2010 AMC 10A Problem 8

Each day Tony works $2$ hours at $\$0.50$ per hour per year of age, so a day's pay is $2 \cdot 0.50 \cdot (\text{age}) = \text{age}$ dollars.

Over $50$ days he earned $\$630$, an average of $\dfrac{630}{50} = 12.6$ dollars per day. Since his daily pay is always a whole number, his age must have changed during the period: he was $12$ for some days and $13$ for the rest.

Let $d$ be the number of days he worked at age $13$. Then
$$12(50 - d) + 13d = 630 \quad\Longrightarrow\quad 600 + d = 630 \quad\Longrightarrow\quad d = 30.$$
So he worked $20$ days at age $12$ and $30$ days at age $13$, and his age at the end of the period is $13$.

The answer is $\boxed{\textbf{(D)}\ 13}$.

Simplify the pay rule first: it collapses to "one dollar per year of age per day." When a computed average of integers is not an integer, the underlying value must have changed, and the average tells you which two consecutive values are involved. The question asks for the age at the end, which is the larger one.

• Rounding 12.6 down to 12 (the age at the start of the period) instead of realizing his birthday fell inside the six months.
• Answering $14$ by assuming he must have been older than the average throughout, or misreading the pay rule as $\$0.50$ per day.