Question

A father is three times as old as his son. After 10 years, his age will be twice as that of his son then. Find their present ages.​

Answer 1

First, Let the ages of father and his son be x and y years respectively.

According to the question,

Case 1 :-

Father's age is three times of son's age.

⇒ Father's age = 3 × son's age

⇒ x = 3y ...(1)

Case 2 :-

After 10 years, Father's age would be twice of son's age.

We must add 10 to both the ages because this case is given after 10 years in the future.

⇒ Father's age + 10 = 2 ( Son's age + 10 )

⇒ x + 10 = 2 ( y + 10 )

⇒ x + 10 = 2y + 20

Substituting x = 3y , from (1), we have

⇒ 3y + 10 = 2y + 20

⇒ y = 10

Now, Substitute y = 10 in (1) to get the age of father,

⇒ x = 3 × 10

⇒ x = 30

Hence, The present age of father is 30 years while the present age of his son is 10 years.

Answer 2

Answer :

Let the ages of father and son be a & b respectively.

So, A.T.Q

→ a = 3b ........(1)

★ After 10 years.

→ a + 10 = 2(b + 10)

→ a + 10 = 2b + 20

→ a - 2b = 10 ........(2)

____________________

From equation (1) put value of a in equation (2).

→ 3b - 2b = 10

→ b = 10

So, son's present age is 10 years

____________________

Put value of b in equation (1).

→ a = 3(10)

→ a = 30

Father's present age is 30 years.