Math, asked by kopparapu12, 11 months ago

The present age of a man is 7 times the present age of his son if two years ago the age of the man was 11 times the age of the son and the present age of man is

Answers

Answered by littyissacpe8b60

Son's age = a

Man's age = 7a

Before 2years

Son's age = a - 2

Man's age = 11(a-2)

7a - 2 = 11a - 22

4a = 20

a = 5

Son's present age = 5years

Man's age = 35years

Answered by AwMAbhishek

Solution: Let present age of son is y years and of man is x years .

=> Present age of man = 7 times of y

=> x = 7y years.

{ Two years ago }

=> Age of man = 11 times the age of son.

=> (x – 2) = (y – 2) × 11

Solving the equation

➥ (x – 2) = 11 × (y – 2)

➥ (x – 2) = 11y – 22

➥ 7y – 2 + 22 = 11y

➥ 20 = 11y – 7y

➥ 20/4 = y

➥ 5 = y

We have got earlier that man's present age is equal to x = 7y years. Then x = 7 × 5 = 35 years.

Required age of man is 35 years.

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