Seven children, each with the same birthday, were born in seven consecutive years. The sum of the ages of the youngest three children is 42 . What is the sum of the ages of the oldest three?
(A) 51
(B) 54
(C) 57
(D) 60
(E) 63
Solution 1

Since the seven children were born in seven consecutive years, then the oldest child is 4 years older than the oldest of the three youngest children, the second oldest child is 4 years older than the second oldest of the three youngest children, and the third oldest child is 4 years older than the youngest child.

Since the sum of the ages of the three youngest children is 42 , then the sum of the ages of the
three oldest children is $42+4+4+4=54$.

## Solution 2

Since the ages of the seven children are seven consecutive integers, let the ages of the youngest three children be $x, x+1$ and $x+2$.

Then $x+x+1+x+2=42$ or $3 x+3=42$ or $x=13$.

So the ages of the seven children are 13, 14, 15, 16, 17, 18, and 19.

Therefore, the sum of the ages of the oldest three children is $17+18+19=54$.

ANSWER: (B)