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>
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== Solution ==
 
== Solution ==
== See Also ==
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<center><asy>
*[[2006 AMC 10A Problems]]
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draw((5,0){up}..{left}(0,5),red);
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draw((-5,0){up}..{right}(0,5),red);
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draw((5,0){down}..{left}(0,-5),red);
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draw((-5,0){down}..{right}(0,-5),red);
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draw((6,0){up}..{left}(0,6),blue);
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draw((-6,0){up}..{right}(0,6),blue);
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draw((6,0){down}..{left}(0,-6),blue);
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draw((-6,0){down}..{right}(0,-6),blue);
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</asy></center>
  
*[[2006 AMC 10A Problems/Problem 13|Previous 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> 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>.
  
*[[2006 AMC 10A Problems/Problem 15|Next Problem]]
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== See Also ==
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{{AMC10 box|year=2006|ab=A|num-b=14|num-a=16}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Revision as of 20:22, 7 September 2020

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

[asy] draw((5,0){up}..{left}(0,5),red); draw((-5,0){up}..{right}(0,5),red); draw((5,0){down}..{left}(0,-5),red); draw((-5,0){down}..{right}(0,-5),red); draw((6,0){up}..{left}(0,6),blue); draw((-6,0){up}..{right}(0,6),blue); draw((6,0){down}..{left}(0,-6),blue); draw((-6,0){down}..{right}(0,-6),blue); [/asy]

Since $d = rt$, we note that Odell runs one lap in $\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}$ minutes, while Kershaw also runs one lap in $\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}$ 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 $\frac{30}{\frac{2\pi}{5}} \approx 23.8$ laps run by both, or $\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 = 47$ meeting points $\Longrightarrow \mathrm{(D)}$.

See Also

2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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