The ages of three contestants in the Cayley Contest are 15 years, 9 months; 16 years, 1 month; and 15 years, 8 months. Their average (mean) age is
(A) 15 years, 8 months
(D) 15 years, 11 months
(B) 15 years, 9 months
(E) 16 years
(C) 15 years, 10 months

## Part B: Each correct answer is worth 6.
The ages of three contestants in the Cayley Contest are 15 years, 9 months; 16 years, 1 month; and 15 years, 8 months. Their average (mean) age is
(A) 15 years, 8 months
(B) 15 years, 9 months
(C) 15 years, 10 months
(D) 15 years, 11 months
(E) 16 years

## Solution 1

Consider one of the ages, say the youngest, as a base age. The other two contestants are one month and five months older respectively. Since $\frac{0+1+5}{3}=2$, this implies that the average age is two months greater than the youngest. This gives an average age of 15 years, 10 months.

## Solution 2

This second solution involves a little more calculation but still gives the same correct answer. Since there are twelve months in a year, the age of the first contestant, in months, is $15 \times 12+9$ or 189 months. Similarly, the ages of the other two students would be 193 and 188 months. The average age would thus be $\frac{189+193+188}{3}$ or 190 months. The average age is then 15 years, 10 months because $190=12 \times 15+10$.

ANSWER: (C)

## Part B: Each correct answer is worth 6.