Question

The ages of two girls are in the ratio 5 : 7. Eight years ago their ages were in the ratio 7 : 13. Find their present ages.

Answer 1

Let the present age of the first girl be 5x and the second girl be 7x.
Now,
there ages 8 yrs ago = (5x-8) / (7x-8) = 7/13
65x - 104 = 49x - 56
16x = 48
x=3
The age of the first girl = 5*3 = 15 yrs
The age of the second girl = 7*3 = 21 yrs. 

Answer 2

Given: Ratio of ages of two gi_rls = 5 : 7

           Ratio of their ages 8 years ago = 7 : 13

To find: Their present ages

Let: Their present ages be $x$  and $y$

Solution: According to the given question and assumption made,

$\frac{x}{y} = \frac{5}{7}$

i.e., $7x - 5y = 0$    ...(1)

also, $\frac{x \ - \ 8}{y \ - \ 8} = \frac{7}{13}$

i.e., $13(x-8) = 7(y-8)$

⇒ $13x - 104 = 7y - 56$

⇒ $13x - 7y = 48$       ...(2)

Using elimination method to solve the equations:

Multiplying equation (1) by 7, we get

$49x - 35y = 0$   ...(3)

Multiplying equation (2) by 5, we get

$65x - 35y = 240$    ...(4)

Subtracting equation (3) from equation (4)

⇒ $(65x - 35y) - (49x - 35y) = 240 - 0$

⇒ $65x - 49x - 35y + 35y = 240$

⇒ $16x = 240$

⇒ $x = \frac{240}{16} = 15$

Putting $x=15$ in equation (1)

⇒ $7 * 15 - 5y = 0$

⇒ $5y = 105$

⇒ $y = \frac{105}{5} = 21$

Hence, present ages of the gi_rls are 15 years and 21 years.