Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.


Given :

Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then.

To find :

We have to find their present ages.

Solution : 

Let the present ages of the son and the father be $x$ and $y$ respectively.

This implies,

Age of the son after 10 years $= x+10$

Age of the father after 10 years $= y+10$.

Age of the son 10 years ago $= x-10$.

Age of the father 10 years ago $= y-10$.

Therefore,

$y+10 = 2(x+10)$

$y+10 = 2x+20$

$y = 2x+20-10$

$y = 2x+10$.....(i)

$y-10=12(x-10)$

$y-10=12x-120$

$(2x+10)-10=12x-120$    (From (i))

$2x+10-10=12x-120$

$12x-2x=120$

$10x=120$

$x=\frac{120}{10}$

$x=12$

$\Rightarrow y=2(12)+10=24+10=34$

The present age of the son is $12$ years and the present age of the father is $34$ years.  

Tutorialspoint
Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

67 Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements