Difference between revisions of "2010 AMC 10A Problems/Problem 8"
(Added Solution 2) |
m (→Solution 3) |
||
(16 intermediate revisions by 6 users not shown) | |||
Line 14: | Line 14: | ||
</math> | </math> | ||
− | + | ==Solution 1== | |
− | + | Tony works <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 during the six month period 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>. | |
− | Tony | ||
− | + | ==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>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> | 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> | 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> | ||
+ | |||
+ | ==Solution 3== | ||
+ | Since Tony worked for <math>50</math> days, he has worked for <math>100</math> hours. Let <math>k</math> be his hourly wage. Then, we know that <math>100k = 630</math>. | ||
+ | |||
+ | Dividing both sides by <math>50</math>, we get that <math>2k</math>(Daily wage, which is equal to his age) <math>\approx</math> <math>12.6</math>. Since we now know that his age was either 12 or 13 during the 6 month span, and we are asked to find his age at the end of this time, the answer is <math>\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}} | {{MAA Notice}} |
Latest revision as of 16:28, 28 July 2023
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 1
Tony works 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 during the six month period 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
Solution 3
Since Tony worked for days, he has worked for
hours. Let
be his hourly wage. Then, we know that
.
Dividing both sides by , we get that
(Daily wage, which is equal to his age)
. Since we now know that his age was either 12 or 13 during the 6 month span, and we are asked to find his age at the end of this time, the answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.