The work is now preserved in the result, and it appears that one part will be of the number to be divided; and the other, of it. This is a rule that will apply to any number.

Suppose a 500 as in the example.

Ans. One part is $125, and the other $375; the same as above.

Suppose it is required to divide $7532 in the same propor

tions.

Ans. One part is $1883, and the other is $5649. 2. A man sold some apples, some pears, and some oranges for a number a of cents, the apples at two cents apiece, the peurs at three cents apiece, and the oranges at five cents apiece. There were twice as many pears as oranges, and three times as many apples as pears. How many were there of each ?

of 184

8

the number

Suppose a 184 cents, then

of oranges; 2 × 8 = = 16 the number of pears; and 6 × 8 48 the number of apples. This is easily proved. 8 oranges, at 5 cents apiece, come to 40 cents; 16 pears, at 3 cents apiece, come to 48 cents; and 48 apples, at 2 cents apiece, come to 96 cents;

40+48 +96=184.

The learner may be curious to know, how it is possible to mal e the examples in such a manner, that the answer may al

ways come out a whole number when it is wished; for if the numbers were taken at random, there would frequently be fractions in the result. The method is to solve it first with a letter, as has been done in the two preceding examples. If any number, which is divisible by 4, be put in the place of a, in the first example, the answer will be in whole numbers. And if any number, which is divisible by 23, be put in the place of a, in the second example, the answer will be in whole numbers.

Let the learner now generalize the examples in Art. I., by substituting a letter instead of the number; and after the result is obtained, put in the numbers again, and see if the answers agree. Let him also try other numbers.

The examples in Art. II. may be generalized in the same

manner.

3. A man being asked his age, answered, that if its half and its third were added to it, the sum would be 88. Required his age.

Instead of 88 put a, and let x = the number required.

Any number that is divisible by 11, being put in the place of a, will give an answer in whole numbers. Let a = 88, then TT of it is 48, agreeing with the answer in Art. II.

In the course of the solution it appears, that a is equal to of x; and the result shows, that x is equal to TT of a. That is, the value of x is found by multiplying a by the fraction inverted.

4. In an orchard of fruit-trees, of them bear apples, of them cherries, and the remainder, which is a, bear peaches. How many trees are there in the orchard?

5 *

Any number that is divisible by 5, may be put in the place If a 15, the answer is 36.

of a.

5. The Sth example of Art. II. is solved as follows:

Instead of 100 put a, and let x = the whole number of

Let a

140-1=39.

135, and find the answer in the same way.

The answer will be 53.

Proof.

53+53 +261 +21=135.

The learner may now generalize the examples in Art. II. The preceding examples admit of being generalized still

more, but the process would be too difficult for the learner at present. The following question admits it more easily.

6. (Art. III. Exam. 1.) Two men, A and B, hired a pasture for $55, and A was to pay $13 more than B. How much did each pay

?

This question is, to divide the number 55 into two such parts, that one may exceed the other by 13.

* Let us represent 55 by a, and 13 by b. The question now is to divide the number a, into two such parts, that one may exceed the other by the number 6: a and b being any two numbers, of which a is the larger.

Let

x= the less part.

x+b= the greater part.
x+x+b= ɑ

Then

And

By transposition,

Dividing by 2,

2x + b = a

2 x = a

2 2

-b, is

When a number, consisting of two or more parts, as a to be divided, it is evident that all the terms must be divided,

α

b

as

2 2

nominator, one numerator may be subtracted from the other

numbers. See below, where 55 and 13 are substituted for a and b.

Hence it appears, that the less part is found by subtracting half of the excess of the greater above the less from half the number to be divided; or by taking half the difference between the number to be divided and the excess.

a

The greater part is equal to x+b; hence if b be added to it will give the greater part :

b

* Whenever the learner finds any difficulty in comprehending the operations in the general solutions, let him first solve the questions with the numbers.

The greater is found by adding half the excess to half the number to be divided; or by taking half the sum of the number to be divided and the excess.

In the above example,

A's part = +

55
2 2

13 55 + 13

or

= 34.

2

Let the learner generalize this question by making x = the greater part. The same results will be obtained.

This is a general rule, and will apply to all questions like it, and should be remembered, for it is frequently useful.

Let the learner find the answers to the 2d, 3d, and 7th examples of Art. III. by this rule. That is, by putting the numbers of those examples in the place of a and b in the formulas. It is easy to see the propriety of the rule. For the formula α ·b 55-13 42 shows, that if the $13 that A pays

or

2

2

=

2'

more than B, be taken out, the remainder is to be paid in equal

shows, that if B were to pay $13 more, he would pay as much as A, and the rent would be paid in equal parts by them.

7. A father, who has three sons (Art. III. exam. 4), leaves them 16000 crowns. The will specifies, that the eldest shall have 2000 crowns more than the second, and that the second shall have 1000 crowns more than the third. What is the share of each?

Let a represent the whole number of crowns, b what the eldest son's share exceeds that of the second, and c what the share of the second son exceeds that of the third.

This question may be expressed in general terms, thus: To divide a given number a, into three such parts, that the great