Father is aged three times more than his son Rony. After 8 years, he would be two and a half times of Rony's age. After further 8 years, how many times would he be of Rony's age?

Question: Father is aged three times more than his son Rony. After 8 years, he would be two and a half times of Rony's age. After further 8 years, how many times would he be of Rony's age?

Solution:
Let, Rony's age is x years and Father is 3x + x = 4x years old

After 8 years, Rony's age is x + 8 years and Father is 4x + 8 years old
4x + 8 = 2.5 (x + 8)
⇒ 4x + 8 = 2.5x + 20
⇒ 1.5x = 12
∴ x = 12/1.5
= 8 year
Father's age (8 × 4) = 32 years

After further 8 years, Rony's age = 8 + 16
= 24 years
Father's age = 32 + 16
= 48 years

∴ Father's age is (48/24) = 2 times his son's age