Question

Two Years ago Dilip was three times as old and his son and
two years hence twice his age will be equal to five times that
of his son. Then the ratio of their present age is (in years)​

Answer 1

Let father age be 'f' and son age be 's'.

Before 2 years :

Father's age = f-2 and son's age = s-2.

Now, f-2 = 3*(s-2) ……………….. eqn 1

After 2 years :

Father's age = f+2 and son's age = s+2.

Here, 2*(f+2) = 5*(s+2) ……………. eqn 2

From eqn 1,

f-2 = 3*(s-2),

f =( 3*(s-2))+2,

f = 3s-6+2,

f = 3s-4 ……………… eqn 3

Sub eqn 3 in 2,

2*(3s-4+2) = 5*(s+2),

2*(3s-2) = 5*(s+2),

6s-4 = 5s +10,

6s - 5s = 10+4,

s = 14.

Since from eqn 3, f = 3s - 4,

f = 3(14)-4,

f = 38.

Let's now check with given cases,

Before 2 years, (38–2) = 3 *(14–2)

And after 2 years, 2*(38+2) = 5*(14+2)

So present age of son = 14

present age of father = 38.

Hey dude,

I hope it will help you.

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!