SUBTRACTION
PROOF.
Add the remainder to the lower line.
EXAMPLES.
. s. d.
From 25(T 15 6
Take 129 12 8 $
31
T. cwt. qr. Ib. oz. dr.
From 246 15 2 18 11 5
Take 89 16 1 24 8 15
Rem. 127 2 9 }
Proof 256 15 6
mi. fur. P. ft. in. be.
From 250 4 24 10 61
Take 125 6 30 5 10 2
D. h. mi. sec.
From 325 18 30 24
Take 236 20 45 50
bn. , P. qt. pt.
From 204 2 6 1
Take 150 3 2
sig. deg. mi. sec.
From 6 16 32 29
Take 3 . 24 16 48
T. hhd. gal. qt. pt.
From 50 2 45 2 1
Take 20 3 60 3
A. R. P.
From 1658 2 Vj
Take 1249 3 ?4
Promiscuous Questions in Compound Addition and
Subtraction.
1. A merchant bought five pieces of linen, containing as
follows: No. 1, 36 yards 3 quarters 2 nails; No. 2, 45
yards 1 quarter 3 nails ; No. 3, 48 yards 2 quarters 1 nail ;
No. 4, 52 yards 3 nails ; No. 5, 64 yards 2 quarters ; how
many yards were in all ? Ans.. 247 yds. 2qr. Ina.
2. Sold 5 head of beef cattle, at the following prices, viz :
the first far 67. 2s. 4d. the second for 5Z. 10s. 9%d. /he third
for 11. the fourth for SI. 10s. 6d. the 5th for Ql. 2s. 6d. and
received. 2.21. 10s. 6d. in ready payment, and a note for the
remainder ; how much did the cattle cost, and for how much
was the note given ?
Ans. The cattle cost 367. 6s. l^d. and the note wastfor
13/. 15*. l%d.
32 COMPOUND MULTIPLICATION.
3. A silversmith bought 26lb. 9oz. lOdwt. of silver, and
. wrought up ISlb. \6dwt. Wgr. how much has he left ?
Ans. Sib. Soz. ISdwt. llgr.
4. A physician bought Qlb. Woz. 6dr. 2sc. (apothecaries'
weight) of medicine, and has used 4Z&. Soz. 4dr. Isc. I7gr.
what quantity has he yet remaining ?
Ans. 2lu. 5oz. 2dr. Osc. 3gr.
5. William was born on the 15th day of January, 1816,
at 6 o'clock in the morning, and Charles was born on the
20th of March, 1817, at 9 in the evening ; how much older
.6 William than Charles 1 Ans. 1 year, 2mo. 5d. 15h.
ft An innkeeper bought four loads of hay, weighing as
lollowing, viz. first load, 19 hundred 2 quarters and 14 Ib.
second load, 16 hundred 3 quarters 18 Ib. third load, 22
hundred and 24 Ib. fourth load, 24 hundred and 1 quarter
how much hay in all ? Ans. 4 tons 2 hundred.
7. From a piece of broadcloth which at first measured
55 yds. I sold to A 5J yds. to B 6, to C 7, to D a quan
tity not recollected, and to E just half as much as to D ; on
Pleasuring the remainder, I found there was 20J yds. left ;
how many yards did D and E each receive 1
Ans. D 10, and E 5 yds.
8. A wine merchant bought 1 pipe 2 hhds. and 3 qr.
casks of wine, each 26 galls. ; of these he sold 1 hhd. and
jr. casks; he also found that the pipe had leaked 17 galls,
the remaining hogshead 11, and the cask 5J ; how many
gallons did he buy, and how many had he left ?
Ans. Bought 330 galls, left 18l galls.
9. Bought 4 pieces of cloth, the two first measured 9 Ells
Fr. 3 qr. 2 na. each, and the two last 8 Ells Fr. 2 qr. 3 na.
each, of these I sold 40 5 yards ; how much have I left ?
Ans. 13y. 2 qr. 2 na.
SECTION 4.
OF COMPOUND MULTIPLICATION.
COMPOUND MULTIPLICATION teaches to multiply any given
quantities or numbers of divers denominations.
Case 1.
When the multiplier does not excer\
12 9j
2 4 2
Ans.
.
11
10
7
s. d.
11 8
2, 84
8i
26 10 6
Case 2.
When the multiplier exceeds 12, uut is the exact produd
of any two factors in the table 1 .
34 COMPOUND MULTIPLICATION.
RULE.
Multiply the given sum by any one of these, in the same
nanner as above, and the product by the oilier.
EXAMPLES.
. s. d. qr.
Multiply 12 8 6 by 18=3x6
3
37 5 6 
6
223 13 4
Application.
1. Multiply 4T. Scwt. Iqr. 16ZZ>. Soz. Wdr. by 36.
Arts. 150 T. 2cwt. Iqr. lib. 6oz. 8dr.
2. 120Z. (5s. 9d. by 24. A/is. 238SZ. 2s. Od.
3. 24 T. 4e7/tf. 2. 4/>. I2?ni. f>v. by 2JK
AH*. 237 />. I/?. 507/w.
COMPOUND DIVISION. 35
Case 4.
When the multiplier exceeds the product of any two fac
tors in the table.
RULE.
Multiply by the units figure, as in case 1, and set down
the amount; again multiply the given sum b}* the figure of
tens, and that product by 10, and place tins amount under
the first ; again multiply by the figure of hundreds, and the
product by 10 and 10, which set down under the other pro
ducts in the same way for thousands, by three tens, &c.
EXAMPLES. 
s. d. 2626
Multiply 2 6 by 245 4 2
5
10 50
12 6 first, or units product 10 10
500 second, or tens do.
25 third, or hundreds do. 500 2100
10
Ans. 30 12 6 total.
25
. s. d. . s. d.
2. Multiply 14 6 by 240 Ans. 174
3. 123 117 130 3 3
4. 126 275 309 7 6
SECTION 5.
OF COMPOUND DIVISION.
COMPOUND DIVISION, teaches to divide any sum or quan
tity of divers denominations.
Case 1.
When the divisor does not exceed 12.
RULE.
Divide the highest, or lefthand denomination; if any re
mains, multiply by that number which will reduce it to the
next highest, add this product to the second, then divide as
Df.'ibre, and so proceed till all are divided.
30 COMPOUND DIVISION.
PROOF By compound multiplication.
EXAMPLES.
. s. d. qr. . s. d. qr
2)465 10 6 i 3)563 15 4
Quotient 232 15 3
2
Proof 465 10 6
T. cwtf. f/r. Ib. yds. ft. in.
6)91 16 1 14 5)960 1 9
T. Mid. gal. qt. . w. d. h. mi. sec.
8)468 1 48 3 10)30 6 18 48 50
Case 2.
When the divisor is the exact product of any two factors
in the table.
RULE.
Divide first by one as above, and the quotient by the
other.
EXAMPLES.
6)224 12 6 by 30=6x5
5) 37 8 9
Quotient
7 9
9
.
s.
d.
.
s.
d.
2. Divide
134
18
S
by 44
Ans. 3
1
4
3.
9H4
144
6
16
8
4.
474
72
6
11
8
Case 3.
When the divisor is not the exact product of any two (ac
tors in the table, or eA,:ee.ds them.
RULE.
Divide the highest denomination in the ^iv<n sum, in the
.s.iuic manner as in case 2, of whole numbers; reduce I lit
remainder, if any, to the next lower denomination, adding it
COMPOUND DIVISION. 37
to the number of the same denomination in the given sum ; di
vide this in the same manner, and so proceed till all are divided.
EXAMPLES.
1. Divide 264Z. 10s. 7$d. by 25.
25)264/. 10s. l^d. (Wl.lls.l^d. Ans.
25
20
25)290(11
25
~40
25
15
12
25 ) 187 ( 7
175
Is
4
25)50(2
50
. s. d. . s. d.
2. Divide 409 13 9 by 345 Ans. 139
3. 232 4 9 524 8 l(5i
4. 3236 12 4J 654 4 l3 llj
5. 132 8 68 1 18 10
Promiscuous Questions, for exercise, in the foregoing rules
of Compound Addition, Subtraction, Multiplication and
Division.
1. What is the value of 672 yards of linen at 2s. 5d. per
yard? Ans. 81Z. 45.
2. A goldsmith bought 11 ingots of silver, each of which
weighed 4db. loz. I5dwt. 22gr. how much do they all
weigh? Ans. 45Z&. loz. 15dwt. 2gr.
3. Bought 8 loads of hay, each weighing 1 ton 2 hundred
3 quarters 16 pounds; how much hay in all?
Ans. 9 ton 3 hun. 161b.
4. Divide 9 ton 3 hundred 161b. into 8 shares.
Ans. 1 ton 2 hun. 3qr. 16lb.
5. Bought 15 tracts of land, each containing 300 acres
2 roods and 20 perches ; what is the amount of the whole ?
Ans. 4509 acres 1 qr. 20 rods.
D
38 COMPOUND DIVISION.
6. Divide a tract of land containing 4509 acres 1 rood
and 20 perches equally among 15 persons; what is each
one's share? Ans. 300 acres 2 roods 20 perches.
7. Bought 179 bushels of wheat for 201 dollars 37^ cts.
what is it per bushel? Ans. I doll. 12^ cents.
8. If a man spends 7 pence per day, how much will it
amount to in a year? Ans. 10Z. 12s. lid.
9. What is the value of 1000 bushels of coal at 10 cents
per bushel? Ans. 105 dolls.
10. Bought 135 gallons of brandy at 1 dollar arid 62
cents per gallon, which was sold for 2 dollars and 5 cents
per gallon ; required the prime cost, what it was sold for, and
the gain? Ans. prime cost 219 dolls. 37^ cts.
sold for 276 dolls. 75 cts. gain 57 dolls. 37 cts.
11. If 27 cwt. of sugar cost 47/. 12s. W$d. what cost 1
cwt. ? * Ans. ll. 15s. 3^d.
12. Suppose a man has an estate of 9708 dollars, which
he divides among his four sons : to the eldest he gives f , and
to the other three an equal share of the remainder ; what is
the share of each?
Ans. eldest son, 3883 dolls. 20 cents, other sons, each
1941 dollars 60 cents.
13. A dollar weighs lldwt. Sgr. what will 45 dollars
weigh at that rate? ^Ans. 39oz.
14. An eagle of American gold coin should weigh lldwt.
tigr. now 150 were found to weigh 84oz. 7dwt. 20gr. how
much was this over or under the just weight ? Ans. &gr. over.
15. What cost 2^ cwt. of sugar, at 13 cents 3 mills per
pound ? Ans. 37 dolls. 24 cts.
16. A merchant deposited in bank 35 twentydollar notes,
63 eagles, 284 dolls. 642 half dolls. 368 qr. dolls. 256
twelve and a half cent pieces ; he afterwards gave a check
to A for 560 dolls, and another to B for 820 dolls. ; what
sum has he still remaining in bank? Ans. 679 dolls.
17. A merchant bought a, piece of broad cloth containing
,o yards, at 4 dolls. 66 cts. per yard ; of this he found 4
yards were so damaged that he sold them at half price ; 8
yards he sold at 5 dolls. 50 cents per yard : on the whole
piece he gained IM) dolls. 56 cents ; at what rate did he sell
the remainder? Ans. (> dollars per yard.
18. Five travellers, upon leaving a tavern in I ho morning,
icund they were charged 12 cents eaeh HM their beds, 4
limes flint sum for their spnpor rind break (hst. 75 eenls fnr
RKDIJCTJOJV. 39
liquor among them all, 25 cents each ibr hay ; the remain
der of their bill, which amounted to 6 dollars, was for oats
at 2 J cents per quart; how many gallons of oats had they,
and how much had each man to pay ?
Ans. 8 gallons ; each paid 120 cents.
19. A laborer engaged to work for 75 cents per day,
working 8 hours each day, or 8 hours for a day's work ; but
being industrious he worked 12 hours 25 minutes each day
for five days, and then 11 hours 30 minutes for 9 days more ;
what sum is he entitled to receive for his services ?
Ans. 15 dollars 52 cents 3 millsf
20. If 25 hhds. contain 1534 galls. 1 qt. and 1 pt. of
brandy, each an equal quantity, how much is there in each
hogshead? Ans. 61 galls. 1 qt. 1 pt.
21. If a man do 114 hours 45 minutes' work in 9 days,
how long did he work each day? Ans. I2h. 45mi.
22. Divide 180 dollars among 3 persons A, B, and C;
give B twice as much as A, and C three times as much as B.
C A 20 dolls.
Ans. I B 40
f C 120
SECTION 6.
OF REDUCTION.
REDUCTION is the changing of numbers from one denomi
nation to another, without altering their real value. Thus.
1 dollar, if reduced to cents, will be 100 cents, which in
their real value are equal to 1 dollar: or, 3 feet reduced
to yards, is one yard, which is still the same length as the
3 feet.
RULE.
If the reduction is from a higher to a lower denomina
tion, multiply ; but if from a lower to a higher denomina
tion, divide by as many of the next less as make one of the
greater ; adding the parts of the same denomination to the
product as it descends ; and setting down the remainders as
il ascends.
Reduction ascending, and descending, mutually prove each
t.lher.
40
REDUCTION.
>>CQd
^ 3 3 ^
&
>>>>>>>> 1
N.
b
S" hrl
Ifli
s r ra
^* fa
o
s*
~3 a
5 30
i^p I*. . >? . ^
S .p
p ^
>t
3 p "p
3
* " p
eta.
* PT* 5'
o
...
^
p^
 ,.
*~
ft
P
XV'^
i.
^
H
t i R*
^
5*
COCO gr
l ~*3
__ ' a
*
000
OOl 1 t 1 i u COH 'iob^
J 55
IS
^
O 1 > 4
Cn
I 1 H* 1 > ^ i 1 "
O5 O5t 'H>v!O5O5tO:o
j JJ , g 1 ^ St'
Cft
1
H*
00000000?
? ^ If ir!
1
000
^HH^H^tO^^.
?^^^ ^',0
P
a s ]1 5 Cfe^ 1
i
t h O5
05
tswsiaoScDooS*
..^l 5" o
s
6C tr^ O
00
oooooooo a.
^*I ! r^^'
3
ff
1
O O O
H H 00
o
00
t H ( 'I * tO Oi i'5 O5 ^
1 ' I H 1
1^ ., L
1"
^1 CO
CO
oooooooo?
1 : ^j^

OO
o
H.M^MW^W^^
^ b3'^ t'l
2
tt )t >
00
t H* 1 *
^ooi^aiCfltooo? 9
2 5 a"^ ^, ! ^^ r*.
o
1
O5 GO O5
^ ^ ^
a ~" ^ o 2 "^
as
'CO
C? C^
r '
OO
OOl H'HOOtO^p
? S 5 > ^
H*
o
O H* ^
en
^JQDI 'h'OOOOO;*
Illl ^
j^
t0 00
^
O5OOCOOOOO .^
? a v 2. 
"^ ^
~~ * *~* "''
i 2 is < 68,56
1000 > ( 6,856
3. Divide 65321 by 23,7 Ans 2756,16 +
4. 234,70525 64,25 3,653
5. 10 3 3,3333 +
6. 9 ,9 10
7. ,00178600398 ,00463 ,385746
DIViSIOiN OF DECIMALS. fj 1
8. Divide ,2327898 by 2,46 Ans. ,09463
9. ' ,2327898 ,09463 2,46
10. * ,000162 ,018 ,009
CONTRACTION IN DIVISION OF DECIMALS.
When only a limited number of decimals in the quotient
is sought for, work by the following
RULE.
1. Take as many figures only on the left hand side of
the divisor, as the whole number of figures sought for in the
quotient, and cut off the rest.
2. Make each remainder a new dividend, and for a new
divisor, point off one figure continually from the right hand
of the former divisor, taking care to bring in the increase,
or carriage of the figures so cut off, as in multiplication.
Note. When the whole divisor does not contain as many figures as
are sought for in the quotient, proceed as in common division, without
cutting off a figure, till the figures in the divisor shall equal the re
maining figures required in the quotient, and then begin to cut off as
above directed.
EXAMPLES.
1. Divide 14169,206623851 by 384,672258, retaining
four decimal places in the quotient, or in all six quotient
figures.
3.8.4,6.7.2258 ) 14169,206623851 ( 36,8345. Ans.
1154017
262903
230803
32100
30774
1326
1154
172
153
19
19
52 REDUCTION OF DECIMALS.
2. Divide ,07567 by 2,32467, true to four decimal places,
or three significant figures, the first being a cipher.
2,3.2467 ) ,07567 ( ,0326. Ans.
697
59
46
13
14
3. Divide 5,37341 by 3,74, true to four decimal places.
3,74 ) 5,37341 ( 1 ,4367. Ans.
374
1633
1496
1374
1122
252
224
28
26
4. Divide 74,33373 by 1,346787, true to three decimal
places. Ans. 55,193.
5. Divide 87,076326 by 9,365407, true to three decimal
places. Ans. 9,297.
6. Divide 32,68744231 by 2,45, true to two decimal
olaces. Ans. 13,34.
7. Divide ,0046872345 by 6,24, true to five decima
places. Ans. ,00075
SECTION 5.
REDUCTION OF DECIMALS.
Case 1.
To reduce a vulgar fraction to a decimal.
REDUCTION OF DECIMALS. 53
RULE.
Annex one or more ciphers to the numerator, and divide
by the denominator; the quotient will be the answer in
decimals.
EXAMPLE.
1. Reduce to a decimal.
4)1,00
,25 Ans.
2. Reduce to a decimal. Ans. ,5
3.  J to a decimal. ,75
4.  J to a decimal. ,875
5.  Jj to a decimal. ,04
6.  j to a decimal. ,95
7.  T 6 3 of a dollar to cents. ,40 cts.
Case 2.
To reduce numbers of different denominations to a deci
mal of equal value.
RULE.
Set down the given numbers in a perpendicular column,
having the least denomination first, and divide each of them
by such a number as will reduce it to the next name, annex
ing the quotient to the succeeding number ; the last quotient
will be the required decimal.
EXAMPLE.
1. Reduce 17s. 8%d. to the decimal of a pound.
4
12
20
3
8,75
17,720166
,8864583 + Ans.
2. Reduce 195. to the decimal of a pound. Ans. ,95
3.  3d. to the decimal of a shilling. ,25
4.  3d. to the decimal of a pound. ,0125
5.  kcwt. 2qr. to the decimal of a ton. ,225
6.  2qr. 14/6. to the decimal of a cwt. ,625
7.  3qr. 3na. to the decimal of a yard. ,9375
E2
54 REDUCTION OF DECIMALS.
Case 3.
To reduce a decimal to its equal value in integers.
RULE.
Multiply the decimal by the known parts of the integer
EXAMPLE.
1. Reduce ,8864583 of a pound to its equivalent value
in integers.
,8864583
20
s. 17,7291660
12
d. 8,7499920
4
qr. 2,9999680
It is usual when the left hand figure in the remaining
decimal exceeds five, to expunge the remainder, and add
one to the lowest integer. Thus, instead of 17s. 8d. 2,999,
die. we may say 17s. 8%d. Ans.
2. What is the value of ,75 of a pound ? Ans. 15s.
3. What is the value of ,7 of a pound troy?
Ans. Soz. Sdwt.
4. What is. the value of ,617 of a cwt. ?
Ans. 2qr. 13Z&. 1 oz. W + dr.
5. What is the value of ,3375 of an acre?
Ans. I rood, 14per.
6. What is the value of ,258 of a tun of wine ?
Ans* Ihhd. 2 {gals.
7. What is the proper quantity of ,761 of a day?
Ans. ISh. 15?ni. 50,4src.
i. What is the proper quantity of ,7 of a Ib. of silver?
Ans. Soz. Sdwt.
: What is the proper quantity of ,3 of a year?
Ans. lQ9d. 13ft
10. What is the difference between ,41 of a day and ,16
of an hour ? Ans. 97*. 40wi. 48sec.
11. What is the sum of ,17T. 19^. ,l7qr. and lib. 1
Ans. Zcwt. 2qr.
DECIMAL KHACT1O.NS. 55
Promiscuous Questions in Decimal Fractions.
1. Multiply ,09 by ,000. Ans. ,00081.
2. In ,36 of a ton (avoirdupois) how many ounces ?
Ans. 12902,402.
3. What is the value of ,9125 of an ounce troy?
Ans. 18dwt. 6gr.
4. Reduce ^j to a decimal. Ans. ,0127 nearly. .
5. Reduce 2oz. I6dwt. 20gr. to the decimal of a pound
troy. Ans. ,2368 f nearly.
6. What is the length of ,1392 of a mile?
Ans. 1 fur. 4 per. 3 yds nearly.
7. What multiplier will produce the same result, as mul
tiplying by 3, and dividing the product by 4 ?
Ans. ,75.
8. What decimal of Icwt. is 6lb. Ans. ,0535714.
9. What part of a year is 109 days 12 hours?
Ans. ,3.
10. In ,04 of a ton of hewn timber, how many cubic
inches? Ans. 3456.
11. What is the value of T 3 T of a dollar divided by 3?
Ans. 6f cents.
12. What is the value of ,875 of a hhd. of wine?
Ans. 55 gal. qt. 1 pt.
13. What divisor, true to six decimal places, will produce
the same result as multiplying by 222 ?
Ans. ,004504.
14. In ,05 of a year, how many seconds, at 365 days 6
hours to the year? Ans. 1577880.
15. What number as a multiplier will produce the same
result as multiplying by ,73 and dividing first by 3, and the
quotient by ,25 ?" Ans. ,973^.
16. What is the difference between ,05 of a year, and ,5
of an hour ? Ans. 2i. 2d~ ISh. 42m.
17. In ,4 of a ton, ,3 of a hhd. and ,8 of a gallon, how
many pints ? Ans. 964.
18. How many perches in ,6 of an acre; multiplied by
X)2? Ans. 1,92.*
19. What part of a cord of timber is 1 cubic inch?
Ans. ,000004 f
20. What part of a circle is 28 deg. 48 minutes ?
ATU. ,08.
66 SINGLE RULE OF THREE DIRECT
PART IV.
PROPORTIONS.
Tins part of arithmetic which treats of proportions is
very extensive and important. By it an almost innumerable
variety of questions are solved. It is usually divided into
three parts, viz. Direct, Inverse, and Compound. The first
of these is called the Single Rule of Three Direct, and
sometimes by way of eminence the Golden Ride. The sec
ond is called the Single Rule of Three Inverse: and the
last is called the Double Rule of Three. In all these, cer
tain numbers are always given, called data* by the multipli
cation and division of which, the answer in an exact ratio
of proportion to the other terms is discovered.
SECTION 1.
SINGLE RULE OF THREE DIRECT
IN this rule three numbers are given to find a fourth, that
shall ha\e the same proportion to the third, as the second
has to the first.
If by the terms of the question, more requires more, or
less requires less y it is then said to be direct, and belongs to
this rule.
In stating questions in this rule, the middle term must
always DC of the same name with the answer required; the
last term is that which asks the question, and that which is
of the same name as the demand, the first. When the ques
tion is thus stated, reduce the first and third terms to the
lowest denomination in either ; and the middle term (if com
pound) to its lowest, and proceed according to the following
RULE.
Multiply the second and third terms together, and divide
the product by the first ; the quotient will be the fourth term,
or answer, in the same name with the second.
SINGLE RULE OF THREE DIRECT. 5?
PROOF.
Invert the question, making the answer the first term ; the
result will be, the first term in the original question.
Note. 1. After division if there be any remainder, and the quotient
be not in the lowest denomination, it must be reduced to the next less
denomination, dividing as before, till it is brought, to the lowest denomi
nation, or till nothing" remains.
2. When any of the terms are in federal money, the operation is con
ducted in all respects as in simple numbers, taking care to place the
separatrix bet'ween dollars and cents, according to what has already
been laid down in federal money and decimal fractions.
EXAMPLE.
1. If 8 yarcls of cloth cost 32 dollars, what will 24 yards
cost ?
Yds. D. Yds. D. Yds. D.
As 8 : 32 :: 24 Proof. As 96 : 24 :: 32
24 32
128 48
64 72
8)768 96)768(8
768
96 Ars.
2. When sugar is sold at 12 dollars 32 cts. per cwt. whni
will Wlb. cost? Ans. I doll. 76 cts.
3. What is the amount of 3 cwt. of coffee at 36 cents per
pound? Ans. 120 dolls. 96 cts.
4. What will 4 pieces of linen come to, containing 23,
24, 25, and 27 yards, at 72 cents per yard ?
Ans. 71 dolls. 28 cts.
5. What will bcwt. 2qr. Sib. of iron come to at 48 cents
for 4/6. ? Ans. 61 dolls. 44 cts.
6. What will V2Slb. of pork come to at 8 cts. per pound?
Ans. 10 dolls. 24 cts.
7. If 9 dozen pair of stockings, cost 68 dollars 40 cent?
what will 3 pair cost ? Ans. 1 doll. 80 cts.
8. If 20 bushels of oats cost 9 dollars 60 cents, what will
three bushels come to? Ans. 1 doll. 44 cts.
9. A merchant bought a piece of cloth for 16 dollars 50
cents, at 75 cents per yard ; how many yards were there iii
the piece ? Ans. 22 yds.
10. If 17 cwt. 3qr. 17 Ib. of sugar cost 320 dollars 80 cts.
what must be paid for 6oz. ? An*. 6 o*nts.
58 SINGLE RULE OF THREE DIRECT
11. If 9,7 Ib. of silver is worth 97 dollars, what is the
value of l,5oz. 1 Ans. 1 doll. 25 cts.
12. If 125,5 acres are sold for 627,5 dollars, what will
4,75 acres cost ? Ans. 23 dolls. 75 cts.
13. If 1,5 gallons of wine cost 4 dollars 50 cents, what
will 1,5 tuns cost? Ans. 1134 dolls.
14. How many reams of paper at 1 dollar 66 cents, 1
dollar 97 cents, and 2 dollars 31 cents per ream may be
purchased for 528 dollars 66 cents, of each an equal num
ber ? Ans. 89 reams of each sort.
15. When iron is sold for 224 dollars per ton, what will
Iqr. 1Mb. cost? Ans. 4* dolls. 20 cts._
16. A merchant paid 1402 dollars 50 cents for flour, at 5
dollars 50 cents per barrel ; how many barrels must he re
ceive? Ans. 255 barrels.
17. A man has a yearly salary of 1186 dollars 25 cents,
how much is it per day ? Ans. 3 dolls. 25 cts.
18. A man spends 2 dollars 25 cents per day, and saves
378 dollars 75 cents at the end of the year, what is his
yearly salary? Ans. 1200 dolls.
19. What will 4T. Wcwt. Iqr. I2lb. of hay come to at
1 dollar 12 cents per cwt.? Ans. 101 dolls. 20 cts.
20. How much will a grindstone 4 feet 6 inches diameter,
stfid 9 inches thick, come to at 1 dollar 10 cents per cubic
foot? Ans. 12 dolls. 53 cts.
21. \\hat will a grindstone 28 inches diameter, and 3,5
inches thick, come to at 1 dollar 90 cents per cubic foot ?
Ans. 2 dolls. 26 cts.
22. At 221. Ss. per ton, what will 203 T. Scwt. 3qr. Mb.
of tobacco come to ? Ans. 4558Z. 3s.
23. If 850 dolls. 50 cents is paid for 18 pieces of cloth
at the rate of 11 dollars 25 cents for 5 yards, how many
yards were in each piece, allowing an equal number to each
piece? Ans. 21 yds.
24. If 124 yards of muslin cost 1Z. 17$. 6rf. what is it
pe> yard? Ans. 3s.
25. If a staff 4 feet long cast a shadow (on level ground)
7 feet long, what is the height of a steeple whose shade at
the same time, is 218 feet 9 inches? A?is. 125 feet.
26. If 4292 dollars 32 cents are paid for 476 acres 3
roods 28 perches of land, how much is it per acre ?
Ans. 9 dollars.
27. If a man's annual income be 1333 dollars, and he
SLN(iLK lUiJJ: OK TUkiaO INVERSK. 59
expend daily 2 dollars 14 cents, how much will lie save at
'the end of the year 1 Ans. 551 dolls. 90 cts.
28. If 321 bushels of wheat cost 240 dollars 75 cents,
what is it per bushel ? Ans. 75 cts.
29. If l yard of cloth cost 2 dollars 50 cents, what will
1 quarter 2 nails come to ? Ans. 62^ cts.
30. Bought 3 pipes of wine, containing 120, 124, and
126 gallons, at 5s. 6d. per gallon ; what do they cost ?
Ans. 102Z. Is. lOicZ.
31. A sets out from a certain place and goes 12 miles a
day ; 5 days after, B sets out from the same place, the same
way, and goes 16 miles a day; in how many days will he
overtake A? Ans. 15 days.
32. If I have owing to me 1000Z. and compound with my
debtor, at, 12s. Qd. per pound, how much must I receive?
Ans. 625Z.
33. If 365 men consume 75 barrels of pork in 9 months,
how many will 500 men consume in the same time ?
Ans. 10241 barrels.
34. How much land at 2 dollars 50 cents per acre, musl
f>e given in exchange for 360 acres at 3 dollars 75 cents 1
Ans. 540 acres.
35. If the earth, which is 360 degrees in circumference,
turns round on its axis in 24 hours, how far are the inhabit
ants at the equator carried in 1 minute, a degree there being
69! miles ? Ans. 17 miles 3 fur.
SECTION 2.
SINGLE RULE OF THREE INVERSE.
IF in any given question, more requires less, or less re
quires more, the proportion is inverse, and belongs to this
rule.
Having stated the question, as in the rule of three direct,
proceed according to the following
RULE.
Multiply the first and second terms together, and divide
the product by the third ; the quotient will be the answer, in
the same name as the second.
60 SINGLE RULE OF THREE INVERSE
EXAMPLE.
1. If 20 men can build a wall in 12 days, how long will
it require 40 men to build the ^ame ?
M. d. M. d. M. d.
As 20 : 12 :: 40 Proo/ As 6 : 40 :: 12
12 40
40 ) 240 ( 6 days. Ans. 12 ) 240 ( '20 men.
240 24
2. If 60 men can build a bridge in 100 days, how long
will it require 20 men to build it ? Ans. 300 days.
3. If a wall 100 yards long requires 65 men 4 days, in
what time would 5 men complete it ? Ans. 52 days.
4. If a barrel of flour will last a family of six persons 24
days, how long would it last if 3 more were added to the
family? Ans. 16 days.
5. If 5 dollars is paid for the carriage of \cwt. weight,
150 miles, how far may Sent* weight be carried for th^
same money ? Ans. 25 miles.
6. If a street 80 feet wide and 300 yards long, can be
paved by 40 men in 20 days, what length will one of 60
feet wide be paved by the same men in the same time ?
Ans. 400 yards.
7. If a field that is 30 rods wide and 80 in length, con
tain 15 acres, how wide must one be to contain the same
quantity, that is but 70 rods long ? Ans. 3412. 4/7. 8&iri.
S. If a board be ,75 of a foot wide, what length must it
be to measure 12 square feet? Ans. 16 feet.
9. How much cloth 1,25 yards wide, can be lined by
42,5 yards of silk that is ,75 of a yard wide ?
Ans. 25,5 yards.
10. If 10 men could complete a building in 4,5 months,
what time would it require if 5 more were employed ?
Ans. 3 months.
11. In what time will 600 dollars gain 50 dollars, when
80 dollars would gain it in 15 years? fins. 2 years.
12. If a traveller can perform a journey in 4 days, when
the days are 12 hours long, what time will lw require when
thf! days arn 1 (> hours long ? _ Ans. 3 days.
I ;*. Suppose 100 nvn in a garrison are supplied
SINGLE RULE Oi TliUKE INVERSE. 61
provisions for 30 days, how many men must be sent out if
they would have the provisions last 50 days ?
Ans. 160 men.
14. Lent a friend 292 dollars for six months ; afterwards
J borrow from him 806 dollars ; how long may I keep it to
balance the favor? Ans. 2 months 5 days.
15. 1200 men stationed in a garrison, have provisions for
9 months, at the rate of 14 ounces per day; how long at
the same allowance will the same provisions last if they are
reinforced by 400 men ? And also what diminution must be
made on each ration, that the provisions may last for the
same time ? Ans. 6 mo. at the same allowance
3J oz. deduction to last for the same time.
16. If a piece of land 40 rods in length and 4 in breadth,
make an acre, how wide must it be if it is but 25 rods long?
Ans. 6 rods.
17. How much in length that is 3 inches broad, will make
a square foot ? Ans. 48 inches.
18. If a pasture field will feed 6 cows 91 days, how long
will it feed 21 cows ? Ans. 26 days.
19. There is a cistern having 1 pipe, which will empty it
in 10 hours ; how many pipes of the same capacity will
empty it in 24 minutes ? Ans. 25 pipes.
20. How many yards cf carpeting that is half a yard
wide, will cover a floor that is 30 feet long and 18 feet wide?
Ans. 120 yards.
21. What is the weight of a pea to a steelyard, which
being suspended 39 inches from the centre of motion, will
l
Ans. 4Z6.
22. A and B depart from the same place, and travel the
same road ; but A goes 5 days before B at the rate of 20
miles a day, B follows at the rate of 25 miles a day ; in
what time, and at what distance, will he overtake A?
Ans. 20 days, and 500 miles.
The following rule, if adopted, will suit for the stating of all questions
in single proportion, whether direct or inverse.
GENERAL RULE,
Place that number for the third term, which signifies the
same kind, or thing, as that which is sought ; arid consider
whether the number sought will be greater or less ; if great
F
equipoise 2QSlb. suspended at the draught end f of an inch ?
f>2 SLXGLK RULF. OF THRRK.
er, place the least of the other terms for the first, but if less
place the greater for the first term, and the remaining one
for the second.
Multiply the second and third terms together, and divide
the product by the first ; the quotient will be the answer
required.
EXAMPLES.
1. If 30 horses plow 12 acres, how many will 40 horses
plow in the same time ?
k. h. acr.
Direct Proportion. 30 : 40 : : 12
12
30)480(16 acres. A us.
2. If 30 horses plow 12 acres in 10 days, in how many
days will 40 horses plow the same quantity ?
h. h. D.
Inverse Proportion. 40 : 30 : : 10
10
40)300(7,5 days. Ans.
3. If 800 soldiers in a garrison have provisions sufficient
for 2 months ; how many must depart that the provisions
may last them for 5 months ? Ans. 480.
1. Bought a hogshead of Madeira wine for 119 dollars,
nine gallons of which leaked out ; what was the remainder
sold at per gallon, to gain 12 dollars on the whole 1
Ans. 2 dolls. 42Jfcts.
5. If 225 pounds be carried 51*3 miles for 20 dollars,
how many pounds may be carried 64 miles for the same
money? Ans. 1800Z6.
6. If 87 dolls. 50 cents be assessed on 1750 dolls, what
is the tax of 10 dolls, at the same rate? Ans. 50 cts.
Promiscuous Questions in Direct and Inverse Proportion.
1. Suppose a man tnivels to market with his wagon
loaded, at the rate of 2 miles .n hour, and returns with \\
empty at the rate of 3 miles an hour ; how long will he be
in performing a journey, going and returning, to a place
123 miles distant? Ana. 84JJ hoars.
SINGLE HULK OF TlfREK. (>3
2. A lent B 1000 dollars fur IS!) days, without interest
how long should B 'end A iif>0 dollars to requite the favor?
Ans. 290fg days.
3. Bought 14 casks of butter, each weighing Icwt. Iqr.
4/6. at 12 dollars 60 cents per cwt. ; what did they come to,
and how much per Ib. ?
Ans. 226 dolls. 80 cts. whole cost ; 1 1 cts 2 \ m. per Ib.
4. Sold 4 chests of tea, each weighing Icwt. Qqr. 14//>.
the first for 80 cents per Ib. the second for 90 cents, the
third for 1 doll. 5 cents, and the fourth for 1 doll. 25 cents ;
how many pounds of tea were there, what was the average
price, and what did the whole come to ?
Ans. 504//?. average 1 doll. come to 504 dolls.
5. When flour is sold at 2 dolls. 24 cents per cwt. what
will be the first cost of one dozen of rolls, each weighing 5oz.
allowing the bread to be in proportion to the flour, as five is
co four ? Ans. 6 cents.
6. If a merchant bought 270 barrels of cider for 780 dol
lars, and paid for freight 37 dolls. 70 cents, and for other
charges and duties 30 dolls. 60 cents ; at what must he sell
it per barrel to gain 143 dolls. Ans. 3 dolls. 67^ 4 T cts.
7. If half a ton of hay was equally divided among 80
horses, how much must be given to 7 ? Ans. 3qr. 14Z&.
8. Suppose the circumference of one of the larger wheels
of a wagon to be 12 feet, and that of one of the smaller
wheels 9 feet 3 inches ; in how many miles will the smaller
wheel make 1000 revolutions more than the larger?
Ans. 7 m. 5 fur. 34 yds. 1 ft. 7 T \ in.
9. If a man perform a journey in 18 days, when the days
are 15 hours long, how many days will it require to per
form the same journey when the days are only 12 hours
long? Ans. 22^ days.
10. A merchant bought a piece of broadcloth measuring
42 yds. for 191 dolls. 25 cents; 15 yards of this being
damaged, he sells it at twothirds of its cost ; the residue he
is willing to sell so as to gain 1 doll, per yard on the whole
piece ; at what rate must he sell the remainder ?
Ans. 6 dolls. 86 T 4 T cents per yd.
11. If 60 yards of carpeting will cover a floor that is 30
feet long and 18 broad, what is the width of the" carpeting ?
Ans. 3 feet.
12. ff a piece of land be 40 rods in length, how wide
must it be to contain 4 arres ? Ans. 1 6 rods.
64 SINGLE RULE OF THREE.
13. Suppose a large wheel, in mill work, to contain 70
cogs, and a smaller wheel, working in it, to contain 52 cogs;
in how many revolutions of the greater wheel, will the les
ser one gain 100 revolutions? Ans. 2'88.
14. The number of pulsations in a healthy person is, say
70 in a minute, and the velocity of sound through the air is
found to be 1142 feet in a second : now I counted 20 pulsa
tions between the time of observing a flash of lightning from
a thunder cloud, and hearing the explosion of the thunder;
what was the distance of the cloud ?
Ans. 3 m. 5 fur. 145 yds. 2j ft.
15. A merchant bought 5 pieces of cloth, of different
qualities, but of equal lengths, at the rate of 5, 4, 3, 2, and 1
doll, per yd. for the different pieces ; the whole came to 532
dolls. 50 cents ; how many yards did each piece contain ?
Ans. 35 yds.
16. What principal will gain as much in 1 month, as 127
dollars would gain in 12 months? Ans. 1524 dolls.
17. If a pair of steelyards be 36 inches in length to the
centre of motion, the pea 5 Ib. and the draught end \ inch
in length, what weight, will they draw? Ans. 360 Ib.
18. Supposing the nbovp steelyards would only Hrow 00
Jb., what is the length of the draught end ?
Ans. 2 inches.
19. If 1 yard of cloth cost 2 dolls. 71 cts. 1J mills, what
will 67i yards come to at the same rate?
Ans. 183 dolls. 4 c. 6 m.
20. If a man's income be 16s. 5d. lJ^qr. per day, what
is it per annum ? Ans. 300Z.
21. How many pieces of wall paper that is 3 qrs. wide.
and 11 yards long, will it require to paper the walls of a
room that is 25 feet long, 15 wide, and 10 high, allowing a
reduction of j\ for doors and windows? Ans. lQf T .
22. The length of a wall being tried by a measuring line,
appears to be 1287 feet 4 inches; but on examination the
line is found to be 50 feet 10 inches in length, instead of
50 feet its supposed length; rrquirod the true length of the
wall? Ans. 1309 feet 10} J inch.
23. If a dealer in liquors use, instead of a gallon, a mea
sure which is deficient by half a pint, what will be the true
measure of 100 of these false gallons? Ans. 93j galls.
DOUBLE RULK OF THREK DIRECT. 66
SECTION 3.
THE DOUBLE RULE OF THREE.
THE DOUBLE RULE OF THREE, or as it is often called,
Compound Proportion, is used for solving such questions as
have five terms given to find a sixth. In all questions be
longing to this rule, the three first terms must be a supposi
tion, the two last a demand.
RULE FOR STATING.
1. Set the two terms of the supposition, one under the
other.
2. Place the term of the same kind with the answer
sought in the second place.
3. Set the terms of the demand, in the third place, ob
serving to place the correspondent terms of the supposition
and demand in the same line. Consider the upper and lower
extremes, with the middle terms, separately, as in the single
rule of three ; if both lines are direct, then the question will
be in direct proportion ; but if either lines are inverse, then
will the question be in inverse proportion.
When the question is in direct proportion, multiply the
product of the two last terms by the middle term for a divi
dend, and multiply the two first terms for a divisor; the quo
tient will be the answer in the same name with the middle
term.
But if the proportion be inverse, transpose the inverse
terms and proceed in the same manner as in direct proportion.
DIRECT PROPORTION.
EXAMPLE.
1. It' 6 men in 8 days earn 100 dollars, how much will
12 men earn in 24 days ?
dolls.
6 men ) , mn 5 12 men
>t \
f '
8 days f ' J ) 24 days
48 288
100
48) 28800 (600 dolls. AM.
288
00
F2
06 DOUBLE RULE OF THREE INVERSE
2. If 10 bushels of oats suffice 18 horses for 20 days,
how many bushels will serve 60 horses 36 days ?
Ans. 60 bushels.
3. If 56 pounds of bread will suffice 7 men 14 days, how
much bread will serve 21 men 3 days ? Ans. 36 pounds.
4. If 8 students spend 384 dollars in 6 months, how much
will maintain 12 students 10 months? Ans. 960 dollars.
5. If 20 hundred weight is carried 50 miles for 25 dol
lars, how much must be given for the carriage of 40 hun
dred weight 100 miles? Ans. 100 dollars.
6. If 14 dollars interest is gained by 700 dollars in t>
months, what will be the interest of 400 dollars for 5 years '.'
Ans. 80 dollars.
7. If 4 men can do 12 rods cf ditching in 6 days, how
many rods may be done by 8 men in 24 days ?
Ans. 96 rods.
INVERSE PROPORTION.
EXAMPLE.
1. If 4 dollars pay 8 men for 3 days, how many days
must 20 men work for 40 dollars ?
days.
As 4 dolls. ( q J 40 dolls.
8 men. \ ) 20 men.
Here the lower line is inverse, which transposed will stand
thus : . at 3Z. 17*. 6d. per cwt.
AIM. 1007. 5*. 3j. at 16 dollars 80 cents per cwt. neat; for which
I gave 18 barrels of flour at 4 dollars 50 cents per bbl. and
I % ton of iron at 120 dollars per ton; what was the balance
still due? Ans. 112 dolls. 65 cts.
5. What is the neat weight of 12 barrels of potash, each
weighing ^cwt. 2qr. 26lb. tare I2lb. per cwt. ; and what
will it come to at 9 dollars per cwt. ?
Ans. SQcwt. 2qr. 23lb. and comes to 456 dolls. 34 cts.
6. Sold a hogshead of sugar, weighing 6cwt. gross, tare
100Z&. tret 4Z&. per 104, for 82 dolls. 50 cents ; what was it
sold for per pound? Ans. 15 cents.
7. In I20cwt. 3qr. gross, whole tare llllb. tret 4Z6. per
104, how much is the neat weight in pounds, and what the
amount at 73 cents per pound?
Ans. 12833,6Z&. and comes to 9368 dolls. 53 cts.
8. Bought 9 hogsheads of sugar, each weighing 6cwt.
'2qr. I2lb gross ; tare lllb. per cwt. what is the neat weight,
and wh?' Joes it amount to at 16 dollars per cwt. ?
Ans. 5Qcwt. Iqr. 22lb. amounts to 807 dolls. 14 cts.
9. Sold 27 bags of coffee, each 2cwt. 3qr. lllb. gross;
ea re l'3ib. per cwt. tret 4Z&. per 104; what is the neat
weight, and what will it co*ne to at 32 cents per pound ?
Ans. 66cwt 2qr. lllb. and comes to 2386 dolls. 88 cts.
SECTION 3.
OF INTEREST.
I NT FOREST is a compensation allowed for the use of money,
ibi a givim time; and is generally throughout the United
States fixed by law at the rate of 6 dollars for every 100,
;K>; annum.
1. Thr sum of w < .ney at interest, is called the Principal.
^. The sum per nt. agreed on, is called the Rate.
3. Tho principal id interest added together, is called the
A/Hf*itn.t. *
Interest is firher. nfflc or compound.
SIMPLE INTEREST. 75
SIMPLE INTEREST.
SIMPLE INTEREST is a compensation arising from the
principal only.
Case 1.
When the given time is one or more years, and the prin
cipal dollars only.
RULE.
Multiply the given sum by the rate per cent. ; the product
will be the interest for one year in cents, which multiplied
by the number of years, will be the answer required.
EXAMPLES.
1. What is the interest of 454 dollars for one year, at 6
per cent. ?
454
6
2724 cents. Ans. 27 dolls. 24 cts.
2. Required the interest of the same sum for 5 years, at
the same rate ?
454
2724
5
13620 cents. Ans. 136 dolls. 20 cts.
3. Required the amount of the same sum for 5 years, at
he same rate ?
454
6
2724
5
13620 interest.
4.5400 principal.
59020 amount. Ans. 590 dofts. 20 cts.
4. What is the interest of 200 dollars for 2 years, at 6
per cent ? Ans. 24 dolls.
76 SIMPLE INTEREST.
5. What is the interest of 1260 dolls, for 4 years, at 7
per cent. ? Ans. 352 dolls. 80 cts.
6. What is the amount of a note for 560 dollars for 3
years, at 8 per cent. 1 Ans. 694 dolls. 40 cts.
7. W T hat sum must be given to discharge a bond given for
4520 dollars, on which there is 6 years interest at 5 per
cent. ? Ans. 5876 dolls.
Note. When the rate per cent, contains a fraction, such as , ^, f
the principal must be multiplied by the fraction, as well as the whole
number : this may be done either by adding the parts of \, i, &.c. of
the principal to the product of the whole number; or reduce the frac
tion to a decimal. See case 1, in reduction of decimals.
8. W r hat is the amount of 400 dollars for 2 years, at 6
per cent. 1 Ans. 452 dolls.
9. What is the interest of 4925 dollars lor y years, at 7^
per cent. ? Ans. 3324 dolls. 37 cts. 5 m.
10. What is the amount of 2500 dollars for 1 year, at 7$
per cent. ? Ans. 2693 dolls. 75 cts.
Case 2.
When the principal is dollars and cents, or dollars, cents,
and mills, and the time years only.
RULE.
Multiply the given sum by the rate per cent, and divide
the product by 100 ; or, what is the same, point off two
figures on the right of the product ; .the quotient, or remain
ing figures, will be the answer, in the same name with the
lowest denomination in the principal.
EXAMPLES.
1. What is the interest of 264 dollars 50 cents for 1 year,
at 6 per cent. ?
264,50
6
cents 1587,00 Ans. 15 dolls. 87 cts.
2. What is the interest of 468 dollars 22 cents and 5
mills for 1 year, M 8 per cent.
468,225
mills 37458,00 Ans. 37 d.ls. 45 cts. 8 m.
SIMPLE INTEREST. ^ 77
3. What is the interest of 364 dollars 50 cents for 5 years,
at 6 per cent, per annum ?
36450 Or, 36450
6 30= product of the rate
and time.
218700 cents 10935,00
5
cents 10935,00 Ans. 109 dollars 35 cents.
4. What is the amount of a note for 1260 dollars 50
cents and 5 mills for 3 years, at 7^ per cent, per annum?
Ans. 1544 dolls. 11 cts. 8f ms.
5. What sum will discharge a bond given for 630 dollars
50 cents, on which there is 5 years interest at 8 per cent,
per annum ? Ans. 882 dolls. 70 cts.
6. What is the difference between the interest of 1274
dollars 64 cents 6 mills for 3 years, at 7J per cent, per an
num, and the interest of 3462 dollars 84 cents, for 4 years,
at 3^: per cent, per annum ?
Ans. The lattep is 163 dolls. 37 c. 3,85 m. the greater.
7. A gave B his bond for 3422 dolls. 25 cents, to be paid
in the following manner, viz. onethird at the end of one
year, onethird at the end of two years, and the remainder
at the end of three years, with interest from the date, at 6
per cent, per annum ; what will be the annual payments,
and what the whole amount ?
Ans. 1st payment 1209 dolls. 19,5 cts. 2d, 1277 dolls. 64
cts. 3d, f346 dolls. 8,5 c. ; whole amount 3832 d. 92 c.
Case 3. .
When the principal is dollars, cents, &c. and the time is
years and months, or months only.
RULE.
Multiply by half the number of months in the given time,
when the rate is 6 per cent, per annum : but if the rate per
cent, be more or less than 6 per cent., multiply the given
number of months by the rate, and divide the product by
12 ; the quotient will be the rate for the time; the principal
multiplied by this rate, will give the interest required.
G2
78 ., SIMPLE INTEREST
EXAMPLES.
1. What is the interest of 650 dollars for 8 months, at 6
per cent, per annum ?
650
4 half the months
cents 2500 Ans. 26 dollars.
2. What 13 the interest of 860 dollars for 1 year and 6
months, at 6 per cent, per annum ?
860
9 half the months
cents 7740 Ans. 77 dolls. 40 cents.
3. What is the interest of 420 dollars for 9 months, at 8
per cent, per annum ?
9 months 420
8 per cent. 6
12 ) 72 2520 Ans. 25 dolls. 20c.
6 the rate for the time.
4. What is the amount of a note for 724 dollars, with 18
months interest due thereon, at 4 per cent, per annum ?
Ans. 767 dolls. 44 cts.
5. What is the interest' of 240 dollars for 15 months, at
7 2 per cent, per annum ? A.ns. 22 dolls. 50 cts.
6. What is the interest of 1260 dollars for 4 months, at
G per cent, per annum ? Ans. 27 dolls. 30 cts.
Case 4.
When the principal is dollars, cents, &c. and the time is
months and days, or days only.
RULE.
Find the interest for the given months by the last case,
find take aliquot parts for the days.
Note. In calculation of interest, 30 days make a month.
Or, multiply the given sum when the rate is 6 per cent.
i)\ the number of days, and divide the product by 60 ; the
<]ii<>;ient is the interest required.
Note. Though both the foregoing methods are considered sufficiently
exart for common business, by merchants and jiccomptants generally,
SIMPLK INTKRKST. 775
9
.55132
1.68948
33
3.22510
4.04898
10
.62889
1.79064
25
3.38635
4.29187
11
.71034
1.89829
26
3.55567
l.r,ii)38
12
.79585
2.01219
27
.T73345
1234
13
1.88565
2.l32;j^
28
3.920 J 3
5.11168
14
1.97993
2.26090
29
4.11613
5.41838
15
2.07892
2.39655
30
4.32194
5.74349
To find the compound interest of any sum by this labic,
multiply the figures opposite the number of years, under the
rate percent, b r 'n primipal ; the product will be
tbt: amount required ; from this subtract the principal, tht.
remainder will b'' th<
INSURANCE COMMISSION. AND BROKAGK. 87
EXAMPLE.
2. What is the compound interest of 1000 crbllars for W
years, at 6 per cent, per annum ?
1,59384 the tabular number for the time
1,000 the principal
1593,84000
1000
593,84 the interest. Ans. 593 dolls. 84 cts.
3. What is the amount of 1500 dollars for 5 years, at 5
pej* cent, per annum? Ans. 1914 dolls. 42 cts.
4. What is the compound interest of 4500 dollars for 16
years, at 6 per cent, per annum ?
Ans. 6931 dolls. 57 cts. 5 m.
5. A has B's note for 650 dollars, payable at the end of
20 years, at 6 per cent, per annum, compound interest ;
what sum will it require to discharge the note, at the expira
tion of the given time ?
Ans. 2084 dolls. 63 cts. 4 m.
6. A father left a legacy of 8000 dollars at compound in
terest, 6 per cent, per annum, to be equally divided among
his three sons, when the youngest, who was 4 years old,
should arrive at the age of 21 ; what will be each one's
share? Ans. 7180 dolls. 72 cts. each share.
SECTION 4.
Insurance, Commission, and Brokage.
INSURANCE is a premium given for insuring the owners of
property against the dangers and losses to which it is liable,
or indemnifying for its loss.
The instrument of agreement by which this indemnity is
secured is termed the policy of insurance.
Commission is a compensation allowed to merchants and
others for buying, selling, and transporting goods, wares,
88 INSURANCE A\V> COMMISSION.
Brokage is an allowance given to brokers for exchanging
money, buying and selling stock, &c.
The method of operation in all these is the same as in
simple interest,
EXAMPLES,
INSURANCE.
1. What is the premium of insuring 1260 dollars, at 5
per cent ?
1260
5
63,00 Ans. 63 dolls.
D. c.
2. 1650 dollars at 15 per cent.  Ans. 255 75
3. 4500 25    1125 00
4. What sum must a policy be taken out for, to cover
900 dollars, when the premium is 10 per cent?
100 policy
10 premium
90 sum covered
As 90 : 100 :: 900 : 1000 dolls. Ans.
5. What sum will it require to cover a policy of insurance
for 4500 dolls, at 25 per cent 1 Am. 6000 dolls.
6. What sum witl.it require to cover a policy of insurance
for 560 dollars, at 9 per cent 7 Ans. 615 dolls, 38$ cts,
COMMISSION.
1. What is the commission on 850 dolls, at 5 per cent.
850
5
42,50 Ans. 42 dolls. 50 cts.
2. What is the commission on 1260 dollars, at 6 per
rent? Ans. 75 dolls. 60 cts.
D. c.
3. 2550 dollars at 4 per cent.   Ans. 102 00
4. 26342 H 790 26
5. 6422 } 48 10 J
BROKAGE Bl'YINU AND SELLING STOCKS. 89
6. A commission merchant receives 1260 dollars to fill
an order, from which he is instructed to deduct his own
commission of 5 per cent, how much will remain to satisfy
the order?
100
5 per cent.
As 105 : 100 :: 1260 : 1200 dolls. Ans.
7. A commission merchant has received 4120 dollars
\vith instructions to vest it in salt at 8 dollars per harrel ; de
ducting from it his commission of 3 per cent, how many
barrels of salt can he purchase ? Ans. 500 barrels.
BROKAGE.
1. What is the brokage on 1000 dollars, at lj per cent?
1000
H
1000
500
15,00 Ans. 15 dolls.
2. What is the brokage on 1625 dollars 50 cents, at 3^
per cent? Ans. 54 dolls. 18 cts.
3. 1868 dollars at 2 per cent. Ans. 46 dolls. 70 cts.
4. 560 6 38 60
SECTION 5.
BUYING AND SELLING STOCKS.
Stock is a fund vested by government, or individuals in a
corporate capacity, in banks, turnpike roads, bridges, &c.
the value of which is subject to rise and fall.
RULE.
Multiply the given sum by the rate per cent, and divide
the product by 100
H2
90 REBATE OR DISCOUNT.
EXAMPLES.
1. What is the amount of 1650 dollars, United States
bank stock, at 125 per cent, or 25 per cent, above par?
1650
125
6250
3300
1650
1,00 ) 2062,50 Ans. 2062 dolls. 50 cts.
Or thus, 25 is i)1650
412 50
2062 50 AM.
D. D. c.
1500 bank stock at 110 per cent. Ans. 1650 00
1686 128 . 2158 08
4. 25000 108   27000 00
5. 1260 90   1134 00
6. 9254 84  7773 36
7. 1518 83 1271 32$
SECTION 6.
REBATE OR DISCOUNT,
Is a reduction made for the payment of money before il
tecomes due. It is estimated in such a manner, as that
the ready payment, if put to interest at the same rate and
time, vvould amount to the first sum. Thus, 6 dollars is the
discount on 106 dollars for 12 months, at 6 per cent, leaving
100 dollars the ready payment, which, if put to interest for
the same rate and time, would regain the 6 dollars discount.
RULE.
As 100 dollars and the interest for the given time, is to
100 dollars, so is the given sum to its present worth. Sub
tract the present worth from tho given sum, and the remain
dor is the discount.
REBATE OR DISCOUNT. 91
^ EXAMPLE.
1. What s the discount of 1696 dollars, due 12 months
hence, at 6 per cent, per annum ?
As 106 : 100 :: 1696 : 1600
1600
96 dolls. Ans.
2. What is the present worth of 2464 dollars, due 1 year
and 6 months hence, discounting at the rate of 8 per cent.
nor annum] Ans. 2200 dolls.
3. A has B's note for 1857 dollars 50 cents, payable 8
months after date ; what is the present worth of said note,
discounting at the rate of 5J per cent, per annum?
Ans. 1791 dolls. 80 cts.
4. What reduction must be made for prompt payment of
a note for 650 dollars, due 2 years hence, 7 per cent, per
annum being allowed for discount ?
AILS. 79 dolls. 83 + cts.
5. What is the present worth of 5150 dollars, due in 4^
months, discounting at the rate of 8 per cent, per annum,
and allowing 1 per cent, for prompt payment ?
Ans. 4950 dolls.
Note. Discount and interest are often supposed to be one and the
same thing; and in business, the interest for the time is frequently
taken for the discount, and it is presumed neither party sustains any
loss. This however is not true, for the interest of 100 dollars for 12
months, at 6 per cent, is 6 dollars, whereas the discount for the same
sum, at the same rate and ti'ne, is only 5 dollars 66 cents, making a
difference of 34 cents for every 100 dollars for 1 year at 6 per cent.
The following 1 examples will show the difference.
EXAMPLES.
1. What is the discount of 1272 dollars, due in 12
months, discounting at 6 per cent, per annum ?
.As 106 : 100 :: 1272 : 1200
discount 72 dolls.
2. What is the interest of the same sum, for the same
time and rate ?
1272 D. c.
6 . Interest 76 32
Discount 72
76,32 interest,
Difference 4 32
92 BANK DISCOUNT.
8. What is the difference between the interest and dis
count on 7280 dollars, for 18 months, at 8 per cent, per an
num? Ans. 93 dolls. 60 cts. difference.
Note. But when discount is made for present payment, without re
gard to time, the interest of the sum as calculated for a year, is the
discount.
EXAMPLE.
1. How much is the discount of 260 dollars at 5 per
cent?
260
5
13,00 Ans. 13 dollars.
2. What is the discount on 1650 dollars, at 3 per cent!
Ans. 49 dolls. 50 cte.
3. What sum will discharge a bond for 2464 dollars, on
which a discount of 8 per cent, is given ?
Ans. 2266 dolls. 88 cts.
SECTION 7.
BANK DISCOUNT.
BANK discount is the interest which banks receive for the
use of money loaned by them for short periods. And as
banks from long established custom, give three days over
and above the time limited by the words of the note, called
days of grace ; and as the day of the date, and the day of
payment are both calculated, which makes the time 4 days
longer than expressed in the note, so interest must be calcu
lated on these days in addition to the regular interest on the
given sum, for the specified time.
RULE.
Add 4 to the number of days specified in the note, multi
ply the given sum by this number, and divide the product by
60. Or,
Multiply the given sum by half t)ie number of days, and
divide bv 30.
EQUATION OF PAYMENTS. f 98
Note. When the cents in the given sum are less than 50, the bank
loses the interest on them, but when they are more than 50 they
charge interest for one dollar.
EXAMPLES.
1. Required the discount of 1500 dollars for 60 days.
1500 Or, 1500
64 32= half the days
6000 3000
9000 4500
6,0)9600,0 3,0)4800,0
16,00 Ans. 16 dolls. 16,00
2. What is the discount of 250 dollars for 30 days ?
Ans. 1 doll. 41 1 cts.
3. What is the discount of 600 dollars for 90 days ?
Am. 9 dolls. 40 cts.
4. What is the discount of 1260 dollars 40 cents for 60
days 1 Ans. 13 dolls. 44 cts.
5. What is the discount of 2649 dolls. 75 cents for 60
days ? Ans. 28 dolls. 26 cts. 4 m.
Form of a note offered for discount.
Pittsburgh, July , 1832.
Dollars
Sixty days after date, I promise to pay A. B. or order, at
the bank of , the sum of dollars, without de
falcation ; value received. J. P.
SECTION 8.
EQUATION OF PAYMENTS.
EQUATION of payments is the finding the mean time, for
the payment of two or more sums of money payable at dif
ferv.~t times.
RULE.
Multiply each sum by its own time. Add the products
94 FELLOWSHIP.
into one sum and divide this amount by the whole debt ; t/ie
quotient will be the mean time.
EXAMPLE.
1. A owes B 600 dollars, of which 200 is to be paid at
4 months, 200 at 8 months, and 200 at 12 months; but they
agree to make but one payment ; when must that paymetit
be made?
200 X 4= 800
200 X 8=1600
200X12 = 2400
600 ) 4800 ( 8 months. Ans.
4800
2. A merchant has owing to him from his friend, the sum
of 3000 dollars, to be paid as follows, viz. 500 dollars at 2
months, 1000 dollars at 5 months, and t le rest at 8 months ;
but they agree to make one payment of the whole ; whit
will be the mean time of payment? Ans. 6 months.
3. A buys of B 50 acres of land, for which he agrees to
pay 1000 dollars at the following times, viz. 200 dollars at
5 months, 300 dollars at 8 months, and the rest at 10 months,
but an equation of payments is afterwards agreed upon ;
when must the payment be made?
Ans. 8 months 12 days.
4. C owes D 1400 dollars, to be paid in 3 months, but D
being in want of money, C pays him 1000 dollars at the ex
piration of 2 months ; how much longer than 3 months may
he in justice defer the payment of the rest ?
Ans. 2% months
SECTION 9.
FELLOWSHIP.
'SHH' loaches to find tho profit or loss arising f o
different partners in trade, in proportion to the cap: J or
stock each has advanced.
Fellowship is either nngle or compound.
SINGLE FELLOWSHIP. 95
SINGLE FELLOWSHIP,
Is when the stocks employed are different, but the time
alike.
RULE.
Find the amount of the whole stock employed ; and then
(by proporfton) as the whole stock is to the whole gain or
loss, so is each partner's stock to his share of the gain or
loss.
EXAMPLE.
1 . Two merchants join their stock in trade ; A puts in
eOO dollars, and B puts in 400 dollars, and they gain 250
dollars ; what part belongs to each ?
A 600 A 1 nnn  n 600 to A's share 150 ^
B 400 ; 400 to B's 100 ! ^^
1000 250 J
2. Three merchants enter into partnership in trade ; A
advanced 7500 dollars, B 6000, and C 4500, with this they
gained 5400 dollars ; what was each partner's share ?
C A 2250 dolls,
Ans. I B 1800
I C 1350
3. A bankrupt is indebted to A 1291 dollars 23 cents, to
B 500 dollars 37 cents, to C 709 dollars 40 cents, to D 228
dollars ; and his estate is worth but 2046 dollars 75 cents ;
how much does he pay per cent, and how much is each
creditor to receive ?
D. c.
(A receives 968 42$
B 375 27 
C 532 05
D 171
4. Three men, A, B, and C, rent a farm containing 585
acres 2 roods and 34 perches, at 600 dollars per year, oJ
which A pays 180 dollars, B 195, and C 225, and they
agree that the larm shall be divided in proportion to the
rents; how many acres must each man have?
A. R. P.
; share is 175 2 34j
Ans. { B's 190 1 17^
219 2 22]
i A's share is 175
, { B's 190
f C's 219
96 S1JNGLE FELLOWSHIP.
5. Three merchants freighted a ship with 2160 barrels of
flour, of which 960 barrels belonged to A, 720 barrels to B,
and 480 barrels to C ; but on account of stormy weather
they were obliged to throw 900 barrels overboard ; how
many barrels did each man lose?
i A lost 400 barrels.
Ans. IE $00
(C 200
6. Three merchants join stock in trade ; A put in 1260
dollars, B 840 dollars, and C a certain sum ; and they gained
825 dollars, of which C took for his part 275 dollars ; re
quired A and B's part of the gain, and how much stock C
put in?
( A gained 330 dolls.
< Ans.
1200 proof.
2. Three merchants enter into partnership for 16 months;
A put into stock at first 600 dollars, and at the end of 8
months, 200 dollars more ; B put in at first 1200 dollars,
but at the end of 10 months, was obliged to take out 600
dollars; C put in at first 1000 dollars, and at the end of 12
months put in 800 more ; with this stock they gained 2300
dollars ; what was each man's share ?
C A's share is 560 dolls.
Ans. 1 B's 780
( C's 960
3. A and B join stock in trade ; A put in 600 dollars or.
the first of January ; B advanced on the first of April a sum
98 COMPOUND FELLOWSHIP.
which entitled him to an equal share of the profit at the end
of the year ; required the sum B put in ?
Ans. 800 dollars.
4. D put in stock 1800 dollars ; E at the end of 4 months
agrees to advance such a sum as at the end of the year wili
entitle him to an equal share of .the profits; what sum must
E advance? Ans. 2700 dollars.
5. Two gentlemen, A and B, hired a carriage in Pitts
burgh to go to Philadelphia, and return, for 160 dollars,
with liberty to take in two others by the way. When at
Philadelphia they took in C, and afterwards, 100 miles from
Pittsburgh, they took in D. Now allowing it to be 300 miles
from Pittsburgh to Philadelphia, and also that each man pays
in proportion to the distance he rode ; it is required to tell
how much each must pay ?
!A pays 60 dolls.
B 60
C 30
D 10
160 proof.
6. Three graziers hired a piece of pasture ground for 145
dolls. 20 cents ; A put in 5 oxen for 4 months, B put in 8
oxen for 5 months, and C put in 9 oxen for 6 months ; how
much must each pay ?
Ans. A pays 27 dolls. B 48 dolls, and C 70 dolls. 20 cts.
7. A, B, and C have received 665 dollars interest; A put
in 4000 dolls, for 12 months, B 3000 for 15 months, and C
5000 for 8 months ; how much is each man's part of the in
terest ?
( A 240 dolls.
Ans. I B 225
f C 200
HI. Three merchants lost by somo dealings 263 dollars 90
outs ; A's stock was 580 dolls, for 6 months, B's 580 dolls,
for 9J months, and C's 870 dolls, frr 8 months ; how much
is each man's part of this loss ?
A's loss 59 dolls. 15 cts.
Ans. B's 86 45
C's 118 30
PROFIT AND LOSS. 99
SECTION 10.
PROFIT AND LOSS.
BY this rule we discover what has been gained or lost
on the purchase and sale of goods, and merchandise of every
kind.
RULE.
Prepare the question by reduction when necessary, and
then work by the Rule of Three or Practice, as the nature
of the question may require.
EXAMPLE.
1. Bought 360 barrels of flour for 6 dollars 25 cents per
barrel, and sold it for 7 dollars 50 cents per barrel; what
is the profit on the whole ?
D. c.
7 50
6 25
1 25 gain per barrel.
B. D.c. B. D.
As 1 : 1 25 :: 360 : 450 Ajis.
2. Bought a piece of cloth for 1 doll, and 20 cents pei
yard, and sold it again for 1 dollar 50 cents a yard ; what
is the gain per cent 1 Ans. 25 per cent.
3. Bought a piece of linen containing 42 yards for 21
dollars, and sold it at 66 cents per yard ; what is the gain
or loss on the whole piece ?
Ans. 6 dolls. 72 cents gain.
4. A merchant bought 6 barrels of whiskey containing
32 gallons each, for 96 dollars ; while in his possession he
l ost 12 gallons by leakage, the residue he sold for such a
sum as gained him 12 dollars on the whole; how much per
gallon did he buy and sell for ?
Ans. Bought for 50 cents, and sold for 60 cents per gall.
5. Bought 120 doz. of knives for 20 cents each knife,
and sold them again for 17 cents each, what was the loss on
the whole? Ans. 43 dolls. 20 cts.
100 PROFIT AND LOSS
6. A merchant gave 149 dollars for 100 yards of cloth ;
at how much per yard must he sell it to gain 51 dollars on
the whole 1 Arts. 2 dollars.
7. Bought a chest of tea at 1 dollar and 25 cents per
pound, but finding it to be of an inferior quality, I am will
ing to lose 18 per cent, by it ; how must I sell it per pound ?
Ans. 1 doll. 24 cents per Ib.
8. A merchant bought 20 dozen of wool hats at 90 cents
per hat ; at what rate must he sell them again to gain 20 per
cent, and how much does he gain on the whole 1
Ans. he must sell at 1 dollar 8 cents per hat, and gains
48 dollars 20 cents.
9. A trader bought a hogshead of rum of a certain proof,
containing 115 gallons, at 1 dollar 10 cents per gallon ; how
many gallons of water must he put into it to gain 5 dollars,
by selling it at 1 dollar per gallon 1
Ans. 16i gallons.
10. A merchant bought 4 hundred weight of coffee for
134 dollars 40 cents, and was afterwards obliged to sell it
at 25 cents per pound ; what was his loss on the whole, and
how much on each pound ?
Ans. 5 cents, loss on each pound, and 22 dollars 40
cents on the whole.
11. If by selling 360 yards of broadcloth for 1728 dollars,
there is gained 20 per cent, profit, what did it cost per yard ?
Ans. 4 dollars.
12. A merchant laid out 1000 dollars on cloth, at 4 dol
lars per yard, and sold it again at 4 dollars 90 cents per
yard ; what was his \yhole gain ? Ans. 225 dolls.
13. A sells a quantity of wheat at 1 dollar per bushel,
and gains 20 per cent. ; shortly after he sold of the same to
the amount of 37 dollars 50 cents, and gained 50 per cent. ;
how many bushels were there in the last parcel, and at what
rate did he sell it per bu >hel ?
Ans. 30 bushels, at 1 doll. 25 cents per bushel.
14. A trader is about purchasing 5000 galls, of whiskey,
which he can have at 48 cents per gallon in ready money,
or 50 cents with two months credit ; which will be the most
profitable, either to buy it on credit, or by borrowing the
money at 8 per cent, per annum, to pay the cash price?
Ans. he will gain 68 dollars by paying the cash.
15. A butcher bought 12 head of beef cattle of equal
BARTER. 101
weight, for 240 dollars, which he sells again for 4 cents per
pound ; what ought each one to weigh, that the butcher may
have the hides and tallow as clear gain ?
Ans. 4:cwt. Iqr. 24Z&.
SECTION 11.
BARTER.
BARTER is the exchanging of one commodity for another
at the rates agreed upon by their owners.
$
RULE.
Proceed by the rules of reduction and proportion, as the
nature of the question may require.
EXAMPLE.
1. How many yards of linen at 50 cents per yard must
be given for 6^ yards of broadcloth, at 4 dollars 50 cents
per yard ?
4,50 dollars
6*
2700
112J
28,12$
c. yd. D. c. yds.
As 50 : 1 :: 28,12* : 56% Ans.
2. A has 320 bushels of salt at 1 dollar 20 cents per
bushel, for which B agrees to pay him 160 dollars in cash
and the rest in coffee at 20 cents per pound ; how much cof
fee must A receive? Ans. 1120 Ib.
3. How much rye at 70 cents per bushel must be given
for 28 bushels of wheat, at 1 dollar 25 cents per bushel ?
Ans. 50 bushels.
4. A barters 319 Ib. of coffee at 23* cents per pound,
with B for 250 yards of muslin ; what does the muslin cost
A per yard 1 Ans. 30 cents nearly.
5. C has flour at 5 dollars per barrel, which he barters
12
102 EXCHANGE.
to D at a profit of 20 per cent, for tea which cost 1 dollar
25 cents per pound ; at what rate must D sell the tea to
make the barter equal? Ans. I doll, 50 cts. per Ib.
6. A has cloth which cost him 2 dollars 50 cents per yard,
but in trade he must have 2 dollars 80 cents ; B has wheat
at 1 dollar 20 cents per bushel ; at how much per bushel
should he sell to A, to make the barter equal ?
Ans. I doll. 34f cents.
7. P has 240 bushels of rye which cost him 90 cents per
oushel ; this he barters with Q at 95 cents per bushel for
wheat which stands Q 99 cents per bushel ; how many bushels
of wheat is he to receive in barter, and at what price, that
their gains may be equal ?
Ans. 218/y bushels, at 1 doll. 4 cts. per bushel.
8. A gives B in barter 26lb. 4oz. of cinnamon, at 1 dol
lar 28 cents per pound, for rice at 6 cents per pound ; how
much rice must A receive ? Ans. 5 cwt.
9. C and D barter ; C has muslin that cost him 22 cents
per yard, and he puts it at 25 cents ; D's cost him 28 cents
per yard ; at what price must he put it to gain 10 per cent,
more than C ? Ans. 34 cents per yard.
10. A buys 250 barrels of flour from B, at 6 dollars 25
cents per barrel, in payment B takes 4 cwt. of coffee at 30
cents per pound, 64 pounds of tea at 1 dollar 75 cents per
Ib. 25 yards of broadcloth at 6 dollars per yard, 206 dollars
10 cents in cash, and the balance in salt, at 8 dollars per
barrel ; how many barrels of salt must B receive ?
Ans. 120 barrels.
SECTION 12.
EXCHANGE.
EXCHANGE is the reducing the money, coin, &c. of one
tate or country to its equivalent in another.
Par is equality in value ; but the course of exchange is
often above or below pcu .
Agiu is a term sometimes used, to express the difference .
between bank and current money.
Case 1.
To change the currency of one state into that of another.
EXCHANGE
RULE.
Work by the Rule of Three ; or by the theorems in the
following table :
* The New England
and Maine.
Note. In some part*
Louisiana, Mississippi, I
GQ
O
f ?
. 
~ p
p
r
OKI
5 2
CL ^
^i 3
o' co
!>
"~i P
CD 3
R?'
^1
II
5* P~*

02
ri
3 p:
Cta
CD
[TABLE, exhibiting
Theorems fo
p
P CD
.0
sr
CD
02
i ,*JL^
K
>l *
Is 3
i
l^ 3
CD o
O 02
P C
CD cT
t
f.^
s ^**
11 B
o '>'
01 ^T
co r
^ 2 s^ s
^^ S
P CD p
* P
cr
jj* Q
P^ O
^. re a ^
Ct5 ^5
0> "
a
<^
p. ^ s" 3"
 1" 3'
50 ^^f
f!
o 4 S

^&^J
a^&'S s
** S 5
j 1
P
t * J^
O5 p
Sf
*m
Ril
ii
*"* JO
0
^ c^
o> i sr
C^
"S _
> .
53 S*
s'S ;
' ll I
BQ
0^ rj (0
QO O^
D
H* p
CJ P
O
g >
New York
and
Vorth Carotin
Z. f?
^ a
si
s>
^ "*
"i
^H
=! CC
;
&1
1 1
E^
^2
.j. ^
? g"
fifi
3
^
i^

oi' 0
8 S
^'^S,
4.
r T
MHP
s'
S
The value of a dollar in any state is found, either opposite
to that state, or under it in the table.
104 EXCHAN^K.
EXAMPLE.
1. What is the value of 480Z. Pennsylvania currency in
North Carolina?
s. d.
s.
.
.
As
7 6
: 8 ::
480
: 512 Ans.
.
Or,
480
Add V
t = 32
512 Ans.
2. What is the value of 256Z. New York currency in
Pennsylvania? Ans. 240Z.
How much South Carolina currency is equal to 1500Z.
of New Jersey? Ans. 933Z. 6s. 3d.
4. What sum New York currency is equal to 180Z. in
Massachusetts? Ans. 240Z.
1. How much Virginia currency will purchase a bill for
280Z. South Carolina ? Ans. 360Z.
6. A bill of exchangj being remitted from Rhode Island
__ Kjouth Carolina for 304Z., what is its value in the currency
of the latter? Ans. 236Z. 85.
Case 2.
To change the currency of the different states to Federal
money.
RULE.
Divide the given sum, reduced to shillings, sixpences, or
pence in a dollar, as it passes in each state.
EXAMPLES.
1. Change 127Z. 12s. New England money to dollars and
cents.
127/. 12s.=2552 shillings.
The dollar, New England, is 6s. ) 2552
425,33
Ans. 425 dolls. 33 cts.
FOREIGN KXCHAIS'GK. I0i
2. Change 37/. 10s. Pennsylvania currency, to dollars.
37 L 10s. 1500 sixpences.
7s. 6d. or 15 sixpences make a dollar Pennsylvania cur
rency ; hence 1500 ~ 15= 100 dolls. Ans.
Or, 37Z. 10s. = 9000 pence= 100,00 cents. Ans.
3. Change 2251. 12s. New York currency to Federal
money.
225L 12s. 4512 shillings h 8 the dollar New York cur
rency=564 dollars. Ans.
4. A bill of exchange for 468Z. 9s. 6d. Virginia curren
cy, is remitted to Philadelphia ; what is its value in Federal
money? Ans. 1563 dollars 25 cts.
5. A merchant deposited in the United States branch
bank at Pittsburgh, the sum of 750Z. 10s. Pennsylvania cur
rency, for what sum may he draw for in Federal money ?
Ans. 2001 dollars 33^ cents.
Note. Federal money being now generally introduced into mrrc;;r
tile business throughout, the United States, has nearly superseded ti":
use of the above rules of exchange between the different States.
Case 3,
FOREIGN EXCHANGE,
Accounts are kept in Kngland, Ireland, and the West
India Islands, in pounds, shillings, pence, and farthings
though their intrinsic value in these places is different.
A TABLE
Of different Moneys^ as they are denominated and valued
in different countries.
REAT BRITAIN, IRELAND, AND THE WEST INDIES.
4 farrhings =1 penny
12 pence 1 shillino
20 shillings I pound
106 FOREIGN EXCHANGE.
FRANCE.
12
Deniers
= 1 Sol
20
Sols
1 Livre
3
Livres
 1 Crown

SPAIN.
4
Marvadies Vellon, or
Marvadies of Plate
= 1 Quarta
8*
34
Quartas, or
Marvadies Vellon
1 Rial Vellon
16
34
Quartas, or
Marvadies of Plate
1 Rial of Plate
8
Rials of Plate
1 Piaster, Pezo, or
Dollar
5
Piasters 
1 Spanish Pistole
2
Spanish Pistoles
 1 Doubloon
ITALY.
12
Deniers
= 1 Sol
20
Sols
1 Livre
5
Livres
1 Piece of Eight at
Genoa
6
Livres
1 Ditto at Leghorn
6
Solidi
 1 Gross
24
Grosses 
1 Ducat
PORTUGAL.
400
Reas
= 1 Crusadoe
1000
Reas
 1 Millrea
HOLLAND.
8
Penning 
= 1 Groat
2
Groats
 1 Stiver =2d.
6
Stivers
I Shilling
20
Stivers
1 Florin, or Guilder
3i
Florins 
1 Rix Dollar
6
Florins
 1 . Flemish
5 Guilders  1 Ducat
DENMARK.
16 Shillings  = 1 Mark
6 Marks ~  1 Rix Dollar
32 Rustics  1 Copper Dollar
6 Copper Dollars   1 Rix Dollar
RUSSIA.
18 Pennins  = 1 Gros
30 Gros  1 Florin
3 Florins  1 Rfx Dollar
2 Rix Dollars  1 Gold Puoai
FOREIGN EXCHANGE ,10?
RULE,
In exchanging of foreign moneys, work by the R lie of
Three, or by Practice ; and for exchanging foreign moneys
to Federal, work by the table in page 40.
EXAMPLES.
1. Philadelphia is indebted to London 1749/. currency,
what sum sterling must be remitted, when the exchange is
65 per cent. ?
. . . .
As 165 : 100 :: 1749 : 1060 sterling. Ans.
2. London is indebted to Philadelphia 1060Z. sterling;
what sum Pennsylvania currency must be remitted, the ex
change being 65 per cent, as above ?
. . . .
As 100 : 165 :: 1060 : 1749 Ans.
Or, 50 J 1060
10 i 530
5  106
53
1749Z. Ans.
3. Baltimore, Oct. 1, 1817.
Exchange for 1260Z. 10s. sterling.
Thirty days after sight of this my first of exchange, sec
ond and third of like tenor and date not being paid, pay to
A. B. or order, twelve hundred and sixty pounds ten shil
lings sterling, value received, and place the same to account,
as per advice from
P S n.
W. L. merchant, London.
What is the value of this bill in Federal money ?
1260Z. 1 Os. = 1260,5.x by 444 cents=5596 dolls. Oa
cents. Ans.
108 FOREIGN EXCHANGE.
4. London, January 1, 1818.
Exchange for 5596 dolls. 62 cts. Federal money.
Thirty days after sight of this my second of exchange,
first and third of the same tenor and date not paid, pay to
J. B. or order, five thousand five hundred and ninety six
dollars sixtytwo cents, value received, and place the same
to account, as per advice from
S.'S.
Mr. T. L. merchant, Baltimore.
How much sterling is the above bill, 4,44 cents to the
pound?
444)5596,62(1260
444
1156
888
2686
2664
222
20
4440(10 Ans. 1260Z. 10*.
4440
5. A merchant of Philadelphia receives from his cor
respondent in Dublin, a bill of exchange for 540Z. 15s. Irish
currency ; what is its value in Federal money ?
Ans. 2217 dolls. 7 cts.
6. A merchant in Philadelphia draws on his correspond
ent in Dublin for the balance of an account amounting to
2217 dolls. 7 cents ; what sum Irish currency mast be re
mitted to satisfy the draft? Ans. 540/. 15s.
Note. In tiPrie last examples tha course of exchange is considered
;IH 1 cin{r at jKir: when the exch.mge is nbove or below par, the per
cent, must be added or subtracted, as the case requires.
7. In a settlement between A of London and B of Phila
delphia, B is indebted to A in the sum of 3207. sterling, what
sum must be remitted by B to A to settle the balance, the
uxv hange being 12J per cent, from the United States to
(iivat Britain? Ans. 1598 dolls. 40 cts.
ALLIGATION. 109
8. C of New Yo] ;v remits 3259 dollars to his correspond
ent in Dublin, to be placed to his account ; for what sum
Irish currency, must he receive credit, the course of ex
change being 8 per cent, in favor of Ireland ?
Ans. 7361. nearly.
Note. The par ?f exchange between the United States of America
and most other trading countries, may be found by the table in page 40.
SECTION 13.
ALLIGATION.
ALLIGATION is a rule for finding the prices, and quantity
of simples in any mixture compounded of those things.*
Case 1.
To find the mean price of any part of the composition,
when the several quantities and prices are given.
RULE.
As the sum of the whole quantity, is to its total value, so
is any part of the composition, to its value.
EXAMPLE.
1. A merchant mixed 2 gallons of wine at 2 dollars per
gallon, 2 at 2 dollars 50 cents, and 2 at 3 dollars ; what is
one gallon of this mixture worth ?
gal.
2 at 2,00^400
2 at 2, 50 = 500
2 at 3,00 = 600
6 1500
G. D. c. G. D.c.
As 6 : 15,00 :: 1 : 2,50 Ans.
2. A grocer mixed 20 Ib. of sugar at 10 cents per Ib. 30
Ib. at 15 cents, and 40 Ib. at 25 cents; what is one pound
of this mixture worth? Ans. 181 cts.
3. A trader mixes 10 bushels of salt at 150 cents, 20 at
K
110 ALLIGATION.
160 cts. and 30 at 170 cts. per bushel; at what rate can he
afford to sell one bushel of this mixture? Ans. 1G3J cts.
4. If 4 ounces of silver at 75 cents per ounce, be melted
with 8 ounces at 60 cents per ounce, what is the value of
one ounce of this mixture ? Ans. 65 cents.
Case 2.
To find what quantity of several simples must be taken
at their respective rates, to make a mixture worth a given
price.
RULE.
Place the rates of the simples under each other, and link
each rate which is less than the mean rate, with one or more
that is greater. The difference between each rate and the
mean price set opposite to the respective rates with which it
is linked, will be the several quantities required.
Note. 1. If all the given prices be greater or less than the mean
rate, they must be linked to a cipher.
2. Different modes of linking will produce different answers.
EXAMPLES.
1. How many pounds of tea at 150, 160, and 200 cents
per pound, must be mixed together, that 1 pound may be
sold for 180 cents?
t 150x 20 at 150 cents }
Mean rate 180 2 160\) 20 at 160 V Ans.
( 200^ 30 + 20=50 at 200 )
2. How many gallons of wine at 3, 5, and 6 dollars per
gallon, must be mixed together, that one gallon may be
worth 4 dollars?
Ans. 3 gallons at 3 dolls. 1 gallon at 5 dolls, and 1
gallon at 6 dollars.
3. How many bushels of rye at 40 cents per bushel, and
corn at 30 cents, must be mixed with oats, at 20 cents, to
make a mixture worth 25 cents per bushel ?
t 20^rx 15 f 5 C 6 bushels of rye
1. Ans. 25 ? 30J ) 5 2. Ans. < 6 do. corn
( 40^ 5 (24 do. oats
4. A grocer has four several sorts of tea, viz. one kind
at 120 cents, another at 110 cents, another at 90 cents
ALLIGATION. Ill
and another at 80 cents per pound, how much of each sort
must be taken to fnake a mixture worth 1 dollar per pound ?
' 2 at 120 cents. ( 3 at 120 cents.
110 2 ^ ns >2 110
(3 80
f 1 at 120 cents.
( 1 80
f 2 at 120 cents.
6. Ans. ) j* *g0
(3 80
Note. From this last example it is manifest that a great many dif
ferent answers may result to the same question, according to the va
rious modes of linking the numbers together.
Case 3.
When the rate of all the simples, the quantity of one of
them, and the compound rate of the whole mixture are given,
to find the several quantities of the rest.
RULE.
Arrange the mean rate, and the several prices, linked to
gether as in case 2, and take their difference.
Then, as the difference of the same name with the quan
tity given,
Is to the rest of the differences respectively :
So is the quantity given,
To the several quantities required.
EXAMPLE.
1. A grocer would mix 40 pounds of sugar at 22 cents
per pound, with some at 20, 14, and 12 cents per pound ;
how much of each sort must he take to mix with the 40
pounds, that he may sell the mixture at 1 8 cents per pound ?
f!2 N 41b.
is) 14 V ~ 2
^ 20_ / / 4
. 22 6 against the price of the given quantity.
As 6 : 40 :: 4 : 26,66 Ib. at 12 cents.)
6 : 40 :: 2 : 13,33 do. 14 } Ans.
and 26,66 do. 20
112 ALLIGATION.
2. How much wheat at 48 cents, rye at 06 cents, and
barley at 30 cents per bushel, must be mixed with 24 bush
els of oats at 18 cents per bushel, that the whole may rate
at 22 cents per bushel ? Ans. 2 bushels of each.
3. How much gold at 16, 20, and 24 carats fine, and how
much alloy must be mixed with 10 ounces of 18 carats fine,
that the composition may be 22 carats fine ?
Ans. 10 oz. of 16 carats fine, 10 of 20, 170 of 24,
and 10 of alloy.
Case 4.
When the price of all the simples, the quantity to be mix
ed, and the mean price are given, to find the quantity of each
simple.
RULE.
Find their differences by linking as before :
Then, as the sum of the differences,
Is to the quantity to be compounded ;
So is the difference opposite to each price,
To the quantity required.
EXAMPLE.
1. How much sugar at 10, 12, and 15 cents per pound,
will be required to make a mixture of 40 pounds, worth 1 3
cents per pound ?
8 sum of the different simples.
As 8 : 40 :: 2 : 10 lb. at 10 els. )
8 : 40 :: 4 : 20 do. 15 \ Ans.
and 10 do. 12 >
2. How much golJ of 15, of 17, of 18, and of 22 carats
fine, must be mixed together to form a mixture of 40 ounces
of 20 carats fine?
Ans. 5 oz. of 15, of 17, and of 1 8, and 25 oz. of 22.
3. How many gallons of water must be mixed wkh wine
at (> dollars per gallon, to fill a vessel of 70 #,i lions, so that
it may he sold without loss at 5 dollars per gallon?
4/7A. 1 1 07) lions of water.
VULGAR FRACTIONS. 1 1 3
PART VI.
VULGAR FRACTIONS.
A VULGAR FRACTION is any supposed part or parts of an
unit, and is represented by two numbers placed one above
the other, with a separating line between them ; thus,  one
fifth, fourninths.
The number above the line is called the numerator, and
that below the line the denominator. Thus,
4 numerator 6
&c.
9denomina. 10
The denominator shows how many parts the unikpr inte
ger is divided into, and the numerator shows how many of
those parts are contained in the fraction.
Vulgar fractions are either proper, improper, compound,
or mixed.
A proper fraction is when the numerator is less than the
denominator, as J, f, j, f , &c.
An improper fraction, is when the numerator is eitner
equal to, or greater than the denominator, as f , f, 2 T , &c.
A compound fraction is a fraction of a fraction, as f of $,
f of/ of jf,&c.
A mixed fraction is a whole number and fraction united,
as 8, 4J, 120f , &c.
SECTION 1.
Reduction of Vulgar Fractions.
Case 1.
To Reduce a Vulgar Fraction to its lowest terms.
RULE.
Divide the greater term by the less, and that divisor by
the remainder, till nothing be left, the last divisor is the
K2
114 REDUCTION OF VULGAR FRACTIONS.
common measure, by which divide both parts of the frac
tion : the quotient will be the answer. Or,
Take aliquot parts of both terms continually, till the frac
tion is in its lowest terms.
Note. 1. 1.' the common measure when found is 1, the fraction ia
already in its lowest terms.
2. Ciphers to the right of both the terms may be cut off thus,
w=f
EXAMPLES.
1. Reduce J to its lowest terms.
36 ) 48 ( i
36
Common measure 1 2 ) 36 ( 3
36
12)36(3
36 3 2 div. 6 div.
Ans. 36 18 3
12)48(4 Or, = = Ans.
48 48 24 4
2.. Reduce ^f to its lowest terms. Ans. f
3. Reduce T 7 ^ to its lowest terms. 
4. Reduce fVVVV to i ts lowest terms. f
5. Reduce T y T to its lowest terms. 
6. Reduce gW/T to its lowest terms. J
Case 2.
To reduce a mixed number to an improper fraction.
RULE.
Multiply the whole number, by the denominator of the
fraction, and add the numerator to the product, for a new
numerator, under which place the given denominator.
EXAMPLE.
1. Reduce 8$ to an improper fraction.
8
4
335
 Ans.
4
REDUCTION OF VULGAR FRACTIONS. 115
2. Reduce 12 T f to an improper fraction. Ans. 2 j~?
3. Reduce 183 / T to an improper fraction. *ff 8
4. Reduce 514 T \ to an improper fraction. 8 T  9
5. Reduce 68425? to an improper fraction. a 7 3 T 7 3
Case 3.
To reduce an improper fraction to a whole or mixed
number.
RULE.
Divide the numerator by the denominator ; the quotient
will be the answer required.
Note. This case and case 2, prove each other.
 EXAMPLE.
1 . Reduce \ 5 to its proper terms.
4)35(8 Ans.
32
3
2. Reduce 3 f T 8 to its proper terms. Ans. 183^
3. Reduce 2 \ 6 5 to its proper terms. 352i
4. Reduce 3 T 6 y to i ts proper terms. 56y
5. Reduce 8 y 9 to its proper terms. 514 T 5
Case 4.
To reduce several fractions to others that shall have one
common denominator, and still retain the same value.
RULE.
Reduce the given fractions to their lowest terms, then
multiply each numerator into all the denominators, but its
own. for a new numerator ; and all the denominators into
each other for a common denominator.
EXAMPLE.
1. Reduce t, , and , to a common denominator.
1X3X412V
2X2X4 16> numerators.
3X2X318)
2X3X424 common denominator.
116
REDUCTION OF VULGAR FRACTIONS.
2. Reduce f , , and , to a common denominator.
Ana. ?&, V&, 
8. Reduce i, , T 4 j, and f , to a common denominator.
Case 5.
To reduce several fractions to others, retaining the same
value, and that shall have the least common denominator.
RULE.
Divide the given denominators by any number that will
divide two or more of them without a remainder ; set the
quotients and undivided numbers underneath ; divide these
numbers in the same manner, and continue the operation,
till no two numbers are left capable of being lessened ; the
product of these remaining numbers, together with the di
visor or jdivisors, will give the least common denominator.
Divide the common denominator, so found, by each par
ticular denominator, and multiply the quotient by its own
numerator for a new numerator, under which place the com
mon denominator.
EXAMPLE.
1. Reduce , J, j, and f , to the least common denom
inator.
3)2 368
2)2 128
111 4 X~2x 3=24 common denominator.
'2)24
Divisors
12X1 = 12
3 8x2=16
6 4X5=20
^8 3X7=21
Then, if, , ff, f i Ans.
2. Reduce f , f , T \, and vV, to the least common donom.
mator. Ans. A n , rV 5 o> T V
REDUCTION OF VULGAR FRACTIONS. _ 117
3. Reduce , f , T 4 5 , and , to the least common denomi
nator. Ans. Jf , H, H> If
Case 6.
To reduce a compound traction to a single one.
RULE.
Multiply all the numerators together for a new numerator,
and all the denominators for a new denominator.
Note. Such figures as are alike in the numerators and denominators
may be cancelled.
EXAMPLE.
1. Reduce f of J of to a single fraction.
2 X 3 X 4 24 2 Or cancelled.
=  Ans. 2 3 4 2
3X4X5=60 5 _ _ _ = _ Ans.
3 4 5 5
2. Reduce J of of T \ to a single fraction.
Ans. fi
3. Reduce of f of J to a single fraction.
Ans. ^
4. Reduce f of ^ of f J to a single fraction.
Ans. T Vo
Case 7.
To reduce a fraction of one denomination to the fraction
.of another, but greater, retaining the same value.
RULE.
Make it a compound fraction, by comparing it with all
the denominations between it and that to which it is to be
reduced ; reduce this fraction to a single one.
EXAMPLE.
1 . Reduce f of a penny to the fraction of a pound.
5 X 1 X 1 5
6 X 12 X 20 1440
2. Reduce of a pennyweight to the fraction of a pound
troy Ans. J T
% Reduce T 9 :i  of a pint of wine to the fraction of a hogs
head. Ans. ^fa
118 REDUCTION OF VULGAR FRACTIONS.
4. Reduce TT of a minute to the fraction of a day.
TJ84
Case 8.
To reduce tne fraction of one denomination to the frac
tion of another, but less, retaining the same value.
RULE.
Multiply the given numerator, by the parts of the denom
inator, between it and that to which it is to be reduced, for a
new numerator, and place it over the given denominate,,,
which reduce to its lowest terms.
1. Reduce TT 5 T g of a pound to the fraction of a penny.
5 X20X12 1200 5
1440X1 X 1 H40 6
2. Reduce $~ of a pound troy to the fraction of a pen
nyweight. Ans. j
3. Reduce 7 ~ of a hogshead to the fraction of a pint.
Ans. ^
4. Reduce yjg4 f a day to tne fraction of a minute.
Ans. VT
Case 9.
To find the value of a fraction in the known paits of an
integer.
RULE.
Multiply the numerator by the known .parts of the inte
ger, and diviilo by the denominator.
EXAMPLE.
1. What is the value of f of a pound sterling?
20 shillings = 1 pound.
2
3 ) 40
13 4 Ans. 13*. 4ct
t. Red are J of a pound troy tc iu proper quantity.
Ans. 7oz. 4dwt.
REDUCTION OF VULGAR FRACTIONS. 11D
3. Reduce f of a mile to its proper quantity.
Ans. 6fur*. Wp.
4. Reduce T 3 F of a day to its proper time.
Ans. 7h. I2min.
5. What is the value of f of a dollar. Ans. 80 cts.
Case 10.
To reduce any given quantity, to the fraction of a greater
Munomination of the same kind.
RULE.
Reduce the given quantity to the lowest denomination
mentioned for a new numerator, under which set the integral
part (reduced to the same name) for a denominator.
EXAMPLES.
1. Reduce 6s. Sd. to the fraction of a pound.
s d. s.
68 20
12 12
80 1 240
= Ans.
240 3
2. Reduce 25 cents to the fraction of a dollar.
25 1
100 4
3. Reduce 31 gallons 2 quarts to the fraction of a hogs
head. Ans. i.
4. Reduce 6 hundred weight 2 quarters 18 pounds to the
fraction of a ton. Ans. J.
Case 11.
To reduce a vulgar fraction to a decimal of the same
value.
RULE.
Add ciphers to the righthand of the numerator, and cfivide
by the denominator.
120 ADDITION OF VULGAR FRACTIONS
EXAMPLE.
' 1. Reduce f o a decimal fraction of the same value.
4 ) 300
,75 Ans.
2. Reduce ^J to a decimal fraction. Ans. ,85
SECTION 2.
ADDITION OF VULGAR FRACTIONS.
Case 1.
To add fractions that have the same common denomi
nator.
RULE.
Add all the numerators together, and divide the amount
by the common denominator.
EXAMPLE.
1. Add T ^, r 5 2, T 7 2> T 9 2 and J together.
numerators.
1
5
7
9
11
common denominator 12 ) 33 ( 2 An*.
24
9 3
12 4
2. Add /y, T 8 T , i j, {, and if together. Anj. 2$
8. Add Jf, H, AJ,and JJ together. 3*.
ADDITION OF VULGAR FRACTIONS. 121
Case 2.
To add fractions having different denominators.
RULE.
Reduce the given fractions to a common denominator, by
case 5, and proceed as in the foregoing case.
EXAMPLE.
1. Add $, J, and T 9 j together.
12
1359 12 18 15 18 18
 =   15
I 4 8 12 24 24 24 *4 18
24)63(2}
48
15
2. Add i, i, and together. Ans* 1 &
3. Add f , f ,3, f and r 8 5 together. 3^
Case 3.
To add mixed numbers.
RULE.
Add the fractions as in the foregoing cases, and the inte
gers as in addition of whole numbers.
EXAMPLES.
I. Add 13^, 9 7 y itself will be equal to the ^iven number.
RUf.R.
1. Point off the given sum into periods of two figures
a_, beginning at the right hand.
SQUARE ROOT.
12!)
2. Subtract from the first period on the left, the greatesi
square contained therein ; setting the root, so found, for the
first quotient figure.
3. Double the quotient for a new divisor, and bring down
trte next period to the remainder lor a new dividual. Try
how often the divisor is contained in the dividual, omitting
the units. figure, and place the {lumber, so found, in the quo
tient, and on the right of the divisor ; multiply and subtract
as in division.
4. Double the quotient for a new divisor ; bring down the
next period, and proceed as before, till all the periods are
brought down. When a remainder occurs, add ciphers for a
new period, the quotient figure of which will be a decimal,
which may be extended to any required degree of exactness.
PROOF. *
Square the root, adding the remainder (if any) to the
product, which will equal the given number.
EXAMPLE,
1. What is the square root of 531441 ?
53144J ( 729 An*.
49
I. double the quotient 14,2 ) 414
284
2. double
do. 144,9) 13041
13041
729
729
6561
1458
5103
Ans. 327
2187.
6561.
19683.
4698.
6031.
1506,23 t
2756,22^4
531441 proof.
2. What is the square root of 106929 ?
3. What is the square root of 4782969 ?
4. What is the square root of 43046721 ?
5. What is the square root of 387420489 ?
6. What is the square root of 22071204 ?
7. What is the square root of 36372961 ?
8, What is the square root of 2268741 ?
9. What is fhe square roof of 7596796 '
130 SQUARK ROOT.
When there are decimals joined to the whole numbers in
/he given sum, make the number of decimals even by adding
ciphers, and point off both ways, beginning at the 'decimal
point.
10. What is the sqiuire root of 9712,718051 ?
Ans. 98,553 +
11. What is the square root of 3,1721812 ?
Ans. 1,78106 +
12. What is the square root of 4795,25731 ?
Ans. 69,247
13. What is the square root of ,00008836 ?
Ans. ,0094
To extract the square root of a vulgar fraction.
RULE.
Reduce the fraction to its lowest term ; then extract the
"Kjuare root of the numerator for a new numerator, and the
scjuare root of the denominator for a new denominator.
Nute. When the fraction is a surd* that is, a number whose exact
ff>ot cannot be found, reduce it to u decimal and extract the root there
from.
EXAMPLES,
1 . What is the square root of Jf A 7 Ans. J
"2. What is th<; square root of Jf A ! f
3. What is the square root of iff f ? ^f
Surds.
4. What is the square root of ffj '! Ans. ,86602 +
5. What is the square root of f Jf ? ,93309 +
6. What is ihr square root of fJ ? ,72414 +
To ertraci thr vjimre root of a mired number.
RUTJv
1. Reduce the fractional part of the mixed number to its
lowest term, and thn mixed number to an improper fraction.
SQT T ARK ROOT. 131
2. Extract the roots of the numerator and denominator,
for a new numerator and denominator.
If the mixed number given be a surd, reduce the frac
tional part to a decimal, annex it to the whole number, and
extract the square root therefrom.
EXAMPLES.
1. What is the square root of 37 ^f? Ans. 6J
2. What is the square root of 27 r 9 F ? 5i
3. What is the square root of 51}? 7}
4. What is the square root of 9ff ? 3i
Surds.
5. What is the square root of 7 T 9 T ? Ans. 2,7961 +
6. What is the square root of 8f ? 2,951 9 +
7. What is the square root of 85ff ? 9,27 f
Any two sides of a right angled triangle given to find the
third side.
RULE.
As the square of the hypothenuse or longest side, is al
ways equal to the square of the base and perpendicular, the
other sides added together ; then it is plain if the length of
the two shortest sides are given, the square root of both
these squared and added together, will be the length of the
third or longest side.
Again, when the hypothenuse, or longest side, and one of
the others are given 5 the square root of the difference of the
squares of these two given sides will be the len ;th of the
remaining side.
132
SQUARE ROOT.
EXAMPLE.
1. The wall of a fortress is 3fi feet high, and the ditch
before it is 27 feet wide: it is required to find the length of
n ladder that will reach to the top of the wall from tho op
posite side of the ditch ? Ans. 45 feet.
27 ft. ditch
2. The top of a castle" from the ground is 45 yards high,
and is surrounded with a ditch 60 yards broad, what length
must a cable be to reach from the outside of the ditch to the
top of the castb? Ans. 75 yards.
8. In a right angled triangle, ABC, the hypothenuse
line A C is 45 feet, the base A B 27 feet; required the length
of the perpendicular line B C? Ans. 36 feet.
B
4. In a right angled triangle, A B C, the line A C is 75
feet, B C 45 feet; required the length of the line A B?
Ans. 60 feet.
To find the side of a square equal in area to any given ./
RULE.
Extract the square root of the content of the given su
H;rfices ; the quotient will give the side of the equal square
sought.
SQUARK ROOT 133
EXAMPLES.
1. If the content oi a given circle be ICO, what is the
side of the square equal? Ans. 12,(>4911{
2. If the area of a circle he '2025, what is the side of the
square equal ? Ans. 45.
3. It* the area of a circle be 750, what is the side of the
square equal? Ans. 27,38612 +
To find tht diameter of a circle of a given proportion
larger or less than a given one.
RULE.
Square the diameter of the given circle, and multiply
(if greater) or divide (if less) the product, by the number of
times the required circle is greater or less than t.\c given
EXAMPLES.
1. There is a circle whose diameter is 4 feet; what is the
diameter of one 4 times as large ? Ans. 8 feet.
2. A has a circular yard of 100 feet diameter, but wishes
to enlarge it to one of 3 times that area; what wall the
diameter of the enlarged one measure? Ans. 173,2 f
3. If the diameter of a circle be 12 inches, what will be
the diameter of another circle of half the size ?
Ans. 8,48 f inches.
The area of a circle given to find the diameter.
RULE.
Multiply the square root of the area by 1,12837, and the
produce will be the diameter.
EXAMPLES.
1. When the area is 160, what is the diameter?
Ans. 1 4,272947 f
2. What length of a halter will be sufficient to fasten a
horse from a post in the centre, so that he may be able to
graze upon an acre of grass, and no more?
Ans. 7,1364 perches, or 117 ft. 9 inches*
M
134 SQUARE ROOT.
* Application.
1. If an army of 20736 men is formed into a square
column ; how many men will each front contain?
Ans. 144 men.
2. How many feet of boards will it require to lay the
floor of a room that is 25 feet square? Ans. 625 feet.
3. A certain square pavement contains 191736 square
stones, all of the same size ; how many are contained in one
of its sides ? Ans. 444.
4. In a triangular piece of ground containing 600 perches,
one of the shortest sides measures 40 perches, and the other
30 ; what is the length of the longest side ?
Ans. 50 perches.
5. Two gentlemen set out from Pittsburgh at the same
time ; one of them travels 84 miles due north, and the other
50 miles due west ; what distance are they asunder ?
Ans. 97^ { miles.
ft. What is the square root of 964,5192360241 ?
Ans. 31,05671.
7. What is the square root of 1030892198,4001 ?
Ans. 32107,51.
As it is probable many teachers rind it difficult to explain
satisfactorily the reasons and principles upon which the rules
for the extraction of the roots are founded, I have subjoined
the following demonstration of the rule for extracting the
square root ; and which will also serve to show the reason
of the rules for extracting the roots of the higher powers.
From what has been already said on this rule, it is sufficiently
evident that the extraction of the square root has always this
operation on numbers, viz. to arrange the number of which
the root is extracted into a square form. Thus, if a car
penter should have 625 feet of dressed boards for laying a
floor ; if he extracts the square root of this number, (625)
he will have the exact length of one side of a square floor,
which these boards will be sufficient to make.
Let this then be the question : Required to find the length
of one side of a square room, of which 625 square feet of
boards will be sufficient to lay the floor.
The first step, according to the rule given, is to point off
the numbers into periods of two figures each, beginning
at the unit's place. This ascertains the number of figures
of which the root will consist, from this principle, that the
product of any two numbers can have, at most, but so many .
places of figures, as there are places in both the factors, and
at least, but one less.
The number (625) will then have two periods, and con
sequently the root will consist of two figures.
Operation. The last, or lefthand period in this
. . number is 6, in which 4 is the greatest
625 ( 2 square, and 2 the root ; hence 2 is the
4 first figure in the root, and as one fig
ure more is yet to be found, we may
225 for the present supply the place of
that figure with a cipher (20) ; then 20
Figure 1. will express the just value of that part
of the root now obtained. But a root
is the side of a square, of equal sides.
A
20
20
Hence, figure 1 exhibits a square,
each side of which is 20 feet, and the
a 400 b area 400, of which 20 is the root now
20 obtained.
As the rule requires, we next subtract the greatest square
contained in the first period, and to the remainder bring
down the next period. 4 is the greatest square contained in
the first period (6), and as it falls in the place of hundreds,
is in reality 400, as may be seen by filling up the places to
the righthand with ciphers ; this subtracted from the (i
leaves 2, as a remainder, to which if the next period is
brought down, the remainder will be 225 ; and the original
number of feet (625) has been diminished by the deduction
of 400 feet, a number equal to the superficial content of the
square A.
Figure 1, therefore, exhibits the exact progress of the
operation, and shows plainly how 400 feet of the boards
have been "disposed in the operation thus far, and also that
225 feet yet remains to be added to this square, by en
larging it in such a manner as not to destroy its quadrate
form, or its continuing a complete and perfect square.
Should the addition be made to one side, only, the figure
would lose its square. The addition must be made to two
sides: accordingly the rule directs to lt double the quotient
136
SQUARE ROOT.
(viz. the root already found) for a new divisor ;" the double
of the root is equal to two sides of the square A, for the
double of 2 is 4, and as this 4 falls in the place of tens,
since the next figure in the root, according to the rule, is to
be placed before it in the place of units, it is in reality 40,
and equal to a b and b c which are 20 each.
Operation continued. Again, as the rule di
. . rects, try how often the di~
625 ( 25 visor is contained in the
4 dividual, omitting the units
figure.
45)225 The divisor is here 4,
225 which, as has already been
shown, is 4 tens, or 40 ;
000 this is to be divided into
the remainder 225 ; omit
n 5 ting the last figure=220.
But 40 the sum of the
two sides b c and c d / to
which the remaining 225
is to be added, and the
square A enlarged, which
omitting the last figure (5)
gives 5 for the last quo
tient figure ; 5 is the breadth
of the two parallelograms
B B, the area of each (5 X
20 b 5 20) is 100.
The rule requires us to omit the last figure in the divid
ual^ and also, to place the quotient figure, when found, on
the right of the divisor ', the reasons for which are, that ad
ditions of the two parallelograms B B ta the sides of the
square A (fig. 2) do not leave it a perfect square, a deficiency
remaining at the corner D ; the righthand figure is omitted
to leave something of the dividend for this deficiency.
And as this deficiency is limited by the two parallelograms
B B, and the quotient figure (5) is the breadth of these,
consequently the quotient 5= the length of each of the
sides of the small square D ; this quotient then being
placed on the right of the divisor and multiplied into itself
gives the area of the square I) ; which being added to the
20
5
B 5
D 5
en
d
100
25
c
A
B 25
20
20
20
5
a
400
100
X
CUR!: ROOT. 137
contents of the two parallelograms B B each (100) 200,
shows that the remaining 225 feet of boards have kasji dis
posed of, in these three additions (B B D) made to the T^st
square A ; whilst the figure is seen to be continued a corrt*
plete square.
Q. E. D.
PROOF.
The square A =400 feet
The parallelograms B B =200
The square D = 25
625 feet.
SECTION 4.
THE CUBE ROOT.
THE cube is the third power of any number, and is found
by multiplying that number twice into itself. As 2 X 2 x 2^.
To extract the cube root, therefore, of any number, is to
find another number, the cube of which will equal the given
number. Thus 4 is the cube root of 64 ; for 4X 4x 4=64.
RULE.
1. Point off the given number into periods of three figures
each, beginning at the units place, or decimal point. These
periods will show the number of figures contained in the
required root.
2. Find the greatest cube contained in the first period,
and subtract it therefrom ; put the root of this cube in the
quotient, and bring down the next period to the remainder
for a new dividual.
3. Square the quotient and multiply it by 3 for a defective
divisor; 2x2x3 = 12. Find how often this is contained
in the dividual, rejecting the units and tens therein, and
place the result in the quotient, and its square to the right of
the divisor. 4 x 4=16 put to the divisor 12 =
M2
138
CUBE TtOOT.
4. Multiply the last tigure in the quotient by the rest, and
the product by 30 ; add this to the defective divisor, ana
multiply this sum by the last figure in the quotient, subtract
that product from the dividual, bring down the next period,
and proceed as before.
Note. When the quotient is 1, 2, or 3, put a cipher in the place of
tens in filling up the square on the right of the divisor.
EXAMPLE.
U What is the cube root of 48228544 1
Greatest cube in 48 is
Operation.
48228544(364 Ans.
27
21228
Square of 3 X by 3=27. 1 def. divis. '
Square of 6 put to 27 =2736
6 last quo. fig. X by the rest
and 30 = 540
Complete divisor 3276 J 19656
Square of 36 X 3=3888. 2 def. divis.
Square of 4 put to 3888 = 388816
4 last quo. fig. X by the rest
and 30 = 4320
Complete divisor
393136
1572544
1572544
2. What is the cube root of 13824 T
3. What is the cube root of 373248 ?
4. What is the cube root of 5735339 ?
5. What is the cube root of 84604519 ?
6. What is the cube root of 27054036008 I
7. What is the cube root of 122615327232 ?
8. What is the cube root of 22069810125 ?
9. What is the cube root of 219365327791 ?
10. What is the cube root of 673373097125 ?
11. What is the cube root of 12,977875 ?
12. What is the cube root of 15926,972504 ?
13. What is the cube root of 36155,027576 ?
Ans. 24
72
179
439
3002
4968
2805
6031
8765
2,35
?Mfi f
33,06 f
CUBE ROOT. 139
14. What is the cube root of ,053258279 ? Ans. ,376 +
15. What is the cube root of ,001906624 ? ,124
16. What is the cube root of ,000000729 ? ,009
17. What is the cube root of 2 ? 1,25 +
To extract the cube root of a vulgar fraction.
RULE.
Reduce the fraction to its lowest terms ; then extract the
cube root of the numerator for a new numerator, and the
cube root of the denominator for a new denominator ; but
if the fraction be a surd, reduce it to a decimal, and extract
the root from it for the answer.
EXAMPLES.
1. What is the cube root of f ff 1 Ans. 4
2. What is the cube root of T 3 /^ ? f
3. What is the cube root of 4f f ? f
Surds.
4. What is the cube root of 4 ? Ans. ,829 f
5. What is the cube root of f ? ,873 f
6. What is the cube root of f ? ,822 f
To extract the cube root of a mixed number.
RULE.
Reduce the fractional part to its lowest terms, and the
mixed number to an improper fraction ; extract the cube
roots of the numerator and denominator for a new numerator
and denominator ; but if the mixed number given be a surd,
reduce the fractional part to a decimal, annex it to the whole
number, and extract the root therefrom.
^ EXAMPLES.
1. What is the cube root of 31 J/j ? Ans. 3
2. What is the cube root of 12if ? 2J
3. What is the cube root of 405 r 2 ^ ? 7f
140 CUBE ROOT.
Surds.
4. What is the cube root of 7} 1 Ans. 1,93f
5. What is the cube root of 8f ? 2,057 +
6. What is the cube root of 9 ? 2,092 +
To find the side of a cube that shall be equal to any given
solid, as a globe, a cone, .
netricnl progression, worh by the folloiring
RULE.
1 . R iise the ratio in the given sum, to that power whose
index 5 hall always bo one less than the number of term?
give n ; multiply the number so found by the first term, and
Tho jrviduct will be the last term, or greater extreme.
GEOMETRICAL PROGRESSION. 145
2. Multiply the last term by the ratio, from that product
subtract the first term, and divide the remainder by the
ratio, less one ; the amount will be the sum of the series, or
of all the terms.
EXAMPLE.
1. Suppose 20 yards of broadcloth was sold at 4 mills
for the first yard, 12 for the second, 36 for the third, &c.
what did the cloth come to, and what was gained by the
sale, supposing the prime cost to have been $15 per yard ?
Note. In this question observe, the first term is 4, the ratio 3, arid
the number of terms 20, consequently the ratio 3 must be raised to
20 1= 19th power. Thus,
g S S 1
! I 1 
& 3 8 I
Indices 1234
Ratio 3 9 27 81
. 81
81
648
6561= the 8th power
6561
6561
39366
32805
39366
43046721= 16th power
27= 3d power
301327047
86093442
1162261467=19th power
X 4 first term
4649045868=20th or last term
X3
13947137604
4 first term
Ratio 31=2)13947137600
69 73568,80,0= sum of the series, or number of
First cost of the cloth 300,00 mills for which the cloth was sold
0=gam,
N
146 GEOMETRICAL PROGRESSION.
2. A father gave his daughter who was married on the
first day of January, one dollar towards her portion, prom
ising to double it on the first day of every month for one
year ; what was the amount of her whole portion 'I
Ans. 4095 dollars.
3. A merchant sold 15 yards of satin ; the first yard for
is. the second for 2s. the third for 4s. &c. in geometrical
progression ; what was the price of the 15 yards?
Ans. 1638Z. 7s.
4. A goldsmith sold 1 pound of gold at 1 cent for the first
ounce, 4 for the second, 16 for the third, &c. ; what did it
come to, and what did he gain, supposing he gave 20 dollars
per ounce ?
Ans. He sold it for 55924 dollars 5 cents, and gained
55684 dollars 5 cents.
5. What sum would purchase a horse with 4 shoes and
8 nails in each shoe, at one mill for the first nail, 2 mills foi
the second, 4 for the third, &c. doubling in geometrical pro
gression to the last ?
Ans. 4294967 dollars 29 cents 5 mills.
t>. What sum would purchase the same horse, with the
same number of shoes, and nails, at 1 mill for the first nail,
3 for the second, 9 for the third, &c., in a triple ratio of
geometrical progression to the last ?
Ans. 926510094425 dollars 92 cents.
7. What sum would purchase the same horse, with the
same number of shoes, and nails, at 1 mill for the first nail,
4 for the second, 16 for the 3d, &c., in a quadruple ratio of
geometrical progression to the last ?
Ans. 6148914691236517 dollars 20 cents 5 mills.
S. Sold 30 yards of silk velvet, at 2 pins for the first
yard, 6 for the second, 18 for the third, &c., and these dis
posed of at 100U for a farthing ; what did the velvet amount
to, and what was gained by the sale, supposing the prime
cost to hav<3 IKJCII 100/. per yard '!
Amount 214469929/. 5s. 3
An *' * Gained 214466929/. 5*. Sjd.
SINGLE POSITION. 147
SECTION 7.
OF POSITION.
POSITION is a rule for finding the true number, by one or
more false or supposed numbers, taken at pleasure.
It is of two kinds, viz. Single and Double.
Single position teaches to resolve such questions as require
but one supposed number.
RULE.
1. Suppose any number whatever, and work in the same
manner with it as is required to be performed in the given
question.
2. Then, as the amount of the errors, is to theN
sum, so is the given number to the one required.
PROOF.
Add the several parts of the result together, and if k
agrees with the given sum, it is right.
EXAMPLES.
1. A schoolmaster being asked how many scholars he had,
said, if I had as many, half as many, and one quarter as
many more, I should have 1 32 ; how many had he ?
Suppose he had 40 As 110: 40 :: 132 : 48 Ana.
as many 40 p roof ^
as many 20 40
\ as many 10 g.
US J?
132
2. It is required to divide a certain sum of money among
4 persons, in such a manner that the first shall have , the
second ^, the third J, and the fourth the remainder, which
is 28 dollars ; what was the sum ?
Suppose 72 As 18 : 72 :: 28 : 112 dolls. AIM.
i is 24 Proof 112
i is 18 , . rzr
1 i 19 * ls d '
* 1S _ i is 28
.54 J is 18f
Rem. 18 84
28 last share.
148 DOUBLE POSITION.
3. A, B, and C, buy a carriage for 340 dollars, of which
A pays three times as much as B, and B four times as much
as C ; what did each pay ? f A paid 240 dolls.
Ans. IE 80
(C 20
4. What is the sum of which 1, ], and , make 148 dol
lars? Ans. 240 dollars.
5. A person having spent ^, and J of his money : had 26f
dollars left; what had he at first? Ans. 100 dollars.
6. A, B, and C, talking of their ages, B said his age was
once and a half the age of A ; C said his was twice and T \
the age of both, and that the sum of their ages was 93 ; what
was the age of * each? i A's age 12 years.
Ans. < B's 18
( C's 63
7. Seveneighths of a certain number exceeds fourfifths
by 6; what is that number? Ans. 80.
8. A gentleman bought a chaise, horse, and harness, for
360 dollars ; the horse came to twice the price of the har
ness, and the chaise to twice the price of the horse and har
ness together ; what did he give for each ?
( 80 dollars for the horse
Ans. < 40 harness
( 240 chaise
9. A gentleman being asked the price of his carriage, an
swered that i, i, i, and of its price was 228 dollars ; what
was the price of the carriage ? Ans. 240 dolls.
10. A saves of his wages, but B, who has the same sal
ary? by spending twice as much as A, sinks 50 dollars a
year; what is their annual salary? Ans. 150 dolls, each.
SECTION 8.
DOUBLE POSITION.
DOUBLE POSITION is making use of two supposed num
bers, to find the true one.
RULE.
1. Take any two numbers, and proceed with them ac
cording to the conditions of the question, noting the errors
DOUBLE POSITION. 149
of the results ; multiply these errors crosswise, viz. the
first position by the last error, and the last position by the
first error.
2. If the errors be alike, that is, both greater, or both less
than the given number, take their difference for a divisor,
and the difference of the products for a dividend : but if they
are unlike, take their sum for a divisor, and the sum of the pro
ducts for a dividend, the quotient will be the answer required
EXAMPLE.
1 . A father leaves his estate to be divided among his three
sons, A, B, and C, in the following manner, viz. A is to
have onehalf wanting 50 dollars, B onethird, and C 10 dol
lars less than B ; what was the sum left, and what was each
son's share?
Operation.
1st. Suppose 240 dollars.
Then 240 f 250=70 A's part
240+ 3= 80 B's part
B's share 80 10= 70 C's part
Sum of all their parts 220
20 er. too little,
2d. Suppose 300 dollars.
Then300f 2 50=100 A's part
300: 3= 100 B's part
B's share 10010= 90 C's part
Sum of all their parts 290
10 er. too little,
errors.
1st. sup. 240^20=6000
2d. sup. 300^10=2400
10)3600(360 An*.
Proof 360; 2 50=130
360^ 3= 120
120 10= 110
360
N2
150 DOUBLE POSITION.
2. A and B have the same income ; A saves the of his,
but B by spending 30 dollars per annum more than A, at the
end of 8 years finds himself 40 dollars in debt ; what is their
income, and how much does each spend per annum ?
( Their income is 200 dolls, per annum
Ans. < A spends 175
( B spends 205
3. A, B, and C, would divide 100 dollars between them,
so as B may have 3 dollars more than A, and C 4 dollars
more than B ; how many dollars must each have ?
( A 30 dollars.
Ans. I B 33
(C37
4. A, B, and C, built a house which cost 10,000 dollars;
A paid a certain sum, B paid 1000 dollars more than A, and
C paid as much as both A and B ; how much did each one
pay?
( A paid 2000 dolls.
Ans. 1 B 3000
I C 5000
5. A gentleman has 2 horses and a saddle worth 50 dol
lars, which saddle if he put on the back of the first horse,
will make his value double that of the second; but if he put
it on the second horse, it will make his value triple that of
the first ; what is the value of each horse ?
A $ First horse 30 dolls.
s ' I Second do. 40
6. The head of a fish is 9 inches long, and its tail is as
long as its head and half its body, and its body is as long as
its head and tail together ; what is its whole length ?
Ans. 6 feet.
7. A laborer hired 40 days upon this condition, that he
should. receive 20 cents for every day he wrought, and for
l<'it 10 cents for every day he was idle; at settlement he re
ceived 5 dollars ; how many days did he work, and how
many was he idle? Ans. Wrought 30 days, idle 10
8. A and B vested equal sums in trade ; A gained a sum
equal to J of his stock, and B lost 225 dollars ; then A's
money was double that of B's ; what sum had each vested ?
Ans. 600 dollars.
9. Divide 15 into two such parts, so that when the greater
PERMUTATION. 151
is multiplied by 4, and the less by 16, the products will be
equal? Ans. Greater 12, less 3.
10. A person being asked in the afternoon what o'clock
it was, answered, that the time past from noon was equal to
fa of the time to midnight ; what o'clock was it ?
Ans. 36 minutes past .1 o'clock.
SECTION 9.
PERMUTATION.
PERMUTATION is the finding how many different ways any
given number of things may be varied in position, or ar
rangement. Thus, 123 are six different arrange
132 ments made upon the three
213 figures 1 2 3.
2 3 1
3 1 2
321
RULE.
Multiply all the terms from an unit up to the given num
ber into one another, and the last product will be the num
ber of changes required.
EXAMPLE.
1. In how many different positions may 5 persons be
placed at a table.
1X2X3X4X5^120 Ans.
2 How many changes may be rung on 12 bells, and how
Jong would they be ringing but once over, allowing 10
changes to be rung in 1 minute, and the year to contain 365
days 6 hours ?
Ans. 479001600 changes, and would require 91 years,
3 weeks, 5 days, and 6 hours.
3 Seven men not agreeing with the owner of a boarding
house about the price of boarding, offer to give 100 dollars
each, for as long time as they can seat themselves every day
differently at dinner ; this offer being accepted, how long
may they stay ? Ans. 5040 days, or 13 years, 295 days.
152 COMBINATION.
4. What number of variations will the 9 digits admit of?
Ans. 362880.
5. How many changes may be made on the 26 letters of
the alphabet ?
Ans. 403,291,461,126,605,635,584,000,000.
Quatril. Trilns. Billions. Millions. Units.
Note. From the answer to this last question, which amounts to a
number, of which we cannot form any conceivable idea, we may dis
cover the surprising power of numbers, and also the endless variety of
ideas that may be distinctly communicated by these 26 simple charac
ters. It will also be evident from the method of notation here used,
that a row of figures of any given length whatever, may be numerated,
though we may be entirely unable to comprehend the amount.
SECTION 10.
COMBINATION.
COMBINATION of quantities, is the showing how often a
less number of things can be taken out of a greater, and
combined or joined tdgether differently.
RULE.
Take a series of 1, 2, 3, &c. up to the number to be com
bined ; take another series of as many places, decreasing by
unity from the number out of which the combinations are to
be made ; multiply the first continually for a divisor, and the
latter fcr a dividend, and the quotient will be the answer.
EXAMPLE.
1. How many combinations may be made of 7 dollars
out of 12?
1x2x3, &c. up to 7 =5040 divisor.
Again, 12 the whole number of terms less 7 = 5
Hence 12 x 11 X 10, &c. down to 5 = 3991680 dividend.
And 5040 ) 3991680 ( 792 Ans.
2. How many combinations can be made of 6 letters out
of 24 of the alphabet ? Ans. 134596.
3. In how many different ways may an officer select 8
men out of 30, so as not to make the same selection twice
Ans. 5H52925
ADDITION OF DUODECIMALS. 153
PART VIII.
MENSURATION.
SECTION L
Duodecimals, or Cross Multiplication.
DUODECIMALS are fractions of a foot, or of an inch, or
parts of an inch, having 12 for their denominator. Inches
and parts are sometimes called primes ('), seconds ("),
thirds ('"), &c.
The denominations are,
12 Fourths ("") make 1 Third
12 Thirds  1 Second
12 Seconds 1 Inch in.
12 Inches 1 Foot ft.
Nole. This rule is much used in measuring 1 and computing the di
mensions of the several parts of buildings ; it is likewise used to find
the tonnage of ships, and the contents of bales, cases, boxes, &c.
ADDITION OF DUODECIMALS.
RULE.
Add as in compound addition, carrying 1 for each 12 u.
the next denomination.
EXAMPLE.
Ft. in. "" '" "" Ft. in. "
25 9 3 5 8 244 6 3 10 5
34 3 9 2 7 355 9851
28 10 4 8 4 559 10 9 5 8
64 11 9 7 2 129 5569
82 7 5 6 8 895 1 10 5 11
15 3 7 9 10 651 1759
44 6 11 2 8 555 9 8 5 5
22 3 6 1 5 388 11 10 10 9
318 8 9 8 4
154 SUBTRACTION OF DUODECIMALS.
SUBTRACTION OF DUODECIMALS.
RULE.
Work as in compound subtraction, borrowing 12 when
accessary*
EXAMPLE.
Ft. in. ' Ft. in. " '" ""
From 125 4 3 8 2 2756 5780
Take 68 9 2 10 1 1839 9 5 11 10
Rem. 56 7 10 1
3. From a board measuring 35 feet, 9 inches, 2 seconds,
cut 24 feet, 10 inches, 5 seconds, and 4 thirds; what is left ?
Ans. Wft. lOin. Ssec. 8'"
4. A joiner having lined several rooms very curiously
with costly materials, finds the amount to be, in square
measure, 803 feet, Z inches, 4 seconds ; but several deduc
tions being to be made for windows, arches, &c. those de
ductions amounted to 70 feet, 3 inches, 7 seconds, 10 thirds,
5 fourths ; how many feet of workmanship must he be paid
for? Ans. 132ft. llin. 8" I"' 7""
MULTIPLICATION OF DUODECIMALS.
Case 1.
When the feet of the multiplier do not exceed 12.
RULE.
Set the feet, or the highest denomination of the multiplier
under the lowest denomination of the multiplicand, and mul
tiply as in compound numbers, carrying 1 for every 12 from
;m.
3. Feet multiplied by seconds, give seconds, &c.
4 Inches multiplied by inches, give seconds.
MULTIPLICATION OF DUODECIMALS. 155
5. Inches multiplied by seconds, give thirds, &c.
6. Seconds multiplied by seconds, give fourths.
7. Seconds multiplied by thirds, give fifths, &c.
PROOF.
Reduce the given sum to a decimal, or work by the rules
of practice.
1. Multiply
EXAMPLES.
Ft. in. " Ft. in.
8 6 9 by 7 3
7 3
Ans.
Ft
59
2
11 3
1 8
3
. 62
11
3
Or practice.
3 is i 8 6 9
7
3
59
2
11
1
3
8
3
62
11
3 A?is.
7 3
Proof decimally.
6 9 = 8,5625
= 7,25
Ft. in.
2. Multiply 9 5
3. 7 10
4. 846
428125
171250
599375
62,078125
12
0,937500
12
11,250000
12
3,000000
Ft. in. "
Ans. 36 10 7
69 10 2
21 10 5
Ft. in.
by 3 11
by 8 11
by 2 74
5. What is the price of a marble slab, whose length is 5
feet 7 inches, and breadth 1 foot 10 inches, at 1 dollar and I
50 cents per foot? Ans. 15 dolls. 35^ cts.
6. There is a house with three tiers of windows, 3 in a
tier, the height of the first tier is 7 feet 10 inches, of the
second 6 feet 8 inches, and of the third 5 feet 4 inches, and
the breadth of each window is ^ ft. 11 inches. ; what will the
glazing come to at 14 cts. per ft. ? Ans. 32 dolls. 62 cts. .
156
MULTIPLICATION OF DUODECIMALS.
Case 2.
When the feet of the multiplier exceeds 12.
RULE.
Multiply by the feet in the multiplier, and take parts for
the inches.
EXAMPLES.
1. Multiply 84 feet 6 inches, by 36 feet 7 inches and 6
seconds.
Operation.
84 6
1
6*cc.
507
6
3042
42 3
3 6
Ans. 3094 9
Ft. in. Ft. in.
2. Multiply 76 7 by 19 10
3. 127 6 by 92 4
4. 184 8 by 127 6
Ft. in. sec.
5. Multiply 311
4 7 by 36 7
6x6=36
Ft. in. sec.
Ans. 1518 10 10
11772 6
23545
Ft. in. sec.
5
(tin. is
II
1868 3
6
11209 9 "
155 8 3 6
25 9
8 7
7 '
10 4
21 987
11402 (I
1 I
MULTIPLICATION OF DUODECIMALS. 157
6. A floor is 70 feet 8 inches, by 38 feet 11 inches ; how
many square feet are therein 1
Arts. 2750ft. lin. 4sec.
7. If a ceiling be 59 feet 9 inches long, and 24 feet 6
inches broad, how many yards does it contain ?
Arts. I62yds. 5ft. W$in.
Note. Divide the square feet by 9, and the quotient will be square
yards.
8. What will the paving of a court yard come to at 15
cents per yard, the length being 58 feet 6 inches, and the
breadth 54 feet, 9 inches ?
Arts. 53 dolls. 38 + cts.
9. What is the solid content of a bale of goods, measur
ing in length 7 feet 6 inches, breadth 3 feet 3 inches, and
depth 1 foot 10 inches?
Ans. 44/fc. 8in. Ssec.
Note, To find the cubic feet, or solid content of bales, cases, boxes,
&c. multiply the length by the breadth, and that product by the thick
ness.
10. A merchant imports from London six bales of the fol
lowing dimensions, viz.
Length. Breadth. Depth.
Ft. in. Ft. in. Ft. in
No. 1. 2 10 24 19
2. 2 10 26 13
3. 36 22 18
4. 2 10 28 19
5. 2 10 26 19
6. 2 11 28 18
What are the solid contents, and how much will the freight
amount to, at 20 dollars per ton of 40 feet ?
Ans. lift. lin. and freight 35 dollars 79 cts.
To find a ship's tonnage by Carpenter's measure.
For single decked vessels, multiply the length, breadth
at the main beam, and depth of the hold together, and di
vide the product by 95 ; but if the vessel be double decked,
take half the breadth of the main beam for the depth of the
hold, and work as for a single decked vessel.
O
158 MULTIPLICATION OF DUODECIMALS
.
EXAMPLES.
1. The length of a single decked vessel is 60' feet, the
breadth 20, and depth 10; what is the tonnage?
Then 60x20x10 = 12000
And 1 2000 ~ 95 = 1 26 fy tons. An*.
Or, as 95 : 20x10 :: 60 : 126 r i ?, &c. signifying
that the inch, 5 inch, &c. are divided into 1:2 parts. These
scales are useful for planning dimensions, that are taken in
feet and inches. Again, the edge of the rule is divided into
inches, and each of these into eight parts, representing half
inches, quarter inches, and half quarters.
In this description the rule is folded ; but when it is opened
and the slider drawn out, the hack part will be ibund divided
like the edge of the rule, so that all together will measure
3 feet or one yard.
USE OF THE CARPENTERS' RULE.
1st. To multiply numbers together.
EXAMPLE.
1. Suppose the two numbers 13 and 24.
Set 1 on B to 13 on A ; then against 24 on B, stands 312
on A, which is the required product of the two given num
bers 13 anu 24.
Note. In any operation when a number runs beyond the end of the
line, seek it on the other radius, or other part of the line ; that is, take
the 10th part of it, or the 100th part of it, &c. and increase the product
of it proportionally 10 fold, or 100 fold, y Iho length of a step is
CARPENTERS' AND JOINERS' WORK. 167
meant the length of the front, and the returns at the two
ends ; and by the breadth, is to be understood the girt of its
two outer surfaces, or the tread and rise.
The rail of a staircase is taken at so much per foot in
length, according to the diameter of the wellhole ; archi
trave string boards, by the foot superficial ; brackets and
strings at so much per piece, according to the workmanship.
Wainscoting is measured by the yard square, consisting
of 9 feet.
Door cases, frame doors, modillion cornices, eaves, frontis
pieces, BIUCKLAYEKS' WORK. *
the mouldings, be 15 feet 9 inches high, and 126 feet 3
in* lies in compass ; how. many yards does that room contain '/
Operation.
By decimals* By duodecimals.
12V5 ' Ft  in 
15,75 Ig 3
63125 63Q
88375 126
63125
12655 189
63 1 6
31 6 9
9)1988,4375 390
Sum 220,8
Sum 220 8
Ans. 220 yds. 8 feet.
11. If a room of wainscot be 16 feet 3 inches high, and
the compass of the room 137 feet 6 inches ; how many
yards are contained in it ? Ans. 248 yards 2 f feet.
12. If the windowshutters about a room be 69 feet 9
inches broad, and 6 feet 3 inches high ; how many yards
are contained therein, at work and half?
Ans. 72,656 yards.
13. What will the wainscoting of a room come to at 80
cents per square yard, supposing the height of the room, in
cluding the cornice and moulding, be 12 feet 6 inches, and
the compass 83 feet 8 inches ; three windowshutters, each
7 feet 8 inches by two feet 6 inches, and the door 7 feet by
3 feet 6 inches ; the shutters and door being worked on both
sides, are reckoned work and half?
Ans. 96 dollars 60J cents.
SECTION 5.
OF BRICKLAYERS' WORK
BKICK WOKK is measure. I ;ind rsfimatr'd in various ways.
!i somn places walls aiv mrasimd !>y fho rod square of
1 i frc'i ; so th..t on' rod in length, and one in breadth
BRICKLAYERS' WORK. 171
contain 272,25 square feet ; in other places the custom is to
allow 18 feet to the rod, that is, 324 square feet.
In other places they measure by the rod of 21 feet long,
and 3 feet high, that is, 62 square feet. Again, in other
places they account 16^ feet long and 1 foot high, that is,
16^ square feet, a rod or perch; or again, by the yard of
9 square feet ; and oftentimes the work is estimated at so
much per thousand bricks.
When brick work is measured by the rod, or perch, it
must be estimated at the rate of a brick and a half thick;
so that if a wall be more or less than this standard thickness,
it must be reduced to it by the following
RULE.
Multiply the superficial content of the wall by the num
ber of half bricks in the thickness, and divide the product
by 3 for the superficial feet in standard thickness.
EXAMPLES.
If a wall be 72 feet 6 inches long, and 19 feet 3 inches
high, and 5 bricks and a half thick ; how many rods of
brick work are contained therein, when reduced to the
standard thickness ?
Operation.
By decimals. By duodecimals.
19,25 = the height Ft. in.
72,5 = the length. 72 6
"9625 19 3
3850 648
13475  72
139^625 1368""
11= the thickness. 18 1 6'
f3)15351,875 9 6
272,25) 5117,291 (18 rods. 1395 7 ^6
239479 i
68,06) 216/79( 3 quarters. 3)15351 10 6
lalT 272)511^ (18 rote.
2397
68) "22f ( 3 quarters
Tffeet.
Note. Observe that 68,06 is the one fourth part of 272,25, and 68 is
only the onefourth part of 272. As the number 272^ is an inconve
nient number to divide* by, the \ is usually ornitfed, and the content in
feet divided only by 272 ; the difference being too trifling to be con
sidercd in practice.
172
BRICKLAYERS' WORK.
To find fixed divisors for bringing the answer into feet or
rods of a standard thickness, without multiplying the su
perficies by the number of half bricks, <$*c.
RULE.
Divide three, the number of half bricks in l, by the
number of half bricks in the thickness, the quotient will be
a divisor, which will give the answer in feet. Or if a divi
sor is sought for, that will bring the answer in rods at once,
multiply 272 by the divisor found for feet, and the product
will be a divisor for rods ; as in the following
TABLE.
1
The thickness of
the wall.
2
Divisors for the
ansiver in feet.
3
Divisors for the
ansiver in rods.
1 brick
1,
408
l brick
1
472
2 bricks
,75
204
2J bricks
,6
163,2
3 bricks
,5
136
3i bricks
,4285
116,6
4 bricks
,375
102
4^ bricks
,3333
90,6
5 bricks
,3
81,6
5J bricks
,2727
74,18
Application of the above Table.
Multiply the length of the given wall bv the breadth , o
serve the number of half bricks it is in thickness ; and op
[>rsite thereto will be found in the second column the di
visor to reduce it to feet ana in the third column the divi
sor for rods. Thus in the above example 72,5 X by 19,25
1395,625.
Ana 1395,620 2727 = 5117+ the number of feet in
standard nnasure.
And i:W5,(>jr>74,18^18,8f th<> num l >cr of rods
BRICKLAYERS' WORK. 173
Or, by the Carpenters' Rule
As the tabular divisor, against the thickness of the wall
: is to the length of the wall : : so is the breadth : to the
content.
As 74,18 on B : 72,5 on A :: 19,25 on B : 18 j on A. Ans.
To find the dimensions of a building, measure half around
on the outside, and half on the inside, for the whole length
of the wall ; this length being multiplied by the height gives
the superficies. All the vacuities, such as doors, windows,
window backs, &c. must be deducted, for materials ; but for
workmanship alone no deductions are to be made, and the
measurement is usually taken altogether on the outside.
This is done in consideration of the trouble of the returns
or angles. There are also some other allowances, such as
double measure for feathered gable ends, &c.
2. How many yards and rods of standard thickness are
contained in a brick wall, wWbse length is 57 feet 3 inches,
and height 24 feet 6 inches ; the wall being 2J bricks thick ?
Ans. 259,74 yards, or 8,58 + rods.
3. If a wall be 245 feet 9 inches long, 16 feet 6 inches
high, and 2 bricks thick; how many rods "of brick work
are contained therein, when reduced to standard thickness ?
Ans. 24 rods 3 quarters 24 feet.
4. A triangle gable end is raised to the hei^ of 15 feet
above the wall of a house, whose width is 45 feet and the
thickness of the wall 2 bricks ; required the contei^ in rods
at standard thickness? Ans. 2 rods 18 feet.
Chimneys by some are measured as if they were solid, de
ducting only the vacuity from the hearth to the mantle, on
account of their trouble.
But by others, they are girt or measured round for their
breadth, and the height of their story, taking the depth of
the jambs for their thickness. And in this case no deduc
tion is made for the vacuity from the floor to the mantletree,
because of the gathering of the breast and wings, to make
oom for the hearth in the next story.
P2
174 MASONS' WORK.
If the chimney back be a party wall, and the wall be
measured by itself, then the depth of the two jambs and
length of the breast is to be taken for the length, and the
height of the story for the breadth, at the same thickness
with the jambs.
Those parts of the chimneyshaft which appear above
the roof are to be girt with a line round about the least place
of them for the length, and take the height for the breadth ;
and if they are 4 inches thick, they are to be accounted as
one brick work, and if they are 9 inches thick, they are to
be taken for l brick work, on account of the trouble of
plastering and scaffolding.
It is customary in most places to allow double measure
for chimneys.
SECTION 6.
OF MASONS' WORK.
MASONS' work is measured sometimes by the foot solid,
sometimes by the foot superficial, and sometimes by the foot
in length. It is also measured by the yard, and mostly by
the rod or perch, which is 16 feet in length, 18 inches in
breadth, and 12 inches in depth.
Walls are measured by the perch; columns, blocks of
stone, or marble, &c. by the cubic foot; and pavements,
slabs, chimneypieces, &c. by the superficial or square foot.
Cubic, or solid measure, is always used for materials, but
square measure generally for workmanship.
In solid measure, the true length, breadth and thickness
are taken and multiplied into each other for the content.
!n superficial measure, the length and breadth of every
part of the projection, which is seen without the general
upright face of the building, is taken for the content.
MASONS' WOHK. 175
EXAMPLES.
1. If a wall be 97 leet 5 inches long, 18 feet 3 inches
high, and 2 feet 3 inches thick, how many solid feel, and
perches, are contained therein ?
Operation.
By decimals. By uuodecimals.
97,417 length Ft. in.
18,25 breadth 97 5
18 3
4870S5
194834 776
779336 97
97417 24 4 3"
600
1777,86025 superficies 1 60
2,25 thickness
1777 10 3
888930125 2
355572050
355572050 3555 8 6
444 5 6
4000,1855625 solidity.
in cubic ft. 4000 2094 u.
4000^24,75^161,616f feet. Ans.
2 How many solid feet and perches are contained in a
wall 53 feet 6 inches long, 12 feet 3 inches high, and 2 feet
thick? Ans. 1310,75 feet, and 52,9595 rods.
3. If a wail be 107 feet 9 inches long, and 20 feet 6 inches
high, how many superficial feet are contained therein ?
Ans. 2208 feet 10 inches.
4. If a wall be 112 feet 3 inches long, and 16 feet 6
inches high, how many superficial rods, each 63 square
feet, are contained therein ? Ans. 29 rods 25 feet.
5. What is a marble slab worth, whose length is 5 feet 7
inches, and breadth 1 foot 10 inches, at 80 cents per fool
superficial 1 Ans. 8 dolls. 19 els*
176 PLASTERERS WOixK
SECTION 7.
OF PLASTERERS' WORK.
PLASTERERS' WORK is principally of two kinds, viz.
first, plastering upon laths, called ceiling ; and second, plas
tering upon walls, or partitions made of framed timber,
called rendering, which are measured separately.
Plasterers' work is usually measured by the yard square,
consisting of 9 square feet ; sometimes it is measured by
the square foot, and sometimes by the square of 100 feet.
Enriched mouldings, cornices, &c. are rated by running,
or lineal measure. In arches, the girt round them multiplied
by the length, is taken for the superficies.
Deductions are to be made for doors, chimneys, windows,
and other large vacuities. But when the windows, or other
openings, are small, they are seldom deducted, as the plas
tered returns at the top and sides are allowed to compensate
for the vacuity.
Whitewashing and coloring are measured in the same
manner as plastering.
EXAMPLES.
1. If a ceiling be 59 feet 9 inches long, and 24 feet 6
inches broad, how many superficial yards of 9 square feet
does it contain ?
Operation.
By decimals. By dtfodecimals.
Ft. in. Ft. in.
59 9=59,75 feet 59 9
24 6= 24,5 do. 24 6
29875 236
23900 118
11950 29 10 6"
18
79)1463,875 feet
Ans. 1 62,65 f yards
r 9) 1463 10 6
162 5 10 6 An*.
\VKHJS WORK 177
2 If the plastered partitions between rooms be 141 feet 6
inches about, and 1 1 feet 3 inches high, how many yards
(Jo tiiey contain? Ans. 176,87 yards.
3. What will the plastering of a ceiling come to at 15
cents per yard, allowing it to be 22 feet 7 inches long, and
13 feet 11 inches Lroad ? Ans. 5 dolls. 20 cts.
4; The length of a room being 20 feet, its breadth 14
feet 6 inches, and height 1 feet 4 inches ; how many yards
of plastering does it contain, deducting a fireplace of 4 feet
by 4 feet 4 inches, and two windows, each 6 feet by 3 feet
2 inches ? Ans. 73^ T yards.
5. The length of a room is 14 feet 5 inches, breadth 13
feet 2 inches, and height 9 feet 3 inches, to the under side
of the cornice, which projects 5 inches from the wall, on the
upper part next the ceiling ; required the quantity of render
ing and plastering ; there being no deductions but for one
door, the size whereof is 7 by 4 feet ?
Ans. 53 yds. 5 ft. of rendering, and 18 yds. 5 ft. ceiling.
6. The circular vaulted roof of a church measures 105
feet 6 inches in the arch, and 275 feet 5 inches in length ;
what will the plastering come to at 12 cents per yard ?
Ans. 387 dolls. 42 cts.
7. What will the whitewashing of a room come to at 2
cents per yard, allowing it to be 30 feet 6 inches long, 24
feet 9 inches broad, and 10 feet high ; no deductions being
made for vacuities? Ans. 4 dolls. 13 cts.
SECTION 8.
OF PAVERS' WORK.
PAVERS WORK is measured by the square yard, consist
ing of 9 square feet. The superficies is found by multiply
ing the length by the I readth.
178 FA1JNTKRS' WORK.
EXAMPLES.
1. What cost the paving of a street 225 feet 6 inches long.
and 60 feet 6 inches wide, at 30 cents per square yard ?
By decimals. By duodecimals.
Ft. in. ^ in.
225 6=225,5 feet
60 6= 60,5 do.
13500
11275 US
13530
9)13642 9
1515 7
30
79 ) 13642,75 superficial ft.
1515,86 yards.
30
26 =the price of 7 ft. 9 in.
Ans. 454,7580 ~454jeT
Ans. 454 dolls. 76 cts.
2. What will the paving of a footpath come to at 28
cents per yard, the length being 35 feet 4 inches, and the
breadth 8 feet 3 inches ? Ans. 9 dolls. 33 cts.
3. What cost the paving of a courtyard at 38 cents per
yard, the length being 27 feet 10 inches, and the breadth 14
feet 9 inches ? Ans. ?7 dolls. 33 cts.
4. What will be the expense of paving a rectangular
yard, whose length is 63 feet, and breadth 45 feet, in which
there is laid a footpath 5 feet 3 inches broad, running the
whole length, with broad stones, at 36 cents a yard ; the
rest being paved with pebbles, at 30 cents a yard ?
Ans. 96 dolls. 70 cts.
SECTION 9.
OF PAINTERS' WORK.
PAINTERS' WORK is computed in square yards of 9 feet
Kvery part is measured where the color lies ; and the mea
suring line is pressed close into all the mouldings, corners,
&ic. over which it passes.
PAINTERS' WORK. 179
Windows, casements, &c. are estimated at so much a
piece ; and it is usual to allow double measure for carved
mouldings, &c.
The value of painting is rated by the number of coats ;
or whether once, twice, or thrice colored over, and the dif
ferent qualities and costliness of the colors.
EXAMPLES.
1, How many yards of painting will a room contain
which (being girt over the mouldings) is 16 feet 6 inches,
and the compass of the room 97 feet 6 inches ?
Operation.
By decimals. By duodecimals.
Ft. in. 97 6
97 6=97,5 16 6
16 6=16,5
4875
5850
975
r 9 ) 1608,75 feet r9 ) 1608
Yards 178,6,75 178,6,9
Ans. 178J yards.
2. A gentleman had a room painted at 8J cents per yard,
the measure whereof is as follows, viz. the height 11 feet
7 inches, the compass 74 feet 10 inches, the door 7 feet 6
inches by 3 feet 9 inches ; five window shutters, each 6 feet
8 inches by 3 feet 4 inches; the breaks in the windows 14
inches deep, and 8 feet high ; the opening for the chimney 6
feel  inches by 5 feet, to be deducted, the shutters and doors
are painted on both sides ; what will the whole come to ?
Ans. 10 dolls. 43 cts.
3. How many yards of painting are there in a room, the
length whereof is 20 feet, its breadth 14 feet 6 inches, and
height 10 feet 4 inches; deducting a fireplace of 4 feet by
4 feet 4 inches, and two windows,  each 6 feet by 3 feet 2
inches? Arts. 73^ T yards.
180 GLAZIERS' WORK.
4. What cost the painting of a room at 6 cents per yard ,
its length being 24 feet 6 inches, its breadth 16 feet 3 inches,
and height 12 feet 9 inches ; also the door is 7 feet by 3
feet 6 inches, and the window shutters of two windows,
each 7 feet 9 inches by 3 feet *6 inches, but the breaks of the
windows themselves, are 8 feet 6 inches high, and 1 foot 3
inches deep ; deducting a fireplace cf 5 feet by 5 feet 6
inches. A?is. 7 dolls. 66 cts. 9 m.
SECTION 10.
OF GLAZIERS' WORK.
GLAZIEKS compute their work in square feet ; and the di
mensions are taken either in feet, inches, and seconds, &c.
or in feet, tenths, hundredths.
Windows are sometimes measured by taking the dimen
sions of one pane, and multiplying its .superficies by the
number of panes. But more generally they measure the
length and breadth of the window over all the panes, and
their frames for the length and breadth of the glazing. And
oftentimes the work is estimated at so much per pane ac
cording to the size.
Circular, or ovtil windows, as fanlights, &c. are measured
as if they were square, taking for their dimensions the
greatest length and breadth, as a compensation for the waste
of glass, and labor in cutting it to the proper forms.
EXAMPLES.
1. I low many square feet are contained in a window,
which is 4 feet 3 inches long, and 2 feet 9 inches broad?
Hv decimals. By duodecimals.
Ft. in." Ft. in.
4 = 4, 25 the length 4 3
2 9 =2,75 the breadth 2 9
2125 8 6
2975 3 2
sf,0
11/H75 foot.
11 S 3 Ans.
MEASUREMENT OF GROUJND 181
2. If a window be 7 feet 3 inches high, and 3 feet 5 inches
broad, how many square feet of glazing are contained
therein ? . An*. 24 feet 9 inches.
3. There is a house with three tiers of windows, 7 in a
tier; the height of the first tier is 6 feet 11 inches, of the
second, 5 feet 4 inches, and of the third, 4 feet 3 inches ; the
breadth of each window is 3 feet 6 inches : what will the
glazing come to at 14^ cents per foot?
Ans. 58 dolls. 61 cts.
4. What will the glazing of a triangular skylight come
to at 10 cents per foot, the base being 12 feet 6 inches long,
and the perpendicular height 16 feet 9 inches?
Ans. 10 dolls. 465 cts.
5. What is the area of an elliptical fanlight of 14 feet
inches in length, and 4 feet 9 inches in breadth ?
Ans. 68 feet 10 inches.
6. There is a house with three tiers of windows, and &
in each tier; the height of the first tier is 7 feet 10 inches,
of the second, 6 feet 8 inches, of the third fret 4 inches
and the common breadth 3 feet 11 inche? , what will the
glazing come to at 14 cents per foot?
Ans. &' s J
Proceeding, i of 1 = ^ A's "
f of !=*'!
Then ^V + rV + A= T^V to be taken
from C's. Thus, ^^
and i of T u_ '=j&C 9 a
And
'>*
Their situation
at the end of
the 3d attack.
200 QUESTIONS FOR EXAMINATIONS,
A. B.
Further, ^ + ^=J^ and f of &=& lost by A and B.
Then, t V of ft + iof & ^^i^A's^ p f
' ' "
Also, A of /, + i of
And * of W + riV =T*tt* C's
i of A+W*
* Of T'O +TWo
* Of * = T ' ?
_.__l_l_ _ _S_4_3J o
l~3*0 Tl 5 26880 ^ f n ff ot j n of
r A got 2863 ^
So that if the } 6335
sugarplums > then J
i
// z.
vQf+4 frr