1.

The sum of ages of a father and son is 45 years. Five years ago the product of their ages was four times the father's age at that time. The twice the difference, of father and son's age is:1. 722. 453. 904. 54

Answer» Correct Answer - Option 4 : 54

Given :

The sum of ages of a father and son = 45 years

Five years ago, the product of their ages was four times the father's age at that time.

calculation :

Let the age of father be (a)years and age of his son be (45 - a) years,

Five years ago,

Age of father = (a - 5) years

Age of his son = (45 - a - 5) years = (40 -a) years

According to the question,

Product of their ages five years ago = 4× father's age five years ago

⇒ (a - 5) (40 - a) = 4(a - 5)

⇒ (a - 5) (40 - a) = 4(a - 5)

⇒ 40 - a = 4

⇒ 40 - 4 = a

⇒ 36 = a

Hence,

Present age of father = a years = 36 years

Present age of his son = (45 - a) = (45 -36) = 9 years

Difference between of father's age and son's age = 36 - 9 = 27

∴The twice the difference, of father and son's age is2× 27 = 54



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