# 2010 AMC 10A Problems/Problem 8

## Problem 8

Tony works $2$ hours a day and is paid $$0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned$$630$. How old was Tony at the end of the six month period?

$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ 14$

## Solution

### Solution 1

Tony worked $2$ hours a day and is paid $0.50$ dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is $12$ years old, he gets $12$ dollars a day. We also know that he worked $50$ days and earned $630$ dollars. If he was $12$ years old at the beginning of his working period, he would have earned $12 * 50 = 600$ dollars. If he was $13$ years old at the beginning of his working period, he would have earned $13 * 50 = 650$ dollars. Because he earned $630$ dollars, we know that he was $13$ for some period of time, but not the whole time, because then the money earned would be greater than or equal to $650$. This is why he was $12$ when he began, but turned $13$ sometime in the middle and earned $630$ dollars in total. So the answer is $13$.The answer is $\boxed{D}$. We could find out for how long he was $12$ and $13$. $12 \cdot x + 13 \cdot (50-x) = 630$. Then $x$ is $20$ and we know that he was $12$ for $20$ days, and $13$ for $30$ days. Thus, the answer is $13$.

### Solution 2

Let $x$ equal Tony's age at the end of the period. We know that his age changed during the time period (since $630$ does not evenly divide $50$). Thus, his age at the beginning of the time period is $x - 1$.

Let $d$ be the number of days Tony worked while his age was $x$. We know that his earnings every day equal his age (since $2 \cdot 0.50 = 1$). Thus, $$x \cdot d + (x - 1)(50 - d) = 630$$ $$x\cdot d + 50x - x\cdot d - 50 + d = 630$$ $$50x + d = 680$$ $$x = \dfrac{680 - d}{50}$$ Since $0 < d <50$, $d = 30$. Then we know that $50x = 650$ and $x = \boxed{(D) 13}$