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:
102 comments Page 2 of 11.

Jawahir Saly said:   4 years ago
3rd person age is = 50/5 = 10.
So, age of 1st person = 10 - 6 = 4 years.
(7)

Dilliganesh said:   2 years ago
@All.

By using using Option Verification method.

Option: A)4.

Step1: consider young child=4
Step2: Try to confirm.

Add the numbers of 5 children in the difference of 3.
child1= (1,2,3) 3 times differ and age is 4
like that, we do all the steps.

Child2= 8, child3= 12, child4= 16, child5= 20.
add=4+8+12+16+20=60.
split 3 times=20+20+20=60.
The difference is =20.
so we have to divide = 20/5=4.
Answer is 4.

Take option: B)8.

Add: 8 + 12 + 16 + 20 + 24 = 80,
can't split 80 in 3 times.
So this option is wrong.

option: C)10
Add: 10+14+18+22+26=90
Split 3 times: 30+30+30=90
The difference is = 30
So we have to divide = 30/5 = 6.
5 is nothing but it is many children.
But in the option, there is no value 6.

So, Finally: option A)4 is correct.
(6)

Kartikey sharma said:   1 month ago
Given:

5 children.
Age gap = 3 years.
Total sum of ages = 50 years.
Assume Youngest Age = x.

Since each child is 3 years older than the previous one:
Ages will be:

x,
x + 3,
x + 6,
x + 9,
x + 12.

From the equation;
Sum = 50.

x(x+3) + (x+6) + (x+9) + (x+12) = 50
5x + (3+6+9+12) = 50.
5x + 30 = 50.
5x = 50 - 30.
5x = 20,
x = 20/5.
x = 4.
Youngest child’s age = 4 years.
(4)

Martin said:   2 decades ago
x+3+x+6+x+9+x+12=30

5x+30=50

5x=50-30

5x=20

x=4.
(1)

MATHAN said:   10 years ago
5x + (3 + 6 + 9 + 12) = 50.

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

x = 4.
(1)

Hariharan said:   6 years ago
@Sonam.

let youngest son Tashi age be x and father's age be y age of Dema be a, And the age of Sangay be b.

y=x+30.
a=x+2.
b=a+3.
a+b+x=52.
x+2+a+3+x=52.
x+2+x+2+3+x=52.
3x+7=52.
3x=45.
x=15.
y=15+30.
y=45.
The age of the father is 45.
(1)

Hari said:   4 years ago
What 50 indicates here? Please explain.
(1)

Prakash said:   4 years ago
How you taken x+3, 6, 9, 12?

Instead we can take x+1, x+2, x+3, x+4, x+5.
(1)

V Seddhartha said:   3 years ago
Let the age of the youngest child is a.

Then the ages of other children are a+3, a+6, a+6, a+9, a+12.

By this first term is a.
The Common difference is 3.

No of the terms is 5.
We know that sum of n terms in AP.
n/2[2a+(n-1)d].
5/2[2a+(4)d] = 50
5/2[2a+12] = 50
5[a+6] = 50
a+6 = 50/5
a+6 = 10
a = 4.

The Age of the youngest child is 4.
(1)

Kartikey sharma said:   1 month ago
Now let’s use your formula:

Youngest = (S-d⋅(n(n-1))/2)/n.
Where:
S = 50.
d = 3.
n = 5.
(50-3⋅(5(4))/2)/5.
(50-3⋅10)/5.
(50-30)/5.
20/5 = 4.
So, the answer is 4.
(1)


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