Ans. A received 575 dollars;

B received 1725 dollars;

C received 3450 dollars;

D received 4025 dollars.

Prob. 53. A farmer has 192 fruit-trees in his or chard, consisting of plum-trees, cherry-trees, peachtrees, and apple-trees. The number of cherry-trees is four times the number of plum-trees; the number of peach-trees is five times the number of plum-trees; and the number of apple-trees is six times the number of plum-trees What is the number of each sort? Ans. There are 12 plum-trees,

48 cherry-trees,

60 peach-trees,

72 apple-trees.

Prob. 54. Four gentlemen entered into a speculation, for which they subscribed 9720 dollars, of which B paid four times as much as A, C paid six times as much as A, and D paid seven times as much as A. What did each pay?

Ans. A paid 540 dollars;

B paid 2160 dollars;

C paid 3240 dollars;

D paid 3780 dollars.

(16.) The pupil can not fail to have remarked a striking similarity between many of the preceding problems. Thus, the first problem requires us to divide the number 6 into two parts, one of which is double the other. Problem 10th requires us to divide the number 36 into two parts, one of which is double the other. It is evident that an infinite number of

QUEST.-In what does Problem 1 differ from Problem 10?

similar problems might be proposed, differing from each other only in the number to be divided. We may discover a general method of solving all these problems by representing the number to be divided by a letter, as a. We shall then have

Thus, in Problem 1, x= one third of 6, which is 2; and in Prob. 10, x equals one third of 36, which is 12.

Again, Prob. 2 requires us to divide the number 60 into two parts, one of which is three times the other Prob. 11 only differs from Prob. 2 in the number proposed to be divided. We can discover a general method of solving all problems of this kind by representing tne number to be divided by a letter, as a. We shall

Thus, in Prob. 2, x is one fourth of 60, or 15; and in Prob. 11, x is one fourth of 72, or 18.

Again, Prob. 3 requires us to divide the number 10 into two parts, one of which is four times as great as the other. Prob. 12 requires us to divide the number 15 in a similar manner. We can solve all problems

QUEST.-In what does Problem 2 differ from Problem 11? In what does Problem 3 differ from Problem 12?

of this kind in a general manner by representing the proposed number by a. We shall then have

A similar method is applicable to problems 4 and 13, 5 and 14, etc.

There is also an analogy between Problems 1, 2, 3, etc. Prob. 1 requires us to divide a number into two parts, one of which is double the other. Prob. 2 requires us to divide a number into two parts, one of which is three times the other. Prob. 3 requires us to divide a number into two parts, one of which is four times the other, etc. All such problems are included in the following more general problem:

Prob. 55. It is required to divide a number a into two parts, one of which shall be m times as great as the other.

This problem may be solved in the following man

ner:

Let x represent one of the parts.

Then m times x, which we will write mx, may represent the other part.

And, by the conditions,

x+mx=a.

We now meet with a difficulty in finding the value of x, because the two terms x and mx can not be united in a single term, as was done in Art. 10. But, since x+mx is equal to x repeated 1+m times, we infer that

and this is a general solution of the first eighteen of the preceding problems.

(17.) It will be readily seen that Problems 19 and 28 differ only in the number proposed to be divided ; and if we represent the proposed number by a, we shall have a general solution of this class of problems. A similar remark is applicable to Problems 20 and 29, Problems 21 and 30, etc.

There is also an analogy between Problems 19, 20, 21, etc. In each of them it is required to divide a proposed number into three parts, such that the second and third shall be multiples of the first. All these problems are included in the following general problem.

Prob. 56. It is required to divide a number a into three parts, the second of which shall be m times as great as the first, and the third n times as great as the first.

This problem may be solved in the following manner: Let x represent the first part.

Then mx will represent the second part,

and nx will represent the third part.

And, by the conditions of the problem,

x+mx+nx=α.

We now encounter the same difficulty as in Prob. 55, because the terms x, mx, and nx can not be united in a single term. Since, however, x+mx+nx is equal to x repeated 1+m+n times, we infer that,

QUEST. In what does Problem 19 resemble Problem 28?

and this is a general solution of all the preceding problems from 19 to 36.

(18.) We shall find in a similar manner that Problems 37, 38, 39, etc., are all included in the following general problem.

Prob. 57. It is required to divide a number a into four parts, the second of which shall be m times as great as the first, the third shall be n times as great as the first, and the fourth p times as great as the first. This problem may be solved in the following man

ner:

Let x represent the first part.

Then mx will represent the second part,

nx will represent the third part,

and pc will represent the fourth part.

And, by the conditions of the problem,

x+mx+nx+px=a.

And, reasoning in the same manner as in Prob. 56, we conclude that

and this is a general solution of all the preceding problems from 37 to 54.

(19.) Problems 1 to 54 are called numerical problems, and are such problems as occur in common arithmetic. Problems 55 to 57 are general problems, and pure Algebra is chiefly confined to problems of this kind, where letters are employed to represent quantities which are supposed to be known, as well as those which are unknown. It becomes necessary, therefore,

QUEST. What are numerical problems? What are general prob lems?