Question

5 year ago , FAther age was 7 times the age of his son . 5 year later ,the father age will be 3 time the age of his son . find their present ages

Answer 1

Given:

5 years ago father was 7 times the age of his son.

After 5 years father's age will be 3 times the age of the son.

Find:

Their present ages

Solution:

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

5 years ago father was 7 times the age of his son.

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

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

So,

(y - 5) = 7(x - 5)

⟶ y = 7x - 35 + 5

⟶ y = 7x - 30 -- equation (1)

And,

After 5 years father's age will be 3 times the age of the son.

Son's age after 5 years = (x + 5) years

father's age after 5 years = (y + 5) years.

So,

⟶ y + 5 = 3(x + 5)

Substitute the value of y from equation(1)

⟶ 7x - 30 + 5 = 3x + 15

⟶ 7x - 3x = 15 + 30 - 5

⟶ 4x = 40

⟶ x = 40/4

⟶ x = 10 years

Substitute the value of x in equation (1).

⟶ y = 7(10) - 30

⟶ y = 70 - 30

⟶ y = 40 years

∴

Son's present age = x = 10 years.

Father's present age = y = 40 years.

I hope it will help you.

Regards.

Answer 2

Answer:

⟶ 7x - 30 + 5 = 3x + 15

⟶ 7x - 3x = 15 + 30 - 5

⟶ 4x = 40

⟶ x = 40/4

⟶ x = 10 years

Substitute the value of x in equation (1).

⟶ y = 7(10) - 30

⟶ y = 70 - 30

⟶ y = 40 years

Step-by-step explanation: