Difference between revisions of "2006 AMC 10A Problems/Problem 15"

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== Problem ==
 
== 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?  
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Odell and Kershaw run for 30 minutes on a [[circle|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?  
  
 
<math>\mathrm{(A) \ } 29\qquad\mathrm{(B) \ } 42\qquad\mathrm{(C) \ } 45\qquad\mathrm{(D) \ } 47\qquad\mathrm{(E) \ } 50\qquad</math>
 
<math>\mathrm{(A) \ } 29\qquad\mathrm{(B) \ } 42\qquad\mathrm{(C) \ } 45\qquad\mathrm{(D) \ } 47\qquad\mathrm{(E) \ } 50\qquad</math>
 
== Solution ==
 
== Solution ==
{{solution}}
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{{image}}
*please note that this is this solution has not been confirmed but is only my own solution to the problem.  
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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>, while Kershaw runs one lap in <math>\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}</math>. 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>23 \cdot 2 + 1 = 47</math> laps (the additional one since the meet at the half-way point of the last lap) <math> \Rightarrow \mathrm{D}</math>.  
  
 
We know that Odell and Kershaw meet after they run a total of one circle together. Odell runs at 250m/min for 30 mins in a circle of length of 100pi, and that  Kershaw runs at 300m/min for 30 mins in a circle 120pi in length. Both Odell and Kershaw run 23.8 laps around their respective circles. together they run 47.6 laps. Therefore, they meet a total of 47 times.
 
 
(D) 47
 
 
== See Also ==
 
== See Also ==
*[[2006 AMC 10A Problems]]
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{{AMC10 box|year=2006|ab=A|num-b=14|num-a=16}}
 
 
*[[2006 AMC 10A Problems/Problem 14|Previous Problem]]
 
 
 
*[[2006 AMC 10A Problems/Problem 16|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 08:19, 15 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?

$\mathrm{(A) \ } 29\qquad\mathrm{(B) \ } 42\qquad\mathrm{(C) \ } 45\qquad\mathrm{(D) \ } 47\qquad\mathrm{(E) \ } 50\qquad$

Solution


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Since $d = rt$, we note that Odell runs one lap in $\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}$, while Kershaw runs one lap in $\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}$. 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 $\frac{30}{\frac{2\pi}{5}} \approx 23.8$ laps run by both, or $23 \cdot 2 + 1 = 47$ laps (the additional one since the meet at the half-way point of the last lap) $\Rightarrow \mathrm{D}$.

See Also

2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions