Question

Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?​

Answer 1

Father age is 3 times more than Son age

Let Ronit Age=X years

Then Father Age=x+3x years
=4x years

After 8 years

Ronit Age=(X+8) years

Father's Age=(4x+8) years

Father will be also 2.5 times Ronit age.

Means

4x+8=2.5(X+8)

4x+8=2.5x+20

4x-2.5x=20-8

1.5x=12

X=12/1.5

X=8

Ronit Age=X=8 years

Ronit's Father age=4x=4×8=32 years

After 8 years

Ronit Age=8+8=16 years

Father Age=32+8=40 years

After Further 8 years

Ronit Age=16+8=24 years

Father Age=40+8=48 years

Father Will be=48÷14=2 Times The Son age.

After Further 8 years Father will be 2 times Ronit age

Answer 2

Answer:

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Step-by-step explanation:

Let ronit age is 'x' years

So his father's age is 3x.

After 8 years,

Therefore,Ronit age is 'x + 8' his fathers age is 2 and half times Ronit age

So ,

3x + 8 = \frac{5}{2}  ( x + 8)

..x = 24 fathers age = 3 (24) = 72

After 8 years ronit age = 32 and fathers age is 80

Again after 8 years ronit age = 40 and fathers age is 88

Therefore fathers age is 2.2 times of ronit age.

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