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3
ROBINSON'S
PROGRESSIVE
Practical Aeithmetic;
CONTAININa
THE THEORY OF NUMBERS, IN CONNECTION WITH CONCISE
ANALYTIC AND SYNTHETIC METHODS OF SOLUTION, AND
DESIGNED AS A COMPLETE TEXT-BOOK ON THIS SCIENCE,
FOR
COMMON SCHOOLS AND ACADEMIES.
BY
DANIEL W. FISH, A.M.,
AUTHOR OF THE TABLB-BOOK, PRIMARY ANO INTELLECTUAL ARITHMETICS,
1CUL.IMENTS, AND THE "SHORTER COURSE."
IVISON, BLAKEMAN, TAYLOR & CO.,
NEW YORK AND CHICAGO.
1881.
mmAnon ube.
ROBINSON'S
Mathematical Series.
traded to the wants of Primary, Intermediate, Grammar,
Normal, and High Schools, Academies, and CollegeB,
•-♦■«
FrogressiTe Table Book.
Progressire Primary Arithmetic.
Progressire Intellectual Arithmetic.
Badiments of Written Arithmetic.
JUNIOB-CLASS ARITHMETIC, Oral and Written. KEW.
ProgrestiTe Practical Arithmetic.
Key to Practical Arithmetic.
ProgresgiTC Higher Arithmetic.
Eey to Higher Arithmetic.
New Elementary Algebra.
Key to New Elementary Algebra.
New Uniyersity Algebra.
Key to New Unirerslty Algebra.
New Geometry and Trigonometry. In one vol.
Qeometry, Plane and Solid. In separate vol.
Trigonometry, Plane and Spherical. In separate voL
New Analytical Geometry and Conic Sections.
New Surreying and Narigation.
New Differential and Integral Calcolns.
UniTorsity Astronomy— Descriptive and Physical.
Key to Geometry and Trigonometry, Analytical Geometry uid Conic Seo>
tiong, Surveying and NaTigatlon.
Entered, according to Act of Congress, in the year 1858, and again in the year 186S, by
DANIEL W. FISH, A.M.,
la }he Clerk'a Office of the District Court of the United States, for the Northern
District of New York.
Co^right^ X877, by DauUl W, Fuk»
Add to Lib.
GIFT
PREFACE.
"PROGRESS and improvement characterize almost every art and
-"- science ; and within the last few years the science of Arithmetic
has received many imjportant additions and improvements, which
have appeared from time to time successively in the different treatises
published upon this subject.
In the preparation of this work it has been the author's aim to
combine, and to present in one harmonious whole, all these modem
improvements, as well as to introduce some new methods and prac-
tical operations not found in other works of the same grade ; in short,
to present the subject of Arithmetic to the pupil more as a science
than an art ; to teach him methods of thought, and how to reason,
rather than whM to do; to give unity, system, and practical utility to
the science and art of computation.
The author believes that both teacher and pupil should have the
privilege, as well as the benefit, of performing at least a part of the
thinking and the labor necessary to the study of Arithmetic ; hence
the present work has not been encumbered with the multiplicity of
** notes," '* suggestions," and superfluous operations so common to
most Practical Arithmetics of the present day, and which prevent the
cultivation of that self-reliance, that clearness of thought, and that
vigor of intellect, which always characterize the truly educated mind.
The author claims for this treatise improvement upon, if not supe.
riority over, others of the kind in the following particulars, viz. : In
the mechanical and typographical style of the work ; the open and
attractive page ; the progremve and scientific arrangement of the
subjects; clearness and conciseness of definitions; fullness and accu^
racy in the new and improved methods of operations and analyses ;
brevity and perspicuity of rvles ; and in th^ very large number of
examples prepared and arranged with special reference to thdr prac-
tical utility, and their adaptation to the real business of active life. Tlip
917
IV PREFACE.
answers to a part of the examples have been omitted, that the learner
may acquire the discipline resulting from verifying the operations.
Particular attention is invited to improvements in the subjects of
Common Divisors, Multiples, Fractions, Percentage, Interest, Pro-
portion, Analysis, Alligation, and the Roots, as it is believed these
articles contain some practical features not common to other authors
upon these subjects.
The improvements in Percentage made necessary by the financial
changes of the last few years are especially noticeable. The different
kinds of United States Securities, Bonds, and Treasury Notes are
described, and their comparative value in commercial transactions
illustrated by practical examples. The difference between Gold and
Currency, and the corresponding difference in prices, exhibited in
trade, are taught and illustrated, and many other things that every
commercial student and business man ought to know and understand.
It is not claimed that this is a perfect work, for perfection is im-
possible ; but no effort has been spared to present a clear, scientific,
comprehensive, and complete system, sufficiently full for the busi-
ness man and the scholar ; not encumbered with unnecessary theories,
and yet combining and systematizing real improvements of a practical
ind useful nature. How nearly this end has been attained the intel-
Igent and experienced teacher and educator must determine.
The Author.
NOTICE.
This book has been newly electrotyped in the latest and best style
of typography. No change has been made in the text, except to cor-
rect positive errors, and drop a few obsolete terms from the tables.
These improvements are designed to give new life to a book that
has proved its real merits by the very large circulation it has obtained.
Bbookltn, Jviyt 18TZ.
CONTENTS.
SIMPLE NUMBERS.
PAGE
DEFunriONS T
Roman Notation 8
Table of Roman Notation . 9
Arabic Notation 10
Numeration Table 15
Laws and Rules for Notation and
Numeration 17
Addition 20
Subtraction 28
Multiplication 35
Contractions 42
PAGS
Division 47
Contractions.. 56
Applications of Preceding PtULES. 60
General Principles op Division ... 64
Exact Divisors 66
Prime Numbers 67
Factoring Numbers 67
Cancellation 69
Greatest Common Divison 73
Multiples 79
Classification op Numbebs 84
COMMON FRACTIONS.
Definitions, etc
General Principles op Fractions.
Reduction op Fractions
Addition of Fractions 96 Promiscuous Examples
Subtraction op Fractions 98
Multiplication op Fractions 101
Division OF Fractions 106
112
DECIMALS.
Decimal Notation and Numera-
tion 116
Reduction of Decimals 121
Addition op Declmals 124
Subtraction op Decimals 126
Multiplication of Decimals 127
Division op Decimals 128
DECIMAL CURRENCY.
Notation and Numeration op
Decimal Currency. . 131
Reduction of Decimal Currency. 132
Addition of Decimal Currency. . . 134
Subtraction of Decimal Currency 135
Multiplication op Decimal Cur-
rency 136
Division of Decimal Currency.. . . 137
Additional Applications 139
When the Price is an Auqttot
Part OF A Dollar 139
To Find the Cost op a Quantity. . 140
To Find the Price of One 141
To Find the Quantity 141
Articles Sold by the 100 or 1000.. 142
Articles Sold by the Ton 143
Bills 144
Promiscuous Examples 146
(v)
VI
CONTENTS.
COMPOUND NUMBERS.
PAGE
Redttction 150
Definitions, BTC 150
English Money 151
Trot Weight 153
Apothecaries' Weight 154
Avoirdupois Weight 155
Long Measure 158
StTRTEYORS* LONG MeASITBB 160
Square Measure 161
Surveyors' Square Measure 164
Cubic Measure 165
Liquid Measure 167
Dry Measure 168
Time 170
Circular Measure 172
Counting ; Paper ; Books, etc 173
PAQH
Reduction of Denominate Frac-
tions 175
Addition op Compound Numbers. . 182
Addition op Denominate Frac-
tions 185
Subtraction 186
To Find the Difference in Dates 188
Table 189
Subtraction of Denominate Frac-
tions 190
Multiplication of Compound Num-
bers 191
Division 193
Longitude AND Time 195
Duodecimals 198
Promiscuous Examples 208
PERCENTAGE.
Definttions, etc 205
Commission and Brokerage 212
Stock- Jobbing 216
United States Securities 220
Gold Investments 225
Profit and Loss . . 227
Insurance 233
Taxes 234
CusTOM-HousE Business 237
Simple Interest 240
Partial Payments or Indorse-
ments 247
Problems in Interest 258
Compound Interest 256
Discount 259
Banking 262
Exchange 266
Equation of Payments 271
RATIO AND PROPORTION.
Ratio 279
Proportion 282
Simple Proportion 283
Compound Proportion 289
Partnership 294
Analysis 298
Alligation Medial 807
Alligation Alternate 808
I Involution 813
Evolution 314
Square Root 815
Cube Root 822
Arithmetical Progression.
Geometrical Progression. .
831
Promiscuous Examples 884
Mensuration 842
The Metric System 845
PEACTICAL AEITHMETIC.
DEFmiTIOl^S. *
1. Quantity is any thing that can be increased, dimin*
ished, or measured.
2. Mathematics is the science of quantity.
3. A Unit is one, or a single thing.
4. A Number is a unit, or a collection of units,
5. An Integer is a whole number.
6. The Unit of a Number is one of the collection of
units forming the number. Thus, the unit of 23 is 1 ; of
23 dollars, 1 dollar ; of 23 feet, 1 foot
7. Like Numbers are numbers that have the same kind
of unit. Thus, 74, 16, and 250 ; 7 dollars and 62 dollars ;
4 feet 6 inches, and 17 feet 9 inches.
8. An Abstract Number is a number used without
reference to any particular thing or quantity. Thus, 17 ;
365; 8540.
9. A Concrete Number is a number used with refer-
ence to some particular thing or quantity. Thus, 17 dollars ;
365 days ; 8540 men.
1. The unit of an abstract number is 1, and is called Unity.
a. Concrete numbers are, by some, called DenomincUe members. DenondnaUon
means the name of the unit of a concrete number.
10. Arithmetic is the Science of numbers, and the Art
of computation.
11. A Sign is a character used to indicate an operation,
or for abbreviating an expression.
13. A Rule is a prescribed method for performing an
operation.
Define quantity. Mathematics. A unit. A number. An integer.
The unit of a number. Like numbers. An abstract number. A
concrete number. The unit of an abstract number. Denominate
numbers. Arithmetic. A sign, or symbol. A rule.
SIMPLE NUMBEES.
H"OTATION AIsTD NUMERATION.
13. Notation is a method of writing or expressing num-
bers by characters.
14. Numeration is a method of readitig numbers ex
pressed by characters.
15. Two systems of notation are in general use — ^th6
Roman and the AraMc.
The Roman Notation is Bnpposed to have been firet used by the Bomans ; hence
its name. The Arabic Notation was introduced into Enrope by the Arabs, by whom
it was supposed to have been invented. But investigations have shown that it was
adopted by them about 600 years ago, and that it has been in use among the Hin-
doos more than 2000 years. From this latter fact it is sometimes called the Indian
Notation.
The Romak NoTATi02<r.
16. Employs seven capital letters to express numbers.
Thus,
Letters. I V X L C D M
Values. one. Five. Ten. Fifty. ^Sed. hSd^d. thJSnd.
17. The Roman Notation is founded upon fL^e principles,
as follows :
1st Repeating a lei tor repeats its yalue. Thus, II repre-
sents two, XX twenty, CCC three hundred.
2d. If a letter of any value is placed after one of greater
value, its value is to be added to that of the greater. Thus.
XI represents eleven, LX sixty, DC six hundred.
3d. If a letter of any value be placed lefore one of greater
value, its value is to be taken from that of the greater.
Thus, IX represents nine, XL forty, CD four hundred.
Define notation. Numeration. What systems of notation are now
in general use? From what are their names derived ? Wliat are used
to express numbers in the Roman notation ? What is the value of
each ? What is the first principle of combination ? Second ? Third?
NOTATION AND NUMEEATION.
9
4th. If a letter of any value be placed between two letters,
each of greater value, its value is to be taken from the su7n
of the other two. Thus, XIV represents fourteen, XXIX
twenty-nine, XCIV ninety-four.
5th. A bar or dash placed over a letter increases its value
one thousand times. Thus, V signifies five, and V five thou-
sand ; L fifty, and L fifty thousand.
.Table of Eoman Notation.
I = One.
XX = Twenty.
n « Two.
XXI " Twenty-one.
Ill ^ Three.
XXX « Thirty.
IV " Four.
XL " Forty.
V " Five.
L " Fifty.
VI " Six
LX " Sixty.
VII " Seven.
T.XX " Seventy.
VIII " Eight.
LXXX " Eighty.
IX « Nine.
XC " Ninety.
X " Ten.
C " One hundred.
IX « Eleven.
CO " Two hundred.
XII " Twelve.
D " Five hundred.
XIII " Thirteen.
DC " Six hundred.
XIV " Fourteen.
M ** One thouf^nd.
XV « Fifteen.
MC " One thousand one hundred.
XVi " Sixteen.
MM " Two thousand.
XVn " Seventeen.
X " Ten thouBand.
SLVlii « Eighteen.
C " One hundred thousand.
XIX •* Nineteen.
M " One million.
The STstem of Roman notation is not well adapted to the purposes of numerical
calculation ; it is principally confined to tiie numbering of chapters and sections of
books, public documents, etc
Express the following numbers by letters :
1. Eleven. Ans, XL
2. Fifteen. . Ans.
. Fourth? Fifth? Repeat the table. What is the value of LVII ?
CLXXm? XCVIII? CDXXXII? XCIX? DCXIX? "VMDCCXLIX?
MDXXVCDLXXXIX? To what uses is the Roman notation now
principally confined 1
10 SIMPLE iqUMBEES.
3. Twenty-five.
4. Thirty-nine.
5. Forty-eight.
6. Seventy-seven.
7. One hundred fifty-nine.
8. Five hundred ninety-four*
9. One thousand five hundred thirty-eight.
10. One thousand nine hundred ten.
11. Express the present year.
The Akabic Notatiok.
18. Employs ten characters or figures to express num^
bers. Thus,
Figures. 0123 456789
Karnes, Cipher. One. Two. Three. Four. Five. Six. Seven. Eight. Nine.
19. The first character, or cipher, is called naught, be-
cause it has no value of its own. The other nine characters
are called significant figures, because each has a value of its
own.
20. The significant figures are also called Digits, a word
derived from the Latin term digitus, which signifies /w^er.
21. The naught or cipher is also called nothing, and zero.
The ten Arabic characters are the Alphabet of Arithmetic,
and by combining them according to certain principles, all
numbers can be expressed. We will now examine the most
impoi-tant of these principles.*
22. Each of the nine digits has a value of its own;
hence any number not greater than 9 can be expressed by
one figure.
* Fractional and decimal notation, and the notation of componnd numbers, -will
be discoBsed in their appropriatai places.
What are used to express numbers in the Arabic notation ? What
is the value of each ? What general name is given to the significant
figures ? Why ? Numbers less than ten, how expressed ?
NOTATION" AND NUMEKATION. 11
33. As there is no single character to represent ten, we
express it by writing the unit, 1, at the left of the cipher, 0,
thus, 10. In the same manner we represent
2 tens,
3 tens,
4 tens.
5 tens.
6 tens.
7 tens,
8 tens.
9 tens.
or
or
or
or
or
or
or
or
Twenty,
Thirty,
Forty,
Fifty,
Sixty,
Seventy,
Eighty,
Ninety,
20;
30;
40;
50;
60;
TO;
80;
90.
24, When a number is expressed by two figures, the right
hand figure is called units, and the left hand figure tens.
We express the numbers between 10 and 20 by writing
the 1 in the place of tens, with each of the digits respec-
tively in the place of units. Thus,
Eleven,
Twelve,
Thirteen,
Fourteen,
Fifteen,
Sixteen,
Seventeen,
Eighteen,
Nineteen,
n,
13,
13,
14,
15,
16,
17,
18,
19.
In like manner we express the numbers between 20 and
30, between 30 and 40, and between any tAVO successive tens.
Thus, 21, 22, 23, 24, 25, 26, 27, 28, 29, 34, 47, 56, 72, 93.
The greatest number that can be expressed by two figures
is 99.
25. We express one hundred by writing the unit, 1, at
the left hand of two ciphers, or the number 10 at the left
hand of one cipher ; thus, 100. In like manner we write
two hundred, three hundred, etc., to nine hundred. Thus,
One Two Three Four Five Six Seven Eight Nine
hundred hundred, hundred, hundred, hundred, hundred, hundred, hundred, hundred,
100, 200, 300, 400, 500, 600, 700, 800, 900.
26. When a number is expressed by three figures, the
right hand figure is called units, the second figure, tens, and
the left hand figure, hundreds.
As the ciphers have, of themselves, no value, but are
always used to denote the absence of value in the places they
Tens, liow expressed ? The right hand figure called, what ? I-.eft
hand figure, what ? What is the greatest numher that can be expressed
hy two figures ? One hundred, how expressed ? When numbers are
expressed by three figures, what names are given to each ?
12 SIMPLE 1?"UMBERS.
occupy, we express tens and units with hundreds, by writ-
ing, in place of the ciphers, the numbers representing the
tens and units. To express one hundred fifty we write 1
hundred, 5 tens, and units ; thus, 150. To express seven
hundred ninety-two, we write 7 hundreds, 9 tens, and 2
units; thus,
M ^ S
w e« t>
7 9 3
The greatest number that can be expressed by three figures
is 999.
Examples for Practice.
1. Write one hundred twenty-five.
2. Write four hundred eighty-three.
3. Write seven hundred sixteen.
4. Express by figures nine hundred.
6. Express by figures two hundred ninety.
6. Write eight hundred nine.
7. Write five hundred five.
8. Write five hundred fifty-seven.
27. We express one thousand by writing the unit, 1, at
the left hand of three ciphers, the number 10 at the left
hand of two ciphers, or the number 100 at the left hand of
one cipher ; thus, 1000. In the same manner we write tAvo
thousand, three thousand, etc., to nine thousand ; thus.
One Two Three Four Five Six Seven Eight Nine
flioaiand, ihoiuand, thoiuand, thoosaud, thousand, thousand, thousand, thousand, thousand.
1000, 2000, 3000, 4000, 5000, 6000, 7000, 8000, 9000.
28. When a number is expressed by four figures, the
places, commencing at the right hand, are unitSf tens, hun-
dreds, thousands.
Use of the cipher, what ? Greatest number that can be expressed by
three figures V One thousand, how expressed ? How many figures
Used 1 J^ames of eaob 'i
NOTATION AND NUMEEATION. 13
To express hundreds, tens, and units with thousands, we
write in each place the figure indicating the number we wish
to express in that place. To write four thousand two hun-
dred sixty-nine, we write 4 in the place of thousands, 2 in
the place of hundreds, 6 in the place of tens, and 9 in the
place of units ; thus,
00 .
73 OD
i 1
i s i n
4 2 6 9
The greatest number that can be expressed hy four figures
is 9999.
Examples for Pbactice.
Express the following numbers by figures :
1. One thousand two hundred.
2. Five thousand one hundred sixty.
3. Three thousand seven hundred forty-one.
4. Eight thousand fifty-six.
5. Two thousand ninety.
6. Seven thousand nine.
7. One thousand one.
8". Nine thousand four hundred twenty-seven.
9. Four thousand thirty-five.
10. One thousand nine hundred four.
Eead the following numbers :
IL 76 ; 128 ; 405 ; 910 ; 116 ; 3416 ; 1025.
12. 2100; 5047; 7009; 4670; 3997; 1001.
39. Next to thousands come iiens of thousands, and next
to these come hundreds of thousands, as tens and hundreds
Come in their order after units. Ten thousand is expressed
by removing the unit, 1, one place to the left of the place
Greatest number expressed by four figures*^ Tens of thousands,
Low expressed ? Hundred,-, of thousands ?
14 SIMPLE NUMBERS.
of thousands, or by writing it at the left hand of fotu
ciphers ; thus, 10000 ; and one hundred thousand is ex-
pressed by removing the unit, 1, still one place further to
the left, or by writing it at the left hand of five ciphers ;
thus, 100000. We can express thousands, tens of thou-
sands, and hundreds of thousands in one number, in the
same manner as we express units, tens, and hundreds in one
number. To express five hundred twenty-one thousand
eight hundred three, we write 5 in the sixth place, counting
from units, 2 in the fifth place, 1 in the fourth place, 8 in
the third place, in the second place (because there are no
tens), and 3 in the place of units ; thus.
^a
o1
a
V.
'O a
o "^
d
§^
§ 3
H 2
^
o
W*"
-B
B*
W
5
2
1
8
Eh P
3
The greatest number that can be expressed by ^i;e figures
ifi 99999 ; and by six figures, 999999.
Examples for Practice.
Write the following numbers in figures :
1. Twenty thousand.
2. Forty-seven thousand.
3. Eighteen thousand one hundred.
4. Twelve thousand three hundred fifty.
5. Thirty-nine thousand five hundred twenty-two.
6. Fifteen thousand two hundred six.
7. Eleven thousand twenty-four.
8. Forty thousand ten.
9. Sixty thousand six hundred.
10. Two hundred twenty thousand.
11. One hundred fifty-six thousand.
12. Eight hundred forty thousand three hundred.
Greatest number expressed by five figures ? Six fi^fures ?
NOTATION AND NUMEEATION
15
13. Five hundred one thousand nine hundred sixty-four.
14. One hundred thousand one hundred.
15. Three hundred thirteen thousand three hundred
thirteen.
16. Seven hundred eighteen thousand four.
17. One hundred thousand ten.
Eead the following numbers :
18. 5006
19. 36741
20. 200200
12304
96071 ;
13061 ;
90402 :
5470
49000 ;
218094 :
203410.
100010.
100101,
400560 ;
75620 ;
For convenience in reading large numbers, point them off,
by commas, into periods of three figures each, counting from
the right hand or unit figure. This pointing enables us to
read the hundreds, tens, and units in each period with
facility.
30. "Next above hundreds of thousands we have, succes-
sively, units, tens, and hundreds of millions, and then fol-
low units, tens, and hundreds of each higher name, as seen
in the following :
Numeration Table.
Periods.
6th.
5th.
4tli.
3d.
2d.
Isfc.
Name. <
i
QQ
§
i
i
• Quad;
!
1
5
1
Orders
OF
Units.
undreds.
ens.
nits.
undreds. ]
ens.
nits. 1
undreds.
ens.
nits.
undreds. ]
ens.
nits. )
undreds. j
ens.
nits.
undreds.
3ns.
nits.
I. W^t)
WEHt)
tXlEnt)
WHti
WEh|:d
WH^
Number.
5 3 0,
4 5,
3 7 0,
3 6,
4 8,
6 0.
The nunc
iber is read 530 quadrillion,
46 trillion, 370 billion, 36
million, 408
thousand 60.
How may figures be pointed off? One million, how expressed?
Next period above millions, what ? Give the name of each successive
period.
16 SIMPLE NUMBERS.
This is called the French method of pointing off the periods, and Is the one in
general use in this country.
31. Figures occupying different places in a number, as
units, tens, hundreds, etc., are said to express different
orders of units.
Simple units are called units of the first order.
Tens '' " '' '' '' second "
Hundreds " " '' '' '' third "
Thousands " " " " '' fourth "
Tens of thousands " '' " " '' fifth "
and so on. Thus, 452 contains 4 units of the third order,
5 units of the second order, and 2 units of the first order.
1,030,600 contains 1 unit of the seventh order (millions), 3
units of the fifth order (tens of thousands), and 6 units of
the third order (hundreds).
Examples for Practice.
"Write in figures, and read the following numbers :
1. One unit of the third order, four of the second.
2. Three units of the fifth order, two of the third, one ol
the first.
3. Eight units of the fourth order, five of the second.
4. Two units of the seventh order, nine of the sixth, four
of the third, one of the second, seven of the first.
5. Three units of the sixth order, four of the second.
6. Nine units of the eighth order, six of the seventh, three
of the fifth, seven of the fourth, nine of the first.
7. Four units of the tenth order, six of the eighth, four
of the seventh, two of the sixth, one of the third, five of the
second.
8. Eight units of the twelfth order, four of the eleventh,
six of the tenth, nine of the seventh, three of the sixth, five
of the fifth, two of the third, eight of the first.
Units of diflferent orders are what ?
NOTATION AND NUMEKATION. 17
33. From the foregoing explanations and illustrations,
several important principles are derived, which we will now
present.
1st. Figures have two values, Simple and Local.
The Simple Value of a figure is its value when taken
alone ; thus, 2, 5, 8.
The Liocal Value of a figure is its value when used with
another figure or figures in the same number ; thus, in 842
the simple values of the several figures are 8, 4, and 2 ; but
the local value of the 8 is 800 ; of the 4 is 4 tens, or 40 ;
and of the 2 is 2 units.
When a figure occupies units' place, its simple and local values are the same,
2d. A digit or figure, if used in the second place, ex-
presses tens ; in the third place, hundreds ; in the fourth
place, thousands ; and so on.
3d. As 10 units make 1 ten, 10 tens 1 hundred, 10 hun-
dreds 1 thousand, and 10 units of any order, or in any place,
make one unit of the next higher order, or in the next
place at the left, we readily see that the Arabic method of
notation is based upon the following
Two General Laws,
L Tlie value of units of the different orders increases from
right to left, and decreases from left to right, in a tenfold
ratio,
II. Every removal of a figure one place to the left, increases
its heal value tenfold; and every removal of a figure oneplac^
io the right diminishes its local value tenfold,
6 is 6 units.
60 is 10 times 6 units.
600 is 10 times 6 tens.
6000 is 10 times 6 hundreds.
60000 is 10 times 6 thousands.
First principle derived ? What is the simple value of a fi^re ? Local?
Second principle ? Third ? First law of Arabic notation ? Second ?
18 SIMPLE NUMBERS.
4th. The local value of a figure depends upon its place
from units of the first order, not upon the value of the
figures at the right of it. Thus, in 425 and 400, the value
of the 4 is the same in both numbers, being 4 units of the
third order, or 4 hundred.
Care should be taken not to mistake the local value of a figure for the value of the
whole number. For, although the value of the 4 (huudreflg) is the same in the two
numbers. 425 and 400, the value of the whole of the first number is greater than that
of the second.
5th. Every period contains three figures (units, tens, and
hundreds), except the left hand period, which sometimes
contains only one or two figures (units, or units and tens).
33. As all the principles upon which the writing and
reading of whole numbers depend have now been analyzed,
we will present these principles in the form of rules.
Rfle for Notation.
I. Beginning at the left hand, write tlie figures belonging
to the highest 2^eriod.
II. Write the hundreds, tens, and units of each successive
period in their order, placing a cipher wherever an order
of units is omitted.
Rule for Numeration.
I. Separate the number into periods of three figures each,
commencing at the right hand.
II. Beginning at the left hand, read each period separately,
and give the name to each period, except the last, or period
of units.
34, Until the pupil can write numbers readily, it may be
well for him to write several periods of ciphers, point them
off, and over each period write its name ; thus.
Trillions, Billions, Millions, Thousands, Units.
000, 000, 000, 000, 000,
Fourth principle ? What caution is given ? Fifth principle ? Rule
for notation ? Numeration ?
KOTATIOK AKD K U M E R ATI K . 19
and then write the given numbers underneath, in their
appropriate places.
Exercises i:n" Notation akd Numeratioi^-.
Express the following numbers by figures: —
1. Four hundred thirty-six.
2. Seven thousand one hundred sixty-four.
3. Twenty-six thousand twenty-six.
4. Fourteen thousand two hundred eighty.
5. One hundred seventy-six thousand.
6. Four hundred fifty thousand thirty-nine.
7. Mnety-five million.
8. Four hundred thirty-three million eight hundred six-
teen thousand one hundred forty-nine.
9. Mne hundred thousand ninety.
10. Ten million ten thousand ten hundred ten.
11. Sixty-one billion five million.
12. Five trillion eighty billion nine million one.
Point off, numerate, and read the following numbers: —
13.
8240.
17.
1010.
21.
370005.
14.
400900.
18.
57468139.
22.
9400706342.
15.
308.
19.
5628.
23.
38429526.
16.
60720.
20.
850026800.
24.
74268113759.
25. Write seven million thirty-six.
26. Write five hundred sixty-three thousand four.
27. Write one million ninety-six thousand.
28. Numerate and read 9004082501.
29. Numerate and read 2584503962047.
30. A certain number contains 3 units of the seventh
order, 6 of the fifth, 4 of the fourth, 1 of the third, 5 of
the second, and 2 of the first ; what is the number ?
31. What orders of units are contained in the number
290648 ?
32. What orders of units are contained in the numbei
1037050 ?
20 SIMPLE NUMBEES.
ADDITION
Mektal Exercises.
35. 1. Henry gave 5 dollars for a vest, and 7 dollars
for a coat ; what did lie pay for both ?
Analysis. He gave as many dollars as 5 dollars and 7 doUarsr
which are 12 dollars.
2. A farmer sold a pig for 3 dollars, and a calf for 8 dol-
lars ; what did he receive for both ?
3. A drover bought 5 sheep of one man, 9 of another,
and 3 of another ; how many did he buy in all ?
4. How many are 2 and 6 ? 2 and 7 ? 2 and 9 ? 2 and 8 ?
2 and 10 ?
5. How many are 4 and 5 ? 4 and 8 ? 4 and 7 ? 4 and 9 ?
6. How many are 6 and 4? 6 and 6? 6 and 9 ? 6 and 7?
7. How many are 7 and 7 ? 7and0? 7and8? 7andlO?
7 and 9 ?
8. How many are 5 and 4 and 6 ? 7 and 3 and 8 ? 6 and
9 and 5 ?
36. From the preceding operations we learn that
Addition is the process of uniting several numbers of
the same kind into one equivalent number.
37. The Sum or Amount is the result obtained by the
process of addition.
38. The sign, +, is called plus, which signifies more.
When placed between two numbers, it denotes that they are
to be added ; thus, 6 -(- 4, shows that 6 and 4 are to be added.
39. The sign, =, is called the sign of equality. When
placed between two numbers, or sets of numbers, it signifies
that they are equal to each other ; thus, the expression
64-4=10, is read 6 plus 4 is equal to 10, and denotes that
the numbers 6 and 4, taken together, equal the number 10.
Define addition. The sum or amount? Sign of addition? Of
equality ?
ADDITIOiq".
21
Case I.
40. "When the amount of each colmnn is less
than 10.
1. A farmer sold some hay for 102 dollars, six cows for
162 dollars, and a horse for 125 dollars; what did he
receive for all ?
OPERATION. Analysis. Arrange the numbers so that
4r^m units of like order shall stand in the same
column. Then add the columns separately,
for convenience commencing at the right hand,
and write each result under the column added.
Thus, we have 5 and 2 and 2 are 9, the sum of
the units ; 2 and 6 are 8, the sum of the tens ;
1 and 1 and 1 are 3, the sum of the hundreds.
Hence, the entire amount is 3 hundreds 8 tens
and 9 units, or 389.
Examples for Pkactice.
Amonnt,
102
162
125
389
(3.)
(3.)
(4.)
(5.)
Pounds.
Rods.
Cents.
Days.
132
245
312
437
243
321
243
140
324
132
412
321
Ans. 699
6. What is the sum of 144, 321, and 232 ? Ans, 697.
7. What is the amount of 122, 333, and 401 ? Ans. 856.
8. What is the sum of 42, 103, 321, and 32 ? Ans. 498.
9. A drover bought three droves of sheep. The first con-
tained 230, the second 425, and the third 340 ; how man;y
sheep did he buy in all ? Ans. 995.
Case H.
41. "When the amount of any column equals or
exceeds 10.
1. A merchant pays 725 dollars a year for the rent of a
Case I is what ? Give explanation. Case II is what ?
OPERATION.
725
475
367
Sum of the units.
17
Sum oi ihe tens,
15
Sum of the hundreds,
14
Total amount,
1567
22 SIMPLENUMBEKS
store, 475 dollars for a clerk, and 367 dollars for other
expenses ; what is the amount of his expenses ?
Analysis. Arranging the num-
bers as in Case I, we first add the
column of units, and find the sum to
be 17 units, which is 1 ten and 7
units. Write the 7 units in the place
of units, and the 1 ten in the place of
tens. The sum of the figures in the
column of tens is 15 tens, which is 1
hundred, and 5 tens. Write the
5 tens in the place of tens, and the
1 hundred in the place of hundreds.
Next add the column of hundreds,
and find the sum to be 14 hundreds,
which is 1 thousand and 4 hundreds. Write the 4 hundreds in the
place of hundreds, and 1 thousand in the place of thousands. Lastly,
by uniting the sum of the units with the sums of the tens and hun-
dreds, we find the total amount to be 1 thousand 5 hundreds 6 tens
7 units, or 1567.
This example may be performed by another methoa,
which is the common one in practice. Thus :
OPERATION. Analysis. Arranging the numbers as before, we add
725 the first column and find the sum to be 17 units ; writ-
A>VK ing the 7 units under the column of units, we add the
o/,« 1 ten to the column of tens, and find the sum to be 16
L tens ; writing the 6 tens under the column of tens, we
1567 add the 1 hundred to the column of hundreds, and find
the sum to bo 15 hundreds ; as this is the last column,
write down its amount, 15 ; and the whole amount is 1567, as before.
1. Units of the same order are written in the same column ; and when the sum
in any column is 10 or more than 10, it produces one or more units of a hiprher order,
which must be added to the next column. This process is sometimes called " car-
rying the tens."
2. In adding, learn to pronounce the partial results without naming the numbers
sepai-ately ; thus, instead of saying 7 and 5 are 12, and 5 are 17, simply pronounce
the results, 7, 12, 17, etc.
Give explanation. Second explanation. What is meant by carry-
ing the tens ?
ADDITIOl^. 23
42. From the preceding examples and illustrations we
deduce the following
EuLE. I. Write the numbers to he added so that the units
of the same order shall stand in the same column j that is,
units under units, tens under tens, etc.
II. Commencing at units, add each column separately, and
write the sum underneath, if it he less than ten.
IIL If the sum of any column he ten or more than ten, write
the unit figure only, and add the ten or tens to the next column,
lY. Write the entire sum of the last colmnn,
Pboof. 1st. Begin with the right hand or unit column,
and add the figures in each column in an opposite direction
from that in which they were first added ; if the two results
agree, the work is supposed to be right. Or,
2d. Separate the numbers added into two sets, by a hori-
zontal line ; find the sum of each set separately ; add these
sums, and if the amount be the same as that first obtained,
the work is presumed to be correct.
By the methods of proof here given, the numbers are united in new combina-
tione, which render it ahnost impossible for two precisely similar mistakes to
occur.
The first method is the one commonly used in business.
Examples for Practice.
(2.)
(3.)
(4.)
(5.)
(6.)
lOles.
Inches.
Tons.
Feet.
Bushels.
24
321
427
1342
3420
48
479
321
7306
7021
96
165
903
6254
327
82
327
278
8629
97
250
1292
1929
22531
10865
Rule, first step ? Second? Third? Fourth? Proof, first method 1
Second ? Upon what principle are these methods of proof founded ?
u
SIMPLE
I^UMBERS.
(7.)
(8.)
(9.)
(10.)
lours.
Years.
Gallons.
Rods.
347
7104
3462
47637
506
3762
863
3418
218
9325
479
703
312
4316
84
26471
4:24.
2739
57
84
11. 42 + 64+98 + 70 + 37 = how many? Ans. 311.
12. 312 + 425 + 107 + 391 + 76 = how many ?
Ans, 1311.
13. 1476 + 375 + 891 + 66 + 80 = how many?
Ans. 2888.
14. 37042 + 1379 + 809 + 127 + 40 = how many?
^/^5. 39397.
15. "What is the sum of one thousand six hundred fifty-
six, eight hundred nine, three hundred ten, and ninety-four?
Ans. 2869.
16. Add forty-two thousand two hundred twenty, ten
thousand one hundred five, four thousand seventy-five, and
five hundred seven. Ans, 56907.
17. Add two hundred ten thousand four hundred, one
hundred thousand five hundred ten, ninety thousand six
hundred eleven, forty-two hundred twenty-five, and eight
hundred ten. Ans. 406556.
18. What is the sum of the following numbers : seventy-
five, one thousand ninety-five, six thousand four hundred
thirty-five, two hundred sixty-seven thousand, one thousand
tour hundred fifty-five, twenty-seven million eighteen, two
hundred seventy million twenty-seven thousand ?
Ans. 297303078.
19. A man on a journey traveled the first day 37 miles,
the second 33 miles, the third 40 miles, and the fourth
35 miles ; how far did he travel in the four days ?
20. A wine merchant has in one cask 75 gallons, in
another 65, in a third 57, in a fourth 83, in a fifth 74, and
in a sixth 67 ; how many gallons has he in all ? Ans- 421
ADDITION. 25
21. An estate is to be shared equally by four heirs, and
the portion to each heir is to be 3754 dollars ; what is the
amount of the estate ? Ans. 15016 dollars.
22. How many men in an army consisting of 52714 in-
fantry, 5110 cavalry, 6250 dragoons, 3927 light-horse, 928
artillery, 250 sappers, and 406 miners ?
23. A merchant deposited 56 dollars in a bank on Mon-
day, 74 on Tuesday, 120 on Wednesday, 96 on Thursday,
170 on Friday, and 50 on Saturday; how much did he
deposit during the week ?
24. A merchant bought at public sale 746 yards of broad-
cloth, 650 yards of muslin, 2100 yards of flannel, and 250
yards of silk ; how many yards in all ?
25. Eive persons deposited money in the same bank ; the
first, 5897 dollars; the second, 12980 dollars; the third,
65973 dollars; the fourth, 37345 dollars; and the fifth as
much as the first and second together ; how many dollars
did they aU deposit ? Ans, 141072 dollars.
26. A man willed his estate to his wife, two sons, and
four daughters ; to his daughters he gave 2630 dollars apiece,
to his sons, each 4647 dollars, and to his wife 3595 dollars ;
how much was his estate ? Ans. 23409 dollars.
(27.) (28.) (29.) (30.) (31.)
476
908
126
443
180
390
371
324
298
976
915
569
503
876
209
207
245
891
569
314
841
703
736
137
563
632
421
517
910
842
234
127
143
347
175
143
354
274
256
224
536
781
531
324
135
245
436
275
463
253
R.P.
26 SIMPLE NUMBERS.
32. A man commenced farming at the west, and raised,
the first year, 724 bushels of corn ; the second year, 3498
bushels; the third year, 9872 bushels; the fourth year,
9964 bushels; the fifth year, 11078 bushels; how many
bushels did he raise in the five years ? Ans. 35136 bushels.
33. A has 3648 dollars, B has 7035 dollars, C has 429
dollars more than A and B together, and D has as many
dollars as all the rest; how many dollars has D? How
many have all ? Ans, All have 43590 dollars.
34. A man bought three houses and lots for 15780 dollars,
and sold them so as to gain 695 dollars on each lot ; for how
much did he sell them? Ans. 17865 dollars.
35. At the battle of Waterloo, which took place June 18th,
18 ^'5, the estimated loss of the French was 40000 men; of
the Prussians, 38000 ; of the Belgians, 8000 ; of the Hano-
verians, 3500; and of the English, 12000; what was the
entire loss of life in this battle ?
36. The expenditures for educational purposes in New
England for the year 1850 were as follows : Maine, 380623
dollars; New Hampshire, 221146 dollars ; Vermont, 246604
dollars; Massachusetts, 1424873 dollars; Rhode Island,
136729 dollars; and Connecticut, 430826 dollars; what was
the total expenditure ? Ans. 2840801 dollars,
37. The eastern continent contains 31000000 square
miles; the western continent, 13750000 ; Australia, Green-
land, and other islands, 5250000 ; what is the entire area of
the land surface of the globe ?
38. The population of New York, in 1850, was 515547 ;
Boston, 136881; Philadelphia, 340045; Chicago, 29963;
St. Louis, 77860; New Orleans, 116375; what was the
entire population of these cities ? Ans, 1216671.
39. The population of the globe is estimated as follows:
North America, 39257819 ; South America, 18373188 ; Eu-
rope, 265368216; Asia, 630671661; Africa, 61688779;
Oceanica, 23444082 ; what is the total population of the
globe according to this estimate ? Ans, 1038803745.
ADDITION
27
40. The railroad distance from New York to Albany is
144 miles ; from Albany to Buffalo, 298 ; from Buffalo to
Cleveland, 183 ; from Cleveland to Toledo, 109 ; from To-
ledo to Springfield, 365 ; and from Springfield to St. Louis,
95 miles ; what is the distance from New York to St. Louis ?
41. A man owns farms valued at 56800 dollars ; city lots
Valued at 86760 dollars ; a house worth 12500 dollars, and
other property to the amount of 6785 dollars ; what is the
entire value of his property ? Ans. 162845 dollars.
(42.)
(43.)
(44.)
(45.)
15038
26881
41919
93808
7404
12173
19577
41371
34971
39665
74736
110525
30359
33249
66768
102936
6293
6318
12673
17087
2875
4318
7193
13251
16660
34705
61365.
112110
64934
80597
155497
220619
80901
95299
183134
225255
7444
8624
16845
68940
57068
53806
111139
176974
17255
18647
. 35902
86590
32543
41609
82182
149162
40022
• 35077
75153
109355
56063
46880
132936
283910
33860
41842
82939
112511
17548
26876
44424
72908
28944
36642
65586
157672
16147
29997
52839
86160
38556
44305
83211
119557
234882
262083
522294
839398
39058
39744
78861
117787
152526
169220
353428
471842
179122
198568
386214
671778
7626
8735
17005
41735
1218099
1395860
28 SIMPLE NUMBEK8.
SUBTEAOTIOTsT.
Mental Exeecises.
43. 1. A farmer, having 14 cows, sold 6 of them ; how
many had he left ?
ANAiiYSis. He had as many left as 14 cows less 6 cows, which are
8 cows.
2. Stephen, haying 9 marbles, lost 4 of them ; how many
had he left ?
3. If a man earn 10 dollars a week, and spend 6 dollars
for provisions, how many dollars has he left ?
4. A merchant, having 16 barrels of flour, seUs 9 of them ;
how many barrels has he left ?
5. Charles had 18 cents, and gave 10 of them for a book;
how many cents had he left ?
6. James is 17 years old, and his sister Julia is 5 years
younger ; how old is Julia ?
7. A grocer, having 20 boxes of lemons, sold 11 boxes;
how many boxes had he left ?
8. From a cistern containing 25 barrels of water, 15 bar-
rels leaked out ; how many barrels remained ?
9. Paid 16 dollars for a coat, and 7 dollars for a vest ;
how much more did the coat cost than the vest ?
10. How many are 18 less 5 ? 17 less 8 ? 12 less 7 ?
11. How many are 20 less 14 ? 18 less 12 ? 19 less 11 ?
12. How many are 11 less 3 ? 16 less 11 ? 19 less 8 ?
20 less 9 ? 22 less 20 ?
44. Subtraction is the process of finding the difference
between two numbers of the same kind.
45. The Minuend is the number to be subtracted from.
46. The Subtrahend is the number to be subtracted.
Define subtraction ? Minuend ? Subtrahend t
SUBTEACTIOK. 29
47. The Diflference or Kemainder is the result ob-
tained by the process of subtraction.
The minuend and subtrahend must be like numbers ; thus, 5 dollars from 9 dol-
lars leave 4 dollars ; 5 apples from 9 apples, leave 4 apples ; but it would be absurd
to say 5 apples from 9 dollars, or 5 dollars from 9 apples.
48. The sign, — , is called minus , which signifies less.
When placed between two numbers, it denotes that the one
after it is to be taken from the one before it. Thus, 8—6
=2 is read 8 minus 6 equals 2, and shows that 6, the sub-
trahend, taken from 8, the minuend, equals 2, the
remainder.
Case I.
49. "When no figure in the subtrahend is greater
than the corresponding figure in the minuend.
1. From 574 take 323.
Analysis. Write the less number un-
der the greater, with units under units,
tens under tens, etc., and draw a line
underneath. Then, beginning at the right
hand, subtract separately each figure of
the subtrahend from the figure above it in
the minuend. Thus, 3 from 4 leaves. 1, which is the difference of the
units ; 2 from 7 leaves 5, the difference of the tens ; 3 from 5 leaves 2,
the difference of the hundreds. Hence, we have for the whole differ-
ence, 2 hundreds 5 tens and 1 unit, or 251.
Examples foe Peactice.
OPERATION.
Minuend,
574
Subtrahend,
323
Remainder,
251
Minuend,
(2.)
876
(3.)
676
(4.)
367
(5.)
925
Subtraliend,
Remainder,
334
542
415
261
152
215
213
712
Case I is wliat ? Give explanation.
30 SIMPLENUMBERS
(6.) (7.)
From 876 732
(8.)
987
(9.)
498
Take 523 522
782
178
10. From 3276 take 2143.
Remainders.
1133.
11. From 7634 take 3132.
4502.
12. From 41763 take 11521.
30242.
13. From 18346 take 5215.
13131.
14 From 397631 take 175321.
222310.
15. Subtract 47321 from 69524.
22203.
16. Subtract 16330 from 48673.
32343.
17. Subtract 291352 from 895752.
604400.
18. Subtract 84321 from 397562.
313241.
19. A farmer paid 645 dollars for a span of horses and
a carriage, and sold them for 522 dollars; what did he
lose?
20. A man bought a mill for 3724 dollars, and sold it for
4856 dollars; what did he gain? Ans, 1132 dollars.
21. A drover bought 1566 sheep, and sold 435 of them ,
how many had he left ? Ans. 1131 sheep.
22. A piece of land was sold for 2945 dollars, which was
832 dollars more than it cost ; what did it cost ?
23. A gentleman willed to his son 15768 dollars, and to
his daughter 4537 dollars; how much more did he will to
the son than to the daughter ? Ans. 11231 dollars.
24. A merchant sold goods to the amount of 6742 dollars,
and by so doing gained 2540 dollars ; what did the goods
cost him ?
25. If I borrow 15475 dollars of a person, and pay him
4050 dollars, what do I still owe him ?
26. In 1850 the white population of the United States
was 19,553,068, and the slave population 3,204,313; what
was the difference ?
27. The population of Great Britain in 1851 was 20,936,468,
and of England alone, 16,921,888 ; what was the difference ?
subtraction. 31
Case IL
50. "When any figure in the subtrahend is greater
than the corresponding figure in the minuend.
1. From 846 take 359.
OPERATION. Analysis. In tMs example we
(7) (13) (16) cannot take 9 units from 6 units.
Minuend, 8 4 6 From the 4 tens we take 1 ten, which
Subtrahend. 3 5 9 equals 10 units, and add to the 6
units, making 16 units ; 9 units from
Eemaiuder, 4 8 7 ^q ^^^^^ ^^^^^ 7 ^i^s, which we
write in the remainder in units' place. As we have taken 1 ten from
the 4 tens, 3 tens onlj are left. We cannot take 5 tens from 3 tens ;
so from the 8 hundreds we take 1 hundred, which equals 10 tens, and
add to the 3 tens, making 13 tens ; 5 tens from 13 tens leave 8 tens,
which we write in the remainder in tens' place. As we have taken
1 hundred from the 8 hundreds, 7 hundreds only are left ; 3 hundreds
from 7 hundreds leave 4 hundreds, which we write in the remainder
in hundreds' place, and we have the whole remainder, 487.
The numbers written over the minuend are used simply to explain more clearly
the method of subtracting ; in practice the process should be performed mentally,
and these numbers omitted.
The following method is more in accordance with prac-
tice.
OPERATION. Analysis. Since we cannot take 9 units from 6
® ^ ^ units, we add 10 units to 6 units, making 16 units ;
§ §1 9 units from 16 units leave 7 units. But as we have
QAf* added 10 units, or 1 ten, to the minuend, we shall
have a remainder 1 ten too large, to avoid which, we
add 1 ten to the 5 tens in the subtrahend, making 6
48 7 *®^s. We can not take 6 tens from 4 tens ; so we
add 10 tens to 4, making 14 tens; 6 tens from 14
tens leave 8 tens. Now, having added 10 tens, or 1 hundred, to the
minuend, we shall have a remainder 1 hundred too large, unless we
add 1 hundred to the 3 hundreds in the subtrahend, making 4 hun-
dreds ; 4 hundreds from 8 hundreds leave 4 hundreds, and we have
for the total remainder, 487, the same as before.
Case n is what ? Give explanation. Second explanation.
32 SIMPLE NUMBERS.
The process of adding 10 to the minuend is sometimes called borrowing 10, and
that of adding 1 to the next figure of the subtrahend, carrying one,
• 51. From the preceding illustrations we have the fol-
lowing
Rule. I. Write the less number under the greater, plac-
ing units of the same order in the same column,
II. Begifi at the right hand, and take each figure of the
subtrahend from the figure above it, and write the result un-
derneath.
III. If any figure in the subtrahend be greater than the
corresponding figure above it, add 10 to that upper figure be-
fore subtracting, and then add 1 to the next left hand figure
of the subtrahend.
Pkoof. Add the remainder to the subtrahend, and if
their sum be equal to the minuend, the work is supposed
to be right.
Examples fob Peactice.
Minuend,
(2.)
873
(3.)
7432
(4.)
1969
(5.)
8146
Subtrahend,
538
6711
1408
4377
Eemainder,
335
From
(6.)
Gallons.
3176
(7.)
Bushels.
9076
(8.)
MUes.
7320
(9.)
Days.
6097
Take
2907
4567
3871
3809
From
(10.)
Dollars.
76377
(11.)
Rods.
67777
(12.)
Acres.
900076
(13.)
Feet
767340
Take
45761
46699
899934
5039
What do we mean by borrowing 10 ? By carrying ? Rule, first step 1
Second? Third? Proof?
SUBTRACTION
33
14. 479—382 = how many ? Ans. 97.
15. 6593—1807 = how many? Ans. 4786.
16. 17380—3417 = how many? Ans. 13963.
17. 80014— 43190 = how many? Ans. 36824.
18. 282731—90756 = how many? Ans. 191975.
19. From 234100 take 9970.
20. From 345673 take 124799.
21. From 4367676 take 256569. Ans. 4111107
22. From 3467310 take 987631. Ans. 2479679.
23. From 941000 take 5007. Afis. 935C93.
24. From 1970000 take 1361111. A7is. 608889.
25. From 290017 take 108045.
26. Take 3077097 from 7045676. Ans. 3968579.
27. Take 9999999 from 60000000. Ans. 50000001.
28. Take 220202 from 4040053. Ans, 3819851.
29. Take 2199077 from 3000001. Ans. 800924.
30. Take 377776 from 8000800. Ans. 7623024.
31. Take 501300347 from 1030810040.
32. Subtract nineteen thousand nineteen from twenty
thousand ten. Ans. 991.
33. From one million nine thousand six take twenty thou-
sand four hundred. Ans. 988606.
34. What is the difference between two million seven
thousand eighteen, and one hundred five thousand seven-
teen?
Examples Combiking Addition and Subtraction.
52, 1. A merchant gave his note for 5200 dollars. He
paid at one time 2500 dollars, and at another 175 dollars ;
what remained due ? Ans. 2525 dollars.
2. A traveler who was 1300 miles from home, traveled
homeward 235 miles in one week, in the next 275 miles, in
.the next 325 miles, and in the next 280 miles ; how far had
he still to go before he would reach home ? Ans. 185 miles.
3. A man deposited in bank 8752 dollars ; hs drew out at
one time 4234 dollars, at another 1700 dollars, at another 962
34 SIMPLE NUMBERS.
dollars, and at another 49 dollars ; how much had he remain-
ing in bank? Ans. 1807 dollars.
4. A man bought a farm for 4765 dollars, and paid 750
dollars for fencing and other improvements ; he then sold
it for 384 dollars less than it cost him ; what did he receive
for it ? Ans, 5131 dollars.
5. A forwarding merchant had in his warehouse 7520 bar
rels of flour ; he shipped at one time 1234 barrels, at anothei
time 1500 barrels, and at another time 1805 barrels ; how
many barrels remained ?
6. A had 450 sheep, B had 175 more than A, and C
had as many as A and B together minus 114 ; how many
sheep had C ? Ans. 961 sheep.
7. A farmer raised 1575 bushels of wheat, and 900 bushels
of com. He sold 807 bushels of wheat, and 391 bushels of
com to A, and the remainder to B ; how much of each did
he sell to B ? Ans. 768 bushels of wheat, and 509 of com.
8. A man traveled 6784 miles ; 2324 miles by railroad,
1570 miles in a stage coach, 450 miles on horseback, 175
miles on foot, and the remainder by steamboat ; how many
miles did he travel by steamboat ? Ans. 2265 miles.
9. Three persons bought a hotel valued at 35680 dollars.
The first agreed to pay 7375 dollars, the second agreed to
pay twice as much, and the third the remainder ; Avhat was
the third to pay ? Ans. 13555 dollars.
10. Borrowed of my neighbor at one time 750 dollars, at
another time 379 dollars, and at another 450 dollars. Having
paid him 1000 dollars, how much do I still owe him ?
Ans. 579 dollars.
11. A man worth 6709 dollars received a legacy of 3000
dollars, ^e spent 4379 dollars in traveling ; how much had
he left?
12. In 1850 the number of white males in the United
States was 10026402, and of white females 9526666 ; of
these, 8786968 males, and 8525565 females were native
bom ; how many of both were foreign bom ? Ans. 2240535.
MULTIPLICATION. 35
M D LTIPLIC ATIOI^.
Mental Exercises.
53. 1. What will 4 pounds of sugar cost, at 8 cents a
pound ?
Analysis. Four pounds will cost as mucli as the price, 8 cents
taken 4 times ; thus, 8 + 8 + 8 + 8=33. But instead of adding, we may
say,— since one pound cost 8 cents, 4 pounds will cost 4 times 8 cents,
or 33 cents.
2. If a ream of paper cost 3 dollars, what will 2 reams
cost?
3. At 7 cents a quart, what will 4 quarts of cherries
cost?
4. At 12 dollars a ton, what will 3 tons of hay cost? 4
tons ? 5 tons ?
5. There are 7 days in 1 week ; how many days in 6 weeks?
In 8 weeks ?
6. What will 9 chairs cost, at 10 shillings apiece ?
7. If Henry earn 12 dollars in 1 month, how much can
he earn in 5 months ? in 7 months ? in 9 months ?
8. What will 11 dozen of eggs cost, at 9 cents a dozen ?
At 10 cents ? At 12 cents ?
9. When flour is 7 dollars a barrel, what must be paid for
7 barrels ? for 9 barrels ? for 12 barrels ?
10. At 9 dollars a week, what will 4 weeks' board cost ?
7 weeks' ? 9 weeks' ?
11. If I deposit 12 dollars in a sa\dngs bank every month,
how many dollars will I deposit in 6 months ? In 8 months ?
In 9 months ?
12. At 9 cents a foot, what will 4 feet of lead pipe cost ?
.feet? 10 feet?
13. When hay is 8 dollars a ton, how much will 3 tons
cost ? 4 tons ? 7 tons ? 9 tons ? 11 tons ?
36
SIMPLE KUMBERS
14 What will be the cost of 11 barrels of apples, at 2
dollars a barrel ? at 3 doUarsi?
15. At 10 cents a pound, what will 9 pounds of sugar
cost? 11 pounds? 12 pounds?
54:, Multiplication is the process of taking one of two
given numbers as many times as there are units in the other.
55» The Multiplicand is the number to be multiplied.
56. The Multiplier is the number by which to multiply,
and shows how many times the multiplicand is to be taken.
57. The Product is the result obtained by the process of
multiplication.
5S, The Factors are the multiplicand and multiplier.
1. Factors are producers, and the multiplicand and multiplier are called factors
because they produce the product.
2. Multiplication is a short method of performing addition when the numbers to
be added are equal.
59. The sign, x , placed between two numbers, denotes
that they are to be multiplied together ; thus, 9 x 6=54, is
read 9 times 6 equals 54.
Multiplication^ Table.
Ix 1= 1
2x 1= 2
3x 1= 3
4x 1= 4
Ix 2= 2
2x 2= 4
3x 2= 6
4x 2= 8
Ix 3= 3
2x 3= 6
3x 3= 9
4x 3 = 12
Ix 4= 4
2x 4= 8
3x 4=12
4x 4=16
Ix 5= 5
2x 5=10
3x 5 = 15
4x 5=20
Ix 6= 6
2x 6 = 12
3x 6 = 18
4x 6=24
Ix 7= 7
2x 7 = 14
3x 7=21
4x 7=28
Ix 8= 8
2x 8=16
3x 8=24
4x 8=3:i
Ix 9= 9
2x 9 = 18
3x 9=27
4x 9=36
1x10=10
2x10=20
3x10=30
4x10=40
1x11=11
2x11=22
3x11=33
4x11=44
1x12 = 12
2x12=24
3x12=36
4x12=48
Define multiplication. Multiplicand. Multiplier. Product. Fac-
tor. Multiplication la a abort method of what ? What is the sign of
multiplication ?
MULTIPLICATION.
37
5x 1= 5
6x 1= 6
7x 1= 7
8x 1= 8
5x 2 = 10
6x 2 = 12
7x 2=14
8x 2=16
5x 3zizl5
6x 3 = 18
7x 3=21
8x 3=24
5x 4=20
6x 4=24
7x 4=28
8x 4=32
5x 5=25
6x 5=30
7x 5=35
8x 5=40
5x 6=30
6x 6=36
7x 6=42
8x 6=48
5x 7=35
6x 7=42
7x 7=49
8x 7 = 56
ox 8=40
6x 8=48
7x 8=56
8x 8=64
5x 9=45
6x 9-54
7x 9 = 63
8x 9 = 72
5x10 = 50
6x10=60
7x10=70
8x10=80
5x11 = 55
6x11 = 66
7x11 = 77
8x11=88
5x12 = 60
6x12 = 72
7x12=84
8x12=96
9x
9x
9x
1 =
2=
3 =
4=
5 =
6 =
7=
X
X
X
X
X 8 =
X 9 =
xlO=
xll =
9x12:
9
18
27
36
45
54
63
72
81
90
99
108
10 X
10 X
10 X
10 X
10 X
10 X
10 X
10 X
10 X
1 =
2 =
3 =
4=
5 =
6 =
7=
10
20
30
40
50
60
70
80
90
9 =
10x10=100
10x11 = 110
10x12 = 120
11 X
11 X
llx
11 X
llx
llx
llx
llx
llx
1 =
2 =
3 =
4=
5 =
6 =
7=
11
22
33
44
55
66
77
88
99
9 =
11x10=110
11x11 = 121
11x12=132
12 X
12 X
12 X
12 X
12 X
12 X
12 X
12 X
12 X
1 =
2=
3 =
4 =
5:=
6 =
7 =
8 =
9=
12
24
36
48
60
72
84
96
108
12x10 = 120
12x11 = 132
12x12=144
OPERATION.
Case I.
60. When the multiplier consists of one figure.
1. Multiply 374 by 6.
Analysis. In this example it is
required to take 374 six times. If we
take the units of each order 6 times,
we shall take the entire number 6
times. Therefore, writing the multi-
plier under the unit figure of the mul-
tiplicand, we proceed as follows: 6
times 4 units are 24 units ; 6 times 7
tens are 42 tens ; 6 times 3 hundreds
are 18 hundreds ; and adding these
partial products, we obtain the entire
product, 2244.
Multiplicand,
Multiplier,
Units,
Tens,
Hundreds,
Product,
374
6
24
42
2244
Case I is what ? Give explanation.
IMPLE NUMBEES.
The operation in this example may be performed in an-
other way, which is the one in common use.
OPEEATION.
874
6
2244
Akalysis. Writing the numbers as before, begin
at the right hand or unit figure, and say : 6 times 4
units are 24 units, which is 2 tens and 4 units; write
the 4 units in the product in units' place, and reserve
the 2 tens to add to the next product ; C times 7 tent
are 42 tens, and the two tens reserved in the last
product added, are 44 tens, which is 4 hundreds and 4 teTis ; write the
4 tens in the product in tens' place, and reserve the 4 hundreds to add
to the next product ; 6 times 3 hundreds are 18 hundreds, and 4 hun-
dreds added are 22 hundreds, which being written in the product in
the places of hundreds and thousands, gives, for the entire product,
2244.
61, The unit value of a number is not changed by re-
peating the number. As the multiplier always expresses
times ^ the product must have the same unit value as the
multiplicand. But since the product of any two numbers
will be the same, whichever factor is taken as a multiplier,
either factor may be taken for the multiplier or multiplicand.
In multiplying, learn to pronounce the partial results, as in addition, without
naming the numbers separately ; thus, in the last example, instead of saying 6 times
4 are 24, 6 times 7 are 42 and 2 to carry are 44, 6 times 3 are 18 and 4 to carry are 22,
pronounce only the results, 24, 44, 22, performing the operations mentally. This
will greatly facilitate the process of multiplying.
Examples for Practice.
(2.)
(3.)
(4.)
Multiplicand,
7324
6812
34651
Multipher,
4
6
5
Product,
29296
40872
173255
(5.)
(6.)
(7.) •
(8.)
82456
92714
28093
46247
3
7
8
9
Second explanation. Repeating a number has what effect on the
unit value ? The product must be of the same kind as what ?
M U L T I P L I C A T I O IsT .
39
9. Multiply 32746 by 5.
10. Multiply 840371 by 7.
11. Multiply 137629 by 8.
12. Multiply 93762 by 3.
13. Multiply 543272 by 4.
14. Multiply 703164 by 9.
Ans,
Ans,
Ans,
Ans,
Ans.
Ans,
163730.
5882597.
1101032.
281286.
2173088.
6328476.
15. Wbat will be the cost of 344 cords of wood at 4 dol-
lars a cord ? Ans. 1376.
16. How much wiU an army of 7856 men receive in one
week, if each man receive 6 dollars ? Ans, 47136 dollars.
17. In one day are 86400 seconds ; how many seconds in
7 days ? Ans, 604800 seconds.
18. What will 7640 bushels of wheat cost, at 9 shillings
a bushel ? Ans. 68760 shillings.
19. At 5 dollars an acre, what will 2487 acres of land
cost ? Ans. 12435 doUars.
20. In one mile are 5280 feet ; how many feet in 8
miles ? Ans. 42240 feet
Case IL
63. "When the miiltiplier consists of two or more
figures.
1. Multiply 746 by 23.
OPERATION.
746
23
Analysis. Writ-
ing the multiplicand
and multiplier as in
Case I, first multiply
each figure in the
multiplicand by the
unit figure of the
multiplier, precisely
as in Case I. Then multiply by the 2 tens. 2 tens times G units, or
6 times 2 tens, are 12 tens, equal to 1 hundred, and 2 tens ; place the
2 tens under the tens figure in the product already obtained, and add
the 1 hundred to the next hundreds produced. 2 tens times 4 tens
are 8 hundreds, and the 1 hundred of the last product added are 9
hundreds ; write the 9 in hundreds' place in the product. 2 tens
Multiplicand,
Multiplier, .
Product,
. j times the mul-
\ tiplicand.
nn ] time? the mul-
•^^ j tiplicand.
go j times the mul-
( tiplicand.
Case n is what t Give explanation.
40 SIMPLE NUMBERS.
times 7 hundreds are 14 thousands, equal to 1 ten thousand and 4
thousands, which we write in their appropriate places in the product.
Then adding the two products, the entire product is 17158.
1. When the multiplier contains two or more figures, the several results obtained
by multiplying by each figure are called particU products.
2. When there are ciphers between the significant figures of the multiplier, pass
over them, and multiply by the significant figures only,
63. From the preceding examples and illustrations w€
deduce the following general
EiTLE. I. Write the multiplier under the multiplicand plac-
ing units of the same order under each other.
II. Multiply the multiplicand ly each figure of the multi-
plier successively f beginning tuith the unitfigure, and torite
the first figure of each partial product imder the figure of the
multiplier used, writing down and carrying as in addition.
III. If there are partial products y add them, and their sum
will he the product required,
64. Proof. I. Multiply the multiplier by the multipli-
cand, and if the product is. the same as the first result, the
work is correct. Or,
2. Multiply the multiplicand by the multiplier diminished
by 1, and to the product add the multiplicand ; if the sum
be the same as the product by the whole of the multiplier,
the work is correct.
Examples for Practice.
Multiply
By
(2.)
4732
36
(3.)
8721
47
(4.)
17605
204
28392
61047
70420
14196
34884
35210
Ans,
170352
409887
3591420
What are partial products ? When there are ciphers in the multi-
plier, how proceed ? Rule, first step? Second? Third? Proof, first
method? Second?
MULTIPLICATION-. 41
(5.)
(6.)
(7.)
7648
81092
37967
328
194
426
8. How many yards of linen in 759 pieces, each piece con-
taining 25 yards ? Ans. 18975 yards.
9. Sound is known to travel about 1142 feet in a second
of time ; how far will it travel in 69 seconds ?
10. A 9ian bought 36 city lots, at 475 dollars each ; what
did they all cost him? Ans, 17100 dollars.
11. What would be the value of 867 shares of railroad
stock, at 97 dollars a share ? Ans. 84099 dollars.
12. How many pages in 3475 books, if there be 362 pages
in each book ? Ans. 1257950 pages.
13. In a garrison of 4507 men, each man receives annu-
ally 208 dollars ; what do they all receive ?
14. Multiply 7198 by 216. Ans, 1554768.
15. Multiply 31416 by 175. Ans. 5497800.
16. Multiply 7071 by 556. Ans. 3931476.
17. Multiply 75649 by 579. Ans. 43800771.
18. Multiply 15607 by 3094. Ans. 48288058.
19. Multiply 79094451 by 76095. A7is. 6018692248845.
20. Multiply five hundred forty thousand six hundred
nine, by seventeen hundred fifty. Ans. 946065750.
21. Multiply four milUon twenty-five thousand three
hundred ten, by seventy-five thousand forty-six.
Ans. 302083414260.
22. Multiply eight hundred seventy-seven million five
hundred ten thousand eight hundred sixty-four, by five
hundred forty-five thousand three hundred fifty-seven.
Ans. 478556692258448.
23. If one mile of railroad require 116 tons of iron, worth
65 dollars a ton, what will be the cost of sufficient iron to
construct a road 128 miles in length ? '
A71S. 965120 dollars.
42 SIMPLE l^UMBEKS,
CONTEAOTIONS.
Case I.
65. "When the multiplier is a composite number.
A Composite Number is one tliat may be produced by
multiplying together two or more numbers ; thus, 18 is a
composite number, since 6 x 3 = 18 ; or, 9 x 2 = 18 ; or,
3x3x2 = 18.
66. The Component Factors of a number are the sev-
eral numbers which, multiplied together, produce the given
number ; thus, the component factors of 20 are 10 and 2
(10 X 2=20) ; or, 4 and 5 (4 x 5=20) ; or, 2 and 2 and 5
(2x2x5=20).
The pnpil mnet not confbtmd the factors with the parts of a Dumber. Thns, the
factors of which 13 is composed, are 4 and 3 (4 x 3=12) ; while the parts of which 12
38 composed are 8 and 4 (8+4=12), or 10 and 2 (10+2=12). Tiie factors are mutti-
pliecl^ while the parts are added^ to produce the number.
1. What will 32 horses cost, at 174 doUars apiece ?
OPERATION. Analysis. The fac-
Ifultlplicand, 174 COSt of 1 horso. tors of 33 are 4 and
l8t factor, 4 ®- If we multiply the
cost of 1 horse by 4,
696 cost of 4 horses. ^e obtain the cost of 4
9d flMstor, 8 horses ; and by multi-
Product, 6568 cost of 32 horses. ?^^°^ *^ t ^'* ^^ . ^
horses by 8, we obtain
the cost of 8 times 4 horses, or 32 horses, the number bought.
6*7, Hence we have the following
KuLE. I. Separate the composite numher into two or more
factors.
II. Multiply tlie multiplicand hy one of these factors, and
What are contractions ? Case I is what ? Define a composite num-
ber. Component factors. What caution is given 1 Give explanation.
Rule, first step ? Second ?
MULTIPLICATION. 43
the product hy another, and so on until all the factors have
been used successively j the last product will be the product
required.
The product of any number of factors wiU be the same in whatever order they are
multiplied. Thus, 4 x 3 >; 5=60, and 5 x 4 x 3= 60.
Examples for Practice.
2. Multiply 3472 by 48=6 x 8. Ans, 166656.
3. Multiply 14761 by 64=8 x 8.
4. Multiply 87034 by 81=3 x 3 x 9. Ans. 7049754.
5. Multiply 47326 by 120=6 x 5 x 4.
6. Multiply 60315 by 96. Ans. 5790240.
7. Multiply 291042 by 125. Ans. 36380250.
8. If a vessel sail 436 miles in 1 day, bow far will she sail
in 56 days ? A7is. 24416 miles.
9. What will 72 acres of land cost, at 124 dollars an acre ?
Ans, 8928 dollars.
10. There are 5280 feet in a mile ; how many feet in 84
miles ? Ans. 443520 feet.
11. What will 120 yoke of cattle cost, at 125 dollars a
yoke ?
Case II.
68. "When the multiplier is 10, 100, 1000, etc.
If we annex a cipher to the multiplicand, each figure is
removed one place toward the left, and consequently the
value of the whole number is increased tenfold ( 32 ). If
two ciphers are annexed, each figure is removed two places
toward the left, and the value of the number is increased
one hundred fold ; and every additional cipher increases
the value tenfold.
69. Hence the following
EuLE. Annex as many ciphers to the multiplicand as there
are ciphers in the multiplier ; the number so formed will be
the product required.
Case II is what ? Giv6 explanation. Rule 2
44 simple kumbees.
Examples for Practice.
1. Multiply 347 by 10. Ans. 3470.
2. Multiply 4731 by 100. Atis, 473100.
3. Multiply 13071 by 1000.
4. Multiply 89017 by 10000.
5. If 1 acre of land cost 36 dollars, what will 10 acres
cost ? Ans. 360 dollars^
6. If 1 bushel of corn cost 65 cents, what will 1000 bushels
cost? A?is, 65000 cents.
Case III.
70. "When there are ciphers at the right hand of
one or both of the factors.
1. Multiply 1200 by 60.
OPERATION. Analysis. Both multiplicand and
Multiplicand, 1200 multiplier may be resolved into their
Multiplier, 60 component factors ; 1200 into 12 and
— 100, and 60 into 6 and 10. If these
Product, 72000 several factors be multiplied together
they will produce the same product
as the given numbers (67). Thus, 12x6=7?, and 72x100=7200,
and 7200 x 10=72000, vehich is the same result as in the operation.
EuLE. Multiply the significant fig wes of the multiplicand
hy those of the multiplier, and to the product annex as many
ciphers as there are ciphers on the right of both factors.
Examples foe Practice.
Multiply
By
(2.)
4720
340000
1888
1416
1604800000
(3.)
10340000
105000
5170
1034
1085700000000
Cafle III is ^'hat ? Give explanation. Rule.
MULTIPLICATION, 45
4. Multiply 70340 by 800400. Ans. 56300136000.
5. Multiply 3400900 by 207000. Ans. 703986300000-
6. Multiply 634003000 by 40020. Ans. 25372800060000.
7. Multiply 10203070 by 50302000.
Ans. 513234827140000.
8. Multiply 30090800 by 600080. Ans. 18056887264000.
9. Multiply eighty million seven thousand six hundred, by
eight million seven hundred sixty. Ans. 640121605776000.
10. Multiply fifty million ten thousand seventy, by sixty-
four thousand. Ans. 3200644480000.
11. Multiply ten million three hundred fifty thousand one
hundred, by eighty thousand nine hundred.
Ans. 837323090000.
12. There are 296 members of Congress, and each one re-
ceives a salary of 3000 dollars a year ; how much do they
all receive ?
Examples Combiking Additioi^", Subteactiok, akd
multiplicatiok.
1. Bought 45 cords of wood at 4 dollars a cord, and 9 loads
of hay at 13 doUars a load ; what was the cost of the wood
and hay ? Ans. 297 dollars.
2. A merchant bought 6 hogsheads of sugar at 31 dollars
a hogshead, and sold it for 39 dollars a hogshead ; what did
he gain ?
3. Bought 288 barrels of flour for 1875 dollars, and sold
the same for 9 dollars a barrel ; what was the gain ?
Ans. 717 dollars.
4. If a young man receive 500 dollars a year salary and
pay 240 dollars for board, 125 dollars for clothing, 75 dollars
for books, and 50 dollars for other expenses, how much will
he have left at the end of the year ? Ans. 10 dollars.
5. A farmer sold 184 bushels of wheat at 2 dollars a
bushel, for which he received 67 yards of cloth at 4 dollars
a yard, and the balance in groceries ; what did his groceries
cost him ?
46 SIMPLE N^UilBEBS.
6. A sold a farm of 320 acres at 36 dollars an acre ; B
sold one of 244 acres at 48 dollars an acre ; which received
the greater sum, and how much? Ans. B, 192 dollars.
7. Two persons start from the same point and travel in
opposite directions, one at the rate of 35 miles a day, and the
other 29 miles a day : how far apart will they be in 16 days ?
Ans, 1024 miles.
8. A merchant tailor bought 14 bales of cloth, each bale
containing 26 pieces, and each piece 43 yards ; how many
yards of cloth did he buy ? Ans. 15652 yards.
9. If a man have an income of 3700 dollars a year, and
his daily expenses be 4 dollars ; what will he save in a year,
or 365 days ? A7is. 2240 dollars.
10. A man sold three houses ; for the first he received
2475 dollars, for the second 840 dollars less than he received
for the first, and for the third as much as for the olher two ;
what did he receive for the three? Ans. 8220 dollars.
11. A man sets out to travel from Albany to Buffalo, a
distance of 336 miles, and walks 28 miles a day for 10 days ;
how far is he from Buffalo ?
12. Mr. bo:ight l4 cows at 23 dollars each, 7 horses at
96 dollars each, 34 oxen at 57 dollars each, and 300 sheep at
2 dollars each ; he sold the whole for 3842 dollars ; what
did he gain ? Ans. 310 dollars.
13. A drover bought 164 head of cattle at 36 dollars a
head, and 850 sheep at 3 dollars a head ; what did he pay
for all ?
14. A banker has an income of 14760 dollars a year ; he
pays 1575 dollars for house rent, and four tiihes as much for
family expenses ; what does he save annually ?
Ans. 6885 dollars.
15. A flour merchant bought 936 barrels of flour at 9 dol-
lars a barrel ; he sold 480 barrels at 10 dollars a barrel, and
the remainder at 8 dollars a barrel ; what did he gain or
lose ? Ans, Gained 24 dollars.
k
DIVISIONe 4:t
DIVISIOK
Meistal Exercises.
71. 1. How many hats, at 4 dollars apiece, can be bought
for 20 dollars ?
Analysis. Since 4 dollars will buy one liat, 20 dollars will buy as
many hats as 4 is contained times in 20, which is 5 times. Therefore,
5 hats, at 4 dollars apiece, can be bought for 20 dollars.
2. A man gave 16 dollars for 8 barrels of apples ; what
was the cost of each barrel ?
3. If 1 cord of wood cost 3 dollars, how many cords can
be bought for 15 dollars ?
4. At 6 shillings a bushel, how many bushels of com can
be bought for 24 shilhngs ?
5. When flour is 6 dollars a barrel, how many barrels can
be bought for 30 dollars ?
6. If a man can dig 7 rods of ditch in a day, how many
days "v^U it take him to dig 28 rods ?
7. If an orchard contain 56 trees, and 7 trees in a row,
how many rows are there ?
8. Bought 6 barrels of flour for 42 dollars ; what was the
cost of 1 barrel ?
9. If a farmer divide 21 bushels of potatoes equally
among 7 laborers, how many bushels will each receive ?
10. How many oranges can be bought for 27 cents, at 3
cents each ?
11. A farmer paid 35 dollars for sheep, at 5 dollars apiece ;
how many did he buy ?
12. How many times 4 in 28 ? In 16 ? In 36 ?
13. How many times 8 in 40 ? In 56 ? In 64 ?
14. How many times 9 in 36 ? In 63 ? In 81 ?
15. How many times 7 in 49 ? In 70 ? In 84 ?
48 SIMPLE NUMBERS.
73. Division is the process of finding how many times
one number is contained in another.
73. The Dividend is the number to be divided.
74. The Divisor is the number to divide by.
75. The Quotient is the result obtained by the process
of division, and shows how many times the divisor is con-
tained in the dividend.
1. When the dividend does not contain the divisor an exact number of times, the
part of the dividend left is called the remainder^ and it must be less than the
divisor.
2. As the remainer is always a part of the dividend, it is always of the same name
or kind.
3. When there is no remainder the division is said to be exact.
76. The sign, -^, placed between two numbers, denotes
division, and shows that the number on the left is to he di-
vided by the number on the rigid. Thus, 20-7-4=5, is read,
20 divided hy 4 is equal to 5.
Division is also indicated by writing the dividend above,
12
and the divisor helow a short horizontal line ; thus, -q-=^>
shows that 12 divided ly 3 equals 4.
Case I.
77. "When the divisor consists of one fignre.
1. How many times is 4 contained in 848 ?
OPERATION. Analysis. After writing the divisor
T,, , , , on the left of the dividend, with a line
Dividend,
. . ^ \ gig between them, begin at the left hand
and say : 4 is contained in 8 hundreds,
Qnotient, 212 3 hundreds times, and write 2 in hun-
dreds' place in the quotient ; then 4 is
contained in 4 tens 1 ten times, and write the 1 in tens' place in the
quotient ; then 4 is contained in 8 units 2 units times ; and writing the
2 in units' place in the quotient, and the entire quotient is 213.
Define division. Dividend. Divisor. Quotient. Remainder. What
is complete division ? What is the sign of division? Case I is what?
Give first explanation.
DIVISION. 49
2. How many times is 4 contained in 2884 ?
OPERATION. Analysis. As we cannot divide 2 thousands by
4 "i 2884 ^' ^^® *^^® *^® ^ thousands and the 8 hundreds to-
gether, and say, 4 is contained in 28 hundreds 7 hun-
*^^ dreds times, which we write in hundreds' place in
the quotient ; then 4 is contained in 8 tens 2 tens
times, which we write in tens' place in the quotient ; and 4 is con-
tained in 4 units 1 unit time, which we write in units* place in th^
quotient, and we have the entire quotient, 721.
3. How many times is 6 contained in 1824 ?
OPERATION. Analysis. Beginning as in the last example, say,
6 ) 1824 6 is contained in 18 hundreds 3 hundreds times,
TTT which write in hundreds' place in the quotient ;
then 6 is contained in 2 tens no times, and write a
cipher in tens* place in the quotient ; and taking the 2 tens and 4 units
together, 6 is contained in 24 units 4 units times, which write in units'
place in the quotient, and we have 304 for the entire quotient.
4. How many times is 4 contained in 943 ?
OPERATION. Analysis. Here 4 is contained in 9
^ \ 9^3 hundreds 2 hundreds times, and 1 hundred
over, which, united to the 4 tens, makes
235 ... 3 Eem. ^4 ^^j^g . 4 in 14 tens, 3 tens times and 3
tens over, which, united to the 3 units
make 23 units ; 4 in 23 units 5 units times and 3 units over. The 3
which is left after performing the division, should be divided by 4 ;
but the method of doing it cannot be explained until we reach
Fractions ; so we merely indicate the division by placing the divisor
under the dividend, thus, f. The entire quotient is written 235|,
which may be read, two hundred thirty-five and three divided by four ^
or, two hundred thirty-five and a remainder of three.
From the foregoing examples and illustrations, we deduce
the following
KuLE. I. Write the divisor at the left of the dividend, with
a line between them.
Second. Third. Rule, first step ?
R. P. 3
60
SIMPLE NUMBERS.
n. Beginning at the left hand, divide each figure of tlie
dividend hy the divisor, and write the result binder the divi"
dend,
in. If there he a remainder after dividing any figure, re-
gard it as prefixed to the figure of the next loioer order in the
dividend, and divide as before.
IV. Should any figure or part of the dividend he less than
the divisor, write a cipher in the quotient, and prefix the
number to the figure of the next lower order in the dividend,
and divide as before.
V. If there he a remainder after dividing the last figure,
place it over the divisor at the right hand of the quotient.
Proof. Multiply the divisor and quotient together, and
to the product add the remainder, if any ; if the result be
equal to the dividend, the work is correct.
1. This method of proof depends on the fact that division is the reverse of mnlti-
plication. The dividend answers to the product^ the divisor to one of the factors,
and the quotient to the other.
2. In multiplication the two factors are ariven, to find the product ; in division, the
product and one of the £Eictors are given to find the other £actor.
Examples for Practice.
1. Divide 7824 by
6.
OPERATION.
PROOF.
Divisor. 6)7824
Dividend.
1304 Quotient.
1304
Quotient
6 Divisor.
7824 Dividend.
(2.)
(3.)
(4.)
4)65432
5)89135
6)178932
(5.)
(6.)
(7.)
7)4708935
8)1462376
9)7468542
Second step? Tliird? Fourth? Fifth? Proof? How does divU
ion differ from multiplication ?
DIVISI02S".
61
8. Divide 3102455 by 5.
9. Di^dde 1762891 by 4.
19. Divide 546215747 by 11.
31. Divide 30179624 by 12.
12. Divide 9254671 by 9.
13. Divide 7341568 by 7.
14. Divide 3179632 by 5.
16. Divide 19038716 by 8.
16. Divide 84201763 by 9.
17. Divide 2947691 by 12.
18. Divide 42084796 by 6.
Sums of quotients and remainders,
Quotients.
Quotients.
620491.
440722f.
49655977.
2514968^.
1028296|.
Rem.
20680083.
28.
19. Divide 47645 dollars equally among 5 men ; what will
each receive? Ans. 9529 dollars.
20. In one week are 7 days ; how many weeks in 17675
days ? A71S. 2525 weeks.
21. How many barrels of flour, at 6 dollars a barrel, can
be bought for 6756 dollars? Ans, 1126 barrels.
22. Twelve things make a dozen ; how many dozen in
46216464 ? Ans. 3851372 dozen.
23. How many barrels of flour can be made from 347560
bushels of wheat, if it take 5 bushels to make one barrel ?
A71S, 69512 barrels.
24. If there be 3240622 acres of land in 11 townships,
bow many acres in each township ?
25. A gentleman left his estate, worth 38470 dollars, to
be shared equally by his wife and 4 children ; what did
each receive ? Ans. 7694 dollars.
Case IL
78. When the divisor consists of two or more
figures.
To illustrate more clearly the method of operation, we will first take an example
usually performed by Short Division.
Case II is what ?
52 SIMPLE i^UMBEKS.
1. How many times is 8 contained in 2528 ?
OPERATION. ANAi/f SIS. As 8 is iiot contained in 2 thou-
8)2528 (316 sands, we take 2 and 5 as one number, and
04. consider how many times 8 is contained in tliis
partial dividend, 25 hundreds, and find that it
1'® is contained 3 hundreds times, and a remainder.
8 To find this remainder, we multiply the divisor.
An 8, by the quotient figure, 3 hundreds, and sub-
tract the product, 24 hundreds, from the par-
tial dividend, 25 hundreds, and there remains
1 hundred. To this remainder we bring down
the 2 tens of the dividend, and consider the 12 tens a second partial
dividend. Then, 8 is contained in 12 tens 1 ten time and a remain-
der ; 8 multiplied by 1 ten produces 8 tens, which, subtracted from
12 tens, leave 4 tens. To this remainder we bring down the 8 units,
and consider the 48 units the third partial di\idend. Then, 8 is con-
tained in 48 units 6 units times. Multiplying and subtracting aa
before, we find that nothing remains, and the entire quotient is 316.
2. How many times is 23 contained in 4807 ?
OPERATION. Analysis. We first find how
Divisor. Divid'd. Quotient. many times 23 is contained in 48,
23 ) 4807 ( 209 the first partial dividend, and place
46 the result in the quotient on the
right of the dividend. We then
multiply the divisor, 23, by the
quotient figure, 2, and subtract the
product, 46, from the part of the
dividend used, and to the remainder bring down the next figure of
the dividend, which is 0, making 20, for the second partial dividend.
Then, since 23 is contained in 20 no times, we place a cipher in the
quotient, and bring down the next figure of the dividend, making a
third partial dividend, 207 ; 23 is contained in 207, 9 times ; multiply-
ing and subtracting as before, nothing remains, and the entire quo-
tient is 209.
1. When the process of dividing is performed mentally, and the results only are
written, as in Case I, the operation is termed Short Division.
2. When the whole process of division ia written, the operation is termed Long
Division,
Give first explanation. Second. What is long division? What ia
short division ? When is each used ?
207
207
DIVISIOK. 53
8. Short Division Is generally used when the divisor is a number that will allow
the process of dividing to be performed mentally.
From the preceding illustrations we derive the following
general
KuLE. I. Write the divisor at the left of the dividend, as
in short division.
II. Divide the least numher of the left hand figures in tho
dividend that will contain the divisor one or more times, and
place the quotient at the right of the dividend, with a line
ietiueen them,
III. Multiply the divisor ly this quotient figure, suUract
the product from the partial dividend used, and to the re-
mainder hring down the next figure of the dividend,
IV. Divide as hefore, until all the figures of the dividend
have teen Ir ought down and divided,
V. If any partial dividend will not contain the divisor,
place a cipher in the quotient, and hring down the next figure
of the dividend, and divide as hefore.
YL If there le a remainder after dividing all the figures
of the dividend, it 7nust be written in the quotient, with the
divisor underneath,
1. If any remainder be egval to, or greater than the divisor, the quotient figure is
too small, and must be increased.
3. If the product of the divisor by the quotient figure be greoUr than the pjirtial
dividend, the quotient figure is too large, and must be diminished.
79. Peoof. 1. The same as in short division. Or,
2. Subtract the remainder, if any, from the dividend, and
di^ade the difference by the quotient ; if the result be the
same as the given divisor, the work is correct.
80. The operations in long division consist of five prin^
cipal steps, viz. :
1st. Write down the numbers.
Rule, first step ? Second ? Third ? Fourth ? Fifth ? Sixth ? First
direction ? Second ? Proof ? Recapitulate the steps in their order.
64 SIMPLE KUMBEB 8.
2d. Find how many times.
3d. Multiply.
4th. Subtract.
5th. Bring down another figure.
Examples for Practice.
3. Find how many times 36 is contained in 11798.
OPEKATION.
Dividend.
Divisor. 36 ) 11798 ( 327 Quotient.
108
PROOF BY MtJl
327
36
[.TIPLICATIQH
Quotient.
Divisor.
99
72
1962
981
278
252
11772
26
Remainder.
26 Remainder.
11798
Dividend.
4. Find how many times 82 ig
! contained in {
B9634.
OPERATION.
B2 ) 89634 ( 1093
82
PROOF BY DIVISION.
89634 Dividend.
8 Remainder.
763 Quotient. :
738
L093 ) 89626 (
8744
82 Divisor
254
246
8
2186
2186
5. Find how many times 154
is contained in
32740.
6. Divide 32572 by 34.
7. Divide 1554768 by 216.
8. Divide 5497800 by 175.
9. Divide 3931476 by 556.
10. Divide 10983588 by 132.
Ans. 958.
Ans, 7198.
Ans, 31416.
Ans. 7071.
Ans. 83209.
Divisioir. 55
11. Divide 73484248 by 19. Ans. 3867592.
12. Divide 8121918 by 21. Ans. 386758.
13. Divide 10557312 by 16. Ans, 659832.
14. Divide 93840 by 63. Ee?n, 33.
15. Divide 352417 by 29. Bern. 9.
16. Divide 51846734 by 102. Bern. 32,
17. Divide 1457924651 by 1204. Bern. 1051
18. Divide 729386 by 731. Bern. 679.
19. Divide 4843167 by 3605. Bern. 1652.
20. Divide 49816657 by 9101. • Bern. 6884.
21. Divide 75867308 by 10115. Bern, 4808.
Quotients. Rem.
22. Divide 28101418481 by 1107. 25385201. 974.
23. Divide 65358547823 by 2789. 23434402. 645.
24. Divide 102030405060 by 123456. 826451. 70404.
25. Divide 48659910 by 54001. 901. 5009.
26. Divide 2331883961 by 6739549. 346. 7.
27. A railroad cost one million eigbt hundred fifty thou-
sand four hundred dollars, and was divided into eighteen
thousand five hundred and four shares ; what was the value
of each share ? Ans. 100 dollars.
28. If a tax of seventy-two million three hundred twenty
thousand sixty dollars be equally assessed on ten thousand
seven hundred thirty-five towns, what amount of tax must
each town pay ? Ans. 6736^^^^ dollars.
29. In 1850 there were in the United States 213 college
Mbraries, containing 942321 volumes ; what would be the
average number of volumes to each library ?
Ans. 4424^ vols.
30. The number of post offices in the United States in
1853 was 22320, and the entire revenue of the post office
department was 5937120 doUars ; what was the average
revenue of each office ? Ans, 266 dollars.
56 SIMPLE NUMBERS.
CONTRACTIONS.
Case L
81. "When the divisor is a composite miniber.
1. If 3270 dollars be divided equally among 30 men, ho-w
many dollars will each receive ?
OPERATION. Analysis. If 8270 dollars be divided
6 ) 3270 equally among 30 men, each man will receive
^rr^77 as many dollars as 30 is contained times in
I 3270 dollars. 30 may be resolved into the
109 Ans, factors 5 and 6 ; and we may suppose the 30
men divided into 5 groups of 6 men each;
dividing the 8270 dollars by 5, the number of groups, we have 654,
the number of dollars to be given to each group ; and dividing the
654 dollars by 6, the number of men in each group, we have 109, the
number of dollars that each man will receive.
EuLE. Divide the dividend iy one of the factors, and the
quotient thus obtained hy another, and so on if there be more
than two factors, until every factor has been made a divisor.
The last quotient will be the quotient required.
Examples for Practice.
2. Divide 3690 by 15=3 x 5.
3. Divide 3528 by 24=4 x 6.
4. Divide 7280 by 35 = 5 x 7.
5. Divide 6228 by 36=6 x 6.
6. Divide 33642 by 27=3 x 9.
7. Divide 153160 by 56=7 x 8.
8. Divide 15625 by 125=5 x 5 x 5.
83. To find the true remainder.
1. Divide 1143 by 64, using the factors 2, 8, and 4, antj
find the true remainder.
Ans,
246.
Ans,
147.
Ans.
208.
Ans,
173.
Ans,
1246.
Ans,
2735.
Ans,
125.
What are contractions 1 Case I is what t Give explanation ? Rule.
DITI8I0K. 67
OPERATION. Analysis. Divid-
2)1143 ing 1143 by 2, we
-TT— haveaquotientofS?!,
^1221 ^ ^^^ and a remainder of
4) 71 3x2=6 '' 1 undivided, which,
"l7. . .3 X 8 X 2=48 << ^^^ I ^.f' f *^^
— given dividend, must
55 true rem. , also be a part of the
true remainder. The
671 being a quotient arising from dividing by 2, its units are 2 times
as great in value as the units of the given dividend, 1143. Dividing
the 571 by 8, we have a quotient of 71, and a remainder of 8 undi-
vided. As this 3 is a part of the 571, it must be multiplied by 2 to
change it to the same kind of units as the 1. This makes a true re-
mainder of 6 arising from dividing by 8. Dividing the 71 by 4, we
have a quotient of 17, and a remainder of 3 undivided. This 3 is a part
of the 71, the units of which are S times as great in value as those of
the 571, and the units of the 5T. i^re 2 times as great in value as those
of the given dividend, 1143 ; therefove, w change this last remainder,
3, to units of the same value as the dividend, we multiply it by 8 and
2, and obtain a true remainder of 48 arising from dividing by 4.
Adding the three partial remainders, we obtain 55, the true remainder.
EuLE. L Multiply each partial remainder, except the fast,
ly all the preceding divisors.
II. Add the several products with the first remainder^ and
the sum will ie the true remainder.
Examples for Practice.
Rem.
2. Divide 34712 by 42=6x7. 20,
a Divide 401376 by 64=8x8. 32.
4. Divide 139074 by 72=3x4x6. 42.
5. Divide 9078126 by 90=3 x 5 x 6. 6.
6. Divide 18730627 by 120=4 x 5 x 6. 67.
7. Divide 7360479 by 96=2x6x8. 63.
8. Divide 24726300 by 70=2 x 5 x 7. 60.
9. Divide 5610207 by 84=7 x 2 x 6. 15.
Explain the process of finding the true remainder when dividing by
Ihe factors of a composite number.
3*
58 SIMPLE ISrUMBERS.
Case IL
83. "When the divisor is 10, 100, 1000, etc.
1. Divide 374 acres of land equally among 10 men ; hew
many acres will each have ?
OPEKATiON. Analysis. Since we have shown,
110)3714 , *^^* ^ remove a figure one place
toward the left by annexing a ciphei
Quotient. 37 ... 4 Rem. increases its value tenfold, or multi.
or, 373V acres. plies it by 10 ( 68 ), so, on the con-
trary, by cutting off or taking away
the right hand figure of a number, each of the other figures is removed
one place toward the right, and, consequently, the value of each is
diminished tenfold, or divided by 10 ( 32 ).
For similar reasons, if we cut ojff two figures, we divide by
100, if three, we divide by 1000, and so on.
Rule. From the right hand of the dividend cut off as
many figures as there are ciphers in the divisor. Under the
figures so cut off, place the divisor, and the whole will form
the quotient.
Examples for Practice.
2. Divide 4760 by 10.
3. Divide 362078 by 100.
4. Divide 1306321 by 1000.
5. Divide 9760347 by 10000.
6. Divide 2037160310 by 100000.
Case III.
84. When there are ciphers on the right hand of
the divisor.
I. Divide 437661 by 800.
OPERATION. Analysis. In this example we
8|00 ) 4376|61 resolve 800 into the factors 8 and
547... 61 Rem.
100, and divide first by 100, by cut-
ting off two right hand figures of the
Case II is what? Give explanation. Rule. Case III is what?
Give explanation.
DIVISION.
S§
dividend ( 83 ), and we have a quotient of 4376, and k remainder of
61. We next divide by 8, and obtain 547 for a quotient; and the
entire quotient is 547/oV'
2. Dmde 34716 by 900.
OPERATION. Analysis. Dividing as
9100 ) 347116 ^ t^6 last example, we
~ „ have a quotient of 38, and
DO Quotient, O, 2d rem, , . , ^r. j t
^ ' two remamders, 16 and 5
5 X 100 + 16 = 516, true rem. Multiplying 5, the last re
38f^, Ans. mainder, by 100, the pre-
ceding divisor, and adding
16, the first remainder (82), we have 516 for the true remainder.
But this remainder consists of the last remainder, 5, prefixed to the
figures 16, cut off from the dividend.
85. When there is a remainder after dividing by the sig-
nificant figures, it must be prefixed to the figures cut off
from the dividend to give the true remainder ; if there be
no other remainder, the figures cut off from the dividend
wiU be the true remainder.
Examples
1 FOR Peactice.
Quotients. Rem.
3.
Divide 34716
by 900.
38.
516.
4.
Divide 1047634
by 2400.
436.
1234.
6.
Divide 47321046
by 45000.
1051.
26046.
6.
Divide 2037903176
by 140000.
63176.
7.
Divide 976031425
by 92000.
3425.
8.
Divide 80013176321
by 700000.
376321.
9.
Divide 19070367428
by 4160000.
4584.
927428
10.
Divide 37902564431S
► by 554000000.
89644319.
11.
The circumference of the earth at the
equate
)r is 24898
iles,
. How many hours would a train of cars
require to
travel that distance, going at the rate of 50 miles an hour ?
A?is. 497|f.
12. The sum of 350000 dollars is paid to an army of 14000
men : what does each man receive ?
Ans, 25 dollars.
How is the true remainder found ?
60 SIMPLE N^UMBERS.
Examples iiq" the Preceding Rules.
1. George Washington was born in 1732, and lived 67
years ; in what year did he die ? A7is, In 1799.
2. How many dollars a day must a man spend, to use an
income of 1095 dollars a year? A^is. 3 dollars.
3. If I give 141 dollars for a piece of cloth containing 47
yards, for what must I sell it in order to gain one dollar a
yard? Ans. 188 dollars.
4. A speculator who owned 500 acres, 17 acres, 98 acres,
and 121 acres of land, sold 325 acres ; how many acres had
he left ? Ans. 41 1 acres.
5. A dealer sold a cargo of salt for 2300 dollars, and gained
625 dollars ; what did the cargo cost him ?
Ans. 1675 dollars.
6. If a man earn 60 dollars a month, and spend 45 dol-
lars in the same time, how long will it take him to save 900
dollars from his earnings ?
7. If 9 persons use a barrel of flour in 87 days, how many
days will a barrel last 1 person at the same rate ?
A71S. 783 days.
8. The first of three numbers is 4, the second is 8 times
the first, and the third is 9 times the second ; what is their
sum? A71S. 324.
9. If 2, 2, and 7 are three factors of 364, what is the
other factor ? Ans. 13.
10. A man has 3 farms ; the first contains 78 acres, the
second 104 acres, and the third as many acres as both the
others ; how many acres in the 3 farms ?
11. If the expenses of a boy at school are 90 dollars for
board, 30 dollars for clothes, 12 dollars for tuition, 5 dollars
for books, and 7 dollars for pocket money, what would be the
expenses of 27 boys at the same rate ? A?is. 3888 dollars.
12. Four children inherited 2250 dollars each ; but one
dying, the remaining three inherited the whole ; what was
the share of each? Ans. 3000 dollars.
PKOMISCUOUS EXAMPLES. 61
13. Two men travel in opposite directions, one at the rate
of 35 miles a day, and the other at the rate of 40 miles a
day ; how far apart are they at the end of '6 days ?
14. Two men travel in the same direction, one at the rate
of 35 miles a day, and the other at the rate of 40 miles a
day ; how far apart are they at the end of 6 days ?
15. A man was 45 years old, and he had been married 19
years ; how old was he when married ? Ans. 26 years.
16. Upon how many acres of ground can the entire popu-
lation of the globe stand, supposing that 25000 persons can
stand upon one acre, and that the population is 1000000000 ?
A?is. 40000 acres.
17. Add 384, 1562, 25, and 946 ; subtract 2723 from the
sum ; divide the remainder by 97 ; and multiply the quo-
tient by 142 ; what is the result ? Ans. 284.
18. How many steps of 3 feet each would a man take in
walking a mile, or 5280 feet ? Ans. 1760 steps.
19. A man purchased a house for 2375 dollars, and ex-
pended 340 dollars in repairs ; he then sold it for railroad
stock worth 867 dollars, and 235 acres of western land val-
ued at 8 dollars an acre ; what did he gain by the trade ?
Ans. 32 dollars.
20. The salary of a clergyman is 800 dollars a year, and
his yearly expenses are 450 dollars ; if he be worth 1350
dollars now, in how many years will he be worth 4500 dol-
lars ? Ans, 9 years.
21. How many bushels of oats at 40 cents a bushel, must
be given for 1600 bushels of wheat at 75 cents a bushel?
A71S. 3000. bushels.
22. Bought 325 loads of wheat, each load containing 50
bushels, at 2 dollars a bushel ; what did the wheat cost ?
23. If you deposit 225 cents each week in a savings bank,
and take out 75 cents a week, how many cents will you have
left at the end of the year ? Ans. 7800 cents.
24. The product of two numbers is 31383450, and one of
the numbers is 4050 : what is the other number?
62 SIMPLE NUMBERS.
25. The Illinois Central Railroad is 700 miles long, and
cost 31647000 dollars ; what did it cost per mile ?
A71S. 45210 dollars.
26. What number is that, which being divided by 7, the
quotient multiplied by 3, the product divided by 5, and this
quotient increased by 40, the sum will be 100 ? Ans. 700.
27. How many cows at 27 dollars apiece, must be given
for 54 tons of hay at 17 dollars a ton ?
28. A mechanic receives 56 dollars for 26 days' work, and
spends 2 dollars a day for the whole time ; how many dollars
has he left ? Ans. 4: dollars.
29. If 7 men can build a house in 98 days, how long would
it take one man to build it ? Ans. 686 days.
30. The number of school houses in the State of New
York, in 1855, was 11,137; if their cash value is 5,301,212
dollars, what is the average value ? A^is. 476 dollars.
31. A cistern whose capacity is 840 gallons has two pipes ;
through one pipe 60 gallons run into it in an hour, and
through the other 39 gallons run out in the same time ; in
how many hours will the cistern be filled ? Ans. 40 hours.
32. The average beat of the pulse of a man at middle age
is about 4500 times in an hour ; how many times does it
beat in 24 hours ? Ans. 108000 times.
33. How many years from the discovery of America, in
1492, to the year 1900?
34. According to the census, Maine has 31766 square
miles ; New Hampshire, 9280 ; Vermont, 10212 ; Massachu-
setts, 7800 ; Rhode Island, 1306 ; Connecticut, 4674 ; and
New York, 47000 ; how many more square miles has all
New England than New York ?
35. What is the remainder after dividing 62530000 by
87900? Ans. 33100.
36. A pound of cotton has been spun into a thread 8 miles
in length ; allowing 235 pounds for waste, how many pounds
will it take to spin a thread to reach round the earth, suppos-
ing the distance to be 25000 miles? Ans. 3360 pounds.
PROMISCUOUS EXAMPLES. 63
37. John has 8546 dollars, which is 342 dollars less than
4 times as much as Charles has ; how many dollars has
Charles? Ans. 2222 dollars.
38. The quotient of one number divided by another is 37,
the divisor 245, and the remainder 230 ; what is the divi-
dend? Ans, 9295.
39. What number multiplied by 72084 will produce
5190048? Ans. 72.
40. There are two numbers, the greater of which is 73
times 109, and their difference is 17 times 28 ; what is the
less number ? Ans. 7481.
41. The sum of two numbers is 360, and the less is 114 ;
what is the product of the two numbers ? Ans. 28044.
42. What number added to 2473248 makes 2568754 ?
43. A farmer sold 35 bushels of wheat at 2 dollars a bushel,
and 18 cords of wood at 3 dollars a cord ; he received 9
yards of cloth at 4 dollars a yard, and the balance in money ;
how many dollars did he receive ? Ans. 88 dollars.
44. A farmer receives 684 dollars a year for produce from
his farm, and his expenses are 375 dollars a year ; what will
he save in five years ?
45. The salt manufacturer at Syracuse pays 58 cents for
wood to boil one barrel of salt, 10 cents for boiling, 5 cents to
the State for the brine, 28 cents for the barrel, and 3 cents
for packing and weighing, and receives 125 cents from the
purchaser ; what does he make on a barrel ? Ans. 21 cents.
46. A company of 15 persons purchase a township of
western land for 286000 dollars, of which sum one man pays
fjOOO dollars, and the others the remainder, in equal amounts ;
what does each of the others pay ? A7is. 20000 dollars.
47. If 256 be multiplied by 25, the product diminished
by 625, and the remainder divided by 35, what will be the
quotient? Ans. 165.
48. Two men start from different places, distant 189 miles,
and travel toward each other ; one goes 4 miles, and the other
5 miles an hour ; in how many hours will they meet ?
64 SIMPLE NUMBEES.
GENEEAL PKINCIPLES OF DIVISION,
86. The quotient in Division depends upon the relative
values of the dividend and divisor. Hence any change in the
value of either dividend or divisor must produce a change
in the value of the quotient. But some changes may be
produced upon both dividend and divisor, at the same time,
that will not affect the quotient. The laws which govern
these changes are called Oeneral Principles of Division,
which we will now examine.
I. 54 -T- 9 = 6.
Multipljdng the dividend by 3, we have
54 X 3 -r- 9 = 162 -^ 9 = 18,
and 18 equals the quotient, 6, multiplied by 3. Hence,
Multiplying the dividend ly any nuniber, multiplies the quo-
tient hy the same number.
II. Using the same example, 54 -h 9 = 6.
Dividing the dividend by 3 we have
■^-i-9=:18-^9 = 2,
and 2 = the quotient, 6, divided by 3. Hence, Dividing the
dividend ly any number, divides the quotient by the same
number.
III. Multiplying the divisor by 3, we have
54 -^ 9 X 3 = 54 -^ 27 = 2,
and 2 = the quotient, 6, divided by 3. Hence, Multiplying
the divisor by any number, divides the quotient by the same
mimber.
IV. Dividing the divisor by 3, we have
54 -J- J = 54 -f- 3 = 18,
Upon what does tlie value of tlie quotient depend ? What is the
first general principle of division 1 Second ? Tliird ? Fourth?
GEKERAL PRINCIPLES OF DIVISION. 65
ftnd 18 = the quotient, 6, multiplied by 3. Hence, Divid-
ing the divisor hy any number, multiplies the quotient hy the
mme number,
V. Multiplying both dividend and divisor by 3, we have
64 X 3 -V- 9 X 3 = 162 -^ 27 = 6.
Hence, Multiplying both dividend and divisor hy the same
mmber, does not alter the value of the quotient,
VL Dividing both dividend and divisor by 3, we have
A^--| = 18--3=:6.
Pence, Dividing both dividend and divisor by the same num*
her, does not alter the value of the quotient.
87. These six examples illustrate all the different changes
we ever have occasion to make upon the dividend and divi-
sor in practical arithmetic. The principles upon which
these changes are based may be stated as follows :
Prin. I. Multiplying the dividend multiplies the qicotient ;
and dividing the dividend divides the quotient, {SQ, I and II.)
Prin. II. Multiplying the divisor divides the quotient;
and dividing the divisor multiplies the quotient. (86. Ill
and IV.)
Prin. III. Multiplying or dividing both dividend and
divisor by the same number, does not alter the quotient, (86.
V and VL)
88. These three principles may be embraced in one
General Law.
A change i7i the dividend produces a like change in the
quotient ; but a change in the divisor produces an opposite
change in the quotient.
I- a number be mnltiplied and the product divided by the same number, the quo-
tient will be equal to the number multiplied. Thus, 15 x 4=60, and 60+4=15.
Fifth? Sixth? Into how many general principles can these br
condensed? What is the first? Second? Third? In what general
law are these embraced ?
66 PROPEETIES OF NUMBERS.
EXACT DIVISOES.
89. An Exact Divisor of a number is one that gives
a whole number for a quotient.
As it is frequently desirable to 'know if a number has an
exact divisor, we will present a few directions that will be
of assistance, particularly in finding exact divisors of large
numbers.
A number whose unit figure is 0, 2, 4, 6, or 8, is called an Even Number. And a
number whose unit figure is 1, 3, 5, 7, or 9, is called an Odd Number^
2 is an exact divisor of all even numbers.
4 is an exact divisor when it will exactly divide the tens
and units of a number. Thus, 4 is an exact divisor of 268,
756, 1284.
5 is an exact divisor of every number whose unit figure is
or 5. Thus, 5 is an exact divisor of 20, 955, and 2840.
8 is an exact divisor when it will exactly divide the hun-
dreds, tens, and units of a number. Thus, 8 is an exact
divisor of 1728, 5280, and 213560.
9 is an exact divisor when it will exactly divide the sum of
the digits of a number. Thus, in 2486790, the sum of the
digits 2 + 4 + 8 + 6 + 7 + 9 + 0=36, and 36-t-9=4.
10 is an exact divisor when occupies units' place.
100 when 00 occupy the places of units and tens.
1000 when 000 occupy the places of units, tens, and hun-
dreds, etc.
A composite number is an exact divisor of any number,
when all its factors are exact divisors of the same number.
Thus, 2, 2, and 3 are exact divisors of 12 ; and so also are 4
(=2x2) and 6 (=2x3).
An even number is not an exact divisor of an odd number.
If an odd number is an exact divisor of an even number.
What is an exact divisor ? What is an even number? An odd num-
ber? When is 2 an exact divisor ? 4? 5? 9? 10? 100? 1000?
When is a composite number an exact divisor ? An even number is
not an exact divisor of what ? An odd number is an exact divisor of
what?
FACTORIl^G NUMBERS.
67
twice that odd number is also an exact divisor of the even
number. Thus, 7 is an exact divisor of 42 ; so also is 7 x 2,
or 14.
PRIME NUMBEES.
90. A Prime K^umber is one that can not be resolved
or separated into two or more integral factors.
For reference, and to aid in determining the prime factors
of composite numbers, we give the following :
Table of Prime Numbers from 1 to 1000.
1
59
139
233
337
439
557
653
769
883
2
61
149
239
347
443
563
659
773
887
3
67
151
241
349
449
569
661
787
907
5
71
157
251
353
457
571
673
797
911
7
73
163
257
359
461
577
677
809
919
11
79
167
263
367
463
587
683
811
929
13
83
173
269
373
467
593
691
821
937
17
89
179
271
379
479
599
701
823
941
19
97
181
277
383
487
601
709
827
947
23
101
191
281
389
491
607
719
829
953
29
103
193
283
397
499
613
727
839
967
31
107
197
293
401
503
617
733
853
971
37
109
199
307
409
509
619
739
857
977
41
113
211
311
419
521
631
743
859
983
43
127
223
313
421
523
641
751
863
991
47
131
227
317
431
541
643
757
877
997
53
137
229
331
433
547
647
761
881
FACTORING NUMBERS.
Case I.
91. To resolve any composite number into its
prime factors.
What is a prime number ? In factoring numbers. Case I is what ?
2
2772
2
1386
3
693
3
231
7
77
11
11
1
68 PROPERTIES OF NUMBERS.
1 What are the prime factors of 2772 ?
OPERATION. Analysis. Divide the given number by 2,
the least prime factor, and the result by 2 ; this
gives an odd number for a quotient, divisible by
the prime factor, 3, and the quotient resulting
from this division is also divisible by 3. The
next quotient, 77, we divide by its least prime
factor, 7, and obtain the quotient 11 ; this being
a prime number, the division can not be carried
further. The divisors and last quotient, 2, 2, 3,
8, 7, and 11 are all the prime factors of the given
number, 2772.
Rule. Divide the given number by any prime factor ;
divide the quotient in the same manner, and so continue the
division until the quotient is a prime number, Tlie several
divisors and the last quotient will be the prime factors re-
quired.
Proof. The product of all the prime factors will be the
given number.
Examples for Practice.
2. What are the prime factors of 1140 ? Ans, 2, 2, 3, 5, 19.
3. What are the prime factors of 29925 ?
4. What are the prime factors of 2431 ?
5. Find the prime factors of 12673.
6. Find the prime factors of 2310.
7. Find the prime factors of 2205.
8. What are the prime factors of 13981 ?
Case II.
92. To resolve a number into all the diflferent sets
of factors possible.
1. In 36 how many sets of factors, and what are they ?
Give explanation. Rule. Proof. Case II is what ?
C A K C E L L A T I 2^
69
^2
3
OPERATION.
xl8
xl2
4
6
2
2
3
2
36 = ^
x9
x6
X 2 X
x3 X
X 3 X
x2 X
Analysis. Writing the 36 at
tlie left of the sign =, arrange
all the different sets of factors into
which it can be resolved under
each other, as shown in the opera-
tion, and we find that 36 can be
resolved into 8 sets of factors.
x3
Examples for Practice.
2. How many sets of factors in the number 24 ? "Wliat
are they? Ans. Q sets.
3. In 125 how many sets of factors ? What are they ?
Ans, 2 sets.
4. In 40 how many sets of factors, and what are they ?
Ans. 6 sets.
5. In 72 how many sets of factors, and what are they ?
Ans. 15 sets.
CANCELLATION.
93. Cancellation is the process of rejecting equal fac-
tors from numbers sustaining to each other the relation of
dividend and divisor.
It has been shown ( 7*7 ) that the dividend is equal to the
product of the divisor multiplied by the quotient. Hence,
if the dividend can be resolved into two factors, one of
which is the divisor, the other factor wiU be the quotient.
1. Divide 63 by 7.
Analysis. We see in
this example that 63 is
composed of the factors 7
and 9, and that the factor
7 is equal to the divisor.
Therefore we reject the factor 7, and the remaining factor, 9, is the
quotient.
OPEKATION.
Divisor, ^ ) /^ X 9 Dividend.
9 Quotient.
Give explanation. What is cancellation ? Upon what principle is
it based ? Give first explanation.
70 PROPEETIES OF NUMBERS.
94. Whenever the dividend and divisor are each com-
posite numbers, the factors common to both may first be
rejected without altering the final result. ( 87, Prin. III.)
2. What is the quotient of 24 times 56 divided by 7 times
48?
OPERATION. Analysis. First
24x66_4x0x;?X$_. J indicate the opera-
~~Z 77^ — I I ^ — ■*' ^ns, ^Jqjj ^q ^q performed
7 X 48 ^ X X ^ V V *i
by writing the num-
<)ers which constitute the dividend above a line, and those which con-
Btitute the divisor below it. Instead of multiplying 24 by 56, in the
dividend, we resolve 24 into the factors 4 and 6, and 56 into the factors
7 and 8 ; and 48 in the divisor into the factors 6 and 8. We next
cancel the factors 6, 7, and 8, which are common to the dividend and
divisor, and we have left the factor 4 in the dividend, which is the
quotient.
When all the factors or numbers in tha divid«ind are canceled, 1 should be
retained.
95. If any two numbers, one in the dividend and one in
the divisor, contain a common factor, we may reject that
factor.
3. In 54 times 77, how many times 63 ?
OPERATION. Analysis. In this example we see that 9 will
Q 11 divide 54 and 63 : so we reject 9 as a factor of 54,
^4- \( "^It ^^^ retain the factor 6, and also as a factor of 63,
and retain the factor 7. Again, 7 will divide 7 in
vP the divisor, and 77 in the dividend. Dividing
'^ both numbers by 7, 1 will be retained in the
divisor, and 11 in the dividend. Finally, the
product of 6 X 11=: 66, the quotient.
4. Divide 25 x 16 x 12 by 10 x 4 x 6 x 7.
OPERATION. Analysis. In this,
5 4^ as in the preceding
t$xHxn 5x4 ,, ^, example, we reject all
= = *il^ = -^-f • the factors that are
40X^X0x7 7 common to both divi-
$ dend and divisor, and
Give second explanation.
CANCELLATION. 71
me have remaining the factor 7 in the divisor, and the factors 5 and 4
m the dividend. Completing the work, we have ^=2f, Arts.
From the preceding examples and illustrations we derive
the following
EuLE. L Write the numbers composing the dividend above
a horizontal line, and the numbers composing the divisor
below it.
II. Cancel all the factors common to both dividend and
divisor.
III. Divide the product of the remaining factors of the
dividend by the product of the remaining factors of the di-^
visor, and the result will be the quotient.
1. Kejecting a factor from any number is dividing the number by that &ctor.
2. When a factor is canceled, the unit, 1, is supposed to take its place.
8. One factor in the dividend will cancel only one equal factor in the divisor.
4. If all the factors or numbers of the divisor are canceled, the product of the
remaining factors of the dividend will be the quotient.
5. By many it is thought more convenient to write the fetors of the dividend on
the right of a vertical line, and tiie fiactors of the divisor on the left,
Examples foe Peactice,
1. What is the quotient of 16 x 5 x 4 divided by 20 x 8 ?
FIRST OPERATION. SECOND OPERATION.
.2
2 .
10X0X^ ^ , ^0
= 2, Ans, rt
A
H'
^
2, Ans,
2. Divide the product of 120 x 44 x 6 x 7 by 72 x 33 x 14.
Rule, first step ? Second ? Third ? What is the effect of rejecting
a factor ? What is the quotient when all the factors in the divisor are
canceled \
n
PEOPERTIES OF KHMBEKS,
FIRST OPERATION.
10.
r2
xMx^xt 10x2
nx$$xU
^ 3 ^
^ = e^Ans.
SECOND OPERATION.
10
^ti
m
Ha
t
3
20
6|,
• 3. Divide the product of 33 >i: 35 x 28 by 11 x 15 x 14.
Ans» 14.
4. What is the quotient of 21 x 11 X 26 divided by 14 x
13 ? Ans. 33.
5. Divide the product of the numbers 48, 72, 28, and 5,
by the product of the numbers 84, 15, 7, and 6, and give
the result. Ans. 9-ij-.
6. Divide 140 x 39 x 13 x 7 by 30 x 7 x 26 x 21.
Ans. 4J.
7. What is the quotient of 66 x 9 x 18 x 5 divided by
22 X 6 X 40. ^ Ans. lOJ.
8. Divide the product of 200 x 36 x 30 x 21 by 270 x
40 X 15 X 14. Ans. 2,
9. Multiply 240 by 56, and divide the product by 60 mul-
tiplied by 28. Ans. 8.
10. The product of the numbers 18, 6, 4, and 42 is to be
divided by the product of the numbers 4, 9, 3, 7 and 6 ;
what is the result ? Ans. 4.
11. How many tons of hay, at 12 dollars a ton, must be
given for 30 cords of wood, at 4 dollars a cord ? Ans. 10 tons.
GREATEST COMMON DIVISOU. 73
12. How many firkins of butter, each containing 56
pounds, at 13 cents a pound, must be given for 4 barrels of
sugar, each containing 182 pounds, at 6 cents a pound ?
Ans. 6 firkins.
13. A tailor bought 5 pieces of cloth, each piece contain-
ing 24 yards, at 3 dollars a yard. How many suits of
clothes, at 18 dollars a suit, must be m^de from the cloth
to pay for it ? Ans. 20 suits.
14. How many days' work, at 75 cents a day, will pay for
115 bushels of corn, at 50 cents a bushel ? Ans. 76f days.
GREATEST COMMON DIVISOR.
96. A Common Divisor of two or more numbers is a
number that will exactly divide each of them.
97. The Greatest Common Divisor of two or more
numbers is the greatest number that will exactly divide
each of them.
Numbers prime to each other are such as have no com-
mon di\dsor.
A common diviBor is sometimes called a Common Measure ; and the greatest
common divisor, the Greatest Common
Case I.
98. "When the numbers are readily factored.
1. What is the greatest common divisor of 6 and 10 ?
Ans. 2.
OPEBATION. Analysis. We readily find by inspection tliat 2 will
6 . . 10 divide both tbe given numbers ; hence 2 is a common
"o K divisor ; and since the quotients 3 and 5 have no com-
mon factor, but are prime to each other, the common
divisor, 2, must be the greatest common divisor.
2. What is the greatest common divisor of 42, 63, and 105 ?
What is a common divisor? The greatest common divisor? A
common measure ? The greatest common measure ? What is Case I.?
Give analysis.
'^^p, 4
3
42 . . 63
.105
7
14 . . 21
. 35
2.. 3.
. 5
74 PROPEBTIES OF UUMBBES.
OPERATION. ANAiiYSis. We observe ttat 3 will
exactly divide each of the given num-
bers, and that 7 will exactly divide
each of the resulting quotients. Hence,
each of the given numbers can be ex-
3 ^ ij'_.21 Ans, actly divided by 3 times 7 ; and these
numbers must be component factors of
the greatest common divisor. Now, if there were any other component
factor of the greatest common divisor, the quotients, 2, 3, 5, would be
exactly divisible by it. But these quotients are prime to each other.
Hence 3 and 7 are all the component factors of the greatest common
divisor sought.
3. What is the greatest common divisor of 28, 140, and
280?
OPERATION. Analysis. We first divide by 4 ;
then the quotients by 7. The re-
sulting quotients, 1, 5, and 10, are
prime to each other. Hence 4 and
7 are all the component factors of
4 X 7=28, A71S. tt^6 greatest common divisor.
From these examples and analyses we derive the following
Rule. I. Write the numbers in a line, with a vertical line
at the left, and divide ly any factor common to all the numbers.
XL Divide the quotients in like manner, and continue the
division till a set of quotients is obtained that have no comm.on
factor.
III. Multiply all the divisors together, and the product wiU
be the greatest common divisor sought.
Examples fob Practice.
1. What is the greatest common divisor of 12, 36, 60, 72?
Ans. 12.
2. What is the greatest common divisor of 18, 24, 30, 36,
42? ^ Ans,^.
Rule, first step ? Second ? Third ?
4
28 . . 140 . . 280
7
7.. 35.. 70
1 . . 5 . , 10
i
GREATEST COMMON DIVISOR. 75
3. What is the greatest common divisor of 72, 120, 240,
384 ? Ans. 24.
4. What is the greatest common divisor of 36, 126, 72,
216 ? Ans. 18.
5. What is the greatest common divisor of 42 and 112 ?
Ans, 14.
6. What is the greatest common divisor of 32, 80, and
256? Ans.ie,
7. What is the greatest common divisor of 210, 280, 350,
630, and 840 ? Ans. 70.
8. What is the greatest common divisor of 300, 525, 225,
and 375 ? Ans, 75.
9. What is the greatest common divisor of 252, 630, 1134,
and 1386? . A7is. 126.
10. What is the greatest common divisor of 96 and 544 ?
Ans. 32.
11. What is the greatest common divisor of 468 and 1184 ?
Ans. 4.
12. Wliat is the greatest common divisor of 200, 625, and
150 ? Ans. 25.
Case II.
99. "When the mimbers can not be readily factored.
As the analysis of the method under this case depends
upon three properties of numbers which have not been in-
troduced, we present them in this place.
I. An exact divisor divides any number of times its divi-
dend.
II. A common divisor of two numbers is an exact divisor
of their sum.
III. A common di^dsor of two numbers is an exact divisor
of their difference.
What is Case II.? What is the first principle upon which it is
founded? Second? Third?
76
PROPERTIES OF Is^ UMBERS.
1. What is the greatest common divisor of 84. and 203 ?
Analysis. Draw two vertical lines, and
place the larger number on tlie right, and
the smaller number on the left, one line
lower down. Then divide 203, the larger
number, by 84, the smaller, and write 2,
the quotient, between the verticals, the
product, 168, opposite, under the greater
number, and the remainder, 35, below.
Next divide 84 by this remainder, writing
the quotient, 2, between the verticals, the
product, 70, on the left, and the new remainder, 14, below the 70.
Again divide the last divisor, 35, by 14, and obtain 2 for a quotient,
28 for a product, and 7 for a remainder, all of which we write in the
game order as in the former steps. Finally, divide the last divisor,
14, by the last remainder, 7, and we have no remainder. 7, the last
divisor, is the greatest common divisor of the given numbers.
\JX
-iSOVA
203
84
2
168
70
2
35
14
2
28
14
2
7, 4'^s,
In order to show that the last di^dsor in such a process is
the greatest common divisor, we will first trace the work in
the reverse order, as indicated by the arrow line below.
OPERATION.
84
70
14
14
7 divides the 14, as proved by
the last division ; it will also di-
vide two times 14, or 28, (1.) Now.
as 7 divides both itself and 28, it
will divide 35, their sum, (11.) It
will also divide 2 times 35, or 70,
(I ;) and since it is a common di-
visor of 70 and 14, it must divide
their sum, 84, which is one of the
given numbers, (II.) It will also
divide 2 times 84, or 168, (I ;) and
since it is a common divisor of 108
and 35, it must divide their sum,
203, the larger number, (II.) Hence 7 is a common divisor of the
given numbers.
Again, tracing the work in the direct order, as indicated below, we
203
168
35
28
■^^
Give analysis.
GREATEST COMMOK DIVISOR.
77
know that the greatest common divisor, whatever it he, must divide
2 times 84, or 168, (I.) Then
U203
84 ^
70
14
168
35
28
since it will divide both 168 and
203, it must divide their differ-
ence, 35, (III.) It will also divide
2 times 35, or 70, (I ;) and as it will
divide both 70 and 84, it must di-
vide their difference, 14, (III.) It
will also divide 2 times 14 or 28,
(I ;) and as it will divide both 28
and 35, it must divide their differ-
ence, 7, (III;) hence, it canirwt he
greater tJian 7
Thus we haye shown,
1st. That 7 is a common divisor of the given numbers.
2d. That their greatest common divisor, whatever it be,
cannot be greater- than 7. Hence it must be 7.
From this example and analysis, we derive the following
Rule. I. Draw two verticals, and write the two numbers,
one on each side, the greater 7iuml)er one line above the less.
II. Divide the greater number by the less, tvriting the quo-
tient between the verticals, the product under the dividend,
and the remainder below,
III. Divide the less number by the remainder, the last di'
visor by the last remainder, and so on, till nothing remains.
The last divisor will be the greatest common divisor sought,
IV. If more than tivo numbers be given, first fi7id the great'
est common divisor of two of them, and then of this divisor
and one of the remaining numbers, and so on to the last;
the last common divisor found will be the greatest common
divisor of all the given numbers,
1. When more than two numbers are given, it is better to begin with the least twe.
2. If at any point in the operation a prime number occur as a remainder, it must
be a common divisor, or the given numbers have no common divisor.
Rule, first step? Second? Third? Fourth? What relation have
numbers when their difference is a prime number ?
78
PKOPEETIES OF NUMBERS.
EXAMPf,ER
FOR
Practice.
. What is the greatest common divisor of 221 and 5512 ?
OPERATION.
5512
221
2
442
1092
4
1
884
208
208
Ans. 13
1
13
78
6
78
I
2. Find the greatest common divisor of 154 and 210.
Ans. 14.
3. What is the greatest common divisor of 316 and 664 ?
Ans. 4.
4. What is the greatest common divisor of 679 and 1869 ?
A}is. 7.
5. What is the greatest common divisor of 917 and 1495 ?
Alls. 1.
6. What is the greatest common divisor of 1313 and 4108 ?
Ans. 13.
7. What is the greatest common divisor of 1649 and 5423 ?
Ans. 17.
The following examples may be solved by either of the
foregoing methods.
8. John has 35 pennies, and Charles 50 : how shall they
arrange them in parcels, so that each boy shall have the same
number in each parcel ? Ans. 6 in each parcel.
9. A speculator has 3 fields, the first containing 18, the
second 24, and the third 40 acres, which he wishes to divide
into the largest possible lots having the same number of
acres in each ; how many acres in each lot? Ans. 2 acres.
MULTIPLES. 79
10. A farmer had 231 bushels of wheat, and 273 bushels
of oats, which he wished to put into the least number of bins
containing the same number of bushels, without mixing the
two kinds ; what number of bushels must each bin hold ?
Ans. 21.
11. A Tillage street is 332 rods long ; A owns 124 rods
front, B 116 rods, and C 92 rods ; they agree to divide theii
land into equal lots of the largest size that will allow each
one to form an exact number of lots ; what will be the width
of the lots? Ans, 4 rods.
12. The Erie railroad has 3 switches, or side tracks, of the
following lengths : 3013, 2231, and 2047 feet ; what is the
length of the longest rail that vail exactly lay the track on
each switch ? Ans. 23 feet.
13. A forwarding merchant has 2722 bushels of wheat,
1822 bushels of corn, and 1226 bushels of beans, which he
wishes to forward, in the fewest bags of equal size that will
exactly hold either kind of grain ; how many bags will it
take ? Ans. 2885.
14. A has 120 dollars, B 240 dollars, and C 384 dollars ;
they agree to purchase cows, at the highest price per head
that will allow each man to invest all his money ; how many
cows can each man purchase ? Ans. A 5, B 10, and C 16.
MULTIPLES.
100. A Multiple is a number exactly divisible by a
given number ; thus, 20 is a multiple of 4.
101. A Common Multiple is a number exactly divisible
by two or more given numbers ; thus, 20 is a common mul-
tiple of 2, 4, 5, and 10.
102. The Least Common Multiple is the least num«
ber exactly divisible by two or more given numbers ; thus,
24 is the least common multiple of 3, 4, 6, and 8.
What is a multiple? A common multiple? The least commpp
multiple?
80 PROPERTIES OF JSTUAIBERS.
103. From the definition ( 100 ) it is evident that the
product of two or more numbers, or any number of times
their product, must be a common multiple of the numbers.
Hence, A co7nmon multiple of two or more numbers may he
found ly multiplying the given numbers together,
104. To find the least common multiple.
First Method.
From the nature of prime numbers we derive the follow-
ing principles :
I. If a number exactly contain another, it will contain all
the prime factors of that number.
II. If a number exactly contain two or more numbers, it
will also contain all the prime factors of those numbers.
III. The least number that will exactly contain all the
prime factors of two or more numbers, is the least common
multiple of those numbers.
1. Find the least common multiple of 30, 42, 66, and 78.
OPERATION. Analysis. The
30 = 2 X 3 x5 number cannot fee
42 = 2 X 3 X 7 ^®^® than 78, since
aa o v^ Q v^ 11 it must contain 78 ;
bb = /i X o X 11 , .^ ^
7Q _ 9 ^ q ^ iq ^e^ce It must con.
2x3x13x11x7x5 = 30030, A7is. 78, viz. :
2 X 3 X 13.
We here have all the prime factors of 78, and also all the factors ol
66, except the factor 11. Annexing 11 to the series of factors,
2 X 3 X 13 X 11,
and we have all the prime factors of 78 and 66, and also all the factors
of 42 except the factor 7. Annexing 7 to the series of factors,
2 X 3 X 13 X 11 X 7,
and we have all the prime factors of 78, 66, and 42, and also all the
How can a common multiple of two or more numbers be found ?
First principle derived from prime numbers ? Second ? Third %
Give anal)'sis.
LEAST COMMOK MULTIPLE. 81
factors of 30 except the factor 5. Annexing 5 to the series of factors,
3 X 3 X 13 X 11 X 7 X 5,
and we have all the prime factors of each of the given numbers ; and
hence the product of the series of factors is a common multiple of the
given numbers, (II.) And as no factor of this series can be omitted
without omitting a factor of one of the given numbers, the product of
the series is the least common multiple of the given numbers, (III.)
From this example and analysis we deduce the following
Rule. I. Resolve the given numbers into their prime
factors.
II. Take all the prime factors of the largest number, and
such prime factors of the other numbers as are not found in
the largest number, and their product will be the least com-
mon multiple.
When a prime factor is repeated in any of the given numbers, it must be used as
many times, as a factor of the multiple, as the greatest number of times it appears
iu any of the given numbers.
Examples for Practice.
2. Find the least common multiple of 7, 35, and 98.
Ans. 490.
3. Find the least common multiple of 24, 42, and 17.
Ans. 2856.
4. What is the least common multiple of 4, 9, 6, 8 ?
Ans. 72.
6. What is the least common multiple of 8, 15, 77, 385 ?
Ans. 9240.
6. What is the least common multiple of 10, 45, 75, 90 ?
Ans. 450.
7. Wliat is the least common multiple of 12, 15, 18, 35 ?
Ans. 1260.
Rule, first step ? Second ? What caution is given ?
4*
2
OPERATION.
4. .6. .9.
12
2
2
..3. .9.
. 6
3
3. .9.
. 3
3
3
82 properties of numbers.
Secoi^d Method.
105. 1. What is the least common multiple of 4, 6, 9,
and 12?
ANAiiYSis. First write the
given numbers in a series, with
a vertical line at the left
Since 2 is a factor of some oi
the given numbers, it must be
a factor of the least common
multiple sought. Dividing as
2x2x3x3 = 36, Ans, many of the numbers as are
divisible by 2, write the quo-
tients and the undivided number, 9, in a line underneath. We no\«
perceive that some of the numbers in the second line contain the
factor 2 ; hence the least common multiple must contain another 2,
and we again divide by 2, omitting to write down any quotient when
it is 1. We next divide by 3 for a like reason, and still again by 3.
By this process we have transferred all the factors of each of the
numbers to the left of the vertical ; and their product, 36, must be the
least common multiple sought, (104, III.)
2. What is the least common multiple of 10, 12, 15, and 75 ?
Analysis. We read-
ily see that 2 and 5 are
among the factors of the
given numbers, and must
be factors of the least
2x5x2x3x5= 300, Ans. ^"^""^^ "'^^^^P^^ ' ^'"'^
we divide every number
that is divisible by either of these factors or by their product ; thus, we
divide 10 by both 2 and 5 ; 12 by 2 ; 15 by 5 ; and 75 by 5. We next
divide the second line in like manner by 2 and 3 ; and afterwards the
third line by 5. By this process we collect the factors of the given
numbers into groups; and the product of the factors at the left of the
vertical is the least common multiple sought.
3. Wliat is the least common multiple of 6, 15, 35, 42,
and 70 ?
2,5
OPERATION.
10 . . 12 . . 15 . . 75
2,3
6.. 3.. 15
5
5
Give explanation.
3,7
2,5
LEAST COMMOIT MULTIPLE. OO
OPEBATiON. Analysis. In this opera-
15 . . 42 . . 70 tion we omit the 6 and 35,
because they are exactly con-
tained in some of the other
5 . . 2 . . 10
3x7x2x5 = 210, Ans, given numbers ; thus, 6 is
contained in 43, and 35 in
70; and whatever will contain 42 and 70 must contain 6 and 35.
Hence we have only to find the least common multiple of the remain-
ing numbers, 15, 42, and 70.
From these examples we derive the following
KuLE. I. Write the numbers in a line, omitting any of the
smaller numbers that are factors of the larger, and draw a
vertical line at the left.
II. Divide hy any prime factor, or factors, that may he con-
tained in one or more of the given numbers, and write the
quotients and undivided numbers i7i a line underneath, omit-
ting the Vs.
III. In nice manner divide the quotients and undivided
numbers, and continue the process till all the factors of the
given numbers have been transferred to the left of the vertical.
Tfien multiply these factors together, and their product will
be the least common multiple required.
Examples for Practice.
4. What is the least common multiple of 12, 15, 42, and
60? Ans. 420.
5. What is the least common multiple of 21, 35, and 42 ?
Ans. 210.
6. What is the least common multiple of 25, 60, 100, and
125? Ans. 1500.
7. What is the least common multiple of 16, 40, 96, and
105? Ans. 3360.
8. What is the least common multiple of 4, 16, 20, 48, 60,
and 72 ? Ans. 720.
9. What is the least common multiple of 84, 100, 224, and
300? Ans. 16800.
Rule, first step ? Second ? Third ?
84 PROPERTIES OF NUMBERS.
10. What is the least common multiple of 270, 189, 297,
243? Ans. 187110.
11. What is the least common multiple of 1, 2, 3, 4, 5, 6,
7, 8, 9 ? Ans. 2520.
12. What is the smallest sum of money for which I could
purchase an exact number of books, at 5 dollars, or 3 dol-
lars, or 4 dollars, or 6 dollars each ? Ans, 60 dollars.
13. A farmer has 3 teams ; the first can draw 12 barrels
of flour, the second 15 barrels, and the third 18 barrels ;
what is the smallest number of barrels that will make full
loads for any of the teams ? Ans, 180.
14. What is the smallest sum of money with which I can
purchase cows at 130 each, oxen at $55 each, or horses at
$105 each? Ans. U310,
15. A can shear 41 sheep in a day, B 63, and C 54 ; what
is the number of sheep in the smallest flock that would
furnish exact days' labor for each of them shearing alone ?
Ans, 15498.
16. A servant being ordered to lay out equal sums in the
purchase of chickens, ducks, and turkeys, and to expend as
little money as possible, agreed to forfeit 5 cents for every
fowl purchased more than was necessary to obey orders. In
the market he found chickens at 12 cents, ducks at 30 cents,
and turkeys at two prices, 75 cents and 90 cents, of which
he imprudently took the cheaper ; how much did he thereby
forfeit ? Ans. 80 cents.
CLASSIFICATION OF NUMBEES.
Numbers may be classified as follows :
106. I. As Even and Odd.
107. II. As Prime and Composite.
What is the first classification of numbers ? What is an even num-
ber? An odd number? Second classification? A prime number?
A composite number ?
.CLASSIFICATION OF KUMBERS. 85
108. III. As Integral and Fractional
An Integral Number, or Integer, expresses whole
things. Thus, 281 ; 78 boys; 1000 books.
A Fractional Number, or Fraction, expresses equal
parts of a thing. Thus, half a dollar ; three-fourths of an
hour ; seven-eighths of a mile.
109. IV. As Abstract and Concrete,
110. V. As Simple and Compound.
A Simple Number is either an abstract number, or a
concrete number of but one denomination. Thus 48, 926 ;
48 dollars, 926 miles.
A Compound Number is a concrete number whose value
is expressed in two or more different denominations. Thus,
32 dollars 15 cents; 15 days 4 hours 25 minutes; 7 miles
82 rods 9 feet 6 inches.
111. YI. As Lihe and Unlike,
Like Numbers are numbers of the same unit value.
If simple numbers, they must be all abstract, as 6, 62,
487 ; or all of one and the same denomination, as 5 apples,
62 apples, 487 apples ; and, , if compound numbers, they
must be used to express the same kind of quantity, as time,
distance, etc. Thus, 4 weeks 3 days 16 hours; 1 week 6
days 9 hours ; 5 miles 40 rods ; 2 miles 100 rods.
Unlike Numbers are numbers of different unit values.
Thus, 75, 140 dollars, and 28 miles ; 4 hours 30 minutes,
and 5 bushels 1 peck. /
What is the third classification ? What is an integral number ? A
fractional number ? What is the fourth classification ? An abstract
number ? A concrete number ? What is the fifth classification ? A
simple number ? A compound number ? Sixth classification ? What
are like numbers ? UnliJie numbers ?
86
FRACTIOifS.
FEACTIOl^S.
Definitions, Notation, and Numeration.
112. If a unit be divided into 2 equal parts, one of the
parts is called one-half.
If a unit be divided into 3 equal parts, one of the parts is
called one-third, two of the parts two-thirds.
If a unit be divided into 4 equal parts, one of the parts
is called one-fourth, two of the parts two-fourths, three of
the parts three-fourths.
If a unit be divided into 5 equal parts, one of the parts
is called one-fifth, two of the parts two-fifths, three of the
parts three-fifths, etc.
The parts are expressed by figures ; thus.
One-half is written
i
One-fifth is written
i
One-third "
i
Two-fifths
1
Two-thirds "
f
One-seventh ^ ''
+
One-fourth "
i
Three-eighths '*
1
Two-fourths "
f
Five-ninths "
1
Three-fourths "
f
Eight-tenths "
A
Hence we see that the parts into which a unit is divided
take their name, and their value, from the number of equal
parts into which the unit is divided. Thus, if we divide
an orange into 2 equal parts, the parts are called halves ;
if into 3 equal parts, thirds ; if into 4 equal parts, fourths,
etc. ; and each third is less in value than each half, and
esLoh fourth less than each third; and the greater the num-
ber of parts, the less their value.
When a unit is divided into any number of equal parts,
one or more such parts is a fractional part of the whole
number, and is called a. fraction. Hence,
113. A Fraction is one or more of the equal parts of a
unit.
Define a fraction.
DEFINITIONS, NOTATION, NUMERATION. 87
114. To write a fraction, two integers are required, one
to express the number of parts into which the whole num-
ber is divided, and the other to express the number of these
parts taken. Thus, if one dollar be divided into 4 equal
parts, the parts are called fourths, and three of these parts
are called three-fourths of a dollar. This three-fourths may
be written
3 the number of parts taken.
4 the number of parts into which the dollar is divided.
115. The Denominator is the number below the line.
It denominates or names the parts ; and
It shows how many parts are equal to a unit.
116. The iN'umerator is the number above the line.
It numerates or numbers the parts ; and
It shows how many parts are taken or expressed by the
fraction.
117. The Terms of a fraction are the numerator and
denominator, taken together.
118. Fractions indicate division, the numerator answer-
ing to the dividend, and the denominator to the divisor.
110. The Value of a fraction is the quotient of the nu-
merator divided by the denominator.
120. To analyze a fraction is to designate and describe
its numerator and denominator. Thus, j is analyzed as
follows : —
4 is the denomin^itor, and shows that the integer is divided
into 4 equal parts ; it is the divisor.
3 is the numerator, and shows that 3 parts are taken ; it
is the dividend, or integer divided.
3 and 4 are the ter7ns, considered as dividend and divisor.
The value of the fraction is the quotient of 3-i-4, or |.
HoTV many numbers are required to write a fraction ? Why ? De-
fine the denominator. The numerator. What are the terms of a frac-
tion ? The value ? Whf>t is the analysis of a fraction ?
88 feacixons.
Examples fok Practice.
Express the following fractions by figures :
1. Seven eighths,
2. Three twenty-fifths.
3. Nine one-hundredths.
4. Sixteen thirtieths.
5. Thirty-one one hundred eighteenths.
6. Seventy-five ninety-sixths.
7. Two hundred fifty-four /owr hundred forty-thirds.
8. Eight nine hundred twenty-firsts.
9. One thousand two hundred thirty-two seventy-five thot^
sand six hundredths.
10. Nine hundred six two hundred forty-three thousand
eighty-seconds.
Eead and analyze the following fractions :
^^' iV J T^'y A? M> TT ? TlTJ WV? iM-
13' ftt; T%^; ^i^; li^ir; ^Hrr; ^m.
131. Fractions are distinguished as Proper and Improper.
A Proper Fraction is one whose numerator is less than
its denominator ; its value is less than the unit, 1. Thus,
tV> a? A> Jf ^^6 proper fractions.
An Improper Fraction is one whose numerator equals
or exceeds its denominator ; its value is never less than tlio
unit, 1 . Thus, ^, f, Jjt, •^-, \^, i^ are improper fractions
123. A Mixed Number is a number expressed by an in-
teger and a fraction; thus, 4J-, 17J|, 9^ are mixed numbers*
123. Since fractions indicate division, all changes in the
terms of a fraction will affect the value of that fraction ac-
cording to the laws of division ; and we have only to modify
the language of the General Principles of Division (87) by
substituting the words numerator, denominator, and fraction.
What is a proper fraction ? An improper fraction ? A mixed num'
ber ? Wliat do fractions indicate ?
REDUCTION. 89
or value of the fraction, for the words dividend, divisor, and
quotient, respectively, and we shall have the following
Gekeeal Principles of Fractions.
124* Prin. I. Multiplying the numerator multiplies the
fraction, and dividing the numerator divides the fraction,
Prin. II. Multiplying the denominator divides the frao-
fion, and dividing the denominator multiplies the fraction,
Prin. III. Multiplying or dividing both terms of the frao*
Hon hy the same number does not alter the value of the fraction.
These three principles may be embraced in one
General Law.
125. A change in the numerator produces a like change
in the value of the fraction ; but a change in the denomina-
tor produces an opposite change in the value of the fraction^
REDUCTION.
Case L
126. To reduce firactions to their lowest terms.
A fraction is in its lowest terms when its numerator and
denominator are prime to each other ; that is, when both
terms have no common divisor.
1. Reduce the fraction f^ to its lowest terms.
FiKST OPERATION. ANALYSIS. Dividing both terms of
I j^=:-|^=^=^, ^^^5. a fraction by the same number does
not alter the value of the fraction or
qnotient, (124, III ;) hence, we divide both terms of f f , by 2, both
terms of the result, f-f, by 2, and both tei-ms of this result by 3. As
the terms of f are prime to each other, the lowest terms of || are |.
We have, in effect, canceled all the factors common to the numerator
and denominator.
First general principle? Second? Third? General law? What
is meant by reduction of fractions ? Case I is what ? What is meant
hj lowest terms f Give analysis.
90 FEACTIONS.
SECOND OPERATION. In this operation we have divided
12 ) |-§-=i^, Ans. hoth terms of the fraction by their
greatest common divisor, (97,) and
thus performed the reduction at a single division.
Rule. Cancel or reject all factors common to both numera-
tor and denommator. Or,
Divide hath terms hy their greatest common divisor.
Examples fob Peactice.
2.
Reduce
Jll^ to its lowest terms.
Ans. \,
3.
Reduce
Iff to its lowest terms.
Ans. ^.
4.
Reduce
j^ to its lowest terms.
Ans, If
6.
Reduce
f^ to its lowest terms.
6. Reduce ^iS% ^^ its lowest terms.
7. Reduce -j^^ to its lowest terms.
8. Reduce -^-^ to its lowest terms.
9. Reduce -f^ff to its lowest terms. Ans. JJ.
10. Reduce Iffl to its lowest terms. Ans. fg- .
11. Reduce ^^% to its lowest terms. Ans. H^.
12. Express in its simplest form the quotient of 441 di*
Tided by 462. Ans. f}.
13. Express in its simplest form the quotient of 189 di-
vided by 273. Ans. A-
14. Express in its simplest form the quotient of 1344 di-
vided by 1536. Ans. }.
Case IL
127. To reduce an improper fraction to a whole
CI mixed number.
1. Reduce ^^ to a whole or mixed number.
OPERATION. Analysts. Since
5JLt = 324 -^ 15 = 2lT«r = 211, Ans. 15 fifteenths equal
1, 324 fifteenths are
equal to as many times 1 as 15 is contained limes in 824, which is
Slf'^y times. Or, since the numerator is a dividend and the denom*
Bule. Case II is what? Give explanation.
REDUCTION". 91
inator a divisor ( 118 ), we reduce the fraction to an equivalent whole
or mixed number, by dividing the numerator, 324, by the denom-
inator, 15.
KuLE. Divide the numerator hy the denominator,
1. When the denominator is an exact divisor of the numerator, the result wiD be
A whole nu.nber.
2. In all answers containing fractions reduce the ftactions to their lowest tenna
Examples foe Peactice.
2. In J^- of a week, how many weeks ? Ans. 1^.
3. In 4^ of ^ bushel, how many bushels ? Ans, 23f .
4. In ^^ of a dollar, how many dollars?
5. In ^f- of a pound, how many pounds ? Ans, 54^.
6. Reduce J-|4^ to a mixed number.
7. Reduce ^f- to a whole number.
8. Change ^|p to a mixed number. Ans. 18f .
9. Change ^^ to a mixed number.
10. Change -^W^ to a mixed number. Ans. 1053f|.
11. Change "^l\l^^ to a whole number. Ans. 7032.
Case III.
128. To reduce a whole mimber to a fraction
having a given denominator.
1. Reduce 46 yards to fourths.
OPERATION. Akalysis- Since in 1 yard there are 4 fourths,
46 in 46 yards there are 46 times 4 fourths, which are
^ 184 fourths = ^^. In practice we multiply 46,
the number of yards, by 4, the given denominator,
•1^, Ans, and taking the product, 184, for the numerator of
a fraction, and the given denominator, 4, for the
denominator, we have ^f^.
Rule. Multiply the tvhole numher ly the given denomina-
tor ; take the product for a numerator, under which write
the given denominator.
Rule. Case III is what ? Give explanation. Rule.
92 FRACTIONS.
A whole number is reduced to a fractional form by writing 1 under it for a de-
nominator ; thus, 9 = ^.
Examples for Practice.
2. Reduce 25 bushels to eighths of a bushel. Ans. -8-g-fi-.
3. Eeduce 63 gallons to fourths of a gallon. A?is, ^^.
4. Reduce 140 pounds to sixteenths of a pound.
5. In 56 dollars, how many tenths of a dollar ? Ans, ^^,
6. Reduce 94 to a fraction whose denominator is 9.
7. Reduce 180 to seventy-fifths.
8. Change 42 to the form of a fraction. Ans. ^,
9. Change 247 to the form of a fraction.
10. Change 347 to a fraction whose denominator shall
be 14. A71S. -^-fl-S-.
Case IY.
129. To reduce a mixed number to an improper
fraction.
1. In 5| dollars, how many eighths of a dollar ?
OPEEATION.
54 Analysis. Since in 1 dollar there are 8 eighths,
Q in 5 dollars there are 5 times 8 eighths, or 40
— eighths, and 40 eighths + 3 eighths = 43 eighths,
4g3., A7IS, or ^K
Rule. Multiply the whole number hy the denominator of
the fraction; to the product add the numerator^ and under
the sum write the denominator.
Examples for Practice.
2. In 4 J dollars, how many half dollars ? Ans. |.
8. In 71f weeks, how many sevenths of a week ?
4. In 341 j- acres, how many fourths ? Ans. -LSj^-^.
5. Change 1^^ years to twelfths.
6. Change 56^^ to an improper fraction. Ans. -V^.
7. Reduce 21^ to an improper fraction. Ans. -^-fF--
8. Reduce 225^ to an improper fraction. A71S. ^l\^ *
* "
Case IV is what ? Give explanation. Rule.
REDUCTION-. 93
9. In 96^<V, how many one hundred twentieths ?
10. In 1297^, how many eighty-fourths ? Ans. ^ ^H^ K
11. What improper fraction will express 400-|4 ?
Case V.
130. To reduce a fraction to a given denominator.
As fractions may be reduced to lower terms by division,
they may also be reduced to higher terms by multiplication ;
and all higher terms must be multiples of the lowest terms.
(103.)
1. Eeduce } to a fraction whose denominator is 20.
OPERATION. Analysis. First divide 20, the required
20 -f- 4 = 5 denominator, by 4, the denominator of the
o ^ K given fraction, to ascertain if it be a mul-
— ^^i^j -4w5. tiple of this term, 4. The division shows
4x5 that it is a multiple, and that 5 is the factor
which must be employed to produce this
multiple of 4. We therefore multiply both terms of f by 5, ( 124,)
and obtain ^f , the desired result.
Rule. Divide the required denominator hy the denomina-
tor of the given fraction, and multiply both terms of the
fraction by the quotient.
Examples for Practice.
2. Reduce f to a fraction whose denominator is 15.
Ans. ^j,
5. Reduce -^ to a fraction whose denominator is 35.
4. Reduce -j^ to a fraction whose denominator is 51.
A71S. If.
6. Reduce %^ to a fraction whose denominator is 150.
6. Reduce J-|f to a fraction whose denominator is 3488.
Ans,im-
7. Reduce -^ to a fraction whose denominator is 1000.
Case V is what? How are fractions reduced to higher terms?
What are all higher terms ? Give analysis. Rule.
94 FEACTI0N8.
Case VL
131. To reduce two or more fractions to a com-
mon denominator.
A Common Denominator is a denominator common
to two or more fractions.
1. Reduce f and | to a common denominator.
OPERATION. Analysis. Multiply the terms of the first
3x5 fraction by the denominator of the second, and
~r K ^^ 2T the terms of the second fraction by the denom-
^ ^ ^ inator of the first, (124.) This must reduce
.2x4 esich. fraction to the same denominator, for
^ zo each new denominator will be the product of
^ '^ * the given denominators.
Rule. Multiply the terms of each fraction by the denom-
inators of all the other fractions.
Mixed numbers must first be reduced to Improper fractions.
Examples for Practice.
2. Reduce |, i, and | to a common denominator. •
^ns, if, ih a
3. Reduce ^ and f to a common denominator.
Ans, f i, ff,
4 Reduce ^, ^, and f to a common denominator.
M.ns. |g5, skof ■fbO'
5. Reduce -|, |, f , and |- to a common denominator.
A71S. -J^, 8T8> 836"^ SSff
6. Reduce ^, ^, and f to a common denominator.
^»«- m, m- A
7. Reduce |^, 2 J, J, and -J^ to a common denominator.
Ans. m, m, Hh ^
8. Reduce IJ, -f^, and 4 to a common denominator.
An s. Jtf , U, W
Case VI is what ? What is a common denominator ? Give analysis.
Rule.
REDUCTIOI?-. 96
Case VII.
132. To reduce fractions to the least common de-
nominator.
The Least Common Denomiinator of two or more
fractions is the least denominator to which they can all be
reduced, and it must be the least common multiple of the
lowest denominators.
1. Eeduce ^, |, and -^ to the least common denominator.
OPERATION. Analysis. First find the
2 3 6 . . 8 . . 12 least common multiple of the
2 2 4 2 given denominators, which.
is 24. This must be the
least common denominator to
which the fractions can be
Ans, reduced, (III.) Then multi-
ply the terms of each fraction
by such a number as will re-
duce the fraction to the denominator, 24. Reducing each fraction to
this denominator, by Case V, we have the answer.
Since the common denominator is already determined, it
is only necessary to multiply the numerators by the multi-
pliers.
Rule. I. Find the least common multiple of the given de-
nominators, for the least common denominator.
II. Divide this common denominator hy each of the given
denominators, and multiply each numerator hy the corres-
.ponding quotient. The products loill be the new numerators.
Examples fok Peactice.
2. Reduce ^, ^, fj, and t^ to their least common de-
nominator. Ans. -^, -^, m, j^.
3. Reduce ^, 4? A> A ^^ their least common denomina-
tor. Ans. m m^ *. m-
What is Case VII ? What must be the least common denominator 1
Give analysis. Rule, first step. Second.
96 FRACTIONS.
4. Beduce f , ^, J, and 6 to their least common denomi-
nator. Arts, ^i^, ^, ill, 1^,
5. Reduce 5^, 2J, and If to their least common denomi-
nator. Arts, ^y ^, ^.
6. Reduce ^, |, 4^, and J- to their least common denomi-
nator. Ans. lii, ffi, III, Hf.
7. Reduce f , ^, f, 2|-, and ^ to their least common de-
nominator. Ans. m, -^y -^, m, ^.
8. Change f , 3^, 3f , 9, and -J to equivalent fractions hav-
ing the least common denominator.
9. Change fi, 1^, -J, -J-J, and 6 to equivalent fractions
having the least common denominator.
10. Change 2yV» |J, 4, If, ^, and | to equivalent frac-
tions having the least common denominator.
11. Reduce f , |, |-, and 3^ to a common denominator.
12. Reduce -J, \, 2f , and | to a common denominator.
13. Reduce {^, -^y f, and 3-^ to equivalent fractions hav-
ing a common denominator. A71S. f|-, -§-}-, H, |-J.
14. Change ^, |, and f to equivalent fractions having a
common denominator. Ans. -f^^, WW> iWlr-
15. Change -j^:? 7-|-, f^, and 5 to equivalent fractions hav-
ing a common denominator. Ans. ff, ^^, ff , -W.
16. Change ^V» ^i, A, 7, 1, and IJ to equivalent fractions
having a common denominator.
ADDITION.
133. 1. What is the sum of J, |, |, and |?
OPEBATION. Analysis. Since the
1 4. I 4_ I 4. I z= Jgt = 2, ^W5. S^^^"" fraxjtions have a
common denominator, 8,
their sum may be found by adding their numerators, 1,3, 5, and 7,
and placing the sum, 16, over the common denominator. We thus
obtain ^^^=2, the required sum.
2. Add ^, A, tV. a. and ^. Ans. 2^.
3. Add ^, rfj, j\, ^, ^y, and {i. Ans. 2 A-
Give first explanation.
ADDITION". 97
4. What is the sum of ^, A, ^, H. if. and H?
6. What is the sum of -^, yV_^ ^^ ^^ and |ff ?
6. What is the sum of ^, ■^, if^, Hi, and f|| ?
134. 1. What is the sum of | and | ?
OPEKATiON. Analysis. In
|+| = || + J^= -^^iffl = H, ^WS. whole numbers
we can add like
numbers only, or those having the same unit value ; so in fractions
we can add the numerators when they have a common denominator,
but not otherwise. As f and f have not a common denominator, first
reduce them to a common denominator, and then add the numerators,
27 + 10=37, the same as whole numbers, and place the sum over the
common denominator.
Rule. I. When necessary, reduce the fractions to their
least common denominator.
II. Add the numerators, and place the sum over the com-
mon denominator.
If the amount be an improper fraction, reduce it to a whole or a mixed number.
Examples for Practice.
2. Add J to f . Ans, ||.
3. Add i- to ii. Ans. 1^,
4. Add }, \, f , and ^. Ans. l-^.
5. Add H, fJ. and ^. Ans. 1^^,
6. Add ^0. A, A. and ^. Ans. f.
7. Add ^, m, U, h and |. Ans. 3^.
8. Add }, h h h h h h h and ^, Ans. 7^^.
9. Add 7i, 5|, and lOf.
OPERATION. Akalysis, The sum of the frac-
•J^ 4- I + } = 1^ tions ^, f , and f is IH ; tlie sum of
7 -i_ 5 _|_ 10 ^ 22 *^® integers, 7, 5, and 10, is 22 ;
and the sum of both fractions and
Ans. 23\i integers is 23^^. Hence,
Give second explanation ? Rule, first step. " Second.
R.P. 5
98 FEACTIONS.
To add mixed numbers, add the fractions and integers
separately, and tlien add their sums.
If the mixed numbers are email, they may be reduced to improper fractions, and
then added after the usual method.
10. What is the sum of 14|, 3^^, If, and IJ ? Ans. 21^.
11. What is the sum of -J, 1^^, 10|, and 5 ? Ans. IS^V
12. What is the sum of 17i, 18^, and 26^?
13. What is the sum of A^ tt^ ^i 3, and iJ ?
14. What is the sum of 125^, 327^, and 25i ? Ans. 478^.
15. What is the sum of ^H, |J, 1^, «, and «|?
Ans. 3ttJ.
16. What is the sum of 3^^, 2|-|, 40|, and 10^ ?
17. Bought 3 pieces of cloth containing 125-J, 96 f, and
48f yards ; how many yards in the 3 pieces ?
18. If it take 5^- yards of cloth for a coat, 3^ yards for a
pair of pantaloons, and J of a yard for a vest, how many
yards will it take for all ? Ans. ^r}^.
19. A farmer divides his farm into 5 fields ; the first con-
tains 26y^ acres, the second 40^ acres, the third 514 acres,
the fourth 59f acres, and the fifth 62|- acres ; how many
acres in the farm ? Ans. 241|^.
20. A speculator bought 175| bushels of wheat for 205 J
dollars, 325^ bushels of barley for 296} dollars, 270^1 bush-
els of corn for 200^ dollars, and 43 7-^^ bushels of oats for
156f|- dollars ; how many bushels of grain did he buy, and
how much did he pay for the whole ? j ^« j 1 2^^ A bushels.
• ( 859}S dollai's.
SUBTKACTION.
135. 1. From ^ take A-
OPERATION. Analysis. Since the given
^1^ — ^ = -j*jj = ^, Ans. fractions have a common denom-
inator, 10, we find the difference
by subtracting 3, the less numerator, from 7, the greater, and write
How are mixed numbers added % Give note.
SUBTRACTION". 99
the remainder, 4, over the common denominator, 10. We thus obtain
y% = f , the required difference.
2. From | take |.
^^5. ^.
3. From ^ take ^.
^/is. 1^.
4. From fO- take ^.
Ans. \\.
5. From f| take ff .
Ans. \^.
6. From -^ take ^i^.
Ans, J.
7. From Ml take iif
Ans. ■^.
136. 1. From f take f.
OPERATION.
Analysis.
4 4 — 42 P- — 32-30 -
- -A- -Xr Atl^
As in whole
¥ t T6 36 3 6
-ir^ — T«> -a./*6.
numbers, we
can subtract like numbers only, or those having the same unit value,
so, we cau subtract fractions only when they have a common denom-
inator. As I and f have not a common denominator, we first reduce
them to a common denominator, and then subtract the less numerator,
30, from the greater, 32, and write the difference, 2, over the common
denominator, 36.
EuLE. I. Wlien necessary y reduce the fractions to a com-
mon denominator.
11. Subtract the numerator of the subtrahend from the
numerator of the minuend, and place the difference over the
co7nmon denominator.
Examples for Peactice.
2. From \ take |. Ans. -fy.
3. From ^ take I . Ans. j^.
4. Subtract -^ from f . Ans. ■^.
5. Subtract -^ from -^. Ans. ^.
6. Subtract f| from ^^. Ans. -^.
7. Subtract ^^^ from f^. Ans. ^^.
8. What is the diSereuce between 9-J and 2} ?
Give explanations. Rule, first step. Second.
100 FEACTIONS.
OPERATION. Analysis. First reduce the fractional
91 z= 9 A- parts, -|- and f , to a common denominator,
03 __ 2_^ ^^' ^^^^® ^® cannot take -f^ from j\, we
add 1 = If to j\, which makes f|, and
6^, Ans. 3»^ from if leaves y'^. We now add 1 to
the 2 in the subtrahend, (50,) and say, 3
from 9 leaves 6. We thus obtain 6xV, the difference required.
Hence, to subtract mixed numbers, we may reduce the
fractional parts to a common denominator, and then subtract
the fractional and integral parts separately. Or,
We may reduce the mixed numhers to improper fractions^
and subtract the less from the greater by the usual method,
9. From 8 J take S-J. Ans. 4ff.
10. From 25|- take 9^. Ans, 16^^.
11. From ^ take \^,
12. Subtract 1^ from 6.
13. Subtract 120^ from 450^. Ans, 330|f.
14. Subtract ^ from 3^V Ans, 3^.
15. Find the difference between 49 and 75 J-.
16. Find the difference between 327| and 196|.
17. From a cask of wine containing 31^ gallons, 17|- gal-
lons were drawn ; how many gallons remained ? Ans. 13 J.
18. A farmer, having 450^ acres of land, sold 304|
acres ; how many acres had he left? Ans. 145^^.
1 9. If flour be bought for 6J- dollars per barrel, and sold
for 7| dollars, what will be the gain per barrel ?
20. From the sum of 4 and 3^ take the difference of 4-J
and 5i. Ans. 3|f .
21. A man, having 25| dollars, paid Q\ dollars for coal,
2^ dollars for dry goods, and J of a dollar for a pound of
tea ; how much had he left ? Ans. Iiejj.
22. What number added to 2|will make 7J? AnsA\^.
23. What fraction added to {^ will make ^ ? Ans. ^.
In how many ways may mixed numbers be subtracted ? Wliat arc
they?
MULTIPLICATION. 101
24. A gentleman, having 2000 dollars to divide among
his three sons, gave to the first 912J dollars, to the second
545^ dollars, and to the third the remainder ; what did the
third receive ? Ans. $542^^.
25. Bought a quantity of coal for 136^ dollars, and of
lumber for 350f dollars. I sold the coal for 184^ dollars,
and the lumber for 41 6f dollars. What was my whole gain ?
Ans. $114^.
MULTIPLICATION.
Case I.
137. To miiltiply a fraction by an integer.
1. If 1 yard of cloth cost f of a dollar, how much will 5
yards cost ?
OPERATION. Analysis. Since 1 yard cost
I X 5 = ^- =- ^i, Ans. 3 fourths of a dollar, 5 yards
will cost 5 times 3 fourtJis of a
dollar, or 15 fourths, equal to 3f dollars. A fraction is multiplied by
multiplying its numerator, (124.)
2. If 1 gallon of molasses cost ^ of a dollar, how much
will 5 gallons cost ?
OPERATION. Analysis. Since 5, the muU
■^ X 5 =: J = If , ^725. tiplier, is a factor of 20, the de-
nominator, of the multiplicand,
we perform the multiplication by dividing the denominator, 20, by the
multiplier, 5, and we have |, equal to 1^ dollars. A fraction is multi-
plied by dividing its denominator, (124.) Hence,
MuUi^jlying a fraction consists in multiplying its nu-
merator, or dividing its denominator.
Always divide the denominator when it is exactly divisible by the multiplier.
Examples for Practice.
3. Multiply f by 5. , Ans.^ = ^.
4. Multiply yV by 7. Ans. l|i.
Case I is what ? Give explanations. Deduction.
102
FRACTIONS.
5. Multiply T^ by 12.
6. Multiply ^ by 63.
7. Multiply 5} by 9.
Ans. 74.
Ans, 15.
OPEKATION.
9 Or,
45 ^x9=^=49i.
Analysis. In multiply,
ing a mixed number, first mul-
tiply the fractional part, and
then the integer, and add the
two products; or, reduce the
mixed number to an improper
fraction, and then multiply it.
8. Multiply 7| by 12.
9. Multiply ^ by 8.
10. Multiply yl^ by 51.
11. Multiply 15| by 16.
12. Multiply m by 22.
Ans. 91|.
Ans. 6-^.
Ans. 2.
Afis. 250.
Ans. 16|-.
13. If a man earn 8^<j dollars a week, how many dollars
will he earn in 12 weeks ?
14. What will 9 yards of silk cost at ^ of a dollar per
yard?
15. What will 27 bushels of barley cost at | of a dollar
per bushel ? Ans. 23| dollars.
Case II.
138. To multiply an integer by a fraction.
1. At 75 dollars an acre, how much will | of an acre of
land cost ?
PIKST OPERATION.
5 ) 75 price of an acre.
1 5 cost of ^ of an acre.
3
Ans, 45 cost of f of an acre.
Analysis. 8 fifths of an acre
will cost three times as much as
1 fifth of an acre. Dividing 75
dollars by 5, we have 15 dollars,
the cost of I of an acre, which
we multiply by 3, and obtain 45
dollars, the cost of f of an acre.
Explain the process of multiplying mixed numbers. What is Case
II ^ Give first explanation.
MULTIPLICATION.
103
6BC0KD OPERATION.
75 price of 1 acre.
3
5 ) 225 cost of 3 acres.
Ans. 45 " " 1^ of an acre.
Or, multiplying tlie price
of 1 acre by 3, we have the
cost of 3 acres ; and as ^ of
3 acres is the same as | of
1 acre, we divide the cost
of 3 acres by 5, and we have
the cost of f of an acre, the
same as in the first opera-
tion. Hence,
Multiplying by a fraction co7isists in multiplying ly the
numerator and dividing ly the denominator of the multiplier,
15
J^^ By using the vertical line and cancellation, we
3
$
Bhall shorten, and combine both operations in
one.
45, Ans,
Examples for Practice.
2. Multiply 3 by f .
3. Multiply 100 by A-
4. Multiply 105 by J|.
5. Multiply 19 by i|.
6. Multiply 24 by 6|,
OPEKA.TION.
24
15 = f of 24 •
^44
Or,
150, Ans,
53
159, Ans,
7. Multiply 42 by 9}.
8. Multiply 80 by 14^.
9. Multiply 156 by f|.
10. At 8 dollars a bushel, what will f of a bushel of clover
seed cost ?
Ans, IJ.
Ans. 64f.
Ans. 85.
Ans, 5if.
Analysis. Mul-
tiply by the integer
and fraction sepa-
rately, and add the
products ; or, reduce
the mixed number
to an improper frac-
tion, and then mul
tiply by it.
Ans, 409f
Ans, 1165.
Ans. 108.
Give second explanation. Note. Deduction.
104 FRACTIONS.
11. If a man travel 36 miles a day, how many miles will
he travel in lOf days : Ans. 384 miles.
12. If a village lot be worth 450 dollars, what is -^ of it
worth? Ans, 262^ dollars.
13. At 16 dollars a ton, what is the cost of 2} tons of hay ?
Case III.
139. To multiply a fraction by a fraction,
1. At f of a dollar per bushel, how much will J of a bushe\
of com cost
9
OPEEATiON. Analysis.
1st step, 1 -^ 4 = y^, cost of } of a busheL Since 1 bush.
SdBtep, yV X 3 = ^% '' '' 1 '' '' " ^1 ^««* t of a
Whole work, 1 X f = A = i, AnS. ^'^^l\ ^ ^^.^^
^ * *•* * bushel will
Or, $
% cost f times f of a dollar, or 3 times
h
^ ^ of f of a dollar. Dividing | of a
^ doll AT by Ai WP! Viavft -jT^^ fhft rnst. nf X
2
1 = J, ^7^5. of a bushel. A fraction is divided by
multiplying its denominator, (124.)
Multiplying the cost of \ of a bushel by 3, we have -^^ of a dollar, the
cost of I of a bushel. It will readily be seen that we have multiplied
together the two numerators, 2 and 3, for a new numerator, and the
two denominators, 3 and 4, for a new denominator, as shown in the
whole work of the operation. Hence, for multiplication of fractions,
we have this general
EuLE. I. Reduce all integers and mixed numlers to im*
proper fractions,
II. Multiply together the numerators for a neio numerator
and the denominators for a new denominator.
Cancel all foctors common to numerators and denominators.
2. Multiply I by f Ans. J.
a Multiply J by |. Ans. ^,
4. Multiply tt by -11. Ans, ■^.
5. Multiply 4 by \. Ans. 3|.
What is Case III ? Give explanation. Rule, first step ? Second !
What shall be done with common factors ?
MULTIPLICATION?^.
105
»^
$
*<r0
7
6
$
$
t
$
t
t
4:
$
$
30
7 = ^.
6. What is the product of ^, f, |, and i ? ^?Z5. ■^.
7. What is the product of 1|, f, 2, and 5-}? J.?is. 11 J^.
8. What is the product of f of ^, | of | of |, and \ of
If?
OPERATION. Or,
— X — X — X — X — X — X— =— , ^ns,
4 10^ Q $ $ t 30
Fractions with the word of between them are sometimes
called compound fractions. The word of is simply an equiv-
alent for the "sign of multiplication, x, and signifies that
th3 numbers between which it is placed are to be multiplied
together.
9. Multiply ^ of 2i by \ of 7^. Ans. lf|.
10. Multiply f of 16 by ^ of 26f. Ans, 85^.
11. What is the product of 3, J of -f-, and \ of 3J ?
12. What is the value of 2 J times f of ^ of 1^ ? Ans, 2.
13. What is the value of J of ^ of If times | of 8 ?
14. What is the product of 12 J multiplied by 5^ times 6f ?
Ans, 464^.
15. At f of a dollar per yard, what will | of a yard of
cloth cost ? Ans. ^oi2i, dollar.
16. If a man own f of a vessel, and sell f of his share,
what part of the whole vessel will he sell ?
17. When oats are worth |- of a dollar per bushel, what is
} of a bushel worth ?
18. What will 7} pounds of tea cost, at | of a dollar per
pound ? Ans. 4J J doUars.
19. What is the product of 9^ by 4| ?
39f product by 4.
Or, 9f x4| = — X— = 46.
Ans. 46
"4|.
What doea ** of" signify when placed between two fractions ? What
a compound fraction t
6*
106 FRACTIONS.
To multiply mixed numbers together, either multiply by
the integer and fractional part separately, and then add
their products ; or, reduce both numbers to improper frac-
tions, and then multiply as in the foregoing rule.
20. Multiply 12f by 8|. A7is. 108|.
21. What cost 6|- cords of wood, at 2f dollars a cord ?
22. What cost f of 2J tons of hay, at 11^ dollars a ton !
Ans. $21^^.
23. What will 8| cords of wood cost, at 2f dollars per
cord? Ans. 22 JJ dollars.
24. What must be paid for | of 6|^ tons of coal, at | of 7}
dollars per ton ?
25. A man owning J of a farm, sold ^ of his share ; what
part of the whole farm had he left ? Ans. ^.
26. Bought a horse for 125} dollars, and sold him for ^ of
what he cost ; what was the loss ? Ans. $25^.
27. A owned f of 123|- acres of land, and sold | of his
share ; how many acres did he sell. Ans. 49^.
28. If a family consume IJ- barrels of flour a month, how
many barrels will five such famiHes consume in 4^ months?
DIVISION.
Case I.
140. To divide a firaction by an integer.
1. If my horse eat ^ of a ton of hay in 3 months, what
part of a ton will last him 1 month ?
OPERATION. Analysis. If he eat i% ot a ton in
_t^ _i_ 3 = rAr, Ans. S months, in 1 month he will eat ^ of
^ of a ton, or -^^ divided by 3. Since
a fraction is divided, by dividing its numerator, (124,) divide the
numerator of the fraction, ^jj, by 8, and have y^^.
2. If 3 yards of ribbon cost -J- of a dollar, what will 1 yard
cost ?
Case I is what ? Give first explanation.
DIVISION. lOT
OPERATION. Analysis. Here we cannot exactly
t -^ 3 =: Jg^ Ans. divide tlie numerator by 3 ; but, since a
fraction is divided by multiplying the
denominator, (124,) we multiply the denominator of the fraction, f ,
by 3, and have ^q, the required result. Hence,
Dividing a fraction consists in dividing its numerator, or
multiplying its denominator.
We divide the numerator when it is exactly divisible by the divisor ; otherwise
w3 multiply the denominator.
Examples foe Pkactice.
3. Divide f by 2. Ans. f
4. Divide ^^ by 3. Ans, \,
5. Divide f| by 5. Ans. ^.
6. Divide ^ by 25.
7. Divide ^ by 14. A?is. ^.
8. Divide f^ by 21. A7is. ^.
9. If 6 pounds of sugar cost | of a dollar, what will 1
pound cost ?
10. At 7 dollars a barrel, what part of a barrel of flour
can be bought for -J of a dollar ? Ans. ^.
11. If a yard of cloth cost 5 dollars, what part of a yard
can be bought for -f of a dollar? Ans. -^.
12. If 9 bushels of barley cost 7-^ dollars, what will 1
bushel cost ?
OPERATION.
-, „ g We reduce the mixed number to an improper
'6 ^~~ & fraction, and divide as before.
^^9=i,Ans.
13. If 12 barrels of flour cost 76|- dollars, what will 1
barrel cost ?
OPERATION. Analysis. We first divide as in simple
12 ) 764 numbers, and have a remainder of 4|.
77 . Reduce this remainder to an improper
6f, Ans. fraction, -«/, which divide (as in Ex. 1),
and annex the result, f , to the partial quotient, 6, and we have 6|, the
required result.
Give second explanation. Deduction.
108 FEACTIONS.
14. How many times will 16 J gallons of cider fill a vessel
that holds 3 gallons ? Ans. 5^.
15. If 9 men consume f of 9| pounds of meat in a day,
how much does each man consume ? Ans. ^ ot s. pound.
16. A man paid I99|| for 4 cows ; how much was that
apiece ? Ans, $24f|.
Case IL
141. To divide an integer by a fraction.
1. At f of a dollar a yard, how many yards of cloth can
be bought for 12 dollars ?
FIRST OPERATiOsr. ANALYSIS. As many yards as | of 8
12 dollar, tlie price of 1 yard, is contained
^ times in 13 dollars. Integers cannot be
divided by fourt/is, because they are not
*^ / ^" of the same denomination. Reducing 12
16 yards. dollars to fourths by multiplying, we have
48 fourths; and 3 fourths is contained in
^fourths 16 times, the required number of yards.
SECOND OPERATION. ANALYSIS. We divide the integer by
3 ) 12 the numerator of the fraction, and multi-
T ply the quotient by the denominator, vv^hich
produces the same result as in the first
^ operation. Hence,
16 yards.
Dividing ly a fraction consists in multiplying by the de^
nominator y and dividing hy the numerator of the divisor.
Examples for Practice.
2. Divide 18 by |. Ans, 48.
3. Divide 63 by -^^ Ans, 117.
4. Divide 42 by ^. Ans, 49.
5. Divide 120 by ■^, Ans. 205f
6. Divide 316 by ^g. Ans. 877f
Case II is what ? Give first explanation. Second. Deduction.
DIVISION. 109
7. How many bushels of oats, worth f of a dollar per
bushel, will pay for f of a barrel of flour, worth 9 dollars a
barrel ? Ans. 15.
8. If f of an acre of land sell for 21 dollars, what will an
acre sell for at the same rate ? Atis. $49.
9. When potatoes are worth |- of a dollar a bushel, and
corn I of a dollar a bushel, how many bushels of potatoes
are equal in value to 16 bushels of corn? Ans. 22-J-.
10. If a man can chop 2f cords of wood in a day, in how
many days can he chop 22 cords ?
OPERATIOK.
3f = ¥
22 Analysis. "We reduce the mixed number
. to an improper fraction, and then divide the
integer in the same manner as by a proper
11 ) 88 fraction.
Ans. 8 days
11. Divide 75 by 13f Ans. 5f| .
12. Divide 149 by 2^. Ans. 6^.
13. A farmer distributed 15 bushels of corn among some
poor persons, giving them If bushels apiece ; among how
many persons did he divide it ?
14. Divide | of 320 by | of 9f Ans. 25f .
15. Bought I of 7^ cords of wood for J of 132 ; how much
did 1 cord cost? Ans. $3-J-.
16. A father divided 183 acres of land equally among his
sons, giving them 45f acres apiece ; how many sons had he ?
Ans. 4.
Case III.
142. To divide a fraction by a fraction.
1. How many pounds of tea can be bought for -}-J of a
dollar, at I of a dollar a pound ?
How divide by a mixed number ? Case Til is what ?
110 FEACTIONS.
OPERATION. ^ Analysis. As
First step, ^x3=f-| many pounds as f
SecondBtep, ||_^2 = |}=:lf. of a dollar is con-
11 2 11 3 11 tained times in f|
Wholework, — -^-=— X^ = -^- = l|, Ans. of a dollar. 1 is
14 6 l^^ 4, b contained in ii, H
times, and i is con-
tained in fl 3 times as many times as 1, or 3 times }|, wliicli is f |
limes, wliicli is the number of pounds that could be bought at |^ of a
dollar per pound ; but f is contained but ^ as many times as J, and
fl divided b/ 3 gives ff, equal to if times, or the number of pounds
that can be bought at | of a dollar per pound.
We see in the operation that we have multiplied the dividend by
the denominator of the divisor, and divided the result by the numer-
ator or the divisor, which is in accordance with 140 for dividing a
fraction. Hence, by inverting the terms of the divisor, the two frac-
tions will stand in such relation to each other that we can multiply
together the two upper numbers for the numerator of the quotient,
and the two lower numbers for the denominator, as shown in the
operation. For division of fractions, we have this general
Rule. I. Reduce integers and mixed numhers to improper
fractions.
II. Invert the terms of tlie divisor, and proceed as in mul-
tiplication.
1. The dividend and divisor may be reduced to a common denominator, and the
numerator of the dividend be divided by the numerator of the divisor ; this will giv«
the same result as the rule.
2. Apply cancellatiou w^here practicable.
Examples foe Peactice.
2. Divide J by J.
3. Divide I by |.
4 Divide { by ■^.
5. Divide ^ by -^.
6. Divide f by fj.
7. How many times is -f- contained in f?
8. How many times is 4 contained in 1-| ?
llule, first step. Second. Wliat other method is mentioned
Ans.
If
Ans.
H-
Ans.
«•
Ans.
W-
Ans.
«•
Ans.
i,V
Ahs.
3t.
DIVISION. Ill
9. How many times is yV contained in fj ? A71S. 2f .
10. How many times is ^ contained in J-f ?
11. How many times is J of } contained in f of 2J?
12. What is the quotient of ^ of 4, divided by f of 3 J ?
13. What is the quotient of ^ of | of 36 divided by 1^
times I ? A71S. di.
14. What is the value of -^
?
OPERATION. , This example
3|__^_7_^35_^ $l_ . is only another
4l~~'^~2 * 8~^^$0~~^^ ^^* ^^^^ ^^^ ®^'
5 pressing divis-
ion of fractions ; it is sometimes called a complex fraction, and the
process of performing the division is called reducing a complex frac-
tion to a simple one.
We simply reduce the upper number or dividend to an improper
fraction, and the lower number, or divisor, to an improper fraction,
and then divide as before.
15. What is the value of || ? Ans. f f.
16. What is the value of ^? Ans. 20.
17. What is the value of g ? Ans. ^.
18. What is the value of ^^ ? Ans. 1.
19. What is the value of } f j; ? Ans. i.
I of 4J^
20. If a horse eat -| of a bushel of oats in a day, in how
many days will he eat 5 J bushels ? Ans. 14.
21. If a man spend If dollars per month for tobacco, in
what time will he spend lOf dollars ? Ans. 6|- months.
What is a complex fraction ?
112 FBACTIOKS.
22. How many times will 4f gallons of camphene fill a
vessel that holds J of f of 1 gallon ? Ans. 10|-.
23. If 14 acre? of meadow land produce 32f tons of hay,
how many tons will 5 acres produce? A^is. 11|.
24. If 2 yards of silk cost $3 J, how much less than $17
will 9 yards cost ? Ans. $2f .
25. If f of a yard of cloth cost 3^ of a dollar, what will
1 yard cost ?
26. A man, having $10, gave f of his money for clover
seed at 13^ a bushel ; how much did he buy ? Ans. 2 bush.
27. How many tons of hay can be purchased for IllQ^ig-,
at $9f per ton? Ans. 12^.
Promiscuous Examples.
1. Eeduce J, f , f , and J to equivalent fractions whose
denominators shall be 24. Ans. Jf , f|, ^, ■^.
2. Change 4 to an equivalent fraction having 91 for its
denominator. Ans. ff*
3. rind the least common denominator of f , If, ^ of f ,
2, i of J- of 1^.
4. Add 4i, 1, f of H, 3, and H.
5. Find the difference between f of 6^ and | of 4,8^.
Ans. liff.
6. The less of two numbers is 475 6|, and their difference
is 128|; what is the greater number? Ans. 4885^.
7. What is the difference between the continued products
of 3, J, I, 4|, and 3^, f , 4, f ? An^. 3^.
4 2i
8. Reduce the fractions - and -j to their simplest form.
9. What number multiplied by i will produce 1825 J ?
10. A farmer had ^ of his sheep in one pasture, J in
another, and the remainder, which were 77, in a third pas-
ture ; how many sheep had he ? Ans. 140.
11. What will 7i cords of wood cost at J of 9^ dollars
per cord? Ans. $24^.
PKOMISCUOUS EXAMPLES. 113
12. At I of a dollar per bushel, how many bushels of ap-
ples can be bought for 5-J dollars ?
13. Paid $183 7| for 7350 J bushels of oats ; how much was
that per bushel ? . Ans. i of a dollar.
14. If 235 J acres of land cost $4725|, what will 628 acres
cost ? Ans. 112601.
15. A man, owning | of an iron foundry, sold ^ of his
share for $540| ; what was the value of the foundry ?
Ans. $4055|.
16. 14f less ^-^^ is f of I of what number ? Ans. 27.
17. A merchant bought 4J cords of wood at $3J per cord,
and paid for it in cloth at -| of a dollar per yard ; how many
yards were required to pay for the wool ?
18. How many yards of cloth, f of a yard wide, will line
20i yards, IJ yards wide ? Ans. 34J-.
19. If the dividend be |, and the quotient ^, what is
the divisor?
20. If the sum of two fractions be |, and one of them be
^, what is the other ? A7is. :^.
21. If the smaller of two fractions be |f , and their differ-
ence -^j what is the greater? Ans. -Jf.
22. If 3f pounds of sugar cost 33 cents, what must be
paid for 65 J- pounds ?
23. If 324 bushels of barley can be had for 259^ bushels
of corn, how much barley can be had for 2000 bushels of
com ? Ans. 2500 bushels.
24. A certain sum of money is to be divided among 5 per-
sons ; A is to have ^J, B ^, 3^, D ^, and E the remain-
der, which is 20 dollars ; what is the whole sum to be di-
Wded? Ans. $50.
25. What number, diminished by the difference between
f and f of itself, leaves a remainder of 34 ? Ans. 40.
26. If I of a farm be valued at $1728, what is the value
of the whole ?
114 FEACTIOITS.
27. Bouglit 320 sheep at $2^- per head ; afterward bought
435 at |1| per head ; then sold | of the whole number at
$1| per head, and the remainder at $2|- ; did I gain or lose,
and how much ? Ans. Lost $44 J.
28. If 5 be added to both terms of the fraction -J, will its
value be increased or diminished ? Ans. Increased ySt-
29. If 5 be added to both terms of the fraction f, will its
value be increased or diminished ? Ans. Diminished -^.
30. How many times can a bottle holding J of | of a
gallon, be filled from a demijohn containing | of 1| gallons ?
Ans. 7i.
31. Bought i of 7i cords of wood for i of $32; what did
1 cord cost ?
•32. Purchased 728 pounds of candles at 16f cents a pound;
had they been purchased for 3-| cents less a pound, how many
pounds could have been purchased for the same money ?
Ans. 953i|.
33. What number, divided by 1-|, will give a quotient of
9i? Ans. 12ii.
34. The product of two numbers is 6, and one of them is
1846 ; what is the other ? Ans. -^.
35. A stone mason worked llf days, and after paying
his board and other expenses with f of his earnings, he had
$20 left ; how much did he receive a day ?
36. If I of 4 tons of coal cost $5|, what will | of 2 tons
cost? A71S. $5.
37. In an orchard f of the trees are apple trees, -^ peach
trees, and the remainder are pear trees, which are 20 morp
than 4 of the whole ; how many trees in the orchard ?
Ans. 800.
38. A man gave 6-| pounds of butter, at 12 cents a pound,
for f of a gallon of oil ; what was the oil worth a gallon ?
Ans. 100 cents.
39. A: gentleman, having 2711- acres of land, sold ^ of it,
and gave | of it to his son ; what was the value of the re*
mainder, at $57^ per acre ? A?is. $4577A-
PEOMISCUOUS EXAMPLES. 115
40. A horse and wagon cost 1270 ; the horse cost IJ times
as much as the wagon ; what was the cost of the wagon ?
41. What number taken from 2 J- times 12| will leave
20f? J-ns. 11 J.
42. A merchant bought a cargo of flour for $2173|^, and
sold it for If of the cost, thereby losing | of a dollar per
barrel ; how many barrels did he purchase ? Ans. 126.
43. A and B can do a piece of work in 14 days ; A can do
J as much as B ; in how many days can each do it ?
Ans, A, 32f days ; B, 24^ days.
44. How many yards of cloth f of a yard wide, are equal
to 12 yards | of a yard wide ? Ans. llj.
45. A, B, and can do a piece of work in 5 days ; B and
C can do it in 8 days ; in what time can A do it ?
46. A man put his money into 4 packages ; in the first ho
put f , in the second ^, in the third -J-, and in the fourth the
remainder, which was $24 more than -^ of the whole ; how
much money had he ? Ans. 1720.
47. If $7} will buy 3 J cords of wood, how many cords can
be bought for 110^ ? Ans. 4f|.
48. How many times is ^ of f of 27 contained in -J of J of
42|?
49. A boy lost J- of his kite string, and then added 30
feet, when it was just ^ of its original length ; what was
the length at first ? Ans. 100 feet.
50. Bought -f of a box of candles, and having used -J of
them, sold the remainder for -J^ of a dollar ; how much
would a box cost at the same rate ? Ans. $5-^.
51. A post stands ^ in the mud, J in the water, and 21
feet above the water ; what is its length ?
52. A father left his eldest son f of his estate, his youngest
son ^ of the remainder, and his daughter the remainder,
who received |1723| less than the youngest son ; what was
the value of the estate? Ans. $21114J|.
116 DECIMALS.
DECIMAL FEACTIONS.
143. Decimal Fractions are fractions which have for
their denominator 10, 100, 1000, or 1 with any number of
ciphers annexed.
1. The word decimal is derived from the Latin decern^ which signifies ten,
2. Decimal fpactions are commonly called decimals.
3. Since -^ =. ^y^, y^ = yuwu* ®*'^-» ^^^ denominators of decimal fractions in-
crease and decrease in a tenfold ratio, the same as simple numbers.
Decimal Notatiok akd Numebation.
144, Common Fractions are the common divisions of
a unit into any number of equal parts, as into halves y fifths,
twenty-fourths, etc.
Decimal Fractions are the decimal divisions of a unit,
thus : A unit is divided into ten equal parts, called tenths ;
each of these tenths is divided into ten other equal parts,
called hundredths ; each of these hundredths into ten other
equal parts, called thousandths ; and so on. Since the de-
nominators of decimal fractions increase and decrease by the
scale of 10, the same as simple numbers, in writing decimals
the denominators may be omitted.
In simple numbers, the unit, 1, is the starting point of
notation and numeration ; and so also is it in decimals. We
extend the scale of notation to the left of units' place in
writing integers, and to the right of units' place in writing
decimals. Thus, the first place at the left of units is tens,
and the first place at the right of units is tenths ; the second
place at the left is hundreds, and the second place at the
right is hundredths ; the third place at the left is thousands,
ftnd the third place at the right is thousandths ; and so on.
What are decimal fractions? How do they diflFer from common
fractions ? How are they written ?
NOTATION AND NUMERATION. 117
The Decimal Point is a period ( . ), wliicli must always be
placed before or at the left hand of the decimal. Thus,
-^ is expressed .6
The decimal point Is also called the Separatrix. This is a correct name for it
only when it stands between the integral and decimal parts of the same numbei*.
.5 is 5 tenths, which = -jig- of 5 units ;
.05 is 5 hundredths, '' z=z^^ofb tenths ;
.005 is 5 thousandths, '^ z=^ofh hundredths.
And universally, the value of a figure in any decimal place
is -^ the value of the same figure in the next left hand place.
The relation of decimals and integers to each other is
clearly shown by the following
Numeration Table.
t
s
1
.
CD
1
i
i
.2
1
<K
O
1
4
1
rn
S
1
■d
usandths
-thousan
dred-tho
1
1
1
'I
i
1
^
1
1
w
1
1
S e^ w
1
9
8
7
6
5
4
3
3
1
.2
3
4 5 6
7
8
9
Integers.
Decimals.
By examining this table we see that
Tenths are expressed by one figure.
Hundredths '' '' '^ two figures.
Thousandths " '' " three ''
Ten thousandths '' " '' four "
And any order of decimals by one figure less than the cor-
responding order of integers.
145. Since the denominator of tenths is 10, of hun-
What is the decimal point ? What is it sometimes called ? Wha*
is the value of a figure in any decimal place ?
118 DECIMALS.
dredths 100, of thousands 1000, and so on, a decimal may be
expressed by writing the numerator only ; but in this case
the numerator or decimal must always contain as many
decimal places as are equal to the number of ciphers in the
denominator ; and the denominator of a decimal will always
be the unit, 1, with as many ciphers annexed as are equal to
the number of figures in the decimal or numerator.
The decimal point must never be omitted.
Examples for Peactice.
1. Express in figures thirty-eight hundredths.
2. Write seven tenths.
3. Write three hundred twenty-five thousandths.
4. Write four hundredths. A?is, .04.
5. Write sixteen thousandths.
6. Write seventy-four hundred-thousandths. Ans. .00074.
7. Write seven hundred forty-five millionths.
8. Write four thousand two hundred thirty-two ten-thou-
sandths.
9. Write five hundred thousand millionths.
10. Eead the following decimals :
.05 .681 .9034 .19248
.24 .024 .0005 .001385
.672 .8471 .100248 .1000087
To read a decimal, first numerate from left to right, and the name of the right
hand figure is the name of the denominator. Then numerate from right to left, aa
in whole numbers, to read the numerator.
146, A Mixed Number is a number consisting of in-
tegers and decimals ; thus, 71.406 consists of the integral
part, 71, and the decimal part, .406 ; it is read the same aa
■^ItWV* '^1 and 406 thousandths.
Examples for Practice.
1. Write eighteen, and twenty-seven thousandths.
2. Write four hundred, and nineteen ten-millionth s.
How many decimal places must there be to express any decimal ?
NOTATIOJ?^ AND NUMERATION. 119
3. Write fifty-four, and fifty-four millionths.
4. Eighty-one, and 1 ten-thousandth.
5. One hundred, and 67 ten-thousandths.
6. Bead the following numbers :
18.027 100.0067 400.0000019
81.0001 64.000054 3.03
75.075 9.2806 40.40404
147. Trom the foregoing explanations and illustrations
we derive the following important
Peinciples of Decimal Notation and !N"umeration.
1. The value of any decimal figure depends upon its place
from the decimal point : thus .3 is ten times .03.
2. Prefixing a cipher to a decimal decreases its value the
same as dividing it by ten ; thus, .03 is ^ the value of .3.
3. Annexing a cipher to a decimal does not alter its value,
since it does not change the place of the significant figures
of the decimal ; thus, -^^ or .6, is the same as 3%, or .60.
4. Decimals increase from right to left, and decrease from
left to right, in a tenfold ratio ; and therefore they may be
added, subtracted, multiplied, and divided the same as whole
numbers.
5. The denominator of a decimal, though never expressed,
is always the unit, 1, with as many ciphers annexed as there
are figures in the decimal.
6. To read decimals requires two numerations ; first, from
units, to find the name of the denominator, and second, tO'
wards units, to find the value of the numerator.
148. Having analyzed all the principles upon which the
writing and reading of decimals depend, we will now pre-
sent these principles in the form of rules.
EuLE FOR Decimal Notation.
I. Write the decimal the same as a ivhole number, placing
What is the first principle of decimal notation ? Second ? Third 1
Fourth? Fifth? Sixth? Rule for notation, first step?
120 DECIMALS.
ciphers where necessary to give each significant figure its true
local value,
IL Place the decimal point before the first figure.
EuLE FOR Decimal Numeration.
I. Numerate frmn the decimal point, to determine the de-
nominator,
II. Numerate towards the decimal point, to determine the
numerator.
III. Bead the decimal as a whole number, giving it the
name or denomination of the right hand figure.
Examples for Practice.
1. Write 425 millionths.
2. Write six thousand ten-thousandths.
3. Write one thousand eight hundred fif tj-nine hundred-
thousandths.
4. Write 260 thousand 8 billionths.
5. Eead the following decimals :
.6321 .748243 .2902999
.5400027 .60000000 .00000006
6. Write five hundred two, and one thousand six mil-
lionths.
7. Write thirty-one, and two ten-millionths.
8. Write eleven thousand, and eleven hundred-thou-
Bandths
9. Write nine million, and nine billionths.
10. Write one hundred two tenths. Ans, 10.2.
11. Write one hundred twenty-four thousand three hun-
dred fifteen thousandths.
12. Write seven hundred thousandths.
13. Write seven hundred-thousandths.
14. Read the following numbers :
12.36 9.052 62.9999
142.847 32.004 1858.4583
1.02 4.0005 27.00045
Second ? Rule for numeration, first step ? Second ? Third ?
EEDUCTION. 121
EEDUCTION.
Case I.
149. To reduce decimals to a common denomina-
tor.
1. Eeduce .5, .375, 3.25401, and 46.13 to their least com-
mon decimal denominator.
OPEEATiON. Analysis. The given decimals must contain
,50000 as many places each, as are equal to the greatest
37500 number of decimal figures in any of the given
onKAoi decimals. We find that the third number con-
tains five decimal places, and hence 100000 must
%0,lo\j0\) jjg ^ common denominator. As annexing ciphers
to decimals does not alter their value, (14:4, 3)
we give to each number five decimal places by annexing ciphers, and
thus reduce the given decimals to a common denominator.
KuLE. Give to each number the same numler of decimal
places, ly annexing ciphers,
1. If the numbers be reduced to the denominator of that one of the given num-
bers having the greatest number of decimal places, they will have their least com-
mon decimal denominator.
2. A whole number may readily be reduced to decimals by placing the decimal
point after units, and annexing ciphers ; one cipher reducing it to tenths, two
ciphers to hundredths^ three ciphers to thousandths, and so on.
Examples fob Practice.
2. Eeduce .17, 24.6, .0003, 84, and 721.8000271 to their
least common denominator.
3. Eeduce 7 tenths, 24 thousandths, 187 millionths, 5
hundred millionths, and 10845 hundredths to their least
common denominator.
4. Eeduce to their least common denominator the following
decimals : 1000.001, 841.78, 2.6004, 90.000009, and 6000.
What is meant by the reduction of decimals? Case I is what?
Give explanation. Eule.
E. P. 6
Ans.
h
Am.
A-
Ans.
m-
Ans.
ih
122 decimals.
Case IL
150. To reduce a decimal to a common fraction.
1. Reduce .75 to ifcs equivalent common fraction.
Analysis. Omit the decimal point, supply
OPERATION. the proper denominator to the decimal, and
.75=:3^=|. then reduce the common fraction thus formed
to its lowest terms.
EuLE. Oinit the decimal point, and supply the proper de-
nominator.
Examples foe Peactice.
2. Eeduce .125 to a common fraction.
3. Reduce .16 to a common fraction.
4. Reduce .655 to a common fraction.
5. Reduce .9375 to a common fraction.
6. Reduce .0008 to a common fraction. Ans, -rhu-
Case III.
151. To reduce a common fraction to a decimal.
1. Reduce J to its equivalent decimal.
FIRST OPERATION. ANALYSIS. First annex the
^:^^^z=zr^^=z.'l[5, Ans, same number of ciphers to
both terms of the fraction;
SECOND OPERATION. this does not alter its value.
4)3.00 Then divide both resulting
~ terms by 4, the significant fig-
ure of the denominator, to ob-
tain the decimal denominator,
100. Then the fraction is changed to the decimal form by omitting
the denominator. If the intermediate steps be omitted, the true
result may be obtained as in the second operation.
2. Reduce ^ to its equivalent decimal.
Case II is what ? Give explanation. Rule. Case III is what ? Ex-
plain first operation. Second.
KEDUCTIOK,
123
THIRD OPERATION.
16 ) 1.0000
.0625, Ans.
Analysis. Dividing as in the former
example, we obtain a quotient of 3 fig-
ures, 625. But since we annexed 4
ciphers, there must be 4 places in the
required decimal; hence we prefix 1 cipher. This is made still
plainer by the following operation ; thus,
i^=iVWft=TtUo=.0625.
From these illustrations we deriye the following
EuLE. I. Annex ciphers to the numerator, and divide by
the denominator.
II. Point off as many decimal places in the result as are
equal to the number of ciphers annexed.
A common fraction can be reduced to an exact decimal when its lowest denomi-
nator contains only the prime factors 2 and 5, and not otherwise.
Examples foe Peactice.
3. Eeduce | to a decimal.
Ans.
.625.
4. Reduce f to a decimal.
5. Reduce ^ to a decimal.
Ans.
.9375.
6. Reduce -J to a decimal.
7. Reduce -^ to a decimal.
Ans.
.08.
8. Reduce -^ to a decimal.
Ans.
.046875.
9. Reduce -| to a decimal.
10. Reduce -^^ to a decimal.
11. Reduce -^ to a decimal
Ans.
.00375.
12. Reduce y|^ to a decimal.
Ans.
.008.
13. Reduce -J- to a decimal.
Ans. .;
33333 + .
The sign, +, in the answer indicates that there is still a remainder.
14. Reduce ^ to a decimal. Ans. .513513 + .
The answers to the last two examples are called repeating decimals ; and the
figure 3 in the 13th example, and the figures 513 in the 14th, are called repetends,
because they are repeated, or occur in regular order.
Third operation. Rule, first step ? Second? When can a common
fraction be reduced to an exact decimal ?
124 DECIMALS.
ADDITION.
152. 1. What is the sum of 3.703, 621.57, .672, and
20.0074?
OPERATION. Analysis. Write the numbers so that figures
3.703 of like orders of units shall stand in the same
g21 57 columns ; that is, units under units, tenths un-
^-,jj der tenths, hundredths under hundredths, etc.
This brings the decimal points directly under
20.0074 each other. Commencing at the right hand, add
645.9524 each column separately, and carry as in whole
numbers, and in the result place a decimal point
between units and tenths, or directly under the decimal point in the
numbers added.
EuLE. I. Write the numbers so that the decimal points
shall stand directly under each other.
II. Add as in whole numbers, and place the decimal point,
in the result, directly under the points in the numbers added.
Examples for Practice.
2. Add .199 3. Add 4.015
2.7569 6.75
.25 27.38203
.654 375.01
Sum, 3.8599 ^-^
Amount, 415.65703
4. Add 1152.01, 14.11018, 152348.21, 9.000083.
Ans. 153523.330263.
5. Add 37.03, 0.521, .9, 1000, 4000.0004.
Ans. 5038.4514.
6. What is the sum of twenty-six, and twenty-six hun-
dredths ; seven tenths ; six, and eighty-three thousandths ;
four, and four thousandths? Ans. 37.047.
Explain the operation of addition of decimals. Qive rule, first step.
Second.
ADDITIOIT. 125
7. What is the sum of thirty-six, and fifteen thousandths ;
three hundred, and six hundred five ten-thousandths ; five,
and three million fchs; sixty, and eighty-seven ten-millionths ?
Ans. 401.0755117.
8. What is the sum of fifty-four, and thirty-four hun-
dredths; one, and nine ten-thousandths; three, and two
hundred seven millionths ; twenty-three thousandths ; eight,
and nine tenths ; four, and one hundred thirty-five thou-
sandths? Ans.n.dddlOH.
9. How many yards in three pieces of cloth, the first piece
containing 18.375 yards, the second piece 41.625 yards, and
the third piece 35.5 yards ?
10. A's farm contains 61.843 acres, B's contains 143.75
acres, O's 218.4375 acres, and D's 21.9 acres; how many
acres in the four farms ?
11. My farm consists of 7 fields, containing 12| acres,
18f acres, 9 acres, 24^ acres, 4^ acres, 8^ acres, and 15J4
acres respectively ; how many acres in my farm ?
Reduce the common fractions to decimals before adding.
A71S, 93.6375.
12. A grocer has 2^ barrels of A sugar, 5f barrels of B
sugar, 3f barrels of C sugar, 3.0642 barrels of crushed
sugar, and 8.925 barrels of pulverized sugar ; how many
barrels of sugar has he ? Ans» 23.8642.
13. A tailor made 3 suits of clothes ; for the first suit he
used 2-J yards of broadcloth, 3^ yards of cassimere, and |
yards of satin ; for the second suit 2.25 yards of broadcloth,
2.875 yards of cassimere, and 1 yard of satin ; and for the
third suit 5-f^ yards of broadcloth, and 1-J- yards of satin.
How many yards of each kind of goods did he use ? How
many-yards of all? Ans. to last, 18.375.
126
DECIMALS.
SUBTEACTION,
153. 1. From 91.73 take 2.18.
OPERATION.
91.73
2.18
Ans. 89.55
2. From 2.9185 take 1.42.
OPERATION.
2.9185
1.42
Ans. 1.4985
3. From 124.65 take 95.58746.
OPERATION.
124.65
95.58746
Ans. 29.06254
Analysis. In each of these
three examples, we write the
subtrahend under the mm-
uend, placing units under
units, tenths under tenths,
etc. Commencing at the
right hand, we subtract as
in whole numbers, and in
the remainders we place the
decimal points directly under
those in the numbers above.
In the second example, the
number of decimal places in
the minuend is greater than
the number in the subtra-
hend, and in the third exam-
ple the number is less. In
both cases, we reduce both
minuend and subtrahend to
the same number of decimal
places, by annexing ciphers;
or Ave suppose the ciphers to
be annexed, before performing the subtraction.
Rule. I. Write the 7ium'bers so that the decimal j)oi7its
shall stand directly under each other.
II. Subtract as in whole numbers, and place the decimal
point in the result directly under the points in the given
numbers,
4. Find the difference between 714 and .916. Ans. 713.084.
5. How much greater is 2 than .298 ? Am. 1.702.
6. From 21.004 take 75 hundredths.
7. From 10.0302 take 2 ten-thousandths. Ans, 10.03.
8. From 900 take .009. Ans. 899.991.
9. From two thousand take two thousandths.
10. From one take one millionth. Ans. .990999.
Expl.iin subtraction of fractions. Give the rule, first step. Second.
I
MULTIPLICATION. 127
11. From four hundred twenty-seven thousandths take
four hundred twenty-seven milHonths. Ans. .426573.
12. A man owned thirty-four hundredths of a township
of land, and sold thirty-four thousandths of the township ;
how much did he still own ? Ans. .306.
MULTIPLICATION.
154. 1. What is the product of .35 multiplied by ,5 ?
OPERATION. Akalysis, We perform the multiplication the
.35 same as in whole numbers, and the only difficulty
5 we meet with is in pointing off the decimal places
in the product To determine how many places to
.175, A?IS. point off, we may reduce the decimals to common
fractions ; thus, .35 = -^^^ and .5 — /^. Perform-
ing the multiplication, we have ^^\ x tu = tuwg' ^^^ t^^is product,
expressed decimally, is .175. Here we see that the product contains
as many decimal places as are contained in both multiplicand and
multiplier.
EuLE. Multiply as in whole numlers, and from the right
hand of the product point off as many figures for decimals as
there are decimal places in both factors.
1. If there be not as many figures in the product as there are decimals in both fiwj-
tors, supply the deficiency by prefixing ciphers.
2. To multiply a decimal by 10, 100, 1000, etc., remove the point as many places to
♦he right as there are ciphers on the right of the multiplier.
Examples for Practice.
"Z. Multiply 1.245 by .27. Ans. .33615.
3. Multiply 79.347 by 23.15. Ans. 1836.88305.
4. Multiply 350 by .7853.
5. Multiply one tenth by one tenth. Ans. .01.
6. Multiply 25 by twenty-five hundredths. Ans. 6.25.
Explain multiplication of decimals. Give rule. If the product
have less decimal places than both factors, how proceed ? How mul-
tiply by 10, 100, 1000, etc?
t28 DECIMALS.
7. Multiply .132 by .241. Ans, .031812.
8. Multiply 24.35 by 10.
9. Multiply .006 by 1000. Ans, 6.
10. Multiply .23 by .009. Ans. .00207.
11. Multiply sixty-four tbousandths by thirteen mil-
liontlis. Ans, .000000832.
12. Multiply eigbty-seTen ten-thousandths by three hun-
dred fifty-two hundred-thousandths.
13. Multiply one million by one millionth. Ans, 1,
14. Multiply sixteen thousand by sixteen ten-thousandths.
Ans. 25.6.
15. If a cord of wood be worth 2.37 bushels of wheat,
how many bushels of wheat must be given for 9.58 cords of
wood? Ans, 22.7046 bushels.
DIVISION.
155. 1. What is the quotient of .175 divided by .5 ?
OPEKATION. Analysis. Performing the division the same as
.5 ) .175 in whole numbers, the only diflBculty we meet with
J rr~ is in pointing oflf the decimal places in the quotient.
* * To determine how many places to point off, we may
reduce the decimals to common fractions ; thus, .175 = xVin7> ^^^
J5=^. Performing the division, we have
35
175 5 m 10 35
X -^ = T7^ ;
1000 * 10 1000 $ 100
»nd this quotient, expressed decimally, is .35. Here we see that the
dividend contains as many decimal places as are contained in both
divisor and quotient.
Rule. Divide as in whole numbers, and from the right
hand of the quotient point off as many places for decimals
(IS the decimal places in the dividend exceed those in the
divisor.
Explain division of decimals. Give rule.
DIVISION". 129
1. If the number of flgnres in the quotient be less than the excess of the decimal
places in the dividend over those in the divisor, the deficiency must be supplied by
prefixing ciphers.
2. If there be a remainder after dividing the dividend, annex ciphers, and continue
the division ; the ciphers annexed are decimals of the dividend.
3. The dividend must always contain at least as many decimal places as the diyi-
gor, before commencing the division.
4. In most business transactions, the division is considered sufliciently exact
when the quotient is carried to 4 decimal places, unless great accuracy is required.
5. To divide by 10, 100, 1000, etc., remove the decimal point as many places to the
left as there are ciphers on the right hand of the divisor.
Examples for Practice.
2. Divide .675 by .15. Ans. 4.5.
3. Divide .288 by 3.6. Ans, .08.
4. Divide 81.6 by 2.5. Ans, 32.64.
5. Divide 2.3421 by 21.1.
6. Divide 2.3421 by .211.
7. Divide 8.297496 by .153. Ans. 54.232.
8. Divide 12 by .7854.
9. Divide 3 by 3 ; divide 3 by .3 ; 3 by .03 ; 30 by .03.
10. Divide 15.34 by 2.7.
11. Divide .1 by .7. Ans. .142857 + .
12. Divide 45.30 by .015. Ans. 3020.
13. Divide .003753 by 625.5. Ans. .000006.
14. Divide 9 by 450. Ans, .0^.
15. Divide 2.39015 by .007. Ans. 341.45.
16. Divide fifteen, and eight hundred seventy-five thou-
sandths, by twenty-five ten-thousandths. Ans, 6350.
17. Divide 365 by 100.
18. Divide 785.4 by 1000. Ans. .7854.
19. Divide one thousand by one thousandth.
A71S. 1000000.
When are ciphers prefixed to the quotient ? If there be a remainder,
how proceed ? If the dividend have less decimal places than the divi-
sor, how proceed ? How divide by 10, 100, 1000, etc. ?
6*
130 decimals.
Pkomiscuous Examples.
1. Add six hundred, and twenty-five thousandths ; four
tenths ; seven, and sixty-two ten- thousandths ; three, and
fifty-eight millionth s ; ninety-two, and seven hundredths.
Ans. 702.501258.
2. What is the sum of 81.003 -f- 5000.4 + 5.0008 -^
73.87563 + 1000 + 25 + 3.000548 + .0315 ?
3. Prom eighty-seven take eighty-seven thousandths.
4. What is the difference between nine million and nine
millionths ? Ans. 8999999.993991.
5. Multiply .365 by .15. Ans. .05475.
6. Multiply three thousandths by four hundredths.
7. If one acre produce 42.57 bushels of com, how many
bushels will 18.73 acres produce ? Ans. 797.3361.
8. Divide .125 by 8000. Ans. .000015625.
9. Divide .7744 by .1936.
10. Divide 27.1 by 100000. Ans. .000271.
11. If 6.35 acres produce 70.6755 bushels of wheat, what
does one acre produce ? Ans. 11.13 bushels.
12. Eeduce .625 to a common fraction. Ans. -|.
13. Express 26.875 by an integer and a common fraction.
Ans. 26|.
14. Eeduce yIt ^^ ^ decimal fraction. Ans. .016.
15. Eeduce -— to a decimal fraction. Ans. .5.
16. How many times will .5 of 1.75 be contained in .25 of
17i? Ans. 5.
17. What will be the cost of 3| bales of cloth, each bale
containing 36.75 yards, at .85 dollars per yard ?
18. Traveling at the rate of 4| miles an hour, how many
hours will a man require to travel 56.925 miles ?
Ans. 12| hours.
NOTATION AND NUMEEATION. 131
DECIMAL OUEEElSrOT.
156. Coin is money stamped, and has a given value es-
tablished by law.
157. Currency is coin, bank bills, treasury notes, etc.,
in circulation as a medium of trade.
158. A Decimal Currency is a currency whose denom-
inations increase and decrease in a tenfold ratio.
The currency of the United States is decimal currency, and is sometimes called
Federal Money ; it was adopted by Congress in 1786.
NOTATION AND NUMERATION.
BY THE " COINAGE ACT OP 1873."
The €oin of the United States consists of gold, silver,
nickel, and bronze.
Tlie Gold Coifis are the double-eagle, eagle, half-eagle,
quarter-eagle, three-dollar and one-dollar pieces.
Tlie Silver Coins are the dollar, half-dollar, quarter-dol-
lar, the twenty-cent, and ten -cent pieces.
The Mckel Coins are the five-cent and three-cent pieces.
The Bronze Coins are the one-cent pieces.
Table.
10 mills (m.) make 1 cent c
10 cents '^ 1 dime d.
10 dimes " 1 dollar $.
10 dollars '' 1 eagle E.
1. The mill is a denomination used only in computations ; it is not a coin.
2. The trade-dollar is designed solely for commerce, and not for currency. Its
weight Is 420 grains. The weight of the currency dollar of 1878 is 412^ grains.
S. The character $ is supposed to be a contraction of U. S. (United States), the
U being placed upon the S.
Wliat is coin? Currency? Decimal currency? Federal money?
What are the gold coins of U. S. ? Silver ? Copper ? What are the
denominations of U. S. currency ? What is the sign of dollars ? From
what de jived ?
132 DECIMAL CURRENCY.
159. The gold dollar is the unit of United States money ;
dimes, cents, and mills are fractions of a dollar, and are
separated from the dollar by the decimal point ; thus, two
dollars one dime two cents five mills, are written $2,125.
By examining the tabUy we see that the dime is a tenth
part of the unit, or dollar; the cent a tenth part of the dime
or a hundredth part of the dollar ; and the mill a tenth part
of the cent, a hundredth part of the dime, or a thousandth
oart of the dollar. Hence the denominations of decimal
currency increase and decrease the same as decimal fractions,
and are expressed according to the same decimal system of
notation ; and they may be added, subtracted, multiplied,
and divided in the same manner as decimals.
Dimes are not read as dimes, but the two places of dimes
and cents are appropriated to cents ; thus, 1 dollar 3 dimes
2 cents, or $1.32, are read one dollar thirty-two cents ; hence.
When the number of cents is less than 10, we write a
cipher before it in the place of dimes.
The half cent is frequently written as 5 mills ; thus, 24^ cents, written $.246.
160. Business men frequently write cents as common
fractions of a dollar ; thus, three dollars thirteen cents are
written ^3^^, and read, three and thirteen hundredths
dollars. In business transactions, when the final result of a
computation contains 5 mills or more, they are called one
cent, and when less than 5, they are rejected.
Examples for Practice.
1. Write four dollars five cents. Ans. $4.05.
2. Write two dollars nine cents.
3. Write ten dollars ten cents.
. 4. Write eight dollars seven mills. Ans. $8,007.
What is the unit of U. S. currency ? What is the general law of
increase and decrease ? In practice, how many decimal places are given
to cents ? In business transactions, how are cents frequently written 1
WTiat is done if the mills exceed 5 ? If less than 5 ?
REDUCTION. 133
5. Write sixty-four cents. Ans. $0.64
6. Write three cents two mills.
7. Write one hundred dollars one cent one milL
8. Read $7.93; $8.02; $6,542.
9. Read $5,272; $100,025; $17,005.
10. Read $16,205; $215,081; $1000.011; $4,002.
REDUCTION.
161. By examining the table of Decimal Currency, we
see that 10 mills make one cent, and 100 cents, or 1000
mills, make one dollar ; hence,
To change dollars to cents, multiply hy 100; that is, annex
two cipliers.
To change dollars to mills, annex three ciphers.
To change cents to mills, annex one cipher.
Examples fob Peactice.
1. Change $792 to cents. Ans. 79200 cents.
2. Change $36 to cents.
3. Reduce $5248 to cents.
4. In 6.25 dollars how many cents? Ans. 625 cents.
To change dollars and cents to cents, or dollars, cents, and mills to mills, remove
the decimal point and the sign, $.
5. Change $63,045 to mills. Ans, 63045 mills.
6. Change 16 cents to mills.
7. Reduce $3,008 to mills.
8. In 89 cents how many mills ?
162. Conversely,
To change cents to dollars, divide ty 100; that is, point
off two figures from the right.
To change mills to dollars, point off three figures.
To change mills to cents, point off one figure.
How are dollars changed to cents ? to mills ? How are cents changed
to mills ? How are cents changed to dollars ? Mills to dollars ? to cents t
134 decimal cukrency.
Examples foe Practice.
1. Change 875 cents to dollars. Ans, 18.75.
2. Change 1504 cents to dollars.
3. In 13875 cents how many dollars ?
4. In 16525 mills how many dollars ?
5. Reduce 524 mills to cents.
6 Eeduce 6524 mills to dollars.
ADDITION.
163. 1. A man bought a cow for 21 dollars 50 cents, a
horse for 125 dollars 37J- cents, a harness for 46 dollars 75
cents, and a carriage for 210 dollars ; how much did he pay
for all ?
OPERATION.
$ 21.50 Analysis: Writing dollars under dol-
125.375 lars, cents under cents, etc., so tliat the
^Q WK decimal points shall stand under each
rt-j r. ri/% other, add and point off as in addition of
Ans. 1403.625
decimals.
Rule. I. Write dollars under dollars, cents under ce7its, etc.
11. Add as in simple numhers, and place the point in the
amount as in addition of decimals.
Examples for Practice.
2. What- is the sum of 50 dollars 7 cents, 1000 dollars 75
cents, 60 dollars 3 mills, 18 cents 4 mills, 1 dollar 1 cent,
and 25 dollars 45 cents 8 mills ? Ans. $1137.475.
3. Add 364 dollars 54 cents 1 mill, 486 dollars 6 cents, 93
dollars 9 mills, 1742 dollars 80 cents, 3 dollars 27 cents 6
mills. A71S. $2689.686.
4. Add 92 cents, 10 cents 4 mills, 35 cents 7 mills, 18 cents
6 mills, 44 cents 4 mills, 12|- cents, and 99 cents. Ans. $3,126.
Explain the process of addition of decimal currency. Rule, first
Btep. Second.
SUBTEACTIOlir. 135
5. A farmer receives 89 dollars 74 cents for wheat, 13
dollars 3 cents for corn, 6 dollars 37^ cents for potatoes,
and 19 dollars 62 J cents for oats; what does he receive for
the whole? Ans. ^US.H'T.
6. A lady bought a dress for 9 dollars 17 cents, trimmings
for 87^ cents, a paper of pins for 6} cents, some tape for 4
cents, some thread for 8 cents, and a comb for 11 cent:;
what did she pay for all? Ans. $10.3375.
7. Paid for building a house 12175.75, for painting the
same $240.3 7|-, for furniture $605.40, for carpets $140. 12^;
what was the cost of the house and furnishing ?
8. Bought a ton of coal for $6.08, a barrel of sugar for
$26,625, a box of tea for $16, and a barrel of flour for $7.40 ;
what was the cost of all ?
9. A merchant bought goods to the amount of $7425.50 ;
he paid for duties on the same $253.96, and for freight
$170.09 ; what was the entire cost of the goods ?
10. I bought a hat for $3.62^, a pair of shoes for $1}, an
umbrella for $lf , a pair of gloves for $.62J, and a cane for
$.87 J; what was the cost of all my purchases ? Ans. $8.25.
SUBTRAOTIOlSr.
164. 1. A man, having $327.50, paid out $186.75 for a
horse ; how much had he left ?
OPERATION. ANAI.YSIS. Writing the less number un-
$327.50 der the greater, dollars under dollars, cents
186.75 under cents, etc., subtract and point oflf
in the result as ia subtraction of deci-
Ans. $140.75 ^^j^
Rule. I. Write the subtrahend under the minuend, dollars
under dollars, cents under cents.
11. SuUract as in simple numbers, and place the point in
the remainder, as in subtraction of decimals.
Explain the process of subtraction. Give rule, first step. Second.
136 decimal currency.
Examples for Practice.
2. From 365 dollars 5 mills take 267 dollars 1 cent 8
mills. Ans. $97,987.
3. From 50 dollars take 50 cents. Ans. $49.50.
4. From 100 dollars take 1 mill Ans. $99,999.
5. From 1000 dollars take 3 cents 7 mills.
6. A man bought a farm for $1575.24, and sold it foi
$1834.16 ; what did he gain? Ans. $258.92.
7. Sold a horse for 145 dollars 27 cents, which is 37 dol-
lars 69 cents more than he cost me ; what did he cost me ?
8. A merchant bought flour for $5.62^ a barrel, and sold
it for $6.84 a barrel ; what did he gain on a barrel ?
9. A gentleman, having $14725, gave $3560 for a store,
and $7015.87^ for goods ; how much money had he left ?
10. A lady bought a silk dress for $13 J, a bonnet for $5 J,
a pair of gaiters for $1-|, and a fan for $J ; she paid to the
shopkeeper a twenty dollar bill and a five dollar bill ; liow
much change should he return to her ? Ans. $3.75.
Reduce the fractions of a dollar to cents and mills. «
11. A gentleman bought a pair of horses for $480, a har-
ness for $80.50, and a carriage for $200 less than he paid
for both horses and harness ; what was the cost of the car-
riage? Ans.$3m.50.
MULTIPLICATIOlSr.
165. 1. If a barrel of flour cost $6,375, what will 85
barrels cost?
OPERATION.
$6,375 Analysis. Multiply as in simple num
85 bers, always regarding the multiplier as an
313'j'5 abstract number, and point off from the right
51000 hand of the result, as in multiplication of
^W5. $541:875 decimals.
^ 1
Give analysis for multiplication in decimal currency.
DIVISION. 137
Rule. Multiply as in simple numbers, and place the point
in the product, as in multiplication of decimals.
Examples for Peactice.
2. If a cord of wood be worth $4,275, what will 300 cords
be worth? ^7Z5. $1282.50.
3. What will 175 barrels of apples cost, at $2.45 per bar-
rel? Ans. $428.75.
4. What will 800 barrels of salt cost, at $1.28 per barrel ?
5. A grocer bought 372 pounds of cheese at $.15 a pound,
434 pounds of coffee at $.12^ a pound, and 16 bushels of
potatoes at $.33 a bushel ; what did the whole cost ?
6. A boy, being sent to purchase groceries, bought 3
pounds of tea at 56 cents a pound, 15 pounds of rice at 7
cents a pound, 27 pounds of sugar at 8 cents a pound ; he
gave the grocer 5 dollars ; how much change ought he to
receive ?
7. A farmer sold 125 bushels of oats at $.37^ a bushel,
and received in payment 75 pounds of sugar at $.09 a pound,
12 pounds of tea at $.60 a pound, and the remainder in
cash ; how much cash did he receive ? Ans. $32. 92 J.
8. A man bought 150 acres of land for $3975 ; he after-
ward sold 80 acres of it at $32.50 an acre, and the remainder
at $34.25 an acre ; what did he gain by the transaction?
Ans. $1022.50.
DIVISION.
166. 1. If 125 barrels of flour cost $850, how much
will 1 barrel cost ?
OPERATION.
125 ) $850.00 ( $6.80, Ans. Analysts. Divide as in sim-
750 pie numbers, and as there is a
^ remainder after dividing the dol-
lars, reduce the dividend to cents,
1000 by annexing two ciphers, and coiif
tinue the division.
Rule. Give rule for division in decimal currency.
138 DECIMAL CURRENCY.
EuLE. Divide as in simple numhers, and place the point
in the quotient, as in division of decimals.
1. In business transactions it is never necessary to carry the division further than
to mills in the quotient.
2. If the dividend will not contain the divisor an exact number of times, ciphers
may be annexed, and the division continued as in division of decimals. In this case
it is always safe to reduce the dividend to mills, or to 3 more decimal places th ffn
the divisor contains^ before commencing the division.
Examples for Practice.
3. If 33 gallons of oil cost $41.25, what is the cost per
gallon ? Ans, 11.25.
3. If 27 yards of broadcloth cost $94.50, what will 1 yard
cost?
4. K 64 gallons of wine cost $136, what wiU 1 gallon
cost? ^»s. $2,125.
5. At 12 cents apiece, how many pine-apples can be bought
for $1.32 ? Ans. 11.
6. If 1 pound of tea costs 54 cents, how many pounds
can be bought for $405 ?
7. If a man earns $180 in a year, how much does he earn
a month ?
8. If 100 acres of land cost $2847.50, what will 1 acre
cost? Ans, $28,475.
9. What cost 1 pound of beef, if 894 pounds cost $80.46 ?
Ans, $.09.
10. A farmer sells 120 bushels of wheat at $1.12^ a bushel,
for which he receives 27 barrels of flour ; what does the flour
cost him a barrel ?
11. A man bought 4 yards of cloth at $3.20 a yard, and
37 pounds of sugar at $.08 a pound ; he paid $6.80 in cash,
and the remainder in butter at $.16 a pound ; how many
pounds of butter did it take ? Ans. b^ pounds.
12. A man bought an equal number of calves and sheep,
paying $166.75 for them; for the calves he paid $4.50 a
head, and for the sheep $2. 75 a head ; how many did he
buy of each kind ? Ans, 23.
APPLICATIONS. 139
13. If 154 pounds of sugar cost $18.48, what will 1 pound
cost ?
14. A merchant bought 14 boxes of tea for $560 ; it being
damaged he was obliged to lose $106.75 on the cost of it ;
how much did he receive a box ? Ans, $32.3 7^.
Additional Applications.
Case I.
167. To find the cost of any number or quantity,
when the price of a unit is an aliquot p2|-rt of one
dollar.
168. An Aliquot Part of a number is such a part as
will exactly divide that number ; thus, 3, 5, and 7i are
aliquot parts of 15.
An cUiqvot part may be a whole or a mixed number, while a factor most be a
whole number.
Aliquot Pakts op Oke Dollak.
50 cents = |- of $1.
33| cents = i ot $1.
25 cents = J of $1.
20 cents = i of $1.
16f cents = i of $1.
12J cents = i of $1
10 cents = ^ of $1
8^ cents = ^ of $1
6| cents = ^ of $1
5 cents = -^^ of $1
1. What will be the cost of 3784 yards of flannel, at 25
cents a yard ?
OPEKATiON. Analysis. If the price were $1 a yard,
4 ) 3784 the cost would be as many dollars as there are
J *Qj.« yards. But since the price is ^ of a dollar a
^ yard, the whole cost will be ^ as many dollars
as there are yards ; or, | of $3784 = $3784 ^ 4 == $946.
KuLE. Take such a fractional part of the given number as
the price is part of one dollar.
Examples foe Practice.
2. What cost 963 bushels of oats, at 33 J cents per bushel ?
Ans. $321.
Case I is what ? What is an aliquot part of a dollar ? Give expla-
nation. Rule.
140 DECIMAL CURRENCY.
3. AYhat cost 478 yards of delaine, at 50 cents per yard ?
4. What cost 4266 yards of sheeting, at 8^ cents a yard ?
A71S. 1355.50.
5. What cost 1250 bushels of apples, at 12|- cents per
bushel? Ans. $156.25.
6. What cost 3126 spools of thread, at 6J cents per spool ?
Ans. $195,375.
7. At 16f cents per dozen, what cost 1935 dozen of eggsr
A71S. $322.50.
8. Wliat cost 56480 yards of calico, at 12|- cents per yard ?
9. At 20 cents each what will be the cost of 1275 salt
barrels? Ans. $255.
Case II.
169. The price of one and the quantity being
given, to find the cost.
1. How much will 9 barrels of flour cost, at $6.25 per
barrel ?
OPERATION. Analysis. Since one barrel cost $6.25, 9
$6.25 barrels will cost 9 times $6.25, and $6.25 x
9 9 = $56.25.
Ans. $56.25
Rule. Multiply the price of one hy tlie quantity.
Examples foe Practice.
2. If a pound of beef cost 9 cents, what will 864 pounds
cost? Ans. $77.76.
3. What cost 87 acres of goyemment land, at $1.25 per
acre?
4. What cost 400 barrels of salt, at $1.45 per barrel ?
Ans, $580.
5. What cost 10 chests of tea, each chest containing 5^
pounds, at 44 cents per pound ?
Case II is what ? Give explanation. Rule.
APPLICATIONS. 141
Case III.
170. The cost and the quantity being given, to
find the price of one.
1. If 30 bushels of corn cost 120.70, what will 1 bushel
cost?
OPERATION. Analysis. If 30 bushels cost $20.70, 1
3|0 ) $2|0.70 bushel wiU cost ^ of $30.70 ; and $20.70 h-
^- 30 = $.69.
EuLE. Divide the cost ly the quantity.
Examples for Peactice.
2. If 25 acres of land cost $175, what will 1 acre cost ?
3. If 48 yards of broadcloth cost $200, what will 1 yard
cost? Ans. $4.16f.
4. If 96 tons of hay cost $1200, what will 1 ton cost ?
5. If 10 Unabridged Dictionaries cost $56.25, what will 1
cost? ^?zs. $5.62J.
6. Bought 18 pounds of tea for $11.70 ; what was the
price per pound ? Ans. $.65.
7. If 53 pounds of butter cost $10.07, what will 1 pound
cost?
8. A merchant bought 800 barrels of salt for $1016 ; what
did it cost him per barrel ?
9. If 343 sheep cost $874.65, what will 1 sheep cost?
Ans, $2.55.
10. If board for a family be $684.37^ for 1 year, how
much is it per day ? Ans, $1.87^.
Case IV.
171. The price of one and the cost of a quantity
being given, to find the quantity.
1. At $6 a barrel for flour, how many barrels can be
bought for $840 ?
Case III is what ? Give explanation. Rule. Case IV is what ?
14^ DECIMAL CUKREKCY.
OPERATION. ANAiiTSis. Sincc $6 will buy 1 barrel
6 ) 840 of flour, $840 will buy ^ as many barrels
as there are dollars, or as many barrels as
Ans, 140 barrels. ^q ig contained times in $840; 840-^6
= 140 barrels.
Rule. Divide the cost of the quantity ly the price of one.
Examples for Practice.
2. How many dozen of eggs can be bought for $5.55, if
one dozen cost $.15 ? Ans. 37 dozen.
3. At $12 a ton, how many tons of hay can be bought for
$216 ? Ans. 18 tons.
4. How many bushels of wheat can be bought for $2178.75,
if 1 bushel cost $1.25? Ans. 1743 bushels.
5. A dairyman expends $643.50 in buying cows at $194-
apiece ; how many cows does he buy? Ans. 33 cows.
6. At $.45 per gallon, how many gallons of molasses can
be bought for $52.65 ?
7. A drover bought horses at $264 a pair ; how many
horses did he buy for $6336 ?
8; At $65 a ton, how many tons of railroad iron can be
bought for $117715 ? Ans, 1811 tons.
Case V.
172. To find the cost of articles sold by the 100,
or 1000.
1. What cost 475 feet of timber, at $5.24 per 100 feet ?
FIRST OPERATION.
$5.24 Analysis. If the price were |5.24 per
475 foot, the cost of 475 feet would be 475 x
$5.24 = $2489. But since $5.24 is the price
of 100 feet, $2489 is 100 times the true
value. Therefore, to obtain the true value,
we divide $2489 by 100, which we may do
100 ) $2489.00 by cutting off two figures from the right,
Ans ^24 89 ^^^ *^® result is $24.89. Or,
Give explanation Rule. Case V is what ? Give first explanation.
2620
3668
2096
APPLICATIONS. 143
SECOND OPEKATION. ANALYSIS. Since 1 foot cost Y^, or .01,
$5.24 of $5.24, 475 feet will cost f^f, or 4.75 times
4.75 ^5.24, which is $24.89.
2620 For the same reasons, when the price is per thxm-
q/j/:.o «awrf, we divide the product by 1000, or, which is more
convenient in practice, we reduce the given quantity
2096 to thousands and decimals of a thousand, by pointing
^ off three figures from the right hand.
$24.8900
EuLE. I. Reduce the given quantity to hundreds and deci-
7nals of a hundred, or to thousands and decimals of a thousand,
II. Multiply the price hy the quantity, and point off in the
result as in multiplication of decimals.
The letter C is used to indicate hundreds, and M to indicate thousands.
Examples for Practice.
2. What will 42650 bricks cost, at $4.50 per M ?
Ans. $191,925.
3. What is the freight on 2489 pounds from Boston to
New York, at $.85 per 100 pounds ? Ans. $21,156.
4. What wiU 7842 feet of pine boards cost, at $17.25
per M ? Ans. $135,274.
5. What cost 2348 pine-apples, at $12^ per 1000 ?
6. A broom maker bought 1728 broom-handles, at $3 per
1000 ; what did they cost him ?
7. What is the cost of 2400 feet of boards, at $7 per M ;
865 feet of scantling, at $5.40 per M ; and 1256 feet of lath,
at $.80 per ? Ans. $31,519.
8. What will be the cost of 1476 pounds of beef, at $4.37i
per hundred pounds ?
Case VL
173. To find the cost of articles sold by the ton
of 2000 pounds.
1. How much will 2376 pounds of hay cost, at $9.50 per
ton?
Give second explanation. Rule, first step. Second. Case VI is what ?
144 DECIMAL CUEREKCY,
OPERATION. Analysis. Since 1 ton, or 2000 pounds, cost
2 ) $9.50 $9.50, 1000 pounds, or ^ ton, will cost |- of $9.50,
or $9.50 H- 2 = $4.75. One pound will cost
■n^, or .001, of $4.75, and 2376 pounds will
cost fro, or 2.376 times $4.75, which is $11,286.
$4.75
2.376
$11.28600
Rule. I. Divide the price of 1 ton ly 2, and the quotient
will he the 'price of 1000 pounds.
II. Multiply this quotient hy the given number of pounds
expressed as thousandths, as in Case V,
Examples for Peactice.
2. At 17 a ton, what will 1495 pounds of hay cost ?
Ans. $5.2325.
3. At $8.75 a ton, what cost 325 pounds of hay ?
Ans. $1,421.
4. What is the cost of 3142 pounds of plaster, at $3.84
per ton? ' Ans. $6,032.
5. What is the cost of 1848 pounds of coal, at $5.60 per
ton?
6. Bought 125 sacks of guano, each sack containing 148
pounds, at $18 a ton ; what was the cost ?
7. What must be paid for transporting 31640 pounds of
railroad iron from Philadelphia to Richmond, at $3.05 per
ton? Ans. $48,251.
BILLS.
174. A Bill, in business transactions, is a written state-
ment of articles bought or sold, together with the prices of
each, and the whole cost.
Find the cost of the several articles, and the amount or
footing of the following bills.
Give explanation. Rule. What is a bill ? Explain the manner of
making out a bill.
BILLS
1^5
ttlr, JoHiq^ EiCE,
(1.)
New Yoek, June 20, 1869.
BoH of Baldwin & Sherwood.
7 yds. Broadcloth, @ 83.60
9 " Satinet, " 1.12^
12 " Vesting, " .90
24 " Cassimere, '' 1.37^
32 '' Flannel, '' .65
Redd Payment,
$99,925
BALDWii?^ & Sherwood.
(2.)
BosT02^r, Jan. 1, 1870.
Daniel Chapmak & Co.,
BoH of Palmer & Brother.
67 pairs Calf Boots, @ $3.75
108
75
27
35
50
'' Thick '
'' Gaiters,
Buskins,
Slippers,
Rubbers,
2.62
1.12
.86
.70
1.04
Re<fd
Payment,
$717.93
Palmer & Brother,
By Geo. Baker.
(3.)
Charlestoi^, Sept. 6, 1875.
G. B. Grai^nis,
BoH of Stewart & Hammokd.
325 lbs. A. Sugar,
@
$.07
148
" B. "
it
.06i
286
" Eice,
(I
.05
95
'' 0. J. Coffee, ''
.12i
60 boxes Oranges,
(I
2.75
75
** Lemons,
a
3.62i
12
*' Raisins,
a
2.85
Eec^d Payment, by note at 4 mo. ^^01.75
r.p. 7 Stewart & Hammond.
J-46 DECIMAL CURRENCY.
(4.)
St. Louis, Oct. 15, 1878.
Messrs. Osborn & Eaton^,
BoH of Rob't H. Carter & Gg^
20000 feet Pine Boards, @ $15.00 per M.
7500 '' Plank, " 9.50 ''
10750 '' Scantling, '' 6.25 "
3960 '' Timber, " 2.62^ "
5287 " " " 3.00 ''
Bec^d Payment,
$464.6935
Rob't H. Carter & Co.
(5.)
CiN^ciirif ATI. Mav 3, 1871.
Mr. J. C. Smith,
BoH of Silas Johnson,
25 lbs.
Coffee Sugar, @
$.11
5 ''
Y. H. Tea,
.62^
26 ''
Mackerel, "
.061
4 gal.
Molasses, "
.42
46 yds.
Sheeting, '^
.09
30 "
Bleached Shirting, '^
.14
6 skeins
3 Sewing Silk, *^
.04
4 doz.
Buttons, "
.12
Chgd. in %. ^^^-^^
Silas Johnson,
Per John Wisk.
Promiscuous Examples.
1. What will 62.75 tons of potash cost, at $124.35 per ton ?
Ans. $7802.9625.
2. What cost 15 pounds of butter, at $.17 a pound?
Ans. $2.55.
3. A cargo of com, containing 2250 bushels, was sold for
$1406.25 ; what did it sell for per bushel ? Ans, $4-
PROMISCUOUS EXAMPLES. 147
4. If 12 yards of cloth cost $48.96, what will one yard
cost?
5. A trayeled 325 miles by railroad, and traveled .45 of
that distance ; how far did travel ? Ans. 146.25 miles.
6. If 36.5 bushels of corn grow on one acre, how many
acres will produce 657 bushels ? Ans. 18 acres.
7. Bought a horse for $105, a yoke of oxen for $125, 4
cows at $35 apiece, and sold them all for $400 ; what was
gained or lost in the transaction ?
8. A man bought 28 tons of hay at $19 a ton, and sold
it at $15 a ton ; what did he lose ? Ans, $112.
9. If a man travel 4| miles an hour, in how many hours
can he travel 34|- miles ? Ans. 7.5 hours.
10. At $.31:J per bushel, how many bushels of potatoes
can be bought for $9 ? Aiis. 28.8 bushels.
11. If a man's income be $2000 a year, and his expenses
$3.50 a day, what will he save at the end of a year, or 365
days?
12. A merchant deposits in a bank, at one time, $687.25,
and at another, $943.64 ; if he draw out $875.29, what will
remain in the bank ?
13. Bought 288 barrels of flour for $1728, and sold one-
half the quantity for the same price I gave for it, and the
other half for $8 per barrel; what did I receive for the
whole ? Ans. $2016.
14. What will eight hundred seventy-five thousandths of
a cord of wood cost, at $3.75 per cord? A^is. $3,281 + .
15. A drover bought cattle at $46.56 per head, and sold
them at $65.42 per head, and thereby gained $3526.82 ; how
many cattle did he buy ? Ans. 187.
16. If 36.48 yards of cloth cost $54.72, what wiU 14.25
yards cost ? Ans. $21. 375.
17. A house cost $3548, which is 4 times as much as the
furniture cost; what did the furniture cost? Ans. $887.
18. How many bushels of onions at $.82 per bushel, can
bo bought for $112.34 ?
148 DECIMAL CURRENCY.
19. If 46 tons of iron cost 13461.50, what will 5 tons cost?
20. A gentleman left his widow one-third of his property,
worth 124000, and the remainder was to be divided equally
among 5 cliildren ; what was the portion of each child ?
Ans. $3200.
21. A man purchased one lot, containing 160 acres of
land, at |>1.25 per acre ; and another lot, containing 80
acres, at $5 per acre ; he sold them both at $2.50 per acre;
what did he gain or lose in the transaction ?
22. A druggist bought 54 gallons of oil for $72.90, and
lost 6 gallons of it by leakage. He sold the remainder at
$1.70 per gallon ; what did he gain ? Ans. $8.70.
23. A miller bought 122-|- bushels of wheat of one man,
and 75 J bushels of another, at $.93} per bushel. He sold
60 bushels at a profit of $12.50 ; if he sell the remainder at
$.81 J per bushel, what will be his entire gain or loss?
Ans. $4,718+ loss.
24. A laborer receives $1.40 per day, and spends $.75 for
his support ; how much does he save in a week ?
25. How many pounds of butter, at $.16 per pound, must
be given for 39 yards of sheeting, at $.08 a yard ?
Ans. 19|- pounds.
26. What cost 23487 feet of hemlock boards, at $4.50 per
1000 feet ? Ans. $105.6915.
27. A man has an income of $1200 a year ; how much
must he spend per day to use it all ?
28. Bought 28 firkins of butter, each containing 56
pounds, at $.17 per pound; what was the whole cost ?
29. A merchant bought 16 bales of cotton cloth, each bale
containing 13 pieces, and each piece 26 yards, at $.07 per
yard ; what did the whole cost ? Ans, $378.56.
30. What cost 4868 bricks, at $4.75 per M ?
31. A farmer sold 27 bushels of potatoes, at $.33|- per
bushel; 28 bushels of oats, at $.25 per bushel ; and 19 bush-
els of corn, at $.50 per bushel ; what did he receive for the
who'.e? Ans. $25.50.
PROMISCUOUS EXAMPLES. 149
32. John runs 32 rods in a minute, and Henry pursues
him at the rate of 44 rods in a minute ; how long will it
take Henry to overtake John, if John have 8 minutes the
start? Ans. 211- mmntes,
33. If 4| barrels of flour cost 132.3, what will 7^ barrels
cost ? Ans. $51.
34. If .875 of a ton of coal cost 15.635, what will 9J tons
cost ? Ans. $59.57.
35. For the first three years of business, a trader gained
$1200.25 a year ; for the next three, he gained $1800.62 a
year, and for the next two he lost $950.87 a year ; supposing
his capital at the beginning of trade to have been $5000, what
was he worth at the end of the eighth year? Ans. $12100.87.
36. What will be the cost of 18640 feet of timber, at $4.50
per 100? Ans. $Sd8.S0.
24
37. Eeduce ^ to a decimal fraction. Ans. .78125.
38. What will 1375 pounds of potash cost, at $96.40 per
ton ? Ans. $66,275.
39. Eeduce .5625 to a common fraction. Ans. -^.
40. Eeduce -^, .62J, .37^^^ f, to decimals, and find their
sum. Ans. 1.464375.
41. A man's account at a store stands thus :
Dr. Cr.
$4,745 $2.76^
2.62^ 1.245
1.27 .62J
.45 3.45
5.28J 1.87i
What is due the merchant ? Ans. $4.41|-.
42. A gardener sold, from his garden, 120 bunches of
onions at $.12^ a bunch, 18 bushels of potatoes at $.62|- per
bushel, 47 heads of cabbage at $. 07 a head, 6 dozen cucum-
bers at $.18 a dozen ; he expended $1.50 in spading, $1.27
for fertilizers, $1.87 for seeds, $2.30 in planting and hoeing ;
what were the profits of his garden ? Ans. $23.68.
150
REDUCTION.
EEDUCTIOW.
175. A Compound Number is a number whose value
is expressed in two or more different denominations.
176. Reduction is the process of changing a number
from one denomination to another without altering its value.
Reduction is of two kinds, Descending and Ascending.
177. Reduction Descending is changing a number of
one denomination to another denomination of less unit
value; thus, $1=10 dimes =100 cents =1000 mills.
178. Reduction Ascending is changing a number of one
denomination to another denomination ot greater unit valtte;
thus, 1000 mills=100 cents=10 dimes=$l.
179. A Scale is a series of units, increasing or decreas-
.ing, according to a certain law.
CURRENCY.
180. I. UiiriTED States Monet.
Table.
10 Mills (m.) make 1 Cent ct. f 10000 m.
10 Cents " IDime d. ^„_J 1000 ct.
10 Dimes " 1 Dollar $. "^ ^— 1 100 d.
10 Dollars " 1 Eagle E. [ 10 1.
Unit Equivalents.
ct.
d. 1 =
$ 1 = 10 =
E. 1 = 10 = 100 =
1 = 10 = 100 = 1000 = 10000
Scale— uniformly 10.
Canada Money.
The currency of the Dominion of Canada is decimal, and
the table and denominations are the same as those of the
United States money.
The currency of the whole Dominion of Canada was made uniform July 1, 1871.
Before the adoption of the decimal eystem, pounds, shillingP, and pence were used.
The Silver Coins are the 50-cent piece, 25-cent piece, 10-cent piece,
and 5-cent piece. The 20-cent piece is no longer coined. The Bronze
Coin is the cent.
The Odd Coin used in Canada is the British Sovereign, worth |4.8C|,
and the Half- Sovereign.
m.
10
100
1000
compoukd numbers. 151
11. English Monet.
181. English Currency is the currency of Great Britain,
Table.
U. S. Value.
4 Farthings (far.) make 1 Penny d $0.0202 +
12 Pence " 1 Shilling s 2433 +
20 ShiUings * * 1 Pound or Sov £., or sov. . . . . $4.8665
Unit Equivalents.
d. far.
8. 1 = 4
£. 1 = 12 = 48
1 = 20 = 240 = 960
Scale ascending, 4, 12, 20 ; descending, 20, 12, 4.
1. Farthings are generally expressed as fractions of a penny ; thus, 1 far., some*
times called 1 quarter (qr.), = |d. ; 3 fer. = |d.
2. The gold coins are the sovereign ( = £1), and the half-sovereign.
3. The silver coin^ are the crown ( = 58.), the half-crown ( = 2s. 6d.), the florin, the
ehilling, and the sixpenny, fourpenny, and threepenny pieces.
4. The copper coins are the penny, halfpenny, and farthing,
5. The guinea ( = 2l8.) and the half-guinea ( = 10s. 6d. sterling), are old gold coins,
and are no longer coined.
6. In France accounts are kept in francs and decimes. A franc is equal to $.198
U. S. money.
Case I.
183. To perform reduction descending.
1. Keduce £21 18s. lOd. 2 far. to farthings.
operation. Analysis. Since in £1 there
£21 IBs. lOd. 2 far. are 20s., in £21 there are 20s. x
20 21 = 420s., and IBs. in the given
number added, makes 438s. in £21
18s. Since in Is. there are 12d.,
in 438s. there are 12d. x 438 =
6266d. 5256d., and lOd. in the given num-
4 ber added, makes 5266d. in £21
18s. lOd. Since in Id. there are
21066 far., Ans. 4 f^r., in 5266d. there are 4 far. x
= 21064 far., and 2 far. in the given number added, makes 21066
far. in £21 18s. lOd. 2 far.
438s.
12
152 REDUCTlOiT.
Rule. I. Multiply the highest dcnominatmi of the given
number hy that number of the scale which will reduce it to
the next loioer denomination, adding to the product the given
number, if any, of that loioer denominatioti.
II. Proceed in the same manner with the results obtained
in each lower denomination, until the reduction is brought to
the denomination required.
Case IL
183. To perform rediiction ascending.
1. Reduce 21066 farthings to pounds.
OPERATION. Analysis. First divide 21066
4 ) 21066 far. far. by 4, since there are \ as
1 o \ "nnnj , o £ many pence as farthings, and we
12 )_o266d. + 2 fax. ^^^ ,^^, ^^^^ ^^^ ^ 52^^^ ^ ^
2|0 ) 43|8s. + lOd. remainder of 2 far. Next divide
r>2i I 1 Qq 5266d. by 12, since there are ^^
J rt^^^^\./s-irt<. ^ many shillings as pence, and
Ans. £n 18s. lOd. 3 far. ^, ^ J ,^,, ^^^ I ^^ ^
lOd. Lastly divide 438s. by 20, since there are ^V ^ many pounds
as shillings, and we find that 438s. = £12 + 18s. The last quotient
with the several remainders annexed in the order of the succeeding
denominations, gives the answer £21 18s. lOd. 2 far.
Rule. I. Divide the given number by that number of the
scale which will reduce it to the next higher deno7nination.
IL Divide the quotient by the next higher number in the
scale ; and so proceed to the highest denomination required.
The last quotient, with the several remainders annexed in a
reversed order, will be the answer.
Reduction descending and redaction ascending mntaally prove eacSi other.
Examples for Practice.
1. In 14194 farthings how many pounds ?
2. In £14 los. 8d. 2 far. how many farthings ?
3. In 15359 farthings how many pounds?
4. In 46 sov. 12s. 2d. how many pence ?
5. In 11186 pence how many sovereigns?
COMPOUND KUMBEES,
153
WEIGHTS.
184. Weight is the measure of gravity, and varies ac-
cording to the quantity of matter a body contains. Three
scales of weight are used in the United States and Great
Britain, namely, Troy, Apothecaries', and Avoirdupois.
I. Teoy Weight.
185. Troy Weight is used in weighing gold, silver, and
jewels, and in philosophical experiments.
Table.
24 Grains (gr.) make 1 Pennyweight pwt.
20 Pennyweights " 1 Ounce oz.
12 Ounces " 1 Pound lb.
Unit Equivalents.
pwt. gr.
oz. 1 = 24
lb. 1 = 20 r= 480
1 = 12 = 240 = 5760
SOALE— ascending, 24, 20, 12 ; descending, 12, 20, 24.
Examples toe Pkactice.
1. How many grains in
141b. 10 oz. 18 pwt. 22 gr.?
OPERATION.
14 lb. 10 OZ. 18 pwt. 22 gr.
12
178 OZ.
20
3578 pwt. ;
3578 pwt. X 24 = 85894 gr.,
Ans.
2. How many pounds in
85894 grains ?
OPERATION.
24 ) 85894 gr.
20 ) 3578 pwt. + 22 gr.
12 )178 oz. + 18 pwt.
14 lb. + 10 oz.
Ans. 14 1b. 10 oz. 18 pwt
22 gr.
3. In 5 lb. 7 oz. 12 pwt. 9 gr., how many grains ?
4. In 32457 grains how many pounds ?
Define weight. Troy weight. Repeat the table. Give the scale.
154
R:feDUCTION
5. Eeduce 41760 grains to pounds. Ans. 7 lb. 3 oz.
6. A miner had 14 lb. 10 oz. 18 pwt. of gold dust ; what
was it worth at $.75 a pwt. ? Ans. $2683.50.
7. How many spoons, each weighing 2 oz. 15 pwt., can be
made from 5 lb. 6 oz. of silver ? Ans, 24.
8. A goldsmith manufactured 1 lb. 1 pwt. 16 grs. of gold
into rings, each weighing 4 pwt. 20 gr. ; he sold the rings for
$1.25 apiece ; what did he receive for them? Ans, $62.50.
II. Apothecaeies' Weight.
186. Apothecaries' Weight is used by Apothecaries
and physicians in compounding dry medicines ; but medi-
cines are bought and sold by Avoirdupois Weight.
Table.
20 Grains (gr. xx) make 1 Scruple sc. or ^.
3 Scruples (3 iij) ** 1 Dram dr. or 3.
8 Drams ( 3 viij) '* 1 Ounce oz. or ^ .
12 Ounces ( § xij) " 1 Pound lb. or a.
Unit EQurvALENTS.
dr.
>z. 1 =
1=8 =
sc.
1 =
3 =
24 =
20
60
480
Tb
1 = 13 = 96 = 288 = 5760
Scale — ascending, 20, 3, 8, 12 ; descending, 12, 8, 3, 20.
Examples for Practice.
1. How many gr. in 12 ft)
8? 33 13 15 gr.
OPERATION.
12ft)8! 3 3 I3l5gr.?
12
152 §
8
1219 3 ;
1219 3 X 3=36583 ;
36583 X 20=73175 gr.,^?i5.
gr.
2. How many ft) in 7317?*
OPERATION.
2|0 ) 7317|5 gr.
3 ) 36583 +15 gr.
8 )1219 3 +13
12 ) 152 g +3 3
12ft) + 8?
Ans. 12 ft) 8? 3 3 l3 15 gr.
Define apothecaries' weight. Repeat the table. Give the scale.
OMPOUND NUMBERS. 155
3. In 16 lb. 11 oz. 7 dr. 2 sc. 19 gr., how many grains ?
4. Eeduce 47ft) 6 § 4 3 to scruples. Ans. 13692 sc.
5. How many pounds of medicine would a physician use
in one year, or 365 days, if he averaged daily 5 prescriptions
of 20 grains each ? Ans. 6ft). 4 | 13.
Til. Avoirdupois Weight.
187. Avoirdupois Weight is used for all the ordinary
purposes of weighing.
Table.
16 Ounces make 1 Pound lb.
100 Pounds " 1 Hundred- weight cwt.
20 Cwt., or 2000 lbs., " 1 Ton T.
Unit Equivalents.
lb. oz.
cwt. 1 = 16
T. 1 = 100 = 1600
1 = 20 = 2000 = 32000
Scale— ascending, 16, 100, 20 ; descending, 20, 100, 16.
The long or gross ton, hundred-weight, and quarter were formerly in common
use ; but they are now seldom used except in estimating English goods at the U. S.
custom-houses, and in freighting and wholesaling coal at the mines.
Long Ton Table.
16 Ounces make 1 Pound, marked lb.
28 Pounds " 1 Quarter, " qr.
4 Quarters " 1 Hundred- weight, " cwt.
20 Cwt. = 2240 lb. « 1 Ton, " T.
The following denominations are also in use :
100 Pounds of Grain or Flour make 1 Cental.
100 " Dry Fish " 1 Quintal,
100 " Nails " 1 Keg.
196 " Flour . " 1 Barrel.
200 " Pork or Beef " 1 Barrel.
280 " Salt at N. Y. S. works " 1 Barrel.
56 " " " " ** 1 Bushel.
240 " Lime " 1 Cask.
32 ♦' Oats " 1 Bushel.
56 " Com " 1 Bushel.
60 " Wheat " 1 Bushel.
Define avoirdupois weight. Repeat the table. Give the scale. The
long ton table. What other denominations are in use ?
156 REDUCTION.
Examples for Practice.
1. In 25 T. 15 cwt. 70 lb,
how many pounds ?
OPERATION.
25 T. 15 cwt. 70 lb.
_20
515 cwt.
100
2. In 51570 pounds how
many tons ?
OPERATION.
100 ) 51570 lb.
2|0 )51|5 cwt. + 70 1b.
25 T. + 15 cwt.
Ans. 25 T. 15 cwt. 70 lb.
51570 lb., Ans.
3. Reduce 3 T. 14 cwt. 74 lb. 12 oz. to ounces.
4. Eeduce 119596 ounces to tons.
5. A tobacconist bought 3 T. 15 cwt. 20 lb. of tobacco, at
22 cents a pound ; what did it cost him ? Ans. $1654.40.
6. What will 115 pounds of hay cost, at $10 per ton?
7. A grocer bought 10 barrels of sugar, each weighing
2 cwt. 17 lb., at 6 cents a pound ; 5 barrels, each weighing
3 cwt. 6 lb., at 7-J cents a pound ; he sold the whole at an
average price of 8 cents a pound ; what was his whole gain?
Ans. $51.05.
8. Paid $360 for 2 tons of cheese, and retailed it for 12 J
cents a pound ; what was my whole gain ? A ns. $140.
9. If a person buy 10 T. 6 cwt. 3 qr. 14 lb. of Enghsh iron,
by the long ton weight, at 6 cents a pound, and sell the same
at $130 per short ton, what will he gain ? Ans. $115.85.
10. A farmer sold 2 loads of corn, weighing 2352 lbs. each,
at $.90 per bu. ; what did he receive ? A^is. $75.60.
11. How many pounds in 300 barrels of flour ?
12. A grocer bought 3 barrels of salt at $1.25 j^er barrel,
and retailed it at | of a cent per pound ; what did he gain ?
Atis. $2.55.
Standard of Weight.
188. In the year 1834 the U. S. government adopted a
uniform standard of weights and measures, for the use of the
custom-houses, and the other branches of business connected
with the general government.
COMPOUND NUMBERS. 157
189. The United States standard unit of weight is the
Troy pound of the mint, which is the same as the imperial
standard pound of Great Britain, and is determined as fol-
lows : A cubic inch of distilled water in a vacuum, Aveighed
by brass weights, also in a vacuum, at a temperature of 62°
Fahrenheit's thermometer, is equal to 252.458 grains, of
which the standard Troy pound contains 5760.
190. Tlie U. 8. Avoirdupois pound is determined from
the standard Troy pound, and contains 7000 Troy grains.
Hence, the Troy pound is 4ofo = ifi of an avoirdupois
pound. But the Troy ounce contains -S^f^ := 480 grains,
and the avoirdupois ounce -2^^ = 437.5 grains ; and an ounce
Troy is 480—437.5=42.5 grains greater than an ounce avoir-
dupois. The pound, ounce, and grain. Apothecaries' weight,
are the same as the like denominations in Troy weight, the only
diif erence in the two tables being in the divisions of the ounce.
191. COMPAEATIVE TABLE OF WEIGHTS.
Troy. Apothecaries'. Avoirdupois.
1 pound = 5760 grains, = 5760 grains, = 7000 grains.
1 ounce = 480 " =480 " = 437.5 "
175 pounds, = 175 pounds, = 144 pounds.
Examples foe Practice.
1. An apothecary bought 5 lb. 10 oz. of rhubarb, by
avoirdupois weight, at 50 cents an ounce, and retailed it at
12 cents a dram apothecaries' weight ; what did he gain ?
2. Change 424 drams apothecaries' weight to Troy weight.
3. Change 20 lb. 8 oz. 12 pwt. Troy weight to avoirdu-
pois weight. Ans. 17-^ lb.
4. Bought by avoirdupois weight 20 lb. of opium, at 40
cents an ounce, and sold the same by Troy weight at 50
cents an ounce ; what did he gain ? Ans. $17.83 J.
What is the U. S. standard of weight ? How obtained ? How is
the avoirdupois pound determined ? How is the apothecaries' pound
determined? What are the values of the denominations of Troy,
avoirdupois, and apothecaries* weight ?
158 EEDUCTION.
MEASUEES OF EXTENSION.
192. Extension has one or more of the dimensions —
length, breadth, and thickness.
A Line has only one dimension — length,
A Surface or Area has two dimensions — length and
Ireadth,
A Solid or Body has three dimensions — length, breadth,
and thickness.
I. Long Measure.
193. Long Measure, also called Linear Measure, is
used in measuring lines and distances.
Table.
13 Indies (in.) make 1 Foot ft.
3 Feet " 1 Yard yd.
5i Yd., or 16^ ft., " 1 Rod rd.
320 Rods ty- yy^t^^H** 1 Statute Mile ... .mi
Unit Equivalents.
ft. in.
yd. 1 = 12
rd. 1 = 3 = 36
mi. 1 = 51 = 16i = 198
1 = 320 = 1760 = 5280 = 63360
Scale— ascending, 12, 3, ^, 320 ; descending, 320, 5^, 3, 13.
The following denominations are also in use :
3 Barleycorns make 1 Inch, Used by shoemakers
, ^ - « w TT J (Used to measure the height or
4 Inches " 1 Hand, \ . . ^-u ^ ^A
( horses at the shoulder.
6 Feet " 1 Fathom, Used to measure depths at sea.
1.152| Statute Mi. " 1 Geog. mile, Used to measure distances at sea.
3 Geographic " *' 1 League = 3.458 st. mi.
60 '* " "I 1 D i^^ latitude on a meridian or of longi-
69.16 Statute " " ) ^^^^® i tude on the equator.
360 Degrees ** the Circumference of the Earth.
How many dimensions has extension ? Define a line. Surface or
area. A solid or body. Define long mensuro. What are the denom-
inations? The value of each. What other denominations are used?
COMPOUND KUMBERS.
159
1. For the purpose of measuring cloth and other goods sold by the yard, the yard
is divided into halves, fourths^ eigfUhs^ and sixteenths. The old table of cloth meas-
ure is practically obsolete.
2. The geographic mile is -^^ of -ji^ or ^ ^^qq of the circumference of the earth.
It is a small fraction more than 1.15 statute miles.
3. The length of a degree of latitude varies, being 68.72 miles at the equator, 68.9
to 69.05 miles in middle latitudes, and 69.30 to 69.34 miles in the polar regions. The
mean or average length is as stated in the table. A degree of longitude is greatest
at the equator, where it is 69.16 miles, and it gradually decreases toward the poles,
where it is 0.
Examples for Practice.
1. In 2 mi. 192 rd. 2 yd.
how many inches ?
2. In 164808 inches how
many miles ?
OPERATION.
operation.
2 mi.
192 rd. 2
yd.
12 ) 164808 in.
320
832 rd.
3 ) 13734 ft.
5J\4578yd.
2 / 2
416
4162
4578 yd.
3
11 ) 9156
32|0)83!2rd. + f yd.=2yd.
2 mi. +192 rd.
13734 ft.
12
Ans. 2 mi. 192 rd. 2 yd.
164808 in.,
Ans,
3. The diameter of the earth being 7912 miles, how many
inches is it ? Arts, 501304320 inches.
4. In 168474 feet how many miles ?
5. In 31 mi. 290 rd. 3 yd. how many feet ?
6. If the greatest depth of the Atlantic telegraphic cable
from l^ewfoundland to Ireland be 2500 fathoms, how many
miles is it ? Ans. 2 mi. 269 rd. 1| ft.
7. If this cable be 2200 miles in length, and cost 10 cents
a foot, what was its whole cost? Ans. $1161600.
160 REDUCTIONS'.
8. A pond of water measures 4 fathoms 3 feet 8 inches in
depth ; how many inches deep is it ? Ans, 332.
9. How many times will the driving wheels of a locomo-
tive turn round in going from Albany to Boston, a distance
of 200 miles, supposing the wheels to be 18 ft. 4 inches in
circumference ? Ans. 57600 times.
10. If a vessel sail 120 leagues in a day, how many statute
miles does she sail? Ans, 414.96 st. mi.
11. How many inches high is a horse that measures 14J
hands? Ans. 5S.
Surveyor's Long Measure.
194, A Gunter's Chain, used by land surveyors, is 4.
rods or 66 feet long, and consists of 100 links.
Table.
7.92 Inches (in.) make 1 Link L
25 Links " 1 Rod rd.
4 Rods, or 66 feet, « 1 Chain ch,
80 Chains " 1 Mile mi.
Unit Equivalents.
1. in.
rd. 1 = 79.3
ch. 1= 35= 198
mi 1 = 4 = 100 = 793
1 = 80 = 320 = 8000 = 63360
Scale— ascending, 7.92, 25, 4, 80 ; descending, 80, 4, 25, 7.93.
Rods are seldom used in chain measure, distances being taken in chains and hun ;
iredths.
Examples for Practice. 1
1. In 3 mi. 61 ch. 73 1. how many links?
2. Reduce 29173 1. to miles.
3. A certain field, enclosed by a board fence, is 17 ch.
31 1. long, and 12 ch. 87 1. wide ; how many feet long is the
fence which encloses it ? Ans. 3983.76 ft.
Repeat the ta])lp of surveyors* long measure. Give the scale.
COMPOUND K UMBERS.
161
12 in.
=
1
fc
1-!
«-"
II
S
f*
-
^
12 in. = 1 ft.
II. Square Measure.
195, A Square is a figure bounded by four equal sides,
and having four right angles.
1 square foot is a figure having
four sides of 1 ft. or 12 in. each, as
shown in the diagram. Its contents
are 12 x 12 = 144 square incheso
Hence,
The contents or area of a square,
or of any other figure having a uni-
form length and a uniform hreadthy
is found hy 7nultiplying the length ly the breadth. Thus, a
square foot is 12 in. long and 12 in. wide, and the contents
are 12^ 12 = 144 square inches. A board 20 in. long and
10 in. wide is a rectangle, containing 20 x 10 = 200 square
inches.
196. Square Measure is used in computing areas or
surfaces ; as of land, boards, painting, plastering, pa^dng, etc.
Table.
144 Square Inches (sq. in.) make 1 Square Foot sq. ft.
9 Square Feet " 1 Square Yard sq. yd.
30|^ Square Yards '* 1 Square Rod sq. rd.
160 Square Rods " 1 Acre A.
640 Acres ** 1 Square Mile sq. mi.
Unit Equivalents.
A.
sq.mi. 1 =
sq. rd.
1 =
160 =
eq. yd.
1 =
30i =
4840 =
sq. ft.
1 =
9 =
2721 =
43560 =
sq. m,
144
1296
39204
6272640
1 = 640 = 102400 = 3097600 = 27878400 = 4014489600
Scale— ascending, 144, 9, 30|, 160, 640 ; descending. 640, 160, 30^,
9,144.
Define a square. How is tlie area of a square or any rectangular
figure found ? For what is square measure used ? Repeat the table.
Give the scale.
162 REDUCTION.
Artificers estimate fheir work as follows :
By the square foot ; glazing and stone-cutting.
By the square yard ; painting, plastering, paving, ceiling, and paper-
hanging.
By the square of 100 feet ; flooring, partitioning, roofing, slating,
and tiling.
Brick-work is generally estimated by the 1000 bricks ; sometimes m
yubic feet.
1. In estimating the painting of moldiogs, cornices, etc., the measuring-line Is
earried into all the moldings and cornices.
2. A brick wall which is a brick and a half thick, is said to be of the standard
thickness. Five courses in the height are called a foot.
Examples for Practice.
1. In 10 A. 65 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in. how
many square inches ? •
OPERATION.
10 A. 65 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in.
160
1665 sq.rd.
30}
416}
49966
50382} sq. yd.
9_
453444} sq. ft.
144
36 = } sq. ft.
1813912 with 136 sq. in.
1813776
453444
65296108 sq. in., Ans,
% In 65296108 sq. in. how many acres ?
How do artisans estimate work ?
COMPOUND Is^UMBERS. 163
OPERATION.
144 ) 65296108 sq. in.
9 ) 453445 sq. ft. + 28 sq. in.
son 50382 sq. yd. + 7 sq. ft.
4 J 4
121 ) 201528 fourths sq. yd.
16|0U66i5 sq. rd. + ^=15| sq. yd.
10 A. + 65. sq. rd.
Ans. 10 A. 65 sq. rd. 15f sq. yd. 7 sq. ft. 28 sq. in.
ilO A. 65 sq. rd. 15 sq. yd. 7 sq. ft. 28 sq. in.
6 sq. ft. 108 sq. in.
Or 10 A. 65 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in.
Analysis. Dividing by the numbers in the ascending scale, and
arranging the remainders according to their order in a line below, we
find the'fequare yards a mixed number, 15|. But f of a sq. yd. = | of
9 sq. ft. = 6| sq. ft. ; and | of a sq. ft. = f of 144 sq, in. = 108 sq. in.
Therefore | sq. yd. = 6 sq. ft. 108 sq. in. ; and adding 108 sq. in. to
28 sq. in. we have 136 sq. in. , and 6 sq. ft. to 7 sq. ft. we have 13 sq.
ft. = 1 sq. yd. 4 sq. ft., and writing the 4 sq. ft. in the result, and
adding 1 sq. yd. to 15 sq. yd. we have for the reduced result, 10 A.
65 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in.
3. Reduce 87 A. 118 sq. rd. 7 sq. yd. 1 sq. ft. 100 sq. in.
to square inches. Ans. 550355068 sq. in.
5. Reduce 550355068 square inches to acres.
5. A field 100 rods long and 30 rods wide contains how
many acres ? Ans, 18 A. 120 sq. rd.
6. How many rods of fence will enclose a farm a mile
square ? Ans. 1280 rods.
7. How much additional fence wiU divide it into four
equal square fields ? Ans. 640 rods.
8. How many acres of land in Boston, at $1 a square
foot, ^vill $100000 purchase ?
Ans. 2 A. 47 sq. rd. 9 sq. yd. 3^ sq. ft.
9. How many yards of carpeting, 1 yd. wide, will be re-
quired to carpet a room 18^ ft. long and 16 ft. wide ?
Ans. 32|^ yards.
164 KEDUCTIOI^r.
10. What would be the cost of plastering a room 18 ft.
long, 16 J ft. wide, and 9 ft. high, at 22 cents a sq. yd. ?
Ans. $22.44.
11. What will be the expense of slating a roof 40 feet
long and each of the two sides 20 feet wide, at $10 per
square? Ans. $160.
Surveyors' Square Measure.
197. This measure is used by surveyors in computing the
Area or contents of land.
Table.
635 Square Links (sq. 1.) make 1 Pole P.
16 Poles " 1 Square Chain sq. ch.
10 Square Chains *' 1 Acre A.
640 Acres " 1 Square Mile sq, mi.
86 Square Miles (6 miles square) " 1 Township Tp.
Unit Equivalents.
P. sq. 1.
sq. ch. 1 = 625
A. 1 = 16 = 10000
sq. mi. 1 = 10 = 160 = 100000
Tp. 1 = 640 = 6400 = 103400 = 64000000
1 = 36 = 23040 = 330400 = 3686400 = 3304000000
Scale— ascending, 635, 16, 10, 640, 36 ; descending, 36, 640, 10, 16,
635.
1. A pqnare mile of land is also called a section.
2. Engineers commonly use a chain, or measuring tape 100 feet long, each foot
divided into tenths.
3. The contents of land are commonly estimated In square miles, acres, and hun-
dredths ; the denomination, rood, is no longer used.
Examples for Practice.
1 . How many poles in a township of land ?
2. Eeduce 3686400 P. to sq. mi.
3. In 94 A. 7 sq. ch. 12 P. 118 sq. 1. how many square
links ?
4. What will be the cost of a farm containing 4550000
square links, at $50 per acre ? Ans. 12275.
Repeat the table of surveyors' square measure. Give the scale.
compou:n"I) numbers.
165
III. Cubic Measure.
198. A Cube is a solid, or body,
ba\ing six equal square sides, or
faces. If each side of a cube be 1
yard, or 3 feet, 1 foot in thickness
of this cube will contain 3x3x1
=9 cubic feet, and the whole cube
will contain 3 x3 x 3 = 27 cubic
feet.
A solid, or body, may have the three dimensions all alike
or all different. A body 4 ft. long, 3 ft. wide, and 2 ft. thick
contains 4 x 3 x 2 = 24 cubic or solid feet. Hence,
The cubic or solid contents of a hody are found hy muUi-
plying the lengtlh breadth, and thickness together,
199, Cubic Measure, also called Solid Measure, is used
in estimatmg the contents of solids, or bodies ; as timber,
wood, stone, etc.
Table.
1728 Cubic Inches (cu. in.) make 1 Cubic Foot en. ft.
27 Cubic Feet
16 Cubic Feet
8 Cord Feet, or
128 Cubic Feet,
24^ Cubic Feet
1 Cubic Yard ... cu. yd.
1 Cord Foot cd. ft.
1 Cord of Wood .Cd.
j Perch of Stone )
\ or Masonry, J
,.Pch.
Scale — ascending, 1728, 27. The other numbers are not in a regular
scale, but are merely so many times 1 foot. The unit equivalents, be-
ing fractional, are consequently omitted.
1. A cubic yard of earth la called a load,
2. Railroad and transportation companies estimate light freight by the ^pace it
occupies in cubic feet, and heavy freight by weight.
3. A pile of wood 8 feet long, 4 feet wide, and 4 feet high, contains 1 cord ; and a
cord foot is 1 foot in length of such a pile.
4. A perch of stone or of masonry is 16 j feet long, 1^ feet wide, and 1 foot high.
Define a cube. How are the contents of a cube or rectangular solid
found ? For what is cubic measure used ? Repeat the table. Give
the scale. How is railroad freight estimated ? What is understood
by a cord foot ? By a perch of stone or masonrj' ?
166 REDUCTIOlf.
6. Joiners, bricklayers, and masons make no allowance for windows, doors, etc.
Bricklayers and masons, in estimating their work by cubic measure, make no allow-
ance for the corners of the walls of houses, cellars, etc., but estimate their worK by
the girt, that is, the entire length of the wall on the outside.
6. Engineers, in making estimates for excavations and embankments, take the
dimensions with a line or measure divided into feet and decimals of a foot. Tho
estimates are made in feet and decimals, and the results are reduced to cubic yards.
Examples for Practice.
1. In 125 cu. ft. 840 cu. in. how many cu. in. ? Ans, 216840.
2. Eeduce 5224 cubic feet to cords. A7is. 40|f.
3. In a solid, 3 ft. 2 in. long, 2 ft. 2 in. wide, and 1 ft.
8 in. thick, how many cubic inches? Ans. 19760.
4. How many small cubes, 1 inch on each edge, can be
sawed from a cube 6 feet on each edge, allowing no waste
for sawing? Ans, 37324:8.
6. In a pile of wood 60 feet long, 20 feet wide, and 15 feet
high, how many cords ? Ans. 140|.
6. How many cubic feet in a load of wood 10 feet long, 3;^
feet wide, and 3^ feet high ? Ans. 113f cu. ft.
7. If a load of wood be 12 feet long and 3 feet wide, how
high must it be to make a cord ? Ans. 3^ ft. high.
8. The gray limestone of Central New York weighs 175
pounds a cubic foot. What is the weight of one solid yard ?
Ans. 2 T. 7 cwt. 25 lb.
9. A cellar wall, 32 ft. by 24 ft, is 6 ft. high and 1 ^ ft. thick.
What did it cost at $1.25 a perch? A}is. $50,909 + .
10. What did it cost to dig the same cellar, at 15 cents a
cubic yard? Ans, $25.60.
11. My sleeping room is 10 ft. long, 9 ft. wide, and 8 ft.
high. If I breathe 10 cu. ft. of air in one minute, in how
long a time will I breathe as much air as the room contains ?
Ans. 72 min.
12. In a school-room 30 ft. long, 20 ft. wide, and 10 ft. high,
with 50 persons breathing each 10 cu. ft. of air in one minute,
in how long a time will they breathe as much as the room
contains? Ans. 12 min.
How are excavations and embankments measured ?
COMPOUND NUMBERS. 167
MEASURES OF CAPACITY.
I. Liquid Measuee.
200. Liquid Measure, also called Wine Measure, is
used in measuring liquids ; as liquors, molasses, water, etc.
Table.
4 Gills (gl.) make 1 Pint pt.
3 Pints " 1 Quart qt.
4 Quarts " 1 GaUon gaL
8H Gallons " 1 Barrel bbl.
3 Barrels, or 63 gal. " 1 Hogshead hlid.
Unit Equivalents.
pt. gL
qt. 1=4
gal. 1=2= 8
bbl. 1 = 4= 8= 82
hhd. 1 = 31i = 126 = 252 = 1008
1 = 2 = 63 = 252 = 504 = 2016
Scale— ascending, 4, 2, 4, 31|, 3 ; descending, 2, 31|^, 4, 2, 4.
The following denominations are also in use :
36 Gallons make 1 Barrel of beer.
54 *' oi-t^ Barrels " 1 Hogshead " "
42 " *• 1 Tierce.
2 Hogsheads, or 130 gallons, " 1 Pipe or Butt
2 Pipes, or 4 Hogsheads, " 1 Tun.
1. The denominations, barrel and hogshead, are used in estimating the capacity
of cisterns, reservoirs, vats, etc.
2. The tierce, hogshead, pipe, butt, and tun are the names of casks, and do not
express any fixed or definite measures. They are usually gauged, and have their
capacities in gallons marked on them.
3. Ale or beer measure, formerly used in measuring beer, ale, and milk, is almost
entirely discarded.
What is liquid measure ? Repeat the table. Give the scale. What
other denominations are sometimes used ? How are the capacities of
cisterns, reservoirs, etc., reckoned ? Of large casks ?
168
REDUCTION.
Examples for Practice.
1. In 2 hhd. 1 bar. 30 gal 2
qt. 1 pt. 3 gi. how many gills ?
OPERATION.
2 hhd. 1 bar. 30 gal. 2 qt.
_2 [1 pt. 3 gi.
5bbl.
31i
185
187i gal-
4
752 qt.
2
1505 pt.
4
2. In 6023 gi. how many
hhds. ?
OPERATION.
4 ) 6023 gi.
2 ) 1505 pt.+3gi.
4) 752qt. + lpt.
31i
2
188 gal.
2
63 ) 376
[gal.
= 30J
2)_5bbl.-f^gal.
2 hhd. + 1 bar.
Ans, 2 hhd. 1 bar. 30J gal.
1 pt. 3 gi.
But -J gal. = 2 qt., making
the Ans. 2 hhd. 1 bar. 30 gal.
6023 gi., Ans. 2 qt. 1 pt. 3 gi.
3. Eeduce 3 hogsheads to gills.
" 4. Reduce 6048 gills to hogsheads.
6. In 13 hhd. 15 gal. 1 qt. how many pints ?
6. In 6674 pints how many hogsheads ?
7. What wiU be the cost of a hogshead of wine, at 6 cents
a gill? Ans. $120.96.
8. A grocer bought 10 barrels of cider, at $2 a barrel ;
after converting it into vinegar, he retailed it all at 5 cents
a quart ; what was his whole gain ? Ans, $43.
9. At 6 cents a pint, how much molasses can be bought
for $3.84? A71S, 8 gal.
10. How many demijohns, that will contain 2 gal. 2 qt. 1 pt
each, can be fiUed from a hogshead of wine ? Ans. 24.
II. Dry Measure.
201. Dry Measure is used in measuring articles not
liquid, as grain, fruit, salt, roots, allies, etc.
What is dry measiL'e ?
COMPOUND NUMBERS. 169
Table.
3 Pints (pt.) make 1 Quart qt.
8 Quarts " 1 Peck pk.
4 Pecks " 1 Bushel bu.
Unit Equivalents.
qt, pt.
pk. 1=3
bu. 1 = 8 = 16
1 = 4 = 32 = 64
Scale— ascending, 3, 8, 4 ; descending, 4, 8, 2.
In England, 8 bu. of 70 lbs. each are called a quarter, used in measaring grain.
The weight of the English quarter is ^ of a long ton.
Examples for Practice.
1. In 49 bu. 3 pk. 7 qt. 1 pt. how many pints ?
2. In 3199 pt. how many bushels ?
3. Reduce 1 bu. 1 pk. 1 qt. 1 pt. to pints.
4. Reduce 83 pints to bushels.
5. An innkeeper bought a load of 50 bushels of oats at
65 cents a bushel, and retailed them at 25 cents a peck ;
how much did he make on the load? Ans, $17.50.
Stan^dard of Exten^siok.
202. TJie U. 8. standard unit of measures of extension,
whether linear, superficial, or solid, is the yard of 3 feet, or
36 inches, and is the same as the imperial standard yard of
Great Britain. It is determined as follows : The rod of a
pendulum vibrating seconds of mean time, in the latitude
Oi London, in a vacuum, at the level of the sea, is divided
into 391393 equal parts, and 360000 of these parts are 36
inches, or 1 standard yard. Hence, such a pendulum rod
is 39.1393 inches long, and the standard yard is |§;§^§ of
the length of the pendulum rod.
203. Hie U. S. standard unit of liquid measure is the
old English wine gallon, of 231 cubic inches, which is equal
to 8.33888 pounds avoirdupois of distilled water at its maxi-
mum density, that is, at the temperature of 39.83° Fahren-
heit, the barometer at 30 inches.
Repeat the table. What is a quarter? Wliat is the TJ. S. standard
unit of measurement of extension ? How is it determined ? What is
the U. S. standard unit of liquid measure ?
n- P. 8
170 REDUCTION^.
304. The U. 8. standard unit of dry measure is the Brit-
fsli Winchester bushel, which is 18^ inches in diameter and
8 inches deep, and contains 2150.42 cubic inches, equal to
77.6274 pounds avoirdupois of distilled water, at its maximum
density. A gallon, dry measure, contains 268.8 cubic inches.
1. The wine and dry measures of the same denomination are of different capaci-
ties. The exact and the relative size of each may he readily seen hy the following
205. COMPARATIVB TABLE OP MEASURES OP CAPACITY.
Cu. in. in On. in. in Cu. in. in Cu. in. in
one gallon. one quart. one pint one gill.
Wine measure, 231 57| 28| 1^
Dry measure, Q pk.,) 268^ 67^ 33f 8f
2. The beer gallon of 282 inches is retained in use only by custom. A bushel is
commonly estimated at 2150.4 cubic iaches.
Examples for Practice.
1. A fruit dealer bought a bushel of strawberries, dry
measure, and sold them by wine measure ; how many quarts
did he gain ? Ans, 5^ quarts.
2. A grocer bought 40 quarts of milk by beer measure,
and sold it by wine measure ; how many quarts did he gain ?
Ans. 844" quarts.
3. A bushel, or 32 quarts, dry measure, contains how
many more cubic inches than 32 quarts wine measure ?
A71S. 302| cu. in.
Time.
306. Time is used in measuring periods of duration, as
years, days, minutes.
Table.
60 Seconds (sec.) make 1 Minute min.
GO Minutes " 1 Hour h.
24Hours ** 1 Day da.
7 Days " 1 Week wk.
365 Days ** 1 Common Year yr.
866 Days '* 1 Leap Year yr.
13 Calendar Months ** 1 Year yr.
100 Years ** 1 Century C.
What is the U. S. standard unit of dry measure? How is it ob-
toined? What is the relative size of the wine and the dry gallon t
What is the size of a beer gallon ? What is time ? Repeat the table.
COMPOUNDS UMBERS. 171
Unit Equivat<f,nt8.
min.
sec
h. 1 =
60
da. 1 = 60 =
8600
wk. 1 = 24 = 1440 =
86400
1 = 7 = 168 = 10080 =
604800
mo. ( 865 = 8760 = 525600 =
12 = "j 366 = 8784 = 527040 =
31536000
31622400
Season.
Names,
Abbreviations.
Winter,
( Istn
\ 2d
lonth
, January,
Jan.
February,
Feb.
( 3d
March,
Mar.
Spring,
■j 4th
April,
Apr.
( 5th
May,
May
( 6th
June,
Jun.
Summer,
\ 7th
July,
July
( 8th
August,
Aug.
9th
September,
Sept.
Autumn,
10th
(nth
October,
Oct.
November,
Nov.
Winter,
12th
December,
Dec.
Scale— ascending, 60, 60, 24, 7 ; descending, 7, 24, 60, 60.
The calendar year is divided as follows ;
No. of days.
31
28 or 29
31
30
31
30
31
31
30
31
30
31
365 or 366
1. The exact length of a solar year is 365 da. 5 h. 48 min. 46 sec. ; but for conve-
nience it is reclioned 11 min. 14 sec. more than this, or 365 da. 6h. = 365j da. This \
day in 4 years makes one day, which, every fourth, bissextile, or leap year, is added
to the shortest month, giving it 29 days. The leap years are exactly divisible by 4,
as 1856, 1860, 1864. The number of days in each calendar month may be easily re-
membered by committing the following lines :
'' Thirty days hath September,
April, JunCj and November ;
All the vest have thirty-one.
Save February, which alone
Hath twenty-eight ; and one day more
We add to it one year in four."
2, In most business transactions 30 days are called 1 month.
Examples fok Pkactice.
1. Reduce 365 da^ 5 h. 48 min. 46 sec. to seconds.
2. Reduce 31556926 seconds to days.
Give the scale. What is the length of each of the calendar months ?
What is the exact length of a solar year? Explain the use of bissextile
or leap year. What is the length of a month in business transactions t
172 REDUCTION.
3. In 5 wk. 1 da. 1 h. 1 min. 1 sec. how many seconds ?
4. In 3114061 seconds how many weeks ?
5. How many times does a clock pendulum, 3 ft. 3 in.
long, beating seconds, vibrate in one day? Ans. 86400.
6. If a man take 1 step a yard long in a second, in how
long a time will he walk 10 miles? Ans. 4 h. 53 min. 20 sec.
7. In a lunar month of 29 da. 12 h. 44 min, 3 sec. how
many seconds? Ans. 2551443.
8. How much time will a person gain in 40 years, by rising
45 minutes earlier every day? Ans. 456 da. 13 h. 30 min.
Circular Measure.
307. Circular Measure, or Circular Motion, is used
principally in surveying, navigation, astronomy, and geogra-
phy, for reckoning latitude and longitude, determining loca-
tions of places and vessels, and computing difference of time.
Every circle, great or small, is divisible into the same num-
ber of equal parts, as quarters, called quadrants, twelfths,
called signs, 360ths, called degrees, etc. Consequently the
parts of different circles, although having the same names,
are of different lengths.
Table.
60 Seconds (' ) make 1 Minute '.
60 Minutes " 1 Degree °.
30 Degrees " 1 Sign S.
12 Signs, or 360°, " 1 Circle C.
Unit Equivalents.
1= 60
S. 1 = 60 = 8600
C. 1 = 30 = 1800 = 108000
1 = 12 = 380 = 21600 = 1296000
Scale— ascending, 60, 60, 30, 13 ; descending, 12, 80, 60, 60.
1. Minutes of the earth's circumference are called geographir or nautical miles.
2. The denomination, signs, is confined exclusively to Astronomy.
Define circular measure. How are circles divided ? Repeat the
table. Give the scale. What is a geographic mile ? What is a sign ?
COMPOUN^D NUMBERS.
173
3. Degrees are not strictly divisions of a circle, but of the space about a point in
any plane.
4. 90° make a quadrant, or right angle, and 60° a sextant, or | of a circle.
Examples for Practice.
1. Eediice 10 S. 10° 10' 10" to seconds.
2. Eeduce 1116610" to signs.
3. How many degrees in 11400 geographic vt nautical
miles? ^ws. 190^
4. If 1 degree of the earth's circumference is 69^ statute
miles, how many statue miles in 11400 geographic miles, or
190 degrees ? Ans. 13148.
5. How many minutes, or nautical miles, in the circum'
ference of the earth ? Ans. 21600' or mi.
6. A ship during 4 days' storm at sea changed her longi-
tude 397 geographic miles; how many degrees and minutes
did she change ?
Ans. 6° 37'.
308. Iif CouinTting.
13 Units or Things make
! 1 Dozen.
12 Dozen
1 Gross.
12 Gross
1 Great Gross.
20 Units
1 Score.
209. Paper.
24 Sheets.. make
....1 Quire.
20 Quires "
1 Ream.
2 Reams
1 Bundle.
5 Bundles **
IBale.
210. Books.
The terms folio, quarto, octavo,
duodecimo, etc., indicate
the number of leaves into which a i
sheet of paper is folded.
A sheet folded in 2 leaves is called a Folio.
A sheet folded in 4 leaves **
a Quarto, or 4to.
A sheet folded in 8 leaves "
an Octavo, or 8vo.
A sheet folded in 12 leaves **
a 12mo.
A sheet folded in 16 leaves *'
a 16mo.
A sheet folded in 18 leaves "
an 18mo.
A sheet folded in 24 leaves
a24mo.
A sheet folded in 32 leaves "
a 82mo.
What is a degree? Repeat the table for counting. For reckoning
paper. For indicating the size of books.
174 REDUCTIOiJ^.
Examples for Practice.
1. If in Birmingham, England, 150 million Gillott pens are
manufactured annually, how many great gross will they
make ? Ans. 86805 great gross 6 gross 8 dozen.
2. In 100000 sheets of paper, how many bales ?
A71S. 20 bales 4 bundles 6 quires 16 sheets.
3. What is the age of a man 4 score and 10 years old ?
4. How many printed pages, 2 pages to each leaf, will
there be in an octavo book, having 8 fully printed sheets ?
Ans. 128 sheets.
5. How large a book will ten 32mo sheets make, if every
page be printed ? Ans. 640 pages.
Promiscuous Examples in Eeduction".
1. How many suits of clothes, each containing 6 yd. Bj qr.,
can be cut from 333 yards of cloth ? Ans. 48.
2. A man bought a gold chain, weighing 1 oz. 15 pwt., at
seven dimes a pennyweight ; what did it cost? Ans. $24.50.
3. A physician, having 2 ft) 3 ^ 5 3 1 3 10 gr. of medicine,
dealt it out in prescriptions averaging 15 grains each ; how
many prescriptions did it make ? A7is. 886.
4. A man bought IT. 11 cwt. 12 lbs. of hay, at IJ cents
a pound ; what did it cost ? Ans. $38.90.
5. What will be the cost of a load of oats weighing 1456
pounds, at 37J- cents per bushel? Atis, $17.0625.
6. If one bushel of wheat will make 45 pounds of flour, how
many barrels will 1000 bushels make ? Ans. 229 bbl. 116 lb.
7. A load of wheat weighing 2430 pounds is worth how
much, at $1.20 a bushel? Ans. $48.60.
8. Paid $12.50 for a barrel of beef ; how much was that
per pound? Ans. 6 J cents.
9. If a silver dollar measure one inch in diameter, how
many dollars, laid side by side on the equator, would reach
round the earth ? A?is. 1577511936.
10. In 10 mi. 7 ch. 4 rd. 20 1., how many links ?
Ans. 808:30 links.
DEIfOMINATE FRACTIONS. 175
11. What is the value of a city lot, 25 feet wide and 100
feet long, if every square inch is worth one cent ? Ans. $3 GOO.
12. How many cords of wood can be piled in a shed 50 ft.
long, 25 ft. wide, and 10 ft. high ? Ans. 97 Cd. 5 cd. ft. 4 cu. ft.
13. A cistern 10 feet square and 10 feet deep, will hold
how many hogsheads of water ? Ans. 118 hhd. 464-^ gal.
14. A bin 8 feet long, 5 feet wide, and 4^ feet high, will
hold how many bushels of grain ? Ans. 144^^ bu.
15. How many seconds less in every Autumn than in
either Spring or Summer ? Ans. 86400 sec.
16. If a person could travel at the rate of a second of dis-
tance in a second of time, how much time would he require
to travel round the earth ? Ans. 15 days.
17. How many yards of carpeting, 1 yd. wide, will be re-
quired to carpet a room 20 ft. long and 18 ft. wide ?
Ans. 40.
18. A printer calls for 4 reams 10 quires and 10 sheets of
paper to print a book ; how many sheets does he call for ?
Ans. 2170.
19. How many times will a wheel, 16 ft. 6 in. in circum-
ference, turn round in running 42 miles ? Ans. 13440.
20. How many days, working 10 hours a day, will it re-
quire for a person to count 110000, at the rate of one cent
each second ? Ans. 27 da. 7h. 46 min. 40 sec.
21. A town, 6 miles long and 4^ miles wide, is equal to
how many farms of 80 acres each? Ans. 216.
22. At $21.75 per rod, what will be the cost of grading
10 mi. 176 rds. of road ? Ans. $73428.
REDUCTION OF DEISTOMINATE FRACTIONS.
Case I.
211. To reduce a denominate fraction from a
greater to a less unit.
1. Reduce -^ of a bushel to the fraction of a piiit.
Case I is what 1
176 REDUCTION.
OPERATION. Analysis. To reduce bushels
bn. pt. to pints, we must multiply by 4,
^Xfxfxf = |, Ans. 8, and 2, tbe numbers in the
Or, scale. And since the given num-
5
1
4 = ^ pt., Ans.
ber is a fraction of a bushel, we
indicate the process as in multi-
plication of fractions, and after
canceling, obtain |.
iluLE. Multiply the fraction of the higher denomination
hy the numbers in the scale successively, between the given
and the required denominations.
Cancellation may be applied wherever practicable.
Examples foe Practice.
2. Reduce xoW of a £ to the fraction of a penny.
Ans. -^ d.
3. Reduce xihr of a week to the fraction of a minute.
Ans. -^ min.
4. What part of a gill is ^^5^3^ of a hogshead ? Ans. i gi.
5. What fraction of a grain is -^q of an ounce? Ans. ^ gr.
6. Reduce ^Q^)i^ft^) of a mile to the fraction of an inch ?
Ans. -gS^in.
7. Reduce | of ^ of 2 pounds to the fraction of an ounce
Troy. Ans. f oz.
8. Reduce -^ of a hogshead to the fraction of a pint.
Ans. fl pt.
9. Reduce y^^ of an acre to the fraction of a rod.
Ans. i rd.
Case II.
212. To reduce a denominate fraction from a less
to a greater unit.
1. Reduce f of a pint to the fraction of a bushel.
Give explanation. Rule. Case IT ia what ?
DENOMINATE FRACTIONS. 177
OPERATION. Analysis. To reduce pints
^1111 to bushels, we must divide
^ ^'2~^~g~^'4"^^ 30^ ^^' by 2, 8, and 4, the numbers
of the scale. And since the
given number of pints is a
fraction, we indicate the pro-
cess, as in division of fractions,
and canceling, obtain -^-q.
Or, 5
2
8
4
80 1 = -gV bu., Ans,
Rule. Divide the fraction of the lower denomination ly
the numbers in the scale, successively, hetioee^i the given and
the required denomination.
The operation will frequently be shortened by cancellation.
Examples for Practice.
2. What part of a rod is ^ of a foot ? Ans. -^ rd.
3. What part of a pound is f of a dram ? Ans. y^^ lb.
4. Reduce i^ of a cent to the fraction of an eagle.
^ns. ^-^ E.
6. A hand is ^ of a foot ; what fraction is that of a mile ?
6. Reduce | of 2 pwt. to the fraction of a pound.
Ans. ^j^^ lb.
7. How much less is f of a pint than -J- of a hogshead ?
Ans. IJf hhd.
8. In I of an inch what fraction of a mile ?
9. f of an ounce Troy is f of what fraction of 2 pounds ?
10. f of an ounce is ^ of what fraction of 2 pounds Troy ?
Case III.
213. To reduce a denominate fraction to integers
of lower denominations.
1. What is the value of | of a hogshead of wine ?
Oive explanation. Rale. Case III is wliat ?
8*
178 REDUCTION.
OPERATION.
I hhd. X 63=2^ gal. = 39| gal.
I gal. X 4 = J^ qt. = If qt. ; I qt. X 2 = I pt. = 1 pt.
Ans. 39 gal. 1 qt. 1 pt.
Analysis, f hhd. = f of 63 gal., or 39 1 gal. ; and f gal. = f of
4 qt., or 1| qt. ; and | qt. = f of 2 pt., or 1 pt.
EuLE. I. Multiply the fraction hy that numlerin the scale
ivhich will reduce it to the next lower denomination, and if
the result he an improper fraction, reduce it to a whole or
mixed number.
II. Proceed with the fractional part, if any, as before,
until reduced to the denominations required.
III. The units of the several denominations, arranged in
their order, will be the required result.
Examples for Practice.
2. Reduce ^ of a month to lower denominations.
Ans. 17 da. 3 h. 25 min. 42^ sec.
3. What is the value of f of a £? Ans. 8s. 6d. 3f far.
4. What is the value of -f of a bushel ?
6. Reduce 4 of 15 cwt. to its equivalent value.
Ans. 12 cwt. 85 lbs. llf oz.
6. Reduce f of f of a pound avoirdupois to integers.
Ans. 4|4 oz.
7. What is the value of \ of an acre ? Ans. 133^ P.
8. Reduce 4| of a day to its value in integers.
Ans. 16 h. 36 min. 55^*^ sec.
9. What is the value of f of a pound Troy ?
10. What is the value of -j- of 5J tons ?
Ans. 4 T. 5 cwt. 55J lb.
11. What is the value of | of 3-| a^jres ? Ans.l A. 60 P.
Case IV.
214. To reduce a compoiind nTimber to a fraction
of a higher denomination.
1. What part of a week is 5 da. 14 h. 24 min. ?
Give explanation. Rule. Case VI is what ?
DENOMINATE FRACTIONS. 179
OPERATION. Analysis. To find
5 da. 14 h. 24 min. = 8064 min. what part one compound
1 Wk. = 10080 min. number is of another,
80 64 — 4 ^k Ans they must be reduced to
TO 08 6 V • ^jjg same denomination.
In 5 da. 14 h. 24 min. there are 8064 minutes, and in 1 week there are
10080 minutes. Since 1 minute is yjj^^^ of a week, 8064 minutes is
im% = I of a week.
Rule. Reduce the given number to its lowest denomination
for the numerator, and a unit of the required denomination
to the same denomination for the denominator of the required
fraction.
If the given immber contain a fraction, the denominator of this fraction must be
regarded as the lowest denomination.
Examples for Practice.
2. What part of a mi. is 266 rd. 3 yd. 2 ft. ? Ans. I mi.
3. What fraction of a £ is 13s. 7d. 3 far. ?
4. Reduce 10 oz. 10 pwt. 10 gr. to the fraction of a pound
Troy ? Ans. \^ lb.
5. Reduce 2 cd. ft. 8 cu. ft. to the fraction of a cord.
Ans. -f^ Cd.
6. Reduce 1 bbl. 1 gal. 1 qt. 1 pt. 1 gi. to the fraction of
a hogshead. Ans, |-^ hhd.
7. What part of 2 rods is 4 yards 1 J feet ? Ans. -^.
8. Reduce l-f pecks to the fraction of a bushel. Ans. f bu.
9. What part of 9 feet square are 9 square feet Y
10. From a piece of cloth containing 8 yd. 3 qr. a tailor
cut 2 yd. 2 qr. ; what part of the whole piece did he take ?
Ans. f.
Case V.
215. To reduce a denominate decimal to integers
of lower denominations.
1. Reduce .78125 of a pound Troy to integers of lower
denominations.
Give explanation. Rule. Case V is what ?
9.37500 oz.
20
180 REDUCTION'.
OPERATION. Analysis. First multiply by 1 2
.78125 lb. to reduce the given number from
12 pounds to ounces, and the result is
9 ounces and the decimal .375 of an
oz. Then multiply this decimal by
20 to reduce it to pennyweights,
7.50000 pwt ^^ S®* "^ P^- ^^^ .5 of a pwt.
nA This last decimal we multiply by
24, to reduce it to grains, and the
12.0000 gr. result is 12 gr. Hence the answer
9 OZ. 7 pwt. 12 gr., Ans, is 9 oz. 7 pwt. 12 gr.
EuLE. I. Miiltiply the given decimal hy that number in the
scale lohich loill reduce it to the next lower denomination, and
point off as in multiplication of decimals.
II. Proceed with the decimal part of the product in the
same manner until reduced to the required denominations.
The integers at the left will he the answer required.
Examples for Practice.
2. What is the value of .217° ? Ans, 13' 1.2".
3. What is the value of .659 of a week?
Ans. 4 da. 14 h. 42 min. 43.2 sec.
4. Eeduce .578125 of a bushel to integers of lower denom-
(nations. Ans. 2 pk. 2 qt. 1 pt.
5. Reduce .125 bbl. to integers of lower denominations.
Ans. 3 gal. 3 qt. 1 pt. 2 gi.
6. What is the value of £.628125 ?
7. What is the value of .22 of a hogshead of molasses ?
Ans. 13 gal. 3 qts. 3.52 gi.
\ What is the value of .67 of a league ?
Ans. 2 mi. 101 rd. 6 ft. 6.24 + in.
^. What is the value of .42857 of a month ?
Ans. 12 da. 20 h. 34 min. 13^^ sec.
IC. What is the value of .78875 of a long ton ?
Ans. 15 cwt. 3 qr. 2 lb. 12.8 oz.
Give explanation. Rule.
DENOMINATE FRACTIONS. 181
11. What is the value of 5.88125 acres ? A^is, 5 A. 141 P.
12. Eeduce .0055 T. to pounds. A7is, 11 lb.
13. Reduce .034375 of a bundle of paper to its value in
lower denominations. Ans. 1 quire 9 sheets.
Case VI.
216. To reduce a compound number to a decimal
of a higher denomination.
1. Reduce 3 pk. 2 qt. to the decimal of a busheL
OPERATION. Analysis. Since 8 quarts make
2.00 qt. 1 peck, and 4 pecks 1 bushel, there
o .^Krvrj 1^ "^1^ ^6 I as many pecks as quarts
— ^ ' (183), and I as many bushels as
.8125 bu., Ans. pecks.
Or, 3 pk. 2 qt. = 26 qt. ^"^ ^'^ ^^^ ^^^^^^ ^ P^- ^ qt- to
^\^ Qo f *^^ fraction of a bushel (as in 214),
^ >< ^^^ ^^ h&ve f I of a bushel, which,
|-| = .8125 bu., Ans. reduced to a decimal, equals .8125.
Rule. Divide the lowest denomination given by that num-
ber in the scale which will reduce it to the next higher, and
annex the quotient as a decimal to that higher. Proceed in
the same manner until the whole is reduced to the denomina-
tion required. Or,
Reduce the given number to a fraction of the required de-
nomination, and reduce this fraction to a decimal.
Examples foe Pkactice.
2. Reduce 3 qt. 1 pt. 1 gi. to the decimal of a gallon.
Ans. .90625 gal.
3. Reduce 10 oz. 13 pwt. 9 gr. to the decimal of a pound
Troy. Ans. .8890625 lb.
4. Reduce 1.2 pints to the decimal of a hogshead.
^?25. .00238+ hhd.
5. What part of a bushel is 3 pk. 1.12 qt. ? Ans. .785 bu.
Case VI is what? Give explanations. Rule.
182 ADDITION.
6. What part of an acre is 132.56 P. ?
7. Reduce 17 yd. 1 ft. 6 in. to the decimal of a mile.
Ans. .00994318+ mi.
8. Reduce .32 of a pint to the decimal of a bushel.
Ans. .005 bu.
9. Reduce 4-J feet to the decimal of a fathom.
Ans. .8125 fathom,
10. Reduce 150 sheets of paper to the decimal of a ream.
Ans. .3125 Rm.
11. Reduce 47.04 lb. of flour to the decimal of a barrel.
12. Reduce .33 of a foot to the decimal of a mile.
13. Reduce 5 h. 36 min. 57^^ sec. to the decimal of a day.
ADDITION.
217. 1. A miner sold at one time 10 lb. 4 oz. 16 pwt.
8 gr. of gold; at another time, 2 lb. 9 oz. 3 pwj.; at another,
11 oz. 20 gr. ; and at another, 25 lb. 16 pwt. 23 gr. ; how
much did he sell in all ?
Analysis. Arranging the num-
bers in columns, placing units of the
same denomination under each other,
we first add the units in the right
hand column, or lowest denomina-
tion, and find the amount to be 51
grains, which is equal to 2 pwt.
Ans. Sd 1 17 3 3 gr. We write the 3 gr. under the
column of grains, and add the 2 pwt.
to the column of pwt. We find the amount of the second column to
be 37 pwt., which is equal to 1 oz. 17 pwt. Writing the 17 pwt. under
the column of pwt., we add the 1 oz. to the next column. Adding this
column in the same manner as the preceding ones, we find the amount
to be 25 oz,, equal to 2 lb. 1 oz. Placing the 1 oz. under the column
of oz., we add the 2 lb. to the column of lb. Adding the last column,
we find the amount to bo 39 lb.
OPERATION.
lb. oz. pwt.
gr.
10 4 16
8
2 9 3
11
20
25 16
23
What is addition of compound numbers ? Give exi)lanation.
COMPOUND liTUMBERS. 183
EuLE. I. Write the numbers so that those of the same unit
value will stand in the same column.
II. Beginning at the right hand, add each denomination as
in simple numbers, carrying to each succeeding denomination
one for as many units as it tahes of the denomination added,
to mahe one of the next higher denomination.
Examples for Practice.
(2.)
<
:3.)
£. s. d.
fe
I
3 3 gr.
48 13 8
12
8
7 2 15
51 6 4
10
4 1 10
67 11 3
15
2 1 19
76 18 10
11
6 12
244 10 1
13
4
4 2
(4.)
(5.)
T. cwt. lb. i
DZ.
bn.
pk. qt. pt.
4 7 18
4
1
3 7 1
15 98
15
3
2 2
3 9 10
6
16 1
1 15
17
5 1
9 12 42
9
45
2 4
6. What is the sum of 4 mi. 150 rd. 2 yd. 1 ft. 10 in.,
5 mi. 258 rd. 1 yd. 2 ft. 6 in., 10 mi. 185 rd. 2 yd. 2 ft.
11 in., and 268 rd. 4 yd. 2 ft. 1 in. ?
7. Find the sum of 197 sq. yd. 4 sq. ft. 104| sq. in., 122
sq. yd. 2 sq. ft. 27} sq. in., 5 sq. yd. 8 sq. ft. 2| sq. in., and
237 sq. yd. 7 sq. ft. 128-^ sq. in. ?
Ans, 563 sq. yd. 4 sq. ft. 118.825 sq. in.
When common fractions occur, they should be reduced to a common denomina-
tor, to decimals, or to integers of a lower denomination, and added according to the
usual method.
Give tbe Rule.
184
ADDITIOI?^.
A.
26
P.
148
(8.)
sq yd. 1
25
sq. ft. sq. In.
8 125
19
118
30
7 150
456
100
16
6 98
603
48
5 85
■-Hi)
(i) = 72
503
48
13
1 13
mi.
1
(9.)
rd. yd.
310 4
ft. in.
2 11
(10.)
hhd. gal. qt pt
27 65 3 2
3
160 2
1 10
112 60 2 3
10
305 1
2 11
60 29 1
16
136 3i
1 8
421 2 3
14 39 1 2
bu.
23
(11.)
pk. qt.
3 7
pt.
1
yr.
25
(12.)
da. hr. min. sec.
300 19 54 35
34
2
1
21
40 12 40 24
42
3 5
3
112 14 15 17
61
1 4
1
6
19 11 45 69
23
3
1
1111
11
3 4
57
109 11 37 16
13. If a printer one day use 4 bundles 1 ream 15 quires
20 sheets of paper, the next day 3 bundles 1 ream 10 quires
10 sheets, and the next 2 bundles 13 sheets, how much does
he use in the three days ?
Ans. 2 bundles 1 ream 6 quires 19 sheets.
14. A tailor used, in one year, 2 gross 5 doz. 10 buttons,
another year 3 gross 7 doz. 9, and another year 4 gross 6
doz. 11 ; how many did he use in the three years ?
Ans. 10 gross 8 doz. 6.
i
COMPOUND NUMBERS. 185
15. A ship, leaving New York, sailed easi the first day
3° 45' 50" ; the second day, 4° 50' 10" ; the third, 2° 10' 55";
the fourth, 2° 39" ; how far was she then east from the place
of starting ? Ans. 12° 47' 34 ".
16. A man, in digging a cellar, removed 127 cu. yd.
20 cu. ft. of earth ; in digging a drain, 6 cu. yd. 25 cu. ft. ;
and in digging a cistern, 17 cu. yd. 18 cu. f t. ; what was
the amount of earth removed, and what the cost at 1 6 cents
a cu. yd. ? Ans, 152| cu. yd. ; $24.37^.
17. A farmer received 80 ce»ts a bushel for 4 loads of
com, weighing as follows : 2564, 2713, 3000, and 3109 lbs.;
how much did he receive for the whole ? Ans. $162,657.
18. A druggist sold for medicine, in three years, at an aver-
age price of 9 cents a gill, the following amounts of brandy,
viz. : 1 bbl. 4 gal. 1 pt. ; 30 gal. 2 qt. 1 gi. ; 2 bbl. 15 gal. ;
how much did he receive for the whole ? Ans, $415.17.
218. To add denominate fractions.
1. Add f of a mile to 13^ rods.
OPERATION. Analysis. Find the value
4 mi. = 266 rd. 11 ft. ' ^^ ^^^^ fraction in integers
mrd.'= 13 rd.' Sift! of less denominations (213).
^ ± — and then add tneir values as
.^715. 280 rd. in compound numbers (217).
Or, 13i rd.-^320=^V mi. ^^' ^^^^^ *^^ ^^^° ^^^^-
, -.K • 91 • nor\ A tions to fractions of the same
A mi. + 1- mi. =14 mi. =280 rd. . . ^. ,«-,ox ^.
2* ' ^ 2* denomination (213), the^
add them, and find the value of their sum in lower denominations (213)
2. Add J of a rod to } of a foot. Ans, 13 ft. 1^ in.
3. What is the sum of -J of a mile, 28^ rods ?
Ans, 308 rd. 2 ft. 9 in.
4. What is the sum of f of a pound and -J of a shilling ?
Ans. 13s. lOd. 2f qr.
5. What is the sum of | of a ton and ^ of 1 cwt. ?
Ans. 12 cwt. 42 lb. 134 oz.
Give explanation of the process of adding denominate fractions-
186 SUBTRACTION.
6. What is the sum of | of a day added to J an hour ?
Ans. 9 h. 30 min.
7. What is the sum of | of a week, J of a day, and ^ of
an hour? • Ans, 1 da. 22 h. 15 min.
8. Add -f of a hhd. to } of a gal.
9. What is the sum of 4^ of a cwt., 8|- lb., and 3^ oz. by
long ton table ? Ans, 73 lb. 1-^ oz.
10. What is the sum of | of a mile, f of a yard, and J of
afoot?
11. Sold 4 village lots; the first contained J of ^ of an
acre ; the second, 60J rods ; the third, -f- of an acre ; and
the fourth, | of | of an acre ; how much land in the four
lots ? Ans. 146 P. 126^ sq. ft.
12. A farmer sold three loads of hay ; the first weighed
1^ T., the second, 1^ T., and the third, 18|- cwt. ; what
was the aggregate weight of the three loads ?
Am. 3 T. 5 cwt. 91 lb. 10| oz.
SUBTRACTION.
319. 1. If a druggist buy 25 gal. 2 qt. 1 pt. 1 gi. of wine,
and sell 18 gal. 3 qt. 1 pt. 2 gi., how much has he left ?
Analysis. Writing tlie subtrahend
under the minuend, placing units of the
same denomination under each other,
we begin at the right hand, or lowest
denomination ; since we cannot take
2 gi. from 1 gi., we add 1 pt. or 4 gi. to
i gi. , making 5 gi. ; and taking 2 gi. from 5 gi., we write the remain-
der, 3 gi., underneath the column of gills. Having added 1 pt. or
4 gi. to the minuend, we now add 1 pt. to the pt. in the subtra-
hend, making 1 pt. ; and 1 pt. from 1 pt. leaves pt., which we write
in the remainder. Next, as v»e cannot take 3 qt. from 2 qt., we add
1 gal. or 4 qt. to 2 qt., making 6 qt., and taking 3 qt. from 6 qt., we
write the remainder, 3 qt., under the denomination of quarts. Add-
ing 1 gal. to 18 gal., we subtract 19 gal. from 25 gal., as in simple
OPERATION.
gal. qt. pt.
25 2 1
18 3
gi.
1
2
Ans. 6 3
3
What is subtraction of compound numbers ? Give explanation.
COMPOUI^^D K UMBERS. 187
numbers, and write the remainder, 6 gal., under the column of gal-
lons.
KuLE. I. Write the subtrahend U7ider the minuend, so that
units of the same denomination shall stand under each other.
II. Beginning at the right hand, subtract each denomina-
tion separately, as in simple numbers.
III. If the number of any denomination in the suhtrahend
exceed that of the same denomination in the miriuend, add to
the number in the minuend as many units as make one of the
next higher denomination, and then subtract; in this case
add 1 to the next higher denomination of the suMrahend
before subtracting. Proceed m the same maimer with each
denomination.
Examples for Practice.
(2.
)
(3.)
lb.
oz.
pwt.
gr-
A.
P.
From 18
6
10
14
25
96.9
Take 10
5
4
6
19
145.14
Rem. 8
1
6
8
5
111.76
(4.)
(5.)
T. cwt.
lb.
yr.
da.
h.
min. Bee.
14 n
69J
38
187
16
45 50
10 12
98|
17
190
20
50 40
20 361 19 55 10
6. A Boston merchant bought English goods to the
amount of £4327 13s. 7id., and he paid £1374 10s. llfd. ;
how much did he then owe ?
7. From 300 miles take 198 mi. 305 rd. 2 yd. 1 ft. 10 in.
Ans. 101 mi. 14 rd. 2 yd. 2 ft. 8 in.
8. "What is the difference in the longitude of two places,
one 75° 20' 30" west, and the other 71° 19' 35" west?
Ans. 4° 55".
9. From lOib 7 1 4 3 1 3 15 gr. take 3ib 8 § 2 3 23 18 gr.
Ans. 6ft) 11 § 13 13 17gr.
Give the Rule.
188 SUBTRACTION.
10. The apparent periodic revolution of the sun is made
in 365 da. 6 h. 9 min. 9 sec, and that of the moon in 29 da.
12 h. 44 min. 3 sec. ; what is the difference ?
Ans, 335 da. 17 h. 25 min. 6 sec.
11. A man, having a hogshead of wine, drank, on an aver-
age, for five years, including two leap years, one gill of wine
a day ; how much remained ? A^is. 5 gal. 3 qt. 1 pt. 1 gi.
12. A section of land containing 640 acres is owned by
four men ; the first owns 196 A. 96 J P. ; the second, 200 A.
60 P. ; the third, 177 A. 36 P. ; how much does the fourth
own ? Ans. 65 A. 127.75 P.
13. From a pile of wood containing 75f Cd. was sold at
one time 16 Cd. 5 cd. ft. ; at another, 24 Cd. 6 cd. ft. 12
cu. ft. ; at another, 27 Cd. 112 cu. ft. ; how much remained
in the pile ? Aiis. 6 Cd. 3 cd. ft. 4 cu. ft.
14. If from a hogshead of molasses 10 gal. 1 qt. 1 pt. be
drawn at one time, 15 gal. 1 pt. at another, and 14 gal. 3 qt.
at another, how much will remain ?
230. To find the difference in dates.
1. What length of time elapsed from the discovery of
America by Columbus, Oct. 14, 1492, to the Declaration of
Independence, July 4, 1776?
FIRST OPERATION. ANALYSIS. Place the earlier date un-
yr. mo. da. der the later, writing first on the left
1776 7 4 the number of the year from the Chris-
1492 10 14 ^^^^ 6ra, next the number of the month,
' ~r~ 7 TT counting January as the first month, and
yioo o <^\} j^g^j. ^i^g number of the day from the
first day of the month. Instead of the number of the year, month,
and day, some use the number of years, months, and days that ham
elapsed since the Christian era, thus :
instead of saying July is the 7th mouth,
we say 6 months and 3 days have
elapsed, and instead of saying October
is the 10th month, we say 9 months and
283 8 20 13 days have elapsed.
How is the difference of dates found ?
SECOND
OPERATION.
yr.
mo. da.
1775
6 3
1491
9 13
COMPOUlN^D NUMBEKS.
189
Both methods will obtain the same result ; the former is generally
used.
1. When hours are to be obtained, we reckon from 12 at night, and if minutes and
seconds, we write them still at the right of hours.
2. In finding the time between two dates, or in computing interest, 12 months are
considered a year, and 30 days a month.
When the exact number of days is required for any period
not exceeding one ordinary year, it may be readily found by
the following
Table,
Shomng the number of days from any day of one month to the same day
of any other month within one year.
raOM ANT DAT
TO THE SAJfE DAT OP THiS NBXT.
or
Jan.
365
Feb.
31
59
Apr.
90
May
Jane
July
Aug.
Sept.
Oct.
Nov.
Dec.
January
120
151
181
212
243
273
304
334
February
a34
365
28
59
89
120
150
181
212
242
273
303
March
306
337
365
31
61
92
122
153
1&4
214
245
275
April
275
306
334
365
30
61
91
122
153
183
214
244
May
245
276
304
335
365
31
61
92
123
153
184
214
J una
214
184
245
215
273
243
304
274
as4
304
365
335
30
365
61
31
92
62
122
92
153
123
183
July
153
August
153
184
212
243
273
304
334
365
31
61
92
122
September
122
153
181
212
242
273
303
334
365
30
61
91
October
92
123
151
182
212
243
273
304
335
365
31
61
November
61
92
120
151
181
212
242
273
304
334
365
30
December
31
62
90
121
151
182
212
243
274
304
335
365
If the days of the different months are not the same, the
number of days of difference should be added when the
earher day belongs to the month /row which we reckon, and
siibtracted when it belongs to the month to which we find
the time. If the 29th of February is to be included in the
time computed, one day must be added to the result.
Examples foe Practice.
2. George Washington was bom Feb. 22, 1732, and died
Dec. 14, 1799 ; what was his age ?
Arts. 67 yr. 9 mo. 22 da.
How can the number of days, if less than a year, be obtained ?
IDO SUBTRACTION".
3. How much time has elapsed since the declaration of
independence of the United States?
4. How many years, months, and days from your birth-
day to this date ; or what is your age ?
5. How long from the battle of Bunker Hill, June 17, 1775,
to the battle of Waterloo, June 18, 1815 ? Ans. 40 yr. 1 da.
6. What length of time will elapse from 20 minutes past
2 o'clock, p. M., June 24, 1856, to 10 minutes before 9 o'clock,
A. M., January 3, 1861 ? Ans. 4 yr. 6 mo. 8 da. 18 h. 30 min.
7. How many days from any day of April to the same day
of August ? of December ? of February ?
8. How many days from the 6th of November to the 15th
of April ? Ans. 160 days.
9. How many days from the 20th of August to the 15th
of the following June ? Ans. 299 days.
231. To subtract denominate firactions.
1. From f of an oz. take J of a pwt.
OPEKATiON. Analysis. We per-
■| oz. =7 pwt. 12 gr. form the same reduc-
<» ^^^^ __ 21 ST. tions as in addition of
denominate fractions,
6 pwt. 15 gr., Ans. ( 2 1 8 ), and then sub-
„ o «^ »„ I tract the less value
Or, |oz.x20 = i^^pwt. from the greater,
iyi- — -J = -5/ pwt.= 6 pwt. 15 gr.
2. What is the difference between ^ rod and | of a foot ?
Ans. 7 ft. 6 in.
3. From £f take | of f of a shilling.
4. From f of a league take i^ of a mile.
Ans. 1 mi. 193 rd. 11 ft. 7.9 in.+
5. From 8^2^^ cwt. take 27h} lb.
Ans. 8 cwt. 62 lb. 9| oz.
6. From -^ of a week take ^ of a day.
A ns. 1 da. 4 h. 48 min.
Give explanation of the process of subtracting denominate fractions.
C0MP0UKD:N^ UMBERS. 191
7. Two persons, A and B, start from two places 120 miles
apart, and travel toward each other ; after A travels f , and
B f, of the distance, how far are they apart?
Ans. 41 mi. 289 rd. 8 ft. 7^ in.
8. From a cask of brandy containing 96 gallons, \ leaked
out, and f of the remainder was sold ; how much still re-
mained in the cask ? Ans. 25 gaL 2 qt. 3-J gi.
MULTIPLICATION.
222, 1. A farmer has 8 fields, each containing 4 A.
107 P. ; how much land in all?
OPEBATION. Analysis. In 8 fields are 8 times as much
A. P. land as in 1 field. We write the multiplier
4 107 under the lowest denomination of the multi-
8 plicand, and proceed thus : 8 times 107 P.
— -~ are 856 P., equal to 5 A. 56 P. ; and we
'^ * ^^ write the 56 P. under the number multiplied.
Again, 8 times 4 A. are 32 A., and 5 A. added make 87 A., which we
write under the same denomination in the multiplicand, and the work
is done.
Rule. I. Write the multiplier under the lowest denomina"
tion of the multiplicand.
II. Multiply as in simple numbers, and carry as in addi-
tion of compound numbers.
Examples
FOR
Practice.
(3.)
bu. pk. qt. pt.
4 2 5 1
2
(3.)
mi rd.
9 180
ft.
13
6
9 13
57 124
12
Multiplication of compound numbers, how perfonned ? Rule.
192
MULTIPLICATION.
(4.)
(5.)
£
s. d.
lb. oz. pwt. gr.
5
18 4
3 4 22
4
7
(6.)
(7.)
T.
cwt. lb.
oz.
14
16 48
12
13° 10' 35"
11
9
8. In 6 barrels of grain, each containing 2 bu. 3 pk. 5 qt.,
how many bushels? Ans. 17 bu. 1 pk. 6 qt.
9. If a druggist deal out 31b4§ I3 23l6gr. of medicine
a day, how much will he deal out in 6 days ?
10. If a man travel 29 mi. 150 rd. 15 ft. in 1 day, how
far will he travel in 8 days ?
11. If a woodchopper can cut 3 Od. 48 cu. ft. of wood in 1
day, how many cords can he cut in 12 days ? Ans. 40^ Cd.
12. What is the weight of 48 loads of hay, each weighing
IT. 3 cwt. 50 1b.?
Analysis. When the multi-
plier is large, and a composite
number, we may multiply by one
of the factors, and that product
by the other. Multiplying the
weight of 1 load by 6, we obtain
the weight of 6 loads, and the
weight of 6 loads multiplied by
8, gives the weight of 48 loads.
13. If 1 acre of land produce 45 bu. 3 pk, 6 qt. 1 pi of
com, how much will 64 acres produce? Ans. 2941 bu.
14. How much will 120 yards of cloth cost, at £1 9s. 8}d.
per yard ?
15. If $80 will buy 4 A. 146 P. 20 sq. yd. 3 sq. ft. of land,
how much will $4800 buy? Ans. 295 A. 10 sq. yd.
16. If a load of coal by the long ton weigh 1 T. 6 cwt. 2 qr.
26 lb. 10 oz., what will be the weight of 73 loads ?
Ans. 97 T. 11 cwt. 3 qr. 11 lb. 10 oz.
OPERATION.
T. cwt. lb.
1 3 50
6
00
8
weight of 6 loads.
56 8 00 weight of 48 loads.
COMPOUND NUMBERS. 193
17. The sun, on an average, changes his longitude 59'
8.33" per day ; how much will be the change in 365 days ?
18. If 1 pt. 3 gi. of wine fill 1 bottle, how much will be
required to fill a great gross of bottles of the same capacity ?
DIVISION.
223. 1. If 4 acres of land produce 102 bu. 3 pk. 2 qt. of
wheat, how much will 1 acre produce ?
OPEKATiON. Analysis. One acre will produce {
bu. pk. qt. pts. as much as 4 acres. Writing the divi-
4 ) 102 3 2 (5or on the left of the dividend, we di-
~ ~ ~ 7~ vide 103 bu. by 4, and obtain a quotient
of 35 bu., and a remainder of 3 bu. We
write the 25 bu. under the denomination of bushels, and reduce the
3 bu. to pecks, making 8 pk., and the 3 pk. of the dividend added
makes 11 pk. Dividing 11 pk, by 4, we obtain a quotient of 3 pk.
and a remainder of 3 pk. ; writing the 3 pk. under the order of pecks,
we next reduce 3 pk. to quarts, adding the 3 qt. of the dividend, mak-
ing 36 qt., which divided by 4 gives a quotient of 6 qt. and a remainder
of 3 qt. Writing the 6 qt. under the order of quarts, and reducing the
remainder, 3 qt. , to pints, we have 4 pt. , which divided by 4 gives a
quotient of 1 pt. , which we write under the order of pints, and the
work is dona
2. A farmer put 132 bu. operation.
1 pk. of apples into 46 bar- ^^* P^-
rels ; how many bu. did he ^^)^^^ 1 ( ^ bu,
put into a barrel ? _
40
4
When the divisor is large, and 161 ( 3 pk.
not a composite number, we di- 138
vide by long division, as shown oQ
in the operation. From these
examples we derive the
8
184 ( 4 qt.
— Ans. 2 hu.S])k. 4. qt
Explain the process of dividing compound numbers.
R. P.
194 DIVISION.
Rule. I. Divide the highest denomination as in simple
numbers^ and each succeeding denomination in the same
manner, if there he no remainder,
II. If there he a remainder after dividing any denomina-
tion, reduce it to the next lower denomination, adding in the
given number of that denomination, if any, and divide as before,
III. The several partial quotients will he the quotient re-
quired.
1. When the divisor is large and is a composite number, we may shorten the work
by dividing by the factors.
2. When the divisor and dividend are both compound numbers, they must both
be reduced to the same denomination before dividing, and then the procesB is the
same as in simple numbers.
Examples
FOE
Practice.
(3.)
(4.)
£
s.
d.
T.
cwt.
lb.
5)25
8
4
7)45
15
25
5
1
8
6
10
75
(5.)
wk. da.
h.
TTiin.
(6.)
4)3 5
22
00
10)25°
42'
40"
6
17
30
2
34
16
7. Bought 6 large silver spoons, which weighed 11 oz.
3 pwt. ; what was the weight of each spoon ?
8. A man traveled by railroad 1000 miles in one day;
what was the average rate per hour ?
Ans. 41 mi. 213 rd. 5 ft. 6 in.
9. If a family use 10 bbl. of flour in a year, what is the
average amount each day? Ans. 5 lb. 5fJ oz.
10. The aggregate weight of 123 hogsheads of sugar is
57 T. 19 cwt. 42 lb. 14 oz.; what is the average weight per
hogshead ? Ans. 9 cwt. 42 lb. 10 oz.
11. How many times are £5 10s. lOd. contained in £537
10s. lOd. ? Ans. 97.
Give the rule. Wlien the divisor is a composite number, how may
we proceed ? When the divisor and dividend are both compound num-
bers, how proceed ?
COMPOCND NUMBERS. 195
12. A cellar 50 ft. long, 30 ft. wide, and 6 ft. deep was
excavated by 5 men in 6 days ; how many cubic yards did
each man excavate daily? Ans. 11 cu. yd. 3 cu. ft.
13. If a town 5 miles square be divided equally into 150
farms, what will be the size of each farm ?
Ans. 106 A. 106 P. 20 sq. yd. 1 sq. ft. 72 sq. in.
14. How many times are 4 bu. 3 pk. 2 qt. contained in
336 bu. 3 pk. 4 qt. ? Ans. 70 times.
15. A merchant tailor bought 4 pieces of cloth, each con-
taining 60 yd. 2.25 qr. ; after selling J- of the whole, he
made up the remainder into suits containing 9 yd. 2 qr.
each ; how many suits did he make ? A7is. 17.
LONGITUDE AND TIME.
234. Every circle is supposed to be divided into 360
equal parts, called degrees.
Since the sun appears to pass from east to west round the
earth, or through 360°, once in every 24 hours, it will pass
through ^ of 360°, or 15° of the distance, in 1 hour ; ana
1° of distance in ^i^ of 1 hour, or 4 minutes ; and 1' of
distance in -^ of 4 minutes, or 4 seconds.
Table of Longitude and Time.
360° of longitude = 24 hours, or 1 day of time.
15° *' " =1 hour '* "
1° '' " = 4 minutes " "
V " " = 4 seconds ** "
Case I.
225. To find the diflference of time between two
places, when their longitudes are given.
1. The longitude of Boston is 71° 3', and of Chicago 87°
30' ; what is the difference of time between these two places ?
Explain how distance is measured by time. Repeat the table ot
lon'^itude and time. Case I is what?
OPERATION.
87°
30'
71°
3'
16°
27'
4
196 LONGITUDE AKD TIME.
Analysis. By subtrRction of
compound numbers we first find
the difllerence of longitude be-
tween the two places, which is
16° 27'. Since 1° of longitude
makes a difference of 4 minutes
1 h. 5 min. 48 sec, Ans. ^^ ^^^^> ^^^ 1' ^^ longitude a
difference of 4 seconds of time,
we multiply 16° 27', the difference in longitude, by 4, and we obtain
the difference of time in minutes and seconds, which, reduced to higher
denominations, gives 1 h. 5 min. 48 sec, the difference in time.
Rule. Multiply the difference of longitude in degrees and
minutes ly 4, and the product will de the differ e7ice of time in
minutes and seconds, which may he reduced to hours.
If one place be in east, and the other in west longitude, the difference of longitude
is found by adding them, and if the sum be greater than 180^, it mubt be subtracted
from 360^
Examples for Peactice.
2. New York is 74° 1' and Cincinnati 84° 24' west longi-
tude ; what is the difference of time ? Ans. 41 min. 32 sec.
jj 3. The Cape of Good Hope is 18° 28' east, and the Sand-
wich Islands 155° west longitude ; what is the difference of
time? Ans. 11 h. 33 min. 52 sec.
4. Washington is 77° 1' west, and St. Petersburgh 30° 19
east longitude ; what is their difference of time ?
Ans. 7 h. 9 min. 20 sec.
5. If Pekin is 118° east, and San Francisco 122° west
longitude, what is their difference of time ?
6. If a message be sent by telegraph without any loss of
time, at 12 M. from London, 0° 0' longitude, to Washington,
77° 1' west, what is the time of its receipt at Washington?
Since the sun appears to move trom east to west, when it Is exactly 12 o'clock at
one place, it will be past 12 o'clock at all places east, and btfore 12 at all places west.
Hence, knowing the difference of time between two places, and the exact time at
one of them, the exact time at the other will be found by adding their difference to
the given time, if it be east, and by subtracting if it be toest.
Ans. 6 h. 51 min. 56 sec. A. m.
Give explanation. Rale.
COMPOUKD KUMBEES. 197
^ 7. A steamer arrives at Halifax, 63° 36' west, at 4 o'clock,
p. M. ; the fact is telegraphed to St. Louis, 90° 15' west,
without loss of time ; what is the time of its receipt at St.
Louis ? Ans. 2 h. 13 min. 24 ^ec, P. m.
<''8. If, at a presidential election, the voting begin at sun-
rise and end at sunset, how much sooner will the polls open
and close at Eastport, Me., 67° west, than at Astoria, Ore-
gon, 124° west ? Ans. 3 h. 48 min.
9. When it was 1 o'clock, A. m., on the first day of Jan-
uary, 1859, at Bangor, Me., 68° 47' west, what was the time
at the city of Mexico, 99° 5' west ?
Ans. Dec. 31, 1858, 58 min. 48 sec. past 10, P. M.
Case IL
226. To find the difference of longitude between
two places, when the difference of time is known.
1. If the difference of time between New York and Cin-
cinnati be 41 min. 32 sec, what is the difference of longitude ?
OPERATION. Analysis. Since 4 minutes of time
min. sec. make a difference of 1° of longitude, and
4 ) 41 32 4 seconds of time, a difference of V of
-.^o nor A longitude, there will be \ as many de-
' ' grees of longitude as there are minutes
of time, and \ as many minutes of longitude as there are seconds of
time.
EuLE. Reduce the difference of time to mimctes and sec-
onds, and then divide hy 4 ; the quotient will he the difference
in longitude, in degrees and minutes.
2. What is the difference of longitude between the Cape
of Good Hope and the Sandwich Islands, if the difference
of time be 11 h. 33 min. 52 sec. ? Ans. 173° 28'.
3. What is the difference of longitude between Washings
ton and St. Petersburg, if their difference of time be 7 h.
9 min. 20 sec? Ans. 107° 20'.
Case II is what ? Give explanation. Rule.
198 DUODECIMALS.
4. When it is half past 4, p. m., at St. Petersburg, 30° 19'
east, it is 32 min. 36 sec. past 8, a.m., at New Orleans,
west; what is the difference of longitude? Ans. 119° 21',
5. The longitude of New York is 74° 1' west. A sea cap-
tain leaving that port for Canton, with New York time,
finds that his chronometer constantly loses time. What is
his longitude when it has lost 4 hours ? 8 h, 40 min.? 13 h.
25 min. ? Ans. 14° 1' west ; 55° 59' east ; 127° 14' east.
6. AVhen the days are of equal length, and it is noon on
the 1st meridian, on what meridian is it then sunrise ? sun-
set? midnight? ^725. 90° west; 90° east; 180° east or west.
DUODECIMALS.
227. Duodecimals are the divisions and subdivisions of
a unit, resulting from continually dividing by 12, as* -j^, y^,
Th'^, etc. In practice, duodecimals are applied to the meas-
urement of extension, the foot being taken as the unit.
If the foot be divided into 12 equal parts, the parts are
called inches, or primes ; the inches divided by 12 give sec-
onds ; the seconds divided by 12 give thirds ; the thirds
divided by 12 give fourths ; and so on.
From these divisions of a foot it follows that
V (Inch or Prime) is -^ of a foot.
1" (Second) or ^ of ^ " ^ir of a foot
1'" (Third) or ^^ ot -^ ot ^. . " ^^ o^ a foot.
Table.
12 Fourtlis, marked (""), make 1 Third marked 1'"
12 Thirds " 1 Second " 1"
12 Seconds ** 1 Prime, or Inch, " 1'
13 Primes, or Inches, " 1 Foot " ' ft.
Scale— uniformly 12.
The marks ', ", '", "", are called indices.
What arc duodecimals? To what applied? Explain the divisions
of. the foot. Repeat the table.
COMPOUND I?^UMBERS. ' 199
Duodecimals are really common fractions, and can always be treated as such ; bat
usually their denominators are not expressed, and they are treated as compoand
numbers.
Addition and Subtraction of Duodecimals
228, We add and subtract duodecimals the same as other
compound numbers.
Examples for Practice.
1. Add 13 ft. 4' 8", 10 ft. 6' 7", 145 ft. 9' 11".
Ans. 169 ft. 9' 2".
2. Add 179 ft. 11' 4", 245 ft. 1' 4", 3 ft. 9' 9".
Ans, 428 ft 10' 5".
3. From 25 ft. 6' 3" take 14 ft 9' 8". A71S. 10 ft 8' 7".
4. From a board 15 ft 7' 6" in length, 3 ft. 8' 11" were
sawed off; what was the length of the piece left?
Ans. 11 ft 10' 7".
Multiplication of Duodecimals.
229. Length multiplied by breadth gives surface, and
surface multiplied by thickness gives sohd contents ( 198 ).
1. How many square feet in a board 11 feet 8 inches long
and 2 feet 7 inches wide ?
Analysis. We first multiply hj the 7'.
7 twelfths times 8 twelfths equals 56 one
hundred forty-fourths, which equals 4
twelfths and 8 one hundred forty-fourths.
We write the 8 144ths — marked with two
indices — to the right, and add the 4 12th6
30 ft 1' 8" ^ *^® ri^^t product. 7' times 11 equals
77', which added to 4' equals 81', equal to
6 feet and 9'. We write the 9' under the
inches, or ; 12ths, and the 6 under the feet, or units. 2 times 8
equals 16', or 1 foot and 4'. We write the 4' under the 9', and
add the 1 foot to the next product. 2 times 11 feet are 22 feet, and
1 foot added make 23 feet, which we write under the 6 feet. Adding
How are duodecimals added and suhtracted ? Give analysis of ex-
ample 1.
OPKRATION
lift
8'
2
7'
6 ft
9'
8"
23
4'
k
200 DUODECIMALS.
these partial products, and we have 30 ft. V and 8" for the entire
product.
It will be seen from the above that the number of indiees to every
product of any two factors is equal to the sum of the indices of those
factors ; thus : 7' x 8'=56" ; 4" x 5"'=20""'.
EuLE. I. Wrile the several terms of the multiplier urder
the corresponding terms of the multiplicand.
II. Multiply each term of the multiplicand hy each term of
the multiplier f heginning with the lowest term in each, and
call the product of any two denominations the denomination
denoted hy the sum of their indices, carrying 1 for every 12.
III. Add the partial products, carrying 1 for every 12 ;
their sum will he the required answer.
Examples for Peactice.
2. How many square feet in a board 13 ft. 9' long and 11
wide ? Ans. 12 ft. 7' 3".
3. How many square feet in a stock of 4 boards, each
11 ft. 9' long and 1 ft. 3' wide ? A7is. 58 ft. 9'.
4. How many square yards of plastering on the walls of a
room 12 ft. 11' square, and 9 ft. 3' bigb, allowing for two
windows and one door, each 6 ft. 2' high and 2 ft. 4' wide ?
Ans, 48 sq. yd. 2 ft. 9'.
5. How many solid feet in a mow of hay 30 ft. 4' long,
25 ft. 6' wide, and 12 ft 5' high ? Ans. 9604 ft. 3' 6".
6. How many cords in a pile of wood 18 ft. 6' long, 12 ft.
wide, and 5 ft. 6' high ? Ans. ^ cords 69 ft.
7. How many cubic yards of earth must be removed in
digging a cellar 36 ft. 10' long, 22 ft. 3' wide, and 5 ft. 2'
deep ? Ans. 156 cu. yd. 22 ft. 3' 7".
8. What would it cost to plaster a wall 32 ft. 8' long and
9 ft. high, at 17 cents per square yard? Ans, $5.55 J.
9. How many yards of carpeting, 27' wide, will be re-
quired to cover a floor 48 ft. long and 33 ft. 9' wide ?
Ans. 240 yards.
Qive the rale.
COMPOUND NUMBEES. 201
Divisioiq- OF Duodecimals.
230. 1. A flagstone, 3 ft. 9' wide, has a surface of 20 ft
11' 3"; what is its length?
OPERATioif. Analysis. We divide
3 ft. 9' ) 20 ft 11' 3" ( 5 ft T. the surface by the width
"l^g 9' to obtain the length. The
— divisor is something more
^ ^3 than 3 ft., and to obtain
2 2 3 the first quotient figure, wo
consider how many times
3 ft. and something more is contained in nearly 21 ft. (20 ft. 11') ; we
estimate it to be 5 times, and multiplying the divisor by this quotient
figure, we have 18 ft. 9', which, subtracted from 20 ft. 11', leaves
2 ft. 2', to which we bring down 3", the last term of the dividend. We
next seek how many times the divisor is contained in this remainder,
and find by trial the quotient 7' ; multiplying the divisor by this figure,
we obtain 2 ft. 2' 3", and there is no remainder.
KuLE. I. Write the divisor on the left hand of the dividend,
as in simple numbers,
II. Find the first term of the quotient either hy dividing
the first term of the dividend by the first term of the divisor,
or hy dividing the first two terms of the dividend by the first
two terms of tlie divisor; m,ultiply the divisor by this term
of the qtwtienty subtract the product from the corresponding
terms of the dividend, and to the remainder bring doicn an-
other term of the divisor.
III. Proceed in like manner till there is no remainder, or
till a quotient has been obtained sufficiently exact.
Examples for Practice.
2. Divide 44 ft 5' 4" by 16 ft. 8'. Ans, 2 ft 8'.
3. The squai'e contents of a walk are 184 ft 3' and tlje
length is 40 ft 11' 4" ; what is the width ? Ans. 4 ft 6'.
4. A blanket whose square contents are 14 ft. 6', is to be
lined with cloth 2 ft 7' wide ; how much in length will be
required ?
Give analysis of example 1. Rule.
9*
202 PKOMISCUOUS EXAMPLES.
5. A block 01 granite contains 64 ft. 2' 5" ; its width is
2 ft. 6', and its thickness 3 ft. 7' ; what is its length ?
Am, 7 ft. 2'.
Since the solid contents are the product of the three dimensions, we divide the
solid contents by any two denominations or by their product, to obtain the other
dimension.
Peomiscuous Examples.
1. In 115200 grains Troy, how many pounds ?
2. In 365 da. 6 h. 48 min. 46 sec, how many seconds ?
Ans. 31556926.
3. A man wishes to ship 1560 bushels of potatoes in bar-
rels containing 3 bu. 1 pk. each ; how many barrels will be
required? Ans, 480.
4. Eeduce 295218 inches to miles.
5. Reduce 456575 grains to pounds, apothecaries' weight.
Ans. 79ib3i I3 l3l5gr.
6. How many sheets in 3 reams of paper ?
7. What is the value of 4 piles of wood, each 20 ft. long,
6 ft. wide, and 10 ft. high, at $3.25 per cord ? Ans. $121.87|.
8. How many bottles, each holding 1 qt. 1 gi., can be filled
from a barrel of cider? A71S, 112.
9. At $26. 40 per sq. rd. for land, what will be the cost of a
village lot SJ rd. long, and ^ rd. wide? Ans. $980.10.
10. Divide 259 A. 50 P. of land into 36 equal lots.
Ans. 7 A. 32^ P.
11. How many times can a box holding 4 bu. 3 pk. 2 qt.
be filled from 336 bu. 3 pk. 4 qt. ? Ans. 70.
12. What is the value of .875 of a gallon ?
13. What part of a mile is 116 rd. 2 yd. ? Ans. •^.
14. What part of 2 days is 13 h. 26 min. 24 sec. ?
' 15. From 26 A. 80 P. of land, 5 A. 120 P. were sold ; what
part of the whole piece remained unsold ? Am. ■^.
16. What is the difference between -J of a pound sterling
and 5|- pence ? ^«*. lis. 6id.
17. What is the sum of ^ of a yard, ^^ of a foot, and \ of
an inch ? An^. 7 inches.
PROMISCUOUS EXAMPLES. 203
18. Reduce 3 cwt. 1 qr. 7 lb. of coal to the decimal of a
long ton. Ans. .165625.
19. Benjamin Franklin was born Jan. 18, 1706, and
George Washington Feb. 22, 1732 ; how much older was
Franklin than Washington ? Ans. 26 jv. 1 mo. 4 da.
20. The longitude of Boston is 71° 4' west, and that of
Chicago 87° 30' west ; when it is 12 m. at Boston, what is the
time in Chicago ? Ans. 10 h. 54 min. 16 sec. A. m.
21. If the difference of time between New York and New
Orleans be 1 h. 4 sec, what is the difference in longitude ?
Ans. 15° 1'.
22. Add f of a mile, ^ of a mile, and -f^ of a rod to-
gether. Ans. 233 rd. 8 ft. 3 in.
23. If a bushel of barley cost $.80, what will 20 bu. 3 pk.
6qt. cost? Ans. $16.75.
24. What is the value of .875 of a gross ? Ans. 10|- doz.
25. How many acres in a field 56^ rods long, and 24.6
rods wide ? Ans. 8 A. 109.9 P.
26. How many perches of masonry in the wall of a cellar
which is 20 feet square on the inside, 8 feet high, and IJ feet
in thickness ? Ans. 44.6 + .
27. A, B, and C rent a farm, and agree to work it upon
shares ; they raise 640 bu. 3 pk. of grain, which they divide
as follows : one-fourth is given for the rent ; of the remain-
der A takes 10^ bu. more than one- third, after which B takes
one-half of the remainder less 7 bushels, and C has what is
left ; how much is C's share ? Ans. 161 bu. 3 pk. 6 qt.
28. What is the value in Troy weight of 13 lb. 8 oz. Avoir-
dupois weight ? Ans. 16 lb. 4 oz. 17 pwt. 12 grt
29. If 154 bu. 1 pk. 6 qt. cost $173.74, how much will 1.5
bushels cost ? Ans. $1,687.
30. What is the value of .0125 of a ton ? Ans. 25 lbs.
31. AYhat fraction of 3 bushels is -^\ of 2 bu. 3 pk. ?
Ans. -^.
32. How many wine gallons in a water tank 4 feet long,
3^ feet wide, and 1 ft. 8 in. deep ; Ans. 174j^.
204 PROMISCUOUS EXAMPLES.
33. How many bushels will a bin contain that is 7| feet
square, and 6 ft. 8 in. deep? Ans. 301.339 4-bu.
34. How much must be paid for lathing and plastering
overhead a room 36 feet long and 20 feet wide, at 26 cents
a square yard ?
35. How many shingles will it take to cover the roof of a
building 46 feet long, each of the two sides of the roof being
20 feet wide, allowing each shingle to be 4 inches wide, and
to lie 5 inches to the weather ? Ans. 13248.
36. John Young was born at a quarter before 4 o'clock, A.
M., Sept. 4, 1836 ; what will be his age at half -past 6 o'clock,
p. M., April 20, 1864 ? Ans. 27 yr. 7 mo. 16 da. 14 h. 45 min.
37. How many cubic yards of earth were removed in dig-
ging a cellar 28 ft. 9' long, 22 ft. 8' wide, and 7 ft. 6' deep ?
A71S. 181-gJj cu. yd.
38. What will 30 bu. 54 lb. of wheat cost, at $1.3 7 J per
bushel? Ans. $42.4875.
39. How many square yards of cai-peting will it take to
cover a floor 24 ft. 8' long and 18 ft. 6' wide ? Ans. 60^.
40. What is the cost of 54 bu. 8 lb. of barley, at 84 cents
per bushel? Ans. $45.60.
41. What is the depth of a lot that has 120 feet front, and
contains 18720 square feet ?
42. How many steps of 30 inches each must a person
take in walking 21 miles ?
43. How long will it require one of the heavenly bodies to
move through a quadrant, if it move at the rate of 3' 12"
per minute ? Ans. 1 da. 4 h. 7 min. 30 sec.
44. How many times will a wheel, 9 ft. 2 in. in circum-
ference, turn round in going 65 miles ?
45. If a man buy 10 bushels of chestnuts, at $5.00 per
bushel, dry measure, and sell the same at 22 cents per quart,
liquid measure, how much is his gain? A7is. $31.92.
46. What will it cost to build a wall 240 feet long, 6 feet
high, and 3 feet thick, at $3.25 per 1000 bricks, each brick
being 8 inches long, 4 inches wide, and 2 inches thick ?
Ans. $379.08.
PERCEKTAGE.
205
PERCENTAGE.
231. Per cent, is a term derived from the Latin words
per centum, and signifies by the hundred, or hundredths,
that is, a certain number of parts of each one hundred parts,
of whatever denomination. Thus, by 5 per cent, is meant 5
cents of every 100 cents, 15 of every $100, 5 bushels of every
100 bushels, etc. Therefore, 5 per cent, equals 5 hundredths
— .05=:yfo=r-^. 8 per cent, equals 8 hundredths =.08=
232. Percentag-e is such a part of a number as is indi-
cated by the per cent.
233. The Base of percentage is the number on which
the percentage is computed.
234. Since per cent, is any number of hundredths, it is
usually expressed in the form of a decimal ; but it may be
expressed either as a decimal or a common fraction, as in
the following
Table.
Decimals. Common Fractions. Lowest Terms.
1 per cent.
=
.01
=
T^iy
=
TcHT
2 per cent.
=
.02
=
TW^
=
^
4 per cent.
=
.04
=
T¥ir
=
1^
5 per cent.
=
.05
=
TlTff
=
1^
6 per cent.
=
.06
=
1%-S
=
t
7 per cent.
=
.07
=
10
=
T^
8 per cent.
=
.08
=
TTTD"
r=
^
10 per cent.
=
.10
=
uny
=
^
16 per cent.
=
.16
=
^%
=
A
20 percent.
=
.20
=
"J?
=
i
25 per cent.
=
.25
=
1^
=
i
50 per cent.
=
.50
=
tVo
=
i
100 per cent.
=
1.00
=
m
=
1
125 per cent.
=
1.25
=
m
=
f
\ per cent.
=
.005
r=
unnr
=
^(T
1 per cent.
=
.0075
=
Tff'imj
=
■^
12|^ per cent.
=
.125
=
^¥^
=
i
16^ per cent.
=
.1625
=
tWA
=
M
What is meant by per cent.? From what is the term derived?
What is percentage? What is the base of percentage? How is per
cent, expressed ?
206 percentage.
Examples for Practice.
1. Express decimally 3 per cent. ; 6 per cent. ; 9 per cent. ;
14 per cent. ; 24 per cent. ; 40 per cent. ; 122 J per cent. ; 150
per cent.
2. Express decimally 6J per cent.; 8f per cent.; 33^ per
cent.; 7i per cent.; lOf per cent.; 9| per cent; 103 J per
cent. ; 225 per cent.
3. Express decimally J per cent.; f per cent.; f per cent.;
f percent.; | per cent; IJ per cent; 2^ per cent; 4^ per
cent; 5} per cent; 7^ per cent; 12^ per cent; 25| per
cent.
4. Express in the form of common fractions, in their low-
est terms, 6 per cent; 8 per cent; 12 percent; 14|^ per
cent; 18-| per cent.; 21f per cent; 31 J- per cent ; 37^ per
cent.; 40f per cent; 112 per cent; 225 per cent
Case I.
235. To find the percentage of any number.
1. A man, having $125, lost 4 per cent of it ; how many
dollars did he lose ?
OPERATION.
$125 AiTALTSis. Since 4 per cent, is yj^ = .04, he lost
Q4 .04 of $125, or $125 x .04 = $5. Or, 4 per cent, is
•^^ T^ir = -iz^ and ^V of $125 ^ $5.
Rule. Multiply the given number or quantity by the rate
per cent, expressed decimally, and point off as in decimals. Or,
Take such a part of the given number as tJie number express-
ing the rate is part of 100.
Examples for Practice.
2. What is 6 per cent of $320 ? Ans. $19.20.
3. What is 8 per cent of $327.25 ? Ans. $26.18.
Case I is what? Give explanation. Rule.
PEECEJfTAGE. 207
4. What is 7} per cent, of $56.75 ? Ans. $4.11^.
5. What is 12J per cent, of 2450 pounds ?
Ans, 306.25 pounds.
6. What is 6| per cent, of 19072 bushels?
Aiis. 1287.36 bushels.
7. What is 33^ per cent of 846 gallons ?
Ans. 282 gallons.
8. What is 9| per cent, of 275 miles? Ans, 26.95 miles.
9. What is 14 per cent, of 450 sheep ?
10. What is 50 per cent, of 1240 men?
11. What is 105 per cent, of $5760 ? Ans, $6048.
12. What is 175 per cent, of S12967?
13. What is 25 per cent, of | ?
25 per 'cent, equals ^^^ = \, and lx^ = -^, Ans.
14. What is 15 per cent, of f ? Ans. y^.
15. What is 2^ per cent, of 6f ? Ans. i.
16. What is 33-J- per cent, of -3^ ? Ans. -^.
17. What is 84 per cent, of 7i? Ans. 6-^^.
18. Find | per cent, of $40.80. Ans. $.306.
19. Find If per cent, of $15.60. Ans. $.26.
20. A farmer, having 760 sheep, kept 25 per cent, of
them, and sold the remainder ; how many did he sell ?
21. A man has a capital of $24500; he invests 18 per
cent, of it in bank stock, 30 per cent, of it in railroad stocks,
and the remainder in bonds and mortgages ; how much does
he invest in bonds and mortgages ? Ans. $12740.
22. A speculator bought 1576 barrels of apples, and upon
opening them he found 12^ per cent, of them spoiled ; how
many barrels did he lose ?
23. Two men engaged in trade, each with $2760. One
of them gained 33-J- per cent, of his capital, and the other
gained 75 per cent. ; how much more did the one gain than
the other? ^ws. $1150.
24. A man, owning f of an iron foundry, sold 35 per cent,
of his share ; what part of the whole did he sell, and what
part did he still own ? Ans. He still owned ^|.
208 PERCENTAGE.
25. A owed B $575.40; he paid at one time 40 per cent,
of the debt ; afterward he paid 25 per cent, of the remain-
der; and at another time 12^ per cent, of what he owed
after the second payment ; how much of the debt did he
stiU owe ? Ans, $226.56|.
Case II.
236. To find what per cent, one number is of
another.
1. A man, having $125, lost $5; what per cent, of his
money did he lose ?
OPERATION. ^ Analysis. Multiply the
5_i_125=.04=:4 per cent. base by the rate per cent.
Or, to obtain the percentage
yf^=-g«^=.04=4 per cent: (235) ; conversely, divide
the percentage by the base
to obtain the rate per cent. Or, since $125 is 100 per cent, of his
money, $5 is yf^, equal to -^^ of 100 per cent., which is 4 per cent.
EuLE. Divide the percentage ly the base, and the quotient
will be the rate per cent, expressed decimally. Or,
Take such a part of 100 as the percentage is part of the
base.
Examples for Practice,
2. What per cent, of $450 is $90 ? Ans. 20.
3. What per cent, of $1400 is $175 ? Ans, 12i.
4. What percent, of $750 is $165?
5. What per cent, of $240 is $13.20 ? Ans. 5f
6. What per cent, of $2 is 15 cents ?
%■ What per cent, of G bushels 1 peck is 4 bushels 2 pecks
6 quarts? ' Ans. 75 per cent
8. What per cent, of 15 pounds is 5 pounds 10 ounces
avoirdupois weight ? -4ws. 37-J- per cent.
9. What per cent, of 250 head of cattle is 40 head ?
Case II is what ? G ive explanation. Rule.
PERCENTAGE. 209
10. From a hogshead of sugar containing 760 pounds, 100
pounds were sold at one time, and 90 pounds at another ;
what per cent, of the whole was sold ?
11. A man, having 600 acres of land, sold J of it at one
time, and -J of the remainder at another time ; what per
cent, remained unsold ? Ans, 50 per cent.
Case III.
237. To find a nnmber when a certain per cent,
of it is given.
1. A man lost $5, which was 4 per cent, of aU the money
he had ; how much had he at first ?
OPEKATiON. Analysis. We are here required to
^5_i_.04=$125. filial the base, of wMcli $5 is the per-
Qp centage. Now, percentage equals base
4^ V 1 00 ^1 25 multiplied by the rate per cent. ; con-
* * Tersely, base equals percentage divided
by rate per cent. Or, $5 Ls 4 per cent, of all he had ; |^ of $5, or f ,
equals 1 per cent, of all he had, and 100 times f equals 100 per cent.,
or all he had.
Rule. Divide the percentage ty the rate per cent., ex-
pressed decimally, and the quotient will be the base, or num-
ber required. Or,
Talce as many times 100 as the percentage is times the rate
per cent.
Examples for Practice.
2. 16 is 8 per cent, of what number ? Ans. 200.
3. 42 is 7 per cent, of what number ?
4. 75 is 12^ per cent, of what number? Ans. 600.
5. 33 is 2| per cent, of what number ? Ans. 1200.
6. $281.25 is 37i per cent, of what sum of money?
Ans. $756.
7. A farmer sold 50 sheep, which was 20 per cent, of his
whole flock ; how many sheep had he at first ?
Case III is what ? Give explanation. Rule.
210 PERCEl^TAGE.
8. I loaned a man a certain sum of money ; at one time
he paid me $59.75, which was 12J per cent, of the whole
sum loaned to him ; how much did I loan him ?
9. A merchant invested $975 in dry goods, which was 15
per cent, of his entire capital ; what was the amount of his
capital ? Ans. $6500.
10. If a man, owning 40 per cent, of an iron foundry,
sell 25 per cent, of his share for $1246.50, what is the value
of the whole foundry ? Ans. $12465.
11. A produce buyer, having a quantity of corn, bought
2000 bushels more, and he found that this purchase was 40
per cent, of his whole stock ; how much had he before he
bought this last lot ? Ans. 3000 bushels.
Case IV.
208. To find a number when the number, in-
creased by a certain per cent, of itself, is given.
1. A man's income this year is $525, which is 5 per cent,
inore than it was last year ; what was it last year ?
OPERATION. Analysis. Since his income
$525 -j- 1.05 = $500. this year is .05 more than it
was hist year, this year's in-
come must be 1.05 times the income of last year ; therefore divide this
year's income by 1.05 and it gives the income of last year.
KuLE. Divide the amount ly 1 plus the rate expressed
decimally, and the quotient will be the base or number re-
quired. Or,
Take as many times 100 as the amount is times 1 plus the
rate per cent.
What is Case IV ? Explanation ? Rule.
peecentage. 211
Examples for Practice.
2. What number increased by 18 per cent, of itseK is
equal to 1475 ? " Ans. 1250.
3. A merchant sells broadcloth for $4 per yard, and there-
by makes 25 per cent. ; what did the cloth cost him ?
Ans. $3.20.
4. A expended a certain sum for a house, and 15 per cent.
of the purchase price on repairs, and then found that the
whole cost was $6900. What was the purchase price ?
5. A certain manufacturing company have sold 1432,250
worth of goods, which is 8^ per cent, more than they sold
last year. How much did they sell last year ? Ans. 1400,000.
6. A merchant bought a stock of goods, and paid 4^ per
cent, of the purchase price for freight, when he found that
the goods cost him $8757. What was the purchase price ?
7. A merchant increased his capital by 20 per cent, each
year for two years, when he found he had $9360 invested.
How much had he at first ? Ans, $6500.
Case Y.
239. To find a number when the number, dimin-
ished by a certain per cent, of itself, is given.
1. A man lost 8 per cent, of his sheep and had 368 left ;
how many had he at first ?
OPERATION. • Analysts. Since the man lost
368 -7- .92 = 400. 8 per cent, of his sheep, he has
92 per cent, left ; hence 368 is .92
times his original flock. We therefore divide 368 by .93 and obtain
the required number of sheep.
Rule. Divide the given mimber hy 1 minus the rate ex-
pressed decimally, and the quotient will he the base or number
required. Or,
Take as many times 100 as the given numler is times 1
miyius the rate.
Wliat is Case V ? Explanation ? Rule ?
^13 rERCENTAGE.
Examples for Practice.
2. What number diminislied by 15 per cent, of itself is
equal to 340 ? Ans. 400.
3. A having a certain sum on deposit drew out 20 per
cent., when he found he had $1000 left. How much had he
on deposit at first ? Ans. $1250.
4. My income this year is $4028, which is 24 per cent,
less than it was last year. How much was it last year ?
5. What number diminished by i per cent, of itself is
equal to 298^^ ? Ans. 300.
6. A sells his horse for $198, which is 10 per cent, less
than his asking price, and his asking price was 10 per cent,
more than he cost him ; what did the horse cost him ?
Ans. $200.
COMMISSION AND BROKEEAGE.
240. An Agent, Factor, or Broker is a person wlio
transacts business for another, or buys and seUs money,
stocks, notes, etc.
241. Commission is the percentage, or compensation
allowed an agent, factor, or commission merchant, for buy-
ing and selling goods or produce, collecting money, and
transacting other business.
242. Brokerage is the fee, or allowance paid to a broker
or dealer in money, stocks, or bills of exchange, for making
exchanges of money, buying and selling stocks, negotiating
bills of exchange, or transacting other like business.
The rates of commiesion and brokerage are not regulated by law, but are nsnally
reckon^ at a certain per cent upon the money employed in the transaction.
Define an agent, factor, or broker. What is meant by commisBion 1
Brokerage ?
COMMISSION AND BROKERAGE. 213
Case I.
243. To find the commission or brokerage on any
Slim of money.
1. A commission merchant sells butter and cheese to the
amount of $1540 ; what is his commission at 5 per cent. ?
OPERATION. Analysts. Since
$1540 X .05 = $77, Ans, the commission on
Or, T^ = ^, and ^ X $1540 = 177. ^^ ^« ^ ^^^« ^^ -^^
it is $1540 X .05 = $77. Or, since 5 per cent, is j^^ = -^^ of the sum
received, the commission is -^ of $1640 = $77.
Rule. Multiply the given sum ly the rate per cent, ex-
pressed decimally, and the result tvill l?e the com?nission or
hrolcerage. Or,
Tahe such a part of the giveii sum as the number express-
ing the per cent, is part of 100.
Examples for Practice.
2. A commission merchant sells goods to the amount of
$G756 ; what is his commission at 2 per cent. ?
Ans. $135.12.
3. What commission must be paid for collecting $17380,
at 3^ per cent. ? Ans, $608.30.
4. An agent in Chicago purchased 4700 bushels of wheat
at 75 cents a bushel ; what was his commission at \\ per
cent, on the purchase money ?
5. A broker in New York exchanged $25875 on the Suf-
folk Bank, Boston, at \ per cent. ; how much brokerage did
he receive ? Ans. $64.6875.
6. An auctioneer sold at auction a house for $3284, and
the furniture for $2176.50 ; what did his fees amount to at
2:J- per cent. ?
7. A broker negotiates a bill of exchange of $2890 for
\ per cent, commission ; how much is his brokerage ?
Ans. $23.12.
Case I is what ? Give explanation. Rule.
214 PERCENTAGE.
8. An agent buys for a manufacturing company 26750
pounds of wool, at 32 cents a pound, and receives a com-
mission of 2f per cent. ; what amount does he receive ?
Ans. $235.40.
9. If I sell 400 bales of cotton, each weighing 570 pounds,
at 9 cents a pound, and receive a commission of 2J per cent,
how much do I make by the transaction ? Ans. $461.70.
10. A commission merchant in New Orleans sells 450
barrels of flour at $7.60 a barrel ; 38 firkins of butter, each
containing 56 pounds, at 25 cents a pound; and 105 cheeses,
each weighing 48 pounds, at 9 cents a pound ; how much
is his commission for selling, at 5^ per cent. ?
Ans. 1242.308.
11. A lawyer collected a note of $950, and charged 6|- per
cent, commission ; what was his fee, and what the sum to
be remitted? Ans, Fee, $61.75 ; remitted, $888.25.
12. An insurance agent's fees are 6 per cent, on all sums
received for the company, and 4 per cent, additional on all
sums remaining, at the end of the year, after the losses are
paid ; he receives, during the year, $30456.50, and pays losses
to the amount of $19814.15 ; how much commission does he
receive during the year ? Ans. $2253.084.
Case IL
244. To find the commission or brokerage, when
it is to be dediicted from the given sum, and the
balance invested.
1. A merchant sends his agent $1260 with which to buy
merchandise, after deducting his commission of 5 per cent ;
what is the sum invested, and how much is the commission?
OPERATION.
$1260 -^ 1.05 = $1200, invested.
$1260 — $1200 = $60, commiesion.
Or, ifj + T*ir = fi ; ^1260 -^ U = $1200, invested ,
And $1260 — $1200 = $60, commission.
Case II is what ? Give explanation. Rule.
COMMISSIOK AND BROKERAGE. 215
Analysis. Since the commission is 5 per cent., the agent must
receive $1.05 for every $1 he expends ; he can invest as many dollars
as ^1.05 is contained times in $1260, which is $1200 ; and the differ-
ence between the given sum and the sum invested is his commission.
Or, the money expended is {^ of itself, the commission is yf^ of
this sum, and the commission added to the sum expended is \^ of
the whole sum. Since $1260 is \^% = U, $1260 -^ |^ = $1200, the
sum expended ; and $1260 — $1200 = $60 the commission.
Rule. I. Divide the given amount by 1 increased by the
rate per cent, of commission, and the quotient is the sum
invested.
II. Subtract tlie investment from the given amount, and
the remainder is the commission.
Examples for Practice.
2. A man sends $3246.20 to his agent in Boston, request-
ing him to lay it out in shoes, after deducting his commis-
sion of 2 per cent. ; what is his commission ? Ans. $63.65.
3. What amount of stock can he bought for $9682, and
allow 3 per cent, brokerage ? Ans. $9400.
4. A flour merchant sent $10246.50 to his agent in Chi-
cago, to invest in flour, after deducting his commission of
3^ per cent. ; how many barrels of flour could he buy at
$5.50 per barrel? Ans. 1800 barrels.
5. An agent receives a remittance of $4908, with which
to purchase grain, at a commission of 4J- per cent. ; what
will be the amount of the purchase ?
6. Remitted $603.75 to my agent in New York, for the
purchase of merchandise, agent's commission being 5 per
cent. ; what amount of broadcloth at $5 per yard should I
receive? Ans. 115 yds.
7. A commission merchant receives $9376.158, with or-
ders to purchase grain ; his commission is 3 per cent. , and
he charges IJ per cent, additional for guaranteeing its de-
livery at a specified time; how much will he pay out, and
what are hid fees ? Ans. Fees, $403,758.
21G PERCE NT AGE.
8. A real-estate broker, whose stated commission is 1|
per cent., receives $13842.07, to be used in the purchase of
city lots ; how much does he invest, and what is his com-
mission? Ans. $13604 invested ; $238.07 commission.
9. A broker received $10650, to be invested in stocks afte?
deducting J per cent, for brokerage ; what amount of stock
did he purchase ?
STOCK-JOBBINa.
245. A Corporation is a body authorized by a general
law, or by a special charter, to transact business as a single
individual.
246. A Charter is the legal act of incorporation, and
defines the powers and obligations of the incorporated body.
24T. A Firm is the name under which an unincorpo-
rated company transacts business,
248. Capital or Stock is the property or labor of an
individual, corporation, company, or firm ; ifc receives dif-
ferent names, as Bank Stock, Kailroad Stock, Government
Stock, etc.
249. A Share is one of the equal parts into which the
stock is divided.
250. Stockholders arc the owners of the shares.
251. The Nominal or Par Value of stock is its first
cost, or original valuation.
The original valuation of a share varies in different companies. A share of bank,
insurance, railroad, or like stock is usually $100.
252. Stock is At Par when it sells for its first cost, or
original valuation.
253. Above Par, at a premium, or in advance, when it
sells for more than its original cost ; and
254. Below Par, or at a discount, when it sells for less
than its original cost.
Define a corporation. A charter. A firm. Capital or stock. Shares.
StQckholders. Par value. At par. Above par. Below par.
STOCKS. 317
255. The Market or Real Value of stock is what it
will bring per share in money.
256. A Dividend is a sum paid to stockholders from
the profits of the business of the company.
357. An Assessment is a sum required of stockholders
to meet the losses or expenses of the business of the company.
25S, Premium or advance, and discount on stock, divi-
dends, and assessments, are computed at a certain per cent,
upon the original value of the shares of the stock.
259. A Stock, Broker is a person who buys and sells
stocks, either for himself, or as the agent of another.
260. The calculations in stock-jobbing are based upon
the following relations :
I. Premium, discount, and brokerage are each a percent-
age, computed upon the par value of the stock as the base.
II. The market value of stock, or the proceeds of a sale,
is the amount or difference, according as the sum is greater
or less than the par value.
In all examples relating to stocks, $100 will be considered a share, unless other-
wise stated.
Case I.
261. To find the value of stock, when at an ad-
vance, or discount.
1. What will $3240 of Bank Stock cost at 8 per cent, ad-
vance, brokerage ^ per cent. ?
OPERATION. Analysis. To find the
$1 + .08 = $1.08. price of stock, we add the
$1.08 + .0025 = $1.0825. rate of advance to, or sub-
$1.0825 X 3240 = $3507.30 tract the discount from, $1,-
to this result add the bro-
kerage, and we have the cost of $1. Hence $3240 stock will cost 3240
times $1.0825.
EuLE. Multiply the cost of $1 by the number indicating
the par value of the stock.
Market value. A dividend. An assessment. Case I is what ? Give
explanation. Rule.
E.P. 10
218 percentage.
Examples for Practice.
2. If the stock of an insurance company sell at 5 per cent,
below par, what will $1200 of the stock cost ? Ans, $1140.
3. What is the market value of 35 shares of New York Cen-
tral Railroad stock, at 15 per cent, below par ? Ans. $2975.
4. What must be paid for 48 shares of Panama Railroad
stock, at a premium of 5^ per cent., if the par value be $150
per share, brokerage i per cent. ? Ans. $7632.
5. What costs $5364 stock in Minnesota copper mines, at
9 per cent, above par, brokerage ^ per cent. ? Ans. $5853.465.
6. A man purchased $6275 stock in the Pennsylvania Coal
Company, at par, and sold the same at a discount of 12 per
cent. ; what was his loss ? Ans. $753.
• 7. What must be paid for 125 shares of United States
stock, at 4J per cent, premium, the par value being $1000
per share, brokerage J- per cent. ? Ans. $131250.
8. Bought 42 shares of Illinois Central Railroad stock, at
14 per cent, discount, and sold the same at a premium of
12-J^ per cent. ; what did I gain ? An^. $1113.
9. What is the market value of 175 shares of stock in the
Suffolk Bank, at f per cent, advance ? Ans. $17631.25.
10. Bought 75 shares of stock in the Bank of New Orleans
at $50 each, at 3 per cent, discount, and sold it at 2J per
cent, advance ; what was my gain ? Ans. $196,875.
11. B. exchanged 28 shares of bank stock, of $50 each,
worth 7 per cent, premium, for 25 shares of railroad stock, of
$100 each, at 12^ per cent, discount, and paid the difference
in cash ; how much cash did he pay? Ans. $689.50.
12. A speculator exchanged $3600 of Railroad bonds, at
5 per cent, discount, for 27 shares of Bank stock, at 3 per
cent, premium, receiving the difference in cash ; how much
money did he receive ? Ans. $639.
13. I bought 120 shares Pacific Railroad stock, at a dis-
count of 2J percent., and sold the same at an advance of 12
per cent.; what was my gain? Afis. $1740.
STOCKS. 219
Case II.
362. To find how much stock may be purchased
for a given sum.
1. How many shares of bank stock, at 3 per cent, advance,
may be bought for $5150 ?
OPERATION. Analysis. Since the stock
$5150 -^ 1.03 = 15000 = is at 3 per cent, advance, $1
50 shares, Ans. ^^ s*«^^ ^* P^^ ^^'^^^ "^^^ ^^-^^ -
and if we divide $5150, the
whole sum to be expended, by 1.08, the cost of $1 of stock, the quo-
tient must be the amount of stock purchased.
Rule. Divide the given sum hy tJie cost of $1 of stoch, and
the quotie7it will he the nominal amount of stock purchased.
2. How many shares of railroad stock, at 5 per cent, ad-
vance, can be purchased for $6300 ? Ans. GO shares.
3. I invested $6187.50, in Ocean Telegraph stock, at 10
per cent, discount ; how much stock did I purchase ?
Ans. S6875.
4. I sent my agent $53500 to be invested in Hlinois Cen-
tral Railroad stock, which was selling at 7 per cent, advance;
what amount did he purchase ? Ans. $50000.
5. Sold 50 shares of stock in a Pittsburgh ferry company,
at 8 per cent, discount, and received $1150 ; what is the par
value of 1 share ? Ans. $25,
Stock Intestmen'ts.
263. The net earnings of a corporation are usually divid-
ed among the stockholders, in semi-annual dividends. The
income of capital stocJc is therefore fluctuating, being depend-
ent upon the condition of business ; while the income arising
from hands, whether of government or corporations, is fixed,
being a certain rate per cent., annually, of the par value, or
face of the bonds.
Case II is what? Give explanation. Explain difference between
income of capital stock, and of bonds.
220 PERCEN^TAGE.
264. Federal or United States Securities are of two
kinds; viz., Bonds and Notes.
Bonds are of two kinds.
First, Those which are payable at a fixed date, and are
known and quoted in commercial transactions by the rate
of interest they bear, thus : U. S. 6's, that is, United States
Bonds bearing Q>% interest.
Second, Those which are payable at a fixed date, but
which may be paid at an earlier specified time, as the Gov-
ernment may elect. These are known and quoted in com-
mercial transactions by a combination of the two dates,
thus : U. S. 5-20's ; or a combination of the rate of inter-
est and the two dates, thus : U. S. 6's 5-20 ; that is, bonds
bearing Q% interest, which are payable in twenty years, but
may be paid in five years, if the Government so elect.
When it is necessary, in any transaction, to distinguish
from each other different issues which bear the same rate of
interest, this is done by adding the year in which they be-
come due, thus : U. S.'5's of 71 ; U. S. 5's of '74 ; U. S. 6'i?
5-20 of '84; U. S. 6's 5-20 of '85.
Notes are of two kinds.
First, Those payable on demand, without interest, known
as United States Legal-tender Notes, or, in common lan-
guage, *^ Greenbacks."
Second, Notes payable at a specified time, with interest,
known as Treasury Notes. Of these there are two kinds, —
Q% Compound-interest Notes, and Notes bearing ^7^% in-
terest, the latter known and quoted in commercial transac-
tions as 7.30's.
The nomenclature here explained is that used in com-
mercial transactions, which involve similar Securities of
States or Corporations.
What are United States Securities composed of ? Explain the dif-
ferent kinds of bonds. Of Notes. In what is the interest on each
payable ?
STOCKS. 221
The interest on all U. S. bonds is payable in gold.
The interest on notes is payable in Legal- tender Notes.
The following are the principal United States Securities :
Bonds.
U. S. 6's of 1867.
U. S. 6's of 1868.
U. S. 6's of 1880.
U. S. 6's of 1881.
U. S. 5's of 1871.
U. S. 5's of 1874.
U. S. 5's (New) of 1881.
U. S. 4i's " of 1886.
U. S. 4's (New) of 1901.
U. S. 5-20's, due in 1882, interest Qfo,
U. S. 5-20's, due in 1884, interest 6%.
U. S. 5-20's, due in 1885, interest 6%.
U. S. 10-40's, due in 1904, interest 5%.
Pacific Railroad 6's of 1895.
Pacific Railroad 6's of 1896.
Notes,
Compound-interest Notes of 1867. I 7.30 Notes of 1867.
Compound-interest Notes of 1808. I 7.30 Notes of 1 868.
1. The 5-20's were issued in 1862, '61, '65, '67, and '70. They bear
interest at 6^, paid semi-annually in gold, except the issue of 1870,
called 5's of '81, which bear interest at 5%, paid quarterly in gold.
2. Bonds issued by States, cities, etc., are quoted in a similar man-
ner. Thus, S. G. 6's are bonds bearing 6% interest, issued by the
State of South Carolina.
Case I.
265. To find what income any investment will
produce.
1. What income will he obtained by investing $6840 in
stock bearing 6^, and purchased at 95^ ?
OPERATION. Analysis. We
$6840 -^ .95 = $7200, stock purchased, divide the invest-
$7200 X .06 = $432, annual income. ^^^t' ^^840, by
the cost of $1, and
obtain $7200, the stock which the investment will purchase (262).
And since the stock bears 6% interest, we have $7200 x .06 = $432,
the annual income obtained by the investment.
Rule. Find hoio much stock the investment toill purchase,
and then compute the income at the given rate upon the par
value.
Name the different kinds of bonds. Of Notes. What is Case I ?
Explanation ? Rule ?
222 PE EC EXT AGE.
2. If I invest $SG7 in U. S. 5-20's of '84 at 102^, what
income will I receive on my investment ? Ans. $51.
3. What will be my yearly income^ if I invest $8428 in
U. S. 10-40's, at 2S% ? Ans. $430.
4. How much stock at a premium of 4^% can be bought
for $10500, brokerage i% ? A7is. $10000.
5. If A invest $4795 in Maryland 5's at 87^, brokerage
i%, what will be his yearly income ? A7is. $274.
6. Having $10476 to invest, I find I can purchase U. S.
6's at 107i^, and U. S. 5-20's of '81 at 96-^^, brokerage i%,
in each instance. How much more will I receive yearly by
investing in the former than in the latter? Ans. $42.
7. A having a farm of 109 acres, which rents for $681.25,
sells the same for $125 per acre, and invests the proceeds in
U. S. 6's at 108}^, brokerage i% for purchasing ; will his
yearly income be increased or diminished, and how much ?
Ans. Increased $68.75.
Case II.
266. To find what sum must be invested to ob-
tain a given income.
1. What sum must be invested in Virginia 6% bonds,
purchasable at 80^, to obtain an income of $600 ?
OPERATION. Analysis.—
$600 -T- .05 = $12000, stock required. Since $1 of the
$12000 X .80 = $9600, cost or investment, stock will ob-
tain $.05 in-
come, to obtain $600 will require $600 h- .05 =$12000 (Case I). Multi-
plying the par value of the stock by the market price of $1, we have
$12000 X .80= $9600, the cost of the required stock, or the sum to he
invested.
Rule. I. Divide the given income by the per cent, which the
stock pays ; the quotient will be the par value of the stock
required.
What is Case II ? Explanation ? Rule ?
STOCKS. 223
11. Multiply the par value of the dock hy the market value
of one dollar of the stock ; the product zvill be the required
investment.
Examples for Practice.
2. If K Y. 6's are 6% below par, wliat sum must be in-
vested in this stock to obtain an income of $840 ?
Ans. $13300.
3. What sum must be invested in U. S. 10-40's at 98J^,
brokerage i% for buying, to secure an annual income of
$1860 ? Ans. 136642.
4. When U. S. 5-20's of '81 are quoted at lOSJ, what sum
must I invest to secure an annual income of $1080, broker-
age i^? Ans. 123436.
5. If I sell $25000 U. S. 5-20's of '82 at 93|^, and invest
a sufficient amount of the proceeds in U. S. 6's, at 109J^ to
yield an annual income of $960, and buy a house with the
remainder, how much will the house cost me ?
A71S. $5957.50.
Case III.
267. To find what per cent, the income is of the
investment, when stock is purchased at a given
price.
1. What per cent, of my investment shall I secure by
purchasing the New York 7's at 105^ ?
Analysis. Since $1 of the stock
OPEKATION. will cost $1.05, and pay $.07, the in-
.07 ^ 1.05 = 6f ^. come is j^j = 6|% of the invest-
ment.
EuLE. Divide the annual rate of income which the stock
dears hy the price of the stock ; the quotient will he the rate
upon the investment.
Examples for Practice.
2. What is the rate of income upon money invested in
Q% bonds, purchased at 87 per cent. ? Ans. 6f| ^.
What is Case III ? Explanation ? Rule ?
324 PERCENTAGE.
3. What per eent. on liis money will a man receive an-
nually if he invest in N. Y. G's at 105^ ? Ans. 5\%.
4. What is the rate of income upon money invested in
Missouri 6's at 76% ? Ans. S%.
5. Purchased U. S. 5-20's of '84 at 107|^, brokerage }^;
what is the income on the investment? Ans. 5|^.
6. Which is the better investment, U. S. 10-40's at 98J-^,
or U. S. 5-20's of '85 at lOSf^, brokerage i% in each?
Case IV.
268. To find the price at which stock must be
purchased to obtain a given rate upon the invest-
ment.
1. At what price must 6% stocks be purchased in order
to obtain 8% income on the investment ?
Analysis. Since $.06, the income
OPERATION. of $1 of the stock, is 8% of the sum
$.06 -^ .08 = 1.75 paid for it, we have (235) $.06^
.08 = 75%, the purchase price.
KuLE. Divide the annual rate of income which the stock
hears hy the rate required on the investment ; the quotient
will he the price of the stock.
Examples foe Practice.
2. What must I pay for Missouri 6's that my investment
may yield 9^ annually ? Ans. 66J^.
3. What rate of premium does 6^ stock bear in the mar-
ket when an investment pays h% ? Ans. %0%.
4. At what rate must I buy U. S. 10-40's that I may re-
ceive Q% on my investment? Ans, 83-J-^.
5. At what rate of discount must U. S. 5-20's of '81 be
purchased that I may secure 7^ annually on the invest-
ment ? Ans. 284^.
What is Case IV ? Explanation ? Rule ?
STOCKS. 225
GOLD INVESTMENTS.
269, Currency is a term used in commercial language.
First, To denote the aggregate of Specie and Bills of Ex-
change, Bank Bills, Treasury Notes, and other substitutes
for money employed in buying, selling, and carrying on ex-
change of commodities between various nations. Second,
To denote whatever circulating medium is used in any
country as a substitute for the government standard. In
this latter sense, the paper circulating medium, when below
par, is called Currency, to distinguish it from gold and
silver. If, from any cause, the paper medium depreciates
in value, as it has done in the United States, gold becomes
an object of investment, the same as stocks. In commercial
language, gold is represented as rising and falling ; but gold
being the standard of value, it can not vary. The variation
is in the medium of circulation substituted for gold ; hence,
when gold is said to be at a premium, the currency, or cir-
culating medium, is made the standard, while it is Adrtually
below par.
Case I.
370. To change gold into currency.
1. How much currency can be bought for 1150 in gold
when gold is at 170^ ?
OPERATION. Analysis. Since a dollar of gold
$1.70 X 150 = $255. is worth $1.70 in currency, there can
be as many times $1.70 of currency-
bought as there are dollars of gold. Therefore, $1.70 x 150 = $255
^a the amount of currency which can be purchased for $150 in gold.
KuLE. Multiply the value of one dollar of gold in currency
hy the number of dollars of gold.
2. What is the value in current funds of $250 gold, when
gold is 147^ ? Ans, $367.50.
3. What is the value in current funds of $320.50, when
goldisl37i^? Ans. $440.68J.
What is Currency? Case I? Explanation? Rule?
10*
226 PEKCEKTAGE.
4. When gold is at a premium of 33%, how much will
$2500 in gold cost? Ans, $3325.
5. A holds $8000 U. S. 10-40's ; what is his annual income
in currency if gold is 138 ? Ajis. $552.
6. What is the yearly income in currency from $9500 of
U. S. 5-20's of '84 when gold is 140 ? Ans, $798.
7. A purchased a house, for which he was to pay $450C
in currency, or $3000 in gold at his option. Will he gain
or lose by accepting the latter offer, gold being 147^%, and
how much in currency? Ans, Gain $75.
Case II.
271. To change currency into gold.
1. How much gold can be purchased for $75 current
funds, gold being at 150^ ?
AifALYSis. A dollar of gold cost $1.50
OPEBATION. in currency, therefore there can be as
$75 _i_ $1.50 = 50. many dollars of gold purchased for $75 in
currency as $1.50 is contained times in $75.
Eule. Divide the amount in currency by the price of gold,
2. What is the value in gold of a dollar in currency, when
gold is quoted at 138^^ ? A7is, $72^^^.
3. Gold being the standard, what is the rate of discount
upon current funds, when gold is at 145, 147, 195|-, 280^ ?
Ans. to last, Q4^%,
4. How much gold can be purchased for $4181 current
funds, when gold is quoted at 148^? Ans, $2825.
5. If I sell prints for 24 cents per yard, in currency, what
is the price in gold, gold being at 160^ ? Ans. 15 cents.
6. Sold $5900 U. S. 10-40's at 90^, and invested the
proceeds in gold at 147J ; how much gold did I purchase ?
Ans. $3G00.
7. What is the value in gold of a dollar, in currency,
when gold is at 145^ ? Ans. $.68f f.
8. I invested $792 of currency in gold, when gold is quoted
at 165^. How much gold did I purchase ? Ans, $480.
What is Case II ? Explanation 1 Rule t
PEOFIT AND LOSS. 227
9. What is gold quoted at, when a dollar in currency is
worth 30 cents in gold ? 45 cts. ? 54 cts. ? 60 cts. ? 74 cts. ?
10. How many yards of cloth at $3.50 in gold can be
bought for $126, currency, when gold is at 140 ? Ans. 25^.
11. Bought flour at $11.75 per barrel in currency, when
gold was at 150^, and afterwards sold it at $10.35 in cur-
rency, when gold was 135^ ; did I gain or lose, and how
much, on a sale of 300 barrels ? Ans. $50.
12. Which is the better investment, a bond and mortgage
at 7%, or U. S. 5-20's of '84 at par, gold being 140^ ; and
what per cent, in gold? Ans. XJ. S. 5-20's 1%.
13. Sold $51100 7-30 Treasury-Notes, at 104^, and in-
vested the proceeds in gold at 146^, with which I bought
U. S. 10-40's at 70^ in gold. Will my yearly income be
increased, or diminished by the transaction, and how much
in gold ? Ans. Increased $45.
PROFIT AND LOSS.
272. Profit and Loss are commercial terms, used to
express the gain or loss in business transactions, which is
usually reckoned at a certain per cent, on the prime or first
cost of articles.
Case I.
273. To find the amoiint of profit or loss, when
the cost and the gain or loss per cent, are given.
1. A man bought a horse for $135, and afterward sold
him for 20^ more than he gave ; how much did he gain ?
OPERATION. Analysis. Since $1
$135 X .20 = $27, Ans, gains 20 cents, or 20%,
Or, A'lr = -J ; ^135 x i = $27. $1^5 will gain $135 x .20
= $27. Or, since 20%
equals yV^ = I, the whole gain will be ^ of the cost.
Rule. Multiply the cost hy the rate per cent, expressed
decimally. Or,
Take such part of the cost as the rate per cent, is part of 100.
What is meant by profit and loss ? Case I. is what ? Give explana-
tion. Rule.
228 percentage.
Examples for Practice.
2. A grocer bought a hogshead of sugar for S84.80, and
sold it at 12 J per cent, profit ; what was his gain ?
3. A miller bought 500 bushels of wheat at $1.15 a bushel,
and he sold the flour at 16| per cent, advance on the cost
of the wheat ; what was his gain? Ans. $95.83^.
4. Bought 76 cords of wood at 13.62-J- a cord, and sold it
so as to gain 26 per cent. ; what did I make ?
5. A hatter bought 40 hats at $1.75 apiece, and sold them
at a loss of 14^ per cent. ; what was his whole loss ?
6. A grocer bought 3 barrels of sugar, each containing 230
pounds, at 8} cents a pound, and sold it at 18^^ per cent,
profit ; what was his whole gain, and what the selling price
per pound? Ans. Gain, $10.35 ; price per lb., 9| cents.
7. A sloop, freighted with 3840 bushels of corn, encoun-
tered a storm, when it was found necessary to throw 37 J per
cent, of her cargo overboard ; what was the loss, at 62|- cents
a bushel ? Ans. $900 loss.
8. A gentleman bought a store and contents for $4720 ;
he sold the same for 12^ per cent, less than he gave, and
then lost 15 per cent, of the selling price in bad debts ; what
was his entire loss ? A7is. $1209.50.
9. A man commenced business with $3000 capital ; the
first year he gained 22^ per cent, which he added to his
capital ; the second year he gained 30 per cent, on the whole
sum, which gain he also put into his business ; the third
year he lost 16| per cent, of his entire capital ; how much
did he make in the 3 years ? Ans. $981.25.
Case II.
274. To find the gain or loss per cent., when the
cost and selling price are given.
1. Bought wool at 32 cents a pound, and sold it for
40 cents a pound ; what per cent, was gained ?
Case II is what ? Givo explanation. Rule,
PROFIT AXD LOSS. 229
OPERATION.
40 — 32 = 8 ; 8 -^ 32 = ^ = .25, Ans.
Or, 40 - 32 = 8 ; 8 -^ 32 = -^ = J ; i X 100 = 25^.
Analysis. Since the gain on 32 cents is 40 — 33 = 8 cents, the
whole gain is -i^ = i of the purchase money ; and ^ reduced to a
decimal is 25 hundredths, equal to 25 per cent. Or, if the gain were
equal to the purchase money, it would be 100 per cent. ; but since the
gain is ^ = ^ of the purchase money, it will be ^ of 100 per cent.,
equal to 25 per cent.
Rule. Make the difference betioeen the purchase andseUi7ig
prices the numerator, and the purchase price the denomiiia-
tor ; reduce to a decimal, and the result luill he the per cent.
Or, Tahe such a part of 100 as the gain or loss is part of
the purchase price.
Examples for Practice.
2. A man oought a pair of horses for 1275, and sold them
for $330 ; what per cent, did he gain ? Ans. 20^.
3. If a merchant buy cloth at $.60 a yard, and sell it for
$.75 a yard, what does he gain per cent. ?
4. A speculator bought 108 barrels of flour at $4.62|- a
barrel, and sold it so as to gain $114.88|- ; what per cent,
profit did he make ? Ans. 23 per cent.
5. Bought sugar at 8 cents a pound, and sold it for
9J cents a pound ; what per cent, was gained ?
6. A drover bought 150 head of cattle for $42 per kead,
and sold them for $5400 ; what was his loss per cent, i
Ans. 14^%.
7. If I sell for $15 what cost me $25, what do I lose per
cent. ? Ans. 40 per cent.
8. Bought paper at $2 per ream, and sold it at 25 cents
a quire ; what was the gain ? Ans. 150^.
9. If I sell ^ of an article for f of its cost, what is gained
per cent. ? Ans. 50 per cent.
10. If f of an article be sold for what ^ of it cost, what is
the loss per cent. ? Ans. 37 J per cent.
230 PERCENTAGE.
11. If I sell 3 pecks of clover-seed for what one bushel
cost me, what per cent, do I gain ? Ans. 33^%.
12. A, having a debt against B, agreed to take $.87J^ on
the dollar ; what per cent, did A lose ?
13. A grocer bought 7 cwt. 20 lb. of sugar, at 7 cents a
pound, and sold 3 cwt. 42 lb. at 8 cents, and the remainder
at 8-J- cents ; what was his gain per cent. ? Ans. 18^ per cent.
14. Bought 2 hogsheads of wine, at $1.25 a gallon, and
sold the same at $1.60 ; what was the whole gain, and what
the gain per cent ? Ans. Gain 28^.
15. A grain dealer bought com at $.55 a bushel and sold
it at $.66, and wheat for $1.10, and sold it for $1.3 7^- ; upon
which did he make the greater per cent. ?
Ans, 5 per cent., upon the wheat.
Case III.
275. To find the selling price, when the cost and
the gain or loss per cent, are given.
1. Bought a horse for $136 ; for how much must he be
sold to gain 25 per cent. ?
OPERATION. Analysts. Since $1 of cost
$1 + . 25 = $1.25. sells for $1.25, $1 3G of cost will
$1.25 X 136 = $170, Ans. ^^^^ *'**^ 1^^ *^^^®s ^^■^^' ^^^^^
Or XM. J- 26 — 12 5 _ 8 ®<l^^^s $170, the selling price.
^h W 1- Tow — ttrtr — Z' Or, since the cost is {U, and
$136 X I = $170, Ans, ^j^g ^^^ ^s^p^^ the selling price
will be \l^ = f of the cost, or
f of $136 = $170. If the horse had been sold at a loss of 25 per cent.,
then $1 of cost would have sold for $1 minus .25, or $.75, etc.
Rule. Multiply $1 increased ly the gain or diminished ly
the loss per cent hy the nuraher denoting the cost. Or,
Talce such a part of the cost as is equal to |^g- increased or
diminished ly the gain or loss per cent.
Case III is what ? Give explanation. Role.
profit and loss. 231
Examples for Practice.
2. If 12^ hundred-weight of sugar cost 1140, how must
it be sold per pound to gain 26% Y Ans. 14 cents.
3. Bought a hogshead of molasses for 30 cents a gallon,
and paid 16f per cent, on the prime cost, for freight and
cartage ; what must it sell for, per gallon, to gain 33^ per
cent, on the whole cost ? A^is. $.46f .
4. For what price must I sell coffee that cost 10^ cents a
pound, to gain 17-J-^ ?
5. If I am compelled to sell damaged goods at a loss of
15 per cent, how should I mark goods that cost me $.62J-?
11.20? I3.87i^? A71S. $.53i; $1.02; $3.29|.
6. A man, wishing to raise some money, offers his house
and lot, which cost him $3240, for 18 per cent, less than
cost ; what is the price ?
7. C bought a farm of 120 acres, at $28 an acre, paid
$480 for fencing, and then sold it for 12^^ per cent, advance
on the whole cost ; what was his whole gain, and what did
he receive an acre ? Ans. $480 gain ; $36 an acre.
8. Bought a cask of brandy, containing 52 gallons, at
$2.60 per gallon ; if 7 gallons leak out, how must the re-
tiiainder be sold per gallon, to gain 87^- per cent, on the
cost of the whole ? Ans, $4.13}.
9. A merchant bought 15 pieces of broadcloth, each piece
containing 23^ yards, for $840, and sold it so as to gain
18| per cent. ; what did he receive a yard?
Case IV.
276. To find the cost, when the selling price and
the gain or loss per cent, are given.
1. A merchant sold cloth for $4.80 a yard, and by so doing
made 33-J per cent. ; how much did it cost ?
OPERATIOJT.
1 + .33i = 1.33J ; $4.80 -^ 1.33^ = $3.60, A7is.
Or, $4.80 = f of the cost ; $4.80 -r- i = $3.60.
Case IV is what?
PERCENTAGE.
Analysis. Since the gain is 83^ per cent, of the cost, $1 of the
cost, increased by 33^ per cent., will be what $1 of cost sold for ;
therefore there will be as many dollars of cost, as 1.33^ is contained
times in $4.80, or $3.60. Or, since he gained 33} per cent. = ^ of the
cost, $4.80 is I of the cost ; $4.80 -i- | = $3.60.
If the rate per cent, be loss, subtract it from 1, instead of adding it.
Rule. Divide the selling price hy 1 increased hy the gain
or diyninished ly the loss per cent., expressed decimally, or in
the form of a common fraction, and the quotient will le the
cost.
Examples for Practice.
2. By selling sugar at 8 cents a pound, a merchant lost 20
per cent. ; what did the sugar cost him? Ans. 10 cents.
3. Sold flour for $6.12|^ per barrel, and lost 12| per cent. ;
what was the cost ? Ans. $7.00.
4. A grocer, by selling tea at $.96 a pound, gains 28 per
cent. ; what did it cost him ? Ans. $.75.
5. Sold a quantity of flour for $1881, which was 18} per
cent, more than it cost : what did it cost ?
6. Sold 25 barrels of apples for $69. 75, and made 24 per
cent. ; what did they cost per barrel ?
7. Sold 9^ cwt. of sugar at $8:1- per cwt., and thereby lost
12 per cent. ; what was the whole cost ?
8. Having used a carriage six months, I sold it for $96,
which was 20 per cent, below cost ; what would I have re-
ceived had I sold it for 15 per cent, above cost ? Ans. $138.
9. B sells a pair of horses to C, and gains 1%^ per cent. ;
C sells them to D for $570, and by so doing gains 18| per
cent. ; what did the horses cost B? Ans. $426.66|.
10. A grocer sold 4 barrels of sugar for $24 each ; on 2
barrels he gained 20 per cent., and on the other 2 he lost 20
per cent. ; did he gain or lose on the whole ? Ans. Lost $4,
11. A person sold out his interest in business for $4900,
which was 40 per cent, more than 3 times as much as he be-
gan ^vith ; how much did he begin with ? Ans. $1166.66J.
Give explanation. Rule.
IKSURAIS^CE. 233
INSUEANCE.
277. Insurance on property is security guaranteed by
one party to another, for a stipulated sum, against the loss
of that property by fire, navigation, or any other casualty.
278. The Insurer or Underwriter is the party taking
the risk.
279. The Insured is the party protected.
280. The Policy is the written contract between the
parties.
281 . The Premium is the sum paid by the insured to
the insurer, and is estimated at a certain rate per cent, of
the amount insured, which rate yaries according to the
degree of hazard, or class of risk.
As a security against fraud, most insurance companies take risks at not more
than two-thirds the ftdl value of the property insured.
282. To find the premium when the rate of in-
surance and the amount insured are given.
1. What must I pay annually for insuring my house to
the amount of $3250, at IJ per cent, premium ?
OPEKATiON. Analysis. Multiply
$3250 X .OIJ or .0125 = 140.625. the amount insured.
Or, IJ per ct. = jf^ = ^ ; $3250, by the rate, 1^
$3250 X A = $40,624. P®^ *^®^*' """^ *^'^ ^^-
®^ ^ suit, $40,635, is the
premium. Or, the rate, 1\ per cent., is -^^ = ^V ^^ the amount in-
sured, and i^ of $3250 is $40.62 1.
Rule. Multiply the amount insured by the rate per cent,
and the product will he the premium. Or,
Take such a part of the amomit insured as the rate is part
of 100.
Define insurance. Insurer, or underwriter. Policy. Premium.
To what amount can property usually be insured ? Give analysLs of
example 1. Rule,
234 perce nt age.
Examples for Practice.
2. What is the premium on a policy for $750, at 4=% ?
Ans. $30.
3. What premium must be paid for $4572.80 insurance,
at 2J per cent. ? Ans. $114.32.
4. A house and furniture, valued at $5700, are insured at
If per cent. ; what is the premium ? Ans. $99.75.
5. A vessel and cargo, valued at $28400, are insured at
3^ per cent. ; what is the premium? Ans. $994.
6. A woolen factory and contents, valued at $55800, are
insured at 2^ per cent. ; if destroyed by fire, what would be
the actual loss of the company ; Ans. $54237.60.
7. What must be paid to insure a steamboat and cargo
from Pittsburg to 'New Orleans, valued at $47500, at f of
1 per cent.? Ans. $356.25.
8. A gentleman has a house, insured for $8000, and the
furniture for $4000, at 2| per cent. ; what premium must
he pay ? Ans. $285.
9. A cargo of 4000 bushels of wheat, worth $1.20 a bushel,
is insured at f of 1^ per cent, on f of its value ; if the cargo
be lost, how much will the owner of the wheat lose ?
Ans. $1636.
10. What will it cost to insure a factory valued at $21000,
at 4 per cent., and the machinery valued at $15400, at |^?
Ans. $264.25.
TAXES.
283, A Tax is a sum of money assessed on the person
or property of an individual, for public purposes.
284. When a tax is assessed on property, it is apportioned
at a certain per cent, on the estimated value.
When assessed on the person, it is apportioned equally
among the male citizens liable to assessment, and is called
a poll tax. Each person so assessed is called a poll.
What is a tax? How is a tax on property apportioned ? On the
person, how ?
TAXES. 235
285. Property is of two kinds — real estate, and per-
sonal property.
286, Real Estate consists of immovable property, such
as lands, houses, etc.
28*7. Personal Property consists of movable property,
such as money, notes, furniture, cattle, tools, etc.
288. An Inventory is a written list of articles of prop-
erty, with their value.
289. Before taxes are assessed, a complete inventory of
all the taxable property upon which the tax is to be levied
must be made. If the assessment include a poll tax, then
a complete list of taxable polls must also be made out.
I. A tax of $3165 is to be assessed on a certain town ;
the valuation of the taxable property, as shown by the
assessment roll, is $600,000, and there are 220 polls to be
assessed 75 cents each ; what will be the tax on a dollar, and
how much will be A's tax, whose property is valued at $3750,
and who pays for 3 polls ?
OPERATION.
$.75 X 220 = $165, amount assessed on the polls.
$3165 — $165 = $3000, amount to be assessed on the property.
$3000 ^ $600,000 = .005, tax on $1.
$3750 X .005 = $18.75, A's tax on property.
$.75 X 3 = $2.25, A's tax on 3 polls.
$18.75 + $2.25 = $21, amount of A's tax.
Rule. I. Find the amount of poll tax, if any, and sub-
tract this sum from the ivhole amount of tax to be assessed.
II. Divide the sum to be raised on property, by the whole
amount of taxable property, and the quotient ivill be the per
cent,, or the tax on one dollar.
III. Multiply each man^s taxable property by the per cent.,
or the tax on $1, and to the product add his poll tax, if any ;
the result ivill be the 'whole amount of his tax.
What is real estate ? Personal |)ropei-ty ? An inventory ? Explain
the process of levying a state or other tax. Rule.
236
P E R C E iq^ T A G E .
Ilaving found the tax on $1, or the per cent., which in the preceding example we
find to be 5 mills, or -^ per cent., the operation of assessing taxes maj' be greatly
facilitated by finding the tax on $2, $3, etc., to $10, and then on $20, $30, etc., to
$100, and arranging the numbers as in the followmg
Table.
Prop.
Tax.
Prop.
Tax.
Prop.
Tax.
Prop.
Tax.
$1 gives
$.005
$10
$.05
$100
$ .50
$1000
$ 5.00
2 **
.01
20
.10
200
1.00
2000
10.
3 *'
.015
30
.15
300
1.50
3000
15.
4 -
.03
•40
.20
400
2.00
4000
20.
5 "
.025
50
.25
500
2.50
5000
25.
6 -
.03
60
.30
600
8.00
6000
30.
7 "
.035
70
.35
700
3.50
7000
35.
8 "
.04
80
.40
800
4.00
8000
40.
9 "
.045
90
.45
900
4.50
9000
45.
Examples for Practice.
2. According to the conditions of the last example, how
much would he a person's tax whose property was assessed
at $3845, and who paid for 2 polls?
Finding the amount from the table,
The tax on $3000 is $15.00
" " '' 800 " 4.00
" ** " 40 " .20
*' " " 5 " .025
*' ** " 2 polls..." 1.50
Total tax is $20,725
3. How much would he Ws tax, who was assessed for 1
poll, and on property valued at $5390 ? Ans, $27.70.
4. A tax of $9190.50 is to be assessed on a certain village ;
the property is valued at $1400000, and there are 2981 polls,
to be taxed 50 cents each ; what is the assessment on a dol-
lar ? what is C's tax, his property being assessed at $12450,
and he paying for 2 polls ?
Ans. $.005 J on $1 ; $69.47^, C's tax.
5. What is the tax of a non-resident, having property in
the same village valued at $5375 ? Ans. $29.5625.
Explain the table and its use.
CUSTOM-HOUSE BUSINESS. 237
6. A mining corporation, consisting of 30 persons, are
taxed $4342.75 ; tlieir property is assessed for $188000, and
each poll is assessed 62J- cents ; what per cent, is their tax,
and how much must he pay whose share is assessed for
$2500, and who pays for 1 poll ? A7is. 2-^% ; $58,125.
7. In a certain county, containing 25482 taxable inhab-
itants, a tax of $103294.60 is assessed for town, county, and
state purposes ; a part of this sum is raised by a tax of 30
cents on each poll ; the entire valuation of property on the
assessment roll is $38260000 ; what per cent, is the tax, and
how much will a person's tax be who pays for 3 polls, and
whose property is valued at $9470 ? A^is. to last, $24,575.
8. The number of polls in a certain school district is 225,
and the taxable property $1246093.75 ; it is proposed to
build a union school house at an expense of $10000 ; if the
poll tax be $1.25 a poll, and the cost of collecting be 2J per
cent., what will be the tax on a dollar, and how much will be
E's tax, who pays for 1 poll, and has property to the amount
of $11500 ? Ans. $.008, tax on $1; $93.25, E's tax.
9. In a certain district the school was supported by a rate-
bill ; the tea<ihei^s wages amounted to $200, the fuel and
other expenses to $75.57 ; the public money received was $98,
and the whole number of day's attendance was 3946; A sent
2 pupils 118 days each ; how much was his rate bill?
Ans. $10.62.
CUSTOM-HOUSE BUSINESS.
290. Duties, or Customs, are taxes levied on imported
goods, for the support of government and the protection of
home industry.
291. A Custom-House is an office established by gov-
ernment for the transaction of business relating to duties.
292. A Port of Entry is a seaport town having a cus-
tom-house.
Define duties. A custom-house.
238 PEKCENTAGE.
293. Tonnage is a tax levied upon a vessel, independent
of its cargo, for the privilege of coming into a port of entry.
294. Revenue is the income to government from duties
and tonnage.
Duties are of two kinds — ad valorem and specific.
295. Ad Valorem Duty is a sum computed on tlie cost
of goods in the country from which they were imported.
296. Specific Duty is a sum computed on the weight
or measure of goods, without regard to their cost.
297. An Invoice is a bill of goods imported, showing
the quantity and price of each kind.
298. By the New Tariff Act, approved March 2, 1857,
all duties taken at the U. S. custom-houses, are ad valorem.
In collecting customs, it is the design of government to
tax only so much of the merchandise as will be available to
the importer in the market. The goods are weighed, meas-
ured, gauged, or inspected, in order to ascertain the actual
quantity and value received in port ; and an allowance is
made in every case of waste, loss, or damage.
299. Tare is an allowance of the weight of the package
or covering that contains the goods. It is ascertained by
actually weighing one or more of the empty boxes, casks,
or coverings. In common articles of importation, it is
sometimes computed at a certain per cent, previously ascer-
tained by frequent trials.
300. Leakage is an allowance on liquors imported in
casks or barrels.
301. Breakage is an allowance on liquors imported in
bottles.
Actual leakage or breakage ie allowed, there being no fixed or legal rate.
302. Gross Weight or Value is the weight or value of
the goods before any allowance has been made.
303. Net Weight or Value is the weight or value after
all allowances have been deducted.
Define Tonnage. Revenue. Ad valorem duty. Specific duty. An
invoice. Tare. Leakage. Breakage. Groas weight or vahio. Not
woiglit or value.
CUSTOM-HOUSE BU S.I NESS. 239
Draft is an allowance for the waste of certain articles, and is made only for statis-
tical purposes ; it does not aflfect the amount of duty. The rates of this allowance are
as follows :
On 1121b lib.
Above 112 lb., and not exceeding 224 lb., 2 lb.
2241b., " " " 336 lb., 3 lb.
3361b., " " " 1120 lb., 4 lb.
" 11201b., *' " *' 2016 lb., 7 lb.
" 2016 lb 9 lb.
What is the duty, at 24 per cent., on 50 gross of London
ale, invoiced at $1.20 per dozen, 2J per cent, being allowed
for breakage '
OPKRATiON. Analysis. First
$1.20 X 12 X 50 = $720, gross yalue. find the cost of the
$720 X .025 = $18, breakage. ^1®' at the invoice
$720 - $18 = $702, net value. T'^^' ^^""^ ^ ^^?'
^.^rs^ ^j ^-ir>r^ Ar^ T , FroHi tMs suui dc-
1702 X .34 = $168.48, duty. ^^^^ ,^^ ^1^^^^^
for breakage, $18, and compute the duty on the remainder.
Rule. Deduct allowances, if necessary, and compute the
duty, at the given rate, on the net value.
In the following examples, the legal rate of duty will be given, according to the
Tariff of 1851.
Examples for Practice.
2. What is the duty at 19 per cent, on 224 yards of plaid
silk, invoiced at $.95 per yard ? Ans. $40.43 + .
3. What is the duty at 24 per cent, on 50 barrels of sperm
oil, each containing originally 31|- gallons, invoiced at $.54
per gallon, allowing 2 per cent, for leakage ? Ans. $200.03 -f.
4. What is the duty at 15 per cent, on 175 bags of Java
coffee, each containing 115 lbs., valued at 15 cents per
pound? Ans. $452.81 J.
5. John Jones imported from Havana 25 hhds. of W. I.
molasses, which was invoiced at 36 cents per gallon ; allow-
ing J per cent, for leakage, what was the duty at 24^ ?
Ans. $135,399,
Define draft. Give analysis. Rule.
240
SIMPLE I2TTEREST.
SIMPLE INTEREST.
304. Interest is a sum paid for the use of money.
305. Principal is the sum for the use of which interest
is paid.
306. Rate per cent, per annum is the sum per cent,
paid for the use of $100 annually.
307. Amount is the sum of the principal and interest.
308. Simple Interest is the sum paid for the use of the
principal only, during the whole time of the loan or credit.
309. Legal Interest is the rate per cent, established by
law. It Taries in different States, as follows :
state.
Lejral
Rate.
Special
Agreement.
state.
^t.
Special
by
Agreement.
Alabama
Arkansas
8%
6%
10%
7%
10%
6%
10%
6%
6%
8%
7%
10%
6%
6%
6%
7%
6%
5%
6%
6%
6%
Any rate.
Any rate.
Any rate.
Any rate.
Any rate.
10%
Any rate.
10%
Any rate.
10%
10%
10%
12%
10%
8%
Any rate.
Any rate.
10%
12%
Mississippi
Missouri
67o
6%
10%
6%
6%
6%
6%
10%
10%
6%
10%
6%
6%
7%
6%
8%
10%
'4
'4
%
10%
10%
California
Montana
Any rate.
' 8 %
12%
Any rate.
8^
12^
Connecticut . .
Colorado
Canada
Dakota
Delaware
New Hampshire. .
New Jersey
New York
North Carolina . . .
Nebraska
Dist. Columbia
Nevada
Florida
Ohio
Georgia
Idaho
Oregon
Pennsylvania
Rhode Island
South Carolina
Tennessee
Texas
Illinois
Indiana
Iowa
Any rate.
Any rate.
10%
Any rate.
Any rate.
12%
10%
Any rate.
Kansas
Kentucky
Louisiana
Utah
Vermont
Virginia
Maryland
Massachusetts .
Michigan
Minnesota
West Virginia. . . .
Wisconsin
Washington Ter. .
England
310. Usury is illegal interest, or a greater per cent, than
the legal rate.
Case I.
311. To find the interest on any sum, at any rate
per cent., for years and months.
PERCEIJTAGE. 241
In percentage, any per cent, of any given number is so
many hundredths of that number; but in interest, any rate
per cent, is confined to 1 year, and the per cent, to be ob-
tained of any given number is greater than the rate per cent,
per annum if the time be more than 1 year, and less than the
rate per cent, per annum if the time be less than 1 year.
Thus, the interest on any sum, at any rate per cent., for 3
years 6 months, is 3^ times the interest on the same sum
for 1 year ; and the interest for 3 months is J of the interest
for 1 year.
1. What is the interest on $75. 19 for 3 years 6 months,
at 6^?
OPERATION.
$75.19
.06 Analysis. The interest on $75.19 for 1 yr.,
rTTTTT at 6 per cent., is .06 of the principal, or $4.5114,
and the interest for 3 yr. 6 mo. is 3y% = 3^ times
the interest for 1 yr., or $4.5114 x 3|, which is
H
22557 $15,789+.
135342
$15.7899, Ans,
Rule. I. Multiply the principal hy the rate per cent,, and
the product will he the interest for 1 year,
II. Multiply this product hy the time in years and frac-
tions of a year, and the result will he the required interest.
Examples foe Pkactice.
2. What is the interest of $150 for 3 years, at 4 per cent?
Ans. $18.
3. What is the interest of $328 for 2 years, at 1% ?
4. What is the interest of $125 for 1 year 6 months, at
6^? Ans. $11.25.
5. What is the interest of $200 for 3 years 10 months, at
7 per cent.? Ans. $53.66 + .
6. What is the interest of $76.50 for 2 years 2 months, at
5^? A71S. $8.287.
Explain the difference between percentage and interest Give
analysis. Rule.
i:.r. 11
242 SIMPLE INTEREST.
7. What is the interest of 11276.25 for 11 months, at 7
percent? Ans, $81.89 + .
8. What is the interest of $2569.75 for 4 years 6 months,
at 6 per cent. ?
9. What is the interest of $1500.60 for 2 years 4 months,
at 6i^? Ans. $218.8375.
10. What is the amount of $26.84 for 2 years 6 months,
at 5 per cent. ? Ans. $30,195.
11. What is the amount of $450 for 5 years, at 7 per cent. ?
12. What is the interest of $4562.09 for 3 years 3 months,
at 3^? Ans. $444.80 + .
13. What is the amount of $3050 for 4 years 8 mouths, at
5{ per cent. ? Ans. $3797.25+.
14. What is the interest of $5000 for 9 months, at 8 per
cent.? Ans. $e'500.
15. If a person borrow $375 at 7 per cent., how much will
be due the lender at the end of 2 yr. 6 mo. ?
16. What is the interest paid on a loan of $1374.74, at 6
per cent., made January 1, 1856, and called in January 1,
1860? Ans. $329,937.
17. If a note of $605.70 given May 20, 1858, on interest at
8 per cent, be taken up May 20, 1861, what amount will then
be due if no interest has been paid ? Ans, $751,068.
Case IL
312. To find the interest on any sum, for any
time, at any rate per cent.
The analysis of our rule is based upon the following
Obvious Relations between Time and Interest
I. The interest on any sum, for 1 year, at 1 per cent., is
.01 of that sum, and is equal to the principal with the
separatrix removed two places to the left.
II. A month being ^ of a year, -^^g^ of the interest on any
sum for 1 year is the interest for 1 month.
What is Case II ? Give the first relation between time and interest.
Second.
PERCENTAGE. 243
III. The interest on any sum for 3 days is -^^ = yig- = .1
of the interest for 1 month, and any number of days may
readily be reduced to tenths of a month by dividing by 3.
IV. The interest on any sum, for 1 month, multiplied by
any given time expressed in months and tenths of a month,
will produce the required interest.
1. What is the interest on 1724.68 for 2 yr. 5 mo. 19 da.,
at 7^?
OPERATION. Analysis. We remove
2 yr. 5 mo. 19 da. = 29.6-J mo. the separatrix in tlie given
12 ^ $7 2468 principal two places to tlie
'- left, and have $7.2468, the
^.o0o9 interest on the given sum for
29.6^ 1 year at 1 per cent. (313,
OA-j o I.). Dividing this by 12, we
have $.6039, the interest for
^"^^^ 1 month, at 1 per cent.
S^^^l (II.) Multiplying this quo-
12078 tient by 29.6^, the time ex-
^17 89557 pressed in months and deci-
-, mals of a month, (III., IV.,)
* we have $17.89557, the in-
$125.26899, Ans. terest on the given sum for
the given time, at 1 per cent.
(IV.). And multiplying this product by 7 (7 times 1 per cent.), we
have $125,268, the interest on the given principal, for the given time,
at the given rate per cent.
KuLE. I. Remove the separatrix in the given principal
tioo places to the left; the result will he the interest for 1
year, at 1 per cent,
II. Divide this interest ly 12 ; the result will he the interest
for 1 month, at 1 per cent.
III. Multiply this interest hy the given time expressed in
months and tenths of a month ; the result ivill he the interest
for the given time, at 1 per cent,
IV. Multiply this interest hy the given rate ; the product
loill he the interest required.
Give the third. Fourth. Give analysis. Rule.
244 SIMPLE INTEREST.
Contractions. After removing the separatrix in the principal, two
places to the left, the result may be regarded either as the interest on
the given principal for 12 months at 1 per cent., or for 1 month at 12
per cent. If we regard it as for 1 month at 12 per cent. , and if the
given rate be an aliquot part of 12 per cent., the interest on the given
principal for 1 month may readily be found by taking such an aliquot
part of the interest for 1 mouth as the given rate is part of 12 per
cent. Thus,
To find the interest for 1 month at 6 per cent., remove the separa-
trix two places to the left, and divide by 2.
To find it at 3 per cent., proceed as before, and divide by 4 ; at 4 per
cent., divide by 3 ; at 2 per cent., divide by 6, etc.
Six Pee Cent. Method.
313. By referring to 309 it will be seen that the legal
rate of interest in 21 States is 6 per cent. This is a sufficient
reason for introducing the following brief method into this
work :
Analysis. At 6 per cent, per annum the interest on $1
For 12 months is $.06.
" 2 months (3% = i of 12 mo.) " .01.
'• 1 month, or 30 days (yV of 12 mo.). . " .00^ = $.005 (^ of $.06).
" 6 days (^ of 30 days) " .001.
" 1 " (^ of 6 da. = ^V of 30 days) . . " .000^.
Hence we conclude that,
1st. The interest on $1 is $.005 per month, or $.01 for
every 2 months ;
2d. The interest on $1 is $.000| per day, or $.001 for
every 6 days.
From these principles we deduce the
Rule. I. To find the rate : — Call every year $.06, every
2 months $.01, every 6 days $.001, and every less number of
days sixths of 1 mill.
II. To find the interest : — Multiply the principal hy the
rate.
1. To find the interest at any other rate per cent, by this method, first find it at
6 per cent., and then increase or diminish the result by as many times itself as the
What contractions are given ? Give analysis of the 6 per cent
method. Rule. Its application to any other rate per cent.
SIMPLE INTEREST. 245
^ven rale is greater or less than 6 per cent. Thus, for 7 per cent, add |, for 4 per
cent, subtract i, etc.
2. The interest of $10 for 6 days, or of $1 for 60 days, is $.01. Therefore, if the
principal be less than $10 and the time less than 6 days, or the principal less than
$1 and the time less than 60 days, the interest will be less than $.01, and may be
disregarded.
3. Since the inte^st of $1 for 60 days, is $.01, the interest of $1 for any number
of days is as many cents as 60 is contained times in the number of days. Therefore,
if any principal be multiplied by the number of days in any given number of months
and days, and the product divided by 60, the result will be the interest in cents.
That is. Multiply the principal by the number of days, divide the product by 60, and
point off two decimal places in the quotient. T/ie result wHl be the interest in the
8am£ denomination as the principal.
Examples for Practice.
2. What is the interest of $100 for 7 years 7 months, at
6 per cent. ? Ans. 145.50.
3. What is the amount of 147.50 for 4 years 1 month, at
9 per cent. ? Ans. 164.956.
4. What is the amount of $2000 for 3 months, at 7 per
cent? A71S. $2035.
5. What is the interest of $250 for 1 year 10 months and
15 days, at 6 per cent. ? Ans. $28. 12 J.
6. What is the interest of $36.75 for 2 years 4 months and
12 days, at 7% ? Ans. $6,088.
7. What is the amount of $84 for 5 years 5 months and
9 days, at 5 per cent. ?
8. What is the interest of $51.10 for 10 months and 3
days, at 4^ ?
9. What is the interest of $175.40 for 15 months and 8
days, at 10 per cent? Ans. $22.31 + .
10. What is the amount of $1500 for 6 months and 24
days, at 7^^? Ans. $1563.75.
11. What is the amount of $84.25 for 1 year 5 months
and 10 days, at 6J per cent. ?
12. What is the interest of $25 for 3 years 6 months and
20 days, at 6 per cent ? Ans. $5.33|.
13. What is the interest of $112.50 for 3 months and 1
day, at 9J^? Ans. $2.70+.
What contractions are given ?
246 PERCENTAGE.
14. What is the interest of $408 for 20 days, at 6 per
cent.? * Ans. $1.36.
15. What is the interest of $500 for 22 days, at 7 per cent. ?
16. What is the amount of $4500 for 10 days, at 10 per
cent.? Ans. $4512.50.
17. What is the amount of $1000 for 1 month 5 days, at
6| per cent. ? Ans. $1006. 56 J.
18. Find the interest of $973.68 for 7 months 9 days, at
19. If I borrow $275 at 7 per cent., how much will I owe
at the end of 4 months 25 days ?
20. A person bought a piece of property for $2870, and
agreed to pay for it in 1 year and 6 months, with Q^ per cent,
interest ; what amount did he pay? Ans, $3149.825.
21. In settling with a merchant, I gave my note for $97.75,
due in 11 months, at 5 per cent. ; what must be paid when
the note falls due ? Ans. $102. 23 + .
22. How much is the interest on a note of $384.50 in 2
years 8 months and 4 days, at S% ?
23. What is the interest of $97.86 from May 17, 1850, to
December 19, 1857, at 7 per cent.? A^is. $51.98 + .
24. Find the interest of $35.61, from Nov. 11, 1857, to
Dec. 15, 1859, at 6 per cent. A^is. $4,474.
25. Required the interest of $50 from Sept. 4, 1848, to
Jan. 1, 1860, at 3^%.
26. Required the amount of $387.20, from Jan. 1, to Oct.
20, 1859, at 7 per cent. Ans. $408,957.
27. A man, owning a furnace, sold it for $6000 ; the terms
were, $2000 m cash on delivery, $3000 in 9 months, and the
remainder in 1 year 6 months, with 7 per cent, interest ;
what was the whole amount paid ? * Ans. $6262.50.
28. Wm. Gallup bought bills of dry goods of Geo. Bliss
& Co., of New York, as follows, viz. : Jan. 10, 1858, $350;
April 15, 1858, $150 ; and Sept. 20, 1858, $550.50 ; he bought
on time, paying legal interest ; what was the whole amount
of his indebtedness Jan. 1, 1859? Ans. $1092.66 + .
PARTIAL PAYMENTS. 247
PARTIAL PAYMENTS OR INDORSEMENTS.
314. A Partial Payment is payment in part of a note,
bond, or other obligation ; when the amount of a payment
is written on the back of the obligation, it becomes a re-
ceipt, and is called an Indorsement,
^^QQQ- Sprikgfield, Mass., Jan. 4, 1857.
1. For value received I promise to pay James Parish, or
order, two thousand dollars, one year after date, with in-
terest. George Joi^es.
On this note were indorsed the following payments:
Feb. 19, 1858 $400
June 29, 1859 $1000
Nov. 14, 1859 $520
What remained due Dec. 24, 1860 ?
OPERATION.
Principal on interest from Jan. 4, 1857 $2000
Interest to Feb. 19, 1858, 1 yr, 1 mo.15 da 135
Amount $2135
Payment Feb. 19, 1858 400
Eemainder for a new principal $1735
Interest from Feb. 19, 1858, to June 29, 1859, 1 yr. 4 mo.
10 da 141.69
Amount $1876.69
Payment June 29, 1859 1000
Eemainder for a new principal $876.69
Interest from June 29, 1859, to Nov. 14, 1859, 4 mo. l5 da. 19.725
Amount $896,415
Payment Nov. 14, 1859 520
Remainder for a new principal $376,415
Interest from Nov. 14, 1859, to Dec. 24, 1860, 1 yr. 1 mo.
10 da 25.0 9
Remains due Dec. 24, 1860 $401,505 +
What is meant by partial payment ? By an indorsement ?
248 PERCENTAGE.
^^^^5.50. jSTew York, May 1, 1855.
2. For value received, we jointly and severally promise to
pay Mason & Bro., or order, four hundred seventy-five dol-
lars fifty cents, nine months after date, with interest.
Jokes, Smith & Co.
The following indorsements were made on this note :
Dec. 25, 1855, received $50
July 10, 1856, '' 15.75
Sept. 1, 1857, '' 25.50
June 14, 1858, '' 104
How much was due April 15, 1859 ?
OPERATION.
Principal on interest from May 1, 1855 $475.50
Interest to Dec. 25, 1855, 7 mo. 24 da 21.63
Amount $497.13
Payment Dec. 25, 1855 50
Remainder for a new pri.ncipal $447.13
Interest from Dec. 25, 1855, to Jime 14, 1858, 2 yr. 5 mo.
19 da 77.29
Amount .- $524.42
Payment July 10, 1856, less than interest then due, ) $15.75
Payment Sept. 1, 1857 S 25.50
Tlieir sum less than interest then due $41.25
Payment June 14, 1858 104
Their sum exceeds the interest then due $145.25
Remainder for a new principal $379.17
Interest from June 14, 1858, to April 15. 1859, 10 mo. 1 da.. 22.19
Balance due April 15, 1859 $401.36 +
These examples have been wrought according to the
method prescribed by the Supreme Court of the U. S., and
are sufificicnt to illustrate the following
PARTIAL PAYMEI^Ta. 249
TJkited States Eule>
I. Find the amount of the given principal to the time of
the first payment, and if this payment exceed the interest
then due subtract it from the amount obtained, and treat
the remainder as a new principal.
II. But if the interest be greater than any payment, cast
the interest on the same principal to a time when the sum of
the payments shall eqtml or exceed the interest due ; subtract-
ing the Slim of the payments from the amount of the princi-
pal, the remainder loill form a new principal, on ivhich inter-
est is to be computed as before.
^514.96. S^i^ Feakcisco, June 20, 1858.
3. Three years after date we promise to pay Ross & TVade,
or order, five hundred fourteen and -f^ dollars, for value
received, with 10 per cent, interest. Wilder & Bro.
On this note were indorsed the following payments : Nov.
12, 1858, $105.50 ; March 20, 1860, $200 ; July 10, 1860,
$75.60. How much remains due on the note at the time of
its maturity ? Ans. $242.12 + ,
^'^QQQ- Charleston-, May 7, 1859.
4. For value received, I promise to pay George Babcocl
three thousand dollars, on demand, with 7 per cent, interest
JoHK May.
On this note were endorsed the following payments :
Sept. 10, 1859, received $25
Jan. 1, 1860, '' 500
Oct. 25, 1860, '' 75
April 4, 1861, '' 1500
How much was due Feb. 20, 1862 ? Ans. $1344.35 -f.
Give the United States Court rule for computing interest where par-
tial payments have been made.
11*
^ /
250 . PERCENTAGE.
$912^^ , ]sfEw Orleans, Aug. 3, 1850.
5. One year after date I promise to pay George Bailey, or
order, nine hundred twelve ^^ dollars, with 5 per cent, in-
terest, for value received. James Powell.
The note was not paid when due, but was settled Sept.
15, 1853, one payment of $250 having been made Jan. 1,
1852, and another of $316.75, May 4, 1853. How much
was due at the time of settlement ? Ans. $467.53+.
^184.56. Cincinnati, April 2, 1860.
6. Four months after date I promise to pay J. Ernst &
Co. one hundred eighty-four dollars fifty-six cents, for value
received. S. Anderson.
The note was settled Aug. 26, 1862, one payment of $50
having been made May 6, 1861. How much was due, legal
interest being 6% ? Ans, $154,188.
a note is on interest after it becomes due, if it contain no mention of interest.
7. Mr. B. gave a mortgage on his farm for $6000, dated
Oct. 1, 1851, to be paid in 6 years, with 8 per cent, inter-
est. Three months from date he paid $500 ; Sept. 10, 1852,
$1126 ; March 31, 1854, $2000 ; and Aug. 10, 1854, $876.50.
How much was due at the expiration of the time ?
Ans. $3284.84+.
315. The United States rule for partial payments has
been adopted by nearly all the States of the Union ; the
only prominent exceptions are Connecticut, Vermont, and
New Hampshire.
Connecticut Rule.
I. Payments made one year or more from the time the in
terest commenced, or from another payment, a7id paymerits
less than the interest due. are treated according to the United
States rule.
(Nearly obsolete. The United States rule is in general use.)
Give Connocticut rule for partial payments.
PARTIAL PAYMEN^TS. 251
II. Payments exceeding the interest due, and made toithin
one year from the time interest commenced, or from a former
fayment, shall draw interest for the balance of the year, pro-
vided the interval does not extend beyond the settlement, and
the amount must be subtracted from the amount of the prin-
cipal for one year ; the remainder will be the new principal.
III. If the year extend beyond the settlement, then find the
amount of the payment to the day of settlement, and subtract
it from the amount of the principal to that day ; the remain-
der luill be the sum due.
$460.
Woodstock, Ct., Jan. 1, 1858.
1. For value received, I promise to pay Henry Bowen, or
order, four hundred sixty dollars, on demand, with interest.
James Makshall.
On this note are indorsed the following payments : April
16, 1858, $148; March 11, 1860, $75; Sept. 21, 1860, $56.
How much was duo Dec. 11, 1860? Ans. $238.15 + .
316. A note containing a promise to pay interest an-
nually is not considered in law a contract for anything
more than simple interest on the principal. For partial
payments on such notes, the following is the
Vermont Eule.
I. Find the amount of the principal from the time interest
commenced to the time of settlement.
II. Find the amount of each payment from the time it was
made to the time of settlement.
III. Subtract the sum of the amounts of the payments from
the amount of the principal, and the remainder loill be the
sum due.
$600.
EuTLAND, April 11, 1856.
1. For value received, I promise to pay Amos Getting, or
order, six hundred dollars on demand, with interest an-
nually. John" Brown.
Give the Vermont rule for partial payments.
252 PEECENTAOE.
On this note were indorsed the following payments : Aug.
10, 1856, $156 ; Feb. 12, 1857, $200 ; June 1, 1858, $185.
What was due Jan. 1, 1859 ? A7is, $105.50 + .
317. In New Hampshire interest is allowed on the an-
nual interest if not paid when due, in the nature of dam-
ages for its detention ; and if payments are made before one
year's interest has occurred, interest must be allowed on
such payments for the balance of the year.
New Hampshiee Eule.
I. Find the amount of the principal for one year, and de-
duct from it the amount of each payment of that year, from
the time it was made up to the end of the year ; the remain-
der will he a neio principal, with ivhich proceed as before,
II. If the settlement occur less than a year from the last
annual term of interest, make the last term of interest a part
of a year, accordingly.
$575. Keei^e, N. H., Aug. 4, 1858.
1. For value received, I promise to pay G-eorge Cooper, or
order, five hundred seventy-five dollars, on demand, with
interest annually. David Geeekmak.
On this note were indorsed the following payments : Nov.
4, 1858, $64 ; Dec. 13, 1859, $48 ; March 16, 1860, $248 ;
Sept. 28, 1860, $60. What was due on the note Nov. 4,
1860 ? • Ans. $215.33.
318. When no payment whatever is made, upon a note
promising annual interest, till the day of settlement, in
New Hampshire the following is the
Court Rule.
Compute separately the interest on the principal from the
time the note is given to the time of settlement, and the inter-
est on each yearns interest from the time it should be paid to
the time of settlement. TJie sum of the hiterests thus obtained,
added to the principal, will be the sum due.
The New Hampshire rule. The New Hampshire court rule.
PARTIAL PAYMENTS. 253
^500. KEEi^E, N. H., Feb. 2, 1855.
1. Three years after date, I promise to pay James Clark,
or order, five hundred dollars, for value received, with in-
terest annually till paid. JoHK S. Briggs.
What is due on the above note, Aug. 2, 1859 ? Ans. $649.40.
Problems ix Ii^terest.
319, In examples of interest there are five parts involved,
the Principal, the Rate, the Time, the Interest, and the
Amount.
Case I.
320. The time, rate per cent., and interest being
given, to find the principal.
1. What principal in 2 years, at 8 per cent., will gain
131.80 interest ?
OPERATION. Analysis. Since $1, in
1.12, interest of $1 in 2 year, at 6^. 2 years, at 6 per cent., will
$31.80 -r- .12 = $265, Ans. ^^ ^-^^ interest, the prin-
cipal that will gain $31.80,
at the same rate and time, must be as many dollars as $.12 is contained
times in $31.80 ; dividing, we obtain $265, the required principal.
EuLE. Divide the given interest hy tlie interest of %1 for
the given time and rate, and the quotient will be the principal.
Examples for Practice.
2. What principal, at 6 per cent., will gain $28.1 2^ in 6
years 3 months ? Ans, $75.
3. What snm, put at interest for 4 months 18 days, at 4
per cent., will gain $9.20? Ans. $600.
4. What sum of money, invested at 7 per cent., will pay
me an annual income of $1260 ? Ans. $18000.
5. What sum must be invested in real estate, yielding 10
per cent, profit in rents, to produce an income of $3370 ?
Ans. $33700.
How many parts are considered in examples in interest
they ? What is Case I ? Give analysis. Rule.
? What are
254 percektage.
Case II.
821. The time, rate per cent., and amount being
given, to find the principal.
1. What principal in 2 years 6 months, at 7 per cent.,
will amount to 188.125?
OPEEATION. AlTALYSIS.
$1,175 Amt. of $1 in 2 years 6 months, at 7?S. Since $1, in
$88,125-^1.175 = 175, Ans, 2 years 6
months, at 7
per cent., will amount to $1,175, the principal that will amount to
$88,125, at the same rate and time, must be as many dollars as $1,175
is contained times in $88,125 ; dividing, we obtain $75, the required
principal.
Rule. Divide the given amount hy the amount of $1 for
the given time and rate, and the quotient will he the principal
Examples for Practice.
2. What principal, at 6 per cent., wiU amount to $655.20
in 8 months ? Ans. $630.
3. What principal, at 5 per cent., wiU amount to $106,855
in 5 years 5 months and 9 days? Ans, $84.
4. What sum, put at interest, at 5 J per cent., for 8 years
5 months, will amount to $1897.545 ? Ans. $1297.09 + .
5. What sum, at 7 per cent., will amount to $221,075 in
3 years 4 months ? Ans. $179.25.
6. What is the interest of that sum, for 11 years 8 days,
at 10|- per cent., which will at the given rate and time
amount to $857.54? ^/^5. $460.04.
Case III.
822. The principal, time, and interest being
given, to find the rate per cent.
1. I lent $450 for 3 years, and received for interest $67.50 ;
what was the rate per cent. ?
Give Case II. Analysis. Rule. Case III.
PROBLEMS IN INTEREST. 255
OPERATION. Analysis. Since at
$ 4.50 1 per cent. $450, in 3
g years, will gain $13.50
interest, the rate per
$13.50, int. of $450 for 3 years, at \%, ^^^^ ^^ ^^^^^ ^1^^ ^^^^
$67.50 -T- 13.50 = h%, A?IS. principal, in tlie same
time, will gain $67.50,
mnst be equal to the number of times $13.50 is contained in $67.50;
dividing, we obtain 5, the required rate per cent.
EuLE. Divide the given interest ly the iiiterest on the prin-
cipal for the given time at 1 per cent., and the quotient will
le the rate per cent required.
Examples for Practice.
2. If I pay $45 interest for the use of $500 for 3 years,
what is the rate per cent. ? Ans. 3.
3. The interest of $180 for 1 year 2 months 6 days is
$12.78? what is the rate %? Ans. 6.
4. A man invests $2000 in bank stock, and receives a
semi-annual dividend of $75 ; what is the rate per cent. ?
5. At what per cent, must $1000 be loaned for 3 years 3
months and 29 days, to gain $183.18 ? Ans. h^.
6. A man builds a block of stores at a cost of $21640, and
receives for them an annual rent of $2596.80 ; what per cent,
does he receive on the investment ? Ans, 12.
Case IV.
323. Principal, interest, and rate per cent, being
given, to find the time.
1. In what time will $360 gain $86.40 interest, at Q% ?
OPERATION. Analysis. Since in
$ 360 1 year $360, at 6 per
QQ cent., will gain $21.60,
the number of years in
$21.60, interest of $360 in 1 year, at 6^. ^liich the same princi-
$86.40 ~- 21.60 = 4 years, Ans. pal, at the same rate,
■will gain $86.40, will be
Analysis. Rulo. Case TV. Analysis.
356 PERCENTAGE.
as many as $21.60 is contained times in $86.40 ; dividing, we obtain
4 years, the required time.
EuLE. Divide the given interest hy the interest on the
principal for 1 year^ and the quotient will he the time re-
quired in years and decimals.
The decimal part of the quotient, if any, may be reduced to months and days (by
215).
Examples for Practice.
2. The interest of 1325 at 6 per cent, is $58.50 ; what is
the time ? Ans. Z years.
3. B loaned $1600 at 6 per cent, until it amounted to
$2000 ; what was the time ? A^is, 4 years 2 months.
4. How long must $204 be on interest at 7 per cent. , to
amount to $217.09 ? Ans. 11 months.
5. Engaging in business, I borrowed $750 of a friend at
6 .per cent., and kept it until it amounted to $942 ; how long
did I retain it ? Ans. 4c years 3 months 6 days.
6. How long will it take $200 to double itself at 6 per
cent, simple interest? Ans. 16 years 8 months.
7. In what time will $675 double itself at 5^?
The time in yearB in which any sum will double itself may be found by dividing
100 by the rate per cent.
COMPOUND INTEREST.
324. Compound Interest is interest on both principal
and interest, when the interest is not paid when due.
The simple interest may be added to the principal annually, semi-annually, or
quarterly, as the parties may agree ; but the taking of compound interest is not
Ugod.
1. What is the compound interest of $200, for 3 years, at
6 per cent. ?
Rule. In what time will any 8um double itself at interest ? What
is compound interest ?
COMPOUND INTEEEST. 257
OPERATION.
$200 Principal for 1st year.
$200 X .06 = 12 Interest for 1st year.
$212 Principal for 2d year. m
$212 X .06 = 12.72 Interest for 2d year.
$224.72 Principal for 3d year.
$224.72 X .06 = 13.483 Interest for 3d year.
$238,203 Amount for 3 years.
200.000 Given principal.
$38,203 Compound interest.
Rule. I. Find the amount of the given principal at the
given rate for one year, and mahe it the principal for the
second year,
II. Find the amount of this new principal, and make it the
principal for the third year, and so continue to do for the
given number of years,
III. Subtract the given principal from the last amount, and
the remainder will be the compound interest,
1. When the interest is payable semi-annually or quarterly, find the amount of the
given principal for the first intei-val, and make it the principal for the second inter-
val, proceeding in all respects as when the interest is payable yearly.
2. When the time contains years, months, and days, find the amount for the
years, upon which compute the interest for the months and days, and add it to the
last amount, before subtracting.
Examples for Practice.
2. What is the compound interest of $500 for 2 years at
7 per cent.? Aiis. $72.45.
3. What is the amount of $312 for 3 years, at 6 per cent,
compound interest ? Ans. $371.59 + .
4. What is the compound interest of 1250 for 2 years,
payable semi-annually, at 6 per cent.? Ans. 131.37 + .
5. What will $450 amount to in 1 year, at 7 per cent, com-
pound interest, payable quarterly ? Ans, $482.33.
6. What is the compound interest of $236 for 4 years 7
months and 6 days, at 6^ ? A7is. $72.66 + .
Explain operation. Give rule.
258
PBBCENTAGB.
7. What is the amount of $700 for 3 years 9 months and
24 days, at 7 per cent compound interest ? A ns. $906.55 + .
A more expeditious method of computing compound, in-
terest than the preceding, is by means of the following
Table,
STwwing the amount o/ $1, <w £1, at 3, 4, 5, 6, and 7 per cent, cott^
pound interest, for any number of years, from 1 to 20.
Yrs.
3 per cent
4 per cent.
5 per cent.
6 per cent.
7 per cent.
1..
2..
3..
4..
5..
1.030,000
1.060,900
1.092,727
1.125,509
1.159,274
1.040,000
1.081,600
1.124,864
1.169,859
1.216,653
1.050,000
1.102,500
1.157,625
1.215,506
1.276,282
1.060,000
1.123,600
1.191,016
1.262,477
1.338,226
1.07,000
1.14,49a
1.22,504
1.31,079
1.40,255
6..
7..
8..
9..
10..
1.194,052
1.229,874
1.266,770
1.304,773
1.343,916
1.265,319
1.315,932
1.368,569
1.423,312
1.480,244
1.340,096
1.407,100
1.477,455
1.551,328
1.628,895
1.418,519
1.503,630
1.593,848
1.689,479
1.790,848
1.50,073
1.60,578
1.71,818
1.83,845
1.96,715
11..
12..
13..
14..
15..
1.384,234
1.425,761
1.468,534
1.512,590
1.557,967
1.539,454
1.601,032
1.665,074
1.731,676
1.800,944
1.710,339
1.795,856
1.885,649
1.979,932
2.078,928
1.898,299
2.012,196
2.132,928
2.260,904
2.396,558
2.10,485
2.25,219
2.40,984
2.57,853
2.75,903
16..
17..
18..
19..
20..
1.604,706
1.652,848
1.702,433
1.753,506
1.806,111
1.872,981
1.947,900
2.025,817
2.106,849
2.191,123
2.182,875
2.292,018
2.406,619
2.526,950
2.653,298
2.540,352
2.692,773
2.854,339
3.025,600
3.207,135
2.95,216
3.15,881
3.37,293
3.61,652
3.86,968
8. What is the amount of $800 for 6 years, at 7 per cent. ?
OPERATION.
From the table $1.50073 Amount of $1 for the time.
800 Principal,
11200.58400, Ans.
9. What is the compound interest of $120 for 15 years, at
6 per cent. ? A ns. $129.47 + .
Of what use is the table in computing compound interest ?
DISCOUNT. 259
10. What is the amount of $.10 for 20 years, at 7 per
cent. ? Ans. $.38696.
DISGOUNT.
325. Discount is an abatement or allowance made for
the payment of a debt before it is due.
326. The Present Wortli of a debt, payable at a future
time without interest, is such a sum as, being put at legal in-
terest, will amount to the given debt when it becomes due.
1. A owes B $321, payable in 1 year ; what is the present
worth of the debt, the use of money being worth 7 per cent.?
OPERATION. Analysis. The
Am't of $1 1.07 ) $321 ( $300, Present value, amount of $1 for
321 1 year is $1.07;
tlierefore tlie
$321 Given sum or debt. present wortli of
300 Present wortli. every $1.07 of
$21 Discount. the given debt is
$1 ; and the pres-
ent worth of $321 will be as many dollars as $1.07 is contained times
in $331. $831-r-1.07=$300, Ans.
Rule. I. Divide the given sum or debt ly the amount of
$1 for the given rate and time, and the quotient will he the
present worth of the delt.
II. SuUract the present worth from the given sum or dedt,
and the remainder will he the discount.
The terms present woi'th, discount^ and debt, are equivalent to 2^nci2)al, interest^
and amount. Hence, when the time, rate per cent, and amount are given, the prin-
cipal may he found by (321) ; and the interest by subtracting the principal from
the amount.
Examples foe Pkactice.
2. What is the present worth of $180, payable in 3 years
4 months, discounting at Q% ? Ans, $150.
Define discount. Present worth. Give analysis. Rule,
260 PERCEKTAQE.
3. What is the present worth of a note for 11315.389, due
in 2 years 6 months, at 7 per cent.? Ans. $1119.48.
4. What is the present worth of a note for $866,038, due
in 3 years 6 months and 6 days, when money is worth 8 per
cent. ? What the discount ? Ans. 1190.15 + , discount.
5. What is the present worth of a debt for $1005, on
which $475 is to be paid in 10 months, and the remainder
in 1 year 3 months, the rate of interest
being e^^^J, w^ ,
lout interest, find tne pres- r
When payments are to be made at different times without
ent worth of each payment separately, and take their sum.
6. I hold a note against C for $529,925, due Sept. 1,
1859 ; what must I discount for the payment of it to-day,
Feb. 7, 1859, money being worth 6%? Ans. $17,425.
7. A man was offered $3675 in cash for his house, or
$4235 in 3 years, without interest ; he accepted the latter
offer ; how much did he lose, money being worth 7 per
cent.? ' A?is. $175.
8. A man, having a span of horses for sale, offered them
for $480 cash in hand, or a note of $550 due in 1 year 8
months, without interest; the buyer accepted the latter
offer ; did the seller gain or lose thereby, and how much,
interest being 6% ? Ans. Seller gained $20.
9. What must be discounted for the present payment of a
debt of $2637.72, of which $517.50 is to be paid in 6 months,
S 7 93. 75 in 10 months, and the remainder in 1 year 6 months,
the use of money being worth 7 per cent. ? Ans. $187.29 -f.
10. What is the difference between the interest and dis-
count of $130, due 10 months hence, at 10;^ ? A71S. $.83 J.
Promiscuous Examples m Percentage.
1. A merchant bought sugar in New York at 6 J- cents
per pound ; the wastage by transportation and retailing was
5 per cent., and the interest on the first cost to the time of
sale was 2 per cent. ; how much must he ask per pound to
gain 25 per cent.? Ans. 8^-1- cents.
PROMISCUOUS EXAMPLES. 261
2. A person purchased 2 lots of land for $200 eacli, and
sold one at 40 per cent, more than cost, and the other at 20
per cent, less ; what was his gain ? Ans. $40.
3. Sold goods to the amount of $425, on 6 months' credit,
which was 125 more than the goods cost ; what was the true
profit, money being worth 6% ? A71S. 112. 62 + .
4. Bought cotton cloth at 13 cents a yard, on 8 months'
credit, and sold it the same day at 12 cents casJi ; how much
did I gain or lose per cent. , money being worth 6 per cent.?
5. A farmer sold a pair of horses for $150 each ; on one
he gained 25 per cent., on the other he lost 25 per cent.; did
he gain or lose on both, and how much ? Ans. Lost $20.
6. A man inyested f of all he was worth in the coal trade,
and at the end of 2 years 8 months sold out his entire inter-
est for $3100, which was a yearly gain of 9 per cent, on the
money invested ; how much was he worth when he com-
menced trade ? Ans. $3750.
7. In how many years will a man, paying interest at 7
per cent, on a debt for land, pay the face of the debt in
interest ? Ans. 14f years.
8. Two persons engaged in trade ; A furnished |- of the
capital, and B | ; and at the end of 3 years 4 months they
found they had made a clear profit of $5000, which was 12|
per cent, per annum on the money invested ; how much cap-
ital did each furnish? Ans. A, $7500; B, $4500.
^9. Bought $500 worth of dry goods, and $800 worth of
groceries ; on the dry goods I lost 20 per cent., but on the
groceries I gained 15 per cent.; did I gain or lose on the
whole investment, and how much ? Ans. Gained $20.
10. What amount of accounts must an attorney collect,
in order to pay over $1100, and retain 8-^ per cent, for col-
lecting? Ans.U200
11. A merchant sold goods to the amount of $667, to be
paid in 8 months ; the same goods cost him $600 one year
previous to the sale of them ; money being worth 6 per cent.,
what was his true gain? Ans. $5,346 + .
262 PERCENTAGE.
12. A nurseiyman sold trees at $18 per hundred, and
cleared J of his receipts ; what per cent, profit did he make ?
Ans. 50^.
13. If f of an article be sold for what -f of it cost, what is
the gain per cent. ? Ans. 40|;^.
14. A lumber merchant sells a lot of lumber, which he has
had on hand 6 months, on 10 months' credit, at an advance of
30 per cent, on the first cost ; if he is paying 5 per cent, inter-
est on capital, what are his profits per cent.? Ans. 21-J^.
15. A person, owning -f of a piece of property, sold 20 per
cent, of his share ; what part did he then own? Ans. J.
16. A speculator, having money in the bank, drew 60 per
cent, of it, and expended 30 per cent, of 50 per cent, of this
for 728 bushels of wheat, at |1.12|- per bushel ; how much
was left in the bank ? Ans. ^SQAO.
17. I wish to line the carpet of a room, that is 6 yards
long and 5 yards wide, with duck f yard wide ; how many
yards of lining must I purchase, if it will shrink 4 per cent,
in length, and 5 per cent, in width ? Ans. 43f f .
18. A's money is 28 per cent, more than B's ; how many
per cent, is B's less than A's ? A7ts. 21^^.
19. A capitalist invested f of his money in railroad stock,
which depreciated 5 per cent, in value ; the remaining f he
invested in bank stock, which, at the end of 1 year, had
gained $1200, which was 12 per cent, of the investment ;
what was the whole amount of his capital, and what was his
entire loss or gain ? Ans. $25000, capital ; $450, gain.
20. O's money is to D's as 2 to 3 ; if I of C's money be
put at interest for 3 years 9 months, at 10 per cent., it will
amount to $1933.25 ; how much money has each ?
BANKING.
327. A Bank is a corporation chartered by law for the
purpose of receiving and loaning money, and furnisliiug a
paper circulation.
What is a bank ?
BAlfKING. 263
328. A Promissory Note is a -WTitteii or printed en-
gagement to pay a certain sum, either on demand or at a
specified time .
329. Bank Notes, or Bank Bills, are the notes made
and issued by banks to circulate as money. They are pay-
able in specie at the banks.
330. The Face of a note is the sum made payable by
the note.
331. Days of Grace are the three days usually allowed
by law for the payment of a note after the expiration of the
time specified in the note.
332. The Maturity of a note is the expiration of the
days of grace ; a note is due at maturity.
333. Notes may contain a promise of interest, which will
be reckoned from the date of the note, unless some other
time be specified.
The transaction of borrowing money at banks is conducted
in accordance with the following custom : the borrower pre-
sents a note, either made or indorsed by himself, payable at
a specified time, and receives for it a sum equal to the face,
less the interest for the time the note has to run. The
amount thus withheld by the bank is in consideration of
advancing money on the note prior to its maturity.
334. Bank Discount is an allowance made to a bank
for the payment of a note before it becomes due.
335* The Proceeds of a note is the sum received for it
when discounted, and is equal to the face of the note less
the discount.
Case L
336. Given the face of a note to find the proceeds.
The law of custom at banks makes the discount of a note
Define a promissory note. Bank notes. The face of a note. Days
of grace. The maturity of a note. Explain the process of discount-
ing a note at a bank. Define bank discount. The proceeds of a note.
W)jatisCaseI?
264 PEECEKTAGE.
equal to the simple interest at the legal rate for the time
specified in the note.
Rule. I. Compute the interest on the face of the note for
three days more than the specified time ; the result will be the
discount.
11. Subtract the discount from the face ofthenote^ and the
remainder will le the proceeds.
Examples for Practice.
1. What is the discount, and what the proceeds, of a note
for $450, at 60 days, discounted at a bank at Q% ?
Ans. Discount, $4,725 ; proceeds, $445,275.
2. What are the proceeds of a note for $368, at 90 days,
discounted at the Bank of New York? Ans. $361,345 + .
3. What shall I receive on my note for $475.50, at 60
days, if discounted at the Crescent City Bank, New Or-
leans? ^W5. $471.33 + .
4. What are the proceeds of a note for $10000, at 90
days, discounted at the Philadelphia Bank ? Ans. $9845.
6. Paid, in cash, $240 for a lot of merchandise. Sold it
the same day, receiving a note for $250 at 60 days, which I
got discounted at the Hartford Bank. 'What did I make
by this speculation ? Ans. $7.37|-.
6. A note for $360.76, drawn at 90 days, is discounted at
the Vermont Bank. Find the proceeds. ^W5. $355,168 + .
7. Wishing to borrow $530 of a western bank which is
discounting paper at 8 per cent., I give my note for $536.75,
payable in 60 days. How much do I need to make up the
required amount ? Ans. $.7645.
1. To indicate the maturity of a note or draft, a vertical line ( I ) is need, with the
day at which the note is nominally due on the left, and the date of maturity on the
right ; thus, Jan. ' 1 1 o-
2. When a note is on inter tst^ payable at a ftiture specified time, the amount is the
foce of the note, or the sum made payable, and must be made the basis of discount.
Give rule.
B A IT K I ]sr G . 265
Find the maturity, term of discount, and proceeds of the
following notes :
^500. BosTOK, Jan. 4, 1859.
8. Three months after date, I promise to pay to the order
of John Brown & Co. five hundred dollars, at the Suffolk
Bank, value received. James Barker.
Discounted March 2. f Due, April*],.
Ans, -l Term of discount, 36 da.
I Proceeds, $497.
^^50. St. Louis, June 12, 1859.
9. Six months after date, I promise to pay Thomas Lee,
or order, seven hundred fifty dollars, with interest, value
received. Bykok QuiiTBY.
Discounted at a broker's, Nov. 15, at 10^.
rDue, Dec. i2|i5.
Ans, -l Term of discount, 30 da.
L Proceeds, $766,434+.
Case IL
337. G-iven the proceeds of a note, to find the
face.
1. I wish to borrow $400 at a bank. For what sum must
I draw my note, payable in 60 days, so that when discounted
at 6 per cent. I shall receive the desired amount ?
OPERATION. Analysis. $400 is the
$1.0000 proceeds of a certain note,
.0105 = disc, on $1 for 63 da. *^e ^8-ce of which we are
required to find. We first
$ .9895 = proceeds of |1. ^^t^in the proceeds of $1
$400 -T- .9895 = $404,244 = by the last case, and then
face of the required note. divide the given proceeds,
$400, by this sum; for, as many times as the proceeds of $1 is con-
tained in the given proceeds, so many dollars must be the face of the
required note.
Give Case II, Analysis.
E.P. 13
HGG PEBCENTAOE.
Rule. Divide the proceeds ly tlie proceeds of $1 for the
time and rate mentioned, and the quotient will be the face of
the note.
Examples fob Pbactice.
2. What is the face of a note at 60 days, which yields
$680 when discounted at a New Hayen bank ?
Ans. $687,215.
3. What is the face of a note at 90 days, of which the
proceeds are $1000 when discounted at a Louisiana bank ?
Ans, $1013.085.
4. Wishing to borrow $500 at a bank, for what sum must
my note be drawn, at 30 days, to obtain the required amount,
discount being at 1% ? Ans, $503.22.
5. James Hopkins buys merchandise of me in New York,
at cash price, to the amount of $1256. Not having money,
he gives his note in payment, drawn at 6 months. What
must be the face of the note ? Ans. $1302.341.
EXCHANGE.
338. Exchange is a method of remitting money from
one place to another, or of making payments by written
orders.
339. A Bill of Exchange is a written request or order
upon one person to pay a certain sum to another person, or
to his order, at a specified time.
340. A Sight Draft or Bill is one requiring payment to
be made " at sight," which means, at the time of its presenta-
tion to the person ordered to pay. In other bills, the time
specified is usually a certain number of days *' after sight."
There are always three parties, and usually four, to .^
transaction in exchange.
341. The Drawer or Maker is the person who signs
the order or bill.
Give the rule. Define exchange. A hill of exchange. A sight
draft. The drawer.
EXCHANGE. 267
342. The Drawee is the person to whom the order is
addressed.
343. The Payee is the person to whom the money is
ordered to be paid.
344. The Buyer or Remitter is the person who pur-
chases the bill. He may be himself the payee, or the bill
may be drawn in favor of any other person.
345. The Indorsement of a bill is the writing upon its
back, by which i\iQ payee relinquishes his title, and transfers
the payment to another. The payee may indorse in Uanh
by writing his name only, which makes the bill payable to
the hearer, and consequently transferable like a bank note ;
or he may accompany his signature by a special order to pay
to another person, who in his turn may transfer the title in
like manner, Indorsers become separately responsible for
the amount of the bill, in case the drawee fails to make pay-
ment. A bill made payable to the dearer is transferable
without indorsement.
346. The Acceptance of a bill is the promise which the
draivee makes when the bill is presented to him to pay it at
maturity ; this obligation is usually acknowledged by writ-
ing the word "Accepted," with his signature, across the
face of the bill.
Three days of grace are usually allowed for the payment of a hill of exchange
after the time specified has expired. But in New York State no grace is allowed on
sight drafts.
From these definitions, the use of a bill of exchange in
monetary transactions is readily perceived. If a man wishes
to make a remittance to a creditor, agent, or any other per-
son residing at a distance, instead of transporting specie,
which is attended with expense and risk, or sending bank
notes, which are liable to be uncurrent at a distance from
the banks that issue them, he remits a bill of exchange,
purchased at a bank or elsewhere, and made payable to the
The drawee. The payee. The buyer. An indorsement. An
acceptance. What of grace on bills of exchange ?
268 PEECENTAGE.
proper person in or near the place where he resides. Thus
a man by paying Boston funds in Boston, may put New-
York funds into the hands of his New York agent.
347. The Course of Exchange is the variation of the
cost of sight hills from their par value, as affected by the rela-
tive conditions of trade and commercial credit at the two
places, between which exchange is made. It may be either at
a premium or discount, and is rated at a certain per cent, on
the face of the bill. Bills payable a specified time after sight
are subject to discount, like notes of hand, for the term of
credit given. Hence their value in the money market is
affected by both the course of exchange and the discount
for time.
348. Foreign Exchange relates to remittances made
between different countries.
349. Domestic or Inland Exchange relates to remit-
tances made between different places in the same country.
An inland bill of exchange is commonly called a Draft.
In this work we shall treat only of Inland Exchange,
Case I.
350. To find the cost of a draft.
^500. Syracuse, May 7, 1859.
1. At sight, pay to James Clark, or order, five hundred
dollars, value received, and charge the same to our account.
M. Smith & Co.
To Messrs. Brown & Foster, )
Baltimore. )
What is the cost of the above draft, the rate of exchange
being 1 J per cent, premium ?
OPERATION. Analysis. Since ex-
$500 X 1.015 = $507.50, Ans. change is at 1^ per cent.
premium, each dollar of
the draft will cost $1,015; and to find the whole cost of the draft.
How is exchange conducted ? Explain course of exchange. For-
eign exchange. Inland exchange. Define a draft. What is Case I ?
Give analysis.
EXCHAKGE. 269
we multiply its face, $500, by 1.015, and obtain $507.50, tbe required
result.
BoSTOK, June 12, 1859.
2. Thirty days after sight, pay to John Otis, or bearer,
four hundred eighty dollars, value received, and charge the
same to account of Amos Teenchakd.
To John Stiles & Co.,
New York.
What is the cost of the above draft, exchange being at a
premium of S% ?
OPEEiATiON. Analysis. Since time
$1.0000 is allowed, the draft must
.0055 = discount for 33 days. Suffer discount in the sale.
The discount of $1, at the
$ .9945 = proceeds of $1. je^al rate in Boston, for
.03 rz: rate of exchange. the specified time, al-
$1.0245 = cost of $1 of the draft ^^^^^ g^^^e, is $.0055,
$480 X 1.0245 = $491.76, Ans. j;^^"^^' ^TZT^ ^'°°'
$1, gives $.9945, the cost
of $1 of the draft, provided sight exchange were at par ; but sight ex-
change being at premium, we add the rate, .03, to .9945, and obtain
$1.0245, the actual cost of $1. Then, multiplying $480 by 1.0245, we
obtain $491.76.
EuLE. I. For sight drafts. — Multiply the face of the draft
dy 1 plus the rate when exchange is at a premium, and by
1 minus the rate when exchange is at a discount,
II. For drafts payable after sight. — Find the proceeds of $1
at lank discount for tlie specified time, at the legal rate where
the draft is purchased; then add the rate of exchange when
at a premium, or subtract it when at a discount, and multiply
the face of the draft by this result.
Examples fou Practice.
3. A merchant in Cincinnati wishes to remit $1000 by
Give analysis. Rrle I ; II.
270 PERCENTAGE.
draft to his agent in New York ; what will the bill cost, ex-
change being at 3 per cent, premium ? Ans. $1030.
4. What will be the cost in Rochester of a draft on Albany
for $400, payable at sight, exchange being at | per cent,
premium? A^is. $403.
5. A merchant m St. Louis orders goods from New York,
to the amount of $530, which amount he remits by draft,
exchange being at 2} per cent, premium. If he pays $20 for
transportation, what will the goods cost him in St. Louis ?
Ans. $564,575.
6. What will be the cost, in Detroit, of a draft on Boston
for $800, payable 60 days after sight, exchange being at a
premium of 2% ? Ans. $806.20.
7. A man in Philadelphia purchased a draft on Chicago for
$420, payable 30 days after sight ; what did it cost him, the
rate of exchange being 1 J per cent, discount ? Ans. $411.39.
8. A merchant in Portland receives from his agent 320
barrels of flour, purchased in Chicago at $10 per barrel ; in
payment for which he remits a draft on Chicago, at 2^ per
cent, discount. The transportation of his flour cost $312.
What must he sell it for per barrel to gain $400 ?
Ans. $12.
Case IL
351. To find the face of a draft which a given
sum will purchase.
1. A man in Indiana paid $369.72 for a draft on Boston,
drawn at 30 days ; what was the face of the draft, exchange
being at 3^ per cent, premium ?
OPERATION. Analysis. We find,
$369.72 -T- 1.027 = $360, Ans. hy Case I, that a draft
for $1 will cost $1,027;
hence the draft that will cost $369.72 must be for as many dollars as
1.027 is contained times in $369.72; dividing, we obtain $360, the
required result.
What is Case II ? Give analysis.
EQUATION OF PAYMENTS. 271
Rule. Divide the given cost hy the cost of a draft for II,
at the given rate of exchange ; the quotient will be the face
of the required draft.
Examples for Practice.
2. What draft may be purchased for $243.60, exchange
being at H per cent, premium? Ans. 1240.
3. What draft may be purchased for $79.20, exchange be-
ing at 1 per cent, discount ? Ans, $80.
4. An agent in Pittsburg holding $282.66, due his em-
ployer in New Haven, is directed to make the remittance
by draft, drawn at 60 days. What will be the face of the
draft, exchange being at 2 per cent, premium ? Ans. $280.
6. An emigrant from Bangor takes $240 in bank bills to
St. Paul, Minn., and there pays ^ per cent, brokerage in ex-
change for current money. What would he have saved by
purchasing in Bangor a draft on St. Paul, drawn at 30 days,
exchange being at IJ per cent, discount? Ans. $5,599.
6. A Philadelphia manufacturer is informed by his agent
in Buffalo that $3600 is due him on the sale of some prop-
erty. He instructs the agent to remit by a draft payable in
60 days after sight, exchange being at | per cent, premium.
The agent, by mistake, remits a sight draft, which, when
received in Philadelphia, is accepted, and paid after the ex-
piration of the three days of grace. If the manufacturer
immediately puts this money at interest at the legal rate,
will he gain or lose by the blunder of his agent ?
Ans. He will lose $8.24 -f-.
EQUATION OF PAYMENTS.
352. Equation of Payments is the process of finding
the mean or equitable time of payment of several sums, due
at different times without interest.
353. The Term of Credit is the time to elapse before a
debt becomes due.
Rule. Define equation of payments. Term of credit.
272 EQUATION OF PAYMENTS.
354. The Average Term of Credit is the time to elapse
before several debts, due at different times, may all be paid
at once, without loss to debtor or creditor.
355. The Equated Time is the date at which the sev-
eral debts may be canceled by one payment.
Case I.
356. When all the terms of credit begin at the
same date.
1. On the first day of January I find that I owe Mr.
Smith 8 dollars, to be paid in 5 months, 10 dollars to be
paid in 2 months, and 12 dollars to be paid in 10 months;
at what time may I pay the whole amount ?
OPEKATION.
$ 8 X 5 = 40
10 X 2 = 20
12 X 10 = 120
30 180 -^ 30 r= 6 mo., average time of credit.
Jan. 1+6 mo. = July 1, equated time of payment
Analysis. The whole amount to be paid, as seen above, is $30 ;
and we are to find how long it shall be withheld, or what term of
credit it shall have, as an equivalent for the various terms of credit
on the different items. Now, the value of credit on any sum is meas-
ured by the product of the money and time. And we say, the credit
on $8 for 5 mo. = the credit on $40 for 1 mo., because 8 x 5 = 40 x 1.
In the same manner, we have, the credit on $10 for 2 mo. =: the credit
on $20 for 1 mo. ; and the credit on $12 for 10 mo. = the credit on $120
for 1 mo. Hence, by addition, the value of the several terms of credit
on their respective sums equals a credit of 1 month on $180 ; and this
equals a credit of 6 months on $30, because
30 X 6 = 180 X 1.
Rule. L Multiply each payment iy its term of credit,
and divide the sum of the products by the sum of the pay-
ments ; the quotient will be the average term of credit.
Average term of credit. Equated time. Give Case I. Analysis.
Rule.
AYERAGIKG CREDITS. 273
II. Add tTie average term of credit to the date at which all
the credits legin, and the result will be the equated time of
payment,
1. The periods of time used as multipliers must all be of the same denomination,
and the quotient will be of the same denomination as the terms of credit ; if these
be months, and there be a remainder after the division, continue the division to
days by reduction, always taking the nearest unit in the last result.
2. The several rules in equation of payments are based upon the principle of bank
discount : for they imply that the discount of a sum paid before it is due equals the
Interest of the same amount paid after it is due.
Examples foe Peactice.
2. On the 25th of September a trader bought merchandise,
as follows : $700 on 20 days' credit ; 1400 on 30 days' credit;
$700 on 40 days' credit : what was the average term of credit,
9,nd what the equated time of payment ?
, ( Average credit, 30 days.
* 1 Equated time of payment, Oct. 25.
3. On July 1 a merchant gave notes, as follows : the first
for $250, due in 4 months ; the second for $750, due in 2
months ; the third for $500, due in 7 months : at what time
may they all be paid in one sum ? Ans, Nov. 1.
4. A farmer bought a cow, and agreed to pay $1 on Mon-
day, $2 on Tuesday, $3 on Wednesday, and so on for a week ;
desirous afterward to avoid the Sunday payment, he offered
to pay the whole at one time : on what day of the week
would this payment come ? Ans. Friday.
5. Jan. 1, 1 find myself indebted to John Kennedy in sums
as follows : $650 due in 4 months ; $725 due in 8 months ; and
$500 due in 12 months : at what date may I settle by giving
my note on interest for the whole amount ? Ans. Aug. 21.
Case II.
357. "When the terms of credit begin at different
dates, and the acconnt has only one side.
358. An Account is the statement or record of mercan-
tile transactions in business fonn.
Give Case II. Define an account.
12*
274
EQUATIOl^ OF PAYMENTS.
359. The Items of an account may be sums due at the
date of the transaction, or on credit for a specified time.
An account may have both a debit and a credit side, the
former marked Dr., the latter Or. Suppose A and B have
dealings in which there is an interchange of money or prop-
erty ; A keeps the account, heading it with B's name ; the
Dr. side of the account shows what B has received from A ;
the Cr. side shows what he has parted with to A.
360. The Balance of account is the difference of the
two sides, and may be in favor of either party.
If, in the transactions, one party has received nothing from
the other, the balance is simply the whole amount, and the
account has but one side. Bills of purchase are of this class.
Book accounts bear interest after the expiration of the term of credit, and notes
after they become due.
361. To Averagre an Account is to find the mean or
equitable time of payment of the balance,
362. A Focal Date is a date to which all the others are
compared in averaging an account.
1. When does the amount of the following bill become
due, by averaging ?
J. 0. Smith,
1859. To C. E. BoRDEif, Dr.
June 1. To Cash $450
'' 12. '' Mdse. on4mos 500
Aug. 16. " Mdse 250
FIKST OPERATION.
SECOND OPERATION.
Due.
da.
Items.
Prod.
Due.
da.
Items.
Prod.
June 1
Oct. 12
Aug. 16
133
76
450
500
250
66500
19000
June 1
Oct. 12
Aug. 16
133
57
450
500
250
59850
14250
1200
85500
1200
74100
85500 -J- 1200 = 71 da.
. ( 71 da. after June 1,
^»'- lor Aug. 11.
Atis.
74100 -4- 1200 = 62 da.
62 da. before Oct. 12,
or Aug. 11.
Define items. Balance. To average an account. A focal date.
AVERAGING ACCOUNTS. 275
Analysis. By reference to the example, it will be seen that the
items are due June 1, Oct. 13, and Aug. 16, as shown in the two opera-
tions. In the first operation we use the earliest date, June 1, as a focal
date, and find the difference in days between this date and each of the
others, regard being had to the number of days in calendar months.
From June 1 to Oct. 12 is 133 da. ; from June 1 to Aug. 16 is 76 da.
Hence the firat item has no credit from June 1, the second item has
133 days' credit from June 1, and the third item has 76 days' credit
from June 1, as appears in the column marked da. After this we pro-
ceed precisely as in Case I, and find the average credit, 71 da., and
the equated time, Aug. 11.
In the second operation, the latest date, Oct. 12, is taken for a focal
date ; the work is explained thus : Suppose the account to be settled
Oct. 12. At that time the first item has been due 133 days, and must
therefore draw interest for this time. But interest on $450 for 133
days = the interest on $59850 for 1 da. The second item draws no
interest, because it falls due Oct. 12. The third item must draw inter-
est 57 days. But interest on $250 for 57 days = the interest on $14250
for 1 day. Taking the sum of the products, we find the whole amount
of interest due on the account, at Oct. 12, equals the interest on $74100
for 1 day ; and this, by division, is found to be equal to the interest on
$1200 for 62 days, which time is the average term of interest. Hence
the account would be settled Oct. 12, by paying $1200 with interest on
the same for 62 days. This shows that $1200 has been due 62 days ;
that is, it falls due Aug. 11, without interest.
EuLE. I. Find the time at tvhich each item becomes due,
ly adding to the date of each transaction the term of credit,
if any he s^iecified, and write these dates in a column.
II. Assume either the earliest or the latest date for a focal
date, and find the difference in days between the focal date
and each of the other dates, and write the results in a second
column.
III. Write the items of the account in a third column, and
multiply each sum by the corresponding member of days in the
preceding column, ivriting the products in a final column.
IV. Divide the sum of the products by the sum of the items.
The quotient will be the average term of credit when the
Give analysis. Rule.
276 EQUATION OF PAi-ME]S"TS.
earliest date is the focal date, or the average term of interest
when the latest date is the focal date j in either case ahvays
reckon, from the focal date toward the other dates, to find the
equated time of payment
Examples for Pkactice.
2. John Brown,
1859. To James Greigg, Dr.
Jan. 1. To 50 yds. Broadcloth, @ $3.00 ^150
'' 16. '^2000 *^ Calico, '' .10 200
Feb. 4. '' 75 '' Carpeting, '' 1.33^ 100
March 3. '' 400 '' Oil Cloth, '' .40 160
If James Greigg wishes to settle the above bill by giving
his note, from what date shall the note draw interest ?
Ans, Jan. 27.
3. Abram Russel,
1859. To Wynkoop & Bro., Dr.
March 1. To Cash $300
April 4. '' Mdse 240
June 18. '' '' on 2 mo 100
Aug. 8. " Cash 400
What is the equated time of payment of the above account ?
Ans. May 26.
4. John Otis,
1858. To James Ladd, Dr.
June 1. To 500 bu. Wheat, @ $1.20 $600
'' 12. ''200'' " " 1.50 300
" 15. "640" " " 1.30 832
" 25. " 760 " " " 1.00 760
" 30. "500" " " 1.50 750
When is the whole amount of the above bill due, per
average? ^W5. June 18.
5. My expenditures in building a house, in the year 1856,
were as follows : Jan. 16, $536.78 ; Feb. 20, $425.36 ; March
4, $259.25 ; April 24, $786.36. If at the last date I agi'oe to
i
AVERAGE ACCOUNTS,
277
sell the house for exactly what it cost, with reference to in-
terest on the money expended, and take the purchaser's note
for the amount, what shall be the face of the note, and
what its date ? ^^^ ( Pace, $2007.75.
t Date, March 8, 1856.
a Thomas Whiting,
1859. To IsEAEL Palmer, Dr.
Jan. 1. To 60 bbls. Flour, @ $7.00 $420
'' 28. '' 90 bu. Wheat,*' 1.50 135
Mar. 15. " 300 bbls. Flour, '' 6.00 1800
If credit of 3 months be given to each item, when will
the above account become due ? Ans. May 30.
Case III.
363. "When the terms of credit begin at different
times, and the account has both a debt and a credit
side.
1. Average the following account:
David Ware.
Dr.
Or
.
185
June
tt
Oct.
3.
1
16
20
To Mdse
*' Draft, 3 mo.
" Cash
400
800
250
00
00
00
185
July
Aug.
Sept.
S.
4 By Mdse
20 " Cash
20 " "
200
150
500
00
00
00
Dr.
OPERATION.
Cr.
Focal
date.
Due.
da.
Items.
Prod.
Due.
da.
108
61
80
Items.
Prod.
June 1
Sept. 19
Oct. 20
141
31
400
800
250
56400
24800
July 4
Aug. 20
Sept. 20
200
150
500
21600
9150
15000
-
1450
850
81200
45750
850
45750
Balances.
600
35450
35450 -^ 600 = 59 da., average term of interest.
Oct. 20 — 59 da. = Aug. 22, balance due.
\Miat is Case III ? Explain operation.
278
EQUATION OF I'AYMENTS,
Analysis. In the above operation we have written the dates,
showing when the items become due on either side of the account,
adding 3 days' grace to the time allowed to the draft. The latest
date, Oct. 20, is assumed as the focal date for both sides, and the two
columns marked da. show the difference in days between each date
and the focal date. The products are obtained as in the last case, and
ft balance is struck between the items charged and the products.
These balances, being on the Dr. side, show that David Ware, on the
iay of the focal date, Oct. 20, owes $600 with interest on $35450 for
1 day. By division, this interest is found to be equal to the interest
on $600 for 59 days. The balance, $600, therefore, has been due 59
days. Reckoning back from Oct. 12, we find the date when the bal-
ance fell due, Aug. 22.
Rule. I. Find the time when each item of the account is
due ; and write the dates, in tivo columns, on the sides of the
account to which they respectively helong.
II. Use either the earliest or the latest of these dates as the
focal date for loth sides, and find the prodiicts as in the last
case,
III. Divide the halance of the products hy the balance of
the account ; the quotient will be the interval of time, which
must be reckoned from the focal date toward the other dates
when both bala^ices are on the same side of the account, but
FROM the other dates when the balances are on opposite sides
of the account.
2. What is the balance of the following account, and when
is it due ?
JOHK WiLSOK.
Dr. Or.
1859.
1
1859. 1
Jan.
1
To Mdse.
448
00
Jan.
20
Feb.
4
" Cash.
364
00
Feb.
16
It
20
€1 it
232
00
tt
25
By Am't bro't forward
" 1 Carriage
" Cash
560
264
900 00
Ans.
j Balance, $680.
/ Due March 13.
3. If the following account be settled by giving a note,
what shall be the face of the note, and what its date ?
Qive anal^'sis. Bule.
RATIO,
279
Isaac Fostee.
Br.
Oi
1858.
1858.
Jan.
1
To Mdse.
on 3 mo.
145
86
May
11
By Cash....
11
00
ti
12
U ti
.. 5 «i
37
48
July
12
« «
15
00
June
3
u «
« 3 "
12
25
Oct.
12
** *
82
00
Aug.
4
(( «*
M O <«
66
48
j«o i $154.07, face of note.
^^' \ Mar. 26, 1858, date.
EATIO.
864. Ratio is the comparison with each other of two
numbers of the same kind. It is of two kinds — arithmetical
and geometrical.
365. Aritlimetical Katio is the difference of the two
numbers.
366. Geometrical Ratio is the quotient of one num-
ber divided by the other.
367. When we use the word ratio alone, it implies geo-
metrical ratio, and is expressed by the quotient arising from
di^dding one number by the other. Thus, the ratio of 4 to 8
is 2, of 10 to 5 is J, etc.
368. Katio is indicated in two ways.
1st. By placing two points between the numbers com-
pared, writing the divisor before and the dividend after the
points. Thus, the ratio of 5 to 7 is written 5:7; the ratio
of 9 to 4 is written 9 : 4.
2d. In the form of a fraction ; thus, the ratio of 9 to 3
is I ; the ratio of 4 to 6 is f .
369. The Terms are the two numbers compared*
370. The Antecedent is the first term.
371. The Consequent is the second term.
372. No comparison of two numbers can be fully ex-
plained but by instituting another comparison ; thus, the
It is thongbt best to omit the qneptione at the bottom of the pages in the remain-
ing part of this work, leaving the teacher to use such as may be deemed appropriate.
280 RATIO.
comparison or relation of 4 to 8 cannot be fully expressed
by 2, nor of 8 to 4 by |^. If the question were asked, what
relation 4 bears to 8, or 8 to 4, in respect to magnitude, the
answer 2, or J, would not be complete nor correct. But if
we make unity the standard of comparison, and use it as
one of the terms in illustrating the relation of the two
numbers, and say that the ratio or relation of 4 to 8 is the
same as 1 to 2, or the ratio of 8 to 4 is the same as 1 to -J,
unity in both cases being the standard of comparison, then
the whole meaning is conveyed.
373. A Direct Ratio arises from dividing the conse-
quent by the antecedent. .
374. An Inverse or Reciprocal Ratio is obtained by
dividing the antecedent by the consequent. Thus, the direct
ratio of 5 to 15 is -1^ = 3 ; and the inverse ratio of 5 to 15
is ^j = h
375. A Simple Ratio consists of a single couplet ; as
3:12.
376. A Compound Ratio is the product of two or more
simple ratios. Thus, the compound ratio formed from the
simple ratios of 3 : 6 and 8:2is|-x| = 3x8:6x2 =
377. In comparing numbers with each other, they must
be of the same kind, and of the same denomination,
378. The ratio of two fractions is obtained by dividing
the second by the first ; or by reducing them to a common
denominator, when they are to each other as their numera-
tors. Thus, the ratio of -^ : | is | -^ ^ = f^ = 2, which is
the same as the ratio of the numerator 3 to the numerator
6 of the equivalent fractions ^ and ^.
Since the antecedent is a divisor and the consequent a divi-
dend, any change in either or both terms will be governed by
the general principles of division, ( 87.) We have only to
substitute the terms antecedent, consequent, and ratio, for
divisor, dividend, and quotient, and these princij^les become
EATIO. 281
GENERAL PRINCIPLES OF RATIO.
Pkik. I. Multiplying the consequent multiplies the ratio;
dividing the consequent divides the ratio.
Pkij^. II. Multiplying the antecedent divides the ratio ;
dividing the antecedent multiplies the ratio.
Prik. III. Multiplying or dividing loth antecedent and
consequent hy the same number does not alter the ratio.
These three principles may be embraced in one
Gen"eeal Law.
A change in the consequeis't produces a like change in
the ratio; tut a change in the antecedent produces an
OPPOSITE change in the ratio.
379. Since the ratio of two numbers is equal to the con-
sequent divided by the antecedent, it follows, that
1. The antecedent is equal to the consequent divided by
the ratio ; and that,
2. The consequent is equal to the antecedent multiplied
by the ratio.
Examples for Practice.
I. What part of 9 is 3 ?
I = -^ ; or, 9 : 3 as 1 : ^, that is, 9 has the same ratio to 3
that 1 has to \.
2. What part of 20 is 5 ? Ans. \.
3. What part of 36 is 4 ? Ans. \.
4. What part of 7 is 49 ? Ans. 7 times.
5. What is the ratio of 16 to 88 ? Ans. h\.
6. What is the ratio of 6 to 8|? Ans. |J.
7. What is the ratio of %\ to 78 ? Ans. 12.
8. What is the ratio of 16 to m ? Ans. 4J.
9. What is the ratio of J to | ? Ans. \.
10. What is the ratio of | to ^^ ? Ans. f .
II. What is the ratio of 3^ to 16|? Ans. 5.
12. What is the ratio of 3 gal. to 2 qt. 1 pt ? Ans. ^.
282 PROPORTION.
13. What is the ratio of 6.3s. to 8s. 6d.? Ans. Iff.
14. What is the ratio of 5.6 to .56 ? Ans. ■^.
15. What is the ratio of 19 Ihs. 5 oz. 8 pwts. to 25 lbs.
11 oz. 4 pwts. ? Ans. IJ.
16. What is the inverse ratio of 12 to 16? Ans. }.
17. What is the inverse ration of f to | ? Ans. ^.
18. What is the inverse ratio of 5J to 17}? Ans. -J.
19. If the consequent be 16 and the ratio 2f, what is the
antecedent? Ans. 7.
20. If the antecedent be 14.5 and the ratio 3, what is the
consequent ? Ans. 43.5.
21. If the consequent be J and the ratio J, what is the
antecedent? Ans. 1|-.
22. If the antecedent be | and the ratio |, what is the
consequent? Ans. ^.
PROPORTION.
380. Proportion is an equaHty of ratios. Thus, the
ratios 6 : 4 and 12 : 8, each being equal to |, form a proportion.
381, Proportion is indicated in two ways :
1st. By a double colon placed between the two ratios;
thus, 2 : 5 : : 4 : 10.
2d, By the sign of equality placed between the two ratios ;
thus, 2 : 5 = 4 : 10.
383. Since each ratio consists of two terms, every pro-
portion must consist of at least fotir terms.
383. The Extremes are the first and fourth terms.
384. The Means are the second and third terms.
385. Three numbers may be in proportion when the first)
is to the second as the second is to the third. Thus, the
numbers 3, 9, and 27 are in proportion since 3 : 9 : : 9 : 27,
the ratio of each couplet being 3.
In such a proportion the second term is said to be a mean
proportional between the other two.
386. In every proportion the product of the extremes is
equal to the product of the means. Thus, in the proportion
3 : 5 : : 6 : 10 we have 3 x 10 = 5 x 6.
SIMPLE PROPORTION-. 283
387. Four numbers that are proportional in the direct
order are proportional by inversion, and also by alternation,
or by inverting the means. Thus, the proportion 2:3::
6 : 9, by inversion becomes 3 : 2 : : 9 : 6, and by alterna-
tion 2 : 6 : : 3 : 9.
388. From the preceding principles and illustrations, it
follows that, any three terms of a proportion being given,
the fourth may readily be found by the following
EuLE. I. Divide the product of the extremes ly one of the
means, and the quotient will he the other mean. Or,
II. Divide the product of the means hy 07ie of the extremes,
and the quotient will be the other extreme.
Examples for Practice.
Find the term not given in each of the following proportions :
1. 48 : 20 : : ( ) : 50.
Ans. 120.
2. 42 : 70 : : 3 : ( ).
Ans, 5.
3. ( ) : 30 : : 20 : 100.
Ans, 6.
4. 1 : ( ) : : 7 : 84.
Ans. 12.
5. 48 yd. : ( ) : : $67.25 : $201.75.
Ans. 144 yd.
6. 3 lb. 12 oz. : ( ) : : $3.50 : $10.50.
Ans, 11 lb. 4 oz
7. ( ) : $38.25 :: 8bu. 2pk. : 76bu.2pk. Ans.U.26.
8. 4|- : 38i : : ( ) : 76^.
Ans. 8^.
9. ( ) :12::i: If
Ans. 7.
10. A: ( )::i:f.
Ans. |.
SIMPLE PROPORTION.
389. Simple Proportion is an equality of two simple
ratios, and consists of four terms, any three of which being
given, the fourth may readily be found.
390. Every question in simple proportion involves the
principle of cause and effect.
391. Causes may be regarded as action, of whatever
kind, the producer, the consumer, men, animals, time, dis-
tance, weight, goods bought or sold, money at interest, etc.
284 PROPORTION.
393. Effects may be regarded as whatever is accom-
plished by action of any kind, the thing produced or con-
sumed, money paid, etc.
393. Causes and effects are of two kinds — simple and
compound.
394. A Simple Cause, or Effect, contains but one ele-
ment ; as goods purchased or sold, and the money paid or
received for them.
395. A Compound Cause, or Effect, is the product of
two or more elements ; as men at work taken in connection
with time, and the result produced by them taken in con-
nection with dimensions, length and breadth, etc.
396. Causes and effects that admit of computation, that
is, involve the idea of quantity , may be represented by num-
bers, which will have the same relation to each other as the
things they represent. And since it is a principle of philoso-
phy that like causes produ'ce like effects, and that effects are
always in proportion to their causes, we have the following
proportions :
1st Cause : 2d Cause : : 1st Effect : 2d Effect.
Or, 1st Effect : 2d Effect ; : 1st Cause : 2d Cause ;
in which the two causes, or the two effects forming one coup-
let, must be like numbers, and of the same denomination.
Considering all the terms of the proportion as abstract
numbers, we may say that
1st Cause : Ist Effect : : 2d Cause : 2d Effect ;
which will produce the same numerical result.
But as ratio is the result of comparing tAvo numbers or
things of the same kind (377), the first form is regarded
as the most natural and philosophical.
397. Simple causes and simple effects give rise to simple
ratios ; compound causes and compound effects to compound
ratios.
398. 1. If 5 tons of coal cost $30, what will 3 tons cost ?
The required term will be denoted by a ( ), and designated " blank."
SIMPLE PROPOHTION. 285
STATEMENT. ANALYSIS. In this
tons. tons. $ $ example an effect is
5 : 3 : : 30 : ( ) required, and 5 tons
Ist cause. 2d cause. Ist effect. 2d effect. must have the same
OPERATION. ratio to 3 tons, as $30,
5x( ) = 3x30 the cost of 5 tons, to
3 >< ^06 (blank) dollars, the
( ) = rt — = ^^^> ^^^- cost of 3 tons.
Since the product
of the extremes is equal to the product of the means (373), and the
product of the means divided by one of the extremes will give the
other ; (blank) dollars will be equal to the product of 3 x 30 divided by
5, which is $18.
2. If 15 barrels of flour cost $90, how many barrels can
be bought for $30 ?
Analysis. In this ex-
ample acaiLse is required,
and the statement may be
read thus : If 15 barrels
cost $90, how many or
(blank) barrels will cost
$30 ? The product of the
extremes, 30 x 15, divided
( ) = 5 bar., Ans, ^^ ^he given mean, 90,
will give the required term, 5, as shown in the operation.
KuLE. I. Arrange the terms in the statement so that the
causes shall compose 07ie couplet, and the effects the other,
putting ( ) in the place of the required term.
II. If the required term he an extreme, divide the product
of the means hy the given extreme ; if the required term he a
mean, divide the product of the extremes hy the given mean.
1. If the terms of any couplet be of different denominations, they must be reduced
to the same unit value.
2. If the odd term be a compound number, it must be reduced to its lowest imit.
3. If the divisor and dividend contain one or more factors common to both, they
should be canceled. If any of the terms of a proportion contain mixed numbers,
they should first be changed to improper fractions, or the fractional part to a decimal.
4. When the vertical line is used, the divisor and the required term are written on
the left, and the terms of the dividend on the right.
STATEMENT.
bar.
bar. $
%
15
• ( ) :: 90
: 30
1st cause.
2d cause. 1st effect.
OPERATION.
2d effect.
00
n
( )
u^
286 PROPORTION".
399. We will now give another method of solving ques-
tions in simple proportion, without making the statement,
and which may be used, by those who prefer it, to the one
already given. We will term it the
SECoiirD Method.
Every question which properly belongs to simple propor-
tion must contain four numbers, at least three of which
must be given (389). Of the three given numbers, one
must always be of the same denomination as the required
number. The remaining two will be like numbers, and
bear the same relation to each other that the third does to
the required number ; in other words, the ratio of the third
to the required number will be the same as the ratio of the
other two numbers.
Regarding the third or odd term as the antecedent of the
second couplet of a proportion, we find the consequent or re-
quired term by multiplying the antecedent by the ratio (379).
By comparing the two like numbers, in any given ques-
tion, with the third, we may readily determine whether the
answer, or required term, will be greater or less than the
third term ; if greater, then the ratio will be greater than
1, and the two like numbers may be arranged in the form
of an improper fraction as a multiplier ; if the answer, or
required term, is to be less than the third term, then the
ratio will be less than 1, and the two like numbers may bo
arranged in the form of a proper fraction, as a multiplier.
1. If 4 cords of wood cost $12, what will 20 cords cost?
OPERATION. Analysis. It will
1^3 X 20 ^^^ be readily seen in this
12 X ^, written — ^ = $60. example, that 4 cords
and 20 cords are the
like terms, and that $12 is the third term, and of the same denomina-
tion as the answer or required term.
If 4 cords cost $12, will 20 cords cost more, or less, than 4 cords 7
evidently more : then the answer or required term will be greater
SIMPLE PBOPORTION-. 287
than fhe third term, and the ratio greater than 1. The ratio of 4
cords to 20 cords is ^^, or 5 ; hence the ratio of $12 to the answer
must be 5, and the answer will be -®^ or 5 times $12, which is $60.
2. If 12 yards of cloth cost $48, what will 4 yards cost ?
OPERATION. Analysis. In this example we
48 X ^ = 116, Ans. see that 12 yards and 4 yards are
the like terms and $48 the third
term, and of the same denomination as the required answer.
If 12 yards cost $48, will 4 yards cost more or less than 12 yards ?
less : then the ratio will be less than 1, and the multiplier a proper
fraction. The ratio of 12 yards to 4 yards is 3% ; hence the ratio of $48
to the answer is j\, and the answer will be j% times $48, which is $16.
Rule. I. With the two given numbers, which are of the
same name or kind, form a ratio greater or less than 1,
according as the answer is to be greater or less than the
third given number.
11. Multiply the third number by this ratio, and the pro-
duct will be the required number or answer.
1. Mixed numbers should first be reduced to improper fractions, and the ratio of
the fractions found according to (378).
2. Reductions and cancellation may be applied as in the first method.
The following examples may be solved by either of the
foregoing methods.
Examples foe Practice.
1. If 48 cords of wood cost $120, what will 20 cords cost ?
Ans. $50.
2. If 6 bushels of com cost $4.75, what will 75 bushels
cost ? Ans. $59.37f
3. If 8 yards of cloth cost $3|^, how many yards can be
bought for $50 ? Ans. 114f yds.
4. If 12 horses consume 42 bushels of oats in 3 weeks, how
many bushels will 20 horses consume in the same time ?
5. If 7 pounds of sugar cost 75 cents, how many pounds
can be bought for $9 ? Ans, 84 lbs.
6. What will 11 lb. 4 oz. of tea cost, if 3 lb. 12 oz. cost
13.50? Ans. $10.50.
288 SIMPLE PROPORTION.
7. If a staff 3 ft. 8 in. long cast a shadow 1 ft. 6 in., what
is the height of a steeple that casts a shadow 75 feet at the
same time? Ans. 183 ft. 4 in.
8. At 12.75 for 14 pounds of sugar, what will be the cost
of 100 pounds ? . Ans. $19.6-4f.
9. How many bushels of wheat can be bought for $51.06,
if 12 bushels can be bought for $13.32 ?
10. What will be the cost of 28J gallons of molasses, if
15 hogsheads cost $236.25 ? A?is. 17.12^.
11. If 7 barrels of flour are sufficient for a family 6 months,
how many barrels will they require for 11 months ?
12. At the rate of 9 yards for £5 12s., how many yards of
cloth can be bought for £44 16s. ? Ans. 72 yds.
13. An insolvent debtor fails for $7560, of which he is
able to pay only $3100 ; how much will A receive, whose
claim is $756 ? Ans. $310.
14. If 2 pounds of sugar cost 25 cents, and 8 pounds of
sugar are worth 5 pounds of coffee, what will 100 pounds
of coffee cost ? Atis. $20.
15. If the moon move 13° 10' 35" in 1 day, in what time
will it perform one revolution ?
16. If 8| bushels of corn cost $4.20, what will be the cost
of 13^ bushels at the same rate ? Ajis. $6.48.
17. If If yards of cotton cloth cost 6J pence, how many
yards can be bought for £10 6s. 8d. ? A71S. 694| yds.
18. If 12i cwt. of iron cost $42^, what will 48| cwt. cost?
19. What quantity of tobacco can be bought for $317.23,
if 8| lbs. cost $1|? Ans. 15 cwt. 22.7+ lbs.
20. If 15| bushels of clover seed cost $156J, how much
can be bought for $95.75 ? Ans. 9 bu. 2 pk. 2| qt
21. If I of a baiTel of cider cost ^^, how much will -J of
a barrel cost ? Ans. ^■^.
22. If a piece of land of a certain length, and 4 rods in
breadth, contain | of an acre, how much would there bo if
it were 11| rods wide? Ans. 2 A. 28 rods.
23. If 13 cwt. of iron cost $42^, what will 12 cwt. cost ?
SIMPLE PKOPORTio:sr. 289
24. A grocer has a false balance, by which 1 pound will
weigh but 12 oz. ; what is the real value of a barrel of sugar
that he sells for $28 ? Ans. $21.
25. A butcher in selling meat sells 144-J- oz. for a pound ;
how much does he cheat a customer, who buys of him to the
amount of $30 ? Ans. $2.46 + .
26. If a man clear $750 by his business in 1 yr. 6 mo.,
how much would he gain in 3 yr. 9 mo. at the same rate ?
27. If a certain business yield $350 net profits in 10 mo.,
in what time would the same business yield $1050 profits?
28. B and have each a farm ; B's farm is worth $25
an acre, and O's $30 J ; if in trading B values his land at $28
an acre, what value should put upon his ? Ans. $34.16.
29. If I borrow $500, and keep it 1 yr. 4 mo., for how
long a time should I lend $240 as an equivalent for the
favor ? Ans. 2 yr. 9 mo. 10 da.
COMPOUND PKOPORTION.
400. Compound Proportion embraces that class of
questions in which the causes, or the effects, or both, are
compound.
The required term may be a cause, or a single element of a
cause ; or it may be an effect, or a single element of an effect.
1. If 16 horses consume 128 bushels of oats in 50 days,
how many bushels will 5 horses consume in 90 days ?
STATEMENT.
let cause. 2d cause. Ist effect. 2d effect.
1 S = 1 9^ == ^^« = ( )
Or, 16 X 50 : 5 X 90 : : 128 : ( )
OPEKATiON. Analysis. In tMs ex-
$ X ^0^ X ^t$^ ample the required term
^0~X~^0 ^ is the second effect ; and
the question may be read.
If 16 horses in 50 days consume 128 bushels of oats, 5 horses in 90
days will consume how many, or (blank) bushels?
These questions are most readily performed by cancellation.
290
PROPORTION.
2. If $480 gain $84 interest in 30 months, what sum will
gain 121 in 15 months ?
1st canse.
480
30
STATEMENT.
2d cause. let effect 2d effect
(480 . j
( 30 • (
( )
15
OPEKATION.
4$0^'
X n
X U
= $240, Ans.
84 : 21
Analysis. The le.
quired term in this ex-
ample is an element of
the second clause : and
the question may be read. If $480 in 30 months gain $84, what prin-
cipal in 15 months will gain $21 ?
3. If 7 men dig a ditch 60 feet long, 8 feet wide, and 6
feet deep, in 12 days, what length of ditch can 21 men dig
in 2| days, if it be 3 feet wide and 8 feet deep ?
STATEMENT.
Ul: 1
21
^
Or, 7 X 12 : 21 X I
reo
8
6
60 X 8 X 6
3
8
( ) X 3 X 8.
^1 X 8 X
OPERATION.
$0» X ^ X
t x:t$ X
Or,
$
8
t
n
n
00»
$
$
3
0»
( )
( )= 80 ft, Ans.
Analysis. In
^ ^ ^ ^ . this example the
X $~x~$ ~ ^^ ^** ^q^^^®^ *^"^ ^
the length of the
ditch, and is an element of the
second effect. The question,
as stated, will read thus : if 7
men, in 13 days, dig a ditch 60
feet long, 8 feet wide, and 6
feet deep, 21 men, in 2f days,
will dig a ditch how many, or
(blank) feet long, 3 feet wide,
and 8 feet deep?
Rule. I. Of the given terms, select tJiose which constitute
the causes, and those which constitute the effects, a7id arrange
them in couplets, putting ( ) in place of the required term.
COMPOUKD PROPORTION-. 291
11. Then, if the Uanh term ( ) occur in either of the ex-
tremes, make the product of the means a dividend, and the
product of the extremes a divisor ; hut if the blank term occur
in either mean, make the product of the extremes a dividend,
and the product of the rneans a divisor,
1. The causes must be exactly alike in the number and Hnd of their terms ; tha
same is trae of the effects.
2. The same preparation of the terms by reduction is to be observed as in simple
proportion.
401. We will now solve an example according to the
Second Method given in Simple Proportion.
1. If 18 men can build 42 rods of wall in 16 days, how
many men can build 28 rods in 8 days ?
Analysis. In this example
OPERATION. n xi *
all the terms appear m coup-
X$^ X X =: 24 men. ^^*^' except one, which is 18
^'^ $ men, and that is of the same
kind as the required answer.
Since compound proportion is made up of two or more simple pro-
portions, if this third or odd term be multiplied by the compound ratio,
or by the simple ratio of each couplet successively, the product will
be the required term.
By comparing the terms of each couplet with the third term we may
readily determine whether the answer, or term sought, will be greater
or less than the third term ; if greater, then the ratio will be greater
than 1, and the multiplier an improper fraction ; if less, the ratio will
be less than 1, and the multiplier a proper fraction.
First we will compare the terms composing the first couplet, 43
rods and 28 rods, with the third term, 18 men. If 42 rods require
\8 men, how many men will 28 rods require? less men; hence the
.•atio is less than 1, and the multiplier a proper fraction, f| ; next, if
16 days require 18 men, how many men will 8 days require ? more
men ; hence the ratio is greater than 1, and the multiplier an improper
fraction, Y-. Regarding the third term as the antecedent of a couplet,
the consequent being the term sought, if we multiply this third term
by the simple ratios, or by their product, we shall have the required
term or answer, thus : 18 x |-| x -^g^ = 24, as shown in the operation.
2. 5 compositors, in 16 days, of 14 hours each, can com-
pose 20 sheets of 24 pages in each sheet, 50 lines in a page.
16
u
^
t
X^
n
u
102
^0
00
m
00
293 COMPOUND PROPORTION.
and 40 letters in a line ; in how many days, of 7 hours each,
will 10 compositors compose a volume to be printed in the
same letter, containing 40 sheets, 16 pages in a sheet, 60
lines in a page, and 50 letters in a line ? Ans. 32 days.
OPERATION.
Days. ^Comp. Hours. Sheets. Pages. Lines. Letters.
16xAxJ^xi*xifxttxtt = 32 days.
BY CANCELLATION. ANALYSIS. The required term or an-
swer is to be in days'; and we see that
all the terms appear in pairs or couplets,
except the 16 days, which is of the same
kind as the answer sought.
We will proceed to compare the terms
of each couplet with the 16 days. First,
if 5 compositors require 16 days, how
many days will 10 compositors require?
less days; hence the multiplier is the
32 days, Ans, proper fraction y^^, and we have 16 x ^^,
Next, if 14 hours a day require 16 days,
how many days will 7 hours a day require ? more days ; hence the
multiplier is the improper fraction \S and we have 16 x ^^^ x \\
Next, if 20 sheets require 16 days, how many days will 40 sheets re-
quire ? more days ; hence the multiplier is the improper fraction f^,
and we have 16 x ^^ x if^ x |§. Pursuing the same method with the
other couplets, we obtain the result as shown in the operation.
Rule. L Of the terms composing each couplet form a
ratio greater or less than 1, in the same manner as if the
answer depended on those two and the third or odd term,
II. Multiply the third or odd term hy these ratios succes-
sively , and the product will be the ayiswer sought.
By the odd term is meant the one that is of the same kind as the answer.
The following examples may be solved by eitlier of tho
given methods :
Examples for Practice.
1. If IG horses consume 128 bushels of oats in 50 days,
how many bushels will 5 horses consume in 1)0 days ?
COMPOUKD PEOPORTIOX. 293
2. If a man travel 120 miles in 3 days when the days are
12 hours long, in how many days of 10 hours each will he
require to travel 360 miles ? Ans. lOf days.
3. If 6 laborers dig a ditch 34 yards long in 10 days, how
many yards can 20 laborers dig in 15 days ? Ans. 170 yds.
4. If 450 tiles, each 12 inches square, will pave a cellar,
how many tiles that are 9 inches long and 8 inches wide will
pave the same ? A7is. 900.
5. If it require 1200 yards of cloth f wide to clothe 500
men, how many yards which is -J wide will it take to clothe
960 men ? Ans, 3291^ yds.
6. If 8 men will mow 36 acres of grass in 9 days, of 9
hours each day, how many men will be required to mow 48
acres in 12 days, working 12 hours each day? Ans, 6 men.
7. If 4 men, in 2J days, mow 6-| acres of grass by work-
ing 8i hours a day, how many acres will 15 men mow in 3|
days by working 9 hours a day ? Ans. 40|^ acres.
8. If, by traveling 6 hours a day at the rate of 4|- miles
an hour, a man perform a journey of 540 miles in 20 days, in
how many days, traveling 9 hours a day at the rate of 4|-
miles an hour, will he travel 600 miles? Ans. 14f days.
9. If 2J yards of cloth If yards wide cost I3.37J, what
cost 36 J yards, 1^ yards wide ? Ans. $52.79+.
10. If 5 men reap 52.2 acres in 6 days, how many men
will reap 417.6 acres in 12 days? Ans, 20 men.
11. If 6 men dig a cellar 22.5 feet long, 17.3 feet wide, and
10.25 feet deep, in 2.5 days, of 12.3 hours, in how many days,
of 8.2 hours, will 9 men take to dig another, measuring 45
feet long, 34.6 wide, and 12.3 deep? Ans. 12 days.
12. If 54 men can build a fort in 24J days, working 12^
hours each day, in how many days will 75 men do the same,
when they work but 10|- hours each day? Ans. 21 days.
13. If 24 men dig a trench 33J yards long, 5f wide, and
3^ deep, in 189 days, working 14 hours each day, how many
hours per day must 217 men work, to dig a trench 23J- yards
long, 3f wide, and 2^ deep, in 5^ days ? Ans. 16 hours.
294 PAETNEESHIP.
PAETNEESHIP.
402. Partnership is a relation established between two
or more persons in trade, by which they agree to sliare the
profits and losses of business.
403. The Partners are the individuals thus associated.
404. Capital, or Stock, is the money or property in-
vested in trade.
405. A Dividend is the profit to be divided.
406. An Assessment is a tax to meet losses sustained.
Case I.
207. To find each partner's share of the profit or
loss, when their capital is employed for equal pe-
riods of time.
1. A and B engage in trade ; A furnishes $300, and B
$400 of the capital ; they gain $182 ; what is each one's
share of the profit ?
OPERATION. Analysis. Since
$300 tlie whole capital
^/^{^^ employed is $300
5r;r7- + $400 = $700, it
$700, whole Btock. .^ ^^^^^^ ^^^^^
^% — f J A's share of thestock. furnishes \l% = f
4^^ = -f, B's share of the stock. of the capital, and
$182 Xf = $78, A'8 share of the gain. B |g§ = | of the
$1 82 X 4 = $104, B's share of the gain. capital. And since
each man's share of
the profit or loss will have the same ratio to the whole profit or loss
that his share of the stock has to the whole stock, A will have f of
the entire profit, and B |, as shown in the operation.
We may also regard the whole capital as the first catise,
and each man's share of the capital as the second cause , the
whole profit or loss as the first effect, and each man's share
of the profit or loss as the second effect, and solve by propor-
tion thus ;
PAETNEKSHIP.
let cause.
3d cause.
Ist effect
ad effect.
$700 :
$300
: : $182
: ( )
$700 :
$400
: : $182
: ( )
»00
$003
tm
**00«
( )
m^
f )
ig^as
( ) = $78, A'8 profit ( ) = $104, B'B profit
Rule. Multiply the whole profit or loss by the ratio of the
whole capital to each man's share of the capital. Or,
The tuhole capital is to each man's share of the capital as the
whole profit or loss is to each man's share of the profit or loss,
2. Three men trade in company ; A furnishes $8000, B
$12000, and $20000 of the capital ; their gain is $1680 ;
what is each man's share ?
Ans. A's $336 ; B's $504 ; C's $840.
3. Three persons purchased a house for $2800, of which
A paid $1200, B $1000, and C $600; they rented it for $224
a year ; how much of the rent should each receive ?
4. A man failed in business for $20000, and his available
means amounted to only $13654 ; how much wiU two of his
creditors respectively receive, to one of whom he owes $3060,
and to the other $1530 ? Ans, $2089.062 ; $1044.531.
5. Four men hired a coach for $13, to convey them to
their respective homes, which were at distances from the
place of starting as follows : A's 16 miles, B's 24 miles, C's
28 miles, and D's 36 miles ; what ought each to pay ?
J j A $2. C $3.50.
^'''- (B$3. D$4.50.
6. A captain, mate, and 12 sailors took a prize of $2240,
of which the captain took 14 shares, the mate 6 shares, and
each sailor 1 share ; what did each receive ?
7. A cargo of com, valued at $3475.60, was entirely lost;
I of it belonged to A, J of it to B, and the remainder to C ;
how much was the loss of each, there being an insurance of
$2512 ? A71S, $120.45, A's. $240.90, B's. $002.25, C's.
296 PARTNERSHIP-
8. Three persons engaged in the lumber trade ; two of the
persons furnished the capital, and the third managed the
business ; they gained $2571.24, of which received $6 as
often as D 14, and E had ^ as much as the other two for
taking care of the business ; how much was each one's share
of the gain? Ans. $12So.62,C's. $857.08, D's. $428.54, E's.
9. Four persons engage in the coal trade ; D puts in
$3042 capital ; they gain $7500, of which A takes $2000, B
$2800.75, and C $1685.25 ; how much capital did A, B, and
C put in, and how much is D's share of the gain ?
.^ (A, $6000. C, $5055.75.
' ( B, $8402.25. D's gain, $1014.
Case IL
408, To find each partner's share of the profit
or loss when their capital is employed for U7iequal
periods of time.
It is evident that the respective shares of profit and loss
will depend upon two conditions, viz. : the amount of cajntal
invested by each, and the time it is employed.
1. Two persons form a partnership ; A puts in $450 for
7 months, and B $300 for 9 months ; they lose $156 ; how
much is each man's share of the loss ?
OPERATION. Analysis. The
$450 X 7 = $3150, A'8 capital for 1 mo. use of $450 capital
$300 X 9 = $2700, B'B " " « for 7 months is the
— same as the use ot
$5850, entire ' 7 tinges $450, or
^44 = tV, A'b Bhare of the entire capital ^^^^^ ^^^ 1 "^o"*^ *>
MM = A B'B " " " - ^^*^ ^^ ^^^ ^«" ^
IfT? 7?' Ar^i months is the same
$156 X A = ^84, A'sloBB. astheuseof 9times
$156 X A = $72, B'B " 1300, or $2700 for
1 month. The en-
tire capital for 1 month is equivalent to $3150 + $3700 = $5850. If
the loss, $156, be divided between the two i)artnerfl, accordinp: to
Case I, the results will be the loss of each as shown in the operation.
PARTNERSniP. 297
Examples of this kind may also be solved by proportion as in Case I,
tbe catises being compounded of capital and tvne ; thus,
$5850 : $3150 : : $156 : ( )
$5850 : $2700 : : $156 : ( )
( )
$1007 $$$0
m^^ ( )
M00^
( ) = $84, A's loss. { ) == ^'^^^ B's loss.
Rule. Multiply each man^s capital hy the time it is em-
ployed in trade, and add the products. Then multiply the
entire profit or loss hy the ratio of the sum of the products to
each product, and the results will he the respective shares of
profit or loss of each partner. Or,
Multiply each man^s capital hy the time it is employed in
trade, and regard each product as his capital, and the sum
of the products as the entire capital, and solve hy proportion,
as in Case I.
Examples for Practice.
2. Three persons traded together ; B put in $250 for 6
months, C $275 for 8 months, and D $450 for 4 months ;
they gained $825 ; what was each man's shai'e of the gain ?
3. Two merchants formed a partnership for 18 months.
A at first put in $1000, and at the end of 8 months he put
in $600 more ; B at first put in $1500, but at the end of 4
months he drew out $300 ; at the expiration of the time
they found that they had gained $1394.64 ; what was each
man's share of the gain? Ans. A's $715.20 ; B's $679.44.
4 Three men took a field of grain to harvest and thresh
for J- of the crop ; A furnished 4 hands 5 days, B 3 hands
6 days, and 6 hands 4 days ; the whole crop amounted to
372 bushels ; what was each man's share ?
5. William Gallup began trade January 1, 1856, with a
capital of $3000, and, succeeding in business, took in M. H.
Decker as a partner on the first day of March following, with
a capital of $2000 ; four months after they admitted J. New-
13*
298 ANALYSIS.
man as third partner, who put in $1800 capital ; they con-
tinued their partnership until April 1, 1858, when they
found that $4388.80 had been gained since Jan. 1, 1856 ;
what was each one's share ? C $2106, Gallup's.
Ans. ^ $1300, Decker's.
L $ 982.80, Newman's.
6. Two persons engaged in partnership with a capital of
$5600 ; A'? capital was in trade 8 months, and his share of
the profits was $560 ; B's capital was in 10 months, and his
share of the profits was $800 ; what amount of capital had
each in the firm? Ans, A, $2613.33^ ; B, $2986.66f.
7. A, B, and C, engaged in trade with $1930 capital ; A's
money was in 3 months, B's 5, and C's 7; they gained $117,
which was so divided that ^ of A's share was equal to J of
B's and to } of C's ; how much did each put in, and what
did each gain ? f A, $700 capital ; $26 gain.
Ans, \ B, $630 " $39 "
L C, $600 " $52 "
ANALYSIS.
409. Analysis, in arithmetic, is the process of solving
problems independently of set rules, by tracing the relations
of the given numbers and the reasons of the separate steps
of the operation according to the special conditions of each
question.
410, In solving questions by analysis, we generally reason
from the given number to unity, or 1, and then from unity,
or 1, to the required number,
4 11 ^ United States money is reckoned in dollars, dimes,
cents, and mills ( 180 ), one dollar being uniformly valued
in all the States at 100 cents.
At the time of the adoption of onr decimal currency by Conprresp, in 1786, the
colonial currency, or billft of credit, ispued by the coloniee, had depreciated in value;
and this depreciation, being unequal in the different colonies, gave riee to the dif-
ferent values of the State currencies.
In many of the States it was customary to pive the retail price of articles in Blii)«
Ungs and pence and the cost of the whole in dollars and cents.
ANALYSIS. 299
This nsage has become nearly If not quite obsolete all over the country ; but as
the matter has an historical interest, it is retained in this new edition, to avoid
derangement with previous editions, and the examples will also afford a pleasant
and profitable exercise for the pupil.
413. In New England, Indiana, )
Illinois, Missouri, Virginia, Kentucky, )• $1 = 6s. = 72d.
Tennessee, Mississippi, Texas )
New York, Ohio, Michigan $1 = 8s. = 96d.
New Jersey, Pennsylvania, Dela- I $i __ 7g ga _ qqa
ware, Maryland )
South Carolina, Georgia ) II = 4s. 8d. = 56d.
The Dominion of Canada ) |1 = 5s. = 60d.
Examples for Practice.
1. What will be the cost of 42 bushels of oats, at 3 shil'
lings per bushel. New England currency ?
OPERATION. Analysis. Since
|4^^ 1 bushel costs 3 sliil-
3 lings, 42 bushels will
x^u -:- u == ^^x. yji, cost 42 times 3s., or
$21, Ans. 42 X 3 = 126s. ; and
as 6s. make 1 dollar
New England currency, there are as many dollars in 126s. as 6 is con-
tained times in 126, or $21.
2. What will 180 bushels of wheat cost at 9s. 4d. per
bushel, Pennsylvania currency ?
OPERATION. Analysis. Multi.
Or,
00
112 $
$224 ^^
plying the number of
28 bushels by the price,
o and dividing by the
value of 1 dollar re-
$224, Ans, duced to pence, we
have $224. Or, when
the pence in the ^ven price is an aliquot part of a shilling, the price
may be reduced to an improper fraction for a multiplier, thus : 9s. 4d.
= 9|s. = -\^-s., the multiplier. The value of the dollar being 7s. 6d. =
7|s. = ^', we divide by -i/, as in the operation.
3. What will be the cost of 3 hhd. of molasses, at Is. 3d,
per quart, Georgia currency ?
800
2
$X
BOO AN-ALYSIS.
OPEBATION.
3 Analysis. In this example we first
0^9 reduce 3 lihd. to quarts, by multiplying
^ by 63 and 4, and then multiply by the
^ f, price, either reduced to pence or to an
improper fraction, and divide by the
405.00 value of 1 dollar reduced to the same
^202 50 denomination as the price.
4. Sold 9 firkins of butter, each containing 56 lb., at Is. 6d.
per pound, and received in payment carpeting at 6s. 9d.
per yard ; how many yards of carpeting would pay for the
butter ?
OPERATION. Analysis. The operation in this is similar
to the preceding examples, except that we di-
gg vide the cost of the butter by the price of a
^ rt 3 unit of the articl e received in payment, reduced
. to the same denominational unit as the price
112 yd. of a unit of the article sold. The result will
be the same in whatever currency.
5. What will 3 casks of rice cost, each weighing 126
pounds, at 4d. per pound, South Carolina currency ?
Ans, 127.
6. How many pounds of tea, at 7s. per pound, must bo
given for 28 lb. of butter, at Is. 7d. per pound ? A7is. 6^.
7. Bought 2 casks of Catawba wine, each containing 72
gallons, for $648, and sold it at the rate of 10s. 6d. per quart,
Ohio currency ; how much was my whole gain ? Ans. $108.
8. What will it cost to keep 2 horses 3 wk. if it cost to
keep 1 horse 1 day, 2s. 6d. Canada currency ?
9. How many days' work, at 6s. 3d. per day, must be given
for 20 bushels of apples at 3s. per bushel ? A7is. 9|.
10. Bought 160 lb. of dried fruit, at Is. 6d. a pound, in
New York, and sold it for 2s. a pound in Pliiladelphia ;
what was my whole gain ? Ans. $!l2.66-|.
11. A merchant exchanged 43| yards of cloth, worth
lOs. 6d. per yard, for other cloth worth 8s. 3d. per yard ;
how many yards did he receive ? Ans. 55^^.
ANALYSIS. 301
12. What will be the cost of 300 bushels of wheat at 9s. 4d.
per bushel, Michigan currency ? Ans. 1350.
13. If f of 4 of a ton of coal cost $2f , how much will 4 of
6 tons cost ?
OPERATION.
^ I 1^2 Analysis. Since | of f = ^ of a ton costs
i:^ U04 $2| = $-V-, 1 ton wiU cost 28 times ^ of $J^,
^ *03 or $-i/xf|; and f of 6 tons = V" toi^» will
— -! cost -^/ times f| of $J^ = $16.
$16, Ans,
14. If 8 men can build a wall 20 ft. long, 6 ft. high, and
4 ft. thick, in 12 days, working 10 hours a day, in how many
days can 24 men build a wall 200 ft. long, 8 ft. high, and 6 ft.
thick, working 8 hours a day ?
OPERATION.
u
X
X
10
^0010
X
1
^ ^0
^ X ^ = 100 da.
Analysis. Since 8 men require 13 days of 10 hours each to build
the wall, 1 man would require 8 times 12 days of 10 hours each, and
10 times (12 x 8) days of 1 hour each. To build a wall 1 ft. long would
require ^^ as much time as to build a wall 20 ft. long ; to build a wall
1 ft. high would require ^ as much time as to build a wall 6 ft. high ;
to build a wall 1 ft. thick, \ as much time as to build a wall 4 ft. thick.
Now, 24 men could build this wall in ^ as many days, by working
1 hour a day, as 1 man could build it, and in J as many days by work-
ing 8 hours a day, as by working 1 hour a day ; but to build a wall
200 ft. long would require 200 times as many days as to build a wall
1 ft. long ; to build a wall 8 ft. high would require 8 times as many
days as to build a wall 1 ft. high ; and to build a wall 6 ft. thick would
require 6 times as many days as to build a wall 1 ft. thick.
15. If 2 pounds of tea are worth 11 pounds of coifee, and
3 pounds of coffee are worth 5 pounds of sugar, and 18
pounds of sugar are worth 21 pounds of rice, how many
pounds of rice can be purchased with 12 pounds of tea ?
302 ANALYSIS.
OPERATION. Analysts. Since 18 lb. of sugar
^1 ' are equal in value to 21 lb. of rice,
5 1 lb. of sugar is equal to y^^ of 21 lb.
-J ^ of rice, or ^ = -g lb. of rice, and 5 lb.
of sugar are equal to 5 times ^ lb.
^^ of rice, or 5^ lb. ; if 3 lb. of coffee
3 I 385 are equal to 5 lb. of sugar, or \^ lb.
. "ToQj. iv. ®^ ^^^^> 1 ^^- ^^ coffee is equal to ^
jlnS, l/i»-^ ID. ^^ 8^5 j^ ^^ ^^^ ^j. 35 j^^ ^^ jj jj^
of coffee are equal to 11 times ff lb. of rice, or -\«/ lb.; if 3 lb. of tea
are equal to 11 lb. of coffee, or ^^- lb. of rice, 1 lb. of tea is equal to ^
of i*y\5- lb. of rice, or ^%^ lb., and 12 lb. of tea are equal to 13 times ^^
lb. of rice, or iJ|^ lb. = 128^ lb.
16. If 16 horses consume 128 bushels of oats in 50 days,
how many bushels will 5 horses consume in 90 days ? Ans. 72.
17. If $10J will buy 4f cords of wood, how many cords
can be bought for $24 J ? Ans, 11.
18. Gave 52 barrels of potatoes, each containing 3 bushels,
worth 33^ cents a bushel, for 65 yards of cloth ; how much
was the cloth worth per yard ? Ans, $.80.
19. If a staff 3 ft. long cast a shadow 5 ft. in length, what
is the height of an object that casts a shadow of 46-| ft. at
the same time of day? Ans. 28 ft.
20. Three men hired a pasture for $63 ; A put in 8 sheep
7|^ months, B put in 12 sheep 4^ months, and put in 15
sheep 6f months ; how much must each pay ?
21. If 7 bushels of wheat are worth 10 bushels of rye,
and 5 bushels of rye are worth 14 bushels of oats, and 6
bushels of oats are worth $3, how many bushels of wheat
will $30 buy? A}is. 15.
22. If $480 gain $84 in 30 months, what capital will gain
$21 in 15 months ? Ans. $240.
23. How many yards of carpeting f of a yard wide are
equal to 28 yards | of a yard wide ? Ans. 31 J.
24. If a footman travel 130 miles in 3 days, when the days
are 14 hours long, in how many days of 7 hours each will he
travel 390 miles ? Ans, 18.
AIS^ALYSIS. 303
25. If 6 men can cut 45 cords of wood in 3 days, how
many cords can 8 men cut in 9 days ? Ans. 180.
26. B's age is 1^ times the age of A, and C's is 2^^ times
the age of both, and the sum of their ages is 93 ; what is
the age of each ? Ans. A^s age, 12 yrs.
27. If A can do as much work in 3 days as B can do in
4|- days, and B can do as much in 9 days as C in 12 days,
and C as much in 10 days as D in 8, how many days' work
done by D are equal to 5 days' done by A ? Ans. 8.
28. The hour and minute hands of a watch are together
at 12 o'clock, M. ; when will they be exactly together the
third time after this ?
OPERATION. Analysis. Since tlie
12 X A- X 3 = 3^^ h. minute hand passes the
Ans. 3 h. 16 min. 21-j9^ sec, p.m. ^^^ ^^^^^ 11 *i"^®s in
12 hours, if both are
together at 12, the minute hand will pass the hour hand the first time
in y\ of 12 hours, or 1-^j hours ; it will pass the hour hand the second
time in -^^ of 13 hours, and the third time in ^^ of 12 hours, or Sf'j
hours, which would occur at 16 min. 21^ sec. past 3 o'clock, p. M.
29. A flour merchant paid 1164 for 20 barrels of flour,
giving $9 for first quality, and $7 for second quality; how
many barrels were there of each ?
OPERATION. Analysis. If aU had been
$9 X 20 = $180 ; first quality, he would have
$180 — $164 = $16. P^i^ ^1^^' o^ ^1^ ™o^® *^^ ^«
Ag $7 = 12 • ^^ ^^^' ^^^^ barrel of sec-
Q , ', , ond quality made a difference
Hy^Z= 8 bbl., 2d quality. .^f ^3 ^ ^^^ ^^^ . ^^^^ ^^^^^
^^ — 8 = 12 bbl., ist quality. were as many barrels of second
quality as $2, the difference in
the cost of one barrel, is contained times in $16, etc.
30. A boy bought a certain number of oranges at the rate
of 3 for 4 cents, and as many more at the rate of 5 for 8
cents ; he sold them again at the rate of 3 for 8 cents, and
304 ANALYSIS.
gained on the whole 108 cents ; how many oranges did he
buy?
OPERATION. Analysis. For
1 + 1= fl; ^^2z=^, average cost tliose he bought
^-U = U = H Cts., gain on each. ** ^^^ ^^^ ^^ ^ ^'^^
1 A? TT JS 4 cents he paid |
lOo -T- 1-V = yO, number of oranges. , x i j
o ' of a cent each, and
for those he bought at the rate of 5 for 8 cents he paid f of a cent
each ; and | + f = ff cents, what he paid for 1 of each kind, which
divided by 2 gives f f cents, the average price of all he bought. He
sold them at the rate of 3 for 8 cents, or | cents each ; the diflerence
between the average cost and the price he sold them for, or | — f f =
^1 = 1^ cents, is the gain on each ; and he bought as many oranges as
the gain on one orange is contained times in the whole gain, etc.
31. A man bought 10 bushels of wheat and 25 bushels of
com for $30, and 12 bushels of wheat and 5 bushels of corn
for 820 ; how much a bushel did he give for each ?
Analysis. We may divide or
multiply either of the expressions
by such a number as will render
one of the commodities purchased,
alike in both expressions. In this
example we divide the first by 5
to make the numbers denoting
the corn alike, (the same result
would be produced by multiply-
ing the second by 5,) and we have
the cost of 2 bushels of wheat and 5 bushels of corn, equal to $6.
Subtracting this from 12 bushels of wheat and 5 bushels of com,
which cost $20, we find the cost of 10 bushels of wheat to be $14 ;
therefore the cost of 1 bushel is yV ^^ $^4, or $1.40. From any one of
the expressions containing both wheat and corn, we readily find the
cost of 1 bushel of corn to be 64 cents.
32. A, B, and agree to build a barn for $270. A and
B can do the work in 16 days, B and C in 13-J^ days, and A
and in llf days. In liow many days can all do it work-
ing together ? In how many days can each do it alone ?
What part of the pay ought each to receive ?
urn
W.
C.
Ist lot, 10
25 $30
2d lot, 12
5 $20
^t ^ 5 = 2
6 $6
10.
$14
1 bu. W.
=z $1.40
1 bu. C.
= $ .64
ANALYSIS,
305
OPERATION.
Yg- = -go", what A and B do in 1 day.
•^ = -^Q, what B and C do in 1 day.
"eV ~~ IsV^ what A and C do in 1 day.
A 4- W + A = it^ what A, B, and C do
in 2 days.
^ -T-2 = -^, what A, B, and C do in 1 day,
1 -T- -^ = S"! days, time A, B, and C will do
the whole work together.
A — A = "8% ? I'^A — ^^ da., C alone.
A~A = A? l-i-^ = 26-|da.,Aalone.
A — A = A J 1 "^ A = ■^^ da., B alone.
^ X 8-| = I", the part of the whole C did.
-^ X 8f = I", the part of the whole A did.
-^ X Sf = f , the part of the whole B did.
$270 Xi = 1120, C's share.
$270x1= $90, A's share.
1270 X f = $60, B'8 share.
Analysis. Since
A and B can do the
work in 16 days,
they can do -^^ of it
in 1 day ; B and C,
in 131 or -^^ days,
they can do -^ of it
in 1 day ; A and C,
in llf or ^ days,
they can do /^ of it
in 1 day. Then A,
B, and C, by work-
ing 2 days each, can
of the work, and by
working 1 day each
they can do | of |f ,
or ^^j of the work ;
and it will take them
as many days work-
ing together to do
the whole work as ^^ is contained times in 1, or 8f days.
Now, if we take what any two of them do in 1 day from what the
three do in 1 day, the remainder will be what the third does ; we thus
find that A does -g%, B -g%, and C /^.
Next, if we denote the whole work by 1, and divide it by the part
each does in 1 day, we have the number of days that it will take each
to do it alone, viz. : A 26| days, B 40 days, and C 20 days. And each
should receive sucb a part of $270 as would be expressed by the part
he does in 1 day, multiplied by the number of days he works, which
will give to A $90, B $60, and C $120.
33. If 6 oranges and 7 lemons cost 33 cents, and 12
oranges and 10 lemons cost 54 cents, what is the price of
1 of each? Ans. Oranges, 2 cents ; lemons, 3 cents.
34. If an army of 1000 men have provisions for 20 days,
at the rate of 18 oz. a day to each man, and they be rein-
forced by 600 men, upon what allowance per day must each
man be put, that the same provisions may last 30 days ?
35. There are 54 bushels of grain in 2 bins ; and in one
bin are 6 bushels less than J as much as there is in the
other ; how many busliels in tlie larger bin ? Aits. 40.
306 ANALYSIS.
36. The sum of two numbers is 20, and their difference
is equal to ^ of the greater number ; what is the greater
number ? Ans. 12.
37. If A can do as much work in 2 days as C in 3 days,
and B as much in 5 days as in 4 days ; what time will B
require to execute a piece of work which A can do in 6
weeks ? Ans. 11 J weeks.
38. How many yards of cloth, | of a yard wide, will line
36 yards 1 J yards wide ? Ans. 60.
39. How many sacks of coffee, each containing 104 lbs.,
at lOd. per pound N. Y. currency, will pay for 80 yards of
broadcloth at $3^ per yard ? A7is. 24.
40. A person, being asked the time of day, replied, the
time past noon is equal to -J of the time to midnight ; what
was the hour ? Ans. 2 p.m.
41. A market woman bought a number of peaches at the
rate of 2 for 1 cent, and as many more at the rate of 3 for 1
cent, and sold them at the rate of 5 for 3 cents, gaining 55
cents ; how many peaches did she buy? Atis. 300.
42. A can build a boat in 18 days, working 10 hours a day,
and B can build it in 9 days, working 8 hours a day ; in how
many days can both together build it, working 6 hours a day ?
43. A man, after spending ^ of his money, and J of the
remainder, had $10 left ; how much had he at first ?
44. If 30 men can perform a piece of work in 11 days,
how many men can accomplish another piece of work, 4
times as large, in ^ of the time ? Ans. 600.
45. If 16i lb. of coffee cost $3 J, how much can be bought
for $1.25? Ans. 6^ lb.
46. A man engaged to write for 20 days, receiving $2.50
for every day he labored, and forfeiting $1 for every day he
was idle ; at the end of the time he received $43 ; how
many days did he labor? Ans. 18.
47. A, B, and C can perform a piece of work in 12 hours ;
A and B can do it in 16 hours, and A and C in 18 hours ;
what part of the work can B and C do in 9| hours ? Ans. 4.
ALLIGATION MEDIAL. 307
ALLIGATION.
413, Alligation treats of mixing or compounding two
or more ingredients of different values. It is of two kinds —
Alligation Medial and Alligation Alternate,
414,, Alligation Medial is the process of finding the
average price or quality of a compound of several simple
ingredients whose prices or qualities are known.
1. A miller mixes 40 bushels of rye worth 80 cents a
bushel, and 25 bushels of corn worth 70 cents a bushel, with
15 bushels of wheat worth $1.50 a bushel ; what is the value
of a bushel of the mixture ?
Analysis. Since 40 bushels
of rye at 80 cents a bushel is
worth $32, and 25 bushels of corn
at 70 cents a bushel is worth
$17.50, and 15 bushels of wheat
at $1.50 a bushel is worth $22.50,
^90 An^ therefore the entire mixture, con-
sisting of 80 bushels, is worth
$72, and one bushel is worth ^V of $72, or 72 -^ 80 = $.90.
EuLE. Divide the entire cost or value of the ingredients
hy the sum of the simples.
Examples for Peactice.
2. A wine merchant mixes 12 gallons of wine, at $1 per
gallon, with 5 gallons of brandy worth $1.50 per gallon, and
3 gallons of water of no value ; what is the worth of one
gallon of the mixture ? Ans. $.975.
3. An innkeeper mixed 13 gallons of water with 52 gallons
of brandy, which cost him $1.25 per gallon ; what is the value
of 1 gallon of the mixture, and what his profit on the sale of
the whole at 6 J cents per gill ? Ans. $1 a gallon ; $65 profit.
4. A grocer mixed 10 pounds of sugar at 8 cts. with 12
pounds at 9 cts. and 16 pounds at 11 cts., and sold the mix-
ture at 10 cents per pound ; did he gain or lose by the sale,
and how much ? Ans. He gained 16 cts.
OPERATION.
80
X 40
= $32.00
70
X 25
= 17.50
1.50
X 15
= 22.50
80
) 72.00
308 ALLIGATIOJq^ ALTERNATE.
5. A grocer bought 7J- dozen of eggs at 12 cents a dozen,
8 dozen at 10^ cents a dozen, 9 dozen at 11 cents a dozen,
and lOJ dozen at 10 cents a dozen. He sells them so as to
make 50 per cent, on the cost ; how much did he receive
per dozen ? Ans. 16-J cents.
6. Bought 4 cheeses, each weighing 50 pounds, at 13 cents
a pound ; 10, weighing 40 pounds each, at 10 cents a pound ;
and 24, weighing 25 pounds each, at 7 cents a pound ; I
sold the whole at an average price of 9|- cents a pound;
what was my whole gain ? Ans. $6.
41 5. Alligation Alternate is the process of finding the
proportional quantities to be taken of several ingredients,
whose prices or qualities are known, to form a mixture of a
required price or quality.
Case I.
416. To find the proportional quantity to be used
of each ingredient, when the mean price or quality
of the mixture is given.
1. What relative quantities of timothy seed worth $2 a
bushel, and clover seed worth $7 a bushel, must be used to
form a mixture worth $5 a bushel ?
OPERATION. Analysis. Since on every ingre-
{214 2 ) dient used whose price or quality
ly \ 1 3 j Ans. is less than the mean rate there will
be a gain, and on every ingredient
whose price or quality is greater than the mean rate there will be a
loss, and since the gains and losses must be exactly equal, the relative
quantities used of each should be such as represent the unit of value.
By selling one bushel of timothy seed worth |2, for $5, there is a gain
of $3 ; and to gain $1 would require ^ of a bushel, which we place
opposite the 2. By selling one bushel of clover seed worth $7, for $5,
there is a loss of $2 ; a 
