1% ^ r^%

IMAGE EVALUATION TEST TARGET (MT-3)

V

//

O

/

.<^^ c^

8

Q-

C/j

fA

1.0

I.I

1.25

IIIIIM IIIIM IlliU 111^

m mil 2.0

1.4

1.8

1.6

V]

^^

'»'

c-i?s

c^

^

^^a

i%

/.

y

^

!^

Photographic

Sdences Corporation

4^

^<^

-^^

\

\

m

23 WEST MAIN STREET

WEBSTER, NY. 14580

(716) 872-4503

6^

'

CIHM/ICMH

Microfiche

Series.

CIHIVI/ICIVIH Collection de microfiches.

Canadian Institute for Historical Microreproductions institut Canadian de microreproductions historiques

1980

Technical and Bibliographic Notes/Notes techniques et bibliographiques

The Institute has attempted to obtain the best original copy available for filming. Features of this copy which may be bibliographically unique, which may alter any of the images in the reproduction, or which may significantly change the usual method of filming, are checked below.

D

D

D D

D D

D

Coloured covers/ Couverture de couleur

I I Covers damaged/

Couverture endommagde

Covers restored and/or laminated/ Couverture restaur6e et/ou pellicul6e

I I Cover title missing/

n

Le titre de couverture manque

Coloured maps/

Cartes gdographiques en couleur

Coloured ink (i.e. other than blue or black)/ Encre de couleur (i.e. autre que bleue ou noire)

□ Coloured plates and/or illustrations/ Planches et/ou illustrations en couleur

Bound with other material/ Relid avec d'autres documents

Tight binding may cause shadows or distortion along interior margin/

La reliure serr^e peut causer de I'ombre ou de la distortion le long de la marge intdrieure

Blank leaves added during restoration may appear within the text. Whenever possible, these have been omitted from filming/ II se peut que certaines pages blanches ajoutdes lors d'une restauration apparaissent dans le texte, mais, lorsque cela 6tait possible, ces pages n'ont pas <§t6 fiimdes.

Additional comments:/ Commentaires suppl^mentaires;

L'Institut a microfilm^ le meilleur exemplaire qu'il lui a 6t6 possible de se procurer. Les details de cet exemplaire qui sont peut-dtre uniques du point de vue bibliographique, qui peuvent modifier une image reproduite, ou qui peuvent exiger une modification dans la mdthode normale de filmage sont indiquds ci-dessous.

□ Coloured pages/ Pages de couleur

□ Pages damaged/ Pages endommagdes

I — I Pages restored and/or laminated/

D D

Pages restaurdes et/ou pellicul^es

Pages discoloured, stained or foxed/ Pages d6color6es, tachet^es ou piqu6es

Pages detached/ Pages d6tach6es

QShowthrough/ Transparence

□ Quality of print varies/ Qualit4 indgale de I'impression

I I Includes supplementary material/

Comprend du materiel supplementaire

Only edition available/ Seule Edition disponible

Pages wholly or partially obscured by errata slips, tissues, etc., have been refilmed to ensure the best possible image/ Les pages totalement ou partiellement obscurcies par un feuillet d'errata, une pelure, etc., ont 6X6 filmdes d nouveau de fa^on d obtenir la meilleure image possible.

This item is filmed at the reduction ratio checked below/

Ce document est film6 au taux de reduction indiqu6 ci-dessous.

10X 14X 18X 22X

26X

SOX

y

12X

16X

20X

24X

28X

32X

ails

du

)difier

une

nage

The copy filmed here has been reproduced thanks to the generosity of:

National Library of Canada

The images appearing here are the best quality possible considering the condition and legibility of the original copy and in keeping with the filming contract specifications.

Original copies in printed paper covers are filmed beginning with the front cover and ending on the last page with a printed or illustrated impres- sion, or the back cover when appropriate. All other original copies are filmed beginning on the first page with a printed or illustrated impres- sion, and ending on the last page with a printed or illustrated impression.

L'exemplaire filmd fut reproduit grdce d la g6n6rosit6 de:

Bibliothdque nationale du Canada

Les images suivantes ont 6t6 reproduites avec le plus grand soin, compte tenu de la condition et de la nettetd de l'exemplaire filmd, et en conformity avec les conditions du contrat de fiimage.

Les exemplaires originaux dont la couverture en papier est imprimde sont filmds en commenpant par le p/emier plat et en terminant soit par la dernidrd page qui comporte une empreinte d'impression ou d'illustration, soit par le second plat, selon le cas. Tous les autres exemplaires originaux sont filmds et> commenpant par la premidre page qui comporte une empreinte d'impression ou d'iiiustration et an terminant par la dernidre page qui comporte une telle empreinte.

The last recorded frame on each microfiche shall contain the symbol — ♦► (meaning "CON- TINUED "I, or the symbol V (meaning "END "), whichever applies.

Maps, plates, charts, etc., may be filmed at different reduction ratios. Those too large to be entirely included in one exposure are filmed beginning in the upper left hand corner, left to right and top to bottom, as many frames as required. The following diagrams illustrate the method:

Un des symboles suivants apparaitra sur la dernidre image de cheque microfiche, selon le cas: le symbole — »- signifie "A SUIVRE", le symbole V signifie "FIN".

Les cartes, planches, tableaux, etc., peuvent dtre filmds d des taux de reduction diffdrents. Lorsque le document est trop grand pour §tre reproduit en un seui clich6, il est film6 d partir de Tangle supdrieur gauche, de gauche d droite, et de haut en bas, en prenant le nombre d'images ndcessaire. Les diagrammes suivants illustrent la mdthode.

rrata :o

pelure.

D

32X

1

2

3

1

2

3

4

5

6

jmn

School Glohet Manufactured hy Dawson Brothers,

THE MACVICAR TELLURIAN GLOBE.

{Patented in Canada.)

This is the latest, simplest, and best device ever presented for illustrating Geography and the Elements of Astronomy. Being the product of long years in the class room, it is thoroughly practical.

The Globe is a working model of the Earth in its relation with the Sun. The horizon and other parts are 80 constructed, that in every illustration the Globe represents the real position and relation of the Eanli to the Sun.

The Tellurian Globe combines in the most con- venient and substantial shape, and at much less expense, the best form of Globe, and the best form of Tellurian. It illustrates also in a more simple and philosophic manner than can be done by the use of both Globe and Tellurian, all problems relating to the phenomena of day and night, change of season, twilight, rising and setting of the sun, the apparent daily motion of the sun across the horizon, the motion of the earth in its orbit.

The Tellurian Globe is made in [two sizes. Eight inch and Twelve inch.

Bt dr. MALCOLM MACVICAR, Ph.D., LL.D., Principal of the State Normal School^ Potsdamj N.Y.j

Handbook of the Mac Vicar Tellurian Jllobe,

for the use of Teachers, Schools and Families ; oon- taimng a complete course of illustrations and problems in Geography and Astronomy.

School Booics Published by Daicson Bros,

By professor ANDREW,

0/ the University of McGill College,

The New Dramatic Beader ; Comprising a Selec- tion of Pieces for Practice in Elocution, with intro- 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- ciples of Contrast and Comparison ; with numerous- Exercises. Price, 30 centta.

By F. C. EMBERSON, M.A.,

Inspector of Schools in the Province of Quebec,

The Art of Teaching: A Manual for the use of

Teachers and School Commissioners. Price 50 cents*

Canadian Elementary School Atlas; For the

use of Junior Classes, containing 16 Maps. Price, 25 cents.

A cheap and yet very complete elementary Atlas. The Maps, which are similar to those in Keith Johnston's ele- mentary Atlases, are clear and attractive.

Lennie's English Grammar

Carpenter's Spelling

The Spencerian Series of Copy Books; A

System of Penmanship very extenbively used in the United States and the Dominion of Canada. The System is comprised in twelve Numbers, and divided into four distinct Series, viz : —

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

No«. 6 & 7* Bunlneso Series.

No«. 8 dc 9, • L.»dle»* Series.

N«s. 10, 11 4c 12 Exerctoo Sorlea.

1^

National Library Bibliotheque nationale of Canada du Canada

0(

D]

^^

A

'J//

/^

/

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)

1.0 ^«

II

|M III 2.5

IM

2.2

1.8

1.25 1.4

1.6

•4 6" —

►

V]

<^

/a

e.

/,

4 >

/A

'♦v^

7

Photographic

Sciences

Corporation

23 WEST MAIN STREET

WEBSTER, NY. MS80

(716) 872-4503

\

^^

^

:\

\

^^

^■^^

%^

9)

o\

I

I*

^

^

■:< M% ids ^ ' ^ \# V i

t4

.^

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

*^

&..

y

<^

/

o^

'h

Photographic

Sciences Corporation

^

#

4t>^

,\

iV

:\

\

^<b

V

<>

->

^'^x

€>

<>

23 WEST MAIN STREET

WEBSTER, NY. 14580

(716) 872-4503

'^<^ ^

.<\

'<>', «y

<\

i ft

i

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 ?

1 ' ■

! i I

it

I!!

f ■ i

If f:^

i

*^T

-.V -

■n>^'

■-••.I'

.|., .;

:#'

J,

'} *•

|i ,

■'!■

V ' ■ '

!■

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 .

i 1

ij.

;

i

i M

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

^q\

•1>^

\\

9)

o^

<>

I i

tl

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

m^r ' r

ik

]-,

tij

••:i

■,'l

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

C/j

o

»' \i

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

*?l '?'

a^

/.. ,

3

tj. ■ 'i

\'^'^ ■• I

'^fcf

A.\

•<l I.

\t'

w

H'

m

[4

I

ScJiool Books Published hy Dawson Bros,

By professor ANDREW, 0/ the University of McGill College,

The New Dramatic Reader : Comprising a i

lion of Pieces for Practice in Elocution, with i ductory hints to Readers. Price, 75 cents.

Seleo- intro-

Br J. D. MORELL, LL.D., H. M. Inspector of Schoohf England.

A Complete Manual of Spelling on the Prin- ciples of Contrast and Comparison ; with numerous Exercises. Price, 30 cents.

By F. C. EMBERSON, M.A

Inspector of Schools in the Province of Quebec,

The Art of Teaching: A Manual for the use of

Teachers and School Commissioners. Price 50 cents.

Canadian Elementary School Atlas ; For the

use of Junior Classes, containing 16 Maps. Price, 25

cents.

A cheap and yet very complete elementary Atlas. The Maps, which are similar to those in Keith Johnston's ele- mentary Atlases, are clear and attractive.

Lennie's English Grammar

Carpenter's bpelling.-.

The Spencerian Series of Copy Books; A

Systtm of Penmanship very extensively used in the . United States and the Dominion of Canada. The System is comprised in twelve Numbers, and divided into four distinct Series, viz: —

Nos. If 2, 3f 4 & 5, Common School 8erle«.

NOS.6&79 Munlneas Serlff*.

Horn. 8^9, I.adles' Series.

Tios- 10, 11 4c 19 ExercUe ISeries.

m

ro8»

ft Seleo- h. intro-

Prin-

imerous

ec.

) use of

»0 cents.

For the rice, 25

s. The on's ele-

^

■MM

■i

in the . The divided

rle«.

ISchool Books Published by Daioson Broiher$.

I — - - ■ — — ■

By dr. DAWSON, F.R.S., F.O.S., &c. Principal of the University of McQill College.

A HandvBook of Zoology: With examples from Canadian Species, Recent ana Fossil. Part 1, Inver- tebrata. Three Hundred Illustrations. Priee, 7ft cents.

Bt henry H. miles, M.A., LL.D.,

Secretary to the Department of Public Instruction for the

Province of Quebec.

The School History of Canada; prepared for

use in the Elementary and Model Schools, with many Wood Engravings, a Map of New France *in(l New England, and a Map of the Operations at the siege of Quebec in 1 759, also Appendices, giving an outline of the Constitution of the Dominion ot Canada, a Table of Chronology and Questions for Examination. Price, 60 cents.

The Child's History of Canada ; prepared for

the use of Elemeutary Schools, and of the Young Reader, with Maps and Illustrations. Price, 30 cents.

Histoire du Canada pour les Enfants ; ^ Tusage

des Ecoles El^mentaires — Traduit de I'edition anglaise, par L. Devisme, B.A., de PUniversit^ de France. Price 25 cents.

The History of Canada under the French Re- gime, 1535-1763. This volume is recommended as a Beading Book in the higher Academies. It is a volume of 535 pages, and is illustrated by Many maps and Plans. An Appendix is given containing Notes and Documents explan- atory of the Text. ^

This series of Histories has been approved by the Council of Public Instruction for use in the English and French Schools.

By T. R. JOHNSON;

A Text Book on Book-keeping ; Donble Eotrj

made easy. Price, $1.25.