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| − | ==Problem== | + | == Problem == |
| − | Tony works <math>2</math> hours a day and is paid $<math>0.50</math> per hour for each full year of his age. During a six month period Tony worked <math>50</math> days and earned $<math>630</math>. How old was Tony at the end of the six month period?
| + | What is <math>\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)</math>? |
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| − | <math> | + | <math>\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020</math> |
| − | \mathrm{(A)}\ 9 | |
| − | \qquad | |
| − | \mathrm{(B)}\ 11 | |
| − | \qquad | |
| − | \mathrm{(C)}\ 12 | |
| − | \qquad | |
| − | \mathrm{(D)}\ 13 | |
| − | \qquad | |
| − | \mathrm{(E)}\ 14 | |
| − | </math> | |
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| − | ==Solution== | + | == Solution == |
| − | ===Solution 1===
| + | <math>20-2010+201+2010-201+20=20+20=\boxed{\textbf{(C)}\,40}</math>. |
| − | Tony worked <math>2</math> hours a day and is paid <math>0.50</math> 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 <math>12</math> years old, he gets <math>12</math> dollars a day. We also know that he worked <math>50</math> days and earned <math>630</math> dollars. If he was <math>12</math> years old at the beginning of his working period, he would have earned <math>12 * 50 = 600</math> dollars. If he was <math>13</math> years old at the beginning of his working period, he would have earned <math>13 * 50 = 650</math> dollars. Because he earned <math>630</math> dollars, we know that he was <math>13</math> for some period of time, but not the whole time, because then the money earned would be greater than or equal to <math>650</math>. This is why he was <math>12</math> when he began, but turned <math>13</math> sometime in the middle and earned <math>630</math> dollars in total. So the answer is <math>13</math>.The answer is <math>\boxed{D}</math>. We could find out for how long he was <math>12</math> and <math>13</math>. <math>12 \cdot x + 13 \cdot (50-x) = 630</math>. Then <math>x</math> is <math>20</math> and we know that he was <math>12</math> for <math>20</math> days, and <math>13</math> for <math>30</math> days. Thus, the answer is <math>13</math>.
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| − | ===Solution 2=== | + | ==Video Solution== |
| − | Let <math>x</math> equal Tony's age at the end of the period. We know that his age changed during the time period (since <math>630</math> does not evenly divide <math>50</math>). Thus, his age at the beginning of the time period is <math>x - 1</math>.
| + | https://youtu.be/KBlf0TKdI4I?si=XFer3NW2lpMxgjW_ |
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| − | Let <math>d</math> be the number of days Tony worked while his age was <math>x</math>. We know that his earnings every day equal his age (since <math>2 \cdot 0.50 = 1</math>). Thus, <cmath>x \cdot d + (x - 1)(50 - d) = 630</cmath> <cmath>x\cdot d + 50x - x\cdot d - 50 + d = 630</cmath> <cmath>50x + d = 680</cmath> <cmath>x = \dfrac{680 - d}{50}</cmath>
| + | ~Charles3829 |
| − | Since <math>0 < d <50</math>, <math>d = 30</math>. Then we know that <math>50x = 650</math> and <math>x = \boxed{(D) 13}</math>
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| | == See Also == | | == See Also == |
| | + | {{AMC12 box|year=2010|before=First Problem|num-a=2|ab=A}} |
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| − | {{AMC10 box|year=2010|ab=A|num-b=7|num-a=9}}
| + | [[Category:Introductory Algebra Problems]] |
| − | {{AMC12 box|year=2010|ab=A|before=First Problem|num-a=2}}
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| | {{MAA Notice}} | | {{MAA Notice}} |
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