Q30 of 186 Page 4

Two years ago, a man was five times as old as his son. Two years later his age will be 8 more than three times the age of the son. Find the present ages of man and his son.

Let the age of a man = x years

And the age of his son = y years


Two years ago,


Man’s age = (x – 2) years


Son’s age = (y – 2) years


According to the question,


(x – 2) = 5(y – 2)


⇒ x – 2 = 5y – 10


⇒ x = 5y – 10 + 2


⇒ x = 5y – 8 ...(i)


Two years later,


Father’s age = (x + 2) years


Son’s age = (y + 2) years


According to the question,


(x + 2) = 8 + 3(y + 2)


⇒ x + 2 = 8 + 3y + 6


⇒ x = 3y + 12 …(ii)


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


5y – 8 = 3y + 12


⇒ 5y – 3y = 12 + 8


⇒ 2y = 20


⇒ y = 10


On putting the value of y = 11 in Eq. (i), we get


x = 5(10) – 8


⇒ x = 50 – 8


⇒ x = 42


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


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