the  expression  that  denotes  one  or 
more  of  the  equal  parts  of  anything. 


REDUCTION". 

PREPARATORY     STEPS. 

222.    Step  I. — A  fraction  is  represented  by  lines  thus :     • 
3  ^^iM  ^■HH  Part  taken. 

3  KMiM^-^a^M       Ttliole. 

Observe  carefully  the  following : 

1.  In  f ,  the  denominator  3  expresses  the  whole,  or  3  thirds, 
and  the  numerator  2  expresses  two  parts  of  the  same  size. 


'ii.ll 


ft 


.  if   'f 

„■'■■!? 


* 


94 


FEA  C  TIONS. 


Hence,  3  equal  lines  for  the  denominator  and  2  equal  lines 
for  the  numerator,  of  the  same  length  as  those  in  the  denom- 
inator, represent  v  orrectly  the  whole,  the  parts  taken,  and  the 
relation  of  the  parts  to  each  other,  as  expressed  by  the  frac- 
tion 


8 
5- 


2.  Represent  by  lines  f ;  f  ;  y?  ;  iV  J  «  ;  i^'a- 
Why  can  the  numerator  and  denominator  of  a  fraction  be 
represented  by  equal  lines  ? 

Step  II. — Show  hy  representing  t?ie  fraction  with  lima  that 
one-half  18  equal  to  two-fourths  ;  XhxxB,  . 

-       %.-\  2        -■ 


Part  takeiit 
Whole, 

1.  By  observing  the  illustration,  it  will  be  seen  that  the  value 
of  the  numerator  and  denominator  is  not  changed  by  making 
ea^h  part  in  each  into  two  equal  parts.  It  will  also  be  seen 
that  when  this  is  done  the  numerator  contains  2  parts  and  the 
denominator  4.     Hence  ^  =  f . 

2.  Show  in  the  same  manner  that  one  half  is  equal  to  three- 
sixths,  four-eighths,  five-tenths,  and  so  on. 

Step  III. — Any  fractional  unit  can,  without  changing  its 
value,  he  divided  into  any  desired  number  of  equal  parts. 

Study  carefully  and  explain  the  following  illustrations : 


'tgeu 


u 


t 


2. 


1 

8 

^  ^^ 

2 
6 

1    ' 

■     sum 

1 

1 

4 

8 
18 

— 

— • 

■         Be 

■      by  tb 

RED  UCTIoy. 


96 


:m 


ORAL    EXAMPLES. 

223.    1.  How  many  tenths  in  I  of  an  orange  ?  How  many 
tifteenths  ?    How  many  twentieths  ?  etc.,  and  why  ? 

2.  How  can  ^  of  a  yard  be  made  into  sixteenths  of  a  yard  ? 

3.  How  many  twelfths  in^?    In^?    In|?    In^? 

4.  Make  |  into  twenty-firsts,  and  explain  the  process. 

5.  Show  by  lines  that  J  =  j*^  ;  that  |  =  tj^  ;  that  i  =  ^. 

6.  Change  |,  without  altering  its  value,  into  a  fraction  con- 
taining 7  equal  part  3 ;  10  equal  parts  ;  25  equal  parts. 


PBINCIPIiES    OP    BEDUCTION. 

224.   Let  each  of  the  following  principles  be  illustrated  by 
the  pupil  with  a  number  of  examples : 

Prin.  I. — The  numerator  and  denominatoi'  of  a  fraction 
represent,  each,  parts  of  the  same  size  ;  thus. 


7 


4 

5 


(1.) 


(3.) 


Observe  in  illustration  (1)  the  denominator  7  represents  the  whole  or 
7  sevenths,  and  the  numerator  3  represents  3  sevenths ;  in  illustration  (2), 
the  denominator  represents  b  fifths^  and  the  numerator  A  fifths.  Hence  the 
numerator  and  denominator  of  a  fraction  represent  parts  of  the  same  size. 

Prin.  H. — Multiplying  both  the  terms  of  a  fraction  by  the 
same  number  does  not  change  the  value  of  the  fraction  ;  thus, 


2 
3 


2x4 
8x4 


8 
12 


Be  particular  to  observe  in  the  illustration  that  the  amount  expressed 
by  the  2  iu  the  numerator  ur  the  3  in  the  denominator  of  |  is  not 


«: 


96 


FRA  CTIO  N8, 


changed  by  making  each  part  into  4  equal  parts;   therefore,  J  and  t\ 
express,  each,  the  same  amoant  of  the  same  whole. 

Hence,  multiplying  the  numerator  and  denominator  by  the  same  number 
means,  so  far  as  the  real  fraction  is  concerned,  dividing  the  equal  partf  in 
ea£h  into  as  many  equal  parts  as  there  are  units  in  the  number  bji  which  they 
are  multiplied. 

Prin.  III. — Dividing  both  terms  of  a  fraction  by  the  same 
number  does  not  change  the  value  of  the  fraction  ;  thus,  , 


12-5-3 


8 

4 


The  amount  expressed  by  the  9  in  the  numerator  or  the  12  in  the  denom- 
inator of  t'?!  Is  not  changed  by  putting  every  3  parts  into  one,  as  will  be 
seen  from  the  illustration. 

Hence,  x"i  and  |  express  each  the  same  amount  of  the  same  whole,  and 
dividing:  the  numerator  and  denominator  by  the  same  number  means  jmt- 
ting  as  many  parts  in  each  into  one  as  there  are  units  in  the  number  by 
which  they  are  divided. 


/»!■ 


hi 


DEFINITIONS. 

*2*2i5.  The   Value  of  a  fraction  is  the  amount  which  it 

represents.  ,  .^ 

226.  Reduction  is  the  process  of  changing  the  terms  of 
a  fraction  without  altering  its  value. 

227.  A  fraction  is  reduced  to  Higher  Terms  when  its 
numerator  and  denominator  are  expressed  by  larger  numbers. 
Thus,  i  =  tV- 

228.  A  fraction  is  reduced  to  Lower  Terms  when  its 
numerator  and  denominator  are  expressed  by  smaller  numbers. 
Thus,  j%  =  I 

229.  A  fraction  is  expressed  in  its  Lowest  Terms  when 
its  numerator  and  denominator  are  prime  to  each  other. 


u 


ILLUSTRATION     OF    PROCESS, 


97 


Thus,  in  f ,  the  numerator  and  denominator  4  and  9  are  prime 
to  each  other;  hence  the  fraction  is  expressed  in  its  lowest 
terms. 

230.  A  Common  Denominator  is  a  denominator  that 
belongs  to  two  or  more  fractions. 

231.  The  Least  Cotnmon  Denominator  of  two  or 

more  fractions  is  the  least  denominator  to  which  they  can  all 
be  red:  V  wed. 

232.  A  Proper  Fraction  is  one  whose  numerator  is 
less  than  the  denominator  ;  as  f ,  f . 

233.  An  Impro2)er  Fraction  is  one  whose  numerator 
is  equal  to,  or  greater  than,  the  denominator  ;  as  f ,  |. 

2*34.  A  3Iixed  Nwinher  is  a  number  composed  of  an 
integer  and  a  fraction ;  as  4f ,  18f . 


ILLUSTRATION    OF    PROCESS. 

235.    Prob.  I.— To  reduce  a  whole  or  mixed  number 
to  an  improper  fraction. 

1.  Reduce  3|  equal  lines  to  fifths. 

WHOLES.  FIFTHS. 


=  IS  fifths  = 


18 


Explanation.— Each  whole  line  is  equal  to  5  J^fths,  as  ehown  in  tiie 
illustration ;  3  lines  must  therefore  bo  equal  to  15 Jlfths.  \hjifths  +  Z fifths 
=  18  fifths.    Hence  in  3J  lines  there  are  V  of  a  line. 

From  this  illustration  we  have  the  following  : 


^H^4 


RULE. 

23C$.  Multiply  the  whole  luimher  by  tlie  given  denominator, 
and  to  the  product  add  the  numerator  of  the  given  fraction,  if 
any,  and  icrite  the  result  over  the  given  denominator. 


98 


FRA  CTIONS, 


EXAMPLES    FOR    PRACTICE. 
237.  Reduce  orally  the  following  : 

1.  In  5  pounds  of  sugar  how  many  fourt?i8  of  a  pound  ? 

Solution.— In  1  pound  of  sugar  there  are  \  fourths ;  hence,  in  5  pounds 
there  are  6  times  i/ourths,  which  Is  \^  of  a  pound. 

2.  In  7  tons  of  coal  how  many  ninths  of  a  ton  ? 

d.  Bow  msLUj  tenths  in  $Qd1    In  42  yards?    In  17  pounds? 

4.  Express  20  slb  fourths.    As  sevenths.    As  hundredths. 

5.  In  $9f  how  many  sevenths  of  a  dollar  ? 

Solution.— In  $1  there  are  7  sevenths.  In  $9  there  must  therefore  be 
9  times  7  sevenths  or  63  sevenths.  63  sevenths  +  3  sevenths  are  equal  to 
66  sevenths.    Hence,  in  $9f  there  are  V  of  a  dollar.  •  '  .s. 

6.  In  12^  acres  how  many  twelfths  of  an  acre? 

7.  How  many  eigJitlis  in  9|  ?    In  11 1  ?    In  7f  ?    In  5|  ? 

Reduce  the  following  to  improper  fractions  : 


8. 

83i. 

12. 

340«. 

16. 

nn 

9. 

45|. 

13. 

462g. 

17. 

3ff. 

10. 

76f. 

14. 

1875. 

18. 

4t^ 

11. 

13H. 

15. 

463^\. 

19. 

3tU- 

238.    Prob.  II.— To    reduce  an   improper   fraction   to 
an  integer  or  a  mixed  number. 


1.  Reduce  9  fourths  of  a  line  to  whole  lines. 
»  =r  9^4  = 


2i 


\il 


Explanation.— A  wholo  line  is  compoeed  of  Af&iiriht<.  Hence,  to  make 
the  ^fourths  of  a  line  into  whole  lines,  we  put  every /ot/r  parts  into  on^,  as 
shown  iu  the  illuntration,  or  divide  the  9  by  4,  which  gives  2  wholes  and  1 
of  ihefourtlis  remaining.    Hence  the  following 


RULE. 

239.  Divide  tlie  numerator  hy  the  denominator. 


EXAMPLES. 


99 


EXAMPLES    FOR    PRACTICE. 


7B  1 


240^  Reduce  and  explain  orally  the  following: 

1.  How  many  bushels  are  -^/  of  a  bushel?    ^■'-  V    y?    y^ 

2.  In  $-"/,  how  many  dollars  ?     In  V  o^  a  yard,  how  many- 
yards  V    In  -^^  of  a  foot,  how  many  feet  ? 

3.  How  many  miles  in  ^^  of  a  mile  ? 


7    • 


Reduce  to  whole  or  mixed  numbers  the  following ; 


4. 

H.I  8 

2S  • 

5. 

508 

Br* 

6. 

w. 

7. 

w. 

8. 

ni 

9. 

^fF. 

10. 

^tW. 

11. 

ntl^ 

12. 

W^ 

13. 

-'bW- 

14. 
15. 
16. 
17. 
18. 


50705 
9  0067 

nsoi 

80«6 
3F~' 


241.    Prob.  III. — To  reduce  a  fraction  to  higher  terms. 
1.  Reduce  f  of  a  line  to  twelfths. 

I  2x4_  8_ 

I  *  3x4~  12 


Explanation.— 1.  To  make  a  whole,  which  is  already  in  thirds,  into  12 
equal  part8,  each  third  must  be  made  miofcmr  equal  parts. 

2.  The  numerator  of  the  given  fraction  expresses  2  thirds,  and  the  denom- 
inator 3  thirds;  making  each  third  in  both  mXo  four  equal  i)arts  (224—11), 
as  shown  in  the  illustration,  the  new  numerator  and  denominator  will  each 
contain  4  times  as  many  parts  as  in  the  given  fraction. 

Hence,  \  of  a  line  is  reduced  to  tice^ths  by  multiplying  both  numerator 
and  denominator  by  4. 

Hence  the  following  rule  for  reducing  a  fraction  to  higher 
terms : 


%l 


RULE.  _.. 

242.  Divide  the  required  denominator  by  the  denominator 
of  the  given  fraction,  and  multiply  the  terms  of  the  given  frac- 
tion hy  the  quotient. 


I 


100 


FEA  CTIONS, 


EXAMPLES    FOR    PRACTICE. 

243.    Reduce  and  explain  orally  the  following  : 

1.  How  would  you  make  liahea  of  an  apple  into  fourths? 
luU)  .sixths?    Into  tenths?    Into  siosteentTis ? 

2.  How  many  twelft/ts  in  f  of  a  rx)rd  o'  wood  ? 

8.  Explain  how  i^,  ^,  and  f  can  be  reduced  to  twentieths. 

4.  Show  by  the  use  of  lines  that  f  =  *  =  ^  =  -^^  =  |§. 

5.  Reduce  |,  f ,  f,  ^,  and  ^f  each  to  forty -seconds. 

6.  In  f  how  many  ninety-eighths  ? 

7.  Change  f ,  f^,  ^,  f ,  and  ^^  each  to  360ths. 

8.  Reduce  f ,  ^,  ^,  ^f ,  and  ft  to  165ths. 


244.     Pros.    IV.— To    reduce    a    fraction   to  lower 

terms. 


Reduce  ^^j  of  a  given  line  to  fourths. 
12^3  "■ 


f 
4 


ExPLAKATiON.— 1.  To  make  into  4  equal  parts  or  fourths  a  whole  which 
is  already  in  12  equal  parte,  or  twelfths,  every  3  of  the  12  parts  must  be 
put  into  one. 

2.  The  numerator  of  the  given  fraction  expresses  9  twelfths,  and  the  de- 
nominator 12  twelfths  ;  putting  every  3  twelfths  into  one,  in  both  (224— III), 
as  shown  in  the  illustration,  the  new  numerator  and  denominator  will  each 
contain  one-third  as  many  parts  as  in  the  given  fraction. 

Hence  tt  of  a  line  is  reduced  to  fourths  by  dividing  both  numerator  and 
denominator  by  3. 

Hence  the  following  rule  for  reducing  a  fraction  to  its 
lowest  terms : 

BULB. 

245.  Reject  from  tJie  terms  of  the  giten  fraction  aU  their 
common  factors.    Or, 

Divide  the  terms  of  the  given  fraction  hy  their  greatest  common 
divisor. 


EXAMPLES. 


101 


EXAMPLES    FOR    PRACTICE. 

24G.    Reduce  and  explain  orally  the  following  : 

1.  In  ^  of  a  bushel,  how  many  thirds  of  a  bushel? 

2.  How  can  twelfths  of  a  bushel  be  made  into  fourths  of 
bushel?     \u\jo  tJiirdaf     Into  halcesf 

3.  Reduce  ^^  of  a  dollar  to  Jijths  of  a  dollar. 

4.  Show  by  the  use  of  lines  that  ^^  =  ig  =  yV  =  I  =  i« 

5.  Reduce  ^\  to  its  lowest  terms.  ■^^.  |g.  ||. 
C.  Express  if  in  parts  8  times  as  great  in  value. 
Reduce  the  following  to  their  lowest  terms  : 

7.  VV\.  10.    Uh  13.    fM.  16. 

8.  ^Vtr-  11.    Uh  14.     ,%¥..  17. 


To* 


9.    Ut 


12. 


847 
88  0 


15. 


fro* 

U^  0  U* 
324 


18. 


61  8  4 


347.    Prob.  V. — To   change   fractions   to    equivalent 
ones  having  a  common  denominator. 

1.  Reduce  f  and  f  of  a  line  to  fractions  having  a  common 
denominator.  *" 

i  3x4  ~  12 


(1.) 


'  '*  fa 

m 


■A' 


"•1 

I- 
ill 


8 

4 


8x3 
4x3 


9^ 
13 


(3.) 


Explanation.— 1.  We  And  the  least  common  multiple  of  the  denomina- 
tors 3  and  4,  which  is  12. 

2.  We  reduce  each  of  the  fi^ctions  to  twelfths  (141),  as  shown  in  illua- 
trations  (1)  and  (2). 

Hence  the  following 


i'';i 


BUTiE. 

248.  Find  the  least  common  multiple  of  all  the  denominators 
for  a  common  denominator;  divide  this  by  each  denominator 


^I'f: 


102 


FRA  CTIONS. 


separately,  and  multiply  the  corresponding  numerator  by  the 
quotient,  and  lorite  the  product  over  the  common  denominator. 

EXAMPLES    FOR    PRACTICE. 

2249.    Reduce  and  explain  orally  the  following  : 

1.  Reduce  |  and  j^  to  sixths,     f  and  f  to  twelfths. 
3.  Change  f  and  f  to  fractions  having  the  same  denominator, 
and  explain  each  step  in  the  process. 

3.  Express  f ,  /^  and  f  as  fortieths. 

4.  What  is  the  least  common  denominator  of  |,  f ,  and  j  ? 

Observe,  fractions  have  a  least  common  denominator  when  their  denom- 
inators are  alike  and  there  is  no  factor  common  to  all  the  nnmerators  and 
the  common  denominator. 

Reduce  the  following  to  their  least  common  denominator  : 

5.  f,  1,1,  and  |.  9.  i,  |,  f ,  A, /f,  and  H- 

6.  I  j%,  and  \l  10.  H.  M»  H.  aiid  ^. 

7.  |,f,f,  and,^.  11.  f,  I,  uV.  FT.  is4^.  and  VW- 

8.  ^y,  Af,andf  13.  h  j^,  \h  ^,  and  ^. 


ADDITION. 

PREPARATORY    PROPOSITIONS. 

250.  Prop.  I. — Fractional  units  of  the  same  kind,  th/xt 
are  fractions  of  the  sams  whole,  a/re  added  in  the  same  manner 
as  integral  units. 

Thus,  f  of  a  yard  can  be  added  to  |  of  a  yard,  because  they 
are  each  fifths  of  one  yard.  But  |  of  a  yard  cannot  be  added 
to  I  of  a  day. 


Solve  orally  the  following  : 

1.  f  +  i  +  f . 

2.  f  +  f  +  f . 

3.  A  +  A  +  H- 


Y   +    T    +   f- 


4. 

5.    {1  +  -h  +  ^z- 

6.       A    +    if    +    yV 


EXAMPLES. 


103 


the 


ator, 


lenom- 
jrs  and 


Lor 


V^- 


manner 


^se  tliey 
added 


7.  In  I  +  i^  +  V  of  a  yard,  how  many  yards  7 

8.  How  many  are  %{^  +  ^^V  +  $t^  +  $ii  +  lA  ^ 

9.  Find  the  sum  of  f  J  +  i  |  +  sV  +  ^V  +  ?t  miles. 

10.  Why  cannot  f  of  a  bushel  and  |  of  a  peck  be  added  as 
now  expressed  ? 

Prop.  II. — Fractions  expressed  in  different  fractional  units 
must  he  changed  to  equivalent  fractions  having  the  same  frac- 
tional unit,  before  tJiey  can  he  added. 

For  example,  f  and  f  of  a  foot  cannot  be  added  until  both 
fractions  are  expressed  in  the  same  fractional  unit.  Thus,  f  of 
a  foot  is  equal  -^^^  of  a  foot,  and  f  of  a  foot  is  equal  -^^^  of  a  foot ; 
A  + 1  ?  ®^  ^  ^^*  —  \h  o'  1 A  f®®**  Hence  the  sum  of  |  +  5  of  a 
foot  —  1^*^^  feet. 

Find  orally  the  sum  of  the  following : 


1. 

l^^ii- 

4. 

tV  +  M- 

7. 

f  +  f  +  A. 

2. 

f  +  f- 

5. 

%   +  ^Z' 

8. 

1  +  A  +  |. 

3. 

h  +  1- 

6. 

^  +  ^' 

9. 

1  +  f  +  ii- 

251.    Prob.  I.— To  find  the  sum  of  any  two  or  more 
given  fractions. 

1.  Find  the  sum  of  |  +  f  +  |. 

ErPLANATioN.  —  1.  We  reduce  the 
fractions  to  the  same  fractional  unit,  by 
reducing  them  to  their  least  common 
denominator,  which  is  72  (247). 

2.  We  find  the  sum  of  the  numera- 
tors, 155,  and  write  it  over  the  common 
denominator,  72,  and  reduce  Y>^  to  2i^. 


4 

32 

y 

-72 

5 

60 

6 

~72 

7 

63 

8 

"72 

155 

72 


=  2M. 


The  sum  of  any  number  of  fractions  may  be  found  in  the 
same  manner ;  hence  the  following 


ii 


m 
'"i  ^i 


BULE. 

/.  Change  the  fractions  to  equivalent  ones  having  the 
host  common  denominator,  then  add  the  numerators,  write  the 


104 


FRA  CriONS. 


remit  over  the  common  denominator^  and  reduce,  when  poss^le, 
to  lower  terms  or  to  a  whole  or  mixed  number. 

II.   When  there  are  mixed  nnmbers  or  integers,  add  the  frac- 
tions and  integers  separately,  then  add  the  results. 


TATRITTEN    EXAMPLES. 


!1 
•I 
-  I 


.'!  i 


253.    Find  the  sum  of  each  of  the  following : 


1. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 


i,f,i?,andH. 

h  h  I.  h  and  \, 
■I,  f ,  J,  \,  and  \. 
I,  \,  and  ,V 
»,  I,  and  \. 


8    1    a 
S>  ¥»  ?> 
1    n 

4)  6« 


and  §. 


and  f . 


8        H         15      anA     6  3 
9>  Iff'  Tff>  »""  'Si' 


10. 
11. 
13. 
13. 
14. 
15. 
16. 
17. 
18. 


a* 


i,  7],  and  8 

I,  T«^,  ^i  and  ^. 

If.  2J,  and  4i. 

8^,  2§,  3§,  and  4f. 

4^,  2},  ^,  and  ^. 

H,4J,T\,andH. 

4|,  lOJ,  and  83-JI. 

8J,  25^,  19,  and  68A- 

68|,  28i.  32?,  7^\,  and  6Bi 


9. 

19.  John  Munro  has  lOJ  acres  of  land  in  one  field,  10^  in 
another,  and  llj^  in  a  third ;  how  many  acres  has  he  in  the 
three  fields  ?  Ans.  32 y^  acres. 

20.  There  are  three  tubs  of  butter,  weighing,  respectively, 
44J  pounds,  56^  pounds,  and  78|  pounds  ;  how  much  butter  in 
the  three  tubs  ?  Ans.  Yl^ ^^  \iowjiAQ. 

21.  I  have  a  board  7f  feet  long,  another  11 1  feet  long,  and  a 
third  9^  feet  long;  what  is  their  united  length  ? 

22.  How  many  yards  in  three  remnants  of  silk,  containing, 
respectively,  2}  yards,  IJ  yards,  and  2|  yards? 

23.  William  earned  3f  dollars,  his  father  gave  him  5^^  dol- 
lars, and  his  brother  gave  him  lyV  dollars  more  than  his  father  ; 
how  much  money  did  he  have  in  all  ?  Ans.  14^  dollars. 

24.  Three  pieces  of  cotton  contain,  respectively,  43f,  54f,  and 
87f  yards  ;  how  many  yards  in  all  ? 

25.  H.  Weston  travelled  42  y^^  miles  on  Monday,  30f  miles  on 
Tuesday,  48J^  miles  on  Wednesday,  and  25 J  miles  on  Thursday ; 
how  far  did  he  travel  during  the  four  days  ? 


EXAMPLES, 


105 


^ie. 


afi- 


in  the 


iCres. 
itively, 
.tier  in 
nds. 
and  a 

gaining, 

ij^^dol- 
I  father; 
lUars. 
^4f ,  and 

lileB  on 
lorsday ; 


SUBTRACTION, 

254.  Prop.  I. — Fractional  units  of  the  same  kind  that 
are  fractions  of  the  same  whole  are  subtracted  iu  the  same  man- 
ner as  integral  units. 

Thus,  7  ninths  —  5  ninths  =  2  ninths,  or  J  —  j  ==  !• 
Perfonn  orally  the  subtraction  in  the  following: 

j» 

Toflf 

Prop.  II. — Fractions  expressed  in  different  fractional  units 
must  be  reduced  to  the  same  fractional  unit  hefm'e  subtracting. 


1.  f-f. 

3.  if  -  tV. 

5.  Jl  -  if. 

7. 

iiJ 

3.  if  -  ^. 

4.  il  -  H- 

ft     io«  40 

"•   TO  8         10  5- 

8. 

87 

TOO 

Thus,  in  I  —  -ji'y  we  reduce  the  |  to  sixteentJis  ;  | 

if  -  Tir  =  T^ir ;  hence,  |  -  A  =  VV- 
Perform  orally  the  subtraction  in  the  following : 


i|,  and 


1.    f-i^. 
3.    ii~M 


4         2    B 
•      -ff  Tff* 

6. 


7       '  — 


_      7 
T5  ~  H* 


8. 
9. 


B 

4 

Tff' 

18    11 

Si  Bii' 


s- 


*1^^,    Prob.   I.— To  find    the  difference  of  any    two 
given  fractions. 

1.  Find  the  difference  between  |  and  ^^. 

7         5         21        10        11  Explanation.  —  1.   We  reduce 

Q        12^^24       24^^24  *^®  given  fractions  to  their  least 

common  denominator,  wliich  is  ^. 

2.  We  find  the  difference  of  the  numerator?,  21  and  10,  and  write  it  over 
the  common  denominator,  giving  Hi  the  required  difference. 

2.  Find  the  difference  between  35f  and  16|. 

35f  =  35^  Explanation.— 1.  We  reduce  the  1  and  \  to  their 

Jl  Qs  __  1  g  »  least  common  denominator. 

*        — ^  2.  i\  cannot  be  taljen  from  A ;  hence,  we  increase 

18ii  the  t's  by  \\  or  1,  taken  from  the  .35.    We  now  sub- 

tract xV  from  ?§,  leaving  H- 

3.  We  subtract  16  from  the  remaining  34,  leaving  18,  which  united  with 
\\  gives  18|},  the  required  difference. 

8 


m 


mm 
if    m 


106 


FRA  VTIONS, 


The  difference  between  any  two  fractions  or  mixed  numbers 
may  be  fuuud  in  the  same  manner ;  hence  the  following 

RULE. 

250.  /.  Reduce  the  given  fractions  to  equivalent  ones  having 
tfie  least  common  denominator ;  then  Jind  t/ie  difference  of  f/ie 
numerators  and  vyrite  it  over  the  common  denominator. 

II.  When  there  are  mixed  numbers,  subtract  the  fraction  JirM, 
then  the  integer. 

If  the  fraction  in  the  minuend  is  smaller  than  that  in  the  sub- 
trahend, increase  it  by  one  from  the  integral  part  of  the  minu- 
end ;  then  subtract. 

WRITTEN     EXAMPLES. 


\i 


257.  Perform  the  following  subtractions  : 

1.    H-f           5.    37j\-33/y. 

9. 

73|  -  29H. 

2.    f-§.             6.     63-4^. 

10. 

84^  -  37f . 

3.     7i-4J.          7.     13  6^-9jV 

11. 

511i  -  34H 

4.    9f-6|.          8.     50ii^-47yV 

13. 

65/^  -  59i§ 

13.  From  a  cask  of  vinegar  containing  31^  gallons,  16g  gal- 
lons were  drawn  ;  how  many  remained?     Ans.  15 J  gallons. 

14.  If  flour  be  bought  for  $9tV  a  barrel,  and  sold  for  $12^, 
what  is  the  gain  per  barrel  ?  Ans.  d^^  dollars. 

15.  If  a  grocer  buy,  4|  and  6^  barrels  of  flour,  and  then  sells 
1^  and  4^  barrels,  how  many  does  he  still  have? 

16.  The  sum  of  two  numbers  is  59§,  and  the  greater  is  30|f  ; 
what  is  the  other  number  ?  Ans.  2S\^. 

17.  P.  Jones  is  to  build  45|  miles  of  railroad,  and  has  com- 
pleted 25 1  miles ;  how  many  miles  has  he  to  build? 

18.  James  found  $2f ,  earned  $1|^,  and  had  |2}  given  him ; 
how  much  more  money  had  he  then  than  George,  who  earned 
$6|  and  spent  $4^? 

19.  I  bought  two  tubs  of  butter,  the  tubs  and  butter  together 
weighing  lllf  pounds,  and  the  tubs  alone  weighing  7f  and  8 
pounds  respectively ;  what  was  the  weight  of  the  butter  ? 


PKUPA  R  ATORr    P  It  O  P  O  S  TTI 0  X  S .      107 


MULTIPLICATION. 


PREPARATORY    PROPOSITIONS. 

258.  Tlie  following  propopitiona  must  be  mastered  por- 
fectly,  to  undcrstaiul  luul  explain  the  process  in  multiplication 
and  division  of  fractions. 

Prop.  I. — Multiplying  the  numerator  of  a  fraction,  while  the 
denominator  remains  unchanged,  multiplies  the  fraction  ;  thus, 

2x4  t 

J  -  5 


gal- 
|ns. 
il2|, 
Lrs. 
sells 

I30II ; 

Hi 

com- 

liim ; 
larned 

retlier 
land  8 


Observe  that  since  the  denominator  is  not  changed,  the  "izo  of  the  jtarts 
remain  the  same.  Hence  the  fraction  ?  is  multiplied  by  4,  as  shown  in  the 
illustration,  by  multiplying  the  numerator  by  4, 

Prop.  II. — Dividing  the  denominator  of  a  fraction  while  the 
numerator  remains  unchanged  multiplies  the  fraction  ;  thus, 

2^  2 

12-1-4  •  3 


(1.) 


(2.) 


Observe  that  In  (1)  the  whole  is  made  into  12  equal  parts.  By  putting 
every  4  of  these  parts  into  one,  or  dividing  the  denominator  by  4,  the  svhole, 
as  shown  in  (2),  is  made  into  3  equal  parts,  and  each  of  the  2  parts  in  the 
numerator  is  4  times  1  ttvdffh. 

Hence,  dividing  the  denominator  of  i\  by  4,  the  number  of  parts  in  the 
numerator  remaining  the  same,  multiplies  the  fraction  by  4. 

Prop.  III. — Dividing  the  numerator  of  a  fraction  ichile  the 
denominator  remains  uncJianged  divides  the  fraction  ;  thus, 

6-^3  _  2 

9  "  9 


(1.) 


^of 


(2.) 


'•••   1-  iiiniMiMMiiM 


108 


FRA  CTIONS. 


.'.Ill 


In  (1)  the  numerator  6  expresses  the  parts  taken,  and  one-third  of  these 
6  partii,  as  shown  by  comparing  (1)  and  (2),  the  denominator  remaining  the 
same,  is  one-third  of  the  value  of  the  fraction.  Hence,  the  fraction  %  is 
divided  by  3  by  dividing  the  numerator  by  3. 

Prop.  IV. — Multiplying  the  denominator  of  a  fraction  ichile 
the  numerator  remains  unchanged  divides  the  fraction ;  thus, 


3 

5x3 


3^ 
10 


(1.) 


(2.) 


In  (1)  the  whole  is  made  into  5  equal  parts ;  multiplying  the  denominator 
by  2,  or  making  each  of  these  5  parts  into  2  equal  parts,  as  shown  in  (2), 
the  whole  is  made  into  10  equal  parts,  and  the  3  parts  in  the  numerator  are 
one-half  the  size  they  were  before. 

Hence,  multiplying  the  denominator  of  |  by  2,  the  numerator  remaining 
the  same,  divides  the  fraction  by  2. 


EXERCISES. 
259.  Show  by  the  use  of  lines  or  objects  that 


^ll 


1. 

2. 

3. 

10. 


4x 

3 

12 

7 

7 

5 

5 

18- 

f-6" 

'3 

14- 

h3 

7 

17 

~ 

"17 

4 

4 

4. 

5x 
8 

3" 

'15' 

7. 

5. 

13- 
3 

4-4 

x4 

=  1. 

8. 

6. 

12 

=  1. 

9. 

^  is  how  many  times  ^ ^ 

7  ^  7x3 


,  and  why  ? 


=  2}. 


9^ 

20-5-5 

15-4-3  _^ 
19       ~  19' 

3x5 
5 


=  3. 


3  3x5 

11.  Why  is  STT    ^  =  KT?       ^    Explain  by  lines. 

20-5-5      20 

7  7 

12.  -T-a  is  how  many  times  it^ — 5 ,  and  why  ? 
lo  lo-5-b 

8  3 

13.  Why  is  ^  greater  than  ^ — r  ?    Explain  by  lines. 


1^1 


EXAMPLES. 


109 


ILLUSTRATION    OF    PROCESS. 

260.    Prob.  L — To  multiply  a  fraction  by  an  integer. 

1.  Multiply  I  by  7. 

Solution. —1.  According  to  (258—1),  multiplying  the  numerator,  the 
denominator  remaining  the  same,  multiplies  the  fraction.    Hence,  7  times 

:i8equalto^'*''  =  ^  =  3S.  - 

2.  According  to  (258—11),  a  fraction  is  also  multiplied  by  dividing  the 
denominator.    Hence  the  following 

RULE. 

201.  Multiply  the  numerator  of  the  fraction  by  th^  given 
integer^  or  divide  the  denominator. 


.  =  ov 


=  3. 


EXAMPLES     FOR    PRACTICE. 

202.  Multiply  orally  the  following,  reduce  the  results  to 
their  lowest  terms,  and  explain  as  above. 

1.  f  X  3.  4.    I  X  9.  7. 

2.  I  X  12.  5.    A  X  6.  8. 


^  X  5. 


3. 


I  X  12. 

f  X  4. 


I  X  9. 


6. 


f  X  8. 


9.    \\ 


14. 


Multiply  the  following  and  reduce.     Cancel  when  possible. 
10.    yVn  ^  8.         13.    Hf  X  48.  16.    T'xmr  x  90. 


11. 
12. 


II  X  9. 
T^  X  10. 


14.  tWt7  X  50. 

15.  ^"A  X  100. 


17.  AViT  X  100. 

18.  Ml  X  75. 


203.  Prob.  II — To  find  any  given  part  of  an  integer. 

1.  Find  ^  of  $395. 

Solution.— 1.  We  find  the  \  of  $395  by  dividing  it  by  5.    Hence  the  first 
step,  $393  -«-  5  =  $79. 

2.  Since  $79  is  1  fifth  of  $395,  four  times  $79  will  be  4  fifths.    Hence  the 
second  step,  $79  x  4  =  $316. 

To  avoid  fractions  until  the  final  result,  wc  multiply  by  the  numerator 
first,  then  divide  by  the  denominator  ;  hence  the  following 

RULE. 

204.  Divide  by  the  denominator  and  multiply  by  the  niimer- 
<Uor  of  the  fraction,  which  indicates  the  part  required. 


'""•h 
'f 


110 


FRA  CTIO  NS. 


EXAMPLES     FOR    PRACTICE. 

1.  How  many  are  ^j^  of  8730  ?  of  57  ?  of  835  ?  of  10  ?  of  100  ? 

2.  Find  |  of  $343  ;  of  $1000 ;  of  $4860  ;  of  $10001. 

3.  A  has  $189,  B  lias  |  of  A's  money,  and  C  has  {^  of  B's 
money  ;  what  is  the  sum  of  their  money  ?  Ana.  $360. 

4.  J.  Moodie  owned  395  acres  of  land,  sold  at  one  time  f  of 
it,  and  at  another  time  ^^^  of  it ;  how  many  acres  does  he  still 
own?  Am.  143f|  acres. 

5.  A  man  having  $1305  gave  f  of  it  to  A  and  f  of  what 
remained  to  B  ;  how  much  had  he  left  ?  Ans.  $580. 

6.  A.  Walker  had  three  pieces  of  cloth  containing  respec- 
tively 187  yards,  163  yards,  and  308  yards,  sold  |  of  the  first 
piece,  I  of  the  second  ;  how  many  yards  has  he  left  ? 

265.    Prob.  III. — To  find  any  given  part  of  a  fraction, 

or,  To  multiply  a  fraction  hy  a  fraction. 

Find  the  f  of  |  of  a  given  line. 

riBST  8TBP. 


1    ^                 8                                     3         3 

of                                                     ss                      .          —    .  . 

3                        4                                       4x3       13 

\,» 

1            ^mm     ^mmm     ^mm                               mhh 

8  ""^ 

8EC0NI)  STEP. 

... 

3x3                                         61 

13                          ~                   13  ~  2 

...  X  2                                         ««-   ... 

ai 


Explanation.— According  to  (258— IV),  a  fraction  is  divided  by  multi- 
plying its  denominator.  Hence  we  find  the  J  of  |  by  multiplying  the 
denominator  4  by  3,  as  shown  in  First  Step. 

Having  found  1  third  of  |,  we  find  2  thirds^  according  to  (268—1),  by 
multiplying  the  numerator  of  t?  by  2,  as  shown  in  Second  Step.  Hence  the 
following 


RULE. 

260-  Divide  hy  the  denominator  and  multiply  hy  the 
ator  of  the  fraction  which  indicates  the  part  required. 


Hmer- 


1 


EXAMPLES, 


111 


EXAMPLES    FOR    PRACTICE. 


267.  Solve  orally  the  followiug,  and  explain  as  above  : 

1.  What  is  f  of  I  ?    fof^?    i\off?    foff? 

2.  What  part  of  1  is  ^  of  f  ?    f  of  i  ?    f  of  ^  V 

3.  If  a  yard  of  cambric  cost  $f ,  what  is  |  of  a  yard  worth  ? 
Find  the  value  of  the  following : 


4.  ^  of  41;  (f +  1)  X  8;  (t-l)  X  I;  (f  ofT«3)-,SV 


6.  What  is  the  cost  of  ^^  of  a  yard  of  cloth,  at  $|  per  yard  ? 

7.  Geo.  Henderson  gave  f  of  his  farm  to  one  son,  and  ^  of  what 
was  left  to  another  ?  what  part  of  the  whole  farm  did  ho  give 
the  second  son  ?  Ans.  f . 


268.    Prob.  IV. — To  multiply  by  a  mixed  number. 

Multiply  372  by  6f . 

873 
__6| 

2332 
265| 

2497f 


ExpLANATioK.— 1.  In  multiplying  by  a  mixed  number, 
the  multiplicand  is  taken  separately  (83),  as  many  times  as 
there  are  units  in  the  multiplier,  and  such  a  part  of  a  time 
as  is  indicated  by  the  fraction  in  the  multiplier ;  hence, 

2.  We  multiply  372  by  6;  by  multiplying  first  by  6,  which 
gives  the  product  2232,  and  adding  to  this  product  |  of  872 
which  is  265J  (363),  giving  2497^,  the  product  of  378  and  6f! 
Hence  the  following 


ilti- 
the 

the 


RUIiE. 

209.  Multiply  first  by  the  integer,  then  by  the  fraction,  and 
add  the  products. 

EXAMPLES     FOR     PRACTICE. 


270.  Multiply  orally  and  explain  the  following : 


1. 
2. 
3. 


13  by  5|. 


4.  20  by  3^.  7. 

5.  100by7|.  8. 

6.  400  by  21.  9. 

Perform  the  multiplication  in  the  following  : 

10.  75x12tV  12.    435x104^  14. 

11.  89x342.     13.  631^x34.      15. 


18  by  2J. 
8  by  9J. 


80  ])y 
36  by 


3|. 

3J. 


34  by  lOf. 

8000x9,«,V 
1000  X  73^ 


\ } 
it 


;fl 


i) 


V: 


■m 


112 


FR  A  CTIO  NS. 


371.  Prob.  v.— To  multiply  when  both  multiplicand 
and  multiplier  are  mixed  numbers. 

Multiply  86f  by  54f. 

(1.)    86|  =  *|a;  54f  =  *fi. 

(3.)    ^^  X  ^^  =  ^^.  =  4741if. 

ExFiiAKATioN.— 1.  We  reduce,  as  shown  in  (1),  both  mnltiplicand  and 
multiplier  to  Improper  fractions. 

2.  We  multiply,  as  shown  in  (2),  the  numerators  together  for  the  numer- 
ator  of  the  product,  and  the  denomitiators  together  for  the  denominator 
of  the  product  (964),  then  reduce  the  result  to  a  mixed  number.  Hence 
the  following 

RULE. 

272.    /.  Reduce  the  mixed  numbers  to  improper  fractions. 

IT.  Multiply  the  nuvierators  together  for  tfw-  numerator  of  the 
product,  and  the  den,ominators  for  the  denominator  of  thf 
product. 

///.  Reduce  the  result  to  a  whole  or  r.iixed  number. 


273.  The  process  in  multiplication  is  shortened  by  cancel- 
lation ;  thus, 


Multiply  32|  by  29,^^. 

(1.)   ^^  =  ^^^ 


4     -T. 


28A  =  W- 


19 


49 


(2.)     ^  X  W  =  19  X  49  =  931.  ,. 

Explanation.— 1.  We  reduce  the  multiplicand  and  multiplier  to  Im- 
proper fractions,  as  shown  in  (1). 

2.  We  indicate  the  multiplication,  and  cancel,  as  shown  in  (2),  the 
factors  common  to  any  numerator  and  any  denominator.     Hence  the 


following 


RULE. 

274.  Indicate  all  the  multiplications,  and  cancel  the  factors 
common  to  any  nnmeratoi'  and  any  denominator. 


$11 


■^ 


EXA3IP  LES, 


113 


EXERCISE     FOR    PRACTICE. 

275.  Multiply  the  following,  cancelling  common  factors: 


1. 
2. 
3. 


iff  X 


n- 


M  > 

T¥ff 


7 

Tff* 


If 


4. 
5. 
6. 


89     V 
"4  5     X 

SOO 


60 
TTT* 


n. 


I!  X  H. 


7. 
8. 
9. 


3  7  4^ 
1  00 

lOOff 


8477' 


TOO' 


Find  the  continued  product  of 

10.  ff .  M,  H,  irW  and  li. 
11-  i»|,  i  6»  I.  f.  and^. 
12.    t.  2|,  3A,  5A,  and  Cxb- 


13. 
14. 
15. 


\\h  /rV.  and  f  f  • 
^\\,  \l,  and  /,V 

31.  4L  and  7^ 


'5»  ^tf» 


25. 


|;6|  a  cord,  and 


16.  A  lady  bought  15  yards  of  silk  at  $2|  a  yard,  and  7| 
yards  of  lace  at  $3|  a  yard ;  what  was  the  cost  of  both  ? 

17.  What  is  the  cost  of  12  cords  of  wood  at 
8  tons  of  coal  at  $11|  a  ton  ?  Am.  $171.50. 

18.  A  province  has  an  area  of  37680  square  miles,  and  the 
average  population  to  a  square  mile,  in  1870,  was  35^*5  J  ^^at 
is  its  population  ? 

19.  What  is  the  cost  of  45^^  tons  of  iron,  at  $27  J  per  ton  ? 

20.  At  $3 1  a  yard,  what  is  the  cost  of  15g  ?    Of  32|  ? 

21.  A  merchant  sold  12|  yards  of  cloth  at  $2i  a  yard  ;  28|  at 
$1| ;  and  52  J  at  |3|  ;  what  did  he  receive  for  the  whole  ? 

22.  Find  the  product  of  f  of  25  J,  and  y\  of  l^A- 

23.  Find  the  product  of  4|  of  f  of  12,  and  7f  of  15. 

24.  Bought  19  pounds  of  butter  at  23^  cents  a  pound,  giving 
in  return  27f  pounds  of  lard  at  15  cents  a  pound,  and  the  rest 
in  cash  ;  what  did  I  give  in  cash  ?  ^n«.  20  cents. 

25.  C.  Smart  has  two  fields  containing  respectively  11 1  acres 
and  21^  acres ;  how  much  hay  will  he  take  from  both  fields,  at 
the  rate  of  If  tons  an  acre  for  the  first,  and  2f  tons  an  acre  for 
the  second  ?  Ans.  69yV  tons. 

26.  Find  the  valae  of  ($37f  -  $13f )  x  (f  of  8  -  2 J). 

27.  Find  the  value  of  (|  of  8)  —  (*  of  9  -  2^). 

28. 


Find  the  value  of  2|  +  3| 


84   4.1 


^ 

M 


i 


m 


'V 


114 


FEA  C  TIO  N S, 


DIVISION". 


ILLUSTRATION    OF    PROCESS. 

276.  Prob.  I.— To  divide  a  fraction  by  an  integer. 

1.  Divide  f  by  4 

ExPLAKATioN.  —  1.  According: 
to  (258—111),  a  fraction  is  divid- 
ed by  dividing  the  numerator. 
Hence  we  divide  {J  by  4,  as  shown 
in  (1),  by  dividing  the  numerator 
8  by  4. 

2.  According  to  (358— IV),  a  fraction  is  divided  by  multiplying  the 
denominator.  Hence  we  divide  %  by  4,  as  shown  in  (2),  by  multiplying  the 
denominator  by  4,  and  reducing  the  result  to  its  lowest  terms.  Hence  the 
following 

EXILE.  ^ 

277.  Divide  the  numerator,  or  multiply  the  denominator^  by 
tlie  given  integer. 


(1.) 

8 

• 

9  ' 

•4 

8-4- 

~9 

4_ 

2 
9 

8 

8 

8 

?, 

(2.) 

9  • 

4- 

~9  X 

4^ 

36  ~ 

9 

EXAMPLES    FOR    PRACTICE. 

278.  Divide  orally  and  explain  the  following.  Dividing  the 
numerator  in  every  case  where  it  can  be  done,  in  preference  to 
multiplying  the  denominator. 

1.  If  -^  4.  3.    H  -^  3.  5.    A  --  4. 

2.  tf  -f-  7.  4.    f  -*-  8.  6.    T^if  -*-  6. 

Perform  the  division  in  the  following : 
57.  9.    tf  H-  25. 


7. 


885 
TS'S 


11. 

12. 


1^^  -  50. 


8.    Ht  -^  32.  10.    11^  ^  75. 

13.  If  7  yards  of  calico  cost  $f ,  what  will  1  yard  cost? 

14.  At  $f  for  4  boxes  of  figs,  what  will  1  box  cost  ? 

15.  Show  that  multiplying  the  denominator  of  f  by  4,  divides 
the  fraction.     Explain  by  lines. 

16.  The  product  of  two  numbers  1^  149 1,  and  one  of  them  is 
S3 ;  what  is  the  other  ? 


EX  A  MP  L  E  S, 


115 


17.  If  a  compositor  earns  $45|^  in  18  days,  how  much  does  he 
earn  in  1  day  ?    In  9  days  V    In  5  days  1    In  27  days  ? 

Find  the  value 

18.  Of  (f  of  f  —  xV)  -5-  8-  20.  Of  (I  of  ^  -  2f )  -5-  12. 

19.  Of  (i^  +  I)  -*-  32.  21.  Of  (f  of  10|  +  if)  -f-  32. 

270.    Prob.  II. — To  divide  by  a  fraction. 

1.  How  many  times  is  f  of  a  given  line  contained  in  twice 
the  same  line  ? 


2  lines 

^^  \ 

z= 

FIRST  STEP. 

10      .       ,. 

-p  of  a  Ime. 
5 

If 

^■B^H 

SECOND  STEP. 

3 

I 

10 
5 

3    -  ^i 

^J 

,1^ 

—  ^i 

ides 


Explanation.— 1.  We  can  And  how  many  times  one  number  is  con- 
tained in  another,  only  when  both  are  of  the  same  denomination  (144). 
Hence  we  first  reduce,  as  shown  in  First  Step,  the  2  lines  to  10  J{fths  of  a 
line ;  the  s&me  fractianal  denomination  as  the  divisor,  Sflffhs. 

2.  The  3  fifths  in  the  divisor,  as  shown  In  Second  Step,  are  contained  in 
the  10  fifths  in  the  dividend  3  times,  and  1  part  remaining,  which  makes  \ 
of  a  time.    Hence  2  equal  lines  contain  I  of  one  of  them  3i  times. 

Observe  the  following  regarding  this  solution : 

(1.)  The  dividend  is  reduced  to  the  same  fractional  denomination  as  the 
divisor  by  multiplying  it  by  the  denominator  of  the  divisor ;  and  when 
reduced,  the  division  is  performed  by  dividing  the  numerator  of  the  divi- 
dend by  the  numerator  of  the  divisor. 

(2.)  By  inverting  the  terms  of  the  divisor  these  two  operations  are 
expressed  by  the  sign  of  multiplication.  Thus,  2-t-3  =  2x  »,  which  means 
that  2  is  to  be  multiplied  by  .5,  and  the  product  divided  by  3. 


< 


m  ii 


m  I 


mmm 


116 


FJRA  CTI0N8, 


2.  How  many  times  is  )  of  a  given  line  contained  in  f  of  it  ? 

FIB8T  STEP. 


2 
3 

4 
6 

— 

1 
2 

8 
"                  6 

■  ■  ■    -t '  ■- 

~ 

4 
3     ~ 

4 

6 

SXCOMD  8TSP. 

3 

-6    = 

Explanation.— 1.  We  reduce,  as  shown  in  First  Step,  the  dividend  | 
and  the  divisor  |  both  to  sixths  (144—1). 

3.  Wo  divide  the  J  by  |  by  dividing  the  numerator  of  the  dividend  by 
the  numerator  of  the  divisor.  The  f  is  contained  in  |,  as  shown  in  Second 
Step,  1|  times.    Hence  \  is  contained  1|  times  in  §. 

280.  When  dividing  by  a  fraction  we  abbreviate  the  work 
by  inverting  the  divisor,  as  follows  : 

1.  In  reducing  the  dividend  and  divisor  to  the  same  fractional 
unit,  the  product  of  the  denominators  is  taken  as  the  common 
denominator,  and  each  numerator  is  multiplied  by  the  denom- 
inator of  the  other  fraction  ;  thus, 

5       2       5x3       8x7       15       14       15    Numerator  of  dividend. 
7*    3~7x3'    3x7~21    *31~14    Numerator  of  divisor. 

2.  By  inverting  the  divisor,  thus,  f  -«-  f  =  f  x  f  =  ||,  the 

numerators  15  and  14  are  found  at  once,  without  going  through 

the  operation  of  finding  the  common  denominator.     Hence 

the  following 

KULE. 

281.  Invert  the  terms  of  the  divisor  and  proceed  as  in  mid- 
iiplication. 


EXAMPLES, 


ii; 


lend, 
tor. 
the 

3nce 


lUlr 


EXAMPLES     FOR    PRACTICE. 

282.  Solve  orally  the  following  and  explain  as  above : 


4 


2.  f  +  f 

3.  8-*-#. 


4.  13  -^  f. 

5.  t-*-f. 

6.  9  -I- 1. 


7.  90 -I- H- 

8.  H  +  f. 

9.  200  4- iM. 


10.  1  is  how  many  times  i ?    i?    ^?    ^?    \t    \t 

11.  At  $J  a  bushel,  how  many  bushels  of  com  c&n  be  bought 
for  |9  ? 

Solution.— As  many  bushels  as  %{  is  contained  times  In  |9.    19  are 
equal  to  $Vi  and  $|  is  contained  in  $V,  10?  times.    Hence,  etc. 

12.  At  $5  a  yard,  how  many  yards  of  serge  can  be  bought 
for|3?    For  $10?    For  $15?    For  $7?    For  $25?    For  $9? 

13.  For  $12  how  many  poimds  of  tea  can  be  bought  at  $f 
per  pound?    At$|?    At$f?    At$f?    At$|?    At$|? 


6  9 


3  9 


7  «; 


? 


10  ' 

IT 


14.  5  are  how  many  times  f  ? 

15.  If  f  of  a  ton  of  coal  cost  $3,  what  will  1  ton  cost  ? 

Solution.— Since  J  of  a  ton  cost  $3,  \  will  cost  \  of  $3,  or  f  J,  and  1  ton, 
or  I,  will  cost  9  times  $|,  or  $V,  equal  to  $6|.    Hence,  etc. 

Or,  1  ton  will  cost  as  many  times  $3  as  |  of  a  ton  is  contained  times  in 
1  ton,    1  ton  +  5  =  s  =  gi.    Hence,  1  ton  will  cost  2i  times  $3,  or  $6*. 

16.  At  $1  for  I  of  a  pound  of  tea,  what  is  the  cost  of  1 
pound  ?    Of  7  pounds  ?    Of  y»j  of  a  pound  ?    Of  |  pounds  ? 

17.  If  |-  of  a  cord  of  wood  cost  $4,  what  will  1  cord  cost  ? 
4  cords  ?    11  cords  ?    |  of  a  cord  ?    ^^  of  a  cord  1 

18.  How  many  bushels  of  wheat  can  be  bought  for  $8,  if  f 
of  a  bushel  cost  $|  ?    If  f  of  a  bushel  cost  $^5  ? 

19.  At  $1  a  yard,  how  much  cloth  can  be  bought  for  %f^  ? 

Solution.— As  many  yards  as  %\  is  contained  times  in  $r"u.    %l  equals 
$^,  and  %^  is  contained  1^  times  in  %^a-    Hence,  etc. 

20.  At  $yV  8-  bushel,  how  many  bushels  of  potatoes  can  be 
bought  for  %l  ?    For  $||  ?    For  $f  ?    For  $^  ?    For  $|  ? 

21.  How  many  pounds  of  sugar  at  $|^  can  be  bought  for  $|? 
For$|?    For$H?    For$||?    For$|? 


5  VH 


ii 


118 


PlfA  CTIO  NS. 


"?«; 


22.  If  3'^  of  a  yard  of  cloth  can  be  bought  for  $  i\,  how  much 
will  1  yard  cost  ?    5  yards  ?    7|  yardw  ? 

23.  Geo.  Graham  expended  f  of  $480  in  purchasing  tea  at  $f 
per  pound,  and  the  balance  in  purchasing  coffee  at  %l  i^er  pound. 
How  many  pounds  did  he  buy  of  each  ? 

Perform  the  division  in  the  follo^ving.  Invert  the  divisor 
and  cancel  common  factors.     (175.) 


24. 

U  +  11- 

29. 

N6       .85 
^iiT   —    if' 

34. 

KH  +  m- 

25. 

.»!•  -*-  M- 

30. 

573  -5-  T^jf. 

35. 

11  -*-  iM. 

26. 

If  +  M- 

31. 

862  -f-  f  4. 

36. 

1000  ^  m- 

27. 

m  +  n. 

32. 

100      .        46 

37. 

3000  ^  f  «t. 

28. 

324  -J-  f . 

33. 

573  -f-  Ii 

38. 

m  -*-  If. 

283.   Prob.  III. — To  divide  when  the  divisor  or  divi- 
dend is  a  mixed  number,  or  both. 

1.  Divide  48  by  4f .  '  t^' 

(1.)  48  -f-  4f  -  48  -^  V-  ExPLANATioN.-l.  We  re- 

duce the  divisor  4^  as  shown 
in  (1),  to  the  improper  ft-ac- 


(2.)  48-4- V  =  48  X  A  =  10? 


2.  We  invert  the  divisor,  as  shown  in  (2),  according  to  (280),  and  mul- 
tiply the  48  by  /„  giving  10?  as  the  quotient  of  48  divided  by  4?. 


2.  Divide  8?  by  3|. 
(1.)  8?  -*-  ^  =  \^ 

(2.)  ¥  -^  V  = 


s. 


'^ 


=  i^  =  2f . 


Explanation.— 1.  We 
reduce  the  dividend  and 
divisor,  as  shown  in  (1), 
to  improper  fractions, 
giving  V  +  V. 
2.  We  invert  the  divisor,  Vi  as  shown  in  (2),  according  to  (280),  and 
cancel  31  in  the  numerator  62  and  denominator  81  (1 76),  giving  V,  or  ^. 
Hence,  8?  -•-  3|  =  2?. 

From  these  illustrations  we  obtain  the  following 

BULE. 

284.  Reduce  mixed  numbers  to  improper  fractions ;  then 
invert  the  divisor  and  proceed  as  in  multiplication,  cancelling 
any  factors  that  are  common  to  any  numerator  and  a  denom- 
inator. 


EXAMPLES, 


119 


mul- 


-l.We 

Indand 

in(l), 

;tlons. 

)),  and 


WRITTEN     EXAMPLES. 

285*  Perform  and  explain  the  division  in  the  following : 


1. 

7«-f-2t. 

8. 

732-i-14f. 

15. 

5,«„»a-^2THff. 

2. 

2A-*-4J.         , 

9. 

36^-^8^. 

16. 

873-*-^?. 

3. 

9i'ir-*-5|. 

10. 

85,\-J-23. 

17. 

302-f-/„V 

4 

89^-7|. 

11. 

37t\j-6tV 

18. 

5. 

8624-421. 

12. 

lOOOA-f-T^B-. 

19. 

H  of  15?  4-5. 

6. 

43TVa-^Ti?in7. 

13. 

'^Tun'^^TTnjTr' 

20. 

f,of5^-^^ 

7. 

100iV-*-5^. 

14. 

936+5x§T,. 

21. 

§  of  28 -f-^  oft. 

22.  At  $f  for  ij^^  of  an  acre  of  land,  what  is  the  cost  of  1  acre  ? 
Of  ViT  of  an  acre  ?    Of  ^  of  an  acre  ?    Of  29^  of  an  acre  ? 

23.  If  a  bushel  of  wheat  cost  $1§,  how  much  can  bo  bought 
for$12|?    For$28«?    For$273|? 

24.  Jas.  Johnston  expended  $597|  in  buying  cloth  at  $2f  a 
yard.  He  afterwards  sold  the  whole  of  it  at  $3f  a  yard  ?  How 
much  did  he  gain  by  the  transaction  ? 


COMPLEX   FKACTIOFS. 

28G.  Certain  results  are  obtained  by  dividing  the  numera- 
tor and  denominator  of  a  fraction  by  a  number  that  is  not  an 
exact  divisor  of  each,  which  are  fractional  in  form,  but  are  not 
fractions  according  to  the  definition  of  a  fraction.  These  frac- 
tional forms  are  called  Complex  Fractions. 

The  following  examples,  which  illustrate  the  three  classes  of 
complex  fractions,  should  be  carefully  studied  : 

2- 
Ex.  1.  Show  that  y\  of  a  line  is  equal  to  -j  of  the  same  line. 


then      1 
eUing      1 
tnom-      1 

8  4-3 
12-^3 

2| 
~                   4 

1 

HHH       MMH       HBB       HI 

■  ■                 ^^     mmm    w^mm    m^ 

m 


ti 
■i 

i 


120 


COMPLEX    FRACTIONS, 


Explanation.— 1.  Dividing  tho  uumorator  and  deiiotuiiintor  of  ,%  by  8 
makcH  every  3  parts  in  each  into  1  part,  as  nbuwu  in  tlio  illuutratlou,  but 
doc8  not  change  tho  value  of  the  fraction  (22-1-III). 

3.  The  denominator  or  whole  contiiinfei  4  of  tbeHu  partn,  and  the  uumcra- 

tor  2  of  them  and  \  of  one  of  them,  ati  will  be  seen  by  the  illuBtration. 

23 
Benco,  {^  of  a  line  itj  equal  to  ~  of  the  same  line. 

4 


Ex.  2.  Show  that  /j  of  a  line  is  expressed  by  — . 


13 


5 
5 


2| 


Explanation.— 1.  Dividing  the  numerator  and  denominator  of  A  ^7  5 
makes  every  5 parts  in  each  into  1  part,  as  shown  in  the  illustration. 

2.  The  denominator  or  whole  contains  two  of  these  parts  uud  }  of  one  of 
them,  and  the  numerator  contains  1  part,  as  shown  in  the  illustration. 

Hence,  t'i  of  a  line  is  represented  by  the  fractional  expression  — . 


Ex.  3.  To  show  that  \%  of  a  line  is  expressed  by 


8f 


lO-i-4 
13 -i- 4 


1$ , 


Explanation.— 1.  Dividing  the  numerator  and  denominator  of  \%  by  4 
makes  every  U  parts  in  each  into  Ipart^  as  shown  in  the  illustration. 

2.  The  denominator  or  whole  contains  3  of  these  parts  and  \  of  one  of 
them,  and  the  numerator  contains  2  of  them  and  \  or  {  of  one  of  them. 

2i 


Hence,  \%  of  a  line  is  represented  by  the  fractional  expression 


3i- 


From  these  illustrations  we  have  the  following  definitions: 

287.  A  Comj)l€X  Fraction  is  an  expression  in  the  form 
of  a  fraction,  having  a  fraction  in  its  numerator  or  denominator, 

^ 

6f 


r        4 

or  in  both ;  thus,  ^,    ^, 

•        Of 


COMPLEX    FRACTIONS, 


121 


288.  A  Slmjde  Fraction  is  a  fraction  having  a  whole 
number  lor  its  numerator  and  for  its  denominator. 


PROBLEMS    IN    COMPLEX    FRACTIONS. 

289.    Pros.  I.— To  reduce  a  complex  fraction  to  a  sim- 
ple fraction. 

4ff 
Reduce  =v  to  a  simple  fraction. 
7f 

4^       4|  X  12      56  Explanation.— 1.  Wo  find  the  least  com- 

7?  ~  7?  X  13  ~  93  ™°°  multiple  of  the  denominatoru  of  the 

partial  fractions  |  and  J,  which  i»  12. 
3.  Multiplying  both  terms  of  the  complex  fraction  by  12  (23'i— II), 
which  is  divisible  by  the  deuominators  of  the  partial  fractions,  %  and  J, 
reduces  each  term  to  a  whole  number.    4]  x  12  =  56 ;  7i  x  12  =  93.    Thcrc- 

41 
fore  jl|  Is  equal  to  the  simple  ft-action  1%.    Hence  the  following 


\%  by  4 

one  of 
them. 


ms: 

form 
Lnator, 


EULE. 

290.  Multiply  both  terms  of  the  complex  fraction  by  the  least 
common  multiple  qfaU  the  denominators  of  the  partial  fractions. 

291.  The  three  classes  of  complex  fraction.s  are  forms  of 
expressing  three  cases  of  division  ;  thus, 


(1.)    ^  =  5|-*-7. 
(2.)    ^  =  32-j-9f 


A  mixed  number  divided  by  an  integer. 
An  Integer  divided  by  a  mixed  number. 


8f- 
2| 


(^•)    of  ~  8f  "*"2|.     A  mixed  number  divided  by  a  mixed  number. 


Hence,  when  we  reduce  a  complex  fraction  to  a  simple  frac- 
tion, as  directed  (290),  we  in  fact  reduce  the  dividend  and 
divisor  to  a  common  denominator,  and  reject  the  denominator 
by  indicating  the  division  of  the  numerator  of  the  dividend  by 
the  numerator  of  the  divisor ;  thus, 

9 


I 


!'>{ 


It;-- 

X-f. 


w 


iW 


122 


FRACTIONS. 


!■ 

.'{ 

^ 

\ 

t 

1 
it 

,     .; 

! ;  %-^ 

r\. 

(1.)  ^ 

(2.) 


5f 

s 

Of 


5fxl2      69  ,.      ^    /o«ON 

=  sl ..  io  =  oT, .  according  to  (289). 


2|  X  12      32 

-21  =  Y-  ^  S.  and  -Y-  -  f  =  f  I 


.13 


f  I,  the  game 


result  as  obtained  by  tbe  method  of  muUiplying  by  the  least 
common  multiple  of  the  denominators  of  the  partial  fractions. 

EXAMPLES    FOR    PRACTICE.  C , 

1292.  Reduce  to  simple  fractions,  and  explain  as  above  : 


1. 


2. 


13| 
16f 


13^ 

23tVi7 


5     1?I?I, 
32^ 


6. 


When  the  numerator  or  denominator  contains  two  or  more 
terms  connected  by  a  sign,  perform  the  operation  indicated  by 
the  sign  first,  then  reduce  to  a  simple  fraction.  -» 

Reduce  the  following  to  simple  fractions  : 


3^  ^  ^ 


8. 


9. 


«l 

+  5f 

4J 

-2i_ 

6? 

-n 

(SI— 

^) 

X 

2 

5 

(O  X  I)  +  (I  of  I) 


10. 


11. 


12. 


(^  of  9)  +  it  of  2) 


fof5 


(?  o^  f )  -  A 
(22  of  2)  -  ^j, 

1000 


293.    Prob.  II. — To  reduce  a  fraction  to  any  given  de- 
nominator. 

1.  Examples  where  the  denominator  of  the  required  fraction  is 
a  factor  of  the  denominator  of  the  give7i  fraction. 

Reduce  |f  to  a  fraction  whose  denominator  ii  8.  • 


17_ 
24  ~ 

fraction. 


17 
24 


Explanation.— We   observe  that  8,  the 
denominator  of  the  required  fraction,  is  a 
factor  of  24,  the  denominator  of  the  given 
Hence,  dividing  both  terms  of  J{  by  3,  the  other  factor  of  24,  the 


3_5| 
3~   8" 


53 


fraction  is  reduced  (224—111)  to  -~,  a  fraction  whose  denominator  is  8. 

o 


3. 
4. 
5. 


COMPLEX    FRACTIONS, 


123 


2.  Examples  wTicre  the  denominator  of  the  nqni  red  fraction  is 
not  a  factor  of  the  denominator  of  the  given  fraction. 


Keduce  j^j  to  a  fraction  whose  denominator  is  10. 


8  _  8    X  10  _  80 
^  ^^    13  ~  13  X  10  ~  130 

80  -T- 13      6fs 


(2)     ^^- 
^   ''    130      130 


13       10 


Explanation.  —  1.  Wc  intro- 

tliicc  the  given  denominator  10  as 

a  factor  into  the  denominator  of 

Vj  by  multii^ying,  a»  t^hown  in 

(1),  both  terms  of  the  fraction  by 

10(224-11). 

2.  The  denominator  130  now  contains  the  factors  13  and  10.    Hence, 

dividing  both  terms  of  the  fraction  ^''s'i,  by  13  (224— III),  as  t^hown  in  (2), 

6  * 
the  result  is    ",  a  fraction  whoso  denominator  is  10. 

From  these  examples  we  obtain  the  following 


RULE. 


jnde- 


Hion  is 


8,  the 
m,  is  a 
[e  given 

24,  the 

Is  8. 


294.  Mvltiply  both  terms  of  the  fraction  by  the  given  denom- 
inator,  and  then  divide  them  by  the  denominator  of  the  fraction. 

Observe  that  when  the  given  denominator  is  a  factor  or  multiple  of  the 
denominator  of  the  fraction,  it  is  not  necessary  to  multiply  by  it,  as  will  be 
seen  in  the  first  example  above. 

EXAMPLES     FOR    PRACTICE. 


295.    1.  Reduce  ^,  ^\,  -j?,  ||,  !|^,  and  fg  each  to  thirdt. 

2.  How  many  sevenths  in  f[  ?    In  ^  ?    In  ^  ? 

3.  In  If  how  many  twentieths?    How  many  ninths? 
4. 


5. 


Reduce  |,  i;, 
Reduce  \,  J, 


10' 


4 


»' 


and  \  each  to  sevenths. 


and  A  to  hundredths. 


6.  Express  as  hundredtlis 

7.  How  many  tenths  in  %  ?    In 

8.  How  many  thousondths  in  ^? 

9.  How  manv   hundredths  in 


5      8       7 


8  7 
•1.V 


and  m. 


13  ? 


In  17?    In 


t'i 


In 


10 


Tn  '>  '1    9 

in  ^^  ,f  T 


In-,^?     In 
In  AS?     In 
In  4^? 


s 
1^.' 


H? 


\V^ 


In 


m 


In7f? 


In9f? 


10.  Reduce  to  hundredths  ? 


43    . 

UITT' 


11.  Reduce  to  hundredths 


2|. 

7' 


5^ 


4     I 

J 


Q4 

t 

8" 


40A 


3* 

05 


i^ 


iii    \- 

■    m 


124 


FRA  C  T I  0  NS, 


REVIEW    EXAMPLES. 


P! 


290.     1.  Reduce  f ,  ^,  -r^,  ^^,  and  ||  each  to  twenty-eighths. 

2.  How  many  thirtieths  in  f ,  and  why  ?    In  |  ? 

3.  Reduce  to  a  common  denominator  — ,  ^,  and  tV. 


4.  State  the  reason  why 


5x4 


(258). 


9-5-4  9 

5.  Redux  i  to  a  fraction  whose  numerator  is  12 ;  is  20 ;  is  2  ; 

is  3;  is  7  (224). 

6.  Find  the  sum  of  |,  y^,  J,  J,  and  | J. 

7.  Reduce  to  a  common  numerator  f  and  f  (241). 

8.  Find  the  value  of  (5  of  A  -  iV)  ^  (|  +  3t)- 

9.  If  ^  of  an  estate  is  worth  $3460,  what  is  i  of  it  worth  ?     . 

10.  $4  is  what  part  of  $8  ?    Of  $12  ?    Of  $32  ?     Of  $48  ? 
Write  the  f?olution  of  this  example,  with  reason  for  each  step. 

11.  If  a  man  can  travel  a  certain  distance  in  150  days,  what  part 
of  it  can  he  travel  in  5  days  ?  In  15  days  ?  In  25  days  ?  In  7^ 
days  ?    In  ^  days?     In  12^  days  ? 

12.  A's  farm  contains  120  acres  and  B's  280  ;  what  part  of  B's 
farm  is  A's  ?  Ans.  |. 

13.  42  is  7^  of  what  number  ? 

Write  the  solution  of  this  example,  with  reason  for  each  step. 

14.  $897  is  I  of  how  many  dollars?  ,  r 

15.  5  of  76  tons  of  coal  is  y^^  of  how  many  tons  ? 

16.  A  piece  of  silk  containing  73  yards  is  I  of  another  piece. 
How  many  yards  in  the  latter  ? 

17.  84  is  {^  of  8  times  what  number  ? 

Write  the  solution  of  this  example,  with  reason  for  each  step. 

18.  Bought  a  carriage  for  $286,  and  sold  it  for  ^  of  what  it 
cost ;  how  much  did  I  lose  ? 


1^! 


EX  A  MP  L  ES. 


125 


trt 


Ice. 


it 


19.  A  lias  $G94  in  a  bank,  which  is  i  of  3  times  the  amount 
B  has  in  the  same  bank  ;  what  is  B's  money  ? 

20.  Two  men  are  86J  miles  apart ;  when  they  meet,  one  has 
travelled  SJ  miles  more  than  the  other ;  how  far  has  each 
travelled  V 

21.  If  -/^  of  a  farm  is  valued  at  $4T32|,  what  is  the  value  of 
the  whole  farm  ? 

22.  The  less  of  two  numbers  is  432|,  and  their  diflFerence 
123  iV .     Find  the  f^reater  number. 

23.  A  man  owning  J  of  a  lot,  sold  f  of  his  share  for  $2800 ; 
what  was  the  value  of  the  lot? 

24.  What  number  diminished  by  f  and  |  of  itself  leaves  a 
remainder  of  32  ?  A  us.  504. 

25.  I  sent  j  of  my  money  to  Quebec  and  sent  ^  of  what  I  had 
left  to  Gait,  and  had  still  remaining  $400.  How  much  had  I  at 
first?  Afi8.  $1800. 

26.  Sold  342  bushels  of  wheat  at  $lf  a  bushel,  and  expended 
the  amount  received  in  buying  wood  at  $4J  a  cord.  How 
many  cords  of  wood  did  I  purchase?  A718.  1232  cords. 

27.  If  5  be  added  to  both  terms  of  the  fraction  f?,  how  much 
will  its  value  be  changed,  and  why  ? 

28.  If  I  of  4  pounds  of  tea  cost  $2i,  how  many  pounds  of  tea 
can  be  bought  for  $7i  ?    For  $12f  ? "  For  $f  ^  ? 

29.  I  exchanged  47J  bushels  of  corn,  at  $5  per  bushel,  for 
24|  bushels  of  wheat  ;  how  much  did  the  wheat  cost  a  bushel  ? 

30.  Bought  f  of  S^  acres  of  land  for  ^  of  $3584;  ;  \vhat  was 
the  price  per  acre  ? 

31.  A  can  do  a  piece  of  work  in  5  days,  B  can  do  the  same 
work  in  7  days  ;  in  what  time  can  both  together  do  it  ? 

32.  A  fisherman  lost  f  of  his  line  ;  he  then  added  8  feet, 
which  was  f  of  what  he  lost ;  what  was  the  length  of  the  line 
at  first?  Anf).  15  feet. 

33.  C.  Poison  bought  cloth  to  the  value  of  $2849?,  and  sold 
it  for  y'^  of  what  it  cost  him,  tliereby  losing  $5  a  yard  ?  How 
many  yards  did  he  purchase,  and  at  what  price  per  yard  ? 


I 


hi 


iiii 


.'*i 


*H 


!*ta 


I 


ili    K.n 


126 


FRA  CTIO  NS. 


-  84.  A  tailor  having  276|  yards  of  cloth,  sold  f  of  it  at  one 
time  and  f  at  another  ;  what  is  the  value  of  the  remainder  at 
$3  a  yard  ? 

35.  A  man  sold  -^^  of  his  farm  at  one  time,  f  at  another,  and 
the  remainder  for  $180  at  $45  an  acre  ;  how  many  acres  were 
there  in  the  farm  ? 

36,  A  merchant  owning  i|  of  a  ship,  sells  ^  of  his  share  to  B, 
and  f  of  the  remainder  to  C  for  $000^  ;  what  is  the  value  of  the 
ship? 


BEVIEW    AND    TEST    QUESTIONS. 

297.  1.  Define  Fractional  Unit,  Numerator,  Denominator, 
Improper  Fraction,  Reduction,  Lowest  Terms,  Simple  Fraction^ 
Common  Denominator,  and  Complex  Fraction. 

2.  What  is  meant  by  the  unit  of  a  fraction  ?  Elustrate  by 
an  example. 

3.  When  may  \  be  greater  than  \  ?    \  than  \1 

4.  State  the  three  principles  of  Reduction  of  Fractions,  and 
illustrate  each  by  lines. 

5.  Illustrate  with  lines  or  objects  each  of  the  following 
propositions:  .  :      v' 

I.  To  diminish  the  numerator,  the  denominator  remaining 

the  same,  diminishes  the  value  of  the  fraction. 
II,  To  increase  the  denominator,  the  numerator  remaining 
the  same,  diminishes  the  value  of  the  fraction. 

III.  To  increase  the  numerator,  the  denominator  remaining 

the  same,  increases  the  value  of  the  fraction. 

IV.  To  diminish  the  denominator,  the  numerator  remaining 

the  same,  increases  the  value  of  the  fraction. 

6.  What  is  meant  by  the  Least  Common  Denominator  ? 

7.  When  the  denoi  linators  of  the  given  fraction  are  prime  to 
each  other,  how  is  the  I^east  Common  Denominator  found, 
and  why  ? 


RE  VIEW. 


127 


8.  State  the  five  problems  in  reduction  of  fractions,  and  illus- 
trate each  by  the  use  of  lines  or  objects. 

9.  Show  that  multii)lying  the  denominator  of  a  fraction  by 
any  number  divides  the  fraction  by  that  numljer  (258). 

10.  Show  by  the  use  of  lines  or  objects  tlie  truth  of  the 
following: 


^ing 
ing 


to 
id. 


(1.)  ^  of  2  equals  f  of  1. 
(3.)  I  of  1  equals  ^  of  3. 


(3.)  I  of  5  equals  |  of  1. 

(4.)  ^  of  9  equals  4  times  \  of  9. 


11.  To  give  to  another  person  f  of  14  silver  dollars,  how 
many  of  the  dollar-pieces  must  you  change,  and  what  is  the 
largest  denomination  of  change  you  can  use  ? 

13.  Show  by  the  use  of  objects  that  the  quotient  of  1  divided 
by  a  fraction  is  the  given  fraction  inverted. 

13.  Why  is  it  impossible  to  perform  the  operation  in  |  +  f , 
or   in  ^  +  f ,   without  reducing  the   fractions  to  a  common 


denominator? 

14.  Why  do  we  invert  the  divisor  when  dividing  by  a  frac- 
tion ?    Illustrate  your  answer  by  an  example. 

5 

15.  What  objection  to  calling  —  a  fraction  (21*>)? 

16.  State,  and  illustrate  with  lines  or  objects,  each  of  the 
three  classes  of  so-caUed  Complex  Fractions. 

17.  Which  is  the  greater  fraction,  5  or  ^}l,  and  how  much  ? 

18.  To  compare  the  value  of  two  or  more  fractions,  what 
must  be  done  with  them,  and  why  ? 

34  4    2i  3     5^  2? 

19.  Compare  -^  and  ^ ;  -j^  and  ^h  5  '?rl  ^^^  51 »  »^d  show  in 

7         o    y         10    I*         oi 

each  case  which  is  the  greater  fraction,  and  how  much. 

30.  State    the   rule    for  working    each    of    the    following 
examples: 


(1.)  3|  +  4f +  8|. 

(3.)  (7|  +  5^)-(8-3i). 

(3.)  5  X I  of  I  of  27. 


(4.)8?x5| 
(5.) 


V  ^  ?• 


(^•)  I'^a"'"^' 


Explain  by  objects. 
Explain  by  objects. 


21.  Illustrate  by  an  example  the  application  of  Cancellation 
in  multiplication  and  division  of  fractions. 


1 

t    I' 


i-.y 


DECIMAL    FRACTIONS. 


DEFINITIONS. 

298.  A  unit  is  separated  into  dec'unal parts  when  it  is 
divide.'  ij>  '    mthd;  thus, 


f  <(i 


UNIT. 


DECIMAL  FARTS. 


399.  A    T}(hnal  FractionaZ   Unit  is  one   of  the 

decirnalpa.'hsui  ^.'-^^Jning. 

300.  By  makiLg  ?,  whole  or  unit  into  decimal  parts,  and 
one  Gi  these  jare  into     '  'loal  parts,  and  so  on,  we  obtain  a 

series  ot  dist'/iict  ct  Ir  .^  v/-.  •  ■••<rr.''l  fractional  units,  each  ^  of 
the  preceding,  having  as  diiaomiuators,  respectively,  10,  100, 
1000,  and  so  on. 

Thus,  sei)arating  a  whole  into  decimal  parts,  we  have,  accord- 
ing to  (235),  1  =  III ;  making  y'^  into  decimal  ])art8,  we  have, 
according  to  (24 1 ),  -^^q  —  .^^^ ;  in  the  same  manner,  j^q  = 
if Sw'  ttjV^  =  ToVo ()»  "^d  so  on.  Hence,  in  the  series  of  frac- 
tional units,  j\,  ,i„,  j^Vtt*  ^^^  ^  on,  each  unit  is  one-tenth  of 
the  preceding  unit. 

301.  A  Deri  in  ff  I  Fi*«cf  eon  is  a  fraction  whose  denom- 
inator is  10,  100,  1000,  etc.,  or  1  with  any  number  of  ciphers 
annexed.     Thus,  j^o,  Tojst  i^^n*  *^re  decimal  fractions. 

302.  The  Dcclinal  Sif/n  (.),  called  the  decimal  point,  is 
used  to  express  a  decimal  fraction  without  writing  the  denom- 
inator, and  to  distinguish  it  from  an  integer. 

Observe  that  a  decimal  fraction  should  be  distinguished  from 
its  corresponding  decimal,  just  as  a  vulgar  fraction  should  be 
distinguished  from  its  quotient.     Thus,  |  =  .35,  and  .35  =  ^^. 


NOTATION    AND     NUMERATION.       129 


NOTATION  AND  NUMEEATIOK 

i503.  Prop.  I. — A  decimal  fraction  is  expressed  without 
toriting  tTie  denominator  by  vsing  the  decimal  point,  and  placing 
the  numerator  at  the  right  of  the  period. 

Thus,  j'^y  is  expressed  .7  ;  yVff  is  expressed  .35. 

Observe  that  the  number  of  figures  at  the  right  of  the 
period  is  always  the  same  as  the  number  of  ciphers  in  the 
denominator ;  hence,  the  denominator  is  indicated,  although 
not  written. 

Thus,  in  .54  there  are  two  figures  at  the  right  of  the  period ; 
hence  we  know  that  the  denominator  contains  two  ciphers,  and 
that  .54  =  -^j^. 

Express  the  following  decimal  fractions  without  writing  the 
denominators: 


Lc- 

of 


Irs 


18 


Im 


1. 
2. 
3. 


9 


4.       #!,. 

5. 
6. 


83 

JOJS- 
97 


7. 
8. 
9. 


4xrt 
1  0  0  ff  • 


10. 

11. 

12. 


58 

706 
1  OTf  (5- 

so  SB 

"I  0  0  u  ff* 


304.  A  decimal  fraction  expressed  without  writing 
the  denominator  is  called  simply  a  Decimal, 

Thus,  we  speak  of  .79  as  the  decimal  seventy-nine,  yet  we 
mean  the  decimal  fraction  seventy -nine  hundredths. 

305.  Prop.  II.— Ciphers  at  the  left  of  significant  figures 
do  not  increase  or  diminish  the  number  expressed  by  these 
figures. 

Thus,  0034  is  thirty-four,  the  same  as  if  written  34  without 
the  tw^o  ciphers. 

From  this  it  will  be  seen  that  the  number  of  figures  in  the 
numerator  of  decimal  fractions  can,  without  chanc:ing  the  frac- 
tion, be  made  equal  to  the  number  of  ciphers  in  the  denomina- 
tor by  writing  ciphers  at  its  left ;  thus,  y^Vir  =  AVtt- 

Hence,  YrnyTr  ^^  expressed  by  using  the  decimal  point  and  two 
ciphers  thus,  .007. 


; ! 


1  s-* 


130 


DECIMAL     FRACTIONS, 


Observe^  that  while  the  number  of  parts  in  the  numerator  is  not  changed 
"by  preflxing  the  two  ciphers,  ye\  the  7  is  moved  to  the  tliird  place  on  the 
right  of  the  decimal  point.  Hence,  according  to  (303),  the  denominator 
is  indicated. 

300.  Prop.  III. —  When  t7ie  fraction  in  a  mixed  number  is 
expressed  decimally,  it  is  loritten  after  the  integer,  wltJi  the  deci- 
mal point  between  them. 

Thus,  57  and  .09  are  written  57.00 ;  8  and  .0034  are  written 
8.0034. 

Express  as  mie  number  each  of  the  following ; 

1.  39  and  .9.  4.     23  and  .1.  7.     303  and  .080. 

2.  42  and  .07.  5.    9  and  .08.  8.     260  and  .008. 

3.  76  and  .507.  6.     7  and  .074.         9.     907  and  .062. 

From  these  illustrations  we  obtain  the  following  rule  for 
writing  decimals: 

RULE. 

307.  Write  the  numerator  of  the  given  decimal  fraction. 
Make  the  number  of  figures  icritten  equal  to  the  number  of 
ciphers  in  the  denominator  by  prefijcing  ciphers.  Place  at  the 
left  the  decimal  point. 


EXERCISE    FOR    PRACTICE. 

308.    Express  the    following   decimal 
writing  the  denominator  : 


1. 
2. 
3. 
4. 
5. 
6. 


8 

TUTS' 


7. 

44«! 

8. 

43 

Twins- 

9. 

TTSOO- 

10. 

ihi^js 

11. 

ToVjTcr 

12. 

?o« 

roouir 

1    fractions  w 

13. 

TinnjW* 

14. 

T^foUTT' 

15. 

T^^FffTT* 

16. 

Tff^C^^' 

17. 

TollloH- 

18. 

TUinfTsinr' 

without 


19.  Three  hundred  seven  hundred-thousandths. 

20.  Six  thousand  thirty-four  millionths. 

21.  Seventy-five  ten-millionths. 

22.  Nine  thousand  sixty-seven  himdred  millionths. 


n 


EXERCISE. 


131 


or 


Express  the  following  by  writing  the  denominator  ;   thus, 
.033  =  xof^j* 


23.  .072. 

24.  .0020. 

25.  .37094. 


26.  .0400302. 

27.  .005062. 

28.  .002000. 


29.  .00974. 

30.  .063. 

31.  .00305. 


;jOt).    Prop.  IV. — Ei'cry  figure  in  the  numerator  of  a  deci- 
mal fraction  represents  a  distinct  order  of  decimal  units. 


Tims,  iVifV  is  Pqiial  ToVff  +  toou  +  lo'ffn-  But,  according 
to  (344),  i%%  =  i'„  and  ^gg^  =  ^-J^.  Hence,  5,  3.  and  7 
each  represent  a  distinct  order  of  decimal  fractional  units,  and 
TnV?T>  or  .537  may  be  read  5  tent/is  3  hundredths  and  7  thou^ 
sandths. 

Analyze  the  following  ;  thus,  .0709  =  y^  +  yi^f  ciy. 


1.  .046. 

2.  .909.    ^' 

3.  .0027. 


4.  .01207. 

5.  .04063. 

6.  .05095. 


7.  .0300702. 

8.  .0003092. 

9.  .0060409. 


s 


't    !' 


at 


15  lO.  Prop.  V. — A  decimal  is  read  correctly  by  reading  it 
as  if  it  were  an  integer  and  giving  the  name  of  the  right-hand 
order. 


900      4.       70         I  6 


TSITQ- 


Hence  is  read,  nine  hun- 


Thus,  .975  —  xwQ^  "^  ToiTTy 
dred  seventy-five  thousandths. 

1.  Observe  that  when  there  are  ciphers  at  the  left  of  the 
decimal,  according  to  (305),  they  are  not  regarded  in  reading 
the  number  ;  thus,  .002  is  read  sixty-two  thousandt?is. 

2.  The  name  of  the  lowest  order  is  found,  according  to 
(303),  by  prefixing  1  to  as  many  ciphers  as  there  are  figures 
in  the  decimal.  For  example,  in  .00209  there  are  five  figures ; 
hence  the  denominator  is  1  with  five  ciphers ;  thus,  10CK)00, 
read  hundred-thousandths. 

From  these  illustrations  we  obtain  the  following 

RULE. 

311.  Read  the  decimal  as  a  ichole  number  ;  then  pronounce 
the  name  of  the  lowest  or  right-hand  order. 


132 


DECIMAL     FRACTIONS. 


Read  the  following: 


1.  .004. 

2.  .00902. 

3.  .4097. 

4.  .0C419. 

5.  .02006. 

6.  .30007. 


7.  .5307. 

8.  .01007. 

9.  .0058. 

10.  .000904. 

11.  .040972. 

12.  .000252. 


13.  .020304. 

14.  .0090409. 

15.  .00030503. 

16.  .00200059. 

17.  .0000007. 

18.  .00034657. 


19.  What  is  the  denominator  of  .000407  ?    How  is  it  found  ? 
What  is  the  numerator,  and  how  read  ? 

20.  What  effect  have  the  ciphers  in  .0085  ? 

21.  Write  a  rule  for  expressing  a  decimal  by  writing  its 
denominator. 

312.  The  relation  of  the  orders  of  units  in  an  integer  and 
decimal  will  be  seen  from  the  following  table : 

♦ 

DECIHAL  NUMERATION  TABLE. 


09 

o 


5 


CD 

a 
o 


S 


P 


5 


5 


a 
a 
m 

o 

I 

« 

(X 

a 

s 

H 
5 


s 

S3 

O 

I 

a 

5 


o 

h 
5 


-a 

a 
o 
K 

5 


n 


-o 

i 

• 

J3 

a 

ai 

6 

T! 

0 

• 

J3 

• 

a 

o 

a 
o 

5 

e 
e 

es 
tt. 

S 

o 
.a 
•^^ 

c 

c 

.a 
*^ 

o 

'a 

m 

0 

MM 

.2 

S 

a 

;^ 

P 

H 

HH 

H 

H 

EC 

Eh 

5 

• 

5 

5 

5 

5 

5 

5 

5 

CO 

g 


'F^        (1> 


3     a 

e 


5 


5 


ORDERS  OP  INTEGERS. 


Y ' 

ORDERS  or  DECUfAL  FRACTIONS. 


Observe  carefully  the  following : 


{-£ 


1.  The  ZTnit  is  the  standard  in  both  cases.  The  integral 
orders  are  multiples  of  one  iiiiit^  and  the  decimal  orders  are 
decimal  fractions  of  one  unit. 

2.  Figures  that  are  equally  distant  from  the  units'  place  on 
the  right  or  left,  have  corresponding  names ;  thus,  tenths  cor- 
respond to  tens,  hundredths  to  hundreds,  and  so  on. 


IS. 


ral 
ire 


EXAMPLES, 


133 


3.  In  reading  an  integer  and  decimal  together,  "  and  "  should 
not  be  lifted  anywhere  but  between  the  integer  and  fraction. 

Thus,  0582.643  should  be  read,  nine  thousand  five  hundred 
eighty-two  and  six  hundred  forty-three  thousandths. 

4.  Dimes,  cents,  and  mUls  being  respectively  tenths,  hun- 
dredths,  and  thousandths  of  a  dollar,  are  written  as  a  decimal. 
Thus,  $.347  is  3  dimes,  4  cents,  and  7  mills.  In  reading  dimes, 
cents,  and  mills,  the  dimes  are  read  as  cents.  Thus,  $62,538  is 
read,  62  dollars,  53  cents,  8  mills. 

EXAMPLES     FOR    PRACTICE. 

313.  Read  the  following : 

1.  $285.56. 
r  2.  $920,905. 

3.  $203.06. 

4.  $70,007. 

5.  $300.02. 

6.  $9,807. 


19. 
20. 
21. 


5 

ruo- 


I^TiHnsTr* 


7.  20040.20104. 

m 

9.00006. 

8.  90309.00703. 

14. 

10.1. 

9.    3001.0201. 

15. 

100.0003. 

10.  50400.000205. 

16. 

35.00045. 

11.    2070.00301. 

17. 

9.30005. 

12.  $9005.009. 

18. 

10.000001. 

g  without  writing  the  denominator  : 

23.     407Tiy^*Wij. 

25. 

10201,^^71. 

23.     703^«^%^. 

26. 

4030,|UiT- 

24.      9Y7y5^J^. 

27. 

lOOjjjijj^^^ny' 

28.  Write  with  figures  :  Eighty-two  thousandths ;  four  hun- 
dred five  millionths  ;  eight  ten- thousandths. 

29.  Three  thousand  six  hundred-millionths ;  ninety-one 
millionths  ;  six  hundred  four  thousand  one  billionths. 

30.  Eighty.four  and  seven  ten-thousandths;  nine  thousand 
six  and  five  hundred  nine  ten-millionths ;  six  and  five  mil- 
lionths. 

31.  Nine  thousand  thirty-seven  and  three  hundred  seven 
billionths  ;  one  million  one  and  one  thousand  one  ten-millionths. 


■'V 


■ 'I 


V    '1 

r 

t  Ml 


'.  -i.t 


m 


134 


DECUfAL     FRACTIONS* 


EEDUCTIOJSr. 


PREPARATORY     PROPOSITIONS. 

The  following  preparatory  proposi  'ions  Bliould  be  very  eare- 
fvlly  studied. 

314.  Prop.  \.— Annexing  a  cipher  or  multiplying  a  num- 
ber by  10  introduces  into  the  number  the  two  prime  factors  2 
and  5. 

Thus,  10  being  equal  2  x  5,  7  x  10  or  70  =  7  x  (2  x  5).  Hence 
a  number  must  contain  2  and  5  as  a  factor  at  least  as  many 
times  as  there  are  ciphers  annexed. 

315.  ^ROP.  II. — A  fraction  in  its  lowest  terms,  whose  de- 
nominat&r  contains  no  other  pnme  factors  than  2  or  5,  can  be 
reduced  to  a  simple  decimal. 

Observe  that  every  cipher  annexed  to  the  numerator  and 
denominator  makes  each  divisible  once  by  2  and  5  (314). 
Hence,  if  the  denominator  of  the  given  fraction  contains  no 
other  factors  except  2  and  5,  by  annexing  ciphers  the  numera- 
tor can  be  made  divisible  by  the  denominator,  and  the  fraction 
reduced  to  a  decimal. 

Thus,  |  =  U^^  (224—11).  Dividing  both  tci-ms  of  the 
fraction  by  8  (224— III),  we  have  l^^  =  y«^Vj  =  .875. 


r  1 


1. 

2. 
3. 


Reduce  to  decimals  and  explain  as  above  : 

¥•  ^      TU'  7.      f. 


4.     1^. 

7. 

5.    H. 

8. 

6.     A. 

9. 

^V 


r¥ir. 


10. 

Ih 

13 

11. 

A. 

14 

12. 

#2. 

15 

49 

m- 

1  s 


1 6.  How  many  ciphers  must  be  annexed  to  the  numerator 
and  denominator  of  f  to  reduce  it  to  a  decimal  ? 

17.  Reduce  -l  to  a  decimal,  and  explain  why  the  decimal 
must  contain  three  places. 

18.  If  reduced  to  a  decimal,  how  many  decimal  places  will  | 
make  ?    Will  ^'V  make  ?    Will  -^^  make,  and  why  ? 


'> 


the 


Iff* 


UtOT 

lal 
HI 


PRKPARATORY    P  R  O  P  0  S ITI 0  XS .      135 

511 0,  Prop.  III.— ^  frnrtion  in  its  lowest  terms,  whose  de* 
nominator  contdina  any  other  prime  factors  ttiau  4  or  5  can  be 
reduced  only  to  a  complex  decimal. 

Observe  that  in  this  case  annexing  ciphers  to  the  numerator 
and  denominator,  wliich  (;J14)  introduces  only  the  factors  2 
and  5,  carmot  make  the  numerator  divisible  by  the  giv»'n  de- 
nominator, which  contains  other  prime  factors  than  2  or  5. 

Hence,  a  fraction  will  remain  in  the  numerator,  after  divid- 
ing the  numerator  and  denominator  by  the  denominator  of  the 
given  fraction,  however  far  the  division  may  be  carried. 

Thus,  \\  =  iUSg  (224—11).    Di\'iding  both  numerator  and 

denominator  by  21,  we  have  ^[^  =  ^^  =  .523H,  a  com- 
plex  decimal. 

Reduce  and  explain  the  following  : 

1 .  How  many  tenths  in  f  V    In  f  ?    In  f  ?    In  |  ?    In  {'^  ? 

2.  Reduce  to  hundredths  f  ;  5  ;  ^\;  ^7^ ;  ^1^  ;  IS- 

3.  How  many  thousandths  in  ^  ?    In  |  ?    Tn  ^  ?    In  j%  ? 

J51  7.  Pkop.  IV. — The  same  set  of  figures  must  recur  indeji- 
niteJy  in  the  same  order  in  a  co'mpi  r  decimal  ichich  cannot  he 
reduced  to  a  simple  decimal. 

Thus    ^~  -  ^^^^  -  ^'^^^'^  -  G363  V 
ihus,  ^^  _  j^^^^  _  ^^^^  -  .WbdfV- 

Observe  carefully  the  following : 

1.  In  any  division,  the  number  of  different  remainders  that 
can  occur  is  1  less  than  the  number  of  units  in  the  divisor. 

Thus,  if  5  is  the  divisor,  4  must  be  the  greatest  remainder 
we  can  have,  and  4,  3,  2,  and  1  are  the  only  possible  different 
remainders  ;  hence,  if  the  division  is  continued,  any  one  of 
these  remainders  may  recur. 

2.  Since  in  dividing  the  numerator  by  the  denominator  of 
the  given  fraction,  each  partial  dividend  is  formed  by  annex- 
ing a  cipher  to  the  remainder  of  the  previous  division,  when  a 


S'-» 


w 


I  If 


; 


^  ' 


136 


DECIMAL     FsRACTIOXS, 


remainder  recurs  the  partial  dividend  must  again  be  the  same 
as  was  used  when  this  remainder  occurred  before ;  hence  the 
same  remainders  and  quotient  figures  must  recur  in  the  same 
order  as  at  first. 

3.  If  we  stop  the  division  at  any  point  where  the  given 
numerator  recurs  as  a  remainder,  we  have  the  same  fraction 
remaining  in  the  numerator  of  the  decimal  as  the  fraction 
from  which  the  decimal  is  derived. 


Thus, 


or  ZTT   = 


7 
11 


11  ' 
70000 


700       63t'i 


1100       1000 


"  =  .63T'r ; 


110000  =  10000  =  •^^^^'^'  ^'^^  ^  ^"- 


Jl'18.  Prop.  V. — The 'value  of  a  fraction  which  can  only 
he  reduced  to  a  complex  decimal  is  expressed,  nearly,  as  a  simple 
decimal,  by  rejecting  the  fraction  from  the  numerator. 

3       27  '- 
Thus,  —  =  ~~  (316).     Rejecting  the  ^  from  the  numer- 
ator, we  have  ^^^,  a  simple  fraction,  which  is  only  y\  of  ^J^ 

27A 
smaller  than  the  given  fraction  ^  or  -jk^^ 

Observe  the  following : 

1.  By  taking  a  suflBcient  number  of  places  in  the  decimal,  the 
true  value  of  a  complex  decimal  can  be  expressed  so  nearly  that 
what  is  rejected  is  of  no  consequence. 

Thus,  -  =  -^yoi^QQQ^;  rejecting  the  -^j  from  the  numer- 
ator, we  have  xVinHrffffVff'  o'  .27272727,  a  simple  decimal, 
which  is  only  ^  of  1  hundred-millionths  smaller  than  the 
given  fraction. 

3.  The  approximate  value  of  a  complex  decimal  which  is 
expressed  by  rejecting  the  given  fraction  from  its  numerator  is 
c  illed  a  Circultding  Decimal,  because  the  same  figure  or  set  of 
figures  constantly  recur. 


Ihe 
lat 


DEFINITIONS. 


137 


319.  Prop.  VI. — Diminishing  the  numerator  and  denom- 
inator by  the  same  fractional  part  of  each  does  not  change  the 
value  of  a  fraction. 

Be  particular  to  .master  tLe  followingf,  as  the  reduction  of 
circulating  decimals  to  common  fractions  depends  upon  this 
proposition. 

1.  The  truth  of  the  proposition  may  be  shown  thus : 

9^_^_tof9_9^-3_6_3 
12  ~  13  — i  of  12  ~  12-4""  8  ~  4* 

Observe  that  to  diminish  the  numerator  and  denominator 
each  by  |  of  itself  is  the  same  as  multiplying  each  by  f .  But 
to  multiply  each  by  f ,  we  multiply  each  by  2  (t224— II),  and 
then  divide  each  by  3  (224 — III),  which  does  not  change  the 
value  of  the  fraction ;  hence  the  truth  of  the  proi>osition. 

2.  From  this  proposition  it  follows  that  the  value  of  a  frac- 
tion is  not  changred  by  subtracting  1  from  the  denominator  and 
the  fraction  itself  from  the  numerator. 


Thus,  ^  =  I      !  =  ^^      Observe 
0       5  —  1        4 


that    1   is  the   1  of   the 


denominator  5,  and  |  is  |  of  the  numerator  3  ;  h«nce,  the 
numerator  and  denominator  being  each  diminished  by  the  same 
fractional  part,  the  value  of  the  fraction  is  not  changed. 


|i 


a. 

le 


lis 
is 
of 


DEFINITIONS. 

320.  A  SimjUe  Decimal  is  a  decimal  whose  numerator 
is  a  whole  number ;  thus,  ^Yff  or  .93. 

Simple  decimals  are  alno  called  Finite  Decimals. 

321.  A  Complex  Decimal  is  a  decimal  whose  numer- 

26* 
ator  is  a  mixed  number:  as  -—  or  .263. 

100  ^ 

10 


138 


DECIMAL     FRACTIONS. 


There  are  two  classes  of  complex  decimals : 

1.  Those  whose  valae  can  be  expressed  as  a  simple  decimal  (315),  as 
.aaj  =  .235 ;  .32:  =  .8275. 

2.  Those  whose  value  cannot  be  expressed  as  a  simple  decimal  (316), 
as  .53J  =  .58333  and  bo  on,  leaving,  however  flar  we  may  carry  the  decimal 
places,  i  of  1  of  the  lowest  order  unexpressed.    See  (3 1 7). 

322.  A  Circulating  Decimal  is  en  approximate  value 
for  a  complex  decimal  which  cannot  be  reduced  to  a  simple 
decimal. 

Thus,  .666  is  an  approximate  value  for  .6662  (318). 

323.  A  Repetend  is  the  figure  or  set  of  figures  that  are 
repeated  in  a  circulating  decimal. 

324.  A  Clrculafiuff  Dechnal  is  exjtressed  hy  writ- 
ing the  repetend  once.  When  the  repetend  consists  of  one 
figure,  a  point  is  placed  over  it ;  when  of  more  than  one  figure, 
points  are  placed  over  the  first  and  last  figures  ;  thus,  .333  and 
so  on,  and  .592592+  are  written  .3  and  .592. 

325.  A  Pure  Circulating  Decimal  is  one  which 
commences  with  a  repetend,  as  .8  or  .394. 

326.  A  Mixed  Circulating  Decimal  is  one  in  which 
the  repetend  is  preceded  hy  one  or  more  decimal  places,  called 
the  finite  part  of  the  decimal,  as  .73  or  .004725,  in  which  .7  or 
.004  is  the  finite  part. 


ILLUSTRATION     OF    PROCESS. 

327.    PuoB.    I.— To  reduce  a  common   fraction   to  a 
decimal. 

Reduce  I  to  a  decimal. 


3 
8 


3000 
800() 


375 


1000 


=  .375. 


Explanation.  —  1.   We   annex 
the   same   number   of  ciphers   to 
both  terms  of  the  fraction  (224— 
Prin.  II),  and  divide  the  resulting  terms  by  8,  the  eigniflcant  figure  In 


..^-.3;.T,>»W':-,VWr*iJ- 


EXE  R  CISE, 


139 


the  denominator  which  must  give  a  decimal  denominator.     Hence,  » 
expressed  decimally  if  .375.' 

2.  In  case  annexing  ciphers  does  not  make  the  numerator  divisible 
(316)  by  the  tiisnificant  flguree  in  the  denominator,  the  number  of  places 
in  the  decimal  can  be  extended  indefinitely. 

In  practice,  we  abbreviate  the  work  by  annexing  the  ciphers 
to  the  numerator  only,  and  dividing  by  the  denominator  of  the 
given  fraction,  pointing  off  as  many  decimal  places  in  the 
result  as  there  were  ciphers  annexed.    Hence  the  following 


Lich 
lied 
or 


a 


RULE. 

328.  /  Annex  ciphers  to  the  numerator  and  divide  by  the 
denominator. 

II.  Point  off  as  many  places  in  th^  result  as  there  are  ciphers 
annexed. 

EXERCISE    FOR    PRACTICE. 


329.  Reoixce  to  simple  decimals; 


813 

ml' 


S7 


1.    If.  8.     /j-  5.     II.  7. 

2  7  4         27  ft  «»  ft 

•'*'"*  Reduce  to  a  complex  decimal  of  four  decimal  places: 

ft        B  11        23  1.^        !«  15 

10      11  19      13  14      ,**  16 

Find  the  repetend  or  approximate  value  of  the  following : 

17.  if.  20.     II.  23.     11.  2C.    8?. 

18.  2.  21.     U.  24.     1%.  27.    ^{^j^. 

19.  i|.  22.     y.  25.     II.  28.    Ui^. 

330.    Prob.  II.— To  reduce  a  simple  decimal  to  a  com- 
mon fraction. 

Reduce  .35  to  a  common  fraction. 

86  7  ExPLANATKN.—Wc  wHte  thc  decimal  with 

the  dcnominiiK  r,  and  reduce  the  fraction  (244) 
to  ite  lo\vei*t  tt-im** ;  hence  the  following 


.35  = 


100      20 


1 


lex 
to 

14- 

In 


RULE. 

331.  Express  the  decimal  by  irHting  the  denominaior,  then 
reduce  the  fraction  to  its  lowest  terms. 


r  r 


f,*-      : 


H 


140 


DECIMAL     FB  ACTIONS. 


mt 


EXAMPLES     FOR     PRACTICE. 

332.  Reduce  to  common  fractions  in  their  lowest  terms: 


1.  .840. 

2.  .215. 

3.  .750. 


4.  .0125. 

5.  .0054. 

6.  .0064. 


7.  .008025. 

8.  .00096. 

9.  .00075. 


10.  .00512. 

11.  .0625. 

12.  .00832. 


333.    Pros.  III.— To  find  the  true  value  of  a  pure  cir- 
culating decimal. 


Find  the  true  value  of  .72. 

72  _  _  7^  _  _  72  _  8 
100  ~  100  -  1  ~"  99  ~  11 


7'2  — 
.1^  — 


ExPLAKATioN.— In  taking 
,72  as  the  approximate  value 
of  a  given  fraction,  we  have 


subtracted  the  given  fraction  from  its  own  numerator,  as*  shown  in  (318— 
V).  Hence,  to  find  the  true  vahie  of  t'u\,  we  must,  according  to  (319— 
VI,  2),  subtract  1  IVom  the  denominator  100,  which  makes  the  denominator 
as  many  9'8  as  there  are  places  in  the  repetend ;  hence  the  following 

RULE. 

334r.  Write  the  figures  in  the  repetend  for  the  numerator 
of  the  fraction,  and  as  many  9's  as  there  are  places  in  the  repe- 
tend for  the  denominator,  and  reduce  the  fraction  to  its  lowest 
terms. 

EXAMPLES     FOR    PRACTICE. 
335*  Find  the  true  value  of 

1.  78.  4.    856.  7.    324.  10.    2718. 

2.  36.  5.    372.  8.    i89.  11.    5368. 

3.  54.  6.    135.  9.    836.  12.    8163. 


Find  tlie  true  value  as  improper  fractions  of 

13.  37.8i.  16.    53.324.  19.  29.i88i. 

14.  9.i08.  17.    89.54.  20.  63.2745. 

15.  3.504.  18.    23.758.  21.      6.636. 


EXAMPLES. 


141 


(1)    .318  r=  .3J?j 


336.  PRO"  IV.— To  find  the  true  value  of  a  mixed  cir- 
culating decimal. 

Find  the  true  value  of  SiS. 

_  3J2  _  315  ^  7 
»0«j  -   10  -  990      22 

Explanation.— 1.  We  find,  according  to  (333)  the  true  value  of  the 
repetend  .Ols,  which  Is  .OiJ.  Annexing  this  to  the  .3,  the  finite  part,  we 
have  .31si  the  true  value  of  .Sis  in  the  form  of  a  complex  decimal. 

gin 

2.  We  reduce  the  complex  decimal  .3JJ,  or  '^*  to  a  eimple  fraction  hy 
multiplying,  according  to  (289),  both  terms  of  the  fraction  hy  99,  giving 
^"*  -  {J5  =  ,^,.    Hence  the  true  value  of  .3*18  is  ^V 


3'* 


(2)     .318     Given  decimal. 
3    Finite  part. 


Abbreviated  Solution.— Obsenr'e 
3iS 


315        m 


■h' 


that  in  simplifying  *^*,  wc  multiplied 

both  terms  by  9ft.  Instead  of  multi- 
plying the  3  by  9ft,  wo  may  multiply 
by  100  and  subtract  8  from  the  product.  Hence  we  add  the  18  to  .300,  and 
subtract  3  from  the  resiUt,  which  gives  us  the  true  numerator.  Hence  the 
following 

RULE. 
337.    I.  Find  the  true  value  of  the  repetend,  annex  it  to  the 
finite  imH,  and  redurs  the  complex  decimal  thus  formed  to  a 
simple  fraction. 

To  abbreviate  the  work  : 

//.  From  the  given  decimal  subtract  the  finite  part  for  a 
numerator,  and  for  a  dcnominatoi'  icrite  as  many  O's  as  there 
are  figures  in  the  repetend,  with  as  many  ciphers  annexed  as 
there  are  figures  in  the  finite  2)art. 


\\ 


;:» 


■r 


EXAMPLES    FOR  PRACTICE. 
338.    Find  the  true  value  of 

1.  .959.       4.  .00641.  7.  .008302685. 

2.  .7i2.        5.  .04328.  8.  .000035739. 

3.  .486.       6.  .03287.  9.  .020734827. 


143 


D  E  CI  MA  L     Fit  ACTIO  AS, 


[:'       '• 


Find  the  true  value,  in  tlie  form  of  an  improper  fraction,  of 

10.     9.753.  12.     7.86.  14.    5.39. 

11     5.328.  13.    2.43.  15.     12.227. 


ADDITIOjN^. 

PREPARATORY     PROPOSITION. 

J^JJl).  Any  two  or  inore  decimals  can  he  reduced  to  a  common 
denomiiuitor  hy  annexing  ciphers. 

Thus,  .7  =  3^.  a"^>  according  to  (224—11),  ^'^  =  ^^^  = 
^'^ooj  =  I'oOflVff.  and  so  on  ;  therefore,  .7  =  .70  =  .700  =  .7000. 
Hence  any  two  or  more  decimals  can  be  changed  at  once  to  the 
same  decimal  denominator  by  annexing  ciphers. 

ILLUSTRATION     OF     PROCESS. 

;54:0.  Find  the  sum  of  34.8,  6.037,  and  27.62. 

Explanation.— 1.  We  arran^  the  num- 
bers BO  that  units  of  the  same  order  stand  In 
the  same  column. 

2.  We  reduce  the  decimals  to  a  common 
denominator,  as  shown  in  (1),  by  annexing 
ciphers. 

.3.  We  add  as  in  integers,  placing  the 
decimal  point  before  the  tenths  in  the  t^um. 

In  practice,  the  ciphers  are  omitted,  as  shown  in  (2),  but  the 
decimals  are  re^rded  as  reduced  to  a  common  denominator. 

Thus  the  3  hundredths  in  the  second  number  and  the  2  hundredths  in  the 
third,  when  added  are  written,  as  shown  in  (2>,  as  50  thousandths ;  in  the 
same  manner,  the  8  tetiths  and  6  tentfis  make  1400  thousandths,  or  1  unit  and 
400  thousandths.  The  1  unites  added  to  the  units  and  the  4  written  in  the 
tenths'  place  as  400  thoii^atidfhs. 

From  this  it  will  be  seen  that  the  addition  of  decimals  is 
subject  to  the  same  laws  (250 — I  and  II)  and  rule  (252)  as 
other  fractions. 


(1.) 

(2.) 

34800 

348 

6.037 

6.03T 

27.620 

27.62 

68.457 

68.457 

I      1 


S  UBTRA  CTION, 


143 


EXAMPLES    FOR    PRACTICE. 

341.  Find  the  sum  of  the  following,  and  explain  as  above 

1.  9.07,  36.000,  84.9,  5.0036,  23.608,  and  .375. 

2.  38.9.  7.05,  59.82,  365.007,  93.096,'  and  8.504. 

3.  $42.08,  $9.70,  $89.57,  ^396.02,  and  $.89. 

4.  395.3,  4.0701,  9.96,  and  83.0897. 

5.  .039,  73.5,  .0407,  2.602,  and  29.8. 

6.  8.0093,  .805,  .03409,  7.69,  and  .0839. 

7.  .80003,  3.09.  13.36.  97.005,  and  .9999. 

8.  $.87,  $32.05.  $9.  $75.09,  $.67,  and  $3.43. 


SUBTRACTIO]^. 


343.  Find  the  difference  between  83.7  and  46.392. 


83.700 
45.392 

38.308 


Explanation.— 1.  We  arrun<,'e  the  numbers  so  that  units 
of  the  same  order  stand  in  the  same  column. 

2.  We  reduce  the  decimals?,  or  regard  them  as  reduced  to 
a  common  denominator,  and  then  subtract  as  in  whole 
numbers. 
The  reason  of  this  course  is  the  same  as  given  in  addition.    The  ciphers 
are  also  usually  omitted. 


EXAMPLES     FOR    PRACTICE. 


343.  Subtract  and  explain  tlie  following  : 

1.  39.073  -  7.0285.  6.    54.5  -  37.00397. 

2.  834.9  -  52.47.  7.    379.000001  -  4.0396. 

3.  $67.09  -  $29.83.  8. 

4.  83.003  -  45.879.  9. 

5.  $95.02  -  $78.37.  10. 


96.03  -  89.09005. 
.7  -  .099909. 
.09  -  .0005903. 


t 


144 


DECIMAL     FRACTIONS. 


11.  A  man  paid  out  of  $3432.95  the  following  sums :  $342.06, 
$593,738,  $729,089,  $1363.43,  $296,085,  $37,507.  How  much 
has  he  left?  ^7W.  $73,091. 

13.  In  a  mass  of  metal  there  are  183.741  pounds ;  ^  of  it  is 
iron,  35.305  pounds  are  copper,  and  3.0009  pounds  are  silver, 
and  the  balance  lead.     How  much  lead  is  there  in  the  mass  ? 

13.  A  druggist  sold  74.53  pounds  of  a  costly  drug.  He  sold 
in  March  10|  pounds,  in  April  25.135,  in  May  31f,  and  the 
balance  in  June.     How  many  pounds  did  he  sell  in  June  ? 


Find  the  decimal  value  of 

14.  {^  -  3|)  +  (7/j  -  If)  -  (9.23  -  8.302). 

15.  {$85J  -  $37|)  +  (f  of  $184.20  - 

16.  $859,085  -  ($138|  +  %^)  + 


>S4 


I/, 


MULTIPLICATIOIT. 


344.  Multiply  3.37  by  8.3. 


(1.)  3.27  X  8.3  =  ^^^x«-^. 


(2.) 


327      83 
100  "^  10 


27141 
1000 


=  37.141. 


Explanation.— 1.  Observe  that  3.27  and  8.3  are  mixed  nnmberei;  hence, 
according  to  (271),  they  are  reduced  before  being  multiplied  to  improper 
^'actions,  as  shown  in  (1). 

2.  According  to  (265),  \U  x  ??,  as  shown  in  (2),  equals  27.141.  Hence, 
27.141  is  the  product  of  3.27  and  8.3. 

The  work  is  abbreviated  thus  : 

(3.) 

3  37  ^®  obeerve,  as  shown  in  (2),  that  the  product  of  3.27  and 

8.3  must  contain  as  many  decimal  places  as  there  are  deci- 
mal places  in  both  numbers.  Hence  we  multiply  the  num- 
bers as  if  integers,  as  shown  in  (8),  and  point  off  in  the 
product  as  many  decimal  places  as  there  are  decimal  places 
in  both  numbers.    Hence  the  following 


a3 

981 
2616 


27.141 


tl 

he 

t0( 

ya 

the 
] 

2 

3 
3 


EXAMPLES. 


146 


RIJIiE. 

345.  Multiply  as  in  integers,  and  from  the  right  of  th£  pro- 
duet  point  off  as  many  fgttres  for  decimals  as  there  are  decimal 
places  in  the  multiplicand  and  multiplier. 


EXAMPLES    FOR    PRACTICE. 


346.  Multiply  and  explain  the  following : 


1.  13.4  X. 37. 
3.  7.3x4.9. 

3.  35.08x6.2. 

4.  183.65  X. 7. 


6.  73. 406  X. 903. 

7.  340007x8.43. 

8.  . 4903 X. 06. 

9.  5.04  X  .072. 


11. 
12. 
13. 


.0007  X  .036. 
.009  X.  008. 
.0405  X .09. 


5.  $97.03x42.   10.  . 935 x. 008. 


14.  . 307 X. 005. 

15.  .00101  X  .001. 


Multiply  and  express  the  product  decimally  ; 

16.  3Jby6|.  m  12 J  by  3|  hundredths. 

17.  $35 1  by  9|.  20.  7f  thousandths  by  f . 

18.  |.05f  by  18 J.  21.  9|  tenths  by  .00031. 


22.  What  is  the  value  of  325.17  pounds  of  iron  at  $.023  per 
pound?  ^n«.  7.47891  dollars. 

23.  A  merchant  sold  86.43  tons  of  coal  at  |9.23  a  ton, 
thereby  gaining  $112.12  ;  what  was  the  cost  of  the  coal  ? 

24.  What  would  12.34  acres  of  land  cost  at  $43.21  per  acre  ? 

25.  A  French  gramme  is  equal  to  15.432  English  grains; 
how  many  grains  are  14|  grammes  equal  to  ? 

26.  A  merchant  uses  a  yardstick  which  is  .00538  of  a  yard 
too  short ;  how  many  yards  will  he  thus  gain  in  selling  438 
yards  measured  by  this  yardstick  ? 

27.  A  metre  is  equal  to  39.3708  inches ;  how  many  inches  are 
there  in  1.325  metres  ? 

Find  the  value  of  the  following  :     . 

28.  $240.09  X  (2.3^ -f  of  If). 

29.  (I  of  12?  -  .9031  +  1.001)  x  375. 

30.  ($375|  -  $87,093)  x  (f  of  36  -  |  of  ^y. 


'■'ill 


I 


146 


DECIMAL     FRACTIONS, 


81.  A  dealer  in  wood  and  hay  boup^ht  2005  tons  of  hay  at 
$14.75  a  ton,  and  2387^  cords  of  wood  at  ^4.50  a  cord ;  how 
much  did  he  pay  for  all  ?  Ana.  $46070. 

32.  Bought  18  books  at  |1.37|  each,  and  sold  them  at  a  gain 
of  .50 J  cents  each  ;  what  did  I  receive  for  the  whole? 

33.  A  boy  went  to  a  grocery  with  a  $10  bill,  and  bought  3J 
pounds  of  tea  at  $.90  a  pound,  7  pounds  of  flour  at  5^.07  a 
pound,  and  4  pounds  of  butter  at  $.35  a  pound  ;  how  much 
change  did  he  return  to  his  father  ?  Ans.  $4.96. 

34.  What  would  15280  feet  of  lumber  cost,  at  $2.37|  for 
each  100  feet  ?  Ans.  $302.90. 


Divisioisr. 


.1 


PREPARATORY    PROPOSITIONS. 

347.  Prop.  I. —  Wlien  the  divisor  is  greater  than  the  divi- 
dend, the  quotient  expi'esses  the  part  the  dividend  is  of  the 
divisor. 

Thus,  4  -*-  6  =  f  =  f.  The  quotient  §  expresses  the  part 
the  4  is  of  6. 

1.  Observe  that  the  process  in  examples  of  this  kind  consists 
in  reducing  the  fraction  formed  by  placing  the  divisor  over  the 
dividend  to  its  lowest  terms.  Thus,  33  -f-  56  =  ^j,  wldch 
reduced  to  its  lowest  terms  gives  4. 

2.  In  caso  the  result  is  to  be  expressed  decimally,  the  j)roces8 
then  consists  in  reducing  to  a  decimal,  according  to  (JJii7),  the 
fraction  formed  by  placing  the  dividend  over  the  divisor.  Thus, 
6  -*-  8  =  ^,  reduced  to  a  decimal  equals  .625. 

Divide  the  following,   and  express  the  quotient  decimally. 

Explain  the  process  in  each  case  as  above. 


1.  7H-20. 

4.  154-32. 

7.  8^-11. 

10.  3-*-20 

2.  3-4-4 

5.  13-J-40. 

8.  5-5-7. 

11.  4-5-13. 

8.  5-*-8. 

6.    9-T-80. 

9.  5-5-6. 

12.  7-5-88 

ILLUSTRATION    OF    PROCESS, 


147 


348.  Prop.  II. — TJie  fraction  remaining  after  the  division 
of  one  integer  by  another  expresses  the  part  the  remainder  is 
of  tJie  divisor. 

Thus,  43  -f- 11  =  3i\.  The  divisor  11  is  contained  3  times 
in  42  and  9  left,  which  is  9  parts  or  ^^  of  the  divisor  11.  Hence 
we  say  that  the  divisor  11  is  contained  3^"^  times  in  42.  Wo 
express  the  -^^  decimally  by  reducing  it  according  to  (3127). 
Hence,  SV'r  =  3.8i. 

Divide  the  following  and  express  the  remainder  decimally, 
carrying  the  decimal  to  four  places  : 

1.  473  -f-  23.  4.    65  -h  17.  7.    3000  -i-  547. 

2.  324  -}-  7.  5.    89  h-  103.  8.    5374  -h  183. 

3.  783  -^  97.  6.    37  -^  43.  9.    1000  h-  101. 

349.  Prop.  III. — Division  is  possible  only  when  the  divi- 
dend and  divisor  are  both  of  the  same  denomination  (144 — I). 

For  example,  fV  -J-  t^tj.  or  .3  -f-  .07  is  impossible  until  the 
dividend  and  divisor  are  reduced  to  the  same  fractional  denom- 
ination ;  thus,  .3  -r-  .07  =  .30  -t-  .07  =  4f  =  4.285714.     , 


ILLUSTRATION    OF    PROCESS. 

360.    Ex.  1.  Divide  .6  by  .64. 

(1.) 
(2.) 


.6  -i-  .64  =r  .60  -*-  .64 
60 


60  -4-  64  =    ^  =  .9375 
04 

Explanation.— 1.  We  reduce,  as  shown  in  (1),  the  dividend  and  divisor 
to  the  same  decimal  unit  or  denomination  (2T9). 

2.  We  divide,  according  to  (279),  as  Bhown  in  (2),  the  numerator  60  by 
the  numerator  M,  which  gives  U-  Reducing  U  to  a  decimal  (32T),  we 
have  .6  -•-  .&4  =  .9375. 


Ex.  2.  Divide  .63  by  .0022. 

.63  -^  .0022  =  .6300  -4-  .0022 
6300  -*-  22  =  286x\  =  386.36 


(1.) 
(2.) 


m 


Ji«i 


148 


DECIMAL     FRACTIONS, 


\^ 


m 


m 


EXPLANATION.—I.  We  rcducc,  as  f«hown  in  (1),  the  dividend  nnd  divisor 
to  the  Humc  decimal  unit  by  auuezin^  ciptiurK  to  the  dividend  (.3311). 

2.  We  divide,  according  to  (270;,  a«  fliowu  iu  (.2),  the  numerator  6300 
by  the  numerator  22,  ^'iving  as  a  quotient  28«,V. 

3.  We  reduce,  according  to  (3'it7),  tho  A  in  the  quotient  to  a  decimal, 
giving  tlie  repetend  .30.    Hence,  .03  +  .0023  =  880.^. 

Ex.  3.  Divide  10.831  by  3.7. 


(1.) 
(3.) 

(8.) 


10.831 
10.831 


3.7  =  10.831  -T-  3.700 


3.700  = 


10831      3700 


1000       1000 
37100  )  1G8|31  (  0.33 
103 

"  63 


54_ 

81 
81 

ExpLANATioN.~l.  We  reduce,  as  shown  in  (1),  the  dividend  and  divlflor 
to  the  same  decimal  unit  by  annexing  ciphers  to  the  divisor  (330). 

2.  The  dividend  and  divisor  each  express  thousandths,  as  shown  in  (2). 
Hence  we  reject  the  denominators  and  divide  as  in  integers  (5170). 

3.  Since  there  are  ciphers  at  the  right  of  the  divisor,  they  may  be  cut 
off  by  cutting  off  the  same  number  of  figures  at  the  right  of  tlie  dividend 
(131).  Dividing  by  87,  we  find  that  it  is  contained  0  times  iu  168,  with  6 
remaining. 

4.  The  6  remaining,  with  the  two  figures  cut  off,  make  a  remainder  of 
621  or  ,Voo.  This  is  reduced  to  a  decimal  by  dividing  both  terms  by  27. 
Hence,  as  shown  in  (3),  we  continue  dividing  by  27  by  taking  down  the  two 
figures  cut  off. 

The  work  is  abbreviated  thus : 

We  reduce  the  dividend  and  divisor  to  the  same  decimal  tmit  by  cutting 
off  from  the  right  of  the  dividend  the  flgurcH  that  express  lower  decimal 
units  than  the  divisor.  We  then  divide  as  shown  in  (3),  prefixing  the 
remainder  to  the  figures  cut  off  and  reducing  the  result  to  a  decimal. 

From  these  illustrations  we  obtain  the  following 

RULE. 

351.  Reduce  the  dividend  and  divisor  to  the  same      4mal 
unit ;  divide  as  in  integers  and  reduce  the  fractional  remaindc 
in  the  quotient,  if  any,  to  a  decimal. 


I 
I 


EXAMPLES, 


14» 


\al 


I 


EXAMPLES     FOR    PRACTICE. 

852.  In  the  following  examples  carry  the  answer  In  each 
case  to  four  decimal  places : 

1.  Divide  27^  by  4.03 ;  by  .72  ;  by  2.3J. 

2.  Divide  53.28  by  3.12 ;  by  7.3  ;  by  9.034. 

8.  Divide  |.93  by  $.847  ;  $73.09^  by  |.75J  ;  |.37J  by  $.74. 

4.  Divide  $726.42  by  $.37;  by  $3.08  ;  by  $.953. 

Wliat  is  the  value  of 

5.  $75.83  -*-  $100.  8.  $10000  -^  $.07. 

6.  {\  of  .73)  -*-  .09.  9.  8.345  -*-  2.0007. 

7.  734|  H-  4.5|.  10.  (8J  +  12.07)  +  (15.03  -  |). 
11.  (fTlKJ  X  64)  -^  (I  of  f  of  12|). 

'   12.  ($354.07- 5  of  $10.84) -i- I  of  $7.08. 

13.  (§  of  $324.18  -  $1)  -i-  $2.0005. 

14.  ($3.052 -*-?)- (I  of  $1.08 -f- ,«). 

15.  At  $2.32,  how  many  yards  of  cloth  can  be  bought  for 
$373.84? 

16.  The  product  of  two  numbers  is  375.04,  and  one  of  them 
is  73.009  ;  what  is  the  other  ?  Ans.  5.1369  + . 

17.  How  much  tea  can  be  bought  for  $134.84,  if  23g  pounds 
cost  $17.70?  ^«.?.  179.786G+  pounds. 

18.  A  farmer  sold  132f  bushels  wheat  at  $1.35  per  bushel, 
and  184  bushels  corn  at  $.78^  per  bushel.  He  bought  coal  with 
the  amount  received,  at  $9.54  a  ton.  How  many  tons  did  he 
buy? 

19.  A  merchant  received  $173.25.  $32.19,  and  $89.13.  He 
expended  the  whole  in  buying  silk  at  $1.37^  per  yard.  How 
many  yards  of  silk  did  he  buy? 

20.  What  decimal  part  of  a  house  worth  $3965  can  be  bought 
for  $1498.77?  Ana. 

21    What  is  the  value  of  27f  acres  of  land  when  .S 
acre      worth  $48  ? 


m 


I 


mi 


150 


D  E  CIMA  L     FRACTIO  XS, 


22.  Geo.  Bain  lost  .47  of  his  capital,  and  had  to  use  .13  more 
for  family  expenses,  and  had  still  remaining  $5380.  What 
was  his  original  capital  ?  Ana.  $13450. 

23.  Henry  Barber  owns  |  of  a  cotton  mill  and  sellc  .8  of  his 
share  for  $1650 ;  what  decimal  part  of  the  mill  does  he  still 
own,  and  what  was  the  mill  worth  ? 


REVIEW    EXAMPLES. 

353.  Answers  involving  decimals,  unless  otherwise  stated, 
are  carried  to  four  decimal  places. 

What  is  the  cost 

1.  Of  4.5  acres  of  land,  if  100  acres  cost  $7385  ? 

2.  Of  .7 J  of  a  pound  of  tea,  if  7  pounds  cost  $6.95? 
8.  Of  9  J  cords  of  wood,  at  $12.60  for  2.8  cords  ? 

4.  Of  13.25  yards  of  cloth,  if  3.75  yards  cost  $9 93 J? 

5.  Of  5384  feet  lumber,  at  $5.75  per  100  feet? 

6.  Of  31400  bricks,  at  $8.95  per  1000  bricks  ? 

7.  Of  158i.  pounds  butter,  if  9.54  pounds  ccst  5^:3.239? 

Reduce  each  of  the  following  examples  to  decimals  : 


8. 
9. 

10. 
11. 


n 

H. 

5 


12. 
13. 

14. 

15. 


5|\— 5|. 
?  of  1|. 
(3g-Hi)x| 

8 


f  of  .3 


8^ 


4.3 


16. 
17. 

18. 

19. 


I  of  ^  of  If 
25 


90.  Seven  car-loads  of  coal,  each  containing^  13.75  tons,  were 
sold  at  $8.53  per  ton.    How  much  was  received  for  the  whole  ? 

21.  Four  loads  of  hay  weighed  respectively  2583.07,  3007f, 
35675,  and  3074^  pounds;  what  wa.^  tht  total  weight? 

22.  At  $1.75  per  100,  wjiat  ia  the  cost  of  5384  oranges? 


REVIEW    EXAMPLES, 


151 


A- 

4 


i) 


^ere 


23.  What  is  the  cost  of  carrying  893850  pounds  of  com  from 
Chicago  to  Montreal,  at  $.35|  per  100  pounds? 

24.  If  freight  from  Sarnia  to  Halifax  is  $.39^  per  100  pounds, 
what  is  the  cobt  of  transporting  3  boxes  of  goods,  weighing 
respectively  783 »,  325 g,  and  286;  pounds? 

25.  A  piece  of  broadcloth  cost  $195.38^,  at  $3.27  per  yard. 
How  many  yards  does  it  contain  ? 

26.  Expended  *460.80  in  purchasing  silk,  .3  of  it  at  $2.25  y>eT 
yard,  J  of  it  at  $1.80  per  yard,  and  tlie  balance  at  $o.45  per 
yard.     How  many  yards  did  I  buy  of  each  quality  of  silk  V 

27.  A  person  Laving  $1142.49f  wishes  to  buy  an  equal  num- 
ber of  bushels  of  wheat,  corn,  and  outs ;  the  wheat  at  $1.37,  the 
corn  at  $.87.^,  and  the  oats  at  $.35J.  How  many  bushels  of 
each  Ciin  he  buy  ? 


28.  What  is  the  value  of  (1^^1—3) 


.48 


29.  A  produce  dealer  exchan^jed  48?  bushels  oats  at  39f  cts. 
por  bushel,  and  13|  barrels  of  apples  at  $3.85  a  barrel,  for 
butter  at  371  cts.  a  pound  ;  how  many  pounds  of  butter  did  he 
receive  ? 

30.  A  fruit  merchant  expended  $523.60  in  purchasing  apples 
at  $3.85  ;i  barrel,  which  he  afterwards  sold  at  an  advance  of 
$1.07  i>er  barrel ;  what  was  his  p^ain  on  the  sale  ? 

31.  A  grain  merchant  bought  1830  bushels  of  wheat  at 
$1.25  a  bushel,  570  bushels  corn  at  731  cts.  a  bushel,  and  468 
bushels  oats  at  35^  cts.  a  bushel.  He  sold  the  wheat  at  an 
advance  of  17i  cts.  a  bushel,  the  com  at  an  advance  of  9^  cts. 
a  bushel,  and  the  oats  at  a  loss  f)f  3  cts.  a  bushel.  How  much 
did  he  pay  for  the  entire  quantity,  and  what  was  his  gain  on 
the  transaction  ? 

32.  The  cost  of  constructing  a  certain  road  was  $5050.50. 
There  were  35  men  employed  upon  it  78  days,  and  each  man 
recoived  the  same  amount  per  day  ;  how  much  was  the  daily 
wages? 


,^ 


w 


152 


DECIMAL     FRACTIONS. 


REVIEW    AND    TEST    QUESTIONS. 

354.  1.  Define  Decimal  Unit,  Decimal  Fraction,  Repetend, 
Circulating  Decimal,  Mixed  Circulating  Decimal,  Finite  Deci- 
mal, and  Complex  Decimal. 

2.  In  how  many  ways  may  \  be  expressed  as  a  decimal  frac- 
tion, and  why  V 

3.  What  effect  have  ciphers  written  at  the  left  of  an  integer? 
At  the  left  of  a  decimal,  and  why  in  each  case  (i505)  ? 

4.  Show  that  each  figure  in  the  numerator  of  a  decimal 
represents  a  distinct  order  of  decimal  units  (iiOO). 

5.  How  are  integral  orders  and  decimal  orders  each  related 
to  the  units  (313)?  Illustrate  your  answer  by  lines  or 
objects. 

6.  Why  in  reading  a  decimal  is  the  lowest  order  the  only  one 
named?    Illustrate  by  examples  (310). 

7.  Give  reasons  for  not  regarding  the  ciphers  at  the  left  in 
reading  the  numerator  of  the  decimal  .000403. 

8.  Reduce  |  to  a  decimal,  and  give  a  reason  for  each  step  in 
the  process. 

9.  When  expressed  decimally,  how  many  places  must  j^/j 
give,  and  why  ?    IIow  many  must  ^^  give,  and  why  ? 

10.  Illustrate  by  an  example  the  reason  why  \\  cannot  be 
expresiised  as  a  simple  decimal  (31(1). 

11.  State  what  fractions  can  and  what  fractions  cannot  be 
expressed  as  simple  decimals  (315  and  316).  Illustrate  by 
examples. 

12.  In  reducing  f  to  a  complex  decimal,  why  must  the  numer- 
ator 5  recur  as  a  remainder  (317 — 1  and  2)  ? 

13.  Show  that,  according  to  (234—11  and  III),  the  value 
oi\\  will  not  be  changed  if  we  diminish  the  numerator  and 
denominator  each  by  |  of  itself. 

14.  Show  that  multiplying  9  by  1|  increases  the  9  by  |  of 
itself. 


b 


RE  VIE  W . 


153 


;r- 


15.  Multiplying  the  numerator  and  denominator  of  \l  each 
by  If  produces  what  change  in  the  fraction,  and  why  V 

16.  Show  that  in  diminishing  the  numerator  of  I  by  I  and 
the  denominator  by  1  we  diminish  each  by  the  same  part  of 
itself. 

IT.  In  taking  .3  as  the  value  of  J,  what  fraction  has  bo<>u 
rejected  from  the  numerator?  What  must  be  rejected  from 
the  denominator  to  make  .3  =  J,  and  why? 

18.  Show  that  the  true  value  of  .81  is  g^.  Give  a  reason  for 
each  step. 

19.  Explain  the  process  of  reducing  a  mixed  circulating 
decimal  to  a  fraction.     Give  a  reason  for  each  step. 

20.  How  much  is  .33333  less  than  |,  and  why? 

21.  How  much  is  .571438  less  than  ^,  and  why? 

22.  Find  the  sum  of  .73,  .0040,  .089,  6.58,  and  9.08703,  and 
explain  each  step  in  the  process  (350 — I  and  II). 

23.  If  tentJis  are  multiplied  by  hundredths,  how  many  deci- 
mal places  will  there  be  in  the  product,  and  why  (IJ44)  ? 

24.  Show  that  a  number  is  multiplied  by  10  by  moving  the 
decimal  point  one  place  to  the  right ;  by  100  by  moving  it  two 
places  ;  by  1000  three  places,  and  so  on. 

25.  State  a  rule  for  pointing  off  the  decimal  ])lnces  in  the 
product  of  two  decimals.  Illustrate  by  an  exauiple,  and  give 
reasons  for  your  rule. 

26.  Multiply  385.23  by  .742,  multiplying  fird  by  the  4  Jiun- 
d/redtJiS,  then  by  the  7  tenths,  and  luat  by  the  2  thoumndtha. 

27.  Why  is  the  quotient  of  an  integer  divided  by  a  pri>per 
fraction  greater  than  the  dividend  ? 

28.  Show  that  a  number  is  divided  by  10  by  moving  the 
decimal  point  one  place  to  the  left ;  by  100  l)y  moving  it  two 
places;  by  1000,  three  places;  by  10000,  four  places. and  so  on. 

29.  Divide  4.9  by  1.305,  and  give  a  rea.son  for  each  step  in 
the  process.    Carry  the  decimal  to  three  i)lace8. 

30.  Give  a  rule  for  division  of  decimals. 

11 


'1 


ill 

m 


•  ill 


'''  'l 

1    II 


DENOMINATE    NUMBERS. 


DEFINITIONS. 

355.  A  Relafed  Unit  is  a  unit  which  has  an  invariable 
relation  to  one  or  more  other  units. 

Thus,  1  foot  =  12  inches,  or  J  of  a  yard ;  hence,  1  foot  has  an 
invariable  relation  to  the  units  inch  and  yard,  and  is  therefore 
a  related  unit. 

356.  A  Detioniinate  Nutnher  is  a  concrete  number 
(1  li)  whose  unit  ( 1 1 )  is  a  related  vnit. 

Thus,  17  yards  is  a  denominate  number,  because  its  unit, 
yard,  has  an  invariable  relation  to  the  units  foot  and  inch, 
1  yard  making  always  3  feet  or  36  laches. 

357.  A   DcHominate    Fraction    is  a  fraction   of  a 

related  unit. 

Thus,  5  of  a  yard  is  a  denominate  fraction, 

358.  The  Orders  of  related  units  are  called  Denom^ 
inntions. 

Thus,  yards,  feet,  and  inches  are  denominations  of  length ; 
dollars  and  cents  are  denominations  of  money. 

351>.  A  Cowpoitnd  Number  consists  of  several  num- 
bers expressing  related  denominations,  written  together  in  the 
order  of  the  relation  of  their  units,  and  read  as  one  number. 

Thus,  23  yd.  3  ft.  9  in.  is  a     »mix)und  number. 

360.  A  Standard  Unit  is  a  unit  established  by  law 
or  custom,  from  which  other  units  of  the  same  kind  are 
derived. 


TABLES, 


156 


iable 


08  an 
efore 


mber 

unit, 
inch, 

[)f   a 


um- 
the 


law 
are 


I 


Thus,  tho  standard  unit  of  nioasurcs  of  extension  is  the 
yard.  By  dividing  the  yanl  into  3  equal  parts,  we  obtain  the 
wxAXfoot ;  into  36  equal  parts,  we  obtain  the  unit  inch  ;  mul- 
tiplyinjf  it  by  5|,  we  obtain  the  unit  rod,  and  so  on. 

301.  Related  units  mav  be  classified  into  dx  kinds  : 


1.  Extension. 

2.  Capacity. 


8.  Weight. 
4.  Time. 


5.  Angles  or  Arcs. 

6.  Money  or  Value. 


.*M>2.  lieditction  of  Denominate  JVttmbers  is  the 

process  of  changing  their  denomination  without  altering  their 
value. 


UNITS    OF    WEIGHT. 

<163.  The  Troy  pound  is,  according  to  law,  the  Stanih 
avil  Unit  of  weight. 


TKOY    WEIGHT. 


TABI.K  OP  TNTTS. 

)i\  gr.  =  1  dwt. 

20  dwt.  =  1  oz. 

12  oz.  -  1  lb. 

3.2  gr.  =  1  carat 


1.  Denotni nations.  —  Grains  (<jr.) 
Pennyweights  (tlwt.),  Ounces  (oz.),  Pounds 
(lb.),  and  Carats*. 

2.  EtjHiralenta.—l  lb.  =  12  oz.  =  240 
dwt.  =  5760  gr. 

3.  Vsr.—[J»cd  in  weighin;?  ?old,  pilver, 


and  precious  stones,  and  in  philosophical  experiments. 

4.  This  weight  was  brought  into  Europe  from  Egypt,  and 
tirst  ado])ted  in  Troyes,  a  city  of  France — whence  its  name. 

5.  A  cubic  inch  of  distilled  water  is  252.458  grains  Troy  with 
the  thermoniPtor  at  02   and  tlie  barometer  at  30  inches. 

6.  The  term  "  carat "  is  applied  to  gold  in  a  relative  sense  ;  any 
quantity  of  pure  gold,  or  of  gold  alloyed  with  some  other  metal, 
being  8U]iposed  to  be  divided  into  24  equal  i)art8  (carats). 
When  gold  is  pure  it  is  said  to  be  24  carats  fine  ;  with  3  parts 
alloy  it  is  said  to  be  22  carats  fine,  etc.  Standard  gold  is 
22  carats  fine ;  jewellers*  gold  is  18  carats  fine. 


t>J 


'*^l 


15G 


D  K  y  QMINATE     .V  V  M  li  KRS. 


AVOIRDUPOIS    WEIGHT. 

1.  DriiomluotioiiH.  —  DramH  (dr.X 
Ounces  (oz.),  PoundH  (lb.>,  QuartcTt»  (qr.j, 
Iluudredwcij^htH  (cwt.),  Tons  (T.). 

2.  KquivnlvHtH.—l  ton  =  20  cwt.  = 
8000  lb.  =  32000  oz. 

3.  l/Jvf .— Used  in  weighing  ^rocerlot*. 
all  coarse  and  heavy  articles,  and  drugs 
at  wholesale.    The  term  aiuitdiijjoiii  is 

derived  from  aroirii  (goods  or  chattels)  and  jxddd  (weight). 

4.  In  wholesale  trausactions  in  coal  and  iron  and  in  the  Custom  IIoubc, 
1  qnarter  =  28  lb.,  1  cwt.  ^  112  lb.,  1  T.  =  2^0  lb.  This  is  usually  called 
till'  iMug  Ton  table. 

5.  In  general,  1  stone  (I  st.)  =  14  lb.  Avoirdupois,  but  for  butchers'  meat 
or  fish,  1  stone  =  8  lb.,  1  firkin  of  butter  =  56  lb.,  1  fodder  of  lead^lSJJ  cwt., 
1  great  pound  of  silk  =  24  ounces,  1  pack  of  wooT  =  240  lb. 


TABLK  or  UNITS. 

10  drams 

=r 

uz. 

lU 

OZ. 

= 

lb. 

25 

lb. 

= 

qr. 

4 

cir. 

= 

cwt. 

20  cvrt. 

^^ 

T. 

APOTHECARIES'    WEIGHT. 


TABLE  OF  UNITS. 

20  Jfl".   =   1  sc.  or 

^. 

3  3=1  dr.  or 

3. 

8    3     =1  oz.  or 

z 

12  oz.   =  1  lb. 

1.  Itvnotnitmtiotta.  —  Oraiiis  (gr.). 
Scruples  (d),  Drams  (3),  Ounces  (5), 
rounds  (lb.). 

2.  KquivnU'Ut.s.—Wi.  \  -  512=  3  06 
=  D  288  -  gr.  5700. 

3.  r/.<»e.— Used  in  medical  prescrip- 
tions. 

4.  Medical  proscriptions  are  usually  written  in  Roman  notation.  The 
nnmbor  is  written  after  the  symbol,  and  the  final  i  is  always  written  j. 
Thus,  5  vij  is  7  ounces. 

.Conijtftfative  Table  of  Units  of  Weight, 

By  Act  of  the  British  Parliament  in  1820,  the  brass  weight  of 
07iC  pound  Troy  of  the  year  1758,  kept  by  the  Clerk  of  the 
House  of  Commons,  was  made  the  unit  or  stundanl  measure  of 
weight,  from  which  all  other  weights  are  derived  and  computed. 
This  hrriAs  weight  havinj^  been  lost  or  destroyed  by  fire  in  1834, 
the  Imperial  standard  pound  is  determined  from  the  weight  of 
a  cubic  inch  of  distilled  water,  as  given  in  3GtS — 5. 

TBOT.  AVOIRDUPOIS.  APOTHECARIKS. 

1  pound  =  5760  grains  =  7000  grains  =  5760  grains. 
1  ounce   =    480      "       =    437.5    "      =480      " 


t:  X  A  M  J'  L  KS. 


157 


PROBLEMS    ON    RELATED    UNITS. 

;{(M:.    Pkob.  I. — To  reduce  a  denominate   or  a  com- 
pound number  to  a  lower  denomination. 


7  oz.  9  dwt.  to  pennyweights. 


Reduce  23  lb 

23  lb.  7  oz. 

12 

283  oz. 

20 

5009  (Iwt. 

o 


uwt. 


Solution.— 1.  Since  12  oz.  make  1  lb.,  in 
any  number  of  poumlH  iben-  are  12  timer-  ao 
many  ounccH  an  pounds.  Ilenrc  we  multi- 
ply the  2;i  lb.  by  12,  and  add  the  7  oz.,K'iving 

sasoz. 

2.  A^ln,  since  20  dwt.  make  1  oz.,  in 
any  number  of  ouncee  there  are  2()  times  n» 
many  peunyweij^hti*  as  ouncet*.    IIcucc  we 
multiply  tlie  283  oz.  by  20,  and  add  the  9  dwt.,  giving  5669  dwt. 


RULE. 

JJO.".  /.  Multiply  the  numher  of  the  highest  denomination 
fficen,  hytJw  number  of  units  of  the  next  lower  lie  nomination  that 
make  I  (f  the  higher,  and  to  the  product  add  the  number  given 
of  thr  loiter  denomination. 

IF.  Proceed  in  this  m<rnner  with  each  surcendre  deiiominntion 
obtained,  until  the  number  is  reduced  to  the  required  denomina- 
iion. 

EXAMPLES     FOR     PRACTICE. 
*$OG.  Reduce  and  explain  orally  the  following; 

1.  How  many  drams  in  3  lb.  7  or,.  ?    In  4  lb.  10  oz.  5  dr.  ? 

2.  IIow  many  grains  in  4  dr.  1  sc?    In  1  oz.  4  dr.  1  so.  ? 

3.  How  many  pennyweights  in  3  lb.  7  oz.?     In  5  oz.  7  dwt.  ? 

4.  How  many  pounds  in  2  T.  132  lb.  ?    In  5  T.  19  lb.  ? 

Reduce 

5.  19  lb.  7  oz.  5  dr.  to  drams.       7.  17  lb.  11  3  23  to  graine. 
(5.  13  T.  17  cwt.  to  pounds.  8.  3  lb.  9  3  5  3  to  grains. 

9.  27  lb.  5  oz.  17  dwt.  to  grains. 
10.  173  T.  5  cwt.  47  lb.  to  pounds. 


n 

m 


m 


^m 


*i-..3 


% 


158 


DEXOMIXA  TE    N  UMB  ERS. 


11.  In  37  lb.  8  oz.  15  dwt.  19  gr.  how  many  graiim? 

12.  Reduce  87  T.  la  cwt.  93  lb.  to  pounds. 

13.  Reduce  23  lb.  11  oz.  9  dwt.  to  pennyweights. 

14.  Reduce  184  T.  to  hundredweights  ;  to  pounds ;  to  ounces. 

15.  How  many  grains  in  1  pound  Apothecaries'  weight  ?    In 
1  pound  Troy  weight V     In  1  pound  Avoirdupois  weight? 

IG.  Reduce  104  lb.  17  dwt.  to  pennyweights. 


I 

i 


rl 


307.    Prob.  II. — To  reduce  a  denominate  number  to 
a  compound  or  a  higher  denominate  number. 

Reduce  7487  sc.  to  a  compound  number. 


3  )7487  sc. 
8  )  2495  dr.  4-  2  sc. 
12  )_311  oz.  +  7  dr. 
25  lb.  11  oz. 


SoLtTTiON.  —  Siucc  3  PC.  make  1  dr., 
7487  ec.  must  umkc  a»  many  drams  as  3  is 
contained  times  In  7487,  or  2495  dr.  +  2  ec. 

2.  Since  8  dr.  make  1  oz.,  24%  dr.  mast 
make  as  many  ouucch  as  8  is  contained 
tirnen  In  24!)5,  or  311  oz.  +  7  dr. 

3.  Since  12  oz.  make  1  lb.,  311  oz.  must 
make  as  many  pounds  as  12  is  contained  times  in  311,  or  25  Ib.  +  lloz. 
Hence,  7487  sc.  are  equal  to  tlie  compound  number  25  lb.  11  oz.  7  dr.  2  ec. 
Hence  the  following 

EULE, 

308.  /.  Divide  the  given  nuviber  by  tJie  number  ofunitt  of 
the  given  denomination  that  make  one  of  the  next  higher  denom- 
ination. 

II.  In  the  same  manner  divide  this  quotient  and  each  9uec£8- 
me  quotient,  omitting  the  remainders,  until  tJie  denomination 
required  is  reached.  The  last  quotient,  with  tJie  remainders 
annexed^  is  the  required  result. 

EXAMPLES    FOR    PRACTICE. 

369.  Reduce  and  explain  orally : 

1.  How  many  pounds  Troy  in  2G  oz.  ?  In  80  oz.  ?  In  64  oz.  t 
In  300  dwt.  ?    In  900  dwt.  ?    In  124  oz.  ? 

2.  How  many  hundredweights  in  1486  lb.  ?    In  1774  lb.  ? 

3.  How  many  tons  in  5800  lb.  t    In  9268  lb.  t 


EXAMPLES. 


159 


4.  In  240  sc.  how  many  drams  ?    How  many  ounces  ? 
Reduce 


5.  876445  lbs.  to  tons. 

C.  389G4  gr.  to  pounds  Troy! 

7.  3  97634  to  pounds. 

8.  503640  oz.  to  tons. 

9.  279647  gr.  to  pounds  Apotb. 


10.  8597  dwt.  to  pounds. 

11.  3468  cwt.  to  tons. 

12.  93875  gr.  to  di-ams. 

13.  534278  gr.  to  pounds  Troy. 

14.  873604  oz.  to  tons. 


15.  In  93645    gr.   how  many  pounds    Troy  ?    How  many 
Avoirdupois  ?    How  many  Apothecaries'  ? 

16.  Reduce  57  lb.  13  oz.  Avoir,  to  Troy  weicrht. 

17.  In  9  lb.  Troy  how  many  pounds  Avoirdupois? 

18.  Reduce  14  lb.  7  oz.  Avoir,  to  Apothecary  weight. 

•     19.  Reduce  5  lb.  10  oz.  17  dwt.  to  Apothecary  weight. 


370.  Prob.  III. — To  reduce  a  denominate  fraction  or 
decimal  to  integers  of  lower  denominations. 

Reduce  §  of  a  ton  to  lower  denominations. 

(1.)  f  T.  =  f  of  20  cwt.  =  §  X  20  =  14  cwt.  +  5  cwt. 
(2.)  f  cwt.  =  f  of  100  lb.  =  S  X  100  =  28  lb.  -♦-  ^  ib. 
(3.)    4  lb.  =  4  of  16  oz.  =  4  X  16  =  91  oz. 

Solution.— Since  80  cwt.  is  equal  1  T.,  ?  of  20  cwt.,  or  14?  cwt.,  equals 
J  of  1  T.  Hence,  to  reduce  the  ?  of  a  ton  to  hundredweif,'ht8,  we  take  f  of 
30  cwt.,  or  multiply,  as  shown  in  (1),  the  \  by  20,  the  number  of  hundred- 
weights in  a  ton. 

In  the  same  manner  we  reduce  the  |  cwt.  remaining  to  pounds,  as*  shown 
in  (2),  and  the  f  lb.  remaining  to  ounces,  as  shown  in  (3).  Ucuce  the  fol- 
lowing 

RULE. 

371 .  /.  Multiply  the  given  fraction  or  decimal  by  the  num- 
ber of  units  in  tlie  next  lower  denomination,  and  reduce  t?ie 
result  to  a  mixed  number,  if  any. 

II.  Proceed  in  the  sam>e  manner  itith  the  fractional  part  of 
each  successive  product. 

III.  The  integral  parts  of  tlie  several  products,  with  the  frac- 
tion, if  any,  in  the  last  product,  arranged  in  proper  order,  is  tlie 
required  result. 


.ii- 

■■'5F 
lift 


■K' 

lit 


IGO 


D  ENO  .V T XA  T E     X  UMBE R S. 


EXAMPLES     FOR     PRACTICE. 


37t2.  Find  the  value  in  lower  denominations  : 


1. 


Of  ,^0  of  a  dram. 


2.  Of  5  of  a  ton. 


Of 


,\  of  a  pound  Troy. 
Of  .0  (»f  a  ixiund  Avoir. 


3. 

4. 

5.  Of  5  of  a  pound  Apotli. 

0.  Of  .85  of  a  ton. 

7.  Of  .73  of  an  ounce  Troy. 

8.  Of  .94  of  a  dram. 


9.  Of  ^  of  a  quarter. 

10.  Of  ^\  of  a  hundredwei|?ht 

11.  Of  tJ  pounds  Troy. 

12.  Of  3.7  hundredweights. 

13.  Of  13?  tons. 

14.  Of  5.1)4  pounds  Apoth. 

15.  Of  .730 1  of  a  i)ound  Troy, 
10.  Of  .9350  of  a  ton. 


17.  In  I  of  a  i)Ound  Avoir,  how  much  Troy  weight? 
IS.  Keduco  .84  of  a  hundredweight  to  Troy  weight. 
19.  How  much  will   ij   of  a  cwt.  make  expressed  in  Troy 
weiglit  V    Expressed  in  Apothecary  weight  ? 


J57IJ.  Prob.  IV. — To  reduce  a  denominate  fraction  or 
decimal  of  a  lower  to  a  fraction  or  decimal  of  a  higher 
denomination.  ,     . 


Reduce  g  of  a  dram  to  a  fraction  of  a  pound. 
(1.)        f  dr.  =     Joz.  X 


■!l 


—  8 

—  Iff 


oz. 


(2.)        :^ijOZ.    =    jljlb.    X    /^    =   3  J,  lb. 

SoLrTioN.— 1.  Since  8  drams  =  1  ounce,  1  dram  ie  equal  J  of  an  oz.,  and 
{  of  a  drain  is  equ.il  |  of  J  oz.    Ilcncc,  as  shown  in  (1),  J  dr.  =  /„  oz. 

2.  Since  12  ounces  =  1  pound,  1  ounce  is  equal  ,»,  of  a  pound,  and,  as 
t*liown  in  (2\  ,",  of  an  ounce  is  equal  ,*„  of  ^^  11>m  or  yJa  lb.    Hence,  |  dr.  = 


1 


lb.    ilencc  the  following 


I 


RULE. 

;J74.  /.  Find  the  part  which  a  unit  of  the  given  denomi- 
nation is  of  a  unit  of  the  next  higher  denomination,  and  multi- 
ply this  fraction  hy  t?i£  given  fraction  or  decimal. 

II.  Proceed  in  the  same  manner  icith  the  result  and  each 
surccmve  result,  until  reduced  to  the  denomination  required. 
Reduce  t?ie  result  to  its  lowest  terms  or  to  a  decimal. 


EXAMPLES, 


161 


EXAMPLES     FOR     PRACTICE, 
375.  Reduce  and  explain  oruUy  : 

1.  f  dr.  to  a  fructiou  ul'u  [xmud. 

2.  ^  Bc.  to  a  t'ractiou  ot'u  i>ouud. 

3.  .7  oz.  to  a  fraction  of  a  pound  Troy. 

4.  .8  lb.  to  a  fraction  of  a  ton. 

5.  .0  dwt.  to  a  fraction  of  a  pound. 

6.  ,''o  H).  to  a  fraction  of  a  hundredweight. 

7.  lleducc  I  dwt.,  ^^  gr.,  \l  oz.,  .JJ2  oz.,  ,74  dwt.,  and  .04  gr. 
each  to  the  fraction  of  a  pound  Troy. 

8.  Ueduce  ^  of  a  scruple  to  a  fraction  of  a  |M)iind. 

J  ExrLANATioN.  —  Since    we 

4  V  J-  ^'  1  V    L    1      IJv  divide  the  j,''veu  fraction  by  the 

f  X  'ft  A  t  X  1*2    —   TSTT  "'•  uiiinberH  in  the  iiHceii<liii>,'»»calo 

•*  KUcceH-ively    (IITH)     between 

the  {riven  and  tljc  required  dCDomination,  we  may  arrange  them  as  shown 

iD  tlie  margin,  and  cancel. 

0.  Reduce  .3  oz.,  .84  lb.,  and  i  cwt.  eacli  to  the  fraction  of 
a  ton. 

10.  What  fraction  of  a  pound  is  fj  of  a  dram  ?    .8  of  a  sc.  ? 

11.  Reduce  to  a  fraction  of  a  pound  Troy  .42  gr. ;  .9()  dwt. 

12.  Reduce  to  the  fraction  of  a  ton  g  cwt. ;  .J)  cwt.  ;  ^  lb.  ; 
.34  cwt.;  .861b.;  .10  oz.  ;  J  oz. 

13.  4  of  an  ounce  ia  what  fraction  of  1  lb.  ?  of  1  cwt.  ? 

370.  Prob.  V. — To  reduce  a  compound  number  to  a 
fraction  of  a  higher  denomination. 

RtKiuce  34  30  D2toa  fraction  of  a  pound. 

(1.)        14  36  92=3116:        lb.  1  =  3288. 
(2.)        Wh  =  n  ;    lience,     5  4  3  6  32  =  lb.  'il. 

Solution.— 1.  Two  numbcrB  can  be  comi)ared  only  when  they  are  the 
same  denomination.  Hence  we  reduce,  as  ehown  In  (1),  the  ^4  36  ^2 
and  the  lb.  1  lo  Hcruplei<,  the  lowest  denomination  mentioned  in  cither 
number. 


m 


'.^ 


162 


DENOMINATE    NUMBERS, 


8.  M  I«  ^aiM'liiKoqHHl  olio,  luid  lb.  1  lu'ln«  niual  oaHH,  54  3  0  Dai* 
the  Huiiu'  part  ol'lb.  1  ait  jllU  iti  uf  j'^  vvUicli  in  ^1|,  or  ||.  Uoucu  J 4 
3  6  02      lb.  Jt.or.OOUm. 

RULE. 

JJ77,  Reduce  the  given  number  t<)  itn  lowest  denomination 
for  the  ninmrato^r  of  the  required  fraction,  and  a  unit  of  the 
required  denoniinntioii  to  the  .same  denomination  for  the  denani' 
inator,  and  reduce  the  fraction  to  its  loircat  terms  or  to  a 
decimal. 


'i 


EXAMPLES    FOR    PRACTICE. 

Si78.   1.  VVhttt  fractiou  of  u  pound  Troy  uro  7  oz.  12  dwt. 
8gr.? 
3.  Reduce  5  cwt.  4  lb.  to  a  fraction  of  one  ton. 
8.  Wbut  fractiou  of  u  hundredweight  are  04  lb.  12  oz.  t 

Reduce  to  the  fractiou  of  a  pound  : 

4.  10  oz.  8  dwt.  10  gr.  7.  5  oz.  4  dr.  3  sc.  20  gr. 

5.  9  oz.  5  dr.  3  8C.  8.  4  oz.  18  dwt.  20  gr. 

6.  6  dr.  1  8C.  18  gr.  9.  11  oz.  19  dwt.  23  gr. 

10.  What  part  of  0  lb.  Troy  are  3  lb.  8  oz.  10  dwt.  1 

11.  Reduce  to  the  fractiou  of  a  ton  8  cwt.  04  lb. 

12.  Wliat  part  of  4  cwt.  are  1  cwt.  35  lb.  V    2  cwt.  50  lb.  V 

13.  Roduct^  8  cwt.  00  lb.  to  the  decimal  of  1  ton  ;  of  8  tons. 

14.  Reduce  8  oz.  10  dwt.  to  the  decimal  of  0  pounds. 

15.  Reduce  8  cwt.  3  qr.  10  lb.  to  the  decimal  of  a  ton. 


Abbreviated  Solution.— Since  the  16  poands 
arc  reduced  to  a  decimal  of  a  quarter  by  redno 
ing  i!  to  a  decimal,  wc  auiiex  two  ciphers  to 
the  16,  ae  ehown  in  the  margin,  and  divide  by 
35,  giviuu  .04  qr. 

To  this  result  wo  prefix  the  3  quarters,  giving 

3  64 
a64  qr.,  which  is  equivalent  to  -'  -  hundredweights ;  hence  wc  divide  by 


25  )  10.00  lb. 
4  )^.U  qr. 
20  )  8.91  cwt. 
.4455  T. 


4,  as  shown  in  the  margin,  giving  .91  cwt. 

To  the  resalt  wc  again  prefix  the  8  cwt.,  giving  8.91,  which  is  eqnlva* 

lent  to  ^;JJJ  of  a  ton,  equal  .4455  T.    Hence,  8  cwt.  3  qr.  16  lb.  =  .4456  T. 


E  X  AM  PL  E  S . 


10» 


10.  IUhIuco  U  oz.  1G  (Iwt.  20  ^r.  to  the  decimal  of  a  pound. 
17.  Whut  decimal  of  31  lb.  Troy  is  2  II).  H  oz.  10  dwt.  ? 
IH.  0  OZ.  10  dwt.  12  ^r.  iin*  wimt  dccimul  of  ii  pound  Y 
lU.  Uiulitce  12  cwt.  2  qr.  IH  lb.  to  tbo  dcciniul  of  u  ton. 

20.  What  (b'cinml  of  a  pound  arc  5  J)  3  5  .j2  t'r.  IHV 

21.  Kt'ducc  8  oz.  0  dr.  2  hc.  to  llir  decinial  of  a  pound. 

22.  Hfduco  7  lb,  5  oz.  Avoir,  to  a  dccintal  of   12  lb.  5  oz. 
:{  dwt.  'i'roy. 

23.  1  lb.  0  oz.  y  dwt.  Im  what  pail  of  3  lb.  Apotli.  Witi^ht? 


)un(ls 

jrs  to 
ie  by 

Iving 
Ide  by 


t*i71>.  Piu»n.  V\. — To  find  the  sum  of  two  or  more  de- 
nominate or  compound  numbers,  or  of  two  or  more  denom- 
inate fractions. 

1.  Find  tbo  Rum  of  7  cwt.  84  lb.  14  oz.,  5  cwt.  97  lb.  8  oz., 
and  2  cwt.  U  lb.  15  oz. 

Solution.—!.  Wo  writ*;  iiuinborH  of  the  Hame 
clonomiiiiitiuii  utulfr<cuch  other,  ua  Hhuwn  In  thu 
mar;;  in. 

3.  We  odd  an  in  Siinj)!*!  NiiiiiIjitc,  cotu- 
uiciicinK  wltli  the  luwuht  licaoiniimlioi).  ThuH, 
15, H,  and  It  ouucch  iiiukc  .'17  oaiiccb,  eciual  2  li>. 
r>  oz.  We  write  the  5  oz.  under  the  uuQceH  aud 
a(l(i  tlic  2  Ih.  to  the  pounds. 
We  proceed  in  the  Hame  manner  with  each  denomination  until  the  entire 
Bum,  15  cwt.  'J3  lb.  5  uz.,  in  luuud. 


cwt. 

lb. 

oz 

7 

84 

14 

6 

97 

8 

2 

9 

15 

15 

92 

6 

2.  Find  the  sum  of  J  lb.,  f  dr.,  and  J  sc. 


Solution.— 1.  According  to  (250)» 
only  rructiunal  unitn  of  the  Hame  kind 
and  of  the  nanie  whole  ran  he  added ; 
hence  we  reduce  I  lb.,  t  dr.,  and  ]  hc.  to 
intej?erg  of  lower  <lenomLnatiouB  (370), 
and  then  add  the  rci^tultH,  as  nhowu  iu 
the  margin.    Or, 

2.  The  given  fractions  may  be  re- 
duced to  fractions  of  the  same  denomination*  and  the  results  added 
according  to  (S50),  and  the  value  of  the  sum  expressed  In  Integeis  of 
lower  denominations  according  to  (3  7  0). 


oz. 

dr. 

BC. 

&r 

Jib. 

=  5 

2 

2 

0 

Hdr. 

~ 

a 

8 

f  sc. 

rr 

15 

5~ 

8 

~a~ 

8 

19  4i 


m 

;tf.3 


164 


DENO MINA  TE     X  U Mil  K R  S, 


EXAMPLES    FOF;    »=»RACTICE. 


;J80.  1.  Add  13  T.  18  cwt.  2  qr  19  lb.  Vi  oz.,  15  cwt.  3  qr. 
20  lb.,  32  T.  19  cwt.,  17  T.  15  cwt.  14  oz.,  and  8  T.  12  cwt. 
13  lb.  15  oz. 

2.  Add  13  lb.  7  oz.  5  ar  2  sc.  9  oz.  7  dr.  15  gr.,  7  lb.  9  oz. 
7  dr.,  11  oz.  5  dr.  2  sc.  19  gr„  and  9  lb.  10  oz.  0  dr.  I  sc.  18  gr. 

3.  Find  the  sum  of  f;  lb.,  5  dwt.,  I  oz.,  uud  ■:  lb. 

4.  A  trader  bought  three  lots  of  sugar,  the  fuRt  containing 
10  cwt.  3  qr.  IT  lb.,  the  second  11  cwt.  3  qr,  27  lb.,  and  the 
third  26  cwt  2  qr.  12  lb. ;  how  much  did  he  buy  ? 

5.  What  is  the  sum  of  Aj  cwt.,  39;  lb  ,  and  14}  oz.  ? 

6.  Find  the  sum  of  3.75  T.,  7.9  cwt.,  and  19^  lb. 

7.  Find  the  sum  of  13.4o  lb.,  8.75  oz.,  and  3.7  dr. 

8.  Find  the  sum  of  .7  oz.,  .5  dwt.,  .75  lb.,  .45  oz.,  .9  dwt. 

9    If  ft  druggist  .lells  in  prescriptions  lb.  4  5  »  3  0  ^2  of  a 
certuiu  drug  in  January,  lb.  8  ^^  7  3  7   D2  in  February,  lb.  10 
3  10  3  2  Dl  in  March,  ?b.  9  51-2   DO  in  April,  and  lb.  7  r  9 
3  3   32  in  May,  how  much  was  sold  during  the  five  months? 

An8.  lb.  40  =10  31. 

10.  A  manufacturing  company  l)ought  4  hurts  of  silver, 
weighing  respectively  11  lb.  8  oz.  10  dwt.  23  gr. ;  10  lb.  8  oz. 
15  dwt.  10  gr. ;  9  lb.  11  oz.  9  dwt  11  gr. ;  and  4  lb.  9  oz.  10  dwt. 
22  gr.  ;  what  was  their  united  weight  ? 

11.  A  wholesale  produce  dea'er  bought  3  T.  9  cwt.  15  lb.  f  f>z. 
ot'])Utter  during  the  spring,  1  T.  12  cwt.  IH  lb.  G  oz.  during  the 
summer,  2  T.  7  c\vt.  10  lb.  6  oz.  during  the  autumn,  and  3  T. 
2  cwt.  98  lb.  15  oz.  during  the  winter  ;  how  much  did  he  buy 
during  the  entire  yeai  ? 

12.  What  is  the  sum  of  8.7  lb.,  3.34  oz.,  and  411  dwt.  ? 

13.  Find  the  sum  of  .8  cwt.  and  .5  <«.  '- 

14.  A  grocer  sold  4  lots  of  tea  containing  respectively  7  cwt. 
391b.  13  oz.,  5  cwt.  84  1b.  15  oz.,  13  cwt.  93  lb.  7  oz.,  and 
7  cwt.  74  1b.  11  oz.;  what  was  the  entire  weight  of  the  tea 
t*old  V 


EXAMPLES. 


165 


3qT. 
5  cwt. 

.  9  oz. 


AiDing 
id  the 


irt. 

)3  of  a 
,11).  10 
).7  rO 
It  lis? 

-01. 

silver, 

).  8  «)/. 
(I  (Iwt. 

lb.  -^  f^7.. 

1111;  the 

bU  3  T. 

le  buy 


i7  cwt. 
and 
llie  tea 


lb. 

oz. 

dwt 

'.»7 

.^ 
4 

15 

13 

9 

18 

JJ81.  PuoB.  VII. — To  find  the  difference  between  any 
two  denominate  or  compound  numbers,  or  denominate 
fractions. 

Find  the  difference  between  27  lb.  7  oz.  M  dwt.  and  IIJ  lb. 
9  02.  18  dwt. 

HoLUTioN.— 1.  Wt'  write  inimbcri*  of  tho  samt' 
(lenomiuntion  uiulcr  each  other. 

a.  VVc  eiubtruct  ae  in  r>iinpl('  luunbiT!*.    Wlu'ii 
the  number  of  auy  deuomiiiatiou  of  the  fiil)tra- 
13         9  17  hcud  cHunot  be  taken   from  the  nnniber  «>!'  the 

same  denomiuatiou  iu  tlie  minuend,  \vc  add  art  in 
ftimple  numbers  (.57)  one  from  tlie  next  hiuh«'r  denomination.  Thun, 
IH  dwt.  cannot  l>e  tiiken  from  15  dwt. ;  we  add  1  of  the  1  oz.  to  tlie  \'i  dwt., 
niakin<;  :io  dwt.  18  dwt.  from  ;^5  dwt.  leaves  17  dwt.,  which  we  write  under 
tl.e  pennywei;rht!'. 

We  tiroceed  in  the  f»ame  manner  witli  each  denomination  until  tlie  entire 
difTerencf.  1.3  lb.  *.»  oz.  17  dwt.,  Is  found. 

To  Hubtract  denominate  fractiouti,  wc  reduce  as  directed  in  addition, 
and  then  subtract. 


EXAMPLES     FOR     PRACTICE. 

JJ812.  Find  the  differei)ce  between 

1.  29  lb.  1  oz.  l;j  dwt.  and  17  lb.  s  oz.  19  dwt.  12  pr. 

2.  lb.  13  =7  :r>   v)l  15  ^^r.  and  lb.  7  :  i)  :G   v)2  12  gr. 

3.  25  T.  10  cwt.  2  qr.  19  lb.  uiid  13  T.  18  cwt.  ^2  lb.  13  oz. 

4.  5;  lb.  and  2:;  dwt.  8.  8.36  T.  and  19.75  cwt. 

5.  IJ  T.  and  2;'  cwt.  9.  9.7  oz.  and  5.3  dwt. 

6.  V,  lb.  and  I  dr.  10.  3«  cwt.  and  7^  lb. 

7.  45  lb.  and  %  oz.  11.  7.5  lb.  and  4.75  sc 

12.  A  druppst  had  13  1b.  4  oz.  5  dr.  of  a  certain  niodirine, 
and  sold  at  one  time  3  lb.  7  oz.  (5  dr.  2  sc.  at  another  5  lb.  9  oz. 
4  dr.  1  sc.  10  pr.     How  much  huH  he  left? 

13.  Out  of  a  stnek  of  ha;'  "ontaininp  10  T.  9  cwt  1  (jr.  12  lb. 
three  loads  were  sold  contninlnp.  respectively,  3T.  A  cwt..  2  T. 
19  cwt.  2  i\r.  9  ib.,  and  3  T.  13  cwt.  14  lb.  Uuw  much  hay  is 
left  in  the  btackV  * 


I 


I 


166 


DEXOMINATE    NUMBERS, 


383*    Pros.  VIII.~To  multiply  a  denominate  or  com- 
pound number  by  an  abstract  number. 

Multiply  18  cwt.  74  lb.  9  oz.  by  6. 


18cwt. 


74  lb.    9  oz. 
6 


5T.    12  cwt.    471b.    6  oz. 


Solution.— We  multiply  as  in 
simple  numbiTH.  commcncinp: 
with  the  loweHt  (Icuoiniiintion. 
Thus,  6  times  9  uz.  equals*  M  oz. 
We  reduce  tlie  .54  oz.  to  pounds 
(36T),  equal  3  lb.  6  oz.  We  write  the  fi  oz.  under  the  ounces,  and  add  the 
.3  lb.  to  the  product  of  the  poands. 

We  proceed  In  this  manner  witli  each  denomination  until  the  entire 
product,  6  T.  12  cwt.  47  lb.  6  oz.  is  found. 


EXAMPLES    FOR    PRACTICE. 

384.  1.  Multiply  7  cwt.  2  qr.  18  lb.  5  oz.  by  9  ;  by  12; 
by  63. 

2.  Multiply  3  lb.  9  oz.  12  dwt.  17  gr.  by  4  ;  by  7. 

3.  Each  of  8  loads  of  hay  contained  2  T.  7  cwt.  19  lb.  ;  what 
is  the  weight  of  the  whole  ? 

4.  A  grocer  sold  12  firkins  of  Imtter.  oach  containing  03  lb. 
13J  oz.     How  much  did  they  all  contain  ? 

6.  A  druggist  bought  25  boxes  of  a  certain  medicine,  each 
l)ox  containing  2  lb.  5  oz.  7  dr.  1  sc.  19  gr.  ;  what  was  tlie 
weight  of  the  whole?  , 

Multiply  and  reduce  to  a  compound  number  : 


6.  8»  lb.  by  14. 

7.  21  lb.  by  9. 

8.  9.50  cwt.  by  7. 

9.  10.95  lb.  by  5. 

10.  13  lb.  7f  oz.  by  8. 

11.  6.84  T.  by  .9. 


12.  2.13  dr.  by  .4. 

13.  7.03  cwt.  by  .34 

14.  S  T.  by  l 

15.  I  lb.  by  .7. 

16.  ^  cwt.  by  ^. 

17.  .9  dwt.  by  .9. 


18.  Tf  a  load  of  coal  by  the  long  ton  weigh  2  T.  8  cwt.  3  qr. 
11  lb.,  what  will  Ik*  the  weight  of  32  loads? 

19.  A  drayman  delivered  on  board  of  a  boat  12  loads  of  coal, 
each  containing  ^  T.  7  cwt.  16  lb.  How  much  coal  did  he  put 
on  board? 


I 


EXAMPLES, 


167 


r  com- 


ply as  in 
mcnclng 
ulimilon. 
ilt*  Muz. 
>  pounds 
1  add  the 

he  entire 


(185.    Prob.  IX.— To  divide  a  denominate  or  compound 
number  by  any  abstract  number. 

Divide  29  lb.  7  ok.  3  dr.  by  7. 


dr. 
7)2U 


oz. 

7 


dr. 
o 


Solution.— 1.  The  object  of  the  division  is 
to  find  \  of  the  compound  number.    ThiA  is 
dune  by  fliidiiiK  the  \  of  each  deuomiuatiou 
4       2       6  separately.    Hence  tlie  procet«8  is  the  same  as 

in  finding  one  of  the  equal  parts  of  a  concrete  numbet*. 

ThiiH,  the  )  of  2*)  lb.  U  1  lb.  and  1  lb.  rcmaininj,'.  We  write  the  4  lb.  in 
the  quotient,  and  reduce  the  1  lb.  to  ounces,  which  added  to  7  oz.  make 
1\«  oz.  We  now  And  the  \  of  the  19  oz.,  and  proceed  as  before  until  the 
entire  quotient,  4  lb.  2  oz.  (3  dr.,  io  found. 


i 


by  12; 


b.  ;  what 


he 


oaib. 


line,  ( 
was 


ach 
the 


|wt.  3  qr. 

of  coal, 
he  put 


I 


EXAMPLES    FOR    PRACTICE. 

1.  If  20  lb.  7  oz.  IG  dwt.  are  made  into  7  equal  parts,  how 
much  will  there  be  in  each  i>art? 

2.  Divide  9  T.  15  cwt.  3  qr.  18  lb.  by  2  ;  by  5  ;  by  8  ;  bj  i2. 

3.  A  drugjjfist  made  12  powders  of  ?!  '  5  of  a  certain  med- 
icino  ;  what  was  th»;  weight  of  each  powder? 

4.  Divide  lb.  3  =7  3  4  32  ])y  4  ;  by  0  ;  by  12  :  by  32. 

5.  The  ig^re^nte  weip^ht  of  43  equal  sacks  of  coffee  is  2  T. 
7  CNNl.  2  ([I    12  lb.  ;  what  is  the  weight  of  each  sack  ? 

C.  Divide  5  T.  U  cwt.  2  qr.  8  lb.  by  3  cwt.  1  (jr.  12  lb. 

Reduce  both  the  dividend  and  divisor  to  the  fianic  denomination,  and 
divide  UK  in  bimple  munbers. 

7.  Divide  lb.  6  5  9  3  7  32  by  3  7  !) 2  gr.  10. 

8.  How  many  boxes,  each  coiitaininj<f  DO  lb.,  will  hold  1  T. 
13  cwt.  2qr.  101b.?  Ana.  35. 

9.  Divide  2  lb.  5  oz.  2  (\\y\.  7  gr.  by  1  oz.  3  dwt.  7  prr. 

10.  Divide  .98  lb.  by  .40  dwt.  ;  |  of  a  ton  by  %  of  a  i)ound. 

11.  A  (IrugL'ist  purchased  154  eqtial  lK)ttle8  of  a  certain  m<Hl. 
icine,  containing  in  the  aggregate  lb.  34  §2  3  5  31;  how 
much  did  each  bottle  contain  ? 

12.  Divide  lb.  75  by  3  .58  ;  .08  T.  by  .0  qr. 


168 


DENOMiy  ATE    iV  UMB  K  li  S. 


fi-  1 


1 


UNITS    OF    LENGTH. 

380.  A  yard  is  the  Standard  Unit  in  linear,  surface, 
and  «(;^ic2  measure. 

LINEAB    MEASURE. 


TABLE  or   UNITS. 

13    in.  =  1  ft. 

3     ft.  =  1  yd. 

5|  yd.  =  1  rd. 

40     rd.  =  1  fur. 

8    fur.  =  1  mi. 

320    rd.  =  1  mi. 


1.  Dvnotninationa. —TncYma  (in.).  Feet 
(ft.),  Yards  (ytl ),  Rods  (rd.),  MIK-h  (ml.). 

2.  Equivalents.— I  mi.  =:  380  rd.  =  5281) 
ft.  =  633<J0  in. 

a.  L'«<'.— Used  In  measuring  lines  and  dis- 
tances. 

4.  In  measuring  tlotli  tlie  yard  i.«  divided 
Into  halves,  fourths,  eighths,  and  sixteenths. 
In  estimating  duties  In  tlie  Custom  House,  it 
is  divided  Into  tenths  and  hundredths. 


Table  of  Special  Denominations, 

00      Geographic  or  ^  ^  "^  h^i\iMi\^  on  a  Mcri 

(iU.lO  Statute  Miles 


\   = 


1  Degree 


dian  or  of  Longitude 

on  the  Equator. 

300      Degrees  =the  Circumference  of  the  Earth. 

^  ^,         ^w.   \  Used  to  measure  dis- 
—  1  Oeog.  Ml.  <      . 

°         (     tances  at  sea. 

(ieographical  Miles^l  League. 

Used  to  measure  depths 

at  sea. 

i  Used    to    measure      the 

Inches  =1  Hand.  \      height  of  horses  at  th 

V     shoulder. 

SURVEYORS'    LINEAR    MEASURE. 


1.10  Statute  Miles 

3 

e      Feet 

4 


=  1  Fathom. 


TABLE  OF  rMT9. 

7.0'2  in.    r^  1  1. 


1.  Ih'notniniition.H,  —  \j\v\\i    (').     i{od 
(rd.>,  Chain  (ch.),  Mile  (ml.). 

2.  Equivalonts.—i  ml.=80ch.-320rd. 
=  8001)1. 

.1  /'Af .— Us'-'d  In  nitasnrlng  roads  and 
boundiiries  i.f  l.'iinl. 

4.  The  t'ltit  of  measure  lu  the  (  unfer'n 
Chain,  v\hlch  CDUiaius  10()  links,  equal  4  rods  or  60  Teet. 


2.1  1.      =  1  rd. 

4  rd.    -   1  rli. 

80  cli.  =  1  mi. 


CLOTH    MEASURE, 


169 


urface, 


n.)»Fect 

ml.)- 
a.  =  5281) 

ft  aiul  <liH- 

is  (llvUU'il 
tixteenfhs. 
nousf,  It 
!/w. 


a  Mcii 

or. 

lurth. 
Hure  (Us- 
ui. 

re  ileptlm 

.ire      the 
('8  (it  the 


rondt*  and 
C  I  inter'' f 


«$87«  The  following  French  measures  are  still  frequently 
used  in  the  Province  of  Quebec : 

1.  The  French  foot  =  1.065765  English,  or  is,  nearly 
enough  for  practical  purix)sts,  thrte-quartera  of  an  inch  longer 
than  the  English  foot. 

3.  The  Avpent,  often  improperly  called  an  aero,  is  a  nwas- 
ure  of  length  equal  to  180  French  I'ect.  The  squart;  Arpiitt 
therefore  contains  32400  Freuch,  or  36801.7  English  8(nnirt< 
feet.  Thirteen  Arpeuts  of  land  are  therefore  very  nearly  equal 
to  eleven  acres. 

3.  The  Toisc!  as  a  measure  of  length  equals  six  French  feet. 
It  is  constantly  used  as  a  measure  of  masonry,  whcMi  it  means 
a  toise  in  length  and  a  toise  in  height  of  a  wall  two  French 
feet  thick  ;  but  as  a  measure  of  quarried  or  broken  stone  it 
means  a  cubic  toise.  Tlie  toise  of  stone  therefore  et^uals  three 
toise  of  masonry,  or  0.684430  cubic  yards  English. 


CLOTH    MEASURE. 
388.  This  measure  is  used  by  linen  and  woolen  drapers. 


2} 

inches  (in. 

)  = 

1  nail, 

marked  na. 

4 

nails 

=z 

1  quarter, 

'•        qr. 

4 

quarters 

= 

1  yard, 

"       yd. 

5 

quarters 

r= 

1  English  ell. 

E.  e. 

6 

quarters 

— 

1  French  ell, 

•*       F.  e. 

8 

(juarters 

=: 

1  Flemish  ell, 

"       Fl.  e 

Tlie  Scotch  ell  contains  4  <iuiirters  li  inch. 

Observe. — The    following    measures    are    used    for   special 
piirposes : 


3  inches 

=r     1  palm. 

18  inches 

=     1  cubit 

3  feet 

=     1  common  i>«co. 

5  feet 

=     1  Roman  nace. 

12 

m 

Pi 

tl 

<     >i 


n 


170 


D£yo  Miy  A  T E   y  um  u  ers. 


f 


EXAMPLES     FOR     PRACTICE, 

380.  Reduce  and  explain  the  following: 

1.  38465  yd.  to  miles.  4.  84  rods  to  linke. 

2.  8  inileH  to  yards.  6.  ^  of  a  rod  to  inches. 
<3.  7  rods  to  inches.                   6.  j^  of  a  ch.  to  links. 

7.  15  degrees  to  statute  miles. 

8.  8.76  geographical  miles  to  statute  miles. 

9.  12  rd.  4  yd.  2  ft.  to  inches. 

10.  210  geographical  miles  to  statute  miles. 

11.  2  mi.  5  ch.  8  rd.  to  links.  ' 

12.  .78  of  a  mile  to  a  compound  number. 
18.  .85  of  a  yard  to  a  decimal  of  a  mile. 

14.  7  yd.  2  ft.  to  a  decimal  of  8  rd.  . 

15.  Find  the  difference  between  3  mi.  5  ch.  2  rd.  13  1.,  and 

-^^  of  (1  mi.  7  ch.  8  rd.  18  1.) 

16.  Find  the  sum  of  ,;  of  a  mi.,  .85  of  a  ch.,  and  3  ch.  2  rd. 

17.  The  f(Mir  sides  of  a  tract  of  land  measure  respectively 
3  mi.  5  ch.  2  rd.,  2  mi.  7  ch.  3  rd.  13  1.,  3  mi.  17  1.,  and  2  mi. 
2  rd. ;  what  is  the  distance  round  it? 

18.  On  a  railroad  182  mi.  234  rd.  4  yd.  2  ft.  long,  tliere  are 
18  stiitioiiH  at  equal  distances  from  each  other.  How  far  are 
the  Htations  ai)art,  there  being  a  station  at  each  end  of  the 
road  ? 

19.  A  '^iiip  moving  r'ue  north  sailed  15.7  degrees.  Ilow  far 
did  she  :;iil  in  statute  miles? 

20.  A  ship  sailing  on  the  equator  moved  45  leagues.  How 
many  degrees  is  she  from  the  place  of  storting,  and  what  is 
the  distance  in  statute  mile.s  V 

21.  ij  of  a  rod  is  what  part  of  3  chains? 

22.  1  link  is  what  decimal  of  1  foot? 

28.  In  125  geog.  miles  how  many  statute  mllea? 

34.  3  ft.  are  what  decimal  of  8  hmIs  ? 

25.  82  fathoms  are  what  decimal  of  a  mile? 


D  EFi  y  I T I  o  ys. 


171 


u:n^its  of  surface. 

;J90.  A  square  yard  is  tho  Sfandat'd  Unit  of  mirface 
measure. 

;5U1.  A  Surface  lias  two  i\\nuinsums-~/cnf;t?i  and  hronlth. 
;iU2.  A  Squni'f  is  a  ;>;^//if  mrface  bounded  by  four  equal 
lines,  and  having  four  right  angles. 

lV,y*\,  A  Jtf'ftattyle  is  any  plane  surface  having  four  sides 
and  four  right  angles. 

«$!>4.  The  Unit  of  Measure  for  surfaces  is  usually  a 
sciuare,  each  side  of  which  is  one  unit  of  u  known  length. 

Thut^,  in  14  ^'q.  ft.,  the  unit  of  measure  in  a  square  foot. 

iJJ)r».  The  Area  of  a  rectangle  is 
the  S'/rfare  included  within  its  boun-  ^/^^^^  '""-• 

darie.s,  and  isexpres.sed  by  the  number 
of  times  it  contains  a  given  unit  of 
meanure. 


Q  Pli.  11 


'.»^<l,  U. 


Than,  Hfnce  a  8qnare  yanl  \»  a  purface 
each  side  of  which  is  3  feet  louy,  it  cun  be  di- 
vided into  .3  rowH  of  square  fei't,  aw  sliown  in 
tho  diagram,  each  row  containini,'  U  fqiiaro 
feet.  Hence,  if  1  sq.  ft.  In  taken  as  the  ('nit 
of  Measurv,  tiie  area  of  a  square  yard  is 
3Hq.  ft.  x8  =  9  8q.  ft. 

The  area  of  any  rectan;,'le  is  fouud  in  tlie  same  manner;  hence  the  fol- 
lowing 

RULE. 

lliHi.  Find  th'  produrt  of  the  numbers  denoting  the  I  nyth 
and  breadth,  expressed  in  the  hurcst  denomination  namtd  in 
either  ;  the  result  w  t/ie  area,  ichieh  can  be  reduced  to  any  re- 
quired denomination. 

To  hnd  either  dimension  of  a  rcctanf?le. 

RULE. 

im7.  Divide  thf  number  trpresxing  the  area  by  the  gieen, 
dimennitm  ;  the  quotient  is  the  oth>  r. 


■lit 

m 


■i%\ 


'■'  j? 


m 


■*f  IT'f 


V  1 


(I  , 


172 


DENOMINATE    NUMBERS. 


SQUARE    MEASUHE. 


TABUS  OP  UNITS. 

144  sq.  iu.  =:  1  sq.  ft. 

9  sq.  ft.  =  1  sq.  yd. 

30  J  »!•  yd.  =  1  sq.  rd.,  or  P. 

IGO  Ki.  ixl.  =  1  A. 

GIO  A.  =  1  sq.  mi. 


1.  Denominations.  —  Square 
lucb  (»q.  iu.),  Squaro  Yard  (ttq.  yd.), 
Bquurc  Uutl  (sq.  nl.),  Acru  (A.), 
Square  Mile  (fq.  ml.). 

2.  IJtiuivalenta.—l  eq.  mi.  = 
WO  A.-10-.MOO  bq.  rd.  =  3UU7tiU>  sq. 
yd.  =  2787*100  eq.  ft.  =  40Ut8»iOO 
eq.  iu. 

8.  A  square  ig  a  four-f>idcd  figure  whot^c  sides  and  angles  are  equal. 
Tbii*  table  in  constructed  Troui  tbo  table  uf  liuoar  mcauore  by  multiplying 
each  dimension  by  itt<t'ir. 

4.  r««*.— Ueed  in  computing  areas  or  surfticee. 

5.  01u7.in<;  and  **tone-cuttlng  arc  L'sfimatcd  by  the  ttquarf  foot ;  planter- 
ing,  paving,  painting,  etc.,  by  the  square  foot  or  square  yard ;  roofing, 
flooring,  etc.,  generally  by  the  square  of  I'iO  square  fett. 

6.  In  laying  Hlungles,  one  thousand^  averaging  4  Inches  wide,  and  laid 
6  inches  to  the  weather,  arc  estimated  to  cover  a  square. 


SURVEYORS'  SQUARE  MEASURE. 


TABLE  OP  UNIT? 
625  sq.  1. 

10  P. 

10  sq.  ch 
C40  A. 


=  1  P. 

=  1  sq.  ch. 

=  1  A. 

=  1  sq.  mi. 


1.  Tienomi  nut  ions,  —  Square  Link 
(nq.  1.),  Poles*  (l*.).  Square  Chain  (;sq.ch.). 
Acres  (A.),  Square  MUe  (sq.  mi.>,  Towu- 
Bhlp  (Tp.). 

2.  i:qnicalentH.—\  Tp.  =  :«  pq.  ml. 
r=  23040  A.  =  2;J0400  eq.  ch.  =  3086400  P. 
=  2304000000  h^q.  1. 

8.  f '«e.— Used  in  computing  the  area 
of  land. 
4.  The  Unit  of  land  meacnre  is  tlic  acre.    The  measurement  of  a  tract 
of  land  is  usually  recorded  in  square  miles,  acres,  and  hundrtdtha  of 
an  acre. 


80  u\.  mi.  =  1  Tp. 


EXAMPLES     FOR    PRACTICE. 
308.  Reduce  and  explain  the  foUowinjj  : 
1.  .83  of  an  A.  to  sq.  yards.        4.  .08  of  an  A.  to  sq.  links. 


2.  5  sq.  mi.  to  sq.  yards. 
8.  8  acres  to  sq.  ft. 


5.  .007  mi.  to  sq.  links. 

6.  3  sq.  mi.  to  sq.  chains. 


E X  A  MP  L  E  S. 


173 


luting  the  area 


7.  35  pq.  yd.  to  a  dpcimal  of  an  acre. 

8.  14  P.  to  t^  decimal  of  a  sq.  mi. 

D.  J  of  a  sq.  mi.  to  a  cnm]>oiind  uumber. 

10.  .0005  of  an  A.  to  sq.  feet. 

11.  5  of  a  Tp.  to  a  compound  number. 

12.  .0008  of  a  Bq.  mi.  to  a  compound  number. 

Find  the  sum  of 

13.  f  of  an  A.,  §  of  a  sq.  rd.,  and  3  A.  158  sq.  rd.  25  sq.  yd. 

14.  I  of  (1  Tp.  18  sq.  mi.  584  A.),  and  ^\  of  (378  A.  9  sq.  ch. 
12  P.) 

15.  Find  tlie  diiferonce  between  (^  of  G  (»i  mi.  +  §  of  an  A.), 
and  (I  of  an  A.  -^  i{  of  a  i)ole). 

16.  Subtract  1  -q.  1.  from  1  aero  ;  from  1  township. 

17.  A  tract  of  land  containing'  084  A.   7  8<|.  ch.  13  P.  was 
divided  into  7  ctjual  farms  ;  what  was  tlie  size  of  each  farm  ? 


What  is  tlio  area  of  rectangles  of  tlie  following  dimensions : 

31.  7.5ch.  by  3ch.  81.? 

23.  4  yd.  2  ft.  4  in.  by  3|  yd.  ? 

33.  3.4  yd.  by  0^  yd.  ? 


18.  15  yd.  by  12  yd.  ? 
10.  10^  yards  square  ? 
20.  93  yd.  by  18|  yd.  ? 


34.  How  many  yards  of  carpeting.  3  ft.  3  in.  wide,  will  be 
required  for  3  rooms  18  ft.  by  34  ft.,  and  4  rooms  12  ft.  by  10  ft. 
C  in.  ?  .1/^^.  309J  yd. 

85.  How  many  boards  13  ft.  long  and  4  in.  wide  required  to 
tloor  a  room  which  is  48  ft.  by  33  ft.?  Ans.  384  boards. 

36.  How  many  square  feet  of  luinb<>r  required  for  the  floors 
of  a  house  containing  3  rooms  15  ft.  by  19  ft.,  5  rooms  14  ft.  by 
IG  ft.,  and  3  rooms  13  ft.  by  15  fr.  ?  Ans.  3230  sq.  ft. 

27.  Find  the  cost  of  carptting  a  house  containing  rooms  as 
follows :  4  rooms  15  ft.  by  10  ft.  6  in.,  carpet  f  yd.  wide  at 
$1.26  iwr  yard  ;  3  rooms  18  It.  by  25  ft.,  carpet  I  yd.  wide  at 
$3.45  per  yard  ;  and  5  rooms  13  ft.  8  in.  by  16  ft.,  carprt  1  yd. 
wide  at  $1.08.  Ans.  $620. 


m 


Ml 


174 


DENO  MIX  A  TE     i\  UM  R  E  h' S » 


28.  Find  tlie  cost  of  glazing  10  windows,  each  9  ft.  10  in.  by 

5  ft.  8  in.,  ftt  ^.1)4  a  square  fcK)t. 

21).  How  many  tiles  10  inches  square  will  lay  a  floor  32  ft. 

6  in.  l)y  28  ft.  0  in.  ?  Ans.  1090.08. 

30.  Tlie  ridge  of  the  roof  of  a  building  is  44  ft.  lonff,  and  the 
diHtance  from  each  cave  to  the  ridge  is  19  ft.  'i  in.  How  many 
shingles  4  in.  wide,  laid  6J  in.  to  the  weather,  will  be  required 
to  roof  the  building,  the  first  row  Ining  doul)le? 

31.  Find  the  cost  of  lathing  and  plastering  a  house  at  ^.52 
per  square  yard,  containing  the  following  rooms,  no  allowance 
being  made  for  doors,  windows,  and  baseboard  ;  3  rooms  14  ft. 
by  18  ft.,  and  2  ro<mi8  12  ft.  by  15  ft.,  height  of  ceiling  11  ft. ; 
4  rooms  12  ft.  by  16  ft. ;  and  2  rooms  12  ft.  by  14^  ft.,  height  of 
ceiling  9  ft.  6  in. 

32.  How  many  flag-stones,  3  ft.  5  in.  by  2  ft.  0  in.  will  bo 
reijuired  to  cover  a  court  125  ft.  by  82  ft.,  and  what  will  be  the 
cost  of  llaggiiig  the  court  at  $187  a  scjuare  yard  V 

33.  What  will  be  the  cost  of  papering  a  room  20  ft.  by  32, 
height  of  ceiling  12  ft.,  with  rolls  of  papc^r  8  yards  long  18 
inclies  wide,  at  $1.63  per  roll,  deducting  132  sq.  ft.  for  doors, 
windows,  and  baseboard  ? 


UNITS    OF    VOLUME. 

JJOO.  A  Solid  or  Volume  has  three  dimensions— ^c/jgr^A, 
t>readtfi,  and  thicknesH. 

400,  A  lU'ctnngulftr  Solid  is  a  body  bounded  by  six 
rectdiiglcs  called  faces. 

4-01.  A  Cube  is  a  rectangular  soUO,  l)ounded  by  six  eijual 
squares. 

4012.  The  Unit  of  Measure  is  a  cube  whoso  edge  is  a 
unit  of  some  known  length. 

403.  The    Volume,  or  Solid  Contents  of  a  body,  is 


CUBIC     3tt:ASUIiL\ 


175 


expressed  by  the  number  of  times  it  (•ontninn  a  given  nnii  of 
vuuAurc.  For  cxumplf,  tin;  contents  of  u  cubic  yurd  is  ex- 
proHstul  us  '^7  cuUicfeet. 

Tlui!',  bIucc  each  face  of  a       m.^    jml'j^ N. 

cuhic  yard  ccu^aIiih  U  »q.  ft., 
if  a  HCi-.tlou  1  ft.  thick  18 
takcu  it  iiiiiKt  COD  tain  3  timcB 
H  cii.  ft.,orUcu.  ft., at)  Hhown 
iu  thu  diuKram. 

And  hIir-u  tho  cubic  yard 
is  8  feet  thick,  it  tniiMt  cou- 
tain  3  HcctiuuH,  oach  cun- 
tainiuK  *J  cu.  ft.,  which  it) 
87  en.  ft. 

lloucu,  the  volume  or  con- 
ten*n  of  A  cubic  yard  uzprcHHcd  in  cubic  fei't  itt  found  by  taking;  tlio  product 
of  iIju  nuinborrt  denoting  itu  3  dlmencioni*  in  font. 

Thu  contents  of  any  rectangular  Holid  in  found  in  the  eame  manner ;  hence 
the  following 

RULE. 

404.  Mnd  the  product  of  the  numbers  denoting  the  three 
dimensions  expressed  in  the  lowed  denomination  named.  This 
result  is  the  voiuniCf  and  can  be  reduced  to  any  required  de- 
nomination. 


Mi 


>■« 


To  find  a  required  dimension  : 


RULE. 

405.  Divide  the  volume  by  the  product  of  the  numbers  de- 
noting the  other  tico  dimensions. 

The  volume,  before  division,  must  bo  rcdnrod  to  a  cubic  unit  correa- 
pouding  with  the  Bquare  unit  of  the  product  of  tlio  two  diinuut-iou.>«. 


V;'\ 


CUBIC    MEASURE. 


1.  />*'n«i»*/»if»fi*oij.«i.— Cubic  Inch 
(cu.  in).  Cubic  Foot  (cu.  ft.),  Cubic 
Yurd  (cu.  yd.). 

a.  l^iiiirtiffHtM — 1  cu.  yd.  =  87 
cu.  ft.  =  4»MW<)cu.  in. 
8.  r'#«».— Used  in  computing  tho  volumo  or  conteut»  of  aollds. 


TABLE  OP  UNITS. 

1728  cu.  in.  =   I  cu.  ft. 
27  cu.  ft.   =  1  cu.  vd. 


r  i . 

■iil 


IMAGE  EVALUATION 
TEST  TARGET  (MT-S) 


1.0 


I.I 


1.25 


m  MM 


1^ 

'^  IM    122. 

!''  IM   '""^ 
.'■:   IM    12.0 


1.8 


U    111.6 


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176 


DENOMINATE    NUMBERS. 


V  I 


EXAMPLES    FOR    PRACTICE. 

406.  Reduce  and  explain  the  following  : 

1.  97  cu.  ft.  to  cu.  in.  3.  4  cu.  yd.  394  cu.  ft.  to  en.  izu 

2.  .09  of  a  cu.  yd.  to  cu.  ft.     4.  .0007  of  a  cu.  yd.  to  cu.  ia. 

5.  Find  the  sum  of  f  of  a  cu.  yd.  and  .625  of  a  cu.  ft. 

6.  .8  of  a  cu.  ft.  to  a  decimal  of  a  cu.  yd. 

7.  f  of  a  cu.  ft.  to  a  decimal  of  a  cu.  yd. 

8.  Find  the  difference  between  |  of  a  cu.  yd.  and  .75  of  a 
cu.  ft. 

Find  the  contents  of  rectangular  solids  of  the  following 
dimensions : 

9.  A  solid  7  ft.  9  in.  long  by  3  ft.  4  in.  by  4  ft.  6  in. 

10.  A  cube  whose  edge.is  3  yd.  2  ft.  8  in. 

11.  A  solid  34  ft.  long  by  1  ft.  6  in.  by  2  ft.  9  in. 

12.  A  solid  12  yd.  1  ft.  9  iu.  long  by  2  yd.  2  ft.  by  2  ft.  8  in. 

13.  How  many  cubic  feet  in  a  stick  of  timber  38  ft.  long  by 
2  ft.  3  in.  by  1  ft.  9  in.? 

14.  A  cistern  9  ft.  sq.  contains  1092  cu.  ft. ;  what  is  its 
depth? 

15.  A  stick  of  square  timber  contains  189  cu.  ft. ;  2  of  its 
dimensions  are  1  ft.  9  in.  and  2  ft.  3  in. ;  what  is  the  other? 

16.  How  many  cubic  yards  of  earth  in  an  embankment 
283  ft.  by  42  ft.  8  in.  by  18  ft.  6  m. ? 

17.  How  many  cubic  feet  of  air  in  a  room  74  ft.  9  in.  long, 
52  ft.  10  in.  wide,  and  23  ft.  6  in.  high? 

18.  A  bin  contains  d2Q\  cu.  ft. ;  2  of  its  dimensions  are  9  ft. 
8  in.  and  7  ft.  6  in. ;  what  is  the  other? 

19.  A  vat  is  7  ft.  2  in.  by  4  ft.  9  in.  by  3  ft.  4  in.    How  many 
cubic  feet  does  it  contain? 

20.  In  digging  a  cellar  48  ft.  6  in.  by  39  ft.  8  in.,  and  8  ft. 
4  in.  deep,  how  many  cubic  yards  of  earth  must  be  removed? 


EXAMPLES. 


177 


21.  What  will  be  the  cost  of  the  following  bill  of  square 
timber,  at  f  .33J  per  cubic  foot : 

(1.)  3  pieces  13  ft.  by  9  in.  by  7  in.  ? 

(2.)  8  pieces  15  ft.  6  in.  by  10  in.  by  8  in.  ? 

(3.)  4  pieces  33  ft.  by  8  in.  by  9  in.  ? 

(4.)  6  pieces  3G  ft.  by  1  ft.  6  in.  by  1  ft.  ? 

(5.)  9  pieces  18  ft.  9  in.  by  1  ft.  3  in.  by  9  in.  ? 

(6.)  12  pieces  15  ft.  by  7^  in.  by  9^  in.  ? 

22.  How  many  perches  in  a  wall  37  ft.  long,  23  ft.  6  in. 
high,  and  2  ft.  G  in.  thick  ? 


Table  of  Units  for  Measuring  Wood  and' Stone* 


16  cu.  ft.  =    1  Cord  Foot  (cd.  ft.)  \  Used  for  measuring 

8cd.ft.or  )   ^     icord(cd.)  [        „,'»"' 

128  cii.  ft.       ^  ;     wood  and  stone. 

2i}l  cu.  ft.  =     1  perch  (pch.)  of  stone  or  masonry. 

1  cu.  yd.  of  earth  is  called  a  load. 


1.  The  materials  for  masonry  are  npually  estimated  by  the  cord  or  perch^ 
the  work  by  the  perch  and  cuMc  foot,  also  by  the  square  foot  and  square 
yard. 

2.  In  estimating  the  mason  work  in  a  building,  each  wall  is  measured 
on  the  outside,  and  no  allowance  is  ordinarily  made  for  doors,  windows, 
and  cornices,  unless  specified  in  contract.  In  estimating  the  material,  the 
doors,  windows,  and  cornices  are  deducted. 

3.  Brickwork  is  usually  estimated  by  the  thousand  bricks,  which  are  of 
various  sizes. 

4.  Excavations  and  embankments  are  estimated  by  the  cubic  yard. 


How  many 


EXAMPLES    FOR    PRACTICE. 

407.  Reduce  and  explain  the  following : 

1.  42  cords  to  cd.  feet.  3.  36  cords  to  cu.  feet. 

2.  64  pch,  to  cu.  feet.  4.  .84  pch.  to  cu.  feet. 


178 


DENOMINATE     NUMBERS. 


n 


I. 


6.  7  of  a  cd.  to  cu.  feet. 

6.  .73  of  a  cu.  ft.  to  a  decimal  of  a  cd. 

7.  f  of  a  cord  to  a  decimal  of  a  cu.  yd. 

8.  ^  of  a  cu.  ft.  to  a  decimal  of  a  pch. 

9.  .85  of  a  cord  to  a  decimal  of  3  cu.  yd. 

10.  I  of  8  cd.  to  a  decimal  of  13  cd. 

11.  Find  the  sura  of  f  pch.,  |  cd.,  and  11  cd.  ft.  38  cu.  ft. 

12.  A  pile  of  wood  containing  84  cd.  7  cd.  ft.  12  cu.  ft.  was 
made  into  5  equal  piles  ;  what  was  the  size  of  each  ? 

13.  How  many  cords  in  a  pile  of  wood  196  ft.  long,  7  ft.  6  in. 
high,  and  8  ft.  wide? 


A  Cord  is  a  pile  of  wood,  stone,  etc.,  8  ft.  long,  4  ft.  wide* 
and  4  ft.  high. 

A  Cord  Foot  is  1  ft.  long,  4  ft.  wide,  and  4  ft.  high,  or  \  of 
a  cord,  aa  shown  in  the  cut. 

14.  What  is  the  cost  of  a  pile  of  ^tone  28  ft.  long,  9  ft.  wide, 
and  7  ft.  high,  at  |3.85  per  cord  ? 

15.  A  load  of  wood  containing  1  cord  is  3  ft.  9  in.  high  and 
4  ft.  wide  ;  what  is  its  length  ? 

16.  How  many  perches  of  masonry  will  18  cd.  5  cd.  ft.  of 
stone  make,  allowing  22  cu.  ft.  of  stone  for  1  perch  of  wall  ? 

17.  How  many  cords  of  stone  will  be  required  to  enclose 
with  a  wall  built  without  mortar  a  lot  28  rods  long  and  17  rods 
wide,  the  wall  being  5  ft.  high  and  2  ft.  9  m.  thick  t 


\n 


EXAMPLES. 


m 


BOARD    MEASURE. 


TABLE  OF  TTNITS. 

12  B.  in.  =  1  B.  ft. 
12  B.  ft.  =  1  cu-  ft. 


408.  A  Bofird  Foot  is  1  ft. 

long,  1  ft.  wide,  and   1  in.  thick. 
Hence,  12  hoard  feet  equals  1  cu.  ft. 


409.  A  Board  Inch  is  1  ft.  long,  1  in.  wide,  and  1  in. 
thick,  or  ^V  of  a  hoard  foot.  Hence,  12  hoa/rd  inches  equals 
1  hoard  foot. 


Observe  carefully  the  following  : 

(1.) 
4  feet  long. 


•a 


« 


Square 
foot. 

1 

1 

1.  Diagram  (1)  represents  a 
l/Ourd  where  botli  dimensions 
are  feet.  Hence  tlie  product  of 
the  two  dimensionB  gives  the 
square  feet  in  surface  (396),  or 
the  number  of  board  feet  when 
the  lumber  is  not  more  than 
1  hich  thick. 

2.  Diagram  (2)  represents  a 
board  where  one  dimension  is 
feet  and  the  other  inches.  It  is 
evident  (408)  that  a  board  1 
foot  long,  1  inch  thick,  and  any 
number  of  inches  wide,  contains 
as  many  board  incites  as  there 
are  inches  in  the  width.  Hence 
the  number  of  square  feet  or 
board  feet  in  a  board  1   inch 

thick  is  equal  to  the  length  in  feet  multiplied  by  the  width  in  inches 
divided  by  12,  the  number  oi  board  inches  in  a  board  foot 

3.  In  case  the  lumber  is  more  than  1  inch  thick,  the  number  of  board 
feet  is  equal  to  the  number  of  square  feet  in  the  surface  multiplied  by  the 
thickness- 


ID 


4  X  2  =  8  sq.  ft.  or  8  B.  ft. 

(2.) 
4  feet  long. 


1ft.  by 
9  in. 

«  4x9=36  B.  In.;  36 B.  in.-i-12=3B.  ft. 


'•">. 


EXAMPLES    FOR    PRACTICE. 

410.  Find  the  contents  of  boards  measuring 

1.  24  ft.  by  13  In.        4.    9  ft.  by  32  in.        7.    5  ft.  by  18  in. 

2.  28  ft.  by  15  in.    5.  13  ft.  by  26  in.    8.  34  ft.  by  15  in. 

3.  18  ft.  by  16  in.    6.  17  ft.  by  30  in.    9.  25  ft.  by  14  in. 


fi-W 


IP! 


180 


DENOMINATE    NUMBERS, 


f'/i 


Find  the  contents  of  boards  measuring 

10.  15  ft.  by  1  ft.  3  in.  12.  19  ft.  by  2  ft.  4  in. 

11.  27  ft.  by  1  ft.  6  in.  13.  23  ft.  by  1  ft.  5  in. 

14.  Find  the  contents  of  a  board  18  ft.  long  and  9  in.  wide. 

15.  How  many  board  feet  in  a  stick  of  square  timber  48  ft. 
long,  9  inches  by  14  inches. 

16.  Find  the  length  of  a  stick  of  timber  8  in.  by  10  in.,  which 
will  contain  20  cu.  ft. 

Opbbation.— (1728  x  20) +(8  x  10)  =  432 ;  432+12  =  3G  ft.,  the  length. 

17.  A  piece  of  timber  is  10  in.  by  13  in.     What  length  of  it 
will  contain  26  cubic  feet  ? 

Find  the  cost  of  the  following : 

18.  Of  234  boards  14  ft.  long  8  in.  wide,  at  $3.25  per  hundred. 

19.  Of  5  sticks  of  timber  27  ft.  long,  ^  in.  by  14  in.,  at  $1.75 
per  hundred  feet  board  measure. 

20.  Of  84  plank  20  ft.  long,  11  in.  wide,  3  in.  thick,  at  $1.84 
per  hundred  feet  board  measure. 


UNITS    OF   CAPACITY. 

411.  The  Standard  Units  of  capacity  are  the  Gallon 
for  Liquid,  pjid  the  Bushel  for  Dry  Measure. 


II 


LIQUID    MEASTTBE. 


1.  Deuoiuinations.—Gil\a  (gi.),  Pints 
(pt.),  Quarts  (qt.),  Gallons  (gal.),  Barrels 
(bbl.). 

2.  Equivalents.— 1  gal.=:4  qt.=8  pt.  = 
32  gi. 

3.  Use.— Used  in  measuring  liquids. 

4.  The  capacity  of  cisterns,  vats,  etc.,  is 
usaally  estimated  by  considering  a  barrel  31J  gal. ;  but  barrels  are  made  of 
various  sizes,  from  30  to  56  gallons.  The  hogshead,  butt,  tierce,' pipe,  and 
tun  are  names  of  casks,  and  have  usually  their  capacity  in  gallons  marked 
upon  them. 


TABLE  OF  UNTTS. 

4    gi.    =  1  pt. 

2    pt.    =1  qt. 

4    qt.    =  1  gal. 

81|  gal.  =  1  bbl. 


DJiT    3IEASURE, 


181 


AFOTHECABIES'    FLUID    MEASURE. 


in. 

in. 

in.  wide. 

mber  48  ft. 

Oin.,wliicli 

}  length, 
ength  of  it 


TABLE  OF  UNITS. 
R  60   =    f    3   1 

f  3    8  =  f  5  1 
f  §  16  =  O.  1 
8  =  Cong.  1 


O. 


1.  Denominations. — Minima  or  drops 
(iTl),  Fluid  Drachm  (f  3),  Fluid  Ounce  (f  5), 
Pint  (O.,  for  octarim,  the  Latin  for  one-eighth 
or  pint),  Gallon  (CoDg.,  for  congius,  the  Latin 
for  gallon). 

2.  JiJqtiivalents,  —  Cong.  1  =  O.  8  = 
f  5  128  =  f  3  lOiM  =  ta  61'M0. 

8.  Use.— JJ^ed  in  prescribing  and  compounding  liquid  medicine. 
4.  The  symbols  precede  the  numbers,  as  in  Apothecaries*  Weight,  as 
shown  in  the  table  of  units. 

DRY    MEASURE. 


ler  hundred, 
in.,  at  $1.75 

ick,  at  $1.84 


TABLE  OP  irniTS. 


tlie  Oallon 


2pt.  = 
8qt.  = 
4pk.  = 


Iqt. 

Ipk. 

Ibu. 


1.  Denotninations.~-riatB  (pt.),  Quarts  (qt.)t 
Pecks  (pk.),  Bushels  (bn.). 

2.  Equivalents*— Ihn.   =  4pk.   =   32  qt.   = 
64  pt. 

8.  Use.— Used,  in  measuring  grain,  roots,  fhiits, 
salt,  etc. 

4.  Heaped  rieamre.,  in  which  the  bushel  is  heaped  in  the  form  of  a  cone, 
is  used  in  measuring  potatoes,  com  in  the  ear,  coarse  vegetables,  large 
fruits,  etc.  Stricken  measure  is  used  in  measuring  grains,  seeds,  and  small 
fruits. 

5.  A  bushel  of  oats  =  34  lb. ;  of  buckwheat,  barley,  timothy  =  48  lb, ;  of 
flaxseed  =  50  lb. ;  of  rye  and  Indian  corn  =  56  lb. ;  of  wheat,  potatoes, 
peas,  beans,  onions,  or  red  clover  seed  =  60  lb. 


bfl  (gl.),  P^nt' 
[(gal.).  Barrels 

:4  qt.=8  pt.  = 

ig  liquids. 
[b,  vatB,  etc.,  is 
els  are  made  of 
h^ce,'iApe,  and 
lallons  marked 


EXAMPLES    FOR    PRACTICE. 

412.  Solve  and  explain  orally  the  following  : 

1.  How  many  gills  in  4  qt.  ?    In  ?  gal.  ?    In  7  qt.  ?    In  8  qt. 
1  pt.  ?    In  3  gal.  3  qt.  ? 

2.  How  many  pints  in  3  bu.  ?    In  3  pk.  5  qt.  ?    In  1  bu.  2  pk. 
7qt.? 

3.  What  is  the  sum  of  O.  5  f  3  12  f  3  7  and  f  §  8  f  3  3  ITilS? 

4.  Multiply  3  pk,  5  qt.  1  pt.  by  3  ;  by  5  ;  by  10 ;  by  7  ;  by  12. 
Reduce 

5.  93584  pt.  to  barrels.  8.  93654  pt.  to  bushels. 

6.  28649  pt.  to  bushels.  9.  57364  gi.  to  barrels. 

7.  TTl  8405  to  gallons.  10.  f  3  7649  to  gallons. 


m 


t,:  * 


. 

1 

I'll 

.  1  *  ! 

1 

182 


DENOMINATE    NUMBERS, 


11.  3  qt.  1  pt.  to  a  decimal  of  a  gallon.  # 

12.  f  of  o  qt.  1  pt.  to  a  decimal  of  2  bushels. 

13.  f  3  7  TT[  15  to  a  decimal  of  Cong.  3. 

14.  A  merchant  bought  5860  bushels  wheat  in  Toronto  at 
$1.25,  and  sold  the  whole  in  Halifax  at  the  same  price.  How 
much  did  he  gain  on  the  transaction  ? 

15.  A  grocer  bought  12  firkins  of  butter,  each  containing 
73  lb.  13  oz.,  at  36  cts.  a  pound ;  7  bu.  3  pk.  clover  seed,  at 
$1.15  a  peck  ;  and  5  loads  of  potatoes,  each  load  containing 
43  bu.  8  pk.',  at  $.32  a  bushel.     How  much  was  the  cost  ? 


> 


Comparative  Table  of  Units  of  Capacity, 


CUBIC  INCHES     CXJBfC  rNCHES       CUBIC  INCHEQ 


IN 
ONE  GAIJ.ON. 

IN 

ONE  QUAKT. 

IN 
ONK  PINT. 

Imperial, 

277.274 

Liquid  Measure  U.  S. 

231 

57f 

m 

Dry  Measure  (|  pk.) 

268| 

671 

m 

1.  The  Imperial  Bushel  of  Great  Britain  contains  2218.192  cu.  in.  and 
the  Standard  Hushel  of  the  United  States  contains  2150.42  en.  in. 

2.  An  English  Quarter  contains  8  imp.  bu.  or  8J  U.  S.  bu.    A  quarter  of 
8  U.  S.  bu.,  or  480  lb.,  is  used  in  shipping  grain  from  New  York. 

3.  A  Register  Ten  is  100  cu.  ft.;  used  in  measuring  the  internal 
capacity  or  t^  nnage  of  a  vessel.    A  Shipping  Ton  Is  40  cu.  ft. 

4.  A  cubic  foot  of  pure  water  weighs  1000  oz.  or  62i  lb.  Avoir. 


\\  if' 


EXAMPLES    FOR    PRACTICE. 

4: 13*    1.  How  many  U.  S.  bushels  in  a  bin  of  wheat  6  ft. 
long,  5  ft.  6  in.  wide,  and  4  ft,  9  in.  deep  ? 

How  many  cubic  feet  in  a  space  that  holds 

2.  1000  U.  S.  bushels?  5.  240  English  quarters? 

3.  1000  imp.  bushels?  6.  18  T.  16  cwt.  of  pure  water?  j 

4.  120  bbl.  water?  7.  804  bu.  3  pk.  U.  S.  bu.? 

8.  A  cistern  containing  5300  gal.  of  water  is  10  ft.  square.  | 
How  deep  is  it?  ^r<«.  7.085 +  . 

9.  How  many  ounces  in  gold  are  equal  in  weight  to  9  pounds] 
14  ounces  of  iron  ? 


|:t;-^ 


TIME    MEASURE, 


183 


Toronto  at 


price. 


How 


;li  containing 
over  seed,  at 
ad  containing 
,lie  cost  ? 

ipacity* 

CUBIC  INCHES 
IM 
ONK  FINT. 

281 
331 

918.192  cu.  in.  and 
)0.42  cu.  in. 

bu.    A  quarter  of 
V  York, 
iring  the  internal 

cu.  ft. 
Avoir. 


of  wheat  6  ft. 


q[uarters  1  . 

.  of  pure  water? 

U.  S.  bu.? 
IS  10  ft.  square.  I 

Ans.  7.085 +  . 
Iglit  to  9  pounds 


TABLE  OP  UNITS. 

60  sec. 

=  1  min. 

GO  min. 

=  Ibr. 

24  hr. 

=  Ida. 

7  da. 

=  1  wk. 

365  da. 

=  1  common  yr. 

366  da. 

=  1  leap  yr. 

100  yr. 

=  1  cen. 

UNITS    OF    TIME. 

414.  The  mean  solar  day  is  the  StiuuUiril  Unit  of  time. 

1.  DenoniinationH.  —  Seconds 
(sec),  Minutes  (min.),  Hours  (hr.), 
Days  (da.),  Weeks  (wk.),  Months 
(mo.),  Years  (yr.),  Centuries  (cen.). 

2.  There  are  12  Caleudur  Months 
In  a  year;  of  these,  April,  June, 
September,  and  November,  have 
80  da.  each.  All  the  other  months 
except  February  have  31  da.  each. 
February,  in  common  years,  has  28  da., 
in  leap  years  it  has  29  da. 

.3.  In  computing  interest,  30  days  are  usually  considered  one  month.  For 
bnsinci's  purposes  the  day  begins  and  ends  at  12  o'clock  midnight. 

415.  The  reason  for  common  and  leap  years  wWl  be  seen 
from  the  following : 

The  true  year  is  the  time  the  earth  takes  to  go  once  arofund  the  sun, 
which  is  365  days,  5  hours,  48  minutes  and  49.7  seconds.  Taking  365  days 
as  a  commcm.  year,  the  time  lost  in  the  calendar  in  4  years  will  lack  only 
44  minutes  and  41.2  seconds  of  1  day.  Hence  we  add  1  day  to  February 
every  fourth  year,  making  the  year  366  days,  or  l^enp  Year.  This  correc- 
tion is  44  rain.  41.2  sec.  more  than  should  be  added,  amounting  in  100  years 
to  18  hr.  37  min.  10  sec.  ;  hence,  at  the  end  of  100  years  we  omit  adding  a 
day,  thus  losing  again  5  hr.  22  min.  50  sec,  which  we  again  correct  by  add- 
ing a  day  at  the  end  of  400  years;  >iLence  the  following  rule  for  finding  leap 
year: 

BULE. 

416.  Enery  year,  except  centennial  years,  exajctly  divisible 
by  4,  is  a  lenp  year.  Every  centennial  year  exactly  divisible  by 
400  is  also  a  leap  year. 

This  will  render  the  calendar  correct  to  within  one  day  for  4000  years. 

417.  Prob.  X. — To  find  the  interval  of  time  between 
two  dates. 

How  many  yr.,  mo.,  da.  and  hr.  from  6  o'clock  P.  M.,  July  19, 
1862,  to  6  o'clock  A.  M.,  April  9,  1876. 


184 


DENOMINATE    NUMBERS. 


i  f 


■■■  t 


yr. 

mo. 

da. 

hr. 

1876 

4 

9 

7 

1862 

7 

19 

18 

13 

8 

19 

13 

Solution.— 1.  Since  the  latter  date 
denutuH  tin;  jjrcuifr  period  of  time,  it 
is  the  minuend,  aud  the  earlier  date, 
the  subtrahend. 

2.  Since  each  year  commonce?  with 
January,  aud  each  day  with  12  o'clock 
midnight,  7  o'clock  a.  m.,  April  9, 1876,  is  the  7th  hour  of  the  9th  day  of 
the  fourth  month  of  1876 ;  and  0  o'clock  p.  m.,  July  19, 18<i2,  is  the  18th  hour 
of  the  19th  day  of  the  seventh  month  of  1862.  Hence  the  minuend  and  sub- 
trahend are  written  as  shown  in  the  mar^^iu. 

8.  Considering  24  hours  as  1  day,  30  days  1  month,  and  18  months  1  year, 
the  subtraction  is  performed  as  in  compound  numbers  (381),  and  13  yr. 

8  mo.  19  da.  13  hr.  is  the  interval  of  time  between  the  given  dates. 

Find  the  interval  of  time  between  tlie  following?  dates  : 

1.  10  P.  M.  October  3, 1812,  and  8  A.  M.  April  17,  1879. 

2.  5  A.  M.  May  19, 1854,  and  7  p.  m.  Sept.  3, 1876. 

3.  March  14,  1776,  and  August  3,  1875. 

4.  7  P.  M.  November  25,  1754,  and  2  a.  m.  May  13, 1873. 

5.  April  19th,  1775,  and  Jan.  20, 1783. 

6.  Washington  died  Dec.  14th,  1799,  at  the  age  of  67  yr. 

9  mo.  22  da.     At  what  date  was  he  born? 


CIBCULAB    MEASUBE. 


'^e/nr-g\t<s^*' 


4 1 8.  A  Circle  is  a  plane 
figure  bounded  by  a  curved 
line,  all  points  of  which  are  | 
equally  distant  from  a  point  j 
within  called  the  centre. 

419.  A  CircunifeV'l 
ence  is  the  line  that  bounds] 
a  circle. 

420.  A  Degree  Is  one! 
of  the  360  equal  parts  into! 
which  the  circumference  of  a| 
circle  is  supposed  to 
divided. 


EXAMPLES, 


185 


the  latter  date 
riod  of  lime,  it 
he  earlier  date» 

omnioncep  with 
r  with  12  o'clock 
t  the  9th  day  of 
5,19  the  18th  hour 

ainucud  and  eub- 

12  months  1  year» 
381),  and  13  yr. 
[;u  dates. 

\gf  dates '. 

17, 1879. 

876. 

y  13, 1873. 

3  age  of  67  yr. 


^irc^e  is  a  plane 
led  by  a  curved 
Us  of  which  are 
[nt  from  a  point! 
the  centre. 

Circtimfer 

lethat  bounds! 

\l)egree  is  onel 

equal  parts  into! 
Ircumferenco  of  a| 
lupposed    to 


421.  The  degree  is  the  Stumlard  Unit  of  circulur 
measure. 

1.  7)e»iomifiaf{on«.— Seconds  (")«  Minutes 
(0,  Degrees  ("),  Si^^ns  (S.),  Circle  (Cir.). 

8.  One-'fuUf  of  a  circun\ference,  or  180°,  as 
shown  by  the  figure  in  the  margin,  i^  called  a 
8emi-cireuTf\ference  ;  Oaefourth,  or  90",  i\  (Quad- 
rant (  One-dxth,  or  60",  a  Sextant ;  and  One- 
twe(fth,  or  ^'\  a  Sign. 

3.  The  length  of  a  degree  varies  with  the  size 
of  the  circle,  as  will  be  seen  by  examining  the  foregoing  diagram. 

4.  A  degree  of  latitude  or  a  degree  of  longitude  on  the  Equator  is  69.16 
statute  miles.  A  minute  on  the  earth's  circumference  is  a  geographical  or 
nautical  mile. 

SPECIAL    UNITS. 


TABLE 

OF 

UNITS. 

60" 

:= 

r 

60' 

= 

1° 

30^ 

:^ 

IS. 

12  S. 

=: 

ICir. 

860° 

=: 

ICir. 

Table  for  Paper, 

24  Sheets   =1  Quire. 
20  Quires   =1  Ream. 

2  Reams  =1  Bundle. 

5  Bundles =1  Bale. 


Table  for  Counting, 

12  Things=l  Dozen  (doz.). 

12  Dozen  =1  Gross  (j?ro.). 

12  Gross   =1  Great  Gross  (G.  Gro.). 

20  Thing8=l  Score  (Sc.). 


EXAMPLES    FOR    PRACTICE. 

422.  Reduce  and  explain  the  following  : 

1.  3  s.  17°  9'  to  seconds.    4.  7°  4'  to  a  fraction  of  a  sign. 

2.  1  cir.  5  s.  to  minutes.     5.  9°  12'  to  a  decimal  of  a  circle. 

3.  5°  27'  43"  to  seconds.    6.  .83  of  a  cir.  to  a  compound  number. 

7.  What  part  of  a  circumference  are  60°  ?  90°  ?  180°  ? 

8.  How  many  degrees,  minutes,  etc.,  in  f  of  a  quadrant  ? 

9.  How  many  sextants  in  120°  ?     In  150°  ?     In  165°  ?    In 
248°?    In  295°? 

10.  In  5  cir.  7  s.  17°,  how  many  sextants  and  what  left? 

11.  Reduce  |  of  a  quadrant  to  a  compound  number. 

12.  America  was  discovered  Oct.  14, 1492.    What  interval  of 

I  time  between  the  discovery  and  July  4, 1876  ? 

13 


i 


i 


I 


i 


r 


186 


DENOMINATE     NUMBERS, 


13.  How  many  dozen  in  7 J  proBB?    In  18 J  jpro.  ? 

14.  How  many  dozen  in  8 J  great  gross  ?    In  15|  ? 

15.  How  many  dozen  in  17|  scores  ?    In  196t^  ?    'o  8^  ^ 

16.  Reduce  ]3  bundles  1  rciira  15  quires  of  })aper  to  sheets. 

17.  18a  sheets  are  what  decimal  of  1  bundle  ?    Of  17  quires  7 


i,  V 


Sf     I 


UNITS    OF   MONET. 

CANADIAN    MONET. 

423.  The  legal  currency  of  the  Dominion  is  composed  of 
dollars,  cents,  and  mills.  The  dollar  is  the  Standard  Unit, 
The  silver  coins  are  the  fifty-cent  piece,  the  twenty-five  cent 
piece,  the  ten-cent  piece,  and  the  five-cent  piece. 

The  following  table  includes  Canadian  and  United  States 
money. 

1.  I>«nom{nae{on«.— Mills  (m.),  Cents  (ct.), 
Dimes  (d.),  Dollars  ($),  Eagles  (B.). 

2.  The  United  States  coin,  as  fixed  by  the 
''New  Coinage  Act"  of  1878,  is  as  follows: 
Gold,  the  double-eagle,  eagle,  half-eagle,  quar- 
ter-eagle, three-dollar,  and  one-dollar ;  Silver 1 1 
the  trade-dollar,  half-dollar,  quarter-dollar,  and  | 

ten-cent ;  NicJcel,  the  five-cent  and  three-cent ;  Bronze^  one-cent. 

3.  Cotnposition  of  Coina.—Gfold  coins  of  Britain  consist  of  22  parts  I 
2nire  gold  and  2  parts  of  copper.    Silver  coins  consist  of  87  parts  pure  silver 
and  8  parts  of  copper.    Gold  coin  of  the  United  States  contains  .9  pure  gold 
and  .1  silver  and  copper.   Silver  coin  contains  .9  pure  silver  and  .1  pure  cop- 
per.   Nickel  coin  contains  .25  nickel  and  .75  copper.    Bronze  coin  contains  | 
.95  copper  and  .06  zinc  and  tin. 

4.  The  7Va«fe-dWtor  weighs  420  grains  and  Is  designed  for  commercial  j 
purposes  solely. 

Observe,  the  Canadian  bronze  cent  is  one  inch  in  diameter, 
and  one  hundred  cents  weigh  one  pound  Avoirdupois.  The 
mill  is  not  coined  but  is  used  in  computation.  Copper  coinage 
is  not  a  legal  tender  for  any  sum  above  30  cents,  nor  silverj 
coinage  for  more  than  $10. 


TABLE  OP  UKITS. 

10  m. 

^^ 

let. 

10  Ct. 

=z 

Id. 

10  d. 

=: 

$1. 

$10 

= 

IE. 

OEUMAN    MONEY, 


187 


ENGLISH    MONEY. 

424.  "Wxeponnd  sterling  is  the  Standanl  Unit  of  English 
luney.    It  is  equal  to  $4.8660  Canadian  money. 


TABUS  OF  UNITB. 


4  far. 
12  d. 

20  e. 

2  8. 

5  s. 


=     Id. 
=     Is. 

1  Sov. 

or  £1. 
=     Ifl. 
=     1  cr. 


=  1 


1.  Denomiiuitiotui.  —  FartliinK*)  (f^r.), 
PcunleB  (d.),  ShiUiugH  («.),  Soverei^'u  (fov.), 
Pound  (£),  Florin  (fl.),  Crown  (cr.). 

2.  The  Coins  in  general  uhc  in  Great  Brltnln 
are  as  followt> :  Gold,  sovereign  uml  liult- 
Bovert'ign  ;  Silver,  crown,  half-crown,  florin, 
BhilllnK,  Blx-penny,  and  three-penny;  Cop- 
per, penny,  half -penny,  and  farthing. 


United  States 


;d  for  commercial 


PEBNCH    MONEY. 

425.  The  silver  franc  is  the  Standard  Unit  of  French 
loney.    It  is  equal  to  $.193  Canadian  money. 


TABLK  OP  UOTTS. 

10  ra.  r=  1  ct. 
10  ct.  =  1  dc. 
10  dc.  =  1  fr. 


1.  DenonUuatt4yHa.  —  Millimes  (m.),  Cen- 
times (ct.),  Decimes  (dc),  France  (fr.). 

2.  Equivalents ,  — \  tr.  =  10  dc.  =  100  ct.  = 
1000  m. 

8.  The  Coin  of  France  is  as  followH :  Gold, 
100,  40,  20,  10,  and  5  francs ;  /Silver,  5,  2,  and 


AW,    ■«/,    «V,    iU,    i»UU    u    uuuvn,     ifttvi-if    c,    •,    at 

franc,  and  50  and  25  centimes ;  Jtronxe,  10,  5,  2,  and  1  centime  pieces. 


GERMAN    MONEY. 

1 426.  The  mark  is  the  Standard  Unit  of  the  New  Oer- 

in  Empire.    It  is  equal  to  23.85  cents  Canadian  money,  and 

divided  into  100  equal  parts,  one  of  which  is  called  a 

\fennig» 

11.  The  C<A,m  of  the  New  Empire  are  as  follows:  Gold,  20,  10,  and 
larks ;  Silver^  2  and  1  mark ;  Nickel,  10  and  5  pfennig. 

1 2.  The  coins  most  frequently  referred  to  in  the  United  States  are  the 
rer  Thaler,  et^jUdX  74.6  cents,  and  the  silver  Oroschen,  equal  2\  cents. 


{.13 


II 


f 


m 


m 


"  ^ 


'  I 


iiii 


188 


DENOMINATE    NUMBERS, 


EXAMPLES    FOR    PRACTICE. 


W-i.:  ■' 


l-i 


4127.  Reduce  and  explain  the  following: 

1.  £3  17s.  to  farthings.  4  $34  to  mills. 

2.  83745  mills  to  dollars.  5.  .7d.  to  a  decimal  of  a  £.| 

3.  5s-  to  a  decimal  of  a  £.  6.  .9s.  to  a  decimal  of  £3. 

7.  8  of  a  £  to  a  compound  number. 

8.  £.84  to  a  compound  number. 

9.  How  many  pounds  sterling  in  $8340  Canadian  money  ? 

10.  In  2368  francs  how  many  dollars  ? 

11.  Remitted  to  England  $436  gold  to  pay  a  debt.     Howl 
much  is  the  debt  in  English  money  ? 

12.  Received  from  Germany  43864  marks.    How  much  isi 
the  amount  in  Canadian  money  ? 

13.  £340  17s.  is  how  much  in  Canadian  money  ?    In  Germanl 
money?    In  French  money? 

14.  Reduce  7  marks  to  a  decimal  of  $4. 

15.  Reduce  12  francs  to  a  decimal  of  $5. 
10.  Exchanged  {i;125  for  French  money.    How  much  FrencLl 

money  did  1  receive? 


THE    METEIO    SYSTEM. 


Decimal  Related  Units. 

4:28.    The  Metric  System  of  Belated  Units  is  formed  according  toj 
the  decimal  scale. 

420.    The  MetrCi  which  is  89.37079  inches  long,  or  nearly  one 
millionth  of  the  distance  on  the  earth's  surface  from  the  equator  to  the  pole| 
is  the  base  of  the  system. 

430.  The  Vrimary  or  Principal  Units  of  the  system  are  th^ 
Metre,  the  Are  (air),  the  Stere  (stair),  the  Litre  (leeter),  and  the  Gramme^ 
All  other  units  are  multiples  and  sub-multiples  of  these. 

431.  The  names  of  Multiple  Units  or  higher  denominations  ar 
formed  by  prefixing:  to  the  names  of  the  primary  tmits  the  Greek  numera 
Deka  (10),  Hecto  (100),  Kilo  (1000),  and  Myria  (10000). 


DECIMAL     RELATED     UNITS. 


189 


4315,    The  names  of  Suh'tnultiple  Tin  ita^  or  lower  denominations, 
[are  formed  by  prefixing  to  the  names  of  the  primary  units  the  Latin 
lumerals,  Deci  (i\,),  Cenli  dio),  and  MiUi  di/oo). 


ladian  money  ? 
y  a  debt.     Howl 


How  much  isl 


ley?    In  Germanl 


UNITS    OF    LENGTH. 

4*J3.    The  Metre  is  tht  principal  tmit  oi  length. 


TABLK  OP  UNITS. 

mm.  =  1  Centimetre  = 

cm.  =  1  Decimetre  = 

dm.  =  1  Metre  = 

M.  =1  Decametre  = 

Dm.  =  1  Hectometre  = 

Hm.  =  1  Kilometre  = 

Km.  =  1  Myriametre  (Mm.)  — 

The  metre  is  used  in  place  of  one  yard  in  measuring  cloth  and  tshort  dis- 
lances.    Long  distances  are  usually  measured  by  the  kilometre. 


10  Millimetrep, 
10  Centimetres, 
10  Decimetres, 
10  Metres, 
10  Decametres, 
10  Hectometres, 
10  KilometrcH, 


.3937079  in. 
3.937079  in. 
89.37079  in. 
32.808992  It. 
19.927817  rd. 

.62l;i824mi. 
'6.213824  mi. 


)w  mucli  FrencLI 


formed  according  toj 


k  or  nearly  one 
[equator  to  the  polel 


(the  system  are  th^ 
]),  and  the  Gramme, 


UNITS    OP    SURFACE. 

434,    The  Square  Metre  is  the  principal  unit  of  surfaces. 

TABLB  OF  UNITS. 

100  Sq.  Millimetres,    sq.  mm.    =     1  Sq.  Centimetre  =  .155+  sq.  in. 

100  Sq.  Centimetres,  sq.  cm.     —    "  Sq.  Decimetre  =  1.5.5+  eq.  in. 

100  Sq.  Decimetres,    eq.  dm.     =     l  Sq,  Metre  (Sq.  M.)  =  1.196+  tq.  yd. 

43*5.    The  Are,  a  square  whose  side  is  10  metres,  is  the  piincipal 
mit  for  measuring  land. 

TABLE  OF  UinTS. 

100  Centiares,    ca.    =    1  Are  =    119.6034  sq.  yd. 

100  Ares,  A.     =    1  Hectare  (Ha.)  =       2.471M  acres. 


UNITS    OF    VOLUME. 

436.    The  Cubic  Metre  is  the  principal  unit  for  measuring  ordinary 
^olids,  as  embankments,  etc. 

TABLE  OF  UNITS. 

1000  Cu.  Millimetres,  en.  mm.  =  1  Cu.  Centimetre  =  .061  cu.  in. 
1000  Cu.  Centimetres,  en.  cm.  =  1  Cu.  Decimetre  =  61.026  cu.  in. 
1000  Cu.  Decimetres,      cu.  dm.     =    1  Cu.  Metre        =    35.316  cu.  ft. 


V 


'h 


i  h 


s^ 


190 


DENOMINATE    NUMBERS, 


m 


m 


437*  The  Stere,  or  Ct/Mc  if(;/r£,  is  the  principal  unit  for  measarl 
wood. 


10  Decisteree,  dst. 
10  Stsbxs,       St. 


TABLB  OF  UNITS. 

=  1  Stere 

=  1  Decastere,  Dst. 


35.316+  cu.  ft. 
13.079+  cu.  yd. 


[V>;'( 


hi 

u 


'4 


:( 


b 


:4: 
■r 


^\  it'^  I 


UNITS    OP    CAPACITY. 

438.  The  Litre  is  the  principal  unit  both   of  Liqnid  and 
Measnre.    tt  is  equal  to  a  vessel  whose  volume  is  equal  to  a  cube  Whc 
edge  is  one-tenth  of  a  metre. 


TABLE  OF  UNITS. 

10  Millilitres, 

ml. 

=  1  Centilitre  =      .6102  cu.  in.  :=:       .338  fl.  oz 

10  Centilitres, 

el. 

=  1  Decilitre    =    6.1022  "     "    =       .845  gill. 

10  Decilitres, 

dl. 

=  1  Zitre         =      .908  qt.         =      1.0567  qt. 

10  LlTBBS, 

L. 

=  1  Dekalitre    =    9.06     "          =     S.6417  gal. 

10  Dekalitres, 

Dl. 

=  1  Hectolitre  =    2.8372+  bu.     =    26.417     " 

10  Hectolitres, 

HI. 

=  1  Kilolitre     =  28.372+      "      =  2ft4.17      " 

10  Kilolitres, 

Kl. 

=  1  Myrialitre  =283.75+       "      =2&41.7        " 

The  Hectolitre  is  used  in  measuring  large  quantities  in  both  liquid  aij 
dry  measure. 


UNITS    OF    WEIGHT. 

430.  The  Oranirne  is  the  principal  unit  of  weight,  and  is  equal  | 
the  weight  of  a  cube  of  distilled  water  whose  edge  is  one  centimetre. 


TABLE  OF  UHITS. 

10  Milligrammes, 

mg. 

= 

1  Centigramme              =       .15432  + 

oz. 

Tro 

10  Centigrammes, 

eg. 

=3 

1  Decigramme               =     1.54S34+ 

ii 

bh 

10  Decigrammes, 

dg. 

= 

1  Granune                   =    15.43248  + 

(i 

(I 

10  Grammes, 

G. 

= 

1  Decagramme               =       .8527  + 

oz. 

Avo 

10  Decagrammes, 

Dg. 

=S 

1  Hectogramme              =      3..527.S9+ 

l( 

u 

10  Hectogrammes, 

Hg. 

= 

1  Kilogramme  or  Kilo.   =      2.20462  + 

lb. 

10  Kilogrammes, 

Kg. 

= 

1  Myriagramme              =    28.04681  + 

»« 

10  Myriagrammes, 

Mg. 

= 

1  Quintal                        =  820.46219+ 

11 

10  Quintals, 

= 

1  Touneau  or  Ton.         =2204.6212  + 

Ik 

l:l'i  i^' 


The  Kilogramme  or  Kilo.,  which  Is  little  more  than  8)  lb.  Ayoir.J 
the  conwKni  weight  in  trade.  Heavy  articles  are  weighed  by  the  To 
neaUf  which  is  804  lb.  tnore  than  a  common  ton. 


#!    Ii 


ER8, 


EXAMPLES. 


191 


wl  unit  formeasari 


Comparative  Table  of  Units, 


1  Inch         =    .0254  metre. 

1  Ca.  Foot     =    .2832  Hectolitre. 

IFoot         =    .3(M3       " 

1  Cu.  Yard      --    .7646  Steres. 

6.316+  cu.  ft. 
3.079+  cu.  yd. 

lYai-d        =    .9144 

1  Cord             =  8.625  Steres. 

IMIle         =  1.0()i)3  Kilometres. 

1  Fl.  Ounce     =    .()29-i8  Litre. 

1  Sq.  Inch  =    .0006452  sq.  metre. 

1  Gallon          =  8.T8<i  Litres. 

1  Sq.  Foot  =     .0929 

1  BuKhel         =    .35--J4  Hectolitre. 

TY. 

1  Sq.  ^-rd  =    .8361       " 

1  Troy  Grain  =    .0648  Gramme. 

1  Acre         =40.47  Arcs. 

1  Troy  Lb.      -     .373  Kilogramme. 

I   of  Liquid  and  I 

1  Sq.  Mile    =    .259  Flectares. 

t  Avoir  Lb.     =    .4536  Kilogramme. 

squal  to  a  cube  whc 

lCu.Inch   =    .01639  Litre. 

1  Ton             =    .9071  Tonneau. 

.  in.  ^       .*W  fl.  oz. 

EXAMPLES    FOR    PRACTICE. 

"    =       .845  giU. 
=      1.0667  qt. 

440.  Reduce 

s      S.&417  gal. 
bu.    =    36.417     " 
"      -  264  17       *' 

1.  84  lb.  Avoir,  to  kilo.         7.  40975  litres  to  cu.  in. 

3.  37  T.  to  tonneau.              8.  31.7718  sq.  metres  to  sq.  yd. 

"      =2641.7        " 

3.  96  bu.  to  hectolitres.        9.  27Si.592  litres  to  bushels. 

4.  75  fl.  oz.  to  litres.           10.  35.808  kilogrammes  to  Troy  gr. 

ties  in  both  liquid  ai 

5.  89  cu.  yd.  to  steres.        11.  133.75  steres  to  cords. 

6.  328  acres  to  ares.            13.  33.307  steres  to  cu.  ft. 

■'Hi 


peight,  and  is  equal  | 
one  centimetre. 


.15432+  oz.  Tro]| 
1.54«W+  " 
15.43248+  "       "' 

.3527  +  oz.  Av( 
3.1527.39+  '* 
2.30462+  lb. 

a.04esi+  '• 

S:M.46919+  " 
204.6212  +  " 

than  9)  lb.  Ayoir., 
Ireigbed  by  the  Iv 


13.  If  the  price  per  gramme  is  $.38,  what  is  it  per  grain  ? 

14.  If  the  price  per  litre  is  $1.50,  what  is  it  per  quart  ? 

15.  At  26.33  cents  per  hectolitre,  what  will  be  the  cost  of  157 
bushels  of  peas  ? 

16.  When  sugar  is  selling  at  2. 168  cents  per  kilogramme, 
what  will  be  the  cost  of  138  lb.  at  the  same  rate  ? 

17.  Reduce  834  grammes  to  decigrammes  :  to  decagrammes. 

18.  In  84  hectolitres  how  many  litres?  how  many  centilitres? 

19.  A  man  travels  at  the  rate  of  28.279  kilometres  a  day. 
How  many  miles  at  the  same  rate  will  he  travel  in  45  days  ? 

20.  If  hay  is  sold  at  f  18. 142  per  ton,  what  is  the  cost  of 
48  tonneau  at  the  same  rate  ? 

21.  When  a  kilogramme  of  coffee  costs  $1.1023,  what  is  the 
cost  of  148  lb.  at  the  same  rate  ? 


I 


-i  f'tf 


Mii 


192 


DENOMINATE    NUMBERS. 


m\ 


IS^ 


DUODECIMALS. 

441.  Duotlecinials  are  equal  parts  of  a  linear,  square 
or  cubic  font,  formed  by  Buccessively  dividing  by  12.  Hence 
the  following: 


TABLE  OF  UNITS. 

/Ills 


12  Thirds  ('")  =  1  Second 
12  Seconds       =  1  Prime 
12  Primes        =  1  Foot  . 


1" 
1' 
ft. 


1.  Observe  that  each  de- 
nomination in  duodecimals 
may  denote  lengthy  surf ace^or 
volume.  Hence  the  bigheet 
denomination  used  must  be 

marked  so  as  to  indicate  whether  the  number  represents  linear ^  surface^  or 

cubic  measure. 

Thus,  if  the  feet  are  marked  ft.,  the  lower  denominations  denote  2«n^/A  / 
if  marked  sq.  ft.,  surface;  if  marked  en.  ft.,  volume. 

2.  Each  of  the  following  definitions  should  be  cartfuUy  studied  by  draw- 
ing a  diagram  representing  the  unit  defined.  The  diagram  can  be  made 
on  the  blackboard  on  an  enlarged  scale. 


I 


1 


4 


!^ ; , 


-  if' 


mi 


442.  A  TAnear  Prime  is  one-twelfth  of  a  foot;  a  TAnear 
Second^  one-twelfth  of  a  linear  prime ;  and  a  Linear  Thirdt  one-twelfth 
of  a  linear  second. 

443.  A  Surface  Prime  is  one-twelfth  of  a  square  foot,  and  Is 
12  inches  long  and  1  inch  wide,  and  is  equal  to  12  square  inches. 

444.  A  Surface  Second  is  one-twe^h  of  a  surface  prime,  and  is 
1  foot  long  and  1  linear  second  wide,  which  is  equal  to  1  square  inch. 
Hence  square  inches  are  regarded  as  surface  seconds. 

445.  A  Surface  Tliird  is  one-twe^h  of  a  surface  second,  and  is 
1  foot  long  and  1  linear  third  wide,  which  is  equal  to  12  square  seconds. 
Hence  square  seconds  are  regarded  as  surface  fourths. 

440.  A  Cubic  Prime  is  one-twelfth  of  a  cubic  foot,  and  is  1  foot 
square  by  1  inch  thick,  and  is  equal  to  a  board  foot. 

447.  A  Cubic  Second  is  one-twetfth  of  a  cubic  prime,  and  is  1  foot 
long  by  1  inch  square,  and  is  equal  to  13  cuHc  inches  or  a  board  inch. 

448.  A  Cubic  Third  is  one-tioelfth  of  a  cubic  second,  and  is  1  foot 
long,  1  inch  wide,  and  1  linear  second  thick,  and  is  equal  to  a  cable 
inch. 


ST. 


LONGITUDE    AND     TIME. 


193 


linea/r,  square 
by  13.    Hence 


've  that  each  de- 
in  duodecimcUs 
length,  mrfaceyOT 
[ence  the  highest 
on  used  must  be 
linear,  surface,  or 

onBdeT^ote  length  ! 

y  studied  by  draw- 
ram  can  be  made 


foot;    a    TAnear 
'hirdt  one-tweyth 


EXERCISE    FOR    PRACTICE. 

4:4<l.    Ulustrate  the  foUowiog  by  diagrams  on  the  blackboard: 

1.  5  feet  multiplied  by  7  in.  equals  35  surf  ace  primes. 

2.  8  ft.  multiplied  by  4"  equals  38  surface  seconds. 

3.  7  feet  multiplied  by  6'"  equals  42  surface  thirds. 

4.  3  in.  multiplied  by  5  in.  equals  15  surface  seconds. 

5.  4'  multiplied  by  3"  equals  12  surface  thirds. 

6.  From  these  examples  deduce  a  rule  for  multiplying  feet,  inches, 
seconds,  etc.,  by  feet,  inches,  seconds,  etc. 

Multiply  and  explain  the  following  : 

7.  17  ft.  5^  8"  by  8  ft.  y  7". 

8.  32  ft.  9^  4"  by  6  ft.  S'  11". 

9.  15  ft.  6'  10"  by  9  ft.  4'  8^'. 

13.  19  ft.  8'  7"  by  2  ft.  y  ^  by  3  ft.  2'  4". 

14.  48  ft.  y  by  1  ft.  7'  9''  by  2  ft.  8'  5". 

Duodecimals  are  added  and  subtracted  In  the  same  manner  as  other  com- 
pound numbers.  Division  being  of  little  practical  utility,  is  omitted.  The 
pupil  may,  if  desired,  deduce  a  rule  for  division  as  was  done  for  multipli- 
cation. 

IiONGITUDE    AND    TIME. 


10.  25  ft.  y  3"  by  14  ft.  7'  2". 

11.  18  ft.  7'  y  by  12  ft.  8'  5". 

12.  34  ft.  8'  by  26  ft.  4'  9". 


lare  foot,  and  is 
inches. 

|face  prime,  and  is 
to  1  square  inch. 


face  second,  and  is 
12  square  seconds. 


foot,  and  is  1  foot 


ime,  and  is  1  foot 
board  inch. 

lond,  and  is  1  foot 
1  equal  to  a  cubic 


450.  Since  the  earth  turns  on  its  axis  once  in  24  hours,  ^ 
of  360',  or  15"  of  longitude,  must  pass  under  the  s  m  in  1  hour, 
and  j/(j  of  15°,  or  15',  must  pass  under  it  in  1  minute  of  time, 
and  ^Q  of  15',  or  15'',  must  pass  under  it  in  1  second  of  time. 
Heuce  the  following 

TABLE  OF  KQUIVALENTS. 

A  difference  of  15''  in  Long,  produces  a  diff.  of  1  hr.  in  time. 
"     ly         "  "  ''        1  min.       " 

»•  "     15"        '•  "  "        1  sec.        " 

Hence  the  following  rule  to  find  the  difference  of  time  be- 
tween two  places,  when  their  difference  of  longitude  is  given  : 

RULE. 

451.  Divide  tTic  difference  of  longitude  of  the  two  places  by 
15,  and  mark  the  quotient  hours,  minutes,  and  6econds,  instead 
of  degrees,  minutes,  and  seconds. 


mi 
m 


m 

J  •-;V«~« 

m 

i                              1 

t    ' 

1 

'    Hi    j 

•Ml 

:f 

'1   .      1 

SI;..' 


"I'H^ 


(-1- 


p 


194 


DENOMINATE    NUMBERS. 


To  find  the  difTerence  of  longitude  when  the  difference  of 
time  is  given. 

BUIiE. 

452.  Multiply  the  difference  of  time  betieeen  the  two  p(.aee% 

by  15 y  and  mark  the  product  degrees ^  minutes ^  and  seconds ^ 

instead  of  hours,  minuteSt  and  secondc. 

Since  the  earth  revolves  trom  west  to  eaut,  time  is  earlier  to  places  west 
and  later  to  places  cast  of  any  given  meriuiao. 

EXAMPLES    FOR    PRACTICE. 

453.  Find  the  difference  in  time  between  the  following : 

1.  Albany  West  Long.  73°  44'  50"  and  Boston  W.  Long. 
71°  3'  30". 

When  the  given  places  are  on  the  same  side  of  the  first  meridian,  the 
difference  of  longitude  is  found  by  subtracting  the  lesser  from  the  greater 
longitude. 

2.  Bombay  East  Long.  73°  54'  and  Berlin  East  Long. 
13°  23'  45". 

3.  New  York  W.  Long.  74°  3'  and  Chicago  W.  Long. 
87°  37'  4'. 

4.  San  Francisco  W.  Long.  122°  and  St.  Louis  W.  Long. 
90°  15'  15". 

5.  Calcutta  E.  Long.  88°  19'  2"  and  Philadelphia  W.  Long. 
75°  9'  54". 

Observe,  that  when  the  given  places  are  on  opposite  sides  of  the  first\ 
meridian,  the  difference  in  longitude  Is  found  by  adding  the  longitudes. 

6.  Constantinople  E.  Long.  28°  59'  and  Boston  W.  Long.] 
71°  8'  30". 

7.  Tlie  difference  in  the  time  of  St.  Petersburg  and  Wash- 
ington is  7  hr.  9  min.  19}  sec.  "What  is  the  difference  in  the 
longitude  of  the  two  places  ? 

8.  When  it  is  12  o'clock  M.  at  Montreal,  what  time  is  it  atj 
a  place  50°  24'  west  ? 


an  VIEW. 


195 


le  difference  of 


9.  In  sailing  from  New  Orleans  to  Albany,  the  chronometer 
lost  1  hr.  5  min.  lOf  sec.  The  longitude  of  Albany  is  73°  44' 
50".    What  ia  the  longitude  of  New  Orleans  ? 

10.  An  eclipse  is  observed  by  two  persons  at  different  points, 
the  one  seeing  it  at  8  hr.  30  min.  p.  m.,  the  other  at  11  hr. 
45  min.  p.  m.    What  is  the  difference  in  their  longitude  ? 


irller  to  places  west 


jrlin  East  Long, 
licago  W.  Long. 


Louis  W.  Long. 


delphia  W.  Long. 


^hat  time  is  it  at 


REVIEW    AND    TEST    QUESTIONS. 

454.  1.  Define  Related  Unit,  Denominate  Number,  De- 
nominate Fraction,  Denomination,  and  Compound  Number. 

2.  Repeat  Troy  Weight  and  Avoirdupois  Weight. 

3.  Reduce  9  bu.  3  pk.  5  qt.  to  quarts,  and  give  a  reason  for 
each  step  in  the  process. 

4.  In  9  rd.  5  yd.  2  ft.  how  many  inchjss,  and  why? 

5.  Repeat  Square  Measure  and  Surveyors'  Linear  Meaanre. 

6.  Reduce  23456  sq.  in.  to  a  compound  number,  and  give  a 
reason  for  each  step  in  the  process. 

7.  Define  a  cube,  a  rectangular  volume,  and  a  cord  foot. 

8.  Show  by  a  diagram  that  the  contents  of  a  rectangle  is 
found  by  multiplying  together  its  two  dimensions. 

9.  Define  a  Board  Foot,  a  Board  Inch ;  and  show  by  diagrams 
that  there  are  12  board  feet  in  1  cubic  foot  and  12  board  inches 
in  1  board  foot. 

10.  Reduce  f  of  an  inch  to  a  decimal  of  a  foot,  and  give  a 
reason  for  each  step  in  the  process. 

11.  How  can  a  pound  Troy  and  a  pound  Avoirdupois  be 
compared  ? 

12.  Reduce  .84  of  an  oz.  Troy  to  a  decimal  of  an  ounce 
Avoirdupois,  and  give  a  reason  for  each  step  in  the  process. 

13.  Explain  how  a  compound  number  is  reduced  to  a  fraction 
or  decimal  of  a  higher  denomination.  Illustrate  the  abbre- 
viated method,  and  give  a  reason  for  each  step  in  the  process. 


i 


i  1  ii 


!:l  i 


if   I 


BUSINESS    ARITHMETIC. 


SHORT    METHODS. 


455.  Praxjtlcal  devices  for  reaching  results  rapidly  are  of 
first  importance  in  all  business  calculations.  Hence  the  fol- 
lowing summary  of  short  methods  should  be  thoroughly 
mastered  and  applied  in  all  future  work.  The  exercises  under 
each  problem  are  designed  simply  to  illustrate  the  application 
of  the  contraction. 

When  the  directions  given  to  perform  the  work  are  not 
clearly  understood,  the  references  to  former  explanations 
should  be  carefully  examined. 


■-a 


«* 


450.    Pkob.  I— To  multiply  by  lo,  lOO,  xooo,  etc. 

Move  the  decimal  point  in  the  multiplicand  as  many  places  to 
the  right  as  there  are  cipJiers  in  the  multiplier,  annexing  ciphers 
wlien  necessary  (82). 


Multiply  the  following : 


1.  84  X  100. 

2.  70  X  1000. 
8.  5.73x100. 


4  8.8097x10000. 

5.  .89753  X  1000. 

6.  3.0084x10000. 


7.  8436x1000. 

8.  7300x100000. 

9.  463x1000000. 


457.  Prob.  II. — To  multiply  where  there  are  ciphers 
at  the  right  of  the  multiplier. 

Move  the  decimal  point  in  th£  multiplicand  as  many  places  to 
the  right  as  there  are  ciphers  at  the  right  of  the  multiplier,  annex- 
ing ciphers  when  necessary,  and  multiply  the  result  by  the  signifi- 
cant figures  in  tlie  multiplier  (84). 


SHORT    METHODS. 


197 


[C. 


lence  the  fol- 
^  tborouglily 
Kcrcises  under 
be  application 

work  are  not 
r   explanations 


000,  etc. 

lany  places  to 
inexing  cipJiers 


3426  X 1000. 
7200  X 100000. 
463x1000000. 

re  are  ciphers 


[many  places  to 
iltiplier,  annex- 
\lt  by  the  signifi- 


Multiply  the  following : 

1.376x800.  4  836.9x2000. 

2.  42.9x420.      5.  7.648x3200. 

3.  500  X  700.      C.  2300  x  5000. 


7.  8800x7200. 

8.  460x900. 

9.  .8725x3600. 


458.    Prob.  III.— To  multiply  by  9,  99,  999,  etc. 

Move  the  decimal  paint  in  tlw  multiplicand  as  many  places  to 
the  HghZ  as  there  are  nines  in  the  7aultiplier,  annexing  ciphers 
when  necessary^  and  subtract  the  given  multiplicand  from  tfte 
result. 

Obperve  that  by  moving  the  decimal  point  as  directed,  wc  multiply  by  a 
number  1  greater  than  the  ^iven  multiplier ;  hence  the  multiplicand  i8  Bub* 
traded  from  the  result.  To  multiply  8,  98, 998,  and  bo  on,  we  move  the 
decimal  point  in  the  same  manner,  and  subtract  ttom  the  result  twice  the 
multiplicand. 


Perform  the  following  multiplication  : 

1.  736458  X  9.  4.  53648  x  990. 

2.  3895x99.  5.  83960x9990. 

3.  87634  X  999.  6.  26384  x  98. 


7.  7364x998. 

8.  6283x9990. 

9.  4397x998. 


459.  Prob.  IV.— To  divide  by  10,  100,  1000,  etc. 

Move  the  decimal  point  in  tTie  dividend  as  many  places  to  the 
left  as  there  are  ciphers  in  the  divisor y  prefixing  ciphers  when 
necessary. 

Perform  the  division  in  the  following  : 

1.  8736^100.  4.  23.97-f-lOOO.  7.  .54-^100. 

2.  437.2-HlO.  5.  5.236-^100.  8.  .07-f-lOOO. 

3.  790.3-1-100.  6.  .6934-s-lOOO.  9.  7.2-t-IOOO. 

460.  Prob.  V. — ^To  divide  v^rhere  there  are  ciphers  at 
the  right  of  the  divisor. 

Move  the  decimal  point  in  the  dividend  as  many  places  to  the 
left  as  there  are  ciphers  at  the  right  of  the  divisoi\  prefixing 
ciphers  when  necessary  (129),  and  divide  the  result  by  tJie  sig- 
nificant  figures  in  the  divisor  (131). 


'!1 


:i1 


pi 


Pi' 


ill 


i 


,ii!lif 


1 1 


M'M: 


ijt*!' 


198 


BUSINESB    ARITHMETIC, 


Perform  the  division  in  the  following : 

1.  785a4-4a  4.  5.2-5-400. 

2.  528.7-1-80.  6.  .96-t-120. 
8.  329.5-f-dOOO.  6.  .06-»-200. 


7.  364.2-t-540. 

8.  978.5-J-360. 

9.  8A57+600. 


46L    Pbob.  VI. — To  mnltiply  one  fraction  by  another. 

Caiieel  aU/raetiona  eomnwn  to  a  numerator  and  a  denomina- 
tor before  multiplying  ( 1 74-— II). 

Perform  the  following  multiplications  by  cancelling  common 
factors : 


1.  H^H. 

4     1  60  V  S-6 


9.  fxIfxT?^. 
10.  il§  X  If  X  T^^f. 


U.  if  X  i§«  X  ,V 

12-inSxT»B«jX/,. 

13. 


80 


81 


14.  ^nxHxi 

15.  ifjxxi^x^ 


5.  H  X  ,vk. 

4!<I2.    Pbob.  VII. — To  divide  one  fraction  by  another. 

Cancel  aU  factors  common  to  both  numerators  or  common  to 
both  denominators  before  dividing  (280).    Or, 
Invert  the  divisor  and  cancel  aa  directed  in  Pr^jb.  VI. 


Perform  the  diTision  in  the  folio wixig,  cancelling  as  directed : 

6.  .28-i-.04.  10.  .63-*- .0027. 

7-  /A-»-^V  *!•  .89-«-.008. 


1.  M-*-?- 

2.  .9-T-.03. 


4.  ^1-^^.  8.  If+if.  12.  iVt^-rHu. 

4^3.    Prob.  VIII.  —To  divide  one  number  by  another. 

Cancel  the  factors  that  are  common  to  the  dividend  and  divisor 
before  dividing  (174 — II). 

Perform  the  following  divisions,  cancelling  as  directed  : 

1.  4635 -s-45.  4.  62500->2500.  7.  75000-i-1500. 

2.  3900-fr-180,  5.  89600-J-800.  «.  32000^400. 

3.  8400-4-30a  6.  3420-H5400.  i9.  9999-*-63. 


SHORT    METHODS. 


199 


)-i-360. 
r-4-600. 

f  anotlicr* 

t  denomina- 

ng  comsaou 


XirV 


xt¥j^^- 


-•J-  X 


11  xi. 


X  j|^  X  y. 

r  Another. 

r  common  to 

VI. 

as  directed : 

10 

^..0027. 
008. 

)y  another. 
d  and  divisor 


reciO 


ted: 


)00-h1500. 
)00^400. 


ALIQUOT    FABTS. 

464.  An  Aliquot  Part  of  a  number  is  any  number, 
integral  or  mixed,  which  will  exactly  divide  it. 
Thus,  3,  2|,  ^,  are  aliquot  parts  of  10 

4:05.  The  aliquot  parts  of  any  number  are  found  by  divid- 
ing; by  3,  8,  4,  5,  and  so  on,  up  to  1  less  than  the  given  number. 

Thus,  100 ^2  =  50;  100-!-3  =  83J;  100-+-4=25.  Each  of 
the  quotients  50,  33},  and  25,  is  an  aliquot  part  of  100. 

460.  The  character  @  is  followed  by  the  prioe  of  a  unit  or 
one  article.  Thus,  7  cords  of  wood  @  $4.50  means  7  cords  of 
wood  at  $450  a  cord. 

467.  Commit  to  memory  the  following  aliquot  parts  of 
100,  1000,  and  $1. 


Table  of  Aliquot  Parts, 


no  =  \ 

25    =  \ 
20    =  \ 

14f  =  \ 
12|  =  i 

10    =tV 


of  100. 


500 

333i 

250 

200 

166|  =  jt  >  of  1000. 


142? 
125 
lllj 
100 


=  i 


50  ct. 
33}  ct. 
25  ct. 
20  ct. 
16|ct 
14|  ct. 
12}  ct. 
lljct. 
10  ct. 


=  Oof$L 


408.    Prob.  IX. — To  multiply  by  using  aliquot  parts. 
t  Multiply  459  hy33|. 

3  )  45900  ExHiAjrArnow.— We  multiply  by  100  by  annexing  two 

cipherB  to  the  multiplicand,  or  by  moving  the  decimal 

15300        point  two  placee  to  the  right.    Bat  100  being  eqoal  to 
8  times  the  multiplier  33},  the  product  46B00  is  8  timeg  m 
large  as  the  required  product ;  hence  wc  divide  by  3. 


f 


1 

v'  ;■' 

'"  J 

::•'•■'' 

i 

f^^4 

. 

11  *                                         u 

1 

r               1  1 
i!                 1 

1 

t 

1 1 

(■'  ' 


{>, 


1:^-; 


•'■1 

';i  i 

M. 

!  I 


!!    I!    •■ 


iU 


.ji^^ 


200 


BUSINESS    ARITHMETIC. 


Perform  the  following  multiplications  by  aliquot  parts : 

2.    805x125.  5.     234x333|.  8.     58.9x250- 

6.  809  X  Hi.  9.     7.63x142? 

7.  73x111^.  10.    4.88x81. 


8.    85.8  xl6«. 


4.    974x50. 

Solve  the  following  examples  orally,  by  aliquot  parts. 

11.  What  cost  48  lb.  butter  @  25  ct.  ?    @  50  ct.  ?    ©  33^  ct.  ? 

Solution.— At  $1  a  pound,  48  would  cost  $48.  Hence,  at  38J  cts.  a 
pound,  which  Ib  i  of  |1,  48  pounds  would  coat  I  of  |48,  which  in  $16. 

12.  What  cost  96  lb.  sugar  @  12^  ct.  ?  @  14?  ct.  ?  @  1G|  ct.  ? 
18.  What  is  the  cost  of  24  bushels  wheat  @  $1.33^? 

Solution.— At  $1  a  bai>hel,  34  bushels  cost  $24 ;  at  33^  ct,  which  Is  \  of 
$1  a  bushel,  34  bushels  cost  $8.  Hence,  at  $1.83|  a  bushel,  34  bushels  cost 
the  sum  of  $24  and  $S,  which  is  132. 

14.  What  cost  42  yards  cloth  @  $1.16|?  @  $2.14f  ? 

15.  What  cost  72  cords  of  wood  @  $4.12J  ?  @  $3.25  ? 

Find  the  cost  of  the  following,  using  aliquot  parts  for  the 
cents  in  the  price. 

16.  2940  bu.  oats  @  33  ct. ;  @  50  ct. ;  @  25  ct. 

17.  100  tons  coal  @  $4.25;  @  $5.50;  @  $6.12^  ;  @  $5.33 J. 

18.  280  yd.  cloth  @  $2.14f ;  @  $1.12^  ;  @  $3.25  ;  @  $2.50. 

19.  150  bbl.  apples  @  $420 ;  @  $4.50 ;  @  $4.33J. 

20.  834  bu.  wheat  @  $1.33^ ;  @  $1.50;  @  $1.25  ;  @  $1.16|. 

21.  896  lb.  sugar  @  12^  ;  @  14f  ;  @  16|. 

22.  What  is  the  cost  of  2960  yd.  cloth  at  37^  ct.  a  yard? 

35  =  I  of  100,  hence  4)2960 


12J=iof  25,  hence  2)  740 

370 


374 


¥ 


Explanation.— At  |1  a  yard, 
2960  yd.  will  cost  $2960.  But 
25  ct.  is  i  of  $1,  hence  \  of  ^2960, 
which  is  $740,  is  the  cost  at 
35  ct.  a  yd. 

2.  Again,  13}  ct.  is  the  ^  of 
25  ct.,  hence  $740,  the  cost  at  25  cts.,  divided  by  2,  gives  the  cost  at  12j  ct., 
which  is  $370.  But  25  ct.  +  12i  ct.  =  37J;  hence,  $740  +  $370,  or  $1110,  is 
the  cost  at  37  J  ct. 


$1110 


SHORT    METHODS, 


201 


t  parts : 

6b.9  X  250. 
7.63  X  142?. 
4.38  x8i. 

parts. 

1    (itiSSict.? 

:e,  at  38  i  cts.  a 
ich  is  $16. 

;.?    @lG|ct.? 

.331? 

ct,  which  iB  \  of 
a4  boBhelB  cost 

14?? 
$3.25? 

t  parts  for  tbe 


;  @  $5.33i. 
;  @  |2.50. 

J5 ;  @  $1.16|. 

a  yard? 

loN.-At»layard, 

Icoet  $2960.  But 
I,  hence  ^  of  $2960, 
\   is  the  cost  at 

Yt\  ct.  iB  the  \  of 

le  cost  at  Vi\  ct, 
I  $370,  or$U10,ts 


23.  495  bu.  barley  @  75  ct.  ;  @  03|  ct. ;  @  87J  ct. 

24.  870  lb.  tea  (t^  GO  ct. ;  %  Q%\  ct. ;  @  80  ct. ;  @  87J  ct. 

25.  4384  yd.  cloth  @  12^  ct. ;  @  16  ct. ;  @  30  ct. ;  ^  86  ct. 
Obsurve^  that  10  ct.  =  i>o  of  100  ct,  and  5  ct.  =  )  of  10  ct. 

20.  680  lb.  coffee  @  87^  ct. ;  @  76  ct.  ;  @  60  ct. 

460*    PnoB.  X.— To  divide  by  using  aliquot  parts. 
1.  Divide  7258  by  83J. 


72.58 
8 

217.74 


ExPLAKATiON.— 1.  We  divide  by  100  by  moving  tbe  deci- 
mal point  two  places  to  the  left. 

8.  Since  100  is  3  times  331,  the  given  divisor,  the  quo- 
tient 72.68  iB  only  i  of  the  required  quotient :  huuco  we 
multiply  the  73.58  by  8,  giving  317.74,  the  required  quotient. 


5. 

894.8+125. 

8. 

460.854-250. 

6. 

98.54-4-50. 

9. 

90.638-J-25. 

7. 

879.0+33i. 

10. 

73090 -^333i. 

Perform  by  aliquot  parts  the  division  in  tbe  following : 

2.  8.375-*-16|. 

3.  9764-5-5. 

4.  8730-S-8J. 

Solve  the  following  examples  orally,  using  aliquot  parts : 

11.  At  33 J  ct.,  how  many  yards  of  cloth  can  be  bought 
)r  $4? 

Solution.— Since  $1,  or  100  ct.,  is  8  times  33i  ct.,  we  can  buy  3  yards  for 
.    Hence  for  $4  we  can  buy  4  times  3  yd.,  wtiich  is  12  yd. 

Observe^  that  In  this  solution  we  divide  by  100  and  mnltiply  by  3,  the 
imber  of  time^  83i,  the  given  price,  is  contained  in  lOOL    Thus,  $4=400  ct., 

-<- 100  =  4,  and  4x3=  13.  In  the  solution,  the  redaction  of  the  $4  to 
Ints  is  omitted,  as  we  recognize  at  sight  tliat  100  ct.,  or  $1,  is  contained 
pmes  in  $4. 

13.  How  many  yards  of  cloth  can  be  bought  for  $8  @  12J  ct.  ? 
14?  ct.?  @33ict.?  @16|ct.?  @25ct.  ?  @10ct.?  @o0ct.? 

j  8  ct.  ?  @  5  ct.  ?  @  4  ct.  ? 

|l3.  How  much  sugar  can  be  bought  at  12^  ct.  per  pound  for 
For  $8?    For|12?    For  $30?    For  $120? 

14.  How  many  pounds  of  butter  @  83J  ct.  can  be  bought  for 
'?    For  $10?    For  $40? 

u 


1 

! 

M. 

:^l 

Jl 

^■- 

m 

,.  /' 

^1 

^^H 

'  "  ( 

i 

m 

m 

In 

4 

' 

■  \ 

..  i 

r  f 

i 

i 

! 

i 

202 


BVSTNESS     ARITHMETIC. 


Solve  the  following,  performing  the  division  by  aliquot 
parts  : 

15.  How  many  acres  of  land  can  be  bought  for  $8954  at  ^25 
per  acre  ?    At  $50  ?    At  $33^  ?    At  $125?  At  $16.-;  ?  At  $250  V 

16.  How  many  bushels  of  wheat  can  be  bought  for  S6354  at 
$1.25  per  bushel  ?     At  $2.50  ? 

Obsen-e,  $1.23  =  |  of  $10  and  |2.50  =  V  of  J'lO.  Hence  by  movino:  the 
decimal  point  one  place  to  the  left,  which  will  give  the  number  of  bu.  at 
$10,  and  multiplying  by  8,  will  give  the  number  of  bu,  at  $1.25.  Multiply- 
ing by  4  will  give  the  number  at  $2.50. 

17.  How  many  yards  of  cloth  can  be  bought  for  $2642  at 
m\i  ct.  per  yard?  At  14f  ct.?  At  25  ct.?  At  $3.33.^?  At 
$2.50?    At$l.lH?    At$1.42f? 

18.  What  is  the  cost  of  138  tons  of  hay  at  $12i  ?  At  $14f  ? 
At  $16|  ?    At  $25  ?    At  $13.50  ?    At  $15. 33^  ?    At  $17.25  ? 


BUSIE'ESS    PEOBL^MS. 

DEFINITIONS. 

70.   Quantity  is  the  amount  of  anything  considered  ii)| 
a  business  transaction. 

47 1 .  Pricey  or  Rate,  is  the  value  in  money  allowed  for  p| 
given   unit,  a  given  number  of  units,  or  a  giv€?t  part  of 
quantity. 

Thus,  in  74  bu.  of  wheat  at  |2  per  buehel,  the  oHce  is  the  value  of  a  uritl 
of  the  quantity ;  in  8735  feet  of  boardb  at  45  ct.  p*  r  100  fe'^t,  the  price  is  thej 
value  of  100  units. 

472.  When  the  rate  is  the  value  of  a  given  number  of] 
units,  it  may  be  expressed  as  a  fraction  or  decimal. 

ThuB,  cloth  at  $3  for  4  yards  may  be  expressed  as  $2  per  yard  ;  7  foij 
every  100  in  a  given  number  may  be  expressed  thti  or  .07.    Hence,  f  of  1 
means  5  for  every  8  in  64  or  5  per  S  of  64,  and  .08  means  8 per  100. 


'j^u.J'S'itKm 


BUSINi:SS     PROBLEMS, 


203 


•r  consWered  in 


473.  Cost  is  the  value  in  money  allowed  for  aL  entire 
quantity. 

Thne,  in  5  barrels  of  apples  at  $4  per  barrel,  $4  is  the  price,  and  $4  <5or 
|20,  tlie  entire  value  of  the  5  barrel:?,  in  the  cost. 

474.  /*er  Cent  meana  l^cr  I  hundred. 

Thus,  8  per  ceut  of  $G00  means  $8  out  of  every  $100,  which  is  $48.    Hence 
a  given  per  cent  is  the  price  or  rale  per  100. 

475.  The  Sign  of  Per  Cent  is  %.    Thus,  8%  is  read, 
ii  per  cent. 

since  per  cent  means  per  hundred,  any  given  ptr  cent  may  bo  expressed 
with  the  sign  jt  or  in  the  form  of  a  decimal  or  common  fraction ;  thus, 

1  per  cent  is  written     1%    or     .01    or    ,i_. 

•07     -     T^a. 
1.00     "     IS". 


7  per  cent  " 

<i 

1% 

It 

100  per  cent  " 

ti 

vm 

it 

135  per  cent  " 

it 

135^ 

ti 

1  per  cent  " 

u 

\% 

tt 

1.35 


135 

loo- 

i 


.OOJ  "     -l-  =  .005. 
100 

476.  Percentage  is  a  certain  number  of  hundredths  of  a 
given  quantity. 

477.  Prnfit  and  Loss  are  commercial  terms  used  to 
express  the  gain  or  loss  in  business  transactions. 

478.  The  Pro/it  or  Gain  is  the  ainount  realized  on  busi- 
ness transactions  in  addition  to  the  amount  invested. 

Thus,  a  man  bought  a  farm  for  $8500  and  sold  it  for  $9200.  The  .?8500 
paid  for  the  farm  is  the  amount  invested,  and  tho  $9200  is  the  whole  sum 
realized  on  the  transaction,  which  is  $700  more  than  what  was  invested ; 
hence  the  $700  is  iheprojit  or  gain  on  the  transaction. 

479.  The  Loss  is  the  amount  which  the  whole  sum 
realized  on  business  transactions  is  less  than  the  au,'0'int 
invested. 

Thus,  if  a  horse  Is  bonght  for  $270  and  sold  again  for  f  170,  there  isa  lofl« 
of  1100  on  the  transaction. 

480.  The  Gain  and  the  Loss  are  usually  expressed  as  a  jjer 
cent  of  the  amount  invested. 


i 


'.-■a^TTOT^ 


J'n! 


llllf'j 


204 


B  USIJV^ESS    ARITHMETIC, 


ORAL    EXERCISES. 


■1 


U'l 


liM  niiiM 


481.  Express  the  following  decimally  : 


1. 

ifc. 

5. 

^%' 

9. 

207%. 

13. 

ifc 

2. 

9%. 

6. 

2J%. 

10. 

145i%. 

14. 

H%. 

8. 

18^. 

7. 

112%. 

11. 

512|%. 

15. 

3|%. 

4. 

26%. 

8. 

i%. 

12. 

1^. 

16. 

TIS%* 

17.  What  is  meant  by  8  %  "?    By  135  %  ?    By  |  %  ? 

18.  What  is  the  difference  in  the  meaning  of  5  per  cent  anc 
6 per  seven? 

19.  How  is  3  per  eight  expressed  with  figures  ?    7  per  Jive , 
13  per  twenty  f    9  per  four  ? 

20.  What  does  ^"^  mean,  according  to  (473)?    What  doee 
f  mean,  according  to  the  same  Art.  ? 

21.  What  is  the  difference  in  the  meaning  of  f  %  and 
of  100? 

22.  What  is  the  meaning  of  .00^  ?    Of.07|?    Of.32J? 

23.  Express  .OOf  with  the  sign  %  and  fractionally. 

24.  Write  in  figures  three  per  cent,  and  nine  per  cent. 


Express  the  following  as  a  "per  cent : 

25.  |.  28.     158.  81.    If 

26.  7f  29.     236.  88.     1. 

27.  ^,  80.    307^.  33.    8. 


84.  100. 

85.  700. 

86.  105. 


482.  In  the  following  problems,  some  already  given  ai 
repeated.    This  is  done  first,  for  review,  and  second,  to  givd 
in  a  connected  form  the  general  problems  that  are  of  con\ 
stant  recurrence  in  actual  business.    Each  problem  should 
fixed  firmly  In  the  memory,  and  the  solution  clearly  unde^ 
stood. 

It  will  be  observed  that  Problems  VIII,  IX,  X,  and  XI,  a^ 
the  same  as  are  usually  given  under  the  head  of  Percent 
uge»    They  are  presented  in  a  general  form,  as  the  solution 
the  same  whether  hund/redtlt>8t  or  some  other  fractionai  pai 
are  used. 


BU/SINESS    PROBLEMS, 


205 


pea  ?    7  per  five 


I  already  given  ai 
id  second,  to  gWJ 
that  are  of  con\ 
Lroblem  should 
Bon  clearly  unde^ 

Ix,  X,  and  XI,  a 
lead  of  Percent 
1,  as  the  Bolution 
sr  fractional  pai 


FBOBLEMS. 

483.    Prob.  I.— To  find  the  cost  when  the  number  of 
inits  and  the  price  of  one  unit  are  given. 

1.  What  is  the  cost  of  35  lb.  tea  @  $^  ? 

Solution.— Since  1  lb.  cost  |f ,  35  lb.  vrili  cost  86  times  $f ,  which  is 
(260)  $25. 

Find  the  cost  and  explain  the  following  orally : 

2.  74  bu.  apples  @  $|.  6.  19  boxes  oranges  @  |4|. 

3.  34  yd.  cloth  @  |2|. 

4.  6f  yd.  cloth  @  ||. 

5.  44i  lb.  butter  @  |f . 


5|- 


7.  17  tons  coal  («) 

8.  98  cords  wood  @  $4y^^. 

9.  9|  yd.  cloth  @  f^V- 


Find  the  cost  of  the  following,  and  express  the  answer  in 
iollars  and  cents  and  fractions  of  a  cent : 


10.  52  yd.  cloth  @  |3J. 

11.  18  bblsi  apples  @  |4f. 

12.  84  bu.  oats  @  $f. 

13.  83  lb.  coffee  @  $f 


14.  32\\  lb.  sugar  @  $gV 

15.  63  iV  lb.  butter  Qi  ^t^. 

16.  169  acr.  land  @  |27i. 

17.  25f  cords  wood  @  $6f 


18.  How  much  will  a  man  earn  in  19|  days  at  $2|  per  day  ? 

19.  Sold  Wra.  Henry  36 J  lb.  butter  @  28|  ct.,  17y<^  lb. 
^offee  @  $.33|,  and  39^1  lb.  sugar  @  $.14|.  How  much  was 
lis  bill? 

20.  A  builder  has  17  carpenters  employed  @  $2.25  per  day. 
[ow  much  does  their  wages  amount  to  for  24|  days  1 

484.  Prob.  II. — To  find  the  price  per  unit  when  the 
^ost  and  number  of  units  are  given. 

1.  If  9  yards  cost  $10.80,  what  is  the  price  per  yard  ? 

Solution.— Since  9  yards  cost  110.80, 1  yard  will  cost  I  of  it,  or  |10.80+ 
I  =  $1.20.    Hence,  1  yard  cost  $1.30. 

Solve  and  explain  the  following  orally : 

2.  If  7  lb.  sugar  cost  f  1.08,  what  is  the  price  per  pound  ? 

3.  At  $4.80  for  12  yards  of  cloth,  what  is  the  price  per  yard? 


!^ 


HI 


-^■• 


,^r 


J  Mili! 


;li 


III 


'ill: 


WA 


1,"       ;  1 

;  1 

206 


BUSINESS    ARITHMETIC, 


4.  If  12  lb.  of  butter  cost  $3.84,  how  much  is  it  a  pound  ? 

5.  Paid  $3.42  for  9  lb.  of  coffee.  How  much  did  I  pay  per 
pound  ? 

Solve  and  explain  the  following  : 

tf.  A  farm  containing  282  acres  of  land  was  sold  for  $22184 
What  was  the  rate  per  acre  ?  Ans.  $78.66  -j- . 

7.  A  piece  of  cloth  containing  348  yd.  was  bought  for  $515.91. 
What  did  it  cost  per  yard?  Ans.  $1.4825. 

8.  Bought  236  bu.  oats  for  $90.80.     What  did  I  pay  a  bu.  ? 

0.  If  85  cords  of  stone  cost  $371,875,  what  is  the  price  per 
cord?  Ana.  $4.?75. 

10.  A  farmer  sold  70000  lb.  of  hay  for  $542.50.  How  much 
did  he  receive  per  ton  ?  Ana.  $15.50. 

11.  A  merchant  bought  42  firkins  of  butter,  each  containing 
63|  lb.,  for  $735.67.     What  did  he  pay  per  pound  ? 

12.  There  were  25  mechanics  employed  on  a  building,  each 
receiving  the  same  wages ;  at  the  end  of  28  days  they  were  paid 
in  the  aggregate  $1925.     What  was  their  daily  wages? 

485.  Prob.  ni. — To  find  the  cost  when  the  number  of 
units  and  the  price  of  any  multiple  or  part  of  one  unit 
is  given. 

1.  What  is  the  cost  of  21  ^b.  sugar  at  15  ct.  for  ^  lb.  ? 

SoLTTTioTir.— Since  I  lb.  cost  15  ct.,  21  lb.  must  cost  as  many  times.l5  ci.  ae 
I  lb.  isi  contained  times  in  it.  Hence,  Fivi^l  utep,  21-;- J  =  27 ;  Second  step, 
$.15  X  27  =  $4.05. 

Find  the  cost  of  the  following  : 

2.  124  acres  of  land  at  $144  for  2f  acres  ;  for  1|  A. 

3.  486  bu.  wheat  at  §11  for  8  bushels  ;  at  $4.74  for  3  bushels  ; 
at  $.72  for  f  of  a  bushel. 

4.  265  cords  of  wood  nt  $21.05  for  5  cords. 

5.  135  yd.  broadcloth  at  $8.97  for  2i^  yd.;  at  $12.65  for 
3|  yd. 


B  USINJSSS     PROBLEMS, 


307 


74  for  3  bushels  ; 


.;  at  $13.65  for 


6.  Wliat  is  the  cost  of  987  lb.  coal,  at  35  ct.  i  er  100  lb.  ? 

Solution.— Ab  the  price  is  per  100  lb.,  we  find  the  number  of  hundreds 
in  087  by  moving  the  decimal  point  two  places  to  the  left.  The  pn'oe  innl- 
tipliod  by  this  result  will  give  the  required  cost.  Hence,  $.35  x  9.87=$3.4545. 
the  cost  of  987  lb.  at  35  ct.  per  100  lb. 

Find  the  cost  of  the  following  bill  of  lumber 

7.  2345  ft.  at  11.35  per  100  (C)  feet;  3«28  ft.  at  $.98  per  C. ; 
1843  ft.  at  $1.90  per  C.  ft. ;  8364  ft.  ai  $2.84  per  C.  ;  4384  ft.  at 
!^27.o0  per  1000  (M)  ft. ;  19364  ft.  at  $45.75  per  M. 

8.  What  is  the  cost  of  84690  lb.  of  coal  at  $6.45  per  ton 
\2000  1b.)? 

Observe,  that  pounds  are  changed  to  tone  by  moviii^  the  decimal  point 
^  places  to  the  left  and  dividing  by  2. 

9.  What  is  the  cost  of  96847  lb.  coal  at  $7.84  per  ton  ? 

486.  Prob.  IV. — To  find  the  number  of  units  when 
the  cost  and  price  of  one  unit  are  given. 

1.  How  many  yards  of  cloth  can  be  bought  tor  $28  @  $f  ? 

Solution.— Since  1  yard  can  be  bought  for  $^,  as  many  yards  can  be 
bonuht  for  $28  as  ^  is  contained  times  in  it.    Hence,  $28-+-$f  -  49  yd. 

Find  the  price  and  explain  the  following  orally  : 

2.  How  many  pounds  of  coffee  can  be  bought  for  $60  at  $^ 
per  ixjund  ?    At  $f  ?    At  $j%  V    Ar,  $|?  ?    At  $.33^  ?    At  $.4  'if 

3.  For  $40  how  many  bushels  of  corn  can  be  bought  at  |^ 
per  bu.  ?    At  $|  ?     At  ^j\  V     At  $f;;  ?    At  .$.8  ?    At  $«  ? 

4.  How  many  tons  of  coal  can  be  bought  for  $56  at  .$4  a  ton  c 
At$7V    At  $8?    At  $14?     At$6V    At$9V    At  $5? 

Solve  the  foUowinff : 

5.  The  cost  of  digging  a  drain  at  $3§  per  rod  is  $187  ;  what 
\s  the  length  of  the  drain  ?  Ans.  51  rd. 

G.  How  many  bur»hols  of  wheat  at  $li)  can  be  purchased  for 
fl840  V    At  $lf-  ?     At  $li  ?    At  $1 1  ?    At  $lf  ? 

7.  The  cost  of  a  piece  of  cloth  is  $480,  and  the  price  per  yard 
^I'l  ;  how  many  yards  does  it  contain  ? 


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208 


BUSINESS    ARITHMETIC. 


8.  A  grain  dealer  purchased  a  quantity  of  wheat  at  $1.2( 
per  bushel,  and  sold  it  at  an  advance  of  9^"^  cents  per  bushel 
receiving  for  the  whole  $616.896 ;  how  many  bushels  did  h( 
purchase? 

9.  A  grocer  purchased  $101.65  worth  of  butter,  at  35f  cents 
a  pound ;  how  many  pounds  did  he  purchase  V     Ans.  285  lb. 

10.  How  many  yards  of  cloth  can  be  bought,  at  $2.75  a  yard, 
for  $1086.25?  Ana.  395. 

11.  A  farmer  paid  $14198  for  his  farm,  at  $65f  per  acre 
how  many  acres  does  the  farm  contain?  Ans.  217  A. 

487.  Pros.  Y. — To  find  the  number  of  units  that  cai 
be  purchased  for  a  given  sum  when  the  cost  of  a  multiply 
or  part  of  one  unit  is  given. 

1.  At  19  ct.  for  f  of  a  yard,  how  many  yards  can  be  bouglit 
for  $8.55? 

SoLUTTON— 1.  Since  §  yd.  cost  19  ct.,  J  must  cost  \  of  19  ct.,  f .  9}  ct.,  an( 
I,  or  1  yard,  mast  cost  3  times  9|  ct.,  or  28|  ct. 

2.  Since  1  yard  cost  28J  ct.,  as  many  yards  can  be  bought  for  $8.55  a 
38}  ct.  are  contained  times  in  it.  Hence,  $8.55  -»-  $.285  =  80,  the  number  o 
yards  that  can  be  bought  for  $8.55,  at  19  ct.  for  %  yd. 

2.  A  town  lot  was  sold  for  $1728,  at  $3  per  8  sq.  ft. 
front  of  the  lot  is  48  ft.    What  is  its  depth  ?         Am.  96  ft. 

3.  How  many  bushels  of  com  can  be  bought  for  $28, 
32  ct.  for  I  of  a  bu.  ?    At  28  ct.  for  |  bu.  ? 

4.  How  many  tons  of  coal  can  be  bought  for  $277.50,  at 
for  I  of  a  ton  ?    At  $8  for  f  of  a  ton?  Ana.  37  T. 

5.  A  piece  of  doth  was  sold  for  $34.50,  at  14  yards  per  $j 
How  many  yards  did  the  piece  contain?  Ana.  483  yd. 

6.  A  cellar  was  excavated  for  $408.24,  at  $4.41  for  evel 
7  cu.  yd.     The  cellar  was  54  ft.  by  36  ft.    How  deep  was  it  ? 

7.  A  pile  of  wood  was  bought  for  $275.60,  at  $1.95  for  3  cor 
feet.     How  many  cords  iw  the  pile  ?  Ana.  53  cd. 

8.  A  drove  of  cattle  was  sold  for  $3738,  at  $294  for  eve^ 
7  head.    How  many  head  of  cattle  in  the  drove  ?      Ana.  89.  \ 


ITIC, 


BUSINESS    PROBLEMS. 


JiOD 


of  wheat  at  |1.20l  488.  Prob.  VI.— To  find  the  cost  when  the  quantity 
%  cents  per  bushelMs  a  compound  number  and  the  price  of  a  unit  of  one 
my  bushels  did  hewenomination  is  given. 


butter,  at  35f  centE 
leY     Ans.  2851b. 
fht,  at  $2.75  a  yardj 
An8.  395. 

.,  at  $65f  per  acre] 
Ans.  217  A. 

of  units  that  caii 
cost  of  a  multiply 


rards  can  be  bouglii 


iofl9ct.,f  .9ict.,  and 

be  bonght  for  $8.55  a] 
=  30,  the  number  < 


3  per  8  sq.  ft. 
?         Ans.  96  ft. 

jbought  for  $28, 

for  $277.50.  at 
Ans.  37  T. 

I  at  14  yards  per  $| 
Ans.  483  yd. 
I  at  $4.41  for  evei 
[ow  deep  was  it  ? 

at  $1.95  for  3  coi 
Ans.  53  cd. 

at  $294  for  evei 
)ve?      Ana.  89. 


1.  What  is  the  cost  of  8  bu.  3  pk.  2  qt.  of  wheat,  at  $1.44  per 
t)ushel? 

Solution.  —  ;.  Since  $1.44  is  the 
price  per  bushel,  $1.44x8,  or  $11.53, 
is  the  cost  of  8  bushels. 

2.  Since  2  pk.  =  J  bu.,  $1.44  -*•  2,  or 
72  cts.,  is  the  cost  of  2  pk.,  and  the  \ 
of  72  ct.,  or  36  ct. .  is  the  cost  of  1  pk. 

3.  Since  there  are  8  qt.  in  1  pk., 
2  qt.  =  J  pk.  Hence  the  cost  of  1  pk., 
36  ct  -•-  4,  or  9  ct.,  Is  the  cost  of  2  qt. 

4.  The  sum  of  the  cost  of  the  parts 
lust  equal  the  cost  of  the  whole  quantity.  Heuce,  $12.69  is  the  cost  of 
I  bu.  3  pk.  2  qt.,  at  $1.44  per  bu. 


2 )  $1.44 
8 

11.52 

2)       72 

4)      36 

9 


Cost  of  8  bu. 
"  ♦*  2  pk. 
*'  "  1  pk. 
"     "  2  qt. 


$12.69,  Ans. 


Find  the  cost  of  the  following  orally : 

2.  7  lb.  8  oz.  sugar,  at  12  ct.  per  pound ;  @  14  ct. ;  @  26  ct. 

3.  7f  yd.  ribbon  @  15  ct. ;  @  40  ct.  ;  @  25  ct. 

4.  19  bu.  3  pk.  6  qt.  of  apples,  @  $1  per  bushel. 

5.  13  lb.  12  oz.  butter,  at  34  ct.  per  pound  ;  at  40  ct. 

Solve  the  following : 

6.  What  will  5  T.  15  cwt.  50  lb.  sugar  cost,  at  $240  per  ton? 

7.  Find  the  cost  of  48  lb.  9  oz.  10  dwt.  of  block  silver,  at  $12 
|er  pound.  Ans.  $585.50. 

8.  Sold  48  T.  15  cwt.  75  lb.  of  hay  at  $15  per  ton,  and  32  bu. 
pk.  6  qt.  timothy  seed  at  $3.50  per  bushel.  How  much  did  I 
sceive  for  the  whole  ?  Ar}S.  $847.9  + . 

9.  How  much  will  a  man  receive  for  2  yr.  9  mo.  25  da.  service, 
\i  $1800  per  year?  Am.  $5075. 

10.  Find  the  cost  of  excavatir.g  240  cu.  yd.  13^  cu.  ft.  of 
|arth,  at  50  cts.  per  cubic  yard. 

11.  How  much  will  it  cost  to  grade  8  mi.  230  rd.  of  a  road, 
It  $4640  per  mile  ?  Ans.  $40455. 


! 

^^H  ) 

J  1 

^^H  ^ 

1  . 

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1 

ij. 

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III 


li^H; 


1!H; 


ill 


210 


BUSINESS     MUTHMETIU, 


489.  Pkob.  Vll. — To  find  what  part  one  number  is  of 
another. 

1.  What  i>art  of  12  is  4? 

Solution.— 1  is  ,'4  ol  !2  and  4  being  4  timcB  1,  is  4  timt!^  ,",  of  12,  whicli 
is  1 9  =  3 ;  lienoe,  4  is  J  of  13. 

Observe,  that  to  atstertuin  what  part  one  number  iw  ofauoiher,  we  may  at 
once  write  the  lormer  a»  the  numenitor  and  the  latter  an  thu  denominator 
of  a  fraction,  and  reduce  the  fraction  to  its  lowest  terms  (5i45). 

3.  What  part  is  15  of  18  V    Of  25  ?    Of  24  ?    Of  45  ? 
8.   What  part  is  3G  of  48  ?    Of  38?    Of  42  V    Of  72  ? 

4.  I  is  what  part  of  f  V 

Solution.— 1.  Only  unltv  of  the  same  integral  and  fractional  denomina- 
tion can  be  compared  (144);  hence  we  reduce  %  and  ?  to  \\  and  J?,  and 
place  the  numerator  14  over  t.'te  numerator  18,  giving  \%  —  l\  hence  J  is  2 
off. 

2.  We  may  express  the  relatioii  of  the  fractions  in  the  form  of  a  complex 
fraction,  and   reduce   the   result   to   a   simple   fraction   (289).     Thus, 


3 


K  =  I.    Hence,  ?  is  J  off. 


5.  7^  is  how  many  times  |  ? 

6.  5y  inches  is  what  part  of  2,',  yards?    (See  370.) 

7.  29  /tj  rods  is  what  part  of  1  mile  ? 

8.  f  is  what  part  of  11  ?    |  is  what  part  of  2  J  ? 

9.  11^  is  how  many  times  22  ? 

10.  What  part  of  a  year  is  24  weeks  ?    8  weeks  10  days  ? 

11.  A  man's  yearly  wages  is  $950,  and  his  whole  yearly  ex- 
penses $590.80.  What  part  of  his  wages  does  he  save  each 
year  ? 

12.  3|%  Is  what  part  of  9%  ?    7]  %  is  what  part  of  8|%  Y 

13.  A  man  owning  a  farm  of  210^  acres,  sold  117^  acres. 
What  part  of  hi^^  whole  farm  has  he  still  left  ? 

14.  4%  is  what  part,  of  12  ;^  ?    8  %  is  what  part  of  14%  "? 

15.  Out  of  |750  I  paid  |240.  What  part  of  my  money  have 
I  still  left  ?  ■     Alls.  11,  or  .68. 


I    !  i 


BUSINESS     PROBLEMS, 


211 


ne  number  is  of 


iiuur  I't  of  12,  which 


partof  8|%r 
3old  117^  acres. 


16.  Illustrate  in  full    the  process  in  the    14th   and    15th 
^amples. 

400.    Prob.  VIII. — To  find  a  given  fractional  part  of  a 
Iven  number. 

1.  Find?  of  238. 

I  SoLtTTioN.— We  find  |  of  2!«  by  dividing  it  by  7  ;  hence  2;«  -*-  7  =  34,  the 
)f  238.    But  I  Is  3  times  \  ;  hence  »4  x  3  =  102,  the  ?  of  238. 

1 2.  Find  ^  of  34() ;  ^^^  of  972  ;  Vtt  of  560. 

k  Find  I  of  48 :  of  96  ;  of  376 ;  of  1035. 

k  Find  \  of  $75  ;  I  of  $824.60  ;  ^^  of  $3.25. 

\Obiierve,  that  |  of  $75  meauB  Buch  n  number  of  dollars  as  will  contain  $4 
every  $5  in  $75 ;  hence,  to  find  the  ]  oi"  |T5,  we  divide  by  5  and  multiply 
quotient  by  4. 

|5.  Find  7%  of  828. 

SoLcnoN.— 1.  1%  means  tJ^.     We  find   ^Jo  by  moving  the  decimal 
^nt  two  places  to  the  left  (459).    Hence  1%  or  j  §7;  of  328  is  equal  to 
X  7  =  22.96. 

2.  We  usually  multiply  by  the  rate  first,  then  poini  off  two  decimal  places 
jlhe  product,  which  divides  it  by  100. 

J.  How  much  is  -?  of  157  acres  ?    -^  of  84  bu  wheat  ? 
5  7 

r.  What  is  8%  of  $736  ?    4%  of  395  lb.  butter  'f 


Find 

8.  ^%  of  278  lb. 

9.  5%  of  300  men. 
10.    7%  of  28  yd. 


Find 

11.  4ff^  of  284  mi. 

12.  12i%of732. 

13.  f%of$860. 


14.  Find  the  amount  of  $832  +  1%  of  itself. 

15.  Find  the  amount  of  $325  +  7%  of  itself. 

16.  A  piece  of  cloth  contained  142  yd. ;  15%  was  sold  :  hoi;^ 
[uy  yards  yet  remained  unsold  ? 

17.  A  firkin  of  butter  contained  72g  lb.  ;  |  of  it  was  sold : 
many  pounds  are  there  left  ? 


'■i     '    ■ 


III 


212 


BUSINESS    ARITHMETIC, 


•'V        f  Jl 


■: 


■'.;( 


,1 1 


18.  J.  Henderson's  fann  contained  284  acres,  and  H.  Lee's 
8J%  less.    How  many  acres  in  H.  Lee's  farm? 

10.  If  tea  cost  96  ct.  per  pound  and  is  sold  at  a  loss  of  12^  fc, 
what  is  the  selling  price  ? 

20.  A  merchant  bought  276  yards  cloth  at  $3.40  per  yard. 
He  sold  it  at  25%  profit.  How  much  did  he  realize,  and  what 
was  his  selling  price  ? 

491.  Prob.  IX.—To  find  a  number  when  a  fractional 
part  is  given. 

1.  FiDd  the  number  of  which  84  is  }. 

Solution  .—Since  84  Is  J  of  the  number,  }  of  84  mnet  be  J ;  hence  84  -♦-  7 
=  13  is  the  i  of  the  required  number.  But  9  times  I  is  equal  to  the  whole ; 
hence,  12  x  9  =  108,  the  required  number. 

8.  Find  the  number  of  yards  of  cloth  of  which  135  yd.  is  /^ . 
8.  $36  is  I  of  how  many  dollars  ?    $49  is  I  of  how  many 
dollars  ? 

4.  John  has  $756,  which  is  |  of  Norman's  money ;  how  many 
dollars  has  Norman  ?  Ans.  $1323. 

5.  The  profits  of  a  grocery  for  one  year  are  $3537,  which  is 
1^  of  the  capital  invested.     How  much  is  the  capital  ? 

6.  Find  the  number  of  dollars  of  which  $296  are  $8,'^ ,  or  .08. 

First  Solution.— Since  $296  are  tSt?  of  the  number,  ^  of  $290,  or  $37,  are 
rio  ;  hence  Igg,  or  the  whole,  is  100  times  $37,  or  $37  x  100  =  $3700. 

Second  Solution.— Since  $296  are  twt  of  the  number,  J  of  $296  is  rh>^ 
and  i  of  too  times  $296  is  fgg,  or  the  required  number.  Hence,  $296  x  100 
=  $29600.  and  i  of  $29600  =  $3700,  the  required  number. 

From  these  solutions  we  obtain  the  following  rule  for  finding  a  number 
when  a  decimal  part  of  it  is  given  : 


I 


BIJIiE. 

492.  Move  the  decimal  point  as  many  places  to  the  right  as 
ihere  are  pku^es  in  the  given  decimal,  annexing  ciphers  if  neces- 
sary, and  divide  the  result  by  the  number  expressed  by  the  sig- 
nificant figures  in  the  given  decimal. 


JUSTNESS    PROBLEMS. 


2ia 


A. 


|8%,or.08. 

j9G,or|37,are 
$3700. 
$296  is  \hr,y 
jce,  $296  X  100 


Find  what  number 

7.  16  is  8%  of. 

8.  34  is  6%  of. 

9.  84  is  7%  of. 

10.  $73  arc  9%  of. 

11.  120  yd.  are  5%  of. 
13.  50  bu.  are  8/o  of. 


find  what  number 

18.  I  is  4%  of. 

14.  $§  is  7^  of. 

15.  \  pk.  is  8%  of. 

16.  .7  ft.  is  5%  of. 

17.  .09  is  4%  of. 

18.  .48  is  13%  of. 


Find  what  number 

19.  8}  is  9%  of. 

30.  $3.10  is  6%  of. 

31.  7|is9%  of. 
23.  27ii8  5%  of 
23.  ^yd.  is  8%  of. 
34.  .96  is  12%  of. 


25.  A  grocer  purchased  186  lb.  butter  on  Saturday,  which  is 
6%  of  the  entire  quantity  purchased  during  the  week.  What 
was  the  week's  purchase  ? 

26.  A  merchant  sells  a  piece  of  cloth  at  a  profit  of  80  ct.  a 
yard,  which  is  30%  of  what  it  cost  him.  "What  was  the  buying 
price  per  yard  ? 

27.  A  man's  profits  for  one  year  amount  to  $2840,  which  is 
8%  of  the  amount  he  has  invested  in  business.  What  is  his 
investment  ? 

38.  A  mechanic  pays  $13  a  month  for  house  rent,  which  is 
16%  of  his  wages.    What  does  he  receive  per  month  ? 

29.  12%  of  f  is  9%  of  what  number? 

30.  If  in  a  certain  town  $3093.75  was  raised  from  a  |%  tax, 
what  was  the  value  of  property  in  the  town  ? 

81.  An  attorney  receives  $1.75  for  collecting  a  bill,  which  is 
2^  per  cent  of  the  bill.     What  is  the  amount  of  the  bill  ? 

32.  A  man  having  failed  in  business  is  allowed  to  cancel  his 
debts  by  paying  30% .  What  does  he  owe  a  man  who  receives 
$270?  Ans.  $1350. 

33.  A  man  sold  his  house  for  $1000,  which  was  13%  of  the 
sum  he  received  for  his  farm.  What  was  the  price  of  the 
farm?  Ans.  $8333.33^. 

34.  How  many  acres  in  a  farm  14%  of  which  contains 
43  acres? 

35.  J.  Simpson  has  35  %  6f  his  property  invested  in  a  house,^ 
10%  in  a  farm,  5%  in  a  bam,  and  the  rest  in  a  grove  worth 
$4800.    What  is  the  amount  of  his  property  ? 


'ft    H  j 


^14 


BUSJ  JV  A'S  a     A  H  J  TH  M  E  Tl  V  . 


ml 


m  I 


49c}.    PiiOB.  X. — To  express  the  part  one  number  is 
of  another  in  any  given  fractional  unit. 

1.  How  many  fftha  of  3  is  8  ? 

Solution.— Since  }  is  J  of  8,  there  must  be  as  xaanyffths  of  8  In  8  as  ^ 


is  contained  times  lu  it.    8  +  J 


13S.    Uence,  8  Ib  1^  of  three. 
5 


Solve  the  folio wiug  orally : 

2.  How  many  fourths  of  0  is  7  ?    Is  5  ?    Is  13  V     Is  20  V 

3.  How  many  hundredths  of  36  is  9  ?    Is  4  ?    In  18?    Is  13  V 

4.  $13  are  how  many  tenths  of  $5  V    Of  !|8  V    Of  $15  ? 

5.  42  yards  are  how  many  sixthB  of  2  yd.  ?    Of  7  yd.  ?    Of 
3  yd.? 

C.  ^^hai  per  cent  of  $11  arc  $3,  or  $8  are  how  many  huu 
dredthsof  $11? 

First  Solution.— Since  ^V  is  pjjj  of  $11,  there  must  be  ab  many  hun- 
dredths of  $11  in  $3  as  ^o\j  is  contained  times  in  $8.    |3  +  tVo  =  8  x  W"-  = 

Vt"  =  27t\.    Hence,  $3  are  ?^I^,  or  27A^  of  $11. 

Second  Solution.— Since  (489)  |8  are  y*,  of  |11,  we  tmve  only  to 
reduce  rV  to  hundredths  to  find  wliat  per  cent  $3  are  of  $11.    A  —  iVJo 

=  Zhl  =  27t\?S.    Hence,  $3  are  27A5«  of  $11. 

From  these  solutions  we  obtain  the  following  rule  for  finding 
what  per  cent  or  what  decimal  part  one  nmubcr  is  of  another : 


BULB. 

494.  Express  the  former  number  as  a  fraction  of  the  lattei 
(489),  and  redu^ce  this  fraction  to  hundredths  or  to  the 
required  decimal  (327). 


Find  what  per  cent 

7.  $36  are  of  $180. 

8.  13  is  of  73. 

9.  16  is  of  64. 

10.  $46  are  of  $414. 

11.  13  oz.  are  of  5  lb. 

12.  7  feet  are  of  8  yards. 


13.  If  is  of  $|;  of  $5;of|2|. 

14.  2  bu.  3  pk.  are  of  28  bu. 

15.  284  acres  are  of  1  sq.  mi. 

16.  31b.  13  oz.  are  of  9  lb. 

17.  48  min.  are  of  3  hr. 

18.  f  of  a  cu.  ft.  is  of  1  cu.  yd. 


H, 


BUSINESS     i'  n  O  B  L  E M  8, 


nb 


nber  is 


a  In  8  aB  ^ 


uany 


liuu 


have  only  to 

»   —  •"-I'- 
ll.     IT  -  It"" 


le  for  finding 
of  another : 


of  the  Uttti 
or   to   the 


^5 ;  of  |2|. 
)f  28  bu. 

\  eq.  mi. 
)f91b. 

lir. 
lof  1  cu.  yd. 


19.  A  morcliant  invoetc^l  ^$3485  in  gomls  which  ho  had  to 
aell  for  ij^JOTS.     Wliat  per  <'eDi  of  lii«  investment  did  he  lose? 

20.  Find  what  i>er  cent  3  bu.  3  i>k.  are  of  8  bu.  3  pk.  5  iit. 

21.  A  man  paid  $24  for  the  i;se  of  |3()0  for  one  year.  What 
rate  per  cent  did  he  pay  ? 

22.  Find  what  por  wnt  |  of  a  t^q.  yd.  is  of  I  of  a  sq.  yd. 

23.  A  dmiorgist  paid  84  ct.  an  ounce  for  a  certain  medicine, 
and  sold  it  at  $1.36  an  ounce.  What  jxr  cent  profit  did  lie 
make? 

SoLUTioN.-tl.86-I.S4=  i.^«;  M  =  JiJ2.!  =  ^^  -=  61Jfj<. 

24.  J.  RoHS  deposited  *2500  ju  a  bank,  and  again  deposited 
enough  to  make  the  whole  amount  to  |2750.  What  per  ceut 
of  the  first  deposit  wad  the  last?  Anti.  10. 

25.  When  a  yard  of  silk  is  bought  lor  |1.20  and  sold  for 
$1.60,  what  per  cent  is  the  profit  of  the  buying  price? 

26.  A  man  owed  me  $350,  but  fearing  he  would  not  [)ay  it  I 
a.j^reed  to  take  .$306.25  ;  what  per  cent  did  I  allow  him  ? 

27.  A  farmer  owning  386  acres  sold  148  acres.  What  pur 
cent  of  his  original  farm  does  he  still  own  ? 

38.  Gave  away  77i^  bushels  of  potatoes,  and  my  whole  crop 
was  500  bushels ;  what  ^  of  the  crop  did  I  give  away  ? 

29.  A  man  pays  $215.34  per  acre  for  4i  acres  of  land,  and 
lets  it  a  year  for  $33,916  ;  what  %  of  the  cost  is  tlie  rent  ? 

495.  Prob.  XI. — To  find  a  number  which  is  a  given 
fraction  of  itself  greater  or  less  than  a  given  number. 

1.  Find  a  number  which  is  §  of  itself  less  than  28. 

SoLTTTiON.— 1.  Since  the  required  number  is  I  of  itself,  and  in  |  of  iteelf 
less  than  28,  hence  28  is  j  +  i  or  j  of  it. 

2.  Since  28  is  I  of  the  number,  |  of  28,  or  4,  is  {.  Hence  |,  or  the  whole 
of  the  required  number,  is  5  times  4  or  20. 

Solve  the  following  orally  : 

2.  What  number  is  |  of  itself  less  than  15  ?  Less  than  40  ? 
Less  than  75  ?    Less  than  26  ?    Less  than  32  ? 


» 


216 


BUSINESS    ARITH3IETIC, 


m 


3.  What  number  increased  f  of  itself  is  equal  100  ?  la  equal 
80  ?  Is  equal  120  ?    Is  equal  13  ?    Is  equal  7  V 

4.  Find  a  number  which  diminished  by  §  of  itself  is  equal 
56.  Is  equal  70.    Is  equal  15.     Is  equal  5. 

Solve  and  explain  the  following  : 

6.  What  number  increased  by  7%  or  il^  of  itself  is  equal 
642? 

Solution.— 1.  Since  a  number  increar-cd  by  7^  or  ,  l„  of  it?elf  iB  \%%  +  iJo 
=  \ll  of  itself,  642  is  \Zl  or  107^  of  the  rt-quired  number. 

2.  Since  642  is  \%l  of  the  required  number,  for  every  107  in  642  there 
must  be  100  in  the  required  number.  Hence,  642-«-107-t;.  and  6  x  100=600, 
the  required  number. 

Observe^  that  642  -»- 1.07  is  the  same  as  dividing  by  107  and  mnltiplyinj," 
by  100  (349);  hence  the  following  rule,  when  unnmbcr  has  been  increased 
or  diminished  by  a  given  pf t  cent  or  any  decimal  of  itself: 


RULE. 

496.  Divide  the  given  number,  according  an  it  is  more  or  less 
than  the  required  number^  by  1  increased  or  diiujiinhed  by  the 
given  decimal. 

6.  A  regiment  after  losing  8%  of  its  number  contained  736 
men  ;  what  was  its  original  number?  Ans.  800. 

7.  A  certain  number  increased  by  80^  of  itseli  is  331.2  ;  what 
is  that  number?  Aiis.  184. 

8.  By  running  155^  faster  than  usual,  a  locomotive  runs  644 
miles  a  day ;  what  was  the  usual  distance  per  day  ? 

9.  What  number  diminished  by  25%  of  itself  i.s  G54? 

10.  What  number  increased  by  15%  of  itself  is  equal  248.40? 

11.  A  man  who  has  had  his  salary  increased  h',(  now  receives 
$1050  a  year  ;  what  was  his  former  salary  ?  Ans.  ^1000. 

12.  A  tailor  wlls  a  coat  for  $8,  thereby  gaining  25%  ;  whai 
did  the  coat  cost  him  ?  Ans.  $6.40. 

13.  A  teacher  lays  up  12|%  of  his  salary,  which  leaves  him 
$1750  to  spend  ;  what  is  his  salary  ?  .1  ns.  $2000. 

14.  T.  Laidlaw  sold  his  farm  for  $3960,  wlilrh  was  10;^  lesj 
than  he  gave  for  it,  and  he  gave  10%  more  than  it  was  worth ; 
what  was  its  actual  value  ?  Ans.  $4000. 


RULES    FOR     PERCENTAGE. 


217 


Is  eqtial 
IB  equal 


:  is  equal 

riBigg  +  iSo 

n  642  there 
6^100=600, 


APPLIOATIOISrS. 

497.  Profit  and  Loss,  Commission,  Insurance,  Stocks, 
Taxes,  and  Duties,  are  applications  of  Business  Problems  VIII, 
IX,  X,  XI.  The  rate  in  these  subjects  is  usually  a  per  cent. 
Hence,  for  convenience  in  expressing  rules,  we  denote  the 
quantities  by  letters  as  follows  : 

1.  /{  repreecnts  the  Base^  or  naraber  on  which  the  percontagc  is 
reckoned. 

2.  li  represents  the  Rate  per  cent  ezprci^sed  decimally. 

3.  I*  represents  the  I*ereentage,  or  the  part  of  the  Base  which  is  de- 
noted by  the  Jiate. 

4.  A  repreeentH  the  Amonnt,  or  sum  of  the  Base  and  Percentage. 
6.  JO  represents  the  Hifferenee,  or  Base  less  the  Percentage. 


FormtUcB,  or  Rules  for  Percentage, 


498.  PaoB.  VIIL    P  =  JB  X  JJ.  Read,  j 

499.  PROB.IX 


500.    PROB.X. 


501.   Pbob.  XL 


B  = 


B  = 


1  +  R 
1—R' 


Read 
Read, I 
Read, \ 


Read 


I'Ae  percentage  is  equal  to  the 
base  multiplied  by  the  rate. 
(  The  base  is  equal  to  the  per- 
'  I     centage  divided  by  the  rate. 
The  rate  is  equal  to  the  per- 
centage divided  by  (he  base. 
The  base  is  equal  to  the  amount 

divided  by  1  plus  the  rate. 

The  base  is  equal  to  the  d\ff"'nce 

divided  by  1  minus  the  rate. 


•I 


502.  Refer  to  the  problems  on  pages  211  to  214  inclusive, 
and  answer  the  following  questions  regarding  these  formula; : 

1.  Wliat  is  meant  hj  B  x  R,  and  why  is  P  =  B  x  II?  Illus- 
trate your  answer  by  an  example,  giving  a  reason  for  each 
step. 

2.  Why  is  P-t-i?  equal  Z??  Give  reasons  in  full  for  your 
answer. 

3.  If  B  is  135%,  which  is  the  greater,  P  or  B,  and  why? 

15 


y-.i. 


218 


BUSINESS    ARITHMETIC, 


f>-  ) 


4.  Why  is  R  equal  to  P  -5-  J5,  and  how  must  the  quotient  of 
P  -*-  JB  be  expressed  to  represent  B  correctly  ? 

5.  If  R  is  248%,  how  would  you  express  R  without  the 

sign  %? 

6.  What  is  meant  by  J.  ?  How  many  times  JR  in  P  (491)  ? 
How  many  times  1  in  ^  ?  How  many  times  \  +  R  must  there 
be  in  A  and  why  ? 

7.  Why  is  B  equal  to  ^  -r-  (1  +  i2)  ?  2)  is  equal  to  B  minus 
how  many  times  B  (490)  ? 

8.  Why  is  B  equal  to  2)  -5-  (1  -  i2)?  Give  reasons  in  full 
for  your  answer. 


¥' 


T 


PEOPIT    AND    LOSS. 

503*  The  quantities  considered  in  Profit  and  Loss  corres- 
pond with  those  in  Percentage  ;  thus, 

1.  The  Costt  or  Capital  invested,  is  the  Base. 

2.  Tlie  Per  cent  of  Profit  or  Loss  is  the  Rate. 

3.  The  Profit  or  Loss  is  the  Percentage. 

4.  The  Sellinfj  Price  when  equal  the  Coat  plus  the  Profit 
is  the  Amount ;  when  equal  the  Cost  minus  the  Loss  is  the 
Difference, 


EXAMPLES     FOR    PRACTICE. 

504.    1.  A  firkin  of  butter  was  bought  for  $19  and  sold  atj 
a  profit  of  16  % .     What  was  the  gain  ? 

Formula,  P  =  B  x  R,    Read,  ProJU  or  Loea  —  Cost  x  Rate  %, 

Find  the  profit  on  the  sale 

2.  Of  84  cd.  wood  bought  @  |4.43^,  sold  at  a  gain  of  20^. 

3.  Of  320  yd.  cloth  bought  @  $1.50,  sold  at  a  gain  of  17%. 

4.  Of  873  bu.  wheat  bought  @  $1.25,  sold  at  a  gain  of  \^%.\ 

Find  the  loss  on  the  sale 

5.  Of  180  T.  coal  bought  @  $7.85,  sold  at  a  loss  of  8J%. 


PROFIT    AND     LOSS, 


219 


tient  of 
out  tlie 

xst  there 


gaud  sold  at] 

;ainof20f.. 
rain  of  17  fc. 
gain  of  mP 

)8 


6.  Of  134  A.  land  bought  @  $84.50,  sold  at  a  loss  of  21^  %. 

7.  If  a  farm  was  bought  lor  $4800  and  sold  for  $729  more 
than  the  coat,  what  was  the  gain  per  cent? 

Formula,  R  =  P+B.    Read,  Hats  %  Gain  =  Profit  -t-  Cost. 

8.  If  ^  of  a  cord  of  wood  is  sold  for  f  of  the  cost  of  1  cord, 
what  is  the  gain  per  cent  ? 

9.  A  piece  of  cloth  is  bought  at  $3.85  per  yard  and  sold  at 
$2. 10  per  yard.    What  is  the  loss  per  cent  V 

10.  Find  the  seUi/ig  price  of  a  house  bought  at  $5385.90,  and 
sold  at  a  gain  of  18  ^ . 

Formula,  A  =  B  x  (1  +  R).    Read,  8eUing  Price  =  Coat  x  (1  +  Rate  %  Gain). 

11.  Corn  that  cost  C5  ct.  a  bushel  was  sold  at  20%  gain. 
What  was  the  selling  price  ?  Ans.  78  ct.  a  bu. 

12.  A  grocer  bought  43  bu.  clover  seed  @  $4.50,  and  sold  it 
ill  small  quantities  at  a  gain  of  40 /t.  What  was  the  selling 
price  per  bu.  and  total  gain  ? 

13.  Bought  184  barrels  of  flour  for  $1650,  and  sold  the  whole 
at  a  loss  of  8  % .    What  was  the  selling  price  per  barrel  ? 

Formula,  D  -  B  (1-R).    Read,  Selling  Price  =  Cost  y.{\— Hate  %  Loss). 

14.  C.  Baldwin  bought  coal  at  $6.25  per  ton,  and  sold  it  at 
a  loss  of  18%.     What  was  the  selling  price  ? 

15.  Flour  was  bought  at  $8.40  a  barrel,  and  sold  so  as  to 
lose  15%.     What  was  the  selling  price? 

16.  Sold  a  house  at  a  loss  of  $879,  which  was  15%  of  the 
cost.    What  was  the  cost  ? 

Formula,  B  =  P-^R.    Read,  CokI  =  Profit  or  Loss  +  Rate  %. 

17.  A  grain  merchant  sold  284  barrels  of  flour  at  a  loss  of 
!fG74.50,  which  was  25%  of  the  cost.    What  was  the  buying 

[and  selling  price  per  barrel? 

18.  A  drover  wished  to  realize  on  the  sale  of  a  Hock  of 
|23f5  sheep  $531,  which  is  30%  of  the  cost.  At  what  price  per 
|liead  must  he  sell  the  flock  ? 

19.  Two  men  engaged  in  business,  each  having  $4380.    A 


^ 


220 


BUSINESS    ARITHMETIC. 


r(... 


If  I 


fh 


!  a 


<jii 


gained  33|%  and  B  75^.    How  much  was  B's  gain  more 

than  A's? 

20.  A  grocer  sells  coffee  that  costs  18  J  cents  per  pound,  for 
lOj  cents  a  ix)mid.     What  is  the  loss  per  cent  ? 

21.  If  I  buy  72  head  of  cattlo  ut  $3G  a.  head,  and  sell  33|-%  of 
them  at  a  gain  of  18 >^,  and  the  remainder  at  a  gain  of  24 ;6, 
what  is  my  gain? 

23.  A  man  bought  24  acres  of  land  at  $75  an  acre,  and  sold  it 
at  a  profit  of  8.V  % .     What  was  his  total  gain  ? 

23.  Fisk  and  Gould  sold  stock  for  $3300  at  a  profit  of  33^%. 
What  was  the  cost  of  it  ? 

24.  A  merchant  sold  tsloth  for  $3.84  a  yard,  and  thus  made 
20  */€ .     What  was  the  cost  price  ? 

25.  Bought  wood  at  $3.25  a  cord,  and  sold  it  at  an  average 
gain  of  30  '/r .     What  did  it  bring  per  cord  ? 

26.  Bought  a  barrel  of  syrup  for  $20  ;  what  must  I  charge  a 
gallon  in  order  to  gain  20%  on  the  whole  ? 

27.  If  land  when  sold  at  a  loss  of  12^%  brings  $11.20  per 
acre,  what  would  be  the  gain  per  cent  if  sold  for  $15.36  ? 


COMMISSION. 

505.  A  Coinmissioit  Merchant  or  Agent  is  a  person 
who  transacts  business  for  anotlier  for  a  percentage. 

50f>.  A  Broker  is  a  person  who  buys  or  sells  stocks,  bills 
of  exchange,  etc.,  for  a  percentage. 

507.  Coniinission  is  the  amount  paid  a  commission mer- 
chiint  or  agent  for  the  transaction  of  business. 

508.  lirohemge  is  the  amount  paid  a  broker  for  the 
transaction  of  business. 

509.  The  JVf  t  Proceeds  of  any  transaction  is  the  sum 
of  money  that  is  left  after  all  expenses  of  commission,  etc., 
are  paid. 


i  more 

ind,  for 

53|%  of 
)f  24^6, 

d  sold  it 

iu8  made 

n  average 

L  cbarge  a 

$11.20  per 
L36? 


is  a  person 

81001^9,131118 

isslon  ineT- 

iter  for  the 

is  the  sum 
Mission,  etc., 


COMMISSION, 


221 


510,  The  quantities  considered  in  Commission  correspond 
with  those  in  Percentage  ;  thus, 

1.  The  amount  of  money  invested  or  collected  is  the  Itaxe. 

2.  The  per  anf  allowed  for  eervicen  in  the  Ratv. 

3.  The  Commission  or  Brokerage  in  the  I'erci'tttnffe, 

4.  The  sum  invested  or  collected,  plus  thccommiusion^  is  the  Amount; 
minus  the  commission  is  the  Uifferenve. 


EXAMPLES    FOR     PRACTICE. 

511,  Let  the  pupil  write  out  the  formulae  for  each  kind  of 
examples  in  Commission  in  the  same  manner  as  they  are  given 
in  Profit  and  Loss. 

What  is  the  commission  or  brokerage  on  the  following : 

1.  The  collection  of  $3462.84,  commission  2^  %  ? 

2.  The  sale  of  484  yds.  cloth  (<h  $2.15,  commission  1^'  %  ? 

3.  The  sale  of  176  shares  stocks  at  $87.50  a  share,  broker- 
age g;i^? 

4.  The  sale  of  85  cords  of  wood  (ib  $4.75,  commission  ^\%  ? 

What  is  the  rate  of  commission  on  the  following : 

5.  Selling  wheat  worth  $1.80  a  bu.,  commission  4  ct.  a 
bushel  ? 

6.  Collecting  a  debt  of  $7500,  commission  $350? 

7.  Selling  a  farm  for  $4800,  commission  $120? 

What  is  the  amount  of  the  sale  in  the  following  : 

8.  The  commission  is  $360,  rate  of  commission  21%  1 

9.  The  brokerage  is  $754.85,  rate  of  brokerage  1:.|  %  ? 

10.  The  commission  is  $26.86,  rate  of  commission  1^/^  ? 

Find  the  amount  of  the  sales  in  the  following  : 

Observe,  that  the  commission  is  on  the  amount  of  the  sales.  Hence  tho 
formula  for  finding  the  amount  of  the  sales  when  the  net  proceeds  are 
given  is  (496) 

Ammtnt  of  sales  =  Net  proceeds  -t-  (1  —  Hate  5"). 

11.  Net  proceeds,  $8360;  rate  of  commission,  3^%. 


Si 


i 


•J"  I 
^1 


If 


n 


222 


B  trSINESS    ARITHMETIC, 


v:-\  •-.5  ji 


ill 


12.  Net  proc^^pds,  $1850;  rate  of  commission,  l^o. 

13.  Net  proceeds,  $3G40 ;  rate  of  commission,  |  ^ . 

Find  the  amount  to  be  invested  in  the  following  : 

Observe,  that  when  an  agent  is  to  deduct  his  commission  from  Ibc 
amount  of  money  in  his  hand  the  formula  is  (495) 

Sum  invented  =  Amount  in  hand  +  (1  +  Hate  50. 

*■>..  ^  :^   <unt  in  band,  $3401.01 ;  rate  of  commission,  Z\%. 

15.  Amount  in  hand,  $60G.43  ;  rate  of  commission,  IJ/^. 

16.  Remittance  was  $393.17  ;  rate  of  commission,  3|%. 

17.  A  lawyer  collects  bills  amounting  to  $492  ;  what  is  his 
corair '    V '    it  ^  %  1  A  ns.  $24.60. 

18.  Alia",  '^^ilws  sent  me  $582.40  to  invest  in  apples,  at 
$5  e  bait  el ;  ao\'^  i-iany  can  I  buy,  commission  being  4^  ? 

19.  Ai-  ar,ent  so;  ''^4  barrels  of  beef,  averaging  202|  lb.  each 
at  9  cenis  a  { v,  m.i  ;        m  was  his  commission  at  2^^  ? 

20.  I  have  remitted  ^i]2<'  to  my  correspondent  in  Lynn  to 
invest  in  shares,  after  deducting  his  commission  of  l|^r  ;  what 
is  his  commission  ?  Ans.  $18.83. 

21.  A  man  sends  $6897.12  to  his  agent  in  New  Orleans, 
requesting  him  to  invest  in  cotton  after  deducting  his  commis- 
sion of  2^  ;  what  was  the  amount  invested? 

22.  An  auctioneer  sold  goods  at  auction  for  $13825.  anrl 
others  at  a  private  sale  for  $12050 ;  what  was  his  commission 
at^%?  Ans.  $129.3750. 


INSUBANCE. 

51 2.  Insurance  is  a  contract  which  binds  one  party  to 
indemnify  another  against  possible  loss  or  damage.  It  is  of 
two  kinds :  insurance  on  property  and  insurance  on  life, 

513.  The  Policy  is  the  written  contract  made  between  | 
the  parties. 

514.  The  Premium  is  the  percentage  paid  for  insurance.  I 


from  till 


INSURANCE. 


223 


515.  The  quantities  considered  in  Insurance  correspond 
with  those  in  Percentage ;  thus. 


1.  Tlie  amcmnt  insured  is  the  liase. 

2.  The  per  cent  of  premium  ia  the  Mate. 
'A.  The  premiom  is  the  l*ercentaye. 


I\ 


hat  is  his 
$34.60. 
apples ,  at 

r4.r/ot 

)2ilb.each 

in  Lynn  to 
t  c;-  ;  what 

$18.83. 

w  Orleans, 

liis  commis- 

^3825.  anf1 
cotnniissiou 
1129.3750. 


one  party  to 
ge.     It  is  of  I 
►n  life. 

tade  between! 


for  insurance.! 


EXAMPLES     FOR    PRACTICE. 

510.  Let  the  pupil  write  out  the  formula)  as  in  Profit  and 

Loss. 

1.  What  is  the  premium  on  a  policy  for  $8500,  at  8%  ? 

2.  My  house  is  insured  for  $7250 ;  what  is  the  yearly  pre- 
mium, at2J%? 

3.  John  Kerr's  house  is  insured  for  $3250  at  3^  per  cent,  his 
furniture  for  $945  at  1|  per  cent,  and  his  barn  for  $1220  at  1| 
l)er  cent ;  what  is  the  amount  of  premium  on  the  whole  prop- 
erty? 

4.  A  factory  is  insured  for  $27430,  and  the  premium  is 
}^685.75  ;  what  is  the  rate  of  insurance  ? 

5.  Goldie's  Mills,  Guelph,  worth  $28000,  being  destroyed  by 
fire,  were  insured  for  f  their  value ;  at  2^  per  cent,  what  is  the 
actual  loss  of  the  insurance  company  ? 

6.  The  premium  on  a  house,  at  f  per  cent,  is  $40 ;  what  is 
the  sum  insured  ? 

7.  It  costs  me  $72  annually  to  keep  my  house  insured  for 
$18000;  what  is  the  rate  ? 

8.  What  must  be  paid  to  insure  from  Montreal  to  Liverpool 
a  ship  valued  at  $37600,  at  f  of  1  %  ? 

9.  My  dwelling-house  is  insured  for  $4800  at  |  %  ;  my  fur- 
niture, library,  etc.,  for  $2500  at  |%  ;  my  horses,  cattle,  etc., 
for  $3900  at  g  %  ;  and  a  carriage  manufactory,  including 
machinery,  for  $4700  at  If  %.     What  is  ray  annual  premium? 

10.  A  cargo  of  800  bundles  of  hay,  worth  $4.80  a  bundle,  is 
insured  at  1}%  on  |  of  its  full  value.  If  the  cargo  be  de- 
stroyed, how  much  will  the  owner  lose  ? 


224 


BUSI.YESS    ARITHMETIC* 


STOCKS. 


i 


r  M 


i 


6 1 7.  A  Corporation  is  a  body  of  individuals  or  company 
authorized  by  law  to  transact  business  as  one  person. 

518.  The  Caintal  Stoeh  is  the  money  contributed  and 
employed  by  the  company  or  corporation  to  carry  on  its 
business. 

The  term  stock  is  also  ueed  to  denote  Government  and  City  bonde,  etc. 

519.  A  Shave  is  one  of  the  equal  parts  into  which  the 
capital  stock  is  divided. 

520.  A  Certificate  of  Stocky  or  Scrip ,  is  a  paper 
issued  by  a  corporation,  securing  to  the  holder  a  given  number 
of  shares  of  the  capital  stock. 

521.  The  I*ar  Value  of  stock  is  the  sum  for  which  the 
scrip  or  certificate  is  issued ;  e.  g.,  $100  stock  sells  for  $100 
money. 

522.  The  Market  Value  of  stock  is  the  price  per  share 
for  which  it  can  be  sold.  Stock  is  at  a  premium  when  it  sells 
at  more  than  its  nominal  value ;  e.g.,  when  $100  stocks  sells  at 
$117  money,  it  is  at  17  per  cent,  premium.  When  it  sells 
below  its  nominal  value,  it  is  at  a  discount. 

523.  The  Pmniiim,  lyiscount,  and  Brokerage  are 

always  computed  on  the  par  value  of  the  stock. 

524.  The  Net  Tlarfiinga  are  the  moneys  left  after  de- 
ducting all  expenses,  losses,  and  interest  upon  borrowed 
ca])ital. 

525.  A  liond  is  a  written  instrument,  securing  the  pay- 
ment of  a  sum  of  money  at  or  before  a  specified  time. 

520.  A  Coupon  is  a  certificate  of  interest  attached  to  a 
bond,  which  is  cut  off  and  delivered  to  the  payor  when  the 
interest  is  discharged. 

527.  Consof.<i  are  a  part  of  the  national  debt  of  Britain. 
Various  annuities  are  consolidated  into  a  joint  3  per  cent  stock 


STOCKS. 


225 


j^ 


mpany 

ed  and 
on  its 

d8,  etc. 
lich  the 


a  paper 
number 


rliich  the 
for  $100 

per  share 

in  it  sells 

:8  sells  at 

In  it  sells 

\rage  are 

after  de- 
borrowed 

the  pay- 

iched  to  a 
when  the 

lof  Britain, 
cent  stock 


—hence  the  name  Consols.  The  value  of  these  does  not  vary 
much,  from  the  fact  that  the  nation  is  regarded  as  willing  and 
able  to  pay  all  interest  upon  her  debt,  which  debt  now 
amounts  to  about  773  million. 

o28.  U,  S,  Bonds  may  be  regarded  as  of  two  classes : 
those  payable  at  a  fixed  date,  and  those  payable  at  any  time 
between  two  fixed  dates,  at  the  option  of  the  government. 

f>2f>.  In  commercial  language,  the  two  classes  of  U.  S. 
bonds  are  distinguished  from  each  other  thus  : 

(1.)  U.  S,  6*M,  bonds  payable  at  a  fixed  time. 

(2.)  U.  s.  6*'«  5'ito,  boHdf  payable,  at  the  option  of  the  Government, 
at  any  time  from  5  to  20  years  from  their  date. 

530.  The  liability  on  bank  stocks  in  Canada  is  limited  to 
double  the  amount  of  capital  subscribed.  On  other  stocks  the 
liability  of  shareholders  is  limited  to  the  amount  of  subscribed 
capital. 

EXAMPLES    FOR    PRACTICE. 

53 .1 .  Let  the  pupil  write  out  the  formula  for  each  class  of 
examples,  as  shown  in  Profit  and  Loss : 

1.  Find  the  cost  of  120  shares  of  the  Toronto  Bank  stock,  the 
market  value  of  which  is  108,  brokerage  1%. 

Solution.— Since  1  phare  cost  108% +  J^,  or  lOSi^  of  $100  =  108.1,  the  coet 
of  1-20  shares  wUl  be  |108i  x  120  =  $13020. 

2.  What  is  the  market  value  of  86  shares  in  the  Freehold 
Loan  Company,  at  3^%  premium,  brokerage  -^  '/c  ? 

3.  Find  the  cost  of  95  shares  bank  stock,  at  G%  premium, 
brokerage  *  ^ . 

4.  How  many  shares  of  the  Dominion  Telegraph  stock  at  S^ip 
discount  can  be  bought  for  f  7030,  brokerage  ^  %  ? 

Solution.— Since  1  share  cost  100;«-8;«+ Jj<,  or  92.\;«  of  $100  =  $92.50,  as 
many  shares  can  be  bought  as  $92.50  are  contained  times  in  $7020,  which 
is  76. 

How  many  shares  of  stock  can  be  bought 

5.  For  $10092,  at  a  premium  of  5;^,  brokerage 


iii 


I 


226 


B  UiSINESS    ARITHMETIC, 


A 


6.  For  $13428,  at  a  discount  oil%,  brokerage  \'/c  ? 

7.  For  $108<i0,  at  a  premium  of  9|  Jo ,  brokerage  |  %  ? 

8.  What  sum  must  be  invested  in  stocks  at  113,  paying  9^, 
to  obtain  a  yearly  income  of  $1260  ? 

Solution.— Siuce  $9  is  the  annual  Income  on  1  ehare,  the  number  of 
Bhareti  must  be  equal  11260  +  $9,  or  140  shares,  and  140  shares  at  $112  a 
share  amount  to  $15080,  the  required  inveetment. 

Find  the  investment  for  the  following  : 

9.  Income  $1800,  stock  purchased  at  109 J,  yielding  12  fo. 

10.  Income  $2601),  stock  purchased  at  105^,  yielding  7^. 

11.  Income  $3900,  stock  purchased  at  93,  yielding  6  % . 

12.  What  must  be  paid  for  stocks  yielding  7%  dividends, 
that  10%  may  be  realized  annually  from  the  investment? 

Solution.— Since  $7,  the  annual  income  on  1  share,  must  be  10^  of  the 
cost  of  1  fhare,  t\j  of  $7,  or  70  ct.,  is  1^.  Hence  100^,  or  70  ct.  x  100=$70,  is 
the  amount  that  must  be  paid  for  the  stock. 

What  must  be  paid  for  stocks  yielding 

18.  5%  dividends  to  obtain  an  annual  income  of  8%  ? 

14.  7%  dividends  to  obtain  an  annual  income  of  12%  ? 

15.  9%  dividends  to  obtain  an  annual  income  of  7%  ? 

16.  How  much  currency  can  be  bought  for  $350  in  gold, 
when  the  latter  is  at  13  %  premium  ? 

Solution.— Since  $1  in  gold  is  worth  11.12  in  currency,  $350  In  gold  are 
equal  $1.12  X  350  =  $392. 

How  much  currency  can  be  bought 

17.  For  $780  in  gold,  when  it  is  at  a  premium  of  9%  ? 

18.  For  $396  in  gold,  when  it  is  at  a  premium  of  IS^  %  ? 

19.  For  $530  in  gold,  when  it  is  at  a  premium  of  13^%  ? 

30.  How  much  is  $507.50  in  currency  worth  in  gold,  the 
latter  being  at  a  premium  of  12^%  ? 

Solution.— Since  $1  of  gold  ie  equal  to  $1.12}  in  currency,  $507.50  in  cur- 
rency muBt  be  worth  as  many  dollars  in  gold  as  $1.12^  is  contained  times 
in  $507.50,  which  is  $451.1i;. 


STOCKS, 


227 


ig  9f<'. 


umber  of 
at  111*  a 


;12%. 

lividends, 

e  m  of  the 
100=170,  is 


0  in  gold, 


in  gold  are 


9%? 

gold,  tlie 


1507.50  in  cur- 
Utained  timeti 


How  much  gold  can  be  bought 

21.  For  f  1053. 17  currency,  when  gold  is  at  a  premium  of 
91  %  ? 

32.  For  $317.47  currency,  when  gold  is  at  a  premium 
of  llg/cV 

23.  For   $418.14   currency,    when   gold   is   at   a  premium 

of  \'il7oi 

24.  Bought  80  shares  in  the  Merchants'  Bank  at  a  discount 
of  2,y/(',  and  sold  the  same  at  an  advance  of  Vi'/h  ;  what  did 
I  pain?  Ans.  JJlllOO. 

25.  An  agent  sells  415  barrels  of  fl(mr,  at  $6  a  barrel,  com- 
mission 5%,  and  invests  the  proceeds  in  stocks  of  the  SuflTolk 
Bank,  Boston,  at  17]  ;^c  discount,  brokerage  |%  ;  how  many 
shares  did  he  buy  V 

26.  Bought  84  shares  in  the  Royal  Canadian  Insurance  Com- 
pany, at  I'/c  discount,  and  sold  them  at  6]'^;  advance;  what 
was  my  profit,  the  brokerage  in  buying  and  selling  being  \  per 
cent  ? 

27.  Bought  bonds  at  70%,  bearing  4^^  interest ;  what  is  the 
rate  of  income  ?  Ana.  %%. 

28.  I  invest  $2397  50  in  a  publishing  company's  stock,  whose 
shares,  worth  $50  each,  are  sold  at  $43.50,  brokerage  |%  ;  What 
annual  income  shall  I  d(>rive,  the  stock  yielding  7%  ? 

29.  0.  E.  Bonney  sold  $0000  Pacific  Railroad  G's  at  107,  and 
with  a  part  of  the  proceeds  bought  St.  Lawrence  County  b(jnds 
at  90,  yielding  6%  dividends  sufficient  to  give  an  annual  income 
of  8180 ;  how  much  has  ho  left  ? 

30.  What  rate  of  income  can  be  derived  from  money  invested 
in  the  stock  of  a  company  paying  a  semi-annual  dividend  of 
5%,  purchased  at  84V %,  brokerajje  i%  ? 

31.  What  must  I  pay  for  bonds  yielding  4J  %  annually,  that 
my  investment  may  ])ay  6^  ? 

32.  What  must  be  paid  for  stocks  paying  5  per  cent,  that  the 
investment  may  return  8^  ? 

33.  A  man  bought  a  farm,  giving  a  note  for  $3400,  payable 
in  gold  in  5  years ;  at  the  expiration  of  the  time  gold  waa 
175%  :  what  did  his  farm  cost  in  currency  ? 


I'V      I 


.«fjs':: 


228 


BUSI^'ESS     ARITHMETIC, 


DUTIES    OR    CUSTOMS. 

5!$2.  Duties  or  Customs  are  taxes  levied  by  the  govern- 
ment upon  imported  goods. 

533.  A  Sjiecific  Duty  is  a  certain  sum  imposed  upon  an 
article  without  regard  to  its  value. 

534.  An  Ad  Valorem  Duty  is  a  per  cent  assessed 
upon  the  value  of  an  article  in  the  country  irom  which  it  is 
brought. 

535.  A  Tariff  is  a  schedule  giving  the  rates  of  duties 
fixed  by  law. 

536.  The  following  deductions  or  allowances  are  made 
before  computing  specific  duties : 

1.  Tare.— An  allowance  for  the  box,  cask,  bag,  etc.,  containing  the 
merchaudiBe. 

2.  T,eakaffe.—An  allowance  for  waste  of  liquors  imported  in  cas^lts  or 
barrels. 

8.  Breakage.— An  allowance  for  loss  of  liquors  imported  in  bottles. 


r"  I! 


EXAMPLES     FOR     PRACTICE. 

537.  1.  What  is  the  duty  on  420  l)oxos  of  raisins,  each 
containing  40  pounds,  bought  for  8  ceyits  a  pound,  at  20  per  cent 
ad  valorem  ? 

2.  Imported  21  barrels  of  wine,  each  containinpr  31  gfallons  ; 
2%  being  allowed  for  leakage,  what  is  the  duty  at  40  cents  per 
gallon  ? 

3.  A  merchant  imported  from  Havana  100  boxes  oranges  @ 
$2.25  per  box ;  75  hogsheads  of  molasses,  each  containing  63  gal., 
@  23  cents  i^er  gal.  ;  50  hogsheads  of  sugar,  each  containing 
340  lb.,  @  6  cents  per  lb.  The  duty  on  the  molasses  was  25%, 
on  the  sugar  30%,  and  on  the  oranges  20%.  What  was  the 
duty  on  the  whole  ? 


UK  VIEW, 


220 


4.   What  i.H  the  duty  on  320  yards  of  cloth,  invoiced  at  |1.15 
|)er  yard,  at  20 ;c  ad  valorem? 


^overn- 

,pon  an 

aBsoBScd 
ich  it  is 

)f  duties 

re   made 

talnlng  tbe 
in  cat^kB  or 
bottles. 


isins,  each 
iO  X)er  cent 

|l  gallons  ; 
cents  per 

oranges  (a 

ling  63  gal., 
[containing 
was  25%, 
lat  was  the 


5. 


At  VZ'/o  ad  valoroiu,  what  is  the  duty  on  100  barrels  of 


kerosene,  invoiced  at  $.18  a  gallon,  2%  leakage? 


BEVIEW    AND    TEST    QUESTIONS. 

5J58.  1.  When  a  fraction  is  to  be  divided  by  a  fraction, 
why  can  the  factors  that  are  common  to  the  denominators  of 
the  dividend  and  divisor  bo  cancelled  ? 

2.  How  does  moving  the  decimal  point  one  or  more  places  to 
tlio  left  or  right  affect  a  number,  and  why? 

3.  Sliow  that  multiplying  by  1000  and  Hubtracting  three 
times  the  multiplicand  from  the  product  is  the  same  as  multi- 
plying by  997  . 

4.  Define  Base,  Percentage,  Amount,  and  D  Terence. 

5.  When  the  amount  and  rate  per  cent  is  given  to  find  the 
base,  why  add  the  rate  expressed  decimally  to  1  and  divide  by 
tlip  result . 

6.  Represent  the  quantities  by  lettt  s  and  ^^  rite  a  formula 
for  solving  each  of  the  following  problems  (41>7) : 

I.  Given,  the  Cost  and  the  Profit,  to  find  the  rate  per 
cent  profit. 
II.  Given,  the  rate  per  cent  profit  and  the  selling  price,  to 
find  the  buying  price. 

III.  Given,  the  amount  of  money  sent  to  an  agent  to  pur- 

chase goods  and  the  rate  per  cent  commissio.i,  to 
find  the  amount  of  the  purchase. 

IV.  Given,  the  rate  at  which  stocks  can  be  purchased,  to 

find  how  much  can  be  secured  for  a  given  sum. 
V.  Given,  the  rate  at  which  stocks  can  be  purchased  and 
the  rate  per  cent  of  dividend,  to  find  the  rate  per 
cent  of  income  on  the  investment. 
VI.  Given,  the  premium  on  gold,  to  find  how  much  can 
be  purchased  for  a  given  sum  in  currency. 


h-  i\ 


If, 


INTEREST. 


i»:?;i 


DEFINITIONS. 

539.  Interest  is  a  sum  paid  for  the  ?/««  of  motiey 

Thus,  I  owe  Wm.  Henry  $200,  which  he  allows  me  to  nee  for  one  year 
after  it  in  due.  At  the  end  of  the  year  I  pay  hira  Ihe  $200  and  $14  for  its 
use.    The  $14  is  called  the  Interest  and  the  $200  the  Principal. 

540.  Principal  is  a  sum  of  money  for  the  use  of  which 
interest  is  paid. 

541.  Rate  of  Interest  is  the  numlierof  units  of  any 
denomination  of  money  paid  for  the  use  of  100  units  of  the 
same  denomination  for  one  year  or  some  given  interval  of  time. 

542.  The  Amount  i?  the  sum  of  the  principal  and  interest. 

543.  Simple  Interest  is  interest  which  falls  due  when 
the  principal  is  paid,  or  when  a  partial  payment  is  made. 

544.  Legal  Interest  is  interest  reckoned  at  the  rate  per 
cent  fixed  by  law, 

545.  Usury  is  interest  reckoned  at  a  higher  rate  than  is 
allowed  by  law. 

540.  The  legal  rate  of  interest  for  Canada  is  six  per  cent ; 
for  England,  five  per  cent ;  and  lor  Ireland,  six  per  rent. 

The  following  tahle  ^ivea  the  lc<;al  rates  of  interest  in  the  different 
States  of  the  U.S. 

Where  two  rates  are  given,  any  rate  between  these  limitK  in  allowed,  if 
apecifled  in  wiiting.  When  no  rate  is  named  in  a  paper  involving  interest, 
the  UgcU  or  lowest  rate  is  alwaye  nndorstood. 


SIMPLE    INTEREST, 


231 


•  one  yew 

iU  for  itB 

jf  whicb 


ts  of  any 
its  of  the 
I  of  time. 

Id  interest. 

due  wben 
[ade. 

le  rate  pet 


ite  than  is 

per  cent ; 
'ent. 
the  different 

u  allowet^  if 
Iving  Interest. 


STATES.    RATE  %.     STATES.   RATB  %. 


Ala..  ..| 

Aik....| 


8i         I 

(i  Any, 

10  Any' 


I 

Cal 

.10  Any  1 

■             iCuiiii... 

'  1        i 

1          c.i.... 

10  Anv 

1 

Dnkota. 

7  Anyl 

1 

Do!  ... . 

6 

1 

D.C.... 

6    10 

■             Flor.  .. 

8  Any 

■             Goo... 

7     10 

1 

Idabc* . . 

10, 

lU 

lud... 
Iowa . . 
Kan... 
Ken. . . 
La..  .. 
Maine. 
Md. . . . 
Ma^s . . 
Mich.. 
Minn.. 
Mibs.  . 


6 
(i 
6 
7 
6 
5 
6 
6 
6 
7 
7 
6 


10  : 

10 
10 
12  i 

10  , 
8  ' 
Any! 
I 
Any  ; 
10 
12 
10  ! 


STATES. 

RATE  %.  \ 

1 

STATES. 

RATEiC 

Mo 

6 

10 

S.C... 

1    7 

Any 

Montana 

10 

,  Tenn. 

I    ^ 

10 

N.H.... 

6 

1  Texas. . 

^   8 

12 

N.  J.... 

7 

Utah... 

no 

Any 

N.  Y.... 

7 

Vt 

6 

X.  C... 

6 

8 

Va 

1 
6 

12 

Neb 

10 

15  1 

i  W.Va.. 

6 

1  Nevada . 

10 

Any| 

W.  T... 

10 

Any 

;01iio....i    6 

8 

Wis.... 

7 

10 

Ore^'on  .,  10  ,  12  , 

\Vy..  .. 

12 

i  Penii....|   6  i    7  1 

|Ri 

1  e 

Any 

54:7.  PiioB.  I. — To  find  the  simple  interest  of  any 
given  sum  for  one  or  more  years. 

1.  Find  the  interest  on  $384  for  5  years,  at  7%. 

Solution.— 1.  Since  the  intcreet  of  $100  for  one  year  is  $7,  the  Interest 
of  $1  for  one  year  is  $.07.  Hence  the  interest  of  $1  for  5  years  is  |,07  x  5 
^  $.35. 

2.  Since  the  interest  of  $1  for  h  yr.  i?  $.35,  the  interest  of  13^4  for  the 
same  time  must  ho  384  times  $.35,  or  J134.40.    Hence  the  following 

RUIiE. 

548.  /.  Fetid  the  interest  of  $1  at  the  given  rate  for  the 
given  time,  and  multiply  this  resnU  by  the  number  of  dollars  in 
(he  glccn  principal. 

II.  To  find  the  amount,  add  the  interest  and  principal. 

EXAMPLES     FOR    PRACTICE. 
540.  Find  the  interest  on  the  tollowing  oraDy  : 


1.  11200  for  3  years  at  d%. 

2.  $800  for  2  years  at  4ji. 
:{.  $200  for  5  years  at  6^. 
4.  $90  for  2  years  at  7%. 
■».  $600  for  4  years  at  5%. 
0.  ?;T0  for4  youwatS^. 


7.  $100  for  12  years  at  9%. 

8.  $400  for  8  years  at  5%. 

9.  $1000  for  5  years  at  8%. 

10.  .*;G00for  lyeursat  10%. 

1 1 .  $500  for  5  yj'ars  at  o  % . 

12.  $20  for  3  years  at  9%. 


"i      -, 


232 


BUSINESS     ARITHMETIC, 


i.'.'i  ■ 


Find  the  interest  on  the  following  : 

13.  $784.25  for  9  years  at  4$^ .  20.  !s:293.50  for  6  years  at  45  % 

14.  $245.36  for  3  years  at  7/o.  21.  .$375.84  for  3  years  at  9J%. 

15.  $836.95  for  2  years  at  ^%.  22.  $600.80  for  9  years  at  8^%. 
IG.  $705.86  for  7  years  at  «  % .  23.  $899.00  for  12  years  at  7|  % . 

17.  $28.95  for  1 A  yr.  at  4?  % .  24.  $50.84  for  5  years  at  1?  % . 

18.  $896.84  for  3^  yr.  at  2^  % .  25.  $262.62  for  6  years  at  6,\  % . 

19.  $414.14  for  4  years  at  |%.  26.  $95.60  for  j  year  at  71%. 


m 


METHOD    BY    ALIQUOT    PABTS. 

550.  Prob.  II. — To  find  the  interest  on  any  sum  at 
any  rate  for  years,  months,  and  days  by  aliquot  parts. 

1.  In  business  transactions  involving  interest,  30  days  are 
usually  considered  one  month,  and  12  months  one  year.  Hence 
the  interest  for  days  and  months  may  be  found  according  to 
(488),  by  regarding  the  time  as  a  compound  number  ;  thus. 

Find  the  interest  and  amount  of  $840  for  2  yr.  7  mo.  20  da., 
at  7%. 

$840    Principal. 
.07    Rate  of  Interest. 


6  mo. 

=1  of  1  yr.,  hence 

2)  58.80 
2 

Interest  for  1  yr. 

117.60 

Int.  for  2  yr. 

1  mo. 

= J  of  6  mo.,  hence 

6)  29.40 

"      6  mo. 

15  da. 

=1  of  1  mo.,  hence 

2)    4.90 

"       1  mo. 

6  da. 

=1  of  15  da.,  hence 

3)    2.45 

"     15  da. 

.811 

••       5  da. 

$155.16f 

2  yr.  7  mo.  20  da 

840.00 

Principal. 

5.16|  Amt.  for  2  yr.  7  mo.  20  da. 

IS51.  The  interest,  by  the  method  of  aliquot  parts,  is  usually 
found  by  finding  first  the  interest  of  $1  for  the  given  time,  and 


SIMPLE    INTEREST . 


233 


i  4^,  % 

,11%. 
It  6^%. 


sum  at 
rts. 

days  are 
..  Hence 
lording  to 
r ;  tiius, 

10.  20  da., 


multiplying  the  given  principal  by  the  decimal  expressing  the 
interest  of  $1 ;  thus, 

Find  the  interest  of  $G80  for  4  yr.  9  mo.  15  da.  at  8^ . 

1.  We  first  find  the  iutereet  of  $1  for  the  given  time ;  thus, 

8  ct.    =•  lut.  of  $1  for  1  yr.,  8  ct.  X 4    =  Int.  for  4  yr.    =32  ct. 

6mo.  =  J  of  1  yr.,  hence,     Jof8ct=        "      6mo.  =     4ct. 

8  mo.  =  J  of  6  mo.,    "  io/'4ct.  =        "      3mo.  =     2  ct. 

15  da.   =iof3mo.,    "  iof2ct.  =        "    15  da.   =  .03^  m. 

Hence  the  interest  on  $1  for  4  yr.  9  mo.  15  da.  =  $.3835. 

2.  The  decimal  .383J  expresses  the  part  of  $1  which  is  the  intercBt  of  |1 
for  the  given  time  at  the  given  rate.  Hence,  $680  x  .3831  =  $260,663,  la  the 
interest  of  $680  for  4  yr.  U  mo.  15  da.,  at  8^ ;  hence  the  following 


RULE. 

552.  /.  Find  by  aliquot  parts  tlie  interest  of  S 1  f 07' t?ie  given 
rate  and  time. 

II.  Multiply  the  principal  hy  the  decimal  expressing  the  inter- 
est jcr  $1,  and  the  product  will  be  the  required  interest. 

II '".  To  find  the  amount,  add  t/ie  interest  to  the  principal. 


It. 


lino. 


20  da. 


'  mo.  20  da. 

L  is  usually 
In  time,  and 


EXAMPLES    FOR    PRACTICE. 
553.  Find  the  interest 

1.  Of  $560.40  for  2  yr.  10  mo.  18  da.  at  7%  ;  at  9%. 

2.  Of  1284  for  3  yr.  8  mo.  12  da.  at  6%  ;  at  8J%. 

3.  Of  $296.85  for  4  yr.  11  mo.  24  da.  at  8%  ;  at  5%. 

4.  Of  $2940.75  for  3  yr.  11  mo.  17  da.  at  7%  ;  at  %\%, 

5.  Of  $860  for  1  yr.  7  mo.  27  da.  at  ^%  ;  at  7^%. 

G.  Find  the  amount  of  $250.70  for  2  yr.  28  da.  at  8%. 

7.  Find  the  amount  of  $38.90  for  3  yr.  13  da.  at  9%. 

8.  Paid  a  debt  of   $384.60,  which  was  upon   interest  for 
11  mo.  16  da.  at  7%.     Wliat  was  the  amount  of  the  payment? 

9.  A  man  invested  $795  at  8%  for  4  yr.  8  mo.  13  da.     How 
much  was  the  amount  of  principal  and  interest  ? 

10.  Find  the  amount  of  $1000  for  9  yr.  11  mo.  29  da.  at  7%. 

16 


234 


BUSINESS     ARITHMETIC, 


W: 


METHOD    BY    SIX    PER    CENT. 

PREPARA.TORY    STEPS. 

554.  Step  I.— To  find  the  interest  for  any  number  of 
months  at  6%. 

1.  Since  the  intereet  of  $1  for  12  months,  or  1  yr.,  at  6%,  is 
6  cents,  the  interest  for  two  months,  which  is  J  of  12  mouths, 
must  be  1  cent,  or  y^  part  of  the  principal. 

2.  Since  the  interest  for  2  months  is  -^l-^  of  the  principal,  the 
interest  for  any  number  of  months  will  be  as  many  times  ^^^ 
of  the  principal  as  2  is  contained  times  in  the  given  number  of 
months.     Hence  the  following 


RULE. 

555.  /.  Move  the  decimal  point  in  the  principal  two 
places  to  tJie  left  (459),  prefixing  ciphers,  if  necessary. 

11.  Multiply  this  result  by  one-half  the  number  of  months. 

Or,  Multiply  y^^^  of  the  principal  by  the  number  of  months  and 
divide  the  result  by  2. 


M 


is 

r 
t 


EXAMPLES     FOR    PRACTICE. 
5.">0.  Find  the  interest  at  6% 

1.  Of  $973.50  for  10  mo. 

2.  Of  !?896  for  8  mo. 

3.  Of  $486.80  for  18  mo. 

4.  Of  .$432.90  for  13  mo. 

5.  Of  $304.40  for  7  mo. 


6.  Of  $398  for  1  yr.  6  mo. =18  mo. 

7.  Of  $750  for  2  yr.  8  mo. 

8.  Of  $268  for  2  yr.  6  mo. 

9.  Of  $186  for  4  yr.  2  mo. 
10.  Of  $873  for  1  yr.  11  mo. 


557.  Step  II.— To  find  the  interest  for  any  number  of  days 
at  0%. 

1.  Since  the  interest  of  $1  for  2  months  at  6%  is  1  cent,  the 
interest  for  1  month,  or  30  days,  must  be  ^  cent  or  5  mills. 
And  since  0  days  are  I  of  30  days,  the  interest  for  6  days  must 
be  1  of  5  mills,  or  1  mill,  which  is  y^'^^  of  the  principal. 


SIMPLE     INTEREST, 


235 


Ur  of 


0, 


IS 


aaouths, 

ipal,  the 
imes  T^TJ 
imber  of 


2.  Since  the  inte.ost  for  6  days  is  y^^^y  of  the  principal,  the 
interest  for  any  number  of  days  will  be  as  many  times  j^Vtt  of 
the  principal  as  6  is  contained  times  in  the  given  number  of 
days.     Hence  the  following 

RDTiE. 

*>58.  /.  Move  the  decimal  point  in  the  principal  three 
2*1(1  cos  to  the  left  (459),  prefixing  ciphers,  if  necessary. 

II.  Multiply  this  result  by  one-sixth  the  number  of  days. 

Or,  Midtiply  jtjVtf  ^f  ^^^  principal  by  the  number  of  days  and 
divide  the  result  by  0. 


EXAMPLES    FOR    PRACTICE. 


ipal  ttvo 

ry. 

onths. 
imths  and 


559.  Find  the  interest  at  6  % 

1.  Of  $384  for  24  da. 

2.  Of  $790  for  12  da. 

3.  Of  $850  for  15  da. 

4.  Of  ^mo  for  16  da. 

5.  Of  $935  for  27  da. 


6.  Of  .$584  for  19  da. 

7.  Of  $809  for  28  da. 

8.  Of  $730  for  22  da. 

9.  Of  $840  for  14  da. 
10.  Of  ^396  for  17  da 


10. =18  mo. 

10. 
10. 
10. 

|mo. 
iberofdays 

1  cent,  the 

[or  5  mills- 

,  days  must 

cipah 


560.  Prob.  III. — To  find  the  interest  on  any  sum  at 
any  rate  for  years,  months,  and  days,  by  the  six  per 
cent  method. 

Find  the  interest  of  $542  for  4  years  9  months  17  days  at  8 
per  cent. 

RoLTTTicN  — 1.  The  interest  of  $542  for  4  years  at  6%,  according  to  (547), 
is  1542  X  .06  X  4  =  $130.08. 

2.  The  interest  for  9  months,  according  to  (554),  is  tJo  of  $542,  or  $5.42 
mnltiplied  by  9.  and  this  product  divided  by  2  =  $24.39. 

3.  The  interest  for  17  days,  according  to  (567),  is  tj^oo  of  $542,  or  $.512 
multiplied  by  17,  and  this  product  divided  by  6  =  $1.535  + . 

Hence  $130.08 +  $24.39  + $1.54  =  $156.01,  the  interest  of  $542  for  4  years 
fl  months  and  17  days. 

4.  Having  found  the  interest  of  $542  at  65«,  to  find  the  interest  at  8;^  we 
have  8^  =  6r;  +  2t,  and  2%Ul  of  6^.  Hence,  $156.01  +  i  of  $156.01  =  $208,013. 
the  interest  of  $542  at  6%  for  4  yr.  9  mo.  17  da. 


■BHI" 


236 


BUSI^'^ESS    ARITHMETIC, 


I't: 


EXAMPLES     FOR     PRACTICE. 
561.  Find  the  interest  by  the  6%  method 

f .  Of  $890.70  for  4  yr.  10  mo.  15  da.  at  7%  ;  at  10%  :  at  4% 

2.  Of  .$384.96  for  2  yr.  8  mo.  12  da.  at  6fc  ;  at  9%  ;  at  8%. 

3.  Of  $280.60  for  11  mo.  27  da.  at  8;^  ;  at  4%  ;  at  7%. 

4.  Of  $890  for  9  mo.  13  da.  at  6i%  ;  at  8^%  ;  at  9^%. 

5.  Of  $480  for  2  yr.  7  mo.  15  da.  at  9%  ;  at  12%  ;  at  4|%. 


a 


METHOD    BY    DECIMALS. 

5G12.  In  tliis  method  the  time  is  regarded  as  a  compound 
number,  and  the  months  and  days  expressed  as  a  decimal  of  a 
year. 

When  the  principal  is  a  small  sum,  suflBcient  accuracy  will 
be  secured  by  carrying  the  riocimal  to  three  places ;  but  when 
a  large  sum,  a  greater  number  of  decimal  places  should  be 
taken. 


^ 


50>{.    Prob.  IV. — To  find  the  interest  on  any  sum  at 
any  rate  for  years,  months,  and  days,  by  decimals. 

What  is  the  amount  of  $450  for  5  yr.  7  mo.  16  da.,  at  6%  ? 

Explanation.  —  1.  We  express, 
according  to  (378—15),  the  days  and 
months  as  a  decimal  of  a  year,  ap 
shown  in  (1). 

2.  We  find  the  interest  on  $450, 
the  given  principal,  for  1  year,  which 
l8  $27,  an  shown  in  (2). 

8.  Since  |97  is  the  interest  on  $450 
for  1  year,  the  Interest  for  5.627  years 
is  5.627  limes  $27,  which  is  $151,929, 
ae  shown  in  (1.) 

4.  The  amount  Is  equal  to  the 
principal  pins  the  interest  (542); 
hence,  $151.93+$450  =  |()01.93iBtb& 
amount.    Hence  the  following 


(1.) 

(2.) 

30  )  16        da. 

$450 

12)    7.533  mo. 

.06 

5.627  yr. 

$37.00 

27 

39  389 

113  54 

$151,929    Interest 

■ 

450 

$601.93      Amount. 


SIMPLE    INTEREST, 


237 


Lt4^ 


RULE. 

564.  /.  To  find  the  interest,  mnltiph/  the  principal  by  the 
rate,  and  this  product  by  the  time,  expressed  in  years  and  deci- 
mals of  a  year. 

II.  To  find  the  amount,  add  the  interest  to  the  principal. 


f  k'.  ■ 


^% 


'0* 


mpound 
aial  of  a 

racy  ^^^ 
put  wlien 
hould  be 


sum  at 

Lis. 

/e  express, 
Jhe  days  and 
a  year,  a? 

Ut  on  $450, 
|year,wWch 

Lrest  on  |450 
Ir  5.627  years 
1  is  $151,921), 

Iqual    to  tlic 
lre«t  (542); 

1601.93 19  the 

lowing 


EXAMPLES    FOR    PRACTICE. 

505.  Find  the  interest  by  the  decimal  method 

1.  Of  $374.05  for  2  yr.  9  mo.  15  da.  at  6%  :  at  9%  ;  at  4%. 

2.  Of  $200  for  I  yr.  8  mo.  12  da.  at  5/«,  ;  at  8%  ;  at  7^. 

3.  Of  $790.80  for  5  yr.  3  mo.  7  da.  at  7%  ;  at  11  %  ;  at  3%. 

4.  Of  $700  for  11  mo.  27  da.  at  ^yo  ;  at  7|%  ;  at  2f  %. 

5.  Of  $460.90  for  3  yr.  5  mo.  13  da.  at  6J  %  ;  at  8]  %  ;  at  d^fo. 

6.  Of  .^890  for  7  yr.  19  da.  nt  iiyi  ;  at  8%  ;  at  5%. 

7.  Of  $580.40  for  17  da.  at  &^'/o  ;  at  9A%  ;  at  5^%. 


EXACT    INTEREST. 

560.  In  the  foregoing  methods  of  reckoning  interest,  the 
year  is  regarded  as  300  days,  wliich  is  5  days  less  than  a  c^ym- 
man  year,  and  0  days  less  than  a  leap  year  ;  hence,  the  interest 
when  found  for  a  part  of  a  year  is  incorrect. 

Thus,  if  tho  interept  of  $100  is  |7  for  a  common  year  or  865  days,  tlie 
interet<t  for  75  dayn  at  the  t^anie  rate  mupt  be  //j  of  $7 ;  but  1)y  the  foreijoinff 
method  aVr.  of  $7  is  taken  as  the  interest,  which  is  too  great. 

Obnerve,  that  in  ueing  3V5  in><tead  of  3'/^,  the  denominator  is  diniinieihed 
s«i  =  A  part  of  itt»clf,  and  consequently  (319)  the  reisult  in  f\  part  of  Itself 
too  j^reat. 

Hence,  when  interest  is  calculated  by  the  forcp^oing  methods,  it  must  be 
diminished  by  ^^  of  itself  for  a  common  year,  and  for  like  reasons  ,'j  of  itself 
for  a  leap  year. 

To  find  the  exact  interest  we  have  the  following 

RULE 

507.    1.  Find  the  interest  for  the  given  nnmber  of  years. 
n.  Find  the  exact  number  of  days  in  the  given  months  and 


S*  1 


If — 


m 


sa 


■J.   ul' 


238 


B  USINESS    ARITHMETIC, 


days,  and  take  audi  a  part  of  the  interest  of  the  principal  for  one 
year,  as  the  whole  number  of  days  is  of  JO  J  days. 

Or,  Find  the  interest  for  tlie  given  months  and  days  by  either 
of  the  foregoing  methods,  then  subtract  ,V  pO'Tt  of  itself  for  a 
common  year,  and  /j  for  a  leap  year. 

III.  Add  tlie  result  to  tlie  interest  for  the  given  number  of 
years. 

EXAMPLES    FOR    PRACTICE. 

5(58.  Find  the  exact  interest  by  both  rules 

1.  Of  $260  for  55  da.  at  8%.  5.  Of  $2360  for  7  da.  at  7^%. 

2.  Of  $836  for  84  da.  at  6% .  6.  Of  $380  50  for  93  da.  at  6^  % . 

3.  Of  $985  for  13  da.  at  9%.  7.  Of  $120  for  133  da.  at  8|%. 

4.  Of  $090  for  25  da.  at  7% .  8.  Of  $260.80  for  17  da.  at  12 % . 

9.  What  is  the  diflference  between  the  exact  interest  of  $896 
at  7J^  from  January  11,  1872,  to  November  19,  1870,  and  the 
interest  reckoned  by  the  six  per  cent  method  ? 

10.  Required  the  exact  interest  of  $385.75  at  7  % ,  from  Jan- 
uary 15,  1875,  to  Aug.  23  following. 

11.  A  note  for  $360.80,  bearing  interest  at  8%,  was  given 
March  1st,  1873,  and  is  due  August  23,  1876.  How  much  will 
be  required  to  pay  the  note  when  due  ? 

12.  What  is  the  exact  interest  of  $586.90  from  March  13  to 
October  23  of  the  same  year,  at  7;^  ? 

5(>0.  Prob.  v.— To  find  the  principal  when  the  inter- 
est, time,  and  rate  are  given. 

Observe,  that  the  interent  of  any  principal  for  a  given  time  at  a  given 
rate,  is  the  interest  of  $1  taken  (547)  as  many  times  as  there  are  dollars  in 
the  principal ;  hence  the  following 


RULE. 

570.  Divide  the  given  interest  by  the  interest  of  $1  for  the 
given  tim£  at  the  given  rate. 


one 

ither 
for  a 

jr  of 


It  12%. 

)f  $89& 
ind  the 


)ni 


Jan- 


is  given 
[icli  will 

li  13  to 


Ic  inter- 


SIM  r  LE     I  y  T  E  li  E  S  T, 


EXAMPLES     FOR     PRACTICE. 


239 


671.  1.  What  sum  of  money  will  gain  $110.25  in  3  yr. 
9  mo.  at  7%  ? 

Solution.— The  interest  of  $1  for  3  yr.  9  mo.  at  1%  is  1.3625.  No\v  since 
$.20-25  \9  the  interest  of  $1  for  the  given  time  at  the  given  rate,  $110.25  is 
the  interest  of  as  many  dollars  for  the  same  time  and  rate  as  $.2()25  is 
contained  times  in  1110.25.  Hence  $110.36  -»-  .SinaSt  =  $420,  the  required 
principal. 

What  principal  or  sum  of  money 

2.  Will  gain  $63,488  in  2  yr.  9  mo.  16  da.  at  8%  ? 

3.  Will  gain  $95,456  in  3  yr.  8  mo.  25  da.  at  7%  ? 

4.  Will  gain  $106,611  in  3  yr.  6  mo.  18  da.  at  6i  %  ? 

5.  Will  gain  $235,609  in  4  yr.  7  mo.  24  da.  at  9%  ? 

6.  Will  gain  $74,221  in  2  yr.  3  mo.  9  da.  at  1\%  ? 

7.  WIU  gain  $30,636  in  1  yr.  9  mo.  18  da.  at  ^%  ? 

672.  Prob.  VI. —To  find  the  principal  when  the 
amount,  time,  and  rate  are  given. 

Observe,  that  the  amount  is  the  principal  plus  the  interest,  and  tliat  the 
interest  contains  the  interest  (547)  of  $1  as  many  times  as*  tliere  are 
dollars  in  the  principal ;  consequently  the  amount  must  contain  (495)  $1 
plus  the  interest  of  $1  for  the  given  time  at  the  given  rate  aa  many  times  as 
there  are  dollars  in  the  principal ;  hence  the  following 


BIJLE. 

673.  Divide  the  amount  by  the  amount  of  SI  for  the  given 
time  at  the  given  rate. 


I! 


i.^ 


It  a  given 
[dollars  in 


for  the 


EXAMPLES    FOR    PRACTICE. 

674.  1.  What  sum  of  money  will  amount  to  $290.50  in 
2yr.  8mo.  12  da.  at  6%  ? 

Solution.— The  amount  of  $1  for  2  yr.  8  mo.  12  da.  at  6;?  is  $1,162.  Now 
since  $1,163  is  the  amount  of  $1  for  the  given  time  at  the  given  rate, 
$290.50  is  the  amount  of  as  many  dollars  a5>  $1.16-2  is  contained  times  in  it. 
Hence,  $290.50  ■*■  $1,162  =  $250,  is  the  required  priucipoL 


240 


BUSINESS    ARITH3IETIC, 


2.  What  is  the  interest  for  1  yr.  7  mo.  13  da.  on  a  sum  of 
money  which  in  this  time  amouuts  to  $487.65,  at  7^  ? 

3.  What  principal  will  amount  to  $310.60  in  3  yr.  6  mo. 
9  da.  at  5%  V 

4.  At  8%  a  certain  principal  in  2  yr.  9  mo.  6  da.  amounted  to 
$09982.     Find  the  principal  and  the  interest. 

5.  What  Bimi  of  money  at  10^  will  amount  to  $436.02  in 
4yr.  8  mo.  23  da.? 

575.  Prob.  VII. — To  find  the  rate  when  the  principal, 
interest  and  time  are  given. 

Observe^  that  the  given  interest  mupt  be  as  many  times  1%  of  the  given 
principal  for  the  given  time  as  there  arc  onito  in  the  rate  ;  hence  the 
following 

BULB. 

57C$.  Divide  the  given  interest  hy  the  interest  of  the  given 
principal  for  the  given  time  at  1  per  cent. 


EXAMPLES    FOR     PRACTICE. 

577.    1.  At  what  rate  will  $260  gain  $45.50  in  2  yr.  6  mo.  ? 

Solution.— The  interest  of  $260  for  2  yr.  6  mo.  at  \%  ie  $6.60.  Now 
since  $6.50  is  \%  of  $250  fur  the  given  time,  $45.50  is  as  many  per  cent 
as  $6.50  is  contained  times  in  $45.50 ;  hence,  $45.50  -«-  $6.50  =  7,  is  the 
required  rate. 


At  what  rate  per  cent 

2.  Will  $524  gain  $206.63  in  5  yr.  7  mo.  18  da.? 
8.  Will  $732  gain  $99,674  in  2  yr.  3  mo.  7  da.? 

4.  Will  $395.80  gain  $53,873  in  2  yr.  8  mo.  20  da.? 

5.  Will  $873  gain  $132.S9  in  1  yr.  10  mo.  25  da.? 

6.  Will  $908.50  gain  $325,422  in  4  yr.  2  mo.  17  da.  ? 

7.  A  man  purchased  a  house  for  $3186,  which  rents  for 
$418  32.  What  rate  per  cent  does  he  make  on  the  invest- 
ment? 

8.  Which  is  the  better  investment  and  what  rate  per  cent 


••  ♦ 


am  of 
5  mo. 
ntedto 
i6.02  in 

incipal, 


the  given 
hence  the 


tU  given 


r.  6  mo.? 

.50.  No^ 
Iny  per  cent 
1=  7,  Is  the 


SIMPLE    INTEREST, 


241 


per  annum,  i?4:3G0  which  yields  in  5  years  $1635,  or  fa860 
which  yields  in  9  years  $2692.45  ? 

9.  At  wliat.  rate  jht  cent  per  annum  will  a  sum  of  money 
donl)l(;  itself  in  7  years  ? 

Solution.— Since  iu  100  years  at  1%  any  Bum  doubles  itself,  to  double 
Ittx'lf  iu  T  yoais  the  rate  per  cent  must  be  a»  many  times  lj{  as 7  is  contained 
times  in  100,  which  is  14j^.    Ilence,  etc. 

10.  At  what  rate  per  cent  per  annum  will  i\ny  sum  double 
itself  in  4,  8,  9,  12,  and  25  years  n^spcclivcly  ? 

11.  Invested  }*^3648  in  a  business  that  yields  $1659.84  in  5 
years.  What  per  cent  annual  interest  did  I  receive  on  my 
investment  ? 

12.  At  what  rate  per  cent  ])er  annum  will  any  sum  triple  or 
quadruple  itself  in  6,  9,  14,  and  18  years  respectively  ? 

r>78.  Prob.  VIII.— To  find  the  time  when  the  princi- 
pal, interest,  and  rate  are  given. 

Observe,  that  the  interest  it?  found  (563)  by  multiplying  the  intercfit  of 
the  given  i)rincipal  for  1  year  at  the  given  rote  by  the  time  expropscd  in 
years ;  hence  the  following 

BULE. 

579.  /.  Divide  the  given  interest  hy  the  interest  of  the  given 
jtrincipal  fur  I  year  at  the  given  rate. 

II.  Rcdi'cc  {"AVI),  when  called  for,  fractions  of  a  year  to 
months  and  days. 


•It 


t 


!'.h      1( 


rents  foT 
le  invest- 

per  cent 


EXAMPLES     FOR    PRACTICE. 
580.     1.  In  what  time  will  $350  gain  $63  at  8%  ? 

Solution.— The  interest  of  $350  for  1  yr.  at  8%  is  $28.  Now  since  $28  is 
tlie  interest  of  ^S.jO  at  B:i  for  1  year,  it  will  take  as  many  years  to  gain  $63 
M  $28  is  contained  times  in  $63  ;  hence  $63  -t-  $28  =  2}  yr.,  or  2  yr.  3  mo., 

tlic  required  time. 

In  what  time  will 

2.  $460  gain  $80.50  at  5%  ? 

3.  $80  jfaiu  $UGat7i%? 

4.  $260  gain  $96.80  at  Bfo  ? 


5.  $690  gain  $301,392  at  7f  %  ? 

6.  $477  f^aia  $152.64  at  12%  ? 

7.  $385  gain  $214.72  at  8f  %  ? 


i 

1 

: 

. 

! 

1 

i 

1 

ml 

hi- 

'  ■.)!.  ■ 


242 


B  USINESS    ARITHMETIC, 


f'l 


8.  My  total  gain  on  an  investment  of  $8G0  at  1%  per  annam, 
is  $455.70.     How  long  has  the  investment  been  made  ? 

9.  How  long  will  it  take  any  sum  of  money  to  double  itself 
at  7%  per  annum? 

SoLUTioN.~At  100;(  any  enm  will  double  itself  In  1  year ;  hence  to  double 
itHuirat  1%  it  will  require  as  many  years  as  1%  is  coutuluud  tiuicb  iu  l(>j;>, 
which  in  14jf.    Hence,  etc. 

Obaervey  that  to  find  how  long  it  will  take  to  triple,  quadruple,  etc.,  any 
earn,  we  muat  take  200^,  300^,  etc. 

10.  At  7%  the  interest  of  $480  is  equal  to  5  times  the  prin- 
cipal.    How  long  has  the  money  been  on  interest  V 

11.  How  long  will  it  take  any  sum  of  moneyut5%,8%,6^%, 
or  9%  per  annum  to  double  itself?    To  triple  itself,  etc.  ? 


R '1      I 


i: 


COMPOUND    INTEREST. 

581.  Cow  pound  Interest  is  interest  upon  principal 
and  interest  united,  at  given  intervals  of  time. 

Observe^  that  the  interest  may  be  made  a  part  of  the  principal,  or  com- 
pounded at  any  interval  of  time  agreed  ujton ;  as,  annually,  semi-annually, 
quarterly,  etc. 

5822.  PROB.  IX. — To  find  the  compound  interest  on  any 
sum  for  any  given  time. 

Find  the  compound  interest  of  $850  for  2  yr.  6  mo.  at  6%. 


$850     Prin.  for  lat  yr. 
1.00 

$901     Prin.  for  2tl  yr. 
1.06 


^ii|! 


$955.06  Prin.  for  6  mo. 
1.03 

$983.71  Total  amount. 

$850  Given  Prin. 

$133.71  Compound  Int. 


Explanation.— Since  at  6^  the  anupmf 
is  1.06  of  the  principal,  we  multiply  $850, 
the  principal  for  the  first  year,  by  1.06,  iri  v 
lug  $901,  the  amount  at  the  end  of  tlie  first 
year,  which  forms  the  principal  for  " 
second  year.    In  the  same  manner  u'«' 
$955.06,  the  amount  at  the        ) 
second  year  which  forms  the         ojpui . 
the  6  mouths. 

2.  Since  Q%  for  one  year  is  3K  to  6 
months,  we  multiply  $955.06,  the  prin*  tl 
for  the  6  months,  by  1.03,  which  gives  the 
total  amount  at  the  end  of  the  2  yr.  6  mc. 


anum, 
)  itself 

)  double 
iu  laj;.', 

etc.,  any 
le  prin- 


principal 


or  com- 
l-aunually. 


t  on  any 


at  6%. 

tho  anuynf 
Itiply  ?850, 
byl.O(i,i:iv- 

pal  for  •' 
\\»  9%  to^  ti 


Ihe  prin< 


A 


Ich  gives  the 
layr.  6mc\ 


COMPOUND     lyTEREST, 


243 


8.  From  the  total  amount  wo  sabtract  $850,  the  fjiven  principal,  which 
prives  im.Tl,  the  compound  Intcrctft  of  |»fiU  for  2  years  6  mouths  at  6,i. 
Hence  the  following 


RULE. 

58;{.  /.  Find  the  amount  of  the  principal  for  the  first 
intercal  of  time  (it  the  end  of  which  interest  is  due,  and  make  it 
the  principal  for  the  second  internal. 

II.  Find  the  amount  of  this  principal  for  the  second  interval 
of  time,  and  so  continue  for  each  successive  interval  and  fraction 
of  an  interval,  if  any. 

III.  Subtract  the  given  principal  from  t?ie  last  amount  and 
the  remainder  will  be  the  compound  interest. 

EXAMPLES    FOR    PRACTICE. 

584.  1.  Find  the  compound  interest  of  $380.80  for  1  year 
at  S^/c ,  interest  payable  quarterly. 

2.  Find  the  amount  of  $870  for  2  years  at  6^  compound 
interest. 

3  What  is  the  compound  interest  of  $650  for  3  years,  at  7^ , 
payable  annually  ? 

4.  What  is  the  amount  of  $1500  for  2  years  9  months  at  8% 
compound  interest,  payable  annually  ? 

5.  What  is  the  difference  in  the  simple  interest  and  compound 
interest  of  $480  tor  4  yr.  and  6  mo.  at  7  %  ? 

6.  What  is  the  amount  of  $600  for  1  year  9  months  at  5  % 
compound  interest,  payable  quarterly? 

7.  What  is  the  annual  income  from  an  investment  of  $2860 
t  7^0  compound  interest,  payable  quarterly  ? 

8.  A  man  invests  $3750  for  3  years  at  7^  compound  intercc.;, 
^livable  semi-annually,  and  the  same  amount  for  the  same  time 
at  7|%  simple  interest.  Which  will  yield  the  greater  amount 
of  interest  at  the  end  of  the  time,  and  how  much? 

9.  Wh  will  be  the  compound  interest  at  the  end  of  2  yr. 
5  mo-  or      note  for  $600  at  7^ ,  payable  semi-annually  ? 


! 

1 


r  1 


.1 ; 


244 


BUSINESS    ARITHMETIC. 


K.I 


INTEREST    TABLES. 

585.  Interest,  both  simple  and  compound,  is  now  almost 
invariably  reckoned  by  means  of  tables,  which  give  the  interest 
or  amount  of  f  1  nt  different  rates  for  years,  months,  and  days. 
The  following  illustrate  the  nature  and  use  of  such  tables  : 


Table  showing  the  simple  interest  of  $1  nt  0,  7,  and  Sfe  ,for 

years,  months,  ami  days. 


w. 

756. 

8J<. 

6<. 

T%. 

8jt. 

Tears. 

1 

Yenra. 

4 

M 

.83 

.06 

.07 

.08 

.38 

2 

.12 

.14 

.16 

5 

.30 

.35 

.40 

3 

.18 

.21 

.24 

6 

.36 

.43 

.48 

3fontha. 

n 

^  * 

.' 

HI  out  ha. 



■_ 



1 

.005 

.00583 

.00006 

7 

.035 

.04088 

.01666 

3 

.01 

.01166 

.01388 

8 

.01 

.04666 

.05333 

3 

.015 

.01750 

.02000 

9 

.045 

.06350 

.06000 

4 

.03 

.02333 

.02666 

10 

.06 

.06833 

.0r>») 

5 

.025 

.02916 

.a3333 

11 

.065 

Jt»l\6 

.07333 

6 
Jiaya. 

1 

m 

.03500 

.04000 

Dftya. 

16 

.00016 

.00019 

.00022 

.ooaw 

.00311 

.00355 

3 

.00033 

.00088 

.00044 

17 

.002S8 

.0»«*l 

.wm 

3 

.00050 

.00058 

.00066 

18 

.00300 

.(nti'4) 

owoo 

4 

.00066 

.00077 

.00088 

19 

.00316 

.003f;9 

.0(K«J 

5 

.00083 

.00097 

.00111 

SO 

.00333 

.00388 

.00444 

6 

.00100 

.00116 

.(X)I33 

21 

.00350 

.0040S 

.00166 

7 

.00116 

.00130 

.00156 

22 

.003(ki 

.mizi 

.0048>^ 

8 

.00138 

.00163 

.00177 

33 

.0038:) 

.00417 

.00511 

9 

.00150 

.00175 

.00200 

21 

.00100 

.00-166 

.00533 

10 

.001  f;6 

.OOIM 

.m-tn 

36 

.00416 

.00180 

.00555 

11 

.00183 

.0021 H 

.00^.14 

36 

.00433 

00505 

.00577 

13 

mim 

.00233 

.0020(5 

1 

37 

.OWflO 

.00525 

.0J«00  i 

18 

.00316 

.00253 

.00288  j 

38 

.0046(( 

.00541 

.00622  1 

14 

.002:» 

.(KI2T2 

.00311 

39 

.00483 

.0066:] 

OOGM  , 

15 

.00250 

.00291 

.0G3{8 

1 

1 

INTEREST    TABLES, 


245 


almost 
intcreBt 
d  days. 


Metliotl  of  using  the  Simple  Interest  Table, 

586.  Find  the  inierest  of  $250  for  5  yr.  9  mo.  18  da.  at  7%. 


1.  Wejlnd  the  interest  0/ %l/or 
the  given  time 


<   .;^6     interest  In  table  for 
[    =   -j  .l>525       *'       "      "     •' 
'  I  .0035       "       "      '*     " 


.;^6     interest  in  table  for  5  yr. 

9  mo. 
18  da. 


Interest  of  |1  for  5  yr.  9  mo.  18  da.  id    .4Uti   of  $1. 

2.  Since  the  interest  of  #1  for  5  yr,  ft  mo.  18  da.  is  .406  of  $1,  the  interest 
of  $^50  for  the  same  time  \^  .406  of  $250. 

Hence,  $230  x  .406  =  $101.50,  tlie  required  interest. 


M. 


.» 

.40 

.48 

-    — 

» 

01666 

p6' 
iO' 

.05333 

.PCOOO 

o' 

.Of.666 

i6 

.07333 

— 

11 

.00355 

*»  1 

.no:n7 

r4> 

owoo 

(19 

.fl01"<2 

SB 

.00144 

OH 

'   .0(M66  \ 

i7 

.00468 

17 

.00511 

bG 

.00533 

80 

.00&55 

05 

.oosn 

•a 

.00600  i 

4-1 

.0*522  1 

(i; 

J       OOOM 

EXAMPLES     FOR     PRACTICH. 

587.  Find  by  using  the  table  the  interest,  at  7%,  of 

1.  $880  for  3  yr.  7  mo.  23  da.  4.  $325.80  for  5  yr.  13  da. 

2.  $438  for  5  yr.  11  nio.  It)  da.        5.  .'?:00.50  for  11  mo.  28  da. 

3.  $283  for  6  yr.  8  mo.  27  da.  6.  $395.75  for  3  yr.  7  mo. 

Table  shomug  the  amount  of  ^i  at  0,  7,  and  S'/c  compound 
interest  from  1  to  t?  years. 


YUS. 

.     1 
2 
3 
4 
5 
6 

w. 

1%. 

9fi. 

;  TBS. 

i 

^%' 

1%. 

8?. 

1.713824 
l.a50930 
1.<KKM)05 
2.1.')8i»25 
2..'«lfi.39 
2518170 

1.060000 
1.123600 
1.191'JIG 
1.36^177 
1.3:18226 
1.41S51U 

1.070000 
1.144900 
1.22oOI3 
1.31(n!W 
1.402552 
1.50O7:J0 

i.oeoooo 

1.16W00 
1269712 
1.. 160 189 
1.469$» 
1.58li874  1 

7 
8 
9 

to 

11 
12 

1.5036.30 

l.5^K^a48 

1.(J8'.M79 
1.7908.18 
1.898299 
2.012197 

1.6a')781 
1.718186 
l.a3S.459 
1.967151 
2.104853 
2.252192 

Method  of  using  the  Compound  Interest  Table. 

588.  Find  the  compound  interest  of  $2800  for  7  years 
at  «%. 

1.  The  amount  of  |1  for  7  year*  at  (H  in  the  table  is  1.5038.3. 

2.  Since  the  amount  of  $1  for  7  yonrs  In  L.^wns,  the  amount  of  12800  for 
the  Pamo  time  mufnt  beVMOtimett  fl..V):)63  ^  |l.SO')6.3  x  3800  =  $4310.161. 
Ilcnce,  $4210.  l&l  —  $2800  =  $1410.1(^  the  rei^ulred  imprest. 


1'- 


'I 


Ji- '■■ 


246 


BUSINESS    ARITIIMErrC, 


EXAMPLES    FOR    PRACTICE. 


If' 

'!■■ 
*'  ■- 


589.  Find  by  using  the 

1.  12000  for  Syr.  at  8%. 

2.  $560  for  9  yr.  at  7%. 

3.  $870  for  11  yr.  at  6%. 

4.  $2500  for  3^  yr.  at  6%. 

5.  $3S00for7yr.  at8%. 

6.  $640for4|yr.  at8%. 

7.  |28ofor9iyr.  at7^. 


table  the  compound  interest  of 

8.  $400  for  4  yr.  7  mo.  at  7%. 

9.  $384.50  for  8  yr.  at  6%. 

10.  $900  for  0  yr.  3  mo.  at  8%. 

11.  $4000  for  9  yr.  2  mo.  at  1%, 

12.  $690  for  12  yr.  8  mo.  at  6%. 

13.  $600  for  11  yr.  6  mo.  at  8;^. 

14.  $3900  for  4  yr.  3  mo.  at  6;^. 


11 


ANNUAL    INTEREST. 

590.  Annual  Interest  is  dmple  interest  on  the  princi- 
pal, and  each  year's  interest  remaining  unpaid. 

Annual  interest  is  allowed  on  promissory  notes  and  othor 
contracts  which  contain  tho  words,  *'  interest  payable  annually 
if  the  interest  remains  unpaid." 

591.  PiiOB.  X.— To  find  the  annual  interest  on  a 
promissory  note  or  contract. 

What  is  the  interest  on  a  note  for  $600  at  7%  at  the  end  of 
3  yr.  6  mo.,  interest  payable  annually,  but  remaining  unpaid  ? 

SoLTTTioN.— 1.  At  1%  the  payment  of  Intcregt  on  $000  due  at  the  end  ol 
each  year  Is  $42,  and  the  simple  interest  for  3  yr.  0  mo.  is  $147. 

8.  The  first  payment  of  |42  of  interest  is  due  at  the  end  of  the  first  year 
and  must  bear  simple  interest  for  2  yr.  6  mo.  The  second  payment  is  due 
at  the  end  of  the  second  year  and  must  bear  simple  interest  for  1  yr.  6  mo., 
and  the  third  payment  being  due  at  tl:e  end  of  ilic  third  year  mut^t  boar 
interest  for  6  mo. 

Hence,  there  is  simple  Interest  on  $42  for  2  yr.  0  mo.  +  1  jnr.  6  mo.  ^ 
6  mo.  =  4  yr.  6  mo.,  and  the  Interest  of  the  $42  for  this  time  at  7,-?  is  ♦13.W. 

3.  The  simple  interest  on  $600  beini;  $147.  and  the  wimple  interest  on  flip 
interest  remaining  unpaid  being  $13.23,  the  total  interest  ou  the  note  at  the 
end  of  the  given  lime  io  $160.23. 


PARTI  A  L     PA  Y^ENTS. 


247 


\^\ 


St  of 
at  7%. 

.  at  8  % . 
o.  at7fo. 
lO.  at  6%. 
lO.  at  8  % . 
10.  at  6  /^ . 


the  princi- 

and  otlior 
ie  annually 


test    on  a 

the  end  of 
Ig  unpaid  ? 

I  at  the  eml  ol 

I  the  first  year 
lyment  l8  tliic 
lir  1  yr.  B  "10.. 
lear  roust  boar 

Il3np.  6  mo.  + 
It  11^  l8  »13.J3. 
Interest  on  t'^" 
Ihe  note  at  ilic 


EXAMPLES    FOR     PRACTICE. 

51)2.  1.  How  much  interest  is  due  at  the  end  of  4yr.  9  mo. 
on  a  note  for  $460  at  ^'/o^  interest  payable  annually,  but  re- 
maining unpaid? 

2.  F.  Clark  has  J.  H.  MacVicar's  note  dated  July  29,  1876, 
for  !?8()0,  interost  payable  annually  ;  what  will  be  due  Novem- 
ber 29,  1880,  at  7%  ? 

8.  Find  the  amount  of  $780  at  7%  annual  interest  for  5^  yr. 

4.  What  is  the  difFerence  between  the  annual  interest  and 
the  compound  interest  of  $1800  for  7  yr.  at  7^  ? 

5.  What  is  the  difference  in  the  simple,  annual,  and  com- 
pound interest  of  $790  for  5  years  at  8^  ? 

6.  Wbat  is  the  annual  interest  of  $830  for  4  yr.  9  mo.  at  8^  ? 


FABTIAL    PAYMENTS. 

59*$.  A  Promissoi'i/  Note  is  a  icrittcn  promise  to  pay  a 
sum  of  money  at  a  specified  time  or  on  demand. 

The  Face  of  a  note  is  the  sum  of  money  made  payable  by  It. 
Tlu;  Maker  or  Drawer  of  a  note  is  the  person  who  piijnH  the  note. 
Tho  raiiee  is  the  pc^rson  to  whom  or  to  whosie  order  iho  money  in  paid. 
An  Indorser  is  a  person  who  signs  his  name  on  the  back  of  the  note, 
and  thud  makes  himself  responsible  for  its  payment. 

594.  A  Negotiable  Note  is  a  note  made  payable  to  the 
bearer,  or  to  some  person's  order. 

When  ft  note  is  so  written  it  can  be  bongbt  and  sold  in  the  same  manner 
as  any  other  property. 

♦*>9,"».  A  Partial  Paffinent  is  a  payment  in  part  of  a 
note,  bond,  or  other  oblif^ation. 

o90.  An  Itulorsement  is  a  written  acknowletlgment  of 
a  i>?trtial  payment,  placed  on  the  back  of  a  note,  bond,  etc., 
("luting  tliu  time  and  amount  of  the  same. 


\\ 


r* 


248 


BUSINESS    A  Ji  ITU  ME  TIC. 


'H' 


1. 


MERCANTILE    KULE. 

597.  The  method  of  reckoning  partial  payments  known  as 
the  Mercantile  Rule  is  very  commonly  used  in  computing 
interest  on  notes  and  accounts  running  for  a  year  or  less.  The 
role  is  as  follows  : 

KULE. 

/>1>8,  /  Find  the  amoitnt  of  the  note  or  debt  frmn  the  time 
it  begins  to  bear  interest^  and  of  each  payment  until  the  date  of 
settlement. 

II.  Subtract  the  sum  of  the  aniountu  of  jyayinents  from,  the 
amount  of  the  note  or  debt ;  the  rtinuindcr  ucUl  be  tJte  hulauce  due. 

Obsen'e,  that  an  accnrate  application  of  the  rule  requiree  that  the  exact 
interest  fhoultl  be  found  according  to  (5GG). 


m 


4-V 


EXAMPLES     FOR    PRACTICE. 

599.     1.  $900.  Woodstock,  Sept.  1st,  1870. 

On  demand  I  promise  to  pay  li.  M.  Mac  Vicar,  or  order,  nine 
hundred  dollars  wUh  interest,  calue  receiced. 

Wakren  Mann. 

Indorsed  as  follows:  Oct.  18th,  1876,  !i;l~)0;  Dec.  23,  1876, 
$200  ;  March  15th,  1877,  $300.  What  is  dne  on  the  note 
July  19th,  1877  ? 

2.  An  accouut  amounting  to  $485  wns  duo  Se[)t.  3,  1875,  and 
was  not  settled  until  Aug.  15,  1876.  The  pnynients  made  upon 
it  wore :  $125,  Dec.  4,  1875 ;  $84,  Jan.  17,  1870 ;  .s95,  June  23, 
1870.  What  was  duo  at  the  time  of  settlement,  allowing  inter- 
est at  7%  ? 

3.  A  note  for  $000  bearing  (uterest  at  Q'/c  from  Jtdy  Ist,  1S74, 
wus  paid  May  16th,  ls75.  The  indorsements  wore:  July  12th: 
1874.  $185;  Sept.  15,  1874,  $70;  Jan.  13.1875,  c^230  ;  i.l.I 
Marc]\  2,  1875,  $115.  What  was  due  on  the  note  at  the  time  of 
payment  ? 


PARTIAL     PAYMENTS, 


240 


own  a& 
aputing 
8.    The 


the  time 
i  date  of 

from  tJie 
aiice  due. 

t  the  exact 


t,  1870. 
\rder,  nine 

Mann. 

22, 187G, 
the  note 

1875,  and 
iiatle  upon 
June  23, 
ring  iater- 

l8t,  lb74, 
[julylStli; 
.230;  11*' 
tlie  time  of 


4  $250.  Hamilton,  ^fnrc?^  25, 1876. 

Niuety-eijfbt  days  after  date  I  promise  to  pay  E.  D.  Brooks, 
or  order,  two  hundred  fifty  dollars  with  interest,  value  received. 

Silas  Jones. 

Indorsements  :  $87,  April  13,  1876 ;  $48,  May  9.    What  is  to 
pay  when  the  note  is  due  ? 


UNITED    STATES    RULE. 

OOO.  The  United  States  courts  have  adopted  the  following 
rule  for  reckoning  the  interest  on  partial  payments.  It  is  also 
very  frequently  adopted  in  Canada. 


BULE. 

601.  /  Find  the  amount  of  the  given  principal  to  the  time 
of  the  first  pat/ment ;  if  the  payment  equals  or  exceeds  the  inter, 
est  then  due,  subtract  it  from  the  amount  obtained  and  regard 
the  remainder  as  the  new  principal. 

II.  If  the  payment  is  less  than  the  interest  due,  find  the 
amount  of  the  given  principal  to  a  time  when  the  sum  of  the  pay- 
ments equals  or  exceeds  the  interest  then  due,  and  subtract  the 
sum  of  the  payments  from  this  amount,  and  regard  the  remain- 
der as  the  new  principal. 

m.  Proceed  with  this  new  principal  and  teUh  each  succeeding 
principal  in  the  same  manner. 

602.  The  method  of  applying  the  above  rule  will  be  seen 
firom  tbe  following  example : 

1.  A  note  for  $900,  dated  Montreal,  Jan.  5th,  1876,  and  paid 
Dec.  20th,  1876,  had  endorsed  upon  it  the  following  pay- 
ments :  Feb.  23d,  1876,  $40  ;  April  26th,  $6;  July  19th,  1870. 
$70.    How  much  was  the  payment  Dec.  20th,  1876,  interest 

at7%? 

17 


I 


Hi 

T 

'   »u  ■ 

BB  If 

■  }i 

HHR  {• 

•    -  '' 

RJ^f 

^ 

W' 

'  ■ 

m'l 

^if'  ■ 

|F^' 

. 

250 


BUSINESS    ARITHMETIC, 


fiOliUTTON. 

FiVHt  Step. 

1.  The  flrRt  principal  in  the  face  of  the  note  . 
3.  We  find  the  interest  from  the  datn  of  the  note  to  the  first  pay 
meiit,  Feb.  ij,  1B7()  (49  da.),  at  7< 

Amount 

3.  The  first  payment,  $40,  being  greater  than  the  interest  then 
due,  iii  subtracted  fl-om  the  amount 

/Second  pnncipal 

Seeotul  Step, 

1.  The  second  principal  la  the  remainder  alter  subtracting  the 

flrot  pajrmcnt  from  the  amount  at  that  date     .... 

2.  The  interest  on  $868.43,  from  Feb.  38  to  April  36, 

187(1  (m  da.),  is $10,463 

8.  Thi»  interest  being  greater  than  the  second  pay- 
ment (fB),  we  find  the  interest  on  $8«8.48 
from  April  26  to  July  1!>,  1876  (84  da.), 
which  is? 


$900 

8.4;j 

$908.43 

40.00 
$8(i8.43 


$868.43 


Interest  from  first  to  third  payment 

Amount.    .... 
4.  The  oum  of  the  second  and  third  pjijmentH  bein^  i;."«* 
the  interest  due,  we  (subtract  it  from  the  amount 

Third  pnncipal 


13.961 
$24,414 

"-  than 


!24.414 

$8«.)2.814 

76 
$816,844 


,1 


Tfilrd  Step, 

We  find  the  interest  on  1816.844,  from  July  19  to  Dec.  20, 1876 

(154  da.),  which  is         .        .  ...  .        24.a')T 

Payment  due  Dec.  50,  1876 I840.M0 

In  the  above  example,  the  interest  has  been  reckomd 
according  to  iiHUi) ;  in  the  following,  360  days  have  been 
regarded  as  a  year. 

EXAMPLES    FOR    PRACTICE. 


IP 


2.  A  note  for  $16l}0  at  8%  interest  was  dated  March  18, 1872, 
and  was  paid  Au^r.  13,  1875.  Tlie  following  sums  were  endorsed 
upon  it:  $160,  Fib.  12,  1873 ;  $48,  March  7,  1874;  and  $.ir>0, 
Aug.  25,  1874.     How  much  was  paid  Aug.  13 1 


^ISC  O  UXT. 


51 

14 

24.414 

$8«.»2.8}4 

\n 

. 

7« 

• 

1816.814 

251 


'^"-  "l-n  i,  Fob.  9,  1870,  iut,.  J,'  afj/^*      '    ^""  '"»<"'  '^'« 
-  follows  :   A,.ri,    3.  ',870%ir    T'f  "^"^  ^«'  '««■  '-"'o^^' 

'•^« ;  1580,  May  7,  ,874    andTn'/r  '''  ""'  ^  *'2''.  A"fc'.  s! 
-  0-  upon  it  Sept.  ,,;Z.  ^Z^::,  t/;7'    «ow  n.tcU 

DISCOITN-T. 

«>04.  The  Present  frovtl,  ->f 

"•  '"""f  '•''•««l  at  interest  nf  a  ,!»     "       'T''  ''  ^'''^  ««""' 
^•ve„  ,„„  „h,„  .,  ^^^^^^^  ^^a  legal  rate,  w,ll  amount  to  the 

"'"■'e^'^aJXafa'rur'im^a'Jr"  '*'"'"''  "^  »"«'  <" 

•imc  and  itB  present  worth 

-  "•    """•  ^'  -^°  «-  fe  present  won.  of  .„, 


^' 


15 


252 


BUSINESS     A  RITIIM  ET/C. 


I- 


.t'i 


EXAMPLES     FOR    PRACTICE. 

007.  What  is  ihe  present  worth 

1.  Of  $360  at  7^ ,  due  in  2  yr.  ?    At  5% ,  due  in  8  mo.  ? 

2.  Of  $800  at  0 J^ ,  due  in  6  mo.  ?    At  8% ,  duo  in  9  mo. ? 

3.  Of  $490  at  8;^,  due  in 42  da.?    At  7 ;^,  due  in  128  da. « 

What  is  the  true  discount 

4.  Of  $580  at  7%,  due  in  90  da.?  At  8i%,due  in  4yr. 
17  da.  ? 

5.  Of  $860  at  7%,  due  in  93  da.  ?  At  12 '/c,  due  in  Syr. 
19  da.? 

6.  Of  $260  at  6J%,  due  in  120  da.?  At  9%,  due  in  2  jr. 
25  da.  ? 

7.  Sold  my  farm  for  $3800  cash  and  a  mortgage  for  $6500 
running  for  3  years  without  interest.  The  use  of  money 
being  worth  7%  per  annum,  what  is  the  cash  value  of  the 
farm? 

8.  What  is  the  true  discount  at  8%  on  a  debt  of  $3200,  due 
in  2  yr.  5  mo.  and  24  da.  ? 

9.  What  is  the  difference  between  the  interest  and  irve  dis- 
count at  7;^  of  $460.  due  8  months  hence? 

10.  A  man  is  offered  a  house  for  $4800  cash,  or  for  $5250 
payable  in  2  yr.  6  mo.  without  interest.  If  he  accepts  the 
former,  how  much  will  he  lose  when  money  is  worth  8^  ? 

11.  A  merchant  buys  $2645.50  worth  of  goods  on  3  mo. 
credit,  but  is  offered  3%  discount  for  cash.  Which  is  the 
better  bargain,  and  how  much,  when  money  is  at  7%  per 
annum  ? 

12.  Which  is  more  profitablo,  and  how  much,  to  buy  wood 
at  $4.50  a  cord  cash,  or  at  $4.66  payable  in  9  months  without 
interest,  money  being  worth  85^  ? 

18.  A  ^rain  dealer  sold  2400  bu.  of  wheat  for  $3600,  for 
which  he  took  a  note  at  4  mo.  without  interest.  W^hat  was  the 
cash  price  per  bushel,  when  money  is  at  6^  ? 


B  A  y  K     DISCO  V  y  T. 


253 


.1 

iO.t 


in  4yr. 

.  in  3  yr. 

le  in  2  yr. 

for  $6500 
of  money 
uo  of  the 

13200,  due 

true  dis- 

for  $5250 
|ccept8  the 

on  3  mo. 
tich  is  the 

It  7fo  per 

buy  woo^ 
|i8  without 

13600,  for 
xat  was  the 


BANK    DlSCOUin'. 

008.  BaHh  lyiscounf  is  the  interest  on  the  Xe.ce  of  a 
note  ftir  tlie  time  it  has  to  run,  including  three  days  grace. 

1.  Thirt  dcdnctiou  iu  made  by  a  bank  for  advancing  the  amount  of  the 
note  before  It  it«  due. 

S.  A  note  to  be  dincounti'd  at  a  bank  must  Ui>nally  be  made  payable  to 
the  order  of  some  person  who  must  endorse  it. 

3.  When  a  note  bcarx  interest,  tlie  discount  ia  computed  on  its  face  plus 
the  interest  for  the  time  it  lias  to  run. 

<JOt).  Days  of  Grace  are  three  days  usually  allowed  by 
law  for  the  i^yment  of  a  note  after  the  expiration  of  the  time 
si)ecified  in  it. 

OlO.  The  3IaturUy  of  a  note  is  the  expiration  of  the 
time  including  days  of  grace. 

61 1.  The  l*roc€e(is  or  Avails  of  a  note  is  the  sum  left 
after  deducting  the  discount. 

0 1 12.  A  Protest  is  a  declaration  in  writinpf  by  a  Notary 
Public,  giving  legal  notice  to  the  maker  and  endorsers  of  a  note 
of  its  non-payment. 

1.  In  Ontario  a  note  must  be  protected  on  tlie  day  of  its  maturity  ;  in  tlio 
i'rovince  of  Quebec  it  may  be  protested  on  thi-  day  it  is  due,  or  tlie  fact  of 
its  maturity  may  be  noted  and  tlie  time  of  protest  extended  to  (be  third  day. 
The  endorsers  are  released  from  all  obli^jation  to  pay  a  note  when  not 
regularly  protested. 

2.  When  a  note  becomes  due  on  Sunday  or  a  legal  holiday,  it  must  be 
paid  on  the  day  following. 

613.  Prob.  XII. — To  find  the  bank  discount  and  pro- 
ceeds of  a  note  for  any  g^ven  rate  and  time. 

observe,  that  the  bank  discount  is  the  interest  on  the  face  of  the  note  for 
the  given  time,  and  the  proceeds  is  the  face  of  the  note  minus  the  bank 
discount.    Hence  the  following 

BULE. 

614.  /  Compute  the  inteirst  for  three  dai/s  more  than  the 
giten  time  on  the  fare  of  the  note  ;  or,  if  the  note  benrs  interest, 
on  its  amount  at  maturity  ;  the  result  is  the  bank  discount. 


W 


• 


254 


D  Um  y  ESS     A  Ji  I  TUME  TI C. 


^f 


,j*  '•: 


//.  Subtract  the  discount  from  tJic  face  of  the  note,  or  ifn 
amount  at  inatarUy  if  it  bearn  interott ;  the  remainder  »«  t/ie 
proceeds. 

EXAMPLES     FOR    PRACTICE. 

Gl/>«  What  are  the  bank  discount  and  proceeds  of  a  note 

1.  Of  1790  for  154  da  at  6%  ?     For  2  nio.  12  da.  at  7%  ? 
a.  Of  $380  for  3  mo.  15  da.  at  7j^c  ?    For  0  nio.  9  da.  at  S%  V 
8.  Of  $1000  for  80  da.  at  7^^  ?    For  140  du.  at  8i  '/c  t 

4.  What  is  the  difference  between  the  bank  and  true  discount 
on  a  note  of  $1000  at  7;^ ,  payable  in  90  days  ? 

5.  Valuing  my  horse  at  $212,  1  sold  him  and  took  a  note  for 
$2.'l5  payable  in  60  days,  which  I  discounted  at  the  bank.  How 
much  did  I  gain  on  the  transaction? 

6.  A  man  bought  130  acres  of  land  at  $16  per  acre.  He  paid 
for  the  land  by  discounting  a  note  at  the  bank  for  $2140.87  for 
90  du.  at  0%.     How  much  cash  has  he  left? 

Find  the  date  of  maturity,  the  time,  and  the  proceeds  of  the 

following  notes  : 

(7.) 
$480  90.  Chatham,  Mar.  16,  187»». 

Seventy  days  after  date  I  promise  to  pay  to  the  order  of 

D.  MacVicar,  ftnir  hundred  eighty  -^^^  dollars,  for  value  received. 

Discounted  Mar.  29.  N.  L.  Sage. 

(8.) 

$590.  Sarnia,  May  13,  1876. 

Three  mouths  after  date  I  promise  to  pay  to  the  order  of 
Wm.  Flint,  five  hundred  ninety  dollars,  for  value  received. 
Discounted  June  2.  Peter  M.vcKenztk. 

(9.) 
$1600.  BELLEvnxE,  Jan.  19,  1876. 

Seven  months  after  date  we  jointly  and  Beverally  agree  to 
pay  James  Richards,  or  order,  one  thousan<l  six  hundred  do). 
lars  at  the  Bank  of  Toronto,  value  received- 
Discounted  May  23.  Robert  Button, 

J.vMEs  Jackson. 


^^.YA'     DISCOUNT, 


255 


01 G.  Pros.  XIII  —To  find  the  face  of  a  note  when  the 
proveedn,  time,  and  rate  are  given. 

Observe^  that  the  ^)roc•ce(l^*  it*  \.\w  face  qf  tKe  itote  nilnuH  the  intertut  on  it 
fur  the  ){ivc'u  time  and  rati*,  and  (■oii^t-tiiicntly  that  thu  proceudH  niuxt  cuu- 
taiu  $1  iniuuH  the  intcront  of  $1  Tor  the  ;:iveii  time  and  rate  ac  many  tituett  aa 
there  are  dollarH  in  the  face  of  the  uute.    Ueucv  the  rulluwiui; 


RULE. 

017.  Divide  tJie  (jicen  procfidn  by  the  proceeds  of  $1  for  the 
given  time  and  rate  ;  the  quotient  in  tfiefice  of  the  note. 


V 


if.- 
ft 


EXAMPLES     FOR    PRACTICE. 

018.  What  must  be  the  face  of  a  note  which  will  give 

1.  For  90  da.,  at  7;^ ,  $450  proceeds  ?    if;  180. 25  ?    $97.32  ? 

2.  For  3  mo.  17  da.  at  C  % ,  J^HfiO  proceedrt  V    |290  ?    f  n:]©.  80  ? 

3.  For  73  da.,  at  8  % ,  $234.60  proceeds  '    $1800  ?    $500.94 ? 

4.  The  avails  of  a  note  for  50  days  when  discoimtcd  at  a 
bank  were  $350.80 ;  what  was  the  face  of  the  note  ? 

5.  What  must  be  the  face  of  a  note  for  80  days,  at  7^,  on 
which  I  can  raise  at  a  bank  $472.86? 

6.  How  much  must  I  make  my  note  at  a  bank  for  40  «la. ,  at 
7^c ,  to  pay  a  debt  of  $296.40  ? 

7.  A  merdiant  paid  a  bill  of  goods  amounting  to  $2850  by 
discounting  three  notes  at  a  bank  at  7^,  the  pn)cee<lH  of  ench 
paying  one-third  of  the  l)ill ;  the  time  of  the  first  note  was 
60  days,  of  the  second  90  days,  and  of  the  third  154  dayn. 
What  was  the/<i<^  of  each  note? 

8.  Settled  a  bill  of  $2380  by  giving  my  note  for  $890  at 
30  days,  bearing  interest,  and  another  note  at  90  days,  which 
when  discounted  at  7Ci  will  settle  the  balance.  What  is  the 
face  of  the  latter  note? 

9.  For  what  sum  must  I  draw  my  note  March  23,  1876,  for 
90  days,  so  that  when  discounted  at  7  '/o  on  May  1  the  proceeds 
may  bo  $490  V 


\ 


M 


EXCHANGE. 


[i  ■ 


u 


61 0.  Exchauf/fi  Is  n  motliod  of  paying  debts  or  otlief 
obligations  at  a  distunco  without  transmitting  the  money. 

Thus,  a  merchant  In  Chicago  deeiring  to  pay  a  debt  of  tiaoo  in  New 
York,  paysa  baulc  in  Chicago  11800,  plus  a  Hmall  per  cent  for  their  trouble, 
and  obtains  an  order  for  this  amount  on  a  bank  in  New  York,  which  ho 
remits  to  his  creditor,  who  receives  the  money  from  the  New  York  bnnk. 

Exchange  between  places  In  the  same  country  is  called  Inland  or  />o- 
mestic  Exchange^  and  between  different  countries  Fotdgn  Exchange. 

«20.  A  Draft  or  Bill  of  Exc/tanf/e  is  a  written  order 
for  the  payment  of  money  at  a  specified  time,  drawn  in  one 
place  and  payable  in  another. 

1.  The  Drawer  of  a  bill  or  draft  is  the  person  who  signs  it ;  the  Drawee^ 
the  person  directed  to  pay  it ;  the  Payee,  the  person  to  whom  the  money  in 
directed  to  be  paid  ;  the  Indoreer,  the  person  who  trutitifers  his  right  to  a 
bill  or  draft  by  indorsing  it ;  and  the  Holder,  the  person  who  has  legal  pos- 
sesHion  of  it. 

2.  A  Sight  Draft  or  BiU  is  one  which  requires  payment  to  be  made  when 
presented  to  the  payor. 

3.  A  Time  Drqft  or  Bill  is  one  which  requires  payment  to  be  made  at  a 
ppecifled  time  after  date,  or  after  Hght  or  \}e\ng  preiented  to  the  payor. 

Three  dnys  of  praco  are  usually  allowed  on  bills  of  exchange. 

4.  The  Acceptance  of  a  bill  or  draft  is  the  agreement  of  the  party  on 
whom  it  i?i  drawn  to  pay  it  at  mnturity.  This  is  indicated  by  writing  the 
wonl  "Accepted"  across  the  face  of  the  bill  and  nigning  it. 

When  a  bill  is  protested  for  non-acceptance,  the  drawer  is  bound  to  pny 
it  immediately. 

5.  Foreign  bills  of  exchange  are  usually  drawn  in  duplicate  or  triplicate, 
and  sent  by  different  conveyances,  to  provide  against  miscarriage,  each 
copy  being  valid  until  the  bill  is  paid. 

These  are  distinguished  from  one  another  by  being  called  the  Jlnt, 
eeoond^  and  third  of  exchange. 


/  A'  L  A  y  I)     E  X  ('  H  A  NG  E, 


257 


•  otlief 

^ 

In  New 
trouble, 
khich  ho 
:  bnnk. 
\d  or  1)0- 
gt. 

•n  order 
i  in  ono 

DrawMy 
Imoney  i«* 
ight  to  a 
jegal  po8- 

Iti  when 
kade  at  a 


[party  on 
king  the 

|d  to  pny 

ipUcate, 
3,  each 

|he  Jlrtt, 


021.  The  Par  of  Exchange  is  the  relative  value  of  the 
coins  of  two  coimtries. 

1.  ThiH,  the  par  of  exchange  between  Canada  and  Britain  Is  the  uumlMjr 
of  (lollnrH,  the  ptandard  unit  nf  Canadiun  money,  which  is  equal  to  the 
pound  Htcrling,  the  etaudard  unit  of  Eu^Hi'li  money. 

*2.  The  itnl  rate  qf  exchange  depends  ou  the  balance  of  trade  aud  is 
calU'd  tho  course  offxchange. 

3.  The  value  of  the  pound  t>tcrling  wac,  by  Act  of  Purliamcnt,  fixed  at  |  IJ. 
ItH  intrinsic  value  Ih  now  fixed  at  $-1,861.  But  raten  of  exchange  are  still 
quoted  In  commercial  papers  at  a  certain  per  cent  on  the  old  jnir  of 
trchautje.  Hence  when  exchange  Is  at  a  premium  of  ft;  per  cent,  it  is  at 
par  between  Great  Britain  and  Canada,  because  $4;  +  U}  per  cent  =  $4,862. 


i 


INLAND    EXCHANGE. 

Ot22.  Inland  Exchange  is  a  method  of  paying  debts  or 
other  obligations  at  distant  places  in  the  same  country,  without 
transmitting  thu  money. 

Form  of  Sight  atid  Time  Draft. 

£2700.  Montreal,  July  25.  1878. 

At  fifteen  days  sight,  pay  to  the  order  of  Taintor  Brothers  <fc 
Co.,  two  thousand  seven  hundred  pounds  sterling,  value  received, 
and  cJuirge  the  same  to  the  account  of 

K.  B.  Jones  &  Co. 

To  Geo.  L.  Batnes,  ) 
Banker,  LONDON.      \ 

1.  This  is  the  usual  form  of  a  draft  drawn  by  a  firm  or  individual  upon  a 
hank.    It  may  also  be  made  payable  at  a  ^'iven  time  after  date. 

2.  All  time  drafts  should  be  presented  for  acceptance  as  soon  as  received. 
Whon  the  cashier  writes  the  word  "  Acceptetl,"  with  the  date  of  occeptanco 
across  the  (hoe,  and  siunn  his  name,  the  bank  is  responsible  for  the  pay- 
ment of  the  draft  when  due. 


V 


«*s 


r^' 


258 


B  aSIAESS    ABITUMETIC, 


i  m 


T 

I 
t 

I 


METHODS  OP  INLAND  EXCHANGE. 

62!S.  FiitST  Method.— 7V/f'  party  drxiring  to  trnntmU 
mimey,  pvrrhmca  a  draft  for  the  a/tj/unt  at  a  imnk,  and  nendM  it 
}ry  vmil  to  Us  destination.    Or, 

If  he  has  a  drponit  aWvady  in  a  hank,  /nihjcH  to  hin  cherk  or 
order,  it  in  cuntoinary  to  send  his  check,  cert  (lied  to  be  fjo<>d  by 
the  cashiir  of  the  bank. 

ObBcrvo  cnrefuUy  the  followinj^  : 

1.  Ill  rnno  ortrannactionH  hi'iwi'cn  dlHtiint  rontitric«,  the  tninoiiortatiou 
ofHpeclu  from  iiit.*  oik;  to  the  otlit  r  would  he  MtteiHlcd  with  (rxp<'Ui»e,  rt«k, 
aud  lotiH  of  InfpnsHt.  To  uvold  tlicnc  ill(•(lIlv<'Ilit•^<•«•^,  Hill-  of  Rxchantto  ar- 
uwd  (610)  In  payliij;  dehtH  rcriprocnily  ditc  In  "iirli  rountrii  -.  A  Himila- 
n)ethu('.  Ih  fruqueutly  fullowud  with  rctipeci  tu  truueuctiouH  \,  Wu  the  miqiu 
cuiintr*  ,  thuH : 

8.  BankH  hcU  draftH  upon  others  in  which  tlicy  have  i\i\toAX»  in 
money  or  equivalent  Hecurity.  ITt^nce  Y)nnk^4  throu;.'hoiit  tlir  country,  in 
order  to  give  them  this  fhclllly,  hrtvo  ;4urh  dt-pofit-'  at  ••(•utre»  of  trad«',  »uch 
an  K   utreal,  Toronto,  Hamilton,  etc. 

8.  A  hank  Draft  will  u  -nally  lie  purchased  by  hanks  in  any  jmrt  of  the 
country,  in  rABo  the  piTHon  oflcrinn  it  it*  fttlly  tdeniifl'-^l  a»  the  party  to 
whom  the  draft  Ih  jjayahhr  Ilencf,  a  debt  or  other  liability  may  !>e  dl-- 
charfi^od  at  any  place  by  u  dral'l  on  a  bank  at  any  centre  of  trade. 

4.  A  draft  may  be  made  payable  to  the  percon  to  whom  It  li«  iieot.  or  f  ■ 
the  person  buylnj,' it.  In  the  latter  ease  the  p<rsun  1)ii>iMt,  it  niu-il  write 
on  the  Imck  "  Pay  to  the  order  of"  (name  uf  party  to  whom  it  \*  sent),  aud 
Hl(:n  his  o-<vn  namu. 

5.  CertiflcateM  of  deposits  and  certlflird  checkM  arc  purrhai"*.-*!  bj  bftjUDl 
In  llie  name  manner  as  Imnk  rraflH, 


Second  MKTiron. —  The  party  deniring  to  tran/tmit  money 
otttaiuH  a  Pont  Office  order  for  the  amount  and  remita  it  ax 
h^ore. 

An  the  amount  th.nt  r.m  be  inrhided  in  oiiO  Po».t  Offl« ••  order  i«  limited, 
thitt  metiiiHl  ia  n^Htricted  In  its  application.  It  iii  usually  employed  iu 
neiulttiim  small  huius  uf  money. 


"*'ni 


IN L  A  y  n    h:  x  c  n  a  no  e. 


25f» 


I* 

\Timnil 
'.ndn  it 

lefk  or 
oi'd  hy 


tortation 
itHi,  riak, 

^  i>imUu- 
the  t-aiuo 

mntry,  In 
ade, «uch 

It  t  of  the 

I»Hrty  t'» 

l>f  dl«- 

jul.  or  t  ' 
lii-l  wrilf 
Lot ).  and 


money 


limited. 
Lloyed  lu 


TniTiD  Method. — 77*/'  pmty  (Imring  to  transmit  mimiy^ 
maki'S  a  draft  or  ti^rder  for  thf  amount  upon  a  party  oiriny  him, 
at  the  pltir,-  irJure  the  muMy  is  to  he  sent,  and  remits  thin  an 
prnioudy  dirertrd. 

1.  By  tli?rt  inofho<I  oin;  |)uri>on  Ih  nuid  to  draw  upon  nnotlior.  Such  (Irafls 
Htiniild  \w  i)r«?«'iii«*d  for  pnynu-nt  an  noon  an  received,  and  if  not  puid  or 
acciptctl  xliould  ho.  protcHic*!  for  non-payment  ininit>(llat<-ly. 

2.  Lnrjfi!  I)^l*ln«•^^^  flrniH  havr  deponitH  in  Imnkn  nt  buHincKH  centrcn,  nnd 
credit  with  otJ^T  bii»«inof»«  flrniH;  h<!nn>,  tludr  drafts  nro  U!*e<l  by  thcni- 
HclvuM  and  uthcrH  thu  Hamu  an  bunk  druflH. 

C(t24.  The  Premium  or  IHwount  on  a  draft  depends  cliirfly 
on  tlio  condition  of  trade  iHjtwccn  tln!  place  where  it  18  pur- 
chuHi'd  and  tlic  place  on  which  it  Ih  nia(l'\ 

Thii»,  for  «'xaniplf,  uiercbuntf*  und  olhrr  bufin('^'rt  men  at  Bufl'ulo,  cou- 
tnict  nion*  obllj^tloni*  In  Now  York,  for  which  they  pay  by  draft,  than  N«!W 
York  buHlneHM  nitm  contract  in  liiitfaio ;  i>onH«>qu«Mitly,  lianki*  at  niitrnio 
nuj-t  actually  >cnd  mon<-y  to  Now  York  by  Kxpro.-*  or  oilit'r  « onveyanc*'. 
He  cc,  for  th<'  (•xp«'n><«;  thun  incurred  and  other  troiMtle  In  handling  thu 
m .  icy,  a  mihuII  premium  \*  charged  at  Bullalo  on  New  York  draftn. 


EXAMPLES     FOR     PRACTICE. 

025.  1.  What  Ih  the  cost  of  a  m^hi  draft  for  *2400.  at  \  % 
prenuuin  ? 

SoLUTlOK.— Cowt    r  $2400      li  of  ♦2100      $2.1ir,. 

2.  What  ie  the  cost  of  a  dnift  for  i|;a*JO().  at  J%  premium? 
Solution.— Co«t  -    |u-floo  .  i«of$:BO()      #.'1201. 

Find  the  cost  of  sight  draft 

;{.  Fo"  $h:14.  pn-mium  2%.  «.  For  %Vm,  disrount  1%. 

4.  For  Y'>'JO(>,  pnmiiun  \'/c.  7.  For  *aH  1.5(1,  dineount  |%. 

T).   Kor  |i:W.HO,  pn'mium  J%.        H.  For  $*21)."».:{0.  dlHrount  li}%. 
U.  Thf  «M»Ht  of  u  i»ij.'ht  di   .'t  piirchaH'(l  at  V/,<   pn  luium  Ib 
$U)3  'iU  ;  what  i8  the  face  of  the  draft  ? 

HoLi'TioN  — At  lj!S  prpaiium,  fl  of  ihe  face  of  tlir  (haft  cont  (l.Ol.'S. 
Ilenci-  tlif  face  of  Ihi-  draft  ia  ai«  nctuy  doilure  au  tl.Ul5  in  cuutuinud  limcH 
in  $4m.«>,  which  Ic  |MW. 


^p 


260 


D  U  SI  NESS     A  li  1  TUM  E  TIC. 


Find  the  face  of  a  draft  which  cost 


h'i'V  ■ 


% 


1 


t|,1 


h 

liil 


(  ■! 


10.  $575.40,  premium  2;%. 

11.  $731.70,  pr-jmium  \\%, 
13.  $483.20,  premium  \^o. 


18.  $810.88,  discount  1%. 

14.  $273,847,  (liHcount  %%. 

15.  $31.>.«5.  discount  \\'/o. 


10.  Wliat  is  the  c<i8t  of  a  drpft  for  $400,  payable  in  3  mo., 
premium  \\%,  the  bank  allowing  intiiest  at  4j^  until  the 
draft  is  paid  ? 

Solution.— A  Klj,'ht  draft  for  $400,  nt  \\%  premium,  coHtx  |4()0,  but  the 
bunk  allows  intorci^t  at  4^  on  the  faco,  $100,  for  3  mo.,  which  iu  %\.  Hence 
tlie  draft  will  cot»t  1406  — $4  =  $402. 

Find  the  cost  of  drafts 

17.  For  $700,  premium  }%,  time  60  da.,  interest  at  3%. 

18.  For  $1600,  premium  Xy/c,  time  50  da.,  interest  at  4%. 

19.  For  $2460,  discount  %%,  time  DO  da.,  interest  at  41  %. 

20.  For  $1800,  discount  •{%,  time  30  da.,  interest  at  5%. 

21.  A  merchant  in  Albany  wishing  to  pay  n  debt  of  $498.48 
in  Chicago,  sends  a  draft  on  New  York,  rxcliange  on  New 
York  being  at  \%  premium  in  Cliicago ;  what  did  he  pny  for 
the  draft? 

Solution.— The  draft  caKhed  in  Chicatro  commandn  n  premium  of  11^  on 
lt>»  face.  The  man  rr([uireH,  therefore,  to  purclmKC  a  draft  whoM»  face  pin** 
%%  of  it  (MinnlH  $4W.4H.  Hence,  accor(llni»  to  (495—5),  the  amount  paid,  or 
face  of  the  draft,  ia  fl'.m.lH  -»-  l.OOS  ~  (4%. 

22.  Excliange  being  nt  98?  (\\'/(  discount),  what  iu  the  cost 
of  a  draft,  time  4  mo.,  interest  at  5%  ? 

23.  The  face  of  n  draft  which  was  ])urcha8ed  at  li%  premium 
is  $2500,  the  time  40  da.,  rate  of  interest  allowed  \'/c  ;  what  was 
its  cost  ? 

24.  My  .agent  in  Halifax  »oJd  a  consignment  of  goods  for 
$8200,  commission  on  the  sale  21  [■',  .  lie  reniitted  the  proceeds 
by  draft  on  Montreal,  at  a  premium  of  i^'/c.  What  is  the 
amount  remitted  ? 


^t^'if 


cost 
lium 

lit  NVllB 

Is  for 
ireeds 
Is  the 


FOREIGN    EXCIIAXOE.  2G1 


FOREIGN    EXCHANGE. 

026.  Foreign  Kxi'hanffv,  is  a  metliml  of  paying  debts 
or  other  obligatious  iu  foreign  couutrieti  without  truuumittiug 
the  money. 

Ob^ert'e,  Xhfii  foreign  exchange  is  bamHl  upon  the  fact  that  diflcrent  coun- 
tries exchange  proiiucts,  Hecurltlc«,  etc.,  with  each  other. 

ThiiH,  Cauada  t^e'ln  v\  heat,  etc.,  to  England,  and  Kut;land  in  return  hcUs 
niuiiufactured  ^uode,  etc.,  to  Canada,  lienct;,  partieH  iu  eacli  country 
Ifcconu  indebted  to  parties  in  the  other.  For  thin  reanon.  a  nuTcliant  in 
Canada  can  pay  for  goodii  purchased  in  England  by  buying'  an  order  upon  a 
firm  in  England  which  id  indebted  to  a  firm  in  Cauada. 

Form  of  a  liill  or  Set  of  Exchange, 

£400.  Ottaw.\,  Ju/y  13,  t870. 

At  Hif/ht  of  this  First  of  Exchange  (second  mtd  third  of 
the  mme  date  and  tenor  unpaid),  pay  to  the  ordtr  of  E,  J), 
lilakvslee  Four  Hundred  Pounds  Sterling,  foi'  value 
''erdced,  and  charge  the  wiine  to  the  account  of 

WiLM.vMs,  BnowN  &  Co. 

To  Martin,  Williams  &  Co.,  London. 

The  per-<on  jiprchnsln)^  the  excliange  receives  three  bllln,  which  he  pcnds 
by  diflcrent  nrnils  to  avoid  inlHcarriaj,'e.  Wlic-u  oi;e  liar*  been  received  aad 
paid,  tlie  othorH  at  *  void. 

The  above  ik  the  form  of  tlie  flrnt  bill.  In  the  Secou'!  Bill  the  wonl 
"FiitsT"  l>«  used  luHtead  of  "Sbconu,"  utul  the  j(an?ntlicr.is  n-adts,  '' MrHt 
and  Third  of  the  Hame  date  and  tenor  unpaid."  A  siuiilar  cliange  is  made 
in  th.'  Third  Bill. 

Oii7.  Exrhnnge  with  Europe  is  conducted  chiefly 
through  prominent  flnancinl  centres,  as  London,  Paris,  Berlin, 
Antwerp,  Amsterdam,  etc. 

iV2H,  ijuotatioits  are  the  puhli.shod  rates  at  which  bills 
of  exchange,  stocks,  bonds,  etc..  are  bouglit  and  sold  in  the 
money  market  from  day  to  day. 

ThcHc  (juot.itlons  give  tin-  ni.irket  ;.'old  value  in  stcrllnjj  money  of  one  or 
more  unltH  of  the  foreiirn  coin 

Thus,  quotJitlons  on  London  give  the  value  of  £1  nterling  in  dollars  ;  on 
Paris.  Antwerp,  and  (jeueva,  the  value  of  ♦!  \i\  francii :  on  IIainbur<;.  B«'r- 
lln,  Bremen,  and  Frankfort,  the  value  uf  4  inarhi  Iu  cetUs  ;  on  Amsterdam, 
Uie  value  of  a  yuildtr  in  centit. 


>.,    I 


2G2 


nirsixESS    arithmetic. 


If*.  ,1 


■1T 


u 
m 


Ot20f  The  following  table  gives  the  par  of  exchange,  or  gold 
value  of  foreign  monetary  units  : 


Table  op  Par  ok  Exchange. 


COUNTRIK8. 


Austria 

I  Bt'lgiuin 

I  Bolivia 

1  Brn/JI 

,  Bci^^ota 

Unified  States 

Cunt. ill  Ainurica.. . 

Chill 

Denmark 

Ecuador 

Egypt 

Franco 

Great  Britain 

Greece 

German  Empire. .. 

Japan  

'  India 

I  Italy 

Liberia 

'  Mexico 

Ni'therlands 

Norway 

'  Pern 

i  Portugal 

RuHHla 

Sandwich  InlandH. . 

Spain 

Sweden 

Swltzi-rlond . 

Trl|.oll 

Tuni*" 

Turkey , 

U.  8.  ofCoIomb'a. 


MONET  ART  UNIT. 


STANDARD. 


Florin 

Franc 

Dollar 

MilrelBoflOOOreis  ... 

PoHO 

Dollar , 

Dollar 

Peso 

Crown 

Dollar 

Pound  of  100  piantrcB . . 

Franc 

Pound  sterling , 

Drachma 

Mark 

Yen  

Rupee  of  16  annas 

Lira 

Dollar 

Dollar 

Florin 

Crown 

Dollar 

Mllrcla  of  1000  reiH  . . . . 
Rouble  of  100  copcckn. 

DolKir 

Peseta  of  100  ccntlmcH 

Crown. 

Franc 

Matibiib  of  20  plantrc .-( 
Pla(»lre  of  iO  caroiibs  . , 

Piastre 

Pt'HO 


Silver 

Gold  and  silver 
(Jold  and  silver 

0(.ld 

Gold 

Gold 

Silver 

Gold 

Gold 

Silver 

Gold      

Gold  and  silver 

Gold 

Gold  and  silver 

G(dd 

Gold 

Silver 

Gold  and  silver 

Gold 

Sliver 

(iold  and  silver 

Gold 

SIher   

Gold 

Silver ... 

Gold 

Gohl  and  sliver 

Gold 

Gohl  and  silver 

SIIVIT     

Silver    

Gold 

Silver 


VAUIK  IN 
CANAniAN 

MCNEY. 


.4ft,  8 
.19,8 
.%,  5 

.%,  6 
11.00 
.'.»1,  8 
.91,2 
.26,8 
.91,8 

4.97,4 
.19,  .3 

4.86.  66 
.19,  3 
.28,8 
.90,7 
.48,6 
.19,8 

1.00 
.99,8 
.38,6 
.26,  8 
.91,  8 

1.08 
.73,4 


1.00 

.19, 

3 

.26, 

8 

.19. 

.\ 

.82, 

St 

.11. 

H 

.04, 

8 

1   ■"'• 

8 

FOREiaX    EXC  H  ANQ  E,  2C3 


■•'•i 


METHODS    OP    DIRECT    EXCHANGE. 

i%*My.  Dh'i'vt  Ejrjhfthf/e  is  a  mcthotl  of  making  pay- 
lutMitH  in  a  torcigD  ooimtry  at  the  quoted  rate  ot  exchange  with 
tliat  country. 

FinsT  MKTnoD. — The  person  deftiring  to  transmit  the  money 
purrhani^a  (t  Set  of  tlrrhange  for  the  amount  on  the  eouutry 
to  ichich  the  money  is  to  he  sent,  and  forirordu  the  three  hUls  by 
different  ukuIs  or  rmttes  to  their  destination. 

Skcoxd  Metthod. — The  person  desirimj  to  transmit  the  money 
instrnrts  his  ci'editor  in  tJie  foreign  eountry  to  tliunv  uponhim, 
that  is,  to  sf'fl  a  set  of  exchange  npon  him,  irhich  he  pays 
in  his  own  country  when  presented. 


!  r 


'fc 


i' 


I'*    t 


EXAMPLES    FOR     PRACTICE. 

031 .  1.  Wliat  is  the  cost  in  currency  of  a  bill  of  excbanffe 
on  Liverpool  for  £285  Os.  6d.,  (>xchange  bein^  quoted  at  $4.88, 
and  gnUi  at  1.12,  brokerage  \'/l  ? 


£285  9s.  fid.  =  £285.475 
$4.88  X    285.475  -  $1393. 118 
$1.1225  X  I.^OJJ.llS  .:  1503.77  4- 


SoLUTioN.— 1.  Wi*  fiMtlure 
the  9t>.  (WI.  to  n  decimsil  of  ^l. 
Ilonro  t'2K'S"N.rKl.  €2Hr..J7.'S. 
2.  Since  tl  ♦•»  SK,  t:*<>.475 
inust  be<'(iiii»l  KRS  v  285.475  ;=:  $i;W3.118,  the  i;oUl  value  of  tin*  hill  with- 
(»ut  hrokora;jc. 

;{.  Hiiicc  $1  iroid  i«  equal  91.12  cnrroncy.  and  tho  hrolcor/ipo  is  J  .thoront 
<>r$l  ^(dd  in  currency  1h  $1.12%.  Hunce  tho  hill  co.xt  lo  curreucy  tl.l225  x 
1W3.118  -  ILVil.T? +• . 

What  is  the  cost  of  a  bill  on 

2.  Loivion  for  .€430  Sj^.  3d.,  sterling  at  4.84|,  brokerage  1%  ? 
8.  Paris  for  4500  francs  at  .198,  brokerage  ^'/y  t 

4.  Geneva,  Switzerlaud.  for  80!H>  francs  at  .189? 

5.  Anfirrrp  for  4000  rriincs  at  .175,  in  currency,  golil  at  1.09  Y 

6.  Amsterdam  for  84^K)  g  lilders  at  41  {,  brokerage  \'/i)  V 

7.  FrnJcfort  for  2500  marks,  quoted  at  .97|? 


264 


BUSINESS    ARITHMETIC, 


p.i^ 


i^.* 


8.  A  merchant  in  Kingston  instructed  his  agent  at  Berlin  to 
draw  on  him  fur  a  bill  of  goods  of  43000  marks,  exchange  at 
24?,.  gold  being  at  1.08^,  brokerage  ]  %  ;  what  did  the  merchant 
pay  in  currency  for  the  goodB  t 


METHODS    OF    INDIRECT    EXCHANGE. 

(iii*2.  Indirect  Ka'chunye  is  a  method  of  making 
payments  in  a  foreign  country  by  taking  advuntuge  of  the  rate 
of  exchange  between  that  country  and  one  or  more  other 
countries. 


Observe  carefully  the  following : 

1.  The  advanta);c  of  indirect  over  direct  exchange  under  certain  finan- 
cial conditions  which  gomctimei*,  owing  to  varlouo  cauttcH,  exitst  between 
different  countrieu,  may  be  ithown  bh  fullowa : 

Suppose  exchau|;e  in  Moutruul  to  be  at  par  on  London,  but  on  Parin  at 
ITcontH  fur  1  franc,  and  at  PariB  on  Ltmdon  at  24  francH  for  jb'l.  Witli 
thcMO  conditions,  a  bill  on  London  for  £10U  will  cobt  in  Moutiual  #-186.(>5  ; 
but  a  bill  on  London  fur  £100  will  co»t  in  Parltt  24  iVancH  x  100  =  3400 
francH,  and  u  bill  on  Pariu  for  3400  francH  will  cost  in  Montreal  17  cents  x 
MM)^  $408. 

Ilencc  €100  can  be  sent  from  Montreal  to  London  by  direct  exchange 
for  f48«.<i5,  an<i  by  Indirect  exchange  or  through  Paris  for  |40H,  giving  an 
advantage  ol  $4H(i.G5  —  $408  =  $7b.tl5  in  fiivor  of  the  latter  method. 

2.  The  pro*  <>f  computing  indirect  exchange  1^^  called  ArbitraiUm  Qf 
ExfhaiKje.  Wli'  there  ih  onlv  one  intermediate  place,  it  is  called  simple 
Arldf ration  ;  v  ,ere  there  are  two  or  more  iutermediate  places  it  is  called 
VotnjtoKnd  Arbitration. 

Eitiier  of  the  following  methods  may  be  pursued  : 

FiiisT  Method. — The  person  deHiring  to  transmit  the  money 
may  buy  n  bUl  of  exchange  for  the  ammint  on  an  ivtermeduitc 
pi^icc,  tehkh  he  nenda  to  Jus  agmt  at  that  plare  itith  instructwus 
to  buy  a  bill  itith  the  yroieetla  on  the  pLiee  to  trhich  the  money  i* 
to  he  aeni,  and  toforicard  it  to  thf  pr<fj)er  party. 


This  in  called  the  method  by  remittance. 


m^ 


FORE  TQ  N    EXCHANGE, 


2G5 


I'm  to 
ge  at 
chant 


E. 

le  rate 
I  otber 


In  flnan- 
betwevu 

Paris  at 

;i.    Wltli 

$486.05; 

3  =  WOO 

ccntB  » 

.'xchangc 
giving  au 

ration  Of 

Ll  sini])l« 

le  cftUod 


money 
edutte 

■Ut'tWftH 

ioney  i* 


Second  Method. — Tfie  pencni  desiring  to  send  the  momy 
inKtructa  hut  creditor  to  draw  for  the  amount  on  hi«  agent  at 
an  inUrmediale  place,  and  his  agent  to  draw  upon  him  for  the 
same  amount. 

This  is  callud  the  method  by  drawing. 

Third  Method.— TA^  person  desiring  to  send  t7ie  moiKi/ 
instructs  his  agent  at  an  iiUei mediate  place  to  draw  upon  him 
for  the  amoufit,  ninl  buy  it  hill  on  t/u  2f^arr  to  which  the  monrifia 
to  be  sent,  andforuard  it  to  the  proper  party. 

This  is  called  the  method  by  drawing  and  remitting. 

Tliese  methods  are  equally  applicable  wlion  the  exchange  is 
made  thi-ough'  two  or  more  intemu'diute  plucus,  and  the  solu- 
tion of  examples  under  each  is  only  an  ai)i)licati()n  of  com|K>und 
numbers  and  business.    Probs.  Vili,  IX,  X.  und  XI. 


EXAMPLES    FOR     PRACTICE. 

0«i3.  1.  Exchange  in  Hamilton  on  London  is  4.8o,  and  on 
Paris  in  London  is  24^  ;  what  is  the  cost  of  transmitting  631)04 
francs  to  Paris  through  London  ? 

Solution.— 1.  We  find  the  copt  of  a  b!U  of  oxclmnsro  In  London  for639ftl 
francs  Since  i\\  franco  r=  £1,  63994 -4-3<i.;  Is  equal  the  rmnihcr  of  £  In 
(^ilKM  fVancB,  which  \%  £9613. 

S.  Wl'  find  the  co8t  of  a  bill  orexchan{,'c  in  Ilumiltou  for  i:2G13.  Since 
£1  =  $4.83,  the  bill  muat  cost  $4.88  x  861$  =  $18615.%. 

2.  A  merchant  in  Quebec  wishes  to  pay  a  debt  in  Berlin  of 
7000  marks.  He  finds  he  can  buy  exchange  on  Berlin  at  .25, 
nnd  on  Paris  at  .18,  and  in  Paris  on  Berlin  al  1  mark  for  1  15 
francs.  Will  ht  gain  or  lose  by  remitting  by  indirect  e.icha.  g, , 
luid  how  much  V 

fi.  What  win  lie  the  cost  t«>  remit  4800  guihlere  from  Mon- 
treal  to  Amsterdam  through  Paris  an<l  IiOnd<tn.  eychang-c  l>eini>; 
(putted  as  follows  :  Montreal  on  Paris,  18/.  ;  at  Paris  <^n  Lon- 
don, 241    francs    to  a  £ ;    and    at    London    on    Amsterdam, 

18 


26G 


BUSI^'£:ss   arithmetic. 


H%, 


i: 


J      1-    1 


> 

". 

t       . 

']^-i 

12  J  guilders  to  the  £.     How  much  more  would  it  cost  by  direct 
exchaugo  at  391  cents  for  1  guilder? 

4.  A  Canadian  residing  in  Berlin  wishing  to  obtain  $0000 
from  Canada,  directs  his  agent  in  I'aris  to  draw  on  Montreal  and 
remit  the  proceeds  by  draft  to  Berlin.  Exchange  on  Montreul 
at  Paris  being  .18,  aud  on  Berlin  at  Paris  1  mark  for  1.2  francs, 
the  agent's  commission  being  \%  hoi\\  for  drawing  and  remit- 
ting, how  much  would  he  gain  by  drawing  directly  on  Canada 
at  34J  cents  i)er  mark  ? 


EQUATION    OF   PAYMENTS, 

($34.  An  Acvomit  is  a  written  statement  of  the  debit  and 
credit  transactions  between  two  persons,  with  their  dates. 

The  debit  or  left-hand  side  of  an  account  (marked  Dr.)  shows  the  Hams 
duo  to  the  Creditor,  or  person  keeping  the  account ;  the  credit  or  right- 
hand  side  (marked  C'r.)  showB  the  Hum«  paid  by  the  Debtor,  or  person 
agaiutit  whuui  the  account  is  made. 

035.  The  lialdiice  of  an  account  is  the  difference  between 
the  sum  of  the  items  on  the  debit  and  credit  sides. 

G'KI.  Equation  of  Paymeuts  is  the  process  of  finding 
a  date  at  which  a  debtor  may  pay  a  creditor  in  one  payment 
several  sums  of  money  due  at  different  times,  without  loss  of 
interest  to  either  party. 

<JJJ7.  The  Kqttaied  Time  is  the  date  at  which  several 
(lel)ta  may  be  eiiuitably  discharged  i>y  one  payment. 

(5;>8.  The  Matuflfy  of  any  obligation  is  the  date  at  which 
it  b(^Ci)nieH  due  or  draws  interest. 

«.'59.  The  Tevm  of  Credit  is  the  interval  of  time  from 
till'  date  a  debt  is  contracted  until  its  maturity. 

040.  The  Averaf/e  Term  of  Credit  is  the  interval  of 
time  from  the  viaturity  of  the  first  item  in  an  account  to  the 
Eqiuitcd  Time. 


■P=T? 


Ei^UATIO  y     OF    PA  I'M  hW  T:i, 


2G7 


lirect 

10000 
i\  and 
lit  real 
rancB, 
remit- 
'auada 


eUt  and 

the  pttTtiB 
or  rigbt- 
>r  person 

jctween 

finding 
taynieni 
loss  ol" 

Beveral 
[t  whicli 
le  from 

>rval  of 
it  to  tliu 


PREPARATORY    PROPOSITIONS. 

041.  The  method  of  ppttling  accounts  hy  tqitdtiini  of  pdy. 
ments  depends  upon  the  following  propositions  ;  hence  they 
shouM  he  carefully  studied  : 

Prop.  1. —  When,  by  agreement,  no  intenst  u<  to  be  paid  on  a 
debt  from  u  ttpecijied  time,  if  any  part  of  the  amount  i-i  paid  by 
the  debtor,  he  is  entitled  to  iitUrcut  untU  the  cjrplrat,  m  of  the 
specified  time. 

Thne,  A  owoh  B  $100,  payable  In  12  months  without  Intorct.  whirh 
means'  that  A  is  entitled  by  agreement  to  the  iice  of  $100  of  B's  mniH-y  l«<r 

12  months.    Hence,  IT  ho  '>nyt«  any  part  of  it  before  the  expiration  of  the 

13  months,  he  lis  entitled       iutcrest. 

Obsertt,  that  when  credit  is  given  without  charging  interest,  the  i)rofltB 
or  advantage  of  the  transaction  are  such  as  to  give  the  creditor  an  equiva- 
lent for  the  loss  of  the  interest  of  his  money. 

Prop.  II. — After  a  debt  is  due,  or  tht  time  expires  for  which 
by  agreement  no  interest  is  charged,  the  creditor  is  entitled  to 
interest  on  the  avMunt  until  it  is  paid. 

Thus,  A  owes  B  $300,  duo  in  10  days.  When  the  10  days  expire,  the  f  :X)0 
should  be  paid  by  A  to  B.  If  not  paid,  B  loses  the  use  of  the  moucy,  aiid  is 
hence  entitled  to  interest  until  it  is  paid. 

Prop.  III.—- W7ien  a  term  of  credit  is  allowed  upon  any  of 
the  items  of  an  account,  the  date  at  irhicJi  such  items  are  due  or 
commence  to  draw  interest  is  found  by  adding  its  term  of  credit 
to  the  date  of  each  item. 

Thns,  goods  purchased  Marrh  10  on  10  days'  credit  would  be  due  or  draw 
interest  March  10  +  40  da.,  or  April  11). 

042.  Prob.  I. — To  settle  equitably  an  account  con- 
taining only  debit  items. 

R.  Bates  bought  merchandipe  of  IT.  P.  Emerson  as  follows  r 
May  IT,  1875,  on  3  montlis'  credit,  $205;  July  11,  on  25  days, 
$4G0 ;  Sept.  15,  on  05  days,  $G50. 


208 


B  US  ly  K  a  a   a  ii  i  tum  e  tic. 


m  1 


Find  the  equated  time  niul  tli<>  ninouut  that  will  equitably 
settle  the  account  at  the  dut«'  when  the  last  Item  is  due,  1% 
intcnst  being  allowed  on  each  item  from  maturity. 


N . 


f 

h  ' 

1 

^^'1 

p  - 

% 

'    .;  ,  » 

1 

,;>( 

* 

¥^ 

JV 

/'. 

i,^'  ■ 

A 


'nraf 


m 


BOLUTIOM  BT  INTKHB8T   MBTUOD. 

1.  Wo  flnd  tbo  datu  of  maturity  of  each  item  thus  : 

$3(i5  on  3  mo.  ix  due  May  17+8  mo.  =  Aug.  17. 
f  t6()  «>ii  S.'i  da.  irt  duo  July  11  +35  da.  -  Au)?.  5. 
$tifiU  on  U5  da.    Ih  duu  Sept.  13  r  65  da.    =  Nov.  19. 

i.  An  the  HnmH  of  the  debt  are  duo  nt  thono  dnt<'«.  It  is  evident  that  when 
thoy  nil  romain  unpaid  until  the  latest  maturity,  II.  V.  Kuxthod  is  entitled 
to  Ic^'al  iuteroBt 

On  laea  from  An;;.  17  to  Nov.  1»    ^    94  do. 
On  |460  from  Au^.    5  to  Nov.  19  ::=  lUii  du. 

The  t050  being  due  Nov.  19  bearti  no  intervnt  before  thiN  date. 

3.  On  Nov.  19,  II.  P.  Ememon  U  entitled  to  receive  |i;n5,  the  Hum  of  the 
IttMii*  of  till!  dfl)t  and  tlie  iutcn-hi  on  |3(j5  fur  94  da.  pluB  the  interest  ou 
$m\  for  l()(i  (!a.  at  "7%,  which  Ih  $14.13. 

Ilcncc  thi'  account  may  bo  eciuifnbly  nettled  on  Nov.  19  by  It.  Baton  pny- 
ing  II.  1*.  KmerHon  |1375  +  |14.19  -  $1389.12. 

4.  Slnco  n.  P.  EmorHon  Ik  entitled  to  rocelvo  Nov.  19,  tvm  ♦•  |14.1'i 
lntol•(•^t,  it  la  evident  that  if  he  Ih  paid  ♦l})7.'i  a  Kuftlelont  tinx-  Ix'forc  Nov.  li» 
to  yii'M  $1 1.13  Interent  ut  thin  (inte,  the  del)t  will  b«'  eiinitably  nettled.  But 
|l.'i7.*>,  accordintf  to  (."STO),  will  yield  $14.13  In  53  +  a  fraction  of  a  day. 

Hence  the  equated  lime  of  hottlomcnt  in  Sept.  80,  which  in  W  day*  pre- 
▼iouH  to  Nov.  19,  the  aHBumed  date  of  settlement. 


SOLUTION  BT  PBOnncT  MSTIIOD. 

1.  Wo  find  in  the  same  manner  as  in  the  Interest  mtthod  the  date*  of 
maturity  and  the  number  of  days  each  item  bear:'  iuterobt. 

3.  A»-nn>in(f  Nov.  19,  the  latent  maturity,  an  the  date  of  settlement.  It  In 
evhiont  that  II.  P.  KmerMon  nhouid  be  paid  nt  tliis  date  flliVS,  the  «um  or 
the  item"  of  the  account  and  the  Interest  on  |t3(S5  for  tM  d.iyH  pluH  the  inter- 
681  on  ^-ttiO  for  llHi  dayx. 

3.  Since  the  interest  on  $3«'»5  for  94  dayn  at  any  fjlven  rate  in  equal  to 
the  Interent  on  ♦3<i6'<9l,  or  f34'.U0,  for  1  day  at  the  came  rate,  and  tin- 
iutereat  on  $100  fur  100  da^tt  i^  equal  to  the  iutcret«t  on  $400x106,  or 


E  QUA  T/0  X     OF    P  A  Y  .V  /;  X  T  S 


200 


5  ^  fl4.1^ 
.r.'  Nov.  !'.• 
WvA.    But 
11  (lay. 
day*  pr»'- 


RULE. 

043.     /.  Find  the  date  of  maturity  of  each  item. 

IT.  AMume  m  the  date  of  Hittlement  the  latest  inatunty,  and 
find  the  nutnber  of  days  from  thi.s  date  to  the  maturity  of  eaeh 
item. 


L'  dfttCB  of 

iu«nt.  It  \* 
I  the  Intor- 

cqnnl  to 

|o.  niifl  the 

I K 106,  or 


In  cft8o  tlu<  linlebto<lne88  is  discharged  at  the  aHt^uim-d  date 
(it'rtettlumont : 

fIL  Find  the  interest  on  eaeh  item  from  its  maturity  to  the 
(late  of  Hettlement.  Thi'  Hitin  of  the  items  plan  thijt  interest  in  ths 
am^iunt  that  mnnt  he  paid  the  creditor. 

In  casu  the  equated  time  or  term  of  credit  U  to  ho  found  and 
tlu'  intlubtodness  discharged  in  one  paymeiit,  cither  hy  cash  or 
note: 


■■•^ 


I 


iltably 

le,  7% 


h»t  when 
«  cntltl«itl 


lium  of  Ibi' 
inuroiit  ou 


$48700,  for  1  day,  the  Intorpnf  dno  TI.  P.  Fmcr!«on  Nov.  19  Id  cqnal  the  Inter- 
<)«t  of  $94910  +  |4K7(tO,  or  fiTStlTO,  n)r  1  day. 

4.  8inco  the  IntttroHt  on  178(170  Tor  1  day  Ih  eqnnl  to  the  Intni^ft  on 
I1HT5  fur  iiD  many  dnyn  uh  |1H75  in  rontainrd  Wnwn  In  $7:J(i70,  whirh  i« 
MiH,  It  In  evident  thai  If  II.  P.  EmerBou  receive  the  ui«e  of  fl.TTS  for 
M  days  previouH  to  Nov.  10,  it  will  \w  equal  to  the  intcreftt  ou  I7W70 
for  1  day  paid  ut  that  date.  CouHequcntly,  U.  Ilutea  liy  |»ttyliu'  11^175 
Sept.  90,  which  I0  &4  dayH  boforo  Nov.  10,  (Hhdiur^em  equitably  the 
indobte<ine«M. 

Hence,  Sept.  80  in  tin*  egiiaftd  flmf,  and  from  Anjj.  5  to  Sept.  96,  or 
.VJ  day»,  Ih  the  arenujt  ttiin  ((f  emtlt 

Ot>Mrv€,  tliat  R.  BatuH  may  dihcbargu  uquital>ly  thu  iuUebtedueHM  iu  OQO 
of  throe  way»  : 

(1.)  liy  paying  Xov.  10,  the  InUst  maturity,  $1875,  the  mm  of  the  ifetru  qf 
the  acemmt,  and  th€  Intermt  (/  $73070/"/  1  day. 

In  thii  case  the  payment  Ih  $1:175  +  $14.1-.J  intcroHt   --  $10K!).i:2. 

(9.)  By  paying  $1075,  tA$  $um  qf  the  itetn*  in  ca*h,  on  ^Sejtt.  '41,  tht 

KqiTATEO  TIME. 

(3.)  Hy  ffiHng  Mji  note  for  81.175,  the  mim  qf  the  itemn  qf  the  account^ 
hi-itring  intere/it  from  Stpt.  ai,  fhf  equated  time. 
ObHorve,  tblb  Ih  equivalent  to  puyiuK  thu  $1075  iu  cai<h  Sept.  96. 

From  these  illustrations  we  ol)tnin  the  following 


j 


M. 


%. 


■.%. 


'V^, 


%^.^^ 


IMAGE  EVALUATION 
TEST  TARGET  {MT-3) 


'y 


A 


,M 


£/  z:^ 


A 


& 
^ 


1.0 


I.I 


1.25 


la  IM  iiiM 

•«  IM    1112.2 
140     III!  2.0 

1-4    IIIIII.6 


V] 


<^ 


/}. 


'm 


•n 


/ 


/A 


'W 


'/ 


Photographic 

Sciences 
Corporation 


33  WEST  MAIN  STREET 

WEBSTER,  N.Y.  14580 

(716)  872-4503 


V 


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270 


B  US  INE  SS     A  li  I  Til  ME  TIC. 


''Kb. 

■     V 


IV.  Multiply  each  item  by  the  number  of  days  from  its  ma^ 
turity  to  the  latest  maturity  in  t/ie  accoujit,  and  dicide  the  sum 
of  these  products  by  the  sum  (f  the  items ;  the  quotient  is  the 
ma/iber  of  days  which  must  be  counted  back  from  the  latest  ma- 
turity to  give  the  equated  time. 

V.  The  first  mMurity  subtracted  from  tJie  equated  time  gives 
the  average  term  of  credit. 


if 


r 


^;f! 


!      • 
j 


EXAMPLES    FOR    PRACTICE. 

044.  1.  Henry  Ross  purchases  Jan.  1,  1876,  $1600  worth 
of  goods  from  James  Mann,  payable  as  follows  :  A])ril  1, 1876, 
$700;  June  1,  1876,  $400;  and  Dec.  1,  1876,  $500.  At  what 
date  can  he  equitably  settle  the  bill  in  one  payment  ? 

When  the  interval  between  the  maturity  of  each  item  and  the  date  of 
settlement  is  months,  as  in  this  example,  the  months  should  not  be  reduced 
todays;  thus, 

Solution.— 1.  Assuming  that  no  pajrment  is  made  until  Dec.  1,  James 
Maun  is  entitled  to  interest 

On  $700  for  8  mo.  =  $700  x  8  or  $5600  for  1  month. 
Ou  1400  for  6  mo.  =  $400  x  6  or  $^400  for  1  mouth. 

Ilence  he  is  entitled  to  the  use  of  $8000  for  1  month. 

2.  $8000 -^  IICOO  =  5,  the  number  of  months  (642—4)  which  must  be 
counted  back  from  Dec.  1  to  find  the  equated  time,  which  is  July  1.  Hence 
the  bill  can  be  equitably  settled  in  one  payment  July  1, 1876. 

2.  Bought  merchandise  as  follows :  Feb.  3,  1875,  $S80  ; 
A])ril  18,  $520 ;  May  18,  $260  ;  and  Aug.  12,  $350,  each  item 
on  interest  from  date.  What  must  be  the  date  of  a  note  fcjr 
the  sum  of  the  items  bearing  interest  which  will  equitably 
settle  the  bill  ? 

3.  A  man  purchased  a  farm  May  23, 1876.  for  $8600,  on  which 
he  paid  ^2600,  and  was  to  pay  the  balance,  without  interest,  as 
follows  :  Aug.  10, 1876,  $2500  :  Jan.  4,  1877,  $1500  ;  and  June 
14,  1877,  $2000.  Afterwards  it  was  agreed  that  the  whole 
should  be  settled  in  one  payment.  At  what  date  must  the 
payment  be  made  ? 


( ma- 
sum 
is  the 
5  ma- 
gives 


worth 

,  1876, 
t  what 

J  date  of 
J reduced 

1,  James 


must  be 
Hence 


$380; 
:h  item 
lote  for 
luitably 

which 

lerest,  as- 

\d  June 

whole 

^ust  the 


H  Q  UA  T 1  0  y    OF    PA  Y  M  EXTS. 


271 


Find  the  date  at  which  a  note  bearing  interest  can  be  given 
as  an  eq uitable  settlement  for  the  amount  of  oacli  of  tlie  fol- 
lowing bills,  each  item  being  on  interest  from  the  date  of 
purchase : 


4.  Purchased  as  foUows  : 
July    9,  1876,  $380 
Sept.  13,     "     $270 
Nov.  24,     "     $840 
Dec.  29,     "     $260. 

6.  Purchased  as  follows : 

April  17,  1877,  $186  ; 
June  24,  "  $250; 
Sept.  13,    "     $462. 


6.  Purchased  as  follows 

May  5, 1876,  $186 
Aug.  10,  "  $230 
Oct.  15,  "  $170 
Dec.  20,    "     $195. 


7.  Purchased  as  follows 

Aug.  25, 1877,  $280; 
Oct.  10,  "  $193; 
Dec.   18,     "     $290. 


8.  Sold  A.  Williams  the  following  bills  of  goods :  July  10, 
$2300,  on  6  mo.  credit ;  Aug.  15,  $900,  on  5  mo. ;  and  Oct.  13, 
$830,  on  7  mo.  What  must  be  the  date  of  note  for  the  three 
amounts,  bearing  interest,  which  will  equitably  settle  the 
account. 

9.  Find  the  average  term  of  credit  on  goods  purchased  as 
follows:  Mar.  23,  $700,  on  95  da.  credit;  May  17,  $480,  on 
45  da. ;  Aug.  25,  $690,  on  60  da. ;  and  Oct.  2,  $380  on  35  da. 


045.  Prob.  II. — To  settle  equitably  an  account  con- 
taining both  debit  and  credit  items. 

Find  the  amount  equitably  due  at  the  latest  maturity  of 
either  the  debit  or  credit  side  of  the  following  account,  and  the 
equated  time  of  paying  the  balance  : 


JDr. 

B.  Whitney. 

( 

7r. 

1877. 

1877. 

Mar.  17 

To  mdse.      .    .    . 

$400 

Apr.   13 

By  cash      .... 

$200 

May  10 

"    at  4  mo. 

aso 

Juno  15 

"  draft  at  30  da. 

250 

Ang.    7 

"       •'    at  2  mo. 

540 

I!/ 

:«!   Ill 


i  i! 


,1  1^! 


':!« 


i^ 


272 


BUSIXESS     ARITHMETIC. 


Before  examining  the  following  solntion,  study  carefully  the  three 
propoBitionB  under  (64 1 ). 


SOLUTION  BY  PBODCCT  METHOD. 


I   •■'>  ' 


Dtie. 

Mar.    17.  400 

Sept.    10.  380 

Oct.       7.  5^ 

Total  debt,  $1320 

Total  paid,  450 

Balance,  |870 


ytmt.    Days.  Products. 


204    = 
27    = 


81600 

$91860 
5&400 


Paid.       Amt.    Days.    Products. 
Apr.    13.        200   X   177    =    35400 
July    15.        250  X     84    =    21000 

450  56400 

Amt.  whose  Int.  for  1  da.  is  due  to  Cred. 
Amt.  whose  Int.  for  1  da.  is  due  to  Debt. 


$35460    Bal.  whose  Int.  for  1  da.  is  due  to  Cred. 


f 


Explanation.— Assuming  Oct.  7,  the  latest  maturity  on  either  side  of 
the  account,  as  the  date  of  settlement,  the  creditor  is  entitled  to  interest 
on  each  item  of  the  debit  side,  and  the  debtor  on  each  item  of  the  credit 
side  to  this  date  (641).  Hence,  we  find,  according  to  (642—3),  the 
amount  whose  interest  for  1  day  both  creditor  and  debtor  are  entitled  to 
Oct.  7. 

2.  The  creditor  being  entitled  to  the  most  interest,  we  subtract  the 
amount  whose  interest  for  1  day  the  debtor  is  entitled  to  from  the  cred- 
itor's amount,  leaving  $35460,  the  amount  whose  interest  for  1  day  the 
creditor  is  still  entitled  to  receive. 

8.  We  find  the  sum  of  the  debit  and  credit  items,  and  subtract  the  latter 
from  the  former,  leaving  $870  yet  unpaid.  This,  with  $6.80,  the  interest 
on  $35460,  is  the  amount  equitably  due  Oct.  7,  equal  $876.80. 

4.  According  to  (641—4),  $35460  +  $870  =  40??,  the  number  of  days 
jrevious  to  Oct.  7  when  the  debt  can  be  discharged  by  paying  the  balance, 
$870,  in  cash,  or  by  a  note  bearing  interest.  Hence  the  equated  time  of 
paying  the  balance  is  Aug.  27. 


Tli3  following  points  regarding  the  foregoing  solution  should 
be  carefully  studied  : 


1.  In  the  given  example,  the  sum  of  the  debit  is  greater  than  the  sum  of 
the  credit  items  ;  consequently  the  balance  on  the  account  is  due  to  the 
creditor.  But  the  balance  of  interest  being  also  due  him,  it  is  evident  that 
to  settle  the  account  equitably  he  should  be  paid  the  $870  before  the 
assumed  date  of  settlement.  Hence  the  equated  time  of  paying  the  balance 
must  be  before  Oct.  7. 

2.  Had  the  balance  of  interest  been  on  the  credit  side.  It  is  evident  the 
debtor  would  be  entitled  to  keep  the  balance  on  the  account  until  the 


e  three 


'Products. 
35400 
21000 

56400 
e  to  Cred. 
3  to  Debt. 

to  Cred. 

er  Bule  of 

0  interest 
the  credit 
i— 3),  the 
jutitled  to 

ibtract  the 

1  the  cred- 
1  day  the 

the  latter 
le  interest 

;r  of  days 
|e  balance, 
}d  time  of 


Bhould 


lie  sum  of 
liie  to  the 
Ident  that 
]>efore  the 

le  balauce 

^ident  the 
until  the 


EQUATION    OF    PAYMEXTS. 


273 


Interest  upon  it  would  be  equal  the  Interest  due  him.    Tlence  the  equated 
time  of  paying  the  balance  woakl  be  after  Oct.  7. 

3.  Had  the  balance  of  the  account  been  on  the  credit  side,  the  creditor 
would  be  overpaid,  and  hence  the  balance  would  be  due  to  the  debtor. 

Now  in  case  the  balance  of  interest  is  also  on  the  credit  side  and  due 
to  the  debtor,  it  is  evident  that  to  settle  the  account  equitably  the  debtor 
should  be  paid  the  amount  of  the  balance  before  the  a^ssumed  date  of  set- 
tlement.   Hence  the  equated  time  would  be  before  Oct.  7. 

In  case  the  balance  of  interest  ia  on  the  debtor  ^^i(le,  it  is  evident  that 
while  the  creditor  has  been  overpaid  on  the  account,  he  is  entitled  to  a 
balance  of  interest,  and  consequently  should  keep  the  amount  he  has  been 
overpaid  until  the  interest  uiwu  it  would  be  equal  to  the  interest  due  him. 
Hence  the  equated  time  would  be  afitr  Oct.  7. 

4.  The  interest  method  given  (642)  can  be  used  to  mlvautage  in  finding 
the  equated  time  when  the  time  is  long  between  the  maturity  of  the  items 
and  the  assumed  date  of  settlement.  In  case  this  method  of  solution  is 
adopted,  the  foregoing  conditions  are  equally  applicable. 

From  these  illustrations  we  obtain  the  following 


HTJLE. 

646.  /.  Find  the  maturity  of  each  item  on  the  debit  and 
credit  side  of  the  accovnt. 

II.  AsfiJfme  as  the  date  of  settlement  the  latest  maturity  on 
either  side  of  the  account,  and  find,  in  the  manner  indicated 
abfwe,  tlie  number  of  days  from  this  date  to  the  maturity  of  each, 
on  both  sides  of  the  account. 

III.  Multiply  each  debit  and  credit  item  by  the  number  of  daps 
from  its  maturity  to  the  date  of  settlement,  and  divide  the  bal- 
ance of  the  debit  and  credit  products  by  the  balance  of  the  debit 
and  credit  items  ;  the  quotient  is  the  number  (fdays  the  equated 
time  is  from  the  assumed  date  of  settlement. 

IV.  In  case  the  balance  of  items  and  balance  of  interest  are  both 
on  the  same  side  of  the  account,  subtract  this  number  of  days 
from,  the  assumed  date  of  settlement,  but  add  this  number  of 
days  in  ease  they  are  on  opposite  sides  ;  the  result  thus  obtainsd 
is  th€  equated  time. 


M 


frw 


'■\ 


if 


w 


274 


BUSINESS    ARITHMETIC, 


EXAMPLES    FOR    PRACTICE. 


647.    1.  Find  the  face  of  a  note  and  the  date  from  which  it 
must  bear  interest  to  settle  equitably  the  following  account : 


Dr. 


jAifES  Hand  in  acct.  vyith  P.  Anstead. 


Cr. 


1876. 

1876. 

Jan.    7 

To  mdee.  on  3  mo. 

$430 

Mar.  15 

By  draft  at  90  da. 

$500 

May  11 

«♦       "     "  2  mo. 

390 

May  17 

*'  cash   .... 

280 

June   6 

"     "  5  mo. 

570 

Aug.    9 

*'  mdse.  on  30  da. 

400 

2.  Equate  the  following  account,  and  find  the  cash  payment 
Dec.  7.1876: 


Dr. 


William  Henderson. 


CV, 


1876. 

1876. 

Mar.  83 

To  mdse.  on  46  da. 

$470 

Apr.  16 

By  cash    .... 

$4!K) 

May  16 

44       4i     ti  25  da. 

380 

June  25 

"  mdee.  on  30  da. 

650 

Aug.    7 

"       "     "  36  da. 

590 

July  13 

"  draft  at  60  da. 

2C0 

8.  Find  the  equated  time  of  paying  the  balance  on  the  follow- 
ing account : 


Dr. 


Hugh  Quthbib. 


Or. 


1876. 

1876. 

Jan.  13 

To  mdse.  on  60  da. 

$840 

Feb.  15 

By  note  at  60  da. 

$700 

Mar.  24 

"      '•  40  da. 

580 

Apr.  17 

"  cash   .    .    .    • 

460 

June   7 

"       "     "  4  mo. 

360 

June  9 

''  draft  at  30  da. 

1150 

July  14 

"     "  80  da. 

730 

1 

4.  I  purchased  of  Wm.  Rodgfers,  March  10,  1876,  $930 
worth  of  goods  ;  June  23,  $G80 ;  and  paid,  April  3,  |870  cash, 
and  gave  a  note  May  24  on  30  days  for  $500.  What  must  be 
the  date  of  a  note  bearing  interest  that  will  equitably  settle 
the  balance  ? 


i,! 


biicli  it 
nt: 

Cr. 


$500 
280 
I.  I    400 

Btynient 
Cr. 


—       ■ 

$4W 

a.        650 

2C0 

s  follow- 

er. 

$700 

4(jO 

1150 

,    $930 

10  cash, 

lUst  be 

settle 


HE  VIEW,  275 


REVIEW    AND    TEST    QUESTIONS. 

048.     1.  Define  Simple,  Compound,  and  Annual  Interest. 

2.  Illustrate  by  an  example  every  step  in  the  six  per  cent 
method. 

3.  Show  that  12%  may  be  used  as  conveniently  as  6J^,  and 
write  a  rule  for  finding  the  interest  for  months  by  this  method. 

4.  Explain  the  method  of  finding  the  e:vnct  interest  of  any 
sum  for  any  given  time.  Give  reasons  for  each  step  in  the 
process. 

5.  Show  by  an  example  the  difference  between  true  and  hiink 
discount.     Give  reasons  for  your  answer. 

6.  Explain  the  method  of  finding  the  present  worth. 

7.  Explain  how  the  face  of  a  note  is  found  when  the  pro- 
ceeds are  given.     Illustrate  each  step  in  the  process. 

8.  Define  Erchange,  and  state  the  difference  between  Inland 
and  Foreign  Exchan<]^e. 

9.  State  the  difference  in  the  three  biUs  in  a  Set  of  Ex- 
change. 

10.  What  is  meant  by  Par  of  Exchange  ? 

11.  State  the  various  methods  of  Inland  Exchange,  and  illus- 
trate each  by  an  example. 

12.  Illustrate  the  method  of  finding  the  cost  of  a  draft  when 
exchange  is  at  a  discount  and  brokerage  allowed.  Give  reasons 
for  each  step. 

13.  State  the  methods  of  Foreign  Exchange. 

14.  Illustrate  by  an  example  the  difference  between  Direct 
and  Indirect  exchan^\ 

15.  Define  Equation  of  Payments,  an  Account,  Equated 
Time,  and  Term  of  Credit. 

16.  Illustrate  the  Interest  Method  of  finding  the  Equated 
Time  when  there  are  but  debit  items. 

17.  State  when  and  why  you  count  forward  from  the  assumed 
date  of  settlement  to  find  the  equated  time. 


i 


If 


i  i 


w 


RATIO. 


PREPARATORY    PROPOSITIONS. 

04:9«  Tico  numbers  are  compared  and  their  relation  deter- 
mined by  dividing  the  first  by  the  second. 

For  example,  the  relation  of  $8  to  $4  is  determined  thus,  |8+$4  =  2. 
Observe,  the  quotient  2  indicates  that  for  every  one  dollar  in  the  $4,  there 
are  two  dollars  in  the  |8. 


I)  "J 


'•■< 


,1 


.»(, 


Be  particular  to  observe  the  following: 

1 .  Wlien  the  greater  of  two  numbers  is  compared  with  the 
less,  the  relation  of  the  numbers  is  expressed  either  by  the 
relation  of  an  integer  or  of  a  mixed  number  to  i\\e  iinit  1,  that 
is,  by  an  improper  fraction  whose  denominator  is  1. 

Thup,  20  compared  with  4  gives  20+4  =  5 ;  that  ie,  for  every  1  in  the  4 
there  are  5  in  the  20,  Hence  the  relation  of  20  to  4  is  that  of  the  integer  5 
to  the  nnit  1,  expressed  fractionally  thus,  |. 

Again,  29  compared  with  4  gives  29+4  =  7^  ;  that  is,  for  every  1  in  the  4 
there  are  7^  in  29.  Hence,  the  relation  of  29  to  4  is  that  of  the  mixed  num- 
h<iT  7i  to  the  unit  1. 

2.  Wlien  the  less  of  two  numbers  is  compared  with  the 
greater,  the  relation  is  expressed  by  a  proper  fraction. 

Thus,  6  compared  with  14  gives  6+14  =  ^\  =  f  (244) ;  that  is,  for  every 
3  in  the  6  there  is  a  7  in  the  14.  Hence,  the  relation  of  6  to  14  is  that  of  3  to 
7,  expressed  fractionally  thus,  f . 

Observe,  ih&t  the  relation  in  this  case  may  be  expressed,  if  desired,  as 

1 
that  of  the  unit  1  to  a  mixed  number.    Thus,  6+14  =  ,",  =  _,-  (244) ;  that 

is,  the  relation  of  6  to  14  is  that  of  the  unit  1  to  2}. 


i,  deter- 


^-|4  =  2 

$4,  there 


itli  the 
by  the 
1,  that 


in  the  4 
linteger  5 

in  the  4 
ied  num- 


ith  the 


for  every 
it  of  3  to 


jired,  ae 
[) ;  that 


BAT  JO. 


EXAMPLES     FOR     PRACTICE. 


277 


650.  Find  orally  the  relation 


1.  Of  56  to  8. 

2.  Of  24  to  3. 

3.  Of  38  to  5. 

4.  Of  70  to  4. 


5.  Of  25  to  100. 

6.  Of  113  to  9. 

7.  Of  13  to  90. 

8.  Of  10  to  48. 


9.  Of  85  to  9. 

10.  Of  42  to  77. 

11.  Of  75  to  300- 

12.  Of  10  to  1000. 


051.  Prop.  II. — iVo  numbers  can  be  compared  but  those 
which  are  of  the  same  denomination. 

Thus,  we  can  compare  $8  with  |2,  and  7  inches  with  2  inches,  bat  we  can- 
not compare  $8  with  2  inches  (144—1). 

Observe  carefully  the  following : 

1.  Denominate  numbers  must  be  reduced  to  the  lowest  de- 
nomination named,  before  they  can  be  compared. 

For  example,  to  compare  1  yd.  2  ft.  with  1  ft.  3  in.,  both  numbers  must 
be  reduced  to  inches.  Thus,  1  yd.  2  ft.  =  60  in.,  1  ft.  3  in.  =  15  in.,  and 
60  in.  -4-  15  in.  =  4 ;  hence,  1  yd.  2  ft.  are  4  times  1  ft.  3  in. 

2.  Fractions  must  be  reduced  to  the  same  fractional  denom- 
ination before  they  can  be  compared. 


For  example,  to  compare  35  lb.  with  }  oz.  we  must  first  reduce  the  3J  lb. 
to  oz.,  then  reduce  both  members  to  the  same  fractional  unit.  Thue, 
(1)  3'  lb.  =  56  oz. ;  (2)  56  oz.  =  ^JQ  oz. ;  (3)  «S?  oz.  -*■  |  oz. 
(279) ;  hence,  the  relation  3 J  lb.  to  |  oz.  is  that  of  70  to  1. 


=  »|°  =  70 


EXAMPLES    FOR    PRACTICE. 


652.  Find  orally  the  relation 

1.  Of  |2  to  25  ct  4.  Of  I  to  II. 

2.  Of  4  yd.  to  3  ft.       5.  Of  f  to  f. 

3.  Of  2\  gal.  to  f  qt.    6.  Of  f  oz.  to  2  lb. 


7.  Of  3  cd.  to  6  cd.  ft. 

8.  Of  11  pk.  to  3  bu. 

9.  Of  If  to  2\. 


Find  the  relation 

10.  Of  6|  to  |. 

11.  Of  105  to  28. 

12.  Of  94  bu.  to  3^  pk. 


13.  Of  $36|  $4?. 

14.  Of  1^  pk.  to  2?j  gal. 

15.  Of2yd.  1?  ft.  to  Jin. 


Ml 


ft 


10^. 


Ji78 


B  USIXI^S S     A  R  1  TU M  ETI  C, 


••■;,,  VI 


1  :,!'«.",i 


fl  ^t 


I 


DEFINITIONS. 

053.  A  Ratio  is  a  fraction  wliicli  expresses  the  relation 
wliicli  the  first  of  two  numbers  of  the  same  denomination  has 
to  the  second. 

Thus  the  relation  of  $6  to  $15  is  expressed  by  g  ;  that  is,  ^6 
is  I  of  $15,  or  for  every  }s^2  in  $G  there  are  $5  in  $15.  In  like 
manner  the  relation  of  $12  to  $10  is  expressed  by  ^. 

054.  The  Special  Sign  of  Ratio  is  a  colon  (:). 

Thus  4  :  7  denotes  that  4  and  7  express  the  ratio  i  ;  hence, 
4  :  7  and  |  are  two  ways  of  expressing  the  same  thing.  The 
fractional  form  being  the  more  convenient,  should  be  used 
in  preference  to  the  form  with  the  colon. 

655.  The  Terms  of  a  Ratio  are  the  numerator  and 
denominator  of  the  fraction  which  expresses  the  relation 
between  the  quantities  compared. 

The  first  term  or  numerator  is  called  the  Anteceflent,  the 
second  term  or  denominator  is  called  the  Consequent. 

656.  A  Simple  Ratio  is  a  ratio  in  which  each  term  is  a 
single  integer.    Thus  9  : 3,  or  f ,  is  a  simple  ratio. 

657.  A  Compound  Ratio  is  a  ratio  whose  terms  are 
fonned  by  multiplying  together  the  corresponding  terms  of  two 
or  more  simple  ratios. 


Thus,  multiplying  together  the  correBpouding  terms  of  the  simple  ratios 
7  :  3  and  5  :  2,  we  have  the  compound  ratio  5  x  7  :  3  x  2  =  85 :  6,  or  ex- 

86 
"6* 


pressed  fractionally  J  x  _  =  __!i_° 
^  8      2      8x2 


Observe,  that  when  the  multiplication  of  the  corresponding  terms  is  per- 
formed, the  compound  ratio  is  reduced  to  a  simple  ratio. 

658.  The  Recij)rocal  of  a  number  is  1  divided  by  that 
number.    Thus,  the  reciprocal  of  8  is  1  -5-  8  =  -J. 


m 


BA  TIO. 


279 


relation 
ion  lias 


It  is,  !J6 
In  like 


;  hence, 

ig.     The 
be  used 


ator  and 
relation 

lent,  the 
t. 

term  is  a 


jrms  are 
IS  of  two 


iplc  ratios 
|:  6,  or  ex- 


IB  is  per- 
by  that 


659,  The  Reciprocal  of  a  Ratio  is  1  divided  by  the 
ratio. 

TbuB,  the  ratio  of  7  to  4  f'^  7  :  4  or  J,  and  Us  reciprocal  1«  1  -•- 1  =  |, 
according  tu  (280).  Hence  the  reciproatl  ul'a  ratio  b  thu  ratio  iDvcrted, 
or  thu  constquerU  divided  by  the  antecedent. 

OHO.  A  Ratio  is  in  its  Sitnptest  Terms  wlien  tlie  ante- 
cedent and  consequent  are  prime  to  each  otlier. 

OOl.  The  Redaction  of  a  Ratio  is  the  process  of  chang- 
ing its  terms  without  changing  the  relation  they  express. 

^tiB  J,  1, 9,  each  expree>s  the  same  relation. 


PROBLEMS    ON    BATIO. 

602.  Since  every  ratio  is  either  a  proper  or  improper  frac- 
tion, the  principles  of  reduction  discussed  in  (224)  apply  to 
the  reduction  of  ratios.  The  wording  of  tlic  principles  must  l)e 
slightly  modified  thus : 

Prin.  I. — TJie  terms  of  a  ratio  must  each  represent  units  of 
the  same  kind. 

Pit  IN.  II. — Multiplying  both  terms  of  a  ratio  by  tJie  same 
number  does  not  change  t?ie  value  of  the  ratio. 

Prin.  III. — Dividing  both  terms  of  a  ratio  by  the  same  num- 
ber does  not  change  the  value  of  the  ratio. 

For  the  illustration  of  these  principles  refer  to  (224). 

063.  Prob.  I.— To  find  the  ratio  between  two  given 
numbers. 

Ex.  1.    Find  the  ratio  of  $56  to  $84. 

Solution.— Since,  accordin":  to  (649),  two  numbers  are  compared  by 
divii'.in^  the  first  by  the  eecoud,  we  divide  |i3G  by  $84,  giving  $56  +  |84  = 
l\  J  that  ie,  $56  is  2'  of  |84.    Hence  the  ratio  of  $56  to  $84  is  ||. 


f 


ii 


m 


280 


Lf  .  , 


■t^' 


I'  f 


r^ 


1  .■ 


i 


-"■ 

: ,  ' 

!  ^ 

;«-  - 

^*t. 

^^B' 

u^u.::-' 

f 

BUSINESS    ARITHMETIC, 


Ex.  2.    Find  th.*  ratio  of  1  yd.  2  ft.  to  1  ft.  3  In. 

Solution.— 1.  Since,  according  to  (051),  only  numbers  of  the  Bamc 
dcnoniliiatlon  can  bo  coniimrotl,  wo  reduce  both  termH  to  Inchec,  jjivlnjr 
6U  In.  and  15  in. 

2.  Dividing  60  in.  by  16  in.  we  havo  60  in.  -«-  16  in.  =  4 ;  that  it),  60  in.  it) 
4  timuH  15  in.    llcnco  Ibo  ratio  of  1  yd.  2  ft.  to  1  It.  3  lu.  it) ;. 


EXAMPLES    FOR    PRACTICE. 

«04.  Find  tlio  ratio 

1.  Of  143  yd.  to  305  yd.  3.  Of  73  A.  to  3G5  A. 

2.  Of  $512  lo  $250.  4.  Of  082  da.  to  21)40  da. 

5.  Of  £41  58.  Od.  to  £2  38.  6d. 

6.  Of  20  T.  0  cwt.  93  lb.  to  25  cwt.  43  lb.  5  oz. 

0(J5.  P«OB.  II.— To  reduce  a  ratio  to  its  simplest 
terms. 

Reduce  the  ratio  Y  ^o  i^^  simplest  tenns. 

Solution.— Since,  according  to  (602— III),  the  value  of  the  ratio  V  i.s 
not  cbanped  by  dividing  both  terms  by  the  pame  number,  we  divide  the 
antecedent  15  and  the  consequent  9  by  3,  their  greatest  common  divisor, 

giving  —  "*"     =  -.    But  having  divided  15  and  9  by  their  greatest  common 
^*'9+33 

divisor,  the  quotients  T>  and  3  must  be  prime  to  each  other.    Ilence  (600) 

1  are  the  simplest  terms  of  the  ratio  V* 


EXAMPLES     FOR    PRACTICE. 
COG.  Reduce  to  its  simplest  terms 


1.  The  ratio  21  :  50. 

2.  The  ratio  0  :  9. 

3.  The  ratio  ^. 


4.  The  ratio  05  :  85. 

5.  The  ratio  |f  f . 

0.  The  ratio  195  :  39. 


Express  in  its  simplest  terms  the  ratio  (see  516) 


7.  Of  I  ft.  to  2  yd. 

8.  Of  90  T.  to  50  T. 


9.  Of  3  pk.  5  qt.  to  1  bu.  2  pk. 
10.  Of8s.  9d.  to£l. 


RATIO. 


281 


1 


667.  Prob.  III.— To  find  a  number  that  has  a  given 
ratio  to  a  given  number. 

How  many  dollars  are  g  of  |72  f 

BoLVTiON.— The  fhiction  }  denotes  the  ratio  of  the  required  nambor  to 
ITS;  uamely,  for  every  $8  iu  f79  there  are  $5  in  the  required  uumbur. 
Connequontly  we  divide  the  $73  by  $8,  and  maltiply  $5  by  the  quotient. 
Honce,  first  step,  173  -»•  $8  =  9;  second  step,  $5x9  =  945,  the  required 
number. 

ObservSt  that  this  problem  is  the  same  as  Pbob.  VIII  (400),  and 
Prob.  n  (963)*  Compare  this  solution  with  the  solution  iu  each  of  these 
problems. 


i 


EXAMPLES    FOR    PRACTICE. 

668.  Solve  and  explain  each  of  the  following  examples, 
regarding  the  fraction  in  every  case  as  a  ratio. 

1.  A  man  owning  a  farm  of  248  acres,  sold  /^  of  it ;  how 
many  acres  did  he  sell  1 

2.  How  many  days  are  -^  of  360  days  ? 

8.  James  has  $706  and  John  has  }  as  much ;  how  much 
has  John? 

4.  Mr.  Jones  has  a  quantity  of  flour  worth  $3140 ;  part  of 
it  being  damaged  he  sells  the  whole  for  ^  of  its  value ;  how 
much  does  he  receive  for  it? 

5.  A  man's  capital  is  $4500,  and  he  gains  ^  of  his  capital ; 
how  much  does  he  gain  ? 

669.  Prob.  IV.— To  find  a  number  to  which  a  given 
number  has  a  given  ratio. 

$42  are  j  of  how  many  dollars  7 

Solution.— The  fraction  |  denotes  the  ratio  of  $43  to  the  required 
number ;  namely,  for  every  |7  In  $43  there  are  $4  in  the  required  number. 
Consequently  we  divide  the  $43  by  $7  and  multiply  |4  by  the  quotient. 
Hence,  first  step^  t4S  -»-  $7  =  6 ;  teoond  step,  $4x6  =  $94,  the  required 
number. 

Observe,  that  this  problem  is  vhe  same  as  Pbob.  IX  (491).  Compare 
the  solutions  and  notice  the  points  of  difference. 

19 


i,'  * 


282 


BUSINESS    ARITHMETIC, 


,1 


EXAMPLES    FOR    PRACTICE. 

670.  Solve  and  explain  each  of  the  following  examples, 
regarding  the  fraction  in  every  case  as  a  ratio. 

1.  I  received  |75,  which  is  f  of  my  wages;  how  much  is 
still  due? 

2.  96  acres  are  |f  of  how  many  acres  1 

3.  James  attended  school  117  days,  or  ^^  of  the  term ;  how 
many  days  in  the  term? 

4.  Sold  my  house  for  |2150,  which  was  ^f  of  what  I  paid 
for  it ;  how  much  did  I  lose  ? 

5.  48  cd.  3  cd.  ft.  of  wood  is  ^j  of  what  I  bought;  how 
much  did  I  buy  % 

6.  Henry  reviewed  249  lines  of  Latin,  or  f  of  the  term's 
work ;  how  many  lines  did  he  read  during  the  term  ? 

7.  Mr.  Smith's  expenses  are  f  of  his  income.  He  spends 
^^1500  per  year  ;  what  is  his  income  ? 

8.  A  merchant  sells  a  piece  of  cloth  at  a  profit  of  $2.o0, 
which  is  ^\  of  what  it  cost  him  ;  how  much  did  he  pay  for  it  ? 

9.  4  gal.  3  qt.  1  pt.  are  /j  of  how  many  gallons? 
10.  7  yards  and  2  ft,  are  f  of  how  many  yards  ? 

671.  Prob.  v.— To  find  a  number  to  -which  a  giren 
number  has  the  same  ratio  that  two  other  given  numbers 
have  to  each  other. 


To  how  many  dollars  have  $18  tlie  same  ratio  that  6  yd. 
have  to  15  yd.  ? 

SoLtrnoN.— 1.  We  find  by  (663—1)  the  ratio  of  (5  yd.  to  15 yd.,  which 
^  r*f  =  i«  a^xording  to  (659). 

8.  SiBoe  f  denotev  the  ratio  of  ikab  fid  to  tito  reqoired  wuDber,  the  |18 
mnat  be  the  antecedent ;  benoe  we  have,  aooordiAg  to  <6^0).  Jin^t  s(ej), 
418  -I-  $3  =  9 ;  aeoond  et^,  $5x9:?  $46,  the  reqaired  number. 

Ob^erre,  that  in  this  problem  we  have  the  antecedent  of  a  ratio  ^ren  to 
And  the  coneaqueut.  In  the  following  we  have  the  consequeot  given  to 
find  the  antecedent 


R  A  TIO. 


283 


mples, 
lucb  is 

tt ;  how 

,t  I  paid 

it;  how 

e  term's 

e  spends 

of  $2.50, 
y  for  it? 


a  given 
lumbers 


It  6  yd. 


[yd. ,  which 
r,  ihe  |18 

lo  piTcn  to 
[t  given  to 


672.  Prob.  VI. — To  find  a  number  that  has  the  same 
ratio  to  a  given  number  that  two  other  given  numbers 
have  to  each  other. 

How  many  acres  have  the  same  ratio  to  12  acres  that  $56 
have  to  $84? 

Solution.— 1.  We  find  by  (663—1)  the  ratio  of  $56  to  $84,  which 
is  IS  =  S,  according  to  (660). 

2.  Since  i  denotes  the  ratio  of  the  required  number  to  12  acres,  the  12 
acres  must  be  the  consequent;  hence  we  have,  according  to  (666".,//>'< 
step,  12  acr.  +  8  acr.  =  4 ;  second  stsp,  2  acr.  x  4  =  8  acres,  the  required 
number. 

fiXAMPLES    FOR    PRACTICE. 

673.  The  following  are  applications  of  Prob.  V  and  VI. 

1.  If  12  bu.  of  wheat  coat  $16,  what  will  42  bu.  cost? 

Regarding  the  solution  of  examples  of  this  kind,  observe 
that  the  price  or  rate  per  unit  is  assumed  to  be  the  same  for  each 
of  the  quantities  given. 

Thus,  since  the  12  bn.  cost  $15,  the  price  per  bushel  or  unit  Is  $1.25,  and 
the  example  aslis  for  the  cost  of  42  bu.  at  this  price  per  bushel.  Cuu- 
eequently  whatever  part  the  12  bu.  are  of  42  bu.,  the  $15,  the  cost  of  12  bu., 
must  be  the  same  part  of  the  cost  of  42  bu.  Hence  we  find  the  ratio  of 
12  bu.  to  42  bu.  and  solve  the  example  by  Prob.  V. 

2.  If  a  man  earn  $18  in  2  weeks,  how  much  will  he  earn  in 
52  weeks  ? 

3.  What  will  16  cords  of  wood  cost,  if  2  cords  cost  $9  ? 

4.  If  24  bu.  of  wheat  cost  $18,  what  will  30  bu.  cost  ? 

5.  If  24  cords  of  wood  cost  $60,  what  will  40  cords  cost  ? 

6.  Bought  170  pounds  of  butter  for  $51 ;  what  would  680 
pounds  cost,  at  the  same  price  ? 

7.  At  the  rate  of  16  yards  for  $7,  how  many  yards  of  cloth 
can  be  bought  for  $100  ? 

8.  Two  numbers  are  to  each  other  as  10  to  15,  and  the  less 
number  is  329  ;  what  is  the  greater? 


P'; 


■>  <.        J' 


^'' 


PROPORTION 


f* 


DEFIinTIONS. 

674,  A  Proportioti  is  an  equality  of  ratios,  the  tenns  of 
the  ratios  being  expressed. 

ThuH  the  ratio  J  is  equal  to  the  ratio  U  ;  hence  i  =  y  is  a  proportion, 
and  is  read,  The  ratio  of  3  to  5  is  equal  to  the  ratio  of  12  to  30,  or  3  is  to  5 
as  12  is  to  20. 

075.  The  equality  of  two  ratios  constituting  a  proportion  is 
indicated  either  by  a  double  colon  (:  :)  or  by  the  sign  (=). 

Thus,  I  =  T*s,  or  8  :  4  =  9  :  19,  or  8  :  4  : :  9  :  12. 

076.  A  Simple  Proportion  is  an  expression  of  the 
equality  of  two  simple  ratios. 

Thus,  t\  =  II,  or  8  :  12  : :  82  :  48,  or  8  :  12  =  32  :  48  is  a  simple  propor- 
tion.   Hence  a  simple  proportion  contains  four  terms. 

C77.  A  Compound  Proportion  is  an  expression  of  the 
equality  of  a  compound  (057)  and  a  simple  ratio  (650). 

2:3) 

Thus,     "     >  : :  48  :  60,  or  S  X  J  =  43,  is  a  compound  proportion.    It  is 
6:5) 

read.  The  ratio  2  into  6  is  to  3  into  5  as  4S  is  to  60. 

078.  A  Proportional  is  a  number  used  as  a  term  in  a 
pro{)ortion. 

Tims  in  the  simple  proportion  3  :  5  : :  6  :  15  the  numbers  2,  5,  6,  and  15 
are  its  terms ;  hence,  each  one  of  these  numbers  is  called  a  proportional, 
and  the  four  numbers  together  are  called  proportionals. 

Wh<^n  three  numbers  form  a  proportion,  one  of  them  is  repeated.  Thus, 
32  :  8  : :  8  : 8. 


SIMPLE    PROPORTION. 


285 


•me  of 


>ortion, 
is  to  5 


rtion  is 


of  tlie 
propor- 
of  tlie 

1.     It  IB 

in  a 


\,  and  15 
)rtion!iK 

Thus, 


670.  A  Mean  Proportional  is  a  number  that  is  the 
Conaequent  of  ono  and  the  Antecedent  of  the  other  of  the  two 
ratios  forming  a  proportion. 

Tbas  in  the  proportion  4  :  8  : :  8  :  16,  the  namber  8  iB  the  consequent  of 
the  first  ratio  and  the  antecedent  of  the  second ;  hence  is  a  mean  propar- 
tional. 

(>80.  The  Aiifeceflents  of  a  proportion  are  the  first  and 
third  terms,  and  the  Consequents  are  the  second  and  fourth 
terms. 

081.  The  Extremes  of  a  proportion  arc  its  first  and 
fourth  terms,  and  the  Means  are  its  second  and  third 
terms. 


SIMPLE    PROPORTION. 

PREPARATORY     STEPS. 

082.  The  following  preparatory  steps  should  be  i^erfectly 
mastered  before  applying  proportion  in  the  solution  of  problems. 
The  solution  of  each  example  under  Step  I  should  be  given  in 
full,  as  shown  in  ((571  and  072),  and  Step  II  and  III  should 
be  illustrated  by  the  pupil,  in  the  manner  shown,  by  a  number 
of  examples.    . 

C583.  Step  I. — Find  by  Prob.  V and  Yl.in  ratio,  the  miss- 
ing term  in  tJiefolloicing  proportions  : 

The  required  term  is  represented  by  the  letter  x. 

1.  24  :  60  : :  r  :  15.  4.  2  yd.  :  8  in.  : :  a;  :  3  ft.  4  in. 


2.  6  :  42  ; 

3.  84  :  X  : 


5  :  X. 
21  :  68. 


5.  5  bu.  2  pk.  :  3  pk.  :  :  ./• :  4  bu. 

6.  a? :  £3  28.  : .  49  T.  :  18  cwt. 


Step  II. — Show  that  the  product  of  tlie  extremes  of  a  propor- 
tion is  eqiuU  to  the  product  of  the  means. 

Thus  the  proportion  2  :  3  : :  6  : 9  expressed  fractionally  gives  1  =  J. 


^86 


BUSINESS    ARITHMETIC. 


El  ■   -J 


Now  if  both  tenne  of  tliis  equality  be  multiplied  by  3  and  by  9,  the 

consoqucntB  of  the  given  ratios,  the  equality  U  not  changed;   hence, 

iix9x8      0x3x9 

_  _ Cancelling  (1 76)  the  factor  3  in  the  left-hand  term 

and  9  in  the  right-hand  term  we  have  2x9  =  6x3.  Bat  2  and  9  are  the 
extremes  of  the  proportion  and  6  and  3  are  the  means ;  hence  the  truth  of 
the  proposition. 


I   ■ 


■| 


1.     14  :  3  : :  3-  :  13. 

4. 

2.    27  :  c  : :  9  :  5. 

6. 

3.    aj :  24  : :  7  :  8. 

6. 

Step  III. — 8?u>w  that,  since  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means ^  any  term  of  a  proportion  can 
he  found  when  the  other  three  are  known. 

Thus  in  the  proportion  8 : »  : :  9  :  15  we  have  known  the  two  extremes 
3  and  15  and  the  mean  9.  Bat  by  Step  II,  3  x  15,  or  45,  U  equal  to 
9  times  the  required  mean  ;  hence  45  +  9  =  6,  the  required  mean.  In  the 
eanic  manner  an*  one  of  the  terms  may  be  found ;  hence  the  truth  of  the 
proposition. 

Find  by  this  method  the  missing  term  in  the  following  : 


$13  :  T  : :  5  yd.  :  3  yd. 
64  cwt.  \x   :  :  $120  :  $15. 
128  bu. :  3  pk.  : :  x  :  ^1.25. 


Solution  by  Simple  Proportion, 

684-.  The  quantities  considered  in  problems  that  occur  in 
practical  business  are  so  related  that  when  certain  conditions 
are  assumed  as  invariable,  they  form  ratios  that  must  be  equal 
to  each  other,  and  hence  can  be  stated  as  a  proportion  thus, 

If  4  yd.  of  cloth  cost  $10,  what  will  18  yd.  cost  ? 

Observe,  that  in  this  example  the  price  per  yard  is  assumed 
to  be  invariable,  that  is.  tlie  price  is  the  same  in  both  cases ; 
consequently  whatever  part  the  4  yd.  are  of  the  18  yd.,  the  $10 
are  the  same  part  of  the  cost  of  the  18  yd.,  hence  the  ratio  of 
the  4  yd.  to  the  18  yd.  is  equal  the  ratio  of  the  flO  to  the 
required  cost,  giving  the  proportion  4  yd. :  18  yd.  : :  $10  :  %x. 


SIMPLE     PROPORTION, 


287 


9,  the 
hence, 

id  term 

are  the 
truth  of 


Ion  can 


sxtrcmes 
equal  to 
.  In  the 
th  of  the 


ng 


S15. 
^1.25. 


)ccur  in 

Editions 

)e  equal 

thus, 


issumed 
cases ; 

[the  $10 
itio  of 
to  the 

lO  :  |2- 


EXAMPLES    FOR     PRACTICE. 

685.  Examine  carefully  the  following  proportions,  and 
state  what  must  be  considered  in  each  case  as  invariable,  and 
why,  in  order  that  the  proportion  may  be  correct. 


1. 


The  number 

of  units 

bought  in 

oue  case 


is  to 


'  The  number  1 

f  The  cost  ] 

of  unit? 
bought  in 

■as  - 

in  the 
first 

is  to 

.  another  case . 

case 

The 
2  Principal 
in  one 
case 


The  number 
of  men 

that  can  do 

a  piece  of 

work  in 

one  case 


The 
Principal 
in  another 

case 


The 

interest  in 

the  first 

case 


f 


8. 


The  number  " 

r     The    1 

of  men 

number 

that  can  do 

of  days 

the  same 

■         •    • 

the 

• 

work  in 

second 

another  case  . 

.    work    , 

'  The  cost 
in  the 
second 
case. 

The 
interest  in 
the  second 

case. 

The 

number 

of  dajs 

the 

first 

work. 


Why  is  the  second  ratio  of  this  proportion  made  the  ratio  of  the 
number  of  days  the  second  work  to  the  number  of  days  the  first  work  f 
Illustrate  this  arrangement  of  the  terms  of  the  ratio  by  other  examples. 

In  solving  examples  by  simple  proportion,  the  following 
course  should  be  pursued  : 

/.  Represent  the  required  term  by  x,  and  make  it  the  last 
extreme  or  consequent  of  the  second  ratio  in  the  proportion. 

II.  Find  the  term  in  the  example  thaJt  is  of  the  same  denom- 
ination as  the  required  term,  and  make  it  the  second  mean  or 
the  antecedent  of  the  second  ratio  of  the  proportion. 

III.  Determine,  hy  inspecting  carefuUy  the  condi'.ions  giren 
in  the  example,  whetTier  x,  the  required  term  of  the  ratio  now 
expressed^  must  he  greater  or  less  than  the  given  term. 

IV.  If  T,  the  required  term  of  the  ratio  expressed,  must  he 
greater  than  the  given  term,  m^ike  the  greater  of  the  remaining 
terms  in  the  example  the  consequent  of  the  first  ratio  of  the 
proportion  ;  if  less,  make  it  the  antecedent. 


% 


288 


BUSINESS    ARITHMETIC, 


m0 


l:\ '« 


* 


t^ 


V.  When  the  proportion  w  stated,  find  the  required  term 
either  as  shown  in  (67 1)  or  in  (672). 

Observe,  that  in  either  way  of  finding  the  required  term,  any  flictor  that 
is  common  to  the  given  extreme  and  either  of  the  given  means  should  t>e 
cancelled,  as  shown  in  (175)* 

4.  If  77  pounds  of  sugar  cost  $8.25,  what  will  84  pounds 
cost? 

5.  How  many  bushels  of  wheat  would  be  required  to  make 
39  barrels  of  flour,  if  15  bushels  will  make  3  barrels? 

6.  If  6  men  pat  up  73  feet  of  fence  in  3  days,  how  many  feet 
will  they  put  up  in  33  days  ? 

7.  I  raised  245  bushels  of  com  on  7  acres  of  land  ;  how 
many  bushels  groW  on  2  acres? 

8.  What  will  168  pounds  of  salt  cost,  if  Sj^  pounds  cost 
37^  cents? 

9.  If  25  cwt.  of  iron  cost  $84.50,  what  will  24|  cwt.  cost? 

10.  Paid  $2225  for  18  cows,  and  sold  them  for  $2675 ;  what 
should  I  gain  on  120  cows  at  the  same  rate  ? 

11.  If  5  lb.  10  oz.  of  tea  cost  $5.25  ;  what  will  7  lb.  8  oz. 
cost? 

12.  My  horse  can  travel  2  mi.  107  rd.  in  20  minutes ;  how 
far  can  he  travel  in  2  hr.  20  min.  ? 

18.  If  a  piece  of  cloth  containing  18  yards  is  worth  $10.80, 
what  are  4  yards  of  it  worth  ? 

14.  If  18  gal.  3  qt.  1  pt.  of  water  leaks  out  of  a  cistern  in 
4  hours,  how  much  will  leak  out  in  36  hours  ? 

15.  Bought  28  yards  of  cloth  for  $20;  what  price  per  yard 
would  give  me  a  gain  of  $7.50  on  the  whole  ? 

16.  My  annual  income  on  U.S.  6%'s  is  $337.50  when  gold 
is  at  112^ ;  what  would  it  be  if  jyold  were  at  $125  ? 

17.  If  I  lend  a  man  $69.60  for  8i^  months,  how  long  should 
he  lend  me  $17.40  to  counterbalance  it  ? 

18.  If  10  bu.  of  apples  cost  $7.50,  what  will  60  bu.  cost? 

Ans.  $45. 

19.  If  a  board  13  ft.  long  cast  a  shadow  of  10  ft.,  what  will 
be  the  height  of  a  tree  which  casts  a  shadow  of  115  ft.  ? 

Ans.  149  It.  6  in. 


COMPOUND    PROPORTION, 


289 


km 


rthflt 
lid  be 


»undB 
make 
y  fe«t 
;  how 
s  cost 

>8t? 

:  wliat 


COMPOUND    PROPOKTION". 

PREPARATORY    STEPS. 

686.  Step.  I. — A  compound  ratio  U  reduced  to  a  simple 
one  hy  multiplying  the  antecedents  together  for  an  antecedent 
and  the  consequents  for  a  consequent  (657). 

ThuB,  the  componnd  ratio  ]  4  !  o  [  J"  reduced  to  a  elmple  ratio  by 

m'iltiplyiiJg  the  antecedents  6  and  4  together,  and  the  consequents  7  and  3. 
Expressing  the  ratios  fractionally  we  have  f  x  j  =  |f  =  ;  (665). 

Obsefve,  that  any  factor  that  Is  common  to  any  antecedent  and  con- 
sequent may  be  cancelled  before  the  terms  are  multiplied. 

Reduce  the  following  compound  ratios  to  simple  ratios  in 
their  simplest  terms. 


1. 


9  : 

25 

15  : 

18 

28  : 

50 

3  : 

7 

2. 


''8^1 
35 

115  J 


8. 


16  :    9 

27  :  15 
28 
8 


Step  II. — A  compound  proportion  is  reduced  to  a  simple 
proportion  by  reducing  the  compound  ratio  to  a  simple  ratio. 

(8*9) 
Thus,  in  the  componnd  proportion  ]  »  ,  4  r  : :  24  :  18,  the  compound 

ratio  2  X  ;  is  equal  the  simple  ratio  ; ;  substituting  this  in  the  proportion 
for  the  compound  ratio  we  have  the  simple  proportion  4  :  3  :  :  24  :  18. 

Observe,  that  when  a  compound  proportion  is  reduced  to  a  simple  pro- 
portion,  the  missing  term  is  found  according  to  (671),  or  (673). 

Find  the  missing  term  in  the  following  : 


(24  :  15) 

I.    <    7  :  16  [• 

(  25  :  21  ) 


24  : 

15 

7  . 

16 

25  : 

21 

23 

12 

8 

32 

36 

9 

(  23  :  12  ) 
i  8  :  32  >■ 
(  36  :    9  ) 


40  :  X, 


75  :  X, 


2. 


(28  :  7) 
i  9  :  36  [• 
(  50  :  10  ) 


28  :  X, 


32  :  X. 


13 


■i. 'vrf 


290 


BUSINESS    ARITHMETIC, 


Solution  by  Cotnpound  Proportion. 

087.  The  following  preparatory  propositions  should  be 
carefully  studied,  and  the  course  indicated  observed  in  solving 
problems  involving  comi)ound  proportion. 

Prop.  I. — There  are  one  or  more  conditions  in  every  example 
incolviug  proportion,  which  must  he  regarded  as  iavuriablc 
in  order  that  a  solution  may  be  given,  thus, 

If  9  lioraes  can  sabsitjt  on  60  bu.  of  oats  for  SO  day^,  how  long  can 
6  burses  subsist  ou  70  bu.  ? 

In  this  example  there  are  two  conditions  that  must  be  considered  as 
invariable  iu  order  to  give  a  solution  : 

1.  The  fact  that  each  horse  subsists  on  the  same  quantity  of  oats 
each  dsjy. 

2.  The  fact  that  each  bushel  of  oats  contains  the  same  amount 
of  food. 

Prop.  II. — To  solve  a  problem  involving  a  compound  pro- 
portion, the  ffect  of  each  ratio,  which  forms  the  compound 
Toiio,  on  the  Required  term,  must  he  considered  separately, 
thus : 

If  5  men  can  build  40  yards  of  a  fence  in  12  days,  how  many 
yards  can  8  men  build  in  9  days  ? 

1.  We  observe  that  the  invariable  conditions  in  this  example 
are 

(1.)  That  each  man  in  both  cases  does  the  same  amount  of  toork  in  the 
same  time. 

(8.)  That  the  sam£  amount  of  work  is  required  in  each  case  to  build  one 
yard  of  the  fence. 

2.  We  determine  by  examining  the  problem  how  the  required 
term  is  affected  by  the  relations  of  the  given  term,  thus  : 

(1.)  We  observe  that  the  5  men  in  12  days  can  build  40  yards.  Now 
since  each  man  can  build  the  same  extent  of  the  fence  in  one  day,  it  is  evi- 
dent that  if  the  8  men  work  12  days  the  same  as  the  5  men,  the  40  yards 
built  by  the  5  men  in  12  days  must  have  the  same  ratio  to  the  number  of 


Id  be 
alving 


cample 
'iable 


ong  can 
iered  as 
of  oats 
amount 


d  pro- 
npaund 
irately, 


many 
tample 

*  in  the 

lUd  one 

luired 


\.    Now 

is>  evi- 
|0  yards 
iber  of 


COMPOUND     PB  O  PORTION. 


291 


yards  that  can  be  built  by  the  8  men  in  12  days  as  5  men  have  to  8  men ; 
hence  the  proportion 

5  men  :  8  men  : :  40  yards  :  x  yards. 

This  proportion  will  give  the  number  of  yards  the  8  men  can  build  in 
12  days. 

(i.)  We  now  observe  that  the  8  men  work  only  9  days  ;  and  since  they 
can  do  the  same  amount  of  work  each  day,  the  work  done  in  12  days  must 
have  the  same  ratio  to  the  work  they  can  do  in  9  days  that  12  days?  have  to 
9  days.    Hence  we  have  the  compound  proportion 

5  men  :  8  men  (..„,, 

We  find  from  this  proportion,  according  to  (686—11),  ttiat  the  8  men 
can  build  48  yards  of  fence  in  9  days. 


EXAMPLES    FOR    PRACTICE. 

688.  1.  If  12  men  can  saw  45  cords  of  wood  in  3  days, 
working  9  hours  a  day,  how  much  can  4  men  saw  iu  18  days^ 
working  12  hours  a  day  ? 

2.  If  it  cost  $88  to  hire  12  horses  for  5  days,  what  will  it  cost 
to  hire  10  horses  for  18  days  ? 

3.  When  the  charge  for  carrying  20  centals  of  grain 
50  miles  is  $4.50,  what  is  the  charge  for  carrying  40  centals- 
100  miles  ? 

4.  If  28  horses  consume  240  bushels  of  corn  in  112  days,  how 
many  bushels  wiU  12  horses  consume  in  lOG  days. 

5.  The  average  cost  of  keeping  25  soldiers  1  year  is  $3000  ; 
what  would  it  cost  to  keep  139  soldiers  7  years  ? 

6.  64  men  dig  a  ditch  72  feet  long,  4  feet  wide,  and  2  feet 
deep,  ill  8  days  ;  how  long  a  ditch,  2|  feet  wide  and  1^  feet 
deep,  can  96  men  dig  in  60  days  ? 

7.  If  1  pound  of  thread  makes  3  yards  of  linen,  1}  yard 
wide,  how  many  pounds  would  make  45  yards  of  linen, 
1  yard  wide  ? 

8.  If  it  requires  8400  yd.  of  cloth  1|  yd.  wide  to  clothe  8500 
8  )ldierH,  how  many  yards  |  wide  will  clothe  6720  ? 


Ill 


I- 


%t:: 


PARTNERSHIP 


u ' 


m 


if 


It-  i 


DEFINITIONS. 

080.  A  Partnership  is  au  association  of  two  or  more 
persons  lor  the  transaction  of  business. 

The  persons  associated  are  called  partners,  and  the  Association  is  called 
a  Vomitany^  Firm^  or  House. 

CM)0.  The  Cajntal  is  the  money  or  other  property  invested 
in  the  business. 

The  Capital  is  aleo  called  the  Investment  or  Joint-stock  of  the 
Company. 

091.  The  Assets  or  Effects  of  a  Company  are  the 
proiM5rty  of  all  kinds  belonging  to  it,  together  with  all  the 
amounts  due  to  it. 

G92.  The  Liabilities  of  a  Company  are  its  debts. 


PREPARATORY    PROPOSITIONS. 

093.  Prop.  I. — The  profits  and  the  losses  of  a  comr 
pany  are  divided  among  the  partners,  according  to  the 
'vcdue  of  eojch  man's  investment  at  the  time  the  division 
is  made. 

Observe  carefully  the  following  regarding  this  proposition  : 

Since  the  use  of  money  or  property  ie  iteelf  value,  it  ia  evident  that 
the  value  of  an  investment  at  any  time  after  it  is  made,  depends  first  upon 
the  amount  invested,  second  on  the  length  of  the  time  the  investment  has 
been  made,  and  third  the  rate  of  interest. 

Thus  the  value  of  an  investment  of  $500  at  the  time  it  is  made  is  just 


P  A  RT  y  E  R  s  n  IP  . 


293 


$500  ;  bnt  at  tfu  end  of  ^  yearn,  reckontn<T  He  ^ue  to  be  worth  1%  per  anntim, 
itt)  value  will  bo  $600  +  $816  =  |815. 

Prop.  II. — The  value  of  any  invefttment  made  for  a  given 
number  of  intervals  of  time,  can  be  represented  by  another 
investment  made  for  one  interval  of  time. 

Tbu8,  for  example,  the  valae  of  an  investment  of  $40  for  5  months  at 
any  f^iven  rate  of  interest  is  the  same  as  the  value  of  5  times  $40,  or  $S00, 
for  one  mouth. 


r  more 

is  called 

invested 

i;   of  the 


are    the 
all  the 


a  com' 

to   the 

Xdinsion 


ition : 

lent  that 

irst  upon 

lent  has 

is  just 


EXAMPLES     FOR    PRACTICE. 

004.  Find  the  value  at  simple  interest 

1.  Of  $350  invested  2  yr.  3  mo.  at  7%  per  annum. 

2.  Of  $800  invested  4  years  at  6%  per  annum. 

3.  Of  $2860  invested  19  months  at  8%  per  annum. 

Solve  the  following  by  applying  (09i$ — II). 

4.  A  man  invests  $600  for  9  months,  $700  for  3  months,  and 
$300  for  7  months,  each  at  the  same  rate  of  interest.  What 
sura  can  he  invest  for  4  months  at  the  given  rate  of  interest,  to 
be  equal  in  value  to  the  three  investments. 

5.  An  investment  of  $200  for  6  months  is  equal  in  value  to 
what  investment  for  4  months  ? 

6.  A  warehouse  insured  for  $35,000  is  entirely  destroyed  by 
fire ;  ^  of  it  belonged  to  A,  \  to  B,  and  the  rest  to  C.  How 
much  of  the  insurance  did  each  receive  ? 


ILLUSTRATION    OF    PROCESS. 

695.  Prob.  I. — To  apportion  gains  or  losses  when 
each  partner's  capital  is  invested  the  same  length  of 
time. 

Observe,  that  when  each  partner^s  capital  is  nsed  for  the  same  length  of 
Hme^  It  is  evident  that  his  share  of  the  gain  or  loss  mast  be  the  same  frac- 
tion of  the  whole  gain  or  loss  that  his  capital  is  of  the  whole  capital.  Hence« 
examples  under  this  problem  may  be  solved— 


i 


u 


( 


.   ■«  ' 

'if 


294 


BUSINESS    ARITIIMKTIC, 


I.  By  Proportion  thus . 


T/ie  whole 

capital 

inveated 


(i  Each  fnan'a  |  |    Whole  \  i         I. 

•   ■<       capital       >  •:  -((^ainor)-    S    ■<  wtOH 
f      invented      1  i     loaa      )  '      or 


Each 
),*a  gain 
loaa. 


II.  By  Percentage  ilius  : 


Find  what  per  cent  (493)  the  whole  gain  or  low  in  of  the 
whole  capital  invested,  and  take  the  same  per  c^ent  of  emh  man's 
investment  as  his  share  of  the  gain  or  loss. 

II.  By  FractioHB  thus  : 

Find  tcJiat  fractional  part  each  man's  investment  is  <f  the 
whole  capital  invested,  and  take  the  same  fractional  part  of  the 
gain  or  loss  as  each  man's  share  of  the  gain  or  loss. 


EXAMPLES    FOR    PRACTICE. 

6S)«.  1.  A  man  failing  in  business  owes  A  $9600,  B  17000, 
and  C  $5400,  and  his  available  property  amounts  to  $5460 ; 
what  is  each  man's  share  of  the  property  ? 

2.  Three  men,  A,  B,  and  C,  form  a  company;  A  puts  in 
$6000  ;  B  $4000  ;  and  C  $5600  ;  they  gain  $4320 ;  what  is  each 
man's  share  ? 

3.  Three  men  agree  to  liquidate  a  church  debt  of  $7890,  each 
paying  in  proportion  to  his  property ;  A's  property  is  valued  at 
$6470,  B'k  at  $3780,  and  C's  at  $7890  ;  what  portion  of  the  debt 
does  each  man  pay  ? 

4.  The  sum  of  $2600  is  to  be  divided  among  four  school 
districts  in  proportion  to  the  number  of  scholars  in  each ;  in 
the  first  there  are  108,  in  the  second  84,  in  the  third  72,  in  the 
fourth  48  ;  what  part  should  each  receive? 

6.  A  building  worth  $28500  is  insured  in  the  ^tna  for  $3200, 
in  the  Home  for  $4200,  and  in  the  Mutual  for  $6500  ;  it  having 
been  partially  destroyed,  the  damage  is  set  at  $10500 ;  what 
should  each  company  pay  ? 


eh 
gain 

taa. 


of  the 
man'n 


of  the 
of  the 


PARTNERSHIP, 


205 


(M)7.  Prob.  II.— To  apportion  gains  or  losses  when 
each  partner's  capital  is  invested  different  lengths  of 
time. 

Observe  carefully  the  following : 

1.  According?  to  (603—11)  wo  can  find  for  each  partner  an  amoont 
whoce  value  inveHtcd  one  interval  of  time  ie  equal  to  the  value  of  liis 
capital  for  the  given  intervalu  of  time. 

2.  Having  found  thiH  we  can,  by  adding  these  amounts,  find  an  amount 
whoHO  value  invested  one  interval  of  time  Ih  eqnal  to  the  total  value  of  the 
whole  capital  invented- 

When  thie  is  done  it  is  evident  that  each  man's  share  of  the  gain  or  IO0B 
must  be  the  same  fraction  of  the  whole  gain  i  r  loss  that  the  value  of  his 
investment  is  of  the  total  value  of  the  whole  capital  invested.  Hence  the 
problem  flrom  this  point  can  be  solved  by  either  of  the  three  methods  given 
under  Prob.  I  (695). 


$7000, 
$5460 ; 

3Tits  in 
I  is  each 

)0,  each 
Llued  at 
|he  debt 

school 
cb;  in 
in  the 

$3200, 

[having 

what 


EXAMPLES    FOR    PRACTICE. 

Of>8.  1.  Three  men  hire  a  pasture  for  $136.50  ;  A  \)\ii%  in 
16  cows  for  8  weeks,  B  puts  in  6  cows  for  12  weeks,  and  C  the 
same  number  for  8  weeks  ;  what  should  each  man  pay  ? 

2.  A  and  B  engage  in  business  ;  A  puts  in  $1120,  for  5  months 
and  B  $480  for  8  months ;  they  gain  $354  ;  what  is  each  man's 
share  of  the  gain  ? 

3.  The  joint  capital  of  a  company  wa«  $7800,  which  was 
docbled  at  the  end  of  the  year.  A  put  in  \  for  9  mo.,  B  ^  for 
8  mo.,  and  C  the  remainder  for  1  year.  What  is  each  one's 
stock  at  the  end  of  the  year  ? 

4.  A  and  B  formed  a  partnership  Jan.  1,  1876.  A  put  in 
^^6000  and  at  the  end  of  8  mo.  $900  more,  and  at  the  end  of 
10  mo.  drew  out  $800 ;  B  put  in  $9000  and  8  mo.  after  $1500 
more,  and  drew  out  $600  Dec.  1 ;  at  the  end  of  the  year  the  net 
profits  were  $8900.    Find  the  share  of  each. 

5.  Jan.  1,  1875,  three  persons  began  business.  A  put  in 
$1200,  B  put  in  $500  and  May  t  $800  more,  C  put  in  $700  and 
July  1  $400  more  ;  at  the  end  of  the  year  the  profits  were  $875 ; 
how  shall  it  be  divided  ? 


m;^ 


iir 


r-    H    -■ 


v.- 1' 


H4 


ALLIGATION. 


ALLIGATION    MEDIAL. 


699*  Alligation  Medial  is  the  process  of  finding  the  mean 
or  average  price  or  quality  of  a  mixture  composed  of  several 
ingredients  of  different  prices  or  qualities. 


■I    i^ 


&  -^ 


EXAMPLES    FOR    PRACTICE. 

TOO.  1.  A  grocer  mixed  7  lb.  of  coffee  worth  30  ct.  a 
pound  with  4  lb.  @  25  ct.  and  10  lb.  (ib  32  ct. ;  in  order  that  he 
may  neither  gain  nor  lose,  at  what  price  must  he  sell  the 
mixture? 


7  lb.  @  30  ct. 

4  lb.  @  25  ct. 

10  lb.  @  32  ct. 

21  lb.  = 

16.30  ^  21  = 


=  $2.10 
=  1.00 
=    3.20 


$6.30 
30  ct. 


Solution.— 1.  Since  the  valne  of 
each  kind  of  coffee  is  not  changed  hy 
mixing,  we  find  the  value  of  the  entire 
mixture  by  finding  the  value  of  each 
kind  at  the  given  price,  and  taking  the 
Bum  of  these  values  as  shown  in  illus- 
tration. 

2.  Having  found  that  the  21  lb.  of  coffee  are  worth  at  the  given  prices 
♦6.30,  it  is  evident  that  to  realize  this  amount  fk-om  the  sale  of  the  21  lb.  at 
a  uniform  price  per  pound,  he  must  get  for  each  pound  ,V  of  $6.30 ;  hence, 
$6.30  -i-  31  =  30  cents,  the  selling  price  of  the  mixture. 

2.  A  wine  merchant  mixes  2  gallons  of  wine  worth  $1.20  a 
gallon  with  4  gallons  worth  $1.40  a  gallon,  4  gallons  worth 
$.90  and  8  gallons  worth  $.80  a  gallon  ;  what  is  the  mixture 
worth  per  gallon  ? 

3.  A  grocer  mixes  48  lb.  of  sugar  at  17  ct.  a  pound  with 
68  lb.  at  13  ct.  and  94  lb.  at  11  ct. ;  what  is  a  pound  of  the 
mixture  worth? 


ALL  IQ  A  Tl  0  N. 


297 


emean 
several 


4.  A  merchant  purchased  60  gallons  of  molasses  at  00  ct. 
per  gallon  and  40  gallons  at  25  cents,  which  he  mixed  with 
8  gallons  of  water.  He  sold  the  entire  mixture  so  as  to  gain 
20  per  cent  on  the  original  cost ;  what  was  his  selling  price 
per  gallon  ? 

5.  A  goldsmith  melts  together  6  ounces  of  gold  22  carats  fino, 
30  ounces  20  carats  fine,  and  12  ounces  14  carats  fine ;  liuw 
many  carats  fine  is  the  mixture? 

6.  A  farmer  mixes  a  quantity  of  barley  at  90  cents  a  bushel 
with  oats  at  37  cents  a  bushel  and  rye  at  65  cents  a  bushel. 
Find  the  price  of  the  mixture. 


50  ct.  a 

that  he 
sell  the 


value  of 
inged  by 
[he  entire 

of  each 
iking  the 

in  illus- 


jn  prices 
21  lb.  at 
:  hence. 


|tll.20  a 

wortli 

lixture 

^d  with 
of  the 


ALLIGATION    ALTERNATE. 

701.  Alligation  Alternate  is  the  process  of  finding 
the  proportional  quantities  of  ingredients  of  different  prices  or 
qualities  that  must  be  used  to  form  any  required  mixture,  when 
the  price  or  quality  of  the  mixture  is  given. 

PREPARATORY    PROPOSITIONS. 

702,  Prop.  I. — In  forming  any  mixture,  it  is  assumed  that 
the  value  of  the  entire  mixture  must  be  equal  to  the  aggregate 
value  of  its  ingredients  at  their  given  prices. 

Thus,  if  10  pounds  of  tea  at  45  ct.  and  5  pound?  at  60  ct.  be  mixed,  the 
value  of  the  mixture  must  be  the  value  of  the  10  pounds  plus  the  vatae  of 
the  5  pounds  at  the  given  prices,  Avhloh  is  equal  $4.50  +  $3.00  ■--  $7.50. 
Hence  there  is  neither  gain  nor  loss  in  forming  a  mixture. 

Prop.  II. — The  price  of  a  mixture  must  he  less  than  tiie 
highest  and  greater  tJian  the  lowest  price  of  any  ingredient  used 
informing  the  mixture. 

Thus,  if  sugar  at  10  ct.  and  at  15  rt.  per  pound  be  mixed.  It  it»  evident  the 
price  of  the  mixture  must  be  lens  than  I.**  cents  and  greater  than  10  cents  ; 
that  u,  It  must  be  some  price  between  10  and  15  cents. 


wr 


298 


BUSIXESS     ARITHMETIC, 


ILLUSTRATION    OF    PROCESS. 

703.  If  tea  at  50  ct.,  60  ct.,  75  ct.,  and  90  ct.  per  pound  be 
mixed  and  sold  at  66  ct.  per  })ound ;  bow  mucb  of  eacb  kind  of 
tea  can  be  put  in  tbe  mixture  ? 

First  Step  in  Solution, 

We  find  the  jmin  or  loss  on  one  unit  of  each  injjredient  thns  : 


66  ct. 
66  ct. 


(1.)    ] 


.     ( 75  ct.  —  66  ct.  =    9  ct.  loss. 
^  ''    <  90  ct.  -  66  ct.  =  24  ct.  loss. 


56  ct.  =  10  ct.  gain. 
60  ct.  =    6  ct.  gaiu. 


Second  Step  in  Solution, 

We  now  lake  an  ingredient  on  which  there  i>*  a  gain,  and  one  on  which 
there  is  a  loss,  and  ascertain  how  much  of  each  must  be  put  in  tbe  mixture 
to  make  the  gain  and  loss  equal ;  thus : 

Producino  Gain.        Gaotbd  and  Lost.         Pbodttcino  Loss. 
(1.)  9  lb.  at  10  ct.  per  lb.  gain.  =  90  ct.  =  10  lb.  at    9  ct.  per  lb.  lojs. 
(2.)  4  lb.  at    6  ct.  per  lb.  gain.  =  24  ct.  =     1  lb.  at  24  ct.  per  lb.  low. 

Hence  the  mixture  must  contain  9  lb.  at  56  ct.  per  pound,  10  lb.  at  75  ct. 
per  poHnd,  4  lb.  at  60  ct.  per  pound,  and  1  lb.  at  90  ct.  per  pound. 

704.  Observe  carefully  the  following : 

1.  The  gain  and  loss  on  any  two  ingredients  may  be  balanced 
by  assuming  any  amount  as  the  sum  gained  and  lost. 

Thus,  ins^tead  of  taking  90  cents,  as  in  (1)  in  the  above  solution,  as  the 
amount  gained  and  lost,  we  might  take  360  cents ;  and  dividing  360  cents 
by  10  cents  would  give  36,  the  number  of  pounds  of  56  ct.  ton  that  would 
gain  this  sum.  Again,  dividing  360  cents  by  9  cents  would  give  40,  the 
number  of  pounds  of  75  ct.  tea  that  would  lose  this  sum. 

2.  To  obtain  intefjrnl  proportional  parts  the  amount  -sumed 
must  be  a  multiple  of  the  gain  and  loss  on  one  unit  of  the 
ingredients  balanced,  and  to  obtain  the  least  integral  propor- 
tional parts  it  must  be  the  least  common  multiple. 


und  be 
kind  of 


I 


ALLIGATION, 


299 


3.  \Mien  a  number  of  ingredients  are  given  on  which  there  is 
a  gain  and  also  on  which  tliere  is  a  loss,  they  may  be  balanced 
with  each  other  in  several  ways ;  hence  a  series  of  diflferent 
mixtures  may  be  formed  as  follows  : 

Taking  the  foregoing  example  we  have 


A  Second  Mixture  thus: 
PBODUCiNa  Gain.  Gained  and  Lost.       Producing  Loss. 

(1.)  24  lb.  at  10  ct.  per  II).  gain.  =  240  ct.  =  10  lb.  at  24  ct.  per  lb.  loss. 
(2.)    9  lb.  at    6  ct.  per  lb.  gain.  =    54  ct.  =    6  lb.  at    9  ct.  per  lb.  loss. 

Ilence  the  mixture  is  composed  of  24  lb.  @  56  ct.,  9  lb.  @  60  ct.,  10  lb. 
@  90  ct.,  and  6  lb.  ®  75  ct. 


on  which 
e  mixture 


Loss. 
tr  lb.  loH. 
T  lb.  lo». 

at  75  ct. 


)alanced 


kn,  as  the 

360  cents 

lut  would 

ire  40,  the 


•sumed 

of  the 

Ipropor- 


A  TJiird  Mixture  thus  : 

Producing  Gain.  Gained  and  Lost.       Pboducino  Loss. 

(1.)  9  lb.  at  10  ct.  per  lb.  gain.  =  90  ct.  =  10  lb.  at  9  ct.  per  lb.  loss. 
(2.)  24  lb.  at  10  ct.  per  lb.  gain.  =  240  ct.  =  10  lb.  at  24  ct.  per  lb.  loss. 
(3.)    9  lb.  at    6  ct.  per  lb.  gain.  =    54  ct.  =    6  lb.  at    9  ct.  per  lb.  loss. 

Observe,  that  in  (1)  and  (2)  we  have  balanced  the  loss  on  the  T.'i  ct.  and 
90  ct.  tea  by  the  gain  on  the  56  ct.  tea;  hence  we  have  9  lb.  +  24  lb.,  or 
33  lb.  of  the  56  ct.  tea  in  the  mixture. 

Observe,  also,  that  in  (3)  we  have  balanced  the  gain  on  the  60  ct.  tea  by  a 
loss  on  the  75  ct.  tea ;  hence  we  have  10  lb.  +  6  lb.,  or  16  lb.  of  the  75  ct. 
tea  in  the  mixture. 

Hence  the  mixture  is  composed  of  33  lb.  @  56  ct.,  9  lb.  @  60  ct.,  16  lb. 
%  75  ct.,  and  10  lb,  @  90  ct. 

4.  Mixtures  may  be  fonned  as  follows  : 

/.  Take  any  pair  of  ingredients,  one  giving  a  gain  and  the 
other  a  loss,  and  find  the  gain  and  loss  on  one  unit  of  each. 

II.  Assume  the  least  common  multiple  of  the  gain  and  loss 
on  one  unit  as  the  amount  gained  and  lost,  by  putting  t?ie  two 
ingredients  in  the  mixture. 

III.  Divide  the  amount  thus  assumed  hy  the  gain  and  then 
by  the  loss  on  one  unit;  the  results  will  be  respectively  the 


i^*;:'f 


:•,  -i-.f 


300 


BUSI^'£SS    ARITHMETIC, 


^f'^^i 


M 


number  of  units  of  each  ingredient  thtt  must  he  in  the  mixture 
that  the  gain  and  loss  viay  balance  ea^h  other. 

IV.  Proceed  in  the  same  manner  with  other  ingredients;  the 
results  will  be  tlie  proportional  parts. 


EXAMPLES     FOR    PRACTICE. 

705.  1.  A  man  wishes  to  mix  sufficieut  water  with  mo- 
lasses worth  40  cents  a  gallon  to  make  the  mixture  worth 
24  cents  a  gallon ;  what  amount  must  he  take  of  each  'I 

2.  How  much  sugar  at  10,  9,  7,  and  5  ct.  will  produce  a 
mixture  worth  8  cents  u  pound  ? 

3.  A  merchant  desires  to  mix  flour  worth  ^6,  $7^,  and  $10  a 
barrel  so  as  to  sell  the  mixture  at  $9  ;  what  proportion  of  each 
kind  can  he  use  V 

4.  A  jeweller  has  gold  16,  18,  22,  and  24  carats  fine;  how 
much  of  each  must  he  use  to  form  gold  20  car»is  fine? 

5.  A  farmer  has  wheat  worth  40,  55,  80,  and  90  cents  a 
bushel ;  how  many  bushels  of  each  must  bo  mixed  with  270 
@  40  ct.  to  form  a  mixture  worth  70  cents  a  bushel? 

Examples  like  this  where  the  quantity  of  one  or  more 
ingredients  is  limited  may  be  solved  thus  : 

First,  we  find  the  gain  or  loss  on  one  unit  as  in  (703). 

Second,  we  balance  the  whole  gain  or  loss  on  an  ingredient 
where  the  quantity  is  limited,  by  using  any  ingredient  giving 
an  opposite  result  thus  : 

Producing  Gain.  Gaiked  and  Lost.       Producing  Loss. 

(1.)  270  bu.  at  30  ct.  per  ba.  gain.  =$81.00=405  bu.  at  20  ct.  per  bu.  loss. 
(2.)      2  bu.  at  15  ct.  per  bu.  gain'=       .30=     3  bu.  at  10  ct.  per  ba.  loss. 

*»72  bu.  +  408  bu.=680  bu.  in  mixtun. 

Observe,,  the  pfaln  on  the  270  bu.  may  be  balanced  with  the  other  injEjre- 
dlent  that  produces  a  loss,  or  with  both  incredieuts  that  produce  a  loss, 
and  these  may  be  put  in  the  mixture  in  different  proportions ;  hence  a 
series  of  different  mixtures  may  thus  be  formed. 


mixture 
its;  the 


ALLIQ  ATIOX, 


301 


ith  mo-  ' 

B  worth 

oduce  a 

[id  $10  a 
L  of  each 

ne;  li3w 

cents  a 
dth  270 

)r  more 

redient 
giving 

jOSS. 

or  bu.  loss, 
her  bu.  loss. 

in  mixtun. 

[er  Inffre- 

te  a  loss, 

hence  a 


6.  I  wish  to  mix  vinegar  worth  18,  21,  and  27  cents  a  gallon 
with  8  gallons  of  water,  making  a  mixture  worth  25  cents  a 
gallon  ;  how  much  of  each  kind  of  vinegar  can  I  use  V 

7.  A  merchant  having  good  flour  worth  $7,  $0,  and  ^12  a 
barrel,  and  240  barrels*  of  a  poorer  quality  worth  $5  a  barrel, 
wishes  to  sell  enough  of  each  kind  to  realize  an  average  ])rice 
of  $10  a  barrel  on  the  entire  quantity  sold.  How  many  barrels 
of  each  kind  can  he  sell? 

8.  A  man  bought  u  lot  of  sheep  at  an  average  price  of  $2 
apiece,  lie  paid  lor  50  of  them  $2.50  per  head,  a«id  for  the 
rest  $1.50,  $1.75,  and  s3.25  i)er  head  ;  how  many  sheej)  could 
there  be  in  the  lot  at  each  price  ? 

9.  A  milkman  mixes  milk  worth  8  cents  a  (juartnyith  water, 
making  24  quarts  worth  6  cents  a  quart ;  how  much  water 
did  he  use  ? 

Examples  like  this,  where  the  quantity  of  the  mixture  la 
limited,  may  be  solved  thus  : 

Solution. — 1.  We  find,  according  to  (T02),  the  smallest  proportional 
parts  that  can  be  ubed,  namely,  3  quarts  of  milk  and  1  quart  of  water, 
making  a  mixture  of  4  quarts, 

2.  Now,  since  in  4  qt.  of  the  mixture  there  are  3  qt.  of  milk  and  1  ql. 
of  water,  in  94  qt.  there  must  be  as  many  times  3  qt.  of  milk  and  1  qt.  of 
water  as  4  qt.  are  contained  times  in  24  qt.  Consequently  we  have  as  the 
Jirst  step  24  qt.  -*-  4  qt.  =r  6,  t^ecoml  itfep  8  qt.  x  ♦»  =  18  qt.  and  1  qt.  x  6 
=  6  qt.  Hence  in  24  qt.  of  the  mixture  there  are  18  qt.  of  milk  and  6  qt. 
of  water. 

10.  A  jeweler  melts  together  gold  14,  18,  and  24  carats  fine, 
80  as  to  make  240  oz.  22  carats  fine;  how  much  of  each  kind 
did  it  require? 

11.  A  grocer  has  four  kinds  of  coffee  worth  20,  25,  35,  and 
40  cents  a  pound,  from  which  he  fills  an  order  for  135  pounds 
worth  32  cents  -^  pound ;  how  may  he  form  tlio  mixture  ? 

12.  I  wish  to  fill  an  order  for  224  lb.  of  sugar  at  12  cents,  by 
forming  a  mixture  from  8,  10,  and  16  cent  sugar;  how  much 
of  each  must  I  take  ? 

13.  How  much  candy  at  35,  39,  and  47  ct.  will  produce 
a  mixture  worth  20  cents  a  pound  ? 


'M 


'A':i:'^ 


■■:■-,    i 


INVOLUTION. 


DEFINITIONS. 


706.  A  Power  of  a  number  is  either  the  number  itself  or 
the  product  obtained  by  taking  the  number  two  or  more  times 
as  a  factor. 

Thus  35  is  the  prodnct  of  6  x  5  or  of  5  taken  twice,  ae  a  factor :  hence  25 1& 
a  power  of  5. 

707.  An  Exjtonent  is  a  number  written  at  the  right  and 
a  little  above  a  number  to  indicate : 

(1.)  The  number  of  times  the  given  number  is  taken  as  a  foctor.  Thus 
in  7"*  the  8  indicates  that  the  7  i^  taken  3  timeb  as  a  fiuitor ;  hence  7'  = 
7x7x7  =  343. 

(2.)  The  degree  of  the  power  or  the  order  of  the  power  with  reference  to 
the  other  powers  of  the  given  number.  Thus,  in  5*  the  4  indicates  that  the 
given  power  is  the  fourth  power  of  5,  and  hence  there  are  three  powers  of 
5  below  5* ;  namely,  5,  5',  and  5\ 

708.  The  Square  of  a  number  is  its  second  power,  so  called 
because  in  finding  the  superficial  contents  of  a  given  square  we 
take  the  second  power  of  the  number  of  linear  units  in  one  of 
its  sides  (395). 

709.  The  Cube  of  a  number  is  its  third  power,  so  called 
because  in  finding  the  cubic  contents  of  a  given  cube  we  take 
the  third  power  of  the  number  of  linear  units  in  one  of  its- 
t^-^.gea  (403). 

710.  Involution  is  the  process  of  finding  any  required 
power  of  a  given  number. 


tseif  or 
e  times 

ence  25  i& 

iglit  and 


ir.    Thus 
!nce  7*  = 

rerence  to 

Is  that  the 

towers  of 


called 
WiTe  we 
one  of 


called 
Kve  take 

le  of  its- 


required 


I NVO  L  UTI  0  X, 


PROBLEMS    IN    INVOLUTION. 


303 


711.  Prob.  I. — To  find  any  power  of  any  given 
number. 

1.  Find  the  fourth  power  of  17. 

Solution.— Since  according  to  (706)  the  fourth  power  of  17  is  the 
product  of  17  talcen  as  a  factor  4  times,  we  have  17  x  17  x  17  >(  17  =  83521,  the 
required  power. 

2.  Find  the  square  of  294.    Of  386.     Of  497.     Of  253. 

3.  Find  the  second  power  of  48.     Of  65.     01  432. 

4.  Find  the  third  power  of  5.    Of  J.    Of  ,;,.    Of  .8. 

5.  Find  the  cube  of  63.     Of  25.    Of  76.    Of  392. 

Observe,  any  power  of  a  fraction  it?  found  by  involving  each  of  its  terms 
separately  to  the  required  power  (256;. 


Find  the  required  power  of  the  following : 

10.  (.3?)'.        12.  (.71)^ 


454. 
2372. 


8.  (H>'.^ 


11.  (.25)*.        13.  (.l/jj)4. 


14.  .0303^ 

15.  (.005;)«. 


712.  Pbob.  II. — To  find  the  exponent  of  the  product 
of  two  or  more  powers  of  a  given  number. 

1.  Find  the  exponent  of  product  of  7^  and  7^ 

Solution.— Since  7*  =  7  x  7  x  7  and  7'  =  7  x  7,  the  product  of  7*  and  7' 
must  be  (7  X  7  X  7)  X  (7  X  7),  or  7  taken  as  a  foctor  as  many  times  as  the  sum 
oi  the  exponents  3  and  2.  Hence  to  find  the  exponent  of  the  product  of 
two  or  more  powers  of  a  given  number,  we  take  the  sum  of  the  given 
exponents. 


Find  the  exponent  of  the  product 


3.  OfCD'^xd)*. 
8.  Of  35*  X  353. 


4.  Of  18*  X 18'. 

5.  of(*rx(jr. 


6.  Of(T-'V)*x(A)». 

7.  Of23'x23^ 


7«. 


8.  Of  (7*)«.     Observe,  (7*)*  =  7*  x  7*  =  7*><« 
Hence  the  required  exponent  is  the  product  of  the  given  exponents. 

9.  Of(12»)*.        10.  Of(9«)\        11.  Of  (16Y.      13.  Of[(i)»]*. 


I 


n 


!!l 


lif 


EVOLUTION. 


DEFINITIONS. 


713.  A  Root  of  a  numl)er  is  either  the  number  itself  or 
one  of  the  equal  factors  or  into  which  it  can  be  resolved, 

ThuB,  since  7  x  7  =  49,  the  factor  7  is  a  root  of  49. 

714:»  The  Second  or  Square  Root  is  one  of  the  two 

equal  factors  of  a  number.     Thus,  5  is  the  square  root  of  25. 

715.  The  Third  or  Cube  Roof  is  one  of  the  three  equal 
factors  of  a  number.     Thus,  2  is  the  cube  root  of  8. 

716.  The  Radical  or  Root  Sign  is  V,  or  a  fractional 
exponent. 

When  the  siffo,  ^,  1b  used,  the  degree  or  name  of  the  root  is  indicated 
by  a  small  figure  written  over  the  sign  ;  when  the  fractional  exponent  is 
used,  the  denominator  indicates  the  name  of  the  root ;  thus, 

/?/9  or  9*  indicates  that  the  second  or  square  root  is  to  be  found. 

yy/27  or  27»  indicates  that  the  third  or  cube  root  is  to  be  found. 
Any  required  root  is  expressed  in  the  same  manner.    The  index  is 
usually  omitted  when  the  square  root  is  required. 

717.  A  Perfect  Power  is  a  number  whose  exact  root 
can  be  found. 

718.  An  Imperfect  Power  is  a  number  whose  exact 
root  cannot  be  found. 

The  indicated  root  of  an  imperfect  power  is  called  a  eurd ;  thus  ^5. 

719.  Evolution  is  the  process  of  finding  the  roots  of 
numbers. 


r  itself  or 
ed. 


of  the  two 
»t  of  25. 

three  equal 

fractional 

iB  indicated 
exponent  is 

ound. 

nd. 

e  index  is 

3xact  root 

lose  exact 

las  Vs. 
roots  of 


nvoLUTioy^,  305 


SQUABE    BOOT. 

PREPARATORY    PROPOSITIONS. 

720.  Prop.  I.  —  Any  perfect  second  power  may  be 
represented  to  the  eye  by  a  square,  and  the  number  of  units  in 
the  side  of  such  square  will  represent  the  second  or  square 
root  of  tlie  given  power. 

For  example,  if  25  is  the  given  power,  we  can  suppose  the 
number  represents  25  small  squares  and  arrange  them  thus : 

1.  Since  25  =  5  x  5,  we  can  arrange  the  35  squares  5  in 
a  row,  making  5  rows,  and  hence  forming  a  equarc  as 
shown  in  the  illustration. 

2.  Since  the  side  of  the  square  is  5  units,  it  represents 
the  pquare  root  of  25,  the  given  power ;  hence  the  truth 
-of  the  proposition. 

731.  Prop.  II. — Any  number  being  given,  by  suppoHvg  it 
to  represent  small  squares,  we  can  find  by  arranging  these 
squares  in  a  large  square  the  largest  perfect  second  power  the 
given  number  contains,  and  hence  its  square  root. 

For  example,  if  we  take  83  as  the  given  number  and  suppose 
it  to  represent  83  small  squares,  we  can  proceed  thus : 


1.  We  can  take  any  number  of  the  83  squares,  as  36, 
that  we  know  will  form  a  perfect  square  (Prop.  I), 
and  arrange  them  in  a  square,  as  t^bown  in  (1),  leaving 
47  of  the  83  squares  yet  to  be  disposed  of. 

2.  We  can  now  place  a  row  of  squares  on  two  adja- 
cent sides  of  the  square  in  (1)  and  a  square  in  the  comer, 
and  still  have  a  perfect  square  as  shown  in  (2). 

8.  Observe,  that  in  putting  one  row  of  small  squares 
on  each  of  two  adjacent  sides  of  the  ^(\xi&re  first  formed, 
we  must  use  twice  as  many  squares  as  there  are  units 
in  the  side  of  the  square. 

4.  Now  since  it  takes  twice  6  or  12  squares  to  put 
one  row  on  each  of  two  a4Jacent  sides,  we  can  put  on 


f!> 


^  r 

I" 

tfelH 

3   _ 

- 

-1-  fe 

fc  P 

s. 

i\tm 

- 

-.}:  r 

p.t.t=j  i_"i: 

(1) 


Bf 


4_i 


1 1 


WW 


306 


BUSINESS    ARITHMETIC, 


m ' 


f^^ 


1 


11 


It 


■^1 


i « 
,1 


(8) 
6x8=18.  8'=9. 


as  many  rows  as  12  is  contained  times  in  47,  the  number  of  squares  remain- 
ing.   Hence  we  can  put  on  8  rows  as  shown  in  (8)  and  have  11  squares  still 

remaining. 

5.  Again,  having  pnt  8  rows  of  squares  on  each 
of  two  adjacent  sides,  it  talces  3  x  8  or  9  squares  to 
fill  the  corner  thus  formed,  as  shown  in  (3),  leaving 
only  2  of  the  11  squares. 

Hence,  the  square  in  (3)  represents  the  greatest 
r-*  perfect  power  in  83,  namely  81 ;  and  9,  the  number 
so  of  units  in  its  side,  represents  the  square  root  of  81. 

•■o  6.  Now  observe  that  the  length  of  the  side  of  the 
square  in  (3)  is  6+3  units,  and  that  the  number  of 
Hmall  squares  may  be  represented  in  terms  of  6  +  3; 
thus. 


6^  =  36. 


(1.)  (6+3)«  =  Q'  +  S^  +  ttoiceQxd  =  36+9  +  36  =  81. 

Again,  suppose  5  units  had  been  taken  as  the  side  of  the  first  square,  the 
number  of  small  squares  would  be  represented  thus : 

(2.)  (5  +  4)'^  =  S'  +  ^'  +  ticke  5x4:  =  25  +  16  +  40  =  81. 

In  the  same  manner  it  may  be  shown  that  the  square  of  the  sum  of  any 
two  numbers  expressed  in  terms  of  the  numbers,  is  the  square  of  each  of 
the  numbers  plus  twice  their  product. 

Hence  the  square  of  any  number  may  be  expressed  in  terms  of  its  tens 
and  units ;  thus  57  =  50+7 ;  hence, 

(3.)  572  =  (50 +  7)"^  =  50^  +  r +  tmce 50x7  =  3249. 

This  may  also  be  shown  by  actual  multiplication.  Tims,  in  multiplying 
57  by  57  we  have,  Jlrst,  57  x  7  =  7  x  7+50  x  7  =  7'  +  50  x  7 ;  we  have,  second^ 
67x50  =  50x7+50x50  =  50x7+50';  hence,  57 »  =  QO^+V+ twice 50x1. 


Find,  by  constructing  a  diagram  as  above,  the  square  root  of 
each  of  the  following : 

Observe,  that  when  the  number  is  large  enough  to  give  tens  in  the  root* 
we  can  take  as  the  side  of  the  first  square  weconstnict  the  greatest  number 
of  tens  whose  square  can  be  taken  out  of  the  given  number. 


1.  Of  144. 

2.  Of  196. 

3.  Of  289. 


4.  Of  520. 

5.  Of  729. 

6.  Of  1089. 


7.  Of  1125. 

8.  Of  584. 

9.  Of  793. 


10.  Of  1054. 

11.  Of  2760. 

12.  Of  3832. 


9  remain- 
laretf  tttill 

)  on  each 
quares  to 
;),  leaving 

3  greatest 
le  number 
root  ol  SI. 

ide  of  the 
lumber  of 
Qsof6+3; 


E  VOL  UTION. 


307 


square,  the 


mm  of  any 
of  each  of 

of  its  ten& 


7S22.  Prop.  Ill— TAd  square  of  any  number  must,  wn- 
tain  twice  as  many  figures  as  the  number,  or  twice  as  many 
less  one. 

This  proposition  may  be  shown  thus : 

1.  Observe,  the  Bqaare  of  either  of  the  digits  1,  2,  8,  is  expressed  by  one 
figure,  and  the  t<quare  of  either  of  the  digits  4,  5,  6,  7,  8,  0,  is  czprenoed 
by  two  figures ;  thus,  2  x  2  =  4,  8  x  3  =  9,  and  4  x  4  =  16,  5  x  6  =  25,  and 
so  on.  • 

2.  Since  10  x  10  =  100,  it  is  evident  the  square  of  any  number  of  tens 
must  have  two  ciphers  at  the  right ;  thus,  20'^  =  20  x  20  =  400. 

Now  since  the  square  of  either  of  the  digits  1,  2,  3,  is  expressed  by  one 
figure,  if  we  have  1, 2,  or  3  tens,  the  square  of  the  number  must  1  e  expressed 
by  3  figures ;  that  is,  one  figure  less  than  twice  as  many  as  are  required  to 
express  the  number. 

Again,  since  the  square  of  either  of  the  digits  4,  5,  6,  7,  8,  9,  is 
expressed  by  two  figures,  if  we  have  4,  5,  6,  7,  8,  or  9  tens,  the  square  of  the 
number  must  contain  four  figures ;  that  is,  twice  as  many  figures  as  are 
required  to  express  the  number.  Hence  it  is  evident  that,  in  the  square  of 
a  number,  the  square  of  the  tens  must  occupy  the  third  or  the  third  and 
fourth  place. 

By  the  same  method  it  may  be  shown  that  the  square  of  hundreds  must 
occupy  the^A  or  the  J^th  and  sixth  places,  the  square  of  thousands  the 
seventh  or  the  seventh  and  eighth  places,  and  so  on ;  hence  the  truth  of  the 
proposition. 


aultiplying 
ve,  second^ 
50x7. 

Ire  root  of 


the  root* 
jst  number 


fl054. 
If  2760. 
If  3833. 


From  this  proposition  we  have  the  following  conclusions : 

723.  I.  If  any  number  be  separated  into  periods  of  tuo 
figures  each,  beginning  with  tlie  units  place,  the  number  of 
periods  wiU  be  equal  to  the  number  of  places  in  the  square 
root  of  the  greatest  perfect  power  which  the  given  number 
contains. 

TL  In  the  square  of  any  number  the  square  of  the  units  are 
fmind  in  the  units  and  tens  place,  the  square  of  the  tens  in 
the  hundreds  and  thousands  place,  the  square  of  the  hun- 
dreds in  tTte  tens  and  hundreds  of  thousands  place, 
and  so  on. 


11 


m 

IP 

Ip 

&*] 

I"    .! 

l^'i 

r\t- 

1  '^':- 

i 

I?   ;■■ 

Is    ^J.  ■ 


1  ■■!  i 

-"■'3        ^ 

^ 

■f 

a'* 

•l.  '  . 

.i 

.  1-i  J 

1 

! 

if- 1 

-',  l 

J  i 

lir' 

308 


BUSINESS    ARITHMETIC, 


ILLUSTRATION    OF    PROCESS. 


724.     1.  Find  the  square  root  of  225. 
(0)10x6=80.       ((<)5«=85. 


■"■ 

^ 

^ 

^ 

" 

" 

— 1 

tn 

4.-     *  -    ..-    . 

+ 

ill. 

s 

II 

^ESSPpe&gs 

r 

7 

;~ 

-_ 

— 

_-■ 

^ 

— z 

^- 

"^ 

E 

u. 

_^ 

^ 

"= 

TZ 

= 

== 

— . 

'z= 

1^ 

o 

a 

^ 

S 

Si 

3 

s 

fc 

_£. 

3 

3 

'3 

^=aE( 

^ 

:ze 

.= 

-1^ 

--,> 

-= 

^ 

.^ 

r 

~ 

:f 

-3 

^ 

s 

^ 

== 

-~ 

ss 

:i=l 

:?. 

-~ 

r 

-~ 

^ 

-3 

:^ 

^ 

^ 

:= 

=a: 

^ 

=ri 

= 

P 

=e 

^ 

^ 

•# 

!?*^ 

* 

^ 

S 

^ 

s 

^ 

^ 

m 

g 

^ 

=^^ 

m. 

(a 

)   1 

0 

i_ 

:1( 

X) 

Id  step. 


id  Step 


\ 


336(10 
10*  B  10  X  10  s  100 

(1)  T.  diviaAQ  %  3=30)  135  (   5 

foo^  15 
S0k5= 

6x6: 


<»>n:rsh 


135 


S 

II 


Explanation.— 1.  We  observe,  as 
ebown  in  (a),  tbat  1  ten  is  tbe  lar^et 
nnmber  of  tens  wboso  sqnara  is  con- 
tained in  335.  Hence  in  let  step  we 
subtract  10*=100  from  836,  leaving  136. 

3.  Having  formed  a  square  wliose 
side  is  10  units,  we  observe,  as  sbown 
1q  {b)  and  (c),  that  it  will  take  tu^ce  ten  to  put  one  row  on  two  acyaoent 
eidesi.    Hence  the  Trial  Divisor  is  10  x  2  =  sa 

8.  We  observe  tbat  SO  is  contained  6  times  In  125,  but  if  we  add  6  unite 
to  the  aide  of  the  square  (a)  we  will  not  have  enough  left  for  the  comer 
((f),  hence  we  add  5  unite. 

4.  Having  added  6  units  to  the  side  of  the  square  (a),  we  observe,  as 
shown  in  (6)  and  (<?),  that  it  requires  twice  10  or  20  multiplied  by  5  plus 
6  X  6,  as  shown  In  ((f),  to  complete  the  square  \  hence  (3)  in  id  step. 


Solution  with  every  Operation  Indicated. 

725.    2.  find  the  square  root  of  466489. 


First  Stkp. 
Sbooni)  Step. 


€00x600 


466489  (  600 
860000       80 


j  (1)  2Via/e;ioi«>r 600x3  =  1300)106489 
•}         i  1300x80=96000 
*  ^  M     80x80=  6400 


f  = 


3 


103400     688  required  root. 


'TmRS  Step, 


1  (1)  2Wa;(fi«i4or 680x3  =  1360)    4089 
•     "j        (  1360x3=4080} 


8x8=      9 


4089 


Bxplanation.— 1.  We  place  a  point  over  every  second  figure  beginning 
with  the  units,  and  thus  find,  according  to  (722),  that  the  root  must  have 
three  places.    Hence  the  first  figure  of  the  root  expresses  hundreds. 


835(10 

.  m 

=20)125 (5 
root  15 

=    1* 

observe,  as 
B  the  Iftr^eat 
aare  is  con- 
1st  «lep  we 
leaving  18S. 

luare  wlioee 
re,  as  pbown 

B  add  6  un\ts 
or  the  corner 

observe,  as 
jed  by  5  plus 
step. 


\ated. 


EVOL  UTION, 


3oa 


8.  We  ob^erro  that  the  square  of  600  is  the  fnreatcst  second  poNor 
of  buudrcd-i  cuuuiiued  iu  4(kM8:).  ileuce  iu  the  tirel  rtep  we  8ublract 
600  X  GOO  -^  360(M)0  from  460^t8*»,  ioavin^  106480 

3.  Wu  now  duubit'  the  600,  the  rout  foaud,  for  a  trial  dlvimor,  accordiug 
to  (744—2).  DlvidiuK  106489  by  1200  we  And,  accordluK  to  (7>i4— 2),  that 
we  cau  odd  80  to  tbe  root.  For  tUiit  additlou  wu  une.  u»  hUowu  Iu  (2), 
teomU  step,  1200  ^m  -  mm  aud  80  X  80  -  0400(7*4-4).  muklii),'  in  all 
10^100.    Subtracting  10^100  from  1»^0,  we  have  atill  rvmainiug  40W. 

4.  We  aguJu  double  680,  the  root  found,  for  a  trial  divisor,  according 
to  (f  24— 3),  and  proceed  in  the  same  manner  as  before,  as  shown  In 
third  vtep. 

obtierve,  from  the  foregoing  it  is  evident  that  one  digit  in  the  root  is 
equivalent  to  two  digits  in  the  square,  and  convorHoly,  two  dlfrltt*  In  the  . 
»quare  give  one  digit  in  the  root.    Flence  by  dividing  thu  giv«n  number 
into  i>erio<iH  of  two  flgurei*  bugiuning  at  the  decimal  points  wc  show  the 
bambcr  of  digits  i^  the  root. 


7tiH,    Contracted    Solution     of    the   foregoing 
Example. 


First  Step. 


6x0  = 


46G4(i9  (  G83 


Ul)        6  X  2=    12)1064 
Second  Step.  ]  ^^^    128,8=  1024 


THIRD  Step     >  ^^^      68x2  =  130)     408J» 
THIRD  STEi .    ^  ^2^  1363  ^  3  ^  40^9 


anired  root, 


beginning 
)t  must  have 
Irede. 


Explanation.— 1.  Observe,  in  the^rs^  step  we  luiow  that  the  square 
600  must  occupy  the  fifth  and  sixth  place  (721).  lleuce  the  ciphers  are 
omitted. 

2.  Observe,  that  in  (1),  second  step,  we  use  6  instead  of  600,  tlius  dividing 
the  divisor  by  100 ;  hence  we  reject  the  tens  and  uniti  firom  the  right  of  the 
dividend  (131). 

3.  Observe,  also.  In  (2),  second  step,  we  unite  in  one  three  operations. 
Instead  of  multiplying  12  by  80,  the  part  of  the  root  found  by  dividing 
1064  by  12,  we  multiply  the  12  flfst  by  10  by  annexing  the  3  to  it  (82),  and 
having  annexed  the  8  we  multiply  the  result  by  8,  which  gives  ua  the- 
prodnct  of  12  by  80,  plus  the  square  of  8.  But  the  square  of  8,  written,  as 
it  is,  in  the  third  and  fourth  place,  is  the  square  of  80. 

Hence  by  annexing  the  8  and  writing  the  result  an  we  do,  we  have 
united  in  one  three  operations  ;  thus,  128  x  8  =  12  x  80  -t-  80*. 


mam 


310 


n  Ua  I N  E  S  S     A  R  I  TUM  E  Tl  C, 


mM 


m 


I 


From  these  illustrations  we  have  the  following 

727.  liuLK. — /.  ISeparate  the  number  into  periods  of  two 
figures  rarh,  by  2d(icin(/  a  point  over  ecery  second  figure,  begin- 
ning with  the  units  figure. 

II.  Mud  the  greatest  square  in  the  left-hand  period  and 
pface  its  root  on  the  right.  (Subtract  this  square  from  the 
ptrioil  and  annex  to  the  remainder  the  next  period  for  a 
dicidend. 

III.  Double  the  j)art  of  the  root  found  for  a  tnal  divisor,  and 
find  flow  many  times  this  divisor  is  contained  in  the  dividend, 
omitting  tlie  right-hand  figure.  Annex  the  quotient  thus  found 
both  to  the  root  and  to  the  divisor.  Multiply  the  ditisor  thus 
completed  by  the  figure  of  the  root  last  obtained,  and  subtract  the 
product  from  tJie  dividend. 

IV.  Tf  there  are  more  periods,  continue  the  operation  in  the 
same  manner  as  before. 

In  applying  this  rule  be  particular  to  observe  : 

1.  When  there  is  a  remainder  after  the  last  period  hae  been  need,  annex 
periodf"  of  ciphers  and  continue  the  root  to  as  many  decimal  places  as  may 
be  required. 

2.  We  eeparatc  a  number  into  periods  of  two  figure?  by  beginning  at  the 
nnltt*  place  and  prococdin/j;  to  the  left  if  the  number  is  an  integer,  and  to 
the  ri{;ht  if  a  decimal,  and  to  the  right  and  left  if  both. 

3.  Mix'-d  numbers  and  fractions  are  reduced  to  decimals  before  extract- 
ins:  the  root.  But  in  case  the  numerator  and  the  denotninator  are  perfect 
powers,  or  the  detwminator  alone,  the  root  may  be  more  readily  fornied  by 
extracting  the  root  of  each  term  separately. 


Tims 


80  on. 


I 


^  _  V49  _  7 


81 


V^i 


9' 


and 


Extract  the  square  root 

1.  Of  ^W  3.  Of 

2.  Of  ,Vj.  4.  Of  f  Jf . 

9.  540131000000. 
10.  191810713444. 


35 
04 


V35   V^S 

-7_  =  -^v,-,  and 

V64     S 


6.  Of  iff. 

7  Of  72p 

6.  Of  ,ViV- 

Q  Of  3  ^  1 

Ans. 

739000. 

Ans. 

43T9G2. 

is  of  two 
re,  begin  - 

riod  nnd 
from  the- 
iod  for  a 

Ivisor,  find 
I  dividend, 
huH  found 
icisor  thua 
ubtract  the 

Uion  in  the 


upcd,  annex 
laces  as  may 

inninir  at  the 
tc'Rer,  aud  to 

eforc  extract- 
r  are  perfect 
y  foriped  by 


Vi^^-,  and 

8   ' 


7.  Of  ,Vo^. 

8.  Of  ^Vj- 


9000. 
79G2. 


E  VO  L  UT  I  ON, 

EXAMPLES    FOR     PRACTICE. 
728.  Extract  the  square  root 


311 


1.  Of  3481. 

2  Of40»0. 

8.  Of  7509. 

4.  Of2l-'()9. 

T).  Of  0210. 

C.  or  8049. 


7.  Of?i2}- 

9.  Of.022.j, 
10    Of  '  "''* 

11.  or  .5770, 

12.  or  .2804 


18.  Of  137041. 

14.  Of  4100.25. 

15.  Of  708427.50. 
10.  or  28022.70. 

17.  Of  57.1530. 

18.  Of  474.8041. 


Find  tlie  square  root  to  three  decimal  i)laces : 


19.  Of  32. 

20.  Of  59. 

21.  or  7. 


22.  or  .93. 

23.  or  .8. 

24.  Of  .375. 


25.  Of  14.7. 
20.  or  80.2. 
27.  Of  5.973. 


28.  Of  I. 

29.  or/g. 

30.  or  A„ 


Perform  the  operations  indicated  in  the  following  : 

34.  ^/558009-<-(4J^)i. 


31.  ^^889-^1024. 
82.  V^209  +  V225. 
33.  V!SF§  ^  ^2209. 


35.  70890^-^^2130. 
30.  (SS?S)^x  131376^. 


87.  What  is  the  length  of  a  square  floor  containing  9025 
square  feet  of  lumber? 

38.  How  many  yards  in  one  of  the  equal  sides  of  a  square 
acre? 

39.  A  square  garden  contains  237109  square  feet ;  how  many 
reet  in  one  or  its  sides? 

40.  A  triangular  field  contains  1900.24  P.  What  is  the 
length  of  one  side  of  a  square  field  of  equal  area  ? 

41.  An  orchard  containing  9210  trees  is  planted  in  the  form 
or  a  square,  each  tree  an  equal  distance  from  another  ;  how 
many  trees  in  each  row? 

42.  Find  the  square  root  of  2,  of  5,  and  of  11,  to  4  decimal 
places. 

43.  Find  the  square  root  of  2,  j'^j,and  of  \l,  to  3  decimal 
places. 


ll 


^*^Tf  :^^'^ 


312 


B  COSINESS     ARITHMETIC, 


M 


..i 


CUBE    ROOT. 

PREPARATORY     PROPOSITIONS. 

729.  Prop.  I.— Any  perfect  third  power  may  be  repre^ 
sented  to  the  eye  by  a  cube,  and  the  number  of  units  in  the  side 
of  such  cube  will  represent  the  third  or  cube  root  of  the 
given  power. 

Represent  to  the  eye  by  a  cabe  343. 

1.  We  can  suppose  the  number  343  to  repre- 
sent small  cubes,  and  we  can  take  2  or  more  of 
these  cubes  and  arrange  them  iu  a  row,  Ub  bhovvu 

2.  Having  formed  a  row  of  5  cube?,  as  shown 
m  (1),  we  can  arrange  5  of  these  rows  side  by  side, 
as  shown  in  (2),  forming  a  square  slab  coiituiiiiug 
5x5  small  cubes,  or  as  many  small  cubes  as  the 
square  of  the  number  of  units  in  the  side  of  the 
slab. 

3.  Placing  5  such  slabs  together,  as  shown  in 
(3),  we  form  a  cube.  Now.  since  each  blab  con- 
tains 5x5  small  cubes,  and  since  5  slabs  are 
placed  together,  the  cube  in  (3)  contain**  5x5x5, 
or  125  small  cubes,  and  hence  represents  the  third 
power  125,  and  each  edge  of  the  cube  represents 
to  the  eye  5,  the  cube  root  of  125. 

We  have  now  remaining  yet  to  be  disposed  of 
»43-125,  or  218  small  cubes. 

4.  Now,  observe,  that  to  enlarge  the  cube  in  (3) 
so  that  it  may  contain  the  343  small  cubes,  wo 
must  build  the  same  number  of  tiers  of  small 
cubes  upon  each  of  three  adjacent  sides,  as  shown 
in  (4). 

Observe,  also,  that  a  slab  of  small  cubes  to 

cover  one  side  of  the  cube  in  (3)  must  contain 

5x5  or  25  small  cubes,  as  shown  iu  (4),  or  as 

many  small  cubes  as  the  square  of  the  number  of  unite  in  one  edge  of  the 

cube  in  (3). 

Hence,  to  find  the  number  of  cubes  necessary  to  put  one  slab  on  each  of 
three  sides  of  the  cube  in  (3),  we  multiply  the  square  of  its  edge  by  3,  giving 
6'  X  3  =  5  X  5  X  3  —  75  small  cubes. 


be  repre' 
)i  the  side 
tot  of  the 


a43  to  repre- 
t  or  more  of 
jw,  as  tiUowu 

ep,  as  shown 
8  side  by  side, 
abcouiaiuiug 
cubes  as  the 
he  side  of  the 

,  as  Bhown  in 
jach  blub  cou- 
5  elabsj  are 
ktains  5x5^5, 
iente  the  third 
jbc  represents 

je  disposed  of 

|the  cube  in  (3) 
lall  cubes,  wo 
[tiers  of  small 
Ides,  as  shown 

Imall  cubes  to 

must  contain 

iu  (4),  or  as 

le  edge  of  the 

Llab  on  each  of 
Ige  by  3,  giving' 


£  VOL  UTI  0  N. 


313 


(5) 


5.  Having  found  that  75  small  cubes  will  put  one  tier  on  each  of  three 
adjacent  sides  of  the  cube  in  (3),  we  divide  218, 
the  number  of  small  cubes  yet  remaining,  by 
75,  and  find  how  many  such  tiers  we  can  form. 
Thus,  218+75  =  2  and  68  remaining.  Hence  we 
can  put  2  tiers  on  each  of  three  adjacent  sides, 
as  shown  in  (5),  and  have  68  email  cubes 
remaining. 

6,  Now,  observe,  that  to  complete  this  cube 
we  must  fill  euch  of  the  three  comers  formed  by 
building  on  three  adjacent  sides. 

Examine  cartfuUy  (6)  and  observe  that  to  fill 
one  of  these  three  comers  we  require  as  many 
email  cubes  as  is  expressed  by  the  square  of 
the  number  of  tiers  added,  multiplied  by  the 
number  of  units  in  the  side  of  the  cube  to  which 
the  addition  is  made.  Hence  we  require  2'  x  5 
or  20  small  cubes.  And  to  fill  the  three  corners 
we  require  3  times  2^  x  5  or  60,  leaving  68—60  or 
8  of  the  small  cubes. 


ffi) 


(7) 


7.  Examine  again  (5)  and  (6)  Andobserve  that 
when  the  three  comers  are  filled  we  require  to 
complete  the  cube  as  shown  in  (7),  another  cube 
whose  side  contains  as  many  units  as  there  are 
units  added  to  the  side  of  the  cube  on  which 
we  have  built.  Consequently  we  require  2*  or 
3x2x2  =  8  small  cubes. 

Hence  we  have  formed  a  cube  containing 
343  small  cubes,  and  any  one  of  its  edges  repre- 
sents to  the  eye  5+2  or  7  units,  the  cube  root  of  343. 

Prom  these  illustrations  it  will  be  seen  that  the  steps  in  finding  the  cube 
root  of  343  may  be  stated  thus  : 


"tt: 


First  Step. 


343 
125 


Skoond  Step. 


We  assume  that  343  represents  small  cubes  and  take 

5  as  the  length  of  the  side  of  a  large  cube  formed 

from  these.    Hence  we  subtract  the  cube  of  5  = 

r  1.  We  observe  it  takes  5'  x  3  =  75  to  put  oiie  tier  on 

three  adjacent  Hides.    Hence  we  can  put  on  75)218(2 

2.  We  have  now  found  that  we  can  add  2  units  to  the 
Bide  of  the  cube.    Hence  to  add  this  we  require 

(1)  For  the  3  sides  of  the  cube         5«  x  2  x  3=  150  ^ 

(2)  For  the  8  corners  thus  formed   2'  x  5  x  3= 

(3)  For  the  cube  in  the  comer  last  formed  2*: 


J=  60  )-  =  218 
»=    8J 


u 


Hence  the  cube  root  of  343  is  5+2  =  7. 


21 


^<''^ 


314 


BUSINESS    ARITHMETIC, 


u  • 


[.f 


■]  ■^ 


730«  Observe^  that  the  number  of  small  cubes  in  the  cube  (7)  in  the 
foregoing  Illustrations,  are  expressed  in  terms  of  6+2;  namely,  the  num- 
ber of  units  in  the  side  of  the  first  cube  formed,  plus  the  number  of  tiers 
added  in  enlarging  this  cube ;  thus  : 


<7) 


(B) 


m 


II 


(5  +  2)8 


5»  +  5«x2x3      +      2«x5x3  +  2» 


In  this  manner  it  may  be  shown  that  the  cube  of  the  sum  of  any  two 
numbers  is  equal  to  the  cube  of  each  number  plus  3  times  the  square  of  the 
first  multiplied  by  the  second  number,  plus  3  times  the  square  of  the  second 
multiplied  by  the^r^^  number. 

Hence  the  cube  of  any  number  may  be  expressed  in  terms  of  its  tens 
and  units ;  thus,  74  =  70+4;  hence, 

(70+4)»  =  70»  +3  times  70"  x  4+3  times  4"  x  70+4*  =  405224. 


Solve  each  of  the  following  examples,  by  applying  the  fore- 
going illustrations : 

1.  Find  the  side  of  a  c  abe  which  contains  739  small  cubes, 
taking  6  units  as  the  side  of  the  first  cube  formed. 

2.  How  many  must  be  added  to  9  that  the  sum  may  be  the 
cube  root  of  4096?    Of  2197?    0/2744? 

8.  Take  20  units  as  the  side  of  the  first  cube  formed,  and  find 
the  side  of  the  cube  that  contains  15625  cubic  units. 

4.  Find  the  jube  root  of  1368.    Of  3405.    Of  2331.    Of  5832. 

5.  ExpresB  the  cabe  of  54,  of  72,  of  95,  of  123,  of  274,  in 
terms  of  the  tens  and  units  of  each  number. 

6.  Express  the  cube  of  83  in  terms  of  80+3. 


be  (7)  in  the 
y,  the  nnm- 
iber  of  tiers 


(fi) 


EVOLUTION, 


315 


II 
5x3+2» 

um  of  any  two 

e  square  of  tbe 
re  of  the  second 

18  of  its  tens 


405224. 

|ing  the  fore- 
small  cubes, 
may  be  the 

jed,  and  find 
Its. 
\\.    Of  5832. 

IS3,  of  274,  in 


731.  Prop.  II. — The  cube  of  any  mimbcr  must  contain 
three  times  as  many  places  as  the  number^  or  three  times  aa 
jnany  less  one  or  two  places. 

This  proposition  may  be  shown  thus  : 

1.  Observe,  V  =  1,  2»  =  8,  3'  =  27,  4*  =  64,  5»  =  125,  and  9»  =  7»; 
hence  the  cube  of  1  and  2  is  expressed  each  by  one  figure,  the  cube  of  3  and 
4  each  by  two  figures,  and  any  number  from  5  to  9  Inclusive  each  by  three 
figures. 

2.  Observe,  also,  that  for  every  cipher  at  the  right  of  a  number  there 
must  (82)  be  three  ciphers  at  the  right  of  its  cube  ;  thus,  10*  =  1,000, 
100'  =  1,000,000.  Hence  the  cube  of  tens  can  occupy  no  place  lower  than 
thousands,  the  cube  of  hundreds  no  place  lower  than  millions,  and  bo  on 
with  higher  orders. 

3.  From  the  foregoing  we  have  the  following : 

(1.)  Since  the  cube  of  1  or  2  contains  one  figure,  the  cube  of  1  or  2  tens 
must  contain /o?/r  places  ;  of  1  or  2  hundreds,  seven  places,  and  so  on  with 
higher  orders. 

(2.)  Since  the  cube  of  3  or  4  contains  two  figures,  the  cube  of  3  or  4  tens 
must  contain ^t'e  places ;  of  3  or  4  hundreds,  eight  places,  and  so  on  with 
higher  orders. 

(3.)  Since  the  cube  of  any  number  from  5  to  9  inclusive  contains  three 
places,  the  cube  of  any  number  of  tens  from  5  to  9  tens  inclusive  must  con- 
tain six  places  ;  of  liuudreds,  from  5  to  9  hundred  inclusive,  ?iin€  places, 
and  so  on  with  higher  orders  ;  hence  the  truth  of  the  proposition. 

Hence  also  the  following  : 

7322.  /  If  any  nuTnber  be  separated  into  periods  of  th/ree 
figures  ea/ih,  beginning  with  the  units  place,  the  number  of 
periods  mil  be  equal  to  the  number  of  pLices  in  tlie  cube  root 
of  the  greatest  perfect  third  power  which  the  given  number 
contains. 

II.  The  cube  of  units  contains  no  order  higher  than 
hundreds. 

III.  The  cube  of  tens  contains  no  order  lower  than  thousands 
nor  higher  than  hundred  thousands,  the  cube  of  hundreds  no 
order  lower  than  millions  nor  ftigh^r  than  hundred  millions,  and 
80  on  with  higher  orders. 


;t! 


^iilll// 


316 


B^/SI^ESS    ARITHMETIC, 


!<■' 


:j 


ILLUSTRATION    OF    PROCESS. 

733.  Solution  ivith  every  Operation  Indicated^ 

Find  the  cube  root  of  92345408. 


FlKST  Stbp. 


400*  =  400x400x400  = 


9234M08(400 
64000000 


(1)  Trial  divisor  400»  x  8  =  480000  )  28M5408  (   50 

Second  Step.  <         /  40O'  x  50  x  3  =  24000000  \ 

(2)  "I  50"^  X  400  X  3  =    3000000  >•    =     27125000 
(  50»  =      125000  J 


Thtpo  \y 


(  (1)  TricU  divisor  450'  x  3  =  607500  )    12204C8  ( _2 

I         I  450"  X  3  X  2  =  1215000  \  Hoot  453 

(  (2;    ■{  2»  X  450  x  3  =       5400 
(  2»  =  8 


\ 


=      1220408 


.  f 


Explanation.— -t.  vV'  j  ...z'.  a  period  over  every  third  fljjure  begln- 
iiiDg  with  the  units,  and  thus  fled,  according  to  (732),  that  the  root 
must  have  three  places.  Hence  the  first  figure  of  the  root  expressee 
huudrcds. 

2.  We  observe  that  400  is  the  greatest  number  whose  cube  is  contained 
k)  the  given  number.  Subtracting  400*  =  64000000  from  92345408,  we  have 
88345408  remaining. 

3.  We  find  a  trial  divisor,  according  to  (729—4),  by  taking  3  times  the 
square  of  400,  as  shown  in  (1),  second  step.  Dividing  by  this  divisor, 
according  to  (729—5),  the  result  is  that  we  find  we  can  add  50  to  the  root 
already  found. 

Observe,  the  root  now  found  is  400+50,  and  that  according  to  (730), 
(400  +  50)»  =  400»  +  400"x50x3  +  50''x400x3+50». 

We  have  already  subtracted  400*  =  640000  from  the  given  number. 
Hence  we  have  only  now  to  subtract, 

400'x50x3  +  50»x400x3+50»  =  27126000, 

as  shown  in  (2),  second  step,  leaving  1220408. 

5.  We  find  another  trial  divisor  and  proceed  in  the  same  manner  to  find 
the  next  figure  of  the  root,  as  shown  in  the  third  8tQ3. 


EVOL  UTION, 


317 


.dicatcd* 

[08(400 
WO 

408  (   50 
)000 

34C8  ( _2 
JiOOt  452 
10408 


i  figure  begin- 
that  the  root 
root  expresses 


)e  iB  contained 
i&408,  we  have 


tag  3  times  the 
ly  this  divisor, 
Id  50  to  the  root 

ig  to  (730), 


given 


number. 


manner  to  find 


734.     Contracted    Solution    of    tJie   foregoing 
Example, 


First  Step. 
Second  Step. 


4»  =  4x4x4  = 


0234M08(452 
64 


i(l)  Trial  divisor  40*  x  3  =     41 
/  40»  X  5  X  3  =  eiooo  \ 
(2)  ^  5»  X  40  X  3  =    3000  >  = 
(  5*  =      125 ) 


Trial  divisor  450*  x  3  =  607500   )    1220408 


Thibd  Stsp. 


ni)  Trials 
}  1 460* 
( (2)  j  2«  X 


X  2  X  3  =  1216000  \ 
450  X  8=       MOOJ-   = 

a»=  8) 


1220406 


Explanation.— 1.  Observe,  in  the  first  step,  we  know  that  the  cube  of 
400  must  occupy  the  seventh  and  eighth  places  (732—111).  Ilence  the 
ciphers  are  omitted. 

2.  Observe,  also,  that  no  part  of  the  cube  of  hundreds  and  tens  is  found 
below  thousands  (732— III).  We  therefore,  in  finding  the  number  of  tens 
in  the  root,  disregard,  as  shown  in  second  step,  the  right-hand  period  in  the 
given  number,  and  consider  the  hundreds  and  tens  in  the  root  as  tens  and 
units  respectively. 

Hence,  in  general,  whatever  number  of  places  there  are  in  the  root,  we 
<lisregard,  in  finding  any  figure,  as  many  periods  at  the  right  of  the  given 
number  as  there  are  places  in  the  root  at  the  right  of  the  figure  we  are 
finding,  and  consider  the  part  of  the  root  found  as  tens,  and  the  figure  we 
fire  finding  as  units,  and  proceed  accordingly. 

From  these  illustrations  we  have  the  following 


BIJIiE. 

735.  /  Separate  the  number  into  periods  of  three  fig^ires 
each,  by  placing  a  point  over  every  third  figure,  beginning  with 
the  units  figure. 

II.  Find  the  greatest  cube  in  the  left-h/ind  period,  and  place 
its  root  on  the  right.  Subtract  this  cube  from  the  period  and 
annex  to  the  remainder  the  next  period  for  a  dividend. 

III.  Divide  this  dividend  by  the  trial  divisor,  which  is  3  time* 
the  square  of  the  root  already  found,  considered  as  tens;  the 
quotient  is  the  next  figure  of  the  root. 


'•>  ■ 


w^ 


(T? 


■«■■" 


318 


BUSINESS    ARITHMETIC, 


r :  . 

if    . 


IV.  fMytractfrom  the  dividend  P>  times  the  square  of  the  root 
before  found,  considered  as  tens,  multiplied  by  the  figure  last 
found,  plus  3  times  the  square  of  the  figure  last  found,  multi- 
plied by  tTie  root  before  found,  plus  the  cube  of  t?ie  figure  last 
found,  and  to  the  remainder  annex  the  next  period,  if  any,  for 
a  new  dividend. 

V.  Jf  there  are  nune  figures  in  the  root,  find  in  the  same 
manner  trial  divisors  and  proceed  as  before. 

In  applying  this  rule  be  particular  to  observe : 

■-  1.  In  dividing  by  the  Trial  Divisor  the  quotient  may  be  larger  than  the 
required  figure  in  the  root,  on  account  of  the  addition  to  be  made,  as 
ehovvn  in  (729—6)  second  step.  In  Buch  case  try  a  figure  1  leps  than  the 
quotient  found. 

S.  When  there  is  a  remainder  after  the  last  period  has  been  used,  annex 
periods  of  ciphers  and  continue  the  root  to  as  many  decimal  places  as  may 
be  required. 

8.  We  separate  a  number  Into  periods  of  three  figures  by  beginning  at 
the  units  place  and  proceeding  to  the  left  if  the  number  is  an  integer,  and 
to  the  right  if  a  decimal,  and  to  the  right  and  left  if  both. 

4.  Mixed  numbers  and  ft'actions  arc  reduced  to  decimals  before  extract* 
ing  the  root.  But  in  case  the  numerator  and  the  denominator  are  perfect 
third  powers,  or  the  denominator  alone,  the  root  m8.y  be  more  readily 
found  by  extracting  the  root  of  each  term  separately. 


EXAMPLES    FOR    PRACTICE. 
736.  Find  the  cubic  root  of 


1.  729. 

2.  216. 

3.  2197. 

4.  1331. 


5.  10648. 

6.  4096. 

7  518 

8.  6859. 


10.  un- 

11.  438976. 

12.  250047. 


13.  47045881. 

14.  24137569. 

15.  113.379904. 

Of  86. 


17 .  Find,  to  two  decimal  places,  the  cube  root  of  11. 
Of  84.    Of  235.    Of^V    Of^^.     Of  75.4.    Of  6.7. 

18.  Find  to  three  decimal  places  the  cube  root  of  3.    Of  7. 
Of  .5.    Of  .04.    Of  009.    Of  2.06. 


EVOL  UTION, 


319 


I  of  the,  root 
figure  last 
mnd,  multi- 
g  figure  last 
,  if  any,  for 

in  the  same 


larger  than  the 

to  be  made,  as 

1  lesB  than  the 

een  used,  annex 
a\  places  as  may 

by  beginning  at 
B  an  integer,  and 

l8  before  extract- 
Inator  are  perfect 
be  more  readily 


13.  47045881. 

14.  24137569. 

15.  113.379904. 

16.  t^VtA^- 
of  11.    0136. 

|6.7. 

ot  ol  3.    0£  7. 


19.  Find  the  sixth  root  of  4096. 

Observe,  the  sixth  root  may  be  found  by  extracting ^r«<  the  square  root, 
then  the  cube  root  of  the  result. 

For  example,  V^M  =  64;  hence,  4096  =  64  x  64.  Now,  if  we  extract 
the  cube  root  of  64  we  will  have  one  of  the  three  equal  factors  of  64,  and 
hence  one  of  the  six  equal  factors  or  sixth  root  of  4096. 

Thus,  <y/64  =  4 ;  hence,  64  =  4  x  4  x  4.  But  we  found  by  extracting 
its  square  root  that  4096  =  64  x  64,  and  now  by  extracting  the  cube  root 
that  64  =  4  X  4  X  4 ;  consequently  we  know  that  4096  =  (4  x  4  x  4)  x  (4  x  4  x  4). 
Hence  4  is  the  required  sixth  root  of  4096. 

In  this  manner,  it  1b  evident,  we  can  find  any  root  whose  index  contains 
no  other  factor  than  2  or  3. 

20.  Find  the  eighth  root  of  43046721. 

21.  Find  the  sixth  root  of  2565726409. 

22.  What  is  the /<>Mr<A  root  of  34012224? 

23.  What  is  the  m»«A  root  of  134217728? 

24.  A  pond  contains  84604519  cubic  feet  of  water;  what 
must  be  the  length  of  the  side  of  a  cubical  reservoir  which 
will  exactly  contain  the  same  quantity  ? 

25.  How  many  square  feet  in  the  surface  of  a  cube  whose 
volume  is  16777216  cubic  inches  ? 

26.  What  is  the  length  of  the  inner  edge  of  a  cubical  cistern 
that  contains  2079  gal.  of  water  ? 

27.  What  is  the  length  of  the  inner  edge  of  a  cubical  bin 
that  contains  3550  bushels  ? 

28.  A  pile  of  cord  wood  is  256  ft.  long,  8  ft.  high,  and  16  ft. 
wide ;  what  would  be  the  length  of  the  side  of  a  cubical  pile 
containing  the  same  quantity  of  wood  ? 

29.  What  is  the  length  in  feet  of  the  side  of  a  cubical 
reservoir  which  contains  1221187.5  pounds  avoirdupois,  pure 
water  ? 

30.  What  are  the  dimensions  of  a  cube  whose  volume  is 
equal  to  82881856  cubic  feet  ? 

31.  Find  the  cube  root  of  843297653. 


'■■% 


*i ' 


PROGRESSIONS 


■I      !■• 


i 


i '  % 


k. 


DEFINITIONS. 

737.  A  Progression  is  a  series  of  numbers  so  related, 
that  each  number  in  the  series  may  be  found  in  the  same  man- 
ner, from  the  number  immediately  preceding  it. 

738.  An  Arithmetical  Progression  is  a  series  of 
numbers,  which  increases  or  decreases  in  such  a  manner  that 
the  difference  between  any  two  consecutive  numbers  is  constant. 
Thus,  3,  7,  11, 15, 19,  23. 

739.  A  Geometrical  Progression  is  a  series  of  num- 
bers, which  increases  or  decreases  in  such  a  manner  that  the 
ratio  between  any  two  consecutive  numbers  is  constant. 

Thus,  6, 10,  20, 40,  80,  is  a  geometrical  progrcsBion. 

740.  The  Terms  of  a  progression  are  the  numbers  of 
which  it  consists.  The  First  and  Last  Terms  are  called  the 
Extremes  and  the  intervening  terms  the  Means. 

741.  The  Comm,on  or  Constant  Difference  of  an 

arithmetical  progression  is  the   difference  between  any  two 
consecutive  terms. 

742.  The  Common  or  Constant  Ratio  or  Multi' 
plier  of  a  geometrical  progression  is  the  quotient  obtained  by 
dividing  any  term  by  the  preceding  one. 

743.  An  Ascending  or  Increasing  Progression 

is  one  in  which  each  term  is  greater  than  the  preceding  one. 

744.  A  Descending  or  Decreasing  Progression 

is  one  in  which  each  term  is  less  than  the  preceding  one. 


PROGRESSION, 


321 


ABITHMETICAL    FBOGBESSION. 

745.  There  are  jive  quantities  considered  in  Arithmetical 
Progression,  which,  for  convenience  in  expressing  rules,  we 
denote  hj  letters,  thus  : 


go  related, 
J  same  man- 


1.  A  represents  the  F%r6t  Term  of  a  progression. 

2.  L  represents  the  Last  Term. 

3.  I>  represents  the  Constant  or  Common  Difference. 

4.  A'  represents  the  Number  qf  Terms. 

5.  S  represents  the  Sum  of  all  the  Terms. 


a  series  of 
manner  that 
8  is  co)i»to^T^^- 


3Ties  of  num- 
ner  that  the 
tant. 

numTsers  of 
ire  called  the 

\rence  of  an 
^een  any  two 

or  ilfiiWi- 

xt  obtained  by 

*ogre8sion 

ling  one. 

Progression 

Ing  one. 


746.  Any  three  of  these  quantities  being  given,  the  other 
two  may  be  found.    This  may  be  shown  thus : 

Taking  7  as  the  first  term  of  an  increasing  series,  and  5  the  constant 
difference,  the  series  may  he  written  in  two  forms  ;  thus  : 


Isi  Term. 

(1)  t 

(2)  7 


M  Term. 
IS 

7+ (5) 


Sd  Term. 
17 

0  ^  \ 

7  +  (5+5) 


Uh  Term. 
22,  and  so  on. 


7+(5+5+5) 


Observe,  in  (2),  each  term  is  composed  of  the  first  term  7  plus  as  many 
times  the  constant  difi'erence  5  as  the  nnmber  of  the  term  less  1.  Thas,  for 
example,  the  ninth  term  in  this  series  would  be  7+5  ;<  (9—1)  =  47. 

Hence,  tram  the  manner  in  which  each  term  is  composed,  we  have  the 
following  formulae  or  rules  : 


1.    A=  L-J)x{N-\).     Read, 


The  first  term  is  equal  to  the  last 
term,  minus  the  common  difference 
multiplied  by  the  number  qf  terme 
less  1. 


2.    X  =  ^  +  2>x(2<r-l). 


8.    2>  = 


T^-A 


{The  last  term  is  equal  to  the  first 
term,  plus  the  common  difference  mul- 
tiplied by  the  number  qf  terms  less  1. 

(The  common  difference  is  equal  to 
the  last  term  minus  the  first  term^ 
divided  by  the  number  of  terms  less  1. 


•:r 


/■  f 


322 


BUsry^ESS   arithmetic. 


il 


m 

r  1 


4.    N—   -  _    +1. 


{The  number  of  teryuH  is  equai  to  (fir 
last  term  minus  the  first  tertn,  iJlvidid 
by  the  common  cUference^  plus  l. 


Observe,  that  io  a  decreasing  ecricp,  tho  flret  term  is  the  largest  ani  tbu 
last  term  the  stnaUest  iu  the  McrU^B.  Uence,  to  make  the  above  formuliu 
apply  to  a  dccrcanlnt,'  Hories,  we  mu'»t  place  L  where  A  la,  and  -i  whore  /> 
la,  and  read  tho  formulee  accordingly. 

747.  To  show  how  to  find  the  sum  of  a  series  let 


(1.)       4       7      10      18      16      19 
(3.)      19  _16_  18_10 7__4 

(8.)      28  +  28  +  28  +  23  +  23  +  28  =  twice  the  sum  of  the  termj. 


be  an  arithmetical  series. 
be  the  same  series  reversed. 


Now,  observe,  that  in  (3),  which  is  equal  to  twice  the  Bum  of  the  ^cric?, 
each  term  is  equal  to  the  first  term  plus  the  last  term ;  hence, 


a=\ot{A  +  L)^y.    Read, 


The  sum  qf  the  terms  of  an  arithmetical 
series  is  equal  to  one-haHf  qf  the  sum  qf  thfi 
first  and  last  term,  multiplied  by  the  number 
qf  terms. 


EXAMPLES     FOR    PRACTICE. 

748.  1.  The  first  round  of  an  upright  ladder  is  12  inches 
from  the  ground,  and  the  nineteenth  246  inches ;  how  far  apart 
are  the  rounds  ? 

2.  The  first  term  of  an  arithmetical  progression  is  4,  the 
common  difference  2 ;  what  is  the  12th  term? 

3.  Weston  travelled  14  miles  the  first  day,  increasing 
4  miles  each  day ;  how  far  did  he  travel  the  15th  day, 
and  how  many  miles  did  he  travel  in  all  the  first 
12  days? 

4.  The  tenth  term  of  an  arithmetical  progression  is  190,  the 
common  difference  20 ;  what  is  the  first  term  ? 

5.  The  first  term  of  an  arithmetical  series  of  100  terms 
is  150,  and  the  last  term  1338 ;  what  is  the  common 
difference  ? 

6.  The  amount  of  $360  for  7  years  at  simple  interest  was 
$486 ;  what  was  tho  yearly  interest  ? 


PROGRESSION, 


32a 


lenn,  dlvidtd 
»,  plus  1. 

argest  ani  tbo 
ibove  formulif 
tnd  ^l  where  /-< 


let 

netical  series, 
series  reversed. 

im  of  Ibe  terms. 

tin  of  the  ecrlc?, 
ice, 

r  an  arithmetiiol 
qf  the  sum  of  the 
led  by  the  number 


er  is  12  inches 
;  Low  far  apart 

Bsion  is  4,  the 

[ay,    increasing 
[the    15th  day, 
all    the    first 

3sion  is  190,  the 

of  100  terms 
the   common 

Lie  interest  was 


7.  A  merchant  bought  10  pieces  of  cloth,  giving  10  cents  for 
the  first  and  $12.10  for  the  last,  the  several  prices  forming  an 
arithmetical  series ;  find  the  cost  of  the  cloth. 

8.  What  is  the  sum  of  the  first  1000  numbers  in  their 
natural  order? 

0.  How  many  less  strokes  are  made  daily  by  a  clock  which 
strikes  the  hours  from  1  to  12,  than  by  one  which  strikes 
from  1  to  24  ? 

10.  A  man  set  out  on  a  journey,  going  6  miles  the  first  day, 
increasing  the  distance  4  miles  each  day.  The  last  day  he 
went  50  miles  ;  how  long  and  how  far  did  he  travel  ? 


GEOMETBICAIi    PROGRESSION. 

749.  There  are  Jive  quantities  considered  in  geometrical 
progression,  which  we  denote  by  lette-s  in  the  same  manner 
as  in  arithmetical  progression  ;  thus  : 

1.  A  =  First  Term.  3.  X  =  Last  Term. 

B.  Jt  =  Conatant  Ratio.  4.  A^  =  Number  of  Terms. 

6.  A  =  the  Sam  <    ull  the  terms. 

750.  Any  t?iree  of  these  quantities  being  given,  the  ot7ier 
two  may  be  found.     This  may  be  shown  thus  : 

Taking  3  as  the  first  term  and  2  as  the  constant  ratio  or  maltiplier,  the 
series  may  be  written  in  three  forms ;  thus  : 


1st  Term. 
(1.)          3 

2d  Term. 
6 

8x2 

8x3 

Sd  Term. 
12 

lUh  Term. 
84 

6f.h  Term. 
48 

(8.)          8 
(3.)          8 

3x(2x2) 
8x8* 

8x(3x3x2) 

8xa» 

3x(2;<2x2xS> 
8x2* 

Observe,  in  (8),  each  term  is  composed  of  the  first  term,  3,  multiplied 
by  the  constant  multiplier  2,  raised  to  the  power  indicated  by  the 
number  of  the  term  less  1.  Thus,  for  example,  the  seventh  term  would 
be3x  2'-'  =3  X  2«  =  192. 


m 


f.ff 


stmm 


m 


Is 


Mr 


.'■s:   ■■■!  I,     :-:i 


I  v  » 

■'  i 


324 


BUSIiVIJSS    ARITHMETIC, 


Hence,  from  the  manner  in  which  each  tenn  is  composed,  we  have 
the  followiug  formulae  or  rules : 

r     T?ie  first  term  is  equal  to  the  last  term^  di- 
\.  A  =  ^^n_i.  Read,  -I  vided  by  the  constant  rnultijTlier  raised  to  the 

1 2^ower  indicated  l/y  the  nunUfer  of  terms  less  1. 

r     The  last  term  is  equal  to  the  first  term,  muUi- 
%.  L  =  A  yi  K°-''    Read,  •{  i^ied  try  the  constant  multiplier  raised  to  the 

I  power  indicated  try  the  number  of  terins  Cess  1. 


n— I  /"T 

■  ij 


8.  «=  V:i 


4.  JM-I  =  ^. 


Read, 


Read, 


The  constant  multiplier  is  equal  to  the  root, 
whose  index  is  indicated  by  the  number  of 
terms  less  one,  of  the  quotient  of  the  last  term 
divided  by  the  first. 

The  number  of  terms  less  ont.  is  equal  to  the 
exponent  of  tfie  power  to  which  the  common 
multiplier  must  6e  reused  to  be  equal  to  the 
quotient  of  the  last  term  divided  by  tlie  first. 


751.  To  show  how  to  find  the  sum  of  a  geometrical  series, 
we  take  a  series  whose  common  multiplier  is  known  ;  thus  : 

^^  =  5  +  15  +  45  +  185  +  405. 


'i    I 
■.  t 


Multiplying  each  term  in  this  series  by  3,  the  common  multiplier,  we 
will  have  3  times  the  sum. 

(1.)  ;S'x3  =  5x3  +  15x3  +  45x3  +  135x3  +  405x3,  or 
(2.)'Sx3  =  lo     +45      +135     +405      +405x3. 


Subtracting  the  sum  of  the  series  from  this  result  as  expressed  in  (2), 
we  liave, 

/Sx3=       15  +  45  +  135  +  405  +  405x3 
S =  5  +  15  +  45  +  135  +  405 

^^x2  =  405x3-5 

Now,  observe,  in  this  remainder  -S  x  2  is  S  x  (ft  -  i),  and  406  x  3  is 
Jj  X  R,  and  5  i&  A.  Ilence,  S  x  {H  —  i)  =  L  x  R  —  A.  And  since 
M  —  \  times  the  Sum  is  equal  to  X  >  iJ  -  ^,  we  have. 


«  = 


Tj  X  R^  A 

R-1      ' 


Read, 


The  sum  of  a  geometrical  series  is  eqtial  to 
the  difference  between  the  last  term  mvlfiplied 
by  the  ratio  and  the  first  term,  diridfd  by  the 
L  ratio  minus  1. 


►sed,  we  have 

last  term^  di- 
raised  to  the 
)f  terms  lt«s  1. 

St  term,  mtUti- 
r  raised  to  the 
of  terms  less  1. 

Ml  to  the  root, 
fhe  nunvber  of 
f  the  last  term 

V  is  eqvcU  to  the 
ch  the  common 
be  eqval  to  the 
i  by  tJie  first. 

etrical  series, 
wn ;  thus : 


1  multiplier,  we 

)5  X  3,  or 
)5  X  3. 

|;xpres8ed  In  (2), 

|3 


and  406  X  3  is 
\A.    And  since 


is  equal  to 
\term  multiplied 
».  diridtd  by  tlu 


PROGRESSION, 


EXAMPLES    FOR    PRACTICE. 


325 


752.  1.  The  first  term  of  a  geometrical  progression  is  1, 
and  the  ratio  2 ;  what  is  the  12th  term  ? 

2.  The  first  term  of  a  geometrical  progression  is  3,  the 
ratio  4  ;  what  is  the  8th  term  ? 

3.  Tlie  extremes  of  a  geometrical  progression  are  2  and  1458, 
and  the  ratio  3  ;  what  is  the  sum  of  all  the  terms  ? 

4.  The  extremes  are  4  and  2916,  and  the  ratio  3  ;  what  is 
the  number  of  terms  ? 

5.  A  man,  coming  from  Winnipeg  to  the  Province  of  Ontario, 
travelled  0  days ;  the  first  day  he  went  5  miles,  and  doubled 
the  distance  each  day ;  his  last  day's  ride  was  160  miles  ;  how 
far  did  he  travel  ? 

6.  The  first  term  is  3,  the  seventeenth  196608  ;  what  is  the 
sum  of  all  the  terms  ? 

7.  The  first  term  of  a  geometrical  progression  is  4,  the  7th 
term  is  2916 ;  what  is  the  ratio  and  the  sum  of  the  series  ? 

8.  Supposing  an  engine,  on  the  Intercolonial  Railway  be- 
tween Quebec  and  Halifax,  should  start  at  a  speed  of  3  miles 
an  hour,  and  the  speed  could  be  doubled  each  hour  until  it 
equalled  96  miles,  how  far  would  it  have  moved  in  all,  and 
how  many  hours  would  it  be  in  motion  7 

ANNUITIES. 

75;j.  An  Annuity  is  a  fixed  sum  of  money,  payable 
annually,  or  at  the  end  of  any  equal  periods  of  time. 

754.  The  Atnount  or  Final  Value  of  annuity  is  the 
sum  of  all  the  payments,  each  payment  being  increased  by  its 
interest  from  the  time  it  is  due  until  the  annuity  ceases. 

755.  The  Present  Worth  of  an  annuity  is  such  a  sum 
of  money  as  will  amount,  at  the  given  rate  per  cent,  in  the 
given  time,  to  the  Amount  or  Final  Value  of  the  annuity. 


1 

II 


IS,  ■■> 


iO 


iT 


L. 


326 


B  USIXESS    ARITHMETIC. 


756.  An  Annuity  at  Simple  Interest  forms  an  arith- 
metical progressUm  whose  common  difference  is  the  interest  on 
the  given  annuity  for  one  interval  of  time. 

Thns  an  annaity  of  $400  for  4  years,  at  7^  eimple  interest,  gives  the  fol- 
lowing progresBion : 


f;  1 

r 


'■{ 


m 


:  i  ■  ■  - 


let  Term,    id  Term, 
a.)       $400       $400 +($38) 
(3.)       $400  $428 


Sd  Term. 
$400 +($28 +  128) 
$466 


lUh  Term. 

$400 -r  ($28 +$28 +128),  or 


Observe,  there  le  no  interest  on  the  laet  payment ;  hence  it  forms  the 
M  Term.  The  payment  before  the  last  bears  one  year's  interest,  hence 
forms  the  Sd  Term ;  and  so  on  with  the  other  terms. 

Hence  all  problems  in  annuities  at  simple  interest  are  solved  by  arith' 
meticcU  progression. 

757.  An  Annuity  at  Compound  Interest  forms  a 
geometrical  progression  whose  common  multiplier  is  represented 
by  the  amount  oi$t  for  one  interval  of  time. 

Thus  an  annuity  of  $300  for  4  years,  at  Z%  compound  interest,  gives  the 
following  progression : 


Ist  Term. 
$900 


id  Term. 
$900x1.06 


Sd  Term. 
$900x1.06x1.06 


kth  Term. 
$900x1.06x1.06x1.06 


Observe  carefully  the  following : 

(1.)  The  last  payment  bears  no  interest,  and  hence  forms  the  1st  Term  of 
the  progression. 

(2.)  The  payment  before  the  last,  when  not  paid  until  the  annuity 
ceases,  bears  interest  for  one  year;  hence  its  amount  is  $800x1.06  and 
forms  the  id  Term. 

(3.)  The  second  payment  before  the  last,  bears  interest  when  the 
annuity  ceases,  for  two  years ;  hence  its  amount  at  compound  interest  ie 
♦900  X  1.06,  the  amount  for  one  year,  multiplied  by  1.06,  equal  $300  x  1.06  x 
1.06,  and  forms  the  Sd  Term^  and  so  on  with  other  terms. 

Hence  all  problems  in  annuities  at  compound  interest  are  solved  by 
geometrical  progression. 


r  RO  G  R  Essioy, 


327 


l! 


ms  an  nritli- 
B  interest  on 


:,  gives  the  fol- 

h  Term. 

28 +$28 +128),  or 
$484. 

;nce  it  forms  the 
s  interest,  hence 

solved  byarf/A- 


^rest  forms  a 

r  is  represented 


aterest,  gives  the 

\kth  Term. 
1.06x1.06x1.06 


18  the  1st  Tenn  of 

intil  the  annnlty 
lis  $300 X  1.06  and 

llerest  when  the 
jpoond  interest  is 
pqual  $300  X  1.06  X 

Bt  are  eolved  by 


EXAMPLES    FOR    PRACTICE. 

758.  1.  A  father  deposits  $150  annually  for  the  benefit  of 
his  son,  beginning  with  his  12fh  birthday ;  what  will  be  the 
amount  of  the  annuity  on  his  2l8t  birthday,  allowing  simple 
interest  at  6%  ? 

2.  What  is  the  amount  of  an  annuity  of  $200  for  6  years  at 
7fc  simple  interest  ? 

3.  What  is  the  amount  of  an  annuity  of  $400  for  4  years  at 
7%,  comi)ound  interest  ? 

4.  What  is  the  present  worth  of  an  annuity  of  $000  for 

5  years  at  8^ ,  simple  interest  ? 

5.  What  is  the  present  worth  of  an  annuity  of  $700  at  B%, 
simple  interest,  for  10  years? 

6.  What  is  the  present  worth  of  an  annuity  of  $100  for 

6  years  at  6%,  compound  interest  ? 

7.  What  is  the  present  worth  of  an  annuity  of  $350  for 
9  years  at  6%,  compound  interest? 

8.  What  is  the  amount  of  an  annuity  of  $600  &t  I'/c,  com- 
pound interest,  for  12  years  ? 

9.  At  what  rate  %  will  $100  amount  to  $119.1016  in  3  years, 
at  compound  interest  ? 

This  example  and  the  fonr  following  should  be  solved  by  applying  the 
formulffi  for  geometrical  progression  on  page  394. 

10.  At  what  rate  %  will  $1000  amount  to  $1500.73  in  6  years, 
compound  interest  ? 

11.  What  sum  at  compound  interest  8  years,  at  7%,  will 
amount  to  $4295.465  ? 

13.  The  amount  of  a  certain  sum  of  money  for  12  years,  at 
7%  compound  interest,  was  $1126.096;  what  was  the  original 
Hum? 

13.  In  how  many  years  will  $20  smount  to  $23.82032,  at  6% 
compound  interest  ? 


'.!"■   'J 


Urn 


m 


f  1 


m 


IM 


MENSURATION 


k*2     f 


' 


'I 

»  "^   i 


GENERAL    DEFINITIONS. 

759*  A  Line  ie  that  which  has  only  IcDgth. 

7 GO.  A  Straight  Line  is  a  Une  which  has  the  same  direction  at 
every  point. 

701.  A  Curved  Utie  la  a  line  which  changes  its  direction  at 
every  point. 

76iS.  Parallel  Line*  are  lines  which  have  the  same  direction. 

703.  An  Angle  is  tlic  opening  between  two  lines  which  meet  in  a 
common  point,  called  the  veitex. 


Angles  are  of  three  Itinde 

, thas : 

(1)                             (2) 

(8) 

(4) 

1 

5 

1                  c 

m    J9                     1 

Honiz 

ONTAL..       "          A 

c 

r — c 

7'wo  Right  Angles.      One  Right  Angle.      Obtuse  Angle.    Acute  Angle. 

7G4:.  When  a  line  meets  another  line,  making,  as  shown  in  (1),  two 
equal  angles,  each  angle  is  a  Right  Angle f  and  the  lines  are  said  to  bo 
perpendicular  to  each  other. 

7GJ^.  An  Obtuae  Angle,  as  shown  in  (3),  is  greater  than  a  right 
angle,  and  an  Acute  Angle^  bl9  shown  In  (4),  is  less  than  a  right  angle. 

Angles  are  read  by  using  letters,  the  letter  at  the  vertex  being  always 
read  in  the  middle.    Thus,  in  (8),  we  read,  the  angle  BAG  or  CAB. 

7C$0.  A  Plane  is  a  surface  such  that  if  any  two  points  in  it  be  joined 
by  a  straight  line,  every  point  of  that  line  will  be  in  the  surface. 


MENSURATIOX. 


329 


me  direction  at 

tB  direction  at 

;  direction. 
wWch  meet  in  a 


(4) 


Acute  Angle. 

lown  In  (1),  two 
are  eaid  to  be 


sr  than  a  right 
right  angle. 

being  always 
CAB. 

In  It  be  joined 
bface. 


7G7«  A   J'lnne  Tiynve  \a  a  plane  bounded  cither  by  straight  or 
curved  lines,  or  by  Oi-w  curved  line. 

70S.  A  Polygon  \»  a  plane  figure  bounded  by  straight  linen.    It  is 
named  by  the  number  of  sides  in  its  boundary ;  thus  : 


Trigon. 


TBtragon, 


Pentagon.       BeKogon,  and  m>  on. 


Observe,  that  a  regular  polygon  Is  one  that  has  all  Its  sides  and  all  its 
angles  equal,  and  that  the  Baae  of  a  polygon  la  the  aide  on  which  it  stands. 

760.  A  Trigon  is  a  ^Are^-sided  polygon.     It  is  usually  called  a 
TYiangle  on  account  of  having  three  angles. 

Triangles  are  of  three  kinds,  thus : 


^        D       c  Ac 

Sight-ana^  Triangle.   Acute-angled  TYiangle.    Obtuse-angled  Trian^ 

Observe^  a  right-angled  triangle  has  okx  right  angle,  an  acntc-angled 
triangle  has  thbeb  acute  angles,  and  au  obtuse-angled  triangle  has  om 
obtuse  angle. 

Observe,  also,  as  ehovm  in  (9)  and  (3),  that  the  Altitude  of  a  triangle  is 
the  perpendicular  distance  from  one  of  its  angles  to  the  side  opposite. 

770.  An  £!quilateral  Triangle  is  a  triangle  whose  three  sides 
are  equal 

771*  An  l80»eele»  Triangle  has  two  of  Us  sides  eqaaL 

772.  A  Scalene  Triangle  luu  all  of  its  sides  unequal. 

773.  A  Tetragon  is  a  four-sided  polygon.  It  is  usually  called  a 
Quadrilateral.  22 


m 


■An 


M'  '1 


ir: 


ill 


^m 


-yrnmavKxm 


330 


D[/SI.yESS    ARITHMETIC, 


Qoadrilaterals  are  of  three  kinds,  thus ; 


*"■ 


Yh>  ) 


ParaUelogram. 


Trapezoid. 


Trapezium. 


f>  S 


1. 


^/  :  \ 


si         <  ' 


'.I 


Observe,  that  a  Parallelogram  has  its  opposite  sides  parallel,  that  a 
Trapezoid  has  only  two  sides  parallel,  and  that  a  Trapezium  has  no  sides 
parallel. 

Observe,  also,  that  the  Diagonal  of  a  qoadrilateral,  as  shown  in  (1),  (3) 
and  (3),  is  a  line  joining  any  two  opposite  angles. 

774.  A  Parallelogram  is  a  quadrilateral  which  has  its  opposite 
Bides  parallel.    Parallelograms  are  of  four  kinds,  thus : 


(I)  (2)  (3) 

ffi      .  J*         »|  -|C  Bm 


Square. 


Rectangle. 


0     A 


— -F     A 


R/iomboid. 


Rhombns. 


Observe,  that  a  Square  has  all  its  sides  equal  and  all  its  angles  right 
angles,  that  a  Rectangle  has  its  opposite  sides  equal  and  all  its  angles  right 
angles,  that  a  Rhomboid  has  its  opposite  ^c^  equal  and  its  angles  a<n//«  and 
obttute,  and  that  a  Rhombus  has  all  its  sides  equal  and  its  angles  acute  and 
obtuse. 

Observe,  also,  that  the  Altitude  of  a  parallelogram,  as  shown  in  (3)  and 
(4),  is  the  perpendicular  distance  between  two  opposite  sides. 

775.  A  Circle  is  a  plane  bounded  by  a  curved  line,  called  the 
circumference,  every  point  of  which  is  equally  distant  from  a  point  within, 
called  the  centre ;  thus  : 

770.  The  Diameter  of  a  circle  is  any  straight 
line,  as  CD,  passing  through  its  centre  and  terminating 
at  both  ends  in  the  circumference. 

777.  The  Itatllus  of  a  circle  is  any  straight  line, 
as  AB,  extending  from  the  centre  to  the  circumference. 


Trapezium. 

parallel,  that  a 
im  bas  no  sides 

hown  in  (.1),  (2) 
has  Its  opposite 


'0 


I  its  angles  right 
I  its  angles  right 
angles  acute  and 
angles  acvi£  and 


;8 


Bhown  in  (8)  and 
des. 

line,  called  the 
tm  a  point  within. 


|e  is  any  straight 
and  terminathig 


I  any  straight  line, 


M  EX  SURA  TION. 


331 


778,  The  Pertmetpr  of  a  polygon  is  the  snm  of  aU  the  lines  which 
form  its  bonndary,  and  of  a  circle  the  circumference. 

770«  The  ^rea  of  any  plane  figure  is  the  surfoce  contained  within 
its  boundaries  or  boondary. 

780,  Mensuration  treats  of  the  method  of  finding  the  lengths  of 
lines,  the  area  of  BarfJacet*,  and  volames  of  HoUds. 


SOLUTION    OP    PBOBLEMS. 

781*  The  eolntiont*  of  problems  in  mensaration  cannot  be  demon- 
strated  except  by  geometry,  but  the  general  principle  which  underlies 
these  ifolntiouij  may  be  stated ;  thus. 

The  contents  of  any  given  mrfaee  or  solid  thai  can  be  measured  can  be 
shown  to  be  equal  to  the  contents  of  a  rectangular  sitrface  or  solid,  whose 
dimensions  are  equal  to  certain  kkown  dimensions  qf  the  given  surface  or 
solidt  thus : 


8  units  long. 


•§ 


1.  Observe,  that  the  nnmber  of  small  squares 
in  (1)  is  eqnal  to  the  prodnct  of  the  numbers 
denoting  the  length  and  breadth.  Thus,  8x5  = 
40  small  f^aares. 

2.  Otfserve,  in  (2),  that  the  plane  bounded  by 
the  lines  FB,  BC,  CE,  and  EF,  is  rectangular 
and  equal  to  the  given  parallelogram  A  BCD. 
bccau-'^  we  have  added  to  the  right  as  much 
surface  as  we  have  taken  oflfat  the  left. 

Hence  the  contents  of  the  parallelogram 
ABCD  Is  found  by  taking  the  product  of  the 
nnmber  of  units  in  the  altitude  CE  or  BP,  and 
in  the  side  BC. 

8.  Observe,  again,  the  diagonal  BD  divides  the  parallelogram  Into  two 
equal  triangles,  and  hence  the  aroa  of  the  triangle  ABD  i.s  one-half  the  area 
of  the  parallelogram,  and  1:*  therefore  found  by  taking  one-half  of  the 
product  of  the  number  of  units  in  the  base  AD  and  in  the  altitude 
BF  or  CE. 

In  view  of  the  flict  tliat  the  aolutions  in  mensuration  depend  upon 
geometry,  and  that  the  pupil  requires  a  considerable  knowledge  of  that 
subject  in  order  to  understand  them,  no  explanations  are  given.  The  rule, 
in  each  cai<e,  must  be  strictly  followed. 


*\ 


'i  ] 
,1 


'II 


n 


n 


332 


BUSINESS     ARITHMETIC. 


\ 

- 

v 

^ 

'*  \ 


FBOBLEMS    ON    TRIANGLES. 

7S12.  Prob.  I.— When  the  base  and  altitude  qf  a  triangk  are  given^ 
to  find  the  area :  Divide  the  product  of  the  babb  and  altitude  by  i. 

Fiud  the  area  of  a  triauglo 

1.  Whose  base  is  8  rd.  and  altitude  2  rd.  7  ft. 
3.  Whose  base  Is  14  ft.  and  altitude  7  tt.  8  in. 

3.  What  is  the  area  of  a  triangular  park  whose  base  is  16.76  chains  and 
altitude  13.4chaiQe? 

4.  Whose  base  is  31  chains  and  altitude  16  chains. 

5.  How  many  stones,  each  2  11.  G  in.  by  1  ft.  9  in.  will  be  used  in  paving 
a  triangular  court  whose  base  is  160  feet  and  altitude  136  feet,  and  what 
will  bo  the  expense  at  $.35  a  square  yard  ? 

6.  How  many  square  feet  of  lumber  will  be  required  to  board  up  the 
gable-ends  of  a  house  30  feet  wide,  having  the  ridge  of  the  roof  17  feet 
higher  than  the  foot  of  the  rafters  ? 

783.  Pbob.  11.— When  the  area  and  one  dimension  are  given,  to 
find  the  other  dinunHon:  Double  the  area  and  divide  by  the  given 
dimension. 

Find  the  altitude  of  a  triangle 

1.  Whose  area  ia  364  square  rods  and  base  24  rods. 

2.  Whose  area  is  75  square  feet  and  base  15  feet. 
8.  Whose  base  is  6  ft.  1  in.  and  area  50  sq.  ft.  100  sq.  in. 

Find  the  base  of  a  triangle 

4.  Whose  altitude  is  3  yd.  2  ft.  and  area  8  sq.  3rd. 

5.  Whose  area  is  3  A.  108  P.  and  altitude  28  rd. 

6.  Whose  area  ia  2  sq.  rd.  19  sq.  yd.  2  sq.  ft.  36  sq.  in.  and  altitude  1  rd 
1  ft.  6  in. 

7.  For  the  gable  of  a  church  76  feet  wide  it  required  250  stones,  each  2 1 
long  and  1  ft.  6  in.  wide ;  what  is  the  perpendicular  distance  from  the  rid^ 
of  the  roof  to  the  foot  of  the  rafters  ? 

784.  Prob.  III.— W^^n  the  three  sides  <tf  a  triangle  are  given, 
find  the  area :  From  haif  the  sum  qf  the  three  sides  subtract  each  tid 
separately.    Multiply  the  half  sum  and  the  three  remainders  together ;  tl\ 
square  root  of  the  product  is  the  area. 

1.  What  is  the  area  of  an  isosceles  triangle  whose  base  is  50  in. 
each  of  its  equal  sides  85  inches  f 

2.  Fiud  the  area  of  a  triangle  whose  sides  arc  15, 30, 35  feet 


c. 


ME^S  UR  A  TI  0  N, 


333 


triangU  are  givetiy 


l8  16.76  chainB  and 


8.  How  many  acres  in  a  triangular  field  whose  fides  meacarc  16,  20, 
30  rode y 

4.  What  is  the  area  of  an  equilateral  triangle  whose  sides  each  measure 
40  foot  ? 

5.  A  piece  of  land  in  the  form  of  an  equilateral  triangle  requires  156  rods 
of  fence  to  enclose  it ;  how  many  acres  are  there,  and  what  ia  the  cost  at 
|40  per  acre  ? 

6.  How  many  square  yards  of  plastering  contained  in  a  triangle  whose 
sides  are  30, 40, 50 ;  and  what  is  the  cost  of  it  at  75  cents  per  sq.  yd.  ? 

Ans.  66;  yds.    Ck>8t  |fiO. 


villbeused  in  pavlnjf  785*     Prob.  Vf.—When  Vie  base  and  perpendicular  are  given  in  a 

iide  126  feet,  and  what  I  right-angled  triangle,  to  find  the  other  side :  Extract  the  square  root  of  Ihs 

mm  qf  the  squares  of  the  base  and  perpendicular. 
,ulredtoboardupthe 

Ige  of  the  roof  H  feet 


llmension  are  given^jo 
U  divide  by  the  given 


The  reason  of  this  rule  and  the  one  in  Prob.  V  will  be  seen  by  examin- 
ing the  diagram  in  the  margin. 


sq.  lo* 


Observe,  that  the  square  on  the  side  AB  opposite 
the  right  angle,  contains  as  many  small  squaren  as 
the  sum  of  the  small  squares  in  the  squares  on  the 
base  AC  and  the  perpendicular  BC.  This  i:^  ebown 
by  geometry  to  be  true  of  all  right-angled  triai)<4les. 

Hence,  by  extracting  the  square  root  of  the  sum 
of  the  squares  of  the  base  and  perpendicular  of  a 
right-angled  triangle,  we  have  the  length  of  the 
Bide  opposite  the  right  angle. 


•a 


1  *■  I 


w. 


The  side  opposite  the  right  angle  is  called  the  Hypothenuse. 


and  altitude  1  tdl   Find  the  hypothenuse  of  a  right-anjjled  triangle 

[   ^  Mn  stones,  each  2  fl  1.  Whose  base  is  15  ft.  and  perpendicular  36  ft. 
t  distance  from  the  rida  2.  Whose  base  is  40  ft.  and  perpendicular  16  ft. 

8.  A  tree  104  ft.  high  stands  upon  the  bank  of  a  stream  76  feet  wide  ; 
rhat  is  the  distance  of  a  man  upon  the  opposite  baulc  from  a  raven  upon  the 
a  triangle  are  given,tov  of  the  tree  f 

gides  subtract  each  sia  4.  What  is  the  length  of  the  shortest  rope  by  which  a  horse  may  be  tied 
emainders  together ;  tl^  a  post  in  the  middle  of  a  field  20  rods  square,  and  yet  be  allowed  to  graze 

'  on  every  part  of  it  f 
6.  A  and  B  start  ftt)m  one  comer  of  a  field  a  mile  square,  travelliner  at  the 
Ivrhoee  base  la  8**  **•  Hme  rate ;  A  follows  the  fence  around  the  field,  and  B  proceeds  directly 

tross  to  the  opposite  comer ;  when  B  reaches  the  comer,  how  far  will 
Is  90, 35  feet.  |  be  trom  A  ? 


li 


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ik 


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tij 


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334 


BUSINESS    ARITHMETIC, 


780.  Proij.  V,— ]rA«n  /A«  6a««  or  perj)endicular  is  to  de  found; 
Extract  the  tquure  root  qfthe  difference  Itetiveen  the  square  qf  the  hypothe- 
nme  and  the  square  of  the  given  aide. 

Flud  tbo  bane  of  a  rigbt-unglcd  triangle 

1.  WhoHe  perpendicular  is  20  feet  and  hypothenase  45  feet. 

2.  Whot>e  hypothenuHe  is  40  feet  and  perpendicular  15  feet. 

3.  Bunker  Hill  monument  is  230  feet  lii^b  ;  a  man  360  feet  from  the  base 
Bhot  a  bird  hovering  above  the  top ;  the  man  wan  423  feet  from  the  bird ; 
how  far  was  the  bird  Arom  the  top  of  the  monument  f 

4.  The  lower  endu  of  two  opposite  rafters  are  48  feet  apart  and  the 
length  of  each  rafter  is  80  feet ;  what  is  the  elevation  of  the  ridge  above 
the  eaves  ? 

5.  A  ladder  36  feet  long  reaches  from  the  middle  of  the  street  to  a  win- 
dow 28  feet  high  ;  how  wide  is  the  street  f 


FBOBLEMS    ON    QUADBILATERALS. 

787.  PnoB.  \l.—To  find  tlie  area  qf  a  parallelogram :  Multiply  the 
base  by  the  altitude. 

1.  now  many  acres  in  a  piece  of  land  in  the  form  of  a  parallelogram, 
whose  base  is  9.86  ch.  and  altitude  7.5  ch.  ? 

2.  Find  the  area  of  a  parallelogram  whose  base  is  3  it.  9  in.  and  altitude 
7  ft.  8  in.  ;  whose  altitude  is  2  yd.  5  in.  and  base  3  yd.  6  in. 

3.  The  base  of  a  rhombus  is  9  A.  8  in.  and  its  altitude  3  ft. ;  how  many 
square  feet  in  its  surface  ? 

4.  How  many  square  feet  in  the  roof  of  a  building  86  ft.  long,  and  whose 
rafters  are  each  16  ft.  6  in.  long  ? 

788.  Prob.  vn — To  find  the  area  qf  a  trapezoid:  Multiply  one^ 
half  of  the  sum  qf  the  parallel  sides  by  the  altitude. 

Find  the  area  of  a  trapezoid 

1.  Whose  parallel  sides  are  8  and  11  inches  and  altitude  6  inches. 

2.  Whose  parallel  sides  are  15  and  25  feet  and  altitude  11  feet. 

3.  One  side  of  a  field  measures  47  rods,  the  side  opposite  and  parallel  to 
it  measures  39  rods,  and  the  distance  between  the  two  sides  is  15  rods ; 
how  much  is  it  worth  at  $40  per  acre  f 

4.  How  many  square  feet  in  a  board  1  ft.  4  in.  wide,  one  side  of  which  i» 
82  ft.  long  and  the  other  side  34  feet  long  f 


to  be  found  ; 
)f  the  hypothe- 


t. 

St. 

it  from  the  base 
from  the  bird; 

t  apart  and  the 
the  ridge  above 

street  to  a  win- 


BALS. 

\m:  Multiply  the 

a  parallelogram, 

in.  and  altitude 

3  ft. ;  how  many 

long,  and  whose 

; ;  Multiply  one^ 


inches, 
feet. 

and  parallel  to 
^es  is  15 rods; 

Bide  of  which  i» 


M  Ey  SUR  A  TIO  N, 


335 


789.  Pbob.  WU..— To  find  the  area  of  a  trapezium :  Multiply  the 
diagoml  by  half  the  turn  of  the  perpe^ndicukvra  to  U  from  the  opposite 
angles. 

Refer  to  diagram  (3>  in  (773)  and  And  the  area  of  a  trapesiom 

1.  Whose  diagonal  ii*  IG  feet  and  perpendiculars  to  this  diagonal  7  feet 
and  6  feet. 

2.  Whose  diagonal  is  45  in.  and  perpendiculars  to  this  diagonal  11  inches 
and  9  inches. 

S.  How  many  acres  in  a  field  in  the  form  of  a  trapezium  whose 
diagonal  is  \  mi.  and  the  perpendiculars  to  this  diagonal  5  ch.  and 
6  ch.? 

4.  Whose  diagonal  is  37  ft.  6  in.  and  perpendiculars  to  this  diagonal  7  ft. 
4  in.  and  8  ft.  8  in. 


790.  Pbob.  VS..— To  find  the  diameter  of  a  circle:  Divide  the  cir- 

eumference  by  S.1U16. 

To  find  the  circumference :  Multiply  the  diameter  by  S.1U16. 

1.  Find  the  circumference  of  a  circle  whose  diameter  is  14  inches; 
whose  radius  is  9  inches. 

2.  Find  the  diameter  of  a  circle  whose  circumference  is  94.248  Inches ; 
whose  ciicumference  is  78-54  feet. 

3.  Ilow  many  miles  does  the  earth  pass  over  in  its  revolution  around 
the  sun,  its  distance  flrom  the  sun  being  95,000,000  miles  ? 

4.  What  will  it  cost  to  fence  a  circular  park  8  rods  in  diameter,  at 
$4.80  per  rod? 

791.  Pbob.  X.— 7b  JInd  the  area  of  a  circle:  Multiply  \  of  it* 
diameter  by  the  drcumference ;  or^  Multiply  the  square  qf  its  diameter 
by  .785k. 

1.  Wliat  is  the  area  of  the  largest  circular  plot  that  can  be  cut  from  a 
field  135  foet  square  ?  How  much  must  be  cut  ofi"  at  the  comers  in  making 
this  plot  ?  How  much  less  will  it  cost  to  fence  this  than  the  square,  at 
$2.50  a  rod  ? 

2.  What  is  the  area  of  a  circle  whose  diameter  in  20  feet?  Whoee 
diameter  is  42  inches  ?    Whose  circumference  is  157.08  feet  ? 

3.  The  distance  around  a  circular  park  is  1 2  miles.  How  many  acres 
does  it  contain  ? 

4.  How  many  square  yards  arc  contained  in  a  circle,  who!«e  diameter 
is  Si  feet?  Ans.  1.069. 


^1 


'SI' 


PM 


'•/',;  i 


tK'  I 


336 


BUSINESS     ARITHMETIC, 


J    /' 


7012.  Pbob.  XI.— To  find  the  diameter  when  the  area  qf  a  circle 
is  given:  Extract  the  tgtuire  root  qf  the  quotient  qf  the  area  divided 
by  .786k. 

Observe,  that  nvben  the  diameter  Is  found,  the  drcnmflDrcnce  can  be 
found  by  multiplyint;  the  diameter  by  8.141(i  (700). 

1.  What  is  the  circumference  of  a  circle  whose  area  is  103.9884 
square  feet? 

2.  What  is  the  diameter  of  a  circle  whose  area  is  50.S666  sq.  ft.  t 

8.  The  area  of  a  circular  lot  is  19.635  square  rods;  what  la  its 
diameter  ? 

4.  The  area  of  a  circle  is  118.0976  sq.  in. ;  what  is  its  circumference  t 

6.  What  is  the  radius  of  a  circle  whose  area  is  804.349U  sq,  in.  ? 

6.  How  many  rods  of  fence  will  be  required  to  enclose  a  circle  whofle 
area  is  2Mfi  square  rods  7 


•'.^  J 


*-r 


FBOBLEMS    ON    SOLIDS    OB    VOLUMES. 

703.  A  Solid  or  Volume  has  three  dimensions:  length,  breadth, 
and  thickness. 

The  boundaries  of  a  solid  are  planes.    They  are  called /(UM,  and  their 
intereectiouB  edgee. 

704.  A  Prism  is  a  solid  or  volume  having  two  of  its  faces  equal  and 
parallel  polygons,  and  its  other  faces  parallelograms. 

Obeerve,  a  prism  is  named  by  the  number  of  sides  in  Its  equal  and 
parallel  foces  or  bases  ;  thus  : 


(1) 


Triangular 
I^ism. 


Quadrangular 
Prism. 


Pent     m- 


Observe,  a  Prism  whose  parallel  faces  or  bases  are  p.   ^Uelogrf    a,  as 
shown  in  (3),  is  called  a  Parol letopipedon. 

Observe,  also,  that  the  Altittide  of  a  prism  is  the  perpendicular  aietance 
between  ita  bases. 


MENSURATIOX. 


337 


"■ea  qf  a  circle 
area  divided 


ierence  can  be 

3a  la   163.9884 

1.  ft.f 

;    what  la  ita 

inference  f 
in.r 
i  circle  whose 


jUMES. 

ngth,  breadth, 

lOM,  and  their 
icee  equal  and 


elogr    id,  aa 
alar  Uietance 


795.  A  CyHnder,  as  ehown  In  (1),  is  a  round  solid  or  volume  having 
Uoo  equal  and  parallel  circles  as  its  ba»e$. 

1.  Obeerve,  that  the  altitude  of  a  cylin- 
der is  the  perpendicular  distance  between 
the  two  circles  forming  Its  b<u«a. 

8.  ObMTve,  also,  that  a  cylinder  is  con> 
coived  to  be  (generated  by  revolving  a 
rectangle  about  one  of  its  sides. 

TOO.  A  Sphere,  au  Mhown  in  (3),  ia  a  solid  or  volume  bounded  by  a 
curved  snrfoce,  such  that  all  points  in  it  are  equally  distafit  fh>m  a  point 
within,  called  the  centre. 

797.  The  Diameter  of  a  sphere  la  a  line,  as  CD  in  (S),  passing 
through  its  centre  and  terminating  at  both  ends  in  the  surface. 

798*  The  Radius  of  a  sphere  is  a  Use  drawn  f^om  the  centre  to  any 
point  in  the  surflice. 

799.  A  Pyratnid,  as  shown  in  (1),  is  a  solid  or  volume  having  as 
its  base  any  polygon,  and  as  its  other/aces  triangles,  which  meet  in  a  com* 
mon  point  called  the  vertex. 


Pyramid. 


Frustum. 


Cone. 


Frustum, 


HOO.  A  Cone^  as  shown  in  (8),  is  a  solid  or  volume  whose  base  is  a 
circle  and  whose  convex  surface  tapers  uniformly  to  a  point,  called  the 
vertex. 

1.  Obfterve,  that  the  Attitude  of  a  pyramid  or  cone  is  the  perpendicular 
distance  between  the  vertex  and  the  bnse. 

2.  Ob/'crtr,  also,  that  the  Slant  Height  of  a  pyramid  is  the  perpen- 
diculjir  (lintance  between  the  vertex  and  one  of  the  sides  of  theiutse; 
a  1  of  a  cone  the  distance  between  the  vertex  and  the  circumference  of 
tJ     base. 

15 


"  ..'I 


m 


_    illL 
Jill 


% 


:  V 


338 


BUSINESS     ARITHMETIC. 


-    <^ 


801.  A  frufiiwtn  of  a  pyramid  or  cone,  as  phown  in  (2)  and  (4), 
is  the  part  which  remains  after  cuttiug  off  the  top  by  a  plane  parallel  to^ 
the  base. 

80t2.     Prob.  XII.—  To  find  the  convex  surface  qf  a  priinn  or  cylinder  : 
Multiply  the  perim etkb  of  the  base  by  the  altitude. 
To  find  tht  ENTIRE  auRFACH  add  the  area  qf  the  bases. 

The  reasion  of  this  rule  may  be  shown  thus  : 

(1)  (2)  (3) 

G  F  e 


Observe,  that  if  the  three  faces  of  the  prism  in  (1)  are  marked  out  side  by 
side,  as  shown  in  (2),  we  have  a  rectangle  which  is  equal  to  the  convex 
surface  in  the  prism. 

Observe,  also,  that  the  surfocc  of  the  cylinder  in  (8)  may  be  conceived  as 
spread  out,  as  shown  in  (2) ;  hence  the  reason  of  the  rule. 

Find  the  area  of  the  convex  surface 

1.  Of  a  cylinder  whose  altitude  is  4  ft.  9  in.  and  the  circumference  of  its 
base  7  ft.  8  in. 

2.  Of  a  prism  whose  altitude  is  8  feet,  and  its  base  a  triangle,  the  sides 
of  whose  base  measures  4  ft.,  3  ft.,  3  ft.  6  in. 

3.  Of  a  prism  whose  altitude  is  9  inches,  and  its  base  a  hexagon,  each 
sido  of  which  is  2}  inches. 

4.  Find  the  entire  surface  of  a  cylinder  9  ft.  high,  the  diameter  of  whose 
base  is  8  ft. 

5.  Find  the  entire  surfiiicc  of  a  parallelopipedon  9  ft  long,  5  (1. 6  in.  wide, 
and  3  ft  high. 

803.  Prob.  'SilW.—  To  find  the  volume  qf  any  prism  or  cylinder: 
Multiply  the  area  of  the  base  by  the  altttude. 

1.  Wtiat  >  the  volume  of  a  triangular  prism  whose  altitude  is  28  ft.,  and 
the  sides  of  its  bnsc  6  ft.,  7  ft,  5  ft.  respectively. 

%.  Find  the  volume  of  a  triangular  prism  wuose  altitude  is  15  ft.  and  the 
tides  of  the  base  each  4  ft. 


in  (2)  and  (4), 
ane  parallel  to 


nn  or  cylinder : 


(3) 


bed  out  side  by 
to  the  convex 


c  conceiTcd  as 


iference  of  its 
gle,  the  side? 
lexagon,  each 
eter  of  whose 
ft.  6  in.  wide, 

or  cylinder  : 

i(<  28  n.,  and 
5  ft.  and  the 


M  E  S  SU  RATION, 


339 


3.  What  is  the  volume  of  a  parallelopipedon  15  ft.  long,  12  ft.  high, 
10  ft.  wide  ? 

4.  Find  the  contents  of  a  cylinder  whose  altitude  is  19  ft.  and  the  diame- 
ter of  its  base  4  ft. 

5.  A  log  is  90  ft.  long  and  its  diameter  is  16  in. ;  how  many  cubic  feet 
does  it  contain  t 

i\.  What  k  the  value  of  a  piece  of  timber  15  in.  square  and  50  feet  long, 
at  40  cents  a  cubic  foot  ? 

804- .     Prob.  XrV.—  To  find  the  convex  mrface  of  a  jryramid  or  cone  : 
MiUti[dy  the  perimeter  of  the  base  by  one-haif  the  slant  height. 

To  find  the  entire  surface,  add  the  area  of  the  base. 

Find  the  convex  surflace  of  a  cone 

1.  Whose   slant  height   is   15  feet,   and   the  diameter  of  the  base 
10  feet. 

2.  Whose   base   is   19  in.    in  circumference,   and   the  slant  height 
12  inchest. 

Find  the  convex  surface  of  a  pyramid 

3.  Whose  base  is  3  ft.  6  in.  square,  and  the  slant  height  6  ft. 

4.  Whose  slant  height  is  19  ft,  and  the  base  a  triangle  whose  sides  are 
12, 14,  8  ft. 

Find  the  entire  surflace  of  a  pyramid 

6.  Whose  slant  height  is  56  in.,  and  its  baao  a  triangle  each  of  whoso 
sides  is  6  in. 

6.  Whose  slant  height  is  45  feet,  and  the  base  a  rectangle  7  ft.  long  and 
8  ft.  wide. 

Find  the  entire  surfbcc  of  a  cone 

7.  Whose  slant  height  is  42  feet,  and  the  circumference  of  the  base 
31.416  ft. 

8.  Whose  slant    height  is   75  in.,   and  the  diameter  of    the  base 
6  inches. 

80R,     Prob.  XV.—  To  find  the  volume  qf  a  pyramid  or  cone :  Multiply 
the  urea  of  the  btise  by  one-third  the  altitude. 

Find  the  volume  of  a  cone 

1.  Whose  altitude  is  24  feet,  and  the  circnmferencc  of  the  base 
6.2832  feet. 


•! 


11 

t  i" 

II 


r'> 


^40 


nusry^Ess   a  rithmei^i c. 


r'1^1 


.*.' 


8.  Whose  altitude  is  12  it,  and  the  diameter  of  the  baee  4  ft. 

Find  (he  volame  of  a  pyramid 

3.  Whose  altitude  is  15  feet,  and  its  base  4  feet  square. 

4.  Whose   altitude   is   45   ft.,    and   its   base   a   rectangle  15   feet  by 
16  feet. 

5.  Whose  altitude  is  18  inches,  and  the  base  a  triangle  8  inches  on 
each  side. 


m 


•( 


:;/■ 


800.  Prob.  XVI.— 7b  find  the  convex  surface  cf  a  frwitum  of  a 
pyramid  or  cone :  Multiply  the  gum  of  the  i)erimeter.'<  or  circumferences 
by  one  ha^  the  slant  height. 

To  find  the  entire  surface^  add  the  area  of  both  the  bates. 

1.  What  is  the  convex  surface  of  a  fhistum  of  a  cone  whose  slant  height 
is  9  inches,  and  the  circumference  of  the  lower  base  17  inches,  and  of  the 
upper  base  18  inchei*  ? 

2.  What  is  the  convex  surftu:e  of  a  frustum  of  a  triangular  pjrramid 
whose  slant  height  is  6  feet,  each  side  of  the  greater  base  3  feet,  and  of  the 
less  base  2  feet? 

3.  Find  the  entire  surftice  of  a  frustum  of  a  pyramid  whose  slant  height 
is  14  feet,  and  its  bases  triangles,  each  side  of  the  larger  base  being  8  feet, 
and  of  the  smaller  base  6  feet. 

4.  Find  the  entire  surfiMse  of  a  frustum  of  a  cone  whose  slant  height  is 
27  feet,  the  circumference  of  the  greater  base  being  87.6992  feet,  and  of  the 
less  base  31.416  feet. 

807.  Pbob.  XVn.— Tto  find  Vie  volume  of  a  frustum  of  a  pyramid 
or  cone:  To  the  mm  of  the  areas  of  both  banes  add  the  square 
root  of  their  product  and  multiply  the  result  by  one-third  of  the 
altitude. 

1.  How  many  cubic  feet  in  a  frurtum  of  a  cone  whose  altitude  Is  9  feet, 
the  diameters  of  its  ba^s  8  feet  and  6  feet  * 

2.  Find  the  volume  of  a  frustum  of  a  square  p3rr«mid  whose  altitude 
is  6  feet,  and  each  side  of  the  lower  base  16  feet,  and  of  the  upper  base 
12  feet. 

8.  How  many  cubic  feet  In  a  section  of  a  tree-trunk  90  feet  long, 
the  diameter  of  the  lower  base  being  IR  inches,  and  of  the  upper  base 
12  inches  ? 

4.  One  of  the  big  trees  of  California  is  32  feet  in  diameter  at  the  foot  of 
the  tree ;  how  many  cubic  feet  in  a  section  of  this  tree  9U  feet  high,  the 
upper  base  being  20  feet  in  diameter  i 


15   feet  by 
i  iuches  on 


-wttum  of  a 
cutnferenceii 


slant  height 
,  and  of  the 

lar  pyramid 
,  and  of  the 


elant  height 
)eiiig  8  feet. 

It  height  is 
and  of  the 


a  ityramid 
the  square 
ird  of  the 


le  is  9  feet, 

pe  altitude 
upper  base 

feet  long, 
ipper  base 

ho  foot  of 
high,  the 


3IEXSURA  TION, 


341 


5.  A  granite  rock,  whose  form  is  a  fhistum  of  a  triangular  pyramid,  is 
40  feet  high,  the  Bides  of  tlic  lower  base  being  30  feel  tach,  and  of  tl»e  upper 
base  16  feet  each.    How  many  cubic  feet  in  the  rock. 

808.  Prob.  XVUl.—  To  find  the  surface  of  a  spJiere :  Multiply  the 
diameter  by  the  circumference  qf  a  great  circle  of  the  given  sphere. 

1.  What  is  the  surface  of  a  globe  9  inches  in  diameter  ? 

2.  Find  the  surfoce  of  a  sphere  whose  diameter  is  8  feet. 

a  How  many  square  feet  In  the  surface  of  a  sphere  45  feet  in  diameter  ? 

4.  How  noAny  square  inches  in  the  surface  of  a  globe  5  inchuH  iu 
diameter  ? 

5.  What  is  the  surface  of  a  globe  whose  radius  is  1  ft.  6  in.  f 


80t).     Pbob.  'KlX.—To  find  the  volume  qf  a  sphere, 
surface  by  one-sixth  of  the  diameter. 


MultliAy  the 


1   How  many  cubic  yards  in  a  sphere  who^e  diameter  is  3  yards  ? 

2.  Find  the  volume  of  a  sphere  whot^c  diameter  is  20  iuches  ? 

3.  How  many  cubic  feet  in  a  globe  9  iuchcB  in  diameter? 

4.  Find  the  solid  contents  of  a  globe  2  ft.  6  in.  in  diameter. 

5.  Find  the  volume  of  a  globe  whose  radius  is  4  inches. 

8  lO.  Pbob.  XX.— To  find  the  capacity  qf  casks  in  galloiui :  MuUijdy 
the  number  of  inches  in  the  length  by  the  square  of  the  number  qf'  inchts  in 
the  mean  diameter^  and  this  product  by  .0031*. 

Observe^  that  the  mean  diameter  is  found  (nearly)  by  adding  to  tltc  head 
diameter  %,  or  if  the  staves  are  but  slightly  curved,  {  of  the  dilTcreuc-e 
between  the  head  and  bung  diameters. 

The  process  of  finding  the  capacity  of  casks  is  called  Gauging, 

1.  How  many  gallons  will  a  cask  hold  whose  head  diameter  is  21  inches, 
bung  diameter  90  inches,  and  length  42  inches  ? 

2.  How  many  gallons  in  a  cask  whose  head  diameter  is  20,  bung  diameter 
26  inches,  and  its  length  80  inches  ? 

8.  A  cask  slightly  curved  is  40  inches  long,  its  head  diameter  being 
22  inches,  and  its  bung  diamet4ir  27  inches  ;  how  many  gallons  will  it 
hold? 

4.  What  is  the  volume  of  a  ca^k  whose  diameters  are  18  and  24  inches 
respectively,  and  the  length  32  inches  ? 


i 

I 


REVIEW  AND  TEST  EXAMPLES. 


!, 


811.  In  using  this  set  of  review  and  test  examples,  the 
following  suggestions  should  be  carefully  regarded  : 

1.  Tlie  examples  cover  all  the  important  subjects  in  arith- 
metic, and  are  designed  as  a  test  of  the  pupil's  strength  in 
solving  difficult  problems  and  of  his  knowledge  of  principles 
and  processes. 

2.  The  teacher  should  require  the  pupil  to  master  the 
thought  expressed  in  each  example  before  attempting  a 
solution. 

To  do  this  he  must  notice  carefully  the  meaning  of  each 
sentence,  and  especially  the  technical  terms  peculiar  to  arith- 
metic ;  he  must  also  locate  definitely  the  business  relations 
involved. 

3.  When  the  soluticms  are  given  in  class,  the  teacher  should 
require  the  pupils  to  state  clearly : 

(a).    Wluit  u  given  and  wJuit  is  required  in  each  example. 
(6).    The  relations  of  the  given  quantities  from  which  what  is 
required  can  he  found. 

(c).  The  xtejis  thtt  must  he  taken  in  their  order,  and  the  pro- 
cesses that  must  he  used  to  nhtuin  the  required  remit. 

In  making  these  three  statements,  no  set  form  should  Oe 
used  ;  each  pupil  should  be  left  free  to  pursue  his  own  course 
and  give  liis  own  solution.  Clearness,  accuracy,  and  brevity 
should  be  the  only  conditions  imposed.  The  pupil  should  be 
allowed  to  make  re)>eated  efforts  until  the  precise  point  of  his 
failure  is  made  apparnt. 

^Vlien  the  work  is  written  on  a  slate,  paper,  or  blackboard, 
neatness  and  a  logical  order  in  arranging  the  steps  in  the 
solution  should  be  invariably  required. 


PLES. 


samples,  the 

i: 

sets  in  arith- 

I  strength  in 

of  principles 

master  the 
ittempting  a 

ning  of  each 
iliar  to  arith- 
less  relations 

acher  should 

rample. 
which  tchat  is 

and  the  pro- 
l. 

m  should  oe 
s  own  course 
and  brevity 
m1  should  be 
point  of  his 

blackboard, 
steps  in  the 


REV  I E  W    E  XA  M  PLES, 


343 


1.  A  gentleman  held  a  note  for  .11643.20,  payable  in  8  mo., 
without  interest.  lie  discounted  the  note  at  8%  for  ready 
cash,  and  invested  the  proceeds  in  stock  at  $104  per  share. 
How  many  shares  did  he  purchase?  Ans.  15  shares. 

2.  Three  daughters,  Mary,  Jane,  and  Ellen,  are  to  share  an 
estate  of  $80000,  in  the  proportion  of  ^,  i|,  and  \,  respectively  ; 
but  Ellen  dies,  and  the  whole  amount  is  to  be  divided  in  a 
proper  proportion  between  the  other  two.  What  share  does 
each  receive  ?  Ans.  Mary,  $48,000 ;  Jane,  $32,000. 

3.  What  must  be  the  dimensions  of  a  rectangular  bin  that 
will  hold  350  bushels  of  grain,  if  its  length  is  twice  its  width, 
and  its  width  twice  its  depth  ? 

Ana.  Length,  15.5  +  ft. ;  width,  7. 75 4- ft.;  depth,  3.87  f  ft. 

4.  A  Montreal  merchant  bought  800  barrels  of  flour  at  $7 
per  barrel,  and  sent  it  to  Halifax,  paying  9%  of  tlie  cost  for 
freight  and  other  charges ;  his  agent  sold  it  at  an  advance  of 
25%  on  the  original  cost  and  charged  3%  commission.  What 
was  the  net  gain  ?  Ah8.  $686. 

5.  What  sum  invested  in  railroad  stock  paying  7%  annually 
willyieldaquarterly  dividend  of  $325.50?       Ana.  $18,600. 

6.  A,  B,  and  C  together  can  dig  a  ditch  in  4  days.  A  can 
dig  it  alone  in  10  days ;  B  can  dig  it  alone  in  12  days.  How 
long  will  it  take  C  to  do  the  work  alone  ?  Ana.  15  days. 

7.  A  person  owning  7^  acres  in  the  form  of  a  rectangle 
3  times  as  long  as  it  is  wide,  wishes  to  tether  his  horw  to  a 
stake  by  the  shortest  rope  that  will  allow  him  to  graze  upon 
any  part  of  the  field.     What  is  the  length  of  rope  required  ? 

Ana.  31.02 +itl, 

8.  A  cubical  block  contains  64  cubic  feet ;  what  is  the  dis- 
tance from  one  corner  to  the  oppf)site  diagonal  corner  ? 

Ana.  6.92+  feet. 

9.  A  farmer  bought  a  horse,  wagon,  and  plough  for  $134 ; 
the  horse  cost  I  as  much  as  the  wagon,  and  the  plough  J  as 
much  as  the  horse.     What  was  the  cost  of  each  ? 

Ana.  Horse,  {^70;  wagon,  $50;  plough,  $14. 


i  u 


Ill  ■■' 


m^ 


y 


■r 


It 


:     . 


t'  ( 


!  1! 


3M 


REVIEW    EXAMPLES, 


10.  A  grocer  mixed  15  pounds  of  Hyson  tea  with  9  pounds  of 
Gunpowder  tea,  and  sold  it  at  $.96  j)er  pound,  thus  gainini;^ 
25%  on  the  original  cost.  If  a  pound  of  the  Gunpowder  cost 
IG  cents  more  than  a  pound  of  the  Hyson,  what  was  the  cost 
of  each  per  pound  ?       An%.  Gunpowder,  $.86  ;  Hysou,  ^.70. 

11.  A  farmer  has  a  cornfield  whose  width  is  to  its  length  as 
8  tu  4,  and  contains  4f  acres.  The  hills  of  corn,  supposing 
them  to  occupy  only  a  mathematical  point,  are  2  feet  apart, 
and  no  hill  is  nearer  the  fence  than  3  feet.  What  must  he  pay 
a  man  to  hoe  his  com,  at  the  rate  of  $.50  per  day,  if  he  hoes 
750  hiUa  in  a  day  ?  Am.  $34.33,\. 

12.  The  duty  at  20%  ad  valorem  on  a  quantity  of  tea  in 
chests,  each  weighing^  75  pounds  gross,  and  invoiced  at  $.70 
per  pound,  was  $6,552,  tare  heing  4%.  How  many  chests 
were  Imported  ?  Ans.  650. 

13.  A  room  is  22  feet  long,  18  feet  wide,  and  14  feet  high. 
What  is  the  distance  from  one  of  the  lowest  corners  to  the 
opposite  upper  comer?  Ans.  Q\. OS  +  ft. 

14.  A  farmer  sold  85  sheep  at  $2,  $2.20  and  $2.80  per  head, 
and  thus  realized  an  average  price  of  $2.40  per  head.  What 
number  of  each  did  the  lot  contain  ? 

Ans.  17  at  $2 ;  84  at  $f  .20 ;  84  at  $2.80. 

15.  If  the  ratio  of  increase  of  a  certain  c\o])  is  3,  and  a  man 
begins  by  planting  5  bushels,  using  all  the  crop  for  seed  the 
next  year,  and  so  on ;  what  will  be  his  crop  the  seventh  year  V 

Ans.  10,935  bushels. 

10.  A  can  do  a  piece  of  work  in  4 J  days  that  requires  B 
G  days  and  C  9  days  to  do  the  same  amount  of  work.  In  how 
many  days  can  they  do  it  working  together  ?      Ans.  2  days. 

17.  A  father  divided  his  property  among  his  wife  and  four 
sons,  directing  that  his  wife  should  have  $8  as  often  as  the 
oldest  son  $6,  the  second  eon  $3  as  often  as  the  wife  $5,  the 
youngest  son  $12  as  often  as  the  third  $14,  the  third  sou  i^o  as 
often  as  the  oldest  $7.  The  youngest  son  received  S4,500; 
what  was  the  value  of  the  father's  property?    Ans.  $32,780. 


1  9  pounds  of 
bus  gaininjor 
ipowder  cost 
was  the  cost 
ysou,  ^.70. 

its  length  as 
rn,  supposing 

2  feet  apart, 
I  must  he  pay 
ly,  if  he  hoes 
J.  $34.23,^. 

tity  of  tea  in 
roiced  at  $.T0 
many  chests 
Ans.  65a 

1 14  feet  high, 
corners  to  tho 
s.  31.08  + ft. 

i3.80  per  head, 
head.    What 

|34at$2.80. 

3,  and  a  man 
for  seed  the 
Kventh  yearV 
)35  bushels. 

lat  requires  B 
rork.  In  how 
[w«.  2  days. 

ife  and  four 

often  as  the 

le  wife  $5,  the 

lird  sou  4^5  at» 

ceived  S4,500; 

^18.  $32,780. 


REVIEW    AXD     TEST    EXAMPLES.    345 

18.  If  72  men  dig  a  trench  20  yd.  long,  1  ft.  G  in.  broad,  and 
4  ft.  deep  in  3  days  of  10  hours  each,  how  many  men  would  be 
required  to  dig  a  trench  30  yd.  long,  2  ft.  3  in.  broad,  and  5  ft. 
deep  in  15  days  of  U  hours  eacli  t  Aii».  45  men. 

19.  Saniuol  Wells  paid  3^  times  as  much  for  a  house  as  for 
a  baru ;  had  the  bam  cost  him  6  %  more,  and  the  house  8  % 
more,  the  whole  cost  would  have  been  $7260.  What  was  the 
actual  cost  ?  Ans.  $G,750. 

20.  Change  ^^5  of —  to  a  simple  fraction,  and  re- 


1  -f- 


3  +  i 


In- 


duce to  lowest  terms.  Ana. 

21.  A  person  sells  out  .*4,500  of  4%  stock  at  95,  and  invests 
the  proceeds  in  bank  stock  at  80,  which  pays  an  annual  divi- 
dend of  2|  % .     How  much  ia  the  gain  or  loss  per  annum  ? 

Am.  $37.50  loss. 

22.  James  Oriswold  bought  f  of  a  ship ;  but  the  property 
having  fallen  in  value  8^,  he  sells  14%  of  his  share  for  $2700. 
What  was  the  value  of  the  ship  at  first  ?  Ans.  $25,000. 

23.  A  gentleman  willed  to  the  youngest  of  his  five  sons 
$2000,  to  the  next  a  sum  greater  by  one-half,  and  so  on,  the 
eldest  receiving  $10,125,  thus  disposing  of  his  entire  estate. 
What  was  the  gentleman's  estate  worth?  Ans.  $20,375. 

24.  I  sent  $7847  to  my  agent  in  New  Orleans,  who  pur- 
chased sugar  at  an  average  price  of  $16  \>et  barrel ;  he  charged 
3|  %  commission.     How  many  barrels  did  he  buy? 

I  Ans.  475. 

25.  Bought  3,000  bushels  of  wheat  at  $1.50  per  bu.shel. 
What  must  I  ask  per  bushel  that  I  may  fall  20%  on  the  asking 
price  and  still  make  16%,  allowing  10 'r  of  the  sales  for  bad 
debts?  Ans.  $241,;. 

26.  Henry  Swift  has  $0,000  worth  of  5%  stock:  but  not 
being  satisfitnl  ^\^th  his  inc-<»me,  lu*  sells  at  96  and  invests  in 
stock  paying  4J%,  which  pves  him  an  income  greater  by 
$45.60.     At  what  price  did  he  purchase  the  latter  strxik? 

^„  Ans.  At  75. 


li 


r 


'.*  ■  ,, 


u.  ,e 


^  ■ 

i  ■ 

(    V 

.  4 

^  ■  ; 

u 

•  ■: 

" 

^1 

j 

ll 

:    1 

i  *  ! 


34G    i?i&r/-Bir    A^D     TEST    EXAMPLES, 

27.  A  drover  bought  a  number  of  horses,  cows,  and  sheep 
for  $3,900.  For  every  horse  he  paid  $75,  for  each  cow  he 
paid  f  as  much  as  for  a  horse,  and  for  each  sheep  ^  as  much  an 
for  a  cow.  He  bought  3  times  as  many  sheep  as  cows,  and 
twice  as  many  cows  as  horses  ;  how  many  did  he  buy  of  each  ? 

Ans.  20  horses  ;  40  cows ;  120  sheep. 

28.  I  shipped  to  my  agent  in  Buffalo  a  quantity  of  flour, 
which  he  immediately  sold  at  $7.50  per  barrel.  I  then  in- 
structed him  to  purchase  goods  for  me  at  a  commission  of 
8^^  ;  he  charged  me  4%  commission  for  selling,  and  received 
as  his  whole  commission  $800.  How  many  barrels  of  flour  did 
I  send  him?  Ans.  1,472. 

29.  Adam  Gesner  gave  his  note  for  $1,250,  and  at  the  end 
of  3  years  4  months  and  21  days  paid  off  the  note,  which 
then  amounted  to  $1504.375 ;  reckoning  only  simple  interest, 
what  was  the  rate  fo  ?  Ana. 


30.  A  hound  in  pursuit  of  a  fox  runs  6  rods  while  tlie  fox 
runs  3  rods,  but  the  fox  had  60  rods  the  start.  How  far  must 
the  hound  run  before  he  overtakes  the  fox  ?     Ans.  150  rods. 

31.  A  man  divided  his  property,  amounting  to  $15,000, 
among  his  three  sons,  in  such  a  manner  that  their  shares  put 
at  6  %  simple  interest  should  all  amount  to  the  same  sum  when 
they  were  21  years  old  ;  the  ages  of  the  children  were  respec- 
tively 6  yr.,  9  yr.,  and  13  yr.     What  was  the  share  of  each  ? 

Ans.  Eldest,  $5683.082+  ;  second,  $4890.094+  ; 
youngest,  $4426.822  + .  ♦ 


a 


32.  A  certain  garden  is  12f  rods  long,  and  9^  rods  wide. 
At  2^  cents  per  cubic  foot,  what  will  it  cost  to  dig  a  ditch  ^ 
around  it  that  shall  be  3 J  feet  wide  and  4  feet  deep  ?  ^^ 

Ans.  $258. 03|. 

33.  A  farmer  sells  a  merchant  40  bushels  of  oats  at  $.60  pe 
bushel  and  makes  20  %  ;  the  merchant  sells  the  farmer  4  yard 
of  broadcloth  at  $3.75  per  yard,  15  yards  of  calico  at  8  cent 
per  yard,  and  40  yards  of  cotton  cloth  at  12  cents  per  ya 


i.^  i 


LE8» 

wB,  and  Bbeep 
each  cow  lie 
)  \  as  much  as 
)  as  cows,  and 
B  buy  of  each  1 
J ;  120  sheep. 

antity  of  flour, 
•el.    I  then  in- 
commission  of 
ig,  and  received 
rrels  of  flour  did 
Ans.  1,472. 


REVIEW    A. YD     TEST    EXAMPLES.     347 

and  makes  a  profit  of  25  ^ .    Which  gains  the  more  by  tbf '  t  rade, 
uud  how  much  ?  Ans.  Merchant  gains  |.20. 

34.  A  triangular  cornfield  consisting  of  146  rows,  has  437  hills 
in  the  longest  row,  and  2  in  the  shortest ;  how  many  corn  hills 


in  the  field  ? 
35.  What  is  the  value  of 

r44|  of  .056  -  3.04  of  t\1 
L    (8-2.4)  +  |of3|     J  ' 


Ans.  82047  hills. 


+  2 


285, 
561 


Ans.  i\. 


36.  Bought  60  barrels  of  flour  at  $8.50  per  barrel,  but  on 
account  of  its  having  been  damaged,  one-half  of  it  was  sold  at 
a  loss  of  10%,  and  the  remainder  at  |9  per  barrel.  W^hat  %  was 
lost  by  the  operation  ?  Ans.  2^  % . 

37.  A  room  22  feet  long,  16  feet  wide,  and  9  feet  high,  con- 
pds  while  the  fox  I  i^jng  4  windows,  each  of  which  is  5 J  feet  high  and  3  feet  wide ; 


^  and  at  the  end 
the  note,  which 
f  simple  interest, 
Ans.  Qfo. 


.t.    How  far  must 
Ans.  150  rods. 

nting  to  $15,000, 
t  their  shares  put 
le  same  sum  when 
iren  were  respec 
share  of  each? 
td,  $4890.094+  ; 

id  9i  rods  wide, 
lost  to  dig  a  dit<" 
let  deep? 

Ans.  $258.03|. 

lof  oats  at  $.60  pe 
4  yard 


the 


farmer 
calico  at 
I12  cents  per  yai 


also  two  doors,  7  feet  in  height  and  3^  feet  in  width.  The 
base-boards  are  |  of  a  foot  wide.  What  will  it  cost  to  plaster 
and  paper  the  room,  if  the  plastering  cost  16  cents  per  square 
yard  and  the  papering  to  cents  ?  Ans.  $21.20^. 

38.  Two  persons,  A  and  B,  each  receive  the  same  salary.  A 
pends  76^  %  of  his  money,  and  B  spends  as  much  as  would 

il  46A  %  of  what  both  received.  At  the  end  of  the  year  they 
both  together  have  left  $276.25 ;  what  part  of  it  belongs  to  A, 
and  what  to  B?  Ans.  A,  $199.75  ;  B,  $76.50. 

39.  Two  persons  280  miles  apart  travel  toward  each  other 
ntil  they  meet,  one  at  the  rate  of  6  miles  jier  hour,  the  other 
t  the  rate  of  8  miles  per  hour.     Ilow  far  docs  each  travel  ? 

Ans.  First,  120  miles  ;  second,  160  miles. 

40.  James  Welch    has  a  debt    in    Ottawa    amounting    to 
4489.32.    For  what  sum  must  a  note  be  drawn  at  90  days,  that 
hen  discounted  at  6%  at  an  Ottawa  bauk,  will  just  pay  the 
bt?  Ans.  $4560. 


i .  I 


1  ! 


«  4    ! 


ill 


348  i?  J?  r  /  A' ir   and    test  examples, 

41.  I  went  to  the  store  to  buy  carpeting,  and  found  that  any 
one  of  three  pieces,  width  rebpectivuly  1^,  1^,  and  2 J  yards, 
would  exactly  fit  my  room  without  cutting  anything  from  the 
width  of  the  carpet.    What  is  the  width  of  my  room  ? 

Aji8.  22^  feet. 

42.  If  10  horses  in  25  days  consume  3^  tons  of  hay,  how  long 
will  6j^  tons  last  0  horsoH,  12  cows,  and  8  sheep,  if  each  cow 
consumes  |  as  much  as  a  horse,  and  each  sbecp  |  as  much  as  a 
cow  ?  Au8.  25  days. 

43.  The  distance  between  the  opposite  corners  of  a  square 
field  is  GO  rods  ;  how  many  acres  in  the  field  ? 

Ans.  11  A.  40  sq.  rd. 

44.  At  |225  per  ton,  what  is  the  cost  of  17  cwt.  2  qr.  21  lb.  of 
sugar?  Ana.  $199.2371. 

45.  A  drover  bought  12  sheep  at  $6  per  head ;  how  many 
must  he  buy  at  $9  and  $15  per  head,  that  he  may  sell  them  all 
at  $12  per  head  and  lose  nothing  ? 

Ans.  1  at  $9, 25  at  $15. 

46.  Three  men  bought  a  field  of  grain  in  circular  form  con- 
taining 9  A.,  for  which  they  paid  $192,  of  which  the  first  man 
paid  !?48.  the  second  $64.  tlie  third  $80.  They  agreed  to  take 
their  shares  in  the  form  of  rini^s  ;  the  first  man  mowing  around 
r,he  field  until  he  got  his  share,  then  the  second,  and  so  on. 
What  depth  of  ring  must  each  man  mow  to  get  his  share  of  the 
grain''  Ans.  1st  man,  2.80+  rd. ; 

2d  man,  4.73  +  rd. ; 
3d  man,  13.81  +  rd. 

47.  A  young  man  inherited  an  estate  and  spent  15%  of  it 
during  the  first  year,  and  30%  of  the  remainder  during  the 
second  year,  when  he  had  only  $9401  left.  How  much  money 
did  he  inherit  ?  Ans.  $15800. 

48.  Mr.  Webster  bought  a  house  for  $6750,  on  a  credit  of 
10  months  ;  after  keeping  it  4  months,  he  sold  it  for  $7000  on 
a  credit  of  8  months.  Money  being  worth  6%,  what  was  his 
net  cash  gain  at  the  time  of  the  sale  ?  Ane.  $177.37  + . 


O^:;   r 


L  E  S, 

nd  that  any 
id  2^  yards, 
ig  from  the 
aiV 
22  J  feet. 

ly,  how  long 
if  each  cow 
18  much  as  a 
.  25  days. 

of  a  square 

40  sq.  rd. 

I  qr.  21  lb.  of 
$199^37^. 

;  how  many 
sell  them  all 

25  at  |15. 

IT  form  con- 
le  first  man 
reed  to  take 
wing  around 
and  so  on. 
sliare  of  the 
J.80+  rd. ; 
73+  rd. ; 
3.81  +  rd. 

t  15%  of  it 
during  the 
luch  money 
$15800. 

a  credit  of 
T  $7000  on 
lat  was  his 
177.37  +  . 


REVIEW    AND     TEST    EXAMPLES,     349 

49.  A  and  B  can  do  a  piece  of  work  in  18  days  ;  A  can  do  | 
418  much  as  B.     In  how  many  days  can  each  do  it  alone? 

Ana.  A,  40J  days  ;  B,  32f  days. 

50.  If  ?  of  a  farm  is  worth  $7524  at  $45  i^er  acre,  how  many 
acres  in  the  whole  I'urm  V  Ana.  195,^^  A. 

51.  A  person  paid  $1450  for  two  building  lots,  the  price  of  one 
being  45^  that  of  the  other ;  he  sold  the  cheai)er  lot  at  a  gain 
of  C0%,  and  the  dearer  one  at  a  loss  of  25%.  What  %  did  he 
gain  or  lose  on  the  whole  transaction  ?  Ana.  \^\  fc  gain. 

52.  A  certain  sum  of  money,  at  8%  compound  interest  for 
10  years,  amounted  to  $2072.568.  What  was  the  amount  at 
interest?  Ana.  $960. 

53.  There  are  two  church  towers,  one  120  feet  high,  and  the 
other  150  feet.  A  certain  object  upon  the  ground  between  thum 
is  125  feet  from  the  top  of  the  first  and  160  feet  from  the  top  of 
the  second ;  how  far  apart  are  their  tope  ? 

Ana.  95.50+  feet. 

54.  A  farmer  sold  to  a  merchant  80  bushels  of  wheat  nt  $1  90 
l^er  bushel,  70  bushels  of  barley  at  $1.10,  and  176  bushels  of 
oats  at  $.75.  He  took  in  payment  a  note  for  5  months,  and 
immediately  got  it  discounted  at  bank  at  6% ;  how  much  money 
did  he  receive  ?  iln«.  $351 .  06  + . 

55.  There  is  a  pile  of  100  railroad  ties,  which  a  man  is  required 
to  carry,  one  by  one,  and  place  in  their  proper  places,  3  feet 
apart ;  supposing  the  first  to  be  laid  3  feet  from  the  pile,  how 
far  will  the  man  travel  in  placing  them  all  ? 

Am.  30300  feet. 

56.  Sound  travels  at  the  rate  of  1142  feet  a  second.  If  a  gun 
be  discharged  at  a  distance  of  4|  miles,  how  much  time  will 
elapse,  after  seeing  the  flash,  before  the  report  is  heard  ? 

Ana.  20J??  sec. 

57.  If  a  company  of  480  men  have  provisions  for  8  months, 
how  many  men  must  be  sent  away  at  the  end  of  6  montlis,  that 
the  remaining  provisions  may  last  6  months  longer  ? 

Alia.  320  men. 


!i;l« 


850    REVIEW    AJSD     TEST    EXAMPLES, 


.^4 


v '  ■  i 


i 


K 


V 


i 


f    r' 


,1^ 


58.  The  first  fear  a  man  was  In  btutineBs  ho  cleared  $300, 
and  each  year  his  profit  increased  by  a  commun  difference  ;  the 
fourteenth  year  he  made  |950.  How  much  did  he  make  the 
third  year  ?  Ana.  $400. 

69.  What  number  is  that,  which  being  increased  by  ^,  |,aQd 

1  of  itself,  and  diminished  by  25,  equals  391  ¥ 

Ans.  180. 

60.  At  what  time  between  5  and  0  will  the  hour  and  minute 
hands  of  a  clock  be  together? 

Ans.  27^  min.  past  five. 

61.  A  field  whose  length  is  to  its  width  as  4  to  3,  contains 

2  A.  2  R.  82  rd. ;  what  are  its  dimeuHions  ? 

Ans.  Length,  24  rd. ;  width,  18  rd. 

62.  Three  persons  formed  a  partnership  with  a  capital  of 
$4600.  The  first  man's  stock  was  in  trade  8  months  and  gained 
$752 ;  the  second  man's  stock  was  in  trade  12  months,  and 
gained  $600 ;  and  the  third  man  had  his  stock  in  16  months, 
and  gained  $640.     What  was  each  man's  stuck  ? 

Ans.  First,  $2350;  second,  $1250  ;  third,  $1000. 

68.  How  many  thousand  shingles,  18  inches  long  and  4  in. 
wide,  lying  ^  to  the  weather,  are  required  to  shingle  the  roof 
of  a  building  54  feet  long,  with  rafters  22  feet  long,  the  first 
row  of  shingles  being  double  ?  Ans.  l^^. 

64.  Employed  an  agent  who  charges  4%  commission  to  col- 
lect a  bill  of  $550.  He  succeeded  in  obtaining  only  85^ ;  how 
much  did  I  receive  ?  Ans.  $448.80. 

65.  A  and  B  entered  into  partnership  and  gained  $4450.50. 
A  put  in  enough  capital  to  make  his  gain  15%  more  than  B's  ; 
what  was  each  man's  share  of  the  gain  ? 

Ans.  A,  $2880.50 ;  B,  $2070. 

66.  A  building  is  75  feet  long  and  44  feet  wide,  and  the 
elevation  of  the  roof  is  14  feet.  How  many  feet  of  boards  will 
be  required  to  cover  the  roof,  if  the  rafters  extend  2  feet 
beyond  the  plates,  and  the  boarding  projects  1\  feet  at  each 
end,  and  ^  allowed  for  waste  ?  Ans.  547497+  feet. 


i'LES, 


REVIEW    AND     TEST    EXAMPLES,    351 


0  cleared  $300, 
difference ;  the 

1  be  make  the 
Am.  $400. 

Bed  by  f  \,  and 

Am.  180. 
)ur  and  minute 

lin.  past  five. 
V  to  8,  contains 

width,  18  rd. 

itb  a  capital  of 
Qths  and  gained 

2  months,  and 
k  in  16  months, 
\ 

third,  $1000. 

long  and  4  in. 

lingle  the  roof 

long,  the  first 

Ana.  14|8. 

nission  to  ool- 
Qly  85%;  how 
rut.  $448.80. 

ined  $4450.50. 
Lore  than  B's ; 

;  B,  $2070. 

nride,  and  the 
>f  boards  will 

)xtend  3  feet 
feet  at  eacb 

4,97+  feet. 


67.  A  circular  court  is  laid  with  19  rows  of  flat  stono^,  each 
row  forming  a  complete  circle ;  the  outside  row  is  89  inches 
wide,  and  the  width  of  each  row  diminishes  2  inches  as  it  uears 
the  centre.     What  is  the  width  of  the  innermost  row  Y 

Ana.  3  inches. 

a  of  fi  of  AJ^ 

68.  Reduce  .    , .    .7^  ^  •  simple  fraction,  and   take  the 

4offofU 

result  from  the  sum  of  10},  ^^j^,  and  7}|.  Ana.  8Jj{. 

69.  Bought  75  yards  of  cloth  nt  10%  less  than  the  first  cost, 
and  sold  it  at  10%  more  than  the  first  cost  and  gained  $25. 
What  was  the  first  cost  per  yard  t  Ana.  |1.66f . 

70.  A  grain  merchant  bought  7500  bushels  of  com  at  $1.85 
per  bushel,  5450  bushels  of  oats  at  $.80,  3250  bushels  of  barley 
at  $.95,  paid  $225  for  freight  and  $170  for  storage  ;  he  immo- 
diately  sold  it  at  an  advance  of  20  %  on  the  entire  cost,  on  a 
credit  of  6  months.  What  %  did  be  gain  at  the  time  of  the 
sale,  money  being  worth  8%  ?  Ana.  15+  %. 

71.  A  farmer  employs  a  number  of  meu  and  8  boys ;  he  pays 
the  boys  .$.05  and  the  men  $1.10  per  day.  The  amount  that  he 
paid  to  all  was  as  much  as  if  each  had  received  $.92  i)er  day  : 
how  many  men  were  employed?  Ana.  12  men. 

72.  S.  Howard  can  mow  6  acres  in  4  days,  and  hib  son  can 
mow  7  acres  in  5  days.  How  long  will  it  take  them  both  to 
mow  49i^  acres  ?  Ana.  11  ^  days. 

78.  I  lent  a  friend  $875,  which  he  kept  1  year  and  4  months. 
Some  time  afterward  I  borrowed  of  him  $350  ;  how  long  must 
I  keep  it  to  balance  the  favor  ?  Ana.  ^  yr.  4  mo. 

74.  Find  the  difference  between  the  surface  of  a  floor  80  ft. 
9  in.  long  and  66  ft.  6  in.  broad,  and  the  sum  of  the  surfaces  of 
three  others,  the  dimensions  of  each  of  which  are  exactly  one- 
third  of  those  of  the  other. 

Ana.  891  sq.  yd.  7  sq.  ft.  12  sq.  in. 

75.  A  tree  broken  off  24  feet  from  the  ground  rests  on  the 
stump,  the  top  touching  the  ground  30  feet  from  the  foot  of  the 
tree.     What  was  the  height  of  the  iree  ?        Ana.  62.41  +  ft. 


X; 

/ 


\ 


}:%  '^:' 


352   BEVJEW    AND     TEST    EXAMPLES, 


'  )'■ ' ' 


"'*  1 


,M! 


I  ( I 


i 


I,! 


Pf» 


76.  Two  persons  entered  into  partnership  for  trading.  A 
put  in  $245  for  375  days  and  rec€ived  ^  of  the  gain  ;  the  num- 
ber of  dollars  that  B  put  in  was  equrl  to  the  number  of  days  it 
was  employed  in  trade.     What  'vas  B's  capital  ? 

Aii8.  $350. 

77.  How  many  square  feet  of  boards  1^  inches  thick  will  lie 
reciuired  to  make  a  box,  open  at  the  top,  whoso  inner  dimen- 
sions are  6  feet  long,  4  feet  wide,  and  3  feet  deep  ? 

AhH.  88,'u  sq.  ft. 

78.  A  farmer  having  80  acres  of  l.ind,  worth  $5.")  an  aero, 
wishes  to  buy  enough  more  at  $50  and  .$G5,  respectively,  so 
that  the  value  of  his  land  shall  averr%'t  $60  an  acre.  How  much 
of  each  must  he  buy?  Am.  1  A.  at  $50  ;  82  A.  at  $65. 

79.  What  is  the  amount  of  an  annuity  of  $700  for  8  years,  at 
6%  comi'ound  inU^rest  ?  Am.  $6928.22J. 

80.  What  must  be  the  price  of  stock  yielding  5}  'v  ,  that  will 
yield  the  same  profit  as  4^%  stock  at  96?  Am.  112. 

81.  A  jjerson  after  sptuiding  \  and  J  of  his  money  an.l  $20, 
had  $80  loft.     What  had  he  at  first  ?  Am.  $240. 

82.  James  Hari>er  has  a  large  jewelry  store,  which  with  its 
contents  he  insun^s  in  the  C'itiz«'ns'  Insurance  Company  for 
*  of  its  estimated  value,  at  3}^.  This  Company  immediately 
insures  \  of  its  risk  in  the  Phoenix  Company,  at  2  J  '/c .  After  two 
years  and  a  half,  the  store  and  Its  contents  w«'ro  destrr)yed  by  fire, 
when  it  was  found  tliat  the  PhoBnix  Company  lost  $2025  more 
than  the  Citizons'  Compaay.  Reckoning  6%  simple  '  iterest 
on  the  premiums  that  the  owner  |>aid,  what  would  1)e  his  entire 
loBsV  Ant.  $78815.75. 

88.  A  drover  sold  42  cows  and  34  oxen  f.)r  $3374,  receiving 
$21  i»er  lu'ad  more  for  the  oxen  than  for  the  cows.  What  aid 
he  receive  for  each  jMir  head  ?  Anit.  $35  for  cows ; 

$56  for  oxen. 

84.  A  certain  room  is  27  ft.  5  In.  long,  14  ft.  7  in.  wide,  and 
12  ft.  10  in.  high.  How  much  paper  ;  of  a  yard  w'uh;  will  he 
required  to  cover  the  walls  ?  Ans.  136  yd.  2  ft.  8  in. 


LES, 

trading.  A 
\ ;  the  num- 
3r  of  days  it 

iii8.  $350. 

hick  will  1)© 
uncr  dimen- 

^8,^  sq.  ft. 

jiS.")  an  iicro, 
ipectively,  so 
.  How  much 
I  A.  at  $63. 

or  8  year«,  at 
$0928. 22  J. 

%,  that  will 
Am.  112. 

,oy  un.l  $20, 
Am.  $240. 

licli  with  its 
'ompuny  for 
immediately 
After  two 
roye*!  by  fire, 
$25)25  moro 
pie  "  itrnist 
Ixi  his  rntire 
^78815.75. 

r4,  rorciving 
What  did 
for  cows; 
for  oxen. 

|n.  wide,  and 
i'mV'!  will  1)6 
2  ft.  8  in. 


REVIEW    AND     TEST     EXAMPLES,     353 

85.  The  area  of  a  triangular  field  is  0  A.  36  rd. ;  the  base  is 
(U  rods.  Whnt  is  the  p<}rpendicular  distance  from  the  base  to 
the  angle  opposite  ?  Aiia.  31  i  rfxls, 

86.  If  the  width  of  a  building  is  50  feet,  and  the  length  of 
the  rafters  30  feet,  what  will  it  cost  to  board  the  gable  ends, 
at  $.18  per  square  yard  ?  Ana.  $16.58  -♦- . 

87.  What  is  the  solidity  of  the  largest  ball  that  can  be  cut 
out  of  a  cubical  bUx:k  whose  sides  are  6  inches  stpiareV 

Am.  113.0976  cu.  in. 

88.  A  privateer  took  a  prize  ivorth  £.348  15s.,  which  was  to 
be  divided  among  1  captain,  3  mates,  and  27  privates,  so  that 
a  private;  should  have  one  share,  a  mate  twice  as  much  as  a 
private,  and  the  captain  6  times  as  much  as  a  mate.  What 
was  the  share  of  each? 

Am.  Private,  £7  ISs. ;  mate,  £15  10s. ;  captain,  £93. 

89.  The'  width  of  a  certain  building  is  38  feet,  and  the 
elevation  of  the  roof  is  16  feet ;  how  many  square  feet  of 
boards  will  be  required  to  cover  the  gable  ends? 

Am.  608  s<i.  ft. 

90.  The  length  of  one  side  of  a  field  in  the  form  of  an  e<iui- 
lateral  triangle  is  40  rods.  How  many  acres  does  the  field 
contain,  and  what  would  it  cost  to  fence  it.  at  $.65  jn-r  rcvi  ? 

Ans.  4  A.  52.8+  sq.  rd.;  $78. 
1       1      1 


91.  Change 


3  "^^  8'*"4 


I 


1 


,-  to  a  simple  fraction,  and  reduce 


2J  "^  3i  "*■  4i 
to  lowest  terms.  A?i8.  IJ;. 

92.  Two  merchants,  Sanford  and  Otis,  Invested  equal  .sums 
In  trade.  Sauford  gainnl  a  sum  e<iual  to  \  of  his  stock  and 
$24  more,  and  Otis  lost  $144  ;  then  Otis  had  just  \  as  uiuch 
money  as  Sauford.     What  did  each  invest?  Aria.  $(175. 

93.  Henry  Norton  sold  his  fann  for  $13270,  $5000  of  which 
was  to  be  pai<l  in  6  mo.  hs,  $4000  in  one  year  and  6  months, 
and  the  rest  in  2  years.  What  was  the  n«»t  cash  value  of  hia 
fann,  money  being  worth  6;^^  ?  Am.  $12336.59  +  . 


rr 


>  '^:* 


354    BEVIEW    AND     TEST    EXAMPLES, 


Li  i 

i      I 


94  At  what  time  between  10  and  11  o'clock  will  the  hands 
be  directly  opposite?  Arm.  21^  min.  past  ten. 

95.  How  much  better  is  it  to  invest  |15000  in  6%  stock,  at 
a  discount  of  25%,  than  to  loan  the  same  sum  at  7%  simple 
interest?  Ans.  $150. 

96.  What  is  the  present  worth  of  an  annuity  of  $550  for 
6  years,  at  8;^  simple  interest?  Ans.  $2675.675  +  . 

jjof9J +  4|of  J\       1.30 


97.  What  is  the  value 


of(« 


5-4f 


') 


.005 

Ans. 


i^oV- 


08.  If  36  men  working  8  hours  a  day  for  16  days  can  dig  a 
trench  72  yd.  long,  18  ft.  wide,  and  12  ft.  deep,  iu  liow  many 
days  will  32  men,  working  12  hours  a  day,  dig  a  trench  04  yd. 
long,  27  ft.  wide,  and  18  ft.  deep  ?  Ans.  24  days. 

99.  Bought  a  piece  of  broadcloth  at  $2.75  per  yard.  At 
what  price  shall  it  be  marked  that  I  may  sell  it  at  5  %  less 
than  the  marked  price  and  still  make  20%  profit  ? 

Ans.  $3.47i'5. 

100.  A  man  hired  a  mechanic  for  85  days,  on  condition  th  -, 
for  every  day  he  worked  he  should  receive  $1.75,  and  to: 
every  day  he  was  absent  he  should  forfe't  $2.50.  At  the  end 
of  the  time  he  received  $40 ;  how  many  days  did  he  work  ? 

Ans.  30  days. 

101.  If  stock  bought  at  25;^  premium  pay  7J;^  on  the  in- 
vestment,  what  %  will  it  pay  if  bought  at  4%  discount? 

Ans.  9g%. 

102.  The  interest  on  a  note  for  2  yr.  3  mo.  18  da.,  at  8%,  was 
$155.02  ;  what  was  the  face  of  the  note  ?  Ans.  $842.50. 

103.  Tlie  distance  on  the  road  anmnd  a  certain  park  is  17 
miles.  If  three  persons  start  from  the  same  point  on  the  road 
at  the  same  time  and  travel  in  the  same  direction  around  tlio 
park,  how  far  will  oach  have  to  travel  before  they  all  come 
together,  if  the  first  travels  5  miles  an  hour,  the  second  6,  and 
the  third  7  miles  an  hour? 

Am.  First,  85  mi. ;  second,  102  mi. ;  third,  110  mi 


LE8. 

ill  the  hands 
k.  past  ten. 
6%  stock,  at 
It  7  %  simple 
Ans.  $150. 

of  $550  for 
i675.675  +  . 

OG 
.005 

Ans.    ^oV- 
ays  can  dig  a 
in  how  many 
trench  04  yd. 
ris.  24  days. 

)er  yard.     At 
it  at  5%  less 

? 

u.  |3.47i'ft. 
londition  th 
.75,  and  to: 
At  the  end 
he  work? 
IS.  30  days. 

^f  on  the  in- 
icount  ? 
Ans.  9g%. 

,  at  8%,  was 
|<w.  1842.50. 

.in  park  is  17 
it  on  the  road 
n  around  the 
jhey  all  come 
lecond  6.  and 

I,  119  mL 


BE  VIEW    ASD     TEST    EXAMPLES.     355 

104.  Two  persons  commence  trade  with  the  same  amount  of 
money ;  the  first  man  spends  48%  of  his  yearly,  and  the  second 
spends  a  sum  tHjual  to  25^^  of  what  both  had  at  first ;  at  the 
end  of  the  year  they  botli  together  had  $8468.  How  much 
liad  each  at  the  end  of  the  year  ? 

Ans.  |17G8,  first  ;  $1700,  second. 

105.  A  rollor  ust^d  for  levelling  a  lawn  being  G  ft.  0  in.  in 
circumference  by  2  ft.  3  in.  in  width,  is  observed  to  make  12 
revolutions  as  it  rolls  from  one  extremity  of  the  lawn  to  the 
other.  Find  the  area  rolled  when  the  roller  has  passed  ten 
times  the  whole  length  of  it.  Ans.  1U5  sq.  yd. 

IOC.  A  and  B  form  a  copartnership ;  A's  stock  is  to  B's  as 
5  to  7.  At  the  end  of  4  months  A  withdraws  f  of  his  stock, 
and  B  \  of  his  .-  their  year's  gain  is  $5650.  How  much  does 
each  receive?  Ans.  A,  $2500;  B,  $3150. 

107.  A  drover  l)ought  a  number  of  sheep,  oxen,  and  cows. 
He  paid  half  as  much  more  for  oxen  as  for  sheep,  and  half  as 
much  more  for  cows  as  for  oxen  ;  he  sold  the  sheep  at  ii 
profit  of  10  ^<.  the  oxen  at  a  i)rofit  of  8f^,  and  the  cows  at  a 
loss  of  i'/e  ;  lie  received  for  the  whole  $8416.  What  did  Ik* 
pay  for  each  lot  ? 

Ans.  $700,  sheep  ;   $1050,  oxen  ;   $1575,  cows. 

108.  A  c "jm mission  merchant,  who  char<^es  IJJ?-,  purchases 
for  me  145  barrels  of  sugar,  pays  for  freight  :rl2.5(),  nuiking 
the  whole  l)ill  ;f»2'i55.07.  If  there  were  100  jMiunds  of  sugar 
in  each  barrel,  what  was  the  i)rice  of  the  sugar  per  pound,  and 
what  was  the  amount  of  commission  ? 

Ans.  $.08  per  pound  ;  $38.57,  com. 

109.  A  merchant  in  Montreal  importi-d  from  England  a 
quantity  of  gomls,  for  wliich  he  had  to  pay  a  duty  of  12 ''^.  On 
account  of  the  drpression  in  trade,  he  is  obliged  to  sell  at  a 
loss  of  7J%  ;  had  he  sold  tliera  two  months  scM)ner,  he  would 
have  received  $896  more  than  he  did,  and  then  would  have 
cleared  3,»  '/i  on  the  transaction.  What  price  did  he  pay  for 
thegtKxls?  ^/w.  $7500. 


i 

1J 


5" 


Vv' 


5.V   '  1 


356      REVIEW   AND    TEST   EXAMPLES, 

110.  How  many  bricks  8  inches  long,  4  inches  wide,  and 
2  inches  thick  will  be  re(;[uired  to  V;uild  a  cubical  cistern,  open 
at  the  top,  that  shall  contain  3000  gallons,  if  the  wall  is  made 
a  foot  thick  and  ^  of  the  entire  wall  is  mortar  ? 

Ans.  5918  bricks. 


111.  What  is  the  value  of  2J  x 


H  + 


41 


Ans.  Ilf 


112.  If  18  men,  working  10  hours  per  day,  can  dig  a  ditch  in 
20  days,  how  long  will  it  take  3  men  and  40  boys,  working 
8  hours  per  day,  to  dig  a  ditch  twice  as  long,  6  men  being 
equal  to  10  boys?  Ans.  o3*  days. 

113.  John  Turner  owes  |350  due  in  7  montlis,  $500  in  3 
months,  and  $650  due  in  5  months,  and  pays  |  of  the  whole  in 
6  months ;  when  ought  the  remainder  to  be  paid  ? 

Ans.  3  months. 

114.  A  and  B  can  do  a  piece  of  work  in  4|  days ;  B  and  C  in 
5i\  days ;  and  A  and  C  in  4|  days.  In  what  time  can  each  do 
the  work  alone  ?  Ans.  A,S  days  ;  B,  10  days  ;  C.  12  da. 

115.  A  and  B  alone  can  do  a  piece  of  work  in  15  and  18  days 
respectively.  They  work  together  on  it  for  3  days,  when  B 
leaves;  but  A  continues,  nnd  after  3  days  is  joined  by  C; 
together  they  finish  it  in  4  days.  In  what  time  could  C  do  the 
piece  of  work  by  himself?  Ans.  24  days. 

116.  Mr.  Smith  paid  'd\  times  as  much  for  a  horse  as  for  a 
harness.  If  he  had  paid  10%  less  for  the  harness  and  7|%  more 
for  the  horse,  they  %vould  together  have  cost  $245.40.  How 
much  did  he  give  for  each  ? 

Ans.  Horse,  $182 ;  harness,  $56. 

117.  James  and  Herbert  are  running  around  a  b^ock  25  rods 
square;  James  runs  around  it  every  7|  minutes,  and  Herbert 
every  8J  minutes.  If  tlu'y  started  together  from  the  same 
point,  how  many  times  muwt  each  ruu  around  the  block  before 
they  will  be  together  ? 

Am.  Jau)e»»  10  limes  ;  HerUrt,  0  times. 


L£8. 

8  wide,  and 
cistern,  open 
rail  is  made 

918  bricks. 
An8,  llf. 

iig  a  ditch  In 
joys,  working 
6  men  being 
J.  o6l  (lays. 

lis,  $500  in  8 
t  the  whole  in 

? 

J.  3  months. 

8 ;  B  and  C  in 
te  can  each  do 
8 ;  C.  12  da. 

.5  and  18  days 
days,  when  B 
joined  by  C; 
:ould  C  do  the 
ins.  24  days. 

lOTse  as  for  a 
and  7^  %  more 
1245.40.    How 


REVIEW  AND    TEST   EXAMPLES,      357 

118.  James  Walker  contracted  to  build  a  stone  wall  ISO  rd. 
long  in  21  days.  He  employed  45  men  12  days,  who  built 
412 \  yards.  How  many  more  men  must  be  employed  to  finish 
the  work  in  the  required  time?  Afis.  '49. 

110.  I  invested  f6345  in  Government  bonds  at  104?i,  broker- 
age 1\%.  How  much  would  I  gain  by  selling  the  same  at  113j[, 
brokerage  1J%  ?  Ans.  $Ji7.j. 

120.  A  grain  merchant  bought  3250  bushels  of  wheat,  at 
i|1.25  per  bushel,  and  sold  it  immediately  at  $1.45  i)er  bushel, 
receiving  in  payment  a  note  due  4  mouths  hence,  which  he  had 
discounted  at  bank  at  6^<> .     What  was  his  gain  ? 

Aii8.  .i;553.a9-»-. 

121.  Two  men  form  a  partnership  for  trading  ;  A's  capital  is 
$3500,  B's  $4800.  At  the  end  of  7  months,  how  much  must  A 
put  in  that  he  may  receive  ^  of  the  year's  gain  ? 

Ans.  $3120. 

122.  A  maft  having  lost  25%  of  his  capital,  is  worth  exactly 
as  much  as  another  who  has  just  gained  15%  on  his  capital ; 
the  second  man's  capital  was  originally  $9000.  What  was  the 
first  man's  capital  V  Ans.  $13800. 

123.  A  merchant  imported  18  barrels  of  syrup,  each  contain- 
ing 42  gallons,  invoiced  at  $.95  per  gallon  ;  paid  .*>b5  for  freight 
ami  a  duty  of  30%.     What  %  will  he  gain  by  selling  the  whole 


lor  $1171.459"^ 


Am. 


15%. 


lamess,  $56. 

I  b^ock  25  rods 
and  Herbert 
■)ni   the  same 
block  before 

rt,  0  times. 


'I 


1  ,  » 


: 


ANSWERS. 


«•  tJH 


Hj 


). 


•t 


!  ii 


The  answers  to  oral  exercises  and  the  more  simple  examples 
have  been  omitted. 

The  answers  for  the  exercises  taken  from  the  Arithmetical 
Tables,  commence  on  page  376. 


Art.  39. 

/.  1213. 
2.   11526. 

3.  vm. 

4.  5726. 

5.  322633. 

6.  1543. 

7.  $4030. 

8.  $795. 
.9.  $2617. 

10.  $1330. 

Art.  45. 

1.  $3553.66. 

2.  16217.66. 
S.   $1004.94. 

4.  IU06.75. 

5.  $75.38. 

6.  $312.09. 
S.  $607.65. 

9.  $17931. 
10  !?(Uy.90. 

11.  $5746.62. 
le    ?1751.'32. 

13.  $228.20. 

14.  $8985.60. 

15.  $70.11. 

Art.  01. 

1.  100  years. 

2.  $UH6. 

3.  6269. 

4.  (i892  feet 

5.  3502. 


6.  6940. 
S.  $550. 
iy.  $41874. 

/ry.  $1074. 
11.  $4820. 
i4^.  $1634. 
13.  $245. 
i4.  $1136. 
75.  1005  ba. 

16.  $262.64. 

17.  $166.38. 

18.  $11.06. 
/!).  $95.12. 
m  $901.92. 
21.   $272.59. 

Art.  81. 

0.   $1995. 

7.  $639. 
^.  i?;:i882. 

.9.  63360  feet. 

iO.  $2556. 

11.  1516  far. 

i.'.  $241. 

!//?.  $311. 

/>^.  $76. 

Art.  90. 

i.  $8:;()0. 
$24630. 
.tl498. 
$72. 
522270  gal. 


J. 

4. 
5. 


n.  14136  bu. 

7.  142692  da. 

8.  8946  trees, 
ry.  $2373. 

10.  $9048. 
ii.  3.')857586. 

12.  $1086800. 

13,  $11.26. 
/4.  $72.25. 

15.  $638.27. 

16.  $6253.04; 


17.  $21.99  gain. 

18.  $69.42. 

19.  $2112; 
$345.60  ga. 

20.  $32.33  gain. 

Art.  113. 

/.  281 ;  302. 

2.  $986. 

/.  210  hours. 

4.  147  barrels. 

5.  121  montlis 

6.  26  we«kB. 

7.  86400  arc. 
cV.  134  baskets 
0.  480. 

/(y.  47. 

//.  59  dozen. 

12.  $178. 

13.  m  sli«M«p. 
i.^.  76  barrels. 


Art.  120. 

7.205. 
2.2X», 

3.  440. 

4.  315. 

5.  882. 

6.  3158. 

7.  1002. 
S.  137. 
.9.445. 

/^.  963. 
11.  4455. 
/J.  375. 
iJ.  4144. 
14.  560. 
/.5.  4661. 
16.  2247. 
/7.  108. 
75.  176. 
7.9.  276. 
20.  5362. 
2U  7967. 
.?i?.  90807. 
?J.  1234. 
U.  4(;834. 
;;A'J.  3147. 
2C.  864;J. 
.^7.  40367. 
/<S\  7967. 
29.  147rt3. 

^(y.  39407. 

31.  50406. 

^^.  105070. 


pie  examples 
Arithmetical 

\vt.  120. 

r.  205. 
^206. 
>.  440. 
815. 
832. 
8158. 
7.  1002. 
S.  137. 
.9.445. 
0.  962. 
i.  44')5. 
J.  375. 
b.  4144. 
U.  560. 

.5.  4561. 
l6\  2247. 

7.  108. 

[5.  176. 

;.9.  2T«. 

[^A  6862. 

;.  7967. 

b.  90807. 

b.  12:54. 

4r»H34. 

8147. 

8<i48. 

40367. 
'S.  7967. 
n.  147^3. 

yy.  89407. 

\l.  50400. 

fi?.  105070. 


Art.  122. 

2.  15712. 

.IT.  43  loads. 

4.  36  acres. 

J.  300  miles. 

C.  2()7  acres. 

7.  758. 

.V.  11200. 

.9.  3600 
W.  ij;T080. 
//.  ♦49151. 
J  J.  275  acres. 

Art.  140. 

2.  $iiiiH4. 
s.  |7;jr>.56. 

4.  :j;767.3«. 
r>.  120502.50. 
a.  :j;658.56. 

7,  ♦632.20. 

5.  $756. 

9.  $598.58. 
JU.  $;i31.89. 

Art.  147. 

S.  $13.  'i  $18000. 

«>.  ^o,      I  .  ♦♦*•'• 

4.  $7.    .V.  $73. 
5. 


$7'.    .V*.  $7J 
$52. 


'     2 


Art.  148. 

$1260. 

.?.  $1148. 
//.  *8172. 
5.  31104  railea. 
a.  $2496. 
7.  ♦162. 

Art.  I4». 

:?.  85  barn^l.s. 

3.  73  horse-s. 

4.  17  weeks. 

5.  \^\\\  acres. 
C.  327  thousand. 

Art.   1 50. 

?.  .{.T.l  T^iinds. 
3.  184  huxes. 


4.  4900  pounds. 
.5.  776  baskets. 
6*.  168  days. 
7.  168  cords. 

Art.  151. 

1.  25  pounds. 
J.  712  corda. 

3.  $2625. 

4.  $3;«24. 

5.  $426l 

6.  $470  gahL 

7.  $1241. 

8.  $504. 

9.  60;  13. 

W.  $2096 ;  $1861. 
//.  4714:  4262. 

12.  $4028 .  $2466. 

13.  50  days. 
/4.  $576. 
75.  $56. 

/6\  149|f  acres ; 
14811  acres. 
/7.  6  bushels. 

Art.  173. 

/.  3,  7,  2,  5,  3. 
.'.  3,  7.  2.  5. 

3.  i,  3. 3,  7. 11. 

4.  2.  3.  7,  11. 

5.  2.  2,  2.  3.  8.  113. 

6.  2,  5.  7.  3.  7. 

7.  7.  7.  81>. 

*.  2.  3. 11, J, J. 
^',  «>,  o,  0,  O,  < )  7. 

/^>.  3,  11,  2.  5,  7,  13. 
//.  5.  2.  5,  2,  2,  73. 
I  /?.  5,  5, 13,  59. 
13.  2,  2,  4007. 
74.  2.  5,  5,  5,  5. 
in.  3,3,31,  37. 

I  m.  3, 5, 2, 2, 2, 2, 2. 7. 

'  77.  2,  5.  2,  5.  2,  2.  2,  2, 
'         2. 2. 

7^.  5.  377,  43. 
/'.  2.  5,  2,  5,  2,  5. 
"'.  2.5.3.  7.  11. 
1  21.  2,  3.  7,  11,  17. 


359 

22.  2, 5, 6, 109. 
»3.  5,  2,  2,  2,  229. 
2Jt.  3,  2,  2,  2,  2.  3,  8, 

8,  8,  3. 
S5.  2,  5,  11,  73. 
Sii.  6,  7,  7.  17. 
.?7.  2,5,2,5,5,5,5,5. 
i?6.  2,5,2,5,2.5,3.8. 

8.3. 
f9.  2,5,2,5,2,5,2,2, 

2,  «,  2,  3. 
SO.  2,  5,  2,  5,  5,  7, 13. 
SU  5,  5,  5.  5,  3,  3,  3. 

52.  2,5,11,13,18. 

53,  3,  5,  2,  3,  8,  7,  13. 


Art. 

177. 

i.  16^. 

f.^' 

2.  \\. 
S.  20. 

5.200. 

4.  15U. 

5.  166j. 

c'y.  20. 

i(>.  2. 

77.  4  barrels. 

12.  42. 

7J.  6  shillings. 

U.  80  weeks. 

7.5.  80  loads. 

76.  915  IM) 

unda 

77.  5  boxes. 

Art. 

185. 

1.  5. 

7^.34. 

2.  15. 

77.  ;^. 

.y.  21. 

12.  35. 

4.  39. 

7.?.  46. 

r>.  18. 

74.  5. 

6-.  4. 

15.  8. 

7.  17. 

7ft\  12. 

S.  81. 

77.  24. 

i/.  75. 

7.V.  46. 

Art. 

192. 

7.  21. 

0'.  5. 

;?.  15. 

7.4. 

^.  22. 

cV.  3. 

4.^. 

.'/.  91. 

5.  35. 

10.^ 

w 


ly 


H' 


111 


\' 


p  ■ 

Iff  1 


,  t 


fi 


360 

Art.  199. 

1.  3  feet. 
^.  8. 

3.  12  inches. 

4.  80  feet. 

5.  13  barrels. 
G.  14  feet. 

7.  14  yards. 

8.  $88; 
214  acres. 
105  acres. 

Art.  20«. 

i.  00  cents. 

^.  2520. 

^.  80  quarts. 

4.  $7i>H. 

5.  U'tH  rows. 

6.  30  feet. 

7.  1050  feet. 

Art.  a  10. 

J.  330. 

i:'.  182. 

5.  390. 

4.  2145. 

5.  5005. 
0.  402. 
7.  0000. 
5.  2548. 

Art.  237. 

S.  fiS". 

i6-.  i^y,K 

17.  \\\ 

Art.  240. 

4.  '6^i\. 


ANSWERS, 


5, 

G. 

7. 

S. 

'J. 
10. 
11. 
U. 
IJ. 

/4. 

15. 
IG. 
17. 
IS. 


12||. 

4.  57 

*i  ore 

8er|. 

80,Vff. 
89  jV 

«3h. 

64J2. 


16  B 
ffOff" 


1  I* 

„7 


Art. 

7. 

O.    -r' 
11.    I?. 


24G. 

/^.  ^|. 

15. 
10 
17 
18.  \ 


'•  Ik. 

.  HI 


Art.  249. 


6. 

G. 
7. 

5. 

'J. 


10. 


11. 


if;  f^ 


u. 

tfC'   BO 
«<» 

4B     . 

T2fl  > 

m- 

mi 

4Hfl   . 
T20  » 

^■5  0 


•iJ 


[J 


i^. « 


OH    . 
DA  • 

Stttl  » 


m 
Ml 
m 

HI 

IHO 


6'. 

ii. 
1^. 
iJ. 

14. 
15. 
10. 
17. 

16'. 
/.''. 
:<f(J. 
?1. 


98H. 

U3a. 

82  ,V. 


4. 
5. 
G. 

7. 

8. 

9. 
10. 
11. 
12. 


15. 
10. 

/;. 

IS. 


179 A  lbs. 
28}  J  feet. 
55!!  yards. 

23.  $142. 

24.  185 [I J  yds. 

25.  14ej»'aV  mi. 

Art.  257. 

1.  fh. 

2.  \\. 

3.  2  f. 

1^ 

8,V. 

43i^ 
40|^. 
17^. 

/5'.  \h\  gallons. 
^4.  $3 A. 

5.',  barrels. 

28U. 

195  niiles. 


Art.  253. 

i.  3 


If 


ni.  4.  a 


87  1 

SO' 


5-  2^5 


/:>.  OOi  pounds, 
Art.  202. 

'//.  35!. 


U.  ISi'o- 
iJ.  50. 

i6\  72. 

17. 10, ni- 

Art.  204. 

1.  7867;  61  A; 

75U;    9; 
90. 

2.  $206; 

$777J : 
$3780 ; 
$7:T85. 

3.  $300. 

4.  142Hi{  acres. 

5.  $580. 

G.   836  j^s  yds. 

Art.  2G7. 

^'  HI;  nu: 

7.  ?• 

Art.  270. 

10.   953 i. 
//.  959|. 
12.   44571 1 . 
7^.  1534. 
i4.  74800. 
15.   73000. 

Art.  275. 

/  1^' 

^'  «• 

4.  t 

T   * 

^'.  li 

"w   4B0 
'•  11*1  IF' 
v   r.  s 


10. 


6  tl  1  • 

"a .'.  • 


i^.  iJ^ 


.  50. 
.  72. 

\rt.  204. 

.  7867;  61A; 

75U;    9; 

90. 
.  $266; 

$7775 ; 
$11780 ; 
$77785. 

.  laoo. 

.  142|;{  acres, 

.  $580. 

.  835A  yds. 

lit.  2G7. 

1^51  I     IS* 

.rt.  270. 

953  i. 

9592. 

44571 I . 

1534. 

74800. 

73600. 

At.  ti75. 

n 

1  r- 

.\. 

4 

«• 
8 
IR" 

,  1  3 

450 

1     "  -' 

hi  .1  • 

M  5- 


ANSWERS, 


i4.   ■ 


i^.  1380104. 

f  i.  126? 

£3.  500it. 
i?4   29  ct. 
S5.  69rV  tone. 

J6\  $10^f. 
S7,  2H. 

^<!'.  aw • 

Art.  278. 


7.    W0OA 

>'!  8540. 


tJi- 


7. 

^-  I  hi' 

i(^'  lif. 

V 


IS- 


53.  825^. 

54.  '  ' 
35. 
36. 
37.  11295. 

Art.  285. 
i.  2H. 

4^  iifif. 

5.  20Ja. 
6\  48071. 
7".  18  |. 

8.  50  . 

9.  4»i  f 
/O.  81  f. 
11   5f 
/^.  iLo.t. 

mil 


16.  6 

^?'.  |2U;  |22| 

$68|. 
^«?-  mil 


Art.  282. 

^4.  1|. 

26.  i|- 
:S?A  Mil. 

^^.'  m'i. 

31.  im^. 


Art.  205. 

7    1  •  A-   ft  • 

-»•    f  •    f  »      g    » 

8|.2|    2J 
3  '  8  •  3 

2    ?!.  ^.^ 

7  '    7  '  7  • 

•^'  ao  •  9* 

7  *    7  *    7  ' 

7'  7'   7* 

^'  'nio  t  TTnF » 
JUL  •  «o  • 
Too  »    Too  » 


361 

100  • 

100*  100  ' 

100  ' 
217,V 

100  • 

Art.  296. 


.J. 


100'  100' 
100'   "^' 

Toy 

100* 

10' 10' 10' 
OJJ 

10' 

S'  tWo;t'<A)%; 


II;  H;  A. 

1.  1 

3f '  ^' 

6.  SU. 

7.  H ;  if. 

.9.'  JI^'tsa. 
^^-  «Hj;  i;  A; 


r!  49. 


1000* 

9  ^m- 


7f. 

13. 

H.  $1614|. 

i5.  40  f  tons. 

16.  12l|7da. 

17.  25» 
IS.  $63|. 
/r>.  $289|. 
^(;.  47 J j  miles; 

38!  X  miles. 


'■>  Too  » 


^'   100  '100' 


.'/.  $8113,',. 
22.  656Ar. 
_  .  „..  .    S3.  $ior>oo. 

J?  .  il*J  .Uj.  11800. 

26.  123?  corrls. 
i'7.  Increased 

by  A. 


7611 
100  ' 


24 


H'. 


n    . 


:  I 


m 


362 

eS.  4|}|  Ibfl.  ; 

SI.  2U  days. 
SS.  15  feet. 
3S.  1106|f$yd& 

5^.  $289i{|. 
^5.  18  U  acres. 

S6.  %rms\. 

Art.  320. 

1.  .6. 
;?.  .4875. 

3.  .36. 

4.  .84875. 

5.  .4625. 
G.  .712. 

7.  .063125. 

8.  .5008. 

9.  .3571f 
m  .42301g. 
11.  .3538  A 
i.?.  .1^571 

13.  .0352 

14.  .0879, 
i5.  .1917tW. 

16.  .niSsYi- 

27.  .019047. 
18.  .42a57i. 
i5.  .5367142^. 
m  .^6410^. 
fSL  .S46153. 
;?5.  .33^92307. 
SS.  ..^02156862- 

7450986. 
f4.  .863. 
S5.  .430. 
i?(>.  8.tl42a5. 
57.  82  4076923. 
28. 24.32i42857. 


ANSWERS, 


i: 


Art.  332. 
1.  H.     ^.  |. 

5.  g 

i^.  iHi- 

Art.  335. 

i?.  A. 


Art.  338. 

2.  M. 


r   _i 


1  !•  a  1 


JO. 

II.  'OT. 

i:\  v^. 
/4.  V. 

15.^. 


Art.  341. 

i.  158.9656. 
J?.  572.877. 
.9.  i58a26. 

4.  492.4198. 

5.  105.9817. 

6.  16.62220. 

7.  115.2549a 

8.  $121.11. 

Art.  343. 

1.  82.0445. 
^.  782.48. 

3.  $87.26. 

4.  87.124. 

5.  $16.65. 

6.  17.49608. 

7.  874.960401. 

8.  6.08995. 

9.  .600091. 

10.  .0894097. 

11.  $72,091. 
if.  94.1881. 
IS.  17.42. 
/4.  5.552. 
15.  $170.75. 
i6.  $788.0225. 

Art.  346. 

/.  4.958. 

2.  85.77. 

3.  217.496. 

4.  $58,555. 

5.  .$4075.26. 
0.  60.285618. 

7.  286  025901. 

8.  .029418. 
5.  .80288. 

70.  .00748. 
//.  .0000352. 
U.  .000072. 
/.?.  .(X)3045. 
/4.  .001535. 
/.I.  .00000101. 
in.  24.17J. 
i7.  1344.13^. 
IS.  $1.06375. 


19.  .46075. 
*o.  .0068. 
il.  .00088575. 
S2.  $7.47891. 

23.  $685.6289. 

24.  $588.2114 

25.  22:.900|  gr. 
2G.  2.85644  yd. 

27.  52.16681  in. 

28.  $271.8017. 
2U.  1908.75. 
SO.  $5260.888  •(- 
31.  $46070. 

?.  $88.88|. 

33.  $4.96. 

34.  $862.90. 

Art.  347. 

1.  .85. 

2.  .75. 
.J.  .625. 

4.  .46875. 

5.  .825. 
6".  .1125. 

7.  .tS. 

5.  .^1428lJ. 

9.  .8^. 
/r;.  .15. 

i/.  .&0769S. 

ii'.  .07954. 

Art.  :)48. 

i.  20.5652  +  . 

2.  46.2857  +  . 
J.  8.0731 +. 

4.  8.8235  +  . 
.';.  .8040  +  . 
(I.  .8604  +  .- 
7.  5.4844  +  . 

5.  29.8661+. 
9.  9.0009  +  . 

Art.  ,352. 

/.  6.854«; 
38.3680. 


16075. 

iX)088576. 
^7.47891. 
|;685.6289. 
1(583.2114. 
•2:.tt0«|  gr. 
J.85644  yd. 
32.10681  in. 
H271.8017. 
1»08.75. 
15260.8884- 
'i46070. 
id.88|. 
4.96. 
;62.90. 

rt.  347. 

.85. 
.75. 
.626. 

.46875. 

.825. 

.1125. 

.'71428b. 

.8^. 
.15. 

.^709i^. 

079M. 

t,  :i48. 

120.5052  +  . 
46.2857  +  . 
8.0721 +  . 
13.8235  +  . 

.8040  +  . 
.8604  +  .- 
i.4844  +  . 
59.8001 +. 
.1)009  +  . 

Irt.  352. 

1854«; 
}8.3080. 


f.  17.0709. 

5.  1.0979. 

4.  1960.5945. 

6.  .7583. 
(;.  6.08JJ3. 

7.  160.6284. 

5.  142857.142a 
9.  41710. 

20.  1.4743. 
J  I.  161.2496. 
1?.  73.3743. 

13.  89.5901. 

14.  |5.8:m2. 

1,5.  101.1879  yd. 
in.  5.1309. 

17.  17!>.78<I0  lb. 

18.  32.9085  long. 

19.  214.2327  yd. 

50.  .37s. 

51.  $2320.4678. 

S^.  $13450. 

Art.  :M3. 

1.  $332,325. 
f.  $.7446. 

3.  $43,875. 

4.  $35.1125. 

5.  $809.58. 
6'.  $281,567. 
7.  $53.8135. 
S.  .8888  +  . 
9.  .9100  +  . 

10.  .4888  +  . 

11.  .5. 

IL^.  .5050+. 
li.  .7714  +  . 

14.  .39. 

15.  .0199  +  . 
KJ.  .3478200+. 
17.  1.7000 +  . 
IS.  .08177  +  . 
19.  .3920  +  . 
SU.  $821.0125. 
SI.  11232.81 10  +  . 
SS.  $94.22. 

S.^.  $3188.005. 
C'4.  $5.5500  +  . 


ANHWERS, 

25.  59.75  vards. 
20.  01.44  vurils; 

82.5800+  yarda; 

48.1):39+  vanls. 
27,  439  bushefa. 
-.V.  lU.    ^ 
29.  100.110  pounds. 
:io.  $14.').52. 
.;/.  $801,785. 
J2,  $1.85. 

Art.  3««. 

5.  1885  drams. 

r>.  27700  pounds. 

7.  103240  grains. 

<v.  21900  grains. 

9.  158328  grains. 
10.  34a'>47  ix)unds. 
//.  217339  graina 

12.  175393  pounds. 
Li.  5749  dwt. 

14.  588S000  ounces. 

15.  7000  grains. 
iC.  39377  dwt. 

Art.  300. 

5.  438  T.     4   cwt. 

451b. 
G.  0  lb.  9  oz.  8  dwt 

12  gr. 

7.  1017  lb.  2  ot. 

8.  16  T.      14    cwt. 
77  lb.  8  oa. 

P.  48  lb.  6  oz.  4  dr. 
2  sc.  7  gr. 

10.  35  lb.  9  oz.  17  dwt 

11.  173  T.  8  cwt 

13.  1504  dr.      1    sc. 

15  gr. 
IS.  92  lb.  9  oz.  1  dwt. 
11  gr. 

14.  27  T.  0  cwt.  4  oz. 

15.  1011).  3oz.  Idwt 
21gr.;13i''^lb. 
10  lb.  8  oz.  2  sc 

5gr. 

16.  70  lb.  3  oz  \  dwt 

23i  gr. 


363 

17.  T,V,  lb. 

IS.  17  lb.  0  oz.  4  dr. 

1  80.  21  gr. 
19.  51b.  310  -0  D2 

8gr. 

Art.  372, 

/.  32  2  gr. 

2.  11   cwt.     11    lb 

1}  oz. 
J.  8  oz.      14    dwt. 

13,Vgr. 

4.  9.0  oz. 

5.  58    34    T)i 
14?  gr. 

6".  17  cwt. 
7.  14  dwt.  14.4  gr. 
S.  £>2  10.4  gr. 
9.  15  lb   10  oz. 

10.  58  lb.  5J  oz. 

11.  7  lb.  5  oz.  0  dwt. 

10  gr. 

12.  8  cwt.  70  lb. 

13.  13  T.      14   t  wt. 

28  lb.  9>  oz. 

14.  lb.   5     3  11     33 

14.4  gr. 

15.  8  oz.     10  dwt. 

17  004  gr. 

16.  18  cwt.    71    lb. 
8.2  oz. 

17.  lib.  15 dwt.  5gr. 

18.  102  lb.  1  oz. 

19.  80  lb.    9  oz 

13  dwt  8  gr.  ; 
lb.80|9  35  31. 

Art.  375. 


0. 


11000ft 


T. 


A'off  '''• 


T. 


.    I 


10.  gj^lb.; 

^^-  sffino  '»*•  >  sift  II*' 
1-'-  ^  T.  ;  ,«„  t.  : 

TTsW  ^'  •  TnOO  ^  •» 
s         T 

iJ.  A  lb. ;  rt^5  cwt. 


''  I 


I! 


■^^ 


IMAGE  EVALUATION 
TEST  TARGET  {MT-3) 


I 


1.0 


I.I 


M    12.0 


12.2 


1.8 


1.25 

1.4      1.6 

■• 6"     

► 

Photographic 

Sciences 

Corporation 


23  WEST  MAIN  STREET 

WEBSTER,  N.Y.  14S80 

(716)  872-4503 


V 


iV 


40^ 


^\ 


.<-*u 


9> 


V 


^^ 


#; 


c^ 


■'^ 


4if     C 


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364 


ANSWERS. 


H 


.1 1 


h 


I..  J 


'J- 


m 

ff! 

|fel . 

yu^.. 

Art.  378. 

i.  IH  lb. 
2.  /A  T. 
5.  m  cwt. 

4.  tJS-  lb. 

5.  Ill  lb. 

6.  ^A  lb. 

7.  if  lb. 
5.  tWs  lb. 

5-  Hie  lb. 

10.  H- 

12.  1. 
i.?.  .05375. 
/^.  .122|. 
15.  .4455  T. 
26.  .9868+  lb. 
17.  .1138|. 
i<9.  .81875  lb. 
19.  .634  T. 
^0.  .8121527J  lb. 
21.  .7361  j  lb. 
^2.  .71498  +  . 
23.  .5944f 

Art.  380. 

1.  74  T.  3  qr.  4  lb. 

10  oz. 

2.  33  lb.  2  oz.  1  sc. 

12  gr. 
S.  1  lb.  3  oz.  18  dwt. 
20f  gr. 

4.  49  cwt.  2  qr.  6  lb. 

5.  5    cwt.     20   lb. 

12^  oz. 
6\  4  T.  3  cwt.  9  lb. 
10  oz. 

7.  Ib.l4  53  34  32 

14  gr. 

8.  10  oz.     4    dwt. 

9.0  gr. 
5.  lb.  40  §10  31. 

10.  87  lb.    2  oz. 

12  dwt.  18  gr. 

11.  10  T.     11   cwt 

42  lb.  15  oz. 


12.  8  lb.    11   oz. 

19  dwt.  15.2  gr. 

13.  80  lb.  .5  oz. 

14.  IT.  14 cwt. 921b. 

14  oz. 

Art.  382. 

1.  11   lb.    4  oz. 
13  dwt.  12  gr. 
;?.  lb.  5  59  3  6  32 

3.  UT.  18cwt.lqr. 

21  lb.  3  oz. 

4.  5  lb.  10  oz.  7  dwt. 

5.  1  T.  18  cwt.  62  lb. 

8  oz. 
G.  5  6  3  7  10  gr. 
7.  4  lb.  6  oz.  11  dwt. 

12  gr. 
.?.  7  T.  7  cwt.  45  lb. 
9.  9  oz.    8  dwt. 

16.8  gr. 

10.  3  cwt.    64  lb. 

12  oz. 

11.  lb.  7  55  3  6  31 

5gr. 

12.  lb.3  511  3132 

10  gr. 

13.  6  T.  12  cwt.  2  qr. 

141b. 

Art.  384. 

1.  3  T.  9  cwt.  14  lb. 

13  oz. ; 

4  T.  12  cwt.  19  lb. 
12  oz. ; 
24  T.  4  cwt.  8  lb. 

11  oz. 

2.  15  lb.    2  o2. 

10  dwt.  20  gr. ; 
261b.  7  oz.  8  dwt. 
23  gr. 

3.  18  T.     17   cwt. 

521b. 

4.  7  cwt.  66  lb.  5  oz. 

5.  62  lb.  4  oz.  7  dr. 

15  gr. 


6.  1  cwt.    19  lb. 
12|  oz. 

7.  25  lb.  14  oz. 

^.  3  T.  6  cwt.  92  lb. 
9.  54  lb.  12  oz. 

10.  1  cwt.    7   lb. 
14f  oz. 

11.  6  T.  3  cwt.  12  lb. 

12.  32  11.12  gr. 

13.  2  cwt.    59  lb. 

6.72  oz. 

14.  9  cwt.    72  lb. 
3f  oz. 

15.  8H  oz. 

16.  50  lb. 

17.  19.44  gr. 

IS.  78T.3cwt.161b. 
19.  40  T.  5  cwt.  92  lb. 

Art.  385. 

1.  4  lb.  a  oz.  16  dwt. 

13|  gr. 

2.  4  T.  17  cwt.  3  qr. 
21  lb.  8  oz.  ; 
IT.  19  cwt.  181b. 


9|  oz. 


1  qr. 


1  T.  4  cwt. 

24  lb.  2  oz. 

16  cwt.  1  qr.  7  lb. 

12  oa. 
5.  3  1  5  gr. 
4.510  3710  gr.; 

57  32  6fgr.; 

53  35  3igr.; 


5. 

6. 

7. 

8. 

9. 
10. 
11. 


5l323213|gr. 
1  cwt.    10  lb. 
IJfloz. 
34, 

I  bo: 


35  boxes. 
25 

5ii,v ;  1920. 

52  3  5  31. 
12.  i2413|f ;  lOf- 

Art.  389. 

i.  21.86611+  mi. 


cwt.    19  lb. 
l^  oz. 

I  lb.  14  oz. 
T.  6  cwt.  92  lb. 
t  lb.  12  oz. 

cwt.    7  lb. 
4|oz. 

T.  3  cwt.  12  lb. 
)2  11.12  gr. 

cwt.    59  lb. 
;.72  oz. 

cwt.    72  lb. 
J|  oz. 

II  oz. 
10  lb. 
.9.44  gr. 
'8T.3cwt.l61b. 

to  T.  5  cwt.  92  lb. 

Art.  385. 

Ub.aoz.l6dwt. 

13?  gr. 

4T.  17  cwt.  3  qr. 

21  lb.  8  oz. ; 
LT.  19  cwt.  181b. 

9|oz.; 

L  T.  4  cwt.  1  qr. 

24  lb.  2  oz. 

6  cwt.  1  qr.  7  lb.     I 

12  oa.  ' 

1  5gr. 

10  3  710gr.; 

7  3  2  6|  gr. ; 

3  3  5  3^  gr. ; 

I3  23213igr. 
cwt.    10  lb. 


{do: 


xes. 
>5. 
)11,^^ ;  1920. 

2  36  31. 

413|i;  lOf. 

,rt.  389. 

51.&65U+  ml* 


2.  .5280  yd. 
J.  1386  in. 

4.  2100  1. 

5.  84f  in. 
e.  5.5f  1. 

7.  1087.5  mi. 

S.  4.3866+   statute 

mi. 
9.  2544  in. 
10.  245  statute  mi. 
//.  165751. 
1^.  233  rd.    3  yd. 

10.8  in. 
J3.  .000482+  mi. 

14.  .4646. 

15.  2  mL  48  ch.  3  rd. 
24 1. 1.08  in. 

16.  66  ch.  2  rd.  7  1. 

1.76  in. 

17.  10  mi.  14  ch.  5  1. 

18.  7  mi.    258  rd, 

2  yd.  2H  ft. 

19.  10a5.916  St.  mi. 

20.  2\  degrees ; 
155.625  St.  mi. 

SI.  w- 
SS.  .66  ft. 

33.  145.83  St.  mi. 

24.  .(m6. 

25.  .036  mi 

Art.  398. 

1.  4017.2  sq.  yd. 

2.  15488000  sq.  yd. 

3.  180680  sq.  ft. 

4.  8000  sq.  I. 

5.  448000  sq.  1. 

6.  19200  sq.  ch. 

7.  .00516528+  A. 

8.  .0001367  + 
sq.  mi. 

9.  284   A.      71    P. 

3  sq.  yd.  3  sq.  ft. 
36  sq.  in. 

10.  21.78  sq.  ft. 


ANSWERS, 

11.  25  sq.  mi.  457  A. 

1   sq.  ch.   6   P. 
535fc  sq.  1. 

12.  81  P.  27  sq.  yd. 
7  sq.  ft.  67.68 
sq.  in. 

IS.  4  A.  68  P.  6  sq. 
ft.  32H  in. 

14.  1  Tp.  12  8q.  mi. 
188  A.  9  sq.  ch. 
IP. 

15.  4  sq.  mi.  319  A. 
9  sq.  ch.  8  P. 
458^  sq.  1. 

IB.  9  sq.  ch.   15  P. 
624  sq.  1. 
35  sq.  mi.  639  A. 
9  sq.  ch.  15  P. 
624  sq.  1. 

17.  140  A.  6  sq.  ch. 

13  P.  89f  sq.  I. 

18.  5  P.  28  sq.  yd.  6 

sq.  ft.  108  sq.in. 

19.  9  P.  5  sq.  yd.  4 
sq.  ft.  108  sq.  in. 

20.  5  P.  29  sq.  yd.  1 

sq.  ft.  18  sq.  in. 

21.  2  A.  3  sq.  ch.  1  P. 

376  sq.  1. 

22.  18  sq.  yd.  1  sq.  ft. 
572  sq.  in. 

23.  1  P.  2  sq.  yd.  8 
sq.  ft.  14.4  sq.  in. 

24.  309i  yds. 

25.  384  boards. 

26.  2230  sq.  ft. 

27.  $620. 

28.  $485,275. 

29.  1096.98  tiles. 

30.  11352  shingles. 

31.  $500.76. 

32.  1200  stones; 
$2120.725. 

33.  $50.53. 

Alt.  406. 

i.  167616  cu.  in. 


365 

2.  2.43  cu.  ft. 

3.  867456  cu.  in. 

4.  32.6592  cu.  in. 

5.  18  cu.  ft.  1080 
cu.  in. 

6'.  .0296  cu.  yd. 

7.  .02*>  cu.  yd. 

8.  9  cu.  ft.  648 
cu.  in. 

9.  4  cu.  yd.  8  cu.  ft. 
432  cu.  in. 

10.  58  cu.  yd.  21  cu. 

ft.  1664  cu.  in. 

11.  3  cu.  yd.  18 

cu.  ft. 

12.  29  cu.  yd.  22  cu. 
ft.  576  cu.  in. 

13.  5cu.yd.l4cu.  ft. 

1080  cu.  in. 

14.  13  ft.  5 J  in. 

15.  48  ft. 

16.  827851  cu.  yd. 

17.  92808U  cu.  ft. 

18.  4  ft.  6  in. 

19.  mil  cu.  ft. 

20.  598JHCU.  yd. 

21.  $233.6244  +  . 

22.  87|«pch. 

Art.  407. 

1.  336  cd.  ft. 

2.  1584  cu.  ft. 

3.  4608  cu.  ft. 

4.  20.79  cu.  ft. 

5.  91?  cu.  ft. 

6.  .005703125  cd. 

7.  2.6337448  + 
cu.  yd. 

8.  .03232  pch 

9.  134.S2098  +  . 

10.  .1111. 

11.  2cd.3cd.ft.8cu. 

ft.  972  cu.  in. 

12.  16  cd.  7  cd.  ft. 
15  cu.  ft.  345^ 
cu.  in. 


366 


AJVSWEBS. 


'1,1  If 


pi 


m 


tn 


13.  91 1  cd. 

14.  $53.0578J. 
i5.  8  ft.  6f  in. 
10.  108,*Tpch. 
17,  mOti  cd. 


i- 


Hi 


Alt.  410. 

1.  26  B.  ft. 

2.  35  B.  ft. 

3.  24  B.  ft. 

4.  24  B.  ft. 

5.  28  B.  ft.  2  B.  in. 

6.  42  B.  ft.  6  B.  in. 

7.  7  B.  ft.  6  B.  in. 

8.  42  B,  ft.  6  B.  in. 

9.  29  B.  ft.  2  B.  in. 

10.  18  B.  ft.  9  B.  in. 

11.  40  B.  ft.  6  B.  in. 
le.  44  B.  ft.  4  B.  in. 
13.  32  B.  ft.  7  B.  in. 
U.  13  B.  ft.  6  B.  in. 

15.  504  B.  ft. 

16.  36  ft. 

17.  31  ft.  2|  in. 

18.  $70.98. 

19.  *24  80625. 
go.  $85,008. 

Art.  412. 

1.  32  gi. :  64  gi. ; 

56  gi.;  28  gi.; 

120  gi. 
e.   128  pt.;  58  pt.; 

110  pt. 
5.  O.  6  f  5  5  f  3  2 

TTV15. 
4.  2  bu.  3  pk.  1  pt. ; 

4  bu.  2  pk.  3  qt. 
1  pt.; 

9  bu.  7  q,t. 


6  bu.  1  pk.  6  qt. 

1  pt. 
11  bu.  2  qt. 
6.   871Hbbl. 

6.  369 1 J  bush. 

7.  Cong.  .13777- 

66  +  . 

8.  1463f^  bu. 
.9.  56|f^  bbl. 

10.  Cong.  7.46972  +  . 

11.  .875  gal. 
IS.   .064453125. 

13.  .00236  +  . 

14.  1523. 21  f. 

15.  $424.52. 

Alt.  413. 

1.   125.958+  bu. 
^.  1244111  cu.  ft. 

3.  1282l|  cu.  ft. 

4.  505  A  cu.  ft. 

5.  2462.431  cu.  ft. 

6.  6011  cu.  ft. 

7.  1001.475 +  CU.  ft. 

8.  7.085+  ft. 

9.  144^*^  oz. 

Art.  417. 

1.  66  yr.  6  mo. 

13  da.  10  hr. 
S.  22  yr.  3  mo. 

14  da.  14  br. 

3.  99  yr.  4  mo.  19  da. 

4.  118  yr.  5  mo. 

17  da.  7  hr. 

5.  7  yr.-  9  mo.  1  da. 

6.  Feb.  22,  1732. 

Art.  422. 

1.  385740". 

2.  30600'. 

3.  19663 '. 

5.  .025  cir. 

6.  9  s.  28'  48'. 

7.  A  ^ 'ir. ;  i  Cir. ; 
i  Cir. 


5.  64"  17'  8|". 
9.  2  sextants;  2i ; 
2| ;  4^  ;  m 

10.  33  sex.  47°. 

11.  2  s.  12". 

12.  883  yr.  8  mo. 

20  da. 

13.  94  doz. ;  165  doz. 

14.  1264  doz.;  2280 

doz. 

15.  29|  doz.;  327| 
doz. ;  141|^  doz. 

IG.   13320  sheets. 

17.   .333. 

Ai-t.  427. 

1.  2736  far. 

2.  $83,745. 

3.  £.0375. 

4.  34000  mills. 

5.  £.002916|. 
G.   .015. 

7.  12s.  6d. 

5.  16s.  9d.  2.4  far. 

9.   £1713  88.  lid. 

10.  $457,024. 

11.  £89  lis.  9id. 

12.  $10461.564. 

13.  $1172.0965}; 
4914  marks 
45.08+  pfennig; 
6073  fr.  3  ct. 

8.9+  m. 

14.  .417375. 

15.  .4632. 

IG.  647  f  r.  6  dc.  6  ct. 
8.3+  m. 


o 

3. 

4- 
6. 
6. 

7. 


Art.  440. 

38.1024  Kg. 
33.5627  Ton. 
33.8304  HI. 
2.2185  L. 
68.0494  St. 
13274.16  A. 
250  cu.  in. 


ANSWFBS. 


367 


If 


eextauts;  2^; 

J  sex.  47°. 

s.  12  . 

33  yr.  8  mo. 

30  da. 

4  doz. ;  165  doz. 

264  doz.;  2280 

doz. 

9|  doz.;  327| 

loz. ;  141}  doz. 

3320  sheets. 

333. 

I.rt.  427. 

5736  far. 

B83.745. 

E.0375. 

34000  mills. . 

£.002916|. 

.015. 

128.  6d. 

16s.  9d.  2.4  far. 

£1713  88.  lid. 

S457.024. 

£89  Us.  8id. 

$10461.564. 
$1172.0965}  ; 
4914  marks 
45.08+  pfennig; 
6073  fr.   3  ct. 

8.9+  m. 
.417375. 
.4632. 
647  f  r.  6  dc.  6  ct. 

8.3+  m. 

Art.  440. 

38.1024  Kg. 
33.5627  Ton. 
33.8304  HI. 
2.2185  L. 
68.0494  St. 
13274.16  A. 
250  cu.  in. 


8.  38  sq.  yd. 

9.  7.7353+  bu. 
10.  652960  gr. 
n.  36.8965+  od. 
J2.  1176.15+  cu.ft 
IS.  $.024624. 
U.  $1.41975. 

15.  $14.5675  +  . 

16.  $1,357  +  . 

17.  8340  dg. 

83.4  Dg. 

18.  8400  L. 

840000  cl. 

19.  790.75+  mi. 
eo.  $960. 
ei.  $74.00048  +  . 

Art.  449. 

7.  153  sq.  ft.  8' 9"  3'" 

8"" 

8.  212  sq.  ft.  9'  11" 

2'"  S"". 

9.  146  sq.  ft.  2'  1" 

W"  S"". 

10.  376  sq.  ft.  2'  2'  8'" 

6"". 

11.  236  sq.  ft.  9'  11" 

2'"  9"".  ' 

12.  915  sq.  ft.  8". 
IS.  156  cu.  ft.  r  7"  7'" 

14'  216  cu.  ft. '8'  11" 

2'"  9"". 


Art.  466. 

i.  8400. 

2.  76000. 

3.  573. 

4.  38097. 

5.  897.52. 

6.  30084. 

7.  3426000. 
\8.  720000000. 

5.463000000. 

Art.  457. 

1.  300800. 

2.  18018. 
S.  350000. 

4.  1673800. 

5.  24473.6. 

6.  11500000. 

7.  27360000. 

8.  414000. 

9.  3141. 

Art.  458. 

1.  6628122. 

2.  385605. 

3.  87546366. 

4.  53111520. 

5.  839516040. 

6.  2585632.  I 

7.  7349272.  I 

8.  62767170. 

9.  4388206. 


4.  .013. 

5.  .008. 
6'.  .0004. 

7.  .674*. 

8.  2.7041*. 

9.  .013921. 

Art.  461. 


Art.  453. 

1.  10  min.  45|  sec. 

5.  3  hr.  58  min.  1  sec 
3.  54  min.  16Y*y  sec. 
4-   2  hr.  6  min.  59  sec. 

6.  10hr.53min.65|j 

sec. 

6.  6  hr.  40  min.  10  sec. 

7.  107°  19'  48 j". 

8.  8hp.38min.24wc 

9.  90°  2'  30". 
10.   43°  45'. 


Art.  459. 

1.  87.36. 

2.  43.72. 

3.  7.903. 

4.  .02397. 

5.  .05236. 

6.  .0006934. 

7.  .0054. 

8.  .00007. 

9.  .0072. 

Art.  460. 

/.  183.8. 

2.  6.54625. 

3.  .1098i. 


Art.  463. 

1.  103. 

2.  30. 

3.  28. 
4'  25. 
5.  112. 

6.  u. 

7.  50. 

8.  80. 

9.  158f. 


Art.  468. 

'.  15300. 

2.  111875. 

3.  596| 

4.  48700. 

5.  78000. 
e.  9655|. 

7.  8000. 

8.  14725. 

9.  1090. 
10.   14.6. 
/e.  $980; 

$1470; 
$735, 

17.  $425 ; 
$550 ; 
$612.50; 
$5334. 

18.  $600; 
1315 ; 
$910; 
$700; 

19.  $630; 
$675; 
$650. 

20.  *1112; 
$1251 : 
$1042.50; 
$973. 

21.  $112; 
$128: 
$149,331. 

22.  $1110. 

23.  $371.25  ; 
$309.37^; 
$433.12  i; 

24.  $522;  ■ 
$543.75; 
$606; 
$761.25. 

25.  $548. 
$657.60, 
$1315.20; 
$1534.40. 

26.  $255; 
$510; 
$408. 


■  \  ■ 


f  *  r      *        '1 


■  ■    ■»' 


rii,'l:l 


a 


;i 


Ti 


H    '• 


368 

Art.  469. 

i.  217.74. 
^.  6025. 
S.  1952.8. 
^.  2619. 
J.  8.1584. 
€.  1.9708. 
7.  11.388. 
S.  1.8434. 
5.  3.62552. 
iO.  219.288. 

15.  358  16  A. 
179.08  A. 
268.62  A. 
8.954  A. ; 
71.632  A.; 
537.24  A. ; 
35.816  A. 

16.  6083.2  bu. ; 
2541.6  bu. 

17.  7926  yd. ; 
18494  yd. ; 
10568  yd. ; 
792.6  vd. ; 
1056.8  yd.  ; 
2377.8  yd.  ; 
1849.4  yd. 

18.  $1725 ; 
$1971f  ; 
$2300; 
$3450; 
$1863  ; 
$2116 ; 
$2380.50. 

Art.  483. 

10.  $14444^. 

11.  $87. 

12.  $56. 
IS.  $35.57|. 
U.  $5.68U. 

15.  $17,791. 

16.  $4656.88f . 

17.  $143.46f . 

18.  $52.66|. 

19.  $19.1  ItV. 
SO.  $940.95. 


ANSWERS, 


o. 


Art.  484. 

6.  $78.66  +  . 

7.  $1.4826. 

8.  $.3811. 
d.  $4,375. 

10.  $15.50. 
U.  $.2758  +  . 
12.  $2.75. 

Art.  485. 

1.  $4.05. 

$7812  ; 

$12361U. 

$668.25 ; 

$7G7.88  ; 

$816.48. 
I^.  $1163.35. 

5.  $465.75 ; 
$465.75. 

6.  $3.4546. 

7.  $1346.2295. 

8.  $273.12525. 

9.  $379.64024. 

Art.  486. 

5.  51  rd. 

6.  626  bu. ; 
516H  bu.  ; 
630  bu. ; 
5341^  bu. ; 
468y^  bu, 

7.  274  yd. 

8.  476  bu. 

9.  285  lb. 

10.  395  yd. 

11.  217  A. 

Art.  487. 

1.  30  yd. 

2.  96  ft. 

3.  38|  bu. ; 
40  bu. 

4.  37T.; 

5.  483yd.  * 

6.  9  ft. 

7.  53  od. 
^.«9. 


Art.  488. 

6.  $1886. 

7.  $685.50. 
*.  $847. 09|. 
9.  $5075. 

iO.  $120.25. 
11.  $40455. 

Art.  489. 

^-  * ;  f ;  ^  f 

4.  I- 

5.  18f. 

10.  -j^  yr. ; 

\l1       898 

^4. 1 ;  f 
i5.  }|,  or  .68. 

Art.  490. 

;?.  20.4;  77.76; 

604. 
^.  42  ;  84  ; 

329 ;  905|. 
-^$60; 

$641.36f : 

5.  22.96. 

6.  78i  A. ; 
82  bu. 

7.  $58.88; 
15.80  lb. 

8.  25.02  lb. 

9.  15  men. 

10.  1.96  yd. 

11.  13.251  mi. 
1^.  91.50. 

13.  $6.45. 

14.  $837.3. 

15.  $347.75. 

16.  120.7  yd. 


17.  244  lb. 
15.  269.86  A. 
19.  84  ct. 
;?0.  $1173 ; 
$4  26  per  yd. 

Art.  491. 

1.  108. 
;?.  324  yd. 

3.  $45 ;  $63. 

4.  $1323. 

5.  $23680. 

Art.  492. 

7.200. 
5.400. 
9.  1200. 

10.  800. 

11.  2400  yd. 

12.  700  bu. 
1^.  20. 

14.  $9H. 

15.  6U  pk. 

16.  14  ft. 

17.  2k 
15.4. 
15>.  42|. 
^6?.  $36. 
21.  874. 
^^.550. 
^^-  10  yd. 
2k.  8. 

^5.  3100  lb. 
26.  $1.50. 
^7.  $35500. 
28.  $75. 
J?5.  1. 

.ro!  $412500. 
31.   $75. 
5iF.  $1350. 
33.   $83381. 
^4.  300  A. 
35.  $8000. 

Art.  494. 

7.  20%. 

8.  16|^. 


r.  24i  lb. 
?.  259.86  A. 
h  84  ct. 
?.  $1173 ; 
(4  25  per  yd. 

Art.  491. 

1.  108. 

2.  324  yd. 

9.  |45 ;  $63. 
^.  $1323. 
5.  $23580. 

Art.  492. 

7.200. 
S.  400. 
9.  1200. 
9.  800. 

1.  2400  yd. 

2.  700  bu. 
S.  20. 

4.  $9H. 

7.2\. 

?.  4. 
^42|. 
^  $36. 

m 


ANSWERS, 


10  yd. 

8. 

3100  lb. 

$1.50. 

$35500. 

$75. 

1. 

$412500. 

$75. 

$1350. 

$8338i. 

300  A. 

$8000. 

Lrt.  494. 

20%. 
16|^. 


9,  716 fe' 

10.  iii%. 

11.  16}%. 

12.  29}%. 

13.  85f  %  ; 

55%; 

10?%. 
U-  95 
15.  44 
26.  42U%. 
i7.  261%. 
18.  liH%. 
i9.  14||^%. 
^0.  39i|% 
^/.  8%. 
J?^.  76A%. 
23.  6l||%. 
;?4.  10%. 
25.  331%. 
^e.  13}%. 

^7.  6im%. 

28.  15|%. 

Art.  496. 

<?.  800  men. 

7.  184 

^.  5(J0  mi. 

9.  872. 
iO.  216. 
ii.  $1000. 
12.   $6.40. 
i5.  $2000. 
U.  $4000. 

Art.  504. 

7.  $3.04. 

2.  $74.48. 
A  $81.60. 
U.  $158.23121. 
6.  $116,574. 

6.  $3278.96^. 

7.  15%. 

8.  20%. 

9.  26A%- 
iO.  $6356.862. 
ii.  $.78. 


IS.  $77.40  ga. 
$6.80   sell- 

ing  price. 
$8.25. 
$5.12^. 
$7.14. 
$5860. 
$9.50    buy. 

ing  price; 
$7.12i  aeU- 

ing  price. 
$9.75. 
$1825. 
20%. 
$570.24. 
$153. 
$2475. 
$3.20  per 

yd. 
$4.23|  per 

cd. 
$.76|j*r. 

20%: 


IS. 

15. 
16. 
17. 


18. 
19. 
20. 
21. 
22. 
23. 
24. 

25. 

26. 
27. 


Art.  511. 

1.  $96.95952. 

2.  $18.3105. 

3.  $57.75. 

4.  $14. 13^. 

5.2|%. 

e.  4|%. 

7.  2|%. 

8.  $14400. 

9.  $47178.12^. 

10.  $1652.92A- 

11.  $8663.21  AV 

12.  $1863.97 +  . 

13.  $3653.70  +  . 
U.  $3286. 

15.  $596. 
i^.  $384048+. 
17.  $24.60. 
18. 112  bbl. 

19.  $876,435. 

20.  $13.82. 

21.  $6761.882  + 

22.  $129,376. 


Art.  516. 

i.  $105. 
2.  $199,371. 
S.  $148.58f. 

4.  \%. 

5.  |22711f 

6.  $6000. 

7.  1%. 

8.  $250. 66|. 

9.  $148.75. 
10.  $1948.80. 

Art.  631. 

2.  $8965.50. 

3.  $10105.82^. 

4.  76  shares. 

5.  96  shares. 
G.  144  shares, 

7.  153  shares. 

8.  $15680. 

0.  $16462.50. 

10.  $40090. 

11.  $59800. 

12.  $70. 

13.  $82.50. 

14.  $58,334. 

15.  $128.57f 

16.  $393. 

17.  $85020. 

18.  $448.47. 

19.  $583.05. 

20.  $451.11^. 

21.  $964. 
?.  $284,982  +  . 

23.  $368. 

24.  $1160. 

25.  281  shares. 

26.  $1639. 

27.  6%. 

28.  $191.80. 

29.  $3730. 

^(?.  iin%. 

5i.  $75. 
32.  $62i. 

34.  $5950. 
JJ.  $680. 


369 

36.  $800,831  +  ; 
$780,821  +  ; 

^  $124,931  +  . 

37.  $8, 

Art.  537. 

/.  $268.80. 

2.  $265,192 

3.  $622.68J. 
4'  $73.60. 
6.  $66.6792. 

Art.  549. 

13.  .V282.38. 

14.  $51.5256. 

15.  $8.3695. 

16.  $41.78265. 

17.  $1.6559  +  . 

18.  $86.8208  +  . 

19.  $18.25248. 

20.  $85,115. 

21.  $107.1144 

22.  $462,016. 

23.  $827.08. 

24.  $4  2301. 

25.  $97.1694. 

26.  $6,736. 

Art.  553. 

1.  $118.1074 ; 
$146.4238. 

2.  $63.048 ; 
$89,318. 

3.  $118.3442 ; 
$73,965  +  . 

4.  $815,976  + 
$1078.254  + 

5.  $641775; 
$106.9625. 

6.  $292.3719. 

7.  $49,529  +  . 

8.  $410,475  +  . 

9.  $1094.096. 
10.  $1699.80  +  . 

Art.  556. 

1.  $48,675. 

2.  $35.84. 


,/'ydlL 


■■p 


§4    't^:' 


iM 


I-  • 


370 

5.  143.812. 

4.  $28.1385. 

6.  $13,754. 

6.  $35.82. 

7.  $130. 

8.  $40.20. 
d.  $46.50. 

i(7.  $100,395. 

Art.  559. 

1.  $1,536. 

2.  $1.58. 

5.  $2,125. 
4.  $1.54|. 

6.  $4.2075. 

6.  $1,849}. 

7.  $3.775|. 
^.  $2.67|. 

9.  $1.96. 
10.  $1,122. 

Art.  561. 

1.  $303.9513; 
$434.216 ; 
$178.6875. 

2.  $62.36352 ; 
$93.54528 ; 
$83.15186. 

3.  $22.2609 ; 
$11.1301 ; 
$19.4783. 

4.  $45.4765; 
$57.7202; 
$66.4656. 

5.  $113.40; 
$151.20 ; 
$56.70. 

Art.  666. 

1.  $63  6533 ; 

$93.9800  ; 

$41.7689. 
S.  $24.65  ; 

$39.44 ; 

$34.51. 
S,  $291.695 ; 

$458.378 ; 

$125.0123. 


ANSWSBS. 


4.  $65.9458; 

$50.3270; 

$19.0895. 
6.  $103.44004 ; 

$131.2892; 

$57.6877. 
6.  $376.6188 ; 

$502.1577; 

$318.8486. 

"1.85 ; 
.6037  ; 
1.5074. 

Art.  668. 

i.  $3.1342. 

2.  $11.5436. 

3.  $3.1574. 

4.  $3.3082. 

5.  $3.3945. 

6.  $6.54407. 

7.  $3.6073. 

8.  $1.4576. 

9.  $.8939. 

10.  $16.2754. 

11.  $461,193. 
U.  $25.2125. 

Art.  571. 

2.  $283,992  +  . 

3.  $364.9987+. 

4.  $462,019. 

5.  $662,984 

6.  $434,994. 

7.  $296. 

Art.  574. 

2.  $49,652  +  . 

3.  $264,998  +  . 

4.  $572,996  +  . 

5.  $295,996  +  . 

Art.  577. 

2.  7%. 

3.  6%. 

4.  6%. 

5.  8%. 

€.  8k^%  nearly. 


11. 
12. 


7.  12%. 
8.\%  better 

2d. 
5.  14f%. 
10.  26 fo;t2^%; 

4%. 

9A%. 
33^%;  60%; 

225%; 
33^%; 
14f%; 
21?%; 
lli%. 
16|%. 

Art.  680. 

i.  2  yr.  3  mo. 

2.  8  yr.  6  mo. 

3.  6yr. 

4.  4  yr.  9  mo. 

5.  5  yr.  7  mo. 

6  da. 

6.  2  yr.  8  mo. 

7.  6  yr.  4  mo. 

24+  da. 
^.  7  yr.  6  mo. 

25  d& 
9.  14f  yr. 

10.  71f  yr. 

11.  20  yr. ; 
12i  yr. ; 
15fs  yr. ; 
Hi  yr. ; 
40  yr.; 
25  yr. ; 
30ft  yr. ; 
22|  yr. 

Art.  684. 

1.  $31.390166  + 

2.  $977,532. 

3.  $146.27795. 

4.  $1864.576. 

5.  $20.0034. 

6.  $647.2378+. 

7.  $205,616+. 


8.  $440824+  gr. 

at  comp.  int. 

9.  $108,595  +  . 

Art.  587. 

1.  $219,658. 

2.  $183.0183. 

3.  $133.55053. 

4.  $1148721672. 
6.  $55.4364 

6.  $992669725. 

Art.  589. 

1.  $1220.2528. 

2.  $469.53704. 

3.  $781.52013. 

4.  $566.8662. 
6.  $2755.8606. 

6.  $248.1272192 

7.  $257.299443. 

8.  $145.728068. 

9.  $228.8346. 

10.  $556.75033. 

11.  $3439.63075. 

12.  $758.952567. 

13.  $854.942736. 

14.  $1097.5152. 

Art.  692. 

1.  $146,004. 

2.  $1071.414. 

3.  $1128.075. 

4.  $23.1868. 

5.  $50.66  dif. be- 

tween Sim. 

and  Annual 

Int. ; 
$4,209  dif.be- 

tween     An. 

and      Com. 

Int.; 
$64.769 +dif. 

between 

Sim.    and 

Com.  Int. 

6.  $868,208. 


.  $440824+  gr. 
at  comp.  int. 
.  $108,595  +  . 

Art.  587. 

.  $219,658. 
.  $183.0183. 
.  $133.55053. 
.  $1148721672. 
.  $55.4364. 
.  $992659725. 

Art.  589. 

1.  $1220.2528. 

2.  $469.53704. 
S.  $781.52013. 

4.  $566.8662. 

5.  $2755.3606. 

6.  $248.1272192 

7.  $257.299443. 

8.  $145.728068. 

9.  $228.3345. 
'0.  $556.75033. 

1.  $3439.63075. 
'2.  $753.952567. 
'3.  $864.942730. 
'4.  $1097.5152. 

Art.  592. 

\i.  $146,004. 

2.  $1071.41*. 

$1128.075. 

$23.1868. 

$50.56  dif. be- 
tween Sim. 
and  Annual 
Int. ; 

$4,209  dif.be- 
tween  An, 
and  Com. 
Int.; 

$54.769+ dif. 
between 
Sim.    and 
Com.  Int. 
I.  $363,208. 


\3 


Art.  599. 

1.  $282.46}. 

2.  $202,793. 

3.  $11,254 

4.  $117,942. 

Art.  602. 

5.  $1491.49  +  . 

3.  $2891.527. 

4.  $420,292. 

5.  $5424.651  +  . 

Art.  607. 

1.  $315,789+; 
$348,387+  ; 

2.  $776,699+  ; 
$754717  +  . 

3.  $485,468+  ; 
$478.10  +  . 

4.  $9,975+; 
$148,456. 

6.  $15.275 ; 
^230.578. 

6.  $5.513 ; 
$40.82  +  . 

7.  $9171.90  +  . 

8.  $530,367. 

9.  $.957. 

10.  $425. 

11.  2d  $33,865  bet- 

ter. 

12.  $.103  more  prof- 

itable at  $4.66. 

13.  $1.47  +  . 

Art.  615. 

1.  $20,671*  B.  Die.; 
$769,328*  Pro.; 
$11.520fB.Dis.; 
$778.479J  Pro. 

5.  $5.88  Bk.  Dis.  ; 
$274.12  Proc'ds; 
$11.94fBk.DiB.; 
$268.05^  Pro. 

S.  $25.82|Bk.  Dis.; 
$1574171  Pro. ; 
$54,022*  B.  Dis.; 
$1545.977J  Pro. 


ANSWERS, 

4.  $.884 
6.  $20.121|. 

6.  $27,194 

7.  DueMav27; 
59  da.  Time ; 
$475,383+  Pro. 

8.  Due  Aug.  16 ; 
75  da.  Time ; 
$581.395J  Pro. 

9.  Due  Aug.  22 ; 
91  da.  Time ; 
$1571.68$  Pro. 

Art.  618. 

7.  $458,287; 
$189.68; 
$99,112. 

2.  $876,061  +  ; 
$295,415+  ; 
$540,713. 

3.  $238.63 ; 
$1830.922. 
$515,648. 

4.  $a54.452. 

5.  $480,616. 

6.  $298,899  +  . 

7.  $961.781, 1st ; 
$967,495,  2d; 
$979,914,  3d. 

8.  $1517.440. 

9.  $495,262. 

Art.  625. 

1.  $2416. 

2.  $3204 

3.  $850.68. 

4.  $6331.50. 

5.  $133,796. 

6.  $1491. 

7.  $382.30284. 

8.  $291,141. 

9.  $486. 

10.  $560. 

11.  $720. 

13.  $824. 
U.  $275.50. 


371 

15,  $821. 

16.  $402. 
n.  $698.25. 

18.  $1615.11J. 

19.  $2415.925. 

20.  $1779. 
21   $496 

^£97A%,op2H% 
dis. 

23.  $2526.38f. 

2lt,  $8013.43  +  . 

Art.  631. 

2.  $2124.99065  +  . 

3.  $893.22}. 

4.  $1642  41. 

5.  $763. 

G.  $3469.83]^. 

7.  $609,375. 

8.  $11456.8126. 

Art.  633. 

1.  $12616.96. 

2.  $301     gain    by 

Ind. 

3.  $124,852  +   less 
by  Ind. 

4  3011.68+  marks. 

Art.  644. 

1.  July  1, 1876. 

2.  April  30, 1876. 

3.  Dec.  27, 1876. 

4.  Oct.  19,  1876. 

5.  Sept.  3, 1876. 

6.  July  21. 1877. 

7.  Oct.  19, 1877. 

8.  Feb.  5. 

9.  60  da. 

Art.  647. 

1.  $210,    Face    of 

note; 
Due  Dec.  12,  '76. 

2.  $100  due; 
Dec.     7,    1876, 

equated  time. 


i'\' 


h'i  \"- 


m 


i  ■' 


lii- 


)i  1 


872 

3.  Apr.  6,  1877. 

4.  March  1, 1876. 

Art.  604. 

of.    •J. 

^. /A- 

Art.  660. 

7.  I. 


4- 

s. 

6. 
7. 
8. 
9.  f  |. 

10.  ^\. 


>.  1  . 


Art.  668. 

i.  46A. 
;?.  150  da. 
^.  f  597. 
^.  $2386.40. 
J.  $2100. 

Art.  670. 

1.  $60. 
;^.  104  A. 

3.  130  da. 

4.  $678}f  • 

5.  177  cd.  3  cd.  ft. 
€.  415  lines. 

7.  $2000. 
S.  $13.50. 
^.  17gal.3qt.lpt. 

Art.  673. 

^.  $72. 
-*.  $27. 


ANSWBJtS, 

5.  $100. 
6-.  $204. 

7.  2284  yd. 

8.  498i. 

Art.  683. 

7.  6. 
e.  35. 
5.  272. 

4.  10  yd. 

5.  29i  bu. 

6'.  £168  158.  6|d. 

1.  56. 
^.  15. 
^.  21. 

4.  $7f . 

5.  8  cwt. 

6.  213i. 

Art.  685. 

4-  $9. 
5.  195  bu. 
6'.  803  ft. 

7.  70  bu. 

5.  183.655. 
10.  $3000. 
ii.  $7. 

i^.  16  mi.  109  rd. 
IS.  $2.40. 

14.  169  gal.  3  qt. 

1  pt. 

15.  $.98-ft-. 

16.  $375. 

i7.  2  yr.  10  mo. 

Art.  686. 

^.  V. 
i.  48. 

2.  23^. 

Art.  688. 

i.  120  cd. 
2.  264. 


5.  $18. 

4.  180  bu. 

5.  $116760. 

6.  1728  ft. 

7.  12  lb. 

8.  23040  yd. 

Art.  694. 

1.  $405,125. 

^.  $992. 

S.  $3222.26}. 

4.  $2400  for  4  mo. 

J.  $800  for  4  mo. 


Art.  606. 


1. 


s. 


$2382.545, 

A'b  share ; 
$1737.272, 

B's  share ; 
$1340.181, 

C's  share. 
$1661.538, 

A's  share ; 
$1107.692, 

B's  share ; 
$1550.769, 

C's  share. 

3.  $2814.128, 

A's  share ; 
$1644.112, 
B's  share ; 
$8431.758, 
C's  share. 

4.  $900,  1st  district ; 
$700.  2d  district ; 

600,  3d  district; 

400, 4th  district. 

1178.947.  ^tna ; 
$1547.363,  Home ; 
$2394.736,     Mu. 
tual. 


4. 


Art.  698. 

1.  $70,451,     A's 
share. 

$39 .629, B's  share; 
$36,419,  C's  share. 


w^ 


■-„.( 


8. 

Obu. 
16760. 
28  ft. 
lb. 
040  yd. 

.rt.  694. 

^.126. 
192. 

!222.26f. 
1400  for  4  mo. 
iOO  for  4  mo. 

.rt.  606. 

1382.545, 
'b  share; 
737.272, 
's  share ; 
340.181, 
's  share. 
661.638, 
's  share ; 
107.692, 
s  share  ; 
50.769, 
share. 
14.128, 
share; 
.112. 
share ; 
1.768, 
share. 
1st  district; 
2d  district ; 
jO,  3d  district ; 
,  4th  district. 
8.947,  iBtna ; 
7.363,  Home ; 
4.736,    Mu. 


).  698. 


ANSWERS, 


n,    A's 


re. 


829,  B'b  share; 
119,  C'b  shure. 


S.  f:!0,  A's  share; 
$144,  B's  share. 

0.  $1940,  A's  stock ; 
13510,  B's  stock ; 
$7150,  C's  stock. 

4.  $8666.06  +  , 

A's  share ; 
$5288.98  +  , 

B's  share. 
.  6.  $385.10  +  , 

A's  profit ; 
$288.56  +  , 

B's  profit ; 
$251.32  +  , 

C's  profit. 

Art.  TOO. 

5.  $1.00. 

5.  $.13^. 

4.  $-31}. 

6.  18}  carats. 

Art.  705. 

1.  3  gal.  of  mo.  to  2 

gal.  of  water. 

2.  1,  3,  2, 1  lb. 

5.  1, 2,  and  6  bbl. 
4.  1  part  of  each. 

6.  2, 1,  and  109  gal. 

7.  2,  2,  604,  and  240 

bbl. 

8.  50,  50,  5,  and  1 
sheep. 

10.  30, 30,  and  180  oz. 
il.  18,  27,  27,  and 

631b. 
i^.  44t,89|,  and89if 

lb. 

Art.  711. 

e.  86436 :  148996 ; 

247009;  64009. 
5.2804;  4225; 

186624 

^*  TijsJ  Fnr »  TTnnF » 
.512. 


5. 

250047;  15625: 

438976 : 

60286288. 

6. 

4100625. 

7. 

66169. 

8. 
9. 

r*fJir- 

10. 

.039304. 

11. 

.00390625. 

12. 

m- 

13. 

.00028561. 

14. 

.00091204. 

16. 

.000000166375. 

Art.  712. 

2. 

7. 

3. 

7. 

4. 

9. 

6. 

15. 

6. 

18. 

7. 

12. 

9. 

12. 

10. 

30. 

11. 

24. 

12. 

12. 

Art.  728. 

1.  59. 

2.  64. 

3.  87. 
4-  53. 

5,  96. 

6.  93. 

5'.  I*. 

S.  If. 

9.  .15. 

m.  u. 

11.  .76. 

12.  .48. 

13.  371. 
14>  64.5. 

15.  876.6. 

16.  167.4 

17.  7.56. 

18.  21.79. 

19.  5.656  +  . 

20.  7.681 +  . 


375 


SI.  2.646+. 

22.  .964  +  . 

23.  .894  +  . 
24-  .612+. 

25.  3.834  +  . 

26.  9.284  +  . 

27.  2.443  +  . 

28.  .881  +  . 

29.  .404  +  . 

30.  .346  +  ' 

31.  51. 

32.  62. 

33.  88*. 
34'  2656. 
36.  6. 

36.  354906 +  . 

37.  95  ft. 

38.  69.57+  yd. 

39.  487  ft. 

40.  44842+  rd. 

41.  96  trees. 

42.  1.4142;  2.2860; 
3.3166. 

43.  .654;  .852;  .785, 


Art.  736. 


m 


flt 


m 


/ 


f^j^(i 


I   ■ 


IV' 


374 

i5.  1.442+ ; 

1.913+  ; 

.798+  ; 

.8414  ; 

.208+  ; 

1.272 +. 
19.  4. 

m  9. 

Sfl.  87. 

;?^.  76.86  +  . 

183.  8. 

;?i  439  ft. 

25.  2730JI  sq.  ft. 

;?6.  78.3+  inches. 

£7.  196.9+  inches. 

2S.  32  feet. 

;?P.  26.9+  feet. 

SO.  436  feet. 

Art.  748. 

1.  18  in. 

;?.  26. 

S.  70 ;  432. 

4.  10. 

5.  12. 

6.  $18. 

7.  $97.60. 
5.  600500. 
9.  144. 

iO.  12  days ;  886  mi. 

Art.  752. 

1.  2048. 

2.  49152. 
5.  2186. 

4.  7. 

5.  815  mi. 

6.  393213. 

7.  8;  4372. 

8.  189  mi. ;  6. 

Art.  758. 

1.  $1905. 
S.  $1410. 
5.  $1775.9772. 
4.  $2485.714+. 


ANSWERS. 

5.  $5288.88  +  . 

6.  $491.73  +  . 

7.  $2380.59  +  . 

8.  $8944.226. 

9.  6%. 
10.  7%. 
ii.  $2500. 
12.  $500. 
i^.  3  years. 


Art.  782. 


i.  3 


rd. 


ch. 


^.  or^  sq 
i?.  53l|  sq.  ft. 
S.  112.292  sq 

4.  168  sq.  ch. 

5.  2160  stones ; 
$367.50. 

6.  510  sq.  ft. 

Art.  783. 

1.  22  rd. 
^.  10  ft. 

3.  16  ft.  8  in. 

4.  6  yd. 
6.  42  rd. 

6.  4  rd.  4  yd.  1  ft. 

9  in. 

7.  20  ft. 

Art.  784, 

1.  612.87  sq.  in. 
;^.  150  sq.  ft. 

3.  .924+  A. 

4.  692.82+  sq.  ft. 

5.  $292.68  +  . 

Art.  785. 

1.  39  ft. 

2.  43.08+  ft. 

3.  128.80+  ft. 

4.  1414+  rd. 

5.  187.45+  rd. 

Art.  786. 

1.  40.31  +  ft. 
S.  37.08+  ft. 


3.  2.10+  ft. 

4.  18  ft. 
^.  42  ft. 

Art.  787. 

i.  7  A.  8  sq.  ch. 
15  P.  125  sq.  1. 

2.  28  sq.  ft.  108  sq. 

in.  ; 

6  sq.  yd.  6  sq.  ft 
138  sq.  in. 

3.  29  sq.  ft. 

4.  2805  sq.  ft. 

Art.  788. 

1.  57  sq.  in. 

2.  220  sq.  ft 

3.  $161.25. 

4.  44  sq.  ft. 

Art.  789. 

1.  104  sq.  ft. 

2.  450  sq.  in. 

3.  11  A. 

4.  800  sq.  ft. 

Art.  790. 

1.  8.9824  in.; 
66.5488  in. 

2.  30  in  ;  25  ft. 

3.  596904000  mi. 

4.  $45,239  +  . 

Art.  791. 

1.  14313.915  sq.ft.; 
3911.085  sq.  ft; 
$17.55+. 

2.  314.16  sq.  ft ; 
1385.4456  sq.  in.; 
1963.50  sq.  ft. 

3.  79.5727+  A. 

Art.  792. 

1.  48.9824  ft 

2.  8  ft. 

3.  5rd. 


.  2.10+  ft. 
.  18  ft. 
.  42  ft. 

Art.  787. 

.  7  A.  8  sq.  ch. 
16  P.  125  sq.  1. 

28  sq.  ft.  108  eq. 
in. ; 

6  sq.  yd.  6  sq.  ft 
188  sq.  in. 

29  sq.  ft. 
2805  sq.  ft. 

Art.  788. 

67  sq.  in. 
220  sq.  ft 
$161.26. 
44  sq.  ft. 

Art.  789. 

104  sq.  ft. 
450  sq.  in. 
11  A. 
300  sq.  ft. 

Art.  790. 

3.9824  in.; 
66.5488  in. 
30  in  ;  26  ft. 
596904000  mi. 
146.239 +  . 

Art.  791. 

1431.3.915  sq.ft.; 
}911.086sq.  ft; 
M7.554. 
J14.16sq.  ft; 
[385.4456  sq.  in.: 
L963.50  sq.  ft. 
r9.5727+  A. 

Vrt.  792. 

3.9824  ft 
ft 
rd. 


-#.  87.6992  in. 
S.  16  in. 
G.  62.882  rd. 

Art.  802. 

i  36,<V  sq.  ft. 
^.  84  sq.  ft. 

5.  185  sq.  in. 

4^  326.7264  sq.  ft 

6.  186  sq.  ft. 

Art.  803. 

i.  411.5132  +  cu.ft. 
^.  103.8  cu.  ft 
3.  1800  cu.  ft. 
f  238.7616  cu.  ft. 
S.  41.888  cu.  ft. 
e.  131.25. 

Art.  804. 

i.  235.62  sq.  ft. 
^.  114  sq.  in. 
3.  35  sq.  ft. 


ANSWERS, 

A.  323  sq.  ft. 

6.  519.68+  sq.  in. 

6.  731  sq.  ft. 

7.  738.276  sq.  ft 

8.  608.686  sq.  in. 

Art.  895. 

i.  26.1828  cu.  ft 
B.  50.2666  cu.  ft 

3.  80  cu.  ft 

4.  3600  cu.  ft. 

6'  136.763+  cu.in. 

Art.  806. 

i'  136  sq.  in. 
^.  45  sq.  ft 

3.  837.29+  sq.  ft. 

4.  1124.6928  sq.  ft. 

Art.  807. 

1.  848.7176  cu.  ft. 
^.  1184cu.ft 
3.  24.871  cu.  ft 


375 

4.  48681.968  cu.  ft. 
^.9446.448+cu.ft: 

Art.  808. 

i.  264.4696  sq.  in. 
^.  201.0624  sq.  ft 
3.  6861.74  8q!*ft. 
4-  78.64  sq.  in. 
S.  4071.5136  sq.  in. 

Art.  809. 

1.  14.1372  cu.  yd. 

2.  4188.8  cu.  in. 

3.  .22089  cu.  ft. 

4.  14137.2  cu.  in. 

5.  268.0832  cu.  in. 

Art.  810. 

1'  104.1012  gal. 
2.  58.752  iraE 
5.  a5gal. 
4.  52.6592  gal. 


B-4  M 


S 


•! 


r 
1^ 


'  •< 


■t ' 


. 


876 


ANSWERS, 


ANSWEBS    FOB    ABITHMETICAIi    TABLES. 

Art.  35. 

Observe^  the  answen  to  examples  taken  from  the  Arithmetical  Tables 
are  in  every  case  arranged  in  the  order  the  pnpil  is  directed  to  take  the 
examples  from  the  Tables.  The  letters  over  the  sets  of  answers  indicate 
the  colamns  of  the  Table  ased,  and  the  black  figures  in  the  margin  the 
number  of  the  answer. 


A,  B,  0. 

B,  C,  D. 

0,  D,  B. 

D,  B, ». 

B, »,  o. 

F,  O,  H. 

O,  H,  I. 

H,I,J. 

1 

7fi7 

1688 

1887 

1886 

877 

1787 

1885 

1868 

2 

1090 

1917 

2183 

1841 

1436 

3281 

1836 

8280 

3 

1489 

2407 

2088 

1908 

3043 

2439 

2407 

9099 

4 

1730 

2320 

2215 

2173 

1751 

2538 

8i^ 

3475 

5 

a021 

2228 

2293 

1948 

1443 

3443 

S445 

^169 

6 

3381 

2331 

3819 

2315 

1171 

2738 

3349 

8507 

7 

2601 

2059 

2604 

9064 

1665 

2673 

2787 

9387 

8 

9741 

2432 

2339 

2416 

218:^ 

2850 

2530 

8317 

9 

2703 

2048 

2501 

2036 

3870 

2720 

3316 

2181 

10 

8730 

2819 

2214 

9161 

2628 

2303 

8039 

3410 

Art.  3(}. 


A,  B,  C,D. 

B,C,D,B. 

C,  D,  B,  F. 

D,  R,F,0. 

E,  F,  O,  H. 

F,  O,  H,  I. 

0,  H,  I,  J. 

1 

13264 

22653 

36551 

25564 

15666 

26680 

36823 

2 

15792 

27940 

29427 

24295 

32977 

29790 

37927 

3 

21276 

32779 

27821 

28237 

22397 

83993 

39961 

4 

25801 

28058 

30610 

26129 

21322 

83240 

82428 

6 

38206 

32077 

30601 

38040 

20129 

84313 

33154 

6 

32928 

29295 

82963 

29860 

18630 

36323 

38366 

7 

35916 

29182 

31853 

28561 

25639 

86115 

84170 

8 

35026 

30286 

32894 

28968 

29706 

37084 

80866 

9 

36916 

29190 

31929 

39317 

83199 

82013 

80169 

Art.  37.    Examples  with  five  numbers  in  each. 


A,  B,0,  D,B. 

181411 

B,  C,  D,  E,F. 

C,D,B,F,0. 

D,B,  F,0,H. 

B,F,0,H,I. 

F,  O,  H,  I,  t. 

1 

314140 

841433 

314362 

243644 

386469 

2 

226631 

336346 

863490 

834985 

249376 

803796 

3 

297622 

876258 

862615 

826186 

261888 

418914 

4 

337843 

378469 

884727 

847308 

273108 

431113 

5 

378053 

380569 

405729 

857326 

273287 

432r3 

6 

427878 

378772 

887766 

877507 

276001 

460039 

7 

435129 

361331 

413346 

833495 

881979 

449619 

8 

448976 

889788 

397915 

879185 

891881 

418844 

,    TABLES. 


ritbmetlcal  Tables 
rected  to  take  the 
'  answers  Indicate 
in  the  margin  the 


a, 


O,  H,  1.  H,  I,  i 


1885 
1836 
3407 
ftd4& 
8446 
9849 
2787 

asao 


8039 


186S 
8280 

aogs 

»175 


8507 
8387 
8217 
2181 
2410 


r,  o,  H,  I.  o,  H,  I,  J. 


26680 

26822 

89790 

27927 

88993 

29961 

83240 

82428 

84318 

33154 

36323 

88256 

86415 

84170 

87084 

80866 

82013 

80169 

in  each. 


O,  H,  I.  »•,  o,  H,  I,  J. 


886460 
893796 
418914 
431113 
4329^3 
460039 
449619 
418844 


1 
2 
3 
4 
5 
6 


ANSWERS, 


Examples  with  six  nnmhers  in  ectch. 


377 


A,B,C,D,B. 

B,C,D,E,F. 

C,  D,  B,  F,0. 

D,  E,  F,0,H. 

E,P,G,H,1. 

r,G,H,i,j. 

1 

250102 

401059 

4Q0(i28 

400320 

2632:«) 

432;«8 

2 

311474 

4147*3 

4t78<)8 

378T24 

287271 

472789 

3 

377407 

474117 

441213 

412172 

83175() 

5n5*«9 

4 

4:ii8l9 

438237 

4SM16 

4»11()5 

34aCH2 

5208()2 

5 

47Wai 

4«<«58 

463G25 

486293 

369966 

5-SMk-rt 

6 

50;58:JO 

*i8250 

482M1 

43M53 

36i5<)5 

WWjSS 

7 

63.'3H1H 

4382-25 

482293 

422974 

429776 

4977f»7 

Examples  with  seven  numbers  in  each. 


A,  B,  C,D,E.  B,  C,  D,  E,P. 


.334915 
391259 
473383 
632397 
552578 
602509 


449496 
312642 
633886 
524026 
525886 
525144 


C,  D,  E,  F,Q, 


493006 
5264(J6 
638902 
540312 
668410 
661488 


D,  E,P,G,H. 


450109 
4(>4710 
489069 
502072 
484149 
614932 


E,  F,G,U,I. 


301125 
ai7139 
3iX)680 
431760 
441629 
449362 


F,G,  H,  I,  J. 

511291 
571434 
<Mr4348 
617645 
615;iS5 
598066 


Art.  58.  Examples 

with  three  figures. 

A,  B,  C. 

B,  C,  D. 

C,  B,  E. 

D,  E,F. 

B,  P,  G. 

P,  O,  H. 

G,  H,  I. 

H,  I,  J. 

1 

96 

88 

881 

188 

115 

153 

471 

298 

2 

257 

571 

288 

123 

23:3 

mi 

265 

;349 

3 

1H2 

:m 

202 

20 

201 

15 

147 

AZi 

4 

70 

299 

8 

82 

182 

\Si 

169 

313 

6 

199 

6 

67 

333 

674 

262 

378 

222 

6 

162 

385 

152 

482 

188 

16!) 

309 

M 

7 

51 

494 

58 

422 

220 

vrt 

27 

2fi8 

8 

162 

381 

191 

91 

91 

89 

106 

64 

9 

26 

260 

398 

21 

207 

70 

206 

34 

10 

226 

263 

369 

311 

111 

111 

114 

134 

11 

227 

274 

258 

416 

162 

377 

2:W 

329 

Art.  59. 

Examples  with  four  fg 

ures. 

A,  B,  C,D. 

962 

B,  C,I),E. 

C,D,E,  p. 

D,  E,  P,G. 

E,P,G,H. 

P,G,H,I. 

1529 

0,  H,  I,  J. 

1 

381 

3812 

1885 

1158 

4707 

2 

2571 

6712 

2877 

1288 

2326 

3266 

2651 

3 

1620 

3798 

2020 

201 

2015 

147 

1473 

4 

701 

2992 

82 

818 

1817 

1831 

l«i88 

5 

1991 

67 

667 

3326 

6788 

2622 

3778 

6 

1616 

3848 

1518 

4817 

1831 

1691 

8084 

17 

506 

4912 

578 

4220 

2197 

1973 

268 

« 

1618 

3S09 

1909 

909 

910 

894 

1064 

0 

260 

2602 

8979 

207 

2070 

704 

29«6 

lO 

2263 

»i31 

8689 

3111 

nil 

1114 

1134 

11 

2274 

2742 

2584 

4162 

1628 

3T67 

2329 

26 


378 


ANSWERS. 


Art.  60.    Examples  with  six  figures. 


A,B,C,D,E,P. 

B,  c,D,B,P,o. 

C,D,E,F,0,H. 

D,  E,F,0,H,I. 

E,  F,  O,  H,  I,J. 

1 

96188 

38115 

881153 

18»i71 

115203 

2 

267123 

671283 

287674 

123265 

282651 

» 

le-iOiO 

379799 

202015 

20147 

201473 

4 

70082 

299182 

8183 

81831 

181688 

it 

199333 

6674 

66738 

33x'622 

673778 

6 

161518 

384817 

151831 

481691 

1830&4 

7 

50578 

494220 

57803 

421973 

219732 

8 

161909 

380909 

190911 

90894 

91064 

9 

26021 

260207 

397930 

20704 

2070:M 

10 

226311 

36:3111 

368889 

311114 

1111:M 

11 

227416 

274162 

258877 

416233 

16232!) 

'  ■■:  I 


Art.  79.    Multiplicand  tliree  figures,  multiplier  one. 


A,  B,  C. 

B,  C,  D. 

C,  D,  E. 

D,  E,  F. 

B,  F,  O. 

F,  G,  H. 

G,  n,  I. 

H,  I,  J. 

1 

1872 

1785 

942 

3565 

414 

3080 

2562 

2710 

2 

690 

2310 

3408 

4200 

1518 

2152 

30fr4 

5010 

3 

2765 

5736 

3384 

5184 

*«)2 

7776 

3240 

4:374 

4 

3899 

2880 

6128 

4008 

6183 

5274 

3180 

6713 

5 

2922 

7876 

758 

5274 

4445 

55(i8 

5784 

5823 

6 

&488 

3476 

2073 

M^ 

1560 

8622 

29:30 

2607 

T 

5936 

3872 

4215 

3933 

3024 

4734 

7160 

4705 

8 

7173 

6846 

4710 

6872 

5382 

6702 

8472 

61  «M 

9 

4795 

4179 

7808 

6912 

41:34 

6279 

7792 

2247 

10 

8865 

3428 

4046 

6312 

4480 

5802 

2712 

7047 

11 

4554 

4752 

8523 

1912 

5'195 

7704 

3^48 

5192 

Multiplicand  four  Jigwes,  multiplier  one. 


A,B,C,I>. 

B,  C,  D,  E. 

C,D,E,F. 

D,  E,  F,G. 

E,F,  G,H. 

F,  G,  H,  I. 

G,  H,  I,  J. 

1 

11735 

6942 

23565 

21414 

11080 

11562 

42710 

2 

8310 

15408 

68:>00 

31518 

10152 

43064 

23010 

3 

23730 

57384 

45184 

45402 

43776 

43240 

5a37t 

4 

27880 

46128 

46008 

60183 

41-^74 

a5180 

55713 

5 

43875 

8758 

68274 

mU5 

69568 

41784 

86823 

6 

27476 

26073 

48433 

73560 

17622 

47930 

17607 

7 

67872 

24215 

759a3 

35024 

227:34 

631«50 

44766 

8 

55848 

58710 

62872 

77382 

4V,m 

85)472 

78165 

9 

67179 

47808 

875)12 

40134 

48279 

7175)2 

29247 

10 

894-^8 

60046 

46312 

39480 

5:38C2 

88712 

61047 

11 

60752 

53523 

87912 

33495 

70704 

59948 

45192 

[.  ;E,F,o,  B,I,J. 


11520;} 
232651 
201473 
1816S8 
673778 
1830&4 
219732 
QIOM 
2070:M 
Illi:i4 
16232!> 


tier  one. 


3,  H,  I. 

H,  I,  J. 

2562 

2710 

30&4 

5010 

3240 

4;i74 

3180 

6713 

5784 

5823 

2930 

2607 

7160 

4705 

8^172 

616.") 

7793 

2247 

2712 

7047 

3W8 

5192 

n,l. 

)62 

G,  H,  I,  J. 

42710 

m 

23010 

MO 

58:nt 

;80 

55713 

m 

8e82;i 

)30 

17607 

m 

44765 

72 

78165 

t>2 

29247 

12 

61047 

»48 

45192 

ANS  WERS. 


Multiplicand  six  figures,  multiplier  one. 


379 


A,B,C,D,B,P. 

B,  C,D,E,F,G. 

c,D,E,r,o,H. 

D,  E,r,Q,H,I. 

E,P,G,  U,I,J. 

1 

117:35«)5 

1041414 

3771080 

2141562 

692710 

2 

1108200 

2911518 

3410152 

4io;jow 

1523010 

3 

3165184 

6()9!>402 

5083776 

3243240 

4378374 

4 

3;«60(» 

51i>0183 

4601274 

2675180 

4815713 

5 

4388274 

4379M5 

6069568 

a52l7»4 

7820823 

6 

48()»433 

6953560 

6227622 

4597930 

587607 

7 

7635!Ki3 

3875024 

50fi27;i4 

3503160 

18947G5 

8 

&382872 

8807382 

5501fM)2 

JM3JM72 

53881«i5 

9 

8<)37912 

8586134 

8838279 

6151792 

2069247 

10 

7886312 

4281>480 

^173802 

3158712 

8071017 

11 

3037912 

416^195 

8530704 

3349948 

6285192 

Art.  89,    Multiplicand  five  figures,  multiplier  three. 


A,  B,  C,  D,  E. 

B,  C,  D,  B,  F. 

C,  D,  E,  F,  O. 

D,  E,  F,  0,H. 

E,  F,  G,  H,  I. 

1 

19997292 

18224326 

11925914 

4M05130 

5306082 

2 

7812528 

24964200 

41432958 

4X392832 

lftl48184 

3 

30300024 

6.*«92864 

38805882 

57015456 

38675160 

4 

42270628 

3379;i448 

66ftll003 

46&47784 

66318380 

5 

33691778 

801915:34 

147M455 

5«i230768 

5(H)60904 

6 

57!)06513 

379a360;i 

26155710 

72554862 

1752(M70 

t 

6(5601755 

416073S3 

50158044 

43175954 

82892860 

8 

778701()0 

75155712 

541.54022 

77129442 

5a311432 

9 

5.>474128 

47156952 

87529»14 

743593W 

4()7frl372 

10 

93a5;«(» 

41077142 

4544*360 

67595752 

50578:392 

11 

52327483 

53172:332 

89761395 

2992:3024 

62615508 

Multiplicand  six  figures,  multiplier  five. 


A,  B,  C,  D,  E,  F. 

B,  C,  D,  E,  F,  G. 

C,  D,  B,  F,  O,  H. 

D,  E,  F,  O,  H,  I. 

1 

9042318325 

29574555914 

24704210975 

18119756082 

2 

13^9639200 

21761400958 

552<»90248:32 

26558882184 

3 

22816228864 

73a50041882 

3777753t>456 

44623739160 

4 

48843909448 

4;3752(}()6003 

450127297t^ 

58161088880 

5 

42:3801875»4 

60605755455 

69768171768 

11496276904 

6 

S3272295603 

7aS109:35710 

3OI001 48862 

^484771 1470 

T 

8302719<5383 

38071142044 

72554010954 

2<)215897860 

8 

47700798174 

955982<i0022 

60439965442 

693086354:33 

9 

82339081154 

34613802344 

77142625399 

68960040378 

10 

5S<'>32758142 

81315762360 

27707044752 

62040262392 

11 

65994081332 

41008041395 

&4813207024 

45360431506 

J  1 


"t; 

t 

,1^1 

Si 


■^il 


nil 


380 


ANSWERS. 


Art.  111.    Dividend  three  figures,  diioiaoi' one. 


A,  B,  C. 

B,  0,  D. 

0,  D,  E. 

D,  K,  P. 

E,  F,  O. 

F,  G,  H. 

O,  H,I. 

H,  I,  J. 

1 

56 

96 

173t 

863 

168» 

55? 

243' 

92 

2 

53 

41 

51| 

159J 

41S 

107f 

183 

82A 

3 

OS 

154 

82 

66} 

127? 

9t>? 

96J 

226S 

4 

288 

192 

111 

283 

68 

1603 

at? 

28 

5 

138 

85} 

20!)J 

41 S 

126:1 

»)!» 

76J 

424 

6 

53 

88 

70? 

185i 

89 

84i 

184  i 

48| 
1131 

7 

159 

403 

329 

97,^ 

121? 

165 

119; 

8 

143 

105? 

m 

im 

42; 

169J 

82 

103 

9 

1«)7 

134^ 

143 

363 

2S4J 

43 

439i 

113? 

10 

237 

242 

Ml 

15Hi 

w» 

76? 

asj 

155i 

11 

74 

232 

40? 

171?, 

1981 

242 

114 

141.] 

Dividend  five  fig wes,  divisor  one. 


A,B,C,]),B. 

B,  0,  D,  E,F. 

C,D,E,F,0. 

D,  B,  r,«,H. 

E,r,a,n,i. 

F,  G,  H,  I,  J. 

1 

3373J 

858611 

144918 

7721  g 

28748} 

6217 

3 

4718J 

61C«i 

51524 

;)017? 

12516i 

12582;; 

3 

18082 

13209 

49271 

11596J 

7096S 

12893 

4 

9614 

25616J 

8568 

14160.1 

7069? 

10(5845 

5 

192091 

7597;-! 

13968.1 

18760 

150761 

2(5924 

6 

(KWO? 

4985  J 

16422J 

11584^ 

6684-J 

8428J 

7 

31829 

6097,^ 

9407 

l!M98i 

106191 

99131    . 

8 

10659i- 

13184'i 

3012S 

12769J 

mb^ 

94863 

9 

13143 

11786 J 

107341 

3265-I 

4(i9ii0] 

5M1{^ 

10 

10538?, 

16158?, 

10694J 

6790} 

887n 

7828. 

11 

8467? 

18571 J 

9582 

21492 

9SM7;i 

161411 

Art.  121.    Dividend  four  figures,  divisor  two. 


A,  B,  0,D. 

B,  0,  D,  B. 

0,  D,  B^. 

D,  B,  F,  G. 

E,  F,  G,  H. 

F,  G,  n,  I. 

O,  H,  I,  J. 

1 

45U 

91H 

14952 

80-5 

137*  J 

58SI 

202|S 
152?! 

ft 

49i',1 

ffi^S 

4i;i 

180A 

^58^^? 

103?  S 

3 

85J* 

14635 

mi 

61 '^1 

llfiJJ 

78*5 

932  S 

4i 

awA 

lOTt-H 

108.^3 

223?~J 

5Tl';i' 

143.U 

79}J 

5 

112jf2 

8i;i 

170U 

as,"* 

121;:? 

216/, 

74 

6 

51(8 

e7i* 

65?  J 

17411 

70>.i 

aQjis 

153J 

T 

1291t 

39Ji' 

263,^:t 

87?? 

116fJ 

130J? 

1121 

1 

8 

ISlJi 

mn 

32r.' 

171,'s 

40J? 

151?!- 

71? 

9' 

146 

12913 

112?^ 

34H 

I9in 

42JI 

325H 

• 

lO 

225?  5 

167,*. 

9US 

125^, 

67^1 

372 

i 

11 

TOa 

1971? 

38,-j 

168  J  S 

i75i: 

210J3 

103S 

w  one. 


G,  U,  I. 

H,  I,  J. 

aWl 

92 

i83 

82* 

96J 

226i 

a>? 

28 

76? 

424 

184i 

48i 

119; 

113J 

82 

103 

439i 

113? 

IB* 

155i 

114 

141:; 

ANSWERS, 


Dividend  six  figures,  divisor  three. 


381 


A,B,C,D,E,F. 

B,  C,D,  E,F,Q. 

C,D,E,F,G,II. 

D,E,r,o,n,i. 

E,  F,  O,  H,  I,J. 

1 

222HI 

289l:3 

lOOOj-H 

1001  „' J? 

looi::; 

2 

611.12? 

933AV 

655i?J 

510^33 

3 

616,Vt 

2412 j?| 

826?SI 
1774i!|5 

8-145  J  i 

1295iJ: 

4 

1!M5:JS 

law.;  51 

w>:.i? 

730.U*; 

o 

905hV's 

1378:  ;.i 

870»n 

600;!  Sj 
2413ij; 

2991 ii; 

3152il 

6 

113:i,,V. 

331  rVs 

915iS| 

7 

687*13 

ISTOj^.-V 

978;;¥S 

m^At 

^206^,?? 

8 

1460^", 

62iai 

553111 

mtn 

r,76?«s 

9 

saira 

2455ii| 
516iU 

504}  U 

597U5 
510'is 

loi2«;f 

10 

32374VS 

2190,St 

930} IS 

11 

1248,V, 

1283yy, 

535? il 

24&4,^j'k 

1280J-I 

Art.  194.    Examples  with  two  numbers. 


»,n,i. 

F,G,H,I,J. 

48} 

6217 

16', 

12582.^, 

W>a 

12893 

m 

10(584; 

T»)1 

2(i924 

^i 

8428J 

'JJ 

9913? 

5:? 

94862 

!9» 

5541? 

u 

7822 

7* 
•  11 

16141]; 

lor  two. 


U  n,  I. 

a,  H,  I,  J. 

e;j 

202IS 
152?,| 

w?S 

81  r, 

932S 

3.H 

79,^J 

6/, 

74 

)j$ 

153J 

)iS 

112J 

us 

71,! 

325^ 

37S 

m 

103^ 

1 

6 

2 

22 

88 

22 

7 

8 

56 

16 

48 

168 

2 

6 

6 

18 

6 

42 

8 

9 

9 

99 

63 

27 

3 

36 

132 

24 

20 

44 

9 

70 

10 

60 

60 

«» 

4 

60 

40 

20 

35 

5 

10 

44 

44 

44 

44 

132 

5 

132 

132 

66 

66 

66 

11 

24 

24 

W 

84 

36 

6 

3r 

105 

12 

3 

45 

1 

12 

156 

26 

26 

91 

39 

Art.  195.    Examples  with  two  nurnbers. 


1 

6 

6 

2 

2 

4 

18 

T 

6 

35 

1 

8 

21 

6 

2 

12 

6 

24 

6 

4 

18 

8 

6 

85 

3 

3 

9 

18 

3 

12 

12 

8 

60 

5 

3 

9 

6 

10 

2A 

4 

60 

6 

4 

12 

24 

4 

42 

1 

3 

lO 

12 

20 

24 

28 

12 

12 

5 

4 

3 

12 

6 

9 

3 

11 

12 

^ 

2 

28 

21 

27 

6 

4 

35 

28 

12 

21 

15 

Art.  196.    Examples  with  two  numbers. 


1 

6 

36 

2S 

16 

44 

im 

6 

2 

105 

3 

3 

68 

9 

2 

12 

30 

8 

6 

7 

3 

7 

6 

TO 

16 

12 

48 

210 

3 

\% 

132 

las 

6 

22 

2ftl 

8 

66 

5 

;« 

77 

8 

66 

4 

20 

15 

20 

12 

35 

195 

9 

42 

10 

24 

60 

m 

90 

o 

12 

8 

4 

6 

6 

24 

1   lo 

12 

4 

2 

28 

3 

8 

|J*.'' 


m 


382 


ANS  WE  Its, 


m 


11^ 


it 


I 


Art.  197,    Examples  with  three  numbers. 


1 

2 

2 

22 

22 

T 

8 

8 

16 

24 

2 

6 

6 

6 

6 

8 

9 

9 

9 

9 

3 

12 

12 

4 

4 

9 

10 

10 

60 

ao 

4 

20 

20 

5 

5 

10 

44 

44 

44 

44 

•■> 

132 

66 

66 

66 

11 

24 

12 

M 

12 

6 

36 

3 

3 

3 

12 

26 

26 

13 

13 

Art.  198.    Examples  with  three  numbers. 


1 

6 

6 

2 

2 

4 

18 

6 

2 

35 

1 

8 

21 

3 

2 

12 

6 

8 

6 

1 

3 

T 

6 

35 

1 

3 

3 

6 

3 

12 

12 

4 

6 

1 

3 

8 

6 

6 

3 

1 

3 

6 

4 

4 

3 

4 

6 

1 

3 

9 

6 

10 

24 

4 

12 

6 

6 

4 

1 

4 

6 

3 

3 

10 

12 

4 

2 

28 

3 

3 

^'1 


f    ■ 


Art.  211.    Examples  with  two  numbers,  according  to  194. 


1 

190 

2772 

1232 

176 

792 

3 

1680 

2520 

216 

756 

252 

3 

2772 

792 

5280 

5280 

3960 

4 

120 

120 

840 

2100 

6825 

5 

264 

792 

2772 

4158 

4752 

6 

420 

420 

5460 

16380 

4095 

T 

3360 

560 

1341 

1344 

1680 

8 

6930 

10395 

2079 

6237 

12474 

9 

1050 

840 

1200 

3600 

5010 

10 

660 

1320 

1408 

4620 

2640 

11 

3360 

10080 

2520 

1260 

3780 

la 

312 

1560 

1820 

1092 

2457 

Art.  211.    Exa/mples  with  two  numbers,  accmding  to  195. 


1 

240 

1260 

5544 

4752 

1848 

1386 

2 

1008 

13860 

792 

4320 

4620 

5544 

3 

1260 

39()0 

1320 

3360 

7700 

51480 

4 

660 

1320 

3960 

4620 

103950 

34320 

5 

4620 

9240 

12860 

11012 

20790 

102960 

6 

8360 

740 

1680 

2496 

4368 

32760 

7 

3168 

2520 

33»}4 

44352 

9072 

83160 

8 

6980 

3150 

237H0 

69300 

45360 

20790 

9 

4620 

770 

2640 

23100 

7920 

55440 

10 

1848 

5280 

5544 

4620 

18860 

23760 

11 

2184 

6210 

32760 

5160 

3276 

7020 

her9. 


mbers. 


16 

a4 

9 

9 

60 

au 

44 

44 

M 

1« 

13 

" 

1 
1 

3 

24 

2 


8 

21 

3 

8 

3 

6 

1 

3 

6 

4 

12 

B 

28 

3 

8 

^•ding  to  194:. 


sioo 

H58 


1260 
1092 


792 
252 
3960 
6825 
4752 
4095 
1680 
12474 
5040 
2640 
3780 
2457 


A  XSWERS. 


383 


Art.  211.    Examples  with  two  numbers,  according  to  15^C5. 


1 

1280 

396 

1848 

5280 

440 

798 

2 

240 

840 

360 

3780 

2100 

8190 

3 

2772 

792 

792 

3(i'.KiO 

5940 

15M 

4 

420 

»K) 

&10 

5-160 

1575 

585 

5 

1056 

8140 

11088 

147M 

»]2W 

18480 

6 

13HtiO 

815 

41580 

36036 

28;J5 

38610 

7 

3:J60 

1400 

1680 

4800 

5040 

2520 

8 

306 

13860 

2376 

2772 

124740 

4752 

9 

840 

16800 

5040 

2100 

5040 

3780 

10 

1716 

17160 

17160 

4004 

60060 

61776 

Art.  211.    Examples  with  three  numbers,  according  to  1?>7. 


1 

140 

210 

84 

48 

48 

2 

90 

90 

90 

72 

72 

3 

1512 

4;« 

48 

48 

168 

4 

1440 

1440 

9<X) 

150 

150 

5 

280 

2610 

360 

360 

72 

6 

878 

1890 

2160 

1680 

16H0 

7 

126 

882 

17W 

392 

168 

8 

160 

224 

2016 

1008 

570 

9 

360 

1080 

1512 

5r.7 

2268 

10 

1890 

1260 

720 

2640 

2(;40 

11 

495 

3465 

1386 

616 

264 

12 

1008 

1680 

3360 

1440 

864 

Art.  211.    Examples  with  three  numbers,  according  to  1 08. 


,ding  to  1U5. 


1848 
4620 
7700 

103950 

[20790 
4368 
9072 

145360 
7920 

118860 
3276 


1386 
6544 

51480 
34320 

102960 
32760 
83160 
20790 
55440 
23760 
7030 


1 

252 

270 

840 

48 

72 

48 

108 

2 

12(50 

864 

1080 

^) 

1800 

120 

420 

3 

420 

30240 

361) 

720 

600 

120 

l'2(iO 

4 

180 

70560 

1080 

m 

2400 

WO 

6;;oo 

5 

2520 

5040 

2016 

10584 

51)4 

3024 

576 

6 

1080 

5040 

1440 

75«0 

5(>10 

14256 

(;;)36 

7 

594 

6930 

39()0 

41580 

55440 

7128 

792 

8 

2376 

18480 

41580 

462C 

5280 

792 

2376 

q 

'^  '■;•! 

•1 

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a^ 

/..    , 

3 

tj.  ■  'i 

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