ARTES
LIBRARY
1817
VERITAS
UNIVERSITY OF MICHIGAN
HEALTHIN. UNDLY
TEEBOR
SCIENTIA
OF THE
QKQUÆRIS PENINSULAM-AMŒNAM.
CIRCUMSPICE
MATAVADA PAŠAKIAIKINIAIO JOKATZIKONZENTRO MIKINYA
THE GIFT OF
Miss Marian Goodrich
Tanto
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PRINCIPLES OF OPERATING BY NUMBERS
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ADAMS'S NEW ARITHMETIC.
ARITHMETIC,
IN WHICH THE
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ANALYTICALLY EXPLAINED,
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ARE
SYNTHETICALLY APPLIED ;
TO BE DERIVED BOTH FROM
THE INDUCTIVE AND SYNTHETIC MODE
OF INSTRUCTING:
THE WHOLE
MADE FAMILIAR BY A GREAT VARIETY OF USEFUL AND INTER-
ESTING EXAMPLES, CALCULATED AT ONCE TO ENGAGE
THE PUPIL IN THE STUDY, AND TO GIVE HIM A
FULL KNOWLEDGE OF FIGURES IN THEIR
APPLICATION TO ALL THE PRAC-
TICAL PURPOSES OF
LIFE.
1
AND
COMBINING THE ADVANTAGES
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THUS
DESIGNED FOR THE USE OF
SCHOOLS AND ACADEMIES
IN THE UNITED STATES.
J
UTICA, N. Y.
PUBLISHED BY HASTINGS & TRACY.
*****
1828.
>
BY DANIEL ADAMS, M. D.
AUTHOR OF THE SCHOLÀR'S ARITHMETIC, SCHOOL GEOGRAPHY, &C.
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A true copy.
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DISTRICT OF NEW-HAMPSHIRE.
District Clerk's Office.
BE IT REMEMBERED, That on the eighteenth day of September, A. D. 1827, in the
fifty-second year of the Independence of the United States of America, DANIEL ADAMB,
of said district, has deposited in this office the title of a book, the right whereof he claims
as author, in the words following, to wit:
}
"ARITHMETIC, in which the Principles of operating by Numbers are analytically ex-
plained, and synthetically applied: thus combining the Advantages to be derived both
from the inductive and synthetic Modo of instructing the whole made familiar by a
great Variety of useful and interesting Examples, calculated at once to engage the Pupil
In the Study, and to give him a full Knowledge of Figures in their Application to all the
practical Purposes of Life. Designed for the Use of Schools and Academies in the United
States. By DANIEL ADAMS, M. D. Author of the Scholar's Arithmetic, School Goog-
raphy, &c."
In conformity to the act of Congress of the United States, entitled, " An Act for
the encouragement of learning, by securing the copies of maps, charts, and books,
to the authors and proprietors of such copies during the times therein mentioned;"
and also to an act, entitled, "An Act supplementary to an act for the encourage-
ment of learning, by securing the copies of maps, charts, and books, to the authors and
proprietors of such copies during the times therein mentioned; and extending the bone-
fits theroof to the arts of dosigning, engraving and etching historical and other printą.",
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Attest, C. W. CUTTER, Clerk.
CHARLES W. CUTTER,
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Stereotyped at the
Boston Type and Stereotype Foundry.
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Clerk of the District of New-Hampshire.
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PREFACE.
5
THERE are two methods of teaching, the synthetic and the analytic.
In the synthetic method, the pupil is first presented with a general
view of the science he is studying, and afterwards with the particulars
of which it consists. The analytic method reverses this order: the
pupil is first presented with the particulars, from which he is led, by
certain natural and easy gradations, to those views which are more
general and comprehensive.
▸
The Scholar's Arithmetic, published in 1801, is synthetic. If that
is a fault of the work, it is a fault of the times in which it appeared.
The analytic or inductive method of teaching, as now applied to ele-
mentary instruction, is among the improvements of later years. Its
introduction is ascribed to PESTALOZZI, a distinguished teacher in
Switzerland. It has been applied to arithmetic, with great ingenuity,
by Mr. COLBURN, in our own country.
The analytic is unquestionably the best method of acquiring know
ledge; the synthetic is the best method of recapitulating, or reviewing
it. In a treatise dosigned for school education, both methods are use-
ful. Such is the plan of the present undertaking, which the author,
occupied as he is with other objects and pursuits, would willingly have
forborne, but that, the demand for the Scholar's Arithmetic still con-
tinuing, an obligation, incurred by long-continued and extended pa-
tronage, did not allow him to decline the labour of a revisal, which
should adapt it to the present more enlightened views of teaching this
science in our schools. In doing this, however, it has been necessary
to make it, a new work,
In the execution of this design, an analysis of each rule is first given,
containing a familiar explanation of its various principles; after which
follows a synthesis of these principles, with questions in form of a sup-
plement. Nothing is taught dognatically; no technical term is used
till it has first boon definod, nor any principle inculcated without a pre-
vious developement of its truth; and the pupil is made to understand
the reason of each process as he proceeds.
The examples under each rule are mostly of a practical nature, be-
ginning wit those that aro very easy, and gradually advancing to
those more ficult, till one is introduced containing larger numbers,
and which is not casily solved in the mind; then, in a plain, familiar
mannor, the pupil is shown how the solution may be facilitated by
figures. In this way he is made to see at once their use and their ap-
plication.
At the close of the fundamental rules, it has been thought advisable
to collect into one clear view the distinguishing properties of those
rules, and to give a number of oxamples involving one or more of them.
These exercisos will prepare the pupil more readily to understand the
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application of these to the succeeding rules; and, besides, will serve
to interest him in the science, since he will find himself able, by the
application of a very few principles, to solve many curious questions.
The arrangement of the subjects is that, which to the author has
appeared most natural, and may be seen by the Index. Fractions have
received all that consideration which their importance demands. The
principles of a rule called Practice are exhibited, but its detail of cases
is omitted, as unnecessary since the adoption and general use of federal
money. The Rule of Three, or Proportion, is retained, and the solu-
tion of questions involving the principles of proportion, by analysis, is
distinctly shown.
The articles Alligation, Arithmetical and Geometrical Progression,
› Annuities and Permutation, were prepared by Mr. IRA YOUNG, a mom-
ber of Dartmouth College, from whose knowledge of the subject, and
experience in teaching, I have derived important aid in other parts of
the work.
{
PREFACE.
•
The numerical paragraphs are chiefly for the purpose of reference:
these references the pupil should not be allowed to neglect. His at-
tention also ought to be particularly directed, by his instructer, to the
illustration of each particular principle, from which general rules are
deduced for this purpose, recitations by classes ought to be instituted
in every school where arithmetic is taught.
1
The supplements to the rules, and the geometrical demonstrations
of the extraction of the square and cubo roots, are the only traits of the
old work preserved in, the new.
DANIEL ADAMS.
Mont Vernon, (N. H.) Sept. 29, 1827.
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Numeration and Notation,
Addition,
Subtraction,
Multiplication,
Division,
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Different Denominations,
Federal Money,
gift
me, Marian Goodrich
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INDEX.
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SIMPLE NUMBERS.
Fractions arise from Division,
Miscellaneous Questions, involving the Principles of the preceding Rules,
COMPOUND NUMBERS.
1
Reduction,
Tables of Money, Weight, Measure, &c.
Addition of Compound Numbers,
Subtraction,
Multiplication and Division,
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to find the Value of Articles sold by the 100, or 1000,
Bills of Goods sold,
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COMMON, or VULGAR. Their Notation,
Proper, Improper, &c.
•
To change an Improper Fraction to a Whole or Mixed Number,
a Mixed Number to an Improper Fraction,
To reduce a Fraction to its lowest Terms,
FRACTIONS.
Greatest common Divisor, how found,
To divide a Fraction by a Whole Number; two ways,
To multiply a Fraction by a Whole Number; two ways,
a Whole Number by a Fraction,
one Fraction by another,
General Rule for the Multiplication of Fractions,
To divide a Whole Number by a Fraction,
one Fraction by another,
General Rule for the Division of Fractions,
Addition and Subtraction of Fractions,
•
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56
57
64
68
69
69-82
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Fage
7
12
19
26
37
42
52
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Common Denominator, how found,
Least Common Multiple, how found,
Rule for the Addition and Subtraction of Fractions,
Reduction of Fractions,
121
124
124
DECIMAL. Their Notation,
132
135
Addition and Subtraction of Decimal Fractions,
Multiplication of Decimal Fractions,
Division of Decimal Fractions,
137
139
To reduce Vulgar to Decimal Fractions,
Reduction of Decimal Fractions,
142
145
146
To reduce Shillings, &c., to the Decimal of a Pound, by Inspection,
the three first Decimals of a Pound to Shillings, &c., by Inspection, 157
89
93
101
102
103
104
105
106
107
110
112
113
114
115
117
118
119
120
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Reduction of Currencies,
To reduce English, &c. Currencies to Federal Money,
Federal Money to the Currencies of England, &c.
one Currency to the Par of another Currency,
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Taxes, Method of assessing,
Alligation,
Duodecimals,
mont blond
•
by Progression,
Equation of Payments,
Ratio, or the Relation of Numbers,
Proportion, or Single Rule of Three,
Same Questions, solved by Analysis, ¶ 65, ex. 1—20.
Compound Proportion, or Double Rule of Three,
Fellowship,
Interest,
Time, Rate per cent., and Amount given, to find the Principal,
Time, Rate per cent., and Interest given, to find the Principal,
Principal, Interest, and Time given, to find the Rate per cent.,
Principal, Rate per cent., and Interest given, to find the Time,
To find the Interest on Notes, Bonds, &c., when partial Payments have
been made,
Compound Interest,
•
Arithmetical Progressiou,
Annuities at Compound Interest,
Practice, 29, cx. 10-19. T 43.
Insurance, ¶ 82.
T
Buying and Selling Stocks, T 82.
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INDEX.
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Involution,
Extraction of the Square Root,
•
Application and Use of the Square Root, sec Supplement,
Extraction of the Cube Root,
•
Application and Use of the Cube Root, see Supplement,
222 Geometrical Progression,
231 Permutation,
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Gauging, ex. 190, 191.
Forms of Notes, Bonds, Receipts, and Orders,
Book-Keeping,
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of a Pyramid, or Cone, ex. 188, 189.
of any Irregular Body, ex. 202, 203.
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187
192
195
197
201
Scale for taking Dimensions in Feet and Decimals of a Foot, 204
205 | Evolution,
207
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MISCELLANEOUS EXAMPLES.
Barter, cx. 21-32.
Position, ex. 89-108.
To find the Area of a Square or Parallelogram, ex. 148–154.
of a Triangle, ex. 155-159.
P),
Having the Diameter of a Circle, to find the Circumference; or, having the
Circumference, to find the Diameter, ex. 171-175.
To find the Arca of a Circle, ex. 176-179.
of a Globe, cx, 180, 181.
To find the Solid Contents of a Globe, ex. 182–184.
of a Cylinder, ex. 185-187.
IN
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Mechanical Powers, ex. 192-201.
•
Commission, T 82; 11 85, ex, 5, 6.
Loss and Gain, ¶ 82; ¶ 88, cx. 1—8.
Discount, ¶ 85, cx. 6-11.
Paro
151
7
153
154
155
156
164
165
166
167
207
212
.. 215
220
225
237
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168
169
229
176
177
179
259
263
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NUMERATION.
or one
T. A SINGLE or individual thing is called a unit,”unity,
one and one more are called two; two and one more
are called three; three and one more are called four; four
and one more are called five; five and one more are called
six; six and one more are called seven; seven and one more
are called eight; eight and one more are called nine; nine
and one more are called ten, &c.
Sy
These terms, which are expressions for quantities, are
called numbers. There are two methods of expressing
numbers shorter than writing them out in words; one called
the Roman method by letters,* and the other the Arabic
method by figures. The latter is that in general use.
In the Arabic method, the nine first numbers have each
an appropriate character to represent them. Thus,
One
Two
Three
Four
Five
Six
Seven
Eight
Nine
Ten
*In the Roman method by letters. I represents one; V, five; X, ten; L, fifty ;
C, one hundred; D, fire hundred and M, one thousand.
As often as any letter is repeated, so many times its value is repeated, unless it
bo a letter representing a less number placed before one representing a greater;
then the less number is taken from the greater; thus, IV represents four, LX, nine;
Sc., as will be seen in the following
Twenty
Thirty
Forly
Fifty
Sixty
Seventy
Eighty
'
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RITHMETIC.
I.
II.
11.
llil. or IV.
V.
VI
TABLE.
Ninety
VII.
VIII.
VIIII. or IX.
X.
- XX.
XXX.
XXXX. & XL.
L.
LX.
LXX.
LXXX.
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One hundred
Two hundred
Three hundred
Four hundred
Five hundred
Six hundred
Seven hundred
Eight hundrod
Nine hundred
One thousand
Five thousand
Ten thousand
Fifty thousand
Hundred thousand
One million
Two million
{
LXXXX. or XC.'
C.
CC.
CCC.
CCCC.
D. or 10.*
DC.
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DCC.
DCCC.
M.
MM.
* 15 is usod instead of D to represont fivo hundrod, and for overy additional Ɔ ar、
noxed at the right hand, the number is increased ten times.
+ CIO is used to represent one thousand, and for evory C. and I put at each end, the
number is increased ton times.
↑ A line over any number increases its value one thousand times
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DCCCC.
M. or CIO.f
100.or V.‡
CC100. or X
1000.
CCCIDO3. or C.
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Four
Five
Six
Seven
Eight
Nine
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A unit, unity, or one, is represented by this character,
Two
Three
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NUMERATION.
P
Tep has no appropriate character to represent it; but is
'considered as forming a unit of a second or higher
order, consisting of tens, represented by the same
character (1) as a unit of the first or lower order,
but is written in the second place from the right
hand, that is, on the left hand side of units; and
as, in this case, there are no units to be written
with it, we write, in the place of units, a cipher, (0,)
which of itself signifies nothing; thus,
Ten
Eleven
Twelve 12.
10.
11.
One ten and one unit are called-
One ten and two units are called
Thirteen 13.
Fourteen 14.
Fifteen· *15
Sixteen 16.
Seventeen 17.
Eighteen 18.
One ten and three units are called
One ten and four units are called
One ten and five units, are called
One ten and six units are called
One ten and seven units are called
One ten and eight units are called
One ten and nine units are called'
Two tens are called
Three tens are called
Four tens are called
Five tens are called
Six tens are called
Seven tens are called
Eight tens are called
Nine tens are called
Nineteen 19.
Twenty 20.
Thirty 30.
Forty
Fifty
40.
50..
60:
Sixty
Seventy
770.
Eighty . 80.
Ninety
90.
Ten tens are called a hundred, which forms a unit of a
still higher order, consisting of hundreds, represented
by the same character (1) as a unit of each of the
foregoing orders, but is written one place further
toward the left hand, that is, on the left hand side
of tens; thus,
One hundred 100.
One hundred, one ten, and one unit, are called
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One hundred and eleven 111..
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12. There are three hundred sixty-five days in a year.
In this number are contained all the orders now described,
viz. units, tens, and hundreds. Let it be recollected, units
occupy the first place on the right hand; 'tens, the second
place from the right hand; hundreds, the third place. This
number may now be decomposed, that is, separated into parts,
exhibiting each order by itself, as follows:-The highest
order, or hundreds, are three, represented by this character,
3; but, that it may be made to occupy the third place, count-
ing from the right hand, it must be followed by two ciphers,
thus, 300, (three hundred.) The next lower order, or tens,
are six, (six tens are sixty,) represented by this character, 6;
but, that it may occupy the second place, which is the place
of tens, it must be followed by one cipher, thus, 60, (sixty.)
The lowest order, or units, are five, represented by a single
character, thus, 5, (five.)
1
NUMERATION.

We may now combine all these parts together, first writing
down the five units for the right hand figure, thus, 5; then
the six téns (60) on the left hand of the units, thus, 65; then
the three hundreds (300) on the left hand of the six tens,
thus, 365, which number, so written, may be read three
hundred, six tens, and five units; or, as is more usual, three
hundred and sixty-five.
T 3. Hence it appears, that figures have a different value
according to the PLACE they occupy, counting from the right hand
towards the left.
Hund.
Tens.
Units.
Take for example the number 3 3 3, made by the same
figure three times repeated. The 3 on the right hand, or in
the first place, signifies 3 units;, the same figure, in the second
place, signifies 3 tens, or thirty; its value is now increased
ten times. Again, the same figure, in the third place, signi-
fies neither 3 units, nor 3 tens, but 3 hundreds, which is ten
times the value of the same figure in the place immediately
preceding, that is, in the place of tens; and this is a funda-
mental law in notation, that a removal of one place towards
the left increases the value of a figure TEN TIMES.
1
Ten hundred make a thousand, or a unit of the fourth
order. Then follow tens and hundreds of thousands, in the
same manner as tens and hundreds of units. To thousands
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succeed millions, billions, &c., to each of which, as to units.
and to thousands, are appropriated three places,* as exhi-
bited in the following examples:
EXAMPLE 1st.
EXAMPLE 2d.
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of Quadrillions.
❤ Units
NUMERATION.
of Trillions.
Hundreds
Tens
A Units
of Billions.
-Hundreds
Tens
Units
J
of Millions.
Hundreds
co Tens
→ Units
co
of Thousands.
Hundreds
→ Tens
∞ Units
A
or Hundreds
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of Units.
- Tens
Units
3 1 7 4 5 9 2 8 3 7 4 6 3 5 1 2
3, 174, 5 9 2, 8 3 7, 4 6 3, 5 1 2,
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To facilitate the reading of large numbers, it is frequently
practised to point them off into periods of three figures each,
as in the 2d example. The names and the order of the pe-
riods being known, this division enables us to read num-
bers consisting of many figures as easily as we can read
three figures only. Thus, the above examples are read 3
(three) Quadrillions, 174 (one hundred seventy-four) Tril-
lions, 592 (five hundred ninety-two) Billions, 837 (eight
hundred thirty-seven) Millions, 463 (four hundred sixty-
three) Thousands, 512 (five hundred and twelve.)
After the same manner are read the numbers contained in
the following
This is according to the French method of counting. The English, after
hundreds of millions, instead of proceeding to billions, reckon thousands, tens and
hundreds of thousands of millions, appropriating six places, instead of three, to
millions, billions, &c.
}
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Hundreds of Millions.
Tens of Millions.
·
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Hundreds of Thousands.
Tens of Thousands.
Millions.
Thousands.
Hundreds.
Tens.
Units.
by
86
4 3 2
7054
86 200
9 0 0 3 7 1
5 0 8 6 0 0 0
1 0 3 0 2 0 7 0
8 0.6 10 5409
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34120*
701602
6539285
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NUMERATION TABLE.
+
NUMERATION.
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Those words at the head of the
table are applicable to any sum or
number, and must be committed per-
fectly to memory, so as to be readily
applied on any occasion:
Note. Should the pupil find any difficulty in reading the
following numbers, let him first transcribe them, and point
them off into periods.
Of these characters, 1, 2, 3, 4, 5,
6, 7, 8, 9, 0, the nine first are some-
times called significant figures, or
digits, in distinction from the last,
which, of itself, is of no value, yet,
placed at the right hand of another
figure, it increases the value of
that figure in the same tenfold pro-
portion as if it had been followed by
any one of the significant figures.
52831209
175264013
3456720834
25037026531
Jurgel,~
11
286297314013
5203845761204
13478120673019
341246801734526
The expressing of numbers, (as now shown,) by figures,
is called Notation. The reading of any number set down in
figures, is called Numeration.
1. Seventy-six.
2. Eight hundred and seven.
After being able to read correctly all the numbers in the
foregoing table, the pupil may proceed to express the fol-
lowing numbers by figures:
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3. Twelve hundred, (that is, one thousand and two hun-
dred.)
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ADDITION OF SIMPLE NUMBERS.
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4. Eighteen hundred,
5. Twenty-seven hundred and nineteen.
6. Forty-nine hundred and sixty.
7. Ninety-two thousand and forty-five.
8. One hundred thousand.
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9. Two millions, eighty thousands, and seven hundreds.
10. One hundred millions, one hundred thousand, one
hundred and one.
T 3, 4,
11. Fifty-two millions, six thousand, and twenty.
12. Six billions, seven millions, eight thousand, and nine
hundred.
13. Ninety-four billions, eighteen thousand, one hundred
and seventeen.
14. One hundred thirty-two billions, two hundred millions,
and nine.
ADDITION
15. Five trillions, sixty billions, twelve millions, and ten
thousand.
16. Seven hundred trillions, eighty-six billions, and seven
millions.
OF SIMPLE NUMBERS.
T4. 1. James had 5 peaches, his mother gave him 3
peaches more; how many peaches had he then ?
2. John bought a slate for 25 cents, and a book for eight
cents; how many cents did he give for both?
3. Peter bought a waggon for 36 cents, and sold it so as
to gain 9 cents; how many cents did he get for it?
4. Frank gave 15 walnuts to one boy, 8 to another, and
had 7 left; how many walnuts had he at first?
5. A man bought a chaise for 54 dollars; he expended 8
dollars in repairs, and then sold it so as to gain 5 dollars;
how many dollars did he get for the chaise ?
6. A man bought 3 cows; for the first he gave 9 dollars,
for the second he gave 12 dollars, and for the other he gave
10 dollars; how many dollars did he give for all the cows?
7. Samuel bought an orange for 8 cents, a book for 17.
cents, a knife for 20 cents, and some walnuts for 4 cents;
how many cents did he spend?
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ADDITION OF SIMPLE NUMBERS.
13
8. A man had 3 calves worth 2 dollars each, 4 calves
worth 3 dollars each, and 7 calves worth 5 dollars each;
how many calves had he?
9. A man sold a cow for 16 dollars, some corn for 20 dol-
lars, wheat for 25 dollars, and butter for 5 dollars; how
many dollars must he receive?
The putting together two or more numbers, (as in the
foregoing examples,) so as to make one whole number, is
called Addition, and the whole number is called the sum, or
amount.
10. One man owes me 5 dollars, another owes me 6
dollars, another 8 dollars, another 14 dollars, and another 3
dollars; what is the amount due to me?
11. What is the amount of 4, 3, 7, 2, 8, and 9 dollars ?
12. In a certain school 9 study grammar, 15 study arith-
metic, 20 attend to writing, and 12 study geography; what
is the whole number of scholars?
2+0=
2+1=
SIGNS. A cross,+, one line horizontal and the other per-
pendicular, is the sign of addition. It shows that numbers,
with this sign between them, are to be added together. It
is sometimes read plus, which is a Latin word signifying
more.
Two parallel, horizontal lines,, are the sign of equality.
It signifies that the number before it is equal to the number
after it. Thus, 53 8 is read 5 and 3 are 8; or, 5 plus
(that is, more) 3 is equal to 8.
1
In this manner let the pupil be instructed to commit the
following
2
OT JA W N
3
1
4
1
ADDITION TABLE.
3+0=
3+1
3+2
3+3
4+0=
441
54+2=
64+
+3
3+4=74 + 4 =
845
4 + 6
4 + 7
+++++
II II
2+2
2+3
+ 4
6
My
3 +
6 8 3 +6
2 +5
2+
2+7=9 3+7
2+8=10 38
114 +8
2 +9 = 11 | 3 + 9 = 12 | 4+9:
B
3
→
1
9
10
4
5
6
* 10 10 10✪
8
=13
5+0
1
5 + 1
5+2
9
10
11
12
5 +8:
13 | 5+ 9
10
5
6
17
8
9
10
11
+3
5 + 4
5+5
5 + 6
5712
13
14
}
?
**
L
7
1
}
S
14
ļ
1
}
K
ADDITION TABLE-CONTINUED,
6 +0
7+0.
7 +1
6 +1
6 +2
myt
6 +3
97
6 + 4 = 10
10 7
6 + 5
11
6 +6
6 + 7
6 +8
69
Banner with
Į
L
J
6
ну
ADDITION OF SIMPLE NUMBERS. T 4,5
15
+ + + + +
123 4D ON∞
ry +
127+ 6
13 77
1478
79
TEET T
:
Spa
718+0
8 + 1
82
8+ 3
8 +4
12
8+ 5
13
8 +6
14
87
158+
16
8
}
9
1.0
{
11
5+9= how many?
87 how many?
4+3+2 how many?
6 + 4+ 5
how many?
M
Pendatang
819+0 = 9
9 9+1=10
109
9+2=11
9+3= 12
9 +4
13
95
14
9 + 6
15
97
16
17
98
9+9
18
11
12
13
14
15
=16
S
8 +9 = 17
2+0+4+6= how many
7+1+0+8= how many
3+0+9+5 how many?
9264+5= how many?
1+3+5+ 7+ 8 how many?
1+2+3+4+5+6 how many?
8+9+0+2+4+5= how many?
6+2+5+0+8+3= how many?
་་་
Joanna
[
1
T 5. When the numbers to be added are small, the addi-
tion is readily performed in the mind; but it will frequently
be more convenient, and even necessary, to write the num-
bers down before adding them.
Pag?
13. Harry had 43 cents, his father gave him 25 cents
more; how many cents had he then?
One of these numbers contains 4 tens and 3 units. The
other number contains 2 tens and 5 units. To unite these
two numbers together into one, write them down one
under the other, placing the units of one number directly
under units of the other, and the tens of one number directly
under tens of the other, thus:
43 cents. Having written the numbers in this man-
25 cents. ner, draw a line underneath.
:
¿
1
T 5.
43 cents.
25 cents.
8
43 cents.
25 cents.
T
ADDITION OF SIMPLE NUMBERS.
15
We then begin at the right hand, and add
the 5 units of the lower number to the 3
units of the upper number, making 8 units,
which we set down in unit's place.
We then proceed to the next column, and
add the 2 tens of the lower number to the
4 tens of the upper number, making 6 tens,
or 60, which we set down in ten's place,
and the work is done.
Ans. 68 cents.
It now appears that Harry's whole number of cents is 6
tens and 8 units, or 68 cents; that is, 43 +25= 68.
14. A farmer bought a chaise for 210 dollars, a horse for
70 dollars, and a saddle for 9 dollars; what was the whole
amount?
Write the numbers as before directed, with units under
units, tens under tens, &c.
OPERATION.
Chaise, 210 dollars.
Horse, 70 dollars.
Saddle, 9 dollars.
}
"
Add as before. The units will
be 9, the tens 8, and the hundreds
2; that is, 210+709289.
Answer, 289 dollars.
Grig
After the same manner are performed the following ex-
amples:
15. A man had 15 sheep in one pasture, 20, in another
pasture, and 143 in another; how many sheep had he in
the three pastures? 15 +20 + 143 how many?
16. A man has three farms, one containing 500 acres,
another 213 acres, and another 76 acres; how many acres
in the three farms? 500+213 +76 how many?
17. Bought a farm for 2316 dollars, and afterwards sold
it so as to gain 550 dollars; what did I sell the farm for?
2316550 how many?
Men dettagl
Hitherto the amount of any one column, when added up,
has not exceeded 9; consequently has been expressed by a
single figure. But it will frequently happen that the amount
of a single column will exceed 9, requiring two or more figures
to express it.
18. There are three bags of money. The first contains
876 dollars, the second, 653 dollars, the third, 524 dollars,
what is the amount contained in all the bags?
1
16
π 5.
Writing down the numbers as
already directed, we begin with the
right hand, or unit column, and
find the amount to be 13, that is,
3 units and 1 ten. Setting down
the 3 units, or right hand figure,
2
in unit's place, directly under the column, we reserve the
1 ten, or left hand figure, to be added with the other
tens, in the next column, saying,, 1, which we reserved, to 2
makes 3, and 5 are 8, and 7 are 15, which is 5 units of its
own order, and I unit of the next higher order, that is, 5 tens
and 1 hundred. Setting down the 5 tens, or right hand figure,
directly under the column of tens, we reserve the left hand
figure, or 1 hundred, to be added in the column of hun-
dreds, saying, 1 to 5 is 6, and 6 are 12, and 8 are 20, which
being the last column, we set down the whole number,
writing the 0, or right hand figure, directly under the column,
and carrying forward the 2, or left hand figure, to the next
place, or place of thousands. Wherefore, we find the whole
amount of money contained in the three bags to be 2053
dollars, the answer.
OPERATION.
First bag,
Second bag,
Third bag,
Amount,
1
ADDITION OF SIMPLE NUMBERS.
}
876
653
524
2053
$1
1
3
PROOF. We may reverse the order, and, beginning at the
top, add the figures downward. If the two results are alike,
the work is supposed to be right.
From the examples and illustrations now given, we de-
rive the following
↓
4
F
J
RULE.
1. Write the numbers to be added, one under another,
placing units under units, tons under tens, &c., and draw a
line underneath.
I
II. Begin at the right hand or unit column, and add to-
gether all the figures contained in that column: if the
amount does not exceed 9, write it under the column; but
if the amount exceed 9, so that it shall require two or more
figures to express it, write down the unit figure only under
the column; the figure or figures to the left hand of units,
being tens, are so many units of the next higher order,
which, being reserved, must be carried forward, and added
to the first figure in the next column.
III. Add each succeeding column in the same manner, and
set down the whole amount at the last column
1
.
}
|
T 5.
10
ADDITION OF SIMPLE NUMBERS.
EXAMPLES FOR PRACTICE.
19. A man bought four loads of hay; one load weighed
1817 pounds, another weighed 1950 pounds, another 2156
pounds, and another 2210 pounds; what was the amount
of hay purchased ?
20. A person owes A 100 dollars, B 2160 dollars, C785
dollars, D 92 dollars; what is the amount of his debts.?
21. A farmer raised in one year 1200 bushels of wheat,
850 bushels of Indian corn, 1000 bushels of oats, 1086 bush-
els of barley, and 74 bushels of pease; what was the whole
amount?
Ans. 4210.
22. St. Paul's Cathedral, in London, cost 800,000 pounds
sterling; the Royal Exchange 80,000 pounds; the Mansion-
House 40,000 pounds; Black Friars Bridge 152,840 pounds;
Westminster Bridge 389,000 pounds, and the Monument
13,000 pounds; what is the amount of these sums?
Ans. 1,474,840 pounds.
23. At the census in 1820, the number of inhabitants in
the New England States was as follows:-Maine, 298,335;
New Hampshire, 244,161; Vermont, 235,764; Massachu-
setts, 253,287; Rhode Island, 83,059; Connecticut, 275,248;
what was the whole number of inhabitants, at that time, în
those States?
Ans. 1,389,854.
25.
2 8 6 3 7 0 5 4 2 1 0 6 1
310742 9 3 1 5 6 3 8
6 2 5 3 0 3 4 7 9 2
247135
8673
1
है
B*
24. From the creation to the departure of the Israelites
from Egypt was 2513 years; to the siege of Troy, 307 years
more; to the building of Solomon's Temple, 180 years; to
the building of Rome, 251 years; to the expulsion of the
kings from Rome, 244 years; to the destruction of Carthage,
363 years; to the death of Julius Cæsar, 102 years; to the
Christian era, 44 years; required the time from the Crea-
tion to the Christian era..
Ans. 4004 years.
17
;
26.
4 3 6 7 5 8 3 0 2 14 6 3
1752349 713 620
608127 5 3 0 6 2 17
5 6 5 2 17 4 6 3 0 1 28
870326 3 4 7 2013
1
f
#
18
27.
5 3 6 4 20 76 3 10 23
2 8 1 2 3 4 5 6 7 2 9 4 8
605 7042087094
3 1 6 2 8 3 5 9 0 6 7 18
7 6 0 4 2 8 6 5 3 7 892
19 C
∞ ∞0,
1
1
SUPPLEMENT TO NUMERATION. AND ADDITION. T 5.
+
I
1
29. What is the amount of 46723, 6742, and 986 dollars?
30. A man has three orchards; in the first there are 140
trees that bear apples, and 64 trees that bear peaches; in
the second, 234 trees bear apples," and 73 bear cherries; in
the third, 47 trees bear plums, 36 bear pears, and 25 bear
cherries; how many trees in all the orchards?
*
}
SUPPLEMENT
TO NUMERATION AND ADDITION.
QUESTIONS.
1. What is a single or individual thing called? 2. What
is notation? 3. What are the methods of notation now in
use? 4. How many are the Arabic characters or figures ?
5. What is numeration? 6. What is a fundamental law in
+
;
28.
9 0 2 3 7 5 4 6 8 2 1 3 5
28 3 4 9 6 7 326 708
9306 34 2 1 6 7 3 2'1
2 3 6 5 4 780 24369
8.0 50 60 70 80 900
1
7
IX.
notation? 7. What is addition? 8. What is the rule
for addition? 9. What is the result, or number sought,
called? 10. What is the sign of addition? 11. of
equality? 12. How is addition proved?
EXERCISES.
¿
1. Washington was born in the year of our Lord 1732;
he was 67 years old when he died; in what year of our
Lord did he die?
{
}
2. The invasion of Greece by Xerxes took place 481 years.
before Christ; how long ago is that this current year 1827 ?
3. There are two numbers, the less number is 8671, the
difference between the numbers is 597; what is the greater
number.
i
1

15
AA
T 5, 6.
19
4. A man borrowed a sum of money, and paid in part
684 dollars; the sum left unpaid was 876 dollars; what was
the sum borrowed?
*
SUBTRACTION OF SIMPLÉ NUMBERS.
5. There are four numbers, the first 317, the second 812,
the third 1350, and the fourth as much as the other three;
what is the sum of them all?
6. A gentleman left his daughter 16 thousand, 16 hun-
dred and 16 dollars; he left his son 1800 more than his
daughter; what was his son's portion, and what was the
amount of the whole estate? Ans. {Son's portion, 19,416.
Whole estate, 37,032.
7. A man, at his death, left his estate to his four children,
who, after paying debts to the amount of 1476 dollars,
received 4768 dollars each; how much was the whole
estate ?
Ans. 20548.
8. A man bought four hogs, each weighing 375 pounds;
how much did they all weigh?
Ans. 1500.
9. The fore quarters of an ox weigh one hundred and
eight pounds each, the hind quarters weigh one hundred
and twenty-four pounds each, the hide seventy-six pounds,
and the tallow sixty pounds; what is the whole weight of
the ox?
Ans. 600.*
10. A man, being asked his age, said he was thirty-four
years old when his eldest son was born, who was then fif-
teen years of age; what was the age of the father?
11. A man sold two cows for sixteen dollars each, twen-
ty bushels of corn for twelve dollars, and one hundred
pounds of tallow for eight dollars; what was his due ?
↓
I
SUBTRACTION
OF SIMPLE NUMBERS.
>
16. 1. Charles, having 18 cents, bought a book, for which
he gave 6 cents; how many cents had he left?
2. John had 12 apples; he gave 5 of them to his brother;
how many had he left ?
3. Peter played at marbles; he had 23 when he began,
but when he had done he had only 12; how many did he
--Jose ?
}
1
1
í
t
20
SUBTRACTION OF SIMPLE NUMBERS.
T 6.
4. A man bought a cow for 17 dollars, and sold her again
for 22 dollars; how many dollars did he gain?
5. Charles is 9 years old, and Andrew is 13; what is the
difference in their ages?
6. A man borrowed 50 dollars, and paid all but 18; how
many dollars did he pay? that is, take 18. from 50, and
how many would there be left ?
7. John bought a book and slate for 33 cents; he gave 8
cents for the book; what did the slate cost him ?
J
}
8. Peter bought a waggon for 36 cents, and sold it for 45
cents; how many cents did he gain by the bargain?
9. Peter sold a waggon for 45 cents, which was 9 cents
more than he gave for it; how many cents did he give for
the waggon?
10. A boy, being asked how old he was, said that he was
25 years younger than his father, whose age was 33 years;
how old was the boy?
+
The taking of a less number from a greater (as in the
foregoing examples) is called Subtraction. The greater
number is called the minuend, the less number the subtra-
hend, and what is left after subtraction is called the differ-
ence, or remainder.
what is
Ans. 5..
12. If the subtrahend be 4, and the minuend 16, what is
the remainder?
13. Samuel bought a book for 20 cents; he paid down 12
cents; how many cents more must he pay?
11. If the minuend be 8, and the subtrahend 3,
the difference or remainder?
J
SIGN. A short horizontal line, -, is the sign of subtrac
tion. It is usually read minus, which is a Latin word signi-
fying less. It shows that the number after it is to be taken
from the number before it. Thus, 8-35, is read 8 mi-
nus or less 3 is equal to 5; or, 3 from 8 leaves 5. The
latter expression is to be used by the pupil in committing
the following
1
1
}
7
3
"
7
'
1
Į
J
:
"
T 6, 7.
·
*?
MPANG GARANTE
22
3—2
4
1
2 2
5—2—3
CO HILO
JOGATJA
4
6 24
P
1
2=5
8-2
2=6
9-2
10-2
proteskakačky
7 - 3
8-5
SUBTRACTION OF SIMPLE NUMBERS.
3 3=0
3=1
5-3-2
LO ECO H
1
9
12
13 - 4
4
3
17
8
SUBTRACTION TABLE.
6-3=3
7-3=4
8-3-5
9-3
103 =
9
10-4
KAŽDÝ
Supported)
4-4
5-4
6-4
17
4
8-4
44
4: 5
6
Splat
how many?
how many?
how many?
how many?
how many
?
OPERATION,
From 237 the minuend,
•
Take 114 the subtrahend,
123 the remainder.
1
17
1
so
2
L
f
5—5—0
6-5- 1
77-5=2
8-5=3
9 —5— 4
10—5—5
6—6=0
7—6—1
8-6=2
9-6=3
10-6 : 4
R
18
28
22
33 5
(A)
I
J
41 - 15
- 15
1
7. When the numbers are small, as in the foregoing
examples, the taking of a less number from a greater is rea-
dily done in the mind; but when the numbers are large,
the operation is most easily performed part at a time, and
therefore it is necessary to write the numbers down before
performing the operation.
14. A farmer, having a flock of 237 sheep, lost 114 of
them by disease; how many had he left?
Here we have 4 units to be taken from 7 units, 1 ten t
be taken from 3 tens, and 1 hundred to be taken from
hundreds. It will therefore be most convenient to write the
less number under the greater, observing, as in addition, to
place units under units, tens under tens, &c. thus:
We now begin with the
units, saying, 4 (units) from
7, (units,) and there remain 3,
(units,) which we set down.
directly under the column in
unit's place. Then, proceed
ng to the next column, we say, 1 (ten) from 3, (tens,) and
here remain 2, (tens,) which we set down in ten's place.
!
htt
{
7-7=0:
8-7:
-7=1
97
7=2
10-7=3
17
how many?
7how many?
13 how many?
how
how many?
how many?
how
8—8=0
9-8=1
10-8=2
9—9—0
10—9—1
21
1
$
F
F
↓
22
T T
*
Proceeding to the next column, we say, 1 (hundred) from 2,
(hundreds,) and there remains 1, (hundred,) which we set
down in hundred's place, and the work is done. It now ap-
pears, that the number of sheep left was 123'; that is,
1
237 — 114 = 123,
SUBTRACTION OF SIMPLE NUMBERS.
After the same manner are performed the following ex-
amples:
15. There are two farms; one is valued at 3750, and the
other at 1500 dollars; what is the difference in the value
of the two farms?
}
16. A man's property is worth 8560 dollars, but he has
debts to the amount of 3500 dollars; what will remain after
paying his debts?
17. James, having 15 cents, bought a pen-knife, for which
he gave 7 cents; how many cents had he left?
OPERATION.
15 cents.
7 cents.
8 cents left.
18. A man bought a horse for 85 dollars, and a cow for
27 dollars; what did the horse cost him more than the
cow ?
A difficulty presents itself here; for we
cannot take 7 from 5; but we can take 7
from 15, and there will remain 8.
OPERATION.
Horse,
Cow, 277
85
1
Difference, 58
The same difficulty meets us here as in
the last example; we cannot take 7 from
5; but in the last example the larger num-
ber consisted of 1 ten and 5 units, which
together make 15; we therefore took 7
from 15. Here we have 8 tens and 5 units. We can now,
in the mind, suppose 1 ten taken from the 8 tens, which
would leave 7 tens, and this 1 ten we can suppose joined to
the 5 units, making 15. We can now take 7 from 15, as be-
T fore, and there will remain 8, which we set down. The
taking of 1 ten out of 8 tens, and joining it with the 5 units,
is called borrowing ten. Proceeding to the next higher or-
der, or tens, we must consider the upper figure, 8, from
which we borrowed, I less, calling it 7; then, taking 2 (tens).
from 7, (tens,) there will remain 5, (tens,) which we set down,
making the difference 58 dollars. Or, instead of making
the upper figure 1 less, calling it 7, we may make the lower
figure one more, calling it 3, and the result will be the same;
for 3 from 8 leaves 5, the same as 2 from 7,
+
ì
>
?
I
1
i
"
1
ཞ༽། 7,
५
8.
SUBTRACTION OF SIMPLE NUMBERS.
19. A man borrowed 713 dollars, and paid 471 dollars;
how many dollars did he then owe?
many?
713-471 how
Ans. 242 dollars.
Ans. 1147.
20. 1612 465
21. 43751
Ans. 36969.
subtraction is
of them; how
Ans. 22 sheep.
how many had
Ans. 40.
{
Cop
}
Baby T
Žema
6782
Kā
}}
how many?
¶ 8. The pupil will readily perceive, that
the reverse of addition.
22. A man bought 40 sheep, and sold 18
many had he left? 40 -18 how many?
23. A man sold 18 sheep, and had 22 left;
he at first? 18+22
18+22= how many?
24. A man bought a horse for 75 dollars, and a cow for
16 dollars; what was the difference of the costs?
how many? Reversed, 59 + 16
75-16
how many?
25. 114—103—how many? Reversed, 11 + 103 = how
many?
26. 143
many?
Belg
}
how many?
23
Hence, subtraction may be proved by addition, as in the
foregoing examples, and addition by subtraction.
To prove subtraction, we may add the remainder to the
subtrahend, and, if the work is right, the amount will be equal
to the minuend.
76= how many? Reversed, 67+76 = how
i
To prove addition, we may subtract, successively, from
the amount, the several numbers which were added to pro-
duce it, and, if the work is right, there will be no re-
mainder. Thus 7+ 8+6 21; proof, 21 — 6 :
15, and
KADET KA
15
-87, and 7—7— 0.
From the remarks and illustrations now given, we deduce
the following
#
RULE.
I. Write down the numbers, the less under the greater,
placing units under units, tens under tens, &c. and draw a
line under them.
II. Beginning with units, take successively each figure in
the lower number from the figure over it, and write the re-
mainder directly below.
III. When the figure in the lower number exceeds the
figure over it, suppose 10 to be added to the upper figure;
but in this case we must add 1 to the lower figure in the
next column, before subtracting. This is called borrowing 10.
↓
8
1
L
1
1
24
مور
*
SUPPLEMENT TO SUBTRACTION.
;
¡
EXAMPLES FOR PRACTICE.
27. If a farm and the buildings on it be valued at 10000,
and the buildings alone be valued at 4567 dollars, what is
the value of the land?
1
}
28. The population of New England, at the census in
1800, was 1,232,454; in 1820 it was 1,659,854; what was
the increase in 20 years?
29. What is the difference between 7,648,203 and
928,671 ?
30. How much must you add to 358,642 to make
1,487,945 ?
31. A man bought an estate for 13,682 dollars, and sold it
again for 15,293 dollars; did he gain or lose by it? and how
much?
32. From 364,710,825,193 take 27,940,386,574.
33. From 831,025,403,270 take 651,308,604,782.
34. From 127,368,047,216,843 take 978,654,827,352.
SUPPLEMENT
TO SUBTRACTION.
1
QUESTIONS.
1
1. What is subtraction? 2. What is the greater number
called? 3. the less number? 4. What is the result
or answer called?
5. What is the sign of subtraction
6. What is the rule? 7. What is understood by borrowing
ten? 8. Of what is subtraction the reverse? 9. How is
subtraction proved? 10. How is addition proved by sub-
traction ?
1
¶ "8%
31
1
EXERCISES.
1. How long from the discovery of America by Colum-
bus, in 1492, to the commencement of the Revolutionary
war in 1775, which gained our Independence?
2. Supposing a man to have been born in the year 1773,
how old was he in 1827?
1
."
3. Supposing a man to have been 80 years old in the year
1826, in what year was he born ?
(
4. There are two numbers, whose difference is 8764; the
greater number is 15687; I demand the less?
→
MAR
•
}
T 8.
5. What number is that which, taken from 3794, leaves
865?
+
3
6. What number is that to which if you add 789, it will
become 6350?
"
7. In New York, by the census of 1820, there were
123,706 inhabitants; in Boston, 43,940; how many more
inhabitants were then in New York than in Boston?
"
8. A man, possessing an estate of twelve thousand dollars,
gave two thousand five hundred dollars to each of his two
daughters, and the remainder to his son; what was his son's
share?
SUPPLEMENT TO SUBTRACTION.
9. From seventeen million take fifty-six thousand, and
what will remain ?
鼻
​10. What number, together with these three, viz. 1301,
2561, and 3120, will make ten thousand?
}
11. A man. bought a horse for one hundred and fourteen
dollars, and a chaise for one hundred and eighty-seven dol-
lars; how much more did he give for the chaise than for
the horse?
25
12. A man borrows 7 ten dollar bills and 3 one dollar
bills, and pays at one time 4 ten dollar bills and 5 one dol-
lar bills; how many ten dollar bills and one dollar bills
must he afterwards pay to cancel the debt?
}
Ans. 2 ten doll. bills and 8 one doll.
13. The greater of two numbers is 24, and the less is 16;
what is their difference?
}
14. The greater of two numbers is 24, and their differ-
ence 8; what is the less number?
15. The sum of two numbers is 40, the less is 16; what
is the greater?
"
16. A tree, 68 feet high, was broken off by the wind; the
top part, which fell, was 49 feet long; how high was the
stump which was left?
17. Our pious ancestors landed at Plymouth, Massachu-
setts, in 1620; how many years since?
18. A man carried his produce to market; he sold his
pork for 45 dollars, his cheese for 38 dollars, and his butter
for 29 dollars; he received, in pay, salt to the value of 17
dollars, 10 dollars worth of sugar, 5 dollars worth of mo-
lasses, and the rest in money; how much money did he
Ans. 80 dollars.
receive ?
19. A boy bought a sled for 28 cents, and gave 14 cents
с
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1
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1
1
1
1
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20
1 84 9.
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to have it repaired; he sold it for 40 cents; did he gain or
lose by the bargain? and how much?
MULTIPLICATION OR SIMPLE NUMBERS.
}
20. One man travels. 67 miles in a day, another man fol-
lows at the rate of 42 miles a day; if they both start from
the same place at the same time, how far will they be apart
at the close of the first day? of the second ?
of the fourth?
of
the third
1
3
会
​"
}
21. One man starts from Boston Monday morning, and
travels at the rate of 40 miles a day; another starts from the
same place Tuesday morning, and follows on at the rate of
70 miles a day; how far are they apart Tuesday night?
Ans. 10 miles.
-
1
1
22. A man, owing 379 dollars, paid at one time. 47 dol-
lars, at another time 84 dollars, at another time 23 dollars,
and at another time 143 dollars; how much did he then
owe?
*!
Ans, 82 dollars.
B
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23. A man has property to the amount of 34764 dollars,
but there are demands against him to the amount of 14297
dollars; how many dollars will be left after the payment of
his debts?
24. Four men bought a lot of land for 482 dollars; the
first man paid 274 dollars, the second man 194 dollars less
than the first, and the third man 20 dollars less than the
second; how much did the second, the third, and the fourth
man pay?
The second paid 80.
Ans. The third paid 60..
The fourth paid 68.
25. A man, having 10,000 dollars, gave away 9 dollars;
how many had he left?
Ans. 9991.
MULTIPLICATION
OF SIMPLE NUMBERS.
1723
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19.1. If one orange costs, 5 cents, how many; cents
must. L give for 2 oranges? - how many cents for 3
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2. One bushel of apples costs 20 cents; how many cents
must I give for 2. bushels ?
for 3 bushels ?
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T 9. MULTIPLICATION OF SIMPLE NUMBERS.
$27
3. One gallón contains 4 quarts; how many quarts in 2
gallons? in 3 gallons? in 4 gallons?
4. Three men bought a horse; each man paid 23 dollars
for his share; how many dollars did the horse cost them?
5. A man has 4 farms worth 324 dollars each; how many
dollars are they all worth?
6. In one dollar there are one hundred cents; how many
cents in 5 dollars?
. How much will 4 pair of shoes cost at 2 dollars a pair?
8. How much will two pounds of tea cost at 43 cents a
pound?
J
9. There are 24 hours in one day; how many hours in 2
days?
in 3 days? in 4 days?
- in 77
days?
10. Six boys met a beggar, and gave him 15 cents each;
how many cents did the beggar receive?
(1
When questions occur, (as in the above examples,) where
the same number is to be added to itself several times, the
operation may be much facilitated by a rule, called Multi-
plication, in which the number to be repeated is called the
multiplicand, and the number which shows how many times
the multiplicand is to be repeated is called the multiplier.
The multiplicand and multiplier, when spoken of collectively,
are called the factors, (producers,) and the answer is called
the product.
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11. There is an orchard in which there are 5 rows of trees,
and 27 trees in each row; how many trees in the orchard?
In the first row, 27 trees.
second 27
•
27
third
fourth 27
fifth
.... 27
...
....
I
In adding, we find
that 7 taken five times
amounts to 35. We write
down the five units, and
reserve the 3 tens; the amount of 2 taken five times is 10,
and the 3, which we reserved, makes 13, which, written to
the left of units, makes the whole number of trees 135.
*******
In the whole orchard, 135 trees.
ㄘ
​In this example, it is
evident that the whole
number of trees will be
equal to the amount of
five 27's added together.
If we have learned that 7 taken 5 times amounts to 35,
and that 2 taken 5 times amounts to 10, it is plain we need
write the number 27 but once, and then, setting the multi-
1
{
1
F
2
1
}
5
}
28
plier under it, we may say, 5 times 7 are 35, writing down
the 5, and reserving the 3 (tens) as in addition. Again, 5
times 2 (tens) are
10, (tens,) and 3,
(tens,) which we
reserved, maké 13,
(tens,) as before.
1
MULTIPLICATION OF SIMPLE NUMBERS ¶ 9,
T 9, 10.
{
}
55555
Multiplicand, 27 trees in each row.
Multiplier,
5 row's.
Product,
137 trees, Ans.
T 10. 12. There are on a board 3 rows of spots, and 4
•
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spots in each row; how many spots on the board?
A slight inspection of the figure will
show, that the number of spots may be
found either by taking 4 three times, (3
times 4 are 12,) or by taking 3 four times,
(4 times 3 are 12;) for we may say there
are 3 rows of 4 spots each, or 4 rows of 3 spots each; there-
fore, we may use either of the given numbers for a multi-
plier, as best suits our convenience. We generally write
the numbers as in subtraction, the larger uppermost, with
units under units, tens under tens, &c. Thus,
1
1
Multiplicand, 4 spots.
Multiplier, 3 rows.
Product,
12 Ans.
Hence, Multiplication is a short way of performing many
additions; in other words,-It is the method of repeating any
number any given number of times.
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​SIGN. Two short lines, crossing each other in the form
of the letter X, are the sign of multiplication. Thus, 3 × 4.
12, signifies that 3 times 4 are equal to 12, or 4 times 3
are 12.
Note. 4 and 3 are the factors,
which produce the product 12.
1
Note. Before any progress can be made in this rule, the
following table must be committed perfectly to memory.
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​¶ 10, MULTIPLICATION OF SIMFLE NUMBERS,
2
3=
4:
2X 5:
2 X 6
2 X
7
2 X 8=16
1
2 times 0 are 04 X 10
2 X 124 X 11
2 X
4 X 12
2 X
2 X
3 X 0
3.X 1
3 X
3 X
3 X
X
3 X
3 X
3 X
3 X
3 X
4 X
4 X
2 X
9
2 X 1020
3
4
5:
1
;;
2 X 11=22
2 X 12
6
7
3 X 10
3 X 11
3 X 12
M
9=
0
1
2
10
∞
8
2 4 6
12
14
18
24
MULTIPLICATION TABLE.
49/10 X
56 10 X
10 10-10-10.10.10 10 10 10 10 10 10 10
0
X10=401 7 x my
44 7X 8
487X9
7 X 10
77 X 11
7 X 12
4
хо
8
5 X X 1
5
5 X. 2
10
315
8 X 0
5 X
X3
5 X 4208X 1=
X 4
5 X X 5
6
5 X
5 X
05 X
5 X
=0
5X
126 X 1
156 X
186 X
=216 X
8246 X
27
96X 0
T
8
9
40
45
1050
50
55
5 X 11
65 × 12 = 60
}
2
3=
4
366 X 9
5
6 X 6
30 6 X 7
33 6 X 848
54
X
256 X
30 8 x
6 X 12
358 X
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S M
0
-6
12
18
24
30
36
42
6 X 1060
6 X 11 =
66
72
Xeya
8 X
8 X
X 4
2 =
3
8 x
8 X
8 ×
8 X 10
8 X 11
8 X 12
9 X
9 ×
5
6
4 X
3127X 0:
0
4 X
4 X 4:
16 7 X
1
ry
4 X 5:
207 X
2 =
14
4 X 6
24 7 X
3
21 10 X
4X my
428 10 ×
287 X
4 X 8=327 × 535 10 X
X
4× 9=367× 6
42 10 X
9 X 0
9 X 1
9 X 2
9 X 3
4
5
9X 6
8=
9:
=
M
11
Peterbata
63 10 X
70 10 X
77 10 X
84 10 X
Wakandanal
0
8
16
24 11 X
32 11 X
40 11 X
48 11 X3
56 11 X
4
5
6
7
8
9
10 X 10
=100
10 X 11 110
10 X 12
120
0
1
2=
330
1
2
Į
0=
0 =
4=
5
64 11 X
72 11 X 6
40
50*
60
= 70
Batignol
80 11 X 7
88 11 X 8
96 11 X 9
011 X 10
911 X 11
18/11 X 12
27 12 X
12 X 0
36 12 X 1=
45 12 X 2
54 12 X 3
A
63 12 X 4
72 12 X 5
81 12 X 6
S
W
The
66
= 77
88
99
110
121
132
11
9 X
9 X
8
9 X 9
9 X 10
90 12 X 77
99 12 X 8
9 X 11
9 X 12108 12 X 9-108
0 12 X 10120
10 12 X 11=132
20 12 X 12144
Copp
29
80
90
II II II
0
11
22
33
44
55
0
12
24
36
48
60
72
84
96
1
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#
9 × 2
4 X 6
8×9
3 x 7
55
how many?
how many?
how many
how many?
how many?
?
MULTIPLICATION OF SIMPLE NUMBERS.
24.
4 X 3 X 2
3 X 2 X 5
how many?
7 × 1×2 = how many?
8 × 3 × 2 how many?
3 X 2 X 4 X 5 how many?
*
2
15. How many inches are there in
being 12 inches?
OPERATION.
253
12
13. What will 84 barrels of flour cost at 7 dollars a bar-
rel?
8
14. A merchant bought 273 hats at
did they cost?
Ans. 588 dollars.
dollars each; what
Ans. 2184 dollars.
253 feet, every foot
1
1
OPERATION.
The product of 12, with each of the signifi-
cant figures or digits, having been commit-
ted to memory from the multiplication table,
it is just as easy to multiply by 12 as by a
single figure. Thus, 12 times 3 are 36, &c.
16. What will 476 barrels of fish cost at 11 dollars a bar-
rel?
Ans. 5236 dollars.
Ans. 3036
1
I
17. A piece of valuable land, containing 33 acres, was
sold for 246 dollars an acre; what did the whole come to?
1
As 12 is the largest number, the product of which, with the
nine digits, is found in the multiplication table, therefore,
when the multiplier exceeds 12, we multiply by each figure
in the multiplier separately. Thus :
J
The multipli-
er consists of 3
ten's and '3 units.
First, multiply-
ing by the 3
units gives 'us
738 dollars, the
price of 3 acres.
We then multiply by the 3 tens, writing the first figure of
the product (8) in ten's place, that is, directly under the figure
by which we multiply. It now appears, that the product by
the 3 tens consists of the same figures as the product by the
three units; but there is this difference-the figures in the
product by the 3 tens are all removed one place further to-
ward the left hand, by which their value is increased ten-
fald, which is as it should be, because the price of 30 acres
pla
246 dollars, the price of 1 acre.
33 number of acres.
1
738 dollars, the price of 3 acres.
738 dollars, the price of 30 acres.
Ans: 8118 dollars, the price of 33 acres.
3
1
1
T 10.
T
Ma
1
5

+
31
T 10.
MULTIPLICATION OF SIMPLE NUMBERS.
is evidently ten times as much as the price of 3 acres, that
is, 7380 dollars; and it is plain, that these two products,
added together, give the price of 33 acres.
These examples will be sufficient to establish the fol-
lowing
RULE.
I. Write down the multiplicand, under which write the
multiplier, placing units under units, tens under tens, &c.,
and draw a line underneath.
II. When the multiplier does not exceed 12, begin at the
right hand of the multiplicand, and multiply each figure con-
tained in it by the multiplier, setting down, and carrying, as
in addition.
"
III. When the multiplier exceeds 12, multiply by each
figure of the multiplier separately, first by the units, then by
the tens, &c., remembering always to place the first figure of
each product directly under the figure by which you multi-
ply. Having gone through in this manner with each figure
in the multiplier, add their several products together, and
the sum of them will be the product required.
1
EXAMPLES FOR PRACTICE.
18. There are 320 rods in a mile; how many rods are
there in 57 miles?
19. It is 436 miles from Boston to the city of Washing-
ton; how many rods is it?
20. What will 784 chests of tea cost, at 69 dollars a
chest?
21. If 1851 men receive 758 dollars apiece, how many
dollars will they all receive?
Ans. 1403058 dollars.
22. There are 24 hours in a day; if a ship sail 7 miles in
an hour, how many miles will she sail in 1 day, at that rate ?
how many miles in 36 days? how many miles in 1 year, or
365 days?
Ans. 61320 miles in 1 year.
23. A merchant bought 13 pieces of cloth, each piece
containing 28 yards, at 6 dollars a yard; how many yards
were there, and what was the whole cost?
Ans. The whole cost was 2184 dollars.
24. Multiply 37864 by 235.
25.
29831 952.
261 ............ 98956 8704.
Product,
8898040.
28399112.
... 817793024.
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...
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1
2
1
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4
5
CONTRACTIONS IN MULTIPLICATION.
I. When the multiplier is a composite number.
11. Any number, which may be produced by the mul-
tiplication of two or more numbers, is called a composite
number. Thus, 15, which arises from the multiplication of
5 and 3, (5 × 315,) is a composite number, and the num-
bers 5 and 3, which, multiplied together, produce it, are called
component parts, or factors of that number. So, also, 24 is a
composite number; its component parts or factors may be 2
and 12 (2 × 12=24;) or they may be 4 and 6 (4×6
24;) or they may be 2, 3, and 4 (2 × 3 × 4
·
24.)
20
3
60
CONTRACTIONS IN MULTIPLICATION,-.
1. What will 15 yards of cloth cost, at 4 dollars a yard ?
15 yards are equal to 5 × 3 yards. The cost of 5
yards would be 5 × 420 dollars; and because 15
yards contain 3 times 5 yards, so the cost of 15 yards
will evidently be 3 times the cost of 5 yards, that is,
20 dollars X 3: 60 dollars.
Ans. 60 dollars.
that get
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1
Wherefore, If the multiplier be a composite number, we may,
if we please, multiply the multiplicand first by one of the com-
ponent parts, that product by the other, and so on, if the com-
ponent parts be more than two; and, having in this way
multiplied by each of the component parts, the last product
will be the product required.
►
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1088 dollars, the price of 8 tons.
12 the other component part, or factor.
[
T11
2. What will 136 tons of potashes come to, at 96 dollars
per ton ?
8 X 1296. It follows, therefore, that 8 and 12 are ·
component parts or factors of 96. Hence,
136 dollars, the price of 1 ton.
8 one of the component parts, or factors.
{
}"
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Ans. 13056 dollars, the price of 96 tons.
3. Supposing 342 men to be employed in a certain piece
of work, for which they are to receive 112 dollars each,
how much will they all receive ?
8X7X2=112.,
Ans. 38304 dollars..
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12, 13. CONTRACTIONS IN MULTIPLICATION.
=
4. Multiply 367 by 48,
5.
853
56.
6............ 1086
$72.
{
...
II. When the multiplier is 10, 100, 1000, &c.
....
1
1
1. 12. It will be recollected, (T3.) that any figure, on be-
ing removed one place towards the left hand, has its value
increased tenfold; hence, to multiply any number by 10, it
is only necessary to write a cipher on the right hand of it.
Thus, 10 times 25 are 250; for the 5, which was units before,
is now made tens, and the 2, which was tens before, is now
made hundreds. So, also, if any figure be removed two places
towards the left hand, its value is increased 100 times, &c.
Hence,
When the multiplier is 10, 100, 1000, or 1 with any number
of ciphers annexed, annex as many ciphers to the multipli-
cand as there are ciphers in the multiplier, and the multi-
plicand, so increased, will be the product required. Thus,
Multiply 46 by 10, the product is 460.
83... 100,
95 ... 1000,
8300.
95000.
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33
Product, 17616.
47768.
78192.
****
**
EXAMPLES FOR PRACTICE.
1. What will 76 barrels of flour cost, at 10 dollars a barrel?
2. If 100 men receive 126 dollars each, how many dol-
lars will they all receive?
1
3. What will 1000 pieces of broadcloth cost, estimating
cach piece at 312 dollars?
4. Multiply 5682 by 10000.
82134 ... 100000.
5:'.
1
T13. On the principle suggested in the last T, it follows,
When there are ciphers on the right hand of the multipli-
cand, multiplier, either or both, we may, at first, neglect
these ciphers, multiplying by the significant fegures only;
after which we must annex as many ciphers to the product
as there are ciphers on the right hand of the multiplicand
and multiplier, counted together.
1
1
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1
1
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34
ને
EXAMPLES FOR PRACTICE.
4. If 1300 men receive 460 dollars apiece, how many
dollars will they, all receive?
OPERATION.
460
1300
6.
7.
3
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1
Ans. 598000 dollars.
138.
46
756
P
CONTRACTIONS IN MULTIPLICATION. T13.
OPERATION.
378
204
1512
000
product, which gives the true answer.
2. The number of distinct buildings in New England,'
appropriated to the spinning, weaving, and printing of cot-
ton goods, was estimated, in 1826, at 400, running, on an
average, 700 spindles each; what was the whole number of
spindles?
3. Multiply 357 by 6300,
4.
... 8600 ...
17.
5.
9340.... 460.
5200 410.
378,.... 204.
1
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....
1
The ciphers in the multiplicand
and multiplier, counted together,
are three. Disregarding these, we
write the significant figures of the
multiplier under the significant fig-
ures of the multiplicand, and multi-
ply; after which we annex three,
ciphers to the right hand of the
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8. Multiply 154326 by 3007.
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In the operation it will be seen, that multi-
plying by ciphers produces nothing. There-
fore,
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1
EXAMPLES FOR PRACTICE.
,
1
1
77112
III. When there are ciphers between the significant figures
of the multiplier, we may omit the ciphers, multiplying by
the significant figures only, placing the first figure of each pro-
duct directly under the figure by which we multiply.
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народ
1
T. 18.
SUPPLEMENT, TO MULTIPLICATION.
9. Multiply 543 by
10.
11:
1
OPERATION.
206.
1620 ... 2103.
........... 36243 32004.
...
t
Product, 464058282
is
1080282
462978
!
154326
3007
'
SUPPLEMENT
TO MULTIPLICATION.
1
!
1
QUESTIONS...
}
1. What is multiplication? 2. What is the number to be
multiplied called? 3. to multiply by called? 4. What
is the result or answer called? 5: Taken collectively, what
are the multiplicand and multiplier called? 6. What is the
sign of multiplication? 7. What does it show? 8. In what
order must the given number be placed for multiplication ?
9. How do you proceed when the multiplier is less than 12?
10. When it exceeds 12, what is the method of procedure?
11. What is a composite number? 12. What is to be under-
stood by the component parts, or factors, of any number?
13. How may you proceed when the multiplier is a compo-
site number? 14. To multiply by 10, 100, 1000, &c., what
suffices? 15. Why? 16. When there are ciphers on the
right hand of the multiplicand, multiplier, either or both,
how may we proceed? 17. When there are ciphers be-
tween the significant figures of the multiplier, how are they
to be treated?
Į
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1
1 !
1
EXERCISES.
1. An army of 10700 men, having plundered a city, took
so much money, that, when it was shared among them, each
man received 46 dollars; what was the sum of money
taken?
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35


{
7
1
1
36
T 13.
E
2. Supposing the number of houses in a certain town to
be 145, each house, on an average, containing two families,
and each family 6 members, what would be the number of
inhabitants in that town?
Ans. 1740.
3. If 46 men can do a piece of work in 60 days, how
many men will it take to do it in one day?
{
\
4. Two men depart from the same place, and travel in
opposite directions, one at the rate of 27 miles a day, the
other 31 miles a day; how far apart will they be at the end
of 6 days?
Ans. 348 miles.
5. What number is that, the factors of which are 4, 7, 6,
and 20?
Ans. 3360.
6. If 18 men can do a piece of work in 90 days, how long
will it take one man to do the same?
SUPPLEMENT TO MULTIPLICATION.
7. What sum of money must be divided between 27
men, so that each man may receive 115 dollars?
8. There is a certain number, the factors of which are 89
and 265; what is that number?
9. What is that number, of which 9, 12, and 14 are
factors?
10. If a carriage wheel turn round 346 times in running
1 mile, how many times will it turn round in the distance
from New York to Philadelphia, it being 95 miles.
Ans. 32870.
11. In one minute are 60 seconds; how many seconds in.
4 minutes ? in 5 minutes ? in 20 minutes?
in 40 minutes?
12. In one hour are 60 minutes; how many seconds in
an hour? in two hours? how many seconds from
nine o'clock in the morning till noon?
13. In one dollar are 6 shillings; how many shillings in
3 dollars?
in 300 dollars?
in 467 dollars?
14. Two men, A and B, start from the same place at the
same time, and travel the same way; A travels 52 miles a
day, and B 44 miles a day; how far apart will they be at
the end of 10 days?
for
15. If the interest of 100 cents, for one year, be 6 cents,
how many cents will be the interest for 2 years ?
4 years? for 10 years? for 35 years?
years?
for 84
16. If the interest of one dollar, for one year, be six cents,
what is the interest for 2 dollars the same time?
5.
dollars ?
8 dollars ? 95 dollars?
7 dollars?
Ka
}
promentumdedertj
↓
T 13, 14.
17. A farmer sold 468 pounds of pork, at 6 cents a pound,
and 48 pounds of cheese, at 7 cents a pound; how many
cents must he receive in pay?
18. A boy bought 10 oranges; he kept 7 of them, and sold
the others for 5 cents apiece; how many cents did he receive?
19. The component parts of a certain number are 4, 5, 7,
6, 9, 8, and 3; what is the number?
!
20. In 1 hogshead are 63 gallons; how many gallons in 8
hogsheads? In 1 gallon are 4 quarts; how many quarts in 8
hogsheads? In 1 quart are 2 pints; how many pints in 8 hogs-
heads?
鬼
​DIVISION OF SIMPLE NUMBERS.
37
DIVISION
OF SIMPLE NUMBERS.
¶ 14. 1. James divided 12 apples among 4 boys; how
many did he give each boy?
2. James would divide 12 apples among 3 boys; how
many must he give each boy?
(
3. John had 15 apples, and gave them to his playmates, who
received 3 apples each; how many boys did he give them to?
4. If you had 20 cents, how many cakes could you buy
at 4 cents apiece?
}
5. How many yards of cloth could you buy for 30 dollars,
at 5 dollars a yard?
6. If you pay 40 dollars for 10 yards of cloth, what is one
yard worth?
7. A man works 6 days for 42 shillings; how many shil-
ings is that for one day?
8. How many quarts in 4 pints? - in 6
in 10 pints?
in 6 pints?
9. How many times is 8 contained in 88?
10. If a man can travel 4 miles an hour, how many hours
would it take him to travel 24 miles?
}
11. In an orchard there are 28 trees standing in rows,
and there are 3 trees in a row; how many rows are there ?
Remark. When any one thing is divided into two equal
parts, one of those parts is called a half; if into 3 equal
parts, one of those parts is called a third; if into four equal
parts, one part is called a quarter or a fourth; if into five,
one part is called a fifth, and so on.
D
2
I
38
DIVISION OF SIMPLE NUMBERS.
1
12. A boy had two apples, and gave one half an apple to
each of his companions; how many were his companions ?-
13. A boy divided four apples among his companions, by
giving them one third of an apple each; among how many
did he divide his apples?
14. How many quarters in 3 oranges?
15. How many oranges would it take to give 12 boys one
quarter of an orange each?
16. How much is one half of 12 apples? ›
£
17. How much is one third of 12?
18. How much is one fourth of 12?
19. A man had 30 sheep, and sold one fifth of them;
how many of them did he sell?
20. A man purchased sheep for 7 dollars apiece, and
paid for them all 63 dollars; what was their number?
12 cents.
First orange, 3 cents.
Ful
21. How many oranges, at 3 cents each, may be bought
for 12 cents?
9
Second orange, 3 cents.
T 14, 15.
}
It is plain, that as many times as 3 cents can be taken
from 12 cents, so many oranges may be bought; the object,
therefore, is to find how many times 3 is contained in 12.
6
Third orange, 3 cents.
1
t
3
}
1
We see in this example, that
we may take 3 from 12 four
times, after which there is no re
mainder; consequently, subtrac
tion alone is sufficient for the ope
ration; but we may come to the
same result by a process, in most
cases much shorter, called Di-
vision.
· 3.
Fourth orange, 3 cents.
0
7 15. It is plain, that the cost of one orange, (3 cents,)
multiplied by the number of oranges, (4,) is equal to the
cost of all the oranges, (12 cents ;) 12 is, therefore, a pro-
duct, and 3 one of its factors; and to find how many times
3 is contained in 12, is to find the other factor, which, mul-
tiplied into 3, will produce 12. This factor we find, by
trial, to be 4, (4 X 312;) consequently, 3 is contained in
12 4 times.
Ans. 4 oranges.
/
}
22. A man would divide 12 oranges equally among 3 chil-
dren; how many oranges would each child have?
Here the object is to divide the 12 oranges into 3 equal
ļ
1

2
邝
​{
¶ 15.
39
parts, and to ascertain the number of oranges in each of those
parts. The operation is evidently as in the last example, and
consists in finding a number, which, multiplied by 3, will pro-
duce 12. This number we have already found to be 4.
Ans. 4 oranges apiece.
As, therefore, multiplication is a short way of performing
many additions of the same number; so, division is a short
way of performing many subtractions of the same number;
and may be defined, The method of finding how many times
one number is contained in another, and also of dividing a num-
ber into any number of equal parts. In all cases, the process
of division consists in finding one of the factors of a given
product, when the other factor is known.
The number given to be divided is called the dividend,
and answers to the product in multiplication. The number
given to divide by is called the divisor, and answers to one of
the factors in multiplication. The result, or answer sought,
is called the quotient, (from the Latin word quoties, how
many?) and answers to the other factor.
4
SIGN. The sign for division is a short horizontal line be-
tween two dots, ÷. It shows that the number before it is
to be divided by the number after it. Thus 27 ÷ 9
93 is
read, 27 divided by 9 is equal to 3; or, to shorten the ex-
pression, 27 by 9 is 3; or, 9 in 27 3 times. In place of the
dots, the dividend is often written over the line, and the di-
visor under it, to express division; thus, 73, read as
before.
1
N NO NO NA MI
I
H| I| -| ||
= ∞ ∞ 10
2***
= 3
12=6
14-7
18
18-9
DIVISION OF SIMPLE NUMBERS.
*
1 $ 1 &
=210=2122
33 183
16 4 20
204244
22 525
355
Le
=624
680
217
by 28
=735
248 32. 840
27936
936 = 942
at at with coko co ctwo
DIVISION TABLE.*
|| || || ||
H
312345
4
•Stre
Kaavat
I
Ħ
|| || ||
7
2 #
21
1 2 3 4 1 C
2
5 30 5 35
6 36
366 6|=6
1742 742=7
848856
94954929
858
4 244 24
*The reading used by the pupil in committing the table may be, 2 by 2 is 1,
4 by 2 is 2, &c. ; or, 2 in 2 one time, 2 in 4 two times, &c.
{
}
{
}
1
+
1
斧
​40
I
1
1
{
↓
DIVISION OF SIMPLE NUMBERS.
M
"}
{
DIVISION TABLE-CONTINUED.
:.1
1 1.1 18
1.
&
162 18248=214=2 44 =2
243273483443| 4=
32=4364 484 444 44 = 4
4054254855 | £} =
486546 486 672
6學
​ff
567 637 78=7H=7| #
648728 1888 =
$$ 8
12=919 | 289 | ft=9|123² =
my
108.
9
42 =
287, or how many?
426, or 42
549, or 54
how many?
how many?
32 8, or 32
3311, or
how many?
how many? 10812, or 108
12
1
}
{
Dividend.
Divisor, 4) 856
Quotient,
214
18=1|H=1|1=1
497, or 42
324, or 32
9911, or
f
84
84 ÷ 12, or $4
រ
M
} I
¶ 16. 23. How many yards of cloth, at 4 dollars a yard,
can be bought for 856 dollars?
✓
ì
Here the number to be divided is 856, which therefore
is the dividend; 4 is the number to divide by, and there-
fore the divisor. It is not evident how many times 4 is con-'
tained in so large a number as 856. This difficulty will be
readily overcome, if we decompose this number, thus:
856800 +40 + 16.
}
Beginning with the hundreds, we readily perceive that 4 is
contained in 8 2 times; consequently, in 800 it is contained
200 times. Proceeding to the tens, 4 is contained in 4 1
time, and consequently in 40 it is contained 10 times.
Lastly, in 16 it is contained 4 times. We now have.
200+10+4=214 for the quotient, or the number of
times 4 is contained in 856.
Ans. 214 yards.
We may arrive to the same result without decomposing
the dividend, except as it is done in the mind, taking it by
parts, in the following manner:
{
For the sake of convenience, we
write down the dividend with the divi-
sor on the left, and draw a line between
them; we also draw a line underneath.
Then, beginning on the left hand,
¶ 15, 16.
'
}
t
how many?
how many?
how many?
how many?
how many?
J
1
T
¶ 16.
41
we seek how often the divisor (4) is contained in 8,
(hundreds,) the left hand figure; finding it to be 2 times,
we write 2 directly under the 8, which, falling in the place
of hundreds, is in reality 200. Proceeding to tens, 4 is con-
tained in 5 (tens) 1 time, which we set down in ten's
place, directly under the 5 (tens.) But, after taking 4 times
ten out of the 5 tens, there is 1 ten left. This 1 ten we join
to the 6 units, making 16. Then, 4 into 16 goes 4 times,
which we set down, and the work is done.
DIVISION OF SIMPLE NUMBERS.
This manner of performing the operation is called Short
Division. The computation, it may be perceived, is carried
on partly in the mind, which it is always easy to do when
the divisor does not exceed 12.
RULE.
From the illustration of this example, we derive this general
rule for dividing, when the divisor does not exceed 12:
I. Find how many times the divisor is contained in the
first figure, or figures, of the dividend, and, setting it direct-
ly under the dividend, carry the remainder, if any, to the
next figuré as so many tens.
༤!
II. Find how many times the divisor is contained in this
dividend, and set it down as before, continuing so to do till
all the figures in the dividend are divided.
PROOF. We have seen, (¶ 15,) that the divisor and quo-
tient are factors, whose product is the dividend, and we
have also seen, that dividing the dividend by one factor is
merely a process for finding the other.
Hence division and multiplication mutually prove each other.
To prove division, we may multiply the divisor by the quo-
tient, and, if the work be right, the product will be the same
as the dividend; or we may divide the dividend by the quo-
tient, and, if the work is right, the result will be the same as
the divisor.
To prove multiplication, we may divide the product by one
factor, and, if the work be right, the quotient will be the other
factor.
EXAMPLES FOR PRACTICE.
24. A man would divide 13,462,725 dollars among 5 men ;
how many dollars would each receive?
D*
+
ĭ
سم
I
1
42
DIVISIÓN OF SIMPLE NUMBERS. ¶ 16, 17.
In this example, as we cannot
have 5 in the first figure, (1,) we
take two figures, and say, 5 in 13
will go 2 times, and there are 3
over, which, joined to 4, the next
figure, makes 34; and 5 in 34 will
go 6 times, &c.
In proof of this example, we mul-
tiply the quotient by the divisor,
and, as the product is the same as
the dividend, we conclude that the
work is right. From a bare in-
spection of the above example and
its proof, it is plain, as before stated, that division is the re-
verse of multiplication, and that the two rules mutually prove
each other.
13,462,725
OPERATION.
Dividend.
Divisor, 5) 13,462,725
Quotient,
2,692,545
PROOF.
Quotient.
2,692,545
5 divisor.
25. How many yards of cloth can be bought for 4,354,560
dollars, at 2 dollars a yard?
át 3 dollars?
4 dollars ?
at 5 dollars ?
at 9?
at 6 dollars?
at 10?
My ?
at 8?
Note. Let the pupil be required to prove the foregoing,
and all following examples.
26. Divide 1005903360 by 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,
and 12.
27. If 2 pints make a quart, how many quarts in 8 pints?
in 12 pints?
in 20 pints?
in 24 pints?
in 248 pints? in 3764 pints? in 47632 pints?
28. Four quarts make a gallon; how many gallons in 8
quarts? in 12 quarts?
in 20 quarts?
quarts?
in 36
in 368 quarts?
in 4896 quarts?
}
Dividend.
Divisor, 5) 86
T17. 5)86
L
17금
​in 5436144 quarts?
29. A man gave 86 apples to 5 boys; how many apples
would each boy receive?
11
Here, dividing the
number of the apples
(86) by the number of
boys, (5,) we find, that
I
Quotient, 17-1 Remainder.
each boy's share would be 17 apples; but there is one apple
left.
J
}
→
K
I
at
at
In order to divide all the apples equal-
ly among the boys, it is plain, we must di-
vide this one remaining apple into 5 equal
}
X
1
t
17.
DIVISION OF SIMPLE NUMBERS.
43
parts, and give one of these parts to each of the boys. Then
each boy's share would be 17 apples, and one fifth part of
another apple; which is written thus, 17 apples.
Ans. 174 apples each.
The 17, expressing whole apples, are called integers, (that
is, whole numbers.) The (one fifth) of an apple, express-
ing part of a broken or divided apple, is called a fraction,
(that is, a broken number.)
Fractions, as we here see, are written with two numbers,
one directly over the other, with a short line between them,
showing that the upper number is to be divided by the
lower. The upper number, or dividend, is, in fractions, call-
ed the numerator, and the lower number, or divisor, is called
the denominator.
*
Note. A number like 174, composed of integers (17)
and a fraction, (,) is called a mixed number.
In the preceding example, the one apple, which was left
after carrying the division as far as could be by whole num-
bers, is called the remainder, and is evidently a part of the
dividend yet undivided. In order to complete the division,
this remainder, as we before remarked, must be divided into
5 equal parts; but the divisor itself expresses the number of
parts.
If, now, we examine the fraction, we shall see, that
it consists of the remainder (1) for its numerator, and the
divisor (5) for its denominator.
}
Therefore, if there be a remainder, set it down at the right
hand of the quotient for the numerator of a fraction, under
which write the divisor for its denominator.
P
Proof of the last example.
17/1/
5
In proving this example, we
find it necessary to multiply
our fraction by 5; but this is
easily done, if we consider, that
the fraction expresses one
86
part of an apple divided into 5 equal parts; hence, 5 times
is 1, that is, one whole apple, which we reserve to be
added to the units, saying, 5 times 7 are 35, and one we re-
served makes 36, &c.
30. Eight men drew a prize of 453 dollars in a lottery;
how many dollars did each receive ?
1
1
{
44
F
Divisor, 8) 453
Quotient, ·56, §
answer 56 dollars
to each man.
Proof.
563
8
Dividend.
453
DIVISION OF SIMPLE NUMBERS.
T 18. Here we may notice, that the eighth part of 5 dol-
lars is the same as 5 times the eighth part of 1 dollar, that
is, the eighth part of 5 dollars is g of a dollar. Hence, §
expresses the quotient of 5 divided by S.
1
TT 18, 19.
Here, after carrying the division as
far as possible by whole numbers, we
have a remainder of 5 dollars, which,
written as above directed, gives for the
and § (five eighths) of another dollar,
p
/
§ is 5 parts, and 8 times 5 is 40, that is, 4º 5,
which, reserved and added to the product of 8 times
6, makes 53, &c. Hence, to multiply a fraction,
we may multiply the numerator, and divide the
product by the denominator.
>
Or, in proving division, we may multiply the whole num-
ber in the quotient only, and to the product add the remain-
der; and this, till the pupil shall be more particularly taught
in fractions, will be more easy in practice. Thus, 56 × 8 =
448, and 448, 5, the remainder, 453, as before.
(http
31. There are 7 days in a week; how many weeks in
365 days?
Ans. 524 weeks.
32. When flour is worth 6 dollars a barrel, how many
barrels may be bought for 25 dollars? how many for 50 dol-
lars? for 487 dollars? for 7631 dollars?
33. Divide 640 dollars among 4 men.
640 ÷ 4, or 549
34. 6786, or how many?
£7 8
35. £que how many?
36. 1234 how many?
37. 2464-how many?
6
38. 2764-how many?
Big Magn
>
39. 40801 how many?
40. 2014012 how many?
}
I
120
W
160 dollars, Ans.
Ans. 113.
T19. 41. Divide 4370 dollars equally among 21 men.
When, as in this example, the divisor exceeds 12, it is
evident that the computation cannot be readily carried on in
the mind, as in the foregoing examples. Wherefore, it is
more convenient to write down the computation at length,
in the following manner:
Ans. 3848.
1
T19.
{
}
{
OPERATION.
Divisor. Dividend. Quotient.
21) 4370 (208.
42
¿
1.
1
170
168
DIVISION OF SIMPLE NUMBERS.
b
'
11
է
>
4
45
We may write the divisor
and dividend as in short di-
vision, but, instead of writing
the quotient under the divi-
dend, it will be found more
convenient to set it to the
right hand.
2 Remainder.
Taking the dividend by
parts, we seek how often we
can have 21 in 43 (hundreds ;) finding it to be 2 times, we
set down 2 on the right hand of the dividend for the high-
est figure in the quotient. The 43 being hundreds, it fol-
lows, that the 2 must also be hundreds. This, however,
we'need not regard, for it is to be followed by tens and units,
obtained from the tens and units of the dividend, and will
therefore, at the end of the operation, be in the place of hun-
dreds, as it should be.
It is plain that 2 (hundred) times 21 dollars ought now
to be taken out of the dividend; therefore, we multiply the
divisor (21) by the quotient figure 2 (hundred) now found,
making 42, (hundred,) which, written under the 43 in the
dividend, we subtract, and to the remainder, 1, (hundred,)
bring down the 7, (tens,) making 17 tens.
We then seek how often the divisor is contained in 17,
(tens;) finding that it will not go, we write a cipher in the
quotient, and bring down the next figure, making the whole
170. We then seek how often 21 can be contained in 170,
and, finding it to be 8 times, we write 8 in the quotient, and,
multiplying the divisor by this number, we set the product,
168, under the 170; then, subtracting, we find the remain-
der to be 2, which, written as a fraction on the right hand
of the quotient, as already explained, gives 208
for the answer.
dollars,
This manner of performing the operation is called Long
Division. It consists in writing down the whole computation.
From the above example, we derive the following
RULE.
I. Place the divisor on the left of the dividend, separate
them by a line, and draw another line on the right of the
dividend to separate it from the quotient.
II. Take as many figures, on the left of the dividend, as
(
46
T 19.
contain the divisor once or more; seek how many times they
contain it, and place the answer on the right hand of the
dividend for the first figure in the quotient.
6
III. Multiply the divisor by this quotient figure, and write
the product under that part of the dividend taken.
IV. Subtract the product from the figures above, and to the
remainder bring down the next figure in the dividend, and
divide the number it makes up, as before. So continue to
do, till all the figures in the dividend shall have been brought
down and divided.
}
1
DIVISIO
{
}
I
Note 1. Having brought down a figure to the remainder,
if the number it makes up be less than the divisor, write
a cipher in the quotient, and bring down the next figure.
Note 2. If the product of the divisor, by any quotient
figure, be greater than the part of the dividend taken, it is an
evidence that the quotient figure is too large, and must be
diminished. If the remainder at any time be greater than
the divisor, or equal to it, the quotient figure is too small, and
must be increased.
SIMPLE NUMBERS.
¿
EXAMPLES FOR PRACTICE.
1. How many hogsheads of molasses, at 27 dollars a hogs-
head, may be bought for 6318 dollars?"
Ans. 234 hogsheads.
2. If a man's income be 1248 dollars a year, how much
is that per week, there being 52 weeks in a year?
Ans. 24 dollars per week.
3. What will be the quotient of 153598, divided by 29?
Ans. 529614.
1
{
4. How many times is 63 contained in 30131?
Ans. 4781 times; that is, 478 times, and 3 of another
time.
5. What will be the several quotients of 7652, divided by
16, 23, 34, 86, and 92?
6. If a farm, containing 256 acres, be worth 7168 dollars,
what, is that per acre?
7. What will be the quotient of 974932, divided by 365 ?
Ans. 2671.
(
8. Divide 3228242 dollars equally among 563 men; how
many dollars must each man receive? Ans. 5734 dollars.
9. If 57624 be divided into 216, 586, and 976 equal parts,
what will be the magnitude of one of each of these equal
parts
+
}
;
N
I
T 20, 21.
47
Ans. The magnitude of one of the last of these equal parts
will be 59%
10. How many times does 1030603615 contain 3215?
Ans. 320561 times.
11. The earth, in its annual revolution round the sun, is
said to travel 596088000 miles; what is that per hour, there
being 8766 hours in a year?
12. 1234567890-how many?
}
13. 1078f8a0= how many?
e
14. 987649931 how many?
CONTRACTIONS IN DIVISION.
+
p
1
CONTRACTIONS IN DIVISION.
I. When the DIVISOR is a COMPOSITE NUMBER.
T20. 1. Bought 15 yards of cloth for 60 dollars; how
much was that per yard?
15 yards are 3 × 5 yards. If there had been but 5 yards,
the cost of one yard would be 12 dollars; but, as there
are 3 times 5 yards, the cost of one yard will evidently be
but one third part of 12 dollars; that is, 4 dollars. Ans.
Hence, when the divisor is a composite number, we may,
if we please, divide the dividend by one of the component
parts, and the quotient, arising from that division, by the
other; the last quotient will be the answer.
2. If a man can travel 24 miles a day, how many days
will it take him to travel 264 miles?
It will evidently take him as many days as 264 contains 24.
OPERATION.
246 × 4.
X
6)264
4)44
11 days.
3. Divide 576 by 48
(8 X 6.)
4. Divide 1260 by 63= (7 × 9.)
5. Divide 2430 by 81.
6. Divide 448 by 56.
or,
{
*
1
1
24)264(11 days, Ans.
24
II. To divide by 10, 100, 1000, &c.
T 21. 1. A prize of 2478 dollars is owned by 10 men;
what is each man's share?
÷
24
24
F
}
í
{
>
+
3
{
L
1
48
CONTRACTIONS IN DIVISION.
Each man's share will be equal to the number of tens con-
tained in the whole sum, and, if one of the figures be cut off
at the right hand, all the figures to the left may be consid-
ered so many tens; therefore, each man's share will be
247 dollars.
**
A
(1
'
It is evident, also, that if 2 figures had been cut off from
the right, all the remaining figures would have been so ma-
ny hundreds; if 3 figures, so many thousands, &c. Hence
we derive this general RULE for dividing by 10, 100, 1000,
&c. Cut off from the right of the dividend so many figures
as there are ciphers in the divisor; the figures to the left
of the point will express the quotient, and those to the right,
the remainder.
*
2. In one dollar are 100 cents; how many dollars in 42400
Ans. 424 dollars.
cents ?
424 00
3. How many dollars in 34567 cents ?
Here the divisor is 100; we therefore cut off 2
figures on the right hand, and all the figures to the
left (424) express the dollars.
المصور
10. In one cent are 10 mills;
mills?
in 400 mills ?
mills?
in 4784 mills?
}
4. How many dollars in 4567840 cents?
1
5. How many dollars in 345600 cents?
Ans. 426.
6. How many dollars in 42604 cents?
7.. 1000 mills make one dollar; how many dollars in 4000
mills?
in 25000 mills?
in 845000?
*
4|0)48|0
8. How many dollars in 6487 mills? Ans. 6487 dollars.
9. How many dollars in 42863 mills? in 368456
mills?
in 96842378 mills?
1
T 21, 22.
12 dolls. Ans.
+
1
2
III. When there are CIPHERS on the right hand of the divisor.
T 22. 1. Divide 480 dollars among 40 men?
OPERATION.
Ans. 345 dollars.
100
mindkettenpla
$
how many cents in 40
in 20 mills?
in 468
in 34640 mills?
In this example, our divisor,
(40,) is a composite number,
(10 X 4 40;) we may, there-
fore, divide by one component
part, (10,) and that quotient by
the other, (4;), but to divide by 10, we have seen, is but to
cut off the right hand figure, leaving the figures to the left
1
,{
#
1
}
Y
W35
l
;
T 22.
40
of the point for the quotient, which we divide by 4, and the
work is done. It is evident, that, if our divisór had been
400, we should have cut off 2 figures, and have divided in
the same manner; if 4000, 3 figures, &c. Hence this gene-
ral RULE : When there are ciphers at the right hand of the di-
visor, cut them off, and also as many places in the dividend;
divide the remaining figures in the dividend by the remain-
ing figures in the divisor; then annex the figures, cut off
from the dividend, to the remainder.
气
​2. Divide 748346 by 8000.
Dividend.
Divisor, 8000) 748|346.
SUPPLEMENT TO DIVISION.
}
Quotient, 93.-4346 Remainder.
3. Divide 46720367 by 4200000.
Dividend.
$
:
47
42
42/00000) 467 20367(11,520867 Quotient.
42
}
1
I
?
ཙྪེཝམཾ ཏཏན?
Ans. 934848.
}
520367 Remainder.
t
4. How many yards of cloth can be bought for 346500
dollars, at 20 dollars per yard?
5. Divide 76428400 by 900000.
6. Divide 345006000 by 84000.
7. Divide 4680000 by 20, 200, 2000, 20000, 300, 4000,
50, 600, 70000, and 80.
SUPPLEMENT TO DIVISION.
QUESTIONS.
1. What is division? 2. In what does the process of di-
vision consist? 3. Division is the reverse of what? 4. What
is the number to be divided called, and to what does it an-
swer in multiplication? 5. What is the number to divile
by called, and to what does it auswer, &c.? 6. What is the
result or answer called, &c.? 7. What is the sign of divi
sion, and what does it show? 8. What is the other way of
expressing division? 9. What is short division, and how is
E
1
1
}
"
{
1
50
T. 22
{
at, performed? 10. How is division proved? 11, How is
multiplication proved? 12. What are integers, or whole
numbers? 13. What are fractions, or broken numbers?
14. What is a mixed number? 15. When there is any thing
left after division, what is it called, and how is it to be
written? 16. How are fractions written? 17. What is.
the upper number called? 18. the lower number?
19. How do you multiply a fraction? 20. To what do the
numerator and the denominator of a fraction answer in di-
vision? 21. What is long division? 22. Rule? 23. When
the divisor is a composite number, how may, we proceed?
24. When the divisor is 10, 100, 1000, &c., how may the
operation be contracted? 25. When there are ciphers at
the right hand of the divisor, how may we proceed?
1
}
EXERCISES.
1. An army of 1500 men, having plundered a city, took
2625000 dollars; what was each man's share?
SUPPLEMENT TO DIVISION,
1
}
?
1
1
1
2. A certain number of men were concerned in the pay-
ment of 18950 dollars, and each man paid 25, dollars; what
was the number of men?
3. If 7412 eggs be packed in 34 baskets, how many in a
basket?
1
++
❞
4. What number must I multiply by 135. that the pro-
duct may be 505710?
5. Light moves with such amazing rapidity, as to pass
from the sun to the earth in about the space of 8 minutes.
Admitting the distance, as usually computed, to be 95,000,000
miles, at what rate per minute does it travel?
6. If the product of two numbers be 704, and the multi-
plier be 11, what is the multiplicand?
Ans. 64.
7. If the product be 704, and the multiplicand 64, what
is the multiplier?
Ans. 11..
8. The divisor is 18, and, the dividend 144; what is the
quotient?
1
9. The quotient of two numbers. is 8, and the dividend
144; what is the divisor'?
1
}
{
10. A man wishes to travel 585 miles in 13 days; how
far must he travel each day?
11. If a man travels 45 miles a day, in how many days
will be travel 585 miles ?
f
*
{
1
1
T-22.
51
12. A man sold 35 cows for 560 dollars; how much was
that for each cow?
SUPPLEMENT TO DIVISION.
13. A man, selling his cows for 1
for all 560 dollars; how many did he sell?
14. If 12 inches make a foot, how many feet are there in
364812 inches?
dollars each, received
15. If 364812 inches are 30401 feet, how many inches
make one foot?
16. If you would divide 48750 dollars among 50 men,
1
how many dollars would you give to each one?
17. If you distribute 48750 dollars among a number of
men, in such a manner as to give to each one 975 dollars,
how many men receive a share?
18. A man has 17484 pounds of tea in 186 chests; how
many pounds in each chest?
}
W
19. A man would put up 17484 pounds of tea into chests
containing 94 pounds each; how many chests must be have?
20. In a certain town there are 1740 inhabitants, and 12
persons in each house; how many houses are there ?— in
each house are 2 families; how many persons in each family?
21. If 2760 men can dig a certain canal in one day, how
many days would it take 46 men to do the same? How
'many men would it take to do the work in 15 days?
in 5 days?
în 20 days?
in 40 days?
in 120 days?
ļ
22. If a carriage wheel turns round 32870 times in run-
ning from New York to Philadelphia, a distance of 95 miles,
how many times does it turn in running 1 mile? Ans. 346.
23. Sixty seconds make one minute; how many minutes
in 3600 seconds?
in 86400 seconds ?
in 604800
seconds ?
in 2419200 seconds?
24. Sixty minutes make one hour; how many hours in
1440 minutes?
in 10080 minutes?
in 40320
minutes?
in 525960 minutes ?
25. Twenty-four hours make a day; how many days in
168 hours?
in 672 hours?
in 8766 hours?
26. How many times can I subtract forty-eight from four
hundred and eighty?
27. How many times 3478 is equal to 47854 ?
28. A bushel of grain is 32 quarts; how many quarts must
I dip out of a chest of grain to make one half (1) of a
bushel? for one fourth (1) of a bushel?
for one
eighth (4) of a bushel?
Ans. to the last, 4 quarts.
STE
1
7
? 2
}
}
C
1
52
29. How many is of 20?,
247?
of 345678?.
30. How many walnuts are one
of 6 walnuts?
nuts?
of 300?
* of 45 ?
of 3456320 ?
JAKĄ
MISCELLANEOUS QUESTIONS.
of '7843 ?
31. What is of 4?
↓
${
of 48?
of 204030648.?
Ans. to the last, 102015324.
third part (3) of 3 wal-
of 12 ?
of 30?
of 478 ?
Ans. to the last, 1152106.
1 of 20 ?
of 320?
Ans. to the last, 19602.
डै
1
MISCELLANEOUS QUESTIONS,
Involving the Principles of the preceding Rules.
Į
Note. The preceding rules, viz. Numeration, Addition,
Subtraction, Multiplication, and Division, are called the Fun-
damental Rules of Arithmetic, because they are the foun-'
dation of all other rules.
T 22, 23.
of
1. A man bought a chaise for 218 dollars, and a horse for
142 dollars; what did they both cost?
2. If a horse and chaise cost 360 dollars, and the chaise
cost 218 dollars, what is the cost of the horse? If the horse
cost 142 dollars, what is the cost of the chaise?
3. If the sum of 2 numbers be 487, and the greater num-
ber be 348, what is the less number? If the less number
be 139, what is the greater number? ·
4. If the minuend be 7842, and the subtrahend 3481,
what is the remainder? If the remainder be 4361, and the
minuend be 7842, what is the subtrahend?
C
I
1 23. When the minuend and the subtrahend are given,
how do you find the remainder?
When the minuend and remainder are given, how do you
find the subtrahend?
When the subtrahend and the remainder are given, how
do you find the minuend?
When you have the sum of two numbers, and one of them
given, how do you find the other?
}
When you have the greater of two numbers, and their
difference given, how do you find the less number?
When you have the less of two numbers, and their differ
once given, how do you find the greater number?
1
f
T 23, 24.
MISCÉLLANEOUS QUESTIONS,
1
5. The sum of two numbers is 48, and one of the numbers
is 19; what is the other?
6. The greater of two numbers is 29, and their difference
10; what is the less number?
7. The less of two numbers is 19, and their difference is
10; what is the greater?
8. A man bought 5 pieces of cloth, at 44 dollars a piece;
974 pairs of shoes, at 3 dollars a pair; 600 pieces of calico,
at 6 dollars a piece; what is the amount?
9. A man sold six cows, worth fifteen dollars each, and a
yoke of oxen, for 67 dollars; in pay, he received a chaise,
worth 124 dollars, and the rest in money; how much money
did he receive?
10. What will be the cost of 15 pounds of butter, at 13
cents per pound?
11. How many bushels of wheat can you buy for 487
dollars, at 2 dollars per bushel?
T 24. When the price of one pound, one bushel, &c. of
any commodity is given, how do you find the cost of any
number of pounds, or bushels, &c. of that commodity? If
the price of the 1 pound, &c. be in cents, in what will the
whole cost be? If in dollars, what? if in shillings?
P
53
1
if in pence? &c.
When the cost of any given number of pounds, or bushels,
&c. is given, how do you find the price of one pound of
bushel, &c. In what kind of money will the answer be?
When the cost of a number of pounds, &c. is given, and
also the price of one pound, &c., how do you find the num-
ber of pounds, &c.
12. When rye is 84 cents per bushel, what will be the cost
of 948 bushels? how many dollars will it be?
13. If 648 pounds of tea cost 284 dollars, (that is, 28400
cents,) what is the price of one pound?
14. What is the product of 754 and 257
E*
When the factors are given, how do you find the product?
When the product and one factor are given, how do you
find the other factor?
When the divisor and quotient are given, how do you
find the dividend?
When the dividend and quotient are given, how do you
find the divisor?
1
1
1
}
† 24, 25.
PS!
15. What number, multiplied by 25, will produce 18850?
16. What number, multiplied by 754, will produce 18850?
17. If a man save six cents a day, how many cents would
he save in a year, (365 days,) ? how many in 45
years? how many dollars would it be? how many cows
could he buy with the money, at 12 dollars each?
Ans. to the last, 82 cows, and 1 dollar 50 cents remainder.
54
1,
MISCELLANEOUS QUESTIONS.
}
}
18. A boy bought a number of apples; he gave away ten
of them to his companions, and afterwards bought thirty-four
more, and divided one half of what he then had among four
companions, who received 8 apples each; how many apples
did the boy first buy?
Let the pupil take the last number of apples, 8, and re-
verse the process.
Ans. 40 apples.
T
$
19. There is a certain number, to which, if 4 be added,
and 7 be, subtracted, and the difference be multiplied by 8,
and the product divided by 3, the quotient will be 64; what
is that number?
Ans. 27.
20. A chess board has 8 rows of 8 squares each; how
many squares on the board?
D
T25. 21. There is a spot of ground 5 rods long, and 3
rods wide; how many square rods does it contain ?
C
Note. A square rod is a
square (like one of those in
the annexed figure) meas-
uring a rod on each side.
By an inspection of the
figure, it will be seen, that
there are as many squares
in a row as rods on one side,
and that the number of rows
A
is equal to the number of rods on the other side; therefore,
5X315, the number of squares.
{
Ans. 15 square rods.
A figure like A, B, C, D, having its opposite sides equal
and parallel, is called a parallelogram or oblong.
M
{
:
5
· 22. There is an oblong field, 40 rods long, and 24 rods
wide; how many square rods does it contain?
23. How many square inches in a board 12 inches long,
Ans. 144.
and 12 inches broad?
Į

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1
¶ 25.
MISCELLANEOUS QUESTIONS.
55
24. How many square feet in a board 14 feet long and 2
feet wide?
25. A certain township is six miles square; how many
square miles does it contain?
Ans. 36.
26. A man bought a farm for 22464 dollars; he sold one half
of it for 12480 dollars, at the rate of 20 dollars per acre; how
many acres did he buy? and what did it cost him per acre?
27., A boy bought a sled for 86 cents, and sold it again for
8 quarts of walnuts; he sold one half of the nuts at 12 cents
a quart, and gave the rest for a penknife, which he sold for
34 cenis; how many cents did he lose by his bargains?
28. In a certain school-house, there are 5 rows of desks;
in each row are six seats, and each seat will accommodate
2 pupils; there are also 2 rows, of 3 seats each, of the
same size as the others, and one long seat where 8 pupils
may sit; how many scholars will this house accommo-
date?
Ans. 80.
29. How many square feet of boards will it take for the
floor of a room 16 feet long, and 15 feet wide, if we allow
12 square feet for waste?
J
30. There is a room 6 yards long and 5 yards wide; how
many yards of carpeting, a yard wide, will be sufficient to cover
the floors, if the hearth and fireplace occupy 3 square yards?
31. A board, 14 feet long, contains 28 square feet; what
is its breadth?
111416 //
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32. How many pounds of pork, worth 6 cents a pound,
can be bought for 144 cents?
33. How many pounds of butter, at 15 cents per pound,
must be paid for 25 pounds of tea, at 42 cents per pound?
34. 4+ 5+ 6+1+8=how many?
35.4310—2—4—6—7—how
many?
of 6?
36. A man divides 30 bushels of potatoes among 3 poor
men; how many bushels does each man receive? What is
of thirty? How many are (two thirds) of 30?
37. How many are one third (1) of 3?
of 9 ?
of 282 ?
of 45674312 ?
38. How many are two thirds (?) of 3 ?
of 9?
of 282 ?
of 45674312?
39. How many are 1 of 40?
60? — § of 60 ?.
of 6?
1 of 80?
4
246876 ?
₫ of 246876 ?
40. How many is of 80?
'
# of 80 ?
of 100?
41. An inch is one twelfth part (T) of a foot; how many
† of 40?
- of 124?
1
of
of
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56
feet in 12 inches?
!
}
in 24 inches?
in 12243648 inches?
42. If 4 pounds of tea cost 128 cents, what does 1 pound
cost?
2 pounds? 3 pounds? 5 pounds?..
100 pounds?
}
43. When oranges are worth 4 cents apiece, how many
can be bought for four pistareens, (or 20 cent pieces?)
44. The earth, in moving round the sun, travels at the
rate of 68000 miles an hour; how many miles does it travel
in one day, (24 hours?) how many miles in one year, (365
days?) and how many days would it take a man to travel
this last distance, at the rate of 40 miles a day? how many
years?
Ans. to the last, 40800 years.
earn in 20 weeks, at 80 cents
45. How much can a man
per day, Sundays excepted ?
46. A man married at the age of 23; he lived with his
wife 14 years; she then died, leaving him a daughter, 12
years of age; 8 years after, the daughter was married to a
man 5 years older than herself, who was 40 years of age
when the father died; how old was the father at his death?
Ans. 60 years.
47. There is a field 20 rods long, and 8 rods wide; how
Ans. 160 rods.
many square rods does it contain ?
48. What is the width of a field, which is 20 rods long,
and contains 160 square rods?
49. What is the length of a field, 8 rods wide, and con--
taining 160 square rods?
}
50. What is the width of a piece of land, 25 rods long,
and containing 400 square rods ?
COMPOUND NUMBERS.
4
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I
1
COMPOUND NUMBERS.
1 26. A number expressing things of the same kind is
called a simple number; thus, 100 men, 56 years, 75 cents,
are each of them simple numbers; but when a number ex-
presses things of different kinds, it is called a compound num-
ber; thus, 43 dollars 25 cents and 3 mills, is a compound
number; so 4 years 6 months and 3 days, 46 pounds 7
shillings and 6 pence, are compound numbers.
Note. Different kinds, or names, are usually called dif
ferent denominations.
¶ 25, 26.
in 36 inches?
}
I
· T 26.
!
FEDERAL MONEY.
Weagles.
dollars.
➡dimes.
Orcents.
Nmills.
FEDERAL MONEY.
Federal money is the coin of the United States. The
kinds, or denominations, are eagles, dollars, dimes, cents,
and mills.
$
10 mills
10 cents, (100 mills,)
10 dimes, 100 cents
10 dollars, (= 100 dimes
SIGN. This character, $, placed before a number, shows
it to express federal money.
1000 mills,)
1000 cents
10000 mills)
57
are equal to
As 10 mills make a cent, 10 cents a dime, 10 dimes a
dollar, &c. it is plain, that the relative value of mills, cents,
dimes, dollars and eagles corresponds to the orders of units,
tens, hundreds, &c. in simple numbers. Hence, they may
be read either in the lowest denomination, or partly in a
higher, and partly in the lowest denomination. Thus:
[
1 cent.
= 1 dime.
1 dollar.
1 eagle."
34652 may be read, 34652 mills; or 3465 cents and 2 mills;
or, reckoning, the eagles tens of dollars, and the dimes tens
of cents, which is the usual practice, the whole may be
read, 34 dollars 65 cents and 2 mills.
For ease in calculating, a point () called a separatrix,†
is placed between the dollars and cents, showing that all the
figures at the left hand express dollars, while the two first
figures at the right hand express cents, and the third, mills.
Thus, the above example is written $34'652; that is, 34
dollars 65 cents 2 mills, as above. As 100 cents make a
dollar, the cents may be any number from
number from 1 to 99, often re-
quiring two figures to express them; for this reason, two
places are appropriated to cents, at the right hand of the
point, and if the number of cents be less than ten, requiring
but one figure to express them, the ten's place must be filled
with a cipher. Thus, 2 dollars and 6 cents are written 2'06.
10 mills make a cent, and consequently the mills never ex-
ceed 9, and are always expressed by a single figure. Only
Hart
* The cagle is a gold coin, the dollar and dime arc silver coins, the cent is a
copper coin." The mill is only imaginary, there being no coin of that denomina-
tion. There are half eagles, half dollars, half dimes, and half cents, real coins:
†The character used for the separatrix, in the “ Scholars' Arithmetic," was
the comma; the comma inverted is here adopted, to distinguish it from the com-
na used in punctuation.
$
}
2
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{
↑
58
one place, therefore, is appropriated to mills, that is, the
place immediately following cents, or the third place from
the point. When there are no cents to be written, it is evi-
dent that we must write two 'ciphers to fill up the places of
1
Six
1
→
REDUCTION OF FEDERAL MONEY.
}
{
cents. Thus, 2 dollars and 7 mills are written 2'007.
cents are written '06, and seven mills are written '007.
75 cents,
24 dollars,
}
1
Note. Sometimes 5 mills a cent is expressed frac-
tionally thus, 125 (twelve cents and five mills) is ex-
'pressed 122, (twelve and a half cents.)
1
Y
17 dollars and 8 mills are written, 17'008
4 dollars and 5 cents,
4'05
675
S
F
1
F!
gd
9 cents,
4 mills,
6 dollars 1 cent and 3 mills,
en
1
1
*
T 27. How many mills in one cent ?
in 3 cents?
in 4 cents?
in' 10 cents ?
cents and 5 mills ?
How many cents in 2 dollars?
8 dollars?
I
lars and 20 cents?
How many dollars in
in 380 cents ?
cents in 1000 mills?
in 3000 mills?
Sta
REDUCTION OF FEDERAL MONEY.
24'
Write down 470 dollars 2 cents; 342 dollars 40 cents
and 2 mills; 100 dollars, 1 cent and 4 mills; 1 mill; 2
mills; 3 mills; 4 mills; cent, or 5 mills; 1 cent and 1 mill;
2 cents and 3 mills; six cents and one mill; sixty cents and.
one mill; four dollars and one cent; three cents; five cents;.
nine cents.
cents?
cents? - in 100 cents, (= 1 dollar) ?
in 3, dollars? in 4 dollars?
in 563 cents? -
109
'004
6'013
in 6 cents?
in 30 cents?
¶ 26, 27.
in 1 cent and 2 mills ?
I
in 2 cents?
- in 9
in 78
in 2 dollars?
in 484 cents ?
in 4
1
+
!
}
in 4 dollars?
in
in 3 dollars and 15 cents? - in 5 dol-
in 4 dollars and 6 cents?
400 cents ?
in 600 cents?
in 40765 cents? How many
How many dollars in 1000 mills?
in 8000 mills?
in 4378
in 846732 mills?
mills?
3
This changing one kind of money, &c. into another kind, with-
out altering the value, is called REDUCTION.
!
£
T 27, 28.
ADDITION OF FEDERAL MONEY,
As there are 10 mills in one cent, it is plain that cents are
changed or reduced to mills by multiplying them by 10, tha
is, by merely annexing a cipher, (T 12.) 100 cents make a
dollar; therefore dollars are changed to cents; by annexing 2
ciphers, and to mills by annexing 3 ciphers. Thus, 16 dollars
= 1600 cents 16000 mills. Again, to change mills back
to dollars, we have only to cut off the three right hand
figures, (21;) and to change cents to dollars, cut off the
two right hand figures, when all the figures to the left will be.
dollars, and the figures to the right, cents and mills.
¿
f
Ans. 3400 cents.
Reduce 34 dollars to cents..
Reduce 240 dollars and 14 cents to cents.
}
,
1
}
3
Reduce 748'143 to mills.
Reduce 748143 mills to dollars.
Reduce 3467489 mills to dollars.
Reduce 48742 cents to dollars.
Reduce 1234678 mills to dollars.
Reduce 3469876 cents to dollars.
Reduce $4867'467 to mills.
Reduce 984 mills to dollars.
Reduce 7 mills to dollars.
Reduce 014 to mills.
Reduce 17846 cents to dollars.
Reduce 984321 cents to mills.
Reduce 96173 cents to dollars.
Ans. $96'17.
Reduce 2064 cents, 503 cents, 106 cents, 9214 cents,
500 cents, 7262 cents, to dollars.
Reduce 86753 mills, 96000 mills, 6042 mills, to dollars.
#
Ans. 24014 cents.
Ans. 748143 mills.
Ans. $748'143,
Ans. 3467'489.
Ans. $487′42,
59%
ADDITION AND SUBTRACTION OF FEDERAL
MONEY.
Ans. $'984
Ans. '007
1 28. From what has been said, it is plain, that we may
readily reduce any sums in federal money to the same de-
nomination, as to cents, or mills, and add or subtract them
as simple numbers. Or, what is the same thing, we may
set down the sums, taking care to write dollars under dollars,
cents under cents, and mills under mills, in such order, that the
separating points of the several numbers shall fall directly under
each other, and add them up as simple numbers, placing the
separatria in the amount directly under the other points.
1
1
1
•
1
60
1
¶ 28%
SUBTRACTION OF FEDERAL MONEY.
What is the amount of $487'643, $132'007, $4'04,
and $264 102 ?
Ans. $887'792.
OPERATION.
487643 mills.
132007 mills.
4040 mills.
264102 mills.
!
Amount, 887792 mills, = $887792.
量
​or,
EXAMPLES FOR PRACTICE.
FOR PRACTICE.
1. Bought 1 barrel of flour for 6 dollars 75 cents, 10
pounds of coffee for 2 dollars 30 cents, 7 pounds of sugar
for 92 cents, 1 pound of raisins for 12 cents, and 2 oranges
for 6 cents; what was the whole amount? Ans. $10'155.
2. A man is indebted to A, $237′62; to B, $350; to C,
$8612; to D, $9'62; and to E, $0'834; what is the
amount of his debts?
Ans. $684'204.
1
3. A man has three notes specifying the following sums,
viz. three hundred dollars, fifty dollars sixty cents, and
nine dollars eight cents; what is the amount of the three
notes ?
Ans. $359'68.
<
1
4. What is the amount of $56'18, $737, $280,
$0'287, $17, and $90'413?
5. Bought a pair of oxen for $76'50, a horse for $85,
and a cow for $1725; what was the whole amount?
1
4750 mills.
2125 mills.*~
2625 mills
A
6. Bought a gallon of molasses for 28 cents, a quarter of
tea for 37 cents, a pound of salt petre for 24 cents, 2 yards
of broadcloth for 11 dollars, 7 yards of flannel for 1 dollar
62 cents, a skein of silk for 6 cents, and a stick of twist for
4 cents; how much for the whole?
OPERATION.
$487'643
132'007
SO T
4'04
$ 264'102
$887 792 Amount.
SUBTRACTION OF FEDERAL MONEY.
7. A man gave 4 dollars 75 cents for a pair of boots, and
2 dollars 12 cents for a pair of shoes; how much did the
boots cost him more than the shoes?
OPERATION.
{
}
or,
$2'625 Ans.
OPERATION.
$4.75
$24125
$2'625 Ans.
T
1
61、.
**T 29.
8. A man bought a cow for eighteen dollars, and sold her
again for twenty-one dollars thirty-seven and a half cents;
how much did he gain?
Ans. $3375.
9. A man bought a horse for 82 dollars, and sold him
again for seventy-nine dollars seventy-five cents; did he gain
or lose? and how much?
Ans. He lost $225.
10. A merchant bought a piece of cloth for $176, which
proving to have been damaged, he is willing to lose on it
$16'50; what must he have for it?
Ans. $159'50.
11. A man sold a farm for $5400, which was $725′37
more than he gave for it; what did he give for the farm?
12. A man, having $500, lost 83 cents; how much had
he left?
wow t
* N
1
15. How much must you add to $16'82 to make $25?
16. How much must you subtract from $250, to leave
$87'14 ?
MULTIPLICATION OF FEDERAL MONEY.
13. A man's income is $1200 a year, and he spends
800'35; how much does he lay up?
14. Subtract half a cent from seven dollars.
17. A man bought a barrel of flour for $6'25, 7 pounds
of coffee for $1'41; he paid a ten dollar bill; how much
must he receive back in change?
Sammy
1
pe = Judge thes
I
}
OPERATION.
4'625
3
Your a
MULTIPLICATION OF FEDERAL MONEY.
129. 1. What will 3 yards of cloth cost, at $4'62) a
yard?
$4'625 are 4625 mills, which
multiplied by 3, the product is
13875 mills. 13875 mills may
now be reduced to dollars by
$13'875, the answer.
placing a point between the third
and fourth figures, that is, between the hundreds and thou-
sands, which is pointing off as many places for cents and
mills, in the product, as there are places of cents and mills
in the sum given to be multiplied. This is evident; for, as
1000 mills make 1 dollar, consequently the thousands in
13875 mills must be so many dollars.
M
2. At 16 cents a pound, what will 123 pounds of butter
cost?
F
Pas
1
F
}
:
}
7
f
62
OPERATION.
J
123, the number of pounds.
16 cents, the price per pound.
7738
123
MULTIPLICATION OF FEDERAL MONEY.
1
1
}
¿
{
T 29.
As the product of
any two numbers
will be the same,
whichever of them
be made the multi-
plier, therefore the
quantity, being the
larger number, is
}
19'68, the answer.
made the multiplicand, and the price the multiplier.
123 times 16 cents is 1968 cents, which, reduced to dollars,
is $19'68.
RULE.
From the foregoing examples it appears, that the multi-
plication of federal money does not differ from the multipli-
cation of simple numbers. The product will be the answer in
the lowest denomination contained in the given sum, which may
then be reduced to dollars.
}
1
EXAMPLES FOR PRACTICE.
3. What will 250 bushels of rye come to, at $0'881 per
bushel?
Ans. $221′25.
4. What is the value of 87 barrels of flour, at $6'37 a
barrel ?
5. What will be the cost of a hogshead of molasses, con-
taining 63 gallons, at 284 cents a gallon? Ans. $17'955.
6. If a man spend 124 cents a day, what will that amount
to in a year of 365 days? what will it amount to in 5
years?
Ans. It will amount to $228'124 in 5 years.
7. If it cost $3675 to clothe a soldier 1 year, how much
will it cost to clothe an army of 17800 men?
Ans. $654150.
8. Multiply $367 by 46..
9. Multiply $0'273 by 8600.
10. What will be the cost of 4848 yards of cafico, at 25
cents, or one quarter of a dollar, per yard? Ans. $1212.
Note. As 25 cents is just of a dollar, the operation in
the above example may be contracted, or made shorter; for,
at one dollar per yard, the cost would be as many dollars as
there are yards, that is, $4848; and at one quarter (1) of a
dollar per yard, it is plain, the cost would be one quarter (1)
as many dollars as there are yards, that is, 44 $1212.
848
ał Janettone
1
1
i
T 29.
MULTIPLICATION OF FEDERAL MONEY.
63
When one quantity is contained in another exactly 2, 3, 4,
5, &c. times, it is called an aliquot or even part of that quanti-
ty; thus, 25 cents is an aliquot part of a dollar, because 4 times
25 cents is just equal to i dollar; and 6 pence is an aliquot
part of a shilling, because 2 times six pence just make 1
shilling. The following table exhibits some of the aliquot
parts of a dollar :
TABLE.
From the illustration of the last
example, it appears, that, when the
price per yard, pound, &c. is one of
these aliquot parts of a dollar, the
cost may be found, by dividing the
given number of yards, pounds, &c.
by that number which it takes of
the price to make 1 dollar. If the
price be 50 cents, we divide by 2;
if 25 cts. by 4; if 12 cts. by 8, &c.
This manner of calculating the cost of articles, by taking
aliquot parts, is usually called Practice.
11. What is the value of 14756 yards of cotton cloth, at
12 cents, or of a dollar, per yard?
By practice.
8)14756
Ans. $1844'50
Cts.
50
of a dollar.
331 of a dollar.
I
of a dollar.
25
20
I of a dollar.
121 of a dollar.
Z
61 of a dollar.
fe
5
of a dollar.
}
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*
By multiplication.
14756
'125
$1844'500 Ans. as before.
12. What is the cost of 18745 pounds of tea, at $50,=
dollar, per pound?
Ans. $9372′50
13. What is the value of 9366 bushels of potatoes, at 33
cents, or of a dollar, per bushel ? 2366 $3122 Ans.
14. What is the value of 48240 pounds of cheese, at
$ '061, — of a dollar, per pound?
15. What cost 4870 oranges, at 5 cents,
apiece?
Ans. $3015.
of a dollar,
Ans. $243'50
of apples, aí 20
Ans. $30204.
16. What is the value of 151020 bushels
of a dollar, per bushel ?
cents,
F
17. What will 264 pounds of butter cost, at 12 cents
per pound?
Ans. $33.
18. What cost 3740 yards of cloth, at $125 per yard?
73780
29512
14756
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64
OPERATION.
865
5
MULTIPLICATION OF FEDERAL MONEY.
4) $ 3740 = cost at
935
cost at $1 per yard.
cost at $ 25 per yard.
OPERATION.
4250
$ 14
Ans. $4675= cost at $125 per yard.
19. What is the cost of 8460 hats, at $1'12 apiece?
;
at $3.20 apiece ?
at
át $150 apiece ?
$4'06 apiece?
Ans. $951750.
9517'50.
1
17000
4250
$59'500
1
$12690.. $27072.
$3436875.
T30. To find the value of articles sold by the 100, or 1000.
1. What is the value of 865
hundred ?
feet of timber, at $5 per
1
}
T
4325= value at $5 per foot.
100 times the true value of the timber; and therefore, if we
divide this number ($ 4325) by 100, we shall obtain the
true value; but to divide by 100 is but to cut off the two
right hand figures, or, in federal money, to remove the separa-
trix two figures to the left.
Ans. $43'25.
/
It is evident, that, were the price so much per thousand,
the same remarks would apply, with the exception of cutting
off three figures instead of two. Hence we derive the
general RULE for finding the value of articles sold by the 100,
or 1000-Multiply the number by the price, and, if it be
reckoned by the 100, cut off the two right hand figures, and
the product will be the answer, in the same kind or denomi-
nation as the price. If the article be reckoned by the 1000,
cut off the three right hand figures.
"
EXAMPLES FOR PRACTICE.
2. What is the value of 4250 feet of boards, at $14 per
1000?
Ans. 59 dollars and 50 cents.
Į
≈
ཟ
¶ 30%
1
Were the price $5
per foot, it is plain, the
value would be 865 X
$5$ 4325; but the
price is $5 for 100 feet;
consequently, $ 4325 is
7
1.
In this example, because the price is at
so much per 1000 feet, we divide by 1000,
or cut off three figures.
The data
1
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DIVISION OF FEDERAL MONEY.
65
3. What will 3460 feet of timber come to, at $4 per
hundred?
4. What will 24650 bricks come to, at 5 dollars per 1000?
5. What will 4750 feet of boards come to, at $12'25 per
1000?
Ans. 58'187.
¶ 30, 31.
6. What will 38600 bricks cost, at $475 per 1000?
7. What will 46590 feet of boards cost, at $ 10'625 per
1000?
8. What will 75 feet of timber cost, at $4 per 100?
9. What is the value of 4000 bricks, at 3 dollars per 1000?
1
DIVISION OF FEDERAL MONEY.
T 31. 1. If 3 yards of cloth cost $5'25, what is that a yard?
OPERATION.
$5'25 is 525 certs,
3)5625
which divided by 3, the
quotient is 175 cents,
which, reduced to dollars,
is $175, the answer.
Answer, 175 cents, = $175.
2. Bought 4 bushels of corn for $3; what was that a
bushel ?
1
OPERATION.
4)3'00
4 is not contained in 3; we may, however, reduce the
$3 to cents, by annexing two ciphers, thus:
*
?
Y
18)42675 (2375 mills,
36.
67
54
Ans. 75 cents.
3. Bought 18 gallons of brandy, for $4275; what did it
cost a gallon?
OPERATION.
135
126
90.
90
1
*
$2'375, the answer.
$ 42'75 is 4275 cents. After bringing
down the last figure in the dividend, and
dividing, there is a remainder of 9 cents,
which, by annexing a cipher, is reduced
to mills, (90,) in which the divisor is con-
tained 5 times, which is 5 mills, and there
is no remainder. Or, we might have re-
duced $42'75 to mills, before dividing, by
annexing a cipher, 42750 mills, which,
divided by 18, would have given the same result, 2375 mills,
which, reduced to dollars, is $2'375, the answer.
F*
¿
300 cents divided by 4, the quotient
is 75 cents, the price of each pair of
shoes.
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66
4, Divide $59′387 by 8.
PUT
OPERATION.
8)59'387
Quotient, 7'423g, that is, 7 dollars, 42 cents, 3 mills, and §
of another mill. The. & is the remainder, after the last di-
vision, written over the divisor, and expresses such fractional
part of another mill. For all purposes of business, it will be
sufficiently exact to carry the quotient only to mills, as the
parts of a mill are of so little value as to be disregarded..
Sometimes the sign of addition (+) is annexed, to show that
there is a remainder, thus, $7423+.
DIVISION OF FEDERAL MONET.
"S
RULE.
From the foregoing examples, it appears, that division of
federal money does not differ from division of simple num-
bers. The quotient will be the answer in the lowest denomina-
tion in the given sum, which may then be reduced to dollars.
J
24
Note. If the sum to be divided contain only dollars, or
dollars and cents, it may be reduced to mills, by annexing
ciphers before dividing; or, we may first divide, annexing
ciphers to the remainder, if there shall be any, till it shall
be reduced to mills, and the result will be the same.
I
I
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1
EXAMPLES FOR PRACTICE.
5. If I pay $46875 for 750 pounds of wool, what is the
value of 1 pound?
Ans. $0'625; or thus, $0'62.
6. If a piece of cloth, measuring 125 yards, cost $181'25,
what is that a yard?
Ans. $1'45.
7. If 536 quintals of fish cost $1913'52, how much is that
a quintal ?
Ans. $357.
8. Bought a farm, containing 84 acres, for $3213; what
did it cost me per acre?
Ans. $38'25.
9. At $954 for 3816 yards of flannel, what is that a yard?
Ans. $0.25.
10. Bought 72 pounds of raisins for $8; what was that
a pound?
how much?
माँ 31.
Portreat the same
е утратили
Ans. $0'111; or, $0'111+.
11. Divide $12 into 200 equal parts; how much is one
of the parts?
how much?
Ans. '0'006.
12. Divide $30 by 750.20
13. Divide $60 by 1200. 28
14. Divide $215 into 86, equal
´one of the parts be? how much?
how much?
parts; how much will
242
how much?
;
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T 31.
67
}
15. Divide $176 equally among 250 men; how much
will each man receive? 178 how much?
}
SUPPLEMENT TO FEDERAL MONEY.
1
!
SUPPLEMENT TO FEDERAL MONEY.
QUESTIONS.
2.
W
1. What is understood by simple numbers?
by compound numbers? 3.
by different denomina-
tions? 4. What is federal money? 5. What are the de-
nominations used in federal money? 6. How are dollars
distinguished from cents? 7. Why are two places assigned
for cents, while only one place is assigned for mills? 8.
To what does the relative value of mills, cents, and dollars
correspond? 9. How are mills reduced to dollars? 10.
17.
multiplication?
18.
-to cents? 11. Why? 12. How are dollars reduced
to cents? · 13.
to mill's? 14. Why? 15. How is
the addition of federal money performed? 16.
subtraction?
divi-
sion? 19. Of what name is the product in multiplication,
and the quotient in division? 20. In case dollars only are
given to be divided, what is to be done? 21. When is one
number or quantity said to be an aliquot part of another?
22. What are some of the aliquot parts of a dollar? 23.
When the price is an aliquot part of a dollar, how may the
cost be found? 24. What is this manner of operating
called? 25. How do you find the cost of articles, sold by
the 100 or 1000 ?
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EXERCISES.
}
1. Bought 23 firkins of butter, each containing 42 pounds,
for 16 cents a pound; what would that be a firkin, and
how much for the whole? Ans. $159'39 for the whole.
{
2. A man killed a beef, which he sold as follows, viz. the
hind quarters, weighing 129 pounds each, for 5 cents a
pound; the fore quarters, one weighing 123 pounds, and the
other 125 pounds, for 4 cents a pound; the hide and tal-
low, weighing 163 pounds, for 7 cents a pound; to what
did the whole amount?
Ans. $35'47.
3. A farmer bought 25 pounds of clover seed at 11 cents
a pound, 3 pecks of herds grass seed for $2'25, a barrel of
flour for $650, 13 pounds of sugar at 124 cents a pound;
for which he paid 3 cheeses, each weighing 27 pounds, at
84 cents a pound, and 5 barrels of cider at $1'25 a barrel.
The balance between the articles bought and sold is 1 cent
is it for, or against the farmer?
L
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168
}
↑
1
7
T 32..
4. A man dies, leaving an estate of $71600; there are
demands against the estate, amounting to $39876 74; the
residue is to be divided between 7 sons; what will each
one receive?
*
}
SUPPLEMENT TO FEDERAL MONEY.
5. How much coffee, at 25 cents a pound, may be had for
100 bushels of rye, at 87 cents a bushel? Ans. 348 pounds.
6. At 12 cents a pound, what must be paid for 3 boxes
of sugar, each containing 126 pounds?
7. If 650 men receive
8675 each, what will they all
receive ?
8. A merchant sold 275 pounds of iron, at 64 cents à
pound, and took his pay in oats, at $0'50 a bushel; how
many bushels did he receive?··
9. How many yards of cloth, at $4'66 a yard, must be
given for 18 barrels of flour, at $9'32,a barrel?
MED SCEN
10. What is the price of three pieces of cloth, the first
containing 16 yards, at $375 a yard; the second, 21 yards,
at $450 a yard; and the third, 35 yards, at $ 5'12} a yard ?.
T 32. It is usual, when goods are sold, for the seller to
deliver to the buyer, with the goods, a bill of the articles
and their prices, with the amount cast up. Such bills are
sometimes called bills of parcels.
Mr. Abel Atlas
}
Bought of Benj. Burdett
124 yards figured Satin, at $250 a yard,
sprigged Tabby,... 1'25
8
Received payment,
5 bags Coffee, 75 lb. each,
I chest hyson Tea, 86 lb..
Received payment,
Boston, January 6, 1827.
Mr. James Paywell
3 hogsheads new Rum, 118 gal: each,
2 pipes French Brandy, 126 and 132 gal.
1 hogshead brown Sugar, 92 cwt.
3 casks of Rice, 269 lb. each,
AANBAA
BENJ. BURDETT.
*
•
Bought of Simeon Thrifty
at $0-31 a gal.
**
:
**
**
For Simeon Thrifty,
******
∙14121..
10'34 cwt.
*
'05 .. lb.
<23
192
.. ....
Salem, June 4, 1827..
*******
$31'20
10'00
PETER FAITHFUL.
41'25.
7
+
$706*521
}
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(
1
$
¶ 32, 33.
}
5682 feet Boards,
2000 .....
Mr. Peter Carpenter
(See T 30.)
800 Thick Stuff,
1500..... Lathing,
650 ..... Plank,
Timber,
879
236 .....
.....
.....
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………..
»
REDUCTION.
1
at $6
..
**
Bought of Asa Falltree
per M.
}
Wilderness, February 8, 1827.
8'34
12'64 ...
4'
·106
...
.....
....
2'50 ..... C.
2475
…………… **
f
Received payment,
$101'849
ASA FALLTREE.
Note. M. stands for the Latin mille, which signifies 1000,
and C. for the Latin word centum, which signifies 100.
REDUCTION.
¶ 33. We have seen, that, in the United States, money
is reckoned in dollars, cents, and mills. In England, it is
reckoned in pounds, shillings, pence, and farthings, called
denominations of money. Time is reckoned in years, months,
weeks, days, hours, minutes, and seconds, called denomina-
tions of time. Distance is reckoned in miles, rods, feet, and
inches, called denominations of measure, &c.
The relative value of these denominations is exhibited in
tables, which the pupil must commit to memory.
1
算
​69
}
ENGLISH MONEY.
The denominations are pounds, shillings, pence, and far-
things.
TABLE.
4 farthings (qrs.) make 1 penny, marked d.
12 pence
1 shilling,
1 pound,
S.
£.
20 shillings
:
Note. Farthings are often written as the fraction of a
penny; thus, 1 farthing is written d., 2 farthings, d., 3
farthings, 2 d.
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$
70
shillings?
penny?
How many farthings in 1
in 2 pence? things?
in 6
Į
8 s. ?
shillings and
in 2 s. 3 d. ?
pence?
in 3 pence?
in 8 pence?
in 9 pence? in 12 pence? things?
in 1 shilling? in 2
in 2 s. 6 d. ?
How many pence in 2 shil-
lings?
in 3 s.?
4 s. ?
in 6 s. ?
in 10 s.?
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1
in 4 s. 3 d. ?
in 3£.
2£. 15 s.?
{
Į
1
2 pence?
How many shillings in 1
in 2£.?
pound?
in 3 £.?
in 4 £. 6 s.?
REDUCTION.
SEKAŽ
in. 4£.?
4
in 2 s. 4 d. ?
in 3 s. 6 d. ?
10 s. ?
T 33.
How many pence in 4 far-
in 8 farthings?
-in 12 farthings? in
24 farthings? in 32 far-
in 36 farthings?
in 48 qrs.? How many
shillings in 48 qrs. ?
96 qrs.?
in
in 6 £.8 s. ?
in
in pence?
in
in 2
}
}
1
How many shillings in 24
in 36 d. ?
in 72 d. ?
in 120 d.?
in 27 d.?
in 30 d. ?
in 51 d. ?
in 48 d. ?
in 96 d. ?
in 26 d. ?
in 28 d.?
in 42 d. ?
It has already been remarked, that the changing of one
kind, or denomination, into another kind, or denomination,
without altering their value, is called Reduction. ( 27.)
Thus, when we change shillings into pounds, or pounds into
shillings, we are said to reduce them. From the foregoing
examples, it is evident, that, when we reduce a denomina-
tion of greater value into a denomination of less value, the
reduction is performed by multiplication; and it is then call
ed Reduction Descending. But when we reduce a denomina-
tion of less value into one of greater value, the reduction is
performed by division; it is then called Reduction Ascending.
Thus, to reduce pounds to shillings, it is plain, we must
multiply by 20. And again, to reduce shillings to pounds,
we must divide by 20. It follows, therefore, that reduction
descending and ascending reciprocally prove each other.
·
86 s.?
70 s. ?
Bad, Bagnatingalė
I
How many pounds in 20 shil-
lings?
in 40 s. ?
in
60 s.?
in 80 s. ?
in
in 128 s. ?
in
in 55 s. ?
1
1
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T 33, 34.
71
1. In 17£. 13 s. 62 d. how 2. In 16971 farthings, how
many farthings?
many pounds?
OPERATION.
OPERATION.
Farthings in a penny, 4)16971 3 grs.
Pence in a shilling,
12)4242 6 d.
Shillings in a pound,
20)35|3 13s.
+
£.
17 13 6 3
20 8.
s. d. qrs
353s. in 17. 13 s.
12 d.
4242 d.
4q.
REDUCTION.
$
17£.
Ans. 17£. 13s. 6d.
Farthings will be reduced
16971 qrs. the Ans.
to pence, if we divide them
In the above example, be- by 4, because every 4 far-
cause 20 shillings make 1 things make 1 penny. There-
pound, therefore we multiply fore, 16971 farthings divided
17. by 20, increasing the by 4, the quotient is 4242
product by the addition of the pence, and a remainder of 3,
given shillings, (13,) which, which is farthings, of the
it is evident, must always be same name as the dividend.
done in like cases; then, be- We then divide the pence
cause 12 pence make 1 shil- (4242) by 12, reducing them
ling, we multiply the shillings to shillings; and the shillings
(353) by 12, adding in the (353) by 20, reducing them
given pence, (6.) Lastly, to pounds. The last quotient,
because 4 farthings make 117 £ with the several re-
penny, we multiply the pence mainders, 13 s. 6 d. 3 qrs. con-
(4242) by 4, adding in the stitute the answer.
given farthings, (3.) We Note. In dividing 353 s. by
then find, that in 17. 13 s. 20, we cut off the cipher, &c.,
62 d., are contained 16971 as taught T 22.
farthings.
""
{
T 34. The process in the foregoing examples, if care-
fully examined, will be found to be as follows, viz.
To reduce high denomina- To reduce low denominations
tions to lower,-Multiply the to higher,-Divide the lowest
highest denomination by that denomination given by that
number which it takes of the number which it takes of the
next less to make 1 of this same to make 1 of the next
higher, (increasing the pro-higher. Proceed in the same
duct by the number given, manner with each succeeding
if any, of that less denomina-denomination, until you have
4
1
+
1
}
|
11
f
72
T. 34.
tion.) Proceed in the same brought it to the denomination
manner with each succeeding required.
denomination, until you have
brought it to the denomination
required.
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3. Reduce 32£. 15 s. 8 d.
to farthings.
}
5. In 29 guineas, at 28 s.
ach, how many farthings?
7. Reduce $163, at 6 s.
each, to pence?
9. In 15 guineas, how
'many pounds?
1
REDUCTION.
EXAMPLES FOR PRACTICE.
A
Jud
eas.
Note.
We cannot reduce guineas directly to pounds, but
we may reduce the guineas to shillings, and then the shil-
lings to pounds.,
↓
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}
}
TROY WEIGHT.
**
By Troy weight are weighed gold, silver, jewels, and all
iquors. The denominations are pounds, ounces, penny-
weights, and grains.
TABLE.
24 grains (grs.) make 1 pennyweight, marked pwt.
20 pennyweights
1 ounce,
1 pound,
12 ounces
↓
4. Reduce 3142 farthings
to pounds.
6. In 38976 farthings, how
many guineas? ·
ད
8. Reduce 11736 pence to
dollars.
10. Reduce 21£. to guin-
}
11. Bought a silver tank-
ard, weighing 3 lb. 5 oz., pay
ang at the rate of $1'08 an
ounce; what did it cost?
13. Reduce 210lb. 8 oz.
12 pwt. to pennyweights.
15. In 7lb. 11 oz.
*
9 grs. of silver, how many to pounds.
grains?
I
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12. Paid $44'28 for a sil-
yer tankard, at the rate of
$1'08 an ounce; what did it
weigh?
14. In 50572 pwt. how
many pounds?
3pwt. 16. Reduce 45681 grains
po
OZ.
lb.
* The fineness of gold in tried by fire, and is reckoned in carats, by which is
understood the 24th part of any quantity; if it lose weaning in the trial, it is said
to be 24 carats fine; if it lose 2 carats, it is then 22 carats fine, which is the stand-
ard for gold,
Silver which abides the fire without loss is said ic ce 12 ounces fine. The
standard for silver coin is 11 oz. 2 pwts. of fine 2.ver, and 18 pwts. of copper
melted together
1
1
1

1 34.
"
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​APOTHECARIES' WEIGHT.
Apothecaries' weight is used by apothecaries and physi-
cians, in compounding medicines. The denominations are
pounds, ounces, drams, scruples, and grains.
REDUCTION.
17. In 9 lb. 83.13.20.
19 grs., how many grains.
TABLE.
1
20 grains, (grs.) make 1 scruple, marked D.
-1 dram,
3.
3 scruples
8 drams
12 ounces
16 drams, (drs.) make
16 ounces
28 pounds
4 quarters
20 hundred weight
1 ounce,
1 pound,
J
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B
Younce,
A pound,
}
島
​AVOIRDUPOIS WEIGHT.†
By avoirdupois weight are weighed all things of a coarse
and drossy nature, as tea, sugar, bread, flour, tallow, hay,
leather, medicines, (in buying and selling,) and all kinds
of metals, except gold and silver. The denominations are
tons, hundreds, quarters, pounds, ounces, and drams.
TABLÉ.
اء
Any
1 quarter,
1 hundred weight,
1 ton,
3.
tb.
18. Reduce 55799 grs. to
pounds.
marked
73
ht
1
1
Oz.
lb.
Note 1. In this kind of, weight, the words gross and net
are used. Gross is the weight of the goods, together with
the box, bale, bag, cask, &c., which contains them. Net
weight is the weight of the goods only, after deducting the
weight of the box, bale, bag, or cask, &c., and all other al-
lowances.
qr.
cwt.
T.
Note 2. A hundred weight, it will be perceived, is 112 lb.
Merchants at the present time, in our principal sea-ports,
buy and sell by the 100 pounds.
* The pound and ounce apothecaries' weight, and the pound and ounce Troy,
are the same, only differently divided, and subdivided..
† 175 oz, Troy =192 oz. avoirdupois, and 175 lb. Troy-144lb. avoirdu-
pois. 1lb. Troy = 5760 grains, and 1 lb. avoirdupois = 7000 grains Troy.
G
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$
4
1
1
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1
1
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↓
74
20. How much sugar, at
19. What will 5 cwt. 3 qrs.
17 lb. of sugar come to, at 12 cents a pound, may be
124 cents a pound.
bought for $82 625?
I
21. A merchant would put 22. In 470 boxes of raisins,
109 cwt. 0 qrs. 12lb. of containing 26 lb. each, how
raisins into boxes, containing many cwt.?
26 lb. each; how many boxes
will it require?
5 quarters,
6 quarters,
REDUCTION.
23. In 12 tons, 15 cwt. 24. In 7323500 drams, how
1 qr. 19lb. 6 oz, 12 dr. how many tons?
many drams?\
25. In 28 lb. avoirdupois,
how many pounds Troy?
27. In 573 yds. 1 qr. 1 na.
ow many nails?
29. In 151 ells Eng. how
many yards?
7
Note. Consult T 34, ex. 9.
}
3
L
TABLE.
1 nails, (na.) or 9 inches, make 1 quarter, marked
1 quarters, or 36 inches,
1 yard,
3 quarters,
1 ell Flemish,
ell English,
Tell French,
B
*
3
wpfafanua ngand, lagler
CLOTH MEASURE.
Cloth measure is used in selling cloths and other goods,
sold by the yard, or ell. The denominations are ells, yards,
quarters, and nails.
}
!
26. In 34lb. 0 oz. 6 pwt.
16 grs. Troy, how many
pounds avoirdupois ?
J
1
3
T 34.
ส
1
marked qr.
yd.
E, FI.
E. E.
E. Fr.
#
28. In 9173 nails, how ma-
ny yards?
LONG MEASURE.
Long measure is used in measuring distances, or othe
things, where length is considered without regard to breadth
The denominations are degrees, leagues, miles, furlongs
rods, yards, feet, inches, and barley-corns.
30. In 1883 yards, how ma-
ny ells English?
}

I
34.
$
3 feet,
5yards, or 16 feet,
S
TABLE.
3 barley-corns, (bar.) make 1 inch,
-12 inches,
1 foot,
1 yard,
1 rod, perch, or pole,
1 furlong,
B
40 rods, or 220 yards,
8 furlongs, or 320 rods, -
3 miles,
REDUCTION.
-
60 geographical, or 691
statute miles,
360 degrees,
691
3240
marked
CA
2160
180 half of the multiplicand.
H
{
*
+
{
Note. To multiply by 2, is Note. The barley-corns be-
to take the multiplicand 2 ing divided by 3, and that
times; to multiply by 1, is to quotient by 12, we have
take the multiplicand 1 time; 132105600 feet, which are to
to multiply by, is to take the be reduced to rods. We can-
multiplicand half a time, that not easily divide by 161 on
is, the half of it. Therefore, account of the fraction ; but
to reduce 360 degrees to stat-16 feet 33 half feet, in 1
ute miles, we multiply first by rod; and 132105600 feet
! the whole number, 69, and to 264211200 half feet, which,
the product add half the multi- divided by 33, gives 8006400
plicand. Thus :
rods.
360
75
in.
ft.
yd.
r. p.
fur.
M.
L.
1 mile,
1 league,
1 degree,
a great circle, or circumfer
ence of the earth.
31. How many barley-corns 32. In 4755801600 barley-
will reach round the globe, it corns, how many degrees ?
being 360 degrees?
deg. or
Hence, when the divisor is
encumbered with a fraction,
2 or 1, &c., we may reduce
the divisor to halves, or fourths,
&c., and reduce the dividend
to the same; then the quo-
25020 statute miles in 360 de-tient will be the true answer.
0
grees.
33. How many inches from 34. In 30539520 inches,
Boston to the city of Wash-how many miles?
ington, it being 482 miles ?
35. How many times will a 36. If a wheel, 16 feet 6
wheel, 16 feet and 6 inches inches in circumference, turn
in circumference, turn round round 12800 times in going
in the distance from Boston to from Boston to Providence,
Providence, it being 40 miles? what is the distance ?
I
1
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}
+
1
1
1
1
$
76
1 yard.
=
3 feet
LAND OR SQUARE MEASURE.
Square measure is used in measuring land, and any other
thing, where length and breadth are considered. The de-
nominations are miles, acres, roods, perches, yards, feet and
inches.
!
¶ 35. 3 feet in length make a yard in long measure; but it
requires 3 feet in length and 3 feet in breadth to make a yard
in square measure; 3 feet in length and one foot wide make
3 square feet; 3 feet in length and 2 feet wide make 2
times 3, that is, 6 square feet; 3 feet in length and 3 feet
wide make 3 times 3, that is, 9 square. feet. This will
clearly appear from the annexed figure.
3 feet
1 yard.
144 square inches
}
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30 square yards
40 square rods,
4
640
Solv
}
{
REDUCTION.
TABLE.
12 × 12; that is,
12 inches in length and 12 inches make 1 square foot.
in breadth
G
9 square feet3x3; that is, 3 feet?
in length and 3 feet in breadth
5X 51, or 2721
J
Va
square feet 16 × 16,
It is plain, also, that a square foot,
that is, a square 12 inches in length
and 12 inches in breadth, must con-
tain 12 × 12 = 144 square inches.
roods, or 160 square rods,
acres,
F
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Jo
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1
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T 35.

GM
Note. Gunter's chain, used in measuring land, is 4 rods.
in length. It consists of 100 links, each link being 7
inches in length; 25 links make 1 rod, long measure, and
625 square links make 1 square rod.
t
1 square yard.
1 square rod,
perch or pole.
1 rood.
1 acre.
1 square mile.
}
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T 35, 36.
}
37. In 17 acres 3 roods 12|
rods, how many square feet? how many acres?
1
yd.=3 feet trick.
Note. In reducing rods to Nate. Here we have 776457
feet, the multiplier will be square feet to be divided by
2721. To multiply by 4, is to 721. Reduce the divisor to
take a fourth part of the mufourths, that is, to the lowest
tiplicand. The principle, is denomination contained in it;
the same as shown 34, then reduce the dividend to
ex. 31.
fourths, that is, to the same
denomination, as shown 132,
ex. 34.
39. Reduce 6 square miles
o square feet/
41. There is a town 6 miles
*
1 yd.3 ft. long.
}
› REDUCTION.
под ку
38. In 776457 square feet,
square; Now many square square miles.
miles in that town?
many acres ?
how
SOLID OR CUBIC MEASURE.
Solid or cubic measure is used in measuring things that
have length, breadth, and thickness; such as timber, wood,
stone, bales of goods, &c. The denominations are cords,
tons, yards, feet, and inches.
T 36. It has been shown, that a square yard contains
3 × 39 square feet. A cubic yard is 3 feet long, 3 feet
vide, and 3 feet thick. Were it 3 feet long, 3 feet wide,
nd one foot thick, it would contain 9 cubic feet; if 2 feet
hick, it would contain 2 × 9: 18 cubic feet; and, as it is
3 feet thick, it does contain 3 × 9 27 cubic feet. This
will clearly appear from the
annexed figure.
1 yd.3 ft. wido.
G*
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40. In 1,784,217,600 square
feet, how many square miles?
42. Reduce 23040 acres to
It is plain, also, that a cubic
foot, that is, a solid, 12 inches
in length, 12 inches in breadth,
and 12 inches in thickness,
will contain 12 X 12 X 12 =
1728 solid or cubic inches.
}
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7
78.
128 solid fect,
TABLE.
1728 solid inches, 12x 12 x 12,
that is, 12 inches in length,
12 in breadth, 12 in thickness,
27 solid feet,
3 X 3 X 3
40 feet of round timber, or, 50 feet)
of hewn timber,
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Ab
8 X 4 X 4, that
is, 8 feet in length, 4 feet in
width, and 4 feet in height,)
Aut
43. Reduce 9 tons of round
timber to cúbic inches.
2
Note. What is called a cord foot, in measuring wood, is
16 solid feet; that is, 4 feet in length, 4 feet in width, and
1 foot in height, and 8 such feet, that is, 8 cod feet make
1 cord.
1
BR
REDUCTION.
1
AN
>
45. In 37 cord feet of wood,
how many solid feet?
47: Reduce 64 cord feet of
wood to cords.
46. In 592 solid feet of
wood, how many cord feet?
48. In 8 cords of wood, how
many cord feet?
50. 2048 solid feet of wood,
49. In 16 cords of wood,
how many cord feet? how how many cord feet? how
many solid feet?
many cords?
WINE MEASURE.
Wine measure is used in measuring all spirituous liquors,
ale and beer excepted; also vinegar and oil. The denomi-
nations are tuns, pipes, hogsheads, barrels, gallons, quarts,
pints, and gills.
·1
M
make
pa
}
TABLE.
4 gills (gi.)
2 pints
4 quarts
31 gallon's
63 gallons
2 hogsheads
1 hogshead,
1 pipe,
1 tun,
2 pipes, or 4 hogsheads
Note. A gallon, 'wine measure, contains 231 cubic inches.
Anak
1
make 1 solid foot.
J
- 1 solid yard.
1 ton or load.
M
1 cord of wood.'
D
↑ 36.

44. In 622080 cubicinches,
how many tons of round tim-
ber?
1 pint, marked
1 quart,
1 gallon,
1 barrel,
1
SON
pt.
J
qt.
gal.
bar.
hhd.
P
T.
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1
↓
f
1
T 36.
51. Reduce 12 pipes of wine|
to pints.
53. In 9 P. 1 hhd. 22 gals.
3 qts. how many gills?
•
55. In a tun of cider, how
many gallons?
/
2 pints (pts.)
4 quarts
36 gallons
54 gallons
M
ALE OR BEER MEASURE.
Ale or beer measure is used in measuring ale, beer, and
milk. The denominations are hogsheads, barrels, gallons,
quarts, and pints...
make
4 pecks
36 bushels
1
ام
2 pints (pts.) make
8 quarts
REDUCTION.
1
TABLE.
52. In 12096 pints of wine,
how many pipes ?
54. Reduce 39032 gills to
pipes.
56. Reduce 252 gallons to
tuns.
1 quart,
1 gallon,
1 barrel,
1 hogshead,
TABLE.
1
Note. A gallon, beer measure, contains 282 cubic inches.
57. Reduce 47 bar. 18 gal.
of ale to pints.
58. In 13680 pints of ale,
how many barrels ?
59. In 29 hhds. of beer,
how many pints?
60. Reduce 12528 pints to
hogsheads.
DRY MEASURE.
Dry measure is used in measuring all dry goods, such as
grain, fruit, roots, salt, coal, &c. The denominations are
chaldrons, bushels, pecks, quarts, and pints.
marked
?
1 quart,
1 peck,
1 bushel,
1 'chaldron,
J
B
marked
79
qt.
gal.
bar.
hhd,
+
Note. A gallon, dry measure, contains 2684 cubic inches.
A Winchester bushel is 18 inches in diameter, 8 inches,
deep, and contains 2150 cubic inches.
•
qt..
pk.
bu.
ch,
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1
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80
61. In 75 bushels of wheat,
how many pints?
63. Reduce 42 chaldrons of
coals to pecks.
60 seconds (s.)
'60 minutes
24 hours
days
4 weeks
B
1
January,
February, 2d,
March, 3d,
April,
4th,
May,
5th,
June,
6th,
7th,
July,
August, 8th,
September, 9th,
October, 10th,
November, 11th,
December, 12th,
Hand
TIME.
The denominations of time are years, months, weeks,
days, hours, minutes, and seconds.
TABLE,
J
REDUCTION.
make
than
I
13 months, 1 day and 6 hours,
or 365 days and 6 hours,
wala
J
T 36, 37.
62. In 4800 pints, how ma-
ny bushels ?
I
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pią
64. In 6048 pecks, how ma-
ny chaldrons?
Ju
K
1st month, has 31 days.
28
31
30
J
HONG
31
30
31
31
30
31
30
31
1 minute, marked
1 hour,
1
1 day,
1 week,
1 month,
1 37. The year is also divided into 12 calendar months,
which, in the order of their succession, are numbered as fol-
lows, viz.
+
1 common, or
Julian year,
www
Bu
Thirty days hath September,
April, June, and November,
February twenty-eight alone;
All the rest have thirty-one.
2
S
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1
}
11
Sa
1
#
m.
h.
d.
W.
mo.
Note. When any year
can be divided by 4 with-
out a remainder, it is call-
ed leap year, in which
February has 29 days.
yr.
The number of days in each month may be easily fixed in.
the mind by committing to memory the following lines;
ܐ
{
81
*
The first seven letters of the alphabet, A, B, C, D, E, F, G,
are used to mark the several days of the week, and they are
disposed in such a manner, for every year, that the letter A
shall stand for the 1st day of January, B for the 2d, &c. In
pursuance of this order, the letter which shall stand for Sun-
day, in any year, is called the Dominical letter for that year.
The Dominical letter being known, the day of the week
on which each month comes in may be readily calculated
from the following couplet:
T 37
¿
REDUCTION....
At Dover Dwells George Brown, Esquire,
Good Carlos Finch And David Fryer.
These words correspond to the 12 months of the year, and
the first letter in each word marks the day of the week on
which each corresponding month comes in; whence any other
day may be easily found. For example, let it be required
to find on what day of the week the 4th day of July falls, in the
year 1827, the Dominical letter for which year is G. Good
answers to July; consequently, July comes in on a Sunday;
wherefore the 4th day of July falls on Wednesday.
Note. There are two Dominical letters in leap years,
one for January and February, and another for the rest of
the year.

65. Supposing your age to
be 15 y. 19 d. 11 h. 37 m.
45 s., how many seconds old
are you, allowing 365 days 6
hours to the year?
66. Reduce 475047465 se-
conds to years.
567. How many minutes from
the 1st day of January to the to days.
14th day of August, inclusive-
ly?
68. Reduce 325440 minutes
70. In 4079160 minutes,
69. How many minutes from
the commencement of the war how many years?
between America and Eng-
land, April 19th, 1775, to the
settlement of a general peace,
which took place Jan. 20th,
1783 ?
"6
i
1
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1
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​82
CIRCULAR MEASURE, OR MOTION.
Circular measure is used in reckoning latitude and longi-
tude; also in computing the revolution of the earth and
other planets round the sun. The denominations are circles,
signs, degrees, minutes, and seconds.
60 seconds (") make
60 minutes
30 degrees
12 signs, or 360 degrees,
SUPPLEMENT TO REDUCTION.
:
古
​{
}
TABLE.
Note. Every circle, whether great or small, is divisible
into 360 equal parts, called degrees.
12 particular things
12 dozen
12 gross, or 144 dozen,
Gay
71. Reduce 9 s. 13° 25′ to 72. In 1020300", how many
seconds.
degrees?
A
The following are denominations of things not included in
the Tables:-
W
1 minute, -
I degree,
}
1 sign,
1 circle of the zodiac.
marked
}
make
20 particular things
6 points make 1 line,
12 lines
1 inch,
4 inches
1 hand, { used in measuring the height of
horses.
1
6 feet
1 fathom, used in measuring depths at sea.
112 pounds
make
1 quintal of fish.
24 sheets of paper make-
20. quires
1 quire.
1 ream.
{;
T 37.
And
S.
1 dozen.
1 gross.
1 great gross.
Also,
make
1 score.
used in measuring the length of
the rods of clock pendulums.
2
I
»
SUPPLEMÉNT TO REDUCTION.
QUESTIONS.
1. What is reduction? 2. Of how many varieties is re-
duction? 3. What is understood by different denominations,
as of money, weight, measure, &c. ? 4. How are high de
}

Get
T 37.
SUPPLEMENT TO REDUCTION.
nominations brought into lower? 5. How are low denomi-
nations brought into higher? 6. What are the denomina-
tions of English money? 7. What is the use of Troy weight,
and what are the denominations? 8.
avoirdupois
weight? the denominations? 9. What distinction do
you make between gross and net weight? 10. What dis-
tinetions do you make between long, square, and cubic
measure? 11. What are the denominations in long mea-
sure ?
in square measure? 13.
in cubic mea-
12.
sure?
14. How do you multiply by ? 15. When the di-
visor contains a fraction, how do you proceed? 16. How is
the superficial contents of a square figure found? 17. How
s the solid contents of any body found in cubic measure?
18. How many solid or cubic feet of wood make a cord?
19. What is understood by a cord foot? 20. How many
such feet make a cord? 21. What are the denominations
of dry measure? 22. of wine measure? 23.
3
'
03
of
time? 24. of circular measure? 25. For what is cir-
cular measure used? 26. How many rods in length is Gun-
ter's chain? of how many links does it consist? how many
inks make a rod? 27. How many rods in a mile? 28. How
many square rods in an acre? 29. How many pounds make
1 cwt. ?
"
EXERCISES.
1. In 46 £. 4 s., how many dollars?
Ans. $154.
2. In 36 guineas, how many crowns, at 6 s. 7 d. each?
Ans. 153 crowns, and 9 d.
3. How many rings, each weighing 5 pwt. 7 grs., may be
made of 3 lb. 5 oz. 16 pwt. 2 grs. of gold?
Ans. 158,
}
4. Suppose West Boston bridge to be 212 rods in length,
how many times will a chaise wheel, 18 feet 6 inches in
circumference, turn round in passing over it?
$
222
Ans. 1891 times.
5. In 470 boxes of sugar, each 26 lb., how many cwt.?
6. In 10 lb. of silver, how many spoons, each weighing
oz. 10 pwt.?
}
+
7. How many shingles, each covering a space 4 inches
one way and 6 inches the other, would it take to cover 1
square foot? How many to cover a roof 40 feet long, and
24 feet wide? (See T 25.) Ans. to the last, 5760 shingles.
8. How many cords of wood in a pile 26 feet long, 4 feet
wide, and 6 feet high? Ans. 4 cords, and 7 cord feet,
5
2
1
}
}
I
1
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i
++
+
t
+
34
}
/ T 37.
9. There is a room 18 feet in length, 16 feet in width,
and 8 feet in height; how many rolls of paper, 2 feet wide,
and containing 11 yards in each roll, will it take to cover the
walls?
Ans. 818.
10. How many cord feet in a load of wood 6 feet long,
2 feet wide, and 5 feet high?
Ans. 4 cord feet.
}
11. If a ship sail 7 miles an hour, how far will she sail,
at that rate, in 3 w. 4 d. 16 h. ?
t
+
唇
​}
J
12. A merchant sold 12 hhds. of brandy, at $275 a gal-
lon; how much did each hogshead come to, and to how
much did the whole amount?
SUPPLEMENT TO REDUCTION.
13. How much cloth, at 7s. a yard, may be bought for
29 £. 1 s.?
14. A goldsmith sold a tankard for 10£. 8 s. at the rate
of 5 s. 4 d. per ounce; how much did it weigh?
15. An ingot of gold weighs 2 lb. 8 oz. 16 pwt.; how
much is it worth at 3 d. per pwt. ?
16. At $0'18 a pound, what will 1 T. 2 cwt. 3 qrs. 16 lb.
of lead come to ?
1
17. Reduce 14445 ells Flemish to ells English.
18. There is a house, the roof of which is 442 feet in
length, and 20 feet in width, on each of the two sides; if
3 shingles in width cover one foot in length, how many
shingles will it take to lay one course on this roof? if 3
courses make one foot, how many courses will there be on
one side of the roof? how many shingles will it take to
cover one side?
to cover both sides?
}
1
pettat
Ans. 16020 shingles.
1
19. How many steps, of 30 inches each, must a man take
in travelling 54 miles?
**
20. How many seconds of time would a person redeem
in 40 years, by rising each morning hour earlier than he
now does?
21. If a man lay up 4 shillings each day, Sundays ex-
cepted, how many, dollars would he lay up in 45 years?
22. If 9 candles are made from 1 pound of tallow, how
many dozen can be made from 24 pounds and 10 ounces?
23. If one pound of wool make 60 knots of yarn, how
many skeins, of ten knots each, may be spun from 4 pounds
6 ounces of wool?
1
•
}
T 38.
ADDITION OF COMPOUND NUMBERS.
ADDITION
OF COMPOUND NUMBERS.
T 38. 1. A boy bought a knife for 9 pence, and a comb
for 3 pence; how much did he give for both? Ans. 1 shilling.
2. A boy gave 2 s. 6 d. for a slate, and 4 s. 6 d. for a book;
how much did he give for both?
3
{
10
3. Bought one book for 1 s. 6 d., another for 2 s. 3 d., an-
other for 7 d.; how much did they all cost? Ans. 4 s. 4 d.
85
4. How many gallons are 2 qts. + 3 qts. + 1 qt.?
5. How many gallons are 3 qts. + 2 qts. + 1 qt. + 3
qts. + 2 qts. ?
6. How many shillings are 2 d. + 3 d. + 5 d. +6 d. +7d.?
7. How many pence are 1 qr. + 2 grs. † 3 qrs. + 2 qrs.
+ 1 qr. ?
8. How many pounds are 4 s. + 10 s. + 15 s. + 1 s. ?
9. How many minutes are 30 sec. + 45 sec. † 20 sec. ?
10. How many hours are 40 min. + 25 min. + 6 min.?
11. How many days are 4 h. +8 h. + 10 h. 20 h.?
12. How many yards in length are 1 f. +2 f. + 1 f. ?
13. How many feet are 4 in. + 8 in. + 10 in. + 2 in.
+1 in. ?
14. How much is the amount of 1 yd. 2 ft. 6 in. +2 yds.
1 ft. 8 in. ?
£.
5.
d.
15 14 6
20 2 8
5
6
15. What is the amount of 2 s. 6 d. 4 s. 3 d. +7s. 8d.?
16. A man has two bottles, which he wishes to fill with
wine; one will contain 2 gal. 3 qts. 1 pt., and the other 3
qts. ; how much wine can he put in them?
17. A man bought a horse for 15£. 14 s. 6 d., a pair of
oxen for 20£. 2 s. 8 d., and a cow for 5£. 6 s. 4 d.; what
did he pay for all ?
4
I
When the numbers are large, it will be most convenient
to write them down, placing those of the same kind, or de-
nomination, directly under each other, and, beginning with
those of the least value, to add up each kind separately.
OPERATION.
Ans. 41 3 6
H
In this example, adding up the
column of pence, we find the amount
to be 18 pence, which being
1 S.
6 d., it is plain, that we may write
down the 6 d, under the column of
pénce, and reserve the 1 s. to be add-
ed in with the other shillings.
1
1
L
86
7
1.
£.
46
16
}
T 38.
/
Next, adding up the column of shillings, together with
the 1 s. which we reserved, we find the amount to be 23 s.
1.3 s. Setting the 3s. under its own column, we add
the 1. with the other pounds, and, finding the amount to be
41., we write it down, and the work is done.
Ans. 41 £. 3 s. 6 d.
Note. It will be recollected, that, to reduce a lower into
a higher denomination, we divide by the number which it
takes of the lower to make one of the higher denomination.
In addition, this is usually called carrying for that number:
thus, between pence and shillings, we carry for 12, and be-
tween shillings and pounds, for 20, &c.
༤
The above process may be given in the form of a general
RULE for the Addition of Compound Numbers:
· ADDITION OF COMPOUND NUMBERS.
I. Write the numbers to be added so that those of the
same denomination may stand directly under each other.
II. Add together the numbers in the column of the lowest
denomination, and carry for that number which it takes of
the same to make one of the next higher denomination.
Proceed in this manner with all the denominations, till you
come to the last, whose amount is written as in simple num-
bers.
Proof. The same as in addition of simple numbers.
EXAMPLES FOR PRACTICE..
λ
74 0
538 19 7 1
S.
d. gr.
11 3 2
اد
1
lb. oz. prot. gr
36 7 10 11
42 6 9 13
81 7 16 15
[
£.
772
18
36 16
d.
9
61
0 104
52
}
$.
I
TROY WEIGHT.
oz. prol. gr.
6 14 9
8
6 16
3
10
11
Į
£.
S.
183 19
4
8
17
10
15 4
d..
Į
oz. prot. gr.
18
13 16
4
3 77
Bought a silver tankard, weighing 2 lb. 3 oz., a silver
cup, weighing 3 oz. 10 pwt., and a silver thimble, weighing
2 pwt. 13 grs.; what was the weight of the whole ?
}
(Th
1
1
+
1
¦
T38.
T.
ADDITION OF COMPOUND NUMBERS.
cwt. gr. lb. Oz. dr.
11 1 16
5
10
9
15
11
9
14
J
25 0 2 11
718
0
25
yds. qr. na.
36 1 2
A man bought 5 loads of hay, weighing as
23 cwt. (= 1 T. 3 cwt.) 2 qrs. 17 lb. ; 21 cwt.
19 cwt. 0 qr. 24 lb. ; 24 cwt. 3 qrs.; 11 cwt.
how many tons in the whole?
41 2 3
65 3 1
1
AVOIRDUPOIS WEIGHT.
?
1
A
CLOTH MEASURE.
E. Fl. qr, na.
41 1 2
3
18 2
57 0 1
1
Deg. mi. fur. r.
59 46 6 29
216 39 1 36
678
53 17
53
2
Pol. ft.
in.
86 179 137
19 248 119
12 96 75
LONG MEASURE.
in. bar.
10
2
ft.
14
7 24 9
cwt. gr. lb.
16 3 18
2 16
22
There are four pieces of cloth, which measure as follows,
viz. 36 yds. 2 qrs. 1 na.; 18 yds. 1 qr. 2 na.; 46 yds. 3 qrs.
3 na.; 12 yds. 0 qr. 2 na.; how many yards in the whole ?
15
6 1
8 1
I
"U
LAND OR SQUARE MEASURE.
oż. dr.
6 14
8
12
11 10
E. En. gr. na.
75 4 2
31 1
28 3
Mi. fur. pol.
3 7
8 6 27
وبه
follows, viz.
1 qr. 16 lb.;
0 qr. 1 lb. ;
A. rood. pol. ft.
56 3 37
245
93
29 1 28
416 2
31
128
87
0
1
in.
228
25
119
J
a
(3 p.
}
+
}
88.
1
T 38.
There are 3 fields, which measure as follows, viz. 17 A.
3 r. 16 p.; 28 A. 5 r. 18 p.; 11 A. Or. 25 p.; how much
land in the three fields?.
Ton. ft.
29 36
36
12
19
8
11
E
1
K
perkaya den
}
in,
1229
64
917
2
ADDITION OF COMPOUND NUMBERS.
d.
Y. mo. 20.
57 11 3 6
0
6 0 5
Hhd. gal. qts. pts.
51 53 1 1
27 29
29
3 0
9
13 0 1
X
Bus. p. qt. pt.
36 2 5 1
19 3 7
0
84 9 2
32
SOLID OR CUBIC MEASURE.
yds. ft. in.
75 22
1412
9 26
195
3 19
1091
*
$
↓
h. m.
23
16
WINE MEASURE.
I
A merchant bought two casks of brandy, containing as
follows, viz. 70 gal. 3 qts.; 67 gal. 1 qt.; how many hogs-
heads, of 63 gal. each, in the whole?
DRY MEASURE.
5 18
{
1
$.
55 11
42 18
5
TIME.
ܐ ܼܼܿܿܿܵ
?
N
3
3
1
1!
!
2
Tun. hhd. gal. qts.
}
37 2 37 2
19 1
,28 2
59 1
00
! +
cords. ft.
37 119
9 110
48 127
"
Ch. bus. p. qts.
48 27 3 5
6 29
1 7
}
Y.
m. 20. d.
40 3 1 5
16 7 0 4
2ry 5 2 0
"
1
1:30. SUBTRACTION OF COMPOUND NUMBERS. $9
SUBTRACTION
1
OF COMPOUND NUMBERS.
T30. 1. A boy bought a knife for 9 cents, and sold it
for 17 cents; how much did he gain by the bargain?
2. A boy bought a slate for 25. 6 d., and a book for 3 s. 6 d.
how much more was the cost of the book thau of the slate ?
3. A boy owed his playmate 2 s.; he paid him 1 1. 6 d. ;
how much did he then owe him?
4. Bought two books; the price of one was 4 s. 6 d., the
price of the other 3s. 9 d.; what was the difference of their
costs?
5. A boy lent 5 s. 3 d.; he received in payment 2 s. 6 d. ;
how much was then dực?
6. A man has a bottle of wine containing 2 gallons and 3
quarts; after turning out 3 quarts, how much remained?
7. How much is 4 gal. less 3 gal. ? 4 gal. (less) 2 qts. ?
4 gal. 1 qt.? 4 gal. 1 gal. 1 qt.? 4 ga!.1 gal. 2 qts.
— ?
4 gal. 1 gal, 3 qts. ? 4 gal.
1 gal. 3 qts.?
Pr
2 gal. 3 qts.?
4 gal. 1 qt.
8. How much is 1 ft.
Kİ,
T
S
Vaging
1 ft. 6 in.? 7 ft. 8 in.
{
pagkakata
M
(less) 6 in. ? 1 ft.
3 in.
10 in. ?
9. What is the difference between 4£. 6 s. and, 1 £. 8 s. ?
10. How much is 3 £.
(less) I s.? 3£. 2 s.? 3£.
2£. 6 s.? 10.£. 4 s.
37 32. 15 s. 3. 4 s.
5£/8.9!
11. A man bought a horse for 30£. 4 s. 8 d., and a cow
for 5£. 14 s. 6d. ; what is the difference of their costs?
OPERATION.
£. S. d.
Minuend, 30 4 8
Subtrahend, 5 14 6
Ans. 24 10 2
:
M
K
8 in.? 6 ft.
4 ft. 2 in.? 7 ft. 8 in. — 5 ft.
J
As the two numbers are large,
it will be convenient to write
them down, the less under the
greater, pence under pence, shil-
lings under shillings, &c. We
may now take 6 d. from 8 d., and
there will remain 2 d. Proceeding to the shillings, we cau-
not take 14 s. from 4 s., but we may borrow, as ia simple num-
bers, 1 from the pounds, 20 s.,.which joined to the 4 s.
makes 24 s., from which taking 14 s. leaves 10 s., which we
set down. We must now carry 1 to the 5., making 6£.,
which taken from 30£. leaves 24 £., and the work is done.
Note. The most convenient way in borrowing is, to sub-
II*
1
{
Y
1
+
I
>
1
"
.
F
90
SUBTRACTION OF COMPOUND NUMBERS. T 39.
tract the subtrahend from the figure borrowed, and add the
difference to the minuend. Thus, in the above example, 14
from 20 leaves 6, and 4 is 10.
The process in the foregoing example may be presented
in the form of a RULE for the Subtraction of Compound Numi-
bers:
#
}
I
I. Write down the sums or quantities, the less under the
greater, placing those numbers which are of the same de-
nomination directly under each other.
II. Beginning with the least denomination, take succes-
sively the lower number in each denomination from the up-
per, and write the remainder underneath, as in subtraction
of simple numbers.
III. If the lower number of any denomination be greater
than the upper, borrow as many units as make one of the
next higher denomination, subtract the lower number there-
from, and to the remainder add the upper number, remem-
bering always to add 1 to the next higher denomination for
that which you borrowed.
Proof. Add the remainder and the subtrahend together,
as in subtraction of simple numbers; if the work be right,
the amount will be equal to the minuend.
EXAMPLES FOR PRACTICE.
1. A merchant sold goods to the amount of 136 £. 7s. 61 d.,
and received in payment 50£. 10 s. 42d; how much re-
mained due?
Ans. - 85£. 17 s. 14 d.
10 s., and, in selling
2. A man bought a farm for 1256 £.
it, lost 87£. 10 s. 6 d.; how much did he sell it for?
Ans. 1168 £. 19 s. 6 d.
}
3. A man bought a horse for 27 £. and a pair of oxen for
19 £. 12 s. 8 d.; how much was the horse valued more than
the oxen?
4. A merchant drew from a hogshead of molasses, at one
time, 13 gal. 3 qts.; at another time, 5 gal. 2 qts. 1 pt.;
what quantity was there left? Ans. 43 gal. 2 qts. 1 pt.
1
5. A pipe of brandy, containing 118 gal. sprang a leak,
when it was found only 97 gal. 3 qts. 1 pt. remained in the
cask; how much was the leakage?
#
6. There was a silver tankard which weighed 3 lb. 4 oz.;
the lid alone weighed 5 oz. 7pwt. 13 grs.; how much did
the tankard weigh without the lid ?
1
1
1
T 39.
grs.
7. From 15 lb. 2 oz. 5 pwt. take 9'oz. 8 pwt. 10 g
8. Bought a hogshead of sugar, weighing 9 cwt. 2 qrs.
17 lb.; sold at three several times as follows, viz. 2 cwt. 1 qr.
11 lb. 5 oz.; 2 qrs. 18 lb. 10 oz.; 25 lb. 6 oz.; what was the
weight of sugar which remained unsold?
SUBTRACTION OF COMPOUND NUMBERS.
Ans. 6.cwt. 1 qr. 17 lb. 11 oz.
9. Bought a piece of black broadcloth, containing 36 yds.
2 qrs.; two pieces of blue, one containing 10 yds. 3 qrs.
2 na.,
the other, 18 yds. 3 qrs. 3 na.; how much more was
there of the black than of the blue?
*
10. From 28 miles, 5 fur. 16 r. take 15 m. 6 fur. 26 r. 12 ft.
11. A farmer has two mowing fields; one containing 13
acres 6 roods; the other, 14 acres 3 roods: he has two
pastures also; one containing 26 A. 2 r. 27 p.; the other,
45 A. 5 r. 33 p.: how much more has he of pasture than of
mowing?
12. From 64 A. 2 r. 11 p. 29 ft. take 26 A. 5 r. 34 p. 132 ſt.
13. From a pile of wood, containing 21 cords, was sold, at
one time, 8 cords 76 cubic feet; at another time, 5 cords 7
cord feet; what was the quantity of wood left?
14. How many days, hours and minutes of any year will
be future time on the 4th day of July, 20 minutes past 3
o'clock, P. M.? `` Ans. 180 days, 8 hours, 40 minutes.
15. On the same day, hour and minute of July, given in
the above example, what will be the difference between the
past and future time of that month?
91
16. A note, bearing date Dec. 28th, 1826, was paid Jan.
2d, 1827; how long was it at interest?
The distance of time from one date to that of another may
be found by subtracting the first date from the last, observing
to number the months according to their order. (¶ 37.)
} OPERATION.
A. D.
}
D.{
1827. 1st m.
1826. 12 ....
Ans.
0
0
4 days.
17. A note, bearing date Oct. 20th, 1823, was paid April
25th, 1825; how long was the note at interest?
18. What is the difference of time from Sept. 29, 1816, to
April 2d, 1819?
Ans. 2 y. 6 m. 3 d.
19. London is 51° 32', and Boston 42° 23′ N. latitude;
what is the difference of latitude between the two places?
Ans. 9° 9'.
´2d day.
28 ...
*****
*
Note. In casting in-
terest, each month is
reckoned 30 days.
1
中
​\
1
5
}
2
毒
​3
Y
I
1
1
"
92
1
T 40.
•
20. Boston is 71° 3', and the city of Washington is 770
43 W. longitude; what is the difference of longitude be-
tween the two places?
· Ans. 6° 40'.
21. The island of Cuba lies between 74° and 85° W. lon-
gitude; how many degrees in longitude does it extend?
1
SUETRACTION OF COMPOUND NUMBERS.
1 40. 1. When it is 12 o'clock at the most easterly ex-
tremity of the island of Cuba, what will be the hour at the
most westerly extremity, the difference in longitude be-
ing 11° ?
}
Note.
The circumference of the earth being 360°, and
the earth performing one entire revolution in 24 hours, it
follows, that the motion of the earth, on its surface, from
west to east, is
14
15° of motion in 1 hour of time; consequently,
1° of motion in 4 minutes of time, and
J
1' of motion in 4 seconds of time.
From these premises it follows, that, when there is a dif
ference in longitude between two places, there will be a
corresponding difference in the hour, or time of the day.
The difference in longitude being 15°, the difference in time
will be 1 hour, the place easterly having the time of the day
1 hour earlier than the place westerly, which must be par-
ticularly regarded.
{
If the difference in longitude be 1°, the difference in time
will be 4 minutes, &c.
1
Hence,-If the difference in longitude, in degrees and
minutes, between two places, be multiplied by 4, the pro-
duct will be the difference in time, in minutes and seconds,
which may be reduced to hours.
We are now prepared to answer the above question.
11°
7
Hence, when it is 12 o'clock at the
most easterly extremity of the island,
it will be 16 minutes past 11 o'clock
at the most western extremity.
44 minutes.
2. Boston being 6° 40′ E. longitude from the city of
Washington, when it is 3 o'clock at the city of Washington,
what is the hour at Boston ?
1
Ans. 26 minutes 40 seconds past 3 o'clock.
3. Massachusetts being about 72°, and the Sandwich
Islands about 155° W. longitude, when it is 28 minutes past
6 o'clock, A. M. at the Sandwich Islands, what will be the
hour in Massachusetts ?.
Ans. 12 o'clock at noon,
¿
}
I
3
1
T 41. MULTIPLICATION OF COMPOUND NUMBERS.
MULTIPLICATION & DIVISION
OF COMPOUND NUMBERS.
¶ 41. 1. A man bought 2 yards of cloth, at 1 s. 6 d. per
yard; what was the cost?
2. If 2 yards of cloth cost 3 shillings, what is that per
yard?
1
7. How many shillings are 3 times 8 d. ?
3 X 10 d. ?
4 X 7 d.?
3. A man has three pieces of cloth, each measuring 10
yds. 3 qrs.; how many yards in the whole?
4. If 3 equal pieces of cloth contain 32 yds. 1 qr., how
much does each piece contain ?
5. A man has five bottles, each containing 2 gal. 1 qt. 1 pt.;
how much wine do they all contain ? ·
6. A man has 11 gal. 3 qts. 1 pt. of wine, which he would
divide equally into five bottles; how much must he put into
each bottle ?
9 d. ?
2 X 3 qrs.?
8. How much is one third
of 2 s. 6 d. ?
To of 7 s. 6 d. ?
3 d. ?
6 d. ?
9. At 1£. 5 s. 8 d. per
yard, what will 6 yards of
cloth cost?
}
OPERATION.
£. s. d.
s. d. gr.
1
5 8 3 price of 1 yard.
6 number of yards.
S
93
7 X 6 d.?
K
5 X 2 qrs.?
of 2 shillings?
of 2 s. 4 d. ?
of 1d.?
↓
Here, as the numbers are large, it will be most convenient
to write them down before multiplying and dividing.
3 X 9 d. ?
10 X
of 2 s.
of 3 s.
of 21 d. ?
OPERATION,
£. s. d. gr.
6)7 14 4 2 cost of 6 yards.
1
Ans. 7 14 4 2 cost of 6 yards.
5 8 3 price of 1 yard.
Proceeding after the man-
6 times 3 qrs. are 18 qrs. ner of short division, 6 is con-
4 d. and 2 qrs. over; we set tained in 7£. 1 time, and 1£.
down the 2 qrs.; then, 6 times over; we write down the
8 d. are 48 d., and 4 to carry quotient, and reduce the re-
makes 52 d. 4 s. and 4 d. mainder (1.) to shillings,
over, which we write down; (20 s.,) which, with the given
again, 6 times 5 s. are 30 s. shillings, (14 s.,) make 34 s. ;
10. If 6 yards of cloth cost
7£. 14 s. 4 d., what is the
price per vard?
}
T
J
i
f
1
**
}
T
94 :
T 41.
r
and 4 to carry makes 34 s. = 6 in 34 s. goes 5 times, and 4 s.
1£. and 14's. over; 6 times over; 4 s. reduced to pence
1£. are 6£., and 1 to carry 48 d., which, with the
makes 7£, which we write given pence, (4 d.,) make 52
down; and it is plain, that the d. ; 6 in 52 d. goes 8 times, and
united products arising from 4 d. over; 4 d. 16 qrs.,
the several denominations is which, with the given qrs.
the real product arising from (2) = 18 qrs. ; 6 in 18 qrs. goes
the whole compound number. 3 times; and it is plain, that
the united quotients arising
from the several denomina-
tions, is the real quotient aris-
ing from the whole compound
number.
by 7.
MULTIPLICATION AND DIVISION **
1
1
11. Multiply 3£. 4 s. 6 d.
12. Divide 22£. 11 s. 6 d.
by 7.
}
13. What will be the cost 14. At 2. 12 s. 6 d. for 5
of 5 pairs of shoes at 10 s. 6 d. pairs of shoes, what is that a
a pair?
pair?
15. In 5 barrels of wheat,
each containing 2 bu. 3 pks.
6 qts., how many bushels?
I
16
#
I
16. If 14 bu. 2 pks. 6 qts.
of wheat be equally divided
into 5 barrels, how many
bushels will each contain ?
18. If 9 coats contain 39
17. How many yards of
cloth will be required for 9 yds. 3 qrs. 3 na., what does 1
coats, allowing 4 yds. 1 qr. coat contain?
3 na. to each ?
19. In 7 bottles of wine, 20. If 5 gal. 1 gill of wine
each containing 2 qts. 1 pt. 3 be divided equally into 7 bot-
gills, how many gallons? tles, how much will each con-
tain?
21. What will be the .22. If 8 silver cups weigh
weight of 8 silver cups, each 3 lb. 9 oz. 1 pwt. 16 grs., what
weighing 5 oz. 12. pwt. 17 is the weight of each?
grs. ?
23. How much sugar in 12 24. If 119 cwt. 1 qr. of su-
hogsheads, each containing gar be divided into 12 hogs-
9 cwt. 3 qrs. 21 lb. ?
heads, how much will each
hogshead contain?
25. In 15 loads of hay, each
weighing 1 T. 3 cwt. 2 qrs.,
how many tons ?
7
>
26. If 15 teams be loaded
with 17 T. 12 cwt. 2 qrs. of
hay, how much is that to each
team ?
$
}
1
14
T 41.
OF COMPOUND NUMBERS.
When the multiplier, or divisor, exceeds 12, the operations
of multiplying and dividing are not so easy, unless they be
composite numbers; in that case, we may make use of the
component parts, or factors, as was done in simple numbers.
S
Thus 15, in the example 15 being a composite num-
above, is a composite number ber, and 3 and 5 its compo-
produced by the multiplica- nent parts, or factors, we may
tion of 3 and 5, (3 × 5
divide 17 T. 12 cwt. 2 qrs. by
15.) We may, therefore, one of these component parts,
multiply 1 T. 3 cwt. 2 qrs. by or factors, and the quotient
one of those component parts, thence arising by the other,
or factors, and that product by which will give the true
the other, which will give the answer, as already taught,
true answer, as has been al-( 20.)
ready taught, (T 11.)
T.
1
3 10 2
OPERATION.
OPERATION.
T. crot. qr.
cul. gr.
3 2
One factor, 3) 17 12 2
3 one of the factors. The other factor, 5) 5 17 2
Ans. 1 3 2
5 the other factor.
17 12 2 the answer.
27. What will 24 barrels
of flour cost, at 2£. 12 s. 4 d.
a barrel?
29. What will 112 lb.. of
sugar cost, at 71 d. per lb. ?
Note. 8, 7, and 2, are fac-lb. ?
tors of 112.
95
{
33. What will 139 yards of
cloth cost, at 3 £. 6 s. 5 d.
per yard?
139 is not a composite num-
ber. We may, however, de-
compose this number thus,
139: 100+30 +9.
28. Bought 24 barrels of
flour for 62 £. 16 s.; how
much was that per barrel ?
30. If 1 cwt. of sugar cost
3 £. 7 s. 8 d., what is that per
31. How much brandy in 32. Bought 84 pipes of
84 pipes, each containing 112 brandy, containing 9468 gal.
gal. 2 qts. 1 pt. 3 g.?
1 qt. 1 pt.; how much in a
pipe?
1
p
34. Bought 139 yards of
cloth for 461. 11 s. 11 d.;
what was that per yard?
When the divisor is such a
number as cannot be produced
by the multiplication of small
numbers, the better way is to
We may now multiply the divide after the manner of
1
is 1
I
96
T 41
price of 1 yard by 10, which long division, setting down
will give the price of 10 yards, the work of dividing and re-
and this product again by 10, ducing in manner as fol
which will give the price of lows:
100 yards.
We may then multiply the
price of 10 yards by 3, which
will give the price of 30 yards,
and the price of 1 yard by 9,
which will give the price of
9 yards, and these three pro-
ducts, added together, will evi-
dently give the price of 139
yards; thus:
£.
· 3
33
1
MULTIPLICATION AND DIVISION
S. d.
6
4
5 price of 1 yd.
10
2 price of 10 yds.
10
£. S. d.
11
139) 461
417
44
20
891 (6 s.
834
57
12
1
695 ( 5 d.
695
}
+
1.
толера
332
99 12
The divisor, 139, is con-
tained in 461 £. 3 times,
8 price of 100 yds. (3£.,) and a、 remainder of
6 price of 30 yds. 44 £., which must now be
29 17 9 price of 9 yds. reduced to shillings, multi-
plying it by 20, and bringing
461 11 11 price of 139 yds. in the given shillings, (11 s.,)
Note. In multiplying the making 891 s., in which the
price of 10 yards (33 £. 4 s. divisor is contained 6 times,
2 d.) by 3, to get the price of (6 s.,) and a remainder of
30 yards, and in multiplying 57 s., which must be reduced
the price of 1 yard (3£. 6 s. to pence, multiplying it by 12,
5 d.) by 9, to get the price of and bringing in the given
9 yards, the multipliers, 3 and pence, (11 d.,) together mak-
9, need not be written down, ing 695 d., in which the di-
}
but may be carried in the visor is contained 5 times,
mind.
(5 d.,) and no remainder.
The several quotients, 3£,
6.8., 5 d., evidently make the
answer.
11 ( 3£ .
#
}
1
↓
I
C
T 41.
The processes in the foregoing examples may now be pre-
sented in the form of a
OF COMPOUND NUMBERS.
3
1
RULE for the Multiplicution of RULE for the Division of Com-
Compound Numbers.
pound Numbers.
tiplier be more than one, mul-
tiply the product of 100 by the
number of hundreds; for the
tens, multiply the product of
10 by the number of tens; for
the units, multiply the multi-
plicand and these several pro-
ducts will be the product re-
quired.
1
!
I. When the multiplier does I. When the divisor does
not exceed 12, multiply suc- not exceed 12, in the manner
cessively the numbers of each of short division, find how
denomination, beginning with many times it is contained in
the least, as in multiplication the highest denomination, un-
of simple numbers, and carry der which write the quotient,
as in addition of compound and, if there be a remainder,
numbers, setting down the reduce it to the next less de-
whole product of the highest nomination, adding thereto the
denomination.
number given, if any, of that
denomination, and divide as
before; so continue to do
through all the denominations,
and the several quotients will
be the answer,
{
Į
97
II. If the multiplier exceed
II. If the divisor exceed 12,
RING
12, and be a composite num- and be a composite, we may di-
ber, we may multiply first by vide first by one of the com-
one of the component parts, ponent parts, that quotient by
that product by another, and another, and so on, if the com-
so on, if the component parts ponent parts be more than
be more than two; the last two; the last quotient will be
product will be the product re- the quotient required.
quired.
1
Gal
III. When the multiplier III. When the divisor ex-
exceeds 12, and is not a com- ceeds 12, and is not a com-
posite, multiply first by 10, posite number, divide after the
and this product by 10, which manner of long division, set-
will give the product for 100; ting down the work of di-
and if the hundreds in the mul-viding and reducing.
}
1
!
The
A
1
98
MULTIPLICATION AND DIVISION, &c.
EXAMPLES FOR PRACTICE.
cloth cost, at 4 s. 7 d. per
yard?
1. What will 359 yards of 2. Bought 359 yards of cloth
for 83 £.0 s. 41 d.; what was
that a yard ?
3. In 241 barrels of flour,
each containing 1 cwt. 3 qr.
9 lb.; how many cwt. ?
4. If 441 cwt. 13 lb. of flour
be contained in 241 barrels,
how much in a barrel ?
5. How many bushels of
wheat in 135 bags, each con-
taining 2 bu. 3 pks. ?
3 X 9 X 5 = 135.
7. What will 35 cwt. of to-
bacco cost, at 3 s. 10
lb. ?!
9. If 14 men build 12 rods
6 feet of wall in one day, how
many rods will they build in
71 days?·
what will 8 lbs.
"
1
}
8. At 759 £. 10 s. for 35
d. per cwt. of tobacco, what is that
per lb. ?
T42. 1. At.10 s. per yard, what will 17849 yards of
cloth cost?
M
6. If 371 bu. 1 pk. of wheat
be divided equally into 135
bags, how much will each bag
contain ?
Note. Operations in multiplication of pounds, shillings,
pence, or of any compound numbers, may be facilitated by
taking aliquot parts of a higher denomination, as already ex-
plained in "Practice" of Federal Money, T 29, ex. 10.
Thus, in this last example, the price 10 s. of a pound;
therefore, of the number of yards will be the cost in
pounds. 17428924 £. 10 s. Ans.
2. What cost 34648 yards of cloth, at 10 s. or £. per
yard?
at 5 s. = 4£. per yard?
at 4 s. =
= ££.
at 3 s. 4 d. £. per yard?
at 2 s.
Ans. to last, 3464 £. 16 s.
per yard?
. per yard?
3. What cost 7430 pounds of sugar, at 6 d.
át 4 d. } s. per lb. ?
¿s. per lb?
at 3 d.
at 2 d. ts. per lb. ?
lb. ?
per lb. ?
帮
​Ans. to the last, 14.30 s. — 928 s. 9 d.
928 s. 9 d.
4. At $1875 per cwt., what will 2 grs.
what will 1
cwt. cost?
10. If 14 men build 92 rods
12 feet of stone wall in 71
days, how much is that per
day?
$
¶ 42.
at 11 d.
qr.cwt. cost?
what will 14 lbs.cwt. cost?
1
Start
Devadattending
P
46 £. 8's. 9 d.
1 cwt. cost?
- what will 16 lb..
سلام
s. per
i s..
P
cwt. cost? Ans. to the last, $ 1339.
$
1
¶ 42.´´
5. What cost 340 yards of cloth, at 12 s. 6 d. per yard ?
12 s. 6 d. 10 s. (.) and 2 s. 6 d. (£.); there-
J
fore,
SUPPLEMENT TO COMPOUND NUMBERS.
340
170 £.
2
42 £. 10 s.
Ans. 212 £ 10 s.
Or,
10s.
£.)340
2s. 6d. of 10 s.) 170 £.
4
cost at 10 s.
at 2 s. 6 d.
at 12 s. 6 d.
per yard.
per yard.
per yard.
at 10 s. per yard.
42 £. 10s. at 2 s. 6 d. per yard.
Ans. 212 £. 10 s. at 12 s. 6 d. per yard.
SUPPLEMENT TO THE ARITHMETIC OF
COMPOUND NUMBERS.
99
QUESTIONS.
4
1. What distinction do you make between simple and
compound numbers? ( 26.) 2. What is the rule for addi-
tion of compound numbers? 3. for subtraction of, &c.?
4. There are three conditions in the rule given for multi-
plication of compound numbers; what are they, and the
methods of procedure under each? 5. The same questions
in respect to the division of compound numbers? 6. When
the multiplier or divisor is encumbered with a fraction, how
do you proceed? 7. How is the distance of time from one
date to another found? 8. How many degrees does the
earth revolve from west to east in 1 hour? 9. In what
time does it revolve 1°? Where is the time or hour of the
day earlier at the place most easterly or most westerly?
10. The difference in longitude between two places being
known, how is the difference in time calculated? 11. How
may operations, in the multiplication of compound num-
bers, be facilitated? 12. What are some of the aliquot parts
of 1 £.?
of 1 s. ?
of 1 cwt. ? 13. What is this
manner of operating usually called?
1
72"
1
3
1
\
100
{m}
1
EXERCISES.
1. A gentleman is possessed of 14 dozen of silver spoons,
each weighing 3 oz. 5 pwt.; 2 doz. of tea spoons, each weigh-
ing 15 pwt. 14 gr. ; 3 silver cans, each 9 oz. 7 pwt. ; 2 silver
tankards, each 21 oz. 15 pwt.; and 6 silver porringers, each
11 oz. 18 pwt.; what is the weight of the whole ?
Ans. 18 lb. 4 oz. 3 pwt.
** [
1
SUPPLEMENT TO COMPOUND NUMBERS.
Note. Let the pupil be required to reverse and prove the
following examples:
鼻
​+
2. An English guinea should weigh 5 pwt. 6 gr.; a piece
of gold weighs 3 pwt. 17 gr.; how much is that short of the
weight of a guinea ?
3. What is the weight of 6 chests of tea, each weighing
3 cwt. 2 qrs. 9 Ib. ?
4. In 35 pieces of cloth, each measuring 2 yards, how
many yards ?
5. How much brandy in 9 casks, each containing 45 gal.
3 qts. 1 pt. ?
6. If 31 cwt. 2 qrs. 20 lb. of sugar be distributed equally
into 4 casks, how much will each contain?
1
占
​CİDD
Note. The pupil will recollect, that 8, 7 and 2 are fac-
•
tors of 112, and may be used in place of that number.
8. If 800 cwt. of cocoa cost 18 £. 13 s. 4 d., what is that
per cwt.? what is it per lb. ?
9. What will 94 cwt. of copper cost at 5 s. 9 d. per lb. ?
10. If 6 cwt. of chocolate cost 72 £. 16 s., what is that
per lb. ?
7. At 43 d. per lb., what costs 1 cwt. of rice ? 2 cwt. ?
3 cwt.?
11. What cost 456 bushels of potatoes, at 2 s. 6 d..
bushel?
per
1
1
13. What cost 7846 pounds of tea, at 7 s. 6 d. per lb..?
at 14 s. per lb. ?
at 13 s. 4 d. ?
14. At $94'25 per cwt., what will be the cost of 2 qrs.
of 3 qrs. ?
of 14 lbs. ?
of 21 lbs.?
of 16 lbs. ?
of 24 lbs, ?
of tea?
{
↑ 42.
(See T42.)
Note. 2 s. 6 d. is † of 1 £.
12. What cost 86 yards of broadcloth, at 15 s. per yard?
Note. Consult T 42, ex. 5. }
Bryophotometers Tveryone
Note. Consult T 42, ex. 4 and 5.
U)
Categ
1
1
1
+
༈༣
1
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at
ܗ
1
*
¶ 42, 43.
101
15. What will be the cost of 2 pks. and 4 qts. of wheat,
1'50 per bushel?
16. Supposing a meteor to appear so high in the heavens
as to be visible at Boston, 71° 3', at the city of Washington,
77° 43', and at the Sandwich Islands, 155° W. longitude,
and that its appearance at the city of Washington be at 7
minutes past 9 o'clock in the evening; what will be the
hour and minute of its appearance at Boston and at the
Sandwich Islands?
1
|^
FRACTIONS.
!
13
FRACTIONS.
1 43. We have seen, (¶ 17,) that numbers expressing
whole things are called integers, or whole numbers; but that,
in division, it is often necessary to divide or break a whole
thing into parts, and that these parts are called fractions, or
broken numbers.
}
+5
It will be recollected, (T 14, ex. 11,) that when a thing
or unit is divided into 3 parts, the parts or fractions are call-
ed thirds; when into four parts, fourths; when into six parts,
sixths; that is, the fraction takes its name or denomination from
the number of parts, into which the unit is divided. Thus,
if the unit be divided into 16 parts, the parts are called six-
teenths, and 5 of these parts would be 5 sixteenths, expressed
thus, The number below the short line, (16,) as before
taught, (T 17,) is called the denominator, because it gives
the name or denomination to the parts; the number above
the line is called the numerator, because it numbers the parts.
The denominator shows how many parts it takes to make
a unit or whole thing; the numerator shows how many of
these parts are expressed by the fraction.
1. If an orange be cut into 5 equal parts, by what frac-
tion is 1 part expressed? 2 parts? 3 parts?
5 parts? how many parts make unity
4 parts ?
or a whole orange?
2. If a pie be cut into 8 equal pieces, and 2 of these
pieces be given to Harry, what will be his fraction of the
pie? if 5 pieces be given to John, what will be his fraction?
what fraction or part of the pie will be left?
It is important to bear in mind, that fractions arise from
division, (¶ 17,) and that the numerator may be considered a
1
1.
'
'
1
24,
V
}
܆ ܐ
102
T. 43%
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1
dividend, and the denominator a divisor, and the value of the
fraction is the quotient; thus, is the quotient of 1 (the
numerator) divided by 2, (the denominator;) is the quo-
tient arising from 1 divided by 4, and is 3 times as much,
that, is, 3 divided by 4; thus, one fourth part of 3 is the
same as 3 fourths of 1.
Hence, in all cases, a fraction is always expressed by the
sign of division.
3 is the dividend, or numerator.
expresses the quotient, of which
4 is the divisor, or denominator.
3. If 4 oranges be equally divided among 6 boys, what
part of an orange is each boy's share?
A sixth part of 1 orange is g, and a sixth part of 4 oranges
is 4 such pieces, = &.
Ans. of an orange.
4. If 3 apples be equally divided among 5 boys, what part
of an apple is each boy's share? if 4 apples, what? if 2
apples, what? if 5 apples, what?
5. What is the quotient of 1 divided by 3?
of 2 by 4?
of 1 by 4?
P
TRACTIONS..
青
​of 2 by 3?
by 7?
of 6 by 8 ?
of 5
of 2 by 14 ?
of 4 by 5?
6. What part of an orange is a third part of 2 oranges?
one fourth of 2 oranges?
t of 3 oranges?
4 of 3?
1 of 3 oranges?
& of 2?
of 4?
f of 2 ?
4 of 5?
/
."
of 3 by 4?
A Proper Fraction. Since the denominator shows the num-
ber of parts necessary to make a whole thing, or 1, it is plain,
that, when the numerator is less than the denominator, the
fraction is less than a unit, or whole thing; it is then called a
proper fraction. Thus, 1, §, &c. are proper fractions.
An Improper Fraction. When the numerator equals or ex-
ceeds the denominator, the fraction equals or exceeds unity, or
1, and is then called an improper fraction. Thus, §, †, 4, 4;
are improper fractions.
1
A Mixed Number, as already shown, is one composed of a
whole number and a fraction. Thus, 142, 137, &c. are mix-
ed numbers.
1
7. A father bought 4 oranges, and cut each orange into 6.
equal parts; he gave to Samuel 3 pieces, to James 5 pieces,
to Mary 7 pieces, and to Nancy 9 pieces; what was each
one's fraction?
}
Was James's fraction proper, or improper? Why?
Was Nancy's fraction proper, or improper? Why?
1
J
103
* 44.
To change an improper fraction to a whole or mixed number.
T44. It is evident, that every' improper fraction must
contain one or more whole ones, or integers.
1. How many whole apples are there in 4 halves (†) of
an apple? in i
20 ?
in &?
in
in 10?
in 984 ?
in 48?
2. How many yards in
in ?
in &?
LfQ?
OPERATION.
4)27
P
P
Sarkanberateraný, aleman
in
in ?
in 48?
15?
in 177?
in 20?
3. How many bushels in 8 pecks? that is, in g of a bushel?
in 19?
in ?
in 18?
-in
in 24 ?
„Ans. 63 oranges.
1?
*
4
1
FRACTIONS.
P
in 120?
P
of a yard?
in?
http
*
This finding how many integers, or whole things, are con-
tained in any improper fraction, is called reducing an impro-
per fraction to a whole or mixed number.
↓
4. If I give 27 children of an orange each, how many
oranges will it take? It will take 27; and it is evident, that
dividing the numerator, 27, (= the num-
ber of parts contained in the fraction,) by
the denominator, 4, (= the number of
parts in 1 orange,) will give the number
of whole oranges.
1
in § of a yard?
- in
Hence, To reduce an improper fraction to a whole or mixed
number,-RULE: Divide the numerator by the denominator;
the quotient will be the whole or mixed number.
EXAMPLES FOR PRACTICE.
Fo
5. A man, spending of a dollar a day, in 83 days would
spend of a dollar; how many dollars would that be?
Ans. $138.
6. In 1411 of an hour, how many whole hours?
The 60th part of an hour is 1 minute: therefore the ques-
tion is evidently the same as if it had been, In 1417 minutes,
how many hours?
Ans. 2387 hours.
7. In 8793 of a shilling, how many units or shillings?
Ans. 730 shillings.
2
8. Reduce 4478 to a whole or mixed number.
9. Reduce 8, 70, £78, 4388, 3485, to whole or mix-
ed numbers.
'
1
1
"
1
1
J
104
↑ 45.
To reduce a whole or mixed number to an improper fraction.
T 45. We have seen, that an improper fraction may be
changed to a whole or mixed number; and it is evident,
that, by reversing the operation, a, whole or mixed number
may be changed to the form of an improper fraction.
J
1. In 2 whole apples, how many halves of an apple? Ans. 4
halves; that is, . In 3 apples, how many halves? in 4
apples? in 6 apples? in 10 apples? in 24? in 60? in
170 in 492 ?
2. Reduce 2 yards to thirds. Ans. §.
thirds. Ans. §. Reduce 3 yards to thirds.
33 yards. 5 yards.
4
OPERATION.
192
5
yards.
3. Reduce 2 bushels to fourths.
61 bushels.
72 bushels.
12
4. In 16
1 make 1 dollar: if, therefore, we multiply 16 by 12, that
is, multiply the whole number by the denominator, the product
will be the number of 12ths in 16 dollars: 16 X 12 192,
and this, increased by the numerator of the fraction, (5,) evi-
dently gives the whole number of 12ths; that is, 197 of a
dollar, Answer.
2
16 dollars.
12
P
FRACTIONS.
Pr
番
​Reduce 23 yards to
3 yards.
6/3/
5 yards.
dollars, how many of a dollar?
f
1
2 bu.- 6 bushels.
252 bushels.
1
1
12ths in 16 dollars, or the whole number.
12ths contained in the fraction.
197 197, the answer.
Hence, To reduce a mixed number to an improper fraction,-
RULE: Multiply the whole number by the denominator of
the fraction, to the product add the numerator, and write
the result over the denominator.
EXAMPLES FOR PRACTICE.
5. What is the improper fraction equivalent, to 2323 hours?
Ans. 1417 of an hour.
6. Reduce 730
shillings to 12ths.
As of a shilling is equal to 1 penny, the question is evi-
dently the same as, In 730 s. 3 d., how many pence?
Ans. 8792 of a shilling; that is, 8763 pence.
요구​요
​/
1
1
T 45, 46.
7. Reduce 118, 1728, 87, 478%, and 7215 to improper
fractions.
8. In 1561 days, how many 24ths of a day?
Ans, 3781 3761 hours.
9. In 342 gallons, how many 4ths of a gallon?
Ans. 1371 of a gallon 1371 quarts.
24
FRACTIONS.
To reduce a fraction to its lowest or most simple terms.
146. The numerator and the denominator, taken to-
gether, are called the terms of the fraction.
T
Indep
If of an apple be divided into 2 equal parts, it becomes 2.
The effect on the fraction is evidently the same as if we had
multiplied both of its terms by 2. In either case, the parts
are made 2 times as MANY as they were before; but they are only
HALF AS LARGE; for it will take 2 times as many fourths to.
make a whole one as it will take halves; and hence it is
that is the same in value or quantity as ..
C3
2 is 2 parts; and if each of these parts be again divided
into 2 equal parts, that is, if both terms of the fraction be
multiplied by 2, it becomes . Hence,=, and the
reverse of this is evidently true, that = 2.
05
It follows therefore, by multiplying or dividing both terms of
the fraction by the same number, we charge its terms without
altering its value.
Thus, if we reverse the above operauon, and divide both
terms of the fraction & by 2, we obtain its equal, ; dividing
again by 2, we obtain 2, which is the most simple form of the
fraction, because the terms are the least possible by which
the fraction can be expressed.
}
Aft
The process of changing into its equalis called re-
ducing the fraction to its lowest terms. It consists in dividing
both terms of the fraction by any “umber which will divide them
both without a remainder, and the quotient thence arising in the
same manner, and so on, till it appears that no number greater
than 1 will again divide them.
*
A number, which will divide two or more numbers with-
out a remainder, is called a common divisor, o common mea-
sure of those numbers. The greatest number that will do
this is called the greatest common divisor.
1

1
1
1
ست
106
1. What part of an acre are 128 rods ?
One rod is
acre.
of an acre, and 128 rods are 18 of an
We
Let us reduce this fraction to its lowest terms.
find, by trial, that 4 will exactly measure both 128 and 160,
and, dividing, we change the fraction to its equal 23. Again,
we find that 8 is a divisor common to both terms, and, di-
viding, we reduce the fraction to its equal 4, which is now
in its lowest terms, for no greater number than 1 will again
measure them. The operation may be presented thus:
4) 1280
8)
128) 160(1
128
32
}
FRACTIONS.
}
160 40 5
32)128(4
128
f
2. Reduce 48,7, 148, and 16 to their lowest terms.
Ans.,,, and . '
of an acre, Answer.
Note. If any number ends with a cipher, it is evidently
divisible by 10. If the two right hand figures are divisible
by 4, the whole number is also. If it ends with an even
number, it is divisible by 2; if with a 5 or 0, it is divisible
by:5.
3. Reduce 488, 4, 145, and 31 to their lowest terms.
I
T 46, 47

T47. Any fraction may evidently be reduced to its lowest
terms by a single division, if we use the greatest common
divisor of the two terms. The greatest common measure of
ny two numbers may be found by a sort of trial easily made.
Let the numbers be the two terms of the fraction 128. The
common divisor cannot exceed the less number, for it must
measure it. We will try, therefore, if the less number, 128,
which measures itself, will also divide or measure 160.
128 in 160 goes 1 time, and 32 re-
main; 128, therefore, is not a divisor of
160. We will now try whether this re-
mainder be not the divisor sought; for if
32 be a divisor of 128, the former divi-
sor, it must also be a divisor of 160,
which consists of 12832. 32 in 128 goes 4 times, with-
out any remainder. Consequently, 32 is a divisor of 128 and
160. And it is evidently the greatest common divisor of
these numbers; for it must be contained at least once more in
160 than in 128, and no number greater than their difference,
that is, greater than 32, can do it.
J
* *
¶ 47, 48.
107
Hence the rule for finding the greatest common divisor of
two numbers :-Divide the greater number by the less, and
that divisor by the remainder, and so on, always dividing
the last divisor by the last remainder, till nothing remain.
The last divisor will be the greatest common divisor required.
/
Note. It is evident, that, when we would find the greatest
common divisor of more than two numbers, we may first find
the greatest common divisor of two numbers, and then of
that common divisor and one of the other numbers, and so
on to the last number. Then will the greatest common
divisor last found be the answer.
1
4. Find the greatest common divisor of the terms of the
fraction, and, by it, reduce the fraction to its lowest terms.
OPERATION.
21)35(1
21
14)21 (1
14
Greatest divis. 7)14(2.
14
↓
FRACTIONS.
1
1
toss
6. Reduce 384 to its lowest terms.
1192
114.
85
7. Reduce 1 to its lowest terms.
8. Reduce 468 to its lowest terms.
1184
9. Reduce 43g to its lowest terms.
Then, 7)21
35
Ans. 1.
5. Reduce to its lowest terms.
96
541
Note. Let these examples be wrought by both methods;
by several divisors, and also by finding the greatest common
divisor.
3
5
!
"
Ans.
To divide a fraction by a whole number.
T 48. 1. If 2 yards of cloth cost of a dollar, what does
1 yard cost? how much is divided by 2?
2. If a cow consume of a bushel of meal in 3 days, how
much is that per day? 3 how much?
4
3. If a boy divide of an orange among 2 boys, how much
will he give each one? 2 how much?
4. A boy bought 5. cakes for 19 of a dollar; what did 1
cake cost? 12÷5 how much?
Ans, f.
Ans.
Ans. 117.
· Ans. 2.
&
4
;
-----
1
{
{
{
i
}
108
48 T
5. If 2 bushels of apples cost of a dollar, what is that
per bushel ?
FRACTIONS.
1 bushel is the half of 2 bushels; the half of & is 1.
(
Ans. dollar.
6. If 3 horses consume 13 of a ton of hay in a month,
what will 1 horse consume in the same time?
}
13 are 12 parts; if 3 horses consume 12 such parts in a
month, as many times as 3 are contained in 12, so many
parts 1 horse will consume.
Ans. of a ton.
7. If g of a barrel of flour be divided equally among 5
families, how much will each family receive?
1
§ is 25 parts; 5 into 25 goes 5 times. Ans. of a barrel.
The process in the foregoing examples is evidently di-
viding a fraction by a whole number; and consists, as may
be seen, in dividing the numerator, (when it can be done
without a remainder,) and under the quotient writing the
denominator. But it not unfrequently happens, that the nu-
merator will not contain the whole number without a re-
mainder.
8. A man divided of a dollar equally among 2 persons;
what part of a dollar did he give to each?
of a dollar divided into 2 equal parts will be 4ths.
Ans He gave
of a dollar to each.
9. A mother divided & a pie among 4 children; what part
of the pie did she give to each ? ÷ 4 how much?
10. A boy divided of an orange equally among 3 of his
companions; what was each one's share? ÷ 3 = how
much?
11. A man divided of an apple equally between 2 chil-
dren; what part did he give to each? divided by 2 =
gov
what part of a whole one?
£
is 3 parts: if each of these parts be divided into 2 equal
parts, they will make 6 parts. He may now give 3 parts to
one, and 3 to the other: but 4ths divided into 2 equal parts,
become 8ths. The parts are now twice so many, but they
are only half so large; consequently, & is only half so much
as 4.
Ans. of an apple.
kat
In these last examples, the fraction has been divided by
multiplying the denominator, without changing the numerator.
The reason is obvious; for, by multiplying the denominator
by any number, the parts are made so many times smaller,
since it will take so many more of them to make a whole
1
I
T. 49.
one; and if no more of these smaller parts be taken than
were before taken of the larger, that is, if the numerator be
not changed, the value of the fraction is evidently made so
many times less.
{
Para Make
·
FRACTIONS.
}
1
T 49. Hence, we have two ways to divide a fraction by
a whole number :-
I. Divide the numerator by the whole number, (if it will
contain it without a remainder,) and under the quotient write
the denominator.—Otherwise,
II. Multiply the denominator by the whole number, and
over the product write the numerator.
jangkiting
109
EXAMPLES FOR PRACTICE.
1. If 7 pounds of coffee cost of a dollar, what is that
per pound? ÷ 7 =
21 7 how much? Ans. of a dollar.
2. It 8 of an acre produce 24 bushels, what part of an
acre will produce 1 bushel?
÷ 24 how much?
of a dollar, what is that a
3. If 12 skeins of silk cost
skein? 12 how much?
19
4. Divide by 16.
Note. When the divisor is a composite number, the in-
telligent pupil will perceive, that he can first divide by one
component part, and the quotient thence arising by the
other; thus he may frequently shorten the operation. In
the last example, 168 x 2, and ÷ 8, and §÷2
§
=78.
Ans. To
5. Divide by 12. Divide by 21. Divide by 24.
28
6. If 6 bushels of wheat cost $43, what is it per bushel ?
Note. The mixed number may evidently be reduced to
an improper fraction, and divided as before.
Ans. 188 of a dollar, expressing the fraction in its
lowest terms. ( 46.)
7. Divide $413 by 9.
8. Divide 129 by 5.
9. Divide 14 by 8.
10. Divide 1841 by 7.
Quot. of a dollar.
Quot. 42=24.
Quot. 137.
Ans. 26T
Note. When the mixed number is large, it will be most
convenient, first, to divide the whole number, and then re-
duce the remainder to an improper fraction; and, after di-
viding, annex the quotient of the fraction to the quotient of
J
1
K
1
}
}
j
L
1
i
1
1
110
T 49, 50,
the whole number; thus, in the last example, dividing 1841
by 7, as in whole numbers, we obtain 26 integers, with 24
remainder, which, divided by 7, gives, and 26 +14
26 Ans.
4
11. Divide 27861 by 6.
Ans. 4648.
12. How many times is 24 contained in 764614?
13. How
many times is 3 contained in 4621?
1
1
+
cost ?
FRACTIONS.
To multiply a fraction by a whole number.
T 50. 1. If 1 yard of cloth cost of a dollar, what will
2 yards cust? × 2 = how much?
1
2. If a cow consume of a bushel of meal in 1 day, how
much will she consume in 3 days? × 3 how much?
1
"
>
3. A boy bought 5 cakes, at of a dollar each; what did
he give for the whole?
× 5
how much?
4. How much is 2 times
times &?
I
1
Ans. 318847.
Ans. 154.
?
5. Multiply by 3.
& by 2.
f by 7.
6. If a man spend & of a dollar per day, how much will
he spend in 7 days?
3 times ?
is 3 parts. If he spend 3 such parts in 1 day, he will
evidently spend 7 times 3, that is, 228 in 7 days.
Hence, we perceive, a fraction is multiplied by multiplying the
numerator, without changing the denominator.
S
But it has been made evident, ( 49,) that multiplying the
denominator produces the same effect on the value of the frac-
tion, as dividing the numerator: hence, also, dividing the de-
nominator will produce the same effect on the value of the
fraction, as multiplying the numerator. In all cases, therefore,
where one of the terms of the fraction is to be multiplied, the
same result will be effected by dividing the other; and where
one term is to be divided, the same result may be effected by
multiplying the other.
This principle, borne distinctly, in mind, will frequently
enable the pupil to shorten the operations of fractions. Thus,
in the following example:
At
of a dollar for 1 pound of sugar, what will 11 pounds
2

Multiplying the numerator by 11, we obtain for the pro-
duct of a dollar for the answer.
?
T 51. ›
111
T 51. But, by applying the above principle, and dividing
the denominator, instead of multiplying the numerator, we at
once come to the answer, &, in its lowest terms. Hence,
there are TWO ways to multiply a fraction by a whole number :
I. Divide the denominator by the whole number, (when it
can be done without a remainder,) and over the quotient
write the numerator. Otherwise,
1
}
II. Multiply the numerator by the whole number, and un-
der the product write the denominator. If then it be an
improper fraction, it may be reduced to a whole or mixed
number.
J
36.
EXAMPLES FOR PRACTICE.
1. If 1 man consume of a barrel of flour in a month,
how much will 18 men consume in the same time?
men ?
9 men ?
2. What is the product of
40 how much?
3. Multiply
by 48.
S
!
1
FRACTIONS..
(
Į
by 12.
by 60.
1
Note. When the multiplier is a composite number, the
pupil will recollect, (T 11,) that he may first multiply by
one component part, and that product by the other. Thus,
in the last example, the multiplier 60 is equal to 12 × 5;
therefore, TX 121, and 1 × 5==55, Ans.
13
An 404.
4. Multiply 52 by 7.
Note. It is evident, that the mixed number may be re-
duced to an improper fraction, and multiplied, as in the pre-
ceding examples; but the operation will usually be shorter,
to multiply the fraction and whole number separately, and
add the results together. Thus, in the last example, 7 times
5 are 35; and 7 times are 151, which, added to 35,
make 401, Ans.
Or, we may multiply the fraction first, and, writing down
the fraction, reserve the integers, to be carried to the product
of the whole number.
L
5. What will 943 tons of hay come to at $17 per ton?
Ans. $164.
6. If a man travel 2 miles in 1 hour, how far will he
Ans. to the last, 1 barrels.
multiplied by 40? 72% X
Ans. 234.
by
by 18.
by 21.
$
1
1
I
1
J
1
112
1
in 8 hours?
L
travel in 5 hours?
in 12 hours?
in 3 days, supposing he travel 12 hours each day?
Ans. to the last, 772 miles.
Note. The fraction is here reduced to its lowest terms;
the same will be done in all following examples.
1
}
FRACTIONS.
To multiply a whole number by a fraction,
C
¶ 52. 1. If 36 dollars be paid for a piece of cloth, what
costs of it? 36 X how much?
4
of the quantity will cost of the price; a time 36 dol-
lars, that is, of 36 dollars, implies that 36 be first divided
into 4 equal parts, and then that 1 of these parts be taken 3
times; 4 into 36 goes 9 times, and 3 times 9 is 27.
Ans. 27 dollars.
¶ 51, 52.
}
From the above example, it plainly appears, that the ob-
ject in multiplying by a fraction, whatever may be the multipli-
cand, is, to take out of the multiplicand a part, denoted by the
multiplying fraction; and that this operation is composed of
two others, namely, a division by the denominator of the
multiplying fraction, and a multiplication of the quotient by
the numerator. It is matter of indifference, as it respects
the result, which of these operations precedes the other, for
36 X 3427, the same as 36 4 X 3 = 27.
}
1


Hence, To multiply by a fraction, whether the multiplicand
be a whole number or a fraction,—
1
RULE.
Divide the multiplicand by the denominator of the multi-
plying fraction, and multiply the quotient by the' numerator;
or, (which will often be found more convenient in practice,)
first multiply by the numerator, and divide the product by
the denominator.
Multiplication, therefore, when applied to fractions, does
not always imply augmentation or increase, as in whole
numbers; for, when the multiplier is less than unity, it will
always require the product to be less than the multiplicand,
to which it would be only equal if the multiplier were 1.
We have seen, (TT 10,) that, when two numbers are multi-
plied together, either of them may be made the multiplier,
without affecting the result. In the last example, therefore,
instead of multiplying 16 by, we may multiply 2 by 16,
(T 50,) and the result will be the same.
J
t
{
T 52, 53.
FRACTIONS.
EXAMPLES FOR PRACTICE.
2. What will 40 bushels of corn come to at & of a dollar
per bushel? 40 X how much?
2
朵
​1
3. What will 24 yards of cloth cost at of a dollar per
yard? 24 X & how much?
fo of 45?
4. How much is of 90?
4
5. Multiply 45 by
To multiply one fraction by another.
53. 1. A man, owning of a ticket, sold
share; what part of the whole ticket did he sell?
how much?
IN
1
}
تر
We have just seen, (T 52,) that, to multiply by a fraction,
is to divide the multiplicand by the denominator, and to multi-
ply the quotient by the numerator. divided by 3, the de-
nominator of the multiplying fraction, (49,) is, which,
multiplied by 2, the numerator, ( 51,) is,
Ans.
The process, if carefully considered, will be found to con-
sist in multiplying together the two numerators for a new nu-
merator, and the two denominators for a new denominator.
& of 369?
Multiply 20 by 1.
1
EXAMPLES FOR PRACTICE,
2. A man, having of a dollar, gave of it for a dinner;
what did the dinner cost him?
Ans. dollar.
Product,
TJ
3. Multiply by . Multiply by .
4. How much is 4 of 3 of 7 of 2 ?
C
4
113
of his
of is
Note. Fractions like the above, connected by the word
of, are sometimes called compound fractions. The word of
implies their continual multiplication into each other.
Ans. 16876.
When there are several fractions to be multiplied con-
tinually together, as the several numerators are factors of the
new numerator, and the several denominators are factors of
the new denominator, the operation may be shortened by
dropping those factors which are the same in both terms, on the
principle explained in T 46. Thus, in the last example,
3, 3, 4, we find a 4 and a 3 both among the numerators and
1
1
1
5
among the denominators; therefore we drop them, multiply-
ing together only the remaining numerators, 2 x 714, for
a new numerator, and the remaining denominators, 5 X 8=
K*
40, for a new denominator, making 13, Ans. as before.
!
1
J
7
{
I
}
114
¶ 53, 54.
5. of of of of of of how much? Ans. 7.
6. What is the continual product of 7, 1, of § and 33?
2
Note. The integer 7 may be reduced to the form of an
improper fraction by writing a unit under it for a denomina-
tor, thus, 7.
Ans. 211'
FRACTIONS.
7. At of a dollar a yard, what will of a yard of cloth
2
cost?
8. At 68 dollars per barrel for flour, what will of a bar-
rel cost?
6; then
× 7=1= $21}}, Ans.
9. At & of a dollar per yard, what cost 7 yards?
10. At $2 per yard, what cost 6ğ yards?
11. What is the continued product of 3, 3,
H of § of $ ?
1
I
T 64. The RULE for the multiplication of fractions may
now be presented at one view :—
{
I. To multiply a fraction by a whole number, or a whole
number by a fraction,-Divide the denominator by the whole
number, when it can be done without a remainder; other-
wise, multiply the numerator by it, and under the product
write the denominator, which may then be reduced to a
whole or mixed number.
+
I
II. To multiply a mixed number by a whole number,-Multi-
ply the fraction and integers, separately, and add their pro-
ducts together.
III. To multiply one fraction by another, Multiply together
the numerators for a new numerator, and the denominators for
a new denominator.
Note. If either or both are mixed numbers, they may first
be reduced to improper fractions.
Aplik
Ans. $611.
Ans. $ 1438.
of 2, 24, and
Ans. 48.
EXAMPLES FOR PRACTICE.
1. At $ per yard, what cost 4 yards of cloth ?
yds. ?
6 yds.? 8 yds. ? 20 yds. ?
2. Multiply 148 by . —by f
Ans. to the last, $15.
by 2. by
Last product, 44.
3. If 2% tons of hay keep 1 horse through the winter,,
++
A
#
?
}
¶ 54, 55.
how much will it take to keep 3 horses the same time?
7 horses?
13 horses?
4. What will 8
bárrel?
FRACTIONS.
15
?
5. At $142 per cwt., what will be the
6. A owned & of a ticket; B owned
ticket was so lucky as to draw a prize of
each one's share of the money
7. Multiply of 3 by of t
8. Multiply 7 by 275
9. Multiply by 23.
10. Multiply 2 of 6 by .
11. Multiply of 2 by of 4.
$244
12. Multiply continually together
and ✈ of 10.
+
13. Multiply 1000000 by §.
Ans. to last, 377 tons.
barrels of cider come to, at $3 per
£
OPERATION:
t
Į
Numerator, 3)40
}
To divide a whole number by a fraction.
*
¶ 55. We have already shown (T 49) how to divide a
fraction by a whole number; we now proceed to show how
to divide a whole number by a fraction.
1. A man divided $9 among some poor people, giving
them of a dollar each; how many were the persons who
received the money? 9 9 how many?
good
1 dollar is 4, and 9 dollars is 9 times as many, that is, 36;
then is contained in 3 as many times as 3 is contained
in 36.
Ans. 12 persons.
&
That is,-Multiply the dividend by the denominator of the
dividing fraction, (thereby reducing the dividend to parts of
the same magnitude as the divisor,) and divide the product
by the numerator.
*
cost of 147 cwt.?
of the same; the
$1000; what was
2. How many times is contained in 8? 8÷÷÷ how
many?
8 Dividend.
5 Dénominator.
}
Product, f
Product, 151.
Product, 21.
Product, 1.
Product, 3.
of 8, 4 of 7, of 9,
Product, 20.
Product, 555555,
"
115.
1
Quotient, 131 times, the Answers
To multiply by a fraction, we have seen, (T 52,) implies
two operations a division and a multiplication; so, also, to
divide by a fraction implies two operations-a multiplication
and a division.
p
3
x
I
..
1
r3
+
116
51,
}
FRACTIONS.
T56. Division is the reverse of multiplication.
To multiply by a fraction, To divide by a fraction,
whether the multiplicand be whether the dividend be a
a whole number or a fraction, whole number or a fraction,
as has been already shown, we multiply by the denomina-
(1 52,) we divide by the de-tor of the dividing fraction,
nominator of the multiplying and divide the product by the
fraction, and multiply the quo- numerator.
tient by the numerator.
A
===
12 multiplied by 2, the pro- |
duct is 9.
I
1
1
Note. In either case, it is matter of indifference, as it
respects the result, which of these operations precedes the
other; but in practice it will frequently be more convenient,
that the multiplication precede the division.
/
}
+
In multiplication, the mul- In division, the divisor be-
tiplier being less than unity, ing less than unity, or 1, will
or 1, will require the product be contained a greater number
to be less than the multipli- of times; consequently will re-
cand, ( 52,) to which it is quire the quotient to be great-
only equal when the multi-er than the dividend, to which
plier is 1, and greater when it will be equal when the di-
the multiplier is more than 1. visor is 1, and less when the
divisor is more than 1.
}
شده با این
*Cou
EXAMPLES FOR PRACTICE.
1. How many times is contained in 7? 7
금
​many?
2. How many times can I draw
of a cask containing 26 gallons?
3. Divide 3 by 2.
4. If a man drink
will 3 gallons last him
5. If 22 bushels of oats sow an acre, how many acres will
22 bushels sow? 22 24 how many times?
Note. Reduce the mixed number to an improper frac--
tion, 24,
Ans. 8 acres.
6. At $42 a yard, how many yards of cloth may be
bought for $37?
Ans. 8 yards.
7. How many times is
contained in 84?
TOS
T 56.
3,
12 divided by 4, the quo-
tient is 16.
10 by 2.
7
6 by 3.
of a quart of rum a day, how long
t
;"
how
01
of a gallon of wine out
↓
3
Ans. 90 times.
1
}
}
1
D
1
1
T 56, 57.
8. How many times is a contained in 6 ?
3
་་
9. How many times is 8g contained in 53 ì
2 X 5
FRACTIONS.
Ans. 644 times.
10. At of a dollar for building 1 rod of stone wall, how
many rods may be built for $87? 87 how many
times?
To divide one fraction by another.
¶ 57. 1. At & of a dollar per bushel, how much rye may
be bought for g of a dollar? is contained in g how many
용
​times ?
Maga
$
1
Had the rye been 2 whole dollars per bushel, instead of a
of a dollar, it is evident, that of a dollar must have been
divided by 2, and the quotient would have been; but the
divisor is 3ds, and 3ds will be contained 3 times where a
like number of whole ones are contained 1 time; conse-
quently the quotient is 3 times too small, and must there-
o
fore, in order to give the true answer, be multiplied by 3,
that is, by the denominator of the divisor;3 times
bush. Ans.
ΤΟ
10'
Ans. § of 1 time.
117
{
The process is that already described, ¶ 55 and T 56. If
carefully considered, it will be perceived, that the numerator
of the divisor is multiplied into the denominator of the divi-
dend, and the denominator of the divisor into the numerator
of the dividend; wherefore, in practice, it will be more con-
venient to invert the divisor; thus, inverted becomes ;
then multiply together the two upper terms for a numerator, and
the two lower terms for a denominator, as in the multiplication
of one fraction by another. Thus, in the above example,
3 X 3
9
as before.
EXAMPLES FOR PRACTICE.
2. At 4 of a dollar per bushel for apples, how many bush-
els may be bought for of a dollar? How many times is
contained in 7?
Ans. 3 bushels.
3. If of a yard of cloth cost of a dollar, what is that
per yard? It will be recollected, (T 24,) that when the cost
any quantity is given to find the price of a unit, we divide
the cost by the quantity. Thus, (the cost) divided by 7:
(the quantity) will give the price of 1 yard.
of
Ans. of a dollar per yard.
}
}
+
1
1
1
}
118
(
}
FRACTIONS.
I
PROOF. If the work be right, (T 16, "Proof,") the pro-
duct of the quotient into the divisor will be equal to the
dividend; thus, X. This, it will be perceived,
is multiplying the price of one yard (3) by the quantity (7)
to find the cost (;) and is, in fact, reversing the question,
thus, If the price of 1 yard be 34 of a dollar, what will of a
금
​yard cost?
Ans. £ of a dollar.
4. How many bushels of apples, at
bushel, may be bought for of a dollar?
1
Note. Let the pupil be required to reverse and prove the
succeeding examples in the same manner.
7. Divide by . Quot. 3.
་
8. Divide 21 by 14.
5. If 4 pounds of butter serve a family 1 week, how
"many weeks will 367 pounds serve them?
Ans. ST
The mixed numbers, it will be recollected, may be re-
duced to improper fractions.
weeks.
Quot. 2.
Quot. 8.
6. Divide by . Quot. 1.
}
Quot. 11.
9. How many times is
10. How many times is
{
}
11. Divide of by of.
2 4
"
T 57, 58.
contained in ?
contained in 47 2
of a dollar per
Ans. 4 bushels.
Divide by 1.
Divide 7 by.
Divide 10g by 21.
1
T 58. The RULE for division of fractions may now be pre-
sented at one view :-
Quot. 414.
Ans. 4 times.
I. To divide a fraction by a whole number,-Divide the
numerator by the whole number, when it can be done with-
out a remainder, and under the quotient write the denomi-
nator; otherwise, multiply the denominator by it, and over the
product write the numerator.
II. To divide a whole number by a fraction,-Multiply the
dividend by the denominator of the fraction, and divide the
product by the numerator.
III. To divide one fraction by another,-Invert the divisor,
and multiply together the two upper terms for a numerator,
and the two lower terms for a denominator.
Note. If either or both are mixed numbers, they may be
reduced to improper fractions.
Ans. 11g times.
Quot. 4.
}
2
T 59. ADDITION AND SUBTRACTION OF FRACTIONS. 119.
•
EXAMPLES FOR PRACTICE.
P
1. If 7 lb. of sugar cost of a
pound?÷7: how much?
ΤΟΥ
2. At $ for & of a barrel of cider, what is that per bar-
rel?
3. If 4 pounds of tobacco cost of a dollar, what does 1,
pound cost?
4. If 4 of a yard cost $4, what is the price per yard ?
5. If 14 yards cost $75, what is the price per yard?
Ans. 5.
barrels of cider, what is that
Ans. $3.
Ans. 1989.
of ..
Quot. 382.
Quot. 18.
Quot. 3.
6. At 31 dollars for 10
per barrel ?
7. How many times is contained in 746 ?
8. Divide of by 2.
1
Divide by
Quot. &.
9. Divide of 4 by § of .
#
10. Divide of 4 by
1-/-/-5
11. Divide 45 by § of 4.
12. Divide of 4 by 48.
$
dollar, what is it per
of is how much?
783%
Septembe
P
1
ADDITION AND SUBTRACTION OF FRACTIONS.
T 59. 1. A boy gave to one of his companions
orange, to another, to another; what part of an
did he give to all? z + ↑ + ž how much?
2. A cow consumes, in 1 month, of a ton of hay; a
horse, in the same time, consumes of a ton; and a pair
of oxen, ; how much do they all consume? how much
more does the horse consume than the cow?
the oxen
4.
T5
than the horse? &+#+
how much?
how much? how much?
☆☆
TS
3
Quot. 22.
Quot. 2.
Be my de
Bangla Val
1
of an
orange
Ans. f.
P
>
f T5
´3. }+{+}= how much?
- how much?
4. 2% +2%+2%+1+2=how much? A =
#
how much?
5. A boy, having of an apple, gave of it to his sister;
what part of the apple had he left?
how much?
120€
1
ADDITION AND SUBTRACTION OF FRACTIONS. T 60.
When the denominators of two or more fractions are alike,
(as in the foregoing examples,) they are said to have a common
denominator. The parts are then in the same denomina-
tion, and, consequently, of the same magnitude or value. It
is evident, therefore, that they may be added or subtracted,
by adding or subtracting their numerators, that is, the num-
ber of their parts, care being taken to write under the re-
sult their proper denominator. Thus, += ; 8 - 4
= f.
6. A boy, having an orange, gave of it to his sister, and
of it to his brother; what part of the orange did he give
away?
4ths and 8ths, being parts of different magnitudes, or value,
cannot be added together. We must therefore first reduce
them to parts of the same magnitude, that is, to a common de-
nominator. are 3 parts. If each of these parts be divided
into 2 equal parts, that is, if we multiply both terms of the
fraction by 2, (T 46,) it will be changed to §; then § and
f are f.
Ans. 1 of an orange.
Here, 3ds cannot be so divided as to become 5ths, nor
can 5ths be so divided as to become 2ds; but if the 3ds be
each divided into 5 equal parts, and the 5ths each into 3
equal parts, they will all become 15ths. The will become
18, and the & will become; then, taken from 12 leaves
TE, Ans.
구
​7. A man had of a hogshead of molasses in one cask,
and g of a hogshead in another; how much more in one
cask than in the other?
1 60. From the very process of dividing each of the
parts, that is, of increasing the denominators by multiplying
them, it follows, that each denominator must be a factor of
the common denominator; now, multiplying all the denomina-
tors together will evidently produce such a number.
Hence, To reduce fractions of different denominators to
equivalent fractions, having a common denominator,-RULE:
Multiply together all the denominators for a common denomi-
nator, and, as by this process each denominator is multiplied
by all the others, so, to retain the value of each fraction,
multiply each numerator by all the denominators, except its
· own, for a new numerator, and under it write the common
denominator,
•
4
I
3
T 60. ADDITION AND SUBTRACTION OF FRACTIONS.
{
EXAMPLES FOR PRACTICE.
1. Reduce, and to fractions of equal value, having a
common denominator.
[
3 X 4 X 560, the common denominator.
X
2 × 4 × 5 = 40, the new numerator for the first fraction.
3 × 3 × 5 = 45, the new numerator for the second fraction.
3 X 4 X 4 = 48, the new numerator for the third fraction.
601
The new fractions, therefore, are 48, 43, and 4. By an
inspection of the operation, the pupil will perceive, that the
numerator and denominator of each fraction have been mul-
tiplied by the same numbers; consequently, (T 46,) that
their value has not been altered.
121
1
8
160
192
2. Reduce,, and to equivalent fractions, having a
common denominator.
Ans. 129, 149, 218, 113.
243,
3. Reduce to equivalent fractions of a common denomi-
nator, and add together, 3, 3, and 4.
Ans. 33 +33 +25=73=133, Amount.
Amount, 117.
4. Add together 2 and §.
5. What is the amount of + ÷ + ÷ + ÷ ?
줌 ​ㅎ
​Ω
T2
Ans. 243 = 1.37.
6. What are the fractions of a common denominator
equivalent to and &? Ans. 18 and 22, or and 12.
We have already seen, (T 59, ex. 7,) that the common
denominator may be any number, of which each given de-
nominator is a factor, that is, any number which may be di-
vided by each of them without a remainder. Such a number
is called a common multiple of all its common divisors, and
the least number that will do this is called their least com-
mon multiple; therefore, the least common denominator of any
fractions is the least common multiple of all their denominators.
Though the rule already given will always find a common
multiple of the given denominators, yet it will not always
find their least common multiple. In the last example, 24
is evidently a common multiple of 4 and 6, for it will exactly
measure both of them; but 12 will do the same, and as 12 is
the least number that will do this, it is the least common
multiple of 4 and 6. It will therefore be convenient to have
a rule for finding this least common multiple. Let the num
bers be 4 and 6.
9
It is evident, that one number is a multiple of another,
when the former contains all the factors of the latter. The
L
I
ご
​1
1
1
{
{
'
›
1
1
惺
​L
}*
3
}
..
122
ADDITION AND SUBTRACTION OF FRACTIONS. T 61.
}
factors of 4 are 2 and 2, (2 × 24.) The factors of 6
are 2 and 3, (2 X 36.) Consequently, 2 X 2 × 3 = 12
contains the factors of 4, that is, 2 X 2; and also contains the
factors of 6, that is. 2 × 3. 12, then, is a common multiple
of 4 and 6, and it is the least common multiple, because it
does not contain any factor, except those which make up.
the numbers 4 and 6; nor either of those repeated more
than is necessary to produce 4 and 6. Hence it follows,
that when any two numbers have a factor common to both,
it
may be once omitted; thus, 2 is a factor common both to
4 and 6, and is consequently once omitted.
2) 4
}
•
T61. On this principle is founded the RULE for finding
the least common multiple of two or more numbers. Write
down the numbers in a lite, and divide them by any number
that will measure two or more of them; and write the quo-
tients and undivided numbers in a line beneath. Divide this
line as before, and so on, until there are no two numbers
that can be measured by the same divisor; then the conti-
nual product of all the divisors and numbers in the last line
will be the least common multiple required.
Let us apply the rule to find the least common multiple
of 4 and 6.
Ś
!
1
2.3
4 and 6 may both be measured by 2; the
quotients are 2 and 3. There is no number great-
er than 1, which will measure 2 and 3. There-
fore, 2 × 2 × 312 is the least common mul-
tiple of 4 and 6.
{
If the pupil examine the process, he will see that the di-
visor 2 is a factor common to 4 and 6, and that dividing 4
by this factor gives for a quotient its other factor, 2. In the
same manner, dividing 6 gives its other factor, 3. Therefore
the divisor and quotients make up all the factors of the two
numbers, which, multiplied together, must give the com-
mon multiple.
7. Reduce ‡, 2, and
least common denominator.
OPERATIÓN,
2) 4. 2. 3. 6
(3) 2
f 3. A
}
2.1.1
•
I
15
1
to equivalent fractions of the
Then, 2 X3 X 2 = 12, least common
denominator. It is evident we need
not multiply by the 1s, as this would
not alter the number.
1
1
I
H
}
}
!
1
T61. ADDITION AND SUBTRACTION OF FRACTIONS. 123
To find the new numerators, that is, how many 12ths
each fraction is, we may take 2, 1, 4 and † of 12. Thus:
of 12
= 9
6
of 12
& of 128
† of 12 = 2
7 = 718
138=138号
​1571521
1
। .
34
348
123 = 1240
{{
+2=3
3 = f
2
Ans. z, fz, and
6 8 2.
15
T
8)
8. Reduce 2, §, and & to fractions having the least com-
mon denominator, and add them together.
Ans. 12 +2 +32=41=144, amount.
9. Reduce and to fractions of the least common de-
nominator, and subtract one from the other.
Ans. 18-18 Ts, difference.
10. What is the least number that 3, 5, 8 and 10 will
measure?
Ans. 120.
*
11
New numerators, which,
written over the com-
mon denominators, give
11. There are 3 pieces of cloth, one containing 7 yards,
another 13 yards, and the other 15 yards; how many
yards in the 3' pieces.
24
Before adding, reduce the fractional parts to their least
common denominator; this being done, we shall have,
Adding together all the 24ths, viz. 18+ 20
+21, we obtain 59, that is, 24 — 211.
We write down the fraction 11 under the
other fractions, and reserve the 2 integers
to be carried to the amount of the other
integers, making in the whole 3711, Ans.
12. There was a piece of cloth containing 343 yards,
from which were taken 124 yards; how much was there
left ?
Ans. 3711
+=1
12
£= 1
}
At
We cannot take 16 twenty-fourths
(14) from 9 twenty-fourths, (;) we
must, therefore, borrow I integer, 24
Ans. 2117 yds.
twenty-fourths, (,) which, with,
makes 2; we can now take 1 from
2, and there will remain 17; but, as we borrowed, so also
we must carry 1 to the 12, which makes it 13, and 13 from
34 leaves 21.
Ans. 2117.
13. What is the amount of of 3 of a yard, of a yard,
and 1 of 2 yards ?
I
Note. The compound fraction may be reduced to a sim-
ple fraction; thus, of; and of 2; then, +
1 }
3 + 4 = 138=1 yds., Ans.
53
→
fi
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F
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124
π 62, 63.
¶ 62. From the foregoing examples we derive the fol-.
lowing RULE:-To add or subtract fractions, add or subtract..
their numerators, when they have a common denominator;
otherwise, they must first be reduced to a common denomi-
nator.
}
Note. Compound fractions must be reduced to simple
fractions before adding or subtracting.
}
}
5. From 14 take 2.
6. From 3 take .
1
REDUCTION OF FRACTIONS.
1. What is the amount of §, 43 and 12?
2. A man bought a ticket, and sold of
of the ticket had he left?
A
1
EXAMPLES FOR PRACTICE.
3. Add together,,, 75, and 1.
4. What is the difference between 14
1
7. From 147 take 484.
8. From of take 1 of 47.
To
9. Add together 112, 311, and 1000₫.
10. Add together 14, 11, 43, and 1.
11. From take. From take 3.
Ans. 17.
of it; what part
Ans.
Amount, 223.
and 167?
•
}
Ans. 118.
Remainder, .
{
and †?
12. What is the difference between
f 륭​? 씁 ​j f
and ? and ? and ? § and 2 ?
&
3
13. How much is 1. #? 1-? 1-2 1-f?.
2-4? 2-4? 24-37 34
21-
To?
Rem. 23.
Rem. 983.
Rem. 476.
?
1000?
i
and ?
REDUCTION OF FRACTIONS.
T63. We have seen, ( 33,) that integers of one denomi-
nation may be reduced to integers of another denomination.
It is evident, that fractions of one denomination, after the
same manner, and by the same rules, may be reduced to
fractions of another denomination; that is, fractions, like
integers, may be brought into lower denominations by mul
tiplication, and into bigher denominations by division.
}
I
1
·
T 63.
REDUCTION OF FRACTIONS.
125
To reduce higher into LOWER To reduce lower into HIGHER
denominations.
denominations.
1
(RULE. See T34.)
1. Reduce
O'
(RULE. See ¶ 34.)
of a pound 2. Reduce & of a penny to
to pence, or the fraction of a the fraction of a pound.
penny.
Note. Division is perform-
Note. Let it be recollect- ed either by dividing the nu-
ed, that a fraction is multiplied merator, or by multiplying the
either by dividing its denomi-denominator.
nator, or by multiplying its nu-|·
12s. ÷ 20—
merator.
Or thus: 5 of 20 of 12 =
of a penny, Ans.
1'
28J
219
§
zło £ × 20 = 14 3. X 12
d. Ans.
"
T280
I
3. Reduce of a pound
to the fraction of a farthing.
1283 £.'X 2028 s. X
12=28 d. X 4 = 28% =
29.
JJ
249
460
1280
1
Tag
Or thus:
Num. 1
20 s. in 1 £.
20
12 d. in 1 s.
240
4 q. in 1 d.
960
1
§ d.
zo £. Ans.
Or thus:
68=zo £. Ans.
2
2
Or thus:
Denom. 4
{
4. Reduce of a farthing
to the fraction of a pound.
20. ÷ 4
a.
32s.+20=40=1280£.
d. ÷ 12
T6
of of %=
4 q. in 1 a.
192
16
12 d. in 1 s.
20 s. in 1 £.
3840
3840
Then, 12'80 £. Ans.
5
2688
Then, 2019. Ans.
5. Reduce of a guinea
to the fraction of a peuny.
7. Reduce 4 of a guinea to
the fraction of a pound.
Consult T 34, ex. 11.
9. Reduce of a móidore, at
36 s. to the fraction of a guinea. to the fraction of a moidore.
11. Reduce of a pound, 12. Reduce of an ounce
10. Reduce of a guinea
37
亨
​>
Troy, to the fraction of an to the fraction of a pound
ounce.
Troy. ••
L*
6. Reduce of a penny to
the fraction of a guinea,
8. Reduce of a pound to
the fraction of a guinea.
1
}
i
I
C
+
1
126
13. Reduce
of a pound, 14. Reduce
avoirdupois, to the fraction of to the fraction
'an ounce.
avoirdupois.
17. A cucumber grew to the
length of ʊ of a mile; what
part is that of a foot?
REDUCTION OF FRACTIONS.
#
T3
15. A man has 7 of a 16. A man has of a pint
hogshead of wine; what part of wine; what part is that of
is that of a pint ?
a hogshead?
18. A cucumber grew to
the length of 1 foot 4 inches
==of a foot; what part
is that of a mile?
16
V
I
[
7
19. Reduce of & of a
pound to the fraction of a shil-what fraction of a pound?
29. 24 of a shilling is & of
I
ling.
非常
​TI
TI
21. Reduce of of 3 22. 18 of a penny is of
pounds to the fraction of a what fraction of 3 pounds?
penny.
182 of a penny is of what
part of 3 pounds? 18 of a
penny is off of how many
pounds?
f
1
T 63, 64.
of an ounce
of a pound
¶ 64. It will frequently be It will frequently be re-
required to find the value of a quired to reduce integers to
fraction, that is, to reduce a the fraction of a greater de-
fraction to integers of less de-
nomination.
nominations.
1
2. Reduce 13 s. 4 d. to the
fraction of a pound.
1. What is the value of
of a pound? In other words,
Reduce & of a pound to shil-
lings and pence.
60
13 s. 4 d. is 160 pence;
there are 240 pence in a
of a pound is 413 shil-pound; therefore, 13 s. 4 d. is
†
lings; it is evident from of 159 of a pound. That
a shilling may be obtained is,-Reduce the given sum or
some pence; of a shilling is quantity to the least denomina-
12 4d. That is,-Multiply tion mentioned in it, for a nu-
the numerator by that mumber merator; then reduce an inte
which will reduce it to the next ger of that greater denomina-
less denomination, and divide tion (to a fraction of which it
the product by the denominator; is required to reduce the giv-
if there be a remainder, multiply en sum or quantity) to the
and divide as before, and so on; same denomination, for a denomi
the several quotients, placed one nator, and they will form the
after another, in their order, fraction required.
[
will be the answer »,
I
A
1
+
1


T 64.
3. What is the value of
of a shilling?
Numer. 3
12
REDUCTION OF FRACTIONS.
EXAMPLES FOR PRACTICE. EXAMPLES FOR PRACTICE.
4. Reduce 4 d. 2 q. to the
fraction of a shilling.
OPERATION.
I
OPERATION.
Denom. 8)36(4 d. 2 q. Ans.
32
4.
16(2 q.
16.
5. What is the value of f
of a pound Troy?
7. What is the value of §
of a pound avoirdupois?
13. Reduce of an acre
to its proper quantity.
15. What is the value of
1
¿
9
17. What is the value of
4 d. 2 q-
4
18 Numer.
A
}
6. Reduce 7 oz. 4 pwt. to
the fraction of a pound Troy.
8. Reduce 8 oz. 143 dr. to
the fraction of a pound avoir-
dupois.
Note. Both the numerator
and the denominator must be
reduced to 9ths of a dr.
9. 4 of a month is how ma-
10. 3 weeks, 1 d. 9 h. 36 m.
ny days, hours, and minutes? is what fraction of a month?
11. Reduce of a mile to
its proper quantity..
1 s.
12
12
18 of a dollar in shillings, the fraction of a dollar.
pence, &c. ?
Note. Let the pupil be required to reverse and
following examples:
21.. What is the value of of a guinea?
"
127
4
48 Denom
12. Reduce 4 fur. 125 yds.
2 ſt. 1 in. 24 bar. to the frac-
tion of a mile.
14. Reduce 1 rood 30 poles
to the fraction of an acre.
of a yard?
19. What is the value of 20. Reduce 4 cwt. 2
鲞
​of a ton?
lb. 14 oz. 12
tion of a ton.
ZE
& Ans.
16. Reduce 5 s. 7 d. to
18. Reduce 2 ft. 8 in. 14 b.
to the fraction of a yard.
dr. to the frac-
qr. 12
prove the
"
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1
7
1
1
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7
لیه السلام اسود
སྙ་
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128
1,64.
22. Reduce 3 roods 17 poles to the fraction of an acre.
23. A man bought 27 gal. 3 qts. 1 pt. of molasse3; what
part is that of a hogshead?
'SUPPLÉMENT TO FRACTIONS.
•
}
24. A man purchased † of 7 cwt. of sugar; how much
sugar did he purchase
25. 13 h. 42 m. 51 s. is what part or fraction of a day?
SUPPLEMENT TO FRACTIONS.
QUESTIONS.
1
1. What are fractions? 2. Whence is it that the parts
into which any thing or any number may be divided, take
their name? 3. How are fractions represented by figures?
4. What is the number above the line called?-Why is it
so called? 5. What is the number below the line called ?
}
}
[
a
-Why is it so called?-What does it show? 6. What is
it which determines the magnitude of the parts?-Why?
7. What is a simple or proper fraction? an improper
fraction? a mixed number? 8. How is an improper
fraction reduced to a whole or mixed number? 9. How is
a mixed number reduced to an improper fraction ?
whole number? 10. What is understood by the terms of the
fraction? 11. How is a fraction reduced to its most simple
or lowest terms? 12. What is understood by a common di-
visor? by the greatest common divisor? 13. How is
it found? 14. How many ways are there to multiply a frac-
tion by a whole number? 15. How does it appear, that di-
viding the denominator multiplies the fraction? 16. How is a
mixed number multiplied? 17. What is implied in multi-
plying by a fraction? 18. Of how many operations does it
consist? What are they? 19. When the multiplier is less
than a unit, what is the product compared with the multi-
plicand? 20. How do you multiply a whole number by
a fraction? 21. How do you multiply one fraction by ano-
ther? 22. How do you multiply a mixed number by a
mixed number? 23. How does it appear, that in multiply-
ing both terms of the fraction by the same number the value
of the fraction, is not altered? 24. How many ways are
there to divide a fraction by a whole number?What are
they? 25. How does it appear that a fraction is divided by
multiplying its denominator? 26. How does dividing by a
}
I
}
T 64, 65.
129
fraction differ from multiplying by a fraction? 27. When
the divisor is less than a unit, what is the quotient compared
with the dividend? 28. What is understood by a common
denominator? the least common denominator? 29.
How does it appear, that each given denominator must be a
factor of the common denominator? 30. How is the com-
mon denominator to two or more fractions found? 31.
What is understood by a multiple? by a common multi-
by the least common multiple?—What is the pro-
cess of finding it? 32. How are fractions added and sub-
tracted? 33. How is a fraction of a greater denomination
reduced to one of a less? — of a less to a greater 34.
How are fractions of a greater denomination reduced to in-
tegers of a less ? integers of a less denomination to the
fraction of a greater?
ple?
}
SUPPLEMENT TO FRACTIONS.
EXERCISÉS.
1. What is the amount of and ?
of 121, 33 and 43 ?
2. To
Note.
of a pound add & of a shilling.
First reduce both to the same denomination.
3. § of a day added to of an hour make how many
hours? what part of a day?
Ans. to the lust, §& d.
1
4. Add lb. Troy to 7 of an oz.
I
Amount, 6. oz. 11 pwt. 16 gr.
5. How much is less??? A-B? 141
44? 648? 128 of of ?
2 3 2
T4
6. From shilling take
7. From of an ounce take
1
of and ?
Ans. to the last, 2011.
Amount, 184 s.
W
of a penny.
of a pwt.
Ans. to the last, 198.
Rem. 51 d.
d. 9
8. From 4 days 7 hours take 1
9. At $§ per yard, what costs
T 65. The price of unity, or 1, being given, to find the
cost of any quantity, either less or more than uuity, multiply
the price by the quantity. On the other hand, the cost of any
quantity, either less or more than unity, being given, to find
the price of unity, or 1, divide the cost by the quantity.
Ans. $ H
Rem. 11 pwt. 3 grs.
h.
TA
Rem. 2 d. 22 h. 20 m. ›
of a yard of cloth?
1
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1
1
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+
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r
130
1. If lb. of sugar cost
a pound cost ?*
This example will require two operations: first, as above,
to find the price of 1 lb.; secondly, having found the price
of 1 lb., to find the cost of of a pound. s. ÷ 11 (11
of 758. ¶ 57) s. the price of 1 lb.
P
70'5
91
18 (13 of s. 153)=4912 s.
Then, s. X
4 d. 34971 q, the
91
8. If
SUPPLEMENT TO FRACTIONS.
¶ 65.
of a shilling, what will 3 of
T5
Answer.
$
T5
T3
13
Or we may reason thus: first to find the price of 1 lb. :
1lb. costs 7 S. If we knew what 7 lb. would cost,
we might repeat this 13 times, and the result would be the
price of 1 lb. 1 is 11 parts. If 1 lb. costs 75 s., it is evi-
dent .lb. will cost of T'S., and th. will cost
13 times as much, that is, s. the price of 1 lb. Then,
43 of Pots. = 793 s., the cost of of a pound. 4873 s.
4 d. 34931 q., as before. This process is called solving
the question by analysis.
୮
T65
T65
1
After the same manner let the pupil solve the following
questions:
A
2. If 7 lb. of sugar cost of a dollar, what is that a
pound? of how much? What is it for 4 lb. ?
for 4 lb. ? 4.
of how much?
&
how much? What for 12 pounds
What for 12 pounds?of= how
much?
Ans. to the last, ^$ 14.
3. If 6 yds. of cloth cost $3, what cost 9
yards?
Ans. $4'269.
4. If 2 oz. of silver cost $2'24, what costs
oz. ?
Ans. $84.
5. If 4 oz. costs $12, what costs 1 oz.? Ans. $1'283.
6. If lb. less by costs 13 d., what costs 14 lb. less by
of 2.1b: ?
Ans. 4£.9 s. 9 d.
yds. cost?
7. If yd. cost $3, what will 40
Ans. $59'062 +.
of her worth?
Ans. $53 785 +.
cost?
6
of a ship costs $251, what
f
9. At 2§£. per cwt., what will 93 lb.
10. A merchant, owning of a vessel,
for $957; what was the vessel worth?
11. If yds. cost £., what will
}
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16
is
"
T
This and the following are examples usually referred to the rule Proportion,
or Rule of Three. See ¶ 95 cx. 35.
36
sold
Ans. 6 s. 3 d.
of his share
Ans. $1794'375.
of an ell Eng, cost?
Ans. 17 s. 1 d. 29 q.
暑
​I
f
>
↓
¶ 65.
131
J
12. A merchant bought a number of bales of velvet, each
containing 12917 yards, at the rate of $7 for 5 yards, and
sold them out at the rate of $11 for 7 yards, and gained
$200 by the bargain; how many bales were there?
First find for what he sold 5 yards; then what he gained
on 5 yards-what he gained on 1 yard. Then, as many times
as the sum gained on 1 yd. is contained in $200, so many
yards there must have been, Having found the number of
yards, reduce them to bales.
Ans: 9 bales.
13. If a staff, 5 ft. in length, cast a shadow of 6 feet, how
high is that steeple whose shadow measures 153 feet?
Ans. 1441 feet.
SUPPLEMENT TO FRACTIONS.
f
14. If 16 men finish a piece of work in 28 days, how
long will it take 12 men to do the same work?
First find how long it would take 1 man to do it; then 12
men will do it in of that time.
Ans. 377 days.
15. How many pieces of merchandise, at 20 s. apiece,
must be given for 240 pieces, at 124 s. apiece? Ans. 149.
16. How many yards of bocking that is 14 yd. wide will
be sufficient to line 20 yds. of camlet that is of a yard
wide ?
2
авто за
P
First find the contents of the camlet in square measure
then it will be easy to find how many yards in length of
bocking that is 11 yd. wide it will take to make the same
quantity.
Ans. 12 yards of camlet.
17. If 14 yd. in breadth require 20 yds. in length to
make a cloak, what in length that is yd. wide will be re-
quired to make the same?
Ans. 34 yds.
18. If 7 horses consume 24 tons of hay in G weeks, how
many tons will 12 horses consume in 8 weeks?
If we knew how much I horse consumed in 1 week, it
would be easy to find how much 12 horses would consume
in 8 weeks.
M
11
4
4
24: LL tons. If 7 horses consume tons in 6 weeks,
1 horse will consume 4 of 4 = 4 of a ton in 6 weeks; and
if a horse consume of a tou in 6 weeks, he will consume
Z old t Ts of a ton in 1 week. 12 horses will consume
183 in 1 week, and in 8 weeks they will
ES
T68
TUS
T6
12 times
consume 8 times 188 LER 63. tons, Ans.
A
C
19. A man with his family, which in all were 5 persons,
did' usually drink 74 gallons of cider in 1 week; how much
will they drink in 22 weeks when 3 persons more are
added to the family?
Ans. 2804 gallons.
{
I
:
1
}
Į
132
DECIMAL FRACTIONS.
T 66, 67.
20. If 9 students spend 10%. in 18 days, how much
will 20 students spend in 30 days.? Ans. 39 £. 18 s. 4 d.
}
}
DECIMAL FRACTIONS.
¶ 66. We have seen, that an individual thing or number.
may be divided into any number of equal parts, and that
these parts will be called halves, thirds, fourths, fifths, sixths,
&c., according to the number of parts into which the thing
or number may be divided; and that each of these parts may
be again divided into any other number of equal parts, and so
on. Such are called common, or vulgar fractions. Their denom-
inators are not uniform, but vary with every varying division
of a unit. It is this circumstance which occasions the chief
difficulty in the operations to be performed on them; for
when numbers are divided into different kinds or parts, they
› cannot be so easily compared. This difficulty led to the in-
vention of decimal fractions, in which an individual thing, or
number, is supposed to be divided first into ten equal parts,
which will be tenths; and each of these parts to be again di-
vided into ten other equal parts, which will be hundredths;
and each of these parts to be still further divided into ten
other equal parts, which will be thousandths; and so on.
Such are called decimal fractions, (from the Latin word decem,
which signifies ten,) because they increase and decrease, in a
tenfold proportion, in the same manner as whole numbers.
¶ 67. In this way of dividing a unit, it is evident, that
the denominator to a decimal fraction will always be 10,
100, 1000, or 1 with a number of ciphers annexed; conse-
quently, the denominator to a decimal fraction need not be
expressed, for the numerator only, written with a point be-
fore it () called the separatrix, is sufficient of itself to ex-
- press the true value. Thus,
ļ
j
I
fo are writtén '6.
427.
27
T00
685
'685.
The denominator to a decimal fraction, although not ex-
pressed, is always understood, and is 1 with as many ci-
phers annexed as there are places in the numerator. Thus,
3765 is a decimal consisting of four places; consequently,
1 with four ciphers annexed (10000) is its proper denomina-
tor. Any decimal may be expressed in the form of a com-
f
.....
1
1
***
T 67..
mon fraction by writing under it its proper denominator.
Thus, '3765 expressed in the form of a common fraction,
is 3765
T0000*
When whole numbers and decimals are expressed to-
gether, in the same number, it is called a mixed number.
Thus, 25'63 is a mixed number, 25', or all the figures on the
left hand of the decimal point, being whole numbers, and
'63, or all the figures on the right hand of the decimal point,
being decimals.
The names of the places to ten-millionths, and, generally,
how to read or write decimal fractions, may be seen from
the following
Numbers.
Whole
M
Decimal Parts.
}
}
Į
3d place.
2d place.
1st place.
f
1
!
M
DECIMAL FRACTIONS.
634
25135
77003000
-8600
50555
1st place.
2d place. O
3d place.
4th place. O
5th place.
6th place.
7th place. O
=
6 3 4 0 0 0 0 0 0 0
A
*****
||
or
6 3
******
.... 634.
25 and 63 Hundredths.
Lad
"
6 0 0 0 0 0
******
TABLE.
0
1JJJJ
6853
…….
8 6 0 0 2
6 8 5 3
t
panelf
******
185
|| || || || || Hundreds.
Tens.
Units.
(
0 5 0
05
150
******
រោ
сл сл
60 5
…………….
75
5 Hundredths.
7 and 9 Millionths.
50 Thousandths.
85002 Hundred-Thousandths.
6853 Ten-Thousandths.,
My hear
}
read 5 Tenths.
}
or Tenths.
Hundredths.
Thousandths.
Ten-Thousandtès.
Hundred-Thousandths.
•
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133
Millionths.
Ten-Millionths.
Į
4
1
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​¿
k
134
T 68, 69.
Į
From the table it appears, that the first figure on the right
hand of the decimal point signifies so many tenth parts of a
unit;
the second figure, so many hundredth parts of a unit;
the third figure, so many thousandth parts of a unit, &c. It
takes 10 thousandths to make 1 hundredth, 10 hundredths.
to make 1 tenth, and 10 tenths to make 1 unit, in the same
manner as it takes 10 units to make 1 ten, 10 tens to make
1 hundred, &c. Consequently, we may regard unity as a
starting point, from whence whole numbers proceed, con-
tinually increasing in a tenfold proportion towards the left
hand, and decimals continually decreasing, in the same pro-
portion, towards the right hand. But as decimals decrease
towards the right hand, it follows of course, that they in-
crease towards the left hand, in the same manner as whole
numbers.
1
DECIMAL FRACTIONS.
}
!
Ad
↓
¶ 68. The value of every figure is determined by its
place from units. Consequently, ciphers placed at the right
hand of decimals do not alter their value, since every signifi-
cant figure continues to possess the same place from unity.
Thus, '5, '50, '500 are all of the same value, each being
equal to, or .
But every cipher, placed at the left hand of decimal frac-
tions, diminishes them tenfold, by removing the significant
figures further from unity, and consequently making each
part ten times as small. Thus, '5, '05, '005, are of different
value, '5 being equal to F, or 1; '05 being equal to 78ʊ, or
z; and '005 being equal to Too, or zoo.
Decimal fractions, having different denominators, are readily
reduced to a common denominator, by annexing ciphers until
they are equal in number of places. Thus, '5, '06, 234 may
be reduced to '500, '060, 234, each of which has 1000 for a
common denominator.
T69. Decimals are read in the same manner as whole
numbers, giving the name of the lowest denomination, or
right hand figure, to the whole. Thus, '6853 (the lowest
denomination, or right hand figure, being ten-thousandths)
is read, 6853 ten-thousandths.
Any whole number may evidently be reduced to decimal
parts, that is, to tenths, hundredths, thousandths, &c. by an-
nexing ciphers. Thus, 25 is 250 tenths, 2500 hundredths,
25000 thousandths, &c. Consequently, any mixed number
1
[
1
[
纂
​T 70. ADDITION AND SUBTRACTION OF DECIMALS.
135
may be read together, giving it the name of the lowest de-
nomination or right hand figure. Thus, 25'63 may be read
2563 hundredths, and the whole may be expressed in the
form of a common fraction, thus, 26.
The denominations in federal money are made to corre-
spond to the decimal divisions of a unit now described, dol-
lars being units or whole numbers, dimes tenths, cents hun-
dredths, and mills thousandths of a dollar; consequently the
expression of any sum in dollars, cents, and mills, is simply
the expression of a mixed number in decimal fractions.
Forty-six and seven tenths 4677% — 46″7.
Write the following numbers in the same manner:
Eighteen and thirty-four hundredths.
Fifty-two and six hundredths.
Nineteen and four hundred eighty-seven thousandths.
Twenty and forty-two thousandths.
One and five thousandths.
135 and 3784 ten-thousandths.
9000 and 342 ten-thousandths.
10000 and 15 ten-thousandths.
974 and 102 millionths.
320 and 3 tenths, 4 hundredths and 2 thousandths.
500 and 5 hundred-thousandths.
47 millionths.
Four hundred and twenty-three thousandths.
}
ADDITION AND SUBTRACTION OF DECIMAL
FRACTIONS.
¶ 70. As the value of the parts in decimal fractions in-
creases in the same proportion as units, tens, hundreds, &c.,
and may be read together, in the same manner as whole
numbers, so, it is evident that all the operations on decimal
fractions may be performed in the same manner as on whole
numbers. The only difficulty, if any, that can arise, must
be in finding where to place the decimal point in the result.
This, in addition and subtraction, is determined by the same
rule; consequently, they may be exhibited together.
1. A man bought a barrel of flour for $8, a firkin of but-
1
f
1
136
ADDITION AND SUBTRACTION OF DECIMALS.
¶ 70.
ter for $350, 7 pounds of sugar for 83 cents, an ounce
of pepper for 6 cents; what did he give for the whole?
1
OPERATION.
8000 mills, or 1000ths of a dollar.
3500 mills, or 1000ths.
835 mills, or 1000ths.
60 mills, or 1000ths.
8
3'50
1
}
'835
'06
ļ
1
Ans. $12'395 = 12395 mills, or 1000ths.
As the denominations of federal money correspond with
the parts of decimal fractions, so the rules for adding and
subtracting decimals are exactly the same as for the same
operations in federal money. (See T 28.)
*
J
2. A man, owing $375, paid $175 75; how much did
he then owe?
}
OPERATION.
$375
37500 cents, or 100ths of a dollar.
175'75 = 17575 cents, or 100ths of a dollar.
1
$199'25 = 19925 cents, or 100ths.
}
The operation is evidently the same as in subtraction of
federal money. Wherefore,-In the addition and subtrac-
tion of decimal fractions, RULE: Write the numbers under
each other, tenths under tenths, hundredths under hun-
dredths, according to the value of their places, and point off
In the results as many places for decimals as are equal to the
greatest number of decimal places in any of the given num-
bers.
I
}
EXAMPLES FOR PRACTICE.
3. A man sold wheat at several times as follows, viz.
13'25 bushels; 8'4 bushels; 23'051 bushels, 6 bushels, and
75 of a bushel; how much did he sell in the whole?
Ans. 51451 bushels.
4. What is the amount of 429, 21, 355, 173ʊ and
17% ?
Ans. 808-143, or 808'143.
5. What is the amount of 2 tenths, 80 hundredths, 89
thousandths, 6 thousandths, 9 teuths, and 5 thousandths?
Ans. 2.
6. What is the amount of three hundred twenty-1-ine, and
seven tenths; thirty-seven and one hundred sixty-two thou-
sandths, and sixteen hundredths?
}
1
*
# 70, 71.
MULTIPLICATION OF DECIMALS,
137
7. A man, owing $4316, paid $376'865; how much did
he then owe?
Ans. $3939'135.
8. From thirty-five thousand take thirty-five thousandths.
Ans. 34999'965.
Ans. 1'5507.
Ans. 234'9925.
9. From 5'83 take 4'2793.
10. From 480 take 245'0075.
11. What is the difference between 179313 and 817
05693 ?
Ans. 976'07307.
σ
12. From 478 take 27%. Remainder, 1985, or 1'98.
13. What is the amount of 297, 374100800T, 97253
315, 27, and 100?
Ans. 942'957009.
9
10002
24
MULTIPLICATION OF DECIMAL FRACTIONS.
¶ 71. 1. How much hay in 7 loads, each containing
23'571 cwt. ?
{
23'571 cwt.
17
OPERATION.
475
25
Ans. 164'997 cwt. — 164997 1000ths of a cwt.
We may here ( 69) consider the multiplicand so many
thousandths of a cit., and then the product will evidently be
thousandths, and will be reduced to a mixed or whole num-
ber by pointing off 3 figures, that is, the same number as are
in the multiplicand; and as either factor may be made the
multiplier, so, if the decimals had been in the multiplier, the
same number of places must have been pointed off for deci-
mals. Hence it follows, we must always point off in the pro-
duct as many places for decimals as there are decimal places in
both factors.
2. Multiply 75 by ‘25.
375
150
1875 Product.
M *
་
OPERATION.
1
23571 1000ths of a cwt.
ry
1
In this example, we have 4 de-
cimal places in both factors; we
must therefore point off 4 places
for decimals in the product. The
reason of pointing off this num-
ber may appear still more plain,
if we consider the two factors as
Jour
!
I
}
11
138 -
T 71.
MULTIPLICATION OF DECIMALS..
on or vulgar fractions. Thus, 75 is 755, and 25 is
1875, Ans. same as be-
common
225: now, 70% x 10's
fore.
συσσ
p
3. Multiply '125 by '03.
OPERATION.
'125
'03
00375 Prod.
them at the left hand.
pear from the following
13: now, 15 × TSU
X
before.
These examples will be sufficient to establish the following
RULE.
In the multiplication of decimal fractions, multiply as in whole
umbers, and from the product point off so many figures for deci-
mals as there are decimal places in the multiplicand and multi-
plier counted together, and, if there are not so many figures in the
product, supply the deficiency by prefixing ciphers.
EXAMPLES FOR PRACTICE.
4. At $5'47 per yard, what cost 8'3 yards of cloth?
1
Here, as the number of significant
figures in the product is not equal to
the number of decimals in, both fac-
tors, the deficiency must be supplied
by prefixing ciphers, that is, placing
The correctness of the rule may ap-
process: 125 is 2%, and '03 is
00375, the same as
{
375
100000
Ans. $45'401.
7
5. At $107 per pound, what cost 26'5 pounds of rice?
Ans. $1855..
6. If a barrel contain 175 cwt. of flour, what will be the
Ans. 11025 cwt..
weight of '63 of a barrel ?
I
7. If a melon be worth $'09, what is "7 of a melon
worth?
Ans. 6 cents.
8. Multiply five hundredths by seven thousandths.
1
+
9. What is '3 of 116 ?
10. What is '85 of 3672 ?
11. What is '37 of '0563 ?
12. Multiply 572 by ‘58..
13. Multiply eighty-six by four hundredths.
'1
Product, ‘00035.
Ans. 34'8.
1
Aris. 3121'2.
Ans. '020831.
Product, 331476.
Product, 3'44..
14. Multiply '0062 by ‘0008.'.
15. Multiply forty-seven tenths by one thousand eighty-.
six hundredths.
}
1
1
¶ 71, 72.
J
嘉
​DIVISION OF DECIMALS.
X
16. Multiply two hundredths by eleven thousandths.
17. What will be the cost of thirteen nudedths of a ton
of hay, at 11 a ton?
18. What will be the cost of three hundred seventy-five
thousandths of a cord of wood, at $2 a cord?
19. If a man's wages be seventy-five hundredths of a dol-
lar a day, how much will he earn in 4 weeks, Sundays ex-
cepted?
1
DIVISION OF DECIMAL FRACTIONS.
139
T 72. Multiplication is proved by division. We have
seen, in multiplication, that the decimal places in the product
must always be equal to the number of decimal places in the
multiplicand and multiplier counted together. The multi-
plicand and multiplier, in proving multiplication, become the
divisor and quotient in division. It follows of course, in di-
vision, that the number of decimal places in the divisor and
quotient, counted together, must always be equal to the number of
decimal places in the dividend. This will still further appear
from the examples and illustrations which follow:
1. If 6 barrels of flour cost $44'718, what is that a bar-
rel?
OPERATION.
6)44 718
Ans. 7'453
. է
By taking away the decimal point, $44718
44'71844718
mills, or 1000ths, which, divided by 6, the quotient is 7453
mills, $7'453, the Answer.
Or, retaining the decimal point, divide as in whole num-
bers.
}
As the decimal places in the di-
visor and quotient, counted toge-
ther, must be equal to the number
of decimal places in the dividend,
there being no décimals in the divisor,-therefore point off
three figures for decimals in the quotient, equal to the number
of decimals in the dividend, which brings us to the same re-
sult as before.
2. At $4'75 a barrel for cider, how many barrels may be
bought for $31?
}
In this example, there are decimals in the divisor, and
none in the dividend. $475 475 cents, and $31, by
annexing two ciphers, 3100 cents; that is, reduce the di
1
1
1
;
J
}
1
J
140
др куп на
vidend to parts of the same denomination as the divisor.
Then, it is plain, as many times as 475 cents are contained in
3100 cents, so many barrels may be bought.
/
250
But the remainder, 250, instead of be-
ing expressed in the form of a common
fraction, may be reduced to 10ths by annexing a cipher,
which, in effect, is multiplying it by 10, and the division con-
tinued, placing the decimal point after the 6, or whole ones
already obtained, to distinguish it from the decimals which
are to follow. The points may be withdrawn or not from
the divisor and dividend.
OPERATION.
ĥ
1
475)3100(6240 barrels, the Answer; that is, 6 barrels and
2850
24% of another barrel.
2500
2375
DIVISION OF DECIMALS.
1250
950
3000
2850
475)31'00 (6'526+ barrels, the Answer; that is, 6 bar-
rels and 526 thousandths of another
barrel.
2850
1
}
150
is of so trifling a value, as not to merit notice.
If now we count the decimals in the dividend, (for every
cipher annexed to the remainder is evidently to be counted
a decimal of the dividend,) we shall find them to be five,
which corresponds with the number of decimal places in the
divisor and quotient counted together.
J
1
By annexing a cipher to the first
remainder, thereby reducing it to
10ths, and continuing the division,
we obtain from it '5, and a still fur-
ther remainder of 125, which, by an-
nexing another cipher, is reduced to
100ths, and so on.
The last remainder, 150, is 1 of
à thousandth part of a barrel, which
¿
3. 'Under T 71, ex. 3, it was required to multiply '125 by
'03; the product was '00375. Taking this product for a
dividend, let it be required to divide '00375 by '125. One
operation will prove the other. Knowing that the number
of decimal places in the quotient and divisor, counted to-
gether, will be equal to the decimal places in the dividend,
we may divide as in whole numbers, being careful to retain
the decimal points in their proper places. Thus,
}
¿
141
The divisor, 125, in 375 goes 3.
times, and, no remainder. We have
only to place the decimal point in
the quotient, and the work is done.
There are five decimal places in the
dividend; consequently there must be five in the divisor and
quotient counted together; and, as there are three in the di-
visor, there must be two in the quotient; and, since we have
but one figure in the quotient, the deficiency must be sup-
plied by prefixing a cipher.
125
The operation by vulgar fractions will bring us to the
same result. Thus, 125 is 7, and 00375 is Too:
now, Too÷TUS = 1938888803, the same
as before.
Q
=
¶
72, 73.
DIVISION OF DECIMALS.
OPERATION.
'125) '00375('03
375
000
125
Τουσ
¶ 73. The foregoing examples and remarks are suffi-
cient to establish the following
RULE.
In the division of decimal fractions, divide as in whole num-
bers, and from the right hand of the quotient point off as many
figures for decimals, as the decimal-figures in the dividend ex-
ceed those in the divisor, and if there are not so many figures in
the quotient, supply the deficiency by prefixing ciphers.
j
If at any time there is a remainder, or if the decimal
figures in the divisor exceed those in the dividend, ciphers.
may be annexed to the dividend or the remainder, and the
quotient carried to any necessary degree of exactness; but
the ciphers annexed must be counted so many decimals of
the dividend.
1
EXAMPLES FOR PRACTICE.
4. If $472'875 be divided equally between 13 men, how
much will each one receive?
Ans. $36'375.
5. At $5 per bushel, how many bushels of rye can be
bought for $141 ?
Ans. 188 bushels,
6. At 12 cents per lb., how many pounds of butter may
be bought for $37?
Ans. 296 b.
7. At 61 cents apiece, how many oranges may be bought
for $8?
Ans. 128 oranges.
rel?
8. If '6 of a barrel of flour cost $5, what is that per bar-
Ans. $8'333 +.
Quot. '037+.
9. Divide 2 by 53'1.
5
"
1
t
142
REDUCTION OF COMMON OR
10. Divide '012 by ‘005.
11. Divide three thousandths by four hundredths.
1.
!
Quot, '075.
12. Divide eighty-six tenths by ninety-four thousandths.
13. How many times is '17 contained in 8?
REDUCTION OF COMMON OR VULGAR FRAC-
TIONS TO DECIMALS.
T74. 1. A man has of a barrel of flour; what is that
4245
expressed in decimal parts?
1
T 73, 74.
Quot. 2'4.
As many times as the denominator of a fraction is con-
tained in the numerator, so many whole ones are contained
in the fraction. We can obtain no whole ones in 4, because
the denominator is not contained in the numerator. We
may, however, reduce the numerator to tenths, (T 72, ex. 2,)
by annexing a cipher to it, (which, in effect, is multiplying
it by 10,) making 40 tenths, or 4'0. Then, as many times as
the denominator, 5, is contained in 40, so many tenths are
contained in the fraction. 5 into 40' goes 8 times, and no
remainder.
Ans. 8 of a bushel.
2. Express of a dollar in decimal parts.
The numerator, 3, reduced to tenths, is 8, 340, which,
divided by the denominator, 4, the quotient is 7 tenths, and
a remainder of 2. This remainder must now be reduced to
hundredths by annexing another cipher, making 20 hun-
dredths. Then, as many times as the denominator, 4, is con-
tained in 20, so many hundredths also may be obtained. 4
into 20 goes 5 times, and no remainder. of a dollar, there-
fore, reduced to decimals, is 7 tenths and 5 huudredths, that
is, 75 of a dollar.
The operation may be presented in form as follows :-
20
20
f
Num.
Denom. 4) 30 (75 of a dollar, the Answer.
28
1
1
י}
1
Sha
{
I
T 74.
66 ) 4′00 ( ‘0606 +, the Answer.
396
400
396
VULGAR FRACTIONS TO DECIMALS.
3. Reduce
to a decimal fraction.
The numerator must be reduced to hundredths, by annex-
ing two ciphers, before the division can begin.
来
​4
t
I
Gambar
}
As there can be no tenths, a cipher must
be placed in the quotient, in tenth's place.
I
f
6
Note. cannot be reduced exactly; for, however long the
division be continued, there will still be a remainder.* It is
sufficiently exact for most purposes, if the decimal be ex-
tended to three or four places.
From the foregoing examples we may deduce the follow-
ing general RULE:-To reduce a common to a decimal frac-
Y
4
A
* Decimal figures, which continually repeat, like '06, in this exam-
ple, are called Repetends, or Circulating Decimals. If only one figure
repeats, as 3333 or 7777, &c., it is called a single repetend. If two or
more figures circulate alternately, as '060606, 234234234, &c., it is
called a compound repetend. If other figures arise before those which
circulate, as 743333, 143010101, &c., the decimal is called a mixed
repetend.
+
143
A single repetend is denoted by writing only the circulating figure
with a point over it: thus, '3, signifies that the 3 is to be continually
repeated, forming an infinite or never-ending series of 3's.
A compound repetcrd is denoted by a point over the first and last re-
peating figure: thus, 234 signifies that 234 is to be continually re-
peated.
may
It not be amiss, here to show how the value of any repetend may
be found, or, in other words, how it may be reduced to its equivalent
vulgar fraction.
фор
If we attempt to reduce & to a decimal, we obtain a continual repe-
tition of the figure 1: thus, '11111, that is, the repetend '1. The value
of the repetend 'i, then, is ; the value of
will evidently be twice as much, that is, 3.
3, and 4 = 4, and §, and so on to 9,
1. What is the value of '8?
“
}
Ans. &.
2. What is the value of '? Ans. §. What is tho' value of “Ž?
of 7?
of 4?
of 15?
of 9 ?
→ of ‹İ ?
If ' be reduced to a decimal, it produces '010101, or the repotend 01.
bo
The repetend '02, being 2 times as much, must be
and 48, being 48 times as much, must be 18, and
and '03 — 5º,
JF, &a.
48
74
=
35
222, &c., the repetend 2,
In the same manner,
which
3
1.
!
1
|
2
Me
!
1
واهر
?
144 REDUCT. OF VULG. FRACTIONS TO DECIMALS. T74.
tion,-Annex one or more ciphers, as may be necessary, to the
numerator, and divide it by the denominitor If then there ha
a remainder, annex another cipher, ana de jure, and
continue to do so long as there shall continue to be a remainder,
or until the fraction shall be reduced to any necessary degree
of exactness. The quotient will be the decimal required,
which must consist of as many decimal places as there are
ciphers annexed to the numerator; and, if there are not so
many figures in the quotient, the deficiency must be sup-
plied by prefixing ciphers.
4. Reduce 2, 4, 1%, and
སྐམ་
Kaplan
1
5. Reduce 3, Tobo, T745, and be to decimals.
TOO,
Ans. '692; '003; 0028; '000183 +
478
6. Reduce 1, 67, 68 to decimals.
EXAMPLES FOR PRACTICE,
Tis's
to decimals.
Ans. “5; 25; '025; '00797 +.
}
1
Đ
7. Reduce †,, 6tu, d, †, T'r, fr, 3 to decimals.
8. Reduce,,t, t, 't, t, t, 25, z's, to decimals.
8, zu,
B
37
If be reduced to a decimal, it produces '001; consequently,
‘002 =55, and ‘037 ², and 42533, &c., As this principle
will apply to any number of places, we have this general Reus for re-
ducing a circulating decimal to a vulgar fraction,--Make tho given
repetend the numerator, and the denominator will be as many s as
there are repeating figures,
3. What is the vulgar fraction equivalent to “704 ?
4. What is the value of '003?
014?
4424?
$2463? ·002103 ?
Ans, to last,
A MARA
O ;
5. What is the valup of '43?
In this fraction, the ropetend begins in the second place, or place of
hundredths. The first figure, 4, is As, and the repetend, 3, is
these two parts anust be added together.
ሊ
3
that is,
1
plaja katta,
}
+
38, Ans. Hence, to find the value of a mixed repetenil,-Find the
valuo of the two parts, separately, and add then together.
•
765 + 580 = 185 = 73, Ans.
ANS. TEST.
6. What is the value of 153?
7. What is the valuo of 0047?
8. What is the value of 128?
48?.
It is plain, that circulatos may be added, subtracted, multiplied, and
divided, by first reducing them to their equivalent vulgar fractions.
(4123?
▼
27
paddy pr
Jadamantan proje
T
UK
Ans. FfF.
01021 ?
EET.
Į
}
of To,
39
ទប
ܝ ܚ ܝ
$
2
་
4
T 75.
}
REDUCTION OF DECIMAL FRACTIONS.
REDUCTION OF DECIMAL FRACTIONS.
¶ 75. Fractions, we have seen, (T 63,) like integers, are
reduced from low to higher denominations by division, and
from high to lower denominations by multiplication.
To reduce a compound num- To reduce the decimal of a
ber to a decimal of the highest higher denomination to integers
denomination.
of lower denominations.
1. Reduce 7 s. 6 d. to the
decimal of a pound.
}
Ans.
2. Reduce 375 £. to in-
tegers of lower denominations.
6 d. reduced to the decimal 375 £. reduced to shillings,
of a shiling, that is, divided that is, multiplied by 20, is
by 12, is 5 s., which annexed 7'50 s.; then the fractional
to the 7 s. making 7'5 s., and part, '50 s., reduced to pence,
divided by 20, is 375 £. the that is, multiplied by 12, is
6 d.
Ans. 7 s. 6 d.
The process may be pre- That is,-Multiply the given
sented in form of a rule, thus:-decimal by that number which
Divide the lowest denomina- it takes of the next lower de-
tion given, annexing to it one nomination to make one of this
or more ciphers, as may be higher, and from the right
necessary, by that number hand of the product point off
which it takes of the same to as many figures for decimals
make one of the next higher as there are figures in the
denomination, and annex the given decimal, and so con-
quotient, as a decimal to that tinue to do through all the de-
higher denomination; so con- nominations; the several num-
tinue to do, until the whole bers at the left hand of the
shall be reduced to the deci- decimal points will be the
mal required.
value of the fraction in the
proper denominations.
1
/14*
·
EXAMPLES FOR PRACTICE.
3. Reduce 1 oz. 10 pwt. to
the fraction of a pound.
OPERATION.
20) 10'0 pwt.
12)1'5 oz.
1
145
'125 lb. Ans.
N
EXAMPLES FOR PRACTICE.
4. Reduce 125 lbs. Troy to
integers of lower denomina-
tions.
OPERATION.
lb. $125
12
oz. 1'500
20
pwt. 10'000. Ane. 1oz. 10pwt.
картина
1
1
!
"
А
146
5. Reduce 4 cwt. 23 qrs, to
the decimal of a ton.
}
REDUCTION OF DECIMAL FRACTIONS.
Note. 232′6.
7. Reduce 38 gals. 3′52 qts.
I
of beer, to the decimal of a hhd. of beer?
hhd.
9. Reduce 1 qr. 2 n. to the
decimal of a yard.
11. Reduce 17 h. 6 m. 43
sec. to the decimal of a day.
13. Reduce 21 s. 10 d. to
the decimal of a guinea.
15. Reduce 3 cwt. 0 qr. 7
lbs. 8 oz. to the decimal of a
ton.
1 76.
6. What is the value of
2325 of a ton ?
/
8. What is the value of "72
Let the pupil be required to reverse and prove the follow-
ing examples:
F
10. What is the value of
$375 of a yard?
12. What is the value of
713 of a day?
14. What is the value of
78125 of a guinea?
16. What is the value of
15334821 of a ton ?
17. Reduce 4 rods to the decimal of an acre.
18. What is the value of "7 of a lb. of silver?
19. Reduce 18 hours, 15 m. 50'4 sec, to the decimal of a
day.
20. What is the value of '67 of a league?
21. Reduce 10 s. 94 d. to the fraction of a pound.
!
176. There is a method of reducing shillings, pence
and farthings to the decimal of a pound, by_inspection, more
simple and concise than the foregoing. The reasoning in
relation to it is as follows:
1
60
1.
7 of 20 s. is 2 s.; therefore every 2 s. is, or '1 £.
Every shilling is, or '05 £. Pence are readily
reduced to farthings. Every farthing is £. Had it so
happened, that 1000 farthings, instead of 960, had made a
pound, then every farthing would have been obʊ, or '001 £.
But 960 increased by part of itself is 1000; conse-
quently, 24 farthings are exactly 30, or '025 £., and 48
farthings are exactly 88, or '050 £. Wherefore, if the
number of farthings, in the given pence, and farthings, be
more than 12, 24 part will be more than ; therefore add 1
to them if they be more than 36, 4 part will be more than
11; therefore add 2 to them then call them so many
thousandths, and the result will be correct within less than
d of robe of a pound. Thus, 17 s. 5 d. is reduced to the
!
1
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J
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147
=
decimal of a pound as follows: 16 s. '8 £. and 1 s. =
'05 £. Then, 54 d. 23 farthings, which, increased by
1, (the number being more than 12, but not exceeding 36,) is
024 £., and the whole is '874 £. the Ans.
Wherefore, to reduce shillings, pence and farthings to the
decimal of a pound by inspection,-Call every two shillings one
tenth of a pound; every odd shilling, five hundredths; and the
number of farthings, in the given pence and farthings, so many
thousandths, adding one, if the number be more than twelve and
not exceeding thirty-six, and two, if the number be more than
thirty-six.
REDUCTION OF DECIMAL FRACTIONS.
¶ 77. Reasoning as above, the result, or the three first
figures in any decimal of a pound, may readily be reduced
back to shillings, pence and farthings, by inspection. Double
the first figure, or tenths, for shillings, and, if the second
figure, or hundredths, be five, or more than five, reckon ano-
'ther shilling; then, after the five is deducted, call the figures
in the second and third place so many farthings, abating
one when they are above twelve, and two when above thir-
ty-six, and the result will be the answer, sufficiently exact
for all practical purposes. Thus, to find the value of '876 £.
by inspection :-
$8 tenths of a pound
'05 hundredths of a pound
'026 thousandths, abating 1, 25 farthings
'876 of a pound
声​越
​Ka
=
1
16 shillings.
1 shilling.
0 s.
64 d.
= 17 s.
17 s.
64 d.
Ans.
EXAMPLES FOR PRACTICE.
1. Find, by inspection, the decimal expressions of 9 s. 7 d.,
and 12 s. 03 d.
Ans. '479£., and '603£.
2. Find, by inspection, the value of '523 £., and '694.
Ans. 10 s. 5 d., and 13 s. 104 d.
3. Reduce to decimals, by inspection, the following sums,
and find their amount, viz.: 15 s. 3 d.; 8 s. 11 d.; 10 s.
61 d.; 1 s. 81 d.; † d., and 21 d. Amount, £1'833.
4. Find the value of '47 £.
Note. When the decimal has but two figures, after taking
out the shillings, the remainder, to be reduced to thousandths,
will require a cipher to be annexed to the right hand, or
supposed to be so.
Ans. 9 s. 4§ d.
I
}
148
π myy.
5. Value the following decimals, by inspection, and find
their amount, viz.: 785 £.; '357 £.; '916 £.; 74 £.;
‘5 £.; ‘25 £.; '09 £.; and ‘008 £. Ans. 3£. 12 s. 11 d.
I
1
SUPPLEMENT TO DECIMAL FRACTIONS.
}
}
SUPPLEMENT TO DECIMAL FRACTIONS.
1
QUESTIONS.
T
1. What are decimal fractions? 2. Whence is the term
derived? 3. How do decimal differ from common frac-
tions? 4. How are decimal fractions written? 5. How
can the proper denominator to a decimal fraction be known,
if it be not expressed? 6. How is the value of every figure
determined? 7. What does the first figure on the right
hand of the decimal point signify?
the second figure ?
third figure? fourth figure? 8. How do ciphers,
placed at the right hand of decimals, affect their value?
9. Placed at the left hand, how do they affect their value?
10. How are decimals read? 11. How are decimal frac-
tions, having different denominators, reduced to a common
denominator? 12. What is a mixed number? 13. How
may any whole number be reduced to decimal parts? 14.
How can any mixed number be read together, and the
whole expressed in the form of a common fraction?
What is observed respecting the denominations in federal
money? 16. What is the rule for addition and subtraction
of decimals, particularly as respects placing the decimal
point in the results? multiplication? division?
17. How is a common or vulgar fraction reduced to a deci-
mal? 18. What is the rule for reducing a compound num-
ber to a decimal of the highest denomination contained in
it? 19. What is the rule for finding the value of any given
decimal of a higher denomination in terms of a lower?
20. What is the rule for reducing shillings, pence and far-
things to the decimal of a pound, by inspection? 21. What
is the reasoning in relation to this rule? 22. How may the
three first figures of any decimal of a pound be reduced to
shillings, pence and farthings, by inspection?
15.
!
I
1 77.
7 yds.
6 &
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∞ ∞ ∞
1. A merchant had several remnants of cloth, measuring
as follows, viz. :
81
.....
37%
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SUPPLEMENT TO DEĆIMAL FRACTIONS. 149
EXERCISES.
How many yards in the whole, and what would
the whole come to at $3'67 per yard?
$133'863+, cost.
2. From a piece of cloth, containing 36g yds., a merchant
sold, at one time, 7 yds., and, at another time, 12g yds.;
how much of the cloth had he left?
Ans. 167 yds..
Note.
Reduce the common fractions to deci-
mals. Do the same wherever they occur in the
examples which follow.
Ans. 36'475 yards.
3. A farmer bought 7 yards of broadcloth for 81 £., a
barrel of flour for 2 £., a cask of lime for 18 £., and 7 lbs.
of rice for £.; he paid 1 ton of hay at 376 £., 1 cow
"
at 6 £., and the balance in pork at £. per lb.; how
many were the pounds of pork ?
Note.
In reducing the common fractions in this example,
it will be sufficiently exact if the decimal be extended to
three places.
Ans. 1084 lb.
4. At 12 cents per lb., what will 373 lbs. of butter cost?
Ans. $4'7182.
5. At $1737 per ton for hay, what will 11 tons cost?
Ans. $201'92§.
6. The above example reversed. At $201'92
of hay, what is that per ton?
for 11½ tons
Ans. $1737.
7. If '45 of a ton of hay cost $9, what is that per ton?
Consult ¶ 65.
Ans. $20.
8. At '4 of a dollar a gallon, what will '25 of a gallon
of molasses cost?
Ans. $1.
9. At $9 per cwt., what will 7 cwt. 3 qrs. 16 lbs. of sugar
cost?
Note. Reduce the 3 qrs. 16 lbs. to the decimal of a cwt.,
extending the decimal in this, and the examples which fol-
low, to four places.
Ans. 71'035+.
10. At $69'875 for 5 cwt. 1 qr. 14 lbs. of raisins, what is
that per cwt.?
Ans. $13.
11. What will 2300 lbs. of hay come to at 7 mills per lb. ?
Ans. $16*10.
at 18 cents per
Ans. $137679
12. What will 7654 lbs. of coffee come to,
lb. ?
N
+
t
5
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{
}
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150
буду курдук
13. What will 12 gals. 3 qts. 1 pt. of gin cost, at 28 cents
per quart?
Note.
Reduce the whole quantity to quarts and the deci-
mal of a quart.
Ans. $14'42.
14. Bought 16 yds. 2 qrs. 3 na. of broadcloth for $100 125;
what was that per yard?
Ans. $6.
1
SUPPLEMENT TO DECIMAL FRACTIONS.
15. At $192 per bushel, how
bought for $72 ?
16. At $9272 per ton, how
chased for $60'268 ?
17. Bought a load of hay for
of $ 16 per ton; what was the weight of the hay?
much wheat may be
Ans. 1 peck 4 quarts.
much irou may be pur-
Ans. 13 cwt,
$9'17, paying at the rate
1
Ans. 11 cwt. 1 qr. 23 lbs.
18. At $3024 per tun, what will 1 hhd. 15 gals. 3 qts.
of wine cost?
Ans. $94'50.
19. The above reversed. At $94′50 for 1 hhd. 15 gals.
3 qts. of wine, what is that per tun?
Ans. $302'4.
1
20. At $1'80 for 31 qts. of
wine, what is that per gal.?
22. If g of a ton of pot-
ashes cost $60'45, what is
that per ton ?
វ
Note. The following examples reciprocally prove each
other, excepting when there are some fractional losses, as ex-
plained above, and even then the results will be sufficiently
exact for all practical purposes. If, however, greater exact-
ness be required, the decimals must be extended to a greater
number of places.
1
Sad
{
21. At $2215 per gal.,
what cost 34 qts. ?
24. If 3 of a yard 25. If a yard of
'8
26. At $25 per ·
of cloth cost $2, cloth cost $2'5, yard, how much
what is that per what will 3 of a cloth may be pur-
yard cost?
chased for $2 ?
yard?
27. If 14 cwt. of 28. If a ton of 29. At 27 £. 10 s.
pot-ashes cost 19 £.pot-ashes cost 27£. a ton for pot-ashes,
5 s., what is that 10 s., what will 14
per ton?
what quantity may
cwt. cost?
be bought for 19 £.
5.s. ?.
23. At $9672 per ton for.
pot-ashes, what will § of a ton
cost?
1
Note. After the same manner let the pupil reverse and
prove the following examples:
31
F
/
1
!
¶ 77, 78.
151
30. At $18'50 per ton, how much hay may be bought
for $12'025 ?
31. What will 3 qrs. 2 na. of broadcloth cost, at $6 per
yard?
32. At $22′10 for transportation of 65 cwt. 46 miles, what
is that per ton?
}
for
33. Bought a silver cup, weighing 9 oz. 4 pwt. 16 grs.
3 £. 2 s. 3 d. 33 q.; what was that per ounce ?
34. Bought 9 chests of tea, each weighing 3 cwt. 2 qrs. 21
lbs. at 4 £. 9 s. per cwt.; what came they to?
35. If 5 acres 1 rood produce 26 quarters 2 bushels of
wheat, how many acres will be required to produce 47
quarters 4 bushels? A quarter is 8 bushels.
Note. The above example will require two operations,
for which consult ¶ 65, ex. 1.
36. A lady purchased a gold ring, giving at the rate of
$20 per ounce; she paid for the ring $125; how much
did it weigh?
REDUCTION OF CURRENCIES.
178. Previous to the act of Congress in 1786 establish-
ing federal money, all calculations in money, throughout the
United States, were made in pounds, shillings, pence and
farthings, the same as in England. But these denominations,
although the same in name, were different in value in dif
ferent countries.
REDUCTION OF CURRENCIES.
England,
Canada and
Nova Scotia,
The New Eng-
land States,
Virginia,
Kentucky, and
Tennessee,
New York,
Ohio, and
N. Carolina,
1
Thus, 1 dollar is reckoned in
4 s. 6 d., called English, or sterling money.
5 s. called Canada currency.
6 s., called New England currency..
8 s., called New York currency..
{
1
1
1
1
1
1
152
New Jersey,
Pennsylvania,
Delaware, and
Maryland,
REDUCTION OF CURRENCIES.
1 dollar is reckoned in
9) 263'040
$
7 s. 6 d., called Pennsylvania currency.
1
S. Carolina and
Georgia,
1. Reduce 6£. 11 s. 61 d. to federal money.
Note. To reduce pounds, shillings, pence and farthings,
in either of the above-named currencies, to federal money,-
First, reduce the shillings, pence and farthings (if any be
contained in the given sum) to the decimal of a pound by in-
spection, as already taught, T 76.
}
4 s. 8 d., called Georgia currency.
54
240
£6576 English money.
40
6£. 11 s. 64 d. = £6'576.
40
English money.—-Now, supposing the above sum to be
English money,-1. is 20 s. = 240 pence, in all the above
currencies. 1 dollar, in English money, is reckoned 4 s. 6 d.
54 pence, that is, of 1 pound. Now, as many
times as, the fraction which 1 dollar is of 1 pound, Eng-
lish money, is contained in £6'576, so many dollars, it is
evident, there must be; that is,-To reduce English to federal
money,-Divide the given sum by, the quotient will be
federal money.
7
29'226 federal money, Answer.
¶ 78
}
Note. It will be
recollected, to di-
vide by a fraction,
we multiply by the
denominator, and
divide the product
by the numerator.
$27′304 federal money, Answer.
Canada CURRENCY.-Supposing the above sum to be Cana-
da currency,-1 dollar, in this currency, is 5 s. = 60 pence,
that is,
of 1 pound. Therefore,-To reduce Cana-
f
da currency to federal money,-Divide the given sum by, and
the quotient will be federal money; or, which is the same.
thing,-Multiply the given sum by 4.
£6'576 Canada currency.
4.
}
1
1
178.
t
;
6 S.
NEW ENGLAND CURRENCY. -1 dollar, in this currency, is
o, or 3 of a pound. There-
fore,-To reduce New England currency to federal money,-Di-
vide the given sum by '3.
72 pence, that is,
3) £.6576 New England currency. "
REDUCTION OF CURRENCIES.
mad Sama
Wapt
ngjaj
153
$21'92 federal money, Answer.
9.6
NEW YORK CURRENCY.-1 dollar, in this currency, is 8 s. =
96 pence, that is, f, or '4 of a pound. Therefore,
To reduce New York currency to federal money,-Divide the
given sum by '4.
'4) £.6'576 New York currency.
$ 16'44 federal money, Answer.
PENNSYLVANIA CURRENCY.-1 dollar, in this currency, is 7 s.
90 pence, that is, Q
of a pound. Therefore,-
6 d.
To reduce Pennsylvania currency to federal money,-Divide by
, that is, multiply the given sum by 8, and divide the pro-
duct by 3.
}
£.6'576 Pennsylvania currency.
8
3)52 608
}
$17'536 federal money, Answer:
=
5.6
To
Georgia curreNCY.-1 dollar, Georgia currency, is 4 s.
8 d. 56 pence, that is, of a pound. Therefore,-
To reduce Georgia currency to federal money,-Divide by or
that is, multiply the given sum by 30, and divide the pro-
duct by 7.
£. 6'576 Georgia currency.
30
7) 197'280
$28'1829 federal money, Answer.
From the foregoing examples, we derive the following
general RULE:— To reduce English money, and the currencies
of Canada and the several States, to federal money,-First, re-
duce the shillings, &c., if any in the given sum, to the deci-
mal of a pound; this being done, divide the given sum by
such fractional part as 1 dollar, in the given currency, is
a fractional part of 1 pound.
7
1
}
}
I
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{
1
1
1
154
1
REDUCTION OF CURRENCIES.
EXAMPLES FOR PRACTICE.
2. Reduce 125 £., in each of the before named currencies,
to federal money.
Answers.
125£. English money,
125£. Canada currency,
125. New England currency,...
125£. New York
125 £. Pennsylvania
125 £. Georgia
$416'6663.
$312'50.
$333'333.
$535 7142.
3. Reduce 1 s. 6 d., in the several currencies, to federal
money.
↓
1
Answers. 1 s. 6 d. '075£. English money, is $ 3331;'
Canada currency, it is $30; New England currency, it is
$25; New York currency, it is $1871; Pennsylvania
currency, it is $20; Georgia currency, it is $321.
4. Reduce 75£. 15 s., in the several currencies, to federal
money.
5. Reduce 18 £.0 s. 8 d., in the several currencies, to
federal money.
6. Reduce 44 d., in the several currencies, to federal
money.
7. Reduce 36 £. 3 s. 7 d., in the several currencies, to
federal money.
Answer.
$118'25, changed to N. York
79. To reduce federal money to any of the before named
currencies, reverse the process in the foregoing operations;
that is,-Multiply the given sum in federal money by such
fractional part as 1 dollar, in that currency to which you
would reduce it, is of 1 pound. The product will be the
answer in pounds and decimals of a pound, which must be
reduced to shillings, pence and farthings, by inspection, as
already taught, ¶ 77.
Pennsylvania
Georgia
}
is $555'5558.
$500.
EXAMPLES FOR PRACTICE.
1. Reduce $118'25 to the several before named cur-
rencies.
00.0
C
£. $. d.
English money, is 26 12 14.
29 11 3..
Canada currency,
N. England currency,... 35
**
ADIDAS á líka e
¶ 78, 79%
***
...
}
...
...
1
35 9 6.
47 6 0.
44 6 104.
27 11 98
¶ 79, 80.
2. Change $250 to the several currencies.
3. Change 56 cents to the several currencies.
4. Change $45'12 to the several currencies.
1
1 80. It may sometimes be required to reduce one cur-
rency to the par, or equality of another currency.
1. Reduce 35£. 6 s. 8 d., English money, to N. England
****
[
currency.
64
$1 is 4 s. 6 d. 54 d. English money. $1 is 6 s. -
*72 d. N. England currency; that is, the value of any number
of pounds, shillings, pence, &c., English money, is 12 405
of the same in N. England currency; consequently,-To re-
duce English money to N. England currency,-Multiply by ,
or, which is the same, increase it by part of itself. Thus,
£. s. d.
3) 35 6 8:
q.
English money, is
11 15 6
2
47
2 2 2 New England currency, Answer.
Hence we have this general RULE for finding a multiplier
to reduce any currency to the par of another :-
Make $1 in pence, of the currency to be reduced, the de-
nominator of a fraction, over which write $1 in pence, of
the currency to which it is to be reduced, for a numerator.
This fraction may then be reduced to its lowest terms be-
fore multiplying.
REDUCTION OF CURRENCIES.
¥
On the same principles, let the pupil form for himself multi-
pliers, by which
LATE
****
To reduce English money to Canada, N. York, Pennsylva-
nia, and Georgia currencies.
155

***
•
N. York currency to English, Canada, N. Eng-
land, Pennsylvania, and Georgia currencies.
Pennsylvania currency to English, Canada, N.
England, N. York, and Georgia currencies.
……..………………... Georgia currency to English, Canada, N. Eng-
land, N. York, and Pennsylvania currencies.
Canada currency to English, N. England, N.
York, Pennsylvania, and Georgia currencies.
N. England currency to Canada, N. York, Penn-
sylvania, and Georgia currencies.
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Livre of France,
Franc
do.
Silver Rouble of Russia,
Florin or Guilder of the United Netherlands,
Rates at which the following foreign coins are estimated at the
Custom Houses of the United States.
}
Sand
t
Mark Banco of Hamburg,
Real of Plate of Spain,
Real of Vellon of do.
Milrea of Portugal,
Tale of China,
Pagoda of India,
Rupee of Bengal,
INTEREST,
July
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HOME
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2. Reduce 8764 livres to federal money.
3. Reduce 10,000 francs to federal money.
4. Reduce 250,000 florins to federal money.
5. In $1000, how many francs?
摹
​2
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1
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f-
T 80, 81.
$ 18.
$
184.
€€€
675.
'40.
$ '331.
'10.
'05.
1
1'24.
1'48.
1'84.
'50.
1
INTEREST.
T 81. Interest is an allowance made by a debtor to a
creditor for the use of money. It is computed at a certain
number of dollars for the use of each hundred dollars, or so
many pounds for each hundred pounds, &c. one year, and
in the same proportion for a greater or less sum, or for a
longer or shorter time.
J
The number of dollars so paid for the use of a hundred
dollars, one year, is called the rate per cent. or per centum;
the words per cent. or per centum signifying by the hundred.
The highest rate allowed by law in the New England
States, is 6 per cent., that is, 6 dollars for a 100 dollars, 6
cents for a 100 cents, 6 pounds for a 100, &c.; in other
words, Too of the sum lent or due is paid for the use of it one
year. This is called legal interest, and will here be under-
stood when no other rate is mentioned.
* In the State of New York, 7 per cent. is the legal interest; in England the
legal interest is 5 per cent.
素
​1
I
T 81.
1
X
7
cent.
cent.
cent.
INTEREST.
Let us suppose the sum lent, or due, to be $1.
100th part of $1, or of a dollar, is 1 cent, and
dollar, the legal interest, is 6 cents, which, written
cimal fraction, is expressed thus,
So of any other rate per cent.
1 per cent., expressed as a common fraction, is
To decimally,
I
cent.
{
per cent. is a half of 1 per cent., that is,
per cent., is a fourth of 1 per cent., that is,
per cent. is 3 times per cent., that is,
2
4
1
Write 24 per cent. as a
2 per cent. is '02, and
Write 4 per cent. as
42 per cent.
8 per cent.
9 per cent.
is ; decimally, ‘10.)
124 per cent.
}
A
Note. The rate per cent. is a decimal carried to two
places, that is, to hundredths; all decimal expressions lower
than hundredths are parts of 1 per cent.
stance, is '625 of 1 per cent., that is, '00625.
per cent., for in-
1
[
decimal fraction.
per cent. is '005.
a decimal fraction.
5 per cent.
82 per cent.
10 per
101 per cent.
15 per cent.
t
'
157
The
80 of a
as a de-
'06.
JA
+
01.
005.
'0025.
'0075.
Ans. '025.
4 per
71 per
9 per
cent. (10 per cent.
11 per
1. If the interest on $1, for 1 year, be 6 cents, what will
be the interest on $17 for the same time?
It will be 17 times 6 cents, or 6 times 17, which is the
same thing:-
$17
'06
1'02 Answer; that is, 1 dollar and 2 cents.
To find the interest on any sum for 1 year, it is evident
we need only to multiply it by the rate per cent. written as a
decimal fraction. The product, observing to place the point
as directed in multiplication of decimal fractions, will be the
interest required.
Note. PRINCIPAL is the money due, for which interest is
paid. AMOUNT is the principal and interest added together.
O
↓
}
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158
T 81, 82.
2. What will be the interest of $32′15, 1 year, at 41 per
cent. ?
per cent. ?
}
i
$32′15 principal.
W
16075
INTEREST.
12860
There being five de-
cimal places in the mul-
tiplicand and multiplier,
five figures must be
pointed off for decimals
from the product, which
gives the answer,-1
Ans. $1'44675
7
dollar, 44 cents, 6 mills, and 75% of a mill. Parts of a mill are
not generally regarded; hence, $1'446 is sufficiently exact
for the answer.
'045 rate per cent.
3. What will be the interest of $11'04 for 1 year, at 3
per cent. ?
per cent. ?
at 5
at 6 per cent. ?
at 74 per cent.?
at 10 per cent. ?
at 81 per cent. ?
at 10
at 112 per cent. ?
at 12 per cent. ?
4. A tax on a certain town is $1627'18, on which the
collector is to receive 2 per cent. for collecting; what will
he receive for collecting the whole tax at that rate?
Ans. $40'679.
Ja
per
at 94 per cent.?
cent. ? at 11
at 12 per cent.?
Note. In the same way are calculated commission, in-
urance, buying and selling stocks, loss and gain, or any
thing else rated at so much per cent. without respect to time.
}
5. What must a man, paying $0'371 on a dollar, pay on
a debt of $132′25 ?
Ans. $49'593.
6. A merchant, having purchased goods to the amount of
$580, sold them so as to gain 12 per cent., that is, 121
cents on each 100 cents, and in the same proportion for a
greater or less sum; what was his whole gain, and what was
the whole amount for which he sold the goods?
Ans. His whole gain was $72'50; whole amount
$ 652'50.
7. A merchant bought a quantity of goods for $763′371 ;
how much must he sell them for to gain 15 per cent. ?
Ans. $ 877'881.
T 82. COMMISSION is an allowance of so much per cent.
to a person called a correspondent, factor, or broker, for as-
sisting merchants and others in purchasing and selling goods.
J
159
T 82.
8. My correspondent sends me word that he has pur-
chased goods to the value of $1286, on my account; what
will his commission come to at 24 per cent.? Ans. $32'15.
9. What must I allow my correspondent for selling goods
to the amount of $2317'46, at a commission of 31 per cent.?
Ans. $75'317.
;
!
INSURANCE is an exemption from hazard, obtained by the
payment of a certain sum, which is generally so much per
cent. on the estimated value of the property insured.
PREMIUM is the sum paid by the insured for the insurance.
POLICY is the name given to the instrument or writing,
by which the contract of indemnity is effected between the
insurer and insured.
INTEREST.
10. What will be the premium for insuring a ship and
cargo from Boston to Amsterdam, valued at $37800, at 41
per cent. ?
Ans. $1701.
11. What will be the annual premium for insurance on a
house against loss from fire, valued at $3500, at 2 per cent. ?
By removing the separatrix 2 figures towards the left, it is
evident, the sum itself may be made to express the premium
at 1 per cent., of which the given rate parts may be taken;
thus, 1 per cent. on $3500 is $35'00, and 2 of $35'00 is
$26'25, Answer.
Y
12. What will be the premium for insurance on a ship and
cargo valued at $25156'86, at per cent.?
at per
at & per cent. ?
at & per cent? at &
Ans. At § per cent. the premium is $157'23.
cent. ?
per cent. ?
STOCK is a general name for the capital of any trading
company or corporation, or of a fund established by govern-
ment.
I
The value of stock is variable. When 100 dollars of
stock sells for 100 dollars in money, the stock is said to be at
par, which is a Latin word signifying equal; when for more,
it is said to be above par; when for less, it is said to be be-
low par.
13. What is the value of $7564 of stock, at 112 per
cent.? that is, when 1 dollar of stock sells for 1 dollar 124
↓
1
!
1
}
160
T 82, 83
cents in money, which is 12 per cent. above par, or 12 per
cent. advance, as it is sometimes called. Ans. $8509'50.
14. What is the value of $3700 of bank stock, at 951
per cent., that is, 4 per cent. below par? Ans. $3533'50.
15. What is the value of $120 of stock, at 921 per cent. ?
at 861 per cent.? at 672 per cent., at 1044
per cent.?
at 1081 per cent.?
at 115 per cent. ?
INTEREST.
Add V
p
at 37 per cent. advance?
1
LOSS AND Gain. 16. Bought a hogshead of molasses for
$60; for how much must I sell it to gain 20 per cent.?
Ans. $72.
17. Bought broadcloth at $2'50 per yard; but, it being
damaged, I am willing to sell it so as to lose 12 per cent.;.
how much will it be per yard?
Ans. $2'20.
18. Bought calico at 20 cents per yard; how must I sell it
to gain 5 per cent.? 10 per cent. ? 15 per cent.?
Ans. to the last, 16 cents per yard.
to lose 20 per cent.?
T 83. We have seen how interest is cast on any sum of
money, when the time is one year; but it is frequently ne-
cessary to cast interest for months and days.
Toda
Now, the interest on $1 for 1 year, at 6 per cent., being
'06, is
'01 cent for 2 months,
'005 mills (or a cent) for 1 month of 30 days, (for so we
reckon a month in casting interest,) and
'001 mill for every 6 days; 6 being contained 5 times in 30,
1
Hence, it is very easy to find by inspection, that is, to cast
in the mind, the interest, on 1 dollar, at 6 per cent. for any
given time. The cents, it is evident, will be equal to half the
greatest even number of the months; the mills will be 5 for
the odd month, if there be one, and 1 for every time 6 is
contained in the given number of the days.
Suppose the interest of $1, at 6 per cent., be required for
9 months, and 18 days. The greatest even number of the
months is 8 half of which will be the cents, '04; the mills,
reckoning 5 for the odd month, and 3 for the 18 (3 times 6
18) days, will be '008, which, united with the cents,
(048) give 4 cents 8 mills for the interest of $1 for 9
months and 18 days.
•
1
1
1
T`83.
161
1. What will be the interest on $1 for 5 months 6 days?
77 months?
8 months
6 months 12 days?
24 days?
11 months 6 days?
months 6 days?
Kādā
1
|
INTEREST.
9 months 12 days?
1
Target lang
¿
Saja Tang
16 months?
ODD DAYS. 2. What is the interest of $1 for 13 months
16 days?
12 months 18 days?
1
✯
The cents will be 6, and the mills 5, for the odd month,
and 2 for 2 times 6 = 12 days, and there is a remainder of
4 days, the interest for which will be such part of 1 mill as 4
days is part of 6 days, that is, & = 3 of a mill. Ans. '0673.
3. What will be the interest of $1 for 1 month 8 days?
2 months 7 days?
3 months 15 days?
months 22 days?
5 months 11 days?
17 days?
7 months 3 days?
1
6 months
8 months 11 days?
10 months 15 days?
12 months 3 days?
9 months 2 days?
11 months 4 days?
10 months?
Note. If there is no odd month, and the number of days be
less than 6, so that there are no mills, it is evident, a cipher must
be put in the place of mills; thus, in the last example,-12
months 3 days,-the cents will be '06, the mills 0, the 3
days a mill.
Ans. '0601
J
4. What will be the interest of $1 for 2 months 1 day?
4 months 2 days? 6 months 3 days?
months 4 days?
10 months 5 days?
for 2 days?
8
for 1 day?
for 5 days?
1) and 1) $56 13 principal.
5. What is the interest of $ 56'13 for 8 months 5 days?
The interest of $1, for the given time, is '0408; therefore,
15
‘040§ interest of $1 for the given time.
224520 interest for 8 months.
2806 interest for 3 days.
1871 interest for 2 days.
2'29197, Ans. $2'291.
for 3 days?
for 4 days?
1
5 days 3 days+2 days. As the multiplicand is taken
$
once for every 6 days, for 3 days take, for 2 days take d
f
F
+
1
1
I
F
}
1
}
162
T 83.
of the multiplicand.+. So also, if the odd days
be 4 2 days+2 days, take of the multiplicand twice;
for 1 day, take f.
I
Note. If the sum on which interest is to be cast be less
than $10, the interest, for any number of days less than 6,.
will be less than 1 cent; consequently, in business, if the sum
be less than $10, such days need not be regarded.
• 2.
1
}
From the illustrations now given, it is evident,-To find the
interest of any sum in federal money, at 6 per cent., it is only
necessary to multiply the principal by the interest of $1 for
the given time, found as above directed, and written as a.
decimal fraction, remembering to point off as many places
for decimals in the product as there are decimal places in
both the factors counted together.
N
EXAMPLES FOR PRACTICE.
6. What is the interest of $87'19 for 1 year 3 months?
Ans. $6'539.
$6751.
$8132.
7. Interest of $116,08 for 11 mo. 19 days?
8.
of $200 for 8 mo. 4 days?
of $0'85 for 19 mo. ?
9.
10.
of $850 for 1 year 9 mo. 12 days?
of $675 for 1 mo. 21 days?
11.
$56737.
12.
of $8673 for 10 days?
13.
$14'455.
$'036.
14.
of $0'73 for 10 mo. ?
of $96 for 3 days?
of $73'50 for 2 days?
Note. The inte-
rest of $1 for 6 days
15.
16.
17.
of $180 75 for 5 days? (being 1 mill, the dol-
of $15000 for 1 day?
lars themselves ex-
press the interest in mills for six days, of which we may take
parts.
MANAG
*10*
UMSHO
INTEREST.
...
1
**•
Thus, 6) 15000 mills,
វ
2'500, that is, $250, Ans. to the last.
When the interest is required for a large number of years,
it will be more convenient to find the interest for one year,.
and multiply it by the number of years; after which find.
the interest for the months and days, if any, as usual.
18. What is the interest of $1000 for 120 years?
Ans. $7200..
19. What is the interest of $520'04 for 30 years and
6 months?
Ans. $951'673..
$'08.
'909.
}
I
ہے
་
T 83, 84.
163
20. What is the interest on $400 for 10 years 3 months
and 6 days?
Ans. $246'40.
for
21. What is the interest of $220 for 5 years?
12 years?
50 years?
Ans. to last, $660.
22. What is the amount of $86, at interest 7 years?
Ans. $122 12.
23. What is the interest of 36 £. 9 s. 61 d. for 1 year?
Reduce the shillings, pence, &c. to the decimal of a pound,
by inspection, (T76;) then proceed in all respects as in
federal money. Having found the interest, reverse the ope-
ration, and reduce the three first decimals to shillings, &c.,
by inspection. (T 77.)
Ans. 2 £. 3 s. 9 d.
24. Interest of 36 £. 10 s. for 18 mo. 20 days? Ans. 3 £.
8 s. 1 d. Interest of 95 £. for 9 mo.? Ans. 4 £. 5 s. 6 d.
25. What is the amount of 18 £. 12 s. at interest 10
months 3 days?
Ans. 19 £. 10 s. 91 d.
26. What is the amount of 100 £. for 8 years?
(dekapotą kepada
$36
'04
Ans. 148 £.
27. What is the amount of 400 £. 10 s. for 18 months?
Ans. 436 £. 10 s. 10 d. 3 q.
28. What is the amount of 640 £. 8 s. at interest for 1
year?
for 2 years 6 months?
for 10 years?
Ans. to last, 1024 £. 12 s. 91 d.
144
'36
2
¶ 84. 1. What is the interest of 36 dollars for 8 months,
at 4 per cent. ?
1
INTEREST.
Note. When the rate is any other than six per cent., first
find the interest at 6 per cent., then divide the interest so
found by such part as the interest, at the rate required, ex-
ceeds or falls short of the interest at 6 per cent., and the
quotient added to, or subtracted from the interest at 6 per
cent., as the case may be, will give the interest at the rate
required.
}
1
}
!
……………………………
41 per cent. is of 6 per cent.; therefore,
from the interest at 6 per cent. subtract ;
the remainder will be the interest at 4 per
cent.
1'08 Ans.
3.
2. Interest of $54'81 for 18 mo., at 5 per ct.? Ans. $4′11.
of $500 for 9 mó. 9 days, at 8 per ct.? $31'00.
of $62′12 for 1 mo. 20 days, at 4 per ct.? $345,
4.
1
}
1
(
164
5. Interest of $85 for 10 mo. 15 days, at 121 per cent.?
}
14
INTEREST.
Ans. $9'295.
6. What is the amount of $53 at 10 per ct. for 7 mo. ?
Ans. $56'091.
The time, rate per cent. and amount given, to find the principal.
¶ 85. 1. What sum of money, put at interest at 6 per
cent., will amount to $61'02, in 1 year 4 months?
The amount of $1, at the given rate and time, is $ 1'08
hence, $61'02 $1'08 56'50, the principal required;
÷
that is,-Find the amount of $1 at the given rate and time, by
which divide the given amount; the quotient will be the princi-
pal required.
Ans. $56'50.
$
2. What principal, at 8 per cent., in 1 year 6 months, will
amount to $85′12 ?
Ans. $76.
3. What principal, at 6 per cent., in 11 months 9 days,
will amount to $99311?
Note. The interest of $1, for the given time, is '0561;
but, in these cases, when there are odd days, instead of
writing the parts of a mill as a common fraction, it will be
more convenient to write them as a decimal, thus, ‘0565;
that is, extend the decimal to four places. Ans. $94.
4. A factor receives $988 to lay out after deducting his
commission of 4 per cent.; how much will remain to be
laid out?
A
T 84, 85
}
It is evident, he ought not to receive commission on his
own money. This question, therefore, in principle, does not
differ from the preceding.
}
Note. In questions like this, where no respect is had to
time, (T 81, ex. 4, note,) add the rate to $1. Ans. $950.
5. A factor receives $1008 to lay out after deducting
his commission of 5 per cent.; what does his commission
amount to?
Ans. $48.
Madagaskar Mada
2
DISCOUNT. 6. Suppose I owe a man $39750, to be paid
in 1 year, without interest, and I wish to pay him now; how
much ought I to pay him when the usual rate is 6 per cent. ?
I ought to pay him such a sum as, put at interest, would,
in 1 year, amount to $397'50. The question, therefore,
does not differ from the preceding.
Ans. $375.
Note. An allowance made for the payment of any sum
L
1
(
A
1
?
1
T. 85, 86.
165
of money before it becomes due, as in the last example, is
called Discount.
{
}
The sum which, put at interest, would, in the time and
at the rate per cent. for which discount is to be made, amount
to the given sum, or debt, is called the present worth.
1
7. What is the present worth of $S34, payable in 1 year
7 months and 6 days, discounting at the rate of 7 per cent.?
Ans. $750.
8. What is the discount on $321'63, due 4 years hence,
discounting at the rate of 6 per cent.? Ans. $62'26.
INTEREST.
9. How much ready money must be paid for a note of
$ 18, due 15 months hence, discounting at the rate of 6 per
cent. ?
Ans. 16'744.
10. Sold goods for $650, payable one half in 4 months,
and the other half in 8 months; what must be discounted
for present payment?
Ans. 18'873
{
11. What is the present worth of $ 56'20, payable in 1
year 8 months, discounting at 6 per cent. ?
at 4 per
"cent.?
7 per cent. ?
at 7 per cent.?
at
at 5 per cent. ?
1
PROTE
at 9 per cent. ?
The time, rate per cent., and interest being given, to find the
principal.
¶ 86. - 1. What sum of money, put at interest 16 months,
will gain $10'50, at 6 per cent. ?
÷
$1, at the given rate and time, will gain '08; hence,
$10'50 $'08 $131'25, the principal required; that
is,-Find the interest of $1, at the given rate and time, by
which divide the given gain, or interest; the quotient will be the
principal required.
Ans. $131′25.
I
Ans. to the last, $48'869.
2. A man paid $4'52 interest, at the rate of 6 per cent.
at the end of 1 year 4 months; what was the principal ?
Ans. $56'50.
3. A man received, for interest on a certain note, at the
end of 1 year, $20; what was the principal, allowing the
rate to have been 6 per cent.?
Ans. $333'3331.
1
I
<
1
2
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1
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1
1
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¦
166
}
J
¶ 87, 88:
The principal, interest, and time being given, to find the rate
per cent.
T 87. 1. If I pay $378 interest, for the use of $36
for 1 year and 6 months, what is that per cent. ?
The interest on $36, at one per cent. the given time, is $ '54;
hence, $378÷54 '07, the rate required; that is,-
Find the interest on the given sum, at 1 per cent. for the given
time, by which divide the given interest; the quotient will be
the rate at which interest was paid.
Ans. 7 per cent.
1
4
2. If I pay $234 for the use of $468, 1 month, what is
the rate per cent. ?
Ans. 6 per cent.
3. At $46'80 for the use of $520, 2 years, what is that
per cent. ?
Ans. 4 per cent.
t
INTEREST.
}
The prices at which goods are bought and sold being given, to
find the rate per cent. of GAIN or Loss.
1
¶ 88. 1. If I purchase wheat at $1'10 per bushel, and
sell it at $1'37 per bushel, what do I gain per cent. ?
(275
This question does not differ essentially from those in the
foregoing paragraph. Subtracting the cost from the price
at sale, it is evident I gain 274 cents on a bushel, that is,
(275
of the first cost. '25 per cent., the Answer. That is,
-Make a common fraction, writing the gain or loss for the numera-
tor, and the price at which the article was
nominator; then reduce it to a decimal.
1410
1410
bought for the de-
Kateg
2. A merchant purchases goods to the
what per cent. profit must he make to gain $66 ?
Ans. 12 per cent.
3. What per cent. profit must he make on the same
purchase to gain $38′50 ? to gain $24'75}
gain $275 P
to
i
Note. The last gain gives for a quotient '005, which is
The rate per cent., it must be recollected, (T 81,
per cent.
note,) is a decimal carried to two places, or hundredths; all
decimal expressions lower than hundredths are parts of 1
per cent.
•
amount of $ 550;
4. Bought a hogshead of rum, containing 114 gallons, at
96 cents per gallon, and sold it again at $1'0032 per gal-
lon; what was the whole gain, and what was the gain per
cent. ?
Ans. 4 gain per cent.
{ $4'924, whole gain.
}
}
1
1
&
¶ 88, 89.
5. A merchant bought a quantity of tea for $365, which,
proving to have been damaged, he sold for $332'15; what
did he lose per cent. ?
Ans. 9 per cent.
-
7. Bought indigo at $1'20 per lb., and
90 cents per lb.; what was lost per cent.?
INTEREST.
6. If I buy cloth at $2 per yard, and sell it for $2'50
per yard, what should I gain in laying out $100 ?
Ans. $25.
S
1
JA
8. Bought 30 hogsheads of molasses, at $600; paid in
duties $20'66; for freight, $40'78; for porterage, $6'05,
and for insurance, $30'84: if I sell them at $26 per hogs-
head, how much shall I gain per cent.? Ans. 11'695 per cent.
14
167
Y
sold the same at
Ans. 25 per cent.
The principal, rate per cent., and interest being given, to find
the time.
F
T 89. 1. The interest on a note of $36, at, 7 per cent.,
was $378; what was the time?
The interest on $36, 1 year, at 7 per cent., is $2'52;
hence, $378 $2'52 1'5 years, the time required; that
is,—Find the interest for 1 year on the principal given, at the
given rate, by which divide the given interest; the quotient will
be the time required, in years and decimal parts of a year; the
latter may then be reduced to months and days.
Ans. 1 year 6 months.
2. If $3171 interest be paid on a note of $226'50,
what was the time, the rate being 6 per cent.?
Ans. 2'331
2 years 4 months.
3. On a note of $600, paid interest $20, at 8 per cent.;
what was the time?
Ans. '416 + 5 months so nearly as to be called 5, and
would be exactly 5 but for the fraction lost.
4. The interest on a note of $21725, at 4 per cent., was
$28'242; what was the time? Ans. 3 years 3 months.
Note. When the rate is 6 per cent., we may divide the
interest by the principal, removing the separatrix two
places to the left, and the quotient will be the answer in
months.
1
мама и на почето
Y
}
1
J
1
+
Ÿ
!
1
'
168
'INTEREST.
To find the interest due on notes, &c. when partial payments
have been made.
!
¡
T 90. In Massachusetts the law provides, that payments
"shall be applied to keep down the interest, and that neither
interest nor payment shall ever draw interest. Hence, if the
payment at any time exceed the interest computed to the
same time, that excess is taken from the principal; but if
the payment be less than the interest, the principal remains
unaltered. Wherefore, we have this RULE :-Compute the
interest to the first time when a payment was made, which,
either alone, or together with the preceding payments, if
any, exceeds the interest then due; add that interest to the
principal, and from the sum subtract the payment, or the
sum of the payments, made within the time for which the
interest was computed, and the remainder will be a new
principal, with which proceed as with the first.
1. For value received, I promise to pay JAMES CONANT, or
order, one hundred sixteen dollars sixty-six cents and six mills,
with interest. May 1, 1822.
$116,666.
SAMUEL ROOD.
On this note were the following endorsements:
Dec. 25, 1822, received $16'666
$1'666
July 10, 1823,
Sept. 1, 1824,
$5'000
June 14, 1825,
$33'333
April 15, 1826,
$62'000
What was due August 3, 1827 ?
1
1
The first principal on interest from May 1, 1822,
Interest to Dec. 25, 1822, time of the first pay-
ment, (7 months 24 days,)
L
T 90,"
Note. In finding the
times for computing the
interest, consult T 40.
Amount,
Payment, Dec. 25, exceeding interest then due,
Ans. $23'775.
$116'666
Remainder for a new principal,
Interest from Dec. 25, 1822, to June 14, 1825,
(29 months 19 days,)
j
4'549
$121215
16'666
104'549
15'490
Amount carried forward, $120'039;
f
1
1
?
T 90, 91.
}
COMPOUND INTEREST.
Payment, July 10, 1823, less than interest
then due,
Payment, Sept. 1, 1824, less than interest
then due,
Payinent, June 14, 1825, exceeding in-
terest then duë,
سے
169
Amount brought forward, $120 039
f
¿
Remainder for a new principal, (June 14, 1825,)
Interest from June 14, 1825, to April 15, 1826,
(10 months 1 day,)
$ 14666
5'000
Payment, April 15, 1825, exceeding interest then
due,
1
33.333
Ad
(
4'015
Amount, $ 84'055
39'999
80'040
Remainder for a new principal, (April 15, 1826,)
Interest due Aug. 3, 1827, from April 15, 1826,
(15 months 18 days,)
19720
Balance due Aug 3, 1827,
$23 775
2. For value received, I promise to pay JAMES. LOWELL, 07
order, eight hundred sixty-seven dollars and thirty-three cents.
with interest. Jan. 6. 1820.
$867'33.
62'000
$22'055
HIRAM SIMSON.
On this note were the following endorsements, viz.
April 16, 1823, received $136'44.
April 16, 1825, received $319.
Jan. 1, 186, received $518'68.
What remained due July 11, 1827 ? Ans. $215'103.
COMPOUND INTEREST.
T 91. A promises to pay B $256 in 3 years, with in-
terest annually; but at the end of 1 year, not finding it con-
venient to pay the interest, he consents to pay interest on
the interest from that time, the same as on the principal.
Note. Simple interest is that which is allowed for the
principal only; compound interest is that which is allowed
P
1
1
{
I
i
1
1
170
T 91.
for both principal and interest, when the latter is not paid at
the time it becomes due.
Compound interest is calculated by adding the interest to
the principal at the end of each year, and making the amount
the principal for the next succeeding year.
1. What is the compound interest of $256 for 3 years,
at 6 per cent.?
(
COMPOUND INTEREST.
t
!
$256 given sum, or first principal.
'06
256400 principal,
to be added together.
'271'36 amount, or principal for 2d year.
'06
16'2816 compound interest, 2d year, added to-
271'36 principal,
do. Sgether.
287'6416 amount, or principal for 3d year.
'06
304'899
256
amount.
first principal subtracted.
*
17'25846 compound interest, 3d year, added to-
287'641
principal,
do.
·
1
鲁
​$
Ans. $48'899
compound intercet for 3 years.
what the amount
2. At 6 per cent., what will be the compound interest, and
what the amount, of $1 for 2 years?
for 3 years?
for 4 years?
for 7 years?
for
6 years?
for 5 years?
for 8 years
?
Ans. to the last, $1'593+.
}
It is plain that the amount of $2, for any given time, will
be 2 times as much as the amount of $1; the amount of
$3 will be 3 times as much, &c.
Hence, we may form the amounts of $1, for several years,
into a table of multipliers for finding the amount of any sum
for the same time.
T 91.
Years. 5 per cont.
1'05
1
2
TABLE,
Showing the amount of $1, or 1£., &c. for any number of
years, not exceeding 24, at the rates of 5 and 6 per cent.
compound interest.
{
304 10
1
5
COMPOUND INTEREST.
6
6 per cent.
1'06
1'1236
{
Years.
13
14
1'1025
1'15762
1'1910115
1'21550 1'26247 16
17
1276281'33822
1340091'41851 || 18
71407101650363+
19
8
9
1'47745 1'59384 20
155132+1'68947+ || 21
1'628891'79084 + || 22
10
11
1710331'89829 +-|| 23
12 1'79585 + 2'01219+|| 24 | 3′22509 +14′04893+
171
6 per cent.
5 per cent.
1'885642'13292 +
1'979932626090 -↓
2'07892 +239655+
2'18287 2454035
2'29201 + 2'69277
2'406612'85433 +
2'52695 3'02559+
2'65329 +3'20713+
Note 1. Four decimals in the above numbers will be suf-
ficiently accurate for most operations.
Note 2. When there are months and days, you may first
find the amount for the years, and on that amount cast the
interest for the months and days; this, added to the amount,
will give the answer.
2'785963'39956 +
2'925263'60353 +
3'071523481974 +
3. What is the amount of $600'50 for 20 years, at 5 per
cent. compound interest? at 6 per cent.?
$1 at 5 per cent., by the table, is $2'65329'; therefore,
2'65329 X 600'50 $1593'30+ Ans. at 5 per cent.; and
3'20713 × 600'50 = $1925'881+ Ans. at 6 per cent.
}
pound interest, for 4 years?
years? for 12 years?
4. What is the amount of $40'20 at 6 per cent. com-
for 10 years?
for 18
for 3 years and 4 months?
for 24 years, 6 months, and 18 days?
Ans. to last, $168'137.
Note. Any sum at compound interest will double itself
in 11 years, 10 months, and 22 days.
From what has now been advanced we deduce the fol-
lowing general
1
RULE.
I. To find the interest when the time is 1 year, or, to find the
rate per cent. on any sum of money, without respect to time, as
1
1
}
COMPOUND INTEREST.
172
T 91.
the premium for insurance, commission, &c., Multiply the
principal, or given sum, by the rate per cent., written as a
decimal fraction; the product, remembering to point off as
many places for decimals as there are decimals in both the
factors, will be the interest, &c. required.
1
II. When there are months and days in the given time, to find
the interest on any sum of money at 6 per cent.,-Multiply the
principal by the interest on $1 for the given time, found by
inspection, and the product, as before, will be the interest
required.
·
III. To find the interest on $1 at 6 per cent., for any given
time, by inspection,-It is only to consider, that the cents will
be equal to half the greatest even number of the months;
and the mills will be 5 for the odd month, (if there be one,)
and 1 for every 6 days.
IV. If the sum given be in pounds, shillings, pence and far-
things,-Reduce the shillings, &c. to the decimal of a pound,
by inspection, ( 76;) then proceed in all respects as in
federal money. Having found the interest, the decimal part,
by reversing the operation, may be reduced back to shillings,
pence and farthings.
V. If the interest required be at any other rate than 6 per
cent., (if there be months, or months and days, in the given time,)
-First find the interest at 6 per cent.; then divide the in-
terest so found by such part or parts, as "the interest, at the
ráte required, exceeds or falls short of the interest at 6 per
cent., and the quotient, or quotients, added to or subtracted
from the interest at 6 per cent., as the case may require, will
give the interest at the rate required.
5
"
Note. The interest on any number of dollars, for 6 days,
at 6 per cent., is readily found by cutting off the unit or right
hand figure; those at the left hand will show the interest in
cents for 6 days.
EXAMPLES FOR PRACTICE.
1. What is the interest of $1600 for 1 year and 3 months?
Ans. $120.
2. What is the interest of $54811, for 1 year 11 months?
Ans. $'668.
3. What is the interest of $2'29, for 1 month 19 days,
at 3 per cent. ?
Ans. $'009.
4. What is the interest of $18, for 2 years 14 days, at 7
per cent. ?
Ans. $2'569.
:
f
1
T 91.
173
5. What is the interest of $1768, for 11 months 28
days?
Ans. $1.054.
6. What is the interest of $200 for 1 day? 2 days?
4 days?
3 days?
5 days?
34
}
SUPPLEMENT TO INTEREST.
cent.?
cent. ?
Žablja
1
Ans. for 5 days, $0‘166.
7. What is the interest of half a mill for 567 years ?
Ans. $0'017.
8. What is the interest of $81, for 2 years 14 days, at 2
per cent. ?
2 per
5 per
cent. ?
₫ per cent. ?
3 per cent.?
6 per cent. ?
& per cent. ?
41 per cent.?
7 per cent.?
9 per cent. ?
1 per
cent. ?
10 per
8 per cent. ?
12 per cent. ?
!
124 per cent.?
Ans. to last, $20'643,
9. What is the interest of 9 cents for 45 years, 7 months,
11 days?
Ans. $0'245.
10. A's note of $175 was given Dec. 6, 1798, on which
was endorsed one year's interest; what was there due Jan..
1, 1803 ?
Note. Consult ex. 16, Supplement to Subtraction of Com-
pound Numbers.
Ans. $207'22.
11. B's note of $5675 was given June 6, 1801, on inter-
est after 90 days; what was there due Feb. 9, 1802?
Ans. $58'19.
12. C's note of $365'37 was given Dec. 3, 1797; June
7, 1800, he paid $97'16; what was there due Sept. 11,
Ans. $328'32.
1800 ?
by commission? 8.
10.
13. Supposing a note of $317'92, dated July 5, 1797, on
which were endorsed the following payments, viz. Sept. 13,
1799, $208′04; March 10, 1800, $76; what was there
due Jan. 1, 1801 ?
Ans. $83'991..
SUPPLEMENT TO INTEREST.
QUESTIONS.
1. What is interest? 2. How is it computed? 3. What
is understood by rate per cent.? 4. by principal?
by legal interest? 7.
insurance? 9. premium?
5.
by amount? 6.
stock?
12. What is under-
above par? 14.
policy? 11.
stood by stock being at par? 13.
P*
1
$
{
f
T 91.
below par? 15. The rate per cent. is a decimal car-
ried to how many places? 16. What are decimal expres-
sions lower than hundredths? 17. How is interest, (when
the time is 1 year,) commission, insurance, or any thing else
rated at so much per cent. without respect to time, found?
18. When the rate is 1 per cent., or less, how may the ope-
ration be contracted? 19. How is the interest on $1, at
6 per cent. for any given time, found by inspection? 20.
How is interest cast, at 6 per cent., when there are months
and days in the given time? 21. When the given time is
less than 6 days, how is the interest most readily found?
22. If the sum given be in pounds, shillings, &c., how is in-
terest cast? 23. When the rate
23. When the rate is any other than 6 per
cent, if there be months and days in the given time, how is
the interest found? 24. What is the rule for casting interest
on notes, &c. when partial payments have been made, and
what is the principle on which the rule is founded? 25.
How may the principal be found, the time, rate per cent.,
and amount being given? 26. What is understood by dis-
count? 27. by present worth? 28. How is the prin-
cipal found, the time, rate per cent., and interest being given?
29. How is the rate per cent. of gain or loss found, the
prices at which goods are bought and sold being given? 30.
How is the rate per cent. found, the principal, interest, and
time being given? 31. How is the time found, the princi-
pal, rate per cent., and interest being given? 32. What is
simple interest? 33. compound interest? 34. How
is compound interest computed?
174
}
1
.
t
SUPPLEMENT TO INTEREST.
EXERCISES.
n
1. What is the interest of $273'51 for 1 year 10 days, at
per cent. ?
Ans. 19677.
2. What is the interest of $ 486 for 1 year, 3 months, 19
days, at 8 per cent. ?
Ans. $50'652.
3. D's note of $203'17 was given Oct. 5, 1808, on inter-
est after three months; Jan. 5, 1809, he paid $50; what
was there due May 2, 1811?
Ans. $17453.
4. E's note of $870'05 was given Nov. 17, 1800, on in-
terest after 90 days; Feb. 11, 1805, he paid $186'06; what
was there due Dec. 23, 1807 ?
Ans. $1041'58.
5. What will be the annual insurance, at § per cent., on
» house valued at $1600 ?
Ans. $10.
}
4
kad
i
$'
1
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1
T 91.
6. What will be the insurance of a ship and cargo, valued
at $5643, at 1 per cent.? at 4 per cent.?
per cent. ?
Note.
at 76
at 11 per cent.?
at per cent. ?
Consult T 82, ex. 11.
SUPPLEMENT TO INTEREST.
$
+
Ans. at per cent. $42'322.
7. A man having compromised with his creditors at 623
cents on a dollar, what must be pay on a debt of $137'46 ?
Ans. $85'912.
8. What is the value of $800 United States Bank stock,
at 112 per cent. ?·
Ans. $900.
9. What is the value of $56075 of stock, at 93 per cent.?
Ans. $521′497
175
10. What principal at 7 per cent. will, in 9 months 18 days,
amount to $422'40?
Ans. $400.
་
11. What is the present worth of $426, payable in 4
years and 12 days, discounting at the rate of 5 per cent.?
In large sums, to bring out the cents correctly, it will
sometimes be necessary to extend the decimal in the divisor
to five places.
Ans. $354'506.
12. A merchant purchased goods for $250 ready money,
and sold them again for $300, payable in 9 months; what
did he gain, discounting at 6 per cent.? Ans. $37'081.
13. Sold goods for $3120, to be paid, one half in 3
months, and the other half in 6 months; what must be dis-
counted for present payment ?
Ans. 68'492.
14. The interest on a certain note, for 1 year 9 months,
was $49'875; what was the principal ? Ans. $475.
15. What principal, at 5 per cent., in 16 months 24 days,
will gain $35?
Ans. $500.
16. If I pay $15'50 interest for the use of $500, 9
months and 9 days, what is the rate per cent. ?
•
17. If I buy candles at $167 per lb., and sell them at
20 cents, what shall I gain in laying out $100 ?
Ans. 1976.
18. Bought hats at 4 s. apiece, and sold them again at 4 s..
9 d.; what is the profit in laying out 100 €.?
+
Ans. 18 £. 15 s.
19. Bought 37 gallons of brandy, at $1'10 per gallon,
and sold it for $40; what was gained or lost per cent.?
20. At 4 s. 6 d. profit on 1 £., how much is gained in laying
out 100 ., that is, how much per cent.? Ans. 22 £. 10 s.
21. Bought cloth at $4'48 per yard; how must I sell it
to gain 124 per cent. ?
Ans. $5'04.
1
1
1
}
1
J
176
T 91, 92
22. Bought a barrel of powder for 4 £.; for how much
must it be sold to lose 10 per cent.?
Ans. 3 £. 12 s.
23. Bought cloth at 15 s. per yard, which
good as I expected, I am content to lose '17
must I sell it per yard?
EQUATION OF PAYMENTS.
ů
24. Bought 50 gallons of brandy, at 92
but by accident 10 gallons leaked out; at
sell the remainder per gallon to gain upon the whole cost at
the rate of 10 per cent. ?
Ans. $1'265 per gallon.
61
25. A merchant bought 10 tons of iron for $950; the
freight and duties came to $145, and his own charges to
$25; how must he sell it per lb. to gain 20 per cent. by it?
Ans. 6 cents per lb.
{
not proving so
per cent.; how
Ans. 12 s. 44 d.
cents per gallon,
what rate must Í
EQUATION OF PAYMENTS.
T 92. Equation of payments is the method of finding the
mean time for the payment of several debts, due at different
times.
1. In how many months will $1 gain as much as 5 dol-.
lars will gain in 6 months?
2. In how many months will $1 gain as much as $40
will gain in 15 months?
Ans. 600.
3. In how many months will the use of $5 be worth as
much as the use of $1 for 40 months?
4. Borrowed of a friend $1 for 20 months; afterwards
lent my friend $4; how long ought he to keep it to become
indemnified for the use of the $1?
5. I have three notes against a man; one of $12, due in
3 months; one of $9, due in 5 months; and the other of
$6, due in 10 months; the man wishes to pay the whole at
once; in what time ought he to pay it?
$12 for 3 months is the same as $1 for 36 months, and
$9 for 5 months is the same as $1 for 45 months, and
$6 for 10 months is the same as $1 for 60 months.
27
141
He might, therefore, have $1 141 months, and he may
keep 27 dollars part as long; that is, = 5 months
6 days, Answer.
1
1
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Į
}
F
1
1
T 93.
RATIO; OR THE RELATION OF NUMBERS. 177
Hence, To find the mean time for several payments,—RULE :
-Multiply each sum by its time of payment, and divide the
sum of the products by the sum of the payments, and the
quotient will be the answer.
Note. This rule is founded on the supposition, that what
is gained by keeping a debt a certain time after it is due, is
the same as what is lost by paying it an equal time before it
is due; but, in the first case, the gain is evidently equal to the
interest on the debt for the given time, while, in the second
case, the loss is only equal to the discount of the debt for that
time, which is always less than the interest; therefore, the
rule is not exactly true. The error, however, is so trifling,
in most questions that occur in business, as scarce to merit
notice.
6. A merchant has owing him $300, tó be paid as fol-
lows: $50 in 2 months, $100 in 5 months, and the rest in
8 months; and it is agreed to make one payment of the
whole in what time ought that payment to be?
Ans. 6 months.
RATIO;
OR
7. A owes B $136, to be paid in 10 months; $ 96, to be
paid in 7 months; and $260, to be paid in 4 months: what
is the equated time for the payment of the whole?
Ans. 6 months, 7 days +.
8. A owes B $600, of which $200 is to be paid at the
present time, 200 in 4 months, and 200 in 8 months; what
is the equated time for the payment of the whole?
Ans. 4 months.
in 3 months,
t
9. A owes B $ 300, to be paid as follows:
in 4 months, and the rest in 6 months: what is the equated
time?
Ans. 4 months.
1
i
THE RELATION OF NUMBERS.
Ans.
T 93. 1. What part of 1 gallon is 3 quarts? 1 gallon is
4 quarts, and 3 quarts is 2 of 4 quarts.
of a gallon.
2. What part of 3 quarts is 1 gallon? 1 gallon, being 4
quarts, is of 3 quarts; that is, 4 quarts is 1 time 3 quarts
and of another time.
Ans. 1.
1
1
!
1
=
I
2
178
RATIO; OR THE RELATION OF NUMBERS.
3. What part of 5 bushels is 12 bushels?
Finding what part one number is of another is the same
as finding what is called the ratio, or relation of one number
to another; thus, the question, What part of 5 bushels is 12
bushels? is the same as What is the ratio of 5 bushels to 12
bushels? The Answer is 12 22.
J
Ratio, therefore, may be defined, the number of times one
number is contained in another; or, the number of times one
quantity is contained in another quantity of the same kind.
4. What part of 8 yards is 13 yards? or, What is the ratio
of 8 yards to 13 yards?
?
13 yards is of 8 yards, expressing the division fractionally.
If now we perform the division, we have for the ratio 18;
that is, 13 yards is 1 time 8 yards, and § of another time.
We have scen, (T 15, sign,) that division may be expressed
fractionally. So also the ratio of one number to another, or
the part one number is of another, may be expressed frac-
tionally, to do which, make the number which is called the
part, whether it be the larger or the smaller number, the nu-
merator of a fraction, under which write the other number for
a denominator. When the question is, What is the ratio, &c.?
the number last named is the part; consequently it must be
made the numerator of the fraction, and the number first
named the denominator.
5. What part of 12 dollars is 11 dollars? or, 11 dollars is
what part of 12 dollars? 11 is the number which expresses
the part. To put this question in the other form, viz. What.
is the ratio, &c. let that number, which expresses the part,
be the number last named; thus, What is the ratio of 12 dol-
lars to 11 dollars?
Ans. 1.
Tapatya Vagyong
6. What part of 1 £. is 2 s. 6 d.? or, What is the ratio of
1 £. to 2 s. 6 d. ?
30 pence; hence,
i £.
-
t
240 pencé, and 2 s. 6 d.
, is the Answer.
24.0
7. What part of 13 s. 6 d. is 1 £. 10 s. ? or, What is the ra-
tio of 13 s. 6 d. to 1 £. 10 s.?
Ans. 20.
8. What is the ratio of 3 to 5?
7 to 19?
15?
1107 ?
of 19 to 7?
of 84 to 160?
of 1107 to 615?
Ja ne dey
jpt the fighting, pagal
!
T 93%
+
https
of 5 to 3?
of 15 to 90?
of 160 to 84 ?
1
of
of 90 to
of 615 to
Ans. to the last, 8.
$
}
1
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T 94,
>
RULE OF THREE.
PROPORTION;
OR
THE RULE OF THREE.
4 21 dollars, Answer.
T 94. 1. If a piece of cloth, 4 yards long, cost 12 dollars,
what will be the cost of a piece of the same cloth 7 yards
long?
I
Had this piece contained twice the number of yards of the
first piece, it is evident the price would have been twice as
much; had it contained 3 times the number of yards, the
price would have been 3 times as much; or had it contained
only half the number of yards, the price would have been
only half as much; that is, the cost of 7 yards will be such
part of 12 dollars as 7 yards is part of 4 yards. 7 yards is
of 4 yards; consequently, the price of 7 yards must be of
the price of 4 yards, or 7 of 12 dollars. of 12 dollars, that
is, 12 × 7 84
1
hours. hows, miles. milos,
6, 11, 30, 55,
2. If a horse travel 30 miles in 6 hours, how many miles
will he travel in 11 hours, at that rate ?
11 hours is of 6 hours, that is, 11 hours is 1 time 6
hours, and of another time; consequently, he will travel, in
11 hours, 1 time 30 miles, and of another time, that is, the
ratio between the distances will be equal to the ratio be-
tween the times.
of 30 miles, that is, 30 x
355 miles. If,
then, no error has been committed, 55 miles must be
30 miles. This is actually the case; for 3⇓.
of
Ans. 55 miles.
1
"/
179
Quantities which have the same ratio between them are.
said to be proportional. Thus, these four quantities,
1
1
written in this order, being such, that the second contains
the first as many times as the fourth contains the third, that
is, the ratio between the third and fourth being equal to the
ratio between the first and second, form what is called a pro-
portion.
It follows, therefore, that proportion is a combination of twą
equal ratios. Ratio exists between two numbers; but pro-
portion requires at least three.
⚫
}
1
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​·
180
T 94, 95.
To denote that there is a proportion between the numbers
6, 11, 30, and 55, they are written thus :-
6. 11 :: 30 : 55
} F
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[
RULE OF THREE.
<\
which is read, 6 is to 11 as 30 is to 55; that is, 6 is the
same part of 11, that 30 is of 55; or, 6 is contained in 11 as
many times as 30 is contained in 55; or, lastly, the ratio or
relation of 11 to 6 is the same as that of 55 to 30.
T 95. The first term of a ratio, or relation, is called the
antecedent, and the second the consequent. In a proportion
there are two antecedents, and two consequents, viz. the an-
tecedent of the first ratio, and that of the second; the con-
sequent of the first ratio, and that of the second. In the
proportion 6 : 11:30:55, the antecedents are 6, 30; the
consequents, 11, 55.
?
6
The consequent, as we have already seen, is taken for the
numerator, and the antecedent for the denominator of the
fraction, which expresses the ratio or relation. Thus, the
first ratio is, the second ; and that these two
§5' 11
ratios are equal, we know, because the fractions are equal.
The two fractions and being equal, it follows that,
by reducing them to a common denominator, the numerator
of the one will become equal to the numerator of the other,
and, consequently, that 11 multiplied by 30 will give the
same product as 55 multiplied by 6. This is actually the
case; for 11 × 30 = 330, and 55 × 6 330. Hence it
follows,-If four numbers be in proportion, the product of the
first and last, or of the two extremes, is equal to the product of
the second and third, or of the two means.
?
Hence it will be easy, having three terms in a proportion
given, to find the fourth. Take the last example. Know-
ing that the distances travelled are in proportion to the times
or hours occupied in travelling, we write the proportion
thus:
1
Drag
1
I
hours. hours. milos. milos.
6: 11 :: 30
/
Now, since the product of the extremes is equal to the
product of the means, we multiply together the two means,
11 and 30, which makes 330, and, dividing this product by
the known extreme, 6, we obtain for the result 55, that is,
55 miles, which is the other extreme, or term, sought.
I
F
181
T 95.
3. At $54 for 9 barrels of flour, how many barrels may
be purchased for $186 ?
In this question, the unknown quantity is the number of
barrels bought for $186, which ought to contain the 9 bar-
rels as many times as $186 contains $54; we thus get the
following proportiòn :
dollars. dollars. barrels. barrels.
54: 186
9
↓
曲
​:: 9:
54
54
RULE OF THREE.
*****
54) 1674 (31 barrels, the Answer.
162
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"
The product, 1674,
of the two means, di-
vided by 54, the
known extreme, gives
31 barrels for the
other extreme, which
is the term sought,
or Answer.
1
A
Any three terms of a proportion being given, the operation
by which we find the fourth is called the Rule of Three.
just solution of the question will sometimes require, that the
order of the terms of a proportion be changed. This may
be done, provided the terms be so placed, that the product
of the extremes shall be equal to that of the means.
4. If 3 men perform a certain piece of work in 10 days,
how long will it take 6 men to do the same?
The number of days in which 6 men will do the work be-
ing the term sought, the known term of the same kind, viz.
10 days, is made the third term. The two remaining terms
are 3 men and 6 men, the ratio of which is .But the
'more* men there are employed in the work, the less time will
be required to do it; consequently, the days will be less in
*The rule of three has sometimes been divided into direct and inverse, a dis-
tinction which is totally useless. It may not however be amiss to explain, in this
place, in what this distinction consists.
The Rule of Three Direct is when more requires more, or less requires less, as
in this example-If 3 men dig a trench 48 feci long in a certain time, how many
feet will 12 men dig in the same time? Here it is obvious, that the more men
there are employed, the more work will be done; and therefore, in this instance,
more requires more. Again:-If 6 men dig 48 feet in a given time, how much
will 3 mon dig in the same time? Hore less requiros less, for the less men there
are employed, the less work will be done.
The Rule of Three Inverse is when more requires less, or less requires more, as
in this example-It 6 mon dig a certain qunutity of trench in 14 hours, how many
hours will it require 12 men to dig the same quantity Here more requires less;
that is, 12 mon being more than 6, will require less time. Again:-If & men per
form a piece of work in 7 days, how long will 3 men be in performing the same
work? Here less requires more; for the number of mon, being less, will require
more time.
Q
S
I
*
→
↓
}
1
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​1
1
182
T 95.
proportion as the number of men is greater. There is still a
proportion in this case, but the order of the terms is inverted; '
for the number of men in the second set, being two times
that in the first, will require only one half the time. The
first number of days, therefore, ought to contain the second
as many times as the second number of men contains the
first. This order of the terms being the reverse of that as-
signed to them in announcing the question, we say, that the
number of men is in the inverse ratio of the number of days.
With a view, therefore, to the just solution of the question,
we reverse the order of the two first terms, (in doing which
we invert the ratio,) and, instead of writing the proportion,
3 men: 6 men, (,) we write it, 6 men : 3 men, (,) that is,
men. men. days. days.
6: 3 :: 10
10 .
C
Note. We invert the ratio when we reverse the order
of the terms in the proportion, because then the antece-
dent takes the place of the consequent, and the consequent
that of the antecedent; consequently, the terms of the frac-
tion which express the ratio are inverted; hence the ratio
is inverted. Thus, the ratio expressed by § 2, being in-
verted, is = {·
Having stated the proportion as above, we divide the pro-
duct of the means, (10 × 3≈ 30,) by the known extreme,
6, which gives 5, that is, 5 days, for the other extreme, or
term sought.
Ans. 5 days.
1
RULE OF THREE.
?
Stat
From the examples and illustrations now given we deduce
the following general
{Manag
RULE.
m
Of the three given numbers, make that the third term
which is of the same kind with the answer sought. Then
consider, from the nature of the question, whether the an-
swer will be greater or less than this term.
If the answer is.
to be greater, place the greater of the two remaining num-
bers for the second term, and the less number for the first
térm; but if it is to be less, place the less of the two re-
maining numbers for the second term, and the greater for
the first; and, in either case, multiply the second and third
terms together, and divide the product by the first for the
answer, which will always be of the same denomination as
the third term.
1
t
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T 95.
RULE OF THREE.
183
Note 1. If the first and second terms contain different de-
nominations, they must both be reduced to the same de-
nomination; and if the third term be a compound number, it
either must be reduced to integers of the lowest denomination,
or the low denominations must be reduced to a fraction of
the highest denomination contained in it.
Note 2. The same rule is applicable, whether the given
quantities be integral, fractional, or decimal.
}
EXAMPLES FOR PRACTICE.
5. If 6 horses consume 21 bushels of dats in 3 weeks,
how many bushels will serve 20 horses the same time?
Ans. 70 bushels.
{
6. The above question reversed. If 20 horses consume 70
bushels of oats in 3 weeks, how many bushels will serve 6
horses the same time?
Ans. 21 bushels.
7. If 365 men consume 75 barrels of provisions in 9
months, how much will 500 men consume in the same time?
Ans. 10244 barrels.
8. If 500 men consume 1024 barrels of provisions in 9
months, how much will 365 men consume in the same
time?
Ans. 75 barrels..
!
9. A goldsmith sold a tankard for 10 £. 12 s., at the rate
of 5 s. 4 d. per ounce; I demand the weight of it.
Ans. 39 oz. 15 pwt.
10. If the moon move 13° 10′ 35″ in 1 day, in what time
does it perform one revolution? Ans. 27 days, 7 h. 43 m.
11. If a person, whose rent is $145, pay $12'63 parish
taxes, how much should a person pay whose rent is $378?
Ans. $32'925.
1
12. If I buy 7 lbs. of sugar for 75 cents, how many pounds
can I buy for $6?
Ans. 56 lbs.
13. If 2 lbs. of sugar cost 25 cents, what will 100 lbs. of
coffee cost, if 8 lbs. of sugar are worth 5 lbs. of coffee?
Ans. $20.
14. If I give $6 for the use of $100 for 12 months,
what must I give for the use of $357'82 the same time ?
Ans. $21'469.
15. There is a cistern which has 4 pipes; the first will
fill it in. 10 minutes, the second in 20 minutes, the third in
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184
T 95.
40 minutes, and the fourth in 80 minutes; in what time will
all four, running together, fill it? -
RULE OF THREE.
20 + 20 + 40+b8 cistern in 1 minute.
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16. If a family of 10 persons spend 3 bushels of malt in
a month, how many bushels will serve them when there are
30 in the family?
Ans. 9 bushels.
K
Japa
Note. The rule of proportion, although of frequent use,
is not of indispensable necessity; for all questions under it
may be solved on general principles, without the formality
of a proportion; that is, by analysis, as already shown, T 65,
ex. 1. Thus, in the above example,-If 10 persons spend
3 bushels, 1 person, in the same time, would spend of 3.
bushels, that is, of a bushel; and 30 persons would spend.
30 times as much, that is, 18
9 bushels, as before.
1
Ans. 5 minutes.
17. If a staff, 5 ft. 8 in. in length, cast a shadow of 6 feet,
how high is that steeple whose shadow measures 153 feet?
Ans. 1444 feet.
18. The same by analysis. If 6 ft. shadow require a staff
of 5 ft. 8 in. 68 in., 1 ft. shadow will require a staff of
of 68 in. or 8 in.; then, 153. ft. shadow will require 153
times as much; that is, 8 X 153 =
1734 in. =
1940 4
1444 ft., as before.
19. If 3 £. sterling be equal to 4. Massachusetts, how
much Massachusetts is equal to 1000 £. sterling?
Ans. 1333 £. 6 s. 8 d.
20. If 1333 £. 6 s. 8 d. Massachusetts, be equal to 1000£.
sterling, how much sterling is equal to 4 £. Massachusetts ?
Ans. 3 £.
21. If 1000 £. sterling be equal to 1333 £. 6 s. 8 d. Mas-
sachusetts, how much Massachusetts is equal to 3 £. ster-
ling?
Ans. 4 £.
22. If 3 £. sterling be equal to 4 £. Massachusetts, how
much sterling is equal to 1333 £. 6 s. 8 d. Massachusetts ?
Ans. 1000 L.
23. Suppose 2000 soldiers had been supplied with bread
sufficient to last them 12 weeks, allowing each man. 14
ounces a day; but, on examination, they find 105 barrels,
containing 200 lbs. each, wholly, spoiled; what must the al-
lowance be to each man, that the remainder may last them
the same time?
Ans. 12 oz. a day.
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T 95.
185
24. Suppose 2000 soldiers were put to an allowance of
12 oz. of bread per day for 12 weeks, having a seventh part of
their bread spoiled; what was the whole weight of their
pread, good and bad, and how much was spoiled?
F
Ans.
1
RULE OF THREE.
The whole weight, 147000 lbs.
{Spoiled,
21000 lbs.
25.--
→ 2000 soldiers, having lost 105 barrels of bread,
weighing 200 lbs. each, were obliged to subsist on 12 oz. a
day for 12 weeks; had none been lost, they might have had
14 oz. a day; what was the whole weight, including what
was lost, and how much had they to subsist on?
Whole weight,
Ans.
147000 lbs.
Left, to subsist on, 126000 lbs.
26.. 2000 soldiers, after losing one seventh part of
their bread, had each 12 oz. a day for 12 weeks; what was
the whole weight of their bread, including that lost, and how
much might they have had per day, each man, if none had
been lost?
Whole weight, 147000 lbs.
Ans. Loss,
21000 lbs.
14 oz. per day, had none been lost.
27. There was a certain building raised in 8 months by
120 workmen; but, the same being demolished, it is required
to be built in 2 months; I demand how many men must
be employed about it.
Ans. 480 men.
28. There is a cistern having a pipe which will empty it
in 10 hours; how many pipes of the same capacity will
empty it in 24 minutes?
Ans. 25 pipes.
29. A garrison of 1200 men has provisions for 9 months,
at the rate of 14 oz. per day; how long will the provisions
last, at the same allowance, if the garrison be reinforced by
400 men?
Ans. 62 months.
30. If a piece of land, 40 rods in length and 4 in breadth,
make an acre, how wide must it be when it is but 25 rode
long?
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31. If a man perform a journey in 15 days
are 12 hours long, in how many will he do it
are but 10 hours long?
32. If a field will feed 6 cows 91 days, how long will it
feed 21 cows ?
Ans. 63 rods.
when the days
when the days
Ans. 18 days.
Ans. 26 days.
33. Lent a friend 292 dollars for 6 months; some time
after, he lent me 806 dollars; how long may I keep it to
balance the favour ?
Ans. 2 months 5 + days.
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T 95..
RULE OF THREE...
34. If 30 men can perform a piece of work in 11 days,
how many men will accomplish another piece of work, 4
times as big, in a fifth part of the time?
Aus. 600 men.
5
35. If 11 lb. of sugar cost of a shilling, what will 24
of a lb. cost?
Ans. 4 d. 3497Z q
Note. See T 65, ex. 1, where the above question is
solved by analysis. The eleven following are the next suc-
ceeding examples in the same 1.
36. If 7 lbs. of sugar cost of a dollar, what cost 12 lbs.?
Ans. $1.
4
37. If 61 yds. of cloth cost $3, what cost 94 yds. ?
38. If 2 oz. of silver cost $224, what costs
1
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39. If 4 oz. cost $12, what costs 1 oz.?
40. If lb. less. by lb. cost. 13
less by of 2 lbs. ?
1/8
41. If yd. cost $7, what will 40
oz.?
Ans. $0'84.
Ans. $1'283.
d., what cost 14 lbs.
Ans. 4 £.9 s. 9% d.
yds. cost?.
?
Ans. $59'062.
of her worth?
Ans. $53'785.
cost?
Ans. 6 s. 3 d.
of his share
I
44. A merchant, owning t of a vessel, sold
for $957; what was the vessel worth ? Ans. 1794 375.-
45. If § yd, cost £., what will of an ell English cost?
Ans. 17 s. 1 ä. 27 q.
46. A merchant bought a number of bales of velvet, each
containing 12017 yds., at the rate of $7 for 5 yds., and sold
them out at the rate of $11 for 7 yds., and gained $200
by the bargain; how many, balcs were there? Ans. 9 bales.
47. At $33 for 6 barrels of flour, what must be paid for
178 barrels ?
Ans. $979.
48. At, $2.25 for 317 cwt. of hay, how much is that per
ton?
Ans. $14'195.
how much will
49. If 25 lbs. of tobacco cost 75 cents,
185 lbs, cost?
42. If of a ship cost $251, what
is
43. At 33. per cwt., what will 9
→
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lbs.
3
Ans. $4'269.
Ans.; $555.
50. What is the value of '15 of a hogshead of lime, at
$2:39 per hhd. ?
Ans. $03585.
I
51, If 15 of a hhd. of lime cost $0.3585, what is it per
bhd. P
Ans. $2439...
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$
COMPOUND PROPORTION.
COMPOUND PROPORTION.
доль
T 96. It frequently happens, that the relation of the
quantity required, to the given quantity of the same kind,.
depends upon several circumstances combined together; it
is then called Compound Proportion, or Double Rule of Three.
18%
1
1. If a man travel 273 miles in 13 days, travelling only
7 hours in a day, how many miles will he travel in 12 days,
if he travel 10 hours in a day?
This question may be solved several ways. First, by analy-
sis:
~~~
If we knew how many miles, the man travelled in 1 hour,
it is plain, we might take this number 10 times, which would
be the number of miles he would travel in 10 hours, or in 1
of these long days, and this again, taken 12 times, would be
the number of miles he would travel in 12 days, travelling
10 hours each day.
TB
ST
If he travel 273 miles in 13 days, he will travel †h of 273
miles; that is, 27 miles in 1 day of 7 hours; and 4 of 273
miles is 3 miles, the distance he travels in 1 hour: then,
10 times 273 27 miles, the distance he travels in 10
hours; and 12 times 2342 22760 360 miles, the dis-
tance he travels in 12 days, travelling 10 hours each day.
Ans. 360 miles.
9.1
But the object is to show how the question may be solved
by proportion:-
First; it is to be regarded, that the number of miles tra-
velled over depends upon two circumstances, viz. the num-
ber of days the man travels, and the number of hours he
travels each day.
………………
We will not at first consider this latter circumstance, but
suppose the number of hours to be the same in each case:
the question then will be,-If a man travel 273 miles in 13
days, how many miles will he travel in 12 days? This will
furnish the following proportion :-
13 days 12 days :: 273 miles : miles
which gives for the fourth term, or answer, 252 miles.
Now, taking into consideration the other circumstance, or
that of the hours, we must say,-If a man, travelling 7 hours
´a day for a certain number of days, travels 252 miles, how fur
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188
T 96.
will he travel in the same time, if he travel 10 hours in a day?
This will lead to the following proportion :—
I
7 hours: 10 hours :: 252 miles :
miles.
This gives for the fourth term, or answer, 360 miles.
We see, then, that 273 miles has to the fourth term, or
answer, the same proportion that 13 days has to 12 days,
and that 7 hours has to 10 hours. Stating this in the form
of a proportion, wè have
13 days
: 12 days
7 hours: 10 hours
·
COMPOUND PROPORTION.
:: 273 miles,:
miles
I
by which it appears, that 273 is to be multiplied by both 12
and 10; that is, 273 is to be multiplied by the product of
12 X 10, and divided by the product of 13 × 7, which, be-
ing done, gives 360 miles for the fourth term, or answer, as
before.
In the same manner, any question relating to compound
proportion, however complicated, may be stated and solved.
2. If 248 men, in 5 days, of 11 hours each, can dig a trench
230 yards long, 3 wide, and 2 deep, in how many days, of 9
hours each, will 24 men dig a trench 420 yards long, 5 wide,
and 3 deep?
Here the number of days, in which the proposed work can
be done, depends on five circumstances, viz. the number of
men employed, the number of hours they work each day,
the length, breadth, and depth of the trench. We will con-
sider the question in relation to each of these circumstances,
in the order in which they have been named:
9 hours 11 hours 5 days:
ฟุ
:
I
3d. Length of the ditches.
+
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1st. The number of men employed. Were all the circum-
stances in the two cases alike, except the number of men and
the number of days, the question would consist only in find-
ing in how many days 24 men would perform the work which
248 men had done in 5 days; we should then have
24 men: 248 men: 5 days:
days.
2d. Hours in a day. But the first labourers worked 11
hours in a day, whereas the others worked only 9; less hours
will require more days, which will give
$
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days.
The ditches being of unequal
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​T 96, 97.
189
length, as many more days will be necessary as the second
is longer than the first; hence we shall have
COMPOUND PROPORTION.
230 length: 420 length: 5 days:
days.
4th. Widths. Taking into consideration the widths, which
are different, we have
3 wide
5 wide: 5 days:
5th. Depths. Lastly, the depths being different, we have
2 deep 3 deep 5 days:
:
I
days.
J
It would seem, therefore, that 5 days has to the fourth
term, or answer, the same proportion
that. 2 depth has to
all which stated in form
Men, 24 : 248
Hours, 9 : 11
Length, 230: 420
Width, 3: 5
Depth, 2: 3
that 24 men has to 248 men,
whose ratio is 248,
that 9 hours has to 11 hours, the ratio of which is,
that 230 length has to 420 length,
that 3 width has to
5 width,
X5 X 3
X3x2=
17186400
298080"
…………….. days.
*******
3 depth,
of a proportion, we have
L
common term.
:: 5 days:
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T 97. The continued product of all the second terms
248 × 11 × 420 × 5 × 3, multiplied by the third term,
5 days, and this product divided by the continued pro-
duct of the first terms, 24 X 9 X 230 X 3X 2, gives
288,84990 days for the fourth term, or answer.
2885.
97
But the first and second terms are the fractions 248, 17,
438, and 2, which express the ratios of the men, and of
'the hours, of the lengths, widths and depths of the two
ditches. Hence it follows, that the ratio of the number of
days given to the number of days sought, is equal to the pro-
duct of all the ratios, which result from a comparison of the
terms relating to each circumstance of the question.
The product of all the ratios is found by multiplying to-
248 × 11X 420
gether the fractions which express them, thus, 24 X 9 X 230
days.
$38,
影
​&;.
17186400
and this fraction, 298080, represents the
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190
compound prOPORTION.
T 97.
ratio of the quantity required to the given quantity of the same
kind. A ratio resulting in this manner, from the multiplica-
tion of several ratios, is called a compound ratio.
From the examples and illustrations now given we de-
duce the following general
RULE
for solving questions in compound proportion, or double
rule of three, viz.-Make that number which is of the
same kind with the required answer, the third term; and,
of the remaining numbers, take away two that are of the
same kind, and arrange them according to the directions
given in simple proportion; then, any other two of the same.
kind, and so on till all are used.
1
Lastly, multiply the third term by the continued product
of the second terms, and divide the result by the continued
product of the first terms, and the quotient will be the fourth
term, or answer required.
EXAMPLES FOR PRACTICE.
1. If 6 men build a wall 20 ft. long, 6 ft. high, and 4 ft.
thick, in 16 days, in what time will 24 men build one 200
ft. long, 8 ft. high, and 6 ft. thick ?
Ans. 80 days.
2. If the freight of 9 hhds. of sugar, each weighing 12
cwt., 20 leagues, cost 16 £., what must be paid for the
freight of 50 tierces, each weighing 24 cwt., 100 leagues?
Āns. 92 £. 11 s. 103 d.
3. If 56 lbs. of bread be sufficient for 7 men 14 days, how
much bread will serve 21 men 3 days?
Ans. 36 lbs.
1
£6
The same by analysis. If 7 men consume 56 lbs. of bread,
1 man, in the same time, would consume of 56 lbs.
lbs.; and if he consume 5 lbs. in 14 days, he would
consume 4 of 26 8 lb. in 1 day. 21 men would con-
sume 21 times so much as 1 man; that is, 21 times
1170 lbs. in 1 day, and in 3 days they would consume 3
times as much; that is, 2838 36 lbs., as before.
=
Ans. 36 lbs.
Note. Having wrought the following examples by the
rule of proportion, let the pupil be required to do the same
by analysis.
4. If 4 reapers receive $11'04 for 3 days' work, how
many men may be hired 16 days for $103'04?
I
Ans. 7 men.
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T 97. SUPPLEMENT TO THE SINGLE RULE OF THREE. 191
5.
If 7 oz. 5 pwt. of bread be bought for 4 d. when corn
is 4 s. 2 d. per bushel, what weight of it may be bought for
1 s. 2 d. when the price per bushel is 5 s. 6 d. ?
Ans. 1 lb. 4 oz. 342 pwts.
6. If $100 gain $6 in 1 year, what will $400 gain in
9 months?
Note. This and the three following examples reciprocally
prove each other.
7. If $100 gain $6 in 1 year, in what time will $400
gain $18?
8. If $400 gain $18 in 9 months, what is the rate per
cent. per annum ?
9. What principal, at 6 per cent. per. ann., will gain $18
in 9 months?'
10. A usurer put out $75 at interest, and, at the end of 8
months, received, for principal and interest, $79; I demand
at what rate per cent. he received interest.
Ans. 8 per cent.
11. If 3 men receive 85 £. for 19 days' work, how
much must 20 men receive for 100 days'?
t
Ans. 305£. 0 s. 8 d.
1
11
SUPPLEMENT TO THE SINGLE RULE OF
THREE.
QUESTIONS.
1. What is proportion? 2. How many numbers are re-
quired to form a ratio? 3. How many to form a proportion?
4. What is the first term of a ratio called? 5. the second
term? 6. Which is taken for the numerator, and which for
the denominator of the fraction expressing the ratio? 7.
How may it be known when four numbers are in proportion ?
8. Having three terms in a proportion given, how may the
fourth term be found? 9. What is the operation, by which
the fourth term is found, called? 10. How does a ratio be-
come inverted? 11. What is the rule in proportion? 12.
In what denomination will the fourth term, or answer, be
found? 13. If the first and second terms contain different
denominations, what is to be done? 14. What is compound
proportion, or double rule of three? 15. Rule?
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FELLOWSHIP.
!
EXERCISES.
}
1. If I buy 76 yds. of cloth for $113'17, what does it
cost per ell English?
Ans. $1'861.
1
2. Bought 4 pieces of Holland, each containing 24 ells
English, for $96; how much was that per yard?
Ans. $0'80.
3. A garrison had provision for 8 months, at the rate of
15 ounces to each person per day; how much must be al-
lowed per day, in order that the provision may last 9
months?
Ans. 1213 oz.
c
4. How much land, at $2'50 per acre, must be given in
exchange for 360 acres, at $375 per acre?
T 97, 98.
J
Ans. 540 acres.
5. Borrowed 185 quarters of corn when the price was
19.s.; how much must I pay when the price is
17 s. 4 d.?
Ans. 20241.
6. A person, owning of a coal mine, sells
for 171.; what is the whole mine worth ?
7. If § of a gallon cost § of a dollar, what
tun ?
}
of his share
Ans. 380£.
8. At 1½ £. per cwt., what cost 34 lbs. ?
9. If 4 cwt. can be carried 36 miles for 35
many pounds can be carried 20 miles for the same money
Ans. 9074 lbs.
10. If the sun appears to move from east to west 360 de-
in
grees in 24 hours, how much is that in each hour?
each minute? in each second?
costs
of a
Ans. $140.
Ans. 10 d.
shillings, how
?
"Ans. lo last, 15" of a deg.
11. If a family of 9 persons spend $450 in 5 months, how
much would be sufficient to maintain them 8 months if 5
Ans. $1120.
persons more were added to the family?
Note. Exercises 14th, 15th, 16th, 17th, 18th, 19th, and
20th, "Supplement to Fractions," afford additional examples
in single and double proportion, should more examples be
thought necessary.
1
FELLOWSHIP.
↑ 98. 1. Two men own a ticket; the first owns 1, and
the second owns & of it; the ticket draws a prize of 40 dol-
lars; what is each man's share of the money?
1
T 98,
193
2. Two men purchase a ticket for 4 dollars, of which one
pays 1 dollar, and the other 3 dollars; the ticket draws 40
dollars; what is each man's share of the money?
f
مگیر
A's stock,
B's stock,
3. A and B bought a quantity of cotton; A paid 100
dollars, and B 200 dollars; they sold it so as to gain 30
dollars; what were their respective shares of the gain ?
The process of ascertaining the respective gains or losses
of individuals, engaged in joint trade, is called the Rule of
Fellowship.
The money, or value of the articles employed in trade, is
called the Capital, or Stock; the gain or loss to be shared is
called the Dividend.
FELLOWSHIP.
It is plain, that each man's gain or loss ought to have the
same relation to the whole gain or loss, as his share of the
stock does to the whole stock.
Partekat
Hence we have this RULE:-As the whole stock: to each
man's share of the stock: the whole gain or loss: his share
of the gain or loss.
*
4. Two persons have a joint stock in trade; A put in
$250, and B $350; they gain $400; what is each man's
share of the profit?
1
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OPERATION.
$250) Then,
$350
#
600: 250 :: 400: 166'6662 dolls. A's gain.
600: 350 :: 400 : 233′3334 dolls. B's gain.
Whole stock, $600
The pupil will perceive, that the process may be contract-
ed by cutting off an equal number of ciphers from the first
and second, or first and third terms; thus, 6: 250 :: 4:
166'666, &c.
It is obvious, the correctness of the work may be ascer-
tained by finding whether the sums of the shares of the gains
are equal to the whole gain; thus, $ 166'666+ $233333
$400, whole gain.
5. A, B and C trade in company; A's capital was $175,
B's $200, and C's $500; by misfortune they lose $250;
what loss must each sustain ?
50', A's loss.
Ans. $ 57'1429, B's loss.
$ 142′8574, C's loss.
6. Divide $600 among 3 persons, so that their shares
may be to each other as 1, 2, 3, respectively.
Ans. $100, $200, and $300.
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194
7. Two merchants, A and B, loaded a ship with 500
hhds. of rum; A loaded 350 hhds., and B the rest; in a
storm, the seamen were obliged to throw overboard 100
hhds.; how much must each sustain of the loss ?
g
Ans. A 70, and B 30 hhds.
8. A and B companied; A put in $45, and took out
of the gain; how much did B put in?
Ans. $30.
Note. They took out in the same proportion as they put
in; if 3 fifths of the stock is $45, how much is 2 fifths.
of it?
1
9. A and B companied, and trade with a joint capital of
$400; A receives, for his share of the gain, as much as B;
what was the stock of each ?
1
FELLOWSHIP.
1
10. A bankrupt is indebted to A $780, to B $460, and
to C $760; his estate is worth only $600; how must it
be divided?
Note. The question evidently involves the principles of
fellowship, and may be wrought by it.
Ans. A $234, B $138, and C $228.
11. A and B venture equal stocks in trade, and clear
$164; by agreement, A was to have 5 per cent. of the
profits, because he managed the concerns; B was to have
but 2 per cent.; what was each one's gain? and how much
did A receive for his trouble ?
}
} .
Ans. A's gain was $117'1429, and B's $46,8574, and
A received $70'2854 for his trouble.
Į
2
T 98
:
12. A cotton factory, valued at $12000, is divided into
100 shares; if the profits amount to 15 per cent. yearly, what
will be the profit accruing to 1 share?
to 2 shares ?
to 5 shares? - to 25 shares?
1
10
A's stock.
Ans.266'666, B's stock.
· Ans. to the last. $450.
13. In the above-mentioned factory, repairs are to be made
which will cost $340; what will be the tax, on each share,
necessary to raise the sum ?
on 2 shares ?.
on 3
shares?
Ans. to the last, $34.
14. If a town raise a tax of $1850, and the whole town
be valued at $37000, what will that be on $1? What
will be the tax of a man whose property is valued at $1780 ?
Ans. $'05 on a dollar, and $89 on $1780.
on 10 shares ?
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T 99.
195
T 99. In assessing taxes, it is necessary to have an in-
ventory of the property, both real and personal, of the whole
town, and also of the whole number of polls; and, as the polls
are rated at so much each, we must first take out from the
whole tax what the polls amount to, and the remainder is to
be assessed on the property. We may then find the tax upon
1 dollar, and make a table containing the taxes on 1, 2, 3,
&c., to 10 dollars; then on 20, 30, &c., to 100 dollars; and
then on 100, 200, &c., to 1000 dollars. Then, knowing the
inventory of any individual, it is easy to find the tax upon his
property.
15. A certain town, valued at $64530, raises a tax of
$2259'90; there are 540 polls, which are taxed $'60
each; what is the tax on a dollar, and what will be A's tax,
whose real estate is valued at $1340, his personal property
at $874, and who pays for 2 polls?
540 X 60 $324, amount of the poll taxes, and
$2259'90 $324 1935'90, to be assessed on property.
$64530: $1935'90 :: $1: '03; or, 1843
1935.90 '03, tax on $1.
TABLE.
dolls. dolls.
dolls.
Tax on 1 is '03 | Tax on 10 is
2.: '06
··
1
}
FELLOWSHIP.
3 ..
'09
4 .. '12
5.. '15
6 .. '18
60 1'80
•
Hy
21
70 .. 2'10
8.. $24
80 .. 2'40
9.. (247 .......... 90
90 .. 2470
J
1
Wh
1
20 ..
30 '90
40
1
Now, to find A's tax, his real estate being $1340, I find,
by the table, that
The tax on
The tax on
The tax on
dolls.
dolls. dolls,
30 | Tax on 100 is 3
'60
6'
200..
300 .. 9
400 ..
12
500 .. 15'
600.. 18'
700 .. 21'
**
Wi
1'20
50.. 1'50
$1000
300
40
- is
3
}
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'
1
Tax on his real estate
In like manner I find the tax on his personal
property to be
2 polls at '60 each, are
-}
800.. 24
900 .. 27′
1000 .. 30°
1
1
1
1
$30
96
1620
$40'20
26'22
1'20
Amount, $6762
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1
1
2
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196
T 99, 100.
16. What will B's tax amount to, whose inventory is 874
dollars real, and 210 dollars personal property, and who pays
for 3 polls?
Ans. $34'32.
17. What will be the tax of a man, paying for 1 poll,
whose property is valued at $3482 ? at $768?
at $940? at $4657? Ans. to the last, $140'31.
18. Two men paid 10 dollars for the use of a pasture 1
month; A kept in 24 cows, and B 16 cows; how much
should each pay?
19. Two men hired a pasture for $10; A put in 8 cows.
3 months, and B put in 4 cows 4 months; how much should
each pay?
FELLOWSHIP.
7
T 100. The pasturage of 8 cows for 3 months is the
same as of 24, cows for 1 month, and the pasturage of 4 cows
for 4 months is the same as of 16 cows for 1 month. The
shares of A and B, therefore, are 24 to 16, as in the former
question. Hence, when time is regarded in fellowship,-
Multiply each one's stock by the time he continues il in trade,
and use the product for his share. This is called Double Fel-
lowship.
Ans. A 6 dollars, and B 4 dollars.
20. A and B enter into partnership; A puts in $100
6-months, and then puts in $50 more; B puts in $200 4
months, and then takes out $80; at the close of the year,
they find that they have gained $95; what is the profit of
´each?
Ans. $43'711, A's share.
$51'288, B's share.
21. A, with a capital of $500, began trade Jan. 1, 1826,
and, meeting with success, took in B as a partner, with a
capital of $600, on the first of March following; four
months after, they admit C as a partner, who brought $800
stock; at the close of the year, they find the gain to be
700; how must it be divided among the partners ?
{
Ans.
$250, A's share.
$250, B's share.
200, C's share.
"
i
常委
​QUESTIONS.
1. What is fellowship? 2. What is the rule for operat-
ing? 3. When time is regarded in fellowship, what is it
called? 4. What is the method of operating in double
fellowship? 5. How are taxes assessed? 6. How is
fellowship proved ?
1
古
​1
7 101, 102.
¿
ALLIGATION.
¶ 101. Alligation is the method of mixing two or more
simples, of different qualities, so that the composition may be
of a mean, or middle quality.
When the quantities and prices of the simples are given,
to find the mean price of the mixture, compounded of them,
the process is called Alligation Medial.
ง
;
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ALLIGATION.
1. A farmer mixed together 4 bushels of wheat, worth
150 cents per bushel, 3 bushels of rye, worth 70 cents per
bushel, and 2 bushels of corn, worth 50 cents per bushel;
what is a bushel of the mixture worth?
It is plain, that the cost of the whole, divided by the num-
ber of bushels, will give the price of one bushel.
T
را
4 bushels, at 150 cents, cost 600 cents.
3
at 70
210
2
at 50
100
9 bushels cost
910 cents.
2. A grocer mixed 5 lbs. of sugar, worth 10 cents per lb.,
8 lbs. worth 12 cents, 20 lbs. worth 14 cents; what is a
pound of the mixture worth?
Ans. 1219.
3. A goldsmith melted together 3 ounces of gold 20
carats fine, and 5 ounces 22 carats fine; what is the fine-
ness of the mixture ?
Ans. 211
4. A grocer puts 6 gallons of water into a cask containing
40 gallons of rum, worth 42 cents per gallon; what is a gal-
lon of the mixture worth?
Ans. 361 cents.
5. On a certain day the mercury was observed to stand in
the thermometer as follows: 5 hours of the day, it stood at
64 degrees; 4 hours, at 70 degrees; 2 hours, at 75 degrees,
and 3 hours, at 73 degrees: what was the mean temperature
for that day?
It is plain this question does not differ, in the mode of its
operation, from the former.
Ans. 69 degrees.
"
197
*****
T 102. When the mean price or rate, and the prices or
rates of the several simples are given, to find the proportions
or quantities of each simple, the process is called Alligation
Alternate: alligation alternate is, therefore, the reverse of
alligation medial, and may be proved by it.
R*
21º 1013 cts. Ans.
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102.
1. A man has oats worth 40 cents per bushel, which he
wishes to mix with corn worth 50 cents per bushel, so that
the mixture may be worth 42 cents per bushel; what pro-
portions, or quantities of each, must he take?
·
$
}
Had the price of the mixture required exceeded the price
of the oats, by just as much as it fell short of the price of
the corn, it is plain, he must have taken equal quantities of
oats and corn; had the price of the mixture exceeded the
price of the oats by only as much as it fell short of
the price of the corn, the compound would have required 2
times as much oats as corn and in all cases, the less the
difference between the price of the mixture and that of one.
of the simples, the greater must be the quantity of that sim-
ple, in proportion to the other; that is, the quantities of the
simples must be inversely as the differences of their prices
from the price of the mixture; therefore, if these differen-
ces be mutually exchanged, they will, directly, express the
relative quantities of each simple necessary to form the com-
pound required. In the above example, the price of the
mixture is 42 cents, and the price of the oats is 40 cents;
consequently, the difference of their prices is 2 cents: the
price of the corn is 50 cents, which differs from the price
of the mixture by 8 cents. Therefore, by exchanging these
differences, we have 8 bushels of oats to 2 bushels of corn,
for the proportion required.
Ans. 8 bushels of oats to 2 bushels of corn, or in that
proportion.
The correctness of this result may now be ascertained by
the last rule; thus, the cost of 8'bushels of oats, at 40 cents,
is 320 cents; and 2 bushels of corn, at 50 cents, is 100
cents; then, 320100420, and 420, divided by the num-
ber of bushels, (8+2,) = 10, gives 42 cents for the price of
the mixture.
ALLIGATION.
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{
2. A merchant has several kinds of tea; some at 8 shil-
lings, some at 9 shillings, some at. 11 shillings, and some
at 12 shillings per pound; what proportions of each must
he mix, that he may sell the compound at 10 shillings per
pound?
Here we have 4 simples; but it is plain, that what has
just been proved of two will apply to any number of pairs,
if in each pair the price of one simple is greater, and that of
the other less, than the price of the mixture required.
Hence we have this
#

T 102.
10s.
RULE.
The mean rate and the several prices being reduced to
the same denomination,-connect with a continued line each
price that is LE3s than the mean rate with one or more that
is GREATER, and each price GREATER than the mean rate
with one or more that is LESS.
88.
9s.
11s.
(12s..
Write the difference between the MEAN rate, or price, and
the price of EACH SIMPLE opposite the price with which it is
connected; (thus the difference of the two prices in each
pair will be mutually exchanged;) then the sum of the differ-
ences, standing against any price, will express the RELATIVE
QUANTITY to be taken of that price.
By attentively considering the rule, the pupil will per-
ceive, that there may be as many different ways of mixing
the simples, and consequently as many different answers, as
there are different ways of linking the several prices.
We will now apply the rule to solve the last question :—
OPERATIONS.
氨
​}
ALLIGATION.
lbs.
Ans.
{
1
Or,
}
10
1
&
1
9
11
12
ļ
(
}
Here we set down the prices of the simples, one directly
under another, in order, from least to greatest, as this is
most convenient, and write the mean rate,, (10 s.) at the
"
left hand. In the first way of linking, we find, that we
may take in the proportion of 2 pounds of the teas at 8
and 12 s. to 1 pound at 9 and 11 s. In the second way,
we find for the answer, 3 pounds at 8 and 11 s. to 1 pound
at 9 and 12 s.
Q
·
199


-2+1=3
]
-1+2=3)
3. What proportions of sugar, at 8 cents, 10 cents, and
14 cents per pound, will compose a mixture worth 12 cents.
per pound?
=1
Ans. In the proportion of 2 lbs. at 8 and 10 cents to 6
lbs. at 14 cents.
Note. As these quantities only express the proportions of
each kind, it is plain, that a compound of the same mean
price will be formed by taking 3 times, 4 times, one half, or
any proportion, of each quantity. Hence,
When the quantity of one simple is given, after finding
Ans.
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6
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200
T.102
the proportional quantities, by the above rule, we may say,
As the PROPORTIONAL quantity is to the GIVEN quantity:
so is each of the other PROPORTIONAL quantities: to the RE-
QUIRED quantities of each,
fanta
4. If a man wishes to mix 1 gallon of brandy worth
16 s. with rum at 9 s. per gallon, so that the mixture
may be worth 11 s. per gallon, how much rum must he
use?
1
ALLIGATION.
Taking the differences as above, we find the proportions
to be 2 of brandy to 5 of rum; consequently, 1 gallon of
brandy will require 24 gallons of rum. Ans. 24 gallons,
5. A grocer has sugars worth 7 cents, 9 cents, and 12
cents per pound, which he would mix so as to form a com-
pound worth 10 cents per pound; what must be the pro-
portions of each kind?
Ans. 2 lbs. of the first and second to 4 lbs. of the third kind.
6. If he use 1 lb. of the first kind, how much must he take
of the others? if 4 lbs., what?
if 6 lbs., what?
if 10 lbs., what? if 20 lbs., what?
Ans. to the last, 20 lbs. of the second, and 40 of the third.
¿
7. A merchant has spices at 16 d. 20 d. and '32 d. per
pound; he would mix 5 pounds of the first sort with the
others, so as to form a compound worth 24 d. per pound;
how much of each sort must he use?
Ans. 5 lbs. of the second, and 74 lbs. of the third.
8. How many gallons of water, of no value, must be
mixed with 60 gallons of rum, worth 80 cents per gallon, to
reduce its valute to 70 cents per gallon? Ans. 84 gallons.
9. A man would mix 4 bushels of wheat, at $1'50
per bushel, rye at $1'16, corn at $75, and barley
at $50, so as to sell the mixture at $ 84 per bushel,
how much of each may he use?
1
10. A goldsmith would mix gold 17 carats fine with
some 19, 21, and 24 carats fine, so that the compound may
be 22 carats fine; what proportions of each must he use?
Ans. 2 of the 3 first sorts to 9 of the last.
11. If he use 1 oz. of the first kind, how much must
he use of the others? What would be the quantity of the
compound?
Ans. to last, 7½ ounces.
I
12. If he would have the whole compound consist of 15
oz., how much must he use of each kind? if of 30
oz., how much of each kind?
if of 373 oz., how much?
Ars. to the last, 5 oz. of the 3 first, and 224 oz. of the last..
St
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201
Hence, when the quantity of the compound is given, we
may say, As the sum of the PROPORTIONAL quantities, found
by the ABOVE RULE, is to the quantity REQUIRED, so is each
PROPORTIONAL quantity, found by the rule, to the REQUIRED
quantity of EACH.
T 102, 103.
1
DUODECIMALS.
1
13. A man would mix 100 pounds of sugar, some at 8
cents, some at 10 cents, and some at 14 cents per pound, so
that the compound may be worth 12 cents per pound; how
much of each kind must he use?
+
We find the proportions to be, 2, 2, and 6. Then, 2 + 2 i
+6 = 10,
10, and
2:
2 :
20 lbs. at 8 cts.
20 lbs. at 10 cts.
Ans.
6
60 lbs. at 14 ets.
14. How many gallons of water, of no value, must be
mixed with brandy at $1'20 per gallon, so as to fill a ves-
sel of 75 gallons, which may be worth. 92 cents per gallon?
Ans. 17 gallons of water to 57 gallons of brandy..
f
10: 100 ::
15. A grocer has currants at 4 d., 6 d., 9d. and 11 d. per
Ib.; and he would make a mixture of 240 bls., so that the
mixture may be sold at 8 d. per lb.; how many pounds of
each sort may he take?
Ans. 72, 24, 48, and 96 lbs., or 48, 48, 72, 72, &c.
Note. This question may have five different answers.
QUESTIONS.
medial? 3.
1. What is alligation? 2.
the
rule for operating? 4. What is alligation alternate? 5.
When the price of the mixture, and the price of the several
simples, are given, how do you find the proportional quanti-
ties of each simple? 6. When the quantity of one simple is
giver., how do you find the others? 7. When the quantity
of the whole compound is given, how do you find the quan-
tity of each simple?
1
DUODECIMALS.
T103. Duodecimals are fractions of a foot. The word
is derived from the Latin word duodecim, which signifies.
twelve. A foot, instead of being divided decimally into ten
equal parts, is divided duodecimally into twelve equal parts,
1
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1
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}
202
T'103.
called inches, or primes, marked thus, ('); Again, each of
these parts is conceived to be divided into twelve other equal
parts, called seconds, ("), In like manner, each second is
conceived to be divided into twelve equal parts, called thirds,
("); each third into twelve equal parts, called fourths,
(); and so on to any extent.
In this way of dividing a foot, it is obvious, that
inch, or prime, is
MULTIPLICATION OF DUODECIMALS.
1
1
of TE,
of a foot.
of 1½ of 12,
Ts of a foot.
second is
1" third is
1" fourth is T of th of th of those of a foot.
1''''' fifth is of of th of th of th='37 of a foot, &c.
ΤΣ
0736
th
A
P
OPERATION.
ft.
Length, 16 7
Breadth, 1 3/
12" fourths make 1 third,
12" thirds
1" second,
12" seconds
1
12 inches, or primes, 1
7
Duodecimals are added and subtracted in the same man-
ner as compound numbers, 12 of a less denomination making
1 of a greater, as in the following
TABLE.
4 1 9/
16 777
Ans. 20 8' 9"
ht
ד
" 2
Note. The marks, ///
"", "", &c., which distinguish the
different parts, are called the indices of the parts or denomi-
nations.
1
MULTIPLICATION OF DUODECIMALS.
Duodecimals are chiefly used in measuring surfaces and
solids.
1. How many square feet in a board 16 feet 7 inches long,
and 1 foot 3 inches wide?
Note. Length × breadth = superficial contents, (¶ 25.)
3
Togg
P
of a foot.
inch or prime,
foot.
=
inches, or primes, of a
Te
foot, and 3 inches of a foot;
consequently, the product of 7 x
of a foot, that' is, 21"
T2
TTT
1 and 9; wherefore, we set
down the 9, and reserve the 1'
to be carried forward to its proper
place. To multiply 16 feet by 3'
}
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1
T 103.
203
is to take of 16 48, that is, 48'; and the 1' which we
reserved makes 49', 4 feet 1'; we therefore set down
the 1', and carry forward the 4 feet to its proper place,
Then, multiplying the multiplicand by the 1 foot in the mul-
tiplier, and adding the two products together, we obtain the
Answer, 20 feet, 8, and 9".
1
The only difficulty that can arise in the multiplication of
duodecimals is, in finding of what denomination is the pro-
duct of any two denominations. This may be ascertained as
above, and in all cases it will be found to hold true, that the
product of any two denominations will always be of the denomi-
nation denoted by the sum of their INDICES. Thus, in the
above example, the sum of the indices of X 3' is "
sequently, the product is 21"; and thus primes multiplied
by primes will produce seconds; primes multiplied by seconds
produce thirds; fourths multiplied by fifths produce ninths, &c.
; con-
It is generally most convenient, in practice, to multiply the
multiplicand first by the feet of the multiplier, then by the
inches, &c., thus :-
Kapland
:
ft.
16 7
1
3/
1
}
MULTIPLICATION OF DUODECIMALS.
16 7
4 1' 9"
20 8' .9"
2. How many solid feet in a block 15 ft. 8' long, 1 ft. 5
wide, and 1 ft. 4' thick ?
OPERATION.
ft.
Length, 15
15 8/
Breadth, 1 5'
15 8'
66' 4"
Thickness, 1 4'
Ans.
22 2/ 44
=
16 ft. X 1 ft. = 16 ft., and 7×
1 ft. 7. Then, 16 ft. × 3′ = 48′
4 ft., and 7' X 3' 21: 1' 9".
The two products, added together, give
for the Answer, 20 ft. 8' 9", as before.
1
22 2′ 4″
7 4' 9" 4"
29 77 14 4
1
7
V
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The length multiplied by the
breadth, and that product by the
thickness, gives the solid con-
tents, (¶ 36.)
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"
MULTIPLICATION OF DUODECIMALS. T 104.
From these examples we derive the following RULE:-
Write down the denominations as compound numbers, and
in multiplying remember, that the product of any two de-.
nominations will always be of that denomination denoted by
the sum of their indices.
EXAMPLES FOR PRACTICE.
{
3. How many square feet in a stock of 15 boards, 12 ft.
8' in length,, and 13 wide?
Ans. 205 ft. 10'.
{
4. What is the product of 371 ft, 2' 6 multiplied by
181 ft. 1/9/?
Ans. 67242 ft. 10.1" 4" 6.
Note. Painting, plastering, paving, and some other kinds
of work, are doue by the square yard. If the contents in
square feet be divided by 9, the quotient, it is evident, will
be square yards.
5. A man painted the walls of a room 8 ft. 2' in height,
1
and 72 ft. 4 in compass; (that is, the measure of all its
sides;) how many square yards did he paint?
Ans. 65 yds. 5 ft. 8' 8".
6. There is a room plastered, the compass of which is
47 ft. 3', and the height 7 ft. 6'; what are the contents ?
Ans. 39 yds. 3 ft. 4' 6".
}
7. How many cord feet of wood in a load 8 feet long, 4
feet wide, and 3 feet 6 inches high?
Note. It will be recollected, that 16 solid feet make a
cord foot.
Ans. 7 cord feet.
8. In a pile of wood 176 ft. in length, 3 ft. 9 wide, and
4 ft. 3' high, how many cords?
Ans. 21 cords, and 7 cord feet over.
9. How many feet of cord wood in a load 7 feet long, 3
feet wide, and 3 feet 4 inches high? and what will it come
to at $40 per cord foot?
Ans. 48 cord feet, and it will come to $175.
10. How much wood in a load 10 ft. in length, 3 ft. 9' in
width, and 4 ft. 8' in height? and what will it cost at $1'92
per cord?
}
Ans. 1 cord and 215 cord feet, and it will come to
$2'62.
T 104. Remark. By some surveyors of wood, dimen-
sions are taken in feet and decimals of a foot. For this pur-
pose, make a rule or scale 4 feet long, and divide it into feet,
and each foot into ten equal parts. On one end of the rule,
Ch
1
1
I
1
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{
T 104, 105..
205
for 1 foot, let each of these parts be divided into 10 other
equal parts. The former division will be 10ths, and the lat-
ter 100ths of a foot. Such a rule will be found very con-
venient for surveyors of wood and of lumber, for painters,
joiners, &c.; for the dimensions taken by it being in feet and
decimals of a foot, the casts will be no other than so many
operations in decimal fractions.
INVOLUTION.
11. How many square feet in a hearth stone, which, by a
rule, as above described, measures 4'5 feet in length, and
2'6 feet in width? and what will be its cost, at 75 cents per
square foot ?.
Ans. 117 feet; and it will cost $84775.
12. How many cords in a load of wood 75 feet in length,
3'6 feet in width, and 4'8 feet in height? Ans. 1 cord,1ft.
13. How many cord feet in a load of wood 10 feet long,
3'4 feet wide, and 3'5 feet high?
Ans. 716.
QUESTIONS.
1. What are duodecimals? 2. From what is the word
derived? 3. Into how many parts is a foot usually divided,
and what are the parts called? 4. What are the other de-
nominations? 5. What is understood by the indices of the
denominations? 6. In what are duodecimals chiefly used
7. How are the contents of a surface bounded by straight lines
found? 8. How are the contents of a solid found? 9. How
is it known of what denomination is the product of any two
denominations? 10. How may a scale or rule, be formed
for taking dimensions in feet and decimal parts of a foot?
!
INVOLUTION.
T 105. Involution, or the raising of powers, is the mul-
tiplying any given number into itself continually a certain
number of times. The products thus produced are called
the powers of the given number. The number itself is called
the first power, or root. If the first power be multiplied by
itself, the product is called the second power or square; if
the square be multiplied by the first power, the, product is
called the third power, or cube, &c.; thus,
5 is the root, or 1st power, of 5.
5×5 25 is the 2d power, or square, of 5,
5X55 125 is the 3d power, or cube, of 5,
5X5×5×5=625 is the 4th power, or biquadrate, of 5, 54.
-52
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The number denoting the power is called the index, or
exponent; thus, 54 denotes that 5 is raised or involved to
the 4th power.
1. What is the square, or 2d power, of 7 ?
2. What is the square of 30?
INVOLUTION.
3. What is the square of 4000 ?
4. What is the cube, or 3d power, of 4 ?
5. What is the cube of 800?
6. What is the 4th power of 60 ?
7. What is the square of 1?
of 4?
8. What is the cube of 1?
of 4?
9. What is the square of ?
}
10. What is the cube of
11. What is the square of ?
de
12. What is the square of 1'5?
kad
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(
18. How much is 27 ?
65?
108 ?
3
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{
Ans. 16000000.
Ans. 64.
Ans. 512000000.
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Ans. 12960000.
of 2?
of 3 ?
Ans. 1, 4, 9, and 16.
of 3?
of 2?
Ans. 1, 8, 27, and 64.
of $?
of
Ans. †, 1§, and 42.
of ?
of it?
Ans. r,, and 348.
the 5th power of ?
27)
Ans. 1, and .
the cube?
Ans. 49.
Ans. 900.
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13. What is the 6th power of 1'2?
14. Involve 24 to the 4th power.
Note. A mixed number, like the above, may be reduced
to an improper fraction before involving: thus, 21; or
it may be reduced to a decimal; thus, 242425.
Ans. $5012518.
Ans. 1821 2342.
15. What is the square of 47?
16. What is the value of 74, that is, the 4th power of 7?
Ans. 2401.
17. How much is 93 ?
361
T 105.
I
65 ?
36 ?
{
104 ?
Ans. 729, 7776, 10000
45 ?
53 ?
Ans. to last, 100000000.
The powers of the nine digits, from the first power to the
fifth, may be seen in the following
TABLE,
;
Ans. 2′25, and 3′375.
Ans. 2'985984.

Roots or 1st Powers 1 | 2 | 3
4.
B1 .6
16425
or 2d Powers|1| 4| 9
343 612
Squares
Cubes for 3d Powers 1 8 27 64| 125 | 216
Biquadrates or 4th Powers|1|16|81| 256 625 |1:296| 2401
| 4096 | 6561
Sursolids
or 5th Powers 1 32 |243 |1024 |3125 7776; 1680732768 159049
16 1
"
71 8
•491
64
9
81
729
T-107
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EXTRACTION OF THE SQUARE ROOT.
}
EVOLUTION.
¶ 106. Evolution, or the extracting of roots, is the me-
thod of finding the root of any power or number.
The root, as we have seen, is that number, which, by a
continual multiplication into itself, produces the given power.
The square root is a number which, being squared, will pro-
duce the given number; and the cube, or third root, is a num-
ber which, being cubed or involved to the 3d power, will
produce the given number: thus, the square root of 144 is
12, because 122² = 144; and the cuhe rool of 343 is 7, be-
cause 73, that is, 7 × 7 × 7, 343; and so of other num-
bers.
207
Although there is no number which will not produce a
perfect power by involution, yet there are many numbers of
which precise roots can never be obtained. But, by the
help of decimals, we can approximate, or approach, towards
the root to any assigned degree of exactness. Numbers,
whose precise roots cannot be obtained, are called surd
numbers, and those, whose roots can be exactly obtained, are
called rational numbers.
The square root is indicated by this character placed
before the number; the other roots by the same character,
with the index of the root placed over it. Thus, the
square
root of 16 is expressed 16; and the cube root of 27 is
expressed 3/27; and the 5th root of 7776,5/7776.
When the power is expressed by several numbers, with
the sign + or between them, a line, or vinculum, is drawn
from the top of the sign over all the parts of it; thus, the
square root of 21 5 is 21
5, &c.
Jolantat
1
EXTRACTION OF THE SQUARE
ROOT.
T 107. To extract the square root of any number is to
find a number, which, being multiplied into itself, shall pro-
duce the given number.
{ {
1. Supposing a man has 625 yards of carpeting, a yard
wide, what is the length of one side of a square room, the
I
दे*
章
​I
1
}
1
>
t
1
,"
208.
EXTRACTION OF THE SQUARE ROOT. T 107.
floor of which the carpeting will cover? that is, what is one
side of a square, which contains 625 square yards?
We have seen, (T 35,) that the contents of a square sur-
face is found by multiplying the length of one side into it-
self, that is, by raising it to the second power; and hence,
having the contents (625) given, we must extract its square
root to find one side of the room.
}
This we must do by a sort of trial: and,
1st. We will endeavour to ascertain how many figures
there will be in the root. This we can easily do, by point-
ing off the number, from units, into periods of two figures
each; for the square of any root always contains just twice as
many, or one figure less than twice as many figures, as are
in the root; of which truth the pupil may easily satisfy him-
self by trial. Pointing off the number, we find, that the
root will consist of two figures,
a ten and a unit.
ď
20
OPERATION.
625 (2
4
225
FIG. I.
A
282812080
20
C
1
•
2d. We will now seek for
the first figure, that is, for
the tens of the root, and it is
plain, that we must extract it
from the left hand period 6,
(hundreds.) The greatest
square in 6 (hundreds) we
find, by trial, to be 4, (bun-
dreds,) the root of which is 2,
(tens, 20;) therefore, we
set 2 (tens) in the root. The
root, it will be recollected, is
one side of a square. Let us,
then, form a square, (A, Fig.
I.) each side of which shall be
supposed 2 tens, 20 yards,
expressed by the root now
obtained.

{
D
The contents of this square are 20 × 20 400 yards, now
disposed of, and which, consequently, are to be deducted from
the whole number of yards, (625,) leaving 225 yards. This
deduction is most readily performed by subtracting the square
number 4, (hundreds,) or the square of 2, (the figure in the
root already found,) from the period 6, (hundreds,) and bring
ing down the next period by the side of the remainder,
making 225, as before.
I
}
1
'
¶ 107.
EXTRACTION OF THE SQUARE ROOT.
209
3d. The square A is now to be enlarged by the addition
of the 225 remaining yards; and, in order that the figure
may retain its square form, it is evident, the addition must
be made on two sides. Now, if the 225 yards be divided by
the length of the two sides, (20+20=40,) the quotient
will be the breadth of this new addition of 225 yards to the
sides c d and b c of the square A.
!
But our root already found, =2 tens, is the length of one
side of the figure A; we therefore take double this root, = 4
tens, for a divisor.
5 yds.
20 yds.
d
j
a
OPERATION-CONTINUED.
625 (25
4
45)225
225
B
FIG. II.
20 yds.
A
1
20
20
400
20
5
100
20 yda
1
C
5 yds.
5
D
1 25
с
20
5
100
5 yds.
5 yds.
}
20 yds.
The divisor, 4, (tens,)
is in reality 40, and we
are to seek how many
times 40 is contained in
225, or, which is the
same thing, we may
seek how many times
4 (tens) is contained in
22, (tens,) rejecting the
right hand figure of the
dividend, because we
have rejected the cipher
in the divisor. We find
our quotient, that is, the
breadth of the addition,
to be 5 yards; but, if
we look at Fig. II., we
shall perceive that this
addition of 5 yards to the
two sides does not com-
plete the square; for
there is still wanting, in
the corner D, a small
square, each side of
which is equal to this last quotient, 5; we must, therefore,
add this quotient, 5, to the divisor, 40, that is, place it at the
right hand of the 4, (tens,) making it 45; and then the whole
divisor, 45, multiplied by the quotient, 5, will give the con-
tents of the whole addition around the sides of the figure A,
which, in this case, being 225 yards, the same as our divi-
dend, we have no remainder, and the work is done. Con-
sequently, Fig. II. represents the floor of a square room, 25
S*
{
Ma
1

}
4
The square A contains 400 yards.
The figure B ..... 100
C
100
D
******
210
EXTRACTION OF THE SQUARE ROOT. T 107.
yards on a side, which 625 square yards of carpeting will
exactly cover.
The proof may be seen by adding together the several
parts of the figure, thus:-
***** 10
.....
۶۳
3
OTATO MA
25..........
}
Or we may prove it
by involution, thus:-
25 X 25=
625, as be-
fore.
Proof, 625 ...
From this example and illustration we derive the following
general
RULE
FOR THE EXTRACTION OF THE SQUARE ROut.
I
I. Point off the given number into pèriods of two figures
cach, by putting a dot over the units, another over the hun-
dreds, and so on. These dots show the number of figures
of which the root will consist.
II. Find the greatest square number in the left hand pe-
riod, and write its root as a quotient in division. Subtract
the square number from the left hand period, and to the re-
mainder bring down the next period for a dividend.
III. Double the root already found for a divisor; seek how
many times the divisor is contained in the dividend, except-
ing the right hand figure, and place the result in the root,
and also at the right hand of the divisor; multiply the di-
visor, thus augmented, by the last figure of the root, and
subtract the product from the dividend; to the remainder
bring down the next period for a new dividend.
IV. Double the root already found for a new divisor, and
continue the operation as before, until all the periods are
brought down.
Note 1. If we double the right hand figure of the last
divisor, we shall have the double of the root.
Note 2. As the value of figures, whether integers or
decimals, is determined by their distance from the place
of units, so we must always begin at unit's place to point off
the given number, and, if it be a mixed number, we must
point it off both ways from units, and if there be a deficiency
in any period of decimals, it may be supplied by a cipher.
1
It is plain, the root must always consist of so many integers
1
1
1
T 108.
EXTRACTION OF THE SQUARE ROOT.
211
and decimals as there are periods belonging to each in the
given number.
EXAMPLES FOR PRACTICE.
2. What is the square root of 10342656 ?
OPERATION.
3
3
t
}
}
!
1
62) 134
124
641 ) 1026
641
3, What is the square root of 43264?
OPERATION.
10342656 (3216, Ans.
9
6426) 38556
38556
¿
}
408) 3264
3264
1
"
43264 (208, Ans.
4
}
1
7. What is the square root of '001296 ?
8. What is the square root of '2916 ?
9. What is the square root of 36372961 ?
10. What is the square root of 164?
}
4. What is the square root of 998001 ?
5. What is the square root of 234'09 ?
6. What is the square root of 964′5192360241 ?
ch
Ň
Į
T
↑ 108. In this last example, as there was a remainder,
after bringing down all the figures, we continued the opera-
tion to decimals, by annexing two ciphers for a new period,
and thus we may continue the operation to any assigned de-
gree of exactness; but the pupil will readily perceive, that
he can never, in this manner, obtain the precise root; for the
last figure in each dividend will always be a cipher, and the
i
Ans. 999.
Ans. 15'3.
1
Ans. 31'05671..
Ans. '036.
Ans. $54.
Ans. 6031.
Ans. 12'8+.
1
1
T
1
$
1
1
1
}
I
212
SUPPLEMENT TO THE SQUARE ROOT. ¶ 108.
↓
Į
last figure in each divisor is the same as the last quotient
figure; but no one of the nine digits, multiplied into itself,
produces a number ending with a cipher; therefore, what-
ever be the quotient figure, there will still be a remainder.
ܕܐ
11. What is the square root of 3?
12. What is the square root of 10?
13. What is the square root of 184′2 ?
14. What is the square root of
? ·
Note. We have seen, (T 105, ex. 9,) that fractions are
squared by squaring both the numerator and the denomina-
tor. Hence it follows, that the square root of a fraction is
found by extracting the root of the numerator and of the de-
nominator. The root of 4 is 2, and the root of 9 is 3. ·
Ans. 3.
Ans. 3.
Ans. fo.
薯
​15. What is the square root of ?
16. What is 'the square root of 10% ?
17. What is the square root of?
18. What is the square root of 201?,
1
}
19. What is the square root of
20. What is the square root of 25?
L
Ans. 173+.
Ans. 3'16+.
Ans. 13'57+.
When the numerator and denominator are not exact
squares, the fraction may be reduced to a decimal, and the
approximate root found, as directed above.
675?
1
Ans. T
2.
Ans. 41.
SUPPLEMENT TO THE SQUARE ROOT.
QUESTIONS.
ΟΙ
a
1. What is involution? 2. What is understood by a
power? 3.
the first, the second, the third, the fourth
power? 4. What is the index, or exponent? 5. How do
you involve a number to any required power? 6. What is
evolution? 7. What is a root? 8. Can the precise root of all
numbers be found? 9. What is a surd number? 10.
rational? 11. What is it to extract the square root of any
number? 12. Why is the given sum pointed into periods of
two figures each ? 13. Why do we double the root for a
divisor? 14. Why do we, in dividing, reject the right hand
figure of the dividend? 15. Why do we place the quotient
figure to the right hand of the divisor? 16. How may we
Ans. '866 +.
Ans. '912+.
1
F
1
T 108.
2
213
prove the work? 17. Why do we point off mixed numbers
both ways from units? 18. When there is a remainder,
how may we continue the operation? 19. Why can we
never obtain the precise root of surd numbers? 20. How
do we extract the square root of vulgar fractions?
鼻子
​SUPPLEMENT TO THE SQUARE ROOT.
EXERCISES.
1. A general has 4096 men; how many must he place in
rank and file to form them into a square?
Ans. 64.
2. If a square field contains 2025 square rods, how many
rods does it measure on each side?
Ans. 45 rods.
t
3. How many trees in each row of a square orchard con-
taining 5625 trees?
Ans. 75.
4. There is a circle, whose area, or superficial contents,
is 5184 feet; what will be the length of the side of a square
of equal area?
/5184 72 feet, Ans..
5. A has two fields, one containing 40 acres, and the other
containing 50 acres, for which B offers him a square field
containing the same number of acres as both of these; how
many rods must each side of this field measure ?
Ans. 120 rods.
J
6. If a certain square field measure 20 rods on each side,
how much will the side of a square field measure, contain-
ing 4 times as much? /20 X 20 X 4 = 40 rods, Ans.
7. If the side of a square be 5 feet, what will be the side
of one 4 times as large?
9 times as large ?
times as large?
large?
25 times as large?
16
36 times as
Answers, 10 ft.; 15 ft.; 20 ft.; 25 ft.; and 30 ft..
8. It is required to lay out 288 rods of land in the form of
a parallelogram, which shall be twice as many rods in length
as it is in width.
Tak ada
Note. If the field be divided in the middle, it will form
two equal squares.
{
Ans. 24 rods long, and 12 rods wide.
9. I would set out, at equal distances, 784 apple trees, so
that my orchard may be 4 times as long as it is broad; how
many rows of trees must I have, and how many trees in
each row?
Ans. 14 rows, and 56 trees in each row.
10. There is an oblong piece of land, containing 192 square
rods, of which the width is as much as the length; re-
quired its dimensions.
Ans. 16 by 12.
1
{
1
I
>
1
1
f
}
1
f
•
214
11. There is a circle, whose diameter is 4 inches; what is
the diameter of a circle 9 times as large?
"
Notë. The areas or contents of circles are in proportion
to the squares of their diameters, or of their circumferences.
Therefore, to find the diameter required, square the given
diameter, multiply the square by the given ratio, and the
square root of the product will be the diameter required.
√√4X4X9= 12 inches, Ans.
12. There are two circular ponds in a gentleman's pleasure
ground; the diameter of the less is 100 feet, and the greater
is 3 times as large; what is its diameter? Ans. 173'2+ feet.
13. If the diameter of a circle be 12 inches, what is the
diameter of one as large?
Ans. 6 inches..
を​興奮
​+
T 109. 14. A carpenter has a large wooden square; one
part of it is: 4 feet long, and the other part 3 feet long; what
is the length of a pole, which will just reach from one end to
the other?
1
"
Note. A figure of 3
sides is called a triangle,
and, if one of the corners
be a square corner, or right
angle, like the angle at B
in the annexed figure, it is
called a right-angled trian-
gle, of which the square
of the longest side, A C,
(called the hypotenuse,)
}
is equal to the sum of the squares of the other two sides, A B
and B C.
T
SUPPLEMENT TO THE SQUARE ROOT. · T 109.
1
Hypotenuse.
{
Base.
'
1
A
ť.
B
I
Perpendicular.
t
4216, and 329; then, 9+16 5 feet, Ans.
15. If, from the corner of a square room, 6 feet be mea-
sured off one way, and 8 feet the other way, along the sides
of the room, what will be the length of a pole reaching from
point to point?
Ans. 10 feet.
16. A wall is 32, feet high, and a ditch before it is 24 feet
wide; what is the length of a ladder that will reach from the
top of the wall to the opposite side of the ditch?
Ans. 40 feet.
17. If the ladder be 40 feet, and the wall 32 feet, what is
the width of the ditch?
Ans. 24 feet.
18. The ladder and ditch given, required the wall.
}
Ans. 32 feet
1
}
1

1
F
T 110.
215
19. The distance between the lower ends of two equal
rafters is 32 feet, and the height of the ridge, above the beam
on which they stand, is 12 feet; required the length of each
rafter.
Ans. 20 feet.
t
·
}
20. There is a building 30 feet in length and 22 feet in
width, and the eaves project beyond the wall 1 foot on every
side; the roof terminates in a point at the centre of the
building, and is there supported by a post, the top of which
is 10 feet above the beams on which the rafters rest; what
is the distance from the foot of the post to the corners of the
eaves? and what is the length of a rafter reaching to the
middle of one side? a rafter reaching to the middle of
one end? and a rafter reaching to the corners of the eaves?
Answers, in order, 20 ft.; 15'62 ft.; 18'86 ft.; and
22'36+ ft.
EXTRACTION OF THE CUBE ROOT."
21. There is a field 800 rods long, and 600 rods wide;
what is the distance between two opposite corners ?
Ans. 1000 rods.
22. There is a square field containing 90 acres; how
many rods in length is each side of the field? and how many
rods apart are the opposite corners ?
Answers, 120 rods; and 1697+rods.
23. There is a square field containing 10 acres; what dis-
tance is the centre from each corner? Ans. 28'28 + rods.
1
EXTRACTION OF THE CUBE
ROOT.
¶ 110. A solid body, having six equal sides, and each of
the sides an exact square, is a CUBE, and the measure in
length of one of its sides is the root of that cube; for the
length, breadth and thickness of such a body are all alike; con-
sequently, the length of one side, raised to the 3d power,
gives the solid contents. (See T 36.)
Hence it follows, that extracting the cube root of any num-
ber of feet is finding the length of one side of a cubic bo-
dy, of which the whole contents will be equal to the given
number of feet.
1. What are the solid contents of a cubic block, of which
each side measures 2 feet? Ans. 23 2 X 2 X 28 feet.
2. How many solid feet in a cubic block, measuring 5 feet
on each side?
Ans. 53 125 feet.
1
http
¿
1
}
E
1
Į
216
EXTRACTION OF THE CUBE ROOT. T 110.
3. How many feet in length is each side of a cubic block,
containing 125 solid feet?
= 5 feet.
Ans. /125
Note. The root may be found by trial.
feet?
216 solid feet?
feet?
4. What is the side of a cubic block, containing 64 solid.
27 solid feet?
512 solid
Answers, 4 ft.; 3 ft.; 6 ft.; and 8 ft.
5. Supposing a man has 13824 feet of timber, in separate
blocks of 1 cubic foot each; he wishes to pile them up in
a cubic pile; what will be the length of each side of such
a pile?
1
It is evident, the answer is found by extracting the cube
root of 13824; but this number is so large, that we cannot
so easily find the root by trial as in the former examples;
We will endeavour, however, to do it by a sort of trial; and,
1st. We will try to ascertain the number of figures, of
which the root will consist. This we may do by pointing
the number off into periods of three figures each ( 107, ex. 1.)
I
Pointing off, we see, the
root will consist of two figures,
'a ten and a unit. Let us, then,
seek for the first figure, or
tens of the root, which must
be extracted from the left
hand period, 13, (thousands.)
The greatest cube in 13
(thousands) we find by trial,
or by the table of powers, to be
8, (thousands,) the root of
which is 2, (tens;) therefore,
we place 2 (tens) in the root.
The root, it will be recollect-
ed, is one side of a cube. Let
us, then, form a cube, (Fig. I.)
each side of which shall be
supposed 20 feet, expressed
by the root now obtained.
The contents of this cube are
20×20×20= 8000 solid feet,
which are now disposed of, and which, consequently, are to
be deducted from the whole number of feet, 13824. 8000
taken from 13824 leave 5824 feet. This deduction is most
readily performed by subtracting the cubic number, 8, or
the cube of 2, (the figure of the root already found,) from
1
B
20
OPERATION.
13824 (2
8
5824
20
20
FIG. I.
C
C
F
Į
20
400
20
8000 feet, Contents.
D
20
E
Ka
1
7

"
T 110.
217
the period 13, (thousands,) and bringing down the next pe-
riod by the side of the remainder, making 5824, as before.
2d. The cubic pile A D is now to be enlarged by the ad-
dition of 5824 solid feet, and, in order to preserve the cubic
form of the pile, the addition must be made on one half of
its sides, that is, on 3 sides, a, b, and c. Now, if the 5824
solid feet be divided by the square contents of these 3 equal
sides, that is, by 3 times, (20 X 20400) 1200, the quo-
tient will be the thickness of the addition made to each of
the sides a, b, c. But the root, 2, (tens,) already found, is
the length of one of these sides; we therefore square the
root, 2, (tens,) 20 × 20 = 400, for the square contents of one
side, and multiply the product by 3, the number of sides,
400 × 3 = 1200; or, which is the same in effect, and more
convenient in practice, we may square the 2, (tens,) and mul-
tiply the product by 300, thus, 2 X 24, and 4 × 300=1200,
for the divisor, as before.
X
?
20
13824 (24 Root.
8
Divisor, 1200)5824 Dividend.
20
n
The divisor, 1200, is con-
OPERATION-CONTINUED. tained in the dividend 4 times;
consequently, 4 feet is the
thickness of the addition made
to each of the three sides, a,'
b, c, and 4 × 1200 = 4800, is
the solid feet contained in
these additions; but, if we
look at Fig. II., we shall per-
ceive, that this addition to the
3 sides does not complete the
cube; for there are deficiencies
in the 3 corners n, n, n.
Now
the length of each of these
deficiencies is the same as the
length of each side, that is, 2
(tens) 20, and their width
and thickness are each equal to
the last quotient figure, (4);
their contents, therefore, or
the number of feet required to
fill these deficiencies, will be
found by multiplying the square
of the last quotient figure, (42)
=16, by the length of all the
deficiencies, that is, by 3 times
20
m
A
7
EXTRACTION OF
TRA
5
mag
20
4800
960
64
5824
0000
FIG. II,
20
N
N
HE CUBE ROOT.

20
1
}
}
218
EXTRACTION OF THE CUPE ROOT. T 110.
1
1
the length of each side, which is expressed by the former
quotient figure, 2, (tens.) 3 times 2 (tens) are 6 (tens)
60; or, what is the same in effect, and more convenient in
practice, we may multiply the quotient figure, 2, (tens,) by
30, thus, 2 X 3060, as before; then, 60 X 16960, con-
×
tents of the three deficiencies n, n, n.
FIG. III.
4
20
4
20
20
f
20 +
Kappa mag ka
+
THE
ፏ
FIG. IV.
24 feet.
4
20
24
4
20
A
1
Looking at Fig. III., we
perceive there is still a de-
ficiency in the corner where
the last blocks meet. This
deficiency, is a cube, each
side of which is equal to the
last quotient figure, 4. The
cube of 4, therefore, (4 X 4
× 464,) will be the solid
contents of this corner, which
in Fig. IV. is seen filled.
►
Now, the sum of these sev-
eral additions, viz. 4800 +
-960645824, will make
the subtrahend, which, sub-
tracted from the dividend,
leaves no remainder, and the
work is done.
24 feel.
given number; or it may be
the contents of all the several parts, thus,
Feet.
8000 = contents of Fig. I.
4800
960
64
1
Pendants Boy
504
Fig. IV. shows the pile
which 13824 solid blocks of
one foot each would make,
when laid together, and the
root, 24, shows the length of
one side of the pile. The
correctness of the work may
be ascertained by cubing the
side now found, 243, thus, 24
× 24 × 24 = 13824, the
proved by adding together
•
addition to the sides a, b, and c, Fig. 1.
addition to fill the deficiencies n, m, n, Fig. II.
addition to fill the corner e, e, e, Fig. IV.
13824 ÷ contents of the whole pile, Fig. IV., 24 feet on
Peach side
►


T 110.
EXTRACTION OF THE CUBE ROOT.
219
From the foregoing example and illustration we derive the
following
६
J
}
RULE
FOR EXTRACTING the CUBE ROOT.
1
I. Separate the given number into periods of three figures
each, by putting a point over the unit figure, and every third
figure beyond the place of units.
II. Find the greatest cube in the left hand period, and put
its root in the quotient.
III. Subtract the cube thus found from the said period,
and to the remainder bring down the next period, and call
this the dividend.
IV. Multiply the square of the quotient by 300, calling it
the divisor.
V. Seek how many times the divisor may be had in the
dividend, and place the result in the root; then multiply
the divisor by this quotient figure, and write the product
under the dividend.
VI. Multiply the square of this quotient figure by the
former figure or figures of the root, and this product by 30,
and place the product under the last; under ail write the
cube of this quotient figure, and call their amount the sub-
trahend.
VII. Subtract the subtrahend from the dividend, and to the
remainder bring down the next period for a new dividend,
with which proceed as before; and so on, till the whole is
finished.
Note 1. If it happens that the divisor is not contained in
the dividend, a cipher must be put in the root, and the
next period brought down for a dividend.
Note 2. The same rule must be observed for continuing
the operation, and pointing off for decimals, as in the square
root.
Note 3. The pupil will perceive that the number which
we call the divisor, when multiplied by the last quotient
figure, does not produce so large a number as the real sub-
trakend; hence, the figure in the root must frequently be
smaller than the quotient figure.
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SUPPLEMENT TO THỂ Cube root.
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EXAMPLES FOR PRACTICE.
6. What is the cube root of, 1860867 ?
OPERATION.
1
300) 860 first Dividend.
600
120
8
12 X 300
22 X 1 X 30
23
i860867(123 Ans.
C
32 X 12 X 30
33 =
ܚܝܚ
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728 first Subtrahend.
122 × 30043200 ) 132867 second Dividend.
129600
3240
27
132867 second Subtrahend.
000000
7. What is the cube root of 373248 ?
8. What is the cube root of 21024576 ?
•
J
T 110
9. What is the cube root of 84'604519 ?
10. What is the cube root of '000343 ?
11. What is the cube root of 2?
12. What is the cube root of?
Note. See T 105, ex. 10, and T 108, ex. 14.
13. What is the cube root of 128?
Τ
14. What is the cube root of 143?
15. What is the cube root of ?
16. What is the cube root of ThE?
Ans. 72.
Ans. 276.
Ans. 4'39.
Ans. '07.
1
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Ans. 1'25+.
Ans.
Ans. &
Ans. z
SUPPLEMENT TO THE CUBE ROOT.
QUESTIONS.
1. What is a cube? 2. What is understood by the
cube root? 3. What is it to extract the cube root?
4. Why is the square of the quotient multiplied by 300
for a divisor? 5. Why, in finding the subtrahend, do
we multiply the square of the last quotient figure by 30
times the former figure of the root? 6. Why do we
cube the quotient figure? 7. How do we prove the
operation?
Ans. 125+.
Ans. f.
}
T 111. SUPPLEMENT TO THE CUBE ROOT.
}
74
1
221
EXERCISES.
Про
1. What is the side of a cubical mound, equal to one 288
feet long, 216 feet broad, and 48 feet high? Ans. 144 feet.
2. There is a cubic box, one side of which is 2 feet; how
many solid feet does it contain ?
Ans. 8 feet.
3. How many cubic feet in one 8 times as large? and
what would be the length of one side ?
/
Ans. 64 solid feet, and one side is 4 feet.
4. There is a cubical box, one side of which is 5 feet;
what would be the side of one containing 27 times as much?
64 times as much?
125 times as much?
Ans. 15, 20, and 25 feet.
5. There is a cubical box, measuring 1 foot on each
side; what is the side of a box 8 times as large?
27
times?
64 times?
Ans. 2, 3, and 4 feet.
T 111. Hence we see, that the sides of cubes are as the
cube roots of their solid contents, and, consequently, their con-
tents are as the cubes of their sides. The same proportion is
true of the similar sides, or of the diameters of all solid figures
of similar forms.
4
6. If a ball, weighing 4 pounds, be 3 inches in diameter,
what will be the diameter of a ball of the same metal, weigh-
ing 32 pounds? 4 32: 33: 63
:
Ans. 6 inches.
7. If a ball, 6 inches in diameter, weigh 32 pounds, what
will be the weight of a ball 3 inches in diameter? Ans. 4 lbs.
8. If a globe of silver, 1 inch in diameter, be worth $6,
what is the value of a globe 1 foot in diameter ?
Ans. $10368.
9. There are two globes; one of them is 1 foot in diame-
ter, and the other 40 feet in diameter; how many of the
smaller globes would it take to make 1 of the larger?
Ans. 64000.
"
10. If the diameter of the sun is 112 times as much as the
diameter of the earth, how many globes like the earth would
it take to make one as large as the sun? Ans. 1404928.
11. If the planet Saturn is 1000 times as large as the
earth, and the earth is 7900 miles in diameter, what is the
diameter of Saturn?
Ans. 79000 miles.
12. There are two planets of equal density; the diameter
of the less is to that of the larger as 2 to 9; what is the ra-
tio of their solidities?
Ans. 785; or, as 8 to 729.
T*
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1
222
ARITHMETICAL PROGRESSION.
PROGRESSION. ¶ 111, 112.
Note. The roots of most powers may be found by the
square and cube root only: thus, the biquadrate, or 4th root,
is the square root of the square root; the 6th root is the
cube root of the square root; the 8th root is the square root
of the 4th root; the 9th root is the cube root of the cube
root, &c. Those roots, viz. the 5th, 7th, 11th, &c., which
are not resolvable by the square and cube roots, seldom oc-
cur, and, when they do, the work is most easily performed
by logarithms; for, if the logarithm of any number be divided.
by the index of the root, the quotient will be the logarithm
of the root itself.
ARITHMETICAL PROGRESSION.
T 112. Any rank or series of numbers, more than two,
increasing or decreasing by a constant difference, is called an
Arithmetical Series, or Progression..
When the numbers are formed by a continual addition of
the common difference, they form an ascending series; but
when they are formed by a continual subtraction of the com-
mon difference, they form a descending serics,
3, 7, 9,
15, is an series.
Thus, { 18, 15, 11, 3; 17, 13, 3, &c. is a descending series.
9, 5,
The numbers which form the series are called the terms
of the serics. The first and last terms are the extremes, and
the other terms are called the means.
There are five things in arithmetical progression, any three
of which being given, the other two may be found:
1st. The first term.
2d. The last term.
3d. The number of terms.
4th. The common difference.
5th. The sum of all the terms.
T
kataka
1. A man bought 100 yards of cloth, giving 4 cents for the
first yard, 7 cents for the second, 10 cents for the third, and
so on, with a common difference of 3 cents; what was the
cost of the last yard?
As the common difference, 3, is added to every yard except
the last, it is plain the last yard must be 99 × 3,,:
cents, more than the first yard.
Ans. 301 cents.
297
1
1
T 112.
223
Hence, when the first term, the common difference, and the
number of terms, are given, to find the last term,-Multiply the
number of terms, less 1, by the common difference, and add
the first term to the product for the last term.
1
ARITHMETICAL PROGRESSION.
2. If the first term be 4, the common difference 3, and
the number of terms 100, what is the last term? Ans. 301.
3. There are, in a certain triangular field, 41 rows of
corn'; the first row, in 1 corner, is a single hill, the second
contains 3 hills, and so on, with a common difference of 2;
what is the number of hills in the last row? Ans. 81 hills.
4. A man puts out $1, at 6 per cent. simple interest,
which, in 1 year, amounts to $1'06, in 2 years to $1'12,
and so on, in arithmetical progression, with a common dif-
ference of $ '06; what would be the amount in 40 years?
Ans. $3'40.
Hence we see, that the yearly amounts of any sum, at
simple interest, form an arithmetical series, of which the
principal is the first term, the last amount is the last term, the
yearly interest is the common difference, and the number of
years is 1 less than the number of terms.
5. A man bought 100 yards of cloth in arithmetical pro-
gression; for the first yard he gave 4 cents, and for the last
301 cents; what was the common increase of the price on
each succeeding yard?
This question is the reverse of example 1; therefore,
301 4 =
297, and 297 ÷ 99 3, common difference..
Kapaļ
Hence, when the extremes and number of terms are given,
to find the common difference,-Divide the difference of the
extremes by the number of terms, less 1, and the quotient
will be the common difference.
}
6. If the extremes be 5 and 605, and the number of terms
151, what is the common difference?
Ans. 4.
7. If a man puts out $1, at simple interest, for 40 years,
and receives, at the end of the time, $3'40, what is the
rate?
If the extremes be 1 and 3'40, and the number of terms
41, what is the common difference?
Ans. '06.
alike; the
what was
Ans. 5 years..
8. A man had 8 sons, whose ages differed
youngest was 10 years old, and the eldest 45;
the common difference of their ages?
1
J
4
}
224
T 112.
9. A man bought 100 yards of cloth in arithmetical series;
he gave 4 cents for the first yard, and 301 cents for the last
yard; what was the average price per yard, and what was
the amount of the whole?
· ARITHMETICAL PROGRESSION.
Since the price of each succeeding yard increases by a con-
stant excess, it is plain, the average price is as much less than
the price of the last yard, as it is greater than the price of
the first yard; therefore, one half the sum of the first and
last price is the average price.
4
301 cts.
One half of 4 cts.
152 cts.. average
price; and the price, 1524 cts. × 100 = 15250 cts.= Ans.
$152'50, whole cost.
pild
Hence, when the extremes and the number of terms are given,
to find the sum of all the terms,-Multiply the sum of the ex-
tremes by the number of terms, and the product will be
the answer.
10. If the extremes be 5 and 605, and the number of
terms 151, what is the sum of the series?' Ans. 46055.
11. What is the sum of the first 100 numbers, in their
natural order, that is, 1, 2, 3, 4, &c. ?
Ans. 5050.
12. How many times does a common clock strike in 12
hours?
Ans. 78.
13. A man rents a house for $50, annually, to be paid at
the close of each year; what will the rent amount to in 20
years, allowing 6 per cent., simple interest, for the use of
the money?
The last year's rent will evidently be $50 without interest,
the last but one will be the amount of $50 for 1 year, the
last but two the amount of $50 for 2 years, and so on, in
arithmetical series, to the first, which will be the amount of
$.50 for 19 years $107.
✔
If the first term be 50, the last term 107, and the number
of terms 20, what is the sum of the series? Ans. $1570.
}
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}
14. What is the amount of an annual pension of $100,
being in arrears, that is, remaining unpaid, for 40 years,
allowing 5 per cent. simple interest?
Ans. $7900.
15. There are, in a certain triangular field, 41 rows of
corn; the first row, being in 1 corner, is a single hill, and
the last row, on the side opposite, contains 81 hills; how
many hills of corn in the field ?
Ans. 1681 hills.
1
·
»
¶ 112, 113. GEOMETRICAL PROGRESSION.
225
16. If a triangular piece of land, 30 rods in length, be 20
rods wide at one end, and come to a point at the other, what
number of square rods does it contain ?.
Ans. 300.
17. A debt is to be discharged at 11 several payments,
in arithmetical series, the first to be $ 5, and the last $75;
what is the whole debt?
the common difference be-
tween the several payments?
Ans. whole debt, $440; common difference, $7.
18. What is the sum of the series 1, 3, 5, 7, 9, &c., to
1001 ?
Ans. 251001.
Note. By the reverse of the rule under ex. 5, the differ-
ence of the extremes 1000, divided by the common difference 2,
gives a quotient, which, increased by 1, is the number of
terms 501.
*
19. What is the sum of the arithmetical series 2, 2, 3,
31, 4, 41, &c., to the 50th term inclusive ?
Ans. 712.
20. What is the sum of the decreasing series 30, 29%, 291,
29, 283, &c., down to 0?
}
Note. 30+ 1
91, number of terms. Ans. 1365.
QUESTIONS.
1
1. What is an arithmetical progression? 2. When is the
series called ascending? 3. when descending? 4. What
are the numbers, forming the progression, called? 5. What
are the first and last terms called? 6. What are the other
terms called? 7. When the first term, common difference,
and number of terms, are given, how do you find the last
term? 8. How may arithmetical progression be applied to
simple interest? 9. When the extremes and number of
terms are given, how do you find the common difference?
10. how do you find the sum of all the terms?
GEOMETRICAL PROGRESSION.
T 113. Any series of numbers, continually increasing by
a constant multiplier, or decreasing by a constant divisor, is
called a Geometrical Progression. Thus, 1, 2, 4, 8, 16, &c.
is an increasing geometrical series, and 8, 4, 2, 1, 1, 1, &c.
is a decreasing geometrical series.
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226
As in arithmetical, so also in geometrical progression,
there are five things, any three of which being given, the
other two may be found :-
t
:
1st. The first term.
2d. The last term.
GEOMETRICAL PROGRESSION.
3d. The number of terms.
4th. The ratio.
5th. The sum of all the terms.
The ratio is the multiplier or divisor, by which the series is
formed.
}
J
1
1. A man bought a piece of silk, measuring 17 yards, and,
by agreement, was to give what the last yard would come
to, reckoning 3 cents for the first yard, 6 cents for the second,
and so on, doubling the price to the last; what did the piece
of silk cost him?
I
3 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2
X2 X2 X2 X2 196608 cents, $1966'08, Answer.
=
T 113.
}
In examining the process by which the last term (196608)
has been obtained, we see, that it is a product, of which the
ratio (2) is sixteen times a factor, that is, one time less than
the number of terms. The last term, then, is the sixteenth
power of the ratio, (2,) multiplied by the first term (3.)
Now, to raise 2 to the 16th power, we need not produce
all the intermediate powers; for 24 2 X 2 X 2 X 2 = 16,
is a product of which the ratio 2 is 4 times a factor; now,
if 16 be multiplied by 16, the product, 256, evidently con-
tains the same factor (2) 4 times + 4 times, = 8 times;
and 256 X 256 65586, a product of which the ratio (2)
is 8 times 8 times, 16 times, factor;
16 times, factor; it is, therefore,
the 16th power of 2, and, multiplied by 3, the first term,
gives 196608, the last term, as before. Hence,
س
When the first term, ratio, and number of terms, are given,
to find the last term,-
|
I. Write down a few leading powers of the ratio with
their indices over them.
1
II. Add together the most convenient indices, to make an
index less by one than the number of the term sought.
III. Multiply together the powers belonging to those in-
dices, and their product, multiplied by the first term, will be
the term sought.
I
}
T 113.
227
2. If the first term be 5, and the ratio 3, what is the 8th
term?
1 2
2 3 + 4 =
my
3, 9, 27, × 81=2187 × 5 first
term, 10935, Answer.
3. A man plants 4 kernels of corn, which, at harvest,
produce 32 kernels; these he plants the second year; now,
supposing the annual increase to continue 8 fold, what
would be the produce of the 16th year, allowing 1000 ker-
nels to a pint?
Ans. 2199023255′552 bushels.
4. Suppose a man had put out one cent at compound in-
terest in 1620, what would have been the amount in 1824,
allowing it to double once in 12 years?
217
131072.
}
Powers of the ratio, with
their indices over them.
1
P
GEOMETRICAL PROGRESSION.
MARIPOSA
Ans: $131072.
5. A man bought 4 yards of cloth, giving 2 cents for the
first yard, 6 cents for the second, and so on, in 3 fold ra-
tio; what did the whole cost him?
2+6+18+ 54 80 cents.
¿
Ans. 80 cents.
In a long series, the process of adding in this manner
would be tedious. Let us try, therefore, to devise some
shorter method of coming to the same result. If all the
terms, excepting the last, viz. 2+6+18, be multiplied by
the ratio, 3, the product will be the series 6 + 18+ 54;
subtracting the former series from the latter, we have, for the
remainder, 54 — 2, that is, the last term, less the first term,
which is evidently as many times the first series (2 +6 +18)
as is expressed by the ratio, less 1: hence, if we divide the
difference of the extremes (54 2) by the ratio, less 1,
(31,) the quotient will be the sum of all the terms, ex-
cepting the last, and, adding the last term, we shall have the
whole amount. Thus, 54 2 52, and 3-1 2; then,
52226, and 54 added, makes 80, Answer, as before.
Maakon
J
Defragt
Hence, when the extremes and ratio are given, to find the
sum of the series,-Divide the difference of the extremes by the
ratio, less 1, and the quotient, increased by the greater term,
will be the answer.
(Calgary)
Magdad de
6. If the extremes be 4 and 131072, and the ratio 8,
what is the whole amount of the series?
131072 4
8-1
+ 131072 = 149796 Answer.
}
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1
228
T 113.
7. What is the sum of the descending series 3, 1, 1, 1,
, &c., extended to infinity?
It is evident the last term must become 0, or indefinitely
near to nothing; therefore, the extremes are 3 and 0, and
the ratio 3.
Ans. 4.
GEOMETRICAL PROGRESSION.
8. What is the value of the infinite series 1+1+&+
64, &c. ?
Ans. 14.
//
9. What is the value of the infinite series, To + 180 t
TOTO + TOOOO, &c., or, what is the same, the decimal
11111, &c., continually repeated?
Ans. f.
10. What is the value of the infinite series, 10 + 10800;
&c., descending by the ratio 100, or, which is the same, the
repeating decimal '020202, &c. ?
Ans..
11. A gentleman, whose. daughter was married on a new
year's day, gave her a dollar, promising to triple it on the
first day of each month in the year; to how much did her
portion amount?
Here, before finding the amount of the series, we must
find the last term, as directed in the rule after ex. 1.
Ans. $265 720
The two processes of finding the last term, and the amount,
may, however, be conveniently reduced to one, thus:-
When the first term, the ratio, and the number of terms, are
given, to find the sum or amount of the series,-Raise the ratio
to a power whose index is equal to the number of terms, from
which subtract 1; divide the remainder by the ratio, less 1,
and the quotient, multiplied by the first term, will be the
answer.
Applying this rule to the last example, 312 531441, and
"
531441
1
X 1265720. Ans. $2654720, as before,
X 1:
Burgenlan
3-1
12. A man agrees to serve a farmer 40 years without any
other reward than 1 kernel of corn for the first year, 10 for
the second year, and so on, in 10 fold ratio, till the end of
the time; what will be the amount of his wages, allowing
1000 kernels to a pint, and supposing he sells his corn at 50
Gents per bushel?
1040 1
10-1
(1,111,111,111,111,111,111,111,111,
111,111,111,111,111 kernels.
Ans. $8,680,555,555,555,555,555,555,555,555,555,555
555.
I
T 113, 114.
229
13. A gentleman, dying, left his estate to his 5 sons, to
the youngest $1000, to the second $1500, and ordered,
that each son should exceed the younger by the ratio of 14;
what was the amount of the estate?
Note. Before finding the power of the ratio 1, it may
be reduced to an improper fraction, or to a decimal, 1'5.
X 1000
5 I
1655 1
X 1000 = $131871; or,
15-1
£ - 1
$13187'50, Answer.
Compound Interest by Progression.
T 114. 1. What is the amount of $4, for 5 years, at 6
per cent. compound interest?
We have seen, (T 92,) that compound interest is that,
which arises from adding the interest to the principal at the
close of each year, and, for the next year, casting the inter-
est on that amount, and so on. The amount of $1 for 1
year is $1'06; if the principal, therefore, be multiplied by
1'96, the product will be its amount for 1 year; this amount,
multiplied by 1'06, will give the amount (compound inter-
est) for 2 years; and this second amount, multiplied by 1'06,
will give the amount for 3 years; and so on. Hence,
the several amounts, arising from any sum at compound in-
terest, form a geometrical series, of which the principal is the
first term; the amount of $1 or 1 £., &c., at the given rate
per cent., is the ratio; the time, in years, is 1 less than the
number of terms; and the last amount is the last term.
The last question may be resolved into this:-If the first
term be 4, the number of terms 6, and the ratio 1'06,
what is the last term?
1'0651'338, and 1'338×4=$5'352+.
GEOMETRICAL PROGRESSION.
Ans. $5'352.
Note 1. The powers of the amounts of $1, at 5 and at 6
per cent., may be taken from the table, under T 91. Thus,
opposite 5 years, under 6 per cent., you find 1'338, &e.
Note 2. The several processes may be conveniently exhi-
bited by the use of letters; thus :—
Let P. represent the Principal.
R.
T.
A
TERRAN
****
CON
bangan) Kateg
the Ratio, or the amount of $ 1, &c. for 1 year.
the Time, in years.
A............. the Amount.
When two or more letters are joined together, like a word,
U
1
1
$
1
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2
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T 114.
they are to be multiplied together. Thus PR. implies, that
the principal is to be multiplied by the ratio.
J
When one
1
!
{
J
letter is placed above another, like the index of a power, the
first is to be raised to a power, whose index is denoted by the
second. Thus RT implies, that the ratio is to be raised to
a power, whose index shall be equal to the time, that is, the
number of years.
}
+
Jedna
GEOMETRICAL PROGRESSION.
1
2. What is the amount of 40 dollars for 11 years, at 5 per
cent. compound interest?
RT. X P.
+
A.; therefore, 1'0511 × 40
68'4.
Ans. $68'40.
3. What is the amount of $6 for 4 years, at 10 per cent.
compound interest?
Ans. $87847%.
4. If the amount of a certain sum for 5 years, at 6 per
cent. compound interest, be $5'352, what is that sum, or
principal?
If the number of terms be 6, the ratio 1'06, and the last
term 5'352, what is the first term?
I
man pe
1'338
/
他
​This question is the reverse of the last; therefore,
A.
5'352
P.; or,
4.
RT.
J
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​Ans. $4.
5. What principal, at 10 per cent. compound interest, will
amount, in 4 years, to $87846 ?
Ans. $6.
6. What is the present worth of $68'40, due 11 years
nence, discounting at the rate of 5 per cent. compound in-
terest?
· Ans, $40.
7. At what rate per cent. will $6 amount to $8'7846 in
·4 years ?
}
If the first term be 6, the last term 87816, and the num
ber of terms 5, what is the ratio?
JA
847846
f 1
Agg
1
P...
1'4641
6
RT, that is,
the ratio; and then, by extracting the
1'10 for the ratio.
{
8. In what time will $6 amount to
·
cent. compound interest?
A.
87816
=RT, that is,
1'46411'10; therefore,
P.
6
if we divide 1'4641 by 1'10, and then divide the quotient
thence arising by 1'10, and so on, till we obtain a quotient
that will not contain 1'10, the number of these divisions will
be the number of years.
Ans. 4 years.
the 4th power of
4th root, we obtain
Ans. 10 per cent.
$8'7846, at 10 per
J
T114, 115. GEOMETRICAL PROGRESSION.
231
{
9. At 5 per cent. compound interest, in what time will
$40 amount to $68'40?
{
11
1
÷
Having found the power of the ratio 1905, as before, which
is 171, you may look for this number in the table, under
the given rate, 5 per cent., and against it you will find the
number of years.
Ans. 11 years.
}
22
10. At 6 per cent. compound interest, in what time will
4 amount to $5352?
Ans. 5 years.
1
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Annuities at Compound Interest.
T115. It may not be amiss, in this place, briefly to show
the application of compound interest, in computing the
amount and present worth of annuities.
An ANNUITY is a sum payable at regular periods, of one
year each, either for a certain number of years, or during the
life of the pensioner, or forever.
When annuities, rents, &c. are not paid at the time they
become due, they are said to be in arrears.
The sum of all the annuities, rents, &c. remaining un-
paid, together with the interest on each, for the time they
have remained due, is called the amount.
1. What is the amount of an annual pension of $100,
which has remained unpaid 4 years, allowing 6 per cent.
compound interest?
The last year's pension will be $100, without interest;
the last but one will be the amount of $100 for 1 year; the
last but two the amount (compound interest) of $100 for
2 years, and so on; and the sum of these several amounts
will be the answer. We have then a series of amounts, that
is, a geometrical series, (T 114,) to find the sum of all the
terms.
1
X 100 = 437'45.
If the first term be 100, the number of terms 4, and the
ratio 1'06, what is the sum of all the terms?
Consult the rule, under T 113, ex. 11.
1'064
'06
Ans. $437'45.
Hence, when the annuity, the time, and rate per cent. are
given, to find the amount,-RAISE the ratio (the amount of
•
}
4
1
1
>
-~1
!
T
!
232
T 115, 116.
1, &c. for 1 year) to a power denoted by the number of
years; from this power subtract 1; then divide the remain-
der by the ratio, less 1, and the quotient, multiplied by
the annuity, will be the amount.
1
Sk
Magdalen
Sagay Citie
Note. The powers of the amounts, at 5 and 6 per cent.
up to the 24th, may be taken from the table, under T 91.
GEOMETRICAL PROGRESSION.
2. What is the amount of an annuity of $50, it being in
arrears 20 years, allowing 5 per cent. compound interest?
Ans. $1653'29.
3. If the annual rent of a house, which is $150, be in
arrears 4 years, what is the amount, allowing 10 per cent.
compound interest?
Ans. $696'15.
4. To how much would a salary of $500 per annum
amount in 14 years, the money being improved at 6 per 7
cent. compound interest?
in 10 years?
in 20
in 24 years?
Ans. to the last, $25407475.
years?
in 22 years?
11
1
}
T 116. If the annuity is paid in advance, or if it be
bought at the beginning of the first year, the sum which
ought to be given for it is called the present worth.
5. What is the present worth of an annual pension of
$100, to continue 4 years, allowing 6 per cent. compound
interest?
The present worth is, evidently, a sum which, at 6 per
cent. compound interest, would, in 4 years, produce an amount
equal to the amount of the annuity in arrears the same time.
By the last rule, we find the amount = $437'45, and by
the directions under T 114, ex. 4, we find the present worth
$346'51.
Ans. $346'51.
}
Hence, to find the present worth of any annuity,-First
find its amount in arrears for the whole time; this amount,
divided by that power of the ratio denoted by the number of
years, will give the present worth.
6. What is the present worth of an annual salary of $100
to continue 20 years, allowing 5 per cent.? Ans. $ 1246′22.
1
I
T 116.
233
The operations under this rule being somewhat tedious,
we subjoin a
1234B67
TABLE,
Showing the present worth of $1, or 1 £. annuity, at 5 and
6 per cent. compound interest, for any number of years
from 1 to 34.
Years.
5
Years.
18.
0'94339
1'83339 19
2'67301 20
3'4651 21
22
4'21236
4'91732 23
5'07569
5678637
5'58238 24
6'46321
6'20979 25
710782
6'80169 26
7672173
736008 27
8'30641
7'88687 28
12
8'86325
8'38384 29
13
9'39357
8'85268 30
14 9'89864
9'29498 31
15
10'37966
9'71225 32
16 10'83777 10'10589 33
11'27407 10'47726 34
17
1
by
8
9
10
11
ܐ
GEOMETRICAL PROGRESSION.
5 per cent.
0'95238
1'85941
2'72325
3'54595
4'32948
years?
&
per cent.
pright @ANNA
\
/

6 per cent.
5 per cent.
11'68958
10'8276
12'08532
11'15811
12'46221 11'46992
12'82115 | 1176407
13'163
12'04158
13'48807
12'30338
12'55035
13 79864
14'09394
12478335
1437518 13'00316
14'64303 | 13′21053
14'89813 | 13'40616
15'14107 13′59072
15'37245
13476483
It is evident, that the present worth of $2 annuity is 2
times as much as that of $1; the present worth of $3 will
be 3 times as much, &c. Hence, to find the present worth
of any annuity, at 5 or 6 per cent.,-Find, in this table, the
present worth of $1 annuity, and multiply it by the given
annuity, and the product will be the present worth.
15'5928113'92908
7. What ready money will purchase an annuity of $150,
to continue 30 years, at 5 per cent. compound interest?
The present worth of $1 annuity, by the table, for
30 years, is $15'37245; therefore, 15'37245 × 150
$2305'867, Ans.
25 years?
U*
15'80268 14'08398
16'00255 14'22917.
16'1929 | 14'36613
8. What is the present worth of a yearly pension of $40,
to continue 10 years, at 6 per cent. compound interest?
to continue 15 years?
at 5 per cent. ?
20
34 years?
Ans. to last, $647716.
1
$
1
1
1
1
>
}
234
GEOMETRICAL PROGRESSION.
116.
When annuities do not commence till a certain period of
time has elapsed, or till some particular event has taken
place, they are said to be in reversion.
1
J
}
9. What is the present worth of $100 annuity, to be
continued 4 years, but not to commence till 2 years hence,
allowing 6 per cent. compound interest?
The present worth is evidently a sum which, at 6 per
cent. compound interest, would in 2 years produce an amount
equal to the present worth of the annuity, were it to commence
immediately. By the last rule, we find the present worth of
the annuity, to commence immediately, to be $346'51, and,
by directions under T 114, ex. 4, we find the present worth
of $346'51 for 2 years, to be $308'393. Ans. $308'393.,
}
,
1
1
Hence, to find the present worth of any annuity taken in
reversion, at compound interest,-First, find the present worth,
to commence immediately, and this sum, divided by the power.
of the ratio, denoted by the time in reversion, will give the.
answer.
;
10. What ready money will purchase the reversion of a
ease of $60 per annum, to continue 6 years, but not to com-
mence till the end of 3 years, allowing 6 per cent. compound
interest to the purchaser?
The present worth, to commence immediately, we find to
295'039
be, $295'039, and
Ans. $24772.
=24772.
1'063
It is plain, the same result will be obtained by finding the
present worth of the annuity, to commence immediately,
and to continue to the end of the time, that is, 3 +69
years, and then subtracting from this sum the present worth
}
طر
$
of the annuity, continuing for the time of reversion, 3 years.
Or, we may find the present worth of $1 for the two times
by the table, and multiply their difference by the given an-
nuity. Thus, by the table,
•
The whole time, 9 years,
The time in reversion, 3 years,
Difference,
Mga
6'80169
2'67301
4'12868
?
60
}
$247472080: Ans..
11. What is the present worth of a lease of $100 to con-
tinue 20 years, but not to commence till the end of 4 years,
3
#
}
dreamed
make
f
t
+
T 117.
allowing 5 per cent.?
8 years?
sion ?
མ
#
and
↑ 117. 12. What is the worth of a freehold estate, of
which the yearly rent is $60, allowing to the purchaser
6 per cent. ?
In this case, the annuity continues forever, and the estate
is evidently worth a sum, of which the yearly interest is equal
to the yearly rent of the estate. The principal multiplied by
the rate gives the interest; therefore, the interest divided
by the rate will give the principal; 60'06
L
Pa
GEOMETRICAL PROGRESSION.
235
what, if it be 6 years in rever-
10 years? 14 years?
Ans. to last, $629′426.
منسيه
ļ
Hence, to find the present worth of an annuity, continuing
forever,-Divide the annuity by the rate per cent., and the
quotient will be the present worth.
Note. The worth will be the same, whether we reckon
simple or compound interest; for, since a year's interest of the
price is the annuity, the profits arising from that price can
neither be more nor less than the profits arising from the an-
nuity, whether they be employed at simple or compound in-
terest.
!
CA
13. What is the worth of $100 annuity, to continue for-
ever, allowing to the purchaser 4 per cent.? allowing
5 per cent.?
15
per cent. ?
10 per cent.?
8 per cent. ?
20 per cent. ?
Ans. to last, $500.
14. Suppose a freehold estate of $60 per annum, to com-
mence 2 years hence, be put on sale; what is its value, al-
lowing the purchaser 6 per cent. ?
Its present worth is a sum which, at 6 per cent. compound
interest, would, in 2 years, produce an amount equal to the
worth of the estate if entered on immediately.
60
$1000
the. worth, if entered on immediately,
'06
1000.
Ans. $1000.
$1000
= $889-996, the present worth.
1'062
The same result may be obtained by subtracting from the
worth of the estate, to commence immediately, the present worth
of the annuity 60, for 2 years, the time of REVERSION. Thus,
by the table, the present worth of $1 for 2 years is 1'83339
X 60
110'0034 present worth of $60 for 2 years,
and $1000 $110'0034 $889'9966, Ans. as before.
THE CONTENT
{
}
1
1
1
+
i
J
236
GEOMETRICAL PROGRESSION.
T 117.
15. What is the present worth of a perpetual annuity of
$100, to commence 6 years hence, allowing the purchaser
5 per cent. compound interest? what, if 8 years in re-
version?
10 years?
15 years?
1
1
Pagkatapos
30 years?
The foregoing examples, in compound interest, have been
confined to yearly payments; if the payments are half year-
ly, we may take half the principal or annuity, half the rate per
cent., and twice the number of years, and work as before, and
so for any other part of a year.
44
1
QUESTIONS.
1. What is a geometrical progression or series? 2. What
is the ratio? 3. When the first term, the ratio, and the num-
ber of terms, are given, how do you find the last term?
4. When the extremes and ratio are given, how do you find
the sum of all the terms? 5. When the first term, the ratio,
and the number of terms, are given, how do you find the
amount of the series? 6. When the ratio is a fraction, how
do you proceed? 7. What is compound interest? 8. How
does it appear that the amounts, arising by compound in-
terest, form a geometrical series? 9. What is the ratio, in
compound interest?
the number of terms?
the
first term?
the lust, term? 10. When the rate, the
time, and the principal, are given, how do you find the
amount? 11. When A. R. and T. are given, how do you find
P.? 12. When A. P. and T. are given, how do you find R.?
13. When A. P. and R. are given, how do you find T. ?
What is an annuity? 15. When are annuities said to be in ar-
rears? 16. What is the amount? 17. In a geometrical series,
to what is the amount of an annuity equivalent? 18. How do
you find the amount of an annuity, at compound interest?
19. What is the present worth of an annuity?
how com-
puted at compound interest ? how found by the table?
20. What is understood by the term reversion? 21. How
do you find the present worth of an annuity, taken in`rever-
sion? by the table? 22. How do you find the present
worth of a freehold estate, or a perpetual annuity? the
same taken in reversion?
14.
by the table?
A
{
Į
JEDNOG OD
1
4 years?
Ans. to last, $4624755.
1
futation, ang mga Am
}
1
1
T 118, 119.
MISCELLANEOUS EXAMPLES.
*
PERMUTATION.
¶ 118. Permutation is the method of finding how many
different ways the order of any number of things may be
varied or changed.
1. Four gentlemen agreed to dine together so long as
they could sit, every day, in a different order or position;
how many days did they dine together?
}
ī
Had there been but two of them, a and b, they could sit
only in 2 times 1 (1 X 2 = 2) different positions, thus,
a b, and b a. Had there been three, a, b, and c, they could
sit in 1 × 2 × 36 different positions; for, beginning the
order with a, there will be 2 positions, viz. a b c, and a c b;
next, beginning with b, there will be 2 positions, b a c, and
bca; lastly, beginning with c, we have c a b, and c b a,
that is, in all, 1 X 2 × 3 = 6 different positions. In the
same manner, if there be four, the different positions will
De 1 X 2 X 3 X 4 = 24.
Ans. 24 days.
237
Hence, to find the number of different changes or permu
tations, of which any number of different things are capable,—
Multiply continually together all the terms of the natural
series of numbers, from 1 up to the given number, and the
last product will be the answer.
2. How many variations may there be in the position of
the nine digits?
Ans. 362880.
3. A man bought 25 cows, agreeing to pay for them I
cent for every different order in which they could all be
placed; how much did the cows cost him?
Ans.
155112100433309859840000.
4. Christ Church, in Boston, has 8 bells; how many
changes may be rung upon them?
Ans. 40320.
Chayking
P
MISCELLANEOUS EXAMPLES.
$
¶ 119. 1. 4+6 × 7—160.
A line, or vinculum, drawn over several numbers, signifies,
that the numbers under it are to be taken jointly, or as one
whole number.
:
1
I
***
}
I
}
1
t
J
}
238
}
2. 9. 8+ 4×8+ 4 -6 how many?
4.
!
MISCELLANEOUS EXAMPLES:
I
3.7 +4 — 2 + 3 + 40 × 5 how many?
S
36 2 X 4 2
2 X 2
how many?
Ans. 31.
5. There are two numbers; the greater is 25 times 78,
1
and their difference is 9 times 15; their sum and product
are required.
5
ppprodatelna
1
Ans. 3765 is their sum; 3539250 their product.
6. What is the difference between thrice five and thirty,
and thrice thirty-five?
35 X 3-5 X 3 + 3060, Ans.
7. What is the difference between six dozen dozen, and
half a dozen dozen?
Ans. 792..
8. What number divided by 7 will fake 6488 ?
9. What number multiplied by 6 will make 2058?
1
4
J
t
T:119.
{
Ans. 30.
Ans. 230.
10. A gentleman went to sea at 17 years of age; 8 years
after he had a son born, who died at the age of 35; after
whom the father lived twice 20 years; how old was the
father at his death?
Ans. 100 years.
11. What number is that, which being multiplied by 15
the product will be ?
÷ 15, Ans:
12. What decimal is that, which being multiplied by 15,
the product will be "75 ?
675 ÷ 15 '05, Ans.
13. What is the decimal equivalent to?
!
1
571
14. What fraction is that, to which if you add g,
will be ?
Ans. '0285714.
the sum
Ans. 13.
15. What number is that, from which if you
굽
​take, the
Ans. 28.
remainder will be ?
+
16. What number is that, which being divided by 2, the
quotient will be 21 ?
Ans. 15.
17. What number is that, which multiplied by
duces ?
pro-
Ans. f.
18. What number is that, from which if you
itself, the remainder will be 12?
take of
Ans. 20.
}
19. What number is that, to which if you add & of & of
itself, the whole will be 20?
20. What number is that, of which 9 is the & part
Ans. 12.
Ans. 13.
21. A farmer carried a load of produce to market: he
sold 780 lbs. of pork, at 6 cents per lb. ; 250 lbs. of cheese,
at 8 cents per lb.; 154 lbs. of butter, at 15 cents per lb. :
1
·
+
1
J
!
T:119.
239
in pay he received 60 lbs. of sugar, at 10 cents per lb.; 15
gallons of molasses, at 42 cents per gallon;
erel, at $375; 4 bushels of salt, at $1′25
the balance in money: how much money did he receive?
barrel of mack-
per bushel; and
t
Ans. $68 85.
142
22. A farmer carried his grain to market, and sold
75 bushels of wheat, at $1'45 per bushel.
64....
... rye,
$495
$ 50
450 ...
...
1
MISCELLANEOUS EXAMPLES.
corn,
Z
In exchange he received sundry articles :-
3 pieces of cloth, each
containing 31 yds., at
2 quintals of fish,
8 hhds. of salt,
****
1
and the balance in money.
How much money did he receive ?
$175 per yd.
$2'30 per quin.
$4'30 per hhd,
Ans. $38'80.
23. A man exchanges 760 gallons of molasses, at 371
cents per gallon, for 664 cwt. of cheese, at $4 per cwt.;
how much will be the balance in his favour? Ans. $19.
24. Bought 84 yards of cloth, at $125 per yard; how
much did it come to? How many bushels of wheat, at
$1'50 per bushel, will it take to pay for it?
Ans. to the last, 70 bushels.
25. A man sold 342 pounds of beef, at 6 cents per pound,
and received his pay in molasses, at 37 cents per gallon;
how many gallons did he receive? Ans. 5472 gallons.
26. A man exchanged 70 bushels of rye, at $92 per
bushel, for 40 bushels of wheat, at $1'37 per bushel, and
received the balance in oats, at $40 per bushel; how
many bushels of oats did he receive?
Ans. 231.
1
27. How many bushels of potatoes, at 1 s. 6 d. per bushel,
must be given for 32 bushels of barley, at 2 s. 6 d. per
bushel?
Ans. 534 bushels.
28. How much salt, at $150 per bushel, must be given
in exchange for 15 bushels of oats, at 2 s. 3 d. per bushel?
Note.. It will be recollected that, when the price and cost
are given, to find the quantity, they must both be reduced to
the same denomination before dividing.. Ans. 34 bushels.
29. How much wine, at $275 per gallon, must be given
in exchange for 40 yards of cloth, at 7 s. 6 d. per yard?
Ans. 18-2 gallons.
}
}
S
240
}
T 119.
•
30. A had 41 cwt. of hops, at 30 s. per cwt., for which
B gave him 20 £. in money, and the rest in prunes, at 5 d.
per lb.; how many prunes did A receive ?
Ans. 17 cwt. 3 qrs. 4 lbs.
31. A has linen cloth worth $30 per yard; but, in bar-
tering, he will have $35 per yard; B has broadcloth worth
$375 ready money; at what price ought the broadcloth
to be rated in bartering with A?
30: 35: 375: $4'375, Ans. Or, of 3975
$4'37, Ans. The two operations will be seen to be ex-
actly alike.
32. If cloth, worth 2 s. per yard, cash, be rated in barter
at 2 s. 6 d., how should wheat, worth 8 s. cash, be rated in
exchanging for the cloth?
Ans. 10 s., or $1'6664.
33. If 4 bushels of corn cost $2, what is it per bushel ?
Ans. $50.
34. If 9 bushels of wheat cost $13'50, what is that per
bushel ?
Ans. $150.
35. If 40 sheep cost $100, what is that
100, what is that per head?
Ans. $2'50.
1
1
1
MISCELLANEOUS EXAMPLES.
I
36. If 3 bushels of oats cost 7 s. 6 d., how much are they
per bushel?
Ans. 2 s. 6 d., $41.
37. If 22 yards of broadcloth cost 21 £. 9 s., what is the
price per yard?
Ans. 19 s. 6 d., $3'25.
38. At $50 per bushel, how much corn can be bought
for $2'00?
Ans. 4 bushels.
39. A man, having $100, would lay it out in sheep, at
$250 apiece; how many can he buy?
Ans. 40.
40. If 20 cows cost $300, what is the price of 1 cow?
of 2 cows? of 5 cows?
of 15 cows?
Ans. to the last, $225.
meat in one week, how
✔
41. If 7 men consume 24 lbs. of
much would 1 man consume in the same time?
5 men ?
10 men?
2 men?
Ans. to the last, 34 lbs.
Note. Let the pupil also perform these questions by the
rule of proportion.
42. If I pay $6 for the use of $100, how much must I
pay for the use of $75 ?
Ans. $4'50.
}
43. What premium must I pay for the insurance of my
house against loss by fire, at the rate of per cent., that is,
dollar on a hundred dollars, if my house be valued at
$2475 ?
Ans. $124375.
*
1
}
$35
Bú
}
ļ
(
t
T 119.
241
44. What will be the insurance, per annum, of a store and
contents, valued at $9876'40, at 1 per centum ?
Ans. $1489146.
+
1
}
¿
45. What commission must I receive for selling $:478
worth of books, at 8 per cent.?
Ans. $38'24.
46. A merchant bought a quantity of goods for $784,
and sold them so as to gain 21 per cent.; how much did he
gain? and for how much did he sell his goods?
}
}
MISCELLANEOUS EXAMPLES.
cost?
Ans. to the last, $.888'14.
47. A merchant bought a quantity of goods at Boston, for
$ 500, and paid $43 for their transportation; he sold them
so as to gain 24 per cent. on the whole cost; for how much
did he sell them ?
Ans. $673'32.
48. Bought a quantity of books for $64, but for cash a
discount of 12 per cent. was made; what did the books
Ans. $56'32.
49. Bought a book, the price of which was marked
$4'50, but for cash the bookseller will sell it at 334 per
cent. discount; what is the cash price ? Ans. $3.00.
50. A merchant bought a cask of molasses, containing 120
gallons, for $42; for how much must he sell it to gain 15
per cent.? how much per gallon?
Ans. to last, $'401.
51. A merchant bought a cask of sugar, containing 740
pounds, for $59'20; how must he sell it per pound, to gain
25 per cent. ?
Ans. $'10.
52. What is the interest, at 6 per cent., of $71'02 for 17
months 12 days?
Ans. $6178 +.
53. What is the interest of $487'003 for 18 months?
Ans. 43'83 +,
54. What is the interest of $8'50 for 7 months?
Ans. $2971.
55. What is the interest of $1000 for 5 days?
Ans. $8331.
56. What is the interest of '50 for 10 years?
Ans. $30.
15 months and 7
Ans. $7486 +.
/
57. What is the interest of $84'25 for
days, at 7 per cent. ?
58. What is the interest of $154'01 for 2 years, 4 months
and 3 days, at 5 per cent. ?
Ans. $18'032.
59. What sum, put to interest at 6 per cent., will, in 2
years and 6 months, amount to $150?
Note. See T 85.
Ans. $130 434 →→→,
60. I owe a man $475'50, to be paid in 16 months with-
X
}
3
***
1
242
T 119.
out interest; what is the present worth of that debt, the use
of the money being worth 6 per cent.? Ans. $440′277 +..
61. What is the present worth of $1000 payable in 4
years and 2 months, discounting at the rate of 6 per cent.?
Ans. $800.
62. A merchant bought articles to the amount of $500,
and sold them for $575; how much did he gain?
What per cent. was his gain? that is, How many dollars
did he gain on each $100 which he laid out? If $500
gain $75, what does $ 100 gain? Ans. 15 per cent.
63. A merchant bought cloth at $3'50 per yard, and sold
it at $425 per yard; how much did he gain per centum?
Ans. 213 per cent.
64. A man bought a cask of wine, containing 126 gallons,
for $283'50, and sold it out at the rate of $275 per gal-
lon; how much was his whole gain? how much per gal-
lon? how much per cent. ?
རྩྭ
1
F
1
MISCELLANEOUS EXAMPLES.
TEOU
Ans. His whole gain, $63'00; per gallon, $50; which
is 223 per centum.
65. If $100 gain $6 in 12 months, in what time will it
gain $4? $10?
$14?
67. 20 men built a certain bridge in 60 days, but, it being
carried away in a freshet, it is required how many men can
rebuild it in 50 days.
1
لی
Ans. to the last, 28 months,
66. In what time will $ 54 50, at 6 per cent., gain $2'18?
Ans. 8 months.
I
days. days. men.
50: 60: 20: 24 men, Ans.
68. If a field will feed 7 horses 8 weeks, how long will it"
feed 28 horses?
Ans. 2 weeks.
69. If a field, 20 rods in length, must be 8 rods in width
to contain an acre, how much in width must be a field, 16
rods in length, to contain the same?
Ans. 10 rods.
70. If I purchase for a cloak 12 yards of plaid § of a yard
wide, how much bocking 1 yards wide must I have to line it?
Ans. 5 yards.
71. If a man earn $75 in 5 months, how long must he
work to earn $460?
Ans. 303 months.
72. A owes B $540, but, A not being worth so much
money, B agrees to take $75 on a dollar; what sum must
B receive for the debt?
Ans. $405.
73. A cistern, whose capacity is 400 gallons, is supplied.
#
!
1
;
T 119.
a
by a pipe which lets in 7 gallons in 5 minutes; but there is
leak in the bottom of the cistern which lets out 2 gallons
in 6 minutes; supposing the cistern empty, in what time
would it be filled?
威
​MISCELLANEOUS EXAMPLES.
243
In 1 minute of a gallon is admitted, but in the same time
of a gallon leaks out.
Ans. 6 hours, 15 minutes.
74. A ship has a leak which will fill it so as to make it
sink in 10 hours; it has also a pump which will clear it in
15 hours now, if they begin to pump when it begins to
leak, in what time will it sink ?
In 1 hour the ship would be
the same time it would be
filled by the leak, but in
emptied by the pump.
Ans. 30 hours.
75. A cistern is supplied by a pipe which will fill it in
40 minutes; how many pipes, of the same bigness, will fill it
in 5 minutes ?
Ans. 8.
76. Suppose I lend a friend $500 for 4 months, he
promising to do me a like favour; some time afterward, I
have need of $300; how long may I keep it to balance the
former favour?
Ans. 63 months.
77. Suppose 800 soldiers were in a garrison with pro-
visions sufficient for 2 months; how many soldiers must de-
part, that the provisions may serve them 5 months?
Ans. 480.
78. If my horse and saddle are worth $84, and my
horse
be worth 6 times as much as my saddle, pray what is the
value of my horse?
Ans. $72.
79. Bought 45 barrels of beef, at $3'50 per barrel, among
which are 16 barrels, whereof 4 are worth no more than 3
of the others; how much must I pay? Ans. $143'50.
+
80. Bought 126 gallons of rum for $110; how much
water must be added to reduce the first cost to $75 per
gallon?
Note. If $75 buy 1 gallon, how many gallons will $110
buy?
Ans. 20 gallons.
81. A thief, having 24 miles start of the officer, holds his
way at the rate of 6 miles an hour; the officer pressing on
after him at the rate of 8 miles an hour, how much does he
gain in 1 hour? how long before he will overtake the thief?
Ans. 12 hours.
}
82. A hare starts 12 rods before a hound, but is not per-
ceived by him till she has been up 14 minutes; she scuds
away at the rate of 36 rods a minute, and the dog, on view,
+
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MISCELLANEOUS EXAMPLES.
T 119.
makes after, at the rate of 40 rods a minute; how long will
the course hold? and what distance will the dog run?
Ans. 144 minutes, and he will run 570 rods.
83. The hour and minute hands of a watch are exactly
together at 12 o'clock; when are they next together?
}
1
In 1 hour the minute hand passes over 12 spaces, and the
hour hand over 1 space; that is, the minute hand gains upon
the hour hand 11 spaces in 1 hour; and it must gain 12
spaces to coincide with it.
Ans. 1 h. 5 m. 27 s.
{
84. There is an island 20 miles in circumference, and
three men start together to travel the same way about it; A
goes 2 miles per hour, B 4 miles per hour, and C 6 miles
per hour; in what time will they come together again?
Ans. 10 hours.
85. There is an island 20 miles in circumference, and
two men start together to travel around it;. A travels 2 miles
per hour, and B 6 miles per hour; how long before they will
again come together?"
B gains 4 miles per hour, and must gain 20 miles to over-
take A; A and B will therefore be together once in every,
5 hours.
-
25
86. In a river, supposing two boats start at the same time
from places 300 miles apart; the one proceeding up stream
is retarded by the current 2 miles per hour, while that mov-
ing down stream is accelerated the same; if both be pro-
pelled by a steam engine, which would move them 8 miles
per hour in still water, how far from each starting place will
the boats meet?·
امی
I
Ans. 112 miles from the lower place, and 187 miles
from the upper place,
87. A man bought a pipe (126 gallons) of wine for
$275; he wishes to fill 10 bottles, 4 of which contain 2
quarts, and 6 of them 3 pints each, and to sell the remainder
so as to make 30 per cent, on the first cost; at what rate
per gallon must he sell it?
Ans. $2'936 +.
88. Thomas sold 150 pine apples at $33 apiece, and
received as much money as Harry received for a certain
number of watermelons at $25 apiece; how much money
did each receive, and how many melons had Harry?
1
Ans. $50, and 200 melons.
89. The third part of an army was killed, the fourth part
taken prisoners, and 1000 fled; how many were in this army?
This and the eighteen following questions are usually
1
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Ť 119.
$
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wrought by a rule called Position, but they are more easily
solved on general principles. Thus, & of the
+ fe
army; therefore, 1000 is of the whole number of men;
and, if 5 twelfths be 1000, how much is 12 twelfths, or the
whole?
Ans. 2400 men.
1
MISCELLANEOUS EXAMPLES.
་
.90. A farmer, being asked how many sheep he had, an-
swered, that he had them in 5 fields; in the first were 1 of his
flock, in the second, in the third, in the fourth, and
in the fifth 450; how many had he?
Ans. 1200.
91. There is a pole, of which stands in the mud, 3 in
the water, and the rest of it out of the water; required the
part out of the water.
Ans. z.
ΤΣ
92. If a pole be in the mud, g in the water, and 6 feet
out of the water, what is the length of the pole? Ans. 90 feet.
93. The amount of a certain school is as follows: of
the pupils study grammar, geography, arithmetic,
T
learn to write, and 9 learn to read what is the number of
each?
5
20
Ans. 5 in grammar, 30 in geography, 24 in arithmetic;
12 learn to write, and 9 learn to read.
94. A man, driving his geese to market, was met by
another, who said, "Good morrow, sir, with your hundred
geese;" says he, "I have not a hundred; but if I had, in ad-
dition to my present number, one half as many as I now
have, and 21 geese more, I should have a hundred:" how
many had he?
100-2 is what part of his present number?
Ans. He had 65 geese.
95. In an orchard of fruit trees, of them bear apples,
pears, plums, 60 of them peaches, and 40, cherries;
how many trees does the orchard contain? Ans. 1200.
96. In a certain village, of the houses are painted white,
red, yellow, 3 are painted green, and 7 are unpainted;
how many houses in the village?
Ans. 120.
97. Seven eighths of a certain number exceed four fifths of
the same number by 6; required the number.
7 -=; consequently, 6 is of the required num--
ber.
Ans. 80.
8
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98. What number is that, to which if of itself be added,
15
the sum will be 30?
Ans. 25.
**
99. What number is that, to which if its and be added,
the sum will be 84 ?
841+1+1=times the required number. Ans. 48.
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MISCELLANEOUS EXAMPLES.
T 119
100. What number is that, which, being increased by and
of itself, and by 22 more, will be made three times as
much?
་
The number, being taken 1, 3, and & times, will make 2
times, and 22 is evidently what that wants of 3 times.
Ans. 30.
101. What number is that, which being increased by ,
and § of itself, the sum will be 2342?
Ans. 90.
age
102. A, B, and C, talking of their ages, B said his
was once and a half the age of A, and C said his age was
twice and one tenth the age of both, and that the sum of
their ages was 93; what was the age of each?
Ans. A 12 years, B 18 years, C 63 years old.
103. A schoolmaster, being asked how many scholars he
had, said, "If I had as many more as I now have, & as many,
as many, and as many, I should then have 435;" what
was the number of his pupils ?
Ans. 120.
3
104. A and B commenced trade with equal sums of
money; A gained a sum equal to g of his stock, and B lost
$200; then A's money was double that of B's; what was
the stock of each ?
4
By the condition of the question, one half of g, that is, g
of the stock, is equal to g of the stock, less $200; conse-
quently, $200 is of the stock.
Ans. $500.
105. A man was hired 50 days on these conditions,—that,
for every day he worked, he should receive $75, and, for
every day he was idle, he should forfeit $25; at the ex-
piration of the time, he received $27'50; how many days
did he work, and how many was he idle?
Guy
+2
Had he worked every day, his wages would have been
$75 X 50 = $37'50, that is, $.10 more than he received;
but every day he was idle lessened his wages $75+ $25
$1; consequently he was idle 10 days.
Ans. He wrought 40, and was idle 10 days.
106. A and B. have the same income; A saves of his;
but B, by spending $30. per annum more than A, at the end
of 8 years finds himself $ 40 in debt; what is their income,
and what does each spend per annum ?
1
Ans. Their income, $200 per annum;. A spends $175,
and B. $205 per annum.
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107. A man, lying at the point of death, left to his three
sons his property; to A wanting $20, to B, and to C
↓
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T 119.
MISCELLANEOUS EXAMPLES.
247
the rest, which was $10 less than the share of A; what
was each one's share?
Ans. $.80, $ 50.and $ 70.
拳
​108. There is a fish, whose head is 4 feet long; his tail
is as long as his head and the length of his body, and his
body is as long as his head and tail; what is the length of
the fish?
The pupil, will perceive, that the length of the body is
the length of the fish.
Ans. 32 feet.
109. A can do a certain piece of work in 4 days, and B
can do the same work in 3 days; in what time would both,
working together, perform it?
Ans. 14 days.
110. Three persons can perform a certain piece of work in
the following manner: A and B can do it in 4 days, B and
C in 6 days, and A and C in 5 days: in what time can they
all do it together ?
Ans. 3 days.
J
111. A and B can do a piece of work in 5 days; A can do
it in 7 days; in how many days can B do it? Ans. 171 days,
112. A man died, leaving, $ 1000 to be divided between
his two sons, one 14, and the other 18 years of age, in such
proportion, that the share of each, being put to interest at 6
per cent., should amount to the same sum when they should ·
arrive at the age of 21; what did each receive?
Ans. The elder, $546'153+; the younger, $ 453'846+.
113. A house being let upon a lease of 5 years, at $60
per annum, and the rent being in arrear for the whole time,
what is the sum due at the end of the term, simple interest
being allowed at 6 per cent.?
Ans. $336.
114. If 3 dozen pair of gloves be equal in value to 40
yards of calico, and 100 yards of calico to three pieces of
satinet of 30 yards each, and the satinet be worth 50
cents per yard, how many pair of gloves can be bought for
$4?
Ans. 8 pair.
I
115. A, B, and C, would divide $100 between them, so
as that B may have $3 more than A, and C $ 4 more than
}
B; how much must each man have?
}
Ans. A $30, B $33, and C$37.
116. A man has pint bottles, and half pint bottles; how
much wine will it take to fill 1 of each sort?
how
much to fill 2 of each sort?
how much to fill 6 of each
sort?
117. A man would draw off 30 gallons of wine into 1
oint and 2 pint bottles, of each an equal number; how
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?
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many bottles will it take, of each kind, to contain the 30
Ans. 80 of each.
gallons?
118. A merchant has canisters, some holding 5 pounds,
some 7 pounds, and some 12 pounds; how many, of each
an equal number, can be filled out of 12 cwt. 3 qrs. 12 lbs.
of tea?
1
Ans. 60.
}
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119. If 18 grains of silver make a thimble, and 12 pwts.
make a teaspoon, how many, of each an equal number, can
be made from 15oz. 6 pwts. of silver? Ans. 24 of each.
120. Let 60 cents be divided among three boys, in such
a manner that, as often as the first has 3 cents, the second
shall have 5 cents, and the third 7 cents; how many cents.
will each receive?
Ans. 12, 20, and 28 cents.
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MISCELLANEOUS EXAMPLES.
121. A gentleman, having 50 shillings to pay among his
labourers for a day's work, would give to every boy 6 d., to
every woman 8 d., and to every man 16 d.; the number of
boys, women, and men, was the same; I demand the number
of each.
Ans. 20.
t
122. A gentleman had 7 £. 17 s. 6 d. to pay among his
labourers; to every boy he gave 6 d., to every woman 8 d.,
and to every man 16 d.; and there were for every boy three
women, and for every woman two men; I demand the num-
ber of each.
Ans. 15 boys, 45 women, and 90 men.
A
}
T 119.
123. A farmer bought a sheep, a cow, and a yoke of oxen
for $82′50; he gave for the cow 8 times as much as for
the sheep, and for the oxen 3 times as much as for the cow;
how much did he give for each ?
Ans. For the sheep 2'50, the cow $20, and the oxen
$60.
124. There was a farm, of which A owned, and B 14,
the farm was sold for $1764; what was each one's share
of the money?
Ans: A's $504, and B's $1260.
{
125. Four men traded together on a capital of $3000, of
which A put in 1, B, C, and D ; at the end of 3 years
they had gained $2364; what was each one's share of the
gain?
}
Ans.
}
126. Three merchants accompanied; A furnished of
the capital, B, and C the rest; they gain $1250; what
A's $1182.
B's $ 591.
C's $ 394.
D's $ 197.
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MISCELLANEOUS EXAMPLES.
part of the capital did C furnish, and what is each one's
share of the gain?
Ans. C furnished
T
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$
of the capital; and A's share of the
gain was $500, B's $46875, and C's $281'25.
9
}
127. A, B, and C, traded in company; A put in $ 500, B
$350, and C 120 yards of cloth; they gained $332'50, of
which C's share was $120; what was the value of C's
cloth per yard, and what was A and B's shares of the
gain ?
}
Note. C's gain, being $ 120, is 13988 of the whole
4
138
gain: hence the gain of A and B is readily found; also the
price at which C's cloth was valued per yard.
Ans.
IC's cloth, per yard, $4.
A's share of the gain, $ 125.
B's do. $87'50.
128. Three gardeners, A, B, and C, having bought a
piece of ground, find the profits of it amount to 120£. per
annum. Now the sum of money which they laid down was
in such proportion, that, as often as A paid 5£., B paid 7£.,
and as often as B paid 4£., C paid 6. I demand how
much each man must have per annum of the gain.
1
Note. By the question, so often as A paid 5£., C paid £ of
7£. Ans. A 26£. 13 s. 4 d., B 37£. 6 s, 8 d., C 56£.
129. A gentleman divided his fortune among his sons,
giving A 9. as often as B 5£., and C 3£. as often as B
7£.; C's dividend was 1537§£.; to what did the whole
estate amount ?
Ans. 11583 £. 8 s. 10 d.
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130. A and B undertake a piece of work for $54, on
which A employed 3 hands 5 days, and B employed 7 hands
3 days; what part of the work was done by A, what part
by B, and what was each one's share of the money?
Ans. A and B ; A's money $22′50, B's $31'50.
131. A and B trade in company for one year only; on
the first of January, A put in $1200, but B could not
put any money into the stock until the first of April; what
did he then put in, to have an equal share with A at the end
of the year?
Ans. $1600.
132. A, B, C, and D, spent 35 s. at a reckoning, and, be-
ing a little dipped, agreed that A should pay, B 1, C †,
and D 1; what did each pay in this proportion?
t
Ans. A 13 s. 4 d., B 10 s., C 6 s. 8 d., and D 5 s.
133. There are 3 horses, belonging to 3 men, employed to
draw a load of plaster from Boston to Windsor for $26′45;
I
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MISCELLANEOUS EXAMPLES.
950
TT 119.
A and B's horses, together are supposed to do of the
work, A and C's fo, B and C's 18; they are to be paid
proportionally; what is each one's share of the money?
A's $1150 18.
Ans. B's 575 = 25.
C's $ 9'20 23
Proof, $26'45.
134. A person, who was possessed of 2 of a vessel, sold
of his share for 375 £.; what was the vessel worth?
A
↑
Ans. 1500£.
7
135. A gay fellow soon got the better of of his for-
tune; he then gave 1500£. for a commission, and his profu-
sion continued till he had but 450 £. left, which he found to
be just of his money, after he had purchased his commis-
sion; what was his fortune at first?
Ans. 3780 £.
136. A younger brother received 1560 £., which was just
of his elder brother's fortune, and 5 times the elder brother's
fortune was as much again as the father was worth; pray,
what was the value of his estate? Ans. 19165£. 14 s. 34 d.
137. A gentleman left his son a fortune, of which he
spent in three months; of of the remainder lasted him
nine months longer, when he had only 537£. left; what
was the sum bequeathed him by his father?
6
1
Ans. 2082£. 18 s. 2 d.
138. A cannon ball, at the first discharge, flies about a
mile in eight seconds; at this rate, how long would a ball
be in passing from the earth to the sun, it being 95173000
miles distant?
Ans. 24 years, 46 days, 7 hours, 33 minutes, 20 seconds.
139. A general, disposing his army into a square battalion,
found he had 231 over and above, but, increasing each side
with one soldier, he wanted 44 to fill up the square; of how
many men did his
men did his army consist?
Ans. 19000.
140. A and B cleared, by an adventure at sea, 45 guineas,
which was 35. per cent. upon the money advanced, and
with which they agreed to purchase a genteel horse and
carriage, whereof they were to have the use in proportion
to the sums adventured, which was found to be 11 to A, as
often as 8 to B; what money did each adventure ?
!
Ans. A 104£. 4 s. 218 d., B 75£. 15 s. 9 d.
141. Tubes may be made of gold, weighing not more
than at the rate of of a grain per foot? what would be
the weight of such a tube, which would extend across the
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T 119.
251
Atlantic from Boston to London, estimating the distance at
3000 miles?
Ans. 1 lb. 8 oz. 6 pwts. 31 grs.
142. A military officer drew up his soldiers in rank and
file, having the number in rank and file equal; on being
reinforced with three times his first number of men, he
placed them all in the same form, and then the number in
rank and file was just double what it was at first; he was
again reinforced with three times his whole number of men,
and, after placing them all in the same form as at first, his
number in rank and file was 40 men each; how many men
had he at first?
Ans. 100 men,
MISCELLANEOUS EXAMPLES.
143. Supposing a man to stand 80 feet from a steeple, and
that a line reaching from the belfry to the man is just 100
fect in length; the top of the spire is 3 times as high above.
the ground as the steeple is; what is the height of the
spire? and the length of a line reaching from the top of the
spire to the man? See T 109.
Arş. to last, 197 feet, nearly.
144. Two ships sail from the same port; one sails direct-
ly east, at the rate of 10 miles an hour, and the other direct-
ly south, at the rate of 74 miles an hour; how many miles
apart will they be at the end of 1 hour? 2 hours?
24 hours? 3 days? Ans. to last, 900 miles.
145. There is a square field, each side of which is 50
rods; what is the distance between opposite corners ?
Ans. 7071 + rods.
146. What is the area of a square field, of which the op-
posite corners are 70'71 rods apart? and what is the length
of each side?
Ans. to last, 50 rods, nearly.
147. There is an oblong field, 20 rods wide, and the dis-
tance of the opposite corners is 334 rods; what is the length
of the field?
its area?
Aga
Ans. Length, 263 rods; area, 3 acres, 1 rood, 134 rods.
148. There is a room 18 feet square; how many yards
of carpeting, 1 yard wide, will be required to cover the floor
of it?
182324 ft. 36 yards, Ans.
149. If the floor of a square room contain 36 square,
yards, how many feet does it measure on each side?
Ans. 18 feet.
When one side of a square is given, how do you find its
area, or superficial contents ?
When the area, or superficial contents, of a square is given,
how do you find one side?
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MISCELLANEOUS EXAMPLES.
¶ 119. -
150. If an oblong piece of ground be 80 rods long and
20 rods wide, what is its area?
E
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Note. A Parallelogram, or Oblong,
has its opposite sides equal and par-
allel, but the adjacent sides unequal.
Thus A B C D is a parallelogram,
and also E F C D, and it is easy
to see, that the contents of both are
equal. Ans. 1600 rods, 10 acres.
151. What is the length of an oblong, or parallelogram,
whose area is 10 acres, and whose breadth is 20 rods?
(not do zincatat Be
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Ans. 80 rods.
152. If the area be 10 acres, and the length 80 rods,
what is the other side?
C
When the length and breadth are given, how do you find
the area of an oblong, or parallelogram?
When the area and one side are given, how do you find
the other side?
F B
153. If a board be 18 inches wide at one end, and 10
inches wide at the other, what is the mean or average width
of the board ?
Ans. 14 inches.
Mean width,
feet, Ans.
>
When the greatest and least width are given, how do you
find the mean width ?
{

154. How many square feet in a board 16 feet long, 1'8
feet wide at one end, and 1'3 at the other?
18 + 13
1'55; and 1'55 X 16 24'8
2
A
155. What is the number of square feet in a board 20
feet long, 2 feet wide at one end, and running to a point at
the other?
Ans. 20 feet.
{
How do you find the contents of a straight edged board,
when one end is wider than the other?
1
If the length be in feet, and the breadth in feet, in what
denomination will the product be?
If the length be feet, and the breadth inches, what parts of
a foot will be the product?
156. There is an oblong field, 40 rods long and 20 rods.
wide; if a straight line be drawn from one corner to the op-
equal right-angled
posite corner, it will be divided into two
triangles; what is the area of each ?
Ans. 400 square rods
2 acres, 2 roods.
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T 119.
MISCELLANEOUS EXAMPLES.
253
157. What is the area of a triangle, of which the base is
30 rods, and the perpendicular 10 rods? Ans. 150 rods.
158. If the area be 150 rods, and the base 30 rods, what
is the perpendicular?
Ans. 10 rods.
159. If the perpendicular be 10 rods, and the area 150
rods, what is the base?
Ans. 30 rods.
When the legs (the base and perpendicular) of a right-
angled triangle are given, how do you find its area?
When the area and one of the legs are given, how do you
find the other leg?
Note. Any triangle may be divided into two right-angled
triangles, by drawing a perpendicular from one corner to the
opposite side, as may be seen by the annexed figure.
C
A
d
}
Here A B C is a triangle, di-
vided into two right-angled trian-
gles, Ad C, and d B C; there-
fore the whole base, A B, multi-
plied by one half the perpendicular
d C, will give the arca of the
whole. If A B = 60 feet, and
Ans. 480 feet.

d C16 feet, what is the area?
160. There is a triangle, each side of which is 10 feet;
what is the length of a perpendicular from one angle to
its opposite side? and what is the area of the triangle?
Note. It is plain, the perpendicular will divide the oppo-
site side into two equal parts. See T 109.
Ans. Perpendicular, 8'66 + feet; area, 43'3 + feet.
161. What is the solid contents of a cube measuring 6
feet on each side?
Ans. 216 feet.
When one side of a cube is given, how do you find its
solid contents?
When the solid contents of a cube are given, how do you
find one side of it?
Y
162. How many cubic inches in a brick which is 8 inches
long, 4 inches wide, and 2 inches thick? in 2 bricks?
in 10 bricks?
feet?
Ans. to last, 640 cubic inches.
163. How many bricks in a cubic foot? in 40 cubic
in 1000 cubic feet?
Ans. to last, 27000.
164. How many bricks will it take to build a wall 40 feet
in length, 12 feet high, and 2 feet thick ?
Ans. 25920.
165. If a wall be 150 bricks, 100 feet, in length, and
1
4 bricks, 16 inches, in thickness, how many bricks will
lay one course ?
10 courses? If the.
2 courses?
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wall be 48 courses,
build it?
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T 119.
8 feet, high, how many bricks will
150 X 4: 600, and 600 × 48 = 28800, Ans.
166. The river Po is 1000 feet broad, and 10 feet deep,
and it runs at the rate of 4 miles an hour,; in what time
will it discharge a cubic mile of water (reckoning 5000
feet to the mile) into the sea? Ans. 26 days, 1 hour.
167. If the country, which supplies the river Po with
water, be 380 miles long, and 120 broad, and the whole
land upon the surface of the earth be 62,700,000 square
miles, and if the quantity of water discharged by the rivers
into the sea be every where proportional to the extent of
land by which the rivers are supplied; how many times
greater than the Po will the whole amount of the rivers be?
Ans. 1375 times.
168. Upon the same supposition, what quantity of water,
altogether, will be discharged by all the rivers into the sea in
a year, or 365 days?
Ans. 19272 cubic miles.
▸
169. If the proportion of the sea on the surface of the
earth to that of land be as 104 to 5, and the mean depth of
the sea be a quarter of a mile; how many years would it
take, if the ocean were empty, to fill it by the rivers running
at the present rate? Ans. 1708 years, 17 days, 12 hours.
170. If a cubic foot of water weigh 1000 oz. avoirdupois,
"and the weight of mercury be 13 times greater than of
water, and the height of the mercury in the barometer (the
weight of which is equal to the weight of a column of air
on the same base, extending to the top of the atmosphere)
be 30 inches; what will be the weight of the air upon a
square foot?*
a square mile? and what will be the
whole weight of the atmosphere, supposing the size of the
earth as in questions 166 and 168?
Ans. 21094375 lbs. weight on a square foot.
mile.
1
MISCELLANEOUS EXAMPLES.
52734375000 ...
44444
1
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*****
10249980468750000000
of the whole atmosphere.
171. If a circle be 14 feet in diameter, what is its circum-
ference?
*****
Note. It is found by calculation, that the circumference of a
circle measures about 34 times as much as its diameter, or,
more accurately, in decimals, 3'14159 times. Ans. 44 feet.
172. If a wheel measure 4 feet across from side to side,
how many feet around it?
Ans. 124 feet.
173. If the diameter of a circular pond be 147 feet, what
is its circumference?
Ans. 462 feet.
1
T 119.
174. What is the diameter of a circle, whose circumfe-
rénce is 462 feet?
Ans. 147 feet.
175. If the distance through the centre of the earth, from
side to side, be 7911 miles, how many miles around it?
77911 X 3'14159 24853 square miles, nearly, Ans.
176. What is the area or contents of a circle, whose diam-
eter is 7 feet, and its circumference 22 feet?
Note. The area of a circle may be found by multiplying
the diameter into the circumference. Ans. 38 square feet.
177. What is the area of a circle, whose circumference is
176 rods?
Ans. 2464 rods.
178. If a circle is drawn within a square, containing 1
square rod, what is the area of that circle ?·
"
Note. The diameter of the circle being 1 rod, the circum-
ference will be 3'14159. Ans. "7854 of a square rod, nearly.
Hence, if we square the diameter of any circle, and multi-
ply the square by 7854, the product will be the area of the
circle.
I
MISCELLANEOUS EXAMPLES.
1
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255
179. What is the area of a circle whose diameter is 10
rods?
102 X 7854 = 78′54.
Ans. 78'54 rods,
180. How many square inches of leather will cover a
ball 34 inches in diameter ?
Note. The area of a globe or ball is 4 times as much as
the area of a circle of the same diameter, and may be found,
therefore, by multiplying the whole circumference into the
whole diameter.
Ans. 384 square inches.
181. What is the number of square miles on the surface
of the earth, supposing its diameter 7911 miles?
7911 X 24853 196,612,083, Ans.
182. How many solid inches in a ball 7 inches in diame-
ter?
Note. The solid contents of a globe are found by multiply-
ing its area by part of its diameter.
1
Ans. 179 solid inches.
183. What is the number of cubic miles in the earth,
supposing its diameter as above?
Ans. 259,233,031,435 miles.
184. What is the capacity, in cubic inches, of a hollow
globe 20 inches in diameter, and how much wine wail it
contain, 1 gallon being 231 cubic inches?
Ans. 4188'8+ cubic inches, and 18413 + gallons.
185. There is a round log, all the way of a bigness; the
areas of the circular ends of it are each 3 square feet;
1
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256
T 119.
how many solid feet does 1 foot in length of this log con-
tain ? 2 feet in length? 3 feet?
10 feet?
A solid of this form is called a Cylinder.
How do you find the solid content of a cylinder, when
the area of one end, and the length are given?
(
186. What is the solid content of a round stick, 20 feet
long and 7 inches through, that is, the ends being 7 inches
in diameter ? i
}
Find the area of one end, as before taught, and multiply it
by the length.
Ans. 5347 cubic feet.
If you multiply square inches by inches in length, what
parts of a foot will the product be? if square inches by
feet in length, what part?
187. A bushel measure is 18'5 inches in diameter, and 8
inches deep; how many cubic inches does it contain ?
Ans. 2150'4+.
It is plain, from the above, that the solid content of all
bodies, which are of uniform bigness throughout, whatever
may be the form of the ends, is found by multiplying the
area of one end into its height or length.
Solids which decrease gradually from the base till they
come to a point, are generally called Pyramids. If the base
be a square, it is called a square pyramid; if a triangle, a
triangular pyramid; if a circle, a circular pyramid, or a cone.
The point at the top of a pyramid is called the vertex, and
a line, drawn from the vertex perpendicular to the base, is
called the perpendicular height of the pyramid.
The solid content of any pyramid may be found by multi-
plying the area of the base by of the perpendicular height.
188.. What is the solid content of a pyramid whose base
is 4 feet square, and the perpendicular height 9 feet?
42 X 9/ 48.
Ans. 48 feet.
189. There is a cone, whose height is 27 feet, and whose
base is 7 feet in diameter; what is its content? ·
Ans. 346 feet.
MISCELLANEOUS EXAMPLES.
1
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190. There is a cask, whose head diameter is 25 inches,
bung diameter 31 inches, and whose length is 36 inches;
how many wine gallons does it contain ?
how many
beer gallons?
Matt)
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Note. The mean diameter of the cask may be found by
adding 2 thirds, or, if the staves be but little curving, 6
tenths, of the difference between the head and bung diame-
↓
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T 119.
257
ters, to the head diameter. The cask will then be reduced
to a cylinder.
Now, if the square of the mean diameter be multiplied by
47854, (ex. 177,) the product will be the area of one end,
and that, multiplied by the length, in inches, will give the
solid content, in cubic inches, (ex. 185,) which, divided
by 231, (note to table, wine meas.) will give the content în
wine gallons, and, divided by 282, (note to table, beer meas.)
will give the content in ale or beer gallons.
In this process we see, that the square of the mean diame-
ter will be multiplied by '7854, and divided, for wine gal-
lons, by 231. Hence we may contract the operation by
only multiplying by their quotient (7840034;) that is,
by '0034, (or by 34, pointing off 4 figures from the product
for decimals.) For the same reason we may, for beer gal-
lons, multiply by (1854 0028, nearly,) 0028, &c.
MISCELLANEOUS EXAMPLES.
Hence this concise RULE, for guaging or measuring casks,-
Multiply the square of the mean diameter by the length; mul
tiply this product by 34 for wine, or by 28 for beer, and, point-
ing off four decimals, the product will be the content in gallons
and decimals of a gallon.
In the above example, the bung diameter, 31 in.
25 in.
the head diameter = 6 in. difference, and 3 of 6 4 inches,
25 in. 4 in. 29 in. mean diameter.
-
Then, 292
841, and 841 X 36 in.
30276.
Then,{
30276 × 34 = 1029384. Ans. 102'9384 wine gals.
30276 × 28 = 847728. Ans. 847728 beer gals.
191. How many wine gallons in a cask whose bung diame-
ter is 36 inches, head diameter 27 inches, and length 45
inches?
Ans. 166'617.
192. There is a lever 10 feet long, and the fulcrum, or
prop, on which it turns, is 2 feet from one end; how many
pounds weight at the end, 2 feet from the prop, will be bal-
anced by a power of 42 pounds at the other end, 8 feet from
the prop
o?
A
Note. In turning around the prop, the end of the lever 8
feet from the prop will evidently pass over a space of 8 inches,
while the end 2 feet from the prop passes over a space of
2 inches. Now, it is a fundamental principle in mechanics,
that the weight and power will exactly balance each other,
when they are inversely as the spaces they pass over. Hence,
in this example, 2 pounds, 8 feet from the prop, will balance
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8 pounds 2 feet from the prop; therefore, if we divide the
distance of the POWER from the prop by the distance of the
WEIGHT from the prop, the quotient will always express the
ratio of the WEIGHT to the POWER; § 4, that is, the weight
will be 4 times as much as the power.
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MISCELLANEOUS EXAMPLES.
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193. Supposing the lever as above, what power would it
require to raise 1000 pounds?
Ans. 1900 250 pounds.
194. If the weight to be raised be 5 times as much as the
power to be applied, and the distance of the weight from the
prop be 4 feet, how far from the prop must the power be
applied?
An's. 20 feet.
195. If the greater distance be 40 feet, and the less of a
foot, and the power 175 pounds, what is the weight?
Ans. 14000 pounds.
i
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196. Two men carry a kettle, weighing 200 pounds; the
kettle is suspended on a pole, the bale being 2 feet 6 inches
from the hands of oue, and 3 feet 4 inches from the hands
of the other; how many pounds does each bear?
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42 X 4 = 168.
197. There is a windlass, the wheel of which is 60
inches in diameter, and the axis, around which the rope
coils, is 6 inches in diameter; how many pounds on the axle
will be balanced by 240 pounds at the wheel?
1
Vote. The spaces passed over are evidently as the diame-
ters, or the circumferences; therefore,
10, ratio.
Ans. 2400 pounds.
198. If the diameter of the wheel be 60 inches, what
must be the diameter of the axle, that the ratio of the weight
to the power may be 10 to 1 ?
Ans. 6 inches.
Note. This calculation is on the supposition, that there
T 119.
Ans. 168 lbs.
Ans. { 1144 pounds.
854
P
is no friction, for which it is usual to add to the power
which is to work the machine.
199. There is a screw, whose threads are 1 inch asun-
der, which is turned by a lever 5 feet,
60 inches, long;
น
what is the ratio of the weight to the power?"
Note. The power applied at the end of the lever will de-
scribe the circumference of a circle 60 X 2 120 inches
in diameter, while the weight is raised 1 inch; therefore,
the ratio will be found by dividing the circumference of a circle,.
whose diameter is twice the length of the lever, by the distance
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between the threads of the screw. 120 X 34 3774 cir-
3774
cumference, and
3774, ratio, Ans.
FORMS OF NOTES;
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200. There is a screw, whose threads are of an inch
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​asunder; if it be turned by a lever 10 feet long, what weight
will be balanced by 120 pounds power? Ans. 30171 pounds.
201. There is a machine, in which the power moves over
10 feet, while the weight is raised 1 inch; what is the
power of that machine, that is, what is the ratio of the
weight to the power ?
Ans. 120.
202. A man put 20 apples into a wine gallon measure,
which was afterwards filled by pouring in 1 quart of water;
required the contents of the apples in cubic inches.
*
"
I
Ans. 1734 inches.
203. A rough stone was put into a vessel, whose capaci-
ty was 14 wine quarts, which was afterwards filled with 2
quarts of water; what was the cubic content of the stone?
Ans. 6644 inches.
FORMS OF NOTES, BONDS, RE-
CEIPTS, AND ORDERS.
Overdean, Sept. 17, 1802.
For value received, I promise to pay to OLIVER BOUNTIFUL,
or order, sixty-three dollars fifty-four cents, on demand, with
interest after three months.
WILLIAM TRUSTY.
Attest, TIMOTHY TESTIMONY.
1
NOTES.
No. I.
:
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No. II.
Bilfort, Sept. 17, 1802.
For value received, I promise to pay to O. R., or bearer,..
dollars
cents, three months after date.
PETER PENCIL..
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FORMS OF NOTES.
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No. III.
By two Persons.
Arian, Sept. 17, 1802.
For value received, we, jointly and severally, promise to
pay to C. D., or order,
cents, on
demand, with interest.
dollars
ALDEN FAITHFUL.
JAMES FAIRFACE.
Attést, CONSTANCE ADLEY..
Observations.
1. No note is negotiable unless the words "cr order,'
wise "or bearer," be inserted in it.
"other-
2. If the note be written to pay him " or order," (No. I.)
then Oliver Bountiful may endorse this note, that is, write
his name on the backside, and sell it to A, B, C, or whom
he pleases. Then A, who buys the note, calls on William
Trusty for payment, and if he neglects, or is unable to pay,
A may recover it of the endorser.
/
3. If a note be written to pay him "or bearer," (No. II.)
then any person, who holds the note, may sue and recover the
same of Peter Pencil.
+
2
4. The rate of interest, established by law, being six per
cent. per annum, it becomes unnecessary, in writing notes, to
mention the rate of interest; it is sufficient to write them
for the payment of such a sum, with interest, for it will be
understood legal interest, wliich is six per cent.
f
5. All notes are either payable on demand, or at the ex-
piration of a certain term of time agreed upon by the parties,
and mentioned in the note, as three months, a year, &c.
Į
6. If a bond or note mention no time of payment, it is
always on demand, whether the words "on demand" be
expressed or not.
7. All notes, payable at a certain time, are on interest as
soon as they become due, though in such notes there be no
mention made of interest.
1
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!
This rule is founded on the principle, that every man
ought to receive his money when due, and that the non-
payment of it at that time is an injury to him. The law,
therefore, to do him justice, allows him interest from the
time the money becomes due, as a compensation for the
injury.
1
8. Upon the same principle, a note, payable on demand,
without any mention made of interest, is on interest after a
1
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261
demand of payment, for upon demand such notes imme-
diately become due.
9. If a note be given for a specific article, as rye, payable
in one, two, or three months, or in any certain time, and the
signer of such note suffers the time to elapse without de-
livering such article, the holder of the note will not be
obliged to take the article afterwards, but may demand and
recover the value of it in money.
FORMS OF BONDS.
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BONDS.
A Bond, with a Condition, from one to another.
Know all men by these presents, that I, C. D. of, &c., in
the county of, &c., am held and firmly bound to E. F., of,
&c., in two hundred dollars, to be paid to the said E. F., or
his certain attorney, his executors, administrators, or assigns,
to which payment, well and truly to be made, I bind myself,
my heirs, executors and administrators, firmly by these
presents. Sealed with my seal. Dated the eleventh day of
, in the year of our Lord one thousand eight hun-
dred and two.
The Condition of this obligation is such, that, if the above-
bound C. D., his heirs, executors, or administrators, do and
shall well and truly pay, or cause to be paid, unto the above-
named E. F., his executors, administrators, or assigns, the
full sum of two hundred dollars, with legal interest for
the same, on or before the eleventh day of
suing the date hereof,-then this obligation to be void, or
otherwise to remain in full force and virtue.
next en- `
Signed, &c.
A Condition of a Counter Bond, or Bond of Indemnity, where
one man becomes bound for another.
The condition of this obligation is such, that whereas the
above-named A. B., at the special instance and request, and
for the only proper debt of the above-bound C. D., together
with the said C. D., is, and by one bond or obligation bear-
ing equal date with the obligation above-written, held and
firmly bound unto E. F., of, &c., in the penal sum of
dollars, conditioned for the payment of the sum of, &c., with
legal interest for the same, on the
day of
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262
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FORMS OF RECEIPTS.
next ensuing the date of the said in part recited obligation,
as in and by the said in part recited bond, with the condition
thereunder written, may more fully appear;-if, therefore, the
said C. D., his heirs, executors, or administrators, do and
shall well and truly pay, or cause to be paid, unto the said
E. F., his executors, administrators, or assigns, the said sum
of, &c., with legal interest of the same, on the said
day of, &c., next ensuing the date of the said in part re-
cited obligation, according to the true intent and meaning,
and in full discharge and satisfaction of the said in part re-
cited bond or obligation,-then, &c.-otherwise, &c..
}
Note. The principal difference between a note and a
bond is, that the latter is an instrument of more solemnity,
being given under seal. Also, a note may be controlled by
a special agreement, different from the note, whereas, in case
of a bond, no spécial agreement can in the least control
what appears to have been the intention of the parties, as
-expressed by the words in the condition of the bond.
1
}
}
Sitgrieves, Sept. 19, 1802.
Received from Mr. DURANCE ADLEY ten dollars in full
of all accounts.
ORVAND CONSTANCE.
}
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}
}
$
Sitgrieves, Sept. 19, 1802.
Received of Mr. ORVAND CONSTANCE five dollars in full
of all accounts. ·
DURANCE ADLEY.
}
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RECEIPTS.
Sitgrieves, Sept. 19, 1802.
Received of Mr. SIMPSON EASTLY (by the hand of Trus
TRUSTY) Sixteen dollars twenty-five cents, which is en-
dorsed on his note of June 3, 1802..
PETER CHEERFUL.
$
Receipt for Money received on a Note.
A Receipt for Money received on Account.
Sitgrieves, Sept. 19, 1802.
Received of Mr. ORAND LANDIKE fifty dollars on ac-
EIDRO SLACKLEY,
count.
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FORMS OF RECEIPTS, &c.-BOOK-KEEPING.
Receipt for Money received for another Person.
Salem, Aug. 19, 1827.
Received from P. C. one hundred dollars for account of
J. B.
ELI TRUMAN.
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Amherst, July 6, 1827.
Received of I. S. thirty dollars, in full of one year's in-
terest of $500, due to me on the
last, on note from the said I. S.
day of
SOLOMON GRAY.
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Receipt for Money paid before it becomes due.
Hillsborough, May 3, 1827.
Received of T. Z. ninety dollars, advanced in full for one
year's rent of my farm, leased to the said T. Z., ending the
first day of April next, 1828.
HONESTUS JAMES.
Receipt for Interest due on a Note.
A
263
Note. There is a distinction between receipts given in
full of all accounts, and others in full of all demands. The
former cut off accounts only; the latter cut off not only ac-
counts, but all obligations and right of action.
1
ORDERS.
Archdale, Sept. 9, 1802.
Mr. STEPHEN BURGESS. For value received, pay to A.
B., or order, ten dollars, and place the same to my account.
SAMUEL SKINNER.
L
Pittsburgh, Sept. 9, 1821.-
་
Mr. JAMES ROBOTTOM: Please to deliver Mr. L. D. such
goods as he may call for, not exceeding the sum of twenty-
five dollars, and place the same to the account of your
humble servant,
NICHOLAS REUBENS.
BOOK-KEEPING.
It is necessary that every man should have some regular,
uniform method of keeping his accounts. What this method
shall be, the law does not prescribe; but, in cases of dis-
pute, it requires that the book, or that on which the charges
were originally made, be produced in open court, when he
will be required to answer to the following questions :---
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Is this your book, and the method in which you keep your accounts?
Did you make the charges, now in dispute, at the time they purport to have been made?
Are they just and true?
*
Have you received pay for them, or any part? if so, how much?
An answer in the affirmative, under oath, to the above questions, (the last only excepted,) is all that -
is required to substantiate his claim.
Dr.
}
It con-
For farmers and mechanics, the following method will be found both convenient and easy.
sists in having one single book, entering the name of the person, with whom an account is to be opened,
at the top of the left hand page, Dr., and at the top of the right hand page, Cr., as follows:-
James Macknight.
1827.
Jan. 5. To 5 cords of Wood, at $1675,
May 16. To one day's work, self and
oxen,
July 23. To 4 bushels of Rye, at 75 cts.
delivered by your order to
C. D.
dolls. cts.
8
75
-1 50
3
13*
00
25
3
James Macknight.
1827.
April 8.
By one Plough,
May 10. By repairing Cart Wheels,
Sept. 12. By Cash to balance,
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Cr.
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9 25
150
2 50
13 25
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