What is the value of 9cwt. of cheese at £1 11s. 5d. per cwt ? Ans. £14 2s. 9d.

Ans. £112.

What is the value of 96 quarters of rye at £ 3s. 4d. per quarter. What is the weight of 7 hhds. of sugar, each weighing 9 cwt. 3qrs. 12lb. Ans. 69. cwt.

In the Lunar circle of 19 years, of 365d. 5h. 48′ 48′′ each, how many days, &c. ? Ans. 6939d. 14h. 27′ 12′′.

Multiply 14f. 9' by 4f. 6'.

Multiply 4f. 7' 8" by 9f. 6'.

Ans. 66f. 4' 6". Ans. 44f. 0′ 10′′.

Required the content of a floor 48f. 6' long and 24f. 3' broad.

Ans. 1176f. 1' 6".

What is the number of square feet &c. in a marble slab,

whose length is 5f. 7' and breadth 1f. 10'?

Ans. 10f. 2' 10."

The result is evidently the may be seen by comparing

highest denomination of the multiplier, and disposing of the several products as in the first example below. same whichever method is pursued, as this example with that of the same question on the right, performed according to the rule in the text. This last arrangement seems to be preferable, as it is more strictly conformable to what takes place in the multiplication of numbers accompanied by decimals.

DIVISION OF COMPOUND NUMBERS.

107. A COMPOUND number may be divided by a simple number, by regarding each of the terms of the former, as forming a distinct dividend. If we take the product found in article 105, namely, £63 126s. 63d. 27q. and divide it by the multiplier 9, we shall evidently come back to the multiplicand, £7 14s. 7d. 3q. We arrive at the same result also, by dividing the above sum reduced, or £69 11s. 9d. 3q. for we obtain one 9th of each of the several parts that compose the number, the sum of which must be one 9th of the whole. But since, in this case, each term of the dividend is not exactly divisible by the divisor, instead of employing a fraction we reduce what remains, and add it to the next lower denomination, and then divide the sum thus formed, by the divisor. The operation may be seen below.

Whence, to divide a number consisting of different denominations by a simple number, divide the highest term of the compound number by the divisor, reduce the remainder to the next lower denomination, adding to it the number of this denomination, and divide the sum by the divisor, reducing the remainder, as before, and proceed in this way through all the denominations to the last, the remainder of which, if there be one, must have its quotient represented in the form of a fraction by placing the divisor under it. The sum of the several quotients, thus obtained, will be the whole quotient required. When the divisor is large and can be resolved into two or more simple factors, we may divide first by one of these factors, and then that quotient by another, and so on, and the last quotient will be the same as that which would have been obtained by using the whole divisor in a single operation. Taking the result of the example in the corresponding case of multiplication, we proceed thus,

By dividing £83 18s. 6d. by 2, we obtain one half of this sum, which being divided by 9, must give one 9th of one half, or one 18th of the whole. The first operation may be considered as separating the dividend into two equal parts, and the second as

distributing each of these into nine equal parts, the number of parts therefore will be 18, and being equal, one of them must be one 18th of the whole.

But when the divisor connot be thus resolved, the operation must be performed by dividing by the whole at once. If the quotient, which we are seeking, were known, by adding it to, or subtracting from it, the dividend a certain number of times, increasing or diminishing the divisor at the same time by as many units, we might change the question into one, whose divisor would admit of being resolved into factors, which would give the same quotient; we should thus preserve the anology which exists between the multiplication and division of compound numbers. But this cannot be done, as it supposes that to be known, which is the object of the operation.

Multiplication and division, where compound numbers are concerned, mutually prove each other, as in the case of simple numbers. This may be seen by comparing the examples, which are given at length to illustrate these rules.

Ans. 37cwt. 3qrs. 18lb.

Divide 1061cwt. 2qrs. by 28.

Divide 375mls. 2fur. 7pls. 2yds. 1ft. 2in. by 39.

Ans. 9mls. 4fur. 39pls. 2ft. 8in.

If 9 yards of cloth cost £4 3s. 74d. what is it per yard?

Ans. 9s. sd. 2q.

If a hogshead of wine cost £33 12s. what is it per gallon?

Ans. 10s. 8d.

If a dozen silver spoons weigh 3lb. 2oz. 18pwt. 12grs. what Ans. 3oz. 4pwt. 11grs.

is the weight of each spoon.

If a persou's income be £150 a year, what is it per day?
Ans. 8s. 24d. nearly.
A capital of £223 16s. 8d. being divided into 96 shares, what

is the value of a share?

Ans. £2 6s. 71.

PROPORTION.

108. We have shown, in the preceding part of this work, the different methods necessary for performing on all numbers, whether whole or fractional, or consisting of different denominations, the four fundamental operations of arithmetic, namely, addition, subtraction, multiplication and division; and all questions relative to numbers ought to be regarded as solved, when, by an attentive examination of the manner in which they are stated, they can be reduced to some one of these operations. Consequently, we might here terminate all that is to be said on arithmetic, for what remains belongs, properly speaking, to the province of algebra. We shall, nevertheless, for the sake of exercising the learner, now resolve some questions which will prepare him for algebraic analysis, and make him acquainted with a very important theory, that of ratios and proportions, which is ordinarily comprehended in arithmetic.

109. A piece of cloth 13 yards long was sold for 130 dollars, what will be the price of a piece of the same cloth 18 yards long.

It is plain, that if we knew the price of one yard of the cloth that was sold, we might repeat this price 18 times, and the result would be the price of the piece 18 yards long. Now, since 13 yards cost 130 dollars, one yard must have cost the thirteenth part of 130 dollars, or 130, performing the divison, we find for the result 10 dollars, and multiplying this number by 18, we have 180 dollars for the answer; which is the true cost of the piece 18 yards long.

A courier, who travels always at the same rate, having gone 5 leagues in 3 hours, how many will he go in 11 hours?

Reasoning as in the last example, we see, that the courier goes in one hour of 5 leagues, or, and consequently, in 11 hours he will go 11 times as much, or of a league multiplied by 11, or 5, that is 18 leagues and 1 mile.

In how many hours will the courier of the preceding question go 22 leagues?

We see, that if we knew the time he would occupy in going one league, we should have only to repeat this number 22 times, and the result would be the number of hours required. Now the