Question

5 years ago a man was 7 times as hos son 5 years hense the father will be 3 times old as his son find their ages​

Answer 1

Answer :-

Age of father is 40 years and age of son is 10 years.

Solution :-

Let the present age of a man be x

Let the present age of son be y

Five years ago

Father's age = x - 5

Son's age = y - 5

Father's age 5 years age = 7 times as his son.

⇒ x - 5 = 7(y - 5)

⇒ x - 5 = 7y - 35

⇒ x = 7y - 35 + 5

⇒ x = 7y - 30--(1)

Five years hence

Father's age = x + 5

Son's age = y + 5

Father's age 5 years hence = 3 times as his son.

⇒ x + 5 = 3(y + 5)

⇒ x + 5 = 3y + 15

⇒ x = 3y + 15 - 5

⇒ x = 3y + 10---(2)

From (1) & (2)

7y - 30 = 3y + 10

⇒ 7y - 3y = 10 + 30

⇒ 4y = 40

⇒ y = 40/4

⇒ y = 10

Substitute y = 10 in (1)

x = 7y - 30

⇒ x = 7(10) - 30

⇒ x = 70 - 30

⇒ x = 40

Therefore age of father is 40 years and age of son is 10 years.

Answer 2

• Let present age of father be M and present age of son be N.

» 5 years ago father was 7 times as his son.

Before 5 years...

Age of father = M - 5

Age of son = N - 5

• A.T.Q.

=> M - 5 = 7(N - 5)

=> M - 5 = 7N - 35

=> M - 7N = - 35 + 5

=> M - 7N = - 30

=> M = 7N - 30 ________ (eq 1)

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» 5 years hence, the father will be 3 times old as his son.

After 5 years...

Age of father = M + 5

Age of son = N + 5

• A.T.Q.

=> M + 5 = 3(N + 5)

=> M + 5 = 3N + 15

=> M = 3N + 15 - 5

=> M = 3N + 10

=> (7N - 30) = 3N + 10 [From (eq 1)]

=> 7N - 30 = 3N + 10

=> 7N - 3N = 10 + 30

=> 4N = 40

=> N = 10

• Put value of N in (eq 1)

=> M = 7(10) - 30

=> M = 70 - 30

=> M = 40

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Present age of father = M = 40 years.

Present age of son = N = 10 years.

_________ [ANSWER]

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