Aptitude - Problems on Ages - Discussion

Discussion Forum : Problems on Ages - General Questions (Q.No. 1)

Problems on Ages

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?

2 times

2	1	times
	2

2	3	times
	4

3 times

Answer: Option

Explanation:

Let Ronit's present age be x years. Then, father's present age =(x + 3x) years = 4x years.

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

8x + 16 = 5x + 40

3x = 24

x = 8.

Hence, required ratio =	(4x + 16)	=	48	= 2.
	(x + 16)		24

Discussion:

319 comments Page 29 of 32.

Uttam said: 1 decade ago

Age of son=x
then,
(x+3x)=4x
after 8 years
(4x+8)=2.5(x+8)
8x+16=5x+40
x=8
after 8 years
(4x+16)=y(x+16)
wkt x=8
s0,
y=(32+16)/(8+16)
y=2.

DINESH REDDY said: 1 decade ago

Ya 4x is absolutely right.

Ashu said: 1 decade ago

Although, I got the solution but cann't we add 16 in the second step, as the fater age is increased by 16 I. E 8 and further 8 years.

M.karthik said: 1 decade ago

How (4x+16) / (x+16) is = 48/24.

Please explain.

Parul said: 1 decade ago

How (x+8) ?

Mahananda said: 1 decade ago

Hi vaishu,

I didn't get the step of adding 8 to 4x+8 &x+8. so pls explain me

Nayana said: 1 decade ago

Hi vaishu,

I didn't get the last step why did you add 8 to that equation 4x+8 and x+8. Can you please explain me that last step?

Nagarjun said: 1 decade ago

Thank you vaishu. I am satisfied after reading your answer.

Manu said: 1 decade ago

I want details answer for 4x+16/x+16 and how did we get answer 48/24 please. Reply me.

Gauri said: 1 decade ago

thanks vaishu, for solving the problem for the last ratio

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