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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
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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
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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(
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A
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/^
/
COMPLETE ARITHMETIC,
ORAL AND WRITTEN.
Designed for the Use of Common and High Schools
AND Collegiate Institutes.
REV. D. H. MACVICAR, LL.D.,
PRLVCIPAL, PrbSBTTBRIAH COLLBOl, MOHTRIAL.
■ « » ■
MONTREAL:
DAWSON BROTHERS, PUBLISHERS
1879.
OA /03
1^31
Entered according to Act of Parliament of Canada, in the year 1879, byl
Dawson Brothers, in the Office of the Minister of Agriculture.
PREFACE.
THIS work is based upon the Complete Arith-
metic prepared two years ago by M. JMacVicar,
Mi.D., LL.D., Principal of the State Normal School,
'otsdam, N. Y., and the undersigned. In the present
vohime the order is changed throughout, and the
work is specially adapted to the wants of the Com-
mon and High Schools, and Collegiate Institutes of
'unada.
Attention is invited to the Properties of Numbers,
rreatest Common Divisor, Fractions, Decimals, Com-
[)ound Numbers, The Metric System, Business Arith-
metic, Ratio and Proportion, Alligation, and Square
md Cube Root, with the belief that the treatment
will be found new and an improvement upon former
uethods.
In every subject the pupil is first required to
baster elements or preparatory steps and propositions
In order to fit him for the more advanced and com-
)lex operations.
A systematic drill is provided on Oral and Written
Exercises and Review and Test Questions, by which
[he principles and processes of numbers, with their
iv
F liEFAC E.
applications to j)racticul busiiiesri, will be permanently
fixed in the mind.
In all cases in which it is possible, each process is
presented objectively, so that the truth is exhibited to
the eye and thus clearly defined in the mind.
The entire drill and discussions are believed to be
so arranged, and so thorough and complete, that by
passing through them the pupil cannot fail to acijuire
such a knowledge of principles and facts, and to
receive such mental discipline, as will prepare him
properly for the study of the higher mathematics.
Invaluable aid in methods of presentation and the
management of class work will be found in the
Teacher's Edition of Dr. M. MacVicar's Complete
Arithmetic, published by Taintor Brothers, Merrill
&Co., N. Y.
D. H. MacVICAE.
Montreal, July, 1879.
CONTENTS.
Page
Notation and Numeua-
TION 1
Roman Notation 9
Revikw and Test Ques-
tions 10
Addition 11
Canadian Money 18
Rkview and Test Ques-
tions 20
Subtraction 21
Review and Test Ques-
tions 27
Multiplication 28
Review and Test Ques-
tions 40
Division 42
Division by Factors ... 55
Review and Test Ques-
tions 60
Properties op Numbers 66
Exact Division 67
Prime Numbers 70
Factoring 72
Cajjcellation 74
Greatest Com. Divisor. 77
Least Com. Multiple. . 85
Pagk
Review and Test Ques-
tions 89
Fractions 90
Reduction 93
Addition 102
Subtraction 105
Multiplication 107
Division 114
Complex Fractions 119
Review and Test Ques-
tions 126
Decimal Fractions 128
Notation and Numera-
tion 129
Reduction 134
Addition 142
Subtraction 143
mui.tiplication 144
Division 140
Review and Test Ques-
tions 152
Denominate Numbers.. 154
Units op Weight 155
Comparative Units op
Weight 156
Units of Length 168
VI
CONTENTS,
Page
Units op Surfaoe 171
Units of Volume. . . . 174
Units of Capacity. . . . 180
Comparative Units of
Capacity 182
Units of Time 183
Units of Money 186
Metuic System 188
Decimal Related Units 188
Comparative Table of
Decimal Units 191
Duodecimals 192
Longitude and Time . . 193
Review and Test Ques-
tions 195
Business Arithmetic... 196
Aliquot Parts 199
Business Problems 202
Applications 217
Profit and Loss 218
Commission 220
Insurance 222
Stocks 224
Duties or Customs 228
Review and Test Ques-
tions 5)29
Interest 230
Method by Aliquot
Parts 232
Method by Six Per
Cent 234
Method by Decimals . . 236
Page
Exact Interest. 237
Compound Interest 242
Interest Tables 244
Annual Interest 246
Partial Payments 247
Discount 251
253
256
257
261
, Bank Discount
'] Exchange
Inland Exchange
Foreign Exchange. . .
Equation of Payments. 266
Review and Test Ques-
tions 275
Ratio 276
Proportion 281
SiMPiJE Proportion 285
Compound Proportion. 289
Partnership 292
Alligation Medial. 296
i Alligation Alternate. 297
Involution 302
j Evolution 304
Progressions 320
Arithmetical Progres-
sion 321
Geometrical Progres-
sion 323
Annuities 325
Mensuration 328
Review and Test Ex-
amples 342
Answers 358
ARITHMETIC.
NOTATION AND NUMERATION.
NUMBERS PROM 1 TO 1000.
Art, 1. Numbers are expressed by means of ten figiirei,
r.».o«,. 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 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.
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 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 MDL XIV ; 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 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 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
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.
5
9
7
G
8
9
7
^
9
I ^
lO.
8
5
7
8
9
G
7
8
S
wk
11.
7
5
9
^
7
8
5
G
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
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 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
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
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
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 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 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 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, 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 + 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 with a remainder 4.
(3.) 10 = + 1 = 1 nine + 1. Hence, 80 = 8 nines + 8, and
is divisible by with a remainder 8.
(4.) From the foregoing it follows that 400 + 80 + 6, or 486, is
divisible by 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 (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
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 P E RTI E S 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. 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
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 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 - = 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
in o1)taincd.
3. Obscrvt; tlial the number of tinu'H 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 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 the snuiUer number. Hence
tlie <i^reatest ronim<m divisor of 2? and 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 (
108
8 ) 2S ( 3
24
4)8(2
•
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 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
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•^,
■-h-
■*m
^
A/.
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82
PROPERTIES OF NUMBERS,
ARITHMETICAL DRILL TABLE NO. 4.
101. Table for Written Exercises in Greatest Common
Divisor and Least Common Multiple.
A.
B.
c.
D.
E.
F.
1.
30
36
154
176
88
198
a.
48
210
72
54
84
126
n.
252
396
264
480
220
792
4.
60
120
40
420
175
195
6.
132
26j^
396
462
594
528
6.
UO
105
420
156
315
585
7.
96
280
112
192
336
840
8.
198
315
297
693
567
594
9.
210
350
240
300
720
630
10.
132
220
264
308
660
528
11.
168
480
504
420
252
540
13.
156
312
130
364
273
351
EXAMPLES FOR PRACTICE.
10i2. Find the greatest common divisor of the following:
1.
357 and 483.
6.
385 and 1085.
3.
195 and 465.
7.
356 and 808.
3.
418 and 330.
8.
195 and 483.
4.
803, 546,- and 124.
9.
546, 4641, and 364.
5.
455 and 1085.
10.
465, 1365, and 215.
19«$. Continue the practice in finding the greatest common
divisor of abstract numbers by taking examples from the
above Arithmetical Tables. Let all the examples taken
e BEATS ST VOilMOy D I VI SOB. 83
I. 4.
Common
F.
198
12G
792
195
52S
58')
8JfO
50^
I GclO
528
5J^0
351
owing :
common
I from the
los token
from Table No. 3 be worked orally and in sets in the same
manner as directed for written exercises.
JExutnples with Two Auinbcrs.
194. FiEST Set. — Take examples for written exercises
from the first line of Table No. 4, thus :
(1.) 30 36
(2.) 36 154
(3.) 154 170
(4.) 176 88
(^.) 88 196
Observe that in each new example, the first number taken
in the last example is omitted and a new number added.
Take in the same manner examples from each line in the
table.
19,1. Second Set. — Take examples from the first and
second lines thus :
(1.) 30 48
(3.) 154
72
(5.) 88 84
(2.) 30 210
(4.) 176
54
(6.) 198 126
To present other examples, omit the first line and use the
second and thii'd, then the third and fourth, and so on to the
bottom of the table.
190. TiiiKD Set. — Take the numbers from the first and
thh'd line, then from the second andfoyrth, then from the tJiird
^nd fifth, etc., to the bottom of the table.
Examples with Uiree Nntnbers.
197. First Set. — Take examples from each line, thus:
(1.) 30 30 154
(2.) 36 154 176
(3.) 154 170 88
(4.) 170 88 198
198. Second Set. — Take the numbers from the first,
second, and third lines, then from the second, third, and fourth,
and so on to the bottom of the table.
>?''>■§
m
■*' ■ - t;
<M
'M.
1 • >!l
84
PROPERTIES OF NUMBERS,
WRITTEN EXAMPLES.
199. 1. I have rooms 12 feet, 15 feet, and 24 feet wide;
what is the width of the widest carpeting that will fit any room
in ray house? A us. 8 feet.
2. Divide the greatest common divisor of 48, 72, UO, and 120
by the greatest common divisor of 21, 30, 39, and 84,
3. In the city of Montreal, some of the sidewalks are
48 inches wide, some CO inches, and others 72 inches ; what
is the widest flagging that can be used in each of these side-
walks without cutting ? Anfi. 12 inches.
4. D. White owns in Hamilton 3 lots of equal depth, the first
having a front of 72 feet, the second 144 feet, and the third
108 feet, which he wishes to divide into as many lots as pos-
sible having equal fronts ; how many feet will each front con-
tain ? Ans. 30 feet.
5. A teamster agrees to cart 132 barrels of flour for a mer-
chant on Monday, 84 barrels on Wednesday, and 108 barrels
on Friday; what is the largest number he can carry at a load,
and yet have the same number in each ? Afis. 12 barrels.
6. I have a lot whose sides measure, respectively, 42 feet,
84 feet, 112 feet, and 126 feet ; I wish to enclose it with boards
having the greatest possible unifonn length ; Avhat will be the
length of each board ? Ano. 14 feet.
7. A merchant has three pieces of cloth containing respec-
tively 42, 98, and 84 yards, which he proposes to sell in dress
patterns of uniform size. What is the largest number of yards
the dress i>atterns can contain so that there may be nothing
left of either piece ?
8. If two farms containing each nn exact number of acres
were purchased for $8132 and $0270 respectively, what is the
highest uniform price per acre th.at could have been paid, and
in this case how many acres in each farm ?
LEAST COJfJlOX MULTIPLE,
85
LEAST OOMMOjS" MULTIPLE.
PREPARATORY PROPOSITIONS.
200. Study carefully each of the following propositions :
Prop. I. — A multiple of a number contaiTis as a factar each
prime factor of the number as muiiy times as it enters into the
number.
Thusi, 60, which Is a multiple of 12, contains 5 times 12, or 5 times
2x2x3, the prime factors of 12. Hence, each of the prime factors of 12
enters as a factor into 60 as many times as it enters into 12.
Prop. ll.— T7ie least common multiple of two or more given
numbers must contain, as a factor, each prime factor in those
numbers the greatest number of times that it enters into any one
of tliem.
Thus, 12 = 2 X 2 X 3, and 9 = 3x3. The prime factors in 12 and 9 are 2
and 3. A multiple of 12, according to Prop. I, must contain 2 as a factor
twice and 3 once. A multiple of 9, according to the same propopition, must
contain 3 as a factor twice. Hence a number which is a multiple of both
12 and 9 must contain 2 as a factor twice and 3 twice, which is equal to
2 X 2 X 3 X 3 = 36. Hence 36 is the least common multiple of 12 and 9.
'•, ?!
'4 ll
11'"
'^i>\
DEFINITIONS.
201. A Multiple of a number is a number that is exactly
divisible by the given number.
Thus, 24 is divisible by 8 ; hence, 24 is a multiple of 8.
202. A Conunon Multifile of two or more numbers is
a number that is exactly divisible by each of them.
Thus, 36 Is divisible by each of the numbers 4, 9, and 12 ; hence, 36 is a
common multiple of 4, 9, and 12.
2015. The Least Common Multiple of two or more
numbers is the least number that is exactly divisible by each
of them.
Thus, 24 is the least number that is divisible by each of the numbers 6
and 8 ; hence, 24 is the least common multiple of 6 and 8.
m
86 PROPERTIES OF NUMBERS,
METHOD BY FACTORING.
ILLUSTRATION OF PROCESS.
204. Prob. I. — To find, by factoring, the least com-
mon multiple of two or more numbers.
Find the least common multiple of 18, 24, 15, and 35.
Explanation.— 1. We observe that
3 is a factor of 18, a4, and 13. Dividing
these numbers by 3, we write the quo-
tients with 35, in the second line.
2. Observing that 2 is a factorof 6 and
8, we divide as before, and find the third
line of numbers. Dividing by 5, we find
the fourth line of numbers, which are priiLe to each other; hence cannot
be further divided.
3. Observe tlie divisors 8, 2, and 5 are all the factors that are common to
any two or more of the given numbers, and th quotients 3. 4, and 7 are the
factors that belong each only to one number. Therefore the divisors and
quotients together contain each of the prime factors of 18, 24, 15, and 35 as
many times as it enters into any one of these numbers. Thus, the divisors
Sand 2, with the quotient 3, are the prime factors of 18; and so with the
other numbers.
Hence, according to (200—11), the continued product of the divisors 3,
2, and 5, and the quotients 3, 4, and 7, which is equal to 2520, is the least
common multiple of 18, 24, 15, and 85.
From this illustration we have the following
3
18
24
15
35
2
6
8
5
35
5
3
4
5
35
3
4
%
RULE.
205. L Write the numbers in a line, and divide by any
prime fci^tor that is contained in any two or more of them,
placing the quotients and the undivided numbers in the line
below.
II. Oj)erate upon the second line ofnumbem in the mme man-
ner, and so on until a line of numbers that are prime to each
other is found.
III. Find the continued product of the divisors used and the
numbers in the last line ; this icill give tlie least common multiple
of th^ given numbers.
LEAST rOJfJfO.Y MULTIPLE,
87
Ml
WRITTEN EXAMPLES.
200. 1. What is the least number of cents that can be
exactly expended in oranges, whether they cost 4, o, or 6 cents
apiece ?
2. What is the least common multiple of the nine digits ?
3. What is the smallest quantity of milk that will exactly
fill either six-quart, nine-quart, or twelve-quart cans?
4. ^Vhat is the smallest sum of money that I can exactly lay
out in calves at 14 dollars each, cowa at 38 dollars each, or
oxen at 57 dollars each ? Ans. 798 dollars.
5. A can lay 42 rows of shingles on my house in a day, and
B can lay 50 rows ; what is the least number of rows that will
give a number of full days' work to either A or B ?
6. What is the width of the narrowest street across which
stepping-stones either 3, 4, or 9 feet long will exactly reach ?
7. Three separate parties are measuring the distance from
the city hall, Kingston, to the University, Toronto ; one party
uses a chain 33 feet long, another a chain 66 feet long, and the
third a chain 50 feet long, marking each chain's length with a
stake ; at what intervals of space will three stakes he driven
at the same place ? Ans. Every 1650 feet.
M
m^
METHOD BY GREATEST COMMON DIVISOR.
ILLUSTRATION OF PROCESS.
207. Prob. II.— To find, by using the greatest com-
mon divisor, the least common multiple of two or more
numbers.
Find the least common multi])le of 195 and 255.
Explanation.—!. We find the greatest common divitor of 195 and 255,
which is 15.
2. The greatest common divipor, 15, according to (182), contains all
the prime factors that are common to 1'15 and 255. Dividing each of these
numbers by 15, we find the factors that are not common, namely, 13 and 17.
l-f
-m
Mm
1..^.
88
PROPERTIES OF N U Jif B E R S .
3. The common divisor 15 and the qnoticnt 13 contain all the prime
factors of 195, and the common divisor 15 and the quotient 17 contain all
the prime factors of 255.
Hence, accor(!'ng to (^200—11), the continued product of the common
divisor 15 and the quotients 13 and 17, which is 3315, is the least common
multiple of 195 and 255.
The least common . uiltiple of any two numbers is found in
the same manner ; hence the following
RULE.
208. /. Find the grefftent common dicisar of the two given
numbers, and divide each of the numbers by this divisor.
II, Find the continued product of the greatest common divisor
and the qvotients ; this will give the least common midtiple of
the two given numbers.
To find the least common multiple of three or more numbers
by this method, we have the foUowinpf
RULE.
200* Find the least common mtdtiple of two of them; then
find the least common multiple of the multiple thus found and
the third number , and so on with four or more numbers. ,
i:XAMPLES FOR PRACTICE.
Ill
m
[!*
210. Find the least common multiple of
1. 110 and 165.
2. 91 and 182.
3. 78 and 195.
4. 143 and 165.
5. 385 and 455.
6. 154 and 231.
7. 462 and 546.
8. 364 and 637.
211. For further practice take examples with two numbers
from Table No. 4, page 82, as directed in (194), (195), and
(196) ; and examples with three numbers from Table No. 3,
as directed in (197) and (198).
Continue to practice with abstract numbers until you can
find the least common multiple of two or more numbers
accurately and rapidly.
LEAST COMMOX MULTIPLE,
89
BEVIEW AND TEST QUESTIONS.
212. 1. Define Prime Number, Composite Number, and
Exact Divisor, and illustrate each by an exami)]e.
2. What is meant by an Odd Number? An Even Number?
3. Show that if an even number is divisible by an odd num-
ber, the quotient must be even.
4. Name the prime numbers from 1 to 40.
5. Why are all even numbers except 2 composite ?
6. State how you would show, in the series of odd numbers,
that every fifth number from 5 is divisible by 5.
7. What is a Factor ? A Prime Factor ?
8. What are the prime factors of 81 ? Of 64 ? Of 125 ?
9. Show that rejecting the same factor from the divisor and
dividend does not change the quotient.
10. Explain Cancellation, and illustrate by an example.
11. Give reasons for calling an exact divisor a measure.
12. What is a Common Measure ? The Greatest Common
Measure ? Illustrate each answer by an example.
13. Show that the greatest common divisor of 42 and 114 is
the greatest common divisor of 42 and the remainder after the
division of 114 by 42.
14. Explain the rule for finding the greatest common divisor
by factoring ; by division.
15. Why must we finally get a common divisor if the greater
of two numbers be divided by the less, and the divisor by the
remainder, and so on ?
16. What is a Multiple ? The Least Common Multiple ?
17. Explain how the Least Common Multiple of two or more
numbers is found by using their greatest common divisor.
18. Prove that a number is divisible by 9 when the sum of
its digits is divisible by 9.
19. Prove that a number is divisible by 11 when the differ-
ence of the sums of the digits in the odd and even places
is zero.
5' '^ -» 1
FRACTIONS.
PREPARATORY PROPOSITIONS.
213. Prop. I. — Any thing regarded as a whole can be
divided Into unequal or equal parts ; thus,
WHOLE.
PARTS.
Hi'
■iH'
(1.)
(2.)
1. Equal parts of a whole are called Fractions.
2. Into what kind of parts can a pear be divided ? A bushel
of wheat? A slate? A garden ? Anything?
3. Make $12 into unequal parts in six ways, and into equal
parts in five ways ?
4. In how many ways can 15 be made into equal parts?
Into unequal parts ?
Prop. II. — The same whole can be divided into equal parts of
different s-izea ; thus, . • •
WHOLE.
EQUAL PARTS.
Halves.
Thirds.
Fourths.
Fifths.
1. Observe, the equal parts are named by using the ordinal
corresponding with the number of parts. Thus, when the
whole is made into three parts, one part is called a third, when
PREPARATORY PROPOSITIONS, 91
into four parts, one part is called a fourth, and so on to any
mimber of parts.
2. When the whole is made into ten equal parts, what is
one part called ? Into sixteen equal parts ? Into twenty-four ?
Into forty-three ?
3. What are the largest equal parts that can be made of a
whole ? The next largest ? The next largest ?
4. What ib meant by one-half of an apple? One-third?
One-fifth ?
5. What is meant by two-thirds of a line? Of an hour?
Of a day?
0. How would you find the fourth of anything? The
seventh? The tenth ?
7. Find the third of 6. Of 12. Of 15. Of 24. Of 48.
8. If a whole is made into twelve equal parts, how would
you name three parts? Seven parts? Five parts? Nine
parts ?
9. How many halves make a whole? How many thirds f
How many sevenths? How many tenths? How many fif-
teenths ?
Prop. III. — Equal parts of a whole, or Fractions, are
expressed by two numbers written one over the other, with a line
between them ; thus,
Numerator, lL Shows the number of equal parts In the fraction.
Dividing Line, ^ Shows that 4 and 5 express a fraction.
Denominator, ^ Shows the number of equal parts in the whole.
Read, Four -fifths.
,' t
1.
Read the following :
3 7 6 9
"5> F» ^» TTF?
I If.
*
2.
What does f signify
? f? If?
23 9
:Vi) •
3.
Express in numbers three-fifths •
nine-
■thirteenths ;
eleven-
titteenths.
4.
Read the following :
fi M-> \l
Uh
ih ni
5.
Write in numbers
eight-twentieths
; twelve-sixteenths; j
fifteen -seventieths ; nine-
■fortieths.
■
92
FHA t'TIONS.
C. What does Numerator mean ? Denominator f Dividing
lino ? Tirniit of a fraction ?
7. How is a fraction oxpreflsod by numbers?
8. Name tlie tenns of I Viy. ifl. n- Vh ill
9. Express in numbers seven-nintlis ? Nineteen forty-fifths.
Prop. IV. — The me or value of the same kind of equal parts
depends upon the size or value of the whale of which they are
parts; tlius,
f
WHOLE.
EQUAL PARTS.
Hnlvpft,
IlalvvH,
1. Tlio equal parts in the illustration, although halves in
both cases, are unequal in size, because the wholes are unequal
in size.
2. Which is the larger, the half of $4 or the half of |6 ?
8. Which is the smaller, the fourth of 13 inches, or the
fourth of 20 inches, and why?
4. If oft'ored the half of either of two farms, which would
you take, and why ?
Prop. V. — The size or value of the equal parts of a whole
diminish as the number of parts increase^ or increase as the
number of parts diminish ; thus.
WHOLE.
EQUAL PABT8.
Tttirda.
Fourths*
Fifths.
1. Which is the greater, one-half or one-third? One-fourth
or one-fifth ? One-sixth or one-ninth, and why ?
2. How much is J of $48 smaller than J of it?
3. Upon what two things does the value of one-half, one-
third, one-fourth, one-fifth, etc., depend? Illustrate your
answer by two examples.
PHE1*AKAT0RY PROPOSITIONS, 93
DEPIinTIONS.
214« A Ft*actional Unit is one of the equal parts of
fiuything regarded as a whole.
iil5. A Fraction is one or more of the equal parts of
anything regarded as a whole.
21($. The Unit of n Fraction is the unit or whole
which is considered as divided into equal parts.
217. The Nnincrator is the number above the dividing
line in the expression of a fraction, and indicates how many
equal parts are in the fraction.
218. The Denominator is the number below the
dividing line in the expression of a fraction, and indicates how
many equal jmrts are in the whole.
210, The Terms of a fraction are the numerator and
denominator.
220. Taken together, the tei^ms of a fraction are called a
Fraction, or Fractional Namher,
221. Hence, the word Fraction, means one or more of the
equal parts of anything, or the expression that denotes one or
more of the equal parts of anything.
REDUCTION".
PREPARATORY STEPS.
222. Step I. — A fraction is represented by lines thus : •
3 ^^iM ^■HH Part taken.
3 KMiM^-^a^M Ttliole.
Observe carefully the following :
1. In f , the denominator 3 expresses the whole, or 3 thirds,
and the numerator 2 expresses two parts of the same size.
'ii.ll
ft
. if 'f
„■'■■!?
*
94
FEA C TIONS.
Hence, 3 equal lines for the denominator and 2 equal lines
for the numerator, of the same length as those in the denom-
inator, represent v orrectly the whole, the parts taken, and the
relation of the parts to each other, as expressed by the frac-
tion
8
5-
2. Represent by lines f ; f ; y? ; iV J « ; i^'a-
Why can the numerator and denominator of a fraction be
represented by equal lines ?
Step II. — Show hy representing t?ie fraction with lima that
one-half 18 equal to two-fourths ; XhxxB, .
- %.-\ 2 -■
Part takeiit
Whole,
1. By observing the illustration, it will be seen that the value
of the numerator and denominator is not changed by making
ea^h part in each into two equal parts. It will also be seen
that when this is done the numerator contains 2 parts and the
denominator 4. Hence ^ = f .
2. Show in the same manner that one half is equal to three-
sixths, four-eighths, five-tenths, and so on.
Step III. — Any fractional unit can, without changing its
value, he divided into any desired number of equal parts.
Study carefully and explain the following illustrations :
'tgeu
u
t
2.
1
8
^ ^^
2
6
1 '
■ sum
1
1
4
8
18
—
— •
■ Be
■ by tb
RED UCTIoy.
96
:m
ORAL EXAMPLES.
223. 1. How many tenths in I of an orange ? How many
tifteenths ? How many twentieths ? etc., and why ?
2. How can ^ of a yard be made into sixteenths of a yard ?
3. How many twelfths in^? In^? In|? In^?
4. Make | into twenty-firsts, and explain the process.
5. Show by lines that J = j*^ ; that | = tj^ ; that i = ^.
6. Change |, without altering its value, into a fraction con-
taining 7 equal part 3 ; 10 equal parts ; 25 equal parts.
PBINCIPIiES OP BEDUCTION.
224. Let each of the following principles be illustrated by
the pupil with a number of examples :
Prin. I. — The numerator and denominatoi' of a fraction
represent, each, parts of the same size ; thus.
7
4
5
(1.)
(3.)
Observe in illustration (1) the denominator 7 represents the whole or
7 sevenths, and the numerator 3 represents 3 sevenths ; in illustration (2),
the denominator represents b fifths^ and the numerator A fifths. Hence the
numerator and denominator of a fraction represent parts of the same size.
Prin. H. — Multiplying both the terms of a fraction by the
same number does not change the value of the fraction ; thus,
2
3
2x4
8x4
8
12
Be particular to observe in the illustration that the amount expressed
by the 2 iu the numerator ur the 3 in the denominator of | is not
«:
96
FRA CTIO N8,
changed by making each part into 4 equal parts; therefore, J and t\
express, each, the same amoant of the same whole.
Hence, multiplying the numerator and denominator by the same number
means, so far as the real fraction is concerned, dividing the equal partf in
ea£h into as many equal parts as there are units in the number bji which they
are multiplied.
Prin. III. — Dividing both terms of a fraction by the same
number does not change the value of the fraction ; thus, ,
12-5-3
8
4
The amount expressed by the 9 in the numerator or the 12 in the denom-
inator of t'?! Is not changed by putting every 3 parts into one, as will be
seen from the illustration.
Hence, x"i and | express each the same amount of the same whole, and
dividing: the numerator and denominator by the same number means jmt-
ting as many parts in each into one as there are units in the number by
which they are divided.
/»!■
hi
DEFINITIONS.
*2*2i5. The Value of a fraction is the amount which it
represents. , .^
226. Reduction is the process of changing the terms of
a fraction without altering its value.
227. A fraction is reduced to Higher Terms when its
numerator and denominator are expressed by larger numbers.
Thus, i = tV-
228. A fraction is reduced to Lower Terms when its
numerator and denominator are expressed by smaller numbers.
Thus, j% = I
229. A fraction is expressed in its Lowest Terms when
its numerator and denominator are prime to each other.
u
ILLUSTRATION OF PROCESS,
97
Thus, in f , the numerator and denominator 4 and 9 are prime
to each other; hence the fraction is expressed in its lowest
terms.
230. A Common Denominator is a denominator that
belongs to two or more fractions.
231. The Least Cotnmon Denominator of two or
more fractions is the least denominator to which they can all
be red: V wed.
232. A Proper Fraction is one whose numerator is
less than the denominator ; as f , f .
233. An Impro2)er Fraction is one whose numerator
is equal to, or greater than, the denominator ; as f , |.
2*34. A 3Iixed Nwinher is a number composed of an
integer and a fraction ; as 4f , 18f .
ILLUSTRATION OF PROCESS.
235. Prob. I.— To reduce a whole or mixed number
to an improper fraction.
1. Reduce 3| equal lines to fifths.
WHOLES. FIFTHS.
= IS fifths =
18
Explanation.— Each whole line is equal to 5 J^fths, as ehown in tiie
illustration ; 3 lines must therefore bo equal to 15 Jlfths. \hjifths + Z fifths
= 18 fifths. Hence in 3J lines there are V of a line.
From this illustration we have the following :
^H^4
RULE.
23C$. Multiply the whole luimher by tlie given denominator,
and to the product add the numerator of the given fraction, if
any, and icrite the result over the given denominator.
98
FRA CTIONS,
EXAMPLES FOR PRACTICE.
237. Reduce orally the following :
1. In 5 pounds of sugar how many fourt?i8 of a pound ?
Solution.— In 1 pound of sugar there are \ fourths ; hence, in 5 pounds
there are 6 times i/ourths, which Is \^ of a pound.
2. In 7 tons of coal how many ninths of a ton ?
d. Bow msLUj tenths in $Qd1 In 42 yards? In 17 pounds?
4. Express 20 slb fourths. As sevenths. As hundredths.
5. In $9f how many sevenths of a dollar ?
Solution.— In $1 there are 7 sevenths. In $9 there must therefore be
9 times 7 sevenths or 63 sevenths. 63 sevenths + 3 sevenths are equal to
66 sevenths. Hence, in $9f there are V of a dollar. • ' .s.
6. In 12^ acres how many twelfths of an acre?
7. How many eigJitlis in 9| ? In 11 1 ? In 7f ? In 5| ?
Reduce the following to improper fractions :
8.
83i.
12.
340«.
16.
nn
9.
45|.
13.
462g.
17.
3ff.
10.
76f.
14.
1875.
18.
4t^
11.
13H.
15.
463^\.
19.
3tU-
238. Prob. II.— To reduce an improper fraction to
an integer or a mixed number.
1. Reduce 9 fourths of a line to whole lines.
» =r 9^4 =
2i
\il
Explanation.— A wholo line is compoeed of Af&iiriht<. Hence, to make
the ^fourths of a line into whole lines, we put every /ot/r parts into on^, as
shown iu the illuntration, or divide the 9 by 4, which gives 2 wholes and 1
of ihefourtlis remaining. Hence the following
RULE.
239. Divide tlie numerator hy the denominator.
EXAMPLES.
99
EXAMPLES FOR PRACTICE.
7B 1
240^ Reduce and explain orally the following:
1. How many bushels are -^/ of a bushel? ^■'- V y? y^
2. In $-"/, how many dollars ? In V o^ a yard, how many-
yards V In -^^ of a foot, how many feet ?
3. How many miles in ^^ of a mile ?
7 •
Reduce to whole or mixed numbers the following ;
4.
H.I 8
2S •
5.
508
Br*
6.
w.
7.
w.
8.
ni
9.
^fF.
10.
^tW.
11.
ntl^
12.
W^
13.
-'bW-
14.
15.
16.
17.
18.
50705
9 0067
nsoi
80«6
3F~'
241. Prob. III. — To reduce a fraction to higher terms.
1. Reduce f of a line to twelfths.
I 2x4_ 8_
I * 3x4~ 12
Explanation.— 1. To make a whole, which is already in thirds, into 12
equal part8, each third must be made miofcmr equal parts.
2. The numerator of the given fraction expresses 2 thirds, and the denom-
inator 3 thirds; making each third in both mXo four equal i)arts (224—11),
as shown in the illustration, the new numerator and denominator will each
contain 4 times as many parts as in the given fraction.
Hence, \ of a line is reduced to tice^ths by multiplying both numerator
and denominator by 4.
Hence the following rule for reducing a fraction to higher
terms :
%l
RULE. _..
242. Divide the required denominator by the denominator
of the given fraction, and multiply the terms of the given frac-
tion hy the quotient.
I
100
FEA CTIONS,
EXAMPLES FOR PRACTICE.
243. Reduce and explain orally the following :
1. How would you make liahea of an apple into fourths?
luU) .sixths? Into tenths? Into siosteentTis ?
2. How many twelft/ts in f of a rx)rd o' wood ?
8. Explain how i^, ^, and f can be reduced to twentieths.
4. Show by the use of lines that f = * = ^ = -^^ = |§.
5. Reduce |, f , f, ^, and ^f each to forty -seconds.
6. In f how many ninety-eighths ?
7. Change f , f^, ^, f , and ^^ each to 360ths.
8. Reduce f , ^, ^, ^f , and ft to 165ths.
244. Pros. IV.— To reduce a fraction to lower
terms.
Reduce ^^j of a given line to fourths.
12^3 "■
f
4
ExPLAKATiON.— 1. To make into 4 equal parts or fourths a whole which
is already in 12 equal parte, or twelfths, every 3 of the 12 parts must be
put into one.
2. The numerator of the given fraction expresses 9 twelfths, and the de-
nominator 12 twelfths ; putting every 3 twelfths into one, in both (224— III),
as shown in the illustration, the new numerator and denominator will each
contain one-third as many parts as in the given fraction.
Hence tt of a line is reduced to fourths by dividing both numerator and
denominator by 3.
Hence the following rule for reducing a fraction to its
lowest terms :
BULB.
245. Reject from tJie terms of the giten fraction aU their
common factors. Or,
Divide the terms of the given fraction hy their greatest common
divisor.
EXAMPLES.
101
EXAMPLES FOR PRACTICE.
24G. Reduce and explain orally the following :
1. In ^ of a bushel, how many thirds of a bushel?
2. How can twelfths of a bushel be made into fourths of
bushel? \u\jo tJiirdaf Into halcesf
3. Reduce ^^ of a dollar to Jijths of a dollar.
4. Show by the use of lines that ^^ = ig = yV = I = i«
5. Reduce ^\ to its lowest terms. ■^^. |g. ||.
C. Express if in parts 8 times as great in value.
Reduce the following to their lowest terms :
7. VV\. 10. Uh 13. fM. 16.
8. ^Vtr- 11. Uh 14. ,%¥.. 17.
To*
9. Ut
12.
847
88
15.
fro*
U^ 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 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 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 ff •
10.
11.
12.
58
706
1 OTf (5-
so SB
"I 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!
•
J3
•
a
o
a
o
5
e
e
es
tt.
S
o
.a
•^^
c
c
.a
*^
o
'a
m
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 S ITI 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
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 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 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 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 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. 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. 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. 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
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. 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 ^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 ; 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. 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. 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)
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Corporation
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23 WEST MAIN STREET
WEBSTER, NY. 14580
(716) 872-4503
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176
DENOMINATE NUMBERS.
V I
EXAMPLES FOR PRACTICE.
406. Reduce and explain the following :
1. 97 cu. ft. to cu. in. 3. 4 cu. yd. 394 cu. ft. to en. izu
2. .09 of a cu. yd. to cu. ft. 4. .0007 of a cu. yd. to cu. ia.
5. Find the sum of f of a cu. yd. and .625 of a cu. ft.
6. .8 of a cu. ft. to a decimal of a cu. yd.
7. f of a cu. ft. to a decimal of a cu. yd.
8. Find the difference between | of a cu. yd. and .75 of a
cu. ft.
Find the contents of rectangular solids of the following
dimensions :
9. A solid 7 ft. 9 in. long by 3 ft. 4 in. by 4 ft. 6 in.
10. A cube whose edge.is 3 yd. 2 ft. 8 in.
11. A solid 34 ft. long by 1 ft. 6 in. by 2 ft. 9 in.
12. A solid 12 yd. 1 ft. 9 iu. long by 2 yd. 2 ft. by 2 ft. 8 in.
13. How many cubic feet in a stick of timber 38 ft. long by
2 ft. 3 in. by 1 ft. 9 in.?
14. A cistern 9 ft. sq. contains 1092 cu. ft. ; what is its
depth?
15. A stick of square timber contains 189 cu. ft. ; 2 of its
dimensions are 1 ft. 9 in. and 2 ft. 3 in. ; what is the other?
16. How many cubic yards of earth in an embankment
283 ft. by 42 ft. 8 in. by 18 ft. 6 m. ?
17. How many cubic feet of air in a room 74 ft. 9 in. long,
52 ft. 10 in. wide, and 23 ft. 6 in. high?
18. A bin contains d2Q\ cu. ft. ; 2 of its dimensions are 9 ft.
8 in. and 7 ft. 6 in. ; what is the other?
19. A vat is 7 ft. 2 in. by 4 ft. 9 in. by 3 ft. 4 in. How many
cubic feet does it contain?
20. In digging a cellar 48 ft. 6 in. by 39 ft. 8 in., and 8 ft.
4 in. deep, how many cubic yards of earth must be removed?
EXAMPLES.
177
21. What will be the cost of the following bill of square
timber, at f .33J per cubic foot :
(1.) 3 pieces 13 ft. by 9 in. by 7 in. ?
(2.) 8 pieces 15 ft. 6 in. by 10 in. by 8 in. ?
(3.) 4 pieces 33 ft. by 8 in. by 9 in. ?
(4.) 6 pieces 3G ft. by 1 ft. 6 in. by 1 ft. ?
(5.) 9 pieces 18 ft. 9 in. by 1 ft. 3 in. by 9 in. ?
(6.) 12 pieces 15 ft. by 7^ in. by 9^ in. ?
22. How many perches in a wall 37 ft. long, 23 ft. 6 in.
high, and 2 ft. G in. thick ?
Table of Units for Measuring Wood and' Stone*
16 cu. ft. = 1 Cord Foot (cd. ft.) \ Used for measuring
8cd.ft.or ) ^ icord(cd.) [ „,'»"'
128 cii. ft. ^ ; wood and stone.
2i}l cu. ft. = 1 perch (pch.) of stone or masonry.
1 cu. yd. of earth is called a load.
1. The materials for masonry are npually estimated by the cord or perch^
the work by the perch and cuMc foot, also by the square foot and square
yard.
2. In estimating the mason work in a building, each wall is measured
on the outside, and no allowance is ordinarily made for doors, windows,
and cornices, unless specified in contract. In estimating the material, the
doors, windows, and cornices are deducted.
3. Brickwork is usually estimated by the thousand bricks, which are of
various sizes.
4. Excavations and embankments are estimated by the cubic yard.
How many
EXAMPLES FOR PRACTICE.
407. Reduce and explain the following :
1. 42 cords to cd. feet. 3. 36 cords to cu. feet.
2. 64 pch, to cu. feet. 4. .84 pch. to cu. feet.
178
DENOMINATE NUMBERS.
n
I.
6. 7 of a cd. to cu. feet.
6. .73 of a cu. ft. to a decimal of a cd.
7. f of a cord to a decimal of a cu. yd.
8. ^ of a cu. ft. to a decimal of a pch.
9. .85 of a cord to a decimal of 3 cu. yd.
10. I of 8 cd. to a decimal of 13 cd.
11. Find the sura of f pch., | cd., and 11 cd. ft. 38 cu. ft.
12. A pile of wood containing 84 cd. 7 cd. ft. 12 cu. ft. was
made into 5 equal piles ; what was the size of each ?
13. How many cords in a pile of wood 196 ft. long, 7 ft. 6 in.
high, and 8 ft. wide?
A Cord is a pile of wood, stone, etc., 8 ft. long, 4 ft. wide*
and 4 ft. high.
A Cord Foot is 1 ft. long, 4 ft. wide, and 4 ft. high, or \ of
a cord, aa shown in the cut.
14. What is the cost of a pile of ^tone 28 ft. long, 9 ft. wide,
and 7 ft. high, at |3.85 per cord ?
15. A load of wood containing 1 cord is 3 ft. 9 in. high and
4 ft. wide ; what is its length ?
16. How many perches of masonry will 18 cd. 5 cd. ft. of
stone make, allowing 22 cu. ft. of stone for 1 perch of wall ?
17. How many cords of stone will be required to enclose
with a wall built without mortar a lot 28 rods long and 17 rods
wide, the wall being 5 ft. high and 2 ft. 9 m. thick t
\n
EXAMPLES.
m
BOARD MEASURE.
TABLE OF TTNITS.
12 B. in. = 1 B. ft.
12 B. ft. = 1 cu- ft.
408. A Bofird Foot is 1 ft.
long, 1 ft. wide, and 1 in. thick.
Hence, 12 hoard feet equals 1 cu. ft.
409. A Board Inch is 1 ft. long, 1 in. wide, and 1 in.
thick, or ^V of a hoard foot. Hence, 12 hoa/rd inches equals
1 hoard foot.
Observe carefully the following :
(1.)
4 feet long.
•a
«
Square
foot.
1
1
1. Diagram (1) represents a
l/Ourd where botli dimensions
are feet. Hence tlie product of
the two dimensionB gives the
square feet in surface (396), or
the number of board feet when
the lumber is not more than
1 hich thick.
2. Diagram (2) represents a
board where one dimension is
feet and the other inches. It is
evident (408) that a board 1
foot long, 1 inch thick, and any
number of inches wide, contains
as many board incites as there
are inches in the width. Hence
the number of square feet or
board feet in a board 1 inch
thick is equal to the length in feet multiplied by the width in inches
divided by 12, the number oi board inches in a board foot
3. In case the lumber is more than 1 inch thick, the number of board
feet is equal to the number of square feet in the surface multiplied by the
thickness-
ID
4 X 2 = 8 sq. ft. or 8 B. ft.
(2.)
4 feet long.
1ft. by
9 in.
« 4x9=36 B. In.; 36 B. in.-i-12=3B. ft.
'•">.
EXAMPLES FOR PRACTICE.
410. Find the contents of boards measuring
1. 24 ft. by 13 In. 4. 9 ft. by 32 in. 7. 5 ft. by 18 in.
2. 28 ft. by 15 in. 5. 13 ft. by 26 in. 8. 34 ft. by 15 in.
3. 18 ft. by 16 in. 6. 17 ft. by 30 in. 9. 25 ft. by 14 in.
fi-W
IP!
180
DENOMINATE NUMBERS,
f'/i
Find the contents of boards measuring
10. 15 ft. by 1 ft. 3 in. 12. 19 ft. by 2 ft. 4 in.
11. 27 ft. by 1 ft. 6 in. 13. 23 ft. by 1 ft. 5 in.
14. Find the contents of a board 18 ft. long and 9 in. wide.
15. How many board feet in a stick of square timber 48 ft.
long, 9 inches by 14 inches.
16. Find the length of a stick of timber 8 in. by 10 in., which
will contain 20 cu. ft.
Opbbation.— (1728 x 20) +(8 x 10) = 432 ; 432+12 = 3G ft., the length.
17. A piece of timber is 10 in. by 13 in. What length of it
will contain 26 cubic feet ?
Find the cost of the following :
18. Of 234 boards 14 ft. long 8 in. wide, at $3.25 per hundred.
19. Of 5 sticks of timber 27 ft. long, ^ in. by 14 in., at $1.75
per hundred feet board measure.
20. Of 84 plank 20 ft. long, 11 in. wide, 3 in. thick, at $1.84
per hundred feet board measure.
UNITS OF CAPACITY.
411. The Standard Units of capacity are the Gallon
for Liquid, pjid the Bushel for Dry Measure.
II
LIQUID MEASTTBE.
1. Deuoiuinations.—Gil\a (gi.), Pints
(pt.), Quarts (qt.), Gallons (gal.), Barrels
(bbl.).
2. Equivalents.— 1 gal.=:4 qt.=8 pt. =
32 gi.
3. Use.— Used in measuring liquids.
4. The capacity of cisterns, vats, etc., is
usaally estimated by considering a barrel 31J gal. ; but barrels are made of
various sizes, from 30 to 56 gallons. The hogshead, butt, tierce,' pipe, and
tun are names of casks, and have usually their capacity in gallons marked
upon them.
TABLE OF UNTTS.
4 gi. = 1 pt.
2 pt. =1 qt.
4 qt. = 1 gal.
81| gal. = 1 bbl.
DJiT 3IEASURE,
181
AFOTHECABIES' FLUID MEASURE.
in.
in.
in. wide.
mber 48 ft.
Oin.,wliicli
} length,
ength of it
TABLE OF UNITS.
R 60 = f 3 1
f 3 8 = f 5 1
f § 16 = O. 1
8 = Cong. 1
O.
1. Denominations. — Minima or drops
(iTl), Fluid Drachm (f 3), Fluid Ounce (f 5),
Pint (O., for octarim, the Latin for one-eighth
or pint), Gallon (CoDg., for congius, the Latin
for gallon).
2. JiJqtiivalents, — Cong. 1 = O. 8 =
f 5 128 = f 3 lOiM = ta 61'M0.
8. Use.— JJ^ed in prescribing and compounding liquid medicine.
4. The symbols precede the numbers, as in Apothecaries* Weight, as
shown in the table of units.
DRY MEASURE.
ler hundred,
in., at $1.75
ick, at $1.84
TABLE OP irniTS.
tlie Oallon
2pt. =
8qt. =
4pk. =
Iqt.
Ipk.
Ibu.
1. Denotninations.~-riatB (pt.), Quarts (qt.)t
Pecks (pk.), Bushels (bn.).
2. Equivalents*— Ihn. = 4pk. = 32 qt. =
64 pt.
8. Use.— Used, in measuring grain, roots, fhiits,
salt, etc.
4. Heaped rieamre., in which the bushel is heaped in the form of a cone,
is used in measuring potatoes, com in the ear, coarse vegetables, large
fruits, etc. Stricken measure is used in measuring grains, seeds, and small
fruits.
5. A bushel of oats = 34 lb. ; of buckwheat, barley, timothy = 48 lb, ; of
flaxseed = 50 lb. ; of rye and Indian corn = 56 lb. ; of wheat, potatoes,
peas, beans, onions, or red clover seed = 60 lb.
bfl (gl.), P^nt'
[(gal.). Barrels
:4 qt.=8 pt. =
ig liquids.
[b, vatB, etc., is
els are made of
h^ce,'iApe, and
lallons marked
EXAMPLES FOR PRACTICE.
412. Solve and explain orally the following :
1. How many gills in 4 qt. ? In ? gal. ? In 7 qt. ? In 8 qt.
1 pt. ? In 3 gal. 3 qt. ?
2. How many pints in 3 bu. ? In 3 pk. 5 qt. ? In 1 bu. 2 pk.
7qt.?
3. What is the sum of O. 5 f 3 12 f 3 7 and f § 8 f 3 3 ITilS?
4. Multiply 3 pk, 5 qt. 1 pt. by 3 ; by 5 ; by 10 ; by 7 ; by 12.
Reduce
5. 93584 pt. to barrels. 8. 93654 pt. to bushels.
6. 28649 pt. to bushels. 9. 57364 gi. to barrels.
7. TTl 8405 to gallons. 10. f 3 7649 to gallons.
m
t,: *
.
1
I'll
. 1 * !
1
182
DENOMINATE NUMBERS,
11. 3 qt. 1 pt. to a decimal of a gallon. #
12. f of o qt. 1 pt. to a decimal of 2 bushels.
13. f 3 7 TT[ 15 to a decimal of Cong. 3.
14. A merchant bought 5860 bushels wheat in Toronto at
$1.25, and sold the whole in Halifax at the same price. How
much did he gain on the transaction ?
15. A grocer bought 12 firkins of butter, each containing
73 lb. 13 oz., at 36 cts. a pound ; 7 bu. 3 pk. clover seed, at
$1.15 a peck ; and 5 loads of potatoes, each load containing
43 bu. 8 pk.', at $.32 a bushel. How much was the cost ?
>
Comparative Table of Units of Capacity,
CUBIC INCHES CXJBfC rNCHES CUBIC INCHEQ
IN
ONE GAIJ.ON.
IN
ONE QUAKT.
IN
ONK PINT.
Imperial,
277.274
Liquid Measure U. S.
231
57f
m
Dry Measure (| pk.)
268|
671
m
1. The Imperial Bushel of Great Britain contains 2218.192 cu. in. and
the Standard Hushel of the United States contains 2150.42 en. in.
2. An English Quarter contains 8 imp. bu. or 8J U. S. bu. A quarter of
8 U. S. bu., or 480 lb., is used in shipping grain from New York.
3. A Register Ten is 100 cu. ft.; used in measuring the internal
capacity or t^ nnage of a vessel. A Shipping Ton Is 40 cu. ft.
4. A cubic foot of pure water weighs 1000 oz. or 62i lb. Avoir.
\\ if'
EXAMPLES FOR PRACTICE.
4: 13* 1. How many U. S. bushels in a bin of wheat 6 ft.
long, 5 ft. 6 in. wide, and 4 ft, 9 in. deep ?
How many cubic feet in a space that holds
2. 1000 U. S. bushels? 5. 240 English quarters?
3. 1000 imp. bushels? 6. 18 T. 16 cwt. of pure water? j
4. 120 bbl. water? 7. 804 bu. 3 pk. U. S. bu.?
8. A cistern containing 5300 gal. of water is 10 ft. square. |
How deep is it? ^r<«. 7.085 + .
9. How many ounces in gold are equal in weight to 9 pounds]
14 ounces of iron ?
|:t;-^
TIME MEASURE,
183
Toronto at
price.
How
;li containing
over seed, at
ad containing
,lie cost ?
ipacity*
CUBIC INCHES
IM
ONK FINT.
281
331
918.192 cu. in. and
)0.42 cu. in.
bu. A quarter of
V York,
iring the internal
cu. ft.
Avoir.
of wheat 6 ft.
q[uarters 1 .
. of pure water?
U. S. bu.?
IS 10 ft. square. I
Ans. 7.085 + .
Iglit to 9 pounds
TABLE OP UNITS.
60 sec.
= 1 min.
GO min.
= Ibr.
24 hr.
= Ida.
7 da.
= 1 wk.
365 da.
= 1 common yr.
366 da.
= 1 leap yr.
100 yr.
= 1 cen.
UNITS OF TIME.
414. The mean solar day is the StiuuUiril Unit of time.
1. DenoniinationH. — Seconds
(sec), Minutes (min.), Hours (hr.),
Days (da.), Weeks (wk.), Months
(mo.), Years (yr.), Centuries (cen.).
2. There are 12 Caleudur Months
In a year; of these, April, June,
September, and November, have
80 da. each. All the other months
except February have 31 da. each.
February, in common years, has 28 da.,
in leap years it has 29 da.
.3. In computing interest, 30 days are usually considered one month. For
bnsinci's purposes the day begins and ends at 12 o'clock midnight.
415. The reason for common and leap years wWl be seen
from the following :
The true year is the time the earth takes to go once arofund the sun,
which is 365 days, 5 hours, 48 minutes and 49.7 seconds. Taking 365 days
as a commcm. year, the time lost in the calendar in 4 years will lack only
44 minutes and 41.2 seconds of 1 day. Hence we add 1 day to February
every fourth year, making the year 366 days, or l^enp Year. This correc-
tion is 44 rain. 41.2 sec. more than should be added, amounting in 100 years
to 18 hr. 37 min. 10 sec. ; hence, at the end of 100 years we omit adding a
day, thus losing again 5 hr. 22 min. 50 sec, which we again correct by add-
ing a day at the end of 400 years; >iLence the following rule for finding leap
year:
BULE.
416. Enery year, except centennial years, exajctly divisible
by 4, is a lenp year. Every centennial year exactly divisible by
400 is also a leap year.
This will render the calendar correct to within one day for 4000 years.
417. Prob. X. — To find the interval of time between
two dates.
How many yr., mo., da. and hr. from 6 o'clock P. M., July 19,
1862, to 6 o'clock A. M., April 9, 1876.
184
DENOMINATE NUMBERS.
i f
■■■ t
yr.
mo.
da.
hr.
1876
4
9
7
1862
7
19
18
13
8
19
13
Solution.— 1. Since the latter date
denutuH tin; jjrcuifr period of time, it
is the minuend, aud the earlier date,
the subtrahend.
2. Since each year commonce? with
January, aud each day with 12 o'clock
midnight, 7 o'clock a. m., April 9, 1876, is the 7th hour of the 9th day of
the fourth month of 1876 ; and 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
.j i^^
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?
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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.
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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.
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B USIJV^ESS ARITHMETIC,
ORAL EXERCISES.
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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?
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BUSINESS ARITHMETIC,
4. If 12 lb. of butter cost $3.84, how much is it a pound ?
5. Paid $3.42 for 9 lb. of coffee. How much did I pay per
pound ?
Solve and explain the following :
tf. A farm containing 282 acres of land was sold for $22184
What was the rate per acre ? Ans. $78.66 -j- .
7. A piece of cloth containing 348 yd. was bought for $515.91.
What did it cost per yard? Ans. $1.4825.
8. Bought 236 bu. oats for $90.80. What did I pay a bu. ?
0. If 85 cords of stone cost $371,875, what is the price per
cord? Ana. $4.?75.
10. A farmer sold 70000 lb. of hay for $542.50. How much
did he receive per ton ? Ana. $15.50.
11. A merchant bought 42 firkins of butter, each containing
63| lb., for $735.67. What did he pay per pound ?
12. There were 25 mechanics employed on a building, each
receiving the same wages ; at the end of 28 days they were paid
in the aggregate $1925. What was their daily wages?
485. Prob. ni. — To find the cost when the number of
units and the price of any multiple or part of one unit
is given.
1. What is the cost of 21 ^b. sugar at 15 ct. for ^ lb. ?
SoLTTTioTir.— Since I lb. cost 15 ct., 21 lb. must cost as many times.l5 ci. ae
I lb. isi contained times in it. Hence, Fivi^l utep, 21-;- J = 27 ; Second step,
$.15 X 27 = $4.05.
Find the cost of the following :
2. 124 acres of land at $144 for 2f acres ; for 1| A.
3. 486 bu. wheat at §11 for 8 bushels ; at $4.74 for 3 bushels ;
at $.72 for f of a bushel.
4. 265 cords of wood nt $21.05 for 5 cords.
5. 135 yd. broadcloth at $8.97 for 2i^ yd.; at $12.65 for
3| yd.
B USINJSSS PROBLEMS,
307
74 for 3 bushels ;
.; at $13.65 for
6. Wliat is the cost of 987 lb. coal, at 35 ct. i er 100 lb. ?
Solution.— Ab the price is per 100 lb., we find the number of hundreds
in 087 by moving the decimal point two places to the left. The pn'oe innl-
tipliod by this result will give the required cost. Hence, $.35 x 9.87=$3.4545.
the cost of 987 lb. at 35 ct. per 100 lb.
Find the cost of the following bill of lumber
7. 2345 ft. at 11.35 per 100 (C) feet; 3«28 ft. at $.98 per C. ;
1843 ft. at $1.90 per C. ft. ; 8364 ft. ai $2.84 per C. ; 4384 ft. at
!^27.o0 per 1000 (M) ft. ; 19364 ft. at $45.75 per M.
8. What is the cost of 84690 lb. of coal at $6.45 per ton
\2000 1b.)?
Observe, that pounds are changed to tone by moviii^ the decimal point
^ places to the left and dividing by 2.
9. What is the cost of 96847 lb. coal at $7.84 per ton ?
486. Prob. IV. — To find the number of units when
the cost and price of one unit are given.
1. How many yards of cloth can be bought tor $28 @ $f ?
Solution.— Since 1 yard can be bought for $^, as many yards can be
bonuht for $28 as ^ is contained times in it. Hence, $28-+-$f - 49 yd.
Find the price and explain the following orally :
2. How many pounds of coffee can be bought for $60 at $^
per ixjund ? At $f ? At $j% V Ar, $|? ? At $.33^ ? At $.4 'if
3. For $40 how many bushels of corn can be bought at |^
per bu. ? At $| ? At ^j\ V At $f;; ? At .$.8 ? At $« ?
4. How many tons of coal can be bought for $56 at .$4 a ton c
At$7V At $8? At $14? At$6V At$9V At $5?
Solve the foUowinff :
5. The cost of digging a drain at $3§ per rod is $187 ; what
\s the length of the drain ? Ans. 51 rd.
G. How many bur»hols of wheat at $li) can be purchased for
fl840 V At $lf- ? At $li ? At $1 1 ? At $lf ?
7. The cost of a piece of cloth is $480, and the price per yard
^I'l ; how many yards does it contain ?
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208
BUSINESS ARITHMETIC.
8. A grain dealer purchased a quantity of wheat at $1.2(
per bushel, and sold it at an advance of 9^"^ cents per bushel
receiving for the whole $616.896 ; how many bushels did h(
purchase?
9. A grocer purchased $101.65 worth of butter, at 35f cents
a pound ; how many pounds did he purchase V Ans. 285 lb.
10. How many yards of cloth can be bought, at $2.75 a yard,
for $1086.25? Ana. 395.
11. A farmer paid $14198 for his farm, at $65f per acre
how many acres does the farm contain? Ans. 217 A.
487. Pros. Y. — To find the number of units that cai
be purchased for a given sum when the cost of a multiply
or part of one unit is given.
1. At 19 ct. for f of a yard, how many yards can be bouglit
for $8.55?
SoLUTTON— 1. Since § yd. cost 19 ct., J must cost \ of 19 ct., f . 9} ct., an(
I, or 1 yard, mast cost 3 times 9| ct., or 28| ct.
2. Since 1 yard cost 28J ct., as many yards can be bought for $8.55 a
38} ct. are contained times in it. Hence, $8.55 -»- $.285 = 80, the number o
yards that can be bought for $8.55, at 19 ct. for % yd.
2. A town lot was sold for $1728, at $3 per 8 sq. ft.
front of the lot is 48 ft. What is its depth ? Am. 96 ft.
3. How many bushels of com can be bought for $28,
32 ct. for I of a bu. ? At 28 ct. for | bu. ?
4. How many tons of coal can be bought for $277.50, at
for I of a ton ? At $8 for f of a ton? Ana. 37 T.
5. A piece of doth was sold for $34.50, at 14 yards per $j
How many yards did the piece contain? Ana. 483 yd.
6. A cellar was excavated for $408.24, at $4.41 for evel
7 cu. yd. The cellar was 54 ft. by 36 ft. How deep was it ?
7. A pile of wood was bought for $275.60, at $1.95 for 3 cor
feet. How many cords iw the pile ? Ana. 53 cd.
8. A drove of cattle was sold for $3738, at $294 for eve^
7 head. How many head of cattle in the drove ? Ana. 89. \
ITIC,
BUSINESS PROBLEMS.
JiOD
of wheat at |1.20l 488. Prob. VI.— To find the cost when the quantity
% cents per bushelMs a compound number and the price of a unit of one
my bushels did hewenomination is given.
butter, at 35f centE
leY Ans. 2851b.
fht, at $2.75 a yardj
An8. 395.
., at $65f per acre]
Ans. 217 A.
of units that caii
cost of a multiply
rards can be bouglii
iofl9ct.,f .9ict., and
be bonght for $8.55 a]
= 30, the number <
3 per 8 sq. ft.
? Ans. 96 ft.
jbought for $28,
for $277.50. at
Ans. 37 T.
I at 14 yards per $|
Ans. 483 yd.
I at $4.41 for evei
[ow deep was it ?
at $1.95 for 3 coi
Ans. 53 cd.
at $294 for evei
)ve? Ana. 89.
1. What is the cost of 8 bu. 3 pk. 2 qt. of wheat, at $1.44 per
t)ushel?
Solution. — ;. Since $1.44 is the
price per bushel, $1.44x8, or $11.53,
is the cost of 8 bushels.
2. Since 2 pk. = J bu., $1.44 -*• 2, or
72 cts., is the cost of 2 pk., and the \
of 72 ct., or 36 ct. . is the cost of 1 pk.
3. Since there are 8 qt. in 1 pk.,
2 qt. = J pk. Hence the cost of 1 pk.,
36 ct -•- 4, or 9 ct., Is the cost of 2 qt.
4. The sum of the cost of the parts
lust equal the cost of the whole quantity. Heuce, $12.69 is the cost of
I bu. 3 pk. 2 qt., at $1.44 per bu.
2 ) $1.44
8
11.52
2) 72
4) 36
9
Cost of 8 bu.
" ♦* 2 pk.
*' " 1 pk.
" " 2 qt.
$12.69, Ans.
Find the cost of the following orally :
2. 7 lb. 8 oz. sugar, at 12 ct. per pound ; @ 14 ct. ; @ 26 ct.
3. 7f yd. ribbon @ 15 ct. ; @ 40 ct. ; @ 25 ct.
4. 19 bu. 3 pk. 6 qt. of apples, @ $1 per bushel.
5. 13 lb. 12 oz. butter, at 34 ct. per pound ; at 40 ct.
Solve the following :
6. What will 5 T. 15 cwt. 50 lb. sugar cost, at $240 per ton?
7. Find the cost of 48 lb. 9 oz. 10 dwt. of block silver, at $12
|er pound. Ans. $585.50.
8. Sold 48 T. 15 cwt. 75 lb. of hay at $15 per ton, and 32 bu.
pk. 6 qt. timothy seed at $3.50 per bushel. How much did I
sceive for the whole ? Ar}S. $847.9 + .
9. How much will a man receive for 2 yr. 9 mo. 25 da. service,
\i $1800 per year? Am. $5075.
10. Find the cost of excavatir.g 240 cu. yd. 13^ cu. ft. of
|arth, at 50 cts. per cubic yard.
11. How much will it cost to grade 8 mi. 230 rd. of a road,
It $4640 per mile ? Ans. $40455.
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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.
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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 ?
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BUSINESS ARITHMETIC,
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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 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
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 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 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 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. 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. 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 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.
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 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 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
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. 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 : 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 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 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
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 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
^ \
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 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
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MENSURATION
k*2 f
'
'I
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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
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^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.
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
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n
332
BUSINESS ARITHMETIC.
\
-
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^
'* \
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 N,
333
triangU are givetiy
l8 16.76 chainB and
8. How many acres in a triangular field whose fides meacarc 16, 20,
30 rode y
4. What is the area of an equilateral triangle whose sides each measure
40 foot ?
5. A piece of land in the form of an equilateral triangle requires 156 rods
of fence to enclose it ; how many acres are there, and what ia the cost at
|40 per acre ?
6. How many square yards of plastering contained in a triangle whose
sides are 30, 40, 50 ; and what is the cost of it at 75 cents per sq. yd. ?
Ans. 66; yds. Ck>8t |fiO.
villbeused in pavlnjf 785* Prob. Vf.—When Vie base and perpendicular are given in a
iide 126 feet, and what I right-angled triangle, to find the other side : Extract the square root of Ihs
mm qf the squares of the base and perpendicular.
,ulredtoboardupthe
Ige of the roof H feet
llmension are given^jo
U divide by the given
The reason of this rule and the one in Prob. V will be seen by examin-
ing the diagram in the margin.
sq. lo*
Observe, that the square on the side AB opposite
the right angle, contains as many small squaren as
the sum of the small squares in the squares on the
base AC and the perpendicular BC. This i:^ ebown
by geometry to be true of all right-angled triai)<4les.
Hence, by extracting the square root of the sum
of the squares of the base and perpendicular of a
right-angled triangle, we have the length of the
Bide opposite the right angle.
•a
1 *■ I
w.
The side opposite the right angle is called the Hypothenuse.
and altitude 1 tdl Find the hypothenuse of a right-anjjled triangle
[ ^ Mn stones, each 2 fl 1. Whose base is 15 ft. and perpendicular 36 ft.
t distance from the rida 2. Whose base is 40 ft. and perpendicular 16 ft.
8. A tree 104 ft. high stands upon the bank of a stream 76 feet wide ;
rhat is the distance of a man upon the opposite baulc from a raven upon the
a triangle are given,tov of the tree f
gides subtract each sia 4. What is the length of the shortest rope by which a horse may be tied
emainders together ; tl^ a post in the middle of a field 20 rods square, and yet be allowed to graze
' on every part of it f
6. A and B start ftt)m one comer of a field a mile square, travelliner at the
Ivrhoee base la 8** **• Hme rate ; A follows the fence around the field, and B proceeds directly
tross to the opposite comer ; when B reaches the comer, how far will
Is 90, 35 feet. | be trom A ?
li
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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
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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
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II
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^40
nusry^Ess a rithmei^i c.
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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.
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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.
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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
r 44| 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.
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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 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.
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850 REVIEW AJSD TEST EXAMPLES,
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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 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
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 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. 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, 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, 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
•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. 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. 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. 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. 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
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
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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.