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

After 6 years, the man’s age will be (x+6) years and son’s age will be (y+6) years. Thus using the given information, we have 

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

`⇒ x+ 6 = 3y +18`

`⇒ x - 3y -12 =0`

Before 3 years, the age of the man was (x-3) years and the age of son’s was (y -3) years. Thus using the given information, we have

` x-3 =9 (y -3)`

`⇒ x -3 = 9y -27`

`⇒ x - 9y +24 =0`

So, we have two equations

` x- 3y-12 =0`

` x- 9y +24=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/((-3)xx24 - (-9)xx(-12))=(-y)/(1xx24-1xx(-12))=(1)/(1xx(-9)-1xx(-3))`

`⇒ x/(-72-108)=(-y)/(24+12)=1/(-9+3)`

`⇒ x/-180=y/36=1/6`

`⇒ x = 180/6,y = 36/6`

`⇒ x =30, y = 6`

Hence, the present age of the man is 30 years and the present age of son is 6 years.