Difference between revisions of "2012 AMC 10A Problems/Problem 16"
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<math> \textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000 </math> | <math> \textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000 </math> | ||
− | == Solution == | + | == Solution 1== |
First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let <math>x</math> be the time these runners run in seconds. | First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let <math>x</math> be the time these runners run in seconds. | ||
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Now we must find a time that is a multiple of <math>1250</math> and results in the 5.0 m/s runner to end up on the start line. Every <math>1250</math> seconds, that fastest runner goes <math>5.0(1250)=6250</math> meters. In <math>2(1250)=2500</math> seconds, he goes <math>5.0(2500)=12500</math> meters. Therefore the runners run <math>\boxed{\textbf{(C)}\ 2,500}</math> seconds. | Now we must find a time that is a multiple of <math>1250</math> and results in the 5.0 m/s runner to end up on the start line. Every <math>1250</math> seconds, that fastest runner goes <math>5.0(1250)=6250</math> meters. In <math>2(1250)=2500</math> seconds, he goes <math>5.0(2500)=12500</math> meters. Therefore the runners run <math>\boxed{\textbf{(C)}\ 2,500}</math> seconds. | ||
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+ | == Solution 2== | ||
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+ | Working backwards from the answers starting with the smallest answer, if they had run <math>1000</math> seconds, they would have run <math>4400, 4800, 5000</math> meters, respectively. The first two runners have a difference of <math>400</math> meters, which is not a multiple of <math>500</math> (one lap), so they are not in the same place. | ||
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+ | If they had run <math>1250</math> seconds, the runners would have run <math>5500, 6000, 6250</math> meters, respectively. The last two runners have a difference of <math>250</math> meters, which is not a multiple of <math>500</math>. | ||
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+ | If they had run <math>2500</math> seconds, the runners would have run <math>11000, 12000, 12500</math> meters, respectively. The distance separating each pair of runners is a multiple of <math>500</math>, so the answer is <math>\boxed{\textbf{(C)}\ 2,500}</math> seconds. | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2012|ab=A|num-b=15|num-a=17}} | {{AMC10 box|year=2012|ab=A|num-b=15|num-a=17}} |
Revision as of 00:39, 9 February 2012
Contents
Problem
Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
Solution 1
First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let be the time these runners run in seconds.
Because is a multiple of 5, it turns out they just meet back at the start line.
Now we must find a time that is a multiple of and results in the 5.0 m/s runner to end up on the start line. Every seconds, that fastest runner goes meters. In seconds, he goes meters. Therefore the runners run seconds.
Solution 2
Working backwards from the answers starting with the smallest answer, if they had run seconds, they would have run meters, respectively. The first two runners have a difference of meters, which is not a multiple of (one lap), so they are not in the same place.
If they had run seconds, the runners would have run meters, respectively. The last two runners have a difference of meters, which is not a multiple of .
If they had run seconds, the runners would have run meters, respectively. The distance separating each pair of runners is a multiple of , so the answer is seconds.
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |