Aptitude - Problems on Ages - Discussion

Discussion Forum : Problems on Ages - General Questions (Q.No. 1)
1.
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
1 times
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 14 of 32.

Indu said:   1 decade ago
If father's present age 37 year then what's the age of son ? please help me.

Sara shree said:   1 decade ago
I think we should take father's age be y.
So x+8 is equal to (3y+8)5/2

Hmdeep said:   1 decade ago
Thanks Jwala.

Adhi said:   1 decade ago
Is there any formula this problem.

Akruti said:   1 decade ago
Father age is 3 times in starting.
After 8 year father age will be 2 and half times of son.


Means after 8 year father age decrease 1/2 times.
i.e. 3-(1/2)=2(1/2)

So, after 16 year father age will be= 2(1/2)-(1/2) = 2 times of son.

Annapoorna said:   1 decade ago
When it is "three times more" why should we take (3x+x)? It can also be (3x+2x) or (3x+.5x)?

Laxman said:   1 decade ago
Hi @Annapoorna,

It is supposed of x years for son and in question father is more than 3 years of son so that it is of (3x+x). Thanks.

Nihal said:   1 decade ago
What is the meaning of last line of Question?

And how it is resolved?

Bhoomika said:   1 decade ago
How do we get (4x + 16) divide by (x + 16) instead of (4x+16)=(x+16)?

Prachi said:   1 decade ago
There are times when we take x & y both variable. So how to identify whether we have to use 1 variable or both?


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