2010 AMC 10A Problem 8

First make the pay rule simpler.

1. Pay for one day. He works $2$ hours, and each hour pays $\$0.50$ for every year of age. So one day pays $2 \times 0.50 = 1$ dollar per year of age. In other words, each day he earns his age in dollars. A 12-year-old earns $\$12$ a day.
2. Average day. He earned $\$630$ in $50$ days, which is $630 \div 50 = \$12.60$ per day on average.
3. What that tells us. His daily pay is always a whole number of dollars (his age). If his age had been the same all period, the average would be a whole number. It is $12.6$, between $12$ and $13$, so he must have had a birthday: some days he was $12$ and later days he was $13$.
4. Check it works. Say he worked $d$ days at age $13$ and $50 - d$ days at age $12$. Then
$$12(50 - d) + 13d = 630.$$
The left side is $600 + d$, so $d = 30$. That means $20$ days at age $12$ and $30$ days at age $13$: $20 \times 12 + 30 \times 13 = 240 + 390 = 630$. It matches.
5. The question asks for his age at the end of the period, which is $13$.

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

Turn a wordy pay rule into "dollars per day" before doing anything else; here it became simply "his age." When an average of whole numbers comes out as a decimal, the numbers must have changed along the way, and the average sits between the old and new values. Always re-read whether the question wants the start or the end.