Difference between revisions of "2010 AMC 10A Problems/Problem 8"
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==Solution== | ==Solution== | ||
+ | ===Solution 1=== | ||
+ | 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>. | ||
− | + | ===Solution 2=== | |
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | 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> | ||
+ | ==Video Solution== | ||
+ | https://youtu.be/P7rGLXp_6es?t=244 | ||
+ | ~IceMatrix | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2010|ab=A|num-b=7|num-a=9}} | {{AMC10 box|year=2010|ab=A|num-b=7|num-a=9}} | ||
+ | {{AMC12 box|year=2010|ab=A|before=First Problem|num-a=2}} | ||
+ | {{MAA Notice}} |
Latest revision as of 06:41, 26 May 2020
Problem 8
Tony works hours a day and is paid $ per hour for each full year of his age. During a six month period Tony worked days and earned $. How old was Tony at the end of the six month period?
Solution
Solution 1
Tony worked hours a day and is paid 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 years old, he gets dollars a day. We also know that he worked days and earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. Because he earned dollars, we know that he was for some period of time, but not the whole time, because then the money earned would be greater than or equal to . This is why he was when he began, but turned sometime in the middle and earned dollars in total. So the answer is .The answer is . We could find out for how long he was and . . Then is and we know that he was for days, and for days. Thus, the answer is .
Solution 2
Let equal Tony's age at the end of the period. We know that his age changed during the time period (since does not evenly divide ). Thus, his age at the beginning of the time period is .
Let be the number of days Tony worked while his age was . We know that his earnings every day equal his age (since ). Thus, Since , . Then we know that and
Video Solution
https://youtu.be/P7rGLXp_6es?t=244
~IceMatrix
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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