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3. In 3 hogsheads of sugar, each containing 10cw 3gr. 5lb., how many hundred weight?

Ans. 32cwt. 1qr. 15lb.

4. How much cloth will it take for 7 suits of clothes, if each suit require 7yd. 3qr. 1na.?

Ans. 54yd. 2qr. 3na.

5. How much wood can a horse draw ir 13 loads, if he

draw 1C. 19S. ft. at each load?

Ans. 14C. 119S. ft. 6. How long will it take a man to saw 6 cords of wood, if he employ 7hr. 30m. 45sec. to saw one cord, allowing 10 working hours for each day?

Ans. 4da. 5hr. 4m. 30sec. 7. The circumference of a wheel is 15 feet 2 inches. What distance will this wheel measure on the ground, if it is rolled over 365 times? Ans. 1mi. 255 ft. 10in.

8. Allowing the year to consist accurately of 365 days, 5 hours, 48 minutes, 49 seconds, what will be the true length of 1843 years? Ans. 673141da. 10hr. 44m. 28 sec.

When the multiplier is a composite number, we may, as in simple numbers, multiply successively by the component parts.

9. What will 35cwt. of cheese cost, at 15s. 6d. per hundred weight?

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10. How much brandy in 84pi., each containing 128gal 2qt. 1pt. 3gi.? Ans. 10812gal. 1qt. 1pt. 11. In 21 loads of wood, each 1C. 1C. ft., how many cords? Ans. 23 C. 5C. ft.

12. Suppose the piston rod of a steam engine to move 3 ft. 4in. 1b. c. at each stroke. Through what distance wil it move in making 1000 strokes? Ans. 3361 ft. lin. b. c. 13. Bought as follows:

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14. What is the amount of the following bill of goods?

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86. Let it be required to divide £100 10s. 3d. equally among 17 men..

14*

EXPLANATION.

OPERATION.

First, we say 17 in £100, is contained 5 times and £15 remaining; and since these £15, as well as the 10s., are yet to be divided among the 17 men, we reduce the pounds to shillings, and add the 10s., making 310s. ; we find 17 to be contained 18 times in 310s. with 4s. remainder. We reduce the 4s. to pence, and add the 3d., making 51d., which divided among the 17 men, gives 3d. each.

17) £100 10s. 3d. (£5

85

15

20

17)310(18s.
17

140

136

4

12

17)51(3d.

51

Collecting, we have

£5 18s. 3d.

NOTE. We do not divide 100 pounds by 17 men, which is impossible; but we separate £100 into 17 equal parts. Each part is expressed by the quotient, and contains £5, (ART. 64. Note.)

Or, adhering to the general definition of Division, (ART. 22 and ART. 64,) we suppose a pound set apart for each man, and then find how many times £17, the number thus set apart, is contained in £100; the quotient will be an abstract number. The answer will, of course, be as many pounds to each man as there are parcels of £17 in £100; that is, as there are units in the quotient.

Had the divisor been one of the nine digits, the work might have been performed by Short Division.

We therefore have this general

RULE.

I. Place the divisor on the left of the dividend, as in Sımple Division. Begin at the left-hand and divide the number of the highest denomination by the divisor.. Reduce the re

mainder, if any, to the next lower denomination, to which add the number of the dividend expressing that denomination, and then divide the sum by the divisor.

II. Proceed in the same way for all the denominations. If there is a remainder after the last division, place it over the divisor, and annex it in a fractional form to the quotient. Each quotient will be of the same denomination as its dividend.

Having placed the divisor as in Simple Division, how do you proceed? When, in dividing any particular denomination, there is a remainder, how do you dispose of it! what denomination will the respective quotients be?

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lb. oz. pwt. gr.

13) 10 8 16 3(076.9oz. 18pwt. 312g7.

12

13) 128(9oz.

117

11

20

13)236(18pwt.

13

106

104

2

24

13)51(311gr.

12 remainder.

(4.)

mi. fur. rd. yd. ft. mi. fur. rd. yd. ft in 23)100 4 30 11 2(4 2 39 3 0 7

92

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5. Divide 10tuns 2hhd. 17gal. 2pt. by 67.

6. Divide 51A. 1R. 11P. by 51.
7. Divide 4gal. 2qt. by 144
8. Divide £113 13s. 4d by 31.

Ans. 39gal. 6pt Ans. 1A. OR 1P. Ans. 1gi.

Ans. £3 13s. 4d.

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