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29. After the earth has passed through 1 s. 15 * 22 of its orbit, what part of its orbit remains for it to pass through to complete a revolution ? Ans. 10 s. 149 38 '.

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33. How much more wine in a cask which contains 73 gal. 1 qt. 3 gl., than in another which contains 51 gal. 1 pt. 1 gl. 2 Ans. 22 gal. 1 pt. 2 gl. 34. A merchant after selling a certain quantity of rum from a hogshead, finds there are 45 gal. 2 qt. 1 pt. remaining. How much did he sell ? * Ans. 17 gal. 1 qt. 1 pt.

, ALE OR BEER MEASURE,

35. If 37 gal. 1 pt. of beer be taken from a cask containing 49 gal., how much will remain 2 Ans. 11 gal. 3 qt. 1 pt.

DRY MEASURE.

36. If 18 bus. 1 p.k. 7 qt. of grain be taken from a bin containing 57 bus., how much will remain 7 Ans. 38 bus. 2 plc. 1 qt. 37. A farmer after selling grain from a bin containing 57 bus. has 38 bus. 2 plc. 1 qt. left; how much did he sell ? Ans. 18 bus. 1 plc. 7 qt.

Questions to be performed by Addition and Subtraction of compound numbers.

1. A merchant imports 75 T. 1 qr. 20 lb. of iron in one vessel, and 60 T. 15 cwt. 3 qr. in another; he then sells 10 T. 3 qr., how much has he left 2 Ans. 125 T. 15 cwt. 1 qr. 20 lb. 2. A father's age is 75 Y. 6 mo. 3 w. ; his eldest son's 50 Y. 5 d., and his youngest son’s 25 Y. 15 h. If the difference between the ages of the sons be added to the age of the father, what will be the sum of the time ! Ans. 100 Y. 6 mo. 3 w. 4 d. 9 h.

Questions to be performed by Multiplication and Subtractions of compound numbers.

1. A gentleman after selling 27 A. 1 R. 14 r. from a farm containing 100 A. 27 ft., has three other farms each equal to what he has left of that from which he sells a part; how much land has he? Ans. 290-A. 2 R. 24 r. 108 ft. 2. A boy was sent to a store with 16.6. 9 s. 1d. 3 qr., to pay a debt of 15 6.9 d. 2 qr. After paying the debt, he takes up goods to 5 times the amount of what money he had left, and leaves the remainder of his money in part pay. What was the new debt 2 Ans. 5 38.13 s. 5d.

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Division of compound numbers is ascertaining how many times a simple number can be taken from the several denominations of a compound number; or it is separating a compound number into several parts.

As the number to be divided is composed of several denominations, the quotient will generally consist of different denominations.

RULE 1.

Divide each denomination of the dividend by the

divisor, beginning with the highest, the same as in simple numbers.

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Illustration of Rule 1.-In the division of simple numbers, 1 is placed in the quotient every time we take the divisor from the dividend. By rule 1, in the division of compound numbers, we divide each denomination as a simple number, consequently, the quotient of each denomination contains 1 as many times as the divisor can be taken from that particular denomination. In the operation of the last question, we divide the price of 4 yards by 4. By taking as many units of any denomination as there are units in the divisor, we take 1 unit from the dividend for each unit in the divisor; but, as we place but 1 unit in the quotient, every time we take the whole divisor from the dividend, thus, 1 set in the quotient, answers to any one in the divisor. 1636. 12 s. 8d. is the price of 4 yards, but as we put but 1 in the quotient of each denomination every time we take 4 from that denomination, the whole quotient can only be a 4th part of the whole dividend; consequently,438. 3 s. 2d. is the price of 1 of the 4 yards.

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If in dividing any denomination except the least there be a remainder, multiply it by as many of the next less denomination as it takes to make 1 in that denomination from which the remainder is derived, and add the number in the less denomination to the product. Then divide the sum thus found by the divisor.

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3 remainder of the quarters.
25 multiply by 25 and add the 20 lb.

75 . . . 20

*

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Illustration of Rule 2. —When a remainder occurs, it is always less than the divisor, so that the divisor cannot be taken from it, and another unit of that denomination cannot be set in the quotient. Yet there being a remainder, the number of units put in the quotient is not the exact quotient, and although we cannot have another unit in the larger denomination, we may have 1 or more units in the less denomination, where each unit dees not express so much. In the last question, each bar will weigh more than 2 cwt. but not so much as 3. Had the weight of the 8 bars been 16 cwt. the weight of each bar would have been just 2 cwt., and if the weight of the whole had been 24 cwt., the weight of each would have been just 3 cwt. ; but as it is more than 16 and less than 24, the weight of 1 bar must be somewhere between 2 and 3 cwt. The 11 qr. express all the quarters in the 2 cwt. which were left and the 3 given quarters; the 95 lb. contain the 3 remaining qr. and the 20 lb. given, and the 120 oz. contain the 7 remaining lb. and the 8 oz. given, for 4 qr, being equal to 1 cwt. by multiplying the 2 cwt. which were left by it, we put 4 into the product for each cwt. remaining, and so of the other denominations, r

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