Question

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.

Answer 1

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

After 10 years, father’s age will be (x+10) years and son’s age will be (y + 10) years. Thus using the given information, we have

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

`⇒ x+ 10 = 2y+ 20`

` ⇒ x - 2y -10 =0`

Before 10 years, the age of father was `(x - 10)` years and the age of son was `(y - 10)` years. Thus using the given information, we have

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

`⇒  x + 10 = 2y +20`

`⇒ x - 12y +110 =0`

So, we have two equations

`x -2y -10 =0`

` x- 12y +110 =0`

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have 

` x / ((-2)xx110 -(-12)xx(-10))=(-y)/(1xx110 -1(-10))= 1/(1xx(-12)-1xx(-2))`

`⇒ x / (-220-120)=(-y)/(110+10)+ 1/(-12+2)`

`⇒ x/-340 =(-y)/120 = 1/(-10)`

`⇒ x/(340)= y/120 = 1/10`

`⇒ x = 340/10,y=120/10`

`⇒ x = 34 , y = 12`

Hence, the present age of father is 34 years and the present age of son is 12 years.