Aptitude - Problems on Ages - Discussion

Son age is "x".
Father age is "3x".

10 year hence (after).
son age will become "x+10".
father age will become "3x+10" but father age is twice of son's age So it will become 2(x+10).


So, the equation will be.
(3x+10) = 2(x+10),
3x+10=2x+20,
3x-2x=20-10.
x=10.

For 10 years after age we need to add 10 in x (the most important).
x=10+10.
x=20.
Now, substitute the value of x in the x+10 equation.
Son age will become "x+10" => 20+10 = 30.
Father age will become "3x+10" => 3*20+10 = 70.

70/30 = 7/3 = 7:3.

@All.

Here's my explanation.

Let's take the present age of
Son= s
Father =f
10 year before,
son= s-10.
father= f-10.
Equation i
3(s-10)=f-10
3s-f=20 --> (i)

After 10 year;
son=s+10
father= f+ 10
equation (ii)
2(s+10) = f+10
2s-f = -10 --> i)
By equation;
S = 30
F = 70.
So, the ratio = 7:3.

Father's age ten years ago: (3x + 10),
Ten years hence: (2x + 10),
Let son be x years:

(x+10) = (2x + 10),
x = 20.
3x+10 : x+10 = 70 : 30 = 7 : 3.

Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10],
3x + 20 = 2x + 40.
x = 20..
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

I think the answer is incorrect because the year is in ago so, we would subtract the 3x-10 but there is 3x+10.
How it is possible?

Good session very clear explanations. Thanks all.

The son age is x.
10 years ago, father is age was 3x-10.
After 10 years, father's age is 2x+10.
then,
3x-10 = 2x + 10.
x = 20.
The required ratio is = (3x+10):(x+10).
= 70:30,
= 7:3.