Question

Ten years hence, a man's age will be twice the age of his son. Ten years ago, the man was four times as old as his son. Find their present ages.

Answer 1

Let the age of a man = x years

And the age of his son = y years

Ten years hence,

Man’s age = (x + 10) years

Son’s age = (y + 10) years

According to the question,

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

⇒ x + 10 = 2y + 20

⇒ x = 2y + 20 – 10

⇒ x = 2y + 10 ...(i)

Ten years ago,

Father’s age = (x – 10) years

Son’s age = (y – 10) years

According to the question,

(x – 10) = 4(y – 10)

⇒ x – 10 = 4y – 40

⇒ x = 4y – 30 …(ii)

From Eq. (i) and (ii), we get

2y + 10 = 4y – 30

⇒ 2y – 4y = – 30 – 10

⇒ – 2y = – 40

⇒ y = 20

On putting the value of y = 20 in Eq. (i), we get

x = 2y + 10

⇒ x = 2(20) + 10

⇒ x = 50

Hence, the age of man is 50 years and the age of his son is 20 years.