Question

9. A father is 4 times as old as his son. 5 years ago, he was 7 times as old as his son was . the present age of son is
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Answer 1

Answer:-

Let the son's present age be x and father's present age be y.

Given:

Father is 4 times son's age.

⟶ y = 4x -- equation (1)

Also given that,

He was 7 times the son's age before 5 years.

• Son's age before 5 years = (x - 5) years.

• Father's age before 5 years = (y - 5) years.

According to the above condition,

⟶ y - 5 = 7(x - 5)

⟶ y - 5 = 7x - 35

Substitute the value of y from equation (1)

⟶ 4x - 7x = - 35 + 5

⟶ - 3x = - 30

⟶ x = - 30/ - 3

⟶ x = 10

∴ The present age of son is 10 years.

Answer 2

Question

A father is 4 times as old as his son. Five years ago, he was 7 times as old as his son was. The present age of son is :

Solution

Let, the present age of son be x

the present age of father be 4x

Five years ago, son's age x-5

Five years ago, father's age be 7(x-5)

ATP

$7(x - 5) = 4x - 5 \\ = ) \: 7x - 35 = 4x - 5 \\ = ) \: 7x - 4x = 35 - 5 \\ = ) \: 3x = 30 \\ = ) \: x = \frac{30}{3} = 10$

Ans:- The present age of son = 10yrs

The present age of father = (10×4)yrs = 40yrs