Aptitude - Problems on Ages - Discussion

Why are you taking odd ones?

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.

@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).

How we get x + 12?

Explain this step.

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.

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

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

x = 4.

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?

Is there any short cut method please?

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.

Take youngest child = x.

So, (x-12) + (x-9) + (x-6) + (x-3) + x = 50.

5x-30 = 50;

5x = 20;

x = 4.

So youngest child is ---> 20;