Question

Five years ago a man was seven times as old as his son. Five years after the father will be three times as old as his son. Find their present ages.

Answer 1

Let $x$ and $y$ be the present age of father and his son.

According to the question,

$x-5=7(y-5)$

$x=7y-30.....\left(1\right)$

Also,

$x+5=3(y+5)$

$x+5=3y+15$

$7y-30+5=3y+15\left(\text{From}\left(1\right)\right)$

$7y-3y=15+25$

$\Rightarrow y=\frac{40}{4}=10$

Substituting the value of $y$ in equation $\left(1\right)$, we have

$x=7\times 10-30=70-30=40$

Hence the present age of father and his son is $40$ and $10$ years respectively.