Question

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

Answer 1

Let son's age 10 years ago be x.
Hence, father's age 10 years ago was 12x.
Also note that their present age will be x+10 and 12x+10 respectively.
10 years later, their ages will be
x+10+10=x+20 and 12x+10+10=12x+20 respectively.
According to question;
2×(x+20) = 12x+20,
=>2x+40=12x+20,
=>10x=20,
=>x=2.
So, son's present age =x+10 =2+10 =12 years,
father's age =12x+10 =12×2+10 =24+10 =34 years
(Answer)