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Difference between revisions of "2010 AMC 12A Problems/Problem 1"

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== Problem ==
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Tony works <math>2</math> hours a day and is paid &#36;<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 &#36;<math>630</math>. How old was Tony at the end of the six month period?
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What is <math>\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)</math>?
 
  
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<math>\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020</math>
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<math>
 +
\mathrm{(A)}\ 9
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\qquad
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\mathrm{(B)}\ 11
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\qquad
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\mathrm{(C)}\ 12
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\qquad
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\mathrm{(D)}\ 13
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\qquad
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\mathrm{(E)}\ 14
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</math>
  
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== Solution ==
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==Solution==
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<math>20-2010+201+2010-201+20=20+20=\boxed{\textbf{(C)}\,40}</math>.
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===Solution 1===
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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===
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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>.
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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>
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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>
  
 
== See Also ==
 
== See Also ==
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{{AMC12 box|year=2010|before=First Problem|num-a=2|ab=A}}
 
  
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[[Category:Introductory Algebra Problems]]
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{{AMC10 box|year=2010|ab=A|num-b=7|num-a=9}}
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{{AMC12 box|year=2010|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:18, 10 May 2020

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

See Also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 10 Problems and Solutions
2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
First Problem 		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

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