Question

10 years ago, the average age of a family of 4 members was 24 years. Two children having been born (with age diference of 2 years), the present average age of the family is the same. The present age of the youngest child is :

A) 1 B) 2 C) 3 D) 4

Answer 1

Let the family members be a,b,c and 10 years ago

So,  a+b+c+d/4= 24

     =a+b+c+d=24x4 = 96......................equation 1

After 10 years, let the two new family members be e and f and let f be younger than e

Age of a= a+10

Age of b= b+10

Age of c= c+10

Age of d= d+10

Age of e= e

Age of f= e-2 (According to Question)

So, new average,

(a+10+b+10+c+10+d+10+e+e-2)/6 =24 (cause it is given that average is same)

= (a+b+c+d) + 10+10+10+10-2+ 2e = 24x 6

= 96 + 38 +2e = 144   ( from equation 1, we know that a+b+c+d= 96)

=2e= 144-96-38

=2e=10

=e=10/2

=e=5

Now, the youngest child was 'f' and 'f' was 2 years younger than 'e'

Since 'e' 5 years old,

'f' will be  e-2

=3 years old

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