Aptitude - Problems on Ages - Discussion

We can take the youngest child as x.

Hence, 1st child = x.

2nd = x+3.
3rd = x+6.
4th = x+9.
5th = x+12.

Therefore, x + x + 3 + x + 6 + x + 9 + x + 12 = 50.

5x + 30 = 50.

5x = 50 - 30 (remember the sign changes when you transpose or change the side).

5x = 20.

x = 20/5 (remember the sign changes when you transpose or change the side).

x = 4.

So, the age of youngest child = 4 years.

Is there any short cut method please?

If we consider the age of younger child x then after 3 years that child will (x + 3) again 3 years 2nd child born and he become (x + 6) like this up to date the child become (x + 12).

Then the age of younger brother should (x + 12) not only x.

Means answer should be (4 + 12). i.e =16.

Then why not getting 4 as the answer. Can anyone explain it?

5x + (3 + 6 + 9 + 12) = 50.

5x + 30 = 50.
5x = 50 - 30.
5x = 20.

x = 4.

Average/total number.

A + B + C + D + E/5 = 50.
50/5 = 10.
C = 10.
10 - 3 = 7.

7 - 3 = 4 -> YOUNGEST ONE'S AGE IS 4yrs.

How we get x + 12?

Explain this step.

@Nikil,

The sum of ages of 5 children and their difference is 3 years.

We consider that 1st children age be "x".

Then the second one be "x + 3".

3rd one be"x + 6".

4th one be "x + 9".

5th one be "x + 12" (because they all are born with three years difference).

Hi guys,

The best easy way to solve this type of question : You can go as per options and we can reduce the time consumption.

Go as per 1st option : for 1st child if 4 years then 2nd child 4+3 =7 years, Similarly 3rd child 10 years, 4th child 13 years and 5th child 16 years.
Then,
Add all the years if u get total 50 years (given in question) you decide your answer is correct. No need to calculate we can reduce our time for this type of easy question.

Why are you taking odd ones?

Can anybody solve the problem? Please.

Three times the present age of a father is equal to eight times the present of his son. Eight years hence the father will twice as old as his son at that time. What are their present age?