8 years ago, the average age of a family of 6 members was 22 years. Two children having been born (with age difference of 3 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:** **(A)** **Explanation:** Total age of 6 members, 8 years ago = (22 x 6) years = 132 years.

Total age of 6 members now = [132 + (8 x 6)] years = 180 years.

Total age of 8 members now = (22 x 8) years = 176 years.

Sum of the ages of 2 children = (180 – 176) years = 4 years.

Let the age of the younger child be x years.

Then, age of the elder child = (x+3) years. So,

=> x+(x+3) = 4

=> x = 1

Age of younger child = 1 year.

Quiz of this Question

## Recommended Posts:

- Algorithms Quiz | SP2 Contest 1 | Question 16
- Algorithms Quiz | SP2 Contest 1 | Question 17
- Algorithms Quiz | SP2 Contest 1 | Question 18
- Algorithms Quiz | SP2 Contest 1 | Question 19
- Algorithms Quiz | SP2 Contest 1 | Question 20
- Algorithms Quiz | SP2 Contest 1 | Question 21
- Algorithms Quiz | SP2 Contest 1 | Question 22
- Algorithms Quiz | SP2 Contest 1 | Question 7
- Algorithms Quiz | SP2 Contest 1 | Question 4
- Algorithms Quiz | SP2 Contest 1 | Question 2
- Algorithms Quiz | SP2 Contest 1 | Question 3
- Algorithms Quiz | SP2 Contest 1 | Question 5
- Algorithms Quiz | SP2 Contest 1 | Question 6
- Algorithms Quiz | SP2 Contest 1 | Question 9
- Algorithms Quiz | SP2 Contest 1 | Question 10
- Algorithms Quiz | SP2 Contest 1 | Question 11
- Algorithms Quiz | SP2 Contest 1 | Question 12
- Algorithms Quiz | SP2 Contest 1 | Question 1
- Algorithms Quiz | SP2 Contest 1 | Question 8
- Algorithms Quiz | SP2 Contest 1 | Question 13