Question

Father is aged three times more than his son Ronit. After 8 yrs, he would be

two two and a half times of Ronit’s age. After further next 8 years, how many

times would he be of Ronit’’s age?

Answer 1

Let the age of Ronit be x.

Then fathers age will be x + 3x = 4x years.

Give, After 8 years he would be  2 1/2 = 5/2of Ronit Age. 

(4x+8) = 5(x+8)/2

2(4x + 8) = 5(x+8)

8x + 16 = 5x + 40

8x - 5x = 40 - 16

3x = 24

x = 8.


After 8 years, The Age of Ronit will be (4x + 16)/(x+16)

                                                               = (4 * 8 + 16)/(8 + 16)

                                                               = (32 + 16)/24

                                                              = 48/24

                                                               = 2.


Hope this helps!

Answer 2

Let the age of Ronit be x.

Then fathers age will be = x + 3x = 4x
after 8 years he would be 2 1/2 = 5/2of Ronit's age

(4x+8) = 5(x+8)/2

2(4x + 8) = 5(x+8)

8x + 16 = 5x + 40

8x - 5x = 40 - 16

3x = 24

x = 8.
After 8 years, The Age of Ronit will be (4x + 16)/(x+16)


= (4 * 8 + 16)/(8 + 16)

= (32 + 16)/24

=48/24

= 2.