Question

The present age of a man is 2 years more than five times the age of his son. Two years hence, the man’* age will be 8 years more than three times the age of his son. Find their present ages.

Answer 1

Let the man’* present age be x years.

Let his son’* present age be y years.

According to the question, we have:

Two years ago:

Age of the man = Five times the age of the son

⇒ (x – 2) = 5(y – 2)

⇒ x – 2 = 5y – 10

⇒ x – 5y = –8 …….(i)

Two years later:

Age of the man = Three times the age of the son + 8

⇒ (x + 2) = 3(y + 2) + 8

⇒ x + 2 = 3y + 6 + 8

⇒ x – 3y = 12 …………(ii)

Subtracting (i) from (ii), we get:

2y = 20

⇒ y = 10

On substituting y = 10 in (i), we get:

x – 5 × 10 = -8

⇒ x – 50 = -8

⇒ x = (-8 + 50) = 42

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