Difference between revisions of "2006 AMC 10A Problems/Problem 15"
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== Solution == | == Solution == | ||
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| − | Since <math>d = rt</math>, we note that Odell runs one lap in <math>\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}</math> minutes, while Kershaw runs one lap in <math>\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}</math> minutes. They take the same amount of [[time]] to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are <math>\frac{30}{\frac{2\pi}{5}} \approx 23.8</math> laps run by both, or <math>\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 = 47</math> meeting points <math> \ | + | Since <math>d = rt</math>, we note that Odell runs one lap in <math>\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}</math> minutes, while Kershaw also runs one lap in <math>\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}</math> minutes. They take the same amount of [[time]] to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are <math>\frac{30}{\frac{2\pi}{5}} \approx 23.8</math> laps run by both, or <math>\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 = 47</math> meeting points <math> \Longrightarrow \mathrm{(D)}</math>. |
== See Also == | == See Also == | ||
Revision as of 15:48, 26 February 2007
Problem
Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
Solution
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Since 
, we note that Odell runs one lap in 
minutes, while Kershaw also runs one lap in 
minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are 
laps run by both, or 
meeting points 
.
See Also
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
