UC-NRLF
735
IN MEMORIAM
FLORIAN CAJOR1
THE
JUVENILE ARITHMETIC!*
AND
SCHOLAR'S GUIDE;
WHEREIN THEORY AND PRACTICE ARE COMBINED AND
ADAPTED TO THE CAPACITIES
OF
YOUNG BEGINNERS;
CONTAINING A DUE PROPORTION OF EXAMPLES
IN
FEDERAL MONEY;
AND THE WHOLE BEING ILLUSTRATED BY NUMEROUS
QUESTIONS SIMILAR TO THOSE
w-
OF
PE&TALOZZL
BY MARTIN RUTER, A. M.
Cfncfwnatf:
PUBLISHED AND SOLD BY N. AND G. GUILFORD
. AND O. FARNSWORTH, JR. PRINTERS
1828,
DISTRICT OF OHIO, TO WIT :
BE IT REMEMBERED, That on the twenty-second day of April, in the
year of our Lord one thousand eight hundred and twenty seven, and in the fifty-
first year of the American Independence, MARTIN RUTER, of said District,
hath deposited in said office the title of a book, the right whereof he claims as au-
thor and proprietor in the words following, to wit:
"THE JUVENILE ARITHMETICS: AND SCHOLAR'S GUIDE:
wherein theory and practice are combined and adapted to the capacities of young
beginners; containing" a due proportion of examples in Federal Money, and the
•whole being illustrated by numerous questions similar to those of PESTALOZ-
ZI,by MARTIN RUTER, A. JI/;"
In conformity to the Act of the 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, of the Act entitled " AH Act supplementary to an Act enti-
tled an Act for the encouragement of Learning, by securing the copies of Man>,
Charts, and Books to the Authors and Proprietors of such Copies during the times
therein mentioned, and extending the benefits thereof to the Arts of designing, en-
graving, and etching historical and other Prints.
WM. KEY BOND,
Clerk of the District of Ohio.
RECOMMENDATIONS.
The following have been selected from the recommendations be*
stowed upon this work.
Messrs. Guilfords, — I have examined hastily the u Juvenile Arithmetick," which
you sent me, and am of opinion that it possesses some advantages over those gen-
erally in use: I particularly refer to the part intended to Cultivate in the learner,
the habit of going through the solutions mentally. Very respectfully yours,
JOHN E. ANNAN,
Professor of Mathematicks and Natural Philosophy in the Miami University.
Oxford, June 5, 1827.
From a hasty review of Dr. Ruter's Arithmetick, I am inclined to think well of
it. The attempt to introduce a rational method of instruction in arty department
of education, is laudable and especially in common schools. This I think the Ju- Ilf
venile Arithmetick is well calculated to do, in that branch of study to which it be- "
longs. The plan of Pestalozzi is excellent, and Dr. Ruter has perhaps imitated it
more successfully (by comprizing more in less space) thanMr. Colburnj between,.
whose Aritfrnietick and this there is however a c
Tour's, &c.
WM. H. M'GUFFY,
Professor of Languages, &c., in the Miami University.
Orford, June 28, 1827.
We havo examined your 'Juvenile Arithmetick,' and feel a pleasure in recom-
nending it to the schools of our country. We think the general arrangement good,
ind have no hesitation in saying, that the questions prefixed and appended to the
•ules, give it superior advantages. Respectfully yours,
JOSEPH S. TOMLINSON,
JOHN P. DURBIN,
March 12, 1823- Professors in Augusta College,-
The Juvenile Arithmetick, from the cursory examination which I have given it,
appears to be a manuel of value for the introduction of youth, into the science of
numbers. In furnishing a second edition, I wish you success.
ELIJAH SLACK.
I concur most cheerfully in th« above opinion. S. JOHNSTON.
Cincinnati, April 2, 1828.
I have used thy compilation of Arithmetick during the last year; and do not hefcP-
fate in recommending it to the publick. The questions pieceding the rules, the
particular attention to fractions, and the sketch of mensuration give it a decidetl
preference to any other here in use. JOHN L. T ALBERT.
Cincinnati, Fourth mo, 5, 1828.
Having examined the above Arithmetick, I cheerfully concur in the foregoing
•pinion of its merits. ARNOLD TKUESDBLI..
M306089
IV RECOMMENDATIONS.
I have carefally inspected "the Juvenile Arithmetick and Scholars Guide," by
Dr. Ruter and am of the opinion, it is well calculated and arranged, to conduct thk
pupil by an easy gradation to a perspicuous conception of the science of numbers-
I therefore recommend it to publick nse, particularly in common schools.
SAMUEL BURR,
September 2, 1827. . Professor of Mathernaticks
A cursory examination of Dr. M. Ruler's Arithmetick, has convinced me, that
the simple and familiar manner in which the learned author unfolds the principles
of this science, and adapts them ttr the understanding of the young learner, can
not fail to give his work a decided preference, for practical purposes, over those
arhhmeticks in common use. In my opinion, teachers who adopt it, as well as
pupils who study it, wilLrealize satisfactory and highly beneficial results.
S. KIRKHAM,
Pittsburgh^ April 2, 1823. Author of Grammar in Familiar Lectures.
I have examined the system of Arithmetick compiled by Dr. Ruter, and am of
opinion that it is well calculated for conveying to youth, a general knowledge of
that science in a shorter time, than any I have seen.
G. GARDNER,
March 29, 1828. Teacher of Mathematicks, Mill-Creek Township-.
Having examined the Juvenile Arithmetick, I have no hesitation in pronouncing
it an excellent elementary School Book. The rules are Judiciously arranged, and
peculiarly well adapted to juvenile comprehension: The work contains multnm in
parvo, and I think its publication will be conducive to publick utility.
Hoping its merits will be duly appreciated, I take great pleasure in recommen-
ding it to the publick patronage. Your's respectfully,
RICHARD MORECRAFT,
Cincinnati, January 2, 1828. Teacher-
From my acquaintance with Ruler's Arithmetick, I am convinced that it is well
calculated to encourage the student, improve his mind, and prepare him for busi-
ness. JOHN LOCKE,
May 16, 1828. Principal of Cincinnati Female Academy.
Gentlemen — I have examined with some attention the Juvenile Arithmetick, &C.
by the Rev. Dr. Ruter, and am decidedly of opinion, that it is admirably calcula
ted for conveying to youth with great facility a general knowledge of that impor
rant science. Tlie ingenious manner in which the compiler has given an elucida-
tion of Vulgar Fractions, together with an exclusion of all extraneous matter,
fenders it in my estimation a treatise of peculiar merit.
Your obedient servant,
JOHN WINRIGHT,
Cincinnati, September 2, 1827 . Teacher
PREFACE.
THIS ARITHMETICK has been compiled with a view
to facilitate the progress of pupils, and lessen the la-
bour of teachers. The questions preceding and fol-
lowing the rules, are designed to lead young learners
into habits of thinking and calculating; and thus,
to prepare them for practical operations. Experi-
ence has demonstrated, that, in the instruction of
children in any science, it is necessary to excite their
entire attention to the subject before them. The la-
tent energies of their minds must be roused up, and
called forth into action. When this can be effectually
done, success is rendered certain. To accomplish
this important object, the best method has been found
in the frequent use of well selected questions.—
Though it is a successful course in all juvenile stu-
dies, it is particularly so in the science of numbers;
and the progress of pupils must be slow without it.
The questions in the following pages are thought to
be sufficiently numerous for the purposes intended;
the rules have been arranged according to the plan
of some of the best authors on this subject, and the
work is offered to the publick with the hope that it
will be useful in the schools of our country.
M. R.
EXPLANATION OF THE CHARACTERS USED IN
ARITHMETICS:.
4- Signifies plus, or addition.
Signifies minus, or subtraction.
X Denotes multiplication.
-T- Means division.
: : ' : Signifies proportion.
= Denotes equality.
Thus, 4+7 denotes that 7 is to be added to 4.
5 — 3, denotes that 3 is to be taken from 5.
8X2, Signifies that 8 is to be multiplied by 2.
9-r3, That 9 is to be divided by 3.
3 : 2 : : 6 : 4, Shows that 3 is to 2 as 6 is to 4.
7-j-9= 16, Shows that the sum of 7 and 9 is equal to 16,
^/ or I/ Denotes the Square Root,
s/ Denotes the Cube Root.
{/ Denotes the Biquadrate Root.
This mark, called a Vinculum, shows that the
several figures over which it is drawn are to
he fcken together as a simple quantity.
ARITHMETICK.
ARITHMETICK is the science which treats of the nature
and properties of numbers: and its operations are con-
ducted chiefly by five principal rules. These are, Nu-
meration, Addition, Subtraction* Multiplication, and Di-
vision.
Numbers in Arithmetick are expressed by the fol-
lowing ten digits or characters, namely: 1 one, 2 two,
3 three, 4 four, 5 five, 6 six, 7 seven, 8 eight, 9 nine,
0 cypher.
An Integer signifies a whole number, or certain quan-
tity of units, as one, three, ten. A Fraction is a broken
number, or part of a number, as ^ one half, f two-thirds,
1 one-fourth, f three-fourths, f five-sevenths, &c.
NUMERATION.
Numeration teaches the different value of figures by
their different places, and to express any proposed num-
bers either by words or characters; or to read and
write any sum or number.
NUMERATION TABLE.
Units*
Tens.
Hundreds.
Thousands,
Tens of thousands.
e< co *5 «3 ^ co w Hundreds of thousands,
Millions.
t-<Ncou5 Tens of millions.
co *- i> co Hundreds of millions,
G< co i> Thousands of millions.
10 °P Ten thousands of millions*
^ Hundred th&usands of millions.,
8 NUMERATION.
Here, any figure in the place of units, reckoning from
right to left, denotes only its simple value; but that in
the second place denotes ten times its simple value; and
that in the third place, one hundred times its simple
value; and so on, the value of any figure in each suc-
cessive place, being always ten times its former value.
Thus in the number 6543, the 3 in the first place denetes
only three; but 4 in the second place signifies four tens,
or 40; 5 in the third place, five hundred; and 6 in the
fourth place, six thousand ; which makes the whole num-
ber read thus — six thousand five hundred and forty three.
The cvpher stands for nothing when alone, or when on
the left hand side of an integer; but being joined OB the
right hand side of other figures, it increases their value
in the same ten fold proportion: thus, 50 denotes five
tens; and 500 is read five hundred.
Though the preceding numeration table contains only
twelve places, which render it sufficiently large for
young students, yet it may be extended to more places
at pleasure.
EXAMPLE.
Quatrillions . Trillions. Billions. Millions. Units.
987,654; 321,234; 567,898; 765,432; 123,456
Here note, that Billions is substituted for millions of
millions: Trillions, for millions of millions of millions:
Quatrillions, for millions of millions of millions of mil-
lions. From millions, to billions, trillions, qwatrillions,
and other degrees of numeration, the same intermediate
denominations, of tens, hundreds, thousands, fyc. are used,
as from units to millions. And thus, in ascertaining the
amount of very high numbers, we proceed from Millions
to Billions, Trillions, Quatrillions, Quintillions, Sextil-
lions, Septillions, Octillions. Nonillions, Decillions, Un-
decillions, Duodecillions, Tredecillions, Quatuordecil-
lions, Quindecillions, Sexdecillions, Septendecillions,
Octodecillions, Novemdecillione, VigintilHons.&c. all of
which answer to millions so often repeated, as their in-
dices respectively require, according to the above pro-
portion.
SIMPLE ADDITION. 0
THE APPLICATION.
~Write down, in figures, the following numbers:
Ten. - 10
Twenty-one. - 21
Thirty-five. - 35
Four hundred and sixty-seven. 467
Two thousand three hundred and eighty-nine. 2389
Thirty-four thousand five hundred and seventy. 34570
Six hundred and three thousand four hundred. 603400
Seven millions eight hundred and four thou- ) 78Q4329
sand three hundred and t wenty-nine. \
Fifty-eight millions seven hundred and thir-j 50739105
ty-two thousand one hundred and five. \ "
Eight hundred and ten millions nine hun-)
d red and two thousand five hundred> 810902512
and twelve. )
Three thousand two hundred and three)
millions six hundred and eight thou-> 3203608999
sand nine hundred and ninety-nine. )
Question 1. What is Arithmetick?
2. What are the ten digits by which numbers
are expressed?
3. What is an integer?
4. What is a fraction?
5. What are the principal rules by which the
operations in Arithmetick are conducted ?
6. What does Numeration teach?
SIMPLE ADDITION.
Simple Addition teaches to put together numbers of
the same demonination into one sum; as 5 dollars, 4
dollars, and 3 dollars, make 12 dollars.
Before the pupil enters upon Addition in the usual way, with
figures, it would be useful for him to learn to perform easy ope-
rations in his mind. For this purpose let him be exercised in
the following questions, or in others which are similar.
10 SIMPLE ADDITION.
1. If you have two cents in one hand and one in the
other, how many hare you in both?
2. If you have three cents in one hand and two in
the other, how many have you in both?
3. If you have five cents in one hand, and two in the
other, how many have you in both?
4. John has six cents, and Robert has three; how ma-
ny have they both together?
5. Charles gave five cents for an orange, and two for
an apple; how many did he give for both?
6. Dick had four nuts, John had three, and David
had two; how many had they all together?
7. Henry had five peaches, Joseph had three and
Tom had two, and they put them all into a basket; how
many were there in the basket?
8. Three boys, Peter, John and Oliver, gave some
money to a beggar. Peter gave seven cents, John four,
and Oliver three. How many did they all give him?
9. A man bought a sheep for eight dollars, and a calf
for seven dollars; what did he give for both?
10. A boy gave to one of his companions eight peaches;
to another six; to another four; and kept two himself;
how many had he at first?
11. How many are two and three? — two and five? —
three and seven? — four and five?
12. How many are twice four? — twice five? — twice six?
twice seven? — twice eight? — twice nine?
13. How many are three and two and one?
14. How many are four and three and two?
15. How many are five and four and three?
16. How many are four and five and two?
17. How many are seven and three and one?
18. How many are eight and four and two?
19. How many are nine and five and one?
20. How many are five and six and seven?
21. Ho iv many are four and three and two and one?
22. How many are two and three and one and four?
23. How many are five and three and two and one?*
* It is expected that m;my of these questions will be varied by the teacher, and
icndered harder, or easier, or others eiibstitutyd, as the capacity of the pupil may
require.
SIMPLE ADDITION. 11
RULE.
Place the figures to be added, one under another, so
that units will stand under units, tens under tens, hun-
dreds under hundreds, &c. Draw a horizontal line un-
der them, and beginning at the bottom of the first column,
on the right hand side, that is. at units, add together the
figures in that column, proceeding from the bottom to
the top. Consider how many tens are contained in their
sum, and how many remain besides the even number of
tens, placing the amount under the column of units, and
carrying so many as you have tens to the next column.
Proceed in the same manner through every column, set-
ting down under the last column its full amount.
PROOF.
Begin at the top of the sum and add the several rows
of figures downwards as they were added upwards, and
if the additions in both cases be coriect the sums will
agree.
EXAMPLES.
I. II. III. IV.
12 321 4000 542210
21 123 3124 135403
34 410 2345 350212
10 203 5234 201304
77 1057 14703 1229129
V. VI. VII.
2 4 0 5 6 7 £. 50678 450789
3540210* 7 6 £'4 3 876543
4321023* 20134 450789
4065243* 56787 876543
2123456 65432 234798
12 SIMPLE ADDITION.
VIII. IX. X.
678987G5 45 20000000
4321234 678 3000000
567898 9876 400000
76543 54321 50000
2123 234567 6000
212 8987654 700
72866775 9287141 23456700
XI.
XII.
XIII.
24681012
54321231
9 1
i 7 6 543 2
42130538
19000310
1 c<
I 3 4 5 5 7 8
71021346
20304986
9 i
J7 6 54 3 2
20324913
19876540
1 S
> 3 4 5 6 7 8
98765432
98755432
9 i
3765432
12345678
12000987
1 <
2345678
APPLICATION.
1. A boy owed one of his companions 6 cents; he owed
another 8, another 5, and another 9. How much did he
owe in all? Ans. 28 cents.
2. A man received of one of his friends 7 dollars, of
another 10, of an another 19, and of another 50. How
many dollars did he receive? Ans. 86 dollars.
3. A person bought of one merchant ten barrels of
flour, and paid 40 dollars; of another twenty barrels of
cider, for which he paid 60 dollars, and twenty barrels
of sugar at 450 dollars; and of another ninety-five bar-
rels of salt at 570. How many barrels did he buy, and
how much money did he pay for the whole?
Ans. 145 barrels, and pqd 1120 dollars.
4. A had 250 dollars; 'B-had 375;*C had 5423; D,
64320; E, 287432, and F, 4321507. ^ow murh would
it all make, if put together? * Ans'.' 4679367 dollars.
Question 1. What does Simple Addition teach?
2. How do you place the numbers to be added?
3. Where do you begin the addition?
4. How do you prove a sum in Addition?
SIMPLE SUBTRACTION. 13
SIMPLE SUBTRACTION.
Simple Subtraction teaches to take a less number from
*i greater of the same denomination, and thus to find the
difference between them.
Questions to prepare the learner for this rule.
k If you have seven cents, and give away two; how
many will you have left ?
2. If you have eight cents, and lose four of them ; how
many will you have left?
3. A boy having ten cents, gave away four of them;
how many had he left?
4. A man owing twelve dollars, paid four of it; how
much did he then owe?
5. A man bought a firkin of butter for fifteen dollars,
and sold it again for ten dollars; how much did he lose?
6. If a horse is worth ten dollars, and a cew is worth
four; how much more is the horse worth than the cow?
7. A boy had eleven apples in a basket, and took out
five; how many were left?
8. Susan had fourteen cherries, and ate four of them ;
how many had she left?
9. Thomas had twenty cents, and paid away five of
them for some plums; how many had he left?
10. George is twelve years old, and William is seven;
how much older is George than William?
11. Take four from eight; how many will remain?
12. Take three from nine; how many will remain?
13. Take five from ten; how many will remain?'
14. Take six from ten; how many will remain?
15. Take six from eleven; how many will remain?
16. Take five from twelve; how many will remain?
17. Take four from thirteen; how many will remain?
18. Take six from fourteen; how many will remain?
19. Take six from fifteen; how many will remain?
20. Take eight from sixteen; how many will remain?
21. Take nine from twelve; how many will remain?
22. Take nine from thirteen; how many will remain?
23. Take three from thirteen; how many will remain?
;4 SIMPLE SUBTRACTION.
24. Talge eight from seventeen ; how many will remain i
25. Take nine from sixteen; how many will remain?
26. Take nine from eighteen; how many will remain?
RULE.
Place the larger number uppermost, and the smaller
one under it, so that units may stand under units; tens
under tens ; hundreds ander hundreds, &c. Draw a line
underneath, and beginning with units, subtract the lower
4rom the upper figure, and set down the remainder. —
But when in any place the lower figure is larger than
the upper, call the upper one ten more than it really is;
subtract the lower figure from the upper, considering it
;is having ten added to it, and setting down the remain-
der add one to the next left figure of the lower line, and
proceed thus through the whole.
PROOF.
Add the remainder and the less line together, and if
the work be right, their sum will be equal to the grea-
ter line.
EXAMPLES.
I. II. III. IV. V.
23 457 54367 73214 84201
11 215 20154 54876 49983
12 242 34213 18338 34218
VI. VII.
98 r 2030405321 700000000000
6054123456789 98765432123
APPLICATION.
1. A man borrowed of his friend four hundred and
eighty dollars; and having afterwards paid one hundred
and sixty-five, how much was still due? Ans. 315 dolls.
2. A owed B 10,000 dollars. He paid at one time
467, and afterwards 297. How much was still due to
B? Ans. 9236 dollars.
SIMPLE MULTIPLICATION. \b
3. B owed C 11,989 dollars, lie paid atone time
2875 dollars; at another, 4243; and afterwards, 3000.
How much did he still owe? Ans. 1871 dollars.
4. A man travelled till he found himself 1300 miles
from home. On his return, he travelled in one week
235 miles; in the next, 275; in the next, 325, and in the
following week 290. How far had h* still to go before
he would reach home? Ans. 175 miles.
Question 1. What does Subtraction teach?
2. How do you place the larger and smaller
numbers?
3. What do you do when the lower number is'
larger than the upper number?
4. How is a sum in Subtraction proved?
SIMPLE MULTIPLICATION.
Simple Multiplication teaches a short method of finding
what a number amounts to when repeated a given num-
ber of times, and thus performs Addition in a very ex-
peditious manner.
1. What will four apples cost, at two cents a piece?
2. What must you give for two oranges, at six cents
a piece?
3. What are two barrels of flour worth, at five do,!-^
lars a barrel?
4. What will three pounds of butter come to, at three
cents a pound?
5. If you can walk four miles in one hour; how far
can you walk in three hours?
6. If a cent will buy five nuts; how many nuts wifl
four cents buy?
7. What are two barrels of cider worth, at three
dollars a barrel?
Before entering upon this Rule, let the pupil so learn the fol-
lowing table, as to answer with readiness any question im
in it 4 after which, he will be able to proceed with facility.
DIMPLE MULTIPLICATION
MULTIPLICATION TABLE.
Twice
3 times
4 times
5 times
6 times
7 times
Imake2
1 make3
1 make 4
1 make5
1 make6
Imake7
2 4
2 6
2 8
2 10
2 12
2 14
3 6
3 9
3 12
3 15
3 18
3 21
4 8
4 12
4 16
4 20
4 24
4 28
& 10
5 15
5 20
5 25
5 30
5 35
6 12
6 18
6 24
6 30
6 36
6 42
7 14
7 2*1
7 28
7 35
7 42
7 49
8 16
8 24
8 32
8 40
8 48
8 56
r 9 is
9 27
9 36
9 45
9 54
9 63
10 20
10 30
10 40
10 50
10 60
10 70
11 22
11 33
11 44
11 55
11 66
11 77
12 24
12 36
12 48
12 60
12 72
12 84
8 times
9 times
10 time?
1 1 times
12 times
1 make 8
1 make 9
1 make 10
1 make 11
1 make 12
2 16
2 18
2 20
2 22
2 24
3 24
3 27
3 30
3 33
3 36
4. 32
4 36
4 40
4 44
4 48
'5 40
5 45
5 50
5 55
5 60
6 48
6 54
6 60
6 66
6 72
7 56
7 63
7 70
7 77
7 84
8 64
8 72
8 80
8 88
8 96
9 72
9 81
9 90
9 99
9 108
10 80
10 90
10 100
10 110
10 120
11 88
11 99
11 110
11 121
11 132
:12 96
12 108
12 120
12 132
12 144
Though the foregoing table extends no farther than
12, it. may be easily continued farther; and if pupils
were to extend it, and commit it to memory, as far as 30
or 40, it would afford them 'great advantage in their pro-
i^rees.
The number to be multiplied is called the multipli-
cand.
The number which multiplies is called the multiplier.*
The number produced by the operation is called the
product.
* The multiplier and multiplicand arc also called farfa' ••
SIMPLE MULTIPLICATION. 17
CASE I»
When the Multiplier is no more than 12.
RULE.
Place the greater number, or multiplicand, uppermost;
set the multiplier under it, and beginning with units,
multiply all the figures of the multiplicand in succession,
carrying one to the next figure for every ten, and setting
down the several products, as in Addition. The whole
of the last product must be set down.
PROOF,
Multiply the sum by double the amount of the multi-
plier, and if the work in both instances be right, the pro-
duct will be double the amount of the former product.*
EXAMPLES.
I. II. Ill* IV. V.
234 3201 51000 43201 354610
2345 6
468 9603 204000 216005 2127660
VI. VII. VIII.
453210 3245Q13 987654321
78 9
3172470 25960104 8888888889
IX. X. XI.
678987654 321234567 898765432
9 11 IS
* Multiplication may be proved by Division; for if the product be divided by tjhe
multiplier, the quotient will be the same as the multiplicand,
2*
18
-SIMPLE MULTIPLICATION.
CASE It.
When the Multiplier is more than 1£.
RULE.
Multiply each figure in the multiplicand by every
figure in the multiplier, and place the first figure of each
product exactly under its multiplier; then a'dd the seve-
ral proddcts together, and their sum will be the answer.
When cyphers occur at the right hand of either of
the factors, omit them in multiplying, and annex them to
the right hand of the product.*
When the multiplier is the product of any two whole
numbers, the multiplication may be performed by mul-
tiplying the sum by one of them, and the preduct by the
other. Thus, if 24 were to be multiplied by 18, (as 6
times 3 make 18.) let it be multiplied by 6, the product
by 3, and the answer will be the same as if multiplied
by 18.
EXAMPLES.
4 3 0 2 1 6 7 ft-
432
86043356
1 29065034
172086712
18585364896
in.
679100
32
13582
20373
2 173 1200
1041 72
78129
87654320
543
262962960
3506 1728O
438271600
47596295760
v.
432000
4300
1296
1728
885462
1857600000
* Multiplying by 10, add a cypher to tbc right hand sride of the sum, and it is done-
Thus, let it be required to multiply 12 by 10, the product will be 120; but if a cy-
pher be added, it will bring the same result. In multiplying by 100, add two cyV
phew: by 1000, three, &c.
SIMPLE MULTIPLICATION". 19
Multiply 18450 by 35.
VI.
18450
7
129150 92250 92250
5 7 55350
645750 645750 645750
9. multiply 420 by 7 product 2940
10. 3240 9 29160
11. 54134 18 974412
12. 37990 24 911760
13. 84522 54 4564188
14. 90203 587 52949161
15. 370456 7854 2909561424
16. 7654876 8765 67094988140
APPLICATION.
1 . A man had 29 cows, and his neighbour had five times
as many. How many had his neighbour? Ans. 145.
2. There are 12 barrels of sugar, each containing 256
pounds. How many pounds did they all contain ?
Ans. 3072.
3. How far will a man travel in a year, allowing the
year to contain 365 days, if he travel 40 miles per day?
Ans. 14600 miles.
4. In one hogshead are 63 gallons ; — how many gallons
are there in 144 hogsheads? Ans, 9072,
I
Q. 1. What does simple Multiplication feach?
2. What is the number to be multiplied, called?
3. What is the number called which is used in mul-
tiplying another number?
4. Are the multiplicand and multiplier called by any
other names?
5. How do you proceed when the multiplier is no
more than 12?
6. When the multiplier is more than 125howdoyou
proceed?
20 SIMPLE DIVISION.
7. What do you do when cyphers occur at the right
hand of either of the factors?
8. How do you proceed when the multiplier is the
product of two other numbers?
9. How may sums in Multiplication be proved?
SIMPLE DIVISION.
Simple Division teaches to find how often one num-
ber is contained in another, and is a concise way of per-
forming several subtractions.
Questions to prepare the learner for this rule.
1. James had 4 apples and John half as many; how
many had John?
2. If two oranges cost 6 cents, what does one cost?
3. If you divide 8 apples equally between 2 boys,
how many will each have?
4. What is one half of eight?
5. If you divide 6 nuts equally among 3 boys, how
many will each have?
6. What is one third of six?
7. If 12 cherries cost nine cents, what will 4 cost?
8. A third of 9 is how many?
9. If you divide 16 nuts equally among 4 boys, how
many will each have?
10. A fourth of 16 is how many?
11. How many times two are there in six?
12. How many times three in six?
13. How m£hy times four in eight?
14. How many times two in twelve?
15. In nine, how many times three?
16. In eight, how many times two?
17. In ten, how many times five?
18. In twelve, how many times three?
19. In twelve, how many times four?
20. In twenty, how many times five?
21. In eighteen, how many times six?
22. In sixteen, how many times two?
SIMPLE DIVISION. 21
23. In thirty, how many times five?
24. In thirty, how many times six?
25. In twenty-one, how many times seven?
26. In twenty-eight, how many times seven?
27. In thirty-six, how many times twelve?
28. In forty-eight, how many times twelve?
29. In forty-eight, how many times sixteen?
30. In fifty-five, how many times eleven?
31. In sixty, how many times twenty?
32. In eighty, how many times twenty?
33. In one hundred, how many times twenty?
34. In one hundred and twenty, how many times thirty ?
35. In ten, how many times four?
Answer. Two times, and two remain.
36. In fourteen, how many times three?
Answer. Four times, and two remain.
37. In twenty-five, how many times four?
Answer. Six, and one remains.
There are in Division four principal parts, *iz:
The dividend, or number to be divided.
The divisor, or number given to divide by.
The quotient, or answer, which shows how many
times the divisor is contained in the dividend.
The remainder, which is any overplus of figures
that may remain after the sum is done, and is
always less than the divisor.
CASE I.
RULE. — First, find how many times, the divisor is con-
tained in as^ many figures on the left hand of the divi-
dend as are necessary for the operation, and place the
number in the quotient. Multiply the divisor by this
number, and set the product under the figures at the left
hand of the dividend before mentioned. Subtract this
product from that part of the dividend under which it
stands, and to the remainder bring down the next figure
of the dividend; but if this will not contain the divisor,
place a cypher in the quotient, and bring down another
figure of the dividend, and so on, until it will contain the
22 SIMPLE DIVISION.
divisor. Divide this remainder (thus increased) in the
same manner as before; and proceed in this manner un-
til all the figures in the dividend are brought down and
used.
PROOF.
Multiply the quotient by the divisor, and to the pro-
duct add the last remainder, if there be any; if the
work is right, the sum will be equal to the dividend.
EXAMPLES.
I.
Divisor. Dividend. Quotient.
3)1439671 82(47989060
Proof 143967182
In this example, I find
2 9 that 3, the divisor, can
2 7 not be contained in the
first figure of the divi-
2 6 dend ; therefore I take
24 two figures, viz: 14, and
inquire how often 3 is
27 contained therein, which
27 I find to be 4 times, and
•— put 4 in the quotient. —
1 8 Then multiplying the
1 8 divisor by it, i set tho
product under the 14,
Remainder. 2 in the dividend, and
find by subtracting that
;here is a remainder of two. To this 2, I bring down
the next figure in the dividend, viz: 3, which increases
the remainder to 23. I then seek how often 3 is con-
tained in 23, and proceed as before. When I bring down
the 1 that is in the dividend, I find that 3 can not be con
tained in it, and therefore place a cypher in the quotient
and bring down the 8, which makes 18. Finding that 3
is contained 6 times in 18, and that there is no remain-
SIMPLE DIVISION. 33
tier, I bring clown the 2; but as 3 can not be contained
in it, I place a cypher in the quotient, and let 2 stand as
the last remainder. In proving the sum by Multiplica-
tion, the 2 is added. This mode of operation is called
LONG DIVISION.
ii. in.
2 ) 3456789 ( 1728394 5 ) 6789876 ( 1357975
2255
14 Proof 3456789 17 Pr. 6789876
14 15
5 28
4 _ 25
16 39
16 35
7 48
6 45
18 37
18 35
9 26
8 25
1 1
IV. V.
42)9870(235 320) 1 286401 608 1 (40200050
84 1280
147 640
126 640
210 1608
210 1600
8i
SIMPLE DIVISION.
VI.
12)301203(25100
24 12
61 Pr. 301203
60
12
12
03
VII.
15)218760(14584
15 15
68
60
72920
1^584
87 218760
75
126
120
60
60
VIII.
€48 ) 2468098 ( 3808
1944
5240
5184
5698
5184
514
Proof
3808
648
30464
15232
22848
514
2468098
1. Divide 87654 by 58 Quo. 1511 Rem. 16
2. 456789 679 672 501
3. 3875642 7898 490 5622
4. 98765432 1234 80036 1008
5 12486240 87654 142 39372
6. 57289761 7569 7569
Note. — When there is one cypher, or more, at the
right hand of the divisor, it may be cut off; but when
this is done, the same number of figures must be cutoff
from the right hand of the dividend; and the figures
thus cut off, must be placed at the right hand of the re-
mainder.
SIMPLE DIVISION. 25
EXAMPLES.
I. II.
^100)567434110(94572 18[000)246864|593(13714
54 18
27 66
24 54
34 128
30 126
43 26
42 18
14 84
12 72
210 Remainder 12593
Note. — In dividing by 10, 100, or 1000, &c. when you
cut off as many figures from the dividend as there are
cyphers in the divisor, the sum is done; for the figures
cut off at the right hand are the remainder, and those
at the left are the quotient, as in the following sums:
III. IV.
Quotient. Quotient.
1 | 0 ) 9 8 7 6 5 ( 4 Rem. 1)00)123456(78 Rem,
Quo. Quo.
1^000)56789(876 Rem. 1 ',0000)8765(4321 Rem.
CASE II.
When the divisor does not exceed 12, seek how often
it is contained in the first figure or figures of the divi-
dend, and place the result in the quotient. Then mul-
tiply in your mind the divisor by the figure placed in the
quotient, subtract the product from the figure under
which it would properly stand in the former case of di-
vision and conceive the remainder, if there be any, to
be prefixed to the next figure. See how often the divi-
sor is contained in these, and proceed, as before, 'till
3
26 SIMPLE DIVISION.
the whole is divided. This operation is called S
DIVISION.
EXAMPLES.
I. II.
4)987654321 8)1 2 3 4567189
Quo. 2469 13580— I Quo. 15432098—6
In the first example, J find that 4 is contained twice
in 9, and that 1 remains. The 1, 1 conceive as prefixed
to the next figure, which is 8, and they become 18. In
18, 1 find 4 is contained four times, and 2 remain. By
prefixing the 2 to the following figure, which is 7, they
make 27. In this manner I proceed, setting the result
of each calculation in the lower line which is the quo-
tient. In the second example, as 8 can not be contained
in 1, I take two figures, and proceed as in the first.
III. IV.
9)1023684200 12)1914678987
v. vi.
11)6789870062 12)1000001246
Note. — When the divisor is of such a number that two
figures being multiplied together will produce it, divide
the dividend by one of those figures, the quotient thence
arising, by the other figure, and it will give the quotient
required. As it sometimes happens that there is a re-
mainder to each of the quotients, and neither of them
the true one, it may be found thus: — Multiply the first
divisor by the last remainder, and to the product add
the first remainder, which will give the true one.
EXAMPLES,
I.
Divide 249738 by 56.
8)249738
8
7|31217—2 4
4459—4 .32 .*
2
64 Remainder.
SIMPLE DIVISION*
The same done by Long Division,
56)249738(4459
224
257
224
333
280
538
504
3 4 Remainder.
n.
Divide 1847562324 by 84.
42)1847562324 " 7)1847562324
7)153963527 12)26393747 4—6
2199478 9—4 2199478 9—6
12 7
4 8 Ren* 4 2
6
Rem. 4 8
3. Divide 84630986 by 72.
4. « 6 7 8 6 0 1 2 1 by 63.
5. " 1 2 4 5 6 7 4 3 by 96.
6. " 34210390 by 81.
7. " 54697283 by 1 0 3.
8. " 75392618 by 1 1 3.
Note. — In all cases in Division, when there is any re-
mainder, the remainder and divisor form a Vulgar
Fraction. Thus, if the divisor be 8 and the remainder
5, they make f or five eights; or, as in one of the pre-
ceding examples, the divisor is 56 and the remainder 34;
which make ff .
28 FEDERAL MONEY.
APPLICATION.
1. If 48672 dollars be equally divided among four song?
how much will each receive? Ans. 12168 dollars,
2. If a field of 32 acres, produces 1 920 bushels of corn,
how much is it per acre. Ans. 60 bushels.
3. Sixty men at a festival, which lasted three days,
spent 240 dollars per day. How much did each man
spend per day, and how much did he spend in the whole?
Ans. 4 dollars per day and 12 in the whole.
4. Divide 151200 Ibs. of meat, equally, among an ar-
my which consists of 27 regiments, each regiment ha-
ving 7 companies, and each company 100 men; and
what would be each man's share? Ans. 8 Ibs,
Q. 1. What does Simple Division teach?
2. What are the four principal parts of Division?
3. How do you proceed when there is one cypher or
more on the right hand of the divisor?
4. How do you proceed in dividing by ten, or a hun-
dred, or a thousand?
5. How do you proceed when the divisor does not
exceed 12?
»). When you divide by any number not exceeding
12, what is the operation called?
7. When the divisor is of such a number that two
figures multiplied together will produce it?
3. What can be vmade by placing the remainder of a
sum over the divisor? Ans. a VulgaRFraction,
9. How is a sum in Division proved?
FEDERAL MONEY.
The denominations of Federal Money, or the money
of the UMTED STATES, are, Eagle, Dollar, Dime, Cent,
>.nd Mill.
TABLE,
10 Mills (in) make 1 Cent, c.
10 Cents - 1 Dime, d.
10 Dimes - 1 Dollar, /). or g.
10 Dollars - 1 Eagle, E.
29
, fn writing Federal Money, it is ctistomary to omit Ea-
gles, Dimes, and Mills, and set down sums in dollars,
cents, and parts of a cent. The parts of a cent gene-
rally used are, halves, thirds, and quarters. Thus, \ is
a half; -J a third; \ a quarter.*
As the column of cents admits of any number under
one hundred, it consists of two rows of figures ; and when
a less number than 10 is written, a cypher is placed to
the left hand of it. In writing a sum in dollars and
cents, if any part of it consist of even dollars without
cents, the place of cents is supplied with two cyphers,
Cents are separated from dollars by a point or period*
Exercises for the learner.
1. How many mills make a cent? — How many half a
cent? — How many a cent and a half? — How many twp
cents?
2. How many halves of a cent make one cent?
3. How many thirds of a cent make a cent?
4. How many fourths of a cent make a half cent?
5. How many fourths make a cent?
6. How many cents make one fourth or quarter of a
dollar.
*7. How many cents make a half dollar?
8. How many cents make three-fourths of a dollar?
9. How many cents make a dollar?
10. How many dollars and cents in one hundred and
ten cents? — How many in two hundred and six cents? —
How many in three hundred and forty-eight cents? —
How many in five hundred and one cents?
11. If you give a dollar for a book, thirty cents for a
slate, and one cent for a pencil; how many cents will
you give for the whole?
12. Write down one dollar and eight c«nts. Two dol-
lars a»d sixteen cents. Twenty dollars and five cents?
13. Write down three hundred dollars and forty cents,
14. Five hundred eighty-four dollars and fifty cents.
* In addition, subtraction, and division of Federal Money, the partp of a cent
less than a fourth are usually omitted. A part greater than a fourth jfe Called fr
half, or three-fourths, according to its proportionate value.
3*
30 FEDERAL MONEY.
15. Eight hundred sixty dollars and sixty-seven cents,
16. Four thousand eight hundred dollars and two cents.
17. Six hundred thirty-one dollars fifty-six and a fourth
cents.
18. Nine hundred and eighty-seven dollars.
19. Thirty-two thousand five hundred dollars eighty-
seven and a half cents.
20. Ten dollars sixty-eight and three-fourths cents.
21. Twelve dollars ninety-three and three-fourths cents,
22. Twenty dollars thirty-seven and a half cents.
23. If hirty- three dollars thirty-three and a third cents.
24. Sixty dollars sixty-six and two third cents.
25. Read the following sums, viz:
$8448.871 $3450.25 $47967.91 $7.10 $115.331
$170.93$ $19.01 $85.061
ADDITION OF FEDERAL MONEY.
RULE.
Begin at the right hand side of the sum, add one row
of figures at a time, and carry one for every ten, from
the lower denomination to the next higher, as in Simple
Addition, until the whole is added. When you come to
the last row on the left hand, instead of setting down
what remains over ten, twenty, or thirty, £,c. set down
the full amount.
Note. — When there are parts of a cent in a sum, such
as halves, &c. find the amount of them in fourths of a
cent; consider how many cents these fourths will make,
and add them to the first row in the column of cents. —
When the parts of a cent are not sufficient to make a
cent, place their amount at the right hand of the column
of cents, as in the first example; and when the parts of
a cent make one cent or more, and some parts remain,
but not enough for another cent, the parts thus remaining
must be set down in the same way, according to the se-
cond example. The proof is the same as in Simple Ad-
dition,
FEDERAL MONEY. % 31
EXAMPLES.
j. ii. in. iv
D. cts. D. cts. D. cts. D. cis.
5432,121 324.871 885.90 98765432
1234.561 987. 43f 125.871 123456.78
7898.76 720.30 440.40 987654.32
5432.12 842. 43J 867.121 123000.45
3456.78 100.621 390.97 678987.65
23454 . 34| 2975 . 671 2710.27 2900753 . 52
APPLICATION.
1. A man bought a farm in five parcels; for the first,
lie gave $250.75; for the second, $350; for the third,
$475.871; for the fourth, $550; and for the fifth, $600.
What was paid for the farm? Ans. 2226.621
2. A merchant, in buying, gave for flour, $325.43£;
for sugar, $854.25; for molasses, $520.621; for coffee,
$944.50; and for cotton, $6427.121. What was the sum
paid? Ans. $9071, 93|.
3. What is the amount of 101 cents; 93| cents; 871
cents; 50 cts.; 311 cts.; 43| cts.; and 11 dollars?
Ans. $14.16f cents.
4. Gave for an Arithmetick 31i cents; fora slate, 371
cents; for quills, 50 cents; for an inkstand, 621 cents;
for a Geography, 1 dollar, and for a History, 87i cents.
How much do they amount to? Ans. $3. 68| cts.
5. Add $75212.50. $90000, $644225.75,
$4587220.50, and $5876432.75.
SUBTRACTION OF FEDERAL MONEY,
RULE.
Place the smaller sum under the larger, setting the
dollars under dollars and cents under cents, and proceed
as in Simple Subtraction. When there is a fraction, or
part of a cent in the upper line of figures, and none m
the lower, set it down at the right of the remainder, as
52 FEDERAL MONfcY,
a part of the answer. When there is a fraction in each
line, and the upper one is the larger, subtract the lower
©ne from it and set down the difference ; but if the lower
one is larger than the upper, subtract it from the nunv
ber that it takes of the fraction to make a cent — add
the difference to the upper one, and set down the amount.
When there is a fraction in the lower line and none in
the upper; subtract the fraction from the number that
it takes of it to make a cent, and set down the remain-
der. In this case, and likewise when the part or frac-
tion below is larger than the upper one, it is necessary
to carry one to the right hand figure of the lower row
of cents.
EXAMPLES.
lit. IV,
D. c. D. c.
687.31 9000.43
599.81 8220.3U
$294.75 $87.50 $780.1 1|
VII. VIII.
D. c. D. c.
5978.311 9810000.12|
4689.93$ 1987654.68|
$237.75 $30.061 $1288.371 $7822345.441
9. Subtract $987.20 from $1000.
10. Subtract $5871.311, from $6430.87J.
11. Take $44.87i, from 300 dollars.
12. Take $11000, from $19876.871.
APPLICATION.
1. Bought goods amounting to $4875.621, and having
paid $2850.93$; how much remains due?
Ans. $2*024.68$.
2. My account against my neighbour amounts to $759.
55; and his account against me js $546. 87-J. How much
does he owe me? Ans, $2 12,37 j.
FEDERAL MONEY. 33
3. Having bought a quantity of goods at $5425, and
sold them at $6932.68£. How much did I make on the
goods? Ans. $1507.68f
4. A owes me $ 11587. 50, but having failed in busi-
ness, he is able to pay $9263.621. How much do I lose?
Ans. $2323.87-1,
5. Subtract $8427.871, from $9000. Ans. $572. 12-J,
MULTIPLICATION OF FEDERAL MONEY.
RULE.
Set the multiplier under the sum, and proceed as in
Simple Multiplication, carrying one for every ten from
a lower to a higher denomination, until Ihe whole is
multiplied. After the sum is done, separate, by a pe-
riod, the two right hand figures of the product for cents,
and the figures at the left hand of the period will be
dollars.
Note. — When the sum to be multiplied contains a frac-
tion, or part of a cent, multiply it by the multiplier, and
consider how many cents are contained in its product. —
Then multiply the first figure of the cents and add to
its product the cents contained in the product of the
fraction, and proceed as directed above. In multiplying1
a fraction, if you find in the product one cent or more,
and a remainder not large enough to make another cent,,
set it down at the right hand of the product, that is un-
der the row of fractions or parts of a cent. When there
is a fraction in jthe sum, and the multiplier exceeds 12,
multiply the sum without the fraction, and afterwards
multiply the fraction and add it to the sum.
EXAMPLES,
I. II.
D. c. D. c.
124.10 830.121
248.20 2490.371 689.20 12258.121 3305.25
FEDERAL MONEY*
VI.
D. C.
164325.11
18
1314600.88
1643251.1
2957851.98
12480.68
62403.4
74884.08
1542.17
6609.3
9*
8151.561
9.
10.
11.
12.
13.
14.
16.
17.
18.
19.
20.
21.
Multiply
$420.50
by
2.
Ans.
$ 84 1.00
519.75
by
3.
Ans.
$1559.25
99.62J
by
4.
Ans.
398.50
75.31|
by
5.
Ans.
376 56|
62.121
by
6.
Ans.
372.75
750.25"
by
7.
Ans.
5251.75
330.124
by
8.
Ans.
2641.00
248.87|
by
9.
Ans.
2239.87^
95.93J
by
12.
Ans.
1151.25
24.17
by
10.
Ans.
241.70*
37.50
by
28.
Ans.
1050.00
58.93|
by
36.
Ans.
2121 75
9876.621 by 208. Ans. 2054338,00
APPLICATION.
1. How much will 18 barrels of flour cost, at 3 dol-
lars per barrel? Ans. 54 dollars,
2. What will 35 pounds of coffee cost, at 20 cents per
pound? Ans 7 dollars.
3. Sold 87 barrels of flour, at $3.121 per barrel.
What was the amount? Ans. $271.871.
4. Bought 160 acres of land, at $1.25 per acre. —
What did the whole cost? Ans. 20Q dollars.
5. What will 225 bushels of apples cost, at 62i cents
per bushel? Ans. 140.621.
6. What will 580 bushels of salt cost, at $1.12} per
bushel? Ans. $652.50
* In multiplying by 10, when there is no fraction in the sum, it is necessary to
add a cypher to tho right hand of the mm, placing the period that separates cents,
t'rom dollars one figure farther towards the ripht hand, and the gurn is done. In
«nu!tiplying by 100, add two cyphers; l>y 1000, three, &c.
FEDERAL MONEY. 35
DIVISION OF FEDERAL MONEY.
RULE.
Proceed as in Simple Division. When the sum con
rsists of dollars and cents, the two right hand figures of
the quotient will be cents. When there is a remainder,
multiply it by 4, adding the number of fourths that are
in the fraction of the sum (if there be any) to its pro-
duct: then divide this product by the divisor, and its
quotient will be fourths, which must be annexed to the
quotient of the sum. When the sum consists of dollars
only, if there be a remainder, add two cyphers to it;
then divide by the divisor as before, and its quotient
will be cents, which must be added to the quotient of
the sum. When the sum is in dollars, and the divisor is
larger than the dividend, add two cyphers to the divi-
dend— then divide, and the quotient will be in cents,
EXAMPLES.
I. II. III. IV.
D. v. D. c. D. c. D. c. D. c,
2)420.50 4)8000.00 5)580.75 7)84.49(12.67
7
210.25 2000.00 116.15
14
14
49
49
V. VI. VII.
D. c. D. c. D. c. &: c. D. c. D. c
9)27.81(3.09 12)144.60(12.05 36)162.36(4.5!
27 12 144
81 24 183
81 24 180
60 36
60 36 *
36 FEDERAL MONE&
VIII. IX.
D. c. D. c. D. c. D. c.
36)1234.72(34.29£-- 44)87654.3*2(1992. ?4i
108 44
436
396
107 405
72 396
352 94
324 88
28 63
4 44
36)112(3 192
108 176
4 16
4
44)64(1
44
20-f
10.
Divide
£640
by
12
11.
u
717. 12£
by
8
12.
a
246.25
by
9
13.
u
687 . 20
by
12
14.
U
980
by
34
15.
u
87654
by
128
16.
u
1284 ,31£
by
112
17.
u
40000
by
188
18.
a
976 . 871
by
225
19.
u
1234.37JL
•by
212
SO.
u
9876 . 44
by
345
21.
a
89876 . 54
by
374
FEDERAL MONEY. 37
APPLICATION.
1. Divide 400 dollars, equally, among 20 persons. —
What will be the portion of each person? Ans. $20.
2. Divide 1728 dollars, equally, among 12 persons.
What does each one of them share? Ans. $144.
3. If 240 bushe's cost 420 dollars,- what is the cost of
one bushel at the same rate? Ans. $1.75.
Promiscuous Examples.
1. What will the following sums amount to, when ad-
ded together, namely: —
$124.621; $248.871; $342.40; $9850.25.
and $20.311? Ans. $10586.461.
2. If my estate is worth 12870 dollars, and I meet
with losses amounting to $4364.50, how much shall I
have left? Ans. $8505.50.
3. A merchant enters into a trade by which he re-
ceives $1324.621 per year, for four years; how much is
his whole gain? Ans. $5298.50.
4. An estate of 98740 dollars is to be divided, equally,
between 8 heirs; what did each receive?
Ans. $12342.50.
A bought of B,
1 barrel of sugar at - - $24.50
1 chest of tea, 60.00
1 hogshead of salt, - • 3.75
20 yards of cloth, - 15.00
1 barrel of flour, - 3.87J
Ans. $107.121.
Q. 1. What are the denominations of Federal Money?
2. How many mills make a cent?
3. How many cents make a dime?
4. How many dimes make a dollar?
5. How many dollars make an eagle?
6. How are the denominations generally used in
writing Federal Money, and in reckoning?
7. Where is Federal Money used as a currency?
Answer. In the United States of North America,
4
TABLE
OF
MONEY, WEIGHTS, MEASURES,
ENGLISH MONEY.
A table of Federal Money has already been given.
The denominations of English Money are pound, shil
ling, penny, and farthing.
4 farthings (qr.) make 1 penny d.
12 pence - 1 shilling s.
20 shillings - 1 pound Jg,
0^- Farthings are written as fractions, thus:
I- one farthing.
A two farthings, or a half-penny.
J three farthings.
PENCE TABLE.
1
d.
«. d.
s.
20 pence make
1 8
20
30 " " -
^ 6
30
40 " "
3 4
40
50 « " r
4 2
50
60 " " -
5 0
60
70 " "
5 10
70
80 « « -
6 8
80
90 « " -
7 6
9Q
100 « « -
8 4
100
110 « «- -
9 2
no
120 « « -
10 0
120
240 <« " -
20 0
130
SHILLING TABLE.
S.
£. *
20 - - -
1 0
30 - - -
1 10
40 - - -
2 0
50 - - -
2 10
60 - - -
3 0
•70 - - -
3 10
80 - > -
4 0
90 - - -
100
4 10
5 0
no -
5 10
120 - - -
6 0
130 - - -
6 10
TABLE OF WEIGHTS AND MEASURES., 39
TROY WEIGHT.
By this weight, jewels, gold, silver, and liquors are
Weighed.
The denominations of Troy Weight are pound, ounce,
pennyweight, and grain.
24 grains (gr.) make 1 pennyweight dwt.
20 pennyweights - 1 ounce - oz.
12 ounces 1 pound Ib.
AVOIRDUPOIS WEIGHT.
By this weight are weighed things of a coarse, dros-
sy nature, that are bought and sold by weight; and all
metals but silver and gold.
The denominations of Avoirdupois Weight are ton,
hundred weight, quarter, pound, ounce, and dram.
16 drams, (dr.) make 1 ounce - oz.
16 ounces - 1 pound - Ib.
28 pounds - 1 quarter of a cwt. qr.
4 quarters, or 112 Ib. 1 hundred-weight cotf,
20 hundred weight - 1 ton T.
APOTHECARIES WEieHT.
By this weight apothecaries mix their medicines, but
buy and sell by Avoirdupois Weight.
The denominations of Apothecaries Weight are
pound, ounce, dram, scruple, and grain.
20 grains (gr.) make 1 scruple 9
3 scruples - 1 dram 3
8 drams - 1 ounce 3
12 ounces 1 pound ft
LONG MEASURE.
Long measure is used for lengths and distances.
The denominations of Long Measure are degree,
league, mile, furlong, pole, yard, fcot, and inch,
40 TABLE OP MEASURES.
12 inches (in.) make 1 foot ft.
3 feet - 1 yard - yd.
5± yards, or 16J feet 1 rod, pole, or perch P.
40 poles (or 220 yds.) 1 furlong - fur.
8 furlongs (or 1760 yds.) 1 mile - M.
3 miles - 1 league - L.
eOgeographick > n d ,
or 69J statute 3
Note. — A hand is a measure of 4 inches, and used in
measuring the height of horses.
A fathom is 6 feet, and used chiefly in measuring the
depth of water.
CUBICK, OR SOLID MEASURE.
By Cubick, or Solid Measure, are measured all (hings
that have length, breadth and thickness.
Its denominations are, inches, feet, ton, or load, and
cord.
1728 inches make 1 cubick foot,
27 feet - 1 yard.
40 feet of round timber)
or 50 feet of hewn£ 1 ton or load,
timber }
128 solid feet, i. e. 8 in)
length, 4 in breadth, > 1 cord of wood,
and 4 in height )
LAND, OR SQUARE MEASURE.
This measure shows the quantity of lands.
The denominations of Land Measure are acre, rood,
square perch, square yard, and square foot.
144 square inches make
9 square feet
rJO-J- square yards
40 square perches
4 roods
'.MO acres
square foot ft.
square yard yd.
square perch 1J.
rood R.
acre A.
mile - m.
TABLE OF MEASURES. 41
CLOTH MEASURE.
By this measure cloth, tapes, &c. are measured.
The denominations of Cloth Measure are English el%
Flemish ell, yard, quaiter of a yard, and nail.
4 nails (na) make 1 quarter of a yard qr.
4 quarters 1 yard - yd.
3 quarters 1 ell Flemish - E. FL
5 quarters - 1 ell English - E. E,
6 quarters 1 ell French - E. F,
DRY MEASURE.
This measure is used for grain, fruit, salt, &c.
The denominations of Dry Measure are bushel, peck7
quart, and pint.
2 pints (pt.) make 1 quart - qt.
8 quarts - 1 peck pe,
4 pecks 1 bushel bit,,-
WINE MEASURE.
By Wine Measure are measured Rum, Brandy, Perry.,
Cider, Mead, Vinegar and Oil.
Its denominations are pints, quarts, gallons, hogsheads^
pipes, &c.
2 pints (pt.) make 1 quart qt.
4 quarts - 1 gallon gal,
42 gallons - 1 tierce - tier.
63 gallons - 1 hogshead hhd.
2 hogsheads - 1 pipe or butt P. or £«
2 pipes - 1 tun - T.
ALE, OR BEER MEASURE*
The deneminations of this measure are pints, quartSj
gallons, barrels, &c,
4*
12 TABLE OF TIME AND MOTION.
2 pints (pt.) make 1 quart - qls.
4 quarts 1 gallon - gal.
8 gallons - 1 firkin of ale - fir.
2 firkins - 1 kilderkin - kil.
2 kilderkins - 1 barrel - bar,
11 barrels, or 54 gallons 1 hogshead of beer hhd.
2 barrels - 1 puncheon - pun.
3 barrels, or 2 hogsheads 1 butt butt.
TIME.
The denominations of Time are year, month, week^
day, hour, minute, and second.
60 seconds (sec.) make 1 minute - mm.
60 minutes - 1 hour - H.
24 hours 1 day - D.
7 days 1 week W.
-52 weeks, 1 day, atid 6 hours J 1 v
or 365 days, and 6 hours \ i ye;
12 months (mo.) 1 year
Note. — The six hours in each year are not reckoned
till they amount to one day: hence, a common year con-
sists of 365 days, and every fourth year, called leap
year, of 366 days.
The following is a statement of the number of days
in each of the twelve months, as they stand in the calen-
dar or almanack:
The fourth, eleventh, ninth, and sixth,
Have thirty days to each affix'd :
And every other thirty-one,
Except the second month alone,
Which has but twenty eight in fine,
Till leap year gives it twenty-nine.
MOTION.
60 seconds make 1 prime minute, "
60 minutes - 1 degree - °
30 degrees - 1 sign - $.
12 signs, or 360 degrees j Thef "*o}% great circle
$ of the Zodiack,
COMPOUND ADDITION.
Compound Addition teaches to add numbers which re
present articles of different value, as pounds, shillings,
pence; or yards, feet, inches, &c. called different de-
nominations. The operations are to be regulated b> the
value of the articles, which must be learned from the
foregoing table.
RULE.
Place the numbers to be added so that those of the
same denomination may stand directly under each other.
Add the figures of the first column or denomination to-
gether, and divide the amount by the number which it
takes of this denomination to make one of the next higher.
Set down the remainder, and carry the quotient to the
next denomination. Find the sum of the next column
or denomination, and proceed as before through the
whole, until you come to the last column, which must
be added by carrying one for every ten as in Simple Ad-
dition.
EXAMPLES.
I. II. III.
£ s. d. qrs. £ s. d. qrs. £ s. d. qrs.
14 10 8 2 19 19 11 3 18 17 11 3
11 16 10 1 10 14 4 1 15 J4 10 3
8 3 11 3 13 13 10 2 17 18 9 2
34 11 6 2 44 8 2 2 62 11 8 0 Ans,
In the first of the above examples, I begin with the
right hand column, or that of farthings; and having ad-
ded it, find that it contains 6. Now, as 6 farthings con
tain 1 penny and 2 over, I set the 2 farthings, under the
column of farthings, and carry the penny to the column
of pence. In the column of pence I find 29, which, with
44 COMPOUND ADDITION.
the one carried from the forth ings, make 30. In 30 pence
1 find there are 2 shillings and 6 pence over: setting the
6 pence under the column of pence, I add the 2 shillings
to the column of shillings. In this column are 29, and the
2 added make 31. Thirty-one shillings contain 1 pound,
and 11 shillings over. The 11 shillings are then placed
under the column of shillings, and the 1 is carried to the
column of pounds. In that column are 33 pounds, which,
with the 1 added, make 34. Thus the amount of the
sum is, 34 pounds, 11 shillings, 6 pence, and 2 farthings;
In all cases in Compound Addition, one must be car-
ried for the number of times that the higher denomina-
tion is contained in the column of the lower denomina-
tion. Thus, in Troy Weight: as 24 grains make one
pennyweight, one from the column of grains is carried
for every 24; in the column of pennyweights, one for
every 20; and in every instance the learner must be
guided by the foregoing table of "Money, Weight?,
Measures, &c."
IV. V.
£ s. d. qrs. £ s. d. qrs.
487 16 11 3 9876 15 4 5
830 10 9 1 2123 14 5 0
500 11 42 6789 18 10 2
620 18 3 3 1234 15 11 1
900 8 10 0 7876 493
,/Vote. — Sums in Compound Addition njay be proved in
the same manner as in Simple Addition.
TROY WEIGHT.
VI. VII.
Ib. 02. dwt. gr. Ib. oz. dwt. gr.
487 10 18 22 6780 11 11 12
500 8 11 10 1100 9 18 22
234 11 10 16 3090 10 17 20
876 3 17 23 2468 8 13 19
COMPOUND ADDITION. 45
AVOIRDUPOIS WEIGHT.
VIII. IX.
Ton. cwt. qr. Ib. oz. dr. Ton. cwt. qr. Ib. Qz.
16 18 2 25 11 14 27 17 3 27 8
97 12 3 17 9 11 98 19 2 11 9
34 11 1 10 10 10 70 11 1 18 7
82 19 2 27 15 13 18 16 0 10 6
APOTHECARIES WEIGHT.
X. XI,
fc 3 3 6 gr. ft 3 3 6 £T-
74 9 7 1 13 20 10 7 1 18
18 11 6 2 17 37 11 5 2 17
91 10 3 0 10 28 9 3 1 15
17 9 5 1 19 14 ft 4 0 11
LONG MEASURE.
XII. XIII.
deg. mil. fur. po. ft. in. mil. fur. po. yd. ft.
118 36 7 19 13 3 976 2 13 4 2
921 15 4 16 10 10 867 6 10 3 I
671 10 6 27 11 11 500 1 11 0 0
643 26 5 15 8 8 123 4 15 3 2
123 14 5 16 7 8 345 6 17 1 0
CUBICK, OR SOLID MEASURE.
XIV. XV. XVI.
Ton ft. in. yd. ft. in. Cord ft. in.
17 10 1229, 29 20 1092 48 120 1630
24 13 1460 11 11 1195 54 110 1500
98 25 1527 18 11 1000 75 88 1264
18 16 1079 27 9 1330 87 113 1128
46
COMPOUND ADDITION.
LAND> OR SQUARE MEASURE.
XV II.
acr. roo. per.
987 2 23
798 3 28
123 2 11
567 1 27
700 0 00
xx.
yd. qr. nL
175 3 3
481 2 1
234 1 2
345 0 1
234 1 2
XXIII.
El. E. qr. nl
87654 1 2
56788 3 1
87654 3 2
12345 0 0
84231 2 3
xxvr.
bush. pk. qt.
187 7 3
290 6 2
185 3 1
.549 1 2
160 5 3
XVIII.
XIX.
acr. roo. per.
8423 1 36
acr. roo. per,
9432 3 24
1234 0 10
4324 2 12
4821 3 11
5678 1 36
6789 2 30
5865 3 11
8000 1 13
8765 2 15
CLOTH MEASURE
XXI.
XXII.
El. Fr. qr. nl.
247 2 3
£J. Fl. qr. nL
9876 2 3
456 1 1
8765 1 2
345 3 0
3456 2 3
236 2 2
4000 0 0
567 0 1
7898 2 3
XXIV.
XXV.
yd. qr. nl.
656547 1 1
yd. qr.
987654321 3
nl.
3
987654 2 0
234567876 0
0
765432 1 3
543212345 3
2
134545 3 2
900087654 1
3
584050 0 1
384563200 3
0
DRY MEASURE.
XXVII.
XXVIII.
busk. pk. qt.
356 3 7
bush. pk. qt.
874 3 6
120 1 6
123 1 2
543 2 1
345 2 5
678 3 5
753 1 7
432 1 :3
936 2 4
COMPOUND ADDITION.
-47
WINE MEASURE.
XXIX.
Tun. hhd. gal. qt. pt.
4820 1 16 2 1
9765 3 18 3 1
8645 2 19 1 0
5432 1 22 3 1
6787 1 10 1 0
XXX.
Tun. hhd. gal. qt. pt.
987654 1 12 1 1
321234 3 15 0 1
125780 2 18 3 1
876531 2 27 1 0
248765 1 49 2 1
ALE OR BEER MEASURE.
XXXI.
hhd. gal. qt. pt.
4820 48 3 1
8765 34 1 1
9877 53 2 1
1234 12 1 0
5678 50 0 1
XXXII.
hhd. gal. qt. pt,
17819174 18 3 1
21350000 27 1 1
12168400 35 0 0
21346870 15 3 1
43212345 50 1 1
TIME.
xxxni.
flu. d, h. m. s.
3 6 23 58 24
3 5 20 49 57
1 4 21 30 30
3 2 13 63 53
1 0 10 10 10
XXXIV.
y. mo. 'os. d. h. m. $.
75 11 3 6 22 50 57
18 10 2 5 16 16 15
84 11 1 4 15 H) 10
40 9 1 0 00 00 00
80 10 1 1 11 11 11
XXXV.
18* 54' 44"
20 25 30
87 30 10
00 11
27 29
11
34
MOTION.
XXXVI.
26° 19' 15"
19 26 20
50 15 19
33 10 11
12 34 31
xxxvir.
10*. 24° 53' .60"
9 0 19 31
3 9 23 42
8 17 44 45
7 10 10 10
48 COMPOUND SUBTRACTION.
Q, 1. What does Compound Addition teach?
2. How do you place the different denominations iu
Compound Addition?
3. How do you proceed after placing the denomina-
tions under each other?
4. What do you observe, in carrying from one de-
nomination to another, that is different from
Simple Addition?
5. How is Compound Addition proved?
COMPOUND SUBTRACTION.
Compound Subtraction teaches to find the difference
between any two sums of different denominations.
RULE.
Place those numbers under each other which are of
the same denomination — the less always being below
the greater.* Begin with the least denomination, and
if it be larger than the figure over it, consider the up-
per one as having as many added to it as make one of
the next greater denomination. Subtract the lower
from the upper figure, thus increased, and set down the
remainder, remembering, that whenever you thus make
the upper figure larger, you must add one to the next
superior denomination.
PROOF,
As in Simple Subtraction.
EXAMPLES.
ENGLISH MONEY.
I. II HI.
£ s. d. qrs. £ s. d. qrs. £ s. d. qr.
460 14 10 3 744 10 81 689 792
320 10 8 2 398 18 10 3 372 18 4 3
140 4 21 345 11 92 316 943
*By this is meant, that the lower line of figures must always be a less sum than
the upper line, though Borne of its smaller denominations may be larger than those
immediately above them, in the upper line.
COMPOUND SUBTRACTION. 49
The first example is, in itself, sufficiently plain. In
the second, finding the upper figure smaller than the
lower one, as it is in farthings, and as four farthings make
a penny, 1 suppose four added to the upper figure, which
makes it 5. Then 1 say, 3 from 5, and 2 remain. Placing
the 2 underneath, 1 add 1 to the next lower figure, name-
ly, the 10, which thus becom* s 1 1 ; and as the 8 standing
above is less, I suppose 12 added to it, which makes it
20. Taking 1 1 from 20, 9 remain. Setting the 9 un-
derneath, and adding one to the 18, it becomes 19; and
as the upper figure is smaller, I suppose 20 added to it,
which makes it 30, I take 19 from 30, and 1 1 remain.
Placing the 1 1 underneath,! carry one to the next figure,
namely, 8; and then proceed as in Simple Subtraction,
TROY
WEIGHT.
IV
V.
Ib.
oz.
dwt.
gr.
Ib.
oz.
dwt.
gr.
947
5
13
16
876543
7
16
11
123
10
18
20
549876
9
17
19
AVOIRDUPOIS WEIGHT.
VI. VII.
"7 bra. .crvt. gr. Ib. Ton. cwt. qr. Ib. * oz. dr.
5 13 1 12 8 16 0 24 11 11
1 H 3 16 6 18 2 26 12 13
APOTHECARIES WEIGHT.
i
VIII. IX.
ft 3 3 6 gr. ft 5 3 6 gr.
44 7 5 1 12 87 4 1 0 10
39 9 6 2 16 48 10 4 1 18
COMPOUND SUBTRACTION.
LONG MEASURE.
X.
deg. mil. fur.po. ft. in.
85 53 7 16 10 iO
60 57 0 27 14 11
XI.
deg. mil. fur. pa
95 10 3 12
79 44 6 13
CUBICK, OR SOLID MEASURE.
XII. XIII. XIV.
Ton ft. in. yd. ft. in. Cord ft. in.
18 17 1040 40 10 940 874 110 112S
11 21 1485 32 16 1080 499 120 1699
LAND, OR SQUARE MEASURE.
XV. XVI. XVII.
acr. roo. per.
987 2 23
798 3 28
acr. roo. per.
8423 1 36
acr. roo. per.
9432 3 12
4123 0 10 7324 2 24
XVIII.
CLOTH MEASURE,
XIX. XX.
XXI.
yd. qr. nl. E. E. qr. nl. E. Fl. qr. nl. E. Fr. qr. hi.
45
29
537
409
2
3
567
389
945
739
XXII.
bush. pk. qt.
74 1 1
42 3 2
DRY MEASURE.
XXIII.
bush. pk. qt.
230 0 0
199 2 1
XXIV.
bush. pk. qt.
56 1 1
28 3 3
COMPOUND SUBTRACTION. M
WINE MEASURE,
XXV. XXVI.
Tun. hhd. gal. qt. pt. Tun. hhd. gal. qt. pt.
482 1 16 1 1 654 2 12 1 0
297 3 22 3 1 276 3 40 2 1
ALE OR BEER MEASURE.
XXVII. XXVITI.
hhd. gal. qt. 'pt. hhd. gal. qt. pt,
8240 12 1 1 11917400 10 0 0
1987 52 2 2 11654000 27 2 2
TIME.
XXIX. XXX.
w. d. h. m. s. y. mo. w. d. h. m. $.
8 2 12 42 30 20 10 1 4 10 27 37
7 1 16 54 40 11 11 3 5 17 40 54
MOTION.
XXXI.
16° 15' 35"
12 45 48
XXXII.
8$. 10° 10' 10"
6 15 50 30
XXXIII.
7s. 8° 37' 47"
4 11 44 55
Application of the two preceding rules.
1. A B & C purchased goods in partnership. A paid
12 pounds, 10 shillings and 8 pence; B paid 124 pounds*
16 shillings; and C paid 87 pounds and 11 pence. —
What was the whole amount paid? Ans. £224 7s. 7d.
2. A merchant has money due him: — from one man,
587 pounds; from another, 420 pounds, 17 shillings and
6 pence; from a third, 200 pounds; and from a fourth,
978 pounds, 16 shillings and 8 pence. How much had
he due in all? Ans, £2186 14s, 2d.
52 COMPOUND MULTIPLICATION.
3. From 20 pounds, take 12 pounds, 19 shillings anct
3 farthings. Ans. £7 Os. lid. Iqr
4. From 22 pounds, take 19 shillings and 1 farthing.
Ans. £21 Os. lid. 3qrs.
6. From 17 pounds, take 9 pounds, 9 shillings and 9
pence. Ans. £7 10s. 3d.
6. A has paid B £7 2s. 3d., £19 J Is. 4d, and £17
18s. 9^d. on account of a debt of £60. How much re-
mains unpaid? Ans. £15 7s. 7 id.
7. A ropemaker received 3 tons, 4 cwt, 2 quarters,
and 5 pounds of hemp; of which he made into cordage
2 tons, 9 cwt, and 1 quarter. How much had he left?
Ans. 15ewt. Iqr. 5lbs.
Q. 1. What does Compound Subtraction teach?
2. How do you set down Compound Subtraction?
3. What do you do when the lower denomination is
larger than the one that is above it?
4. How is Compound Subtraction proved?
COMPOUND MULTIPLICATION.
Compound Multiplication teaches how to find the value
of any given number of different denominations, re-
peated a certain number of times. It is of great use in
finding the value of goods, which is generally done by
multiplying the price by the quantity.
CASE I.
When the quantity or multiplier does not exceed 12.
Set down the price of 1, and place the multiplier un-
der the lowest denomination; and in multiplying by it,
observe the same rules for carrying from one denomina-
tion to another as in Compound Addition.
PROOF.
Double tbe multiplicand, and multiply by half the
multiplier: or, divide the product by the multiplier.
COMPOUND MULTIPLICATION* 53
EXAMPLES.
ENGLISH MONEY.
I. What will 7 yards of cloth cost, at £1 12s. lOfd
per yard? 7
£11 10s. 3id,
In this example, I say 7 time 3 make 21— -that is, 21
farthings, equal to five pence and one farthing. I set
down the one farthing under the place of farthings, and
carry the five pence to the place of pence saying, 7 times
10 arc 70, and 5 make 75 pence — equal to 6 shillings and
3 pence. I set down the 3 pence under the pence in the
sum and carry the 6 shillings, saying, 7 times 12 are 84,
and 6 make 90 shillings, equal to 4 pounds and 10 shillings.
Setting down the 10 shillings under the shillings, I carry
the 4 pounds, saying 7 times 1 are 7, and 4 make 1 1
pounds, making the answer to the question 1 1 pounds,
10 shillings, 3 pence and 1 farthing,
II. III. IV.
£ s. d. £ s. d. £ *, A
Multiply 4 14 lOf 7 12 9 14 15 9J
by 2 4 8
V. VI. VII.
£ s. d. £ s. d. £ s. d.
Multiply 14 17 81 24 16 101 50 15 5|
by 9 7 12
TROY WEIGHT.
VIII. IX.
Ib. oz. dwt. gr. Ib. oz. dwt. gr,
Multiply 11 9 16 10 17 5 12 6
by 4 5
5*
54 COMPOUND MULTIPLICATION.
AVOIRDUPOIS WEIGHT.
Mult. 3
by
x. xi.
Ton. cwt. qr. Ib. oz. dr. Ton. cwt. qr. Ib. oz. dr.
. 3 11 3 10 5 4 6 17 3 13 2 15
fi ft
APOTHECARIES WEIGHT.
XII. XIII.
ft 3 3 6 gr. fe 3 3 6 g-r.
Mult. 43 10 6 2 10 4 10 7 2 13
by 5 6
LONG MEASURE.
XIV. XV.
cleg. in. fur. p. yd. ft. in. L. m. fur. p.
Mult. 7 22 6 20 2 2 10 15 2 7 30
by 26
CUBICK, OR SOLID MEASURE.
XVI. XVII. XVIII.
Ton. ft. in. yd. ft. in. Cord ft. in.
Mult. 10 16 15 14 2 19. 24 13 18
by 2 4 6
LAND, OR SQUARE MEASURE.
XIX. XX. XXI.
A. R. P. A. R. P. A. R. P.
Mult. 20 3 12 37 2 15 47 1 18
by 2 4 6
COMPOUND MULTIPLICATION.
55
CLOTH MEASURE.
XXII. XXIII. XXIV. XXV.
yd. qr. nl ELFr.gr. nl El.Fl.qr.nl. El.E. qr.nl
Mult. 17 3 3 32 2 1 42 3 1 53 2 1
by 4 6 8 9
XXVI.
bush.pk. qt.
Mult. 637
by 5
DRY MEASURE.
XXVII.
bush. pk. qt.
14 3 2
6
XXVIII.
bush. pk. qt,
34 2 3
8
WINE MEASURE.
XXIX. XXX.
Tun. hhd. gal. qt. pt. Tun. hhd. gal. qt. pt.
Mult. 1 2 12 3 1 2 3 40 3 1
by 4 10
ALE, OR BEER MEASURE.
XXXI.
XXXII.
hhd. gal. qt. pt.
Mult. 3 12 2 1
by 5
hhd. gal. qt. pt.
4 15 3 1
XXXIII.
y. mo. w. d.
Mult. 7735
by 9
XXXV.
Mult. 24° 19' 11"
by 10
TIME.
XXXIV.
y< mo. w. dt h. m. s.
8 5 3 6 20 32 10
MOTION.
XXXVI.
Ws. 30° 17' lOf"
12
tOMPOUND MULTIPLICATION.
CASE II.
When the multiplier or quantity exceeds 12. and is the pro
duct of two factors in the Multiplication Table- that is,
of two numbers which being multiplied together, amount
to the same as the multiplier.
Multiply the sum by one of the two numbers, and then
multiply the product by the other.
EXAMPLES.
I. II.
£ s. d. £ s. d.
Multiply 8 18 llfbylS. 13 12 9J by 27,
6 9
53 13 10i
3
161 1 7J
£ s. d.
122 15
368
£ *. d.
3. Multiply 10 10 10 by 14. Product 147 11 8
4.
5.
6.
7.
8.
9.
10.
11.
12,
£<
a
11 11 11 by 15.
12 12 9 by 24.
513 41 by 28.
4 15 10" by 42.
7 17 7| by 64.
6 10 3 by 72.
9 19 llf by 81.
10 15 9| by 84.
3 11 7£ by 96.
173 18 9
303 6 0
158 14 6
201 5 0
504 9 4
468 18 0
809 16 7 A
906 8 3
343 16 0
CASE III.
When the quantity, of multiplier, is such a number that no
two numbers in the Multiplication Table will produce it.
Multiply the sum by two numbers whose product will
amount to nearly the same as the multiplier; then mul-
tiply the sum by the number which will make the pro-
duct of the two numbers equal to the multiplier, and
add its product to the sum produced by the two num-
bers.
COMPOUND MULTIPLICATION. 57
EXAMPLES.
I.
£ s. d. £ s. d.
Multiply 7 10 5 7 10 5
by 62 10 2
75 4 2 15 0 10
6
451 50 Here note, I multiply by 10, then
15 0 10 by 6, because 10 times 6 make 60;
then I multiply the same sum by
466 5 10 2, that [ multiplied, first, by 10,
— and add its product to the other
product, which makes the amount ®f the answer.
£ *. d. £ s. d.
2. Multiply 2 10 10 by 31. Product 78 15 10
3. " 3 II 11 by 38. " 136 12 10
4. " 411 21 by 68. « 310 0 9
5. " 1 8 8 by 26. « 37 5 4
6. " 13 3£ by 47. « 54 14 81
7. « 12 41 by 83. « ^92 17 li
CASE IV.
When the multiplier is greater than the product of any two
numbers in the Multiplication Table.
Multiply the sum by 10, and that product by 10, which
is equal to multiplying by 100; then multiply the pro-
duct by the number of hundreds in the multiplier, and if
the sum be even hundreds, the product will be the answer.
If there be odd numbers over even hundreds, as 70, 80, or
87, &c., multiply the amount or product of the first mul-
tiplication by 10, by the number of tens over 100; thus,
if the»-e be 70 over, multiply by 7. If, in addition to
ten«. there are smaller numbers, as 7, 8, 5, &c,, the sum
must be multiplied by such number; and the amount of
aU the multiplications being then added together, their
sum will be the answer,
58 COMPOUND MULTIPLICATION*
EXAMPLES.
Multiply
by 4321
£
s.
d.
i
2
4
10
11
3
4 amount of
10.
10
Ill 13 4 amount of 100,
10
1116 13 4 amount of 1000,
4
4466 13 4 amount of 4000.
335 00 0 amount of 300.
23 9 0 amount of 21,
4825 2 4 Answer.
In the foregoing example, 1 first multiply by 10, three
times, which gives the amount of the sum multiplied by
1000; then by 4, which gives the amount of 4000. The
sum is yet to be multiplied by 32 1 . To do this, I take the
product of the sum multiplied by 100, namely, 111/. 5s.
and multiply it by 3, which gives the product of the
sum by 300. But as there are 21 to multiply by, 1 take
the original sum and multiply it by 7, and then by 3;
and then adding the products together, I obtain the an-
swer.
£. s. d. £ s. d.
-2. Multiply 1 4 by 190. Product 12 13 4
3. « 1 2 3 by 430. " 478 7 6
4. « 7 6 by 506. " 189 15 0
5. " 88 by 684. « 296 8 0
6. « 1 3 9 by 375. " 445 6 3
7. « 12 by 3456. « 201 12 0
APPLICATION.
1. What do B4 pounds of sugar cost at 9d. per pound?
COMPOUND MULTIPLICATION. 59
2. What do 18 yards of cloth cost at 19s. per yard?
3. Sold 7 tons of iron at £32 10s. per ton: how much
'is the amount? Ans. £227 10s.;
4. What is the weight of 4 hogsheads of sugar, each
weighing 7 cwt. 3qns. 191b? Ans. 31cwt. 2qrs. 201bs.
5. What is the weight of 6 chests of tea, each weigh
ing 3cwt. 2qrs. 91bs? Ans. 2 Icwt. Iqr. 261bs.
6. What is the value of 79 bushels of wheat, at 11s.
5fd. per bushel? Ans. £45 6s. 101.
7. What is the value of 94 barrels of cider, at 12s.
2d. per barrel? Ans. £57 3s. 8d.
8. What is the value of 114 yards of cloth at 15s.
3fd. per yard? Ans. £87 5s. 7Jd.
9. What is the value of 12 cwt. of sugar, at £3 7s.
4d. per cwt.? Ans. £40 8s.
10. What is the worth of 63 gallons of oil at 2s. 3d.
per gallon? Ans. £7 Is. 9d.
11. What is the amount of 120 days wa§:es at 5s. 9d.
per day? Ans. £34 10ss
12. What is the worth of 144 reams of paper at 13s.
4d. per ream? Ans. £96.
J3. What will Icwt. of sugar cost, at lOfd. per lb.?*
Ans. £5 Os. 4d.
14. If I have 9 fields, each containing 12 acres, 2 roods
and 25 poles; how many acres have I in the whole?
Ans. 113A. 3R. 25P.
Q. 1. What does Compound Multiplication teach?
2. In what is it particularly useful?
3. Which is made the multiplier — the price or the
quantity?
4. How do you proceed when the multiplier does not
exceed 12?
5. How Ho you proceed when the multiplier exceeds
12?
6. When the multiplier consists of no two component
numbers, as in case third, how do you proceed?
7. How do you proceed in case fourth?
8. How is Compound Multiplication proved?
*ft must be recoHected that Icwt. is 1121bs.
£0 COMPOUND DIVISION.
COMPOUND DIVISION.
Compound Division teaches the manner ©f dividing
numbers of different denominations.
CASE I.
When the divisor does not exceed 12.
Begin at the highest denomination, and after dividing
that, if any thing remain, reduce it to the next lower
denomination, adding it to that denomination in the sum,
and proceed in this manner until the whole is divided.
If the number of either denomination should be too
small to contain the divisor, reduce it to the next lower
denomination, and add it thereto, asxlirected in case of
a remainder. The denominations in the quotient must
be kept separate.
PROOF.
Multiply the quotient by the divisor, and the product,
if right, will be equal to the dividend.
EXAMPLES.
I. II. III.
£ s. d. £ s. d. £ s. d.
Divide 2)6 88 4)3 3 10 5)7 2 3
344 15 111 is 51+3
IV. V. VI.
£ s. d. £ s. d. £ s. d.
5)6 17 11 6)9 9 9 12)21 16 Hi
177 1 11 7i 1 16 4f-flO
In doing the 6th sum, which is divided by 12. I find
the divisor is contained once in 21; and setting down 1
1 find 9 pounds remaining; which, reduced to shillings,
and added to the 16 shillings in the sum, make 196
Shillings. The divisor being contained 16 times in 196,
with 4 remaining, I set down 16, and reducing the 4
shillings to pence, and adding them to the 1 1 pence, in
the sum, the amount is 59 pence. The divisor is con-
COMPOUND DIVISION. 61
lamed 4 times in 59, leaving 1 1 pence remaining. I sef
down 4, and the remaining 1 1 pence reduced to farthings
and added to the half penny or 2 farthings in the sum,
make 46 farthings; and as the divisor is contained 3
times in 46, leaving a remainder of 10, 1 set down £ and
place the final remainder at the right hand of the sum.
£
s.
d.
£ *.
d.
7.
Divide
12
10
10
by
5.
Quotient
2
10
2
8.
u
13
13
9
by
4.
"
3
8
5J
9.
u
2
18
H£
by
3.
u
19
7f
10.
u
7
7
7
by
4,
u
1
16
10 J
11.
"
177
19
llf
by
12.
« 14
16
7f
CASE II.
ie divisor exceeds 12, cmrf w tfta product of two
numbers multiplied together.
Divide by one of the numbers: then divide the quo-
tient by the other.
EXAMPLES.
Divide £5 10s. 6d. by 48,
£ \ d.
6)5 10 6
8) 18 5
2 3+5 Answer.
Note. — If there be any remainder in the first opera-
tion, and not any in the second, it is the true one. When
there is a remainder in the second operation, multiply it
by the first divisor, and add it to the first remainder, if
there be any, and it forms the true remainder.
£ s. d. £ s. d.
2. Divide 240 12 10 by 16. Quotient 15 0 9^+4
3. « 88 13 11 by 21. « 44 5|+2
4. « 90 15 41 by 32. « 2 16 8f
5. « 450 8 8 by 42. " 10 14 5|+26
6. , « .789 19 9 by 64. " • 12 6
7. « 840 4 3£by 72. « 11 13
62 COMPOUND DIVISION,
CASE III.
When the divisor is more than 12, and can not be produced
by multiplying any two numbers together.
Divide after the manner of Long Division, reducing
from higher to lower denominations, as in the following
EXAMPLES,
Divide £61 12s. by 23.
i.
£ *•
23)61 12(£2 13s. 6d. 3qrs.-f-3 Ans.
46
— Divide £14 10s. llfd. by 95.
15 ii.
20 X £ s. d.
95)14 10 11£(£0 3s. 0£d.+2 Ans,
312 20X
28
-*- 290
82 285
69
— 5
13 12X
12X
71
156 4-X
138
287
18 285
4X
2
72
69
3
Note. — In the second example, 1 tind the divisor grp a-
ter than the number of pounds in the dividend. I there-
fore set down a cypher in the place of pounds in the
quotient, then reduce the 14 pounds in the sum, inte
COMPOUND DIVISION. 83
(•.hillings, at the same time adding the 10 shillings in the
sum, to the amount, which therehy becomes 290. In
290 the divisor is contained 3 times, and 5 over. This
6 shillings 1 multiply by 12, to reduce it to pence, adding
to it the 1 1 pence in the sum; and the amount being still
smaller than the. divisor, 1 set down a cypher in the
place of pence, in the quotient, and reduce it to far-
things ;% which, with the 3 farthings in the sum, amounts
to 287 farthings. In 287 the divisor is found three
times, and there is a remainder of 2. The quotient,
therefore, contains a cypher in the place of pounds; 3,
in the place of shillings; a cypher in the place of pence,
and 3 in the place of farthings.
Though this operation is longer, it is, perhaps, less
liable to error than either of the preceding cases.
£ ,. a. £. s. d.
3. Divide 20 10 8 by 17. Quotient 1 4 l£+9
4. " 27 18 7 by 29. " 0 19 3-»+6
5. « 147 4 4 by 65. « 25 31+18
6. " 581 19 ll£by 73. « 7 19 51+49
7. " 77 3 3f by 19. « 41 21+17
8. « 319 7 101 by 29. « 11 0 3i+l
APPLICATION.
1. If 42 cows cost £126 16s. 6d; what was the price
of each? Ans. £3 Os. 4id,
2. If £1000 be divided, equally, among 40 men;—
what will each receive? Ans. £25*
3. Five men bought a quantity of hay, weighing 21
tons. 13 hundred, and 3 quarters; which they divided,
equally among them. What was the share of each£
Ans 4 tons, 6cwt. 3qrs.
4. A farmer had 3 sons, to whom he gave a tract of
land containing 520 acres, 3 roods, 29 perches; and the
land was too be divided,, equally among them. What
was the portion of each? Ans. 173A. 2R. 23P:
5. Divide 375 miles, 2 furlongs, 7 poles, 2 yards, 1
foot, 2 inches, by 39. Ans. 9M. 4fur. 39P. Oyd. 2ft. 8m.
6. Divide 571 yards, 2 quarters, 1 nail by 47.
Ans, 12yds. Oqr. 2na,
64 REDUCTION.
7. Divide 120 months, 2 weeks, 3 days, 5 hours, 20
minutes, by 111. Ans. Imo. OW. 2D. 10H. 12 min,
8. Divide 54 dollars, 54 cents, 4 mills, among 3 girls
and 2 boys; and give to each girl twice as much as to
each boy. What does each girl receive?
Ans. $13 63c. 6m,
9. Divide $20 among 8 persons; and give the first
JOc. more than the second; the second lOc. more than
the third, Sue. ; what sum does the eighth person receive?
Ans. $2.15.
Q. 1. What does Compound Division teach?
2. How do you proceed when the divisor does not
exceed 12?
3. What do you do when the number of either de--
nomination is too small to contain the divisor?
4. What do you eto when the divisor exceeds 12, and
is the product of two numbers multiplied to-
gether?
5. How do you proceed when the divisor is more
than 12, and can not be produced by multih
plying any two numbers together?
$. How is Compound Division proved?
REDUCTION.
Reduction teaches to change numbers of one denomfc
nation into those of other denominations, retaining the
same value. Its operations are performed by Multipli-
cation and Division. When performed by Multiplica-
tion, it is called Reduction Descending, when performed
by Division, it is called Reduction Ascending.
1 . How many farthings will it take to make two pence ?
How many pence to make two shillings? — How many
shillings to make two pounds?
2. [low many gills to make three pints? — How many
pints to make three quarts? — How many quarts to make
three gallons?
3. How many quarts to make four pecks? — How ma-
ny pecks to make four bushels?
REDUCTION. 65
4. How many pence are there in eight farthings? —
How many shillings in twenty- four pence? — How many
pounds in forty shillings?
5. How many pints in twelve gills? — How many quarts
in six pints? — How many gallons in twelve quarts?
6. How many pecks in thirty-two quarts? — How ma-
ny bushels in sixteen pecks?
7. How many pounds and shillings in thirty shillings?
How many shillings and pence in thirty pence?
DEDUCTION DESCENDING,
RULE.
Multiply the numbers in the highest denomination
given, by the number that it takes of the next less de-
nomination to make, one of that greater; and thus pro-
ceed until you shall have multiplied each higher deno-
mination by the number that it takes to form the next
lower, until you come to the lowest of all.
PROOF.
Descending and Ascending Reduction prove each
other.
SIMPLE EXAMPLES.
r.
Reduce 25 pounds to shillings. Ans. 500 shillings*
25
20 shillings in a pound.
500 shillings,
Reduce 50 shillings to pence, Ans £00 pence
50
12 pence in a shilling.
600 pence.
6*
66 REDUCTION.
III.
Reduce 15 pence to farthings. Ans. 60 farthings
1 o
4 farthings in a penny,
60 farthings.
IV.
Reduce 10 tons to hundred weights, Ans. 200cwt-
10
20 hundred in a ton*
* 200 hundred.
v.
Reduce 36 peunds to ounces. Ans. 576 ounces;
36
16 ounces in a pound,
216
36
576 ounces.
6. Reduce 70 miles to furlongs, Ans. 560 fur,
7. Bring 30 furlongs to rods. Ans. 1200 rods.
8. Bring 20 rods to feet. Ans. 330 feet
9. Bring 24 feet to inches. Ans. 288 inches.
10. Reduce 32 acres to roods. Ans. 128 roods.
11. Bring 24 square perches to square yards.
Ans. 726 square yards.
12. Reduce 10 hogsheads to gallons. Ans. 630gal.
13. Bring 25 gallons to pints. Ans. 200 pints.
14. Reduce 23 bushels to pecks. Ans. 92 pecks.
15. Bring 12 pecks to pints. Ans. 192 pints.
16. Reduce 15 years to months. Ans. 180 months.
17. Bring 75 days to hours. Ans. 1800 hours.
18. Bring 24 hours to minutes. Ans. 1440 minutes,
19. Bring 10 signs to degrees. Ans. 600 degrees
REDUCTION. 67
COMPOUND EXAMPLES,
I.
£ s. d. qfs.
In 15 17 11 3 how many farthings,?
20 shillings in a pound.
317 shillings.
12 pence in a shilling.
3815 pence.
4 farthings in a penny.
15263 farthings.
Note. — In multiplying by 20, I added in the 17 shil-
lings, by 12, the 11 pence; and by 4, the 3 farthings^
and this must be observed in all similar cases.
To prove this sum, let the order of it be changed,
and it will stand thus: in 15263 farthings, how many
pounds?
4)15263
12)3815+3 quarters.
210)31)7+11 pence.
£15 17s. lid. 3qrs. Ans.
In reducing Federal Money from a higher to a lower
denomination, it is only necessary to annex as many cy-
phers as there are places from the denomination given
to that required; or, if the given sum be of different
denominations, annex the figures of the several denomi-
nations in their order, and continue with cyphers, whep
the sum requires it, to the denomination intended.
2. Thus, in 7 eagles, 3 dollars, how many mills?
Ans. 73000.
3. In 85 dollars, how many mills? Ans. 85000.
4. In 574 eagles, how many dollars? Ans. 5740.
5. In 469 dollars, bow many cents? Ans. 48900,
68 REDUCTION.
6. In 844 dollars, 75 cents, how many mills?
Ans. 844750
7. In 10QO dollars, how many mills? Ans. 1000000,
8. In 25 dollars, 47 cents, 8 mills; how many mills?
9. In 29 guineas at 28s. each, how many pence t'
Ans. 97*4.
10. In 20 acres, 29 poles, or perches, how many square
perches? Ans. 3229,
11. How many solid feet in 30 cords of wood?
Ans. 3840.
12. How many grains in 100 Ibs.— Troy Weight?
Ans. 576000.
13. How many Ibs. in a ton :— -Avoirdupois Weight?
Ans. 2240.
14. In 27 Ibs. — Apothecaries Weight; how many grains?
Ans. 155520.
15. In 30 yards, how many nails? Ans. 480.
16. In 360 degrees, being the distance round the
world, how many inches, allowing 69J miles to a de-
gree? Ans. 1,587,267,200.
17. How many pints are there in one tun of wine?
Ans. 2016.
18. How many half pints in one hogshead of beer?
19. How many pints in 400 bushels? Ans. 25600.
20. How many seconds in 80 years?
Ans. 2,524,554,960;
21. How many yards in 4567 miles? Ans. 8037920.
22. In £20 17s., how many pence and half pence?
Ans. 5004 pence, and 10,008 half pence,
REDUCTION ASCENDING.
RULE.
Divide the figure or figures in the lowest denomina
tion, by so many of that name as make one of the next
higher; and continue the division until you have brought
it ioto that denomination which your question requires.
In reducing Federal Money from a lower to a higher
denomination, nothing more is necessary than to cut off
REDUCTION. 69
so many places on the right hand side of the sum, as
there are denominations lower than the one required,
Thus, 98765 mills are reduced to dollars, cents, and
mills, by cutting off one. figure for mills, two more for
cents, and the remaining figures being dollars, the
amount is $98|76|5 — or ninety-eight dollars, seventy-six
cents, five mills.
SIMPLE EXAMPLES.
1. How many dollars are there in 8000 mills?
8|00|0 Ans.' 8,
2. In 487525 cents, hpw many dollars and cents?
4875125 Ans. $4875.25.
3. In 999888 mills, haw many dollars, cents, and mills ?
999J88J8 Ans. jj999.8a.87
4. In 19200 farthings, how many pounds?
4)19200
12)4800
20)400
Ans. 20 p
5. In 480 nails, how many yards?
4)480
4)120
30 Ans.
COMPOUND EXAMPLES.
6. la 52300 farthings, how many pounds?
4)52300
12)13075
2|0)108|9+7
Ans. £54 9s. 7d,
7. In 8428 Ibs. Avoirdupois Weight, how many tons?
Ans. 3 tons. 18cwt. 3qrs. 81bs,
. 0 REDUCTION.
8. In 524 Ibs. Avoirdupois Weight, how many cwt. &,c,
Ans. 4cwt. 2qrs. SOlbe,
9. In 125440 grains, Troy Weight, how many Ibs?
Ans. 44,
10. In 155520 grains, Apothecaries Weight, how many
pounds? Ans. 27.
1 1 . How many miles are there in 1,585,267,200 inches?
Ans. 25020.
12. In 4000 nails, how many yards? Ans. 250,
13. In 8000 square rods, how many acres? Ans. 50.
14. In 2016 pints of wine, how many tuns?. Ans. 1,
15. How many bushels are there in 80,000 quarts?
Ans. 2500.
16. In 2,524,554,960 seconds, how many years? Ans. 80.
17. In 3840 solid feet, how many cords? Ans. 30.
18. [n 1728 half pints of beer, how many hogsheads?
Ans. 2.
19. Bring 240,000 pence to pounds. Ans. £1000,
20. Bring 112 quarters to cwt. Ans. 28 cwt.
21. Bring 120 miles into leagues. Ans. 40L,
22. Bring 1280 poles into furlongs. Ans. 32 fur.
23. Reduce 960 nails to quarters. Ans. 240 qrs.
24. Reduce 17280 cubick, or solid inches, to solid feet,
Ans. 10 solid feet.
25. In 768 pints, how many bushels? Ans. 12.
26. In 1890 gallons, how many hogsheads? Ans. 30,
Q, 1. What does Reduction teach?
2. By what rules are its operation performed?
3. When performed by multiplication, what is it
called?
4. What is your rule for Reduction Descending?
5. When performed by Division what is it called?
6. What is your rule for Reduction Ascending?
7. How do you reduce Federal Money from a lower
to a higher denomination?
3. How is Reduction proved ?
EXCHANGE. 71
EXCHANGE.
- Exchange teaches to change a sum of one kind of
money to a given denomination of another kind.
T*o reduce the currency of each of the United States to dol-
lars and cents, or Federal Money.
RULE.
Reduce the sum to pence ; to the pence annex two
cyphers; then divide by the number of pence in a dol-
lar, as it passes in each State, the quotient or answer
will be in cents, which may be easily reduced to dollars*
Note. — This rule applies to the currency of any other
country, if its currency be in pounds, shillings, pence, &c.
EXAMPLES.
1. Reduce 621 pounds, New England, Virginia, and
Kentucky currency, to dollars and cents j a dollar being
72 pence,
£
621
20
12420
12
72)14904000(^2070.00
144
504
504
000
2. Reduce 12 pounds, 3 shillings, and 9 pence to dol-
dollars and cents. Ans. $40.621,
3. Reduce 30 pounds and 3 shillings to dollars and
cents. Ans. $100.50
4. In £763 New England and New York currencies,
how many dollars, cents, and mills?
Ans. $2543.33cts. 3m. N. E. cur.
$1907,50cts. N, Y. cur.
72 EXCHANGE.
5. In 9 pounds and 16 shillings in New York and
North Carolina currency, how many dollars and cents,
reckoning 96 pence to a dollar?
£ s.
9 16
20
196
12
3X12=96 8)235200
12)29400
$24.50 Answer,
6. In 30 pounds, how many dollars and cents?
Ans. $75.00.
7. In 27 pounds, 2 shilling, how many doilars and
cents? " Ans. $67.75.
8. fn 942 pounds of New Jersey. Pennsylvania, Dela-
ware, and Maryland currency ; how many dollars and
Qents,.a dollar being 90 pence?
•*" 942
20
18840
12
9|0)2260800jO
$2512.00 Answer,
"0. In 12 pounds how many dollars and cents?
Ans. $32.
10. In £86 6s. 5£d. how many dollars, cents and mills?
Ans. $230.19cts. 4m.
11. In 21 pounds, South Carolina and Georgia, curren-
cy, how many dollars and cents, there being 56 pence
in a dollar?
EXCHANGE
£
21
20
420
12
7X8=56) 7)504000
8)720.00
$90.00 Answer.
12. In 56 pounds, how many dollars, &c.? Ans. $240.
13. In 108 pounds, Canada and Nova Scotia currency,
iiow many dollars, &c., there being 60 pence in a dollar?
£
108
20
2160
12
6|0)25920.0|0
$432.00 Answer.
14. In 460 pounds and 16 shillings sterling, or EnglisL
money, how many dollars, &c.3 there being 54 pence ip
a dollar?
£
460 16
20
9216
12
9X6=54 9)110592.00
6)12288.00
$2048.00 Answer,
7
74 EXCHANGE.
15. Reduce 16 pounds, 6 shillings, and 3 pence, English
money, to dollars and cents.
£ 9. d.
16 6 3
20
326
12
9)3915.00
6)435.00
$72.50 Answer.
To bring Federal Money into pounds, shillings and pence,^
RULE.
Multiply the dollars, or dollars and cents, by the num-
ber of pence in a dollar of the currency to which you
wish to change the given sum; — the answer will be in
pence, which can then be reduced to shillings and pounds.
When there are cents in the sum to be reduced, two fi-
gures must be cut off from the right of the product, be-
fore bringing it into pounds.
Note. — This rule applies to the currency of any coun-
try whose currency is in pounds, shillings, &c.
EXAMPLES.
1. In $16.50 how many pounds and shillings, in sterr
ling, or English Money, a dollar being four shillings and
six pence, or 54 pence?
$16.50
9X6=54 9
12)891.00
2|0)7|4.-}-3
£3 14s, 3d.
EXCHANGE. "75
-- 2. In 33 dollars, how many pounds, &c.'?
$33
9
297
6
12)1782
2|0)14|8— 6d.
£7. 8s. 6d, Answer.
3. In U)00000 dollars, how many pounds sterling?
$ 1000000
9
9000000
6"
12)54000000
2|0)450000|0
£225.000 Answer,*
4. Reduce 432 dollars into the currency of Canada
and Nova Scotia, a dollar being equal to five shillings,
or 60 pence.
$432
60
12)25920
3|0)216|0
£108 Answejr.
5. In $490,50 how many pounds, shillings, &c.?
Ans. £122 12s. 6d^
* Federal Money may, also, be changed into English Money, by multiplying the
dollars by 9, and dividing the product by 40.
T6> VULGAR FRACTIONS*
6. Bring $150.25, into the currency of New England,
Virginia, and Kentucky, a dollar, being equal to 7£
pence.
$150.25
9
9X8=72 135225
8
12)10818|00
2|0)90l I— 6
£45. Is. 6d. Answer.
Q-. 1. What does Exchange teach?
2. How do you reduce the currency of any one of
the United States to Federal Money?
3. Does this rule apply to the currency of any other
country ?
4. How do you change Federal Money into pounds,
shillings, and pence of any state, or country?
6. Among the various kinds of money, what kind is
the most easily reckoned £
VULGAR FRACTIONS.
^-^ Fractions are broken numbers, expressing any assigna-
ble part of an unit, or whole number. They are repre-
sented by two numbers placed one above another, with
aline drawn between them; thus, f, f, &c. signifying
4wo fifths and five eights.
The figure above the line is called the numerator,
and that below the line, the denominator. The denomi-
nator shows into how many equ.il parts the whole quan-
tity is divided, and represents the divisor in division. —
The numerator shows how many of those parts are ex-
pressed by the fraction; being the rfr», .i^c'er after di-
vision. Both these numbers are sometimes calied the
terms of the fraction.
VULGAR FRACTIONS. !fl
Questions to prepare the learner for this rule.
1. If a pear be cut into two equal parts, what is one
of those parts called? Ans. a half.
2. If you cut a pear into three equal parts, what is-
one of these parts called? Ans. one third,
3. How many thirds of any thing make the whole?
4. If a pear be cut into four equal parts, what is one
of those parts called? Ans. one fourth. What are two
of the parts called ? Ans. two fourths. What are three
of them called? Ans. three fourths.
5. How many fourths of a thing make the whole?
6. If an orange be cut into rive equal parts, what i's
one of the parts called? Ans. one fifth. What are two
of the parts called? Ans. two fifths. What are three
of them called? Ans. three fifths. What are four of
them called? Ans. four fifths.
7. How many fifths of a thing make the whole?
8. If you cut a pear into six equal parts, what is one
of the parts called? What are two of them called?- —
What are three of them called ? What are four of them
called?
9. How many thirds are there in one? How many
fourths? If four fourths make the whole, what part are
two fourths? What part of three is one? What part
of four are two? What part of six are two? What
part of eight are two? What part of eight are six? —
What part of 9 are 6? What part of 10 are 2? What
part of 10 are 4? What part of 10 are 7? What part
of 12 are 6? What part of 12 are 4? What part of
12 are 3? What part of 12 are 2?
10. How many are two fourths of 12? How many are
three fourths of 12? Two thirds of 12, are how many?
How many are 5 times 8? In one eighth of 40, tow
many ? In three eights of 40, how many ? Four eights
of 40, are how many ? Then £ of any number, or of
any thing amount to how many, or how much? How
many are f of 30? How many are f of 30? How ma-
ny in £ of 60? In £ of 60, how many? In J of 60,
78 VULGAR FRACTIONS.
how many? In ^ ->f 60, how many? How many are
^ of 60? Hew many are £ of 60? How many are |
of 60? In 2 and }, how many fifths? In 5 and | how
rtiany fifths? In 1 of 100 how many? In £ of 100
cents, or 1 dollar, how much? How much are f and i?
How much are J and I? How much are f and -J? How
much are f ? How much are ^ and |? How much are
f, J, and i? In ^ how many? In f, how many? If
you take £ from one dollar, how much will remain? If
you take J from one dollar, how much will remain? If
you take J from a pound, how much will remain? If
you take £ from one, kow much will remain? How
many fourths are 2 times £? How many are 5 times f ?
How many are 3 times f ? In \-, how many? In J^,
how many?
11. If one half, three fourths, and a quarter, be added,
how much will be their amount?
12. If you take two eights from eleven eights, how
much will remain?
13. What is a proper fraction?
Ans. When the numerator is less than the deaomina^
tpr, as i, or | , &c.
14. What is an improper fraction?
Ans. It is that in which the numerator is equal, or
superior to the denominator; as f , or f , or J, &c.
45. What is a simple fraction?
Ans. It is a fraction expressed in a simple form; as,
*» 4, f -
16. What is a compound fraction?
Ans. It is the fraction of a fraction, or several frac-
tions connected together with the word of between
them; as |, of |, of |, or f of ^ &G., which are read
thus, one half of two thirds. &e.
17. What is a mixed number?
Ans. It is composed of a whole number and a frac-
tion; as 3J. or 12£.
1 8. What is the commen measure of two or more num-
bers?
Ans. It is that number which will divide each of them
without a remainder; thus, 5 is the common measure of
DEDUCTION OF VULGAR FRACTIONS. 79
10. 20, and 30, and the greatest number that will do this,
is called the greatest common measure.
19. What is meant by the common multiple?
Ans. Any number which can be measured by two or
more numbers, is called the common multiple of those
numbers; and if it be the least number that can be so
measured, it is called the least common multiple; thus,
40, 60, 80, 100, are multiples of 4 and 5; but their
least common multiple is 20.
20. When is a fraction said to be in its lowest terms?
Ans. When it is expressed by the smallest numbers
possible.
21. What is meant by a prime number?
Ans. It is a number which can only be measured by
itself, or, an unit.
22. What is meant by a composite number?
Ans. That number, which is produced by multiplyingr
several numbers together, is called a composite number.
23. What is a perfect number?
Ans. A perfect number is one that is equal to the sum
of its aliquot parts.*
REDUCTION OF VULGAR FRACTIONS.
Reduction of Vulgar Fractions, is the bringing of
them out of one form into another, in order to prepare
them for Addition, Subtraction, Multiplication, &c.
CASE I.
To reduce a fraction to its lowest terms.
RULE.
Divide the greater term by the less, and that divisor
by the remainder, and so continue till nothing be left;;
* The following perfect numbers are all which are, at present, known1,
6 8589869056
28 137438691328
496 2305843008139952128
8128 24178521639228158837784576
33550336 9903520314282971830448816128
80 REDUCTION OP VtfLGAB FRACTIONS.
the last divisor will be the common measure; then divide
both parts of the fraction by the common measure, and
the quotients will express the fraction required,
c Note. — If the common measure happen to be 1, the
fraction is already in its lowest term. Cyphers, on the
right hand side of both terms, may be rejected ; as ||£ ? •
FXAMPLES.
1. Reduce if 4. to its lowest terms.
144)240(1 48)144(3
144 144
96)144(1 I Ans,
96 48)240(5
Greatest common — 240
measure 48)96(2
96
Thus 48 is the greatest common measure, and the
true answer is obtained by dividing the fraction by it.
This reduction may be performed, also, by another
rule, thus: — Divide the numerator and denominator of
the fraction by any number that will divide them both
without a remainder; divide the quotients in the same
manner, and so on, 'till no ir.tmber will divide them both?
and the last quotients will express the fraction in its
lowest terms.
The same sum done by this method : —
12)m(i* ' 4>2L!(f Answer.
2. Reduce -f^ to its lowest terms. Ans. J.
3. Reduce iff to its lowest terms. Ans. -i.
4. Reduce |f | to its lowest term?. Ans. T^.
5. Reduce }Jf£ to its lowest terms. Ans. }.
CASE II.
To reduce a mixed number to an improper fraction.
RULE.
Multiply the whole number by the denominator of
the fraction, and add the numerator to the product; then
set that sum, namely, the whole product, above the de-
nominator for the fraction required.
REDUCTION OF VULGAR TRACTIONS. &1
EXAMPLES.
1. Reduce 23f to an improper fraction.
5
117 ' y Answer.
2. Reduce 12 J to an improper fraction. Ans. 1%S,
3. Reduce 14 T7¥ to an improper fraction. Ans. l£?.
4. Reduce 1 03 -/j- to an improper fraction. Ans. 3|f 8.
CASE III.
To reduce an improper fraction to a whole or mixed num-
ber.
RULE.
Divide the numerator by the denominator, and the
quotient will be the whole or mixed number sought.
EXAMPLES.
1. Reduce ^ to its equivalent number.
3)12(4 Answer.
12
2. Reduce V5 to its equivalent number.
7)15(21 Answer.
14
1
3. Reduce 7T4T9 to its equivalent number. Ans. 44 y1^
4. Reduce 5T6 to its equivalent fraction. Ans. 8.
5. Reduce 'f| 2 to its equivalent fraction. Ans. 54 \\.
6. Reduce 9Ty to its equivalent number. Ans. 171 T|.
CASE IV.
To reduce a whole number to an equivalent fraction, having
a given denominator.
RULE.
Multiply the whole number by the given denomina-
tor; then set the product above for a numerator, and
the given denominator below, and they will form the
fraction required.
EXAMPLES.
1. Reduce 9 to a fraction whose denominator shall
be 7. 9X7=63, then 6T3 is the answer,
2 Reduce 13 to a fraction whose denominator shall
be 12. Ans, ^ .
82 REDUCTION OF VULGAR FRACTION^.
3. Reduce 27 to a fraction whose denominator shaU
be 11. Ans. Yy7-
CASE V.
To reduce a compound fraction to a simple one.
RULE.
Multiply all the numerators together for a new nu
merator, and all the denominators for a new denomina*
tor? then reduce the fraction to its lowest term.
EXAMPLES.
1. Reduce \ of f of f to a single or simple fraction-.
1X2X3 6 1
.. ._ __ _. — _. _ Answer.
2X3X4 24 4
2. Reduce | of f of \\ to a single fraction. Ans. T4T.
3. Reduce ^ of £ to a single fraction. Ans. ||.
4. Reduce f of A Of \\ to a simple fraction. Ans. ££$.
CASE VI.
To reduce fractions of different denominations to others
of the same value, and having a common denominator.
RULE.
Multiply each numerator into all the denominators ex-'
eept its own, for a new numerator, and all the denomi^
nators into each other for a common denominator.*
EXAMPLES.
1. Reduce i, f, and f to a common denominator.
1X3X4 = 12 the numerator for |.
2X2X4=16 the numerator for f.
3X2X3 = 18 the numerator for f.
*The least common denominator, or multiple, of two or more numbers, may be
found thus: Divide the given denominators by any number that will divide two or
more of them without a remainder, and set the quotients and undivided numbers
underneath. Divide these quotients by any number that will divide two or more
of them as before, and thus continue, 'till no two numbers are left, capable of be-
ing lessened. Then multiply the last quotients, and the divisor, or divisors together,
tiii'l 'li" product will be the answer.
What is the least common multiple of f, J, TGT, and T^?
8)9 8 15 16
3)9 1 15 2
3152
3X1X5X2=30X3X8=720, Ans.
REDUCTION" OF VULGAR TRACTIONS. 83
Deneminator 2X3X4=24 the common denominator.
Therefore, the results are ij, |f and i|.
Or the multiplications may be performed mentally,
and the results given -'-, |, & =i|, if, i|.
2. Reduce f and f to a common denominator.
Ans. |f and f f .
3. Reduce f , f , and J to a common denominator.
Ans. |£, |f, |f.
4. Reduce £, f and J to fractions of a common denomi-
nator. Ans ,yL, Jfo and }££,
CASE VII.
To reduce a fraction of one denomination to the fraction
of another, but greater, retaining the same value.
RULE.
Make the fraction a compound one, by comparing it
with all the denominations between it and that denomi-
nation to which you would reduce it; then reduce that
compound fraction to a simple one.
EXAMPLES.
1. Reduce 4 of a cent to the fraction of a dollar. By
comparing it, it becomes 4 °f yV °f TO* which being re^
duced by case five, will be 4X1X1=4 and 7X10X10
=700. Ans. ^ D.
2. Reduce | of a mill to the fraction of an eagle.
Ans. jof o-o-
3. Reduce f of a penny to the fraction of a pound.
3X1 X 1 = 3 1
|0fTVof¥V - — — =£ Ans'
5X12X20=1200 400.
4. Reduce -| of an ounce to the fraction of a pound,
Avoirdupois Weight. Ans. JT Ib,
5. Reduce | of a dwt. .to the fraction of a pound,
Troy Weight. Ans. T/^ Ib,
6. Reduce }| of a minute to the fraction of a day.
Ans. TTV? day.
CASE VIII.
To reduce the fraction of one denomination to the fraction
of another, but less, retaining the same value.
84 REDUCTION OF VULGAR FRACTieNS.
RULE.
Multiply the given numerator by the parts in the de-
nomination between it and that to which you would re-
duce it, and place the product over the given denomi-
nator.
EXAMPLES.
1. Reduce ,^j of a dollar to the fraction of a cent.
The fraction is ,^ of V° of V; then,
J_X10XU)_100 and this reduced, is equal to
175X ] X J ~~175 Ans. 4 c.
2. Reduce T^£ 77 of an eagle to the fraction of a mill
Ans. |.
3. Reduce ^i^- of a pound to the fraction of a penny.
An?, f .
4. Reduce ^ of a pound Avoirdupois, to the fraction
of an ounce. Ans. £.
5. Reduce l¥7^ of a pound Troy, to the fraction of a
pennyweight. Ans. £ dwt.
6. Reduce , JJ¥ of a day to the fraction of a minute.
Ans. |£ of a min.
CASE IX.
To find the value of the fraction in the known parts of the
integer; or, to reduce a fraction to its proper value.
RULE.
Multiply the numerator by the known parts of the in-
teger, end divide by the denominator.
EXAMPLES.
1. What is the value of | of a pound?
2 thirds of a pound.
20
3)40 thirds of a shilling.
13s.+l third of a shilling.
12
3)12 thirds of a penny.
4d. An8. 13s. 4d.
REDUCTION OF VULGAR FRACTIONS. 85
5. Reduce f of a shilling to its proper value
2 fifths of a shilling.
12
5)24(4d.
20
4 fifths of a penny.
4
6)16 fifths of a farthing.
3 qr.-f 1 fifth. Ans. 4d. 3qr. £.
3. Reduce | of a Ib. Troy, to its proper quantity.
Ans. 7 oz. 4dwt
4. Reduce | of a mile to its proper quantity.
Ans. 6 fur. 16 poles.
5. Reduce T5^ of a cwt. to its proper quantity.
Ans. 2 qrs.
6. Reduce f 0f an acre to its proper value.
Ans. 2R. 20P.
7. Reduce fV of a day to its proper value.
Ans. 7 hours 12 min,
CASE X.
To reduce any given quantity to the fraction of a greater
denomination of the same kind.
RULE.
Reduce the given quantity to the lowest denomina-
tion mentioned for a numerator, and the integer into the
same denomination for a denominator.
EXAMPLES.
1. Reduce 16s. 8d. to the fraction of a pound;
16 8 Integer £1
12 20
Numerator 200 20
= £ Ans. !«--
Deno«unator240
— r 240 Denominator^
8
ot> ADDITION OF VULGAR FRACTIONS.
2. Reduce 6 furlongs and 16 poles to the fraction ot
a mile. Ans. £ .
3. Reduce f of a farthing to the fraction of a pound.
Ans. TT'¥Tr.
4. Reduce £ dwt. to the fraction of a pound Troy.
Ans. ^.
5. Bring 80 cents to the fraction of a dollar.
A dollar is 100 cents, then 80 cents are equal to T8/ff
of a dollar; which, being reduced, is equal to f Ans.
6. Bring 16 cents 9 mills to the fraction of an eagle,
16 cents 9 mills = 169
1 eagle = 10000
7. Bring 2 quarters 3£ nails to the fraction of an ejl<
English.*
2 quarters 3i nails.
4
11
•9
Numerator 100
Denominator 9 of -J of J = ||£ = A Ans.
ADDITION OF VULGAR TRACTIONS.
CASE I.
To add fractions having a common denominator.
PtULE.
Add all the numerators together, and place the sum
over the common denominator, which will give the sum
of the fractions required.
EXAMPLES.
1. Add f , f and f together.
!+!+! = *= H Answer.
* When the sum contains a fraction, as in the 7th example, multiply both p.-uts
ot the sum by the denominator thereof, and to the numerator add the nunn r-itoi
of the given fraction.
ADDITION OF VULGAR TRACTIONS. 87
&. Add j, 5, % and -f together.
' Y+f+f+f =' V = H Answer.
CASE II.
7o add fractions having different denominators.
RULE.
Find the common denominator by Case VI, in Reduce
lion; then add, as in the preceding example?.
EXAMPLES.
1. Add | and f together.
4X5= 20)
• numerators.
47 sum.
4X9 = 36 com. denom. f J ±= l£.i Ans.
2. Add f a«d T5T together. Ans. 1§r.
CASE III.
To add mixed numbers.
RULE.
Add the fractions as in Case I, in Addition, and the
whole numbers as in Simple Addition; then add the frac-
tions to the sum of the whole numbers. If the fractions
have different denominators, reduce them to a common
denominator, and then add the fractions to the integers
or whole numbers.
EXAMPLES.
1. Add 13TL, 9T45 and 3T^ together.
13-f9-f-3 = 25 whole numbers.
TV+T45+T7s = H = f Thus> 25f Ans.
2. Add 5|, 6f and 4i together.
5-J-6+4 =15 whole numbers.
Then, 2X8X2 = 32
7X3X2 = 42
1X3X8= 24
98 sum of the numerators.
3X8X2 = 48 common denominator.
Then, f| = 2^. Thus, 15+2^ = 17 JT Answer,
''}. Add If, 2} and 3f together. Ans. 7}$&
38 ADDITION OF VULGAR
CASE IV.
To- add compound fractions.
RULE.
Reduce them to simple ones, and proceed as before
EXAMPLES.
1. Add i of I of £, to I of J of }f
—
2X3X4 = ^^ simple fraction.
2 X3 X 10 —
and 3X5XH = A°5 = T4i simple fraction.
Then find a common denominator.
- ., 4X 4= 16
for J, ^ thus, i x j j _ || numerator.
27 sum of the nuirieratop$.
4XH = 44 common denominator.
Therefore f J is the Answer,
2. Add -f. 9i and | of i together.
jVo^e> — The mixed number of 9J «= 4/ ; the compound
fraction f of -|- •= f. Then the fractions are, f, y and
|; which must be reduced to. fractions of a common de-
nominator and added. Ans.
CASE V.
When the given fractions are of several denominations.
RULE.
Reduce them to their proper values, or quantities,
and add them according to the following examples,
EXAMPLES.
1. Add | of a pound to | of a shilling.
Thus, | of a pound = 13s. 4d.
and f of a shilling = Os. 4d.
13s. 8d. SJqr.
2. Add £ of a pound and ,3- of a shilling together.
Ans. 15s. 10 ,^d.
3. Add i of a week. -]- of a day, and ^ of an hour to-
gether, Ans. 2d.
SUBTRACTION OF VULGAR FRACTIONS. 89
4. Add £ of a yard, J of a foot, and £ of a mile to-
gether. Ans. 1100yds. 2ft. 7in.
5. Add J of a dollar, | of a cent, ,3¥ of a cent, and, J
of a mill together. Ans. 20c. Urn.
6. Add J of pound, ^ of a shilling, and £ of a penny
together. Ans. 2s.
SUBTRACTION OF VULGAR FRACTIONS',
CASE I.
When the fractions have a common denominator.
RULE.
Subtract the less numerator from the greater, and se!
the remainder over the common denominator, which
will show the difference of the given fractions,
EXAMPLES.
1. Subtract f from f. Ans. f
2. What is the difference between f and f . Ans. f ={•,
3. Take ,\ from ^ • Ans. &==£.
4. Take f from f Ans. f=i
CASE II.
When fractions, or mixed numbers, are to be subtracted
from whole numbers.
RULE.
Subtract the numerator from its denominator, and un-
der the remainder place the denominator; then carry
One to be deducted from the whole number.
EXAMPLES.
J. Take | from 12.
Thus, 12
llf Answer,
2. Subtract 27f$ from 32. Ans. 4J$ ,
3. From 10, take TV Ans. 9&\
4. From 9, take 5i. Ans. 3£.
5. From 26, take 24TV Ans. T\= J-,
90 SUBTRACTION OF VULGAR FRACTIONS,
CASE III.
To subtract fractions having different denominators.
RULE,
Reduce the fractions to a common denominator, by
Case VI in Reduction, and subtract the less numerator
from the greater — the difference will be the answer.
EXAMPLES.
1. What is the difference between ij- and ff ?
Thus, -J-a. and f f are equal to |£i, TW,
And 88 from 171, leaves 85. Ans. TW-
2. From ^ take f . Ans. JT.
3. Take JL from f . Ans. |.
4. Subtract T7^ from ^. Ans. ^
CASE IV.
To distinguish the largest of any two fractions.
RULE.
Reduce them to a common denominator, and the one
that has the larger numerator is the larger fraction.
EXAMPLE.
Which is the greater fraction, }J, or ff ?
Thus, 192 common denominator.
12X15 = 180 numerator.
16X11= H6 numerator.
4 numerator.
Then, Tf 2 = TV
Therefore, if is the greater fraction by ?L., Ans
CASE V.
To subtract one mixed number from another, when thefra&
tion to be subtracted is greater than that from which the
subtraction is to be made.
RULE.
Reduce the fractions to a common denominator; sub-
tract the numerator of the greater from the common
denominator, and add to the remainder the less numera-
tor; then set the sura of them over the common denomi-
nator, and carry one to the whole number, and subtract
as'io Simple Subtraction.
MULTIPLICATION OF VULGAR FRACTIONS, 91
EXAMPLES.
1. From 12£ subtract 8}f.
Thus. | reduced to a common denominator, = T5T7?,
and || reduced to a common denominator, = y7^-.
Then 72 taken from 1 14, leaves 42; which, added to
57, the less numerator, makes 99 for the numerator in
the answer. Then carrying 1 to the whole number,
namely, 8 makes it 9; and taking 9 from 12 leaves 3.
Therefore, the answer is
2. From 10^, take 1TV Ans.
CASE VI.
When fractions are of different denominations.
RULE.
Reduce them to their proper values, or quantities,
and subtract as in Compound Subtraction,.
EXAMPLES.
1. From f of a pound, take ^ of a shilling.
Thus, | of a pound = 17s. 6d.
And | of a shilling =0 4
17s. 2d. Answer.
2. From f of a ton take T97 of a cwt.
Ans. 14cwt. Oqr. lllb. 3oz. 3}dr,
3. From J of a pound, take £ of a shilling, and what
will be the remainder? Ans. 14s. 3d.
4. From | of a pound, Troy Weight, take £ of an
ounce. Ans. 8oz. 16dwt. 16gr,
MULTIPLICATION OF VULGAR FRACTIONS.
RULE.
Reduce compound fractions to simple ones, and mixed
numbers to equivalent fractions; then multiply all the
numerators together for a numerator, and all the de-
nominators together for a denominator which will give
the product required,
92 DIVISION OF VULGAR FRAGTIO^
EXAMPLES,
1. Multiply f by f.
Here, f Xf = A = £» the answer.
2. Multiply f by f. Aus. ^.
3. " A by A- Ans.Ty
4. " £ of 7 byf Ans. If.
5. " 6| by f Ans. ||.
6. « 414. by 3f . Ans. 14i|^
7. <{ 41 byf Ans.T%,
DIVISION OF VULGAR FRACTIONS.
RULE.
Reduce compound fractions to simple ones, and mixed
numbers to equivalent fractions; tben multiply the nu-
merator of the dividend by the denominator of the di-
visor, for a new numerator, and the denominator of the
dividend by the numerator of the divisor, for the de-
nominator; the fractions thus formed will be the answer'
EXAMPLES.
1. Divide 4 by f.
Thus, 4 numerator of the dividend.
3 X denominator of the divisor,
12 numerator.
Then 7 denominator of the dividend-,
2 X numerator of the divisor.
14 denominator.
Therefore, }| = & is the answer^
2. Divide f by f .
Thus,
'
A
9 Answer,
3. Divide Jf by f Ans. fv
5. " |- by y. Ans. TV
6. « Hbyf AnM*
7. " f by f Ans. -if
S. « ^^
DECIMAL TRACTIONS. 93
9. Divide f by £. Ans. ^ .
10. " 71 by 9£» Ans. ff .
11. " | of i by -f of 7f . Ans. Tfp
12. What part of 33 JT, is 28J--J. Ans. f .
Q. 1. What are Vulgar Fractions?
2. How are they represented in figures'?
3. What is the upper figure called?-
4. What is the lower figure called?
5. What does the denominator show?
6. What does the numerator show?
7. What are the two numbers of a fraction some**
times called?
DECIMAL FRACTIONS.
Decimal Fractions are parts of whole numbers, and
lire separated from them by a point, thus, 8.5; which is
read, eight and five tenths, or 8 T5^. All the figures on
the left of the point are whole numbers; those on the
tight are fractions. An unit is supposed to be divided
into ten equal parts, and the figure at the right of the
point expresses the number of those parts. Decimals
decrease in a tenfold proportion, as they depart from
the separating point. Thus, .5 is 5 tenths, or one half;
.57 is 57 hundredths; .05 is 5 hundredths; and .005 is B
thousandths. Cyphers placed at the right hand of de-
cimals do not alter their value; thus, .5 or T\; .50 or
To\i -500 or T5oW? are a^ °^ t*16 same value, and equal
to |. The first place of decimals is called tenths; the
second, hundredths, &c.
DECIMAL JVtIMERATION TABLE.
1 1
765 4 321 . 654321
94 ADDITION or DECIMALS.
ADDITION OF DECIMALS.
RULE.
Place the figures according to their values — units un-
der units, tenths under tenths, &c., and add as in Simple
Addition of whole numbers; observing to place the
point in the sum under those in the given numbers.
EXAMPLES.
1. Add together the following sums, viz: 252. 25?
343.5, 17.85, 1244,75 and .425.
Thus, 252.25 Ab?e.—The answer to this sum
343.5 is read thus: One thousand eight
17.85 hundred and fifty-eight, and seven
1244.75 hundred and seventy -five thou'
.425 sandths.
1858.775 Answer,
ii. in.
87654.321 987654.3
23456.78 212345.67
98765.4 898765.432
209876.501 2098765.402
4. Add 420.4, 38.05, 54.9, 27.003 and 29.384.
Ans. 569.737.
5. Add 376.25, 86.125, 6.5, 41.02 and 358.865.
Ans. 868.760.
6. Add .64, .840. .4, .04, .742, .86, .99 and .450.
Ans. 4.962.
Note. — Dimes, cents and mills are decimals of a dol-
lar. A dime is one tenth, a cent is one hundredth, a
mill is one thousandth; which shows that the addition
of Federal Money is the addition of decimals. Thus,
5 tenths of a dollar is the same as 50 hundredths, or 50
cents; and 25 hundredths of a dollar is equal to 25
cents, &c. It may he likewise added, that .5, or .60, or
.500, being equal to one half, .25 equal to one quarter,
and .75 equal to three quarters or three fourths, so .7,
or .35, or any intermediate fractions, have a propor-
tionate value.
SUBTRACTION AND MULTIPLICATION OF DECIMALS.
SUBTRACTION OF DECIMALS.
RULE.
Write the larger number first, and the smaller one
under it; then subtract as in Simple Subtraction; obser-
ving, that the dividing point in the answer, or remain
der, must be placed under those in the sum.
EXAMPLES.
From
take
i.
91.73
2.138
n.
2.73
1.9185
89.592
0.8115
in.
214.81
4.90142
209.90858
IV.
From 1.5
take .987654321
0.512345679
v.
.8234567890
.5987654329
.2246913561
MULTIPLICATION OF DECIMALS.
RULE.
Place the multiplier under the multiplicand, and mul-
tiply as in Simple Multiplication; then point off as many
places for decimals as there are decimals in the multi-
plicand and multiplier. If there be not so many figures
in the product as there are decimals in both factors, the
deficiency must be supplied by prefixing cyphers.
EXAMPLES.
Multiply 24.85
by 6.25
12425
4970
14910
155.3125
228375
3197*5
365400
3996.5625
in.
79.347
23.15
396735
79347
238041
158694
1836.88305
96 DIVISION OF DECIMALS.
IV. V. VI. VII.
Multiply .63478 .567 .285 .25
by .8994 5 .003 .25
253912 2.835 .000855 125
571302 50
571302
507824 .0625
.570921132
£. Multiply .63478 by .8204. Ans. .520773512.
'9. Multiply .385746 by .00464. Ans. .00178986144.
DIVISION OF DECIMALS.
RULE.
Divide as in Simple Division, and point off as many
ugures from the right hand of the quotient, for decimals,
as the decimal figures in the dividend exceed in number
those in the divisor. When there are not so many fi-
gures in the quotient as this rule requires, the deficiency
must be supplied by prefixing cyphers to the left of the
quotient. When there are more decimal figures in the
divisor than in the dividend, place as many cyphers to
the right of the dividend as will make them equal.—
When the number of decimals in the divisor, and the
number in the dividend are equal, the quotient will al-
ways be in whole numbers, unless there should be a re-
mainder after the dividend is all brought down. When
there is a remainder, cyphers must be annexed to it
and the division continued and the quotient thence ari-
sing will be decimals.
^DIVISION OP DECIMALS.
EXAMPLES.
II.
6.4)128.64(20.1 324.8)9876.5(30.4079
128 9744
64
64
IV.
,5).75(1.5
5
25«.
25
13250 cypher
12992 annexed.
25800
22736
30640
29232
1408+
in.
.48)65.88(137
48
178
144
348
336
12rem
v.
VI.
179).48624097(.0027I643
358
1282
1253
.2685)27.0000(100.55865
26.85
15000 cyphers
13425 annexed.
294
179
1150
1074
35750
13425
23250
21480
769
716
537
537
7. Divide 234.70525 by 64.25.
8. Divide 14 by .7854.
8. Divide 2175.68 by 100.
9
17700
16110
15900
13425
2475
Ans. 3,653
Ans. 17.825.
Ans. 21.7568.
98 REDUCTION OF DECIMALS*
REDUCTION OF DECIMALS.
CASE I.
To reduce a vulgar fraction to a decimal.
RULE.
Place cyphers to the right of the numerator until you
can divide it by the denominator; and divide 'till noth-
ing remains; or, if it be a number that will not divide
without a remainder, then divide until you get three or
more figures for the quotient. The quotient will be thf,
vulgar fraction expressed in decimals.
EXAMPLE^.
1. Reduce J to a decimal.
Thus, 2)1.0(.5
10
2. Reduce \ and | to decimals.
4)1.00(.24 4)S.OO(.75
8 28
20 20
20 20
3. Reduce £ to a decimal.
3)1.0Q(.333 Answer,
10
9
10
9
4. Reduce £ to a decimal. Ans. .376.
CASE II.
To reduce any sum, or quantity, to the decimal of any
given denomination.
REDUCTION OP DECIMALS. 9§
RULE.
Reduce the quantity to the lowest denomination, and
reduce the proposed integer to the same denomination;
then divide the quantity by the amount of the integer,
and the quotient will be the answer.
EXAMPLES.
1. Reduce 3s. 9d. to the fraction of a pound.
One pound reduced to pence makes 240; and 3s. 9dr
reduced to pence makes 45.
Then, 240)45.0000(.1875 Answer:
240
2100
1920
1800
1680
1200
1200
The same sum may be done by writing the given num-
bers from the least to the greatest in a perpendicular
column, and dividing each of them by such number as
tvill reduce it to the next denomination, annexing the
quotient to the succeeding number.
Thus—
12 9.00
3.750|0
.1875 Answe3r.
2. Reduce 7 drams to the decimal of a pound, Avoir-
dupois Weight. Ans. .02734375.
3. Reduce 14 minutes to the decimal of a day.
Ans. .009722.
4. Reduce 21 pints to the decimal of a peck.
Ans. .013125.
5. Reduce 15s. 6d. to the decimal of a pound.
Ans. .775.
6. Reduce 56 gallons 3 quarts 1 pint to the decimal
of a hogshead. Ans. .9027:
IOQ REDUCTION OF DECIMALS.
7. Reduce 12dwts. 16grs. to the decimal of a poand>,
Troy Weight. Ans, .0527-
8. Reduce 4 mills to the decimal of a dollar Ans, .004.
9. Reduce 7 cents to the fraction of a dollar. Ans. .07=
Note. — In doing sums in this rule, it will be necessary
to keep in mind the tables of the different weights,,
measures, money, &c.
CASE III..
To find the value of any decimal fraction.
RULE.
Multiply the decimal by the number of parts in the
next lower denomination; point off as many figures for
decimals as is required by the rule in nnulti plication of
decimals; then multiply the decimal by the number of
parts in the next lower denomination, and so on, to the
last. The figures on the left of the points will show the
value of the decimal in the different denominations.
EXAMPLES.
L What is the value of .775 of a pound?
£.776
20
5.15.500
12
d.6.000 Answer 15s. 6d.
, What is the value of .625 of a cwt.i
4
4000
1000
14.000 Ans. 2qr. 14ib.
3. What is the value of .625 of a shilling? Ans. 7-J-d
4. What is the value of .4694 of a pound, Troy
Weight? Ans, 5oz, 12dwts, I5,744gr?,
DUODECIMALS. 101
5. What is the value of .6875 of a yard ? Ans. 2qrs. 3na,
6. What is the value of .3375 of an acre ? Ans. 1 R . 1 4P.
7. What is the value of .0008 of an Eagle? Ans. 8m,
Q. 1. What are decimal Fractions?
2. How are they separated from whole numbers?
3. In what manner do they decrease as they depart
from the separating point?
4. In the table of numeration, what is the first place
called?
5. What money, or currency, is reckoned after the
manner of Decimal fractions?
DUODECIMALS.
Duodecimals are fractions of a foot or of an inch, or
parts of an inch, and have 12 for their denominator. —
They are useful in measuring' planes, or surfaces, and
solids. In adding, subtracting, and multiplying by Duo-
decimals, it is necessary to carry one for twelve.
The denominations are feet, inches, seconds, third?
and fourths.
12 fourths "" make 1 third
12 thirds - 1 second ".
12 seconds - 1 inch /.
12 inches - 1 foot Ft,
ADDITION OF DUODECIMALS,
RULE.
$5- Add as ip Compound Addition,
I. EXAMPLES. II.
Ft. I. " '" Ft. I. " '" "h
24 10 11 10 ' 80 11 %10 10 11
18 9 8 3 25 4 3 2 1
12 10 1 7 75 10 11 11 10
56 6 9 8 182 3 2 0 10
9*
102 DUODECIMALS.
SUBTRACTION OF DUODECIMAL^
RULE.
(ffir Subtract as in Compound Subtraction.
EXAMPLES.
1. II.
Ft. I. " '" Ft. I. " '" ""
18 9 8 3 80 1 2 4 6
12 10 11 1.0 39 11 10 10 8
5 10 8 5 40 1 3 5 10
MULTIPLICATION OF DUODECIMALS.
RULE.
Set down the different denominations, one under the
other, so that feet stand under feet, inches under inches?
seconds under seconds, &c. Multiply each denomina-
tion in the sum, by the feet in the multiplier, and set the
result of each under its corresponding term, observing
to carry one for every 12 from one denomination to
another. Then multiply the sum by the inches in the
multiplier, and set the result of each term one place
removed to the right of those in the sum; and in like
manner, multiply the sum or multiplicand by seconds^
fhirds, &c., if there be aoy in the multiplier.
Or, instead of multiplying by inches, &,c., take such
parts in the multiplicand, as these are of a foot.
Add the amount of the multiplications together, and
#ieir sum will b* the answer.
EXAMPLES.
i.
Ft. I.
Multiply 4 7
by 6 4
27 6
1 6 4
29 0 4" 66 4 6"
DUODECIMALS*
103
Ft.
Multiply 8
by 4
in.
/. "
4 2
2
///
10
IV.
Ft. I.
11 10
10 9
33
1
4 11
4 8
4
5 8
118 4
8 10
34
7 9 8"" 127 2 6"
JVote. — In doing the third sum, I begin with 4, which
stands under the 8, and multiply the sum, beginning with
the right hand figure which is 10; saying 4 times 10
are 40. In 40, I find there are 3 times 12 and 4 over.
Setting down 4, 1 multiply the next figure, adding three
to it, which makes 11, and thus multiply the whole sum.
Then taking the 2 for the multiplier, I say 2 times 10
are 20. In 20 I find 12 is contained once, and 8 over.
Setting down 8 one place farther to the right, I say 2
times 2 are 4, and one to carry makes 5; and after this
manner multiply all the figures in the sum. Then ad-
ding the two rows of figures together, 1 obtain the an-
swer.
Method of doing the same sum by taking the frac?
tional parts.
Ft. I. " "'
2 inches =
8"" Answef,
In this last example, I multiply the sum by 4, as in the
former case. Then, as 2 inches make i of a foot, I di-
vide the Sjum by 6, which I had multiplied by 4, divi-
ding it after the manner of Compound Division, multi-
plying each remainder by 12, and adding it to the next
lower denomination; and setting the result under the
1
8
4
2
10
4
2
33
4
11
4
1
4
8
5
34
9
7
9
104 DUODECIMALS.
amount of the multiplication. Then I add the two sums
as before,
5. What are the solid contents of a cubick block that
is 4 feet 4 inches in length, 3 feet 8 inches in breadth,
and 2 feet 8 inches in thickness?
Ft. L
4 4
3 8X
13 0
2 10 8
15 10 8
2 8X
31 9 4
10 714
42 4 5" 4'" Answer.
6. What is the product of 12 feet 9 inches, multiplied
by 6 feet 4 inches. Ans. SOFt. 91.
7. What is the product of 3 feet 2 inches 3" multi-
plied by 3 feet 2 inchesS"? Ans. lOFt. II. 11"0'" 9"".
8. What is the price of a marble slab, whose length
is 5 feet 7 inches, and breadth 1 foot 10 inches, at one
dollar per foot? Ans. $ 10.23,
Q. 1. What are Duodecimals?
2. In what are they useful?
3« In adding, subtracting, and multiplying Duodeci-
mals, what d© you observe in carrying from
one denomination to another?
4. What are the denominations used in Duodecimals?
5. Repeat the rule for Multiplication of Duodecimals9.
SINGLE RULE OF THREE* 105
SINGLE RULE OF THREE.
The Rule of Three, which is sometimes called the
Rule of Proportion, teaches how to find a fourth pro-
portional to three numbers given. As it has three terms
given to find a fourth, it is generally called the Rule of
Three.
Questions to prepare the learner for this rule.
1. If 2 apples cost 3 cents, how much will 4 applet
cost at the same rate?
2. If you give 2 cents for 4 nuts, how many cents*
must you give for 8 nuts?
3. If a pound of butter cost 8 cents, how much will
4 pounds cost?
4. A boy has 20 melons to sell, and asks 10 cents for
two, how much will they all come to at the same rate?
5. If 6 men can reap a field of wheat in 4 days, how
long will it take 12 men to reap the same field?
6. If 4 yards of cloth cost 1 dollar, how much will 2
yards cost?
7. How much will a gallon of milk come to, at four
cents a quart?
8. How milch will a bushel of peaches come to, at
25 cents a peck?
9. If 2 cents will buy 3 apples, how many apples will
9 cents buy?
10. If a boy can run 2 miles in one hour, how far can
he run in 4 hour*?
RULE.
Set the term in the third place, which is of the same
kind with that in which the answer is required. Then
determine \vhether the answer ought to be greater or
less than the third term. If the answer ought to be
greater than the third term, set the greater of the other
two numbers on the left for a second or middle term;
and the less number on the left of the second term, for
a first term. If the answer ought to be less than the.
third term, the less of the two other numbers must be
the middle term, and the greater must be the first ternu
106 SINGLE RULE OF THREE.
After thus stating the sum, proceed to do it in the fol-
lowing manner, viz : Reduce the third term to the lowest
denomination mentioned in it. Reduce, likewise, the
first aod second terms to the lowest denomination that
either of them hag. Then multiply the second and
third terms together, and divide their product by the
first term. The quotient thus obtained will be the an*
swer.
It will not be necessary to distinguish between direct
and inverse proportion, because the foregoing rule is
calculated for both.
PROOF.
By reversing the statement.
EXAMPLES.
1. If 3 pounds of sugar cost 2& cents, what will IB
pounds cost at the same rate?
Ibs. Ibs. cts.
Thus, 3 : 18 : : 25
18
200
25
3)450
$1.50 Answer.
2. If 7 pounds of coffee cost 87| cents, what must f
pay for 244 pounds?
Ibs. Ibs. cts.
7 : 244 : : 87
1708
1952
122
7)21350
$30.50 Answer
SINGLE RULE OF THREE* 107
3. If 450 barrels of flour cost $ 1350, what will 8 bar-
rels cost?
bbls. bbls. $ $
Thus— As 450 : 8 : : 1350 : 24, Answer:*
4. If 15 yards of cloth cost £6, what number of
yards may be bought for £125?
£ £ yds. yds.
As 6 : 125 : : 15 : 312| Answer.
5. If 12 men can do a piece of work in 20 days, ip
what time will 18 men do it?
m. m. d. d.
As 18 : 12 : : 2Q : 13| Answer.
6. What will be the cost of 17 tons of lead, at 22£
dollars 66 cents per ton?
T. T. D. cts. D. cts.
As 1 : 17 : : 223.66 : 2802.22 Ans.
7. WTmt will 72 yards of cloth cost at the rate of 9
yards for £5J2s.
~yds. yds. £ s. £ s.
As 9 : 72 : : 5 12 : 44 16 Ans,
8. If 750 men require 22500 rations of bread for a
month, what will a garrison of 1200 require? Ans, 36000.
9. What must be the length of a board that is 9 inches
in width, to make a surface of 144 inches, or a square
foot? Ans. 16 inches.
10. How many yards of a matting 2 feet 6 inches broad,
will cover a floor that is 27 feet long, and 20 broad?
Ans. 72 yards.
11. If a person's annual income be 520 dollars, what
is that per week? Ans. 10 dollars,
12. If a pasture be sufficient for 3000 horses for 18
days, how Jong will it be sufficient for 2000?
H. H. D. D.
As 2000 : 3000 : : 18 : 27 Ans.
13. What must be the length of a piece of land 13|
rods in breadth, to contain one acre?
Ans. 11 rods, 4yds. 2ft. Oifin.
* The sum in the third example is read thus:— As 450 is to 8, so is 1350 to the
answer This is the manner of reading all sums when stated in the Rule of Three.
108 SINGLE RULE OF THREE.
14. If 8 men can build a tower in 12 days, in what
time can 12 build it?
M. M. D. D.
As 12 : 8 : : 12 : 8 Answer.
15. If a piece of land be 5 rods in width, what must
be its breadth to make an acre?
R. R. R. PC.
As 5 : 160 : : 1 : 32 Answer.
16. How much carpeting that is IJ yards in breadth,
will cover a floor 7\ yards in length, and 5 yards in
breadth?
By Decimal Fractions,
yds. yds. yds.
As 1.5 : 5 : : 7.5 : 25 Answer.
17. What will one quart of wine cost at the rate of 12
dollars for 16 gallons?
gals. qts. qt. D. cts.
As 16 or 64 : 1 •. : 12.00 : 18J Answer.
18. If 10 pieces of cloth, each piece containing 42
yards, cost 531 dollars 30 cents, what does it cost per
yard? Ans. $ 1.261.
19. If a hogshead of brandy cost 78 dollars 75 cents,
what must be given for 5 gallons at the same rate?
Ans. $6.25.
20. If a staff 4 feet in length, cast a shade on level
ground, 8 feet in length, what is the height of a tower
whose shade, at the same time, measures 200 feet?
ft. ft. ft. ft.
As 8 : 200 : : 4 : 100 Answer.
21. I lent my friend 350 dollars for 5 months, he pro-
mising to do me the same favour; but when requested,
he could spare only 125 dollars. How long ought 1 to
keep it to balance the favour?
D. D. M. M.
As 125 : 350 : : 5 : 14 Answer.
22. If 7 oxen be worth 10 cows, how many cows will
-21 oxen be worth?
Ox. Ox. C. C. .
As 7 : 21 : : 10 : 30 Answer
SINGLE RULE OF THREE. 109
23. If 48 men can build a fortification in 24 days, how
many men can do the same in 192 days?
D. D. M. M.
As 192 : 24 : : 48 : 6 Answer.
24. A certain piece of work was done by 120 men in 8
months, how many men will it take to do another piece
of work of the same magnitude in 2 months? Ans. 480.
25. A merchant failing in trade, owes 29475 dollars:
he delivers up his property which is worth 21894 dol-
lars 3 cents; how much does this sum pay on the dollar
towards what he owes? Ans. 74cts. 2m.-|~
26. If a tax of 30,000 dollars be laid on a town in
which the ratable property is estimated at 9,000,000 dol-
lars, what will be the tax of one of the citizens whose
ratable estate is reckoned at 750 dollars.
D. D. D. D. cts.
As 9,000,000 : 30,000 : : 750 : 2 . 50 Ans*
27. If property rated at $28, pay a tax of $21, how
much is that on the dollar?
D. D. D. cts.
As 28 : 21 : : 1 : 75 Answer.
28. How far are the inhabitants on the equator carried
in a minute, allowing the earth to make one revolution
in 24 hours; and allowing a degree to contain 691 miles?
The earth being divided into 360 degrees, allowing
69^ miles to a degree, makes the distance round it to be
25020 miles; — the number of minutes in 24 hours is
1440: then,
min. min. miles, miles, fur.
As 1440 : 1 : : 25020 : 17 . 3 Answer.
29. There is a cistern having 4 spouts; the first will
empty it in 15 minutes, the second in 30 minutes, the
* In making taxes in a due proportion, according to the value of each man's ra-
table estate, proceed in the following manner. Make the amount of ratable pro
perty the first term; make the sum to be raised the second term; and one dollar the
third term; and the number arising from this operation will be the amount to be
raised on the dollar. From this, make a tax table from one dollar to 30, or any
amount necessary. In the same manner find what is to be paid on a cent of rata-
ble estate; and from this, mako a table from 1 to 99 cents; then, from these tableg,
take each man's tax. Thus, if the tax were 75 cents on the dollar, and you would
know what a portion of property pays, that is rated at $28.80, the tables will show
the amount to be $21, for the dollars, and 60 cts., for the cents. In estimating pro-
perty for making taxes, it is customary to rate it much lower tha» its rea! value.
10
i'10 SINGLE RULE OF THREE.
third in 45 minutes, and the fourth in 60 minutes: in
what time would the cistern be emptied, if they were
all running together?
As 15 : 1 : 90 : 6
30 : 1
45 ; 1
60 : 1
90 : 3
90 : 2
90 : li
12J
cisterns, cist. min. min.sec.
Then, decimally, as 12.5 : 1 : : 90 : 7 . 12 Ans.
30. If a ship's company of 15 persons have a quanti-
ty of bread, sufficient to afford to each one 8 ounces per
day, during a voyage at sea, what ought to be their al-
lowance, under the same circumstances, if 5 persons be
added to their number. Ans. 6 ounces.
Note. — As the Rule of Three in Vulgar and Decimal
Fractions require the same statements as in whole num-
bers, and is performed by multiplication and division
after the sanae manner of other sums in the Rule of
Three, it is deemed unnecessary to give any examples.
When the pupil understands Fractions and the Rule of
Three, he will find no difficulty with the Rule of Three
in Fractions.
Ct- 1. What is the Rule of Three sometimes called?
2. What does it teach?
3. Which of the terms must be set in the third place?
4. How do you ascertain which ought to be the first
term, and* which the second?
5. If the third term consist of different denomina-
tions, what do you do with them?
6. What do you do if the first and second terms are
of different denominations?
7. After stating the sum, and reducing, when neces
sary, the terms to similar denominations, how
do you proceed to do the sum?
3. How are sums in the Single Rule of Three proved •>
DOUBLE RULE OF THREE* 1H
DOUBLE RULE OF THREE.
The Double Rule of Three is that in which five or
more terms are given to find another term sought.
RULE.
Set the term which is of the same denomination as
the term sought, in the third place; then consider each
pair of similar terms separately, and this third one, as
making the terms of a statement in the Single Rule of
Three, setting the similar terms in the first or second pla-
ces, according to the rule of the Single Rule of Three.
After stating the question in this manner, and reducing,
if necessary, the similar terms to similar denominations,
then multiply the terms in the second and third places
together for a dividend, and the terms in the first place
together for a divisor — the quotient, after dividing, will
be the term sought.
Sums in this rule may also be done by two or more
statements in the Single Rule of Three.
PROOF.
By inverting the statement, or, more easily, by two
statements in the Single Rule of Three.
EXAMPLES.
1. If 8 men, in 16 days, can earn 96 dollars, how rritich
can 12 men earn in 26 days?
men 8 : 12 : : ) ^
days 16 : 26 : : \ ™
128 312
96X
1872
2808
D.
128)29952(234 Anfewer.
256
435
384
512
512
112 DOUBLE RULE OF THREE*
F $100 gain $6 in 12 months w
Miths?
months 12 : 9 : : \ $400
2. If $100 gain $6 in 12 months what will $400 em
in 9 months?
1200 54
400 X
12|00)216[00
Answer.
3. If 16 men can dig a trench 54 yards in length in
o days, how many men will be necessary to complete
one, 135 yards in length in 8 days?
By two statements in the Single Rule of Three*
yds. yds. men. men.
As 54 : '135 : : 16 : 40.
days. ds. men. men.
Then, as 8 : 6 : : 40 : 30 Answer.
4. If $100 in one year gain $5 interest, what will be
*he interest of $750 for 7 years? Ans. $262.50.
5. If 9 persons expend $120 in 8 months, how much
will 24 persons spend in 16 months at the same rate?
AHS. $640.
6. If 54 dollars be the wages of 8 men for 14 days,
what must be the wages of 28 men for 20 days at the
same rate? Ans. $270,
7. If a horse travel 130 miles in 3 days, when the
days are 12 hours in length, in how many days of 10
hours each can he travel 360 miles? Ans. 9|| days.
8. If 60 bushels of corn can serve 7 horses 28 days3
how many days will 47 bushels serve 6 horses?
Ans. 51?8j days.
9. If a barrel of beer serve 7 persons for 12 days,
how many barrels will be sufficient for 14 persons for a
year, or 365 days? Ans. 60 J barrels.
10. If 8 men spend 32 dollars in 13 weeks, what will
24 men spend in 52 weeks?
* When a sum in the Double Rule of Throe appears difficult to be stated for owe
operation, it may uiways be done with ease by two statements in the Single Rwle
•>f Tljp'e, us in the above example.
PRACTICE. 113
By two statements in the Single Rule of Tliree.
men. men. D. D.
As 8 : 24 : : 32 : 96
weeks, weeks. D. D.
Then, as 13 : 52 : : 96 : 384 Answer.
Q. 1. How many terms are generally given in the Dou-
ble Rule of Three?
2. Which of the terms must be set in the third place?
3. How do you ascertain which of the other terms
should be placed in the first, and which in the
secend place?
4. Which of the terms do you multiply together for
a dividend?
5. How do you form a divisor?
6. How do you proceed when the terms consist of
different denominations?
7. By what other method may sums be done in the
Double Rule of Three, besides the one first
given?
8. How is a sum in the Double Rule of Three proved?
PRACTICE.
Practice is a short method of doing all sums in the
Single Rule of Three, that have one for their first term,
and is of great use among merchants.
It may be proved by Compound Multiplication, or by
the Single Rule of Three.
Questions to prepare the learner for this rule.
1. What will 50 yards of tape cost at £ of a cent per
yard?
2. What will 40 pounds of beef come to at } of a
eent per pound?
3. What will 100 figs come to at f of a cent a piece?
4. How many pence will 40 peaches come to at one
farthing a piece?
5. How many shillings and pence will 50 peaches come
to at one farthing a piece?
10*
114
PRACTICE.
PRACTICE TABLE, OR TABLE OF ALIQUOT
cts.
dols. s.
d.
£- d.
^.
50
= i
10
0
2"
1 _.-
JL
25
o
***
6
8
o
li
V
20
5"
P0
'5
0
I
2
121
•1
Q
4
0
A
P
3
1
4
4
rV
P
3
4
6"
. O
Q
4
5
iV
2
6
-I
D
6
JL
4
AJ
1
8
TV
P*
qrs.
/^.
m.
cis.
1
0
A J
2 or
56
5
i ) 2,
0r.
rf.
1
28
2
i «»
2
* I d
16
1
rV) a
1
i 1
14
8
7
cwt.
TV
CASE I.
« price o/" owe yard, pound, fyc. is in farthings,
RULE.
Divide by the aliquot parts of a penny, and the an-
swer will be in pence, which reduce to shillings, pounds,
&c.
EXAMPLES.
1. What is the value of 2. What is the value of
380 at one farthing each? 744 at 3 farthings each?
380 2 1 i 1 744
12)95 pence. 1
7s. lid. Ans.
3. What is the value of
460 at 2 farthings each?
460
12)230 pence.
19s. 2d. Answer
372
186
12)558
20)46—6
£2 6s. 6d. Am*
>ftACTlCE.
4. What is the worth of 298 at id.? Ans. 6s. 2-£d.
5. What is the worth of 586 at id.? Ans. £1 4s. 5d,
6. What is the worth of 964 at Jd.? Ans. £3 Os. 3d,
CASE II.
When the price is any number of pence less than 12.
RULE.
Divide by the aliquot parts of a shilling, and the an-
swer will be in shillings, which may be reduced to
pounds.
EXAMPLES.
1
i.
TV
II.
672 at Id.
20)56
£2 16s. Ans.
444 at
20)74
£3 14s. Ans.
3. What is the value of 237 at 3d.?
4. What is the value of 594 at 4d,?
5. What is the value of 868 at 6d.?
6. What is the value of 988 at 5d.?
£ s. d.
Ans. 2 19 3
Ans. 9 18 0
Ans. 21 14 0
Ans. 20 1 1 8
7, What is the value of 1049 at 8d.? Ans. 34 19 4
8. What is the value of 1294 at 10d.? Ans. 53 18 4
4d.
Id.
988
at 5d.
20)411
£20 11s. 8d. Answer.
Note. — In this last example, as 5d is not an aliquot
part of a shilling, I take 4 pence, which is one third of
a shilling, and after dividing by that, I take one penny,
which is one fourth of four, and divide the first product
by it. Then, adding them together, and reducing them
to pounds, &c, I obtain the answer,
116
PRACTICE.
CASE III.
When the price in pence exceeds the number of 12.
RULE.
Consider the number given in the sum as containing
so many shillings. Then divide by such aliquot parts
as may be formed by the pence over a shilling, adding
the product to the sum. The answer will be in shillings,
EXAMPLES.
600 at 13|d.
75
20
675
£33 ISs. Answer.
Note. — In this example, I consider the sum as 600
shillings. Then, as the given price is lid. over a shil-
ling, which makes | of a shilling, I divide the sum by 8,
and add the quotient to the given sum j which makes 675
shillings, or £33 15s.
£ s. d.
2. What is the worth of 450 at 14d.? Ans. 26 5 0
3. What is the worth of 570 at 16d.? Ans. 38 0 0
CASE IV.
When the price is any number of shillings under 20.
RULE.
Divide by the aliquot parts of a pound, and the an-
swer will be in pounds. Or, consider the sum as being
so many shillings, then multiply the sum by the number
of shillings in the price. The product will be the an-
swer in shillings; which reduce to pounds.
EXAMPLES.
5s.
1296 at 5s.
£324 Ans.
n.
723 at 12s.
12X
20)8676
£433 16s. Answer.
PRACTICE*
117
The second example is done by the second method,
which is thought by many to be the easier way.
£
3. What is the value of 1128 at 3s.? Ans. 169
4. What is the value of 889 at 4s.? Ans. 177
5. What is the value of 1616 at 9s.? Ans. 727
s.
4
16
4
S. What is the value of 2868 at 18s.? Ans. 2581 4
CASE V.
When the price is in pounds, shillings and pence.
RULE.
Multiply the gum or quantity by the number of pounds
in the price, then divide by the aliquot parts of shillings
and pence, and add the quotients to the product — their
sum will be the answer.
EXAMPLES.
I. II.
10
448 at £41 Os. 6d.
4
4
i
5
5678 at £7
4s. 9d,
7
39746
6
1_
1135 12
3
JL
2
141 19
70 19
6
1792
224
11 4
£2027 4s. Ans.
£41094 10s. 6d.
Note. — In the second example, after multiplying the
sum by the number of pounds, as 4s. is i of a pound, I
divide by 5, which gives 1135. pounds in the quotient;
and leaving a remainder of 3 pounds, which reduced to
shillings and divided by 5 give 12s. Then, as 6 and 3
make 9, the number of pence, and as 6d. is | of 4s., I
divide the quotient by 8, which gives 141 pounds with
a remainder of 7; this being reduced to shillings, and
the 12 shillings above added to it make 152, which di-
vided still by the 8 give 19 shillings. And as 3 is \ of
6, or its aliquot part, I divide the last quotient by 2. —
This Drives 70 pounds and a remainder of one, which is
20 shillings; and adding it with 19 shillings above, the
amount is 39 shillings. This divided by the 2 gives 19
118
PRACTICE.
shillings and a remainder of 1 shilling, or 12 pen<ff. '
which, divided still by the 2, makes 6d. And thus the
answer is obtained.
3. What is the amount of 288 at £5 10s. 4d.?
Ans. £1588 16s,
4. What is the amount of 642 at £9 4s. 6d.?
Ans. £5922 9s.
5. What is the amount of 734 at £12 2s. 8d.?
Ans. £8905 17s. 4cl
CASE VI.
IVken the quantity consists of different denominations, and
the price is in pounds? shillings^ <$'C.
RULE.
Multiply the price of the highest denomination given,
by the whole of the highest denomination, then divide
by aliquot parts of each of the lower denominations in
the sum. Add the results together, and their sum will
be the answer.
EXAMPLES.
£
at 4
6
2 per cwt
i
S
12
18
6
1
2
3
1
10
9*
3cwt. 2qrs. 14 Ibs.
2 qre. are £ of a cwt.
14 Ibs. are £ of 2 qrs.
£15 12s. 4£d. Ans.
2. 4 cwt. 3 qrs. 12 Ibs. at £8 4s. 4d. per cwt.
Ans. £39 18s. 2d.
3. 5 cwt. 3 qrs. 4 Ibs. at £9 6s. 8d. per cwt.
Ans. £51 18s. lOJd,
4. 7 cwt. 0 qr. 14 Ibs. at £2 3s. 4d. per cwt.
Ans. £15 8s. 9d.
5. 8 cwt. 3 qrs. 24 Ibs. at £1 2s. 3d. per cwt.
Ans. £9 19s. 5]d,
6. 9 cwt. 1 qr. 18 Ibs. at £3 10s. lOd. per cwt.
Ans. £33 6s. 7d,
7. 10 cwt. 2 qrs. 10 Ibs. at £4 4s. 6d. per cwt.
Ans. £44 14s, 9]d
3PBACT1CE.
119
FEDERAL MONEY.
CASE I.
When the price is -j, 1, or | of a cent.
RULE.
Divide the sum by the even parts of a cent, and the
answer will be cents.
EXAMPLES.
1. What is the worth of 452 Ibs. at \ cent per lb.?
i -
L 452 at £ cent.
226 cents,
or $2.26. Answer.
II.
in.
1
4
2468 at i cent each? 1
JL
2
987654 at f ct.?
617 cents.
or $6.17 Ans.
|
\
493827
2469131
$7407.401 Ans,
CASE II.
When the price is in cents.
RULE.
Divide the sum by the aliquot parts of a dollar, and
the answer will be in dollars.
EXAMPLES.
1. What is the worth of 2345 yards at 20 cents per
yard?
20
2345
Answer.
25 i 348 at 25 cents.
$87— Ans.
CASE III.
When the price is dollars and cents.
RULE.
Multiply the quantity by the dollars, then work for
the cents as in the last case, add the products together
for the answer.
EXAMPLES.
- 1. What is the worth of 5220 at three dollars and
fwenty cents each?
120
PRACTICE.
20
i
52.
20 II.
3 50
i1
8.48 at
$5.50.
—
5
"
156.
60
10.
44
42
.40
: ,
., —
4
.24
$ 167.04
—
—
$46.64 Answer
CASE IV.
When the price is no aliquot part of a dollar.
RULE.
Divide by two or more numbers, whose sum will make
the number wanted.
EXAMPLES.
1. W^hat will 9754 Ibs. cost at 30 cents per lb.?
" 9754
> cts.
10 cts. i of 20
1950.80
975
$2925.80 Answer.
2. What will 642 Ibs. cost at 60 cents per lb.?
Ans. $385.20.
CASE V.
When there are several denominations in the quantity , and
the price is dollars and cents.
RULE.
Multiply the dollars in the price by the number of
the highest denomination in the quantity; work for the
cents by the rules in the preceding cases; for the parts
in the quantity, take aliquot parts of each lower denomi-
nation, and add the products together.
EXAMPLES.
1. What is the value of 20cwt. 3 qrs. 14 Ibs. at 12
dollars and 25 cents per cwt.?
25 cts. = I i I 20 cwt.
I 12 dollars.
240 price at 12 dollars.
5 price at 25 cents.
$245 price of 20 cwt, at $12.25.
FELLOWSHIP. 121
To obtain the parts in the quantity.
cwt.
2 qrs. = i 12.25
1 qr. =
14 Ibs. =
6.121 price of 2 qrs.
3.061 price of 1 qr.
1.53 price of 14 Ibs.
10.71| price of 3 qrs. 14 Ibs.
245.00 price of 20 cwt.
$255.7 If Answer.
2. What is the worth of 64 cwt. 2 qrs. 141bg. at 16
dollars and 20 cents per cwt.? Ans. $1046.921 .
Q. 1. What is "practice?
2. Wherein is it particularly useful?
3. Repeat the table of aliquot parts.
4. How many cases are there in pounds, shillings
&c.?
5. Repeat the rule of each different case.
6. What number of cases do you find in Federal
Money?
7. Repeat the rule of each case.
8. How are sums in Practice proved?
FELLOWSHIP.
Fellowship is an easy rule by which merchants or
other persons in company, are enabled to make a just
division of the gain or loss in proportion to each per-
son's share. Sums in Fellowship are generally done by
the Rule of Three.
CASE T.
When the several shares are considered without regard to
time.
RULE.
As the sum of all the stock is to each person's parti-
cular share of the stock, so is the sum of all the gain 01
loss, to the gain or loss of each person.
11
13-2 FELLOWSHIP.
PROOF.
Add together all the shares of gain or loss, and if it
be right, the sum will be equal to the whole gain or loss.
EXAMPLES.
1. A and B purchase certain goods amounting to $580.
of which A pays $350 and B $230. They gain 262;—
what is each man's share of the gain?
A $350
B $230
A's share, gain. $ cts.
As 580 : 350 : : 262 : 158. lOf £ A?s gain.
B's share, gain.
Then, as 580 : 230 : : 262 : 103.89f| B's gain.
2. A, B and C formed a company. A put in $40, B
60 and C 80. They gained $72 : — what was each man's
share? Ans. A gained $16, B 24 and C 32.
3. A, B and C lose a quantity of property worth
$2400; of which A owned |, B -J- and the remainder to
C; what does each lose?
Ans. A loses $600, B 800 and C 1000.
4. A and B have gained $800, of which A was to re-
ceive 10 per cent, more than B; what did each receive?
Ans. A received $440 and B 360.
5. A and B purchase goods worth $80, of which A
pays 30 and B 50. They gain $20; — what is the gain
of each? Ans. A's gain is $7.50 and B's 12.50.
6. Four men formed a capital of $3200. They gained
in a certain time $6560. A's stock was $560, B's 1040,
C's 1200 and D's 400. What did each gain?
Ans. A's gain was $1 148, B's 2132, C's 2460 and D's 820.
CASE II.
When the different stocks in company are considered in re-
lation to time.
RULE.
Multiply each man's stock by the time it has been a
part of the whole stock: then, as the sum of the pro-
ducts is to either single product, so is the whole sum of
gain or loss to the gain or lass of each man.
TARE AND TRET. 123
EXAMPLES.
1. A, B and C hold a pasture in common, for which
they pay $40 per annum. A put in 9 cows for five weeks;
B, 12 cows for 7 weeks; and C, 8 cows for 16 weeks.*--
What must each man pay of the rent?
9X 5 = 45
12X 7 = 84
8X16= 128
Dolls. Dolls.
As 257 : 45 : : 40 : 7^ A'a part.
As 257 : 84 : : 40 : 13^ B's part.
As 257 : 128 : : 40 : 19ffX C's part.
2. A with a capital of £1000, entered into business
<jn the first of January. On the first of March follow-
ing he took in B as a partner, who brought with him a
capital of £1500; and three months after they are
joined by C with a capital of £2800. At the end of the
year, they find they have gained £1776 10s. How
must it be divided among them?
Ans. A's part will be £457 9s. 4£d.
B's part will be £571 16s. 8£d.
C's part will be £747 3s. Hid.
Q. 1. What is Fellowship?
2. By what rule are sums in Fellowship usually done?
3. How do you proceed when the shares are consi-
dered without regard to time?
4. How do you proceed when the shares are consi-
dered in relation to time?
5. How are sums in Fellowship proved?
TARE AND TRET.
Tare and Tret are certain allowances made by mer-
chants in selling their goods by weight.
Tare is an allowance made for the weight of the bar-
rel, bag, &c., that contains the article or commodity
bought.
Tret is an allowance of 4 Ibs. in every 104 Ibs. f&f-
waste, dust, &c.
124 TARE AND TRET*
Gross weight is the weight of the goods, together
with the barrel, box, or whatever contains them. When
the tare is deducted from the gross, what remains is
called suttle.
Neat weight is the weight of articles after all allow-
ances are deducted.
CASE I.
When the fare is so much per hhd. on any given quantity,
RULE.
Subtract the tare from the quantity — the remainder
will be the neat weight.
EXAMPLE.
In 6 hhd. of sugar, each weighing 9 cwt. 2 qrs. 10 Ibs,
gross, tare 25 Ibs. per hhd. how much neat weight?
cwt. qr. Ib. cwt. qrs. Ibs.
25X6=1 1 10 tare. 9 2 10
6X
57 2 4 gross
1 1 10 tare.
56 0 22 An«,
CASE IT.
When the tare is at so much per cwt.
RULE.
Divide the gross weight by the aliquot parts of a cwt,
then subtract the quotient from the gross, and the re-
mainder will be the neat weight.
EXAMPLES.
1. In 129 cwt. 3 qrs. 161bs. gross, tare 14 Ibs. per
cwt. what neat weight?
14 Ibs.
129 3 16 gross.
16 0 261
113 2 17-J Answer.
2. In 97 cwt. 1 qr. 7 Ibs. gross, tare 20 Ibs. per cwt..
what neat weight?
TARE AND TRET. 125
16 Ibs.
4 Ibs.
Subti
-act
97
1
7
gross.
J Add.
[- tare.
13
3
3
1
25
17
1
14,
79
3
20| Answe
./Vote. — When the tare per cwt. is not an aliquot part,
the tare may be found by the Rule of Three, thus — As
112 is to the number of pounds gross, so is the rate per
cwt., to the tare required.
3. What is the neat weight of 38 cwt. 0 qr. 4 Ibs. tare
at 1 1 Ibs. per cwt.
cwt. qr. Ibs.
38 0 4 = 4260 pounds.
Ibs. Ibs. Ibs.
Then, as 112 : 4260 : : 11 : 4 18T4TV Answer.
4260
418/ft
cwt. qrs. Ibs.
3841T6T83 neat = 34 1 5TfiT^ Answer,
CASE III.
When tare and tret are allowed.
RULE.
Find the tare according to the preceding rules, sub-
tract it from the gross, and the remainder will be suttle;
then divide the suttle by 26, and the product will be the
tret, which, subtract from the suttle — the remainder will
be the neat.
Note. — As 4 pounds on the 104 Ibs. is the customary
allowance for tret, we divide by 26, because 4 is -^ of
104.
EXAMPLES.
1. In 247 cwt. 2 qrs. 15 Ibs. gross, tare 28 Ibs. per cwt,
and tret 4 Ibs. for every 104 Ibs. how much neat?
1 26
28 Ibs. =
4lbs.= | aVof 104
247 2 15
61 3 17 12 tare subtract.
185 2 25 4
7 0 16 0
Ans. 178 2 9 4 neat.
2. In 9cwt. 1 qr. lOlbs. gross, tare 28 Ibs. per cvt
and tret 4 Ibs. for every 104 Ibs. how much neat?
Ans. 6cwt. 2 qrs. 26ilbsL
3. A merchant purchased 4 hhds. of tobacco, weigh-
ing as follows: — The first 5 cwt. 1 qr. 12 Ibs. gross, tare
65 Ibs. per hhd.; the 2d. 3 cwt. Oqr. 19 Ibs. gross, tare
75 Ibs.; the 3d. 6 cwt. 3 qrs. gross, tare 49 Ibs.; the 4th
4 cwt. 2 qrs. 9 Ibs. gross, tare 35 Ibs. and allowing tret
to each at the rate of 4 Ibs. for every 104 Ibs. What
was the neat weight of the whole?
Ans. 17 cwt. 0 qr. 19 Ibs. 2 oz:
Q. 1. What do you understand by Tare and Tret 2
2. What is tare?
3. What is tret?
4. What is gross weight?
5. What is neat weight?
6. What is called suttle?
BARTER,
Barter is the exchange of one commodity for another,
and leaches merchants to proportion their quantities
without loss.
Questions in Barter are solved either by the Rule of
Three, or by Practice.
When a quantity of one commodity is to be bartered
fcr a quantity of another, first find the value in money
of the quantity to be exchanged, then find what quanti-
ty of the other may be had for that amount.
EXAMPLES.
1. How much flour at g3 per barrel must be given in
exchange for 100 hhds. of salt worth $4.80 cts. per hhd
SIMPLE INTEREST. 127
$4.80
100 hhd.
$480.00 price of the salt.
dolls. bar. dolls. bar.
Then, As 3 : 1 : : 480 : 160 Answer.
2. Two merchants wish to make an exchange, A ha"fc
30 cwt. of cheese, at £l. 3s. 6d. per cwt. and B has 9
pieces of cloth, at £3. 15s. per piece — which must re-
ceive money, and how much? Ans. B must pay A £1 10s.
3. A has 150 bushels of wheat at $1.25 per bushel,
for which B gives 65 bushels of barley, worth 62^ cents
per bushel, and the balance in oats at 37 J cts. per bushel ;
what quantity of oats must A receive from B? Ans. 39 If,
SIMPLE INTEREST.
Interest is a premium paid for the use of money. In
calculating interest on money, four things are necessary
to be considered, viz. the principal, the time, rate per
cent., and amount.
The principal is the money lent for which interest is
to be received.
The rate per cent, per annum (by the year) is the in-
terest for 100 dollars or 100 pounds for one year.
The time is the number of years, months or days, for
which interest is to be calculated.
The amount is the sum of the principal and interest^
when added together.
Questions to prepare the learner for this rule.
1. If you give $6 for the use of $100 for a year; how
much must you give for the use of $50?
2. If you give $6 for the use of $100 for a year; how
much must you give for the use of it for six months?
3. How much for 3 months? — How much for 4 months?
How much for 8 months? — How much for 9 months?
4. If the interest of $200 be one dollar for a month;
fcow much will it be for 15 days?-^-How much for 10
days? — How much for 20 days?
128 SIMPLE INTEREST.
CASE I.
When the time is one year, and the rate per cent, is any
number of dollars, pounds, #c.
RULE.
Multiply the principal by the rate per cent , divide the
product by 100, and the quotient will be the interest for
one year.
EXAMPLES.
1. What is the interest of 328 dollars for one year at
6 per cent.?
328 In this example, as cutting off
6 the two right hand figures is the
same as dividing by 100, the di-
$19.|68cts. Ans. vision is omitted.
2. What is the interest of $9876 for one year at 6
percent.? 6
$592|56 cts. Answer.
When the sum is in pounds, if there be a remainder
after dividing, or after cutting off the two right hand
figures, the remainder, or figures cut off must be redu-
ced to shillings; and if there be still a remainder after
dividing the shillings, it must be reduced to pence, &c.
3. What is the interest of £573 13s. 9id. at 6 per
cent, per annum?
£573 13s. 9id. ./Vote.— When the interest is
6 for more than one year, mul-
tiply the interest for one year
£34|42 2 9 by the number of years. To
20 obtain the amount, the interest
must be added to the princi-
8|42 pal.
12
5)13 £34. 8s. 5d. Answer.
SIMPLE INTEREST, 129
4. What is the interest of £40. 19s. 1 Id. 3qrs, for one
vear, at 6 per cent, per annum?
£40 19s. lid. 3qrs.
6
2|45 19 10
20
9)19
12
2|38
4
1|52 remain. Ans. £2. 9s. 2d. Iqr.
5. What is the interest of 87 dollars for one year, at
6 per cent, per annum? Ans. $5.22.
6. What is the interest of 143 dollars for one year at
7 per cent, per annum? Ans. $ 10.01.
When the rate per cent, consists of a whole number
and a fraction, as 6£, 6|, or 6J, multiply the principal
by the whole number, to the product add £, or i, as the
case may be, of the principal and then divide by 100, or
cut off the two right hand figures as before.
7. What is the interest of 228 dollars for one year, at
6J- per cent, per annum?
1368
57
$14|25cts. Answer.
When the principal consists of dollars and cents, mul-
tiply by the rate per cent, without any reference to the
separating point; then from the product cut off the first
right hand figure as a fraction or remainder, the next
figure will be mills, the two next cents, and the other
figures, that is, those on the left of the cents, will be
dollars,
13® SIMPLE INTEREST.
8. What is the interest of $98.79' for one year, at 6
per cent, per annum? 6 '
5|92|7|4 fraction.
Ans. $5 92c. 7m.
9. What is the interest of 432 dollars 73 cents for 4
years, at 6 per cent, per annum?
$432.73
6 rate per cent.
259638
4 number of years.
103|85|5|2 frac. Ans. $103 85c. 5m,
10. What is the interest of $8420.82 for three years,
at 8 per cent, per annum? Ans. $2020. 99c. 6m.
11. What is the interest and amount of $7462.131 for
four years at 7 per cent, per annum?
Ans. Interest, $2089.39c. 7m. Am't. $9551.53c. 2m,
CASE II.
To find the interest when the given time is months or days.
RULE.
Find the interest for one year, then say — as one year
is to the given time, so is the interest of the sum for one
year, to the interest for the time required. Or, instead
of the Rule of Three, it may be done by Practice, thus:
For the number of months, take aliquot parts of a year;
and for days, the aliquot parts of 30.*
EXAMPLES.
1. What is the interest of $98.50 for 9 months and 18
days, at 6 per cent, per annum?
$98.50
6
$5.91|0|0 for one year.
year. mo. days. $ cts. $ cts. m.
Then, as 1 : 9 18 : : 5 91 : 4 72 8 Ans.
* In those calculations, a j'ear is reckoned at 360 (Jays, and a month at 30 days.
SIMPLE INTEREST. 131
in this sum, the year is reduced to SCO days, the 9
months and 18 days to 288 days, and the third term
stands as 591 cents.
The same is done by Practice, thus —
$98.50 ./Vote.— Interest may
6 be calculated in the
following manner.viz :
mo.
6, | of a year.
3. -i of 6 mo.
15d. iof 3 mo.
3, i of 15 ds.
5.91.010 When rate per cent, is
9? multiply the prin-
2.95.5 cipal by Jof the given
1.47.71 number of months; —
24.61 when it is 8, multiply
4.9i the principal by | of
the number of months:
Ans. $4.72.8 when the rate is 6,
multiply by 1 the number of months; when it is 4, multi-
ply by i; when it is 3, by i; and when the rate is 2,
multiply by i — the product in any of those cases will
show the answer.
2. What is the interest of $120.60 for one year and
three months, at 6 per cent, per annum ? Ans. $9. 04c. 5m.
3. What is the interest on $187.061 for 10 months, at
6 per cent, per annum? Ans. $9.35c. 3m.
4. What is the interest and amount of 640 dollars for
4 years and 7 months, at 5 per cent, per annum?
Ans. g 146.66| interest. Am't. $786.66|.
5. What is the interest of $300 for 4 years, 4 months,
and 20 days, at 81 per cent, per annum? Ans. $1 11.91|.
6. What is the~interest of $5420 for 17 months at 4
per cent, per annum? Ans. $307.131.
7. What is the interest of $7200 for 14 months at 6
per cent, per annum? Ans. $504.
8. What is the interest of $8050.871 for 3 years and
11 months at 6 per cent, per annum? 4ns. $1891.95c. 5m.
9. What is the interest of $948.621 for 8 months, at
8 per cent, per annum? Ans. $50.59c. 3m.
10. What is the interest of £421 16s. 9d., for 2 years
and 8 months, at 5 per cent, per annum?
Ans. £56 4s.
|32 SIMPLE INTEREST.
11. What is the interest of 580 pounds for 5 years, %
months and 10 days, at 7 per cent, per annum?
Ans. £210. 17s. lO^d.
12. What is the interest of $36 for 1 month at 8 per
cent, per annum? Ans. 24 cents.
Bank interest is generally reckoned by days only ; and
to find the interest for any number of days at 6 per cent,
as computed at banks, multiply the dollars by the num-
ber of days, and divide by 6; — the quotient will be the
interest in mills.
Note.— The interest of any number of dollars for 60
days, will be exactly the number of cents, thus — $80
for 60 days, at 6 per cent, is 80 cents.
CASE III.
The amount, time, and rate per cent, given to find the
principal.
RULE.
Find the amount of 100 dollars at the rate and time
given; then say, as the amount of 100 dollars, is to the
amount given, so are 1 00 dollars tr the principal required.
EXAMPLKS.
1. What principal at interest for two years, at 6 per
cent, per annum, will amount to $134.40?
$100
6
6.00
2
12.00
100.00
$112 amount of 100 for two years.
dolls. $ cts. dolls, dolls.
Then, as 112 : 134.40 : : 100 : 120 Ans.
2. What principal at interest for 5 };ears, at 6 per
cent, will amount to $780? Ans. $600.
3. What principal at interest for 4 years and 3 njonf hs
at 6 per cent, will amount to $1192.25. Ans. $950?
SIMPLE INTEREST. 133
CASE IV.
To find the rate per cent, when the amount, time and prin-
cipal are given.
RULE.
Take the principal from the amount, the remainder
will be the interest for the given time; then, as the
principal is to one hundred dollars, so is the interest of
the principal for the given time, to the interest of 100
dollars for the same time. Divide the interest of 100
dollars thus found, by the time, and the quotient will be
the rate per cent.
EXAMPLES.
1. At what rate per cent, will $500 amount to $650
in three years. 650 Amount.
500 Principal.
150 Interest for the time,
D. D. D D.
As 500 : 100 : : 150 . 30 Interest of 100.
Then divide by the same 3)30(10 Ans. per cent
30
2. At what rate per cent, per annum will $1850 dou-
ble in 5 years? Ans. 20 per cent.
CASE V.
To find the time when the principal, amount, and rate per
cent, are given.
RULE.
Find the interest of the principal for one year; find
the interest of the principal for the whole time, by sub-
tracting the principal from the amount; then divide the
whole interest by the interest for one year — the quotient
will show the time required.
EXAMPLES.
1. In what time will $300 amount to $1000 at 5 per
cent, per annum?
800 1000 Then, 4|0)20|0
5 800 5
$40|00 200 Whole InH. Ans. 5 years.
12
134 COMPOUND INTEREST.
2. In what time will $80 amount to $182.40 at 8 per
cent, per annum? Ans. 16 years,
COMPOUND INTEREST.
Compound Interest is that which arises from the in-
terest being added to the principal, and becoming a part
of the principal, at each time of payment.
RULE.
Find the amount of the principal, for the time of the
first payment, by Simple Interest; this amount, contain-
ing the principal and interest for the first year, will be
the principal for the second year; and the amount of
this principal, which consists of the principal and inte-
rest for the second year, will be the principal for the
third year, and so on, for any number of years. From
the last amount, subtract the given principal, and the
remainder will be the compound interest.
EXAMPLES.
1. What is the compound interest of $8000 for two
years, at 6 per cent, per aunum?
$8000
6
Interest for the first year 480100
Principal 8000
Amount 8480
6
Jfi't. for the second year 508.180
Principal 8480.00
8988.80
Subtract 8000.00
$988. 80c. Answer.
2. What is the compound interest of $554 for 3 years,
at 8 per cent, per annum? Ans. $143.88.
3. What is the compound interest of $744 for 2 years,
at 7 per cent, per annum? Ans. $107.80c. 5m.
4. What is the compound interest of $50 for 8 years,
$t 8 per cent. per annum? Ans. $42.54c. 6m,
INSURANCE, COMMISSION AND BROKERAGE. 133
5. What is the compound interest of £48 5s. for 3
years, at 6 per cent, per annum? Ans. £9 4s. S^cL
Q. \. What is interest?
2. What are the four things considered in calculating
interest?
3. What is the principal ?-What is the rate per cent.?
What is the time? — What is the amount?
4. How do you proceed in the first case?
5. How do you proceed in pounds, shillings, &c.?
6. How do you proceed when the rt'te per cent, con-
sists of a whole number and a fraction?
7. How do you proceed when the principal is in dol-
lars and cents.?
8. How do you calculate interest for more than a
year? — How, when the time is in months?
9. What other method is there for calculating inte-
rest, besides the method of multiplying the
sum by the rate p*r cent.?
10. How is bank interest reckoned? — What is the rule
for casting it?
11. Do you understand all the cases and rules of in-
terest?
12. What is Compound Interest?
13. Repeat the rule for calculating Compound Interest?
INSURANCE, COMMISSION AND BROKERAGE.
Insurance, Commission and Brokerage, are premiums
allowed to insurers, factors and brokers at a certain rate
per cent.; and is obtained after the manner of the first
case in Simple Interest.
EXAMPLES.
1. What is the insurance of $4500, at 2£ per cent.*?
9000
2250
$112l50c. Answer,
1 36 DISCOUNT*
2. What is the commission on a sale of goods amotm
ting to $1184 at 5 per cent.? Ans. $59.2O,
3. What is the brokerage of $987 at 3 per cent.?
Ans. £99.61.
4. What is the commission on a sale of goods amoun-
ting to $4820 at 4i per cent.? Ans. $21.6.90.
DISCOUNT.
Discount is an allowance made for the payment of
any sum of money before it becomes due, and is the dif-
ference between that sum, due some time hence, and its
present worth. RULE.
As the amount of $100 at the given rate and time is
to $100, so is the given sum or debt to the present worth.
Subtract the present worth from the given sum, and the
remainder will be the discount.
EXAMPLES.
1 . What is the present worth of $500 due in 3 yearsr
at 6 per cent, per annum?
$ $ $
$100 118 : 100 : : 500
6 100
$ c. m,
6100 118)50000(423.72.8
3 472
18 280
100 236
118 amount of $100. 440
354
86.00|72c,
82.6
340
236
104 remainder.
EQUATION. 1 37
2. What is the present worth of $350 payable in 6
months, discounting at 6 per cent, per annum?
Ans. $339 80c. 5m.
3. What is the discount on $1000 due in one year, at
6 per cent, per annum? Ans. $56.60c. 3m.
4. What is the present worth of £65 due in 15 months
at 6 per cent, annum? Ans. £60 9s. 3£d.
5. What sum will discharge a debt of $1695 due af-
ter 5 months and 20 days at 6 per cent, per annum?
Ans. $1541.32c. 6m.
6. What is the present worth of $426.55, at 6 per
cent, per annum, due in 8 months? Ans. $410.14c. 5m.
Note. — When discount is made without regard to time,
it is found as the interest of the sum would be for one
year.
EQUATION.
Equation is the method for finding a time to pay at
once, several debts due at different times.
RULE.
Multiply each payment by the time at which it is due,
and divide the sum of the products by the sum of all the
payments — the quotient will be the time required.
EXAMPLES.
1. A owes B $480 to be paid in the following manner,
viz: $100 in 6 months, $120 in 7 months, and $260 in 10
months; what is the equated time for payment of the
whole debfc?
100X 6= 600
120X 7= 840
260X10= 2600
480 )4040(8T5g months,
3840
138 LOSS AND GAIN.
2. A owes B $1100, of which 200 is to be paid in 3
months, 400 in 5 months, and 500 in 8 months — what is
the equated time for payment of all? Ans. 6 months.
3. C is indebted to a merchant to the amount of $2500;
of which $1000 is payable at the end of 4 moMths,$800
in 8 months, and 700 in 12 months — when ought pay-
ment to be made, if all are paid together?
Ans. 7 months 15J days.
LOSS AND GAIN.
Loss and Gain is a rule by which persons in trade are
able to discover their profit or loss; and to increase or
lessen the prices of their goods so as to gain or lose on
them to any given amount.
Questions in Loss and Gain are solved by the Rule of
Three, or by Practice.
EXAMPLES.
1. A merchant bought 100 yards of silk at 75 cents
per yard, what will be his gain in the sale, if he sell it
at 90 cents per yard.
75 cents.
yard, yards. cts. Dolls.
15 gain per yard. As 1 : 100 : : 15 : 15 Ans.
2. If a grocer buy 250 Ibs. of tea, at $225, and sell
the whole at $1.25 per Ib. what will be his gain by the
transaction?
$1.25 $312.50
250 225.00
G250 $87.50
250
$312.50
3. If a yard of calico cost 28 cents, and is sold for 31
cents, what is the gain on 293 yards? Ans. $8.79.
4. Bought 420 bushels of corn at 25 cents per bushel,
and sold the same at 38 cents per bushel; what was the
amount gained? Ans. $54.60.
INVOLUTION. 139
5. A merchant bought 12 cwt. of coffee at 26 cents
per Ib, and afterwards obliged to sell it 20 cents per Ib.
what was his loss? Ans. $00.64.
6. It a merchant gain $80 on $560, what is that per
cent.? Ans 14f per cent.
7. If a yard of vel vet be bought for 1 6s. and sold again
for 12s. what is the loss per cent.? Ans. 25 per cent,
INVOLUTION,
OR THE RAISING OF POWERS*
The product of any number multiplied by itself any
given number of times, is called its power, as in the fol-
lowing example.
Thus, 2X2 = 4 the square, or second power of 2.
2X2X2 = 8 the cube, or third power of 2.
2X2X2X2= 16 the biquadrate, or 4th power of
2.* Hence, 3 raised to the 4th power makes 81. The
number which denotes a power is called the index, or
exponent of that power.
When a power of a vulgar fraction is required, it is
only necessary to raise, first the numerator, and then the
denominator to the given power, and place the product
of the one over the product of the other; thus, the 3d
2X2X2 ==
power of * 3X3X3 = *'
EXAMPLES.
1. What is the square of 4567? Ans. 20857489.
2. What is the cube of 567? Ans. 182284263,
3. What is the biquadrate of 67? Ans. 20151121.
4. What is the ninth power of 2? Ans. 512.
5. What is the cube of J? Ans. |ff.
6. What is the cube or third power of .13?
Ans. .002197.
7. What is the sixth power of 5.03?
Ans. 16196.005304479729,
* Any given number is considered the first power of itself, and when multiplied
by itself the product is the second power, &c.
HO E.V0LUT10N.
TABLE OF THE FIRST IVINE POWERS.
?
o
C/3
-0
fi
1
1
O
c
cr
O>
£»
S"
T3
0
<T>
in
cr
"3
0
<T>
^t
a*
i
o
-*
cr
T3
0
<B
CO
tr
*d
0
n
CO
&
1
0
p
1
1
1
16
1
1
1
1
1
2
4
8
«->!i!.
64
128
256
512
n
O
9
27
SI
241'
729
2187
6561
19683
4
16
64
256
1024
4096
16384
65536
262144
5
25
E
f,
125
625
3125
15625
78125
390625
1953125
6
7
8
9
216
1296
7776
1680?
46656
279936
1979616
10077696
342?
5401
117649
823543
5764801
40353607
64
512
4096
-J2768
>621 44 2097152 16777216
134217728
81
72C
6561
o904fJ
531441 14782969S43046721
38742048S
EVOLUTION,
OR THE EXTRACTION OF ROOTS.
The root of a number, or power, is any number, which
Ueing multiplied by itself a certain number of times,
will produce that power; and is called the square, cube,
biquadrate root, &c. according to the power to which it
belongs. Thus, 3 is the square root of 9, because when
multiplied by itself, it produces 9; and 4 is the cube
root of 64, because 4X4X4 = 643 and so of any other
number.
THE SQUARE ROOT.
Extracting the square root of a number, is the taking
a smaller number from a larger, and *uch as will, being
multiplied by itself, produce the larger number,
EVOLUTION. 141
RULE.
1. Separate the sum into periods of two figures each,
beginning at the right hand figure.
2. Seek the greatest square number in the left hand
period; place the square, thus found, under that period,
and the root of it in the quotient. Subtract the square
number from the first period; to the remainder bring
down the next period, and call that the resolvend.
3. Double the quotient, and place it on the left hand
of the resolvend for a divisor. Seek how often the di-
visor is contained in the resolvend, omitting the units
figure, and set the answer in the quotient, and also on
the right hand side of the divisor. Then multiply the
divisor, including the last added figure, by that figure,
that is, by the figure last placed in the quotient; place
the product under the resolvend, subtract it, and to the
remainder bring down the next period, if there be any
more, and proceed as already directed. If there be a
remainder after the periods are all brought down, annex
cyphers, two at a time, for decimals, and proceed till
the root is obtained with sufficient exactness.
Note. — When a sum in the Square Root consists of
whole numbers and decimals, point off the whole num-
bers as above directed, then point the decimal part,
commencing at the decimal point and forming periods of
two figures each towards the right, observing when
there is only one figure left for the last period, to add a
cypher to the right of it, to make an even period. —
When the sum consists entirely of decimals, separate
the periods after the same manner. If it be required
to extract the square root of a vulgar fraction, reduce
it to its lowest terms; then extract the root of the nu-
merator for the numerator vf the answer, and the root
of the denominator for the denominator of the answer.
If the fraction be a surd, that is, a number whose root
can never be exactly found, reduce it to a decimal, and
then extract the root from it; and if the sum be a mixed
number, the root may be obtained in the same way.
PROOF.
Square the root, adding in the remaider, (if any,) and
the result will equal the given number.
142
EXAMPLES.
1. Whaft is the square root of 20857489"?
.... Root.
20857489(4567 Answer,
16
divisor 85)485 resolvend.
425
divisor 906)6074 resolvend.
5436
divisor 9127)63889 resolvend.
63889
2. What is the square root of 294849? Ans. 54$.
3. What is the square root of 41242084? Aris. 6422,
4. What is the square r^ot of 17.3056? Ans. 4.16,
5. What is the square root of .000729? Ans. .027,
6. What is the square root of 5? Ans. 2.23606,
7. What is the square root of //T? Ans. ^.
8. What is the square root of 17f ? Ans. 4.168333.
9. A general has an arrny of 7056 men; how many
must he place on a side to form them into a compact
square? Ans. 84.
10. If the area of a circle he 184.125, what is the side
of a square that shall contain the same area?
Thus, ^184.125= 13.569-|-Answer.
11. If a square piece of land contain 61 acres and 41
square poles, what is the length of one of its sides?
A. P.
Thus, 61 41 =9801 square poles.
Then, ^9801 = 99 rods, or poles in length, Answer.
12. There is a circle whose diameter is 4 inches; what
is the diameter of a circle 3 times as large?
Thus, 4X4= 16; and 16X3= 48 and ^48= 6.928
-{-inches. Ans.
13. There is a circle whose diameter is 8 inches; what
is the diameter of a circle which is only one fourth as
farge. 8X8=64; and 64-7-4=16; and ^16=4 inches,
Ans, 4 inches,
THE CUBE ROOT. 143
The square of the longest side of a right angled trian-
rJe, is equal to the sum of the squares of the other two sides;
therefore, the difference of the squares of the longest side,
and either of the other sides, is the square of the remaining
side.
14. The wall of a certain city is 20 feet in height, it
is surrounded by a ditch 20 feet in breadth; what must
be the length of a ladder, to reach from the outside of
the ditch to the top of the wall? Ans. 28i feet.
THE CUBE ROOT.
The cube root of a given number, is such a number
as being multiplied by itself, and then into that product,
preduces the given number.
RULE.
1. Point off the sum into periods of three figures each,
beginning with units.
2. Find the greatest cube in the left hand period,
place the root of it in the quotient, subtract the cube
from the left hand period, and to the remainder bring
down the next period for a resolvend.
3. Square the quotient, and multiply the square by 3
for a defective divisor.
4. Seek how often the defective divisor is contained
in the resolvend, omitting the units and tens, or two
right hand figures. Place the result in the quotient, and
its square to the right of the divisor, supplying the place
of tens with a cypher, whenever the square is less than
ten.
5. Multiply the last figure of the quotient or root by
all the figures in it previously ascertained ; multiply that
product by 30, and add their product to the divisor, to
complete it.
6. Multiply and subtract as in Simple Division, and
to the remainder bring down the next period, for a new
resolvend. Find a divisor as before, and thus proceed
imtil all the periods are brought down.
}44 THE CUBE ROOT.
Note. — The cube root of a vulgar fraction is found by
reducing it to its lowest terms, and extracting, as in the
square root; and if the fraction be a surd, reduce it to
a decimal, and then extract the root.
In extracting the cube root, if the sum be in part de-
cimals, or if the whole be decimals, point the figures as
in the square root, observing to have three figures in a
period instead of two; and in all cases in the cube root,
when there is a remainder, if it be required to obtain
decimal figures to the root, proceed as directed in the
square root, only add three cyphers, in place of two, to
the remaider.
PROOF.
Involve the root to the third power, adding the re-
mainder, (if any,) to the result.
EXAMPLES.
1. What is the cube root of 182284263?
. . . Root.
182284263(567 Answer.
125
f Defective divisor and square of 6. -
Complete divisor — 8436)50616
Defective 56X56X3 =940849),™™^
divisor. 7X56X30= 11760J6^8263 new resolv.
Complete divisor 952609)6668263
2. What is the cube root of 48228 544? Ans. 36.4.
3. What is the cube root (or 3d root) of 2?
Ans. 1.259921.
4. What is the cube root of 132651 ? Ans. 51.
5. What is the cube root of 4173281? ADS. 161.
6. What is the cube root of .008649? Ans. .2052-f.
7. What is the cube of iff? Ans. f.
8. If the contents of H globe amount to 5832 cubick
inches, what are the dimensions of the side of a cubick
block containing the same quantity ? Ans. 1 8 in. square
BIftUADRATE ROOT. 14'i>
THE BiaUADRATE ROOT.
To extract the biquadrate root, is to find out a number
which being involved 4 times into itself, will produce
the given number, that is the fourth power.
RULE.
Extract the square root of the sum, then extract the
square root of that root, and the last root will be the
answer.
EXAMPLES.
1 What is the biquadrate root (or 4th root) of 531441?
Root.
531441 ) 729 ( 27 Answer,
49 4
142)414 47)329
284 329
1449)13041
13041
2. What is the biquadrate root of 4096? Ans. *.
3. What is the biquadrate root of 146841? Ans. 11.
Mote. — The roots of several other powers may be ob-
tained by means of the foregoing rules, thus —
To obtain the root of the 6th power, extract the square
root of the cube root.
For the 8th, take the square root of the biquadrate
root. For the 9th, take the cube root of the cube root.
For the 12th root, take the cube root of the biquadrate
root.
Questions concerning the powers and roots.
1. WThat is called a power?
2. AVhat power is the square? Ans. The 2d. power
3. What is the cube of a number called?
4. AVhat is the biquadrate?
5. How do you raise the pow^r of a vulgar fraction I
6. What is the root of a power?
7. What is meant by extracting the square root'?
8. Repeat the rule for doing it?
146 ALLIGATIONS7.
9. How do you proceed when the sum consists in patt;
or altogether, of decimals?
10. How do you extract the square root of a vulgar
fraction ?
11. How do you proceed when the fraction is a surd?
12. What do you understand by the cube root?
13. Repeat the rule for extracting it?
14. How do you extract the cube root of a vulgar frac-
tion?
15. What do you understand by extracting the biqua-
drate root?
16. Repeat the rule for extracting it?
17. How are sums in the square roet proved?
18. How are sums in the cube root proved.?
ALLIGATION.
Alligation is a rule for mixing simples of different
qualities, in such a manner that the composition may be
of a mean or middle quality.
CASE I.
To find the mean price of any part of the mixture, when
the quantities and prices of several thing* are given.
RULE.
As the sum of the quantities is to any part of the com-
position, so is the price of the quantities to the price of
any particular part.
EXAMPLES.
1. A trader mixes 60 gallons of wine at 100 cents per
gallon; 40 gallons at 80 cents, and 30 gallons of water.
What should be the price per gallon?
gals. cts. %
Wine 60 at 100=60.00
Wine 40 at 80 = 32.00
Water 30
gals. gal. $
J30 1 : : 92.0*
ALLIGATION. 1 4?
$. A trader mixes a quantity of tea as follows, viz:—
*lbs. of tea at 42 cents per lb.; 6 Ibs. at 33 cents; 121bs,
75 cents, and 15 IDS. at 30 cents. What can he sell it
for per lb.? Ans.*66ff cents,
3. A farmer mixes 20 bushels of wheat at 5s. per
bushel, with 36 bushels of rye at 3s., and 40 bushels of
of barley at 2s, per bushel; how much is a bushel of
*,he mixture worth? Ans. 3s,
CASE II.
When the prices of several simples are given to find what
quantity of each, at their respective prices, must he taken
to make a compound at a proposed price.
RULE.
Set the prices of the simples in a column under each
other. Connect with a continued line, the rate of each
simple which is less than that of the compound, with one
or any number of those that are greater than the com-
pound, and each greater rate, with one or more of the
less. Place the difference between the mixture rate,
and that of each of the simples, opposite to the rates
with which they are linked. Then, if only one differ-
ence stand against any rate, it will be the quantity be-
longing to that rate; but if there be several, their sum
will be the quantity. Different modes of linking will
produce different answers.
EXAMPLES.
1. A merchant would mix wines at 17s. 18s. and 22s,
per gallon, so that the mixture may be worth 20s. per
gallon: what quantity of each must be taken?
Ans. 2 gallons at 17s., 2 at 18s., and 5 at 22s.
2. How much barley at 40 cents, corn at 60, and wheat
at 80 cents per bushel, must be mixed together, that the
compound may be worth 62| cents per bushel?
Ans. 17J- bush, of barley, 17|- of corn, and 25 of wheat;
14$ ALLIGATION-
CASE III.
When the prices of all the simples, the quantity of one of
them, and the mean price of the mixture, are given, to
.find the quantities of the other simples.
RULE.
Find an answer as before, by connecting; then, as the
difference of the same denomination with the given
quantity, is to the differences respectively, so is the given
quantity, to the different quantities required.
EXAMPLES.
1. How much gold of 15, 17, 18, and 22 carats fine
must be mixed together to form a composition of 40 oz>
of 20 carats fine?
f!5 ^ . . 2
Mean or Mixture j 17^ 2
rate. 20 | IS\ \ 2
L22;J 5+3+2=10
16
then as 16 : 2 : : 40 : 5)
and as 16 : 10 : : 40 : 25J AnSW6r'
Answer 5 oz. of 15, 17, and 18 carats fine, and 25 QXI
of 22 carats fine.
2. A grocer has currents at 4d., 6d., 9d., and lid.,
perlb. and he would make a mixture of 240 Ibs. that
might be sold at 8d. perlb.j how much of each kind-
must he take?
Ans, 72 Ibs. at 4d., 24 at 6d., 48 at 9d. and 96 at lid,
CASE IV.
When the prices of the simples, the quantity to be mixed,
and the mean price are given, to find the quantity of each
simple.
RULE*
Connect the several prices, and place their differen-
ces as before; then, as the sum of the differences thus
given, is to the difference of each rate, so is the quantl
ty to be compounded, to the quantity required
POSITION* 149
EXAMPLES.
1. How much sugar at 9 cents, 11 cents and 14 cents
per Ib. will be necessary to form a mixture of 20 Ibs,
worth 12 cents perlb.?
(91 2
12 Iliv | 2
(14/J 3+1 = 4
8
Then, as 8 2 : : 20 : 5 Ibs. 9 cents.)
2 : : 20 : 5 Ibs. 11 cents.) Answer.
4 * : 20 : 10 Ibs. 14 cents.)
2. A grocer has sugar at 24 cents perlb. and at 13
cents perlb.; and he wishes so to mix 2cwt. of it, that
he may sell it at 16 cents perlb.; how much of each
kind must he take? Ans. 162j£ Ibs. of that at 13 cents,
and 61TVlbs. of that at 24 cents.
3. How many gallons of water must be mixed with
wine worth 60 cents per gallon, so as to fill a vessel of
80 gallons, that may be sold at 41 £ cents per gallon?
Ans. 18f gallons of water, and 61 A of wine>
POSITION.
Position is a rule for solving questions, by one or more
supposed numbers. It is divided into two parts, namely
single and double,
SINGLE POSITION.
Single position teaches to solve questions which re-
quire but one supposition.
RULE.
Suppose a number, and proceed with it as if it were
the real one, setting down the result — Then, as the re-
Suit of that operation, is to the number given, so is the
supposed number, to the number sought,
IS*
1 50
EXAMPLES.
1. What number is that, which being multiplied by
and the product divided by 6, will give 14 for the
flentf
Suppose 18
7
6)126
Theft, as 21 : 14 : : 1&
18
112
14
£ 1)252(12 Answer
21
42
42
& What number is that, of which one half eiceeefe
^ne third by 15?
Suppose 60— Then £ | 60 | i|60
30 20
Subtract 20
10
rtfhen, as 10 : 15 : : 60 : 90 Answer.
3. What number is that, which being increased by |?
•J. and J of itself, the sum will be 125? Ans. 60,
4. A schoolmaster being asked how many scholars he
foad, answered, that if | of his number were multiplied
by 7, and | of the same number added to the product,
the sum would be 292. What was his number? Ans. 60.
5. A schoolmaster being asked what number of schol-
ars he had, said, if I had as many, half as many, and
fourth as many, I should have 99. What was hrs
Ans, -36
POSITION. 1^>
b. A person, after spending i and £ of his money. v
§30 left; what had he at first? Ans. $180.
7. Seven eighths of a certain numher exceed four
fifths by 6. What is that number? Ans. 80.
8. A certain sum of money is to be divided among 4
persons, in such a manner that the first shall have i of
it, the second |, the third £, and the fourth the remain-
der, which is $28; what is the sum? Ans. $112.
9. What sum, at 6 per cent, per annum, will amount
t® £860. in 12 years? As.
DOUBLE POSITION.
Double Position teaches to resolve questions by meaifs
of two supposed numbers.
RULE.
Suppose two convenient numbers, and proceed with
each according to the condition of the question, and set
down the efrours of the results. Multiply the errours
into their supposed numbers, crosswise ; that is, multiply
the first supposed number by the last errour, and the
last supposed number by the first errour.
If the errours be alike, that is, both too much, or both
too little, divide the difference of their products by the
difference of the errours — the quotient will be the an-
swer. But if the errours be unlike, that is, one too-
large and the other too small, divide the sum of the
products by the sum of the errours.
EXAMPLES.
1. What number is thatj whose 1 part exceeds the |
part by 16?
Suppose 24; and as 1 of 24 is 8, and 1 of it is 6, it is
evident that the third part exceeds the fourth part by 2f
instead of 16; and therefore the errour is 14 too small,
Again, suppose 48; and i of 48 being 16, and i being 12,
it is manifest that the third part exceeds the fourth by
4, instead of 16; hence the errour is 12 too small,—*--
Then, the errours being alike, proceed thus— »
POSITION,
er.
1. supposition 24 \ / 14 too small,
er.
2. supposition 48 / \ 12 too small,
14 672 product. 288 product,
12 288
2 dif.of er.2)384 difference of the products.
192 Answer.
2. A son asking his father how old he was, received
this answer: Your age is now '- of mine; but 5 years
ago, your age was | of mine. What are their ages?
Ans. 20 and 80.
3. Two persons, A and B, have each the same income,
A saves \ of his; but B, by spending 50 dollars per an-
num more than A, finds himself at the end of 4 years
one hundred dollars in debt. What was their income,
and what did each spend?
Ans. Their income was $125 per annum for each,- A
spends $100 and B spends $150 per annum.
4. What number, added to the sixty-second part of
7626, will make the sum of 200? Ans. 77.
5. A man being asked how many sheep he had in his
drove, said, if 1 had as many more, half as many more,
one fourth as many more, and 12|, I should have 40. —
How many had he? Ans. 10.
6. An officer had a divison, £ of which consisted of
English soldiers, \ of Irish, £ of Canadians, and 50 of
Ihdians. How many were there in the whole? Ans. 600.
7. A servant being hired for 30 days, agreea to re-
ceive 2s. 6d. for every day he laboured, and to forfeit
Is. for every day he played. At the end of the term
his pay amounted to £2. 14s. How many of the days
did he labour? Ans. 24.
8. What number is that, which being multiplied by 6,
the product increased by adding 18 to it, and the sum
divided by 9, the quotient will be 20? Ans, 27,
ARITHMETICAL PROGRESSION". 153
ARITHMETICAL PROGRESSION.
Arithmetical Progression is a series of numbers in<
creasing or decreasing by a common difference; as, 1^
2, 3, 4, 5; 1, 3, 5, 7, 9; 5, 4, 3, 2, 1; 9, 7, 5, 3, 1, &c.
The numbers in a series are called terms — the first and
last terms are called extremes, and the common differ-
ence is the number by which the terms in a series differ
from each other; as in 2, 5, 8, 11, &c. — the common dif-
ference is 3.
In any series in Arithmetical Progression, the sum of
the two extremes is equal to the sum of any two terms,
equally distant from them, or equal to double the mid-
dle term when there is an uneven number of terms in
the series. Thus, in the series 2, 4, 6, 8, 10, 12, — the
extremes are 2 and 12, equal to 14, and if you add 10
and 4, or 8 and 6, the result will be the same; and in the
series 2, 4, 6, 8, 10, the extremes are 10 and 2, and as
the number of terms is uneven 6 is the middle one,
which, when doubled makes 12, and the extremes whea
added together make the same amount,
CASE I.
The first term, common difference, and" number of terms,
being given, to find the last term and sum of all the
terms.
RULE.
Multiply the common difference by one less than the
number of terms, and to the product add the first term,
the sum will be the last. Add the first and last terms
together, multiply their sum by the number of terms,
and half the product will be the sum of all the terms.
EXAMPLES.
1. The first term in a certain series is 3, the common
difference 2, and the number of terms 9; to find the last
ferm, and the sum of all the terms,
154 ARITHMETICAL PROGRESSION
One less than the number of terms is 8.
2 common difference.
8 number of terms less oiie
16 product.
3+first term
19 last term.
3+first term.
22
9 X number of terms.
2)198
Answer 99 sum of all the terms.
£. A person sold 80 yards of cloth at 3 cents for th€
Urst yard. 6 for the second, and thus increasing 3 cents
every yard : what was the whole amount? Ans. $97.20.
3. How many times does a clock usually strike in 12
hours? Ans. 78.
4. A man on a journey travelled 20 miles the first
day. 24 the second, and continued to increase the num-
ber of miles by every clay for 10 day. How far did he
travel? Ans. 380 miles.
5. A farmer bought 20 cows, and gave 2 dollars for
the first, 4 for the second, and so on, giving in the same
proportion from the first to the last. What did he give
for the whole? Ans. $420.
CASE II.
When the two extremes and the number of terms are given
to find the common difference.
RULE.
Subtract the less extreme from the greater, and di-
vide the remainder by one less than the number of
terms — the quotient will be the common difference.
EXAMPLES.
1. The extremes being 3 and 19, and the number of
ferms 9, what is the common difference?
GEOMETRICAL PROGRESSION. ^5
9 19 II.
13 13 number "\
1 of terms 80 ' p .
8 8)16 - 2Qf Extremes.
12 -J
Ans. 2 12)60 difference of
— extremes.
Common difference 5 Answer.
3. If the extremes be 10 and 70, and the number of
terms 21, what is the common difference, and the sum
of the series? Ans. com. diff. 3, and the sum, 840.
4. A certain debt ctm be paid in one year, or 52
weeks, by weekly payments in Arithmetical Progres-
sion, the first payment being 1 dollar, and the last 103
dollars. What is the common difference of the terms?
Ans. $2.
5. A debt is to be discharged at 16 several payments
in Arithmetical Progression; the first payment to be 20
dollars, and the last 1 10 dollars. What is the common
difference. Ans. $6,
GEOMETRICAL PROGRESSION.
Geometrical Progression is the increase of any series
of numbers by a common multiplier, or the decrease of
any series by a common divisor; as 3, 6, 12, 24, 48;
and 48, 24, 12, 6, 3. The multiplier or divisor by which
any series is increased or decreased, is called the ratio.
CASE I.
To find the last term and sum of the series,
RULE.
Raise the ratio to a power whose index is one less
than the number of terms given in the sum. Multiply
the product by the first term, and the product of that
multiplication will be the last term: then multiply the
last term by the ratio, subtract the first term from the
product, and divide the remainder by a number that is
one less than the ratio — the quotient will be the sum of
^he series.
156 GEOMETRICAL PROGRESSION.
EXAMPLES.
1. Bought 12 yards of calico, at 2 cents for the first
yard, 4 cents for the second, 8 for the third, &c. : what
was the whole cost?
NOTE. — The number of terms 123 and the ratio 2.
1st. term 2 1st. power.
2
4 2d. power,
2
8 3d. power,
2
16 4th. power,
2
32 5th. power
32
1024 10th. power.
2
2048 llth. power, or one less than the
2 1st. term. [number of terms.
4096
2 the ratio.
8192
2 subtract the 1st. term.
1)8190 1, is one less than the ratio.
$81 90 Answer.
£. Bought 10 Ibs. of tea, and paid 2 cents for the first
pound, 6 for the second, 18 for the third, &c. What did
tne whole cost? Ans. $590.48.
SEOMETRICAL PROGRESSING 157
3. The first term in a sum is 1, (he whole number of
terms 10, and the ratio 2; what is the greatest term,
and the sum of all the terms?
Ans. The greatest term is 512, and the sum of the
terms 1023.
4. What debt may be discharged in 12 months, by
paying 1 dollar the first month, 2 dollars the second
month, 4 the third month, and so on, each succeeding
payment being double the last; and what will be the
amount of the last payment?
Ans. the debt is $4095, and the last payment $2048.
5. A father whose daughter was married on a new
3rear'$ day, gave her one cent, promising to triple it on
the first day of each month in the year: what was the
amount of her portion? Ans. $2657.20.
6. One Sessa, an Indian, having invented the game of
chess, shewed it to his prince, who was so delighted with
it, that he promised him any reward he should ask; —
upon which Sessa requested that he might be allowed
one grain of wheat for the first square on the chess
board, 2 for the second. 4 for the third, and so on, doub-
ling continually, to 64, the whole number of squares. —
JNTow, supposing a pint to contain 7680 of these grains,
and one quarter or 8 bushels to be worth 27s. 6d., it is
-required to compute the value of all the wheat?
£64481488296.
7. What sum would purchase a horse with 4 shoes, and
eight nails in each shoe, at one farthing for the first nail,
& halfpenny for the second, a penny for the third, &c.,
doubling to the last? Ans. "£4473924. 5s. 3ieJ.
8. A merchant sold 15 yards of satin, the first yard
for Is. the second for 2s. the third for 4s. the fourth for
8s. &c.; what was the price of the 15 yards?
Ans. £1638. 7s.
9. Bought 30 bushels of wheat, at 2d. for the first
bushel, 4d. for the second, 8d. for the third, &c. ; what
does the whole amount to, and what is the price per
bushel on an average?
\ £8947848. 10s. 6d. Amount.
'• | £298261. 12s. 4d. per bushel.
14
158 PERMUTATION.
PERMUTATION.
Permutation is used to show how many ways things
may be varied in place or succession.
RULE.
Multiply all the terms of the series continually, from
1 to the given number inclusive; and the last product
will be the answer required.
EXAMPLES.
1. How many changes can be made with 8 letters of
the alphabet?
1X2X3X4X6X6X7X8=40320 Answer.
2. In how many different positions can 12 persons
place themselves round a table?
1X2X3X4X5X6X7X8X9X10X11X12 =
479001600 Ans.
3. How many changes may be made with the alpha-
bet? Ans. 620448401733239439360000,
SKETCH OF MENSURATION,
OF PLANES AND SOLIDS.*
Planes, surfaces, or superficies, are measured by the
inch, foot, yard, &c., according to the measures used by
different artists. A superficial foot is a plane or surface
of one foot in length and breadth, without reference to
thickness. Solids are measured by the solid inch, foot,
yard, &c.; thus, 1728 solid inches, that is 12X12X12
make one cubicle or solid foot. Solids include all bodies
which have length, breadth and thickness.
ARTICLE I.
To measure a square having equal sides.
RULE.
Multiply any one side of the square by itself, and the
product will be the area, or superficial contents, in feet,
yards, or any other measure, according to the measure
used in measuring the sides.
* Planes are the same as superficies, or surfaces.
SKETCH OF MENSURATION.
159
EXAMPLE.
Let A, B, C and D represent a square, having equal
sides each measuring 20 feet. Multiply the length of
one side by itself, thus —
20 feet.
20 feet.
Ans. 400 square feet.
ARTICLE II.
To measure the plane or surface of a parallelogram.
RULE.
Multiply the length hy the breadth — the product will
be the superficial contents.
EXAMPLE.
Let A, B, C and D represent a parallelogram whose
length is 40 yards, and breadth 15 yards.
yards.
breadth 15
length 40 yards.
Ans. 600 square yards. C
Note. — The contents of boards and other articles
which, are measured by feet, &c., may be easily found
by Duodecimal fractions.
ARTICLE III.
To measure the plane or surface of a triangle.
RULE.
Multiply the base by half the perpendicular, if it be
a right angled triangle, and the product will be the area,
or superficial contents; or multiply the base and perpen-
dicular together, and half the product will be the area.
But if it be an oblique angled triangle, multiply half
the length of the base by a perpendicular let fall on the
base from the angle opposite to it, and the product will
be the area.
160
SKETCH OF MENSURATION,
EXAMPLES.
1. Let C, H and G represent a right angled triangle,
having the right angle at G ; the base C G being 40
feet, and the perpendicular H G, 28 feet.
No. 1.
14 feet, or half the perpendicular.
40 feet, or the base.
560 feet — the area.
2. Let B, C and D represent an oblique angled tri-
angle; the length of the base B D being 80 feet, and
the perpendicular C E, 28 feet.
No. 2.
28 the perpendicular. C
40 half the base.
1120 Answer.
Mote. — Right angled triangles are such as have one
angle like the corner of a square, and which is called
the right angle, containing 90 degrees; as the angle G
in the triangle. No. 1. — Oblique angled triangles are
such as have each of the angles, either more or less
than 90 degrees, as in the triangle, No. 2.
ARTICLE IV.
To measure a circle.
Note. — Circles are round figures, bounded every where
by a circular line, called the periphery, and also the cir-
cumference. A line passing through the centre is cal-
led the diameter. Half the length of the diameter is
called the radius.
The diameter may be found by the circumference,
-thus — As 22 is to 7 so is the circumference to the diam
eter; and in like manner may the circumference be
SKETCH OF MENSURATION. 161
found by the diameter; for, as 7 is to 22, so is the di-
ameter to the circumference.
ARTICLE V.
To find the superficial contents, or area, of a circle.
RULE.
Multiply half the circumference by half the diame-
ter, and the product will be the answer. Or, multiply
the square of the diameter by .7854; or multiply tlie
square of the circumference by .07958, and in either
case the product will be the answer.
EXAMPLE.
How many square feet are contained in a circle whose
circumference is 44 feet, and whose diameter is 14 feet?
22 half the circumference.
7 half the diameter.
154 square feet. Answer.
The same may be done by multiplying the diameter
find circumference together, and dividing the product
by 4, thus, 44X 14=61 6-;-4= 154. Answer.
ARTICLE VI,
To measure the surface of a globe or sphere.
RULE.
Multiply the circumference by the diameter, the pro-
duct will be the surface, or area.
EXAMPLES.
1. What are the superficial contents of a globe whose
circumference is 220 feet, and whose diameter is 70
feet ? 220 X 70= 1 5400 square feet. Ans.
2. How many square miles are contained on the sur-
face of the whole earth, or globe, which we inhabit?
The circumference of the earth is estimated to be
25020 miles, and the diameter, 7964, nearly.
Then, 25020X7964=199259280.
Ans. 199259280 square miles.
14*
162 SKETCH OF MENSURATION,-
ARTICLE VII.
To find the solid contents of a cube*
RULE.
Multiply the length of one side by itself, and multiply
the product by the same length, that is, by the same
multiplier; the last product will be the solid contents of
the cube.
EXAMPLES.
1. How many solid feet are contained in a cube, or
solid block of 6 equal sides, each side being 3 feet in
length, and 3 in breadth?
3X3X3=27 solid or cubick feet. Ans.
When the contents are required of right angled solids,
whose length, breadth, &c., are not equal: multiply the
length by the breadth, and that product by the thick-
nes* — the product will be the answer
2. Required the contents of a load of wood, whose
length is 8 feet, breadth 4 feet, and height or thickness
4 feet. 8X4X4=128 solid feet, or 1 cord. Ans.
3. Required the contents of a stone i6£ feet in length,
\\ in hrf.adth, and 1 foot in thickness.
16.5X1.5X1 = 24.75 solid feet, or 1 perch. Ans,
Note, — Solids whose dimensions are in feet and inches,
are more easily measured by Duodecimals.
ARTICLE VIII.
To find the contents of a prism.
A prism is an angular figure, generally ®f three equal
sides, whose ends are in the form of triangles. It re-
sembles a fife of three sides, whose whole length is of
equal bigness.
RULE.
Find the area or superficial contents of one end as of
any other triangle, then multiply the area by the length
of the prism, and the product will be the solidity.
EXAMPLE.
What are the solid contents of a prism, the sides of
the triangles of which measure 13 inches, the perpen
* A cwbe » a solid body of equal sides, each of which is an exact square.
SKETCH OF MENSURATION. 163
dicular extending from one of its angles to its opposite
side, 12 inches, and its length 18 inches?
13X12=156-^-2=78X18—1404 cubick inches. Ans,
ARTICLE IX.
To find the contents of a cylinder.
A cylinder is a long round body, all its length being
of equal bigness, like a round ruler.
RULE.
Find the area of one end, by the rule for finding the
area of a circle, then multiply it by the length, and the
product will be the answer.
EXAMPLE.
What is the solidity of a cypher, the area of one end
of which contains 2.40 square feet, and its length being
12.5 feet? 2.40 X 12.5=30 solid feet. Ans.
ARTICLE X.
To find the solid contents of a round stick of timber, which
is of a true taper from the larger to the smaller end.
RULE.
Find the area of both ends; add the two areas to^
gether, and reserve the sum; multiply the area of the
larger end by the area of the smaller end, extract the
square root of the product, add the root to the reserved
sum, then multiply this sum by one third the length of
the stick, and the product will be the solidity.
Note. — As this method requires considerable labour,
the following has been preferred for common use, though
not quite so accurate.
RULE.
Girt the stick near the middle, but a little nearer to
the larger than to the smaller end; this will give the
circumference at that place. Find the diameter by the
circumference; multiply the circumference and diame-
ter together; then multiply one fourth of the product
by the length, and the answer will be nearly the solid
contents.
EXAMPLE.
What is the solidity of a round stick of timber that
164 PRACTICAL QUESTIONS.
is 10 feet long, and its circumference near the middle is
2.61 feet?
As 22 : 7 : : 2.61 : .83 diameter,
cir. diam. length, feet.
2.16 X. 83=2. 1663-^4^.5415X10=^ 5 41 50.
Ans. 5.4 160 solid feet.
ARTICLE XI.
To find the solid contents of a globe.
RULE.
Multiply the cuhe of the diameter by .5236, the pro-
duct will be the solid contents. Or, multiply the super-
ficial contents, or surface, hy one sixth part of the sur-
face. Or, multiply the cube of the diameter by 11.
and divide the product by 21 — in either case the pro-
duct will be the solidity.
EXAMPLES.
1. What are the solid contents of a globe whose di-
ameter is 14 inches?
14X 14 X 14=2744X. 5236= 1436.7584
cubick inches. Ans.
2. How many solid miles are contained in the earth,
or globe, which we inhabit?
Suppose the diameter to be 7954 miles; then, 7954X
7954X7954=503218686664 thecubenf the earth'saxis,
or diameter; then,
503218686664X.5236=263485304337
cubick miles. Ans.
Note. — The solidity of a globe may be found by the
circumference, thus — Multiply the cube of the circum-
ference by .016887 — the product will be the contents.
PRACTICAL QUESTIONS.
1. A cannon ball goes about 1500 feet in a second of
time. Moving at that rate, what time would it take in
going from the earth to the sun; admitting the dis-
tance to be 100 millions of m'les, and the year to con-
tain 365 days, 6 hours? Ans.
PRACTICAL QUESTIONS. iGc/
2. A young- man spent j of his fortune in 8 months, f
of the remainder in 12 months more, after which he
had £410 left. What was the amount of his fortune?
Ans. £966 13s. 4d.
3. What number is that, from which if you take % of
J, and to the remainder add •£$ of -^ , the sum will be
10? Ans. KVAV-
4. What part of 3, is a third part of 2? Ans. f.
5. If 20 men can perform a piece of work in 12 days,
how many will accomplish another thrice as large, in
one fifth of the time? Ans. 300.
6. A person making his will, gave to one child ±% of
his estate, and the re*t to another. When these lega-
cies were paid, the one proved to be £600 more than
the other. What was the worth of the whole estate?
Ans. £2000.
7. The clocks of Italy go on to 24 hours; how many
strokes do they strike in one complete revolution of the
index? Ans. 300.
8. What quantity of water must be added to a pipe
of wine, valued at £33. to bring the first cost to 4s. 6d.
per gallon? Ans. 20| gallons.
9. A younger brother received £6300, which was J
of his elder brother's portion. Wrhat was the whole es-
tate? Ans. £14400.
10. WThat number is that which being divided by 2, or
3, 4, 5, or 6, will leave 1 remainder, but which if di-
vided by 7 will leave no remainder? Ans. 72L
11. What is the least number that can be divided by
the nine digits without a remainder? Ans. 2520.
12. How many bushels of wheat, at $1.12 per bushel,
can I have for $81.76? Ans. 73,
13. What will 27 cwt. of iron come to, at $4.56 per
cwt.? Ans. $123.12.
14. When a man's yearly income is 949 dollars, how
much is it per day? Ans. $2.60,
15. My factor sends me word he has bought goods to
the value of £500. 13s. 6d. upon my account,* what will
his commission come to at 3£ per cent.?
Ans. £17. 10s. 5Jd
166 PRACTICAL QUESTIONS.
16. How many yards of cloth, at 17s. 6d. per yard,
can I have for 13 cwt. 2 qrs. of wool, at 14d. per lb.?
Ans. 100 yards. 3} qrs.
17. There is a cellar dug that is 12 feet every way,
in length, breadth, and depth; how many solid feet of
earth were taken out of it? Ans. 1728.
18. If | of an ounce cost £ of a shilling, what will £
of a lb. cost? Ans. 17s. 6d.
19. If | of a gallon cost f of a £. what will f of a tun
cost? Ans. £105.
20. If | of a ship be worth £3740, what is the worth
of the whole? Ans. £9973. 6s. 8d.
21. What is the commission on $2176.50, at 21 per
cent.? Ans. $54.411.
22. In a certain orchard | of the trees bear apples, i
pears, £ plums, 60 of them peaches, and 40 cherries;
how many trees are in the orchard? Ans. 1200.
23. If A travel by mail at the rate of 8 miles an hour,
and when he is 50 miles on his way, B start from the
same place that A did, and travel on horseback the
same road at 10 miles an hour, how long and luow far
will B travel to come up with A?
Ans. 25 hour?, and 250 miles.
24. Bought a quantity of cloth for 750 dollars, & of
which I found to be inferior which 1 had to sell at 1 dol-
lar 25 cents per yard, and by this I lost 100 dollars:
what must I sell the rest at per yard that 1 shall lose
nothing by the whole? Ans. $3.15if .
25. If the Earth goes round the sun in 365 days, 5
hours, 48 minutes, 49 seconds, and its distance from the
sun 95000000 miles, what must be the distance of the
planet Mercury from the Sun, admitting the time of its
revolution round the Sun to be 87 days, 23 hours, 15
minutes, 40 seconds?
Note. — The planets descrihe equal areas in equal
time?: therefore, as the square of the time of the revo-
lution of one planet, round the Sun, is to the square of
the time of the revolution of any other planet, so is the
cube of the distance of one planet from the Sun, to the
cube of the distance of any other from the Sun,
PLATE
10 BE USED IN STUDYING VULGAR FRACTIONS.
I I
In using; the Fractional Plate, the student must count the white
spans, and not the black lines. The first row of squares, or white
spaces, at the top, are whole numbers ; the second row is divided
into halves ; the third, into thirds, and so on from the top to the
bottom. Thus it may be shown at one glance, that 7 halves
.make three a;id a half, or that 8 thirds make 2 and 2 thirds, &e.
CONTENTS.
Numeration, - - -
Simple Addition, -------
Simple Subtraction,
Simple Multiplication, ------ 15
Simple Division, ------- 20
•Federal Money, 28
Tables of Money, Weights, Measures, £c., -
Compound Addition, ------ 43
Compound Subtraction, ------ 48
Compound Multiplication, ----- 52
Compound Division, ------ 60
Reduction, -------- 64
Exchange, -------- 71
Vulgar Fractions, ------- 76
Decimal Fractions, ------ 93
Duodecimals, ------- 101
Single Rule of Three, 105
Doable Rule of Three, - - - - - - 111
Practice, 113
Fellowship, - 121
Tare and Tret, 123
Barter, 126
Siuinle Interest, 127
Compound Interest, - - - - - - 134
Insurance, Commission and Brokerage, - - - 135
Discount, -------- 136
Equation, 137
Loss and Gain, - - 138
Involution, - - - - - - - - 139
Evolution, -------- 140
.Square Root, -------- ib.
Cube Root, 143
Biquaratc Root, 145
Alli'/Jition, 146
Position, 149
Double Position, 151
Arithmetical Progression, - 153
Geometrical Progression, ----- 155
fVni; itation, -- 158
Sketch of Mensuration, ib.
Practical Questions, .----- 164
02426
iv?308O89
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