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:
325 comments Page 33 of 33.

Gora4663 said:   5 years ago
Let son age = x
Father age = y.

X + 3x =y( acc to 1st statemnt) ---> eq1
5/2(X+8) = y+8
5x+40 = 2y+16
5x-2y=-24 ---> eq2.

Solve eq 1 and 2 we get;
X=8 then y = 24,
After 16 year,
Son age 24 & father age 48.
Therefore 2 times.

Chinnu said:   4 years ago
If we take 2.5 or 5/2 will we get the same answer? Anyone explain.

Chinnu said:   4 years ago
Anyone, Can you explain this sum by taking 2.5?

Hamza said:   4 years ago
@All.

Please explain, father is two and a half mean 5/2. Then why we write 5/2 With son's age?

Because when any question ask father is 5 times of son then we put father is 5x and son is x.

Anyone, please explain this.

Pranali said:   16 hours ago
Yes, I agree with you @Janarthanan.


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