34. Fifteen years ago, the average age of a husband and a wife was 20 years. The average remains the same today, even though they now have two children. What is the present age of the youngest child if the children differ in age by 2 years?

(a) 5 years
(b) 6 years
(c) 4 years
(d) 7 years



Answer :

Sure, let's solve this problem step by step.

1. Initial Information 15 Years Ago:
- The average age of the husband and wife 15 years ago was 20 years.
- Therefore, the sum of their ages 15 years ago was [tex]\( 2 \times 20 = 40 \)[/tex] years.

2. Total Age Increase Over 15 Years:
- Each year both the husband and the wife age by 1 year. So, in 15 years, both of them will have aged by [tex]\( 15 \times 2 = 30 \)[/tex] years.
- Therefore, the total of their ages now is [tex]\( 40 + 30 = 70 \)[/tex] years.

3. Current Average Age with Children:
- Currently, the couple plus their two children have the same average age of 20 years.
- There are 4 people in total. Hence, the sum of their current ages is [tex]\( 4 \times 20 = 80 \)[/tex] years.

4. Total Age of the Children:
- We already calculated that the combined age of the husband and wife today is 70 years.
- Therefore, the combined age of their two children is [tex]\( 80 - 70 = 10 \)[/tex] years.

5. Ages of the Children:
- Let the age of the youngest child be [tex]\( x \)[/tex] years.
- Since the children differ in age by 2 years, the age of the oldest child will be [tex]\( x + 2 \)[/tex] years.
- The sum of their ages is given by [tex]\( x + (x + 2) = 10 \)[/tex].

6. Solve for [tex]\( x \)[/tex]:
- Simplify the equation: [tex]\( 2x + 2 = 10 \)[/tex].
- Subtract 2 from both sides: [tex]\( 2x = 8 \)[/tex].
- Divide both sides by 2: [tex]\( x = 4 \)[/tex].

Therefore, the present age of the youngest child is [tex]\( 4 \)[/tex] years.

So the correct answer is:
(c) 4 years