Aptitude - Problems on Ages - Discussion

Discussion Forum : Problems on Ages - General Questions (Q.No. 2)
2.
The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
4 years
8 years
10 years
None of these
Answer: Option
Explanation:

Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

5x = 20

x = 4.

Age of the youngest child = x = 4 years.

Discussion:
99 comments Page 5 of 10.

Durgesh said:   1 decade ago
Avg. of age of 5 children = 50/5 = 10

Now there are 5 no. having sum 50 and avg. as 10 so no. can only be
= (10-6),(10=3),(10),(10+3),(10+6)
=4 , 7 ,10 ,13 ,16

Mitali said:   1 decade ago
I have a other way to solve it.
Avg of 50 = 50/5 = 10.
Hence, the middle No. In the series i.e. x+6=10 (middle no is the avg value of the series).
So x = 10-6= 4.

Vineshgujjari said:   1 decade ago
Since given 5 children=50
let one children be x
given interval 3 each
so x+(x+3)+(x+6)+(x+9)+(x+12)=50
5x+30=50
x=20/5
x=4..

PRATEESH K said:   2 years ago
sum = 50
x,x+3,x+6,x+9,x+12 => eq1.
eq 1 is in arithmetic mean so the average is x+6
average of 5 children is 50/5 = 10,
So, x+6 = 10,
x = 4.
(14)

Prasad Baanu said:   1 decade ago
Hai Nandish..., They have already gave in the question that the sum of ages of those boys are 50:-)
so x + (x+3)+ (x+6)+ (x+9)+(x+12) = 50

Jinju said:   1 decade ago
We can easily understand if the avg is 8 then youngestg should be less thgan 8. Then after we calculate the validity of remaining values.

Akhil said:   7 years ago
According to me;

The youngest children age = x+12.
= 4+12.
= 16.

Basavaraj Marapur said:   10 years ago
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;

Sudheer said:   1 decade ago
Let x+(x-3)+(x-6)+(x-9)+(x-12)=50
5x-30=50
5x=80
x=16
is this correct or not?
if it is wrong then explain the reason..

Ishu Subham said:   1 decade ago
Let the ages be (x-6), (x-3), (x), (x+3), (x+6).
then sum of ages = 5x
=> 5x = 50
x = 10
youngest = x-6
=4


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