Question

43. Two years ago, a man was five times as old as his son. Two years later,
will be 8 more than three times the age of his son. Find their
present ages.
[CBSE 2008]
is added to the father's age, the sum is​

Answer 1

Answer:

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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