# 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?

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

5*x* = 20

*x* = 4.

Age of the youngest child = *x* = 4 years.

Discussion:

98 comments Page 1 of 10.
Dilliganesh said:
3 weeks 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.

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.

Giri said:
3 months ago

Total sum 50.

Total children 5.

So, 50/5 =10.

Interval each 3,

So, the youngest age 4.

Total children 5.

So, 50/5 =10.

Interval each 3,

So, the youngest age 4.

(2)

PRATEESH K said:
5 months 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.

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.

(7)

Abhishek said:
6 months ago

Short trick.

5 children born at interval of 3 years in 50 years.

1st child. 1+3 = 4 year.

2nd child. 2+3 = 5 year.

3rd child. 3+3 = 6 year.

4th child. 4+3 = 7 year.

5th child 5+3 = 8 year.

=>who is youngest child of all above this?

=> Ans= 4 years.

5 children born at interval of 3 years in 50 years.

1st child. 1+3 = 4 year.

2nd child. 2+3 = 5 year.

3rd child. 3+3 = 6 year.

4th child. 4+3 = 7 year.

5th child 5+3 = 8 year.

=>who is youngest child of all above this?

=> Ans= 4 years.

(38)

Yahaya said:
6 months ago

Let the ages of the 5 children be a, b, c, d, e.

Intervals of 3

i.e if a=3,

b= a+3,

c= a+3+3,

d=a+3+3+3 and e= a+3+3+3+3.

Therefore their sum becomes:

a+a+3+a+3+3+a+3+3+3+a+3+3+3+3 = 50

Then;

5a + 3(10 times) = 50,

5a + 30 = 50.

5a = 50 - 30.

5a =20.

Divide both sides by 5.

5a/5 = 20/5,

a = 4 years.

Intervals of 3

i.e if a=3,

b= a+3,

c= a+3+3,

d=a+3+3+3 and e= a+3+3+3+3.

Therefore their sum becomes:

a+a+3+a+3+3+a+3+3+3+a+3+3+3+3 = 50

Then;

5a + 3(10 times) = 50,

5a + 30 = 50.

5a = 50 - 30.

5a =20.

Divide both sides by 5.

5a/5 = 20/5,

a = 4 years.

(3)

V Seddhartha said:
7 months 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.

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)

Raju Sampang Ample said:
9 months ago

Let the X be the first or Oldest Child born to Uncle “Bhokal”.

Since All the child is born at 3 years later than their elder Siblings

Therefore sequence will be

X( eldest Child)

X-3

X-6

X-9

X-12( youngest Child)

So, finding X value does not mean the youngest child rather it’s the age of the eldest child

Now, as per the question,

Total age of Children = 50

X+X-3+X-6+X-9+X-12=50

Total X’s + Total Numbers= 50

5X-30=50

5X= 50+30

X=80/5

X=16 (Eldest Son Age)

Since Youngest Child came 12 years later than his /her eldest sibling.

So youngest child= X-12= 16-12=4 years

Since All the child is born at 3 years later than their elder Siblings

Therefore sequence will be

X( eldest Child)

X-3

X-6

X-9

X-12( youngest Child)

So, finding X value does not mean the youngest child rather it’s the age of the eldest child

Now, as per the question,

Total age of Children = 50

X+X-3+X-6+X-9+X-12=50

Total X’s + Total Numbers= 50

5X-30=50

5X= 50+30

X=80/5

X=16 (Eldest Son Age)

Since Youngest Child came 12 years later than his /her eldest sibling.

So youngest child= X-12= 16-12=4 years

(8)

Alisha Mugale said:
1 year ago

Consider A, B, C, D, E are 5 children.

Let the lowest value as 4 for the youngest child.

Start from 4 and add the difference of 3 to each value.

4 + 7+ 10+ 13+ 16 = 50.

Let the lowest value as 4 for the youngest child.

Start from 4 and add the difference of 3 to each value.

4 + 7+ 10+ 13+ 16 = 50.

(27)

Arpitha said:
1 year ago

Here, How come 30? Please explain me.

(9)

Gopinath said:
1 year ago

First, you notice every 5 children birth in 3 years cap in 50 years what's is the age in the youngest child.

X, (x+3), (x+6), (x+9), (x+12) years.

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

Then, here it's using the method in adding x in out 5X +30 = 50.

5X= 50-30.

5X= 20, divisible by 5,

X= 4.

X, (x+3), (x+6), (x+9), (x+12) years.

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

Then, here it's using the method in adding x in out 5X +30 = 50.

5X= 50-30.

5X= 20, divisible by 5,

X= 4.

(26)

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