2010 AMC 10A Problem 8

Tony worked $2\cdot50=100$ hours, so his average hourly pay was $\$630\div100=\$6.30.$ Because his hourly pay is $\$0.50$ times his age in full years, his average age on the days he worked was $6.30\div0.50=12.6$ years. During a six-month period his age can change by at most one, so he must have worked some days at age $12$ and some at age $13.$ Thus he was $13$ at the end of the period. Thus, D is the correct answer.