Question

13. Four years ago a father was six times as old as his son. Ten years later, the father will be two and a half times as old as his
son Find the present age of father and his son.
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Answer 1

Answer:

• Present age of son is 10 yrs and
• Present age of Father is 40 yrs.

Explanation:

Let, present age of Father be x and

present age of son be y

then,

Considering Case I

• Four years ago Father was six times as old as his son

so, four years ago age of father will be ( x - 4 ) and age of son will be ( y - 4 )

hence,

→ ( x - 4 ) = 6 ( y - 4 )

→ x - 4 = 6 y - 24

→ x = 6 y - 24 + 4

→ x = 6 y - 20  ____equation (1)

Considering case II

• Ten years later, the father will be two and half times (means 5/2 times ) as old as his son

so, ten years later age of Father will be ( x + 10 ) and of son will be ( y + 10 )

hence,

→ ( x + 10 ) = 5/2 ( y + 10 )

→ x + 10 = 5 y/2 + 25

multiplying by 2 both sides

→ 2 x + 20 = 5 y + 50

using equation (1)

→ 2 ( 6 y - 20 ) + 20 = 5 y + 50

→ 12 y - 40 + 20 = 5 y + 50

→ 12 y - 5 y = 50 + 40 - 20

→ 7 y = 70

→ y = 10

→ y = 10 yrs

putting value of y in equation (1)

→ x = 6 y - 20

→ x = 6 ( 10 ) - 20

→ x = 60 - 20

→ x = 40 yrs

therefore,

• Present age of son is 10 yrs and
• Present age of Father is 40 yrs.