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lO 1 ^ 1/3 I
PREFACE. ^£/^c.,
Progress and improvement characterize almost every art and
science ; and within the last few years the science of Arithmetic has
received many miportant additions and improvements, which have
appeared from time to time successively in the different treatises pub-
lished upon this subject.
In the preparation of this work it has been the author's aim to com-
bine, and to present in one harmonious whole, all these modem im-
provements, as well as to introduce some new methods and practical
operations not foimd 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 what to do; to give imity, 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 Qxidi the labor necesseiry 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 superi-
ority over, others of the kind in the follo'W'ing particulars, viz. : In
the mechanical and typographical style of the work; the open and
attractive page ; the progressive and scientific a)rangement of the
subjects ; clearness and conciseness of definitio7is ; fullness and accu-
racy in the new and improved methods of operations and analyses;
brevity and pei'spicuity of rules; and in the very large number of
(in)
ir
PREFACE.
examples prepared and Arranged wiOi apeeial reference to their practical
utility, and their adaptation to the real business of active life. The an-
swers 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 Com-
mon Divisors, Multiples, Fractions, Percentage, Interest, Proportion, An-
alysis, Alligation, and the Eoots, 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 noticable. The ^different
kinds of United States' Securities, Bonds, and Treasury Notes are de-
scribed, and their comparative value in commercial transactions illus-
trated 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.
There has also been added a full and practical presentation of the
Metric System of Weights and Measures, containing many new and orig-
inal improvements in the arrangement, notation, and applications, not
before presented to the public, and which greatly simplify and adapt the
system to general use.
It is not claimed that this is a perfect work, for perfection is impossi-
ble; but no effort has been spared to present a clear, scientific, compre-
hensive, and complete system, suflSciently full for the business man and
the scholar; not encumbered with unnecessary theories, and yet com-
bining and systematizing real improvements of a practical and useful
nature. How nearly this end has been attained the intelligent and ex-
perienced teacher and educator must determine.
The Author.
CONTENTS.
SIMPLE NUMBERS.
PAGE
Definitions, 7
Roman Notation 8
Table of Roman Notatiott, 9
Arabic Notation, 10
Nuineratiun Table, 15
Iaws and Rules for Notation and
Numeration, IT
Addition, 20
Subtraction, 28
Multiplication, 35
Contractions^ 42
Piioa
Division, 4T
CJontractions, 56
Applications of preceding Rules, 60
General Principles of Division, 64
Exact Divisors, 66
Prime Numbers, 6T
Factoring Numbers, 6T
Cancellation, 69
Greatest Common Divisor, 73
Multiples 79
Classiflcatiou of N umbers, 84
COMMON FRACTIONS.
Definitions, &c., 86
General Principles of Fractions,. .... 89
Reduction of Fractions, 83
Addition of Fractions, 96
Subtraction of Fractions, 9S
Multiplication of Fractions, 101
Division of Fractions, 108
Promiscuous Examples, 112
DECIMALS.
Decimal Notation and Numeration,. 116
Reduction of Decimals, 121
Addition of Decimals, 124
Subtraction of Decimals, 12S
Multiplication of Decimals, 127
Division ofDccimals, 12*
DECIMAL CURRENCY.
Notation and Numeration o. Decimal
Currency, 131
Reduction of Decimal Currency,. . . . 132
Addition of Decimal Currency, 134
Subtraction of Decimal Currency,... 135
Multiplication of Decimal Curren-
cy, 136
Division of Decimal Currency, 137
Additional Applications, J39
When the Price Is an Aliquot Part
of a Dollar, ....\ 139
To find the Cost (Ai Quantity, 140
To find the Price of One, 141
To find the Quantity, 141
Articles sold by the 100 or 1000, 143
Articles sold by the Ton, 143
Bills, 144
Promiscuous Examples, 148
W
Tl
CONTENTS.
COMPOUND NUMBERS.
PAGB
Reduction, 150
Definitions, Ac.,. 150
English Money, 151
Troy Welglit, 153
Apothecaries' Weight, 154
Avoirdupois Weight, 155
Long Measure, 153
Surveyors' Long Measure, 160
Square Measure, 161
Surveyors' Square Measure, 164
Cubic Measure, 165
Liquid Measure, 167
Dry Measure, 168
Time, 170
Circular Measure, 172
TAOm
Counting ; Paper ; Books, &c^ 173
Reduction of Denonainate Fractions, 175
Addition of Compound Numbers,. .. 189
Addition of Denominate Fractions, . 185
Subtraction, 186
To find the Diflference in Dates, 188
Table, 189
Subtraction of Denominate Frac-
tions, 190
Multiplication of Compound IS um-
bers, 191
Division, , 193
Longitude and Time, 195
Duodecimals, 193
Promiscuous Examples, 203
PERCENTAGE.
Definitions, Ac, 205
Commission and Brokerage, 212
Stock- Jobbi ng, 216
United States' Securities, 220
Gold Investments, 225
Profit and Loss, 227
Insurance, 283
Taxes, 234
Custom House Business, 28T
Simple Interest...... 240
Partial Payments or Indorsements,. 247
Problems in Interest, 253
Compound Interest, 256
Discount, 259
Banking, 262
Exchange, 266
Equation of Payments, 271
RATIO AND PROPORTION.
Katio, 279
Proportion,... 282
Simple Proportion, 283
Compound Proportion, 289
Partnership, 294
Analysis, 298
Alligation Medial, 807
Alligation Alternate, 808
Involution, 818
Evolution, 814
Square Boot, 816
Cube Boot, 822
Arithmetical Progression, 328
Geometrical Progression, .'. 881
Promiscuous Examples, 884
Mensuration, 842
The Metric System, 845
PEACTICAL ARITHMETIC.
DEFINITIONS.
1. ftnantity is any thing that can be increased, dimmished,
or measured.
3. 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 same kind or
name as the number. Thus, the unit of 23 is 1 ; of 23 dollars,
1 dollar ; of 23 feet, 1 foot.
7, Like Numbers have the same kind of unit. Thus, 74,
16, and 250 ; 7 dollars and 62 dollars ; 19 pounds, 320 pounds,
and 86 pounds ; 4 feet 6 inches, and 17 feet 9 inches.
8, An Abstract Number is a number used without refer-
ence to any particular thing or quantity. Thus, 17; 365 ; 8540.
O, A Concrete Number is a number used with reference
to some particular thing or quantity. Thus, 17 dollars ; 365
days ; 8540 men.
Notes. 1. The unit of an abstract number is 1, and is called Unity.
2. Concrete numbers are, by some, called Denominate Nufnbers,
Denomination 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 indicating an operation to be
performed.
12. A Rule is a prescribed method of performing an op-
eration.
Define quantity. Mathematics. A imit. A number. An integer.
The tmit 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 NUMBERS.
KOTATION AND NUMERATION.
13. ITotation is a method of writing or expressing numbers
by characters ; and,
14. Numeration is a method of reading numbers,expressed
by characters.
15. Two systems of notation are in general use — the
Roman and the Arabic.
Note. The Roman Notation is supposed to have been first used
by the Romans; hence its name. The Arabic Notation was intro-
duced into Europe 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 ROMAN NOTATION
16. Employs seven capital letters to express numbers,
thus :
Letters, I V X L C D M
one
hundred, hundred, thousand.
Values, one, five, ten, fifty, . «3« , . «!«
17. The Roman notation is founded upon five principles,
as follows :
1st. Repeating a letter repeats its value. Thus, II repre-
sents two, XX twenty, CCC three hundred.
2d. If a letter of any value be placed after one of greater
value, its value is to be united to that of the greater. Thus,
XI represents eleven, LX sixty, DC six hundred.
3d. If a letter of any value be placed before 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. WTiat systems of notation are now
in general use ? From what are their names derived ? What are used
to express numbers in the Roman notation ? What is the value of each?
What is the first princijjle of combination ? Second ? Third ?
NOTATION AND NUMERATION. '9
4th. If a letter of any value be placed between two letters,
each of greater value, its value is to be ta/cen from the united
value of the other two. Thus, XIV represents fourteen,
XXIX twenty-nine, XCIV ninety-four.
5 th. A bar or dash placed over a letter increases its value
one thousand fold. Thus, V signifies five, and V five thou-
sand ; L fifty, and L fifty thousand.
TABLE OF ROMAN NOTATION.
I
is
One.
XX
is
Twenty.
II
«
Two.
XXI
«
Twenty-one,
III
4(
Three,
XXX
«
Thirty.
IV
t(
Four.
XL
«
Forty.
V
«
Five.
L
ti
Fifty.
VI
«
Six. •
LX
it
Sixty.
VII
«
Seven.
LXX
it
Seventy.
VIII
«
Eight.
LXXX
it
Eighty.
IX
<i
Nine.
XC
«
Ninety.
X
((
Ten.
C
tt
One hundred.
XI
((
Eleven.
CC
«
Two hundred.
XII
<(
Twelve.
D
((
Five hundred.
XIII
«
Thirteen.
DC
«
Six hundred.
XIV
,«
Fourteen.
M
ti
One thousand, {dred.
XV
«
Fifteen.
MC
«
One thousand one hun-
XVI
«
Sixteen.
MM
((
Two thousand.
XVII
it
Seventeen.
X
«
Ten thousand.
XVIII
li
Eighteen.
C
«
One hundred thousand.
XTX
((
Nineteen.
M
tt
One million.
Note. The system of Roman notation is not well adapted to the
purposes of numerical calculation ; it is principally confined to the
numbering of chapters and sections of books, public documents, &c.
Express the following numbers by letters :
1. Eleven. Ans,
2. Fifteen. Ans.
XL
Fourth ? Fifth ? Repeat the table. What is the value of LYH ?
CLXXIII? XCVIII? CDXXXII? XCIX? DCXIX?
VMDCCXLIX? MDXXVCDLXXXIX ? To what uses is tha
Roman notation now principally coniined ?
1*
10 SIMPLE NUMBERS.
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 ARABIC NOTATION
18. Employs ten characters or figures to express numbers.
Thus,
Kgures, 0123456789
Names and ') "*"gh' °^^> *wo, three, four, five, six, Beven, eight, nine-
values, 5 ^.jph'er,
10, The first character is called naughty because it has no
value of its own. The other nine characters are called signif-
icant figures, because each has a value of its own.
20, The significant figures are also called Digits, a word de-
rived from the Latin term digitus, which signifies^n^er.
31 • 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
important 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.
* Fractfonal and uecimal notation, and the notation of compound nombers, will bft
discussed in ttieir appropriate places.
What are used to express numbers in the Arabic notation ? "\Miat
is the value of each ? What general name is given to the significant
figures ? Wliy ? Numbers less than ten, how expressed ?
NOTATION AND NUMERATION. 11
S3. As we have 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;
70;
80; 90.
34.
"When a number is e
;xpress€
d by two 1
Sgures, 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 respectively
in the place of units. Thus,
eleven, twelve, thirteen, fonrteen, fifteen, sixteen, serenteen, eighteen, nineteen.
11, 12, 13, 14, 15, 16, 17, 18, 19.
In like manner we express the numbers between 20 and
SO, between 30 and 40, and between any two 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 hun-
dred, three hundred, &c., 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 unitSy 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 occupy.
Tens, how expressed ? The right hand figure called what ? Left
hand figure, what ? What is the greatest number that can be expressed
by two figures ? One hundred, how expressed ? When numbers are
expressed by three figures, what names are given to each }
12 SIMPLE KUMBERS.
we express tens and units with hundreds, by writing, 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,
I
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.
5. 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 two
thousand, three thousand, &:c., to nine thousand ; thus,
one two three four five six seven eight nine
thousand, thousand, thousand, thousand, thousand, thousand, thousand, thousand, thousand.
1000, 2000, 3000, 4000, 5000, GOOO, 7000, 8000, 9000.
38. When a number is expressed by four figures, the
places, commencing at the right hand, are units, tens, hundreds^
thousands.
Use of the cipher, what ? Greatest number that can be expressed by
three figures ? One thousand, how expressed ? IIow many figures
used ? Names of each ?
NOTATION AND NUMERATION. 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 <^
units J thus,
i -3
5 c 2 .1^
2 3 C3 d
4 2 6 9
The greatest number that can be expressed hj four figures
is 9999.
EXAMPLES FOR PRACTICE.
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.
Read the following numbers : —
11. 76; 128; 405; 910; 116; 3416; 1025.
12. 2100; 5047; 7009; 4670; 3997; 1001.
20. Next to thousands come tens 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 ?
expressed ? Hundreds of thousands ?
14 SIMPLE NUMBERS.
of thousands, or by writing it at the left hand of four ci-
phers; thus, 10000; and one hundred thousand is expressed
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 thousands, 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,
«M
O . . •
^ M CO CO •
<u 5 § a ■"
la a -5 I I -I
5 2 18 3
The greatest number that can be expressed by five figures
is 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.
S. 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 niunber expressed by five figures ? Six figures ?
NOTATION AND NUMERATION.
15
13.
14.
15.
teen.
16.
17.
Five hundred one thousand nine hundred sixty-four.
One hundred thousand one hundred.
Three hundred thirteen thousand three hundred thir-
Seven hundred eighteen thousand four.
One hundred thousand ten.
Read the following numbers : —
18. 5006; 12304; 96071,' 5470
19. 36741 ; 400560 ; 13061 • 49000
20. 200200; 75620; 90402; 218094
203410.
100010.
100101.
Por convenience in reading large numbers, we may point
them oif, by commas, into periods of three figures each, count-
ing 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 follow
units, tens, and hundreds of each higher name, as seen in the
following
NUMERATION TABLE.
S .2
i
■s
s
OOOOOOOO
'^^^ ro^^r^ /'xjv^/^ r^-A^^ /v^\>^ /'n^a-^ z'^ov-^ /•>^A^r^
I I I I I I I
j§5i§5i§2|§=ii-ii=iiiii
S 8,7 65, 43 2, 10 9, 8 76, 5 5 6, 78 9, 01 2,
v^v>^v>'v'>^ v.>-\rvy \-/-v"v>' v^'v'vy v-or^^ v^>-v>^ v>-y"v^
ninth eighth seventh sixth fifth fourth third second
period, period, period, period* periodi period, period, period.
o
a « -is
3 S a
345
first
period.
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.
Note. Thi? is called the French method of pointing off the peri-
ods, and is the one in general use in this country.
31* Figures occupying different places in a number, as
units, tens, hundreds, &c., are said to express different orders
of units.
Simple units are called units of the Jirst order.
Tens " " " " " second "
Hundreds « « « " " third "
Thousands " " « « « fourth "
Tens of thousands " « « « " //^A «
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 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 of 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 sec-
ond.
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 different orders are what }
NOTATION AND NUMERATION. 17
32. From the foregoing explanations and illustrations, we
derive several important principles, which we will now pre-
sent.
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 Local Value of a figure is its value when used with an-
other 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.
Note. "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, expresses
tens ; in the third place, hundreds ; in the fourth place, thou-
sands ; and so on.
Sd. 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.
I. The different orders of units increase from right to lefty
and decrease from left to right, in a tenfold ratio.
II. Every removal of a figure one place to the left, increases
its local value tenfold ; and every removal of a figure one place
to the right, diminishes its local value tenfold^ .
Thus,
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 figure ? Local r
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 hoth numbers, being 4 units of the third order, or 4
hundred.
Note. 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 (hundreds) is the same in the two numbers, 425 and 400, the value
of the whole of the first nvmaber is greater than that of the second.
5th. Every period contains three figures, (units, tens, and
hundreds,) except the left hand period, which sometimes con-
tains only one or two figures, (units, or units and tens.)
33. As we have now analyzed all the principles upon
which the writing and reading of whole numbers depend, we
will present these principles in the form of rules.
RULE FOR NOTATION.
I. Beginning at the left hand^ write the figures belonging to
the highest period.
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,
over each period write its name, thus,
Trillions, Billions, Millions, Tliousands, Unttn.
000, 000, 000, 000, 000
Fourth principle ? What caution is given ? Fifth principle ? Rule
for notation ? Niuneration ?
NOTATION AND NUMERATION. 19
and then write the given numbers underneath, in their appro-
priate places.
EXERCISES IN NOTATION AND NUMERATION.
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. Ninety-five million.
8. Four hundred thirty-three million eight hundred sixteen
thousand one hundred forty-nine.
9. Nine 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 number
1037050?
20 SIMPLE NUMBERS.
ADDITION.
MENTAL EXERCISES.
35. 1. Henry gave 5 dollars for a vest, and 7 dollars for
a coat ; how much did he pay for both ?
Analysis. He gave as many dollars as 5 dollars and 7 dollars,
which are 12 dollars. Therefore he paid 12 dollars for both.
2. A farmer sold a pig for 3 dollars, and a calf for 8 dol-
lars ; how much 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 ? G and 6 ? 6 and 9 ? 6 and 7 ?
7. How many are 7 and 7? 7 and 6? 7 and 8? 7 and 10?
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 perceive 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 -f- 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
6 -f- 4 z= 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 ?
ADDITION. 21
CASE I.
40, When the amount of each column is less
than 10.
1. A farmer sold some hay for 102 dollars, six cows for
162 dollars, and a horse for 125 dollars ; how much did he re-
ceive for all ?
OPERATION. Analysis. We arrange the numbers so
4 .^ that units of Hke order shall stand in the
Jl'g same column. We then add the columns
102 separately, for convenience commencing at
IQ2 the right hand, and write each result under
■125 *^® column added. Thus, we have 5 and 2
and 2 are 9, the sum of the units ; 2 and 6
Amount, 389 are 8, the sum of the tens ; 1 and 1 and 1
are 3, the sum of the hundreds. Hence, ♦^he
entire amomit is 3 hundreds 8 tens and 9 units, or 389, the Answer,
EXAMPLES FOR
PRACTICE.
(2.)
pounds.
132
(3.)
rods.
245
(4.)
cents.
312
(5.)
days.
437
243
321
243
140
324
132
412
321
Ans. G99
G. What is the sum of 144, 321, and 232 ? Aiis, 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 many
sheep did he buy in all ? Ans. 995,
CASE II.
41. Wlien 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 ?
Sum of the units,
17
Sum of th^tens,
15
Sum of the hundreds,
14
22 SIMPLE NUMBERS.
store, 475 dollars for a clerk, and 3G7 dollars for other ex-
penses ; what is the amount of his expenses ?
OPERATION. Analysis. Arranging the num-
-S^Ǥ hers as in Case I, we first add the
^11 column of units, and find the sum
725 to be 17 units, which is 1 ten and
475 7 units. We write the 7 units in
gQj 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. We write the 5 tens in the
„ ^ , ^ 1 Kon place of tens, and the 1 hundred in
Total amount, 1567 f, , e\ i i ^ir
the place oi hundreds. We next
add the column of hundreds, and find the sum to be 14 hundreds,
which is 1 thousand and 4 hundreds. We w rite 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 hundreds, we find the total amount to be 1 thousand 5 hundreds
6 tens and 7 units, or 1567.
This example may be performed by another method, which
is the common one in practice. Thus :
OPERATION. Analysis. Arranging the numbers as before, we
725 add the first column and find the sum to be 17 units ;
475 writing the 7 units under the column of units, we add
Qg7 the 1 ten to the column of tens, and find the sum to be
16 tens ; writing the 6 tens under the column tens, wo
1567 add the 1 hundred to the column of hundreds, and find
the sum to be 15 hundreds ; as this is the last column,
we write down its amount, 15 ; and we have the whole amount, 1567,
as before.
Notes. 1. Units of the same order arc written in the same column ;
find when the sum in any column is 10 or more than 10, it produces
one or more units of a higher order, which must be added to the next
column. This process is sometimes called «* carrying the tens."
2. In adding, learn to pronounce the partial results without naming
the numbers separately ; thus, instead of saying 7 and 5 are 12, and
6 are 17, simply pronounce the results, 7, 12, 17, &c.
Give explanation. Second explanation. 'What is meant by carry-
ing the tens ?
^
ADDITION. 23
4:'3, From the preceding examples and illustrationrf we
deduce the following
Rule. I. Write the numbers to he added so that all the units
of the same order shall stand in the same column ; that is, units
under units, tens under tens, S^c.
II. Commencing at units, add each column separately^ and
write the sum underneath, if it he less than ten.
III. 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.
IV. Write the entire sum of the last column.
Proof. 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.
Note. By the methods of proof here given, the numbers are iinited
in new combinations, which render it almost impossible for two pre-
cisely similar mistakes to occur.
The first method is the one commonly used in business,
EXAMPLES FOR PRACTICE.
(2.)
(3.)
(4.)
(5.)
(6.)
miles.
inches.
tons.
feet.
bushels.
24
S21
427
1342
3420
48
479
321
7306
7021
96
165
903
5254
327
82
327
278
8629
. 97
250
1292
1929
22531
10865
Rule, first step ? Second ? Third ? Fourth ? Proof, first method ?
Second ? Upon what principle are these methods of proof founded ?
24
(7.) (8.) (9.) (10.)
hours. years. gallons. rods.
347 7104 3462 47637
506 3762 863 3418
218 9325 479 703
312 ' 4316 84 26471
424 2739 57 84
11. 42 + 64 H- 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 ?
Ans. 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 thou-
sand one hundred five, four thousand seventy-five, and five
hundred seven. Ans. 56907.
17. Add two hundred ten thousand four hundred, one hun-
dred 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 thir-
ty-five, two hundred sixty-seven thousand, one thousand four
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? A7is. 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 40() miners ?
23. A merchant deposited 56 dollars in a bank on Monday,
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. Five persons deposited money in the same bank ; the
first, 5897 dollars; the second, 12980 dollars; the third,
65973 dollars; the fourth, 37345 dollars j and the fifth as
much as the first and second together ; how many dollars did
they all 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
RP
26 SIMPLE NUMBEBS,
, 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 dol-
lars 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,
1815, 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 en-
tire population of these cities ? Ans. 1216671.
39. The population of the globe is estimated as follows :
North America, ?.9257819 ; 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 Toledo to Spring-
field, 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
51365
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
571778
7626
8735
17005
' 41735
1218099
1395860
BIMPLE NUMBERS.
SUBTRACTION.
MENTAL EXERCISES.
4:3, 1. A farmer, having 14 cows, sold 6 of them; how
many had he left ?
Analysis. He had as many left as 14 cows less 6 cows, which
are 8 cows. Therefore, he had 8 cows left.
2. Stephen, having 9 marbles, lost 4 of them ; how many
had he left?
3. If a man earn 10 dollars a week, and spend G dollars for
provisions, how many dollars has he left ?
4. A merchant, having 16 barrels of Hour, sells 9 of them ;
liow many has he left ?
5. Charles had 18 cents, and gave 10 of them for a book ;
how many 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. IIow many are 20 less 14 ? 18 less 12 ? 19 less 11?
12. How many are 11 less 3? 16 less 11? 19 less 8f
20 less 9 ? 22 less 20 ?
44. Subtraction is the process of determining the diftcr-
cnce, between two numbers of the same unit value.
4:5. The Minuend is the number to be subtracted from.
40. The Subtrahend is the number to be subtracted.
Define subtraction. Minuend, Subtrahend.
' JSUBTRACTION. 29
47. The Difference or Eemainder is the result obtained
by the process of subtraction.
Note. The minuend and subtrahend must be like numbers ; thus,
5 dollars from 9 dollars leave 4 dollars ; 5 apples from 9 apples leave 4
apples ; but it would be absurd to say 5 apples from 9 dollars, or a
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 G equals 2, and denotes that 6, the subtrahend,
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.
OPERATION. Analysis. We write the less num-
Minuend, 574 ^^^ under the greater, with units under
„ . , , , QOQ units, tens under tens, &c., and draw
Subtrahend, 6z6 ,. , , „,' ,' . .
a Ime underneath. Ihen, begmnmg at
Pemainder, 251 the right hand, we 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 difference, 2 hundreds 5 tens and 1 unit, or 251.
EXAMPLES FOR PRACTICE.
(2.) (3.) (4.) (5.)
Minuend, 876 676 367 925
Subtrahend, 334 415 152 213
Remainder, 542 261 215 712
Case 1 is what ? Give explanation.
SIMPLE NUMBERS.
(G.) (7.)
(8.)
(9.)
From
876 732
987
498
Take
523 522
782
178
X ^ r
Remainders.
10.
From 3276 take 2143.
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 ; how much did he
lose ?
20. A man bought a mill for 3724 dollars, and sold it for
4856 dollars ; how much did he gain ? Arts. 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, how much 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; how much
was the difference ?
27. The population of Great Britain in 1851 was 20,936,468,
and of England ulonc; 16,921,888} what was the difiercnce?
- SUBTRACTION. ^l
CASE II.
50. When any figure in the subtrahend is greater
than the corresponding figure in the minuend.
> 1. From 846 take 359.
OPERATION. Analysis. In this example we
(7) (13) (If;) 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
Remainder, 4 8 7 Units, making 16 units ; 9 units from
16 units leave 7 units, which we
write in the remainder in units' place. As we have taken 1 ten
from the 4 tens, 3 tens only 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 remaindei' in tens' place.
As we have taken 1 hundred from the 8 hundreds, 7 hundreds otily
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.
Note. The numbers written over the minuend are used simply to
explain more clearly the method of subtracting ; in practice the pro-
cess 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
•|„-i units, we add 10 units to 6 units, making 16 units;
^^ 5 9 units from 16 units leave 7 units. But as we have
o'*" added 10 units, or 1 ten, to the minuend, we shall
359 have a remainder 1 ten too large, to avoid which, we
An J add 1 ten to the 5 tens in the subtrahend, making 6
tens. 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 11 is what ? Give explanation. Second explanation.
32
SIMPLE NUMBERS.
Note. Tho 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 examples and illustrations we
have the following general
Rule. I. Write the less number under the greater^ placing
units of the same order in the same column,
II. Begin at the right hand^ and take each figure of the sub-
trahend from the figure above it, and write the result under-
neath,
III. If any figure in the subtrahend be greater than the cor-
responding 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.
Proof. Add the remainder to the subtrahend, and if their
sum be equal to the minuend, the work is supposed to be right.
EXAMPLES FOR
PRACTICE.
Minnend,
(2-)
873
(3.)
7432
(4-)
1969
(5.)
8146
Subtrahend
538
6711
1408
4377
Remainder,
335
From
(6.)
gallons.
3176
(7.)
Uusheln.
9076
(8.)
inilei^.
7320
(9.)
days.
5097
Take
2907
4567
3871
3809
From
(10.)
dollars.
76377
(11.)
rodfl.
67777
(12.)
acres.
900076
(13.)
feot.
767340
Take
45761
46699
899934
5039
"VVTiat do we mean by borrowing 10 ? By carrying ? Ride, first step ?
Second? Third? Proof?
SUBTRACTION.
14 479 — 382 = how many ?
15. 6593 — 1807 — how many ?
16. 17380 — 3417 z= how many ?
17. 80014 — 43190 = how many?
18.
19.
20.
21.
22.
23.
24.
25.
2G.
27.
28.
29.
30.
31.
32.
Ans. 97.
Jns. 4786.
A71S. 13963.
Ans, 36824.
Ajis. 191975.
Ans.
A71S.
Ajis.
Ans.
4111107.
2479679.
935993.
608889.
Ans. 3968579.
Ans. 50000001.
Ans. 3819851.
Ans. 800924.
A?is. 7623024.
282731 — 90756 = how many!
From 234100 take 9970.
From 345673 take 124799.
From 4367676 take 256569.
From 3467310 take 987631.
From 941000 take 5007.
From 1970000 take 1361111.
From 290017 take 108045.
Take 3077097 from 7045676.
Take 9999999 from 60000000.
Take 220202 from 4040053.
Take 2199077 from 3000001.
Take 377776 from 8000800.
Take 501300347 from 1030810040.
Subtract nineteen thousand nineteen from twenty thou^
sand ten. Ans. 991.
33. From one milHon 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 COMBINING ADDITION AND SUBTRACTION.
SfB, 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 ; he drew out at
one time 4234 dollars, at another 1700 dollars, at another 962
2*
84 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 ; how much 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 1224 barrels, at another
time 1500 barrels, and at another time 1805 barrels; how
many barrels remained ?
G. 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. 9G1 sheep.
7. A farmer raised 1575 bushels of wheat, and 900 bushels
of corn. He sold 807 bushels of wheat, and 391 bushels of
corn to A, and the remainder to B ; how much of each did
lie sell to B ? Ans. 7G8 bushels of wheat, and 509 of .corn.
8. A man traveled G784 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. 22G5 miles.
9. Three persons bought a hotel valued at 35G80 dollars.
The first agreed to pay 7375 dollars, the second agreed to
pay twice as much, and the third the remainder ; how much
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 C709 dollars, received a legacy of 3000
dollars. He spent 4370 dollars in traveling ; how much had
he left ?
12. In 1850 the number of wnite males in the United
States was 1002G402, and of white females 9526GG6; of
these, 878G9G8 males, and 85255G5 females were native
born; how many of both were foreign boru? Ans. 2240535.
MULTIPLICATION. 85
MULTIPLICATION.
MENTAL EXERCISES.
5S, 1. What will 4 pounds of sugar cost, at 8 cents a
pound ?
Analysis. Four pounds will cost as much as the price, 8 cents
taken 4 times ; thus, 8-[-8-|-8-[-8zzi32. But instead of adding,
we may say, — since one pound costs 8 cents, 4 pounds will cost 4
times 8 cents, or 32 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 wuU 3 tons of hay cost? 4
tons? 5 tons?
0. 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, how much 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 savings 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 ?
7 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 ?
8Q
SIMPLE NUMBERS.
14. What ■will be the cost of 11 barrels of apples, at 2 dol.
lars a barrel ? at 3 dollars ?
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 othen
55 » The Multiplicand is the number to be taken.
5G, The Multiplier is the number which shows how
many times the multiplicand is to be taken.
57, The Product is the result obtained by the process of
multiplication.
58, The Factors are the multiplicand and multiplier.
KoTES. 1. Factors are producers, and the multiplicand and mul-
tiplier 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 n^ 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
4 X 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
1 X 8=1 8
2X 8 = 16
3 X 8 = 24
4X 8 = 32
IX 9= 9
2X 9 = 18
3X 9 = 27
4 X 9 = 36
1 X 10=10
2 X 10 = 20
3 X 10 = 30
4X 10 = 40
1 X 11 = 11
2X 11=22
3X 11=33
4X11=44
1 X 12 = 12
2X 12 = 24
3X 12 = 36
4X 12 = 48
Define multiplication. Multiplicand. MultipHer. Product. Fac-
tors. Multiplication is a short method of what ? What is the sign of
multiplication ?
MULTIPLICATION.
37
5X
5X
5X
5X
5X
5X
6X
5X
5X
5X 10
5X11
5X12
: o
:10
:lo
:20
:25
:30
35
40
45
50
55
60
6X
6X
6X
6X
6X
6X
6X 7
6X 8
6X 9
6X 10
6X 11
6X 12
: 6
:12
:18
:24
:30
:36
42
48
54
GO
G6
72
7X
7X
7X
7X
7X
7X
7X
7X
7X
7X
TXll
7X 12
: /
:14
:21
:28
35
42
49
56
63
70
77
84
8X
8X
8X
8X
8X
8X
8X
8X
8X
8X
1= 8
2:^16
3 = 24
4 = 32
5 = 40
6 = 48
7 = 56
8 = 64
9 = 72
10 = 80
8X 11 = 88
8X 12 = 96
9X 1= 9
10 X 1= 10
IIX 1= 11
12 X 1= 12
9X 2= 18
10 X 2= 20
11 X 2= 22
12 X 2= 24
9X 3= 27
10 X 3= 30
11 X 3= 33
12 X 3= 36
9X 4= 36
10 X 4= 40
11 X 4= 44
12 X 4= 48
9X 5= 45
10 X 5= 50
1 1 X o — 55
12 X 5= 60
9X 6= 54
10 X 6= 60
11 X 6= 66
12 X 6= 72
9X 7= 63
10 X 7= 70
11 X 7= 77
12 X 7= 84
9X 8= 72
10 X 8= 80
11 X 8= 88
12 X 8= 96
9X 9= 81
10 X 9= 90
11 X 9= 99
12 X 9 = 108
9X 10= 90
10X10 = 100
11 xio=iio
12X10 = 120
9X 11= 99
10X11 = 110
11 Xll = 121
12X11 = 132
9X 12 = 108
10X12 = 120
11 X 12 = 132
12X12 = 144
CASE I.
60, When the multiplier consists of one figure.
1. Multiply 374 by 6.
Analysis. In this example it i.>»
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-
pHer 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.
— n >
Case I is what ? Give explanation.
OPERATION.
III
Slultiplicand,
374
Multiplie
[•»
6
units,
24
tens,
42
hundreds,
18
Product,
2244
83 SIMPLE NUMBERS.
The operation in this example may be performed in another
way, which is the one in common use.
OPERATION. Analysis. Writing the numbers as before, we
374: begin at the right hand or unit figure, and say: G
P times 4 units are 24 units, which is 2 tens and 4
imits ; write the 4 units in the product in units*
2244 place, and reserve the 2 tens to add to the next prod-
uct ; 6 times 7 tens are 42 tens, and the two tens re-
served in the last product added, are 44 tens, which is 4 hundreds
and 4 tens ; 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 hundreds added are 22 hundreds, which, being
written in the product in the places of hundreds and thousands,
gives, for the entire product, 2244.
Gl, The unit value of a number is not changed by re-
peating the number. As the multiplier always expresses
iimeSj the product must have the same unit value as the mul-
tiplicand. But, since the product of any two numbers will be
the same, w^hichever factor is taken as a multiplier, either
factor may be taken for the multiplier or multiplicand.
Note. 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, G 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.
(3.)
G812
6
Multiplicand,
Multiplier,
(2.)
7324
4
Product,
29296
(5.)
S2456
3
(6.)
927U
7
40872
(7.)
28093
8
Second explanation. Repeating a number has what effect on the
unit value ? The product must be of the same kind as what ?
MULTIPLICATION.
S9
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.
Ajis,
Ans.
A71S.
Ans.
Ans.
163730.
5882597.
1101032.
281286.
2173088.
6328476.
15. What will be the cost of 344 cords of wood at 4 dol-
lars a cord ? Ans. 1376.
16. How much will an army of 7856 men receive in one
week, if each man receive 6 dollars ? A7is. 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 dollars.
20. In one mile are 5280 feet ; how many feet in 8 miles ?
Ans. 42240 feet.
CASE II.
63. When the multiplier consists of two or more
figures.
1. Multiply 746 by 23.
Multiplicand,
Multiplier,
Product,
OPERATION.
746
23
2238
1492
17158 23
. < times the tnnl-
l tiplicand.
) < times the mul-
( tiplicand.
times the mul-
tiplicand.
Analysis. Writ-
ing the multiplicand
and multiplier as in
Case I, we first mul-
tiply each figure in the
multiplicand by the
unit figure of the mul-
tiplier, precisely as in
Case I. We then multiply by the 2 tens. 2 tens times 6 units, or 6
times 2 tens, are 12 tens, equal to 1 hundred, and 2 tens ; we 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 ; we write the 9 in hundreds' place in the product. 2 tens
Case II is what ? 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, we have the entire product, 17158.
Notes. 1. "SVhen the multiplier contains two or more figures, the
several results obtained by multiplying by each figure are called partial
products.
2. When there are ciphers between the significant figures of the
multiplier, pass over them, and multiply by the significant figoires only.
63. From the preceding examples and illustrations we
deduce the following general
Rule. I. Write the midtiplier under the multiplicand, placing
units of the same order under each other.
II. Multiply the multiplicand hy each figure of the multi-
plier successively, beginning with the unit figure, and write the
first figure of each partial product under the figure of the mul-
tiplier used, writing down and carrying as in addition.
III. If there are partial products, add them, and their sum
^fr-^ill bs the product required.
, f
64:. Proof. 1. 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.
(2.)
(3.)
(4.)
Multiply
4732
8721
17605
By
36
47
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 ? Kule, first step ? Second ? Third ? . Proof,
first metliod ? Second ?
MULTIPLICATION. 4]
(o.) (6 ) (7.)
7648 81092 37967
328 194 426
8. How many yards of linen in 759 pieces, eacli piece con-
taining 25 yards ? A71S. 18975 yards.
9. Sound is known to travel about 1142 feet in a second oi
time ; how far will it travel in 69 seconds ?
10 A man bought 36 city lots, at 475 dollars each ; how
much 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. 8 1099 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 annually
208 dollars ; how much do they all receive ?
14. MuUiply 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. MuUiply 15607 by 3094. Ans. 48288058.
19. Multiply 79094451 by 76095. Ans. 6018692248845.
20. Multiply live hundred forty thousand six hundred nine,
by seventeen hundred fifty. Ans. 946065750.
21. Multiply four million twenty-five thousand three hun*
dred ten, by seventy-five thousand forty-six.
Ans. 302083414260.
22. Multiply eight hundred seventy-seven miUion five hun-
dred 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? Ans. 965120 dollars.
42 SIMPLE NUMBERS.
CONTRACTIONS.
CASE I.
65. When the multiplier is a composite number.
A Composite Number is one that may be produced by
multiplying together two or more numbers ; thus, 18 is a com-
posite number, since 6 X 3 z:z 18 ; or, 9 X 2 ^ 18 • or, 3 X
3X 2:^:18.
66. The Component Factors of a number are the sev-
eral numbers which, multiphed 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, (2 X 2 X 5 = 20 )
Note. The pupil must not confound the faclots with the parts of a
number. Thus, the factors of which 12 is composed, are 4 and 3,
(^4X3= 12 ;) while the parts of which 12 is composed are 8 and 4,
(8 -f- 4 = 12,) or 10 and 2, (10 -f 2 = 12.) The factors arc multiplied,
while the parts are added^ to produce the number.
I. AVhat will 32 horses cost, at 174 dollars apiece?
orER.\TioN. Analysis. The fac-
MuitipiicanJ, 174 cost of 1 liorsc. tors of 32 are 4 and
1st factor, 4 ^- If we multiply the
< cost of 1 horse by 4,-
C9G cost of 4 horses, ^e obtain the cost of 4
2d factor, 8 horses ; and by multi-
rroduot, 5568 cost of 32 horses, f^"^^
horses by 8, we obtain
the cost of 8 times 4 horses, or 32 horses, the number bought
OT. Hence we have the following
Rule. I. Separate the composite number into two or more
factors.
II. Multiply the multiplicand by one of these factors^ and
"What are contractions? Ca.se I is what? Define a composite
number. Com])onent factors. What caution is given? Give ex-
planation. Rule, first step ? Second ?
MULTIPLICATION. 43
that product hy another, and so on vntil all the factors have
been used successively ; the last product will be the product re-
quired.
Note. The product of any number of factors will be the same in
"whatever order they are multiplied. Thus, 4 X 3 X 5 = 60, and
5X4X3 = 60.
EXAMPLES FOR PRACTICE.
2. Multiijly 3472 by 48 = 6 X 8. Ans, IGCGoG.
3. Multiply 14761 by 64 = 8 X 8.
4. Multiply 87034 by 81 — 3 X 3 X 9. Ans. 70497o4.
5. Multiply 47326 by 120 m 6 X 5 X 4.
6. Multiply 60315 by 96. Ans. 5790210.
7. Multiply 291042 by 125. Ans. 36380250.
8. If a vessel sail 436 miles in 1 day, how far will she sail
in 56 days ? Ans. 24416 miles.
9. How much will 72 acres of land cost, at 124 dollars an
acre ? Ans. 8928 dollars.
10. There arc 5280 feet in a mile; how many feet in 84
miles? Ans. 443520 ^qX.
11. What will 120 yoke of cattle cost, at 125 dollars a
yoke ?
CASE II.
68. When the multiplier is 10, 100, 1000, &c.
If we annex a cipher to the multiplicand, each figure is re-
moved one place toward the left, and consequently the value of
the whole number is increased tenfold, (3^.) 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
Rule. Annex as many ciphers to the multiplicand as there
are ciphers in the multiplier j the number so formed will be
the product required.
Case II 13 what ? Give explanation. Rule ?
44 SIMPLE NUMBERS.
EXAMPLES FOR PRACTICE.
1.
Multiply 347 by 10. Ans.
3470.
2.
Multiply 4731 by 100. Ans.
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 wi
11 1000 bushels
cost?
Ans.
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
-J ... .. nrx component factors ; 1200 into 12 and
u ip ler, ^^^^ ^^^ ^^ .^^^ ^ ^^^ ^^ j^ ^^^^^
Product, 72000 several factors be multiplied together
they will produce the same product as
the given numbers, (67.) Thus, 12 X 6 = 72, and 72 X 100 =
7200, and 7200 X 10 = 72000, which is the same result as in the
operation. Hence the following
Rule. Multiply the significant figures 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.
Multiply
By
EXAMPLES
(2.)
4720
340000
1888
1416
1604800000
FOR
PRACTICE
(3.)
10340000
105000
5170
1034
1085700000000
1
Case III is what ?
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
hundi-ed, 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 COMBINING ADDITION, SUBTRACTION, AND
MULTIPLICATION.
1. Bought 45 cords of wood at 4 dollars a cord, and 9 loads
of hay at 13 dollars 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 ; how much
did he gain?
3. Bought 288 barrels of flour for 1875 dollars, and sold
the same for 9 dollars a barrel ; how much was the gain ?
Ans. Ill 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; how much did his gro-
ceries cost him ?
46 SIMPLE NUMBERS.
G. 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 ; wdiat will he save in a year, or
365 days ? Ans. 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 other two ;
how much 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. C bought 14 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 lor 3842 dollars ; how
much 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 ; how much 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 times as much for
family expenses ; how much 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 ; how much did he gain or
lose ? Ans. Gained 24 dollars.
DIVISION. 47
DIVISION.
MENTAL EXERCISES.
yi, 1. How many hats, at 4 dollars apiece, can be bought
for 20 dollars ?
Analysis. Since 4 dollars will buy one hat, 20 dollars will buy
as many hats as 4 is contained times in 20, which is 5 times. There-
fore, 5 hats, at 4 dollars apiece, can be bought for 20 dollars.
2. A man gave 1 6 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 corn can
be bought for 24 shillings ?
o. 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 will 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 iriany 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 apit^ce
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. rivision 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 contained
in the dividend.
Notes. 1. "Wlien 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 remainder is always a part of the dividend, it is always
of the same name or kind.
5. When there is no remainder the division is said to be complete.
7G. The sign, -^, placed between two numbers, denotes
division, and shows that the number on the left is to be divided
by the number on the right. Thus, 20 -J- 4 = 5, is read, 20
divided by 4 is equal to 5.
Division is also indicated by writing the dividend ahove^ and
12
the divisor helow a short liorizontal line ; thus, — = 4, shows
that 12 divided by 3 equals 4.
CASE I.
77. When the divisor consists of one figure.
1. How many times is 4 contained in 848 ?
OPERATION. Analysis. After writing the divisor
on the left of the dividend, with a line
A \ oTq' between them, we begin at the left hand
Divisor, 4 ) 848 , . . ^ • J • o I A A
^ and say : 4 is contamed m 8 hundreds,
Quotient, 212 2 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, we have the entire quotient, 212.
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 ?
OPERATiox. Analysis. As we cannot divide 2 thousands by
4)2884 ^' ^^ ^^^^ ^^^ ^ thousands and the 8 hundreds to-
gether, and say, 4 is contained in 28 hundreds 7 hun-
'^^1 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 the
quotient, and we have the entire quotient, 721.
3. How many times is 6 contained in 1824 ?
OPERATIOX. Analysis. Beginning as in the last example, we
6) 1824 say, 6 is contained in 18 hundreds 3 hundreds times,
which we write in hundreds' place in the quotient ;
^'^^ then 6 is contained in 2 tens no times, and we 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 we
write in units' place in the quotient, and we have 304 for the entire
quotient.
4. How many times is 4 contained in 943 ?
OPERATIOX. Analysis. Here 4 is contained in 9
4x9^3 hundreds 2 hundreds times, and 1 hundred
over, which, united to the 4 tens, makes
235 ... 3 Rem. 14 tens ; 4 in 14 tens, 3 tens times and 2
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, |. The entire quotient is written 23o|,
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
Rule. I. Write the divisor at the left of the dividend, with
a line between them.
Second. Third. Riile, first step?
T?.P 8
50
SIMPLE NUMBERS.
II. Beginning at the left hand, divide each figure of the
dividend hy the divisor, and write the result under the divi-
dend,
III. If there he a remainder after dividing any figure, re-
gard it as prefixed to the figure of the next lower 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 num-
ber to the figure of the next lower order in the dividend, and
divide as before,
V. Jf there be 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 2)roduct add the remainder, if any ; if the result be equal
to the dividend, the work is correct.
Notes. 1. This method of proof depends on the fact that division is
the reverse of multiplication. The dividend answers to the product^ the
divisor to one oi the factors^ and the quotient to the other.
2. In multiplication the two factors are given, to find the product :
in division, the product and one of the factors are given to find the
other factor.
EXAMPLES FOR PRACTICE.
1. Divide 7824 by 6.
OPERATION.
Divisor. 6)7824 Dividend.
1304 Quotient.
PROOF.
1304 Quotient.
6 Divisor.
7824 Dividend.
(2.)
4)65432
(3.)
5)89135
(4.)
6)178932
(5.)
7)4708935
(6.)
8)1462376
(7.)
9)7468542
Second step ? Third ? Fourth ?
ion differ from multiplication ?
Fifth ? Proof ? How does divis-
DIVISION.
il
8. Divide 3102455 by 5.
9. Divide 1762891 by 4.
10. Divide 546215747 by 11.
11. Divide 30179624 by 12.
12. Divide 9254671 by 9.
Quotients.
Quotients.
620491.
440722f.
49655977.
2514968y8^,
1028296^
Kem.
13. Divide 7341568 by 7.
14. Divide 3179632 by 5.
15. Divide 19038716 by 8.
16. Divide 84201763 by 9.
17. Divide 2947691 by 12.
18. Divide 42084796 by 6.
Sums of quotients and remainders, 20680083. 28.
19. Divide 47645 dollars equally among 5 men ; how
much will each receive ? Ans. 9529 dollars.
20. In one week are 7 days; how many weeks in 17675
days ? Ans. 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 ?
Ans. 69512 barrels.
24. If there be 3240622 acres of land in 11 townships,
how many acres in each township ?
25. A gentleman left his estate, worth 38470 dollars, to be
shared equally by his wife and 4 children ; how much did
each receive ? Ans. 7694 dollars.
CASE II.
78. When the divisor consists of two or more figures.
Note. To illustrate more clearly the method of operation, we will
first take an example usually performed by Short Division.
Case n is what ?
52 SIMPLE NUMBERS.
1. How many times is 8 contained in 2528 ?
OPERATION. Analysis. As 8 is not contained in 2 thou-
8 ) 2528 C 316 sands, we take 2 and 5 as one number, and con-
n . sider how many times 8 is contained in this
partial dividend, 25 hundreds, and find that it
12 is contained 3 hundreds times, and a remainder.
8 To find this remainder, we multiply the divisor,
"TT 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 multipHed 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 dividend. Then, 8 is con-
tained in 48 units 6 units times. Multiplying and subtracting as
before, we find that nothing remains, and we have for the entire
quotient, 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 tho
^ ' quotient figure, 2, and subtract the
product, 46, from the part of tho
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 ; multi-
plying and subtracting as before, nothing remains, and we have for
the entire quotient, 209.
Notes. 1. 'WTicn the process of dividinpj is performed montally, and
the results only are written, as in Case I, the operation is temied Short
Division.
2. When the whole process of division is written, the operation is
termed LoTig Division.
Give first explanation. Second. What is long division ? What ia
short division ? When is each used ?
DIVISION. 53
3. 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
Rule. I. Write the divisor at' the left of the dividend, as
in short division.
II. Divide the least number of the left hand figures in the
dividend that will contain the divisor one or more times, and
place the quotient at the right of the dividend, with a line be-'
tween them.
III. Multiply the divisor by this quotient figure, subtract
the product from the partial dividend used, and to the remain-
der bring down the next figure of the dividend.
IV. Divide as before, until all the figures of the dividend
have been brought down and divided.
Y. If any partial dividend will not contain the divisor^
place a cipher in the quotient, and bring down the next figure
of the dividend, and divide as before.
VI. If there be a remainder after dividing all the figures of
the dividend, it must be written in the quotient, with the divi-
sor underneath.
Notes. 1. If any remainder be eqtial to, or greater than the divisor,
the quotient figure is too stnall, and must be increased.
2. If the product of the divisor by the quotient figure be greater
than the partial dividend, the quotient figure is too large^ and must be
diminished.
70. Proof. 1. The same as in short division. Or,
2. Subtract the remainder, if any, from the dividend, and
divide 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.
54
SIMPLE NUMBERS.
2d. Find how many times.
3d. Multiply.
4th. Subtract.
6th. Bring down another figure.
EXAMPLES FOR PRACTICE.
3. Find how many times 36 is contained in 11798.
OPERATION. PROOF BY MULTIPLICATION.
DividcHd.
11798 (
327 Quotient.
327
Quotient.
108
36
Divisor.
99
1962
72
981
278
11772
252
26
Remainder.
26
Remainder.
11798
Dividend.
4. Find how many times 82 is contained in 89634.
OPERATION.
82 ) 89634 ( 1093
82
PROOF BY DI'S^SI0N.
89634 Dividend.
8 Eeraaindor,
763
738
Quotient.
1093 ) 89626 ( 82
8744
Divisor.
2186
2186
254
246
^ 8
5. Find how many times 154 is contained in 32740.
6. Divide 32572 by 34. Ans. 958.
7. Divide 1554768 by 316. Ans. 7198.
8. Divide 5497800 by 175. Ans, 31416.
9. Divide 3931476 by 556. Ans. 7071.
10. Divide 10983588 by 132. Ans. 83209.
DIVISION. 5o
11. Divide 73484248 by 19.
12. Divide 8121918 by 21.
13. Divide 10557312 by 16.
14. Divide 93840 by 63.
15. Divide 352417 by 29.
16. Divide 51846734 by 102.
17. Divide 1457924651 by 1204.
18. Divide 729386 by 731.
19. Divide 4843167 by 3605.
20. Divide 49816657 by 9101.
21. Divide 75867308 by 10115.
22. Divide 28101418481 by 1107.
23. Divide 65358547823 by 2789.
24. Divide 102030405060 by 123456.
25. Divide 48659910 by 54001.
26. Divide 2331883961 by 6739549.
27. A railroad cost one million eight hundred fifty thousand
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. 6736yVTT?V dollars.
29. In 1850 there were in the United States 213 college
libraries, containing 942321 volumes; what would be the
average number of volumes to each library ?
Ans. 44242-f 7 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 dollars; what was the average
revenue of each office ? Ans. 266 dollars.
Ans. 3867592.
Ans. 386758.
Ans. 659832.
Rem.
33.
Rem.
9.
Rem.
32.
Rem.
1051.
Rem.
579.
Rem.
1652.
Rem.
6884.
Rem.
4808.
Quotients.
Rem.
25385201.
974.
23434402.
645.
826451.
70404.
901.
5009.
346.
7.
56 SIMPLE NUMBEKS.
CONTRACTIONS.
CASE I.
81. When the divisor is a composite number.
1. If 3270 dollars be divided equally among 30 men, how
many dollars will each receive ?
OPERATION. Analysis. If 3270 dollars be divided
5)3270 equally among 30 men, each man will receive
as many dollars as 30 is contained times in
6)654 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 3270 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. Hence,
Rule. Divide the dividend hy one of the factors, and the
quotient thus obtained by 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 0.
7. Divide 153160 by 56 = 7 X 8.
8. Divide 15625 by 125 = 5 X 5 X 5.
82. To find the true remainder.
1. Divide 1 1 43 by 64, using the factors 2, 8, and 4, and fiud
the true remainder.
What are contractions ? Case I is what ? Give explanation. Rule;
Ans.
246.
Ans.
147.
Ans.
208.
Ans.
173.
Ans.
1246.
Ans.
2735.
Ans.
125.
OPERATION.
2)1143
8)571 -
1 rem.
4)71-
3X2= 6 "
17-
- 3 X 8 X 2 = 48 "
55 true rem.
57
Analysis. Divid-
ing 1143 by 2, we
have a quotient of
571, and a remainder
of 1 undivided, which,
being a part of the
given dividend, must
also be a part of the
true remainder. The
571 being a quotient arising from dividing by 2, its units are
2 times as great in value as the units of the ^ven dividend, 1 143.
Dividing the 571 by 8, we have a quotient of 71, and a remainder
of 3 undivided. 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 remainder of 6 arising from dividing by 8. Dividing the 7 1 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 8 times as great in value
as those of the 571, and the units of the 571 are 2 times as great
in value as those of the given dividend, 1143 ; therefore, to 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. Hence,
Rule. I. Multiply each partial remainder, except the first,
hy all the preceding divisors.
II. Add the several products with the first remainder, and
the sum will be the true remainder.
EXAMPLES FOR PRACTICE.
Rem.
2.
Divide 34712 by
42 =: 6 X 7.
20.-
3.
Divide 401376 by
64 = 8 X 8.
32.V
4.
Divide 139074 by
72 = 3 X 4 X 6.
42.-
5.
Divide 9078126 by
90 = 3 X 5 X 6.
6i
6.
Divide 18730627 by
120 r= 4 X 5 X 6.
67.
7.
Divide 7360479 by
96 = 2X6X8.
63v
8.
Divide 24726300 by
70 r= 2 X 5 X 7.
60i.
9.
Divide 5610207 by
84 = 7 X 2 X 6.
15r/
Explain the process of finding the true remainder when dividing by
the factors of a composite number.
3*
58 SIMPLE KITMBERS.
CASE IT.
83. When the divisor is 10, 100, 1000, &c.
1. Divide 374 acres of land equally among 10 men ; how
many acres will each have ?
OPERATION. Analysis. Since we have shown,
110)3714 that to remove a figure one place
— - — toward the left by annexing a cipher
Quotient. 37 - - - 4 Rem. increases its value tenfold, or multi-
or, 37^ 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 off two figures, we divide by
100, if three, we divide by 1000, and so on. Hence the
Rule. I'rom 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.
1. Divide 437661 by 800.
OPERATION. Analysis. In this example we
8100)4376161 resolve 800 into the factors 8 and
100, and divide first by 100, by cut-
547 - - - 61 Rem. ting off two right hand figures of the
Caco II is what ? Give explanation. Rule. Case III is whati
Give explanatioii.
DIVISION-.
59
dividend, (83,) and we have a quotient of 4376, and a remainder of
61. We next divide by 8, and obtain 547 for a quotient; and the
entire quotient is 547^^.
2. Divide 34716 by 900.
OPERATION.
9]00 )347|16
38 Quotient. 5, 2d rem.
5 X 100 + 16 == 516, true rem.
38|^^, Ans,
ruuy
Analysis. Dividing
as in the last example, we
have a quotient of 38, and
two remainders, 16 and
5. Multiplying 5, the
last remainder, by 100,
the preceding divisor, and
adding 16, the first remainder, (82,) we have 516 for the true re-
mainder. But this remainder consists of the last remainder, 5, pre-
fixed to the figures 16, cut off from the dividend. Hence,
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 oflf from the dividend will, be the
true remainder.
Quotients. Rem.
3. Divide 34716 by 900. 38 516-
4. Divide 1047634 by 2400. 436 1234v
5. Divide 47321046 by 45000. 1051 ^26046
6. Divide 2037903176 by 140000. ' '.63176
7. Divide 976031425 by 92000. -3425 -
8. Divide 80013176321 by 700000. i.376321
9. Divide 19070367428 by 4160000. 4584 C927428
10. Divide 379025644319 by 554000000. 89644319
1 1 . The circumference of the earth at the equator is 24898
miles. How many hours would a train of cars require to travel
that distance, going at the rate of 50 miles an hour ?
Ans. 497 1§.
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 NUMBEBS.
EXAMPLES IN THE PRECEDING RULES.
^1. George Washington was born in 1732, and lived 67
years ; in what year did he die ? Ans. in 1799.
2. How many dollars a day must a man spend, to use an
income of 1095 dollars a year ? Ans. 3 dollars.
3. If I give 141 dollars for a piece of cloth containing 47
yards, for how much 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 ? Alls. 411 acres.
5. A dealer sold a cargo of salt for 2300 dollars, and gained
625 dollars ; what did the cargo cost him ?
A?is. 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 jflour in 87 days, how many-
days will a barrel last 1 person at the same rate ?
Ans. 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
sura? Ans. 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 ? Ans. 3888 dollars.
12. Four children inherited 2250 dollars each; but one
dying, the remaining three inherited the wliole ; what was the
share of each ? Ans. 3000 dollars.
PROMISCUOUS 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 ? A)is. 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?
Ans. 40000 acres. S^
/l7. Add 884, 1562, 25, and 946; subtract 2723 from the
sum ; divide the remainder by 97 ; and multiply the quotient
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 ; how much 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 dol-
lars now, in how many years will he be worth 4500 dollars ?
Ans. 9 years.
21. How many bushels of oats at 40 cents a bushel, must
be given for 1600 bushels of wheat at 75 cent« a bushel ?
Ans. 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 j 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 ?
Ans. 45210 dollars.
2C. What number is that, which being divided by 7, the
quotient multiphed hy 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 bay 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 ; suppose their cash value to have
been 5,301,212 dollars, what would be the average value?
Ans. 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
tlie 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 ninny 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? A71S. 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? Aiis. 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?
Ans. 95506.
43. A farmer sold 35 bushels of wheat at 2 dollars a bush-
el, 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 ; how many
dollars 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 packing barrel, and 3
cents for packing and weighing, and receives 125 cents from
the purchaser ; how many cents 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
6000 dollars, and the others the remainder, in equal amounts ;
how much does each of the others pay ? Ans. 20000 dollars.
47. If 256 be multiplied by 25, the product diminished by
625, and the remainder divided by 35, what will be the quo-
tient? 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 NUMBERS.
GENERAL PRINCIPLES OF DIVISION.
80. 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 General Principles of Division, which we
will now examine.
L 54^9 = 6.
If we multiply 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 hy any number, multiplies the quotient
by the same number.
II. Using the same example, 54 -f- 9 =: 6.
If we divide the dividend by 3 we have
Y-j-9 = 18-^9:z=2,
and 2 = the quotient, 6, divided by 3. Hence, Dividing the
dividend by any number, divides the quotient by the same
number.
III. If we multiply the divisor by 3, we have
54 _^ 9 X 3 = 54 -^ 27 m 2,
and 2 = the quotient, 6, divided by 3. Hence, Multiplying
the divisor by any number, diindes the quotient by the same
number.
V. If we divide the divisor by 3, we have
54 -^ § = 54 ~- 3 = 18,
Upon what does the vnhie of the quotient depend ? "NMiat is the
first general principle of division J {Second ? Third ? Fourth }
GENERAL PRINCIPLES OF DIVISION. 65
and 18 =: the quotient, 6, multiplied by 3. Hence, Dividing
ike divisor by any number^ multiplies the quotient by the same
number.
V. If we multiply both dividend and divisor by 3, we have
54 X 3 -^ 9 X 3 :=: 162 ^ 27 = 6.
Hence, Multiplying both dividend and divisor by the same num-
ber, does not alter the value of the quotient.
VI. If we divide both dividend and divisor by 3, we have
Y-M=:18-^3 = 6.
Hence, Dividing both dividend and divisor by the same num-
ber, 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 divisor
in practical arithmetic. The principles upon which these
changes are based may be stated as follows :
Prin. I. Multiplying the dividend multiplies the quotient ;
and dividing the dividend divides the quotient. (80. 1 and 11.)
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 j
divisor by the same number, does not alter the quotient. (86. i
V and VI.)
88. These three principles may be embraced in one
GENERAL LAW,
A change in the dividend produces a like change in the I
quotient ; but a change in the divisor produces an opposite /
change in the quotient.
Note. If a number be multiplied and the product divided by the
same number, the quotient will be equal to the number multiplied.
Thus, 15 X 4 = 60, and 60 -5- 4 = 15.
Fifth ? Sixth ? Into how many general principles can these be con-
densed ? What is the first ? Second ? Third ? In what general law
are these embraced ?
66 PROPERTIES OF NUMBERS.
EXACT DIVISORS.
I89o 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 wdll present a few directions that will be of assistance,
particularly in finding exact divisors of large numbers.
Note. A number whose unit figure is 0, 2, 4, 6, or 8 is called an
Eve7i Number, And a number whose unit figure is 1, 3, 6, 7, or 9, is
called an Odd Number,
. f 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, 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 + = 36, and 36-^9 = 4.
' 10 is an exact divisor when occupies units' place.
' 100 when 00 occupy the places of units and tens.
V 1000 when 000 occupy the places of units, tens, and hun-
dreds, &c.
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
(r= 2 X 2) and 6 (= 2 X 3).
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 ? WTiat is an even number ? An odd num-
ber ? WTicn 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 ? Att odd niuabej is an eacact divisor of
what?
FACTORINO NUMBERS.
67
twice that odd number is also an exact divisor of the even
number. Thus, 7 is an exact divisor of 42 j^so also is 7 X 2,
or 14.
PEBIE NUMBEES.
'S 00. A Prime Number 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
m
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
rACTORINQ NUMBERS.
CASE I.
91. To resolve any composite number into its
prime factors.
What is a prims niunber ? In fhctoruig 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. We 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
mxt quotiont, 77, we divide by its least prime
factor, 7, and we obtain the quotient 11 ; this be-
ing a prime number, the division can not be car-
ried further. The divisors and last quotient, 2,
2, 3, 3, 7, and 11 are all the prime factors of the
given number, 2772. Hence the
Rule. Divide the given number hy any prime factor ; di^-
vide the quotient in the same manner, and so continue the
division until the quotient is a prime number. The several
divisors and the last quotient will be the prime factors required.
Proof. The product of all the prime factors will be tha
given number.
EXAMPLES FOR PRACTICE.
2. What are the prime factors of 1 1 40 ? 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 IT.
02. To resolve a number into all the different sets
of factors possible.
1. In 36 how many sets of factors, and what are they ?
Give explanation. Rule. Proof. Case 11 is what ?
30=^
CANCELLATION. 69
OPERATION. Analysis. Writing the 36 at
"2 X 18 ^^^ ^^^^ °^ ^^^ ^^S^ — ^' ^^ arrange
Q w 1 2 ^^ the different sets of factors into
, q which it can be resolved under
each other, as shown in the opera-
'^ tion, and we find that 36 can be
2X2X9 resolved into 8 sets of factors.
2X3X6
3X3X4
12X 2 X3 X3
EXAMPLES FOR PRACTICE.
2. How many sets of factors in the number 24 ? What
are they ? Ans. 6 sets.
3. In 125 how many sets of factors ? What are they ?
A71S. 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 factors
from numbers sustaining to each other the relation of dividend
and divisor.
It has been shown (77) 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 will be the quotient.
1. Divide 63 by 7.
OPERATION. Analysis. We see in
DiTisor, ;^);5f X 9 Dividend. this example that 63 is
composed of the factors 7
9 Quotient. ^nd 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.
Give explanation. What is cancellation ? Upon what principle is
it based ? Give first explanation.
70 PROPERTIES OF NUMBERS.
04:. Whenever the dividend and divisor are each composite
numbers, the factors common to both may first be rejected
without altering the final result. ( 8^5 Prin. Ill, )
2. What is the quotient of 24 times 56 divided by 7 times
OPERATION. Analysis.
24 X 56 4 X X :^ X $ . ^^ ^^f '''-
=■ "^4, u4.ns. dicate the op-
7x48 J^X0X0 eration to be
performed by
writing the numbers which constitute the dividend above a line, and
those which constitute 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 fac-
tors 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.
Note. "When all the factors or niunbers in the dividend are can-
celed, 1 should be retamed.
05 • 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
6 II divide 54 and 63 ; so we reject 9 as a factor of 54,
fj-y Hr4 ^'^^ retain the factor 6, and also as a factor of 63,
p^ X 7IJ ^^^ retain the factor 7. Again, 7 will divide 7 in
0^ the divisor, and 77 in the dividend. Dividing
r^ 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
54^ ^^^1 as in the pre-
n-KU^X^ 5X4 ,^ ^^ ceding example, we
=r = ^^ = 2f . reject all the fac-
a'0 X ji X X 7 7 tors that are com-
^ mon to both divi-
dend and divisor,
Give second explanation.
CANCELLATION. 71
and we have remaining the factor 7 in the divisor, and the factors 5 and
4 in the dividend. Completing the work, we have K^- ==: 2f , Ans,
From the preceding examples -and illustrations we derive
the following
Rule. I. 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 div-
idend by the product of the remaining factors of the divisor,
and the result will be the quotient.
Notes. 1. Rejecting a factor from any number is dividing the number
by that factor.
2. When a factor is canceled, the unit, 1, is supposed to take its
place.
3. 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 factors of
the dividend on the right of a vertical line, and the factors of the divisor
on the left.
EXAMPLES FOR PRACTICE.
1. What is the quotient of 16 X 5 X 4 divided by 20 X 8 ?
FIRST OPERATION. SECOND OPERATION.
^ ha
4
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 ? WTiat is the effect of rejecting
a factor ? What is the quotient when all the factors in the divisor are
canceled ?
72
TROPERTIES OP NUMBERS.
FIRST OPERATION.
10
Tt^X^^XXi 3
3^
^ij> = 6|, Am,
SECOND OPERATION.
10
>
i^0
ii
It
3
20
6|, ^715.
3. Divide the product of 33 X 35 X 28 by 11 X 15 X 14.
Am, 14.
4. What is the quotient of 21 X H 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, 9f.
6. Divide 140 X 39 X 13 X 7 by 30 X 7 X 26 X 21.
Ans. 4 J.
7. What is the quotient of 66 X 9 X 18 X 5 divided by
22 X 6 X 40 ? Ans, 10|.
8. Divide the product of 200 X 36 X 30 X 21 by 270 X
40 X 15 X 14. Ans, 2.
9. Muhiply 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 dollai-s a cord ? Ans. 10 tons.
GREATEST COMMON DIVISOR. 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 containing
24 yards, at 3 dollars a yard. How many suits of clothes, at
18 dollars a suit, must be made 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, 76§ days.
GREATEST COMMON DIVISOIl.
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 num-
bers is the greatest number that will exactly divide each of
them.
(^Numbers prime to each other are such as have no common
divisor.)
Note. A common divisor is sometimes called a Common Measure ;
and the greatest common divisor, the Greatest Common Measitre.
CASE I.
98. When the numbers are readily factored.
1. What is the greatest common divisor of 6 and 10 ?
Ans. 2.
OPERATION. Analysis. We readily find by inspection
6 . . 1 that 2 will divide both the given numbers ;
■~ ~ hence 2 is a common divisor; and since the
• • i_ quotients 3 and 5 have no common 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.
RP. 4
3
42 . . 63 . . 105
7
14 . . 21 . . 35
2.. 3.. 5
74 PROPERTIES OF NUMBERS.
OPERATION. Analysis. We observe that 3
■will exactly divide each of the given
numbers, and that 7 will exactly
divide each of the resulting quo-
tients. Hence, each of the given
numbers can be exactly 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 ;
28 . . 1 40 . . 280 t^^n the quotients by 7. The re-
2 ~ suiting quotients, 1, 5, and 10, are
7 . . do . . /O prime to each other. Hence 4 and
1 . . 5 . . 10 ^ ^^^ ^^^ t^^ component factors of
the greatest common divisor.
4 X 7 = 28, A71S,
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 by any factor common to all the numbers.
II. Divide the quotients in like manner, and continue the
division till a set of quotients is obtained that have no common
factor.
III. Multiply all the divisors together, and the product will
he the greatest common divisor sought,
EXAMPLES FOR 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. 6.
Jiule, first step ? Second ? Third ?
GREATEST COMMON DIVISOR. 76
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. 16.
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? Ans. 126.
10. What is the greatest common divisor of 96 and 544 ?
A71S. 32.
11. What is the greatest common divisor of 468 and 1184?
Ans. 4.
12. What is the greatest common divisor of 200, 625, and
150? Ans. 25.
CASE n.
99. When the numbers can not be readily factored.
As the analysis of the method under this case depends upon
three properties of numbers which have not been introduced,
we present them in this place.
I. An exact divisor divides any number of times its dividend.
II. A common divisor of two numbers is an exact divisor
of their sum.
III. A common divisor of tvvc numbers is an exact divisor
of their difference.
"VNTiat is Case U ? WTiat is the first principle upon which it is
founded ? Second ? Third ?
76
PROPERTIES OF NUMBERS.
1. What is the greatest common divisor of 84 and 203 ?
OPERATION. Analysis. Wc draw two vertical
203
84
2
70
2
14
2
•14
2
168
35
28
7, Ans,
lines, and place the larger number on
the right, and the smaller number on
the left, one line lower down. We
then divide 203, the larger number, by
84, the smaller, and write 2, the quo-
tient, between the verticals, the prod-
uct, 168, opposite, under the greater
number, and the remainder, 35, below.
We 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. We 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 same order as in
the former steps. Finally, dividing 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.
In order to show that the last divisor 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.
84
70
14
14
OPERATION.
A
2
7 divides the 14, as proved
by the last division ; it will
also divide two times 14, or 28,
(I.) Now, as 7 divides both
itself and 28, it will divide 35,
their sum, (XL) It will also
divide 2 times 35, or 70, (I ;)
and since it is a common
divisor 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 168 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, wc
203
168
35
28
Give analysis.
GREATEST COMMON DIVISOR. 77
84
70
U
168
28
7
know that the greatest common divisor, whatever it he, must divide
w 2 times 84, or 168, (I.) Then
V 203 since it will divide both 168
and 203, it must divide their
diflference, 35, (III.) It will
also divide 2 times 35, or 70,
(I ;) and as it will divide both
35 70 and 84, it must divide 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
difference, 7, (III;) hencey it
cannot be greater than 7.
Thus we have 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 number one line above the less.
II. Divide the greater number by the less, writing 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 divisor
hy the last remainder, and so on, till nothing remains. The
last divisor will be the greatest common divisor sought.
IV. If more than two numbers be given, first find the greatest
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.
Notes. 1. Wlien more than two numbers are given, it is better to
begin with the least two.
2. If at any point in the operation a priyne number occur as a re-
mainder, it must be a common divisor, or the given nmnbers have no
common divisor-
Rule, first step ? Second ? Third ? Fourth ? What relation have
nmnbers when their difference is a prime number ?
78
PROPERTIES OF NUxMBERS.
EXAMPLES FOR PRACTICE.
1. What is the greatest common divisor of 221 and 5512 ?
OPERATION.
221
2
4
208
1
Am. 13
1
6
5512
442
1092
884
208
13
78
78
2. Find the greatest common divisor of 154 and 210.
Ans. 14.
-^ 3. "What is the greatest common divisor of 316 and 664?
^-"^ A)is. 4.
, 4. "What is the greatest common divisor of 679 and 1869 ?
Ans. 7.
5. "What is the greatest common divisor of 917 and 1495 ?
Ans. 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 fore-
going methods.
8. John has 35 pennies, and Charles 50 : bow shall they
arrange them in parcels, so that each boy shall have the same
number in each parcel? Ans. 5 in each parcel.
9. A speculator has 3 fields, the first containing 18, tho
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 village street is 332 rods long; A owns 124 rods
front, B 116 rods, and C 92 rods; they agree to divide their
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
fallowing lengths: 3013, 2231, and 2047 feet; what is the
length of the longest rail that will 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
takfe? 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 multiple
of 2, 4, 5, and 10.
109. The Least Common Multiple is the least number
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 rommoa
multiple ?
80
PROPERTIES OF NUMBERS.
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 common multiple of two or more numbers may be found by
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, Q&, and 78.
OPERATION. Analysis. The
SO = 2 X 3 X 5 number cannot be
42 = 2 X 3 X 7 ^^^^ ^^^^ ^^' ^^"^^
66 = 2X3X11 ^^ TCi\xs,i contain 78 ;
yg 2y3VlS hence it must con-
tain the factors of
2X3X 13 X 11X7X5 = 30030, ^ws.*^^' '''^* ;
2 X 3 X 13.
We here have all the prime factors of 78, and also all the factors of
QQ, except the factor 11. Annexing 11 to the series of factors,
2X3X 13X 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,
2X3X 13X 11 X7,
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 analysis.
LEAST COMMON MULTIPLE. 8i
factors of 30 except the factor 5. Annexing 5 to the series of factors,
2 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 Jiumber, and
such prime factors of the other numbers as are not found in the
largest number, and their product will be the least cotnmo7i
midtiple.
Note. 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 niunber of times it appears in 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.
5. 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. What is the least common multiple of 12, 15, 18, 35 ?
Anx. 1260.
Rule, first step ? Second ? What caution is given i
4*
82
PROPERTIES OP NUMBERS.
2
4.
. 6.
.9.
.12
2
2.
.3.
.9.
. 6
3
3.
.9.
. 3
3
3
SECOND METHOD.
lOS, 1. What is the least common multiple of 4, 6, 9,
and 12 ?
, OPERATION. Analysis. We first write
the given numbers in a series,
with a vertical line at the left.
Since 2 is a factor of some of
the given numbers, it must be
a factor of the least common
multiple sought. Dividing as
^^g^ many of the numbers as are
divisible by 2, we >vrite the
quotients and the undivided number, 9, in a line underneath. We
now perceive that some of the numbers in the second line contain
the factor 2 ; hence the least common multiple must contain another
2, and Ave again divide by 2, omitting to write down any quotient
when it is 1. We next divide by 3 for a Hke 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 ?
2X2X3X3=
OPERATION.
2,5
10.
. 12.. 15.. 75
2,3
6.. 3.. 15
5
5
2X5X2X3X5 = 300, Ans.
Analysis. We read-
ily see that 2 and 5 are
among the factors of the
given numbers, and must
be factors of the least
common multiple ; hence
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 i)rocess 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. What is the least common multiple of 6, 15, 35, 42,
and 70?
Give explanation.
2,5
LEAST COMMON MULTIPLE. 83
OPERATION. Analysis. In this oper-
15 . . 42 . . 70 ation 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 42, 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 re-
maining numbers, 15, 42, and 70.
From these examples we derive the following
KuLE. I. Write the niimhers 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 in a litie underneath, omitting the Vs.
III. In like manner divide the quotients and undivided num-
bers, and continue the process till all the factors of the given
numbers have been transferred to the left of the vertical. Then
miiltiply 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? ^ A?is. 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 mgney for which I could
purchase an exact number of books, at 5 dollars, or 3 dollars,
or 4 dollars, or 6 dollars each ? A7is. 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 $30 each, oxen at $oo each, or horses at
$105 each? Ans. $2310.
15. A can shear 41 sheep in a day, B G3, 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 im-
prudently took the cheaper; how much did he thereby for»
feit? Ans. 80 cents.
CLASSIFICATION OF NU^IBERS.
Numbers may be classified as follows :
106. I. As Bven 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 OP NUMBERS. 85
10§. 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.
no. 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. VI. As Like 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 ap-
ples, 487 apples; and, if compound numbers, they must be
used to express the same kind of quantity, as time, distance,
&c. 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 ? >Vhat is the fourth classification ? An abstract
number ? A concrete number ? \Vhat is the fifth classification ? A
simple number ? A compound nxmiber ? Sixth classification ? What
are like numbers ? Unlike nvunbers ?
86 FRACTIONS.
FRACTIONS.
DEFINITIONS, NOTATION, AND NUMERATION.
lis. 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 07ie 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 ffths, three of the parts
three fifths, &:c.
The parts are expressed by figures ; thus,
One half is written
^
One fifth is
written
i
One third
u
^
Two fifths
a
1
Two thirds
a
ii
One seventh
((
+
One fourth
«
\
Three eighths
a
t
Two fourths
«
i
Five ninths
a
i
Three fourths
«
i
Eight tenths
it
A
Hence we see that the parts into which a unit is divided take
their name, and their value, from the 7iumber 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, &c. ; and each
third is less in value than each half, and each fourth less than
each third; and the greater the number 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
ll«l. A Fraction is one or more of the equal parts of a
unit.
Define a fraction.
DEFINITIONS, NOTATION, AND NUMERATION. 87
114:, To write a fraction, two integers are required, one
to express the number of parts into which the whole number
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 Numerator 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 de-
nominator, taken together.
118. Fractions indicate division, the numerator answering
to the dividend, and the denominator to the divisor. Hence,
119. The Value of a fraction is the quotient of the nu-
merator divided by the denominator.
ISO. To analyze a fraction is to designate and describe
its numerator and denominator. Thus, f is analyzed as fol-
lows : —
4 is the denominator, and shows that the unit 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 terms, considered as dividend and divisor.
The value of the fraction is the quotient of 3 -^ 4, or f .
How many numbers are required to write a fraction ? Why ? Do-
fine the denominator. The numerator. What are the terms of a frac-
tion ? The value ? >Vhat is the analysis of a fraction ?
88 FRACTIONS.
EXAMPLES FOR 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.
G. Seventy-five ninety-sixths.
7. Two hundred fifty-four /oz/r hundred forty-thirds.
8. Eight nine hundred twenty-firsts.
9. One thousand two hundred thirty-two seventy-five thou-
sand six hundredths.
10. Nine hundred six two hundred forty-three thousand
eighty-seconds.
Read and analyze the following fractions :
11. -1^; i^', /^; ^1; jf ; -^fe; ^V^; i||.
12. T^^; tW^; -t^^i tB^; ^U^i ,VVV
1^' tVi J T^iTtrJ fI if T 5 iSflfjF*
121 • 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, /j* t&>
^175 -hi ^^® proper fractions.
An Improper Fraction is one whose numerator equals or
exceeds its denominator ; its value is never less than the
unit, 1. Thus, ^, §, -\S -3^, ^g, -^j^ are improper fractions.
122. A Mixed Number is a number expressed by an in-
teger and a fraction; thus, 4^, 17^|, Dj^jy are mixed numbei-s.
123. Since fractions indicate division, all changes in the
terms of a fraction will affect the value of that fraction according
to the laws of division ; and we have only to modify the lan-
guage of the General Principles of Division (87) by substi-
tuting the words numerator ^ denominator , and fraction, or value
"What is a proper fraction ? An improper fraction ? A mixed
number ? What do firactions indicate ?
EEDUCTION. 89
of the fraction^ for the words dividend^ divisor, and quotient,
respectively, and we shall have the following
GENERAL PRINCIPLES OF FRACTIONS.
1124:. Prin. I. Multiplying the numerator multi'plies the
fraction^ and dividing the numerator divides the fraction.
Prin. II. Multiplying the denominator divides the fraction,
and dividing the denominator multiplies the fraction.
Prin. III. Multiplying or dividing both terms of the frac-
tion hy the same number does 7iot alter the value of the fraction.
These three principles may be embraced in one
GENERAL LAW.
13^. A change in the numerator produces a like change
in the value of the fraction ; but a change in the denomina-
tor j9roc?wces an opposite change in the value of the fraction.
REDUCTION.
CASE I.
1^6* To reduce fractions to their lowest terms.
A fraction is in its lowest terms when its numerator and de-
nominator are prime to each other ; that is, when both terms
have no common divisor.
1. Reduce the fraction |J to its lowest terms.
FIRST operation. ANALYSIS. Dividing both
4S:ir:§^zz:i2r=^, Aus. terms of a fraction by the same
number does not alter the value
of the fraction or quotient, (124, III ;) hence, we divide both
term^ of |f, by 2, both terms of the result, |-^, by 2, and both terms
of this result by 3. As the terms of | are prime to each other, the
lowest terms of |f are |. We have, in effect, canceled all the fac>
tors common to the numerator and denominator.
First general principle ? Second ? Third ? General law ? WTiat
is meant by reduction of fractions? Case I is what? "N\Tiat is
meant by lowest terms ? Give analysis.
Ans.
i-
Ans.
t-
Ans.
fi-
90 FRACTIONS.
SECOND OPERATION. In this operation we have divided
12) 4^-::= * Ans, ^^^^ terms of the fraction by their
greatest common divisor, (97,) and
thus performed the reduction at a single division. Hence the
Rule. Cmicel or reject all factors common to both numerG'
tor and denominator. Or,
Divide both terms by their greatest common divisor,
EXAMPLES FOR PRACTICE.
2. Reduce |^J to its lowest terms.
3. Reduce §|| to its lowest terms.
4. Reduce ||^ to its lowest terms.
5. Reduce f || to its lowest terms.
6. Reduce g^VeV ^^ ^^^ lowest terms.
7. Reduce fi^j to its lowest terms.
8. Reduce -jVi^ff to its lowest terms.
9. Reduce f^f 8" ^^ i^^ lowest terms. Ans. ^J.
10. Reduce If |^ to its lowest terms. Ans. ^%,
11. Reduce t^feVo- to its lowest terms. Ans. ^^\.
12. Express in its simplest form the quotient of 441 divided
by 462. Ans. |f
13. Express in its simplest form the quotient of 189 di-
vided by 273. • Ans. ^^.
14. Express in its simplest form the quotient of 1344 di-
vided by 1536. Ans. |.
CASE II.
127. To reduce an improper fraction to a whole
or mixed number.
1. Reduce ^^ to a whole or mixed number.
OPERATION. Analysis. Since
Mr^. — 324— 15 = 21^ = 21-^, vl?2S. 15 fifteenths equal
^ • ^ 1,324 fifteenths are
equal to as many times 1 as 15 is contained times in 324, which is
21^ times. Or, since the numerator is a dividend and the denom-
Bulc. Case II is what } Give explanation.
REDUCTION. 91
mator a divisor, (118,) we reduce the fraction to an equivalent
whole or mixed number, by dividing the numerator, 324, by the
denominator, 15. Hence the
Rule. Divide the numerator hy the denominator.
Notes. 1. "When the denominator is an exact divisor of the numer-
ator, the result will be a whole number.
2. In all answers contaimng fractions reduce the fractions to their
lowest terms,
EXAMPLES FOR PRACTICE.
2. In X^- of a week, how many weeks ? Ans. 1 f .
3. In -1^^ of a bushel, how many bushels ? Ans. 23|
4. In ^ |-S^ of a dollar, liow many dollars ?
5. In ^^ of a pound, how many pounds ? Ans. 54^.
6. Reduce -f ^^ to a mixed number.
7. Reduce ^^- to a whole number.
8. Change -'-Ip to a mixed number. Ans, 18|.
9. Change ^f f ^ to a mixed number.
10. Change ^^2^-^ to a mixed number. Ans. 1053f f.
11. Change ^^§^^^^ to a whole number. Ans. 7032.
CASE in.
IS8. To reduce a whole number to a fraction hav-
ing a given denominator.
1. Reduce 46 yards to fourths.
OPERATION. Analysis. Since in 1 yard there are 4 fourths,
4g in 46 yards there are 46 times 4 fourths, which are
4 184 fourths = J-l^. In practice jve multiply 46,
the number of yards, by 4, the given denominator,
f-*-, Ans. ^jj(j taking the product, 184, for the numerator of a
fraction, and the given denominator, 4, for the de-
nominator, we have 1|^. Hence we have the
Rule. Multiply the whole number by the given denominator ;
take the product for a numerator^ under which write the given
denominator.
Rule. Case III is what ? Give explanation. Rule.
92 FRACTIONS.
Note. A whole number is reduced to a fractional form by writing
1 mider it for a denominator ; thus, 9 n: ^.
EXAMPLES FOR PRACTICE.
2. Reduce 25 bushels to eighths of a bushel. A7is. ^^oa
3. Reduce G3 gallons to fourths of a gallon. Ans. '|2.
4. ReduceU40 pounds to sixteenths of a pound.
5. In 56 dollars, how many tenths of a dollar ? A71S. ^^^p■,
G. 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. Ans. ^ff^.
CASE IV.
139. To reduce a mixed number to an improper
fraction.
1, In 5| dollars, how many eighths of a dollar ?
OPERATION.
5 3 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 z=z 43 eighths,
^g^j ^W5. or ^. From this operation we derive the following
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 4i dollars, how many half dollars ? Ans. |.
3. Iii 71^ weeks, how many sevenths of a week ?
4. In 341 J acres, how many fourths? Ans. ^^-^.
5. Change 12/^ years to twelfths.
6. Change 5Gr^^ to an improper fraction. Ans. -W^.
7. Reduce 21 5?)^ to an improper fraction. Ans. -^i^"^*
8. Reduce 225 1| to an improper fraction. Ans. ^|§*.
Case IV is what ? Give explanation. Rule.
HEDUCTION. 93
9. In 9Gy*j^(y, how many one hundred twentieths ?
10. In 1297^4, how many eighty-fourths ? Ans. ^Q||^J-.
11. What improper fraction will express 400§2 ?
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. Reduce f to a fraction whose denominator is 20.
OPERATION. Analysis. We first divide 20, the
20 -L. 4, ^^ ^ required denominator, by 4, the denomi-
nator of the given fraction, to ascertain
_2 X 5 __ , 5 ^;25, if it be a multiple of this term, 4. The
4 \/ 5 division shows that it is a multiple, and
that 5 is the factor which must be em-
ployed to produce this multiple of 4. We therefore multiply both
terms of f by 5, (124,) and obtain -^f , the desired result. Hence the
Rule. Divide the required denominator hy the denominator
of the given fraction^ and multiply both terms of the fraction hy
the quotient,
examples for practice.
2. Reduce 5 to a fraction ■whose denominator Is 15.
Ans. f^.
3. Reduce f to a fraction whose denominator is 35.
4. Reduce |J to a fraction whose denominator is 51.
Ans. ^f.
5. Reduce §g to a fraction whose denominator is 150.
6. Reduce |§| to a fraction whose denominator is 3488.
Ans. '^l%%,
7. Reduce ^J^ to a fraction whose denominator is 1000.
Case V is what ? How are fractions reduced to higher terms ?
What are all higlier terms ? Give analysis. Rule,
94 FRACTIONS.
CASE VI.
131, To reduce two or more fractions to a com-
mon denominator.
A Common Denominator is a denominator (fommon to two
or more fractions.
1. Reduce J and f to a common denominator.
OPERATION. Analysis. We multii)ly the terms of the
3X5 first fraction by the denominator of the second,
~ ^ =z ^^ and the terms of the second fraction by the
^ ^ ^ denominator of the first, (124.) This must re-
\( A ^^c^ ^^c^ fraction to the same denominator,
— ^:^ 8 for each new denominator will be the product
5X4 of the given denominators. Hence the
Rule. Multiply the terms of each fraction hy the denomina-
tors of all the other fractions.
Note. Mixed numbers must first be reduced to improper fractions.
EXAMPLES FOR PRACTICE.
2. Reduce §, ^, and J to a common denominator.
Ans. i|,^f,M.
3. Reduce f and J to a common denominator.
^ns. il.'il'
4. Reduce |, -^^^ and |^ to a common denominator.
5. Reduce |, |, §, and ^ to a common denominator.
Ans. m^n. Uh Ml-
G. Reduce j^g, ^, and | to a common denominator.
Ans. i%h-nhN2'
7. Reduce ^, 2|, J, and ^ to a common denominator.
8. Reduce 1§, y^o» ^^d 4 to a common denominator.
Ans. -V^, §*, \^^-.
Cnse VI is whnt ? "NVlmt is n pominon denominator ? Give analysis,
llulp.
OPERATION.
2,3
6. .8. .12
2,2
4.. 2
REDUCTION. 95
CASE VII.
13S. To reduce fractions to the least common de-
nominator.
The Least Common Denominator 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 denom-
inators.
1. Reduce ^, f , and ^^ to the least common denominator.
Analysis. AVe first find
the least common multiple
of the given denominators,
which is 24. This must be
2X3X2X2=^24: the least common denom-
1 __ 4. ■^ inator to which the frac-
3 _- Is [. Ans. tions can be reduced. (III.)
J_ ^J I ' We then multiply the terms
1 j 2t J Qf gg^f.]^ fraction by such a
number as will reduce the fraction to the denominator, 24. Re-
ducing 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 multipliers.
Hence the following
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 correspond-
ing quotient. The products will he the new numerators.
EXAMPLES FOR PRACTICE.
2. Reduce /^, fjj, |^, and y*^ to their least common de-
nominator. Ans. ^\%, rV^, \ih, tI^.
3. Reduce ^, ^, -^^, /^ to their least common denominator.
Ans. ^7j5, ^Tj^, Tj^/^'^g-, gjg^.
"What is Case VII ? What must be the least common denominator ?
Give analysis. Rule, first step. Second.
96 FRACTIONS.
4. Reduce f , gV? h ^"^ ^ ^^ their least common denomi-
nator. Ans. ^5';j, 2^5^^, ^11, Vb'/.
5. Reduce 5^, 2^, and 1| to their least common denomina-
tor. A71S. V-,Y-,¥-
6. Reduce y\, f , f , and J to their least common denomi-
nator. Ans. m^Pih Ml Mh
7. Reduce f , |, f , 2f, and ^ to their least common de-
nominator. A?is. ifg, i^e^g, tVf» til' iVs-
8. Change |, j\, 3|, 9, and |^ to equivalent fractions hav-
ing the least common denominator.
^ 9. Change f|^, 1^, |, xi)^"<i ^ to equivalent fractions hav-
ing the least common denominator.
10. Change 2/^, |J, 4, 1§, ^^, and | to equivalent frac-
tions having the least common denominator.
11. Reduce |, §, ^, and /g^ to a common denominator.
12. Reduce ^, f , 2 J, and ^ to a common denominator.
13. Reduce |f, /jj, §, and 3^ to equivalent fractions hav-
ing a common denominator. Ans. ||, §^, f g, |5'
14. Change ^j, §, and J to equivalent fractions having a
common denominator. Ans. fV%, ^tjVtt' tWjt*
15. Change ^, 74^, §^, and 5 to equivalent fractions hav-
ing a common denominator. Ans. §|, 4^^-, |g, ^5*^.
16. Change /^, 6^, -^^jy, 7, f, and 1^ to equivalent fractions
having a common denominator.
ADDITION.
133. 1. What is the sum of f f, I, and I ?
OPERATION. Analysts. Since the
| + |-|-| + ^ — J^ = 2, Ans. ^^'^^ fractions have a
common denominator, 8,
their sum may be found by addlnjir 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 f^, tV, T-V, ^0, and ^^. Ans. ^.
8. Add j^, T^, l2y -i^» yV» and j^J. Ans. 2^.
Give first explanation.
ADDITION. 97
4. What IS the sum of /^, ^% /^, ^f , ^f , and f ^ ?
5. What is the sum of ^V^, j%\, /^V, f/^, and -i^l ?
6. What is the sum of jV\-, ^^A. il^^ Mi' and ||§ ?
134. 1. What is the sum of f and f ?
OPERATION. Analysis. In
% + §=U + hi'=^ ^'-ts^^- = n^ ^^s, 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 | and f have not a common denominator,
w^e first reduce them to a common denominator, and then add the nu-
merators, 27 -{- 10 := 37, the same as whole numbers, and place the
sum over the common denominator. Hence the following
Rule. I. When necessary, reduce the fractions to a corn^
mon or to their least common denominator.
II. Add the num^ratorsy and place the sum over the common
denominator.
Note. If the amomit be an improper fraction, reduce it to a whole
or a mixed nimiber.
EXAMPLES FOE PRACTICE.
2. Add f to f. Ans. f |.
3. Add I to \l, Ans. If^.
4. Add f, ^, f, and ^. Ans, I^Vf-
\5. Add ^A, 2^, and ^^. Ans, l/^V
N6. Add ,V(T. T% ?V. and ^V- ^^«- f-
v7. Add ^i, it^, f I, ^, and f . • Ans, 3^^.
^8. Add f , J, f, I, t, f , I, I, and ^. Ans, 1^^^^%
9. Add 7^, of, and lOJ.
OPERATION. Analysis. The sum of the fraer
^_|_|i 3— ij^ tions \, |, and f is 1^^ ; the sum of
y I ^ I jQ __ 22 " the integers, 7, 5, 'and 10, is 22 ;
' ' and the sum of both fractions
Ans, 23 II- and integers is 231^. Hence,
Give second explanation ? Rule, first step. Second.
R.P 5
98 FRACTIONS,
To add mixed numbers, add the fractions and integers sep-
arately, and then add their sums.
Note. If the mixed numbers are small> they may be reduced to im-
proper fractions, and then added after the usual method.
10. Whatisthesumof 14|,3/ff,l§,andfg? Ans, 2Hg.
11. What is the sum of |, 1/2^, 10|, and 5 ? Ans. IS^^,
12. What is the sum of 17f, ISj^, and 2G.j\ ?
13. What is the sura of /g, ^, 1 ^, 3, and ^| ?
14. What is the sum of 125f, 327 rV, and 25^? Ans. 4785^.
15. What is the sum of ^fg, |J, l^^g, i^, and |ga ?
Ans. 3|gJ.
16. What is the sum of 3^9^, 2i|, 40f, and 10^^ ?
17. Bought 3 pieces of cloth containing 125|, 96|, and
48 1 yards ; how many yards in the 3 pieces ?
18. If it take 5^ yards of <;loth for a coat, 3^ yards for a
pair of pantaloons, and ^ of a yard for a vest, how many yards
will it take for all ? Ajis. d^^.
19. A farmer divides his farm into 5 fields; the first con-
tains 26/j acres, the second 40j^f acres, the third 51 f acres,
the fourth 59| acres, and the fifth G2§ acres ; how many acres
in the farm? Ans. 241 1 J.
2C. A speculator bought 175| bushels of wheat for 205^
dollars, 325f bushels of barley lor 2903 dollars, 270|f bush-
els of corn for 200|4 dollars, and 437 j\ bushels of 6at3 for
15Gf J dollars ; how many bushels of grain did he buy, and how
much did he pay for the whole ? J i ^^^^ti\ bushels,
^** (85911 dollars.
SUBTRACTION.
135. 1. From /^ take ^.
oPEiiATiON. Analysis. Since the given
^^ — -^ = -jS^ rr: I, A7is. fractions have a common denom-
inator, 10, we find the din'crcnce
by subtracting 3, the less numerator, from 7, the greater, and write
How arc mixed numbers added ? Give note.
SUBTRACTION. 99
the remainder, 4, over the common denominator, 10. We thus
obtain -^\ ::= |, the required difference.
2. From | take |. Ans. ^.
3. From if take | j. Ans. i.
4. From J^ take /y. ^W5. ^f .
5. From A§ take f §. A7is. ^.
6. From -j^^^ take ■^^\. Ans. ^.
7. From ^|| take ^J|. ^«5. ^V
136. 1. From f take |.
OPERATION. Analysis.
I - 1 = f f - f g zzr 3_2_- 3j) ^ ^2^ ^ _i^, ^^s. As in whole
numbers, we
can subtract like numbers only, or those haiving the same unit value,
so, we can subtract fractions only when they have a common de-
nominator. As f and | 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 wTite the difference, 2,
over the common denominator, 36. Vie thus obtain ^ = ^^, the
required difference. Hence the following
Rule. I. When necessary, reduce the fractions to a
common denominator.
II. Subtract the numerator of the subtrahend from the
numerator of the minuend, and place the difference over the
common denominator.
T^r
EXAMPLES FOR PRACTICE.
2. From ^ take f . . Ans.
3. From ^^ take f . A?is. /g.
4. Subtract j\ from |. Ans. y^\.
5. Subtract ^^ from -j^^jy. Ans. i^.
6. Subtract ^a from jfao. Ans. f^\.
7. Subtract -^-^J^ from |^. Ans. ^%Y^.
8. What is the difference between 9^ and 2 J ?
Give explanations. Rule, first step. Second.
100 FRACTIONS.
OPERATION. Analysis. We first reduce the frac-
9^ z= 9^ tional parts, ^ and |, to a common denom-
23 — . 2-9 inator, 12. Since we cannot take ^ from
^, we add 1 ziz i^ to y^, which makes \^,
6 /^ Ans. and j\ from i| leaves y^. We now add 1
to the 2 in the subtrahend, (50,) and say»
3 from 9 leaves 6. We thus obtain 6^'^, 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 numbers to improper fractions,
and subtract the less from the greater by the usual method*
-9. From 8J^ take 3^. Ans. 4||.
10. From 25| take 9^. Ans. 16-^.
11. From 4f take {|. ^ J^
12. Subtract If from 6.
13. Subtract 120jf^ from 450^. Ans. 330||.
14. Subtract -^\ from 3/^. Ans. Z^f-^.
15. Find the difference between 49 and 75^.
16. Find the difference between 227|^ and 196f.
^ 17. From a cask of wine containing 31^ gallons, 17| gal-
lons were drawn ; how many gallons remained ? Ans. 13|.
18. A farmer, having 450-j7g acres of land, sold 304 j acres ;
how many acres had he left ? Ans. 145;^^.
19. If flour be bought for G^ dollars per barrel, and sold
for^7§ dollars, what will be the gain per barrel ?
20. From the sum of f and 3J^ t|ike the difference of 4^
^nd 5^. Ans. 3||.
21. A man, having 251 dollars, paid 6^ dollars for coal, 2^
dollars for dry goods, and } of a dollar for a pound of tea ;
how much had he left? Ans. $16|J.
22. What number added to 2| will make 7^ ? Ans. 4§|.
23. What fraction added to |J will make ^§ ? Ans. ^.
In how many ways may mixed numbers be subtracted ? ^Vhat are
they ?
MULTIPLICATION. 101
24. A gentleman, having 2000 dollars to divide among his
three sons, gave to the first 912^ dollars, to the second 545^
dollars, and to the third the remainder ; how much did
the third receive ? Ans. $542y^^.
25. Bought a quantity of coal for 136 j-^^ dollars, and of
lumber for 350f dollars. I sold the coal for 184^ dollars, and
the lumber for 41 6| dollars. How much was my whole gain?
Atis. §114jV
MULTIPLICATION.
CASE I.
137. To multiply a fraction by an integer.
1. If 1 yard of cloth cost J of a dollar, how much will 5
yards cost ?
OPERATION. Analysis. Since 1 yard cost
3 >^ 5 — . j^. — - 33 ^yig^ 3 fourths of a dollar, 5 yards
will cost 5 times 3 fourths of a
dollar, or 15 fourths^ equal to 3| dollars. A fraction is multipHed
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
J^^J• X 5 = T = 1 -, Ans. multiplier, is a factor of 20, the
denominator, of the multipli-
cand, we perform the multiplication by dividing the denominator,
20, by the multipHer, 5, and we have |, equal to If dollars. A
fraction is multiphed by dividing its denominator, (124.) Hence,
Multiplying a fraction consists in multiplying its numerator^
or dividing its denominator.
Note. Always divide the denominator when it is exactly divisible
by the multiplier.
EXAMPLES FOR PRACTICE.
3. Multiply ^ by 5. Ans. y == 2f
4. Multiply fV by 7. Ans. 1-fft.
Case I is what ? Give explanations. Deduction.
102 FRACTIONS.
5. Multiply j\ by 12. Ans. 7f
6. Multiply /t- by G3. Ans. 15.
7. Multiply 5J- by 9.
OPERATION. Analysis. In multiply-
5^ ing a mixed number, we first
9*^ Qj. multiply the fractional part,
51 zzz -WL ^^^ ^^^'^ ^^^ integer, and
^i J I y "gj. 4q 1 ^^^ the two products ; or we
45 ^ ^ ^* reduce the mixed number to
49J.
an improper fraction, and
then multiply it.
8. Multiply 7f by 12. Ans. 9 If
^ 9. Multiply j\\ by 8. A?is. 5^.
\0. Multiply y^2- by 51. Ans. 2.
11. Multiply 15 1 by 16. Ans. 250.
12. Multiply |g^ by 22. Ans. 16|.
13. If a man earn S^^jy 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 f of an acre of
land cost ?
FIRST OPERATION.
Analysis. 3 fifths of an
5 ) 75 price of an aero.
acre will cost three times as
much as 1 fifth of an acre.
15 cost of -^ of an aero.
Dividing 75 dollars by 5, we
3
have 15 dollars, the cost of
Ans. 45 « « J« " "
^ of an acre, which we mul-
tiply by 3, and obtain 45
dollars, the cost of f of an acre.
Explain the process of multiplying mixed niunbers. AVhat is Case
II ? Give first explanation.
MULTIPLICA-riON.
103
SECOND OPERATION.
75 price of 1 acre.
3
Or, multiplying the 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
operation. Hence,
Muitijplying hg a fraction consists in multiplying by the nu-
merator and dividing by the denominator of the muUipUer*
5 ) 225 cost of 3 acres.
Ans. 45 «
■3. of an acre.
15
3
Note. By using the vertical line and
cancellation, we shall shorten, and com-
bine both operations in one.
45, Ajis.
EXAMPLES FOR PRACTICE.
2. Multiply 3 by |.
3. Multiply 100 by yV
4. Multiply 105 by |^.
5 Multiply 19 by ^f
a Multiply 24 by 6|.
Ans. 1^.
Ans. 64f.
Ans. 85.
Ans. 5]f.
OPERATION.
24
15 = 1 of 24; Or, ^
53
Ans.
ai
Analysis. We
multiply by the in-
teger and fraction
separately ,and add
the products; or,
144
159, Ans,
159,
reduce the mixed
number to an im-
proper fraction,
id then multiply by it.
7. Multiply 42 by 9f .
8. Multiply 80 by 14^^
9. Multiply 156 by f^.
10. At 8 dollars a bushel, ^
ivhat will f
Atis. 409^.
Ans. 1165.
Ans. 108.
of a bushel of clover
seed cost ?
Give second explanation. Note. Deduction.
104 FRACTIONS.
11. If a man travel 36 miles a day, how many miles will
he travel in 10§ days ? Aiis. 384 miles.
12. If a village lot be worth 450 dollars, what is yV 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 § of a dollar per bushel, how much will J of a bushel
of corn cost ?
OPERATION. Analysis.
1st atep, I _t. 4 ^^ ^, cost of ^ of a bushel. Since 1 bush-
2(J .tep, T-^ X 3 = A, " " ^ " " " ®^ COS* i 0^ ^
... , V /C3 6 1 J dollar,! of a
Whole wor., f X f = ^ = ^ ^n5. ^^^^^J ^ ^ .^j
cost I times f of a dollar, or 3 times
1^ of f of a dollar. Dividing f of a
dollar by 4, we have ^, the cost of
Or,
o 1 1 A ^ ^ of a bushel. A fraction is di-
^' ' vided by multiplj-ing its denomina-
tor, (124.) Multiplying the cost of ^ of a bushel by 3, we have ^\
of a dollar, the cost of | 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 denom-
inator, as shown in the whole work of the operation. Hence, for
multiplication of fractions, we have this general
KuLE. I. Reduce all integers and mixed numbers to
Improper fractions.
II. Multiply together the numerators for a new numerator,
and the denominators for a neii) denominator.
Note. Cancel all factors common to numerators and denominators.
EXAMPLES FOR PRACTICE.
2. Multiply f by |.
3 Multiply I by |.
4. Multiply ^l by f ^.
5. Multiply 4^ by f .
AVhat is Case III ? Give explanation. Kule, first step ? Second ?
"What shall be done with common factors ?
Ans.
i-
Ans.
-ft-
Ans.
A-
Ans.
3?.
MULTIPLICATION.
105
6. What is the product of y^^, f, ^j and ^ ? Ans. j'g*
7. What is the product of If, f, 2, and 5^ ? ^W5. 11||.
What is the product of f of -/j, f of f of ^, and |^ of
8.
n ^7 $
— X— X— X
4 ^0.6
OPERATION. Or,
^xIx^X^=:L, Ans.
^ $ K $ ZQ
Note. Fractions with the word of between them
are sometimes called compoimd fractions. The word
of is simply an equivalent for the sign of multiplica-
tion, and signifies that the numbers between which
it is placed are to be multiplied together.
i
^
X0
7
6
$
^
^
%
7t
4
$
$
3017:
Ans.
Ans.
HI-
85^.
9. Multiply -r\ o^ 2^ by ^ of 7^.
10. Multiply f of 16 by -/^ of 262.
11. What is the product of 3, J- of f , and f of 3|- ?
12. What is the value of 2^ times f of | of 1^ ? Ans. 2.
13. What is the value of | of ^ of 1| times f of 8 ?
14. What is the product of 12^ multiplied by 5^ times 6 J ?
Ans. 464y^.
15. At I of a dollar per yard, what will | of a yard of
cloth cost ? Ans. ^ o£ a. 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
f of a bushel worth ?
18. What will 7} pounds of tea cost, at f of a dollar per
pound ? Ans. 4^§ dollars.
19. What is the product of ^ by 4§ ?
9f
4§ - 23 2
39? product by 4. Or, 9,- X4? = —X— = 46.
__6f « " f. ' ' ;sf ^
Ans. 46 " « 4f .
What does " o/"' signify when placed between two fractions ?
is a compound fraction ?
What
6«
106 FRACTIONS.
To multiply mixed numbers together "wc may either mul-
tiply by the integer and fractional part separately, and then
add their products ; or, we may reduce both numbers to
improper fractions, and then multiply as in the foregoing rule.
20. Multiply 12| by 81. Ans. 108^.
21. What cost 6 1 cords of wood, at 2| dollars a cord ?
22. What cost J of 2 J- tons of hay, at 11^^ dollars a ton ?
A?is. $21-f5V.
23. What will 8f cords of wood cost, at 2^ dollars per
cord ? Ans. 22 j J dollars.
24. What must be paid for ^ of 6^ tons ef coal, at f of 7^
dollars per ton ? ^^ -^ "
25. A man owning ^ of a farm, sold ^ of his share ; what
part of the whole farm had he left ? A7is. ^^.'
26. Bought a horse for 125| dollars, and sold him for |- of
what he cost ; how much was the loss ? Ans, $25^^^.
27. A owned f of 123f acres of land, and sold ^ of his
share ; how many acres did he sell ? A7is. 49y^^.
28. If a family consume 1|- barrels of flour a month, how
many barrels will five such families consume in 4:^fj montlis ?
DIVISION.
CASE I.
140. To divide a fraction by an integer.
1. If my horse cat -^j of a ton of hay in 3 months, what
part of a ton will last him 1 month ?
OPERATION. Analysis. If he eat ^^^ of a ton in
^ -1- 3. =r -j^, Ans. 3 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,) we divide
the numerator of the fraction, ^, by 3, and we have ^, the answer.
2. If 3 yards of ribbon cost | of a dollar, what will 1 yard
cost?
Case I is what ? Give first explanation.
Ans.
f
Ans.
+•
Ans.
H-
Ans.
^■
Ans.
ri^'
DIVISION. 107
OPERATION. Analysis. Here we cannot exactly
s _i_ ^ _- _5 Jins, divide the numerator by 3 ; but, since a
fraction is divided by nxiltiplying the
denominator, (124,) we multiply the denominator of the fraction,
|, by 3, and we have ^, the required result. Hence,
Dividing a fraction consists in dividing its numerator^ or
multiplying its denominator.
Note. We divide the numerator when it is exactly divisible by the
divisor ; otherwise we multiply the denominator,
EXAMPLES FOR PRACTICE,
3. Divide f by 2.
4. Divide /y by 3.
6. Divide || by 5.
6. Divide {•^ by 25.
7. Divide \^ by 14,
8. Divide f^ bj 21.
9. If 6 pounds of sugar cost f of a dollar, how much will
1 pound cost .''
10. At 7 dollars a barrel, what part of a barrel of flour can
be bought for | 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, how much will I
bushel cost ? •
OPERATION. ;n'ote. We reduce the mixed number
71 r= S6. to an improper fraction, and divide as
4^9'= I, Ans. ^^^«'^^-
13. If 12 barrels of flour cost 76 J dollars, how much will
1 barrel cost ?
OPERATION. Analysis. Her^ we first divide as in
12 ) 76^ simple numbers, and we have a remainder
of 4^. We reduce this remainder to an
b^, Ans. improper fraction, ^, which we divide (as
in Ex. 1,) and annex the result, |, to the partial quotient, 6, and
we have 6|, the required result.
Give second explanation. Deduction,
J08 FRACTIONS.
14. How many times will 16f gallons of cider fill a vessel
that holds 3 gallons? Aiis. 5-^^.
15. If 9 men consume f of 9f pounds of meat in a day,
how much does each man consume ? Ans, J of a pound.
16. A man paid $99f^ for 4 cowsj how much was that
apiece? Ans, $24f|.
CASE II.
141. To divide an integer by a fraction.
1. At I of a dollar a yard, how many yards of cloth can be
bought for 12 dollars?
riRST OPERATION. ANALYSIS. As many yards as | of a
12 dollar, the price of 1 yard, is contained
A times in 12 dollars. Integers cannot be di-
Tided by fourths, because they are not of
^ ) 48 the same denomination. Reducing 12 dol-
16 yards lars to /bwr^A^ by multiplying, we have 48
fourths ; and 3 fourths is contained in 48
fourths 16 times, the required number of yards.
SECOND OPERATION. ANALYSIS. Here we divide the integer
3 \ 1 2 ^y the numerator of the fraction, and mul-
tiply ^p^ quotient by the denominator,
^ which produces the same residt as in the
4 fiist operation. Hence,
1 6 yards.
Dividing hy a fraction consists in multiplying hy the denom
inatory and dividing hjj the numerator of the divisor.
EXAMPLES FOR
PRACTICE,
2. Divide 18 by f.
Ans, 48.
3. Divide 63 by ^7^. ''
Ans, 117.
4. Divide 42 by f.
Ans. 49.
5. Divide 120 by ^7^.
Ans. 205f.
6. Divide 316 by ^3.
Ans. 877J.
Case II is what ? Give first explani
Ation. Second
. Deduction.
DIVISION. 109
7. How many bushels of oats, worth f of a dollar per bushel,
will pay for § of a barrel of flour, worth 9 dollars a barrel ?
Ans. 1 5.
8. If ^ of an acre of land sell for 21, dollars, what will an
acre sell for at the same rate ? Ans. $49.
9. When potatoes are worth f of a dollar a bushel, and
corn I of a dollar a bushel, how many bushels of potatoes are
equal in value to 1 6 bushels of corn ? Ans. 22^.
10. If a man can chop 2^ cords of wood in a day, in how
many days can he chop 22 cords ?
OPERATION.
22 Analysis. "VVe reduce the mixed number
^ to an improper fraction, and then divide the
integer in the same manner as by a proper
11J_88 fraction.
Ans, 8 days.
11. Divide 75 by 13f. Ans. 5^f.
12. Divide 149 by 24|. Ans. 6x%.
13. A farmer distributed 15 bushels of com among some
poor persons, giving them If bushels apiece; among how
many persons did he divide it ?
14. Divide | of 320 by ^ of 9^. Ans. 25^.
15. Bought ^ of 7^ cords of wood for J of $32 ; how much
did 1 cord cost ? Ans. $3|.
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.
14^. To divide a fraction by a fraction.
1. How many pounds of tea can be bought for -f^ of a dol-
lar, at § of a dollar a pound ?
How divide by a mixed number ? Case III is what }
110 FKACTIONS.
OPERATION. . Analysis. As
First stop, 1 1- X 3 ^ f f ™any pounds as f
Socoud step, -fj-^2 = §| = lf. of a dollar is con-
11 ^11 ^ 11 tained times in i^
Whole work. TZ'^Z'^TZ^ -= -=lf , ^/i^.of a dollar. 1 is
12 S X^^ 2 8 contained in \^, \}^
times, and | is con-
tained in \^ 3 times as many times as 1, or 3 times \^, which is ||
times^ which is the number of pounds that could be bought at | of
a dollar per pound ; but f is contained but ^ as many times as ^,
and f I divided by 2 gives ||, equal to 1| times, or the number of
pounds that can be bought at f 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 of the divisor, which is in accordance with 140 for dividing a
fraction. Hence, by invertiiig the terms of the divisor, the two
fractions will stand in such relation to each other that we can mul-
tiply together the two upper numbers for the numerator of the quo-
tient, and the two lower numbers for the denominator, as shown in
the operation. For division of fractions, we have this general
Rule. I. Heduce integers and mixed numbers to improper
fractions.
II. Invert the terms of the divisor, and proceed as in multi-
plication.
Notes. 1. The dividend and divisor may be reduced to a common
denominator, and the numerator of the dividend be divided by the nu-
merator of the divisor ; this will give the same result as the rule.
2. Apply cancellation where practicable.
EXAMPLES FOR PRACTICE.
2.
Divide | by |.
Ans, H.
3.
Divide f by ^.
Ans, 3^.
4.
Divide \ by y^(j«
Ans, 15.
5.
Divide ^ by j^j.
Ans, if.
6.
Divide | by f |.
Ans. ff.
7.
How many times
is*
contained
in
1?
Ans, l^V-
8.
How many times
is^
contained
in
13?
Ans. 8|.
■■' >j
■\ I
<" '
•4'
J '.
7 "
l; ", S
" c
^ '^
^
^- 1
Rule, first step. Second.
What other method
is mentioned ?
DIVISION. Ill
\9.'"How many times is -^g contained in |^ ? Ans, 2f.
10. How many times is i\ contained in ^§ ?
11. How many times is ^ of f contained in f of 2j^?
12. What is the quotient of j% of 4, divided by | of 3| ?
13. What is the quotient of ^ of J of 36 divided by 1|
times f ? Ans. 3^.
3^
14. What is the value of — ?
OPERATION.
3i.__ I- ___7 , 35 ^ $ This example
7~ — — — TT"^ "TT ^^ — ^ — ^^ t> Ans, is only another
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 — ?
8f
16. AThat is the -value of — -?
f
17. What is the value of -^^P
^' .
18. AVhat is the value of ^i^-^?
i off ,
I of 4^'
20. If a horse eat f of a bushel of oats in a day, in how
many days will he eat 51 bushels? Ans. 14.
21. If a man spend If dollars per month for tobacco, in
what time will he spend 10| dollars ? A71S, 6| months.
What is a complex fraction
19. What is the value of
Ans.
ft.
Ans.
20.
Ans,
m-
Ans.
1,
Ans.
h
112 FRACTIONS.
22. How many times will 4| gallons of camphene fill a
vessel that holds ^ of ^ of 1 gallon ? Ans. lO^.
23. If 14 acres of meadow land produce 32 f tons of hay,
how many tons will 5 acres produce ? Ans. 11§.
24. If 2 yards of silk cost $3^, how much less than $17
will 9 yards cost ? Ans. $2|.
25. If f of a yard of cloth cost -^^j of a dollar, how much
will 1 yard cost ?
26. A man, having $10, gave f of his money for clover
seed at $3^ a bushel; how much did he buy? A}is. 2 bush. .
27. How many tons of hay can be purchased for $119^^5,
at $9^ per ton? Ans. I2/5.
PROMISCUOUS EXAMPLES.
1. Reduce J, f, f, and ^ to equivalent fractions whose de-
nominators shall be 24. Ans. ^f , |^, /j, ^•.
2. Change j- to an equivalent fraction having 91 for its
denominator Ans. |f.
3. Find the least common denominator of J, If, ^ of |, 2,
^ofioflA.
4. Add 4i, J, t of 1|, 3, and U.
5. Find the difference between f of 6/^ and | of 4j\.
^^15. li§|.
/ 6. The less of two numbers is 475 6 1, and their difference
is 128£ ; w^hat is the greater number? Ans. 4885 j''^.
7. What is the difference between the continued products of
3,^,?, 4f,and3§, §, 4,f? Ans. 3J5,
4 2^
8. Reduce the fractions — and — to their simplest form.
* 11
9. What number muhiplied by f will produce 1825| ?
Ans. 3043 1.
10. A farmer had -J of his sheep in one pasture, ^ in an-
other, and the remainder, which were 77, in a third pasture ;
how many^heep had he ? Ans. 140.
11. What will 7 1 cords of woo'd cost at ^ of 9^ dollars per
cord? ^ns. $24^^.
PROMISCUOUS EXAMPLES. 113
12. At ^ of a dollar per bushel, how many bushels of apples
can be bought for 5| dollars ?
13. Paid $1837| for 7350^ bushels of oats ; how much was
that per bushel ? Ans. | of a dollar.
14. If 235^ acres of land cost $4725f , how much will 628
acres cost ? ^ns. $12601.
15. A man, owning | of an iron foundery, sold ^ of his share
for$540f; what was the value of the foundery ? Ans. $4055 1.
16. 14f lessA^^ is f of J of what number?
^^\ Ans. 27.
17. A merchant bought 4| cords of wood at $3^ per cord,
and paid for it in cloth at f of a dollar per yard ; how many
yards were required to pay for the wood ?
18. How many yards of cloth, f of a yard wide, will line
20^ yards, 1^ yards wide ? Ans. 34^.
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
j^jj, what is the other ? Ans. ^V
21 If the smaller of two fractions be f f, and their differ-
ence ^T^j what is the greater ? Ans. ^§.
22. If 3f pounds of sugar cost 33 cents, how much must be
paid for 65.^ 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
corn ? Ans. 2500 bushels.
24. A certain sum of money is to be divided among 5 per-
sons ; A is to have ^, B -^, C -jJ^, D g^, and E the remainder,
which is 20 dollars ; what is the whole sum to be divided ?
Ans. $50.
25. What number, diminished by the difference between |
and f of itself, leaves a remainder of 34 ? Ans. 40.
26. If f of a farm be valued at $1728, what is the value of
the whole ?
114 FRACTIONS.
27. Bought 320 sheep at $2 J per head ; afterward bought
435 at $1| per head ; then sold f of the whole number at Slf
per head, and the remainder at $2^; did I gain or lose, and
how much ^ Aiis. Lost $44^.
28. If 5 be added to both terms of the fraction |, will its
value be increased or diminished? Arts. Increased y^^.
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 ^ of § of a gal-
lon, be filled from a demijohn containing J of If gallons ?
A?is. 7J-.
31. Bought ^ of 7^ cords of wood for ^ of $32 ; how much
did 1 cord cost ?
32. Purchased 728 pounds of candles at 1 6f cents a pound j
had they been purchased for 3| cents less a pound, how many-
pounds could have been purchased for the same money ?
Ans. 953^|.
33. What number, divided by If, will give a quotient of
9|? Ans. 12§f.
34. The product of two numbers is 6, and one of them is
1846; what is the other? Ans. gfj.
35. A stone mason worked 11§ days, and after paying his
board and other expenses with ^ of his earnings, he had $20
left ; how much did he receive a day ?
36. If f of 4 tons of coal cost $5^, what will f of 2 tons
cost? Ans. $5.
37. In an orchard f of the trees are apple trees, .yV peach
trees, and the remainder are pear trees, which are 20 more than
•J 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 ^ of a gallon of oil ; how much was the oil worth a gal-
lon ? Ans. 100 cents.
39. A gentleman, having 271 J 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 ? Ans, $4577^^.
PROMISCUOUS EXAMPLES. 115
40. A horse and wagon cost $270 ; the horse cost 1| times
as much as the Avagon ; what was the cost of the wagon ?
41. What number taken from 2^ times 12| will leave
20f? Jns. llf
42. A merchant bought a cargo of flour for $2173^, and
sold it for f f of the cost, thereby losing | of a dollar per bar-
rel; how many barrels did he purchase ? Ans. 126.
43. A and B can do a piece of work in 14 days ; A can do
I as much as B ; in how many days can each do it ?
Ans. A, 32| days ; B, 241 days.
44. How many yards of cloth f of a yard wide, are equal
to 12 yards f of a yard wide ? A7is. 11^.
45. A, B, and C 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 he
put |, in the second J, in the third ^, and in the fourth the re-
mainder, which was $24 more than ^^ of the whole ; how much
money had he ? Ans. $720.
47. If $71 will buy 3^ cords of wood, how many cords can
be bought for $10^ ? Ans. 4|^.
48. How many times is ^ of | of 27 contained in | of J^ of
42§?
49. A boy lost ^ 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 ^ of
them, sold the remainder for ^| of a dollar ; how much would
a box cost at the same rate ? Ans. $5^.
51. A post stands ^ in the mud, ^ in the water, and 21 feet
above the water ;"what is its length ?
5-2. A father left his eldest son ^ 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. $21114JJ.
116 DECIMALS.
DECIMAL FRACTIONS.
14:3. Decimal Fractions are fractions which have for
their denominator 10, 100, 1000, or 1 with any number of
ciphers annexed.
Notes. 1. The word decimal is derived from the Latin decern,.
which signifies ten.
2. Decimal fractions are commonly called decimals.
3. Since -^ = y^, j^ =^ T^|t' &c., the denominators of decimal
fractions increase and decrease in a, tenfold ratio, the same as simple
numbers.
DECIMAL NOTATIOK AND NUMERATION.
144. Common Fractions are the common divisions of a
unit into any number of equal parts, as into halves, fifths,
twenty-fourths, &c.
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 hun-
dredths ; each of these hundredths into ten other equal parts,
called thousandths; and so on. Since the denominators of
decimal fractions increase and decrease by the scale of 10, the
same as simple numbers, in writing decimals the denomina-
tors 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,
and the third place at the right is thousandths ; and so on.
"What are decimal fractions ? How do they differ from common
fractions ? How are they written ?
T^aV "
((
xWiT "
il
NOTATION AND NUMERATION. 117
77ie Decimal Point is a period ( . ), which must always be
placed before or at the left hand of the decimal. Thus,
■j^g- is expressed .6
.54
.279
Note. The decimal point is also called the Separairix. This is a
correct name for it only when it stands between the integral and deci-
mal parts of the same nizmber.
.5 is 5 tenths, which = J^ of 5 units ;
.05 is 5 hundredths, " z= j'^ of 5 tenths ;
.005 is 5 thousandths, " rr y\y of 5 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 clear-
ly shown by the following
NUMERATION TABLE.
£ S « I
475 3.6 2418695
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 corre-
sponding order of integers.
14^. Since the denominator of tenths is 10, of hun-
What is the decimal pomt ? What is it sometimes called ? What is
the value of a figiire 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 PRACTICE.
1. Express in figures thirty -eight hundredths.
2. Write seven tenths.
3. "Write three hundred twenty-five thousandths.
4. Write four hundredths. Ans. .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. Read the folio winjr decimals :
.05
.681
.9034
.19248
.24
.024
.0005
.001385
.672
.8471
.100248
.1000087
Note. To read a decimal, we first numerate from left to right, and
the name of the right hand figure is the name of the denominator. We
then numerate from right to left, as in whole nvmibers, to read the
numerator.
14®. A mixed number is a number consisting of integers
and decimals; thus, 71.406 consists of the integral part, 71,
and the decimal part, .406; it is read the same as 71^^®o,
71 and 406 thousandths.
EXAMPLES FOR PRACTICE.
1. Write eighteen, and twenty-seven thousandths.
2. Write four hundred, and nineteen ten-millionths.
How many decimal places must there be to express any decimal ?
NOTATION 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. Eead the following numbers :
18.027 100.0067 400.0000019^
81.0001 54.000054 3.03 "
75.075 9.2806 40.40404
147. From the foregoing explanations and illustrations
we derive the following important
PRINCIPLES OF DECIMAL NOTATION AND NUMERATION.
1. The value of any decimal figure depends upon \\s, jplace
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, -f^^ or .6, is the same as jV^j 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, /ro?7z
units, to find the name of the denominator, and second, towards
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 present
these principles in the form of rules.
RULE FOR DECIMAL NOTATION.
I. Write the decimal the same as a whole number, 'placing
"What is the first principle of decimal notation ? Second ? Thu-d ?
Fourth ? fifth ? Sixth ? Rule for notation, first step ?
120 DECIMALS.
ciphers where necessary to give each significant figure its true
local value,
II. Place the decimal point before the first figure,
RULE FOR DECIMAL NUMERATION.
I. Numerate from the decimal point, to determine the de-
nominator,
II. Numerate towards the decimal point, to determine the
numerator,
III. Head 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.
o. Write one thousand eight hundred fiiiy-nine hundred-
thousandths.
4. Write 260 thousand 8 billionths.
5. Read the following decimals :
.6321 .748243 .2962999
.5400027 .60000000 .00000006
6. Write five hundred two, and one thousand six milUonths.
7. Write thirty-one, and two ten-millionths.
8. Write eleven thousand, and eleven hundred-thousandths.
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 * 82.004 1858.4583
1.02 4.0005 27.00045
Second ? Rule for numeration, first step ? Second ? Third ?
REDUCTION. 121
KEDUCTION.
CASE I.
149. To reduce decimals to a common denomina-
tor.
1. Reduce .5, .375, 3.25401, and 46.13 to their least com-
mon decimal denominator.
OPERATION. Analysis. The given decimals must contaia
.50000 ^s many places each, as are equal to the greatest
S7500 number of decimal figures in any of the given
q t)KA(\i decimals. "Wo find that tho third number con-
. * Q/xAA t^^T^s five decimal places, and hence 100000 must
^rb.loUOU Yye a common denominator. As annexing ciphers
to decimals does not alter their value, (144., 3) we give to each number
five decimal places by annexing ciphers, and thus reduce the given
decimals to a common denominator. Hence,
Rule. Give to each number the same number of decimal
places, by annexing ciphers.
Notes. 1. If the numbers be reduced to the denominator of that
one of the given numbers having the greatest nimiber of decimal places,
they will have their least common decimal denominator.
2. A whole number may readily be reduced to decimals by placing
the decimal point after units, and annexing ciphers ; one cipher re-
ducing it to tenths, two ciphers to hundredths, three ciphers to thou-
sandths, and so on.
EXAMPLES FOR PRACTICE.
2. Reduce .17, 24.6, .0003, 84, and 721.8000271 to their
least common denominator.
3. Reduce 7 tenths, 24 thousandths, 187 millionths, 5 hun-
dred millionths, and 10845 hundredths to their least common
denominator.
4. Reduce 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. Rule.
R.P 6
122 DECIMALS.
CASE 11.
150. To reduce a decimal to a common fraction.
1. Reduce .75 to its equivalent common fraction.
OPERATION. Analysis. We omit the decimal point,
^75 irz: J^ ^ nr ^. supply the proper denominator to the deci-
mal, and then reduce the common fraction
thus formed to its lowest terms. Hence,
Rule. Omit the decimal point, and supply the proper
denominator,
EXAMPLES FOR PRACTICE.
2. Reduce .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.
CASE III.
151. To reduce a common fraction to a decimal.
1. Reduce f to its equivalent decimal.
FIRST OPERATION. ANALYSIS. We first annex
3 — - 3 — - 7 5 — - ^75 jIyis^ the same number of ciphers
to both terms of the fraction ;
SECOND OPERATION. this does not alter its value.
A\or\(\ We then divide both resulting
^ — '- — terms by 4, the significant fig-
.75 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 n is what ? Give explanation. Rule. Case III is what ?
Explain first operation. Second.
Ans.
h
Ans.
^'
Ans.
m-
Ans,
u-
Ans.
TzVu*
INDUCTION.
123
THIRD OPERATION. ANALYSIS. Dividing as In the former
16) 1.0000 example, we- obtain a quotient of 3 fig-
7 ures, 625. But since we annexed 4
.Ob2o, Ans. 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,
1 10000 62_5 — 0(\9^
From these illustrations we derive the following
Rule. I. Annex ciphers to the numerator, and divide hj
the denominator.
II. Point off as many decimal places iii the result as are
equal to the number of ciphers annexed.
Note. A common fraction can be reduced to an exact decimal when
its lowest denominator contains only the prime factors 2 and 5, and
not otherwise.
EXAMPLES FOR PRACTICE.
A71S.
Ans.
.625.
.9375.
3. Reduce |^ to a decimal. •
4. Reduce f to a decimal.
5. Reduce J^| to a decimal.
6. Reduce |^ to a decimal.
7. Reduce /^ to a decimal.
8. Reduce -^-^ to a decimal.
9. Reduce f to a decimal.
, 10. Reduce -^jy to a decimal.
11. Reduce ^§(j to a decimal.
12. Reduce -j^^ to a decimal.
13. ' Reduce ^ to a decimal.
Note. The sign, -|-i hi the answer indicates that there is still a
remainder.
Ans.
.08.
Ans.
.046875.
Ans.
.00375.
Ans.
.008.
Ans.
.33333+.
14. Reduce ^f to a decimal.
Ans. .513513+.
Note. 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 iii
regular order,
Third operation. Rule, first step ? Second ? "When can a common
fraction be reduced to an exact decimal ?
124 DECIMALS.
ADDITION.
1.13. 1. What is the sum of 3.703, 621.57, .672, and
20.0074?
OPERATION. Analysis. We write the numbers so that fig-
3.703 ^^^s of Hke orders of units shall stand in the same
621.57 columns; that is, units under units, tenths under
/.„2 tenths, hundredths under hundredths, &c. This
on' An" 1 brings the decimal points directly under each
____1___ other. Commencing at the right hand, we add
645.9524 each column separately, and carry as in whole
numbers, and in the result we place a decimal
point between units and tenths, or directly under the decimal point
in the numbers added. From this example we derive the following
Rule. I. Write the numbers so that the decimal points
shall stand directly under each other.
n. Add as in whole \umbers, and place the decimal pointy
in the result, directly under the points in the numbers added.
2. Add
Sum,
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. Give rule, first step.
Second.
EXAMPLES
FOR
PRACTICE.
.199
3.
Add
4.015
2.7569
6.75
.25
27.38203
.654
375.01
3.8599
2.5
ADDITION. 125
7. What is the sum of thirty -six, and fifteen thousandths ;
three hundred, and six hundred five ten-thousandths ; five,
and three miUionths ; -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 hun-
dred seven millionths ; twenty-three thousandths ; eight, and
nine tenths ; four, and one hundred thirty-five thousandths ?
Ans. 71.399107.
^ 9. How many yards in three pieces of cloth, the first piece
containing 18.375 yards, the second piece 41.G25 yards, and
the third piece 35.5 yards ?
10. A's farm contains 01.843 acres, B's contains 143.75
acres, C'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 12f acres, 18f
acres, 9 acres, 24| acres, 4|| acres, 81?^ acres, and 15=^§ acres
respectively ; how many acres in my farm ?
Note. Keduce the common fractions to decimals before adding.
Ans. 93.6375C>
12. A grocer has 2^ barrels of A sugar, 5f barrels of B
sugar, 3 1 barrels of C sugar, 3.0642 barrels of crushed
sugar, and 8.925 barrels of pulverized sugar ; how many bar-
rels of sugar has he ? Ans. 23.8642.
v^l3. A tailor made 3 suits of clothes; for the first suit he
used 2^ yards of broadcloth, 3^^ yards of cassiraere, 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-i^ yards of broadcloth, and I4 yards of satin.
How many yards of each kind of goods did he use ? How
many yards of all ? Ans. to last, 18.375.
O
126 DECIMALS.
SUBTRACTION.
153. 1. From 91.73 take 2.18. Analysis. In each of these
three examples, we write the
OPERATION. subtrahend under the minu-
91.73 end, placing units under
2.18 units, tenths under tenths,
A on ^- &c. Commencing at the
right hand, we subtract as
2. From 2.9185 take 1.42. \ ^'^^l^. numbers, and in
the remainders we place the
OPERATION. decimal points directly under
2.9185 those in the numbers above.
2 42 In the second example, the
number of decimal places in
Ans, 1.4JoO \^Q minuend is greater than
•^ ^,^K , ^K^^-ri« the number in the subtra-
3. From 124.05 take 95.58746. j^end, and in the third exam-
OPERATION. pie the number is less. In
124.65 '^o\}a. cases, we reduce both
95 58746 minuend and subtrahend to
'- the same number of decimal
Ans. 29.06254 places, by annexing ciphers;
or we suppose the ciphers to
be annexed, before performing the subtraction. Hence the
Rule. 1. \Yrite the numbers so that the decimal points
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.
6. How much greater is 2 than .298 ? Ans. 1.702.
^6. From 21.004 take 75 hundredths.
7. From 10.0302 take 2 ten-thousandths. A7is. 10.03.
8. From 900 take .009. Ans. 899.991.
9. From two thousand take two thousandths.
^ 10. From one take one millionth. Ans. .999999.
£xplain subtraction of fractions. Give the rule, first step. Second.
MULTIPLICATrON. 127
11. From four hundred twenty-seven thousandths take
four hundred twenty-seven millionths. 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.
1^4. 1. What is the product of .35 multiplied by .5 ?
OPERATION. Analysis. We perform the multipHcation 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
.1 /Oj^ns. point off, we may reduce the decimals to common
fractions; thus, .35 z=:^^ and .5:=z^^-^. Perform-
ing the multiplication, and we have -^-^ X t¥ =^ TV?nr» ^^^ this
product, expressed decimally, is .175. Here we see that the prod-
uct contains as many decimal places as are contained in both mul-
tiplicand and multiplier. Hence the following .
Rule. Multiply as in whole numbers, and from the right
hand of the product point off as many figures for decimals as
there are decimal places in both factors.
Notes. 1. If there be not as many figtires in the product as there
are decimals in both factors, supply the deficiency by prefixing ciphers,
2. To multiply a decimal by 10, 100, 1000, &c., remove the point as
many places to the right as there are ciphers on the right of the multi-
plier.
EXAMPLES.
2. 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 mtiltiplication of decimals. Give rule. If the product have
less decimal places than botli factors, how proceed ? IIo^v M.ultiply by
10, 100, 1000, &c. ?
128 DECIMALS.
7. Multiply .132 by .241. Ans. .031812.
a Multiply 24.3,5 by 10.
9. Multiply .006 by 1000. Ans, 6.
..' 10. Multiply .23 by .009. Ans. .00207.
~ 11. Multiply sixty-four thousandths by thirteen railliontha
Ans, .000000832.
12. Multiply eighty-seven 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.
YSSo 1. "What is the quotient of .175 divided by .5 ?
OPERATION. Analysis. We perform the division the same as
.5 ) .175 ^^ whole numbers, and the only difficulty we meet
with is in pointing off the decimal places in the quo-
AnS' cOO tient. To determine how many places to point off,
we may reduce the decimals to common fractions; thus, .175 =
•j^j^, and .5 zzz ^-^. Performing the division, and we have
175 5 ;i'^$ 10 35
X — =
1000 10 1000 $■ 100
and 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. Hence the following
Rule. Divide as in whole numbers, and from the right
hand of the quotient point off as many places for decimals
as ike decimal places in the dividend exceed those in the
divisor.
Explain division of decimals. Give rule.
DIVISION. 129
Notes. 1. If the number of figures 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 divisor, before commencing the division.
4. In most business transactions, the division is considered suffi-
ciently exact when the quotient is carried to 1 decimal places, imless
great accuracy is required.
5. To divide by 10, 100, 1000, &c., 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. .02.
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.
Ans. 1000000.
When are ciphers prefixed to the quotient ? If there be a remainder,
liow proceed ? If the di\ndend have less decimal places than the divi-
sor, how proceed ? How divide by 10, 100, 1000, &c, ?
6*
130 DECIMALS.
PROMISCUOUS EXAMPLES.
1. Add six hundred, and twenty-five thousandths; four
tenths ; seven, and sixty-two ten-thousandths ; three, and fifty-
eight millionths ; ninety-two, and seven hundredths.
Ans. 702.501258.
2. What is the sura of 81.003 -f 5000.4 -f 5.0008 -[-
73.87563 + 1000 + 25 + 3.000548 + .0315 ?
3. From eighty-seven take eighty-seven thousandths.
4. What is the difference between nine million and nine
millionths? Ans. 8999999.999991.
5. Multiply .365 by .15. Ans. .05475.
G. Multiply three thousandths by four hundredths.
7. If one acre produce 42.57 bushels of corn, 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. Reduce .625 to a common fraction. Aiis. f.
13. Express 26.875 by an integer and a common fraction.
Ans. 26|.
14. Reduce yf^ to a decimal fraction. Ans. .016.
15. Reduce -|| to a decimal fraction. Ans. .5.
16. How many times will .5 of 1.75 be contained in .25 of
17^? 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 4f miles an hour, how many
hours will a man require to travel 56.925 miles.
Ans. 12§ hours.
NOTATION AND NUMERATION. 131
DECIMAL CURRENCY.
1^0. Coin is money stamped, and has a given value es-
tablished by law.
\37, Currency is coin, bank bills, treasury notes, &c., in
circulation as a medium of trade.
158 . A Decimal Currency is a currency whose denom-
inations increase and decrease in a tenfold ratio.
Note. The currency of the United States is decimal currency, and
is sometimes called Federal Money ; it was adopted by Congress iii 1786.
NOTATION AND NUMERATION. "^
The Coin of the United States consists of gold, silver,
nickel, and bronze.
2Vie Gold Corns are the double-eagle, eagle, half-eagle,
qu;irter-eagle, three-dollar and one-dollar pieces.
The Silver Coins are the dollar, half-dollar, quarter-
dollar, dime, half-dime, and three-cent pieces.
The Nickel Coins are the five-cent and three-cent pieces.
The Bronze Coins are the two-cent and 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.
■5
UNIT EanVALENTS.
Mills. Cents.
1^ =1 Dimes.
100 =10 =1 ^,„,,.
1000 =100 =10 =1
Eagle.
10000 = 1000 = 100 r= 10 = 1
Note. The character $ is supposed to be a contraction of U. S.,
(United States,) the U being placed upon the S.
'S\Tiat is coin ? Currency ? Decimal currency ? Federal money ?
What arc the gold coins of U. S. ? Silver ? Copper ? What are the
denominations of U, S. currency ? What is the sign of dollars ? From
what derived ?
132 DECIMAL CURRENCY.
1^9* The dollar is the unit of United States money;
dimes, cents, and mills are fractions of a dollar, and are sepa-
rated from the dollar by the decimal point ; thus, two dollars
one dime two cents five mills, are written $2,125.
By examining the table, we see that the dime'i^ 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 part 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.
Note. The half cent is frequently written as 5 mills ; thus, 24^ cents,
written $.245.
160. Business men frequently write cents as common
fractions of a dollar ; thus, three dollars thirteen cents are
written $3y\y3^, 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.
\Miat is the unit of IT. S. currency ? \Miat is the p;encral law of
increase and decrease ? In practice, how many decimal places are given
to cents ? In business transactions, how are cents frequently written ?
What is done if the mills exceed 5 ? If less than 5 ?
REDUCTION. 13S
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 seir
that 10 mills make one cent, and 100 cents, or 1000 mills,
make one dollar ; hence.
To change dollars to cents, multiply ^ 100 ; that is, annex
ttvo ciphers.
To change dollars to mills, annex three ciphers.
To change cents to mills, annex one cipher,
EXAMPLES FOR PRACTICE.
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.
Note. 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 ?
16S. Conversely,
To change cents to dollars, divide ^ 100 ; that is, point off
two figures from the right.
To change mills to dollars, point off three Jigures,
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 ?
)34 DECIMAL CURRENCY.
EXAMPLES FOR PRACTICE.
1. Change 875 cents to dollars. Ans. $8.75.
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. Reduce 6524 mills to dollars.
ADDITION.
103. 1. A man bought a cow for 21 dollars 50 cents, a
horse for 1 25 dollars 37| cents, a harness for 46 dollars 75 cents,
and a carriage for 210 dollars ; how much did he pay lor all ?
OPERATION.
S 21 50 Analysis. Writing dollars under dol-
^^'^^ lars, cents under cents, &c., so that the
decimal points shall stand under each
other, we add and point off as in addition
^^^•QQ of decimals. Hence the following
Ans, $403,625
Rule. I. Write dollars under dollars, cents under cents, Sfc.
II. Add as in simple numbers, aiid 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. Ans. $2689.686.
4. Add 92 cents, 10 cents 4 mills, 35 cents 7 mills, 18 cents
G mills, 44 cents 4 mills, 12^ cents, and 99 cents. Ans. $3,126.
Explain the process of addition of decimal currency. Rule, first step.
Second.
SUBTRACTION. 135
5. A farmer receives 89 dollars 74 cents for wheat, 13 dol-
lars 3 cents for corn, 6 dollars 37^ cents for potatoes, and 19
dollars 62^- cents for oats; what does he receive for the
whole? Ans. $128.77.
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 1 1 cents ; what
did she pay for all ? A?is. $10.3375.
7. Paid for building a house $2175.75, for painting the
same $240.37^, 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 $3.62j, a pair of shoes for $1J, an
umbrella for $1§, a pair of gloves for $.62|, and a cane for
$.871 J what was the cost of all my purchases ? Ans. $8.25.
SUBTRACTION.
164. 1. A man, having $327.50, paid out $186.75 for
a horse ; how much had he left ?
OPERATION. Analysis. "Writing the less number un-
$327.50 ^^^ the greater, dollars under dollars, cents
186 75 under cents, &c., we subtract and point oflE
'- — in the result as in subtraction of decimals.
Ans. $140.75 Hence the following
Rule. I. Write the svhtrahend under the minuend, dollars
under dollars^ cents under cents, ^c.
11. Subtract 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 ifought a farm for $1575.24, and sold it for
$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 ; how much 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 ?
'-f 10. A lady bought a silk dress for $13^, a bonnet for $5^, a
pair of gaiters for $lf, and a fan for $| ; she paid to the shop-
keeper a twenty dollar bill and a five dollar bill ; how much
change should he return to her ? Ans. $3.75.
Note. 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 j what was the cost of the carriage?
Ans. $360.50.
MULTIPLICATION.
MS, 1. If a barrel "^of flour cost $6,375, what will 85
barrels cost.'' \
operation:
$6,375 Analysis. We multiply as in simple
85 numbers, always regarding the multii)Her
_ _- as an abstract number, and point oft' from
^^ the right hand of the result, as in multipli-
^^^^^ cation of decimals. Hence the following
Ans. $541,875
Give analysis for multiplication in decimal currency.
DIVISION. 137
HuLE. Multiply as in simple numbers, and place the point
in the product, as in multiplication of decimals.
EXAMPLES FOR PRACTICE.
2. If a cord of wood be worth $4,275, what will 300 cords
be worth? -^ns. $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 o-rocer bought 372 pounds of cheese at $.15 a pound,
434 pounds of coffee at $.12i a pound, and 16 bushels of pota-
toes 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^.
^. A man bought 150 acres of land for $3975 ; he after-
ward sold 80 acres of it at $3^.50 an acre, and the remainder
at $34.25 an acre j how much 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. Analysis. We divide as in
125 ) $850.00 ( $6.80, Ans. ^^"^P^^ numbers, and as there
rj^Q is a remainder after dividing
the dollars, we reduce the div-
1000 idend to cents, by annexing two
1000 ciphers, and continue the di-
~ vision. Hence the following
Bule. Give rule for division in decimal currency.
138 DECIMAL CURRENCY.
Rule. Divide as in simple numbers, and place the point in
the quotient, as in division of decimals.
Notes. 1. In business transactions it is never necessary to carry
the division further than to mills in the quotient.
. 2. K the dividend will not contain the divisor an exact number of
times, ciphers may be annexed, and the division continued as in divis-
ion of decimals. In this case it is always safe to reduce the dividend
to mills, or to 3 more decimal places than the divisor contains, be-
fore commencing the division.
EXAMPLES FOR PRACTICE.
2. If 33 gallons of oil cost $41.25, what is the cost per gal-
lon? Ans. $1.25.
3. If 27 yards of broadcloth cost $94.50, what will 1 yard
cost?
4. If 64 gallons of wine cost $136, what will 1 gallon cost?
Ans. $2,125.
5. At 12 cents apiece, how many pine-apples can be bought
for $1.32? Ans. II.
6. If 1 pound of tea cost 54 cents, how many pounds can
be bought for $405 ?
7. If a man earn $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.
— tf). 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. 56 pounds.
1 2. 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 $5 GO; it being
damaged he was obhged to lose $106.75 on the cost of it;
how much did he receive a box ? Ans. $32.37J-.
Additional Applications.
CASE I.
167. To find the cost of any number or quantity,
when the price of a unit is an aliquot part of one dollar.
168* An Aliquot Part of a number is such a part as will
exactly divide that number; thus, 3, 5, and 7^ are aliquot
parts of 15.
Note. An aliquot part may be a whole or a mixed number, while a
factor must be a whole nvmiber.
ALIQUOT PARTS OF ONE DOLLAR.
50 cents = ^ of 1 dollar.
S3 J cents = ^ of 1 dollar.
25 cents == |^ of 1 dollar.
20 cents = i of 1 dollar.
16§ cents i= ^ of 1 dollar.
12^ cents = ^ of 1 dollar.
10 cents = y'g^ of 1 dollar.
8^ cents = jV ^^ ^ dollar.
6^ cents = j'^ of 1 dollar.
5 cents z= ^\j of 1 dollar.
1. \yhat will be the cost of 3784 yards of flannel, at 25
cents a yard ?
OPERATION. Analysis. If the price were $1 a yard,
4 ") 3784 ^^® ^^^^ would be as many dollars as there are
yards. But since the price is ^ of a dollar a
Ans. $946 yard, the whole cost will be \ as many dollars
as there are yards ; or, J of 3784 = 3784 -^- 4 = $946. Hence the
Rule. Take such a fractional part of the given number as
the price is part of one dollar,
EXAMPLES FOR PRACTICE.
2. What cost 963 bushels of oats, at 33^ cents per bushel?
Ans. $321.
Case I is what ? What is an aliquot part of a dollar ? Give ex-
planation. Rule.
140 DECIMAL CURRENCY.
3. What cost 478 yards of delaine, at 50 cents per yard ?
4. What cost 4266 yards of sheeting, at 8^ cents a yard?
Ans. $355.50.
5. What cost 1250 bushels of apples, at 12| cents per
bushel? A71S. $156.25.
6. What cost 3126 spools of thread, at 6| cents per spool?
Ans. $195,375.
7. At 16§ cents per dozen, what cost 1935 dozen of eggs?
Ans. 322.50.
8. What cost 56480 yards of calico, at 12^ 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
g 9 — $56.25. Hence
Ans. $56.25
Rule. Multiply the price of one hy the quantity.
EXAMPLES FOR PRACTICE.
2. If a pound of beef cost 9 cents, what will 864 pounds
cost? Ans. $77.76.
3. What cost 87 acres of government land, at $1.25 per
acre ?
4. What cost 400 barrels of salt, at $1.45 per barrel ?
Ans. $580.
5 What cost 16 chests of tea, each chest containing 52
pounds, at 44 cents per pound?
Case n is what ? Give explanation. Rule.
APPLICATIONS. 141
CASE III.
170 . The cost and the quantity being given, to find
tlie price of one.
1. If 30 bushels of corn cost $20.70, what will 1 bushel
cost?
OPERATION. Analysis. If 30 bushels cost $20.70, 1
310 ) S'^IO 70 bushel will cost ^ of $20.70; and $20.70-^
' ^-^^- 30 = $.69. Hence,
$.69
Rule. Divide the cost hy the quantity.
EXAMPLES FOR PRACTICE.
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. $4.1 6§.
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? Ans. $5.62^.
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 $684.37J^ 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 ?
142 DECIMAL CURRENCY.
OPERATION. Analysis. Since $6 will buy 1 barrel
6 ) 840 of flowi'j $840 will buy | as many barrels
as there are dollars, or as many barrels as
$6 is contained times in $840; 840-^-6
r= 140 barrels. Ilence,
AjIS. 140 barrels.
Rule. Divide (he cost of (he quantity hy 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 $19J.
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 y.
17S. To find the cost of articles sold by the 100,
1000, &c. '
1. What cost 475 feet of limber, at $5.24 per 100 hci ?
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 z= $2489. But since $5.24 is the
2620 price of 100 feet, $2489 is 100 times the true
3668 value. Therefore, to obtain the true value,
2096 ^^ divide $2489 by 100, which we may do
bv cutting: off two fii^ures from the right, and
100) $2489.00 the result is $24.89: Or,
Ans. $24.89
Give explanation. Rule. Case V is what ? Give first uxi)lauutioiu
APPLICATIONS. 143
SECOND OPERATION. ANALYSIS. Since 1 foot costs y^, or. 01,
$5.24 o^ $5.24, 475 feet will cost 4^, or 4.75 times
4*75 $5.24, which is $24.89.
2g20 Note. For the same reasons, when the price
or^po is per thousand, we divide the product by 1000,
*^""^ or, which is more convenient in practice, we re-
2096 duce the given quantity to thousands and deci-
mals of a thousand, by pointing off three figures
$24.8900 from the right hand. Hence the
Rule. I. Reduce the given quantity to hundreds and deci*
mdls of a hundred^ or to thousands and decimals of a thousand*
II. MidtipJy the price hy the quantity, and point off in the
result as in multiplication of decim,als.
Note. The letter C is used to indicate himdreds, and Mto indicate
thousands,
EXAMPLES FOR PRACTICE.
^ 2. What will 42GoO 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 will 7842 feet of pine boards cost, at $17.25
perM? Ans. $135,274+.
5. What cost 2348 pine-apples, at $12^ per 100 ?
6. A broom maker bought 1728 broom-handles, at $3 per
1000 ; how much did they cost him ?
7. What is the cost of 2400 feet of boards, at $7 perM;
865 feet of scantling, at $5.40 per M ; and 1256 feet of lath, at
$.80 per C? Ans. $31,519.
8. What will be the cost of 1476 pounds of beef, af*$4.37^
per hundred pounds ?
CASE VI.
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 CURRENCY.
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 -^ 2 = $4,75. One pound will cost
$4.75
2.376
T1jW» ^^ '^^^* ^^ $4.75, and 2376 pounds will
cost 11^, or 2.376 times $4.75, which is $1 1.286.
$11.28600 ^^^^^
Rule. I. Divide the price of 1 ton by 2, and the quotient
will be the price of 1000 pounds.
II. Multiply this quotient by the given number of pounds
expressed as thousandths, as in Case V,
EXAMPLES FOR PRACTICE.
2. At $7 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. 145
(1.)
Mr. John Rice, ^^^ York, June 20, 1859.
BoH, o/ Baldwin & Sherwood,
7 yds. Broadcloth, ^$3.60 XS.3lO
9 " Satinet, « 1.124- iOAlS^
12 " Vesting, « .90" I %^0
24 « Cassimere, « 1.37^ ^XW
32 " Flannel, « .65 J/?,;^/
Rec^d Payment, ^ ^^^-^^^
Baldwin & Sherwood.
(2.)
Daniel Chapman & Co., Boston, Jan. 1, 1860.
BoH, o/ Palmer & Brother.
67 pairs Calf Boots, (a) $3.75 -o^^/. 3. ^
108 « Thick « « 2.62:»<^S2.76
75 « Gaiters, « 1.12- S!^. tfZ?
27 « Buskins, « .86 - l^X^^
35 « Slippers, « .70 - -^^, ^^
50 « Rubbers, « 1.04 ^ X^ ^/)
-KecW. Payme^at, $717.93 cZ^i^-U/^^/
Palmer & Brother,
By Geo, Baker.
(3.)
G. B. Grannis, Charleston, Sept. 6, 1859.
Bo't. of Stewart & Hammond,
325 lbs. A. Sugar, <a) $.07
148 " B. « « .06^
286 « Rice, « .05
95 « O. J. Coffee, « .12J.
50 boxes Oranges, " 2.75
75 " Lemons, « 3.62^
12 " Raisins, " 2.85 „
7> ,_. r. 1501.75
/fee rf. Payment, hy note at 4t mo.
R p 7 Stewart «k Hammond.
146
DECIMAL CURRENCY.
(4-)
Messrs. Osboen & Eaton, ^''^ ^<'""' 0«'- ^^' ' 858.
Bo't. of Rob't. H. Carter & Co,
20000 feet Pine Boards (5) $15 per M. ;rxi,
7500
« Plank,
ii
9.50 "
ll'^'ir
10750
" Scantling,
((
6.25 "
:-iyi
3960
« Timber,
i(
2.62^ "
M^
5287
(( u
ii
3.00 «
H.^' ^ '
$464.6935
Rec'd, Payment,
Rob't. H. Carter & Co.
(5.)
Mr. J. C. Smith, Cincinnati, May 3, 1861.
Bo't. of Silas Johnson,
25 lbs. Coffee Sugar,
5 " Y. H. Tea,
26 « Mackerel,
4 gal. Molasses,
46 yds. Sheeting,
SO « Bleached Shirting,
6 skeins Sewing Silk,
4 doz. Buttons,
Chgd.{n %
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 corn, containing 2250 bushels, was sold for
$1406.25 ; what did it sell for per bushel ? Ans, $|.
(a)
$.11 y.)0
.62^ y^"^'
Mi ,--^
.42
.09
ing, "
.14 U,^
.04 ^.^^
.12
$18.24
Silas Johnson,
Per John Wise.
PROMISCUOUS EXAMPLES. 147
4. If 12 yards of cloth cost $48.96, what will one yard
cost .'*
5. A traveled 325 miles by railroad, and C traveled .45 of
that distance ; how far did C 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 ; how much
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 ; how much did he lose ? Ans. $112.
9. If a man travel 4f miles an hour, in how many hours
can he travel 34^ miles ? Ans. 7.5 hours.
10. At $.31:^ per bushel, how many bushels of potatoes
can be bought for $9 ? Ans. 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?
1 2. A merchant deposits in a bank, at one time, $687.25,
and at another, $943.64 ; if he draw out $875.29, how much
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 ; how much did I receive for the whole ?
A?is. ^i#*^^^#t7.,
14. What will eight hundred seventy-five thousandths of a 1^0 ^ ^
cord of wood cost, at $3.75 per cord ? Ans. $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 will 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, caa
be bought for $112.34?
148 DECIMAL CURRENCY.
19. If 4G tons of iron cost $3461.50, what will 5 tons cost?
-- 20. A gentleman left his widow one third of his property,
worth $24000, and the remainder was to be divided equally
among 5 children ; how much was the portion of each child ?
Ans. $3200.
-21. A man purchased one lot, containing 1 60 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 ; how much did he gain ? A)is. $8.70.
23. A miller bought 122^ bushels of wheat of one man,
and 75|- bushels of another, at $.93| per bushel. He sold 60
bushels at a profit of $12.50; if he sell the remainder at
$.81|- 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 ?
Alls. 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 j what did he receive for the
whole? 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 o\'^ertake John, if John have 8 minutes the start ?
Ans. 21^ minutes.
33. If 43. barrels of flour cost $32.3, what will 7^ barrels
cost? Ans. $51.
34. If .875 of a ton of coal cost $5,635, what will 9^ 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. $838.80.
37. Reduce ^ to a decimal fraction. Ans. .78125.
^i
38. What will 1375 pounds of potash cost, at $96.40 per
ton? Ans. $66,275.
39. Reduce .5625 to a common fraction. Ans. ■^^.
40. Reduce ^^^, .62^, .37 j^, §, 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 .62^
.45 3.45
5.28^ 1.87J
What IS due the merchant ? Ans. $4.41 J.
42. A gardener sold, from his garden, 120 bunches of on-
ions at $.12J 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.
KEDUCTION.
1 75. A Compound Number is a concrete number whoso
value is expressed in two or more different denominations.
I'T'O. Reduction is the process of changing a number from
pne 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 oiless unit value ; thus,
$1 = 10 dimes =100 cents = 1000 mills.
178. Reduction Ascending" is changing a number of one
denomination to another denomination of greater unit value ;
thus, 1000 mills = 100 cents = 10 dimes = $1.
170. A Scale is a series of numbers, descending or as-
cending, used in operations upon compound numbers.
CURRENCY.
180. I. United States Money.
TABLE.
1 10 mills (m.) make 1 cent, ct.
10 cents " 1 dime, <.].
10 dimes " 1 dollar, $.
10 dollars " 1 eagle, E.
UNIT EQUIVALENTS.
Ct. m.
d. 1 = 10
$ 1 = 10 = 100
E. 1 = 10 = 100 = 1000
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.
Note Tlie currency of the whole Dominion of Canada was made nnlform
July 1, 1871. Before the adoption of the decimal system, pounds, shillings, and
pence were used.
The Silver Coins are tlic 50-ccnt piece, 25-cent piece, 10-cent piece,
and 5 cent piece. The 20-ceiit piece is no longer coined. The Bronze
Coin is the cent.
The Gold Coin used in Canada is the British sovereign, worth
$4.86^, and the half-sovereign _
COMPOUND NUMBERS. 151
n. English Money.
181. Englisll Currency is the currency of Great Britain.
' 4 farthings (far. or qr.) make 1 penny, d.
12 pence " 1 shilling, s.
20 shillings " 1 pound or sovereign, ,,.£, or sov.
UNIT EQUIVALENTS.
X, or sov. 1= 12= 48 /^"'^-Vl. ^7^0-*^
1 = 20 = 240 = 960 1a ^ \b pAlh r^
Scale ascending, 4, 12, 20 ; descending, 20, 12, 4. ' ^ ^ \4^ >' '^^^ %
Note. 1. Parthings are generally expressed as fractions dfa penny;
thus, 1 far., sometimes called 1 quarter, (qr.), = |^d ; 3 far. = f d.
2. The gold coins are the sovereign ( = £1), and the half sovereign,
S. The silver coins are the crown (= 5s.), the half-crown ( = 2s. 6d.), the
florin, the shilling, and the sixpenny, foiirpenny, and threepenny pieces.
4. The copper coins are the penny, halfpenny, and farthing.
5. The guinea (= 21s.) 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
18.6 cents U. S. money.
CASE I.
182. To perform reduction descending.
1. Reduce 2l£ 18 s. 10 d. 2 far. to flirthin^s.
OPERATION. Analysis. Since in £1 there
21 £ 18 s 10 d 2 far ^^^ ^^ ^'' ^^ '^^ ^ there are 20 s. x
2Q * * * 21 = 420 s., and 18 s. in the given
-— — number added, makes 438 s. in 21£
18 s. Since in 1 s. there are 12 d,,
if in 438 s. there are 12 d. x 438 =
^266 d. 5256 d., and 10 d. in the given
number added, makes 5266 d. in
21066 far. Ans, 21£ 18s. lOd. Since in 1 d. there
are 4 far., in 5266 d. there are 4 far.
X 5266 = 21064 far., and 2 far. in the given number added, makes
21066 far. in the given number. Hence,
152 REDUCTION.
PvULE. I. Multiply the highest denomination of the given
number by that number of the scale which will reduce it to the
next lower denomination^ and add to the product the given
number^ if any^ of that lower denomination.
II. Proceed in the same manner with the results obtained in
each lower denomination^ until the reduction is brought to the
denomination required,
CASE II.
183. To perform reduction ascending.
I. Reduce 21066 farthings to pounds.
OPERATION. Analysis. We first divide
4 ) 21066 far. ^^^ 21066 far. by 4, because
there are \ as many pence as.
12)5266d.+ 2 far. farthings, and we find that
2|0 ) 43|8 s. + 10 d. * 21066 far. = 5266 d. + a re-
^ £_\-\o^ mainder of 2 far. We next
divide 5266 d. by 12^ because
Ans. 21 £ 18 s. 10 d. 2 far. ^^^eve Sive J^ as many shiUings
as pence, and we find that 5266 d. = 438 s. + 10 d. Lastly we
divide the 438 s. by 20, because there are 2V ^3 many pounds as
shillings, and we find that 438 s. =2l£ + 18s. The last quotient
with the several remainders annexed in the Order of the succeeding
denominations, gives the answer 21 £ 18 s. 10 d. 2 far. Hence,
HuLE. I. Divide the given number by that number of the
scale which will reduce it to the next higher denomination.
II. 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.
Note. Reductiou descending and reduction ascending mutually
prove each other.
EXAMPLES FOR PRACTICE.
1. In 14194 farthings how many pounds?
2. In 14 <£ 15 s. 8 d. 2 far. how many farthings ?
3. In 15359 farthings how many pounds?
4. In 46 sov. 12 s. 2 d. how many pence i
6. In 11186 pence how many sovereigns?
COMPOUND NUMBERS.
153
AVEIGHTS.
184:. Weight is a measure of the quantity of matter a
body contains, determined according to some fixed standard.
Three scales of weight are used in the United States
and Great Britain, namely, Troy, Apothecaries', and Avou*-
dupois.
I. Troy Weight.
18^. Troy Weight is used in weighing gold, silver, and
jewels ; in philosophical experiments, &;c.
T\BLE.
24 grains (gr.) make 1 pennyweight,, .pwt or dwt
20 pennyweights " 1 ounce, oz.
12 ounces " 1 pound, lb.
UNIT EQUIVALENTS.
lb.
1
pwt. gr.
OZ. 1 = 24
1 z= 20 = 480
= 12 =: 240 = 5760
Scale — ascending, 24, 20, 12; descending, 12, 20, 24.
EXAMPLES FOR PRACTICE.
1. How many grains in 14 lb.
10 oz. 18 pwt. 22 gr.?
OPERATION.
141b. 10 oz. 18 pwt. 22 gr.
12
178 oz.
20
3578 pwt.
24
14334
7156
2. How many pounds in
85894 grains ?
OPERATION.
24 ) 85894 gr.
20) 3578 pwt. +22gr.
12 )178 oz. + 18 pwt.
141b. + 10 oz.
141b. 10 oz. 18 pwt
Ans.
22 gr.
85894 gr., Ans.
3. In 5 lb. 7 oz. 12 pwt. 9 gr., how many grains ?
4. In 32457 grains how many pounds ?
Define weight.
Troy weight.
7*
Repeat the table. Give the scale.
154
REDUCTION.
5. Reduce 41 7 CO grains to pounds. Ans. 7 lb. 3 oz.
^. A miner had 141b. 10 oz. 18 pwt. of gold dust; how
much was it worth at $.75 a pwt. ? Ans. $2683.50.
V 7. How many spoons, each weighing 2 oz. 15 pwt., can be
made from 5 lb. 6 oz. of silver ? Am. 24.
V8. A goldsmith manufactured 1 lb. 1 pwt. 1 G grs. of gold into
rings, each weighing 4 pwt. 20 gr. ; he sold the rings for $1.25
apiece ; how much did he receive for them ? Ans. $62.50.
II. Apothecaries' Weight.
186. Apothecaries' Weight is used by apothecaries and
physicians in compounding medicines; but medicines are
bought and sold by avoirdupois weight.
table.
Vv
20 grains (gr.) make 1 scruple, sc. or 9.
3 scruples " 1 dram, di*. or 5«
8 drams " 1 ounce, oz. or
12 ounces " 1 pound, lb. or lb.
UNIT EaUIVALENTS.
sc. gr.
dr. 1 = 20
oz. 1 = 3 = 60
lb. 1 r= 8 = 24 = 480
1 := 12 = 96 := 288 = 5760
Scale — ascending, 20, 3, 8, 12 ; descending, 12, 8, 3, 20.
examples for practice.
2. How many lb in 73175
1. How many gr. in 12 lb
8§35 19 15 gr.?
operation.
12ib 8S3 3 19 15gr.
12
152 S
8
1219 5
3
3658 3
20
73175 gr., Ans.
operation.
2|0 ) 7317|5 gr.
3 ) 3658 9 + 15 gr.
8 ) 1219 5 + 13
12) 152 § + 3 3
12lb + 8i
Ans. 12lb8i3 3 19 15gr.
Define apothecaries' weight. Repeat the table. Give the scale.
k 16
COMPOUND NUMBERS. 155
3. In 161b. 11 oz. 7 dr. 2 sc 19 gr., how many grains?
4. Reduce 47 ib 6 S 4 5 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 ? J.7is. 6 tb. 4 i 1 9.
in. Avoirdupois Weight.
18T. Avoirdupois Weight is used for all the ordinary pur-
poses of weighing.
TABLE.
drams (dr.) make 1 ounce, oz.
16 ounces " 1 pound, lb.
100 lb. " 1 hundred weight, . cwt.
20 cwt., or 2000 lbs., « 1 ton, T.
UNIT EQUIVALENTS.
OZ. dr.
lb. 1 z=z 16
cwt. 1 = 16 = 256
T. 1 = 100 r= 1600 = 25600
1 =z 20 == 2000 = 32000 = 512000
Scale— ascending, 16, 16,100, 20 ; descending, 20, 100, 16, 16.
Note. The lo7icf or ffroSs ton, hundred weight, and quarter were
formerly in common use ; ^but they are now seldom used except in
estimating EngKsh goods at the U. S. custom-houses, and in freighting
and wholesaling coal from the Pennsylvania mines.
LONG TON TABLE.
28 lb. make 1 quarter, marked qr.
4 qr. = 112 lb. " 1 hundred weight, " cwt.
20 cwt. = 2240 lb, <«• 1 ton, «« T.
Scale — aseending, 28, ^,20; descending, 20, 4, 28.
The following denominations are also in use.
56 pounds make 1 firkin of butter.
100 " " 1 quintal of dried salt fish.
100 " " 1 cask of raisins.
196 « " 1 barrel of flour.
200 " « 1 " " beef, pork, or fish.
280 « « 1 « « salt at the N. Y. State salt works.
56 " " 1 bushel" " " " " " «
32 « « 1 « « oats.
48 « « 1 " « barley.
56 « « 1 « « corn or rye.
--60 « « 1 « " wheat.
Define avoirdupois weight. Repeat the table. Give the scale. The
long ton table. What other denominations are in. use ? What is ths
value of each ?
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.
21075115 cwt. + 70 lb.
25 T. + 15 cwt.
Ans. 25 T. 15 cwt. 70 lb.
515701b., Ans.
3. Keduce 3 T. 14 cwt. 74 lb. 12 oz. 15 dr. to drams.
'4. Reduce 1913551 drams to tons.
5. A tobacconist bought 3 T. 15 cwt. 20 lb. of tobacco, at
22 cents a pound ; how much did it cost him ? Ans, $1 654.40.
6. How much will 115 pounds of hay cost, at $10 per ton ?
7. A grocer bought 10 barrels of sugar, each weighing
2 cwt. 171b., at 6 cents a pound; 5 barrels, each weighing
3 cwt. 6 lb., at 7^ cents a pound ; he sold the whole at an
average price of 8 cents a pound ; how much was his whole
gain? A?is. $51.05.
8. Paid $360 for 2 tons of cheese, and retailed it for 12^
cents a pound ; how much was my whole gain ? Ans. $140.
*Q. If a person buy 10 T. 6 cwt. 3 qr. 14 lb. of English iron,
by the long ton weight, at 6 cents a pound, and sell the same
at $130 per short ton, how much 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 ? Ans. $75.60.
1 1 . How many pounds in 300 barrels of flour ?
"12. A grocer bought 3 biu-rels of salt at $1.25 per barrel,
and retailed it at f of a cent per pound ? what did he gain ?
A?is. $2.55.
STANDARD OP WEIGHT.
188. In the year 1834 the U. S. government adopted a
uniform standard of weights and measures, for the use of tho
custom houses, and the other branches of business connected with
the general government. Most of the States which have adopt-
ed any standards have taken those of the general government.
COMPOUND NUxMBERS. 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, weighed
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. The U. S. Avoirdupois pound is determined from
the standard Troy pound, and contains 7000 Troy grains^
Hence, the Troy pound is ^l^% = ^f ^ of an avoirdupois
pound. But the Troy ounce contains ^^J-^ = 480 grains, and
the avoirdupois ounce ^^^ =: 437.5 grains ; and an ounce Troy
is 480 — 437.5 =. 42.5 grains greater than an ounce avoirdu-
pois.^ The pound, ounce, and grain, Apothecaries' weight,
are the same as the like denominations in Troy weight, the only
difference in the two tables being in the divisions of the ouno
191* COMPARATIVE TABLE OF WEIGHTS.
Troy. Apothecaries'. Avoirdupois.
1 pound := 5760 grains, = 5760 grains, := 7000 grains.
1 ounce z:^ 480 " = 480 " =437.5 «
175 pounds, z=z 175 pounds, =z 144 pounds.
->■ EXAMPLES FOR PRACTICE.
vl. 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 ; how much did he gain ?
Ans. $33.75.
2. Change 424 drams apothecaries' weight to Troy weight.
Ans, 4 lb. 5 oz.
3. Change 20 lb. 8 oz. 12 pwt. Troy weight to avoirdu-
pois weight. Ans. 17^\*-5 lb.
f 4. Bought by avoirdupois weight 20 lb. of opium, at 40
cents an ounce, and sold the same by Troy v/eight at 50 cents
an ounce ; how much was gained or lost ? Ans. $17.83^.
::j^
"What is the V. S. standard of weight ? How obtained ? How is
'v the avoirdupois pound determined ? How is the apothecaries' pound
determined ? "SMiat are the values of the denominatioiis of Troy, avoir-
dupois, and apothecaries' weight?
158 REDUCTION.
MEASURES OF EXTENSION.
19^. Extension has three dimensions — length, breadth,
and thickness.
A Line has only one dimension — length.
A Surface or Area has two dimensions — length and breadth.
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 or distances.
TABLE.
M) . . ^
12 inches (in.) make 1 foot, ft.
3 feet " 1 yard, yd.
5^ yd., or 16^ ft, « 1 rod, rd.
40 rods " 1 furlong fur.
8 furlongs, or 320 rd., ^' 1 statute mile,, .mi.
UNIT EQUIVALENTS.
ft. in.
yd. 1 = 12
rd. 1 r= 3 = 36
far. 1 r= 5^ z=z 16^ =: 198
„,i. 1 = 40 z=z 220 =: 660 = 7920
1 — 8 = 320 = 1760 =z 5280 =z 6336P
Scale — ascending, 12, 3, 5^, 40, 8 ; descending, 8, 40, 5-^, 3, 12,
The following denominations are also in use : —
3 barleycorns make 1 inch, J "i^^^ ^^ shoemakers in measuring
' ; the length of the foot.
4 inches « 1 hand, \ ^^^ in measuring the height of
I horses du'ectly over the fore feet.
^ feet " 1 fathom, used in measuring depths at sea.
1.1521 statute m. « l geographic mile, \ f^^ ^" measuring dis-
° *' ^ 'I tances at sea.
3 geographic « « 1 league.
60 " « " ) .^^^^ 5 of latitude on a meridian or of
69.10 statute « « $ ^ ^^^^^^ J longitude 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 measure. What are the denom-
inations ? The value of each. AVTiat other denominations are used ?
COMPOUND NUMBERS.
159
Notes. 1. For the purpose of measuring cloth and other goods sold
by the yard, the yard is divided into halves, fourths, eighths, and six-
teenths. The old table of cloth measure is practically obsolete.
2. The geographic mile is g\ of -j^-g^ or ^ ^q( ^ (^ of the distance round
the center 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 mOes 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 polesj where it is 0.
EXAMPLES FOR PRACTICE.
1. In 2 mi. 4 fur. 32 rd.
2 yd. how many inches ?
OPERATION.
2 mi. 4 fur. 32rd. 274.
8
20 fur.
40
832 rd.
416
4162
4578 yd.
3
2. In 164808 inches how
many miles ?
OPERATION.
1 2) 164808 in.
3) 13734 ft.
5^ ^ 4578 yd.
2" j 2
11)9156
4| 0)83|2 rd.+^yd. — 2yd.
8)20fur.+32rd.
2 mi. + 4 fur.
Ans, 2 ml 4 fur. 32 rd. 2 yd.
13734 ft.
12
164808 in., ^Tis.
3. The diameter of the earth being 7912 miles, how many-
inches is it? Ans. 501304320 inches.
4. In 168474 feet how many miles ?
N^ 5. In 31 mi. 7 fur. 10 rd. 3 yd., how many feet ?
^^. If the greatest depth of the Atlantic telegraphic cable
from Newfoundland to Ireland be 2500 fathoms, how many
miles is it ? Ans. 2 mi. 6 fur. 29 rd. 1^ ft.
160 REDUCTION.
^ 7. If this cable be 2200 miles in length, and cost 10 cents
a foot, what was its whole cost? Ans. $1161600.
8. A pond of water measures 4 fathoms 3 feet 8 inches in
depth ; how many inches deep is it ? Ans. 332.
V 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 cir-
cumference? Ans, 57600 times.
VI 0. If a vessel sail 120 leagues in a day, how many stat-
ute miles does she sail ? Aiis. 414.
„11. How many inches high is a horse that measures 14^
hands ? A7is. 58.
surveyors' 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, .... 1.
25 links " 1 rod, rd.
4 rods, or 66 feet, " 1 chain ..ch.
80 chains " 1 mile, . . mi.
UNIT EQUIVALENTS.
1. in.
rd. 1 = 7.92
ch.
1 = 25 = 198
r.1. T=: A=z 100= 792
1 = 80 r= 320 = 8000 = 63360
Scale — ascending, 7.92, 25, 4, 80 ; descending, 80, 4, 25, 7.92.
KoTE. Hods are seldom used in chain measure, distances being
taken in chains and links.
EXAMPLES FOR PRACTICE.
1. In 3 mi. 51 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 j how many feet long is the fence
Avhich encloses it? ^««- 3983.76 ft.
Kepeat the table of surveyors' long measure. Give the scale.
COMPOUND NUMBERS.
161
IT. Square Measure.
10^. A Square is a figure having four equal sides, and
four equal angles or corners.
12 in. =1 ft. 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 inches. Hence
The contents or area of a square, or
of any other figure having a uniform
length and a uniform breadth, is found
by multiplying the length by the breadth.
Thus, a square foot is 12 in. long and 12 in. wide, and the con-
tents are 12 X 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 sur-
faces ; as of land, boai'ds, painting, plastering, paving, &c.
t*
~
~
~
~
,
e
r"
II
.-
W
' '
^
12 in. = 1 ft.
TABLE.
144 square inches (sq. in.) make 1 sqnare foot, marked sq. ft.
9 square feet " 1 sq iare yard, « sq. yd.
?jfn oniio^Q TrovrJo « 1 squaic xod, " sq. rd.
1 rood,
1 acre,
1 square mile,
9 square feet
30^ square yards
40 square rods
4 roods
640 acres
sq. rd.
R.
A.
sq. mi.
UNIT EQUIVALENTS
sq.mi.
1 =
R.
1 =
4 =
sq. rd.
40 =
160 =
pq. yd.
1 =
301 =
1210 =
4840 =
sq.ft.
1 =
9 =
272| =
10890 =
43560 =
sq. in.
144
1296
39204
1568160
6272640
1 = 640 = 2560 = 102400 = 3097600 = 27878400 = 4014489600
Scale — ascending, 144, 9, 30^, 40, 4, 640; descending, 640, 4,
40, 30^, 9, 144.
Define a square. How is the 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 their work as follows :
By the square foot : glazing and stone-cutting.
i3y the square yard: painting, plastering, paving, ceiling, and
paper-hanging.
By the square of 100 feet : flooring, partitioning, roofing, slating,
and tiling.
Brick-laying is estimated by the thousand bricks j also by the
square yard, and the square of 100 feet.
Notes. 1. In estimating the pauiting of moldings, cornices, &c., the
measuring-line is carried into all the moldings and cornices.
2. In estimating brick- laying by the square yard or the square of
100 feet, the work is xmderstood to be l^ bricks, or 12 inches, thick.
EXAMPLES FOR PRACTICE.
1. In 10 A. 1 R. 25 sq. rd. 16 sq. yd. 4 sq. fl. 136 sq. in.
now many square inches ?
OPERATION.
10 A. IR. 25 sq. rd. 16 sq.yd. 4 sq. ft. 136 sq. in.
-i
41 IL
40
1665 sq.rd.
30i
416J
49966
50382| sq.yd.
9
453444^ sq.ft.
144
36 = I sq. ft.
1813912 with 136 sq. in.
1813776
453444
65296108 sq. in., Ans.
2. In 65296108 sq. in. how many acres?
How do artisans estimate work f
CX)MPOUND NUMBERS. 163
OPERATION.
144 ) 65296108 sq. in.
9 ) 453445 sq. ft. + 28 sq. in.
30^^ 50382 sq. yd. + 7 sq. ft.
4 J 4
121 ) 201528 fourths sq. yd.
4! 0)166[5 sq. t(\.-\-^— 15| sq. yd.
4)41R.+25sq.rd.
10 A. -|- 1 R.
10 A. 1 R. 25 sq. rd. 15f sq. yd. 7 sq. ft. 28 sq. in.
10 A. 1 R. 25 sq. rd. 15 sq. yd. 7 sq. ft. 28 sq. in.
6 sq. ft. 108 sq. in.
10 A. 1 R. 25 sc;. rd. IG sq. yd. 1 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 square yards a mixed number, 15|. But f of a sq. yd.
= I of 9 sq. ft. z= 6| sq. ft. ; and | of a sq. ft. =: | of 144 sq. in.^=
108 sq. in. Therefore | sq. yd. =z 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. 1 R. 25 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in.
3. Reduce 87 A. 2 R. 38 sq. rd. 7 sq. yd. 1 sq. ft. 100 sq.
in. to square inches. -Ans. 550355068 sq. in.
4. Reduce 550355068 square inches to acres.
5. A field 100 rods long and 30 rods wide contains how
many acres? -^ns. 18 A. 3'R.
6. How many rods of fence will enclose a farm a mile
square ? -^^^s. 1280 rods.
7. How much additional fence will divide it into four equal
square fields ? Ans. 640 rods.
8. How many acres of land in Boston, at $1 a square foot,
will $100000 purchase ?
Ans. 2 A. 1 R. 7 sq. rd. 9 sq. yd. 3^ sq. ft.
9. How many yards of carpeting, 1 yd. wide, will be required
to carpet a room 18J ft. long and 16 ft. wide ? Ans. 82| yd.
164 REDUCTION.
10. What would be the cost of plastering a room 18 ft. long,
16^ ft. wide, and 9 ft. high, at 22 cts. a sq. yd. ? Ans. $22.44.
"yll- 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.
S" 625 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.
36 square miles (6 miles square) " 1 township,. Tp.
UNIT EQUIVALENTS.
P. sq. 1.
Bq. Ch. 1 = 625
A 1= 16=: 10000
eq m\ 1= 10 = 160 := 100000
Tp 1 = 640 = 6400 = 102400 = 64000000
1 = 36 r= 23040 — 230400 — 3686400 = 2304000000
Scale — ascending, 625, 16, 10, 640, 36; descending, 36, 640,
10, 16, 625.
Notes. 1. A square mile of land is also called a section.
2. Canal and railroad engineers commonly use an engineers* chain,
which consists of 100 links, each 1 foot long.
3. The contents of land arc commonly estimated in square miles,
acres, and himdredths ; the denomination, rood, is fast going into dis-
use.
EXAMPLES FOR PRACTICE.
1. How many poles in a township of land ?
2. Reduce 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. $2275.
Repeat the table of siurveyors' squaie measure. Give the scale.
COMPOUND NUMBERS.
165
^-~ y / A
y^ y y gm
y^ — -7 — T^ ^^.^JSm.
!
j^l;
1
1 .
,::„ W
III. CuBJC Measure.
198. A Cube is a solid, or body,
liaving 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
z=z 9 cubic feet,and the whole cube will
contain 3 X 3 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 we see that
The cubic or solid contents of a hody are found by mvltif ly-
ing the lengthy breadth, and thickness together.
100. Cubic Measure, also called Solid Measure, is used
in estimating the contents of solids, or bodies ; as timber^vood,
tftone, &c.
TABLE.
1728 cubic inches (cu. in.) make 1 cubic foot, cu. ft.
27 cubic feet
16 cubic feet
8 cord feet, or
128 cubic feet
:1
1 cubic yard, . . . . cu. yd.
1 cord foot, cd. ft.
:4| cubic feet
" 1 cord of wood, .... Cd.
« 1 S perch of stone ) p ,
\ or masonry, ^
Scale — ascending, 1728, 27. The other numbers are not in a
regular scale, but are merely so many times 1 foot. The unit
equivalents, being fractional, are consequently omitted.
Notes. 1. A cubic yard of earth is called a load.
2. Railroad and transportation companies estimate light freight by
} the space it occupies in cubic feet, and hea\'y freight by weight.
'*T*^5. A pile of wQod 8 feet long, 4 feet Avide, 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^ 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 cubte measure \ised? Repeat the table.
Give the scale. How is railroad freight estimated ? ^\Tiat is under-
istood by a cord foot ? Ry a perch of ston^ or masonry ?
166 REDUCTION.
5. Joiners, bricklayers, and masons make no allowance for windows,
doors, &c. Bricklayers and masons, in estimating their work by
cubic measure, make no allowance for the corners of the walls of
houses, cellars, &c., 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 deci-
mals of a foot. The estimates are made in feet and decimals, and the
results are reduced to cubic yards.
EXAMPLES FOR PRACTICE.
V 1. In 125 cu. ft. 840 cu. in. how many eu. in. ? Ans. 21 6840.
2. Reduce 5224 cubic feet to cords. Ans. 40||.
- 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
7 5- y)2^^^^^ from a cube 6 feet on each edge, allowing no waste for
eawing? Ans. 373248.
5. 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 J- feet high ? Ans. 113^ 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.
7^ ^S. The gray limestone of Central New York weighs 175
-"^ I)ounds 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.
How much did it cost at $1.25 a perch? ~X Ans. $50,909+
'10. How much did it cost to dig the same cellar, at 15
cents a cubic yard ? -^^s. $25.60.
V^ i V 11. My sleeping room is 10 ft. long, 9 ft. wide, and 8 h. high.
70 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 ? -4?^s. 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 embanknienta measured ?
( ^-^ rOMPOTTND NUMBERS. 167
MEASURES OF CAPACITY.
^^ I. Liquid Measure.
^^ 900. Liquid Measure, also called Wine Measure, is used
^S^in measuring liquids ; as liquors, molasses, water, &c.
^ ^\) TABLE.
J ^ 4 gills (gi.) make 1 pint, pt.
2 pints " 1 quart, qt.
4 quarts " 1 gallon, ga L
3U gallons " 1 barrel, .... .bbl.
2 barrels, or 63 gal. " 1 hogshead,, .hhd.
UNIT EQUIVALENTS.
pt. gi.
qt. 1=4
g,i. 1=2= 8
w.i. 1 = 4 = 8 = 32
i.hd. 1 = 3U = 126 = 252 = 1008
1 = 2 = 63 = 252 = 504 = 2016
Scale — ascending, 4, 2, 4, 31^, 2 ; descending, 2, 31^, 4, 2, 4.
The following denominations are also in use :
36 gallons make 1 barrel of beer.
54 " or IJ barrels " 1 hogshead " "
42 " "1 tierce.
2 hogsheads, or 120 gallons, " 1 pipe or butt.
2 pipes or 4 hogsheads, " 1 tun.
Notes. 1. The denominations, barrel and hogshead, are^sed in es-
timating the capacity of cistern s, r eservoi rs, vats, &c. ''^
2. The tierce, hogshead, pipe, biitT, 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 measiuing beer, ale, and
milk, is almost entirely discarded.
"What is liquid measure ? Repeat the table. Give the scale. "VMiat
other denominations are sometimes used ? How are the capacities of
cisterns, reservoirs, &c., 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 [lpt.3gi.
5 bbl.
31^
185
187J-gal.
4
752 qt
2
1505 pt.
4
2. In G023 gi. how many
hhds..?
OPERATION.
4 ) 6023 gi.
2 ) 1505 pt. + 3 gi.
4 ) 752 qt. + 1 pt.
31^ 188 gal.
2 J 2
[gal.
63 )376
2)_5bbl. + -V-gal.=r30^
2 hhd. + 1 bar.
^ Ans. 2 hhd. 1 bar. 30^ gal.
pipt. 3gi.
But ^ gal. = 2 qt., making
the Ans. 2 hhd. 1 bar. 30 gaL
2 qt. 1 pt. 3 gi.
6023 gi., Ans.
3. Reduce 3 hogsheads to gills.
4. Reduce 6048 gills to hogsheads.
5. In 13 hhd. 15 gal. 1 qt. how many pints ?
6. In 6674 pints how many hogsheads ?
^ 7. What will 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 atr 5 cents a
quart ; how much 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 filled from a hogshead of wine ? Ans. 24.
II. Dry Measure.
SOI, Dry Measure is used in measuring articles not
liquid, as grain, fruit, salt, roots, ashes, &c.
What is dry measure i
COMPOUND NUMBERS, 169
TABLE.
2 pints (pt.) make 1 quart, qt.
8 quarts " 1 peck, pk.
4 pecks " 1 bushel, . bu. or bush.
UNIT EQUIVALENTS,
qt. pt.
pk. 1 = 2
. i,„. 1 =r 8 == 16
1 r= 4 =: 32 = 64
Scale — ascending, 2, 8, 4 ; descending, 4, 8, 2.
y^NoTE. In England, 8 bu, of 70 lbs. each are called a quarter, used in
measuring grain. The weight of the English quarter is -|^ of a long ton.
EXAMPLES FOR PRACTIQE.
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.
N^ 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.
STANDARD OP EXTENSION.
SOS. The U. S, standard unit of measures of extensioriy
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 of
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 tf^^ta ^^ the length of
the pendulum rod.
S03. The 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 maximum
density, that is, at the temperature of 39.83° Fahrenheit, the
, barometer at 30 inches;.
Repeat the table. What is a quarter ? What is the IT. S. standard
unit of measurement of extension ? How is it determined S What is
the U. S. standard unit of liquid measure ?
E.P. 8
170 REDUCTION.
304: • The U. S. standard unit of dry measure is the Brit-
ish 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, drj measure, contains 2G8.8 cubic inches.
Note. 1. The wine and dry measures of the same denomination
are of different capacities. The exact and the relative size of each may
be readily seen by the following
205, COMPARATIVE TABLE OF MEASURES OF CAPACITY^
Cu. in. in | Cu. in. in Cu. in. in Cu. in. in
one gailonJ one quart. one pint. one gill.
XWine measure, 231 / 57| 28| 7^
i)ry measure, (^ pk.,) 268| 67| 33| Sf
^'7 2. The beer gallon of 282 inches is retained in use only by custom
^'i;:^ bushel is commonly estimated at 21504 cubic inches.
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-^f quarts.
2. A grocer bought 40 quarts of milk by beer measure, and
sold it by wine measure ; how many quarts did he gain ?
A7is. 8f f quarts.
3. A bushel, or 32 quarts, dry measure, contains how many
more cubic inches than 32 quarts wine measure ?
Ans. 302f cu. in.
Time.
206. Time is used in measuring periods of duration, as
years, days, minutes, &c.
TABLE.
60 seconds (sec.) make 1 minute, min.
60 minutes " 1 hour, h.
24 hours " 1 day, da.
7 days " 1 week, wk.
365 days " 1 common year,. . .yr.
366 days " 1 leap year, yr.
12 calendar months " 1 year, yr.
100 years " 1 century, C,
What is the U. S. standard unit of dry measure ? How is it ob-
tained ? What is the relative size of the wine and the dry gallon ?
>Vh^t is tjie size pf & beer gallon ? AVhat is time ? Repeat the table.
COMPOUND NUMBERS. 171
UNIT EQUIVALENTS.
mjn. sec.
h. 1 =: 60
da. 1 =: 60 =r 3600
wk.
1 == 24 =r 1440 == 86400
1 =r:
7 z=z 168 = 10080 = 604800
yr
mo. '
{ 365 = 8760 =z 525600 = 31536000
1
= 12 z=;
I 366 — 8784 = 527040 =z 31622400
Scale — ascending, 60, 60, 24, 7 ; descending, 7, 24, 60, 60.
The calendar year is divided as follows : —
No. of mo.
Season.
Names.
Abbreviations.
No. of days.
1
Winter,
^ January,
; February,
Jan.
31
2
((
Feb.
28 or 29
3
Spring,
C March,
Mar.
31
4
a
<. April,
Apr.
30
5
«
^May,
31
6
Summer,
C June,
Jun.
30
7
«
^July,
31
8
((
( August,
Aug.
31
9
Autumn,
(T September,
Sept.
30
10
a
< October,
Oct.
31
11
li
( November,
Nov.
30
12
Winter,
December,
Dec.
31
365 or 366
Notes. M. The exact leng^of a solar year is 365 da. 5 h. 48 min. 46
sec. ; but for conveni&Jiee it is reckoned 11 min. 14 sec. more than this,
or 365 da. 6 h. = 365^ da. This \ day in 4 years makes one day,
which, every fourth, bissextile, or leap year, is added to the shortest
month, gi\'ing 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 remembered by committing the ifoUo wing lines i-/^
" Thirty days hath September,
April. June, and November ;
All the rest 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 FOR PRACTICE.
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 ?
172 REDUCTION.
3. In 5 wk./ 1 da. 1 h. 1 min. 1 sec. how many seconds ?
4. In 3114061 seconds how many weeks?
^^. 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
loner a time will he walklO 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 prin-
cipally in surveying, navigation, astronomy, and geography,
for reckoning latitude and longitude, determining locations 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, &c. Consequently the
parts of different circles, although having the same names, are
of different lengths.
TABLE.
GO seconds (") make 1 minute,...'.
60 minutes " 1 degree, . . .
30 degrees " 1 sig-n, S.
12 signs, or 360°, " 1 circle, C.
•UNIT EQUIVALENTS.
60
S. 1 = 60 — 3600
C. 1 z= 30 = 1800 = 108000
1 == 12 = 360 = 21600 =: 1296000
Scale — ascending, 60, 60, 30, 12; descending, 12, 30, 60, 60.
Notes. 1. Minutes of the earth's circumference are called geo-
graphic or nautical miles.
2. The denomination, signs, is confined exclusively to Astronomy.
Define circular measure. How are circles divided ? llopcat the
table. Give the scale. What is a geographic mile? What is a
bign?
COMPOUND NUMBERS. 17^
8. 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 1 of a
circle.
EXAMPLES FOR PRACTICE.
' 1. Reduce 10 S. 10° 10' 10'' to seconds.
2. Reduce 1116G10" to signs.
3. How many degrees in 11400 geographic or nautical
miles? ' ' Ans. 190°.
4. If 1 degree of the earth's circumference is 69^ statute
miles, how many statute 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 h^r longitude
397 geographical miles ; how many degrees and minutes did
she change ? Ans. 6° 37'.
S08, In Counting.
12 units or things .make. . . .1 dozen.
12 dozen " 1 gross.
12 gross " 1 great gross.
20 units " 1 score.
309. Paper.
24 sheets make 1 quire.
20 quires " 1 ream.
2 reams " 1 bundle.
5 bundles " 1 bale.
310. Books.
The terms folio, quarto, octavo, duodecimo, Sec, indicate
the number of leaves into which a 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 " a 24mo.
A sheet folded in 32 leaves " a 32mo.
"WTiat is a degree ? Repeat the table for coxmting. For reckoning
paper. For indicating the size of books.
174 REDUCTION.
EXAMPLES FOR PRACTICE.
1. If in Birmingham, England, 150 million Gillott pens are
maniifactured annually, how many great gross will they make ?
Ans. 8G805 great gross 6 gross 8 dozen.
2. In 100000 sheets of.paper, how many bales?
Ans. 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 pages.
5. How large a book will ten 32mo. sheets make, if every
page be printed? ^^-^ Ans. 640 pages.
PROMISCUOUS EXAMPLES IN REDUCTION.
1. How many siUts of clothes, each containing 6 yd. of 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.
X 3. A physician, having 2 lb 3i 55 IB 10 gr. of medicine,
dealt it out in prescriptions averaging 15 grains each ; how
many prescriptions did it make ? Ans. 886.
4. A man bought 1 T. 11 cwt. 12 lbs. of hay, at 1^ 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 37^ cents per bushel ? Ans. $17.0625^"^^
^'6. If one bushel of wheat will make 45 pounds of flour, how
many barrels will 1000 bushers make ? Ans. 229 bbl. 116 lb.
7. A load of wheat weighing 2430 pounds is worth how
much, at $1.20 a bushel? u'/y /.,.'> A?is. $48.60.
8. Paid $12.50 for a barrel of beef; how much was that
per pound ? Ans. 6^ 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 ? ^ Ans. 1577511936.
10. In 10 mi. 7 ch. 4 rd, 20 1., how many hnks ?
Ans. 80820 hnks.
DENOMINATE 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. $3600.
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.
s^l3. A cistern 10 feet square and 10 feet deep, will hold
how many hogsheads of water? Ans. 118 hhd. 46|-^ gal.
14. A bin 8 feet long, 5 feet wide, and 4| feet high, will
hold how many bushels of grain ? Ans. 1441^^ 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 circumfer-
ence, 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 $10000, at the rate of one cent
each second? Ans. 27 da. 7h. 46min. 40 sec.
21. A town, 6 miles long and 4i 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 DENO^nXATE FRACTIONS.
CASE I.
Sll. To reduce a denominate fraction from a
greater to a less unit.
1. Reduce ^^ of a bushel to the fraction of a pint
Case I is what ?
176 KEDUCTION.
OPERATION.
^1^ X t X f X f = I, Ans, Analysis. To reduce bushels
Q to pints, we must multijjly by 4,
' 8, and 2, the numbers in the
1 scale. And since the given num-
4 ber is a fraction of a bushel, we
^ indicate the process as in multi-
g plication of fractions, and after
— canceling, obtain |, the Answer.
4 = 1 pt., Ans. Hence,
HuLE. Multiply the fraction of the higher denomination hy
the numbers in the scale successively, between the given and the
required denominations.
Note. Cancellation may be applied wherever practicable.
EXAMPLES FOR PRACTICE.
2. Reduce y^V?j of a £ to the fraction of a penny.
Alls. ^^ d.
3. Reduce jii^yu of a week to the fraction of a minute.
Ans. ^jj min.
4. What part of a gill is ^xy^S" ^^ ^ hosghead ? Ans. ^ gi.
5. What fraction of a grain is ^^^ of an ounce ? Ans. ^ gr.
6. Reduce lUTj^utjo- of a mile to the fraction of an inch.
Ans. ^V¥r in-
7. Reduce f 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. f^pt.
9. Reduce T^iy of an acre to the fraction of a rod.
Arhs. ^ rd.
CASE IT.
313. 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. Bule. Case It is what ^
DENOMINATE FRACTIONS. 177
OPERATION. . ^
Analysis. To reduce
_^J;_^1 ^ J pints to bushels, we must
a^S^^SO' ^^''^^^ ^y 2. 8, and 4, the
Or, 5
2
8
4
80
numbers of the scale. And
4 smce the given number of
pints is a fraction, we indi-
cate the process, as in divis-
ion of fractions, and cancel-
— ing, obtain -gL, the Answer.
1 = ^V bu., Ans.
Rule. Divide the fraction of the lower denomination hy the
numbers in the scale, successively, between the given and the
required denomination.
Note. 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 fSf a dram ? Ans. xjfjt lb.
4. Reduce ^ of a cent to the fraction of an eagle.
5. A hand is ^ of a foot ; what fraction is that of a mile ?
^ ^ns. ^-^ij^mi.
6. Reduce •f of 2 pwt. to the fraction of a pound. Ans. -^^j^ lb.
7. How much less is f of a pint than ^ of a hogshead ?
Ans. H§hhd.
8. In -I of an inch what fraction of a mile ? Ans. y^ 5^55(7 ^i»
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.
313. To reduce a denominate fraction to integers
of lower denominations.
1. What is the value of f of a hogshead of wine ?
Give explanation. Rule, Case III is what ?
8*
178 REDUCTION.
OPERATION.
I bhd. X C3 = ^^^ gal. = 39f gal.
fgal. X4:=J^-qt. = l|qt.; | qt. X 2 nr f pt. == 1 pt.
A71S. 39 gal. 1 qt. 1 pint.
Analysis. | hhd. = | of 63 gal., or 39| gal. ; and | gal. = | of
4 qt., or 1| qt. j and | qt. =-| of 2 pt., or 1 pt. Hence,
Rule. I. Multiple/ the fraction hy that number in the scale
which 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. 42f sec.
3. What is the value of ^ of a £ ? Ans. 8 s. 6 d. 3^ far.
4. What is the value of | of a bushel ?
5. Reduce f of 15 cwt. to its equivalent value.
Ans. 12 cwt. 85 lbs. y oz. 6f dr.
6. Reduce f of | of a pound avoirdupois to ihtegers.
Ans. 4oz. llff dr.
7. What is the value of ^ of an acre ? Ans. 3 R. 13^ P.
8. Reduce ^| of a day to its value in integers.
Ans. 16 h. 36 min. 55^^ sec.
9. What is the value of ^ of a pound Troy ?
10. What is the value of I of 51 tons ? Ans. 4 T. 5 cwt. 55|. lb.
11. What is the value of f of 3§ acres ? Ans. 1 A. 1 R. 20 P.
CASE IV.
^14, To reduce a compound number to a fraction
of a higher denomination.
1. What part of a week is 5 da. 14 h. 24 min.?
Give explanation. Rule. Case IV 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,
_8 64 — 4^vk Ans. they must be reduced to
itjosff 5 •' • tiig game denomination.
In 5 da. 14 h. 24 min. there are 8064 minutes, and in 1 week there
are 10080 minutes. Since 1 minute is yo^ of a week, 8064 min-
utes is -j8j)^^rzz^ of a week. Hence,
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.
Note. If the given number 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 6 fur. 26 rd. 3 yd. 2 ft. ? Ans, ^ mi.
3. What fraction of a £ is 13 s. 7 d. 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. -^^ 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 14 feet ? Ans. -j^.
8. Reduce If pecks to the fraction of a bushel. Ans. | bu.
9. What part of 9 feet square are 9 square feet ?
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.
31^. To reduce a denominate decimal to integers
of lower denominations.
1. Reduce .78125 of a pound Troy to integers of lower de-
nominations.
Give explanation. Rule. Case V is what ?
180 REDUCTION.
OPERATION. Analysis. We first multiply
.78125 lb. ^y ^^ ^° reduce the given number
22 froni pounds to ounces, and the
result is 9 ounces and the decimal
9.37500 oz. .375 of an oz. We then multiply,
20 this decimal by 20 to reduce it to
7.50000 pwt. pennyweights, and get 7 pwt. and
2 i '5 of a pwt. This last decimal we
multiply by 24, to reduce it to
12.0000 gr. grains, and the result is 12 gr.
f. rr . ^ct A Hence the answer is 9 oz. 7 pwt.
9 oz. 7 pwt. 12 gr., Ans. ^^ ^^ ^
Rule. I. Multiply the given decimal hy that number in the
scale which will 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 in-
tegers at the left will he the answer required.
EXAMPLES FOR PRACTICE.
2. What is the value of .217° ? Ai^s. 13' 1.2'^
3. What is the value of .659 of a week ?
Ans. 4 da. 14 h. 42 min. 43.2 sec.
4. Reduce .578125 of a bushel to integers of lower denom-
inations, ^ns. 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 Jrogshead of molasses ?
Ans. 13 gal. 3 qts. 3.52 gi.
8. What is the value of .67 of a league ?
/ Ans. 2 mi. 3 rd. 1 yd. 3f in.
9. What is the value of .42857 of a month ?
Ans. 1 2 da. 20 h. 34 min. 13^ sec
10. 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 ? Ans. 5 A. 3 R. 21 P.
12. Reduce .0055 T. to pounds. Ans. 11 lb.
13. Reduce .034375 of a bundle of pajDcr to its value hi
lower denominations. Ans. 1 quire 9 sheets.
CASE VI.
316. 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
"TTTTr , will be | as many pecks as quarts
d.ZoOO pk. (183), and J as many bushels as
.8125 bu., ^725. pecks.
Or we may reduce 3 pk. 2 qt. to
. Or, 3 pk. 2 qt. = 26 qt. ^^^ ^^^^^^^^ ^^ ^ ^^^^^^ ^^^ ^^ ^^^^
1 bu. — 32 qt. and we have ||^ of a bushel, which,
f^ = .8125 bu., Ans. reduced to a decimal, equals .8125.
Hence the
Rule. Divide the lowest denomination given hy that num-
her in the scale which will reduce it to the next higher, and an-
nex the quotient as a decimal to that higher. Proceed in the
same manner until the whole is reduced to the denomination
required. Or,
Reduce the given number to a fraction of the required de-
nomination, and reduce this fraction to a decimal.
EXAMPLES FOR PRACTICE.
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. .88906251b.
4. Reduce 1.2 pints to the decimal of a hogshead.
Ans. .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 3 R. 12.56 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.
A?is. .005 bu.
9. Reduce 4^ 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. 36mia. 57-^^ sec. to the decimal of a day.
ADDITION.
317. 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 pwt. ; at another, 1 1 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 oth-
er, we first add the units in the
light hand column, or lowest de-
nomination, and find the amount to
be 51 grains, which is equal to 2
Ans. 39 117 3 pwt. 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 col-
umn 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, wo
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 be 39 lb. Ilcnco
the following
What is addition of compoimd numbers ? Give explanation.
OPERATION.
lb.
07.. pwt. pr.
10
4 16 8
2
9 3
11 20
25
16 23
COMPOUND NUMBERS. 183
HuLE. I. Write the numbers so that those of (he same unit
value will stand in the same column.
II. Beginning at the right hand, add each denomination cus
in simple numbers, carrying to each succeeding denomination
one for as many units as it takes of the denomination addedy to
make one of the next higher denomination.
EXAMPLES FOR PRACTICE.
(20
£. s. d.
43 13 8
lb.
12
1
8
(3.)
3. 9. gr.
7 2 15
51 6 4
10
4 1 10
G7 11 3
15
00
2 1 19
76 18 10
11
6 12
244 10 1
13
4
4 2 00
(40
T. cwt. lb. oz.
4 7 18 4
dr.
10
bu.
1
(5.)
pk. qt. pt.
3 7 1
15 98 15
5
3
2 2
S 9 10 6
15
1 6 1
1 15
4
17
5 1
9 12 42 11
2
45
2 4
G. What is the sum of 4 mi. 3 fur. 30 rd. 2 yd. 1 ft. 10 in.,
5 mi. 6 fur. 18 rd. 1yd. 2 ft. 6 in., 10 mi. 4 fur. 25 rd. 2 yd.
2 ft. 11 in., and 6 fur. 28 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. 27f sq. in., 5 sq. yd. 8 sq. ft. 2§ sq. in., and 237
sq.yd. 7 sq.ft. 128-isq.in.?
Ans. 563 sq. yd. 4 sq. ft. 118.825 sq. in.
Note. When common fractions occur, they should be reduced to a
common denominator, to decimals, or to integers of a lower denomi-
nation, and added according to tlie usual method.
Give the Rxile.
184
ADDITION.
A.
26
3
(8.)
P. sq. yd. sq.ft.
28 25 8
sq. in.
125
19
2
38 30 7
150
456
2
20 16 6
98
603 1 8 12Q)5 85
(i) = Hi)
ffl=72
503 1 8 13 1 13
(9.)
(1
0.)
mi.
fur.
rd.
yd,
ft.
in.
hhd.
gal.
qt.
pt.
1
7
30
4
2
11
27
65
3
2
3
4
00
2
1
10
112
60
2
3
10
7
25
1
2
11
50
421
29
00
2
1
16
3
16
3.
^
1
8
3
14
39
1
2
(1
1.)
(12.)
bu.
pk.
qt.
pt.
T^'
da.
h. :
min.
sec.
23
3
7-
1
25
300
19
54
35
34
2
1
21
40
40
24
42
3
5
3
112
15
17
51
1
4
1
6
19
45
59
23
3
1
1
1
1
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 bales 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 J how many did he use in the three years?
A71S. 10 gross 8 doz. 6.
COMPOUND NUMBERS. 185
15. A ship, leaving New York, sailed east the first day 3^
45' 50" ; the second day, 4° 50' 10" ; the third, 2° 10' 55" ;
the fourth, 2° 39" ; how I'ar 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. yds. 18 cu. ft. ; what was the amount
of earth removed, and what the cost at 16 cents a cu. yd.?
Ans. 152^ cu. yds.; $24.37^.
17. A farmer received 80 cents a bushel for 4 loads of
corn, weighing as follows: 2564, 2713, 3000, and 3109 lbs.;
how much did he receive for the whole ? Ans, $162.6574"
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. 4gal. Ipt.; 30 gaL 2 qt. 1 gi. ; 2 bbh 15 gal;
how much did he receive for the whole ? Ans. $415.17.
318. To add denominate fractions.
1. Add 1^ of a mile to ^ of a furlong.
Analysis. We find the
value of each fraction in in-
tegers of less denominations
(213), and then add their
values as in compound num-
bers (SIT).
, ". 1 t . /i* • rr n Or, we mav reduce the
2V mi. -f- 4 mi. == 41 nil. = 7 fur. . ' . "^ i. .- r
** * ' *' "^ * given fractions to fractions 01
the same denomination (212), then add them, and find the value
of their sum in lower denominations (213).
2. Add I of a rod to f of a foot. Ans. 13 ft. 1^ in.
3. What is the sum of ^ of a mile, f of a furlong, and ^ of
a rod ? Ans. 7 fur. 27 rd. 8 ft. 3 in.
4. What is the sum of f of a pound and f of a shilling ?
Ans. 13 s. 10 d. 2f qr.
5. What is the sum of | of a ton and f of 1 cwt. ?
Ans. 12 cwt. 421b. 13f oz.
Give explanation of the process of adding denominate fractions.
1
i
mi. =
fur.=
Ans.
r, 1 fur
6 fur. 26 rd.
13 rd.
11
ft
ft.
7 fur. 00
mi.
3 S8 SUBTRACTION.
6. What is the sum of f of a day added to J an hour ?
Ans. 9 h. 30 min.
7. What is the sura of ^ of a week, f 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, 8f lb., and 3^^ oz. by
long ton table ? Ans. 73 lb. 1 oz. 3|i dr.
10. What is the sum of f of a mile, f of a yard, and f of
a foot ?
11. Sold 4 village lots; the first contained ^ of ^ of an
acre ; the second, 60f rods ; the third, f of an acre ; and the
fourth, f of f of an acre ; how much land in the four lots ?
Ans. 3 R. 26 P. 126-J-V5- 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 ?
Ans. 3 T. 5 cwt. 911b. lOfoz.
SUBTRACTION.
219. 1. If a druggist buy 25 gal. 2 qt. 1 pt. 1 gi. of
wine, and sell 18 gal. 3qt. 1 pt. 2 gi., how much has he left?
OPERATION. Analysis. Writing the subtrahend
gal. qt. pt. gi. under the minuend, placing units of the
25 2 1 1 same denomination under each other,
18 3 2 we begin at the right hand, or lowest
A ~A ^ J\ ^~ denomination ; since we cannot take
MS. K> 6 V o 2 gi. from 1 gi., we add 1 pt. or 4 gi. to
1 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 we 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
What is subtraction of compound numbers ? Give explanation.
COMPOUND NUMBERS.
187
numbers, and write the remainder, 6 gal., under the column of gal-
lons. Hence the following
Rule. I. Write the subtrahend under the minuend, so that
units of the same denomination shall stand under each other,
II. Beginning at the right hand, subtract each denomination
separaiely, as in simple numbers.
III. ]f the number of any denomination in the subtrahend
exceed that of the same denomination in the minuend, 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 subtrahend before
subtracting. Proceed in the same manner with each denomi-
nation.
EXAMPLES FOR PRACTICE.
(2.)
lb. oz. pwt.
From 18 6 10
14
(3.)
A. R. P.
25 2 16.9
Take 10 5
4
6
19 3 25.14
Rem. 8 1
6
8
5 2 31.76
(4.)
T, cwt. lb.
14 11 69f
yr.
38
da.
187
(5.)
h. min. sec,
16 45 50
10 12 98f
17
190
20 50 40
20 361 19 55 10
6. A Boston merchant bought English goods to the amount
of 4327 £ 13 s. 7|d., and he paid 1374£ 10s. 11} d.; how
much did he then owe ?
7. From 300 miles take 198 mi. 7 fur. 25 rd. 2 yd. 1ft.
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 ?
Jns. 4° 55'^
9. From 10 ft 7 § 4 3 1 9 15 gr. take 31b8§2329
18 gr. Ans. 6 ft 11 i 1 3 1 9 17 gr.
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 ? Ans. 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. 2 R. 16^ P. ; the second, 200
A. 1^ R. ; the third, 177 A. 36 P. ; how much does the fourth
own ? Ans, 65 A. 3 R. 7.75 P.
13. From a pile of wood containing 75^ Cd. was sold
at one time 16 Cd. 5 cd. ft.; at another, 24 Cd. 6cd.ft. 12
cu. ft. ; at another, 27 Cd. 112 cu. ft. ; how much remained in
the pile ? Ans. 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 ?
S30. 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 orERATiON. ANALYSIS. "We place the earlier date
yr. mo. da. under the later, writing first on the left
1* • " • 4 the number of the year from the Chris-
1492 10 14 tian era, next the number of the month,
ooo ^ ^ counting January as the first month, and
next the 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
SECOND OPERATION. ?«^^ ^^^f ^ '^'^ ^.^^ ^'^"f ^'i ?'''' '^" T '
^^^ mstead of saymg July is the /th month,
2jy^ 6 3^® ^^y ^ months and 3 days have
■lAqi f) no 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 ?
COMPOUND NUMBERS.
189
Both methods will obtain the same result ; the former is generally
used.
Notes. 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,
Showing the number of days from any day of one month to the same day
of any otlier month within one year.
FB.OM ANY
TO THE SAME DAY OF THE NEXT.
DAY OF
Jan.
365
Feb.
31
Mar.
59
Apr.
90
May.
120
June
151
July
181
Aug.
212
Sept.
243
Oct.
273
Nov.
304
Dec.
January
334
February . .
334
365
28
59
89
120
150
181
212
242
273
303
March ....
306
337
365
31
61
92
122
153
184
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
June
214
245
273
304
334
365
30
61
92
122
153
183
July
184
215
243
274
304
335
365
31
62
92
123
153
August . . .
153
184
212
243
273
3C4
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
33.5
365
If the days of the different months are not the same, the
number of days of difference should be added when the earlier
day belongs to the month from which we reckon, and subtracted
•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 FOR rRACTICE.
2. George Washington was born Feb. 22, 1732, and died
Bee. 14 1799 ; what was his age ? Ans. 67 yr. 9 mo. 22 da.
How can the number of days, if less than a year, be obtained ?
190 SUBTRACTION.
3. How much time has elapsed smee the declaration o^
independence of the United States ?
4. How many years, months, and days from your birthday
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 ? A7is. 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.
SSI. To subtract denominate fractions.
1. From f of an oz. take |^ of a pwt.
OPERATION. Analysis. We per-
foz. nz 7 pwt. 12gr. form the same reduc-
1 p^vt. =r 21 gr. tions as in addition of
denominate fractions,
6 pwt. 15 gr., Ans. . (^jg )^ ^nd then sub-
tract the less value from
Or, f oz. X 20 = -%^- pwt. the greater.
^^- — i = ^^ pwt. =z 6 pwt. 15 gr.
2. What is the difference between J- rod and } of a foot ?
Ans. 7ft. 6 in.
3. From ^ £ take § of f of a shilling.
4. From § of a league take -/^ of a mile.
Ans. 1 mi. 2 fur. 16 rd.
5. From 85^^ cwt. take 1 qr. 2f lb.
Ans. 8 cwt. 2 qr. 14 lb. 5 oz. 157^\dr.
6. From -^ of a week take ^ of a day.
Ans. 1 da. 4 h. 48 min.
Give explanation of the process of subtracting dcDominate fractions.
COMPOUND NUMBERS. 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 •^, of the distance, how far are they apart ?
Ans. 41. mi. 7 fur. 9 rd. 8 ft. 7^ in.
8. From a cask of brandy containing 96 gallons, -i leaked
out, and f of the remainder was sold ; how much still remained
in the cask ? ' Ans, 25 gal. 2 qt. 3^ gi.
MULTIPLICATION.
2S2. 1. A farmer has 8 fields, each containing 4 A. 2 R.
27 P. ; how much land in all ?
Analysis. In 8 fields are 8 times as much
land as in 1 field. "VVe write the multiplier
under the lowest denomination of the mul-
tiplicand, and proceed thus ; 8 times 27 P«
„ are 216 P., equal to 5 R. 16 P.; and we
^ write the 16 P. under the number multiplied.
Then 8 times 2 R. are 16 R., and 5 R added make 21 R., equal to
4 A. 1 R. ; and we write the 1 R. under the number multiplied.
Again, 8 times 4 A are 32 A., and 4 A. added make 36 A., which Me
write under the same denomination in the multipHcand, and the
work is done. Hence,
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.
(2.) (3.)
tn. pk. qt. pt. mi. fur. rd. ft.
4 2 5 1 • 9 4 20 13
2 6
OPERATION.
A.
K.
p.
4
2
27
8
9 13 57 3 4 12
Multiplication of compoimd numbers, how performed ? Rule.
192
MULTIPLICATION.
(4.)
£. s. c\
5 18 4
(5.)
lb. oz. pwt. gr.
3 4 22
4
7
(6.)
T. cwt. lb.
14 16 48
oz.
12
(7.)
13° 10' 35"
11
9
8. In 6 barrels of grain, each containing 2 bu. 3 pk. 5 qt.,
how many bushels ? -^ws. 17 bu. 1 pk. 6 qt.
9. If a druggist deal out 3 lb 4 § 1 5 2 9 16 gr. of med-
icine a day, how much will he deal out in 6 days ?
10. If a man travel 29 mi. 3 fur. 30 rd. 15 ft. in 1 day,
how far will he travel in 8 days ?
11. If a woodchopper can cut 3 Cd. 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
1 T. 3 cwt. 50 lb. ?
OPERATION. Analysis. "When the multi-
T. cwt. lb. plier is large, and a composite
1 3 50 number, we may multiply by one
6 of the factors, and that product
~ ~ ~ by the other. Multiplyinsr the
7 1 00 welghtoteioad. weightoflloadby6,weobtaia
the weight of 6 loads, and the
56 8 00 weight of 48 loads. 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 pt. of
corn, how much will 64 acres produce ? Ans. 2941 bu.
14. How much will 120 yards of cloth cost, at 1 £ 9 s. 8^ d.
per yard ?
15. If $80 will buy 4 A. 3 R. 26 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. 1 1 lb. 10 oz.
COMPOUND NUMBERS. 193
17. The sun, on an average, changes his longitude 59' 8.33"
per (lay ; how much will be the change in 365 days ?
18. If 1 pt. 3 gi. of wine fill 1 bottle, how much will be re-
quired to fill a great gross of bottles of the same capacity ?
DIVISION.
!3S3. 1. If 4 acres of land produce 102 bu. 3 pk. 2 qt. of
wheat, how much will 1 acre produce ?
OPERATION. Analysis. One acre will produce \
pt. bu. pk. qt. pts. as much as 4 acres. Writing the divi-
4)102 3 2 sor on the left of the dividend, we divide
9 - 2 n 1 1^2 bu. by 4, and we obtain a quotient of
25 bu., and a remainder of 2 bu. We
write the 25 bu. under the denomination of bushels, and reduce the
2 bu. to pecks, making 8 pk., and the 3 pk. of the dividend added
makes 11 pk. Di\dding 11 pk. by 4, we obtain a quotient of 2 pk.
and a remainder of 3 pk. ; writing the 2 pk. under the order of
peeks, we next reduce 3 pk. to quarts, adding the 2 qt. of the
dividend, making 26 qt, which divided by 4 gives a quotient of 6 qt.
and a remainder of 2 qt. Writing the 6 qt. under the order of
quarts, and reducing the remainder, 2 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 done.
2. A farmer put 132 bu. operation.
1 pk. of apples into 46 barrels ; '^"* p*^-
how many bu. did he put into ^^ ) 1^2 1(2 bu.
a barrel ?
40
4
When the divisor is large, and
not a composite number, we di- Ibl ( o pfc.
vide by long division, as shown ^^^
in the operation. From these 23
examples we derive the o
184(4qt
Arts. 2 bu. 3 pk. 4 qt
jj p Explain the process of dividing compound numbers.
if
194 DIVISION".
Rule. I. Divide the highest denomination as in simple
numbers, and each succeeding denomination in the same man-
ner, if there be no remainder.
II. If there be 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 be-
fore.
III. The several partial quotients will be the quotient re-
quired.
Notes. 1. When the divisor is large and is a compcstie 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 process is the same as in simple numbers.
EXA
MPLE
S FOR
PRACTICE.
(3.)
£. s. d.
5 ) 25 8 4
T.
7)45
(4.)
cwt.
15
lb.
25
5 18
6
10
75
(5.)
wk. da. h.
4)3 5 22
min.
00
10 ) 25°
(6.)
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. 5 fur. 13 rd. 5 fi. G in.
9. If a family use 10 bbl. of flour in a year, what is the
average amount each day ? Ans. 5 lb. 5 oz. 14^^ dr.
10. The aggregate weight of 123 hogsheads of sugar is
57 T. 19 cwt. 42 lb. 14 oz. ; what is the average weight per
hogsheaf? ? Ans. 9 cwt 42 lb. 10 oz.
11. IIovv many times are 5 £ 10 s. 10 d. contained in 537 £
10 s. 10 d.? Ans. 9 7.
Give the rule. When the divisor is a composite number, how may
we proceed? When the divisor and dividend are both compound
numbers, how proceed ?
COMPOUND NUMBERS. 195
12. A cellar 50 ft. long, 30 ft. wide, and 6 ft. deep was ex-
cavated by 5 men in 6 days ; how many cubic yards did each
man excavate daily? An9. 11 cu.yd. 3 cu.ft.
13. If a town 5 miles sqLuare be divided equally into 150
farms, vvhat will be the size of each farm ?
Ans. 106 A. 2 R. 26 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. 3pk. 4qt.? Ans. 70.
15. A merchant tailor bought 4 pieces of cloth, each con-
taining 60 yd. 2.25 qr. ; after selling ^ of the whole, he made
up the remainder into suits containing 9 yd. 2 qr. each ; how
many suits did he make ? Ans* 17.
LONGITUDE AND TIME.
394. 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 ; and 1° of
distance in ^^ of 1 hour, or 4 minutes; and 1' of distance in
■^\j of 4 minutes, or 4 seconds.
TABLE OF LONGITUDE AND TIME.
360° of longitude z= 24 hours, or 1 day of time.
15° " '' := 1 hour " "
1° " " =3 4 minutes « «
1' " " =: 4 seconds " «
CASE I.
225, To find the diiference 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 of
longitude and time. Case I is what }
OPERATION.
87°
30'
71
3'
16°
21'
4
196 LONGITUDE AND TIME. .
Analysis. By subtraction of
compound numbers we first find
the difference of longitude be-
tween the two places, which is
16° 27'. Since 1° of longitude
makes a difference of 4 minutes
1 h. 5mm.4Tsec., Ans. ^J^^' '"'^ / "^ '""fc'''"*?^ "
' 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 difierence in
time. Hence the
Rule. Multiple/ the difference of longitude in degrees and
minutes hy 4, and the product will be the difference of time in
minutes and seconds, which may he reduced to hours.
Note. If one place be in east, and the other in west longitude, the
difference of longitude is found by adding them, and if the svun be
greater than 180°, it must be subtracted from 360°.
EXAMPLES FOR PRACTICE.
2. New York is 74° 1' and Cincinnati 84^ 24' west longi-
tude ; what is the difference of time ? Ans. 41 min. 32 sec.
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. Petersburg 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 ?
Note. Since the sun appears to move from east to west, when it is
exactly 12 o'clock at one place, it will be past 12 o'clock at all places
east, anf' befyre 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 bo found by adding their difference to the
given time, if it be east, and by subtracting if it be icest.
Ans. 6 h. 51 min. 56 sec, A. M.
Give explanation. Rule.
COMPOUND NUMBERS. 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 sec, P. M.
8. If, at a presidential election, the voting begin at sunrise
and end at sunset, how much sooner will the polls open and
close at Eastport, Me., 67^ west, than at Astoria, Oregon, 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 II.
3^6. 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 Cincin-
nati 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 1' of
ITT 77^ . longitude, there will be ^ as many de-
, ^fis. grees of longitude as there are minutes
of time, and \ as many minutes of longitude as there are seconds of
time. Hence,
Rule. Reduce the difference of time to minutes and sec-
onds^ and then divide by A:', 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 Washington
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 rain. 36 sec. past 8, A. M., at New Orleans, west ;
what is the difference of longitude ? Ans. IIO" 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 longi-
tude 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. When the days are of equal length, and it is noon on
the 1st meridian, on what meridian is it then sunrise? sun-
set ? midnight ? Ans. 90^ west ; 90 ' east ; 180 ' east or west
DUODECIMALS.
22T. Duodecimals are the divisions and subdivisions of
a unit, resulting from continually dividing by 12, as 1, y^, y^j,
ttVs' ^^' ^^ 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 di-
vided by 12 give fourths; and so on.
From these divisions of a foot it follows that
1' (inch or prime) is yV of a foot.
1'' (second) or -j^ of ^, . " ^^^ of a foot.
r' (third) or yV of yV of yJj, . . « y^^ of a foot, &c.
TABLE.
12 fourths, mai'ked {f"'), make 1 third marked V"
12 thirds " 1 second, " 1"
12 seconds « i prime, or inch, " 1'
12 primes, or inches, " 1 foot, « ft.
Scale — 'jn'^ormly 12.
The marks ', '', ''\ "\ are called indices.
What are duoHecimals ? To what applied ? Explain the divisions
of the foot. Repeat the table.
COMPOUND NUx^IBERS. 199
Note. Duodecimals are really common fractions, and can always
be treated as such ; but usually their denom'jiators are not expressed,
and they are treated as compound numbers.
Addition and Subtraction of Duodecimals.
SS§. We add and subtract duodecimals the same as other
compound numbers.
examples.
1. Add 13 ft. 4' 8'', 10 ft. G 7", 145 ft. 9' 11".
Ans. 169 ft, 9' 2".
2. Add 179 ft. IV 4", 245 ft. 1' 4", 3ft. 9' 9".
Ans. 428 ft. 10' 5".
3. From 25 ft. 6' 3" take 14 ft. 9' 8". Ans. 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. lift. 10' 7".
Multiplication of Duodecimals.
3^9. Length multiplied by breadth gives surface, and
surface multiplied by thickness gives solid contents (108).
1. How many square feet in a board 11 feet 8 inches, long
and 2 feet 7 inches wide?
Analysis. We first multiply by the 7'.
7 twelfths times 8 twelfths equals 56 one
hundred forty-fourths, which equals 4
twelfths and 8 one hundred forty-fourths.
We WTite the 8 144ths — marked with two
indices — to the right, and add the 4 12ths
30 ft. 1' 8" ^° *^^ next product 7' times 11 equals
77', which added to 4' equals 81', equal to
6 feet and 9'. We %vrite 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 1 1 feet are 22 feet, and
1 foot added make 23 feet, which we write under the 6 feet. Add-
operation.
lift. 8'
2 7'
6 ft.
23
9' 8''
4'
How are duodecimals added and subtracted ? Give analysis of ex-
dmple 1.
200 DUODECIMALS.
ing these partial products, and we have 30 ft. 1' and 8" for the
entire product.
It will be seen from the above that the number of indices to every
product of any two factors is equal to the sum of the indices of those
factors ; thus 7' X 8' z=: 56" ; 4'^ X ^'" — 20'^'". Hence the
HuLE. I. Write the several terms of the multiplier under
the corresponding terms of the multiplicand.
II. Multiply each term of the multiplicand hy each term of
the multiplier, beginning with the lowest term in each, and call
the product of any tiuo denominations the denomination denoted
hy the sum of their indices, carrying 1 for every 1 2.
III. Add the partial products, carrying 1 for every 12 ;
their sum will be the required answer,
EXAMPLES FOR PRACTICE.
2. How many square feet in a board 13 ft. 9' long and IV
wide? Ans. 12ft. 7' 3".
3. How many square feet in a stock of 4 boards, each 1 1 ft.
9' long and 1 ft. 3' wide ? Ans. 58 ft. 9'.
4. How many square yards of plastering on the walls of a
room 12ft. 11^ square, and 9 ft. 3-^ high, allowing for two win-
dows 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 18ft. 6' long, 12ft.
wide, and 5 ft. 6' high ? ^ns. 9 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. 156cu.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^.
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.
Give the rule.
18
9'
2
2
%' 3"
2' 3"
COMPOUND NUMBERS. 20!
Division op Duodecimals.
S8©. 1. A flagstone, 3 ft. 9' wide, has a surface of 20 ft
ir 3''; what is its length?
OPERATION. Analysis. We divida
3 ft. 9' ) 20 ft. 11' 3" ( 5 ft. T, the surface by the width
to obtain the length. The
divisor is something more
than 3 ft., and to obtain
the first quotient figure, we
consider how many times
3 ft. and something more is contained in nearly 21ft. (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 do^vn 3", the last term of the divi-
dend. "We next seek how many times the divisor is contained in
this remainder, and find by trial the quotient 7 ; multiplying the
diviaor by this figure, we obtain 2 ft. 2' 3", and there is no remain-
der. Hence the
Rule. I. Write the divisor an the left hand of the dividend,
as in simple numbers,
II. J'ind the first term of the quotient either hy dividing the
first term of the dividend hy the frst term of the divisor, or by
dividing the frst two terms of the dividend by the frst two
terms of the divisor ; multiply the divisor by this term of the
quotient, subtract the product from the corresponding terms of
the dividend, and to the remainder bring down another term of
the dividend.
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 16ft. 8'o Ans. 2 ft. 8\
3. The square contents of a walk aro 184 ft. 3', and the
length is 40 fl. 11' 4" ; what is the width? Ans. 4fl. 6'.
, 4. A blanket whose squaro contents are 14 fl. 6', is to be
lined with cloth 2 ft. 1' wide ; how much in length will be re-
quired ?
Give analysis of example 1. Rule.
9*
202 PROMISCUOUS EXAMPLES.
5. A block of granite contains 64 ft. 2' 5" ; its width is
2 ft. 6', and its thickness 3 ft. T ; what is its length ?
Note. Since the solid contents arc the product of the three dimen-
sions, we divide the solid contents by any two dimensions or by their
product, to obtain the other dimension.
Ans, 7 ft. 2'.
PROMISCUOUS EXAMPLES.
1. In 115200 grains Troy, how many pounds.^
2. In &65 da. 5 h. 48 min. 46 sec, how many seconds ?
A71S. 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. Reduce 295218 inches to miles.
5. Keduce 456575 grains to pounds, apothecaries' weight
Ans. 79 1b 3 S 13 IB 15 gr.
6. How many sheets in 3 reams of paper ?
7. What is the value of 4 pilei 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? Ans. 112.
9. At $2G.40 per sq. rd. for land, what will be the cost of a
village lot 8^ rd. long, and 4^ rd. wide ? Ans. $980.10.
10. Divide 259 A. 1 R. 10 P. of land into 36 equal lots.
Ans. 7 A. 32^ P.
-41. How many times can a box holding 4 bu. 3 pk. 2 qt. be
miedfrom 336bu. 3 pk. 4qt.? Ans. 70.
12. What is the value of .875 of a gallon ?
13. What part of a mile is 2 fur. 36 rd. 2 yd. ? Ans. -^.
14. What part of 2 days is 13 h. 26 min. 24 sec. ?
15. From 26 A. 2 R. of land, 5 A. 3 R. were sold j what
part of the whole piece remained un.«old ? Ans. ■^^.
16. What is the difference bet wren f of a pound sterling
and 5^ pence? Ans. 11 s. 6 J d.
17. "V^Tiat is the sum of f of a yard, ^ of a foot, and -f of
an inch ? Ans. 7 inches.
PROMISCUOUS EXAMPLES. 203
• 18. Reduce S cwt. Iqr. 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 g how much older was Franklin
than Washington? A7is. 26 yr. 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 furlong, and r\ of a rod to-
gether. Ans. 5 fur. 33 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. 2 R. 29.9 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 1 J feet
in thickness ? i^ J a&fC^ ^ns. J^^. (j / 7
27. A, B, and C rent a farm, and agree to woi-k it upon '
shares ; they raise 640 bu. 3 pk. of gmin, which they divide
as follows : one fourth is given for the rent ; of the remainder
A takes l^ 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. 11.4 dr.
avoirdupois weight? Ans. 161b. 5 ozo 10 pwt. 11.7 +gr.
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. What fraction of 3 bushels is -/_ of 2 bu. 3 pk. ?
Ans. yW
32. How many wine gallons in a water tank 4 feet long,
3^ feet wide, and 1 ft. 8 in. deepP Ans. 174y\.
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 -[- hu.
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 yaixi ?
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 ?
Ans. 181-5^" cu. yd.
38. What will 30 bu. 54 lb. of wheat cost, at $1.37^ per
bushel? Ans. $42.4875.
39. How many square yards of carpeting will it take to
cover a floor 24 ft. 8' long and 18 ft. 6' wide ? Ans. 504f .
40. What is the cost of 54 bu. 8 lb. of barley, at 84 cents
per bushel ? Ans. $45.50.
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? Ans. $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.
PERCENTAGE.
205
PEROExVrAGE.
231 • Per cent, is a term derived from the Latin words 'per
centum^ and signifies hy the hundred, or hundredths, that is, a cer-
tain number of parts of each one hundred parts, of whatever de-
nomination. Thus, by 5 per cent, is meant 5 cents of every 100
cents, $5 of every $100, 5 bushels of every 100 bushels, &c.
Therefore, 5 per cent, equals 5 hundredths = .05 = y^^ ziz ^V
8 per cent, equals 8 hundredths z=i .08 =: yf g- z= ^^,
2S2, Percentage 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.
334:« 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 TemiB.
1 per cent. -
= .01
r=
TTiT =
= tk
2 per cent. '
* .02
«
tIt *
A
4 per cent.
' .04
«
ITW
^
5 per cent.
* .05
11
ih •
A
6 per cent.
' .06
tIt
*
7 per cent.
' .07
IrtO
Tinr
8 per cent.
' .08
lod
A
10 per cent.
' .10
T% '
A
16 per cent.
' .16
T% '
ih
20 per cent. *
' .20
m '
i
25 per cent.
* .25
m •
i
50 per cent.
' .50
T% '
i
100 per cent.
' 1.00
m '
1
125 per cent.
' 1.25
m ;
i
^ per cent.
* .005
T¥oT
"Soir
f per cent.
" .0075
TlFoo^
^
12^ per cent.
.125
t'/.V '
' i
16J per cent.
' .1625
^VA •
H
^V^lat is meant by per cent. ? From what is the term derived ?
^Vhat is percentage ? What is the base of percentage ? How is per
cent, expressed?
206 PERCENTAGE.
EXAMPLES FOR PRACTICE.
14
per
1. Express decimally 3 per cent. ; 6 per cent. ; 9 per cent. ;
: per cent. ; 24 per cent. ; 40 per cent. ; 112^ per cent. ; 150
jt^^r cent.
2. Express decimally 6 J- per cent. ; 8 J per cent. ; 33^ per
cent. ; 7^ per cent. ; lOf per cent. ; 9f per cent. ; 103^ per
cent. ; 225 per cent.
3. Express decimally ^ per cent. ; f per cent. ; § per cent.;
•| per cent. ; | per cent. ; 1^ per cent. ; 2f per cent. ; 4^ per
cent.; 5f per cent.; 7|- per cent.; 121- per cent.; 25 f per
cent.
4. Express in the form of common fractions, in their lowest
terms, 6 per cent. ; 8 per cent. ; 12 per cent. ; 14^ per cent. ;
18 1 per cent. ; 21 1 per cent. ; 31^ per cent. ; 37^ per cent. ;
40| 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 Analysis. Since 4 per cent, is 3-^ = .04, he lost
^04 .04 of $125, or $125 X .04 = $5. Or, 4 per cent.
'- is ^±^ =z ^, and ^V of $ 1 25 = $5. Hence the
$5.00
Rule. Multiply the given number or quantity by the rate
per cent, expressed decimally, and point off a^ in decimals. Or,
Take such a part of the given number as the number ex-
pressing 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,
PERCENTAGE. 207
4. What IS 7i per cent, of $56.75 ? Ans. $4.11/5.
5. What is 12^ per cent, of 2450 pounds ?
Ans. 306.25 pounds.
6. What is 6f per cent, of 19072 bushels ?
Ans. 1287.36 bushels.
7. What is 33^ per cent, of 846 gallons ?
Ans. 282 gallons.
8. What is 9| per cent, of 275 miles? A)is, 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 $12967 ?
13. What is 25 per cent, of ^ ?
25 per cent.- equals -^\% r= J, and | X i = -^, A71S.
14. What is 15 per cent, off? , Ans. y^.
15. What is 2^ per cent, of 6| ? Ans. |.
16. What is 33^ per cent, of y9^ ? Ans. -^\.
17. What is 84 per cent, of 7^? Ans. 6yV
18. Find f 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^ per cent, of his capital, and the other gained
75 per cent. ; how much more did the one gain than the other ?
Ans. $1150.
24. A man, owning ^ of an iron foundery, 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 ^l
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 remainder ;
and at another time 12^ per cent, of what he owed after the
second payment; how much of the debt* did he still owe ?
Ans. $226.56J.
CASE n.
S36. To find what per cent, one number is of an-
other.
1. A man, having $125, lost $5 ; what per cent, of his
money did he lose ?
OPERATION. Analysis. We multi-
5 -r 125 = .04 rr: 4 per cent. ply the base by the rate
Qj. per cent, to obtain the
tJ J = 5V = -04 = i per cent P^'-'^'^ntege (235) ; con-
^^^ '''' ^ versely, we divide the per-
centage by the base to obtain the rate per cent. Or, since $125 is
100 per cent, of his money, $5 is ^^, equal to t^ of 100 per cent,
which is 4 per cent. Hence the
Rule. Divide the percentage hy the base, and the quotient
will he 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. 12^
4. What per cent, of $750 is $1 65 ?
5. What per cent, of $240 is $13.20 ? Ans. 5^
6. What per cent, of $2 is 15 cents ?
7. What per cent, of 6 bushels 1 peck is 4 busheU 2 pecks
6 quarts ? Ans. 75 per cent
8. What per cent, of 1 5 pounds is 5 pounds 1 ounces
avoirdupois weight ? Ans. 37^ per cent,
9. What per cent, of 250 head of cattle is 40 head ?
Case II is what ? Give explanation, Bule.
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 ^ of it at one
time, and ^ of the remainder at another time ; what per cent,
remained unsold ? Ans. 50 per cent.
CASE III.
337. To find a number when a certain per cent, of
it is given.
1. A man lost $5, which was 4 per cent, of all the money
he had ; how much had he at first ?
OPERATION. Analysis. We are here required to
$5 _^ .04 = $125. find the base, of which $5 is the per-
Qp centage. Now, percentage equals base
4 V 100 ^125 multiplied by the rate per cent; con-
■* ' versely, base equals percentage divided
by rate per cent. Or, $5 is 4 per cent, of all he had ; \ of $5, or J,
equals 1 per cent, of all he had, and 100 times f equals 100 per
cent., or all he had. Hence the
Rule. Divide the percentage ly the rate per cent., ex-
pressed decimally, and the quotient will he the base, or number
required. Or,
Take 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 2f per cent, of what number ? Ans. 1200.
6. $281.25 is 37^ per cent, of what sum of money ?
Ans. $750.
7. A farmer sold 50 sheep, which was 20 per cent, of his
whole flock ; how many sheep had he at first ?
Case in is what ? Give explanation. Rule.
210 PERCEirrAGE.
8. I loaned a man a certain sum of money ; at one time
he paid me $59.75, which was 12^ 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 foundery, sell
25 per cent, of his share for $1246.50, what is the value of
the whole foundery ? 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 i Ans. 3000 bushels.
CASE IV.*
238. To find a number when the number, increased
by a certain per cent, of itself, is given.
1. A man's income this year is $525, which is 5 per cent,
more than it was last year ; what was it last year ?
OPERATION. Analysis. Since his income
$525 — 1.05 = $5.00. this year is .05 more than it
was last year, this year's income 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. Hence the
Rule. Divide the amount hy 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.
* The changes that have been wrought in financial and commercial operations
the past few years, require some new methods and applications in the subject of
percentage. Several pages have therefore been inserted in this worlc, including U. 8.
Securities, Stoclc, and Gold Investments, and their comparative values in commer-
cial transactions sliown by practical examples, &c.
All the changes tliat have been made, and the new matter inserted, are contained
in the following fdxtfen pages, except the repaging, and renumbering the Articlea,
which is continued througli the remainder of tlie book.
What is Case IV. ? Explanation ? Rule ?
PERCEinrAGE. 211
EXAMPLES FOR PRACTICE.
2. What number increased by 18 per cent, of itself is
equal to 1475? Ans. 1250.
3. A merchant sells broadcloth for $4 per yard, and thereby
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 ?
6. A certain manufacturing company have sold $432,250
worth of goods, which is 8 Jg per cent, more than they sold
last year. How much did they sell last year ? Ans. $400,000.
6. A merchant bought a stock of goods, and paid 4| per
cent, of the purchase price for freight, when he fonnd that the
goods cost him $8757. What was the purchase price ?
7. A merchant increased his capital by 20 per cent, each
yeai for two years, when he found he had $9360 invested.
How much had he at first ? Ans. $6500.
CASE V.
339. To find a number when the number, diminished
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. Analysis. Since the man lost
368 -i- .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 .92 and obtain the required number of sheep.
Hence the
Rule. Divide the given number by 1 minus the rate ex-
pressed decimally, and the quotient will be the base or number
required. Or,
Take as miny times 100 as the given number is times 1
minus tne rate.
What i3 Case Y. ? Explanation ? Rule ?
212 " PERCENTAGE.
EXAMPLES FOR PRACTICE.
2. What number diminished 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 ^ 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 BROKERAGE.
340. An Agent, Factor, or Broker is a person wiio
transacts business for another, or buys and sells money, stocks,
notes, <fec.
341. Commission is the percentage, or compensation
allowed an agent, factor, or commission merchant, for buying
and selling goods or produce, collecting money, and transact-
ing other business.
943. 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.
Note. The rates of commission and brokerage are not regulated
by law, but are usually reckoned at a certain per cent, upon the
money employed in the transaction.
Define an agent, factor, or broker. What is meant by commis-
gion ? Brokerage ?
COMMISSION AND BROKERAGE. 213
CASE I.
343. To find the commission or brokerage on any
sum of money.
1. A commission merchant sells butter and cheese to the
amount of $1540 ; what is his commission at 5 per cent. ?
oPERATiox. Analysis.
1540 X -05 :i= $77, Ans. Since the com-
Or, j^,j = ^L , and g-V X 1540 =z $77. mission on $1 is
5 cents or .05 of
a dollar, on $1540 it is $1540 X .05=: $77. Or, since 5 per cent
is j^z=z-^^ of the sum received, the commission is -j^ of $1540
z=: $77. Hence the
Rule. Multiply the given sum ly the rate per cent, ex-
pressed decimally, and the result will be the com,mission or bro-
kerage. Or,
Take such a part of the given sum as the number expressing
the per cent, is part of 100.
EXAMPLES FOR PRACTICE.
A commission merchant sells goods to the amount of
$6756 ; 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 14- 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^ per cent. ?
7. A broker negotiates a bill of exchange of $2890 for f
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 commis-
sion of 2 J per cent. ; what amount does he receive ?
A71S. ^235.40.
9. If I sell 400 bales of cotton, each weighing 570 pounds,
at 9 cents a pound, and receive a commission of 2^ per cent.,
how much do I make by the transaction ? Ans. $461.70.
10. A commission merchant in New Orleans sells 450 bar-
rels of flour at $7.60 a barrel ; 38 firkins of butter, each con-
taining 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 5i per centl ? Ans. $242,308.
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.
A^r'-rAn 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 II.
344. To find the commission or brokerage, when
it is to be deducted from the given sum, and the bal-
ance 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 -7- 1.05 = $1200, Invested.
$1260 — $1200 = $60, commission.
O/, m + ihu = U ; ^1260 -^§h = S1200, inveeted;
And $1260 $1200 = $60, commission.
Case 11 is what ? Give explanation. Rule.
COMMISSION 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
tlie difference between the given sum and the sum invested is his
commission.
Or, the money expended is 1^ of itself, the commission is ^^ of
this sum, and the commission added to the sum expended is if| of
the whole sum. Since $1260 is \^ = f ^,$1260 -^ U = ^^200,
the sum expended; and $1260 — $1200=: $60 the commission.
Hence the
Rule. T. Divide (he given amount hy 1 increased by the rate
jper cent, of commission, and the quotient is the sum invested.
II. Subtract the 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 commission
of 2 per cent; how much is his commission? Ans. $63.65.
3. What amount of stock can be bought for $9682, and al-
low 3 per cent, brokerage ? Ans. $9400.
4. A flour merchant sent $10246.50 to his agent at Chica-
go, 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 4^ per cent. ; what will be
the amount of the purchase ?
6. Remitted $603.75 to my agent in New York, for the pur-
chase 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 1^ per cent, additional for guaranteeing its delivery at
a specified time ; how much will he pay out, and what are
his fees ? Ans. Fees, $403,758.
216 PERCEin^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 commis-
sion ? Ans. $13604 invested; $238.07 commission.
9. A broker received $10650, to be invested in stocks after
deducting | per cent, for brokerage; what amount of stock
did he purchase ?
STOCK-JOBBING.
24:5, A Corporation is a body authorized by a general
law, or by a special charter, to transact business as a single
individual.
340. A Charter is the legal act of incorporation, and de-
fines the powers and obligations of the incorporated body.
247. A Firm is the name under which an unincorporated
company transacts business.
248. Capital or Stock is the property or labor of an in-
dividual, corporation, company, or firm ; it receives different
names, as Bank Stock, Railroad Stock, Government Stock, &c.
249. A Share is one of the equal parts into which the
stock is divided.
250. Stockholders are the owners of the shares.
251. The Nominal or Par Value of stock is its first cost,
or original valuation.
Note. The original value 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.
Stockholders. Par value. At par. Above par. Below par.
STOCKS. 217
355. The Market or Real Value of stock is what it will
bring per share in money.
256. A Bividend is a sum paid to stockholders from the
profits of the business of the company.
257. An Assessment is a sum required of stockholders to
meet the losses or expenses of the business of the company.
St58. 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.
So9. A Stock Broker is a person who buys and sells stocks,
either for himself, or as the agent of another.
SOO. The calculations in stock-jobbing are based upon the
following relations :
I. Premium, discount, and brokerage are each a percentage,
computed upon the par value of the stock as the base,
II. The market value of stock, or the proceeds of a sale, is
tbe amount or difference, according as the sum is greater or less
than the par value.
Note 1.— In all examples relating to stocks, $100 will be considered a share, unless
otherwise stated.
CASE I.
361. To find the value of stock, when at an advance,
or discount.
1. What will $3240 of Bank Stock cost at 8 per cent,
advance, brokerage \ per cent. ?
OPERATION. Analysis. To find the
$1 + .08 = $1.08. P"'" f 'f "^' "^^ ^^^ ^^^
rate of advance to, or sub-
11.08 4- .0025 = $1.0825. ^ract tlie discount from, $1;
$1.0825 X 3240 = $3507.30. to this result add the
brokerage, and we have the cost of $1. Hence $3240 stock will cost
8240 times $1.0825. Hence the
Rule. 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.
R.P. 10
218 PERCEin:AGE.
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
• Central 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 1150
per share, brokerage ^ per cent ? Ans. $7632.
5. What costs $5364 stock in the 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 4| per cent, premium, the par value being $1000 per
share, brokerage | 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^
per cent.; how much did I gain? A7is. $1113.
9. What is the market value of l75 shares of stock in the
Suftblk Bank, at | per cent, advance? Ans. $17631.25.
10. Bought 75 shares of stock in the Bank of New Orleans
of $50 each, at 3 per cent, discount, and sold it at 2 1 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 12i 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 2^ per cent., and sold the same at an advance of 12
per cent, ; what was my gain ? Ans. $1-740.
STOCKS. 219
CASE II.
SOS. 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 85150 ?
OPERATION. Analysis. Since the stock
$5150-4- 1.03 = $5000 = is at 3 per cent, advance, $1
50 shares, Ans, ^^ «*°^^ ^* P^' ^^^^ ^^^ ^^-^^ J
and if we divide $5150, the
whole sum to be expended, by 1.03, the cost of $1 of stock, the
quotient must be the amount of stock purchased. Hence the
Rule. Divide the given sum hy the cost of ^l of stocky
and the quotient will be the nominal amount of stock purchased.
2. How many shares of raih'oad stock, at 5 per cent, ad-
vance, can be purchased for $6300 ? Ans. 60 shares.
3. I invested $6187.50, in Ocean Telegraph stock, at 10
per cent, discount ; how much stock did I purchase ?
Ans. $6875.
4. I sent my agent $53500 to be invested in Illinois 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 INVESTMENTS.
^03* The net earnings of a corporation are usually divid-
ed among the stockholders, in semi-annual dividends. The
income of capital stock is therefore fluctuating, being depend-
ent upon the condition of business ; while the income arising
from bonds, 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 PERCENTAGE.
^64. Federal or United States Securities are of two
kinds ; viz., Bonds and Notes.
Bonds are of two kinds.
Firsts Those which are payable at a fixed date, and are
known and quoted in conimercial transactions by the rate of
interest they bear, thus : U. S. 6's, that is, United States
Bonds bearing 6 ^ interest. *
Second^ Those which are payable at a fixed date, but which
may be paid at an earlier specified time, as the Government
may elect. These are known and quoted in commercial transac-
tions by a combination of the two dates, thus : U. S. 5-20's ;
or a combination of the rate of interest and the two dates,
thus : U. S. 6's 5-20 ; that is, bonds bearing 6 % 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 diff'erent issues which bear the same rate of interest,
this is done by adding the year in which they become due,
thus : U. S. 6's of '71 ; U. S. 5's of '74 ; U. S. 6's 5-20 of
'84 ; U. S. 6'8 5-20 of '85.
Notes are of two kinds.
Firsty Those payable on demand, without interest, known
as United States Legal-tender Notes, or, in common language,
" Greenbacks."
Second^ Notes payable at a specified time, with interest,
known as Treasury Notes. Of these there are two kinds, — .
6 % Compound-interest Notes, and Notos bearing *J-^^ ^ in-
terest, the latter known and quoted in commercial transac-
tions as 7.30's.
The nomenclature here explained is that used in com-
* The character % generally employed ia business transactions signifios per
eenU; tlius 6 % signifies 6 per cent.
What are United States Securities composed of? Explain the
different kinds of bonds. Of Notes. In what is the interest on
each payable ?
STOCKS. 221
mercial transactions, which involve similar Securities of States
or Corporations.
The interest on all bonds is payable in gold.
Tbe interest on notes is payable in Legal-tender Notes.
When Bonds or Stocks are sold, a revenue stamp must be
used equal in value to one cent on each $100, or fraction of
$100, of their currency value. If sold by a broker, this is
chai'ged to the person for whom they are sold.
The following are the principal United States Securities: —
BONDS.
U. S. 4's (New) of 1901.
U. S. 5-20's, due in 1882, interest 6%,
U. S. 5-20's, due in 1884, interest 6%.
U. S. 5-20'8, due in 1885, interest 6%.
U. S. 10-40's, due in 1904, interest 5fo.
Pacific Railroad G's of 1895.
Pacific Raihroad 6's of 1896.
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. 5'8(New)ofl881.
U.S. 4^8 " of 1886.
NOTES.
Compound-interest Notes of 1867. I 7.30 Notes of 1867.
Compound-interest Notes of 1868. | 7.30 Notes of 1868.
CASE I.
36«5. To find what income any investment will produce.
1. What income will be obtained by investing $6840 in
stock bearing 6 ^, and purchased at 95 ^?
OPERATION. Analysis. We
$6840 ~ .95 = $7200, stock purchased. <iivide the invest-
$7200 X .06 = $432, annual income. ^^^*' ^^^f?' ^^
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. Hence the
Rule. F'ind how much stock the investment will purchase^
and then compute the income at the given rate upon the par
value.
What is the value of stamps to be attached to any bond when
sold? Name the different kinds of bonds. Of Notes. What is
Case I. ? Explanation ? Rule ?
222 PERCEin:AGE.
2. If I invest $867 in U. S. 6-20's of 84 at 102 %, what
income will I receive on my investment? Ans. $51.
3. What will be my yearly income, if T invest $8428 in
U. S. 10-40's, at 98 ^? Ans. $430.
4. How much stock at a premium of 4| % can be bought
for $10500, brokerage \ % ? Ans. $10000.
5. If A. inv.est $4795 in Maryland 5's at 87 %, broker-
age i %i what will be his yearly income ? Ans. $274.
6. Having $10476 to invest, I find I can purchase U. S. 6's
at 107^ %, and U. S. 5-20's of '81 at 96 A %, brokerage ^ %, in
each instance. How much more will I receive yearly by in-
vesting 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 \ % for purchasing ; will his
yearly income be increased or diminished, and how much?
Ans, Increased $68.75.
CASE II.
366. To find what sum must be invested to obtain a
given income.
1. What sura must be invested in Virginia 6 ^ bonds, pur-
chasable 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 obtain
$.05 income, to
obtain $G00 will require $G00 ^ .05 = $12000 (Case 1). Multiply-
ing 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
bo invested. Hence the
Rule. I. Divide the given income by the per cent, which the
stock pays ; the quotient will be the par value of the stock re-
quired.
What is Case II ? Explanation ? Rule ?
STOCKS. 223
11. Multiply the par value of the stock by the market value
of one dollar of the stock ; the product will he the required
investment,
EXAMPLES FOR PRACTICE.
2. If N. Y. 6's are 5 % below par, what sum must be in-
vested in this stock to obtain an income of |840 ? Ans, $13300.
3. What sura must be invested in U. S. 10-40's at 98 1 ^,
brokerage \ % for buying, to secure an annual income of
$1860? Ans. $36642.
4. When U. S. 5-20's of *81 are quoted at 108j, what sum
must I invest to secure an annual incomeof $1080, brokerage
\%\ ^715. $23436.
5. If I sell $25000 TJ. S. 5-20's of '82 at 93| %, and
invest a sufficient amount of the proceeds in U. S. 6's, at
109| % to yield an annual income of $960, and buy a house
with the remainder, how much will the house cost me?
Ans, $5957.50.
CASE III.
367. To find what per cent, the income is of the in-
vestment, 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
OPERATION. will cost $1.05, 2iii6. pay $.07, the in-
.07 -T- 1.05 = 6| %. CO™® is T^5 = 6| ^ of the invest-
ment. Hence the
Rule. Divide the annual rate of income which the stock
bears by the price of the stock ; the quotient will be the rate
upon the investment.
EXAMPLES FOR PRACTICE.
2. What is the rate of income upon money invested in
6 % bonds, purchased at 87 per cent.? Ans. 6§| ^.
What is Case III? Explanation? Rule?
\
224: PEBCENTAGE.
3. What per cent, on his money will a man receive an-
nually if he invest in N. Y. 6's at 105 % ? Ans, ^ %,
4. What is the rate of income upon money invested in
Missouri 6's at 75 ^ ? Ans, 8 %.
5. Purchased U. S. 6-20's of '84 at 107| %, brokerage \ %\
what is the income on the investment ? Ans. of %,
6. Which is the better investment, U. S. 10-40's at 98| ^,
or U. S. 5-20's of '85 at 108| %, brokerage \%m each?
CASE IV.
268 c To find the price at which stock must be pur-
chased to obtain a given rate upon the investment.
1. At what price must 6 % stocks be purchased in order to
obtain 8 % income on the investment ?
OPERATION. Analysis. Since $.06, the income
|.06 -^ .08 = $.75 o* $1 of the stock, is 8 .^ of the sum
paid for it, wo have (235) $.06 -4-
.08 = 75 %y the purchase price. HencO;
Rule. Divide the annual rate of income which the stock
bears hy the rate required on the investment ; the quotient will
be the price of the stockr
V
EXAMPLES FOR PRACTICE.
2. What must I pay for Missouri C's that my investment
may yield 9 ^ annually ? Aris. 66 ^ %.
3. What rate of premium does 6 % stock bear in the market
when an investment pays 5 ^ ? Ans. 20 %,
4. At what rate must I buy TJ. S. 10-40'8 that I may re-
ceive 6 ^ on my investment? Ans. 83] ^.
5. At what rate of discount must U. S. 5-20's of '81 be
purchased that I may secure 7 % annually on the investments
Ans. 28;J ^.
What is Case IV? ExplanaUon? Rule?
STOCKS. 22S
GOLD INVESTMENTS.
369. Currency is a terra 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 exchange
of commodities between various nations. Second, To denote
•whatever circulating medium is used in any country as a sub-
stitute for the government standard. In this latter sense, the
paper circulating medium, when below par, is called C urrency ,
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 circulating medium, is made the
standard, while it is virtually below par.
CASE I.
270. To change gold into currency.
1. How much currency can be bought for $150 in gold
when gold is at 170 ^?
OPERATION. Analysis. Since a dollar of gol^
$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
is the amount of currency which can be purchased for $150 in gold.
Rule. Multiply the value of one dollar of gold in currency
by the number of dollars of gold.
2. What is the value in current funds of $250 gold, when
gold is 147 %'i Ans. $367.50.
3. What is the value in current funds of $320.50, when
gold is 137^ %'i Ans. $440.68|.
What is Currency ? Case I. ? Explanation ? Rule ?
10*
aJeS PERCENTAGE.
4. When gold is at a premium of 33 ^, how much will
2500 ^n gold cost ? Ans. $3325.
5. A holds $8000 U. S. 10-40's ; what is his annual income
in currency if gold is 138 ? Ans. $552.
6. What is the yearly income in currency from $9500 of
U. S. 5-20's of '84 when gold is 140 I Ans. $798.
7. A. purchased a house, for which he was to pay $4500 in
currency, or $3000 in gold al his option. Will he gain or
lose by accepting the latter offer, gold being 147^ %y and how
much in currency ? Ans, Lose $75.
CASE II.
371* To change currency into gold.
1. How much gold can be purchased for $75 current funds,
gold being at 150 ^1
Analysis. A dollar of gold cost f 1.50
OPERATION. in currency, therefore there can be as
^75 -L. $1.50 = 50. many dollars of gold purchased for $75 in
currency as $1.50 is contained times in $75.
Hule. 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^ % ? Ans. %.1'U\^^.
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., 64f %,
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 147|"; how much gold did I purchase?
Ans. $3600.
7. What is the value in gold of a dollar, in currency,
when gold is at 145 %'i Ans. $.68§§.
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 ? Rule ?
PKOFIT AKD 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? ^cJT^ ^2z_^<^
11. Bought flour at $11.75 per barrel in currency,/^hen
gold was at 150 ^, and afterwards sold it at $10.25 in gold,
when gold was 135 %\ did I gain or lose, and how much, on
a sale of 300 barrels ? Ans. ^^ ^jfez IT y 2* 2
12. Which is the better investment, a bond and mortgage /c^-^*^
at 7 %, or U. S. 5-20's of '84 at par, gold being 140 ^ ; and
what per cent, in gold ? Ans. U. 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
TJ. S. 10-40's at 70 % in gold. Will my yearly income be
increased, or diminished by the transaction, and how much
in gol4 ? Ans. Increased $45.
PROFIT AND LOSS.
3TS» Profit and Loss are commercial terms, used to ex-
press the gain or loss in business transactions, which is usually
reckoned at a certain oer cent, on the prime or first cost of
articles.
CASE I.
373. To find the amount 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 20 _ 1 . d^TOK y 1 __ ^oH $135 will gain $135 X .20
^r, T^^ - ^, $135 X ^ - $27. ^ ^27. Or, since 20 %
equals -^^ = i, the whole gain will be ^ of the cost. Hence the
following
Rule. Multiply the cost hy the rate per cent, expressed
decimally. Or,
Take such part of the cost as the rate per vent, 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 $84.80, and sold
it at 12^ 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 1 6 J per cent, advance on the cost of
the wheat; what was his gain? Ans. $95.83^.
4. Bou ght 76 cords of wood at $3.62^^ 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 14f per cent. ; what was his whole loss ?
6. A grocer bought 3 barrels of sugar, each containing 230
pounds, at S^ cents a pound, and sold it at 18/^ per cent, profit ;
what was his whole gain, and what the selling price per pound ?
Ans. Whole gain, $10.35 ; price per pound, 9^ cents.
7. A sloop, freighted Avith 3840 bushels of corn, encoun-
tered a storm, when it was found necessary to throw 37^ per
cent, of her cargo overboard ; what was the loss, at 62^ cents
a bushel ? Aits. $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 ? Aiis. $1209.50.
9. A man commenced business with $3000 capital; the
first year he gained 22^ per cent., which he added to his capi-
tal ; 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 16f per cent, of his entire capital; how much did he make
in the 3 years ? Ans. $981.25. , .
CASE II. ^^'^
!5574. 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 n is what ? Give explanation. Kule.
^
PROFIT AND LOSS. 229
OPERATION.
40 — 32 = 8 ; 8 -r- 32 = ^8_ _ .25, Ans.
Or, 40 — 32 = 8 ; 8 ^ 32 =\^^ := i ; -^ X 100 = 25 ^.
Analysis. Since the gain on 32 cents is 40 — 32 i::r: 8 cents, the
■whole gain is -^2 ^^ i °^ ^^^ purchase money ; and J 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 ^^ iz= ^ of the purchase money, it will be J of 100 per
cent., equal to 25 per cent. Hence the following
Rule. Ifake the difference between the purchase and selling
prices the numerator^ and the purchase pries the denominator ;
reduce to a decimal, and the result will be the per cent. Or,
Take such a part of 100 as the gain or loss is part of the
purchase price,
EXAMPLES FOR PRACTICE.
2^ A man bought a pair of horses for $275, 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.^2J a
barrel, and sold it so as to gain $114.88|i-; what per cent,
profit did he make? Ans. 23 per cent.
5. Bought sugar at 8 cents a pound, and sold it for 9^ cents
a pound ; what per cent, was gained ?
6. A drover bought 150 head of cattle for $42 per head,
'd sold them for $5400 ; w^hat was his loss per cent. ?
Ans. 14f^.
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; \vhat was the gain ? Ans. 150^.
9. If I sell ^ of an article for J of its cost, what is gained
per cent. ? Ans. 50 per cent.
10. If I of an article be sold for what J- of it cost, what is
the loss fo. Ans. 37^ 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 $.87 J- 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^ cents ; what was his gain per cent. ? Ans. 182^3- 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 corn at $.55 a bushel and sold
it at $.66, and wheat for $1.10, and sold it for $1.37^; upon
which did he make the greater per cent. ?
Ans. 5 per cent., upon the wheat.
CASE III.
375. 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. Analysis. Since $1 of cost
$1 + .25 ~ $1.25. sells for $1.25, $136 of cost will
$1.25 X 136 = $170, Ans. sell for 136 times $1.25, which
Or 1 00 I 2s — j2ft _ 5 equals $170, the selling price.
$136 X ^ = $170, A71S. ,, "'' !^"^^;^^/^««t ^^. tiro' and
^ '^4 V J o ^j^g g^jj^ ^^^ ^Yie selling price
will be \^^ = I of the cost, or
•I 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,
&c. Hence,
RuLEl Multiply $1 increased hy the gain or diminished hy
the loss per cent, hy the number denoting the cost. Or,
Take such a part of the cost as is equal to \%% increased or
diminished hy the gain or loss per cent.
Case III is what ? Give explanation. Riile.
^ i
PROFIT AND LOSS. 231
%.
EXAMPLES FOR PRACTIC^^,
2. If 12 J- hundred weight of sugar cost $140, how must
it be sold per pound to gain 25 ^ ? A7is. 14 cents.
3. Bought a hogshead of molasses for 30 cents a gallon,
and paid 16§ per cent, on the prime cost, for freight and cart-
age ; how much must it sell for, per gallon, to gain 33^ per
cent, on the whole cost? Arts. $.46§.
4. For what price must I sell coffee that cost 10^- cents a
pound, to gain 17^ ^ ?
5. If I am compelled to sell damaged goods at a loss of 15
per cent., how should I mark goods that cost me $.62^?
$1.20? $3.87^? Ans. $.53^; $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 remain-
der be sold per gallon, to gain 37^ 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 18f
per cent. ; how much did he receive a yard ?
(5ase rv.r- ^^
376. 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^ per cent. ; how much did it cost ?
OPERATION.
1 + .331 =1.33^; $4.80 -^ 1.33^ r= $3.60, Jns.
Or, $4.80 = f of the cost ; $4.80 -M- 2= $3.60.
^ — . — ^ , -^^ =.^-
Case IV is what ? X
232 PERCENTAGE.
Analysis. Since the gWi is 33^ per cent, of the cost, $1 of the
cost, increased by '3'S^ per cent., will be what $1 of cost sold for :
therefore there will be as many dollars of cost, as 1.33^ is con-
tained times in $4.80, or $3.60. Or, since he gained '33^ per cent.
=r ^ of the cost, $4.80 is |- of the cost ; $4.80 -^ | = $3.60.
Note. If the rate per cent, be loss, we subtract it from 1, instead
of adyi^g it. Hence the following
Rule. Divide the selling price hy 1 increased hy the gain
or diminished hy the loss per cent., expressed decimally , or in
the form of a common fraction, and the quotient will be 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. ; how much did it cost him ? Ans. $.75.
5. Sold a quantity of flour for $1881, which was 18f per
cent, more than it cost ; how much did it cost ?
6. Sold 25 barrels of apples for $69.75, and made 24 per
cent. ; how much did they cost per barrel ?
7. Sold 9^ cwt. of sugar at $8|- per cwt., and thereby lost
12 per cent. ; how much 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 received
had I sold it for 15 per cent, above cost? Ans. $138.
9. B sells a pair of horses to C, and gains 12^ per cent. ;
C sells them to D for $570, and by so doing gains 18 J per
cent. ; how much 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 w^hole ? 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 began
with ; how much did he begin with ? Ans. $1166.66§.
Give explanation. Rule.
INSURANCE. 233
INSURANCE.
Syy. 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.
28 1 1 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 varies according to the degree of
hazard, or class of risk.
Note. As a security against fraud, most insurance companies take'
risks at not more than two thirds the full value of the property
insured.
282. To find the premium when the rate of insur-
ance and the amount insured are given.
1. What must I pay annually for insuring my house to the
amount of $3250, at 1;^ per cent, premium ?
OPERATION. Analysis. We
$3250 X .0-1 i- or .0125 = $40,625. multiply the amount
Or, 1^ per ct. = ^^^ = -gi^ ; insured, $3250, by
$3250 X gV= ^40.624. *^^ rate IJ per
^ '^ cent., and the result,
$40,625, is the premium. Or, the rate, \\ per cent, is ^^=z^ of
the amount insured, and -^-^ of $3250 is $40.62^. Hence the
Rule. Multiply the amount insured hy the rate per cent.,
and the product will be the premium. Or,
Take such a part of the amount insured as the rate is part
of 100.
Define insurance. Insurer, or underwriter. Policy. Premium.
To what amount can property usually be insured ? Give analysis of
example 1. Rule.
234 PERCENTAGE.
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
2^ 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 fur-
niture for $4000, at 2f 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 fof 1^ per cent on f of its value ; if the cargo be
lost, how much will the owner of the wheat lose ? A7is. $1 636.
10. What will it cost to insure a factory valued at $21000,
at ^ per cent ; and the machinery valued at $15400, at ^ ^ ?
A?i$, $264.25. ..
^ TAXES.
383. A Tax is a sum of money assessed on the person
or property of an individual, for public purposes.
2 8 4:. 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.
"V\Tiat 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 personal
property.
380* Real Estate consists of immovable property, such
as lands, houses, &c.
387. Personal Property consists of movable property,
such a3 money, notes, furniture, cattle, tools, &c.
28 8» An Inventory is a written list of articles of proper-
ty, with their value.
S89* 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 as-
sessment roll, is $600,000, and there are 220 polls to be as-
sessed 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 z=i $18.75, A's tax on property.
$.75 X 3 zz: $2.25, A's tax on 3 polls.
$18.75 '-{■ $2.25 — $21, amount of A's tax.
Hence the following
Rule. I. Find the amount of poll tax, if any, and subtract
this sum from the whole 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 will be the per
cent., or the tax on one dollar.
III. Multiply each marCs taxable property by the per cent.,
or the tax on %\, and to the product add his poll tax, if any ;
the result will be the whole amount of his tax.
"What is real estate ? Personal property ? An inventory ? Explain
the process of levying a state or other tax. Rule.
236
PERCENTAGE.
Note. Ha\Hng found the tax on $1, or the per cent., which in the
preceding example we find to be 5 mills, or i per cent., the operation
of assessing taxes may be greatly facilitated by finding the tax on $2,
$3, &c., to $10, and then on $20, $30, &c., to $100, and arranging
the numbers as in the following
TABLE.
Prop.
Tax.
Prop.
Tax.
Prop.
Tax.
Prop.
Tax.
$1 give?
$.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 «
.02
40
.20
400
2.00
4000
20.
5 "
.025
50
.25
500
2.50
5000
25.
6 •«
.03
60
.30
600
3.00
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.
. $15.00
4.00
.20
.025
1.50
EXAMPLES FOR PRACTICE.
2. According to the conditions of the last example, how
much would be 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 . ,
" " " 800 ... . "
" " " 40 .... "
** " " 5 . . . . **
" " " 9. polls "
Total tax is $20,725
3. How much would be "VVs 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 dollar ?
what is C's tax, his property being assessed at $12450, and ho
paying for 2 polls ? Ans. $.005^ on $1 ; $09,474-, 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; their property is assessed for $188000, and
each poll is assessed 62^ 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 ? Ans. 2x^0 ^- 5 $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 as-
sessment 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 ? Ans. 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 24- 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. $.00a, tax on $1 ; $93.25, E's tax.
9. In a certain district the school was supported by a rate-
bill ; the teacher's wages amounted to $200, the fuel and other
expenses to $75.57 ; the public money received was $98, and
the whole number of days' attendance was 3946 ; A sent 2
pupils 118 days each; how much was his rate-bill ? Ans. $10.62
CUSTOM HOUSE BUSINESS.
390* 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 govern-
ment for the transaction of business relating to duties.
S9S* A Port of Entry is a seaport town having a cus-
tom house.
Define duties. A custom house.
238. PEIipENTAGE. -
S03. Tonnage is a tax levied upon a vessel, independent
of its cargo, for the privilege of coming into a port of entry.
294L» 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 the cost
of goods in the country from which they were imported.
S90* Specific Duty is a sum computed on the weight or
measure of goods, without regard to their cost.
29T, An Invoice is a bill of goods imported, showing
the quantity and price of each kind.
298. By the New Tarifi* 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, measured,
gauged, or imported, 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 ascertained by
frequent trials.
300. Leakage is an allowance on liquors imported in
casks or barrels.
301. Breakage is an allowance en liquors imported in
bottles.
Note. Actual leakage or breakage is allowed, there being no fixed
or legal rate.
303. Gross Weight or Value is the weight or value of
the goods before nny allowance has been njade.
303. Net Weight or Value is the weight or value after
all allowances have been deducted.
Define Tonnage. Revemie. Ad valorem duty. Specific duty. An
invoice. Tare. Leakage. Breakage. Gross weight or value. Net
weight or value.
CUSTOM HOUSE BUSINESS. 239
Note. Draft is an allowance for the waste of certain articles, and
is made only for statistical purposes ; it does not affect the amount of
duty. The rates of this allowance are as follows :
On 112 1b lib.
Above 112 lb. and not exceeding 224 lb., 2 lb.
" 224 lb. '' '« " 336 lb., 3 lb.
" 336 1b. " " " 1120 1b., 4 1b.
«' 11201b. " " ♦* 20161b., 7 1b.
" 2016 1b 9 1b.
1. "What is the duty, at 24 per cent., on 50 gross of London
ale, invoiced at 81.20 per dozen, 2i per cent, being allowed
for breakage ?
OPERATION. Analysis. We
$1.20 X 12 X 50 = $720, gross value, first find the cost of
$720 X .025 = $18, breakage. ^^e ale, at the in-
$720 — $18 = $702, net value. voice price, which is
$702 X .24 = $168.48, duty. ^-"^O. From this
sum we deduct the
allowance for breakage, $18, and compute the duty on the re-
mainder. Kence the following
Rule.. Deduct allowancesj if necessary, and compute the
duty, at the given rate, on the net value.
Note. — 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 -f-.
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 + .
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. $452,811.
5. John Jones imported from Havana 25 hhds. of W. I.
molasses, which was invoiced at 36 cents per gallon; allowin*^
\ per cent, for leakage, what was the duty atPA^?
Ans. $135,399 + .
Define draft. Give analysis Rule.
240
SIMPLE INTEREST.
SIMPLE INTEREST.
304. Interest is a sum paid lor 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 varies in different States, as follows:
>
1 ^
f
§
1 n
r«
p
^2"
t
^
«
Alabama
Arkansas
California
Connecticut
Colorado
Canada
Dakota
Delaware
Dist, Columbia. .
Florida
Georgia ........
Idaho
Illinois
Indiana
Iowa
Kansas
Kentucky
Louisiana
Maine
Maryland
Massachusetts . .
Michigan
Minnesota
8%
6%
Any rate.
10%
Any rate.
H
Any rate.
10%
Any rate.
(i%
....
10%
Any rate.
m
....
6^
10%
m
Any rate.
1%
10^
lO^g
Any rate.
0^
10^
6^
10^
^%
10;^
1%
12%
Q%
10^
5^
8%
6^
Any rate.
Any rate.
H
10^
I 1%
12%
Mississippi
Missouri
Montana
New Hampshire.
New Jersey
New York .*
North Caiolina..
Nebraska
Nevada
Ohio
Oregon
Pennsylvania .. .
Rhode Iglfiud. . .
South Caiolina .
Tennessee
Texas
Utah
Vermont
Virginia
West Virginia. .
Wisconsin
Washington Ter.
England
K%
C^
"*%
e^
10%
10%
K%
(i%
10^
6^
10%
5%
30%
Any late.
12%
Any rate.
H
12%
Any late.
Any rate.
Any late.
Any late.
'12%
VoV
Any rate.
3 10. 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.
PERCENTAGE. 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 obtained
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 3i times the interest on the same sum for 1 year ;
and the interest for 3 months is ^ of the interest for 1 year.
1. What is the interest on $75.19 for 3 years 6 months, at
OPERATION.
875.19
.06 Analysis. The interest on $75.19, for 1 jt.,
r~T7T~r at 6 per cent., is .06 of the principal, or $4.51 14,
^ * and the interest for 3 yr. 6 mo. is 3^^ =: 3^ times
^ the interest for 1 yr., or $4.5114 X 3^, which is
22557 $15.789 -}-> the Ans. Hence, the following
135342
$15.7899, Ans.
KuLE. 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 fractions
of a year, and the result will he the required interest.
EXAMPLES FOR PRACTICE.
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 7 ^ ?
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. AVhat is the interest of $76.50 for 2 years 2 months, at
5^^? ' Ans. $8,287 + .
Explain the difference between percentage and interest. Give
analysis. Rule.
HP.
11
\
242 SIMPLE INTEREST. \
7. Wliat is the interest of $1276.25 for 11 months, at 7
percent.? Ans. $81.89 + .
8. What is the interest of $25 GO. 7 5 for 4 years G months,
at 6 per cent. ?
9. What is the interest of $1500.C0 for 2 years 4 months,
at G^^? Ans. $218.8375.
10. What IS the amount of $26.84 for 2 years 6 months, at
5 percent.? 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 -f- .
13. AVhat is the amount of $3050 for 4 years 8 months, at
6^ per cent. ? Ans. $3797.25 + .
14. What is the interest of $5000 for 9 months, at 8 per
cent.? Ans. $300.
15. If a person bonow $375 at 7 per cent., how much will
be due the lender at the end of 2 yr. 6 mo. ?
IG. 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, oh 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 IT.
SlSu To Imd tlie interest on any sum, for any
time, at any rate per cent.
The analysis of our rul^is based upon the following
Obvious Itelutions heti^een Time and Interest.
T. The interest on any sum, for 1 year, at 1 per cent., is
.01 of that sum, andjis equal to the principal with the separatrix
removed two ])laces to the left.
II. A month being -^^ of a year, ^ of the interest on any
Bum for 1 year is the interest for 1 month.
»■ - — ■—■■■
"What is Case II ? Give the first relation between tinie and interest.
8ecoud.
/••
PERCENTAGE. 243
III. The interest on any sum for 3 days is tj^^. = j\^ z= .1
of the interest for 1 month, and any number of days may
readily be reduced to^fUhs 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.
I. What is the interest on $724.68 for 2 yr. 5 mo. 19 da.,
at 7 ^ ?
OPERATION. Analysis. We remove
2yr. 5 mo. 19 da. = 29.6^ mo. the separatrix in the given
1 rt V oy 94pQ principal two places to the
^-^-— ^-^ left, and we have $7.2468,
$.6039 the interest on the given sum
29.6^ for 1 year at 1 per cent.
— — (313 L). Dividing this oy
^^^^ 12, we have $.6039, the inter-
36234: est for 1 month, at 1 per cent.
54351 (II.) Multiplying this
12078 quotient by 29.6|, the time
OQ--7 expressed in months and deci-
$17.89007 j^^jg Qf. ^ ^^^^^^ (HI. IV.,)
we have $17.89557, the in-
$12o.2689|p, 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. Hence,
Rule. I. Remove the separatrix in the given principal
two places to the left / the result will he the interest for 1 year,
at 1 per cent.
II. Divide this interest hyl2; 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 will he the interest
for the given time, at 1 per cent.
IV. Multiply this interest hy the given rate ; the product
will he the interest required.
Give the third. Fourth, Give analysis. Rule.
244: SIMPLE INTEREST.
Contractions. After removing the separatrix iii the principal
two places to the left, the result may be regarded either as the in-
terest on the given ^)riucipal for 12moi^^j^t 1 per cent., or for I
month at 12 per cei^ If we rcgtrftTll^^oW month at 12 per
cent., and if the givo^rate be an aliquot^iart 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 month as the given
rate is part of 12 per cent. Thus,
To find the interest for 1 month at 6 per cent., remove the sep-
aratrix two places to the left, and divide by 2.
To find it at 3 per cent., ])roceed as before, and divide by 4 ; at 4
per cent., divide by 3 ; at 2 per cent., divide by G, &c.
SIX PER 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 G per cent- per annum the interest on $1
For 12 months is $.0G.
" 2 months {-^=1- of 12 nro.) '' .01.
" 1 month, or 30 days(^V of 12 mo.) " .00J=$.005 ( jL of $.06).
" 6 days (I- of 30 days)' " .001.
" 1 " (]ofGda.=5Jjof30days) " .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
C days.
From these principles we deduce the
EuLE. I. To find the rate : — Call every year $.0G, every
2 months $.01, every G days $.001, and any less number of days
&ixlhs of 1 mill,
II. To find the interest : — Multiply the principal hy the rate,
IIOTES. — 1. To find the interest nt any other rate per cent, by this
method, first find it at G per cent., atid then increase or diminish the
result by as many times itself as the given rate is greater or less than G
per cent. Thus, for 7 per cent, add i, for 4 per cent, subtract i, &c.
What contractions are given ? Give analysis of the 6 per cent, method.
Rule. Its application to any other rate per cent,
SIMPLE INTEREST. 245
r
2.' JFhe 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 titDe less than 60 days, the
interest will be less than $.01, and may be disregarded.,;
3. Since the interest of $1 for 60 days i.s $.01, the interest of $1 fo.r
any number of days is as many cents as 00 is contained times in tho
number of days. Therefore, if any principal be multiplied by the num-
ber of d.iys in any given number of months and d.ays, and the product
divided by 60, the result will be the interest in cents. That is. Multi-
ply the principal by the number of day Sy. divide the product by 60, and point
off two decimal places in the quotient. The result will be the interest in the
tame denomination as the principal
EXAMPLES FOR PRACTICE.
2. What is the interest of $100 for 7 years 7 months, at 6
p«r cent. ? Ans, $45.50.
3. What is the amount of $47.50 for 4 years 1 month, at 9
per cent. ? Ans. $64,956 + .
4. What is the amount of $2000 for 3 months, at 7 per
cent.? Ans, $2035.
5. What is the interest of $250 for 1 year 10 months and
15 days, at 6 per cent. ? Ans. $28,124^.
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 6|- 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, at9i ^? 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 daysf at 10 per
cent.? Ar^. $^12.50.
17. What is the amount of $1000 for 1 month 5 days, at
6^ per cent. ? 4fis. $n)06.56^.
18. P'ind the interest of $973.68 for 7 monthl 9 days, at
19. If I borrow $275 at 7 per cent., how much will I owe
he 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 64- per cent
interest; what anSount did he pay ? A7is. $3149.825.
21. In setthng 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.? A?is, $102.23 +.
22. How much is the interest on a note cf $384.50 in 2
years 8 months and 4 days, at 8 ,^ ?
23. What is the interest of $97.86 from May 17, 1850, to
December 19, 1857, at 7 per cent. ? Ans. $51.98 +.
24. Find the interest of $35.61, from Nov. 11, 1857, to
Dec. 15, 1859, at 6 per cent. Ans. $4,474.
25. Required the interest of $50 from Sept. 4, 1848, to
Jan. 1, 1860, at 31^ •?
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 in cash on delivery, $3000 in 9 months, and the
remainder in 1 year 6 months, with 7C.per cent, interest;
what was the whole apount paid ? Ans. $6262.50.
28. Wm. GalluiNSu^t bills of dry goods of Geo. Bliss
& Co., of New YorkTas follows, viz. : J^ui. 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 receipt, and
is called an Indorsement,
$2000. SrRiNGFiELD, Mass., Jan. 4, 1857.
1. For value received I promise to pay James Parish, or
order, two thousand dollars, one year after date, with interest.
George Jones. ^^^
On this note were indorsed the following payments ;
Feb. 19, 1858, $400 // / ^
June 29, 1859, $1000 / _
4^
Nov. 14, 1859, $520
What remained due Dec. 24, 1860? y/ ^ j' v'
OPERATION.
Principal on interest from Jan. 4, 1857, $200(V '~~^
Interest to Feb, 19, 1858, 1 yr. 1 mo. 15 da. *135 1 i ^5^
Amount, $'2135
Payment Feb- 19, 1858 .^ 400
Remainder for a new principal, J^^1735
Interest from Feb. 19, 1858, to June 29, 1859, 1 >t.
4 mo. 10 da., 141.69
■=*«
Amount, $1876.69
Payment June 29, 1859, 1000.
Remainder for a new principal, $876.69
Interest from June 29, 1859, to Nov. 14, 1859, 4 mo.
15 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., - 25.09
Remains due Dec. 24, 186Q, $401,505-4-
"What is meant by partial payment ? By an indorsement J
248 PERCENTAGE.
^^75.50. jq-Ew 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.
Jones, S^iith & 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
PajTnent Dec. 25, 1855, 50.
Remainder for a new principal, $447.13
Interest from Dec. 25, 1855, to June 14, 1858, 2 yr.
5 mo. 19 da., 77.29
Amount, $524.42
PajTnent July 10, 1856, less than interest ^
then due, > $15.75
Payment Sept. 1, 1857, J 25.50
Their 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 -f-
These examples have been wrought according to the method
prescribed by the Supreme Court of the U, S., and are suf'
ficient to illustrate the following
PARTIAL PAYMENTS. 249
United States Rule.
L 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 he greater than any payment, cast -the
interest on the same principal to a time when the sum of the
payments shall equal or exceed the interest due ; subtracting the
sum of the payments from the amount of the principal, the re-
mainder will form a new principal, on which interest is to be
computed as before*
^^^^•^^- San Francisco, June 20, 1858.
3. Three years after date we promise to pay Ross &
Wade, or order, five hundred fourteen and -^^^ dollars, for
value received, with 10 per cert, interest. Wilder & Bro.
On this note were indorse 1 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 -f-.
^^^^^' Charleston, May 7, 1859.
4. For value received, I promise to pay George Babcock
three thousand dollars, on demand, with 7 per cent, interest.
John May.
On this note were indorsed 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 + .
Give the United States Court rule for computing interest where
partial payments have been made.
11*
250 PERCENTAGE.
$912y'^oV j^g^y Orleans, Aug. 3, 1850.
5. One year after date I promise to pay George Bailey, or
order, nine hundred twelve ■^jf'jj 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 + .
^^^^•^^- 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 -|- .
Note. 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, interest.
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 -|- .
SIS, 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, and payments less
than the interest due, are treated according to the United States
rule.
(Nearly obsolete. The United States rule is In general upe.)
Give Connecticut rule for partial payments.
PARTIAL PAYMENTS. 251
II. Payments exceeding the interest due, and made within
one year from the time interest commenced, or from a former
payment, 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 principal for
one year ; the remainder will be the new principal,
III. Jf 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 remainder
will be the sum due.
^^^^- 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 Marshall.
On this note are indorsed the following payments : April
16, 1858, $148 ; March 11, 1860, $7.5 ; Sept. 21, 1860, $53.
How much was due Dec. 11, 1860 ? Ans. $238.15+.
310* A note containing a promise to pay interest an-
nually is not considered in law a contract for any thing more
than simple interest on the principal. For partial payments
on such notes, the following is the
• Vermont Rule.
I. Find the amount of the principal from the time interest
commenced to the time of settlement,
II. Fitid 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 will be the
sum due.
$600,
Rutland, April 11, 1856.
1. For value received, I promise to pay Amos Cotting, or
order, six hundred dollars on demand, with interest annually.
John Brown.
Give the Vermout rulo for partial payments.
258 PERCENTAGE.
On this note were indorsed the following payments : Aug.
10, I80G, $I5G ; Feb. 12, 1857, $200 • June 1, 1858, $185.
What was due Jan. 1, 1859 .? Ans. $105.50+.
317, In New Hampshire interest is allowed on the an-
nual interest if not paid when due, in the nature of damages
for its detention ; and if payments are made before one year's
interest has occurred, interest must be allowed on such pay-
ments for the balance of the year. Hence the following
New Hampshire Rule.
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 remainder will
he a new principal, with which proceed as before.
II. Jf the settlement occur less than a year from the last an-
nual term of interest, make the last term of interest a part of a
year, accordingly.
^^^^- Keene, N. H., Aug. 4, 1858.
1. For value received, I promise to pay George Cooper, or
order, five hundred seventy-five dollars, on demand, with in-
terest annually. David Greenman.
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 interest
on each year's interest from the time it should he paid to the
time of settlement. The sum of the interests thus obtainedy
added to the principal, will he the sum due.
The New Hampshire rule. The New Hampshire court rule.
PARTIAL PAYMENTS. 253
^^^^- Keene, 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 interest
annually till paid. John S. Briggs.
What is due on the above note, Aug. 2, 1859 ? A7is. $649.40.
Problems in Interest.
319. In examples of interest there are five parts involved
the Principal, the Eate, the Time the Interest, and the
Amount.
CASE I.
330. The time, rate per cent., and interest being
given, to find the principal.
1. What principal in 2 years, at 6 per cent., will gain
$31.80 interest ?
OPERATION. Analysis. Since $1, in
$.12, interest of $1 iu 2 years at 6 ^. 2 years, at 6 per Cent., will
$31.80 — .12 — $265, Ans. gain $.12 interest, the prin-
cipal that will gain $31.80,
at the same rate and time, must be as many dollars as $.12 is con-
tained times in $31.80; dividing, we obtain $265, the required
principal. Hence,
Rule. Divide the given interest hy the interest of $1 for
the given time and rate, and the quotient will he the principal,
examples for practice.
2. What principal, at per cent., will gain $28.12^ in 6
years 3 months ? Ans. $75.
3. What sum, 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? What
are they ? What is Case I ? Give analysis. Rule,
254 PERCENTAGE.
CASE II.
321. The tiniG. 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 $88,125?
OPERATION. Analysis.
$1,175 Amt. of $1 in 2 years 6 months, at^. Since $1, in
$88,125 ^ 1.175 = $75, Ans, ^ ^!^''\ t
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. Hence the
Rule. Divide the given amount by the amount of $1 for
the given time and rate, and the quotient tvill be the principal
required.
EXAiTPLES FOR rRACTICE.
2. What principal, at G per cent., will amount to $G55.20
in 8 months ? Ans. $630.
3. What principal, at 5 per cent., will amount to $106,855
in 5 years 5 months and 9 days ? Ans. $84.
4. What sum, put at interest, at 54- per cent., for 8 years
5 months, will amount to $1897.545 ? Ans. $1297.09+.
5. AVhat 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? Ans. $460.04.
CASE III.
339. The principal, time, and interest being given,
to find *iie rate per cent.
1. I lent $450 for 3 years, and received for interest $67.50 ;
what was the rate per cent ?
Give case 11, Analysis, Rule, Case III.
p
i^^ 75 - *^
A
yx PROBLEMS IN INTEREST. 255
OPERATION. Analysis. Since at
$ 4.50 1 per cent. $450, in 3
3 years, will gain $13.50
~r~rTZ ^ . interest, the rate per
$13.00, int. of $450 for 3 years at 1 %. ^^^^^ ^^ ^^^^^^ ^^^ J^^
$67.50 -T- 13.50 = 5 ^. Ans, principal, in the same
time, will gain $67.50,
must be equal to the number of times $13.50 is contained in $67.50;
dividing, we obtain 5, the required rate per cent. Hence the
Rule. Divide the given interest hy the interest on the prin-
cipal for the given time at 1 per cent., and the quotient will be
the rate per cent, required.
EXAMPLES FOR PRACTICE.
2. If I pay $45 interest for the use of $500 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 fo ? Ans. 6.
4. A man invests $2000— i»>, 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. 5^.
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.
333. Principal, interest, and rate per cent, being
given, to find the time.
1. In what time will $360 gain $86.40 interest, at 6^^?
OPERATION. Analysis. Since in
$ 360 1 year $360, at 6 per
.00 cent., will gain $21.60,
the number of years in.
which the same princi-
pal, at the same rate,
will gain $86.40, will be
$21.60 Interestof$360inlycarat6,f. ^.j^j^j^ ^^^ ^^^^ ^^-^^^^^
$86.40 -i- 21.60 =: 4 years, ^Tis. pal, at the same rate.
Analysis. Rule. Case IV, Analysis.
256 PERCENTAGE.
as many as $21.00 is contained times in $86.40 j dividing, we ob-
tain 4 years, the required time. Hence the
Rule. Divide the given interest hy the interest on the prin-
cipal for 1 year, and the quotient will be the time required in
years and decimals.
Note. The decimal part of the quotient, if any, may be reduced to
months and days (by 215).
EXAMPLES FOR PRACTICE.
2. The interest of $325 at 6 per cent, is $58.50 ; what is
the time ? Ans. 3 years.
3. B loaned $1600 at 6 per cent, until it amounted to
$2000 ; what was the time ? Ans. 4 years 2 months.
4. How long must $204 be on interest at 7 per cent., to
amount to $217.09 ? t ' ^ns. l) months.
5. Engaging in business, I borrowed $750 of a friend at 6
per -Wt., and kept it until it amounted to $942 ; how long did
I retam it ? Ans. 4 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 ^ ?
Note. The time in years in which any sum will double itself may
be fomid by dividing 100 by the rate per cent.
ND^INTEREST. '
COMPOUND
334* Compound Interest is interest off both principal and
interest, when the interest is not paid when due.
Note. 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 ler/al.
1. What is the compound interest of $200, for 3 years, at
6 per cent, r
Rule. In what time will any sxuu double itself at interest ? "What
*« compound mterest i
COMPOUND INTEREST. 257
OPERATION.
$200 Principal for 1st year.
$200 X '06 == 12 Interest for 1st year.
$212 Principal for 2d year.
$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 make 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.
Notes. 1. "When the interest is payable semi-annually or quar-
terly, find the amount of the given principal for the first interval, and
make it the principal for the second interval, 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 amoimt, before subtracting.
EXAMPLES FOR PRACTICE.
2. What is the compound interest of $500 for 2 years at 7
per cent. ? Ans. $72.45.
3. What is the amount of $312 for 3 years, at 6 per cent,
compound interest ? Ans. $371,594-.
4. What is the compound interest of $250 for 2 years,
payable semi-annually, at G per cent.? Ans. $31.37-}-.
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 ^? Ans. $72.66-|-.
Explain operation. Give rule.
^Z'.
258
PERCENTAGE.
7. What is the amount of $700 for 3 years 9 months and
24 days, at 7 per cent, compound interest ? Ans. $90G.55-[-.
A more expeditious method of computing compound interest
than the preceding, is by means of the following
TABLE,
Showing the amount of$l, or £1, at 3, i, 5, 6, and 7 per cent., compound
interest, for any number of years, from 1 to 20-
Yrs.
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
3 per cent.
1.030,000
1.060,900
1.092,727
1.125,509
1.159,274
1.194,052
1.229,874
1.266,770
1.304,773
1.343,916
1.384,234
1.425,761
1.468,534
1.512.590
1.557,967
1.604,706
1.652,848
1.702,433
1.753,506
1.806,111
4 per cent.
1.040,000
1.081,600
1.124,864
1.169,859
1.216,653
1.265,319
1.315,932
1.368,569
1.423,312
1.480,244
5 per cent.
1.539,454
1.601,032
1.665,074
1.731,676
1.800,944 2.078,928
1.050,000
1.102,500
.L157,625
1.215,506
1.276,282
1.340,096
1.407,100
1.477,455
1.551,328
1.628,895
1.710,339
1.795,856
1.885,649
1.979,932
1.872,981
1.947,900
2.025,817
2.106,849
2.182,875
2.292,018
2.406,619
2.526,950
2.191,12312.653,298
6 per cent. 7 per cent.
1.060,000; 1.07,000
1.123,600; 1.14,490
1.191,016j 1.22,504
1.262,4771.31,079
1.338,2261.40,255
1.418,519 1.50,073
1.503,630' 1.60,578
1.593,8481 1.71,818
1.689,479jl.83,84a
1.790,848 1.96,715
1.898,299 2.10,485
2.012,196'2.25,219
2.132,928'2.40,984
2.260.904 2.57,853
2.396,558 2.75,903
2.540,352
2.692,773
2.854,339
2.95,216
3.15,881
3.37,293
3.025,60013.61,652
3.207,135.13.86,968
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.
$1200.58400, Ans,
9. What is the compound interest of $120 for 15 years, at
5 per cent.? Ans. $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.? Arts. $.38696.
DISCOUNT.
3^d. Discount is an abatement or allowance made for
the payment of a debt before it is due.
3^6. The Present Worth 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 pres-
ent worth of the debt, the use of money being worth 7 per
cent. ?
OPERATION. Analysis. The
Am'tof $1, 1.07) $321 ($300, Present value, amount of $1 for
321 1 year is $1.07;
therefore the pres-
$321 Given sum or debt. e^t worth of every
300 Present worth. $1.07 of the given
$21 Discount. ^ebt ^s $1; and
the present worth
of $321 will be as many dollars as $1.07 is contained times in $321.
$321 -f- 1.07 =z $300, Ans. Hence the following
E-ULE. I. Divide the given svm or debt by the amount of
$1 for the given rate and time, and the quotient will be the pres-
ent worth of the debt.
11. Subtract the present worth from the given sum or debt,
and the remainder will be the discount.
Note. The terms present worthy discount y and debt, are equivalent
to principal, interest, and amount. Hence, when the time, rate per
cent., and amount are given, the principal may be found by (321) ;
and the interest by subtracting the principal from the amount.
EXAMPLES FOR PRACTICE.
2. What is the present worth of $180, payable in 3 years
4 months, discounting at 6 ^ ? Ans. $150.
Define discount. Present worth. Give analysis. Kule.
260 PERCENTAGE.
3. "VYhat is the present worth of a note for $1315.389, due
in 2 years 6 months, at 7 per cent.? Ans. $1 111). 48.
4. AVhat is the present worth of a note for S8GG.038, due
in 3 years 6 months and 6 days, when money is worth 8 per
cent.? What the discount ? Ans. $190.15+, 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 G ^ ?
Note. When payments are to be made at different times without
interest, find the present worth of each payment separately, and take
their sum.
Ans. $945.40-}-.
6. I hold a note against C for $529,925, due Sept. 1, 1859 ;
what must I discount foi; 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
oflfer ; how much did he lose, money being worth 7 per cent. ?
I Ans. $175.
h-^n^' A man, having a span of horses for sale, offered thorn
^^^—mr $480 cash in hand, or a note of $550 due in 1 year 8
f^ months, without interest ; the buyer accepted the latter offer ;
did the seller gain or lose thereby, and how much, interest be-
ing 6 ^ ? Ans. Seller gained $20.
9. What must be discounted for the present payment of a
debt of $2G37.72, of which $517.50 is to be paid in 6 months,
$793.75 in 10 months, and the remainder in 1 year G months,
the use of money being worth 7 per cent. ? Ans. $187.29 -[-.
10. What is the difference between the interest and discount
of $130, due 10 months hence, at 10 ^ ? Ans. $.SS^.
PROMISCUOUS EXAMPLES IX PERCENTAGE. ^^^
1. A merchant bought sugar in New York at 6|^ 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. ? ? / P.'^f r ; ; ., ^||r Ans, 8^+ cents.
I
i
i
^
PROMISCUOUS EXAMPLES. 261
2. A person purchased 2 lots of land for $200 each, and
sold one at 40 per cent, more than cost, and the other at 20
per cent, less ; how much did he gain ? Ans. $40.
3. Sold goods to the amount of $425, on 6 months* credit,
■which Avas ^25 more than the goods cost; what was the true
profit, money being worth G^^? Aiis. $12.62 -\-.
4. Bought cotton cloth at 13 cents a yard, on 8 months*
credit, and sold it the same day at 12 cents cash; how much
did I gain or lose per cent., money being worth 6 per cent. ?
Ans. Lost 4 ^ ?
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 invested § of all he was worth in the coal trade,
and at the end of 2 years 8 months sold out his entire interest
for $3100, which was a yearly gain of 9 per cent, on the
money invested ; how much was he worth when he commenced
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. \Af years.
8. Two persons engaged in trade ; A furnished f 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 collect-
ing? /I ^^ Ans. $1200.
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 ? ? •'••'^ -'■ft At^- $5,346 +.
262 ' PERCENTAGE.
12. A nurseryman sold trees at $18 per hundred, and
cleared ^ of his receipts ; what per cent, profit did he make ?
Ans. 50 fc ?
13. If 1^ of an article be sold for what f of it cost, what is
the gain per cent. ? Ans. 40 1.
14. A lumber merchant sells a lot of lumber, which he has
h.ad 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. 211.
15. A person, owning | of a piece of property, sold 20 per
cent, of his share ; what part did he then own ? Ans. 4-
16. A speculator, having money in the bank, drew 60 per
cent, of it, and expended 30 per cent, of 50 ^er cent, of this for
728 bushels of wheat, at $1.12iper bushel; how much was
left in the bank ? Ans. $3640.
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. 43|f .
18. A's money is 28 per cent, more than B's ; how many
per cent, is B's less than A's ? Ans. 21|^.
19. A capitalist invested ^ of his money in railroad stock,
which depreciated 5 per cent, in value ; the remaining f he in-
vented in bank stock, which, at the end of 1 year, liad 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. C's money is to D's as 2 to 8 ; if ^ 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 ?
Ans, C, $2812 ; D, $4218.
BANKING.
3h27« a BaJlk is a corporation chartered by law for the
purpose of receiving and loaning money, and furnishing a
paper circulation.
is a bank }
BANKING. 263
338. A Promissory Note is a written or printed engage-
ment to pay a certain sum, either on demand or at a specified
time.
339. Bank Notes, or Bank Bills, are the notes made
and issued by banks to circulate as money. They are payable
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.
333* 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.
33d« 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 I.
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 Ox a note. Days
of grace. The maturity of a note. Explain the process of discounting
a note at a bank. Define bank discount. The proceeds of a note.
AVhat is Case I ?
264 PERCENTAGE.
equal to the simple interest at the legal rate for the time spe-
cified in the note. Hence the
Rule. I. Compute the interest on the face of the note for
three days more than the specijied time ; the result will he the
discount.
II. Subtract the discount from the face of the note, and the
remainder will he the proceeds.
EXAMPLES FOR PRACTICE.
1. What is the discount, and what the proceeds, of a note
for $450, at GO days, discounted at a bank at 6^?
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 GO
days, if discounted at the Crescent City Bank, New Orleans ?
Ans. $471.33+.
4. What are the proceeds of a note for $10000, at 90 days,
discounted at the Philadelphia Bank ? Ans. $9845.
5. 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. AVhat 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. Ans. $355.1 68 -|-.
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.
Notes. 1. To indicate the maturity of a note or draft, a vertical
line ( I ) is used, with the day at which the note is nominally due on
the left, and the date of maturity on the right ; thus, Jan. ' | lo .
2. AVTien a note is on interest, payable at a future specified time, the
amount is the face of the note, or the sum made payable, and must be
made the basis of discount.
Give rule.
BANKING. 265
Find the maturity, term of discount, and proceeds of the
followinor notes : —
$oOO. Boston, 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. ( Due, April ^l^.
A71S. < Term of discount, 36 da.
( Proceeds, $497.
$750. 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 re-
ceived. Byron Quinby.
Discounted at a broker's, Nov. 15, at 10 ^.
[ Due, Dec. i2|^^.
Ans. < Term of discount, 30 da.
(Proceeds, $766,434+.
CASE II.
337. Given the proceeds of a note, to find the
face.
1. I wish to borrow $100 at a bank. For what sum must
I draw my note, payable in CO 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 =z disc, on $1 for €3 da. ^he face of which we are
required to find. We first
$ .9895 = proceeds of $1. obtain the proceeds of $1
$400 -^ .9895 = $404,244 = by the last case, and then
face of the required liote. 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. Hence the
r.p.
Give Case II. Analysisj
12
2G6 PERCENTAGE.
Rule. Divide the proceeds by the proceeds of %\ for the
time and rate me7itioned, and the quotient will be the face of
the note.
EXAMPLES FOR PRACTICE.
2. What is the face of a note at 60 days, which yields
$680 when discounted at a New Haven bank ?
Ans. $687,215+.
3. "What is the face of a note at 90 days, of which the pro-
ceeds 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 7 ^ ? 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. AVhat
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 sura to another person, or to
his order, at a specified time.
34:0. 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 siglit."
Tlierc are always three parties, and usually four, to a trans-
action in exchange :
341. The Drawer or Maker is the person who signs the
order or bill.
Give the rule. Define exchsingc. A bill of exchange. A sight draft.
Xhe drawer.
EXCHANGE. 2G7
34^* The Drawee is the person to whom the order is
addressed.
•^43. The Payee is the person to whom the money is or-
dered to be paid.
344. The Buyer or Remitter is the person who purchases
the bill. He may be himself the payee, or the bill may be
drawn in favor of any other person.
34^. The Indorsement of a bill is the writing upon its
back, by which the payee relinquishes his title, and transfers
the payment to another. The payee may indorse in blank 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 payment.
A bill made payable to the hearer is transferable without in-
dorsement.
346. The Acceptance of a bill is the promise which the
drawee makes when the bill is presented to him to pay it at
maturity ; this obligation is usually acknowledged by writing
the word " Accepted," with his signature, across the face of
the bill.
Note. Three days of grace are usually allowed for the payment of
a bill of exchange after the time specified has expired. But in Xew
York State no grace is allowed on sight drafts.
From these definitions, the use of a bill of exchange in mon-
etary transactions is readily perceived. If a man wishes to
make a remittance to a creditor, agent, or any other person
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
issued them, he remits a bill of exchange, purchased at a bank
or elsewhere, and made payable to the proper person in or
The drawee. The payee. The buyer. An indorsement. An
acceptance. What of grace on bills of exchange ?
268 PERCENTAGE.
near the place where he resides. Thus a man bj paying
Boston funds in Boston, may put New York funds into tlic
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 pre-
mium 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 be-
tween 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.
$^^00. 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.
To M. Smith & Co,
Messrs. Brown & Foster, )
Baltimore. *
What is the cost of the above drafl, the rate of exchange
being 1^ per cent, premium?
operation. Analysis. Since ex-
$500 X 1.015 = $507.50, Ans. ^^^"^e is at \\ per cent
premium, each aoUar ot
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.
EXCHANGE. 269
>ve multiply Its face, $500, by 1.015, and obtain $507.50, the re-
quired Ans.
$480. Boston, 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 Trenchard.
To John Stiles & Co.
New
feCo.,|
York. )
What is the cost of the above draft, exchange being at a
premium of 3 ^ ?
OPERATION. Analysis. Since
$1.0000 time is allowed, the
.0055 n: discount for 33 days. draft must suffer dis-
r~~~7r count in the sale. The
$ .9945= proceeds of $1. ^.^^^^^^ ^^ ^^^ ^^ ^^^
.VO z=: rate of exchange. , i ^ • d . x-
legal rate in Boston, for
$1.0245 r= cost of $1 of the draft. the specified time, al-
$480 X 1.0245 = $491.76, Ans, lowing grace, is $.0055,
which, subtracted from
$1, gives $.9945, the cost of $1 of the draft, provided sight ex-
change were at jmr ; but sight exchange 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, the Ans.
From these examples we derive the following
EuLE. I. For sight drafts. — Multiply the face of the draft
hy 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 hanh discount for th^ 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 FOR PRACTICE.
3. A merchant in Cincinnati wishes to remit $1000 by
vjrivc analysis. Rule I; IE.
270 PERCENTAGE.
draft to his agent in New York ; what will th^ bill cosf, 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 f per cent, pre-
mium? Ans. $403.
5. A merchant in St. Louis orders goods from New York,
to the amount of $530, which amount he remits by draft, ex-
change being at 2f per cent, premium. If he pays $20 for
transportation, what will the goods cost him in St. Louis ?
Ans. $5G4.575.
6. What will be the cost, in Detroit, of a draft on Uoston
for $800, payable 60 days after sight, exchange being at a pre-
mium 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^ 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 ? A?is. $12.
CASE 11.
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 — 1.027 = $360, Ans. ^V ^^^^ ^' ^^^^ ^ ^^'^^
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
Ans. From this example and analysis we derive the following
What is Case II ? Give analysis.
EQUATION OF PAYMENTS. 271
Rule. Divide the given cost hy the cost of a draft for $1,
at the given rate of exchange ; the quotient will he the face of
the required draft.
EXAMPLES FOR TRACTICE.
2. What draft may be purchased for $243.G0, exchange
being at 1 ^ per cent, premium ? Ans. $240.
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.
5. An emigrant from Bangor takes $240 in bank bills to
St. Paul, Min., 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 1^ per cent, discount? Ans. 85.599+.
0. A Philadelphia manufacturer is informed by his agent in
Buffalo that $3600 is due him on the sale of some property.
He instructs the agent to remit by a draft payable in 60 days
after sight, exchange being at J per cent, premium. The agent,
by mistake, remits a sight draft, which, when received in Phila-
delphia, is accepted, and paid after the expiration 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-(-.
EQUATION OF PAYMENTS.
359. Equation of Payments is the process of finding the
mean or equitable time of payment of several sums, due at
different times without interest.
35«S. Tlie 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.
3«54:. The Average Term of Credit is the time to elapse
before several debts, due at diiferent times, may all be paid
at once, without loss to debtor or creditor.
3SS» The Equated Time is the date at which the several
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 ?
OPERATION.
$ 8 X 5 = 40
10 X 2 =r 20
12 X 10 = 120
SO 180 -^ 30 = 6 mo., average time of credit.
Jan. 1. -f- 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 Avithheld, 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. =zthe credit on $40 for 1 mo., because 8 X 5 := 40
XI- 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 lOmo.izz
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, be-
cause
30 X 6 = 180 X 1.
Rule. I. Multiply each payment by its term of credit^ and
divide the sum of the products by the sum of the payments ; the
quotient will be the average term of credit.
Average term of credit. Equated time. Give Case I. Analysis.
Bule.
AVERAGING CREDITS. 273
II. Add the average term of credit to (he date at which all
the credits begin, and the result will he the equated time of
•payment.
Notes. 1. The periods of time used as multipliers must all be of
the same denomination, and the quotient will be of the same denomi-
nation as the terms of credit ; if these be months, and there be a re-
mainder 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
jjaid before it is due equals the interest of the samQ amount paid after
it is due.
EXAMPLES FOR PRACTICE.
2. On the 25th of September a trader bought merchandise,
as follows : $700 on 20 days' credit; $400 on 30 days' credit ;
$700 on 40 days' credit : what was the average term of credit,
and what the equated time of payment ?
. ( Average credit, 30 days.
( 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, I find myself indebted to John Kennedy in sums
as follows : $650 due in 4 montks ; $725 due in 8 months ; and
$500 due in 12 monttis : at what date may I settle by giving
my note on interest for the whole amount? Ans. Aug. 21.
CASE IT.
357. When the terms of credit begin at different
dates, and the account has only one side.
3^8* An Account is the statement or record of mercantile
transactions in business form.
Give Case II. Define an account.
12*
274
EQUATION OF PAYMENTS.
3^0. 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 Cr. 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 Balanca 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.
Note. Book accounts bear interest after the expiration of the term
of credit, and notes after they become due.
361. To Average an Account is to find the mean or
equitable time of payment of the balance.
369. 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. C. Smith,
1859. To C. E. Borden, Dr.
June 1. To Cash, $450
« 12. " Mdse. on 4 mos., . . ; 500
Aug. 16. « Mdse., 250
FIRST OPERATION.
SECO
STf 0]
PERATIO
N.
Duo.
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-^1200=1 71 da.
. S 71 da. after June 1,
^'**- J or Aug. 11.
Ans.
74100-^-1200 = 62 da.
J 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. 12, and Aug. 16, as shown in the two
operations. 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 cal-
endar months. From June 1 to Oct. 12 is 133 da. ; from June 1 to
Auo-. 16 is 76 da. Hence the first 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 proceed 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. 1 2. The third item
must draw interest 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 terra 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 Au^. 11,
without interest. Hence the following
Rule. I. Find the time at which each item becomes due^
by adding to the date of each transaction the term of credit, if
any be specified^ 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 number of days in the
preceding column, writing 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. Bule.
276 EQUATION OP PAYMENTS.
earliest date is the focal date, or the average term of interest
when the latest date is the focal date ; in either case always
reckon from the focal date toward the other dates, to find the
equated time of payment.
examples for practice.
2. John Brown,
1859. To James Greigg, Dr.
Jan. 1, To 50 yds. Broadcloth, rS) $3.00, . . . $150
" 16. " 2000 " Calico, " .10, o . o 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,
1859o To Wtnkoop & Bro., Dr.
March 1. To Cash, . „ , r c » c » o . . . » . c , c . . $300
April 4. " Mdse., .oooe.o.oooocoococ. 240
June 18. " " on 2 mo., o « c o o o o ^ o c o > 100
Aug. 8. « Cash, . , c « o c c „ o c . . . « c . 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, . . , - o . $600
« 12. " 200 " " « 1.50, 300
« 15„ " 640 « « " 1.3Q, .o 832
« 25. ** 760 « « " 1.00, .0. ^. 760
" 30. " 500 " « « 1.50,..oc .. 750
Wlien is the whole amount of the above bill due, per
average ? Ans. June 18.
5. My expenditures in building a house, in the year 1856,
were as follows : Jan. 1 6, $536.78 ; Feb. 20, $425.36 ; March 4,
$259.25 ; April 24, $786.36. If at the last date I agree to
AVERAGING ACCOUNTS.
277
sell the house for exactly what it cost, with reference to interest
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 ? ^^^^ I Face, $2007.75.
"** ( Date, March 8, 1856.
6. Thomas Whiting,
1859. To Israel Palmer, Br.
Jan. 1, To 60 bbls. Flour, rS) $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
•
1858.
1 1858.
June
1
To Mdse
400
00
1 July
4
By Mdse
200
00
((
16
« Draft, 3 mo. .
800
00
Aug.
20
" Cash
150
00
Oct.
20
" Cash,
250
00
Sept.
20
<( <(
500
00
Pocal
date.
2)r.
OPERATION.
Cr.
Due
da.
Items.
Prod.
1 Due
1 July 4
Aug, 20
Sept. 20
da.
Items.
Prod.
June 1
Sept. 19
Oct. 20
141
31
400
800
250
56400
24800
108
61
30
200
150
500
850
21600
9150
15000
45750
1450
850
81200
45750
35450
Balances.
600
35450 -^ 600 =: 59 da., average term of interest
Oct. 20 — 59 4a. = Aug. 22, balance due.
What is Case III ? Explain operation.
/
278
EQUATION OF PAYMENTS.
Analysis. In tlie above operation we have written the dates,
showing when the items become due on either side of the ac-
count, 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 a balance is struck. between the items charged and the
products. These balances, being on the Dr. side, show that David
Ware, on the day 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,
lias been due 59 days. Reckoning back from Oct. 12, we find the
Hence the following
data when the balance fell due, Aug. 22.
Rule. I. Find the time when each item of the account is
due ; and write the dates, in two columns, on the sides of the
account to ichich they respectively belong.
II. Use either the earliest or the latest of these dates as the
focal date for both sides, and find the products as in the last
case.
III. Divide the balance of the products by 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 balances 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?
John Wilson,
Dr.
Cr.
1859. 1
Jan.
1
Feb.
4
t(
20
To Mdse.
" Cash.
1859. 1
448
00
Jan.
20
364
00
Feb.
16
232
00
(<
2.5
By Am't bro't forward | 560
" 1 Carriage 264
'♦ Cash 900
Ans.
S Balance, $680.
I 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?
Give analysis. Knle.
RATIO.
279
Isaac Foster.
Dr
Cr.
1858o
1858.
Jan. 1
To Mdse. on 3 mo.
145
86
May
11
By Cash
11
00
12
u i< .. 5 a
37
48
July
12
" " ....
15
00
June 3
<( « « 3 ((
12
25
Oct.
12
« ((
82
00
Aug. 4
« <( << 2 <'
66
48
Ans,
^ $154.07, face of note.
I Mar. 26, 1858, date.
RATIO.
304:* Batio is the comparison with each other of two num-
bers of the same kind. It is of two kinds — arithmetical and
geometrical.
365. Aritlmietical Ratio is the difference of the two
numbers.
366. Geometrical Ratio is the quotient of one number
divided bj the other.
36T. When we use the word ratio alone, it implies geo-
metrical ratio, and is expressed by the quotient arising from
dividing one number by the other. Thus, the ratio of 4 to 8
is 2, of 10 to 5 is ^, &c.
368. Ratio is indicated in two ways.
1st. By placing two points between the numbers compared,
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 f ;
the ratio of 4 to 6 is |.
369. The Terms are the two numbers compared.
370. The Antecedent is the first term.
37 li The Consequent is the second term.
37^. No comparison of two numbers can be fully ex-
plained but by instituting another comparison ; thus, the com-
NoTE. It is thought best to omit the questions at the bottom of the pan-es. in the
remaining part of this work, leaving the teacher to use such aa may be deemed ap-
propriate. "^ *^
280 RATIO.
parison 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 ^, would not be complete nor correct. But if we make
iinifi/ the standard of comparison, and use it as one of the
terms in illustrating the relation of the two numbers, and
gay 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-, U7i{ty in both cases
being the standard of comparison, then the whole meaning i3
conveyed.
373* A Direct Hatio arises from dividing the consequent
by the antecedent.
374. An Inverse or Eeciprocal Ratio is obtained by di-
■viding the antecedent by the consequent. Thus, the direct
ratio of 5 to 15 is -^ = 3 ; and the inverse ratio of 5 to 15 is
IF — 3-
375* A Simple Ratio consists of a single couplet; as
3: 12.
376. A Compoimd Ratio is the product of two or more
simple ratios. Thus, the compound ratio formed from the
simple ratios of3:Gand8:2isfXf = 3X8:6X2 =
377. In comparing numbers with each other, they must
be of the same kindj and of the same denomination.
37 8« The ratio of two fractions is obtained by dividing
the second by the first ; or by reducing them to a common de-
nominator, when they are to each other as their numerators.
Thus, the ratio of tIt • f is f "T" tV = ^^ — 2, which is the
same as the ratio of the numerator 3 to the numerator 6 of
the equivalent fractions f'^ 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 divi'
«or, dividend, and quotient, and these principles become
I
RATIO. 281
GENERAL PRINCIPLES OF RATIO.
Prix. I. Multiplying the consequent multiplies the ratio;
dividing the consequent divides the ratio.
Prin. II. Multiplying the antecedent divides the ratio ; di-
viding the antecedent multiplies the ratio.
Prix. III. Multiplying or dividing both antecedent and con-
sequent by the same number does not alter the ratio.
These three principles may be embraced in one
GENERAL LAW.
A change in the consequent produces a like change in the
ratio ; but 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.
1. What part of 9 is 3?
I = ^ ; or, 9 : 3 as 1 : ^, that is, 9 has the same ratio to 3 that 1
h«s to \.
2. What'part of 20 is 5 ? Ans. f
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. b\.
6. What is the ratio of 6 to 8| ? Ans. |J.
7. What is the ratio of 6^ to 78 ? Ans. 12.
8. What is the ratio of 16 to 66? Ans. 4|.
9. What is the ratio of f to f ? Ans. \.
10. What is the ratio off to -rV? ^««- §•
11. 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 6.3 s to 8 s. 6 d. ? Ans. 1^
14. What is the ratio of 5.6 to .56 ? Ans. ^^.
15. What is the ratio of 19 lbs. 5 oz. 8pwts. to 25 lbs. 11
oz. 4'pwts. ? Ans, 1^.
1 6. What is the inverse ratio of 1 2 to 16.-^ Ans. f .
17. What is the inverse ratio of f to |? Ans, ■^^,
18. What is the inverse ratio of 5| 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 f, what is the an-
tecedent ? Ans. 1^.
22. If the antecedent be f and the ratio ^, what is the
consequent ? A?is. -^q.
PROPORTION.
380. Proportion is an equality of ratios. Thus, tlie ratios
6 : 4 and 12:8, each being equal to f , 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 =1 4 : 10.
38S. Since each ratio consists of two terms, every pro-
portion must consist of at least four terms.
383. The Extremes are the first and fourth terms.
384. The Means are the second and third terms.
38«S* Tiiree numbers may be in proportion when the first
is to the second as the second is to the third. Thus, the num-
bers 3, 9, and- 27 are in })roportion since 3 : 9 : : 9 : 27, the
ratio of each coui)let being 3.
In such a proportion the second term is said to be a mean
proportional between the other two.
380. In every proportion the product of the extremes is
equal to the product of the means. Thus, in the proportion
3 : 5 : : : 10 we have 3X10=5X6.
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 alternation 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
Rule. I. Divide the product of the extremes hi/ one of the
means, and the quotient will be the other mean. Or,
II. Divide the product of the means hy one 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:: (oc) : 50^ Ans. 120.
2. 42 : 70::3 : (J"^. ^^^^' A7is. 5,
3. (O :30 :: 20 : 100. ' ' Ans. 6.
4. 1 : (/-u) :: 7 : 84. Aiis. 12.
5. 48 yd. :( ):: $G7.25 : $201.75. Ans. 144 yd.
6. 3 lb. 12 oz. : ( ) : : $3.50 : $10.50. Ans. 11 lb. 4 oz.
7. ( ) : $38.25 : : 8 bu. 2 pk. : 76 bu. 2 pk. %is. $4.25.
8. 4i- : 38^ : : ( ) ; 761. Ans. 8^.
9. ( ) : 12 : : f : If Ans. 7.
r
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.
3110. 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, distance,
weight, goods bought or sold, money at interest, &c.
393« Effects may be regarded as whatever is accom-
284 PROPORTION.
plished by action of any kind, the thing produced or consumed,
money paid, &c.
3i$3. Causes and effects are of two kinds — simple and
compound.
394. A Simple Cause, or Effect, contains but one element ;
as goods purchased or sold, and the money paid or received
tor 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 connection
with dimensions, length and breadth, &c.
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 produce 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 num-
bers, we may say that
1st Cause : 1st Effect : : 2d Cause : 2d Effect,
which will produce the same numerical result.
But as ratio is the result of comparing two 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 co^t?
Note. The required term will be denoted by a ( ), and designated
♦« blank."
SIMPLE PROPORTION,
285
STATEMENT.
tons. tons. $ $
5 : 3 : : 30 : ( )
Ist cause. 2d cause. 1st effect. 2d effect.
OPERATION.
5x( )==3X30
3X30^ 01Q ^
/■ \ _- ^ ^z= $18, Ans,
Analysis. In this
example an eject is
required, and 5 tons
must have the same
ratio to 3 tons, as
$30, the cost of 5
tons, to (blank) dol-
lars, the cost of 3
tons.
bar.
15
t cause.
STATEMENT,
bar. $
: ( ) : : 90
2d cause. 1st effect.
%
30
2d effect.
OPI
00
:ration.
( )
Xt>^
( ) = 5 bar.. Am.
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, Ans.
2. If 15 barrels of flour cost $90, how many barrels can be
bought for $30 ?
Analysis. In this ex-
ample a cause 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, di-
vided by the given mean,
90, will give the required
term, 5, as shown in the operation. Hence we deduce the following
Rule. I. Arrange the terms in the statement so that the
causes shall compose one couplet^ and the effects the other, put-
ting ( ) 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 hj the given mean.
Notes. 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 unit.
3. K 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 frac-
tions, 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.
286 PROPORTION.
390. 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
Second Method.
Every question which properly belongs to simple propor-
tion must contain four numbers, at least three of which must
be given (S89). 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 num-
ber ; in other words, the ratio of the third to the required
number will be the same as the ratio of the other tv/o nuniT
bers.
Regarding the third or odd term as the antecedent of the sec-
ond couplet of a proportion, we find the consequent or required
term by multiplying the antecedent by the ratio (S79).
By comparing the two like numbers, in any given question,
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 tlie third term, then the ratio will be less than 1,
and the two like numbers may be arranged in the form of a
proper fraction, as a multiplier.
1. If 4 cords of wood cost $12, what will 20 cords cost?
orERATioN. Analysis. It will
X'^'^ X SO te readilv seen in this
IS X •^, written z=z $G0. example,' that 4 cords
^ and 20 cords are the
like terms, and that
$12 is the third term, and of the Fame denomination as the answer
or required term.
If 4 cords cost $12, will 20 cords cost more, or less, than 4 cords?
evidently more : then the answer or required term will be greater
SIMPLE PROPORTION. 287
than the 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 T5^^ $16, 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 ^\ ; hence the ratio of
$48 to the answer is ^%, and the answer will he\\ times $48, which
is $16. Hence the following
Rule. I. With the two given mtmbers, which are of the
same name or kind, form a ratio greater or less than 1, accord-
ing as the answer is to he greater or less than the third given
number.
II. MultijpJy the third nuwher hy this ratio, and the product
will be the required number or answer.
Note. 1. Mixed numbers shouW first be reduced to improper frac-
tions, and the ratio of the fractions found according to (378 )•
2. Reductions and cancellation may be applied as in the Urst method.
The following examples may be solved by either of the
foregoing methods.
EXAHrPLES FOR PRACTICE.
1. If 48 cords of wood cost $120, how much will 20 cords
cost? ' -Ans. $50.
2. If 6 bushels of corn cost $4.7rs how much will 75 bush-
els cost ? Ans. $59.37^-.
3. If 8 yards of cloth cost $3J-, 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
$3.50? Ans. $10.50.
288 SIMPLE PROPORTION.
7. If a staff 3 ft. 8 in. long cast a shadow 1 ft. G 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 $2.75 for 14 pounds of sugar, what will be the cost
of 100 pounds? A71S. $19.64f.
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 2 8 J- gallons of molasses, if 15
hogsheads cost $236.25 ? Ans. $7.1 2 J.
11. If 7 barrels of flour are sufficient for a family 6 months,
how many barrels will they require for 1 1 months ?
12. At the rate of 9 yards for £5 12 s., 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 ? Ans. $20.
15. If the moon move 13° 10' 35' in 1 day, in what time
will it perform one revolution ?
16. If 8 J bushels of corn cost $4.20, what will be the cost
of 13^ bushels at the same rate ? Ans. $6.48.
17. If 1 J yards of cotton cloth cost 6]- pence, how many
yards can be bought for £10 6 s. 8 d. ? Ans. 694f yds.
18. If 12j- cwt. of iron cost $42^, how much will 48| cwt.
cost? Ans. $163.50-1-.
19. What quantity of tobacco can be bought for $317.23,
if 8f lbs. cost $1} ? Ans. 15 cwt. 22.7-|- lbs.
20. If 15| bushels of clover seed cost $156|, how muclj
can be bought for $95.75 ? Ans. 9 bu. 2 pk. 2f qt.
21. If I of a barrel of cider cost $f J, how much will I of
a barrel cost? Ans. $^j.
22. If a piece of land of a certain length, and 4 rods in
breadth, contain f of an acre, how much would there be 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 PROPORTION. 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 14|^- 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 C have each a farm ; B's farm is worth $25
an acre, and C's $30^ ; if in trading B values his land at $28
an acre, what value should C 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 $240 as an eouivalent for the favor ?
A71S. 2 yr. 9 mo. 10 da.
COMPOUND PROPORTION.
400. Compound Proportion embraces that class of ques-
tions in which the causes, or the effects, or both, are compound.
The required term may be a cause, or a single element of a
cause ; or it may be an effect, or a single element of an effect.
1. If 16 horses consume 128 bushels of oats in 50 days,
how many bushels will 5 horses consume in 90 days ?
STATEMENT.
1st can«»e.
2d cause.
1st effect. 2d effect
1 '^ :
( 50
1 90
:: 128 : ( )
Or, 16 X 50 :
5 X 90
: : 128 : ( )
OPERATION.
Analysis. In this ex-
^ X 003 X X^$^
ample the required term
72 bu.
is the second effect; and
HX0 ~
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 ?
Note. These questions are most readily performed by cancellation,
R.p. 13
290
PROPORTION.
2. If $480 gain $84 interest in 30 months, vfhat sum will
gain $21 in 15 months?
STATEMENT.
1st cause. 2d cause. 1st effect. 2d effect.
(480
I 30 •
OPERATION.
( )
15
84 : 21
Analysis. The re-
quired term in this ex-
$240, Ans, ample is an element of
^4 X X^ the second cause ; and
the question may be
read, If $480 in 30 months gain $84^ what principal 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.
UlA
21
'^ 3
Or, 7X 12 : 21 X f :
OPERATION.
GO
8: -s
G ( 8
C0X8XG:( )X3X8
Analysis. In
^i X 8 X 00-^ X $ X 0^
^ X t^X ^ X$X^
Or, n 8
( )
0^
( ) = 80 ft., Ans.
this example the
:: 80 ft., Ans. required term is
the length of the
ditch, and is an element of the
second effect. The question,
as stated, will read thus : if 7
men, in 12 days, dig a ditch 60
feet long, 8 feet Mide, 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 ?
Hence we have the following
Rule. T. Of the (jiven terms, select those tvhich constitute
the causes, and those which constitute the effects, and arrange
them in couplets, putting ( ) in place of the required term.
COMPOUND PROPORTION. 291
II. Then, if the hlanh term ( ) occur in either of the ex-'
tremes, make the product of the means a dividend, and the
product of the extremes a divisor ; but if the blank term occur
in either mean, make the product of the extremes a dividend,
and the product of the means a divisor.
Notes. 1. The causes must be exactly alike in the member and kind
of their terms ; the same is true of the effects.
2. The same preparation of the terms by reduction is to be observed
as in simple proportion.
4:01. 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 23 rods in 8 days ?
OPERATION. Analysis. We see in
^S^ X^^ this example that all the
jt^^ X — X =24. men. terms appear in couplets,
4-'^ P 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
proportions, if this third or odd term be multiplied by the compound
ratio, or by the simple ratio of each couplet successively, the prod-
uct 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, 42
rods and 28 rods, with the third term, 18 men. If 42 rods require
18 men, how many men will 28 rods require ? less men ; hence the
ratio is less than 1, and the multiplier a proper fraction, || ; 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, '^-^. 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 |f X \^ ^== 24, as
shown in the operation.
2. 5 compositors, in 16 days, of 14 hours each, can compose
20 sheets of 24 pages in each sheet, 50 lines in a page, and
X0
^0
$0
40
292 ^ COMPOUND PROPORTION.
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, GO lines in a
page, and 50 letters in a line ? Ans. 32 days.
OPERATION,
days. comp. hours, sheets, pages, lines, letters.
16X T^ X Jy^ X la X ^f X f a X U= 32 days.
BY CANCELLATION. ANALYSIS. The required term or an-
1 6 swer is to be in days ; and we see that
g all the terms appear in pairs or couplets,
^ . except the 16 days, which is of the same
^ kind as the answer sought.
40 We will proceed to compare the terms
^09 of each couplet with the 16 days. First,
^w if 5 compositors require J16 days, how
. many days will 10 compositors require ?
^^ less days ; hence the multiplier is the
32 days, A71S. proper fraction ^\, and we have 16 X iV
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 ^, and we have 16 X x\ X V*
Next, if 20 sheets require 16 days, how many days will 40 sheets
require ? more days ; hence the multiplier is the improper fraction
1^^, and we have 16 X yu X V X -f^- Pursuing the same method
with the other couplets, we obtain the result as shown in the opera-
tion. Hence we have the following
R.ULE. I. 0/ 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 successivehjy
and the product will he the answer sought.
Note. By the odd term is meant the one that is of the same kiiid
as the answer.
The following examples may be solved by either of the
given methods.
EXAMPLES FOR PRACTICE.
1. If 16 horses consume 128 bushels of oats in 50 days,
how many bushels will 5 horses consume in 90 days ?
COMPOUND PROPORTION. 293
f >
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 oGO miles? A7is. 10^ 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 ? A?is. 900.
5. If it require 1200 yards of cloth |- wide to clothe 500
men, how ipany yards which is ^ wide will it take to clothe
960 men? Ans. 329U yds.
G. 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 ? A?is. 6 men.
7. If 4 men, in 2^- days, mow 6| acres of grass by work-
ing 8^ hours a day, how many acres will 15 men mow in 3f
days by working 9 hours a day ? Ans, 40 j^ 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 4J
miles an hour, will he travel 600 miles ? Ans. 14f days.
9. If 21 yards of cloth If yards wide cost 83.371, what
cost 36J^ yards, Ij 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 241 days, working 12^
hours each day, in how many days will 75 men do the same,
when they work but lOi hours each day? Ans. 21 days.
13. If 24 men dig a trench 33^ yards long, 5| wide, and
31 deep, in 189 days, working 14 hours each day, how many
hours per day must 217 men work, to dig a trench 23^ yards
long, 3 1 wide, and 2^ deep, in 5^ days ? Ans, 16 hours.
294 PARTNERSHIP.
• PARTXERSHIP.
4:0S. PartnersMp is a relation established between two
or more persons in trade, bj which they agree to share the
profits and losses of business.
403. The Partners are the individuals thus associated.
4:04:* Capital, or Stock, is the money or property invested
in trade.
405. A Dividend is the profit to be divided.
406* An Assessment is a tax to meet losses sustained.
CASE I.
407. To find each partner's share of the profit or
loss, when their capital is employed for equal periods 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 the whole capital
%^{)Q employed is $300
-^^— -f $400 = $700, it
$700, whole stock. is evident that A
53-§ =z ^, A's share of tho stock. fumishcs 4^^ i= ^
40.0 ;— ; 4 jj.g « « « of the capital, and
$182 X ? = $78, A's share of the gain. ^ t^f —^ f .^^"^
^ ^ . ^ ^ capital. And since
$182Xf = $104,D-, 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 Avhole stock, A will have
4 of the entire profit, and B 4, as shown in the operation.
We may also regard the whole capital as the Jirst causCy
and each man's share of the capital as the second causCy the
whole profit or loss as ihafrst effect, and each man's share of the
profit or loss as the second effect, and solve by proportion thus :
PARTNERSHIP.
1st cause.
2d cause. 1st effect.
2(1 cffecU
$700
S300 :-; $182
= ( )
$700 ,
$400 : : §182
= ( )
t00
^00^
;?00
400*
C )
xn''^
( )
j'S^ss
295
( ) — ^'^8, A's profit.
Hence we have the following
( } — 'i{5l^'*> L's profit.
EuLE. Multiply the whole profit or loss hj the ratio of the
whole cajntal to each marCs share of the capital. Or,
Tlie whole 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 C 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; thej 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 will 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?
(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 ; how much did each receive ?
7. A cargo of corn, valued at $3475.60, was entirely lost ;
^ of it belonged to A, ^ of it to B, and the remainder to C ;
how much was the loss of each, there being an insurance of
$2512? ^ws, $120.45, A's. $240.90, B's. $602.25, C's.
296 PARTNERSHIP.
8. Three persons engaged in the lumber trade ; two of the
persons furnished the capital, and the third managed the busi-
ness; they gained $2571.24, of which C received $6 as often
as D $4, and E had ^ as much as the other two for taking care
©f the business ; how much was each one's share of the gain ?
Ans. $1285.G2, C's. $857.08, D^s. $42^.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 ^jiuch is D's share of the gain ?
^^^•{b;
$6000. C, $5055.75.
$840
CASE II.
$8402.25. D's gain, $1014.
4:08. To find each partner's share of the profit or
loss when their capital is employed for unequal periods
of time.
It is evident that the respective shares of profit and loss will
depend upon two conditions, viz. : (he amount of capital in-
vested 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's capital for 1 mo. use of $450 capital
$300 X 9 = $2700, Kb « « « for '^ months is the
~ — - — same as the use of
$5850, entire « « " 7 tj^es $450, or
m^ = /jj, A's share of the entire capital. ^^^^^ ^or 1 month;
Zt(i(^=6 r,,, « „ „ u and of $300 for 9
lr^Vr Ky 7 ' «^o i months is the same
$loG X -i^Tr=:$84, A'sloss. *!, en*'
^ ITT "^^^j '" =• as the use of 9 times
$150 X 1% = S72, B'8 «' $300, or $2700 for
1 month. The en-
tire capital for 1 month is equivalent to $3150 -[-$2700= $5850.
If the loss, $156, be divided between the two partners, according^ to
Case I, the results will be the loss of each as shown in the operation.
PARTNERSHIP. 297
Examples of this kind may also be solved by proportion as in
Case I, the causes being compounded of capital and time ; thus,
$5850 : $3150
: $156 : ( )
$5850 : $2700
: $156 : ( )
$^$0
U$0'
$$$0 \ m00^
( )
U$'^-
( ) t$^'^
= $8-^
r, A's loss.
( ) = $12, WbIoss.
(
Hence the following
Rule. Multiply each man's capital hj the time t't is em-
ployed in trade, and add the products. Then multiply the
entire pro/it or loss by 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 by 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 by 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 ; how much was each man's share of the
gain ?
3. Two merchants formed a partnership for 1 8 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 ; how much 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 C 6 hands 4 days ; the whole crop amounted to
372 bushels ; how much was each one's share ?
5. AVilliam 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
13*
•298 ANALYSIS.
a capital of $2000; four months after they admitted J. New-
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; how
much was each one's share ? ( $2106, Gallup's.
Ans.-\ $1300, Decker's.
( $ 982.80, Newman^s.
6. Two persons engaged in partnership with a capital of
$5600 ; A's 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 ? A7is. 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 j how much did each put in, and what
did each gain? ( A, $700 capital ; $26 gain.
^ws.^B, S630 " $39 «
/ C, $600 « $52 «
ANALYSIS.
4:09, 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.
4: 1 O* In solving questions by analysis, we generally reason
from the ffiven number to unity ^ or 1, and then from unity, or
1, to the required number.
411. United States money is reckoned in dollars, dimes,
cents,andmilLs(180),one dollar being uniformly valued in all
the States at 100 cents ; but in most of the States money is
sometimes still reckoned in pounds, sWllings, and pence.
Note. At the thne of the adoption of our decimal currency by
Cono;ress, m 1786, the colonial cutrmc;/, or bills of credit, issued by the
colonies, had depreciated in value, and this depreciation, bcin^ inicqual
in the different colonies, gave rise to the different values of the State
currencies ; and this variation continues wherever the denominations of
shillings and pence are in use.
OPERATION.
4^^
42X3=rl26s. ^
8
26-^-6z=$21 Or, —
ANALYSIS. 299
4H3. In New England, Indiana, \
Illinois, Missouri, Virginia, Kentucky, > $1 r=: G s. =: 72 d.
Tennessee, Mississippi, Texas, , , . ^ ,)
New York, Oiiio, Michigan, $1 = 8 s. = 96 d.
New Jerse3% Pennsylvania, Bela- )
TVT 11 t$l=-7s. 6d.z:=90d
Ware, Maryland, )
South Carolina, Georgia, ) $l=4s. 8d. = 56d.
The Dominion of Canada, / $1 = 5 s. = 60d.
EXAMPLES FOR PRACTICE.
1. What will be the cost of 42 bushels of oats, at 3 shillings
per bushel, New England currency ?
Analysis. Since
1 bushel costs 3 shil-
lings, 42 bushels will
cost 42 times 3 s., or
121, Ans. 42X3 = 126 s.; and
as 6 s. make 1 dollar
New England currency, there are as many dollars in 126 s. as 6 is
contained times in 126, or $21.
2. What will 180 bushels of wheat cost at 9 s. 4d. per
bushel, Pennsylvania currency ?
OPERATION. Analysis. Multi-
plying the number of
bushels by the price,
and dividing by the
value of 1 dollar re<.
duced to pence, we
we have $224. Or,
when the pence in the
given price is an aliquot part of a shilling, the price may be reduced.
to an improper fraction for a multiplier, thus : 9 s. 4 d. = 9^ s. =:
^- s., the multipher. The value of the dollar being 7 s, 6 d. = 7^ s.
=^ J^, we divide by ^, as in the operation.
3. What will be the cost of 3 hhd. of molasses, at 1 s. 3 d.
per quart, Georgia currency ?
00
112
Or,
28
$224
2
1224, Am.
30(^ ANALYSIS.
m
2
$t
OPERATION.
3 Analysis. In this example we first
^0^ reduce 3 hhd. to quarts, by multiplying
V by 63 and 4, and then multiply by tho
price, either reduced to pence or to an
^'^ improper fraction, and divide by the
405 00 "value of 1 dollar reduced to the same
'. denomination as the price.
8202.50
4. Sold 9 firkins of butter, each containing 56 lb., at 1 s. 6 d.
per pound, and received in payment carpeting at 6 s. 9 d. 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-
K/> vide the cost of the butter by the price of a
.g miit of the article received in payment, reduced
^ to the same denominational unit as the price
112 yd. ^^ ^ ^^^^^ °^ ^^® article sold. The result will be
the same in whatever currency.
5. What will 3 casks of rice cost, each weighing 126
pounds, at 4 d. per pound, South Carolina currency ? A?is. $27.
6. How many pounds of tea, at 7 s. per pound, must be
given for 28 lb. of butter, at 1 s. 7 d. per pound ? Ans. 6 J.
7. Bought 2 casks of Catawba wine, each containing 72
gallons, for $648, and sold it at the rate of 10 s. 6 d. per quart,
Ohio currency ; how much was my whole gain ? Ans. $108.
8. "What will be the expense of keeping 2 horses 3 weeks
if the expense of keeping 1 horse 1 daf be 2 s. 6 d., Canada
currency ? Ans. $21.
9. How many days* work, at 6 s. 3 d. per day, must be
given for 20 bushels of apples at 3 s. per bushel ? A7is. 9^.
10. Bought 160 lb. of dried fruit, at Is. 6d. a pound, in
New York, and sold it for 2 s. a pound in Philadelphia ; how
much was my whole gain ? Ans. $12.66§.
11. A merchant exchanged 43^ yards of cloth, worth 10 s.
6 d. per yard, for other cloth worth 8 s. 3 d, per yard ; how
many yards did he receive ? Ans. 55^.
ANALYSIS. 301
12. What will be the cost of 300 bushels of wheat at 9 s.
4 d. per bushel, Michigan currency ? Arts. $350.
13. If J of f of a ton of coal cost $2f , how much will f of
6 tons cost ?
OPERATION.
$ I t^^ Analysis.. Since | of ^ z= -|| of a ton costs
1$ ' ^$^ ^^ = ^¥' 1 ton will cost 28 times -^-^ of $^,
rf I Hrio or $i/ X ft ; and 4 of 6 tons = ^ tons, will
"^ ' ^^ . cost Y times fi of $ ^ — $i6.
116, 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.
1^ $ 10 ^0010 $
-- X — X — X —7— X — X— = 100 da.
I j^4 $ ^0 4
Analysis. Since 8 men require 12 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 i 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 1 as many days by working 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 coffee, 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 su-
^Z^ gar are equal in value to 21 lb. of
5 rice, 1 lb. of sugar is equal to -^-^
11 of 21 lb. of rice, or M =| lb. of
y^ • rice, and 5 lb. of sugar are equal
to 5 times | lb. of rice, or ^^^ lb. ;
3 I 385 if 3 lb. of coffee are equal to 5 lb.
A i.;)Qi 11 ^^ sugar, or \^ lb. of rice, 1 lb. of
* * coffee is equal to ^ of \^ lb. of
rice, or f| lb., and 11 lb. of coffee are equal to 11 times f | lb. of
rice, or -^Z lb. ; if 2 lb. of tea are equal to 11 lb. of coffee, or ^^^ ^b.
of rice, 1 lb of tea is equal to -| of ^^s/ lb. of rice, or W^- lb., and 12
lb. of tea are equal to 12 times \\^ lb. of rice, or ^^ 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 4| cords of wood, how many cords
can be bought for $24f ? A?is. 11.
18. Gave 52 barrels of potatoes, each containing 3 bushels,
worth 33^ cents a bushel, for Qo 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 46j 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 C put in 15
sheep 6§ 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? A71S.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 § of a yard wide are
equal to 28 Awards ^ of a yard wide?. Ans. 314.
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. IS,
ANALYSIS. 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 2y\j 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
1 2 o'clock, M. ; when will they be exactly together the third
time after this ?
OPERATION. Analysis. Since
12 X t't X 3 := '^fr^' ^^^ minute hand pass-
Ans. 3h. IGmin. 21^^ sec, P. M. ^s the hour hand 11
times in 12 hours, if
both are together at 12, the minute hand will pass the hour hand
the first time in J^ of 12 hours, or l^j hours ; it Avill pass the hour
hand the second time in ^^ of 12 hours, and the third time in j\ of
12 hours, or 3^^ hours, which would occur at 16 min. 21^^^ sec.
past 3 o'clock, P. M.
29. A flour merchant paid $164 for 20 barrels of flour,
giving $9 for first quality, and $7 for second quality ; how
many barrels were there of each ?
orERATiox. Analysis. If all had been
$9 X 20 ziz $180 ; first quality, he would have paid
^IgQ $164 = $16. $180, or $16 more than he did
CQ CT CO • ^^^^'' -^^'^^T harrel of second
, - ^ ^ , , ', quality made a difference of $2
IG -1- 2 = 8 bbl., 2d qualily. j^ ^^^ ^^^^ . ^^^^^ ^j^^^.^ ^^.^^^
20 — 8 — 12 bbl., 1st « j^s many barrels of second qual-
ity as $2, the difference in the
cost of one barrel, Is contained times in $16, &c.
SO. 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 ;
lie sold them again at the rate of 3 for 8 cents, and gained on
the whole 108 cents; how many oranges did he buy?
304 ANALYSIS.
OPERATION. Analysis. For
^ -j- I z=L 4i; ^1 -^2=^ f I, average cost. those he bought
I — Zl=z i| z= li cts., gain on each. at the rate of 3 for
108 -~ U = 00, number of oranges. V^"^' ^^ ^'^"^ ^
of a cent each, and
for those he bought at the rate of 5 for 8 cents he paid | of a cent
each; and |-|-|=i4| cents, what he paid for 1 of each kind,
which divided by 2 gives \ | cents, the average price of all he bought.
He sold them at the rate of 3 for 8 cents, or | cents each ; the dif-
ference between the average cost and the price he sold them for, or
f — if:= 15==^ ^i 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, &c.
31. A man bought 10 bushels of wheat and 25 bushels of
corn for $30, and 12 bushels of wheat and 5 bushels of corn
for $20 ; 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 corn, which
cost $20, we find the cost of 10 bushels of wheat to be $14 ; there-
fore the cost of 1 bushel is -^^ of $14, 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 C agree lo build a barn for $270. A and
B can do the work in 16 days, B and C in 18^ days, and A
and C in 11^ days. In how many days can all do it working
together ? In how many days can each do it alone ? AVhut
part of the pay ought each to receive ?
OPERATION.
W.
C.
1st lot, 10
25 $30
2d « 12
5 $20
l8t-^5=:2
5 $6
10.
....$14
Ibu. W,
.^ $1.40
1 bu. C.
= $ .64
ANALYSIS.
305
•&V — To »
■eV + A +
OPEEATTON.
^ = -^, what A and B do in 1 day.
■i^ = -,% " J^-^<^ " "
A and C " "
z= i|, -what A, B, and C do in
2 days.
18 _i_ 2 r=r -g^Q, what A, B, and C do in 1 day.
1 -1- ^^ zzz 8|- days, time A, B, and C, will do the
whole work together.
Pj = 8^ ; 1 -^ -io = 20 da., C alone.
ss
Po=^'l
■^«0 = 26|da.,A"
^-sV==8^0-l-^A=^40da.,B «
^^ X 8| =r A, the part of the whole C did.
■ioX^=h " " " ^ "
$270 X I = 8120, C8 share.
$270 X I = $90, A-s «
$270 XI ==$60, Bs «
Analysis. Since
A and B can do the
■work in 16days,they
can do ^\ of it in h
day; B and C, in
131 or Y days» they
can do ^^ of it in 1
day; AandC, inll|
or ^ days, they can
do -^ of it in 1 day.
Then A, B, and C,
by working 2 days
each, can do ^-j-
^«o + 8V = i!o^'the
work, and by work-
ing 1 day each they
can do J of if, or ^
of the work ; and it
will take them as
many days working
together to do the
T\hole work as g\ 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 -^, B -^q, and C -g^.
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 26f days, B 40 days, and C 20 days.
And each should receive such a part of $270 as would be ex-
pressed 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 1 2 oranges
and 10 lemons cost 54 cents, what is the price of 1 of each ?
A71S. 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 reinforced
by GOO men, upon what allowance per day must each man be
put, that the same provisions may last 30 days ? Ans. 7^ oz.
35. There are 54 bushels of grain in 2 bins ; and in one bin
are 6 bushels less than i as much as there is in the other;
how many bushels in the larger biu ? Ans. 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? jins. 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 C in 4 days ; what time will B
require to execute a piece of work which A can do in G
weeks.? Jns. 11^ weeks.
38. How many yards of cloth, f of a yard wide, will line
36 yards 1^ yards wide ? Ans. GO.
39. How many sacks of coffee, each containing 104 lbs ,
at 10 d. per pound N. Y. currency, will pay for 80 yards of
broadcloth at $3^ per yard ? Ans. 24.
40. A person, being asked the time of day, replied, the time
past noon is equal to | of the time to midnight ; what was
the hour ? Jns. 2, P. M.
41. A market woman bought a number of poaches 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 J- of his money, and ^ of the
remainder, had $10 left ; how much had he at first ?
44. If 30 men can perform a piece of work in 1 1 days, how
many men can accomplish another piece of work, 4 times as
large, in -^ of the time ? Ans. GOO.
45. If IC-} lb. of coffee cost $3|-, how much can be bought
for $1.25? Ans. G^ lb.
46. A man engaged to write for 20 days, receiving $2.50
for every day he labored, and foifciting $1 for every day ho
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 9f hours ? Ans. f .
ALLIGATION MEDIAL. 307
ALLIGATION.
413. Alligation treats of mixing or compounding two or
more ingredients of different values. It is of two kinds — Alli-
gation Medial and Alligation Alternate.
4: 1 4:« Aliigation Medial is the process of finding the aver-
age price or quality of a compound of several simple ingredi-
ents 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 ?
OPERATION. Analysis. Since 40 bushels
80 X 40 = $32.00 of rye at 80 cents a bushel is
70X25:= 17.50 worth $32, and 25 bushels of com
1 KQ v^ 15 __ 22 50 ^^ "^^ cents a bushel is worth
— $17.50, and 15 bushels of ^Yheat
80 ) 72.00 at $1.50 a bushel is worth $22.50,
therefore the entire mixture, con-
sistino: of 80 bushels, is worth
$.90, Ans.
$72, and one bushel is worth -^^ of $72, or 72 -^ 80 = $.90.
Hence the following
Rule. Divide the entire cost or value of the ingredients
hy the sum of the simples.
EXAMPLES FOR PRACTICE.
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 gal-
lon 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|- 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 10 pounds at 11 cts., and sold the mixture
at 10 cents per pound; did he gain or lose by the sale, and
how much? Ans. He gained 16 cts.
308 ALLIGATION ALTERNATE.
6. A grocer bought 7^ dozen of eggs at 12 cents a dozen,
8 dozen at 10^ cents a dozen, 9 dozen at 11 cents a dozen,
and 10^ 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? A?is. IG^ 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 91 cents a pound ; how much
was my whole gain ? A7is. $Q.
415* 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.
410. 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 ?
OPERATiox. Analysis. Since on every in-
("21^12) gredient used whose price or qual-
j 7 JL 3 j ^ws. jjy jjj i^gg ^Ijj^j^ ^j^jj mean rate there
will be a ga{?i, and on every in-
gredient whose price or quality is g)*ea{er 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 repre-
sent the unit of t'aliie. 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 ; and to lose $1
would require |f of a bushel, which wo place opposite the 7.
In every case, to find the unit of value we must divide $1 by the
gain or loss per bushel or pound, &c. Hence, if, every time we take
J^ of a bushel of timothy seed, we take ^ of a bushel of clover seed,
the gain and loss will be exactly equal, and wc shall have ^ and -^
for the proportional quantities.
ALLIGATION ALTERNATE.
309
1
2
13
4
5
3
i
4
4
1 ^
i
1
1
7
1
2
2
I 10
i
3
3
If we wish to express the proportional numbers in integers, we
may reduce these fractions to a common denominator, and use their
numerators, since fractions having a common denominator are to
each other as their numerators. ( 37§ ) thus, |- and ^ are equal to
I and |, and the proportional quantities are 2 bushels of timothy
seed to 3 bushels of clover seed.
2. What proportions of teas worth respectively 3, 4, 7 and
10 shillings a pound, must be taken to form a mixture worth
6 shillings a pound ?
OPERATION. Analysis. To preserve the
equality of gains and losses, we
must always compare two prices
cr simples, one greater and one
less than the mean rate, and
treat each pair or couplet as a
separate example. In the given
example we form two couplets,
and may compare either 3 and 10, 4 and 7, or 3 and 7, 4 and 10.
"We find that ^ of a lb. at 3 s. must be taken to gain 1 shilling,
and ^ of a lb. at 10 s. to lose 1 shiUing ; also ^ of a lb. at 4 s. to gain
1 shilhng, and 1 lb. at 7 s. to lose 1 shilHng. These proportional
numbers, obtained by comparing the two couplets, are placed in
columns 1 and 2. If, now, we reduce the numbers in columns 1 and
2 to a common denominator, and use their numerators, we obtain
the integral numbers in columns 3 and 4, which, being arranged in
column 5, give the proportional quantities to be taken of each.*
It will be seen that in comparing the simples of any couplet, one
of which is greater, and the other less than the mean rate, the pro-
portional number finally obtained for either term is the difference
between the mean rate and the other term. Thus, in comparing 3
and 10, the proportional number of the former is 4, which is the
difference between 10 and the mean rate 6 ; and the proportional
number of the latter is 3, which is the difference between 3 and the
mean rate. The same is true of every other couplet. Hence, when
the simples and the mean rate are integers, the intermediate steps
taken to obtain the final proportional numbers as in columns 1, 2, 3,
and 4, may be omitted, and the same results readily found by taking
the difference between each simple and the mean rate, and placing
it opposite the one with which it is compared.
* Prof. A. B. Canfield, of Oneida Conference Seminary, N. Y., used this method of
Alligation, essentially, in th© instruction of his classes as early as ISIG, and he wag
doubtless the author of it.
310 ALLIGATION ALTERNATE.
From the foregoing examples and analyses we derive the following
Rule. I. Write the several prices or qualities in a column,
and the mean price or quality of the mixture at the left,
IL Form couplets hy comparing any price or quality lees,
with one that is greater than the mean rate, placing the part
which must he used to gain 1 of the mean rate opposite the less
simple^ and the part that must be used to lose 1 opposite the
greater simple, and do the same for each simple in every couplet.
111. If the proportional numbers are fractional, they may be
reduced to integers, and if two or more stand in the same hori-
zontal line, they must be added; the final results will be the pro-
portional quantities required.
Notes. 1. If the numbers in any couplet or colimin have a com-
mon factor, it may be rejected.
2. We may also multiply the numbers in any couplet or colmnn by
any multiplier we choose, without affecting the equality of the gains
and losses, and thus obtain an indefinite number of results, any one of
which being taken will give a correct final result.
EXAMPLES FOR PRACTICE.
3. A grocer has sugars worth 10 cents, 11 cents, and 14
cents per pound ; in what proportions may he mix them to
form a mixture worth 12 cents per pound?
Ans, 1 lb. at 10 cts., and 2 lbs. at 11 and 14ct8.
4. What proportions of water at no value, and wine worth
$1.20 a gallon, must be used to form a mixture worth 90 cents
a gallon ? Ans. 1 gal. of water to 3 gals, of wine.
5. A farmer had sheep worth $2, $2J, $3, and $4 per
liead ; what number could he sell of each, and realize an
average price of 82| per head ?
Ans. 3 of the 1st kind, and 1 each of the 2d and 3d,
and 5 of the 4th kind.
6. What relative quantities of alcohol 80, 84, 87, 94, and
96 per cent, strong must be used to form a mixture 90 per
cent, strong ?
Ans. 6 of the first two kinds, four of the 3d, 3 of the 4th,
and 16 of the 5th
ALLIGATION ALTERNATE. 311
CASE II.
417. When the quantity of one of the simples is
limited.
1. A miller has oats worth 30 cents, corn worth 45 cents,
and barley worth 84 cents per bushel ; he desires to form a
mixture worth GO cents per bushel, and which shall contain 40
bushels of corn ; how many bushels of oats and barley must
he take ?
OPERATION. Analysis. By
( 30 g^tj 4 4 20 ) the same process
60 < 45 ^V 8 8 40 [ Ans. ^^ ^^ ^^^^ ^ we
( 84 2^ ^V 5 5 10 50 ) fi^^ ^^^ P^-°P°^-
tional quantities
of each to be 4 bushels of oats, 8 of corn, and 10 of barley. But
we wish to use 40 bushels of corn, which is 5 times the propor-
tional number 8, and to preserve the equality of gain and loss we
must take 5 times the proportional' quantity of each of the other
simples, or 5 X 4 •.= 20 bushels of oats, and 5 X 10 :=: 50 bushels
of barley. Hence the following
Rule. Find the proportional quantities as in Case I,
Divide the given quantity by the proportional quantity of the
same ingredient, and midtipty each of the other proportional
quantities hy the quotient thus obtained.
EXAMPLES FOR PRACTICE.
2. A merchant has teas worth 40, GO, 75, and 90 cents per
pound ; how many pounds of each must he use with 20 pounds
of that worth 75 cents, to form a mixture at 80 cents.
Ans. 20 lbs. each of the first three kinds, and 130 lbs. of
the fourth.
3. A farmer bought 24 sheep at $2 a head ; how many
must he buy at $3 and $5 a head, that^he may sell the whole
at an average price of $4 a head, without loss ?
Ans. 24 at $3, and 72 at $5.
4. How much alcohol worth 60 cents a gallon, and how
much water, must be mixed with 180 gallons of rum worth
$1.30 a gallon, that the mixture may be worth 90 cents a
gallon ? Ans, 60 gallons each of alcohol and water.
312
ALLIGATION ALTERNATE.
5. How many acres of land worth 35 dollars an acre must
be added to a larm of 75 acres, worth $50 an acre, that the
average value may be $40 an acre ? Ans. 150 acres.
6. A merchant mixed 80 pounds of sugar worth 6^ cents
per pound with some worth 8^ cents and 10 cents per pound,
so that the mixture was worth 7J- cents per pound ; how much
of each kind did he use ?
CASE III.
418. When the quantity of the whole compound is
limited.
1. A grocer has sugars worth G cents, 7 cents, 12 cents,
and 13 cents per pound. He wishes to make a mixture of
120 pounds worth 10 cents a pound ; how many pounds of
each kind must he use ?
OPERATION. Analysis. By Case
I we find the propor-
tional quantities of each
to be 3 lbs. at 6 cts., 2
lbs. at 7 cts., 3 lbs at 12
cts., and 4 lbs. at 13 cts.
By adding the propor-
tional quantities, we find
that the mixture would be but 12 lbs. while the required mixture is
120, or 10 times 12. If the whole mixture is to be 10 times as much
as the sum of the proportional quantities, then the quantity of each
simple used must be 10 times as much as its respective proportion-
al, which would require 30 lbs. at 6 cts., 20 lbs. at 7 cts., 30 lbs. at
12 cts., and 40 lbs. at 13 cts. Hence we deduce the following
Rule. Find the proportional numbers as in Case I. Z)i-
vide the given quantity/ hy the sum of the proportional quaji-
titles, and midtiphj each of the proportional quantities hy the
quotient thus obtained.^
EXAMPLES FOR PRACTICE.
2. A farmer sold 170 sheep at an average price of 14
shillings a head ; for some he received 9 s., for some 12 s., for
some 18 s., and for others 20 s.; how many of each did he
sell ? Ans. 60 at 9 s., 40 at 12 s., 20 at 18 s., and 50 at 20 s.
i
!
3
I
3
30
\
2!
2
20
i
3
3
30
i
4
4
40
1
il2
120
INVOLUllON. 313
3. A jeweler melted together gold 16, 18, 21, and 21:
carats fine, so as to make a compound of 51 ounces 22 carats
fine ; how much of each sort did he take ? Aiis. 6 ounces
each of the first three, and 33 ounces of the last.
4. A man bought 210 bushels of oats, corn, and wheat, and
paid for the whole $178.50 ; for the oats he paid $^, for the
corn $2, and for the wheat $1^ per bushel ; how many bush-
els of each kind did he buy ? Ans. 78 bushels each of oats
and corn, and 54 bushels of wheat.
5. A, B, and C are under a joint contract to furnish 6000
bushels of corn, at 48 cts. a bushel ; A's corn is worth 45 cts.,
B's 51 cts., and C's 54 cts. ; how many bushels must each put
into the mixture that the contract may be fulfilled ?
6. One man and 3 boys received $84 for 56 days' labor; the
man received $3 per day, and the boys S^, $|, and $1^ re-
spectively ; how many days did each labor ? Ans. The man
16 days, and the boys 24, 4, and 12 days respectively.
INVOLUTION.
419. A Power is the product arising from multiplying a
number by itself, or repeating it several times as a factor;
thus, in 2 X 2 X 2 =: 8, the product, 8, is a power of 2.
4^0. The Exponent of a power is the number denoting
how many times the factor is repeated to produce the power,
and is written above and a little tc the right of the factor; thus,
2 X 2 X 2 is written 2^, in which 3 is the exponent. Exponents
likewise give names to the powers, as will be seen in the
following illustrations :
3 r= 3^ =: 3, the first power of 3 ;
3X3 == 32 =: 9, the ^cond power of 3
3X3X3 r= 33 = 27, the third power of 3.
4^1. The Sq[liare of a number is its second power.
49^2. The Cube of a number is its third power.
4^3. Involution is the process of raising a number to a
given power.
R.P. 14
314 EVOLUTION.
4^4. A Perfect Power is a number that can be exactly
produced by the involution of some number as a root ; thus, 25
and 32 are perfect powers, since 25 = 5 X 5, and 32 i= 2 X
2X2X2X2.
L What is the cube of 15?
OPERATION. Analysis. We multiply
15 X 15 X 15 :r: 3375. Ans. ^^ ^y 1^' ^""^ ^^^, Product
by 15, and obtain 3375,
which is the 3d power, or cube of 15, since 15 has been taken 3
times as a factor. Hence, we have the following
E-ULE. Multiply the number by itself as many times, less 1,
cts there are units in the^ exponent of the required power,
EXAMPLES FOR PRACTICE.
2. What is the square of 25 ? Ans. 625.
3. AVhat is the square of 135 ? Ans. 18225.
4. What is the cube of 72 ? ' Ans. 373248.
5. What is the 4th power of 24 ? Ans. 331776.
6. Raise 7.2 to the third power. Ans. 373.248.
7. Involve 1.06 to the 4th power. Ans. 1.26247696.
8. Involve 12 to the 5th power. Ans. .0000248832.
9. Involve 1.0002 to the 2d power. Ans. 1.00040004.
10. What is the cube off?
OPERATION.
A A 2 2X2X2 _ 2^ 8_
y^ 5 ^ 5 "5X^5X5" 53"~125
It is evident from the above operation, that
A common fraction may be raised to any power, by raising
each of its terms, separately, to the required power.
11. What is the square of f ? Ans. ^.
12. What is the cube of |f ? Ans. f f |J.
13. Raise 24f to the 2d power. Ans. 612-j%.
EVOLUTION.
4^5, A Root is a factor repeated to produce a power ;
thus, in the expression 5 X 5 X 5 = 125, 5 is the root from
which the power, 125, is produced.
SQUARE ROOT. 315
4L20, Evolution is the process of extracting the root of a
number considered as a power, and is the reverse of Involution.
437. The Radical Sign is the character, ^J , which, placed
before a number, denotes that its root is to be extracted.
4:S8. The Index of the root is the figure placed above the
radical sign, to denote what root is to be taken. When no
index is written, the index 2 is always understood.
4:30. A Surd is the indicated root of an imperfect power.
430. Roots are named from the corresponding powers, as
will be seen in the following illustrations :
The square root of 9 is 3, written ^^9 z=z 3.
The cube root of 27 is 3, written .5/'27 =: 3.
The fourth root of 81 is 3, written ^Sl =z 3.
431* Any number whatever may be considered a power
whose root is to be extracted ; but only the perfect powers can
have exact roots.
SQUARE ROOT.
433. The Square Boot of a number is one of the two
equal factors that produce the number ; thus the square root
of 49 is 7, for 7 X 7 =r 49.
433. In extracting the square root, the first thing to be
determined is the relative number of places in a given number
and its square root. The law governing this relation is exhib-
ited in the following examples : —
Roots.
Squares.
Eoots.
Squares.
1
1
1
1
9
81
10
1,00
99
98,01
100
1,00,00
999
99,80,01
1000
1,00,00,00
From these examples we perceive
1st. That a root consisting of 1 place may have 1 or 2
places in the square.
2d. That in all cases the addition of 1 place to the root
adds 2 places to the square. Hence,
316 EVOLUTION.
If we 'point off a number into two-figure periods, commen'
cing at the right hand, the number of fall periods and the left
hand full or partial period will indicate the number of places
in the square root ; the highest period answering to the highest
figure of the root.
434. 1. What is the length of one side of a square plat
containing an area of 5417 sq. ft. ?
OPERATION. Analysis. Since the given figure is
54,17|73.6 a square, its side will be the square root
49 of its area, which we will proceed to com-
7~~ ~~ pute. Pointing off the given number, the
2 periods show that there will be two in-
l4o 42 J tegral figures, tens and units, in the root.
146.0 88.00 '^^^ ^^^^ ®^ ^^^ ^<^°* must be extracted
146 6 87 96 i^am the first or left hand period, 54 hun-
dreds. The greatest square in 54 hun-
4 dreds is 49 hundreds, the square of 7 tens ;
we therefore write 7 tens in the root, at
the right of the given number.
Since the entire root is to be the side of a square, let us form a
Fig I. square (Fig. I), the side of which is 70 feet long.
The area of this square is 70 X 70 z= 4900 sq. ft.,
which we subtract from the given number. This
is done in the operation by subtracting the
square number, 49, from the first period, 54,
and to the remainder bringing down the sec-
ond period, making the entire remainder 517.
If we now enlarge our square (Fig. I) by the addition of 517
square feet, in such a manner as to preserve the square form, its
size will be that of the required square. To preserve the square
form, the addition must be so made as to extend the square equally
in two directions ; it will therefore be composed of 2 oblong figures
at the sides, and a little square at the corner (Fig. II). Now, the
width of this addition will be the additional length to the side of the
square, and consequently the next figure in the root. To find width
we divide square contents, or area, by length. But the length of
one side of the little square cannot be found till the width of the
addition be determined, because it is equal to this width. We will
therefore add the leni^ths of the 2 oblong figures, and the sum will
be sufficiently near the whole length to be used as a trial divisor.
SQUARE ROOT.
317
Fig. II.
^_3
70
c
rial DiviFor = 140
Complete Divisor = 143
Each of the oblong figures is equal in length to the side of the
square first formed ; and their united length
is 70 + 70 =: 140 ft. (Fig. III). This num-
ber is obtained in the operation by doubling
the 7 and annexing 1 cipher, the result being
written at the left of the dividend. Dividing
517, the area, by 140, the approximate length,
we obtain 3, the probable width of the addi-
tion, and second figure of the root. Since 3 is
also the side of the little square, we can now
find the entire length of the addition, or the complete divisor, which
is 70 -|- 70 + 3 =1 143 (Fig. III).
'^' ^^^* This number is found in the oper-
ation by adding 3 to the trial di-
visor, and writing the result un-
derneath. Multiplying the com-
plete divisor, 143, by the trial
quotient figure, 3, and subtracting
~ the product from the dividend, we
obtain another remainder of 88 square feet. -With this remainder,
for the same reason as before, we must proceed to make a new
enlargement ; and we bring down two decimal ciphers, because the
next figure of the root, being tenths, its square will be hundredths.
The trial divisor to obtain the width of this new enlargement,
or the next figure in the root, will be, for the same reason as
before, twice 73, the root already found, with one cipher annexed.
But since the 7 has already been doubled in the operation, we have
only to double the last figure of the complete divisor, 143, and
annex a cipher, to obtain the new trial divisor, 146.0. Dividing, we
obtain .6 for the trial figure of the root ; then proceeding as before,
we obtain 146.6 for a complete divisor, 87.96 for a product ; and
there is still a remainder of .04. Henee, the side of the given
square plat is 73.6 feet, nearly. From this example and analysis
we deduce the following
Rule. I. Point off the given number into periods of two
figures each, counting from unit's place toward the left and
right.
II. Find the greatest square number in the left hand period,
and write its root for the first figure in the root ; subtract the
square number from the left hand period^ and to the remainder
^ring down the next period for a dividend.
318 EVOLUTION.
III. At the left of ike dividend write twice the Jirst figure of
the root, and annex one cipher, for a trial divisor ; divide the
dividend hy the trial divisor, ayid write the quotient for a trial
figure in the root.
IV. Add the trial figure of the root to the trial divisor for a
complete divisor ; multiply the complete divisor hy the trial
figure in the root, and subtract the product from the dividend,
and to the remainder bring down the next period for a new
dividend.
y. To the last complete divisor add the last figure of the
root, and to the sum annex one cipher, for a new trial divisor,
with which proceed as before.
Notes. 1. If at any time the product be greater than the dividend,
diminish the trial figure of the root, and correct the erroneous work.
2. If a cipher occur in the root, annex another cipher to the trial
divisor, and another period to the dividend, and proceed as before.
'EXAMPLES FOR PRACTICE.
2. What is the square root of 406457.2516?
OPERATJON^ 1 ^
40,64,57.25,16 I 637.54, ^ws.
36
Trial di^^so^,
120
464
CompJete «
123
369
Trial «
1260
9557
Complete "
1267
8869
Trial «
1274.0
688.25
Complete "
1274.5
637.25
Trial
1275.00
51.0016
t!ompleto "
1275.04
51.0016
Notes. 3. The decimal points in the work may bo omitted, care
being taken to point off in the root according to the nimiber of deci-
mal periods used.
4. The pupil will acquire greater facility, and secure greater accura-
cy, by keeping units of like order under each other, and each divisor
opposite the corresponding dividend, by the use of the lines, as shown
in the operation.
3. "What is the square root of 576 ? Ans. 24.
SQUARE ROOT. 319
4. "What is the square root of 6561 ? Ans. 81.
5. What is the square root of 444889 ? Ans. 667,
6. What is the square root of 994009 ? Ans. 997.
7. What is the square root of 29855296 ? A?is. 5464.
8. What is the square root of 3486784401 ? Ans. 59049.
9. What is the square root of 54819198225 ?
Note. The cipher in the trial divisor may be omitted, and its place,
after division, occupied by the trial root figure, thus forming in suc-
cession only complete divisors.
10. What is the square root of 2 ?
2. I 1.4142+ , Ans.
1
24
100
96
281
400
281
2824
11900
11296
28282
60400
56564:
11. Extract the square roots of the following numbers:
V3 = 1.7320508 +
V5 = 2.2360679 +
V6 = 2.4494897 +
V7 = 2.6457513 +
V8 =: 2.8284271 +
VIO = 3.1622776 +
12. What is the square root of .00008836 ? Ans. .0094.
13. What is the square root of .0043046721 ? Ans. .06561.
Notes. 5. The square root of a common fraction may be obtained
by extracting the square roots of the numerator and denominator
separately, provided the terms are perfect squares; otherwise, the
fraction may first be reduced to a decimal.
6. Mixed numbers may be reduced to the decimal form before ex-
tracting the root ; or, if the denominator of the fraction is a perfect
square, to an improper fraction.
14. Extract the square root of ^Wx* Ans. f|.
15. Extract the square root of |jf f- Ans. ^.
16. Extract the square root of |. Ans. .816496 +.
17. Extract the square root of 17f. Ans. 4.1683 +.
320
EVOLUTION.
thus, the two lines, A B and A C, meeting.
APPLICATIONS.
4LS5» An Angle is the opening between two lines that
meet each other
form an angle at A.
436. A Triangle is a figure having three
sides and three angles, as A, B, C.
437. A Right-Angled Triangle is a tri-
angle having one right angle, as at C.
438. The Base is the side on which it
stands, as A, C.
439. The Perpendicular is the side
forming a right angle with the base, as B, C.
440. The Hypotenuse is the side opposite the right angle,
as A, B.
441. Those examples given below, which relate to trian-
gles and circles, may be solved by the use of the two following
principles, which are demonstrated in geometry.
1st. The square of the hypotenuse of a right-angled triangle
is equal to the sum of the squares of the other two sided.
2d. The areas of two circles are to each other as the squares
of their radii, diameters, or circumferences.
1. The two sides of a right-angled triangle are 3 and 4
feet ; what is the length of the hypotenuse ?
Analysis. Squaring
OPERATION.
32 rr= 9, square of one side.
42 1= 1 6, square of the other side*
25, square of hypotenuse.
^25 = 5, Ajis.
the two sides and add-
ing, we find the sum to
be 25 ; and since the sum
is equal to the square of
the hypotenuse, we ex-
tract the square root, and
obtain 5 feet, the hypot-
enuse. Hence,
To find the hypotenuse. Add the squares of the two sides,
and extract the square root of the sum.
To find either of the shorter sides. Subtract the square of
the given side from the square of the hypotenuse, and extract the
square root of the remainder.
SQUARE ROOT. 321
EXAMPLES FOR PRACTICE.
2. If an army of 55225 men be drawn up in the form of a
square, how many men will there be on a side ? A?is. 235.
3. A man has 200 yards of carpeting 1^ yards wide ; what
is the length of one side of the square room which this carpet
will cover ? Ans. 45 feet.
4. How many rods of fence will be required to inclose 10
acres of land in the form of a square ? Ans. 1 60 rods.
5. The top of a castle is 45 yards high, and the castle is sur-
rounded by a ditch 60 yards wide ; required the length of a
rope that will reach from the outside of the ditch to the top
of the castle. Ans. 75 yards.
6. Required the height of a May-pole, which being broken
39 feet from the top, the end struck the ground 15 feet from
the foot. Ans. 75 feet.
7. A ladder 40 feet long is so placed in a street, that
without being moved at the foot, it will reach a window^ on
one side 33 feet, and on the other side 21 feet, from the
ground ; what is the breadth of the street ? Ans. 56.64 -|- ft.
8. A ladder 52 feet long stands close against the side of a
building ; how many feet must it be drawn out at the bottom,
that the top may be lowered 4 feet ? Ans. 20 feet.
9. Two men start from one corner of a park one mile
square, and travel at the same rate. A goes by the walk
around the park, and B takes the diagonal path to the opposite
corner, and turns to meet A at the side. How many rods
from the corner wdll the meeting take place ? Ans. 93.7 -|- rods.
10. A room is 20 feet long, 16 feet wide, and 12 feet high ;
what is the distance from one of the lower corners to the op-
posite upper corner? Ans. 28.284271 -[-feet.
11. It requires 63.39 rods of fence to inclose a circular
field of 2 acres ; what length will be required to inclose 3
acres in circular form ? Ans. 77.63+ rods.
12. The radius of a certain circle is 5 feet; what will be
the radius of another circle containing twice the area of the
first? . Ans. 7.07106 + feet.
14* '
322 EVOLUTION.
CUBE ROOT.
443. The Cube Root of a number is one of the three
equal factors that produce the number. Thus, the cube root
of 27 is 3, since 3 X 3 X 3 = 27.
443. In extracting the cube root, the first thing to be
determined is the relative number of places in a cube and its
root. The law governing this relation is exhibited in the fol-
lowing examples : —
Roots.
Cubes.
Roots.
Cubes.
1
1
1
1
9
729
10
1,000
99
907,299
100
1,000,000
999
997,002,999
1000
1,000,000,000
From these examples, we perceive,
1st. That a root consisting of 1 place may have from 1 to
3 places in the cube.
2d. That in all cases the addition of 1 place to the root
adds three places to the cube. Hence,
If we 'point ojf a number into three-figure periods, com-
mencing at the right hand, the number of full periods and the
left hand full or partial period will indicate the number of
places in the cube root, the highest period answering to the
highest figure of the root,
444. 1. "What is the length of one side of a cubical block
containing 413494 solid inches ?
OPERATION — COMMENCED. ANALYSIS. SinCO the block IS a
413494 I 74 cube, its side will be the cube root of
343 its solid contents, which we will pro-
14700 70404 ^^^^ ^° compute. Pointing off the
14:/ UU /U^y-l given number, the two periods show
that there will be two figures, tens and
units, in the root. The tens of the root must be extracted from the
first period, 413 thousands. The greatest cube in 413 thousands is
343 thousands, the cube of 7 tens ; wc therefore write 7 tens in the
root at the right of the given number.
CUBE ROOT.
323
Ti-. I.
Since the entire root is to be the side of a cube, let us form a
cubical block (Fig. I), the side
of which is 70 inches in length.
The contents of this cube are
70 X 70 X 70 = 343,000 solid
inches, which we subtract from
the given number. This is done
in the operation by subtracting
the cube number, 343, from the
first period, 413, and to the re-
mainder bringing down the sec-
ond period, making the entire
remainder 70494.
If we now enlarge our cubical
block, (Fig. I), by the addition of 70494 solid inches, in such a
manner as to preserve the cubical form, its size will be that of
the required block. To preserve the cubical form, the addition
must be made upon three adjacent sides or faces. The addition
will therefore be composed of 3 flat blocks to cover the 3 faces,
(Fig. II) ; 3 oblong blocks to fill the vacancies at the edges,
(Fig. Ill) ; and 1 small cubical block to fill the vacancy at the cor-
ner, (Fig. IV). Now, the thickness of this enlargement will be the
additional length of the side of the cnhe, and, consequently, the
second figure in the root. To find thickness, we may divide soUd
I'ig- II- contents by surface, or area.
But the area of the 3 oblong
blocks and little cube cannot
be found till the thickness of
the addition be determined, be-
cause their common breadth is
equal to this thickness. "We will
therefore find the area of the
three flat blocks, which is suffi-
ciently near the whole area to be
used as a trial divisor. As these
are each equal in length and
breadth to the side of the cube
whose faces they cover, the whole
area of the three is 70 X 70 X
3 = 14700 square inches. This number is obtained in the operation
by annexing 2 ciphers to three times the square of 7 ; the result
being written at the left hand of the dividend. Dividing, we obtain
324
EVOLUTION.
Fig. Ill
OPERATION — CONTINIIEI).
41349 4 I 74
T. IT. 343
"4, the probable thickness of the addition, and second figure of the
root. With this assumed figure,
we -will complete our divisor by
adding the area of the 4 blocks,
before undetermined. The 3 ob-
long blocks are each 70 inches
long ; and the little cube, being
equal in each of its dimensions
to the thickness of the addition,
must be 4 inches long. Hence,
their united length is 70 -j- 70
-|- 70 -[- 4 = 214. This number
is obtained in the operation by
multiplying the 7 by 3, and an-
nexing the 4 to the product, the
result being written in column
I, on the next line below the
trial divisor. Multiplying 214,
the length, by 4, the common
width, we obtain 856, the area of
the four blocks, which added to
14700, the trial divisor, makes
15556, the complete divisor ; and
multiplying this by 4, the second
figure in the root, and subtract-
ing the product from the divi-
dend, we obtain a remainder of
0270 solid inches. With this re-
mainder, for the same reason as
before, we must proceed to make
a new enlargement. But since
we have already two figures in
the root, answering to the two
periods of the given number,
the next figure of the root must
be a decimal ; and we therefore
annex to the remainder a period
of three decimal ciphers, mak-
ing 8270.000 for a new dividend.
The trial divisor to obtain the
thickness of this second enlarge-
ment, or the next figure of the root, will be the area of three new flat
blocks to cover the three sides of the cube already formed j and this
214 856
14700
15556
70494
62224
8270.000
CUBE ROOT.
325
surface, (Fig. IV,) is composed of 1 face of each of the flat blocks
already used, 2 faces of each of the oblong blocks, and 3 faces of
the little cube. But we have in the complete divisor, 15556, 1
face of each of the flat blocks, oblong blocks, and little cube ;
and in the correction of the trial divisor, 8o6, 1 face of each of
the oblong blocks and of the little cube; and in the square of
the last root figure, 16, a third face of the little cube. Hence, 16
-|- 856 -\- 15556 i= 10428, the significant figures of the new trial
divisor. This
OPERATION — CONTINUED.
413494 I 74.5
I. IT. 343
214
856
14700
15556
70494
62224
222.5 111.25
1642800
16539.25
8270.000
8269.625
.375
number is ob-
tained in the
operation by
adding the
square of the
last root fig-
ure mentally,
and combin-
ing units of
like order.
thus : 16, 6, and 6 are 28, and we write the unit figure in the new
trial divisor ; then 2 to carry, and 5 and 5 are 12, &c. We annex
2 ciphers to this trial divisor, as to the former, and dividing, obtain
5, the third figure in the root. To complete the second trial di-
visor, after the manner of the first, the correction may be found by
annexing .5 to 3 times the former figures, 74, and multiplying this
number by .5. But as we have, in column I, 3 times 7, with 4
annexed, or 214, we need only multiply the last figure, 4, by 3,
and annex .5, making 222.5, which multiplied by .5 gives 111.25,
the correction required. Then we obtain the complete divisor,
16539.25, the product, 8269.625, and the remainder, .375, in the
manner shown by the former steps. From this example and analysis
we deduce the following
Rule. I. Pohit off the given number into periods of three
figures each, counting from units' place toward the left and right.
II. Find the greatest cube that does not exceed the left hand
period, and write its root for the first figure in the required
root ; subtract the cube from the left hand period, and to the
remainder bring down the next period for a dividend.
III. At the left of the dividend write three times the square
of the first figure of the root, and annex two ciphers, for a trial
divisor ; divide the dividend by the trial divisor, and write the
quotient for a trial figure in the root.
326 EVOLUTION. J'vUX Hi^ fl .^ y^^
IV. Annex the trial figure to three times the J^ mem fyuio ^
and write the result in a column marked I, one line below the
trial divisor ; multiple/ this term by the trial figure, and write
the product on the same line in a column marked II ; add this
term as a correction to the trial divisor, and the I'esult will be
the complete divisor.
V. Multiply the complete divisor by the trial figure, and
subtract the product from the dividend, and to the remainder
bring down the next period for a new dividend.
VI. Add the square of the last figure of the root, the last
term in column II, and the complete divisor together, and annex
two ciphers, for a new trial divisor ; with which obtain an-
other trial figure in the root.
VII. Multiply the unit figure of the last term in column I
by 3, and annex the trial figure of the root for the next term
of column I ; multiply this result by the trial figure of the root
for the next term of column II ; add this term to the trial
divisor for a complete divisor, with which proceed as before.
Notes. 1. If at any time the product be greater than the dividend,
diminish the trial figure of the root, and correct the erroneous work.
2. If a cipher occur in the root, annex two more ciphers to the
trial divisor, and another period to the dividend ; then proceed as be-
fore with column I, annexing both cipher and trial figure.
EXAMPLES FOR TRACTICE.
1. What is the cube root of 79.112 ?
OPERATIOK.
79.112 I 4.2928 + , Ans.
C)4.
H dy*
122
244
1209
11421
12872
25744
128768 1030144
4800
5044
15112
10088
529200
540021
5024000
4805589
55212300
55238044
158411000
110476088
5520379200 47934912000
5527409344 44219274752
3714637248 rem.
CUBE ROOT. 327
2. What is the cube root of 84604519 ? Ans. 439.
3. What is the cube root of 2357947691 ? Ans. 1331.
4. What is the cube root of 10963240788375 ? Ans. 22215.
5. What is the cube root of 270671777032189896 ?
Ans. 646866.
6. What is the cube root of .091125 ? Ans. .45.
7. What is the cube root of .000529475129 ? Ans. .0809.
8. What is the approximate cube root of .008649 ?
A71S. .2052 + .
Extract the cube roots of the followinsr numbers : —
72 =: 1.259921+
^3 = 1.442249+
^4 z= 1.587401+
^5 = 1.709975+
-^6== 1.817120+
-^7 =z 1.912931+
APPLICATIONS IN CUBE ROOT.
1. What is the length of one side of a cistern of cubical
form, containing 1331 solid feet? Ans. 11 feet.
2. The pedestal of a certain monument is a square block of
granite, containing 373248 solid inches ; what is the length
of one of its sides ? Ans. 6 feet.
3. A cubical box contains 474552 solid inches ; what is
the area of one of its sides ? Ans. 42 J sq. ft.
4. How much paper will be required to make a cubical
box which shall contain ||- of a solid foot ? Ans. f of a yard.
5. A man wishes to make a bin to contain 125 bushels, of
equal width and depth, and length double the width; what
must be its dimensions ? Ans. Width and depth, 51.223 +
inches; length, 102.446 + inches.
Note. Spheres are to each other as the cubes of their diameters or
circumferences
6. There are two spheres whose solid contents are to each
other as 27 to 343 ; what is the ratio of their diameters ?
Analysis. Since spheres are to each other as the cubes of their
diameters, the diameters will be to each other as the cube roots of
the spheres ; and ^27 = 3, ^343 = 7 j hence the diameters required
are as 3 to 7.
328 ARITHMETICAL PROGRESSION.
7. The diameter of a sphere containing 1 solid foot is 14.9
inches ; what is the diameter of a sphere containing 2 solid
feet ? Ans. 1 8.7 + inches.
8. If a cable 4in. in circumference, will support a sphere 2ft.
iu diameter, what is the diameter of that sphere which will be
supported by a cable 5in. in circumference? Ans. 2.32 +ft.
ARITHMETICAL PROGRESSION.
445. An Arithmetical Progression, or Series, is a series
of numbers increasing or decreasing by a common difference.
Thus, 3, 5, 9, 11, &c., is an arithmetical progression with an
ascending series, and 13, 10, 7, 4, «&;c., is an arithmetical pro-
gression with a descending series.
440. The Terms of a series are the numbers of which it
is composed.
447. The Extremes are the first and last terms.
448. The Means are the intermediate terms.
44D. The Common Difference is the difference between
any two adjacent terms.
4o0. There are Jiiw parts in an arithmetical series, any
three of which being given, the other (wo may be found.
Tliey are as follows : the Jirst term^ last term, common differ'
ence, number of terms, and sum of all the terms,
CASE I.
451. To find the last term when the first term,
common diirerencc, and number of terms are given.
Let 2 be the first term of an ascending series, and 3 the
common difference ; then the series will be written, 2, 5, 8, 11,
14, or analyzed thus : 2, 2 + 3, 2 + 3 + 3, 2-|-3 + 3 + 3,
2 + 3 + 3 + 3 + 3.
Here we see that, in an ascending series, we obtain the
second term by adding the common difference once to the first
term ; the third term, by adding the common difference twice
to the first termj and, in general, we obtain any term by
ARITHMETICAL PROGRESSION. 329
aflding the common difference as many times to the first term
as there are terms less one.
Note. The analysis for a descending series would be similar.
Hence,
Rule. Multiply the common difference hy the number of
terms less one, arid add the product to the first term, if the
series be ascending, and subtract it if the series be descending,
EXAMPLES.
1. The first term of an ascending series is 4, the common
difference 3, and the number of terms 19; what is the last
term ? Ans. 58.
2. "What is the 13th term of a descending series whose first
term is 75, and common difference 5 ? Ans. 15,
3. A boy bought 18 hens, paying 2 cents for the first, 5
cents for the second, and 8 cents for the third, in arithmetical
progression ; what did he pay for the last hen ?
4. What is the 40th term of the series ^, f, 1, 1^, &c. ?
Ans. lOj.
5. A man travels 9 days ; the first day he goes 20 miles,
the second 25 miles, increasing 5 miles each day; how far
does he travel the last day of his journey ? Ans. 60 miles.
6. What is the amount of $100, at 7 per cent., for 45
years ? $100 + $7 X 45 = $415, Ans.
CASE II.
459. To find the common difference when the
extremes and number of terms are given.
Referring to the series, 2, 5, 8, 11, 14, analyzed in 4.38,
we readily see that, by subtracting the first term from any
term, we have left the common difference taken as many times
as there are terms less one ; thus, by taking away 2 in the
fifth term, 2 + 3 + 3 + 3 + 3, we have 3 taken 4 times.
Hence,
Rule. Divide the difference of the extremes by the number
of terms less one.
330 ARITHMETICAL PROGRESSION.
EXAMPLES.
1. The first term is 2, the last term Is 17, and the number
of terms is 6 ; what is the common difference ? Ans. 3.
2. A man has seven children, whose ages are in arithmetical
progression; the youngest is 2 years old, and the eldest 14;
what is the common difference of their ages ? Ans. 2 years.
3. The extremes of an arithmetical series are 1 and 50^,
and the number of terms is 34 ; what is the common difference ?
4. An invalid commenced to walk for exercise, increasing
the distance daily by a common difference ; the first day he
walked 3 miles, and the 14th day 9^ miles ; how many miles
did he walk each day ?
Note. When we have found the common difference we may add it
once, twice, &c., to the fiist term, and we have the series, and conse-
quently the meatis,
Ans. 3, ^, 4, 4^, 5, 5^, &c.
CASE III.
453. To find the number of terms when the ex-
tremes and common difference are given.
Examining the series, 2, 5, 8, 11, 14, analyzed in 438,
we also see that after taking away the first term from any
term, we have left the common difference taken as many
times as the number of terms, less 1. Hence,
Rule. Divide the difference of the extremes by the common
difference J and add 1 to the quotient.
examples.
1. The extremes are 7 and 43, and the common difference
is 4; what is the number of terms ? Ans. 10.
2. The first term is 2^, the last term is 40, and the common
difference is 74; what is the number of terms ? Ans. 6.
3. A laborer agreed to build a fence on the following con-
ditions : for the first rod he was to have 6 cents, with an
increase of 4 cents on each successive rod ; the last rod came
to 226 cents ; how many rods did he build ? Ans, bQ rods.
GEOMETRICAL PROGRESSION. 331
CASE IV.
454. To find the sum of all the terms when the
extremes and number of terms are given.
To deduce a rule for finding the sum of all the terms, we
will take the series 2, 5, 8, 11, 14, writing it under itself in an
inverse order, and add each term ; thus,
2 -)- 5 + 8 + 11 + 14 = 40, once the sum.
14-1-11-1-' 8+ 5+ 2=^=40, " « "
16 _|_ 16 -|- IG + 16 + 16 = 80, twice the sum.
Here we perceive that 16, the sum of the extremes, multi-
plied by 5, the number of terms, equals 80, which is twice the
sum of the series. Dividing 80 by 2 gives 40, which is the
sum required. Hence,
Rule. Multiply the sum of the extremes hy the number of
terms, and divide the product hy 2.
EXAMPLES.
1. The extremes are 5 and 32, and the number of terms 12 ;
what is the sum of all the terras ? Ans. 222.
2. How many strokes does a common clock make in 12
hours ? Ans. 78 strokes.
3. What debt can be discharged in a year by weekly pay-
ments in arithmetical progression, the first being $24, and the
last $1224? Ans. $32448.
4. Suppose 100 apples were placed in a line 2 yards apart,
and a basket 2 yards from the first apple ; how far would a
boy travel to gather them up singly, and return with each
separately to the basket ? Ans. 20200 yards.
GE03^IETRICAL PROGRESSION.
455. A Geometrical Progression is a series of numbers
increasing or decreasing by a constant multiplier.
When the multiplier is greater than a unit, the series is
332 GEOMETRICAL PROGRESSION.
ascending I thus, 2, 6, 18, 54, 162, is an ascending series, in
which 3 is the multiplier.
When the multiplier is less than a unit, the series is descend-
ing \ thus, 162, 54, 18, 6, 2, is a descending series, in which ^
is the multiplier.
4^0* The Ratio is the constant multiplier.
4t57 • In every geometrical progression there are five
parts to be considered, any three of wiiich being given, the
other two may be determined. They are as follows: The^Vs^
term, last term, ratio, number of terms, and the sum of all the
terms.
The first and last terms are the extremes, and the interme-
diate terms are the means.
CASE I.
458. To find any term, the first term, the ratio,
and number of terms being given.
The first term is supposed to exist independently of the
ratio. Using the ratio once as a factor, we have the second
term ; using it twice, or its second power, we have the third
terra ; using it three times, or its third power ^ we have the
fourth term ; and, in general, the power of the ratio in any
term is one less than the number of the term. Tlie ascending
series, 2, 6, 18, 54, may be analyzed thus: 2, 2 X 3, 2 X
3X3, 2X3X3X3.
In this illustration we see that
1st term, 2, is independent of the ratio.
2d " 6 = 2X3 = the first term into the 1st power of
fhe ratio.
3d term, 18 = 2 X 32 = the first term into the 2d power
of the ratio.
4th term, 54 = 2 X 3 ^ = the first term into the 3d power
of the ratio. Hence
RuLK. Mdtiply the first term hy that power of the ratio
denoted hy the number of terms less 1.
GEOMETRICAL PEOGRESSION. 333
EXAMPLES. X
1. The first term of a geometrical series is 4, the ratio is 3 ;
what is the 9th term ? Ans. 4 X 38 =r 26244.
2. The first term is 1024, the ratio I, and the number of
terms 8 ; what is the last term ? Ans. y^^.
3. A boy bought 9 oranges, agreeing to pay 1 mill for the
first orange, 2 mills for the second, and so on ; what did the
last orange cost him ? Ans, $.256.
4. The first term is 7, the ratio ^, and the number of terms
7 ; what is the last term ? Ans. T^iuy
5. What is the amount of $1 at compound interest for 5
years, at 7 per cent, per annum ? Ans. $1.40255 -|-.
Note. In the above example the first term is $1, the ratio is $1.07,
and the ntmiber of terms is 6.
6. A drover bought 7 oxen, agreeing to pay $3 for the first
ox, $9 for the second, $27 for the third, and so on ; what did
the last ox cost him ? Ans. $2187.
CASE II.
459, To find the sum of all the terms, the ex-
tremes and ratio being given.
If we take the series 2, 8, 32, 128, 512, in which the ratio
is 4, multiply each term by the ratio, and add the terms thus
multiplied, we shall have
8 + 32 + 128 + 512 + 2048 = 2728 = {,^-- ttTe.^.""
But 2 + 8 + 32 -f 128 + 512 = 682 = j t^'Ums!""" ''^ ^"
Hence, by subtracting, we get 2048 — '2 — 2046 — { ^Z'thTtlrml!''"'
Dividing by 3, the ratio less one, 2046 -^ 3 =: 682 = { Se'terms.'"'"' ""^ ""^
The subtraction is performed by taking the lower line or
series from the upper. All the terms cancel except 2048 and
2. Taking their difference, which is 3 times the sum, and di'^
viding by 3, the ratio less one, we must have the sum of all
the terms. Hence
334 PROMISCUOUS EXAMPLES.
Rule. Multiply the greater extreme hy the ratio, subtract
the less extreme from the product, and divide the remainder
by the ratio less 1.
Note. Let every decreasing series be inverted, and the first term
called the last ; then the ratio will be greater than a unit. K the series
be infinite^ the first term is a cipher.
EXAMPLES.
1. The first terra is 2, the last term 512, and the ratio 3;
what is the sum of all the terms ? Ans. 1^1.
2. The first term is 4, the last term is 262144, and the
ratio is 4; what is the sum of the series ? Ans. 349524.
3. The first term of a descending series is 162, the last
term 2, and the ratio ^ ; what is the sum ? Ans. 242.
4. What is the value of |, ^^, y^, &c., to infinity ? Ans. \.
Note. In the following examples we first find the last term by the
Rule under Case I.
5. What yearly debt can be discharged by monthly pay-
ments, the first being $2, the second $6, and the third $18,
and so on, in geometrical progression ? Ans. $531440.
6. If a grain of wheat produce 7 grains, and these be sown
the second year, each yielding the same increase, how many
bushels will be produced at this rate in 12 years, if 1000 grains
make a pint? Ans. 252315 bu. 4^ qt.
7. Six persons of the Morse family came to this country
200 years ago; suppose that their number has doubled every
20 years since, what would be their number now ?
Note. The other cases in Progression will be found in the Higher
Arithmetic.
lO°
ilr^
PROMISCUOUS EXAJ^IPLES. J^^-^
1. One hdf the sum of two numbers is 800, and one half tho
difference of the same numbers is 200 ; what are the numbers ?
Ans. 1000 and 600.
2. What number is that to which, if you add -f of A of itself,
the sum will be 61 ? Ans. 55.
3. What part of a day is 3 h. 21 min. 15 sec. ? Ans. ^lA^.
PROMISCUOUS EXAMPLES. 335
4. A commission merchant received 70 bags of wheat, each con-
taining 3 bu. 3 pk. 3 qt. ; how many bushels did he receive ?
5. Four men, A, B, C, and D, are in possession of $1100; A
has a certain sum, B has twice as much as A, C has $300, and D
has $200 m.ore than C ; how many dollars has A ? Aiis. $100.
6. At a certain election, 3000 votes were cast for three candi-
dates, A, B, and C ; B had 200 more votes than A, and C had 800
more than B ; how many votes were cast for A ? A?is. 600.
7. What part of 17^ is 31 ? ^ Ans. if ^
8. The difference between -| and -J of a number is 10 ; what is
the number ? Ans. 560.
9. A merchant bought a hogshead of rum for $28.35 ; how
much water must be added to reduce the first cost to 35 cents per
gallon ? Ans. 18 gal.
10. A and B traded with equal sums of money ; A gained a sum
equal to 1 of his stock ; B lost $200, and then he had -^ as much as
A ; how much was the original stock of each ? A7is. $500.
11. A farmer sold 17 bushels of barley, and 13 bushels of wheat,
for $31.55 ; he received for the wheat 35 cents a bushel more than
for the barley ; what was the price of each per bushel ?
Ans. Barley, $.90; wheat, $1.25.
12. What is the interval of time between March 20, 21 minutes
past 3 o'clock, P. M., and April 11th, 5 minutes past 7 o'clock,
A. M. ? ^ Ans. 21 da. 15 h. 44 min.
13. What o'clock is it when the time from noon is ^^ of the
time to midnight ? Ans. 5 o'clock 24 min. P. M.
14. Wliat is the least number of gallons of wine that can be ship-
ped in either hogsheads, tierces, or barrels, just filUng the vessels,
without deficit or excess ? Ans. 126 gal.
15. A ferryman has four boats ; one will carry 8 barrels, another
9, another 15, and another 16 ; what is the smallest number of bar-
rels that will make full freight for any one, and all of the boats ?
16. A and B have the same income ; A saves 1 of his, but B,
by spending $30 a year more than A, at the end of four years finds
himself $40 in debt j what is their income, and how much does
each spend a year? C Income, $160.
Ans. < A spends $140.
( B spends $170.
17. If a load of plaster weighing 1825 pounds cost $2.19, how
much is that per ton of 2000 pounds ? Ans. $2.40.
18. If 2^ yards of cloth If yards wide cost $3.37|, what will be
the cost of 36^ yards U yards wide? Ans. $52,779.
19. I lend my neighbor $200 for 6 months ; how long ought he
to lend me $1000 to balance the favor ?
20. Bought railroad stock to the amount of $2356.80, and found
that the sum invested was 40 per cent of what I had left ; what
sum had I at first ? Ans. $8248.80.
21. 20 per cent, of | of a number is what per cent, of | of it ?
Ans. 12^.
336 PROMISCUOUS EXAMPLES.
22. Divide a prize of $10200 among 60 privates, 6 subaltern
officers, 3 lieutenants, and a commander, giving to each subaltern
double the share of a private, each lieutenant 3 times as much as
the subaltern, and to the commander double that of a lieutenant ;
how much is each man's share? Ans. Com. $1200; each man, $100.
23. A is 51 miles in advance of B, who is in pursuit of him ; A
travels 16 miles per hour, and B 19 ; in how many hours will B
overtake A ?
24. How much wool, at 20, 30, and 54 cents per pound, must be
mixed with 95 pounds at 50 cents, to make the whole mixture
^vorth 40 cents per pound ?
Ans. 133 lb. at 20 ; 95 lb. at 30 ; 190 lb. at 54 cents.
25. If 240 bushels of wheat are purchased at the rate of 18
bushels for $22^, and sold at the rate of 22| bushels for $33.75,
"what is the profit on the whole ? Ans. $60.
26. My horse, wagon, and harness together are worth $169 ; the
wagon is worth 4 times the harness, and the horse is worth double
the wagon ; what is the value of each ? C Horse, $104.
Ans. ^ Wagon, $ 52.
( Harness, $ 13.
27. The shadow of a tree measures 42 feet ; a staff 40 inches in
length casts a shadow 18 inches at the same time ; what is the
height of the tree ? Ans. 93 1 ft.
28. If a piece of land 40 rods long and 4 rods wide make an
acre, how wide must it be to contain the same if it be but 25 rods
long ? Ans. 6| rods.
29. A, B, and C are employed to do a piece of work for $26.45 ;
A and B together are supposed to do | of it, A and C ^^, and B
and C 4|, and paid proportionally ; how much must each receive ?
30. It 12 ounces of wool make 21 yards of cloth that is 6 quar-
ters wide, how many pounds of avooI will it take for 150 yards of
clath 4 quarters wide ?
31. Six persons, A, B, C, D, E, and F, are to share among them
$0300 : A is to have | of it, B|, C |, D is to have as much as A
and C together, and the remainder is to be divided between E and
F in the proportion of 3 to 5 ; how much does each one receive ?
32. What is the amount of $200 for 8 years at 6 per cent, com-
pound interest ? Ajis. $318,769.
33. A garrison, consisting of 360 men, was provisioned for 6
months ; but at the end of 5 months they dismissed so many of the
men that the remaining provision lasted 5 months longer; how
many men were sent away?
84. A certain princi|)ul, at compound intere«:t for 5 years, at 6
percent., will amount to $0(31). 118; in what time will the same
principal amount to the same sum, at G per cent, simple interest?
Ajis. 5 yr. 7 n)0. 19.3-|-da.
85. Paid $148,352 for 9728 feet of pine lumber ; how much waa
that per thousand 1
PROMISCUOUS E^MPLES. / C"^ 337
36. Comparing two numbers, 483 was ^und to be tbeir least
common multiple, and 23 their greatest common divisor ; what
is the product of the numbers compared ? Ans. 11,109.
37. Eight workmen, laboring 7 hours a day for 15 days, were
able to execKte |^ of a job ; in how many days can they complete the
residue, by working 9 hours a day, if 4 workmen are added to their
number ? Ans. 15| days.
38. K a hall 36 feet long and 9 feet wide require 36 yards of
carpeting 1 yard wide to cover the floor, how many yards 1^ yards
wide will cover a floor 60 feet long and 27 feet wide ?
Ans. 144 yards.
39. A, B, and C traded in company ; A put in $1 as often as B
put in $3, and B put in $2 as often as C put in $5 ; B's money was
in twice as long as C's, and A's twice as long as B's ; they gained
$52.50 } how much was each man's share of the gain ? C A's, $12.
Ans.^B's, $18.
(C's, $22.50.
40. A and B found a watch worth $45, and agreed to divide the
value of it in the ratio of f to | ; how much was each one's share ?
. 5 $20, A's.
^'^^- } $25, B's.
41. A man received $33.25 interest on a sum of money, loaned
5 years previous, at 7 per cent. ; what was the sum lent ?
Ans. $95.
42. The diameter of a ball weighing 32 pounds is 6 inches ;
•what is the diameter of a ball weighing 4 pounds ? Ans. 3 inches.
43. Divide $360 in the proportion of 2, 3, and 4,
Ans. $80, $120, $160.
44. If by working 6f hours a day a man can accomplish a job in
12^ days, how many days will be required if he work 8^ hours per
day? Ans. 9^^^ days.
45. An open court contains 40 square yards ; how many stones,
9 inches square, will be required to pave it ? Ajis. 640.
46. A drover paid $76 for calves and sheep, paying $3 for calves,
and $2 for sheep ; he sold \ of his calves and f of his sheep for
$23, and in so doing lost 8 per cent, on their cost ; how many of
each did he purchase ? Ans. 12 calves ; 20 sheep.
47. If a cistern, 17^ feet long, 10^ broad, and 13 deep, hold 546
barrels, how many barrels will that cistern hold that is 16 feet long,
7 broad, and 15 deep ? Ans. 384 bbls.
48. If 12 men, working 9 hours a day, for 15|- days, were able to
execute f of a job, how many men may be withdrawn, and the resi-
due be finished in 15 days more, if the laborers are employed only
7 hours a day ? Ans. 4 men.
49. A general formed his men into a square, that is, an equal
number in rank and file, and found that he had 59 men over ; and
increasing the number in both rank and file by 1 man, he wanted 84
men to complete the square ; how many men had he ? Ans. 5100.
R.p. 15
338 PROMISCUOUS EXAMPLES. .
50. Bought wheat at $1.50 per bushel, corn at $.75 per bushel,
and barley at $.60 per bushel ; the wheat cost twice as much as the
corn, and the corn twice as much as the barley ; of the sum paid,
$243 and \ of the whole was for wheat, and $153 and ^V of the
whole was for the corn ; how many bushels of grain did I purchase ?
Alls. 756.
51. Divide $630 among 3 persons, so that the second shall have
■| as much as the first, and the third ^ as much as the other two ;
what is the share of each ? C 1st, $240.
Ans.\2([, $180.
(3d, $210.
52. Bought a hogshead of molasses for $28, and 7 gallons
leaked out ; at what rate per gallon must the remainder be sold to
gain 20 ^ ?
63. 20 per cent, of ^ of a number is how many per cent, of 2
times f of li times the number ? Ans. 7^.
54. B andf C, trading together, find their stock to be worth $3500,
of which C owns $2100 ; they have gained 40 per cent, on their first
capital ; what did each put in ? ^ S B, $1000.
^''^' i C, $1500.
55. If the ridge of a building be 8 feet aoove the beams, and the
building be 32 feet wide, what must be the length of rafters ?
66. If 12 workmen, in 12 days, working 12 hours a day, can
make up 75 yards of cloth, f of a yard wide, into articles of clothing :
how many yards, 1 yard wide, can be made up into like articles, by
10 men, working 9 days, 8 hours each day ? Ans. 23^,
57. A grocer sells a farmer ICO pounds of sugar, at 12 cents a
pound, and makes a profit of 9 per cent. ; the farmer sells him 100
pounds of beef, at 6 cents a pound, and makes a profit of 10 per
cent. ; who gains the more by the trade, and how much ?
Ans. The grocer gains $.446 -|- more.
58. In 1 yr. 4 mo. $311.50 amounted to $336.42, at simple
interest ; what was the rate ^ ? * Ans. 6.
59. Three persons engage to do a piece of work for $20 ; A and
B estimate that they do | of it, A and C that they do | of it, and B
and C that they do | of it ; according to this estimate, what part of the
$20 should each man receive ? Ans. A's, $114 ; li's, $5f ; C's, $2f
60. Paid $375, at the rate of 2^ per cent., for insurance on a
cotton factory and the machinery ; for what amount was the policy
given ?
61. A merchant bought goods in Boston to the amount of $1000,
and gave his note, dated Jan. 1, 1857, on interest after 3 months;
six months after the note was given he paid $560, and 5 months
after the first payment he paid $406 j what was due Aug 23, 1859 ?
Ans. $06.68-h
62. If * of A's money be equal to f of B's, aiid f of B's be equal
to j of C^s, and 4 of C's be equal to § of D's, and D has $45 more
than C, how much has each ? ^„. S A, $378 ; C, $360 ;
A}is. <
B, $336 ; 1), $405.
PROMISCUOUS EXAMPLES. 339
-t
63. A owed B $900, to be paid in 3 years ; but at the expiration
of 9 months A agreed to pay $300 if B would wait long enough for
the balance to compensate for the advance ; how long should B
wait after the expiration of the 3 years ? Ans. 13^ mo.
64. A certain clerk receives $800 a year ; his expenses equal -^^
of what he saves ; how much of his salary does he save yearly ?
65. A merchant sold cloth at $1 per yard, and made 10 per cent,
profit ; what would have been his gain or loss had he sold it at $.87-^
per yard ? Ans. Loss, 3| ^.
21-9
GG, AVhat is the cube of —j- Ans. |J.
IT ^o
63
07. What is the cube root of — - A7is. |.
68. A miller is required to grind 100 bushels of provender worth
50 cents a bushel, from oats worth 20 cents, corn worth 35 cents,
rye worth 60 cents, and wheat worth 70 cents per bushel j how
many bushels of each may he take ?
69. A man owes $6480 to his creditors ; his debts are in arith-
metical progression, the least being $40, and the greatest $500;
required the number of creditors and the common difference between
the debts. j ^24 creditors.
^'^^' I $20 difference.
70. Two ships sail from the same port ; one goes due north 128
miles, and the other due east 72 miles ; how far are the ships from
each other ? Ans. 146.86 -|- miles.
71. If 10 pounds of cheese be equal in value to 7 pounds of
butter, and 1 1 pounds of butter to 2 bushels of corn, and 14 bushels
of com to 8 bushels of rye, and 4 bushels of rye to 1 cord of wood ;
how many pounds of cheese are equal in value to 10 cords of wood ?
Ans. 550.
72. A and B traded until they gained 6 per cent, on their stock ;
then f of A's gain was $18 ; if A's stock was to B's as f to ^, how
much did each gain, and what was the original stock of each ?
. 5 A's gain, $45 ; stock, $750.
^"^•^B's " $37.50; " $625.
73. If 20 men, in 21 days, by working 10 hours a day, can dig a
trench 30 ft. long, 15 ft. wide, and 12 ft. deep, when the ground is
called 3 degrees of hardness, how many men, in 25 days, by work-
ing 8 hours a day, can dig another trench 45 ft. long, 16 ft. wide,
and 18 ft. deep, when the ground is estimated at 5 degrees of
hardness ? Ans. 84.
74. Wishing to know the height of a certain steeple, I measured
the shadow of the same on a horizontal plane, 27^ feet; I then
erected a 10 feet pole on the same plane, and it cast a shadow of 2|
feet ; what was the height of the steeple ? Ans. 103^ ft.
75. A can do a piece of work in 3 days, B can do 3 times as
much in 8 days, and C 5 times as mu<;h in 12 days ; in what time
can they all do the fii-st piece of work ? Ans. f da.
340 PROMISCUOUS EXAMPLES. J
I ^
76. A person sold two farms for $1890 each ; for one ho received
25 per cent, more than its true value, and for the other 25 per cent,
less than its true value 5 did he gain or lose by the sale, and how
much .P Ans. Lost $252.
77. Three men paid $100 for a pasture ; A put in 9 horses, B 12
cows for twice the time, and C some sheep for 2-^ times as long-
as B's cows ; C paid one half the cost ; how many sheep had he,
and how much did A and B each pay, provided 6 cows eat as much
as 4 horses, and 10 sheep as much as 3 cows ? T C had 25 sheep.
Ans. \ A paid $18.
( B « $32.
78. A man purchased goods for $10500, to be paid in three equal
installments, without interest ; the first in 3 months, the second in
4 months, the third in 8 months ; how much ready money will pay
the debt, money being worth 7 %'t Ans. $10203.94-}-.
79. A farmer sold 50 fowls, consisting of geese and turkeys; for
the geese he received $.75 apiece, and for the turkeys $1.25 apiece,
and for the whole he received $52.50 ; how many were there of
each ? Ans. 20 geese, 30 turkeys.
80. There is an island 73 miles in circumference, and 3 footmen
start together and travel around it in the same direction ; A goes 5
miles an hour, B 8, and C 10 ; in what time will they all come
together again if they travel 12 hours a day ? Ans. 6 da. 1 h.
81. A, B and C are to share $100000 in the proportion of 4, {,
and \, respectively ; but C dying, it is required to divide the whole
sum proportionally between the other two ; how much is each one's
share .P . $ A's, $57142.85f ^
^^^-}B% $42857. 14fM^
82. A, B, and C have 135 sheep ; A's plus B's are to B's plus
C's as 5 to 7, and C's minus B's to C's plus B's as 1 to 7 ; how many
has each ? Ans. A, 30 ; B, 45 ; C, 60.
83. A man sold one hog, weighing 250 ])ounds, at 4 cents per
pound ; a second, weighing 300 pounds, at 4i cents ; and a thn-d,
weighing 369 pounds, at 5 cents ; what was the average price per
pound for the whole ? Ans. 4|^| cents.
84. In a certain factory are employed men, women and boys ;
the boys receive 3 cents an hour, the women 4, and the men 6 ; the
boys work 8 hours a day, the women 9, and the men 12 ; the boys
receive $5 as often as the women $10, and for every $10 paid to the
women, $24 are paid to the men ; how many men, women, and
boys are there, the whole jiumber being 59 ?
Ans. 24 men, 20 women, 15 boys.
85. A fountain has 4 receiving pipes. A, B, C, and I) ; A, B, and
C will fill it in 6 hours, B, C, and 13 in 8 hours, C, D, and A in 10
hours, and D, A, and B in 12 hours; it has also 4 discharging
pipes, \V, X, Y, and Z ; W, X, and Y will empty it in 6 hours. A, Y,
and Z in 5 hours, Y, Z, and W in 4 hours, and Z, W, and X in 3
hours ; suppose the pipes all open, and the fountain full, in what
time would it be emptied ** • Ans, G-^ h.
PROMISCUOUS EXAMPLES. 34X
86. How many building lots, each 75 feet by 125 feet, can be
laid out or. 1 A. 1 R. 6 P. 18^ sq. yd. ? Ans. 6.
87. A man bought a house, and agreed to pay for it $1 on the
first day of January, $2 on the first day of February, $4 on the
first day March, and so on, in geometrical progression, through
the year ; what was the cost of the house, and what the average
time of payment ? A ^ $4095.
^^' I Average time, Nov. 1.
88. A man sold a rectangular piece of ground, measuring 44
chains 32 links long by 3(3 chains wide ; how many acres did it
contain ? Ans. 159 A. 2 R. 8.32 P.
89. What number is that which being increased by its half, its
thud, and 18 more, will be doubled ? Aiis. 108.
90. A merchant has 200 lb. of tea, worth $.62-^ per pound, which
he will sell at $.56 per pound, provided the purchaser will pay in
cofiee at 22 cents, which is worth 25 cents per pound ; does the
merchant gain or lose by the sale of the tea, and how much per
cent. ? Ans. gained lA- %.
91. A man owes a debt to be paid in 4 equal installments at 4,
9, 12, and 20 months, respectively; discount being allowed at 5 per
cent., he finds that $750 ready money will pay the debt ; how much
did he owe? Ans. $784.74+.
92. A and B traded upon equal capitals ; A gained a sum equal
to f of his capital, and B a sum equal to ^^ of his ; B's gain was
$500 less than A's ; what was the capital of each ? Ans. $4000.
93. I purchase goods in bills as follows : June 4, 1859, $240.75 ;
Aug. 9, 1859, $137.25; Aug. 29, 1859, $65.64; Sept. 4, 1859,
$230.36; Nov. 12, 1859, $36. If the merchant agree to allow
credit of 6 mo. on each bill, when may I settle by paying the whole
amount? Atis. Feb. 1, 1860.
94. A young man .inherited a fortune, ^ of which he spent in 3
months, and ^- of the remainder in 10 months, when he had only
$2524 left ; how much had he at first ? Ajis. $5889.33 +.
95. A man bought a piece of land for $3000, agreeing to pay 7
per cent, interest, and to pay principal and interest in 5 equal an-
nual insttdlments ; how much was the annual payment ?
Ans. $731.67 -|-.
96. I have three notes payable as follows : one for $200. due
Jan. 1. 1859, another for $350, due Sept. 1, and another for $500,
due April 1, 1860 ; what is the average of maturity ?
V Ans. Oct. 24, 1859.
97. A man held three notes, the first for $600, due Julv 7, 1859 ;
the second for $530, due Oct. 4, 1859 ; and the third for $400, due
Feb. 20, 1860 ; he made an equitable exchange of these with a
speculator for two other notes, one of which was for $730, due Nov.
15, 1859 J Avhat was the face of the other, and when due ?
. 5 Face, $800.
"^'*^' } Due Aug. 29, 1859.
342 MENSURATION.
MENSURATION OF LINES AND SUPERFICIES.
460. In taking the measure of any line, surface, or solid, we are
always governed by some denomination, a unit of which is called the
Unit of Measure. Thus, if any lineal measure be estimated in feet,
the unit of measure is 1 foot ; if in inches, the unit is 1 inch. If any
superficial measure be estimated in feet, the unit of measure is 1
square foot ; if in yards, the unit is 1 square yard.
461. If any solid or cubic measure be estimated in feet, the
unit of measure is 1 cubic foot ; if in yards, the unit is 1 cubic yard.
462. The area of a figure is its superficial contents, or the
surface included within any given lines,
without regard to thickness.
463. An Oblique Angle is an angle
greater or less than a right angle ; thus,
ABC and C B D are oblique angles.
CASE I.
464. To find the area of a square or a rectangle.
465. A Square is a figure having four equal sides and four
right angles.
466. A Rectangle is a figure having four right angles, and
its opposite sides equal.
Rule. Multiply the length hj the breadth^ and the product will
he the square contents.
EXAMPLES FOR PRACTICE.
1. How many square inches in a board 3 feet long and 20 inches
wide? Ans. 720.
2. A man bought a farm 198 rods long and 150 rods wide, and
agreed to give $32 an acre ; how much did the farm cost him ?
Ans. $5940.
3. A certain rectangular piece of land measures 1000 links by
100 ; how many acres does it contain ? Ans. 1 A.
CASE II.
467. To find the area of a rhombus or a rhomboid.
46§, A Rhombus is a figure having four equal sides and four
oblique angles.
469. A Rhomboid is a figure having its opposite sides equal
and parallel, and its angles oblique.
Note. The square, rectangle, rhombus, and fhomboid, havinej their op-
posite sides parallel, are called by the general name, parallelogram.
It is proved in geometry that any ])arallelogram is equal to a rec-
tangle of the same length and width ; hence the
MENSURATION. 343
Rule. Multiply the length hy the shortest or perpendicular dis-
tance between two opposite sides.
EXAMPLES FOR PRACTICE.
1. A meadow in the form of a rhomboid is 20 chains long, and
the shortest distance between its longest sides is 12 chains ; how
many days of 10 hours each will it take a man to mow the grass on
this meadow, at the rate of 1 square rod a minute ? Ans. 6 da. 4 h.
2. The side of a plat in the form of a rhombus is 15 feet, and a
perpendicular drawn from one oblique angle to the side opposite,
will meet this side 9 feet from the adjacent angle ; what is the area
of the plat? Ans. 180 sq. ft.
CASE III.
470. To find the area of a trapezoid.
471. A Trapezoid is a figure having j-?
four sides, of which two are parallel. • /
The mean length of a trapezoid is one /
half the sum of the parallel sides ; hence the [}
IlULE. Multiply one half the sum of the parallel sides hy the per-
pendicular distance between them.
EXAMPLES FOR PRACTICE.
1. AVhat are the square contents of a board 12 feet long, 16
inches wide at one end, and 9 at the other ? Ans. 12^ sq. ft.
2. AVhat is the area of a board 8 feet long, 16 inches wide at
each end, and 8 in the middle ? Ans. 8 sq. ft.
3. One side of a field is 40 chains long, the side parallel to it is
22 chains, and the perpendicular distance between these two sides
is 25 chains ; how many acres in the field ? Ans. 77 A. 5 sq. ch.
CASE IV.
472, To find the area of a triangle.
473, The Base of a triangle is the side on which it is supposed
to stand.
474, The Altitude of a triangle is the perpendicular distance
from the angle opposite the base to the base, or to the base produced
or extended.
475, A Triangle is one half of a parallelogram of the same
base and altitude ; hence the
Rule. Multiply one half the base by the altitude, or one half the
altitude by the base. Or, Multiply the base by the altitude, and
divide the product by 2.
EXAMPLES FOR PRACTICE.
1. How rnany square yards in a triangle whose base is 148 feet,
and perpendicular 45 feet ? Ans. 370 yds.
344 MENSURATION.
2. The gable ends of a barn are each 28 feet wide, and the per-
pendicular height of the ridge above the eaves is 7 feet ; how many
feet of boards will be required to board up both gables ?
Ans. 196 feet.
CASE V.
476. To find the circumference or the diameter of a circle.
477. A Circle is a figure bounded by one
uniform curved line.
47 §. The Circumference of a circle is the
curved line bounding it.
479. The Diameter of a circle is a straight
line passing through the center, and termina-
ting in the circumference.
It is proved in geometry that in eveiy circle the ratio between the
diameter and the circumference is 3.1416 -}-. Hence the
Rule. I. To find the circumference. — Multiply the diameter
hrj 3.1416.
II. To find the diameter. — Multiply the circumference by .3183.
EXAMPLES FOR PRACTICE.
1. AVTiat length of tire will it take to band a carriage wheel 5
feet in diameter ? Aiis. 15 ft. 8.4 -|- in.
2. What is the circumference of a circular lake 721 rods in
diameter ? Ans. 7 mi. 25 rds. 1.54 -|- ft.
3. What is the diameter of a circle 33 yards in circumference ?
A71S. 10.5 -|- yards.
CASE VI.
4S0. To find the area of a circle.
From the principles of geometry is derived the following
Rule. I. When both diameter and circumference are given ; —
Multiply the diameter by the circumference, and divide the product
II. When the diameter is given ; — Multiply the square of the
diameter by .7854.
III. When the circumference is given ; — Multiply the square of
the circumference by .07958.
EXAMPLES FOR PRACTICE.
1. The diameter of a circle is 113, and the circumference 355;
what is the area ? ^ ^W5. 10028.75.
2. What is the diameter of a circular island containing 1 square
mile of land ? Ans. 1 mi. 41 rd. 1.4 + ft.
3. A man has a circular garden requiring 84 rods of fencing to
inclose it j how much land in the garden ? Ans. 3 A. 81.5-j- P.
THE METRIC SYSTEM
OF
WEIGHTS AND MEASURES.*
INTRODUCTION.
The metric system of weights and measures — so called, because
the metre is the unit from which the other units of the system are
derived — had its origin in France during the Revolution, a time when
all regard for institutions of the past was repudiated. In the year
1790, the French government resolved to introduce a new system ;
and, in order that it might he received with general favor, other
countries were invited to join with it in the choice of new units.
In response to this invitation, a large number of scientific men, com-
missioned by various countries, met in Paris, in consultation with the
principal men of France. In the year 1791, a commission, nomi-
p3 nated by the Academy of Sciences, was appointed by the Government
. to prepare the new system. The first work of the commission was to
select a standard of lengths from which the system of units adopted
might at any time be restored if from any cause the original unit
.'/' should be lost. A quadrant of the earth's meridian was chosen as
-, the standard, and the ten-millionth part of it taken as the unit of
lengths, which was called a metre. In 1795, this standard and a
provisional metre whose length was determined from measurements
* 3r. Mc Vicar, A.M., rrincipal of the State Normal and Training School at
Broekport, Jf .Y., a most thorough and critical scholar as well as teacher, prepared
this article, which contains many practical improvements in Notation, Nomencla-
ture, and Applications, not before presented to the public.
Entered, according to Act of Congress, in the year 1867, by D. W. Fisn, A.M., in the Clerk's
Office of the District Court of the United States for the Southern District of New York.
15*
346 THE METRIC SYSTEM.
of the earth's meridian, which had already been made, was adopted
by the government.
In the meantime, two eminent astronomers, Mechain and Delambre,
were engaged in determining the exact length of the arc of the meri-
dian between Dunkirk in the north of France, and Barcelona in
Spain. At a later period, Biot and Arago measured the prolonga-
tion of the same meridian as far as the island of Formentara. From
these measurements, together with one formerly made in Peru, they
deduced, as they supposed, the exact distance from the equator to
the pole, which differed slightly from the standard assumed in 1795.
In 1799, a law was passed changing the length of the metre adopted
in 1795 so as to conform with this difference. The metre thus de-
termined was marked by two very fine parallel lines drawn on a pla-
tinum bar, and deposited for preservation in the national archives.
While a part of the commission were engaged in establishing the
exact length of the metre, other members pursued a course of inves-
tigation for the purpose of determining a unit of weights, which would
sustain an invariable relation to the unit of lengths. As the result
of their investigations, the weight of a cube of pure water whose edge
was one-hundredth part of a metre was the unit chosen. The water
was weighed in a vacuum, at a temperature of 4° C, or 39.2° F.,
which was supposed to be the temperature of greatest density. This
weight was called a gramme; and a piece of platinum weighing one
thousand gi*ammes was deposited as the standard of weights in the
national archives.
Had the work of the commission ended in determining tlieso
standards of lengths and weights, their labor would have been futile.
For, while the conception of basing their system upon an absolute
standard in nature was good, the execution proved a failure. Later
investigations have shown that the metre is less than the ten-millionth
part of the earth's meridian ; consequently the metric system of
weights and measures is referable not to an invariable standard in
nature, but to the platinum metro deposited in the national archives
of France. The great benefits which result from i\\Q labors of the
commission arise from the adoption of the decimal scale of units, and
a simple yet general and expressive nomenclature. The amount of
THE METRIC SYSTEM. 347
time and money used in carrying on exchanges between different coun*
tries, which would be saved by the universal adoption of t^is system,
is incalculable. The system was declared obligatory throughout the
whole of France after Nov. 2, 1801 ; but, owing to the prejudices
of the people in' favor of established customs, and the confusion con-
sequent upon the use of the new measures, the Government, in 1812,
adopted a compromise, in the systeme usuel, whose principal units
were the new ones, while the divisions and names were nearly those
formerly in use, ascending commonly in the ratios of two, three, four,
eight, or twelve. In 1837, the government abolished this system,
and enacted a law attaching a penalty to the use of any other than
the metric system after Jan. 1, 1841. Since that time, the system
has been adopted by Spain, Belgium, and Portugal, to the exclusion
of other weidits and measures. In Holland, other weiojhts are used
only in compounding medicines. In 1864, the system was legalized
in Grreat Britain ; and its use, either as a whole or in some of its parts,
has been authorized in Greece, Italy, Norway, Sweden, Mexico,
Guatemala, Venezuela, Ecuador, United States of Columbia, Brazil,
Chili, San Salvador, and Argentine Republic. In 1866, Congress
authorized the metric system in the United States by passing the fol-
lowing bills and resolution : —
An Act to authorize the use of the Metric System op TVeights
AND Measures.
Be it enacted hy the Senate and House of Representatives of the United States
of America in Congress assembled, That, from ami after the passage of this
Act, it shall be lawful throughout the United States of America to employ
the Weights and Measures of the Metric System : and no contract or dealinjr,
or pleading in any court, shall be deemed invalid, or liable to objection, be-
cause the -weights or measures expressed or referred to therein are weights or
measures of the Metric System.
Section 2. And he it further enacted, That the tables in the schedule
hereto annexed shall be recognized in the construction of contracts, and in
all legal proceedings, as establishing, in terms of the Aveights and measures
now in use in the United States, the equivalents of the weights and measures
expressed therein in terms of the Metric System ; and said tables may be
lawfully used for computing, determining, and expressing in customary
weights and measures, the Aveights and measures of the ^.Ictric System.
348
THE METRIC SYSTEM.
A Bill to authorize the Use in Post Offices of the Wkights
OF the Denomination of Grammes.
Be it enacted by the Senate and House of Representatives of the United States
oj" America in Congress assembled, That the Postmaster General be, and he is
hereby, authorized and directed to furnish to the post-offices exchanging
mails with foreign countries, and to such other offices as he shall think expe-
dient, postal balances denominated in grammes of the metric system ; and,
until otherwise provided by law, one-half ounce avoirdupois shall be deemed
and taken for postal purposes as the equivalent of fifteen grammes of tho
metric weights, and so adopted in progression ; and the rates of postage shall
be applied accordingly.
Joint Resolution to enable the Secretary of the Treasury
TO furnish to each State one set of the Standard Weights
AND Measures of the Metric System.
Be it resolved by the Senate and House of Representatives of the United States
of America in Congress assembled, That the Secretary of tlie Treasury be, and
he is hereby, authorized and directed to furnish to each State, to be delivered
to the governor thereof, one set of the standard weights and measures of tho
metric system, for tlie use of the States respectively.
TABLES AUTHORIZED BY CONGRESS.
MEASURES OF LENGTHS.
Metric Dehominations and Values.
Equivalents in Denominations in uso.
Myriametre, . . .
10,000 metres,
6.2137 miles.
Kilometre,
1,000 metres,
0.62137 miles, or 3280 feet, 10 inches.
Hectometre , . . .
100 metres,
32 S feet and 1 inch.
Decametre, . . .
10 metres,
393.7 inches,
Metre,
1 metre,
39.37 inches.
Decimetre,
^^g- of a metre, . .
3.937 inches.
Centimetre, , . .
W'-^ of a metre, . .
0.3937 inch.
Millimetre,
TWTT 0^ ^ "^«^^^' •
0.0394 inch.
MEASURES OF SURFACES.
Metric Denominations and Values.
Equivalents in Denominations in use.
Hectare,
Are,
Centiare,
10,000 square metres,
100 square metres,
1 square metre,
2.471 acres.
119.0 square yards.
1550 square inches.
THE METRIC SYSTEM.
349
MEASURES OF CAPACITY.
Metric Denominations and Values.
Equivalents in Denominations in use.
Names.
No. of
Utres.
Cubic Measure.
Dry Measure.
Liquid or wine
measure.
Kilolitre, or stere,
Hectolitre,
Decalitre.
Litre,
1000|1 cubic metre,
100 j^^ of a cubic metre, . . .
lOjlO cubic decimetres,. . .
1.308 cubic yd.
2 bu. 3.35 pk.. .
9.08 quarts,
0.908 quart, . . .
6.1022 cubic in.
0.6102 cubic in.
0.061 cubic in. .
264.17 gallon.
26.417 gallon.
2.6417 gallon.
1.0567 quart.
0.845 gill.
0.338 fluid oz.
0.27 fluid dr.
Decilitre,
Centilitre,
Millilitre,
TTTo
-^Q of a cubic decimetre,
10 cubic centimetres, . .
1 cubic centimetre,
WEIGHTS.
Metric Denominations and Values.
Equivalents in De-
nominations in use.
Names.
Number of
grammes.
Weight of what quantity of water
at maximum density.
Avoirdupois weight.
Millier, or tonneau, .
Quintal,
1,000,000
100,000
10,000
1,000
100
10
i
1000
1 cubic metre,
2204.6 pounds.
220.46 pounds.
22.046 pounds.
2.2046 pounds.
8.5274 ounces.
0.3527 ounce.
15.432 grains.
0.5432 grain.
0.1543 grain.
0.0154 grain.
1 hectolitre,
Myriagramme,
Kilogramme, or kilo.
Hectogramme,
Decagramme,
Gramme,
10 litres,
1 litre, * .
1 decilitre,
10 cubic centimetres,
1 cubic centimetre,
1-10 of a cubic centimetre,
10 cubic millimetres,
1 cubic millimetre,
Decigramme,
Centigramme,
Milligramme
Note. — The spelling in the above tables is not the same as in
the tables in the schedule annexed to the report of the committee of
the House of llepresentatives on weights and measures. The change
is not made to indicate any preference for any standard upon this
subject ; but to carry out what the author believes to be an essential
condition to the utility and success of the system.
As remarked by a distinguished senator when the tables were
adopted by Congress, "77?e names are cosmopolitan ;" and to re-
tain this character fully , the spelling must also be cosmopolitan.
The French introduced the nomenclature and spelling ; and, so
long as the names remain unchanged, the spelling should be retained.
R.P. 16
850
THE METRIC SYSTEM.
NOMENCLATUKE AND TABLES.
There are eight kinds of quantities for which tables are usually
constructed; viz., Lengths, Surfaces, Volumes or Solids, Capacities,
Weights, Values, Times, and Angles or Arcs. The table for Times
is the same in the metric as in the ordinary system. The table for
Angles is constructed upon a centesimal scale. The tables for the
other six kinds of quantities are constructed upon a decimal scale.
Ill each of the tables for Lengths, Surfaces, Volumes, Capacities, and
Weights, there are eight denominations of units, — one principal and
seven derivative. The principal units are the Tnetre, which is the
base of the system, and those derived directly from it. The two
following tabular views present the facts regarding the principal and
derivative units, which should be fixed in the memory.
r I. Metre,
II. Are, .
III. Stere,
IV. Litre,
V. Gramme,
1. Principal unit of Lengths.
2. The base of the metric system, and nearly
one ten-millionth part of a quadrant of
the earth's meridian.
^ 3. Equivalent, 39.3708 inches.
1. Principal unit of surfaces.
2. A square whose side is ten metres.
3. Equivalent, 119.6 square yards.
1 . Principal unit of volumes or solids.
2. A cube whose edge is one metre.
3. Equivalent, 1.308 cubic yards.
' 1. Principal unit of capacities.
2. A vessel whose volume is equal to a cube
whose edge is one-tenth of a metre.
3. Equivalent, .908 quart dry measure, or
1.0567 quarts wine measure.
1. Principal unit of weights.
2. The weight of a cube of pure water whoso
edge is .01 of a metre.
3. The water must be weighed in a vacuum
4° C, or 39.2° F.
4. Equivalent, 15.432 grains.
THE METRIC SYSTEM.
351
a 2
C m
O S
. o
1. Three orders of smaller units, or submultiples of eacb
kind, are formed by dividing eacb of the principal units
into tenths, hundredths, and thousandths.
2. Four orders of larger units, or multiples of eacb kind,
are formed by considering as a unit ten times, one
hundred times, one thousand times, and ten thousand
times, eacb of the principal units.
" The names of derivative units are formed by
attaching a prefix to the name of the princi-
pal unit from which they are derived, which
indicates their relation to the principal unit.
1. Millesimus, one thousandth, contracted
2U
s.
3 3
Milli. Example, Millilitre
8 millilitres = i-i^js of a litre.
roV(jofalitre;
2. Centesimus, one hundredth, contracted
centi. Ex., Centiare = yic? of an are; 4
centiares = y-*-o of an are.
3. Decimus, tenth, contracted deci. ^.r.. De-
cimetre = i^jj metre ; 3 decimetres = -^^
metre.
1. Deca, ten. Example, Decametre = 10
metres ; 5 decametres = 50 metres.
2. Hecaton, one hundred, contracted hecto.
Ex., Hectolitre = 100 Utres; 7 hectolitres
= 700 litres.
3. Kilioi, one thousand, contracted kilo. Ex.
Kilogramme = 1000 grammes.
4. Myria, ten thousand. Ex., Myriastere =
10,000 stores; 3 myriasteres =30,000 steres.
5. The a in deca and myra, and the o in hecto
and kilo, are dropped when prefixed to are.
The tables being constructed upon a decimal scale, ten
units of a lower order .make one of the next higher,
thus : 10 millimetres = 1 centimetre ; 10 centimetres
= 1 decimetre; 10 decimetres = 1 metre; 10 me-
tres = 1 decametre, &c.
"2 •
g OJ
352
THE METRIC SYSTEM.
The facts in the preceding views being mastered, the tables can be
constructed by the pupil at sight. For example : The names of the
derivative units are formed by attaching the seven prefixes, in their
order, to the principal units of the tables. The order of progression
being ten, the table of capacities will be written thus : —
10 Millilitres = 1 Centilitre.
10 Centilitres = 1 Decilitre.
10 Decilitres = 1 Litre.
10 Kilolitres
10 Litres = 1 Decalitre.
10 Decalitres = 1 Hectolitre.
10 Hectolitres = 1 Kilolitre.
1 Myrialitre.
All the tables peculiar to the Metric System are presented together
in a convenient form in the two following tables : —
TABLE OF SUBMULTIPLES AND PRINCIPAL UNITS.
Names ok Units.
T*nr>XTTvr"T \TTr»v
Symbols.
PREFIX.
BASE.
JL XVVy^^ i^ ^1 \^ X^L X XV/*> .
P Metre
Miir-e-mee'-ter
8^
10 MilH-
Are
Mili'-e-are
3^ .
Equal
Sterc
MiU'-e-ster
sS
1 Centi-
Litre
Miir-c-li'-ter
8^^
8^
. Gramme
Miir-e-gram
r Metre
Sent'-e-mee'-ter
^M
10 Centi-
Are
Scnt'-e-&re
A
Equal
Stere
Sent'-e-ster
2S
1 Dcci-
Litre
Sent'-c-li'-tcr
^
. Gramme
Scnt'-e-gram
fi
- Metre
Des'-e-mee'-tcr
,M
10 Dcci-
Are
Dcs'-c-are
A
Equal
Stere
Dcs'-e-ster
iS
1 Principal Unit.
Litre
Dcs'-c-li'-ter
JL
. Gramme
Des'-e-gram
fi
- Metre
Mee'tcr
M
10 Principal Units
Are
Are
A
E(iiial
Store
Stor
S
1 Dcca-
Litre
Li'-tcr
L
. Gramme
Gram
G
THE METRIC SYSTEM.
353
TABLE or MULTIPLES.
Names oi
<- Units.
Peonuxciatiox.
PREFIX.
BASK.
r Metro
Dck'-a-mee-ter
^\I
10 Deca-
Arc
Dck'-are
^A
Equal -
Store
Dck'-a-ster
^S
1 Hecto-
Litro
Dek'-a-li'-ter
'l
. Gramme
Dck'-a-gram
'G
- Metre
Hec'-to-mee-tcr
hi
10 Hecto-
Are
Hec'-tare
'a
Equal -
Sterc
Ilec'-to-ster
's
1 Kilo-
Litre
Ilec^-to-li'-tcr
'l
- Gramme
Ilcc'-to-gram
'g
r Metre
Kiir-Q-mee-ter
hi
10 Kilo-
Are
Kiir-are
'a
Equal -
Stere
Kill'-o-ster
's
1 Myria-
Litro
Kiir-o-li'-ter
'l
> Gramme
Kill'-o-gram
'g
- Metre
Mir'-e-a-mee-tcr
'm
Are
Mir'-e-are
\
Myria- -
Stere
Mir'-e-a-ster
*s
Litre
MiZ-e-a-li'-ter
*L
^ Gramme
Mir'-e-a-gram
'G
ABBREVIATED NOMENCLATURE.
To secure the fullest advantage to business men by the universal
adoption of the new system of weights and measures, it is necessary
that the names used should be short and easy to write and pronounce,
that they should express clearly the relation of the different denomi-
nations of the same table to each other, and that they should be
identical in all languages.
The last two of these requirements would be secured by the uni-
versal use of the nomenclature adopted by the French. It is cosmo-
politan in its character : it belongs to their language no more than to
any other. The former, however, is not secured. It is evident to
all, that, for business purposes, the long names of the metric system
are inconvenient, and that to shorten them would prove a great
354
THE METRIC SYSTEM.
advantage. Efforts have been macje to introduce short names;
but these efforts have invariably sacrificed their universal and expres-
sive character, which is of more importance to the business world
than their shortness.
The only true course which seems to be open, is to abbreviate the
names already introduced, in such a way as to retain their peculiar
characteristics.
To secure this, the following plan of abbreviation is suggested : —
First. Let the prefixes be abbreviated thus : Myr, kil, hect, dec,
des, cent, mil.
Second. Let the initial letter of the names of the five principal
units be used, instead of the names themselves, thus : For metre, use
a capital M ; for are, use a capital A ; for store, a capital S ; for
litre, a capital L ; and, for gramme, a capital G.
Third. For the names of multiples and sub-multiples, attach to
these initial capital letters the abbreviated prefixes, thus : Kil M, pro-
nounced kill-em' ; Kil S, pronounced kill-ess', &c.
By this method of abbreviation, the elements of the original terms
are retained in such a form that each part is clearly indicated. The
capital letter used after the prefix will always point to the base-word
of which it is the initial, although the pronunciation is changed.
TABLES WITH ABBREVIATED NOMENCLATURE.
MEASURES OF LENGTHS.
Written.
rronounccd.
10 Mil ]M,
Mill-em', make 1 Cent M
10 Cent M,
Centrem',
1 Des M.
10 Des M,
Des-eni' '
1 M.
10 M,
Em
1 Dec M.
10 Dec M,
Dek-cm',
1 Hect M.
10 Hect M,
Ilect-em', *
1 Kil M.
10 Kil M,
Kill-em',
1 Myr M.
Myr M,
Mir-em'.
THE METRIC SYSTEM.
355
MEASURES OF SURFACES.
Written.
Pronounced.
10 Mil A,
Mill-ii',
make
1 Cent A
10 Cent A,
Cent-a',
1 Des A.
10 Des A,
Des-a',
1 A.
10 A,
A,
1 Dec A.
10 Dec A,
Dek-a',
1 Hect A,
10 Hect A,
Hect-a',
1 Kil A.
10 Kil A,
Kill-a',
1 Myr A.
Mjr A,
Mir-a'.
MEASURES OF VOLUMES, OR SOLIDS.
Written.
Pronounced.
10 Mil S,
Mill-ess',
make
1 Cent S.
10 Cent S,
Cent-ess\
1 Des S.
10 Des S,
Des-ess',
IS.
10 S,
Ess,
1 Dec S.
10 Dec S,
Dek-ess',
1 Hect S,
10 Hect S,
Hect-ess',
1 Kil S.
10 Kil S,
Kill-ess',
IMyrS.
Myr S,
Mir-ess'.
MEASURES OF CAPACITT.
Written.
Pronounced
10 Mil L,
Mill-eir,
10 Cent L,
Cent-eir,
10 Des L.
Dess-eir
10 L,
Ell,
10 Dec L,
Dek-eir,
10 Hect L,
Hect-eir,
10 Kil L,
Kill-ell',
Myr L,
Mir-eir.
make
1 Cent L.
1 Des L.
1 L.
1 Dec L.
1 Hect L.
1 Kil L.
1 Myr L.
356 THE METBIC SYSTEM.
MEASURES OF WEIGHTS.
Written.
Pronounced.
10 Mil G,
Mill-gee',
make
1 Cent G.
10 Cent a.
Cent-gee',
1 Des G.
10 Des G,
Dee-gee',
IG.
10 G,
Gee,
1 Dec G.
10 Dec G,
Dek-gee',
1 Hect G.
10 Hect G,
Ilect-gee',
1 Kil G.
10 Kil G,
Kill-gee',
1 Myr G.
Myr G,
Mir-gee'.
NOTATION AND NUMERATION.
In the practical application of the metric system, it is not always
convenient to use the principal units as the unit of number. For
example : Should the gramme, the principal unit of weight, be used
as the unit of number, in the grocery or any similar business, small
quantities would be expressed by inconveniently large numbers.
Example : 386 lbs. are expressed by 175,000 grammes. To avoid
this inconvenience, the higher denominations are used as the unit of
number when large quantities are measured.
No general system of notation is yet agreed upon. The same
quantity is written in various ways by different authors. Example :
42 metres, 8 decimetres, and 5 centimetres, are written
m cm
42.85 M. 42? 85. 42.85. M 42.85. &c.
Inasmuch as the principal units of measure arc not always used as
the unit of number, it is important that a system of notation be adopt-
ed, wliich will apply equally well to both principal and derivative
units.
It is believed that the system given below, while simple and con-
venient, expresses clearly the relation of the unit of number to the
principal unit of measure ; and, hence, has an advantage over any
contractions of the names of the derivative units or arbitrary signs
which might be adopted.
the metric system. 357
General principles of notation.
I. The scale in the metric system being decimal, the consecutive
denominations are expressed by the consecutive orders of units in a
number. Thus, 78642.358 metres is an expression for 7 myria-
metres, 8 kilometres, 6 hectometres, 4 decametres, 2 metres, 3 deci-
metres, 5 centimetres, 8 millimetres.
II. Whichever one of the eight denominations of units of measure
is used as the unit of a number, the higher denominations are ex-
pressed as tens, hundreds, and so on ; and the lower as tenths, hun-
dredths, and so on. Example : 784.56 decametres. Here the unit
of the number is a decametre ; consequently the tens and hundreds
are, respectively, hectometres and kilometres, and the tenths and
hundredths are metres and decimetres.
From these principles and illustrations, we derive the following rule
for notation : —
KuLE. Write the consecutive denominations in their order, com-
mencing with the higher, and placing a cipher wherever a denomi-
nation is omitted, and the decimal point after the denomination
which is the unit of the number.
RULES FOR IXDICATOG THE DENOiC^fATION.
Rule I. When a principal unit of measure is the unit of num-
ber, place the initial letter of the unit used before the number, thus :
M 342.5. Read, three hundred and forty-two and five-tenths
metres ; or, 3 hectometres, 4 decametres, 2 metres, 5 decimetres.
EXAMPLES FOR PRACTICE.
Write the numbers which represent the following quantities, con-
sidering the principal unit of measure the unit of number.
1. Seven myriametres, 4 hectometres, three decametres, and eight
centimetres. Ans. M 70430.08.
2. Thirty-four kilometres and forty-three millimetres.
Ans. M 34000.043.
3. Eighty-seven hectogrammes and fifty-nine centigrammes.
Ans. G 8700.59.
358 THE METRIC SYSTEM.
4. Thirty-two myriagrammes, forty-eight decagrammes, five milli-
grammes. Ajis. G 320480.005.
6. Three hundred and two kilares, eight hundred and seven ccn-
tiares. A7is. G 302008.07.
6. Four myrialitres, sixty-two decalitres, five millilitres.
Ans. L 40620,005.
7. Four hundred and thirty-three kilosteres, nine hundred and
eighty four hectosteres, seven thousand two hundred and three centi-
steres. - Ans. S 53147203.
KuLE II. When a multiple of a principal unit of measure is the
unit of number ; — First, Place before the number (he initial letter
of the principal unit from which the multiple is derived. Second,
Indicate the order of multiple used h^j a small figure placed to the
left and above the letter prefixed to the number, (See symbols in
table of multiples.)
Example. 42.5 kilometres, is written ^M 42.5.
The M before the number indicates that the metre is the unit of
measure from which the unit of the number is derived. The small
3 indicates that the third order of multiple, or kilometre, is the unit
of number.
EXAMPLES FOR PRACTICE.
Write the numbers which represent the following quantities, con-
sidering the denomination named as the unit of number ; —
Unit of Number, Kilogramme.
1. 43 myriagrammes, 7 decagrammes, 5 grammes.
Ans. «G 430.075.
2. 8 kilogrammes and 3 centigrammes. Ans. ^G 8.00003.
3. 736 hectogrammes, 243 centigrammes, and 4 milligrammes.
Ans. »G 73.602434.
4. 2009 hectogrammes and 3 centigrammes.
Ans. ^G 200.90003.
Unit of Number, Decalitre.
5. 254 litres and 43 millilitres. Ans. ^L 25.4043.
THE METRIC SYSTEM. 359
6. 364 mjrialitres, 47 litres, 384 millilitres.
Am. ^L 304004.7384.
7. 243 decalitres, 47 centilitres. Ans. ^L 243.047.
Unit of Niimher^ Second Order of Multiples.
8. 23 myriametres, 72 millimetres. Ans. ^M 2300.00072.
9. 4000007 steres and 2 millisteres. Ans, ^S 40000.07002.
10. 3 kilares and 43 centiares. Ans. ^A. 30.0042.
Unit of Number, Myriametre,
11. 3 hectometres and 2 centimetres. Ans. *M .030002.
12. 6 millimetres. Ans. *M .0000005.
13. 3 decametres and 2 centimetres. Ans. *M .003002.
KuLE III. When a submultiple of a principal unit of measure
is the unit of number ; — First, Place before the number the initial
letter of the principal unit from which the submidtiple is derived.
Second, Indicate the order of submultiple used by a small figure
placed to the left and below the letter prefixed to the number. (See
symbols in table of submultiples.)
EXAMPLES FOR PRACTICE.
Write the numbers which represent the following quantities, con-
sidering the denomination named as the unit of number.
Unit of Number, 3Iillimetre.
1. 32 decametres and 2 decimetres. ^ws. 5M 320200.
2. 7002 hectometres. Ans. gM 700200000.
3. 7 myriametres and 5 metres. Ans. 3M 70005000.
4. 3 kilometres and 2 decametres. Ans. 3M 3020000.
Unit of Number, Second Order of Submultiples.
5. 5 kilogrammes and 9 grammes. Ans. gG- 500900.
6. 302 myriasteres, 5 decasteres, and 3 centisteres.
Ans. 2S 302005003.
7. 4009 kilolitres and 5 litres. Ans. gL 400900500.
8. 2 hectares and 2 centiares. Ans. jA 20002.
360 THE METKIC SYSTEM.
Unit of Number, Decilitre.
9. 3002 hectolitres and 4 millilitrcs. Ans. jL 3002000.04.
10. 6 myrialitres and 1 decalitre. Ans. iL GOO. 100.
11. .404 millilitres. Ans. jL .00004.
REDUCTION.
Rule for Reduction Descending. Multiply the given quantity
hy the number of the required denomination which makes a unit of
the given denomination.
Since the multiplier is always 10, 100, 1000, &c., the operation
is performed by removing the decimal point as many places to the
right as there are ciphers in the multiplier, annexing ciphers when
necessary.
EXAMPLES FOR PRACTICE.
1. Reduce ^M 32.58 to millimetres.
2. Reduce *M 5 to decimetres.
3. Reduce G402 to milligrammes.
4. Reduce ^A42.3 to centiares.
5. Reduce ^L 93.2 to decilitres.
G. Reduce *S 895 to decasteres.
7. Reduce 'A 903.2 to milliares.
8. Reduce 'G 539 to centisrammes.
Rule for Reduction Ascending. Divide the given quantity
hy the number of its own denomination which makes a unit of the
required denomination.
Since the divisor is always 10, 100, 1000, &c., the operation is
performed by removing the decimal point as many places to the left
as there are ciphers in the divisor, prefixing ciphers when necessary.
1. Reduce 2 A 5 to myriares.
2. Reduce gM 403 to kilometres.
3. Reduce iS42.3 to hectosteres.
4. Reduce 3 A 7.2 to decares.
EX.iMPLES FOR PRACTICE.
5. Reduce sG 3 to kilogrammes.
6. Reduce ^^ 5.7 to hectolitres.
7. Reduce 3M 9 to myriametres.
8. Reduce 084 7.3 to decasteres.
MEASURES OF SURFACES.
RELATIONS OF UNITS OF SURFACE TO UNITS OF LENGTH.
Decimilliare = One square decimetre = 100 square centimetres.
^.,.. _ f -^^ square decimetres, or a plane figure wfioso
( length is one metre and breadth one decimetre.
Centiare = One square metre = 100 square decimetres.
THE METRIC SYSTEM.
361
Declare
Are
Decare
Hectare
Kilare
Myriare
={
10 square metres, or a plane figure wbose length is one
decametre and breadth one metre.
One square decametre = 100 square metres.
ilO square decametres, or a plane figure whose length
is one hectometre and breadth one decametre.
One square hectometre = 100 square decametres.
(10 square hectometres, or a plane figure whose length
I is one kilometre and breadth one hectometre.
One square kilometre = 100 square hectometres.
NUMERAL EXPRESSION FOR SURFACE.
Tlie contents of a plane figure is expressed numerically by giving
the number of times Ifc contains some given area, which is assumed as
the unit of surface.
The following illustrations will show how the various denomina-
tions of the table are used in numerical expressions of surface : —
ILLUSTRATION FIRST.
r
Length 6 metres.
It win be seen, by examining this figure, that the lines drawn
parallel to the sides, at the supposed distance of a metre from each
other, divide the surface into square metres, and that there are as
many rows of square metres as there are metres in the breadth, each
row containing as many square metres as there are metres in the
length. Hence the number of square metres in the area of the figure
is equal to the product of the two numbers which indicate the length
and breadth ; and A 0.21 is a numerical expression for its contents.
16
362
THE METRIC SYSTEM.
ILLUSTRATION SECOND.
1=
C metres.
In this figure, the lines drawn parallel to the sides divide the
figure into 36 milliares, or oblongs, whose length is one metre and
breadth one decimetre. It is evident that ten of these oblongs put
together will constitute a centiare, or square metre. Hence the ex-
pression, 36 milliares, may be written 3.6 centiares; and read, three
and six tenths centiares, or three centiares and six milliares.
By reducing the length to decimetres, the numerioal expression of
the contents will be, by Illustration First, 60 x 6, or 360 dccimilliares
or square decimetres.
ILLUSTRATION THIRD.
Length 1 decametre, 2 metres, and 1 decimetre.
1
1
1
Deciare.
sV
/•
Dcciarp.
^.v
Milliare. Dccimilliare,
In this figure, we have illustrated the relations of different denomi-
nations of units in expressing the contents of a given surface.
THE METRIC SYSTEM. 363
In the following analysis, each part of the contents is presented
separately, as it would be obtained by multiplying the length by the
breadth. The learner should carefully note each part, and analyze a
sufficient number of examples to fix the principles in the mind.
ANALYSIS.
Jq ( One decimetre = 1 decimilliare = A 0.0001
One decimetre x \ Two metres = 2 milliares = A 0.002
( One decametre = 10 milliares = 1 centiare = A 0.01
{One decimetre =s 2 milliares = A 0.002
Two metres =s 4 centiares = A 0.04
One decametre == 2 declares = A 0.2
iOne decimetre =s 10 milliare = 1 centiare = A 0.01
Two metres =r 2 declares = A 0.2
. , One decametre = 1 are or square metre = A 1.
X =
1—1
CI
^M 1.21 X ^M 1.21 = A 1.4641
From these illustrations, we derive the following rule for finding a
numerical expression for a given surface of uniform length and
breadth : —
Rule. Reduce the length and hreadth to the same denomination ;
find the product of the two dimensions after reduction, and point
off as many decimal places in this product as there are decimal
places in the two dimensions.
The unit of the numerical expression thus found will be a decimil-
liare when the unit of length is a decimetre, a centiare when the
unit of lenojth is a metre, an are when the unit of leno;th is a deca-
metre, a hectare when the unit of length is a hectometre, and a
myriare when the unit of length is a kilometre.
EXAMPLES FOR PRACTICE.
1. How many ares in a floor M 1.25 long, and M 8.7 wide?
Ans. A. 10875.
2. How many centiares, how many kilares, and how many hec-
tares in the same floor? Ans. gA 10.875.
3. How many ares in a board M 5.32 by gM 47. ?
Ans. A .025004.
4. How many milliares, how many myriares, and hectares in the
same board ?
5. How many metres of a carpet nine decimetres wide will cover
364 THE METRIC SYSTEM.
a floor six metres long and five and four-tenths metres wide * ^d
what would be the cost of the carpet, at $2.50 a centiare ?
Ans. M 3G. $};0.
G. In a farm consisting of four fields of the following dimensions,
bow many hectares ? First field, length M 342, breadth M 273 ;
second field, length M 634, breadth M 350 ; third field, length
M 450, breadth M 329 ; fourth field, length M 730, breadth M 632.7.
Ans. ^A 92.5187.
7. A pile of lumber was found to contain 150 boards M 4 long
and iM4. wide, 225 boards M 6.2 long and gM 52. wide, and 642
boards M 5.2 long and gM 43 wide. How much was it worth, at $42.
per are, face measure. Ans. $1008.38 -\-.
8. How many bricks iM2.2 X iM 1.1 would pave a side-walk
M 842.6 long and M 2.2 wide? and what would be the whole cost
at 82 cents per centiare. Ans. 76600 bricks. $1520.05 +.
MEASURES OF VOLUMES, OR SOLIDS.
RELATIONS OF UNITS OF VOLUMES TO UNITS OF LENGTHS.
Millistere = A cubic decimetre = 1000 cubic centimetres.
r 10 cubic decimetres, or a volume, or solid, whose
Centistere = ■< length is one metre, and breadth and thickness one
( decimetre.
f 10 centisteres = 100 cubic decimetres, or a volume
Decistere = •< whose length and breadth is one metre, and thick-
C ness one decimetre.
^ _ ( ^ cube metre = 10 decisteres = 100 centisteres =
( 1000 millisteres or cubic decimetres.
_ (10 cubic metres, or a volume whose length is one
Decastere ^^ •%
(. decametre, and breadth and thickness one metre.
/" 10 decasteres = 100 cubic metres, or a volume whoso
Ilcctostere = -< length and breath is one decametre, and thickness
(. one metre.
Kilostere = A cubic decametre = 1000 cubic metres.
r 10 kilosteres, or a volume whoso length is one hecto-
Myriastere = •< metre, and breadth and thickness each one deca-
( metro.
THE METRIC SYSTEM.
365
NUMERICAL EXPRESSION FOR YOLUME, OR SOLIDITY.
The solidity, or contents, of a volame is expressed numerically by
giving the number of times it contains some given solid as the unit
of volume.
The following illustrations will show how the various denominations
of the table are used in numerical expressions of volume.
MiUistere, or Cubic Decimetre.
10 millisteres, placed side by side, make a volume whose length
is one metre, and breadth and thickness each one decimetre, thus, —
10 centistere, placed side by side, make a volume whose length
and breadth is each one metre, and thickness one decimetre, thus, —
Decistre = 10 Centisteres = 100 3Iillisteres.
10 decisteres, placed face to face, make a cnbe whoso edge is one
metre, thus, —
Stere = 10 Decisteres = 100 Centisteres = 1000 Millisteres,
From these illustrations, it is evident that the contents of a cubic
metre may be expressed numerically, as S 1, iS 10, 2S 100, 3S 1000.
366
THE METRIC SYSTEM.
The followlnoj fiojures illustrate the use of
the same four denominations in expressing
the contents of a cubic volume whose edge
is one metre and one decimetre. The sur-
face of one face of the volume contains
one centiare, two milliares, and one deci-
milliare, thus, —
Centiare.
Milliaif.
jr
Taking a slab of the face one decimetre thick, thus, —
and we have one decistere, two
centisteres, and one millistere.
But the volume is eleven deci-
metres thick ; therefore we have iiliilH
eleven such slabs, or eleven times one decistere, two centisteres, and
one millistere.
r 11 millisteres = 1 centistere and 1 millistere = S 0.011
= K 22 centisteres = 2 decisteres and 2 centisteres = S 0.22
( 11 decisteres = 1 stere and 1 decistere = S 1.1
Ml.l X Ml.l X Ml.l = S1.331
From these illustrations, we derive the following rule for finding a
numerical expression for a given volume of uniform length, breadth,
and thickness : —
Rule. Reduce the length, breadth, and thickness to the same
denomination ; find the product of the three dimensions, after re-
duction, and point off as many decimal places in this product as
there are decimal places in the three dimensions.
The unit of the numerical expression thus found will be a millistere
when the unit of length is a decimetre, a stere when the unit of length
is a metre, a kilostere when the unit of leno:th is a decametre.
EXAMPLES FOR PRACTICE.
1. How many steres in a wall twenty-four metres long, eight and
five-tenth metres higli, and fifty-two centimetres thick? And what
would be the cost of building it, at $4.25 a store?
Ans. S 106.08. Cost, $450.84.
THE METRIC SYSTEM. 367
2. Wbat would be the cost of a pile of wood fifteen and seven-
tenths metres long, three metres high, and seven and fifty- two hun-
dredths metres wide, at $1.50 a stere? Ans. $531.29—.
3. What would be the cost of excavating a cellar eighteen and
three-tenths metres long, ten and seventy-three hundredths metres
wide, and three and four-tenths metres deep, at 15 cents per sterc?
A71S. S100.14+.
4. How deep must a box be, whose surface is thirty-two milliarcs,
to contain seven and thiiiy-six hundredths stores? Ans. iM 23.
5. How many stores in five sticks of timber of the following di-
mensions : First, iM 5.2 by jM 7.3, and M 13 long; second, gM 43.
by 2M 65, and M 17.5 long; third, ^M 5.3 by ^M 3.7, and M 15.42
long; fourth, 2M 39 by gM 56, and M 14 long; fifth, iM 4.52 by
iM 3.78, and M 15 long. Ans. S 18.470352.
6. What must be the height of a load of wood, M 3.2 long and
M 1.1 wide, to contain S 4.0128. A}is. M 1.14.
MEASUREMENT OF ANGLES.
In the ordinary or sexagesimal system, a right-angle, which is used
as the measure of all plane angles, is divided into 90 equal parts,
called degrees; a degree is divided into 60 equal parts, called
minutes ; and a minute into 60 equal parts, called seconds.
In the centesimal or French system, a right-angle is divided into
100 equal parts, called grades; a grade into 100 equal parts, called
minutes; and a minute into 100 equal parts, called seconds.
The former is called the sexagesimal system, on account of the
occurrence of the number sixfy in forming the subdivisions of a de-
gree ; and the latter centesimal, on account of the occurrence of the
number one hundred.
Grades, minutes, and seconds are usually written thus : 35^ 42^
24^^ ; read, thirty-five grades, forty-two minutes, twenty-four seconds.
Since the scale is centesimal, minutes may be expressed as hun-
dredths, and seconds as ten-thousandths ; hence any number of grades,
minutes, and seconds may be expressed decimally thus : 73= 4569 ;
read, seventy- three ^ades, forty-five minutes, sixty-nine seconds.
368 THE METRIC SYSTEM.
In a right-angle, there are 100 grades, or 90 degrees; benco, for
every 10 grades there are 9 degrees. Dividing the 10 gi-ades into
9 equal parts or degrees, each part will contain 1^ grades; therefore
a degree s equal to 1 1 grades. Ilcncc, in any number of grades
there are as many degrees as 1^ is contained times in the given
number of grades ; and, conversely, in any number of degrees there
are 1^ times as many grades as there arc degrees. Hence the fol-
lowing rules : —
TO CHAJJ^GE THE CENTESIMAL MEASUxlE TO THE SEXAGESIMAL.
Rule. Express the minutes and seconds as a decimal of a
grade ^ divide hy \\'. the quotient will express the number of de-
grees and decimals of a degree in the given number of grades, min-
utes, and seconds.
EXAMPLES.
Change the following quantities from the centesimal measmi-e to
the sexagesimal.
Ans, 22° 48' 35.208".
Ans. 31' 16.932^
Ans, 74° 49^ 29.388".
Ans. 33° 17' .06".
Ans. 12° 44' 25.44".
Ans. 81° 49' 5.16".
Ans. 10° 39' 8.1".
TO CHANGE THE SEXAGESIMAL MEASITIIE TO THE CENTESiaLVL.
Rule. Reduce the minutes and seconds to a decimal of a de-
gree ; multiply the degrees and decimal of a degree by 1 ^ : the pro-
duct is the number of grades, minutes, and seconds in the given
number of degrees, minutes, and seconds.
EXAMPLES.
Change the following quantities from the sexagesimal measure to
the centesimal.
1. 36° 18' 27". Ans. 40« 34^ Uf\
2. 56' 54". Ans. 1« 5^ Zl^''.
1.
25^ 34^ 42^\
2.
5r 93 \
3.
83" 13^ 87^\
4.
36S 98^ lb'\
5.
14» 15' 60'^
6.
90" 90' W\
7.
18^ 50' 25''.
8.
27° 36' 45''.
4.
189° 15' 20".
5.
63° 14' 58".
G.
147° 24' 48".
7.
117° 36' 54'.
THE METRIC SYS-TEM. 369
Ans. 30« 68^ 5f ^
Arts. 210^ 28^ 39|r-
Ans. 70s27''71|f-\
Ans. 163^ 79^ 25|f \
Ans. 130^ 68^- 33-^^\
TO CUAXGE THE METPwIC TO THE COMMON SYSTEM.
Rule. Reduce the given quantity to the denomination of the
principal unit of the table ; multiply by the equivalent y and reduce
the product to the required denomination.
1. 3M3.6, how many feet?
OPERATiox. Analysis. — The metre is
^M 3.6 X 1000 = M 3600 ^^^ principal unit of the table ;
39.37 in. X 3600 = 141732 in. ^^"^^ ^« ^^^"^^ ^^^ ^^^«"^^-
141732 in. -f- 12 in. = 11811 ft. *^^' *^ °'^*''''- . ^'""^^^ ^^'^f^
are 39.37 inches in a metre, in
3600 metres there are 3600 times 39.37 inches; and since there are
12 inches in a foot, there are as many iceH as 12 inches is contained
times in 141732 inches. Therefore ^M3.6 is equal to 11811 ^nQt,
EXAMPLES FOR PRACTICE.
2. How many feet in 472 centimetres ? Ans. 15.4855^ ft.
3. How many cubic feet in 2 kilosteres? Ans. 70632 cu. ft.
4. How many gallons, wine measure, in 325 decilitres ?
Ans. 8 gals. 2.343— qts.
5. How many gallons in 108.24 litres ? Ans. 28.594 -j- gals.
6. How many bushels in 3262 kilolitres ?
Ans. 92559.25 bush.
7. How many square yards in 436 ares ?
Ans. 52145.6 sq. yds.
8. In 942325 centilitres, how many bushels ?
Ans. 267.3847 + bush.
9. In 436 mjTiagrammes, how many pounds ?
Ans. 9611.9314 lbs.
370 THE METRIC SYSTEM.
TO CHANGE FROM THE COMMON TO THE METRIC SYSTEM.
Rule. Iteduce the given quantity to the denomination in which
the equivalent of the principal unit of the metric table is expressed ;
divide hj this equivalent, and reduce the quotient to the required
denomination.
1. In 10 lbs. 4 oz. liow many myriagrammes ?
OPERATION. Analysis. — •
10 lbs. 4 oz. = 10.25 lbs. T^e gramme,
10.25 lbs. X 7000 = 71750 gr. *^® principal
71750 gr.-h 15.432 gr. = 04649.43— ""'^ °^ ^^^
G 4649.43 j- 10000 = ^G .464943 — Ans. *^^^^' '^ ^"^
pressed in
grains hence we reduce the pounds and ounces to grains. 15.432
grains make one gramme; hence there are as many grammes in 71750
grains as 15.432 grains is contained times in 71750 grains. And since
there are 10000 grammes in a myriagramme, dividing G4G49.43 — by
10000 will give the myriagrammes in 10 pounds 4 ounces. Therefore,
10 lbs. 4 oz. is equal to *G 464943 —
EXAMPLES FOR PRACTICE.
2. In 6172.8 pounds, how many decagrammes ?
Ans. ^0 280000.
3. How many hectares in 2392 square yards ? Afis. ^A .2.
4. How many ares in a square mile ?
Ans. A 25899.665552—.
5. How many millistercs in 18924 cubic yards?
Ans. 3S 14467889.9082 +.
6. In 892 grains, how many hectogrammes ?
Ans. 2G. 578019.
7. In 2 miles, 6 furlongs, 39 rods, and 5 yards, how many
kilometres? Ans. «M 4.620416 +.
8. Bought 454 bush, wheat at $3, and sold the same at $8.75
per hectolitre ; how many hectolitres did I sell ? Did 1 gain or lose,
and how much ?
Ans. ^L 160. Gain, $38.
THE METRIC SYSTEM.
371
MISCELLANEOUS EXASIPLES.
Kequired the footings of the following bills : —
(!•)
^Y. J. Milne,
New York, April 23, 1867.
Bo't of L. CooLEY & Son.
M 122 Broadcloth,
@ $6.00
" 320 Bid. Shirting,
" 230 White Flannel,
.35
.30
- 206.5 Ticking,
.31
" 107.9 Blk. Silk,
" 2.40
Hec'd Payment,
Ans. $1235.975
L. Cooli:y & Son.
(2.)
Buffalo, May 1, 1867.
CiiAS. D. McLean,
Bo't of AYm. Benedict.
40 chests Tea, each
«G30.5 @ $ 2.50
12 sacks Java Coffee,
'* 40.00
25 bbls. Coffee Sugar, each
^GllO - .32
10 *' Crushed "
^a 95 " .38
30 boxes Eaisins, "
«a 12 *' .50
Jxecd Payment^
Ans. $4951.00
Wm. Benedict.
3. A man bought a lot of land ^M 40 long and ^M 20 wide, and
sold one-third of it. How many ares had ho left, and what was the
cost of the lot, at $100 per acre ?
Ans. to first, A 53333.33^-. Ans. to second, $197685.95.
4. A fai-mer sold ^L 540 of wheat at $6, and invested the pro-
ceeds in coal at $3 per ton. How many myriagrammes of coal did
he purchase? Ans. ''G 36741.835147 +.
5. What will be the cost of a pile of wood M 42.5 long, M 2. high,
M 1.9 wide, at $2 per stere ? Ans. $323.
372 THE METRIC SYSTEM.
6. How many metres of shirting, at $.25 per metre, must bo
given in exchange for ^L 300 oats, at $1.20 per hectolitre?
Ans. M 1440.
7. A grocer buys butter at $.28 per lb., and sells it at $.60 per
kilogramme. Does he gain or lose, and what per cent. ?
Ans. Lost 2^1-^- %.
8. A bin of wheat measures M 5 square, and M 2.5 deep. How
many hectolitres will it contain, and what will be the cost of the
wheat, at $2 per bushel ? Ans. ^L(j25. $3546.875.
9. What price per pound is equivalent to $2.50 per ^G?
Ans. $11.34.
10. A merchant bought M 240 of silk at $2, and sold it at $1.95
per yard. Did he gain or lose, and how much ?
Ans. Gain $31.81.
11. Find the measure of 1^ 5^^ in decimals of a degree.
A71S. .00945.
12. A merchant shipped to France 50 bbls. of coffee sugar, each
containing 250 lbs., paying $2 per hundred for transportation. He
sold the sugar at $.34 per kilogramme, and invested the proceeds in
broadcloth at $4 per metre. How many yards of broadcloth did he
purchase? Ans. 458.71 -j- ytls.
13. The difference between two angles is 10 grades, and their
sum is 45°. Find each angle. Ans. 18° and 27°.
14. Determine the number of degrees in the unit of angular
measure when an angle of 66§ grades is represented by 20.
A71S. 3°.
15. How many centiares of plastering in a house containing six
rooms of the following dimensions, deducting one-twelfth for doors,
windows, and base ? and what would bo the cost of the work at 38
cents per centiare? First room, M 6.2 X M4.7; second room,
M 4.52 X M 4 ; third room, M 6 X M 5.2 ; fourth room, M 3.82 X
M3.82; fifth room, M7 X M6.2; sixth room, M4.5 X M4.25.
Height of each room, M 3.8. Ans. jA 562.039 — . $213.57 +.
i^iy^.
^ r^/^L ^Jj^ ^-^mii^/
/,
/£
/ 'f
2
--f.
^ »'' ,Lz^^^;?JJ'H,^^^/-
iZi^^^JEi
0-<rn/,',-'i
.^^-t' f'ii-
DISTILLING WATER.
r6194.]— Beixg an amateur photographer In a
country town, where the chemist has the impudence
to charge me the same
price for distilled
water that I pay for
my ale, I determined
to distil for myself.
The following simple
apparatus, modified
from a drawing in
the Photographic News,
works admirably, and
I find it very conve-
nient. I can now use
distilled water for all
my solutions, and for"
other purposes. A is
a common tin sauce-
pan, with a small hole
in the side, for a tobac-
co-pipe; B a " steamer,"
on top, with a bottom
like an inverted cone, an inch of wire being soldered at
the apex. A gas jet (Bunsen's, if possible) boils the
water in the saucepan ; the ascending steam Is con-
densed on the lower surface of the steamer, runs
down to the point of the wire, down the pipe into the
bottle. A small jet of cold water keeps B cool.
v.^:
X
,4^
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■'J t^^-?
m.
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m^ 15 1969
G 5REC'D-12iyi
LD 21A-30m-6.'67
(H24728l0)476
General Library
University of California
Berkeley
HEADERS.
-ectura. No. a.
AUBRARY
"he Inorganic Chem-
:ments of Common
ptive of The Amer-
tooKs, and The Edu-
l information, mailed
g CO.,
iT,NEW YORK.
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THE
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SANDERS' UNION READERS and SPELLERS. New and beav
IVISON, BLAKEN^
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ROBINSON'S MATHEMATICS ' ,'.; lete Series or" ARiiHiiKj
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WEBSTER'S SCHOOL i- Srand.ird. '
DANA'S MANUAL, ANn 1
Of world-wide repui..t;-ir:.
SILLIMAN'S PHYSICS, a;
guyh and practical.
I '
WILLSONS HISTOAIES. InM
WELL^' SCIEN - . CompU
HRAY'S BOt.i: . ;:^
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