Aptitude - Problems on Ages - Discussion

Let Father = F and Son = S.

10 Years Ago:

F-10 = 3(S-10).
F = 3S-20 ------ (i).

10 year hence:

F+10 = 2(S+10) ------ (ii).

Replace (i) in (ii).

3S-20+10 = 2S+20.
3S-10 = 2S+20.
S = 30 <----- Son's Age.

Again using (i).

F = 3(30)-20.
F = 70.

So Ratio is 70:30 ----> 7:3.

What is the meaning of fathers age will be twice that of his son.

2(3x+10+10) = (x+10+10).

Why are you taking?

(3x+10+10) = 2((x+10)+10).

Please tell me.

10 years ago means 3x - 10. But how it become 3x + 10?

Father's age is twice of his son's age. 8 years ago. He was thrice as old as his son. So what will be d son's age at present.

@Javeed:

The age is calculating with respect to the present age. So ten years ago means 3x-10. So the present age will b 3x+10.

Now after ten years 3x+10+10.

Thanks to all. Now I got some idea.

@Arshad.

We can calculate from present age either hence or later and also ago, before, after in any age problems.

The age of the son and the father are in the ratio 3:10. If the father have 52 years, how old is the son?

Let's present ages of father and son will be x and y.
Before 10 yearss, it will be (x - 10) = 3(y - 10).

Ten years hence , the equation will be x + 10 = 2(y + 10).
Solve these two equations, we get x = 70 years and y =30 years.

Hi.

Let as consider
Father's age = x.
Son's age = y.

=> 10 year Ago Father's age is x - 10 = 3(y-10) -----> Equation 1.
=> After 10 years Father's age x + 10 = 2(y+10) ------> Equation 2.

Solve 2 equation......
=> x - 10 = 3y - 30.
x - 3y = - 20 --> Equation 3.
=> x - 2y = 10 --> New Equation 4.
=> Substract Equation 3 from Equation 4.
x - 3y = - 20
x - 2y = 10
----------------
- y = - 30.

So, y = 30 --> age of son.

Then put y =30 into any equation.

x - 2(30) = 10
x - 60 = 10.
x = 70 --> age of father.

x = 7.
y = 3.

I hope, this will help you.