Question

Six years hence a man's age will be three times the age of his son and three years ago, he was nine times as old as his son. Find their present ages.

Answer 1

Let the age of man be $x$

and the age of his son be $y$

According to the question$:$

After $6$ years $:$

$(x+6)=3(y+6)$

$\Rightarrow x-3y=12⟶\left(i\right)$

Three years ago $:$

$x-3=9(y-3)\Rightarrow x-3=9y-27\Rightarrow x-9y=-24⟶\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right),$

$6y=36$

$\Rightarrow y=6$

Putting value of $y$ in equation $\left(i\right),$

$x-3xy=12$

$\therefore x=12+18$

$x=30$

$\therefore$ Age of man $=30$ years$.$

Age of son $=6$ years$.$