Question

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

Answer 1

Let the man’s present age be x years.
Let his son’s 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.