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0cric3 of ^rational Scljool Books.
TREATISE ON ARITHMETIC,
IN
THEORY AND PEACIICE.
FOR
^i):^ M^^ (91! giijj)aiiyasj
Aathorh'^.d by the Council of Public I?islru<:tio?i
for Upper Canada.
T O R O N T O :
rUfU.ISIIED BY ROBHRT MciMIAlL,
Qb, King Street E^rsT.
1860.
^ %
\ ^
\
%^ ii^^'d
PREFACE
In the present edition a vast number of exercises have
been added, that no rule, however trifling, might be left
without so many illustrations as should serve to make it
sufficiently familiar to the pupil. ^Vnd when it was feared
that the application of any rule to a particular class of ca-ses
might not at once suggest itself, some question calculated
to remove, or diminish the difficulty has been introduced
among the examples.
A considerable space is devoted to the "nature of num
bers," and " the principles of notation and numeration ;"'
for the teacher may rest assured, that the facility, and even
the success, TN-ith which subsequent parts of his instruction.
will be sonveyed to the mind of the learner, depends, in a
great degree, upon an adequate acquaintance with them.
Hence, to proceed without securing a perfect and practical
knowledge of this part of the subject, is to retard, rather
than to accelerate improvement.
The pupil, from the very commencement, must be made
•perfedlif familiar with the terms and signs which are intro-
duced. Of the great utihty of technical language (accu-
rately understood) it is almost superfluous to say anything
here : we cannot, however, forbear, upon this occa-sion, re-
calling to remembrance what is so admirably and so effec-
tively inculcated in the '• Easy Lessons on Reasoning."
** Even in the common mechanical arts, something of a
technical language is foimd needful for thos<» w-ho are learn-
VI PREFACE
ing or exercising them. It would be a very great lu
convenienpe, even to a common carpentw, not to have a
precise, well understood name for each of the several opera-
tions he performs, such as chiselling, sawing, planing, ^^c.,
and for the several tools [or instruments] he works with.
And if we had not such wox'ds as addition, subtraction,
multiplication, division, &c., employed in an exactly defined
sense, and also fixed rules for conducting these and other
arithmetical processes, it would be a tedious and uncertain
work to go through even such simple calculations as a child
Tery soon learns to perform with perfect ease. And after
all there would be a fresh difficulty in making other per-
sons understand clearly the correctness of the calculations
made.
" You a,re to observe, however, that technical language
and rules, if you would make them really useful, must be not
only distinctly understood, but also learned and remembered as
familiarly as the alphabet, and employed coiistantUj, and
with scrupulous exactness ; otherwise, technical language will
prove an encumbrance instead of an advantage, just as a
suit of clothes would be if, instead of putting them on and
wearing them, you were to carry them about in your hand."
Page 11.
What is said of technical language is, at least, equally true
of the signs and characters by which we still further facihtate
the conveyance of our ideas on such matters as form the
subject of the present work. It is much more simple to put
down a character which expresses a process, than to write
the name, or description of the latter, in full. Besides, in
glancing over a mathematical investigation, the mind is
able, with greater ease, to connect, and understand its dif-
ferent portions when they are briefly expressed by familiar
signs, than when they are indicated by words wliich have
nothing particularly calculated to catch the ei/e, and which
cannot even be clearly understood without considerable
attention. But it must be bortie in mind, that, while such
«• treatise as the present, will seem easy and intelligible
PREFACE Vll
■>
enough if the signs, which it contains in ahnost every page,
are as familiar as they should be, it must nece&?arily appear
more or less obscure to those who have not been habituated
to the use of them. They are, however, so few and so simple,
that there is no excuse for their not being perfectly under-
stood— ^particularly by the teacher of arithmetic.
Should peculiar circumstances render a different arrange-
ment of the rules preferable, or make the omission of any
of them, for the present at least, advisable, the judicious
master will never be at los^ aow to act — there may be
instances in which the shortness of the time, or the limited
intelligence of the pupil, will render it necessary to confine
his instruction to the more important branches. The teacher
should, if possible, make it an inviolable rule to receive
no answer unless accompanied by its explanation, and its
reason. The references which have been subjoined to the
different questions, and which indicate the paragraphs where
the answers are chicjly to be obtained, and also those refer-
ences which are scattered through the work, vrill, be found
of considerable assistance ; for, as the most intelligent pupil
will occasionally forget something he has learned, he may
not at once see that a certain principle is applicable to a
particular case, nor even remember where he has seen it
explained.
Decimals have been treated of at the same time as integers,
because, since both of them follow precisehj the same laws,
when the rules relating to integers are fully understood,
there is nothing nevj to be learned on the subject — particu-
larly if what has been said with reference to numeration and
notation is carefully borne in mind. J:^hould it, however, in
any case, be preferred, what relates to them can be omitted
until the learner shall have made some further advance.
The most useful portions of mental arithmetic have been
introduced into " Practice" and the other rules with which
they seemed more immediately connected.
The different rules should be very carefully impressed on
the mind of the learner, and when he is found to have been
Vm PRKFACE.
guilty of any inaccuracy, he should be made to correct him
self by repeating each part of the appropriate rale, and
exemplifying it, until he perceives his error. It should be
continually kept in view that, in a work on such a subject
as arithmetic, any portion must seem difficult and obscure
without a knowledge of what precedes it.
The table of logarithms and article on the subject, also
the table of squares and cubes, squar*:> roots and cube roots
of nimibers, which have been introduced at the end of the
work, will, it is expected, prove very acceptable to the more
advanced arithmetician.
CONTExNTS
PART I.
Multiplication Table, ....
Tables of money, weights, and measures,^
Definitions, .....
Section I. — Notation and Numeration,
Arabic system of numbers,
Roman notation, . . " .
Skction II.— Simple Additi(»i,
To prove Addition,
Addition of quantities containing decimals-,
Simple Subtraction,
To prove Subtraction,
Subtraction of quantities containing decimals.
Simple ^Multiplication, .
To multiply when neither multiplicand nor mul
tiplier exceeds 12, .
When the multiplicand exceeds 12,
To prove Multiplication,
To multiply quantitie.s, when tliere are cyphers
or decimals, ....
When lx>tli multiplicand and multiplier ex
cee.-^. 12, ....
T(j prove Multiplication,
To prove MultipKcation, by casting out the nines
'Jo multiply, so as to have a certain number o;
decimal places in the product.
To multiply by a composite number,
by a number not composite,
To multiply by a number consisting of nines.
Simple Division. ....
a 2
Pare
I
3
12
14
18
20
33
B(i
37
45
40
49
53
57
57
CO
Gl
65
CG
C7
70
72
73
74
78
./
CONTENTS
NTS f
To dlvi^ Avlien the divisor does not exceed 12,
nor tli«dividend 12 times the divisor,
When tlie divisor does not exceed 12, but
the dividend is more than 12 times the divisor,
To prove Division, ....
To divide when the dividend, divisor, or both,
contain cyphers or decimals,
When the divisor exceeds 12,
To prove Division,
To abbreviate the process of division, when there
are decimals, . . . . .
To divide by a composite number,
To divide by a number but little less than one
expressed by unity and one or more cyphers,
To find the greatest common measure of numbers,
To find their least common multiple,
SfiCTiGN III. — Reduction Descending,
Reduction Ascending, .
To prove Reduction, . t.
The Compound Rules, .
Compound Addition,
Compound Subtraction,
Compound Multiplication, when the multiplier
does not exceed 12, .
When the multiplier exceeds 12 and
composite, ....
When the multiplier is the sum of compo-
site numbers, ....
When the multiplier is not composite.
Compound Division, when the divisor is abstract,
and does not exceed 12,
When the divisor exceeds 12 and is cc
posite, ....
When the divisor exceeds 12 and is not com-
posite, ....
When the divisor and dividend are both ap-
plicate, but not of the same denomination ; or
more than one denomination is found in either
or both, .....
Pag«
80-
85
86
89
94
95
96
101
104
107
109
110
113
114
123
i2e»
123
12a
130
132
134
134
1»»
CONTEBPTS
\ f^' , V rf/
?5!cc:tion IV. — Vulgar Fractions,
v, To reduce au improper fraction to a mixed n^
To reduce an integer to a fraction,
To reduce fractions to lower terms,
To find the value of a fraction in terms of a lo-^er
denomination, ..... 143
To express one quantity as the fraction of another, 143
To add fractions having a common denominator, 144
To add fractions when their denominators are
different and prime to each other, . . 145
To add fractions ha^-ing different denominators,
not all prime to each other, . . . 146
To reduce a mixed number to an improper frac-
tion, ...... 147
To add mixed numbers, .... 148
To subtract fractions which have a commoti
denominator, ..... 149
To subtract fractions which have not a common
denominator, . . . . 150
To subtract mixed numbers, or fractions from
mixed numbers, .... 150
To multiply a fraction and whole number together, • 152
To multiply one fraction by another, . . 153
To multiply a fraction, or mixed number by a
mixed number, .... 154
To divide a fraction by a whole number, . 156
To divide a fraction by a fraction, . . 156
To di\T.de a whole number by a fraction, . 158
To divide a mixed number by a whole number or
fraction, ..... 158
To divide an *iteger by a mixed number, . 159
To divide a fraction or mixed number by a mixed
number, ..... 159
>Vheu the divisor, dividend, or both, are com-
pound, or complex fractions, . . . 160
Decimal Fractions, .... 162
To reduce a vulgar fraction to a decimal or to 3
decimal fraction, . . . , 163
To reduce a decimal to a lower denomination, , 163
JUl CONTENTS.
VAg*
To find at once the decimal equivalent to any
number of shillings, pence, &c., . . 164
To find the number of shillings, &c., equivalent
to a decimal, ....
Circulating Decimals,
To change a circulating decimal into its equi
valent vulgar fraction,
When a vulgar fraction will give a finite decimal
The number of decimal places in a finite decimal,
The number of digits in the period of a circulate
When a circulating decimal will contain a
finite part, . ...
Contractions in multiplication and division, de
rived from the properties of fractions,
Section V. — Proportion,
Nature of ratios, ....
Nature of Proportion,
To find the arithmetical mean of two quantities, 181
To find a fourth proportional, when the first term
is unity, ..... 184
When the second or third term is imity, . 184
To find the geometrical mean of two quantities, 184
Properties of a geometrical proportion, . . 185
Rule of Simple Proportion, . . . 185
When the first and second terms are not of
the same, or contain different denominations, 190
When the third term contains more than one
denomination, ..... 192
If fractions or mixed numbers are found in
any of the terms, . . . .195
Rule of Compound Proportion, . . . 202
To abbreviate the process, ' . . , 204
166
16»
167
170
170
171
172
173
176
17T
179
PART n.
Skction VI. — Practice, .... 209
To find aliquot parts, . . . . 209
To find the price of one denomination, that of a
higher being given, , . . . 211
CONTENDS. xiii
Page
To find the price of more than one lower denomi-
nation, ...... 212
To find the price of one higher denomination, . 213
To find the price of more than one higher deno-
mination, ..... 213
Given the price of one denomination, to find that
of any number of another, . . . 214
When the price of any denomination is the aliquot
part of a shilling, to find the price of any num-
ber of that denomination, . . . 215
When che price of any denomination is the ali-
quot part of a pound, to find the price of any
nimiber of that denomination, . . . 215
When the complement of the price, but not the
price itself, is the aliquot part or parts of a
pound or shilling, . . . .216
When neither the price nor its complement is the
aliquot part or parts of a pound or shilling, . 217
When the price of each article is an even number
of shilLi-\i3, to find the price of a number of
articiet^, .... 218
When the price is an odd number and lesa
than 20, . . . . .219
To find the price of a quantity represented by a
mixed number, .... 220
Given the price per cwt., to find that of CTvt.,
qrs., &c., 221
Given the price per pound, to find that of cwt.,
qrs., &c., 222
Given the price per pound, to find that of a ton, 223
Given the price per ounce, to find that of ounces,
pennyrveights, &c., .... 223
Given the price per yard, to find that of yards,
qrs, &c., 225
To find the price of acres, roods, &c., . . 226
Given the price per quart, to find that of a hogs-
head, 226
Given the price per quart, to find that of a ferai, 227
Xiy CONTENTS
Page
Given the pri-^e of one article in pence, to find
that of any number, .... 227
Given wages per day, to find their amount per
year, ...... 228
Bills of parcels, ..... 229
Tare and Tret, ..... 233
Section VII. — Simple Interest,
To find the simple interest on any sum, for a year, 237
When the rate per cent, consists of more
than one denomination, . . . 233
To find the interest on any sum for years, . 239
For years, months, &c., . , . 239
To find the interest on any sum, for any time,
at 5, 6, &c., per cent., .... 240
When the rate, or number of years, or both,
are expressed by a mixed number, . . 242
To find the interest for days, at five per cent., . 243
To find the interest for days, at any other ratQ, . 244
To find the interest for months, at 6 per cent., 244
To find the interest of money left after one or
more payments, .... 246
Given the amount, rate, and time — to find the
principal, ..... 248
Given the time, rate, and principal — to find the
amoimt, ..... 249
Given the amount, principal, and rate — to find
the time, . . . . .249
Given the amount, principal, and time, to find
the rate, 250
Compound Interest — gi\en the principal, rate,
and time — to find the amount and interest, . 251
To find the present worth of any sum, . . 256
Given the principal, rate, and amomit — to find
the time, ..... 257
Discount, . . . . .260
To find discount, . . . . . 261
To compute Commission, Insurance, Brokerage, 263
CCNTENT8.
To find wLat insurance must be paid that^ if the
goods are lost, both their value and the insur
ance paid may be recovered, .
Purchase of Stock,
Equation of Payments,
SECTION VIII. — Exchange, .
Tables of foreign money.
To reduce bank to current money^
To reduce current to bank money.
To reduce foreign to British money,
To reduce British to foreign money,
To reduce florins, &c., to pounds, &c., Flemish,
To reduce pounds. &c., Flemish, to florins, &c.,
Simple Arbitration of Exchanges,
Compound Arbitration of Exchanges,
To estimate the gain or loss per cent.,
Profit and Loss, ....
To find the gain or loss per cent..
Given the cost price and gain — to find the sellin
price, .....
Given the gain or loss per cent., and the sellin
price— to find the cost price, .
Simple Fellovrship,
Compound Fellowship,
Barter, .....
Alligation ■Medial,
Alligation Alternate,
When a given amount of the mixture is re
quire^, .....
When the amount of one ingredient is given
SECTION IX. — Involution,
To raise a number to any poTver,
To raise a fraction to any power.
To raise a mixed number to any power, .
Evolution, ....
To find the square root, .
When the square contains decimals,
To find the square root of a fraction,
Page
XVI
CONTENTS.
To find the square root of a mixed number,
To find the cube root,
When the cube contains decimals,
To find the cube root of a fraction.
To find the cube root of a mixed number,
To extract any root whatever,
To find the squares and cubes, the square and
cube roots of numbers, by the table, .
Logarithms, ....
To find the logarithm of a given number by the
table, .....
To find the logarithm of a fraction.
To find the logarithm of a mixed number,
To fijid the number corresponding to a given
logarithm, . , , , ,
If the given logarithm is not in the table.
To multiply numbers by means of their loga
rithms, ....
To divide numbers by means of their logarithms.
To raise a quantity to any power by means of its
logarithm, .....
To evolve a quantity by means of its logarithm,
Section X, — ^On Progression,
To find the sum of a series of terms in arithmeti-
cal progression, . .
In an arithmetical series given the extremes and
number of terms — to find the common difier-
encc, ......
To find any number of arithmetical means ,.be-
tween two given numbers.
To find any particular term of any arithmetical
series, ......
In an arithmetical series, given the extremes and
common difi'erence — to find the number of terms,
Given the sura of the series, the number of
terms, and one extreme — to find the other,
To find the sura of a series of terms in geometri-
oal progression, ....
rage
313
313
315
31(3
31G
317
318
319
321
323
323
324
324
325
326
327
327
32a
329
330
331
332
332
333
3*4
CONTEN'W.
In a geometrical series, given the extremes and
number of terms — to find the common ratio,
To find any number of geometrical means between
two quantities, ....
To find any particular term of a geometrical series,
In St geometrical series, given the extremes and
common ratio — to find the number of terms, .
In a geometrical series, given the common ratio,
the nimiber of terms, and one extreme — to find
the other ....
Annuities, ....
To find the amount of a certain nmnber of pay
ments in arrears, and the interest due on them
To find the present value of an annuity,
When it is in perpetuity, .
To find the value of an annuity in reversion,
Position, .
Single Position, .
Double Position, .
Miscellaneous exercises,
Table of Logarithms,
Table of squares and cubes, and of square and
cube roots, ....
Table of the amounts of £1, at compound interest
Table of the amounts of an annuity of £1,
Table of the present values of an annuity of £1,
Irish converted into British acres,
■ Value of foreign money in British,
rag©
335
336
336
337
O O (J
OOO
340
340
342
343
344
345
346
347
355
361
377
385
385
386
386
386
T5(EATISi: ON ARITHMETIC
THEORY AND PRACTICE.
ARITHMETIC
PART I.
TABLES.
MULTIPLICATION TABLE.
1
7 times |
1 are 7
2 — 14
3 — 21
4 — 28
5—35
6—42
7—49
8 — 56
9 — 63
10 — 70
11 — 77
112 — 84
Twice
3 times
4 times
5 times
6 times
1 are 2
1 are 3
1 are 4
1 are 5
1 are 6
2—4
2—6
2—8
2 — 10
2 — 12
3—6
3—9
3 — 12
3 — 15
3 — 18
4—8
4 — 12
4 — 16
4 — 20
4-24
1 5 - 10
5 — 15
5—20
5 — 25
5 — 30
6 — 12
6 — 18
6 — 24
6 — 30
6 — 36
7 - 14
7 — 21
7—28
7 — 35
7—42
8 — 16
8 — 24
8 — 32
8 — 40
8 — 48
i 9 — 18
9 — 27
9 — 36
9 — 45
9 — 64
llO — 20
10 — 30
10 — 40
10 — 50
10 — 60
111 — 22
11 — 33
11 — 44
11 — 55
11 — 66
jl2 — 24
12 — 36
12 — 48
12 — 60
12 — 72
8
!
times 1
1
are
8
2
16
3
24
4
32
5
40
6
48
7
56
8
—
64
9
—
72
10
—
80
11
—
88
12
—
96
t
9 times
1 are 9
2—18
3—27
4—36
5—45
6—54
7—63
8—72
9—81
10 — 90
11 — 99
12 — 108
10 times
1 are 10
2—20
3—30
4—40
50
60
70
6 —
6 —
7 —
8—80
9—90
10 — 100
11 — 110
12 — 120
11 times
1
are
11
2
22
3
33
4
44
5
55
6
—
m
7
—
77
8
—
88
9
99
10
110
11
121
12
—
132
12 times
1 are 12
2—24
3 — 36 I
4 — 48 I
5 — 60 I
6 — 72 I
7 — 84 .
8—96
9 — 108 ,
10 — 120 ;
11 — 132 I
12 — 114
It appears from this table, that the multiplication of tho
same two numbers, in whatever order taken, produces the
"signs used in this treatise.
+ the sign of addition; as 5+7, or 5 to be addc*
to 7.
— the sign of subtraction ; as 4— 3, or 3 to be sub-
tracted from 4.
X the sign of multiplication ; as 8X9) or 8 to bo
multiplied by 9.
-h the sign of division ; as IS -r 6, or 18 to be divided
by 6.
the vinculum, which is used to show that all
the quantities united by it are to be considered as but
one. Thus 4+3 — 7X6 means 4 to be added to 3, 7
to be taken from the sum, and 6 to be multiplied into
the remainder — the latter is equivalent to the whole
quantity under the viPxCulum.
= the sign of equality ; as 5 + 6 = 11, or 5 added to
6, is equal to 11.
I >^, and |<|, mean that | is gi-eater than |, and
that I IS less than f.
: is the sign of ratio or relation ; thus 5:6, means
the ratio of 5 to 6, and is read 5 is to 6.
: : indicates the equality of ratios ; thus, 5 : 6 : : 7 : 8,
means that there is the same relation between 5 and 6
as between 7 and 8 ; and is read 5 is to 6 as 7 is to 8.
^ the radical sign. By itself, it is the sign of the
square root ; as ^o^ which is the same as 5^, the square
root of 5. ^6, is the cube root of 3, or 3*. 7^4, is
the 7th root of 4, or 4^, &c.
Example.—a/ 8 -3 + 7 X 4 h- 6 + 31 X ^ 9 -f- 10^ X 52=
641-31, &c. may be read thus : take 3 from 8, add 7 to the
difference, multiply the sum by 4, divide the product by 6
take the square root of the quotient and to it add 31, then
multiply the sum by the cube root of 9. divide the product
by the square root of 10, multiply the quotient by the
eqnare of 5, and the product will be equal to 641-31, &c.
These signs arc fidbj explained in their proper places.
2 MULTIPLICATION TAULE.
eaixie result ; thus 5 times G, and G times 5 are 30 : — tho
reason will be cx|)laiued when we treat of multiplication.
There are, therefore, several repetitions, which, although
many persons conceive them unnecessary, are not, perhaps,
quite unprofitable. The following is free from such an
objection : —
f
Twice 2
» 3
„ 4
» 6
M 6
„ 7
,. 8
„ 9
are 4
— 6
— 8
— 10
— 12
— 14
— 16
— 18
5 times 7 are 35
» 8-40
,. 9—45
10 times 8
» 9
., 10
» 11
are 80
— 90
— 100
— 110
6 times 6—36
» 7-42
„ 8-48
,, 9—64
11 times 2
„ 3
„ 4
„ s
„ 6
„ 7
» 8
„ 9
— 22
— 33
— 44
— 55
— 66
— 77
— 88
— 99
8 times 3
» 4
» 5
» 6
„ 7
,, 8
„ 9
— 9
— 12
— 15
— 18
— 21
— 24
— 27
7 times 7 — 49
„ 8-56
„ 9—63
8 times 8 — 64
„ 9—72
12 times 2
,, 3
„ 4
., 5
" 5
>> 7
., 8
„ 9
» 10
„ 11
„ 12
— 24
— 36
- 48
60
— 72
— 84
— 96
— 108
— 120
— 132
— 144
1
4 times 4
„ 5
,, 6
>> 7
M 8
» 9
— 16
— 20
— 24
— 28
— 32
— 36
9 times 9 81
10 times 2 are 20
„ 3—30
„ 4—40
„ 6—60
„ 6—60
„ 7—70
5 times 5
„ 6
I
— 25
— 30
" Ten," or " eleven times," in the above, scarcely requires
to be committed to memory; since we perceive, that to
multiply a number by 10, we have merely to add a cypher to
the right hand side of it : — thus, 10 times S are 80 ; and to
multiply it by 11 we have ordy to set it down twice : — thus,
11 timea 2 ciro 23.
TABLE OF MONEY.
The following tables are required for reduction, the
compound rules, &c., and may be committed to memory
TABLE OF MONEY.
A farthing is the smallest coin generally used in this
Symbols.
country, it is represented by . . . . ^
Farthings
make 1 halfpenny, h
1 penny, d,
. 1 shilling, s
hah'pcnce
4 or
.
48
24 or
480
504
pence
12
9G0
1,008
240 or
252 or
shillings
20
21
1 pound,
1 guinea,
The symbols of pounds, shillings, and pence, are placed
£ s. d.
over the numbers which express them. Thus, 3 „ 14 „ G,
means, three pounds, fourteen shillings, and sixpence. S(jme-
times only the symbol for pounds is used, and is placed
s. d.
before the whole quantity ; thus, £3 „ 14 „ 6. 3 9i meana
three shillings and ninepence halfpenny. 2s. 6|f?. means two
shillings and sixpence three farthinfrs,' &c.
AVhen learning the above and following tables, the pupil
sh'ould be ri^uired, at first, t<<) commit to memory only those
portions which are over the thick angular lines ; thus, in the
one just given: — 2 farthings make one halfpenny: 2 half-
pence one penny : 12 pence one shilling ; 20 shillings one
pound ; and 21 shillings one guinea.
\y i, ^, reaUy mean the quarter, half, and three quarters
of a penny. (/. is used as a symbol, because it is the first
letter of ■' denarius," the Latin word signifying a penny ; s.
was adopted for a similar reason — '• solidus,"' meaning, in
tlie same language, a shilling ; and <£ also — '• Libra," signify*
ing a pound. ,
2 6 make one half Crown.
5 0 one Crown.
13 4 one Mark.
WEIGHTS,
AVOIRDUPOISE WEIGHT.
Its name is derived from French — and ultimately
from Latin words signifying "to have weight." It ia
used in weighing heavy articles
SymboU
make 1 ounce, oz.
. 1 pound, lb.
. 1 quarter, q.
. 1 hundred,cwt.
hundreds
20 . 1 ton, t.
14 lbs., and in some cases 16 lbs., make 1 stone.
20 stones ... 1 barrel.
TROY WEIGHT.
It is so called from Troyes, a city in France, where
it was first employed ; it is used in philosophy, in
weighing gold, &,c.
Symbols.
Grains ■ , . . , . . grs.
24 . . . make 1 pennyweight, ^wt.
pennyweights
20 . . 1 ounce, ot.
Drams
16
ounces
16
pounds
28
,
256 or
.
7,168
448 or
1,792
35,840
.
28,672
112 or
2,240
quarters
4
573,440
80 or
480 or
5,760
ounces
240 or I 12 . 1 pound, . ft).
A grain was originally the weight of a graii of corn,
taken^^from the middle of the ear ; a pennyweight, that of
the silver penny formerly in use.
APOTHECARIES WEIGHT.
In mixing medicines, apothecaries use Troy weight,
hut subdivide it as follows : —
Grains Symbols
20 . . . . . make 1 scruple, 9
1 dram, 3
1 ounce, 5
ounces
12 . 1 pound, lb.
The " Carat," Avhich is equal to four grains, is used in
weigliing diamonds. The term carat is also applied in
estimating the tinuucss of gold ; the latter, when jjerfeetjy
00 or
scruples
3
480
24 or
288
drams
8
5,700
90 or
MEASimSS.
pure, 13 said to be " 24 carats fine."' If there are 23 parts
gold, and one part some other material, the mixture is said
to he "23 carats fine ; " if 22 parts out of the 24 are gold,
it is " 22 carats fine," &c. ; — the whole mass is, in aU cases,
supposed to be divided into 24 parts, of which the number
consisting of gold is specified. Our gold coin is 22 carats
fine : pure gold being very soft would too soon wear out.
The degree of fineness of gold articles is marked upon them
at the Goldsmith's Hall; thus we generally perceive " 18 " on
the cases of gold watches ; this indicates that they are " 18
carats fine" — the lowest degree of piu-ity which is stamped.
grs.
A Troy ounce contains . 480
An avoirdupoise ounce . 4374
A Troy pound . . 5,760'
An avoirdupoise pound . 7,000
A Troy poimd is equal to 372- 965 French grammes.
175 Troy pounds are equal to 144 avoirdupoise ;
175 Troy are equal to 192 avoirdupoise ounces.
CLOTH MEASURE.
Inches
2i
.
aails
9 or
4
86
16 or
27
12 or
45
20 or
54
24 or
quarters
4
3
5
0
make 1 nail.
1 quarter.
1 yard.
. 1 Flemish ell
. 1 English ell.
1 French ell
LONG MEASURE.
(It is used to measure Length.)
'. . make 1 inch.
1 foot.
144 or
inches.
12
432
36 or
198
252
7,920
10,080
63.860
80.640
feet
o
2,376
3,024
IGiOT
21 or
660
840
5,280
6,720
yards
5i
7
•
95.040
120,960
220 or
280 or
1,760
2,210
perches
40
40
700,320
l'C7,6S0
320 or
320 or
1 yard.
1 English perch
1 Irish perch.
1 English furlong
1 Irish furlong,
furlongs
8 1 English mile.
8 1 Irish mile
a
6
MEASURE?
Three miles make one league. CO^U English miles make
60 nautical, or geographical miles ; which are equal to one
degree, or the three hundred and sixtieth part of the cir-
fluiufcrence of the globe — as measured on the equator.
4 inches make 1 hand (used in measuring horses).
inches
3 palms
18 laches
5 feet
6 feet
120 fathoms
1 palm.
1 span.
1 cubit
1 pace.
1 fathom.
1 cable's length.
100 links, 4 English perches (or poles), 22 yards, GG feet,
or 792 inches, make one chain. Each link, therefore, is
eqtlal to 7/o\ inches. 11 Irish are equal to 14 English
miles. The Paris foot is equal to 12792 Enghsh inches;
the Koman foot to 11-604 3 and the French metre to 39-383.
MEASURE OF SURFACES.
A surface is called a square when it has four equal
sides and four equal angles. A square inch, therefore,
is a surface one inch long and one inch wide ; a squar«
foot, a surface one foot long and one foot wide, &.c.
Square inches
144
1,296
39,204
63,504
1,568,160
9,640,160
C,272,640
10,160,640
square feet
9
441
10,890
17,640
43,560
70,560
4,014,439,600 27,878.400
6,e02,809,600i25,io8,400
sq. yards
30i
49
1,210 or
1,960 or
4,840
7,840
3,097,600
jq. perehea.
40
40
102,400 2,660 or!
.5,017,60oll02,400 I2.06O or!
irake 1 sq. foot.
1 square yard.
1 sq. Eng. perch.'
1 sq. Irish pereh.
I sq. Eng. rood. '
I sq. Irish rood.
1 statute acre.
1 plantation acre.
1 sq. Eng. mile.
1 sq. Irish mile.
640
640
The English, called also the statute acre, consists of 10
square chains, or 100,000 square links.
The English acre heing 4,840 square yards, and the Irish,
or plantation acre, 7,840 ; 19G square English are equal to
121 square Irish acres.
Tlie English square mile being 3.007,G00 square yards,
and the Irish 5,017,600; 196 English square miles are
equal tc 121 Irish : — we have seen, however, that 14 English
are equa.1 to 11 Irish Ihicar mileg
MEASURES.
MEASURE OF SOLIDS.
The teacher will explain that a cube is a solid havin|^
six equal squai-e surfaces ; and \7ill illustrate this by
models or examples — the more familiar the better. A
cubic inch is a solid, each of whose ji^:^ sides or faces is
a square inch ; a cubic foot a solid raoh of whose SM.
sides is a square foot ^ &:c.
Cubic inches
1,728
1 cubic feet
27
nifrks 1 cub''? fQivV
cubio r^<^
WINE MEASURE.
Gills or nagglns
4
8 or
320
576
1,344
2,016
2,688
4,032
8,064
pints
quarts
8 or
4
.
gallons
80
40 or
10
144
72
18
336
168
42
504
252
63
672
336
84
.
hogsheads
1,008
504
1,008
126 or
252
2
2,016
4 or 1
1 gaJoA
1 anker.
1 runlet.
1 tierce.
1 hogshea«
1 puncheon
1 pipe or butt
1 tun.
Jn some places a gill is equal to half a pint.
Foreign wines, &:c., are ofi»en suld by measures differing
from the above.
ALE MEASURE.
Gallons
16 or
firkins
2
32
4 or
6
8
12
kilderkins
0
48
64
96
3 or
4 or
6 or
barrels
u
0
i
make 1 firkin.
1 kilderkin.
1 barrel.
1 hogshead.
1 puncheon.
1 butt.
MEASURES.
BEER MEASURE.
•f ~llon3
9
18 or
54
72
108
firkins
2
4 or
6
8
12
kilderkins
2
3 or
4 or
6 or
barrels
o
make 1 fiikin.
1 kilderkin
1 barrel.
1 hogshead.
1 puncheon.
1 butt.
DRY MEASURE.
(It is used for vrheat, and other dry goods.)
Pints
4 or
8
IG
61
1&2
256
576
512
2,048
2,560
5,120
quarts
2
4 or
32
96
12J^^
288
25G
1,024
1,280
2,560
pottles
9
4 or
16
48
64
144
128
512
640
1,280
gallons
24
32
72
64
256
320
640
pecks
4
12 or
16or
36 or
32
128
160
320
bushels
3
4
9
coombs
2
make 1 pottle.
1 gallon.
1 peck.
1 bushel.
1 sack.
1 coomb.
1 vat.
8 or
32
40
80
« or
10 or
20
1 quarter.
quarters
4
5
1 chaldron
1 wey.
weys
10 or I 2 'l last.
The pint dry measure contains about 34| cubic inches;
277i cubic inches was made the standard gallon for both
lir{uid and dry goods, by an Act of Parliament which came
into operation in 182G.
Coals are now sold by weight; 140 pounds make one bag;
IG bags one ton.
TIME.
Thirds
MEASURE OE TIME.
Syaaols
make 1 second "
ECConJs
3600, or tji)
5,184,000
1 0.288,000
. io,15i000
l,»9-i, 160.000
1,897,344,000
1,892,160,000
3600 or
86,400
1,440 or
lO.OSO
604,800
2,419,200 40,320
31,536.000 J525.G00
31,622,400 ,527,040
31,536,000 ,.525,600
i
hours
24
163 or
672 or
3,7G0or ."iGS
1 hear h.
1 daj d
1 week w.
1 lunar month.
1 common year
1 leap year,
calendar mon."!
lunar months f ^ J'^^'"'
13 j
The following will exemplify the use of the above symbols : — •
The solar year consists of 3G5 d. 5 h. 48' 45" 30'": read "three
hundred and sixty-five days, five liours, forty-eight minutefu
forty-five second.s. and thirty third:?.
The number of days in each of the twelve calendar months
will be easily remembered by means of the well known lines,
'■ Thirtv- days hath September,
April, June, and November,
February twenty-eight alone
And all the rest thiriy-ene."
The following table will enable us to fijid how many daya
there are from any day in one month to any day in another.
Fr
3M A>-V DaV ].V
1
2
z
<
1 ._
Jan.|Feb.i.Mar
April,
275I
306j
334I
May
June
214
245
273
July; Aug
184! 153;
215i 184;
243J 212|
Sept.!
1221
18l|
Oct. jXOT
92i 61
123| 92
loli 120
Dec
31
62
90
Jan.
Feb.
365! 334; 306
31; 365^ 337
245
276
304
Mar.
59| 2^; 366
April
90
59; 31
365J
335
304
274! 243;
2I2I
182J 151
121
May
120] 89J 61
30,
365
334
304| 273
242 j
2121 181
151
June! 151 1 120j 92
61j
31
365
3351 304|
273i
243| 212
182
Julyj 181; 1.50: 122
Aug.i 212! 161 ; 153
9lj
1221
61
9-2
30
61
365
334!
303j
3.^1
273i 242
304' 273
212
31
365'
!
243
Sept.j 243J 212', 184
Oct. 1 273! 242 214
Nov.j 304! 273 245
1.53;
163,
2141
123
92
122
1.53
62
92
31
61
365:
30|
6,|
335
304
334
274
153
365
3W
184
123} 92|
31 1 365
3.85
Dec.l 334! 303' 275
1 ! 1
2n|
!
214
183
153: 122!
i !
"1
61
30
Z
10 TIME.
To find by this table tlic distance between any two
days in two different months :
Rule. — Look along that vertical row of figures at
the head of which stands the fii'st of the given months ;
and also along the horizontal row which contains the
second ; the number of days from any day in the one
month to the same day in the other, will be found where
these two rows intersect each other. If the given day
in the latter month is earlier than that in the former,
find by how much, and subtract the amount from the
number obtained by the table. If, on the contrary, it
is later, ascertain by how much, and add the amount.
When February is included in the given time, and
it is a leap year, add one day to the result.
Example 1. — How many days are there between the
fifteenth of INIaroh and the fourth of October '? Looking
down the vertical row of figures, at the head of which INIarch
is placed, and at the same time, along the horizontal row at
the left hand side of which is October, we perceive in their
intersection the number 214 : — so many days, therefore, in-
tervene between the fifteenth of jNIarch and the fifteenth of
October. But the fourth of October is eleven days earlier
than the tifteentii ; we therefore subtract 11 from 214, and
obtain 203. tlie number required.
Example 2. — How many days are there between the
third of January and the nineteenth of JNIay ? Looking as
before in tlie table. Ave find that 120 days intervene between
the tliird of January and the third of May 5 but as the nine-
teenth is sixteen days later than the third, we add 16 to 120
and obtain 136, the number required.
Since February is in this case included, if it were a leap
year, as that month would then contain 29 days, we should
add one to the 136, and 137 would be the answer.
During the lapse of time, the calendar became iuaccu-
cate : it was corrected by Pope G-regory. To understand
Uow this became necessary, it must be borne in mind that
the Julian Calendar, formerly in use, added one day every
fourth year to the month of February ; but this being
(iomewhat too much, the days of the months were thrown
i)ut of then- proper places, and to such an extent, that
«ach had become ten days too much in advance. Pope
txregory, to remedy this, ordained tIi:U wliat, according
TIME
n
to the Julian style, would have been the 5th of October
1582, should be considered hs the 15th ; and to prevem
the recuiTence of such a mistake, he desired that, in
place of the last year of every century being, as hitherto.
a leap year, only the last year of every foui-th centur;y
should be deemed such.
The " New Style," as it is called, was not introducea
into England until 1752, when the error had become
eleven days. The Gregorian Calendar itself is sHghtly
inaccurate.
To find if any given year be a leap year. If not the
last year of a century :
Rule. — Divide the number which represents the
given year by 4, and if there be no remainder, it is a
leap year. If there be a remainder, it expresses how
long the given year is after the preceding leap year.
Example 1. — 1840 was a leap year, because 1840 divided
by 4 leaves no remainder.
Example 2. — 1722 was the second year after a leap year,
because 1722 divided by 4 leaves 2 as remainder.
If the given year be the last of a centui-y :
^lULE. — Divide the number expressing the centuries
by 4, and if there be no remainder, the given one is a leap
year ; if there be a remainder, it indicates the number
of centuries between the given and preceding last year
of a century which was a leap year.
Example 1. — 1600 was a leap year, because 10, being
divided by 4, leaves nothing.
Ex.ajmple 2. — 1800 was two centm-ies after that last year
of a century which was a leap year, because, divided by 4,
it leaves 2. •
DIVISION OF THE CIRCLE.
Thirds
60
3600 c
216,000
77,760,000
seconda
60
3,600 or
mmutes
60
make 1 second "
1 minute '
1 decree °
degrees
1,296,000 I 21,600 or | 360 1 circumference.
Evory circle is supposed to be divided into the samo
w^/^sr of degrees, minutes, &c. ; the greater or less, there-
12 DEFINITIONS.
fore, the circle, the greater or less each of these will be. The
following will exemplify the applications of the symboU
60° 5' 4" 6'" 5 which m " '
eecondi, and six thirds.
DEFINITIONS
1. Arithmetic may be considered either as a science
or as an art. As a science, it teaches the properties of
numbers ; as an art, it enables us to apply this know-
ledge to practical purposes ; the former may be called
theoretical, the latter practical arithmetic.
- 2. A Unit, or as it is also called. Unity, is one of tho
individuals under consideration, and may include many
units of another kind or denomination ; thus a unit of
the order called " tens" consists of ten simple units. Or
it may consist of one or more parts of a unit of a higher
denomination ; thus five units of the order of " tens" are
five parts of one of the denomination called " hundreds ;"
three units of the denomination called " tenths" are
three parts of a unit, which we shall presently term 'the
" unit of comparison."
3. Number is constituted of two or more units ;
strictly speaking, therefore, unity itself cannot be con-
sidered as a number.
4. Abstract Numbers are those the properties of
which are contemplated without reference to their appli-
cation to any particular purpose — as five, seven, &c. ;
abstraction beiij^ a process of the mind, by which it sepa-
rately considers those qualities which cannot in reality
exist by themselves ; thus, for example, when we attend
only to the length of anything, we are said to abstract
from its breadth, thickness, colour, &c., although these
are necessarily found associated with it. There is nothing
inaccurate in this abstraction, since, although length
cannot exist without breadth, thickness, &c., it has pro-
perties independent of them. In the same way, five, seven,
&c., can be considered only by an abstraction of the
mind, as not applied to indicate some particular things.
5. Applicate Numbers are exactly the reverse of
D^,Fl^•ITig^'s. 13
abstract, being applied to indicate particular objects —
as five men, six houses.
6. The Unit of Comparison. In every number
tbere is some unit or individual which is used as a
standard : this we shall henceforward call -the " unit
of comparison." It is by no means necessary that it
should always be the same ; for at one time we may
speak of four objects of one species, at another of four
objects of another species, at a third, of four dozen, or
four scores of objects ; in all these cases four is tho
number contemplated, though in each of them the idea
conveyed to the mind is different — this difference arising
from the different standard of comparison, or unity
assumed. In the first case, the " unit of comparison"
was a single object ; in the second, it was also a single
object, but not of the same kind ; in the third, it became
a dozen ; and in the fourth, a score of objects. Increas-
ing the " unit of comparison" evidently increases the
quantity indicated by a given number ; while decreas-
ing it has a contrary effect. It will be necessary to
bear all this carefully in mind.
7. Odd Numhirs. One, and every succeeding alter-
nate number, are termed odd ; thus, three, five, seven, &c.
8. Even Numbers. Two, and every succeeding alter-
nate number, are said to be even ; thus, four, sLx, eight,
&c. It is scarcely necessary to remark, that after tak:ing
away the odd numbers, all those which remain are even,
and after taking away the even, all those which remain
are odd.
We shall introduce many other definitions when treat-
ing of those matters to which they delate. A clear
idea of what is proposed for consideration is of the
greatest importance ; this must be derived from the
definition by which it is explained.
Since nothing assists both the understanding and tho
memory more than accurately dividing the subject of
instruction, we shall take this opportunity of remarking
to both teacher and pupil, that we attach much impor-
tance to the divisions which in future shall actually be
mad/^, or shall be implied by the order in which tho
different heads will be examined.
b2
14
SECTION I.
ON NOTATION AND NUMERATION.
1. To avail ourselves of the properties of numbers,
R-e must be able both to form an idea of them om-selves,
and to convey this idea to others by spoken and by written
language ; — that is, by the voice, and by characters.
The expression of number by characters, is called
notation^ the reading of these, nuimratian. Notation,
therefore, and numeration, bear the same relation to
each other as writing and readings and though often
confounded, they are in reality perfectly distinct.
2. It is obvious that, for the pm-poses of Arithmetic,
we require the power of designating all possible num-
bers ; it is equally obvious that we cannot give a dif-
ferent name or character to each, as their variety Ls
boundless. We must, therefore, by some means or
another, make a limited system of words and signs
suffice to express an unlimited amount of numerical
quantities : — ^with what beautiful simplicity and clear-
ness this is effected, we shall better understand presently.
3. Two modes of attaining such an object present
themselves ; the one, that of comhining words or cha-
racters already in use, to indicate new quantities ; the
other, that of representing a variety of different quan-
tities by a single word or character, the danger of
mistake at the same time being prevented. The Romans
simplified then' «ystem of notation by adopting the prin-
ciple of comhination ; but the still greater perfection of
ours is due also to the expression of many numbers by
the same character.
4. It will be useful, and not at all difficult, to explain
to the pupil the mode by which, as we may suppose, an
idea of considerable numbers was originally acquned,
and of which, mdeed, although unconsciously, we still
avail om'selves ; we shall see, at the same time, how
methods of simplifying both numeration and notation
were naturally suggested.
NOTATION A-ND NUMERATION. 15
1^' -as suppose no system of numbers to be as yei con-
etructed, iiud that a lieap, foi' example, of pebbles, Ls
placed before us. that we may discover their amount,
[f tiiis is considerable, we cannot ascertain it by look-
ing at them all together, nor even by separat-cly in-
specting them ; we must, therefore, have recourse to
that contrivance which the mind alwa^'s uses when it
desires to grasp what, taken as a whole, is too great for
its powers. If we examine an extensive landscape, as
the eye cannot take it all in at on^ view, we look suc-
cessively at its different portions, and form our judg-
ment upon them in detail. We must act similarly with
reference to large numbers ; since we cannot compre-
hend "them at a sinijle ijlance, we must divide them into
a sufficient number of parts, and, examining these in
succession, acquire an indirect, but accurate idea of
the entue. This process becomes by habit so rapid,
that it seems, if carelessly observed, but one act, though
it is made up of many : it is indispen.sable, whenever we
desire to have a dea-r idea of numbers — which is not,
however, every time they are mentioned.
5. Had we, then, to form for ourselves a numerical
systeni, we would naturally divide the individuals to l^e
reckoned into equal groups, each group consisting of
Bnme number quite within the limit of our comprehen-
sion ; if the groups were few, our object would be attained
without any further effort, since we should have acquired
an accurate knowledge of the number of groups, and of
the number of individuals in each group, and therefore
a satisfactory, although indirect estimate of the whole.
We ought to remark, that different 'j^ersons have
very different limits to their perfect comprehension of
number ; the intelligent can conceive with ease a com-
paratively large one ; there are savages so rude as to be
incapable of forming an idea of one that is extremely
email.
6. Let us call the number of indi-\*iduais that we choose
to constitute a group, the ratio ; it is evident that the
larger the ratio, the smaller the number of groups, anc
the smaller the ratio, the larger the number of gi'oupa -
but the smaller the number of groups the bc-\ter.
16 NOTATION AND NUMERATION.
7. If the groups into Tvliich wg have divided the
objects to be reckoned exceed in amount that number
of which we have a perfect idea, we must continue the
process, and considering the groups themselves as indi-
viduals, must form with them new groups of a higher
order. We must thus proceed until the number of our
highest group is sufficiently small.
8. The ratio used for groups of the second and higher
orders, would naturally, but not necessarily, be the same
as that adopted for the lowest ; that is, if seven indi-
viduals constitute a group of the first order, we would
probably make seven groups of the first order constitute
a group of the second also ; and so on.
y. It might, and very likely would happen, that wo
should not have so many objects as would exactly form
a certain* number of groups of the highest order —
Bome of the next lower might be left. The same might
occur in forming one or raoi'e of the other groups. We
might, for example, in reckoning a heap of pebbles,
have two grouj^s of the fourth order, three of the third,
none of the second, five of the first, and seven indi-
viduals or "units of comparison."
10. If we had made each of the first order of groups
consist of ten pebbles, and each of the second order
consist of ten of the first, each group of the third of ten
of the second, and so on with the rest, we had selected
the dediiuil system, or that which is not only used at
present, but which was adopted b}^ the Hebrews, Greeks,
Romans, &c. It is remarkable that the language of
every civilized nation gives names to the diiierent
groups of this, but not to those of any other numerical
i5ystem ; its very general diffusion, even among rude
and barbarous people, has most probably arisen from
the habit of counting on the fingers, which is not
altogether abandoned, even b}' us.
11. It was not indispensable that we should have
ased the same ratio for the groups of all the diilereut
orders ; we might, for example, have made four pebbles
form a group of the first order, twelve groups of the
first order a group of the second, and twenty groups
of the second a group of the third order : — in such a
NOTATION AND NinlERATIOX. 17
caso we had adopted a system exact Ij like tliat to be
iound in the table of money (page 3), in which four
farthings make a group of the order jieTKe^ twelve
pence a gwup of tlie order s/iU!i?igs^ twenty shillings a
group of the order pounds. While it must be admitted
that the use of the same system for applicate, as for
abstract numbers, would greatly simplify our arithmetical
processes — as will be very evident hereafter, a glance
at the tables given already, and those set down in treat-
ing- of exchange, will show that a great variety of systems
have actually been constructed.
12. When we use the same ratio for the groups of all
the orders, we term it a common ratio. There appeai-s to
have been no particular reason why ten should have been
selected {is a '^ common ratio" in the system of numbers
ordinarily used, except that it was suggested, as already
remarked, by the mode of counting on the fingers ; and
that it is neither so low as unnecessarily to increa.se
the number of orders of groups, nor so hvih. as to exceed
tlie conception of any one for whom the system was
intended.
13. A sy.stem in which ten is the " common ratio"
is called decimal^ from " decem," which in Latin signifies
ton : — ours is, therefore, a " decimal .^^ystera" of num])ers.
If the common ratio were sixty, it would be a sexoged-
vw.l system ; such a one was formerly used, and is still
retained — ^as will be perceived by the tables already
given for the measurement of arcs and angles, and of
lime. A quinary system w6uld have five for its " com-
rnon ratio ;" a diwdecimal^ twelve ; a vigesimal^ twenty,
14. A little reflection will show that it was useless
to give different names and characters to any numbers
except to those whicji are le.ss than that which consti-
tutes the lowest group, and to the different orders of
groups ; because all pcssible numbers must consist of
individuals, or of groups, or of both . individuals and
groups : — in nci'jier case would it be refjuircd to specify
more than the number of individuals, and the numbei
of each species of group, none of which numbers — as is
evident — can be greater thiin the coaiinon ratio. Thij»
Nnmci.
Cluira^ttit.
fOiie
1
Two
2
Three .
3
Four
4
Five
5
Six .
6
Seen .
7
F-i.^bt .
8
Nit.e
0
Ten
10
Hundred
100
Thousand
1,000
Ten thonsand
10,000
Hundred thousand lOO.COn
Million .
1,000,000
18 NOTATION AND NUMERATION.
is just what we Lave clone in our nuraerical Rystem,
except that we have formed the names of some of the
groups by combination of those ahcady used ; thus,
" tens of thousands," the group next higher than thou-
sands, is designated by a combination of words already
applied to express other groups — which tends yet further
to simplification.
15, ARABIC SYSTEM OF NOTATION!
Units of domparison,
First group, or units of the second order,
Second group, or units of the third orde-r,
Third group, or units of the fourth order,
Fourth group, or units of the fifth order,
Fifth group, or units of the sixth order.
Sixth group, or units of the seventh order,
16. The characters which express the nine first num-
bers are the only ones used ; they are called digits^ from
the custom of counting them on the fingers, already
noticed — " digitus" meaning in Latin a finger ; thoy are
also called significant figures^ to distinguish them from
the cypher, or 0, which is used merely to give the digits
their proper position with reference to the decimal point.
The pupil will distinctly remember that the place where
the "- units of comparison" are to be found is that imme-
diately to the left hand of this point, which, if not ex-
pressed, is supposed to stand to the right hand side of
all the digits — thus, in 468-76 the 8 expresses " units
of comparison," being to the left of the decimal point ;
in 40 the 9 expresses " units of comparison," the doci-
inal point being understood to the right of it.
17. We find by the tabic just given, that after tlie
nine first numbers, the same digit is constantly repeated,
its position with reference to the decikial point being,
however, changed : — that is. to hidicate each succeeding
group it is moved, Ijy means of a cypher, one place
farther to the loft. Any of the digits miiy be used to
NOTATION AND NUMERATION 19
express its respective number of any of the groups : — •
tlius 8 would be eight '' units of comparison ;" 80,
eight groups of the first order, or eight " tens" of
simple units ; 800, eight groups of the second, .or unita
of the third order ; and so on, AVe might use any of
the digits with the different groups ; thus, for example,
6 for groups of the third order, 3 for those of the second,
7 for those of the first, and 8 for the " units of compari-
son ;" then the whole set down in full would be 5000,
300, 70, 8, or for brevity sake, 5378 — for we never uso
the cypher when we can supply its place by a significant
figure, and it is evident that in 5378 the 378 keeps the
5 four places from the decimal point (understood), just as
well as cyphers would have done ; also the 78 keeps the
3 in the third, and the 8 keeps the 7 in the second place,
18. It is important to remember that each digit has
two values, an absolute and a relative ; the absolute
value is the number of units it expresses, whatever these
units may be, and is unchangeable ; thus 6 always
means six, sometimes, indeed, six tens, at other times
six hundred, &c. The relative value depends on the
orcUr of units indicated, and on the nature of the " unit
of comparison."
19. What has been said on this very important sub-
ject, is intended principally for the teacher, though an
ordinary amount of industry and intelligence will be
quite suuicient for the purpose of explaining it, even to
a child, particularly if each jpoint is illustrated by an
appropriate example ; the pupil may be made, for in-
stance, to arrange a number of pebbles in groups, some-
times of one, sometimes of another, and sometimes of
several orders, and then be desired to express them by
figures — the " unit of comparison" being occasionally
changed from individuals, suppose to tens, or hundreds, or
to scores, or dozens, &c. Indeed the pupils must be well
acquainted with these introductory matters, otherwise
they will contract the habit of answering without any
very definite ideas of many things they will be called
upon to explain, and which they should be expected
perfectly to understand. Any trouble bestowed by the
teacher at thi^ period will be well repaid by the ease
20 NOTATION AND NUMLIRATIU.V.
and rapidity with wiiieli tlic scliolar will afterwards
ads'anco ; to be assured of this, he has only to recol-
lect that most of his future reasonings will be derived
from, and his explanations grounded on the very prin-
ciples we have endeavoured to unfold. It may be taken
as an important truth, that what a child learns without
understanding, he will acquire with disgust, and will
5JOon cease to remember ; for it is with childi'en as with
persons of more advanced years, when we appeal suc-
cessfully to their understanding, the pride and pleasure
they feel in the attainment of knowledge, cause the
labour and the weariness which it costs to be under-
valued, or forgotten.
20. Pebbles will answer well for examples ; indeed,
their use in computing has given rise to the t^rm ailcio-
lation., '' calculus" being, in Latin, a pebble : but while
the teacher illustrates what he says by groups of par-
ticular objects, he must take care to notice that hi^
remarks would be equally true of any others. He must
also point out the difference between a group and its
equivalent unit, which, from their perfect eciuality, are
generally confounded. Thus he may show, that a penq^,
while equal to, is not identical vrith four farthings. Thi.s
seemingly unimportant remark will be better appre-
ciated hereafter ; at the same time, without inaccuracy
of result, we may, if we please, consider any group
either as a unit of the order to which it belongs, or so
many of the next lower as are equivalent.
21. lloman Notation. — Our ordinary numerical cha-
racters have not been always, nor every where used tc
express numbers ; the letters of the alphabet naturally
presented themselves for the purpose, as being abeady
familiar, and, accordingly, were very generally adopted —
for example, by the Hebrews, Greeks, Ilomans, &e.,
each, of course, iLsing their own alphabet. The pupil
should be acquainted with the Roman notation on
account of its beautiful simplicity, and its being still
employed in inscriptions, &;c. : it is found in the foUow-
inor table : —
NOTATION AND NUMERATION.
21
ROMAN NOTATION,
Anticipated cliam
Cliange
Cfiavaclcrs.
I.
ir.
III.
Anticipated change IIII. or IV.
(Jhange . .V.
Yl. .
VII. .
VIII. .
IX.
X.
XI.
XII. .
XIII. .
XIV. .
XV. .
XVI. .
XVII. .
XVIII
XIX. .
XX. .
XXX., &c
Anticipated change XL.
Change . . L.
LX., &c.
Anticipated cliaage XO.
Change . . C.
CC, &c.
Anticipated change CD.
Change . . D. or I3.
Anticipated change CM.
Change . . M. or CIq.
V. or I,),-).
X. or CCIoo
Nuinhers Exprrtsed.
, One.
. Two.
. Three
. Four.
. Five.
. Six.
. Seven.
. Eight.
. Nine.
. Ten.
. Eleven.
. Twelve.
. Thirteen.
. Fourteen.
. Fifteen.
. Sixteen.
. Seventeen.
. Eighteen.
. Nineteen.
. Twenty.
, Thirty, &c.
. Forty.
. Fifty.
. Sixty, &c.
. Ninety.
. One liundred.
. Two hundred, e'^c
. Four hundred.
. Five hundred, ,S:c
. Nine hundred.
■. One thousand, ko,.
. Five thousand, &c.
. Ten thousand, &c.
. Fifty thousand, &o.
. One hundred thousand, ka
CCCIooo.
22. Thus Tve fiad that the liomaas used very few
characters — fewer, indeed, than we do, although our
system is still more simple and eiTeclive, from our apply-
ing-the principle of " position," unkuovf^n to them.
They expressed all numhcrs by the following symLols,
or combinations of them : I. V. X. L. C. D. or Iq. M.,
or CL3. In constructing their system, they evidently
had a quinary in view ; that is, as we have said, one in
which five would be tlie c/)m?non ratio ; for we find that
they changed their character, not only at ten, ten times
ApfO NUMERATION.
ten, &c., but also at five, ten times five, &c. : — a purely
•.leciinal system would suggest a change only at ten, ten
times ten, &c. ; a pm-ely quinary, only at five, five times
five, &c. As far as notation was concerned, what they
adopted was neither a decimal nor a quinary system,
nor even a combination of both ; they appear to have
supposed two primary groups, one of five, the other of
ten " units of comparison ; " and to have formed all the
other groups from these, by using ten as the common
ratio of each resulting series.
23. They anticipated a change of character ; one
unit before it would natui-ally occur — that is, not one
" unit of comparison," but one of the units under consi-
deration. In this point of view, four is one unit before
five; forty, one unit before fifty — tens being now the
units under consideration ; foui* hundred, one unit before
five hundi-ed — hundi-eds having become the units con-
templated.
24. When a lower character is placed before a
higher its value is to be subtracted from, when placed
after it, to be added to the value of the liigher ; thus,
IV. means one less than five, or four ; VI., one more
than five, or six.
25. To express a number by the Roman method of
notation : —
iR-ULE. — Find the highest number within the .given
one, that is expressed by a single character, or the
" anticipation " of one [21] ; set down that character,
or anticipation — as the case may be, and take its value
from the given number. Find what highest number
less than the remainder is expressed by a single charac-
ter, or " anticipation ; " put that character or " anticipa-
tion " to the right hand of what is already wi'itten, and
take its value from the last remainder : proceed thus
until nothin<]j is left.
Example. — Set down the present year, eighteen h'mdred
and forty-four, in Roman characters. One thousap.d, ex-
pressed by M., is the highest number within the givnn one,
indicated by one character, or by an anticipation ; we pu fc do^vn
M,
and take one thousand from the given number, which loavea
NOTATIO!* AND NUMERATION. 23
eight hundred and forty-four. Five hundr« n thG
highest number within tho last remainder (e tji* f<«iidr<jd
and furty-fom-) expressed Lj one character, or an " antici-
pation :" Tve set down D to the right hand of M,
3ID,
and take its value from eight hundred and forty-four, which
leaves three hundred and forty-four. In this the highest
number expressed by a single character, or an " anticipa-
tion," is one hundred, indicated by C ; which we set down j
and for the same reason two other Cs.
]MDCCC.
This leaves only forty-four, the highest number withic
which, expressed by a single character, or an -anticipation,''
is forty, XL — an anticipation ; we set this down also,
MDCCCXL.
Four, expressed by IV.. still remains; which, being al8«
added, the whole is as follows : —
JSIDCCCXLIV.
26. Fosilian. — The same ^jharacter may have dif-
ferent values, according to ^e flacz it holds with refer-
ence to the decimal point, or, perhaps, more strictly,
to the "unit of comparison." This is the principle of
'positirm.
27. The places occupied by the units of the different
orders, according to the Arabic, or ordinary notation
[15] , may be described as follows : — units of comparison,
one place to the left of the decimal point, expressed^
or understood ; tens, two places ; hundreds, three places,
&c. The pupil should be made so familiar with these,
as to be able, at once, to name the " place" of any order
of units, or the " units" of any place,
28. When, therefore, we are desked to write any
number, we have merely to put down the digits expres-
sing the amounts of the different units in then* proper
places, according to the order to which each belongs.
If, in the given number, there is any order of which
there are no units to be expressed, a cypher must be set
down in the place belonging to it ; the object of which
is, to keep the significant figures in their own 'posi-
tions. A C3^pher produces no effect when it is not
between significant figures and the decimal point ; thus
0536, 536-0, and 536 would mean the same thing — \\\q.
24 NOTATION AND NUMERATION.
second is, however, the correct form. 536 and 5360 dre
diiierent ; in the latter case the cypher aflfects the value,
because it alters the position of the digits.
Example. — Let it be required to set down sis hundred
and two. The six must he in the third, and the two in the
first place ', for this purpose we- are to put a cypher between
the 6 and 2 — thus, 602 : without the cypher, the six would
be in the second place — thus, 62 ; and would mean not six
hundreds, but six tens.
29. In numerating, we begin with the digits of the
highest order and proceed downwards, stating the num-
ber which belongs to each order.
To facilitate notation and numeration, it is usual to
divide the places occupied by the different orders of
units into periods ; for a certain distance the English and
French methods of division agree ; the English billion
is, however, a thousand times greater than the French.
This discrepancy is not of much importance, since we
are rarely obliged to use so high a number, — ^we shall
prefer the French method. To give some idea of the
amount of a billion, it is only necessary to remark, that
according to the English method of notation, there
has not been one billion of seconds since the bu-th of
Christ. Indeed, to reckon even a million, counting on
an average three per second for eight houi-s a day,
would require nearly 12 days. The following are the
two methods.
ENGLISH METHOD.
Trillions. Billions. Millions. Units.
000-000 000-000 COO -000 000-000
FRENCH METHOD.
Billions. Millions. Thousands. Units.
auiiclietls. 'I'eiis. Unita. Hunil. Tens. Units. Hiind. Tens. Units. Hnnd. Trus. Unit*.
0 00 000 000 000
30. Use of Periods. — Let it be required to read off
llie follovring number, 576934. We put the first point
to the left of the hundreds' place, and find that there are
exactly two periods — 576,934 ; this does not always
occur, as the highest period is often imperfect, consisting
only of ono or two digits. Dividing the number thus
NOTATION AND NUMERATION. 25
into parts, shows at once that 5 is in the third phice of
the second period, and of course in the sixth place to
the 4eft hand of the decimal point (understood) ; and,
therefore, that it expresses hundreds of thousands. The
7 being in the fifth place, indicates tens of thousands ;
the 6 in the fourth, thousands ; the 9 in the third, hun-
dreds ; the 3 in the second, tens ; and the 4 in the first,
units (of" comparison"). The whole, therefore, is five
hundreds of thousands, seven tens of thousands, six
thousands, nine hundreds, three tens, and fom- units, — •
or more briefly, five hundred and seventy-six thousand,
nine hundred and thirty-four,
31. To prevent the separating point, or that which
divides into periods, from being mistaken for the decimal
point, the former should be a comma (,) — the latter a
full stop (•) Without this distinction, two numbers
which are very different might be confounded : thus,
498-763 and 498,763, — one of which is a thousand
times greater than the other. After a while, we may
dispense with the separating point, though it is conve-
nient to use it with considerable numbers, as they are
then read with greater ease. "•
32. It will facilitate the reading of large numbers
not st^parated into periods, if we begin with the units of
comparison, and proceed onwards to the left, saymg at
the first digit " units," at the second "• tens," at the
third ''hundreds," &c., marking in our mind the deno-
mination of the highest digit, or that at which we stop.
We then commence with the highest, express its number
and denomination, and proceed in the same way with
each , until we come to the la.st to the risiht hand.
Rv'AAiPLE. — Let it be required to read off 6402, Looking
at the 2 (or pointing to it), we say ••units:" at the 0, -tens;"'
At the 4. '"hundreds;"' and at the 6, ••thousands."'^ The
latter, therefore, being six thousands, the next digit is four
iiundreds, kc. Consequently, six thousands, four hundreds,
no tens, and two units : or, briefly, six thousand four hun-
dred and two, is the reading of the given number,
33. Periods may be used to facilitate notation. The
pupil will fixst wiit^ down a number of periods of cyphers
26 NOTATION AND NUMERATION.
to represent tbe places to be occupied by tlie various
orders of units. He will then put the digits express-
ing the different denominations of the given number,
under, or instead of those cyphers which are in corres-
ponding positions, with reference to the decimal point — ■
beginning with the highest.
Example. — Write down three thousand six hundred and
fifty-four. The highest denomination being thousands, will
occupy the fourth place to the left of the decimal point. It
will be enough, therefore, to put down four cyphers, and
under them the corresponding digits — that expressing the
thousands under the foiu-th cypher, the hundreds under the
third, the tens under the second, and the units under the
first; thus
0,000
3,654
A cypher is to be placed under any deaomination in
which there is no significant figure.
Example. — Set down five hundred and seven thousand,
and sixty-three.
000,000
a07,063
After a little practice the periods of cyphers will
become unnecessary, and the number may be rapidly
put down at once.
34. The units of comparison are, as we have said,
always found in the first place to the left of the
decimal point ; the digits to the left hand progressively
increase in a tenfold degree — those occupying the first
place to the left of the units of comparison being ten
times greater than the units of comparison ; those occu-
pying the second place, ten times greater than those
which occupy the first, and one hundred times gi-eater
than the units of comparison themselves ; and so on.
jMoving a digit one place to the left multiplies it by
ten, that is, makes it ten times greater ; moving it two
places multiplies it by one hundred, or makes it one
hundred times greater ; and sc of the rest. If all the
digits of a quantity be moved one, two, &c., places to
the left, the whole is increased ten, one hundred, &c.,
times — as the case may be. On the other hand, moving
NOTATION AND NUMERATION. 27
a digit, or a quantity one place to the right, divides it,
by ten, that is, makes it ten timr^s smaller than before ;
moving it two places, divides it by one himdi-ed, or
makes it one hundred times smaller, &;c.
35. We possess this power of easily increasing, or
diminishing any number in a tenfold, &c. degree, whether
the digits are all at the right, or all at the left of the
decimal point ; or partly at the right, and partly at the
left. Though we have not hitherto considered quantitiea
to the left of the decimal point, their relative value will be
very easily understood from what we have already said.
For the pupil is now aware that in the decimal system
the quantities increase in a tenfold degree to the left,
and decrease in the same degree to the right ; but
there is nothing to prevent this decrease to the right
from proceeding beyond the units of comparison, and
the decunal point ; — on the contrary, from the very
natm-e of notation, we ought to put quantities ten times
kss than units of comparison one place to the right of
them, just as we put those which are ten times less tliaa
hundreds, &c., one place to the right of hundreds, &:c
AVe accordingly do this, and so continue the notation
not only upwards, but downwards, calling quantities te
the left of the decimal point integers, because none of
them is less than a tchole " unit of comparison ;" an(*
those to the right of it decimals. When there are deci-
mals in a given number, the decimal poiat is actually
expressed, and is always found at the right hand side-
of Ihe units of comparison.
3G. The quantities equally distant from the unit of
comparison bear a very close relation to each other
v>'hich is indicated even by the similarity of then* names ;
those which are one place to the left of the units of com-
parison are called " tens," being each identical with, or
equivalent to ten units of comparison ; those which are
one place to the right of the units of comparison an3
called " tenths," each being the tenth part of, that is, ten
times as small as a unit of comparison ; quantities two
places to the left of the units of comparison are called
" hundreds," being one hundi-ed times greater ; and
those two places to the rig/itj " huncbedths," being one
2S NOTATION AND NUMERATION
hundred times less than the units of coniparison ; and so
of all the others to the right and left. This will be more
evident on in?pccting the following table : — •
Asccn'.ling Scries, or Integers. I Desccndin? Series, or Decimals.
One Unit . . 1;1' One Unit.
Ten . . 101 -1 Tenth.
Hundred . . 100 j -01 Hundredth.
Thousand . . 1.0001-001, Thousandth.
Ten thousand . 10,000| -000.1 Ten-thousandth.
Hundred thousand 100,0001 -000,01 Hundred- thousandth.
&c. 1 &c.
We have seen that when we divide integers into periods
[29] , the first separating point must be put to the right
of the thousands ; in dividing decimals, the first point
must be put to the right of the thousandths.
37. Care must be^ taken not to confound what we
now call " decimals," with what we shall hereafter design
nate " decimal fractions ;" for they express equal, but
not identically the same quantities — the decimals being
what shall be termed the " quotients" of the corres-
ponding decimal fractions. This remark is made here to
anticipate any inaccurate idea on the subject, la those
who already know something of Arithmetic.
38. There is no reason for treating integers and deci-
mals by different rules, and at different times, since they
follow precisely the same laws, and constitute parts of
the very same series of numbers, Besides, any quantity
may, as far as the decimal point is concerned, be ex-
pressed in different ways ; for this purpose we have
merely to change the unit of comparison. Thus, let it
be required to set down a number indicating five hun-
dred and seventy-four men. If the " unit of compari-
son" be oiie man^ the quantity would stand as follows,
574. If a band of ten men, it would become 57*4 — foi*,
as each man would then constitute only tlie tenth part
of the " unit of comparison," four men would be only
four-tenths, or 0*4 ; and, since ten men would form but
one unit, seventy men would be merely seven units of
comparison, or 7 ; &c. Again, if it wore a band of one
hundred ??ic?i, the number must be Avrittcn 5'74 ; and
lastly, if a band of a thmvsand niaij it would be 0*574
NOTATION AND NUMERATION. 29
Should the " unit " be a band of a dozen, or a score
men, the change would be still more complicated ; as,
not only the position of the decimal point, but the very
digits also, would be altered.
39. It is not necessary to remark, that moving the
decimal point so many places to the left, or the digits
an equal number of plaees to the right, amount to the
same thing.
Sometimes, in changing the decimal point, one or
more cyphers are to be added ; thus, when we move 42' G
three places to the left, it becomes 42600 ; when wo
move 27 five places to the right, it is '00027, &c.
40. It follows, from what we have said, that a deci-
mal, though less than what constitutes the unit of com-
parison, may itself consist of not only one, but several
individuals. Of course it will often be necessary to indi-
cate the " unit of comparison," — as 3 scores, 5 dozen, 6
men, 7 companies, 8 regiments, &c. ; but its nature does
not affect the abstract properties of numbers ; for it is
true to say that seven and five, when added together,
make twelve, whatever the unit of comparison may be : —
provided, however, that the sainc standard be applied to
both ; thus 7 men and 5 men are 12 men ; but 7 men
and 5 horses are neither 12 men nor 12 horses,; 7 men
and 5 dozen men are neither 12 men nor 12 dozen men.
When, therefore, numbers are compared, &c., they must
have the same unit of comparison, or — without altering
their value — they must be reduced to those which have.
Thus we may consider 5 tens of men to become 50
indiiidual men — the unit of comparison being altered
from ten men to one man^ without the value of the
quantity being changed. This principle must be kept
in mind from the very commencement, but its utility
will become more obvious hereafter.
EXAMPLES IN NUMERATION AND NOTATION.
jV^otation.
1. Put down one hundred and four . . 104
2. One tliousand two hundred and forty . 1,240
3. Twenty tliousand, three hundred and forty-five 20,345
c
30 NOTATION AND NUMERATION.
4. Two hundred and tliirty-four thousand, fivo '^^.i.
hundred and sixty-seven . . , 234,507
5. Three hundred and tAventv-nine thousand,
seven hundred and seventy-nine . . 329,779
6. Seven hundred and nine thousand, eight hun-
dred and twelve .... 709,812
7. Twelve hundred and forty-seven thousand,
four hundi-ed and fifty-seven . . 1,247,457
8. One million, three hundred and ninety-seven
thousand, four hundred and seventy-live . 1,397,475
9. Put down fifty-four, seven-tenths . . 54-7
10. Ninety-one, five hundredths . . . 9105
11. Two, three-tenths, four thousandths, and four
hundred-thousandths . .. . 2*30404
12. Nine tliousandths, and three hundred thou-
sandths ..... 000903
13. Make 437 ten thousand times greater . 4,370,000
14. Make 27 one hundred times greater . 270
15. Make 0056 ten times greater . . 056
16. Make 430 ten times less ... 43
17. Make 2-75 one thousand times los^i . . 000275
Numeracion
1. read 132
2. — 407
3. — 2760
4. — 5060
5. — 37654
0. — 8700002
7. read 8540326
8. — 5210007
9. _ 6030405
10. _ 560075
11. — 3000006
12. — 0- 0040007
13. Sound travels at the rate of ahout 1142 feet in a
second : light moves ahout 195.000 miles in the same time.
14. The sun is estimated to he 886,149 miles in diameter;
its size is 1,377,613 times greater than that of the earth.
15. The diameter of the planet mercury is 3,108 miles,
and his distance from the sun 36,814,721 miles.
16. The diameter of Venus is 7,498 miles, and her dis-
tance from the sun 68,791,752 miles.
17. The diam.etcr of the earth is ahout 7,964 miles: it is
95,000,000 miles from the sun. and travels round the latter
at the rate of upwards of 68.000 miles an hour.
18. The diameter of the moon is 2,144 miles, and her dis-
tance from tlic earth 236.847 miles.
19. The diameter- of Mars is 4,218 miles, and his distance
from the sun 144.907,630 miles.
20. The diameter of Jupiter is 89,069 miles, and his dis-
tance from the sun 494,499,108 miles.
NOTATION AND NUMERATION. 31
21. The diameter of Saturn is 78,730 miles, and his dis-
tance from the siin 907,089, 032 miles.
2'2. Tlie length of a pendulum which would vibrate
seconds at the equator, is 39011,084 inches; in the latitude
of 45 degrees, it is 39-116.820 inches; and in the latitude of
90 degrees, 39-221,950 inches.
23. It has been calculated that the distance from the
earth to the nearest fixed star is 40.000 times the diameter
of the earth's orbit, or annual path in the heavens ; that is,
about 7,600,000,000,000 miles. Now suppose a cannon
ball to fly from the eai-th to this star, with a uniform velocity
equal to that with which it first leaves the mouth of the
gun — say 2,500 feet in a second — it would take nearly
1,000 years to reach its destination.
24. A piece of gold equal in bulk to an ounce of water,
would weigh 19-258 ounces ; a piece of iron of exactly the
same size, 7-788 ounces: of copper, 8- 788 ounces; of lead,
11-352 ounces; and of silver, 10-474 ounces.
Note. — The examples in notation Diay be made to answer
for numeration ; and the reverse.
QUESTIONS IN NOTATION AND NUMERATION.
[The references at the end of the questions show in -what
paragraphs of the preceding section the respective answers
are principally to be found.]
1. What is notation .^ [1].
2. What is numeration ? [1].
3. How are we able to express an infinite variety of
numbers by a few names and characters .' [3] .
4. How may we suppose ideas of numbers to have
been originally acquired .' [4, &c.].
5. What is meant by the commo-n ratio of a system
of numbers.^ [12].
6. Is any particular number better adapted than
another for the common ratio t [12].
7. Are there systems of numbers without a common
ratio.? [11].
S. What is meant by quinary, decimal, duodecimal,
vigesimal, and sexagesimal systems t [13].
9. Explain the Arabic system of notation .? [15].
10. What are digits.? [16].
11. How are they made to express all numbers } [17] .
32 NOTATION AND NUMERATION.
12. What is meant by their absolute and rclativo
/alues? [18].
13. Are a digit of a higher, and the equivalent num-
-^er of units of a lower order precisely the same thing .'
c20].
14. Have the characters we use, always and every
♦7here been employed to express numbers r [21].
15. Explain the Eoman method of notation.? [22, &c.].
16. ^Yhat is the decimal point, aad the position of
the different orders of units with reference to it ? [26
and 27] .
17. When and how do cyphers affect significant
figures.? [28].
18. What is the difference between the English and
French methods of dividing numbers into periods } [29] .
19. What is the difference between integers and
decimals r [35] .
20. "Wliat is meant by the ascending and descending
series of numbers ; and how are they related to each
other.? [36].
21. Show that in expressing the same quantity, we
must place the decimal point differently, according to
the unit of comparison we adopt .? [38] .
22. What effect is produced on a digit, or a quantity
by removing it a nmnber of places to the right, or left,
or similarly removing the decimal point ? [34 and 39]
33
SECTION II.
THE SIMPLE RULES. ' ■
SIMPLE ADDITION.
1. If numbers are changed by any arithmetical pro-
cess, they are either increased or diminished ; if in-
creased, the effect belongs to Addition; if diminished,
to Subtraction. Hence all the rules of Arithmetic are
ultimately resolvable into either of these, or combina-
tions of both.
2. When any number of quantities, either different^
or repetitions of the same, are united together so as
to form but one, we term the process, simply, " Addi-
tion." When the quantities to be added are the savm^
but we may have as many of them as we please, it is
called " Multiplication ;" when they are not only the
same., but their number is indicated by one of them., the
process belongs to "Involution." That is, addition re-
stricts us neither as to the kind, nor the number of the
quantities to be added ; multiplication restricts us as to
the kind, but not the number ; involution restricts us
both as to the kind and number : — all, however, are
really comprehended under the same rule — addition.
3. Simple Addition is the addition of abstract num-
bers ; or of applicate numbers, containing but one deno-
mination.
The quantities to be added are called the addeiids ;
the result of the addition is termed the sum.
4. The process of addition is expressed by + , called
the plus, or positive sign; thus 8 + 6, read S plus 6,
means, that 6 is to be added to 8. When no sign ia
prefixed, the positive is understood.
The equality of two quantities Is iDdicated by =,
thus 9-h7=:16, means that the sum of 9 and 7 is equal
to 16.
34 ADDITION.
Quantities connected bj the sign of addition, or that of
equality, may be read in any order ; thus if 74-9=16, it
is true, also, that 9 + 7=16, and that 16=7+9, or 9+7.
5. Sometimes a single horizontal line, called a vhi-
ciihiju, from the Latm word signifying a bond or tie,
is placed over several numbers ; and shows that all the
quantities under it are to be considered, and treated as
^ut one ; thus in 4 + 7=11, 4 + 7 is supposed to form
but a single term. However, a vinculum is of little
consequence in addition, since putting it over, or remov-
ing it from an additive quantity — that is, one which has
the sign of addition prefixed, or understood — does not
in any way alter its value. Sometimes a parenthesis () is
used in place of the vinculum; thus 5 + 6 and {d + 6)
mean the same thing.
6. The pupil should be made perfectly familiar with
these symbols, and others which we shall introduce as
we proceed ; or, so far from being, as they ought, a
great advantage, they will serve only to embarrass him.
There can be no doubt that the expression of quantities
by characters, and not by words written in full, tends
to brevity and clearness ; the same is equally true of the
processes which are to be performed — the more con-
cisely they are indicated the better.
7. Arithmetical rules are, naturally, divided into two
parts ; the one relates to the setting down of the quan-
tities, the other to the operations to be described. We
shall generally distinguish these by a line.
To add Numhers.
Rule. — I. Set down the addends under each other,
so that digits of the same order may stand in the same
vertical celumn — units, for instance, under units, tens
under tens, &c.
II. Draw a line to separate the addends from their
III. Add the units of the same denomination together,
focginninw at the right hand .'^de.
IV. a the sum of any column be less than ten, set it
down under that column ; but if it be greater, for evei7
ADDITION'. 35
ten it contains, carry one to tlie next column, and put
down only what remains after deducting the tens ; if
*^othing remains, put down a cypher.
V. Set down the sum of the last column in full.
8. Example.— Find the sum of 542+375-f-984—
542 )
375 > addends.
984 \
1901 sum.
We have placed 2, 5. and 4, which belong to* the order
''units," in one column; 4, 7, and 8, which are "tens," in
another; and 5, 3, and 9, which are '• hundreds," in another.
4 and 5 units are 9 units, and 2 are 11 units — equivalent
to one ten and one unit : we add, or as it is called, •• carr}''"
the ten to the other tons found in the next column, and set
down the unit, in the units' place of the " sum."
The pupil, having learned notation, can easily find
how many tens there are in a given number ; since all
the digits that express it, except one to the right hand
side, will indicate the number of "tens" it contains;
thus in 14 there are 1 ten, and 4 units ; in 32, 3 tens,
and 2 units ; in 143, 14 tens, and 3 units, &c.
The ten obtained from the sum of the units, along with 8,
7, and 4 tens, makes 20 tens ; this, by the method just men-
tioned, is found to consist of 2 tens (^of tens), that is, two of
tlie next denomination, or hundreds, to be carried, and no
units (of tens) to be set down. We •• carry," 2 to the hun-
dreds, and write down a cypher in the tens' place of the
'"sum."
The two hundreds to be "carried," added to 9, 3, and 5,
hundreds, make 19 hundreds ; which are equal to 1 ten (of
Imndreds) : or one of the next denomination, and 9 imits (of
hundreds) ; the former we "carry" to the tens of hundreds,
or thousands, and the latter we set down in the hvmdreds'
place of the "sum."
As there are no thousands in the next column — that is,
nothing to which we can "carry" the thousand obtained
by adding the hundreds, we put it down in the thousands'
[ilace of the "sum ;" in other words, we set down the sum of
the last column in full.
9. Reason of I. (the first part of the rule). — We put
units of the same denomination iu the same vertical column.
36 ADDITION.
that we may easily find those quantities which are to be added
together ; and that the value of each digit may be more clear
from its being of the same denomination as those which are
under, and over it.
Reason of II. — "We use the separating line to prevent the
gum from being mistaken for an addend.
PtEAsoN OF III. — We obtain a correct result only by adding
units of the same denomination together [Sec. I. 40] : — hun-
dreds, for instance, added to tens, would give neither hnndi-eda
nor tens as their sum.
We begin at the right hand side to avoid the necessity of
more than one addition ; for, beginning at the left, the process
would be as follows —
542
375
984
1,700
190
n
1,000
800
100
1
1,901
The first column to the left produces, by addition, 17 hun-
dred, or 1 thousand and 7 hundred ; the next column 19 tens,
or 1 hundred and 9 tens; and the next 11 units, or 1 ten and
1 unit. But these quantities are still to be added : — beginning
again, therefore, at the left hand side, we obtain 1000, 800, 100,
and 1, as the respective sums. These being added, give 1,901
as the total sum. Beginning at the right hand rendered the
successive additions unnecessary.
Reason of IY. — Our object is to obtain the sum, expressed
in the highest orders, since these, only, enable us to represent
any quantity with the lowest numbers ; we therefore consider
ten of one denomination as a unit of the next, and add it to
those of the next which we already have.
After taking the " tens " from the sums of the diJBferent
columns, we must set down the remainders, since they arc
parts of the entire sum; and they are to be put under the
columns that produced them, since they have not ceased to
belong to the denominations in these columns.
Reason- of V. — It follows, that the sum of the last column
is to be set down in full ; for (in the above example, for in-
stance,) there is nothing to be added to the tens (of liundreds)
it contains.
10. Proof of Addition. — Cut off the upper addend,
by a separating line ; and add tlic sum of the quantities
/ ADDITIOW 87
under, to what is above this line. If all the additions
have been correctly performed, the latter sum vdll be
equal to the result obtained by the rule : thus —
5,673
4,632
8,697
2,543
21,545 sum of all the addends.
15,872 sum of all the addends, but one.
5,673 upper addend.
21,545 same as sum to be proved.
Til is mode of proof depends on the fact that the whole is
equal to the sum of its parts, in whatever order they are
taken; but it is liable to the objection, that any error com-
mitted in the first addition, is not unlikely to be repeated in
the second, and the two errors would then conceal each other.
To prove addition, therefore, it is better to go through
tho process again, beginning at the top, and proceeding
doTvnwards. From the principle on which the last mode of
proof is founded, the result of both additions — the direct
and reversed — ought to be the same.
It should be remembered that these, and other proofs of
the same kind, afford merely a high degree of probability,
since it is not in any case quite certain, that two errors cal-
culated to conceal each other, have not been committed.
11. To add QiMiitities containing Decimals. — From
what has been said on the subject of notation (Sec. I.
35), it appears that decimals, or quantities to the right
hand side of the decimal point, are merely the continu-
ation, dopmwards, of a series of numbers, all of which
follow the same laws ; and that the decimal point is
intendf'd, not to show that there is a difference in the
nature of quantities at opposite sides of it, but to mark
whe^-e the " unit of comparison" is placed. Hence the
rule for addition, already given, appHes at whatever side
all, or any of the digits in the addends may be found
It is necen.'ary to remember that the decimal point in
the sum, should stand precisely under the decunal points
of the ?\arle.nds ; since the digits of the sum must be, from
the very nature of the process [9] , of exactly the same
value"., respectively, as the digits of the addends under
38 ADDITION.
wliicli tliey are ; and if set down as tliey should be, their
denominations are ascertained, not only by their position
with reference to their own decimal point, but also by
their position with reference to the digits of the addends
above them.
Example.
263-785
460-502
637-008
626-3
1887-595
It is not necessary to fill up the columns, by adding
cyphers to the last addend ; for it is sufficiently plain
that we are not to notice any of its digits, until we come
to the third column.
12. It follows from the nature of notation [Sec. I.
40] , that however we may alter the decimal points of
the addends — provided they are all in the same vertical
column — the digits of the sum will continue unchanged ;
fhus in the followino^ : —
4785 478-5 47-85 -4785 -004785
3257 325-7 32-57 -3257 -003257
6546 654-6 65-46 -6546 -006546
14588 1458-8 145-88 1-4588 -014588
EXERCISES.
(A
dd the following numbers.)
Addition.
(1) (2) (3)
4 8 3
6 4 9
3 7 7
6 6 6
7 2 5
Multiplication. Involution.
(4) (5) (6) (7) (8) (
6 4 9 (3 f4
6 4 9 CO ■<3 ^ J 4
6 4 9 ^3 "^^ 14 o •
6 4 9 — [4
6. 4 9 _ —
9)
5
5
5
.5
(10)
6763
2341
5279
(11)
3707
2465
5678
(12) (13) (14) (
2367 6978 6767 7
3246 3767 4579 1
1239 1236 1236 3
15)
647
239
789
AD'oliiOlf.
3
(16)
(IT)
(18)
(19)
(20)
(21)
5673
3767
3001
5147
34567
7345G
123?
4507
2783
3745
47891
4567>^
2315
1:434
4507
0789
41234
9123J
(22)
(23)
(24)
(25)
(26)
(27)
7G789
34567
78789
34076
73412
36707
40707
89123
01007
78707
70700
46770
1247G
45678
34057
45079
47076
30707
(28)
(29)
(30)
(31)
(32)
(33)
45607
76767
23456
45078
23745
87967
37076
45677
78912
91234
67891
32785
36707
76988
34507
50789
23456
64127
(31)
(35)
(30)
(37)
(38)
(39)
30071
45676
37645
47656
76767
45676
45667
37412
67456
12345
12345
34567
12315
37373
12345
07891
37676
12345
47676
45674
67891
10707
71267
67891
(40)
(41)
(42)
(43)
(44)
(45)
71234
19123
93456
45678
45679
76756
1241)8
67345
13767
34507
34507
34567
91379
67777
37124
12345
12345
12345
9:^456
88899
12456
99999
76767
67891
(46)
(47)
(48)
(49)
(50)
(51)
37676
78967 •
34507
47076
67678
57667
12677
12345
12345
12345
12345
34567
88991
73707
77700
6767 1
67912
23456
23478
12071
67345
- 10070 -
40707
76799
..
39
40
ADDITION.
(52)
76769
12345
70775
45666
(53)
57567
19807
34076
13707
(54)
767346
476734
467007
123456
(55)
473894
767367
412345
671234
(56)
376767
123764
345678
912345
(57)
576
4589
87
84028
(58)
74564
7674
376
6
(59)
5676
1567
63
6767
(60)
76746
71207
100
56
(61)
67674
75670
36
77
(62)
42-37
56-84
27-93
62-41
(63)
0-87
5-273
8-127
?.5-63
(64)
03-785
20-766
00-253
10-004
(65)
85-772
6034-82
57-8563
712-52
(66)
•00007
•06236
-0572
•21
(67)
5471-3
563-47
21-502
0-00007
(08)
81-0235
376-03
4712-5
6-53712
(69)
0-0007
5000-
427-
37-12
(70)
8456-5
-37
8456-302
•007
(71)
576-34
4000-005
213-5
2753-
72.
73.
74.
76.
/7.
£7654 4- £50121 + £100 + £76767 + £675
=:£135317.
£10 + £7676 + £97674 + £676 + £9017
=£115053.
£971 + £400 + £97476+£30+£7000+£76734
=£182611.
10000 + 76567 + 10 + 76734 + 6763 + 6767 + 1
=176842.
1 + 2 + 7676 + 100 + 9 + 7707 + 67=15622.
76 + 9970 + .33 + 9977+100 +67647 + 676760
=704563.
ADDITION. 41
78. -75 + -6 4- -756 -f •7254 + -345 + -5 + *005 + -07
=3-7514.
79. ■4+74-47 + 37-007 + 7505 + 747077=:934-004.
80. 56-05 + 4-75 + -007 + 36-14+4-672=101-619.
81. •76 + -0076 + 76 + -54-5-f--05.r=82-3176.
82. 'd-{--OD-{--OOo-\-D-\-oO-{-DOO=DDD-odo.
83. •367 + 56-74-762 + 97-6 + 471=13S7-667.
84. l + -l + 104--0H-160 + '001rrrl7Mll.
85. 3-76+44-3-f 476-1 + o-;3=529-66.
86. 36-774-4-42-1- 1-1001 + -6=42-8901.
87. A merchant owes to A. £1500 ; to B. £408 ; to
G. £1310 ; to D. £50 ; and to E. £1900 ; what is the
sum of all his debts r An^. £5168.
8S. A merchant has received the following sums : —
£200, £315, £317, £10, £172, £513 and £9 ; what id
the amount of all ? Aiis. £1536.
89. A merchant bought 7 casks of merchandize. No.
1 weighed 310 ib ; Xo. 2, 420 lb ; No. 3, 338 lb ; No.
4, 335 ib ; No. 5, 400 ib ; No. 6, 412 lb ; and No. 7
429 ib : what is the weight of the entire r
Ans. 2644 &.
90. What is the total weight of 9 casks of goods : —
Nos. 1, 2, and 3, weighed each 350 ib ; Nos. 4 and 5,
each 331 ib ; No. 6, 310 ib ; Nos. 7, 8, and 9, each
342 ib .' A?is. 3048 ib.
91. A merchant paid the following sums : — £5000,
£2040, £1320, £1100, and £9070; how much was
the amount of all the payments ? Ans. £18530.
92. A linen draper sold 10 pieces of cloth, the first
c mtained 34 yards ; the second, third, fourth, and fifth,
each 36 yards ; the sixth, seventh, and eighth, each 33
yards ; and the ninth and tenth each 35 yards ; how
many yards were there in all .' Aii3. 347.
93. A cashier received six bacrs of money, the first
held £1034 ; the second, £1025 ;lhe third, £2008 ; the
fourth, £7013 ; the fifth, £5075 ; and the sixth, £89 :
how much was the whole sum ? Ans. £16244.
94. A vintner buys 6 pipes of brandy, containing as
follows:— 126, lis, 125, 121, 127, and 119 gallons;
how many gallons in the whole .' Ans. 736 gals.
95- What is the total weight of 7 cai^ks, No. 1, con-
42 ADDITION.
taining, 960 lb ; No. 2, 725 fib ; No. 3, 830 ib ; No. 4,
798 ib ; No. 5, 697 ib ; No. 6, 569 fib ; and No. 7,
987 lb ? Ans. 5566 ib.
96. A merchant bouglit 3 tons of butter, at £90 per
ton ; and 7 tons of tallow, at £40 per ton ; how much
is the price of both butter and tallow ? Ans. £550.
97. If a ton of merchandize cost £39, what will 20
tons come to } Ans. £780.
98. IIow much are five hundi-ed and seventy-three ;
eight hundred and ninety-seven ; five thousand six hun-
dred and eighty-two ; two thousand seven hundred and
twenty-one ; fifty-six thousand seven hundred and seventy-
one ? Ans. 66644.
99. Add eight hundred and fifty-six thousand, nme
hundred and thirty-three ; one million nine hundred and
seventy-six thousand, eight hundred and fifty-nine ; two
hundred and three millions, eight hundred and ninety-
five thousand, seven hundi*ed and fifty-two.
Ans. 206729544.
100. Add three millions and seventy-one thousand ;
four millions and eighty-six thousand ; two millions and
fifty-one thousand ; one million ; twenty -five milUons and
six ; seventeen millions and one ; ten millions and two ;
twelve millions and twenty-three ; four hundred and
seventy-two thousand, nine hundred and twenty-three ;
one hundred and forty-three thousand ; one hundred and
forty- three millions. Ans. 217823955.
101. Add one hundred and thirty-three thousand;
seven hundred and seventy thousand ; thirty-seven tliou-
sand 5 eight hundred and forty-seven thousand ; thu'ty-
three thousand ; eight hundred and seventy-six thousand ;
four hundred and ninety-one thousand. Ans. 3187000.
102. Add together one hundred and sixty-seven thou-
sand ; three hundred and sixtj^-seven thousand ; nine hun-
dred and six thousand ; two hundred and fort3'-seven
thousand ; ten thousand ; seven hundred thousand ; nine
hundred and seventy-six thousand ; one hundred and
ninety-five thousand ; ninety-seven thousand.
yi7w\ 3665000.
103. Add three tcn-thousandtlis ; forty-four, five
tenths ; five hundredths ; six thousandths, eight tcn-tliou-
ADDiriOM. 43
^andths ; four thousand and forty 'one ; twcnt_y-tAvo, one
tenth; one ten-thousandth. ' Af is. 4107 -6^72.
104. Add one thousand ; one ten-thousandth ; five hun-
dredths ; fourteen hundi-ed and forty ; two tenths, three
ten-thousandths ; five, four tenths, four tliousandths.
Ans. 2445-6544.
105. The circulation of promissory notes for the four
weeks ending February 3, 1S44, was as follows : — Bank
of England, about £21,228,000 ; private banks of Eng-
land and ^\'ales, ^£4,980,000 ; Joint Stock Banks of
l!:ngland and Wales, £3,446,000 ; all the banks of Scot-
land, £2,791,000 ; Bank of Ireland, £3,581,000 ; all the
other banks of Ireland, £2,429,000 : what was the total
circulation ? Ans. £38,455,000.
106. Chronologers have stated that the creation of
theVrorld occurred 4004 years before Christ ; the deluge,
2348 ; the call of Abraham, 1921 ; the departure of t1ie
Israelites, from Egypt, 1491 ; the foundation of Solomon^s
temple, 1012 ; the end of the captivity, 536. This being
the year 1844, how long is it since each of these events ?
Ans. From the creation, 5848 years ; from the deluge,
4192 ; from the call of Abraham, 3765 ; from the de-
parture of the Israelites, 3335 ; from the foundation of the
temple, 2856 ; and from the end of the captivity, 2380
107. The deluge, according to this calculation, occur-
red 1656 years after the creation; the call of Abraham
427 after the deluge ; the departure of the Israelites,
430 after the call of Abraham ; the foundation of the
temple, 479 after the departure of the Israelites ; and
the end of the captivity, 476 after the foundation of the
temple. How many years from the first to the last ?
Ans. 3468 years.
108. Adam lived 930 years ; Seth, 912 ; Enos, 905 ;
Cainan, 910 ; Mahalaleel, 895 ; Jared, 962 ; Enoch, 365 ;
Methuselah, 969 ; Lamech, 777 ; Noah, 950 ; Shem, 600 ;
Arphaxad, 438 ; Salah, 433 ; Heber, 464 ; Peleg, 239 ;
Eeu, 239 ; Serug, 230 ; Nahor, 148 ; Terah, 205^ Abra-
ham, 175 ; Isaac, 180 ; Jacob, 147. What is the sum of
all their ages } • Ans. 12073 years
13. The pupil should not be allowed to leave addition,
44 ADDITION.
until he can, with great rapidity, continually add any of
the nine digits to a given quantity ; thus, beginning with
9, to add 6, he should say: — 9, 15, 21, 27, 33, &c.,
without hesitation, or further mention of the numbers.
For instance, he should not be allowed to proceed thus :
9 and 6 are 15 ; 15 and 6 are 21 ; &c. ; nor even 9 and
6 are 15 ; and 6 are 21 ; &c. He should be able, ulti-
mately, to add the following —
5638
4756
9342
19786
in this manner : — 2y 8 ... 16 (the sum of the column ;
of which 1 is to be carried, and 6 to be set down) ; 5,
10 ... 13; 4, 11 ... 17; 10, 14 ... 19.
QUESTIONS TO BE ANSWERED BY THE PUPIL.
1. To how many rules may all those of arithmetic be
reduced.^ [1].
. 2. What is addition r [3].
3. What are the names of the quantities used in addi-
tion .^ [3].
4. What are the signs of addition, and equality t [4].
6. What is the vinculum ; and what are its effects on
additive quantities ? [5] .
6. What is the rule for addition r [7].
7. What are the reasons for its different parts ? [9] .
8. Does this rule apply, at whatever side of the deci-
mal point all, or any of the quantities to be added ars
found? [11]. _
9. How is addition proved .? [10]-
10. What is the reason of this proof .^ [10] •
SUBTRACTION. 45
SIMPLE SUBTRACTION.
14. Simple subtraction is confined to abstract numbers,
and applicate which consist of but one denomination.
Subtraction enables us to take one number called the
subtrahend^ from another called the miniteiul. If any-
thing s left, it is called the excess ; in commercial con-
cerns, it is termed the remainder ; and in the mathema-
tical sciences, the difference.
15. Subtraction is indicated by — , called the minus,
or negative sign. Thus 5 — 4=1, read five minus four
equal to one, indicates that if 4 is substracted from 5,
unity is left.
Quantities connected by the negative sign cannot be
taken, indifi"erently, in any order ; because, for example,
5 — 4 is not the same as 4 — 5. In the former case the
positive quantity is the greater, and 1 (which means
+ 1 [4] ) is left ; in the latter, the negative quantity
is the greater, and — 1, or one to be subtracted, still
remains. To illustrate yet further the use and nature
of the signs, let us suj^posc that we hmx five pounds,
and owe four ; — the five pounds we ho.te will be repre-
sented by 5, and our debt by — 4 ; taking the 4 from
the 5, we shall have 1 pound ( + 1) remaining. Next
let us suppose that we have only four pounds and owe
five ; if we take the 5 from the 4 — that is, if we pay
as ftbr as we can — a debt of one pound, represented by
— 1 , will still remain ; — consequently 5-— 4=1 ; but
4— 5=— 1.
16. A vinculum placed over a subtractive quantity,
or one having the negative sign prefixed, alters its
value, unless we change all the signs but the first : —
thus 5 — 3+2, and 5 — 3+2, are not the same thing;
for 5—3+2=4 ; but 5—3+2 (3 + 2 being considered
now as but one quantity) =0 ; for 3+2=5 ; — therefore
— 3+2 is the same as 5 — 5, wh5(.^ leaves nothing ; or,
in other words, it is equal to 0. If, however, we change
all the signs, except the first, the valuo of the quantity ia
46 SUF.TRACTION.
uot altered by the vmculuin ; — thus 5 — 3-r2=4j and
f)— 3 — 2, also, is equal to 4.
Again, 27-44-7-3=27.
27-4+7-3=19.
Rnt 97 4 7_i_'^ (chansring all the signs of the ? 97
but Zi — ±— / -^6 ori-iual quantitiesrbut the first) ^ — ^' •
The following example will show how the vinculum
affects numbers, according as we make it include an
additive or a subtractiv(3 quantity : —
48+7-3-8+7-2=49.
40 i^-'_q_ oj^-T 9 4Q . what is under the vinculum being
lo-f* o o < z, — ly, additive, it is not necessary to
change any signs.
,10 I 7 Q I Q TTo /to. it is now necessary to change all the
45_|- / — i+0^+^=4y , gjg^g ,^,j^^^ ti^g vinculum.
48+7 — 3 — 8—7+2=49; it is necessary in this case, also,
AQ '^ o o ^ o in to change tlift signs.
48 + / — o — 0+/ — 2=49 ; it is not necessary in this case.
In the above, we have sometimes put an additive, and
sometimes a subtractive quantity, under the vinculum ;
in the former case, we were obliged to change the signs
of all the terms connected by the vinculum, except the
first — that is, to change all the signs undtr the vin-
culum ; in the latter, to preserve the original value of
the quantity, it was not necessary to change any sign.
To Subtract Nwmhers.
17. Rule. — I. Place the digits of the subtrahend
under those of the same denomination in the minuend —
units under units, tens under tens, &c.
II. Put a line under the subtrahend, to separate it
from the remainder.
III. Subtract each digit of the subtrahend from the
one over it in the minuend, beginning at the right hand
side.
ly. If any order of the minuend be smaller than the
quantity to be subtracted from it, increase it by ten ; and
cither consider the next order of the minuend as lessened
by unity, or the next order of the subtrahend as in-
creased by it.
V. After subtracting any denomination of the sub-
SUBTRALTK)^. 47
trahend from the correspond lug part of the minuend,
set down what is left, if any thing, in the place whicii
belongs to the same denomination of the ** remainder."
VlT Bat if there is nothing left, put dovrn a cypher —
provided any digit of the " remainder" will be more dis-
tant from the decimal point, and at the same side of it.
18.. EXAMPLE 1. — Subtract 427 from 702.
iv2 mimiend.
427 subtrahend.
305 remainder, ditferenee, or execs?.
We cannot take 7 units from 2 units : but ''boriowing." as
it is called, on-c of the 9 tons in the minuend, and consider-
ing it as {en unit.s. Ave add it to the 2 units, and then have
12 units : taking 7 from 12 units, 5 are left : — avc put 5 in
the units place of the *• remainder." We may consider the
0 tens of the minuend (one having been taken away, or
borrowed) as 8 tens; or, which is the same thing, may
suppose the 9 tens to remain as they were, but the 2 tens
of the subtrahend to have become 3 ; then, 2 tens from 8
tens, or 3 tens from 9 tens, and G tens are left: — we put G
in the tens" place of the •• remainder." 4 hundreds, of the
subtrahend, taken from the 7 hundreds of the minuend,
leave 3 hunilreds— which we pat in the hundreds" place of
the -remainder."'
Example 2. — Take 504 from 708.
708
504
204
Wlien 6 tens are taken from 6 tens, nothing is loft : we
therefore put a c_v|)her in the tens' place of the ••remainder."'
Ejlamfle 3. — Take 537 from 594.
504
537
~57'
When 5 hundreds are taken from 5 hundreds, notliing
remains : but we do not here set down a cypher, since no
sigtiificant S.L'ure in the remainder is at the same side of,
and farther from the decimal point, than the place which
would be occupied by this cypher.
19. Reason of I. — We put digits of the sar.ic denomina-
tions in the sf.aie vertical column, that the different parts
48 SUBTRACTION.
of the Rubtraliend may be near those of the minuend from
which they are to be taken ; we are then sure that the coy-res-
pojiding portions of the subtrahend and minueud may be
easily found. By tliis arrangement, also, "we remove any
doubt as to the denominations to which the digits of the ^uh-
trahend belong — their value? being rendered more certain, by
their position with reference to the digits of the minuend.
Keasox of II. — The separating line, though convenient, 13
not of such importance as in addition [9] ; sinpe the " remain-
der " can hardly be mistaken for another quantity.
IXeasox of III. — "When the numbers are considerable,
the subtraction cannot be effected at once, from the limited
powers of the mind; we therefore divide the given quantities
into parts ; and it is clear that the sum of the dififerenccs of
the corresponding parts, is equal to the difference between
the sums of the parts : — thus, 578 — 327 is evidently equal
to 600— 300-f-70— 2D-f 8— 7, as can be shown to the child by
pebbles, &c. We begin at the right hand side, because it may
be necessary to alter some of the digits of the minuend, so as
to make it possible to subtract from them the corresponding
ones of the subtrahend ; but, unless we begin at the right hand
side, we cannot know what alterations may be required.
llEAso^r OF IV. — If any digit of the minuend be smaller
than the corresponding digit of the subtrahend, we can proceed
in cither of two ways. First, we may increase that denomina-
tion of the minuend which is too small, by borrowing one from
the next higher, (considered as teji of the lower denomination,
or that which is to be increased,) and adding it to those of the
lower, already in the minuend. In this case we alter the
form, but not the value of the minuend ; which, in the exam-
ple given above, would become —
Hundreds. tens. units.
7 8 12 = 792, the minuend.
4 2 7 = 427, the subtrahend.
3 G 5 ^ 365, the difference.
Or, secondly, we may add equal quantities to both minuend
and subtrahend, which will not alter the difference ; then we
■would have
Hundreds. tens. units.
7 9 2 -j- 10 = 702 -f- 10, the minuend + 10.
4 2-1-1 7 =427-1-10. tlie subtrahend -f- 10.
3 6 5= 365 -\~ 0, the same dilTcrence.
In this mode of proceeding we do not use the given minuend
and subtrahend, but others which produce the same remainder.
Reason of V. — The remainders obtained by subtracting,
successively, the different denominations of the subtrahend
froaa those which correspond in the minuend arc tlie parts of
SUBTRAC'lION. 40
tbe total remainder. Tliey are to be set down under the deno-
minations^ft'hich produced them, since they belong to these
denouunations.
Kkason of VI. — Unless there is a significant figure at the
same side of the decimal point, and more distant from it thar.
the cypher, tlie latter — not being between the decimal point
and a significant figure — will be useless [Sec. 1. 28], and may
therefore be omitted.
20. Proofs of Suit radian. — Add tog'.^lher the re-
mainder and subtrahend ; and the sum should be equal
to the minuend. For, the remainder expresses by how
much the subtrahend is smaller than the minuend ;
adding, therefore, the remainder to the subtrahend,
should make it equal to the minuend ; thus
8754 minuend.
5S39 subtrahend. ^
2915 difference. 5
Sun\ of difference and .subtrahend, 8754=minuend.
Or ; subtract the remainder from the minuend, and
what is left sliould be equal to the subtrahend. For
the remainder is the excess of the minuend abo^x; the
Rubtraherid ; therefore, taking away this excess, should
leave both equal ; thus
8634 minuend. Proof : 8634 minuend.
7985 subtrahend. 649 remainder.
649 remainder. New remainder, 7985=subtrahend.
In practice, it is sufficient to set down the quantities
once ; thus
8034 minuend.
7085 subtrahend.
649 remainder.
Difference between remainder and minuend, 7985=subtrahend.
21. To Subtract, ichen the quantities contain Deci-
mals.— The rule just given is applicable, at whatever
side of the decimal point all or any of the digits may
bo found; — this follows, as in addition [11], from the
very nature of notation. It is necessary to put the
decimal point of the remainder under the decimal points
of the minuend and subtrahend ; otherwise the digits
of the remainder will not, as they ought, have the same
value as the digits from which they have been derived.
60
SUBTRACTION.
Example.— Subtract 427-85 from 5G304.
663-04 *
427-85
135-19
Since the digit to the right of the decimal point in the
remainder, indicates what is left after the subtraction of the
tenths, it expresses so many tentiis : and since the digit to
the left of the decimal point indicates wimt remains after
the subtraction of the units, it expresses so many units; —
ail this is shown by the position of the decimal point.
22. It follows, from the principles of notation [Sec. I.
40], that however we may alter the decimal points of
the minuend and subtrahend, as long as they stand in
the same vertical column, the digits of the difference
are not changed ; thus, in the following examples, the
eame dioits are found in all the remainders : —
4302
3547
436-2
354-7
43-62
35-47
■4362
•3547
-0815
•0004362
-0003547
815
81-5
8-15
•0000S15
EXERCISES IN SUBTRACTION.
(1)
(2)
(3)
(4)
(5)
(6)
From
1909
7432
9076
8140
3176
70377
Take
1408
6711
4567
4377
2907
45761
(7)
(8)
(9)
(10)
(11)
(12)
From
86167
67777
71234
900076
376704
745674
Take
61376
46699
43412
899934
297610
376789
(13)
From 67001
Take 35090
(14)
9733376
4124767
(15)
567074
476476
(16)
473(576
321799
(17) (18)
6310756 376576
3767016 240940
SUBTRACTI0f7.
51
(19^ (20) (21) (22) (23) (24)
From 345670 234100 4367676 345078 70101076 67360000
Take 1799 990 250569 124799 37691734 31237777
(25) (26) (27) (28) (29) (30)
From 1970000 7010707 67345001 1074561 14767674 4007070
Take 1361111 3441216 47134777 1123640 7476909 3713916
From
Take
(31)
7045676
3077097
(32)
37670070
26716615
(33)
70000000
9999999
(34)
70040500
56767767
(35)
50070007
41234016
(38)
8000800
377776
From
Take
(36)
I 1000000
9919919
(37)
3000001
2199077
(39)
8000000
62358
(40)
4040053
220202
(41)
85-73
42-10
From
Take
(42)
805-4
73-2
(47)
874-32
5-63705
(43)
594-763
85-600
(44)
47-630
0-078
(45)
52-137
20-005
(49)
'632-
0-845003
From
Take
(46)
0-00063
0-00048
(48)
57-004 4-
2-3
(50)
400-327
0-0006
51.
52. 560789—75674=501115.
53. 941000—5007=935993.
54. 97001—50077=46924.
55. 76734—977=75757.
56 . 56400—100=50300 .
57 . 700000—99=699901 .
58. 5700—500=5200.
5';). 9777-89=9688.
60. 76000—1=75999.
01. 90017—3=90014.
•45676— 567456=178220.1 02. 97777—4=97773.
63. 60000—1=59999.
64. 75477—76=75401.
65. 7-97-1-05=6-92.
66. 1-75— -074=1-676.
67. 97-07—4-769=92-301.
68. 7-05—4-776=2-274.
09. 10-761—9-001=1-76.
70. 12-10009—7-121=4-97909
71. 176-1 — -007=176-093.
I 72. 15-00—7-863=7-197.^
62 SUBTRACTION.
73. What number, added to 0700, will make it 10001
Ans. 1192.
74. A vintner bought 20 pipes of brandy, containing
2450 gallons, and sold 14 pipes, containing 1680 gal-
lons ; how many pipes and gallons had he remaining f
Ans. 6 pipes and 779 gallons.
75. A merchant bought 564 hides, weighing 16S00
lb, and sold of them 260 hides, weighing 7S09 lb ; how
many hides had he unsold, and what was their weight .'
A.m. 304 hides, weighing 8991 lb.
76. A gentleman who had 1756 acres of land, gives
250 acres to his eldest, and 230 to his second son ; how
many acres did he retain in his possession .' Ans. 1276.
77. A merchant owes to A. £800 ; to B. £90 ; to C.
£750 ; to D. £600. To meet these debts, he has but
£971 ; how much is he deficient ? Ans. £1269.
78. Paris is about 225 English miles distant from
London ; Eome, 950 ; Madrid, 860 ; Vienna, 820 ;
Copenhagen, 610 ; G-eneva, 460 ; Moscow, 1660 ; Gib-
raltar, 1160; and Constantinople, 1600. How much
more distant is Constantinople than Paris ; Eome than
Madrid ; and Vienna than Copenhagen. And how much
less distant is G-eneva than Moscow ; and Paris than
Madrid? Ans. Constantinople is 1375 miles more- dis-
tant than Paris ; Eome, 90 more than Madrid ; and
A'ienna, 210 more than Copenhagen. Geneva is 1200
miles less distant than 3Ioscow ; and Paris, 635 less
than Madrid.
79. How mucb was the Jewish greater than the
English mile ; allowing the former to have been 1*3817
miles English } Ans. 0-3817.
80. How mucli is the English greater than the Eomau
mile ; allowing the latter to have been 0"915719 of a
mile English .' ^?zs. 0-084281
8 1 . What is tbe value of 6 - 3 -f- 1 5 — 4 ? A^is. 14
82. Of 43 + 7-3— 14. => Ans. 33
83. Of47-6— 2+1— 244-16 — -34.' A713. 5294
84. What is tbe difference between 15 + 13—6—81 +
()2, and 15 + 13—6—81 + 62.? Ans. 38.
23. Before leaving this rule, the pupil should be able
MULTIPLICATION. 53
to take any of the nine digits contin\ially from a given
number, without stopping or hesitating. Thus, sub-
tracting 7 from 94, he should say, 94, 87, 80, Sec. ; and
should proceed, for instance, with the following example
5376
4298
1078
in this manner: — 8, 16.. .8 (the difference, to be set
down) ; 10, 17.. .7 ; 3, 3...0 ; 4, 5...1.
QUESTIONS TO BE ANSWERED EV THE PUPIL.
1 . What is subtraction ? [14j.
2. What are the names of Ae terms used in subtrac-
tion .' t"14].
3. What is the sia-n of subtraction .' [15].
4. How is the yinculum used, with a subtractive
quantity? [16].
5. AVhat is the rule for subtraction } [17].
6. What are the reasons of its different parts? [19],
7. Does it apply, when there are decimals ? [21].
8. How is subtraction proved, and why ? [20] .
9. Exemplifv a brief mode of performing subtrac-
tion ? [23].
SIMPLE MULTirLICATION.
24. Simple multiplication is confined to abstract
numbers, and appiicate vrhich contain but one denomi-
nation.
Multiplication enables us to add a quaniity, called the
multijplicand^ a number of times indicated by the multi-
plier. The multiplicand, therefore, is the number mul-
tiplied ; the multiplier is that by Vrhich we multiply :
the result of the multiplication is called the product.
It follows, that what, in addition, vrould be called m
" addend," in multiplication, is termed the " raultipli-
cand ; " and what, in the former, would b,e oaiied tho
" sum," in the latter, i.? designated tiie " product." Tho
quantities v/hichj when muip.idjcd together, give tho
54 MULTIPLICATION.
product, are called also factors^ and, when they ara
integers, suhmuUiples. There may be more than two
factors ; in that case, the multiplicand, multiplier, or
both, will consist of more than one of them. Thus, if 5
6, and 7, be the factors, either 5 times 6 may be con-
sidered as the multiplicand, and 7 as the multiplier — oi
5 as the multiplicand, and 6 times 7 as the multiplier.
25. Quantities not formed by the continued addition
of any number, but unity — that is, which are not the
products of any two numbers, unless unity is taken as
one of them — are called jprime numbers : all others are
termed amiposiie. Thus 3 and 5 are prime, but 9
and 14 are composite numbers ; because, only three,
multiplied by c?/^, will p^duce " three," and only^re,
multiplied by one^ will produce '^ five," — but, three
multiplied by t/irec will produce " nine," and seven mul-
tiplied by two will produce " fourteen,"
26. Any quantity contained in another, some number
of times, expressed by an integer — or, in other words,
that can be subtracted from it without leaving a re-
mainder— is said to be a measure, or aliquot part of
that other. Thus 5 is a measure of 15, because it is
contained in it three times exactly — or can be sub-
tracted from it a number of times, expressed by 3, an
integer, without leaving a remainder ; but 5 is not a
measure of 14, because, taking it as often as possible
from 14, 4 will still be left;— thus, 15—5=10, 10- 5r=
5, 5 — 5=0, but 14 — 5r=9, and 9 — 5r=:4. Measure,
submultiple, and aliquot part, are synonymous.
27. The common measure of two or more quantities
is a number that will measure each of them : it is a
measure common to them. Numbers which have no
common measure but unity, are said to be prime to each
other ; all otliers are composite to each other. Thus 7
and 5 are prime to each other, for unity alone will
measure both ; 9 and 12 are composite to each other,
because 3 will measure either. It is evident that two
•privie numbers must be prune to each other ; thus 3
and 7 ; for 3 cannot measm-e seven, nor 7 three, and —
except unity — there is no other number that will mea-
sure either of them.
MUI.TIPLICATIOX. 65
Two riuiiiltcrs may bo omnposito to each other ^ and
yet one of them ma}- be a irr'une niiiiiber ; thus 5 and 25
ai-e both measured by 5, still the former is 'prime.
Two numbers may bo camposite^ and yet prime to
each other ; thus 9 and 14 are both composite numbers,
yet they have no cmmnoii measure but unity.
28. The greatest common measure of two or more
numbers, is the greatest number which is their common
measure ; thus 30 and 60 are measured by 5, 10, 15,
and 30 ; therefore each of these is theii* common mea-
sure ; — but 30 is their greatest common measure. When
a product is formed by factors which are integers, it is
measured by each of them.
29. One number is the multiple of another, if it
contain the latter a number of times expressed by an
integer. Thus 27 is a multiple of 9, because it con-
tains it a number of times expressed by 3, an integer.
Any quantity is the multiple of its measure, and the
measure of its multiple.
30. The common multiple of two or more quantities,
is a number that is the multiple of each, by an integer ; —
thus 40 is the common multiple of 8 and 5 ; since it is a
multiple of 8 by 5, an integer, and of 5 by 8, an integer.
Thj least common multiple of two or more quantities,
is the least number which is their common multiple ; — •
thus 30 is a common multiple of 3 and 5 ; but 15 ia
their least common multiple ; for no number smaller
than 15 contains each of them exactly.
31. The equimultiples of two or more numbers, are
their products, when multiplied by the same number ; —
thus 27, 12. and 18, are equimultiples of 9, 4, and 6 ;
because, midtiplying 9 by /^res, gives 27, multiplying 4
by three., gives 12, and multiplying 6 by three., gives 18.
32. Multiplication greatly abbre\4ates the process of
addition ; — for example, to add 68965 to itself 7000 times
by *' addition," would be a work of great labour, and con-
sume much time ; but by " multiplication," as we shall find
presently, it ca^i be done with ease, in less than a minute.
33. At first it may seem iiiaecm-ate, to have stated
[2] that multiplication is a species of addition ; since we
can know the product of two quantities without having
56 MULTIPLICATION.
recourse to that rule, if they are found in the multipli-
cation table. But it must not be forgotten that the mul-
tiplication table is actually the result of additions, long
since made ; without its assistance, to multiply so sunple
a number as 4 by so small a one as five, we should be
obliged to proceed as follows,
4
4
4
4
4
20
performing the addition, as with any other addends.
The multiplication tabl^is due to Pythagoras, a cele-
brated Grreek philosopher, who was born 590 years
before Christ.
34. We express multiplication by X ; thus 5X7=
35, means that 5 multiplied by 7 are equal to 35, or
that the product of 5 and 7, or of 5 ly 7, is equal to 35.
When a quantity under the vinculum is to be multi-
plied by any number, each of its parts must be multi-
plied— for, to multiply the whole, we must multiply
each of its parts :— thus, 3X 7 + 8—3=3 X 7 + 3 X S—
3X3; and 4 + 5X8 + 3 — 6, means that each of the
terms under the latter vinculum, is to be multiplied by
each of those under the former.
35. Quantities connected by the sign of multiplication
may be read in any order; thus 5x6=6X5. This
will be evident from the following illustration, by which
it appears that the very same number may be considered
either as bX^-, or QX^^ according to the view we take
of it :—
« 6
Quantities connected by the sign of multiplication,
MULTIPLICATION. 57
are multiplied if we multiply one of the factors ; thus
6X7X3 multiplied by 4=6X7 multiplied by 3X4.
36. To prepare him for multiplication, the pupil
should be made, on seeing any two digits, to name then*
product, without mentioning the digits themselves. Thus,
a large number having been set down, he may begin
with the product of the first and second digits ; and
then proceed with that of the second and thu'd, &c".
Taking
5876349258G7
for an example, he should say : — 40 (the product of 5
and 8) ; 56 (the product of 8 and 7) ; 42 ; 18 ; &c., as
rapidly as he could read 5, 8, 7, &c.
To Multij)Iy Nuvibers.
37. When neither multiplicand, nor multiplier ex-
ceeds 12 —
Rule. — Find the product of the given numbers by
the multiplication table, page 1.
The pupil should be perfectly familiar with this table.
Example. — What is the product of 5 and 7 1 The mul-
tiplication table shows that 5x7=35, (5 times 7 are 35).
38. This rule is applicable, whatever may be the
relatiiie values of the multiplicand and multiplier ; that
is [Sec. I. IS and 40], whatever may be the kiiid of
units expressed — provided their ahsolute values do not
exceed 12. Thus, for instance, 1200x90, would come
under it, as well as 12x9 ; also '0009 XO'S, as well as
9X8. We shall reserve what is to be said of the man-
agement of cyphers, and decimals for the next rule ; it
will be equally ti-ue, however, in all cases of multiplica-
tion.
39. When the multiplicand does, but the multiplier
does not exceed 12 —
Rule. — I. Place the multiplier under that denomi-
nation of the multiplicand to which it belongs.
II. Put a line under the multiplier, to separate it from
the product.
III. Multiply each denomination of the multiplicand
by the multiplier — beginning af the right hand side.
58 MULTIPLICATION-.
IV. If tlie product of the multiplier and any digit
of the multiplicand is letis tlian ten, set it down under
that digit ; but if it be greater, for every ten it contains
carry one to the next product, and put down only what
remams, after deducting the tens ; if nothing remains,
put down a cypher.
V. Set down the last product in full.
40. Example. 1.— What is the product of 897351x4 ?
897351 multiplicand.
4 multiplier.
3589404 product.
4 times oue unit are 4 imits ; since 4 is less than ten, it
gives nothing to be '• carried," we, therefore, set it down in
the units' place of the product. 4 times 5 are twenty (tens)-;
which are equal to 2 tens of tens, or hundreds to be carried,
and tw units of tens to be set down in the tens' place of
the product — in which, therefore, we put a cypher. 4
times 3 are 12 (hundreds), which, with tlie 2 hundreds to be
carried from the tens, make 14 hundreds ; these are equal
to one thousand to be carried, and 4 to be set down in the
thousands' place of the product. 4 times 7 are 28 (thou-
sands), and 1 thousand to be carried, are 29 thousands ; or
2 to be carried to the next product, and 9 to be set down
4 times 9 are 36, and 2 are 38 ; or 3 to be carrried, and 8 to
be set down. 4 times 8 are 32, and 3 to be carried are 35 ;
which is to be set do^vn. since there is nothing in the next
denomination of the multiplicand.
Example 2.— rNlultiply 80073 by 2.
80073
2
16014G
Twice 3 units are G imits : G being less than ten, gives
nothing to be carried, hence we put it down in the units'
place of the quotient. Twice 7 tens are 14 tens ; or 1 hundred
to be carried, and 4 tens to be set down. As there are no
hundreds in the multiplicand, w^e can have none in the pro-
duct, except that which is derived from the multiplication
of the tens : w^e accordingly put the 1, to be carried, in the
hundreds' place of the product. Since there are no thou-
sands in the multiphcaud. nor any to be carried, we put a
cypher in that denomination of the product, to keep any
significaiit figures that follow, in their proper places.
MULTIPLICATION. 59
41. Reason of I. — The multiplier is to be placed under that
denomination of the multiplicand to which it belongs; since*
there is then no doubt of its value. Sometimes it is necessary
(o add cyphers in putting down the multiplier ; thus, »
Example 1. — 178 multiplied by 2 hundred —
478 multiplicand.
200 multiplier.
Example 2. — 5b9 multiplied by 3 ten- thousandths —
530 • multiplicand.
0-0003 multiplier.
Reason of II. — It is similar to that given for the separating
line in subtraction [19].
Reason of III. — When the multiplicand exceeds a certain
amount, the powers of the mind are too limited to allow us
to multiply it at once ; we therefore multiply its parts, in suc-
CQSsion, and add the results as we proceed. It is clear that
the sum of the products of the parts by the multiplier, is equal
to the product of the sum of the parts by the same multi-
plier :— thus, 537 X 8 is evidently equal to 500 x8-f-o0x8-|-7 X8
For multiplying all the parts, is multiplying tlie whole ; since
the whole is equal to the sum of all its parts.
We begin at the right hand side to avoid the necessity of
aftiiti'ards adding together the subordinate products. Thus,
taking the example given above ; were we to begin at the left
hand, the process would be —
897351
4
3200000=800000X4
360000= 90000x4
28000= 7000X4
1200= 300X4
200= 50X4
4= 1X4
3589404=sum of products.
Reason of IV. — It is the same as that, of tlie fourth part of
the rule for addition [0] ; the product of the multiplier and
ony denomination of the multiplicand, being equivalent to the
hum of a column in addition. It is easy to change tlie given
cxauipie to an exercise in addition; for'&97o5lx4, is the same
thing !i3
897351
897351
897351
897351
3589404
60 MULTIPLICATION.
Reason of V. — It follows, that the last product is to be set
clown in full ; for the tens it contains will not be increased :
they may, therefore, be set down at once.
This rule includes ail cases in wbi?-li the absohUe
value of the digits in the multiplier docs not exceed
12. Their relative value is not material ; for it is as
easy to multiply by 2 thousands as by 2 units.
42. To prove multiplication, when the mnltiplier does
not exceed 12. Multiply the multiplicand by the mul-
ti2)lier, minus one ; and add the multiplicand to the pro-
duct. The sum should be the same as the product of
the multiplicand and multiplier.
Exa:\iple. — IMultiply 6432 by 7, and prove the i^vault.
6432 multiplicand.
6=7 (the multiplier)—!
6432 38592 multiplicand X 6 .
7 (=6+1) 6432 multiphcaadx 1.
45024 = 45024 multiplicand multiplied by G \ 1=:7.
We have multiplied by 6, and by 1, and added the results ;
but six times the multiplicand, plus once the nmltlplicand,
is equal to seven times the multiplicand. ^Vhat we oHain
from the two processes should be the same, for we )mve
merely used tAvo methods of doing one thing.
EXERCISES FOR THE PUPIL.
Multiply
By
(1)
76762
2
(2)
67456
2
(6)
456769
7
(3)
78976
6
(4)
57346
5
Multiply
By
(5)
763452
6
0)
354709
8
(8^
45678?
IMultiply
(9)
806342
11
(10)
738579
12
(11)
476387^
11
(12)
faa2976.?
12
MULTIPLICATION. 6i
43. To Multiply tchen the Qimniities contain Cyphers.
or Decimals. — The rules already given are applicable ;
those which follow are consequences of them.
When there are cjpliers at the erid of the multipli-
cand (cyphers in the middle of it, have been already
noticed [40])—
Rule. — Multiply as if there were none, and add to the
product as many cyphers as have been neglected. For
The greater the quaaitity multiplied, the greater ought to
be the product.
Example.— Multiply 5G000 by 4.
56000
4
224000
4 times 6 units in the fourth place from the decimal point,
are e^-idently 24 units in the same place ; — that is, 2 in the
fifth place, to be carried, and 4 in the fourth, to be set down.
That we may leave no doubt of the 4 being in the fourtli
Elace of the product, we put three cyphers to the right
and. 4 times 5 are 20. and the 2 to be carried, make 22.
44. K the multiplier contains cyphers —
Rule. — Multiply as if there were none, and add to
the product as many cyphers as have been neglected.
The greater the multiplier, the greater the number of times
the multiplicand is added to itself; and, therefore, the greater
the product.
Example.— Multiply 567 by 200.
200
113400
From what we have said [35], it follows that 200x7 is
the same as 7x200 : but 7 times 2 hundred are 14 hundred ;
and, consequently, 200 times 7 are 14 hundred : — that is, 1
in the fouiih place, to be carried, and 4 in the third, to be set
down. We add two cyphers, to show that the 4 is in the
third place.
45. If both multiplicand and multiplier contain
cyphers —
Rule. — Multiply as if there were none in either, and
add to the product as many cyphers as are found in
both.
d2
62 MULTIPLICATION.
Each of the quantities to be multiplied adds cyphers to the
product [43 aud 4-i].
Example.— IMultiply 46000 by 800.
4G0ud
800
^6800000
8 times G thousand Trould be 48 thousand : but 8 Jtundrcd
times six thousand ought to produce a number 100 times
greater — or 48 hundred thousand ; — that is. 4 in the sevcntk
phice from the decimal point, to be carried, and 8 in the
sixth place, to be set do^Yn. But, 5 cyphers are required,
to keep the 8 in the sixth place. After ascertaining the
position of the iirst digit in the product — from what the
pupil already knows— there can be no difficulty with the
other digits.
46. When tliere are decimal places in the multipli-
cand—
Rule. — Multiply as if tbere were none, and remove tbe
product (by means of the decimal point) so many places
to the right as there have been decimals neglected.
The smaller tlie quantity multiplied, the less the product
Example. — JMultiply 5-G7 by 4.
5'67
4
22-68
4 times 7 hundredths are 28 hundreths :— or 2 tentlis, to
be cjirried, and 8 hundredth*— or 8 in the sscunil place, to
the right of the decimal point, to be set doAvn. 4 times 6
tenths are 24 tenths, which, with the 2 tenths to be carried,
make 2G tenths : — or 2 units to be carried, and G tenths to
be set doAvn. To show that tlie 6 represents tenths, we put
the decimal point to the left of it. 4 times 5 unit^ are 20
units, which, with the 2 to be carried, make 22 units.
47. "When there arc decimals in the multiplier —
Rule. — Multiply as if there were none, and remove
the product so many places to the right as there are
decimals in the multiplier.
The smaller the quantity by which we multiply, the less
must be the result.
MULTIPLICATiON. 63
Example. — ^lultiuly 563 by -07
503
007
39-41
3 multiplied hy 7 himdrcdths, is the same [35] as 7 hun-
dredths multiplied by 3 : which is equal to 21 hundredths : —
or 2 tenths to be carried, and 1 hundredth — or 1 in the
second place to the right of the decimal point, to be set dovm.
Of course the 4, derived from the next product, must be one
place from the decimal poirtt, 6:c.
48. When there are decimals iu both multiplicand
and multiplier —
Rule. — Multiply as if there were none, and mOYC
the product so many places to the right as there are
decimals in both.
In this case the product is diminished, by the smallnees of
both multiplicand and multiplier.
Example 1.— Multiply 56-3 by -08.
563
•08
4-504
8 times 3 tenths are 2-4 [46] ; consequently a quantity
one hundred times less than 8 — or -OS, multiplied by three-
tenths, will give a quantity one hundred times less than 2-4 —
or -024 ; that is, 4 in the third place from the decimal point,
to be set dovNTi, and 2 in the second place, to be carried.
Example 2.— :Multiply 5 63 by 0-00005.
5-63
0-00005
0-0002815
49. "When there are decimals in the multiplicand, and
cyphers in the multiplier ; or the contrary —
Rule. — 31ultiply as if there were neither cyphers
nor decimals ; then, if the decimals exceed the cyphers,
move the product so many places to the right as will be
equal to the excess ; but if the cyphers exceed the deci-
mals, move it so many places to the left as will be
equal to the excess. '
The cyphers move the product to the left, the decimals to
the right ; the effect of both together, therefore, will be equal
to the difference of their separate effects.
(J4
MULTIPLICATION.
Example 1.— Multiply 4600 by -06.
4600
006 2 cyphers and 2 decimals ; excess = 0
276
Example 2.— Multiply 47-63 by 300.
47-63
300 2 decimals and 2 cyphers j excess = 0.
14289
Example 3.— Multiply 85-2 by 7000.
85-2
7000 1 decimal and 3 cyphers j excess =2 cyj>ieri
596400
Example 4.— Multiply 578-36 by 20.
578-36
20 2 decimals and 1 cypher; excess =1 decimal
11567-2
Multiply
By
EXERCISES
(13)
48960
5
FOR THE
(14)
75460
9
PUPIL
(15)
678000
8
(16)
57ci00
6
•
Multiply
By
(17)
7463
80
(21)
743560
800
(18)
770967
900
(19)
147005
4000
(20)
561*76748
30000
Multiply
By
(22)
534900
30 000
(23)
50000
300
(24)
86000
5000
Multiply
By
(25)
52736
2
(26)
8-7563
4
(27)
•21375
6
(28)
0-0007
8
MULTIPLR.ATIOX. 65
(20) (30) (31) (32)
Multiply 50341 85037 72108 2170-38
By 0-0003 0-005 0-0007 0-06
Multiply
By
(33)
875-432
0-04
(34)
78000
0-3
(35)
61-721
6000-
(36)
32
0-00007
-00224
In the last example we are obliged to add cyphers to the
product, to make up the required number of decimal places.
50. When both multiplicand and multiplier exceed
12—
Rule. — ^I. Place the digits of the multiplier under
those denominations of the multiplicand to which they
belong.
II. Put a line under the multiplier, to separate it from
the product.
III. Multiply the multiplicand, and each part of the
multiplier (by the preceding rule [39]), beginning ^th
the digit at the right hand, and taking care to move the
product of the multipKcand and each successive digit
of the multiplier, so many places more to the left, than
the preceding product, as the digit of the multiplier
which produces it is more to the left than the signifi-
cant figure by which we have last multiplied.
IV. Add together all tlie products ; and then- sum
will be tlie product of the multiplicand and multiplier.
51. EXA3IPLE.— :Multiply 5034 by 8073.
5G34
8073
16902=prodact l)y 3.
39438 =product by 70.
45072 =product by 8000.
45483282=product by 8073.
The pr'xiuct of the multiplicand by 3. requires no e»^^.^
nation.
66 MULTIPLICATION.
7 tens times 4. or [35] 4 times 7 tens are 28 tens : — 2 hun-
dreds, to be carried, and 8 tens (8 in the second place from
the decimal point) to be set down, &:c. 8000 times 4, or 4
times 8000, are 32 thousand : — or 3 tens of thousands to be
carried, and 2 thousands (2 in the fourth place) to bo set
down. &c. It is unnecessary to add cyphers, to show the
values of the first digits of the different products: as they
are sufficiently indicated by the digits above. The products
by 3, by 70, and by 8000. are added together in the ordinary
way.
52. Reasons of I. and II. — They are the same as those
given for corresponding parts of the preceding rule [41].
Reason of III. — We are obliged to multiply successively
by the parts of the multiplier ; since we cannot multiply by
the luhole at once.
Reason of IV. — The sum of the products of the multipli-
cand by the parts of ihQ multiplier, is evidently equal to the
product of the multiplicand by the whole multiplier ; for, in
the example just given, 5634 X 8073 =5634 X 8000 4-70 + 3=
[34] 5634X8000+5634X70+5634X3. Besides [35], we may
consider the multiplicand as multiplier, and the multiplier as
nmltiplicand ; then, observing the rule would be the same
thing as multiplying the neAV multiplier into the different
parts of the new multiplicand ; which, we have already seen
[41], is the same as multiplying the whole multiplicand by
the multiplier. The example, just given, would become
8073X5634.
8073 new multiplicand
5634 new multiplier.
We are to multiply 8, the first digit of the multiplicand, by
5634, the multiplier; then to multiply 7 (tens), the second
digit of the multiplicand, by the multiplier ; &c. When the
multiplier was small, we could add the different products as
we proceeded; but we now require a separate addition, — which,
however, does not affect the nature, nor the reasons of the
process.
5-3. To jprove multiplication, when the multiplier ex-
ceeds 12 —
Rule. — Multiply the multiplier by the multiplicand ;
and the product ought to be the same as that of the
multiplicand by the multiplier [35] . It is evident, that
we could not avail ourselves of this mode of proof, in the
last rule *[42] ; as it would have supposed the pupil to
be then able to multiply by a quantity greater than 12
MULTIPLICATION. 67
54. We may prove rauitiplication by wliat is called
" casting out the nines."
PiULE. — Cast the nines from the sum of the digits of
the multiplicand and multiplier ; multiply the remain-
ders, and cast the nines from the product : — what is now
left shoukl be the same as what is obtained, by cast-
ing the nines, out of the sum of the digits of the product
of the multiplicand by the multiplier.
Example 1. — Lot the quantities multiplied be 942G and
3785.
Taking the nines from 9426, we get .3 as remainder.
And from 3785, we get 5.
47130
75408 3x5=15, from which 9
65982 being taken,
28278 • 6 are left.
Taking the nines from 35677410, 6 are lefc.
Tlie remainders being equal, we are to presume the
multiplication is correct. Tlie same result, however, would
have been obtained, even if we had misplaced digits, added
or omitted cyphers, or fallen into errors which had counter-
acted each other : — with ordinary care, however, none of
the.se is likely to occur.
Example 2. — Let the numbers be 76542 and 8436.
Taking the nines from 76542, the remainder is 6.
Taking them from 8436, it is 3.
459252
229626 6x3=18, the
306168 remainder from which is 0.
612336
Taking the nines from 645708312 also, the remainder is 0.
The remainders being the same, the multiplication may
be considered right.
Example 3. — Let the numbers be 463 and 54.
From 463, the remainder is 4.
From 54, it is 0.
1852 4x0=0 from which the remainder is 0.
2315
From 25002 the remainder is 0.
68 MULTIPLICATION.
Tlie remainder, being in each case 0, Tve are to suppose
that the multiplication is correctly performed.
This proof applies -whatever he the position of the
decimal point in either of the given numbers.
55. To understand this rule, it must be known that
"a number, from which 9 is taken as often as possible,
will leave the same remainder as will be obtained if 9
be taken as often as possible from the sum of its digits."
Since the pupU is not supposed, as yet, to have learned
division^ he cannot use that rule for the purpose of
casting out the nines ; — nevertheless, he can easily
effect this object.
Let the given number be 5C3. The sum of its digits is
5-I-G4-3, while the number itself is 500+60-}-3.
First, to take 9 as often as possible from the sum of its
digits. 5 and 6 are* 11; from which, 9 being taken, 2 are
left. 2 and 3 are 5, which, not containing 9, is to be set
down as the remainder.
Next, to take 9 as often as possible from the number itself-.
563_=500 + G0-f3=5xl00+Gxl0+3=5x9H^+Gx
9-j-l-]-3,= (if we remove the -vinculum [34]"), 5x99-|-5-f-
Gx9-j-6-f-3. But any number of nines, mil be found to be
tlie product of the same number of ones by 9 : — thus 999=
111x9: 99=11x9: and 9=1x9. Hence 5x99 expresses
a certain number of nines — being 5x11x9 ; it may, there-
fore, be cast out: and for a similar reason, 6x9: after which,
there will then be left 5-j-6-f-3 — from which the nines are
still to be rejected; but, as this is the sum of tlie digits, we
must, in casting the nines out of it, obtain the same remain-
der as before. Consequently "we get the same remainder
whether we cast the nines out of the number itself, or out
of the sum of its digits."'
Neither the above, nor the following reasoning can
oflfer any difficulty to the pupil who has made himself
as famiiiiu- with the use of the signs as he ought: —
they will both, on the contrary, serve to show how much
simplicity, is derived from the u^e of characters express-
ing, not only quantities, but processes ; for, by means
of such characters, a long series of argumentation ma^
be seen, as it were, at a single glance.
56. "Casting the nines from the f:\ctors, multiplying th(>
resulting remainders, and casting tlic nines from this product,
MULTIPLICATION. 69
will leave the same remainder, as if the nin^s were cast from
the product of the factors,'" — prowled the multiplicatiun
has been rightly performed.
To show this, set down the quantities, and take away the
nines, as before. Let the factors be 573x-l'J-l-
Casting the nines from 5-f-7-f-3 (which we have just seen
is the same as casting the nines from 573), we obtain 6 aa
remainder. Casting the nines fi-om 4-|-C-|-4, we get 5 aa
remaiiuler. Multiplying 6 and 5 we obtain 30 as product ;
which, being equal to 3x10=3x94-1=3x94-3, will, when
the nines are taken away, give 3 as remainder.
We can show that 3 will be the remainder, also, if we
cast the nines from the product of the factors : — which is
effected by setting down this product : and taking, in suc-
cession, quantities that are equal to it — as follows,
573x-lC-i: (the product of the factors) =
5xl'^04-7xi04^3 X 4x1004-6x104-4=
5 X 094-1 H-"X 04-1+3 X 4 X 994-14-6 x94-l-h4=
5x99+5-1-7x94-74-3 X 4x994-44-6x94-64-4.
5x99, as we have seen [55], expresses a number of nines;
it will continue to do so, when multipHed by all the quan-
tities under the second vinculimi. and is, therefore, to be
cast out; and. for the same reason, 7x9. 4x99 expresses
a number of nines ; it will continue to do so when multiplied
by the quantities under the first vinculum, and is, therefore,
to be cast out ; and, for the same reason, 6x9. There will
then be left, only 5 4-74-3x44-64-4, — from which the nines
are still to be (?ast out, the remainders to be multiplied together,
and the nines to be cast from their product ; — but we have
done all this already, and obtained 3. as the remainder.
EXERCISES FOR THE PUPIL.
Multiply
(37)
765
765
(38)
732
456
(39)
997
345
(40)
767
347
Products
Multiply
By
(41)
657
789
(42)
456
791
(48)
767
789
(44)
745
741
Products
70
MULTIPLICATIOX.
57. If there -are cyphers, or decimals in the multipli-
cand, multiplier, or both ; the same rules apply as when
the multiplier does not exceed 12 [43, &c.].
(1)
4600
57
(2)
2784
620
EXAMPLES.
(3) (4)
32-68 7856
26- 0-32
(5)
87-96
220-
(6)
482000
0-37
262200
1726080
849-68 2513-92
19351-2
178340
Contradioiis in Midtiplication.
58. When it is not necessary to have as many deci-
mal places in the product, as are in both multiplicand
and multiplier —
KuLE. — Reverse the multiplier, putting its units' _p^^ce
under the jplace of that denomination in the multipli-
cand, which is the lowest of the required product.
Multiply by each digit of the multiplier, beginning
with the denomination over it in the multiplicand ; but
adding what would have been obtained, on multiplying
the preceding digit of the multiplicand — unity, if the
number obtained would be between 5 and 15 ; 2, if
between 15 and 25 ; 3, if between 25 and 35 ; &c.
Let the lowest denominations of the products, arising
from the diflFerent digits of the multiplicand, stand in
the same vertical column.
Add up all the products for the total product ; from
which cut off the required number of decimal places.
59. Example 1.— Multiply 5-6784 by 9 7324, so as to
have four decimals in tlte product.
Short Method.
Ordinary Method
56784
5-67F t
42379
9/324
511056
22J7136
39740
1131568
1703
170352
113
39748;8
22
511056
55-2643
55'2644'GOia
MULTIPLICATION. 71
9 in tlie multiplier, expresses units ; it is therefore put
under the foiirtli decimal place of the multiplicand — that being
the place of the lowest decimal required in the product.
In multiplying by each succeeding digit of the multiplier,
we neglect an additional digit of the multiplicand ; because,
as the multiplier decreases, the number multiplied must in-
crease— to keep the lowest denomination of the different pro-
ducts, the same as the lowest denomination required in the
total product. In the example given, 7 (the second digit of
the multiplier) multiplied by 8 (the second digit of the mul-
tiplicand), will evidently produce the same denomination as 9
(one denomination higiier than the 7), multiplied by 4 (one
donouiination lower than the 8). Were we to multiply the
lowest denomination of the multiplicand by 7, we should get
[4i3] a result in the fifth place to the right of the decimal point ;
which is a denomination supposed to be, in the present in-
stance, too inconsiderable for notice — since we are to have
only foUr decimals in tlie product. But we add unity for
every ten that would arise, from the multipl cation of an addi-
tional digit of the multiplicand ; since every such ten consti-
tutes ane, in the lowest denomination of the required product.
When the multiplication of an additional digit of the multi-
plicand would give more than 5, and less than 15 ; it is nearer
to the truth, to suppose we have 10, than either 0, or 20 ; and
therefore it is more correct to»add 1, than either 0, or 2. When
it would give more than 15, and less than 25, it is nearer to
the truth to suppose we have 20, than either 10, or 30; and,
therefore it is more correct to add 2, than 1, or 3 ; &c. We
may consider 5 either as 0, or 10 ; 15 either as 10, or 20 ; &c.
On inspecting the results obtained by the abridged,
and ordinary methods, the difference is perceived to be
inconsiderable. When greater accuracy is desired, we
should proceed, as if we intended to have more decimals
in the product, and afterwards reject those which are
unnecessary.
Example 2. — Multiply 8-76532 by -5704, so as to hav*
3 decimal places.
8-76532
4075
4383
613
52
3
6-051
^
72 MULTIPLICATION.
There are no units in the multiplier; but, as the Yvle
directs, we put its units' place under tho. third decimal place
of the multiplicand. In multiplying by 4, since there is no
digit over it in the multiplicand, we merely set down what
would have resvilted from multiplying the preceding deno-
mination of the multiplicand.
Example 3.— Multiply -4737 by -6731 so as to have 6
decimal places in the product.
•47370
1376
284220
33159
1421
47
•318847
ifVe have put .he units' place of the multiplier under the
siith decimal place of the multiplicand, adding a cypher, or
BUjitposing it to be added.
Example 4.— Multii^ly 84G732 by •0050, sc as to have
four decimal places.
84-0732
65
4234
508
•4742
Example 5 .—Multiply -23257 by -243, so as to have four
decimal places.
2.^257
342
465
7 41b
•0505
^Ve are obliged to place a cypher in the product, to make
up the required niunbcr of decimals.
GO. To multiply by a Composite Number—
lluLC. — Multiply, Pucccissivoly, by its fiictors.
MULTIPLICATION. 73
Example.— Multiply 732 by 9G. 96 = 8x12- thcreforo
32x06 = 732x8x12. [35].
5856, product by 8.
70272, product by 8 x 12, or 96.
If we multiply by 8 only, we multiply by a quantity 12
times too small ; and, therefore, the product will be 12 times
los.s than it should. "We rectify this, by making the product
12 times greater — that is, we multiply it by 12.
61. V\'hen the multiplier is not exactly a Compo.sit^
Number —
Rule. — Multiply by the factors of the nearest com-
posite ; and add to, or subtract from the last product,
so many times the multiplicand, as the assumed compo-
site Is less or greater than the given multiplier
Example 1. — r^Iultiply 927 by 87.
87 = 7x124-3: therefore 927 X 87 = 027 X 7x12 + 3 =
927x7x12 + 927x3. [34].
027
7
6^9=927 X 7.
12
77868=927x7x12.
2781=927x3.
80649 = 927 X 7 x 12 + 927 x 3, or 927 x S7.
If we multiply only by 84 (7 X 12), we take the number to
be multiplied 3 times less than we ought ; this is rectiiied, by
abiding 3 times the multiplicand.
Example 2.— Multiply 432 by 79. 79 = 81-2 = 9 x 9-2;
therefore 432 X 79=432 X 9 X 9-2 = 432 x 9 x9-432x 2.
432
9
3888 = 432x9.
9
34J02 = 432x9x9.
»04 = 432x2.
34128 = 432 x 9 x 9-432 x 2, or 432 x 70.
/4 ilULTIPLICA/ION.
In multiplying by 81, the composite number, we have taken
the number to be multiplied twice too often ; but the inaccu-
racy is rectified by subtracting twice the multiplicand from
the product.
62. This method is particularly convenient, when the
multiplier consists of nines.
To Multiply by any Xumber of Nines, —
lluLE. — llemove the decimal point of the multipli-
cand so many places to the right (by adding cyphers if
necessary) as there are nines in the multiplier ; and
subtract the multiplicand from the result.
.Example.— ^Multiply 7347 by 999.
7347 X 999 = 7347000-7347 = 7339653.
We, in such a case, merely multiply by the next higher
convenient comj^osite number, and subtract the multiplicand
so many times as we have taken it too often ; thus, in the
example just given —
7347x999=7347x1000-1=7347000-7347=7339653.
63. We may sometimes abridge multiplication by
considering a part or parts of the multiplier as pro-
duced by multiplication of one or more other parts.
Example.— ]Multiply 57839208 by 62421648. The mid*
tiplier may be divided as follovrs : — 6, 24, 216, and 48.
6 = 6
24 = 6x4
216=24x9
48 = 24x2
57839268, multiplicand
62421648, multiplier.
347035608: : : product by 6 (60000000).
1388142432 : : product by 24 (2400000).
12493281888 : product by 216 (21600).
2776284864 product by 48.
3010422427673664 product by 62421648.
The product by 6 when multiplied by 4 will give the pro-
duct by 24 ; the product by 24, multiplied b}-^ 9, will give the
product by 216 — and, multiplied by 2, the product by 48.
64. There can be no difficulty in finding the places of
the first digits of the difi*ercnt products. For when tlicro
are neither cyphers nor decimals in the multiplicand —
and during multiplication, we may suppose that there aro
neither [48, &c.] — the lowest dcuoiaination of each pro-
MULTIPLICATION. 75
duct, will be the same as tlie lowest denomination of the
multiplier that produced it ; — thus 12 units multiplied
by 4 units will give 48 units ; 14 units multiplied by 4
tens will give 56 tens ; 124 units midtiplied by 35 units
will be 4340 units, &c. ; and, therefore, the beginning of
each product — if a significant figure — must stand under
the lowest digit of the multiplier from which it arises.
When the process is finished, cyphers or decimals, if
necessary, may be added, according to the rides already
given.
The vertical dotted lines show that the places of the lowest
digits of the respective multipHers. or those parts into which
the whole multiplier has been divided, and the lowest digits
of their resulting products are — as they ought to be — of the
same denomination.
48 being of the denomination units, when multiplied into
8 units, will produce units: the first digit, therefore, of the
prixluct by 48 is in the units' place. 216, being of the deno-
mination hundreds when multiplied into units will give hun-
dreds ; hence the first digit of the product by 216 will be in
the hundreds' place, &c. The parts into which the multi-
plier is divided are, in reahty,
60000000 ^
.^^21600 1^=62421648, the whole multiplier.
48]
"We shall give other contractions in multiplication
hereafter, at the proper time.
EXERCISES.
45. 745X4561^39720.
46. 476X767=365092.
47. 345X579=199755
48. 476X479=228004.
49. 897X979=878163.
50. 4 -59X705=3235 -95.
51. 767X407=312169.
52. -457 X- 606= -276942.
53. 700x810=567000.
60. 707X604=427028.
61. 777 X -407=316-239.
62. 7407X4404=32620428.
63. 5767X1307=7537469.
64. 67 -74X -1706=11 -556444
65. 4567X2002=9143134.
66. 7-767x301-2=2339-4204
67. 9600x7100=68160000.
68. 7800X9100=70980000.
54. 670X910=609700. ; 69. 6700x6700=44890000.
55. 910X870=791700.
56. 5001-4x70=350098.
57. 64 -001X40=2500 -04.
58. 91009X79=7189711.
70. 5000x7600=38000000.
71. 70-814x90l07=63808-37098.
72. 97001X76706=7440558706.
73. 95400X67407=6295813800.
59. 40170X80=3213600. I 74. -56007x45070=25242-35490
7f) MULTIPLICATION.
75. How many shillings in £1395 ; a pound being
20 shillings ? Ans. 27900.
76. In 2480 pence how many farthings ; four far-
things being a penny? ^7^. 9920.
77. If 17 oranges cost a shilling, how many can be
had for 87 shillings ? Ans. 1479.
78. How much will 245 tons of butter cost at £2d a
ton.? Ans. G125.
79. If a pound of any thing cost 4 pence, how much
will 112 pounds cost ? Ans. 448 pence.
80. How many pence in 100 pieces of coin, each of
which is worth 57 pence } Ans. 5700 pence.
81. How many gallons in 264 hogsheads, each con-
taining 63 gallons ? Ans. 16632.
82. If the interest of £1 be ^60-05, how much will
be the interest of £376 ? Ans. £18-8.
83. If one article cost £0-75, what will 973 such
cost.? Ans. £729 -Id.
84. It has been computed that the gold, silver, and
brass expended in building the temple of Solomon at
Jerusalem, amounted in value to £6904822500 of our
money ; how many pence are there in this sum, one
pound containing 240 ? Ans. 1657157400000.
85. The following are the lengths of a degree of the
meridian, in the following places : 60480-2 fathoms in
Peru ; 60486-6 in India ; 60759-4 in France ; 60836-6
in England ; and 60952-4 in Lapland. 6 feet being a
fathom, how many feet in each of the above .? Ans.
362881-2 in Peru; 362919-6 in India; 364556-4 in
France ; 365019-6 in England ; and 365714-4 in Lapland.
86. The width of the Menai bridge between thi5
points of suspension is 560 feet ; and the weight between
these two points 489 tons. 12 inches being a foot, and
2240 pounds a ton, how many inches in the former,
and pounds in the latter ?
Alls. 6720 inches, and 1095360 pounds.
87. There are two minims to a semibreve ; two
crotchets to a minim ; two quavers to a crotchet ; two
semiquavers to a quaver : and two dcmi-semiquavers to
a semiquaver : how many demi-semiquavcrs arc equal
to seven semibreves ? Ans. 224
MULTI PLICATION. 77
8S. 32,000 soc'Jis have been couuted in a single poppy ;
how many Avonld be found in 297 of these ? Ans. 9504000.
89. 9,344,000 eggs have been found in a single cod
fish ; how many would there be in 35 such .'
Ans. 327040000.
65. When the pupil is familiar with multiplication,
in working, for instance, the following example,
897351, multipHcand.
4, multiplier.
3589404, product.
He should say : — 4 (the product of 4 and 1), 20 (the pro-
duct of 4 and 5), 14 (the j^roduct of 4 and 3 plus 2, to be
carried), 29, 38, 35 ; at the same time putting do"wn
the units, and carrying the tens of each.
QUESTIONS TO BE ANSWERED BY THE PUPIL.
1 . What is multiplication } [24] .
2. What are the multiplicand, multiplier, and pro-
duct.' [24].
3. What are factors, and submultiples .' [24].
4. WTiat LS fhe difference between prime and compo-
site numbers [25] ; and between those which are prime
and those which are composite to each other ? [27] .
5. What is the measure, aliquot part, or submultiple
of a quantity } [26] .
6. What is a multiple .' [29].
7. What is a cmnmon measure } [27] .
8. What is meant by the greatest common measure .'
9. W hat is a c<?//i?7zo/i multiple .'' [30].
10. What is meant by the least common multiple ?
[30].
11. What are e^zamultiples : [31].
12. Does the use of the multiplication table prevent
multiplication from being a species of addition .' [33] .
13. Who first constructed this table t [33] .
14. What is the sign used for multiplication } [34] .
15. How are quantities imder the vinculum affect^?;
by the sign of multiplication } [34] .
16. Show that qiiantities connected by the sign o/
multipHcation may be read in any order t [35] .
78 DITI.SION.
17. Wliat is tlio rulo for multiplicatiou, when neither
multiplicand nor niultiplier exceeds 12 ? [37J.
18. AVhat is the rule, when only the multiplicand
exceeds 12? [39].
19. AVhat is the rule when both multiplicand and
multiplier exceed 12 t [50].
20. What are the rules when the multiplicand, mul-
tiplier, or both, contain cyphers, or decimals ? [43, &c.] :
and what are the reasons of these, and the preceding
rules.? [41,43, &c., 52].
21. How is multiplication proved .? [42 and 53].
22. Explain the method of proving multiplication,
by " casting out the nines [54] ;" and show that we can
co.st the nines out of any number, without supposing a
knowledge of division. [55] .
23. How do we multiply so as to have a required
number of decimal places .'' [58].
24. How do we multiply hy a composite number [60] ;
or by one that is a little more, or less than a composite
number ? [61].
25. How may we multiply by any number of nines t
[62].
26. How is multiplication very briefly performed r
[65].
SIMPLE DIVISION.
66. Simple Division is the division of abstract num-
bers, or of those which are applicate, but contain only
one denomination.
Di\dsion enables us to find out how often one number,
called the divisor ^ is contained in, or can be taken from
another, termed the dividend ; — the number expressing
how often is called the quotient. Division also enables
us to tell, if a quantity be divided into a certain number
of equal parts, what will be the amount of each.
When the divisor is not contained in the dividend
any number of tunes exactly, a quantity, called the
remainder, is left after the division.
67. It will help us to understand how greatly di^q-
sion abbreviates subtraction, if we consider how long a
process would be required to discover — by actually sub-
tracting it — liow often 7 Is contained in 8563495724,
wliile, a^ we shall find, the same thing can be effected
by dicisityn^ in less than a minute.
C)S. Division is expressed by -f-, placed between the
dividend and divisor; or by putting the divisor undf.H-
th J dividend, with a separating line between : — thus
6-h3=:2, or— =2 (read 6 divided by 3 is equal to 2)
means, that if 6 is divided by 3, the quotient will be 2.
69. "When a quantity under the vinculum is to bo
didded, we must, on removing the vinculum, put the
divisor under each of the terms connected by the sign
of addition, or subtraction, otherwise the value of what
was to be di\'ided will be changed; — thus 5-1-6 — 7h-3=
'— -f- — — ; for we do not divide the whole unless
ij o o
we divide all its parts.
The line placed between the dividend and divisor occa-
sionally assumes the place of a vinculum ; and there-
fore, when the quantity to be divided is subtractive, it
will sometimes be necessary to change the signs — as
already directed [16] : — thus — -! ~^^ 9 ''
27 15— 6-f9 27—15 + 6—0 ,,
but — = . lor when, as
o S o
in tliese cases, all the terms are put under the vinculum,
the effect — as far as the subtractive signs are concerned —
is the same as if the vinculum were removed altogether ;
and then the signs should be changed hack again to
what they must be considered to have been before the
vinculum was a5i:^ed [16].
When quantities connected by the sign of multiplica-
tion are to be divided, dividing any one of the factors,
will be the same as dividing the product ; thus, 5X lOX
25-i-5= - - X 10X25 ; for each is equal to 250.
5
To Divide QuaiUiiies.
70. "When the di\'isor does not exceed 12, nor the
dindond 12 times the divisor
80 DIVISION.
BuLE. — I. Find by the multiplication table tlic
greatest number wliich, multiplied by the divisor, will
give a product that does not exceed the dividend : this
vnll be the quotient requu-ed.
II. Subtract from the dividend the product of this
number and the divisor ; setting down the remainder, if
any, with the divisor under it, and a Line between them.
EsL\3iPLE. — Find how often 6 is contained in 58; or, in
other words, what is the quotient of 58 divided by G.
We learn from the multiplication table that 10 times 6
are 60. But 60 is greater than 58 ; the latter, therefore, does
not contain 6 10 times. We find, l)y the same table, that 9
times 6 are 54, which is less than 58 :— consequently 6 is con-
tained 9, but not 10 times in 58 : hence 9 is the quotient ;
and 4 — the diiFerence between 9 times 6 and the given num-
ber— is the remainder.
4 4 58 4
The total quotient is 9-f-77, or 9 ; that is, -^z=9rT.
If we desire to carry the division farther, we can effect it
by a method to be explained presently.
71. Reason- of I. — Our object is to find the greatest num-
ber of times the divisor can be taken from the dividend ; that
is, the greatest multiple of 6 which will not exceed the num-
her to be tlivided. The mnltiplicatiou table sho^vs the pro-
ducts of any two numbers, neither of which exceeds 12 ; and
therefore it enables us to obtain the product we require ; this
must not exceed the dividend, nor, being subtracted from it,
leave a number equal to, or greater than, the divisor. It is
hardly necessary to remark, that the divisor would not have
been subtracted as often as possible from the dividend if a
number equal to or greater than it were left ; nor would the
quotient answer the question, hotv ofttn the divisor could be
taken from the dividend.
Reason- of II. — We subtract the product of the divi.sor
and quotient from the dividend, to learn, if there be any
remainder, what it is. Wlien there is a remainder, we in
reality suppose the dividend divided into two parts ; one of
these is equal to the product of the divisor and quotient — and
this we actually divide ; the other is the difference between
that product and the given dividend — this we express, by the
notation alreadv explained, as still to he divided, la the ex.am.-
. 58 '5J-f4 54 ,4 ^ , 4
pie given, -=-^=-+^=9+-
72. When the divisor docs not exceed 12, buV the
dividend exceeds 12 times the divisor —
DIVISION. 81
KuLE. — I. Set down the dividend with a line under
it to separate it from the futui-e quotient : and put the
divisor to the left hand side of the dividend, with a line
between them.
II. Divide the divisor into all the denominations of
the dividend, beginning with the highest.
III. Put the resulting quotients under those deno-
minations of the dividend which produced them.
IV. If there be a remainder, after subtracting the pro-
duct of the divisor and any denomination of the quotient
from the corresponding denomination of the dividend,
consider it ten time.s as many of the next lower deno-
mination, and ;idd to it the next digit of the dividend.
V. If any denomination of the dividend (the preced-
ing remainder, when there is one, included) does not
contain the divisor, consider it ten times as many of
the next lower, and add to it the next digit of the
dividend— putting a cypher in the quotient, under the
digit of the dividend thus reduced to a lower denomi-
nation, unless there are no significant figures in the
quotient at the same side of, and farther removed fioiu
the decimal point.
VI. If there be a remainder, after dividing tlie
" units of comparison," set it down — as already directed
[70] — ^with the divisor under it, and a separating line
between them ; or, writing the decimal point in the
C[uotient, proceed with the division, and consider each
\emainder ten times as many of the next lower deno-
mination ; proceed thus until there is no remainder, or
imtil it is so trifling that it may be neglected without
inconvenience.
73. Ex.A3iPLE. — V\'hat is the quotient of 04450-^-7 I
Divisor 7)G445G dividend.
9208 quotient.
C tens of thousands do not contain 7, even oiice ten thou-
sand times ; for ten thousand times 7 are 70 thuusand, which
is greater tlran GO thousand ; there is. therefore, no digit
t<) he put in the teii-tliousands' place of the quotient^ — we
do not, however, put a eyp"icr in that place, since no digit
82 DIVISION.
of the quotient can be further removed from the decimal
point than this C3'pher : for it would, in such a case, prodiice
no effect [Sec. 1. 28]. Considering the G tons of thousands
as 00 thousands, and adding to these the 4 tliousands already
in the dividend, we have 64 tliousands. 7 Anil " go"' into
(that is, 7 can be taken from) 64 thousand, 9 thousand times ;
for 7 times 9 thousand are 03 thousand — which is less tlian
64 thousand, and therefore is not too large ; it does not leave
a remainder equal to tlie divisor — and therefore it is not too
small : — 9 is to be set down in the thousands' place of the
quotient; and the 4 already in the dividend being added to
one thousand (the dilference between 64 and 03 thousand)
considered as ten times so many Irandreds, we have 14 hun-
dreds. 7 will go 2 hundred times into 14 liundreds, and leave
no remainder ; for 7 times 2 liundreds are exactly 14 hun-
dreds : — 2 is, therefore, to be put in tiie hundreds' place of
the quotient, and there is nothing to be carried. 7 will not
go into 5 tens, even once ten times: since 10 times 7 are 7
tens, which is more than 5 tens. But considering the 5
tens as 50 units, and adding to them the other 0 units of the
dividend, we have 56 units. 7 will go into 56, 8 times, leav-
ing no remainder. Ad the 5 tens gave no digit in tiie tens'
place of the quotient, and there are significant figures farther
removed from the decimal point than this denomination of
the dividend, we have been obliged to use a cypher. The
division being finished, and no rem.ainder left, the required
quotient is found to be 9208 exactly ; that is, — y — ==^208.
74. Example 2.— What is the quotient of 73208, divided
by 6?
6)73268
We may set down the 2 units, which remain after the
units of the quotient are found, as represented ; or we may
proceed with the division as follows —
6)73208
12211-333, &c.
Considering the 2 units, left from the units of the divi-
dend, as 20 tenths, we perceive that 6 will go into them
three tenths times, and leave 2 tonths — since 3 tenths times
6 (=0 times 3 tenths [35]) are lii tenths: — Ave put 3 in the
tenths' place of the quotient, and consider the 2 tenths re-
maining, as 20 hundredths. For similar reasons, 0 win go
into 20 hundredths 3 huadredths times, and leave 2 lum-
DIVISION'. 83
(Irodtlis. Considering these 2 hundredths as 20 thousandtli?,
they win give 3 thousandths as quotient, and 2 thousandths
as remainder, &c. The same remainder, constantly recur-
ring, ATill evidently produce the same digit in the successive
denominations of the quotient ; ■we may, therefore, at once
put down in the quotient as many threes as will leave the
linal remainder so small, that it may he neglected.
75. Example 3. — Divide 473G5 by 12.
12)47365
• 3947-0«, &c.
In this example, the one unit left (after obtaining the 7 in
the quotient) even when considered as 10 tenths, does not
contain 12 : — there is, therefore, nothing to he set down in
the tenths' place of the quotient — except a cypher, to keep
the following digits in their proper places. The 10 tenths
are by consequence to be considered as 100 liimdredths ,
12 will go into 100 hundredths 8 hundredths times, &c.
This may be applied to the last rule [70], when we desire
to continue the division.
Example. — Divide 8 by 5.
8-^5 = 1^, or 1-37, ko.
76. When the pupil fully understands the real deno-
minations of the dividend and quotientj he may proceed,
for example, with the following
5)46325
In this manner : — 5 will not go into 4. 5 into 46, 9 times
and 1 over (the 46 being of the denomination to which 6
belongs [thousands], the first digit of the quotient is to be
put under the 6 — tiiat is. under the denomination which
produced it). 5 into 13, twice and 3 over. 5 into 32, 6
times and 2 over. 5 into 25, 5 times and no remainder.
When the divisor does not exceed 12, the process is
called short (Hvision.
77. Reasox of I. — In this arrangement of the qitantities —
■which is merely a matcer of couvenieuce — the values of the
digits of the quotient are ascertained, -both by their position
•with reference to the digits of the dividend, and to their own
decimal point. The separating lines prevent the dividend,
divisor, or quotient from being in any way mistaken.
Reason of II. — Vir'e divide the divisor successively into all
the parts of the dividend, because we cannot divide it at once
into the whole : — the sum of the numbers of times it can be
subtracted from these parts is evidently equal to the number
y4 DIVL^iON,
of times it c.in be subtracted from tlieir sum. Thus, if 5 goes
into Goo. 100 times, into 50, 10 times, and into 5, once; it
Tvill go into 500-1-50-1-5 (=555), 100-f-lO-J-l (=111) times.
Tlie pupil perceives by the examples given above, that, ia
dividing the divisor successively into the parts of the dividend,
each, or any of these parts does not necessarily consist of one
or more digits of the dividend. Thus, in finding, for example,
the quotient 61458-J-7, we are not obliged to consider the parts
as 60000, 4000, 400, 50, and G :— on the contrary, to render the
dividend suited to the process of division, we alter its form,
while, at the same time, we leave its value unchanged; it be-
comes
Thousands. Hundreds. Tens. Units.
63 -f 14 -f 0 -f- 50 (=64456).
Each part being divided by 7, the different portions of the
dividend, with their respective quotients, will be,
Thnutjands. Hundreds. Teas. Units.
7i63 14 0 56 = 64466.
9 2 0 8 = 9208.
. "We begin at the left hand side, because what remains of the
higher denomination, may still give a quotient in a lower ;
and the question is, how often the divisor will go into the
dividend — its different denominatix)ns being taken in any con-
venient way. We cannot know how many of the higher we
ehall have to add to the lower denominations, unless we begin
with the higher.
Reaso.v of III. — Each digit of the quotient is put under
that denomination of the dividend which produced it, because
it belongs to that denomination ; for it expresses what number
of times (indicated by a digit of that denomination) the divisor
can be taken from the corresponding part of the dividend : —
thus the tens of the quotient express how many tens of times
the divisor can be taken from the tens of the dividend : the
kundreds of the quotient, how many hundreds of times it can
be taken from the hundreds, &c.
Reason of IV. — Since what is left belongs to the total re-
mainder, it must be added to it ; but uuless considered as of a
lower denomination, it will give nothing farther in the quotient.
PvEASox OF V. — We are to look upon the remainder as of
the highest denomination capable of giving a quotient; and
though it may not contain the divisor a number of times ex-
presse^i. by a digit of one denomination, it may contain it some
number of times expressed by one that is lower.
The true remainder, after subtracting each product, is the
%'hole remainder of the dividen<l; but we " bring down" only
gii muclj of it as is necessary for our present object. Thus, in
iooking for a digit in the hundreds' jjlace of the quotient, it
will not be necessary to take into account the lens, or units
of the dividend ; since they cannot add to the number tif luiw
drcds of times the divisor may be taken from the dividend.
Divisio:^. 85
A cypher mnst he added [Sec. I. 28], -when it is required,
to give significant figures their proper value — which is never
the cxae, except it comes between them and the decimal point.
Reaso.v of VI. — We may continue tlie process of division,
if we please, as long as it is posible to obtain quotients of a/i,y
denoDiinatioa. Quotients will be produced although there are
no longer any significant figures in the dividend, to which we
can add the successive retuairiders.
78. The smaller the divisor the larger the quotient—
for, the smaller the parts of a given quantity, the greater
their number will be ^ but 0 Is the least possible divi-
sor, and therefore any quantity divided by 0 will give the
largest possible quotient — which is injinitij. Hence,
though any quantity multiplied by 0 is equal to 0, any
number diWded by 0 is equal to an infinite number.
It appears strange, but yet it is true, that-=- ; for
each is equal to tlie grmlcst possible number, and one,
therefore, cannot be greater than another — the appa-
rent contradiction arises from our being unable to form
a true conception of an infinite quantity. It is neees.sary
to bear in mind also that 0, in this case, indicates a
quantity infinitely small, rather than a'nsolutely nothing.
7T). To 'prove Division, — Multiply the quotient by
the divisor ; the product shoidd be equal to the divi-
dend, minus the remainder, if there is one.
For, the dividend, exclusive of the remainder, contains the
divisor a number of times indicated by the quotient ; if, there-
fore, the divisor, is taken that number of times, a quantity
equal to the dividend, minus the remainder, will be producetL
It follows, that adding the remainder to the product of the
divisor and quotient should eive the dividend.
"6832
Example 1. — Prove that — r— =1708.
4
4)6H32 ruGOF. 1708, (quotient.
1708 -^ d\\-i&jx.
C832, product of divi-
For and quotient, equal to the dividend.
Example 2. — Prove t!iat " _ ''t^ 12234 i.
I 7
Proof. Proof.
1223i or 122^4
_I ^
8o638='l;yWeml n.inu. 5, the tcinaia V. 85G38-|-6=di»!.|cnd
86 DIVISION.
EXERCISES.
(1) (2) (3) (4)
2)78345 8)91234 3)67859 9)71234
(5) (6) (7) (8)
4)96707 10)134567 5)767456 11)37067
(9) (10) (11) (12)
6)970763 12)876967 7)891023 9)763457
80. When the dividend^ divisor^ or hoth contran
cyphers or decimals. — The rules already given are appli •
cable : those which follow are consequences of them.
When the dividend contains cyphers —
Rule. — Divide as if there were none, and remove
the quotient so many places to the left as there havo
been cyphers neglected.
The greater the dividend, the greater ought to be tlie
quotient; since it expresses the number of times the divisor
can be subtracted from the dividend. Hence, if 8 will go into
56 7 times, it will go into 5600 (a quantity 100 times greater
than 56) 100 times more than 7 times — or 700 times.
Example 1.— What is the quotient of 568000-i-4 T
^-f=U2:therefore»=142000.
4 4
Example 2.— What is the quotient of 40G0000-f-5 ?
^—81-2; therefore ^'^^^^'^'^=812000 [Sec. I. 39.].
81. When the divisor contains cyphers —
Rule. — Divide as if there were none, and move the
quotient so many places to the right as there are cyphers
in the divisor.
The greater the divisor, the smaller the number of times it
can be subtracted from the dividend. If, for example, 6 can
be taken from a quantity any number of times, 100 times 6
can be taken from it 100 times less often.
56
Example. — What is the quotient of kkr ?
L^=7i therefore ^=-07.
DIVISION. S7
82. If both diviJend and divisor contain cj'pber.s —
Rule. — DiHide as if there were none, and move the
quotient a number of places equal to the difierence
between the numbers of cyphers in the two given quan-
tities : — if the cyphers in the dividend exceed those in
the divisor, move to the left ; if the cyphers in the
divisor exceed those in the dividend, move to the right.
We have seen tliat the effect of cyphers in the dividend is
to move the quotient to the left and of cyphers in the divisor,
to move it to the right ; when, therefore, both causes act
together, their effect must be erjual to the difference between
their separate effects.
E5:amples.
(1)
7)63
"9
(2)
7)6300
900
(3) (4)
70)63 70)6300
0-9 90
(5)
700)630
0-9
(6)
700)6300
9
In the sixth example, the diffei-ence between the numbers
of cyphers being = 0, the quotient is moved neither to the
right nor the left.
83. K there are decimals in the dividend —
Rule. — Divide as if there v/ere none, and move the
quotient so many places to the right as there are deci-
mals.
The smaller the dividend, the less the quotient.
ExAiiPLi:. — What is the quotient of -048-7-8 ?
48. ,. . -048 __
■^=6, therefore -Tr-=- 005.
84. If there are decimals in the divisor —
Rule. — Divide as if there were none, and move the
quotient so many places to the left as there are deci-
mals.
The smaller the divisor, the greater the .quotient.
Ex^vMPLE. — What is the quotient of 54-h-006 "?
54 54
-7r=9. therefore-7r7^=9000.
G ■ -006
85. If there are decimals in both dividend and di-
visor—
Rule. — Divide as i^ there were none, and move the
quotient a number of places equal to the diiferenco
88 DivrsfoN.
between the numbers of decimals in the two given quan-
tities : — if the decimals in the dividend exceed those in
the divisor, move to the right ; if the decimals in the
divisor exceed those in the dividend, move to the left.
We have seen that decimals in the dividend move the
quotient to the right, and that decimals in the divisor move
it to the left ; when, therefore, both causes act together, the
effect must be equal to the difference between their separate
effects.
Examples.
(1)
5)45
9
(2)
5) -45
•09
(3) (4)
•05)45 •5)-045
900 -09
(5)
•005) 450
90000
(6)
•05) -45
9-00
86. If there are cyphers in the dividend, and deci-
mals in the divisor —
lluLE. — Divide as if there wore neither, and move
the quotient a number of places to the left, equal to
the number of both cyphers and decimals.
Both the cyphers in the dividend, and the decimals in the
divisor increase the quotient.
Example. — What is the quotient of 270-f-'03 :
?Z=9, therefore, 270^-03 = 9000.
3
87. If there are decimals in the dividend, and cyphers
in the divisor —
Rule. — Divide as if there were neither, and move
the quotient a number of places to the right equal to
the number of both cyi^hers and decimals.
Both the decimals in the dividend, and the cyphers in tlio
divisor diminish the quotient.
Example. — What 'is the quotient of -18-^20 T
1?=9, therefore ■-=•009.
2 20
The rules which relate to the management of cyphers
and decimals, in multiplication and in division — though
niunorous — will be very easily remembered, if the pupil
merely considers what ov^ht to bo the cHuvt vt' cither
DIVISIOX.
89
EXERCISES.
(13) (14) (15) (16)
8)10000 IDIGOOO 3)70170 0)68530
(17)
20)36526
(18)
3000)47865
(1^)
40)56020
(20)
80)75686
(21)
12)63-075
(22) (23) (24)
10) -08756 -07)54268 •09)57-368
(25)
•0005)60300
(20) (27) (28)
700) -03576 -008)57-362 400)63700
(29)
110)97-634
88. When tlie divisor exceeds 12 —
The process used is called Icrng division ; that is, we
perform the multiplications, subtractions, &c., in full,
and not, as before, merely in the mind. This will be
undei-stood better, by applying the method of long divi-
sion to an example in which — the divisor not being
gi- ater than 12 — it is unnecessary.
Short Division
8)5763472
720434
the same by Lon^ Division.
8)5763472(720404
56
16
10
27
24
In the second method, we multiply the divisor by the
different parts of the quotient, and in each case set down
90 DIVISION'
the product, subtract it from tlie correspondm^ portion of
the dividend, u'rite the remainder, and bring dovm the re-
quiired digits of the dividend. All this must be done when
the divisor becomes large, or the memory would be too
heavily burdened.
89. Rule — I. Put the divisor to the left of the divi-
dend, with a separating line.
II. Mark oif, by a separating line, a place for the
quotient, to the right of the dividend.
III. Find the smallest number of digits at the left
hand side of the dividend, which expresses a quantity
not less than the divisor.
IV. Put under these, and subtract from them, the
greatest multiple of the divisor which they contain ;
and set down, underneath, the remainder, if there is
any. The digit by which we have multiplied the divisor
is to be placed in the quotient.
y. To the remainder just mentioned add, or, as it is
said, '' bi'ing down" so many of the next digits (or
cyphers, as the case may be) of the dividend, as are
reqmred to make a quantity not less than the divisor ;
and for every digit or cypher of the dividend thus
brought down, except one^ add a cypher after the digit
last placed in the quotient.
VI. Find out, and set down in the quotient, the
mimlcr of times the divisor is contained in this quan-
tity ; and then subtract from the latter the product of
the divisor and the digit of the quotient just set down.
Proceed with the resulting remainder, and with all that
succeed, as with the last.
VII. If there is a remainder, after the units of the
dividend have been " brought down" and divided, either
place it into the quotient with the divisor under it, and
a separating line between them [70] ; or, putting the
decimal point in the quotient — and adding to the re-
mainder as many cyphers as will make it at least equal
to the divisor, and to the quotient as many cyphers
minus one as there have been cyphers added to the
remainder — proceed with the division.
DIVISION." 91
00. Example 1.— Di\dde 7832582G by 82.
82)78325826(955103
7o»
7G2
738
240
24G
82 will not go into 7 ; nor into 78 : Ijut it vrill go 0 times
into 783 : — 9 is to be put in the quotient.
The values of the higher denominations in the quotient
ynW be sufficiently marked by the digits which succeed
them — it will, liowever, sometimes be proper to ascertain,
if the pupil, as he proceeds, is acquainted with the orders
of units to which they belong.
9 times 82 are 738. "which, being put under 783. and sub-
tracted from it. leaves 45 as remainder : since this is less than
the divisor, the digit put into the quotient is — as it ought to
be [71] — the largest possible. 2, the next digit of the divi-
dend, being brought down, we have 452, into which 83 goes
5 times : — 5 being put in the quotient, we subtract 5 times
the divisor from "452, which leaves 42 as remainder. 42,
with 5, the next digit of the dividend, makes 425. into which
82 goes 5 times, leaving 15 as remainder : — we put another
5 in the quotient. The last remainder, 15, with 8 the next
digit of the dividend, makes 158, into which 82 goes once,
leaving 76 as remainder ; — 1 is to be put in the quotient. 2,
the next digit of the dividend, along with 76. makes 762,
into which the di^-isor goes 9 times, and leaves 24 as remain-
der ; — 9 is to be put in the quotient. The next digit being
brought down, we have 246, into which 82 goes 3 times
exactly: — 3 is to be put in the quotient. This 3 indicates
3 units, as the last digit brought do^vn expressed units-.
Therefore 1.5^^=955193.
02 DIVISION.
Example 2.— Divide G4212S4 by G42.
G42)G421284(10002
G42
1284
1284
G42 goes once into G42; and leaves no remainder. Brinj;-
ing down the next digit of the dividend gives no digit in
the quotient, in which, therefore, we put a cypb.er after tlie
1. I'he next digit of the dividend, in the same way, gives
no digit in the quotient, in which, consequently, we put
another cypher : and. for similar reasons, another in Lringing
down the next; hut the next digit makes the quantity
])rought down 1284, which contains the divisor twice, and
gives no remainder : — we put 2 in the quotient.
91. When there is a remainder, we may continvie the
diTision, adding decimal i^daces to the quotient, as follows —
Example 3. — Divide 79G347 by 847.
847)790347 (94019, &c.
7C23
3404
3388
1670
847
8230
7G23
92. The learner, after a little practice, will guess
pretty accurately what, in each case, should be the next
digit of the quotient. He has only to multiply iu his mind
the last digit of the divisor, adding to the product what
he would probably have to carry from the multiplica-
tion of the second last : — if this sum can be taken from
the corresponding part of what is to be the minuend,
leaving little, or nothing, the assumed number is likely
to answer for the next digit of the quotient.
93. Rkapoi^ of I. — This arrangement is merely a matter of
convenience; some put the divisor to the riglit of the dividend,
and immediately over the quotient — believing Miat it is more
corrvenient to have two quantities which are to be multiplied
together as near to each other as possible. Thus, in dividing
6125 by 54 —
DIVISIO^^ 93
6425 • 54
^ Ul8, &c.
102
54_
485
432
53, &c
Rkasoni of II. — This, also, is only a matter of convenience
Reason of III. — A smaller part of the dividend -would give
no digit in the quotient, and a larger would give more than
one.
Reason- of IV. — Since the numbers to be multiplied, and
the products to be subtracted, are considerable, it is not so
convenient as in short division, to perform the multiplications
and subtractions mentally. The rule directs us to set do-wu
each multiplier in the quotient, because the latter is the sum
of the multipliers.
Reason of V. — One digit of the dividend brought do-«*n
■would make the quantity to be divided one denomination lower
than the preceding, and the resulting digit of the quotient
also one denomination lower. But if -we are obliged to bring
down two digits, the quantity to be divided is tico denomi-
nations lower, and consequently the resulting digit of the quo-
tient is two denominations lower than the preceding — which,
from the principles of notation [Sec. I. 28], is expressed by
using a cypher. In the same way, bi'inging down three
figures of the dividend reduces the denomination three places,
and makes the new digit of the quotient three denominations
lower than the last — two cyphers must then be used. The
same reasoning holds for any number of characters, whether
significant or otherwise, brought down to any remainder.
Reason of VI. — We subtract the products of the different
parts of the quotient and the divisor (these different parts of
the quotient being put down successively according as they
are found), that we may discover what the remainder is from
which we are to expect the next portion of the quotient. From
what we have already said [77], it is evident that, if there are
no decimals in the divisor, the quotient figure will always be
of the same denomination as the lowest in the quantity from
which we subtract the product of it and the divisor.
Reason of VII. — The reason of this is the same as what
was given for the sixth part of the preceding rule [77].
It is proper to put a dot over each digit of the divi-
dend, as we bring it down ; this will prevent our forget-
ting any one, or bringing it down twice.
94. TVTien there are cyphers, decimals, or both, tho
rules already given [8), &,c.] are applicable.
94 DIVISION.
95. To prove the. Division. — Multiply the quotient
by the divisor ; the product should be equal to the divi-
dend, minus the remainder, if tliere is any [79] .
To prove it by the method of " casting, out the
nines" —
Rule. — Cast the nines out of the divisor, and the
quotient ; multiply the remainders, and cast thp nines
from their product : — that which is now left ought to
be the same as what is obtiiiiied by casting the nines
out of the dividend minus the remainder obtained from
the process of division.
Example. — Prove that —^^=1181/4.
Considered as a' question in multiplication, this -becomes
1181 X 54 = 63776—2 = 63774. To try if this be true,
Castino; the nines from 1181, the remainder is 2. ) 9 ^ A_n
„ „ from 54, „ isO. j-^^^-^'
Casting the nines from 63774, the remainder is . .0
The two remainders are equal both being 0 : hence the
multipUcation is to be presumed right, and, consequently,
the process of division \Yhich supposes it.
The division involves an example of multiplication ; since
the product of the divisor and quotient ought to be equal_ to
the dividend minus the remainder [7'J]. Hence, in proving
the multiplication (supposed), as already explained [54], we
indirectly prove the division.
(30)
24)7654
318ff
EXERCIS
(31)
15)6783
452,^,
ES.
(32) .
16)5674
3541^
(33)
-17)4675
275
(34)
18)7831
435,',
(35)
19)5977
314f^
(36)
21)6783
323
1400)6
(37)
22)9767
^3fi
(38)
23)707500
(39)
300)5807
(40) ,
767600
33360i|
14-8807
4635-3425
(41)
250)77676700
303424 -G094
DIVISION.
(42)
67-1) -1342
-002
(48)
•153) -829749
5-4232
(44)
64-25)123-70536
2-2803
(45)
14-35)269-0625
18-75
(46)
•0037) 555
150000
In example 40 — and some of those which follow — after
obtaining as many decimal places in the quotient as ai'c
deemed necessary, it will be more accurate to consider the
remainder as equal to the divisor (since it is more than one
half of it), and add imity to the last digit of the quotient.
CONTRACTIONS IN DIVISION.
96. We may abbreviate the process of division when
there are many dechnals, by cutting oflF a digit to the
right hand of the divisor, at eacli new digit of the
quotient; remembering to carry what would have been
obtained by the multii)lication of the figure neglected —
unity if this multiplication would have produced more
than 5, or less than 15 ; 2 if more than 15, or less than
25, &c. [59].
Example.— Divide 754-3.37385 by 61-347.
Ordinary Method. Contracted Method.
61-3n)754-33 7385(12-296 61-347)754- 337385(12-296
G1347 61347
14086 7 14086
12269 4 " 12260
181733 1817
1226 94 1227
590 398 590
552 123 552
38 2755 38
36,8082 37
146730
96 L4 VI. Si ON.
According ns the denominations of the quotient become
small, their products by the lower denomination of the divisor
become inconsiderable, and may be neglected, and, conse-
quently, the portions of the dividend from which they would
liave been subtracted. "What should h^ve been carrifd ironi
the multiplication of the digit neglected — since it belongs to a
higher denomination than what is neglected, should still be
retained [59].
97. "We may avail oui'selves, in division, of contri-
vances very similar to those used in multiplication
[60].
To divide by a composite number —
Rule. — Divide successively by its factors.
Example. -Divide 98 by 49. 49=7x7.
7)98
7)14
2=98-^7x7, or 49.
y
Dividing only by 7 we divide by a quantity 7 times too
small, for we are to divide by 7 times 7 ; the result is, therefore,
7 times too great : — this is correct^egl if we divide again by 7
98. If the divisor is not a composite number, we
cannot, as in multiplication, a,bbreviate' the process,
except it is a quantity whicli is but little less than a
number expressed by unity and one or more cyphers
When this is the case —
Rule. — Divide by the nearest higher number, ex-
pressed by unity and one or more cyphers ; add to re-
mainder so many times tlie quotient as the assumed
exceeds the given divisor, and divide the sum by the
preceding divisor. Proceed thus, adding to the remain-
der in each case so many times tlie foregoing quotient
as the assumed exceeds the given divisor until the exact,
or a sufficiently near approximation to the exact quotient
is obtained — the last divisor must be the given, and not
the as.sumed one. The last remainder will be the true
one ; and the sum of all the quotients will be the true
quotient
DIVISION. 97*
EiAMpy.E. -Divide 9876G3425 by 908.
087CG3.v425=087GC3425-^-100U.
1'J75a751=967G03x 2-1-420 H- 1000.
4,701=fcl97ox2Hh7oI-hl000-
0- 7.v090=4 x2-f-701 -^ lOOD.
001a040=-7x24-'J^1000.
0-000^420=Ulx2-f--4-^1000. .^' -:
00004.v0208=-01x24--4-j-998
tliat irs, th© last quotient is 0-0004, and -0203 is the last
remainder.
9876G3
1975
all the quotients are \ .-..-
1 0-01
I 0-0004 • .- ,
The true quotient is 089642-7104, or the sum of the quotients.
And the true remainder 00208, or the last remaindej^ .
Unless Tve add twice the preceding quotient to each succes-
Bive remainder, we shall have subtracted from the lUvideml,
or the part of it j.u.st divided, 1000, and not 998 times the
quotient — in T.'hich case the remainder would be too small to
the amount of twice the quotient. — W§ have used (a) to sepa-
rate the quotients from the remainders.
There can be no difficulty when the learner, by this
process, comes to the decimals of the quotient. Thus in
the third line, 4701 gives, when divided by 1000, 4 units
as quotient, and 701 units stiil to he divided — that is,
701 as remainder. 4*701 would expre.ss 4701 actually
divided by 1000. A number occupying four places, aU
to the left of the decimal point, when divided by 1000,
gives units as quotient ; but if, as in 709 '0 (in the next
line), one is a decimal place, the quotient must be of
a lower denomination than before — that is, of the order
tenths ; and in 01 0*40 (next line), since two out of the
four places are decim-als, the quotient mast be hun-
dredths, Sec.
In adding the necessary quantities, we must carefully
bear in mind to what denominations the quotient multi-
plied, and the remainder to which the product is to be
added, belonir
9S DIVISION.
EXERCISES.
47. 56789-^741=76i|f
48. 478907 -^971=493|0f
49. 977076 -^47600=20f4I|.
50. 567897 -f-842=G74f|i'.
51. 7867674-f-971 2=810 J^W
52. 307O7U0h-467000=G'7193.
53. 6765158-4-7894=857.
54. 67470-f-3900=17-3.
55. 69000-^47600=l•4496.
56. 767674-40700=1-8862.
57. 6114592-^764324=8.
58. 9676744-^910076=10-6329.
69. 740070000^741000=998-7449.
60. 9410607111-f-45673=206043-1132.
V 61. 454076000-^400100=1134-9063.
62.- 7376476767-^345670=21339 -649.
68. 47>6782975-^26•175=l-8177.
64. 47-655-^4-5=10-59.
65. 756-98-^-76-73612=9-866.
66. 7o-3470-^3829=196-7798.
67. 0-1^7-6345=0-0000131.
68. o378-^0 -00096=5602083 -33, &a
69. If ^67500 were to be divided between 5 persons,
how much ought each person to receive .? Ans. £1500.
70. Divide 7560 acres of land between 15 persons.
Ans. Each will have 504 acres.
71. Di\T[de £2880 between 60 persons.
Ans. Each will receive £48.
72. What is the ninth of £972 r Ans. £108,
73. What is each man's part if £972 be divided
among 108 men ? Ans. £9.
74. Divide a legacy of £8526 between 294 persons.
Ans. Each will have £29.
75. Divide 340480 ounces of bread between 1792
persons. Ans. Each person's share will be 190 ounces.
76. There are said to be seven bells at Pekin, each
of which weighs 120,000 pounds ; if they were melted
up, how many such as great Tom of Lincoln, weighing
9894 pounds, or as the great bell of St. Paul's, in
London, weighing 8400 pounds, could be made from
them ? Ans. 84 like great Tom of Lincoln, with 8904
pounds left ; and 100 like the great bell of St. Paid's.
77. Mexico produced from the year 1790 to 1830 a
DIVISION. 99
quantity of gold which was worth iB6,436,443, or
6', 178,985,280 farthings. How many dolkrs, at 207
farthiiigj! each, are in that sum.' Ans. 29850170 nearl
78. A single pound of cotton has been spun into
thread 76 miles in length, and a pound of wool into a
thread 95 miles long ; how many pounds of each would
be rcquii-ed for 'threads 5854 miles in length } Aiis.
77-0263 pounds of cotton, and 61-621 pounds of wooli^
79. The earth travels round its orbit, a space equal
to 567,019,740 miles, in about 365 days, 8765 hours,
525948 minutes, 31556925 seconds, and 1S93415530
thirds ; supposing its motion uniform, how mucli would
it travel per day, hour, minute, second, and third ? ^*.
About 1553480 miles a day, 64691 an hour, 1073 a
minute, 18 a second, and 0*3 a third.
80. All the ii'on produced in Great Britain in^.e
year 1740 was 17,000 tons from 59 furnaces ; and'm
1827, 690,000 from 284. What may be considered a;?^'
the produce of each fui'nace in 1740, one with anoiif^';
and- of each in 1827. Ans. 288-1356 in 1740; and
2429-5775 in 1827.
81. In 1834, 16,000 steam ejigines in Great Britain
saved the laboui* of 450,000 horses, or 2 millions and a
half of men ; to how many horses, and how many men,
may each steam engine be supposed equivalent, one
with another.' Ans. About 28 horses; and 156 men.
99. 3efore the pupil- leaves division, he should be
able to carry on the process as follows : —
Example.— Divide 84380848 by 87532.
87532)84380848(964
560204
350128
He will say (at first aloud) 4 (the digit of the dividend to
■ -^ brought do^vn). 18 (9 times 2) : 0 (the remainder after
r-'.ihtracting the right hand digit of 18 from 8 in the di^-idend) .
28 (VI times 3 -f the 1 to be carried from the 18) : 2 (the
remainder after subtracting the right hand digit of 28 from
0. or rather 10 in the dividend). 48 (9 times 5 + the 2 to
l.»e carried from 2S. arid 1 to compensate for what we bor-
rowed when we con^^idered 0 in tlie dividend as 10) } 0 (tha
100 Divisior^.
remainder -when ^e subtract the right hand digit of 48 from
8 in the dividend). 67 (9 times 7 -f- the 4 to be carried
from the 48) ; 6 (the remainder after subtracting the right
hand digit of 67 from 3. or ratlier 13 in the dividend). 79
(0 times 8 -f- the 6 to be carried from 'the 67 -}- the 1, for
what we borrowed to make 3 in the dividend become 13) ;
0 (the remainder after subtracting 79 frojn 84 in the divi-
dend).
As the parts in the parentheses are merely explanatory,
and not to be repeated, the whole process Avould be,
First part, 4. 18: 6. 28; 2. 48: 0. 67: 6. 79; 5.
Second part, 8. 12; 2. 19 ; 1. 32; 0. 45; 5. 53 ; 3.
Third part, 8 ; 0. 12 ; 0. 21 ; 0. 30 ; 0. 35 ; 0.
The remainders in this case being cyphers, are omitted.
All this will be very easy to the pupil who has prac-
tised what has been recommended [13, 23, and 65].
The chief exercise of the memory will consist in recol-
lecting to add to the products of the diiferent parts of the
divisor by the digit of the quotient under consideration,
what is to be carried from the preceding product, and
unity besides — when the preceding digit of the dividend
has been increased by 10 ; then to subtract the right
hand digit of this sum from the proper digit of the
dividend (increased by 10 if necessary),
QUESTIONS FOR THE PUPIL.
1 . Y\"hat is division } [66] .
2. AYhat are the dividend, divisor, quotient, and re-
mainder r [66] .
3. What is the sign of division ? [6S].
4. How are quantities under the vinculum, or united
by the sign of multiplication, divided ? [69] .
5. What is the rule when the divisor does not exceed
12, nor the dividend 12 times the divisor.^ [70].
6. Give the rule, and the reasons of its different
paris, when the divisor does not exceed 12, but the
dividend is more than 12 times the divisor ? [72 and 77] .
7. How is division proved ? [79 and 95] .
8 What are the rules when the dividend, divisor, or
both contain cyphers or decimals ? [^0].
9. Wliat is the rule, and what are the reasons of its
different parts, when the divisor exceeds 12 .' [89 and 93] .
GREATEST COMMON MEASURE. 101
10. What Is to be done with the remainder ? [72
and 89] .
1 1 . How is division proved by casting out the nines ?
f9-5]- ... . •
12. How may division be abbreviated, when there are
decimals? [96].'
13. How is division performed, when the divisor is
u composite number ? [97] .
14. How is the division performed, when the divisor
is but little less than a number which may be expressed
by unity and cyphers .^ [98]-
15. Exemplify a very brief mode of performing divi*
sion. [99]. -
THE GREATEST COMMON MEASURE OF NUMBERS.
100. To find the greatest common measure of two
quantities —
Rule. — Divide the larger by the smaller ; then
the divisor by the remainder ; next the preceding
divisor by the new remainder : — continue this process
until nothing remains, and the last divisor will be the
greatest common measure. If this be unity, the given
numbers s.r<-i prime to mc/i other.
Example —Find the greatest common measure of 3252
and 4248.
'^25 9)4248(1
8252
"996)3252(3
2988
264)996(3
792
204)264(1
204
60)204(3
180
24)60(^2
48
12)24(2
24
102 GREATEST COMMON .MEASURE.
996, the first remainder, becomes the second divisor 264,
the second remainder, becomes the third divisor, &c. 12,
the kst divisor, is the required greatest common measure.
101. PtKAsox OF THK RuLK. — Bcfove we prove the correct-
ness of the rule, it will be necessary for the pupil to be satis-
tied that. " if any quantity measures another, it Tvill measure
any multiple of that other ; " thus if 6 go into 80, 5 times, it
vrill evidently go into 9 times 30, 9 times 5 times.
Also, that " if a quantity measure two others, it will measure
their sum, and their diiference." First, it will measure their
sum, for if 6 go into 24, 4 times, and into 36, 6 times, it will evi-
24 36
dently go into 24-f-36, 4-f-6 times :— that is, if — =4, and— =
24 ,36 , , . 6 6
Secondly, if 6 goes into 36 oftener than it goes into 24, it is
because of the difference between 36 and 24 ; for as the differ-
ence between the numbers of times it will go into them is
due to this difference, 6 must be contained in it some number
-^. ^, ^ . . 36_. ,24 ,36 24/ 36— 24\
of times : — that is, since— =6, and ^=4, I ^^ i
6 6 6 6 \"^ 6 /
=6 — 4=2, a whole number [26] — or, the difference between
the quantities is measured by 6, their measure.
This reasoning would be found equally correct with any
other similar numbers.
102. Next ; to prove the rule from the given example,
it is necessary to prove that 12 is a common measure ;
and that it is the greatest common measure.
It is a com}7i(yn measure. Beginning at the end of the process,
we find that 12 measures 24, its multiple ; and 48, because it is
a multiple of 24 ; and their sum, 24-J-48 (because it measures
each of them) or 60 ; and 180, because it is a multiple of 60 ;
and 180-|-24 (we have also just seen that it measui^es each of
these) or 204: and 204+60 or 264; and 792, because a multi-
ple of 264 ; and 792-f-204 or 996 ; and 3988, a multiple of 996 ;
and 2988-|-264 or 3252 (one of the given numbers) and 3252-}-
996 or 4248 (the other given number). Therefore it measures
each of the given numbers, and is their commu7i measure.
- 103. It is also their s;reatest common measure. If not, let
some other be greater; then (beginning now at the top of the
process) measuring 42-18 and 3252 (this is the supposition), it
measures their difference, 996; and 2988, because a multiple
of 996 ; and, because it measures 3252, and 2988, it measures
their difference, 261 ; and 792, because a multiple of 264 ; and
the difference between 996 and 792 or 204 ; and the difference
between 2(54 and 204 or 00 : and 180 because a multiple of 60 ;
and the difference between 204 and 180 or 24; and 48, because
a multiple of 24 ; and the difference between 60 and 48 or 12.
But measuring 12, it cannot be greater than 12.
GREATEST COMMON MEASURE. 103
In the same way it could be shown, that any other common
measure of the given numbers must be less th.-in 12 — and con-
sequently that 12 is their grtattst common measure. As the
rule miglit be proved from any other example equally well,
it is true in all cases.
104. "We may here remark, that the measure of two
or more quantities can sometimes be found by inspection •
Any quantity, the digit of wliose lowest denomination
is an even number, is divisible by 2 at least.
Any number ending in 5 is divisible by 5 at least.
Any number ending in a cypher Is divisible by 10 at
least.
Any number which leaves nothing when the threes
are cast out of the sum of its digits, is divisible by 3 at
least ; or leaves nothing when the nines are cast out of
the sum of its digits, is divisible by 9 at least.
EXERCISES.
1 . "What is the greatest common measure of 464320
and 18945 } Am^b.
2. Of 63S296 and 33SS8 .? Ans. 8.
3. Of 1S996 and 29932 ? A?is. 4.
4. Of 260424 and 54423 f Ans. 9.
5. Of 143168 and 2064SSS } Ans. 8.
6. Of 1141874 and 19823208 } Am. 2.
105. To find the greatest common measure of more
than two numbers —
Rule. — Find the greatest common measure of two of
them ; then of this common measure and a third ; next,
of this last common measure and a fourth, &c. The
last common measure found, will bo the greatest common
measure of all the given numbers.
ExAMPi.E 1. — Find the greatest common measure of 679,
5901, and 6734.
By the last rule we learn that 7 is the greatest common
Measure of 679 and 5901 : and by the same rule, that it. the
greatest common measure of 7 and 6734 (the remaining
number), for 6734 -^ 7 = 962, with no remain^r. Therefore
7 is the required number. ^^
ExiMPLE 2. — Find the ijrcatest common mea.sm-e of 936,
736, and 142.
104 LEAST CO.>f!\rON MULTIPLE.
Tlic greatest common measure of 93G and 73G is 8, and
the common measure of 8 and 142 is 2 ] therefore 2 is the
greatest common measure of the given numbers.
lOG. Reasox of the Rule. — It may be shown to be correct
in the same way as the last ; except that in proving the num-
ber found to be a common measure, we are to begin at the end
of all the processes, and go through all of them in succession ;
and in proving that it is the greatest common measure, we
are to begin at the commencement of the first process, or that
used to find the common measure of the two first numbers,
hud proceed successively through all.
EXERCISES.
7. Find the greatest common measui'e of 29472,
176832, and 1074. Ans. 6.
8. Of 648485, 10810, 3672835, and 473580. Atis. 5.
9. Of 16264, 14816, 8600, 75288, and 8472. Ans 8.
THE LEAST COMMON MULTIPLE OF NUMBERS.
107. To find the least common multiple of t-^vo quan-
tities—
Rule. — Divide their product by their greatest com-
mon measure. Or ; divide one of them by their greatest
common measure, and multiply the quotient by the
other — the result of either method will be the required
least common multiple.
Example. — Find the least common multiple of 72 and 84.
12 is their greatest common measure.
"79
_=G, and G X 84 = 504, the number sought.
108. Reason of the Rule. — It is evident that if we mul-
tiply the given numbers together, their product will be a
multiple of each by the other [30]. It will be easy to find
Uie smallest part of this product, which will still be their
ioinmon multiple. — Thus, to learn if, for example, its nine-
teenth part is such.
From wha^p have already seen [69], each of the factors
of any prodi^pivided by any number and multiplied by the
product of the other factors, is e<iual to the produ«*,t of all the
f-ic-^-o'u 4ividcel by the s.irac number. Hence, 72 and 81 being
U • • '-*.n nuHjbers —
LEAST COMMON MULTIPLE. 105
^1^ -^ (the nineteenth part of their proauct)==_X84, or 72 X
}9 19
84 T'' 84 7'^
_. Now if — and — be equiyalent to intcsrers, -^X84 will be a
ly 19 19 ^ 19
multiple of 84, and'^ix72, will be a multiple of 72 [29];
and I?^_, -X84, and 72 X— will each be the common
19 19 19
multiple of 72 and 84 [30]. But unless 19 is a common measure
of 72 and 84, '— and — cannot be both equivalent to intef^ers.
19 19 ^
Tlierefore the quantity by which we divide the product of the
given numbers, or one of them, before we multiply it by the
other to obtain a new, and less multiple of them, must be the
common measure of both. And the multiple we obtain will,
evidently, be the least, when the divisor we select is the
greatest quantity we can use fur the purpose — that is, the
greatc.)!t common measure of the given numbers
It follows, that the least comuion multiple of two
numljurs, prinie to each other, is theu- product.
EXERCISES.
1. Find the least common multiple of 7S and 93.
Ans. 2418.
2. Of 19 and 72. Ans. 136S.
3. Of 464320 and 1S945. Ans. 1759308480.
4. Of 638296 and 33888. Ans. 2703821856.
5. Of 18996 and 29932. Ans. 142147068.
6. Of260424 and 54423. Ans. 157478392R.
109. To find the least common multiple of three or
more numbers —
lluLE. — Find the least common multiple of two of
them ; then of this common multiple, and a third ; next
of this last common multiple and a fourth, &,c. The
\;\?t common multiple found, will be the least common
multiple sought.
Example. — Find the least common multiple of 9, 3, and 27.
% is the greatest common measure of 9 and 3 ; therefore
9
- X 3, or 9 is the least common multiple of 9 and 3,
• 9 is the greatest common measure of 9 aiSRT ; therefore
'^ X 0, or 27 is the required least common multiple.
106 LEAST COMMON MULTIPLE.
110. TtEAPOKT OF THE RuLE. — Bj the last rule it is evident
tliat 27 is the least common multiple of 9 and 27. But since
9 is a multiple of 3, 27, which is a multi2)le of 9, must also be
a multiple of 3 ; 27, therefore, is a multiple of each of the
given numbers, or their common multiple.
It is likewise their least common multiple, because none
that is smaller can be common, also, to both 9 and 27, since
they >vere found to have 27 as theii* least common multiple.
EXERCISES.
7. Find the least common multiple of 18, 17, and 43.
Ans. 13158.
8. Of 19, 78, 84, and Gl. Am. 1265628.
9. Of 51, 170832, 29472. and 5862. Am. 2937002688.
10. Of 537842, 16819, 4367, and 2473.
Am. 8881156168989038.
11. Of 2163G, 241816, 8669, 97528. and 1847.
Am. 152SS35550537452616.
QUESTIONS
1. IIow is the gi'eatest common measure of two quan-
tities found.? [lUUJ.
2. What principles are necessary to prove the correct-
ness of the rule ; and how is it proved ; [101, &c.].
3. How is the greatest common measui-e of thi-ee, or
more quantities found .? [105].
4. How is the rule proved to be correct } [106].
5. How do we find the least common multiple of two
numbeis that are composite t [107].
6. Prove the rule to be correct [108].
7. How do we find the least common multiple of two
prime numbers } [lOS.]
S. How Ls th<3 least common multiple of three or
more numbers found } [109].
9. Prove the 7ule to be correct [110].
In future it will be taken for granted that the pupi^
is to be agkcd the reasons for each rule, &c.
107
SECTIOxN III
REDUCTION AND THE CO,A [POUND RULES.
The pupil should now be made familiar with most of
the tables given at the commencement of this treatise.
"'^ .REDUCTION.
1. Reduction enables us to change quantities from
one '^nomination to another without altering their
value. Taken in its more extended sense, we have often
practised it already : — thus we have changed units into
tens, and t^ns into units, Sec. ; but, considered as a
separate rule, it is restricted to applicate numbers, and
is not confined to a change from one denomination to
"the next higjier, or lower
2. Reduction i» either descending, or ascending. It
is rcdudimi descending when the quantities are changed
from a higher to a lower denomination ; and reduction
asixiiding when from a lower to a higher.
Reduction Descending.
3. Rule. ---Multiply the highest given denomination
by that quantity which expresses the number of the
next lower contained in one of its units ; and add to
the product that number of the next lower denomina-
tion which is found in the quantity to be reduced.
Proceed in the same way with the result ; and continue
tlie process until the requii-ed denomination is obtained.
ExA3iPLE. — Reduce £6 16.s. Q\d. to farthings.
£ 5. d.
6 „ 16 „ 01
_^
136 shillings =£6 „ 16.
12
1632 pence = £6 „ 16 „ 0.
4
C529 farthings = £6 „ 16 „ 0\.
108 REDUCTION.
We multiply the pounds by 20, and at the saine time add
the shillings. Since multiplying by 2 tens (20) can give no
units in the product, there can be no units of shillings in
it except those derived from the 6 of the IG.s. : — Ave at once,
therefore, put down 6 in the shillings' place. l\rice (2 tens'
times) 6 are 12 (tens of shillings), and one (ten shillings), to
be added from the 16.s., are 13 (tens of shillings) — which we
put down. £(j 16s. are, consequently, equal to 1365.
12 times Gd. are 12cl. : — since there are no pence in the
given quantity, there are none to be added to the 12(1. — we
put down 2 and carry 7. 12 times 3 are 36, and 7 are 43.
12 times 1 are 12, and 4 are 16. £Q l(js. are, therefore,
equal to 1632 pence.
4 times 2 are 8, and i (in the quantity to l)e reduced) to
be carried are 9, to be set down. 4 times 3 are 12. 4 times
6 are 24, and 1 are 25. 4 times 1 are 4. and 2 are 6. Hence
£(5 IQs. Oirf. are equal to 6529 farthings.
4. Reasoi^ts of the Rule. — One pound is equal to 20^. ;
therefore any number of pounds is equal to 20 times as many
shillings ; and any number of pounds and shillings is equal
to 20 times as many shillings as there are pounds, plus the
shillings.
It is easy to multiply by 20, and add the shillings at the
same time ; and it shortens tlie process.
Shillings are equal to 12 times as many pence ; pence to
4 times as many farthings ; hundreds to 4 times as many
quarters; quarters to 28 times as many pounds, &c.
1. How many farthings in 23328 pence .'' Ans.
83312.
2. How many shillings in £348 > Ans. 6960.
3. How many pence in £38 10^. ? Ans. 9240.
4. How many pence in £58 135. ? Ans. 14076.
5. How many farthings in £58 135. f Ans. 56304.
6. How many farthings in £59 135. 6^d. ? An^.
57291.
7. How many pence in £63 05. 9d. } A71S. 15129.
8. How many pounds in 16 cwt., 2 qrs., 16 lb. ?
Ans. 1864.
9. How many pounds in 14 cwt., 3 qrs., 16 lb. ?
Ans. 1668.
10. How many grains in 3 lb., 5 oz., 12 dwt., 16
grains.? Ans. 19984.
KEDUCTION. 109
11. How many grains in 7 lb., 11 oz., l.i>dwt.,'14
grains ? Avs. 45974.
12. How many hom-s in 20 (common) years .^ Ans.
175200.
13. How many feet in 1 English mile ? Ans. 52S0.
14. How many feet in 1 Irish mile .' A^is. 6720.
15. How many gallons in Go tuns r Ans. 16380.
16. How many minutes in 46 years, 21 days, 8 hours,
56 minutes (not taking leap years into account) ? Aiis.
24208376.
17. How many square yards in 74 square English
perches ? AlTis. 2238-5 (2238 and one half).
IS. How many square inches in 97 square L-ish perch-
es .? Ans. 6159888.
19. How many square yards in 46 English acres, 3
roods, 12 perches .? Ans. 226633.
20. How many square acres in 767 square English
miles ? Ans. 490880.
21. How many cubic inches in 767 cubic feet ? Atis.
1325376.
22. How many quarts in 767 pecks ? Aiis. 6136.
23. How many pottles in 797 pecks ? Atis. 3188.
Redudio-n Ascending.
5. Rule. — Divide the given quantity by that number
of its units which is required to make one of the next
higher denomination — the remainder, if any, will be of
the denomination to be reduced. Proceed in the same
manner until the highest required denomination is
obtained.
Example. — Reduce 856347 farthings to pounds, kc.
4)856347
12)2140861
20)17840 „ ^
892 '„ 0 „ Gf-=856347 farthings.
4 divided into 856347 farthings, gives 214080 pence and
3 farthings. 12 divided into 214086 pence, gives 17840
shillings and 6 pence. 20 divided into 17840 shillings, gives
£892 and no shillings ; there is, therefore, nothing in the
Bhillings' place of the result.
f2
no
KKDUCrriON.
\Vf' (llvi.lo by 20 if we diviae hy 10 and 2 [Sec. If. 97J.
To divide b}'- 10, vre iuvve merely to cut off the units, if
any. [Sec. I. 34], wliich will then be the units of shillings
in "the result : and the quotient will be tens of shillings : —
di^iding the latter by 2, gives the pounds as quotient, and
tlie tens of shillings, if there are any in the required qiian-
tity, as remainder.
6. Reasons of the Rule. — It is evident that every 4
farthings are equivalent to one penny, and every 12 pence to
one shilling, &c. ; and that what is left after taking away 4
farthings as often as possible from the farthings, must be
farthings, what remains after taking away 12 pence as often,
as possible from the pence, must be pence, &c.
7. To prove Reduction. — Reduction ascending and
descending prove each other.
Example.— £20 17.s.
farthings=£20 lis. 21.
Reduction
l^ 4)20025
f 12)50061
Proof ^ 20)417 „ 2
5i€Z.=20025 farthings J and 20025
(1. farthings.
2} r 4)20025
Reduction-^ 12)5006
Proof
I £20 „ 17
[ 20)417
£20,
20
417
12
"5006
4
21
20025 farthings.
EXERCISES.
24. How many pence in 93312 farthings } Ans.
23328.
25. How many pounds in 6960 shillings } Ans. £348.
26. How many pounds, &c. in 976 halfpence } Ans.
£2 05. 8^.
27. How many pounds, &c. in 7675 halfpence ? Atis.
£15 19s. 9U.
28. How many ounces, and Dounds in 4352 drams ?
Afis. 272 oz., or 17 ib.
REDUCTION. Ill
29. How many cwt., qrs., and pounds iu 1864 pounds ?
^7^5. 16 cwt., 2 qrs., 16 ib.
30. How many hundreds, &c., in 16oS pounds. Aiis.
14 cwt., 3 qrs., 16 lb.
31. How many pounds Troy in 115200 grains.'
Am. 20.
32. How many pounds in 107520 oz. avoirdupoise .'
Ans. 6720.
33. How many hogsheads in 20C58 gallons ? Ans.
J27 hogsheads, 57 gallons.
34. How many days in 8760 hours .- Ans. 365.
35. How many Irish miles in 1834560 feet f A7ts.
173.
36. How many English miles in 17297280 inches ?
-ins. 273.
37. How many English miles, &c. in 4147 yards ?
Ans. 2 miles, 2 furlongs, 34 perches.
38. How many Irish miles, &c. in 4247 yards ? Aiis.
I mde, 7 furlongs, 6 perches, 5 yards.
3l). How many English ells iu 576 nails ? Ans. 28
cils, 4 qrs.
40. How many English acres, &c. in 5097 square
yards } An^. 1 acre, 8 perches, 15 yards.
41. How many IrLsh acres, &.c. in 5097 square yards ?
Ans. 2 roods, 24 perches, 1 yard.
42. How many cubic feet, &c., in 1674674 cubic
inches f Ans. 969 feet, 242 inches.
43. How many yards in 767 Flemish ells } Ans.
575 yards, 1 quarter.
44. How many French ells in 576 English ? Ans. 480.
45. Pteduce £46 14^. 6d., the mint value of a pound
of gold, to farthings .' Ans. 44856 farthings.
46. The force of a man has been estimated as equal
to what, in turning a winch, would raise 256 ft), in
pumping, 419 lb, in ringing a bell, 572 lb, and in row-
ing, 608 tb, 3281 feet in a day. How many hundreds,
quarters, &c., in the sum of all these quantities ? Ans
16 cwt., 2 qrs., 7 lb.
47. How many lines in the sum of 900 feet, tho
112 REDUCTION.
length of the temple of the sun at Balbec, 450 feet its
breadth, 22 feet the circumference, and 72 feet the
height of many of its columns ? Ans. 207936.
48. How many square feet in 760 English acres, the
inclosure in which the porcelain pagoda, at Nan-King,
in China, 414 feet high, stands ? Aiis. 33105600.
49. The great bell of Moscow, now lying in a pit
the beam which supported it having been burned, weighs
360000 ft), (some say much more) ; how many tons, &,c.,
in this quantity ? Ans. 160 tons, 14 cwt., 1 qr., 4 ft).
QUESTIONS FOR THE PUPIL.
1 . What is reduction ? [1] .
2. "What is the difference between reduction descend-
ing and reduction ascending r [2] .
3. What is the rule for reduction descending .'' [3]
4. What is the rule for reduction ascending ? [5] .
5. How is reduction proved r [7] .
Questions fomided on the Talk page 3, c,-c.
6. How are pounds reduced to farthings, and farthings
to pounds, &c. ?
7. How are tons reduced to drams, and drams t€
tons, &c. ?
8. How are. Troy pounds reduced to grains, and
grains to Troy pounds, &c. ?
9. How are pounds reduced to grains (apothecaries
weight), and grains to pounds, &c. .'
10. How are Flemish, English, or French ells, re-
duced to inches ; or inches to Flemish, English, or French
ells, &c. >
11. How are yards reduced to ells, or ells to yards,
&c..?
12. How arc Irish or English miles reduced to lines,
or lines to IrLsh or English miles, &c. ?
13. How are Irish or English square miles reduced
to square inches, or sijuare inches to Irish or English
square miles, &.c. ?
COMPOUND RULES. 113
14. II ow are cubic feet reduced to cubic inches, or
cubic inches to cubic feet, &c. ?
15. How are tuns reduced to nacrgins, or naggins to
tuns, &c. ^
16. How are butts reduced to gallons, or gallons to
butts, &:c. r
17. How are lasts (dry measure) reduced to pints,
and piuts to lasts, &c, ?
IS. How are years reduced to thii'ds, or thirds to
years, &:c, ?
19. How arc degrees (of the circle) reduced to thirds,
or thirds to degrees, &:c. .'
THE COMPOUND RULES.
S. The Compound Rules, are those which relate to
applicate numbers of more than one denomination.
If the tables of money, weights, and measures, were
constructed according to the decimal system, only the
rules for Simple Addition, &c., would be required.
This would be a considerable advantage, and greatly
tend to simplify mercantile transactions. — If 10 far-
things were one penny, 10 pence one shilling, and 10
sliillings one pound, the addition, for example, of £1
95. 8f i. to £Q 8s. 6^d. (a point being used to separate
a pound, then the " unit of comparison," from its parts,
and 0'005 to express ^ or 5 tenths of a penny), would
be as follows —
£
1-9S3
6-865
Sum, 8-&48
The addition might be performed by the ordinary
rules, and the sum read off as follow.s — " eiglit pounds,
eight shillings, four pence, and eight forthings." But
even Avith the present arrangement of money, weights,
and measures, the rules already given for addition, sub-
traction, Sec, might easily have.b<?en made to include
the addition, subtraction, &c., of applicate numbers
consisting of more than one denomination ; since the
114 COMPOUND ADDITION.
principles of both simple unci compound rules are pre-
cisely the same — the only thing necessary to bear
carefully in mind, being the number of any one de-
nomination necessary to constitute a unit of the next
higher.
COMPOUND ADDITIOX.
9. Rule. — I. Set down the addends so that quanti-
ties of the same denomination may stand in the same
vertical column — units of pence, for instance, under
units of pence, tens of pence under tens of pence, units
of shillings under units of shillings, &c.
II. Draw a separating line under the addends.
III. Add those quantities which are of the same
denomination together — farthings to farthings, pence to
pence, &c., beginning with the lowest.
IV. If the sum of any column be less than the num-
ber of that denomination which makes one of the next
higher, set it down under that column ; if not, for each
time it contains that number of its own denomination
which makes one of the next higher, carry one to the
latter and set down the remainder, if any, under the
column which produced it. If in any denomination
there is no remainder, put a cypher under it in the
sum.
10. Example.— Add together £52 17s. S^tf., £47 5^. QUI,
and £(JQ Us. 2ld.
£ s. d.
52 17 3-f )
47 5 6;^- [ addends.
66 14 2} )
166 17 0^
I and i- make 3 farthings, which, with J, make 6 far-
things; these are equivalent to one of tlie next denomina-
tion, or that of pence, to be carried, and two of tlie present,
or one half-penny, to be set do^v^^. 1 penny (to be carried)
and 2 are 3, and 6 are 9, and 3 are 12 pence — equal to one
COMPOUND ADDITION,
115
of the next denomination, or that of shillings, to be carried,
Jind no pence to be set down; we therefore put a cypher
in the pence" place of the sum. 1 sliilling (to be carried)
and 14 are 15, and 5 are 20. and 17 are 57 shillings — equal
to one of the next denomination, or that of pounds, to be
carried, and 17 of the present, or that of shillings, to be
set down. 1 pound and 6 are 7, and 7 are 14, and 2 are
16 pounds — equal to 6 units of pounds, to be set down, and
I ten of pounds to be carried ; 1 ten and 6 are 7 and 4 are
II and 5 are 16 tens of pounds, to be set down.
11. This rule, and the reasons of it, are the same as
those already given [Sec. 11. 7 and 9]. It is evidently
not SO necessary to put a cypher where there is no
remainder, as in Simple Addition.
12. When the addends are very numerous, we may
divide them into parts by horizontal lines, and, adding
each part separately, may afterwards find the amount
of all the sums.
Example :
£ s.
d.
57 14
21
32 16
4
£
s.
d.
19 17
6
> = 151
7
ir
8 14
o
32 5
Jj
£ s. d.
= 404 11 10
47 6
4]
32 17
2
56 3
27 4
9
2
. = 253
3
11
52 4
4
37 8 2j
13. Or, in adding each column, we may put down
a dot as often as we come to a quantity which is at
least equal to that number of the denomination added
which is required to make one of the next — carrying
forward what is above this number, if anything, and
putting the last remainder, or — when there is nothing
left at the end — a cypher under the column : — we carry
to the next column one for every dot. Using the same
example —
116' CO.NiPOUND RULES.
£,
s.
d.
57
•14
2
32
10
4
19
•17
■G
8
•14
2
32
5
•9
47
•6
4
32
17
2
56
•3
•9
27
4
2
52
4
4
37
8
2
i04
11
To
2 pence and 4 are 6, and 2 are 8, and 9 are 17 pence —
equal to 1 shilling and 5 pence ; Ave put down a dot and carry
5. 5 and 2 are 7, and 4 are 11, and 9 are 20 pence — equal
to 1 philling and 8 pence ; we put down a dot and carry 8.
8 and 2 are 10 and 6 are 16 pence — equal to 1 shilling and
4 pence : we put down a dot and carry 4. 4 and 4 are 8 and
2 are 10 — which, being less than 1 shilling, we set down
under the column of pence, to which it belongs, &c. We find,
on adding them up, that there are three dots ; we therefore
carry 3 to the column of shillings. 3 shillings and 8 are 11,
and 4 are 15, and 4 are 19, and 3 are 22 shillings — equal to
1 pound and 2 shilKngs ; we put down a dot and carry 1.
1 and 17 are 18, &,c.
Care is necessary, lest the dots, not being distinctly marked,
may be considered as either too few, or too many. Thia
method, though now but little used, seems a convenient one.
14. Or, lastly, set down the sums of the farthings,
shillings, &c., under their respective columns ; divide
the farthings by 4, put the quotient under the sum of the
pence, and the remainder, if any, in a place set apart
for it in the sum — under the column of farthings ; add
together the quotient obtained from the farthings and
the sum of the pence, and placing the amount under
the pence, divide it by 12 ; put the quotient under the
sum of the shillings, and the remainder, if any, in a
place allotted to it in the sum — under the column of
pence ; add the last quotient and the sum of the shil-
lings, and putting under them their sum, divide the
latter by 20, set down the quotient under the sum of
COMPOUND ADDITION. 117
the pound?', and put the remainder, if any, in the sum —
under the column of shillings; add the last quotient
and the sum of the pounds;, and put the result under
the pounds. Using the folio ft'ing example —
£ s. d.
47 9 21
362 4 11»
51 16 2|
97 4 6
541 13 2i
475 6 4
6 11 Hi
72 19 9r-
13 mrthincrs.
1051
82
47
4
4
3
86
50
1055 C 2]
The sum of the farthings is 13. which, divided by 4, give.i
3 as quotient (to be pat down vmder tiie pence), and one
farthing as remainder (to be put in the sum total — imder
the farthings). 3.-./. (tlie quotient from the farthings) and
47 (the sum of the pence) are 50 pence, which, being put
duwn and divided by 12, gives 4 shillings (to be set down
under the shillings), and 2 pence (to be set down in tho
sura total — under the pence). 45. (the quotient from the
pence) and S2 (the sum of the shillings) are 86 shillings,
which, being set down and divided by 20, gives 4 pounds
(to be seWo^vn under the pounds), and 6 shillings Tto be
set do^vn in the smn total — under the >;liillings) . £A (the
quotient from the shillings) and 1651 (the sum of the
pounds) are 1655 pounds (to be set down in the sum total —
under the pounds). The sum of the addenda is, therefore,
found to be £1655 6.s. 2\d.
15. In proving the compound rule?, wc can generally
avail ourselves of the methods used \7itii tlie sin.pl*? ♦ulf>*
[Sec. II. 10- k-r
118 COMPOUND ADDITION.
EXERCISES FOR THE PUPIL
Money.
(1)
(2)
(3)
(4)
£ s. d.
£ s.
d.
£ s. d.
£ s. d.
76 4 6
58 14
t
75 14 7
84 3 2
57 9 9
69 15
6
67 15 9
96 4 Oi
49 10 8
72 14
8
76 19 10
41 0 6
183 4 11
(5)
(6)
(7)
(8)
£ s. d
£ s.
d.
£ 5. d.
£ s. d.
674 14 7
767 15
6
567 14 7
327 8 6
456 17 8
472 14
6
476 16 6
501 2 111
676 19 8
567 16
7
547 17 6
864 0 6
527 4 2
428 3 10
527 14 3
121 9 84
(9)
(10)
—
(11)
(12)
£ s. d.
£ 5.
a
£ s. d.
£ s. d
4567 14 6
76 14
7
3767 13 11
5674 17 61
776 15 7
667 13
6
4678 14 10
47G7 16 lU
76 17 9
67 15
7
767 12 9
3466 17 104
51 0 10
5 4
2
10 11 5
5984 2 24
44 5 6
5 3
4
3 4 11
8762 9 9
(13)
(14)
—
(15)
(16)
£ s. d.
£ s.
d.
£ s. d.
£ 5. d.
9767 0 61
6767 11
6k
5764 17 64
634 7 114
7649 11 2k
7676 16
94
74.57 16 5
65 7 7
4767 16 lOa
5948 17
81;
6743 18 04
7 12 lOJ
164 1 1
5786 7
6
67 6 6i
5678 18 8
92 7 2i
6325 8
24
432 5 9
439 0 0
(17)
(18)
—
(19)
(20)
£ s. d.
£ s.
rf.
£ s. d.
£ s. d.
0 14 71
5674 16
7^
5674 1 94
4767 14 7i
677 1 0
4767 17
04
4767 11 104
743 13 74
5767 2 6
1545 19
7^
78 18 Hi
7074 14 6i
3697 14 7 i
3246 17
6
0 19 104
7 13 34
5634 0 0^
4766 10
54
5044 4 1
750 6 4
COMPOUND ADDITION. 119
(21)
(22)
(23)
(24)
£ s. d.
£ ...
d.
£ s. d.
£ s.
d.
674 11 Wk
476 14
7
674 13 3i
674 17
6i
667 14 101
576 15
6.?
45 15 74
123 12
2
476 4 11
76 17
74
476 4 61
567 0
7A
347 15 01
676 11
8
577 16 04
579 18
94
476 13 9i
463 14
94
578 6 31
476 6
64
(2.5)
(26)
—
(27)
(28)
£ s. d.
£ s.
d.
£ s. d.
£ s.
d
57G 4 Ik
549 4
6i
876 0 3
219* 0
6
7 7 6
7 19
91
0 5 0
32 11
84
732 19 04
0 16
64
56 11 11
0 0
04
567 0 9^
734 19
9^
123 5 24
127 8
2
754 2 6i
566 14
44
12 0 0
29 6
5i
Avoirdtipoise Weight.
(29) (30) (31) (32)
cwt. qrs. R) cwt. qrs. R) cwt. qrs. lb cwt. qrs. tt)
6 3 14 44 1 16 14
37 2 ■ 15 56 3 11
6i
14 1 11 47 1 16 4<
3
17
56
3
14
1
16
57
1
17
2
27
58
2
26
128 3 12
(33) (34) (35) (36)
cwt. qrs. lb cwt. qrs. lb cwt. qrs. ft) cwt. qrs. B)
76 1 ^ 19 88 2 17 476 3 15 567 2 19
56
3
13
59
2
20
764
1
7
4
1
20
47
2
17
0
3
0
6
3
14
67
3
2
81
2
14
67
1
15
0
1
18
767
1
11
(37) (38) (39) (40)
cwt. qrs. lb cwt. qrs. ft) cwt. qrs. ft) cwt. qrs. ft)
767
1
16
476
1
24^,
447
1
7
14
12
12
44
1
17
756
3
214
576
1
6
3
4
7
567
3
13
767
1
16
467
1
7i
0
5
15
576
1
0
567
2
15
563
1
6
7
0
3
341
2
11
973
1
12
428
0
04
0
0
14
130 COMPOUND ADDITION.
Troy
TVeight.
(41)
(42)
(43;
tb
oz.
dwt. grs.
tb
oz.
dwt.
p-s.
tb
0!!. dwt. gra.
{
0
5
9
5
9
7
0
88
7 9 8
0
6
6
7
0
0
6
7
80
9 8 6
9
0
6
8
8
7
6
4
0
8 7 5
21
11
18
0
(44)
(45)
(46)
tb
oz.
dwt
• grs.
tb .
oz.
dwt.
grs.
tb
oz. dwt. grs.
57
9
12
14
87
3
7
12
57
10 14 11
67
9
11
11
0
11
12
3
0
0 11 10
66
8
10
5
0
0
16
14
46
9 9 8
74
6
5
3
44
12
10
13
22
8 7 5
12
3
5
4
67
8
9
10
11
10 13 14
~~-^
Cloth Measure.
(47)
(48)
(49)
(50)
yds.
qrs
. nls.
yds,
. qrs.
nls
. yds
. qrs
. nls
1. yds. qrs. nig.
99
a
1
176
3
3
37
3
2
0 2 1
47
1
3
47
0
o
0
2
3
5 3 2
76
3
2
7
3
3
0
0
2
0 0 3
224
0
2
^51)
(52)
(53)
(54)
yds.
qrs
. nls,
. yds.
qrs.
nls
. y.ls.
qrs.
nls.
yds. qrs. nls.
567
8
2
147
3
3
157
2
1
156 1 1
476
1
0
173
1
0
143
3
2
176 3 1
72
3
3
148
2
1
0
1
2
54 1 0
5
2
1
92
3
9
54
0
3
573 2 3
IVine
Measure.
"*~~
(55)
(56)
(57)
ts.
hhds.
pis.
ts
hhds.
gls.
ts. hhds. gls.
^9
3
9
89
1
3
3
76 3 4
SO
0
39
7
3
4
67 3 44
98
3
46
76
1
56
0 1 56
87
2
27
44
2
7
5 3 4
41
1
26
54
2
17
6
02 0 27
407
3
sn
yrs.
99
88
77
(58)
ds. hrs.
859 9
0 8
120 7
OC
ms.
56
57
49
>MPOUND ADDITION.
Time.
(59)
yrs. ds. hrs. ms. yrs.
60 90 0 50 59
6 76 1 57 0
0 0 3 58 76
6 12 0 6
(60)
ds. Lrs.
127 7
120 9
121 11
47 3
9 11
121
ms.
50
44
44
41
17
265
115 2
42
61. "WTiat is the sum of the following: — three hun-
dred and ninety-six pounds four shillings and two pence ;
five hundred and seventy-three pounds and four pence
halfpenny ; twenty-two pounds and three halfpence ;
four thousand and five pounds six shillings and three
farthings.? Ans. £4996 105. 8|^.
62. A owes to B £567 165. 7irf. ; to 0 £47 165. ;
and to D £56 Id. How much does he owe in all }
Ans. £671 125. S\d.
63. A man has owing to him the following sums : —
£3 105. Id. ; £46 l\d. ; and £52 145. Qd. How much
is the entu-e .? Ans. £102 55. ^\d.
64. A merchant sends ofi" the following quantities of
butter: — 47 cwt., 2 qrs., 7 ib ; 38 cwt., 3 qrs., 8 lb ;
and 16 cwt., 2 qrs., 20 lb. How much did he send off
in all > Ans. 103 cwt., 7ib.
65. A merchant receives the following quantities of
tallow, viz., 13 cwt., 1 qr., 6 ft) ; 10 cwt., 3 qrs., 10 lb;
and 9 cwt., 1 qr., 15 ft). How much has he received in
all } Ans. 33 cwt., 2 qrs., 3 ft).
66. A silversmith has 7 ft), 8 oz., 16 dwt. ; 9 ft), 7
oz., 3 dwt. ; and 4 ft), 1 dwt. What quantity has he ?
Alts. 21 ft), 4 oz.
67. A merchant sells to A 76 yards, 3 quarters, 2
nails ; to B, 90 yards, 3 quarters, 3 nails ; and to C, 190
yards, 1 nail. How much has he sold in all r Ans. 357
yards, 3 quarters, 2 nails.
68. A wine merchant receives from his conespondeut
4 tuns, 2 hogsheads ; 5 tuns, 3 hogsheads ; and 7 tuns,
1 hogshead. How much is the entire ? Ans. 17 tuns,
2 hogsheads.
122 COMPOUND ADDITION.
^69. A man lias three farms, the first contains 120
acres, 2 roods, 7 perches ; the second, 150 acres, 3
roods, 20 perches ; and the third, 200 acres. How much
land does he possess in all ? Ans. All acres, 1 rood, 27
perches.
70. A servant has had three masters ; with the first
he lived 2 years and 9 months ; with the second, 7
years and 6 months ; and with the third, 4 years and 3
months. What was the servant's age on leaving his
last master, supposing he was 20 years old on going
to the first, and that he went du-ectly from one to the
other .'' Ans. 34 years and 6 months.
71. How many days from the 3rd of March to the
23rd of June i Ans. 112 days.
72. Add together 7 tons, the weight which a piece
of fir 2 inches in diameter is capable of supporting ; 3
tons, what a piece of iron one-thu'd of an inch in
diameter will bear ; and 1000 lb, which will be sustained
by a hempen rope of the same size. Ans. 10 tons, 8
cwt., 3 quarters, 20 ft).
73. Add together the following: — 2^., about the
value of the Eoman sestertius ; ifd.., that of the dena-
rius ; li^., a G-reek obolus ; 9c?., a drachma; £3 155.
a mina ; ^£225, a talent ; Is. Id.^ the Jewish shekel ; and
£342 35. 9^., the Jewish talent. Ans. £bl\ 2s.
74. Add together 2 dwt. 16 grains, the Grreek drachma;
1 ft), 1 oz., 10 dwt., the mina ; 67 ft), 7 oz., 5 dwt., the
talent. Ans. 68 ft), 8 oz., 17 dwt., 16 grains.
QUESTIONS FOR THE PUPIL.
1. What is the difference between the simple and
compound rules r [8] .
2. Might the simple rules have been constructed so
as to answer also for applicate numbers of different
denominations } [8] .
3. What is the rule for compound addition } [9].
4. How is compound addition proved } [15].
5. How are we to act when the addends are numei>
ous ? [12, &c.]
COMPOUND SUBTRACTION. ^ 123
COMPOUND SUBTPwACTION. •
16. Rule — ^I. Place the digits of the subtrahend
under those of the same denomination in the minuend — ■
farthings under farthings, units of pence under units of
pence, tens of pence under tens of pence, &c.
n. Draw a separating line.
III. Subtract each denomination of the subtrahend
from that which corresponds to it in the minuend —
beginning with the lowest.
IV. If any denomination of the minuend is less than
that of the subtrahend, which is to be taken from it,
add to it one of the next higher — considered as an equi-
valent number of the denomination to be increased ;
and, either suppose unity to be added to the next deno-
mination of the subtrahend, or to be subtracted from
the next of the minuend.
V. If there is a remainder after subtracting any
denomination of the subtrahend from the correspond-
ing one of the minuend, put it under the column which
produced it.
YI. If in any denomination there is no remauider,
put a cypher under it — ^unless nothing is left fi-om any
higher denomination,
17. Example.— Subtract £56 13.s. 4|(f., from £0Q 7s. 6|(i.
£ s. d.
96 7 6]^. minuend.
56 13 4|; subtrahend.
39 14 l\, difference.
We cannot take ^ from |, but — borrowing one of the
pence, or 4 farthings, we add it to the \, and then say 3 far-
things from 5, and 2 farthings, or one halfpenny, remains :
we set down 4- under the farthings. 4 pence from 5 (we
have borrowed one of the C pence), and one penny re-
mains : we set down 1 under the pence (1^/. is read •' three
halfpence"). 13 shillings cannot be taken from 7, but (bor-
rowing one from the pounds, or 20 shillings) 13 shillings
from 27, and 14 remain : we set do"wn 14 in the shillings'
place of the remainder. 6 pounds cannot be taken from 5
(we have bon'owed one of the 6 pounds in the mirwieaid)
124
COMPOUND SUBTRACTION.
but 6 from 15, and 9 remain : vre put 0 under the units of
pounds. 5 tens of pounds from 8 tens (we have borrowed
one of the 9), and 3 remain: we put 3 in the tens of poimds'
place of the remainder,
18. This rule and the reasons of it are substantially the
same as those already given for Simple Subtraction [Sec. II.
17, &c.] It is evidently not so necessary to put down cyphers
where there is nothing in a denomination of the remainder.
19. Compound may be proved in the same way as simple
Bubtraction [Sec. II. 20].
(1)
£ s. d.
From 1098 12 6
Take 434 15 8
EXERCISES.
(2) (3) (4)
£ s. d. £ s. d. £ s.
767 14 8 76 15 6 47 16
486 13 9 0 14 5 39 17
(5)
d. £ s. d.
7 97 14 6
4 6 15 7
663 16 10
From
Take
(6)
£ s. d.
98 14 2
77 15 3
(7)
£ s. d.
47 14 6
88 19 9
£
97
88
(8) (9) (10)
s. d. £ s. d. £ s. d.
16 6 147 14 4 560 15 6
17 7 120 10 8 477 17 7
From
Take
(11)
£ s. d.
99 13 3
47 16 7
(12)
£ s.
767 14
476 6
d.
7i
(13)
£ s.
891 14
677 15
d.
U
61
(I'i)
£ s. d.
576 13 71
407 14 95
From
Take
(15)
£ s. d.
567 11 51
479 10 lOi
(16)
£ s.
971 0
0 0
d.
Oi
7
£ s.
437 15
0 11
d.
0
(18)
£ s. d.
478 10 0
47 11 0^
From
Take
(19)
cwt. qrs. lb
200 2 20
99 3 15
Avoirdupoise
(20)
cwt. qrs. lb
275 2 15
27 2 7
1 Weight.
^21)
cwt. qrs.
9004 2
9074 0
lb
25
27
(22)
cwt. qrs. t>
554 0 0
476 3 5
100 3 11
COMPOUND SUBTRACTION.
125
ft)
c'rom 554
fake 97
(23)
oz. dwt.
9 19
0 16
Troy Weight.
(24)
gr. It) oz. dwt. gr.
4 946 0 10 0
15 0 0 17 23
(24)
ft) oz. dwt. gr.
917 0 14 9
798 0 18' 17
457 9
13
Wine Measure.
(26) (27) (28) (29)
ts. hlids. gla. ts. hhds. gls. ts. hhds. gls. ts. hhds. gls.
From 31 3 15 54 0 27 304 0 64 56 0 1
Take 29 2 26 0 3 42 100 3 51 27 2 25
2 0 52
Time.
(30) (31) (32)
yrs. ds. lis. ms. yrs. ds. hs. ms. yrs. ds. hs. nis
From 767 131 6 30 476 14 14 16 567 126 14 12
Take 476 110 14 14 160 16 13 17 400 0 15 0
291 20 16 16
33. A shopkeeper bought a piece of cloth containing
42 yards for £22 105., of which he sells 27 yards for
£\b lbs. 5 how many yards has he left, and what have
they cost him } Aiis. 15 yards ; and they cost him
£6 1.55.
34 A merchant bought 234 tons, 17 cwt., 1 quarter,
23 lb, and sold 147 tons, 18 cwt., 2 quarters, 24 ib ; how
much remained unsold .? Aiis. 86 tons, 18 cwt., 2 qrs.
27 ft).
35. If from a piece of cloth containing 496 yards, 3
quarters, and 3 nails, I cut 247 yards, 2 quarters, 2 nails,
what is the length of the remainder } Ans. 249 yards,
1 quarter, 1 nail.
36. A field contains 769 acres, 3 roods, and 20 perches,
of which 576 acres, 2 roods, 23 perches are tilled ; how
much remains untilled ^ Ans. 193 acres, 37 perches.
37. I owed my friend a bill of £76 165. 9|^., out of
which I paid £b'^ lis. lOf r/. ; how much remained due >
Ans. £\Q 185. lO^d.
126 COMPOUND MULTIPLICATION.
38. A merchant bought 600 salt ox hides, weighing
561 cwt., 2 ib ; of which he sold 250 hides, Weighing
239 cwt., 3 qrs., 25 lb. How many hides had he left,
and what did they weigh r Ans. 350 hides, weighing
321 cwt., 5 lb.
39. A merchant has 209 casks of butter, weighing
400 cwt., 2 qrs., 14 lb ; and ships off 173 casks,
weighing 213 cwt., 2 qrs., 27 lb. How many casks has
he left ; and what is their weight .' Ans. 36 casks,
weighing 186 cwt., 3 qrs., 15 lb.
40. AVhat is the difference between 47 English miles,
the length of the Claudia, a Roman aqueduct, and 1000
feet, the length of that across the Dee and Vale of
Llangollen .^ Av.s. 247160 feet, or 46 miles, 4280 feet.
41. What is the difference between 980 feet, the
width of the single arch of a wooden bridge erected at
St. Petersburg, and that over the Schuj^lkill, at Phila-
delphia, 113 yards and 1 foot in span.? Ans. 640 feet
QUESTIONS FOR THE PUPIL.
1. What is the rule for compound subtraction } [16].
2. How is compound subtraction proved } [19].
COMPOUND MULTIPLICATION.
20. Since we cannot multiply pounds, &g., by pounds,
&c., the multiplier must, in compound multiplication,
be an abstract number.
21. AYhen the multiplier does not exceed 12 —
Rule — I. Place the multiplier to the right hand
side of the multiplicand, and beneath it.
II. Put a separating line under both.
III. Multiply each denomination of the multiplicand
by the multipUer, beginning at the right hand side.
IV. For every time the number required to make
one of the next denomination is contained in any pro-
duct of the multiplier and a denomination of the multi-
plicand, carry one to the next product, and set down the
remainder (if there is any, after subtracting the number
equivalent to what is carried) under the denomination
CO.MPOU-ND MULTirLICATION. 12/
to which it belongs ; but should there be no reuiaiiider,
put a CA^er in that denomination of the product.
22. ExAMPus.— :Multiply £62 17.s. lOd. by 6.
£ s. d.
62 17 10, multiplicand.
6, multiplier.
377 7 0, product.
Six times 10 pence are 60 pence ; these are equal to '»
fihiUings (5 times 12 pence) to be carried, and no pence to
be set down in the product — we therefore -wTite a cypher in
the pence place of the product. 6 times 7 are 42 shillings,
and the 5 to be carried are 47 shillings— we put down 7 in
the units" place of shillings, and carry 4 tens of shiUings.
6 times 1 (ten shillings) are 6 (tens of shillings), and 4 (tens
of shillings) to be carried, are 10 (tens of shillings), or o
pounds (5 times 2 tens of shillings) to be carried, and
nothing, (no ten of shillings) to be set down. 6 times 2 pounds
are 12, and 5 to be carried are 17 pounds— or 1 (ten pounds)
to be carried, and 7 (units of poimd.s) to be set down.
6 times 6 (tens of pounds) are 36, and 1 to be carried arc
37 (tens of pounds).
23. The reasons of the rule will be very easily understood
from what we have already said [Sec. II. 41]. But since, in
compound multiplication, the value of the multiplier has no
connexion with its position in reference to the multiplicand,
where we set it down is a mere matter of convenience ; neither
is it go necessary to put cyphers in the product in those deno-
minations in which there are no significant figures, as it is in
simple multiplication.
24. Compound multiplication may be proved by re-
ducing the product to its lowest denomination, dividing
by the multiplier, and then reducing the quotient
Example.— IMultiply £4 3^. M. by 7.
£ s. d. ■ Proof :
4
3
8
7
29
20
5 8
29
5
8, product.
585
12
7)7028,
12)1004
20)83
produ(
8
;t reduced
quotient reduced 4 3 8=lmdtiplicand.
123 COMPOUND MULTIPLICATION.
£29 5s. Sd. are 7 times the multiplicand ; if, tWefore, the
process has been rightly performed, the seventh part of this
should be equal to the multiplicand.
The quantities are to be " reduced," before the division by 7,
since the learner is not supposed to be able as yet to divide
£29 5s. 8d.
EXERCISES.
£ s. d. £ s. d.
1. 76 14 7^X 2= 153 9 3.
2. 97 13 6^X 3= 293 0 7i
3. 77 10 7iX 4= 310 2 5.
4. 96 11 7hX 6= 482 18 li
5. 77 14 6|X 6= 466 7 Ih.
6. 147 13 3iX 7=1033 13 Oh.
7. 428 12 7^ X 8=3429 1 0.~
8. 572 16 6 X 9=5155 8 6.
9. 428 17 3 X 10=4288 12 6.
10. 672 14 4 X 11=7399 17 8.
11. 776 15 5 X 12=9321 5 0.
12. 7 R) at 55. 2id. #", will cost £1 16s. Sid.
13. 9 yards at lO-s. Hid. W, will cost £4 ISs. 5i</.
14. 11 gallons at 135. 9d. W, will cost £7 II5. 3d.
15. 12 lb at £1 3s. 4:d. W, will cost £14.
25. When the multiplier exceeds 12, and is a com-
posite number —
Rule. — Multiply successively by its factors
Example 1.— IMultiply £47 13s. 4d. by 56.
£ s. d.
47 13 4
7
5G=7x8 £ s. d.
333 13 4=47 13 4x7.
2669 6 8=47 13 4x7x8, or 56.
Example 2.— Multiply 14s. 2d. by 100.
i\ d.
14 2
10
100=10x10 s. d.
£7 1 8=14 2x10.
10
X70 16 8=14 2x10x10, or 100.
COMPOUND MrLTrri.fCATlO.V. 129
F.xAMJip: 3.— Multiply £8 2s. 4d. bv 700.
X s. d.
8 2 4
81
3
4
10
=8
811
13
4
7
=8
o
4x10.
4x10x10, or 100.
568113 4=8 2 4x10x10x7, or 700.
The reason of this rule has been already giyen [Sec. II. 60].
26. "When the multiplier is the sum of composite
numbers —
Rule. — Multiply by each, and add the results.
Example.— .INIultiply £3 14.5. (jd. by 430.
£ .^. d.
3 14 6
10
f. < fl p. *
,]
o 1
5 0 x 3=1 11 15 0. or 3 14
10
Cx30.
72
10 0x4=1400 0 0. or 3 14
6x400.
IGOl 15 U, or 3 14 6x430.
The reason of the rule is the same as that already givea
[^^ec. II. 52]. The sum of the products of the multiplicand by
the parts of the multiplier, being equal to the product of the
multiplicand by the -vrhole multiplier.
EXERCISES.
£ .v. d. £ s. d.
16. 3 7 6 x 1S= GO 15 0.
17. 4 16 7 X 20= m 11 8.
18. 5 14 6kX 22=125 19 11.
19. 2 17 6 X 8r3=103 10 0.
20. 3 16 7 X 56=214 8 8.
21. 2 3 6 X 64=139 4 0.
22. 3 4 7 X 81=261 11 3.
23. 0 9 4 X 100= 46 18 4.
24. 0 16 4 X1000=S16 13 4
25. 100 yards at O.s. 4hd. W, will cost £46 17 6.
26. 700 gallons at 135. 4d. ip, will cost 466 13 4.
27. 240 gallons at 6s. Sd. W, will cost 80 0 0.
28. 360 yards at 13s, id. 4f% will cost 210 0 0.
130 COMPOUND MULTIPLICATION.
27 If the multipliGr is not a composite num^r—
jj,ULE. — Multiply successively by the factors of the
nearest composite, and add to or subtract from the pro-
duct so many tunes the multiplicand as the assumed
composite number is less, or greater than the giv^'X
multiplier.
Example 1 —Multiply £62 12.s. Od. by 7G.
£> s. d.
62 12 6
8
76=8x9+4
501 0 0
9
4509
250 10 0=62 12 6X-1.
4759 10 0=62 12 6x8x9+4, or 76.
Example 2.— Multiply £42 os. M. by 27.
£ s. d.
42 3 4
27=4x7-1
4
168 13 4
7
£ s. d.
1180 13 4=42 3 4x4x7, or 28.
42 3 4=42 3 4x1.
1138 10 0=42 3 4x4x7-1, or 27.
The reason of the rule is the same as that already given
[Sec. II. 61].
EXERCISES.
£
s.
d. £
s.
d.
29.
12
2
4 X 83= 1005
13
8.
30.
15
0
0iXl46= 2193
3
0^.
31.
122
5
0 Xl02= 12469
10
0.
32.
963
0
01X999— 9G2040
2
6i.
S-8. When the multiplier is large, we may often con-
reniently proceed as follows —
Kule. — Write once, ten times, &c.,the multiplicand,
jind, multiplying these respectively by the units, tens
&c., of the multiplier, add the results.
£ .9. d. £ f.
, 47 IG 2x3= 143 8
10
(J.
0.
478 1 8x8= 3824 13
10
4.
4780 10 8x7= 334G5 10
10
8.
47808 0 8x5 = 230041 13
4.
COMPOUND MULTIPLICATION. 131
Example.— I\rultiply £47 lOs. 2'/. by 5783.
5783 = 5 X 1000 + 7 X lOO-j-8 x 10 + 3 x 1-
Units of the multiplicand,
Tens of the multiplicand,
Hundreds of the multiplicand,
Thousands of the multiplicand, 4( 808
Product of multiplicand and multiplier = 27047,5 11 10.
EXERCISES.
£ s. d. £ s. d.
33. 76 14 4 X 92= 7057 18 8.
34. 974 14 2 X 76 = 74077 16 8.
35.780 17 4 X 92 = 71839 14 8.
36. 73 17 7^X122= 9013 10 3.
37. 42 7 7iXl62= 6865 11 lOi.
38. 76 gallons at £0 13 4 #", vill cost £50 13 4.
39. 92 gallons at 0 14 2 #*, will cost 65 3 4.
40. "What is the difierencG between the price of 743
ounces of gold at £3 17s. lO^d. per oz. Troy, and that
of the same weight of silver at 62d. per oz. .' Ans.
£2701 2.9. S^d.
41. In the time of King John (money being then more
valuable than at present) the price, per day, of a cart
v/ith three horses was fixed at Is. 2d. ; what would be
the hue of such a cart for 272 days ? Ans. £15 175. 4d.
42. Veils have been made of the silk of caterpillars,
a square yard of which would weigh about 4 grains ;
what would be the weight of so many square yards of
this texture as would cover a square English mile ?
Ans. 2151 ft), 1 oz., 6 dwt., 16 grs., Troy.
QUESTIONS TO BE ANSWERED BY THE PUPIL.
1. Can the multiplier be an applicate number } [20J.
2. What is the rule for compound multiplication
when the multiplier does not exceed 12 ? [21].
.3. What is the rule when it exceeds 12, and is a
composite number .' [25].
132 COMPOUND DIVISION.
4. When it is the simi of composite numbers ? [26].
5. When it exceeds 12, and not a composite number ?
6. How is compound multiplication proved ? [24].
co:mpound division.
29. Compound Division enables us, if we divide an
applicate number into any number of equal parts, to
ascertain what each of them will be ; or to find out
how many times one applicate number is contained in
another. -
If the divisor be an applicate, the quotient will be an
abstract number — ^for the quotient, when multiplied by
the divisor, must give the dividend [Sec. II. 79] ; but
two applicate numbers cannot be multiplied together
[20] . If the divisor be abstract, the quotient will be
applicate — for, multiplied by the quotient, it must give
the dividend — an applicate number. Therefore, either
divisor or quotient must be abstract.
30. When the divisor is abstract, and does not ex-
ceed 12 —
Rule — I. Set down the dividend, di\nsor, and sepa-
rating line — as directed in simple division [Sec. II. 72].
H. Divide the divisor, successively, into all the deno-
minations of the dividend, beginning with the highest.
III. Put the number expressing how often the divisor
is contained in each denomination of the dividend under
that denomination — and in the quotient.
IV. If the divisor is not contained in a denomina-
tion of the dividend, multiply that denomination by the
number which expresses how many of the next lower
denomination is contained in one of its units, and add
the product to that next lower in the di^^[dend.
V. " Reduce" each succeeding remainder in the same
way, and add the product to the next lower denomi-
nation in the dividend.
AQ. If any thing is left after the quotient from the
lowest denomination of the dividervd Is obtained, put i*
COMPOUND DIViBION. 13'
down, TTitli the divisor under it, and a separating line
between : — or omit it, and if it is not less tlian Lalt*
the divisor, add unity to the lowest denomination of the
quotient.
31. Example 1.— Divide £72 Gs. 9^1 by 5.
£ s. d. '
5)72 6 n
14 9 41
5 will go into 7 (tens of pounds) once (ten time.-), aivl
leave 2 tens. 5 will go into 22 (units of pounds) 4 times, and
leave two pounds or 40s. 40s. and Gs. are 4G.s., into which 5
will go 0 limes, and leave one shilling, or 12//, 12'i. and ihl.
are 21d., into which 5 will go 4 times, and leave 1(/., or 4
farthings. 4 farthings and 2 farthings are 6 farthings, into
which 5 will go once, and leave 1 farthing — still to be divided ;
this would give i, or the fifth part of a farthing as quotient,
which, being less than half the divisor, may be neglected.
A knowledge of fractions will hereafter enable us to
understand better the natui-e of these remainders.
Example 2.— Divide £52 4s. IM. by 7.
7)52 4 If
~7 9 2
One shilling or 12</. are left after dividing the shillings,
which, with the Id. already in the dividend, make 13(7. 7
goes into 13 once, and leaves 6c?.. or 24 farthings, which,
with ^. make 27 farthings. 7 goes into 27 3 times and G
over; but as 6 is more than the half of 7. it may be consi-
dered, with but little inaccuracy, as 7 — which will add one
farthing to the quotient, making it 4 farthings, or one to
be added to the pence.
32. This rule, and the reasons of it, are substantially the
game as those already given [Sec. II. 72 and 77]. The remain-
der, after dividing the farthings, may, from its insignificance,
bo neglected, if it is not greater than half the divisor. If it is
greater, it is evidently more accurate to consider it as giving
one farthing to the quotient, than 0, and therefore it is proper
to add a farthing to the quotient. If it is exactly half the
divisor, we may consider it as equal either to the divisor, or 0.
33. Compound division may be proved by multipli-
cation— since the product of the quotient and divisor,
plus the remainder, ought to be equal to the dividend
[Sec. II. 79].
134
COMPOUND DIVISION.
EXERCISES.
£ S.
d. £ s.
fl.
1.
9G 7
6-
- 2=48 3
9.
2.
76 14
7-
- 3=25 11
61
3.'
47 17
G-
- 4=U 19
4h.
4.
98 19
4-
- 5=19 7
10 i.
5.
77 16
7_
- G=12 19
51
6.
32 12
2-
- 7= 4 13
2.
7.
44 16
7 —
- 8= 5 12
1."
8.
97 14
3-
_ 9=10 17
11.
9.
147 14
G-
-10=14 15
5i
0.
157 16
7-:
-11=14 6
Hi
1.
176 14
6-
-12=14 14
6i
The above quotients are true to the nearest fit
ehinor.
34. Wkon the divisor exceeds 12, and is a composite
number —
Rule. — ^Divide successively by the factors.
Examplj:.— Divide £12 Us. 9d. by 36.
3)12 17 9
12)4 5 11
36=-^Xl2 7~2"
This rule will be understood from Sec. II 97.
EXERCISES
£ s. J.
£ s.
12.
24
17
6-j- 24= 1
0
81
13.
576
13
3-j- 36=16
0
4^
14.
447
12
2-1. 48= 9
6
6.
15.
547
12
4-4- 56= 9
15
7.
16.
9740
14
6-rl20=81
3
5^
17.
740
13
4-j. 49=15
2
31
35. "When the divisor exceeds 12, and is not a com-
posite number —
Rule. — Proceed by the method of long division ;
but in performing the multiplication of the remainders
by the numbers which make them respectively a deno-
mination lower, and adding to the products of that next
lower denomination whatever is already in the dividend,
set down the multipliers, &c. obtained. Place the quo-
tient as directed in long division [Sec. TI. S9].
COMPOUXD DIPISiOX. 135
Example.— Divide £87 1G..\ 4<7. hy C2.
£, s. d. £ ^. il.
62)87 IG 4 (1 8 4.
62
25
20 multiplier.
vhillings 51G(=2.5x20-f-lG)
40G
"20
12 multiplier.
pence ^(=20xl2-f4)
186
4 multiplier
farthings 232 (=58x4)
18G
"46
62 goes into £87 once (that is. \t gives £1 in the nuotionr'),
and leaves £25. £25 are equal to 500.?. (25x20j, which,
with IGs. in the dividend, make 51.6.S. C2 goes into 516<. 8
times rthat is, it gives 8.5. in the quotient), and leaves 205.,
or 240r7. (20x12) as remainder. t«2 goes into 240, ka.
Were we to put | in the quotient, the remainder would be
46. which is more than half the divisor: we consider the
quotient, therefore, as 4 farthings, that is. we add one pennj?
to (3) the pence supposed to be aJready in the quotient.
£1 85. Ad. is nearer to the true quotient than £1 8s. 3^d.[32].
This is the same in principle as the mle given above [30] —
bui since the numbers are large, it is more convenient actually
to set down the sums of the diiierent denorainations of the divi-
dend and the preceding remainders (reduced), the products ol
the divisor and quotients, and the numbers bv which we multi-
ply for the necessary reductions : this prevents the memory
from being tc-o much burdened [Sec. II. 93].
36. Y\"hen the divisor and dividend are both applicate
numbers of one and the same denomination and no
reduction is required —
PtULE. — Proceed as already dkected [Sec. II 70,
72, or S9].
136
COMPOUND DIVISION. '
Example. — Divide £45 l3Y £5.
£5)45
That is £5 is the ninth part of £45 ,
37. When the divisor and dividend are applicate, but
not of the same denomination ; or more than one de-
nomination is found in either, or both —
EuLE. — Reduce both divisor and dividend to the Iot^
est denomination contained in either [3], and then pru
ceed with the division.
Example.— Divide £37 5s. 9ld. by 35. O^d.
s. d. £ s. d.
3 61 37 5 91
12
20
42
4
170 farthinors.
745
12
8949
4
170)35797(211
340
179
170
"~97
Therefore os. G\u. i.s Hie
211th pai't of £37"5s. 9] J.
97 not being less than the half of 170 [32], vre consider it
PS equal to the divisor, and therefore add 1 to the 0 obtained
as the last quotient.
EXERCISES.
£ s.
d.
£ s. d.
18.
17G 12
2 -
- 191= 0 18 6.
19.
134 17
8 -:
- 183= 0 14 9.
20.
4736 14
7 —
_ 443=10 13 lOi
21.
73 16
7 -
- 271= 0 5 5^.
22.
147 14
w -
- 973= 0 8 0^
23.'
157 16
7 -
- 487= 0 6 5|.
24.
58 15
o _
- 751= 0 1 6|
25.
62 10
6i-
- 419= 0 2 Ul
26.
8764 4
Oi-
- 408=18 14 6^.
27.
4728 1)
2 -
- 317=14 18 4i.
28.
8234 0
54-
- 261=31 10 Hi
29.
5236 2
71-
- 875= 5 19 84.
80.
4593 4
2 -
^9812= 0 9 44.
COMPOUND DIVISION. 137
31 . A cubic foot of distilled water weighs 1000 ounces
wliat will be the weight of one cubic inch ? Arts
253'1S29 grains, nearly.
32. How many Sabbath days' journeys (each 1155
yards) in the Jewish days' journey, which was equal to
33 miles and 2 furlongs English } Ans. 50'66, &c.
33. How many pounds of butter at llf^. per lb
would purchase a cow, the price of which is £14 155. ?
Am. 301 -2706.
QUESTIONS FOR THE PUPIL.
1. What is the use of compound division } [29].
2. What kind is the quotient when the divisor is an
abstract, and what kind is it when the divisor is au
applicate number ? [29] .
3. What are the rules when the divisor Is abstract,
and does not exceed 12 ? [30] ;
4. When it exceeds 12, and is composite .^ [34J ;
5. When it exceeds 12, and is not composite ? [35] ;
6. And when the divisor is an apphcate niunber ? [36
and 37|.
133
SECTION IV.
FRACTIONS.
1. If one or more units are divided into equal parts,
and one or more of these parts are taken, we have what
is called a fraction.
Any example in division — before the process has been
performed — may be considered as affording a fraction : — ■
thus I (which means 5 to he divided by 6 [Sec. II. 68] )
is a fraction of 5 — its sixth part ; that is, 5 being divided
into six equal parts, f will express one of them ; or (as
we shall see presently), if unity is divided into six equal
parts, five of them will be represented by f.
2. When the dividend and divisor constitute a frac-
tion, they change their names — the former being then
termed the numerator ^ and the latter the denomin/itor ;
for while the denominator tells the denomination or
kind of parts into which the unit is supposed to be
divided, the numerator numerates them, or indicates the
number of them which is taken. Thus f (read three-
sevenths) means that the parts are " sevenths," and that
" three" of them are represented. The numerator and
denominator are called the terms of the fractions.
3. The greater the numerator, the greater the value
of the fraction — because the quotient obtained when we
divide the numerator by the denominator is its real
value ; and the greater the dividend the larger the
quotient. On the contrary, the greater the denomina-
tor the less the fraction — since the larger the divisor
the smaller the quotient [Sec. II. 78] : — hence -f is
greater than f — which is expressed thus, ■f]>4 ; but |
is less than | — which is expressed by f <^-f--
4. Since the fraction is equal to the quotient of its
numerator divided by its denominator, as long as tbis
quotient is unchansed, the value of the fraction is the
■^ \VA can
•ne
FRACTIONS.
139
to increase or diminisli both the dividoiiJ and divLsor —
which does not affect the quotient.
5. The following will represent unity, seven-sevenths,
and five-sevenths.
Unity.
.1 ^ I
The very faint lines indicate wliat -f wants to make
it equal to unity, and idcnticil icith ^. In the diagrams
which are to follow, we shall, in this manner, generally
subjoin the difference between the fraction and unity.
The teacher should impress on the mind of the pujiil
that he might have chosen any oiker unity to exemplify
the nature of a fraction.
6. The following will show that -f may be considered
as either the 4 of 1, or the -i of 5, both — though not
identical — being perfectly equal.
I of 5 units.
2 of 1 unit.
Unity.
m
J 1 11 1 1
'III IT
"mil
' mil
c^
In the one case we may suppose that the five parts
belong to but one unit ; in the other, that each of the
five belongs to different units of the same kind.
^ Lastly, 4 may be considered as the 4 of one unit fivo
times ri.<s large as the former ; thus —
4 of 1 unit. 4 of 5 units.
equal to
« ___^___^_^_
140
FRACTIONS.
7. If its numerator is equal to, or greater than its
denominator, the fraction is said to be imjproper ; be-
cause, although it has the fi*actional form, it is equal
to, or greater than an integer. Thus | is an improper
fraction, and means that each of its seven parts is equal
to one of those obtained from a unit divided into iivc
equal parts. When the numerator of a proper fraction
is divided by its denominator, the quotient will be ex-
pressed by decimals ; but when the numerator of an
improper fraction is divided by its denominator, part,
at least, of the quotient will be an integer.
It is not inaccurate to consider | as a fraction, since
it consists of " parts " of an integer. It would not,
however, be true to call it jpart of an integer ; but this
is not requu-ed by the definition of a fraction — ^which,
as we have said, consists of " part," or " parts " of a
unit [1].
8. A mixed number is one that contains an integer
and a fraction ; thus If — which is equivalent to, but
not identical with the improper fraction |. The fol-
lowing will exemplify the improper fraction, and its
equivalent mixed number —
Unity.
d
cniD
-i 1 -; 1
Unity -f-
9. To reduce an improper fraction to a mixed number
An improper fraction is reduced to a mixed number if
we divide the numerator by the denominator, and, after
the units in the quotient have been obtained, set down
the remainder with the divisor under it, for denominator ;
thus I is evidently equal to If — as we have already
noticed when we treated of di^asion [Sec. II. 71].
10. A simjph fraction has reference to one or more
integers ; thus ^ — which means, as we have seen [6] ,
the ^ug-sevenths of oiic unit, or the (?7i€-scventh of fivt
units.
FRACTIONS
141
11. A compound fraction supposes one fraction to
refer to another ; thus a of f — represented also by | X ^
(three-fourths multiplied by four-ninths), means not
the four-ninths of unity, but the four-ninths of the
three-fourths of unity : — that is, unity being chvided into
four parts, three of these are to be divided into nine
parts, and then foui' of these nine are to be taken ; thus —
3
4
M I i
■^ I I
■H I I
I I
I I
I I
12. A complex fraction has a fraction, or a mixed
2.
number in its numerator, denominator, or both ; thus ^,
which means that we are to take the fourth part, not
of unity, but of the | of unity. This will be exem-
plified by —
I f
Unity.
I
8 I 1
7' ~5
— 5-^, are complex fractions, and will be better
5 8 ^ -^6
understood when we treat of the divLsion of fractions.
13. Fractions are also distinguished by the nature of
their denominators. When the denominator is unity^
followed by one or more cyphers, it is a decimal frac-
tion— ^thus, /„-, xo'Vo ) ^c- 5 ^U other fractions are vulgar
—thus, f , I, ^f ^, &c.
Arithmetical processes may often be performed with
fractions, without actually dividing the numerators by
the denominators. Since a fraction, like an integer,
may be increased or diminished, it is capable of addi-
tion, subtraction, Sec.
142 FRACTIONS
14. To rediK-e an iut(^gcr to a fraction of any deno-
niinatiou.
An integer may be considered as a fraction if we make
unity its denominator : — thus f may be taken for 5 ;
since | = 5.
We may give an integer any denominator we please
if we previously multiply it by that denominator ;
^, .25 30 35 25 5X5 5 ^
thus, o^-, or -, or y, &c, for -=^—^=- = 5 ;
,30 5X6 5
EXERCISES.
1. Reduce 7 to a fraction, having 4 as denominator
Ans. V-
2. Reduce 13 to a fraction, having 16 as denomina-
tor. Ans. Yf •
3. 4=V- I 4. 19 = V. I 5. 42 = V2'. | 6. 71 = 6^i4.
15. To reduce fractions to lower terms.
Before the addition, &lc., of fractions, it will be often
convenient to reduce their terms as much as possible.
For this purpose —
Rule. — Divide each term by the greatest common
measure of both.
40 5 ^ 40 40-^8 5
Ex.MPLE.-^=^-. For^=^^^g=g-.
We have abeady seen that we do not alter the quotient—
which is the real value of the fraction [4] — if we multiply or
divide the numerator and denominator by the same number.
What has been said, Sec XL 104, will be usefully remem-
bered here.
EXERCISES.
Reduce the following to tlieir lowest terms.
7
8
9
10
11
12. T.
574
— 237
Tn?0
— J40'
41U_
_4 1
TWO
-T¥-
9"6_
-976
TT3 —
-743-
549
_ 183
TT43
— 2 3^T
240_
_12n
Tfl-
-jirr-
1Q 6 3_7
• tXB— Ta-
le: 39 13
^^- 92 37-
1A 43 4
■'■"• ^0 ~5-
neo 5
• ns-
18. J^— 7
20. ,,,^
19.
20.
Ol 5131)0 1120
-^- -^l-^J — ISTx
99 42 5 8 5
90, 412 .2 0 8
24 5J.2_25J5
■"^' Bl4 307*
In the answers to questions given as exercises, we shall,
in future, generally reduce fractions to their lowest deno-
minations
143
16. To find the value of a fraction in terms of a
lower denomination —
RuLE.^ — Ketiuee the numerator by the rule already
given [Sec. III. 3], and place the denominator under it.
ExAMPLR. — What is th(? value, in shillings, of J of a pound ?
£,Z reduced to shillings=GOs. ; therefore £>j reduced to shil-
liugS=V'-5.
The renson of the rule is the same as that already given
[See. III. 4]. The I of a pound becomes 20 times as much if
the " unit of comparison" is changed from a pound to a shilling.
We may, if we please, obtain the value of the result-
ing fraction by actually performing the diviaiou [9] ;
thus \°s.=^\ds. : — hence £^^^Ids.
EXERCISES.
25. £?-^=14.s. Gd.
20. £|5=17.?. 4c/.
27. £i['=19c?.
28. £?=15.?.
29. £-r\=5.^.
30. £^U=^cf-
17. To express one quantity as the fraction of an-
other—
Rule. — Reduce both quantities to the lowest deno-
mination contained in either — if they are not already
of the same denomination ; and then put that which is
to be the fraction of the other as numerator, find the
remaining quantity as denominator.
E.VAMPLE. — "What fraction of a pound is 2^d. ? X1=0G0
farthings, and 2|J.=9 farthings , "
quired fraction, that is, 2|;d.=£^^^f.
Reason of the Rule. — One pound, for example, contains
9G0 farthings, therefore one farthing is £g j-^- (the 9G0th pare
of a pound), and 9 times this, or 2]-, is •£9Xg]o=ofo.
EXERCISES.
31. What fraction of a pound is 14s. ed. .? Ans. ff.
32. What fraction of £100 is 17;?. 4d. r Aiis. yif ^.
33. What fraction of £100 is £32 105. .? Ans. \l.
34. What fraction of 9 yards, 2 quarters is 7 yards,
3 quarters } Ans. ^\.
35. What part of an Iri.sh is an English mile } Ans. \{.
36. What fraction of 6^. ^d. is 25. 1^. } Ans. j\ .
37. What part of a pound avoirdupoise is a pound
Troy .' Alls, -j-ff .
14^ VULGAR FRACTIONS.
QUESTIONS.
1. What is a fraction ? [1].
2. When the divisor and dividend are made to con-
stitute a fraction, what do their names become ? [2] .
3. What are the effects of increasing or diminishing
the numerator, or denominator ? [3] .
4. Why may the numerator and denominator bo mul-
tiplied or divided by the same number without altering ,
the value of the fraction ? [4] .
5. What is an improper fraction } [7].
6. What is a mixed number .? [8].
7. Show that a mixed number is not identical with
the equivalent improper fraction r [8] .
8. How is an improper fraction reduced to a mixed
number ? [9] .
9. What is the difference between a simple, a com-
pound, and a complex fraction ? [10, 11, and 12] ;
10. Between a vulgar and decimal fraction ? [13].
11. How is an integer reduced to a fraction of any
denomination? [14].
12. How is a fraction reduced to a lower term ?
[15].
13. How is the value of a fraction found in terms of
a lower denomination ? [16] .
14. How do we express one quantity as the fi-action
of another.? [17].
VULGAR FRACTIONS.
ADDITION.
18. If the fractions to be added have a common
denominator —
Rule. — Add all the numerators, and place the com-
mon denominator under then* sum.
Example. — f -f- 1 = y .
Eeason- of the Rule. — If ^ve add together 5 and 6 of any
kind of individuals, their sum must be 11 of the same kind
of individuals— siiica the process of additiou has not changed
VULGAR FRACTIONS.
145
their nature. But tlic units to be added were, iu the present
instance, sevenths ; thereibre their sum consists of sevenths.
Addition may be illustrated as follows : —
6
1
Unity
U-l.
iLij
-f
KXEnCISES.
1. 3_^t_|_3^1.:J=15.
O J_L^ I ^ ■'•
^ ll_l_10 |_ri 30 O 4_
"^^ is~r~V3'T-f3 — 13 — -"3"
A I '_L 3 _l_ 5 -'5 1 I I
5.
-K^-frk
6. l+?4-H5=U-
7 I s_i_i 7_i_i :! +y O ''
8- H+}H-H=H=^.^-
n 13|14J_11 33 O
2 3 ^23"
Si-
ll T4 I fi I r2_:T7
f^;*-- 2_3"r2rT^?3 — Lf3 — :r23
li'
1-1
15
IG.
_ •: I !_ ii
23 I 23 1" 23 23'
i* I 1 » I 2 fi :: i
rT"rTT~rTT — fT-
19. If the fractions to be added have not a common
denominator, and ail the denominators are prime to each
other —
Rule. — Multiply the numerator and donominator of
each fraction by the product of the denominators of all
the others, and then add the resulting fractious — by
the last rule.
Ex.\MPLK. — What is the sum of §4-|+f ^
2 3 4 2x4x7 . 3x3x7 4x3x4^56 63 4S_1G7
3+4+7~3x4x7"^4x3x7"*"7x3x4 84"^84"^84~ 84
Having fuund the denominator of one fraction, we may at
once put it as the common denominator : since the same
factors (the given denominators) must necessarily produce
the same product.
20. IxEAJ'ON- OF THE EcLE. — To bring the fractions to a
common denominator v<'e have merely multiplied the nume-
rator and denominator of e.ich by the same number, "svhich
[4] does not alter the fraction. It is necessary to find a
common denominator ; for if ve add the fractions with-
out so doing, Tvfe cannot put tlie denominator of any one
o{ them as the denominator of their stun; — thus ""^ '^
3
for instance, would not be correct — since it would suppose all
the quantities to be thirds, while some of them are fourtlis
an'l sevenths, which are less than thirds ; neither would
24-3+4 ,
— ij be correct — since it would suppose aU c/ them to ba
146
VULGAR FRACTIONS.
Bevenths, altliougli some of thein are thirds and foxirthS;
which are greater than sevenths.
21. In altering the denominators, we have only changed
the parts into whicli the unit is sujiposed to he divided, to an
equivuleut number of others which are smaller, it is neces-
Barj to diminish the size of tliese parts, or each fraction would
not be exactly equal to some number of them. This will
be more evident if we take only two of the above fractions.
Thus, to add f and |,
2 3 2x4 3x3 8^ , 9 _1 7
3~''4~'8 X 4 4 X 3~12'^12~1 2
These fractions, before and after they receive a common
denominator, will be represented as follows : —
Uuitj.
3
4
equal to
equal to
Ti \
"We have increased the number of the parts just as much
as we have diminished their size ; if we had taken parts larger
than twelfths, we could not have found any numbers of them
exactly equivalent, respectively, to both i and |.
EXERCISES.
17 14_2_|_4 59 129
18.
+f+^
21.
22
23. To-rjT"r-yF.T — -^2ft7T3 0'
9 I 83 1 91 I 47 0272933
.3 3 7 112 7
'2 1 0 ^2 I 0-
' - .{■-"-' -1- 4 9_573
2 0 1^2 1 T^^ "1 iS4?r-
1 7 1421 45 1 134937
3i>T-jT"ryF.T — -'^r ~
^°- 3 + 4-r:^ rTiT
10 2_l_2_l_2 142 "I 37
■•"X* 3i^5-ri — Too— '•ToTT-
90 3 I 2_I5 2ol 112 1
^^- 4-rj-t-T— T40— '■T40-
22. If the fraction.s to be added have not a coininon
denominator, and all the denominators arc not primo
to each other —
Proceed as directed by the last rule ; or —
Rule. — Find the least common nmltinle of all the
denominators [Sec. II. 107, &c.], this will be the common
denominator; multiply the numerator of each fraction
VULGAR FRACTION'S. i4?
int<> the qnotient obtained on dividing the common mul-
tiple by its denominator — this will give the new nume-
vators ; then add the numerators as already directed [IS].
Example.— Add A + A -J- -\- —3 is the least common
.5 4 3 2S8-r-82X'5
multiple of 32, 48, and 72 ; therefore so+^+f^^ 2^ "
289^48X4 288-^72 X3_ 45 24 "j^_81*
+ 2>5 + 2bS ==28S+2fe3+2S8— 288*
23. Reason- of the Rule. — "We have multiplied each nume-
rator and denominator by the same number (the least common
multiple of the denominators [4] J — since 5 X 288 -7-32
288 ^^•'^
in3tance)= _p^-^^ For we obtain the same quotient, whether
324-2^8"
we multiply the 'livisor or divide the dividend by the same
number — as in both cases we to the very same amount,
diminish the number of time^ the one can be subtracted from
the other.
When the denominators are not prime to each other the frac-
tions we obti^in hare lower terms if we make the least common
multiple of the denominators, rather than the product of the
denominators, the common denominator. In the present in-
stance, had we proceeded accorcling to the last rule [19], w«
,, , .,583 17230 1S432 4608
would have found _ i __i__ t — _ — _l.-— -^
32^^48 n'2~110.3y2^11Uo92 ' 110.5y2
40320 , . 40320
1 105F* ' 11059^ ^ evidently a fraction containing larger
terms than ii_
288-
EXERCISES.
1 2
32. H4l4-|=if^=2
O 1 5 (41! i'-' 1 3
^^- Ti-t-T-i-D— 4 2— ^sT-
•^5 21514 3 5 11?
^■-'- rTT-TQ — r^ — ^Tf-
'\Ci 1 1 3 I 4 I 5 3 3 2
3^- i+H-^-H-^-^^'^^
0240
"^9 2 4 0*
OQ 5 15 15 3 I- 5 0 29
O^l ±_l_2l i IJT 1_5_
Qil 2_l 2_1_5 5? 1 i>
0 1 15 117 15 2C1T O 6 01
^^- 19 Pisi"-: — TooB 'TcToT-
24. To reduce a mixed number to an improper frac-
tion—
Rule. — Change the integral pai-t into a fraction,
having the same denominator as the fractional part
[14], and add it to the fractional part.
Example. — What fraction is equal to 4| ? 45 = ^-(-.|=s
8r. I 5 — 4_j
V ^ ti « •
1.48
VULGAR FRACTIONS.
25. Reason" of the Rxjle. — We have already seen that an
integer may be expressed as a fraction having any denomi-
nator we please : — the reduction of a mixed number, there-
fore, is really the addition of fractions, previously reduced to
a common denominator.
EXERCISES.
38.
16^=U3.
30.
18|='|^
40.
41.
47i='r.
42.
74i=^r-
43.
95W^«.
44.
00 1 1 0 9
1 1 I 1
45.
12^=VV
4G.
i'H=V-^
47.
4Gf=^|3.
48.
iH='r-
49.
2TH=W
26. To add mixed numbers —
KuLE. — Add together the fractional parts ; then, if
the sum is an improper fraction, reduce it to a mixed
number [9] , and to its integral part add the integers in
fthe given addends ; if it is not an improper fraction, set
it down along with the sum of the given integers.
Example 1. — What is the sum of 4|-j-18g ?
: I 5 I 2 1 4
5"r8 — ^ — -^0
u
18:
sum 23 1
7 eig;hths are 12 eighths ; but, as 8 eightha
5 eighths and
make one unit, 12 eighths are equal to one unit and 4
eighths — that is. one to be carried, and | to be set down. 1
and 18 are 19, and 4 are 23.
Example 2.— Add 12i and 2911.
sum 421^
In this case it is necessary, before performing the addition
[19 and 22], to reduce the fractional parts to a comnitin
denominator.
27. Reason of the Rule. — The addition of mixed num-
bers is performed on the same principle as simple addition
but, in the first example, for instance, eight of one denomina-
tion is equal to one of the next — -while in simple addition [Sec
II. 3], ten of one denomination is equal to one of the next.
EXERCISES.
50. 4i-f3?=8|. 55. 3i+lU+14|!=29f3i,
51. 8U+2;ii=lli|i. 56. 40,^-f38i-f-40i=119|.
52. 19^3,-+7|=26:|i. 57. 81?-f-6'^+ll=99J^.
53. 10^-fllVW22^V S8. -- - '■
54. 1U+8]=193. . 59.
92^^-+37^+7|=i37f-5^
17.V2-F«^4-91|H273f^l-
VULGAR FRACTIONS.
149
QUESTIONS.
1. WJiat is tlio rule for adding fractions which have
A commoQ dcnomiuator ? [18].
2. How are fractions brought to a common denomi-
nator ? [19 and 22].
3. What is the rule for addition when the fractions
have different denominators, all prime to each other .''
[19].
4. What is tho rule when the denommators are not
the same, but are not all prime to each other .' [22] .
5. How is a mixed number reduced to an improper
fraction t [24] .
6. How are mixed numbers added t [26].
SUBTRACTION.
28. To subtract fractions, when thcj have a common
leuominator —
Rule. — Subtract the numerator of the subtrahend
iTom that of the minuend, and place the common deno-
jninator under the difference.
Example. — Subtract |- from |.
7 4 7-4 3
9~0"~ 0 "~0'
20. Reasox of the Rule. — If we take 4 individuals of any
kind, frum 7 of the same kind, three of them will remain, lu
the example, we take 4 (ninths) from 7 (.nintlis), and 3 are left —
which must be ninths, since the proce.ss of subtraction cannot
have changed their nature. The following will exemplify the
eubtractiou of fractions : —
Unity
U
I!!ii
s
- 1 ^
j !
1 i
i
160 VULGAR FRACTIONS.
EXEUCISES.
■*•• 12 12 i- I "-2 3 2 3
2 15 "f 1 71U 8
• T¥ T^ — 2- '• ar 2T—
•i i9._i7 J_ I Q '?_l 3
^- 20 20 10* I "• ^ 1 4*
4. 17 5 2 Q "7 4 _
5. 2J. 7___JL. I 10.
22 11* • 27 2l 9"
30. If tlie subtrahend and minuend have not a com-
mon denominator —
Rule. — Reduce them to a common denominator [19
and 22] ; then proceed as directed by the hist rule.
Example. — Subtract | from |.
7 5 6 3 4 0 2 3
■& 9 T2 T2 T2'
81. ReasojV of the Rule. — It is similar to that already
given [20] for reducing fractions to a common denominator,
previously to adding them.
exercises.
11.
1-1=^.
15.
m-m=^%
12.
Ii-T^^=f|.
16.
u-^^=mi
13.
7 3 I
^~"4 ?•
17.
3ti 43 5
14.
i|-|t=rtT-
18.
32. To subtract mixed numbers, or fractions from
mixed numbers.
If the fractional parts have a common denominator —
Rule — I. Subtract the fractional part of the subtra-
hend from that of the minuend, and set down the differ-
ence with the common denominator under it : then
subtract the integral part of the subtrahend from the
integral part of the minuend.
II. If the fractional part of the minuend is less than
that of the subtrahend, increase it by adding the com-
mon denominator to its numerator, and decrease the
integral part of the minuend by unity.
Example 1. — 4| from 9|.
9 1 minuend.
4| subtrahend,
5^ difference.
3 eighths from 5 eighths and 2 eighth.8 (==]) remain. 4
from 9 and 5 remain.
VULGAR FRACTIONS. 151
ExAMPLK 2.— Subtract 12? from 18].
181 minuend.
12 J subtrahend.
51 difference.
o fourths cannot be taken from 1 fojirth ; but (borrowing
one from the next denomination, considering it as 4 fourths,
and adding it to the 1 fourth) 3 fourths from 5 fourths and
2 fourths (==^) remain. 12 from 17, and 5 remain.
If the minuend is an integer, it may be considered as
a mixed number, and brought under the rule.
Example 3. — Subtract 34 from 17.
17 may be supposed equal to 17^; therefore 17—34 =
ll's-H- But, by the rule, 175-34 = lGf-34 = 13i.
33. Reason of the Rule. — The principle of this rule is
the same as that already given for simple subtraction [Sec
II. 19] : — but in example 3, for instance, five of one denomina-
tion make orie of the next, while in simple subtraction ten of
one, make otie of the next denomination.
34. If the fractional parts have not a common deno-
minator—
Rule. — Bring them to a common denominator, and
then proceed as du-ected in the last rule.
Example 1.— Subtract 42i from 561.
56|=56y*^, minuend.
subtrahend.
35. Reason of the Rule. — We are to subtract the dif-
ferent denominations of the subtrahend from those which cor-
respond in the minuend [See. II. 19] — but we cannot subtract
fractions unless they have a common denominator [30].
EXERCISES.
19. 27f— 31=241.
20. 15|-7|=7f.
21. 12#-12'=|.
22. 8411—11=84.
23. 147if — if=1473
24. 821^1— 7|if=74if4.
25 76'— 709=0,39 *^^
26. 671 — 34^=32i|.
27. 971—3211=643^
28. 604— 41f?j=19i.
29. 92^— 90-fV=2-fi;
30. 1001— 9I=90|.'*
31. 60— .3 =59^,-.
32. 121—101=11.
152 X TULGAR FRACTIONS.
QUESTIONS.
1. "What is the rule for the subtraction of fractions
when they have a common denominator ? [28] .
2. What is the rule, when they have not a common
denominator ? [30] .
3. How are mixed numbers, or fractions, subtracted
from mixed numbers, or integers ? [32 and 34] .
MULTIPLICATION.
36. To multiply a fraction by a whole number; or
the contrary —
Rule. — Multiply the numerator by the whole number,
and put the denominator of the fraction under the pro-
duct.
Example.— Multiply f by 5.
7X0 y .
87. Reason- of the Rule. — To multiply by any number,
we are to add the multiplicand [Sec. II. 33] so many times
as are indicated by the multiplier ; but to add fractions having
a common denominator we must add the numerators [18], and
put the common denominator under the product. Hence —
4 vr, _ 4 , 4 , 4 , 4 , 4_4-}-4-|-4-M-H_4x5_20
7 7/777 7 77
We increase the nnmher of those " parts" of the integer
which constitute the fraction, to an amount expressed by the
multiplier — their size being unchanged. It would evidently
be the same thing to increase their size to an equal extent
•without altering their nnmher — tliis would be effected by
dividing the denominator by the given multiplier ; thus
-j^-j X 5=-*. This will become still more evident if we reduce
the fractions resulting from both methods to others having a
common denominator — for — (=__), and - ( =-= — -\
15 \ 15 / 3 \ 15-7-5/
will then be found equal.
As, very frequently, the multiplier is not contained in the
denominator any number of times expressed by an integer,
the method given in the rule is more generally applicable.
The rule will evidently apply if an integer is to be multi-
plied by a fraction — since the same product is obtained in
whatever order the factors are taken [Sec. II. 35].
VULGAR FRACTIONS.
153
38. The integi-al quantity whicli is to form one of
the factors may consist of more than one denomination
EX.A.MPLE.— What is the 4 of £5 2s. 9^/. ?
s. d. d
2 9x|J
d.
9x2
1. |X2=lf.
2. 5x8=6f.
3. ^x 12=104.
4. ^Xl2=9i.
5. v.x30=i4.
EXERCISES.
27xt=12.
YVxl8=3f.
fix8=7i.
21x5=9.
10. 15x*,=3.
11. j|x<eG=34.
12. i|x20=19.
13. 22xf=4i.
14. AX17=1V^.
15. 143x?=61|-
Ahs
16. How much is yVe of 26 acres 2 roods ?
20 acres 3 roods.
17. How much is ^f of 24 hours 30 minutes } Ans
7 hours.
18. How much is ^W\ of 19 cwt., 3 qrs., 7 Eb ? Ans
7 cwt., 3 qrs., 2 ib.
19. How much is if of £29 } Ans. £\V=£8 195
6^d.
39. To multiply one fraction by another —
Rule. — 3Iultiply the numerators together, and under
then- product place the product of the denominators.
Example.— ^vlultiply | by f .
4x5^20
9x6 54'
40. Reason of the Rule. — K, in the example given, -we
were to multiply ^ hy 5, the product (-/) would be 6 times
too great — since it was by the sixth part of 5 (f), we should
have multiplied. — But the product will become what it ought
to be (that is, 6 times smaller), if we multiply its denominator
by 6, and thus cause the size of the parts to become 6 times less.
We have already illustrated this subject when explain-
ing the nature of a compound fraction [11].
EXERCISES.
25 . ^ X f XyV=ftt-
26. 1x4=^.
07 3 14 V 1T7__ 93P3 01 3 v'ti
■^ ' ■ 4 J3 -^ 3TT — 233.fji ^^- IT'^S TS-
is th(
9^6
7 \/ 5 3 5
1 4 \r 5 7
20
21
22. ^xtX^=2
32. How much
33
28.
29.
30.
31.
ToXjt — 3.
1 V ' 1
_4 w _T_ 1
iT-^ 2 4-^1 5-
e|off
H)w much is the i- of i-
A?is.
An^.
154 VULGAR FRACTIONS.
41 . When wc multiply one proper fyaction by another,
we obtain a product smaller than either of the factors. —
Nevertheless such multiplication is a species of addi-
tion ; for when we add a fraction oiice, (that is, when
we take the whole of it,) we get the fraction itself as
result ; but when we add it less than once, (that is, take
so muck of it as is indicated by the fractional multiplier,)
we must necessarily get a result which is less than when
we took the whole of it. Besides, the multiplication
of a fraction by a fraction supposes multiplication
by one number — the numerator of the multiplier, and
(which will be seen presently) division by another — thft
denominator of the multiplier. Hence, when the division
exceeds the multiplication — which is the case when the
multiplier is a proper fraction — the result is, in reality,
that of division ; and the number said to be multiplied
must be made less than before.
42. To multiply a fraction, or a mixed number by a
mixed number.
Rule. — Reduce mixed numbers to improper fractions
[24], and then proceed according to the last rule.
Example 1. — Multiply | by 4|.
4f=V ; therefore f x4f=|xV=V¥-
Example 2.— Multiply 5| by 6f .
5i=V, and Gf=3^2 ; therefore 51x6f=V X ^=^1^.
43. Reasojt of the Rule. — "We merely put the mixed
numbers into a more conYcnient form, without altering their
value.
To obtain the required product, we might multiply each
part of the multiplicand by each part of the multiplier. — Thus,
taking tke first example.
EXERCISES.
34. 8fxl=7H.
35. 5^x?=2ii.
36. 4ix7JrX3=101l.
o/. ToXoj-XttXyo — ^32-
38. 5^-xl0xlOi=880«t.
39. 3^xl9iX|=50if
40. 6|x|X|Xf=23-V
41. 12ixl3i-x6f=1097^
43. 14xl5yVx3f=:749-,VTr.
44. What is the product of 6, and the | of 5 .'
Ans. 20.
45. What is the product of | of f , and |- of 3f ?
Ans. If
. ULGAR FRACllONS. 155
44. If vre perceive the numerator of one fraction to
be the same as the denominator of the other, we may,
to perform the multiplication, omit the number which
is common. Thus f Xf = t-
This is the same as dividing beth the numerator and deno-
minator of the product by the same number — and therefore
does not alter its value ; since
5 6_5x6_5x6j-6_5
6 ^ 9^6x9 ""Cx^TB—y*
45. Sometimes, before performing the multiplication,
we can reduce the numerator of one fraction and the
denominator of another to lower terms, by dividmg
both by the same number : — thus, to multiply f by 4.
Dividing both 8 and 4, by 4, we get in their places,
2 and 1 ; and the fractions then are -| and |, which,
multiplied together, become |Xy = t4-
This is the same as dividing the numerator and denomina-
tor of the product by the same number ; for
3 4 3x4-^4_3xl /_3 1\ 3^
8^7~6xT-r-4'~2x7 \~2^7) ~U'
QUESTIONS.
1 . How is a fraction multiplied by a whole number
or the contrary ? [36] .
2. Is it necessary that the integer which constitutes
one of the factors should consist of a single denomina-
tion ? [3S].
3. What is the rule for multiplying one fi-action by
another? [39].
4. Explain how it is that the product of two proper
fractions is less than either } [41].
5. What is the rule for multiplying a fraction or
a mixed number by a mixed number .' [42].
6. How may fractions sometimes be reduced, ^before
they are multiplied .' [44 and 45] .
156
VULGAR FRACTIONS.
DIVISION.
46. To divide a vulgar fraction by a whole number — •
Rule. — Multiply the denominator of the fraction by
the whole number, and jDut the product under lis nu-+
merator.
2
EXA.MPLE. ~f -f- 4 = ^ : = -p^.
^ 3x4 '-
47. Reaso:v of the Rule. — To divide a quantity by 3,
for instance, is to make it 3 times smaller than before. But
it is evident that if, while we leave the fiumber of the parts the
same, we make their size 3 times less, wo make the fraction
itself 3 times less — hence to multiply the denominator by 3,
is to divide the fraction by the same number.
A similar effect will be produced if we divide the numera-
tor by 3 ; since the fraction is made 3 times smaller, if,
while we leave the size of the parts the same, we make their
8_^ 4 _^Tf _?
numerator is not always exactly divisible by the divisor, tho
method given in the rule is more generally applicable.
The division of a fraction by a whole number has
been already illustrated, when we explained the nature
of a complex fraction [12].
-=-. Cut since the
!9 . IQ 1
1 . o 1
EXERCISES.
1 1 . Q 1 1
r-.'k ~''^'
_9 1.3___3_
0. u-^^=^
10. t-Ml=
11. ^1,-42=,^^
12. ^-14=3V-
19'
48 It follows from what we have said of the multi-
plication and division of a fraction by an integer, that,
when we multiply or divide its numerator and denomi-
nator* by the same number, we do not alter its value —
sinco we then, at the same time, equally increase and
decrease it.
49. To divide a fraction by a fraction —
Rule. — Invert the divisor (or suppose it to be in-
verted), and then proceed as if the fractions were to be
multiplied.
VULGAR FRACTIONS.
157
Example. — Divide f by f .
5 3
4-7^3-7x3-
5x4_20
/ 4 I o "7x3 21'
Reasox of the Rule. —If, for instance, in the example
just given, we divide | by 3 (the numerator of the divisor),
■we use a quantity 4 times too great, since it is not by 3, but
the fourth part of 3 (J) we are to divide, and the quotient
(^_5_^ is 4 times too small. — It is, however, made what it ought
to be, if we multiply its numerator by 4 — when it becomes
1^, which was the result obtained by the rule.
50. The division of one fraction by another may be
illustrated as follows —
-i
i^i
i
—
' ' '. \ '
Unity.
«*-
._Li_.__
J J. ,
■ 1
The quotient of 4-i-| must be some quantity, which,
taken three-fourth times (that is, multiplied by :^), will
be equal to 4 of unity. For since the quotient multiplied
by the divisor ought to be equal to the dividend [Sec
II. 79] , 4 is f of the quotient. Hence, if we divide the
five-sevenths of unity into three equal parts, each of
these will be one-iouith of the quotient — that is, precisely
what the dividend wants to make it foui'-fom-ths of the
quotient, or the quotient itself.
51. "When we divide one proper fraction by another,
the Cjuotient is greater than the dividend. Xevertheless
such division is a species of subtraction. For the quo-
tient expresses hoic often the divisor can be taken fi-om
the dividend; but were the fraction to be divided by
unity, the dividend itself would express how often the
divisor could be taken from it ; when, therefore, the
divisor is less than unity, the number of times it can be
taken from the dividend must be expressed bv a quantity
greater than the dividend [Sec. 11. 76"!. Besides, diA^d-
ing one fraction by another supposes the multiplication
of the dividend by one number and the division of it by
another — but when the multiplication is by a grea';er
-42
168
VULGAR FRACTIONS.
number than the division, the result is, in reality, that
of multiplication, and the quantity said to be divided
must be increased.
13. i-f-|=l^
14. l~-^=h
15.
,1 l3
EXERCISES.
16. *~I=P
17.
18.
1^- II-tV=1i^
Ol 1 . i 2
52. To divide a whole number by a fraction —
KuLE. — Multiply the whole number by the denomi-
nator of the fraction, and make its numerator the deno-
minator of the product.
Example. — Divide 5 by f .
5 ■ 3_5X7^35
• 7 3 3*
This rule is a consequence of tlie last ; for every whole num-
ber may be considered as a fraction haviiig unity for deno-
minator [14]; hence 5-f-?=f-H7=f X^=V-
It is not necessary that the whole number should consist of
but one denomination [38].
Example.— Divide 17s. old. l>y I.
Us. 3ic/.^2=17.s. 3i:(/.x^=£l Ss. OUL
EXERCISES.
22. 3^1=6?.
23. Il-f-f=i94.
24. 42-f-yi-,=864
26.
27.
=20.
Ans.
Ans.
Ans.
28 8-^^4=8*.
20. 14-^^=38.
30. 16-v-i=32.
^17 Us. A^d
£10 85.
£b 95. 2^d.
31. Divide £1 IQs. 2d. by a.
32. Divide £8 13^. Ad. by f .
33. Divide £b Os. Id. by |i.
53. To divide a mixed number by a whole number
or a fraction —
PtULE. — Di^dde each part of the mixed number accord-
ing to the rules already given [46 and 49], and add the
quotients. Or reduce the mixed number to an improper
fraction [24], and then divide, as already directed [40
and 49] .
Example 1. — Divide 9^ by 3.
9^^3=9^34-?-^3=3-fl=3^.
Example 2. — Divide 14y\ by 7.
14yT=-iT ; therefor
e 14/V
= Vt X
3_^1 250
VULGAR FRACTION?. 159
54. IIea?ox of the Rulk. — Ijj the first cxntuplc we have
divided each part of the dividfii'l by the divisor and added
the results — which [Sec. II. 77] is the same as dividing the
^hole dividend by the divisor.
In the second example we have put the mixed number into
a more convenient form, without altering its value.
34. 8|^17=f|.
35. 51f-^3=17#^
36. lS7f>^-i-jK=^mi
37. 19U^4:l=jU-'
QQ -laX'X) . 4 3 if 17
EXERCISES.
30
5_^41 'iJ5450
40. 84li^2l;3lll"""
42. ioB.|y-i:^=iy6M'.
43. 184-j.ll=l|i.
55. To divide an integer by a mixed number —
Rule. — Reduce the mixed number to an improper
fraction [24] ; and then proceed as already directed
(52].
KxLVMPLE. — Divide 8 by 4§.
45=2^^ therefore 8-f-4f=8-^2^^=8xf3=lH-
Reason- of the Rule. — It is evident that the improper
fraction which is equal to the divisor, is contained in the divi-
dend the same number of times as the divisor itself.
EXERCISES.
44. 5-^3i=lf. I 46 14-^14=7-=:^.
45. 16^1li|=liif I 47 21~14,-^=lf,3,.
4S. Divide £7 IGs. Id. by 3i. Ans. £2 6s. ll^d.
49. Divide £3 35. 3d. by 4^. Ans. 14s. O^d.
56. To di^ide a fraction, or a mixed number, by a
mixed number —
RuL.^.— Reduce mixed numbers to improper fractions
[24] ; and then proceed as already directed [49] .
Example 1. — Divide f by 5}.
51^^-2 thereforp ^ i.51 — 3_:^53Jl.3y 9 2?
Ex-^MPLE 2.— Divide 8-,^ by 7f .
^8^=fi, and 7|=V, therefore 8^--7|=fi^V=H X
4T — JTl'
47 Reasox of the Rule.— "\Ye (as in the last rule) merely
change the mixed numbers into others more conveniently
divided — without, however, altering their value
160
VI/LGAR FRACTIONS.
EXERCISES.
50. ^-55=^.
ejO 3 . Q 5 213
^^- So-^^TT — Silo-
t\'i la . 14 2 5
;u,
57. k-^s^^m-
58. 13H-2i-{-5i-3Hr^o
59. 2i-.^+Hl/r-
58. When the divisor, di^ddend, or both, are com-
pound, or complex fractions —
Rule. — Reduce compound and complex to simple
fractions — by performing the multiplication, in those
which are compound, and the division, in those which
are complex ; then proceed as already dh-ected [49, &c.]
Example 1. — Divide 4 of | by f .
5 r.f fi 3 0 rQQl +liArpfnrP Sv^ • 3 3 0 . 3 3 o v * 1 ^0
T 01 ^=jiB- L^^J' tnereiore yX^-t-j — 3^—4 — 56^X3—1^8.
4
Example 2. — Divide -tt by |.
1=4% [46], therefore ^^l=^^l=^x^=i^.
EXERCISES.
\J\J. iX^-T-g 2 7-
61- 4ii--y^yXA=50^.
3
G2.
5 _:_ -1. — 22
a'i 22.2V'' '17
"^- Qy~3'*^T3 42^?
Q 3
27
1 9
fiA J. . 3v5 ^2 2 1
QUESTIONS.
1 . How is a fraction dived by an integer ? [46] .
2. How is a fraction divided by a fraction } [49].
3. Explain how it occurs that the quotient of two
fractions is sometimes greater than the dividend ? [51].
4. How is a whole number divided by a fraction }
5. What is the Ailc for dividing a mixed number by
an integer, or a fraction .' [53] .
6. What are the rules for dividing an integer, a frac-
tion, or mixed number, by a mixed number ? [55 and
66].
7. What is the rule when the divisor, dividend, or
both arc compound, or complex fractions ? [58].
VULGAR FRACTIONS. 161
MISCELLANEOUS EXERCISES IN VULGAR FRACTIONS,
1 . How mucli is ^ of 1 S6 acres, 3 roods ?' Ans.
20 acres, 3 roods.
2. How much is | of 15 hours, 45 minutes ? Ans.
7 hours.
3. How mucli is jVtV of 19 cwt., 3 qrs., 7 lb .^ Aiis.
7 cwt., 3 qrs., 2 tb.
4. How much is ^\%% of i^lOO f Aiis. £3G 9s.
a If one farm contains 20 acres, 3 roods, and
another 26 acres, 2 roods, what fraction of the former
is the latter ? Ans. jW.
6. "What is the simplest form of a fraction express-
ing the comparative magnitude of two vessels — the one
containing 4 tuns, 3 hlids., and the other 5 tuns, 2
hhds. r Ans. if.
7. What is the sum of | of a pound, and |- of a
shillino: ? Ans. 13s. lO^d.
8.
What is
the sum (
.f|...
ai
ad j'^d.
? Ans. 7
T\d.
9.
What is
the sum
of
^'■.
h 1^-)
and Y^od. I
' A.
ns
35. 1
10
more
. Suppose
j what is
! I have
my entii-
^of
e shii
a
Lre
ship, and that I
f Ans. ii.
buy
A
11 . A boy dinded his marbles in the following manner :
he gave to A J- of them, to B yV, to C I, and to D }y
keeping the rest to himself ; how much did he give
away, and how much did he keep f An^s. He gave away
j\\ of them, and kept y^o •
12. What is the sum of 4 of a yard, ^ of a foot, an()
4 of an inch r A.ns. 7 inches.
13. What is the diiSference between the | of a pound
and D}d. } Ans. lis. 6^d.
14. If an acre of potatoes yield about 82 barrels of
20 stone each, and an acre of wheat 4 quarters of 460
ft> — but the wheat gives three times as much nourish-
ment as the potatoes ; what will express the subsistence
given by each, in terms of the other ? Ans. The pota-
toes will give 4^^- times as much as tlic wheat ; and the
wheat the //y part of what is given by the potatoes.
15. In Fahrenheit's thermometer there are 'SO de-
grees between the boilhig and freezing points , in that
162 DECIMAL FRACTIOXS.
of Reaiimar only 80 ; wliat fraction of a degree in the
latter expresses a degree of the former r Ans. |-, .
16. The average fall of rain in the United Kingdom
is about 34 inches in depth during the year in the plains ;
but in the hilly countries about 50 inches ; Tvhat fraction
of the latter expresses the former ? Ans. ^l.
17. Taking Chimborazo as 21,000 feet high, and
Purgeool, in the Himalayas, as 2-2,480 ; what fraction
of the height of Pui'geool expresses that of Chimborazo .'*
A 'lis A 2 5.
IS. Taking 4200 feet as the depth of a fissure or
crevice at Cutaco, in the Andes, and 5000 feet as the
depth of that at Chota, in the same range of mountains ;
how will the depth of the former be expressed as a
fraction of the latter .' Ans. ^.i.
DECIMAL FRACTIONS.
59. A decimal fraction, as already remarked [13],
has unity with one, or more cyphers to the right hand,
for its denominator ; thus, y/o o" ^^ ^ decimal fraction.
Since the division of the numerator of a decimal fraction
by its denomiuiitor — from the very nature of notation
[Sec. I. 34] — is performed by mo^dng the deoimal
point, the quotient of a decimal fraction — the equi-
valent decimal — is obtained with the greatest facility.
Thus yo\o-r=:-005 ; for to divide any quantity by a
thousand, we liave only to move the decimal point three
places to the right.,
60. It is as inaccurate to confound a decimal fraction
with the corresponding decimal, as to .confound a vulgar
fraction with its quotient. — For if 75 is the qitoiicni
of ^f °, or of VoV) ^^^ i'^ distinct from either ; so also
is '75 the quotient of f or of yYu-, and equally distinct
from either.
61. A decimal is changed into its correspondmg deci-
mal fraction by putting unity with as many cyphers as
it contains decimal' places, under it, for denominator —
having first taken away its decimal point. Thus "5646=
tnnnn ? \J\JO T ?. iV .T '. t^^
DECIMAL FRACTI0.V3. 163
62. Decimal fractions follow esaeti/ uic same rules
as vulgar fractious. — It ls, however, ^euerally uioro
convenient to obtain tlieir quotients [5G], and then per-
form on them the required processes of addition, &c.,
by the methods already described [Sec. II. 11, &:c.]
63. To reduce a vulgar fraction to a dccmmlj or to a
decimal fraction —
Rule, — Divide the numerator by the denominator — •
this will give the required decimal ; the latter may be
changed into its corresponding decimal fraction — as
already icicribed [61].
Example 1. — Reduce | to a decimal fraction.
4)3
0-75=^.
Example 2. — ^^Vhat decimal of a pound is l\d. *
7^/.= [17] Xj^gL; but £-JV',=£0032, &c.
This rule requires no explanation.
|=-G2o. I 9. //.-=-9047G. kc.
53=-973&c. ! 10. 4=-8.
1.
2
1=375.
5.
6.
s!
JV=-36.
7-
4.
1^ 25
4 — Too-
8.
l=-5. I 11. i=-5625
-■'>'
iiz.;
^^-3125. 1 12. |i=-5375.
13. Reduce 12^. Od. to the decimal of a pound. Ans
C-j.o.
14. Reduce 15^. to the decimal of a pound. Ans. -75
15. Reduce 3 quarters, 2 nails, to the decimal of a
yard. Ans. -875.
16. Reduce 3 cwt., 1 qr., 7 lbs, to the decimal of a
ton. A.ns. -165625
64. To reduce a decimal to a lower denomination —
Rule. — ^Reduce it by the rule already given [Sec.
III. 3] for the reduction of integers.
r.xAMPLK 1. — Express £-0237 in terms of a shilling
•6237
20
Answer. 124740 shiHiDcr5=£C237
164 DECIMALS.
Example 2. — Reduce £-9734 to shillings, &;c.
•9734
20
19-4G80 8iiimngs=£-9734.
12
5-6160 penee=-4G8.5
4
2-4640 farthmgs=-G16^Z
Answer, £-9734=195. 5^1.
G5. This rule is founded on the same reasons as were given
for the mode of reducing integers [Sec. IIL 4].
■Multiplying the decimal of a pound by 20, reduces it to shil-
lings and the decimal of a shilling. Multiplying the decimal
of a shilling by 12, reduces it to pence and the decimal of a
penny. Multiplying the decimal of a penny by 4, reduces it
to farthings and the decimal of a farthing.
EXERCISES
23. What is the value of <£-S6S75 .' Ans. 17s. 4\d
24. What is the value of i2-5375 } Ans. IQs. 9d.
25. How much is -875 of a yard } Ans. 3 qrs., 2 nails.
26. How much is '165625 of a ton.' Ans. 3 cwt.,
1 qr., 7 ft.
27. What is the value of £-05 t Ans. \s.
28. How much is '9375 of a cwt. 't Ans. 3 qrs.,
21 lb.
29. What is the value of £'95 } Ans. 19^.
30. How much is '95 of an oz. Troy r Ans. 19 dwt.
31. How much is "875 of a gallon } Ans. 7 pints.
32. How much is '3945 of a day.' Ans. 9 hours,
28', 4', 48'".
33. How much is '09375 of an acre ^ Ans. 15
perches.
66. The following will be found useful, and — being
intimately connected with the doctrine of fractions — •
may be advantageously introduced here :
To find at once what decimal of a pound is equiva-
lent to any number of shillings, pence, &c.
AVhon there is an even number of shillings —
Rule. — Consider them to be half as many tenths of
a pound.
DECIMALS. 165
Example. — 16s.=£-8.
Eyery two sliillings are equal to one-tenth of a pound; there-
fore 8 times 25. are equal to 8 tenths.
67. When the number of shillings is odd —
Rule. — Consider half the next lower even number,
as so many tenths of a pound, and with these set down
5 hundredths.
Example. — 15.s.=£- 75 .
For, 15.-.=145.+ls\ ; but by the last rule 14.s.=£-7 ; and
since 2^.=1 tenth — or, as is evident, 10 hundredths of a
pound — 16.=5 hundredths.
68. When there are pence and farthings —
Rule. — If, when reduced to farthings, they exceed
24, add 1 to the number, and put the sum in the second
and thu'd decimal places. After taking 25 from the
number of farthings, divide the remainder by 3, and put
the nearest quantity to the true quotient, in the fourth
decimal place.
If, when reduced to farthings, tlicy are less than 25,
set down the number in the third, or in the second and
third decimal places : and put what is nearest to one-
third of them in the fourth.
Example 1. — AVhat decimal of a pound is equal to Sid. 1
8^=35 farthings. Since 35 contains 25. we add one tc
the number of farthings, which makes it 36 — we put 3G in
the second and third decimal nlaces. The number nearest
to the third of 10 (35-25 farthings) is 3— we put 3 in the
fourth decimal place. Therefore, S|=£03G3.
Example 2. — What decimal of a pound is equal to l^d. ?
If =7 fiirthings : and the nearest number to the third of
1 is 2. Therefore 1|J.=£0072.
Example 3. — What decimal of a poimd is equal to 5ld. T
5]J.=21 farthings ; and the third of 21 is 7. Therefore
. 5,^^.=£0217.
69. Reasox of the PitrLE.-^We consider 10 farthings as
the one hunch'edth, and one farthing as the one thousandth of
6 pound — because a pound consists of nearly one thousand
farthings. This, however, in 1000 farthings (taken as so
many thousandths of a pound) leads to a mistake of about 40 —
since £l=(not 1000, but) 1000—40 farthujgs. Hence, to a
thousand farthings (considered as thousandtlis o*" a pound).
1C6 CIRCULATi.XG JjEtJIMALS.
forty, or one in 25, must be added; that is, about the «ne-
thirtieth of the number of farthings. It is evident tliat, aa
those above 25 have not been allowed for when we added one
to the farthings, one-thirtieth of t/ieir number, also, must be
added — or, which is the same thing, one-third of their number,
in the f OK rih or next lower decimal jjlace.
If tiie farthings are less than 25, it is evident that the
correction should still be about the thirtieth of their number,
or one-third of it, in the fourth decimal place.
EXERCISES
17. 195. ll|^.=£-9977.
18. 7|^Z.=£-0322.
19. £27 55. 10d.=£27-2915.
20. 14s. 31^. =£7155.
21. 19s. ll|J.==£-9987.
22. £42 lis. 6irf.=£ •42-577,
70. To find at once tlie number of sliillings, pence,
&.C., in any decimal of a pound —
Rule. — Double the number of tenths for shillings—
to which, if the hundredths are not less than 5, add one.
Consider the digit in the second place (after subtracting
5, if it is not less than 5), as tens, and that in the thu-d
as units of farthings ; and subtract unity from the result
if it exceeds 25.
Example.— £-6874=135. 9d.
6 tenths are equal to tivelve shillings ; as the hundredths
are not less than 5, there is an additional shilling— which
makes 13a\ Subtracting 5 from tlie hundredths and adding
the remainder (reduced to thousandths) to the thou-
sandths, we have 37 thousandths from which — since they
exceed 25, we subtract unity ; this leaves 3G as the number
of farthings. £-6874, therefore, is equal to 13s. and 3G
farthings — or 13s. 9d.
This rule follows from the last three — being the reverse of
them.
CIRCULATING DECIMALS.
71. We cannot, as already noticed [Sec. II. 72],
dlwjjys obtain an exact quotient, when we divide one
number by another : — in such a case, what is called an
in-1-p.rmiiuite or (because the same digit, or digits, con-
Btautly recur, or circulate") a recur ring y or drculaling
LI KLU LATIN G BKCLMALS. !•)/
decimal is produced. — Tho decimal is said to be termi-
nate if there is an exact quotient — or one which leaves
no remainder.
12. An interminate decimal, in which only a single
figure is repeated, is called a repetend ; if two or more
digits constantly reclir, they form a periodical decimal.
Thus -77, &c., is a repetend ; but '597597, &c. is a
periodical. For the sake of bre^^ty, the repeated digit,
or period is set down but once, and may be marked as
fallows, -5' ( = -555, ka.) or -^93' ( = -493493493, &c.)
The ordinary method of marking the period is some-
what different — what is here given, however, seems
preferable, and can scarcely be mistaken, even by those
in the habit of using the other.
When the decimal contains only an iitfimte part —
that is, only the repeated digit, or period — it is a pure
repetend, or a ;)?trc_periodical. ' But when there is both
a finite and an infinite part, it is a inixed repetend or
viixed circulate. Thus
•3' (=-o33. &c.) is a pure repetend.
■578' (==-57888, &c.) is a mixed repetend.
'397' (=-397397397. &:c.) is a pure circulate.
805^64271' (=-865G427164271C427Lkc)i3 a mixed circulate
73. The number of digits in a period must always ba
less than the divisor. For, diflfercnt digits in the perioci
suppose different remainders during the division ; but
the number of remainders can never exceed — nor even
be equal to the divisor. Thus, let the latter be seven : the
only remainders possible are 1, 2, 3, 4, 5, and 6 ; any
other than one of these would contain the divisor at
least once — which would indicate [Sec. II. 71] that the
quotient figure is not sufficiently large.
74. It is sometimes useful to change a decimal into
its equivalent vulgar fraction — as, for instance, when in
adding, &c., those which circulate, we desire to obtain
an exact result. For this purpose —
Rule — I. If the decimal is a pure repelevd, put the
repeated digit for numerator, and 9 for denominator.
II. If it is a pure periodiadj put the period for
numerator, and so many nines as there are digits in the
period, for denominator.
1(38 CIRCULATIMG DFXIMALS.
Example 1. — What vulgar fraction is equivalent to -3' 1
Ans. f . * . . .
Ex.\MPLE 2. — What vulgar fraction is eqxiivalent to
•^TS54' I Ati6. ^^|.
75. Reason of I. — i will be found equal to '111, &c. — or
•1'; therefore! (=3x^)=-333, &c.=(3Xin, &c.) For i^
we multiply two equal quantities by the same, or by equal
quantities, the products will still be equal.
In the same way it could be shoAvu that any other digit
divided by 9 would give that other digit as a repetend. — And,
consequently, a repetend of any digit will be equal to a vulgar
fraction having the same digit for numerator, and 9 for deno-
minator.
Reason of II. — gg- will give "0101, &c. — or -^01' as quotient.
For before unity can be divided by 99, it must be considered
as 100 hundredths ; and the quotient [Sec. II. 77] will be one
hundredth, or '01. One hundredth, the remainder, must be made
100 ten thousatidths before it will contain 99 ; and the quotient
will be one ten thousandth, or -0001. One ten thousandth, the
remainder, must, in the same way, be considered as ten ynilHon-
eths ; and the next quotient will be one miHioneth, or -QOOOOl —
and so on with the other quotients, which, taken together,
will be •Ol-fOOOl-f-OOOOOl-f&G., or -010101, &c.— represented
by •^0l'.
- fl (=37XsV=37X.^01') will give -373737, &c.— or -^37' as
quotient. Thus
010101, &c
37
70707
30303
373737, &c.=37X"01'.
In the same way it could be shown that any other two digits
divided by 99 would give those other digits as the period of a
circulate. — And, consequently, a circulate having any two
digits as a period, will be equal to a vulgar fraction having the
same digits for numerator, and 2 nines for denominator.
For similar reasons ^^g will give -001001, &c., or '^001' aa
quotient. But 001001, &c., X (for instance) 563=-oG3563, &c
Thus 001001001, &c.
503
3003003003
6006006006
5005005005
563563563563, &c.=563x-^001 .
In the same way it could be shown that any other three digits
divided by 999 would give a circulating decimal havinsj these
CIRCLhATING DECIMALS. 169
C'gits as n period. — And, consequently, a circulating decimal
liaving any three digits as period will be equal to a vulgar
traction having the same digits for numerator, and 3 nines
tor denominator.
We might, in a similar vraj, sho^sv that any number of digits
divided by an equal number of nines must give a circulate,
each period of which would consist of those digits. — And,
Consequently, a circulate whose periods would consist of any
digits must be equal to a vulgar fraction having one of : s
i-eriods for numerator, and a number of nines equal to t.ie
number of digits in the period, for denominator.
76. If tlie decimal is a mixed repetend or a mixed
circulate —
Rule. — Subtract the finite part from the Tvhole, anrl
set down the difference for numerator ; put for deno-
minator so many cyphers as there are digits in the Jinife
part, and to the left of the c}-]?hers so many nines as
there are digits in the infinite part.
ExA?j?LE. — What is the vulgar fraction equivalent to
•97^8734' ?
There are 2 digits in 07, the finite part, and 4 in 8734,
the infiaite part. Therefore
978734-97 978G37 . ^ • , , ^ ..
-999300-=999900' ^^' '^^ ^''"^^^"'^ ^^^^^"^ ^'^^'*^^^'^-
77. Reason of "fhe Rule. — If, fiDr example, we multiply
•97^87S4' by 100, the product is 97 •8734=97-f--8734. This (by
the last -rule) is equal to 97-j-f^||. which (as we multiplied by
100) is one hundred times greater than the original quantity —
but if we divi.le it by 100 we obtain iVo+jtHJ^tj "which is
equal the original quantity. To perform the addition of ~^^
oTi
uenomin=*^tor — when they become
97X99^00. 873400 97X9999 8734 _ . qqqq __
99990000 ' 99990000~ 999900 ' 999300 ^^^^^® ^^^^ ~
lOOOn-n ^"XlOOOO-1 8734 _97xl0000-97 8784
' 999900 ^"999900" 999900 +999900""
970000—97 8734 978734-97 978637 .
999900 "^99y900~ 999900 ~"999900' ^ ^^^ ^^ exactly the
result obtained by the rule. The same reasoning would hold
with any other example.
EXERCISES.
7. •574'=4^|.
8. •83^25'=i|i|.
9. •147^658'=i±t,fVo-
10. -432^0075 '=i||^*-i
11. 875 -49^05 '=875 1 ^^-.
12. 301-S2^756'=30]i^f'^.
1.
-5'=|.
2.
•^8'=i.
3.
• 73'=X?y.
4.
5.
.v057'=3U,.
6.
•^45632'=MIM.
170 CIRCULATING DECIxMALS.
78. Except where great accuracy is required, it is not
necessary to reduce circulating decimals to their equi-
valent vulgar fractious, and we may add, and subtract
them, &c., like other decimals — merely taking care to
put down so many of them as will secure sufficient
accuracy.
79. It may be here remarked, that no vulgar fraction
will give a finite decimal if, when reduced to its lowest
terms, the denominator contains any prime factors (fac-
tors that are prime numbers — and all the factors, can
be reduced to such) except twos or fives. For neither
10, 100, 1000, &c., nor any multiples of these — as
30, 400, 5000, &c., nor the sum of any of their multi-
ples—as 6420'(5000 + 400+20), &c., will exactly con-
tain any prime numbers, but 2 or 5. Thus | (consi-
, - 30 tenths \ .„ . . .
dered as = \ will give an exact quotient; so also
(. , , 70 tenthsx ^ .„
considered as ^ )• -^^^ t '"'"^ '^^ S^^^^
10 tenths 100 hundredths,
one ; for | (considered as = j or ;z
&c.) does not contain 7 exactly.
For a similar reason -f icill not give an exact quo
. , , 40 tenths, 400 hundredths,
tient ; since -t- (considered as = or ^
&c.) does not exactly contain 7.
80. A finite decimal must have so many depimal places
as will bs equal to the greatest number of twos, or fives,
contained as factors in the denominator of the original
vulo;ar fraction, reduced to its lowest terms.
Thus -i- will give one decimal place; for 2 (found
once in its denominator) is contained in 10 {5X2) ; and
10 tenths , , .„ . ,..,.,
therefore 5 (=2) ^^^^ g^"^*^ ^^"^^ ^"S^^ \^^ ^^^
tenths' place [Sec. II. 77]), that is, one decimal as
quotient.
I (=^- — -J will give two decimal places ; because
2 being found twice as a fac+.':^ in its denominator,
\t will not be enough to consid^ir the nuincratoi- as so
CIRCULATING DECIMALS. 171'
30 tenths
niany tenths ; for 7 (=f ) cannot give an exact
quotient — 30 being equal to 3X2X5, which contains 2,
but not 2x2. It will, however, be sufficient to reduce
, , ' , 300 hundredths
the numerator to hundredths : because ;
w'dl give an exact quotient — for 300 is equal to 3X2X
2x^X0, and consequently contains 2X2. Eut 300
hundredths divided by an integer will give huiulredths —
or two decimals as quotient. Hence, when there are two
twos found as factors in the denominator of the vulgar
fraction, there are also two decimal places in the quotient.
4^0 (=.-,^r)wos^ - J contains 2 repeated three times
> '^X'^X'^X'^
as a factor, in its denominator, and will give thrcf
decimal places. For though 10 tenths — and therefore
t^iXlO tenths — contains 5, one of the factors of 40, \\
does not contain 2X2X2, the others ; consequentlj*
it will not give an exact quotient. — Nor, for the same
reason, will 6x100 hundredths. 6x1000 thousandths
6 X 1000 thousandths
will giYe one — that is, -7^ (=4'V) '^^
leave no remamder ; for 6x1000 (=6 X2X2X2X5X
5X5) contains 2X2X2X5. Bnt Qx'i-000 thausandths
divided by an integer will give thousandths — or three
decimals as quotient. Hence, when there are three twos
found as factors in the denominator of the vulgar frac-
tion, there are also three decimal places in the quotient.
SI. Were the Jives to constitute the larger number of
factors — as, for instance, in j\ jf ^, &c., the same reason
ing would show that the number of decimal places would
be equal to the number of ftves.
It might also be proved, in the same way, that were
the greatest number of twos or fives, in the denominator
of the vulgar fraction, any other than one of those num-
bers given above, there would be an equal number of
decimal places in the quotient.
82. A pure circulate wiU have so many digits in its
period as will be equal to the lea^t nimiber of nines, which
would represent a quantity measured by the dGnoraina-
172 CIRCULATING DECIMALS.
tor of tlie original vulgar fraction, reduced to its lowest
terms. For we have seen [74] that such a circulate will
be equal to a fraction haA^ng some period for its nume-
rator, and some number of nines for its denominator —
that is, it will be equal to some fraction, the numerator
of which (the period of the circulate) will be as vmny
limes the numerator of the given vulgar fraction, as the
quantity represented by the nines is of its denominator.
For if a fraction having a given denominator is equal to
another which has a larger, it is because the numerator
of the latter is to the same amount larger than that of
the former — in which case the increased size of the nu-
merator counteracts the effect of the increased size of the
denominator. Thus -|-=|f ; because, if the numerator
of If is 5 times greater than that of |, the denominator
of If, also, is five times greater than that of f.
Let the given fraction be y\. Since f'3=*^3846ir/ ;
and •\SS461o' 3 8.4.6.15. . _5_ oUr) is PrniM] to 3 8.4.6.15. . _^
and, therefore, whatever multiple 384615 is of 5, 0991 '99
ts the same of 13. — But 999999 is the least multiple of
13, consisting of nines. If not, let some other be less.
Then take for numerator, such a multiple of 5, as that
lesser number of nines is of 13 — and put that lesser
number of nines for its denominator. The numerator of
this new fraction will [75] form the period of a circulate
equal to the original fraction. But as this new period Is
different from 384615 (the former one), the circulate of
which it is an element, is also different from the former
circulate ; there are, therefore, two different circulates
equal to y\ — that is two different values, or quotients
for the same fraction — which is impossible^ Hence it
is absurd to suppose that any less number of nines is a
multiple of 13.
83. The periodical obtained does not contain a finite
part, when neither 2 nor 5 is found in the denominator
of the vulgar fraction — reduced to its lowest terms.
For [76] a finite part would add cyphers to the right
hand of the nines in the denominator of the vulgar
fraction, obtained from the circulate. But cyphers v;ould
suppose the denominator of the original fraction to
contain twos, or fives — since uo other prime factors
CIRCLL.^ liNCi DI-Xl.MALS. 173
could give cyphers in tlicir multiple — the denomiuatvyr
of the vulgar fraction obtained from the circulate.
84. If there is a finite part in the decimal, it will
contain as many digits as there are units in the greatest
number of twos or fives found in the denominator of the
original vulgar fraction, reduced to its lowest terms.
Let the origiual fraction be J-^. Since 56=2 X^X
2X7, the equivalent fraction must have as many nines as
■will just contain the 7 (cyphers would not cause a number
of nines to be a multiple of 7), multiplied by as many
tens as form a product which will just contain the twos as
factors. But we have seen [SO] that one ten (which adds
one cypher to the nines) contains one two, or five ; that
the product of two tens (which add two cyphers to the
nines), contains the product of two twos or fives ; that
the product of three tens (which add three cyphers to the
nines), contains the product of three twos or fives, &c.
That is, there will be so many cyphers in the denomi-
nator as will be equal to the greatest number of twos or
fives, found among the factors in the denominator of the
oriijinal vulo;ar fraction.
But as the digits of the finite part of the decimal add
an equal number of cyphers to the denominator of the
new vulgar fraction [75 j , the cyphers in the denominator,
on the other hand, evidently suppose an equal number of
places in the finite part of a circulate : — there will there-
fore be in the finite part of a circulate so many digits
as will be equal to the greatest number of twos or fives
found among the factors in the denominator of a vulgar
fraction containing, also, other factors than 2 or 5.
85. It follows from what has been said, that there is no
number which is not exactly contained in some quantity
expressed by one or more nines, or by one or more nines
followed by cyphers, or by unity followed by cyphers.
Contractions in Mitltiplicatign and Division
(derived from the properties of fractions.)
86. To multiply any number by 5 —
Rule. — Remove it one place to the left hand, and
divide the result by 2
174 C!1xti:.\';tion's.
ExAMT'i.F.-^7:',GxO-^"V=:-o80.
Reason,— ')=^V • tlierefore 7oGx5=7oGx y^="V=36/<0.
ST. To inultiplY by 2o—
Kui.K. — rteuiovG the ({iiantir^f two places to t\.e left, and
divide by 4.
Example.— 6732x25=^ '3^2oo^-^C^30n.
IIeason.— 25=',^'^: therefore G732x25=6732x'r' •
b^. To multiply by 125—
Hulk. — Uemove the quantity three places to the loft, and
divide the result by 8.
Example.— 7cSG5 X 1 25= ' ^ ^- -^ ^ o o^c g:ji25 .
Reason.— 125="^'' ^' ; therefore 78G5x 125=7865=' "'/^
89. To miiltiply by 75 —
Rule — Remove the quantity two places to the left, then
multiply the result by 3. and divide the product by 4.
ExAMPLE._G85x75=-'"^^r—=2''*J"'' =51375.
Reason.— 75 = " '^" = 100 X 4 : therefore 685 X 75 = 685 x
lOOx^
00. To multiply by 35—
Rule. — To the multiplicand removed two places to the
left and divided by 4, add tlie multiplicand removed one
place to the lefr.
Example 1.-67800x35 = g-s^moo _|_ 678960 = 1G97400
4-678900=2376300.
Reason.- 35='^«+10; therefore 67896x35 =67896x
Many similar abhreviatious y^ill easily suggest tliemselves to
both pupil aud teacher.
01. To divide by any one of the multipliers —
Rule. — Multiply by the equivalent fraction, inverted.
Example.— Divide 847 by 5. 847-^5=847-^ '^"=847 X
,-=,=169-4.
IiEAsox. — We divide by any number -when we divide hy the
fraction equivalent to it ; but wo-divide by a frnction wlieh we
invert it, and then consider it as .a multipher [49].
92. Sometimes what is convenient as a multiplier will
not be equally so as a divisor; thus 35. For it is not so
easy to divide, as to multiply by '^"-f-10, its equivalent
mised number. ^
DECIMALS. 175
QUESTIONS FOR THE PUPU..
1 Show that a decimal fraction, and the corrcspoad-
hig decimal arc not identical [59] .
2. How is a decimal changed into a decimal frac
iion? [61].
3. Ai-e the methods of adding, &c., vulgar and deci-
*mal fractions different r [62] .
4. How is a vulgar reduced to a decimal fraction ?
[63].
5. How is a decimal reduced to a lower denomina-
tion ? [64] .
6. How are pounds, shillings, and pence changed, at
oiuXj into the corresponding decimal of a pound r [66,
67, and 6S].
7. How is the decimal of a pound changed, at ona^
into shillings, pence, &c. r [70].
S. What are terminate and circulating decimals ?
9. ^^ hat are a repetend and a perio .ical, a puro
and a mixed circulate ? [72] .
10. Why cannot the number of digits in a 'neiiod bo
equal to the number of units contained in the divisor .'
[73].
11. How is a pure circulate or pure repetend changed
into an equivalent vulgar fraction .' [74] .
12. How is a mixed repetend or mixed circulate
reduced to an equivalent vulgar fraction } [76].
13. What kind of vulgar fraction can produce no
equivalent finite decimal r [79] .
14. What number of decimal places must necessarily
be found in a finite decimal .- [80] .
15. How many digits must be found in the periods
of a pure circulate ? [fc2] .
16. When is no finite part found in a repetend, or
circulate .' [S3] .
17. How many digits must be found in the finite part
of a 7/iixed circulate r [84] .
IS. On what principal can we use the properties of
fractions as a means of abbreviating the processes of
multiplication and division ? [S6, ko.]
176
SECTION V.
PROPORTION.
1. The rule of Proportion is called also the golden
ruky from its extensive utility ; in some cases it is termed
the ?'ule of three — because, by means of it, when three
numbers are given, a fourth, which is unknown, may be
found.
2. The rule of proportion is divided into the simpie^
and the compound. Sometimes also it is divided into
the direct^ and inverse — which is not accurate, as was
shown by Hatton, in his arithmetic published nearly one
hundred years ago.
3. The pupil to have accurate ideas of the rule of
proportion, miT-t be acquainted with a few simple but
important pri .ciples, connected with the nature of ratios^
and the dorJ nne of proportion.
The following truths are self-evident : —
If the same, or equal quantities are addcl to equal
quantities, the sums are equal. Thus, if we add the scwu.
quantity, 4 for instance, to bX^ and 3X10, which are
equal, we shall have 5 X 6 + 4=3 X 1 0 + 4.
Or if we add equal quantities to those which are
equal, the sums will be equal. Thus, since
5X6=3X10, and 2-|-2=4
5x64-2x2=3x10-1-4.
4. If the same, or equal quantities are subtracted
from others which are equal, the remainders will be
equal. Thus, if we subtract 3 from each of the equal
quantities 7, and 5+2, we shall have
7-3=5+2-3.
And since S=G + 2, and 4=3 + 1.
8-4=6+2-3+1.
5. If eqi^l quantities are mnUiplied by the same, or
by equal quantities, the products will be equal. Thus
PROPORTION. 177
if wc multiply tliG equals 5-|-C, aud 10-f-l l>y 3, we
sliull have
64^3=10+1x3.
Aud siuce 4 -{- 9= 1 H , aud 3 X 6= 1 S .
444^X3X6=13X18.
6. If equal quantities are divided by the sauic, or by
equal quantities, tlie quotients will be equal. Thus if
we divide the eimals 8 and 4-}- 4 by 2, we shall have
8_4-}-4
2 2
And since 20=17 + 3, and 10=2X">.
20_17+3
io~~ 2X0
7. Rntio is the relation which cxi.-ts between two
quantities, and is expressed by two dots ( : ) placed be-
tween them — thus 5 : 7 (read, 5 is to 7) ; which means
that 5 has a certain relation to 7. The former quantity
is called the antecedent^ and the latter the consequent.
S. If we invert the terms of a ratio, we shall have
their inverse ratio ; thus 7 : 5 is the inverse of 5 : 7.
9. The relation between two cpantities may consist
in one being greater or less than the other — then the
ratio is termed arithnetiad ; or in one being some mid-
ti.ple or part of the other — and then it is geometriad.
If two quantities are equal, the ratio betv>'een them
is said to be that of equality ; if they are unequal it is
a ratio of greater inequalUy when the antecedent is
greater than the consequent, and of lesser inequality
when it is less.
10. As the afithmetical ratio between two quantities
is measured by their dij/erenc^^ so long as this difference
is not altered, the ratio is unchanged. Thus the ratio
of 7 : 5 is equal to that 15 : 13 — for 2 is, in each case,
the diiSference between the antecedent and consequent.
Hence we may add the same quantity to both the
antecedent and consequent of an arithmetical ratio, or
may subtract it from them, without changing the ratio.
Thus 7:5, 7 + 3:5 + 3, aud 7 — 2:5-2, are eipial
arithmetical ratios.
l»ut we cannot multiply or divide the terms of an arith-
178 PROPORTION.
metical ratio by the same number. Thus 12x2 : 10X2,
12H-2 : IOh-2, and 12 : 10 i^-e not equal arithmatical
ratios; for 12X2— 10X2=:4, 12-^2— 10-^ 2=1, and
12-10=2.
11. A geometrical ratio is measured by the quotient
obtained if we di\dde its antecedent by its consequent ; —
therefore, so long as this quotient is unaltered the ratio i»
not changed. Hence ratios expressed by equal fractions
are equal ; thus 10 : 5=12 : 6, for V= V • — Hence, also,
we may multiply or diride both terms of a geometrical
ratio by the same number without altering the ratio;
thus 7X2 : 1^X2=7 : 14— because _I^^ J.
14X2 14
But we cannot add the same quantity to both terms
of a geometrical ratio, nor subtract it from them, with-
out altering the ratio.
12. When the pupil [Sec. IV. 17] was taught how
to express one quantity as the fraction of another, he
in reality learned how to discover the geometrical ratio
between the two quantities. Thus, to reiDcat the ques-
tion formerly given, " VVhat fraction of a pound is
2^d. .?" — which in reality means, " What relation is
there between 2\d. and a pound ;" or '' What must we
consider 2\d.^ if we consider a pound as unity ;" " or,"
in fine, " Yv^hat is the value of 2^ : 1" —
We have seen [Sec. I. 40] that the relation between
quantities cannot be ascertained, unless they are made
to have the same " unit of comparison :" but a farthing
is the only unit of comparison which can be applied to
hoth 2\d. and £1 ; we must therefore reduce them to
farthings — when the ratio of one to the other will be-
come that of 9 : 960. But we have also seen that a
geometrical ratio is not altered, if we divide both its
terms by the same number ; therefore 9 : 960 is the same
ratio as ^f ^ : f|f , or ^f^ : 1. — That is, the ratio between
2\d. and £1 may be expressed by 2\d. : iEl, or 9 : 960,
*^i' 96 0 • 1 7 or, the pound being considered as unity, the
farthing will be repi-esented by ^f „.
13. The geometrical ratio between two numbers is the
same as that which exists between the quotient of the
fraction which represents their ratio, and unity. Tb\is,
PROPORTION. 179
:n the last example 9 : 900 aiul ^f- : 1 are equal ratios.
It is not necessary that we should be able to express by
integers, nor even bj a finite decimal, what part or mul-
tiple one of the terms is of the other ; for a geometrical
ratio may be considered to exist between any two quan-
tities. Thus, if the ratio is 10 : 2, 5 ( y) is the quantity
by which we must multiply one term to make it equal
to the other ; if 1 : 2, it is 0'5 (^), a finite decimal ; but
if 3 : 7, it is M2S571' (|), an infunile decimal — in which
case we obtain only an approxunation to the value of
the ratio. But though the measure of the ratio is ex-
pressed by an infinite decimal, when there is no quantity
which will exactly serve as the multiplier, or divisor of
one quantity so as to make it equal to the other — since
we may obtain as near an approximation as we please — ■
there is no inconvenience in supposing that any one
number is some part or multiple of any other ; that is,
that any number may be expressed in terms of another —
or may form one term of a geometrical ratio, unity
being the other.
14. PrffporiioUy or analogy^ consists in the equality
of ratios, and is indicated by putting =, or : :, between
the equal ratios ; thus 5 : 7==9 : 11, or 5 : 7 : : 9 : 11 (read,
5 is to 7 as 9 : 11), means that the two ratios 5 : 7 and
9:11 are equal ; or that 5 bears the same relation to 7
that 9 does to 1 1 . Sometimes we express the equality
of more than two ratios ; thus 4:S::6:12::1S: 36,
(read, 4 is to 8, as 6 is to 12, as IS is to 36), means
there is the same relation between 4 and 8, as between
6 and 12 ; and between IS and 36, as between either 4
and 8, or 6 and 12 — it follows that 4 : S : : 18 : 36 — ^for
two ratios which are equal to the same, are equal t^
each other. When the equal ratios- are arithmetical, the
constitute an arithmcliad proportion ; when geometri
cal, a geometrical proportion
15. The quantities which form the proportion are
called proportionah ; and a quantity that, along with
three others, constitutes a proportion, is called a fourth
proportional to those others. In a proportion, the two
outside terms are called the extremes^ and the two middle
terms the Tnmns ; thus in 5 : 6 : : 7 : 8, 5 and S are the
ISO PKOPORTIO.V.
extremes, 6 and 7 the mcrins. When ilie same qiiantity
is found in hoik means, it is called i/ie mean of the
extremes ; thus, since 5:6: : 6 : 7, 6 is Me mea.n of 5 and
7. When the proportion is arithmetical, the mean of
two quantities is called their arithmdical mean ; when
the proportion is geometrical, it is termed their geome-
trical mean. Thus 7 is the arithmetical mean of 4
and 10; for, since 7— 4=10 — 7, 4 : 7: :7 : 10. And 8 is
the geometrical mean of 2 and 32 ; for, since f =^\,
2 : 8 : : 8 : 32.
16. In an arithmetical proportion, " the skvi of the
means is equal to the sum of the extremes." Thus, since
11 : 9: : 17 : J 5 is an arithmetical proportion, 11 — 9=
17—15 ; but, adding 9 to both the equal quantities, we
have 11-9 + 9=17 — 15 + 9 [3] ; and, adding 15 to
these, we have 11-9 + 9 + 1.5=17—15 + 9 + 15; but
11 — 9 + 9 + 15 is equal to 11 + 15 — since 9 to be sub-
tracted and 9 to be added =0 ; and 17—15 + 9 + 15=
17+9 — since 15 to be subtracted and 15 to be added :=0 :
therefore 11 + 15 (the sum of the extremes) =17 + 9
(the sum of the means). — The same thing might be
proved from any other arithmetical proportion ; and,
therefore, it is true in every case.
17. This equation (as it is called) , or the equality which
exists between the sum of the means and the sum of the
extremes, is the test of an arithmetical proportion : — that
is, it shows us whether, or not, four given quantities
constitute an arithmetical proportion. It also enables us
to find a fourth arithmetical proportional to three given
numbers — since any mean is evidently the difference
between the sum of the extremes and the other mean ;
and any extreme, the difference between the sum of the
means and the other extreme —
For if 4 : 7: :8 : 11 be the arithmetical proportion,
44-11=7 + 8 [16] ; and, subtracting 4 from the equals,
we have 11 (one of the extremes) =7 + 8—4 (the sum of
the means, minus the other extreme) ; and, subtracting 7,
we have 4+11 — 7 (the sum of the extremes minus one
of the means) :r=8 (the other mean). We might in the
j*ame way find the remaining extreme, or the remaining
mean. Any other arithmetical proportion would liavo
PROPORTION. 18l
answered just as well — hence what we have said is true
in all cases.
18. ExA.MPLE. — Find a fourth proportional to 7, 8, 5.
Iklaking the rer|iiired number one of the extremes, and
putting the note of interrogation in the place of it, we have
7 : 8 : : 5 : ? ; then 7 : 8 : : 5 : 8-f-5— 7 (the sum of the means
minus the given extreme, =6) ; and the proportion com-
pleted will be
7 : 8 :: 5 : 6.
Making the required number one of the means, we shall
have 7 : 8 : : '? : 5, then 7:8:: 7+5-8 (the sum of the
extremes minus the given mean, =4) :5j and the proportion
completed will be
7 : 8 :: 4 : G.
As the sum of the means will be found equal to the sum
of the extremes, we have, in each case, completed the pro-
portion.
19. The arithinetiad mean of two quantities is half
ihe sum of the extremes. For the sum of the means is
equal to the sum of the extremes ; or — since the means
are equal — twice one cf the moans is equal to the sum
of the extremes ; consequently, half the sum of the
means — or one, of them, will he equal to half the sum of
the extremes. Thus the arithmetical mean of 19 and
1 *-) ' 27
27 is — ~ — (=23) ; and the proportion completed is
10 : 23 :: 23 : 27, for 19 + 27=23 + 23.
20. If with any four quantities the sum of the means
Is equal to the sum of the extremes, these quantities aro
in arithmetical proportion. Let the quantities be
8 G 7 5.
As the sum of the means is equal to the sum of the
extremes
8 + 5 = 6 + 7.
Subtracting 6 from each of the equal quantities, wo
have 8 + £. — 6 = 6 + 7— 6 ; and subti-acting 5 from each
of these, we have 8 + 5 — 6 — 5=6 + 7 — 6 — 5. But
8+5 — 6 — 5 is equal to 8 — 6, since 5 to be added
and 5 to be subtracted are ?==0 ; and +6+7—6—0=
7 — 5, since 6 to be added and 6 to V-o subtracted =0 ;
»82 ' PROPORTION. '
therefore 8 + 5 — 6 — 5=6+7 — 6 — 5 is the same as
8—6=7—5 ; but if 8—6=7—5, 8 : 6 and 7 : 5, are
two equal arithmetical ratios ; and if they are two equal
ftrithmetical ratios, they constitute an arithmetical pro-
portion. It might in the same way be proved that
»ny other four quantities are in arithmetical proportion,
«f the suiii of the means is equal to the sum of the
extremes.
21. In a gecmietHcal proportion, ^' the product of
he means is equal to the product of the extremes."
fhus, since 14 : 7 : : 16 : 8 is a geometrical proportion,
^*=^Y [1^] j ^'^^'> multiplying each of the equal quanti-
ses by 7, we have (yx7)=yx7; and multiplying
iachoftheseby8,wehavel4X8=16x7(VX7x8) : —
out 14X8 is the product of the extremes; and 16X7
<s the product of the means. The same reasoning would
nold with any other geometrical proportion, and there-
fore it is true in all cases.
22. This equation (as it is called), or the equality of
the product of the means and the product of the extremes,
IS the test of a geometrical proportion : that is, it shows
as whether or not fom* given quantities constitute a
geometrical proportion. It also enables us to find a
fourth geometrical proportional to three given quanti-
ties— which is the object of the rule of three ; since any
mean is, evidently, the quotient of the product of the
extremes divided by the other mean ; and any extreme,
is the quotient of the product of the means divided by
the other extreme.
For if 7 : 14 :: 11 : 22 be the geometrical proportion,
7X22=14X 11 ; and, dividing the equals by 7, we have
14X11
22 (one of the extremes) = — zrz — (^^^ product of the
means divided by the other extreme) ; and, dividing these
7X22
by 11, we have — ri — (the product of the extremes di-
vided by one mean)=14 (the other mean). We might
in the same way find the remaining mean or the remain-
ing extreme. Any other proportion would have answered
just as well — and therefore what we have said is true
in every case.
PROPORTION.
tS3
23. Example. — Find a fourth proportional to 8, 10, and 14.
Making the required quantity one of the extremes, we shall
lUxU
have 8 : 10 : : 14 : ? ; and 8 : 10 : : 14 : — —-(the product
o
i>f the means divided by the given extreme, =17-5).
And the proportion completed will be
8 : 10 : : 14 : 175.
MiUdng the required number one of the means, we shall
8x14
have 8 : 10 : : ? : 14; and 8 : 10 : :——-( the product of
the extremes divided by the given mean, =11-2) : 14.
And the proportion completed wiU be
8 : 10 : : 11-2 : 14.
EXERCISES.
Find fourth proportionals
1. To 3, 6, and 12 . ^ns. 24.
4.
16.
8.
1020.
10.
3,
6
8
G,
12
4
10,
150
„ 08
1020,
08
„ 150
150,
10
„ 1020
68,
1020
„ 10
150.
24. If with any four quantities the product of the
means is equal to the product of the extremes, these
quantities are in geometrical proportion. Let the
Quantities be
5 20 6 24,
As the product of the means is equal to the prod, ct
of the extremes,
5x24=20x6.
5X24 20X6
Dividing the equals by 24, we have ~^^ — = 24 ">
5X24 20X6
and, dividing these by 20, we have 20X24 20X24-
^ ^ 5X24 5^ 20X6 6_ 5 _6 ^
^ 20X24~20 5 and 20X24~24 5 therefore 20~24 '
consequently the geometrical relation between 5 and 20
is the same as that between 6 and 24 ; hence there are
two equal geono'^trical ratios — or a geometrical propor-
184 PROPORTION.
tion. It might, in the same way, be proved that amy
other four quantities are in geometrical proportion, if
the product of the means is equal to the product of the
extremes.
25. When the first term is unity, to find a fourth
proportional —
KuLE. — Find the product of the second and third.
KxAMPLE. — What is the fourth proportional to 1, 12, and
1 : 12 : : 27 : 12x27=324
We are to divide the product of the means by the given
extreme ; but we may neglect the divisor when it is unity —
since dividing a number by unity does not alter it.
EXERCISES.
Find fourth proportionals
9. To 1, 17, and 8 . Ans. 136.
10. „ 1, 23 „ 20 . . 460.
11. „ 1, 100 „ 73 . . 7300.
12. „ 1, 63 „ 110 . . 5830.
13. „ 1, 15 „ 1234 . . 18510.
26. When either the second, or third term is unity —
Rule. — Divide that one of them which is not unity^
by the fii'st.
Example. — Find a fourth proportional to 8, 1, and 5.
8 : 1 : : 5 : |.
We are to divide the product of the means by the given
extreme : but one of the means may be considered as the
product of lx)th, when the other is unity. For, since multi-
plication fjy unity produces no efiect, it may be omitted.
EXERCISES.
Find fourth proportionals.
M. To 5, 20, and 1 . Ans. 4
15. „ 5, 1 „ 20
16. „ 7, 21 „ 1
17. „ 8, 24 „ 1
18. „ 6, 1 „ 50
19. „ 17, 1 „ 68
20. „ .?00, 1000 „ 1
21. „ 200, 1 „ 1000
27. When the means are equal, each is said to be
iJie gfcomotrieil mean of the extremes ; and tlie product
4.
8i.
4.
6.
5.
RULE OF PROPORTION. 185
of the extremes is equal to the Tnmn multiplied by itself.
Hence, to discover the geometrical vwin of two quan-
tities, we have only to find some number which, multi-
plied by itself, wiU be equal to their product — that is,
to find, what we shall term hereafter, the square root
of their product, ^hus 6 is the geometrical mean of 3
and 12 ; for 6X6=3X12. And 3 : 6 : : 6 : 12.
2S. It will be useful to make the pupil acquainted with
the following properties of a geometrical proportion —
We may consider the same quantity either as a mean,
or an extreme. Thus, if 5 : 10 : : 15 : 30 be a geometrical
proportion, so also will 10:5: : 30 : 15 ; for we obtain the
same equal products in both cases — in the fomier, 5X
30=10X15 ; and in the latter, 10X15=5X30 — which
are the same thing. This change in the proportion is
called inversion.
29. The product of the means will continue equal to
the product of the extremes — or, in other words, the
proportion will remain imchanged —
If we alternate the terms ; that is, if we say, " the
first is to the third, as the second is to the fourth" —
If we '* multipbjy or dirule the first and second, oi
the first and third terms, by the same quantity" —
If we " read the proportion bacJcivards''^ —
If we say " the first term pliLS the second is to the
second, as the third plus the fourth is to the fourth" —
If we say "• the first term plus the second is to the
first, as the third plus the fourth is to the thud" — &c.
RULE OF SIMPLE PROPORTION.
30. This rule, as we have said, enables us, when thre?
quantities are given, to find a fourth propoi-tional.
The only difficulty consists in stating the question ;
when this is done, the required term is easily found.
In the rule of simple proportion, two ratios are given,
the one perfect, and the other imperfect.
31. PtULE — I. Put that given quantity which belongs
to the imperfed ratio in the thud place.
II. If it appears from the nature of tlie question that
the required quantity mu^t bo greater than the other,
\S6 RULE OF PROPORTION.
or given term of the same ratio, put the larger term
of the perferi ratio in the second, and the smaller ia
the fii'st place. But if it appears that the required
quantity must be less, put the larger term of the perfcci
ratio in the first, and the smaller in the second place.
III. Multiply the second and third terms together,
and divide the product by the first.^— The answer will
be of the same kind as the third term.
32. Example 1. — If 5 men build 10 yards of a wall in one
day, how many yards would 21 men build in the same time ?
ft will facilitate the stating, if the pupil puts down the
question briefly, as follows — using a note of interrogation to
represent the required quantity —
5 men.
10 yards.
21 men.
^ yards.
10 yards is the given term of the imperfect ratio — it must,
therefore, be put in the third place.
5 men, and 21 men are the quantities which form the
•perfect ratio ; and, as 21 will build a greater number of yards
than 5 men, the required number of yards will be greater
than the given number — hence, in this case, we put the larger
term of the perfect ratio in the second, and the smaller in
the first place —
5 : 21 : : 10 : ?
And, completing the proportion,
5 : 21 : : 10 : ?i^i^42, the required number.
0
Therefore, if 5 men build 10 yards in a day, 21 men will
build 42 yards in the same time.
33. Example 2. — If a certain quantity of bread is sufficient
to last 3 men for 2 days ; for how long a time ought it to
last 5 men '? This is set down briefly as follows :
3 men.
2 days.
5 men.
'? days.
2 days is the given terra of the imperfect ratio — it must,
therefore, be put in the third place.
The larger the number of men, the shorter the time a given
quantity of bread will last them j but this sJiotler time is the
RULE OF PROPORTION, 187
required quantity — hence, in this case, the greater term of
the perfect ratio is to be put in tlie first, and the smaller in
the second place—
5 : 3 :: 2 : ?
And, completing the proportion,
3x2
5 : 3 : : 2 : — ^— =li, the required term.
34. Example 3.— If 25 tons of coal cost X21, ^vhat will
be the price of 1 ton '?
Iy21 21
25 : 1 : : 21 : ~~ pounds £25=^^"'- ^2'^'
It is necessary in this case to reduce the pounds to lowep
denominations, in order to divide them by 25 : this causes
the answer, also, to be of different denominations.
35. Reason- of I. — It is convenient to make the required
quantity the fourth term of the proportion — that is, one of the
extremes. It could, however, be found equally well, if consi-
dered as a mean [23].
Reason of II. — It is also convenient to make quantities of
the same kind the terms of the same ratio ; because, for in-
stance, we can compare men with men, and days with days —
but we cannot compare 7»(?n with dai/s. Still there is nothing
inaccurate in comparing the number of one, with the 7inmber of
the other ; nor in comparing the number of men with the quan-
tity of work they perform, or with the number of loaves they
eat ; for these things are proportioned to each other. Hence we
ehall obtain the same result whether we state example 2, thus
6 : 3 :: 2 : .?
or thus 5 : 2 : : 3 : .'
"When diminishing the kind of quantity which is in the per-
fect ratio increases that kind which is in the imperfect — or the
reverse — the question is sometimes said to belong to the inverse
rule of three ; and diflFerent methods are given for the solution
of the two species of questions. But liatton, in his Aritli-
metic, (third edition, London, 1753,) suggests the above gene-
ral mode of solution. It is not accurate to say " the inverse
rule of three" or " inverse rule of proportion ;" since, although
there is an inverse ratio, there is no \ji.ye.T&Q proportioji.
Reason of III. — "We multiply the second and third terms,
and divide their product by the first, for reasons already given
[22].
The answer is of the same kind as the third term, since
neither the multiplication, nor the division of this term has
changed its nature ;— 2O5. the payment of 5 days divided by 5
188 RULE OF PROPORTION.
20s. • 205.
gives -p~ as tlio payment of one tlay; and —;^, the payment
of one (lay multiplied by 9 gives II—' X 9 as tlie payment of 9
5
days.
If the fourth term -were not of the same kind as the third,
it would not complete the imperfect ratio, and therefore it
would not be the required fourth jjroportional.
36. It will often be convenient to divide the first and
second, or first and third terms, by their greatest com-
mon measure, when these terms are composite to each
other [29].
Example. — If 36 cwt. cost £24, what will 27 cwt. cost 1
36 : 27 : : 24 : ? ,
Dividing the first and second by 9 we have
4 : 3 : : 24 : ?
And, dividing the first and third by 4,
1 : 3:: 6 : 3x6=kl8.
EXERCISES FOR THE PUPIL.
Find a fourth proportional to
1. 5 pieces of cloth : 50 pieces : : £27. Ans. £270
2. 1 cwt. : 215 cwt. : : 50s. Ans. 10750s.
3. 10 lb : 150 ib : : os. Ans. Ids.
4. 6 yards : 1 yard : : 21s. Ans. As. Gd.
5. 9 yards : 36 yards ; : IBs. Ans. 12s.
6. 5 lb : 1 lb : : 155. Ans. 3s.
7. 4 yards : IS yards : : \s. Ans. 4s. Gd.
8. What will 17 tons of tallow come to at £'25 per
ton ? Ams. £425.
9 If one piece of cloth cost £27, how much will 50
pieces cost } Ans. £1350.
10. If a certain quantity of provisions would last 40
men for 10 months, how long would they suffice for 32 :
Ans. \2\ months.
11. What will 215 cwt. of madder cost at 50s. pei
cwt. } Ans. 10750s.
12. I desire to have 30 yards of cloth 2 yards wide,
with baize 3 yards in breadth to line it, Iiow much of
the latter shall I require } Ans. 20 yards.
RULE OF PROPORTION. 189
13. At 10?. per barrel, what will be the price of 130
barrels of barley ? Ans. £65.
14. At 5s. per lb, what will be the price of 150 lb of
tea ? Ans. 750s.
16. A merchant agreed with a carrier to brinof 12
cwt. of goods 70 miles for 13 crowns, but his waggon
being heavily laden, he was obliged to unload 2 cwt. ;
how far should he carry the remainder for the same
money ? Ans. 84 miles.
16. What will 150 cwt. of butter cost at £3 per cwt. ?
Ans. £450.
17. K I lend a person £400 for 7 mouths, how much
ought he to lend me for 12 t Ans. £233 6s. Sd.
18. How much will a person walk in 70 days at the
rate of 30 miles per day ? Ans. 2100.
19. If I spend £4 in one week, how much will I
spend in 52 ? Ans. £20S.
20. There are provisions in a town sufficient to sup-
port 4000 soldiers for 3 months, how many must bo
sent away to make them last 8 months .'' Ans. 2500.
21. What is the rent of 167 acres at £2 per acre }
Ans. £334.
22. If a person travelling 13 hours per day would
finish a journey in 8 days, in what time will he accomplish
it at the rate of 15 hours per day ? Ans. 6if days.
23. What is the cost of 256 gallons of brandy at 12s.
per gallon ? Ans. 3072s.
24. What will 156 yards of cloth come to, at £2 per
yard ? Ans. £312.
25. If one pound of sugar cost 8^/., what will 112
pounds come to ? Ans. S96d.
2b. If 136 masons can bmld a fort in 28 days, how
many men would be required to finish it in 8 days f
Ans. 476.
27. If one yard of calico cost 6J., what will 56 yards
come to } Ans. 336d.
28. What will be the price of 256 yards of tape at
2d. per yard } Ans. b\2d.
29. If £100 produces me £6 interest in 365 dayy,
what would bring the same amount in 30 days } Ans
£1216 13s. Ad.
I'.^O RULE OK PRorORTIOX.
30. "W^bat sliall I receive for 157 pair of gloves, ai
lO^Z. per pair ? Ans. 157 Od.
31. What would 29 pair of shoes come to, at 95. per
pair } Ans. 26ls.
32. If a farmer lend his neighbour a cart horse which
draws 15 cwt. for 30 days, bow long should he have a '
horse in return which draws 20 cwt. ? Ans. 224- days.
33. What sum put to interest at £6 per cent, would
give £6 in one month .'■ Ans. J21200.
34. Ifl lend £400 for 12 months, how long ought £150
be lent to me, to return the kindness r Ans. 32 months.
35. Provisions in a garrison are found suificient to
last 10,000 soldiers for 6 months, but it is resolved to
add as many men as would cause them to be consumed
in 2 months ; what number of men must be sent in ?
Ans. 20,000.
36. If S horses subsist on a certain quantity of hay
for 2 months, how Ions; will it last 12 horses } Ans
li months.
37. A shopkeeper is so dishonest as to use a weight
of 14 for one of 16 oz. ; how many pounds of just will
be equal to 120 of unjust weight ? Aiis. 105 lb.
38. A meadow was to be mowed by 40 men in 10
days; in how many would it be finished by 30 men?
Ans. 13J- days.
37. When the first and second terms of the proportion
are not of the same denomination ; or one, or both of
them contain difi"erent denominations —
PtULE. — Reduce both to the lowest denomination con-
tained in either, and then divide the product of the
second and third by the first term.
Example 1. — If three ounces of tea co.st loci, what will 87
pounds cost ]
The lowest denomination contained in either is ounoes.
oz. It) d. i3oo^-|.5 d.
3 : 87 : : 15 : n — =60r)0=£29.
16
1302 ounces.
There is evidently the same ratio between 3 oz and 87 R)
as between 3 oz. and 1392 oz. (the equal of 87 lb).
RULE OF rROPORXION. 191
Example 2. — If 3 yards of any thing cost 4^. 0^(/., what
n be hough t fur £-1 '.
The lowest denomination in either is farthings.
4 9? : f::3: 1920^3
12 ' 20 231
yds. q nls.
=24 3 3.
57 pence. 40 shiUino;s.
4 12
231 farthings. 480 pence.
1920 farthings.
There is evidently the same ratio between 45. 9 id. and £2,
as between the numbers of farthings they contain, respectively
For there is the same ratio between any two quantities, as
between two others which are equal to them.
RvAMPLE 3. — If 4 cwt., 3 qrs., 17 lb, cost £19, how much
will 7 cwt. 2 qrs. cost ^
The lowest denomination in either is pounds.
cwt. qr. lb cwt. qr. £ 840x19
4 3 17 : 7 2 : : 19 : ■ . „> =£29 Is. 5d.
4 4 ^^-^
19 qrs. 30 qrs.
28 28
649 lbs. 840 lbs.
EXERCISES.
Find fourth proportionals to
39. 1 cwt. : 17 tons : : '^£5. A7is. £1700.
40. OS. : £20 : : 1 yard. Ans. 80 yards.
41. SO yards : 1 qr. : : 4005. A7is. Is. 3d.
42. 35. 4d. : £1 10s. : : 1 yard. Am. 9 yards.
43. 3 cwt. 2 qrs. : 8 cwt. 1 qr. : : £2. Am. £4.
44. 10 acres, 3 roods, 20 perches : 21 acres 3 roods :
£60. Am. £120.
45. 10 tons, 5 cwt., 3 qrs., 14 lb : 20 tons, 11 cwt ,
3 qrs. ; : £840. Am. £1680.
192 RULE OF PROPORTION.
46. What is the price of 31 tuns of wine, at £18 per
hhd. Ans. £22'32.
47. If 1 ounce of spice costs 45., what will be the
price of 16 Yb? Ans. £51 4s.
48. What is the price of 17 tons of butter, at £5 per
cwt. } Ans. £1700.
49. If an ounce of silk costs 4^., what will bo the
price of 15 Jb .^ Ans. £4.
60. What will 224 ft) 6 oz. of spice come to, at 35.
per oz. ? Ans. £538 IO5.
51. How much will 12 !b 10 oz. of silver come to, at
55. per oz. t Ans. £38 IO5.
52. What will 156 cwt. 2 qrs. come to, at 7d. per
fi).? J.715. £511 45. 8^.
53. What will 56 cwt. 2 qrs. cost at IO5. 6d. per
qr. .? ^715. £118 135.
54. If 1 yard of cloth costs £1 55., what will 110
yards, 2 qrs., and 3 nails, come to ? Ans. £138 75. 2\d.
55. If 1 cwt. of butter costs £6 65., how much will
17 cwt., 2 qrs., 7 ft), cost ? Ans. £110 125. lO^-^.
56. At 155. per cwt., what can I have for £615 155. >
Ans. 821 cwt.
57. How much beef can be bought for £760 125., at
325. per cwt. Ans. 475 cwt., 1 qr., 14 ft).
58. If 12 ft), 6 oz., 4 dwt., cost £150, what will 3 ft),
1 oz., 11 dwt., cost.^ Ans. £37 IO5.
59. If 10 yards cost 175., what will 3 yards, 2 qrs.
cost.? Ans. 55. 11^^.
60. If 12 cwt. 22 ft) cost £19, what will 2 cwt. 3
qrs. cost } Ans. £4 55. 8^d.
61. If 15 oz., 12 dwt., 16 grs., cost 195., what will
13 oz. 14 gi'S. cost? Ans. 15s. lOd.
38. If the third term consists of more than one deno-
mination—
Rule. — Reduce it to the lowest denomination which
it contains, then multiply it by the second, and divide
the product by the first term. — The answer will be of
that denomination to which the third has been reduced
and may sometimes be changed to a higher [Scc-
m. oj.
RULE OF PROPORTION. 19S
E.TAJVIPLE 1. — If 3 yards cost 95. 2|c/., wliat will 327 yards
eobt?
The lowest denomination in the third term is farthings.
yds. yds. 5. d. 327x441 ^ ^- ^•
3 : 327 : : 9 2| : ^ farthings=50 1 5^
12 ^
110 pence.
4
441 farthings.
Example 2. — If 2 yards 3 qrs. cost Hid., -what will 27
yards, 2 qrs., 2 nails, cost 1
The lowest denomination in the first and second is nails,
and in the third farthings.
yds. qr. yds. qr. n. d. 44*^x45
2 3 : 27 2 2 : : lU : — ^ farthing8=9*-. 5d,
4 4 4
11 qr. 110 qr. 45 farthings.
4 4
44 nails. 442 nails.
Reducing the third term generally enables us to perform the
required multiplication and division with more facility. — It iai
Bometimes, however, unnecessary.
Example.— If 3 lb cost £3 lis. 4^., what will 96 Tb cost ?
ft) lb £ s. d. ^ s. d. £ s. d. £ s. d.
3 11 4^x96
3: 96:: 3 11 4| : 7r^^—=Z 11 4? x 32=114 4 8
EXERCISES.
Find fourth proportionals to
62. 2 tons : 14 tons : : £28 10^. Aiu. 199 10^.
63. 1 cwt. : 120 cwt. : : 18^. 6^. Ans. iSlll.
64. 5 barrels : 100 barrels : : 65. Id. Ans. M Us. Sd
65. 112 S) : 1 lb : : £3 10s. Ans. l\d.
66. 4 lb : 112 ft) : : b\d. Ans. \2s. 3d.
67. 7 cwt., 3 qrs., 11 lb : 172 cwt., 2 qrs., 18 & : : ^£3
9s. 4\d. Ans. i:S7 55. 4d.
194 RULE OF PROPORTION.
68. 172 c^t., 2 qrs., IS !b : 7 cwt., 3 qrs., 11 lb : : £87
65. 3d. Ans. £3 19s. A\cl.
69. 17cwr.,2cirs., 14 lb : 2 cwt., 3 qrs., 21 lb : : £73
Ans. £12 35. Ad.
70. £87 Qs. 3d. : £3 19s. AU. . : 172 cwt., 2 qrs., 18
lb. Ans. 7 cwt., 3 qrs., 11 lb"
71. £3 195. 4irZ. : £87 65. 3d. : : 7 cwt., 3 qrs., 11 lb.
Ans. 172 cwt., 2 qrs., IS lb.
72. At I85. Qd. per cwt., what will 120 cwt. cost.'
Anis. £111.
73. At 3\d. per pound, what will 112 lb come to .^^
Ans. £1 105. Ad.
74. What will 120 acres of land come to, at 145. Qd.
per acre } Ans. £87.
75. How much would 324 pieces come to, at 25. Q\d.
per piece } Ans. £43 175. Qd.
76. Wliair is the price of 132 yards of cloth, at I65.
4f/. per yard .? Ans £107165.
77. If 1 ounce of spice costs 35. Ad.., what will 18 ib
10 oz cost.? Ans. £49 135. Ad.
78. K 1 ft) costs 65. 8rZ., what will 2 cwt. 3 qrs. come
to .? Ans. £102 135 Ad.
79. If £1 25. be the rent of 1 rood, what will be the
rent ( i 156 acres 3 roods } Ans. £689 145.
80. At IO5. Qd. per qr., what will 56 cwt. 2 qrs. be
worth.? Ans. £118 135.
81. At 155. Qd. per yard, what will 76 yards 3 qrs
come CO } Ans. £59 95. l\d.
82 What will 76 cwt. 8 ib come to, at 25. Qd. per
ib? ^725. £1065.
83 At 145. Ad. per cwt., what will be the cost of 12
cwt. 2 qrs. .? Ans. £8 195. 2d.
84. How much will 17 cwt. 2 qrs. come to, at 195.
IQd. -oer cwt. Ans. £17 75. Id.
85- If 1 cwt. of butter costs £6 65., what will 17 cwt ,
2 qrs , 7 lb, come to .? Ans. £102 125. lOi^.
86 If 1 qr. 14 ft) cost £2 155. 9^., what will be the
cost of 50 cwt., 3 qrs., 24 lb .^ .l?i5. £378 I65. S{d.
RULE OF PROPOBTION. I&5
ST. If the shilling loaf weigh 3 ft 6 oz., when flour
sells at £1 13?. 6^/. per cwt., what should be its weight
when flour sells at £1 7s. Gd ? Ans. 4 lb Iff oz.
88. If 100 lb of anything cost £25 Gs. Sd.j what will
be the price of 625 lb .- A/is. £158 4^. 0-^d.
89. If 1 ft) of spice cost 105. Sd., what is half an oz.
worth ? Ans. Ad.
90. Bought 3 hhds. of brandy containing, respectively,
Gl gals., 62 gals., and 62 gab. 2 qts., at Qs. Sd. per
gallon; what is their cost.^ Aiis. £61 I6s. Sd.
39. K fractions, or mixed numbers are found in ono
or more of tlie terms —
KuLE. — Having reduced them to improper fractions,
if they are complex fractions, compound fractious, or
mixed numbers — multiply the second and thud terma
together, and divide the product by the first — according
to the rules already given [Sec. lY. 36, &c., and 46,
kc] for the management of fractions.
Example. — If 12 men build oZ yards of wall in f of a
week, how long will they require to build 47 yards 1
3 1 yards=-,^ yards, therefore
3 v-i7
2 5 . 47 . . 3 . i_ii — =9i weeks, nearlv.
7
40. — If all the terms are fractions —
lluLE. — Invert the fii-st, and then multiply all the
terms together.
Example. — If ^ of a regiment consume f| of 40 tons of
flour in i of a year, how long will f of the same regiment
take to consume it "?
f : ^: I : |Xi-|=lx|Xf=V^=2C2-8 days.
This rule follows from that -whicli was given for tlie division
of one fraction by another [Sec. IV. 49].
41. If the first and second, or the first and third
terms, are fractions-^—
KuLE. — Reduce them to a common denominator
(should they not have one already), and then omit the
denominator:^
196 RULE OF PROPORTION.
Example. — If f of 1 cwt. of rice costs £2, what "will -j^ of
a cwt. cost 1
3 • To • • -^ • •
Reducing the fractions to a common denominator, we have
2^0 . 2 7 . . O . 1
30 • 30 • • ^ ' •
And omitting the denominator,
20 : 27 : : 2 : ?^=£2-7=£2 Us.
This is merely multiplying the first and second, or the first
and third terms by the common denominator — which [30] doea
not alter the proportion.
EXERCISES.
91. What will f of a yard cost, if 1 yard costs 135
6d. ? A71S. 10s. \\d.
92. If 1 ib of spice costs f^., what will 1 lb 14 oz.
cost.? Ans. Is. A^d.
93. If 1 oz. of silver costs 5|5., what will f oz. cost ?
Ans. As. Sd.
94. How much will \ yard come to if |- cost |5. >
Ans. -^jS.
95. If 2i yards of flannel cost 3^^., what is the price
of 4f yards } Ans. Qs. Ad.
96. What will 3f oz. of silver cost at Q\s. per oz. }
Ans. £\ \s. A\d.
97. If f'g- of a ship costs £273}, what is ^-^ of her
worth } Ans. £221 \2s. Id.
98. If 1 lb of silk costs 16|5., how many pounds can
I have for 37is. .? Ans. 2\ ft).
99. What is the price of 49tV yards of cloth, if 7f
cost £1 185. Ad. } Ans. £51 35. l^^d.
100. If £100 of stock is worth £98f , what will £362
85. l\d. be worth .? Ans. £358 75. Id.
101. If 945. is paid for 4f yards, how much can be
bought for £2y=V ? Ans. 24 yards, nearly.
MISCELLANEOUS EXERCISES IN SIMPLE PROPORTION.
102. Sold 4 hhds. of tobacco at 10irZ. per Ib : No. 1
weighed 5 cwt., 2 qrs. ; No. 2, 5 cwt., 1 qr., 14 lb ; No.
3, 5 cwt., 7 lb ; and No. 4, 5 cwt., 1 qr., 21 lb. What
was their price } Ans £104 145. 9^/.
RULE OF PROPORTIO!T. 197
103. Suppose that a bale of merchandise weighs 300 fc,
and costs £15 4s. 9d. ; that the duty is 2d. per pound ;
that the freight is 25s. ; and that the porterage home
is IS. 6d. : how much does 1 ib stand me in ?
£ s.
15 4
2 10
1 5
d.
9 cost.
0 duty.
0 freight.
;00 : 1 :: 19 1
20
6 porterage.
3 entire cost.
"12
300)4575
15 Id. Answer.
104. Received 4 pipes of oil containing 4S0 gallons
which cost OS. b^d. per gallon ; paid for fi-eight As. per
pipe; for duty, 6^. per gallon; for porterage, \s. per
pipe. What did the whole cost ; and what does it stand
me in per gaUon .- Ans. It cost £144, or 6s. per gallon
105. Bought three sorts of brandy, and an equal
quantity of each sort : one sort at 55. ; another at 6s. ;
and the third at 7s. What is the cost of the whole —
one gallon with another ? Ans. 6s.
106. Bought three kinds of vinegar, and an equal
quantity of each kind : one at 3^^. ; another at Ad. ;
and another at A^d. per quart. Ha\ing mixed them
I wish to know what the mixture cost me per quart }
Alls. Ad.
107. Bought 4 kinds of salt, 100 barrels of each ;
and the prices were 14s., 16s., 17s., and 19s. per barrel.
If I mix them together, what will the mixture have cost
me per barrel : Ans. 16s. M.
108. How many reams of paper at 9s. 9^., and
12s. 2>d. per ream shall I have, if I buy £55 worth of
both, but an equal quantity of each .' Ans. 50 reams
of each.
109. A ^-intner paid £171 for three kinds of wine :
one kind wa.^ £S 10s. ; another £9 5j. ; and the third
198 RULE OF PROPORTION.
£10 loSi per Llid. He had of each an equal quantity,
the auiomit of which is required.
£ .*.
8 10
9 5
10 15
£
28
28 10,
s. £
10 : 171 :
10, the price of three hogsheads of each.
£171x3
£28 10=^^ ^^*^^-
110. Bought three kind.s of salt, and of each an equal
quantity ; one was 14^,, another 165., and the third 195.
the barrel ; and the whole price was £490. How many
barrels had I of each ? Atis. 200.
111. A merchant bought ceriain goods for £1450,
with an agi-eement to deduct £1 per cent for prompt
payment. What has he to pay ? Ans. £1435 10^.
112. A captain of a ship is pro^-ided with 24000 lb
of bread for 200 men, of which each man gets 4 ft) per
week. How long will it last r Ans. 30 weeks.
113. How long would 3150 ft) of beef last 25 men, if
they get 12 oz. each three times per week r Ans. 56
weeks.
114. A fortress containnig 700 men who consume
each 10 lb per week, is provided with 184000 lb of
provisions. How long will they last r Ans. 26 weeks
and 2 da3^6.
115. In the copy of a work containing 327 pages, a
remarkable passage commences at the end of the 156th
page. At what page may it be expected to begin in a
copy containing 400 pages ? Ans. In the 191st page.
116. Suppose 100 cwt., 2 qrs., 14 lb of beef for
ship's use were to be cut up in pieces of 4 lb, 3 ib, 2 lb,
1 ftD, and -i- ft) — there being an equal number of each.
How many pieces would there be in all ? Ans. 1073 ;
and 3i ib'left.
117. Suppose that a greyhound makes 27 sprmga
while a hare makes 25, and that their springs are of
equal length. In how many springs will the hare be
overtaken, if she is 50 springs before the hound .?
RULE OF PROPOrvTIOX. tj9
The time taken by the greyhound for one spring ia to
that required by the hare, as 25 : 27, as 1 : §7, or aa
1 : ItTj [12]. Tlie greyhound, therefore, gains" {r^ of a
spring during every spring of the hare. Therefore
2-- : 50 : : 1 spring : 50-r-/5=G75, the number of springs
the hare will make, before it is overtaken.
118. If a ton of tallow costs d£35, and is sold at the
rate of 10 per cent, profit, v.hat is the selling price ?
Ans. £38 lUs.
119. If a ton of tallow costs d237 105., at what ratt
must it be sold to gain by 15 tons the price of 1 ton >
Ans. XMO.
120. }3ought 45 barrels of beef at 21s. per barrel ;
among them are 16 barrels, 4 of which would be worth
only 3 of the rest, liow much must I pay > Ans.
£43 Is.
121. If 840 eggs are bought at the rate of 10 for a
penny, and 240 more at S for a penny, do I lose or gain
if I sell all at IS for 2d. ? Ans. I gain 6d.
122. Suppose that 4 men do as much work as 5
women, and that 27 men reap a quantity of corn in 13
days. In how many days would 21 women do it } Ans.
The work of 4 men=that of 5 women. Therefore (dividing
each of the equal quantities by 4, they will remain equal),
4 mens work the work of 5 women
■ -T (one man's work) == . — . Con-
bcquently 1| times the work of one woman =^1 man's work : — •
that is. the Avork of one man, in terms of a woman's work,
is Ij ; or a womahs work is to a mans work : : 1 : 1|.
Hence 27 men's Avork = 27xl]- women's work; then, in
place of saying—
21 women : 27 men : : 13 days : ?
say the work of 21 women : the work of 27 xl]- (=33^
333 y 13
women : : 13. : %-^ =20^1 days.
123. The ratio of the diameter of a circle to its
circumference being that of 1 : 3' 14159, what is the
circumference of a circle v/hose diameter is 47'36 feet?
Ans. 148-78618 feet.
124. If a pound (Troy Avoight) of silver is worth GCs.,
200 RULE OF PROPORTION.
what is the value of a pound avoirdupoise ? Ans. £4
05. 2ld.
125. A merchant failing, owes £40881871 to his
creditors ; and has property to the amount of £12577517
105. 11^. How much per cent, can he pay ? Ans. £30
Ids. S^d.
126. If the digging of an English mile of canal costs
£1347 7s. 6d., what will be the cost of an Irish mile?
Ans. £1714 165. 9^d.
127. K the rent of 46 acres, 3 roods, and 14 perches,
is £100, what will be the rent of 35 acres, 2 roods, and
10 perches.? Ans. £75 185. 6^d.
128. When A has travelled 68 days at the rate of
12 miles a day, B, who had travelled 48 days, overtook
him. How many miles a day did B travel, allowing
both to have started from the same place ? Ans. 17.
129. If the value of a pound avohdupoise weight be
£4 05. 2^d., how many shillings may be had for one
pound Troy ? Ans. €65.
130. A landlord abates ^ in a shilling to his tenant ;
and the whole abatement amounts to £76 35. 4^d,
What is the rent .? Ans. £228 IO5. Id.
131. If the third and tenth of a garden comes to £4
IO5., what is the worth of the whole garden.? Ans.
£10 75. 8\d.
132. A can prepare a piece of work in 4^ days ; B
in 6i days ; and C in 8^ days. In what time would all
three do it ? Ans. 2yH^.
4tk days : 1 day : : 1 whole of the work : f part of the whole —
or what A would do in a day.
65 days : 1 day : : 1 whole of the work : ^V part of the whole —
or what B would do in a day.
8i days : 1 day : : 1 whole of the work : ^ part of the whole —
or what C would do in a day.
|-|-p__^-2^=i^4^=what all would do in a day.
Then the if f4 part of' the work : 1 whole of the work : : 1
day (the time all would require to execute ^|^4 of the work) :
2-|.|^ days, the time all would take to do the wliole of it.
133. A can trench a garden in 8| days; B in 51-
days ; but when A, B, and C work together, it will be
finished in 1^^ days. In how many days would C be
able to do \i by huuself .? Ans. 2if f days.
RULE OF PROPORTION. 201
t^% vvi A, B, and C's work in one day =5 of the whole=l^J
Subtract- ( A*s work in 1 day=/^ j _i ,o ^f the whole- ^^o
ing j B"s work in 1 day=/^ j "^ST ^1 tiie wnole— j-,-^,
C's work in one day remains equal to . . . j\Vi
Then y"^%^^ (C's work in one day) : 1 whole of the work : : 1
day : 2 ^^, the time required.
134. A ton of doals yield about 9000 cubic feet of
c^ ; a street lamp consumes about 5, and an argand
uurner (one in which the air passes through the centre
of the flame) 4 cubic feet in an hour. How many tons
of coal would be required to keep 17493 street lamps,
and 192724 argand burners in shops, &.C., lighted for
1000 hours } Ans. 95373^.
135. The gas consumed in London requires about
50,000 tons of coal per annum. For how long a time
would the gas this quantity may be suppos'ed to pro-
duce (at the rate of 9000 cubic feet per ton), keep one
argand light (consuming 4 cubic feet per hour) con-
stantly burning .' Ans. 12S42 years and 170 days.
136. It requires about 14,000 millions of silk worms
to produce the silk consumed in the United Kingdom
annually. Suppo.sing that every pound requires 3500
worms, and that one-fifth is wasted in throwing, how
many pounds of manufactured silk may these worms
be supposed to produce .' Ans. 14SS tons, 1 cwt., 3 qrs.,
17 1b.
137. If one fibre of silk will sustain 50 grams, how
many would be required to support 97 lb f Ans 13580.
138. One fibre of silk a mile long weighs but 12
grains ; how many miles would 4 millions of pounds,
annually consumed in England, reach ?
Ans. 23333333331 miles.
139. A leaden shot of 4^ inches in diameter weighs
17 ib ; but the size of a shot 4 inches in diameter, is to
that of one 4| inches in diameter, as 64000 : 91125 :
what is the weight of a leaden ball 4 inches in diameter ?
Afis. 11-9396.
140. The sloth does not advance more than 100
yards in a day. How long would it take to crawl from
Dublin to Cork, allowing the distance to be 160 English
miles? An.<(. 2S16 da3's; or 8 years, nearly.
202 COMPOUXD PROPORTION.
141. Euglisli race horses have been known to go at
the rate of 58 miles an hour. In what time, at this
velocity, might the distance from Dublin to Cork be
travelled over ? Ans. 2 hours, 45' 31" 2 '
142. An acre of coals 2 feet thick yields 3000 tons ;
and one 5 feet thick 8000. How many acres of 6 feet
thick would give the same quantity, as 48 of 2 feet
thick } Ans. 18.
143. The hair-spring of a watch weighs about the
tenth of a grain ; and is sold, it is said, for about ten
shillings. How much would be the price of a pound of
crude iron, costing one halfpenny, made into steel, and
then into hair-springs — supposing that, after deducting
waste, there are obtained from the iron about 7000
f^jrains of steel } Ans. ^£35000.
CO:vIPOUXD PROPORTIOx\.
42. Compound proportion enables us, although two
or more proportions are contained in the question, to
obtain the required answer by a single stating. In
compound proportion there are three or more ratios, one
of them imperfect, and the rest perfect.
43. Rule — I. Place the quantity belonging to the
imperfect ratio as the third term of the proportion.
II. Put down the terms of each of the other ratios
iu the first and second places — in such a way that the
antecedents may form one column, and the consequents
another. In setting down each ratio, consider what
oftect it has upon the answer — if to increase it, set down
the larger term as consequent, and the smaller as ante-
cedent ; if to diminish it, set down the smaller term as
consequent, and the larger as antecedent.
III. ]^Iultiply the quantity in the thu-d term by tin?
product of all the quantities in the second, and divide
the result by the product of all those in the first.
44. EsAMPLE 1. — Tf 5 men build 16 yards of a wall in 20
days, in how many days would 17 men build 37 yards ?
The question briefly put doAvn [32], v>-ill ba as foUuws *
COMrOUND PROrORTION. 203-
TP 1 > conditions which jrivo 20 days,
lb yards ) o j
20 days imperfect ratio.
1 days, the number sought.
17 men
'7 yards
conditions which give the vequired number oi Ja'
'ilie imperfect ratio consists of days — therefore Ave are to
put 20, the given number of day.s, in the third place. Two
ratios remain to be set down — that of numbers uf men, and
that of numbers of yards. Taking the former first, we a.sk
ourselves how it affects the answer, and find tliat the more
men there are, the smaller the required number will be — since
the greater the number of men, the shorter the time required
to do the work. We, therefore, set down 17 as antecedent,
and 5 as consequent. Next; considering the ratio consisting
of yards, we find that the larger the number of yards, the
longer tho time,- before they are built — therefore increasing
their number increases the quantity required. Hence we
put 37 as consequent, and 16 as antecedent: and the whole
will be as follows : —
17 : 5 : : 20 : )
10
o/
'i ^0v.iv.'>7
And 17 : 5 :: 20 : ~\.^\'.. =13G duvs. n.::ir;v.
10 : 37 ^'><^^
45. The result obtained by the rule is the same as -svould lie
found by taking, in succession, the two proportions suppo.'iod
by the question. Thus
If 5 men would build 16 yards in 20 days, iu how many
days would they build 37 yards '
16': 37 :: 20 : — — number of davs which 5 men would
16
require, to build 37 yards.
If 5 meu would build 37 yards iu" ^' days, in how many
16
days would 17 men build them ?
,-.-.. 20X37 . 20x37^r . .-_ 20x5x37 ,, ,
17 : 0 : : — — — : — — — Xo-;-li= ^^ ;\ , the number
16 16 17X16
of days found by the rule.
40. Example 2. — If 3 men in 4 days of 12 working hours
each build 37 perches, in how ma'ny days of 8 working
hours ought 22 men to build 070 perches '?*
204 COMPOUND PROPORTION.
3 men.
4 days.
12 hours.
37 perches.
"? days.
8 hours.
22 men.
970 perches.
3x12x970x4
22 : 3 : : 4 : no^A^or =2U days, nearly.
8 : 12
37 : 970
22X8X37
The number of days is the quantity sought ; tlierefore 4
days constitutes the imperfect ratio, and is put in the tliird
place. The more men the fewer the days necessary to per-
form the work ; therefore, 22 is put first, and 3 second. The
smaller the number of working hours in the day, the larger
the number of days ; hence 8 is put first, and 12 second.
The greater the number of perches, the greater the number
of days required to build them : consequently 17 is to be
put first, and 970 second.
47, The process may often be abbreviated, by divid-
ing one term in the first, and one in the second place ;
or one in the first, and one in the third place, by the
same number.
Example 1. — If the carriage of 32 cwt. for 5 miles costs
8s., how much will the carriage of 160 cwt. 20 miles cost ?
32:160::8:i^^5^«=160
5 : 20 '^^^^
Dividing 32 and 160 by 32 we have 1 and 5 as quotients.
Dividing 5 and 20 by 5 we have 1 and 4 ; and the propor-
tion will be —
1 : 5 :: 8 : 5x4x8=160
1 : 4
48. "We arc to continue this kind of division as long
as possible — that is, so long as any one number will
measure a quantity in the first, and another in the second
place ; or one in the first and another in the thu'd place
This will in some instances change most of the quantities
into unity — which of course may be omitted.
COMPOUND PROPORflON. 205
Example 2. — ]f 28 loads of stone of loj^vrt. each, build a
wall 20 feet lon^ and 7 feet high, how many loads of 10 cwt.
will build one 323 feet long and 9 feet high 1
19 : 15 : : 28 : 15x323x9x28_^,^
20 : 323 19x20x7
7 : 9
Dividing 7 and 28 by 7, we obtain 1 and 4. — Substituting
these, we have
19 : 15 : : 4 : T
20 : 323
/ 1:9
Dividing 20 and 15 by 5, the quotients are 4 an/i 3 :
19 : 3 : : 4 : ?
4 : 323
1 : 9
Dividing 4 and 4 by 4, the quotients are 1 and 1 :
19 : 3 : : 1 : '?
1 : 323
1 : 9
Dividing 19 and 323 by 19, the quotients are 1 and 17 :
1 : 3 :: 1 : 3x17x9=459.
1 : 17
1:9
In this process we merely divide the first and second, or
first and third terms, by the same number — which [29] does
not alter the proportion. Or we divide the numerator and
denominator of the fraction, found as the faurth term, by the
game niuuber — which [Sec. IV. 15] does not alter the quo-
tient.
EXERCISES IN COMPOUNB PROPORTION.
1. If £240 in 16 months gains ^£64, how much will
£60 gain in 6 months } Ans. £6.
2. "With how many pounds sterling could I gain
£5 per annum, if with ^50 I gain £30 in 16 months ?
Ans. £100.
3. A merchant agrees with a carrier to bring 15 cwt
of goods 40 miles for 10 crowns. How much ought h«
to pay, in proportion, to have 6 cwt. carried 32 miles t
Ans. 165.
K 2
'«?')0 COMPOUND PllOPORTIOX.
4. If 20 cwt. rfre carried the distance of 50 miles for
£dj how much will 40 c\vt. cost, if carried 100 milos ?
Am. i320.
5. If 200 ib of merchandise are carried 40 miles
for 35., how many pounds might be carried 60 miles
for £22 145. 6d. Am. 20200 Bd.
6. If 286 It) of merchandise are carried 20 miles
for 35. , how many miles might 4 cwt. 3 qrs. be carried
for ie32 6s. 8d. } Ans. 2317-627.
7. If a wall of 28 feet high were built in 15 days
by 68 men, how many men would build a wall 32 feet
high in 8 days } Ans. 146 nearly.
8. If 1 ft) of thread make 3 yards of linen of li
yards wide, how many pounds of thread would be required
to make a piece of linen of 45 yards long and 1 yard
wide-? Ans. 12 ft).
9. If 3 ft) of worsted make 10 yards of stuff of 1^
yards broad, how many pounds would make a piece 100
yards long and 1^ broad r Ans. 25 ft).
10. 80000 cwt. of ammunition are to be removed
from a fortress in 9 days ; and it is found that in 6 days
18 horses have carried away 4500 cwt. How many horses
would be required to carry away the remainder in 3
days ^ Ans. 604.
11. 3 masters who have each 8 apprentices earn £36
in 5 weeks — each consisting of 6 working days. How
much would 5 masters, each having 10 apprentices,
earn in 8 weeks, working 5^ days per week — the wages
being in both cases the same .^ Am. £110.
12. If 6 shoemakers, in 4 weeks, make 36 pair of
men's, and 24 pair of women's shoes, how many pair of
each kind would IS shoemakers make in 5 weeks }
Ans. 135 pair of men's, and 90 pair of women's shoes.
13. A waU is to be built of the height of 27 feet ;
and 9 feet high of it are built by 12 men in 6 days.
How many men must be employed to finish the remain-
der in 4 days r Ans. 36.
14. If 12 horses in 5 days draw 44 tons of stones,
how many horses would draw 132 tons the same dis-
tance in 18 days } Am. 10 horses.
15. If 275. are the wages of 4 men for 7 days,
COMPOUND PROPORTION. 207
what will be the wages of 14 men for 10 days? Ans.
£6 Ids.
16. If 120 busliels of corn last 14 horses 56 days,
Low many days ayiII 90 biLshels last 6 horses ? Ails.
98 days.
17. K a footman travels 130 miles in 3 days when
the days are 14 hours long, in how many days of 7 hours
each will he travel 390 miles r Aiis. IS.
18. If the price of 10 oz. of bread, when the corn
is 4s. 2d. per bushel, be od., what must be paid for 3 lb
12 oz., when the corn is os. bd. per bushel ? Aiis. Zs. 3d.
19. 5 compositors in 16 days of 14 hours long can
coraposo 20 sheets of 24 pages in each sheet, 50 lines
in a page, and 40 letters in a line. In how many
days of 7 hours long may 10 compositors compose a
volume to be printed in the same lett<ir, containing 40
sheets, 16 pages in a sheet, 60 lines in a page, and
50 letters in a line r Ans. 32 days.
20. It has been calculated that a square degree (about
69X69 square miles) of water gives off by evapora-
tion 33 millions of tons of water per day. How much
may be supposed to rise from a square mile in a week ^
Alls. 48519-2187 tons.
21. When the mercury in the barometer stands at
a height of 30 inches, the pressure of the aii* on every
square inch of surface is 15 ib. Y/hat will be the pres-
sure on the human body — supposing its whole surface
to be 14 square feet ; and that the barometer stands at
31 inches .' Ans. 13 tons 19 cwt.
QUESTIONS IX RATIOS AND PROPORTION.
1 . What is the rule of proportion ; and is it ever
called by any other name .'' [1].
2. What is the difference between simple and com-
pound proportion .^ [30 and 42] .
3. What is a ratio .' [7].
4. What are the antecedent and consequent } [7] .
5. What is an inverse ratio i [8] .
6. What is the difference between an arithmetical
and a geometrical ratio ? [9] .
208 COMPOUND PROPORTION.
7. fiow can we know whether or not an arithmetical
or geometrical ratio, is altered in value ? [10 and 11].
8. How is one quantity expressed in terms of an
other? [12].
9. What is a proportion, or analogy .'' [14].
10. What are means, and extremes ? [15].
1 1 . What is the arithmetical, or geometrical mean of
two quantities ? [19 and 27].
12. How is it known that four quantities are in arith-
metical proportion ? [16].
13. How is it known that four quantities are in geo-
metrical proportion ? [21].
14. How is a foui'th proportional to three quantities
found .^ [17 and 22].
15. Mention the principal changes which may be
made in a geometrical proportion, without destroying
it.? [29].
16. How is a question in the simple rule of three to
be stated, and solved .? [31].
17. Is it necessary, or even correct, to divide the
rule of three into the dhect, and inverse ? [35] .
18. How is the question solved, when the first oi
second terms are not of the same denomination ; or one^
or both of them contain different denominations ? [37]
19. How is a question in the rule of proportion solved,
if the thkd term consists of more than one denomina-
tion .? [38] .
20. How is it solved, if fractions or mixed numbers
are found in the first and second, in the first and third,
or in all the terms ? [39 and 40] .
21. How is a question in the rule of compound pro-
portion stated, &c. ? [43] .
22. Can any of the terms of a question in the rulo
of compound proportion ever be lessened, or altogeth>?r
banished ? [47 and 48] .
I
209
A R I T II M E T I C
PART II.
SECTIOx^f VI
PrxACTICE.
1. Practice is so called from its being the method
of calculation practised by mercantile men : it is an
abridged mode of performing processes dependent on
the rule of three — particularly when one of the terms
is unity. The statement of a question in practice, in
gencrai terms^ would be, " one quantity of goods is to
another, as the price of the former is to the price of
the latter."
The simplification of the rule of three by means of
practice, is principally effected, either by dividing the
given qiL illy into " parts," and finding the sum of
the prices ^.C these parts ; or by dividing the price into
^' jparts," and finding the sum of the prices at each of
these parts : in either case, as is evident, we obtain the
required price.
2 Parts are of two kinds, " aliquot" and " aliquant."
The aliquant parts of a number, are those which do
not measure it — that is, which cannot be multiplied by
any integer so as to produce it ; the aliquot parts are,
Jis we have seen [Sec. II. 26],, those which measure it.
3. To find the aliquot parts of any number —
Rule. — Divide it by its least divisor, and the result-
ing quotient by its least divisor : — proceed thus until
the last quotient is unity. All the divisors are the priim
aliquot parts ; and the product of every two, every three,
&.C., of them, are the co??ipound aliquot parts of the
given number.
210
PRACTICE.
4. Example.— What are the prime, and compound aliquot-
parts of 84 ?
2)84
2)42
3)21
'^)"
The prime aliquot parts
are 2, 3, and
7 : and
2x2= 4]
2x3= 6
2x7=14
;> are the comp
3x7=21
ound aliquot parts.
2x2x3=12
2x2x7=28
2x3x7=42 J
All the aliquot parts, placed in order,
are 2, 3, 4, 6, 7, 12,
14, 21. 28, and 42.
5. We may apply this rule to appUmte numbers. — Let it
be required to find the aliquot parts of a
pound, in shillings
and rjencc. 240rf.— £1.
2)240
2)120
2)60
2)30
3)15
5)5
1
The prime aliquot parts of a pound
are, therefore, 2r/.,
3</., and bd. : and the compound,
d.
2x2= 4
2x3= G
2x5= 10
2x2x2= 8 s.
d.
2x2x3- 12- 1
0
2x2x5= 20= 1
8
2x3x5= 30= 2
6
2x2x2x2= 16= 1
4
2x2x2x3= 24= 2
0
2x2x2x5= 40= 3
4
2x2x3x5= 60= 5
0
2x2x2x2x3= 48= 4
0
2x2x2x2x5= 80= 6
8
2x2x2x3
X 5=120=10
0
PRACTICE.
211
And placed in order —
£ d.
£ d. s. d.
rh=^
tV=1G=1 4
Vo=3
^= 20= 1 8
S=4
X=24=2 0
.\=5
'\= 30= 2 6
^V=6
^= 40= 3 4
S=8
j= 48= 4 0
A=10 5. d.
j= 00= 5 0
iV=12=l 0
^= 80= 6 8
1=120=10 0
Aliquot parts of a shilling, obtained in the Banie way —
s. d.
s. d.
s. d.
s=i
tt
a
%=i
i=2
1=6
Aliquot parts of avoirdupoise weight —
Aliquot parts of a ton.
Aliquot parts of a cwt.
Aliquot parts of a quarter
ton cw-t. qr.
CWt. H)
qr. ib
A= t = 2
3V=2
t\-= 2
^V=I = 4
o'.= 4
^=4
tV= 1|= 5
r6= '
i=7
tV= 2 = 8
T^i=8
1=14
1= 2^=10
i=14
1= 4'=16
1=16
1= 5 =20
1=28
J=10 =40
^=
=56
Aliquot parts may, in the same manner, be easily
obtained by the pupil from the other tables of weights
and measures, page 3, &c.
6. To find the price of a quantity of o?ie denomina-
tion— the price of a " higher" being given.
Rule. — Divide the price by that number which ex-
presses how many times we must take the lower to
make the amount equal to one of the higher denomina-
tion.
Example. — What is the price of 14 lb of butter at 72j.
per cwt. 1
We must take 14 lb, or 1 stone 8 times, to make 1 cwt.
Therefore the price of 1 cwt. divided by 8. or 72s.-f-8=9d\,
is the price of 14 ib.
The table of aliquot parts of avoirdupoise weight showa
that 14 R) is the | of a cwt. Therefore its price is the ] of
the price of 1 cwt.
212 PRACTICE.
EXERCISES.
"What is the price of
1. I cwt., at 29^\ 6d. per cwt. ? Ans. Is. A\d.
2. J- a yard of clotli, at 85. Gd. per yard ? Ans. 4s. 3d
3. 14 ft) of sugar, at 45^. 6d. per cwt. ? Ajis. ds. S^d
4. What is the price of ^ cwt,, at 50^. per cwt. ?
£ s. d.
50s.=2 10 0
qrs. cwt. £ s.
The price of 2=i is 15 0=2 104-2
of l=|-^2 is 0 12 6=1 5-^2
Therefore the price of 2-|-l qrs.(=| cwt.) is 1 17 6
J cwt., or 3 qrs.=2-f-l qrs. But 2 qrs.=| cwt. ; and its
price is half that of a cwt. 1 qr.=j cwt.-^2; and its price
is half the price of 2 qrs. Therefore the price of f cwt. is
half the price of 1 cwt. plus the half of half the price of
one ewt.
"What is the price of
5. |- oz. of cloves, at 95. 4d. per ib ? Ans. 3^d.
6. 1 nail of lace, at 155. 4d. per yard .^ Ans. ll^d.
7. i ft), at 235. 4d. per cwt. ? Ans. l^d.
8. I lb, at 185. 8^. per cwt. ? Ans. l^d.
7. When the price of more than one "low3r" deno-
mination is required —
!KuLE. — Find the price of each denomination by the
last rule ; and the sum of the prices obtained will be
the requii-ed quantity.
Example. — What is the price of 2 qrs. 14 Vb of sugar,
at 455. per owt. ?
5. d.
45 0 price of 1 cwt.
[or ^ of 1 cwt.
cwt. And 22 6, or 455.-^-2, is the price of 2 qrs.,
2 qrs.=i 5 7-^-, or'455.-^8=22<. 0(i.-i-4, is the
14 lb=^, or I of 2 qrs. "' price of 1411), the ^ of 1 cwt.,
■ or. the \ of 2 qrs.
And 28 1^ is the price of 2 qrs. 14 lb.
2 qr8.=:^ of 1 cwt. Therefore 455. (the price of 1 cwt.) --%
or 265. 6(/., is the price of 2 qrs.
PRACTICE. 213
14 lb is the | of 1 cwt., or the | of 2 qrs. Therefore
45s.-f-8, or 22s-. 6fZ.-^4=5s. 7^d., is the price of 14 lb.
And 22>. 6</.+5.'?. 7^<l., or the price of 2 qrs. plus the price
of 14 lb, is the price of 2 qrs. 14 lb.
EXERCISES.
What is the price of
9. 1 qr., 14 ft) at 465. 6d. per cwt. ? Ans. I7s. 5\d.
10. 3 qrs. 2 nails, at 175. 6d. per yard .' Ans.
Ids. 3|fZ.
11.5 roods 14 perches at 3s. IGd. per acre .' Aiis.
5s. l^d.
12. 16 dwt. 14 grs., at £4 4s. 9d. per oz. ? Ans.
£3 105. 3id.
13. 14 & 5 oz., at 255. 4d. per cwt. ? A^is. 3s. 2^d.
8. When the price of omi "higher" denomination is
rtiquired —
kuLE. — Find whaC niimber of times the lower deno-
mination must be taken, to make a quantity equal to
one of the given denomination ; and multiply the price
by that number. (This is the reverse of the rule given
above [G]).
Example. — What is the price of 2 tons of sugar, at 505.
per cwt. *
1 cwt. is the jY of 2 tons ; hence the price of 2 tons \vill
Ikj 4U times the price of 1 cwt. — or 50.9.x40=£100.
50.';. the price of 1 cwt. multiplied
by 40 the number of hundreds in 2 tons,
gives 2000.<:.
or £100 as the price of 40 cwt., or 2 tuns.
EXERCISES.
What is the price of
14. 47 cwt., at l5. Sd. per lb ?. Ans. £488 V3s. 4d
15. 36 yards, at 4d. per nail r Ans. £9 125.
16. 14 acres, at 55. per perch } Ans. £d.f>0.
17. 12 lb, at If^Z. per grain ? Ans. £504.
18. 19 hhds., at 3d. per gallon r Ans. £14 195. 3^/.
9. When the price of more th(iu one "higher" dono=
minatio^i is required —
214 PRACTICE
KuLE. — Find the price of each by the last, and add
the results together. (This is the reverse of the rule
given above [7] ) .
Example.— What is the price of 2 cwt. 1 qr. of flour,
at 2s. per stone 1
1 stone is the yV of 2 cwt. Therefore
2s., the price of one stone,
multiplied by 16, the number of stones in 2 cvrt.,
gives 325., the price of 16 stones, or 2 cwt.
There are 2 stones in 1 qr, : therefore 25. (the price of 1
stone) x2=45. is the price of 1 qr. And 325.-}-45.=36.?.=
£1 I65., is the price of 2 CTvt. 1 qr.
EXERCISES.
What is the price of
19. 5 yards, 3 qrs., 4 nails, at 4fi?, per nail.? Ans.
£1 12s.
20. 6 cwt. 14 ft), at 3d. per lb } Ans. £8 II5. Qd.
21. 3 ft) 5 oz., at 2\d. per oz. .? Ans. 95. \\\d.
22. 9 oz., 3 dwt., 14 grs., at ^d. per gr. } Ans.
£13 155. ^d.
23. 3 acres, 2 roods, 3 perches, at bs. per perch }
Ans. £\A0 155.
10. When the price of one denomination is given, to
find the price of any number of another —
Rule. — Find the price of one of that other denomi-
nation, and multiply it by the given number of the
latter.
Example. — What is the price of 13 stones at 255. per
cwt.?
1 8tone=-j cwt. Therefore
8)255., the price of 1 c^^'t. divided by 8,
gives 3 1|, the price of 1 stone, or | of 1 cwt.
Multiplying this by 13,' the number of stones,
we obtain £2 0 7^ as the price of 13 stones.
1 stone is the | of 1 cwt. Hence 2o5.-j-8=35. l|rf., is the
price of one stone ; and Zs. I|<i.xl3, the price of 13 stones.
PRACTICE. 31^
EXERCISES.
What is the price of
24. 19 lb, at 2d. per oz. } Am. £.2 IQs. Sd.
25. 13 oz., at \s. Ad. per tb } Atis. \s. \d.
26. 14 ft), at 2s. 6d. per dwt. ? Ans. £420.
27. 15 acres, at 185. per perch ? Ans. £2160.
28. 8 yards, at 4d. per nail r Ans. £2 2s. Sd.
29. 12 hhds., at od. per pmt ? Am. £126.
30. 3 quarts, at 91s. per hhd. } Am. Is. Id.
11. When the price of a given denomination is the
aliquot part of a shilling, to find the price of any num-
ber of that denomination —
Rule. — Divide the amount of the given denomina-
tion by the number expressing what aliquot part the
given price is of a shilling, and the quotient will be the
required price in shillings, &c.
Example. — What is the price of 831 articles at 4d. per }
3)831
277s.=£13 175., is the required price.
M. is the I of a shilling. Hence the price at 4rf. is ^ of
what it would be at Is. per article. But the price at 1$. per
article would be 831.s.:— therefore the price at -id. is So1.s.h-3
or 277s.
EXERCISES.
W^hat is the price of
31. 379 ft) of sugar, at 6d. per ft> > Am. £9 9s. 6d.
32. 5014 yards of calico, at 3d. per yard ? Am.
£62 135. 6d.
33. 258 yards of tape, at 2d. per yard } Am. £2 Ss.
12. When the price of a given denomination is the
aliquot part of a pound, to find the price of any number
of that denomination —
KuLE. — Divide the quantity whose price is sought
by that number which expresses what aliquot part the
given price is of a pound. The quotient will be the re-
quired price in pounds, &c.
216 PRACTICE.
Example. — What is the price of 1732 ft) of tea, at 5i
per ft) ?
5.S. is the | of £1 ; therefore the price of 1732 ft) is th^
^ of what it would be at £1 per Ih. But at £1 per ft) it
would be £1732; therefore at 5s. per ft) it is £1732-f.4=
£433.
EXERCISES.
What is the price of
34. 47 cwt., at 6s. Sd. per cwt. ? Ans. £15 13s. 4d.
35. 13 oz., at 4s. per oz. ? Ans. £2 12s.
36. 19 stones, at 2s. 6d. per stone ? Ans. £2 7s. 6d.
37. 83 ft), at Is. Ad. per ib ? ^7i5. £5 10s. 8^.
38. 115 qrs., at ^d. per qr. } Ans. iS3 16s. Sd.
39. 976 ftj, at 10s. per ib .? ^tis. £488.
40. 112 ft), at bd. per ft) .? ^7is. £2 Qs. Sd.
41. 563 yards, at 10^. per yard ? Ans. £23 9s. 2d.
42. 112 ft), at 5s. per ft) .? ^7is. £28.
43. 795 ft), at Is. 8^. per ft) .? .A?zs. £66 5s.
44. 1000 ft), at 3s. 4d. per ft) f Ans. £166 13s. 4^^.
13. The complermni of the price is what it wants of a
pound or a shilling.
When the complement of the price is the aliquot part
or parts of a pound or shilling, but the price is not —
Rule. — Find the price at £1, or Is. — as the case
may be — and deduct the price of the quantity calculated
at the complement.
Example. — What is the price of 1470 yards, at 13s. 4(Z.
per yard 1
6s. Sd. (the complement of 13s. 4J.) is ^ of £1.
From £1470, the price at £1 per yard,
subtract 490, the price at Gs. Sd. (the complement)
per yard,
and the difference, 980, will be the price at 13s. Ad. per yard.
1470 yards at 13s. 4tZ., plus 1470 at Gs. Sd., are equal to
1470 at 13s. 4(Z.4-6.<J. 8f/., or at £1 per yard. Hence the
price of 1470 at 13s. 4f/.=the price of 1470 at £1, minus
.the price of 1470 at Gs. Sd. pcfr yard.
PRACTICE. 217
EXERCISES.
"WTiat 13 the price of
45. 51 ft), at 175. 6d. per ft) .? tI?^^. £44 125. 6a
46. 39 oz., at 7d. per oz. .? ^?Z5. £1 2s. 9d.
47. 91 ft), at lOcZ. per ft) .? J.715. £3 155. lOd.
48. 432 cwt., at 165. per cwt. ? Aiis. £345 125.
14. When neither the price nor its complement is
the aliquot part or parts of a pound or shilling —
Rule 1. — Di\dde the price into pounds (if there are
any), and aliquot parts of a pound or shilling; then
find the price at each of these (bj preceding rules) : —
the sum of the prices will be what is required.
Example.— What is the price of 822 lb, at £5 19s. 3|J.
per lb ? £d 19s. 3fJ.=£5-f-195. 2>ll.
s. d. £
10 0 =i
6 8 =1
But 195. Z^d.=
6=1-
'TB'fT*
0 0i=i|^-j-6=^, OT^ of the last
Hence the price at £5 195. 3|J. is equal to
£ £ s. d. £ s. d.
822x5 =4110 0 0, the price at 5 0 0 per lb.
«|- =411 0 0 „ £1 or 0 10 0
8f 2 = 274 0 0 „ £l or 0 6 8
«|2 = 102 15 0 „ £» or 0 2 6
^lfer-H20)= 5 29 .,,£jhoTO on „
,.2yB22_ 6) = 0 17 1-^ „ £^^^ or 0 0 0^
And £4903 14 lOi is the price at £5 19 3| „
The price at the whole, is evidently equal to the sum of
the prices at each of the parts.
If the price were £o 19s. S^d. per lb, we should sub-
tract, and not add the price at }d. per lb ; and we then
would have j£4902 05. 7^d. as the answer.
15. Rule 2. — ^Find the price at a pound, a shilling,
a penny and a farthing ; then multiply each by their
218 PRACTICE.
respective numbers, in the given price ; and add the
products. Using the same example —
£ s. d. £ s. d.
20)822 0 0 (the price at £1)X 5=4110 0 0 the price at £5
12)41 2 0 (the price at l5.)X 19= 780 18 0 „ 19«.
4)3 8 6 (the price at lc?.)X 3= 10 5 6 „ Zd
17 14(the price at ic?.)X 3= 2 11 4i „ id.
And the price at £5 19^. Zld. is £4903 14 lOi
16. Rule 3. — Find the price at the next number of
the highest denomination ; and deduct the price at the
difference between the assumed and given price.
Using still the same example —
£6 is next to £5 — the highest denomination in the given
price.
£ s. d. £ s. d
From the price at £6 0 0 . . . . or 4932 0 0
Deduct the price ( the price at 8c?. =:27 80 ) no k n
atsirf. "^ I ,, id.= o 17 nr'^ ^^ ^ ^^
The difference Tvill be the price at £5 lO*. 3| or £4903 14 10^
17. Rule 4. — Find the price at the next higher
ahquot part of a pound, or shilling ; and deduct the price
at the difference between the assumed, and given price
Example. — What is the price of 84 It), at 6s, per lb I
6s.=6s. 8d. minus 8J.=iminus|-i-10.
£> £> s. d.
Therefore 84-i-3=28 0 0 is the price at Gs. 8d. per It).
Deducting f^ of this=2 16 0 the price at 8(Z.,
we have £25 4 0, the price at 6s.
EXERCISES.
AVhat is the price of
49. 73 ft), at 135. per ft) ? Ans. £47 95.
50. 97 cwt., at 155. 9^. per cwt. ? Ans. £76 7s. 9d.
51. 43 ft), at 35. 2d. per ft) ? Arts. £6 16s. 2d.
52. 13 acres, at £4 5s. Ud. per acre .^ Am. £55
05. lid.
53. 27 yards, at75.'5|^. per yard.? Ans. £10
l5. llirf.
18. When the price is an even number of shillings,
and less than 20.
PRACTICE. 319
Rule. — Multiply the number of articles by half the
number of shillings ; and consider the tens of the pro-
duct as pounds, and the units dmibled^ as shillings.
Example.— What it the price of C4G 11), at IQs. per lb "?
646
8
516;8
2
£516 16.5.
29. being the tenth of a pound, there are, in the price,
half as many tenths as shillings. Therefore half the number
of .shillings, multipUed by the number of articles, will express
the number of tenths of a poimd in the price of the entire.
The tens of these tenths \rill be the number of pounds ; and
the units (beijg tenths of a pound) Tvill be half the required
number of shillings — or, multiplied by 2 — the required num-
ber of shillings.
In the example, I65., or £>S. is the price of each article.
Therefore, since there are 646 articles, 64Gx£"8=£516-8
is the price of them. But 8 tenths of a pound (the wiits^in
the product obtained), are twice as many shillings; and hence
we are to multiply the imits in the pro'luct by 2.
EXERCISES.
What is the price of
54. 3215 ells, at 65. per ell } Ans. £964 lOs.
55. 7563 lb, at Ss. per To r A113. £3025 45.
56. 269 cwt., at 16s. per cwt. > Ans. £21^ 4s.
57. 27 oz., at 45. per oz. } Ans. £d 8s.
58. 84 gallons, at 14s. per gallon ? Ans. £58 16^.
19. When the price is an odd number of shillings,
and less than 20 —
Ruj.E. — Find the amount at the next lower even
number of shillings ; and add the price at one shilling.
Example. — What is the price of 275 lb. at 175. per lb T
275
8
The price at I65. (by the last rule) is 220 0
The price at I5. is 275.s.=: . . 13 15
Hence the price at 165.-f-l5 , or 17s., is £233 155.
220 PRACTICE
The price at 17^-. is equal to the price at IG5., plus the
price at one aiiiliing.
EXERCISES.
59. 86 oz., at 55. per oz. ? Ans. £2\ lOs.
60. 62 cwt., at 195. per cwt. ? Ans. £bS I85.
61 14 yards, at 17^. per yard } Ans. £\\ IS5.
62. 439 tons, at II5. per ton? Ans. £2A\ 95.
63. 96 gallons, at 7s. per gallon } Ans. £33 125.
20. When the quantity is represented by a mixed
number —
KuLE. — Find the price of the integral part. Then
multiply the given price by the numerator of the frac-
tion, and divide the product by its denominator — the
quotient vrill be the price of the fractional part. The
sum of these prices will be the price of the whole quan-
tity.
Example.— What is the price of 8| lb of tea, at 55. per
£ s. d.
The price of 8 lb is 8x05.= 2 0 0
The price of | lb is — r — = 0 3 9
And the price of 8| Ih is . .239
The price of f of a pound, is evidently f of the price of a
pound.
EXERCISES.
T\^hat is the price of
64. b\ dozen, at 35. 3^. per dozen } Ans. lis. 10-iJ.
65. 273| lb, at 2s. 6d. per ft) ? Ans. £34 3s. Ud.
66. 530f ft), at 145. pei: ft) .? Ans. 371 IO5. Qd.
67. 178f cwt., at 175. per cwt. .? Ans. £151 125
A^d.
68. 762f cwt., at £1 125. Qd. per cwt. t Ans. £1239
45. 6d.
69. 817y\ cwt., at £3 75. 4d. per cwt. .^ Ans.
£2751 ll5. 6^d.
PRACTICE. 221
21. The rules for finding the price of several deno-
minations, that of one being given [7 and 9], may be
abbreviated by those which follow —
Avoir dujpoise Weight. — Given the price per cwt., to
find the price of hundreds, quarters, &c. —
Rule. — Having brought the tons, if any, to cwt.,
multiply 1 by the number of hundreds, and consider the
product as pounds sterling ; 5 by the number of quar-
ters, and consider the product as shillings ; 2J-, the
number of pounds, and consider the product as pence : —
the sum of all the products will be the price at £\ per
cwt. From this find the price, at the given. number of
pounds, shillings, &c.
Example. — What is the price of 472 cwt., 3 qrs.. IG lb,
at X5 9.S. Gd. per cwt. ?
£ s. d.
1 5 2|
Multipliers 472 3 16
472 17 lOi is the price at £,1 i)er cwt.
2364
212
i64 9 Z\ the price, at £5 per cwt.
>12 16 0| the price, at 9^. (£J^x9.^
11 16 51 the price, at U. (£"^^^2:)
2589 1 9} the price, at £5 9«. U.
At XI per cwt., there will be £1 for every cwt. We mul-
tiply the qrs. by 5, for shillings ; because^ if one cwt. costs £1,
the fourth of 1 cwt., or one quarter, will cost the fourth of
a pound, or bs. — and there will be as many times 5^. as there
are quarters. The pounds are multiphed by 2i ; because if
the quarter costs 5c>., the 28th part of a quarter, or 1 lb,
must cost the 28th part of 5s., or 9.\d. — and there will be as
many times 2^(/. as there are pounds.
EXERCISES.
AVhat is the price of
70. 499 cwt., 3 qrs., 25 lb, at 2bs. \ld. per cwt. }
Am. £647 175. 7^d.
71. 106 cwt., 3 qrs., 14 ft), at IS^. 9^. per cwt. }
Ans. ieiOO 3s. lO^d.
aXZ PRACTICE.
72. 2061 cwt., 2 qrs., 7 ib, at 16^. 6fZ., per cwt. ?
Ans. £1700 155. ^\cl.
73. 106 cwt., 3 (-[I'S., 14 H), at 9^. Ad. per c^t. } Aiis.
^9 175. 6^.
74. 26 cwt., 3 qrs., 7 ft), at 155. 9^^. per cwt. .? ^?i5.
£21 2s. S^d.
75. 432 cwt., 2 qrs., 22 lb, at 185. 6d. per cwt. ?
Am. £400 45. lOi^?.
76. 109 cwt., 0 qrs., 15 ft), at 195. 9(r/. per cwt. .? Ans.
£107 155. 4|c?.
77. 753 cwt., 1 qr., 25 lb, nt 155. 2d. per cwt. t
Am. £571 75. 8^.
78. 19 tons, 19 cwt., 3 qrs., 27ift), at £19 195. 11|J.
per ton .? Am. £399 195. 6^.
22. To find the price of cwt., qrs., S:c., the price of
a pound being given —
EuLE. — Having reduced the tons, if any, to cwt.,
multiply 95. 4d. by the number of pence contained in
the price of one pound : — this will be the price of one
cwt. Divide the price of one cwt. by 4, and the quotient
will be the price of one quarter, &c.
Multiply the price of 1 cwt. by the number of cwt. ;
the price of a quarter by the number of quarters ; the
price of a pound by the number of pounds ; and the sum
of the products will be the price of the given quantity.
Example.— What is the price of 4 cwt., 3 qrs., 7 lb, at
Sd. per lb. '?
s. d,
9 4
d.
4)74 8 the price of 1 csvt. X4, will give 298 8 the price of 4 cwt.
28)18 8 the price of Iqr. XSjTvillgive 56 0 the price of 3 qrs.
8 the price of lib X7, will give 4 8 the price of 7 lb.
20)359 4
And the price of the whole will be £17 19 4
At Id. per lb the price of 1 cwt. would be 112^/. or 95. 4d. : —
therefore the price per c%vt. will be as many times Os. 4<Z. as
there are pence in the price of a pound. The price of a
quarter is -]- the price of 1 cwt. : and there will be as many
times the price of a quarter, as there are quarters, &:c.
PRACTICE. 223
EXERCISES.
"What is the price of
79. 1 cwt., at 6d. per ib ? Ans. £2 i6s.
80. 3 cwt., 2 qrs., 5 ft), at 4d. per ft) ? Ans. £6
12s. 4d.
81. 51 cwt., 3 qrs., 21 ft), at 9d. per ft) .^ -4?w. £218
25. 9d.
82. 42 cwt., 0 qrs., 5 ft), at 25c?. per ft) .^ ^tw. £490
10s. 5d.
83. 10 cwt., 3 qrs., 27 ft), at 51^. per ft) ? ^Itw.
£261 ll5. 9^.
23. G-iven the price of a pound, to find that of a ton —
Rule. — Multiply £9 6^. Sd.' by the number of pence
contained in the price of a pound.
Example. — ^What is the price of a ton, at Id. per ft) ?
£ s. d.
9 6 8
65 6 8 is the price of 1 ton.
If one pound cost Id., a ton will cost 2240rf., or £9 6s-. Sd.
Hence there will be as many times £9 6s. Sd. in the price
of a ton, as there are pence in the price of a pound.
EXERCISES.
AVhat is the price of
84. 1 ton, at 3d. per ft) > Aits. £28.
85. 1 ton, at 9d. per ft) ? Atls. £84.
86. 1 ton, at 10 J. per ft) ? Am. £93 6s. Sd.
87. 1 ton, at 4d. per ft) ? Ans. £37 65. Sd.
The price of any number of tons -will be found, if we mul-
tiply the price of 1 ton by that number.
24. Troy Weight. — Griven the price of an ounce — to
find that of ounces, pennyweights, &c. —
Rule. — Having reduced the pounds, if any, to ounces,
set down the ounces as pounds sterling ; the dwt. as
shillings ; and the grs. as halfpence : — this will give the
price at £1 per ounce. Take the same part, or parts,
&c., of this, as the price per ounce is of a pound.
224 PRACTICE.
Example 1. — What is the price of 538 oz., 18 dwt , 14
grs.. at lis. 6d. per oz. ?
£ s. d.
2)538 18 7 is the price, at £1 per ounce.
10)269 9 3i is the price, at IO5. per ounce.
2) 26 18 11^ is the price, at Is. per ounce.
13 9 5| is the price, at 6d. per ounce.
And 309 17 8^ is the price, at lis. 6d. per ounce.
14 halfpence are set down as 7 pence.
If one ounce, or 20 dwt. cost £1, 1 dwt. or the 20th part
of an ounce will cost the 20th part of £1 — or Is. ; and tho
24th part of 1 dwt., or 1 gr. will cost the 24th part of
Is. — or irf.
Example 2. — What is the price of 8 oz. 20 grs., at X3
2s. Qd. per oz. ?
£ s. d.
8 0 10 is the price, at £1 per ounce.
3
24 2 6 is the price, at £2> per ounce.
Price at £1-7-10= 0 16 1 is the price, at 2s. per ounce*
Price at 2s. -v- 4= 0 4 0^ is the price, at Qd. per ounce.
And £25 2 7^ is the price, at £3 2s. 6^. per oz.
EXERCISES.
What is the price of
88. 147 oz., 14 dwt., 14 grs., at 7s. 6d. per oz. }
Ans. £55 7s. ll^d.
89. 194 oz., 13 dwt., 16 grs., at lis. 6d. per oz. ?
Ans. ieill 18s. lO^d.
90. 214 oz., 14 dwt., 16 grs., at 12s. 6d. per oz. }
Ans. £134 4s. 2d.
91. 11 ft), 10 oz., 10 dwt., 20 grs., at 10s. per oz. .?
Ans. £71 5s. 5d.
92. 19 ft), 4 oz., 3 grs., at £2 5s. 2d. per oz. ? Ans,
£523 18s. 111^.
93. 3 oz., 5 dwt., 12 grs., at £1 65. 8^. per oz. ?
Ans. £4 7s. 3^d.
PRACTICE. 225
25. Cloth Measure. — Given the price per yard — to
find the price of yards, Cjuarters, &c. —
Rule. — Multiply £>\ by the number of yards ; 55. by
the number of quarters ; Is. 3d. by the number of nails ;
and add these together for the price of the quantity at
£1 per yard ? Take the same part, or parts, &c., of this,
as the price is of £1.
Example 1. — What is the price of 97 yards, 3 qrs., 2
nails, at 8s, per yard 1
£1 5s. Is. Zd.
MultipHers 97 3 2
2)97 17 6 is the price, at £1 per yard.
5)48 18 9 is the price, at 10s. per yard.
From this subtract 9 15 9 the price, at 2s. per yard.
And the remainder 39 3 0 is the price, at 8.s. (IO5.— 2s.)
If a yard costs £1, a quarter of a 3-ard must cost 5.s. ; and
a nail, or the 4th of a yard, will cost the 4th pai-t of 5s. or
Is. Zd.
Example 2. — What is the price of 17 yards, 3 qrs., 2
nails, at £2 5s. 9d. per yard ?
£1 5s. Is. 3 J.
MultipHers 17 3 2
17 17 6 is the price, at £1 per yard
o
35 15 0 is the price, at £2 per yard.
The price at £1h- 4=4 9 4^ is the price, at 5s.
The price at 5s. -i- 10=0 8 llf is the price, at 6d.
The price at (jd.-~ 2=0 4 5^ is the price, at Zd.
And £40 17 9^ is the price, at £2 5s. 9^.
EXERCISES.
What is the price of
94. 176 yards, 2 qrs., 2 nails, a 15s. per yard } Ans.
JS132 9s. 4^d.
95. 37 yards, 3 qrs., at £1 5s. per yard ? Ans. £41
3s. 9d.
96. 49 yaj-ds, 3 qrs., 2 nails, at £1 10s. per yard.?
Ans. £74 16s. 3d.
97. ^ yards, 3 qrs., 1 nail, at £\ 15s. per yard .=>
Ans. igl72 185. b^d.
226 PRACTICE.
98. 3 yards, 1 qr., at 17^. Qcl. per yard .^ Am £2
165. 101^.
99. 4 yards, 2 qrs., 3 nails, at £1 2s. 4d. per yard ?
Am. £5 4s. S^d.
26. Land Measure. — Rule. — Multiply £1 by the
number of acres ; 5.5. by the number of roods ; and l^d.
by the number of perches : — the sum of the products will
be the price at £1 per acre. From this find the price,
at the given sum.
Example. — What is the rent of 7 acres, 3 roods, 16
perches, at j£3 8s. per acre ?
£ s. d.
1 5 11
Multipliers 7 3 16
Sum of the products 7 17 0, or the price at £1 per acre.
3
23 11 0 the price at £Z per acre.
3 18 6 the price at 10s. per acre.
27 9 6 the price at £3 lO.s. per acre.
Subtract 0 15 8^ the price at 2s. per acre.
And 26 13 S^ is the price at £3 85.
If one acre costs £1, a quarter of an acre, or one rood, must
cost 55. ; and the 40th part of a quarter, or one perch, must
cost the 40th part of 5s.— or Ihd.
EXERCISES.
What is the rent of
100. 176 acres, 2 roods, 17 perches, at £d 6s. per
acre .? Ans. ^£936 Os. 3d.
101. 256 acres, 3 roods, 16 perches, at £6 6s. 6d,
per acre ? Am. iei624 Us. e^d.
102. 144 acres, 1 rood, 14 perches, at £0 6s. 8d. per
acre .? Am. ^£769 I65
103. 344 acres, 3 rtjods, 15 perches, at £4 Is. Id.
per acre ? Am. £1398 Is. Id.
27. Witie Measure. — To find the price of a hogs-
head, when the price of a quart is given —
Rule. — For each hogshead, reckon as many pounds,
and shillings as there a?'c pence per quart.
PRACTICE. 227
LxAMTLT. — What is tho price of a hogshead at Od. per
quart T Aiis. £9 9s. j
One hogshead at Id. pei* quart would be G3x4, since there
are 4 quarts in one gallon, nnd C3 gallons in one hhd. But
G;iX4rf.=2o2(/.=£l Is. ; and, therefore, the price, at dd. per
quart, "will be nine times as much — or 9XJ£1 ls.=£d 9s.
EXERCISES.
AVhat is tho price of
104. 1 hhd. at IS^. per quart ? Ans. £18 ISs.
105. 1 hhd. at I9d. per quai-t .? Ans. £19 I9s.
106. 1 hhd. at 20d. per quart.? Ans. ^£21.
107. 1 hhd. at 2s. per quart ? Ans. £25 4s.
108. 1 hhd. at 2s. 6d. per quart ? Ans. £31 10s
When the price of a pint is given, of course we know that
of a quart.
28. Given the price of a quart, to find that of a tun —
Rule. — Take 4 times as many pounds, and 4 times
fts many shillings as there are pence per quart.
Example. — What is the price of a tun at lid. per quart "^
£ s.
11 11
4
40 4 is the price of a tun.
Since a tun contains 4 hogsheads, its price must be 4 tiiupji
the price of a hhd. : that is, 4 times as many pounds and shil-
riugs, as pence per quart [27].
EXERCISES.
What is the price of
109. 1 tun, at I'Jd. per quart.? Ans. £79 IGs.
110. 1 tun, at 20d. per quart ? Ans. £84.
111. 1 tun, at 2s. per quart ? Ans. £100 16^.
112. 1 tun, at 2s. 6d. per quart.? Ans. £126.
113. 1 tun, at 2s Sd. per quart.? Ans. £134 8^.
29. A nuviber of Articles. — Given the price of 1
article in pence, to find that of any number —
Rule. — Divide the number b.y 12, for shillings and
228 PRACTICE.
pence ; and multiply the quotient by the number of
pence in the price.
Example. — What is the price of 438 articles, at Id eaci'
12)438
365. 6fZ., the price at Id. each.
7
20)255~~6
£12 15 6 the price at 7d. each.
438 articles at Id. each will cost 438tZ.=36y. 6d. At 7d. each ,
they will cost 7 times as much— or 7XoQs. Gd.=2oo5. Gd.T^
£12 105. ed.
EXERCISES.
"What is the price of
114. 176 a, at 3d. per ft) .? Ans. £2 4s.
115. 146 yards, at 9^. per yard ? Ans. £5 9s. 6d
116. 180 yards, at lO^d. per yard ? Ans. £7 17s. 64
117. 192 yards, at 7if/. per yard ? Ans. £6.
118. 240 yards, at 8^d. per yard t Ans. £8 10s
30. Wages. — Having the wages per day, to find
their amount per year —
Rule. — Take so many pounds, half pounds, and 5
pennies sterling, as there are pence per day.
Example. — What are the yearly wages, at 5J. per day 1
£> s. d.
1 10 5
5 the number of pence per day.
7 12 1 the wages per year.
One penny per day. is equal to SQod.=2i0d.-\-120d.-{-5d.=:
£l-{-}Qs.-\-dd. Therefore any number of pence per day, must
be equal to £1 10*. 5d. multiplied by that number
What is the amount per year, at
119. 3^. per day.? Ans. £4 lis. 3d.
120. 7d. per day.? Ans. £10 125. lid.
121. 9d. per day .= Ans. £13 13^. 9^.
122. 14^. per day.? Ans. £21 bs. lOd.
123. 2s. 3d. per day .? Ans. £41 Is. 3d.
124 S^d. per day .? Ans. £12 18s. ^d.
PRACTICE.
229
BILLS OF PARCELS.
Mr. John Day
Dublin, im April, 1844.
Bought of Richard Jones.
15 \
24 y
127 \
ir. V
12 Y
ards of fine broadcloth, at 13 G per yard
ai\U of superfine ditto, at 18 9 „
ards of yard wide ditto, at 8 4 „
ard.s of drugget, at . .Go „
ard.s of serge, at .- . 2 10 .,
yards of shalloon, at . . 18 ,,
£ s.
d.
10 2
6
22 10
0
11 5
0
5 0
0
1 14
0
2 13
4
£53 4 10
.^^^. James Paul,
0 pair of worsted stocking?, at 4
0 pair of silk ditto, at . . 15
17 pair of thread ditto, at . 5
23 [^air of cotton ditto, at . 4
14 pair of yarn ditto, at . 2
18 pair of womens silk gloves, at 4
10 yards of flannel; at . . 1
Dublin, m May, 18-14.
Bought of Thomas Norton.
s. d.
per pair
per yard
Aris. £23 15 4,>
Mr. James Gor
40
34
31
2'.)
1
ells of dowlas, at
ells of diaper, at
ells of Holland, at
yards of Irish elotli
vards of muslin, at
at
Did)lin, 11th May, 1844.
Bought of John Walsh &: Co
d.
C per ell
8 .
4 per yard
]3| j-ards of cambric, at lU G
5'^ yards of printed calico, at 1 2
Ans. £84 5 lOJ
I. 2
:230 PRACTICE
Dahlia, 20th May, 1844.
Lady Denny,
Bought of Richard Mercer
s. d.
91 yards of silk, at . . 12 9 per yard
13"' yards of flowered do., at 15 G „'
llf yards of histring, at . G 10 „
14 yanls of brocade, at . 11 3 „
12] yards of satin, at . 10 8 „
11| yardi? of velvet, at . iJi 0 „
Aas. xiTislo
Dublin, 2lst May, 1844.
Bought of William Roper,
•s. d.
I5i lb of currants, at . .04 per lb
17i flj of JNlalaga raisins, at . 0 5^ ,,
19^ lb of raisins of the sun, at . 0 G" „
17 lb uf riee, at . . .03^.,
81 lb of pepper, at . . . 1 G^ ,.
3 loaves of sugar, weight 32^ lb. at 0 8i „
13 oz. of cloves, at . . .09 j)er oz.
Am. £3 13 Qi
Dublin, 21th June, 1844.
Mr. Thomas Wriglit,
Bought of Stephen Brown & Co.
5. d.
252 gallons of jn-ime whiskey, at G 4 per gallon
252 gallons of old malt, at .68
252 gallons of old malt, at .80 ,,
Ans. £264 12 0
MISCELLANEOUS EXERCISES.
What is the price of
1. 4715 yards of tape, at }d. per yard .^ Ans.
£4 18.V. 2f«'.
2. 3o4 lb, at 11^/. per fo ? Ans. £1 I6s. lO^d.
3. 4756 lb of sunjar, at 12i^/. per lb f Ans. £242
15.S-. U.
4. 425 pair of silk stockings, at Gs. per pair .' Aiis
£127 lOi.
ruAcj'ire.
231
5. 37ol pair of gloves, at 2s. Cjd. ? Ans. £469 ds
6. 3520 pair of gloves, at 3^. 6d. ? A)is £616.
7. 7341 cwt., at £2 6s. per cwt. ? Ans. £16884 6s.
H. 435 cwt. at £2 Is. per cwt. .' Aiis. £1022 os.
9. 4514 cwt., at £2 Us. 7^d. per cwt. } Ans.
£13005 19a-. 3d.
10. 3749f cwt., at £3 I5s. Gd. per cwt. } Ans.
£14153 17i\ Ofri.
11. 17 cwt., 1 qr., 17 ib, at £1 4s. 9d. per cwt. ?
£21 105. Sid.
12. 78 cwt., 3 qrs., 12 !b, at £2 17.?. 9^/. per cwt. ?
Aiui. £227 14.V.
13. 5 oz., 6 dwt., 17 grs., at os. IQd. per oz. .' Ans
£1 Us-. \y.
14. 4 3-ards, 2 (/rs., 3 nails, at £1 2.?. 4d. per j'ard .?
A7?5. £5 4s. S\d.
15. 32 acres, 1 rood, 14 perches, at £1 16s. per
acre .' Ans. £58 4s. l|fZ.
in. 3 jrallons, 5 pints, at 7^. 6d. per gallon.' Ans.
£1 7.V. 2Jr/.
17. 20 tons, 19 cwt., 3 fp'S., 27^ ft), at £10 lOs
per ton .- Ans. £220 Qs. 1 l^d. nearly.
18. 219 toas, 16 cwt., 3 qrs., at £11 7^. 6d. per
ton .' Ans. £2500 135. 0^^/.
QUESTIOXS IN PRACTICE.
1. What is practice ? [1"|.
2. AVhy is it so called.- [1].
3. What is the difference between aliquot, and aliquant
parts .^ L^].
4. How are the ali!|iiot parts of afjstract, and of
applicate numbers found ? [3] .
5. Wliat is. the diiFerence between prime, and com-
pound aliquot parts c [3] .
6. ITow is the price of any denomination found, that
of another beinp: given ? [6 and 8].
7. How is the pi-ice of two or -more denominations
found, that of one being ;i;iven ? [7 aud 9j.
S. The price of one dcnoniinatlon l);}ing given, how
do we dud tlut of any number of anolher .- [10].
232 PRACTICE
9. When tlie price of any denomination is the aliquot
part of a shilling, how is the price of any number of that
denomination found ? [11].
10. When the price of any denomination is the
aliquot part of a pound, how is the price of any num-
ber of tiiat denomination found r [12].
1 1 . What is meant by the complement of the price r
[13].
12. When the complement of the price of any deno-
mination is the aliquot part of a pound or shilling,
but the price is not so, how is the price of any number
of that denomination found ? [13].
13. When neither the price of a given denomination,
nor its complement, is the aliquot part of a pound or
shilling, how do we find the price of any number of
that denomination ? [14, 15, 16, and 17].
14. How do we find the price of any number of
articles, \A\en the price of each is an even or odd num-
ber of sliillings, DD.d less than 2G f [IS and 19].
15. How is the price of a (quantity, represented by a
mixed number, found ? [20] .
Uj. How do we find the price of cwt., qrs., and lb,
when tlie price of 1 cwt. is- given .' [21].
17. How do we find the price of cwt., qrs., and lb,
when the price of 1 lb is given 'f [22] ,
IS. }lo\7 is tlie price of a ton found, when the price
of 1 ir. is given ? [23].
19. How do we find the price of oz., dwt., and grs.
wh?ii the price Cf an ounce is given r [24].
20. How do we find the price of yards, qrs., and nails,
when the price of a yard is given .' [25].
21. How do we liiid the price of acre^, roods, and
perches f [P.t)].
22. How may the price of a hhd, or a tun be found,
when the nri'jo of a quart is given ^ [27 and 2S].
23. Ho^ may the price of any number of articles bo
r\)und, tlie price of each in pence being given t [29],
24. How are wages per year found, those per day being
given ^ [30]
TARE AND TRET. 233
TAKE AND TRET.
3\ TLe --rToss weight is tlie weight both of the
goo(^i, ami ot the bag, &:c., in which they are.
Tare is an allowance for the bag, &c., which contains
the article.
Sultle is the weight which remains, after deducting
the tare.
Tret Is, usually, an allowance of 4 lb in every 104 lb,
or Vg of the weight of goods liable to waste, after the
trire has been deducted.
Clcff is an allowance of 2 ib in every 3 cwt., after
both tare and tret have been deducted.
Y/hat rem:iins after making ail deductions is called
the wf/, or neat weight.
Diffjrent allowances are made in different places,
nnd for different goods ; but the mode of proceeding is
in -all cases very simple, and may be understood from
the following —
EXERCISE?.
1. Bought 100 carcasses of beef at 15^. M. per cwt.;
grosa weight 4.50 cwt., 2 fjrs., 23 ^ ; tret 8 lb per cai--
ea.ss. V/hat is to be paid for them .'
cwt. qrs. lb. 100 carcasses.
Gruss 4-30 2 2?> 8 lb per carCviss
'i'ret 7 0 Vo cwt. qrs. Hj
Tret, on the entire, 800 1T)=7 0 10
443 2 7 at 18.'. Gi. per cwt.=X410 Ss. 10 ■a\
2. What is the price of 400 raw hides, at 195. 10^/.
per cwt. ; tlie gross weight being 306 cwt., 3 qrs., l.o
lb ; and the tret 4 ft) per hide .' ^Am. £290 3.5. 2^d.
3. If 1 cwt. of butter cost £3, what will be the price
of 2.>0 fiikins; gross weight 127 cwt., 2 qrs., 21 lb;
tare 11 lb per iiikin r Am. £309 S.?. 0^^. .
4. AVhat is the price of S cwt., 3 <|rs., 11 lb, at 15.?. \
C>il. per cwt., allovring the usual tret.' An^. £(') lis.
103^.
23-1 TARE AND TRET.
per
5. Vriiat is the price of S cwt. 21 lb, at 18^. A^d.
^,-1- cwt., allowing the usual tret ? Ans. £7 4s. S^d.
6. Bought 2 hhds. of tallow ; No. 1 weighing 10 cwt.,
1 qr., 1 1 ib, tare 3 qrs., 20 lb ; and No. 2, 11 cwt., 0 qr.,
17 ft), tare 3 qrs., 14 lb; tret 1 lb per cwt. What do
thej come to, at SO^-. per cwt. r
cwt. qrs. lb. cwt. qrs. K).
Gross weight of No. 1, 10 111 . Tare 0 3 20
Gross weight of No. 2, 11 0 17 . Tare 0 3 14
(4ross weight, . . 21 2 0 13 6
Tare, . . .13 0
Suttle. . . . 19 2 22
Tret 1 lb per cwt. . 0 0 19j
Net weight, 19 2 2i^. The price, at 305. per
cwt., is £29 5s. l^lld.
It is evident that the' tret may he found by the following
proportion. —
cwt. cwt. qrs. ft. ft. ft.
1 : 19 '2 22 :: 1 : 19ff.
7. What is the price of 4 hhds. of copperas ; No. 1,
weighing gross 10 cwt., 2 qrs., 4 lb, tare 3 qrs. 4 ib ;
No." 2, 11 cwt., 0 qr., 10 lb, tare 3 qrs. 10 lb ; No. 3,
12 ewt., 1 qr., tare 3 qrs. 14 ib ; No. 4, 11 cwt., 2
qi's., 14 Ib, tare 3 qrs. IS lb; the tret being 1 lb per
cwt. ; and the price lO^. per cwt. r Ans. £20 I7s.
•■ 7 8 4 ' •
y. What will 2 bags of merchandise come to ; No. 1,
weighing gross 2 cwt., 3 qrs., 10 ib ; No. 2, 3 cvN't.,
3 qrs., iO ib ; tare, 16 ft) per bag ; tret 1 ft) per cwt. ;
and at 1.?. Sd. per ib .' Ans. £59 2^. 8\d.
9. A merchant has sold 3 bags of pepper; No. 1,
weighing gross 3 cwt. 2 qrs. ; No. 2, 4 cwt., 1 qr., 7 lb ;'
No. 3, 3 cwt., 3 qrs., 21 lb ; tare 40 ib per bag ; tret
1 lb per cwt. ; and the price being 15^. per ib. What
do they come to .' Ans. £74 Is. 7|f^.
10. ]5ought 3 packs of wool, weighing. No. 1, 3 cwt.,
1 qr., 12 lb ; No. 2, 3 cwt., 3 qrs., 7 tb ; No. 3, 3 cwt.,
2 qrs., lo ib ; tare 30 lb per pack ; tret S lb for every
'"> stone; and at 105. Sti. per stone. What do thei
amount to ''
iakl: a^d tkkt
23.:
cwt. qrs. lb.
Vo. 1, 3 1 12
i\o. 2, 3 3 7
No. 3, 3 2 15
l\iYe 30
Tare 30
Tare 30
Gross, 10 3 G
Tare, 0 3 G
St.
1
90=3 qrs. G lb
Sattle, 10 0 0=70 stones.
St. St. lb. lb.
20 : 70 :: 8 : 28 =
St. Y\i.
Suttle, 70 0
Tret, 1 12
ft.
12
Net weight, G8 4, at 10.^\ 6d. per stone=£35 I65. 7^d.
1 1 . Sold 4 packs of wool at 95. 9d. per stone ; weigh-
ing, No. 1, 3 cwt., 3 qrs., 27 lb. ; No. 2, 3 cwt., 2 qrs.,
16 lb. ; No. 3, 4 cwt., 1 qr., 10 ft). ; No. 4, 4 cwt., 0
qr., 6 ft) : tate 30 ft) per pack, and tret 8 lb for every
20 stone. What is the price ? Ans. £49 Ids. 2^^jd.
12. Bought 5 packs of wool ; weighing, No. 1, 4 cwt.,
2 qrs., 15 lb ; No. 2, 4 cwt., 2 qrs. ; No. 3, 3 cwt.,
3 qrs., 21 lb ; No. 4, 3 cwt., 3 qrs., 14 ft) ; No. 5, 4
cwt., 0 qr., 14 ft) : tare 2'S lb per pack ; tret 8 ft) for
every 20 stone ; and at lis. M. per stone. What is
the price } Ans. £11 15s. 8}^^.
13. Sold 3 packs of wool ; weighing gross, No. 1, 3
cwt., 1 qr., 27 lb ; No. 2, 3 cwt., 2 qrs.,"l6 ftj ; No. 3,
4 cwt., 0 qr., 21 lb : tare 29 tb per pack ; tret 8 ft) for
every 20 stone ; and at Us. Id. per stone. What is the
price i Ans. £41 13s. 7f f i^.
14. Bought 50 casks of butter, weighing gi'oss, 202
cwt., 3 qrs., 14 lb ; tare 20 ft) per cwt. What is the
net weight }
qrs.
cwt.
1
cwt. qrs. ft.
202 3 14
2t)
4040 ft,
10
cwt. qrs. ft.
Gross weight, 202 3 14
Tare, . . 30 0 254
il=!
Net weight, 16G 2 161
5 = 1 of the last, [ =the tare on 3 qr. 14 lb.
2| = I of the last, ^
Tare, 40571 ft = 3G cwt., 0 qr., 25] ft.
236 TARE AND TKET.
15. The ffross wcierlit of ten hlids. of tallow is 104
cwt., 2 qrs., 25 ib ; and the tare 14 ib per cwt. WKat
is the net weight ? Aiis. 91 cwt., 2 qrs., 14|- ib.
16. The gross weight of six butts of currants is 58
ewt., 1 qr., 18 ib ; and the tare 16 tb per cwt. What is
the net weight ? Ans. 50 cwt., 0 qr., 7^ lb.
17. What is the net weight of 39 cwt., 3 qrs., 21 ib ;
the tare being 18 lb per cwt. ; the tret 4 lb for 104 tb ;
and the cloff 2 tb for every 3 cwt. .^
cwt. qrs. tb.
39 3 21
lb. ib.cwt. Tare, . . 0 1 13
1 1
cwt. ars.
lb.
<
39 3
21
Gross weight,
ib. cwt.
Tare,
16=}
5 2
23
2=K8
0 2
24
Suttle, .
Tret=2Vth, or
Tare, 6 1 13
2 lb in 3 cwt. is the ,4^th part of 3 cwt. 32 0 2o
Hence the cloflFof 32 CAVt. 26 lb is its yg-jth part, or 0 0 22.
Net weight, 32 0 4
18. What is the net weight of 45 hhds. of tobacco ;
weighing gross, 224 cwt., 3 qrs., 20 tb ; tare 25 cwt.
3 qrs. ; tret 4 lb per 104 ; cl-jif 2 lb for every 3 cwt. .'
Aiis. 190 cwt., 1 qr., 14^\ ib.
19. What is the net weisrht of 7 hhds. of suo;ar,
weighing gross, 47 cwt., 2 qrs., 4 lb ; tare in the wholo,
10 cwt., 2 qrs., 14 lb ; and trot 4 lb per 104 lb ? Afi.s.
35 cwt., 1 qr., 27 lb.
20. In 17 cv/t., 0 qr.. 17 lb, gross weight of galls,
how ranch net ; allowing 18 ib per cwt. tare ; 4 lb per
104 lb tret ; and 2 lb per 3 cwt. cloff.? Ans. 13 cwt.,
3 qrs., 1 ib nearly.
QUESTIONS.
1. What is the gross weight ? [31].
2. What is tare ? [31].
3. What is suttle.' [31].
4 What is tret.? [31].
5. What is cloff.? [31].
6. What is the net weight > [31].
7. Are the allowances made, always the same ? [31].
237
SECTION VII.
INTEREST, &c.
1. Interest is the price which is allowed for the use of
money ; it depends on the plenty or scarcity of the latter,
and the rLsk which is rim in lending it.
Interest is cither simple or compound. It is simjih
when the interest due is not added to the sum lent, so
as to bear interest.
It is compound when, after certain periods, it is made
to bear interest — being added to the sum, and considered
as a part of it.
The money lent is called the principal. The sum
allowed for each hundred pounds " per annum" (for a
year) Ls called the " rate per cent." — (per iiilOO.) The
amount is the sum of the principal and the interest due.
SIMPLE INTEREST.
2. To find the interest, at any rate per cent., on any
fc;um, for one year —
liuLE I. — Multiply the sum by the rate per cent.,
and divide the product by 100.
Example. — What is the interest of £672 14s. od. for one
year, at 6 per cent. (£6 for every £100.)
£ s. d.
672 14
6
40-36 5 6
20
7-25 The quotient, £40 Is. Zd.. is the iLterefit required.
¥06
We have diviaed by ICK), by merely altering the decimal
point [Sec. I. 34].
-i-'C> IMERll.ST.
ir tliu interest were 1 percent., it would be the liimdredt-h
part of the principal— or the principsil multiplied by yn-^ ; but
being G per cent., it is G times its much — or tlie principal mul-
tiplied by ylo .
3. Rule II. — Bivido the interest into parts of £100 ;
and take corresponding parts of the principal.
Example.— What is the interest of £32 4s. 2d., at 6 per
cent. ?
£G = oG5-j-£l, or £_--phis£ — ^-^-5. Therefore the in-
terest is the -^V of the principal, plus the I of the 75L.
£ s. d. ^
20)32 4 2
5) 1 12 2i is the interest, at 5 per cent.
0 6 5i is the interest, at 1 per cent.
And 1 18 7| is the interest, at 6 (5+1) per cent.
EXERCISES.
1. What is the interest of i£344 175. Qd. for one year,
at C per cent. : Ans. £20 135. lO^^Z.
2. What is the interest of .£600 for one 3'ear, at 5 per
cent. } Ans. £30.
3. What is the interest of £480 155. for one year, at
7 per cent, t Ans. £33 135. OfrZ.
4. What is the interest of £240 IO5. for one year, at
4 per cent. : Ans. £9 125. Aid.
4. To find the interest when the rate jier cent, con-
sists of more than one denomination —
iluLE. — Find the interest at the highest denomina-
tion ; and take parts of this, for those which are lower.
The sum of the results will be the interest, at the given
rate.
Example. — What is the interest of £97 8s. 4d., fo^ one
year, at £5 lOs. per annum ?
£5 = £VV; and 10s. = £-^.
£ s. d.
20)97 8 4
10)4 17 5 is the interest, at 5 per cent.
0 9 9 is the interest, at lOs, per cent.
And 5 7 2 is the int^jrcst, at £5-f-10«. per cent.
INTEKLST. 239
At 5 per cent, the interest is the t^V of the principal ; at
lOv. per cent, it is tlie y^ of wliat it is at 5 per cent. There-
fore, fit £o lOi-. per cent., it is the sum of both.
5. What is the interest of £371 19^. 7U. for one
year, at £3 155. per cent. .' Ans. j£13 ISs.'ll^-d.
tJ. What is the interest of i^84 lis. 10}jd. for one
year, at £4 5s. per cent, t Ans. JC3 lis. lOfd.
7. What is the interest of j291 0^. 3}d. for one year,
at £6 I2s. 9d. per cent. ? Ans. £6 Os. I0\d.
S. What is the interest of ^'968 55. for one year, at
£o 145. (jd. per cent. ? Ans. £55 85. Sd.
5. To find the interest of any sum, for several
years —
KuLE. — ^Multiply the interest of one 3'ear by the num-
ber of years.
Example. — WJiat is the interest of £32 145. 2d. for 7
yravs, at 5 per cent, l
£ 5. d.
20)32 14 2
1 12 8^ is the interest for one 3'ear, at 5 per cent.
And 11 8 11^7 is the interest for 7 years, at 5 per cent.
This rule requires no explanation.
EXERCISES.
9. What is the interest of £14 25. for 3 years, at 6
per cent. .- Ans. £2 lOs. 9d.
10. What is the interest of £72 for 13 years, at £6
105. per cent. .- .4??5. £60 I65. 9*r/.
11. Wliat is the interest of £853 O5. 6y. for 11
years, at £4 125. per cent. ? Ans. £431 125. 7'^d.
6.^ To find the interest of a given sum for years,
months, Sec. —
lluLE. — Having found the interest for tlie yeai-s, as
already directed [2, &c.], take parts of the interest
of one year, for that of the months, &c. ; and then add
the results.
;^iO INTEREST.
Example. — What is tbe ir^tercst of £86 Ss. 4J. for 7 years
and 5 months, at 5 per cent. >
£ s. cl.
20)86 8 4
4 6 5 is the interest .^^ I r^a*. ai^ 5 jxv • cer
£> .'?. d. 30 4 11 is the interest for 7 years.
4 0 5 ~-o= I 8 9| is the interest for 4 month?*.
1 8 9f-i-4= 0 7 2i is the interest for 1 month.
And 32 0 Hi is the required interest.
EXERCISES.
12. What is the interest of £211 55. for 1 year and
6 months, at 6 per cent. .' Ans. £19 05. Sd.
13. What is the interest of £514 for 1 year and 7^
months, at 8 per cent. } Ans. £66 165. 4Ui
14. What is the interest of £1090 for 1 year and 5
months, at 6 per cent. ? Ans. £92 13s.
15. What is the interest of £175 IO5. 6d. for 1 year
and 7 monMis, at 6 per cent, r Ans. £16 135. OjWd.
16. AVhat is the interest of £571 155. for 4 years
and 8 months, at 6 per cent. .' Ans. £160 I5. 9ff/.
17. What is the interest of £500 for 2 years and 10
months, at 7 per cent. ? Ans. £99 35. 4d.
IS. What is the interest of £93 175. 4d. for 7 years
and 11 mouths, at 6 per cent. .' Ans. £44 II5. 7^d.
19. What is the interest of £84 95. 2d. for 8 years
and S months, at 5 per cent. ? Ans. £36 11 5. lli^.
7. To find the interest of any sum, for any time, at
5, or 6, &c., per cent.
At 5 per cent. — *
KuLE. — Consider the years as shilhngs, and the
months as pence ; and find what alitjuot pai;t or parts
of a pound these are. Then take the same part or parts
of the principal.
To find the interest at 6 per cent., find the interest
-at 5 per cent., and to it a^Id its fiftli part, &c.
The interest at 4 per cent, will be the interest at
C'per cent viiniis its fifth part, &c.
intkrp:6T. 2^1
8. EXA3IPLE 1.— What ia the interest of X427 5.s. Od. for
C years and 4 months, at 5 per cent. 1
G 3'ears and 4 months are represented by 65. 4d. ; but
G.>-. 4,-/.=o.s.-}-1.5.-f 4c/ — 1+3'^- of a pound + the }j of tha ^V-
£ s. d.
4)427 5 0
5)100 16 5] is the ] of principal.
3)21 7 ? -" -
_I ^ ''' '' ^^'"^ ^s ^^ ""^^^^ of principal.
' And i;)j G 1;' is the required interest.
The interest of £1 for 1 3'car, at 0 per cent., would be l.f.
fur 1 month !</. ; for any number of years, the same number
of shillings ; for any number of months, the same number of
pence ; and for years and mouths, a corresponding number of
shillings (7/1(^7 ponce. But whatever pai't, or parts, tliese shil-
lings, and peuce arc of a pound, the interest of any other sum,
for the same time and rate, nmst be the same part or parts of
tliat other sum — since the interest of any sum is proportional
to the interest of £1.
Example 2.— AVhat is the interest of £14 2s. 2d. for 0
years and 8 months, at G per cent. ?
G.?. Sd. is the ^ of a pound.
£ .s. d
3)14__2^_
5)4 14 0^ is the interest, at 5 per cent.
0 18 Oj is the interest, at 1 per cent.
5 12 liH i'S the interest, at G (5-f-l) per cent.
EXERCISES.
20. Find the interest of £1090 17.?. 6d. for 1 year
and 8 months, at 5 per cent. .^ Ans. JgOO 18;?. l^d.
21. Find the interest of ^2976 14.?. 7d. for 2 years
and 6 months, at 5 per cent, f Ans. £122 \s. Ofd.
22. Find the interest of £780 175. 6d. for 3 years
and 4 months, at G per cent. ? Ans. £156 35. 6d.
23. AVhat is the interest of £197 1 15. for 2 years
and 6 months, at 5 per cent. ? Jins. £24 13.5. l()\d.
24. What is the interest of £279 lis. for 7^ month.s,
at 4 per cent. .' Ans. £6 195. 9f',yfi?.
25. What is the int<ircst of £790 105. for 6 yeai
and S months, at 5 per cent. ? A%s. £263 12?.
242 INTKKEST.
26. What is the interest of iS124 2s. 9rZ. for 3 yearg
aod 3 months, at 5 per cent. } Ans. £,20 35. of ^^.
27. What is the interest of ^21837 45. 2d. for 3 year a
and 10 months, at 8 per cent. .' Am. £563 85. 3rZ.
9. Vrhen the ratt., or number of years, or both of
them, are expressed by a mixed number —
KuLE. — Find the interest for 1 year, at 1 per cent.,
and multiply this by tlie number of pounds and the frac-
tion of a pound (if there is one) per cent. ; the sum uf
these products, or one of them, if there is but one, will
give the interest for one year. Multiply this by the
number of years, and b}^ the fraction of a 3-ear (if there
is one) ; and the sum of these products, or one of them,
if there is but one, will be the required interest.
Example 1. — Find the interest of i^21 2f. M. for 3| years
at 5 per cent. %
£21 2<. 6.'^-^ 100=45. 2ld. Therefore
£ 5. d.
0 4 25 is the interest for 1 year, at 1 per cent.
5
1 1 If is the interest for 1 year, at 5 per cent.
3 3 5 5^' is the interest for 3 years, at do.
0 15 10} istheiuterestfor^ofayear (£1 l5.1jc/.X4')>atdo.
3 19 3^ is the interest for 3^- years, at do.
Exa:mple 2. — What is the interest of £300 for 5j years,
at 3J per cent. T
£ 5. (/.
£300^-100=3 0 0 is the interest for 1 year, at 1 per cent.
3
9 0 0 is the interest for 1 Tear, at 3 per cent.
2 5 0 is the interest for 1 year, at £i (£3 xf)
11 5 0 is the interest for 1 year, at 3| per cent.
5
56 5 0 is the interest for 5 years, at 3 J percent
5 12 G is the interest for ^ year (£11 5.s^-^2)
2 IG 3 is the do. for l yeaf (£5 125. G|r/.-^2)
And G4 13 0 is the interest for 5^ years, at 3^ do.
INTEREST. 243
EXCRCISF.S.
28. What is tlic interest of £379 25. Gd. for 4^ years,
at 5| per cent. .' Ans. £91 5.5. od.
29. V/bat is the interest of £640 105. 6d. for 2 J
years, at 4i per cent. ? Aiis. £72 l5. 2^\d.
30. What is the interest of £600 105. Gd. for 3^
years, at 5f- per cent. ? An.s. £115 25. Q^\d.
31. What is the interest of £21? 85. 1^. for 6?j
years, at 5f per cent. } Ans. £81 85. 5ffZ.
10. To find the interest for daysj at 5 per cent. —
Rule. — Multiply the principal by the number of days,
and divide the product by 7300.
F.XAMPLE. — What is the interst of £20 4.s. 2J. fur 8 days ?
£ s. d.
20 4 2
200 13 4
20
4193
12
rSOO). 50320(6 ^-|5J.
•43800
6520
The required interest is 0|?&, or Id. — since the remainder
is greater than half the divisor.
The interest of £1 for 1 year is £^V' "^^ ^'^^ ^ daj^V-v-365=
0(3^3^5=7300; that is, the ToOOth part of the principal.
Therefore the interest of any other sum for one clay, is the
7300tli part of that sura; and foi' any number of days, it is
that number, multiplied by the ToOOtlx part of the principal —
or, which is the same thing, the principal multiplied by the
number of days, and divided by 7300.
EXERCISES.
32. Find the interest of £140 105. for 76 days, at 5
per cent. Ans. £1 95. S-f-^^d.
33. Find the interest of £300 for 91 days, at 5 per
cent. Ans. £3 145. 9^d.
34. What is the interest of £800 for 61 days, at 5
per cent. ? Aiu. £6 135. 3%^d.
244 iNTr.UKST.
1 1. To find tlio interest fcir (hrj/s^ at cnnj other rate —
JLluLE. — Find the interest at 5 per cent., and take
parts of this for the remainder.
ExA.MPLE.— What is the interest of £3324 Gs. 2cL for 11
days, at £Q 10s. per cent. "?
X3324 6s. 2;7.xll-^T300=£5 Os. 21(7. Therefore
£ .s. d.
5)5 0 2\ is the interest for 11 days, at 5 per cent.
2)1 0 Oi is the interest for 11 days, at 1 per cent.
0 10 0 is the interest for 11 days, at 10.!>. per cent.
And 6 10 2| is the interest for 11 days, at £6 lOs. (£5-f
i:i+10<f.)
This rule requires no explanation.
KXERCISE."?.
35. What is the interest of £200 from the 7th May
to the 26th September, at 8 per cent. } Ans. £Q As.
36. What is the interest of £150 15^. 6^. for 53
days, at 7 per cent. } Ans. £1 105. l^d.
37. What is the interest of £371 for 1 year and 213
days, at 6 per cent, t Ans. £35 55. Qd.
38. What is the interest of £240 for 1 year and 135
days, at 7 per cent. .? Ans. £23 05. md.
Sometimes the number of days is the aliquot part of
a year ; in which case the process is rendered more easy.
Example. — What is the interest of £175 for 1 year and
73 days, at 8 per cent. ?
1 year and 73 days=li year. Hence the required intereat
is the interest for 1 year -{-its lifth part. But the interest of
£175 for 1 year, at the given rate is £14. Therefore its
interest for the given time is £14-j-£y=£14-f-£2 lG.s.=:
£1G 1G5.
12. To find the interest for months., at 6 per cent — '
EuLE. — If the number expressing tlie months is erm,
multiply the principal by half the numler of months
and divide by 100. But if it is odd., multiply by the
half of one less than the numhcr of months ; divide the
result by 100 ; and add to the quotient what will be
obtained if we divide it by one less than the number of
months.
INTEREST 245
EXAMPLK. 1. —
months, at G per
72
What
cent ?-
5. r/.
G 4
4
is the interest
required intere
of £72 6.-.
St is £2 I7s
4J.
. 10
for 8
£289
20
17-855.
12
5 4
The
^Id.
10-2-W,
4
0-'JG=ld. nearly.
Solving the question by the rule of three, we shall have —
£100 : £72 Os. M. ': £G : £72 6.s. 4r/.x8x6
12 : 8 100^2 = ('li^'i^-
ing l«>th numerator and denominator hv G [Sec. IV. 4]).
£72Cw. 4//.x8x6-f-G .£72 Gy. 4./.X8 ,,..,. , ^
. 100x12:^0 = n.u^2 = (dividing both
£72 G5. 4</.x8-^2
oj^merator and denominator by 2) — ~ioTk7^-t--^ ' "^^
£72 as. 4c/. x4
100
— that is. the required interest is equal t-«o the given sum,
multiiilied Ity half the number which expres;>es the month.-*,
an.l divided by lUU.
Ex.\MPLE 2.— What is the interest of £84 C>s. 2^1. fur U
months, at 6 per cent, t 11=104-1 10-f-2==5.
£ s. d.
84 6 2
^) One less than the given number of
•;. . ^. , , , month ==10.
£4-21 10 10
20
£ «. d.
4 oOs. 10)4 4 3| is the interest for 10 nioiUbs, at fi per cent.
3-70r/. And 4 12 9 U the iutcrest for 11 (10-f 1) raontLs. st U •♦
4
2-^0f.=2a. nearly.
The interest fur 11 months is cviden^^ly the inteves:
\T ^ month, plus the uilevo^t of 1 1^- 1 month -^11—1.
.M
246 INTEREST.
EXERCISES.
39. What is the interest of £250 175. 6d. for 8
months, at 6 per cent. ? Ans. dElO Qs. 8|^'.
40. What is the interest of £571 155. for 8 months,
at 6 per cent. ? Ans. £22 175. 4^d.
41. What is the interest of £840 for 6 months, at 6
per cent. } Ans. £25 45.
42. What is the interest of £3790 for 4 months, at
6 per cent. ? Ans. £75 165.
43. What is the interest of £900 for 10 months, at
6 per cent. ? Ans. £45.
44. What is the interest of £43 25. 2d. for 9 months,
at 6 per cent. ? Ans. £1 185. 9^^.
13. To find the interest of money, left after one or
more payments —
KuLE. — If the interest is paid by days^ multiply the
sura by the number of days which have elapsed before
any payment was made. Subtract the first payment,
and multiply the remainder by the number of days
which passed between the first and second payments.
Subtract the second payment, and multiply this remain-
der by the number of days which passed between the
second and third payments. Subtract the third pay-
ment, &c. Add all the products together, and find the
interest of their sum, for 1 day.
If the interest is to be paid by the week or mouthy
substitute weeks or months for days., in the above rule.
Example. — A person borrows £117 for 94 days, at 8 per
cent., promising the principal in parts at his convenience,
and interest corresponding to the money left unpaid, up
to the difierent periods. In 6 days he pays £17 : in 7
days more £20 : in 15 more £32 ; and at the'end of the 94
days, all the money then due. What does the interest
come to '?
£ days. £ day.
117x 6= 702x1)
100 X 7= 700x1 1^.770
80x15=1200x1 f— ^^''"•
48x6G=31C8xlJ
The interest on 5770 for 1 day, at 5 per cent., is 155. 9|ti
Therefore
INTEREST. 2A /
5)0 15 9| is the intorost at 5 pei cent.
0 3 2 is the interest, at 1 per cent.
o)0 18 11^ is the interest, at G per cent.
0 6 4 is the interest, at 2 per cent.
And 1 5 3| is the interest, at 8 per cent., for th*-* given
sums and times.
If the entire sum were 6 days unpaid, the interest would be
the same as that of 6 times as much, for 1 day. Next, £100
due for 7 days, should produce as much as £700, for 1 day,
&c. And all the suras due for the different periods should
produce as much as the sura of their equivalents, in 1 day.
EXERCISES.
45. A merchant borrows £250 at 8 per cent, for 2
years, with condition to pay before that time as much
of the principal as he pleases. At the expiration of 9
months he pays £89, and 6 months after £70 — leaving
the remainder for the entire term of 2 years. How
much interest and principal has he to pay, at the end
of that time ? Ayis. £127 I6s.
46. I borrow £300 at 6 per cent, for IS months,
with condition to pay as much of the principal before
the time as I please. In -3 months I pay £60 ; 4 months
after £100 ; and 5 months after that £75. How much
principal and interest am I to pay, at the end of IS
mouths ? ^725. £79 Ids.
47. A gives to B at interest on the 1st November,
1804, £6000, at 4^ per cent. B is to repay him with
interest, at the expiration of 2 years — having liberty to
pay before that time as much of the principal as he
pleases. Now B pays
£
The 16th December, 1804, . . 900
The 11th March, 1805, . . 1260 ,
The oOth :\Iarch. ... 600
The 17th August, . . . 800
The 12th February, 1806, . . 1048
How much principal and interest is he to pay on th»
1st November, 1S06 ? Ans. £1642 9^. 2{f^d.
48. Lent at interest £600 the 13th May, 1833, for
24R I-^TKREST.
1 year, at 5 per cent. — with condition that the receive'*
may discharge as much of the principal before the time
as he pleases. Now he pays the 9th July £200 ; and
the 17th September £150. How much principal and
interest is he to pay at the expiration of the year ?
Ans. £26Q 13s. b-^\d.
14. It is hoped that the pupil, from what he has
learned nf the properties of proportion, will easily un-
derstand the modes in which the following rules are
proved to be correct.
Of the principal, amount, time, and rate — given any
three, to find the fom'th.
Given the amoimt, rate of interest, and time ; to find
the principal —
Rule. — Say as £100, plus the interest of it, for the
given time, and at the given rate, ls to i£100 ; so is the
given amount to the principal sought.
Example. — What will produce £862 in 8 years, at 5 per
cent.'?
^ £40 (=£5x8) is the interest for £100 in 8 years at the
given rate. Therefore
£140 : £100 : : £862 : ^^^.^q ^^ =£615 Us. 3iJ.
When the time and rate are given —
£100 : any other sum : : interest of £100 : interest of
that other sum.
By alteration [Sec. V. 29], this becomes—
£100 : interest of £10'/ : : any other sum : interest of
that sum.
And. saying " the first + the second : the second,"' &c.
[Sec. V. 29] we have—
£100 -f- its interest : £100 : : any other sum -|- its in-
terest : that sum — which is exactly the rule.
EXERCISES.
49. What principal put to interest for 5 years will
nmount to £402 105., at 3 per cent, per annum ? Ans.
£350.
50. What principal put to interest for 9 years, at 4
per cent., will amount to £734 S^. } Ans. £540.
INTEREST. 249
51. The amount of a certain principal, bearing inter-
est for 7 years, at 5 per cent., is i£334 165. What is
the principal ? Ans. i£24S.
15. Given the time, rate of interest, and principal —
to find the amount —
KuLE. — Say, as £100 is to £100 plus its interest for
the given time, and at the given rate, so is the givea
sum to the amount requu-ed.
Example. — What will £272 come to, in 5 years, at 5 per
cent. ?
£125 (=£1004-.£5x5) is the principal and interest of
£100 for 5 years ; then —
272x125
£100 : £125 : : £272 : — ^^^=£340, the required
amount.
We found by the last rule that
£100-j-its interest : £100 : : any other sum -f-its interest :
that sum.
Inversion [Sec. V. 29] changes this into,
£100 : £100-|-its interest : : any other sum : that other
Bum+its interest — which is the pi«sent rule.
EXERCISER.
52. What will £350 amount to, in 5 3'ears, at 3 per
cent, per annum .' Ans. £402 10^.
53. What will £540 amount to, in 9 years, at 4 per
cent, per annum } Ans. £734 Ss.
54. What will £24S amount to, in 7 years, at 5 per
cent, per annum ? Ans. £334 16^.
55. What will £973 4^. 2cl. amount to, in 4 years
and 8 months, at 6 per cent. } Ans. £1245 14^. \^d.
56. What vrill £42 3^. Q\d. amount to, in 5 years
and 3 months, at 7 per cent. .' Atis. £57 13^. \Q\d.
16. Given the amount, principal, and rate — to find
the time —
Rule. — Say, as the interest of the given sum for 1
year is to the given interest, so is 1 year to the re-
piired time.
250 INTEREST.
Example. — When would £281 135. 4d. become £338, at
'6 per cent. ?
£14 Is. Sd. (the interest of £281 13^. 4d. for 1 year [2]) :
.^56 Os. Sd. (the given interest) : : 1 : ^p 1 4 i q^ ==^> the
jequl^ed number of years.
17. Iience briefly, to find the time — Divide tho
interest of the given principal for 1 year, into the entu*e
interest, aud the quotient will be the time.
It is evidort, the principal, and rate being given, tho
interest is prcponional to the time ; the longer the time, the
more the interest, »md the reverse. That is —
The interest for one time : the interest for another : :
the former time : ths latter.
Hence, the interest of the given sum for one year (the
interest for one time) : the given interest (the interest of
the same sum for ancther time) : : 1 year (the time which
produced the former) : the time sought (that which pro-
duced the latter) — which is the rule.
EXERCISES.
57. la what time would £300 amount to £372, at 6
per cent. .? Ans. 4 years.
58. In what time would £211 5^. amount to £230
5j. 3^., at 6 per cent. .? Ans. In 1 year and 6
months.
59. "WTien would £561 lo^. become £719 Qs. 9^^.,
at 6 per cent, r Ans. In 4 years and 8 months.
60. When would £500 become £599 35. 4^., at 7 per
cent, i Ans. In 2 years and 10 months.
61. When will £436 95. Ad. become £571 85. \\d.,
at 7 per cent, i Ans. In 4 years and 5 months.
18. Griven the amount, principal, and time — to find
the rate —
Rule. — Sa}--, as the principal is to £100, so is the
given interest, to the interest of £100 — which will give
the interest of £100, at the same rate, and for the same
time. Divide this by the time, and the quotient yhM be
the rate.
INTEREST. !Xil
Example.— At -what rate will £350 amount to £402 10s
ia 5 years ?
£350 : £100 : : £52 10.9. : "^^^^^'-^^i^^. =£15, the in
350
terest of £100 for the same time, and at the same rate
Then '/=3, is the required number of years.
We have seen [14] that the time and rate being the same,
£100 : any other sum : : the interest of £100 : interest
of the other sum.
This becomes, by inversion [Sec. V. 20] — •
Any sum : £100 : : interest of the former : interest of
100 (for same number of years) .
But the interest of £100 divided by the number of years
which produced it, gives the interest of £100 for 1 year —
or, in other words, the rate.
EXERCISE.S.
62. At what rate will £300 amount in 4 years to
ie372 ? Ans. 6 per cent.
63. At what rate will £248 amoimt in 7 years to
£334 16s. f Ans. 5 per cent.
64. At what rate will £976 14.5. 7d. amount in 2 years
and 6 months to £1098 IGs. 4^d. r A/is. 5 per cent.
Deducting the 5th part of*the interest, will give the in-
terest of £076 l-is. Id. for 2 years.
65. At what rate will £780 175. 6d. become £937
Is. in 3 years and 4 months } Ans. 6 per cent.
66. At what rat^ will £843 55. ^d. become £1047 1^.
7|c?., in 4 years and 10 months } Ans. At 5 per cent.
67. At what rate will £43 2s. 4^d. become £60 75
4i^., in 6 years and 8 months .' Ans. At 6 per cent.
68. At what rate will £473 become £900 135. e\d
in 12 years and 11 months .' Ans. At 7 per cent.
COMPOUND INTEREST.
19. Given the principal, rate, and time — to find the
amount and interest —
KuLE I. — Find the interest due at the first time of
payment, and add it to tlic principal. Find the interest
252 INTEREST.
of that sum, considered as a new principal, and add it
to what it would produce at the next payment. Con-
sider that new sum as a principal, and proceed as
before. Continue this process through all the times of
payment.
Example. — What is the compound interest of £07, for 4
years, at 4 per cent, half-yearly ?
£ .1. d.
97 0 0
3 17 11 is the interest, at the end of 1st half-year.
100 17 7} is the amount, at end of 1st half-year.
4 0 83- is the interest, at the end of 1st year.
104 18 3^ is the amount, at the end of 1st year.
4 3 11 Ms the interest, at the end of 3rd half-year.
109 2 3 is the amount, at the end of 3rd half-year.
4 7 3Ms the interest, at the end of 2nd year.
113 9 C} is the amount, at the end of 2nd year.
4 10 0\ is the interest, at the end of 5th half-year.
118 0 4 is the amount, at the end of 5tli half-year.
4 14 5 is the interest, at the end of 3rd year.
122 14 9 is the amount, at the end of 3rd year.
4 18 2]- is the interest,* at the end of 7th half-year.
127 12 11| is the amount, at the end of 7th half-year.
5 2 IMs the interest, at the end of 4th year.
132 15 0} is the amount, at the end of 4th year.
97 0 0 is the principal.
And 35 15 0| is the comj)Ound interest of <£97, in 4 years.
20. This is a tedious mode of proceeding, particularly
when the times of payment are numerous ; it is, there-
fore, better to use the following rules, which will be
found to produce the same result —
Rule II. — Find the interest of £l for one of the
payments at the given rate. Find the product of so
•many factors (ocach of them d61+its interest for one
payment) as there are times of payment ; multiply this
product by the given principal ; and the result will bo
the principal, plus its compound interest for the given
INTEREST. 25^-
time. From tills subtract the principal, and the remain-
der will be its compound interest.
Example 1. — What is the compound interest of £237 for
3 years, at 6 per cent. 1
£06 is the interest of £1 for 1 year, at the given rate;
and there are 3 payments. Therefore £1-06 (£l-f-£OG) is
to be taken 3 times to form a product. Hence lOGxlOGx
106x£237 is the amount at the end of three years: and
1-OG X lOGx 1-06 X £237— £237 is the compound^interortt.
Tlie following is the process in full —
£
1-06 the amount of £1, in one year.
1-06 the multiplier.
1-1236 the gmount of £1, in two years
106 the multiplier.
1-191016 the amount of £1. in three years;
iSIulti plying by 237, the principal,
£ s. d.
vre find that 232-270792=282 5 5 is the amount •
and subtracting 237 0 0, the principal,
we obtain 45 5 5 as the compound interests
Example 2. — What are the amount and compound inte-
rest of £79 for 6 years, at 5 per cent. '?
The amount of £1 for 1 year, at this rate would be £105.
^ Therefore £105xl05xl05xl05xl05xl05x79 is the
amount. &c. And the process in full will be —
£
1-05
1-05
1-1025 the amount of £1, in two years.
11025
1-21551 the amount of £1, in four years.
1-1025
1-34010 the amount of £1. in six years.
'1 & ,,. ,.
£105-86790=105 17 4| is the required amount.
79 0 0
And 26 17 4] is the required interest
M 2
2^4 INTEKKST
Example o. — What are the ainoimt, and compound interest
of £27, fur 4 3'ears. at £2 lOs. per cent, half-yearly.
The amount of £1 for one pnvraent is £1025. Therefore
£1023 X 1025 X 1025 x I02o x 1025 x 1025 x 1025 x
l025 x27 is the amount, &c. And tlie process in full will be
£
1025
1025
1050G3 the amount of £1, in one year,
105063
1- 10382 the amount of £1. in two xears.
110382
1-21842 the amount of £1, in four years.
£ s. ar
£32-89734=32 17 11^ is the required amount.
27 0 0
And 5 17 11 1 is the required interest.
21. EuLE III.— Find by tho interest tabic (at the end
of the treatise) the amount of £1 at the given rate, and
for the given number of payments 5 multiply this by the
given principal, and the product will be the required
amount. From this product subtract the principal, and
the remainder will be the required compound interest.
Example. — What is the amount and compound interest
of £47 10s. for 6 years, at 3 per cciit.. half-yearly '?
£47 106\=£47-5.
We find by the table that
£1-42576 is the amount of £1. fur the given time and rate.
47- 5 is the multiplier.
£ s. (J.
67-7236=67 14 5? is the required amount.
47 10 0'
And 20 4 5^^ is the required interest.
22. Rule i. requires no explanation.
Rkason- of Rulk II. — When the time and rate are the
game, two principals are proportional to their corresponding
amounts. Tlicrefore
£1 (one principal) : £1 03 (its corresponding amount) :
£10G (.another principal) : £1-06 X lOG (its corresponding
amount).
rXTF.REST. 255
Hence the fimouut of £1 for two years, is £1 06xl"06 —
or tlie prorluct of two factors, each of them the amount of £1
for one year.
Again, for similar reasons,
£1 : £1'06 :: £l-06Xl-0G : £1 -OGXl-OGXl-OO.
Hence the amount of £1 for three years, is £l-00xl'00Xl'0G —
or the product of three factors, each of them the amount of
£1 for one year.
The same reasoning would answer for any number of pn^'-
ments.
The amount of any principal will be as much greater than
the amount of £1, at the same rate, and for the same time, as
the principal itself is greater than £1. Hence we multiply
the amount of £1, by the given principal.
Rule III. requires no explanation.
2-3. Wlieu the decimals become numerous, we may
proceed as already directed [Sec. II. 58].
We may also shorten tlie process, in many cases, if
we remember that the product of two of the factors
multiplied by itself, is equal to the product of four of
tliem ; that the product of four multiplied by the pro-
duct of two is equal to the product of six ; and that the
product of four multiplied by the product of four, is
equal to the product of eight, &c. Thus, in example 2,
11025 (=l-05xl-05) xl-1025=105xl-05xl-05xl-05.
EXERCISES.
1 . What are the amount and compound interest of
£91 for 7 years, at 5 per cent, per annum .' Ans. £12S
Os. lid. is the amount; and £37 0^. lld.j the com-
pound interest.
2. What are the amount and compound interest of
£142 fur 8 years, at 3 per cent, half-yearly.' A?is.
£227 17s. 4^d. is the amount ; and £85 17^. 4|^., the
compound interest,
3. What are the amount and compound interest of
£63 OS. for 9 years, at 4 per cent, per annum : An-s.
£90 05. 5^d. is the amount; and £26 15^. o^d.^ thw
compound interest.
4. What are the amount and compound interest of
£44 55. 9^. for 11 years, at 6 per cent, per annum.'*
256 INTEREST.
Ans. £84 Is. bd. is the amount; and £39 155. 8(£. ,
the compound interest.
5. What are the amount and compound interest of
£32 4.9. 9ffi?. for 3 years, at £2 105. per cent, half-
yearly .' Ans. £37 Is. S\d, is the amount ; and £5
25. lOifZ., the compound interest.
6. What are the amount and compound interest of
£971 05. 2\d. for 13 years, at 4 per cent, per annum ?
Alls. £1616 155. llffZ. is the amount; and £645 155
9^^., the compound interest.
24. Given the amount, time, and rate — to jQnd the
principal ; that is, to find the present worth of any sum
to be due hereafter — a certain rate of interest being
allowed for the money now paid.
lluLE. — Find the product of as many factors as there
are times of payment^ — each, of the factors being the
aviount of £1 for a single payment ; and divide this
product into the given amount.
Example. — ^What sum would produce £834 iu 5 years,
at 5 per cent, compound interest '?
T]io amount of £1 for 1 year at the given rate is £1-05 :
and the product of this taken 5 times as a factor lOox
105xl05xl'05xl"05. which (according to the table) is
1-27628. Then
£834-M-27G28=£653 9s. 2ld., the required principal.
25. Reason ok the Eitl,e. — "We have seen [21] that the
amount of any sum is equal to the amount of £1 (for the same
time, and at the same rate) multiplied by the principal ; that is,
The amount of the given principal=the given principal X
the amount of £1.
If we divide each of these equal quantities by the same
number [Sec. V. 6], the quotients will be equal. Therefore —
The amount of the given principalH-the amount of £l=the
given principalXthe amount of £l-f-the amount of £1. That
is, the amount of the given principal (the given amount)
divided by the amount of £1, is equal to the principal, or
quantity requli^ed — which is the rule.
EXERCISES.
7. What ready money ought to be paid for a debt of
£629 175. l-o}^-, to bo due 3 years hence, allowing
5 per cent, compound interest ^ Ans. £500.
INT r REST. 257
8. TVIiat principal, put to interest for G years, would
amount to £268 05. 4^^., at 5 per cent, per annum?
Ans. ^200.
9. V/hat sum would produce £742 195. ll^d. in 14
years, at 6 per cent, per annum } Ans. X'328 125. 7c/.
10. What Is X^495 195. llfrZ., to be due in IS years,
at 3 per cent, half-yearly, worth at present. Ans.
£171 25. S^d.
26. Given the principal, rate, and amount — to find
the time —
KuLE I. — Divide the amount by the principal; and
into the quotient divide the amount of £1 for one pay-
ment (at the given rate) as often as possible — the number
of times the amount of £1 has been used as a divisor,
will be the rerpiired number of payments.
Example. — In Avhat time will £02 amount to £100 135.
O'^d.. at 3 per cent, half-yearly ?
£106 135. 0|(/.-h £02=1- 15027. The amount of £1 for
one payment is £103. But 1-15027 -^ 1-03 = 11255 ;
11255 -J- 1-03 = 1-00272 : 1-00272 -^ 103 = 1-OGOO : and
10609-f-103--103; l-03-f-103=l. _ We have used 103
as a divisor 5 times : therefore the time is 5 payment^ or
21 years. Sometimes there will be a remainder after divid-
ing; by 1-03. kc. as often as possible.
In explaining tlie method of finding the powers and roots
of a given quantity, we shall, hereafter, notice a shorter
method of ascertaining liow often the amount of one pound
can be used as a divisor.
27. Rule II. — Divide the given principal by the
given amount, and ascertain by the interest table in how
many payments £1 would be equal to a quantity nearest
to the quotient^ — con.sidered as pounds : this will be the
required time.
Example. — In what time will £50 become £100, at G
per cent, per annum compound interest 1
£100-^50=2.
We find by the tables that in 11 years £1 will become
£1-8983. -^vhich is less : and in 12 years that it ^vill become
£20122, which is greater than 2. The answer nearest t«
the truth, therefore, is 12 years.
ift S INTEREST.
28. Reason of Rule I. — The given amount is [20] equal
to the given principal, multiplied by a pi'oduct -which contains
as mauv factors as there are times of payment— each factor
being the amount of £l,f«jr one paj-ment. Hence it is evi-
dent, that if we divide the given amount by the given prin-
cipal, we must have the product of these factors ; and that, if
vre divide this product, and tlie successive quotients by one
of the factors, we shall ascertain their number.
Reason of Rule II. — We can find the required number
of factors (eacli the amount of £1), by ascertaining how often
the amount of <£! may be considered as a factor, without
forming a product much greater or less than the quotient
obtained when we divide the given amount by the given
principal. Instead, however, of calculating for ourselves, we
may have recourse to tables constructed by those who have
already made the necessary multiplications — which saves much
trouble.
29. TVlien the quotient [27 j is greater than any
amount of £1^ at the given rate, in the table, divide it
by the greatest found in the table ; and, if necessary,
divide the resulting quotient in the same way. Continue
the process until the quotient obtained is not greater
than the largest amount in the table. Ascertain vrhat
number of j>aijmcnts corresponds to the last quotient,
and add to it so many times the largest number of pay-
fjients in the table, as the largest amount in the table
has been used for a divisor
Example. — When would £22 become £535 12s. 0^(Z.,
at 3 per cent, per annum 1
£535 129. 0J(/.-^ 22=24-34500, -^-hich is greater than any
amount of £1, at the given rate, contained in the table.
24-34500-^4-3830 (the greatest amount of £1, at 3 per cent.,
found in the table) =5 -55330 : but this latter, also, is greater
than any amount of £1 at the given rate in the tallies.
6-55339-r-4-3830=l-20077, which is found to be the amount
of £1, at 3 per cent, per payment in 8 payments. We
have divided by the highest amount for £1 in the tables, or
that corresponding to fifty payments, twice. Therefore, the
required time, is 50-f-50-f-8 payments, or 108 years.
EXERCISES.
11. When would £14 Qs. Sd. amount to JglS 2s. 8^d.
at 4 per cent, per annum, compound interest .- Ans.
In 6 years.
INTEREST. 259
12. When would ^£54 2^. 8^. amount to £76 35. 5^.,
at 5 per cent, per annum, compound interest ? Ans.
lu 7 years.
13. In what time would £793 0.?. 2]^/. become £1034
13s. lOi/'Z., at 3 per cent, half-yearly, compound interest ?
Ans. In 4^^ years.
14. ^\'hen would £100 become £1639 75. 9^7., at 6
per cjr.t. half-yearly, compound interest .'* Ans. In 24
v"ears.
QUESTIONS.
1. What is interest .' [1].
2. What is the difference between simple and com-
pound interest r [1].
3. What are the principal, rate, and amount ? [1].
4. How is the simple interest of any sum, for 1 year,
found? [2 &c.].
5. How is the sunple interest of any sura, for several
years, found ? [5] .
6. How is the interest found, wlicn the rate consists
of more than one denomination .' [4] .
7. How is the simple interest of any sum, for years,
months, &c., found .' [6].
8. How is the simple interest of any sum, for any
time, at 5 or 6, &c. per cent, found ? [7].
9. How is the simple interest found, when the rate,
number of years, or both are expressed by a mixed
number ? [9] .
10. How is the simple interest for days, at 5 per cent.,
found .5 [10].
11. How is the simple interest for days, at any other
rate, found.? [11].
12. How is the simple interest of any sum, for months
at 6 per cent., found r [12].
13. How is the interest of money, left after one or
more payments, found r [13] .
14. How is the principal found, when the amount,
rate, and time are given .'' [14].
lo. How is the amount found, when the time, rato,
and principiil are given .' [15].
2G0 DISCOUNT.
IG. How ia tlie tiniG found, when tlie amount, prin
cipal, and rate are given ? [IG].
17. How is the rate found, when the amount, princi
pal, and time are given ? [IS].
18. How are the amount, and compound interest found^
when the principal, rate, and time are given ? [lU].
19. How is the present worth of any sum, at com-
pound interest for any time, at any rate, found ? [24].
20. How is the time found, when the principal, rate
of compound interest, and amount are given } [^Q]-
DI3C0UXT.
30. Discount is money allowed for a sura paid before
it is due, and should be such as would be produced by
what is paid, were it put to interest from the time the
payment w, until the time it ought to he made.
The 'present loorth of any sura, is that which
wouldj at the rate allowed as discount, produce it, if
put to interest until the sum becomes due.
31. A bill is not payable until three days after the
time mentioned in it ; these are called days of grace.
Tlius, if the time expires on the 11th of the month, the
bill will not be payable until the 14th — except the latter
falls on a Sunday, in which case it becomes payable on
the preceding Saturday. A bill at 91 days will not be
due until the 94th day after date.
32. When goods are purchased, ascertain discount is
often allowed for -prompt (immediate) payment.
The discount generally taken is larger than is sup-
posed. Thus, let what is allowed for paying money
one year before it is due be 5 per cent. ; in ordinary
circumstances £95 would be the payment for £100.
But £95 would not in one year, at 5 per cent., produce
more than £99 15.s., which is less than £100 ; the error,
however, is inconsiderable when the time or sura is small
Hence to find the discount and present worth at any
rate, we may generally use the following —
DISCOUNT. 261
33. Rule. — Find the interest for the sum to be paid,
at the discount allowed ; consider this as discount, and
deduct it from what is due ; the remainder will be the
required present worth.
Example. — £02 will be due in 3 months : what should be
allowed on immediate pa^onent, the discount being at the
rate of 6 per cent, per annum ?
The interest on £02 for 1 year at 6 per cent, per annum
is £3 14.S. 4^</. : and for 3 months it is I85. l^d. Therefore
£02 minus I8.5. l^d.=£Ol h. 4^J., is the required present
worth.
34. To find the present worth a^r.iuTJely —
liuLE. — Say, as £1-00 plus its mter^si for the given
time, is to i^lOO, so is the given sum to the required
present worth.
Example. — What would, at present, pay a v?e>t of .£142
to be due in 0 months. 5 per cent, per annum discoirn*. being
allowed \
£
£ £ s. £ £ 1 oQ ^ 119. £ »'. d.
102-5 (100+2 10) : lOO : : 142 ; — 1^^=138 10 8
Tliis is merely a question in a rule already given [14].
EXERCISES,
1. TVTiat is the present worth of i6850 15.?., payable in
one year, at 6 per cent, discount .' Ans. £802 II5. lOf^
2. What is the present worth of £240 IO5., payable
in one year, at 4 per cent, discount .' A)is. £231 55.
3. What is the present worth of £550 IO5., payable
in 5 years and 9 months, at 6 per cent, per an. discount }
Alls. £409 55. lOii.
4. A debt of £1090 will be due in 1 year and 5
months, what is its present worth, allowing 6 per cent.
per an. discount } Ans. £1004 12^. 2d.
5. What sum will discharge a debt of £250 17^. 6c?.,
to be due in 8 months, allowing 6 per cent, per an.
discount .- Atis. £241 4^. ^d.
6. What sum will discharge a debt of £840, to be
due in 6 months, allowing 6 per cent, per an. discount }
Ans. £815 IO5. SH
262 DISCOUNT.
7. What ready money now will pay a debt of £200,
to be due 127 duys hence, discounting at 6 per cent,
per an. .' Arts. £l9d ISs. 2-}d.
8. What ready money now will pay for £1000, to be
due in 130 days, allowing 6 per cent, per an. discount }
Ans. £979 Is. Id.
9. A bill of £150 105. will become due in 70 days,
what ready money will now pay it, allowing 5 per cent,
per an. discount } Ans. £149 Is. 5d.
10. A bill of £140 10^. will be due in 76 days, what
ready money will now pay it, allowing 5 per cent, per
an. discount f Ans. £139 1^. O^d.
11. A bill of £300 will be due in 91 days, what will
now pay it, allowing 5 per cent, per an. discount } Ans.
£296 6^. l^d.
12. A bill of £39 55, will become due on the first
of September, what ready money will pay it on the
preceding 3rd of July, allowing 6 per cent, per an. ?
Ans. £38 185. l^d.
13. A bill of £218 35. Sid. is drawn of the 14th
August at 4 months, and discounted on the 3rd of Oct. ;
what is then its worth, allowing 4 per cent, per an.
discount .? Aiis. £216 85. Ud-
14. A bill of £486 185. Sd. is drawn of the 25th
March at 10 months, and discounted on the 19th June,
what then is its worth, allowing 5 per cent, per an.
discount.^ Ans. £472 9s. U^d.
15. What is the present worth of £700, to be due in
9 months, discount being 5 per cent, per an. .' Ans.
£674 135. md.
16. What is the present worth of £315 125. 4}d.,
payable in 4 years, at 6 per cent, per an. discount .''
Ans. £254 10s. l^d.
17. What is the present worth and discount of £550
105. for 9 months, at 5 per cent, per an. ? A^is. £530
125. O^d. is the present worth; and £19 175. lli^.
s the discount.
18. Bought goods to the value uf £35 135. Sd. to be
Daid in 294 days ; what ready money are they now
ivorth, 6 per cent, per an. discount being allowed .'*
Ans. £34 05. O^d. i
COMMISSION. 263
19. If a legacy of £600 is left to me on the 3rd of
May, to be paid on Christmas day following, what must
I receive as present papnent, allowing 5 per cent, per
an. discount.' Ans. JJ5S1 4s. 2^d.
20. AVhat is the discount of £i756, the one half pay-
able in 6, and the remainder in 12 months, 7 per cent,
per an. being allowed .' Ans. £37 145. 2^d.
21. A merchant owes £110, payable in 20 months,
and i-224, payable in 24 months ; the first he pays in 5
mouths, and the second in one month after that. What
did lie pay, allowing S per cent, per an. .' Ans. £300.
QUESTIONS rOR THE PUPIL.
1 . Wliat is discount .- [30] .
2. What is the present u-orth of any sum .' [30].
3. Vsh^t ara days of grace 1 [31].
4. How is dL^icount ordinarilj/ calculated ? [33]
5. How is it accu.ratdy calculated ? [34].
C0}4MISSI0X, .^c
35. CovimissiGn is an allowance per cent, made to a
person called an agent, who is employed to sell goods.
Insurance is so mucli per cent, paid to a person who
undertakes that if certain goods are injured or destroyed,
he will give a stated sum of money to the owner.
Brokerage is a small allowance, made to a kind of
agent called a hroker^ for assi.sting in the disposal of
goods, negotiating bills, &c.
36. To compute commission, ecc. —
Rule. — Say, as £100 is to tlie rate of commission, so
is the given sum to the corresponding commission.
Example. — "What will be the commission on goods worth
£437 bs. 2(L at 4 per cent. 1
4 V :f 4*^7 '^v- ^ J
£100 : £4 : : £437 5s. 2cl. : ^qq " " = ^1" ^^^
9^d., the required commisi^ion.
37. To find what insurance mu.st be paid so that, if
the goods are lost, both their value and the insuranoa
paid may be recovered —
264 COMMISSION^
Rule. — Say, as £100 minus the rate per cent, is to
£100, so is the value of the goods insured, to the
required insurance.
Example. — What sum must I insure that if goods worth
£400 are lost, I may receive both their value and the in-
surance paid, the latter being at the rate of 5 per cent. ?
£05 : £100 :: £400 : ^^'-^'^ X 400^ ^^^^ ^^
If £100 Ts-ere insured, only £95 would be actually received,
since £5 was paid for the £100. In the example, £421 Is. OgtZ.
are received; but deducting £21 Is. Oid.y the insurance, £400
remains.
EXERCISES.
1. What premium must be paid for insuring goods
to the amount of £900 155., at 2\ per cent, t A7is.
£22 105. A{cl.
2. What premium must be paid for insuring goods
to the amount of £7000, at 5 per cent. ? Ans. £350.
3. W^hat is the brokerage on £976 175. dd., at 55.
per cent. > Ans. £2 Ss.-lOld.
4. W^hat is the premium of insurance on goods worth
£2000, at 7i per cent. ? Ans. £150.
5. What is the commission on £767 145. 7d.j at 2^
per cent. ? ui7is. £19 35. lOf J.
6. How much is the commission on goods worth
£971 145. 7d.y at 55. per cent, r Ans. £2 85. 7j\d.
7. What is the brokerage on £3000, at 25. 6d. per
cent. ? Ans. £3 155.
S How much is to be insured at 5 per cent, on goods
worth £900, so that, in case of loss, not only the value
of the goods, but the premium of insurance also, may be
repaid ? Ans. £947 75. 4j\d.
9. Shipped off for Trinidad goods worth £2000, how
much must be insured on them at 10 per cent., that in
case of loss the premium of insurance, as well as their
value, may be recovered ? Ans. £2222 45. o^d.
QUESTIONS FOR THE PUPIL.
1 . What is commission ? [35]
2. What is insurance ? [35].
3. What is brokerage ? [35]
PURCHASE OF STOCK. 265
4. How are commission, insurance, &c., calculated?
[36].
5. How is insurance calculated, so that both the in-
surance and value of the goods may he received, if the
latter are lost .- [37].
PURCHASE OF STOCK.
33. Stock is money borrowed by Government from
individuals, or contributed by merchants, &c., for the
purpose of trade, and bearing interest at a fixed, or
variable rate. It is transferable either entirely, or in
part, according to the pleasure of the owner.
If the price per cent, is more than £100, the stock in
question is said to be alom^ if less than £100, lelow " par."
Sometimes the shares of tradiu'r companies are only
gradually paid up ; and in many cases the whole price
of the share is not demanded at all — they may be £50,
£100, &:c., shares, while only £5, £10, &c., may have
been paid on each. One person may have many shares
When the intesest per cent, on the money paid ls con-
siderable, stock often sells for more than what it origi-
nally cost ; on the other hand, when money becomes
more valuable, or the trade for which the stock was
Contributed is not prosperous, it sells for less.
39. To find the value of any amount of stock, at any
rate per cent. —
Rule. — Multiply the amount by the value per cent.,
and divide the product by 100.
ExA^iPLE. — When £60^ will purchase £100 of stock, what
will purchase £042 ?
£042x001
— j-3--=£443 15.. lid.
It is evitlent that £100 of stock is to any other ambnnt of
it, as the price of the former is to that of tlie latter. Thixs
£100 : £612 : : £69'= : ^^_X-^
100
EXERCISES.
1. What must be given for £750 16.. in the 3 per
cent, annuities, when £64]- will purchase £100 .^ Ans.
£4S1 95. OVV^i-
266 EQUATION OF PAYMENTS.
2. What must be given for £1756 Is. ^d. India stock,
when <£196i will purchase dBlOO r Ans. £3446 17^. 'S\d.
3. What is the pui-chase of £9757 bank stock, a"i
£125|- per cent. ? Ans. £12257 45. l^d.
QUESTIONS.
1. What is stock } [38].
2. When is it ahovt., and when helow " par" ? [38] .
3. How is the value of any amount of stock, at auj
rate per cent., found ? [39].
EQUATION OF PAYMENTS.
40. This is a process by which we discover a time,
when several debts to be due at different periods may be
paid, at onre.^ without loss either to debtor or creditor.
lluLE. — Multiply each payment by the time which
should elapse before it would become due ; then, add
the products together, and divide their sum by the sum
of the debts.
ExAMPi.K 1. — A person owes another £20. payable in 6
months ; £50, payable in 8 months ] and £90, payable in
12 monthg. At what time may all be paid together, without
loss or gain to either party ?
£ £
20x G= 120
50 X 8= 400
90x12=1080
100 1(jO) 1600(10 the required number of mouths.
160
Example 2.— A debt of £450 is to be paid thus : £100
immediately, £300 in four, and the rest in six months. AVlien
BhoulJ it be paid altogether T
£ £
100 X 0= 0
300 X 4=1200
50 X 6= 300
450 450)1500(31 monthf?
1350
~150
EQUAliOiS" OF P.WMKNio. 207
41. We have (according to a principle formerly u^cd
[13]) reduced each debt to a sum which would bring the
same interest, iu one month. For G times £20, to he due
in 1 month, should evidently produce the same as X20, to
be due in 6 months — and so of the other debts. And the
interest of £1600 for the smaller time, -vrill just be equal to
the interest of the smaller sum for the larger time.
EXERCISES.
1. A owes B jeeOO, of which £200 is payable in 3
months, £150 in 4 months, and the rest in 6 months ;
but it is agreed that the whole sum shall be paid at
once. When should the payment be made r Ans. In
4^ months.
2. A debt is to be discharged in the following man-
ner : 1 at present, and i every three months after until
all is paid. What is the equated time .'' Aiis. 4^
months.
3. A debt of £120 will be due as follows : £50 in
2 months, £40 in 5, and the rest in 7 months. When
may the whole be paid together ? Ans. In 4} months.
4. A owes B £110, of which £50 is to be paid at
the end of 2 years, £40 at the end of 3^, and £20 at
the end of 4^ years. When should B receiye all at
once f Ans. In 3 years.
5. A debt is to be discharged by paying ^ in 3 months,
i- in 5 months, and the rest in 6 months. What is the
equated time for the whole ? Ans. 4^ months.
QUESTIONS.
1. What is meant by the equation of payments }
2. '\\ hat is the rule for discovering when money, to
be due at different times, may be paid at once .' [40].
2C8
SECTION VIII.
EXCHANGE, &c.
1. Exchange enables iis to find what amount of the
iiionej of one country is equal to a given amount of the
money of another.
Money is of two kinds, real — or coin, and imaginary —
or money of exchange, for which there is no coin ; as,
for example " one pound sterling."
The par of exchange is that amount of the money
of one country actually equal to a given sum of tho
money of another ; taking into account the value of
the metals they contain. The course of exchange h
that sum which, in point of fact, would be allowed
for it.
2. "When the course of exchange with any place is
above " par," the balance of trade is against that place.
Thus if Hamburgh receives merchandise from London
to the amount of £100,000, and ships off, in return, goods
to the amount of but £50,000, it can pay only half what
it owes by bills of exchange, and for the remainder must
obtain bills of exchange from some place else, giving
for them a premium — which is so much lo?.t. But the
exchange cannot be much above par, since, if the pre-
mium to be paid for bills of exchange is high, the
merchant will export goods at less profit ; or he will
pay the expense of transmitting aixl insuring coin, or
bullion.
3. The nominal value of commodities in these countries
was from four to fourteen times less formerly than at
present ; that is, the same airount of money would then
buy much more than now. We may estimate the value
of money, at any particular period, from the amount of
corn it would j)^ii'chase at that time. The value of
money fluctuates from ihc nature of the crops, the state
of trade, &o.
EXCHANGE.
209
la exchange, a variable is given for a fixed sura ; thus
London receives different values for £1 from different
countries.
Agio is the difference which there is in some places
between the cwrrent or cash moncj, and the cx/urnge
or haiik money — which is finer.
The following tables of foreign coins are to be mad^^
familiar to the pupil.
FOREIGN MONEY.
MONEY OF AMSTERDAM.
Flemish Money.
make 1 grote or penuy.
1 stiver.
Peanings
8
16 or
320
800
1920
grotes
40 or
100
stivers
20
50 or
210 120 or
guilders
2i
6
1 rixdollar.
1 pound.
rfcnnings
6
3I0NEY OF HAMBURGH.
Flemish Money.
grotes
12
72 or
1440
pfennings Pence
12 or 2
llings
make 1 grote or penny
1 skill! ng.
1 pound.
192
384
676
I skil
240 orf 20
Ha ?n h ii> rgh Mo ney .
make 1 schilling, equal to 1 stivor
1 mark.
32 or
64
96
schiilinga
16
marks
9
48
1 dollar of exchange.
1 rixdollar.
We find that 6 schillings=l skilling.
Hamburgh money is distinguished hy the word " Hambro."
" Lub," from Lubec, where it was coined, was formerly used
for this purpose ; thus, " one mark Lub."
Wo exchange with Holland and Flanders by the pound
Billing.
N
'^70 EXClIANfJE.
FRE.NCIJ MONKV.
Accounta were formerly kept ia livres, &c.
Dernicrs
12 . . . . . make 1 sou.
sous
'^lO or I 20 . . . 1 livre.
livres
720 I 60 or I 3 . . lecuorcrown
Accounts are now kept in francs and centimes.
Centimes
10 .... . make 1 decime.
decimes
100 or I 10 . . . . 1 franc.
81 Iivres=80 francs.
PORTUGUESE MONEY.
Accounts are kept in milrees and rees.
Tvces
400 ..... make 1 crusado.
" I criisados
1000 or 2k . . . . 1 milree.
4800 I 12 . . . . 1 moidere.
SPANISH MONEY.
Spanish money is of two kinds, plate and vellon ; the latter
being to the former as 82 is to 17. Plate is used in exchange
with us. Accounts are kept in piastres, and maravedi.
Maravedie
34
s
reals
8
make 1 real.
272 or
1 piastre or piece of eight
1088
375
1 piastres
32 or 1 4 .
1 pistole of exchange.
1 ducat.
AMERICAN MONEY.
In some parts of the United States accounts are kept in
dollars, dimes, and cents.
Cents
10 . .... make 1 dimei
dimea
100 or I 10 1 dollar.
In other parts accounts are kept in pounds, shillings, and
pence. These are called cun-ency, but they are of much lesi
-value than with ua, paper maaey being used.
E5CIIANGE
DAMSEl MONEY.
Ff-nnings
12
Ekillings
16 .
102 or
l' j marks
1152 jOG or I 6
6 Danisii^iy Hamburgh marks.
VENETIAN MONEY.
Dnnnii (the plural of denaro)
12 . . . , make 1 soldo.
soldi
20 .
lire soldi
121 or 1-6 4
:?7l
make 1 skilliu^.
1 mark.
1 riidollar
:10 or
l^RS
1920
Pfennings
4
1 lira.
160
AUSTRIAN MONEY.
210 or
SOO
Grains
10
croutzers
60 .
florins
90 or I Ih
NEAPOLITAN MONEY.
I carl ins
10
1 ducat current.
1 dut^t elFcctive
make 1 creutzer
1 florin.
1 rixdollar.
make 1 carlin.
1 ducat re^fA?
MONEY OF GENOA.
Lire soldi
4 and 12 make 1 scudo di cambio, or cro^rn of exchange.
10 and 14 1 scudo d'oro, or gol 1 crown.
OF GENOA AND LEGHORN.
Denari di pe/.za
12
soldi di pe/.za
210 or I 20
Denari di lira
12
I soldi di lira
240 or I 20
1380 lloorj 5|
SWEDISH MONEY.
Fcnnings, or oers
12 ....
make 1 soldo di pozza.
1 pezza of 8 reals,
make 1 soldo di lira.
Iskillings
43
1 lira.
1 pezza of 8 reals
make 1 skilling.
1 rixdollai
872 EXCHANGE.
RUSSIAN MONEY.
Copccs
100 . ^ . make 1 ruble.
EAST INDIAN MONEY.
Cowries
2560 . . . make 1 rupee.
E-upees
100.000 ... 1 lac.
10,000,000 . . 1 crore.
The cowrie is a small shell found at the Maldives, and near
Angola : in Africa about 5000 of them pass for a pound.
The rupeQ has different valijes : at Calcutta it is Is. ll^d.
the Sicca rupee is 2s. O^d. ; and the current rupee 2*. — if we
divide any number of these by 10, we change them to pounds
of our money; the Bombny rupee is 2s. od., &c. A sum of
Indian money is expressed as follows; 5 '38220, which means
6 lacs and ^8220 rupees.
4. To reduce bank to current money —
KrLE. — Say, as £100 is to iSlOO + the agio, so is
the given amount of bank to the requii'ed amount of
current money.
Example. — Ho"^ many guilders, current money, are equal
to 463 guilders, 3 stivers, and 13f| pennings banco, agio
ling 4^ 1
100
7
700
65
: 104f
733
IMultiplyi:
will p-ivfi
:: 463 g. 3 st. 13^.^ P- : "?
20
02G3 stivers.
10
45500
148221 pennings.
ag by 65, and adding 04 to the
96344-^0 product,
jMultiplying by 733
and dividing by 45500)7062036457
will give 155200 pennings.
16)155209
20)0700 9
And 485 g. 0 st. 9||fU p. is the amount sought.
5. We multiply the first and second terms by 7, and add tha
numerator of the fraction to one of tlie products. This is the
same thing as reducing these terms to fractious having 7 for
their denominator, and then multiplyijig them by 7 [Sec. V. 29].
For the same reason, and in tlie same way, we multiply the
first and third terms by 65, to banish the fraction, without
deitroying the pruportlou.
EXCHANGE. 273
The remainder of the process is according to the rule of
proportion [Sec. V. 31]. We reduce the ansvrer to pennings,
stivers, and guildeis. •
EXERCISES.
1. Reduce 374 guilders, 12 stivers, bank money, to
current money, agio being 44 per cent. ? Ails. 3l'2 g.,
5 St., SyW p-
2. Keduce 437S guilders, *^ stivers, bank money, to
current monov, agio being 4f per cent. ? Avs. 4577 g. ,
17st.,3/.VP-
3. Ileduce S73 guilders, 11 stivers, bank money, to
current money, agio being 4| per cent, f Aiis. 91G g.,
2 St., lUf P-
4. Reduce 1642 guilders, bank money, to current
money, agio beimj 4|i per cent. } Ans. 1722 g., 14st.,
lO/j p.
6. To reduce current to bank money —
Rule. — Say, as £100-^ the agio Ls to £100, so is tlie
given amount of current to tlie reqiiii^ed amount of
bank moae}'.
Example. — How much bank money is there in 485 guil-
ders aud ^-^ijiil pennings, agio being 4f'?
^. St. p.
104^. : 100 :: 4S5 0 911^5 - •
7 7-20
733 700 9700
45500 10
33351500 155200
Multiplying by 45500 the denominator,
7062009500
and addino; 25057 tlie numerator,
we get 7062035457
700
33351500)4043424810900
Quotient 14!
16) 148221 ff
20)0263
4C3 3 13^i is the amount sought
274 EXCHANGE.
EXERCISES.
5. Reduce 58734 gi^lders, 9 stivers, 1 1 penningfl,
current luoney, to Lank jnoiicy, agio being 4f per cent. .?
Ans. 560P6 g., 10 st.. 1 l^fi p. "
6. E educe 4326 guildt;rs, 15 punnings, current money,
to bank money, agio being 44 per cent. ? Ans. 4125 ff.,
13 St., 2.i|^ p.
7. Reduce 1186 guilders, 4 stivers, 8 pennings, cur-
rent, to bank money, agio being 4f per cent, t Ans
1136 g., 10 St., Oiff p."
8. Reduce 8560 guilders, S stivers, 10 pennings,
current, to bank money, agio being 4f per cent. .
./17Z5. 8183 g., 19 St., 5f if p.
7. To reduce foreign money to British, &c. —
Rule. — Put the amount of British money considered
in the rate of exchange as third term of the proportion,
its value in foreign money as first, and the foreig-u
money to be reduced as second term.
Example 1. — llemislt Money. — How much British money
is equal to 1054 guilders, 7 itivers, the exeliangc being 336'.
Acl flemish to £ L British ?
335. 4/. : 1054 g. 7 st. : : £1 : '?
12 20
400 pence. 21087 stivers.
400)42174 Flemish pence.
£105-435 = £105 85. ^<l.
£1, the amount of British money considered "in the rate,
is put in the third term ; 33.5. 4c/., its value in foreign money,
in the first : and 1054 g. 7 st., the money to be reduced,
in the second.
9. How many pounds sterling in 1680 guilders, at
335. 3d. Flemish per pound sterling ^ Ans. £\QH Ss.
10. Reduce 6048 guilders, to British' money, at 33.9.
lid. Flemish per pound, British .^ Ans. i,594 7s.
11. Reduce 2048 guilders, 15 stivers, to British
money, at 345. bd. Flemish per pound sterling ? Aiu
£198 85. G^l^d.
EXCHANGE. 27b
\U,. np\7 many pounds sterling in 1000 guilders, 10
stivers, exchange being at 33s. 4</. per p'oimd sterling }
Ans. i,10(j Is.
Example 2. — Hamburgh Jloneu. — How much British
money is equivalent to 476 marks. 'J skillings, the exchange
being 33i\ 6d. Flemish per pound British ?
s. d. m. 8.
33 6 : 476 Of : : £1 : ?
12 32 2
402 grotes. 15232-f 19'=15251l grotes.
402)152.511
£37-9386=X37 ISs. Od.
Multiplying the schillings by 2. and the marks by 32,
reduces both to pence.
13. How much Briti.sh money is equivalent, to 3083
marks, 12| schillings Hambro', at 32^. 4d. Flemish per
pound sterling .' Ans. £254 6s. Sd.
14. How much English money is equal to 5127 marks,
5 Schillings, Hambro' exchange, at 365. 2d. Flemish
per pound sterling r Ayis. £378 Is.
15. How many pounds sterling in 244^ marks, 9^
schillings, Hambro', at 32s. 6d. Flemish per pound ster-
ling .^ \4.ns. £200 10s.
16. Reduce 7S54 marks, 7 schillings Hambro', to
British money, exchange at 34s. 11^. Flemish per
poimd sterling, and agio at 21 per cent. .' Ans. £495
15s. 0^6?.
Example 3. — French Money. — Reduce 8G54 francs, 42
centimes, to Brirish money, the exchange bemg 23f., 50c.,
per £\ British.
f. c. f. c. 8C54-42
23 50 : 8C54 42 : : 1 : ^g.-^^=i=£358 bs. b\d.
42 centimes are 042 of a franc, since 100 centimes make
1 franc.
17. Reduce 17969 francs, 85 centimes, to British
money, at 23 francs, 49 centimes per pound sterling ^
Ans. £765.
18. Reduce 7672 francs, 50 centimes, to British
money, at 23 francs, 25 centimes per pound sterling ?
Ans. £330.
276 EXCHANGE.
19. E educe 15647 francs, 36 centimes, to_ British
money, at 23 francs, 15 centimes per j)Ouiid sterling .''
Alls. ^£675 ISs. 2ld. *
20. Reduce 450 francs, 5S^- centimes, to British
money, at 25 francs, 5 centimes per pound sterling ?
Alls. £\1Q 145.
Example 4. — Portuguese Money. — How much British
money is equal to 540 milrees, 420 rees, exchange being at
5s\ (jcl. per mikee T
m. m. r. 5. d.
1 : 540-420 : : 5 G : 540-420x5.s. 6d.=£US 12s. 3,^^.
In this case the British money is the variable quantity,
and 5s. (jd. is that amomit of it which is considered in the
rate.
The rees are changed into the decimal of a milree by
putting them to the right hand side of the decimal point,
since one ree is the thousandth of a milree.
21. In S50 milrees, 500 rees, how much British
money, at 55. 4d. per milree ? Ans. £226 IGs.
22. Reduce 2060 mili-ees, 380 rees, to English money,
at OS. 6^d. per milree ? Ans. ^£573 05. 10^^.
23. In 1785 milrees, 581 rees, how many pounds
sterlinf;:, exchange at 64|- per milree r Ans. £A7d
17s. 6d.
24. In 2000 milrees, at 55. S^d. per milree, how
many pounds sterling.? Ans. i£570 165. Sd.
ExAJiPLE 5. — Spanish Money. — Reduce 84 piastres, 6 reals,
19 maravedi, to British money, the exchange being 40(/. the
piastre.
p. p. r. m. d.
1 : 84 6 19 : : 49 : ?
8 8
8
34
678 reals.
34
>72
23052 mara^
49
'2)1129548
•edi,
EXCHANGE. 277
EXERCISES.
25. Reduce 2448 piastres to British money, exchange
at oOd. sterling per piastre .' Ans. ^510.
26. Reduce 30000 piastres to British money, at 40^.
per piastre .' Ans. i^oOOO.
27. Reduce l.'>25 piastres, G reals, 22|f-^ maravedi, to
British money, at :^9^d. per piastre ? Ans. £167 155. Ad.
Example G. — American Money. — Reduce 3765 dollars to
British money, at 4.s. per dollar. 45.=£l; therefore
5)3765 ' dol. dol. s. £
753 is the required sum. Or 1 : 37G5 : : 4 : 753
28. Reduce £'292 3<f. 2^d. American, to British money,
at 66 per cent. ? Ati'i. £176.
29. Reduce 5611 dollars, 42 cents., to British money,
at 4s. D^d. per dollar t Ans. £1250 175. 7d.
30. Reduce 274f5 dollars, 30 cents., to British money,
at 45. 3^^/. per dollar ? Ans. £589 C5. 2^.
From these examples the pupil will very easily under-
stand how any other Ivind of foreign, may be changed
to British money.
8. To reduce British to foreign money —
Rule. — Put that amount of foreign money which is
considered in the -rate of exchange as the third term,
its value in British money as the first, and the British
money to be reduCvid as the second term.
Example 1. — Flemis'i Money. — How many guilders, &:c.,
in i^23G 149. 21. British, the exchange being o-U. 2/1. Flemi.-h
to £1 Britisli ?
£ £ s. d. $. d.
1 : 23G U 2 : : 34 2 : T
20 20 12
20 4734 410 ponce.
12 12
i40 568 lOJ.
410
240)23202100
T2)97050-4, &c.
20)8087 6
jW04~7''(H Fhniik.
278 EXCHANGE.
We nil
^ht take
part
s for the 345. 2d.—
34.^
2d.-.
=£1 -f 105.-
f45.-f2J.
£
£
5.
d.
£1 =
= 1
236
14
2
105.=
- i
118
7
1
45.=
= 1
47
G
10
2d.=
'rh
(oVofi)l
19
H
£404
7
61 Flemish
EXERCISES.
31. In £100 Is.y how much Flemish money, exchange
at 335. 4d. per pound sterling ? Ans. 1000 guilders,
10 stivers.
3.2. Reduce £168 85. 5jl^d. British into Flemish,
exchange being 335. Sd. Flemish per pound sterling .''
Ans. 1680 guilders.
33. In £199 ll5. lO^^-^d. British, how much Flemish
money, exchange 345. ^d. per pound sterling } Ans.
2080 guilders, 15 stivers.
34. Reduce £198 85. Q\\^d. British to Flemish
money, exchange being 345. bd. Flemish per pound
sterling } Ans. 2048 guilders, 15 stivers.
Example 2. — Hamburgh Money. — How many marks, &c.,
in £24 65. British, exchange being 33.5. 2d. per XI British '?
£1 : £24 65. : : 335. 2d. : ]
20 20 12
20 486 398 grates.
398
20)193428
2)9671 8 pence.
16)4835 schillings, 1 penny.
302 marks, 3 schillings, 1 penny.
35. Reduce £254 65. Sd. English to Hamburgh
money, at 325. Ad. per pound sterling } VI715. 3083
marks, 12| stivers.
36. Reduce £378 I5. to Hamburg money, at 365
2d. Flemish per pound sterling } Ans. 5127 marks^
5 schillings.
37. Reduce £536 to Hamburgh money, at S65. Ad.
per pound sterling } Ans. 7303 marks.
EXCHANGE. 279
3S. Keduce £495 Ids. 0}d. to Hamburg currency,
at 345. 11^. per pound sterling ; agio at 21 per cent. ?
Ans. 7854 marks 7 schillings.
Example S. — French Moncij. — How much French money
is equal in value to £83 2>\ 2d.^ exchange being 23 francs
25 centimes per £1 British '?
£ £ s. d. f.
1 : 83 2 2 : : 23-25 : ?
20 20
20 1662
12 12
240 "19946
23-25
240)463744-50
19322-7, or 19322f. 70c. is the required sum
39. Reduce £274 5^. 9d. British to fi-ancs, &c., ex-
change at 23 francs 57 centimes per pound sterling r
Ans. 6464 francs 96 centimes.
40. In £765, how many francs, &c., at 23 franca
49 centimes per pound sterling .' Ans. 17969 franca
85 centimes.
41. Reduce £330 to francs, &c., at 23 francs 25 cen-
times per pound sterlijjg ? Ans. 7672 francs 50 cents.
42. Reduce £734 As. to French money, at 24 francs
1 centime per pound sterling.^ Ans. 1769 francs 42-}
centimes.
Example 4. — PoHuguese Money. — How many milrees and
rees in £32 65. British, exchange being bs. 9d. British pt
milree ?
s. d. £ s.
5 9 : 32 0 : : 1000 : %
12 20
69 646
12
7752
1000
,«4uired sum
69)7752000
112348 rees=112 milrees 348 rees, is tne
280 EXCHANGE.
43. Reduce £226 16s. to milrees, &c., at 5^. 4d. per
mili-ee .'' Ans. 850 milrees 500 rees.
44. Reduce £479 175. 6d. to milrees, &c., at 64^(1.
per milree r Ans. 1785 milrees 581 rees.
45. Reduce £570 16s. 8d. to milrees, &c., at 55. S^d.
per milree ? Ans. 2000 milrees.
46. Reduce £715 to milrees, &c., at 55. 8^. per mil-
ree ? Ans. 2523 milrees 529y\ rees.
Example 5. — Spanish Money. — How many piastres, &c.,
in £62 British, exchange being 50^/. per piastre '?
d. £
50 : 62 : : 1 : ?
20
1240 p. r. m.
22 297 0 32|f , is the required sum.
50)14880
297*6 piastres.
48 reals.
34
50)16
_32
32|| maravedis.
-ii
t7. How many piastres, &c., shall I receive for £510
sterling, exchange at 50d. sterling per piastre ? Ans.
2448 piastres.
48. Reduce £5000 to piastres, at 40<^. per piastre ?
Ans. 30000 piastres.
49. Reduce £167 155. 4d. to piastres, &:c., at S9\d.
per piastre ? Ans. 1025 piastres, 6 reals, 22|4-f mara-
vcdis.
50. Reduce £809 95 8^. to piastres, &c., at 40f J. per
piastre ? Ans. 4767 piastres, 4 reals, 2^^-^ maravedis.
Example 6. — American Money. — Reduce £170 British to
Aiuerican currency, at 66 per cent.
£ £ £
100 : 176 :: 100 : :
106
100)29216
£292 35. 2» J., is the required sum.
EXCHANGE. 281
EXERCISES.
61. Reduce £753 to dollars, at 45. per dollar ? Ans.
3765 dollars.
b2. Reduce £532 4ts. Sd. Britisli to American money,
at 64 per cent. } Ans, £872 175. 3d.
53. Reduce £1250 175. Id. sterling to dollars, at
45. b^d. per dollar ^ Ans. 5611 dollars 42 cents.
54. Reduce £589 Qs. 2^-^d. to dollars, at 45. 3^^.
per dollar } Ans. 2746 dollars 39 cents.
55. Reduce £437 Britisli to American money, at 78
per cent. } Ans. £777 175. 2^d.
9. To reduce florins, &c., to pounds, &c., Flemish —
Rule. — Divide the florins by 6 for pounds, and —
adding the remainder (reduced to stivers) to the stivers
— divide the sum by 6, for skillings, and double the
remainder, for grotes.
Example. — How many pound.s, skillings, and grotes, in
1G5 florins 19 stivers ?
f. St.
6)165 19
£27 13s. 2rf., the required sum.
6 will go into 165, 27 times — leading 3 florins, or 60 stivers,
which, with 19, make 79 stivers ; 6 will go into 79, 13 times —
leaving 1 ; twice 1 are 2.
10. Reason- of the Rule. — There are 6 times as many
florins as pounds ; for we find by the table that 240 grotes
make £1, and that 40 (^f) grotes make I florin. There are
6 times as many stivers as skilling-^ ; since 96 pennings make
1 skilling, and 16 {^^) pfennings make one stiver. Also* since
2 grotes make one stiver, the remaining stivers are equal to
twice as many grotes.
Multiplying by 20 and 2 would reduce the florins to grotes ;
and dividing the grotes by 12 and 20 would reduce them to
pounds. Thus, using the same example —
f. St.
165 19
20
3319
2
12)6638 .
20)553 2
£27 Vos. 2i/., as before, is the result.
282 EXCHANGE.
EXERCISES.
56. In 142 florins 17 stivers, how many pounds, &c.,
Atis. £23 165. 2d.
57. In 72 florins 14 stivers, how many pounds, &c.,
Ans. £12 2s. 4d.
58. In ISO florins, how many pounds, &c. ? Ans. £30
11. To reduce pounds, &c., to florins, &c. —
Rule. — Multiply the stivers by 6 ; add to the product
half the number of grotes, then for every 20 contained
in the sum carry 1, and set down what remains above
the twenties as stivers. Multiply the pounds by 6, and,
adding to the product what is to be carried from the
stivers, consider the sum as florins.
Example. — How many florins and stivers in 27 pounds,
13 skillings, and 2 grotes 1
£ s. d.
27 13 2 .
6
IGofl. 19st., the required sum.
6 times 13 are 78, which, with half the number (f ) of
grotes. make "^9 stivers — or 3 florins and 19 stivers (3 twenties,
and 19) ; putting down 19 we carry 3. 6 times 27 are 162,
and the 3 to be carried are 165 florins.
This rule is merely the converse of the last. It is evident
that multiplying by 20 and 12, and dividing the product by 2
and 20, would give the same result. Thus
£ s. d.
27 13 2
20
553
. 12
2)6638
20)33T9_
16511. 19st., the same result as before.
EXERCISES.
59. How many florins and stivers in 30 pounds, 12
skillings, and 1 grote } Ans. 183 fl., 12 st., 1 g.
60. How many florins, &c., in 129 pounds, 7 skil-
lings .? Ans. 776 fl. 2 st.
61. In 97 pounds, 8 skillings, 2 grotes, how many
florins, &c. t Ans. 584 fl. 9 st.
ARBITRATIOX OF EXCHANGES. 283
QUESTIONS.
1. What is exchange ? [1].
2. What is the difference between real and imagin-
ary money? [1].
3. What are the jpar and course of exchange ? [1].
4. What is agio 1 [3] .
5. AVhat is the difference between current or cash
noney and exchange or bank money ? [3] .
6. How is bank reduced to current money ? [4].
7. How is current reduced to bank money ? [6] .
8. How is foreign reduced to British money ? [7].
9. How is British reduced to foreign money ? [8].
10. How are florins, &,c., reduced to pounds Flemish,
fee? [9].
11. How are pounds Flemish, &g., reduced to florins,
fcc. ? [11].
ARBITRATIOX OF EXCHANGES.
12. In the rule of cxdiange only two places arc con-
ce.med ; it may sometimes, however, be more beneficial
10 the merchant to draw through one or more other
places. The mode of estimating the value of the money
of any place, not drawn directly, but through one or
more other places, is called the arbitration of exchanges,
and is either simjph or compound. It is " simple "
when there is only one intermediate place, " compound "
when there are more than one.
All questions in this rule may be solved by one or
more proportions.
13. Simple Arbitration of Exchanges. — Given the
course of exchange between each of two places and
a third, to find the par of exchange between the
former.
KuLE. — Make the given sums of money belonging to
the third place the first and second terms of the propor-
tion ; and put, as third term, the equivalent of what is
in the first. The fourth proportional will be the value
of what is in the second term, in the kind of meney
contained in the third term.
284 ARBITRATION OF EXCHANGES.
Example. — If London exchanges with Paris at lOd. per
franc, and -witli Amsterdam at 34s. 8t/. per £,1 sterlinj^. what
ought to be tho course of exchange, between Paris and
AuiSterdam, that a merchant may without lo.-^3 roniit-from
London to Amsterdam through Paris '?
£1 : 10<:/. : : 34.s. Sd. (the equivalent, in Flemish money,
of £1) : 1 the equivalent of iOd. British (or of a franc) in
Fleinish money.
oil o J V 1 0
Or 240 : 10 : : 34.s. 8d. : ^40 =^''¥-^ ^^^e re-
quired value of 10 J. British, or of a franc, in Flemish money.
£1 and 10:/. are the two given sums of English money, or
that which belongs to the third place; and 345. Sd. is the
given equivalent of £1.
It is evident that, llid. (Flemish) being the value of lOd.,
the equivalent in British money of a franc, when more than
llid. Flemish is given for a frnnc, the merchant will gain if
he remits througli Paris, since he will thu.s indirectly receive
\nore than I7^d. for lOd. sterling — that is, more than its equi-
valent, in Flemish money, at the given course of exchange
between London and Amsterdam. On the other hand, if less
than 17id. Flemish is allowed for a franc, he will lose by
remitting though Paris ; since he will receive a franc for lOd
(British) ; but he will not i-eceive Hid. for the franc : — while,
had he remitted lOd., the value of the franc, to Amsterflam
directly, he would have been allowed Hid.
EXERCISES.
1. If the exchange between London and Amsterdam
is 335. 9d. per pound sterling iind the exchange be-
tween London and Paris 9^d. per franc, what is the
par of exchange between Amsterdam and Paris ? Ans.
Nearly 16d: Flemish per franc.
2. London is indebted to Pctersburgh 5000 rubles ;
while the exchange between Petersburgh and London
is at oOd. per ruble, but between Petersburgh and
Holland it is at 90^. Flemish per ruble, and Holltmd
and England at 365. 4d. Flemish per pound sterling.
Which will be the more advantageous method for Lon-
don to be drawn upon — the direct or the indirect ? Ans.
London will gain £i) 11.?. lj%\d. if it makes payments
by way of Holland.
5000 rubles=£l041 13*. Ad. British, or £1875 Flemish,
but £1875 Flcmish=£1032 2s. 2i^^d. British.
ARBITRATIOX OF EXCHANGES. 2S5
14. Compound Arhif. ration of Exchanges. — To find
what should be the course of exchange between two
places, through two or more others^ that it may be on a
par with the course of exchange between the same two
places, diredly —
Rule. — Having reduced monies of the same kind to
the same denomination, consider each course of exchange
as a ratio ; set down the difierent ratios in a vertical
column, so that the antecedent of the second shall be
of the same kind as the consequent of the first, and the
antecedent of the third, of the same kind as the conse-
quent of the second — putting down a note of interroga-
tion for the unknown term of the imperfect ratio. Thei*
divide the product of the consequents by the product of
the antecedents, and the quotient will be the value of the
given sum if remitted thi-ough the intermediate places.
Compare with this its value as remitted by the direct
exchange.
15. Example. — £824 Flemish being due to me at Am-
sterdam, it is remitted to France at IGa. Flemish per frane;
from France to Venice at 300 francs per 60 ducats : from
Venice to Hamburgh at lOOcZ. per ducat : from Hamburji;h
to Lisbon at 5 Or/, per 400 rees ; and from Lisbon to En, gland
at 5*. 8J. sterling per milree. Shall I gain or lo.?e. and how
much, the exchange between England and Amst<}rdam being
345. -id. per £\ sterling ?
15'^/. : 1 franc.
300 francs.: 60 ducats.
1 ducat : 100 pence Flemish.
50 pence Flemish : 400 rees.
1000 rees : 68 pence British.
-? : £824 Flemish.
1x00x100x400x68x824 ...
>— -; — — — — - — — — -— — — =(ii we reduce the terma
loxoOOxlXoOxlOOO ^
17xS'^4
[Sec. V. 47]) ,^, =£560 6s. 4iJ.
But the exchange between England and Amsterdam f«.f
£824 Flemish is £480 sterling.
Since 345. 4cZ. : £824 : : £1 : ^^'^^ =£480
34.^\ 4cf. ^
I gain therefore by the circular exchange £560 6^;, 4|<i'.
minus £480=£80 65. 4W.
286 ARBITRATION OF EXCHANGES.
If commission is cliarged in any of tlie places, it must
bo deducted from the value of the sum which can bo
obtained in that place.
The process given for the compound arbitration of ex-
change may be proved to be correct, b}' putting down the
different proportions, and solving them in succession. Thus,
if lijd. are equal to 1 franc, wha,t will 300 francs (=00
ducats) be worth. If the quantity last found is the value of
GO ducats, what will be that of one ducat (=100d.), &c. 1-
EXERCISES.
3. If London would remit £1000 sterling to Spain,
the direct exchange being 42^d. per piastre of 272
maravedis ; it is asked whether it will be more profit-
able to remit directly, or t-o remit first to Holland at
35.S. per pound ; thence to France at I9^d. per franc ;
thence to Venice at 300 francs per 60 ducats ; and
thence to Spain at 360 maravedis per ducat ? Ans.
The circular exchange is more advantageous by 103
piastres, 3 reals, 19|^ maravedis.
4. A merchant at London has credit for 6S0 piastres
at Leghorn, for which he can draw directly at 50^. per
piastre ; but choosing to try the circular way, they are
by his orders remitted first to Venice at 94 piastres per
1 00 ducats ; thence to Cadiz at 320 maravedis per
ducat ; thence to Lisbon at 630 rees per piastre of 272
maravedis ; thence to Amsterdam at 5 W. per crusade
of 400 rees ; thence to Paris at lS|f/. per franc ; and
thence to London at lOl-d. per franc ; how much is the
circular remittance better than the direct di-aft, reckon-
ing i per cent, for commission ? Ans. £14 125. 7\d
16. To estimate the gain or loss per cent. —
lluLE. — Say, as the par of exchange is to the course
of exchange, ao is £100 to a fourth proportional. From
this subtract £100.
Example. — The par of exchange is found to be \Sld.
Flemish, but the course of exchauge is 19J. per franc j
what is the gain per cent. 1
ISUL : m. :: £100 : — V.^--=X104 7*. llcZ.
18i
PROFIT AND L0S5. 287
Tliu.s the gaiu per cent.=:C104 7s. lid. minus £100=
X4 7s. lid. if the merchant remits through Paris.
If in reuiitting through Paris commission must be
paid, it is to be deducted from the gain.
EXERCISES.
5. The par of exchange is found to be I8^d. Flemish,
but the course of exchange is IQ^t/., whatis the gain per
cent. ? Ans. £4 ISs. 2\d.
6. The par of exchange is ]7fJ. Flemish, but the
course is ISffZ., what is the gain per cent. ? Ans. £4
6s. 11 ifZ.
7. The par of exchange is IS^d. Flemish, but the
com'se of exchange is 17|f fZ., what is the loss per cent. .'*
Ans. £1 165. 2d.
QUESTIONS,
1. What is meant by arbitration of exchanges ? [12].
2. What Ls the difference between simple and com-
pound arbitration ? [12].
3. What is the rule for simple arbitration ? [13].
4. What is the rule for compound arbitration ? [14].
5. How are we to act if commission is charged in
any place ? [15].
6. How is the gain or loss per cent, estimated } [16].
PROFIT AND LOSS.
17. This rule enables us to discover how much we
gain or lose in mercantile transactions, when we sell at
certain prices.
Given the prime cost and selling price, to find the
gain or loss in a certain quantity.
Rule. — Find the price of the goods at prime cost
and at the selhng price ; the difiference wiU be the gain
or loss on a given quantity
Example. — What do I gain, if I buy 460 ft) of butter at
GJ., and sell it at 7d. per ft) '?
The total prime cost is 460(7. x6=2760f/.
The total selling price is 4C,Od.x7=2>220d.
The total gain is Z220d. minus 27G0cZ.=460(/.=Xl 18s. 4d.
28S PROFIT AND LOSS.
EXERCISES.
1. Boiiglit 140 ft) of butter, at 10^. per ib, and sokl
it at 146?. per ft) ; what was gained ? yhis. £2 6s. iyd.
2. Bought 5 cwt., 3 qrs., 14 ft) of cheese, at £2 \2s.
per cwt., and sold it for £2 185. per cwt. What was
the gain upon the whole ? Ans. £1 Ids. Sd.
3. Bought 5 cwt., 3 qrs., 14 ft) of bacon, at 345. per
cwt., and sold it at 365. 4d. per cwt. What was the
gain on the whole .'' Ans. ISs. 8^d.
4. If a chest of tea, containing 144 ft) is bought
for 6s. Sd. per ft), what is the gain, the price received
for the whole being ^£57 105. ? Ans. £9 lOs.
18. To find the gain or loss per cent. —
Rule.: — Say, as the cost is to the selling price, so is
dBlOO to the requu-ed sum. The fourth proportional
minus £100 will be the gain per cent.
Example 1. — What do I gain per cent, if I buy 1460 R)
of beef at 2>d.^ and sell it at 2>Ul- per ib '^
3f/.xl4G0=4380J., is the cost price.
And 3i(/.xl4G0=5110rf., is the selling price.
Then 4380 : 5110 :: 100 : ^^g^j — =£116 135. 4d.
Am. £116 139. 4d. minus £100 (=£16 135. 4J.) is the gain
per cent.
Eeasoiv of the Rule. — The price is to the price plus the
gain in one case, as the price (£100) is to the price plus the
gain (£100-t-the gain on £100) in another.
Or, the price is to the price plus the gain, as any multiple
or part of the former (£100 for instance) is to an equimultiple
of the latter (£l00-|-the gain on £100).
ExAMPLK 2. — A person sells a horse for £40, and loses 9
per cent., while he should have made 20 per cent. AVhat is
his entire loss '?
£100 minus the loss, per cent., is to £100 as £40_(what
the horse cost, minus what he lost by it) is to what it cost.
91 : 100 : : 40 : — j^— =£43 195. 1^., what the horse cost.
But the person should have gained 20 per cent., or |
of the price ; therefore his profit should have been
«3_105^2M-^£ij 15,, 9:,,,. A„d
PROFIT AXD LOSS. 2S9
£ ,t. (I.
3 19 1.^ is the difference Letween cost and selling price.
8 15 9} is what be should have received ahove cost.
12 14 11^ is his total loss.
f:XERCiSES.
5. Bought beef at 6d. per ft), and sold it at 8^.
What what was the gain per cent. : Ans. 33|.
6. Bought tea for' os. per lb, and sold it for 3s.
What was the loss per cent. ? Aiis. 40,
7. If a pound of tea is bought for 6s. 6d., and sold
for 7s. 4^/., what is the gain per*cent. ? Ans. 12f|.
S. If 5 cvrt., 3 qrs., 26 lb, are bought for £9 85.,
and sold for £\l 185. ll(f., how much is gained per
cent. } Ans. 27 j^-.
9. When wine is bought at 175. 6d. per gallon, and
sold for 27.V. 6^/., what is the gain per cent. } Aiis. 57-i.
10. Bought a quantity of goods for i260, and sold
them for £7o ; what was the gain per cent. ? Ans. 25.
il. Bought a tun of wine for £dO, ready money, and
sold it for £.54 IOj., payable in 8 months. How much
per cent, per anauni is gained by tliat rate .? Ans. \S^.
12. Having sold 2 yards of cloth for ll5. 6^., I
gained at the rate of 15 per cent. What would I have
gained if I had sold it for 12s. r Ans. 20 per cent.
13. If when I sell cloth at 75. per yard, I gain 10
per cont. ; what will I gain per cent, when it is sold for
85. f}d. } Ans. £33 ll5. b^d.
'7s. : 8.S'. 6(1. ■: XllO : £133 Us. 5ld. And £133 II5.
5]//. — £100:=£33 Us. 5ld.y is the required gain.
19. Given the cost price and gain, to find the selling
price —
EuLE. — Say, as £100 is to £100 plus the gain per
cent., so. is the cost price to the requii'cd selling price.
Example. — At what price per yard must I sell 427 yards
of cloth which I bought fijr 195. per yard, so that I may
gain S per cent. 1
108x195.
100 : 108 :: 10... : — ^^,^=£1 Os. Old.
This result may be proved by the la;^t rule.
290 PROFIT AND LOS?.
EXERCISES.
14. Bought velvet at 4s. 8d. per yard ; at what price
tnust I sell it, so as to gain 12^- per cent. ? Ans. At
bs. 3d.
15. Bought muslin at os. per yard ; how must it he
sold, that I may lose 10 per cent. } Ans. At 4s. 6d.
16. If a tun of hrandy costs £40, how must it be
sold, to gain 6^ per cent. ? Ans. For i£42 lOs.
17. Bought hops at £4 165. per cwt. ; at what rate
must they be sold, to lose 15 per cent. ? Ans. For £4
Is. 7ld.
IS. A merchant receives ISO casks of raisins, which
stand him in I65. each, and trucks them against other
merchandize at 28^. per cwt., by which he finds he has
gained 25 per cent. ; for what, on an average, did he sell
each cask f Ans. 80 lb, nearly.
20. Given the gain, or loss per cent., and the selling
price, to find the cost price —
Rule. — Say, asiSlOO plus the gain (or as i2100 minus
the loss) is to £100, so is the selling to the cost price.
Example 1. — If I sell 72 lb of tea at 6.s. per lb, and gain
9 per cent., what did it cost per lb '?
£100x0
109 : 100 : : G : — ^^^^=5s. (Jd.
What produces £109 cost £100 ; therefore wliat pro-
duces Gs. must, at the same rate, cost 5^. 6^/.
Example 2. — A merchant buys 97 casks of butter at o0.<.
each, and selling the butter at £4 per cwt., makes 20 per
cent. ; for how much did he buy it per cwt. 1
30.s\x97=2910.s., is the total price.
*^910y y I'^O
Then 100 : 120 : : 2910 : -- — -j^ — ^=34929., the
3492^^/ 3492s. N ,, ,_ ,
Belling price. And -g^(^=— ^- ^=43Go, is the number
of cwt. ; and n77-=50|^i lb, is the uverage weight of each
cask.
lb lb s. 1 1 o V q
Then 50m : 112 : : 30 : ibL><^=GG;. M. = £^ 6s.
oOlSj
8(i., the required cost pricC; per cwt.
FELLOWSHIP. 291
EXERCISES.
19. Raying sold 12 yards of cloth at 205. per yard,
and lost 10 per cent., what was the prime cost .^ Ans.
22s. 2%d.
20. Having sold 12 yards of cloth at 20^. per yard,
and gained 10 per cent., what was the prime cost .'' Ans.
185. 2yV/.
21. Having sold 12 yards of cloth for £5 14^., and
gained S per cent., what was the prime cost per yard .^
Ans. Ss. y^^.
22. For vi^hat did I buy 3 cwt. of siigar, which I
sold for i^6 35. , and lost 4 per cent. } Ans. For £Q
Ss. nd.
23. For what did I buy 53 yards of cloth, which I
sold for £25, and gained £>b \0s. per cent. } Ans. For
£23 135. 111^.
QUESTIONS.
1. "What is the object of the rule ? [17].
2. Given the prime cost and selling price, how ia
tiie profit or loss found .' [17].
3. How do we find the profit or lo.ss per cent } [18].
4. Given the prime cost and gain, how is the selling
price found .' [19].
5. Given the gain or loss per cent, and selling price,
how do we find the cost price } [20] .
FELLOWSHIP.
21. This rule enables us, when two or more persons
a-e joined in partnership, to estimate the amount of
profit or loss which belongs to the share of each.
Fellowship is either single (simple) or double (com-
pound). It is single, or simple fellowship, when thd
d.iffereut stocks have been in trade for the saim time.
It is double, or compound fellowship, when the different
^ stocks have been employed for different times.
This rule also enables us to estimate how much of a
bankrupt's stock is to be given to each creditor.
293 FELLOWSHIP.
22. Single Felloicship. — Rule. — Say, as the wliolo
stock is to the whole gain or loss, so is each person's
contribution to the gain or loss v/hich belongs to him.
Example. — A put £720 into trade, B £340, and C
£9G0 ; and they gained £47 by the traffic. AVhat is B'a
share of it 1
£
720
S40
9G0
£47x340
2020 : £47 : : £340 : ~~202()~""^^ ^^^' ^^
Each persons gain -or loss must evidently be proportionai
to his contribution.
EXERCISES.
1. B and C buy certain merchandizes, amounting
to £S0, of which B pays £30, and C £dQ ; and they
gain £20. How is it to be divided ? Ans. B £7 10s. ,
andC £12 lOs.
2. B and C gain by trade £182 ; B put in £300,
and (J £400. What is the gain of each r Ans. B £78,
and C J^104.
3. 2 persons are to share £100 in the proportions
of 2 to B and 1 to C. AYhat is the share of each ?
A'fis. B £66|, C £33|-.
4. A merchant failing, owes to B £500, and to G
£900; but has only £1100 to meet these demands.
How much should each creditor receive ? Aiis. B £392-^,
and C £707|.
5. Three merchants load a ship with butter ; B
gives 200 casks, C 300, and J) 400 ; but when they are
at sea it is found necessary to throw ISO casks over-
board. How much of this loss should fall to the share
of each merchant ? Ans. B should lose 40 casks, C
60, and D SO.
6. Three persons are to pay a tax of £100 accord-
ing to their estates. B's yearly pro,perty is £800, C's
£600, and D's £400. How much is each i^erson's share ?
Ans. B's is £44-^, C's £33^, and D\s £22|.
7. Divide 120 into three such parts as shall be to
each other as 1,2, and 3 } Ans. 20, 40, and GO.
FELLOWSHIP. 203
^. A' sliip worth £900 is entirely lost ; } of it be-
Wged to B, i to C, and the rest to D. What should
be the loss of each, £540 being received as insurance ?
Ans. B £45, C £90, and D £225.
9. Three persons have gained £1320 ; if B were to
take £G, C ought to take £4, and D £2. What is each
person's share"? A^is. B's £660, C's £440, and D'a
£220.
10. B and C have gained £600 ; of this B is to
have 10 per cent, more than C. How much will each
receive r Ans. B £314f , and C £2S54.
11. Three merchants form a company; B puts in
£150, and C £260 ; D^s share of £62, which they gained,
comes to £16. How much of the gain belongs to B,
and how much to C ; and what is D's share of the stock ?
Ans. B's profit is £16 16^. 7^\d.^ C's £29 3s. 4^^d. ;
and J) put in £142 12^. 222-^.
12. Three persons join ; B and C put in a certain
stock, and D puts in £1090 ; they gain £110, of which
B takes £35, and G £29. How much did B and C put
in ; and what is B's share of the gain .' Ans. B put
in £S29 6s. llii^., C £687 3^'. o^^d. ; and B's part of
the profit is £46.
13. Three farmers hold a farm in common ; one pays
£97 for his portion, another £79, and the third £100.
The county cess on the farm amounts to £34 ; what is
each person's share of it .' Ans. £11 18^. ll^f^. ; £9
14s. 7^d. ; and £12 6s. A^^d.
23. Compound Fellowship. — Bule. — Multiply each
person's stock by the time during which it has been in
trade ; and say, as the sum of the products is to the whole
gain or loss, so is each person's product to his share of
the gain or loss.
Example. — A contributes £30 for 6 months, B £84 for
11 months, and C £96 for 8 months; and they lose £14,
What is C's share of this loss ?
30 X 6=180 for one month. )
84x11=024 for one month. [ =£1872 for one month.
90 X 8=708 for one month. )
1872 : £14 : : £708 : ^^^^'^^ =£0 Is. 4W., C's bhare
1872 ^
294 FELLOWSIJiP.
•
24. Rkason of the Rulk. — It is clear that £30 cotflributed
for 6 months are, as far as the gain or the loss to be derived
from it is concerned, the same as 6 times £30— or £180 con-
tributed for 1 month. Hence A"s contribution may be taken
as £180 for 1 month ; and, for the same reason, B's as £924
for tlie same time ; and C"s as £7G8 also for the same time
This reduces the question to one in simple fello-wship [22].
EXERCISES.
14. Three merchants enter into partnership ; B puts
in £89 05. for 5 months, G £'92 15^. for 7 months, ana
J) £38 105. for 11 months; and they gain £86 165.
What should be each person's share of it ? Am. B's
£25 105., C's £37 2s., and B's £24 45.
15. B, C, and D pay £40 as the j^ear's rent of a farm.
B puts 40 cows on it for 6 months, C 30 for 5 months,
and J) 50 for the rest of the time. How much of the
rent should each person pay } Ayis. B £21y\, C £13/y,
and D £4^^.
16. Three dealers, A, B, and C, enter into partnership,
and in a certain time make £291 135. 4d. A's stock,
£150, was in trade 6 months ; B's, £200, 3 months ; and
C's, £125, 16 months. "What is each person's share of
the gain } Aiis. A's is £75, B's £50, and C'5 £166
135. 4d.
17. Three persons have received £665 interest ; B
had put in £4000 for 12 months, C £3000 for 15 months,
and D £5000 for 8 months ; how much is each person's
part of the interest.^ Am. B's £240, C's £225, and
B's £200.
18. X, Y, and Z form a company. X's stock is in
trade 3 months, and he claims Jj of the gain ; Y's
stock is 9 months in trade ; and Z advanced £756 for
4 months, and clahns half the profit. How much did
X and Y contribute ? Ans. X £168, and Y £280.
It follows that Ys gain was /_. Then i : ^ : : i:756x4 :
504=Xs product, which, being di^ided by his number of
months, will give XI 08. as his contribution. Ys share of
the stock rnay be foimd in the same way.
19. Three troops of horse rent a field, for which they
pay £80 ; the fii'st sent into it 56 horses for 12 days, tho
FELLOWSl^JIP. 295
second 64 for 15 days, and tlic third SO for IS days.
What mast each pay ? A/is. The first must pay £17
105., the second JL;25, and the third ii;37 105.
20. Three merchants are concerned in a steam vessel ;
the first, A, puts in £240 for 6 months ; the second, 13,
a sum unknown for 12 months ; and the third, C, £160,
for a time not known when the accounts were settled. A
received £300 for his stock and profit, B £600 for his,
and C £260 for his ; wliat was B's stock, and C's time ?
Ans. B's stock was £400 ; and C's time was 15 months.
If £300 arise from £240 in 6 months, £000 (B's stock and
profit) will be found to arise from £iOO (B's stock) in 12
months.
Then £400 : £1G0 :: £200 (the profit on £400 in 12
months) : £80 (the profit on £100 in 12 months). And £160-f
80 (£1(30 with its i-rofit for 12 months) : £260 (£160 with
its profit for some other time) : : 12 (the number of months
in the one case) : if,- . „j^ (the number of months in the other
ibU-j-bU
caso)=13, tlie niunber of months reqxzired to produce the
difference between £160, C's stock, and the £260, wiiich he
received.
21. In the forecroing question A's gain was £60
daring 6 months, B's £200 during 12 months, and C'a
£100 during 13 months; and the sum of the- products
of their stocks and times is 8320. What were their
stocks ? Ans. A's was £240, B's £400, and C's £160.
22. In the same question the sum of the stocks ia
£800 ; A's stock was in trade 6 months, B's 12 months,
and C's 15 months ; and at the settling of accounts,
A is paid £60 of the gain, B £200, and C £100.
What was each person's stock t Ans. xi's was £240,
B's £400, and C's £160.
QUESTIONS.
1. What is fellowship .' [21].
2. What is the diflerence between single and double
fellowship ; and are these ever called by any other
names.' [21].
3. What are the rules for single, and double fellow-
ship ? [22 and 23].
296 BARTER.
BAETER.
25. Barter enables the merchant to exchange one
commodity for another, without either loss or gain.
Rule. — Find the price of the given quantity of one
kind of merchandise to be bartered ; and then ascertain
how much of the other kind this price ought to pur-
chase.
ExA>[rLE 1. — How much tea, at 8.?. per lb, ought to be
vlvcn for 3 cwt. of tallow, at £1 IGs. Sd. per cwt. 7
£ s. d.
1 16 8
5 10 0 is the price of 3 cwt. of tallow.
And £5 106\-r-8.5.=13|, is the number of pounds of tea
tvhich £5 10s.. the price of the tallow, would purchase.
There must be so many pounds of tea, as will be equal to
the number of times that 8s. is contained in the price of the
tallow.
Example 2. — I desire to barter 96 lb of sugar, which
cost me Sd. per lb, but which I sell at 13J., giving 9
months' credit, for calico which another merchant sells for
lid. per yard. gi\"ing 6 months' credit. How much calico
ought 1 to receive ?
I first find at what price I could sell my sugar, were I to
give the same credit as he does —
If 9 months give me 5t/. profit, wha1; ought 6 months to
give 1
9 : 6 : : 5 : ^=J^=3.kZ.
0 9 '
Hence, were I to give 6 months' credit, I should charge
U^d. per lb. Next—
As my selling price is to my buying price, so ought his
soiling to be to his buying price, both giving the same credit.
in- : 8 : : 17 :5^=12J.
11}
The price of my sugar, tliorerorc, is 96><8J., or 7G8<7. ;
ind of his cahco, V2d. per yard.
Hence YV=G4, is the required number of yards.
BARTER 397
EXERCISES,
1 . A merchant has 1200 stones of tallow, at 2s. S^d.
the stone ; B has 110 tanned hides, weight 3994 lb, at
o}d. the ft) ; and they barter at these rates. How much
money is A to receive of B, along with the hides ? Ans.
£40 ll5. 2hL
2. A has silk at 14s. per ft) ; B has cloth at 12-s, 6d.
which cost only 105. the yard. How much must A charge
for his silk, to make his profit equal to that of B r Ans.
I7s. 6d.
3. A has coffee which he barters at lOd. the lb more
than it cost him, against tea which stands B in 10.s\,
but which he rates at 125. 6d. per. ft). How much did
the coffee cost at first } Ans. 3s. Ad.
4. K and L barter. K has cloth worth Ss. the yard,
which he barters at 95. 3«. with L, for linen cloth at
3.V. per yard, which is worth only 25. Id. Who has the
advantage ; and how much linen does L give to K, for
70 yards of his cloth .- Ans. L gives K 215f yards ;
and L has the advantage.
5. B has five tons of butter, at £2o 105. per ton, and
101 tons of tallow, at £33 155. per ton, which he barters
with C ; agreeing to receive £150 l5. Qd. in ready
money, and the rest in beef, at 2l5. per barrel. How
many barrels is he to receive i Ans. 316.
6. I have cloth at S^^. the yard, and in barter charge
for it at 13<;7., and give 9 months' time for payment;
another wierchant has goods which cost him 12^. j)er
lb, and with v/hich he gives 6 months' time for payment.
How high must he charge his goods to make an equal
barter .' Ans. At \ld.
7. I barter goods which cost ^d. per ft), but for
which I charge 13c?., giving 9 months' time, for goods
which are charged at 17^., and with which 6 months'
time are given. E.equu-ed the cost of what I receive }
Ans. \2d.
8. Two persons barter ; A has sugar at S^. per lb,
charges it at 13^., and gives 9 months time ; B has
at \2d. per lb, and charges it at 17f/. per lb. How
time must B give, to make the barter equal .'*
6 months.
298 ALLIGATION.
QUESTIONS.
1. What is barter? [25].
2. What is the rule for barter > [25] ,
ALLIGATION.
.26. This rule enables us to find what mixture -will be
produced bj the union of certain ingredients — and then
it is called alligation medial; or what ingredients will
be required to produce a certain mixture — when it is
termed alligation alternate ; further division of the
subject is unnecessary : — it is evident that any change
in the amount of one ingredient of a given mixture
must produce a proportional change in the amounts
of the others, and of the entire quantity.
27. Alligation Medial. — Given the rates or kinds
and quantities of certain ingredients, to find the mixture
they will produce —
Rule. — Multiply the rate or kind of each ingredient
by its amount ; divide the sum of the products by the
number of the lowest denomination contained in the
whole quantity, and the quotient will be the rate or
kind of that denomination of the mixture. From this
may be found the rate or kind of any other denomination.
Example 1. — What ought to be the price per lb, of a
mixture containing 08 lb of sugar at 9c/. per lb, 87 lb at
OJ.. and 34 lb at Grf.
d. d.
9x08 = 882
5x87 = 435
6x34 = 204
219 219)1521
Alls. Id. per lb, nearly.
Tlie price of each eugnr, is the number of pence per pound
multiplied by the number of pounds ; and the price of the
whole is the sum of the prices. But if 219 lb of sugar have
cost ]ij'2\d., one lb, or the 219th part of thi.s, must cost the
219th part of 1521c/., or ^-^rdd. =7d., nearly.
ALLIGATION. 29^
Example 2. — What will be the price per lb of a mixture
containing 9 lb 6 oz. of tea at 5.s. 6r/. per lb. 18 lb at 6.!..
per lb; and 4G lb 3 oz. at 95. 4irf. per lb 1
R) oz. s. d. £ s. d.
9
18
46
6at5
0 C
3 9
6 per lb= 2 11
0 per lb= 5 8
41 per lb=21 13
0
73
JG
9
1177)29 12
Ans
. Gd.
per oz. nearly
1177 ounces.
And 6c/.xlG=8s., is the price per pound.
In this case, the lo^^vest denomination being ounces, vre
reduce the whole to ounces ; and having found the price of aa
ounce, "we multiply it by 16, to find that of a pound.
Example 3. — A goldsmith has 3 lb of gold 22 carats line,
and 2 lb 21 carats fine. What will be the fineness of tha
mixture ?
In this case the value of each kind of in:^redient is rcpro-
sented by a number o? carats —
lbs
3x22 = GG
2x21 = 42
5 5ym
The mixture is 212 carats fine.
EXERCISES.
1. A vintner mixed 2 gallons of wine, at 14.?. per
gallon, with 1 gallon at 125., 2 gallons at 95., and 4
gallons at 8s. What is one gallon of the mixture worth ?
Ans. 105.
2. 17 gallons of ale, at 9d. per gallon, 14 at 7^c?., 5
at 9|^., and 21 at 4^d.^ are mixed top:ether. How
much per gallon is the mixture worth ? Ans. 7j\d.
3. Having melted together 7 oz. of gold 22 carats
fine, 12^ oz. 21 carats fine, and 17 oz. 19 carats fine, I
wish to know the fineness of each ounce of the mixture .'*
Ans. 20|| carats.
28. Alligation Alternate. — Given the nature of the
mixture, and of the ingredients, to find the relative
amounts of the latter —
Rule. — Put down the quantities greater than the
given mean (each of them connected with the difierence
300 ALLIOATION
between it and the mean, by the sign — ) in one column ;
put the differences between the remaining quantities
and the mean (connected with the quantities to which
they belong, by the sign + ) in a column to the right
hand of the former. Unite, by a line, each plus with some
minus difference ; and then each difference will express
how much of the quantity, with whose difference it is
connected, should be taken to form the required mixture.
If any difference is connected with more than one
other difference, it is to be considered as repeated for
each of the differences with which it is connected ; and
the' sum of the differences with which it is connected is
to be taken as the required amount of the quantity
whose difference it is.
Example 1. — How many pounds of tea^ at 5s. and 85. per
K), would form a mixture worth 75. per ib ?
Price. Diflerences. Price. '
5. S. .*. ■<?.
The mean=8— 1 2-\-o=thQ mean.
1 is connected with 2s., the difference between the mean
and 55. : hence there must be 1 lb at 5s. 2 is connected
with 1, the difference between 85. and the mean ; hence there
must be 2 lb at 85. Then 1 ib of tea at 55. and 2 lb at 8.s-.
per ft), will form a mixture worth 75. per lb — as may be
proved by the last rule.
It is evident that any equimultiples of these quantities
would answer equally well ; hence a great number of answers
may be given to such a question.
Example 2. — How much sugar at 0(/., Id., 5d., and 10'/ ,
will produce sugar at 8d. per lb ?
Prices. Piftercnces. Prices.
d. d.''^"'!. d^.
The mean= j io~o "si's j =^^^ ^^^•
1 is connected with 1, the difference between Id. and the
mean ; hence there is to bo 1 11) of sugar at Id. per lb. 2 is
connected with 3, the difference between 5d. and the mean ;
hex\ce there is to be 2 lb at 5d. 1 is connected with 1, the
difference between 9(Z. and the mean \ lience there is to be
1 Yo at M. And 3 is connected Avith 2,'the difference between
10(/. and the meanj hence there arc to h^ 3 ib at lOci.
per lb.
Al.LIGAIION. 301
Consequently we are to take 1 lb at Id., and 2 lb at o</.,
1 lb at 0</., and 3 Hj at 10//. If we examine wliat mixture
these will give [27], we shall find it tu be the given mean.
Example 3. — What quantities of tea at 4s., 65., Ss., and
ds. per lb, will ja'oduce a mixture worth os. '?
Trices. Diflerences. Prices.
The mean:
:the mean.
3, 1, and 4 are connected with 1?., the differece between
ii. and the mean : therefore we are to take o lb -f- 1 Ih -f- 1
lb of tea, at 4s. per lb. 1 is connected with os., Is., and 4>\,
the differences between 8.s.. Gs., and 9s., and the mean;
therefore we are to take 1 lb of tea at 8s., 1 lb of tea at Qs.,
and 1 lb of tea at Os. per lb.
We find in tliis example that 8s., 6s.. and Os. are all con-
nected with the same 1 : this shows that 1 lb of each will
be required. 4s. is connected with 3, 1, and 4; there must
be, therefore, 3-f-l-{-4 lb of tea at 4.s.
Example 4. — How much of anything, at 3s., 4s., 5s., Is.,
8s., 9s., lis., and 12s. per lb, would form a mixture worth
Gs. per lb ]
Prices. Dilference?. Prices.
1 lb at 3.?.. 2 R) at 4s.. 3 lb at 7s., 2 It) at 8s.. 3-f-5-f.6 (14)
lb at 5s., 1 lb at 9s., 1 lb at lis., and 1 lb at 12?. per lb, will
form the required mixture.
29. Reason of. the Rule. — The excess of one ingredient
above the mean is made to counterbalance what the other
wants of being equal to the mean. Thus in example 1, 1 lb
at OS. per lb gives a deficiency of 2s. : but this is corrected by
25. excess in the 2 lb at 8.?. per lb.
_ In example 2, 1 lb at Id. gives a deficiency of Icf., 1 lb at 9^/.
gives an excess of Id. ; but the excess of Id. and the deficiency
of \d. exactly neutralize each other.
Again, it is evident that 2 lb at 5r/. and 3 lb at lOd. are
worth just as much as 5 lb at Sd. — that is, 8f/. will be tha
average price if we mix 2 lb ai od. with o lb at lOcZ.
o 2
■)2 ALI.IOATIOX.
EXERCISE.-
4. How raucli wine at 85. Qd. and 95. per gallon will
make a mixture worth 85. \0d. per gallon ? Ans. 2
gallons at 85. 6^., and 4 gallons at 9^. per gallon.
5. now much tea at 65. and at 3^. 8d. per fe, will
make a mixture worth 45. 4d. per tb ? Ans. 8 fe at
65., and 20 ft) at 35. 8^. per ft).
6. A merchant has sugar at 5d., lOd., and 12d. per
\h. How much of each kind, mixed together, will be
worth Sd. per ft) ? Ans. 6 lb at 5^., 3 lb at 10^., and
3 lb at I2d.
7. A merchant has sugar at 5cZ., 10c?., I2d.j and 16d
per ft). How many "lb of each will form a mixtm'e worth
lid. per lb? Ans. 5 ft) at od., 1 lb at 10^., 1 lb at
12tZ., and 6 fta at I6d.
8. A grocer has sugar at 5^?., 7^., 12^., and 13^.
per ft). How much of each kind will form a mixture
v/orth lOd. per lb ? Ans. 3 lb at od., 2 lb at 7^., 3 ft)
at 12d.j and 5 lb at I3d.
30. When a given amount of the mixture is required,
to find the corresponding amounts of the ingredients —
Rule. — Find the amount of each ingredient by the
last rule. Then add the amounts together, and say, as
their sum is to the amount of any one of them, so is the
requhed quantity of the mixture to the corresponding
amount of that one .
Example 1. — What must be the amount of tea at 4s. per
It), in 736 Jb of a mixtm-e worth 55. per lb, and containing
tea at Cs., 85., and 95. per lb 1
To produce a mixture worth 55. per lb, we require 8 lb
at 45., 1 at 85., 1 at 65., and 1 at 95. per ib._[28]._ But all
of these, added together, will make 11 lb", iu which thero
are 8 lb at 45. Therefore
lb lb lb 8x736 1^ ^^■
11 : 8 : : 736 : .-. =526 4/y, the, required quantity
of tea at 45.
That is, in 736 ft) of the mixture there will be 536 lb 4y*y
oz. at 45. per lb. The amount of each of the other ingre-
dients may be found in the game way.
ALUGATlOlN. 303
Example 2. — Ilicro, king of Syraeui^e, gave a certain
quantity of gold to form a crown ; but when he received it,
piispecting tiiat the goldsmith luid taken some of the gold,
au(l sup})l!cd its phico by a ba^er metal, he commisdioued
Arcbiinedo^^, the celebrated n:ulhematiciaii of .Syracuse, to
a>>certain if his su-spicioa was well founded, and to what
extf'nt. Archimedes was for some time unsuccessful in hi^
researclics, until one da}', going into a bath, lie remarked
that he displaced a quantity of water equal to his own bulk.
Seeing at once that the same weight of diflercnt bodies
would, if immersed in water, displace very diifcront quan-
tities of the fluid, he exclaimed with delight that he had
found the desired solution of the problem. Taking a mass
of gold e((ual in weiglit to vdiat was given to the goldsmith,
he found thnt it displaced Ic.'^s Avater than the crown ; which,
tlierefore, was made of a lighter, bccauso a more bulky
metal — and, consequently, was an allou of gold.
Now supposing copper to have been the substance with
which the crovrn was adulterated, to find its amount —
Let the gold given by Hiero have w«ig]ied 1 lb, this
would displace about -052 lb of water ; 1 lb of copper v.'ould
displace about -1124 lb of water; but let the crown have
displaced only -072 lb. Then
Gold diiFers from -072, tlio mean, by — •02i).
Copper diifers from it by . . -[-■ 040-1.
Copper. Difiereaces. Gol-L
Hence, the mean=r=- 1124 -0404 •020-f--052=thc mean.
Therefore -020 lb of copper and -0404 lb of gold would
produce the alloy in the crown.
But the crown was supposed to weigh 1 lb ; therefore
•0G04 lb (•0204--0404) : -0404 lb : : 1 lb : l^lil^i+ll^
= GGO lb, the quantity of gold. And 1— •G69=-331 lb is
the quantity of copper.
EXERCISES.
'J. A druggist is desirous of producing, from medicine
at 55., Gi-., S.S., and 9i\ per Tb, 1^ cwt. of a mixture
Avorth Is. per lb. How much of eacli kind must lie
use for the purpose t Ans. 2S lb at 5.?., 56 lb at 6^.,
56 ft) at 85., and 2S lb at 95. per lb.
10. 27 ft) of a mixture worth 45, Ad. per ib are re-
<][uired. It is to eontaiu tea at 55. and at 35. Q>d. per
304 ALLIGATION.
lb. How much of crich must bo used ? J\ns. 15 lb at
OS., and 12 ib at 35. 6d.
11. How much sugar, at 4d., 6d., and S^. per lb,
must there be in 1 cwt. of a mixture worth 7d. per lb .-'
Ans. 1S| ft) at 4^., ISf lb at 6«., and 74| lb at 8^.
per lb.
12. How much brandy at 12^., 135., 145., and 145.
Gd. per gallon, must there be in one hogshead of a mix-
ture worth 135. 6d. per gallon } Ans. 18 gals, at 125.,
9 gals, at 135., 9 gals, at 145., and 27 gals, at 145. 6d.
per gallon.
31. When the amount of one ingredient is given, to
find that of any other —
EuLE. — Say, as the amount of one ingi-edient (found
by the rule) is to the given amount of the same ingredient,
so is the amount of any other ingredient (found by the
rule) to the req^dred quantity of that other.
Example 1. — 29 lb of tea at 4s. per To is to be mixed with
teas at Gs., 85.. and 9.5. per R), so as to produce what will be
"vorth -S."?. per tb. What quantities must be used 1
8 }h of tea at 45., and 1 lb at G5., 1 lb at 85., and 1 lb at
9.S., will make a mixture worth 5s. per lb [27]. There fere
8 lb (the quantity of tea at 45. per lb, as found by the rule) .
29 lb (the given quantity of the same tea) : : 1 lb (the
quantity of tea at 65. per lb, as found by the rule) : ——^ —
8
(the quantity of tea at G5., which corresponds with 29 ib at
45. per lb) =3-^ ib.
We may in the same manner find what quantities of tea at
85. and 9s. pcrib corre.-pund with 29 lb — or the given amount
of tea at 45. per lb.
Example 2.— A refiner has 10 oz. of gold 20 carats fine
and melts it with 16 oz. 18 carats fine. AVliat must be
added to make the mixture 22 carats fine 1
10 oz. of 20 carats nne=10x20 = 200 carats.
16 oz. of 18 carats fine=lGxl8 = 288
26 : 1 :: 488 : 18 };[ carats, the
fineness of the mixture.
24 — 22=2 carats baser metal, in a mixture 22 carats fine.
24 — 18||=5jHj carats baser metal, in a mixture 18 j^
carats fine.
Then 2 carats : 22 carats : : S/'j : 57j"j carats of pure
ALLIGATION. 30")
^irold— required to change 5,-^3 carats baser metal, into i\
mivturo 22 carats line. But there arc ah'cady in the mixture
IH\'^ carats goM : therefore oTy'.r— 18}f^=o8|| carats gold
iire to bo a<ldod to every ounce. There are 20 oz. : therefore
20x38J;^=1008 carats of gold are -wanting. There arc
24 carats (page 5) in every oz. ; tliercfore '^^ carats=42
oz. of gold must 1)0 added. There will then be a mixture
containing
oz. car. car.
10X20 = 2'.)0
Mix 18 = 288
42x24 =- 1008
08 : 1 oz. : : 14'JG : 22 carats, the required fiaencss.
EXERCISES.
13. How much tea at 65. per lb must be mixed with
12 tb at 35. i"d. per fo, so that the mixtui-e may be
worth 4s. Ad. per ib ." Ans. 44 ib.
14. How mucli brass, at 14<^. per lb, and pewter, at
lO^d. per Bj, must I melt with 50 lb of copper, at 16r/.
per lb, so as to make the mixture worth Is. per lb ?
uhis. 50 Ijj of bra.ss, aud 200 ib of pewter.
15. How much gold of 21 aud 23 carats fine must
be mixed vrith 30 oz. of 20 carats fine, so that the mix-
ture may be 22 carats fine .' Ans. 30 of 21, and 90
of 23.
16. How much wine at Is. 5rZ., at 5^. 2d.., and at
As. 2d. per gallon, must be miyjcd with 20 gallons at
C$. Sd. per gallon, to make the mixture worth 6.?. per
gallon } Alls. 44 gallons at 7s. vd., 16 gallons at 55
2d., and 34 gallons at 45. 2d.
QUESTIONS.
1 . AVhat is alligation medial ? [26] .
2. Wliat is the rule for alligation me lial ? [27].
3. Wliat is alligation alternate .? [261.
4. What is the rule for alligation altcvnate .? [28].
5. AYliat is the rule, when a certain a?"»iount of tl <j
mixture is required ? [30] .
6. What is the rule, when the amouu'
of the iugredients Ls given ? [31].
306
SECTION IX.
INVOI.UTION AND EVOLUTION, kc.
1. Involution. — A quantity wHich Ls the product of
two or more footers, each of tlieiu the same uumber, is
termed a power of that number ; and the nuiribcr, mul-
tiplied by itself, is said to be invclced. 'J'hus 5x5x5
(=125) Ls a " poTver of 5 ;" and 125, is 5 '' involved."
A power obtains its denomination from the number of
times the root (or quantity involved) is taken as a factor.
Thus 25 (r=5x5) is the seanid power of 5. — The
second power of any number is also called its square ;
because a square surface, one of whose sides is expressed
by the given number, will have its area indicated by the
igecond power of that number ; thus a square, 5 inches
every w^ay, will contain 25 (the square of 5) square
inches ; a square 5 feet every way, v/ill contain 25
scfuare feet, &c. 216 (6X0X6) i^ the l/iird power of
6. — The third power of any number is also termed its
mibc ; because a cube, the length of one of whose sides
is expressed by the given number, will have its solid
contents indicated by the thiid power cf that number.
Thus a cube 5 inches every way, will contain 125 (the
cube of 5) cubic, or. solid inehes ; a cube 5 feet every
way, will contain 125 cubic feet, ^:c.
2. In place of setting dovvn all the factors, we put
down only one of thera, and mark how often they are
sv/pposed to be set down by a small figure, which, since
it points out the number of the factors, is called the
index^ or crpoTient. Thus 5^ is the abbreviation for
5 X 5 : — and 2 is th>5 index. 5^ means 5 X 5 X 5 X 5 >^ 5,
or 5 in the fifth power. 3^ means 3X3X3X3, or 3 in
the fourth power. S"^ means SxSXSXJ^X^XSXS,
or 8 in the seventh power, &c.
3. Sometimes the vinculum [Sec. II. 5] is used in con-
junction with the index ; thus 5 + 82 means that the sum
of 5 and 8 is to be raised to the second power — this
INTOLUTIO!f. 307
is very cluTerent from n^-f S-, wliicli means tlio sum of
the squares of 5 and S : 5 + 8= being 169 ; 'n-liile 5^ -\- 8^
is only 89.
4. In multiplication the multiplier may be considered
as a species of index. Thus in 187x5, 5 points out
how often 187 should be set down as an addend ; and
187X5 is merely an abbreviation for 187+187+187 +
187 + 187 [Sec. II. 41]. In 187^, 5 points out how
often 187 should be set down as a factor ; and 187^ is
an abbreviation for 187X187X187X187X187 : — that
is, the " multiplier" tells the number of the addends, and
the " index" or " exponent," the number of the fadors.
5. To raise a number to any power —
Rule. — Find the product of so many factors as the
index of the proposed power contains units — each of the
factors being the number which is to be involved.
ExAJiPLE 1. — ^Vhat is the 5 th power of 7 ?
7=- =7x7x7x7x7=16807.
Example 2. — What is the amount of £1 at compound
interest, for 6 years, allowing 6 per cent, per annum '?
The amount of £1 for 6 years, at 6 per cent, is —
_10Gxl0Gxl-0Gxl-06xl-06xl-06 [Sec. YII. 20], or
Hrd'=l-41852.
We, as already mentioned [Sec. VII. 23], may abridge
<he process, by using one or more of the products, already
•obtained, as factors.
"^ EXERCISES.
1. 3'=243.
2. 20"'=10240000000000.
3. 3^=2187.
4. 105'=134009o64032^5.
5. 105''=1-340095640G25.
-6. To raise a fraction to any power — '
Rule. — Raise both numerator and denominator to
that power.
EXAMPLE.-(5)=^=|X|X|=|I.
To involve a fraction is to multiply it by itself. But to
multiply it by itself any number of times, we must multiply
its numerator by itself, and .*\lso its denominator by itself, tlia(
number of times [Sec. IV. 39].
iOS EVOLUTION.
EXFRCrSKS
Q / 0 \ 5 :! 1 2.'. ^
7. To raise a mixed number to any power —
lluLE. — lleduce it to an improper fraction [Sec. IT
24] ; and then proceed as directed by the last rule.
ESAMI'LK.— (2i)*=(f)*=\-^^
EXERCISES.
10. (11D3^13JL_19 3
11. C3i)^=«t
0 4:'.^
F.I
1Q fA^Sl 42G18442W77
8. Evolution is a process exactly opposite to involution ,
since, by means of it, we find what number, raised to a
given power, would produce a given quantity — the num-
ber so found is termed a root. Thus we " evolve " 25
y\'hen we take, for instance, its square root ; that is, when
we find what number, multiplied by itself, will produce
25. Roots, also, are expressed by erpomnts — but as these
exponents are fractions, the roots are called '■''fractional
powers." Thus 4^ means the square root of 4 ; 4^ the
cube root of 4 ; and 4 ^ the seventh root of the fifth power
of 4. Eoots are also expressed by ^, called the radiad
sign. When used alone, it means the square root — thus
^3, is the square root of 3 ; but other roots are indicated
by a small figui-e placed within it — thus ^b ; which
means the cube root of 5. ,^1" (7t), is the cube root
of the square of 7.
9. The fractional exponent, and radical sign are some-
times used in conjunction with the vinculum. Thua
4—3-, is the square root of the difference between 4
and 3 ; ^5-f 7, or 5 + 7^, is the cube root of the sura
of 5 and 7.
iO. To find the square root of any number —
EuLE — I. Point off the digits in pans, by dots ; put-
ting one dot over the units' flace^ and then another dot
over every second digit hofJi to the right and left of
the units' place — if there are digits at loth sides of the
decimal point.
EVOLUTION. 309
II. Find the liighost number the square of wliich
will not exceed the araount of the highest period, or
that which is at tlie- extrcuie left — this number will be
the first digit iu the requh'ed squ:ire root. Subtract its
square from the highest period, and to the remainder,
considered as hundreds, add the next period.
III. Find the highest tligit, which being multiplied
into twice the part of the root already found (consi-
dered as so many tens), and into itself, the 5?/??^ of the
products will uot exceed the sum of the last remainder
and the period added to it. Put this digit in the root
after the one last found, and subtract the former siu/i
from the latter.
IV. To the remainder, last obtained, bring down
another period, and proceed as before. Continue this
process until the exact square root, or a sufficiently
n?ar approximation to it is obtained.
11. Example. — What is the square root of 22420225 ^
22420225(4735, is tlie required root.
IC
87)642
GOO
943)S3iJ2
2829
47325^
22 ivS the highe.st period; and 4- is the highest square wliiek
doe.s not exceed it — we put 4 in the root, and subtract 4-^,
or 16 from 22. This leaves G. which, along with 42. the next
period, makes G42.
Vv'e subtract 87 (twice 4 tens-}-7. the highest digit which
we can now put in the root) X 7 from 042. This leaves
33, which, along with 02, the next period, makes 3302.
"We subtract 943 (twice 47 tens -f-^- the nest digit of the
root) x3 from 3302. This leaves 473. which, aloilt* Y'lt'i
25. the only remaining period, makes 47325.
We subtract 9405 ('twice 473 teas -j.n, the ne-j dl-nt of
the root) x5. This leaves rt.-^ ■rema,inder. °
^ The given numbev, therefore, is exactly t- gquai-e : and
its square root is 4735. . i /
12. Heasojv of I.-^Yg point off -,o ^j^tg of the givea
Bquare m pairs, and con.si.ier the ^^^^J^^ ,^%,, ,3 i.^diclting
310 EVOLUTION
the number of digits in the root, since neither one nor two
digits in the square can give more or less than one in the root ;
neither three nor four digits in the square can give more or
less than two in the root, &c. — which the pupil may easily
ascertain by experiment. Thus 1, the smallest single cligit,
will give one digit as its square root ; and 99, the largest pair
of digits, can give only one — since 81, or the square of 9, is
the greatest square which does not exceed 99.
Pointing off the digits in pairs shows how many should be
brought down successively, to obtain the successive digits of
the root — since it will be necessary to bring down on® period
for each new digit ; but more than one will not be required.
PiEASOjf OF II. — ^Ye subtract from the highest period of the
given number the highest square which does not exceed it,
and consider the root of this square as the first or highest
digit of the required root ; because, if we separate any number
into the parts indicated by its digits (5t3o, for instance, into
500, 60, and 3), its square will be found to contain the square
of each of its parts.
Reasozv of III. — Y*'e divide twice the quantity already in
the root (considered as expressing tens of the next denomina-
tion) into what is left after the preceding subtraction, &c., to
obtain a new digit of the root; because the square of any
quantity contains (besides the square of each of its parts)
twice the product of each part multiplied by each of the other
parts. Thus if 14 is divided into 1 ten and 4 units, its square
will contain not only 10^ and 4-, but also twice the product
of 10 and 4. — We subtract the square of the digit last put in
the root, at the same time that we subtract twice the product
obtained on multiplying it by the part of the root wliicli pre-
cedes it. Thus in the example which illustrates the rule,
AV'hen we subtract 87X7, we really subtract 2x40x7+7^.
It will be easily to show, that the square of any quantity
contains the squares of the parts, along with twice the pro-
duct of every two parts. Thus
22420225=4735^=40004-700-1-30-1-5^.
4000^=16000000
330225
2X4000X30-|-2XT0OX30-|-3O2= 282Q00
47325
2X'1000X5-|-2X700X5-h2X30X5-f-52=:47325
PtEAsox OF IV. — Dividing twice the quantity already in
the root (considered as expressing tens of the next denomi-
nation) into the remainder of the given number, &c., givea
the next digit ; because the square contains the sum of
twice the products (or, what is tlie same thing, the product
EVOLUTlOJi; 311
of twice the fiwrn) of tlio part? of the root rJreafly fonnJ,
multiplied by tlie new digit. Thus 22420225, the square of
4735, contains 4000--f-700--f30--f5- : aud also twice 4000X
700 -f twice 4000 X 30-}- twice 40uOx5 : plus twice 700x30-f-
twice 700x5: plus twice 80x5: — that is. the square of each
of its parts, with the sum of twice the product of every two of
them (which is the same as each of them multiplied by twice
the sum of all the rest). This would, on examination, be
found the case with the square of any other number.
If we examine the example given, we shall find that it will
not be necessary to bring down more than one period at a
time, nor to add cyphers to the quantities subtracted.
13. When the given square contains decimals —
If any of the periods cousi.st of decimals, the digits
in the root obtained on bringing down these periods to
the remainders will also be decimals. Thus, taking the
example just given, but altering the decimal point, we
shall have ^•224202-2d=473-d] V2242-022.5=47-35 ;
^22 •420225 = 4-735 ; ^/ -22420225 = -4735 ; and
^-0022420225 = -04735, &c. : this is obvious. If there
is an odd number of decimal places in the power, it
must be made even by the addition of a cypher. Using
the same figures, ^2242022-5 = 1497-338, &c.
2242022- 50 (1407-838, &c
1
24)124
289)2820
2601
2987)21^22
20909
29943)101350
89829
299403)1152100
898389
2994668)-2537n00
23957344
141375G
In this case the highest period consists but of a single digit
and the given number is not a perfect square.
There must be an even number of decimal places ; since nf
number of decimals in the root will produce an odd number
in th<^ square [Sec. II. 4S] — as may be proved by experimeni
312 EVOLUTION.
EXERCISES.
14. yi95364=442
15. ^328329=573
16. y- 0(37 6= -26
17. ^87 -65=9 -3622
18. y861=:29-3428
19. ^984064=992
20. y5=2- 23607
21. ^-5= -707 106
22. ^91 -9681=9 -50
23. ^238144=188
24. ^32 -3761=5 -69
25. V''331776=-576
14. To extract the square root of a fraction —
EuLE. — Having reduced the fraction to its lowest
terms, make the square root of its numerator the nume-
rator, and the square root of its denominator the deno-
minator of the required root.
KXAMPLE.— y|=2-.
15. Reason of the Rule. — The square root of any quan-
tity must be such a number as, multiplied by "itself, will pro-
duce that quantity. Therefore | is the square root of A ; for
|Xf=|-- The same might be shown by any other example.
Besides, to square a fraction, we must multiply its numera-
tor by itself, and its denominator by itself [6] ; therefore, to
take its square root — that is, to bring back both numerator
and denominator to what they were before — we miLSt take the
square root of each.
16. Or, when the numerator and denominator are
not squares —
E,ULE. — Multiply the numerator and denominator
together ; then make the square root of the product the
numerator of the require I root, and the given denomi-
nator its denominator ; or make the square root of the
product the denominator of the required root, and the
given numerator its numerator.
Example. — What i.s the square root of 4 ? (1)^ =
^/1>^ or ^==4-472136-4-5 = -894427.
0 5X4
17. We, in this case, only multiply the numerator and
denominator by the same number, and then extract the square
4 4x5 4x4 • /4\^
root of each product. For -=- — -, or - — -. Therefore ( = )*
^ 5 5X0 5X4 \5y
/4x5y_y4x5 /4x4y_ _4__
~V5X5/ — 6 ' \5X4/ "^ybxi :
EVOLUTION. 313
IS. Or, lastly—
Rule. — Reduce the given fraction to a decimal
rSeo lY. 63], and extract its square root [13]
EXERCIs^ES.
28-o:^9G8o2
.0 (!).=
27 /14U 14
9066295
28. /3 \^ 6-244998
(
13/ "~ 13
29.
30.
(|)'=.7453a6
(j5)---8fl602.34
3. ^^^,
■8451542
19. To extract the square root of a mixed number —
Rule. — Reduce it to an improper fraction, and then
proceed as ah-eady directed [14, &c.]
Example.— y2]-=yl=^=li.
EXERCISES.
32.
yolfH"? 1 35. yl7i=41GS3
33.
y27^=5A 30. y 6^=2 5203
34.
yl8^,=101858 37. yl3i=3G332
20. To find the cube root of any number —
Rule — I. Point off the digits in threes, by dots —
putting the first dot over the units' place., and then
proceeding both to the right and left hand, if there are
digits at both sides of the decimal point.
II. Find the highest digit -whose cube will not ex-
ceed the highest period, or that which is to the left hand
side — this will be the highest digit of the required root ;
subtract its cube, and bring down the next period to
the remainder.
III. Find the highest cHgU, which, being multiplied
by 300 times the square of that part of the root,
already found — being squared and then multiplied by
30 times the part of the root abeady found — and being
multiplied by its own square — the siu/i of all the pro-
ducts will not exceed the sitm of the last remainder and
the period brought down to it. — Put this digit in the
root after what is already there, and subtract the former
$um from the latter.
IV. To what now remains, bring; down the next
314 EVOLUTION.
period, and proceed as before. Continue tLis process
until the exact cube root, or a sufficiently near approxi-
mation to it, is obtained.
ExA^MPLE.— What is the cube root of 179597069288 ?
179597009288(5642, the required root.
125
54597
= 506re
300x5=x6
30x5x6*
6'X6 ) ^r-.r-^-
300x56^x4 ) o9blOt)y
30x50x4^ = 3790144
300x564^X2 190925288
30x564x2^ }= 190925288
2''X2 )
We find (by trial) that 5 is the first. 6 the second, 4 the
third, and 2*^the last digit of the root. And the given
number is exactly a cube.
21. Eeasox of I. — We point off the digits in threes, for a
reason similar to that which caused us to point them off in
t-fros, vrhen extractiag the square root [12].
Reasox of II. — Each cube will be found to contain the
cube of each part of its cube root.
Reasox of III. — The cube of a number divided into any
two parts, will be found to contain, besides the sum of the
cubes of its parts, the sum of 3 times the product of «ach
part by the other part, and 3 times the product of each part
by the' square of the other part. This will appear froiP the
following : —
179597069288
6000'=1 25000000000
54597069288
2 X GOOCxOOO-fS Xo000x600=-}-600'= 50616000000
3981069288
3 X 5000' X 40-f3 X 5600 X 4v=-f40'= 3790144000
1909252ns
3 X 5640^ X 2-1-3 X 5640 X 2*4-2'= 190925283
Hence, to find the second digit of the root, we must find by
trial some number which — being multiplied by 3 times the
square of the part of the reot already found — its square being
315
multiplied by 3 times the part of the root already found— and
bciug multiplied by the square of itself — the sum of the pro-
ducts will not exceed what remains of the given number.
Instead of considering the part of tlie ruot already found as
60 many tc7is [12] of the denomination next fullowing (as it
really is), which would add one cypher to it, and two cyphers
to its square, we consider it as so many units, and multiply
It, not by 3, but by 30, and its square, not by 3, but by 300.
For 300 X o'' X 6 -j-30 X o X G--f-6"^XG is the same thing as
8Xi>0"XB-f-oX-50xO"-f-t3"X6 ; since we only change the jiosi-
tion of the factors 100 and 10, wliich does not alter the product
[Sect. II. 35].
It is evidently unnecessary to bring* down more than one
period at a time ; or to add cyphers to the subtrahenJs.
Reason of IV. — The portion of the root already found may
be treated as if it were a single digit. Since into whatever
two parts we divide any number, its cube root will contain
the cube of each part, with 3 times the square of each multi-
plied into the other.
22. When there are dechrials in the given cube —
If any of the periods consist of decimals, it is e^^dent
that the digits found on bringing down these periods
must be decimals. Thus ^179597-0692SS = 56-42, &c.
When the decimals do not form complete periods, the
periods are to be completed by the addition of cyphers.
Example. — What is the cube root of -3 1
6-306(-609, &c.
21G
300X6^X6 ) 84000
30X6X6- > =71496
GX6^
300X66^X9 ) 12504000
30X66X9^ =11922309
9X9' ) 581691, &c.
^•3='669, &c. And -3 is not exactly a cube.
It is necessary, in this case, to add cyphers : since one decimal
in the root will give 3 decimal places in the cube; two decimal
»^laces in the root will give six in the cube, o:c. [Sec. 11. 48.]
EXERCISES.
38. ^|3=^3 -207534
39. ^39=3-391211
40. ;/2T2=5 -962731
41. ^123505992=498
42. ^190109375=575
43. ^458314011=771
44. ^483-736625=7-85
45. ^^0056= -86
46. ^999=0-996866
47. ^-979140657= -993
/
316 EVOLUTION.
23. To extract the cube root of a fraction —
Rule. — JIa\4ng reduced the given fraction to its
lowest terms, make the cube root of its numerator the
numerator of the required fraction, and the cube root
of its denominator, the denominator.
2-1. IXeason- of the Rule. — The cube root of any number
must be such as that, taken three times as a factor, it will
produce that number. Tlierefore f is the cube root of y4 -j ;
/or I X f X I = yf-. — Tlie same thing might be showu by any
other example.
Besides, to cube a fraction, we must cube both numerator and
denominator; therefore, to take its cube root — that is to reduce
it to what it was before — we must take the cube root of both.
25. Or, when the numerator and denominator are
not cubes —
Rule. — Multiply the numerator by the square of the
denominator ; and then divide the cube root of the pro-
duct by the given denominator ; or divide the given
numerator by the cube root of the product of the given
denominator multiplied by the square of the given
numerator.
Example. — What is the cube root of f ?
(r.Y = ^^J: or -^ = 5-277632 ~- 7 = •753047.
^'■' 7 ^/7x3'
This rule depends on a principle already explained [10].
26. Or, lastly—
Rule. — Reduco the given fraction to a decimal
[See. ly. 63] , and extract its cube root [2.2] .
8-Go.340;
EXERCISES.
51.
52.
53.
(^|-y=- 941036
(0^=-o6O9O7
(l4)'='4'21G3
27. To find the cube root of a niLxed number —
Rule. — Reduce it to an improper fraction ; and thea
proceed as already directed [23, &c.]
ExAMPLi:.— ;/3||=^"^=l-54.
EVOLUTION. 317
EXERCISKS.
54. (283-) ^=3-0035
55. (7i)'=103098
50. (0^)i=20928
57. (71|)i=41553
58. (32^«y)'=31087
59. (5i)^^l-7592
28. To extract any root whatever —
Rule. — When the index of the root is some power
of 2, extract the square root, when it is some power of 3,
extract the cube root of the given number so many time.'*,
successiyely, as that power of 2, or 3 contains unity.
r
Example 1.— The 8th root of G553C=>/ ^ >v/6o536=4.
Since 8 is the third power of 2, we are to extract the
square root three times, successively.
Example 2.— 134217728'"=yVl34217728=8.
Since 9 is the second power of 3, we are to extract the
cube root twice, successively.
29. In other cases we may use the following (Hutton
Mathemat. Diet. vol. i. p. 135).
Rule. — Find, by trial, some number which, raised
to the power indicated by the index of the given root,
will not be far from the given number. Then say,
as one less than the index of the root, multiplied by the
given number — plus one more than the index of the root,
multiplied by the assumed number raised to the power
expressed by the index of the root : one more than the
index of the root, multiplied by the given number —
plus one less than the index of the root, multiplied by
the assumed number raised to the power indicated by
the index of the root, : : the assumed root : a still
nearer approximation. Treat the fourth proportional
thus obtained in the same way as the assumed number
was treated, and a still nearer approximation will be
found. Proceed thus until an approximation as near as
de.'^irable is discovered.
Example.— What is the 1 3th root of 923 %
Let 2 be the assumed root, and the proportion will bo
12x9234-14x2^3 . Ux9234-12x2»3 .. 2 : a nearer
approximation. Substituting this nearer approximation for
2. in the above proportion, we get another approximation,
which we may treat in the same v»'ay.
p
318 EVOLUTION
EXKRCISES.
CO. (9GG98)'=6-7749
61. (GG457)TT=2-7442
G2. (23G5)l=olo85
au
68. (8742G)t=5084-29
G4. (80G5) '=1-368
6-5. (■075426)t4=-04G988
)0. To find the squares and cubes, the square and
cube roots of numbers, by means of the table at the end
of the treatise —
This table contains the squares and cubes, the square
wid cube roots of all numbers which do not exceed 1000
hut it will be found of con.siderable utility eyen when very
high numbers are concerned — provided the pupd bears
in mind that [12] the square of ariV number is equal to
the sum of the squares of its parts (which may be found
by the table) plus tvi-ice the product of each part by the
sum of all the others; and that [21] the cube of a
number divided into any two parts is equal to the sum
of the cubes of its parts (v^•hich may be found by the
tiible) plus three times the product of each part multi-
plied by the square (found by means of the table) of
the other. One or two illustrations will render this
sufficiently clear.
Example 1. — Find the square of 873450.
87345G nuij'-be divided into two parts, 873 (thousand) and
45G (units). But we find by the table that 873'=7G2120 and
4o~g'=207'J36.
Therefore 762129000000=873000'
700176000=873000 X twice -I'jO
207936=450-'
HxAMPLK 2. — Fipd the cube of 864379. rnvidin^ this into
8G£ (thousand )_and 379 (units), we find 864W44972544
Bu4'=746496, 379'=;34439939, and 379 =143641
Therefore 644972544000000000=864000'
848765952000000=3 x 8G4O0O' X 379
37231 7472000=3 x 8G4000 x 37lj'
And G45821G82323911939=86437U'
LOQARITHMS. 319
It
6\ In fiu']ing the square and cube roots of larger numbers,
we obtain their thret highest digits at once, if we look in the
tuble for the Jiighest cube or square, the highest perii;d of
which (^the required cyphers being .idded) does not exceed the
highest period of the given number. The remainder of the
process, also, may often be greatly abbreviated by means of
the table. •
QUESTIONS.
1. "V\Tiat are involution and evolution t [1].
2. What are a power, index, and exponent } [1 & 2J.
3. What is the meaning of square and cube, of th'3
square and cube roots .' [1 and 8].
4. What is the difference between an integral and a
fractional index .- [2 and 8] ,
5. How is a number raised to any power .' [5].
6. What is the rule for finding the square root .' [10].
7. What L3 the rule for finding the cube root .^ [20].
8. How is the square or cube root of a fraction or
of a mixed number found ? [14, &c., 19, 23, &c., 27].
9. How is any root found .' [28 and 29] .
10. How are the squares and cubes, the square roots
and cube roots, of numbers found, by the table .' [30].
LOG.AJHIHMS.
32. Logarithms are a set of artijidal numbers, which
represent the ordinary or twJural numbers. Taken
along with what is called the base of the system to
which they belong, they are the equals of the corres-
ponding natural numbers, but without it, they are
merely their representatives. Since the base is un-
chan^-eable, it is not written alons; with the losrarithm.
The logarithm of any number is that power of the base
which is equal to it. Thus 10^ is eqiuil to 100 ; 10 is
the hase^ 2 (the index) is the logarithm^ and 100 is the
corresponding natural number. — Logarithms, therefore,
are merely the indices "which designate certain powers
of some base.
33. Logarithms afford peculiar facilities for calcu-
lation. For, as we shall see presently, the multiplica-
tion of numbers is performed by the addition of their
330 LOGARITHMS.
logarithms ; one number is divided by another if we
subtract the logarithm of the divisor from that of the
dividend ; numbers are involved if we multiply their
logarithms by the index of the proposed power ; and
evolved if we divide their logarithms by the index of
the proposed root. — But it is evident that addition and
subtraction are much easier than multiplication and
division ; and that multiplication and division (particu-
larly when the multipliers and divisors are very small)
are much easier than involution and evolution.
34. To use the properties of logarithms, they must be
exponents of the same base — that is, the quantities raised
to those powers which they indicate must be the saine.
Thus 104 X 123 is neither 10' nor 12- , the former being
too small, the latter too great. If, therefore, we desire
to multiply lO"* and 12 ^ by means of indices^ we must
find some power of 10 which will be equal to 12^, or
some power of 12 which will be equal to 10^, or finally,
two powers of some other number whicb will be equal
respectively to lO'^ and 12^, and then, adding these
powers of the same number, we shall have that power
of it which will represent the product of lO"* and 123.
This explains the necessity for a taUe of logarithms —
we are obliged to find the powers of some one base which
will be either equal to all possible numbers, or so nearly
equal that the inaccuracy is not deserving of notice. The
base of the ordinary system is 10 ; but it is clear that
there may be as many difierent systems of logarithms
as there are difi'erent bases, that is, as there are different
numbers.
35. In the ordinary system — which has been calcu-
lated with great care, and with enormous labour, 1 is
the logarithm of 10 ; 2 that of 100 ; 3 that of 1000, &c.
And, since to divide numbers by means of these loga-
rithms (as we shall find presently), we are to subtract
the logarithm of the divisor from that of the dividend,
0 is the logarithm of 1, for 1=1^— 10^-'=10» ; — 1 is
10
the logarithm of -1, for •l=l=ii''=10o-'=10-' : and
. 10 io>
for the same reason, —2 is the logarithm of '01 ; —3
tJuit of -001, &e.
LOGAKITHitS. 52l
36. The logarithms of numbers Idiceen 1 and 10,
must' be more than 0 and less than 1 ; that is, must be
some decimal. The logarithms of numbers between 10
and 100 must be more than 1, and less than 2 ; that
is, unity with some decimal, &c. ; and the logarithms of
numbers between '1 and "01 must be — 1 and some deci-
mal ; between '01 and '00 1, —2 and some decimal, &c.
The decimal part of a logaritlim is always positive.
37. As the integral part or characteristic of a posi-
tive logarithm is so easily found — being [35] one l^ss
than the number of integers in its corresponding num-
ber, and of a negative logarithm one more than thi
number of cyphers j^rejixed in its natural number,
ifc is not set down ia the fcibles. Thus the logarithm
corresponding to the digits 9S72 (that ls, its decimal
part) is 99440-^ ; hence, the logarithm of 9872 Ls 3
•994405 ; that of 9S7-2 is 2-994405 ; that of 9-S72 is
0-994405 ; that of -9872 is- 1-994405 (since there is no
integer, nor prefixed cypher) ; of '009872— 3'994405,
&c. : — The same digits, whatever may be their value.
Lave the same decimals in their logarithms ; since it
is the integi-al part, only, which changes. Thus the
logarithm of 57864000 is 7-762408 ; that of 57S64, is
4-7G240S ; and that of -0000057864, i&— 6-762408.
3S. To find the loixarithrn of a given number, by the
table—
The iutcgi-al part, or charactenstic, of the logarithm
may be found at once, frora what has been just said [37] —
When tlie number is not greater than 100, it will be
found in the column at the top of which is N, and the
decimal part of its logarithm immediately opposite to it
in the next column to the riglit hand.
If the number is greater than 100, and less than
1000, it will also be found in the column marked N,
and the- decimal part of its logarithm opposite to it, iii
the column at the top of which is 0.
K the number contains 4 digits, the first three of
them will be found in the column under N, and th«
fourth at the top of the page ; and then its logarithm
iu ihe same horizontal line as the three first digits of
the given number, and in the same eolunm as its fourth
322 LOGARITHMS.
If the number contains more than 4 digits, find the
logarithm of its first, four, and also the dilference be-
tween that and the logarithm of the next higher nimi-
ber, in the table ; multipi}' this diiference by the remain-
ing digits, and cutting off from the pr(^luct so many
digits as were in the multiplier (but at the same time
adding unity if the highest cut off is not less than 5), add
it to th(? logarithm corresponding to the four fii'st digits.
Example 1. — The logarithm of 59 is 1-770852 (the charac-
teristic being positive, and one less than the number oi integers) .
KxAMPLE 2.— The logarithm of 338 is 2-528817.
Example 3.— The logarithm of -0004587 is — 4 (;G1529
(the characteristic being negative, and one more than the
number of prefixed cyphers) .
Ex-^MPLE 4.— The logarithm of 28434 is 4-453838.
For, the difference between 453777 the logarithm of 2843,
the four first digits of the given nmuber, and 453030 the
logarithm of 2844. the next number, is 153 ; -whicli, multi-
plied by 4, the remaining digit of the given number, pro-
duces 012: then cutting off one digit from this (since wo
have multiplied by only one digit) it becomes 61, which being
added to 453777 (^the logaritlim of 2844) makes 453838, and,
with the characteristic, 4-453838, the required logarithm.
Example 5. — The logarithm of 873457 is 5-941242.
For, the difference between the logarithms of 8734 and
8735 is 50, which, being multiplied by 57, the remaining
digits of the given number, makes 2850; from this we cut
off two digits to the right (since we have multiplied by two
digits), when it becomes 28: but as the highest digit cut
off is 5, we add unity, which makes 29. Then 5-941213 (the
logarithm of 8734) -{-29=5 -9412425 is the required logarithm.
39. Except when the logarithms increase very ra-
])idly — that is, at the commencement of the table — the
differences may be taken from the right hand column
(and opposite the three first digits of the given number)
where the mean differences will be found.
Instead of multiplying the mean difference by the
remaining digits (the fifth, &c., to the right) of the given
number, and cutting off so many places from the product
ys are C(|ual to (he number of digits in the multipli-jr,
to ublaiu the prqwrllutial pari — or what is to be added
L00APJTHM6. 3l83
to tlie logarithm of tlie first four digits, we may tako
the ^'oportional part corresponding to each of the re-
maining digits from that part of the column at the hft
hand side of the page, "which is in tlie same horizontal
division as that in which the fii-st three digits of the
given number have been found.
Example.— What is the logarithm of 839785 ?
The (decimal part of the) logarithm of 839700 is 924124.
Opposite to 8. in the same horizontal division of the page,
we tind 42. or rather, (since it is 80) 420, and opposite to
5, 2u. Hence the reiiuired logarithm is 024124-j-420-f-20=
924570 ; and, with the characteristic, 5-924570.
40. The method given for finding the proportional part — or
what is to be added to the next lower logarithm, in the table —
arises from tlie di.feroucc of numbers being proportional to the
ditfereuce of their logarithms. Hence, using the last example,
100 : So : : 52 (9:i4176, the logarithm of 839800—924124,
the logarithm of 839700) : ""Tqo"' °^' *'''^ difference (the mean
difference may generally be used'iXby the remaining digits of
the given number -r- 100 (the division being performed by cut-
ting off two digits to the right). It is evident that the number
of digits to be cut off depends on the number of digits in the
multiplier. The logaritiim found is not exactly correct, be-
cause numbers are not ea:ac//y proportional to the differences
of their logarithms.
The proportional parts set down in the left hand column,
have been calculated by making the necessary multiplica-
tions and divisions.
41. To find the logarithim of a fraction —
Rule. — Find the logarithms of both numerator and
denominator, and then subtract the former from the
latter ; this will give the logarithm of the quotient.
P:xample.— Log. i| is 1-672098 - 1-748187= - 1-023010.
We find that 2 is to be subtracted from 1 (the character-
istic of the numerator) : Imt 2 from 1 leaves 1 still to bo
8u])tractcd. or [Sect. II, 15] — 1, the characteristic of the
quotient.
We shall find presently that to divide one quantity bj
anotlior. wc have merely to subtract the lop:aritbm of the lattei
from that of the former.
42. To find the logarithm of a mixed number —
liuLE. — ^Reduce it to an improper fraction, and pro
ced as directed by the last rule.
324 LOGARITHMS.
43. To nud the numljcr wliicli corresponds to a given
logaritlim — •
If the logarithm itself is found in the tahle —
ItULE. — Take from the tahle the number "which cor-
responds to it, and place the decimal point so that there
may be the requisite number of integral, or decimal
places — according to the characteristic [37].
Example. — What number corresponds to the logarithm
4-214314 I
We find 21 opposite the natural number 103 ; and look-
i.'g along the horizontal lino, we find the rest of the logarithm
under the figure 8 at the top of the page: therefore the digits
of tlie required number are 1038. But as the characteristic
is 4. there must in it be 5 places of integers. Hence the
required nmnber is 10380.
44. If the given logarithm is not found in the table — •
KuLE. — Find that logarithm in the table which is
next lower than the given one, and its digits will be
the highest digits of the required number ; find the
difference between this logarithm and the given one,
annex to it a cypher, and then divide it by that differ-
ence in the table, which corresponds to the four highest
digits of the required number — the quotient will be the
next digit ; add another cypher, divide again by the
tabular difference, and the quotient will be the next
digit. Continue this process as long as necessary.
Example. — What number corresponds to the logarithm
5-054329 ?
G54273. which corresponds with the natural number 4511,
is the logarithm next less than the given one : therefore the
first four dio:its of the required number are 4511. Adding
a cypher to 50, the difference between 054273 and the given
logarithm, it becomes 500, which, being divided by 90, the
laJnjJar difference corresponding; with 4511, gives 5 as quo-
tient, and 80 as remainder. Therefore, the first five digits
of the required number are 45115. Adding a cypher to 80,
it becomes 800 ; and, dividing this by 96, we obtain 8 as
the next digit of the required number, and 32 as remainder.
I'lie integers of the required number (one more than 5, the
characteristic) are, therefore, 451158. We may obtain the
decimals, by continuing the addition of cyphers to the re-
mainders, and the division by 90.
LOGARITHMS. 325
45. We arrive at tlie same lesult, Ly subtracting
from -the difference between the given logarithm and
the next less in the table, the highest (which does not
exceed it) of those proportional parts found at the right
hand side of the page and in the same horizontal divi-
sion vv'ith the first three digits of the given number —
continuing the process by the addition of cyphers, until
nothing, or almost nothing, remains.
Example. — Using the last, 4511 is the natural number
corresponding to the logarithm G54273, which differs from
the given logarithm by 50. The proportional parts, in the
same horizontal division as 4511, are 10. 19, 29, 38, 48. 58,
67, 77, and 80. The highest of these, contained in 56, is
48. which we find opposite to, and therefore corresponding
witli. the natural number 5: hence 5 is the next of the
required digits. 48 subtracted from 50, leaves 8 : this, when
a cypher is added, becomes 80, which contains 77 (corres-
ponding to the natui'al number 8) ; therefore 8 is the next
of the required digits. 77, subtracted from 80, leaves 3 ;
this, when a cypher is added, becomes 30, &c. The inte-
gers, therefore, of the required number, are foimd to be
451158, the same as those obtained by the other method.
The rules for finding the numbers corresponding to
given logarithms are merely the converse of those used
for finding the logarithms of given numbers.
Use of Logarithms in Arithmetic.
46. To multiply numbers, by means of their loga-
rithms—
Rule. — Add the logarithms of the factors ; and the
natural number corresponding to the result will be the
required product.
Example.— 87 x24=l-939519 (the log. of 87) -f- 1-380211
(the log. of 24)=3-319730; which is foxmd to correspond
with the natural number, 2088. Therefore 87x24=2088.
Reason of the RrLiE. — This mode of multiplication arises
from the vei-y nature of inuices. Thus 5^Xo*=5x5XjXo
multiplied oXoXoXoXoXoXoXo ; and the abbreviation for
thJs [2] is 5'-. Bat 12 is equal to the sum of the indices
(logarithms). The rule might, in the same way, be proved
correct by any otlicr example.
326 LOGARITHM?
47. yriien the characteristics of the logarithms^ to be
added are both positive, it is evident that their sum will
be positive. AVhcu they are both negative, their sum
(diminished by what is to be carried from the sum of
the positive [3oJ decimal parts) vrill be negative. Wlieu
one is negative, and the other positive, subtract the less
from the greater, and prefix to the difierence the sign
belonging to the greater — bearing in mind what has
been already said [Sec. II. 15] with reference to the
subtraction of a greater from a less quantity.
48. To divide numbers, by means of their logarithms —
Rule. — Subtract the logarithm of the divisor from
that of the dividend ; and the natural number, corres-
ponding to the result, wiU be the required quotient.
Example.— 1134-^-42 ==3- 0546 13 (the log. of 1134) —
1 623240 (the log. of 42) = 1-431364, which is found to
correspond vdth the natural number. 27. Therefore 1134-r-
42=27.
Reason' of the Rule. — This mode of division ax'ises from
tlie nature of indices. Thus 4'-f-4'=[2] 4X'1x4x4x4h-4X
^ ^ 4X4X4X4X4 ^ ^ 4x4x4 ^ , ,
4X4= — 4^^4^4 — =^X4X^^^^^=4X4, the abbreviation
for -which is 4'^ But 2 is equal to the index (logarithm) of
the dividend minus that of the divisor. The rule might, in
the same "way, be proved correct by any other example.
49. In subtracting the logarithm of the divisor, if it
is negative, change the sign of its characteristic or inte-
gral part, and then proceed as if this wore to be added
to the characteristic of the dividend ; but before making
the characteristic of the divisor positive, subtract what
was borrowed (if any thing), in subtracting its decimal
part. For, since the decimal part of a logarithm is
positive, what is horrowed^ in order to make it possible
to subtract the decimal part of the logarithm of the
divisor from that of the dividend, must be bo much
taken oway from what is positive, or added to what is
negative in the remainder.
Wo change the si^n of tho negative eharactoristic, and
then add it; for. adding a positive, is the same as taking
away a negative quantity.
LOGARITHMS. 827
50. To raise a quantity to any powor, by means of
it*- logarithm —
lluLE. — Multiply the logarithm of tlie quanity by
the index of the pov/er ; and the natural number cor-
responding to the result will be the requu-ed power.
Example. — RaLso 5 to the 5th power.
The If.garithm of 5 is 069897, which, multiplied by 5,
give.s 3-49485, the logarithm of 3125. Therefore, the 5th
power of 5' is 3125.
Keason of the Rule. — Tliis rale also follows from tlie
nature of indices, o" raised to the 5th povrei- is oX-J multiplied
by 6x^ iiiultiplied by 5x5 -multiplied by 5x5 multiplied
by 5X5, or 5x5x5x5x5x5x5x5x5x5, the abbreviation
for •W'liicli is [2] 5'^ But 10 is equal to 2, the index (logarithm)
of the quantity, nuiltiplied by 5, that of the power. The
rnle might, in the same Avay, be proved correct by any other
example.
51. It follows from what has been said [47] that when
a negative characteristic is to be multiplied, the product
in negative ; and that what is to be carried from the
nfiultijKJvcation of the decimal part (always positive) is
to be suhtrddcd from tlii.s negative result.
52. To evolve any quantity, by means of its loga-
rithm—
lluLE. — Divide the logarithm of the given quantity
by that number which expresses the root to be taken ;
and the natural number corresponding to the result will
be the required root.
Example.— What is the 4th root of 2401.
The logarithm of 2401 is 3-380392, which, divided by 4,
the number expressing the root, gives -845098, the logarithm
of 7. Therefore, the fourth root of 2401 is 1 -.
Reaso.v of the Pvule. — This rule follows, likewise, from
the nature of indices. Thus tlie 5tli ruot of 16" is such a
number as, raised to the 5th power — that is, taken 5 times as
q, factor — would produce 16^°. But 16 oj taken 5 times as a
factor, would produce 16^". The rule might be proved correct,
equally ^vell, by any other example.
53. "Wlien a negative characteri.^Mic is to be divided —
E-Tl;: I. — If tlie cliaractcristic is cxadJy divi.-^ible by
the divisor, divide in the ordinary way, but make tho
characteristic of the quotient negative.
o2S LOGARITHMS.
TI. — If ilic negative characteristic is not exactly
divisible, add what will make it so, both to it and to the
decimal part of the logarithm. Then proceed with the
division.
Example. — Divide the logarithm —4-837564 by 5.
4 Avants 1 of being divisible by 5 ; then —4- 837504-4-5=
— 5-fi-837564-^5==l-367513, the rec^uired logarithm.
Reason of I. — The quotient multiplied by the divisor must
give the dividend; but [51] a negative quotient multiplied by
a positive divisor will give a negative dividend.
Keasox of II. — In example 2, -we have merely added -f- 1
{ind — 1 to the same quantity — which, of course, does not
.alter it.
QUESTIONS.
1 . What are logarithms .? [32] .
2. How do they facilitate calculation .' [33].
3. Why is a table of logarithms necessary } [34].
4. What is the characteristic of a logarithm ; ant?
how is it found .^ [37] .
5. How is the logarithm of a number foundry th<»
table.? [38].
6. How are the "differences," given in the table
used.? [39].
7. What is the use of ** proportional parts .?" [39].
S. How is the logarithm of a fraction found } [41].
9. How do we find the logarithm of a mixed num-
ber } [42] .
10. How is the number corresponding to a given
logarithm found } [43] .
1 1 . How is a number found when its corresponding
logarithm is not in the table .' [44] .
12. How are multiplication, division, involution and
evolution effected, by means of logarithms .? [46, 48,
50, and 52] .
13. Yfhen negative characteristics arc added, what
is the sign of their sum } [47] .
14. What is the process for division, when the cha-
racteristic of the divisor is negative } [49] .
15. How is a negative characteristic multiplied : [51].
IG. How id a uogat/ve charactoristic divided } [53]
329
SECTION X.
PROGRESSION, &c.
1. A progression consists of a niiniber of quantities
increasing, or decreasing by a certain law, and forming
^'liat are called coiitmued proporiionals. When the
terms of the series constantly increase, it is said to
be an ascending^ but when they decrease (increase to
the left), a descending series.
2. In an cqiiidijj'erent or arithmeiical progression, the
quantities increase, or decrease by a cvmmon difference.
Thus 5, 7, 9, 11, &c., is an ascending, and 15, 12, 9, 6,
&c., is a descending arilhrneiical series or progression.
The common difference in the former is 2, and in the
latter 3. A continued proportion may be formed out
'r)f such a series. Thus —
5:7::
9 : G, &c.
15 : 12 ::
3. In a geometrical or eq id rational progression, the
quantities increase by a common ratio or multiplier.
Thus 6, 10, 20, 40, &c. ; -and 10000, 1000, 100, 10, &c.,
are geometrical scries. The common ratio in the former
case is 2, and the quantities increase to the right ; in
the latter it is 10, and the quantities increase to the
left. A continued proportion may be formed out of
such a series. Thus —
5 : 10 : : 10 : 20 : : 20 : 40, &c. ; and 10000 : 1000 : :
1000 : 100 : : 100 : 10, &c. Or we may say 5 : 10 : : 20 :
40 : : &c. ; and 10000 : 1000 : : 100 : 10 : : &c.
4. The first and last terms of a progression are called
its extremes^ and all the intermediate terms its means.
5. Arithmeiical l^rogression. — To find the sum of a
series of terms in arithmetical progression —
Rule. — Multiply the sum of the extremes by half
tlic number of terms.
7 :9 ::9: 11, &c.; and 15
: 12 :
: 12 :
: 9 : :
Or we may say 5 : 7 : ; 9
: 11 :
: &c.
; and
9 : 6 : : &c.
330 PROGRESSION.
Example. — What is the sum of a series of 10 terms, tho
first being 2, and last 20 '? Ans. 2-|-20x ^=110.
G. Reason of the Kulk. — Tliis rule can be easily proved.
For this purpose, set down the progression twice over — but
in such a way as that the last term of one shall be under the
first term of the other scries.
Then, 24+21-1-18+154-12-1- 0=the sum.
9+12-f-15-j-18+21-|-24=the sum. And,
adding the equals, 33+33-j-33+33+33+33=twice the sum.
That i.s, twice t!ie sum of the series will be equal to the sum
of an many quantities as there are terms in the series — each
of tlie quantities being equal to the sum of the extremes.
And tlie sum of the series itself will be equal to half as much,
or to the sum of tlie extremes taken ha/f as many times as
there are terms in the series. The rule might be proved
correct by any other example, and, therefore, is general.
EXERCISES.
1. One extreme is 3, the other 1.5, and the number
of terras is 7. YVhat is the sum of the series ? Ans. 63.
2. One extreme is 5, the other 93, and the number
of terms is 49. What is tke sum ? Ans. 2401.
3. One extreme is 147, the other |, and the number'
of terms is 97. V/hat is the sum ? Ans. 7165-875.
4. One extreme is 4f , the other 143, and the num
ber of terms is 42. What is the sum ? Ans. 3094-875
7. Given the extremes, and number of terms — to find
the common diiference —
lluLE. — Find the diiference between tho given ex-
tremes, and divide it by one less than the number of
terms. The quotient will be the common difference.
Example. — In an arithmetical series, the extremes are 21
and 3, and the number of terms is 7. What is the common
diiference ?
21 — 3-f-7 — 1 = 18-J-6 = 3, the required number.
8. Reason of the Rule. — The difference between the
greater ami lesser extreme arises frcm the common differcnco
being added to the lesser extreme once for every term, ex-
cept tbe lowest ; that is, the greater contains the lesser extreme
plus tlie common difference taken once less than the number
of terms. Therefore, if we subtract the lesser from the greater
extreme, the difference obtaiued will be equal to the common
difference multiplied by one less than the number of terms.
And if we divide the difference by one less than the number
of terms we will have the cummun diiference.
TROCRESSIO-^ 331
EXERCISES.
5. The extremes of an arithmetical series are 21
and 497, and the number of terms is 41. What is the
common difference r Ans. 11-9.
C. The extremes of an arithmetical series are 127||
and 9|, and the, number of terms is 26. What is the
common difference ? Ans. 4f .
7. The extremes of an arithmetical series are 77||
and f , and the number of terms is 84. What is the
common difference } Ans. \^.
9. To find a?nj numhcr of arithmetical means between
two given numbers —
Rule. — Find the common difference [7] ; and, ac-
cording as it is an ascendimj or a descendiuix series, add
it to, or subtract it from the first, to form the second
term ; add it to, or subtract it from the second, to form
the third. Proceed in the same way with the remain-
ing terms.
We must remember that one less than the number of
terms is one more than the number of means.
Example 1. — Find 4 arithmetical means between 6 and
21. 21—6 = 15. TTT=3) the common difference. And
the series is —
6 . 6+3 . 6-f-2x3 . 64-3x3 . 6-{-4x3 . 6-|-5x3.
Or6 . 9 . 12 . 15 . 18 . 21.
Example 2. — Find 4 arithmetical means between 30 and
20
10. 30 — 10=20. 1 , 1=^' the common difference. And
"• ^
the series is — ■
30 . 26 . 22 . 18 . 14 . 10
This rule is eTideut.
EXERCISES.
8. Find 11 arithmetical means between 2 and 26
Am. 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, and 24.
9. Find 7 arithmetical means between 8 and 32
..4715. 11, 14, 17, 20, 23, 26, 29.
10 Find 5 arithmetical means between 4^, and 13^
Ans. 6, 7
\
332 PROGRESSION.
10. Given tlie extremes, nnd tlie number of terms —
to find any term of an arithmetical progression —
Rule, — Find the common diiference by the last rule,
and if it is an ascending series, the required term wiU
be the lesser extreme fliis — if a descending series, the
greater extreme minus the common difference multiplied
by one less than the number of the term.
Example 1. — What is the 5th term of a series containing
9 terms, the first being 4. and the last 28 1
28-4
— Q — =3, is the common difference. And 4-f-3x5 — 1=
16, is the required term.
Example 2. — What is the 7th term of a series of 10 terms,
the extremes being 20 and 2 T
20-2 —
— q— =2, is the common difference. 20 — 2x7—1=8.
is the required term.
11. Reasojv of the Kxtle. — In an ascending series th?
required term is greater than the given lesser extreme to the
amount of all the differences found in it. But the number of
differences it contains is equal only to the number of terms
■which p7-ec€de it — since the common difference is not found in
the^ir*-^ term.
In a descending series the required term is less than the
given greater extreme, to the amount of the differences sub-
tracted from the greater extreme — but one has been subtracted
from it, for each of the terms y^hich precede the required term.
EXERCISES,
11. In an arithmetical progression the extremes are
14 and 86, and the number of terms is 19. What is
the 11th term f Ans. 54,
12. In an arithmetical series the extremes are 22 and
4, and the number of terms is 7. What is the 4th
t^rm .' Ans. 13.
13. In an arithmetical series 49 and f are the ex-
tremes, and 106 is the number of terms. What is the
94th term ? Ans. 6-2643.
12. Given the extremes, and common difference — to
find the number of terms —
Rule. — Divide the difference bet^vcen the given ex-
tremes by the common difference, and the quotient plus
unity will bs the number of terms.
I'KOGRESSrON. 'S'S'i
ExAMPi.F. — How many terms in tin arithmetical scries of
whicli the extremes arc 5 and 20, and the common ditfer
ence 3 ?
20 -5
— r; — =7. And 7-j-l=8, is the nmnbcr of terms.
o
13. PtKAsorf OF THE RuLE. — The greater differs from tho
lesser extreme to the amount of tlie dili'crcnces found in all the
terms. But th» common ditierence is found in all the terms
except the lesser extreme. Therefore the difference between
the extremes contains the common difference once less than
will be expressed by the number of terms.
EXERCISES.
14. In an arithmetical series, the extremes are 96
and 12, and the common difference is 6. What is the
number of terms ? Ans. 15.
15. In an arithmetical series, the extremes are 14
and 32, and the common difference is 3. What is the
number of terms ? Ans. 7.
16. In an arithmetical series, the common difference
is |-, and the extremes are 14f and xl. What is the
number of terms } Ans. S.
14. Given the sum of the series, the number of terms,
and one extreme — to find the other —
IvULE. — Divide twice the sum by the number of
terms, and take the given extreme from the quotient
The difference will be the required extreme.
Example. — One extreme of an arithmetical series is 10
the number of terms is 6, and the sum of the series is 42
^Vhat is the other extreme 1
2x42
— T. — —10 = 4, is the required extreme.
15. Reason of the Rule. — We have seen [5] that 2 X the
sum = sum of the extremes X the number of terms. But if we
divide each of these equal quantities by the number of terms,
■we shall have
2 X the sum sura of extremes X the number of terms
the number of terms the number of terms
2 X the sura „ , . , ,
Or IT— i n-i = sum of the extremes. And sub-
tlie number oi terms
tracting the same extreme from each of these equals, tvc shall
have
.."^^..-il^'" — oneuxtvomc=tlie sum of the extremes
the number of terms
the same extreme.
twice the sum
Or 7i ^r IT- minus one extreme = the other cx-
the number ot terms
.I'cme.
EXERCISES.
17. One extreme is 4, the number of terms is 17,
/ind the sum of the series is 884. What is the other
extreme ? Am. 100.
18. One extreme is 3, the number of terms is 63,
and the sum of the series is 252. What is the other
extreme ? Ans. 5.
19. One extreme is 27, the number of terms is 26,
and the sum of the scries is 1924. What is the other
extreme ? Ans. 121.
16. Geometrical Progressian. — Given the extremes
and common ratio — to find the sum of the series —
EuLE. — Subtract the lesser extreme from the product
of the greater and the common ratio ; and divide the
difference by one less than the common ratio.
ExAj.iPLE. — In a geometrical progression, 4 and 312 are
the extremes, and the common ratio is 2. What is the sum
of the series.
312 X 2 4
^ ^ = 620, the required number.
17. Reason of the Rule. — The rule may be proved by
setting down the series, and placing over it (but in a reverse
order) the product of each of the terms and the common ratio.
Then
Sum X common ra'fto = 8 -f- 16 + 32, &c. . + 312 + 624
Sum= 4+8 + 16-1-32, &c. . +312 .
And, subtracting the lower from the upper line, we shall have
Sum X common ratio — Sum = 624 — 4. Or
Common ratio — 1 X Sum = 624 — 4.
And, dividing each of the equal quantities by the common
ratio minus 1
642 (last term X common ratio)— 4 (the first term)
Sum = Ti — ■ — ^ ~~
common ratio — 1
Which \6 the rule.
PROfiRE.SSION » 335
EXERCISES.
20. The extremes of a geometricTil scries are 512 and
2, and the common ratio is 4. What is the simi .'
Alls. 682.
21. The extremes of a geometrical series are 12 and
175692, and the common ratio is 1 1. VV^hat is the sum }
Ans. 193260.
22. The extremes of an infinite geometrical series
are yV ^^^ ^j ^^'^ tV ^'^ ^'^^ common ratio. ^Vhiit is
the sum.' Ans. ±. [Sec. lY. 74.]
Since the series is infinite, tlic lesser extreme=0
23. The extremes of a geometrical series are "3 and
937*5, and the coinmon ratio is 5. Yv^hat is tlie sum ?
Alls. 1171-875.
18. Given the extremes, and number of terms in a
geometrical seiies — to find the common ratio —
HuLE. — Divide the greater of the given extremes
by the lesser ; and take that root of the quotient v/hicii
is indicated by the number of terms minus 1. This will
be the required number.
Example.— 3 and 80 are the extremes of a geometrical
progression, in which there are 5 terms. What is the eo»v-
men ratio T
80
7r-==lG. And ^'10=2, the required common ratio.
19. RcAsoiv OT THE RuLE. — The greater extreme is e.<inal
to the lesser multiplied by a product which has for its factors
tlic comuioa ratio tuken once less than the number of terms —
since the common ratio is not found in the J??-.?/! term. That
is, the greater extreme contains the common ratio raised to a
j.Dwer indicated by 1 less th:)n tlie number of terras, ami mul-
tipliol by the lesscr extreme. Conseqncutlj if, after dividing;
by the lesser extreme, we take tliat root of the quotient, which
is indicated by one less than the number of terms, we shall
obtain the coaimon ratio itself.
EXERCISES
24. The extremes of a geometrical series are 491 02
and 3, and the number of terms is 8. What is the
common ratio .- Al7is. 4.
25. The extremes of a geometrical series arc 1 and
336 TROGRKSSIOX.
!I5625, and the number of terms is 7 What is the
common ratio ? Ans. 5.
26. The extremes of a geometrical series arc
20176S035 and 5, and the number of terms is 10
^Yhat is the common ratio ? Ans. 7.
20. To find 0711/ number of geometrical means be
twcen two quantities —
Rule. — Find the common ratio (by the last rule)^
and — according as the series is ascending, or descend-
ing— multiply or divide it into the first term to obtain
the second ; multiply or divide it into the second ta
obtain the third ; and so on with the remaining terms.
We must remember that one less than the number
of terms is one more than the number of means.
Example 1. — Find 3 geometrical means between 1 and
81.
^-j-=3; the common ratio. And 3. 9, 27, arc the re-
quired means.
Example 2. — Find 3 geometrical means between 12")f
and 2.
1250 , 1250 1250 1250 ^. ^ ^, ^,
4/-^=5. And -J- ^^ 5-^^5^, or L.O. 50, }•;
are the required means.
This rule requires no explauaMon.
EXERCISES.
27. Find 7 geometrical means between 3 and ly 683 '
Ans. 9, 27, SI, 243, 729, 2187, 6561.
28. Find S geometrical means between 4096 and 8 ?
Ans. 2048, 1024, 512, 256, 128, 64, 32, and 16.
29. Find 7 geometrical means between 14 and
23514624.? Ans. 84, 504, 3024, 18144, 108864,
653184, and 3919104.
21. Given the first and last term, and the number of
terms — to find any term of a geometrical series —
Kule. — If it bo an ascending series, multiply, if a
descending series, divide the first term b;y that })Owof
of the common ratio which is indicated by the numoei
of the term minus 1.
PROGRESSmiV. 337
Example 1.— Find tlie 3itl term of a geometrical scries,
of Avhich the fir^t term is 0, the last 1456. and the number
of terms G.
Tlie common ratio is ^-.:~=Z. Therefore the required
term is Cx 3^=54.
E3L4MPLE 2. — Find the 5th term of a series, of which tli*
extremes are 524283 and 2. and the number of terms is 10.
5242S8 5242.^^
The common ratio ^—7^ — =4. And — ,4 " = 20rl8,
is the required term.
22. Rkasox of the Ritlk.— In .an ascending series, anj
term is the product of the tirst and the coiumou ratio taken
as a factor so many times as there are preceding terms — sirwe
it is not found in the first term.
la a descending series, any terra is equal to the first term,
divided by a product containing the common ratio as a factor
so many limt-s as there are preceding terms — since every term
bat that which is required adds it once to the factors which
constitute the divisor.
EXERCISES.
30. What is the 6th term of a series having 3 and
5859375 as extremes, and containincc 10 terms r A as.
9375.
31. Given 39366 and 2 as the extremes of a series
Laving 10 terms. What is the 8th term ? Ans. 18.
32. Given 1959552 and 7 as the extremes of a series
having 8 toims. AVhat is the 6ih term ? A71S. 252.
23. Given the extremes and common ratio — to find
the number of terms —
Rule. — Divide the greater by the lesser extreme,
and one more than the number expressing Tvhat pov/er
of common ratio is equal to the quotient, will be tho
requh'ed quantity.
Example. — Wow rmmj terms in a series of wluch tho
extremes are 2 and 25G. and the common ratio is 2 ?
256
-o-=128. But 2"=128. There are, therefore, 8 termc
The common ratio is fonnd as a ftictor (in the quotient of
the greater dividcil by the lesser extreme) once less than the
number of terms.
838 PROGRESSION.
EXERCISES.
33. IIow many terms lu a series of wliicU the first is
78732 and the last 12, and tlie cuninion ratio i.s 9 ^
Ans. 5.
34. How many terms in a series of wLicli the ex-
tremes and common ratio are 4, 470.596, and 7 ? Aiis. 7.
35. How many terms in a series of which the ex-
tremes and common ratio, are 19G608, 6, and 8 .' Aiis. 6.
24. Given the common ratio, number of terms, and
one extreme — to find the other —
lluLE. — If the lesser extreme is given, multiply, if
the greater, di\dde it by the common ratio raised to a
power indicated by one less than the number of terms.
Example 1. — In a geometrical series, the le.sser extreme
is 8, the number of terri\s is 5. aiul the common ratio is G;
what is the other extreme '? Ans. 8xG'~'=10oG8.
ExA?iPLE 2. — In a geometrical serie.-^, the greater extreme
is 6561, the number of terms is 7, and the conmion ratio is
S ; M'hat is the other extreme '? Am. 65G1h-3'~'=0.
This rule does not require any explanation.
EXERCISES.
36. The common ration is 3, the number of terms is 7,
and one extreme is 9 ; what is the other } Ans. 6561.
37. The common ratio is 4. the number of terms is
6, and one extreme is 1000 ; what is the other : Aiis.
1024000.
38. The common ratio is 8, the number of terms i*
10, and one extreme is 402653184 ; what is the other >
Ans. 3.
In progression, as in many other rules, the application of
algebra to the reasoning would greatly simplify it.
MISCELLANEOUS EXERCISES IN PROGRESSION.
1 . The clocks in Venice, and some other places strike
the 24 hours, not bcoinning again, as ours do, after 12.
How many strokes do they give in a day t Ans. 300.
2 A butcher bought 100 sheep; for the first ho
gave l5., and for the lust X'O Yds. AVhat did ho pny for
PROGRESSIOJf. 339
all, supposi ig tlicii- prices to form an arithmetical scries ?
Alls. £di)0.
3. A person bought 17 yards of clotli ; for the fii-st
yard he gave 25., and for the last 105. AVhat was the
price of all .' Ans. £5 2s.
4. A person travelling into the country went 3 miles
the first day, S miles the second, 13 the thu-d, and so
on, until he went 53 miles in one day. How many days
did he travel } Ans. 12.
5. A man being asked how many sons he had, said
that the youngest was 4 years old, and the eldest 32,
and that he had added one to his family every foui'th
year. How many had he ? Ans. 8.
6. Find the sum of an infinite series, i, i, Jy, &c.
Ans. \.
7. Of what value Is the decimal -463' ? -4?^. ^f f .
S. What debt can be dLscharged in a year by montlily
payments in geometrical progression, the first term
being £1^ and the last ^22048 ; and what will be the
common ratio .' Ans. The debt will be £4095 ; and
the ratio 2.
9. What vrill be the price of a horse sold for 1 far-
thing for the first nail in his shoes, 2 farthings for the
second, 4 for the third, &c., allowing S nails in each
shoe } Ans. £4473924 5^. ^d.
10. A nobleman dying left 11 sons, to whom he be-
queathed his property as follows ; to the youngest he
gave £1024; to the next, as much and a half; to the
next, 1^ of the preceding son's share ; and so on. What
was the eldest son's fortune ; and what wa.s the amount
of the nobleman's property .' Ans. The eldest son re-
ceived £59049, and the father was worth £175099.
QUESTIONS.
1. T^Tiat is meant by ascending and descending
series? [1].
2. What is meant by an arithmetical and geome-
trical progression ; and are they designated by any other
names .' [2 and 3] .
3. What are the common dllierence and common
ratio } [2 and 3] .
540 ANNL'IT{i:S -
4 b).-:}N tlizU a contiiuiofl proportion mny ]>g f.irraed
froTi. a Scries of either kind : [2 and 3] .
o. What are means and extremes ? [4].
6. ilow is the sum of an arithmetical or a geome-
trical series found? [5 and 16].
7. How is the common difference or common ratio
found? ['/and 18].
8. How is any number of arithmetical or geometrical
means fourd ? [9 and 20].
9. How is any particular arithmetical or geometrical
mean found ? [10 and 21 ] .
10. How is the number of terms in an arithmetical
or geometrical series found ? [12 and 23].
11. How is one extreme of an arithmetical or geome-
trical scries found ? [14 and 24].
ANNUITIES.
25. An annuity is an income to be paid at stated
times, yearly, half-yearly, &c. It is either in possession^
that is, entered upon already, or to be entered upon
inniiediately ; or it is in reversion^ that ls, not to com-
mence until after some period, or after something has
occurred. An annuity is certain when its commence-
ment and termination are assigned to definite periods,
anitingtnt when its beginning, or end, or both are
uncertain ; is in arrears when one, or more payments
are retained after the}^ have become due. The amount
of an annuity is the sum of the payments forborne (in
arrears), and the interest due upon them.
When an annuity is paid off at once, the price given
for it is called its present zcortk^ or valne — ^^vhich ought
to be such as would — if left at compound interest until
the annuity ceases— produce a sum equal to what would
be due from the annuity left unpaid until that time.
This value is said to \q so many years'' purchase; thai
is, so many annual payments of the income as would be
just equivalent to it.
26. To find the amount of a certain number of pay-
ments in arrears, and the interest due on them —
AX^•UITIES• 341
Hule. — Find the interest due on each pn3micnt ; then
the sum of the payments and interest due on them, will
be the rec|uk^d amount.
Example 1. — What will he the amount of cGl .per annum,
unpaid for G years, 5 per cent, simple interest being allowed I
The last, and preceding payments, Avith the interest due on
them, form the arithmetical series £l-j-<£05 x 5 . £ l-|-£05 x
4 . . £lx£05 £1. And its sum is £l+£l+£U5x5x
t=£2-|-£-25x3=£6-75=£G 15.s., the required amount.
ExA.AiPLE 2. — If the rent of a farm wortli £00 per annum
is mipaid for 19 years, how much does it amount to, at 5
per cent, per an. compound interest '.■
In this case the series is geometrical ; and the last payn^ent
with its interest is the amount of £1 for 18 (19 — 1) yepra
luultiplied by the given annuity, the preceding payment
with its interest is the amount of £1 for 17 years multiplied
];y tlic given annuity. Sec.
The amount of £1 (as wc find by the table at the end of
the treatise) fur 18 years is £2-40G62. Then the sum of
tlie series is—
£2-40GG2xl-05xC0-C0
Tpr 1 [lGJ=1832-4, the required amount.
llie amount of £1 for 18 years multiplied by 105 is the
same as the amount of £1 for 19, or the given numl^n- of
years, which is found to be £2527. And 105 — 1, the divi-
sor, is equal to the amount of £1 for one payment minus
£1 ; that is, to the interest of £1 for one payment, llcnoe
£2-527x00 — 00
the required sum will be :r—~ = £1832-4.
It would evidently be the same thing to consider the
annuity as £i, and then multiply the result by CO. Thus
?^^^ X 60 = £1832-4. For an annuity of £00 ought
to be 00 times as productive as one of only £1.
llcnco, briefly, to find the amount of any number of
payments in arrears, and the coivpowral interest due on
them —
Subtract £1 from the amount of £1 for the given
number of payments, and divide the difference by 4hq
interest of £1 for one payment ; tlicn multiply the qn.j -
tii-nt by the given sum. '
.__ Q
ANNUITIES.
27. Reason ok thk Hule. — Each payment, with its inte-
resc, eviih'Uilj constitute a separate amount ; Qui the sum dae
must be the sum of tlie.se amounts — which form a decreasing
series, because of the decreasing interest, arising from the
decreasing number of times of payment.
When simple interest is allowed, it is evident that what is
due will be the sum of an arithmetical series, one extreme of
VfhJch is the first payment plus tlie interest due upon it at the
time of the last, the other the last payment ; and its common
difference the interest on one payment due at the next.
But when compound interest is allowed, what is due will be
the sum of a geometrical series, one extreme of which is the
first payment plus the interest due on it at the last, the other
the last payment; and its common ratio £1 plus its interest
for the interval between two payments. And in each case the
interest due on the first payment at the time of the last will
be the interest due for 07ie less than the number of payments,
since interest is not due on the first until the time of the 8econ<i
payment.
EXERCISE.S.
1. "What is tlie amount of i£37 per annum unpaid
for 11 years, at 5 per cent, per an. simple interest.^
Am. £508 155.
2. What is the amount of an annuity of iSlOO, to
continue 5 years at 6 per cent, per an. compound inte-
rest ?^ Ans. £563 145. 2\d.
3. 'What is the amount of an annuity of £356, to
continue 9 years, at 6 per cent, per an. simple interest .'*
Ans. £3972 195. 2\d.
4. What is the amount of £49 per annum unpaid
for 7 years, 6 per cent, compound interest being allowed }
Ans. £411 55. 11^^.
28. To find the present value of an annuity —
Rule. — Find (by the last rule) the amount of the
given annuity if not paid up to the time it will cease.
Then ascertain how often this sum contains the amount
of £1 up to the same time, at the interest allow«^d.
Example. — What is the present worth of an annuity of
£12 per annum, to be paid for 18 years, 5 per cent, com-
pound interest being allowed 1
An annuity of £12 unpaid for 18 year.s would amount t«
ANN-L'iT[i:o. 343
Bat CI put to interest for 18 yeai'3 at the same rate
would amouiit to X2-40o02. Tlieretbre
£oo7-5S85G «-. ,^ ^ ^ , . , . -, ,
— aThr-o =£140 5.S. Gd. is the required value.
The sum to be paid for the annuity should evidently be such
as would produce the same as the annuity itself, in the same
time.
EXERCISES.
5. "What is the present worth of an annuity of jg27,
to be paid for 13 years, 5 per cent, compound interest
being allowed ? Ans. £>2d3 I2s. d^d.
6. What is -the present worth of an annuity of iS324,
to be paid for 12 years, d per cent, compound interest
being allowed .' Atis. i:2S71 135. lO^d.
7. What is the present worth of an annuity of £22,
to be paid for 21 years, 4 per cent, compound interest
being allowed ? Ans. £308 12s. lOd.
29. To find the present value, when the annuity is
in perpetuity —
IluLE. — Divide the interest which £1 would produce
in perpetuity into £1, and the quotient will be tn^ sum
required to produce an annuity of £>! per annum in
perpetuit}'. Multiply the quotient by the number of
pounds in the given annuity, and the product vVill be
the requii-od present worth.
Example. — What is the value of an income of £17 for ever *
Let us suppose that £100 would produce £5 per C(;nt. per
an. for ever : — -then £1 would produce £-05. Therefore,
to produce £1, we require as many pounds as will be equal
£1
to the number of times £05 is contained in £1. But 777^=*
£20, therefore £20 would produce an annuity of £1 for
ever. And 17 times as much, or £20x17=340, which
would produce an annuity of £17 for ever, is the required
present value.
EXERCISES.
S. A small estate brings £25 per annum ; whit is
its present worth, allowing 4 per cent, per annum ir\te-
rest ? Ans. £625.
9. What is the present worth of an income of £347
344 ANNUITIES.
in perpetuity, allowing G per cent, interest ? Ans
it35783 Qs. ScL
10. What is the value of a perpetual annuity oi £46,
allowing 5 per cent, interest ? Ans. .£920.
30. To find the present value of an annuity in rever-
sion—
Rule. — Find the amount of the annuity as if it were
forborne until it should cease. Then find- what sum,
put to interest now, would at that time produce the
same amount.
ExAiNiPLE. — What is the value of an annuity of £10 per
annum, to continue for 6, but not to commence for 12 years,
5 per cent, compound interest being allowed ?
An annuity of XIO for 6 years if left unpaid, would be
worth £68-0191 ; and £1 would, in 18 years, be worth
£1108959. Therefore
£080191
11-68959 ~^'^^ ^^' ^^^■' ^^ ^^^ required present worth.
EXERCISES
11. what is the present worth of £75 per annum,
which is not to commence for 10 years, but will con-
tinue 7 years after, at 6 per cent, compound interest ?
Ans. £155 9.9. 7^d.
12. The reversion of an annuity of £175 per annum,
to continue 11 years, and commence 9 years hence, is to
be sold ; what is its present worth, allowing 6 per cent,
per annum compound interest ? Ans. £430 7s. Id.
13. What is the present worth of a rent of £45 per
annum, to commence in 8, and last for 12 years, 6 per
cent, compound interest, payable half-yearly, being
allowed ^ Ans. £117 2s. S^d.
31 When the annuity is contingent, its value depend,^
on the probability of the contingent circumstance, or
circumstances.
A life annuity i.^ equal to its amount multiplied by
the value of an annuity of £1 (found by tables) for the
given age. The tables used for the purpose are calcu-
lated on principles derived from the doctrine of chances,
observations on the duration of life in different circum-
stances, the rates of compound interest, &.c.
POSITION^ 345
QUESTIONS.
1 . Whnt is an annuity ? [25] ,
2. AVhat is an annuity in possession — in reversion —
certain — contingent — or in arrears ? [25] .
3. What is meant by the present worth of an an-
nuity ? [25] .
4. How is the amount of any number of payments
in arrears found, the interest allowed being simple or
■compound ? [26] .
5. How is the present value of an annuity in posses-
sion found .' [28] .
6. How is the present value of an annuity in per-
petuity found ? [29] .
7. Hovr is the present value of an annuity in rever-
sion found ? [30] .
POSITION.
32. Position, called also the " rule of false," is a rule
which, by the use of one or more assumed, but false
numbers, enables us to find the true one. By means of
it we can obtain the answers to certain questions, which
we could not resolve by the ordinary direct rules.
When the results are really proportional to the sup-
position— as, for instance, when the number sought is
to be muliijplied or divided by some proposed number ;
or is to be increased or diminished by itself^ or by some
given mnltiiph or jtart of itself — and when the question
contains only one pj-Gposition, we use what is called
single position, assmning only one number ; and the
quantity found is exacihj that which is required. Other-
wise— as, for instance, when the num.ber sought is to bo
increased or diminLshed by some absolute number, which
is not a known multiple, or part of it — or when two
propositions, neither of which can be banished, are con-
tained in the problem, we use double position, assuming
two numbers. If the number sought is, dm-ing the
process indicated by the question, to be involved or
evolved, we obtain only an approxinuition to the quan-
tity required.
346 POSITION.
33. Shigle Position. — Rule. Assume a number, and
perform with it the operations described in the question ;
then say, as the result obtained is to the number used^
so is the true or given result to the number required.
Example. — What number is that which, being multiplied
by 5, by 7, and by 9, the sum of the results shall be 231 '?
Let us assume 4 as the quantity sought. 4xo-|-4x7-(-
4x0=84. And 84 : 4 :: 231 : i^l=^=ll, the required
o4
num.ber.
31. REAso?f OF THE RuLE. — It is evident that two num-
bers, miiliijjfied or divided by the same, should produce pro-
portionate results. — It is otherwise, however, when the a^me
quantity is added to, or subtracted from them. Thus let the
given question be changed into the following. What number
is that which being multiplied by 5, by 7, and by 9, the sum
of the pj-oducts, plus 8, shall be equal to 239 .'
Assujiiing 4, the result will be 92. Then we cannot say
92 (814-8) : 4 : : 239 (231-f-8) : 11.
For though 81 : 4 : : 231 : 11, it does not follow that
84-J-8 : 4 :: 231-f8 : 11. Since, while [Sec. V. 29] we may
multiply or divide the first and third terms of a geometrical
proportiun by the same number, we cannot, without destroy-
ing the proportion, add the same number to, or subtract it
from them. The question in this latt-jr form belongs to the
rule of double position.
EXERCISES.
1. A teacher being asked how many pupils he had,
replied, if you add i, ^, and J- of the number together,
the sum will be 18 ; what was their number } Ans. 24.
2. What number is it, which, being increased by I,
J-, and \ of itself, shall be 125 } Ans. 60.
3. A gentleman distributed 78 pence among a nuni-
ber of poor persons, consisting of men, women, and chil-
dren ; to each man he gave 6^/.^ to each woman. Ad..,
and to each child, 2d. ; there were twice as many
women as men, and three times as many children as
women. How many were there of each ^ Ans. 3 men,
6 women, and 18 children.
4. A person bought a chaise, horse, and harness, for
£60 ; tlie liorse came to twice the price of the harness,
and the chaise to twice the price of the horse and bar-
POSITION. 347
dess. Wliat did he give for each ? Ans. He gave for
the harness, £6 I3s. 4d. ; for the horse, £\3 65. 8^. ;
and for the chaise, X40.
5. A's acre is double that of ]]'s ; E's is treble that
of C's ; and the sum of all their ages is 140. What is
the age of each ? Ans. A's is S4, 13's 42, and C's 14.
6. After paying away J- of my money, and then 4 of
the remainder, I had 72 guineas left. What had I at
first ? Ans. 120 guineas.
7. A can do a piece of work in 7 days ; B can do the
same in 5 days ; and C in G days. In what time will
ttll of them execute it .' Ans. in Ijf-f days.
8. A and B can do a piece of work in 10 days ; A
by himself can do it in 15 days. In what time will B
do it ? Ans. In 30 days.
9. A cistern has three cocks ; when the first is opened
all the water runs out in one hour ; when the sscond is
opened, it runs out in two hours ; and when the thhd i?
opened, in three hours. In what tune wiU it run out, if
all the cocks are kept open together .' Ans. In -y^y hours.
10. "What is that number whose i, ^, and | parts,
taken together, make 27 ? Ans. 42.
11. There are 5 mills; the fii-st grinds 7 bushels of
corn in 1 hour, the second 5 in the same time, the third
4, the fourth 3, and the fifth 1. In what time will the
five grind 500 bushels, if they work together } A71S.
In 2d hours.
12. There is a cistern which can be filled by a cock
in 12 hours ; it has another cock in the bottom, by
which it can be emptied in IS hours. In what time will
it be filled, if both are left open } Ans. In 36 hours.
35. Double Position. — Eule I. Assume two con-
venient numbers, and perform upon them the processes
supposed by the question, marking the error derived
from each with -f- or — , according as it is an error of
e.r-ce55, ©r of defect. Multiply each assumed number into
the error which belon*gs to the other ; and, if the errors
are hotk plus, or hath minus, divide the difference of the
products by the difference of the errors. But, if one is
a plus, and the otjicr is a minus error, divide the sum of
34S POSITION.
tlie prodacts by the sum of the errors. In cither case
the result will be the uuiriber soui^'ht, or ati approxi
Illation to it.
" ExAMVLi: 1. — If to 4 times the price of my horse XIO is
added, the sum will be £100. ^Vhat; did it cost ?
Assuming numbers which give two errors of excess — •
First, let 28 be one of them,
jNIultiply by 4
112
Add 10
From 122, the result obtained,
sul^tract 100, the result required,
and the remainder. +22, is an error of excesa.
Multiply by 31, the other assumed numl^er
and G82 will be the product.
Next, let the assumed number be 31
Multiply by 4
124
Add 10
From 134, the result obtained,
subtract 100, the result reqiured,
^ and the remainder, -f-^-l? is an error of erccs's.
jMuitiply by 28, the other assumed sium.
and 052 will be the prodm't.
From this subtract 682, the product found altove,
divide by 12)270
and the required quantity is 22o=£22 I'^.s.
Difference of errors=34 — 22=12, the number b^ which
we have divided.
36. PuEAsoTV OF THE PtCLK. — Wlieu ill cxnmple 1, we mul-
tiply 28 and 31 by 4, we-iaultiply the error belongim. to eacii
by 4. Hence 122 and 184 are, respectively, equal to the true
result, plus 4 times one of the errors. Subtracting lOO, thy
true result, from each of them, we obtain 22 (4 times I \e error
in 2S) au'l 34 (4 times the error in 31).
But, as numbers are proportional to their eyj/an.lt-iples
the error in 28 : the error in 31 : : 22 (a multiple of the for-
mer) : 34 (an equimultiple of the latter).
And from the nature of proportiuu [Sec. V. 21] —
POSITION. 349
The error in 28x34=the error in 31x22.
But 682=the error in 31-f-tlie required number X 22.
And 952=the error in 28-|-tJie required number X 34.
Or, since to multiply quantities under the vinculum [Sec
[I. 34], we are to multiply each of them —
G82=22 times the error in 31-J-22 times the required number.
952=34 times the error in28-f-34 times the required number.
Subtracting the upper from the lower line, we shall have
952 — 682=34 times the error in 28 — 22 times the error in
31-J-34 times the required number — 22 times the required
number.
But, as we have seen above, 34 times the error in 28=:22
times the error in 31. Therefore, 34 times the error in 28 — 22
times the ei-ror in 31=0; that is, the two quantities cancel
each other, and may be omitted. We shall then have
952 — 682=34 times the required number — 22 times the re-
quired number; or 270=34—22 (=12) times the required
number. And, [Sec. V. 61 dividing both the equal quanti-
ties by 12,
270 34-22
'^(22 5)= — T^ — times (once) the required number.
37. Example 2. — Using the same example, and assuming
numbers which give two errors of defect.
Let them be 14, and 10 —
14 16
4 4
50 04
10 10
eOj the result obtained, 74, the result obtained,
100', the result required, 100, the result required,
— 34, an error of defect. — 20, an error of defect.
10 14
544 364
304 Difference of errors = 34 — 20 = 8.
8)180
22-5 =£22 lOs., is the required quantity.
In this example 34=:four times the error (of defect) in 14;
and 26 = four times the error (of defect) in 16. And, tince
aurabers are proportional to their e(iuimultixjle3,
The error in 14 : the error in 16 : : 34 : 26. Therefore
The error in 14x26=the error in 16X34.
But 544=the required number — the error in 16X34
Ando64=the requu-ed number — the error in 14X26
n 2
? posrnox.
If wc subtract tlio lower from the uppei' line, we shall havo
r)l4 — oG4=(romoviiig tiie vinculum, and ch.-ingiiig the «igu
[Sec. 11. IH]) 8-4 times the requiretl number — 26 times tlie
required number — 34 times the error in 1G-}-2*d times the error
in 14.
But we found above that 34 times the error in 16='i6 times
the error in 14. Therefore — 34 times the ei-ror in 10, ;in(l+2o
times the error in 14=0, and miiy be omitted. We will tiien
have 544 — 364=34 times the required nuuiber — 26 times the
required number; or 180=8 times the required number; and,
dividing both these equal quantities by 8,
180 8
-Q- (22'5) =r^ times (once) the required number.
38. Example 3. — Using still the same exanij/lc, and as-
suming numbers which will give an error oi" ej:ccss, and an
error of difed.
Let them be 15., and 23 —
lo 23
4 4
GO 02
10 10
70, the result obtained. 102, the result obtained.
100, the result required. 100, the result required.
— 30, an error of defect. ^ +2, an error of eocecsn.
23 15
G90
30
;2)72b
■22-5
Sum of err 01 s = 30 -f 2 =
£22 10.s\, the required quantity.
In this example 30 is 4 times the error (of defect) in 15 ;
and 2, 4 times the error (of excess) 'm 23. And, since numbers
are proportioned to the equimultiples,
The error in 23 : the error in 15 : : 2 : 30. Therefore
The error in 23x30=the error_m 15X2^_
But 090=the required number+the error in 23X'^0-
And 30=tlieTequired number — the error in 15X2.
If we add these two lines together, we siiall have 600+30=
(removing tlie vinculum) 30^ times the required number+
twice tlie required number -{- 30 times the error in 23 — twice
the error in 15.
But we found above that 30xtbe error in 2or=2xt}ie error
in 15. Therefore 3CVKthe error in 23 --2 X the error in 15=0
POSITION, 351
and PAfxy be omitie«l. We shall then hare G90-f-30=the re-
quired number X 30 4- the required number X 2 ; or 720=;32
times? the required number. And dividing each of these equal
quantities by 32.
720 32
-^(22-5)=— times (once) the required number.
The given questions might be clianged into one belonging
to single position, thus —
Four times the price of my horse is equal to £100 — £10 ;
or four times the price of my horse is equal to £90. What did
it cost? This change, however, supposes an eflfort of the mind
not required when the question is solved by double position.
39. Example 4. — What is that number which is equal to
4 times its square root +21 ?
Assume 64 anri
81-
y64=8
y81= 9
4
4
32
36
21
21
53.
result
ohtf
inrd.
57. result obtained
64,
result
reqi
lircd.
81, result required
-11
-24
81
64
891
1536
891
13)645
The first
approximation is 49- 6154
It is evident that 11 and 24 are not the errors in the assumed
numbers multiplied or divided by the same quantity, and
therefore, as the reason upon which the rule is founded, does
not apply, we obtain only an approximation. Substituting
this, however, for one of the assumed numbers, we obtain a
Btill nearer approximation.
40. Rule — II. Find the errors by the last rule ; then
divide their difference (if they are both of the same
kind), or their sum (if they are of different kinds), into
the product of the difference of the numbers and one of
the errors. The quotient ^vill be the correction of that
error which has been us<3d as midtiplier.
352 rosiTioN.
Example. — Taking the same as in the last rulc; and s^
Bumiiig 10 and 2-3 as the required number.
19 25
4 4
76 Too
10 10
80 the result obtained. 110 the result obtained.
100 the result required. 100 the result required.
—14. is error o? defect. -\-^0, is error of excess.
The errors are of cliffe rent kinds ; and their sum is 14-f
Z0=24 ; and the difference of the assumed numbers is 25 —
19=6. Therefore
14 one of the errors,
is multiplied by 6, by the difference of the numbers. Then
divide by 24)84
and 35 is the correction for 19, the number
which gave an error of 14.
19-f-(the error being one of defeat, the correction is to be
added) 3 5=22-5=£22 10s. is the required quantity.
41. XIeasoist of the Rule. — The difference of the results
arising from the use of the different assumed numbers (tho
difference of the errors) : the difference between the result ob-
tained by using one of the assumed numbers and that obtained
by using the true number (one of the errors) : : the difference
between the numbers in the former case (the difference between
the assumed numbers) : the difference between the numbers
in the latter case (the difference between the true number, and
that assumed number which produced the error placed in the
third term — that is the correction required by that assumed
number).
It is clear that the difference between the numbers used
produces a proportional difference in the results. For the
results are different, only because the diffox'ence between the
assumed numbers has been multiplied, or divided, or both —
in accordance Avith the conditions of the question. Thus, in
the present iufftance, 25 produces a greater result than 19,
because G, the difference between 19 and 25, has been multi-
plied by 4. For 25X'4=;=19x4-f-6x4. And it is this 6x4
■which makes up 24, the 7-eal difference of the errors. — The
diifercnce between a negative and positive result being the
sum of the differences between each of them and no result.
Thus, if I gain lO^., I am richer to the amount of 245. than if
1 lose ll.v.
PosrwcN. 333
EXERCISES.
13. What number is it which, being multiplied by 3,
the product boiug increased by 4, and the sum divided
by 8, the quotient will be 32 ? Ans. 84.
14. Aifon asked his father how old he was, and re-
ceived the following answer. Your age is now } of
miue, but 5 years ago it was only 4. What are thtir
ages .' Alls. 80 and 20
15. A workman was hired for 30 days at 2^. 6d. for
every day he worked, but with this condition, that for
every day he did not work, he should forfeit a shilling.
At the end of the- time he received £2 146\, how many
days did he work .' Ans. 24.
16. Kequii'ed what number it is from which, if 34
be taken, 3 times the remainder will exceed it by \ of
itself.' Ans. 5Sf.
17. A and B go out of a town by the same road. A
goes S miles each day ; B goes 1 mile the fii-st day,
2 the second, 3 the thii-d, &c. V.'hen will B ovt-r-
takc A .=
A.
B.
A.
B.
Suppose 5
8
1
2
Suppose
7
8
1
2
40
15
3
4
5
5G
2«
o
4
5
--5
15
7
)28
-4
G
7
7
5
28
35
20
2'J
1)15
5-4=1
Wc divide tlie entire
error
by
the munber of
djiys
c-.se, which gives the error ia one day.
each
IS. A gentleman hires two labourers; to the one ho
gives 9d. each day; to the other, on the first day, 2d.y
on the second day, 4r/., on the third day, 6^., &,c. In
how many days will they earn an cqunl sum ? A?ts. In 8.
19. What are tho.ve nusubers which, when addrd,
J554 POSITION.
mate 25 ; but when one is halved and the other douhled,
give equal results ? uins. 20 and 5.
20. Two contractors, A and B, are each to Kiild a
wall of equal dimensions ; A employs as many men as
finish 22| perches in a day ; 13 employs the first day as
many as finish 6 perches, the second as many as finish
9, the third as many as finish 12, &c. In what time
will they have built an e(j[uul number of perches .'
Ans. In 12 days.
21. What is that number whose i, |, and |, multi-
plied together, make 24 .'
Suppose 12 Suppose 4
1=3 1=1
Product=18 Product=:2
3_41 3 11
81 result obtained. 3 result obtained.
24 result required. 24 result required.
+57 =21
64, the cube of 4. 1728, the cube of 12.
3618, product.
36288 To this product
3648 is added.
57+21=78
57-21=78.
78)39036 is the sum.
And 512 the quotient.
-2/512=8, is tlie required number.
miiltiplv the alternate error by the cube of the supposed
er, beciiuse the errors belong to the g^^th part of the cube
We
mimbei
of the assumed numbers, and not to the numbers themselves ;
for, in reiility, it is the cube of some number that is required
— since, 8 bo'iag f^ssumed, according to the question we have
22. What namber is it whose i, i, j, and J-, multi-
plied together, will produce 699Sf t Ans. 36.^
2.3. A said to B, give me one of your shillings, and
I shall have twice as many as you will have left. B
ans^vered, if you give me 1.?., I shall have as many a.s
yen. How many had each r Avs. A 7, and B 5.
FOSITION. 355
24. There are two numbers which, when added to«
gothcr, make 30; but the i, ■}, and J^, of the grcat^ir
are equal to i, -|, and i, of the lesser. What are they ?
Ans. 12 and IS.
25. A gentleman has 2 horses and a saddle worth
jS50. The saddle, if set on the back of the first horse,
will make his value double that of the second ; but if
set on the back of the second horse, it will make his
value treble tliat of the first. What is the value of
each horse ? Ans. £30 and £40.
26. A gentleman finding several beggars at his door,
gave to each 4d. and had vd. loft, but if he had given
(id. to each, he would have had 12^/. too little. How
many beggars were there ? Ans. 9.
It is so likely tliat those who are desirous of studying
this subject furthc^r will be acquainted with the method
of treating algebraic equations — which in many case?
afFoi-ds a so much simpler and easier mode of solviu^g
questions belonging to position — that we do not deem
it necessary to enter further into it.
QUESTIONS.
1. What is the dificreuce between single and double
position ? [32] .
2. In what cases may we expect an exact answer by
^,hesc rules r [32] .
3. What is the rule for single position .^ p3].
4. What are the rules for double position ? [35 and
40].
MISCELLANEOUS EXERCISES.
1. A father being ^asked by his son how old he was;
replied, your age is now \ of mine ; but 4 years ago
It was only i of what mine is now ; v/hat is the age of
each f A as. 70 and 14.
2. Find two numbers, the difference of which is 30,
and the relation between them as 7^ is to 3^ .'' A')is.
58 and 28.
3. Find two numbers whose sum and product are
equal, neither of them being 2 - Ans. 10 and 1^.
356 EXERCISES.
4. /^L.person being asked tlio hour of tlic day, answered,
It is between 5 and 6, and Ijoth the hour and minute
bands are together. Requii-ed what it was.? Ans.
27j\ minutes past 5.
5. What is the sura of the series i, i, |, S:c. } Ans. 1.
6. What is the sum of the series f , yj? i^d t'/u ^^- •''
Ans. 11.
7. A person had a sakry of £75 a year, and let it
remain unpaid for 17 years. How much had he to
receive at the end of that time, allowing 6 per cent,
per annum compound interest, payable half-yearly ?
Ans. £204 lis. 10}d.
8. Divide 20 into two such parts as that, when the
greater is divided by the less, and the leys by the greater,
and the greater quotient is multiplied by 4, and the less
by 64, the products shall be equal.? An.'?. 4 and 16.
9. Divide 21 into two such parts, as that when the
less is divided by the greater, and the greater by the
lov's, and the greater quotient is multiplied by 5, and
th? less by 125, the products shall be equal ? Ans.
3i and 171.
J" A, B, and C, can finish a piece of work in 10
days ; B and C will do it in 16 days. In what time will
A do it by himself ? Ans. 26 1 days.
1. A can trench a garden in 10 daj^s, B in 12, and
C in 14. In what time will it bo done by the three if
they work together t Ans. In SyVi tlays.
\2. What number is it which, divided by 16, will
leave 3 ; but which, divided by 9, will leave 4 .' Ans.
67
13. What number is it which, divided by 7, will
leave 4 ; but divided by 4, will leave 2 .? Ans. IS.
14. If £100, put to interest at a certain rate, wib,
at the end of 3 years, be augmented to £115"7625
(compound interest being allowed), what principal and
interest will b^ due at the end of the first year .? Ans.
£105.
15. An eldoi'ly person in trade, desirous of a little
respite, proposes to admit a sober, and industrious young
person to a share in the business ; and to encourage
him, lie olVors, t'aat if lii> circumstances all.)w him to
EXERCISES. 357
advance £100, liis salary shall be £40 a year ; that if
he is able to advance £200, he shall have £55 ; but
that if he can advance £300, he shall receive £70
annually. In this proposal, what was allowed for his
attendance simply ? Aiis. £25 a year.
16. If 6 apples and 7 pears cost 33 pence, and 10
apples and 8 pears 44 pence, what is the price of one
apple and one pear .^ Ans. 2d. is the price of an apple,
and 3d. of a pear.
17. Find three such numbers as that the first and ^
the sum of the other two, the second and i the sum of
the other two, the third and J- the sura of the other
two vrill make 34 f A?is. 10, 22, 2G.
18. Find a number, to which, if you add 1, the sum
win be divisible by 3 ; but if you add 3, the sum will
be divisible by 4 .' A/is. 17.
19. A market woman bought a certain number of
eggs, at two a penny, and as many more at 3 a penny ;
and having sold them all at the rate of five for 2d.., slie
found she had lost fourpence. How many eggs did she
buy > A.ns. 240.
20. A person was desirous of giving 3d. a piece to
some beggars, but found he had Sd. too little ; he there-
fore gave each of them 2d.j and had then 3d. remain-
ing. Requh-ed the number of beggars.^ Ans. 11.
21. A servant agreed to live with his master for £8
a year, and a suit of clothes. But being turned out
at the end of 7 months, he received only £2 135. 4d.
and the suit of clothes ; what was its value f Ans,
£4 IGs.
22. There is a number, consisting of two places of
figures, which is equal to four times the sum of its
digits, and if 18 be added to it, its digits wiU be in-
verted. Wiiat is the number .' Ans. 24.
23. Divide the number 10 into three such parts, that
if the first is multiplied by 2, the second by 3, and the
third by 4, the three products will be equal ? Ans.
4-3_ "^J- 2JL
^ 1 3 ' '-' 1 3_> f' 1 3 •
24. Divide the number 90 into four such parts that,
If the first is increased by 2, the second diminished by
2, the third inultiplitjd by 2, and the fourth divided by
S5S EXERCISES,
2, the sura, difFercnce, product, and quotient will bo
equal r Ans. IS, 22, 10, 40.
25. Wliat fraction is that, to the numerator of which,
if ] is added, its value will be i ; but if 1 be added to
tlie denominator, its value will be i ? Ans. j*j.
26. 21 gallons were drawn out of a cask of wine,
which had leaked away a third part, and the cask
being then guaged, was found to be half full. How
much did it hold ? Ans. 126 gallons.
27. There is a number, ^ of which, being divided by
6, I of it by 4, and \ of it by 3, each quotient will
be 9 ? Ans. 103.
28. Having counted my books, I found that when I
multiplied together i, i, and | of their number, the
product was 162000. How many had I ? An^. 120.
29. Find the sum of the series l+^ + j + i, &c. .'*
Aiu. 2.
30. A can build a wall in 12 days, by getting 2 days'
assistance from B ; and B can build it in 8 days, by
getting 4 days' assistance from A. In what time will
both together build it ? Ans. In 6f days.
31. A and B can perform a piece of work in 8 days,
when the days are 12 hours long ; A, by himself, can
do it in 12 days, of 16 hours each. In how many days
of 14 hours long will B do it f Ans. 13-f .
32. in a mixture of spirits and water, ^ of the whole
plus 25 gallons was spirits, but i of the whole minus 5
gallons was water. How many gallons were there of
each ? Ans. 85 of spirits, and 35 of water.
33. A person passed ^ of his age in childhood, ^^ of
it in youth, \ of it +5 years in matrimony; he had
then a son whom he survived 4 years, and who reached
only \ the age of his father. At what age did this per-
son die } Ans. At the age of 84.
34. What number is that whose i exceeds its \ by
72 } Ans. 540.
35. A vintner has a vessel of wine containing 500
gallons ; dravang 50 gallons, he tlicn fills up the cask
with water. After doing this five times, how much
wine and how much water arc in the cask } Ans
295 3V0 gallons of wine, and 204 {'Vo g'l-Hons of water.
EXERCI^S. 359
4j. A raothor and two (lauglit<?rs working together
•,a»4 spin 3 lb of flax in one day ; the mother, by herself,
ian do it in 2^ days ; and the eklest daughter in 2i
<iays. In what time can the youngest do it ? Ans.
In 6^ days.
37. A merchant loads two vessels, A and B ; into
A he puts 150 hogsheads of wine, and into B 240 hogs-
heads. The ships, having to pay toll, A gives 1 hogs-
head, and receives 12.?. ; ]^ gives 1 hogshead and 30.?.
besides. At how much was each hogshead valued ?
Am. £4 12.?.
3.S. Three merchants traffic in company, and their
stock is £400 ; the money of A continued in trade 5
months, that of B six months, and that of C nine
months ; and they gained £375, wliich they divided
equally. "What stock did each put in.' Ans. A £167^-^-,
B £13yf f , and C £93j\.
39. xi fountain has 4 cocks, xV, B, C, and D, and
under it stands a cistern, which can be filled by A in 6,
by B in 8, by C in 10, and by D in 12 hours ; the
cistern has 4 cocks, E, F, G, and II ; and can be
emptied by E in G, by F in 5, by Gr in 4, and by H in
3 hours. Suppose the cistern is full of water, and that
the S cocks are all open, in what time will it be emptied ?
Ans. In 2/g hours.
40. What is the value of '2^97' ? Ajis. i-l.
41. AYhat is the value of -5416' ? Ans.''^.
42. What is the value of -0^76923' : Aiis. j\.
43. There are three fishermen, A, B, and C, who
have each caught a certain number of fish ; when xV's
fish and B's are put together, they make 110; when
B's and Cs are put together, they make 130 ; and when
A's and C's are put together, they make 120. K the
fish is di\dded equally among them, what will be each
man's share ; and how many fi.sh did each of them
catch .' Ajis. l^ach man had 60 for his share ; A caui^ht
50, B 60, and C 70.
44. There is a golden cup valued at 70 crowns, and
two heaps of crowns. The cup and first heap, are worth
4 tinifs the value of the second Iioap ; but the cup and
siJcoud heap, arc v>'orth double the value of the first
3^10 EXERCISES.
heap. How many croTvns are tlicre in eacli lieap ? Ans
50 in one, and 30 in another.
45. A certain number of horse and foot soldiers are
to be ferried over a river ; and they agree to pay 2^d.
for two horse, and 3ld. for seven foot soldiers ; seven
foot always followed two horse soldiers ; and when they
were all over, the ferryman received £2o. How many
horse and foot soldiers were there ? Ans. 2000 horse,
and 7000 foot.
46. The hour and minute hands of a watch are to-
gether at 12 ; when will they be together again .' Ans.
at 5y\ minutes past 1 o'clock.
47. A and B are at opposite sides of a wood 135
fathoms in compass. They begin to go round it, in the
same direction, and at the same time ; A goes at the
rate of 11 fathoms in 2 minutes, and B at that of 17
in 3 minutes. How many rounds will each make, before
one overtakes the other t Ans. A will go 17, and B
16i.
48. A, B, and C, start at the same time, frem the
same point, and in the same direction, round an island
73 miU'S in circumference ; A goes at the rate of 6,
B at the rate of 10, and C at the rate of 16 miles per
day. In what time will they be all together again ?
Ans. in 36^ days
M^ATIIEMATICAL TABLES
LOGATUTIIMS OF NUMBERS FKOM 1 TO 10,000, AVITH
DIFFERENCES AND PROPOimoXAL PARTS.
362
LOGARITHMS.
pp
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LOGAlMTiiMS.
363
pp
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7033
750
8016 Sa62l 81091
4919
5390
6360
6329
6799
7267
77351
8203
4966J 5013 5061
5437 54.84I 5531
5907 5954! GOOl
6376 64231 6470
6345' 6392J 6939
7314I 73511 7408
7782, 73-29* 7875
8-24!*! S-296' 8343
745!
79-22
8390
S43C
968483:9do530;963576 968523
89501 8996j 9043| 9090
9416i 9463 9509: 9556
9332i D9-2.3I 9975 970021
4 970347 '970393!970440i 0486
0312
1-278
1740
2203
2566
03581
13-221
1786
2249
2TI-2
0904! 0951
1369! 1415
133-21 1379
22951 2342
•2758 2304
968670 9o37!6 908763 96331 0l96S356:9639O3
9136
91831 92-29 9276
9323 9360
9602
9549! 9595' 9742
9789 9835
0088
97011 4'9701 61 970207
970254 970300
0533
0579! 0826; 0672
0719 0765
0997
1044J 1090- 1137
1183 1-2-29
1461
1503! 1554: 1601
1647 1692
19-25
1971' 2018. 2064
2110 2157
2388
2434I 2481 1 2527
2573 2619
2851
2397 2943! 2989
3035 30^2
J76
LOGARITHMS.
pp
5
940
1
0 I 1 1 2
3 1 4
5 1 S 1 7 1
' \^±J
D.
9731-2S
3500
973174 973220
3636 3682
973266 973313 973359 973405|97345l|
973497;
973543!
4005
46
46
372S
3774 3320; 3366 j
39131
3959|
9
2
4051
-40971 4143
4189
4235
4281
4327
4374|
4420 4466|
46
14
a
4512
45581 4604
4660
4696
4742
4788
4334
4880
4926
46
18
4
4972
501SI 5064
5110
5156
5202
&US 5294
5340
5386
46
5
5432
6478 6524
6570
5616
6662
57071 5753
5799
5846
46
'iS
G
5391
5937! 6983
6029
6075
6121
61671 6212
6258
6304
46
'i-2
7
6350
6396| 6442
6488
6533
6579
6625 1 6671
6717
6763
46
37
8
6303
6854i 6900
6946
6992
7037
7083| 7129
7175
7220
46
4]
9
900
7265
7312; 7358 7403
7449
7495
7541 1 7586
7632
7678
46
077724
977769:977615 977861i977906;977952;97799s'97S043!978039!978135
46
6
1
8181
8226! 8272
8317
8363 8409
8454 8500
8546i
8591
46
9
2
8637
86a3 87-:?8
8774
8819 8865
8911
8956
9002
9047
4S
14
3
9093
9138 9184
9230
9275 9321
9366
9412
9457
9503
46
ks
4
954S
9594 9639
9685
97301 9776
9821
9367
9912
9953
46
■2-i
5
980003
980049 980094
9S0140
9801851930231 9802761980322
980307 9804121
45
6
0458
05031 0549
0594
06401 0685
0730 0776
0821
0867
45
3-2
7
0912
09571 1003
1048
1093 1139
1184 1229
1275
1320
45
M
8
1366
1411 1456
1501
1547 1592
1637 1683
1728
1773
45
4i
9
901/
I3i9| 18641 . 1909
1954
2000 2045
2090 2135
2181
2226
45
)S2271
982316 982362
982407 982452;982497 982543;982583i982633:982678
45
5
J
2723
27691 2814
2859 29041 2949 2994
3040
30851 3130
45
9
2
3175
3220! 3265
3310 3356 1 3401 1 3446
3491
3536
3581
45
14
3
3626
3671! 3716
3762
3307 3S52
3397
3942
3987
4032
45
Iri
4
4077
4122
4167
4212
4257
4302
4347
4392
4437
4482
45
•23
0
4527
4572
4617
4662
4707
4752
4797
4842 4887
4932
45
07
b
4977
5022
6067
5112
5157
5202
5247
6292
5337
5382
45
32
5426
5471
5510
55S1
6606
66511 5696
5741
5786
5830
45
3o
8
5875
5920
5905
6010
6055
61001 6144
6139
C234
6279
45
41
'->
6324
6369
fr413
6458
6503
6543 6593 1 6637
6882
6727
45
9868171936861
966906-936951 !936996'98704o'987035i9S7r3U;987175
45
970
J86772
5
1
7219
7264
7309
7353 73981 7443 | 743S| 753«
7577
7622
45
!.'
7666
7711
7756
7300
7845
7890 7934 7979
8024
8068
45
14
3
8113
S157
8202
8247
8291
83361 8031 8425
8470
8514
45
IS
4
8559
8604
8648
8693
S737
8782| ^326 8371
8916
8960
45
:23
■^
9005
0049
9094
9138
9133
92271 92721 9316
9361
9405
45
27
6
9450
9494
9539
9583
9328 9072 1 9717
9761
9806
9350
44
3-2
7
9S95
9939
9983
990028;990072 990117:980161
990206990250
990294
44
:i-6
8
990339i0903rf3
39042b
04721 0516 0561 1 0605
0650i 0694
0733
44
4]
9
:)8..
0733J 0827
0871
0916J 0960 10041 1049
10931 1137
1182
44
iJ9 1226^991 270
991315
991359!991403 99J44B 991492|991536,99]580;991626
44
4
1
1669! 1713
1758
I8O2; lB46i 18901 19351 1979! 2023
2067
44
9
2
2ilJ
2150
2200
2244
2288 23331 2377
2421
2165
2509
44
13
3
2554
! 259S
2642
2686
2730 2774 2819
2363
2907
2951
44
'l.S
4
2995
1 3039
308?
3127
3172 32161 S260
3304
3343
3392
44
■2-
0
3136
3480
3524
3568
36131 36571 3701
3745
3739
3333
44
•2(i
e
3377
3921
3965
4009
4053! 40971 4141
4185
4229
4273
44
31
4317
4361
4405
4449
44931 4537
4581
4625| 4669
4713
44
3;
4757j 4H0I
4S4r
4839
4933 4977
5021
5065 6108
5152
44
4i.
390
51961 6240
5284
5328; 5372! 5416
5460
1
5504 5547
5591
44
995635i995679|995723
995767 995311 995854 995893;996942|995986i 996030
44
4
6074| 6il7
6161
6205
, 6249
1 6293; 6337
6380
6424
6468
44
9
0
6)12 6555
6599
6643
1 6687
1 67311 6774
681b
6862
6906
44
13
3
6.149 6993
7037
7030
7124
1 7168
7212
7255
7299
7343
44
1^
4
73S0 7 130
7474
7517
7561
1 7605
7648
7692
7'^36
7779
44
V--
7S23 7867
7910
7954
1 7998
: 8041
8035
8129
' 817-..
8216
44
n
8259 S303
8347
8390
! 8434
j 8477
6521
8564
860b
8652
44
31
7
8695 8739', 878^
8826
8869
8913
8956
9000
9043
9087
44
3 1 8
9131 9174 921S
9'J61
9305
9348
9392
9435
9479
9522
44
40J ^
9565 96091 9652' 9696
9739
9783
0826
9870
99i3{ 9957
43
A TABLE OF SQUARE?, cOUES, AND ROOTS.
377
No. ■
Sqanre.
Cu!-?. 1
Sq. K'jot. Cu'ae Root
No.
Square •
CuLe. j .«<}. Root. Cube Root
1
1
1-0000000 1-000000
u
40SS
202141' 8-000000014-000000
41
8
1-4142136 1-2-59921
65
4225
2746-25i 8-06-22.577 4^020726
i
9l
27 j
1-73-20.508 1-44-2-2.3U
66
4356
287496 8 • 1240394 14- 041 240
4
16;
64;
2-0000000 1-587401
•67
4439;
306763: 8-18.33.5-28|4-061548
5
25
125 j
2-23606S0 1-709976
63
4.-524'
3144321 8-2462113 4-0S1656
6
36
2161
2-4494897 1-817121
69
4,-61'
328509: 8 -3066239i4- 101566
7
49
343!
264-37513 1-91-2931
70
4t>00
34.3000! 3-3656003!4- 121-285
8
64;
5!2|
2-S234271 2-OOOOOU
71
5041
3.379il! 8-426149314- 149318
9
81
729
3-0000000 2 -0300S4
72
51»4
373-248 8-485-2314;4- 160168
lo!
1001
1000
3- 16-2-2777 2- 154435
73
5329
339017 8-5440037 4-179339
11
121!
1331
3-3166-248 2 -2-239f'0
74
.5476
40.5224 8-60-23-253;4- 193336
1-2
144;
17^2S;
3-4641016 2-289423
75
5625
421875 8-6602-340! 1-217163
13
169!
2197,
3-C0.55513 2-351.3;i5
76
5776
438976 8-7177979i4 -23-5324
14
196;
£744;
3-7416574 2-410142
77
59-29
45653.31 8-7749644'4-254321
Id
2251
3375
3-37-298.33 2-466-212
79
6084
47435-2! 8-8317609 4-272659
16
256',
4096 i
4-OU00000 2-51984-2
79
0-241
49.3039i 8-8831944U-290841
17
289:
49131
4-1-231056 2-6712^2
80
64<!0
51-2090! 8-944-2719;4-303870
IS
S24i
58321
4-2426407 2-620741
SI
6.561
531441; 9-0000000:4-3-26749
19
861 :
6359! 4-3.53S939 2^5684n2|
82
67-24
551363 9-0.553851 ;4- 344481
■20
400;
8000 1
4-4721360 2-714418!
83
6339
571787: 9-1104336 4-36-2071
•21
441 !
9-261;
4-532-57.57 2-7.53924
84
70-56
592704: 9-1651514 4-379519
•22
484'
10648]
4-6304158 2-80-2039
35
7-225
814!25! 9-2195445 4 -.396330
23
529i
12167!
4-7958315 2-843367
86
7396
636056; 9-273613514-414005
•24
576
133241 4 •S<»39795 2-334499!
87
7569
6.53.503: 9-3273791,4-431047
•25
625;
1.56-25 1
6-0000000 2-9-24013
83
7744
631472: 9-330331514-447960
:i6
676
17.576!
5-0990195 2-962496
93
7921
704969 9 • 433931 1 !4 - 464745
27
■729;
19633 i
5- 1961 5243 -000000
90
8100
7-29OO0, 9-4863320 4-481405
28
7»4i
21952
O-29150-2D 3-035589
91
8-28 i
7-53.371
9-539.3920 4-497941
29
30
841!
24339*
o- 3851 648 3-07-2317
92
8464
778633
9-5916630 4-514357
9ca;
27000'
5-477-2256 3-107-232
93
8649
e94.')57
9-6436506 4-530655
il
961
•29791
5-5677644 3-141381
94
8s36
830584
9-69.53.597 4-5463.36
3-2
10-24'
3-2768
5-6568542 3-174802
95
90-25
857 375 9 • 7467943 4 • 562903
33
1QS9:
^Mr.il
5-74156-26 3-207.534
96
9216
88-1736 9 • 7979590 4 - .573857
3-1
1156,
39;VJi
a-830S5I9 3-239612
97
9409
91 267-31 9-84^3578 1- 594701
35
1225=
4-2875
5-916a7-?3 3-2710C6
93
9504
94119-2! 9-S994949 4-510436
S6
1296!
46656
6-0000000 3-3019-27
99
9301
970:i«9i 9-9493744 4-626065
3;
136J:
506:o
6-03-27625 3-3322-22
100
lOUOO
lC0O-Ot)Oi 10-0000000 4-6415S9
3S
1444
5487-2
0-1644140 3-361975
101
10201
1030301 llO- 04937.56 4-657010
39
. 1-521,
59319
6-2449980 3-391211 102
10404
1061-203!lO-0'H9.3049.4-67-2329
40
ISOO
64000
6-32155.53 3-419952 103
10609
109-27-27 10-1433913 4-687.543
■SI
IGoi
6892 i
6-4031242 3-448217'l04
10316
1 1 24364] 10-1 9.-10390 4 - 702669
4-2
17G4
74033
6-4807107 3- -176027
6 -d574335;3- 508393
195
110-25
1 1 57625! 1 0 - 2469503 4 - 7 1 7694
13
13*9
79507
106
li-236
1191016! 10-29.3630 1 4 - 7.3-2624
44
i 1936
85184
6 •6332496 3-530348
i07
11149
1 225043! 10 ■ 3449804 4 - 747459
1 2025'
9«i25
6 •708-2033 '3 -5-50S93
103
11664
12537 12 10- 3523043.4 - 762203
Iti
1 2116
973.36
6 •782.3300 3-583048
109
lloSl
1295029 10-4403065 4-7768.56
4i
1 2209
103823
6-8556546 3-608820
110
12100
1331000|10-4S308S5,4-7914-20
43
1 2304
11059-2
6-i.'23-2032 3-634-24i
ill
12321
1 36763 1 i 1 0 • 5356.53ti 4 • 80-5896
49
1 2401
117619
7 -0000000, 3 -6-39306
112
1-2541
1404928; 10 •58;J00.52 4 •820-284
00
1 23U0
lioOOG
7 -07 10678:3 -634031
113
1-2769
1442897110-6.301453 4-834.>38
5i
1 2-501
132651
7-14U-284 3-7e84:i(.«
114
12996
14i>1544 10-6770783 4-843308
02
1 2704
UOoOa
7-2ili026;3-73-2511
115
116
13-225
152037511 0-7-2330-53 4-862944
S3
: 2ri03
14d877
7-2S01099;3-7.56286
13456
1 JS0!*ii6. 10-7 7 J32ao 4 - 876999
d4
2i>'16
1574o4
7 -3434692;3 -779763
117
13639
160!6i3:10-816o538 4890973
da
3025
166375
7-4131935;3-S0-29.53lll8
i 13924
1-643032110-8827800 4-90486.-
Oil
3i36
1756J6
7-483314S;3-S-2.5862
^19
J-20
: 1-161
1 685 1 59' i 0 • DOS 7 12 i 4 - 9 1 3635
57
3i49
155 193
7-5498344,3-843501
; 14100
i 728000! 1 0 - 954451 2 4 - 932424
o;i
3;364
195112
7-6157731 I3-S70377
121
i 11841
1771.56i!ll-00O0OO0 4-943058
oa
Wt5l
205379
7-53114-37 3- 3;i29y6
!i22
j Ui84
; l8i5843;ll-04.53610 4-959675
&u
i 3600
2161)00
7-7459S57'3-9148o7
U-23
1 15129
13e0367ill -0905363 4-973190
61
Hiii
22i>J8i
7 -8102497 ;3-936497jl24
7 -8740079 3-957632|l-25
13376
l;;06624
11-1355287 4-9cio631
6i.
3844;
238328
156-20
1953125
11 •1303399 5-000000
62
3969'
250047
7-9372539:3-97O0.57|l'26
10376
2000376
11-2-2497-22.5-013-298
1
SQUARES, CUBES, AND ROOTS.
Root No
1-29 i letjiij
16900
17161
174:241
176891
179561
131
13-2
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
160
151
152
lo3
154
155
156
157
158
159
160
161
162
163
164j
105
1
167
1S225
18496!
187691
19044
193.>11
196001
198311
'20164'
2U449
20736
21025
21315
21609
21004
22201
22500
22501
23104
2340;)
23716
24025
24336
24649
24964
25281
25600
25921
26244
26569
2639 J
27J25
27555
2788-.)
168 28224
169 28561
170 28900
171 29241
29584
29929
30276
30625
3vi976
... 31329
178! 31684
179i 3v;041
1801 3J400
181 32,- 61
182! 33124
183 33189
184 33850
185 34225
1861 34.".96
187 34969
188l 35344
189' 35721
204S3S3! 11 •2694277:5 -026526
20971521 11 -3137085 :5-0396Sl
2146689111 •3578167|5-052774
172
173
!74
175
176
r
2197000
2248091
2299968
2352637
2400104
2460375
2515456
2571353
2628072
ll-4017543i5-055797
ll-445523i;5-078753 194
ll-4891263]5-09i64
ll-53256-26:5-10
4469
11 -575836915 -117230
11 -6189500'5- 129928
11-6619038;5-142503
ll-7046999;5-I5513
11- 747344415 -167649
2685619 11 -7898261^5- 18010
2741000 11 •8321596'5-19249
280322 111 S74342 1 5 - 204828
'i863288iir9l63753'5 -217103
292i-207ill-95S-2607'|5-2-29321
•2985981 12- 0000000'5 -241 433
3048825 12 - 0415946' 5 -2535SS
311-2136 12-08304:;' -,■.■-
3176523 i2-124;r;v : , .,
' Sq. Root. Cuba Root
3211792 12- 1655-.!.,. .
3307949 12-20655i>-).j-.
3375000 12-247448715-313-293
.3442951 12-2882056:5-325074
35 1 1 808 1 2 • 32S82S0 5 - 336803
35S1577 12-369316f> 5-31S1-?!
365-2264 12-409673u 5-36lil0^
3723875 12 - 4498996 i 5 - 37 1 68.J
3796416 1 2-4899960 ;5- 383213
3869893 1 2 - 5299641 ! 5 - 394691
3944312 1:2-5698051 !5-4G6l20
4019679 12-6095-202,5-417501
4096000 12 - 6491 106 \o- 428835
417328112- 6885775^5 -44012-
4251523:12-7279221 |5 -451362
4330747 1 2 - 7671453 ' 5 - 462556
4410944 12 •80624S5;5- 473704
4 1921251 1 2 - 8452326 ! 5 - 484806
45742.J6i2-8S409o7 '5-495865
46574o3| 12 - 92284S0 j5 - 50CS79
474t632[l2-9614814'5^517848
482680913 ■O000000i5-62S776
4913000ii3-038404S'5 -639658
6000211 13 -07669615,5 -550499
6088448'l3- 1148770:5- 5612!;
61777171 13 -1529464 5- 5720^
5265024' 13- 1909G60i5 -582770
5359375, 1 3 - •2287565;5 - 593445
5451 77613 - 2664992: 5 • 604079
5545233! 13 -3041347:5 - 614673
563:-)7o2: 1 3 - 34! 6341 5 - 625226
5735339,13-3790892 5- 63574i|
5832(J00'l 3 -4164079 5-046216
0929 74113- 4536 240; 5 - 6.->665 1
6028568') 3 -4907376 5-667051
6 128187 ; 1 3 - 5-277493 5 - 6774 1 1
6-229.;0 l' 1 3 - 5646600: 5 • 687734
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■548-2048; 9
■56.57137|9
■68321199
■6006993 9
■6181760J9
113781
117793
121801
125805
1-29306
133803
137797
14178S
145774
1497.57
153737
1-57714
161686
165656
1696-22
173.5a5
177544
•181-500
•135453
-189402
•19.3:J47
•197^289
•201229
••205164
■209096
•213025
•2169-50
•2-20873
• -224791
-223707
•23-2619
•237.52S
•240433
•244335
•2482.34
•2-521.30
•256022
•2-59911
■263797
-2676S0
- -271539
■275435
•279303
-28317S
■2S7044*
•290907
•-294767
•298624
■30-2477
■306323
•S10175
•314019
•317360
•321697
•325532
■3-29363
•333192
■337017
•340833
•344657
•34S473
•3-52236
S«609d
820 67f240e 551368000 28'
321 674041 .553337661 ^28
322 675634 55-5412248 -28
323 6773-29 5.57441767:23
3-24 678976:5.594762-24 -28 •
325 630625j5615156^25r23
826 632276, -563559976;-28
827 6839-29; 56-3609283/28
3-23 685584:567663.552'28
329 687241 !5697^227S9 23
330 688900,'571787000;^28
331 690.36115738.56191128
332 6922-24'57593036S'29'
S33 093339 573909.537 128 ■
8.34 695556 .580093704 28
835 6972-2-5 ; .582 1 82875 128 •
336 69SS96'5S4277056:-23
837 700369 536.376-253 '28 ■
333 702244 5SS4><0472;-28'
339 70.3921 ;. 590.5897 19 >28'
340 70.5600 592704000123 ■
341 707231 .5948-2.3.321 -29 ■
842 708964 59694768S'29-
843 710649 599077107:29'
344 71-2336 601211534,-29'
845 714025 603351125!-29'
546 715716 605495736!29-
347 717409 607645423^29 •
343 719104 609800192'-29'
349 720301 61 1960049r29'
350,722500 614125000|-29-
351 724201 616-29.30.51-29-
3-52 725904 613470-203 -29'
353 727609 6206-50477 29 ■
«i.54 729316 6-2-2335864I-29'
355 731 025 6-25026375'29 -
35673-2736 6-27-2-22016:-29 •
3-57 734449 629422793:29 ■
358 736 164 63 16237 12-29 ■
3.5'3 7.37331 6.3.3339779 '29-
360739600 6.36056000 '-29 •
361J741321 633277331-29 •
362 743044 640o0.3925;-29-
363 744769 642735547|^29-
364;746496 644972.544 29-
S65'743-225 64721 46-2o'-29-
366 74.1955 649461 S96 29-
;67 751639 651714363-29-
363753424 6.53972032;-29-
369;755161 6-56-234909,-29-
0;756»00 65350300029 •
117.58641 66077631 1;-29 •
21760334 6630.54343 '29 •
3!7621-29 6653386I7;^29^
4;763376'667627624l29-
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6j767376'67-2221376-29-
7;7691-29'674526133;-29-
3!770384 676836152|29-
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7923601 9
8097206 9
8-270706 9
8444102 9
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8963666,9
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9309523 9
94Si2297 9
9654967 9
9827535 9
0000000 9
017-2363 9
0344623 9
05167819
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1-204.396 9
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457800
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■494919
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■•520730,
■5-24406
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■531749
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564556
563293
571938
57557*
■579203)
582840 !
58646S!
•590094!
!
S4
SQ^fcRES, CUBES, AND ROOTS.
o. Squar
.H87
7796S9' 688465387
7814o6J6aOS07104
7S322.5|693154125
784996
788541
790321
792100
793SS1
795664
797449
394 799236
395801025
1896 802816
1897 804609
18981806404
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! 900 810000
! 901 811801
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907 822649
908 824464
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91Q82S100
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912831744
9131833569
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9171840889
Cube Root Wo. Square
695506456
697S64103
00227072
702595369
04969000
07347971
709732288
12121957
14516984
716917375
193231
21734273
724150792
26572699
729000000
731431701
733870808
36314327
3S763264
741217625
743677416
46142643
48613312
51089429
753571000
756058031
75a55052S
76104849
763551944
766060375
758575296
771095213
9181842724 773620632
29-7153159 9-593716
29-7321375 9-597337
29-7489496 9-600955
29-7657521 9-604570
29-7825452 9-608182
29-7993-289 9-611791
29-8161030
29-8328673
•29-8496231
29-8663690
29-8831056
29-8998328
29-9165506
29-9332591
29-9499583
29-9666481
29-9833287
30-0000000
30-U166620
30-0333148
30-0499584
30-0665928
30-0832179
30-0998339
30-116440'
30-1330383
30'- 1496269
30-166206319-090521
30-182776519-694069
30-1993377I9-697615
30-2158899'9-701158
30-232432919-704699
30-2489869 9-708-23
30-2d54919'9-711772
30-282007919-715305
30 -298514819 -71S335
30-3150123 9-722363
9-615398
9-619002
9-622603
9-626201
9-6-29797
633390
9-636981
9-640569
9-644154
9-64773
9-651317
9-654394
9-658468
9-662040
9-665609
9-669176
9-67-2740
9-676302
9-679860
9-683416
9 -686970
9191844561776151559
920!s46400i7786SSOOO 30-3315018 9-725883
n2l!8;^24l]78122996l'30-3479S18 9-7-29411
922S500S4!783777448'j30-3644529J9-732931
923'851929!786330467!30 • 330915r9 • 736448
924:S53776i788889024i30-39736S3'9- 739963
925 855625 791453125130 - 4138127 9 - 743476
926'857476 79402-2776' 30 - 4302481 !9 - 746936
927 8593-29 796597983, 30-4466747|9-750493
923 861184 799i78752i30-4G30924|9-75399S
929,863041 8017650S9',30-4795013l9-757500
930 864900 804357000130-4959014 9-761000
931 866761 806954491 130 -6122926,9 -76449'
932 868624;S09557568: 30 -528675019 -767992
933 870469, 8121662S7i30-o450487|9 -771434
934 87235C;b14780504;30-5614136;9-774974
935 874-225;8i7400375, 30 -6777697|9- 77846:
936!676096,S20025856;30-6941171 9-782946
9371877969 322656953|30-6104557!9 •785429
938'879844j3-25-293672 30-6267857 9-
939,88172ll827936019,30-6431069'9-79-2386
P40 883600:S30584000l30-6594194'9-79o861
4l!88d48i;83.'i22
( 621 30 • 6757233 9 - 799334
967
963
969
9701
971
972
973
974
975
976
977
978
979
930
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
887364
889249
891136
893025
894916
896809
898704
900601
902500
904401
906304
908209
910116
912025
913936
915849
917764
919681
921600
923521
925444
927369
929296
9312-25
933166
935039
937024
938961
940900
942841
944784
S4672
948676
950625
95257
954529
956484
958441
960400
962361
964324
968289
963256
9702'
972196
974169
976144
978121
980100
98-2081
984064
986049
988030
990025
99-2016
994009
996604
998001
1000000
635896888
838561807
841232384
843908625
846590536
849278123
851971392
854670349
857375000
860085351
862801408
865523177
868250664
870983875
67372-2816
876467493
879217912
881974079
Sq. Root. Cube Kool
30-6920185
30-7083051
30-7245830
30-7408523
30-7571130
30-7733651
30-7896086
30-8058436
30-8220700
30-8382379
30-8544972
30-8706981
30-9030743
30 •'"192497
3' 9354166
30-9515751
30-9677251
8S4736000 30-9838668
837503681
890277128
893056347
895841344
31-0000000
31-0161248
31-032-2413
31-0483494
898632125 31-0644491
901428696 31-0805405
904231063131
907039232 ;31 -1126984
909853it«'9;3l- 1287648
91-2673('"0 31-1448230
915493&,1
918330048
921167317
9-24010424
926859375
929714176
9-80-3804
9 •806-271
9-809736
9-813199
9-816655
9-8-20117
9-823572
9-8-27025
9-830476
9-833924
9 •837369
9-840313
9-844254
9-847692
9-851128
9-854562
9-857993
9-861422
9-864548
9-868272
9-871694
9-875113
9-878530
9-881945
9-88535'
9-892175
9-895580
9-898
9-902383
9-906782
9-909178
9-912571
9-915962
9-919351
9-922738
9-9-26122
9-9-29504
31-1608729
31-1769145
31-19-29479
31-2089731
31-2249900
,31-2409987
932574833131 -2569992
935441352131-27-29915
938313739 31-28397571
94119-2000J31- 30495171 9-932884
94407614131-3209195 9-936261
946966168131-3368792 9-939636
949362087 31 - 35283081 9 - 943009
95-2763904131 -3687743: 9-946380
955671625'31 -3847097 9-949748
958535256:31-4006309 9-953114
961504803131 -4165561] 9-956477
964430-272131 -43246731 9-959839
967361669131-44837041 9-963198
970-299000|31 '4642654' 9-966555
973242-271131-48015251 9-969909
976191483,34-4960315 9-973262
979146657J31 -61190251
9821077S4;31 -62776551
935074875,31-54362061
938047936 31-5594677
9910'26973;3r 5763063
994011992 31-5911380
997002999 31-6069613!
9-97661
9-979960
9-983305
9 •986649
9-989990
9-9933-29
9-996660
1000000000,31 • 6227766^10 - 000000
TABLES.
33<^
TABLE OF THE AMOUNTS OF £1 AT COMPOLTST) I>7TEREST.
3 per cent
1-030001
l-0<i090]
1-09273!
1 - 1-2551
1-159.>7
1-19405
l-22Ps7
r2ti677
1-30477
1-343:^2
l-aS4-23
1-42576
1 -45S53
1-51-259
1-55797
1-60471
1-65236
S
9
10
11
12
13
14
15
16
17
IS j 1-70-243'
10 I 1-75351
20 j 1-806U
21 l-doO-29
22 1-91610
23 1-97359
24 2-03-279
•25 2-09378
1-04000
1-08160
1-1-24S6
1-16986
1-21665
1 -26532
1-81593
1-36S57
1-4-2331
l-45i0-24
1-53945
1-60103
1-66507
1-7316S
1-80094
1- 87-298
1-94790
2 -02582
2-10685
2-19112
2-27877
2-35992
2-464
2-66584
5 por cent
1-05000
1-10-250
1-15762
1-21551
1-276-28
1-34010
1-40710
1-47745
1-55133
1-62S33
1-71034
1 -79586
1-88565
1-97993
2-07893
2 18-287
2-29202
2-40662
2-5-2695
2-65330
2-7S596
2-925-26
2-07162
3-2-2510
3-3S635
No. ofl
6 per cent
Par-
ments
1-06000
26
1-12360
27
1-19102
28
1-2624S
29
1-338-23
30
1-41852
31
1-50363
32
1-593S5
33
1-68948
34
1-79035
35
1-89830
36
2-01220
37
2-13-293
38
2-26090
39
2-39656
40
2-54035
41
2-69277
42
2-85401
43
3-02560
44
3-20713
45
3-39956
46
3-60354
47
3-81975
48
4-04893
49
4-29187
50
3 per cent
2-15659
2 --22 129
2-28793
2-35657
2-4-27-26
2-50008
2-57503
2-65-233
2-73190
2-81386
2-89828
2-9352
3 07478
3-15703
3-26-204
3 • 36990
3-46070
3-56452
3-67145
3-78160
3-89504
401190
4-13-225'
4-256-22!
4-3S391i
4 per cent
2-7724
2-8833
2-99870
3-11865
3-24340
3-37313
•50806
64838
79432
■94609
3
3
3
3
4
4-26809
4-43331
4-61637
4-80102
4-99306
5-19278
5-40049
5-61651
5-84118
6-07432
6-31732
6-57053 10
6 -83335! 10
7-10663ill
55567
73346
9-2013
11614
32194
53304
76434
00319
25335
51601
79182
03141
33543
70475
03999
39199
76159
14967
55715
93501
43426
9059-
401-2-
92133
46740
6 per ceul
61933
82235
11169
41839
74349
03S10
45339
sioyj
25102
14725
6:i6o-J
15425
70351
28572
902S6
55703
25045
93543
76461
59049
46592
393S7
37750
4-2015
TABLE OF THE AMOUNTS OF AN ANNUITY OF £1.
Ko.ol, I
Pay- 3 per cent 4 per
per cetit 6 per eent
No.ofi
Pay. ;
menlsj
1-oonooi 1
2-OHOOOJ 2
3-09t»90 i
4-133ti3' 4
5-30913' 5
6-46341} 6
7 -662461 7
S-89i34i 9
10-15911 10
ll-463{r8!l-i
l2S0779ll3
14- 19-2031 16
15-61779|16
17 -086321 IS
IS • 5939! i -20
■21 -Tn 150-23
■23-41 4 13 25
•25-lUJ>7i-27
•26-S70371-29
28 6T643;31
:10- 53678134
32-46233U6
34- 4-2647 !S9
36-4d9-26'41
00000 I !•
040001 2-
12160: 3
-24646' 4
41632' 4'
63-297 1 6
89329 8
214-23 9
58279 ill
0061L12
43635 :i4'
0-25801 15
6-2684 !l 7
291Si:i9*
0-2359 21
8-2453 '23'
69751 '25
64541-28
671-2330
77308 33
9«920 35
24797.38
61789 41
Oo-260 44
646t»l 47
00000
05000
15250
3 1 012
62563
e0191
14-201
54911
02656
57789
20679
91713
71298
09S63 21
67858 -23
00000
06000
65749
84037 28
1 3-233
53900
06595
71925
6U621
43047
50200
72710
37462
63709
97532
30384
89747
49131
1S079
97164
8699
SS214
01506
•2759
67253
2123
9t>56
75999
78559
99273
S9229
9958:
8155S
86451
4 per cewl 5 per ceut
553041 44
56
66
>ol
70963;
93092
21885
57541
00268 59
50276! 62
07784]
73018
4620S lo
27594; 77
17422i 81'
15945; 85
23423 90
401261 95
66330; 99
023-20:104
43389110
04341 115
71986 121'
50146 1-26
39650 132
10339,139
54065 145
79667!l5*<i'
31174
08421
96768
96629
08494
32S33
70147
20953i 80
85791 j 85
65222 90
598311 95
702251101
97034! 107
40915!1U
02551! 120
82654! 127
81960:135
01-238:142
41233!l51
0-2939; 159
87057U68
94539' 178
■20321 ■ 138
.33373' 193
6670c>.:209
6 [cr cent
•11345 59
66913| 63
40256 j es
32-271 73
43886 79
76079 81
29829 90
063771 97
066961104
3203l!lll
83631^^ IIS
62814 J 27
70954 135
09502 145
79977 154
83970 165
-231/
.1 -:
99334
14300
70015
68516'
11942
0-253-^
42666
»479'!
-1563c
-70576
-52311
-6393(»
-05819
-S016b
-8397c
•34316
-18375
-43478
-1-2037
■2631-:
-904-29
-05346
-76196
-04768
•95054
-50758
-75303
•74351
■5831-.
-09361
-56453
-95840
•33590
386 TABLES.
TABLK OF THE PRESEXT VALUT.S OF AN" AXXUITY OF £1.
r.vo. oi
N'o. od
3 per cent ^ [Jtr cent
Sptrceiit
5 per cent
Psv- 3 per cent
menlsl
4 per cent
5 per cent
6 per cent
1
0-970&7
0-9615^1
0-95238
0-94.340
26
17-87684
16-98277
14-37618
13-00316
2
1-91347
1-Sti619
1-85941
1-83339
27
18-3-2703
16-3-2968
14-64303
13-21053
• 3.
2-828{Jl
2-77519
2-75325
2-67301
28
18-76411
16-66306
14-89812
13-40616
4
3-71710
3-6^999
3-54595
3-46510
29
19-18846
16-98371
15-14107
13-59072
5
4-57971
4-45182
4-32948
4-21-236
30
19-60044
17-29203
15-37-245
13-76483
6
5-41719
5 --242 14
5-07569
4-91732
31
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15-80-267
14-08404
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7-01969
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6-46.321
6-20979
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20-76579
18-14764
16-00255
14-23023
9
7-78611
7-43533
7-10732
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31
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18"2559-2| 15-76186
IRISH CONVERTED IN
TO STATUTE ACRES.
Irith.
Statute.
Irish.
Statute.
Irish,
StaluU.
R.
?.
A. R. P. Y.
A.
A.
R. P. Y.
A. A. B. P. T
0
1
0 0 1 18|
0 0 3 7|
1
1
2 19 5L
20
32 1 23 141
48 2 16 6,
0
2
2
3
0 38 104
3 17 15|
30
0
3
0 0 4 26
3
4
40
64 3 6 2>l
0
1
0 0 6 14^
4
6
1 36 21
50
80 3 38 20;
0
5
0 0 8 3
5
8
0 15 26j
2 35 1}
100
161 3 37 10]
323 3 34 21]
0 1
0
0 0 16 6
6
9
200
0 -J
0
0 0 32 12
7
11
1 14 6.^
3 33 111
300
485 3 32 2
1
0
0 1 24 24
8
12
400
647 3 29 123
■2
0
0 3 9 173
9
14
2 12 17
500
809 3 26 23
3
0
1 0 34 Hi
10
16
0 31 -2^
lOOO
1619 3 13 16J
VALUE OF FOREIGN MONEY IN BRITISH,
Silver being 5.». per ounce
*. d.
1 Florin is worth
16 Schillings (Hamburg)
1 ?vlark (Frankfort) .
1 Franc
1 Milree (Lisbon)
8 Reals
1
1
1
0
4 8
3 U
1 Dollar (New York)
96 Skillings (Copenhagen)
1 Lira (Venice)
1 Lira (Genoa)
1 Lira (Leghorn) .
1 Ruble
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