Question

$5$ years ago a man was $7$ times as old as his son. After $5$ years he will be thrice as old as his son. Find their present ages.

Answer 1

Let the present age of the son be $x$ years and that of the father be $f$ years.
$5$ years back, the father was $7$ times as old as his son.
$\Rightarrow (f-5)=7(x-5)$
$f=7x-35+5$
$\therefore f=7x-30$ ...... (1)
After $5$ years, ages of the father and son will be $f+5$ and $x+5$, respectively.
After $5$ years, the father will be three times older
than his son.
$\Rightarrow f+5=3\times (x+5)$
$7x-30+5=3x+15$ [inserting the value of $f$ from equation 1]
$7x-25=3x+15$
$7x-3x=25+15$
$4x=40$
$x=\frac{40}{4}=10$
Therefore, the present age of the son is $10$ years.

Father's present age $=(7x-30)=7\left(10\right)-30$
$=70-30=40$years