Aptitude - Problems on Ages - Discussion

3rd person age is = 50/5 = 10.
So, age of 1st person = 10 - 6 = 4 years.

@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.

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.

x+3+x+6+x+9+x+12=30

5x+30=50

5x=50-30

5x=20

x=4.

How you taken x+3, 6, 9, 12?

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

What 50 indicates here? Please explain.

@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.

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

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

x = 4.

How we get 5 in this?

How you took that values x, x+3, x+6, x+9, x+12.

Please reply.