Difference between revisions of "2010 AMC 10A Problems/Problem 8"

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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 $x = \boxed{(D) 13}$

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 10 Problems and Solutions

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AMC10 box|year=2010|ab=A|num-b=7|num-a=9}}
+{{AMC12 box|year=2010|ab=A|before=First Question|num-a=2}}
 {{MAA Notice}}

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