Question

two years ago dilip was three times as old as his son and two years hence, twice his age will be equal five times that of his son. Then the present age of dilip is what ​

Answer 1

Step-by-step explanation:

let present age of Dilip be x

present age of son be y

age of Dilip 2 years ago = x - 2

age of son 2 years ago = y - 2

Dilip was 3 times the age of son

x - 2 = 3(y - 2)

x - 2 = 3y - 6

x - 3y = - 6 + 2

x - 3y = - 4 equ (1)

two years later Dilip age will be x + 2

two years later son's age will be y + 2

twice Dilip age will be 5 times his sons age

2(x + 2) = 5(y + 2)

2x + 4 = 5y + 10

2x - 5y = 10 - 4

2x - 5y = 6. equ (2)

multiply equ (1) with 2

2x - 6y = - 8. equ (3)

subtract (2) - (3)

2x - 5y - (2x - 6y) = 6 - (-8)

2x - 5y - 2x + 6y = 6 + 8

y = 14

substitute y= 14 in equ (2)

2x - 5y = 6

2x - 5(14) = 6

2x - 70 = 6

2x = 6 + 70

2x = 76

x = 76/2

x = 38

Dilip present age = 38 years

son's present age = 14 years

hope you get your answer

Answer 2

Given :-

→ Let Dilip's son's age 2 years ago be x years.

Then, Dilip's age 2 years ago = ( 3x ) years

Thus , the son's age 2 years hence (x + 4) years.

Dilip's age 2 years hence (3x + 4) years.

→ 2 ( 3x + 4 ) = 5 ( x + 4 )

→ 6x + 8 = 5x +20

→ x = 12.

∴ Dilip's son's age 2 years ago = 12 years and Dilip's age 2 years ago = 36 years.

∴ son's present age = 14 years, and

Dilip's present age = 38 years.

Thank You!