Aptitude - Problems on Ages - Discussion

At present, F and R.

R=R,

Father's age would 3 times more than Ronit's age
F=3R+R -> F=4R -----> (1)

After 8 years, F+8 and R+8.

Father would be 2 and half a times of Ronit's age
F+8=2 1/2(R+8).

Mixed fraction 2 1/2 = ((2*2)+1)/2 = 5/2.
F + 8=5/2(R+8),
2F+16 = 5R + 40,
2F-5R = 40 - 16,
2F-5R = 24 -------> (2)

Sub F=4R in (2).
2(4R)-5R = 24,
8R-5R = 24,
3R= 24.
R = 8 years, F = 32 years.
After further 8 years means;
F + 16 = 32 + 16= 48,
R + 16 = 8 + 16= 24,
Required ratio= 48/24 = 2.
Hence, After further 8 years, 2 times he would be of Ronit's age.

If we will take Rohit's age I think it should be 3x (x+3).
If we will take Rohit's age as 1 it can be 4 only.

Correct me, If I am wrong.

@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.

Anyone, Can you explain this sum by taking 2.5?

If we take 2.5 or 5/2 will we get the same answer? Anyone explain.

It should be 3x + 8 = 5/2 (x+8).

It should be 3x + 8 = 5/2 (x+8).

The right answer is 3.

At first it is;
1 : 4.

then;
2 : 5.

Here difference after 8 years is 1.
1 part = 8 years.

Again we increase 8 years;
3 : 6.

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

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.