Question

Two years ago, a father was five times as old as his son. Two year later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.

Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of his son. 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 2 years, father’s age will be (x + 2) years and the age of son will be (y + 2) years.

Thus using the given information, we have

x + 2 = 3(y + 2) + 8

⇒ x + 2 = 3y + 6 + 8

⇒ x – 3y – 12 = 0

Before 2 years, the age of father was (x – 2) years and the age of son was (y – 2) years.

Thus using the given information, we have

x – 2 = 5(y – 2)

⇒ x – 2 = 5y – 10

⇒ x – 5y + 8 = 0

So, we have two equations

x – 3y – 12 = 0

x – 5y + 8 = 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) xx 8 - (-5) xx -12) = (-y)/(1 xx 8 - 1 xx (-12)) = 1/(1 xx (-5) - 1 xx (-3))`

⇒ `x/(-24 - 60) = (-y)/(8 + 12) = 1/(-5 + 3)`

⇒ `x/(-84) = (-y)/20 = 1/(-2)`

⇒ `x/84 = y/20 = 1/2`

⇒ `x = 84/2, y = 20/2`

⇒ x = 42, y = 10

Hence, the present age of father is 42 years and the present age of son is 10 years.