THE LIBRARY
OF
THE UNIVERSITY
OF CALIFORNIA
LOS ANGELES
; /■■ - -•»> '- ■■ » • * m/.
'V* ^
ARITHMETIC SIMPLIFIED,
PREPARED FOR THE USE OP
PRIMARY SCHOOLS, FEMALE SEMINARIES,
AND
HIGH SCHOOLS,
IN THREE PARTS;
ADAPTED TO CLASSES OP DIFFERENT AGES, AND OF DIFFERENT
DEGREES OF ADVANCEMENT.
BY CATHARINE E. BEECHER,
LATE PRINCIPAL OF THE HARTFORD FEMALE SEMINARY.
HARTFORD :
PUBLISHED BY Df F. ROBINSON & CO.
1832.
Entered according to Act of Congress, in the year 1832, by D. F.
Robinson, & Co. in the Clerk's office of the District Court of Con.
necticut.
P. CANCTELD, PRINTER.
HARTFORD.
PREFACE.
\o\
— T539cu
The public have this claim upon any author, who offers
a new school book, that such a work shall contain some es-
sential advantages, which are not to be found, in any othei
work already in use. If a writer cannot sustain such a
claim, the public are needlessly taxed, for an article which
is not wanted.
It therefore seems proper, that a statement should be
made of what are supposed to be, the peculiar advantages
and improvements in this work.
The writer, for several years, has been engaged in in-
struction, and has either used, or examined, all the most
popular works on Arithmetic. The following are the de-
ficiencies, which have been experienced, and which it is the
aim of this work to supply. It should, however, be pre-
viously remarked, that all these difficulties have not been
experienced in every work of the kind, heretofore examined ;
but some have existed in one, and some in another, and no
one work, yet known to the writer, obviates them all.
1. The first difficulty, for which a remedy is here at-
tempted, originates from the fact, that in every school, there
is such a variety of age, intellect, and acquisition, that no
(me book is fitted for them all. If a work is found adapted
to advanced classes, it is too difficult for the younger and
less advanced. If it is fitted to these last, it is too easy for
the others.
To remedy this, the following work is divided into Three
Parts. The First Part is adapted to the comprehension of
young children. The Second Part is fitted to older classes.
The Third Part completes an entire system of Arithmetic,
containing all that is required of students on entering col-
lege. The whole work embraces every thing of any con-
sequence, that can be found in the most complete and ex-
tended works ever used, and yet so simplified as to occupy
much less space, and to demand much less labor. There
are enough subjects, of difficulty, to call forth mental in-
dustry and effort, without allowing those which are plain
to remain involved in needless difficulties.
2. Another difficulty which this arrangement remedies.
1*
865520
Vi PREFACE.
arises from the fact, that in most works of this kind, owing
to the length of the various exercises under each head, the
pupils lose the general principles they gain in one part, before
they reach another. Thus, before "Reduction is attained,
the principles employed in Addition and Subtraction, are
partially forgotten, and the pupils do not gain a clear and
general view of the whole science.
But in the First Part of this work, in the compass of
twenty pages, the pupil gains all the fundamental principles
of the science, which in the succeeding parts, are developed
in more minute particulars.
To aid in the same object, and to secure other advanta-
ges, a new method of classification has been adopted. Tin
chiefbenefit aimed at, by this new arrangement, is to sim-
plify the science, by leading the pupils to understand, that
all the various exercises of Arithmetic are included under
the same general principles of Addition, Subtraction. Mul-
tiplication, Division, and Reduction.
The common method of teaching the Simple Rules first,
and then of introducing Vulgar and Decimal Fractions,
has a tendency to render the science much more compli-
cated and perplexing. Thus the pupil is first taught the
process of Simple Addition. Then follows the exercises of
the other simple rules, and by the time the pupil has some-
what forgotten the principles of Simple Addition, comes
Compound Addition, which seems to the child as much on
a new principle, as Multiplication, or Division. Then,
after another interval, comes Decimal Addition, and then
the Addition of Vulgar Fractions.
But a child who is first taught the system of numeration,
as including whole numbers and fractions, and the nature of
each of these modes of expressing numbers, can immedi-
ately commence the operation of Addition, in all its various
particulars, and recognize the general principle, that runs
through the whole, and at the same time the peculiarity
which distinguishes each.
The difficulties arising from the common mode of ar-
rangement, are particularly felt, when the processes of
multiplying and dividing by fractions, are introduced. In all
previous operations, the pupil has found that multiplication
increases a number, and division diminishes it. But when
fractions are introduced, a new science seems to commence,
in which multiplication lessens and division increases a number,'
and all heretofore learned, seems to be contradicted and'
undone.
But if, at the commencement of the science, the pupil under-
PREFACE. Vll
stands the peculiar character of Fractions, and then finds
them arranged with whole numbers, so as to be able to
compare and distinguish the same general principles of the
various exercises, and at the same time the specific differ-
ence, the perplexity arising from an apparent multiplicity
of operations, and their seemingly contradictory nature, is
avoided.
When this plan was first attempted, some difficulty was
felt from the necessity of the operation of Division in the
previous operation of multiplying by a fraction. But this dif-
ficulty has been obviated by having the First Part precede,
in which Multiplication and Division are explained, with-
out entering minutely into the various processes of Frac-
tional Multiplication.
It will be found that pupils, by learning the Division Ta-
ble, and the First Part, can perform all the exercises in
Simple and Fractional Multiplication, without any other
knowledge of the rule of Division. There are two or three
exceptions, however, where some exercises for the slate are
introduced, where the rule of Division must beemployed.
These are intended for older pupils, who are supposed to
understand the process of Division, and may be omitted by
new beginners, until a review.
The writer has herself employed this method of classifi-
cation, in teaching, and it has been used by the teachers in
the institution under her care, for several years ; and thus
exjierience has enforced the conviction, which at first was
the result of reasoning, that this mode of classification will
better secure the benefits sought for, in all attempts at gen-
eralization. It certainly attains advantages, and avoids
difficulties, much more than the common method.
3. Another difficulty experienced in using some of the
most popular works of this kind, has arisen from the fact,
that the mental and ivritten exercises have been entirely
separate ; in some cases being placed in different books.
Thus the pupil, after completing an Arithmetic designed
for mental exercise alone, will often be found repeating ex-
actly the same processes in written Arithmetic, without
recognizing the principles, which, in mental operations,
have been constantly employed. To remedy this, both
mental and written exercises are placed together under
every general rule.
4. Another defect in teaching this science, has arisen
from a want of some method of stating and explaining the
rationale, of each arithmetical process. In many Arithme-
tics, a mechanical method is presented, of performing cer-
Vlll PREFACE.
tain operations according to rule, without any exhibition of
the reason for such operations. Thus, in Subtraction, ichy
one is carried, and ten borrowed ; or, in Multiplication, why
the figures are placed in a certain method ; or, in Division,
why multiplication and subtraction are performed, is never
explained or illustrated. To a child, they are a sort of ca-
balistical process, which he finds will bring the right answer,
and this is all he can know from any thing he gains from
the book. To remedy this, in the following work, every
rule is accompanied by a full explanation of the reason, for
each process employed. In the mental operations also, a
proper mode of stating each process is given. The defini-
tions, rules, and explanations, will be found to be more simple
and concise than in many works of this kind, and perhaps
may be considered as improvements.
As the writer has been in a situation, in which she has
had to employ various teachers, of different qualifications,
it has been one great aim to furnish a work, by which new
and inexperienced teachers could avail themselves of the ex-
perience of others. This work has been maturing for
several years, and the results of the experience of several
able and ingenious teachers employed by the writer, have
been added to her own. It is believed that any teachers with
common talents and industry, can, with the aid of this
work, do all for their pupils in this science, which needs to
be done, in order to make them thorough and expert pro-
ficients.
For the purpose of perfecting such a work as this, and
to make a fair trial of the several improvements contem-
plated, a small work on this plan was printed some years
ago, for the use of the pupils of the writer. But as it was
intended for an experiment, and was necessarily very im-
perfect and incomplete, it was never published. Yet, as in
several cases, those who have been teachers and pupils in
this institution, have introduced it into their schools, it may
be proper to add, that this work is very different from the
former, being much easier, much more extensive and com-
plete, and improved in several respects which it is unne-
cessary to mention.
The writer does not lay claim to any great originality,
in these various particulars, but has aimed to unite in one
work the various excellencies, which might be otherwise
scattered among a variety.
Hartford Female Seminary, Jan. 1, 1S32.
TO TEACHERS.
It is very desirable that new beginners should review the
First Part, till it is very thoroughly understood. It will save
much trouble to both teacher and pupil.
It will be found advantageous, to require older pupils te
study the First Past, previous to commencing the Second ;
for though some of the exercises are very simple, there are
some important explanations and illustrations, not found in
the Second Part.
It is very desirable that pupils should become thorough
and expert in Numeration, especially in Decimal Numera-
tion, before taking the next lessons, and one or two reviews
are recommended previous to proceeding.
In Compound Addition and Fractional Midtiplicalion, if the
pupil has never practised Simple Division, omit those exer-
cises which require this rule, till a review.
The Second Part should be reviewed, before commencing
the Third Part.
If any teachers have a preference for the common
method of classification, it is very easy to direct the pupils
to learn the Simple Rules first. But every pupil will find it
advantageous at least to review on the plan of arrangement
adopted in this work.
When young beginners take the Second Part, it is re-
commended, that they take the easiest exercises, and re-
serve the more difficult, till a review.
ERRATA.
N. B. Pupils are requested to make these corrections with a pen, before
prooeeding to study the book.
Page 39, line 13 from tire bottom, for ascending read descending. P. 53, line
12 from the bottom, for orders read periods. P. 82, line 14 from the bottom,
change the signs from multiplication to addition. P. 92, line 10, for gr.
read qrs. P. 98, line 8 from the bottom, for 2 read 8. P. 110, line 1 for 10,
read 18 ; in the answer to the fifth sum, for 5 read 11 ; the answer to the
seventh sum should be 256 yds. ; line 8 from the bottom, for qrs. read roods.
P. 118, line 6 from the bottom, for 4 read 12. P. 169, lines 6 and 14, for 3 read 3
12 7 «'
P. 173, line 5, for lb. read £. P. 200, at the end of line 3 from the bottom, insert
at 6 per cent. P. 201, at line 15, insert in 1 yr. 4 mo. P. 233, lines 4 and 5 from
the bottom, for seconds and thirds, read twelfths and seconds.
In many places in the first and second parts of the work, an error will be per-
ceived in the manner of expressing decimals, tens and hundreds being used in-
stead of tenths and hundredths ; thus, tens of thousandths, instead of tenths of
thousandths, or ten thousandths.
INDEX.
ARITHMETICAL TABLES.
Addition Table, .... Page 13
Subtraction Table, ----- 13
Multiplication Table, ■ - - - - . 14
Division Table ------ 15
Table of Weights and Measures, - - - 16
Table of Foreign Coins in Federal Money, - - 19
Table of Scripture Weights, Measures, and Coins, - 19
ARITHMETIC.— PART FIRST.
Addition, ---... 25
Subtraction, ------ 26
Multiplication, ...... 28
Division, ...... 31
Reduction, ...... 37
PART SECOND.
Numeration. *
Numeration of Whole Numbers, 42
Numeration of Vulgar Fractions, 55
Decimal Numeration, ----- 58
Addition, ...... 66
Simple Addition, ..... 67
Decimal Addition, ..... 73
Compound Addition, ..... 78
Addition of Vulgar Fractions, .... 82
Subtraction.
Simple Subtraction, ..... 83
Decimal Subtraction, ..... 86
Compound Subtraction, ..... 91
Subtraction of Vulgar Fractions, .... 93
Multiplication.
Simple Multiplication, ..... 94
Decimal Multiplication, .... 103
Compound Multiplication, .... 109
Multiplication of Vulgar Fractions, - - - 111
Division, ...... 123
Simple Division, ..... 124
Compound Division, ..... 132
INDEX. XI
Division of Vulgar Fractions, .... 135
Decimal Division, .... - 144
Reduction.
Reduction Ascending and Descending, - • • 154
Reduction of Fractions to Whole Numbers, - - 158
Reduction of Whole Numbers to Fractions, - - 159
Reduction of Vulgar to Decimal Fractions,
Reduction of Fractions to a Common Denominator, - 162
Reduction of Fractions to the Lowest Terms, - - 167
Reduction of Fractions from one order to another, - 170
Reduction of Fractions of one order, to Units of another order, 171
Reduction of Units of one order to Fractions of another order, 172
Reduction of a Compound Number to a Decimal,
Reduction of a Decimal to Units of Compound Orders, - 175
Reduction of Currencies, -
Reduction of Different Currencies to Federal Money, - 179
Reduction of Federal Money to Different Currencies, 180
Reduction from one Currency to another, - 181
PART THIRD.
Numeration.
Roman Numeration, .... 184
Other methods of Numeration, .... 185
Common, Vulgar, and Decimal Numeration, - 186
Addition.
Simple, Vulgar, and Decimal Addition, - - 187
Subtraction.
Simple, Vulgar, and Decimal Subtraction, - - 188
Multiplication.
Simple, Vulgar, Compound, and Decimal Multiplication, 189
DrvisioN.
Simple, Vulgar, Compound, and Decimal Division, - 191
Exercises in Reduction ..... 193
Interest, - - . . . - 194
Simple Interest, ..... 196
To find the Interest on Sterling Money, ... 200
Various Exercises in Interest, .... 200
Endorsements, ...... 202
First Method, ...... 203
Massachusetts Rule, ..... 204
Connecticut Rule, ..... 205
Compound Interest, . . .. . . 208
Discount, ...... 210
Stock, Insurance, Commission, Loss and Gain, Duties, . 210
Equation of Payments, .... 214
Ratio, ....... 215
Xll
INDEX.
Proportion, ...... 216
Simple Rule of Three; or Simple Proportion, . - 218
Double Rule of Three ; or Compound Proportion, - 221
Fellowship, ...... 225
Alligation, ...... 228
Duodecimals, ...... 232
Involution, - - - - . - 235
Evolution, ...... 237
Extraction of the Square Root, .... 239
Extraction of the Cube Root, .... 243
Arithmetical Progression, .... 248
Geometrical Progression, .... 252
Annuities, ...... 254
Permutation, - - - - ' - - 259
Miscellaneous Examples, - 260
Position, Ex. 40—50. Barter, Ex. 51—58.
To find the Area of a Square, Ex. 69 — 71.
„ „ of a Parallelogram, Ex. 72 — 74.
„ „ of a Triangle, Ex. 75, 76.
To find the Solid Contents of a Cube, Ex. 77 — 79.
To find the Circumference, Diameter and Area of a Circle,
Ex. 80—84.
To find the Area of a Globe or Ball, Ex. 85.
To find the Solid Contents of a Globe or Ball, Ex. 86.
„ „ „ of a Cylinder, Ex. 87.
„ „ „ of a Pyramid, Ex. 88, 89-
Forms of Notes, Receipts, and Orders, - - - 269
Book-Keeping, ...... 272
ARITHMETICAL TABLES.
ADDITION TABLE.
■!
and
are 2
4 and
are 4 6
and
are 6
8
and
are a
2
1
V
1
1
7
8
1
9
2
2
44
o
6 6
2
8
a
2
10
1
3
54
3
3
9
■<
3
11
s
4
6 4
4
86
4
10
8
4
12
2
5
7
4
5
g s
5
11
8
5
IS
S
6
8
4
6
iim;
6
■ 12
8
6
14
3
7
9
4
7
n r.
7
13
8
7
15
2
8
10
4
8
12(1
8
14
8
8
16
o
9
11
4
9
1316
9
15
8
9
17
:<
and
are 3
5 and
are 5,7
and
are 7
9
and
are 9
3
1
4 5
1
eb
I
8
9
1
10
3
2
5
5
2
" ■
n
9
9
2
11
3
3
6
5
3
87
3
10
9
3
12
3
4
7
5
4
9 7
4
11
9
4
13
S
5
8
5
5
10 7
5
12
9
5
14
3
6
9
5
G
11 7
6
13
ii
6
IS
3
7
10
5
7
I -J 7
7
14
9
7
16
3
8
11
5
8
13 :
8
15
9
8
17
3
9
12
5
9
14 7
9
16
9
18
SUBTRACTION TABLE.
from
1
leaves
4
from
4 leaves 7
from
7
leaves
2
1
4
5
1
7
8
1
3
2
4
G
o
7
9
2
4
3
4
7
3
7
10
3
5
4
4
8
4
7
11
4
6
5
4
9
5
7
12
T:
7
6
1
10
G
7
13
6
8
7
I
11
7
7
14
7
9
8
4
12
8
7
15
8
10
9
4
13
9
7
16
g
2
from
2
leaves
5
from
5 leaves
8
from
8
leaves
8
3
1
5
6
1
8
9
i
2
4
2
5
7
2
,-■
10
s
!
5
3
5
8
3
8
11
3
2
6
1
5
9
4
8
12
4
2
7
5
5
10
5
8
13
5
S
8
6
5
11
6
8
14
G
2
9
7
5
12
7
8
15
7
g
10
*
8
5
13
88
16
8
<>
11
9
.5
14
9
8
17
9
a
from
3
leaves
6
from
G
9
fran
9
leaves
3
4
1
6
7
I
9
10
1
3
5
2
6
8
2
9
11
3
3
6
3
6
9
3
'J
12
3
3
7
4
6
10
4
9
13
4
3
8
5
6
11
5
'.)
14
5
3
9
B
6
12
6
9
15
6
3
10
7
6
13
7
9
16
7
3
11
8
6
14
89
17
a
3
12
9
6
15
9
9
18
g
14
MULTIPLICATION TABLE.
MULTIPLICATION TABLE.
2 times ;
ire
5 times are
3 times
ire
1 1 times
are
2
x 1
= 2
•"> X 1
= 5
8 X
1
= 8
11
X i =
= 11
2
fit
4
5 2
10
8
2
16
11
2
22
2
3
6
5 3
15
8
3
24
11
3
33
2
4
8
5 4
20
8
4
32
u
4
44
2
5
10
5 5
25
8
5
40
11
5
55
2
6
12
5 6
30
8
6
48
11
6
66
2
7
14
5 7
35
8
7
56
11
7
77
2
8
16
5 8
40
8
8
64
11
8
88
2
9
18
5 9
45
8
9
72
11
9
99
2
10
20
5 10
50
8
10
80
11
10
110
2
11
22
5 11
55
8
11
88
11
11
121
2
12
24
5 12
60
8
12
96
11
12
132
3 times
are
6 times
are
9 times
are
12 i
imes
are
3
X I
= 3
6 X 1
= 6
9 X 1
= 9
12
x i
= 12
3
2
6
6 2
12
9
2
18
12
2
24
3
3
9
6 3
18
9
3
27
12
3
36
3
4
12
6 4
24
9
4
36
12
4
48
3
5
15
6 5
30
9
5
45
12
5
60
3
6
18
6 6
36
,9
6
54
12
6
72
3
7
21
6 7
42
9
7
63
12
7
84
3
8
24
6 8
48
9
8
72
12
8
96
3
9
27
6 9
54
9
9
81
12
9
108
3
10
30
6 10
60
9
10
90
12
10
120
3
11
33
6 11
66
9
11
99
12
11
132
3
12
36
6 12
72
9
12
108
12
12
144
4 times
are
7 times
are
| tirnesO are
1 3 times
are
4
X 1
= 4
7 X 1
= 7
10
X i
= 10
13
X 1 =
= 13
4
2
8
7 2
14
10
2
20
13
2
26
4
3
12
7 3
21
10
3
30
13
3
39
4
4
16
7 4
28
10
4
40
13
4
52
4
5
20
7 5
35
10
5
50
13
5
65
4
6
24
7 6
42
10
6
60
13
6
78
4
7
28
7 7
49
10
7
70
13
7
91
4
8
32
7 8
56
10
8
80
13
8
104
4
9
36
7 9
63
10
9
90
13
9
117
4
10
40
7 10
70
10
10
100
13
10
130
4
11
44
7 11
77
10
11
110
13
11
143
4
12
48
7 12
84
10
12
120
13
12
156
DIVISION TABLE.
15
DIVISION TABLE.
2 in 2
1-|
6 in 6
11
10 in
10
11
2 4
2
6 12
2
10
20
2
2 6
3
6 18
3
10
30
3
2 6
4
~
G 24
4
~
10
40
4
*■*■
2 10
5 !
. 3
CD
6 30
5
CD
10
50
5
. 3
2 12
6
CO
6 36
6
CO
10
60
6
CO
2 14
7
6 42
7
10
70
7
2 16
8
6 48
8
10
80
8
2 18
9j
6 54
9J
10
90
9J
3 in 3
11
7 in 7
11
11 in
11
11
3 6
2
7 14
11
22
2
3 9
3
7 21
3
11
33
3
3 12
4
-.
7 28
4
a 11
44
4
<-»
3 15
5
TO
7 35
5
■ 3 11
55
5
CD
3 18
6
QD
7 42
6
tn
11
66
6
CO
3 21
7
7 49
7
11
77
7
3 24
8
7 56
8
11
88
8
3 27
9J
7 63
9.
11
99
9,
4 in 4
1^
8 n 8
11
12 in
12
1^
4 8
2
8 16
2
12
24
2
4 12
3
8 24
3
12
36
3
4 16
4
8 32
4
r*
12
48
4
** *
4 20
5
> 5
CD
8 40
5
•I
12
60
5
> 3
CO
4 24
6
Cfl
8 48
6
QQ
•
12
72
6
QQ
4 28
7
8 56
7
12
84
7
4 32
8
8 64
8
12
96
8
4 36
9.
8 72
9.
12
108
9
5 in 5
I*
! 9 in 9
1-
13 in
13
1-
5 10
2
9 18
2
13
26
2
5 15
3
9 27
3
13
89
3
5 20
4
r*-
9 36
4
13
52
4
r*'
5 25
5
J'
9 45
5
i 3
• CD
13
65
5
► 3
f CD
5 30
6
00
9 54
6
05
13
78
6
CO
•
6 35
7
9 63
7
I l3
91
7
5 40
8
9 72
8
1 13
104
8
5 45
9.
9 81
9
1 13
117
9,
16 WEIGHTS AND MEASURES.
1. Troy Weight.
24 grains (gr.) make 1 penny-weight, marked pwl~
20 penny-weights, 1 ounce, oz.
12 ounces, 1 pound, lb.
2. Avoirdupois Weight.
16 drams {dr.) make 1 ounce, oz.
16 ounces, 1 pound, lb.
28 pounds, 1 quarter of a hundred weight, qr.
4 quarters, 1 hundred weight, cwt.
20 hundred weight, 1 ton. T.
By this weight are weighed all coarse and drossy goods,
grocery wares, and all metals except gold and silver.
3. Apothecaries Weight.
20 grains (gr.) make 1 scruple, 9
3 scruples, 1 dram, 3
8 drams, 1 ounce, g
12 ounces, 1 pound, ft
Apothecaries use this weight in compounding their
medicines.
4. Cloth Measure.
4 nails (na.) make 1 quarter of a yard, qr.
4 quarters,, 1 yard, yd.
3 quarters, 1 Ell Flemish,. E. Fl.
5 quarters, 1 Ell English, E. E.
6 quarters, 1 Ell French, E. Fr.
5. Dry Measure.
2 pints (pL) make 1 quart, qt.
8 quarts, 1 peck, pk.
4 pecks, 1 bushel, bu.
This measure is applied to grain, beans, flax-seed, salt,
oats, oysters, coal, &c.
6. Wine Measure.
4 gills (gi.) make 1 pint, pt.
2 pints, 1 quart, qt.
4 quarts, 1 gallon, gal.
3l£ gallons, 1 barrel, bL
WEIGHTS AND MEASURES. 17
42 gallons, 1 tierce, tier.
63 gallons, 1 hogshead, hhd.
2 hogsheads, 1 pipe, p.
2 pipes, 1 tun, T.
All brandies, spirits, mead, vinegar, oil, &c. are meas-
ured by wine measure. Note. — 231 solid inches, make
a gallon.
7. Long Measure.
3 barley corns (&. c.) make 1 inch, marked in.
12 inches, 1 foot, ft.
3 feet, 1 yard, yd.
5| yards, 1 rod, pole, or perch, rd.
40 rods, 1 furlong, fur.
8 furlongs, 1 mile, m.
3 miles, 1 league, lea.
691 statute miles, 1 degree, on the earth.
% 360 degrees, the circumference of the earth.
The use of long measure is to measure the distance of
places, or any other thing, where length is considered,
without regard to breadth.
N. B. In measuring the height of horses, 4 inches make
1 hand. In measuring depths, six feet make one fathom or
French toise. Distances are measured by a chain, four
rods long, containing one hundred links.
8. Land, or Square Measure.
144 square inches make 1 square foot.
9 square feet, 1 square yard.
30| square yards, or ) , ,
o-ro e t. l ! square rod.
272£ square wet, ) ^
40 square rods, 1 square rood.
4 square roods, 1 square acre.
640 square acres, 1 square mile.
Note. — In measuring land, a chain, called Gunter's chain, 4 rods
in length, is used. It is divided into 100 links. Of course, 25 links
make a rod, and 25 times 25=625 square links make a square rod.
In 4 rods, there are 792 inches. Of course, 1 link is 7 |-|._
9. Solid, or Cubic Measure.
1728 solid inches make 1 solid foot.
2*
18
WEIGHTS AND MEASURES.
40 feet of round timber, or ) , , ,
50 feet of hewn timber, $
128 solid feet or 8 feet long, > rf f ^
4 wide, and 4 high, $
All solids, or things that have length, breadth and depth,
are measured by this measure. N. B. The wine gallon con-
tains 231 solid or cubic inches, and the beer gallon, 282..
A bushel contains 2150,42 solid inches.
10. Time.
60 seconds (S.) make 1 minute, marked M.
60 minutes, 1 hour, h.
24 hours, 1 day, d.
7 days, 1 week, w.
4 weeks, 1 month, mo.
13 months, 1 day and 6 hours, 1 Julian year, yr.
Thirty days hath September, April, June, and Novem-
ber, February twenty-eight alone, all the rest have thirty-
one.
N. B. In bissextile or leap-year, February hath 29
days.
11. Circular Motion.
60 seconds (") make 1 minute,
60 minutes, 1 degree,
30 degrees, 1 sign, S,
12 signs, or 300 degrees, the whole great circle of the
Zodiac.
12 units make A Dozen.
12 dozen A Gross.
144 dozen A Great Gross.
20 units A. Score.
24 sheets of paper A Quire.
20 quires A Ream.
VALUE OF FOREIGN COINS.
19
Value of Foreign Coins in Federal Money.
Shilling Sterling, «0.222 '
Crown 5s. 1.111
Sovereign, (a gold coin, - £,) 4.444
Guinea, (21s. nearly out of ;
use in England,) $
Livre of France, 0. 185-j-
Franc " 0.1875—
Pistole* 10 livres" 1.852—
Louis d'or, " 4,444-f
Five franc piece, " 0.937
Real of Plate, of Spain, 0.100
RealofVellon, " 0.050
Pistole, " 3.60
Dollat; " 1.00
Re. of Portugal, $0.0012-|-
Testoon, " 0.125
Milre,* " 1.250
Moidore, " 6.000
Joanese, " 8.000
Marc Banco of Hamburgh, 0.333-f-
Pistole of Italy, 3.200
* Those denominations which have the asterisk, (as the Pistole of France,
and the Milre of Portugal,) are merely nominal; that is, they are represented
by no real coin. In this respect, they are like the Mill in Federal Money.
Rix Dollar of Austria,
0.778—
Rix Dollar of Denmark ^
1.000
and Switzerland, J
Rix Dollar* of Sweden,
1.037
Rix Dollar of Prussia,
0.778—
Florin, "
0.259-j-
Ducat of Sweden and )
Prussia, )
2.074
Piaster of ex, of Spain,
0.80
Ducat of ex, » "
1.102—
Stiver of Holland,
0.019-4-
Guilder or Florin, "
0.388
Rix Dollar, "
0.970
Ducat, "
2.079
Gold Ducat, "
8.000
Ducat of Denmark,
8.833+
Ruble, of Russia,
1.000
Zervonitz, "
2.000
Tale, of China,
1.480
Pagoda, of India,
1.840
Rupee, of Bengal,
0.500
Xeriff, of Turkey,
2.222
A TABLE OF SCRIPTURE WEIGHTS, MEASURES, AND MONEY.
A Cubit,
A Span, half cubit,
A Hand breadth,
A Finger,
A Fathom,
Ezekiel's reed,
The measuring line,
Sabbath day's journey,
Eastern mile,
Stadium, or Furlong,
Day's journey,
S OF LENGTH.
feet.
inches.
• • ■
1
9,88
• . .
10,94
...
3,68
...
0,91
. • .
7
3,55
• • •
10
11,32
. • •
145
11,04
miles* furlongs.
roa
s. feet.
5
21
H
i
3
2
3
o
1
4
3
33
1
12
6
20
SCRIPTURE WEIGHTS AND MEASURES.
MEASURE OF LIQUIDS.
The Homer or Cor.
gall.
75
pints, sol. inch.
5 7,6
The Bath,
7
4
15,2
The Hin,
1
2
2,5
The Log,
24,3
The Firkin, .
7
4,9
MEASURE OF
THINGS.
The Homer,
bushels, pecks.
8
pints.
1,6
The Lethech,
4
0,8
The Ephah,
The Seah,
3
1
3,4
1,1
The Orner,
5,1
The Cab,
2,9
WEIGH
TS.
A Shekel,
lb.
oz.
9
pwt.
9
gr-
2,0
The Maneh,
2
3
6
10,3
A Talent,
113
10
1
10,3
MONE
Y.
dolls.
cents.
mills.
A Shekel,
50
5
The Bekah, (half Sheh.) .
25
3
The Zuza,
12
5
TheGerah,
02
5
Maneh or Mina,
25
29
6
A Talent of Silver,
1,H
85
7
A Shekel of Gold,
8
09
4
A Talent of Gold,
24,2£
71
4
Golden Daric or Drachm,
4
85
7
dolls. ■
cents.
mills.
Piece of Silver, (Drachm)
Tribute money, (Didrachm)
Piece of Silver, (Stater)
Pound, (Mina)
14
14
28
57
35
3
7
4
1
Penny, (Denarius)
14
3
Farthing, (Assarium)
00
6
Farthing, (Quadrands) .
Mite,
00
00
3
1
ARITHMETIC.
PART FIRST.
Arithmetic is the science of numbers.
A unit is a whole thing of any kind.
A fraction is a part of a thing.
Thus a dollar is a unit ; a man is a unit; a picture is
a unit ; a bushel of apples is a unit, &c.
A half of an apple is a fraction ; a quarter of a dollar
is a fraction ; a third of a loaf of bread is a fraction, &c.
Let the pupil mention other units and fractions.,
If an apple is cut into two equal parts, each part is
called one half of the apple. If it is cut into three equal
parts, each part is called one third. If it is divided into
four equal parts, each part is called one fourth. If it is
divided into jive equal parts, each part is called one fifth, <SfC.
If a unit is divided into six equal parts, what is one of
those parts called ? If a unit is divided into seven equal
parts, what is one of those parts called ? If a unit is divi-
ded into eight equal parts, what is one of those parts
called ? Into nine ? Into twenty ? Into an hundred ?
Into fifteen ? Into twenty-two ?
How many halves make one unit ? How many thirds
make one unit ? How many fourths ? How many fifths?
How many sixths ? How many sevenths ? How many
eighths? How many ninths ? How many tenths ? How
many twentieths ? How many hundredths ?
For illustrating the exercises which immediately follow,
the teacher should be provided with a proper number of
the several coins of the U. S. viz : eagles, dollars, dimes,
cents and mills. As mills have never been coined, round
bits of stiff paper may be employed to represent them.
The pupil should first see the several coins and learn the
value of them.
Ten mills are one cent.
Ten cents are one dime.
Ten dimes are one dollar.
Ten dollars are one eagle.
22
ARITHMETIC. FIRST PART.
How many mills make a cent ? What part of a cent is
one mill ? What part of a cent is two mills ? What part
of a cent is three mills? What part of a cent is four
mills ? What part of a cent is five mills ? Six mills ?
Seven mills? Eight mills ? Nine mills?
How many cents make a dime ? One cent is what
part of a dime ? Two cents is what part of a dime ?
Three cents is what part of a dime 1 Four cents is what
part of a dime? Five cents ? Six cents ? Seven cents?
Eight cents ? Nine cents ?
How many dimes make a dollar ? What part of a dol-
lar is one dime ? What part of a dollar is two dimes ?
What part of a dollar is three dimes ? Four dimes ? Five
dimes? Six dimes? Seven dimes? Eight dimes?
Nine dimes ?
How manv dollars make an eagle ? One dollar is what
part of an eagle ? Two dollars is what part of an eagle ?
Three dollars ? Four dollars ? What part of an eagle is
five dollars ? Six dollars ? Seven dollars ? Eight dol-
lars ? Nine dollars?
The same thing, may be considered sometimes as a
unit and sometimes as a fraction — thus, one dollar is a
unit or whole thing of the kind or order called dollars, and
one dollar is also the tenth -part of an eagle, or the fraction
of an eagle. One cent is a unit or whole thing, of the or-
der of cents, and one cent is also the tenth part of a dime,
or the fraction of a dime. One mill is a unit of the order
of mills, and one mill is the tenth part of a cent, or the
fraction of a cent. One day is a unit or whole thing of
the order of days, and one day is also the seventh, fart of
a week, or the fraction of a week. One week is a unit or
whole thing of the order of weeks, and one week is the
fourth part of a month, or the fraction of a month. One
month is -a. unit or whole thing of the order of months, and
one month is also the twelfth part of & year, or the fraction
of a year.
Of what order is one dollar a unit ? Of what order is it
a fraction ? Of what order is one cent a unit ? Of what
order is it a fraction ? Of what order is one week a unit ?
Of what order is it a fraction ? Of what order is one foot
a unit? Of what order is it a fraction? One day is a
unit, of what order, and a fraction of what order ? &c.
ARITHMETIC. FIRST PART. 23
What is half of four cents ? What is a third of six
cents ?
Let the pupil take six cents, and divide them into three
equal portions, and then tell what is one of these parts ?
What is & fourth of eight cents? Let the pupil divide
eight cents into^owr equal portions, and tell how many in
each portion.
There are twice six cents in twelve cents, what part of
twelve is six cents ?
There are three times four cents in twelve cents, what
part of twelve is four cents ?
There are three times five cents in fifteen cents, what
part of fifteen is five cents ?
There are three times three in nine, what part of nine
is three ?
There are two times three in six, what part of six is
three ?
There are four times two in eight, what part of eight is
two 1 ?
There are four times three in twelve, what part of twelve
is three ?
There are^ve times six in thirty, what part of thirty is
six?
There are three times seven in twenty-one, what part
of twenty-one is seven ?
There are four times six in twenty.four, what part of
twenty-four is six 1
There are six times seven in forty-two, what part of
forty-two is seven ?
What part of twelve is three ? Is four ?
What part of nine is three ?
What part of fifteen is three ? Is five ?
What part of sixteen is four 1
What part of eighteen is three 1 Is six ?
What part of twenty. one is three ? Is seven ?
What part of twenty-four is six ? Is four ?
What part of twenty-eight is seven ? Is four 1
What part of thirty-two is eight? Is four ?
What part of thirty -six is nine ? Is four ?
If an apple is cut into two equal parts, what is each part
24 ARITHMETIC. FIRST PART.
called? If it is cut into three equal parts, what is each
part called ?
The more parts a thing is divided into, the smaller these
parts must be. If one thing is divided into twice as many
parts as another thing, each part is twice as small.
If one apple is cut into twice as many pieces as another,
how much smaller is each piece ? How much larger is a
half than a fourth ? Ans. There are twice as many fourths
as halves in a thing, therefore a half is twice as large as
a fourth:
If one apple is cut into four pieces, and another into
eight pieces, how much larger are the fourths than the
eighths ? Ans. As there are twice as many pieces when
there are eighths, as when there are fourths, an eighth is
tioice as small as a fourth.
If one apple is cut into twelve parts and another into six
parts, which has the most parts and which has the largest
parts ? How much larger is a sixth than a twelfth ? Ans.
Twelve is twice as many as six, therefore a sixth is twice as
large as a twelfth.
Which is the largest, a fifth or a tenth ? How much
larger is a fifth than a tenth ? ,
Which is the largest a seventh or a fourteenth?
How much smaller is a fourteenth than a seventh ?
Which is the largest a third or a fifth ?
Which is the smallest a half or a fourth ?
Which is the smallest a third or a half? Ans. The
more pieces there are the smaller they must be, therefore
a third must be smaller than a half.
If one apple was cut into four pieces, and another into
six pieces, which would be the largest a fourth or a sixth ?
Which is the largest a sixth or a ninth ?
Which is theiargest a fifth or a fourth 1
Which is the smallest a twelfth or a tenth ?
Which is the smallest a seventh or a ninth ?
Which is the smallest an eighth or a seventh ?
Which is the smallest a fifteenth or a fifth ?
Which is the largest an eighth or a sixteenth?
Which is the largest a fifth or a half?
If an apple is divided into four pieces, what is each
piece ? If it is divided into twice as many and twice as
ADDITION. 25
small pieces, how many are there, and what are they
called I
If an apple is divided into thirds, what would you change
them to, to make them twice as many and twice as small !
Make two fourths twice as small and twice as many
pieces and what is the answer ?
What part of a thing is twice as small as a half? As
a third ? As a fourth ? As a fifth 1 As a sixth ? As a
seventh ? As an eighth ? As a ninth 1 As a tenth 1 As
an eleventh ? As a twelfth ? ' .
What part of a thing is twice as large as a fourth? A-
a sixth ? As an eighth ? As a tenth ? As a twelfth 1
As a fourteenth ? Asa sixteenth ? As an eighteenth ?
As a twentieth ?
ADDITION.
Two cents, and four cents, and six cents, and nine
cents are how many ?• Sixteen cents, and twelve cents,
are how many ?
Five dollars, and four dollars, andkiine dollars, are how
many ? %
Four halves of %.n apple, and six" halves, and nii>
halves, arc how many halves ?
Five sixths of an 'apple, and four sixths, and nine sixths,
are how manv sixths ?
J
Three fifths of an orange, and four fifths, and nine fifths,
and twelve fifths, are how many fifths ?
Addifmi is uniting several numbers in one.
When whole numbers are added, it is Simple Addition.
When fracf ions are added, it is Fractional Addition.
Six dimes, five dimes, and four dimes are how many ?
Seven dollars, ei^ht dollars, and nine dollars are how
many ?
Nine cents, three cents, twelve cents, and ten cents are
how many ?
Four, three, and seven are how many ?
Eight, five, and three are how many ?
Nine, six, and two are how many ?
3
26 ARITHMETIC. FIRST PART.
Seven, five, and six are how many 1
Eight, nine, and two are how many?
Seven, eight, and one are how many?
Eleven, five, and six are how many?
Ten, seven, and three are how many?
Ten twentieths, six twentieths, and five twentieths are-
how many twentieths ?
One thirteenth of a unit, four thirteenths, and seven
thirteenths are how many thirteenths ?
One fifth of a dollar, three fifths, and eight fifths are
how many fifths ?
One ninth of an orange, four ninths, and six ninths are
how many ninths ?
Seven tenths of an eagle, two tenths, and five tenths
are how many tenths ?
Three eighteenths, nine eighteenths, and four eight-
eenths are how many eighteenths ?
Ten thirtieths, six thirtieths, and five thirtieths are how
many thirtieths ?
Two fourths, six fourths, nine fourths, ten fourths, and
five fourths, are how many fourths ?
Sixteen halves, five halves, nine halves, and six halves,
are how many halves ?
Six eighths, four eighths, seven eighths, sixteen eighths,
are how many eighths?
The number made by adding several numbers together,
is called he sum.
What is the sum of four, six, nine and five ?
What is the sum of four tenths, six tenths, and nine
tenths ?
SUBTRACTION.
If you take two cents from three cents, how many re-
main ?
If you take three dollars from six dollars how many re-
main ?
If you take four dollars from seven dollars how many
remain ?
SUBTRACTION'. 27
If you take five eagles from nine eagles how many re-
main ?
If you take six dimes from ten dimes how many re-
main ?
If two tenths are taken from four tenths how many re-
main ?
If four ninths are taken from eight ninths how many re-
main ?
If two tenths are taken from seven tenths how many re-
main ?
Subtraction is taking one number from another, to find the
remainder.
When whole numbers are subtracted it is Simple Sub-
traction. When fractions are subtracted, it is Fractional
Subtraction.
What is the remainder, when four cents are taken from
nine cents ?
What is the remainder, when three mills are taken
from eight mills ?
What is the remainder, when seven dimes are taken
from twelve dimes ?
What is the remainder, when five dollars are taken
from ten dollars ?
Five from eleven ? Seven from thirteen ? Eight from
twelve ? Five from fourteen ? Nine from sixteen .' Five
from twelve ? Eight from thirteen ? Ten from Twenty ?
What is the remainder, when two sevenths of an apple,
are taken from eight sevenths ? When four sevenths of
a dollar are taken from six sevenths? Eight twelfths
from ten twelfths? Three ninths from eight ninths?
Ten twentieths from twelve twentieths ? Six elevenths
from ten elevenths ? Seven twelfths from twelve twelfths ?
Eight ninths from thirteen ninths ? Three sevenths from
nine sevenths ? Four eighths from eleven eighths ? Four
thirds from twelve thirds ? Five twentieths from seven
twentieths ?
The number which has a number subtracted from it, is
called the minuend.
The number which is to be subtracted from another num-
ber is called the subtrahend.
If eight is subtracted from twelve, what is the subtra-
hend and what is the minuend ?
28 ARITHMETIC. FIRST PART.
If four tenths, is .subtracted from nine tenths, what is
the subtrahend and what the minuend ?
If ten cents be taken from thirteen cents, what is the
subtrahend, and what the minuend ?
MULTIPLICATION.
If you take two cents, three times, what is the amount
of the whole 1
If you take three dollars, four times, what is the amount
of the whole ?
If you take half of an apple three times, what is the
amount ?
If you take two thirds of a dollar four times, what is the
amount ?
If 3 r ou take two fourths of an eagle, six times, what is the
amount 1
Multiplication is repeating a number as often as there
are units in another number.
If you take five dollars four times, what is the amount 1
If you repeat four dollars five times, what is the amount 1
If you take six dollars five times, what is the amount 1
If you repeat six dollars six times, what is the amount ?
Seven times ? Eight times ?
If you take seven dollars three times, what is the amount ]
If you repeat seven, four times, what is the amount ?
Five times ? Six times ? Seven times ?
If you repeat eight twice, what is the amount ?
If you repeat eight three times, what is the amount ?
Four times ? Five times ? Six times ? Seven times 1
Eight times ?
If you repeat nine three limes, what is the amount ? &c.
If you take one fifth of a dollar six times, what is the
amount ? Seven times 1 Eight times ? Nine times?
Tfyou repeat two sixths of a dollar three times, what is
the amount ?
MULTIPLICATION.
M
If you repeat two sixths of a thing four times, what is
the amount ? Five times ? Six times ? Seven times ?
Eight times ?
What is the amount, if four sevenths be repeated
four times ? Five times ? Six times ? Seven times ?
Eight times ?
What is the amount if Jive ninths be repeated eight
times ? Nine times ? Ten times ? Eleven times ?
What is the amount, if eight twentieths be repeated
seven times ? Nine times ? Eight times ? &c.
The number to he repeated, is the multiplicand.
The number which shows how often the multiplicand is to
bo repeated, is called the multiplier.
The midtiplier and multiplicand together, are called the
factors.
The answer obtained is called the product.
Height is repeated four times what is the product ?
What is the multiplier 1 The multiplicand ? The fac-
tors ?
If three sixths are repeated/owr times what are the fac-
tors ? The multiplier ? The multiplicand ?
If you take a. fourth of twelve and repeat it three times,
what is the multiplicand ? The multiplier ? The pro-
duct ?
If you take a sixth of eighteen and repeat it three times,
what is the product ? factors 1 multiplier ? multipli-
cand ?
Simple Multiplication is where both factors are whole
numbers.
Fractional Multiplication is where one or both factors
are fractions.
If twelve is repeated four times, is it simple or fraction-
al multiplication ?
If one fourth of twelve is repeated three times, is it sim-
ple or fractional multiplication ? If one sixth is repeated
seven times, which kind of multiplication is it ?
Exercises in Simple Multiplication.
1. If a man spends three dollars a week, how much
does he spend a month ?
3*
30 ARITII31ETIC. FIRST PART.
Let the pupil state the sum in this manner.
As there are four weeks in a month, a man will spend
four times as much in a month, as in a week ; four times
three is twelve. He will spend twelve dollars.
Let all the following sums be stated in the same way.
Both teachers and pupils will find great advantage in be-
ing particular to follow this method of stating.
2. If a man spend five dollars a month, how much
does he spend in a year ?
3. If a man can make eight pens in a minute, how
many can he make in ten minutes ?
4. If one orange cost six cents, what costs eight oran-
ges ?
5. Eight beys have seven cents apiece, how much have
all?
6. There is an orchard in which there are six rows of
trees, and seven in each row, how many trees in the or-
chard ?
7. The chess board has eight rows of blocks, and eight
blocks in each row, how many blocks in the whole ?
8. Twelve young ladies have each five books apiece,
how many have they all ?
9. If a young lady spends six cents a week, how much
does she spend in a month ?
10. There are nine desks in a school room, and six
scholars at each of the desks, how many are in the room?
11. There are in a window five rows of panes of glass,
and seven panes in each row, how many in the whole ?
12. If one lemon cost four cents, how much will twelve
lemons cost ?
EXERCISES IN FRACTIONAL MULTIPLICATION.
Multiplication of a fraction by whole numbers.
1. If vou repeat one half four times what is the pro-
duct ?
'I. If you multiply three fourths by seven, what is the
product ?
3. What is two thirds multiplied by eight ?
DIVISION. 31
4. If a man spend two twelfths of a dollar a day, how
many twelfths does he spend in a week ?
Ans. As there are seven days in a week, a man spends
seven times as much in a week as in one day. Seven
times two twelfths is fourteen twelfths. He spends four-
teen twelfths of a dollar in a week.
5. If a man gives two eighths of a pound of meat to six
persons, how many eighths does he give away ?
6. If a boy gives two fourths of an orange to seven of
his companions, how many fourths does he give away 1
7. If a man drinks three jourths of a pint of brandy a
clay, how many fourths does he drink in a week?
8. What is three times three eighths ? Six times si.)
sevenths ?
9. If a man lays by two eighths of a dollar a day, how
much does he save in a week 1
10. If there are two thirds of a pound of meat for each
one in a family of seven, how much is there in the whole ?
11. What is six times four tenths ?
12. What is nine times two thirds ?
18. What is seven times four ninths?
14. What is eight times six tenths ?
15. What is twelve times two fourths ?
16. What is nine times three tenths?
17. What is five times three sixteenths?
18. What is six times seven twentieths ?
The multiplication of whole numbers by fractions, is defer-
red to the Second Part, because it involves the process of
Division, which must first be explained.
DIVISION.
How many two cents are there in four cents ?
How many two cents in six cents ?
How many two cents in eight ?
How many two cents in ten ?
How many two cents in twelve ?
How manv three cents are there in six cents ? How
many in nine ? How many in twelve ?
32 ARITHMETIC. FIRST PART.
How many four cents are there in eight ? How many
in twelve ?
How many five cents are there in ten ?
What part part of two cents is one cent ?
What part of four cents is two ? What part of six is
two ? What part of eight is two ? What part of ten is
two ? What part of twelve is two ?
Three cents is what part of six ? Three is what part of
nine ? Of twelve 1
What part of eight is four ? What part of twelve
is four ?
What part of five cents is one ? What part of five is
two ? What part of five is three ? Four ? Five ?
Six? &c.
How many two sixths are there in four sixths ?
How many three fourths are there in six fourths ?
How many four twelfths in eight twelfths ?
What part of two twelfths is one twelfth ?
What part of four twelfths is two twelfths ?
What part of nine twelfths is three twelfths?
Division is finding how often one number is contained in
another, and thus finding ivhat part of one number is another
number.
How many times is six contained in twelve ? In
eighteen ?
What part of twelve is six ? What part of eighteen is
six ?
How many times is five contained in ten ? In fifteen ?
Five is what part often ? Of fifteen ?
How many times is seven contained in fourteen ? In
twenty-one ?
What part of fourteen is seven 1 What part of twen-
ty-one is seven ?
How many times is nine contained in eighteen ?
How many times is ten contained in twenty? In thir-
ty? In forty ?
What part of sixteen is four 1
What part of eighteen is six ?
What part of sixteen is eight ?
One is what part of thirty ? Two is what part of thir-
DIVISION. 33
ty ? Three is what part of thirty ? Six ? Eight ? Eleven ?
Fourteen? Twenty is what part of thirty ? &c.
How many two sevenths are there in ten sevenths ?
' How many three eighths arc there in nine eighths ?
How many six tenths in eighteen tenths ?
How many seven ninths in twenty-one ninths 1
How many five elevenths in twenty elevenths ?
How many three eighteenths arc there in twelve eight-
eenths ?
Two sixths is what part of four sixths ?
Two sevenths is what part often sevenths ?
Three eighths is what part of nine eighths ?
What part of eighteen tenths is six tenths?
What part of fourteen ninths is seven ninths ?
What part of fifteen elevenths is five elevenths?
What part of twelve eighteenths is three eighteenths ?
The number which is divided is called the Dividend.
The number by which you divide is called the Divisor.
The answer is called the Quotient.
If you find how many times three there are in twelve.
which is the Divisor ? The Dividend ? The Quotient ?
If twelve is divided by six, which is the Dividend ? The
Divisor ? The Quotient ?
When whole numbers are divided by whole numbers, it
is called Simple Division.
When either the divisor or dividend is a fraction, it is
called Fractional Division.
Exercises in Simple Division.
1. If you divide 12 cents equally among three boys, how
many will each one have ?
Ans. Each one will have as many as there are threes
in twelve ; ox four cents.
2. If there are forty-eight panes of glass in a window,
and there are eight panes in each row, how many rows
are there ?
Ans. As many as there are eights in forty-eight ; ov
six rows.
3. How much broadcloth, at six dollars a yard, can
you buy for twenty-four dollars ?
34 ARITHMETIC. FIRST PART.
4. How many hours would it take you to travel twen-
ty-one miles, if you travelled three miles an hour 1
5. If you divided thirty-six apples equally among four
hoys, how many would you give them apiece ?
6. How many pounds of raisins, at nine cents a pound,
can you buy for sixty-three cents ?
7. How many reams of paper, at seven dollars a ream,
can you buy for forty-nine dollars ?
8. A man agreed to work eight months, for seventy-two
dollars, how much did he receive a month ?
9. If you buy a bushel of pears for forty-eight cents,
how much is it a- peck ?
10. If there are six shillings in a dollar, how many dol-
lars in thirty-six shillings ?
11. Four men bought ahorse for forty-eight dollars,
what did each man pay ?
12. A man gave sixty-three cents for a horse to ride
nine miles, how much was that for each mile ?
13. A man agreed to pay eight cents a mile for a horse,
and he paid sixty-four cents, how many miles did he go ?
14. A man had forty-two dollars, which he paid for
wood, at seven dollars a cord, how many cords did he
buy?
15. Two boys are running, and are forty-eight rods
apart. The hindermost boy gains upon the other, three
rods a minute, in how many minutes will he overtake the
foremost boy 1
16. A vessel contains sixty-three gallons, and dischar-
ges seven gallons an hour, in how many hours will it be
emptied ?
17. If you wish to put sixty-four pounds of butter in
eight boxes, how many pounds would you put in each
box?
EXERCISES IN FRACTIONAL DIVISION.
Division of whole numbers by Fractions.
1. How many halves are there in six oranges?
2. How many thirds are there in four apples ?
DIVISION. 35
Ans. One apple has three thirds, four apples haxc four
times as many, or twelve thirds.
3. How many fourths are there in three oranges ?
4. How many fifths are there in four apples ?
5. How many sixths are there in two oranges ?
C. How many half dollars are there in four dollars?
7. How many quarters of a dollar in five dollars 1
8. How many half eagles in eight eagles 1
9. In two dollars how many thirds of a dollar ?
10. If there are six one thirds in two dollars, how many
two thirds are there ?
Ans. There are only half as many too thirds as there
are one thirds, or three two thirds.
11. In two dollars, how many one sixths? How main
two sixths ?
12. A man divided two dollars anions his workmen,
and gave them a third of a dollar apiece, how many work-
men had he ?
13. A man divided four dollars equally among his chil-
dren, and gave them each two thirds of a dollar, how ma-
ny children had he ?
Ans. As many children as there are two thirds in four
dollars. In four dollars there are twelve one thirds.
There are half as many two thirds, or six. He had six
children.
14. If a man gave two "sevenths of a dollar to each of
his servants, and gave away in the whole four dollars,
how many servants had he ?
15. How many two sixths in four ?
16. How many two eighths in four ?
17. How many two thirds in eight ?
18. How many two ninths in six?
19. How many two twelfths in two ?
20. How many two twelfths in four ?
Dicis ion of Fractions by whole numbers.
In dividing fractions by whole numbers, we do not find
how many times a whole thing is contained in a part of the
same thing, for that would be absurd ; but we find what
part of once, a whole number is contained in a fraction.
Thus if we wish to divide one halj by one, we say, one
unit is contained in one half, not once, but one half of once.
36 ARITHMETIC. FIRST PART.
1. One is contained in one fourth, what part of once ?
2. One is contained in one fifth, what part of once ?
3. One is contained in one sixth, what part of once 1
4. One is contained in one seventh, what part of once ?
5. One is contained in one eighth, what part of once ?
6. One is contained in one ninth, what part of once 1
7. One is contained in one tenth, what part of once ?
8. One is contained in one eleventh, what pail of once 1
9. One is contained in one twelfth, what part of once 1
10. If you divide one fourth, by one, which is the divi-
sor ? The dividend ? What is the quotient ?
11. If you divide one sixth by one, what is the quotient ?
The divisor ? The dividend ?
12. If you divide one third by one, what is the quo.
tient 1 The divisor ? The dividend ?
13. If one fourth contains one, a fourth of once, what
part of once does two fourths contain it?
Ans. Twice as much, or two fourths of once.
14. If two sixths is divided by one, what is the answer ?
Ans. Tifo sixths of once.
15. Two eighths contain one, what part of once ? Six
eighths contain one, what part of once ?
16. Two twelfths contain one, what part of once ?
Four twelfths contain one, what part of once ?
17. Eight twelfths contains one, what part of once ?
"18. Six twelfths contains one, what part of once ?
If six twelfths contains one, six twelfths of once, it would
contain two, only half as often, or three twelfths of once.
19. Four eighths contains one, what part of once ?
Contains two, wiiat part of once? It contains two, only
half as often, or two eighths of once.
20. Six tenths contains one, what part of once 1 Con-
tains two, what part of once ?
21. Eight tenths contains one, what part of once ?
Contains two, what part of once ?
22. Four eighths contains one, what part of once ? —
Contains two, what part of once?
23. Six elevenths contains one, what part of once 1 —
Contains two, what part of once ?
24. Eight twelfths contains one, what part of once ?
Contains two, what part of once ?
REDUCTION.
37
REDUCTION.
One dime is how many cents ? How many mills ?
One unit of the order of dollars, is how many units of
the order of dimes? How many of the order of cents?
How many of the order of mills ?
One eagle is how many dollars ? How many dimes ?
Cents ?
One unit of the order of dimes is how many units of the
order of cents ?
Reduction is changing units of one order, to those of another.
A unit of the order of eagles is how many units of the
order of dollars ? Of dimes?
Two eagles are how many dollars ? How many dimes ?
How many dollars in two hundred cents ?
How many dollars in twenty dimes ?
Thirty units of the order of dimes, is how many units
of the order of dollars ?
Two pints are one quart.
Eight quarts are one peck.
Four pecks are one bushel.
Two units of the order of quarts, are how many units
of the order of pints ?
Eight pints are how many quarts ?
Two bushels how many pecks ?
Eight pecks how many bushels ?
Three barley-corns are one inch.
Twelve inches are one foot.
Three feet are one yard.
One inch how many barley-corns ? Two inches are
how many ?
Twelve barley-corns how many inches ?
One foot how many inches ? Three feet how many 1
One yard is how many feet ? How many inches ?
How many barley-corns ?
Two yards are how many feet ? How many inches ?
How many barley-corns ?
Three yards are how many feet ? How many inches ?
How many barley-corns ?
4
38 ARITHMETIC. FIRST PART.
How many feet are there in five yards ? How many
inches in five yards ? How many barley-corns 1
How many barley-corns are there in seven yards ?
From the preceding exercises, you learn that a unit of
one order may contain several units of another order.
What do you learn from the preceding exercises ?
How many units of the order of cents, are there in one
unit of the order of dimes ?
How many units of the order of dollars, are there in one
unit of the order of eagles ?
How many units of the order of mills, are there in one
unit of the order of cents ?
How many units of the order of pints, are there in one
unit of the order of quarts ?
How many units of the order of pecks, are there in one
unit of the order of bushels 1
How many units of the order of barley-corns, are there
in one unit of the order of inches ? N
How many units of the order of feet, are there in one
unit of the order of yards ?
How many units of the order of da3"s, are there in one
unit of the order of weeks 1
How many units of the order of weeks, in one unit of
the order of months ?
Change two units of the order of dimes, to units of the
order of cents.
Change twenty units of the order of cents, to units of
the order of dimes.
Change three units of the order of yards, to units of
the order of feet.
Change nine units of the order of feet, to units of the
order of yards.
Change ten units of the order of pints, to units of the or-
der of quarts.
Change five units of the order of quarls, to units of the
order of pints.
Change twenty-one units of the order of days, to units
of the order of weeks, &c.
When units of one order are changed to units of a high-
REDUCTION.
39
er order, the process is called Reduction ascending ; and
when units of one order are changed to those of a lower
order, the process is called Reduction descending.
If twenty cents are changed to dimes, which kind of
reduction is used ?
Iftwenty cents are changed to mills, which kind of re-
duction is used ?
If four gallons are changed to pints, which reduction is
used?
If eight feet are changed to inches, which kind of re-
duction is used ?
In changing twelve barley-corns to inches, which kind
of reduction is used ?
In changing fourteen days to weeks, which reduction is
used?
In chancins five hours to minutes, which reduction
DO '
is used ?
In changing one hundred and twenty minutes to hours,
which reduction is used?
Reduce three dimes to cents ; to mills. Which kind
of reduction is it ?
Reduce three hundred mills to cents ; to dimes ; and
which kind of reduction is it ?
Reduce ihree hundred mills to dollars, and which kind
of reduction is it ?
Reduce two halves to quarters, and which kind of re-
duction is it ?
Ans. As a half is of more value, it is a higher order than
a quarter, therefore it is reduction ascending.
In performing this last exercise, the pupil will find the
necessity for the following distinction in regard to units.
A unit has been defined as " any whole thing of a
kind," and a fraction is defined as " a part of a thing."
But it is very often the case, that fractions are consid-
ered as units. Thus when we reduce quarters to halves,
and halves to quarters, we change units of the order called
quarter, to units of the order called half.
When we say a whole quarter of an apple, and a half a
quarter of an apple, we think of a quarter as a whole thing
of its kind.
The difference between the two kinds of units is this :
40 ARITHMETIC. FIRST PART.
When we think of a whole quarter, we think of another
thing of which the quarter is a part. We think of it as a
whole thing in one respect, and as a part of a thing in
another respect. But when we think of a whole apple, we
do not necessarily think of another thing of which it is a
part.
When we think of a half of a loaf of bread, do we think
of something of which the half is a part?
When, we think of a biscuit, do we necessarily think of
something of which it is a part ?
When we think of a third of an orange, do we necessa-
rily think of something of which it is a part ?
When we think of a house, do we necessarily think of
any thing of which it is a part ?
Those units which do not require us to think of any oth-
er thing of which they are parts, are called whole numbers,
and those units which do require us to think of other
things of which they are parts, are called fractions.
What is the difference between units that are whole
numbers, and units that are fractions ?
Reduce two yards to quarters, and which kind of re-
duction is it ?
Reduce twenty-four inches to feet, and which kind of
reduction is it ?
Reduce three feet to inches, and which kind of reduc-
tion is it ?
Which is of highest value, a half or a quarter ?
Reduce eight quarters to halves, and which kind of re-
duction is it ?
Reduce two halves to quarters, and which kind of re-
duction is it ?
Reduce sixteen quarters to halves, and which kind of
reduction is it 1
Reduce two fifths to tenths ; six tenths to fifths ; eight
tenths to fifths ; twelve tenths to fifths ; three fifths to
tenths ; six fifths to tenths.
Reduce one seventh to fourteenths ; four fourteenths
to sevenths ; four sevenths to fourteenths ; eight four-
teenths to sevenths.
SUMMARY OF DEFINITIONS. 41
Reduce two sixths to twelfths ; four twelfths to sixths ;
eight twelfths to sixths ; five sixths to twelfths ; four
twelfths to sixths.
SUMMARY OF DEFINITIONS.
A unit is any whole thing of a kind.
A fraction is a part of a thing.
Addition is uniting several numbers in one.
Subtraction is taking one number from another, to find
the remainder.
The largest number is the minuend, the smallest num.
ber is the subtrahend.
Multiplication is repeating one number as often as
there are units in another number.
The multiplicand is the number to be repeated ; the
multiplier is the number which shows how often the multi-
plicand is to be repeated ; the factors are both the multi-
plier and multiplicand ; and the product is the number ob-
tained by multiplying.
Division is finding how often one number is contained
in another number, and thus finding what part of one num.
ber, is another number.
The dividend is the number to be divided. The divi-
sor is the number by which you divide. The quotient is
the answer obtained by dividing.
Reduction is changing units of one order, to units of
another order.
Reduction ascending, is changing units of a lower, to a
higher order.
Reduction descending is changing units of a higher, to
a lower order.
Note to Teachers. — A review of this First Part, will
be found more useful than an increased number of ex-
amples.
4*
ARITHMETIC.
SECOND PART.
NUMERATION.
Numeration is the art of expressing numbers by words,
or by figures.
Figures are sometimes called numbers, because they
are used to represent numbers. Thus the figure 4, is oft-
en called the number four, because it is used to represent
that number.
There are thirty-five words, that are commonly used in
numeration ; viz : one, two, three, four, five, six, seven,
eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, six-
teen, seventeen, eighteen, nineteen, twenty, thirty, forty, fif-
ty, sixty, seventy, eighty, ninety, hundred, thousand, million,
billion, trillion, quadrillion, quintillion, sextillion.
Those words ending in tetn, are the words two, three,
four, &c with teen, which signifies and ten, added to
them.
What is the meaning of fourteen ? Ans. Four and ten.
What is the meaning of thirteen? of nineteen? of seven-
teen ?
Those ending in ty, are the words two, three, four, &c.
with ty, which means tens, added to them.
What is the meaning of sixty ? of seventy ? of eighty ?
of twenty? of thirty?
The words of spoken numeration would be more uni-
form, it eleven and twelve, had been called oneleen and tivo-
teen.
The Latin and Greek numerals are so often used in the
various sciences, that it is important for pupils to learn
their names. They are therefore put down with the fig-
ures, and the English names. The figures are called
Arabic, because first introduced into Europe from Ara-
bia.
NUMERATION.
43
ENGLISH, LATIN, AND GREEK NUMERALS.
Greek Names.
Eis.
Duo.
Treis.
Tessares.
Pente.
Hex.
Hepta.
Okto.
Ennea.
Dcka.
Endcka.
Doileka.
Dekatreis.
Dekatessares.
Dekapente.
Dekaex.
Dekacpta.
Dekaocto.
Dekaennea.
Eikosi.
Triakonta.
Tesserakonta.
Pentakonta.
Hexakonta.
Hebdomekonta.
Ogdoekonta.
Ennenekonta.
Hckaton.
Chilio.
Billion, Trillion, (Quadrillion, Quintillion, Sextillion, &c. are
made, by adding ciphers to 1.
If any higher number than sextillion is to be expressed,
the names are made by the Latin numerals, with illion
added to them ; as seplillion, octillion, &c.
A unit has been defined as " a single thing of any
kind."
But a unit of one kind, maybe made up of several units
of another kind. Thus the unit one dollar is made up of
ten units, of the kind, ox order called dimes; and one
dime is made up often units of the order called cents.
A unit which is of the most value, is called a unit of a
higher order.
Arabic Figures.
English Name,
s. Latin Names.
1
One.
Unus.
2
Two.
Duo.
3
Three.
Tres.
4
Four.
Quatuor.
5
Five.
Q,uinque.
6
Six.
Sex.
7
Seven.
Septem.
8
Eight.
Octo.
9
Nine.
Novem.
10
Ten.
Decern.
11
Eleven.
Undecim.
12
Twelve.
Duodecim.
13
Thirteen.
Tredecim.
14
Fourteen.
Quatuordecim
15
Fifteen.
Quindecim.
16
Sixteen.
Sexdecim.
17
Seventeen.
Septendecim.
18
Eighteen.
Oetodecim.
19
Nineteen.
Novemdecini.
20
Twenty.
Viginti.
30
Thirty.
Triginta.
40
Forty"
duadraginta.
50
Fifty.
Q,uinquaginti.
00
Sixty.
Sexaginta.
70
Seventy.
Septuaginta.
HO
Eighty.
Octoginta.
90
Ninety.
Nonaginta.
100
Hundred.
Centum.
1000
Thousand.
Mille.
1000000
Million.
44 ARITHMETIC. SECOND PART
Which unit is of the highest order, a dollar or a cent?
How many units of the order of dimes, are there in one
unit of the order of dollars ?
How many units of the order of mills, make one unit of
the order of cents?
How many units of the order of cents, make one unit of
the order of dimes ?
Every figure represents a certain number ; but the num-
ber it represents, depends upon the order in which it is
placed.
If the figure (2) stands alone, it represents two units,
and is said to be in the^rs^or unit order.
But if it has a figure to the right of it, thus (20) it rep-
resents two tens, or twenty, and is in the second order, or the
order of tens.
The cipher is put to the right, to make the 2 stand in
the order of tens, and to show that there are no units of
the unit order. If some figure was not placed there, the 2
would be in the unit order.
If the figure* 2 has two figures to the right of it, thus
(20G) it represents two hundreds, and stands in the third,
order, or the order of hundreds.
From this it appears, that in numeration, (he number ex.
pressed by any figure, depends upon the order in which it
stands.
The number which any figure expresses when it is con-
sidered alone, is called its simple value. The number it
expresses when placed with other figures, is called its lo-
cal value.
When 2 is considered alone, what is its simple value ?
When it is considered as in the order of tens, what is its
local value 1 When in the order of hundreds, what is its
local value '!
Questions. — What does every figure represent ? What
does the number which any figure represents depend up-
on ? If a figure stands alone, in what order is it ? If it
has one figure at the right of it, in what order is it ? If it has
two figures tit the right of it, in what order is it? In this
number, (234) in what order is the 2 ? the 3 ? the 4 ?
Write one ten. — Why is the cipher used ? What would
the number be., if the cipher were removed ?
NUMERATION. 45
Write one ten and one unit. What is the name of this
number? Ans. Eleven.
Write one ten and two units. What is the name of this
number?
Write one ten and three units. What is the name 1
Write one ten and four units. What is the name ?
Write one ten and Jive units. What is the name ?
Write one ten and six units. What is the name ?
Write one ten and seven units. What is the name ?
Write one ten and eigjit units. What is the name 1
Write one ten and nine units. What is the name ?
Write two tens. What is the name ? Ans. Twenty.
Write three tens. What is the name ?
W rite four lens ; Jive tens ; six tens; seven tens; eight
tens ; nine tens ; and tell their names.
Write one of the order of hundreds.
Write two of the order oHiundreds ; one of the order of
lens ; and Jour of the order of units.
Write two of the order of hundreds ; no tens ; four
units.
Write 4 hundreds, no tens, no units.
Write two hundreds, eight tens, and nine units. Sev-
en hundreds, six tens, and three units. Two tens, and
two units. Nine tens, and six units. Four hundreds,
six tens, and four units. Five hundreds, five tens, and
five units. Nine hundreds, seven tens, and three units.
Four hundreds, eight tens, and four units. Eight hun-
dreds, nine tens, and nine units. Two hundreds, six tens,
and three units. One hundred, two tens, and three
units. Two hundreds, five tens, and seven units. One
ten, and three units. Seven tens, and three units. Nine
hundreds, nine tens, and nine units.
In readin« numbers, we can either mention each order
separately, or simply mention the names of the numbers.
Thus we can call this number, (21) either two tens, and
one unit, or twenty-one.
This number (305) can be read, 3 hundreds ; tens ;
5 units ; or it can be called three hundred and five.
The following numbers are read both ways, thus ;
10 One ten ; no units ; or ten.
1 1 One ten ; one unit ; or eleven.
46 ARITHMETIC. SECOND PART.
208 Two hundreds ; no tens ; eight units ; or two hun-
dred and eight.
40 Four tens ; no units ; or forty.
Let the pupil read the following numbers both ways.
111. 203. 41. 37. 542. 1. 11. 12. 60.
300. 101. 639. 700. 305.
In this number, (20U) why is the cipher put in? What
would the number be if it were left out?
In numeration, every unit of one order, is considered
as composed of ten units of a lower order ; just as in the
coins of this country, ten units of the order of cents, make
one unit of the order of dimes, and ten units of the order
of dimes, make one unit of the order of dollars.
So in numeration, ten units of the order of units, make
one ten ; ten units of the order of tens, make one unit of
the order of hundreds ; ten hundreds, make one unit of
the order oUhousands ; ten thousands make one of the or-
der of tens of thousands; ten tens of thousands, make one
of the order of hundreds of thousands ; ten hundreds of
thousands, make one of the order of millions, &c.
Wherever there are nine units of any order, if there is
another added, the number becomes one unit of the next
higher order.
If we had nine cents, and should add another, instead of
calling the amount ten cents, we could call it one dime ;
and so when ten units are added together, we can call
them one unit of the order of tens, instead of ten units of the
unit order ; and when we have ten units of the order of
tens, we can call them one unit of the order of hundreds.
Questions. — If nine cents have one more added, in
what order do they become a unit ?
If nine dimes have another added, in what order do
they become units 1
Ten units of the order of dollars, make one unit of what
order ?
Ten tens, make one unit of what order ?
Ten units, make one unit of what order ?
Ten hundreds make one unit of what order ?
The following are the names of the orders.
First order, Units.
NUMERATION.
47
Second order,
Third order,
Fourth order,
Fifth order,
Sixth order,
Seventh order,
Eighth order,
Ninth order,
Tenth order,
Eleventh order,
Twelfth order,
Thirteenth order,
Fourteenth order,
Fifteenth order,
Sixteenth order,
Seventeenth order,
Eighteenth order,
Nineteenth order,
Twentieth order,
Tens.
Hundreds .
Thousands.
Tens of thousands.
Hundreds of thousands.
Millions.
Tens of millions.
Hundreds of millions.
Billions.
Tens of billions.
Hundreds of billions.
Trillions.
Tens of trillions.
Hundreds of Trillions.
Quadrillions.
Tens of Quadrillions.
Hundreds of Quadrillions.
Quiutillions.
Tens of Quiutillions.
Hundreds of Quintillions.
Sextillions.
Twenty-first order,
Twenty-second order,
Sextillions are as high as there is ordinarily any need
of writing or reading.
In all the above orders, " Ten units of one order, make
one unit of the next higher order.
If a figure 2 stands in thefrst order, what number does
it express? What number does it express, if it stands in
the fourth order? In the second order ? In the fjth or-
der ? In the sixth ? seventh? eighth?
Let the pupil write the following :
1.
Five units.
13.
One hundred and sixty.
2.
Three tens ; two units.
14.
One hundred, and six tens.
3.
Thirty-two.
15.
Two hundred, two tens.
-1.
Three and ten, or thirt<
:n.
16.
Two hundred and twenty.
5.
Four and ten.
17.
Two hundred and thirty.
6.
Four tens, or forty.
18.
Two tens and two units.
7.
Six and ten.
19.
Twenty-two.
8.
Six tens.
20.
Two hundreds and two units.
9.
Sixteen.
21.
Five tens and two units.
10.
Sixty.
22.
Five hundreds.
11.
One hundred and sixteen.
23.
Five tens.
12
One hundred, one ten,
and six.
24.
Fifty.
48 ARITHMETIC. SECOND PART.
25. Five hundred, and five units.
26. Five and ten.
27. Fifteen.
28. Fifty seven.
29. Four hundreds, six tens.
30. Four hundred and sixteen.
31. Four hundreds, one ten, and six.
32. Four hundred, and six.
33. Two hundred and sixty-six.
34. Three hundred, ten, and one.
35. Three hundred and eleven.
3G. Three hundred, ten and two.
37. Three hundred and twelve.
38. Four hundred and one.
39. One hundred and forty-two.
40. Two hundreds, two tens.
Let the pupil write the following :
1. One unit of the fourth order. What number is if. ?
Which orders have ciphers in them ?
2. Two units of the fourth order ; one unit of the sec-
ond order, and one unit of the first order. What number
is it ? What order has a cipher in it ?
3. Two thousands ; one hundred ; five tens ; six units.
4. Twenty-one hundreds ; five tens ; six units.
Is there any difference between the two last numbers ?
5. Three thousands, four hundreds, six tens and three
units.
0. Thirty-four hundred, and sixty-three.
Is there any difference in the two last numbers ?
7. Three thousands and three units.
Which orders have ciphers placed in them ?
8. Three thousands, six hundreds.
Which orders have ciphers placed in them ?
9. Thirty-six hundred.
What two ways of reading this last number?
10. Twenty thousand.
11. Two tens of thousands.
Is there any difference between these two last num-
bers?
12. Twentyifour thousand.
What two ways of reading this last number?
13. One hundred thousand, two tens of thousands, five
thousands, six hundreds, four tens, and three units.
14. One hundred and twenty-five thousand, six hun-
dred and forty-three.
Is there any difference between these two last numbers ?
15. Two tens of thousands, one thousand, four hun-
dreds, six tens, five units.
What two ways of reading this number ?
16. Four hundred and sixty-two thousand, five hundred
and six.
NUMERATION. 49
What two ways of reading this last ?
17. Forty-four thousand, four hundred and forty-four.
What two ways of reading this last ?
18. Four hundreds of thousands, five thousands, six
hundreds, two tens, five units.
What two ways of reading this last numher ?
19. Two hundred thousand, two thousand, two units.
What orders have cyphers placed in them ?
20. Twenty thousand, and two units.
21. Two hundred and six thousands, four hundred and
six.
22. Sixty-four thousand and three.
23. Sixteen thousand.
24. Fourteen thousand and seven.
25. Five tens of thousands, and six units.
26. Two hundreds of thousands, two hundreds, two
units.
27. Two hundred and sixty-four thousand, and six.
28. Four thousand, and five units.
29. One hundred thousand, and three.
30. Sixteen thousand, six hundred and six.
31. Twenty-four thousand and three.
In order to read and write numbers more conveniently,
they are divided into periods of three figures each, by
means of commas, thus :
870,409,764,256,622,895,946,852.
Thefirst right hand period is called lhe unit period ; and
contains the orders called units, fens and hundreds
The second period, is called the thousand period ; and
contains the orders called thousands, tens of thousands, and
hundreds of thousands.
The third period is called the million period, and con-
tains the orders called millions, tens of millions, and hun-
dreds oj millions.
The fourth period is called the billion period; and con-
tains the orders called billions, tens of billions, and him-
dreds of billions.
The fifth period is called the trillion period; and con-
tains the orders called trillions, tens of trillions, and hun-
dreds of trillions.
5
50 ARITHMETIC. SECOND PART.
The sixth period is called the quadrillion period ; and
contains the orders called quadrillions, tens of quadrillions,
and hundreds of quadrillions.
The seventh period is called the quintillion period ; and
contains the orders called quintillions, tens of quinlillions,
and hundreds of quintillions.
The eighth period is the sextillion.
The following are the periods which must be learned
in succession, beginning with the highest, as well as with
the lowest ; thus,
First Period Unit. Eighth Period, Sextillion.
Second Period, Thousand. Seventh Period, Quintillion.
Third Period, Million. iSixth Period, Quadrillion.
Fourth Period, Billion,
Fifth Period, Trillion.
Sixth Period, Quadrillion.
Seventh Period, Quintillion
Fifth Period, Trillion.
Fourth Period, Billion.
Third Period, Million.
Second Period, Thousand.
Eighth Period, Sextillion. iFirst Period, Unit.
What is the first period ? ' the third ? the fifth ? the
second ? the fourth ? the seventh ? the sixth ? the
eighth 1
The pupil may write the names over the periods until
accustomed to reading them ; thus,
Trill. Bil. Mil. Thous. Units.
32 427 983 254 693
The above may be read in the following manner :
The first left hand period is read, 3 tens of trillions ; 2
units of trillions : or thirty-two trillions.
The next period is read, 4 hundreds of billions ; 2 tens
of billions ; 7 units of billions ; or four hundred and twen-
ty-seven billions.
The next period is read, 9 hundreds of millions; 8 tens
of millions ; 3 units of millions, or nine hundred and eigh-
ty-three millions.
The next period is read, 2 hundreds of thousands ; 5
tens of thousands ; 4 units of thousands ; or two hundred
and fifty-four thousand.
The next period is read, 6 hundreds ; 9 tens ; 3 units;
or six hundred and ninety-three.
The following is a number in which several orders are
omitted, having ciphers in place of numbers.
NUMERATION.
51
Quin. Quad. Tril. Bil. Mil. Th. U.
33 067 004 803 064 000 400
Let the pupil first tell what periods and what orders are
omitted, having ciphers instead ofnumhers.
The above number may be read thus :
Begin at the left and read ; 3 tens of quintillions, and
3 units of quintillions ; or thirty three quintillions.
The next period is, no hundreds of quadrillions ; 6 tens
of quadrillions ; and seven units of quadrillions ; or sixty-
seven quadrillions.
The next period is, no hundreds of trillions ; no tens of
trillions; 4 units of trillions ; or four trillions.
The next period is, 8 hundreds of billions; no tens of
billions ; 3 units of billions ; or eight hundred and three
billions.
The next period is, no hundreds of millions ; 6 tens of
millions ; 4 units of millions ; or sixty-jour millions.
The next period, as it has no hundreds, tens, or units
of thousands, may be omitted entirely, when reading.
The next period is, 4 hundreds ; no tens ; no units ; or
four hundred.
The best and most common way of reading, is that in
the italics, and then all together, it reads thus :
Thirty-three quintillion ; sixty-seven quadrillion ; four
trillion ; eight hundred and three billion ; sixty-four mil-
lion ; four hundred.
Let the pupil read the following sum in both ways :
Quin. Quad. Tril. Bil. Mil. Th. Un.
607 300 000 763 490 068 002
RULE* FOR READING WHOLE NUMBERS.
Point off into periods of three figures earh, beginning at
the right. Read each period as if it stood alone, and then
add the name of the period.
Note. — When a period or order is omitted, it is not ne-
cessary to mention it at all.
Before reading, let the pupil tell what periods and or-
ders are omitted, and represented by ciphers.
52
I
ARITHMETIC
SECOND PART.
Let the pupil point oft', and re
ad the following figures
1
2
31
304
300046
200200200
111
24
40
600
300005
2030003000
100
136
400
611
1200437
311001300
101
3024
4040
693
1200039
60009090
1011
2002
6000
4004
4960004
100100001
2002
46900
40640
103006
1430096
2071113603
3041
60021
600003
1063007
6000007
1000673
201
62003
100014
103964
86004369
101700013
2010
6040064
600436
140001
20064000
600040006
3004
46923
64003
400006
400400400
300010000
227034293
9623000062
200004900
10043259054
3690200000
43600078609
30006340200
459643723007
602030004296
612942004000040367
40000643209437
3907650060042300000
237
6000964
300600C
)0
396770000543965000076
It is necessary for the pupil to understand, that the
French and English arithmeticians use different methods
of numeration.
The English have their periods contain six orders, and
the French only three.
This makes no difference till we come to hundreds of
millions. After that, it makes a great difference, as will
be seen by the following comparison.
It must be noticed, that the same figures are used in both.
English Method.
Trillions.
579364,
Billions.
028635,
Millions.
419763,
Units.
215468.
French Method.
Sext. Quin. Qua. Trill. Bill. Mill. Th. Units.
579, 364, 028, 635, 419, 763, 215, 468.
From the above it can be seen, that all the orders above
hundreds of millions, in both methods, give the same name,
to a very different value.
Thus, the orders of thousands of millions, tens of thous-
ands of millions, and hundreds of thousands of millions, in
the English method, would be read as billions, tens of bill-
ions and hundreds of billions, in the French method.
NUMERATION.
53
Billions, tens of billions, and hundreds of billions, in the
English method, are equivalent to trillions, tens of trillions,
and hundreds of trillions, in the French method.
Five trillion, in the French method, would be read five
billion, in the English ; and five trillion, in the English
method, would be read five quadrillion, in the French.
Questions. — How would a billion, in the English meth-
od, be read in the French?
How would one hundred billion, in the English method,
be read in the French ?
How would one billion, in the French method, be read
in the English ?
How would six hundred billion in the French method,
be read in the English ?
The French method is adopted in this woik, because
it is both the most convenient, and the most common.
But the pupil needs to understand the difference be-
tween the two modes, and the teacher should make the
class point off and read numbers by both.
Point oft' and read the following numbers, first by the
French, and then by the English method.
7G543217G50U431
9870000654321765432
32698000000040000360093
436789643645964379629364
In order to write numbers correctly, the pupil must
learn thoroughly, the succession of the orders beginning
at the left. Thus, SextiUion, Quintillion, Quadrillion,
Trillion, Billion, Thousand and Unit.
Rule for Writing Whole Numbers.
Begin with the highest period, and write first the hundreds,
then the tens, and then the units of that period. Proceed thus,
until all the periods are written. Place a comma between
each period. If any period or order is omitted, place ciphers
in their place.
Note. — Ciphers prefixed to a whole number, have no
effect upon the value. A number, therefore, should nev-
er be begun with a cipher.
5*
54
ARITHMETIC. SECOND PART.
Write two thousand and two. What orders are omit,
ted?
Write two million, two thousand, and four. What or-
ders are omitted ?
Write Three hundred and twenty. four. What period
and orders in this number ?
Write Two hundred thousand and four. What orders
omitted in this last number ?
Write ; Two million and six ? What period omitted ?
what orders omitted ?
Write, Six million ; two hundred and three. Which
period and what orders are omitted ?
Write, Twenty. lour million ; three hundred. Which
period and what orders are omitted ?
Write the following sums and mention the periods and
orders which, are omitted.
1. One billion; twenty-four million; three thousand
and three.
2. Four hundred and sixty-nine billion ; forty. four
thousand ; and seventeen.
3. Fifty billion ; three hundred million ; four hundred
and fifty thousand ; and nineteen.
4. Fifty billion, and seven.
5. Four hundred and thirteen million, and two thous-
and.
6. Nineteen billion, and one million.
7. Six trillion ; nine thousand, and ten.
8. Seven trillion ; nineteen billion ; ten thousand, and
four hundred.
9. Four hundred and nine trillion ; sixteen million ;
eleven thousand and forty.
10. Fifteen billion ; two hundred and four million ; six
thousand, and twenty-one.
11. Sixty-four million ; Tour hundred thousand ; three
hundred.
12. Sixteen million ; five hundred thousand, and six.
13. Three trillion ; fourteen million ; seven thousand.
14. Two hundred and sixteen million.
15. Two billion ; sixteen million, and sixteen.
16. Three hundred and six trillion ; four thousand, and
six.
NUMERATION.
55
17. Two quintillion ; six quadrillion and five.
18. Three hundred and sixty-four thousand.
19. Three million and six.
20. Fourteen trillion ; three hundred.
21. Sixteen trillion, four million, two hundred and four
thousand, seven hundred and one.
22. Three sextillion, one hundred quadrillion, fourteen
trillion, two hundred and sixty billion, four hundred mill-
ions, sixteen thousand, four hundred and one.
23. Five million, two hundred thousands, and sixty-two.
24. Two hundred and five millions, and seventy-four.
25. Twelve hundred and six billions, four millions, and
six thousand.
26. Two hundred sextillions, four hundred millions,
three hundred and four thousand, two hundred and six.
27. Fifteen quintillion, six quadrillions, one hundred
trillions, forty-four billions, two millions, and forty-nine.
28. Fifty "quadrillions, six hundred trillions, forty-three
millions, two thousands four hundred and six.
29. Two hundred and six trillions, forty-three billions,
four hundred and nine millions, sixty-four thousands, four
hundred and ninety-six.
30. One hundred and four billions, six millions, forty-
nine thousands, four hundred and ninety-six.
31. Thirteen millions, four hundred "thousands, six hun-
dred and forty-nine.
3x5. Six sextillions, five quintillions, four quadrillions,
three trillions, two billions, and one million.
NUMERATION OF VULGAR FRACTIONS.
Figures are of two kinds, — 'Figures for a number of
whole things, and figures for a number of parts of things.
A unit is a whole thing of any kind.
A fraction is a part of one thing ; or a part of several
things.
Figures may therefore be divided into fractional and
unit figures.
The following is the mode of showing when the num-
lg 13 lin, «.vw VI ^v,„.w &
56 ARITHMETIC. SECOND PART.
bers represented are several whole things, and when they
are several parts of things.
When there are two whole things, their number is ex-
pressed thus, (2). This is called a unit figure.
But if a whole thing is divided into three parts, and we
wish to express two of these, by figures, we write one fig-
ure, to show into how many parts the whole thing is divi-
ded, and then above it, write the number of parts we wish
to express ; thus, (|).
This is called a fractional figure. The lower figure
shows into how many parts the whole thing is divided,
and the upper figure shows how many of these parts are
expressed.
In f , into how many parts is the whole thing divided,
and how many of these parts are expressed?
In f, into how many parts is the whole thing divided,
and how many parts are expressed ?
In 4? In a? In*] In f ? In T \? In & ?
Fractional figures show into how many parts one whole
thing is divided, and how many of these parts are expres-
sed. Besides this, they can show what part is taken from
several whole things. Thus f shows that one thing is divi-
ded into four parts, and three of them are taken ; or that
three whole things, have a fourth taken from each of them.
For, three fourths of one whole thing, is the same quantity
as one fourth of three whole things.
If you have three apples, and take one fourth out of
each, how much will you have, and how will you express
it in figures 1 If you divide one apple into four parts, and
take three of these parts, how do you express the quantity
taken ?
If you have two apples, and take one sixth from each,
how much will you have, and how will you express it in
figures ?
If you divide an apple into six parts, and take two of
these parts, how much will you have, and how will you
express it in figures ?
If an apple is divided into eight parts, and you take
three of them, how much wdl you have, and how will you
express it in figures 1
If the fraction is considered as showing how many parts
NUMERATION OF VULGAR FRACTIONS- 57
are taken from one unit, then the lower figure shows into
how many farts a unit is divided, and the upper figure
shows how many of these parts are taken. But if the frac-
tion is considered as showing what part is taken out of
several units, then the upper figure shows the number of
uni&s, and the lower figure shows what part is taken from
each.
Thus the fraction § may be considered as expressing,
two sixths oCone thing, or as one sixth of two things.
-j 3 2 is either one twelfth of three things, or three twelfths
of one thing.
£ is either four fifths of one thing, or one fifth of four
things.
| either shows that one ninth is taken out of two things ;
or that two ninths are taken out of one thing.
If | is considered as showing how many parts are taken
out of one thing it is four sevenths of one unit. If it is con-
sidered as showing what part is taken out of several
things, it is one seventh of four units.
If | shows how many parts are taken out of one thing, it
is two thirds of one thing. If it shows what part is taken
out of several things, it is one third of two things.
If | is considered as showing how many parts are taken
out of one unit, what does the 8 show, and what does the
7 show?
If it is considered as expressing what part is taken out
of several units, what does the 7 show, and what do.es the
8 show ?
If * is considered as expressing how many parts are ta-
ken out of one unit, what does the 6 show, and what does
the 4 show 1 If it is considered as expressing what part
is taken out of several units, what does the 4 show, and
what does the 6 show 1
Whenever the numerator is larger than the denominator ,
the fraction is called an improper fraction, and always is
to be considered as expressing what part is taken out of
several units'.
Which of the following are improper fractions ?
6. 7 _8_ 9. 1_2 1_4 6. 4. R.
3 4 14 6^2 ,5 9 13*
What does an improper fraction show 1
58
ARITHMETIC. SECOND PART.
RULE FOR READING VULGAR FRACTIONS.
Read the number of parts expressed by the numerator, and
tlien the size oftheparts expressed by the denominator ; or
Read the part expressed by the denominator, and then the
number of units, expressed by the numerator.
Read the following fractions in both ways, thus : £ is
either three fourths of one thing, or one fourth of three
things.
§ is either three fifths, or one fifth of three.
4. 6. 2. 13. _S_ 6_ _9_
6 9 6 13 10 12 18
RULE FOR WRITING VULGAR FRACTIONS.
Write the number of parts into which a unit is divided,
and draw a line above it. Over it ivrite the number of
parts which are to be expressed; or
Write the whole numbers which have a certain part taken
from them, and draio a line wider. Beneath it write the
figure which expresses the part which is to be taken out of
each of the units above.
Let the. pupil write the following :
If a man divided an apple into eight parts, and gave
away five of these parts, how do you express the quantity
he gave away, and the quantity he kept?
If a man had three apples, and cut out a fourth part of
each, and gave it away, how do you express what he gave
away I
If a man had twelve oranges, and one sixth of each was
decayed, how do you express the quantity of decayed or-
anges he had ?
If a man had five casks of wine, and a twelfth part leak-
ed out of each, how do you express what he lost ?
DECIMAL NUMERATION.
There is another mode of writing fractions, in which the
numerator only, is written. The denominator, although
not written, is always understood to be 1, and a certain
number of ciphers.
These fractions are called Decimals.
NUMERATION OF VULGAR FRACTIONS. 50
Thus in writing decimals, if we are to express two tenths,
instead of writing it thus j\, the numerator only is written,
and a comma, called a separatrix, is placed before it,
thus ,2.
The following is the rule, by which it is known what is
the denominator.
The, denominator of a decimal is always I , and ax many
cyphers as there are figures in the numerator, or decimal.
What is the denominator of this decimal, ,2 ?
Ans. 1 and one cypher.
How many cyphers in the denominators of these deci-
mals, ,34. ,000. ,3240. ,50945. ,3694. ?
If the decimal has one figure, it expresses tenths. Thus
,2 is two tenths.
If it has two figures, it expresses hundredths. Thus ,02
is two hundredths.
If it has three figures, it expresses thousandths. Thus
,002 is two thousandths.
If it has four figures, it expresses tens of thousandths.
Thus ,0002 is two tens of thousandths.
If it has five figures, it expresses hundreds of thous-
andths Thus ,00002 is two hundreds of thousandths.
What does this decimal express, ,3?
Ans. Three tenths.
What does this decimal express, ,30 ?
Ans. Thirty hundredths.
What does this decimal express, ,003 ?
Ans. Three thousandths.
What does this decimal express, ,0003 ?
Ans. Three tens of thousandths.
What does this decimal express, ,5 ] Ans. 5 tenths.
What does this decimal express, ,15 ? Ans. Fifteen
hundredths.
What does this decimal express, ,110?
What does this decimal express, ,2000?
W T hatdoes this decimal express, ,00002?
Ans. Two hundredths of thousandths.
A decimal must always have the number of figures in
the numerator, equal to the number of ciphers in the de-
nominator ; therefore it is necessary to learn how many
ciphers there are in each kind of denominator.
GO ARITHMETIC. SECOND PART.
If the decimal is tenths, there is one cipher in the de-
nominator ; if hundredths, there are two ciphers ; if
thousandths, there are three ciphers ; if tens of thous-
andths, there are four ciphers ; if hundreds of thous-
andths, there are five ciphers, &c.
Of course in writing decimals, if tenths are to be ex-
pressed, there must be only one figure in the numerator, or
decimal ; if hundredths, there must be two figures ; if
thousandths, there must be three figures ; if tens of thous-
andths, there must be four figures ; if hundreds of thous-
andths, there must be Jive figures, &c.
If you are to write two tenths, how many figures must
there be in the numerator or decimal, and how many ci-
phers are understood to be in the denominator ? Write
two tenths. (,2).
If you are to write two hundredths how many ciphers
are understood to be in the denominator, and how many
figures must there be in the numerator 1
Write, two hundredths.
In writing this last, the pupil must first write the 2, and
then as there must be as many figures in the numerator,
as there are ciphers in the denominator, a cipher is placed
before the 2, and then the separatrix is prefixed thus, ,02.
If the cipher were placed after the 2, how would it
read ?
Ans. Twenty hundredths, instead of two hundredths.
If the cipher were not placed before the 2, how would
it read ? Ans. Two tenths.
If another cipher is placed before the ,02 thus, ,002 how
does it read 1
What does the. denominator express, when there are
three figures in the decimal. Ans. Thousandths.
What does it express when there are four figures in the
decimal ?
Let the pupil write the following.
1. Two tenths.
2. Two hundredths.
3. Two thousandths.
4. Two tens of thousandths.
5. Two hundreds of thousandths.
6. Five tenths.
DECIMAL NUMERATION. 61
7. Fifteen hundredths.
8. Fifteen thousandths.
9. Fifteen tens of thousandths.
10. Fifteen hundreds of thousandths.
11. One tenth.
12. Eleven hundredths.
13. One hundred and fifteen thousandths.
14. Five tenths.
15. Fifty-five hundredths.
16. Five hundred thousandths.
17. Five hundred and five thousandths.
18. Fifteen thousandths.
19. Five thousandths.
20. Two hundred thousandths.
21. Twenty-nine thousandths.
22. Five hundredths.
23. Forty hundredths.
24. Nine tews o/" thousandths.
25. Nineteen tens of thousandths.
26. Nine hundred tens of thousandths.
27. Two thousand tens of thousandths.
28. Two thousand and two tens of thousandths.
29. Three thousand three hundred tews o/ thousandths.
30. Thirty-two hundred te»s o/* thousandths.
31. Six tens of thousandths.
32. Four hundreds of thousandths.
33. Fourteen hundreds of thousandths.
34. Four hundred hundreds of thousandths .
35. .Two thousand and six hundreds of thousandths.
36. Sixty-four thousand hundreds of thousandths.
37. Sixteen thousand and four hundreds of thousandth's.
38. Four thousand and nine hundreds of thousandths.
39. Six hundreds of thousandths.
40. Five thousand and four hundreds of thousandths.
41. Sixty-five thousand hundreds of thousandths.
42. Nine hundred and one hundreds of thousandths.
43. Twenty-nine hundred hundreds of thousandths.
44. Twelve tens of thousandths.
45. Fifteen hundredths.
46. Sixty-four thousandths.
47. Nine hundred and one tens of thousandths.
6
62 ARITHMETIC. SECOND PART.
Decimals can be read in two different ways.
Thus, 21 can be read, either as two tenths, and one
hundredth ; or as twenty-one hundredths.
This can best be illustrated, by the coin of the United
States. Thus, 2 dimes, 1 cent, can be read, either as
twenty-one cents, or as two dimes and one cent.
Thus again, 1 dollar, 3 dimes, and 2 cents, can be
called, either 132 cents ; or 13 dimes, 2 cents ; or 1 dol-
lar, 3 dimes, and 2 cents.
In like manner, decimals may be read in different ways.
Thus, 234 can be read either as 234 thousandths ; or 2
tenths, 3 hundredths, and 4 thousandths ; or 23 hun-
dredths, and 4 thousandths ; or 2 tenths, and 34 thou-
sandths.
Write two tenths.
Write twenty hundredths.
,2 is how many hundredths ?
Ans. There are ten times as many hundredths as there
are tenths in a thing. Therefore ,2 is ten times as many
hundredths, or 20.
Is there any difference in the value of ,2 and ,20 ? What
is the difference between them ?
Ans. The ,20 has ten times more pieces, and each
piece is ten times smaller than the ,2 ; but there is no
difference in the value.
,3 is how many hundredths ? ,4 is how many hun-
dredths ?
,30 is how many tenths ? ,40 is how many tenths ?
Write two tenths, and four hundredths. In this sum
how many hundredths ?
Write thirty-four hundredths. In this sum how many
tenths ?
Write 2 tenths, hundredths, or twenty-six hundredths.
Write 4 tenths, 9 hundredths, and read it both ways.
Write 6 tenths, 7 hundredths, five thousandths, or six
hundred and seventy-five thousandths.
Write 6 tenths, 4 hundredths, and 5 thousandths.
Write nine tenths, six hundredths, and six thousandths,
and read them both ways.
Write seven tenths, six hundredths, five thousandths,
and nine tens of thousandths, and read them both ways.
DECIMAL NUMERATION. 63
Write nine tenths, no hundredths, six thousandths, no
tens of thousandths, and five hundreds of thousandths, and
read it both ways.
Write six tenths, no hundredths, no thousandths, and
five tens of thousandths, and read it both ways.
Write six thousand four hundred and thirty-six, tens of
thousandths, and tell how many tenths, hundredths, and
thousandths there are.
Write four hundred and seventy. nine thousandths, and
tell how many tenths, and hundredths there are.
Write five hundred and six thousandths, and tell how
many tenths there are.
Write five hundred and ninety-six hundreds of thou,
sandlhs, and read it both ways.
From the above it appears, that in decimals, the order
next to the separatrix is tenths ; the second order from the
separatrix is hundredths ; the third order is thousandths ;
the fourth order is tens of thousandths ; the fifth order is
hundreds of thousandths, &c.
Questions. — In decimals what is the first order, at the
right of the separatrix ? What is the second order? What
is the fourth order ? What is the third ? the fifth ?
Decimals, are often written with whole numbers. Thus,
2,5. 36,349.
Whole numbers and decimals together, are called mixed
decimals.
Write twenty-four whole numbers, and twenty-four
hundredths. Two hundred whole numbers, and five
tenths. What are the mixed decimals ?
Rule for reading decimals.
Read the numerator, as if it were whole numbers, and then
add the name of the denominator ; or, Read the number of
each separate order, and follow it with the name of the order
in which it stands.
Read the following decimals both ways.
,11. ,020. ,5005. ,32568. ,0505. ,521. ,43002.
24,690. 6,40043. 6,4000. 69,9604. 86,0092. 2,002.
16,00020.
In writing decimals from the dictation of the teacher,
64 ARITHMETIC. SECOND PART.
the pupil needs to understand the two methods very clearly.
Thus for example, he may have this decimal, ,00205,
dictated in two ways, viz. : 205 hundreds of thousandths,
or 2 thousandths, and 5 hundreds of thousandths.
In the first mode of dictation, he must write the 205 as
if it were whole numbers, and then prefix ciphers to
make the figures of the numerator equal to the ciphers of
the denominator.
In the second mode of dictation, he must put a cipher
in each order which is not mentioned ; viz. : in the orders
tenths, hundredths, and tens of thousandths, and a 2
in the order of thousandths, and a 5 in the order of hun~
dreds of thousandths.
Let the pupil write the following in both methods of dic-
tation.
8 hundredths, 6 tens of thousandths ; or 806 tens of
thousandths.
2 tenths, 4 tens of thousandths ; or 2004 tens of. thou-
sandths.
2 thousandths, 5 tens of thousandths ; or 25 tens of thou-
sandths.
3 hundredths, 6 thousandths, 5 tens of thousandths ; or
365 tens of thousandths.
RULE FOR WRITING DECIMALS.
Write the numerator as if it were whole numbers, and
tlten prefix a separatrix. If the figures of the decimal, do
not equal in number the ciphers of the denominator, prefix
ciphers to make them equal, before placing the separatrix ; or
Write each order separately, placing ciphers in the orders
omtted.
Write the following :
1. Two hundred and ten thousandths.
2. Two tenths, five thousandths, six tens of thousandths.
Here the order of hundredths is omitted, and has a cipher
put in it.
3. Two hundred and four hundreds of thousandths.
4. Two thousandths ; four hundreds of thousandths.
What orders are omitted ?
5. Sixteen tens of thousandths.
DECIMAL NUMERATION.
65
6. One thousandth, six tens of thousandths. What or-
ders are omitted ?
7. Four hundred and five thousandths. What orders
omitted 1
8. Four tenths, five thousandths. What orders are
omitted ?
9. Three hundred and sixty-five tens of thousandths.
What order has a cipher placed in it ?
10. Four hundredths, five tens of thousandths. What
orders are omitted ?
11. Twenty-six thousand, nine hundred and forty-six
hundreds oj thousandths.
12. Two tenths, six hundredths, nine thousandths, four
tens of thousandths, six hundreds of thousandths.
In mixed decimals, it will be seen, that the orders are
reckoned from the separatrix, both ways.
Thus in 98423,40795, the first order at the right of the
separatrix is tenths, and thefirst order at the left is units.
What is the second order at the right, and the second or-
der at the left of the separatrix ?
What is the third order at the right, and at the left of
the separatrix ?
What is the fourth order at the right, and at the left of
the separatrix ?
What is the fifth order at the right, and at the left of the
separatrix ?
If you have the decimal ,2, and place a cypher at the
right, thus ,20, what does it become ? Is the value alter-
ed ? How is it altered 1
Ans. The parts are made ten times smaller, and there
are ten times more of them, so that the value remains the
same.
If you place a cypher at the left of ,2 thus, ,02, what
does it become ? How much smaller is a hundredth, than
a tenth ?
How much smaller does it make a decimal to prefix a
cipher to it ?
If you put ttco ciphers at the right of ,2, what effect is
produced ? If you put them at the left of it, what effect is
produced ?
The following principle is exhibited above :
Ciphers placed at the right of decimals, change their
names but not their value.
6*
66 ARITHMETIC. SECOND PART.
Ciphers placed at the left of decimals, diminish their vah
tie ten times, for every cipher thus prefixed.
Prefix a cipher to ,91 and read it. Annex a cipher to
,91 and read it.
Prefix a cipher to ,20 and read it. Annex a cipher to
,20 and read it.
Signs and Abbreviations used in Arithmetic.
The following signs are used instead of the words they
represent.
-f- signifies plus or added to.
— signifies minus or lessened by.
X signifies multiplied by.
-H- signifies divided by.
= signifies equals.
E. signifies Eagles.
$ signifies Dollars.
d. signifies Dimes.
cts. signifies cents.
m. signifies mills.
ADDITION.
Addition is uniting several numbers in one.
There are four different processes of addition.
The first is Simple Addition, in which ten units of one
order make one unit of the next higher order. Thus, ten
units make one of the order of tens — Ten tens make one
of the order of hundreds — Ten hundreds, make one of the
order of thousands, &c.
The second is Decimal Addition, in which decimal frac-
tions are added to each other. Thus, ,5 ,50 ,505 are ad-
ded together.
The third is Compound Addition, in which other num-
bers besides ten, make units of higher orders. Thus,
four units of the order of farthings, make one unit of the
order of pence. Twelve units of the order of pence,
make one of the shilling order. Twenty of the shilling
SIMPLE ADDITION. 67
order, make one of the pound order, &c.
The fourth is the Addition of Vulgar Fractions, in which
parts of units are added to each other. Thus \ i and f
are added to each other.
SIMPLE ADDITION.
If 8 units are added to 9 units, how many are there
of the order of tens ?
Write the 8 under the 9, and draw a line under. Place
the units of the answer, under the figures added, and set
the 1 ten before them.
If 13 apples are added to 25 apples, how many are
there in the whole ?
Write the units under units, and tens under tens. Add
the units first, and place the answer under the unit column.
Then add the tens in the same way.
Add 12 cents to 5 cents.
Add 13 apples lo 14 apples.
Add 14 dollars to 19 dollars.
Add 5 and 2 and 12 together.
Add 13 and 12 and 14~together.
Let the pupil add small sums, which do not amount to
ten of any order, till it can be done quickly and with a full
understanding of the process.
In the next process let the coins be used to illustrate.
If 25 cents be added to 16 cents, how many cents are
there ?
Let 2 dimes be laid on the table, and 5 cents placed
at the right of them. Under the 2 dimes place 1 dime,
and under the 5 cents place G cents. Let the child then
add the to the 5, and the answer will be 11 cents.
Eleven cents are 1 dime and 1 cent. Let him leave 1
cent under the column of cents, and substitute 1 dime for
the 10 cents. Let him place this dime with the 2 dimes,
and his answer will be 3 dimes 1 cent. Ask how many
cents in 3 dimes 1 cent, and the answer will be 31 cents.
Thus his answer will be either 3 dimes 1 cent, or 31 cents.
If the pupil thus sees the principle once illustrated, by
68 ARITHMETIC. SECOND PART.
a visible process, the method will be much more readily
understood and remembered. Let the following sum also,
be done by the coins.
Add $1,36 to $2,97.
Add 2$. 6d. 8 cts. to 3$. 8d. 9 cts.
Add 7 E. 2$. 5d. 6 cts. to 4 E. 8$. 6d. 4 cts.
Add 5d. 6 cts. 7m. to 8d. 4 cts. 9m.
Add 4 E. 0$. 6d. 5 cts. to 5 E. 0$. 4d. 6 cts.
Let the teacher dictate such simple sums until the pro-
cess of writing and adding is well understood, and can be
done with rapidity and accuracy.
Note to teachers.
It is very desirable that pupils should, be required to
write their figures with accuracy and neatness, and learn
to place them in strait lines, both perpendicular and hori-
zontal. Also that they learn to add by calculation, and
not by counting, as young scholars are very apt to do. If
a teacher will but be thorough, at the commencement, in
these respects, much time and labor will be saved.
Mary has 4 apples, James 5, and Henry 7, how many
have all together 1
One boy has 6 marbles, another 4, and another 9, how
many have all together ?
A man gave 9 cents to one boy, 8 to another, and 11
to another, how many did he give to all 1
10 and 11 and 9 are how many?
12 and 7 and 4 are how many ?
4 and 5 and 7 are how many ?
One man owns 6 horses, another 8, and another 9, how
many have they all 1
In a school, 10 study history, 11 geography, and 15
grammar, how many scholars in the whole ?
One house has 10 windows, another 7, and another 12,
how many are there in all ?
James lent one boy 8 cents, another 6, and another 17,
how many did he lend them all ?
If a lady pays 7 dollars for a veil, 9 dollars for a dress,
and 3 dollars for a necklace, what amount does she
spend ?
6 and 9 and 18 are how many 1
SIMPLE ADDITION.
69
1 and 5 and 7 are how many ?
8 and 11 and 14 are how many?
Let the pupil be taught to add using the signs.
the last sum. 8 + 11 + 14 = 33
Thus
Rule for Simple Addition.
Place units of the same order in the same column, and draw
a line under-. Add each column separately, begim&ng at the
right hand. Place the units of the amount, under the column
tovMch they belong, and carry the tens to tlw next higher or-
der.
Add 2694 and 3259 and 6438.
Placing units of the same order in the same column,
they stand thus.
2694
3259
6438
12391
Let the pupil at first learn to add in this manner. 8
units added to 9, are 17, and 4 are 21 units, which is 1 of
the unit order, to be written under that order, and 2 of the
order of tens, to be carried to that order. 2 tens carried
to 3 tens, are 5, and 5 are 10, and 9 are 19 tens ; which
is 9 of the order of tens, to be written under that order,
and 1 of the order of hundreds, to be carried to that order.
Thus through all the orders.
Add the following numbers.
(1)
22321
(2)
23432
(3)
110331
(4)
222311
41332
42212
224212
131232
12123
13124
103123
101221
13220
21101
220320
234031
88996
99S69
657986
688795
(5)
275496
(6)
456789
(?)
369543
(8)
4976432
8732
654321
695432
4976432
70
ARITHMETIC
FIRST PART.
54976
456789
567897
6325498
843215
654321
432591
5192346
7621
543219
526387
8763945
49673
345678
489549
763497
1239713
3111117
3081399
30998150
(9)
(10)
(11)
(12)
30648
30430
764325
29367
46469
25895
70504
29367
74057
57644
98469
29367
63396
72919
57157
29367
55275
3647
46946
29367
90534
57246
3284
29367
8953
30142
363
247781
176202
1041 04ft
399474
Let the pupil now learn to place units of the same order
in the same column, by the following examples.
Let the teacher dictate the following. The pupils
should be required previously to . attempt writing them,
while studying their lesson.
1
One million, four hundred and sixty thousand, and two.
Twenty- four million, six hundred and one.
Three hundred and sixty thousand, four hundred and
six.
Ninety-four million, five hundred and seventy-eight
thousand, three hundred and forty-one.
Six million, seven thousand, and forty-three.
2
Two hundred and six thousand, five hundred and forty-
two.
One million, one thousand, and one.
Nine hundred and ninety million, nine hundred and
ninety-nine.
Eighty-eight thousand, eight hundred and eighty-eight.
Ninety-nine million, seven hundred and sixty-five thou-
sand.
SIMPLE ADDITION. 71
3.
Two hundred and six million, five thousand, four hun-
dred and one.
Fifty-six million, four hundred thousand, five hundred
and six.
Three billion, ninety-nine thousand, and four.
Five hundred million, thirty thousand, four hundred and
forty.
Seven million, six hundred and fifty-four thousand,
three hundred and seventeen.
4.
Four million, four hundred and thirty-two thousand, one
hundred and seventy-six.
Forty-nine thousand, and three.
Nineteen million, seven hundred and sixty-five thou-
sand, nine hundred and eighty-four.
Five hundred and ninety-one.
Seven hundred and sixty-three thousand, nine hundred
and forty-three.
Ninety-nine million, nine thousand and ninety.
5.
Four hundred and four.
Five million, six hundred and forty-three thousand, two
hundred and seventeen.
One million, and two.
Nine thousand, and ninety-nine.
Four million, five hundred and seventy-six thousand,
three hundred and eighty-four.
Forty-four million, three hundred and twenty-one thou-
sand, seven hundred and four.
6.
One hundred million, one thousand, and ten.
Nine billion, eight hundred thousand, nine hundred and
forty.
Four hundred and eighty-eight million, nine hundred
and five thousand.
Eighty-eight million, seven hundred and seventy-seven
thousand, and nine.
Nine hundred and ninety-nine.
72 ARITHMETIC. SECOND PART.
7.
Ninety-nine million, eight, thousand, and four.
Five hundred and eighty-seven million, six hundred
and forty-nine thousand.
Twenty-eight thousand, eight hundred and ninety-nine.
Four hundred thousand, eight hundred and seven.
One billion, fifty-nine million, four thousand and eighty-
seven.
8.
Seven hundred million, ninety-nine thousand, and sev-
enty-nine.
Fifty-five thousand, seven hundred and forty-four.
Nine million, eight hundred thousand, eight hundred.
Eight thousand, eight hundred.
Seven billion, and seventeen.
9.
Eighty-four thousand, and nineteen.
Nine million, fifty-four thousand, seven hundred.
Seven hundred and sixty-eight thousand, eight hundred
and four.
Four billion, twenty million, ten thousand and fifty.
Sixty million, two hundred thousand.
Eleven hundred and forty-two.
10.
Forty thousand, and twelve.
Nine billion, eight thousand.
Sixty million, seven hundred thousand, and ten.
Nine billion, ninety million, eighty thousand, and sev-
enty-eight.
Sixty-five million, and four hundred.
One billion, and four.
11.
Nine hundred thousand.
Four million, fifty-five thousand, and eighty.
Three hundred and sixty-four thousand, seven hundred
and thirty-eight.
Forty million, four hundred and four.
Six hundred and thirty thousand.
12.
Ten million, four hundred.
DECIMAL ADDITION.
73
Seventy-six thousand, three hundred and twenty-one.
Eight million, forty-two thousand, six hundred and
seventy-three.
One thousand, four hundred.
Sixty-four thousand, three hundred and twenty.
One billion, and seventy-three.
DECIMAL ADDITION.
Rule for adding decimals.
Place figures of the same order under each other. Add
each column, as in Simple Addition, and in the answer place
a separatrix between the orders of units and tenths.
Example.
What is the sum of 234,406. 4,6490. 13,234. 2,2.
3650,4002. 990,4699.
Placing units of the same order under each other, they
stand thus : —
234,406
4,6490
13,234
2,2
3650,4002
999,4699
4904,3591
Let the pupils proceed as in Simple Addition, calling
the names of each order, thus : —
9 tens of thousandths added to 2, are 11 tens of
thousandths ; which is 1 ten of thousandths, to be written
under that order ; and 1 of the order of thousandths, to be
carried to that order.
1 thousandth carried to 9, is 10, and 4 are 14, and 9
are 23, and 6 are 29 thousandths ; which is 9 thousandths,
to be written under that order, and 2 hundredths, to be
carried to the next order.
Thus through the other orders, observing to place a
separatrix between the orders of units and tenths.
Arrange the following mixed decimals according to
their orders, and then add them.
7
74 ARITHMETIC. SECOND PART.
(1)
306,42001. 20,3391. 3246,42. .39,4695. 634,001.
84,6302,
(2)
99,987. 65432,02564. 64,65. 596,32. 87632,-
51739. 36,50. 51639,2154.
63,204. 6359,42591. 8642,39. 86423,2915. 68,241.
(4)
63,9876. 59432,1103. 95,02. 876,3254. 8634,251.
3426,549.
Let the pupil write and add the following sums in De-
cimals.
1.
Four units, six tenths, four hundredths, five thou-
sandths.
Two tens, four units, six hundredths.
Three tens, two units, two hundredths, seven thou-
sandths.
Six units, five tenths, seven hundredths, four thou-
sandths, three tens of thousandths.
One unit, three tenths.
2.
Forty-two units, sixteen thousandths.
Five units, sixty-three hundreds of thousandths.
Seventy. four units, seven thousand five hundred and
fifty-three tens of thousandths.
Two units, five hundred and sixty tens of thousandths.
3.
Two hundred and forty-three units, two hundred and
forty-three thousandths, seventeen units, nine hundred
and seventy-three tens of thousandths.
Fifty units, six thousand seven hundred and forty-three
hundreds of thousandths.
Five units, eight thousandths.
One thousand units ; one thousand tens of thousandths.
4.
One thousand and one units ; one thousand and one
hundreds of thousandths.
DECIMAL ADDITION.
75
Nine hundred and ninety-nine units, nine thousand
nine hundred and thirty hundreds of thousandths.
Four units, thirty tens of thousandths.
Five units, fifty-five thousand and forty-three millionths.
5.
Sixteen units, seven hundred and sixty-four thousandths.
Two units, forty-five hundreds of thousandths.
Fifty units, forty-two millionths.
Seven units, nine hundred and ninety -eight tens of thou-
sands.
Six units, five hundred and forty-nine millionths.
6.
Four thousand units, four thousand thousandths.
Forty-one units, four thousand, four hundred and nine
hundreds of thousandths.
Seven units, eighty-seven tens of thousandths.
Four hundred and forty-one units, ninety-nine hun-
dredths.
Four units, four hundreds of thousandths.
7.
Seventeen units, nine thousand eight hundred and sixty
hundreds of thousandths.
Nine units, sixteen tens of thousandths.
Four units, fifty-five hundredths.
Sixty-three units, ninety-nine millionths.
One unit, seventy-four thousandths.
8.
Five hundred and forty-four units, eight thousand seven
hundred and fifty-five millionths.
Ninety-nine units, four hundred hundreds of thou,
sandths.
Six units, eight hundred and eighty-eight thousandths.
Eight thousand units, seventy -four tens of thousandths.
Six units, eighty-eight hundredths.
9.
Seventeen units, forty thousandths.
Five units, ninety-three millionths.
Forty -four units, eighty-seven hundredths.
Six units, nine hundred and ninety-nine thousandths.
Four hundred and twelve units, seventy-five tens of
thousandths.
76 ARITHMETIC SECOND PART.
10.
Seventy-eight units, four thousand and five tens of thou-
sandths.
Two units, five hundred hundreds of thousandths.
Seven units, eighty-nine millionths.
Five hundred and seventy-two units, seventy-six thou-
sand, eight hundred and sixty-four hundreds of thou-
sandths.
Nine thousand and fifty units, nine thousand and fifty
millionths.
11.
Five hundred and eighty-seven units, twenty-nine hun-
dred tens of thousandths.
Forty units, five hundred and sixteen millionths.
Eight units, four hundred and ninety-six thousand mill-
ionths.
Five hundred and forty-two units, two thousand hun-
dreds of thousandths.
Seventeen units, nine thousand nine hundred hundreds
of thousandths.
12.
Sixty-five units, sixty-five hundreds of thousandths.
One hundred and eighty units, one hundred and eighty
tens of thousandths.
Twenty-four units, twenty-four millionths.
Sixteen units, sixteen hundredths.
Five units, five thousandths.
Fifty units, fifty hundreds of thousandths.
13.
One hundred and seventy-six units, one hundred and
seventy-six hundreds of thousandths.
Four units, two thousand four hundred and seventy-five
tens of thousandths.
Eighty-four units, seven hundred and sixty-three mil-
lionths.
Two hundred units, two thousand and forty tens of
thousandths.
Seventeen units, four thousand and four millionths.
14.
Seventy-four units, nine hundred and eighty millionths.
Four units, four hundreds of thousandths.
DECIMAL ADDITION. 77
Eighty-one units, nine thousand four hundred hundreds
<of thousandths.
One unit, ninety thousand and one millionths.
Eleven units, one hundred tens of thousandths.
15.
Seventy units, seventy thousandths.
Five units, four hundred and forty hundreds of thou-
sandths.
Four hundred units, seven thousand and forty-three
millionths.
Nineteen units, eighty thousand and nine millionths.
Six units, one hundred and one hundreds of thou-
sandths.
16.
Nine tenths, four hundredths, three tens of thousandths.
Five tens, sixteen thousandths, four millionths.
Forty units, one hundredth, ten tens of thousandths.
Seven units, five tens of thousandths, three millionths.
Six units, four tenths, two hundredths.
17.
Two teas, two units, nine tens of thousandths.
One unit, four tenths, two hundredths, seven millionths.
Eight tens, two hundredths, six hundreds of thousandths.
Four hundreds, fourteen millionths.
Six units, forty thousand hundreds of thousandths.
Fifty-nine units, fifty-nine thousand millionths.
18.
Eighteen units, four hundred and sixty-three thou-
sandths.
Nine units, eight hundred and forty-three millionths.
Twenty-two units, eleven thousand and one hundreds
of thousandths.
Nine units, ninety-nine hundreds of thousandths.
Eighty-eight units, nine millionths.
Four units, eight hundred and eighty-eight thousandths.
METHODS OF PROVING ADDITION.
1. Commence at the top instead of the bottom of the sev-
7*
78 ARITHMETIC. SECOND PART.
cral columns, and if the same answer is obtained, it may
be considered as right.
2. Draw a line "nd cut off the upper figure of all the
orders. Add the leniainder which is not cut off. Then
add the sum of this remainder to the figures cut off, and
if the answer is the same as the first answer, it may be
considered as right.
COMPOUND ADDITION.
In order to understand the following sums, the pupil
must commit to memory the tables inserted in the com-
mencement of the book.
Sums for Menial Exercise.
If a man has 2 lbs. 1C oz. of beef, and buys 6 lbs. 8
oz. more, how much has he in the whole ?
The answer will be 8 lbs. 18 oz. In 18 oz. how many
pounds, and how many ounces over ? Set down the oun-
ces that are over, and add the number of lbs. to the 8 lbs.
and what is the answer ?
A boy has 3 yards 2 quarters of cloth, and buys 2 yards
and 3 quarters more, how much has he in the whole ?
One man buys 3 bushels and 2 pecks of grain, another
buvs 2 bushels and 3 pecks, how much do both together
buy ?
If you have 1 quart and 1 pint of milk, and buy 2 quarts
and 1 pint more, how much will you have ?
One rope is 3 feet, 7 inches ; another is 4 feet, 6 inch-
es ; how many feet are there in both together ?
If 2 weeks 4 days, be added to Iweek 5 days, how ma-
ny weeks will there be in all ?
If 6 pounds 9 oz. be added to 5 pounds 8 oz. how ma-
ny pounds will there be in all ?
If 3 bushels 2 pecks, be added to 4 bushels 3 pecks,
how many bushels will there be ?
If 7 yards 2 quarters, be added to 8 yards 3 quarters,
how many yards will there be ?
COMPOUND ADDITION. 79
RULE FOR COMPOJNP ADDITION.
Place units of the same order in the same column. Find
the sum of each order. Find how m> y units of the next
higher order are contained, in the svu, and carry them to that
order. Set the remainder under th'' order added.
EXAMPLE.
£. *. d.
5 „ 6 „ 8
4 „ 9 „ 9
9 „ 9 „ 5
19„5„10
Let the pupil add thus : 5 pence added to 9 are 14, and
8 are 22 pence. This sum contains 1 of the order of shil-
ling *o he carried to that order, and 10 to bo written un-
der ,<i order added. One shilling carried to 9 makes
10, and 9 are 19, and 6 are 25 shillings. This sum con-
tains 1 of the order of pounds, to be carried to that order,
nnd 5 of the order of shillings, to be written under that or-
der. 1 pound carried to 9 makes 10, and 4 are 14, and 5
are 19 pounds, which are written under that order.
Accu&tom the pupils to add in this manner ; also require
them to separate their orders in Compound Addition by
double commas, as in the above sum.
Add the following sums :
STERLING
MONEY.
£. s. d.
£. s. d.
12 „13 „ 10
n „i3„ ii
14 „ 9„ 9
13 „ 10 „ 2
16 „ 6„ 5
10 „ 17 „ 3
18 „ 12 „ 11
8 ,, 8 „ 7
TROY
WEJOHT.
lbs. oz. put.
oz. pwt. gr.
16 „ 11 „ 19
11 „19„21
4 „ 4 „ 16
10 „ 16 „ 8
8 „ 8 „ 19
8 „ 17 „ 21
6 „ 9 „ 14
6 „ 8 „ 23
80 ARITHMETIC. SECOND PART.
AVOIRDUPOISE WEIGHT.
cwt. qr. lb. lb. oz. dr.
2„3.'27 24 „ 13 „ 14
1 „1„1^ 17 „ 12„11
4„2„26 26 „ 12 „ 15
6„ 1„13 16 „ 8„ 7
APOTHECARIES WEIGHT.
3. 9- gr. 3- 3. B.
9 „ 1 „17 10 „7 „ 2
3 „ 2 „ 9 6 „ 3 „
6 „ 1 „ 22 7 „ 6 „ 1
4 „ „ 16 9 „ 5 „ 2
CLOTH MEASURE.
yd. qr. na. E. E. qr. na.
71„3„3 44„3„2
13 „ 2 „ 1 ' 49„4„3
16 „ „ 1 06 „ 2 „ 3
42 „ 3 „ 3 84 „ 4 „ 1
DRY MEASURE.
pk. qu. pt. bit. pk. qt.
1 „ 7 „ I 17 „ 2 „ 5
2 „ 6 „ 34 „ 2 „ 7
1 „ 5 „ 13 „ 3 „ 6
2 „ 4 „ 1 16 „ 3 „ 4
WINE MEASURE.
gal. qt. pt. hhd. gal. ql
39 „ 3 „ 1 42 „ 61 „ 3
17 „ 2 „ 1 27 „ 39 „ 2
24 „ 3 „ 9 „ 14 „
19 ., „ 16 „ 24 „ 1
COMPOUND ADDITION. 81
LONG MEASURE.
yds. ft. in. m. fur. po.
4 „ 2 „ 11 46 „ 4 „ 16
3 „ 1 „ 8 58 „ 5 „ 23
1 „ 2 „ 9 9 „ 6 „ 34
6 „ 2 „ 10 17 „ 4 „ 18
LAND, OR SQUARE MEASURE.
acres, roods, rods. s <j-fl- S <1- in.
478 „ 3 „ 31 13 „ 1446
816 „ 2 „ 17 16 „ 1726
49 „ 1 „ 27 3 „ 866
63 „ 3 „ 34 14 „ 284
SOLID
MEASURE.
ton. ft.
cords, ft.
41 „ 43
3 „ 122
12 „ 43
4 ,,114
49 „ 6
7„ 83
4„27
10 „ 127
TIME
i •
y- m. to.
h. tnin. sec.
W» 11 „ 3
23 „ 54 „ 32
3„ 9„ 2
12 „ 40 „ 24
29 „ 8, ,2
14 „ 00 „ 17
46 „ 10 „ 2
6„16„13
CIRCULAR.
MOTION.
*■ o .
O ' H
3 „ 29 „ 17
29 „ 59 „ 59
1 „ 6 „ 10
00 „ 40 „ 10
4„ 18„17
4 „ 10 „ 49
6 „ 14 „ 18
11 „ 6„ 10
82 ARITHMETIC. SECOND PART.
ADDITION OF VULGAR FRACTIONS.
Sums for Mental Exercise.
If one boy has one half an orange, and another three
halves, and another four halves, how many halves are
there in all ?
If one third of a dollar, five thirds, and six thirds, be ad-
ded together, how many are there in all 1
One man owns four twentieths of a building, another
six twentieths, and another eight twentieths, how many
twentieths do all own ?
Seven thirtieths, nine thirtieths, and six thirtieths, are
how many 1
Eight twenty-fifths, four twenty-fifths, and seven twen-
ty-fifths, are how many ?
RULE FOR ADDING VULGAR FRACTIONS, WHEN ALL HAVE
THE SAME OR A COMMON DENOMINATOR.
Add the numerators, and place their sum over the com-
mon denominator.
EXAMPLE.
Add ft ft ft and ft.
The sum of the numerators is 15, which being placed
over the common denominator, gives the answer if.
Add the following sums, using the signs, thus :
Add ft ft and ft. Ans. ft X ft X ft - |f.
Add ft ft and |f Add ft ft ft and ft.
Add ft ft and ft. Add f f f.
When fractions having a different denominator, are ad-
ded, it is necessary to perform a process which will be ex-
plained hereafter.
Those fractions which have the numerator larger than
the denominator, are called improper fractions, thus :
V I-
When we use the expression seven halves, we do not
mean seven halves of owe thing, because nothing has more
than two halves. But if we have seven apples, and take a
half from each one, we shall have seven halves; and they
are halves of seven things, and must be written as above.
SIMPLE SUBTRACTION. 83
SUBTRACTION.
There are four kinds of Subtraction.
The first is Simple Subtraction, in which the minuend
and subtrahend are whole numbers, and ten units of one
order, make one unit of the next higher order.
The second is Decimal Subtraction, in which the minu-
end and subtrahend are Decimals.
The third is Compound Subtraction, in which other num-
bers beside ten, make units of a higher order.
The fourth is Subtraction of Vulgar Fractions, in which
the minuend and subtrahend are vulgar fractions.
SIMPLE SUBTRACTION.
If 8 cents are taken from 12 cents, what will remain ?
If 9 apples are taken from 14 apples, how many will re-
main?
If 12 guineas are taken from 20 guineas, how many
will remain ?
If from 18 books, 12 be taken, how many will remain ?
Let the following examples be illustrated bv the coin of
the U. S.
If $ 2, 5d. 6 cts. be taken from $3, 6d. 7 cts., how much
will remain ? Which is the subtrahend, and which the
minuend ?
Place $3, 6d. 7 cts. on a table, side by side, and let the
pupil take the amount of the subtrahend from them.
Subtract $3, 4d. 5 cts. from $G, 7d. 7 cts.
Subtract 3d. 4 cts. 2 m. from 5d. 6 cts. 8 m.
Subtract 8d. 7 cts. 5 m. from 9d. 9 cts. 9 m.
Let the teacher place on the table the coins, thus :
S3, 4d. 6 cts.
Under this place for the subtrahend, the following, so
that the coins shall stand under others of the same order.*
$2, 2d. 4 cts.
What is the remainder, when the value expressed by
the subtrahend, is taken from the minuend ?
Now if 10 cents be added to the 6 cents of the min-
uend, and 1 dime be added to the 2 dimes of the subtra-
* The pupil must understand that the subtrahend shows how many of
the same kinds of coin, are to be taken from the minuend.
84 ARITHMETIC. SECOND PART.
hend, will there be any difference in the answer. Let
the pupil try it and ascertain.
If 10 dimes be added to the 4 dimes of the minuend,
and 1 dollar be added to the 2 dollars of the subtrahend,
will there be any difference in the answer ?
Let this process be continued until every member of the
class fully understands it, and then let them commit to
memory this principle.
" If an equal amount be added to the Minuend and the
Subtrahend the Remainder is unaltered.
Let the following coins be placed as minuend and sub-
trahend.
$ d. cts.
2 13 Minuend.
14 5 Subtrahend.
Which is the largest sum taken as a whole, the minuend
or subtrahend ?
If each order is taken separately, in which orders is the
minuend the largest, and in which the smallest ?
Can you take 5 cents from 3 cents 1
If you add 10 cents to the 3 cents, you can subtract 5
from it, but what must be done to prevent the Remainder
from being altered ?
$ d. cts. tn.
From 4 3 2 4
Subtract 14 5 6
In which orders are the numbers of the subtrahend
larger than those of the minuend ?
Can 6 mills be taken from 4 mills ?
What can you do in this case ?
If 10 mills be added to the 4 mills of the minuend, why
must 1 cent be added to the 5 cents of the subtrahend ?
From 6432, subtract 3256.
Can 6 units be taken from 2 units ?
What must be done in this case ?
simple subtraction. 85
Rule for Simple Subtraction.
Write the subtrahend under the minuend, placing units of
the same order under each other, and draw a line under.
Subtract each order of the subtrahend, from the same order
of the minuend, and set the remainder under. If any order
of the subtrahend is greater than that oj the minuend, add
ten units to the minuend, and one unit to the next higher or-
der of the subtrahend. Then proceed as before.
EXAMPLE.
Subtract 4356
From 2187
21 09
Let the pupil subtract thus :
Seven units cannot be taken from 6 ; therefore add 10
to the minuend, which makes 16. 7 from 16 leaves 9.
As 10 units have been added to the minuend, the same
amount must be added to the subtrahend. 1 of the order
of tens is the same amount as 10 units, we therefore add
1 to 8 tens, making it 9 tens. We cannot subtract 9 tens
from 5 tens, we therefore add 10 to the minuend, which
makes 15. 9 tens from 15 leaves 6 tens. As 10 tens
have been added to the minuend, the same amount must
be added to the subtrahend — 1 of the order of hundreds is
the same amount as 10 tens ; we therefore add 1 to 1
hundred, which makes 2 hundred. This subtracted from
3 hundred leaves 1 hundred.
Thus through all the orders.
Mode of Proof.
A sum in Subtraction is proved to be right, by adding
the remainder to the subtrahend ; and if the sum is the
same as the minuend, the answer may be considered as
right.
Let the following sums be explained as above.
Subtract
34695
from
56943
<(
653215
cc
956432
cc
500032
cc
867200
cc
6291540
cc
8732418
cc
354965
8
cc
5360025
86 ARITHMETIC. SECOND PART.
« 7985430 « 989763
" 3542685 « 6542169
5321543 « 7954324
" 1223345 « 8500642
" 1549768 « 3895463
3543257 » 6385241
" 2006935 « 5000623
The pupil should learn to subtract by the use of the
signs, thus :
Subtract 5 from 7. Ans. 7 — 5=2.
Subtract8 from 11. Ans. 11 — 8=3.
Subtract the following numbers in the same way. 8
from 17. 9 from 14. 6 from 20. 40 from 85. 800
from 950. 1000 from 2744. 85 from 760. 95 from 700.
440 from 763.
DECIMAL SUBTRACTION.
If 2 tenths, 4 hundredths of a dollar, be taken from 4
tenths, 6 hundredths, what will remain ?
If 3 hundredths, 5 thousandths of a dollar, be taken
from 5 hundredths, 7 thousandths, what will remain ?
If 5 dimes, 6 mills, be taken from 7 dimes, 8 mills, how
much will remain 1
If 4 dimes, Scents, be taken from 7 dimes, 9 cents,
how much will remain ?
If 4 units, 6 tenths, be taken from 6 units, 8 tenths, how
much will remain ?
In simple subtraction, if the number in any order of the
minuend, was smaller than the one to be subtracted, what
did you do ?
The same is to be done in Decimal Subtraction.
Take 4 tenths, 7 hundredths of a dollar, from 6 tenths,
5 hundredths.
In which order is the number of the subtrahend the lar-
gest ?
Can 7 hundredths be taken from 5 hundredths ? What
must be done in this case ?
Take 5 dimes, 6 cents, from 8 dimes, 9 cents.
DECIMAL SUBTRACTION. 87
la which order is the number of the subtrahend the lar-
gest ?
Can 9 cents be taken from 6 cents ? What must you
do in order to subtract ?
Subtract 7 hundredths, 8 thousandths of a dollar, from
8 hundredths, 7 thousandths.
Can 8 thousandths be subtracted from 7 thousandths ?
What must be done in this case 1
Rule for Decimal Subtraction.
Proceed by the rule for common Subtraction, and in the
answer -place a separatrix between the orders of units and
tenths.
EXAMPLE.
Subtract 2,56 from 24,329. Placing the subtrahend
under the minuend, so that units of the same order stand
in the same column. They stand thus :
24,329
2,56
21,769
Let the pupil learn to subtract in this manner :
Nothing from 9 thousandths, and 9 remains to be set
down. 6 hundredths cannot be taken from 2 hundredths ;
we therefore add 10 to the minuend, which makes 12. 6
taken from 12 leaves 6. As 10 was added to the minu-
end, an equal quantity must be added to the subtrahend.
1 of the order of tenths is the same as 10 hundredths, we
therefore add 1 to the 5 tenths, making it 6 tenths. 6
tenths cannot be taken from 3 tenths, we therefore add
10 to the minuend, which makes 13. 6 taken from 13,
leaves 7. As 10 was added to the minuend, an equal
amount must be added to the subtrahend. 1 of the order
of units is the same as 10 tenths, we therefore add 1 to the
2 units, making it 8 units.
Proceed thus through all the orders, remembering to
place a separatrix between the orders of units and tenths.
Let the following sums be arranged and subtracted in
the same way :
88 ARITHMETIC. SECOND PART.
Subtract 25,25 from 62,904
« 790,4 " 96,409
« 2,4693 « 354,268
« 5,34689 " " 40,62
« 6,6543 « 23,3291
« 432,54916 " 542,65329
« 53,00300 « 646,01201
832,2 « 9988,659
" 51,895 « 64,59432
8,4156 « 400,21
« 321,01013 " 4333,0063
" 659,09543 « 679,2941
1.
Subtract two tens, four units, three tenths, five hun-
dredths, and four thousandths ; from four tens, two tenths,
five hundredths, and four thousandths.
2.
Subtract two tens, three units, six tenths, nine hun-
dredths, and three thousandths, from four tens, four units,
three thousandths, and five tens of thousandths.
3.
Subtract two units, four thousand three hundred and
seventy-four tens of thousandths ; from
Twentv-three units, seven thousand five hundred tens
of thousandths.
4.
Subtract ninety-eight units, two thousand nine hundred
and eighty-seven tens of thousandths ; from
Seven hundred and seventy-seven units, four thousand
three hundred and twenty-six tens of tliousandths.
5.
Subtract seven units, six thousand five hundred and for-
ty-three tens of thousandths ; from
Three hundred and sixty-nine units, forty-two hun-
dredths.
6.
Subtract seventy-seven units, twenty- four tens of thou-
sandths : from
DECIMAL SUBTRACTION. 89
Two hundred and twenty-five urfits, seven thousand six
hundred and fifty-four tens of thousandths.
7.
Subtract twelve units, one millionth ; from
Thirty units, ten thousandths.
8.
Subtract one hundred units, eleven tens of thousandths ;
from
Three hundred units, one tenth.
9.
Subtract five hundred and fifty millionths ; from
Ninety-five hundredths.
10.
Subtract ninety-eight units, fifty-four thousand tens of
thousandths ; from
Eight hundred and eighty-seven units, thirty-four thou-
sand tens of thousandths.
11.
Subtract twenty units, seven thousand three hundred
and twenty-one tens of thousandths ; from
Thirty-nine units, eighty-four thousand, three hundred
and twenty-one hundreds of thousandtlis.
12.
Subtract forty units, twenty-five thousand, nine hun-
dred and eighty-three hundreds of thousandths ; from
Eight hundred and forty-one units, six hundred and for-
ty-three tens of thousandths.
13.
Subtract eight units, forty-one tens of thousandths ; from
Seventy, seven units, forty-three thousand and eleven
millionths.
14.
Subtract eight units, one thousand and fourteen mil.
Months ; from
Eight hundred units, twenty-one tens of thousanths.
8*
SO ARITHMETIC. SECOND PART*
15.
Subtract four hundred units, sixty hundredths ; from
One thousand units, three tenths.
16.
Subtract fifteen hundred millionths ; from
Eighteen hundreds of thousandths.
17.
Subtract eighty units, eighty thousandths ; from
Eight hundred units, and eighty millionths.
18.
Subtract two units, seventy-six thousand and eight mil*
lionths ; from
Nine hundred and eighty-seven units, forty-four hun»
dreds of tliousandths.
COMPOUND SUBTRACTION.
A man has 5 yds. 3 quarters of cloth, and cuts off 2
yds. 1 qr. how much is left ?
A man has 6 lbs. 3 oz. of beef, and sells 4 lb. 2 oz.
how much is left ?
If 4 bushels, 3 pecks, are taken from 8 bushels, 5
pecks, how many remain ?
A man has 12 bushels, 6 pecks of grain and sells 7
bushels 5 pecks, how many will remain ?
If 4 yards, 3 quarters, 2 nails, be taken from 6 yds. 4
qrs. 3 nails, how many will remain ?
If 4j£ „ 3s. „ 4d. be subtracted from 6£ „ 8s. „ bd.
how many will remain ?
If the same quantity be added to the minuend and sub-
trahend, is the remainder altered ?
Can you add a certain quantity to the minuend in one
order, and the same quantity to the subtrahend in another
order? Give an example.
If you wish to subtract 1 yd. 3 quarters, from 5 yds.
2 qrs. can you subtract the 3 qrs. from the 2 qrs. ?
What can you do to get the right answer ?
COMPOUND SUBTRACTION. 91
If 4 shillings 4 pence, be taken from 6 shillings 3
pence, how many will remain ?
In which order is the subtrahend larger than the minu-
end ? Can 4 pence be taken from 3 pence ? What must
you do in order to subtract ?
From 10 lbs. 8 oz. subtract 9 lbs. 9 oz.
In which order is the subtrahend larger than the minu-
end ? What must be done in this case ?
From 7 feet 4 inches, subtract 5 feet 6 inches.
In which order is the subtrahend larger than the minu-
end ? What must be done in this case 1
Rule for Compound Subtraction.
Write the subtrahend under the minuend, placing units of
the same order under each other. Subtract each order of
the subtrahend, from the same order of the minuend, and
set the remainder under. If in any order the subtrahend is
larger than the minuend, add as many units to the minuend
as make one of the next higher order ; then add one unit to
the next higher order of the subtrahend.
Example.
Subtract 29£ 19s. Sd. from 36£ \5s. Id.
Placing them according to rule thev stand thus.
<£. s. d.
36 „ 15 „ 7
29 „ 19 „ 8
6 « 15 „ 11
Subtract thus : 8 shillings cannot be taken from 7 ;
therefore add as many units of this order to 7, as are re-
quired to make one unit of tbe next higher order ; that is
12 (as 12 pence make 1 shilling). 12 added to 7 are 19.
Subtract 8 from 19, and 11 remain to be set down.
As 12 pence have been added to the minuend, an equal
quantity must be added to the subtrahend ; therefore car.
ry 1 shilling to the 19 which makes 20. This cannot be
subtracted from 15 ; therefore add to the 15 as many of
02 ARITHMETIC SECOND PART.
this order, as are required to make one unit of the next
higher order ; that is 20. This being added to 15 makes
35. Subtract 20 from 35, and 15 remain to be set down ;
as 20 shillings have been added to the minuend, 1 pound
must be carried to the subtrahend of the next higher or-
der, which makes it 30 ; and this subtracted from 36,
leaves 6 to be written under that order.
Let the following sums be explained as above.
Sterling Money.
£. s. d. s. d. gr.
44 „ 10 „ 2 16 „ 8 „ 2
36 „ 11 „ 8 10 „ 7 „ 4
Troy Weight.
lb. oz. pwt. oz. pwt. gr.
6 „ 11 „ 14 4 „ 19 „ 21
2 „ 3 „ 16 2 „ 14 „ 23
Avoirdupois Weight.
c. qr. lb. lb. oz. dr.
7 „ 3 „ 13 8 „ 9 „ 12
5 „ 1 „ 15 6 „ 12 „ 9
Apothecaries Weight.
3 9 qr- 3 3 &
4 „ 1 „ 17 10 „ 3 „ 1
1 „ 2 „ 15 7 „ 6 „ 1
Cloth Measure.
na.
yd. qr. na. E. E. qr.
35 „ 1 „ 2 67 „ 3 „ 1
19 „ 1 „ 3 21 „ 3 „ 2
Dry Measure.
hi. pk. qt. pk. qt. pt.
65 „ 1 „ 7 2 „ 3 „
14 „ 3 „ 4 1 „ 6 „ 1
SUBTRACTION.
Wine Measure.
93
gal. at. pt. hhd. gal. qt.
21 „ 2 „ tt 13 „ „ 1
14 „ 2 „ 1 10 „ 60 „ 3
Long Measure.
yd. jt. in. m. fur. po.
4 „ 2 „ 11 41 „ 6 „ 22
2 „ 2 „ 11 10 „ 6 „ 23
Land or Square Measure.
A. roods, rods. A. r. po.
29
24
» 1
„ 1
„ 10
„ 25
29
17
>j 2 „
55 * 5>
17
36
Solid Measure.
tons.
116 „
109 „
ft.
24
39
cords
72
41
. ft.
„ 114
„ 120
Time.
yrs.
54
43
mo.
„ 11
» 11
we.
„ 3
„ 3
20 „
min.
41 „
49 „
sec.
20
19
Circular Motion.
9 „ 23
3 „ 7
! „ 45
„40
o
29
19
„ 34 „
„ 40 „
it
54
36
SUBTRACTION OF VULGAR FRACTIONS.
If a boy has 6 ninths of an apple, and gives away 4
ninths, how much remains ?
If he has 8 ninths, and gives away 5 ninths, what re.
mains ?
94 ARITHMETIC. SECOND PART.
If he has 7 twelfths, and gives away 4 twelfths, what
remains ?
In doing those sums let the pupil tell first which is the
minuend and which the subtrahend.
A man has 9 twentieths of a dollar and loses 5 twenti-
eths, how much remains ?
If he has 11 twentieths and loses 7 twentieths, what re-
mains ?
If he has 8 sixteenths, and loses 5 sixteenths, what re-
mains 1
Subtract T 3 5 from r \. Subtract -^ from if.
Rule for Subtracting Vulgar Fractions.
Subtract the numerator of the subtrahend, from the nu-
merator of the minuend, and place the remainder over the
common denominator.
Let the pupil in doing the sums, use the signs in this
way.
Subtract f of a dollar from f .
Alia. 8 8 — 8 .
Subtract ^ from ¥ \. Subtract f f from f f .
(( _6_4_ It ?LO.L t< 2.10. <C .32.0.
900 900* 480 48 0*
It X&. tl JL9 It JL6 It 20_
8 8 0' 3 3 0"
(( _£6_ (C _12J)_ (« _4 <t J.2
5000 5 0* 20 II*
A man owns f of a pasture, and sells f , how much re-
mains his own ?
A boy has if of a guinea, and gives away %%, how
much has he left?
T 9 o from |f, are how many 1 f £ from §f are how
many ?
if from |f are how many 1 ^ from if ? T \ from }f ?
SIMPLE MULTIPLICATION. x
Midtiplication is repeating a number, as often as there
are units in another number.
The number to be repeated, is called the multiplicand.
SIMPLE MULTIPLICATION. 95
The figure expressing the number of times the multipli-
cand is to be repeated, is called the multiplier.
The answer is called the product, because it is the sum
produced by multiplication.
The multiplier and multiplicand are called the factors,
from the Latin word factum, (made,) because they are the
numbers by which the product is made.
There are four processes of multiplication.
The first is Simple Multiplication, where the factors are
whole numbers, and ten units of one order make one unit
of the next higher order.
The second is Decimal Multiplication, where one, or
both the factors are decimals.
The third is Compound Multiplication, where the multi-
plicand consists of orders, in which other numbers be-
sides ten, make units of a higher order.
The fourth is the multiplication of vulgar fractions,
where one, or both the factors, are vulgar fractions.
A boy gives 8 apples to each of 7 companions, how
many does he give to them all ?
A man travels 7 miles an hour, how far will he travel in
9 hours ?
If one pound of raisins cost 11 cents, how much will
6 pounds cost ?
One boy has 7 cents, and another twelve times as many,
how many has the last 1
At six cents apiece, how much will 9 lemons cost 1
At 12 cents a dozen, how much will 8 dozen marbles
cost ?
One pound of sugar costs 9 cents, how much will 5
pounds cost ? 8 pounds ? 11 pounds ? 12 pounds ?
Multiplication has been defined as repeating, or talcing
one number as often as there are units in another num-
ber. Let this process be illustrated by the coins ; thus,
$ d. cts.
2 " 4 " 3
Let the multiplier be 2.
Nov/ the pupil is to take 3 cents, as often as there are
units in 2, and give the answer. Then he is to take 4
dimes as often as there are units in 2, and then 2 dollars
in like manner.
96 ARITHMETIC. SECOND PART.
Let the following sum be"done by the coins.
$ d. cts.
2 " 4 " 4
Multiplied by 3
When the pupil has taken 4 cents three times, he will
have 12 cents. Let a dime be substituted for ten of these
cents, to be caried to the next product, and there remain
two cents, to be placed in the order of cents. Then let 4
dimes be taken 3 times, which make 12, and the one dime
of the other product is added, making 13 dimes. Let a
dollar be substituted for ten of the dimes, and carried to
the next product, and three dimes will remain to be placed
in the order of dimes. Two dollars taken three times,
will make 6 dollars, and adding the one dollar of the other
product, the amount is 7 dollars, to be placed in the order
of dollars.
The pupil should practice in this way until the principle
is fully understood.
Rule for multiplying, when the multiplicand has
several orders, and the multiplier does not exceed
TWELVE.
Place the multiplier below the multiplicand. Beginning
at the right, multiply each order of the multiplicand, by the
multiplier. Place the units of the product, under the order
multiplied, and carry the tens to the next product. Write
the whole of the last product.
Let the pupils at first be exercised thus : —
Example.
249
8
1992
Eight times 9 units are 72 units ; which is 2 units to
be written under that order, and 7 tens to be carried to
the next product. Eight times 4 tens, are 32 tens, and
the 7 tens carried, make 39 tens, which is 9 of the order
of tens, to be written under that order, and 3 hundreds to
be carried to the next product. Eight times 2 hundreds,
are 16 hundreds, and the 3 hundreds carried, make 19
hundreds, which are written down.
SIMPLE MULTIPLICATION. 97
Examples.
ultipl
y 348
by
4.
Multiply
2469
by
6.
u
728
«
5.
14
6923
«
7.
a
4693
C(
6.
u
4593
«
8.
(«
2914
((
7.
a
12468
«
9.
<<
3403
«
8.
<(
42469
«c
10.
<<
6798
((
9.
u
532"; 3
«<
5.
<c
5124
<(
10.
<<
65492
(<
8.
c<
8763
((
11.
When the multiplier consists of several orders, another
method is adopted. For example^**
Multiply 324 by 67.
The 324 is first to be multiplied by the 7 units, accord-
ing to the former rule, and the figures stand thus,
324
67
2268
The 324 is now to be multiplied by the 6 ; what is the
number represented by the 1 Ans. 60 or 6 tens.
If 4 is multiplied by 6 tens, the answer is 24 (ens, or
240. The 4 is to be written in the order of tens, under
the 6, and the 2 (which is 200) is to be carried to the next
product. See below.
324
67
2268
1944
21708 Ans.
The 2 tens, or (20) are next multiplied by the 6 tens,
(or 60) and the answer is 12 hundreds, (1200) and the 2
hundreds to be carried to it, make 1400. The 4 is writ-
ten in that order, and the 1 carried to the next product.
Next the 3 hundreds are multiplied by the 6 tens, and the
answer is 18 thousands, (18000) and the 1 to be carried
Q
98 ARITHMETIC. SECOND PART.
to it, make 19 thousand, which are placed in their orders.
Then the two products are added together, and the an-
swer is obtained.
Let the pupil answer the following questions on the
above sum.
What number does the 6 of the multiplier, represent ?
What number does the 2 represent 1 If they are multi-
plied together, as if they were units, what is the product?
How many ciphers must be added, to express the true
value of 2 tens, multiplied by 6 tens ? How many fig-
ures are at the right hand of both the factors, 2 tens and 6
tens ? Is the number of ciphers added, the same as the
number of figures at'the right hand of both the factors?
What is the answer if the 3 hundreds be multiplied by
6 tens, as if they were units ? How many ciphers must
be added, to make the product express the true value ?
Does the number of ciphers added, correspond to the
number of figures, at the right of both factors ?
By answering the above questions, the pupil will un-
derstand the following principle ?
Figures of any order may be multiplied together like
units, and the true value is found, by annexing as many ci-
phers, as there are figures at the right of both the fac-
tors.
Let the following questions be answered.
Multiplicand r*G9
Multiplier '237
What number is represented by 6 ? by 3 ?
If the 6 is multiplied by the 3, what is the answer, if
the factors are considered as units ? What is the true an-
swer ?
If the 2 is multiplied by 3, what is the answer if they
are considered as units ? what is the true answer ?
What number is represented by 2 ? by 8 f If the 2 is
multiplied by 8, what is the answer if they are considered
as units 1 What is the true answer ?
Let the pupil now learn to multiply the above sum, and
place the figures in the orders to which they belong ;
thus,
SIMPLE MULTIPLICATION, 99
869 Multiplicand.
237 Multiplier.
6083
2607
1738
205952 Answer.
The multiplicand is first multiplied by the 7 of the mul-
tiplier, and the product is 6083.
Then the 3 tens (or 30) are multiplied into the 9 units,
and the answer is 270 ; which is 7 tens to be set in the
order of tens, and 2 hundreds to be carried to the next
product.* Then the 6 tens (or 60) are multiplied by 3
tens, and the product is 1800, and the 2 that were to be
carried make 2000 ; which is 2 of the order of thousands
to be carried to the next product, and to be set in the or-
der of hundreds. Then the 8 hundreds are multiplied by
3 tens, and the answer is 24000, and the 2 to be carried
make 26000 ; which is 6 to be set in the order of thou-
sands, and 2 in the order of tens of thousands.
Next take the 2 hundred as multiplier, and multiply 9
units by it, and the answer is 1800 ; which is 8 to be set
in the order of hundreds, and 1 to be carried to the next
product.
Proceed thus, till all the orders have been multiplied by
the 2 hundred. Then add the several products and the
answer is obtained.
Rule for Simple Multiplication, when the mul-
tiplier HAS SEVERAL ORDERS.
Place the multiplier below the multiplicand, so that units of
tlie same order, may stand in the same column. Multiply
by each order of the multiplier. Write the units of each
product, in the order to which they belong, and carry the
tens to the next product. Add tlie products of the several
orders, and the sum is the answer.
* The cipher is omitted, because as the figure is set under the 8,
we can tell what order it belongs to, without the cipher.
100 arithmetic. second part.
Example.
826
234
3304
2478
1652
193284
Multiply by the 4 units according to the other rule.
Then multiply each order of the multiplicand by the 3
tens (or 30) thus : 6 units multiplied by 3 tens are 18 tens,
which is 8 tens to be written in that order, and 1 of the
order of hundreds to be carried to the next product. 2
tens, (or 20) multiplied by 3 tens (or 30) are 600, and
the 100 carried, makes 700, which is 7 to be written in
the order of hundreds. 8 hundreds multiplied by 3 tens,
(or 30) is 24000 ; which is 4, to be written in the order of
thousands, and 2 tens of thousands to be set in that order.
Lastly, multiply each order of the multiplicand by the 2
hundreds. 6 units multiplied by 2 hundreds, are 12 hun«
dreds, which is 2 hundred to be written in that order, and
1 thousand to be carried to the next product. 2 tens (or
20) multiplied by 2 hundreds, are 4000, and the 1000 car.
ried makes 5000, which is 5 to be placed in the order of
thousands. 8 hundreds multiplied by 2 hundreds, are
160,000, which is 6 tens of thousands, to be written in that
order, and 1 hundred of thousands, to be written in the
order of hundreds of thousands.
Add all the orders of the products, by the rule of com-
mon addition, and the sum is the answer.
Examples.
Multiply 256 by 26 Multiply 4567 by 234
3639 " 329 « 4654 " 496
4638 « 462 « 6789 " 596
5943 « 567 " 5432 « 281
2345 « 234 « 4568 " 362
7892 " 456 « 8382 " 945
If five stands alone (5) of what order is it ? If a ci-
SIMPLE MULTIPLICATION. 101
ipher is affixed, oi what order is it ? How much larger is
the sum, than it was before ? By what number was it
multiplied when the cipher was added ?
If two ciphers are added to the 5, in what order will it
stand ? How much larger is the sum than it was before ?
By what number was it multiplied when the ciphers were
added ?
If three ciphers are added to 5, in what order will it
stand 1 How much larger is the sum than it was before ?
By what number was it multiplied when the ciphers were
added ?
If you wish to multiply 5, by 10, what is the shortest
way? If you wish to multiply 5, by 100, what is the
shortest way ? If you wish to multiply 5, by 1000, what
is the shortest way ?
If you wish to multiply 50, by 2, how would you do it 1
Would it make any difference if you should multiply
the 5 first, and then affix a cipher to the answer 1
If you are to multiply 5000, by 2, can you begin by mul-
tiplying the 5 first ?
If you are to multiply 35000, by 2, can you multiply
the 5 first, and then the 3, and afterwards affix the three
ciphers ?
If you are to multiply 20 by 30, can it be done by mul-
tiplying (he 3 and 2 together, and then affixing 2 ciphers
■to the product ?
Multiply 200 by 20 in the same way.
Rule for Multiplying when the factors are ter-
minated BY CirHKRS.
Multiply the significant figures together, and to their ■pro-
duct annex as many ciphers as terminate both theJactor&.
Note. — All figures are called significant, except ciphers.
ultiply
«
30
3000
by 20
" 9
400
96
by
00
30
<<
200
6
4400
4t
90
<(
2000
" 40
«
100
" 100
•
(44
2400
160
" 2000)
•" 4200)
9*
102 ARITHMETIC. SECOND PART.
When any number is made by multiplying two numbers
together, it is called a composite number.
Thus 12 is a composite number, because it is made by
multiplying 3 and 4 together.
Is 18 a composite number? What two numbers multi-
plied together make 18 1
Is 14 a composite number ? Is 13 a composite num-
ber ? Is 9 a composite number ?
If 12 is multiplied by 8, what is the product? What
are the factors which compose 8 ?
If you multiply 12 by one of these numbers, and the
product by the other, will the answer be the same as if
you multiplied 12 by 8 ?
Let the pupil try and see.
What are the numbers that compose 18 ?
Multiply 123 by 18. Multiply it by one of the num-
bers that compose 18, and the product by the other num-
ber, and what is the result 1
Rule for multiplying, when the multiplier ex-
ceeds 12, AND IS A COMPOSITE NUMBER.
Resolve the multiplier into the factors which compose it,
end multiply the multiplicand by one, and the product by the
other.
Let the following sums be done by the above rule.
Multiply 33 by 20 Multiply 587 by 16
268
329
426
7G54
6543
49 " 6543 " 24
54 " 521 " 27
32 « 72 « 30
96 " 793 " 36
64
40
In multiplication it makes no difference in the product,
which of the factors is used for multiplier or multiplicand ;
for 3 times 4, and 4 times 3, give the same product, and
thus with all other factors. It is in most cases most con-
venient to place the largest number as multiplicand.
DECIMAL MULTIPLICATION. 103
DECIMAL MULTIPLICATION.
In explaining decimal multiplication, it is needful to un-
derstand the mode of multiplying and dividing by the sep-
aratrix.
If we have 2,34 we can make it ten times greater, by
moving the separatrix one order to the right, thus, 23,4.
For 23 units, 4 tenths, is ten times as much as 2 units, 34
hundredths.
It is therefore multiplied by 10.
We can multiply it by 100 by removing the separatrix
entirely, thus, 234, for the 2 units and 34 hundredths, be-
come 234 units, and are thus multiplied by 100.
Whenever therefore we wish to multiply a mixed or
pure decimal, by any number composed of 1 and ciphers,
we can do it by moving the separatrix as many orders to
the right, as there are ciphers in the multiplier.
EXAMPLES.
Multiply 402,5946
" 2,6395
« 4,03956
« 54,0329
« 4,6930
" 3694
« 4,6934
But if the decimal has not as many figures at the right,
as are needful in moving the separatrix, ciphers can be ad-
ded thus. Multiply 2,5 by 1000. Then in order to mul-
tiply by a cipher, it is necessary to move the separatrix
as many orders to the right, as there are ciphers in the
multiplier, 1000, in order to do this, two ciphers must be
added thus,
2500,
Here 2 units, and 5 tenths, are changed to 2 thousands
and 5 hundreds, and of course are made 1000 times lar-
ger, or multiplied by 1000.
In the following examples, in order to multiply by mo-
ving the separatrix, it is necessary to add ciphers to the
right of the multiplicand.
3 y
100
(<
1000
K
10000
K
10
C<
1000
((
100
cc
10000
104 ARITHMETIC. SECOND PART.
EXAMPLES.
Multiply 3,7 by 100
« 2,35 " 1000
2,5 " 10000
" 34,200 « 100000
Multiply 5,2 by 100
30,3 « 1000
3,869 « 10000
« 5,6469 " 100000
Division also, can be performed on decimals, by the use
of the separatrix.
Whenever we divide a number, we make it as much
smaller, as the divisor is greater than one.
If we divide by 10, as 10 is ten times greater than one,
we make the number 10 times smaller.
If we divide by 100, we make the numbers 100 times
smaller.
If therefore we make a number 10 or 100 times smaller,
we divide by 10 or 100.
If we make it 1000 times smaller, we divide by 1000,
&c.
If then we are to divide 323,4 by 10, we must make it
10 times smaller. This we can do by moving the sepa-
ratrix one order to the left, thus, 32,34. If we are to di-
vide by 100, we can do it by moving the separatrix two or-
ders to the left, thus, 3,234.
If we are to divide by 10,000, we can do it by moving
the separatrix 4 orders to the left, thus, ,3234.
Whenever therefore, we wish to divide a pure or mixed
decimal, by a number composed of 1 and ciphers, we can
do it by moving the separatrix as many orders to the left,
as there are ciphers in the divisor.
EXAMPLES.
ivide
32,5 bv 10
Divide 32,69 by
10
it
342,6 " 100
3269,1 "
100
a
469,3 " 1000
2390,4 «
1000
a
46936,7 « 10000
" 12346,95 "
10000
it
23469,8 « 100000
" 15463,96 «.
100000
But if the decimal has not enough figures to enable the
separatrix to be moved, according to the rule, ciphers
must be prefixed.
Thus if we wish to divide 3,2 by 100, we do it thus,
,032. Here the 3 is changed from 3 units, to 3 hundredths.
and of course made 100 times less.
DECIMAL MULTIPLICATION.
105
EXAMPLES.
vide 2,4 by
100
Dividi
3 23,4 by
10000
32,4 "
1000
M
246,4 "
100000
'< 932,5 «
10000
((
293,6 "
100000
21,6 "
100000
U
546,9 "
100000
" 600,7 "
1000000
It
32,3 "
1000000
« 286,9 "
10000000
(<
100,4 "
10000000
" 542,8 "
100000000
M
3094,9 "
1000000
A decimal can also be multiplied, by expunging the sep-
aratrix.
Thus 2,4 is multiplied by 10, by expunging the separa-
trix, thus, 24.
2,56 is multiplied by 100, by expunging the separatrix,
thus, 256.
In all these cases, the decimal is multiplied by a num-
ber composed ofl, and as many ciphers as there are deci-
mals at the right of the separatrix which is expunged.
If you expunge the separatrix of the following decimals,
by what number are thev multiplied ?
2,46. 3,295. 54,6823.
54,63. 89,46321. 5,6432.
How can you multiply 3,1 by 10 ? What is it after
this multiplication ?
How do you multiply 3,12 by 100 ?
this multiplication?
How do you multiply 9,567 by 1000 ?
this multiplication ?
If the separatrix is expunged from 2,52, by what is it
multiplied ?
If the separatrix is expunged from 2,56934, by what is
it multiplied ?
If the separatrix is removed from 5,94321 6, by what is
it multiplied 1
If the separatrix is removed from 3,462*1 , by what is it
multiplied ?
If the separatrix is removed from 3,5 ; by what is it
multiplied ?
If a man supposes he owes $54,23, and finds he owes
10 times as much, what is the sum he owes] How do
you perform the multiplication with the separatrix ?
What is it after
What is it after
106 ARITHMETIC. SECOND PART.
What does the number become after being thus multipli-
ed?
Multiply in the above mode $244,635 by 10, by 100,
and by 1000. What does the sum become, by each of
these operations ?
Divide $244,635 by 10, by 100, and by 1000, with the
separatrix. What does the sum become by each of these
operations ?
Divide and multiply, with a separatrix, $2556,436, by
10, by 100, and by 1000.
If before multiplying, the multiplicand is made a certain
number of times larger, the product is made as much lar-
ger. If the multiplicand is made too large, the product is
as much too large.
For example ;
If we wish to find how much twice 2,3 is, we can change
it to whole numbers, and multiply it by 2, and we know
the answer is 10 times too large. For 23 is 10 times lar-
ger than 2,3, and therefore when it is multiplied by 2, its
product is 10 times too large. If then we make it 10
limes smaller, we shall have the right answer. When-
ever therefore, we wish to multiply a decimal, we can
change it to whole numbers, and multiply it by the rule
for common multiplication. We then can make the pro-
duct as much smaller, as we made the multiplicand larger,
by changing it to whole numbers.
For instance, if we wish to multiply 3,6 by 3, we can
expunge the separatrix, and the multiplicand becomes 10
times too large. We then multiply it as in whole num-
bers thus,
36
3
108
This product is also 10 times too large, and we find the
right answer, by placing a separatrix so as to divide it by
10, thus making it ten times smaller.
In like manner, if the multiplier is increased a certain
number of times, the product is increased in the same pro-
portion.
DECIMAL MULTIPLICATION.
107
If we are to multiply 32 by 2,3, and should by expung-
ing the separatrix, change the multiplier to whole num.
bers, it would make the product 10 times too large, and to
obtain the right answer we must divide the product by 10
with a separatrix, thus making it 10 times smaller.
Multiply 2,5 by 4.
By what number do you multiply, when you expunge
the separatrix of the decimal ?
What is the product of the multiplication after the sep-
aratrix is expunged ? How much too large is this pro-
duct 1
How do you divide this product by the same number as
;ou multiplied the decimal ?
Explain each process as above.
Multiply
u
((
c<
(«
((
It
Multiply
12,46
18,23
,846
36,2
25,36
44,429
92,1234
329
426
362
4689
4693
32678
by
8
9
7
5
4
7
by
2,4
3,5
' 39,5
• 2,36
< 5,462
< 6,H246
Multiply
3,2 1
oy 6
it
52,23
7
«
286,45
« 8
«<
123,678
« 9
<(
32,92
" 12
u
64,64
" 11
((
988,931
9
Multiply
764 by
8,925
(«
2875 "
72,63
c<
30021 "
984,4
cc
8643 "
6,529
u
2875 «
,462
"
7628 "
,3596
Let the multiplier be 2, 4, and the multiplicand is 3,6.
Changing the multiplier to whole numbers, would make
the product ten times too large. Should the multiplicand
be changed to whole numbers, the product would again
be made ten times larger, so that it would be made 100
times too large. Therefore to bring the answer right, we
must divide it by 100, thus making it 100 times smaller.
This is done by the use of a separatrix. 3,6 and 2,4,
when changed to whole numbers and multiplied together,
are 864. This is 100 times too large, and is brought right,
by dividing it by 100 thus, 8,64.
108 ARITHMETIC. SECOND PART.
RULE FOR DECIMAL MULTIPLICATION.
Change the Decimals to whole numbers by expunging the
separatrix. Multiply as in whole numbers. Divide the an-
swer by the product of the two numbers by which the j actors
were multiplied, in expunging the separatrix.
EXAMPLE.
Multiply 8,61 by 4,7.
Change these to whole numbers, and they become 861
and 47. (Here the multiplicand, in expunging the sepa-
ratrix, is multiplied by 100, and the multiplier by 10.)
Multiplying them together, they produce 40467. The
product of the two numbers bv which the factors were
multiplied, (10 and 100), is 1000. Dividing 40467 by it,
gives the answer 40,467.
EXAMPLES.
Multiply 2,37 by 4,6.
By what do you multiply each factor when you remove
the separatrix ? What is the product of the two numbers
by which you multiplied the factors ?
How do you divide by this product ?
Multiply 2,64 by 3,8
" 362,68 " 48,72
6895,40 « 3,651
334,02 « 28,54
2195,334 » 3,2
" 3456,567 " ,51
937,8 " ,84
1234,636 « 36,4
" 765,3 " 1,23
" 89123,002 « ,591
The following common rule for decimal multiplication,
includes all the others, and may be used after understand-
ing the preceding.
RULE FOR DECIMAL MULTIPLICATION.
Multiply as in whole numbers, and then point off in the
product, as many orders of decimals, as are found in both
the factors.
COMPOUND MULTIPLICATION. 109
EXAMPLES.
•ply
3,69
by 3,8
Multiply
,12
by
4,6
<(
18,600
•< 5,9
u
1,94
cc
,600
K
224,7
« 2,3
M
351,9
a
6
«(
9,427
" 3,4
(C
,658
U
,236
COMPOUND MULTIPLICATION.
If 4 grains, 3 penny-weights, are repeated 3 times,
what is the product ?
If 3 yards, 1 quarter, be repeated 3 times, what is the
product ?
If 5 feet, 2 inches, be repeated 4 times, what is the
product ?
If 2 hogsheads, 5 gallons, be repeated 5 times, what is
the product?
If 4 drams, 2 ounces, be repeated 3 times, what is the
product ?
What is 4 times 2 days, 7 hours ?
What is 5 times 3 months, 4 days ?
RULE FOR COMPOUND MULTIPLICATION.
Place the multiplier below the multiplicand. Multiply
each order separately, beginning with the lowest. In the
product of each order, find how many units there are of the
next higher order. Carry these units to the next product,
and set the remainder under the order multiplied.
£. s. d.
1 « 9 " 6
4
5 " 18 "0
Proceed thus : — Four times six pence are 24 pence,
which is 2 units of the next higher order, (or shillings,)
to be carried to that order ; and as no pence remain, a ci-
pher is to be placed in the order of pence. Four times
9 shillings are 36 shillings, and the 2 carried make 38
shillings, which is 1 pound, to be carried to the next pro-
10
110 ARITHMETIC. SECOND PART.
duct, and 10 shillings to be written in the shilling order.
Four times 1 pound is 4 pounds, and the 1 carried, makes
5, which is to be written in the order of pounds.
Let the pupil do the following sums, stating the process
while doing it, as above.
What cost 9 yards of cloth, at 5s. 6d. per yard ?
What cost 5 cwt. of raisins, at £1, 3s. 3d. per cwt. ?
What cost 4 gallons of wine, at 8s. 7d. per gallon ?
What is the weight of 6 chests of tea, each weighing 3
cwt. 2 qrs. 9 lbs. ?
What is the weight of 7 hogsheads of sugar, each
weighing 9 cwt. 3 qrs. 12 lbs. ?
How much brandy in 9 casks, each containing 41 gals.
3 qts. 1 pt. ?
yds.
1. Multiply 14
hhd.
2. Multiply 21
le.
3. Multiply 81
a.
4. Multiply 41
yr.
5. Multiply 20
S,
6. Multiply 1
yds.
7. Multiply 3
1. In 35 pieces of cloth, each measuring 27 J yds. how
many yards? Ans. 971 yds. 1 qr.
2. In 9 fields, each containing 14 acres, 1 rood, and 25
poles, how many acres ? Ans. 129 a. 2 qrs. 25 rods.
3. In 6 parcels of wood, each containing 5 cords and
96 feet, how many cords ? Ans. 34 cords, 64 feet.
4. A gentleman is possessed of 1| dozen of silver
spoons, each weighing 2 oz. 15 pwt. 11 grs., 2 dozen of
tea-spoons, each weighing 10 pwt. 14 grs., and 2 silver
tankards, each 21 oz. 15 pwt. Pray what is the weight of
the whole ? A7is. 8 lb. 10 oz. 2 pwt. 6 grs
ANSWERS.
. qr. na.
yds. qr. na.
3 2 by 11
163 2 2
:. g. qt. pt.
hhd. g. qt. pt.
15 2 1 by 12
254 61 2
m. fur. po.
le. m. fur. po.
2 6 21 by 8
655 1 4 8
r. p
a. r. p.
2 11 by 18
748 38
m. w. d.
yr. m. w. d.
5 3 6 by 14
286 5 2
' II
S. ° ' ''
15 48 24 by 5
7 19 2
• ft-
yds. ft.
87 by 8
29 56
MULTIPLICATION OP VULGAR FRACTIONS. Ill
MULTIPLICATION OF VULGAR FRACTIONS.
MULTIPLICATION WHEN ONLY THE MULTIPLICAND IS A
FRACTION.
A man gave one child three quarters of a dollar, and
another four times as much, how much did he give the
last?
A man has 12 barrels of wine, and takes a half pint
from each 3 times, how many half pints does he take ?
If a man has an ounce of silver, and takes 2 sixteenths
from it 6 times, how many sixteenths does he take 1
How much is 4 times two sixths ?
How much is 5 times two sixths ? 6 times ? 7 times ?
From the above examples it appears, that we can mul-
tiply a fraction by a whole number, by multiplying its nu-
merator.
Let the pupil perform the following sums, first mental-
ly, and then on the slate.
1. What is 9 times ^7
2. What is 3 times T \ ?
3. What is 6 times ^ ?
4. What is 7 times T % 1
5. What is 8 times ^ ?
G. What is 7 times T W ?
7. What is 3 times J T ?
8. What is 5 times ^ ?
9. What is 8 times / T ?
10. What is 4 times j\ 1
11. What is 6 times ^ ?
12. What is 5 times £ ?
13. What is 4 times r \ 1
14. What is 8 times ^ ?
15. What is 5 times JL ?
16. What is 6 times WV t
17. What is 4 times /_?
18. What is 3 times ^ ?
19. What is 9 times / ¥ ?
20. What is 6 times ft 1
In performing these sums on the slate, let the pupil use
the signs, thus :
112 ARITHMETIC. SECOND PART.
Two twentieths multiplied by nine, equals eighteen
twentieths ; and is expressed by signs as follows :
_2_ v Q 18
2 ^ " 2
There is another method, by which the value of a frac-
tion is multiplied, by increasing the size of the parts ex-
pressed by the denominator.
For example, when we wish to multiply T 4 ? by 2, the
most common way is to multiply the numerator by 2,
thus :
jl v 1 8_
12 1 2
But the same effect is produced, if we divide the denom-
inator by 2, thus :
1 2 a ■* — 6
It will easily be seen, that T 8 2 and | are the same quan-
tity. The only difference is, that in one case the unit is
divided into 12 parts and 8 are expressed, and in the oth-
er case, the unit is divided into 6 parts, and 4 are ex-
pressed. In one case, we make twice as many pieces,
and in the other we make them twice as large.
When we multiply the numerator, the number of parts is
multiplied, and when we divide the denominator the size
of the parts is multiplied.
If we multiply T 4 2 by 3, in what two ways can it be
done ?
If we multiply the numerator, what is it that is multi-
plied 1
If we divide the denominator, what is it that is multi-
plied ?
Multiply f by 3 in both ways, and tell what each
method multiplies.
Rule for multiplying when only the multiplicand
is a fraction.
Multiply the numerator, or divide the denominator by the
multiplier.
Let the following sums be performed, and explained as
above.
(<
4
18
(«
((
4
24
(C
H
S
27
((
l<
5
30
<«
((
6
35
it
tt
9
4
((
a
6
4
((
«
_4
4 2
(1
((
6
8 1
M
Multiply |
by
2
«
3
42"
«
7
(i
6
49
«
7
(«
4
64
«
8
a
5
1 5
(C
3
u
'6
8
((
10
u
7
5
«
5
<(
3
56
((
8
«
3
4 8
((
6
((
4
M
11
MULTIPLICATION OF VULGAR FRACTIONS. 113
Multiply T \ by 4
9
6
9
10
7
5
8
6
9
Multiplication where only the multiplier is a
Fraction.
1. If you bave twelve cents, and give away a sixth of
them, to each of four children, how many cents do you
give away ?
Ans. A sixth of twelve cents is two cents. Two cents
given to each of Jour children would be eight cents given
away.
2. If a man has fifteen cents, and gives a fifth of them,
to each of three children, how many does he give away ?
Ans. One fifth of fifteen is three. Three times three
is nine. He gives away nine cents.
From the above examples it appears that when we mul-
tiply by a fraction, we take a part of the multiplicand, and
repeat it a certain number of times. In the last case the
man had fifteen cents, which is the multiplicand. We
take a fifth of it and repeat it three times.
3. If a man had eighteen cents, and gave a ninth of
them, to six different boys, how many cents did he give
away ?
In the above question, what is the multiplicand ? What
part are you to take from it, and how often are you to re-
peat it ?
4. If a man has twelve dollars, and gave a fourth of
them to three different workmen, how many did he give
awav 1 What is the multiplicand ? What part are you
to take from it, and how often are you to repeat it ?
5. How do you multiply twelve by three fourths 1
Ans. We take a fourth of twelve and repeat it three
10*
114 ARITHMETIC. SECOND PART,
times. One fourth of twelve is three. Three fourths are
three times as much. Three times three is nine.
6. How do you multiply eightby three fourths ?
7. How do you multiply eighteen by three ninths ?
8. If you multiply twelve by three, do you make it larger
or smaller 1 If you multiply it by three fourths, do you
make it larger or smaller ?
Why is the multiplicand made smaller when you multi-
ply by three fourths ?
Ans. Because we do not repeat the whole number, but
only a fourth part of it ; and this is repeated only three
times, which does not make it as large a number as the
multiplicand.
9. If you multiply eight by three, do you make it larger
or smaller ? If you multiply it by three fourths, do you
make it larger or smaller ? Why ?
10. Multiply jJ/ieen by two thirds.
] 1. Multiply twenty-four by five sixths.
12. Multiply thirty-two by three eighths.
13. Multiply fourteen by three sevenths.
14. Multiply sixteen by two eighths.
15. Multiply twenty-four by five sixtJis.
Multiplication has been defined, as repeating a number,
as often as there are units in another number.
In multiplying by a fraction, we take such a part of a
number, as is expressed by the denominator, and repeat it
as often as there are units in the numerator.
Thus in multiplying 12 by | we take a sixth part of 12,
and repeat it 4 times, and the answer is 8.
Note. — The propriety of calling the number in the
numerator units, is explained on page 40, where the dis-
tinction is shown between units that are whole numbers,
and units that are fractions. It is shown also on page 57,
where it appears that the numerator may be considered
as whole numbers, divided by the denominator.
In multiplying let the pupil use the signs thus : —
Multiply 12 by f.
12 — 6 = 2
2x3=6. Answer.
16
by
i
4
by
3
4
by
8
4
18
it
a"
<<
3.
6
u
6
24
a
i
8
u
7
8
<<
1 3
2
36
«
1
9
a
1
8
c«
8
1 1
42
U
1
7
a
4
7
<<
V
63
<<
1
9
a
5
9
«
8
a
MULTIPLICATION OP VULGAR FRACTIONS. 115
In doing the above sum what part of 12 is taken ? How
often is it repeated 7
Is the product larger or smaller than the multiplicand ?
Multiply 12 by f .
Is | a proper or improper fraction ?
Is there a whole unit in f ?
Is the product larger or smaller than the multiplicand,
when 12 is multiplied by § ?
Why is it larger when multiplied by g and smaller when
multiplied by J ?
Let the following sums be stated thus ; 16 X f « One
fourth of 16 is 4, and two fourths, is twice as much, or 8.
Multiply
(C
a
a
K
Examples for Mental Exercise.
If you have 14 apples, and give one seventh of them
to each of four boys, how many do you give away ?
Ans. A seventh to 1 boy, would be 2, and four
sevenths, would be four times as much, or 8.
What is a of 14 ?
If you have 48 cents, and give a twelfth of them, to
«ach of two boys, how many do you give away?
What is T \ of 48 ?
A man has 35 sheep, and sells four fifths of them, how
rnanv does he sell ?
A boy has 40 marbles, and loses f of them, how many
does he lose ?
What is 40 multiplied by f ?
What is 36 multiplied by T *V ?
What is f of 21 ? f of 24 1 f of 81 ! 4 of 49 ? f of 64?
I of 16? 4 of 40? I of 45? T \of60? T <W of 96? J of 24?
i of 30? '
What is ft of 18 ? A of 100 ? f of 40 ? f of 28 ? f of
27 ? -A- of 33 ? f of 48 ? f of 81 ? -^ of 144 ? }£ of 99 ?
What is f- of 54? a of 49 ? f of 32 ? f of 81 ? T % of
70 ? -pV of 88? -a of 96 ? f of 16? f of 12 ? f of 18 ! T %
©f24? I of 15? ' r \ of 36?
11g arithmetic second part.
Examples for the slate.
1122 x A
144 X T 6 2
2608 X U
720 X 2V
1335 x ||
578 x A
1912 x if
1357 x \l
545 x U
722 X A
304 X if
420 X T 9 j
If a number is to be both multiplied and divided by two
figures, it makes no difference which is done first, provi-
ded the same figures are used as multiplier and divisor.
For example, let it be multiplied by 2, and dividedby 9.
We can divide first by 9, and then multiply the quo-
tient by 2 ; or we can multiply first by 2, and then divide
the product by 9, and the answer is the same.
Thus 18 multiplied by 2, is 36 ; and this divided by 9
is 4.
Again 18 divided by 9, is 2 ; and this multiplied by 2
is 4.
If then we multiply 12 by f we divide by 4, to find one
fourth of 12, and multiply by 3, to obtain three fourths,
and the answer is 9. But if we should multiply 12 by 3,
and then divide the product by 4, the answer would be
the same. Thus 12 x 3 = 36 and 36 -r- 4 == 9. Thus
9 is the same answer as is obtained by dividing 12 by the
denominator, and multiplying the answer, by the numera-
tor. What are the two ways in which 18 can be multi-
plied by a < What will be the answer, if it is divided by 6
first, and the quotient multiplied by 4 ? What will be the
answer, if it is multiplied by 4 first, and then the product
divided by 6 ?
KuLE FOR MULTIPLYING WHEN ONLY THE MULTIPLIER
IS A FRACTION.
Divide by the denominator, to obtain one part, and multi-
ply by the numerator, to obtain the required number of parts.
But in case this division should, leave a remainder;
Multiply by the numerator first, and then divide the pro-
duct by the denominator.
MULTIPLICATION OF VULGAR FRACTIONS.
117
EXERCISES FOR THE SLATE.
In all these cases it is best to multiply by the numerator
first, and then divide by the denominator. If any remains
after division, place the divisor under it, for a fraction.
Multiply,
1309 by f
II
tt
5463
3204
a
a
5
8
2
s
it
43256
a
6
It
86432
a
8
4
it
3549
a
8
9
tt
54683
a
5
B
ltiply
4081
by
2
4
cc
3042
a
8
ii
5903
a
6
2
it
40938
a
5
9
a
63921
a
A
a
26438
a
4
5
tt
39021
it
S.
8
EXAMPLES FOR MENTAL EXERCISE.
1. If 15 is Jive eighths of some number, what part of
15 is one eighth of that number?
2. If 12 is four sixths of some number, what part of
12 is one sixth of that number ?
3. If 18 is six ninths of some number, what part of 18
is one ninth of that number?
4. If 15 is | of a number, what is } of that number?
5. If 15 is | of some number, what is that number ?
Let such exercises be stated thus.
6. If 15 is Jive eighths, a fifth of 15 is one eighth. A
fifth of 15 is 3. If 3 is one eighth then the whole is 8
times as much, or 24.
7. 24 is f of what number ?
8. 30 is | of what number ?
9. 42 is I of what number ?
10. If a man can do | of a piece of work in 12 days,
how long would it take him to do | of it ?
(Ans.) It would take him only one sixth of the time, to
do one seventh that it does to do f . £ of 12 is 2. It
would take him 2 days.
Let the remaining sums be stated as above.
11. If a man bought 2. G f a barrel of wine for 18 dol-
lars, how much will } cost ?
12. How much will the whole cost?
118 ARITHMETIC. SECOND PART.
13. Bought | of a chaldron of coal for 24 shillings,
how much will i cost ? How much will the whole cost ?
14. If 15 is f of some number, what is one eighth of
that number ?
15. What is the whole of that number? If 23 is £ of
one number, what is that number ?
16. If a man bought ■* of a cask of brandy for 42 dol-
lars, what is i worth ? what is the whole worth ?
17. If a of a month's board cost 12 dollars what is it a
month 1
18. If | of a cord of wood cost 16 shillings, what
would i cost and what would the whole cost ?
19. 28 is | of what number 1
20. 48 is | of what number ?
21. 56 is | of what number ?
22. 32 is T V of what number ?
23. How many times is 4 contained in 5 ?
(Ans.) Once and one over.
24. What is i of 1 ? What is } of 1 ?
25. What is i of 1 ? What is fof 2 ? What is § of 2 ?
26. What is f of 12 ? What is i of 1 ? What is ± of 2 ?
What is i of 4? What is § of 4?
27. What is § of 4? What is £ of 2? What is f of 1?
28. How many units in £ of 2?
29. How many units in £ of 3 ?
30. How many units in | of 5 ?
31. How many units in ^ of 11 ?
32. How many units in f of 6?
33. How many units in | of 18 1
34. How many units in | of 16 ?
35. How many units in l f of 21 ?
It will be seen that in fractions, as in whole numbers, it
makes no difference in the product, which factor is used
as multiplier.
For 12 X f = 9
And f X T V= V = 9
Here when the whole number is used as multiplier, the
answer is an improper fraction, which, if changed to whole
numbers is 9.
Multiply 18 by | and | by 18, and tell in what respects
the answers differ
MULTIPDICATION OE VULGAR FRACTIONS. 119
Multiply ^ by 14, and 14 by ^.
Is there any difference in the value of the answers 1
In what respect do they differ ?
Multiplication where both factors are frac-
tions.
1. If we had i an orange and should give away half of
this i what part of an orange should we give away ?
How much is \ of \ ?
2. If we have \ of an orange, and should give away
I of it, what part "of a whole orange should we give
away ?
(Ans.) If the two halves of any thing be divided into
4 pieces each, the whole is divided into 8 pieces. Ta-
king i of one of these halves then, is taking | of the
whole.
iofiis}
3. If we have j of an orange, and give away half of it,
what part of the whole orange do we give away 1
Ans. If an orange is divided into 4 pieces, and each of
these pieces are halved, the orange is divided into 8 pieces,
and each piece is i of the whole.
iofi is 1.
4. If you receive i of an orange, and you give { ot it
away, what part of the whole orange do you give away ?
Ans. The orange is divided into 3 parts ; if each of
these parts is divided into 4 parts, the whole orange would
be divided into 12 parts, and each part is T ' f of the
whole.
i nf 1 i<! JL
4 Ul 3 l ° T2
5. If you have an apple and it is cut into 5 equal parts,
what part of ihe apple is each piece ? If each piece is
cut into 3 equal parts, what part of the whole apple is
each piece ?
Ans. If an apple is cut into 5 equal parts, each part is
one fifth of the whole, and if each of these pieces is divi-
ded into 3 parts, each part is ,l of the Avhole.
120 ARITHMETIC. SECOND PART.
6. If you have an orange and it is divided into 3 equal
parts, each part is one third, if each i is divided into 6
equal pieces, what part of the i is each piece ?
7. What part of the whole orange is each piece ?
8. If a loaf of hread is cut into 4 equal parts, each part
is \. If each I is divided into 5 equal pieces, each piece
is i of the i, and J^ of the whole loaf, \ of \ then is J^.
9. If a sheet of paper is cut into 5 pieces, each piece
is ^. If each | is cut into 3 equal pieces, each piece is
i of the i, and r l j of the whole, i of i then is -Jj.
10. If a yard of cloth is cut into 8 equal pieces, and
each piece is then cut into 3 equal parts, what part of the
whole is each piece ?
11. If a bushel of apples is divided into fourths of a
bushel, and each fourth is divided into 6 equal portions,
what part of the whole is'each portion 1
12. If you divide a pine apple into 3 equal parts, and
each of those parts into 6 equal pieces, what part of the
whole is each piece ?
13. If you have \ of a dollar and wish to give | of it
to each of 7 children, what part of the whole dollar do
you give to each ?
14. If you have 1 of a lb. of raisins and wish to divide
it equally between 3 children, what part of a lb. do you
give to each ?
15. If you have J- of a yard of muslin, and divide it in-
to 8 equal pieces, what part of a is each piece, and what
part of the whole yard is each piece ?
16. What part of a unit is £ of •*■ ?
Ans. If a unit is divided into 6 parts, and each of these
parts into 8, the unit would be divided into 48 parts, and
each part is J ? of the whole.
Let the following sums be stated in the same manner.
17. What part of a unit is i of } 1
18. What part of a unit is i of £?
19. What part of a unit is \ of i ?
20. What part of a unit is | of i 1
21. What part of a unit is i of | ?
22. What part of a unit is £ of } ?
23. What part of a unit is £ of T ' T ?
24. What part of a unit is \ of \ ?
MULTIPLICATION OF VULGAR FRACTIONS. 121
25. What part of a unit is 1 of T V ?
26. What part of a unit is i of } 1
27. What part of a unit is } of T ^ ?
28. What is i of \1
29. What is J of T V 1
30. What is } of T \ 1
31. What is i of 4 ?
32. What is i of i?
33. What is j of |? What is i of i ?
34. What is i of i n What is i of \ ?
35. What is | of i ? What is \ of ~\ ?
36. What is i of i ? What is \ of \ ?
37. What is \ of T \ ? What is Jg. of \ ?
38. What is i of i ? What is \ of \ ?
39. What is a of T V ? What is T V of \ %
40. What is \ of £l What is \ of J ?
41. What is £ of T ' T ? What is'i of T S T ?
42. What is T ' r of \ 1 What is \ of T V ?
43. What is \ of t l ? What is | of i f
After finding \ of one third we knoiv that ± of two thirds
is twice as much.
1. What i of £ ? What is i of § ?
2. What is i of i ? What is i of f ?
3. What is a of £ 7 What is j of £ ?
4. What is \ of i ? What is | of f ?
5. What is i of i ? What is f of f ?
6. What is i of x ? What is i of f ?
7. What is i. of | ? What is i off'?
8. What is i off?
9. What is J of £ ? What is J- of f ? What is | of $ 1
What is | of f ? What is \ of f ?
10. What is f of i ? What is a of f ?
11. What is J- of i ? What is A of f ?
12. What is a of T V 1 What is a of , \ ?
13. What is i of T \ ? What is A of T 9 /?
14. What is } of f ? What is i of ^ ?
15. What is 4 of f ? What is T V of | ?
16. What is T V off ? What is \ of f ?
17. What is i of ft ? What is T \ of ^ ?
18. What is -' of }i ? What is l of jf ?
19. What is $ of ^ ? What is § of f ?
11
122 ARITHMETIC SECOND PART.
20. What is i of | ? What is i of T \ ?
21. What is } of f ? What is | of $ 7
After finding one part of a fraction, we find the other
parts by multiplication.
Thus after finding what one fourth of a fraction is, we
can find three fourths by multiplying by 3.
Thus i of | is ^j, therefore £ of § is 3 times as much
1. What is i of f? What is | of f ? What is £ of f 7
2. What is i of f ? What is £ of f ? What is § of f ?
3. What is £ of f ? What is § off ? What is $ of $ 7
Let the pupil reason thus : What is f of f 1 One sixth
of one third is T \. One sixth of two thirds is -fj. Four
sixth of two thirds is 4 times as much, or T \.
4. What is £ off? Whatisioff? What is § of £?
5. What is | of |? What is f of £ ? What is f of f ?
What is f of | ? What is -f of £ ?
6. What is i of & 7 What is ^ of j4 ?
7. What is 11 of|? What is ]-# off?
8. What T % off ? What is f of f ? What £ of $ ? What
is £ off? What is f off ?
In multiplying one fraction by another, we are to take
a certain part of one fraction, as often as there are units
in the numerator of the other fraction.
Thus, if we are to multiply £ by f we are to take a sixth
of £ four times.
To explain the rule for multiplying, when both factors
are fractions, take an example.
What is f of f 7
One fifth off is ^\, and this is made by multiplying the
denominator 6, by the denominator 5.
Three fifths of f is three times as much or ||, and this
is made by multiplying the numerator 4, by the numera-
tor 3.
Therefore multiplying the denominators together ob-
tained one fifth off, and multiplying the numerators togeth-
er, obtained three fifths.
SIMPLE DIVISION. 123
Rule for Multiplying when both factors are
fractions.
Multiply the denominators together to obtain one part, and
the numerators together to obtain the required number of
parts.
In performing these sums upon the slate, let the pupil
use the signs thus :
Multiply | by ft. f X ft = U
Exercises for the Slate.
What is ft of ft ? What is f of £ 1 What is f of % 1
What is f of | ?"What is f of Jf ? What is ft of £f 1
What is if of U » What is 3 « of i|| ? What is ^
of 2.4-6 o ?
Multiply H by j f . Multiply ||f by fft.
Multiply Hi by Jft. Multiply Iff by Hf.
DIVISION.
Division is finding how often one number is contained
in another, and thus finding what part of one number is
another number.
The number to be divided is called the Dividend.
The number by which we divide is called the Divisor.
The answer is called the Quotient.
What is left over, after division, is called the Remainder.
There are four kinds of division.
The first is Simple Division, in which both the dividend
and divisor are whole numbers, and ten units of one
order, make one unit of the next higher order.
The second is Compound Division, in which other num-
bers besides ten, make units of higher orders.
The third is Division of Vulgar Fractions, in which the
dividend or divisor (or both) are Vulgar Fractions.
The fourth is Decimal Division, in which the dividend,
or divisor, (or both) are decimal fractions.
124 ARITHMETIC. SECOND PART.
SIMPLE DIVISION.
How many 9 cents are there in 63 cents ?
What part of 63 cents is 9 cents 1
How many times is 8 contained in 56 ?
If 8 is contained 7 times in 56, what part of 56 is 8 ?
If 56 is divided by 8, how much smaller is the quotient
than the dividend ?
How many 7 dollars are there in 42 dollars ?
What part of 42 dollars is 7 dollars ?
How many times is 6 contained in 66 ?
If 6 is contained 11 times in 66, what part of 66 is 6?
If 66 is divided by 6, how much smaller is the quotient
than the dividend ?
There are many numbers which cannot be divided into
equal parts, without making a fraction. For example, if
we wish to divide 7 apples into two equal portions, we
should have for answer 3 apples and i of an apple.
If we had 13 apples, and wished to give a third of them
to each of 3 friends, we should divide the 13 by 3, and the
answer would be 4, and 1 left over. That is, we could
give 4 apples to each of the 3 friends, and one would be
left to divide among them. This divided by 3, (or into 3
equal parts) would give a third to each one. 13 then, di-
vided by 3, gives 4 and i as answer.
If you are to divide 7 apples equally among 3 persons,
how many ivhole apples would you give to each, and what
would remain to be divided ?
If you had 14 oranges, and wished to divide them equal-
ly among 6 persons, how many whole oranges would you
give each ?
How w.ould you divide the two that remained ?
Ans. Divide each into 6 equal parts, and give one of
the parts of each orange to the 6 persons. Each person
would then have 2 oranges and f- .
If you have two apples, each cut into 12 parts, and take
4 of these parts from each apple, how much do you take ?
Ans. T V For yL from each apple makes T \ in the whole.
If we take 9 twelfths from each of the two divided ap-
ples, we shall have {§ in the whole.
Now this is not }§ of one apple, for nothing has more
SIMPLE DIVISION.
125
than 12 twelfths. Whenever therefore we find an improper
fraction, we know that more than one unit has been divided.
What part of 13 apples is 3 apples and \ of an apple ?
Ans. It is a fourth of 13, because 4 times 3 and \ make
13.
What part of 5 is 1 ? is 2 ? is 3 ? is 4 ? is 6 ?
In the last question we reason thus : If 1 is one fifth of
5, 6 must be 6 times as much, or f of 5.
What part of 8 is 1 ? is 4 ? is 7 ? is 9 ?
What part of 15 is 1 ? is 2 ? is 3 ? is 14 ? is 19 ?
What part of 10 is 1 ? is 2 ? is 5? is 9? is 11 ? is 20 ?
What is a sixth of 19 1 What is a fourth of 21 ?
What is an eighth of 26?
If you had 19 pears, and divided them equally among
6 persons, how much did you give to each ?
What part of 19 is 3 and i.
Ans. As there is 6 times 3 and | in 19, it is ± of 19.
When one number is placed over another, it signifies
that the upper number is divided by the lower.
Thus, | signifies that the 3 is divided by 4. For a
fourth of three things is 3 fourths, and f signifies either
3 fourths of owe thing, or a fourth of 3 things.
If you wish to divide 3 dollars into 5 equal parts, what
would it be necessary to do, before you could divide them !
Ans. Change them to dimes.
What would be the answer ?
If you wished to divide 4 dimes into 10 equal parts,
what would it be necessary to do before you could divide
them?
What would be the answer ?
(Let this be shown by the coins.)
How can 3 dollars be divided so as to give ten of the
class, each an equal part ?
Ans. Change the dollars to dimes, and then dividing
them into ten equal parts, there will be 3 dimes for each
of the ten.
Divide $1,2 so as to give 6 scholars, each an equal part.
Divide $2,4 so as to give 8 scholars, each an equal
part ?
Divide 1 dime equally between two scholars.
Divide 1 dime 5 cents equally between 3 scholars.
11* "
126 ARITHMETIC. SECOND PART.
If 1 dime 8 cents are divided by 6, what is the answer?
If 3 dimes 9 cents are divided by 6, what is the quo-
tient, and what the remainder ?
If 5 dimes 6 cents are divided by 7, what ate the quo-
tient and remainder?
If 4 dimes 7 cents are divided by 6, what are the quo-
tient and remainder ?
In the above sums, it will be seen that when one order of
the dividend will not contain the divisor once, it is reduced,
and added to the next lower order, and then divided.
Thus when 4 dimes, 6 cents were to be divided by 6, the 4
dimes were changed to cents, and added to the 6 cents,
and then divided.
It will also be seen, that the quotient and the remainder
are always of the same order as the dividend.
Thus if 4 dimes 7 cents are divided by 6, the 4 dimes
are reduced, and added to the cents, and the quotient is 7
cents, and the remainder is 5 cents.
Thus, also, if 17 thousands are divided by 5, the quo-
tient is 3, and 2 remainder. The 3 is 3 thousands, and the
2, is 2 thousands.
If the order of the dividend were millions, the quotient
and remainder would also be millions.
If the order were tens the quotient and remainder
would also be tens.
If we divide 8 tens by 3, the quotient is 2 tens, and the
remainder 2 tens.
When the dividend has several orders, we divide each
order separately, beginning with the highest orders. This
is called Short Division.
If there is any remainder, after the division of each or-
der, it is changed to the next lower order, added to it, and
then divided.
For example. Let 9358 be divided by 4.
We first divide the 9 thousands by 4, add the remain-
der to the 3 hundreds and divide that. Then divide the
tens and units.
Place them thus : 4)9358
2339|
SIMPLE DIVISION. 127
The 9 thousands is first divided. In 9 units there would
be 2 fours, and 1 remainder. But as this is 9 thousands,
the quotient and remainder must be the same order as the
dividend, and the 2, is 2 thousand fours, and is set under
the 9 in the thousands order. The remainder also is 1
thousand, and is changed to hundreds and added to the 3,
making it 13 hundred. This is then divided by 4. The
quotient is 3 hundreds, which is put under that order, and
the 1 hundred that remains, is changed to tens and added
to the 5 tens, making 15 tens. This is divided by 4, and
the quotient is 3 tens, which is set in that order. 3 tens
remain which, changed to units and added to the 8, make
38 units. This is divided by 4, and the quotient is 9 units,
which is put in that order. 2 units remain, which are di-
vided by the 4 thus f .
9358, then, contains 4, 2 thousands of times, 3 hundreds
of times, 3 tens of times, and 9 units of times. The 3 left
over, is f of another time.
Let the pupil in performing each operation on the slate,
explain it thus :
7)2496
366 f
7 is contained in 24 units 3 times, in 24 hundreds, 3
hundred times, which are set in the order of hundreds.
3 hundred are left over, which changed and added to the
9 tens, make 39 tens.
7 is contained in 39 tens, 5 tens of times, which are set
in the order of tens. 4 tens are left over, which, changed
and added to 6, make 46 units.
Divide 46 units by 7, and the answer is 6 unite, which
are set in that order, and 4 remain, which have the 7 set
under them, to show that they are divided by 7.
Rvle for Short Division.
Divide the highest order, and set the quotient under it.
If any remains, reduce and add it to the next lower order,
and divide as before. If the number in any order, is less
iiian the divisor, place a cipher under it in the quotient ; then
128
ARITHMETIC. SECOND PART.
reduce and add it to ike next lower order, and divide as be-
fore. If any remains when the lowest order is divided,
place the divisor under it as a fraction.
Examples.
Divide
3694 by
3
Divide 3456 by
3
4329 «
4
7892 "
4
6548 "
5
3456 "
5
3621 "
6
7892 "
6
4638 «
7
1234 "
7
29639 "
8
5678 «
8
36964 "
9
91234 "
9
24697 «
10
56789 "
10
36941 "
11
12345 «
11
1263 "
12
67891 "
12
When both the divisor and dividend, have several orders,
another method is taken called Long Division. Let
6492 be divided by 15. In performing the operation de-
scribed below, we set the figures thus.
Dividend.
Divisor 15)6492(432 }§ Quotient.
60
49
45
42
30
12
We first take as many of the highest orders as would,
if units, contain the divisor once and not more than 9
times. In this case we take 64 hundreds. Now we can-
not very easily find exactly how many times the 15 is con-
tained in 64 hundreds. But we can find how many hun-
dreds of times it is contained thus. As 15 would be con-
tained 4 units of times, in 64 units, it is contained 4 hun-
dreds of times, in 64 hundreds. Which 400 is to be set
in the quotient, (omitting the ciphers.)
As we have found that the dividend contains 15, 4 hun-
dreds of times, we subtract 4 hundred times 15 from the
dividend, to find how often 15 is contained in what re-
SIMPLE DIVISION. 129
mains. 400 times 15 is 60 hundreds (6000) which sub-
tracted from the 64 hundreds, leaves 4 hundreds.
This 4 hundreds changed to tens, and the 9 tens of the
dividend put with it, make 49 tens. We now find how
many tens of times the 15 is contained in the 49 tens, thus :
as 15 would be contained 3 units of times in 49 units, it is
contained 3 tens of times in 49 tens, which 3 tens is set in
the quotient. We now subtract 3 tens of 15 (or 45 tens)
from the 49 tens, and 4 tens remain. These are changed
to units and have the 2 units of the dividend put with
them, making 42 units. 15 is contained in 42 units 2
units of times, which is set in the quotient. Twice 15
from 42 units, leave 12, which is \\ of another 15. The
15 then, is contained in the dividend, 4 hundreds of times,
3 tens of times, 2 waits of times, and }§ of another time,
or 432 times, and }| of another time.
Again, divide 6998 by 24.
To do it we first find how many hundreds of times, the
dividend contains the divisor, and subtract these hundreds ;
then how many tens of times, and subtract these tens ;
then how many units of times and subtract these units ;
and then what remains has the divisor set under it.
Let the pupil in doing sums explain them as below.
24)6998(201 if
48
219
216
38
24
14
24 is contained in 69 units, 2 times ; in 69 hundreds, 2
hundred times. 2 hundred times 24 is 48 hundred, which
subtracted from 69 leaves 21 hundred.
130 " ARITHMETIC. SECOND PART.
21 hundreds are 210 tens, and the 9 tens of the dividend
brought down, make 219 tens.
24 is contained in 219, 9 times ; in 219 tens, 9 tens of
times. 24 multiplied by 9 tens, is 216 tens, which sub-
tracted from 219 tens leaves 3 tens.
3 tens are 30 units, and the 8 units of the dividend
brought down make 38 units. 24 is contained in 38
units once, and 14 over, which is if of another time.
The dividend then contains the divisor 2 hundreds of
times, 9 tens of times, 1 unit of times, and if of another
time, or 291 times and if of another time.
Thus it appears, that in Long Division, each quotient
figure, when set down, does not show the exact number of
times the divisor is contained in the order which is divided ;
but it shows, that the divisor is contained so many times
as the quotient figure expresses, and then, a process fol-
lows for discovering how many more times it is contained.
Let the pupil do the following sums, and explain them
as above, until perfectly familiar with the mode.
Divide 2479 by 14 Divide 3568 by 16
" 1954 " 18 « 5896 " 23
" 36964 " 17 « 38907 " 21
" 29006 " 28 " 46032 " 36
Rule for Long Division.
Place the divisor at the left of the dividend, and draw a
line between. Take as many of the highest orders as would,
if units, contain the divisor once, and not more than 9
times. Divide the orders so taken, as if they were units.
Place the quotient fgxire at the right of the dividend, and draw
a line between. Multiply the quotient and the divisor to-
gether, and subtract them from the part of the dividend al-
ready divided. To the remainder, add as many of the next
undivided orders of the dividend as would enable it, if units,
to contain the divisor once, and not more than 9 times, and
then divide as before.
If it is needful to add more than one order of the dividend
to any remainder, (to enable it to contain the divisor) put one
cipher in the quotient for every additional order. If any
SIMPLE DIVISION. 131
remains after dividing the unit order, put the divisor under
it for a fraction.
Examples.
Divide 2649
by
12
Divide
3294
by
14
tt
2468
«
16
M
64329
«
16
tt
1234
a
17
(<
5678
a
18
tt
56789
tt
19
tt
8234
n
36
It
35673
n
59
it
76542
a
41
((
45678
it
256
tt
96743
tt
348
((
912345
it
481
It
59624
n
562
It
678122
tt
984
n
23864
ti
541
It
34568
a
639
It
35469
it
856
tt
543219
a
656
a
1459862
a
942
It
678912
a
9481
tt
724368
tt
2586
tt
9876533
it
6002
it
159864
n
2851
Examples foe Mental Exercises.
1. Bought 12 pounds of raisins for 3 shillings a pound,
how many dollars did they cost ?
State the process thus. If one pound cost 3 shillings,
12 pounds cost 12 times as much, or 36 shillings. As
there are 6 shillings in a dollar, they cost as many dollars
as there are sixes in 36.
Let the following sums be stated in the same manner.
2. Bought 5 bushels of peaches at 4 shillings a bushel,
how many dollars did they cost ?
3. How many peaches at 4 cents each must you give
for 9 oranges at 5 cents apiece.
State the last sum thus. If one orange cost 5 cents, 9,
cost 9 times as much, or 45 cents. As each peach is
worth 4 cents, you must give as many peaches as there
are fours in 45.
4. If you buy 10. yards of cotton, at 5 shillings a yard,
and pay for it with butter at 2 shillings a pound, how ma-
ny pounds will pay for it ?
5. How many apples at 4 cents each, must you give for
3 pine apples at 12 cents each 1
6. If you buy 48 bushels of coal for 12 cents per bush-
132 ARITHMETIC. SECOND PART.
el, and pay for it with cheese at 10 cents per lb. how ma-
ny pounds do you give ?
7. How much rye at 5 shillings a bushel must you give
for 12 bushels of wheat at 8 shillings a bushel.
8. How much cloth worth 9 shillings a yard must you
give for a firkin of butter worth 12 dollars ?
(Change the dollars to shillings.)
9. How many dozen of eggs at 9 cents per dozen must
be given, for 3 yards of cotton worth 20 cents per yard?
10. If you have 8 pine apples worth 9 cents each, and
your companion has 9 quarts of strawberries worth 8 cents
a quart, which he gives to buy the same worth of pine ap-
ples, how many pine apples must you give him ?
COMPOUND DIVISION.
Divide £4 „ 8*. by 2.
Divide £6 „ 12s. by 3.
If 2 dresses contain 24 } ds. 2 qrs. how much in each
dress ?
If 3 silver cups weigh 9 lbs. 6 oz. what is the weight
of each ?
In division we find how often one number is contained
in another, and thus what part of one number is another.
Thus if we divide 8 lbs. 16 oz. by 4, we can either say
how many times is 4 contained in 8 and in 16, or we can
say what is one fourth of 8 lbs. and 16 oz.
If there is any remainder in dividing one order, it must
be changed to unit3 of the next lower order and added to
it and then divide again.
In doing the sum we place the figures thus.
£ s. d.
3)4 " 18 " 9
1 «' 12 « 11
We proceed thus in explaining the process.
A third of 4£ is 1£ which is set under that order, and
there is 1£ remaining which is changed to 20 shillings
and added to the 18 making 38. A third of 38 shillings is
SIMPLE DIVISION. 133
12 shillings, which are set under that order. 2 shillings
remain which are changed to 24 pence and added to the
9d. making 33 pence ; a third of 33 pence is 11 pence
which are set in that order.
Let the following sums be performed and explained as
above.
Divide 22£ 11*. 6d. by 6.
At 2£ 8s. 6d. for 6 pair of shoes what is that a pair ?
If 9 silver cups weigh 31b. 6oz. 8pwt. 3qr. what is the
weight of each ?
If 8 dresses contain 59 yds. 3 qrs. 2n. how much in
each dress ?
If the divisor exceeds 12 and is a composite number di-
vide the sum by one of the factors as above and the an-
swer by the other.
Examples.
Divide 2£ « 8s. " lid. « 4qr. by 44.
If 18 gal. " Cqr. " 4g. of brandy be divided equally
into 28 bottles how much does each contain ?
If 24 coats contain 62 yds. 3 qrs. 4 na. how much does
each contain ?
If 32 teams be loaded with 40T. 16 cwt. 3 qrs. how
much is that for each team ?
If the divisor exceeds 12 and is not a composite number
the following method is used.
Let the figures be placed thus.
£. s. d. £. s. d.
139)461 « 11 « 11( 3" 6 " 5
417
44
20
891
834
» 57
12
695
695
12
134 ARITHMETIC. SECOND PART.
We first divide the pound order and 3 is the quotient
figure, which is of the pound order because the dividend is
pounds. This is put in the quotient with the £ put over
it to indicate its order.
In order to find the remainder we subtract the product
of the quotient and divisor from the 4G1.
The remainder is 44£. This must be changed to
shillings which is done by multiplying it by 20 and then
the 11 shillings of the dividend are added.
This sum is then divided by 139 and the quotient fig-
ure is 6, which is of the shilling order and must be put in
the quotient under that sign. Proceed as before till the
orders are all thus divided.
Let the following examples be performed and explain-
ed as above.
Divide 239£ " 16s. " 4d. « Sqr. by 123.
If 230 yds. of cloth cost 49=£ 19s. lid. what was that
per yard 1
Note. — Change the pounds to shillings first.
If 349 cwt. 3 qrs. 12 lbs. is contained in 204 barrels
how much is in each barrel.
If 42 cwt. of tobacco cost 826^ 18s. 9d. what is that
per lb.
Rule for Compound Division.
If the divisor does not exceed 12, divide each order sepa-
rately, beginning with the highest, remembering to make the
quotient figure of the same order as the dividend.
Whenever there is a remainder change it and add it to
the next lower order and divide as before.
If the divisor exceeds 12, either resolve it into factors and.
divide first by one and then by the other, or proceed after
the manner of long Division.
T. cwt. lb. oz. dr.
Divide
29 « 13 « 25 " 12 " 13 by 6
lb. oz. pwt. grs.
Divide
7 « 10 " 15 " 2 by 5.
DIVISION OF VULGAR FRACTIONS. 13.")
yds. qrs. na.
Divide 76 "3 "2 by 4.
deg. m. fur. pol. ft. in. bar.
Divide 97 " 55 " 7 « 35 « 4 " 2 1
by 7.
£ s. d. qrs.
Divide .25 " 16 " 10 " 3 by 9.
Division where only the Divisor is a Fraction.
If we have 3 apples, how many -i in the whole ? Ans.
In one apple there are two halves, and in three apples there
are three times as many, or six halves.
If we have 6 dollars how many i in the whole ? Ans.
In one dollar there are three thirds and in six dollars
there are six times as many, or eighteen thirds.
If we have 9 apples how many i ?
In 8 apples how many J- ?
In 12 apples how many | 1
In 7 apples how many jj ?
It thus appears that when we divide by a fraction (un-
less it be an improper fraction) the quotient is larger than
the dividend.
Thus 12 divided by \ is 48, for there are 48 one fourths
in 12 units.
Again 9 divided by i is 27, for there are 27 one thirds
in 9 units.
How many i in 8 ?
How many £ in 12 ?
Divide 7 by } Divide 6 by i Divide 12 by i Divide
lObyi Divide 8 by i Divide 11 by i Divide 12 by
T ! 2 Divide 9 by £.
How many + in 8 ? How many + in 7 ?
If you divide 8 by 1 the answer is 24, for there are 24
one thirds in 8. But if we are to divide 8 by | there will be
but half as many. For there is but half as many two
thirds as there one thirds in a number. Therefore if 8 di-
vided by i is 24, when divided by ^ it is half as much,
or 12.
How many f in 3 ?
136
ARITHMETIC. SECOND PART.
Ans. In 3 there are 18 one sixths and half as many two
sixths, or 9.
How many f in 12 ?
How many f in 2 ?
How many | in 4 1
How many | in 6 ?
How many £ in 3 ?
Divide 4 by f Divide 5 by £ Divide 3 by f Divide
3 by | Divide 2 by § Divide 7 by § Divide 5 by T %
Divide 12 by f .
If you have 12 yards of long lawn and wish to cut a
number of handkerchiefs of f of a yard each, how many
can you make from the whole piece 1
If you have 4 oranges and wish to give f of an orange
to your mates, to how many could you give them ?
If you have 4 pounds of rice to distribute to the poor,
and are to give f of a pound to each person, to how many
persons can you give 1
If a reservoir is filled by a spout in | of an hour, how
many limes would the cistern be filled in 9 hours ?
If a pound of raisins can be bought for f of a dollar,
how many pounds can you buy for 4 dollars ?
If | of a barrel of flour will last a family one week,
how long will 6 barrels last ?
If a cow eats | of a ton of hay a month, how long
would 4 tons last her ?
ii £ of a barrel of flour last a family one week, how
long will 10 barrels last ?
It is seen by the preceding examples, that when a num-
ber is to be divided by a fraction, it is multiplied by its de-
nominator, and divided by its numerator.
Thus if we are to divide 2 by f we multiply by the de-
nominator 4 to change 2 into fourths and then divided by
the 3 to find how many three fourths there are.
Divide 3 byf.
Why do you multiply by the denominator ? Why do
you divide by the numerator ?
division of vulgar fractions. 137
Rule for Fractional Division where only the di-
visor is a fraction.
Multiply the dividend by the denominator, and divide the
product by the numerator.
EXAMPLES for
THE 1
SLATE
•
Divide
23
by
4
7
Divi
de
364
by
<«
25
II
6
1 2
<<
24
t<
«
32
a
8
1 2
(<
21
si
<(
325
a
9
1 5
ci
486
ii
u
9470
(I
8
ci
381
ii
«
342
((
IS
8 4
(i
542
ii
<<
G81
<(
JUL
4 9 8
II
232
II
«
3292
K
JUL
5 i a
((
4285
II
9
a
4
7
9
IS.
4 1
I 8
s_2
34_
2 1 5
6 5
12.
1 9
Division where the Dividend only is a Fraction.
When the dividend only is a fraction, and we divide it
by a whole number, we are to find how many parts of a
time, a certain number is contained in certain parts of a
unit.
Thus if we divide \ by 2, we know that \ does not con-
tain 2 units once, but we can find what part of one time the
£ contains the 2.
If \ is divided by 1 unit, we find that it contains it, not
once, but i of once. It can contain two units but half as
many times as one unit. Therefore ^ contains 1 one half
a time, and it contains 2 just half as often, or i of a time. £
divided by 2 then is i. If i is to be divided by 3, we
reason in the same way. \ contains 1, i a time.
It contains 3 only a third as often, i of { is i, and there-
fore i contains 3, £ of a time.
Again if \ is to be divided by 4, we reason thus :
If 4 contains 1, \ a time, it contains 4 only ^ as often.
\ of i is }. Then i contains 4 not one time, but \ of one
time.
Again let \ be divided by 4, and we reason thus :
If i is divided by 1, it contains it not 1 time, but \
12*
138 ARITHMETIC. SECOND PART.
of one time. But it can contain 4 only a as often, a of
s T v.
I 2
The dividend | contains the divisor 4, not one time, but
'-, of one time.
Divide i, £, j, }, £, ,V. 3 each by one.
Proceed thus : a contains 1 not one time, but a of one
time. £ contains 1 not onetime, but a of one time, &c.
Divide a by 2
««
" i « 6
8 "
c( i <( 7
1 2 '
i o u
Divide } by 3
11 4 " 5
" 4 " 10
<< i_ « 7
l o '
TT J
How often is 2 contained in a ?
(Ans.) As 1 would be contained a of a time, 2 is con-
tained half as often, or T 'g of one time.
How often is 3 contained in A ?
How often is 5 contained in a ?
How often is 6 contained in a ?
How often is 7 contained in r \ 1
How often is 8 contained in Jj- ?
How often is 9 contained in jL ?
How often is 10 contained in \ 1
How often is 11 contained in a 1
How often is 12 contained in a 1
How often is 9 contained in a ? •
How often is 8 contained in A ?
How often is 9 contained in a ?
After finding how often 4 is contained in one part, we
find by multiplying, how often it is contained in a given
number of parts. For instance, 4 is contained in one fifth
■J„ of one time. In /wo fifths it would be contained twice
as often, or ^ of one time.
Again, let f be divided by 4, and we reason thus : 4 is
contained in one seventh one fourteenth of one time, in 2
sevenths it is contained twice as often, or two fourteenths
of one time.
How often is 3 contained in £ ?
(Ans.) 3 is contained in a one eighteenth of one time.
In a it is contained 4 times as often, ox four eighteenths
of one time.
DIVISION OF VULGAR FRACTIONS. 139
How often is 4 contained in f ?
How often is 5 contained in f ?
How often is 6 contained in a ?
Divide T 5 o by 3. Divide ^ by 5.
Divide | by 6. Divide f by 7.
Divide T 3 2 by 8. Divide § by 9.
Divide | by 11. Divide £ by 8.
How many times is 6 contained in -f 7
How many times is 4 contained in f ?
How many times is 7 contained in f 1
How many times is 8 contained in j; ?
How many times is 9 contained in T 3 2 ?
In all the above cases it will be observed that the an-
swer is obtained by simply multiplying the denominator of
the fraction by the divisor.
Thus f is divided by 4 thus. 4 is contained in £ Jfc of
one time, and in f twice as often, or ^ of one time. It
can be seen that the answer is obtained by multiplying the
denominator of f by the divisor 4. This is a method
which can always be pursued in dividing any fraction by a
whole number, viz : " multiply the denominator by the di-
visor."
But there is another method which is sometimes more
convenient.
Let A- be divided bv 4.
ID J
Now the quotient of 8 units divided by 4, is 2 units. Of
course the quotient of 8 sixteenths divided by 4, is 2 six-
teenths. In this case we have divided the numerator by
the divisor 4. This can be done in all cases where the
numerator can be divided without remainder.
But when a remainder would be left, it is best to di-
vide, by multiplying the denominator. The answer is of
the same value either way, though the name is different.
For example ; in dividing £■ by 2, we are to find how
many times 2 is contained in *-. Divide by multiplying
the denominator hy 2, and we find that it is contained not
once, but T 4 F of once. By dividing the numerator by 2, we
find also that it is contained not once, but f of once. Now
| and T \ is the same value, by a different name. For if
a thing is divided into eighteen parts, and we take four of
140
ARITHMETIC. SECOND PART.
them, we have the same value as if it were divided into
nine parts and we tookfrco of them.
Divide the following by both methods, and explain them
as above.
Divide
I by
6
8
12
1 6
15
1 7
16
s o
40
9
114
2 5
3
Divide
4
1 2
by
2
4
Cf
1 s
20
6
5
(«
2 '
54
7
8
(«
2 7
6 5
9
10
u
_7 7
10
11
12
a
8 1
e o o
9
Rule for Division where the Dividend is a
Fraction.
Divide the numerator of the Fraction by the Divisor, or,
(if this would leave a remainder,) multiply the denominator
by the Divisor.
Examples for the Slate.
In the following examples, divide the numerator by the
divisor.
vide
1 2
I S
25
8
c<
32
6 1
(<
5
!
a
1 44
2 5
by
4
Divide
1 32
5 2 9
5
«
4 6 8
360
8
<c
S 6
36 9
10
(,'
15
5 5 9
12
by
11
16
7
75
In the following examples multiply the denominator by
the divisor.
Divide
1
by
4
D
ivide
8
9
by
5
it
5
8
(c
6
«<
6
g
<«
7
(<
6
9
«
8
«(
7
1
«
9
(C
3
1
u
12
«<
9
1 l
<«
12
a
1 5
9 1
((
24
<c
72
8
«
61
DIVISION OF VULGAR FRACTIONS. 141
In the following examples divide the numerator by the
divisor.
Divide
a
a
ti
30
4 2
by
3
D
ivide
3 2
a
8
(<
1 2
64
9 ) 2
a
8
<(
43
ii
7
a
3 2
2 8
it
4
a
54
2 4
8
_3_6_
4 6
8.1
9 2
4.2
6 S
36
5
by
6
a
12
a
9
a
6
it
9
Examples for Mental Exercises.
1 . If you have £ of an orange, and wish to divide it
equally between 2 children, what part do you give each '?
2. If you have f of a load of hay, and divide it equally
among 6 horses, how much do you give each ?
3. If you have T \ of a yard of muslin, and divide it into
3 equal parts, what part of a yard is each part ?
4. If you have ±% of an ounce of musk, and divide it
into 12 equal portions, what part of an ounce is each por-
tion ?
5. If you divide i§ of a dollar into 4 equal parts, what
part of a dollar will each part be ?
6. If a man owns if of a cargo, and divides it equally
among 4 sons, how much does he give each ?
Division of one Fraction by another.
When one fraction is to be divided by another, the
same principle is employed, as when whole numbers are
divided by a fraction.
For example, if the whole number 12 is to be divided
by 5, we first multiply by the denominator 4, to find how
often one fourth is contained in 12, and then divide by 3,
to find how often three fourths are contained in it.
In like manner, if we wish to find how many times, or
parts of a time, f is contained in ^, we first find how often
one fourth is contained jn it, by reasoning thus :
One unit would be contained in -f^, two twelfths of
one time.
One fourth would be contained four times as often, or
T \ of one time.
142 ARITHMETIC. SECOND PART.
We thus find how often one fourth is contained in ^, by
multiplying it by 4, thus :
_2_ y 4 — _8_
But three fourths would be contained onlv one third as
often, and we find a third of T 8 2 by multiplying its deno-
minator by 3. For when we wish to divide a fraction
by 3, we multiply its denominator, and thus make the
parts represented by the denominator, three times smaller,
thus :
12~ • U ■3 6*
Here the twelfths are changed to thirty-sixths ; and a
thirty-sixth is a third of one twelfth.
Again let f be divided by §.
It will be found by examining the foregoing process,
that in dividing one fraction by another, the fraction which
is the dividend has its numerator multiplied by the denom-
inator of the divisor, and its denominator multiplied by the
numerator of the divisor.
Let another example be taken and observe thus.
Let I be divided by f .
I if divided by one unit would contain it not once but |
of once. But if divided by one sixth it would contain it
6 times as often or 6 times f which is ' ¥ 2 .
Here the numerator of the dividend (f) has been multi-
plied by the denominator of the divisor (£), and we have
thus found how often one sixth is contained.
Four sixths would be contained only one fourth as
often, and we therefore divide \f by 4 by multiplying its
denominator and the answer is if, and here the denomina-
tor of the dividend has been multiplied by the numerator
of the divisor (|).
We therefore multiplied the numerator of the dividend
by the denominator of the divisor to find how often one sixth
was contained, and multiplied the denominator of the divi-
dend by the numerator of the divisor to find how often jfowr
sixths were contained.
Let the following be performed and explained as
above.
Divide I by f Divide £ by
3
7
a 5, a 6. a _3_ u _9_
9 8 12 10
(( _3_ tl S. " JL3. << _e_
12 9 2 9 1 i
DIVISION OF VULGAR FRACTIONS. 143
This process corresponds with that used in dividing a
whole number by a fraction.
For if we divide 12 by f we first multiply it by 4 to find
how many one fourths there are in 12, and then divide the
answer by 3 to find how many three fourths there are.
So in dividing f by f we first multiply it by 4 to find
how many times one jourth is contained thus (§), and then
divide it by 3 to find how many times three fourths are con-
tained thus, ( T 8 g).
ExAMl'LKS.
Divide
2
by ■»
Divide
3
8
by
3
1 2
j
1
6
1 2
<< 6
n
" 9
1 2
3
1 3
6
8
«
4
t>
5
U
We invert a fraction when we exchange the places of
the numerator and the denominator.
Thus \ inverted is \, and | inverted is f and 4g in-
verted is f £ &c.
Now it appears, as above, that if we wish to divide 3
by | we are to multiply its numerator (3) by the denomi-
nator (6) and its denominator (4) by the numerator (2).
This is more easily done, if we invert the divisor f , thus #.
When the divisor is thus inverted we can multip/y the
numerators together for a new numerator and the denomi-
nators for a new denominator and the process is the same.
Thus let us divide a by |.
Inverting the divisor § the two fractions would stand to-
gether thus £ f . We now multiply the numerators and
denominators together and the answer is }| and it is the
same process, as if we had not inverted the divisor, but
multiplied the numerator of the dividend by the denomi-
nator of the divisor and its denominator by the numerator
of the divisor.
This method therefore is given as the easiest rule, but
it must be remembered that in this process we always mul-
tiply the dividend by the denominator of the divisor and di-
vide it by the numerator, as we do in case of whole
numbers.
144 ARITHMETIC. SECOND PART.
Common rule for dividing one fraction by another.
Invert the Divisor, and then multiply the numerators and
denominators together.
Examples for the slate.
Divide |f by j\.
Invert the divisor and the fractions stand thus ££ y .
Multiply them together, and the answer is £i|.
Divide f| by 44 I Divide || by |f
ii 32. (( 5.6. it 'J.4. (( 1.4
4 9 3 9 1 1 1 9 1 T
«( 3.8 << .32 I (I _5 6 «< _9_3_
56 2 1 I 412 506
il 6_5_ it 3 4.X I : ' ! 6. " 9_3_
138 302 I 49 102
DECIMAL DIVISION.
In order to understand the process of Decimal Division,
it is needful to recollect the method of dividing and multi-
plying, by ciphers and a separatrix.
If we wish to multiply a number by a sum composed of
1 with ciphers added to it, we add as many ciphers to the
multiplicand, as there are ciphers in the multiplier. Thus
if we wish to multiply 61 by 10, we do it by adding one
cipher, 040. If we are to multiply by 100, we add two
ciphers thus, 6400, &c
Examples.
Multiply 3 by 100 Multiply 46 by 100
« 19 " 1000 " 2 " 100000
If we wish to multiply a decimal by any number com-
posed of 1 with ciphers annexed, we can do it by removing
the separatrix as many orders to the right, as there are ci-
phers in the multiplier.
Thus if ,2694 is to be multiplied by 10, we do it thus ;
2,694. If it is to be multiplied by 100, we do it thus ;
DECIMAL DIVISION.
145
26,94. If it is to be multiplied by 1000 we do it thus ;
269,4. But to multiply by a million, we must add ci-
phers also, in order to be able to move the separatrix as
far as required, thus ; 209400,.
Examples.
Multiply 2,64
« 36,9468
3,2
by
10
100
" 1000
Multiply 6,4
1,643
3,2
by 10000
« 10
« 1000000
The same method can be employed in dividing deci-
mals, by any number composed of 1 and ciphers an-
nexed.
The rule is this. Remove the separatrix as many or-
ders to the left, as there are ciphers in the divisor.
Thus if we wish to divide 23,4 by 10 we do it thus ;
2,34.
If we wish to divide it by 100 we do it thus, 234. But
if we wish to divide it by a thousand it is necessary to pre-
fix a cipher thus ,0234. If we divide it by 10,000 we do
it thus ,00234.
Examples.
Divide 2,4
2,46
3,2
« 2,4
Multiply 2,4
Divide 328,94
Mult'y. 326,94
by
(<
100
10
1000
10
10
100
100
Divide 24,3
246,9
" 2,3
34,26
Mult'y. 34,26
Divide 3,2
Multiply 3,2
by
«
10
100
100000
1000
1000
10000
10000
It is needful to understand that a mixed decimal, can
be changed to an improper decimal fraction.
For example, if we change 3,20 to an improper decimal
fraction, it becomes 320 hundredths (f§£), which is an
improper fraction, because its numerator is larger than
the denominator.
But we cannot express the denominator of 320 hun-
dredths, by a separatrix in the usual manner, for the rule
requires the separatrix to stand, so that there will be as
13
I
146 ARITHMETIC. SECOND PART.
many figures at the right of it, as there are ciphers in the
denominator.
If then we attempt to write 320 hundredths in this way,
it will stand thus 3,20, which is then a mixed decimal, and
must be read three units and 20 hundredths. If it is writ-
ten thus, 2-f-j}, it is then a vulgar and not a decimal frac-
tion.
But it is convenient in explaining several processes in
fractions, to have a method for expressing improper deci-
mal jr actions, without writing their denominator. The fol-
lowing method therefore will be used.
Let the inverted separatrix be used to express an im-
proper decimal fraction. Thus let the mixed decimal 2,4
which is read two and four tenths, be changed to an im-
proper decimal thus, 2'4 which may be read twenty-Jour
tenths .
The denominator of an improper decimal, (like that of
other decimals) is always 1 and as many ciphers as there
are figures at the right of the separatrix. It is known to
be an improper decimal, simply by having its separatrix
inverted.
Thus 24'69 is read, two thousand four hundred and six-
ty. nine hundredths. 239'6 is read, two thousand three
hundred and ninety-six tenths, &c.
Examples.
Change the following mixed decimals to improper deci-
mals, and read them.
246,3 24,96 32,1
326,842 3,6496 49,2643
8,4692 368,491 26,3496
Rule for writing an Improper Decimal.
Write as if the numerator were whole numbers, and place
an inverted separatrix, so that there will be as many figures
at the right, as there are ciphers in the denominator.
Write the following improper decimals.
Three hundred and six tenths.
Four thousand and nine hundredths.
DECIMAL DIVISION. 147
Two hundred and forty-six thousand, four hundred and
six tenths.
Three millions, five hundred and forty-nine tens of thou-
sandths.
Two hundred and sixty-four thousand, five hundred and
six thousandths.
Five hundred and ninety. six teiiths.
Decimal Division when the Divisor is a whole number.
The rules for Decimal Division are constructed upon
this principle, that any quotient figure must always be put
^ in the same order as the lowest order of that part of the
dividend taken.
Thus if we divide ,25 (or two tenths", five hundredths,)
by 5, the quotient figure must be put in the hundredth or-
der, thus, (,05) because the lowest order of the dividend is
hundredths.
Again, if ,250 is divided by 50, the quotient figure must
be 5 thousandths, (,005) for the same reason.
Let us then divide ,«56 by 2. We proceed exactly as
in the Short Division of whole numbers, except in the use
of a separatrix.
Let the pupil proceed thus :
2),256
,128
2 tenths d'vided by 2, gives 1 as quotient, which is 1
tenth, and is set under that order with a separatrix before
it. 5 hundredths divided by 2, gives 2 as quotient, which
is 2 hundredths, and is set under that order.
1 hundredth remains, which is changed to thousandths,
and added to the 6, making 10 thousandths.
This, divided by 2, gives 8 thousandths as quotient, which
is placed in that order.
If the divisor is a whole number, and has several or-
ders in it, we proceed as in Long Division, except we use
a separatrix, to keep the figures in their proper order.
Thus if we divide 15,12 by 36, we proceed thus :
•
148 ARITHMETIC. SECOND PART
36)15,12(,42
14,4
,72
,72
,00
We first take the 15,1 and divide it, remembeiing that
the quotient figure is to be of the same order as the lowest
order in the part of the dividend taken, of course the quo-
tient 4 is 4 tenths (,4) and must be written thus in the
quotient.
We now subtract 30 times ,4 which is 14,4. (See
rule for Decimal Multiplication page 108) from the part
of the dividend taken and 7 tenths (.7) remain.
To this biing down the 2 hundredths. Divide, and the
quotient figure is 2 hundredths which must be set in that
order in the quotient.
Subtract 36 times ,02 (or ,72) from the dividend and
nothing remains.
Let the following sums be performed and explained as
above.
Divide
76,8 by 24
Divide
37,8 by 21
«
94,6 « 43
«(
85,8 " 20
Sometimes ciphers must be prefixed to the first quotient
figure, to make it stand in its proper order.
For example, let ,1512 be divided by 36, and we pro-
ceed thus,
36),1512(,0042
,144
,0072
,0072
0000
I
. DECIMAL DIVISION. 149
We take ,151 first, which is 151 thousandths (for the
denominator of any decimal is always of the same order
as the lowest order taken).
This divided by 36 gives 4 as quotient. This 4 is 4
thousandths, because the lowest order in the part of the
dividend taken is thousandths. Therefore when it is put
in the quotient it must have two ciphers and a separatrix
prefixed thus ,004.
We now subtract from the dividend 36 times, ,004 or
,144. (See rule for Decimal Multiplication.)
It is desirable in such cases to place ciphers and a sep-
aratrix in the remainders, to make them stand in their
proper orders.
To the remainder (,007) bring down the 2 tens of
thousandths making 72 tens of thousandths.
This divided by 36 gives 2 tens of thousandths as quo-
tient which is set in that order. 36 times 2 tens of thou,
sandths (or ,0072) being subtracted, nothing remains.
Sometimes we must add ciphers to the dividend before we
can begin to divide.
For example, let ,369 be divided by 469, and we pro-
ceed thus,
460),3690(,00078
,3283
,04070
,03752
,00318 z±%
I
We find that ,369 cannot be divided by 469, so we add
a cipher to it, making it 3690 tens of thousandths.
This divided by 469 gives 7 as quotient, which is 7 tens
of thousandths, (,0007) because the lowest order of the
dividend is of that order.
We now subtract 469 times ,0007 (which is ,3283) from
the dividend, and ,0407 remain.
13*
150 ARITHMETIC. SECOND PART.
To this remainder we add a cipher, and change it from
407 tens of thousandths to 4070 hundreds of thousandths.
This divided hy 469 gives 8 as quotient, which is 8 hun-
dreds of thousandths, hecause the lowest order in the div-
idend is hundreds of thousandths.
We now subtract 469 times 8 hundreds of thousandths
(or ,03752) from the dividend and ,00318 remain.
We could continue dividing, by adding ciphers to the
remainders, but it is needless. Instead of this we can set
the divisor under the remainder as in common division,
thus -3J.1
It is not needful to retain the separatrix and ciphers
when thus writing a remainder, because when put in the
quotient, it is not considered as the |£| part oi' a whole
number, but as a part of the lowest order in the decimal,
by which it is placed.
Thus when this is put with the above quotient, we read
the answer thus 78 hvndreds of thousandths, and |£| of
another hundred of thousandth.
Let the following sums be performed and explained as
above.
Divide 42869 by 95
" 3,69428 « 49
« ,269 " 482
« 481,4 " 81
28,1 " 15
Decimal Division when the Divisor is a Decimal.
When the divisor is a decimal, we proceed as in divi-
ding by a Vulgar Fraction, viz.
We multiply by the denominator, and divide by the nu-
merator.
Thus if we are to divide 24 by ,4, we are to find how ma-
ny 4 tenths there are in 24.
We first multiply 24 by the denominator 10, to find how
many one tenths there are, and then divide by the numer-
ator 4, to find how many 4 tenths there are. 24 is multi-
plied by ten, thus ; 24'0, and has the inverted separatrix,
to show that it is not 240 whole numbers, but tenths.
We now have found that in 24 there are 240 one tenths,
Divide
3,694
by 84
cc
,36946
« 841
(<
3,26
" 589
(<
32,4
" 386
a
364,6
" 99
*
DECIMAL DIVISION.
151
we now divide by 4, to find how many 4 tenths there are.
The answer is 60, which according to the rule, must be of
the same order as the lowest order in the dividend, or 60
tenths, and must be shown by the inverted separatrix thus
(0'0.) This may be changed to whole numbers by revert-
ing the separatrix thus (6,0.)
When the dividend is a decimal, we can multiply by re-
moving the separatrix.
Thus let 8,64 be divided by ,36.
Here we are to multiply by 100, to find how many one
hundredths there are in the dividend, and then divide by
36 to find how many 36 hundredths there are.
We multiply by 100, by removing the separatrix two
orders toward the right, and then dividing by 36, we have
"24 as answer, which is 24 units, because the dividend is
units, as appears below.
36)864,(24
72
000
If the divisor is a mixed decimal, we change it to an im-
proper decimal, and then proceed as before, multiplying
by the denominator and divide by the numerator.
Thus let 10,58 be divided by 4,6.
We first change the divisor into an improper decimal
thus, 4'G (46 tenths.)
We now are to multiply the 10,58 by 10, to find how
many one tenths there are, and then divide by 46, to find
how many 46 tenths there are.
We multiply by 10 by removing the separatrix thus,
105,8, and proceed as follows.
152
ARITHMETIC. SECOND PART.
46)105,8(2,3
92
13,8
13,8
000
Here we divide 105 units by 46, and the quotient fig-
ure is 2 units.
We then subtract 46 times 2 units from the dividend,
and 13 units remain. To this bring down the 8 tenths.
This is divided as if whole numbers, but the quotient 3 is
3 tenths, because the lowest order in the dividend is tenths.
It is set in the quotient with the separatrix before it, and
then 46 times ,3 (or 13,8) is taken from the dividend, and
nothing remains.
Let the following sums be performed, and explained as
above.
Divide
a
The following then is the rule for Decimal Division.
Rule for Decimal Division.
If the divisor is a ivhole number, divide as in common di'
vision, placing each quotient figure in the same order as the
lowest order of the dividend taken.
If the divisor is a decimal, multiply by the denominator,
and divide by the numerator, placing each quotient figure in
the same order as the lowest order of the dividend taken.
If the divisor is a mixed decimal, change it to an improper
decimal, and then proceed to midtiply by the denominator
and divide by the numerator.
N. B . The ride for multiplying and dividing Federal
Money, is the same as for Decimals.
46,4
by 3,6
Divide
891,6 by
,2
,431
2,41
(<
8,964 "
8,6
4,56
" 3,64
CI
89,96 "
4,861
464,92
« 3,2649
re
8,641 »
,4169
DECIMAL DIVISION. 153
Examples.
How many times is $2,04 contained in $9,40 ?
Divide $2,04 by $,84
02 « 8,41
" 2,41 « 19,24
324,07 " 64,81
20,46 " ,49
As it is found to be invariably the case that the decimal
orders in the divisor and quotient always equal those of the
dividend, the common rule for decimal division, is formed
on that principle, and may now be used.
Common Rule for Decimal Division.
Divide as in whole numbers. Point off in the quotient
enough decimals to make the decimal orders of the divisor
and quotient together equal to those of the dividend, counting
every cipher annexed to the dividend, or to any remainder, as
a decimal order of the dividend. If there are not enough
figures in the quotient prefix ciphers.
In pointing off by the above rule, let the teacher ask
these questions.
How many decimals in the dividend ? How many in
the divisor ? How many must be pointed off in the quo-
tient, to make as many in the divisor and quotient, as there
are in the dividend ?
Examples.
At $,75 per bushel, how many bushels of oats can be
bought for $14,23/
How much butter at 16 cents a pound, can be bought for
$20?
A half cent can be written thus, $,005 (for 5 mills is
half a cent, or 5 thousandths of a dollar.)
A quarter of a. cent can be written thus, $,0025 (for { of
a cent is 25 tens of thousandths of a dollar.)
At 12£ cents per hour, in how much time will a man
earn #46" ?
154 ARITHMETIC SECOND PART.
At 6^ cents per pint, how much molasses may be bought
for $2 ?
At $,06 an ounce, how much camphor can be bought
for $3 ?
At$,l2i a bushel, how much coal could be bought for
$5? > V*
Divide ,032 by ,005.
Exercises in Decimal Multiplication and Division.
Multiply ,25 by ,003. Divide ,25 by ,003.
Multiply 3,4 by 2,68. Divide 3,4 by 2,68.
Multiply ,005 by ,005. Divide ,004 by 16,4.
If you buy 24 bushels of coal, at $,09 per bushel, what
does the whole cost?
If a man's wages be fifty hundredths of a dollar a day,
what will it be a month ?
What will be the cost of 25 thousandths of a cord of
wood, at $2 a cord 1
What will be the cost of twelve hundredths of a ton of
hay, at $11 a ton?
If a man pays a tax of two mills on a dollar, how much
must he pay if he is worth $350 ?
If a man pays $,06 a year for the use of each dollar he
borrows of his neighbor, how much must he pay in a year
if he borrows 264 dollars ? How much in two years ?
REDUCTION.
Reduction is changing units of one order, to units of
another order.
Reduction Ascending, is changing units of a lower to a
higher order.
Reduction Descending, is changing units of a higher to a
lower order.
Examples for Mental Exercise.
In 4 gallons how many quarts ?
Note. Let each sum be stated thus. One gallon con.
REDUCTION. 155
tains four quarts, and four gallons four times as much. 4
times 4 is 16.
In 4 gallons how many pints ?
In 8 yds. 3 qrs. how many quarters ?
In 8 feet how many inches ?
In 4 bushels how many quarts ?
In 5 hours how many minutes ?
Are the above sums in Reduction Ascending or De-
scending ?
In 32 quarts how many gallons ?
Let such sums be stated thus. One gallon contains 4
quarts. In 32 quarts therefore, there are as many gal.
Ions as there are 4's in 32.
In 42 pints how many gallons ?
In 49 quarters how many yards ?
In 50 nails, how many quarters and how many yards ?
In 64 inches how many feet ?
In 36 barley corns how many inches 1
In 96 quarts how many bushels ?
In 120 minutes how many hours ?
In 48 feet how many yards 1
In 94 inches how many feet ?
In 3 yards how many inches 1
In 4 gallons how many pints ?
In 32 quarts how many gallons ?
In 80 penny weights how many ounces ?
In 24 ounces how many penny weights I
In 8 pounds how many shillings ?
In 40 shillings how many pence ?
In £2, 9s. 6d. 3 qrs. how many farthings ?
In doing this sum we proceed in the following manner :
£. s. d. qr.
2 « 9 '« 6 " 3
20
49 shillings.
594 pence.
4
2379 farthings.
156 ARITHMETIC. SECOND PART.
We first change the pounds to shillings, by multiplying
by 20, and add the 9 shillings to them, making 49 shil-
lings.
We then change the 49 shillings to pence, by multiply-
ing by 12, and add the 6 pence to them, making 594
pence.
We then change the 594 pence to farthings, by multi-
plying by 4, and add the 3 qrs. and thus we obtain the an-
swer 2379 qrs
This is Reduction Descending, because we have chan-
ged units of a higher order to those of a lower.
Why did we multiply by 20, 12, and 41
Let us now reverse the process, and change 2379 far-
things to pounds.
We proceed thus :
£. s. d. qr.
4)2379(2 "9 "6 " 3.
12)594
20)49
2
We first change the 2379 farthings to pence, by divi-
ding by 4, and the answer is 594 pence, and 3 farthings
(or qr.) over, which is put in the quotient with qr. over it.
We then change the 594 pence to shillings, by dividing
by 12, and the answer is 49 shillings, and six pence over,
which is put in the quotient with d. written over.
We next change the 49 shillings to pounds, by dividing
by 20, and find there is £2 and 9s. over, which are both
put in the quotient with their signs written over them.
Why did we divide by 4, 12, and 20 1
Let the following sums be performed and explained in
the same way.
Change 2486 farthings to pounds.
Change £2 18s. 4d. 2qr. to farthings.
Change 241 shillings to pounds.
Change 249 pence to shillings and pounds.
Change £21 2s. to farthings. t
Change 361 pounds to pence.
Change 35 shillings to pounds.
REDUCTION. 157
Rule for Reduction.
To reduce from a higher to a lower order.
Multiply the highest order by the number required of tin
next lower order, to make a unit of this order. . Add the
next lower order to this product, and multiply it by the num.
ber required of the next lower order, to make a unit of this
order, adding as before. Thus through all the orders.
To reduce from a lower to a higher order.
Divide the amount given, by the number required to make,
a unit of the next higher order. Divide the answer in the
same way, and continue thus till the answer is in units of the
order demanded. The remainders are of the same order
as the dividend, and are to be put as a part of the answer.
Exercises.
Bought a tankard of silver weighing 5 lb. 3 oz. for
which I paid $1,12 an oz. how much did it cost ?
Reduce 2 lb. 8 oz. 11 pwt. to grains.
In 81b. 93.43. 29. 16grs. how many grains?
In 11924 grains how many pounds ?
What cost 4 cwt. 3 qrs. 17 lb. of sugar, at 12a cents
per lb?
In 436 boxes of raisins, each containing 24 lbs. how
many cwt. ?
In 63469542 drams, how many tons ?
In 546 yards how many nails ?
In 5486 nails how many yards 1
In 118J yards, how many Ells Flemish ?
How many barley corns will reach round the globe, it
being 360 degrees ?
How many miles in 836954621 barley corns ?
In 18 acres, 3 roods, 12 rods, how many square feet 1
How many square feet in 16 square miles ?
In 9269546231 square feet how many square miles ?
In 37 cords of wood how many solid feet ?
In 20486 solid feet how many cords ?
In 4 pipes of wine how many pints ?
In 9120854 pints how many pipes ?
14
158 ARITHMETIC. SECOND PART.
In 464 bushels how many quarts ?
In 964693 pints how many bushels ?
REDUCTION OF FRACTIONS TO WHOLE
NUMBERS.
1. In ten fifths, how many units ?
2. In fourteen sevenths, how many units ?
3. Change fifteen fifths to units.
4. Change thirteen fourths to units, and what is the an-
swer ?
5. Change eighteen fourths to units, and what is the an-
swer?
6. Change fourteen sixths to units.
It will be perceived, that in answering these questions,
the pupil divides the numerator by the denominator. Thus
in changing twelve fourths to units, the numerator twelve,
is divided by the denominator four. The above sums are
to be performed mentally first, and the answers given, and
then they are to be written, thus,
7. Change fourteen sixths to units.
Ans. L- 4 = 14 -r- 6 = 2 §
Let the pupil be required to perform all the above sums,
in this manner.
Rule for Reducing Fractions to Whole Numbers.
Divide the numerator by the denominator ; write the re-
mainder, if there be any,- over the denominator, and annex
the fraction, thus formed, to the quotient.
Examples.
1. Reduce 3 T 9 to a whole or mixed number. Ans. 9|.
2. Reduce V- Ans - 9 I- ¥• Ans - H- V- Ans -
151. y. Ans. 2f.
REDUCTION OF FRACTIONS. 159
3. Reduce 'j 8 . Ans. 52§. ays. Ans. 565. 21 f3s.
Ans. 2425.
4. Reduce 8, £*'. 62 g 7 6. 5 18_4 32. 9 15_8 7 3. 132|9 6S #
5. Reduce 08765.4321. 700070007, 600344002.
6. Reduce vii23_45499 49563J0217 33322.UU6
9 • 1 ' 6 *
59248 32 1768
REDUCTION OF WHOLE NUMBERS TO
FRACTIONS.
1. In three units, how many fourths, and how is the
answer expressed in figures ?
2. How many fifths is three units and two fifths, and
how is the answer written ?
3. Reduce nine units to sixths.
4. Reduce seven units and two twelfths to twelfths.
Rule for reducing Whole Numbers to Fractions.
Multiply the whole number, by the denominator of the
fraction to which it is to be reduced, and place the product
over this denominator. If there is with the units, a fraction
of the same denominator, add the numerator of this fraction
to the product, before placing it over the denominator.
Examples.
1. How many 4ths. in 1 ? How many in U f In la t
In If ? ^ 4 4 •
2. How many 5ths. in 1 ? In 5? In 14 ? In 1^ 1 In
74? 55
3. How many 7ths. in 7 ? In 8 ? In 12 ? In 73 t In
4. How many 12ths. in 9 A ? In 7 T % ? In 3 A 1 In
5 - 5 - ' In 8 ' 1
18* 12"
5. How many 6ths. in 3 ? In 4 ? In 5 # ? In 7 « ? In
8? In 9 i? In 12?
160 ARITHMETIC. SECOND PART.
6. How many 27ths. in 3 ? In 2 ? In 5 ft ? Ans. |i.
5 4 I_4_4
27* 2
7. How many 19ths. in 15 ? In 13 T \ 1 In 17 }f ?
Ans 2 - 8 - 5 . 2 - s -° 3-Aj
REDUCTION OF VULGAR TO DECIMAL
FRACTIONS.
Decimal Fractions are generally used in preference to
Vulgar, because it is so easy to multiply and divide by
their denominators.
Vulgar fractions can be changed to Decimals by a pro-
cess which will now be explained.
In this process, the numerator is to be considered as
units divided by the denominator.
Thus £ is 3 units divided by 4, for | is a fourth of 3 units.
We can change these 3 units to an improper decimal
ihus, 3'0 (30 tenths), and then divide by 4 ; remembering
that the quotient is of the same order as the dividend.
4)3'0(,75
2'8
,20
,20
Thus the 30 tenths are divided by 4, and the answer is 7
tenths, which is placed in the quotient, with a separatrix
prefixed. 4 times 7 tenths (or 28 tenths) are then sub-
tracted, and the remainder is ,2. This in order to divide
it by 4, must have a cipher annexed, making it 20 hund-
redths. The quotient of this is 5 hundredths, and no re-
mainder.
(In performing this process, particular care must be
taken in using the separatrix, both for proper and improper
decimals.)
Let | be reduced in the same way.
The two units are first changed to an improper decimal
thus :
REDUCTION OF FRACTIONS. 161
8)2<0(,25
1<6
,40
,40
00
We proceed thus. 20 tenths divided by 8, is 2 tenths,
which is placed in the quotient. 8 times ,2, or 16 tenths
(1*6) is then subtracted, and ,4 remain.
This is changed to 40 hundredths (,40) by adding a
cipher, and then divided by 8. The quotient is 5 hund.
redths, which is put in the quotient and there is no re-
mainder.
Note. After 3 or 4 figures are put in the quotient, if
there still continues to be a remainder, it is not needful to
continue the division, but merely to put the sign of addition
in the quotient to show that more figures might be added.
Examples.
Reduce T 3 ^ to a decimal, and explain as above.
Reduce $~ f $ T 2 T T 8 3 4 f £ eacn t0 a decimal of the
same value.
Let the pupil be required to explain sums of this kind
as directed above, until perfectly familiar with the
principle.
When fractions of dollars and cents are expressed, their
decimal value is found by the same process.
For example, change \ a dollar to a decimal.
Here the 1 of the numerator, is one dollar, divided by
2. By adding a cipher to this 1 and using the inverted
separatrix, the dollar is changed to 10 dimes, and when
this is divided by 2, the answer is 5 ; which being of the
same order as the dividend is 5 dimes.
The answer is to be written with the sign of the dollar
before it, thus $0,5.
The only difference between the answer when \ is re-
duced to a decimal, and when \ a dollar is reduced to a
decimal, is simply the use of the sign of a dollar ($) and
a cipher in the dollar order.
162 ARITHMETIC. SECOND PART.
1. Reduce \ to a decimal. Ans. ,5.
2. Reduce | a dollar to a decimal. Ans. $0,5.
3. Change \ of a dollar to a decimal. Ans. $0,125.
4. Change t l of a dollar to a decimal. Ans. $0,0625.
In this last sum there must be two ciphers added to the
numerator, changing the 1 dollar to cents, instead of
dimes ; and in this case a cipher is put in the order of
dimes, and the quotient (being of the same order as the
dividend) is placed in the order of cents.
5. Reduce \ of a dollar to a decimal. Ans. $0,2.
6. Reduce f of a dollar to a decimal. Ans. $0,625.
7. Re-duce T 3 g of a dollar to a decimal. Ans. $0,1871.
8. Reduce J^ to the decimal of a dollar. Ans. $0,01.
Rule fok the reduction of Vulgar to Decimal
fractions.
Change the numerator to an improper decimal, by annex-
ing ciphers and using an inverted separatrix. Divide by
the denominator, placing each quotient figure in the same or-
der as the lowest order of the part divided.
1. Reduce ^i^ to a decimal. Ans. .0016.
2. Reduce 5 | u to a decimal. Ans. .028.
3. Reduce z f 7 to a decimal. Ans. .05625.
4. Reduce i to a decimal. Ans. .3333333-f
Note. We see here, that we may go on forever, and
the decimal will continue to repeat 33, &c. therefore, the
sign of addition + in such cases may be added, as soon
as it is found that the same number continues to recur
in the quotient.
REDUCTION OF FRACTIONS TO A COMMON
DENOMINATOR.
Before explaining this process, it must be remembered
that | | | | &c. or a fraction which has the numerator
and denominator alike, is the same as a unit. If therefore
we take a fourth of £ it is the same as taking a fourth of
B
REDUCTION OF FRACTIONS. 163
one. If we take a sixth of £ it is the same as taking a
sixth of one.
If we take § of £ it is the same as taking f of one.
Whenever therefore we wish to change one fraction
to another, without altering its value, we suppose a unit to
be changed to a fractional form, and then take such apart
of it, as is expressed by the fraction to be changed.
For example, if we wish to change \ to twelfths, we
change a unitto twelfths and then take i of it, and we have
i of }f, which is the same as ± of one.
If we wish to change \ to eightJis, we change a unit to
| and then take i of it, for \ of f is the same as i of one.
Change | to twelfths, thus, a unit is }§. One third
of is. is T V Two thirds is twice as much, or ^, Then
I are T \.
Change j to twentieths. A unit is §£. One fifth of |§
is 5 4 ¥ . Four fifths is four times as much, or ^.
Change the following fractions and state the process in
the same way.
Change f to twenty fourths.
Change £ to twelfths.
Reduce § to twenty sevenths.
Reduce f to sixty fourths.
Reduce | to twenty fifths.
Reduce £ to twenty sevenths.
Reduce f- to thirty sixths.
Reduce £ to forty ninths.
Reduce j\ to thirty sixths.
Reduce £ to sixteenths.
Reduce fe to fortieths.
Reduce }£ to thirty thirds.
Reduce # to thirty sixths.
Reduce f and f each to twelfths.
Reduce | and T \ each to twentieths.
Reduce A f and £ each to twelfths.
Reduce a T 2 g and ^ each to fortieths.
Reduce i § t \ and ¥ 8 T each to sixty fourths.
Reduce f £ T 8 T5- and if- each to forty eighths.
In the above examples it is seen that when several frac-
tions are to he reduced to a common denominator, a unit is
changed first to a fractional form with the required deno-
164 ARITHMETIC. SECOND PART.
minator. Then it is divided by the denominator of each
fraction, to obtain one part, and multiplied by the numera-
tor, to obtain the required number of parts.
Thus changing | and f each to twelfths, we first change
a unit to a fraction with the required denominator 12 ;
thus, ||. We then divide it by the denominator of f , to
obtain one fourth, and multiply the answer by 3, to obtain
three fourths. In like manner with the §. We divide }§
by the denominator 6, to obtain one sixth, and multiply by
the numerator to obtain two sixths.
In changing fractions to common denominators then, the
unit must be changed to that fractional form which will
enable us to divide it by all the denominators of the frac
tions (which are to be reduced) without remainder.
Thus if we wish to reduce ^ and f to a common denom-
iuator, we cannot reduce them to twelfths, because || can-
not be divided by either the denominator 5, or 7, without
remainder. We must therefore seek a number that can
be thus divided, both by 7 and 5. 35 is such a number.
We now take | of ff and f of f f and the two fractions
are then reduced to a common denominator.
One mode of reducing fractions to a common deno.
MINATOR.
Change a unit to a fraction whose denominator can be
divided by all the denominators of the fractions to be redu-
ced, without remainder. Divide this fraction by the deno-
minator of each fraction to obtain one part, and multiply by
the numerator to obtain the required number of parts.
FURTHER examples for mental exercise.
Reduce § | and J to a common denominator.
Let the unit be reduced to f £ .
Reduce f | f to a common denominator. Let the unit
be reduced to ||.
Reduce f f \ and T 4 F to a common denominator.
Reduce § £ f to a common denominator.
Reduce £ f £ to a common denominator.
Reduce rV£ f f £ *° & common denominator.
But there is another method of reducing fractions to a
REDUCTION OF FRACTIONS. 165
common denominator which is more convenient for opera-
tions on the slate. When a fraction has both its terms
(that is its numerator and denominator) multiplied by the
same number, its value remains the same.
For example ; multiply both the numerator and deno-
minator of | by 4, and it becomes T 8 ¥ . But § and T 8 5 are
the same value, with different names.
The effect, then, of multiplying both terms of a fraction
by the same number is to change their name, but not their
value.
If therefore we have two fractions, and wish to change
them so as to have both their denominators alike, we can
do it by multiplication.
For example ;
Let §- and £ be changed, so as to have the same deno-
minator. This can be done by multiplying both terms of
the | by 9, and of •* by 3. The answers are i-f and £f,
and the value of both fractions is unaltered.
In this case both terms of each fraction were multiplied
by the denominator of the other fraction.
Let the following fractions be reduced to a common de-
nominator in the same way.
1. Reduce f and f to a common denominator. Multiply
'the | by the denominator 7, and the f by the denomina-
tor 5.
. 2. Reduce § and f to a common denominator.
3. Reduce f and £ to a common denominator.
4. Reduce ' T \ and £ to a common denominator.
The same course can be pursued, where there are sev-
eral fractions, to be reduced to a common denominator.
Thus if i | and £ are to be reduced to a common deno-
minator, we can multiply both terms of the \ first by the
denominator 3, and then multiply both terms of the answer
by the denominator 4, and it becomes \\ and its value re-
mains unaltered. For \ and \\ have the same value with
a different name.
Then we can multiply both terms of the § first by the
denominator 2, and then by the denominator 4, and it be-
comes i| and its value remains unaltered.
Then f may be multiplied, first by the deneminator 2,
166 ARITHMETIC SECOND PART.
and then by the denominator 3, and it becomes i| and its
value is unaltered.
The three fractions J § and £ are thus changed to \\
\% and \\ which have a common denominator, and yet
their value is unaltered.
But instead of multiplying each fraction, by each sepa-
rate denominator, it is a shorter way to multiply by the
product of these denominators.
Thus in the above example, instead of multiplying the
\, first by 3, and then the answer by 4, it is shorter to mul-
tiply by 12 (the product of 3 and 4), and the answer will
be the same.
In like manner, if we were to reduce £ £ and | toa
common denominator, we should multiply both terms of
each fraction by the denominators of all the other frac-
tions. But instead of each denominator separately, as
multiplier, we can take the product of them for the mul-
tiplier.
Reduce f £ and i to a common denominator.
Here both terms of the § are first multiplied by the pro-
duct of the other two denominators (which is 12). Then
both terms of £ are multiplied in the same way by the pro-
duct of the other two denominators (15). Then both
terms of \ are multiplied by the product of the other two
denominators (20).
Rule for reducing fractions to a common denomi-
nator.
Multiply both terms of each fraction by the product of all
the denominators except its own.
Reduce \ £ f to a common denominator.
Reduce £ f\ and \\ to a common denominator.
An<t 1A» *&*- and !££..
Reduce iff and £ to a common denominator.
/two 144 J.92 2.4 0. anf ] 252
• /A/W " 288 288 288 aI1U 2 8 8'
Reduce £ ^ and T \ to a common denominator.
Reduce £ £ and 121 to a common denominator.
Art? 54 60 8 88
Jllld. -ijj -sj-g if 2 •
Reduce | £ and f of ji to a common denominator.
4«« 16J_ 2 5.92 13.8
■«•'"»• 15 4 5 6 34 56 34 56'
reduction of fractions. 167
Reduction of Fractions to their Lowest Terms.
What is the difference between a and \ ?
Ans. They express the same value, by different names.
Which fraction has the smallest numbers employed to
express its value ?
In the two fractions f and T \ is there any difference in
the value ?
Which fraction has its value expressed by the smallest
numbers ?
A fraction is reduced to its lowest terms, when its zalue
is expressed by the smallest numbers which can be used, to
express that value.
For example, £ is reduced to its lowest terms, because
no smaller numbers than 3 and 4 can express this value.
The value of a fraction is not altered if both terms of it
are divided by the same number.
Thus if | has both its terms divided by 2, it becomes }
and the value remains the same. If it is divided by 4, it
becomes \ and its value remains unaltered.
When it was divided by 2, it was not reduced to its low.
est terms, because smaller numbers can express the same
Rvalue as f . But when it was divided by 4, it was reduced
to its lowest terms, because no smaller numbers than 1
and 2 can express its value.
The shortest way to reduce a fraction to its lowest
terms is, to divide it by the largest number which will di-
vide both terms, without a remainder.
Any number which will divide two or more numbers
without a remainder is called a common measure, and the
largest number which will do this, is called the greatest
common measure.
In many operations it saves much time to have a frac-
tion reduced to its lowest terms. Thus for example, if we
are to multiply 3429 by || it would be much easier to re-
duce the fraction to £ (which are its lowest terms) and
then multiply.
There are many fractions which can be reduced to their
lowest terms without much trouble. For example let the
pupil reduce these fractions.
168 ARITHMETIC. SECOND PART.
Reduce f £ T 5 ^ r \ ^ to their lowest terms.
But there are many fractions, which it is much mor>„
difficult to reduce. Thus if we wish to reduce yWe" to ^ ts
lowest terms, we could not so readily do it.
In such a case as this there are two ways of doing it ;
the first is as follows.
Rule for reducing a fraction to its lowest terms.
Divide the terms of the fraction by any number that will
divide both, without a remainder. Divide the answer ob-
tained in the same way. Continue thus, till no number can
be found, that will divide both terms without a remainder.
Thus,
Reduce T 2 g 3 3 4 F to its lowest terms.
N. B. The brackets at the right of the fractions show
that both terms of the fraction are to be divided by the di-
visor, and not (he fraction itself, as in the division effrac-
tions.
234 \_i_3 _7_8_
18 36^ • " 6 I 2
6 1 2/^^ 3 6
3 3 o 9 a)^-3= T V 3 2 Answer.
In the above process, both terms of the fraction T 2 g 3 g 4 g
are divided by 3 ; the answer is divided by 2 ; and this
answer again is divided by 3.
The last answer is T \\ which cannot have both terms
divided by any number without a remainder.
The other method of reducing a fraction to its lowest
terms, is first to find the number which is the greatest
common measure, and then to divide the fraction by this
number.
The following is the method of finding the greatest
common measure, and reducing to the lowest terms.
Reduce §4- to its lowest terms.
The denominator is first placed as a dividend, and the
numerator, as a divisor; (below.) After subtracting, the
remainder (14) is used fnr the divisor, and the first divisor
(21) is used for the dividend. This process of dividing
REDUCTION OF FRACTIONS. 16 ( .)
the last divisor by the last remainder is continued till
nothing remains. The last divisor (7) is the greatest
common measure.
We then take the fraction §•]• and divide both terms by
7, the greatest common measure, and it is reduced to its
lowest terms, viz. |.
21)35(1
21
14)21(1
14
7)14(2
14
00
S-M-T-7:
35/ '
Rule for finding the greatest common measure of
a Fraction and reducing it to its lowest terms.
Divide the greater number by the less. Divide the divi-
sor by the remainder, and continue to divide the last divisor
by the last remainder, till nothing remains. The last divisor
is the greatest common measure, by which both terms of the
fraction are to be divided, and it is reduced to its lowest
terms.
Reduce the following Fractions to their lowest terms.
486 • 144 • 324. 14 29. 16. 4.4. . _4J>_8_ . _4_7_4_S_ . . S_ 0_S__ .
9120 ' T72 J ' 64 8' 28S8 ' 2192' 1184' 38433' 42315'
Hff ; f&H&s- Ans - 2V ; tV ; I ; \ ; 1 5 H£ ; ,¥8?,
_3_ • 3 . X.
RpHiipp thp fnllnwinir • 516 . 4932 . _i_2_3JL5_ . 2yj7 .
JXeUUUt; Hie lOUUVVlUg . -943- , 8^64 ' 678910 ' 342954 '
8.31^8 • 3 9_9_72_ • _9.9.8.811 • JULSJLSJL
95636' 812322' 9998881T' 7 3 28472*
15
170 ARITHMETIC. SECOND PART.
REDUCTION OF FRACTIONS FROM ONE ORDER
TO ANOTHER ORDER.
It will be recollected that in changing whole numbers
from one order to another, it was done by multiplication
and division.
Thus, if 40 shillings were to be changed to pounds, we
divided them by the number of shillings in a pound, and if
£2 were to be reduced to shillings, we midtiplied them by
the number of shillings in a pound.
The same process is used in changing jr actions of one
order to fractions of another order.
Thus, if we wish to change -^^ of a £ to a fraction of
the shilling order, we multiply it by 20, making it ^Vo*
For 2V0 °f a shilling is the same as ^£0 of a pound.
Ifwewishto change ^W °f a shilling, to the same
value in a fraction of the pound order, we divide ^Vo D y
20, making it ^^. (This could also be divided by multi-
plying its denominator by 20.)
If then we wish to change a fraction of a lower order to
the same value in a higher order, we must divide the frac-
tion, by multiplying the denominator, by that number of units
(of the order to which the fraction belongs) which make
a unit of the order to which it is to be changed.
Thus if we wish to change | of a penny to the same
value in the fraction of a shilling, we multiply its denomi-
nator by 12, making it g\ of a shilling. If we wish to
change this to the same value in a fraction of the pound
order, we must now multiply its denominator by the num-
ber of shillings which make a pound, making it T? 2 5? of a
pound. It must be remembered that multiplying the deno-
minator of a fraction, is dividing the fraction.
If, on the contrary, we wish to change a fraction of a
higher order to one of the same value in a lower order, we
must multiply.
Thus, to change T | T of a shilling to the penny order,
we must multiply it by 12. This we do, by multiplying
Us numerator by 12, and the answer is ^\ . For as there
are 12 times as many whole pence in a whole shilling, so
there are 12 times as many T | ? of a penny in T | T of a
shilling.
reduction of fractions. 171
Rule for reducing fractions of one order to an-
other order.
To reduce a fraetion of a higher to one of a lower
order.
Multiply the f'action by that number of units of the next
lower order, which are required to make one unit of the order
to which the fraction belongs. Continue this process till the
fraction is reduced to the order required.
To reduce a fraction of a lower to one of a higher
order.
Divide the fraction (by multiplying the denominator) by
the number of units which are required to make one unit of
the next higher order. Continue this process till the frac-
tion is reduced to the order required.
Examples.
Reduce jo\i °? a guinea, (or of 28 shillings,) to the
traction of a penny.
Reduce a of a guinea to the fraction of a pound.
Reduce ^\ of a pound Troy, to the fraction of an
ounce.
Reduce T 3 7 of an ounce to the fraction of a pound Troy.
Reduce -^ of a pound avoirdupoise to the fraction of an
ounce.
A man has F | T of a hogshead of wine, what part of a
pint is it ?
A vine grew T / T ^ of a mile, what part of a foot was it ?
Reduce | off of a pound to the traction of a shilling.
Reduce § off of 3 shillings, to the fraction of a pound.
REUCTION OF FRACTIONS OF ONE ORDER,
TO UNITS OF A LOWER ORDER.
It is often necessary to change a fraction of one order,
to units of a lower order. For example, we may wish to
change | of a unit of the pound order, to units of the
shilling order.
This | of a £ is 2 pounds divided by 3. These 2 pounds
172 ARITHMETIC. SECOND PART
are changed to shillings, by multiplying by 20, and then
divided by 3, and the answer is 13^ shillings. This i of
a shilling maybe reduced to pence in the same way, for ^
of a shilling is 1 shilling divided by 3. This 1 shilling can
be changed to pence, and then divided by 3, the answer is
4 pence.
Rule for finding the Value of a Fraction in units
of a lower order.
Consider the numerator as so many units of the order in
which it stands, and then change it to units of the order in
which you wish to find the value of the fraction. Divide
by the denominator, and the quotient is the answer, and is of
the same order as the dividend.
Examples.
1. How many ounces in f of a lb. Avoirdupoise ?
2. How many days, hours and minutes, in f of a month ?
3. What is the value of f of a yard ?
4. What is the value of -^ of a ton ?
5. How many pence in § of a lb. ?
6. How many drams in f of a lb. avoirdupoise ?
7. How many grains in f of a lb. Troy weight ?
8. How many scruples in | of a lb. Apothecaries
weight ?
9. How many pints in f of a bushel ?
REDUCTION OF UNITS OF ONE ORDER TO
FRACTIONS OF ANOTHER ORDER.
It is necessary often to reverse the preceding process,
and change units, to fractions of another order. For ex-
ample, to change 1 3s. 4d. to a fraction of the pound or-
der.
To do this we change the 13s. 4d. to units of the lowest
REDUCTION OF FRACTIONS.
173
order mentioned, viz. 160 pence. This is to be the numer-
ator of the fraction. We then change a unit of the pound
order to pence (240) and this is the denominator of the
fraction. The answer is -i££ of a pound.
For if 13s. 4d. is 160 pence, and a lb. is 240 pence,
then 13s. 4d. is |££ of a pound.
RULE FOR REDUCING VISITS OF ONE ORDER TO FRACTIONS
OF ANOTHER ORDER.
Change the given sum to units of the lowest order men-
tioned, and'make them the numerator.
Change a unit of the order to which the sum is to he re-
duced, to units of the same order as the numerator, and place
it for the denominator.
Examples.
Reduce 6oz. 4pwt. to the fraction of a pound Troy.
Reduce 3 days, 6 hours, 9 minutes to the fraction of a
month.
Reduce 2cwt. 2qrs. 161bs. to the fraction of a ton.
Reduce 21b. 4oz. to the fraction of a cwt.
REDUCTION OF A COMPOUND NUMBER TO A
DECIMAL FRACTION.
It is often convenient to change a compound number, to a
decimal fraction.
Thus we can reduce loz. lOpwt. to a decimal of the
pound order.
Let the figures be placed thus, and the process will be
explained below. The 10 pwts. are first written, and then
the 1 oz. set under.
20)10'0 pwt.
12) l'5oz.
<125lb.
15*
174 ARITHMETIC. SECOND PART,
We first change the lowest order (10 pwts.) to an im-
proper decimal, thus lO'O. Now as 20 pwts. make an oz.,
there are but one twentieth as many ounces in a sum as
there are penny weights.
For the same reason, in any sum there are but one twen-
tieth as many tenths of an ounce as there are tenths of a
penny weight.
As there are then 100 tenths ofapwt. in this sum, if we
take one twentieth of them, we shall find how many tenths
of an oz. there are.
We therefore divide the lO'O pwts. by 20, and the
amount is ,5. This ,5 is placed (beside the 1 oz. of the
sum) under the lO'O pwts., and thus, instead of reading
the sum as loz. 10 pwts., we read it as 1,5 oz., or loz.
and 5 tenths of an oz.
As the pwts. are thus reduced to the decimals of an oz.
we now reduce the l,5oz. to the decimal of a lb. in the
same way.
We make the 1,5 an improper decimal, thus 1*5 (15
tenths) of an oz.
Now as there are 12oz. in a lb., there are but one twelfth
as many tenths of a lb. in a sum, as there are tenths of an
oz. We therefore divide the 15 tenths of an oz. by 12,
and the answer is ,1 of a lb. and 3 left over. This 3 is re-
duced to hundredths by adding a cipher and dividing it
again. The quotient is 2 hundredths. The next remain-
der is changed to thousandths in the same way, and the
answer is ,125 of a £.
Rule for changing a compound number to a deci-
mal.
Change the loxcest order to an improper decimal. Divide
it by the number of units of this order ; which are required,
to make a unit of the next higher order, and set the answer be-
side the units of the next higher order. Repeat this process
till the sum is brought to the order required.
Examples.
Reduce 10s. 4d. to the decimal of a lb.
REDUCTION OF FRACTIONS. 175
Reduce 8s. 6d. 3qrs. to the decimal of a lb.
Reduce 17hrs. 16min. to the decimal of a day.
Reduce 3qrs. 2na. to the decimal of a yd.
Reduce 32gals. 4qts. to the decimal of a hogshead.
Reduce lOd. 3qrs. to the decimal of a shilling.
REDUCTION OF A DECIMAL, TO UNITS OF
COMPOUND ORDERS.
The preceding process can be reversed, and a decimal
of one order, be changed back to units of other orders.
Thus, if we have ,125 of a lb. Troy, we can change it
to units of the oz. and pwt. order.
In performing the process, we place the figures thus.
,1251b.
12
l,500oz.
20
I0,000pwt.
We reason thus. In ,125 of a lb. there must be 12
times as many thousandths of an oz. (for 12 oz. = 1 lb.
We therefore multiply by 12, and point off according to
rule, and the answer is 1 oz. and 500 thousandths of
an oz.
Now as we have found how many oz. there are, we
must find how many pwts. there are in the ,500 of an oz.
There must be 20 times as many thousandths of a pwt. as
there are thousandths of an oz. therefore multiply the de-
cimal only, by 20, and point off according to rule, and we
find there are 10 pwts.
We have thus found that in ,125 of a lb. there are loz.
and lOpwts.
176 arithmetic. second part.
Rule for changing a decimal of one compound or-
der, TO UNITS OF OTHER ORDERS.
Multiply the decimal by the number of units of the next
lower order which are required to make one unit of the order
in which the decimal stands.
Point off according to rule, and multiply the decimal part
of the answer in the same way, pointing off as before. Thus
till the sum is brought into the order required. The units
of each answer make the final answer.
In ,1257 of a £ how many shillings, pence and farth- *
ings ?
What is the value of ,2325 of a ton ?
What is the value of ,375 of a yard ?
What is the value of ,713 of a day ?
What is the value of ,15834821 of a ton ?
REDUCTION OF CURRENCIES.
There are few exercises in Reduction, of more prac
tical use than the Reduction of Currencies, by which a
sum in one currency is changed to express the same val-
ue in another currency.
An example of this kind of reduction occurs, when the
value of $1 is expressed in British currency thus, 4s. 6d.
The necessity for using this process in this country, re-
sults from the following facts.
Before the independence of the U. States, business was
transacted in the currency of Great Britain. But at vari-
ous times, the governments of the different States, put
bills into circulation, which constantly lessened in value,
until they became very much depreciated. For example,
a bill which was called a pound or twenty shillings, British
currency, was reduced to be worth only fifteen shillings,
in the New England states.
This depreciation was greater in some states than it
was in others, and the result is, that pounds, shillings, and
pence have different values in different states.
12 pence make a shilling, and 20 shillings make a
pound, in all cases, but the value of a penny, a shilling,
or a pound, depends upon the currency to which it be-
longs.
REDUCTION OF CURRENCIES. 177
The following table shows the relative value of the sev-
eral currencies, by showing the value oi one dollar in each
of the different currencies.
VALUE OF ONE DOLLAR IN EACH OF THE DIFFERENT CUR-
RENCIES.
equals 6*. New England currency.
" 8s. New York currency.
" 7s. 6d. Pennsylvania currency.
" 4s. 8d. Georgia currency.
" 4s. Gd. Sterling money, or Eng. currency.
" 5s. Canada currency.
" 4s. I0±d. Irish currency.
" £2. 14s. Scotch currency.
VALUE OF ONE FOUND OF EACH OF THE DIFFERENT CUR-
RENCIES, EXPRESSED IN FEDERAL MONEY.
£1 N. England currency equals $3,333 A-
£1 N. York currency " $2,50
£1 Pennsylvania currency " $2,66(5
£1 Georgia currency " $4,2855
£1 Sterling money " $4,444f
£1 Canada currency " $4,00
£1 Irish currency « $4,10|
£1 Scotch currency " $0,370fe
The following sums for mental exercise, will be found
of much practical use, and should be practised till they can
be readily answered.
Examples in N. England currency for mental ex-
ercise.
1. If 6 shillings equal a dollar or 100 cts. how many
cents in 3 shillings ? in 2 shillings ? in 1 shilling ? in 4
shillings 1 in 5 shillings ?
2. If 1 shilling is 16| cts. how many cents in 6 pence ?
in 3 pence ? in 9 pence 1 in 4 pence ? in 7 pence? in
8 pence ? in 11 pence ?
3. How many cents in Is. 6d. ? in Is. 9d. ? in Is. 3d.?
in 2s. 6d ? in 2s. 9d. ? in 3s. 4d. ? in 5s. 6d. ? in 7s. 6d.?
178 ARITHMETIC. SECOND PART.
in 8s. 6d. ? in 9s. 1 in 9s. 6d. ? in 10s. 6d. ? in lis. ? in
Us. 6d. ? in 12s. ?
4. If 6d. is 8i cts. how many cents is 3d. ? how many-
is Id. ? now many is 2d. ?
5. If you buy 8 yds. of ribbon at Is. 6d. yer yd. how
much will the whole cost ?
6. If you buy 2f yds. of muslin at 2s. 6d. per yd. how
much will it cost in dollars and cents ?
7. If you buy 31 yds. of ribbon at Is. 9d. per yd. how
much will it cost ?
8. If you buy a brush for 2s. 3d. and a penknife for
4s. 6d. and a comb for Is. 6d. how much is given for the
whole ?
9. If you pay 3s. 6d. for scissors, 2s. 4d. for a thimble,
and Is. 9d. for needles, how much will the whole cost ?
10. If linen is 4s. 6d. per yd. how much will 4f yds.
cost ?
11. If a piece of calico is 2s. 3d. per yd. how much will
Q\ yds. cost ?
12. If muslin is 4s. 6d. per yard, what will 2f yds cost ?
13. How much is lHd. ? 10id. ? 9|d. ? 8£d. ? 7id. 1
12£d.? 16id. ?
Examples in N. York ovrrency for mental exercise.
1. If a dollar in N. York currency is 8s. how many
cents in 4s. ? in 2s. ? in Is. ? in 5s. i in 6s. ? in 7s. ? in
9s. ? in 10s. ? in lis. ? in 12s. ? in 13s. ? in 14s. ? in
15s. ? in 16s, ?
2. If one shilling is 12| cts. how many cents in 6d. ? in
3d. ? in Id. ? in 2d. 1 in 4d. 1 in 7d. ? in 8d. ? in 9d. ?
in lOd. ? in lid.?
3. How many cents is Is. 6d. N. York currency ? is
2s. 6d. ? is 3s. 6d. ? is 5s. 3d. ? is 6s. 9d. 1 is 4s. 8d. ?
Questions can be asked in the other currencies in the
same manner.
REDUCTION OF CURRENCIES. 179
REDUCTION OF CURRENCIES TO FEDERAL
MONEY.
Sums of this kind, which are too complicated to be
done mentally, may be performed on the slate, by the
following rules.
To REDUCE BRITISH CURRENCY TO FEDERAL MONEY.
Reduce the sum to a decimal of the pound order, and di-
vide the answer by -fa.
The. reason of this rule is that a dollar is JL. of a £ of
this currency, and therefore there are as many dollars in
the sum as there are ^ in it.
Note. Before reducing any currency to Federal mon
ey, the sum must be reduced to a decimal of the pound or'
der. After this process the following rules may be used.
To reduce Canada currency.
As a dollar is £ of a £ in this currency, there will be as
many dollars as there are } in the sum. Therefore,
Reduce the sum to the decimal of a £ and divide itby\.
To reduce New England Currency.
As 1 dollar is ,3 of a pound in this currency, so there
are as many dollars in a sum of N. England currency as
there are ,3 in it. Therefore
Reduce the sum to the decimal of a £ and divide it by ,3.
To reduce New York Currency.
As 1 dollar is ,4 of a pound in this currency, there will
be as many dollars in a sum of New York currency, as
there are ,4 in it. Therefore
Reduce the sum to the decimal of a £ and divide it by ,4.
To reduce Pennsylvania Currency.
As 1 dollar is f of a £ in this currency there are as
many dollars in the sum as there £ contained in it.
Therefore
Reduce the mm to the deciinal of a £ and divide it by %.
180 ARITHMETIC. SECOND PART.
To reduce Georgia Currency.
As 1 dollar is -fa of a pound in this currency there are
as many dollars in the sum as there ^ contained in it.
Therefore
Reduce to the decimal of a £ and divide the sum by £^.
1
REDUCTION OF FEDERAL MONEY TO THE
SEVERAL CURRENCIES.
To change a sum in federal money to the different cur-
rencies, the preceding process is reversed, and the sum is
to be multiplied [instead of divided) by the several frac-
tions. The answer is found in pounds and decimals of a
pound. The decimal can be reduced to units of the shil-
ling and pence order by a previous rule. (p. 176.)
Examples.
1. Reduce Is. 6d. in the several currencies to Federal
money.
Answers.
Of Canada Currency, it is $,30
British,
N. England,
N. York,
Penn.
Georgia,
$,333i
$,25
$,187i
$,20 '
$,321f
2. Reduce 4id. of the several currencies to Federal
money.
3. Reduce 4s. 6d. of the several currencies to Federal
money.
4. Reduce 35£ 3s. 7{d of the several currencies to
Federal money.
5. Reduce $118,25 to the several currencies.
REDUCTION OF CURRENCIES. 181
I
Ansicers to the lasi
!.
p
£ s. d.
In Canada (
British,
currency,
it is
29 " 11 " 3
26 " 12 " H
N. Eng.
N. York,
M
((
35 " 9 " 6
47 « 6 "
Penn.
u
44 " 6 « 10i
Georgia,
<(
27 " 11 " 9a"
4
Reduce 2s. Dd. of N. England currency to the same
value in all other currencies.
Reduce 4s. 6d. N. York currency to the same value in
all the other currencies.
REDUCTION FROM ONE CURRENCY TO AN-
OTHER.
The following table will enable the pupil to reduce a
sum from one currency to another, with more facility than
by any other method. Each fractional figure shows the
relative value of a sum in one currency to the same sum
in another currency.
For example, the £ in the second perpendicular and the
fourth horizontal column, shows that£l sterling is £ of the
number which expresses the same value in New England
currency. Thus £6 sterling is £ of the number which
expresses the same value in New England currency. That
is, £6 is £ of the answer to be obtained when the same
value is expressed in New England currency. To find
the answer, we reason thus. If 6£ is three fourths, £2 is
one fourth, and 8£ is the answer. Thus dividing by |.
Rule for changing a sum in one currency, to the
same value in another currency.
To change a sum in a currency written in the upper space
to one written in the right hand space, divide by the fraction
that stands where both spaces meet.
If there are shillings, pence and farthings in the sum,
first reduce them to the decimal of a £.
16 .
182
ARITHMETIC. SECOND PART.
TABLE
EXHIBITING THE COMPARATIVE VALUES OF THE 3EVERAL CURRENCIES.
ANY SUM EXPRESSED IN
m
a
k
it
on
CD
it
O
CD
O
ft
1— 1
H
t— •
CO
ft
o
p
P
•
ft
H
ft
CD
13
3
ft
as
ft
CO
o
o
o
IS
1
27
1
1 2
7
8 I
2 5
2 7 6 9
5
54
i
9
5
36
4
27 ,1
£ Scot.
o
•}
K
K
>
3
K
Of
e
S
a
X
»
B
w
M
O
z
1
4
9
I 6
7
1 2
112 5
184 6
5
8
3
4
1 6
1 5
27
4
£N.Y.
8
3
3
5
2 8
4 5
6
9 2 3
2
3
4
5
1 6
1 5
36
5
£ Pen.
1
3
3 i
4 9
75
92 3
5
6
5
4
4
3
9 limes
£N.E.
4 times
9
1
I 4
1 5
9
9 2 3
6
5
3
2
8
5
5 4
5
£ Can.
92.3
22 5
92 3
i o c n
64 6 1
6 7 5
9 2 3
9
92 3
7 5
92 3
6
18 4 6
112 5
2 7 6 9
2 5
£ Irish.
3
7
2 7
2 8
6 7 5
6 4 6 1
15
1 4
9
7
4 5
2 8
7
8 1
7
£ Geo.
4
9
I 2 8
1 27
10
9 2 3
1
9
4
3
5
3
1 6 ! ' 2
-a 1 times
1
£ Ster.
9 7
4 1 3
22 5
9 2 3
1
4
3
1
3
8
4 2 7
10 10
$F.M.
EXAMPLES FOR PRACTICE.
1. Reduce £4 N. E. to F. M.
2. Reduce 2£ 3s. 9d. N. E. to F. M
8. Reduce £6 N. Y. to F. M.
4. Reduce £8 ; 4 ; 9 N. Y. to F. M.
5. Reduce £3 ; 2 ; 3 Penn. to F. M.
6. Reduce $152.60 to N. E.
7. Reduce $196.00 to N. E.
8. Reduce $629.00 to N. Y.
Ans. $13.3331.
Ans. $7,2913.
Ans. $15.00.
Ans. $20.593f .
Ans. $8.30.
Ans. £45 ; 15 ; 7.2.
Ans. $58 ; 16.
Ans. 251 ; 12.
9! Reduce £35 ; 6 ; 8 sterling to N. E
Ans. £47 ; 2 ; 2 ; 2|
10. Reduce £120 N. E. to Can. Ans. £100
11. Reduce £155 ; 13 N. E. to Sterling.
Ans. £116; 14; 9
REDUCTION OF CITRKENCIK3. 183
12. Reduce £104 ; 10 Can. to N. Y. Ans. £167 ; 4.
13. Reduce £300 ; 10 ; 4 ; 2 Can. to Penn.
Ans. £450 ; 15 ; 6 ; 3.
14. Reduce £937 ; 18; 11 ; 1 N. E. to Geo.
Ans. £721; 14; 8; 3.
15. Reduce $224 ; 60 to Can. Ans. £56 ; 3.
16. Reduce £225 ; 6 N. E. to F. M. Ans. $752.00.
17. Reduce £880 15 ; 11 ; 1 Perm, to Sterling.
Ans. 528 9 ; 6 ; 3.
18. Reduce £6,750 Irish to Geor. Ans. £6,461.
19. Reduce £1,846 Ster. to Irish. Ans. £2,000.
20. Reduce £1,722 ; 18 ; 9 ; 3 N. E. to N. Y.
Ans. £2,298 ; 5 ; 1.
21. Reduce £2,114 ; 1 ; 3 Can. to F. M.
Ans. $8,456.25.
22. Change £784 ; 5 ; 6 ; 2 Penn. to Geor.
Ans. £487; 19; 10 ; 2ff.
23. Change £923 Sterling to Irish.
24. Change £,4000 Irish to Sterling.
25. Change £157 ; 8 ; 3 ; 3 N. Y. to N. E.
26. Change £1,654 ; 3 ; 8 ; 1 Penn. to N. E.
27. Change £ 947 ; 9 ; 4 ; 2 N. E. to F. M.
28. Change $1,444.66 to N. E. To N. Y. To Penn.
29. Change $945.22 to N. Y. To Geor. To Can.
30. Change £1,846 ; 15 ; 4 N. E. to F. M. To Penn
To Georgia.
31. Change $4,444,444f to Sterling.
32. Reduce £1,000,000 Sterling to F. M.
ARITHMETIC.
THIRD PART.
NUMERATION.
In the following, Third Part, there will be a reveiw of
the preceding subjects, embracing the more difficult ope-
rations. The rules and explanations will not be repeated,
as the pupils can refer to them in the former part.
ROMAN NUMERATION.
Before the introduction of the Arabic figures, a method
of expressing numbers by Roman Letters was employed.
As this method has not entirely gone out of use, it is im-
portant that it should be learned. The following letters
are employed to express numbers.
I. One. X. Ten.
II. Two. L. Fifty.
III. Three. C. One Hundred.
IIII. or IV. Four. D. Five Hundred.
V. Five. M. One Thousand.
The above letters, by various combinations, are made
to express all the numbers ever employed in Roman Nu-
meration.
RULE FOR WRITING AND READING ROMAN NUMBERS.
As often as a letter is repeated, its value is repeated.
When a less number is put before a greater, the less number
is subtracted. But when the less number is put after the great-
er, it is added to the greater.
Examples. In IV. the less number I. is put before the
greater number V. and is to be subtracted, making the
number four.
In VI. the less number is put after the greater, and it is
to be added, making the number six.
NUMERATION.
J8. r i
In XL. the ten is subtracted from the fifty.
In LX. the ten is added to the fifty.
The following is a table of Roman Numeration :
TABLE.
LXXXX. orXC
C.
CC.
CCC.
CCCC.
D. or Io*
DC.
DCC.
DCCC.
DCCCC.
M. or Cly.t
IQO. or v.t_
CtJIOO- or X.
iooo-
ICCCIooO-orC.
M
MM
\q. is used instead of D. to represent five hundred, and for every additional
Q. annexed at the right hand, the number is increased ten times.
t C13. is used to represent one thousand, and for every C. and 3. put at each
end, the number is increased ten times,
t A line over any number increases its value one thousand timts.
One
I.
Ninety
Two
II.
One hundred
Three
in.
Two hundred
Four
IIII. or IV.
Three hundred
Five
V.
Four hundred
Six
VI.
Five hundred
Seven
VII.
Six hundred
Eight
VIII.
Seven hundred
Nine
Villi, or IX.
Eight hundred
Ten
X.
Nine hundred
Twenty
XX.
One thousand
Thirty
XXX.
Five thousand
Forty
XXXX.orXL.
Ten thousand
Fifty
L.
Fifty thousand
Sixty
LX.
Hundred thousa
Seventy
LXX.
One million
Eighty
LXXX.
Two million
Write the following numbers in Roman letters :
5. 7. 3. 9. 8. 10. 4. 14. 5. 1*5. G. 16.
26. 36. 30G. 1. 11. 111. 7. 17. 77. 777.
1800. 1832. 1789.
Read the following Roman numbers :
VI. XIX. XXIV. XXXVI. XXIX. LV. XLI.
LXIV. LXXXVIII. XCIX. MDCCCXVIII.
OF OTHER METHODS OF NUMERATION.
By the common method of numeration, ten units of out
order, make one unit of the next higher order. But it is
equally practicable, to have any other number than ten.
to constitute a unit of a higher order. Thus we might
have six units of one order make one unit of the next
higher order. Or twelve units of one order might make
one of the next higher order.
The number which is selected to constitute units of the
higher orders, is called the radix of that system of nu-
meration. 16.
186 ARITHMETIC. THIRD PART.
The radix of the common system is ten, and this num-
ber it is supposed was selected, because men have ten
fingers on their hands, and probably used them in express-
ing numbers.
Before the introduction of the Arabic figures, Ptolemy
introduced a method of numeration, in which sixty was
the radix. The Chinese and East Indians use it to this
But in Ptolemy's system there were not sixty different
characters employed. Instead of this, the Roman method
of numeration was used for all numbers as far as sixty,
and then for the next higher orders the same letters were
used over again, with an accent (') placed at the right.
For the third order two accents (") were used, and for the
fourth order three accents ("').
To illustrate this method by Arabic figures, 31' 23
signifies 31 sixties and 23.
We have some remnants of this method in the division
of time into 60 seconds for a minute, and 60 minutes for
an hour, and also the division of the degrees of a circle,
into 60 seconds to a minute, and 60 minutes to a degree.
EXERCISES IN NUMERATION, COMMON, VULGAR, AND
DECIMAL.
(See rides on pages 53, 58, and 64.)
1. Two million, four thousand, one hundred and
six.
2. Two hundred thousand, and six tenths.
3 Twenty six billion, six thousand, and fifteen thou-
sandths.
4. Two hundred and sixty thousand millionths.
5. One sixth of two apples how much and how written ?
6. One ninth of twenty o?'anges, how much, and how
written ?
Is it a proper or improper fraction ?
7. One sixth of four bushels how much? how written ?
is it a proper, or improper fraction ?
8. One tenth of forty bushels, how much ? how written ?
is it a proper or improper fraction ?
9. One tenth of three oranges, how much ? how express-
ed ?
ADDITION. 187
10. Three tenths of three oranges, how much 1 how ex-
pressed.
11. Four sixths of twelve apples, how much ? how ex-
pressed ?
12. Three thousand tenths of thousandths.
13. Four billions, six thousand, and five ten thousandths.
14. Sixteen billions, three hundred and six millions,
five hundred thousand, and six tenths of millionths.
15. Five trillion, five million, five units, and three
hundred and sixty five millionths.
It!. Sixteen hundred and twenty four, and four tenths of
billionths.
ADDITION.
Let the pupil add the following numbers :
1
Two hundred and six million ; twenty four thousand, five
hundred and six.
Thirty seven billion, twenty six thousand and three.
Four hundred and seventy nine billion, six hundred and
sixty seven million, nine hundred and eighty four thou-
sand, six hundred and ninety nine.
Fifteen million, seventy seven thousand, nine hundred.
Thirty six trillion, four hundred million, and six.
Four quadrillion, seventeen million, three hundred and
six.
Six quadrillion, fourteen trillion, seventeen million, four-
teen thousand, three hundred and nine.
Twenty four sextillion, five hundred million and nine.
2
Sixteen thousand, four hundred and sixty four, and nine
tenths.
Two hundred and sixty nine million, fourteen hundred
and three, and thirteen hundredths.
Forty four million three thousand and six, and twenty
thousandths.
Five hundred million, nine hundred and ninety nine
thousand, eight hundred and seventy nine, and two hun-
dred and sixty four tenths of thousandths.
188 ARITHMETIC. PART THIRD.
Six hundred and seventeen thousand, four hundred and
sixty eight, and five hundred and seventy nine hundredths
of thousandths.
Forty six million, nine thousand, and seventy millionths.
3
Add two twelfths, three fourths, and four sixths. (See
page 166.)
Add twenty four fiftieths, sixteen tenths, and twenty
halves.
5
Add forty nine eightieths, seventy nine fortieths, and two
hundred thousandths.
6
Add nine twenty sevenths, thirteen forty fourths, and
twenty nine seventieths .
SUBTRACTION.
1
From,
Three hundred and sixty nine million, four hun-
dred twenty seven thousand, three hundred seventy six,
Subtract,
Two hundred and ninety three million, four hun-
dred and eighty three thousand, nine hundred and
eighty seven.
From,
Twenty four billion six hundred and thirteen mil-
lion, four hundred and forty four thousand, eight hundred
and eighty six, and twenty nine hundredths,
Subtract,
Sixteen billions, twenty four thousand and sixteen, and
four hundred and six thousandths.
3
From,
Sixty four sextillion, ninety trillion, seven billion, twen-
ty nine million, forty thousand three hundred and six, and
twenty nine tenths of millionths,
MULTIPLICATION. 189
Subtract,
Fourteen quintillions, nine quadrillions, seven trillions,
fourteen thousand and eighty, and seven hundredths of
milliontJis.
4
From nine twelfths, subtract two fifths. (See page
, 166.)
5
From thirteen twenty sevenths, subtract three twenty
fourths.
6
From threejijths, subtract twenty nine seventy sevenths.
7
From,
Twelve hundred and six, four hundred and twentieths,
Subtract,
Four hundred and nine, nine hundred and ninetieths.
MULTIPLICATION.
1. Multiply 32694302 by 365.
2. Multiply 24,2 by 27 (See page 108.)
3. Multiply 321,92 by 236.
4. Multiply 236,49 by 2,4.
5. Multiply 47,2935 by 2,08432.
6. Multiply 870,24 by 32,94.
7. Multiply 14 yds. 3 qrs. 2 na. by 28.
8. Multiply 8 le. 2 m. 6 fur. 22 po. by 362.
9. Multiply 2 bu. 3 pk. 1 qr. 1 pt. by 172.
10. Multiply | by 3 (Seepage 112.)
11. Multiply ^ bv 48.
12. Multiply i| by 32.
13. Multiply 12 by § (See page 116.)
14. Multiply 24 by f.
15. Multiply 324 by T V
16. Multiply 2342 by £&.
17. Multiply f by | (Se'e page 123.)
18. Multiply f by f .
19. Multiply f by f
20. Multiply T <V by ||.
21. Multiply Iff by if
190 ARITHMETIC. THIRD PART.
SUMS FOR MENTAL EXERCISE.
Multiply 5 and § by f .
Let such sums be stated thus :
One fourth of 5 is 1 unit, and 1 remains.
This remaining 1 is changed to sixths and added to the
| making f
One fourth of one sixth would be 5 ' ¥ , therefore one
fourth of eight sixth is ^\.
In the above operation we find that one fourth of 5 is 1
and 1 remains. This remainder is changed to sixths and
added to the fraction §, and then is divided by 4. The an-
swer is 1 and J» T .
1. If a yard of muslin cost 2^, what will a a yard cost ?
What is i of 2± ?
2. If a barrel of wine cost lOi dollars, what cost \ a
barrel ? What is \ of 10i ?
3. If 4 bushels of rye cost 8 dollars and § , what cost 2
bushels ? What is \ of 8f ?
4. If you have 21 oranges, and givei away, how much
do you keep ? What is \ of 2£ ? What is £ of 8} ?
5. If 9 bushels of wheat cost 18f dollars, how much is
that a bushel ? What is 1 of 18a ?
6. If 12 pieces of linen cost 16| dollars, how much is
that by the piece 1
7. If 8 gallons of brandy cost 14| dollars, how much
is that a gallon ?
8. If 8 yards of broadcloth cost 28| dollars, how much
is that a yard ?
9. How much would 4 yards cost ?
10. If a man bought 8 barrels of cider for 25| dollars,
how much is one barrel ?
11. How much is 9 barrels ?
12. If 12 yards of linen cambric cost 42| dollars, what
would 7 yards cost ?
13. If you have 12 dollars and f and lose 3i times as
much, how much do you lose ?
We first multiply the 12| by 3 and then by \.
14. 3 times 12 is 36 and 3 times f is | or 1, which
added to 36 is 37.
15. 12 and | multiplied by \ is 6 and £ which added to
the 37 makes 43 and i.
16. Multiply 8 and f by 4 and i.
DIVISION. 191
In doing this sum, first multiply the 8 and then the frac-
tion by 4, and add the products together. Then multi-
ply the 8, and the fraction by |, and add these to the
former products.
Thus 4 times 8 is 32. Four times § is f , which is 1
and f . This added to 32 is 33 and f .
One third of 8 is 2, and 2 remains. Add 2 to the 33
making 35. Change the remainder to fifths and add the
f making l 3 . One third of one fifth would be T ' T , there-
fore i of '/ is j-f, which added to 35 and § makes 35 and
=-1, which equals 36 and T \.
17. Multiply 5 and £ by 2 and i.
18. Multiply 12 and £ by 2 and }.
19. Multiply 9 and T \ by 6 and f .
20. Multiply 7 arid f by 4 and f .
21. Multiply 11 and f by 3 and ±.
22. Multiply 8 and g by 8 and |.
23. Multiply 10 and | by 7 and |.
24. If you buy 9 and | gallons of wine and return 2f
times as much, how much do you return ?
25. If one boy takes 12 apples and J and another takes
5£ times as many, how many does the last take ?
26. If one room requires 12 and -•}- yards of carpeting,
and another requires 3 and | times as much, how much is
required ?
DIVISION.
1. Divide 9123648 bv 79632.
2. Divide 246,2 bv 23 (See page 152.)
3. Divide 2394,609 by 235.
4. Divide 3246,9214 by 39.
5. Divide 32.4 by 9,4 (See page 152.)
6. Divide 3294 by 2,79.
7. Divide 324,976 by 2,4 (See page 152.)
8. Divide 329,42 by 3,24.
9. Divide 329021,4639 by 296,029.
10. Divide 112£. 12s. 7d. 4qrs. by 38.
11. Divide 29 yds. 2 qrs. 3 na. by 39.
12. Divide 2 m. 5 fur. 17 po. 3 yds. by 91.
192 ARITHMETIC. PART THIRD.
13. Divide 12 by J (Seepage 137.)
14. Divide 128 by f .
15. Divide 418 by t *l.
16. Divide 324 by fr
17. Divide 3297 by if.
18. Divide if by 6 (See page 140.)
19. Divide ff by 16.
20. Divide fff by 27.
21. Divide ftfi by 361.
22. Divide 5 |§f 5 by 249.
23. Divide f by f (See page 144.)
24. Divide y % by f
25. Divide^ by f.
26. Divide |f| by iif .
EXAMPLES FOR MENTAL EXERCISE.
1. Divide i by}. Divide ^bv y L (See page 164.)
2. Divide i by r \. Divide j by JL.
3. Divide § by T \. Divide i by i.
4. Divide § by §•. Divide f by t l.
5. Divide £ by T 4 g . Divide T 8 ¥ by j.
6. Divive T 8 2 by §. Divide £ by T V
7. How many times is i contained in T 9 ^ ?
8. How many times is f contained in J ?
9. How many times is -2 contained in -| ?
10. How many times is § contained in | ?
11. If beef is i of a dollar a pound, how much can be
bought for i of a dollar.
12. If a yard of muslin cost ^ °f a dollar, how much
can be bought for i of a dollar ?
In case the divisor and dividend have whole numbers
with the fractions, the whole numbers must be reduced also,
with the fractions, to a common denominator.
Thus if we wish to find how many § there are in 4 and
J, we must change the 4 and f to twelfths, and the § to
twelfths also, and then divide as before. Thus; 4 and |
is ff , and | is ^.
In 57 twelths, there are 7 times 8 twelfths, and one
twelfth left over. This one twelfth, is one eighth of the
divisor ^ .
REDUCTION. 193
The answer then is 7 and }. That is, 4f contains f ,
just 7 times and i of another time.
Again ; how often is 2| contained in 5f ? First, reduce
the divisor and dividend to fractions of a common deno-
minator.
2f is f §, and 5f is ff.
Divide 69 twelfths by 32 twelfths, and the answer is 2
and 5 twelfths left over.
This 5 twelfths is 5 thirty secondths of the divisor. For
Jive twelfths is ^ of 32 twelfths.
1. How many times is If contained in 8|?
2. How many times is 2f contained in 5£ ?
3. How many times is 9| contained in 16| ?
4. If you distribute 13£ lbs. of flour among a certain
number of persons, and give 2$ lbs. to each, to how many
persons do you give ?
5. If 4| bushels of wheat last a family one week, how
long will 12| bushels last them?
6. If 5| tons of hay will keep a horse 6 months, how
many horses will 12f tons keep during the same time?
7. If a cistern is filled in 3f of an hour, how many times
will the cistern be filled in 12§ hours?
8. If you distribute 18f dollars among the poor, and
give 2f dollars to each person, to how many do you give ?
9. At 3] dollars a lb. how many pounds of gum can be
bought for 24f dollars 1
10. How many times is f contained in 2f ?
11. How many times is 5| contained in Si 1
12. How many times is 2| contained in 14| ?
13. How many times is 3| contained in 7|?
14. How many times is 5f contained in 12f ?
REDUCTION.
1. In 29 gallons how many quarts ? (See page 157.)
2. In 65 pints how many gallons ?
3. In 2£. 14s. 9d. 3qrs. how many farthings ?
4. In 923469 farthings, how many pounds, shillings,
and pence ?
5. Reduce ^" to a decimal. (Seepage 162.)
17
194 ARITHMETIC. PART THIRD.
6. Reduce §£ to a decimal.
7. Reduce ^f § to a decimal.
8. Reduce § % and T 4 g to a common denominator. (See
page 166.)
9. Reduce T 4 ¥ f§ ^ T to a common denominator.
10. Reduce T \ ^_ jl t a common denominator.
11. Reduce J^- to its lowest terms. (Seepage 169.)
12. Reduce a4|| to its lowest terms.
13. Reduce || to its lowest terms.
14. Reduce | of a guinea to the fraction of a pound.
(See page 171.)
15. Reduce xgf^g to the fraction of a foot.
16. Reduce § of f of f of a pound to the fraction of a
shilling.
17. Reduce | of £ of 3 shillings to the fraction of a
pound.
18. What is the value of | of a ton in lbs ? (See page
172.)
19. How many ounces in £ of a lb. Apothecary's
weight ?
20. How many pints in T \ of a bushel ?
21. Reduce 8 oz. 6 pwts. to the fraction of a lb. Troy.
(See page 173.)
22. Reduce 4 days 16 hours to the fraction of a year.
23. Reduce 36 gals. 4 qts. to the decimal of a hogs-
head. (See page 174.)
24. Reduce lid. 3qrs. to the decimal of a shilling.
What is the value of ,169432 of a ton ? (See page 176.)
25. What is the value of ,24694 of a £ ?
What is the value of ,396 of an hour 1
26. Reduce 7s. 8d. of each of the different currencies
to the same value in Federal money. (See page 179.)
27. Reduce $6, 29 to the same value in each of the
different currencies. (See page 180.)
INTEREST.
In conducting business, men often find it necessary to
borrow money of each other, and it is customary to pay
those who lend, for the use of their money until it is re-
turned.
INTEREST.
195
The sum of money lent, is called the principal.
The sum paidjor the use of money, is called interest.
Amount is the principal and interest added together.
Per annum signifies by the year.
It is customary to pay a certain sum for every hundred
dollars, pounds, &c. Thus in New England six dollars a
year is paid for the use of every hundred, and in New
York seven dollars for every hundred that is borrowed.
The expressions six per cent , seven per cent , &c signify
that six or seven dollars are paid for every hundred bor-
rowed. Per signifies for and cent, is the abreviation of
centum, the Latin word for hundred. Rate per cent., then,
signifies rate by the hundred. When a man borrows a
sum of money he gives to the one of whom he borrows a
writing in this form :
$500 ,00. Hartford, April 1, 1832.
On demand I promise to pay D. F. Robinson or order,
five hundred dollars with interest, value received.
Samuel Jones.
This is called a note and is said to be on interest.
In this case the borrower, Samuel Jones, is obligated to
pay six dollars a year for each hundred dollars, till the
$ 500 are returned.
In Connecticut, the law does not permit men to receive
any more than 6 per cent, interest ; in New York it allows
7 per cent., and the rate by law varies in the different
states. When the rate per cent, is not mentioned, it is
always to be understood that the interest is what is allowed
by the laws of the state where the note is given.
Usury is taking more interest than the law allows.
Legal interest is the rate allowed by law.
In all notes on interest, if no particular rate per cent.
is mentioned, it is always understood to be legal interest
that is promised. In this work 6 per cent, will be under-
stood when no rate per cent, is mentioned.
Sometimes it occurs that when a man has borrowed a
sum of money, after a time he wishes to pay a part of the
debt.
In this case, when the payment is made, the note which
was given to the lender is taken, and an endorsement is
written on it, stating that such a part of the note was paid
196 INTEREST.
at a particular time. After this the borrower only pays
interest for that part of the debt which remains unpaid.
Notes are given either with or without interest. If the
words " with interest" are not written, a note is under-
stood to be without interest. If a note is given without in-
terest, promising to pay at a certain time, after that time
has expired, the note draws interest from that time.
Notes are given sometimes, promising to pay the inter-
est annually, but oftener the interest is not to be paid until
the note is paid.
When interest is paid only upon the sum lent, it is call-
ed simple interest.
But when the yearly interest is added each year to the
principal, and then interest is taken upon both principal and
interest, it is called compound interest.
The laws of the several states forbid taking compound
interest ; but a man who has lent money, can collect the in-
terest every year, and put it out at interest, and thus gain
compound interest.
But when a man borrows, if the creditor does not collect
the interest every year, he cannot be compelled to pay in-
terest on the interest.
In calculating interest, the rate per cent, is a certain
number of hundredths of the sum lent. Thus if 1 per cent,
is paid for $100, it is T ^ part of the sum lent. If 6 per
cent, is paid, it is the T f „ part of the sum lent.
For this reason all calculations in interest are sums in
decimal multiplication. We divide by the denominator to
find one hundredth, by means of the separatrix, and multi-
ply by the numerator to find the required number of hun-
dredths. For example, if we wish to find the interest of
$263 for one year, at 6 per cent, we must obtain the T f ^
part of the $263. This is done by dividing by the denomi-
nator 100, by means of a separatrix, and multiplj-ing by
the numerator 6. In this case the multiplication is done
first.
$263
6
$15,78
INTEREST. 197
The rale per cent, therefore, may always be written as
a decimal fraction of the order of hundredths.
1 per cent, is written ,01
2 per cent. " ,02
i per cent. " ,005
i per cent. " ,0025
£ per cent. " ,0075
Write 2i per cent, as a decimal fraction.
2 per cent, is ,02, and a per cent, is ,005. Ans. ,025.
Write 4 per cent, as a decimal fraction. 4i per
cent. 4f per cent. 5 per cent. 7£ per
cent. — — 8 per cent. 8f per cent. 9 per
cent. 9i per cent. 10 per cent. (10 per cent.
is T \\; decimally, ,10.) 10i per cent. 11 per
cent. 121 per cent. 15 per cent.
1. If the interest on $l,for 1 year, be 6 cents, what will
be the interest on $ 17 for the same time ?
It will be 17 times 6 cents, or 6 times 17, which is the
same thing : —
$17
,06
1,02 Answer ; that is, 1 dollar and 2 cents.
To find the interest on any sum for 1 year, it is evident
we need only to multiply it by the rate per cent, written as a
decimal fraction. The product will be the interest re-
quired.
What is the interest of $121 at 3f per' cent. ? at2£ per
cent. ? at 8i per cent. ? at 9f per cent. ? at 4i per cent. ?
When we wish to obtain the interest for several years,
we have only to multiply the interest of one year by the num-
ber of years.
Examples.
What is the interest of $214 for 4 years at 2£ per cent ?
for 3 yrs. ? for 9 yrs. ? for 24 yrs. ?
What is the interest of $364,41 for 8 yrs. at 61 per ct. ?
What is the interest sf $1000 for 120 yrs. ?
Ans. $7200.
It may often be needful to calculate the interest on a
sum, for a less time than a year.
17*
198 INTEREST.
When this is needful the following mode is the most
simple and expeditious.
Let the interest be at 6 per cent, as that H& the most
common rate.
At 6 per cent, each dollar gains 6 cents a year, (or 12
mo.) 6 cents for 12 mo. is i a cent (or 5 mills) for 1 month.
As 30 days is called a month, in calculating interest, 5
mills a month, is 1 mill for every 6 days.
Interest at 6 per cent then gains on each dollar,
$,06 a year
$,005 a month
$,001 for every 6 days, and | of a mill for each day.
Whenever therefore we wish to calculate the interest of
any sum for less than a year, we can first calculate the in-
tcrest on one dollar for the given time, calculating 5 mills
for every month, 1 mill for every 6 days, and J- of a mill for
each odd day.
After finding the interest for one dollar we can multiply
this interest by the number of dollars in the sum.
Examples.
What is the interest of $36 at 6 per cent for 9 mo. 12
days? for 6 mo. 3 days ? for 8 mo. 18 days ?
Note. The fractions of a mill had better be changed to
decimals. Thus instead of writing 5£ mills we can write
} 0055— 5| mills can be written ,0053+. (The sign of ad-
dition is added to the last because there are more decimal
orders that may be added.)
What is the interest of $334 for 4 mo. 2 d. ? for 9 mo.
5 d. ? for 7 mo. 4 d. ?
What is the interest of $826 for 2 d. ? for 5 mo. 3 d. ?
for 16 d. ? for 9 mo. 16 d. ?
If it is wished to obtain the interest of any sum for less
than a year, at any other than 6 per cent, the method is, to
find the interest at 6 per cent, and then take such parts of
it, as the rate mentioned, is parts of 6 per cent.
Thus if we wish to find the interest of $560 for 4 mo.
8 d. at 5 per cent, we first find the interest at 6 per cent.
INTEREST. 199
for that time, and then subtract £ of the sum from itself.
• For the interest at 5 per cent is £ less than the interest at
ty per cent.
Thus if the rate is 3 per cent, we must take i of the in-
terest at 6 per cent.
If it is 4 per cent, we must take f (or |) of the interest
at 6 per cent.,&c.
What is the interest of $241,62 cents for 8mo. 6d. at 2
per cent. 1 at 3 per cent. ? at 4 per cent. ? at 9 per cent. ?
at 12i per cent ? at 15 per cent. ?
What is the interest of $54.81 for 18mo. at 5 per cent. ?
Ans. $4.11.
What is the interest of $500 for 9mo. 9d. at 8 per cent. ?
Ans. 31.00.
What is the interest of $62.12 for lmo. 20d. at 4 per
cent. ? Ans. $0,345.
What is the interest of $85 for lOmo. 15d. at 12£ per
cent. ? Ans. $9,295.
Rules for calculating Interest.
To find the interest for years.
Multiply the sum by the rate per cent, as a decimal of the
order of hundredths, and the interest for one year is found.
Multiply this answer by the number of years.
To find the Interest for months and days.
Calculate the interest on one dollar for the given time,
thus ; calculate 5 mills for every month, 1 mill for every
six days, and a of a mill for each odd day. Add these to.
getlier and multiply the answer by the number of dollars and
cents in the sum, pointing off decimals according to rule.
If the rate is any other than 6 per cent., calculate the in.
teresl at 6 per cent., and then add to, or subtract from the
sum such parts of itself, as the rate per cent, is parts of 6
per cent.
200 arithmetic. part third.
Examples.
What is the interest of $116,08 for llmo. 19d. ?
Ans. $ 6,422
of $200 for 8mo. 4d. ? 8,132
of 0,85 for 19mo.? 0,08
of 8,50 for lyr. 9mo. 12d. ? 0,909
of 675 for lmo. 2 Id. ? 5,737
of 8673 for 10d.? 14,455
of 0,73 for lOmo. ? 0,36
Rule for Sterling Money.
When the principal is pounds, shillings, and pence, reduce
the sum to the decimal of a £ (see page 174), and proceed
as in federal money. The answer is in decimals of a £,
and must be changed back to units (see page 176).
What is the interest of £36 ; 9s. 6id. for lyr. ?
Ans. £2.3s. 9£d.
What is the interest of £36 ; 10s. for 18mo. 20d. ?
Ans. £3.8s. lid.
What is the interest of £95 for 9mo. ? Ans. £4.5s. 6d.
Find the interest on <£13 ; 3 ; 6 for 1 yr. A. 15s. 9id.
Find the interest on £13 ; 15s. 3id. for lyr. 6mo.
A. £1 ; 4 ; 9id.
Find the interest on £75 ; 8 ; 4 for 5yrs. 2mo.
A. £23.7s. 7d.
Find the interest on £174 ; 10 ; 6 for 3yrs. 6mo.
A. £36. 13s.
Find the interest on £325 ; 12 ; 3 for 5yrs.
A. £97.13s.8d.
Find the interest on £150 ; 16 ; 8 for 4yrs. 7mo.
A. £41. 9s. 7d.
VARIOUS EXERCISES IN INTEREST.
To find the Principal, when the Time, Rate and
Amount are known.
If in lyr. 4mo. the interest and principal of a sum
amount to $61,02 what is the principal ?
We first find what will be the amount of a dollar with its
INTEREST. 201
interest, for the given time. This amounts to $1,08. Now
as every dollar in the original sum gained 8 cents interest,
there were as many dollars as there are $1.08 in $61,02.
Ans. $56,50.
Rule.
Find the interest of $lfor the given time and add to it.
Divide the sum given by this amount.
Examples.
What principal at 8 per cent, will amount to $85,12 in
lyr. 6 mo.? Ans. $76.
What principal at 6 per cent, will amount to $99,311
in llmo. 9d. ? Ans. $94.
To find the Principal, when the Time, Rate and
Interest are known.
What sum put at interest at 6 per cent, will gain $10,50 1
One dollar put at interest for that time, would gain $,08
and therefore it requires as many one dollars as there are
$,08 in $10,50. Ans. $131,25.
Rule.
Find tJiv interest of $1 for the given rate and time. Di-
vide the interest given by this, and the quotient is the principal.
Examples.
A man paid $4,52 interest at the rate of G per cent, at
the end of lyr. 4mo. ; what was the principal 1
A. $56,50.
A man received $20 for interest on a certain note at
the end of lyr. at the rate of 6 per cent. ; what was the
principal ? Ans. $333,3331.
To find the Rate, when the Principal, Interest,
AND TlME ARE KNOWN.
If $3,78 is paid for using $54, lyr. 6mo. what is the
rate per cent. ?
If this sum were at interest at one per cent, it would
produce $.54.
As many times therefore as $,54 is contained in $3,78
so much more than 1 per cent, is the rate.
202 arithmetic. third part.
Rule.
Divide the given interest by what would be the interest of
the same sum at 1 per cent.
If $2,34 is paid for the use of $468 for lmo. what is the
rate percent. ? Ans. 6 per cent.
At $46,80 for the use of $520 for 2yrs. what is it per
cent. ? Ans. 4£ per cent.
To find the Time, when the Principal, Rate and
Interest are known.
What is the time required to gain $3,78 on $36 at 7
per cent. ?
We first find what would be the interest on that sum for
one year, at 7 per cent.
This would be $2,52. As many times as this sum is
contained in the interest mentioned in the sum, so much
more time than one year is required.
Rule.
Divide the interest given, by the interest lohich the princi-
pal would gain rj the same rate, in one year.
Paid §20 for the use of $600 at 8 per cent. ; whut was
the time ? Ans. 5mo.
. Paid $28,242 for the use of $217,25 at 4 per cent ;
what was the time ? Ans. 3yrs. 3mo.
ENDORSEMENTS.
In transacting business, it is often necessary to calculate
interest upon notes, where partial payments have been
made, and endorsed upon the note. For example, a man
borrows $500, and gives his note promising to repay it
with interest.
Two years after, he pays $150, and has it endorsed.
Then two years after, he pays $75, and has it endorsed.
At the end of six years he is ready to settle the note, and
the question is how much interest he shall pay.
There are different modes established by the laws of
different states on this subject.
INTEREST. 203
The three following are the most common. The first
is the one which formerly was most commonly used.
First Method.
Find the amount of the principal for the%>liole time.
Find the amount of each payment to the time of settlement.
Add the amounts of the payments, and subtract them from
ike amount of the principal.
Example.
On April 1st, 1825, 1 gave a note to A. B. promising to
pay him $300 for value rec'd. and interest on the same at -
6 per cent, till settlement.
Oct. 1, 1825, I paid #100.
April 16, 1826, paid $50.
Dec. 1, 1827, paid $120.
What do I owe on April 1st, 1828 ?
$ cts. m.
300,00,0 principal dated April 1, 1825. yrs. mo. da.
54,00,0 interest up to April 1st, 1828. 3. 0. 0.
354,00,0 amount of principal.
100,00,0 1st payment, Oct. 1, 1825.
15,00,0 interest up to April 1st, 1828. 2. 6. 0.
1 15,00,0 amount of 1st payment.
50,00,0 2nd payment, April J 6th, 1820.
5,87,5 interest up to April 1st, 1828. 1. 11. 15.
55,87,5 amount of second payment.
120,00,0 3rd payment, Dec. 1st, 1827.
2,40,0 interest up to April 1st, 1828. 0. 4. 0.
122,40,0 amount of 3rd payment.
55,87,5 " 2nd payment.
115,00,0 " 1st payment.
293,27,5 total amount of payments.
204 ARITHMETIC. THIRD PART.
354,00,0 amount of principal.
293,27,5 total amount of payments subtracted.
A. 60,72,5»remains due April 1st, 1828.
Rule in Massachusetts.
Find the Amount of the Principal to the time when one
payment, or several payments together, exceed the interest due.
From this subtract the payments and the remainder will be a
new Principal. Proceed thus till the tune of settlement.
Examples.
For value received I promise to pay James Lawrence
$116,666 with interest.
May 1st, 1822.
$116,666. John Smith.
On this note were the following endorsements.
Dec. 25, 1822, received $16,666.
July 10,1623, " $ 1,666.
Sept. 1, 1824, " $ 5,000.
June 14, 1825, " $33,333.
April 15, 1826, " $62,000.
What was due August 3, 1827 ?
Ans. $23,775.
The first principal on interest from May 1,
1822, $116,666
Interest to Dec. 25, 1822, time of the first
payment (7 months 24 days), 4,549
Amount, $121,215
Payment, Dec. 25, exceeding interest then due, 16,666
Remainder for a new principal, 104,549
Interest from Dec. 25, 1822, to June 14,
1825 (29 months, 19 days), 15,490
Amount, $120,039
INTEREST. 205
Payment. July 10, 1823, less than interest
then due, $ 1,606
Payment, Sept. 1, 1824, less than
interest then due, 5,000
Payment June 14, 1825, exceed,
ing interest then due, 33,333
$39,909
Remainder for a new principal (June 14,
1825), 80,040
Interest from June 14, 1825, to April 15,
1826 (10 months 1 day), 4,015
Amount, $84,055
Payment, April 15, 1825, exceeding inte-
rest then due, 62,000
Remainder for a new principal (April 15,
1826), $22,055
Interest due Aug. 3, 1827, from April 15,
1826 (15 months 18 days), 1,720
Balance due Aug. 3, 1827, $23,775
The Rule now adopted in Connecticut, is founded on
the principle that interest is to be paid by the year, so that
if a man pays before a year is ended, he receives interest
on all he pays, from the time he pays it, to the end of the
year when the interest is due.
Rule in Connecticut.
If the payment be made at the end of a year or more, add
the interest due on the whole sum, at this time, to the princi-
pal, and subtract the payment.
Whenever other payments are made, proceed in the same
manner, calculating interest on the principal from the time
of the last payment.
If payment is made before a year has elapsed (from the
date of the note, or jrom the last payment), find the amount
of the principal for one year. Find also the amount of the
payment from the time of payment to the end of the year when
18
206 ARITHMETIC. PART THIRD.
the interest would he due, and subtract the latter from the
former. If hotoever a year extends beyond the time of settle-
ment, find the amount up to that time, instead of for a year.
If any remainder after subtraction, be greater than the
preceding principal, then the preceding principal is to be con-
tinued as the principal for the succeeding time instead of the
remainder, and the difference to be regarded as so much
unpaid interest.
Let interest on the following note be calculated by the
three different rules.
A note for 820,000 is given July 1st, 1825.
1st payment, January 1st, 1826, $1400
2d do. : January 1st, 1827, 2000
3d fajM. September 1st, 1827, 5000
Settlement. January 1st, 1829.
What is due on the note ?
Ansu'ers.
By the common rule, $14,90^,00
By the Massachusetts rule, 15,212,96
By the Connecticut rule, 15,209,47
Let the following be calculated by the Connecticut rule.
$1000,00 Hartford, Jan. 4, 1826.
On demand I promise to pay James Lowell, or order,
one thousand dollars with interest ; value received.
Hiram Simpson.
On this note were the following endorsements.
Feb. 19, 1827, received
June 29, 1828, "
Nov. 14, 1828,
Dec. 29, 1831,
$200.00
500.00
260.00
25.00
What is the balance, June 14,
1832
?
Answer $204.49
Find the balance due on the following note by the Mas-
sachusetts rule.
v
$500.00. Hartford, Feb. 1, 1820.
Value received 1 promise to pay A. B. or order five
hundred dollars with interest. Samuel Jones.
INTER Ei
IT.
2
Endorsements.
May 1, 1820, received,
$40.00
Nov. 14, 1820, "
8.00
April 1, 1821,
12.00
May 1,1821,
30.00
How much remains Sept. 16,
1821 ?
Am. $445.57
207
Find the balance due on the following note by the Con-
necticut rule.
For value received I promise to pay (i. B. or order,
eight hundred and seventy-rive dollars, with interest.
$875.00.
S \MUEL J^MiS.
Hartford, Jan. 10, 1821.
Endorsements.
Aug. 10, 1824, received
S260.00
Dec. 16, 1825,
300.00
March 1, 1826, "
50.00
July 1, 1827, «
150.00
What was due Sept. 1, 1828 ?
Ans. $474.95.
The three rules used above, are all considered as objec-
tionable.
By the first rule, when a man pays a part of his debt,
his payments are not applied to discharging the interest,
but entirely to lessening the principal. By this rule, if a
man should borrow a sum and promise to pay it, with the
interest, in twenty-five years, if he should simply pay what
would be the yearly interest, and have it endorsed, at the
end of 25 years the debt would be entirely extinguished.
Whereas if he should wait till the end of the time agreed
upon, he would have to pay the original sum borrowed,
and the yearly interest upon it also.
The objection to the other two rules is, that the man
who makes payments before the time of settlement, actu-
ally is obliged to pay more than one who pays nothing be-
fore that time. Thus the most punctual man is obliged to
pay more than the negligent.
Compound Interest is the only method, which will do
exact justice to both creditor and debtor. For a man who
208 ARITHMETIC. PART THIRD.
lends money is fairly entitled to receive interest at the end
of each year ; and then by investing the interest in other
stock, he can obtain compound interest. The borrower,
therefore, who detains this yearly interest, ought, in jus-
tice, to pay what the creditor could gain, if the debtor
were punctual.
COMPOUND INTEREST.
Compound Interest is an allowance made for the use of
the sum lent, and also for the use of the interest when it is
not paid.
Rule.
Calculate the Interest, and add it to the principal at tlie
eyul of a year. Make the Amount a new principal for the
next year, with which proceed as before, till the time of set-
tlement.
1. What is the compound interest of $256 for 3 years,
at 6 per cent. ?
$256 given sum, or first principal.
,6
15,36 interest ) be added t ther .
256,00 principal, $ to
271,36 amount, or principal for 2d year.
,00
16,2816 compound interest, 2d year, > added
271,36 principal, do. $ together.
287,6416 amount, or principal for 3d vear.
,06
17,25846 compound interest, 3d year, ) added
287,641 principal, do. $ together.
304,899 amount.
256 first principal subtracted.
A $4^,899 compound interest for 3 years.
INTEREST.
209
3. At 6 per cent, what will be the compound interest,
and what the amount, of $ 1 for 2 years ? what the
amount for 3 years ? for 4 years ? for 5 years ?
for 6 years ?
for seven years ? for 8
Ans. to the last, $1,593+
years ?
It is plain that the amount of $2 for any given time,
will be 2 times as much as the amount of $ 1 ; the amount
of $3 will be 3 times as much, &c.
Hence, we may form the amounts of $1, for several
years, into a table of multipliers for finding the amount of
any sum for the same time.
TABLE,
Showing the amount of $1, or 1£, &c. for any number
of years, not exceeding 24, at the rates of 5 and 6 per
cent, compound interest.
Y'rs.
1
2
3
4
5
5 per cent.
1,05
1,1025
1,15762 +
1,21550+
1,27628+
6| 1,34009+
1,40710 +
1,47745 +
1,55132 +
1,62889+
1,71033+
12 1,79585 +
7
8
9
10
11
6 per cent.
1,06
1,1236
1,19101 +
1,26247 +
1,33822 +
1,41851 +
1,50363 +
1,59384 +
1,68947+
1,79084 +
1,89829 +
2,01219 +
Y'rs. | 5 per cent.
13 1,88564+
14|1,97993 +
152,07892 +
162,18287 +
172,29201 +
132,40661+
192,52695
202,65329+
2112,78596 +
222,92526 +
233,07152 +
6 per cent.
2,13292+
2,26090+
2,39655+
2,54035+
2,69277+
2,85433+
3,02559+
3,20713-|-
3,39956 +
3,60353 +
3,81974 +
24 3,22509+14,04893+
will be
Note 1. Four decimals in the above numbers
sufficiently accurate for most operations.
Note 2. When there are months and days, you may
first find the amount for the years, and on that amount cast
the interest for the months and days ; this, added to the
amount, will give the answer.
3. What is the amount of $000,50 for 20 years, at 5 per
cent, compound interest ? at 6 per cent. ?
$1 at 5 per cent., by the table, is $2,65329 ; therefore,
2,65329 X600,50=$1593,30+ Ans. at 5 per cent. ; and
3,20713x600,50=^1925,881+ Ans. at 6 per cent.
18*
4
210 ARITHMETIC. PART THIRD.
4. What is the amount of $40,20 at 6 per cent, com-
pound interest, for 4 years ? for 10 years ?
for 18 years ? for 12 years ? for 3 years and
4 months 1 for 24 years, 6 months, and 18 days ?
Ans. to last, $168,137
DISCOUNT.
Discount is a deduction made from a debt, for paying it
before it is due.
If, for example, I owe a man $300 two years hence,
and am willing to pay him now, I ought to pay only that
sum, which, with its interest, would in two years, amount
to $300.
The question then is, what sum, together with its inte-
rest at 6 per cent., would, in two years, amount to $300 ?
Such operations are performed by the rule for finding
the principal, when the time, rate, and amount are given
(see page 201).
The sum which, in the time mentioned, would, by the
addition of its interest, amount to the sum which is due, is
called the present worth.
What is the present worth of $834, payable in 1 yr. 7 mo.
6 days, discounting at the rate of 7 per cent. 1
Ans. $750
What is the discount on $321,63, due 4 years hence,
at 6 per cent. 1 Ans. $62,26
What principal, at 8 per cent., in 1 yr. 6 mo. will amount
to $85,12 ? Ans. $76
What principal, at 6 per cent, in 1 1 mo. 9 d. will amount
to $99,311 ? Ans. $94
How much ready money must be paid for a note of $18,
due 15 months hence, discounting at the rate of 6 per
cent. ? Ans. $16,744
STOCK, INSURANCE, COMMISSION, LOSS AND
GAIN, DUTIES.
Stock is a name for money invested in banks, in trade,
in insurance companies, or loaned to a national govern-
ment, for the purpose of receiving interest.
STOCK, INSURANCE, &C. 211
Persons who invest money thus, are called stockholders.
When stockholders can sell their right to stock, for more
than they paid, it is said that stock has risen, and when
they cannot sell it for as much as they paid, it is said that
stock has fallen.
Stock is bought and sold in shares, of from $50 to $100
a share.
The nominal value of a share is the amount paid, when
the stock was first created.
The real value is the sum for which a share will
sell.
When stock sells for its nominal value, it is said to be at
par.
When it sells for more than its nominal value, it is said
to be above par, and when for less it is below par.
When stock is above par it is said to be so much per
cent, advance.
An Insurance Company, is a body of men, who in re-
turn for a certain compensation, promise to pay for the
loss of property insured.
The written engagement they give, is called a Policy.
The sum paid to them for insurance, is called Premium.
Commission, is a certain sum paid to a person called a
correspondent, agent, Jactor, or broker, for assisting in trans-
acting business.
Loss and Gain refer to what is made or lost, by mer-
chants, in their business.
The calculations relating to stock, insurance, commis-
sion, loss and gain, and duties, are performed by the rule
for calculating interest, when the time is one year ?
Rule.
Multiply the sum given, by the rate per cent, as a deci-
mal. (See page 199.)
Examples.
Stock.— 1. What is the value of $350.00 of stock at
105 per cent, that is, at 5 per cent, advance ?
Ans. $367.50
The rate here is 105 per cent —105 hundredths. The
question, then, is, what is 105 hundredths of 350 ; or, mul-
tiply 350 by 1.05.
'212 ARITHMETIC. PART THIRD.
2. What is the value of 35 hundred dollar shares of
stock, at | per cent, advance? Rate 1.0075
Ans $3,526.25
3. At I12i per cent, what must I pay for $7,564.00 of
stock? Rate"l. 125. Ans. 8,509.50
4. What is the value of $615.75 of stock, at 30 per
cent, advance ? Ans. $800,475
5. What is the value of $7,650.00 of stock at 119i per
cent. ? Ans. $9,141.75
6. What is the value of $1,500.00 of stock at 110 per
cent. ? Ans. $1,650.00
7. What is the value of $3700 bank stock at 95i per
cent., that is at 4| per cent, below par? Ans. $3,533.50
Insurance. — 1. What premium must be paid for the
insurance of a vessel and cargo, valued at $123,425.00,
at 15i per cent. ?
15i per cent. =.155, and the question is, what is .155
of 123,425. Ans. $19,130,875
2. What must I pay annually for the insurance of a
house worth $3,500.00, at If per cent. ? Ans. $61.25
3. What must be paid for the insurance of property, at
6 per cent., to the amount of $2,500.00 ? Ans. $150.00
4. What insurance must be paid on $375,000-00, at 5
per cent. ? . Ans. $18,750.00
5. What premium must be annually paid for the insur-
ance of a house worth $10,050.00, at 3 per cent.; and a
store worth $15,875.00, at 4 per cent. ; and out houses
worth $3,846 00, at 5 per cent. ?
6. What premium must be annually paid for the insur-
ance of a Factory worth $30,946.00, at 10 per cent. ; and
7 duelling houses, worth 875.00 each, at 8 percent. ; and
3 grist mills, worth $1,930.00, apiece, at 7 per cent. ; and
one storeing house, worth $9,859.00, at 6 per cent. ? Also,
what is the average rate of insurance on the whole ?
7. If I pay $930.00 annually for insurance, at 5 per
cent., what is the value of the property insured?
Here 930 is .05 of the answer ; 930 ~. 05=$ 18,600 An.
Profit and Loss. — 1. Sold a bale of goods at $735.00,
by which I gain at the rate of 6 per cent. What sum do
I gain? Ans. $44.10
2. In selling 50 hhds. of molasses, at 38 dollars a hhd.,
I gain 10 per cent. What is my gain ? Ans. $190.00
DUTIES. 213
3. In selling 25 bales of cloth, each containing 27
pieces, and each piece 50 yards, a merchant gained 20
per cent, on the cost, which was 10 dollars a yard. What
did he gain, and what did he sell the whole for ?
Ans. Gain $07,500.00. Whole $405,000.00
4. A merchant gained at the rate of 15 per cent, in
selling the following articles : 6 hhds. of brandy for which
he paid $1.50 per gal. ; 7 barrels of flour, cost 11 dollars
a barrel ; 2 quintals of fish, cost 4 cents a pound; 16
hhds. of molasses, cost 56 cents per gal. and 25 bis. of
sugar, containing each 175 lbs., cost 9 cents per lb. What
was his gain on the whole, and what did he receive in all ?
Commission. — 1. If my agent sells goods to the amount
of $2,317.46, what is his commission at 3i per cent. ?
Ans. $75.31745
2. What commission must be allowed for a purchase of
goods to the amount of $1,286.00, at 2~ per cent.
Ans. $32.15
3. What commission shall I allow my correspondent for
buying and selling on my account, to the amount of
$2,836.23, at 3 per cent. ?
4. A merchant paid his correspondent $25.00 commis-
sion on sales to the amount of $1,250.00. At what per
cent, was the commission ?
He paid him T ||^= J '-^ T |^=.02=2 per cent. Ans.
Duties. — Duty is a certain sum paid to government
for articles imported.
When duty is at a certain rate on the value, it is said to
be ad valorem, in distinction from duties imposed on the
quantity.
An Invoice is a written account of articles sent to a pur-
chaser, factor, or consignee.
In computing duties, ad valorem, (or ad vol. as it is
commonly written,) it is usual in custom houses to add one
tenth to the invoice value, before casting the duty. This
makes the real duty one tenth greater than the nominal du-
ty. It will be equally well to make the rate one tenth
greater, instead of increasing the invoice.
1. Find the duty on a quantity of tea, of which the in-
voice is $215.17, at 50 per cent.
Ans. $118.3435=$! 18.343X
214 ARITHMETIC. PART THIRD.
In this example we may add, as directed above, one
tenth of 215.17, to 215.17. Thus, 215.17+21.517=
236 687. Then 236.687X50=$ 118.3435. Or we may
add to the rate .50, one tenth of itself=.05 : thus, .50-j-
.05=55. Then, 215.17 x-55=$l 18.3435, as before.
2. Find the duty on a quantity of hemp at 131 per cent.,
of which the invoice is $654.59. The second of the
above modes is recommended. Another might be used,
viv. : to find, first, the duty on the invoice at the given
rate, and add to it one-tenth of itself. Thus, 654.59 X
13-' =$88. 36965. Ans. 897.2066 15
3. What is the duty on a quantity of books, of which the
invoice is $ 1,670.33," at 20 per cent. ? Ans. $367.4726
EQUATION OF PAYMENTS.
Equation of payments is a method of finding a time for
paying several debts due at different times, all at once ;
and in such a way that both creditor and debtor will have
the same value, as if the debts were paid at the several
times promised.
For if a man pays a debt before it is due, the creditor
gains ; if he pays it after it is due, the debtor gains.
In how many months will $1 gain as much at interest
as $8 will gain in 4 months ? Now as the $1 is 8 times
less than 8, it will require 8 times more time, or 8x4=32
months.
In how many months will the interest on $9 equal the
interest on $1 for 40 months ?
Supposing a man owes me $12 in 3 months, $18 in 4
months, and $20 in 9 months. He wishes to pay the
whole at once ; in what time ought he to pay ?
$12 for 3 months=$l for 36 months.
$18 for 4 months=$l for 72 months.
$20 for 9 months=$l for 180 months.
$50 288 months.
Now it appears that it will be the same to him to have
$1 for 36, for 72, and for 180 months, as it would to have
the 12, the 18, and the 20 dollars for the number of
months specified.
RATIO. 215
He might therefore keep $1 just 288 months, and it
would be the same as keeping the $50 for the number of
months specified. But as the whole sum of money lent
was -$50, he may keep this only one fiftieth (j\) of the time
he might keep $1. Therefore divide the 288 months by
the 50, and the answer is 5|| months.
Rule for finding the mean time of several payments.
Multiply each sum by the time of its payment. Divide the
sum of these products by the sum of the payments, and the
quotient is the mean time.
A man is to receive $500 in 2mo. ; $100 in 5 mo. ;
$300 in 4 mo. If it is paid all at once, at what time should
the payment be made 1
A man owes me $300, to be paid as follows ; i in 3
months ; } in 4 months ; and the rest in G months ; what
is the mean time for payment ? Ans. 4| months.
RATIO.
The word ratio means relation ; and when it is asked
what ratio one number has to another, it means in what
relation does one number stand to another.
Thus, when we say the ratio of 1 to 2 is ^, we mean
that the relation in which 1 stands to 2 is that of one half
to & whole.
Again, the ratio of 3 to 4 is £ , that is, 3 is J of 4, or
stands in the relation of £ to the 4. So also the ratio of
4 to 3 is | ; for the 4 is 4 thirds of 3, and stands to it
therefore in the relation of |.
What is the relation of 11 to 12 1 of 12 to 11 ?
When therefore we find the ratio of one number to an-
other, we find what part of one number another is.
Then the ratio of 4 to 6 is £ ; that is, 4 is 4 sixths of 0.
The ratio of one number to another, then may always
be expressed by a fraction in which the first number (called
the antecedent) is put for numerator, and the second number
(called the consequent) is put for denominator. Thus the
ratio of 8 to 4 is f. This is an improper fraction, and,
changed to whole numbers, is 2 units. The ratio of 8 to
216 ARITHMETIC. PART THIRD.
4, then, is 2. That is, 8 is twice 4, or stands to 4 in the
relation of a duplicate or double*
PROPORTION.
When quantities have the same ratio, they are said to be
proportional to each other. Thus the ratio of 2 to 4 is £,
and the ratio of 4 to 8 is i ; that is, 1 has the same relation
to 2, that 4 has to 8, and therefore these numbers are called
proportionals. Again, 4 is the same portion or part of 8,
as 10 is of 20, and therefore these numbers are called
proportionals. A proportion, then, is a combination of equal
ratios.
Points are used to indicate that there is a proportion
between numbers. Thus 4:8::9:18 is read thus ; 4 has
the same ratio to 8, that 9 has to 18. Or more briefly, 4
is to 8, as 9 to 18.
It will always be found to be the case in proportionals,
that multiplying the two antecedents into the two consequents,
produce the same product.
Thus, 2 : 4 : : 6 : 12
Here let the consequent 4 be multiplied into the ante-
cedent 6, and the product is 24 ; and let the atecedent 2
be multiplied into the consequent 12, and the product also
is 24.
If then we have only three terms in a proportion, it is
* The pupil needs to be forewarned that there is a difference be-
tween French and English mathematicians in expressing Ratio.
The French place the antecedent as denominator, and the conse-
quent as numerator. The English, on the contrary, place the ante-
cedent as numerator, and the consequent as denominator. It seems
desirable that there should be an agreement on this subject, in school
books at least. Two of the most popular Arithmetics now in use,
have adopted the French method, viz. Colburn and Adams. It seems
needful to mention this, that pupils may not be needlessly perplexed,
if called upon to use different books.
The method used here, is the English ; as the most common, and
as most consonant with perspicuity of language. For there seems
to be no propriety in saying that the relation of 2 to 4 is A. The
ratio between these two numbers maybe either | or |, but the rela-
tion of 2 to 4, to use language strictly can be nothing but a.
PROPORTION. 217
easy to find the fourth. For when we have multiplied one
antecedent into one consequent, we know that the term
left out, is a number that, multiplied into the remaining
term, would produce the same product.
Thus let one term be left out of this proportion.
8 : 4 : : 12 :
Here the consequent is gone from the last ratio. We
multiply the antecedent 12 into the consequent 4, and the
answer is 48. We now know that the term left out, is a
number which, multiplied into the 8, would produce 48.
This number is found by dividing 48 by 8, the answer is 6.
Whenever, therefore, a term is wanting to any propor-
tion, it can be found by multiplying one of the antecedents
by one of the consequents, and dividing the product by
the remaining number.
What is the number left out in this proportion?
3 : 12 : : 24 :
What is the number left out in this proportion ?
9 : 8 : : 27 :
In a proportion, the two middle terms are called the
means, and the first and last terms are called the extremes.
Rule for finding a fourth term in a Proportion.
Multiply the means together, and divide the product by
the remaining number.
It is on this principle, that what is commonly called the
" Rule of Three," is constructed. By this process, we
find a fourth term when three terms of a proportion are
given.
Such sums as the following are done by this rule.
If 4 yards of broadcloth cost $12, what cost 9 yards ?
Now the cost is in proportion to the number of yards ;
that is, the same ratio exists between the number of yards,
as exists between the cost of each.
Thus, — as 4 yards is to 9 yards, so is the cost of 4 yards
to the cost of 9 yards. The proportion, then, is expressed
thus :
yds. yds. $
4 : 9 : : 12 :
Here the term wanting, is the cost of 9 yards ; and if
we multiply the means together, and divide by the 4, the
19
218 ARITHMETIC. PART THIRD,
answer is 27 ; which is the other term of the proportion;
and is the cost of 9 yards.
Again, if a family of 10 persons spend 3 bushels of malt
a week, how many bushels will serve at the same rate
when the family consists of 30 ?
Now there is the same ratio between the number of
bushels eaten, as between the numbers in the family.
That is, as is the ratio of 10 to 30, so is the ratio of 3 to
the number of bushels sought.
Thus, 10 : 30 : : 3 :
Rule of Proportion ; or Rule of Three.
When three numbers are given, place thai one as third term,
which is of the same kind as the answer sought. If the an-
swer is to be greater than this third term, place the greatest of
the remaining numbers as the second term, and the less num-
ber as first term. But if the answer is to be less, place the
less number as second term, and the greater as first.
In either case, multiply the middle and third terms toge-
ther, and divide the product by the first. The quotient is thf
answer, and is always of the same order as the third term.
Note. This rule may be used both for common, com-
pound, and decimal numbers. If the terms are compound,
they must be reduced to units of the lowest order men-
tioned.
Many of the sums which follow will be better understood
if performed by the mode of analysis, which has been
explained and illustrated in a former part.
For example, we will take the first sum done by the rule
of proportion.
If 4 yards of broadcloth cost $12, what cost 9 yards !
We reason thus, — If 4 yards cost $12, one yard must
cost a fourth of $12. Therefore, divide $12 by 4, and we
have the cost of one yard. Multiply this by 9, and we
have the cost of 9 yards.
(It is usually best to multiply first and then divide, and
it has been shown that this is more convenient, and does
not alter the answer.)
PROPORTION. 219
Lot the following sums be done by the Rule of Propor-
tion, and then explained by analysis.
1 . If the wages of 15 weeks come to 64 dols. 19 cts. what
is a year's wages at that rate ? Ans. $222, 52 cts. 5m.
2. A man bought sheep at 81.11 per head, to the amount
of $51.6 ; how many sheep did he buy ? Ans. 46
3. Bought 4 pieces of cloth, each piece containing 31
yds. at 1 6s. 6d. per yard, (New England currency,) what
does the whole amount to in federal money ? Ans. 8341
When a tun of wine cost 8140, what cost a quart ?
Ans. 13 cts. S^m.
4. A merchant agreed with his debtor, that if he would
pay him down 65 cents on a dollar, he would give him up
a note of hand of 2 19 dollars 88 cts. I demand what the
debtor must pay for his note ? Ans. $162.42 cts. 2m.
5. If 12 horses eat 30 bushels of oats in a week, how
many bushels will serve 45 horses the same time ?
Ans. 112^ bushels.
6. Bought a piece of cloth for 848.27 cts. at 8 1.19 cts.
per yard ; how many yards did it contain ?
Ans. 40 yds. 2 qrs. ^
7. Bought 3 hhds. of sugar, each weighing 8cwt. lqr.
12 lb. at $7.26 cts. per cwt. ; what come they to ?
Ans. $182.1 ct. 8 in.
8. What is the price of 4 pieces of cloth, the first piece
containing 21, the second 23, the third 24, and the fourth
27 yards, at 81.43 cts. a yard ?
Ans. $135.85 cts. 21+23+24+27=95 yds.
9. Bought 3 hhds. of brandy, containing 61, 62, 62£
gals, at 81 38 cts. per gallon. I demand how much they a-
mount to ? Ans. $255.99 cts.
10. Suppose a gentleman's income is $1836 a year, and
he spends $3.49 cts. a day, one day with another, how much
will he have saved at the year's end? Ans. $562.15 cts.
11. A mereh't bought 14 pipes of wine, and is allowed 6
months credit, but for ready money gets it 8 cents a gallon
cheaper ; how much did he save by paving ready money ?
" Ans. $141.12 cts
12. Sold a ship for 537Z. and I owned f of her ; what
was my part of the money? Ans. £201.7s. 6d,
220 ARITHMETIC. PART THIRD.
13. If T 5 ? of a ship cost $718. 25 cents, what is the whole
worth ? 5 : 781,25 : : 16 : $2500 Ans.
14. If I buy 54 yards of cloth for £31. 10s. what did it
cost per Ell English 1 Ans. 14s. 7d.
15. Bought of Mr. Grocer 11 cwt. 3 qrs. of sugar, at
$8,12 per cwt. and gave him James Pay well's note for
£ 19.7s. (New England currency) the rest I pay in cash ;
tell me how many dollars will make up the balance.
Ans. $30,91
16. If a staff 5 feet long casts a shade on level ground 8
feet, what is the height of that steeple whose shade at the
same time measures 181 feet 1 Ans. I13i ft.
17. If a gentleman has an income of 300 English guineas
a year, how much may he spend, one day with another, to
lay up 500 dollars at the year's end ? Ans. $2,46cts. 5m.
18. Bought 50 pieces of kerseys, each 34 Ells Flemish, at
8s. 4d. per Ell English ; what did the whole cost ? £425
19. Bought 200 yards of cambric for £90, but being
damaged, I am willing to lose £7. 10s. by the sale of it ;
what must I demand per Ell English ? Ans. 10s. 3f d.
20. How many pieces of Holland, each 20 Ells Flemish,
may I have for £23.8s. at 6s. 6d. per Ell English ?
Ans. 6 pieces.
21. A merchant bought a bale of cloth containing 240 yds.
at the rate of $7i for 5 yards, and sold it again at the rate
of $11| for 7 yards ; did he gain or lose by the bargain,
and how much ? Ans. He gained $25,71 cts. 4m. +
22. Bought a pipe of wine for 84 dollars, and found it had
leaked out 12 gallons ; I sold the remainder at 12i cents
a pint ; what did I gain or lose ? Ans. I gained $30
23. A gentleman bought 18 pipes of wine at 12s. 6d. (N.
Jersey currency) per gallon ; how many dollars will pay
the purchase ? Ans. $3780
24. Bought a quantity of plate, weighing 15 lb. 11 oz. 13
pwt. 17 gr. how many dollars will pay for it, at the rate of
12s. 7d. (New York currency,) per ounce ?
Ans. $301,50 cts. 2 T \ m.
25 A factor bought a certain quantity of broadcloth and
drugget, which together cost £81 per yard, the quantity
of broadcloth was 50 yards, at 18s. per yard, and for every
5 yards of broadcloth he had 9 yards of drugget ; 1 de.
PROPORTION. 221
mand how many yards of drugget he had, and what it cost
him per yard ? Ans. 90 yards at 8s. per yard.
26. If I give 1 eagle, 2 dollars, 8 dimes, 2 cents and 5
mills, for 675 tops, how many tops will 19 mills buy ?
Ans. 1 top.
27. If 100 dollars gain 6 dollars interest in a year, how
much will 49 dollars gain in the same time 1
Ans. $2,94 cts.
28. If 60 gallons of water, in one hour, fall into a cistern
containing 300 gallons, and by a pipe in the cistern, 35 gal-
lons run out in an hour ; in what time will it be filled ?
Ans. in 12 hours.
29. A and B depart from the same place and travel the
same road ; but A goes 5 days before B, at the rate of 15
miles a day ; B follows at the rate of 20 miles a day ;
what distance must he travel to overtake A ?
Ans. 300 miles.
COMPOUND PROPORTION.
Compound proportion, is a method of performing such
operations in proportion, as require two or more Stat-
ings. It is sometimes called Double Rule of Three, be-
cause its operations can be performed by two operations of
the Rule of Three.
For example : If 56 lbs. of bread are sufficient for 7 men
14 days, how much bread will serve 21 men 3 days?
Here the amount of bread consumed depends upon two
circumstances, the number of days, and the number of men.
We will first consider the quantity of bread as depend-
ing upon the number of men, supposing the number of days
to be the same.
The proportion would then be this ;
7 men : 21 men : : 56 lbs. to the number of lbs. re-
quired.
Here we multiply the means together, and divide the
answer by 7, and the answer is 168. That is, if the time
was the same, viz. 14 days, the 21 men would eat 168 lbs,
in that time.
We now make a second statement thus :
19*
. 222 ARITHMETIC. PART THIRD.
14 days : 3 days : : 168 lbs. : number of lbs. requir-
ed.
The result of this statement is 36 lbs. which is the an-
swer.
In performing this operation, let the pupil notice that in
the first statement, the 56 was multiplied by the 21 and
the answer divided by 7. This gives the same answer as
would be given, did we divide first, and then multiply.
That is, 56 multiplied by 21, and the product divided by
7, is the same as 56 divided by 7 and the quotient multi-
plied by 21.
We divide by 7, to find how much one man would eat in
the same time, or 14 days, and multiply by 21, to find
how much 21 men would eat.
When we make the second statement, as we have found
how much 21 men would eat in 14 days, we divided the
quantity (168 lbs.) by 14, to find how much they would
eat in one day, and then multiply by 3, to find how much
they would eat in 3 days. But in this case also, the mul-
tiplication is done first.
Let the pupil also notice that the 56 lbs. was multiplied
by 21 and divided by 7, and then that the answer to this
(168 lbs.) was multiplied by 3 and divided by 14. Here
21 and 3 are used as multipliers, and 14 and 7 are used
as divisors.
The answer will be the same (as may be found by trial)
if 56 is multiplied by the product of these multipliers, and
the answer divided by the product of the divisors.
It is on this principle that the common rule in compound
proportion is constructed, which is as follows.
Rule of Compound Proportion.
Make the number which is of the same kind as the answer
required, the third term.
Take any two numbers of the same kind, and arrange
them in regard to this third term, according to the rule of
proportion. Then take any other two numbers of the same
kind, and arrange them in like manner, and so on till all the
numbers are used.
Then multiply the third term, by the product of the second
PROPORTION. 228
terms, and divide the answer by the product of the first terms.
The quotient is the answer.
Examples.
1. If a man travel 273 miles in 13 days, travelling only
7 hours a day, how many miles will he travel in 12 days at
the rate of 10 hours a day ?
Here the number, which is of the same kind as the
answer required, is the 273 miles, and this is put as third
term.
We now take two numbers of the same kind, viz. 13
days and 12 days, and placing them according to the rule
of simple proportion, the question would stand thus.
13 : 12 : : 273 :
We next take two other numbers of the same kind, viz.
10 hours, and 7 hours, and arrange them under the former
proportion according to the same rule, thus :
7 : 10$ ' * *
We now multiply the 273 by the product of 12 and 10,
and divide by the product of 13 and 7 and the quotient is
the answer.
We can explain this process analytically, thus.
We divide by 13, to find how much the man would tra-
vel in one day, at the rate of 7 hours per day.
We multiply by the 12, to find how much he would
travel in 12 days, at the same rate.
We divide by 7 to find how much he would travel in one
hour, and multiply by 10 to find how much he would trave 1
in 10 hours.
Let the pupils explain the following in the same man-
ner.
Examples.
2. If £100 in one year gain £5 interest, what will be
the interest of £750 for 7 years ? Ans. £262. 10s.
3. What principal will gain £262. 10s. in 7 years, at 5
per cent, per annum ? Ans. £750.
4. If a footman travel 130 miles in 3 days, when the
days are 12 hours long ; in how many days, of 10 hours
each, may he travel 360 miles ? Ans. 9|f days.
224 ARITHMETIC. PART THIRD.
5. If 120 bushels of corn can serve 14 horses 56 days,
how manv days will 94 bushels serve 6 horses ?
Ans. 102if days.
6. If 7 oz. 5 pwts. of bread be bought at 4fd. when corn
is at 4s. 2d. per bushel, what weight of it may be bought
for Is. 2d. when the price of the bushel is 5s. 6d. ?
Ans. 1 lb. 4 oz. 3££§ pwts.
7. If the carriage of 13 cwt. 1 qr. for 72 miles be £2.
10s. Gd. what will be the carriage of 7 cwt. 3 qrs. for
112 miles ? Ans. £2.5s. lid. l^q.
8. A wall, to be built to the height of 27 feet, was raised
to the height of 9 ft. by 12 men in 6 days ; how many men
must be employed to finish the wall in 4 days at the same
rate of working ? Ans. 36 men.
9. If a regiment of soldiers, consisting of 939 men, can
eat 351 quarters of wheat in 7 months ; how many soldiers
will eat 1464 quarters in 5 months, at that rate 1
Ans. 5483fl&.
10. If 248 men, in 5 days of 11 hours each, dig a trench
230 yards long, 3 wide and 2 deep ; in how many days
of 9 hours each, will 24 men dig a trench of 420 yards
long, 5 wide and 3 deep ? Ans. 2883 S eV
11. If 6 men build a wall 20 ft. long, 6 ft. high, and 4 ft,
thick, in 16 days, in what time will 24 men build one 200
ft. long, 8 ft. high, and 6 ft. thick ? Ans. 80 days.
12. If the freight of 9 hhds. of sugar, each weighing 12
cwt., 20 leagues, cost £16, what must be paid for the
freight of 50 tierces, each weighing 2| cwt., 100 leagues 1
Ans. £921. Is. 10|d.
13. If 4 reapers receive $11.04 for 3 days' work, how
many men may be hired 16 days for $103.04 ? Ans. 7 men.
14. If 7 oz. 5 pwt. of bread be bought for 4fd. when
corn is 4s. 2d. per bushel,what weight of it may be bought
for Is. 2d. when the price per bushel is 5s. 6d. ?
Ans. 1 lb. 4 oz. 3f|f pwts.
15. If 8100 gain $6 in 1 year, what will 400 gain in
9 months ?
16. If $100 gain $6 in 1 year, in what time will $400
gain $18 ?
17. If $400 gain $18 in 9 months, what is the rate per
cent, per annum 1
FELLOWSHIP. 225
18. What principal, at 6 per cent, per arm., will gain $18
in 9 months ?
19. A usurer put out $75 at interest, and, at the end of 8
months, received, for principal and interest, $79 ; I demand
at what rate per cent, he received interest. Ans. 8 per ct.
20. If 3 men receive £8 T \ for 19^ days work, how much
must 20 men receive for 1001 days' ? Ans. £305 0s. 8d.
21. If 40 men in 10 days, can reap 200 acres of grain,
how many acres can 14 men reap in 24 days ?
Ans. 168 acres.
22. If 14 men in 24 days, can reap 168 acres of grain ;
how many acres can 40 men reap in 10 days ?
Ans. 200 acres.
23. If 16 men in 32 days, can mow 256 acres of grass ; in
how many days will 8 men mow 96 acres ? Ans. 24 days.
24. If 4 men mow 96 acres in 12 days ; how many acres
can 8 men mow in 16 days ? Ans. 256
25. if a family of 16 persons spend $320 in 8 months ;
how much would 8 of the same family spend in 24 months ?
Ans. 8480
26. If a family of 8 persons in 24 months spend $480 ;
how much would they spend, if their number were doub-
led, in 8 months ? Ans. $320
27. If 12 men build a wall 100 feet long, 4 ft. high, and
3 ft. thick, in 40 days ; in what time will 6 men build one,
20 ft. long, 6 ft. high, and 4 ft. thick ?
FELLOWSHIP.
The Rule of Fellowship, is a method of ascertaining the
respective gains or losses of individuals engaged in joint
trade.
Let the pupils perform the following sums as a mental
exercise.
1. Two men own a ticket; the first owns i, and the
second owns f of it ; the ticket draws a prize of 40 dollars ;
what is each man's share of the money ?
2. Two men purchase a ticket for 4 dollars, of which
one man pays 1 dollar, and the other 3 dollars ; the ticket
draws 40 dollars ; what is each man's share of the money ?
226 ARITHMETIC. PART THIRD.
3. A and B bought a quantity of cotton ; A paid $100,
and B $200 ; they sold it so as to gain $30 ; what were
theirrespective shares of the gain ?
The value of what is employed in trade is called the
Capital, or Stock. The gain or loss to be shared is called
the Dividend.
Each man's gain or loss is always in proportion to his
share of the stock, and on this principle the rule is made.
Rule.
As the whole stock is to each man's share of the stock, so
is the whole gain or loss, to his share of the gain or loss.
4. Two persons have a joint stock in trade ; A put in
$250, and B $350 ; they gain $400 ; what is each man's
share of the profit ?
Operation.
A's stock, $250 1 Then,
B's stock, $350 j^ 60() . 250 . . 400 . $ 166#666 i A ' s gam .
Whole stock $000 j 600 : 350 : : 400 : $233,333^ B's gain.
The pupil will perceive that the process may be con-
tracted by cutting off an equal number of ciphers from the
first and second, or first and third terms ; thus, 6 : 250 : :
4 : 166.666|, &c.
It is obvious the correctness of the work may be ascer-
tained by finding whether the sums of the shares of the
gains are equal to the whole gain ; thus, $166.666§-f-
$233.333i=$400, whole gain.
5. A, B, and C, trade in company : A's capital was $175,
B's 200, and C's $500 ; by misfortune they lose $250 ;
what loss must eaeh sustain ? ( $ 50., A's loss.
Ans. 1$ 57.1421, B's loss.
($142.857i, C's loss.
6. Divide 600 among 3 persons, so that their shares may
be to each other as 1, 2, 3, respectively.
Ans. $100, $200, and $300
In assessing taxes, it is customary to obtain* an inven-
tory of every man's property, in the whole town, and also
a list of the number of polls. Each poll is rated at a tax
FELLOWSHIP.
^1
of a certain value. From the whole tax to be raised is
taken out what the tax on polls amounts to, and the re-
mainder of the tax is to be assessed on the property in the
town.
We may then find the tax upon 1 dollar, and make a
table containing the taxes on 1,2, 3, dzc. to 10 dollars ;
then on 20, 30, &c. to 100 dollars ; and then on 100, 200,
&c. to 1000 dollars. Then, knowing the inventory of
any individual, it is easy to find the tax upon his proper-
ty-
I. A certain town, valued at $64530, raises a tax ot
$2259.90; there are 540 polls, which are taxed $,60
each ; what is the tax on a dollar, and what will be A's
tax, whose realestate is valued at $1340, his personal pro-
perty at 8874, and who pays for 2 polls ?
540 x ,60 = $324, amount of the poll taxes, and
$2259,90,— $324=1935,90, to be assessed on property.
$645301: $1935,90 : : $1 : ,03 ; or,»|£g tV=> 03 > tax on 81
TABLE.
dolls.
,30
,60
30
dolls, dolls
Tax on 1 is ,03
« 2 " ,06
« 3 " ,09
« 4 " ,12
<«
,15
6 « ,18
7 « ,21
8 " ,24
9 « ,27
dolls.
Tax on 10 is
« 20 "
dolls. dolls.
Tax on 100 is 3,
" 200 « 0,
" 300 " 9,
" 400 " 12,
" 500 « 15,
" 600 " 18,
" 700 " 21,
« 800 « 24,
" 900 « 27,
'« 1000 " 30,
Now, to find A's tax, his real estate being $1340, I
find by the table, that
The tax on $1000 - is - 30,
The tax on 300 . 9,
The tax on 40 1,20
40
50
60
70
80
90
,90
1,20
1,50
1,80
2,10
2,40
2,70
Tax on his real estate - - - $40,20
In like manner I find the tax on his personal > ^g oo
property to be $ '
2 polls, at ,60 each, are 1,20
Amount, $67,62
228 ARITHMETIC. PART THIRD.
2. What will B's tax amount to, whose inventory is 874
dollars real, and 210 dollars personal property, and who
pays for 3 polls ? Ans. $34.32
3. What will be the tax of a man, paying for 1 poll,
whose property is valued at $3482 ? at $768 ?
Ans. to the last, $140.31
Two men paid 10 dollars for the use of a pasture 1
month ; A kept in 24 cows, and B 16 cows ; how much
should each pay ?
4. Two men hired a pasture for $10 ; A put in 8 cows
3 months, and B put in 4 cows 4 months; how much
should each pay ?
The pasturage of 8 cows for 3 months is the same as oi
24 cows for 1 month, and the pasturage of 4 cows for 4
months is the same as of 16 cows for 1 month. The shares
of A and B, therefore, are 24 to 16, as in the former ques-
tion. Hence, when lime is regarded in fellowship, —
Multiply each one's stock by the time he continues it in trade,
and use the product for his share. This is called Double
Fellowship. Ans. A 6 dollars, and B 4 dollars.
5. A and B enter into partnership ; A puts in $100
6 months, and then puts in $50 more ; B puts in $200
4 months, and then takes out $80 ; at the close of the
year they find that they have gained $95 ; what is the
profit of each ? . ( $43,711, A's share.
Ans ' I $51,288, B's share.
6. A, with a capital of $500, began trade, Jan. 1, 1826,
and, meeting with success, took in B as a partner, with a
capital of 600, on the first of March following ; four
months after, they admit C as a partner, who brought $800
stock ; at the close of the year, they find the gain to be
$700 ; how must it be divided among the partners ?
$250, A's share.
Ans. ^ $250, B's share.
$200, C's share.
ALLIGATION.
The rule of Alligation teaches how to gain the mean
value of a mixture that is made by uniting several articles
of different values.
ALLIGATION. 229
Alligation Medial, teaches how to obtain the value, (or
mean price,) of a mixture, when the quantities and prices
of the several articles are given.
Rule.
As the whole mixture is to the whole value, so is any par'
of the composition, to its mean price.
Examples.
1. A farmer mixed 15 bushels of r) r e, at 64 cents a
bushel, 18 bushels of Indian corn, at 55 cts. a bushel, and
21 bushels of oats, at 28 cts. a bushel ; I demand what a
bushel of this mixture is worth ?
bu. cts. $ cts.
bu.
$ cts. bu
15 at 64=9,60
As 54 :
25,38 : : 1
18 55=9,90
1
21 28=5,88
cts.
— 54)25,38(.47 Ans.
54 25,38
2. If 20 bushels of wheat at 1 dol. 35 cts. per bushel,
be mixed with 10 bushels of rye at 90 cents per bushel,
what will a bushel of this mixture be worth ?
Ans. 91,20 cts.
3. A tobacconist mixed 36 lb. of tobacco, at Is. 6d.
per lb., 12 lb. at 2s. a pound, with 12 lb. at Is. JOd. per
lb. ; what is the price of a pound of this mixture ?
Ans. Is. 8d.
4. A grocer mixed 2 C. of sugar at 56s. per C. and 1
C. at 43s. per C. and 2 C. at 50s. per C. together ; I de-
mand the price of 3 cwt. of this mixture? Ans. £7. 13s.
5. A wine merchant mixes 15 gallons of wine at 4s.
2d. per gallon, with 24 gallons at 6s. 8d. and 20 gallons
at 6s. 3d. ; what is a gallon of this composition worth ?
Ans. 5s. lOd. 24| qrs.
Alligation Alternate, teaches how to find the quantity of
each article, when the mean price of the whole mixture,
and also the prices of each separate article are known.
Rule.
Reduce the mean price and the prices of each separate
article to the same order.
20
230 ARITHMETIC. PART THIRD.
Connect with a line each price that is less than the mean
price, with one or more that is greater ; and each price
greater than the mean price, with one or more that is less.
W?ite the difference between the mean price, and the price
of each separate article, opposite the price with which it is
connected ; then the sum of the differences, standing against
any price, will express the relative quantity to be taken oj
that price.
Examples.
1. A merchant has several kinds of tea ; some at ti
shillings, some at 9 shillings, some at 11 shillings, and
some at 12 shillings per pound ; what proportions of each
must he mix, that he may sell the compound at 10 s.
per pound ?
The pupil will perceive, that there may be as many
different ways of mixing the simples, and, consequently
as many different answers, as there are different ways of
linking the several prices.
Operations.
lbs.
8s. ,-2 \ Or,
lit
Here the prices of the simples, are set one directly
under another, in order, from least to greatest, and the
mean rate, (10s.) written at the left hand. In the first way
of linking, we take in the proportion of 2 pounds of the
teas at 8 and 12s. to 1 pound at 9 and lis. In the second
way, we find for the answer, 3 pounds at 8 and lis. to 1
pound at 9 and 12s.
2. What proportions of sugar, at 8 cents, 10 cents, and
14 cents per pound, will compose a mixture worth 12
cents per pound ?
Ans. In the proportion of 2 lbs. at 8 and 10 cents, to f>
lbs. at 14 cents.
Note. As these quantities only express the proportions
of each kind, it is plain, that a compound of the same mean
price will be formed by taking 3 times, 4 times, one half,
or any proportion, of each quantity. Hence,
ALLIGATION. 231
When the quantity of one simple is given, after finding
the proportional quantities, by the above rule, we may say,
As the proportional quantity ; is to the given quantity : .
>«> is each of the other proportional quantities : to the re-
quired quantities of each.
3. If a man wishes to mix 1 gallon of brandy worth
16s. with rum at 9s. per gallon, so that the mixture may
be worth lis. per gallon, how much rum must he use ?
Taking the differences as above, we find the propor-
tions to be 2 of brandy to 5 of rum ; consequently, 1 gal-
lon'of brandy will require 21 gallons of rum.
Ans. 2i gallons.
4. A grocer has sugars worth 7 cents, 9 cents, and 12
cents per pound, which he would mix so as to form a com-
pound, worth 10 cents per pound ; what must be the pro.
portions of each kind ?
Ans. 2 lbs. of the first and second, to 4 lbs. of the 3d kind.
5. If he use 1 lb. of the first kind, how much must he
take of the others? if 4 lbs., what ? if 6 lbs.,
what ? if 10 lbs., what ? if 20 lbs., what ?
Ans. to the last, 20 lbs. of the 2d, and 40 of the 3d.
6. A merchant has spices at 16d. 20d. and 32d. per
pound ; he would mix 5 pounds of the first sort with the
others, so as to form a compound worth 24d. per pound ;
how much of each sort must he use ?
Ans. 51bs. of the second, and 1\ lbs. of the third.
7. How many gallons of water, of no value, must be
mixed with 60 gallons of rum, worth 80 cents per gallon,
to reduce its value to 70 cents per gallon ? Ans. 8± galls.
8. A man would mix 4 bushels of wheat, at $1,50 per
bushel, rye at $1,16, corn at 8,75, and barley at 8,50, so
as to sell the mixture at 8,84 per bushel ; how much of
each may he use ?
When the quantity of the compound is given, we may
say, As the sum of the proportional quantities, found by
the above rule, is to the quantity required, so is each pro-
portional quantity, found by the rule, to the required
quantity of each.
9. A man would mix 100 pounds of sugar, some at 8
cents, some at 10 cents, and some at 14 cents per pound,
232 ARITHMETIC. PART THIRD.
so that the compound may be worth 12 cents per pound ;
how much of each kind must he use ?
We find the proportions to be, 2, 2, and 6. Then, 2-f-2
+6= 10, and C 2 : 20 lbs. at 8 cts. )
10 : 100 : : 1 2 : 20 lbs. at 10 cts. V Ans.
(6 : 60 lbs. at 14 cts. )
10. How many gallons of water, of no value, must be
mixed with brandy at $1,20 per gallon, so as to fill a ves-
sel of 75 gallons, which may be worth 92 cents per gal. ?
Ans. 17i gallons of water to 57i gallons of brandy.
11. A grocer bas currants at 4d., 6d., 9d., and lid. per
lb. ; and he would make a mixture of 240 lbs., so that
the mixture may be sold at 8d. per lb. ; how many pounds
of each sort may he take 1
Ans. 72, 24, 48, and 96 lbs., or 48, 48, 72, 72, &q.
Note. This question may have five different answers.
DUODECIMALS.
Duodecimal is derived from the Latin word duodecim,
signifying twelve.
They are fractions of a foot, which is supposed to be
divided into twelve equal parts called primes, marked thus,
('). Each prime is supposed to be subdivided into 12 equal
parts called seconds, marked thus, ("). Each second is
also supposed to be divided into twelve equal parts called
thirds, marked thus ('"), and so on to any extent.
It thus appears that
1 an inch or prime is t l of a foot.
1" a second is T \ of T ' 5 or T ^ of a foot.
V" a third is T \ of T » 5 of t l, or T ^j of a foot, &c.
Whenever therefore any number of seconds (as 5") are
mentioned, it is to be understood as so many y i T of a foot,
and so of the thirds, fourths, &c.
Duodecimals are added and subtracted like other com-
pound numbers, 12 of a less order making 1 of the next
higher, thus,
12"" fourths make 1 third 1"'.
12"' thirds make 1 second 1".
12" seconds make 1 prime or inch 1'.
DUODECIMALS. 233
12' inches or primes, make 1 foot.
The addition and subtraction of Duodecimals is the
same as other compound numbers.
These marks ' " '" "" are called indices.
Multiplication of Duodecimals.
Duodecimals are chiefly used in measuring surfaces and
solids.
How many scpiare feet in a board 16 feet 7 inches long,
and 1 foot 3 inches wide ?
Note. The square contents of any thing are found by
multiplying the length into the breadth.
The following example is explained above.
Examples.
16 r
1 3'
16 7'
4 1' 9
20 8' 9"
It is generally more convenient to multiply by the higher
orders of the multiplier first.
Thus we begin and multiply the multiplicand first by the
1 foot, and set down the answers as above.
We then multiply by the 3' or T 3 2 of a foot. 16 is chan-
ged to a fraction, thus y, and this multiplied by T 3 ^ is f£,
or 48', which is 4 feet, (for there are 12' in every foot,)
and is set under that order.
We now multiply 7' (or T \) by 3' (or T 3 ^) and the answer
is T V T or21".
This is 1' to set under the order of seconds, and 9" (pf T )
to be set under the order of thirds.
The two products are then added together, and the
answer is obtained, which is 20 feet 8 primes 9 seconds.
Another example will be given in which the cubic con-
20*
234 ARITHMETIC. PART THIRD.
tents of a block are found by multiplying the length, breadth
and thickness together.
How many solid feet in a block 15 ft. 8' long, 1 ft. 5'
wide, and 1 ft. 4' thick ?
Length,
Breadth,
i
15
1
Operation.
8'
5'
4"
15
6
8'
6'
Thickness
22
1
2'
4'
4"
22
7
2'
4'
4"
9"
4'"
Ans.
29
r
1"
4'"
Let this example be studied and understood before the
rule is learned. If any difficulty is found, let both multi-
plier and multiplicand be expressed as Vulgar Fractions,
and then multiply.
In duodecimals it is always the case that the 'product of
two orders, will belong to that order which is made by ad-
ding the indices of the factors.
Rule.
Write the figures as in the addition of compound numbers.
Multiply by the higher orders of the multiplier first, remem-
bering that the product of two orders belongs to the order de-
noted by the sum of their indices.
If any product is large enough to contain units of a higher
order, change them to a higher order, and place them where
they belong.
Examples.
How many square feet in a pile of boards 12 ft. 8' long,
and 13' wide ?
INVOLUTION. 235
What is the product of 371 ft. 2' 6" multiplied by lttl
a. 1' 9" ? Ans. 07242 ft. 10' 1" 4'" 6"".
If a floor be 10 ft. 4' 5" long, and 7 ft. 8' 6" wide, what
is its surface ? Ans. 79 ft. 11' 0" 6'" 6"".
What is the solidity of a wall 53 ft. 6' long, 10 ft. 3'
high, and 2 ft. thick ? Ans. 1096| ft.
INVOLUTION.
When a number is multiplied into itself, it is said to be
involved, and the process is called Involution, .
Thus, 2X2x2 is 8. Here the number 2 is multiplied
into itself twice.
The product which is obtained by multiplying a number
into itself, is called a Power.
Thus, when 2 is multiplied into itself once, it is 4, and
this is called the second power of 2. If it is multiplied into
\tse\t' twice (2x2x2=8) the answer is 8, and this is called
the third power.
The number which is involved, is called the Root, or
first power.
Thus, 2 is the root of its second power 4, and the root of
its third power 8.
A power is named, or numbered, according to the number
of times its root is used as a factor. Thus the number 4
is called the second power of its root 2, because the root
is twice used as a factor ; thus, 2x2=4.
The number 8 is called the third power of its root 2 ;
because the root is used three times as a factor ; thus,
2X2x2=8.
The method of expressing a power, is by writing its
root, and then above it placing a small figure, to show the
number of times that the root is used as a factor.
Thus the second power of 2 is 4, but instead of writing
the product 4, we write it thus, 2 2 .
The third power of 2 is written thus, 2 3 .
The fourth power of 2 is 10, and is written thus, 2 4 .
The small figure that indicates the number of times that
the root is used as a factor, is called the Index, or Expo-
nent.
236 ARITHMETIC. PART THIRD.
The different powers have other names beside their
numbers.
Thus, the second i>ower is called the Square.
The third power is called the Cube.
The fourth power is called the Biquadrate.
The fifth power is called the Sursolid.
The sixth power is called the Square-cubed.
Powers are indicated by exponents. When a power is
actually found by multiplication, involution is said to be
performed, and the number or root is involved.
Rule of Involution.
To involve a number, multiply it into itself, as often as
there are units in the exponent, save once.
Note. — The reason why it is multiplied once less than
there are units in the exponent, is, that the first time the
number is multiplied, the root is used twice as a factor ;
and the exponent shows, not how many times we are to
multiply, but how many times the root is used as a factor.
1. What is the cube of 5 ? Ans. 5x5X5=125
2. What is the 4th power of 4 ? Ans. 256
3. What is the square of 14? Ans. 196
4. What is the cube of 6 ? Ans. 216
5. What is the 5th power of 2 ? Ans. 32
6. What is the 7th power of 2 ? Ans. 128
7. What is the square of \ 1 Ans. J
8. What is the cube of § ? Ans. JL
A Fraction is involved, by involving both numerator and
denominator.
9. What is the fourth power off ? Ans. g%L
10. What is the square of 51 ? Ans. 30£
11. What is the square of 30i ? Ans. 915 T ' F
12. Perform the involution of 8s. Ans. 32,768
13. Involve ^ }i, and f to the third power each.
An<S 64_. 1331. 5 12
^"''5 8 319 ' T7 2?' ?29
14. Involve 21 1 3 . Ans. 9,393,931
15. Raise 25 to the fourth power. Ans. 390,625
16. Find the sixth power of 1.2. Ans. 2.985,984
EVOLUTION. 237
EVOLUTION.
Evolution is the process of finding the root of any num-
ber; that is, of finding that number which multiplied into
itself, will produce the given number.
The Square Root, or Second Root, is a number which be-
ing squared (i. e. multiplied once into itself) will produce
the given number. It is expressed either by this sign, put
before a number, thus \/4, or by the fraction 1 placed
i
above a number thus, 4 2 .
The Cube Root, or Third Root, is a number, which be-
ing cubed, or multiplied by itself twice, will produce the
3 •
given number. It is expressed thus, </12 ; or thus, 12 3 .
All the other roots are expressed in the same manner.
4
Thus the fourth root has this sign v put before a number,
or else £ placed above it.
6
The sixth root has •/ before it, or £ above it, &c.
There are some numbers whose roots cannot be pre-
cisely obtained ; but by means of decimals, we can ap-
proximate to the number which is the root.
Numbers whose roots can be exactly obtained, arc
called rational numbers.
Numbers whose precise roots cannot be obtained, are
called surd numbers.
When the root of several numbers united by the sign
-f- or — is indicated, a vinculum, or line is drawn from
the sign of the root over the numbers. Thus, the square
root of 36 — 8 is written ^/SG— 8.
The root of a rational number, is a rational root, and the
root of a surd number, is a surd root.
It is very necessary for practical purposes, to be able to
find the amount of surface there is in any given quantity.
For instance, if a man has 250 yards of matting, which is
2 yards wide, how much surface will it cover ?
The rule for finding the amount of surf ace, is to multiply
the length by the breadth, and this will give the amount of
square inches, feet, or yards.
It is important for the pupil to learn the distinction be-
tween a square quantity, and a certain extent that is in the
238
ARITHMETIC. PART THIRD.
—
form of a square. For example, Jour square inches, and
Jour inches square are different quantities.
A Four square inches may be represented
in Fig. A. In this figure there are four
square inches, but it makes a square which is
only two inches on each side, or a two inch
square.
A Jour inch square may be re-
presented by Fig. B.
Here the sides of the square are
four inches long, and it is called
a Jour inch square. But it con-
tains sixteen square inches. For
when the four inch square is cut
into pieces of each an inch square
it will make sixteen of them.
A Jour inch square then, is a square whose sides are four
inches long.
Four square inches are four squares that are each an
inch on every side. .
When we wish to find the square contents of any quan-
tity, we seek to know how many square inches, ovjeet, or
yards, there are in the quantity given, and this is always
found by multiplying the length by the breadth.
When the length and breadth of any quantity are given,
we find its square contents, or the amount of surface it will
cover, by multiplying the length by the breadth.
What are the square contents of 223 yds of carpeting }
wide ?
What are the sq. contents of 249 yds of matting f wide ?
If any quantity is placed in a square form, the length oj
one side is the square root of the square contents of this
figure. Thus in the preceding example, B, the square
contents of the figure are 16 square inches. The side of
the square is 4 inches long ; and 4 is the square root of 16.
The square root, therefore, is the length of the sides of a
square, made by the given quantity.
If we have one side of a square given, by the process of
Involution, we find what are the square contents of the quan-
tity given.
EXTRACTION OP THE SQUARE ROOT. 239
If on the contrary, we have the square contents given,
by the process of Evolution, we find what is the length of
one side of the square, which can be made by the quantity
given.
Thus if we have a square whose side is four inches, by
Involution we find the surface, or square contents to be 16
square inches.
But if we have 16 square inches given, by Evolution we
find what is the length of one side of the square made by
these 16 inches.
EXTRACTION OF THE SQUARE ROOT.
Extracting the square root is finding a number, which,
multiplied into itself, will produce the given number ; or,
it is finding the length of one side of a certain quantity,
when that quantity is placed in an exact square.
It will be found by trial, that the root always contains
just half as many, or one figure more than half as many
figures as are in the given quantity. To ascertain, there-
fore, the number of figures in the required root, we point
off the given number into periods of two figures each, be-
ginning at the right, and there will always be as many
figures in the root as there are periods.
1. What is one side of a square, containing 784 square
feet ?
784(2 Pointing off as above, we find that the root will
4 consist of ttvo figures, a ten and a unit.
We now take the highest peri-
384 od 7 (hundreds), and ascertain
how many feet there will be in
Fig. 1. the largest square that can be
made of this quantity, the sides
of which must be of the order of
tens. No square larger than 4
° (hundreds) can be contained in 7
g 5 (hundreds), the sides of which
will be each 20 feet (because 20x
20=400). These 20 feet (or 2
tens) being sides of the square
B
20
20
400
20 feet. are placed in the quotient as the
first figure of the root.
240
ARITHMETIC. PART THIRD.
This square may be represented by Fig 1.
We now take out the 400 from 700, and 300 square feet
remain. These are added to the next period (84 feet),
making 384, which are to be arranged around the square
B, in such a way as not to destroy its square form; conse-
quently the additions must be made on two sides.
To ascertain the breadth of these additions, the 384
must be divided by the length of the two sides (20+20),
and as the root already found is one side, we double this
root for a divisor, making 4 tens or 40, for as 40 feet is
the length of these sides, there will be as many feet in
breadth, as there are forties in 384. The quotient arising
from the division is 8, which is the breadth of the addition
to be made, and which is placed in the quotient, after the
4 tens. . .
784(28 Root.
4
48
384
384
000
Fig. 2.
20 feet.
8 feet.
—
C
E
20X8=160
8x8=64
*o
B
o
X
<2
II D
o
o
20X20=400
©
GO
cd
CD
O
a
CD
20 feet.
8 feet.
But it will be seen by Fig. 2, that to complete the square,
the corner E must be filled by a small square, the sides of
which are each equal to the width of C and D, that is, 8
EXTRACTION OF THE SQUARE ROOT. 241
feet. Adding this to the 4 tens, or 40, we find that the
whole length of the addition to be made around the square
B, is 48 feet, instead of 40. This multiplied by its breadth,
8 feet (the quotient figure), gives the contents of the whole
addition, viz. 384 feet.
As there is no remainder, the work is done, and 28 feet
is the side of the given square.
The proof may be seen by involution, thus ; 28X28=
784 ; or it may be proved, by adding together the several
parts of the figure, thus;
B contains 400 feet.
C " 160 "
D " 160 "
E " 64 "
Proof 784
If, in any case, there is a remainder, after the last period
is brought down, it may be reduced to a decimal fraction,
by annexing two ciphers for a new period, and the same
process continued.
Whenever any dividend is too small to contain the divi-
sor, a cipher must be placed in the root, and another
period brought down.
From the above illustrations, we see the reasons for the
following rule.
Rule for Extracting the Square Root.
1. Point off the given number, into periods of two figures
each, beginning at the right.
2. Find the greatest square in the first left hand period,
and subtract it from that period. Place the root of this
square in the quotient. To the remainder bring dozen the
next period for a dividend.
3. Double the root already found (understanding a cipher
at the right) for a divisor. Divide the dividend by it, and
place the quotient figure in the root, and also in the divisor.
4. Multiply the divisor, thus increased, by the last figure
of the root, and subtract the product from the dividend. To
the remainder bring down the next period, for a new divu
21
242
ARITHMETIC. PART THIRD.
dend. Double the root already found, for a new divisor.
and proceed as before.
Examples.
What is the square root of 99800 i ?
998001(999 Rooi.
81
189)1880
1701
1989)17901
17901
000
Find the sq. root of 784. A. 28. Of 070. A. 20.
Of 625. A. 25. Of4S7,204. A. 698. Of 038,401.
A. 779. Of 556,510. A. 746. Of 441. A. 21.
1024. A. 32. Of 1444. A. 38. Of 2916. A.
Of 6241. A. 79. Of 9801. A. 99. Ot 17,956.
134. Of 32,761. A. 181
488,601. A. 699.
Find the sq. root of 69.
Of 39,601. A. 199.
9.1104336.
17.2916165.
16.7928556.
18.7349940.
31.2889757.
31.6069613.
26.2106848.
Of 97.
Of 222.
Of 394.
Of 699.
Of 989.
Of 397.
Of 892.
A.
A.
A
A.
A.
A.
A.
A. 8.3066239.
9.8488578.
14.8996644.
. 19.8494332.
26.4380081.
31.4483704.
19.9248588.
29.8063090.
Of 83.
Of 299:
Of 282.
Of 351.
Of 979.
Of 999.
Of 687.
Of
54.
A.
Of
A.
A.
A.
A.
A.
A.
A.
It was shown in the article on Involution, that a fraction
is involved hy involving hoth numerator and denominator,
hence to find the root of a fraction, extract the root loth of
numerator and denominator. If this cannot be done, the
fraction may be reduced to a decimal, and its root ex-
tracted.
What is the square root of §£ ? A. |
A 4 01 Of S.3 7i«a. ? A A.S7. Of 43 3 35 ?
-'*• • ? (I 9 • v/1 480249 • *"■* 69 3 yJl 483025 !
KJl 61 7 70 6 • "■*• 786* V1 942841 • ■*• 9Tj*
Of J-SJ18 01 1
1 24 9 1 -
65 C
EXTRACTION OF THE CUBE ROOT. 243
Find the sq. root of £. A. .6660254. Of ft. A.
15497. O
J3649167.
288617394+
.645497. Of 17f. A. 4.168333. Of &. A.
.193649167. Of f%. A. .83205. Of ft. A
EXTRACTION OF THE CUBE ROOT.
A Cube is a solid body, having six equal sides, each ol
which is an exact square. Thus a solid, which is 1 foot
long, 1 foot high, and 1 foot wide, is a cubic joot ; and a
solid whose length, breadth, and thickness are each 1 yard,
is called a cubic yard.
The root of a cube is always the length of one of its
sides ; for as the length, breadth, and thickness of such a
body are the same, the length of one side, raised to the
third power, will show the contents of the whole.
Extracting the Cube Root of any quantity, therefore, is
rinding a number, which multiplied into itself, twice, will
produce that quantity ; — or it is finding the length of one
side of a given quantity, when that quantity is placed in
an exact cube.
To ascertain the number of figures in a cube root, we
point off the given number, into periods of three figures
each, beginning at the right, and there will be as many
figures in the required root as there are periods.
1. What is the length of one side of a cube, containing
32768 solid feet? . .
32768(3
27
5768
Pointing off as above, we find there will be two figures
in the root, a ten and a unit.
214
ARITHMKTIC. PART THIRD,
We now take the highest
period, 32 (thousands), and
ascertain what is the largest
cube that can be contained
in this quantity, the sides of
which will be of the order of
tens. No cube larger than 27
(thousands) can be contained
in 82 (thousands). The sides
of this are 3 tens or 30 (be-
cause 30x30x30=27,000)
which are placed as the first figure of the root.
This cube mav be represented by Fig 1.
We now take the 27000 from 32000, and 5000 solid feet
remain. These are added to the next period (768), ma-
king 5768, which are to be arranged around the cubic
figure 1, in such a way as not to destroy its cubic form ;
consequently the addition must be made to three of its
sides.
We must now ascertain, what will be the thickness of
the addition made to each side. This will of course de-
pend upon the surface to be covered. Now the length of
one side has been shown to be 30 feet, and, as in a cube,
the length and breadth of the sides are equal, multiplying
the length of one side into itself will show the surface of
side, and this multiplied by 3, the number of sides, gives
the contents of the surface of the three sides. Thus 30x
30=900, which multiplied by 3=2700 feet.
Now as we have 3768 solid feet to be distributed upon a
surface of 2700 feet, there will be as many feet in the
thickness of the addition, as there are twenty-seven hun-
dreds in 3768. 2700 is contained in 3768 twice ; there-
fore 2 feet is the thickness of the addition made to each of
the three sides.
By multiplying this thickness, by the extent of surface
(2700x2) we* find that there are 5400 solid feet contained
in these additions.
EXTRACTION OF THE CUBE ROOT.
245
32768(32
27
2700)5708
5400
360
8
5768
0000
But if* we examine Fig. 2 we shall
tind that these additions do not com-
plete the cube, for the three corners
a an need to be tilled by blocks of the
same length as the sides (30 feet) and
of the same breadth and thickness as
the previous additions (viz. 2 feet).
Now to find the solid contents of
these blocks, or the number of feet
required to fill these corners, we multi-
ply the length, breadth, and thickness
of one block together, and then multiply
this product by 3, the num-
ber of blocks. Thus, the
breadth and thickness of
each block has been shown
to be 2 feet ; 2X2=1, and
this multiplied by 30 (the
|30 length)=120, which is the
solid contents of one block.
But in three, there will be
three times as many solid
feet, or 300, which is the
number required to fill the
deficiences.
In other words, we square
the last quotient figure (2)
multiply the product by the
first figure of the quotient
(3 tens) and then multiply
the last product by 3, the
number of deficiencies.
But by examining Fig. 3,
it appears that the figure is
not yet complete, but that a
small cube is still wanting,
where the blocks last added
meet. The sides of this small cube, it will be seen, are
each equal to the width of these blocks, that is, 2 feet.
If each side is 2 feet long, the whole cube must contain 8
21*
246
ARITHMETIC. PART THIRD.
Fig. 4.
32 feet.
solid feet (because 2x2x2
=8), and it will be seen
by Fig. 4, that this just fills
the vacant corner, and
completes the cube.
We have thus found,
that the additions to be
made around the large
cube (Fig. J) are as fol-
lows.
32 teet.
5400 solid feet upon three sides, (Fig. 2).
3G0 " " to fill the corners a a a.
8 " •' to fill the deficiency in Fig. 3.
Now if these be added together, their sum will be 576S
solid feet, which subtracted from the dividend leave no
remainder and the work is done. 32 feet is therefore the
length of one side of the given cube.
The proof may be seen by involving the side now found
to the third power, thus ; 32x32x32= 32768 ; or it may
be proved by adding together the contents of the several
parts, thus,
27000 feet=contents of Fis. 1.
5400 " ^addition to three sides.
360 " =addition to fill the corners a a a.
8 " —addition to fill the corner in Fie. 3.
32768 Proof.
From these illustrations we see the#easons for the fol-
lowing rule.
Rule fob extracting the Ctbe Root.
1 Point off the given number, into periods of three figures
each, beginning at the right.
2. Find the greatest cube in the left hand period, and
subtract it from that period. ' Place the root in the quotient,
and to the remainder bring down the next period,for a divi-
dend.
3. Square the root already found [understanding a cipher
at the right) and multiply it by 3 for a divisor.
EXTRACTION OF THE CUBE ROOT.
247
Divide the dividend by the divisor, and place the quotient
for the next figure of the root.
4. Multiply the divisor by this quotient figure. Multiply
this quotient figure by the former figure or figures of the root.
Finally cube this quotient figure, and add these three results
together for a subtrahend.
5. Subtract the subtrahend from the dividend. To the
remainder bring down the next period, for a new dividend,
and proceed as before.
If it happens in any case, that the divisor is not con-
tained in the dividend, or if there is a remainder after the
last period is brought down, the same directions may be
observed, that were given respecting the square root.
(See page 241.)
Examples.
What is the cube root of 373248 ?
373248(72
343
70s X 3=14700)30248(First Dividend.
29400
2 2 X70x3= 840
2 3 = 8
30248 Subtrahend.
0000
Find the
958,565,256.
494,913,671.
196,122.941.
57,512,450.
39,651,821.
510,082,399.
of 941,192,000.
Of 478,21 1,768.
Of 445,943,744.
Of 204,336,469.
Of 6,751,269.
Of 42,508,549.
Of 469,097,433.
A. 980.
A. 782.
A. 764.
A. 589.
A
cube root
A. 986.
A. 791.
A. 581.
A. 386.
A. 341.
A. 799.
Find the cube root of 7. A. 1.912933
3.448217. Of 49. A. 3.659306. Of 94. A. 4.546836.
Oi'97. A. 4.610436. Of 199. A. 5.838272. Of 179.
A. 5.635741. Of 389. A. 7.299893. Of 364. A.
189.
A. 349.
A. 777.
Of 41.
Oi
Of
Of
Of
Of
Of
248 ARITHMETIC. PART THIRD,
7.140037. Of 499. A. 7.931710. Of 699. A.
8.874809. Of 686. A. 8.819447. Of 886. A. 9.604569.
Of 981. A. 9.936261.
The cube root of a fraction, is obtained by extracting
the root of numerator and denominator, but if this cannot
be done, it may be changed to a decimal, and the root ex-
tracted.
Find the cube root of T |^ T . A. T 3 T . Of ^f-fff. A.
2 4 Of 450533 A 7 7. Of 7_3_0J_3_8 4_ A~ 19.4 Of
3 1' wl 770295' ■"■• 9 9* "' ^6130899' ■"■* 2 9 9' v "'*
2 34 6.111 \ 273.
0257077 3" ' 39T
Find the cube root of £-. A. .8549879. Of &. A.
.5593445. Of J^o- A - -4578857. Of 5 VV A -
.4562903. Of 4-14. A. .9973262.
ARITHMETICAL PROGRESSION.
Any rank, or series of numbers, consisting of more than
two terms, which increases or decreases by a common dif-
ference, is called an Arithmetical series, or progression.
When the series increases, that is, when it is formed by
the constant addition of the common difference, it is called
an ascending series, thus,
1, 3, 5, 7, 9, 11, &c.
Here it will be seen that the series is formed by a con-
tinual addition of 2 to each succeeding figure.
When the series decreases, that is, when it is formed by
the constant subtraction of the common difference, it is
called a descending series, thus,
14, 12, 10, 8, 6, 4, &c.
Here the series is formed by a continual subtraction of
2, from each preceding figure.
The figures that make up the series are called the
terms of the series. The first and last terms are called
the extremes, and the other terms, the means.
From the above, it may be seen, that any term in a se-
ries may be found by continued addition or subtraction,
but in a long series this process would be tedious. A much
more expeditious method may be found.
1. The ages of six persons are in arithmetical progres-
sion. The youngest is 8 years old, and the common dif-
ARITHMETICAL PROGRESSION. 249
ference is 3, what is the age of the eldest ? In other words,
what is the last term of an arithmetical series, whose
first term is 8, the number of terms 6, and the common
difference 3 ?
8, 11, 14, 17, 20, 23.
Examining this series, we find that the common differ-
ence. 3, is added 5 times, that is one less than the number
of terms, and the last term, 23, is larger than the first
term, by five times the addition of the common difference,
three ; Hence the age of the elder person is 8+3x5=23.
Therefore when the first term, the number of terms,
and the common difference, are given, to find the last
term,
Multiply the common difference into the number of terms,
less J, and add the product to the first terra.
2. If the first term be 4, the common difference 3, and
the number of terms 100, what is the last term ?
Ans. 301.
3. There are, in a certain triangular field, 41 rows of
corn ; the first row, in 1 corner, is a single hill, the second
contains 3 hills, and so on, with a common difference of 2 ,
what is the number of hills in the last row ? A. 81 hills
4. A man puts out 81, at 6 per cent, simple interest,
which, in 1 year, amounts to 81,06 in 2 years to 81,12,
and so on, in arithmetical progression, with a common
difference of 80 ,06 ; what would be the amount in 40
years 1 A. 83 ,40.
Hence we see, that the yearly amounts of any sum, at
simple interest, form an arithmetical series, of which the
principal is the first term, the last amount is the last term,
the yearly interest is the common difference, and the num-
ber of years is 1 less than the number of terms.
It is often necessary to find the sum of all the terms, in
an arithmetical progression. The most natural mode of
obtaining the amount would be to add them together, but
an easier method may be discovered, by attending to the
following explanation.
1. Suppose we are required to find the sum of all the
terms, in a series, whose first term is 2, the number of
terms 10, and the common difference 2.
250 ARITHMETIC. PART THIRD.
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
20, 18, 16, 14, 12, 10, 8, 6, 4, 2
22, 22, 22, 22, 22, 22, 22, 22, 22, 22,
The first row of figures above, represents the given se-
ries. The second, the same series with the order inverted,
and the third, the sums of the additions of the correspond-
ing terms in the two series. Examining these series, we
shall find that the sums of the corresponding terms are the
same, and that each of them is equal to the sum of the ex-
tremes, viz. 22. Now as there are 10 of these pairs in
the two series, the sum of the terms in both, must be 22x
10=220.
But it is evident, that the sum of the terms in one series,
can be only half as great as the sum of both, therefore, if
we divide 220 by 2, we shall find the sum of the terms in
one series, which was the thing required. 220—2=110,
the sum of the given series.
From this illustration we derive the following rule ;
When the extremes and number of terms are given, to
find the sum of the terms,
Multiply the sum of the extremes by the number of terms,
and divide the product by 2.
2. The first term of a series is 1, the last term 29, and
the number of terms 14. What is the sum of the series ?
A. 210.
3. 1st term, 2, last term, 51, number of terms, 18. Re-
quired the sum of the series. A. 477.
4. Find the sum of the natural terms 1, 2, 3, &c. to
10,000. A. 50,005,000.
5. A man rents a house for $50, annually, to be paid
at the close of each year ; what will the rent amount to in
20 years, allowing 6 per cent., simple interest, for the use
of the money ?
The last year's rent will evidently be $50 without in-
terest, the last but one will be the amount of $50 for 1
year, the last but two the amount of $50 for 2 years, and
so on, in arithmetical series, to the first, which will be the
amount of $50 for 19 years = $107.
If the first term be 50, the last term 107, and the num-
ber of terms 20, what is the sum of the series ? A. 1570.
ARITHMETICAL PROGRESSION. 251
6. What is the amount of an annual pension of $100,
being in arrears, that is, remaining unpaid, for 40 years,
allowing 5 per cent, simple interest ? A. $7900.
7. There are, in a certain triangular field, 41 rows of
corn ; the first row, being in 1 corner, is a single hill, and
the last row, on the side opposite, contains 81 hills ; how
many hills of corn in the field? A. 1681.
The method of finding the common difference, may be
learned by what follows.
1. A man bought 100 yards of cloth in Arithmetical
progression : for the first yard he gave 4 cents, and for the
last 301 cents, what is the common increase on the price
of each yard ?
As he bought 100 yards, and at an increased price upon
every yard, it is evident that this increase was made 99
times, or once less than the number of terms in the series.
Hence the price of the last yard was greater than the first,
by the addition of 99 times the regular increase.
Therefore if the first price be subtracted from the last,
and the remainder be divided by the number of additions
(99), the quotient will be the common increase ; 301 — 4=
297 and 297-^-99 = 3, the common difference.
Hence, when the extremes and number of terms are
^iven, to find the common difference,
Divide the difference of' the extremes, by the number of
terms less 1 .
2. Extremes 3 and 19 ; number of terms 9. Required
the common difference. A. 2.
3. Extremes 4 and 56 ; number of terms 14. Required
the common difference. A. 4.
4. A man had 15 houses, increasing equally in value,
from the first, worth $700, to the 15th, worth $3500.
What was the difference in value between the first and
second? A. 200.
In Arithmetical progression, any three of the following
terms being given, the other two maybe found. 1. The
first term. 2. The last term. 3. The number of terms.
4. The common difference. 5. The sum of all the
terms.
252 ARITHMETIC. PART THIRD.
GEOMETRICAL PROGRESSION.
Any series of numbers, consisting of more than two
terms, which increases by a common multiplier, or decrea-
ses by a common divisor, is called a Geometrical Series.
Thus the series 2, 4, 8, 16, 32, &c. consists of
terms, each of which is twice the preceding, and this is
an increasing or ascending Geometrical series.
The series 32, 16, 8, 4, 2, consists of numbers,
each of which is one half the preceding, and this is a
decreasing or descending Geometrical series.
The common multiplier or divisor is called the Ratio,
and the numbers which form the series are called Terms.
As in Arithmetical, so in Geometrical progression, /if
any three of the five following terms be given, the other
two may be found.
1. The first term. 2. The last term. 3. The number
of terms. 4. The common difference. 5. The sum of
all the terms.
1. A man bought a piece of cloth containing 12 yards,
the first yard cost 3 cents, the second 6, the third 12, and
so on, doubling the price to the last ; what cost the last
vard ?
*3x2x2x2x2x2x2x2x2x2x2x2=3X2' '=6144, Ans.
In examining the above process, it will be seen, that the
price of the second yard is found by multiplying the first
pavment into the ratio (2) once ; the price of the third
yard, by multiplying by 2 twice,<kc, and that the ratio (2)
is used as a factor eleven times, or once less than the num-
ber of terms. The last term then, is the eleventh power of
the ratio (2) multiplied by the first term (3).
Hence the first term, ratio, and number of terms, being
given, to find the last term.
Multiply the first term, by that potcer of the ratio, whose
index is one less than the number of terms.
Note. In involving the ratio, it is not always necessary
to produce all the intermediate powers ; the process may
often be abridged, by multiplying together two powers al-
ready obtained, thus,
The 11th power = the 6th power X the 5th power, &c.
GEOMETRICAL PROGRESSION. 2 r )3
2. If the first term is 2, the ratio 2, and the number of
terms 13, what is the last term ? A. 8,192.
3. Find the 12th term of a series, whose first term is 3,
and ratio, 3. A. 531,441.
4. A man plants 4 kernels of corn, which, at harvest,
produce 32 kernels ; these he plants the second year ;
now, supposing the annual increase to continue 8 fold,
what would be the produce of the 16th year, allowing 1000
kernels to a pint? A. 2199023255.552 bushels.
5. Suppose a man had put out one cent at compound
interest in 1620, what would have been the amount in
1824, allowing it to double once in 12 years ?
2^=131072. A. 1310.72.
The most obvious method of obtaining the sum of the terms
in a Geometrical series, might be by addition, but this is
not the most expeditious, as will be seen.
1. A man bought 5 yards of cloth, giving 2 cents for the
first, 6 cents for the second, and so in 3 fold ratio ; what
did the whole cost him ?
2, 6, 18, 54, 162
6, 18, 54, 162, 486
The first of the above lines, represents the original
series. The second, that series, multiplied by the ratio 3.
Examining these series, it will be seen that their terms
are all alike excepting two : viz. the Jirst term of the first
series, and the last of the second series. If now we sub-
tract, the first series from the last, we have for a remainder
486 — 2=484, as all the intermediate terms vanish in the
subtraction.
Now the last series is three times the first, (for it was
made by multiplying the first series by 3,) and as we have
already subtracted once the first, the remainder must of
course be twice the first.
Therefore if we divide 484 by 2, we shall obtain the
sum of the first series. 484-r-2=242 Ans.
As in the preceding process, all the terms vanish in the
subtraction, excepting the first and last, it will be seen,
that the result would have been the same, if the last term
only, had been multiplied, and the first subtracted from
the product.
22
254 ARITHMETIC. PART THIRD.
Hence, the extremes and ratio being given, to find the
sum of all the terms,
Multiply the greater term by the ratio, from the product
subtract, the least term, and divide the remainder by the ratio
less 1.
2. Given the first term, 1 ; the last term, 2,187 ; and
the ratio, 3 ; required the sum of the series. A. 3,280.
3. Extremes, 1 and 65,536 ; ratio 4 ; required the sum
of the series. A. 87,381.
4. Extremes, 1,024 and 59,049 ; required as above.
A. 175,099.
5. What is the sum of the series 16, 4, 1, \, Jg, ^\,
and so on, to an infinite extent ? A. 21^.
Here it is evident, the last term is 0, or indefinitely near
to nothing, the extremes therefore are 16 and 0, and the
ratio 4.
ANNUITIES.
An annuity is a sum payable periodically, for a certain
length of time, or forever.
An annuity, in the proper sense of the word, is a sum
paid annually, yet payments made at different periods, are
called annuities. Pensions, rents, salaries, &c. belong to
annuities.
When annuities are not paid at the time they become
due, they are said to be in arrears.
The sum of all the annuities in arrears, with the interest
on each for the time they have remained due, is called the
amount.
The Present worth of an annuity, is the sum which should
be paid for an annuity yet to come.
When an annuity is to continue forever, its present
worth is a sum, whose yearly interest equals the annuity.
Now as the principal, multiplied by the rate, will give
the interest, the interest, divided by the rate, will give the
principal.
Hence to find the present worth of an annuity, continuing
forever,
Divide the annuity by the rate per cent.
ANNUITIES. 255
1. What is the worth of 8100 annuity, to continue for-
ever, allowing to the purchaser 4 per cent. ? allowing 5
per cent. ? 8 per cent. ? 10 per cent. ? 15 per cent. ?
20 per cent. ? Ans. to last, $500.
2. What is an estate worth, which brings in $7,500 a
year, allowing G per cent. ? A. $125,000.
ANNUITIES AT COMPOUND INTEREST.
It has been shown (page 208) that Compound Interest
is that which arises from adding the interest to the principal
at the close of each year, and making the amount a new
a new principal. The amount of $1 for one year at 6
percent, is $1.06, and it will be found, that if the princi-
pal be multiplied by this, the product will be the amount
for 1 year, and this amount multiplied by LOG, will be the
amount for 2 years, aud so on. Hence we see that any
sum at compound interest, forms a geometrical series, of
which the ratio is the amount of $1 at the given rate per
cent.
1. An annuity of $40 was left 5 years unpaid, what was
then due upon it, allowing 5 per cent, compound interst ?
It is evident that for the fifth or last year, the annuity
alone is due ; (or the fourth, the amount of the annuity for
1 year ; for the' third the amount of the annuity for 2 years,
and so on ; and the sum of these amounts will be the
answer, or what is clue in 5 years.
-From this we find that the amount of an annuity in ar-
rears, forms a geometrical progression, whose first term is
the annuity, the ratio, the amount of $1 at the given rate,
and the number of terms, the number of years.
The above example, then, may be resolved into the fol-
lowing question. What is the sum of a geometrical series
whose first term is $40, the ratio 1.05, and the number of
terms 5 ? First find the last term, by the first rule in Geo-
metrical progression, and then the sum of the series by
the second rule. The answer will be found to be $221.02.
Hence, to find the amount of an annuity in arrears, at
compound interest,
Find the sum of a Geometrical series, whose first term is
256 ARITHMETIC. PART THIRD.
the annuity, whose ratio, the amount of $1 at the given rate
per cent., and whose number of terms is the number of years.
Note. A table, showing the amount of $1 at 5 and 6
per cent., compound interest, for any number of years not
exceeding 24, will be found on page 209.
2. What is the amount of an annuity of $50, it being
in arrears 20 years, allowing 5 per cent, compound inte-
rest? A. $1653,29.
3. If the annual rent of a house, which is $150, be in
arrears 4 years, what is the amount, allowing 10 per cent,
compound interest ? A. $696,15.
4. To how much would a salary of $500 per annum
amount in 14 years, the money being improved at 6 per
cent., compound interest 1 in 10 years 1 in 20 years ?
in 22 years ? in 24 years ?
Ans. to the last, $25,407,75.
5. Find the amount of an annuity of $150, for 3 years,
at 6 per cent. A. $477,54.
A rule has been given, for finding the present worth of
an annuity, to continue forever ; but it is often necessary
to find the present worth of an annuity, which is to con-
tinue for a limited number of years ; thus,
6. What is the present worth of an annual pension of
$100 to continue 4 years, allowing 6 per cent, compound
interest?
The present worth is evidently a sum, which, at com-
pound interest, would in 4 years produce an amount equal
to the amount of the annuity, for the same time.
Now to find a given amount, at compound interest, we
multiply a sum by the amount of $1 at the given rate per
cent, as many times successively as there are years.
Hence to find a sum, which will produce a given
amount in a certain time, we must reverse this process and
divide by the amount of $1 for the given time.
Applying this to the above example, we find by the pre-
ceding rule, that the amount is $437,46 Dividing this by
the amount of $1 for 4 years, we find the present worth,
437,46-f-l,26247=$346,511, Ans,
ANNUITIES.
257
Hence to find the present worth of an annuity,
Find the amount in arrears for the whole time, and divide
it by the amount oj $1 at the given rate per cent., for the
given number of years.
The operations under this rule, will be facilitated by the
TABLE,
following
showing the present worth of $1, or £1 annuity, at 5 and
6 per cent, compound interest, for any number of years
from 1 to 34.
Fears.
5 per cent.
1
0,95238
2
l,8. r )!lll
3
2,72325
4
3,54595
5
4,32948
6
5,07569
7
5,78637
8
6,46321
9
7,10782
10
7,72173
11
8,30641
12
8,86325
13
9,39357
14
9,89864
15
10,37966
16
10,83777
17
11,27407
6 per cent.
0,94339
1,83339
2,67301
3,4651
4,21236
4,91732
5,58238
6,20979
6,80169
7,36008
7,88687
8,38384
8,85268
9,29498
9,71225
10,10589
10,47726
Years.
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
5 per cent.
11,68958
12,08532
12,46221
12,82115
13,163
13,48807
13,79864
14,09394
14,37518
14,64303
14,89813
15,14107
15,37245
15,59281
15,80268
16,00255
16,1929
6 per cent.
10,8276
11,15811
11,46992
11,76407
12,04158
12,30338
12,55035
12,78335
13,00316
13,21053
13,40616
13,59072
13,76483
13,^2908
14,08398
14,22917
14,36613
It is evident, that the present worth of $2 annuity is 2
times as much as that of $1 ; the present worth of $3 will
be 3 times as much, &c. Hence, to find the present worth
of any annuity, at 5 or 6 per cent, — Find, in this table, the
present worth of $1 annuity, and multiply it by the given
annuity, and the product wdl be the present worth.
7. Find the present worth of a $40 annuity, to continue
5 years, at 5 per cent. A. $173,173.
8. Find the present worth of $100 annuity, for 20 years,
at 5 per cent. A. $1,246.22.
9. Find the present worth of an annuity of $21,54 for
7 years at 6 per cent. A. 120.244+
22*
258 ARITHMETIC. PART THIRD.
10. Find the present worth of an annuity of $100, to
continue J 2 years, at 6 per cent. A. $838,384.
11. Find the present worth of an annuity of $936, for
20 years, at 5 per cent. A. $11,664.629 —
As the present worth of any annuity may be found, by
multiplying the annuity by one of the numbers, in the
above table, it is plain that if any present worth be divided
by the same number, it will give the annuity itself.
Hence to discover of what annuity any given sum is the
present worth, we may use the above, as a table of divi-
sors, instead of multipliers.
What annuity to continue 19 years, will $6,694,866
purchase, when money will bring 6 per cent. 1 A. $600.
An annuity is said to be in reversion, when it does not
commence until some future time.
12. What is the present worth of $60 annuity, to be con-
tinued 6 years, but not to commence till 3 years hence,
allowing 6 per cent, compound interest ?
The present worth is evidently such a sum as would in
3 years, at six per cent., compound interest, produce an
amount, equal to the present worth of the annuity, were it
to commence immediately.
We must therefore first find the present worth of an an-
nuity of $60 to commence immediately, according to the
last rule. This we shall discover to be $295,039.
We now wish to obtain a sum, whose amount in 3 years
will equal this present worth. This may be found by di-
viding the $295,039 by the amount of $1 for 3 years thus,
$295.039-r-l, 19101=247.72.
Ans. $247,72.
Hence to find the present worth of any annuity taken in
reversion, at compound interest,
Find the present worth to commence immediately, and this
sum divided by the amount of $1 for the time in reversion,
will give the answer.
13. If an annuity of $100 be 14 years in reversion, to
continue 20 years afterwards, what is its present worth,
discounting at 5 per cent. ? A. $629,426.
14. What is the present worth of a lease of $100 to
continue 20 years, but not to commence till the end of 4
PERMUTATION. 259
years, allowing 5 per cent. ? what if it be 6 years in re-
version ? 8 years ? 10 years ? 14 years ?
Ans. to last, $629,426.
15. What is the present worth of $100 annuity, to be
continued 4 years, but not to commence till 2 years hence,
allowing 6 per cent, compound interest ? A. $308,393.
PERMUTATION.
Permutation is the method of finding how many chan-
ges may be made, in the order in which things succeed
each other.
What number of permutations may be made on the let-
ters A and B ? They may be written A B, or B A.
What number on the letters ABC?
Placing A first, A B C, or A C B.
Placing B first, B A C, or B C A.
Placing C first, C A B, or C B A.
From these examples it will be seen, that of two things
there mav be 2 changes, (1x2=2,) and of 3 things there
may be 6 changes. (1X2X3=6.)
Hence, to find the number of different changes, or per-
mutations, of which any number of different things are
capable,
Find the continual ■product of the natural series of num-
bers, from 1 to the given number .
1. Four gentlemen agreed to remain together,' as long
as they could arrange themselves differently at dinner.
How many days did they remain ? A. 24 days.
2. 10 gentlemen made the same agreement, but they
all died before it could be fulfilled. The last survivor lived
53 yrs. 98 days, after the agreement. How much did the
bargain then want of being fulfilled, allowing 365 days to
the year ? A. 9,888 yrs. 237 d.
3. How many years will it take to ring all the possible
changes on 12 bells, supposing that 10 can be rung in a
minute, and that the year contains 365 d. 5 h. 49 m ?
A. 91 yrs. 26 d. 22 h. 41m.
4. How many variations may there be in the position of
the nine digits \ Ans. 362880
260 ARITHMETIC. PART THIRD.
5. A man bought 25 cows, agreeing to pay for them 1
cent for every different order in which they could all be
placed ; how much did the cows cost him 1
Ans. $1551 12100433309859840000.
MISCELLANEOUS EXAMPLES.
Many of these sums are designed for mental exercise.
In solving the first 50, the pupil should not be allowed to
use the slate.
1. If two men start from the same place and travel in
opposite directions, one at the rate of 4| miles an hour,
and the other at the rate' of 3§ miles an hour, how far will
they be apart in 6 hours 1
2. If 6 bushels of oats will keep 3 horses a week, how
manybushels will be required to keep 12 horses the same
time?
3. If you give 5 men 3a bushels of corn apiece, how
much do you give the whole ?
4. If 8 dollars worth of provisions will serve 9 men 5
days, how many days will it serve 12 men ? how many
days would it serve 3 men ?
5. If $6 worth of provision will serve 5 men 8 days,
how many days would it serve 9 men 1 how many days
would it serve 3 men ?
G. If $12 worth of provision would serve 5 men 7 days,
how many men would it serve 9 days ?
7. If one peck of wheat afford 9 six penny loaves, how
many ten penny loaves would it afford ?
8. If a man paid $60 to his laborers, giving to every
man 9d. and to every boy 3d. if the men and boys were
equal in number, how many were there of each ?
9. Two men bought a barrel of flour together, one paid
$3 and the other paid $5 ; what part of the whole did each
pay, and what part of the barrel ought each to have ?
10. Three men hired a field together, A paid $7, B
paid $3, and C paid $8, what part of the whole did each
pay, and what part of the produce ought each to have 1
11. Three men bought a lottery ticket together, A paid
$6, B paid $4, and C paid $10. They drew a prize of
$150, what was each man's share ?
MISCELLANEOUS EXAMPLES. 261
12. Three men hired a pasture together for $60. A
put in 2 horses, B 4 horses, and C 6 horses, how much
ought each to pay ?
13. Three men commenced trade together, and ad-
vanced money in this proportion — For every $5 that A
put in, B put in 3, and C put in $2, they gained $100,
what was each man's share ?
14. Two men hired a pasture for $32. A put in 3
sheep for 4 months, and B put in 4 sheep for 5 months,
how much ought each to pay ?
Note. 3 sheep for 4 months is the same as 12 sheep
for one month, and 4 sheep for 5 months is the same as 20
sheep for one month.
15. A and B traded together and invested money in the
following proportions, A put in 10 for 2 months, and B
put in $ 5 for 3 months. They gained $70 ; what was
each man's share ?
16. Three men traded in company, and put in money in
the following proportions. A put in 4 dollars as often as
B put in 3, and as often as C put in 2. A's money was in
2 months, B's 3 months, and C's 4 months. They gained
$100 ; what was each man's share ?
17. Two men traded in company. A put in $2 as often
as B put in $3. A's money was employed 7 months, and
B's 5 months. They gained 58 dollars. What was each
man's share ?
18. If A can do i of a piece of work in 1 day, and B
can do ^ of it in one day, how much would both do in a
day ? How long would it take them both together to do the
whole ?
19. If 1 man can do a piece of work in 2 days, and
another in 3 days, how much of it would each do in a day ?
How much would both together do 1 How long would it
take them both to do the whole ?
20. A cistern has 2 cocks ; the first will fill it in 3
hours, the second in 6 hours ; how much of it would each
fill in an hour ? How much would both together fill ? How
long would it take them both to fill it 1
21. A man and his wife found by experience, that, when
when they were both together, a bushel of meal would
last them only 2 weeks ; but when the man was gone, it
262 ARITHMETIC. PART THIRD.
would last his wife 5 weeks. How much of it did both
together consume in 1 week ? What part did the woman
alone consume in 1 week ? What part did the man alone
consume in 1 week ? How long would it last the man
alone ?
22. If 1 man could build a piece of wall in 5 days, and
another man could do it in 7 days, how much of it would
each do in 1 day ? How many days would it take them
both to do it ?
23. A cistern has 3 cocks ; the first would fill it in 3
hours, the second in 6 hours ; the third in 4 hours ; what
part of the whole would each fill in 1 hour ? and how long
would it take them all to fill it, if they were all running at
once?
24. A and B together can build a boat in 8 days, and
with the assistance of C they can do it in 5 days ; how
much of it can A and B build in 1 day? How much of it
can A, B, and C, build in 1 day? How much of it can C
build alone in 1 day ? How long would it take C to build it
alone ?
25. Suppo.se T would line 8 yards of broadcloth that is
li yard« wide, with shalloon that is f of a yard wide ; how
many yards of the shalloon will line 1 yard of the broad-
cloth ? How many yards will line the whole ?
26. If 7 yards of cloth cost 13 dollars, what will 10
yards cost ?
27. If the wages of 25 weeks come to 75 dollars, what
will be the wages of seven weeks ?
28. If 8 tons of hay will keep 7 horses three months,
how much will keep 12 horses the same time ?
29. If a stafF4 feet long cast a shadow 6 feet long, what
is the length of a pole that casts a shadow 58 feet at the
same time of day ?
30. If a stick 8 feet long cast a shadow 2 feet in length,
what is the height of a tree which casts a shadow 42 feet
at the same time of day ?
31. A ship has sailed 24 miles in 4 hours ; how long
will it take her to sail 150 at the same rate ?
32. 30 men can perform a piece of work in 20 days ;
how many men will it take to perform the same work in
8 days ? '
MISCELLANEOUS EXAMPLES. 263
33. 17 men can perform a piece of work in 25 days ;
in how many days would 5 men performme same work ?
34. A hare has 76 rods the start of a greyhound, but
the greyhound runs 15 rods to 10 of the hare ; how many
rods must the greyhound run to overtake the hare ?
35. A garrison has provision for 8 months, at the rate
of 15 ounces per day ; how much must be allowed per
day, in order that the provision may last 1 1 months ?
36. If 8 men can build a wall 15 rods in length in 10
days, how many men will it take to build a wall 45 rods
in length in 5 days?
37. A man being asked the price of his horse, an-
swered, that his horse and saddle together were worth
100 dollars ; but the horse was worth 9 times as much as
the saddle. What was each worth 1
38. A man having a horse, a cow, and a sheep, was
asked what was the value of each. He answered that the
cow was worth twice as much as the sheep, and the horse
3 times as much as the sheep, and that all together were
worth 60 dollars. What was the value of each ?
39. If 80 dollars worth of provision will serve 20 men
24 days, how many days will 100 dollars worth of provi-
sion serve 30 men ?
40. The third part of an army was killed, the fourth
part taken prisoners, and 1000 fled ; how many were in
this army ?
This, and the following 10 questions, are usually classed
under the rule of Position, bat they may be solved in a much
more simple and easy manner. Thus, i-f-|= T 7 5 of the army.
Now as there are 12 twelfths in the whole, 1000 must be
the remaining 5 twelfths. If 1000 is 5 twelfths of the ar-
my, 1 fifth of 1000, or 200, will be 1 twelfth ; and if 200
is 1 twelfth, the whole, or 12 twelfths will be 12 times as
much, or 2400.
41. A farmer being asked how many sheep he had, an-
swered, that he had them in 4 pastures ; in the first he
had i of his flock ; in the second i ; in the third | ; and in
the fourth 15 ; how many sheep had he ?
42. A man driving his geese to market, was met by
another, who said, good morrow, master, with your hun-
dred geese ; says he, I have not a hundred ; but if I had
264 ARITHMETIC. PART THIRD.
half as many more as I now have, and two geese and a
half, I should have a hundred ; how many had he ?
43. What number is that, to which if its half be added
the sum will be GO 1
44. What number is that, to which if its third be added
the sum will be 48 ?
45. What number is that, to which if its 5th be added
the sum will be 54 ?
46. What number is that, to which if its half and its
third be added the sum will be 55 ?
47. A man being asked his age, answered, that if its
half and its third were added to it, the sum would be 77 ;
what was his age ?
48. What number is that, which being increased by its
half, its fourth, and eighteen more, will be doubled?
49. A boy being asked his age, answered, that if i and
i of his age, and 20 more were added to his age, the sum
would be 3 times his age. What was his age ?
50. A man being asked how many sheep he had, an-
shered, that if he had as many more, \ as many more,
and 2i sheep, he should have 100. How many had he?
51. A farmer carried his grain to market, and sold
75 bushels of wheat, at $1,45 per bushel,
64 „ „ rye, „ $ ,95 „ „
142 „ „ corn, „ $ ,50 „ „
In exchange he received sundry articles : —
3 pieces of cloth, each
containing 31 yds, at $1,75 per yd.
2 quintals offish, „ $2,30 per quin.
8 hhds. of salt, „ $4,30 per hhd.
and the balance in money.
How much money did he receive ? Ans. $38,80
52. A man exchanges 760 gallons of molasses, at 37£
cents per gallon, for 66i cwt. of cheese, at $4 per cwt. ;
how much will be the balance in his favor ? Ans. $19
53. Bought 84 yards of cloth, at $1,25 per yard ; how
much did it come to ? How many bushels of wheat, at
$1,50 per bushel, will it take to pay for it ?
Ans. to the last, 70 bushels.
MISCELLANEOUS EXAMI>LES. 265
54. A man sold 342 pounds of beef, at 6 cents per
pound, and received his pay in molasses, at 37^ cents per
gallon ; how many gallons did he receive ?
Ans. 54,72 gallons.
55. A man exchanged 70 bushels of rye, at -$,92 per
bushel, for 40 bushels of wheat, at $1,37} per bushel, and
received the balance in oats, at $,40 per bushel ; how
many bushels of oats did he receive ? Ans. 23^
56. How many bushels of potatoes, at 1 s. 6 d. per
bushel, must be given for 32 bushels of barley, at 2 s. 6 d.
per bushel ? Ans 53^ bushels.
57. How much salt, at $1,50 per bushel, must be given
in exchange for 15 bushels of oats, at 2 s. 3 d. per bushel ?
Note. It will be recollected that, when the price and
cost are given, to find the quantity, they must both be re-
duced to the same denomination before dividing.
Ans. 3£ bushels.
58. How much wine, at $2,75 per gallon, must be
given in exchange for 40 yards of cloth, at 7 s. 6 d. per
yard ? Ans. 1 8 T 2 T gallons.
59. There is a fish, whose head is 4 feet long ; his tail
is as long as his head and \ the length of his body, and his
body is as long as his head and tail ; what is the length of
the fish ?
The pupil will perceive that the length of the body is
i the length of the fish. Ans. 32 feet.
60. A gentleman had 7 £. 17 s. 6 d. to pay among his
laborers ; to every boy he gave 6 d., to every woman 8d.,
and to every man 16 d. ; and there were for every boy
three women, and for every woman two men ; I demand
the number of each. Ans. 15 boys, 45 women, and 90 men.
61. A farmer bought a sheep, a cow, and a yoke of
oxen for $82,50 ; he gave for the cow 8 times as much as
for the sheep, and for the oxen 3 times as much as for the
cow ; how much did he give for each ?
Ans. For the sheep $2,50, the cow $20, and the oxen
$60.
62. There was a farm, of which A owned |, and B if ;
the farm was sold for $1764; what was each one's share
of the money 1 Ans. A's $504, and B's $1260
23
266 ARITHMETIC. PART THIRD.
63. Four men traded together on a capital of $ 3000, of
which A put in \, B \, C }, and D t l ; at the end of 3 yrs.,
thev had gained $2364 ; what was each one's share of the
gain? rA's$1182
A I B's $ 591
Ans> ) C's $ 394
' D's $ 197
64. Bought a book, the price of which was marked
$4,50, but for cash the bookseller would sell it at 33^ per
cent, discount ; what is the cash price ? Ans. $3,00
65. A merchant bought a cask of molasses, containing
120 gallons, for $42 ; for how much must he sell it to gain
15 per cent. ? How much per gallon ? Ans. to last, $,40i
Q6. A merchant bought a cask of sugar, containing 740
pounds, for $59,20 ; how must he sell it per pound to
gam 25 per cent ? Ans. $,10
67. What is the interest, at 6 per cent., of $71,02 for
17 months 12 days ? Ans. $6,178+
08. What is the interest of $487,008 for 18 months ?
Ans. $43,83+
It has been shown that the length of one side of a
square multiplied into itself, will give the square con-
tents.
Hence to find the area, or superficial contents of a
square when one side is given,
Multiply the side of the square into itself.
69. There is a room 18 feet square ; how many yards
of carpeting 1 vard wide will cover it ?
Ans. 182=324 ft.=36 yards.
70. The length of one side of a square room is 31 feet;
how many square feet in the whole room 1 Ans. 961
71. If the floor of a square room contain 36 square
yards, how many feet does it measure on each side ?
Ans. 18 feet.
Note. This answer is obtained by finding the square
root of the area 36 feet.
A parallelogram, or oblong, is a four sided figure, ha-
ving its opposite sides equal and parallel.
To find the area of a parallelogram,
Multiply the length by the breadth.
MISCELLANEOUS EXAMPLES.
267
72. A garden in the form of a parallelogram is 96 feet
long and 54 wide ; how many square feet of ground are
contained in it 1 Ans. 5184 sq. ft.
73. What is the area of a parallelogram 120 rods long
and 60 wide ? Ans. 7200 sq. rods.
74. If a board be 21 feet long, and 18 inches broad,
how many square feet are contained in it ?
Ans. 31} sq. feet.
A triangle is a figure bounded by three lines.
If a line be drawn from one corner of a parallelogram
to its opposite, (as in the Fig. A B,) it will divide it into two
B equal parts of the same
length and breadth as
the parallelogram, but
containing only half its
surface. These two
parts are triangles. —
Now supposing the
length of this parallelo-
gram to be 6 feet, and its breadth* 2, the area would be
12 feet. But the triangle will contain only half the sur-
face, or 6 feet.
Hence to find the area of a triangle,
Multiply the length by half the breadth, or tJie breadth by
half the length.
75. In a triangle 32 inches by 10, how many square
inches ? Ans. 160 sq. inches.
76. What is the area of a triangle whose base is 30
rods and the perpendicular 6 rods ? Ans. 90 rods.
It has been shown that the length of one side of a cube
raised to its third power will give the solid contents of the
cube.
Hence to find the solid contents of a cube, when one
side is given,
Multiply the given side into itself twice, or raise it to ita,
third power.
77. The side of a cubic block is 12 inches ; how manv
solid inches does the block contain ? Ans. 12 3 =1728
78. One side of a cube is 59 feet ; what are its solid
contents ? Ans. 205379
268 ARITHMETIC. PART THIRD.
79. If a cube contains 614,125 cubic yards, what is the
length of one side ? Ans. 85 yards.
Note. This answer is obtained bv finding the cube
root of 614125.
A circle is a figure contained by one line called the cir.
eumference, every part of which is equally distant from a
point within called the centre.
The diameter of a circle, is a line drawn through the
centre, dividing it into two equal parts.
It is found by calculation, that the circumference of a
circle measures about 3| times as much as its diameter, or
more accurately in decimals, 3,4159 times.
Hence to find the circumference of a circle when the
diameter is known,
Multiply the diameter by 3j.
To find the diameter when the circumference is known,
Divide the circumference by 3{.
To find the area of a circle,
Multiply i the diameter into \ the circumference.
80. If the diameter of a wheel is 4 feet, what is its cir-
circumference 1 Ans. 12^ feet.
81. What is the circumference of a circle, whose di-
ameter is 147 feet ? Ans. 462 feet.
82. What is the diameter of a circle, whose circum-
ference is 462 feet ? Ans. 147 feet.
83. What is the area of a circle, whose diameter is 7
feet, and its circumference 22 feet ? Ans. 38i sq. feet.
84. What is the area of a circle, whose circumference
is 176 rods ? Ans. 2464 rods.
The area of a globe, or ball, is 4 times as much as the
area of a circle of the same diameter.
Hence, to find the area of a globe,
Multiply the wliole circumference into the whole diameter.
85. What is the number of square miles on the surface
of the earth, supposing its diameter 7911 miles ?
Ans. 7911x24853=196,612,083.
To find the solid contents of a globe, or ball,
Multiply its area by J- part of its diameter.
86. How many solid inches in a ball 7 inches in diame-
ter ? Ans. 179$.
FORMS OF NOTES, RECEIPTS, &C. 269
A cylinder is a round body, whose ends are circles, and
which is of equal size from end to end.
To find the solid contents of a cylinder,
Multiply the area of one end by the length.
87. There is a cylinder 10 feet long, the area of whose
ends is 3 square feet ; how many solid feet does it con-
tain ? Ana. 30.
Solids which decrease gradually from the base till they
come to a point, are called pyramids. The point at the
top of a pyramid is called the vertex. A line drawn from
the vertex perpendicular to the base, is called the perpen-
dicular height of the pyramid.
To find the solid contents of a pyramid,
Multiply the area of the base by i of the perpendicular
height.
88. There is a pyramid whose height is 9 feet, and
whose base is 4 feet square ; what are its contents ?
Ans. 48 feet.
89. There is a pyramid, whose height is 27 feet, and
whose base is 7 feet in diameter ; what are its solid con-
tents ? Ans. 346^ feet.
FORMS OF NOTES, RECEIPTS, AND
ORDERS.
When a man wishes to borrow money, after receiving
it, he gives his promise to repay it, in such forms as those
below.
NoTE-
No. 1.
Hartford, Jan. 1, 1832.
For value received, I promise to pay D. F. Robinson,
or order, two hundred sixty four dollars, twenty-five cents,
on demand, with interest. John Smith.
No. 2.
New York, Jan. 15, 1332.
For value received, I promise to pay William Dennis,
or bearer, twenty dollars, sixteen cents, three months after
date. George Ellis.
23*
270 ARITHMETIC. PART THIRD.
No. 3.
Philadelphia, July 6, 1831.
For value received, we, jointly, and severally, promise
to pay to Henry Reddy, or order, one hundred dollars,
thirteen cents, on demand, with interest.
James Barnes.
Attest. James Cook. William Hedge.
Remarks.
1. The sum lent, or borrowed, should be written out in
words, instead of using figures.
2. When a note has the words " or order," or " or
bearer," it is called negociable ; that is, it may be given or
sold to another man, and he can collect it.
If the note be written, to pay him " or order," (see No.
I,) then D. F. Robinson can endorse the note, that is,
write his name on the back of it, and then sell it to any one
he chooses. Whoever buys the note, demands pay from
the signer, John Smith.
3. If the note be written, " or bearer," (see note 2,)
then whoever holds the note can collect it of the signer.
4. When no rate of interest is mentioned, it is to be
understood at the legal rate in the state where the note is
given.
5. All notes are payable on demand, unless some par-
ticular time is specified.
6. All notes draw interest after the time of promised
payment has elapsed, even if there is no promise of inte-
rest in the note.
7. Notes that are to be paid on demand, draw interest
after a demand is made.
8. If a man promises to pay in certain other articles, in-
stead of money, after the time of promised payment has-
elapsed, the creditor can claim payment in money.
Receipts.
Hartford, June 16, 1831.
Received of Mr. Julius Peck, twelve dollars, in full of
all accounts. John Osgood.
FORMS OF NOTES, RECEIPTS, &C 271
Receipt for money on a note.
Hartford, June 18, 1831.
Received of John Goodman, (by the hand of Willliam
Smith,) twenty dollars, sixteen cents, which is endorsed
on his note of July 6, 1829. John Reed.
Receipt j or money on account.
Hartford, April 6, 183L
Received of Albert Jones, forty dollars, on account.
Peter Trusty.
Receipt of Money for another Person.
Hartford, June 1st, 1831.
Received of A. B. one hundred and six dollars, for
I. C. Samuel Wilson.
Receipt for Interest due on a Note.
Hartford, Aug. 1, 1832.
Received of W. B. thirty dollars in full of one
year's interest of $500, due to me on the day of
last, on note from the said W. B.
William Gray.
Receipt for Money paid before it is due.
Newport, June 1, 1829.
Received of A. F. sixty dollars advanced, in full for one
year's rent of my house, leased to said A. F. ending the
first day of September next, 1829.
John Graves.
Note. — If a receipt is given in full of all accounts, it
cuts off only the claims of accounts. But " in full of all
demands" cuts off all claims of every kind.
Orders.
New York, June 9, 1830.
Mr. John Ayers. For value received, pay to N. S. or
order, fifty dollars, and place the same to my account.
Solomon Green.
New York, July 9, 1831.
Mr. William Redfield,— Please to deliver Mr. L. D.
such goods as he may call for, not exceeding the sum of
one hundred dollars, and place the same to the account
of vour humble servant. Stephen Birch.
272
ARITHMETIC. PART THIRD*
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This book is DUE on the last
date stamped below.
UC SOUTHERN REGIONAL LIBRARY FA
A 000 257 693 2
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