Aptitude - Problems on Ages - Discussion

it is in arithmetic progression with a common difference of 3.

So,
Sn = n/2 [2a + (n-1) d],
50 = 5/2 [2a + (5-1)3],
10 = 1/2[2a + 12],
20 = 2a + 12,
8 = 2a.
a = 4.
Where a is the first term or the age of the youngest child.

@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.

Total sum 50.
Total children 5.
So, 50/5 =10.
Interval each 3,
So, the youngest age 4.

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.

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.

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.

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.

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

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.

Here, How come 30? Please explain me.