Wherefore by transposition = (23-9=) 14, the answer required.

2. What number is that, from which 27 being subtracted, the remainder is 41?

Let the number required be called x.

From which subtracting 27, the remainder will be x-27.
This, by the problem, equals 41, whence x-27=41.

Therefore, by transposition x= (41+27) 68, the answer required.

3. What number is that, which being multiplied by 4, and 5 being subtracted from the product, the remainder will be 6? Let the required number be x.

This multiplied by 4, is 4x.

From which subtracting 5, the remainder is 4x-5.

This remainder by the problem equals 6, wherefore 4x-5-6.
By transposition 4x= (6+5) 11.

And by division x= (

11

4

:) 23, the answer '.

4. What number is that, which being divided by 7, with 8 added to the quotient, the sum will be 9?

Let the required number be called x.

7

Which sum, by the problem, equals 9, whence —+8=9.

11

• Proof of prob. 3. The required number is 23, that is

For if it be multiplied by 4, the product is 11.

From this subtracting 5, the remainder is 11-5=6, according to the problem.

5. What number is that, from which 12 being subtracted, one fourth of the remainder will be 22 ?

Let the number required be represented by x.

From which subtracting 12, the remainder is x—12.

One fourth of this remainder is

x-12

4

X- -12

22.

4

This by the problem equals 22, whence

Wherefore by multiplication x-12=88.

And by transposition x= (88+12=) 100, the answer *. 6. What number is that, to which 7 being added, two thirds of the sum will be 8?

By multiplication 2x+14=24.

By transposition 2x= (24-14=) 10.

And by division x=5, the answer y.

7. What number is that, which being added to 16, and subtracted from 20, the remainder will be two sevenths of the sum ?

To which adding 8, the sum is I +8=9, according to the problem.

X Proof. The number required is 100.

From which subtracting 12, it becomes 100-12=88.

Also subtracting x from 20, the remainder is 20—x.

2x+32

Wherefore by the problem 20-x=' 7

This by multiplication becomes 140-7x=2x+32.

And by transposition 9x=108.

Whence by division x=12, the answer 2.

8. What number is that, of which its one-fourth part exceeds its one-fifth part by 4.

By multiplication 5x=4x+80.

Whence by transposition xSO, the answer ".

9. A lady being asked her age, replied, If you add,, and of my age together, the sum will be 18:' how old was she?

2 Proof. The required number is 12; for by adding 16 to it, the sum is 12+16=28; also subtracting the given number 12 from 20, the remainder is

a Proof. The number to be found is 80; for one-fourth part of it is

4

80

20, and one-fifth part

5

16: now 20 exceeds 16 by 4, which was to be shewn,

Whence by multiplication by 12 (4x+3x+2x=) 9x=216.
And by division x=24 years, the answer ".

10. Divide 17 shillings between two persons, so that one may have 4 shillings more than the other.

Let the less share be called x.

Then will the greater be x+4.

And both shares added together, will be 2x+4.
This by the problem equals 17, or 2x+4=17.
Whence by transposition 2x= (17—4=) 13.

13
2.

And by division x= (—=6}=) 6s. 6d. = the least share;

subtract this from 17, and (17—Cs. 6d.=) 10s. 6d.= the greater share c.

Or thus,

Let the greater share be x.

Then the less will be x-4.

The sum of these equals 17, viz. 2x−4=17.

Whence 2x=21, x=10s. 6d. and x—4=6s. 6d. as before.

11. Three persons, A, B, and C, rent 140 acres of land between them, of which A has twice as much as B, and B thrice as much as C; how many acres has each ?

Let C's number be called x.

Then B's will be (=thrice C's, or) 3 x.

And A's (=twice B's, or) 6 x.

And the sum of these, by the problem, 10x=140.

Whence x=14 C's share, 3x= (3 × 14=) 42=B's share, and 6x= (6×14=) 84=A's share a.

12. If the half, third, and fourth parts of my number of shillings be added together, the sum will be one shilling more than I have; how many shillings have I got?

Let x=the number required, then will —=the half, ~———

2

3

the fourth part; also x+1=one more than

x

I have; whence by the problem + + =x+1; this equation

2. 3 4

cleared of fractions, becomes (6x+4x+3x=) 13x=12x+12 ; wherefore x=12, the number required.

13. A legacy of 120l. was left between A and B, in such sort, that of A's share was equal to of B's; what sum did each receive?

Let 8x=A's share, then will 7x=B's, and their sum=

(8x+7x=) 15x=120 by the problem; whence x= (

consequently 8x=641.=A's share, and 7 x=561.=B's.

120

15

=) 8:

14. A post is in the mud, in the water, and 11 feet above the water; required the length of the post?

Let its length be called x, then the part will be

7, and the

4'

+11=x: this equation

x

cleared of fractions, &c. we shall have x=20 feet; ==5 feet in

d Proof. A's, B's, and C's shares added together, viz. 84 + 42 + 14–140, which is one condition of the problem.

Also A's share is double of B's, for 84-2 × 42; and B's share is triple of C's, for 42-3 × 14, which is the other condition.

e

Proof. The number obtained by the solution is 12; the half of which is 6, the third 4, and the fourth 3: now 6+4+3=13=12+1; that is, the sum of the half, third, and fourth, exceeds the number I have by 1, which was to be shewn. In like manner the truth of the conclusions in all the following problems may be demonstrated.