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41. How many solid feet in a stick of timber 19 ft. long, 1 ft. 3 in. wide, and 1 ft. thick?

Ans. 23 ft. 9 in.

42. How many solid feet in a stick of timber 23 ft. long, 1 ft. 4 in. wide, and 11 inches thick.

Ans. 28 ft. 1 in. 4".

43. How many solid feet in a stick of timber 30 ft. 6 in. long, 2 ft. wide, and 1 ft. 10 in. thick?

Ans. 111 ft. 10 in.

44. How many solid feet in 7 sticks, each 56 ft. long, 11 in. wide, and 10 inches thick?

Ans. 299 ft. 5 in. 4".

Note 7.--To find how many tons of timber, or how many cords of wood, are contained in any number of feet, we have only to divide the given number of ft. by the number of solid ft. in a ton or cord.

45. How many tons of hewn timber in 24 sticks, each 72 ft. 8 in. long, 2 ft. wide, and 1 ft. 10 in. thick ? Ans. 127 T. 44 ft. 8 in. 46. How much wood in a load 8 ft. long, 4 ft. high, and 4 ft. wide? Ans. 1 cord. 47. How much wood in a load 11 ft. 4 in. long, 3 ft. 7 in. high, and 4 ft. wide?

Ans. 1 C. 34 ft. 5 in. 4".

16 solid feet are frequently called 1 foot of wood, and 8 feet of wood make 1 cord, because there are 8 times 16 in 128, the number of solid feet in a cord; to find how many feet of wood there are in any number of solid feet, we have only to divide by 16.

The answers in this work will be given in cords and solid feet.

48. How much wood in a load 9 ft. 10 in. long, 4 ft. 9 in. high, and 4 ft. wide, allowing 3 in. in the height for loose package ? Ans. 1 C. 49 ft. 49. How many cords in a pile of wood 37 ft. 10 in. long, 5 ft. 3 in. high, and 2 ft. 4 in. wide?

Ans. 3 C. 79 ft. 5 in. 6".

50, How many solid or cubick feet in a bale of

goods, 6 ft. 2 in. long, 3 ft. 3 in. wide, 2 ft. 8 in. deep? Ans. 53 ft. 5 in. 4".

Diggers work by the square of 6 ft. long, wide, and deep, which contains 216 solid feet.

51. How many squares of the above dimensions must he dig to make a canal 400 feet long, 30 feet wide, and 4 feet deep? Ans. 222,48.

Remark. When we multiply feet and inches by feet and inches once, the inches in the product do not express square inches, for it has been proved that each inch is a twelfth part of a foot, and there being 144 square inches in a square foot, each inch in the product will represent 12 square inches. When we multiply the 2d time by ft. and in. to get the solid contents, the inches in the last product become twelfths of a solid foot, consequently each inch must represent 144 solid inches, which is the twelfth part of 1728, the number of solid inches in a solid foot.

When we say, a board contains 9 ft. 6 in., the 6 in. represent a part of the board equal to a board 12 in. long and 6 in. wide. So in a stick of timber measuring 9 ft. 6 in., the 6 in. represent a part of the stick equal to a stick 12 in. long, 12 in. wide, and 6 in. thick.

Questions to be performed by the addition and multiplication of compound numbers.

1. A teamster loaded his waggon with 1 hhd. of molasses weighing 9 cwt. 1 qr. 21 lb. ; 2 bales of cotton, each 1 cwt. 3 qr. 6 oz.; 3 bar. of flour, each 2 cwt. 1 lb.; and 4 bars of iron, each 1 cwt. 3 lb. What was the weight of the load?

Ans. 1 T. 3 cwt. 8 lb. 12 oz. 2. What is the whole weight of a ship's cargo consisting of 103 hhd. of salt, each weighing 4 cwt. 1 qr. 21 lb.; 35 hhd. of wine, each 4 cwt. 3 qr. 2 lb.; 106 hoxes of lemons, each 2 qr. 16 lb.; 126 casks of

raisins, each 2 qr. 16 lb. 10 oz.; and other articles to the amount of 65 T.?

Ans. 103 T. 13 cwt. 3 qr. 3 lb. 12 oz.

3. How many feet of boards will it require to cover a building, 60 ft. 6 in, long, 40 ft. 3 in. wide, 22 ft. high, and each side of the roof 24 ft. 2 in.; allowing 523 ft. 3 in.. for the gable ends, and making no deduction for doors or windows?

Ans. 7880 ft. 5 in.

SECTION V.

SUBTRACTION OF COMPOUND NUMBERS.

Subtraction of compound numbers is taking one compound number from another, or a part of a compound number from itself, to find the difference.

RULE 1.-Write the least number under the greater in such a manner that every denomination may stand under one of its own name.

Commence subtracting with the least denomination, and set the difference between the two numbers in each denomination in a line below.

Question 1.

ENGLISH MONEY.

What is the difference between

35 £. 15 s. 8 d. 3 qr., and 20 £. 8 s. 3 d. 1 qr. ?

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Illustration.-The 15 . is the difference between the pounds in the minuend and the pounds in the subtrahend; 7 s. is the difference of the shillings, 5 d. the difference of the pence, and the 2 qr. the differ

ence of the farthings; therefore the compound number, 15£. 7 s. 5 d. 2 qr. is the difference between the compound numbers given. As each denomination by itself is a simple number, when the under numbers in all the denominations are the smallest, subtracting by the above method will give the difference between any numbers, for it is only subtracting as many simple numbers as there are denominations. 2. If 99 £. 1 s. 10 d. 1 qr. be taken from 100 £. 19 s. 11 d. 3 qr., what will remain ?

Ans. 1 £. 18 s. 1 d. 2 qr.

3. Required the difference between 19 s. 3 d. 2 qr. Ans. 19 s. 1 d. 1 qr.

and 2 d. 1. qr.

TROY WEIGHT.

4. From 9 lb. 11 oz. 19 pwt. 23 gr. take 6 lb.

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5. How much heavier is a gold necklace, which weighs 5 pwt. 19 grs. than one which weighs 4 pwt. Ans. 1 pwt. 18 grs.

1 gr?

APOTHECARIES WEIGHT.

6. Subtract 5 lb. 7 oz. 6 dr. 19 grs. from 11 lb. 10 oz. 7 dr. 2 sc. 19 grs.

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7. Take 1 lb. 6 oz. 2 grs. from 1 lb. 7 oz. 3 dr. 1 sc. 17 grs? Ans. 1 oz. 3 dr. 1 sc. 15 g

grs.

RULE 2. When the under number in any denomination is greater than the upper one, add as many units to the upper one as it takes of that denomination to make 1 in the next greater, and subtract the under number from that sum; then add 1 to the lower number in the next higher denomination.

AVOIRDUPOIS WEIGHT.

8. Required the difference between 18 T. 3 qr. 19 lb. 12 oz. 5 dr. and 12 T. 14 cwt. 22 lb. 3 oz. 11 dr.

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Explanation. We cannot subtract 11 dr. from 5 dr., we therefore add 16 dr. (which are equal to 1 in the denomination of oz.) to the 5 drs. and subtract 11 from 21, the sum of 16 and 5, and add 1 to the 3 in the next place. the denomination of lb., because 28 lb. equal 1 qr., we add 28 to 19, and subtract 22 from 47, the sum. In the next denomination, the under place being occupied by a cipher, we suppose 1 qr. placed where the cipher is, and subtract it from the 3 qr. to balance the 28 lb., which we added to the 19 lb. As there are no cwt. in the minuend, we suppose 1 T., that is, 20 cwt., to stand in the place of cwt., and subtract the 14 from it, and then add 1 T. to the 12 T.

9. Bought 12 cwt. 2 qr. of fish; how much remained after selling 9 cwt. 27 lb. ?

Ans. 3 cwt. 1 qr. 1 lb.

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