In following the calculation of the work, we find, that after having obtained 10 in the quotient, the division could not be continued, we then multiplied the remainder 9461 by 6, to reduce it to feet, and in that we acted upon the same principle as in the preceding example; having obtained 4 feet for the quotient, and 4350 for the remainder, we multiplied this remainder by 12, to reduce it to inches.

For the proof, which is here brought forward, below the quotient, we have taken the given divisor for multiplicand, and before we added the remainder of the division to the product, we divided it by 72,. because this remainder, not being part of the quotient in the proof, must be considered as a number of pence, resulting from the first remainder of the division 9461d. which has been successively multiplied by 6 and by 12.

ALLIGATION.

WE call Alligation, the mixture which we make of several commodities of different values, to compose a whole of the same number of parts, equal to one another, and of a mean rate.

Suppose we have, for example, two kinds of wine, one at 15d. and the other at Sd. a bottle, and wish to mix them together; we shall form a mixture, the price of which per bottle, will evidently be the mean rate between 15d. and 8d.

The necessity there is in commerce of making mixtures, gives rise to two kinds of questions, which

F

form two different cases, which the rule of Alligation shews how to resolve.

First Case. To determine the price of each of the equal parts of the mixture, when we know the value of each of the mixed parts, and the number of the parts of the mixture.

The method to be pursued in this case, is to find the total price of the mixture, divide it by the num ber of mixed parts, and the quotient gives the price of each of the equal parts of the mixture.

QUESTION. To form of three kinds of wine, a mixture composed of 150 bottles: we have taken 96 bottles of wine at 16d, 33 at 13d, and 21 at 10d. it is required to find the price of one bottle of this mixture.

It is evident, that if we knew the total price of the mixture, we should only have to divide it by the number of bottles which compose it, then we should have the price of a bottle of this mixture. Now, to have the whole value of the mixture, we must follow this reasoning. Since one bottle of the first wine costs 16d. the 96 bottles will cost 96 times 16d. that is to say 1536 pence the 33 bottles at 13d. each, will be worth the 21 bottles at 10d. each, will be worth therefore the total price of the mixture amounts to

429

210

2175 pence.

Dividing this number by that of the bottles, which is 150, it is very plain that the quotient will be the price of a bottle of the mixture, which we find to he 14d.

Second Case. To determine the parts we are to take, of each commodity to be mixed, when we know

the price of each of the commodities, and the value of the mixture.

First Question. We wish to compose a wine at 15d. per bottle, with two sorts of wine, one at 20d. and the other at 12d. the bottle, how many are to be taken of each kind of wine? nothing is easier to be found.

20..S

12..5

We always dispose the numbers which point out the price of the wines, as we see it done here; we then take the difference be- 15 tween the highest price and the mean rate, that is to say, between 20 and 15d. and write this difference, which is 5, on the same line with the lowest price; we afterwards take, in like manner, the difference between the mean rate and the lowest price, that is, between 15 and 12, and we write this difference, on the same line with the highest price; then, if we mix together 3 bottles of wine at 20d. and 5 bottles of that at 12d, the mixture will be composed of wine at 15d. the bottle.

For, by each bottle of wine at 20d. which we put in to form the mixture, we lose 5d, since we are to sell each bottle of the mixture for only 15d. then the 3 bottles will make together a loss of 3 times 5d or 158; but, on the other hand, for each bottle of wine at 12d. which we put in to form the mixture, we gain 3d. since we are to sell it at 15d; thus, for the 5 bottles, we shall gain 5 times 3d. or 15d. whence it is evident, we gain as much on one side as we lose on the other; therefore, the mixture is such as it ought to be.

We have solved the question generally, without paying attention to the total quantity of the wine we should wish to have in the mixture; but if this quantity was limited, supposing, for example, we had only 30 bottles of wine of the first quality, and the

question was then, to find how many we ought to take of the second, to form with these 30 bottles, a mixture at the price demanded; it is very easily per ceived, that we ought to take 5 bottles of wine of the second quality as many times, as we have put in 3 bottles of wine of the first quality; that is, we are to multiply 5 by the number of times, that 3 is comprehended in 30, or by the quotient of 30 divided by 3; thus we are to put in 50 bottles of wine of the second quality.

The better to comprehend the meaning of this method, let us take another example, and suppose we wish to put into the mixture, only two bottles of wine of the first quality; we must then, according to what we have just said, divide 2 by 3, which gives, and then, multiplying 5 by, we shall have

3, which will be the number of bottles of wine of the second quality, which we are to join to the two bottles of wine of the first quality, to form a mixture at the price of 15. the bottle.

There may be occasion for mixing together more than two commodities, as in the following problem.

Second Question. We wish to make a composition of coffee at 21d. the pound, with three sorts of coffee, 1st at 27d. a pound, 2nd at 25d. and the 3d at 18. how many pounds of each sort must we take?

We dispose the numbers as in the preceding example, and suppose at first, that we have to mix only two kinds of coffee, one at 27d. and the other at 18d. the pound; proceeding as in the first question, I write the difference of 27 and 21 by the side of 18, and the difference of 21 and 18 be

side 27; it is evident then,

27 3

25 3

21

that if we take 3 pounds

of the coffee at 27d. and 6 pounds of the coffee at

15d. we shall have coffee at 21d...Suppose the mixture made; the question, at present, is to bring in the coffee at 254. without augmenting the value of the mixture; to effect this, I proceed in the same manner as above, considering only the two prices 25d. and 18d, I write, on the same line with 18, and by the side of 6, the difference between 25 and 21, and I also write by the side of 25 the difference between 21 and 18; then I say, that to form the mixture demanded, we must take 3 pounds of the coffee at 27d. 3 pounds of the coffee at 25d. and 10 pounds of that at 187. for, we shall see, that the loss on one side is equal to the gain on the other; in effect, the 3 pounds at 27d. put in the mixture, will make a loss of 18d. and the 3 pounds at 25d. will make another loss of 12d. which makes the total loss 30d. which is precisely equal to the gain, which the 10 pounds of coffee at 18d. will produce, since each of them, as we see, gives a gain of 3d. It is by following the same method, that we can form mixtures with any number of commodities whatever.

The necessities of commerce give rise to a great variety of questions on mixtures; the principles we have just laid down will suffice to solve them; but we shall, however, here discuss an example which might embarrass the learner, and which is of great use in the coining of money, and in a goldsmith's operations.

Third Question. A goldsmith has 25 marks of silver, of 12 penny-weights fine, and wishes to make jewels of them of 10 penny-weights fine; what quantity of other matter, of brass for example, must he add to the 25 marks, to accomplish his end?

The question, therefore, at present, is, to form a mixture with two matters, the value of one of which