Question

02) A father is three times as old as his son. 8 years ago the father
was 7 times as old as his son. Find their present ages.​

Answer 1

Answer:

Let the present age of the son be x and that to of his father which is 7 times =7x

Then, 2 years ago their ages :

(x-2) of son

(7x-2) of father

Given, 13(x−2)=7x−2

Then,

13x−26=7x−2

=13x−7x=26−2

6x=24

or x=4 and that of his father = 28 years

ageofson=x=4years

ageoffather=7x=28years

Answer 2

Let us consider

Son’s age = X

Two years ago son’s age = X – 2

His Father’s age at that time = 3(X – 2)

Present age of Father = 3X – 6 + 2

The present age of Father = 3X – 4

Two years hence father’s age = 3X – 4 + 2

Two years hence father’s age = 3X – 2

Two years hence son’s age = X + 2

Given:

5 × (X + 2) = 2 × (3X – 2)

5X + 10 = 6X – 4

X = 14

Son’s present age X = 14 years.

Father’s present age = (3 × 14) – 4

Father’s present age = 38 years.