Difference between revisions of "2006 AMC 10A Problems/Problem 15"
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== Problem == | == Problem == | ||
− | Odell and Kershaw run for <math>30</math> minutes on a [[circle|circular]] track. Odell runs clockwise at <math>250 m/min</math> and uses the inner lane with a radius of <math>50</math> meters. Kershaw runs counterclockwise at <math>300 m/min</math> and uses the outer lane with a radius of <math> | + | Odell and Kershaw run for <math>30</math> minutes on a [[circle|circular]] track. Odell runs clockwise at <math>250 m/min</math> and uses the inner lane with a radius of <math>50</math> meters. Kershaw runs counterclockwise at <math>300 m/min</math> and uses the outer lane with a radius of <math>60</math> meters, starting on the same radial line as Odell. How many times after the start do they pass each other? |
<math>\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad</math> | <math>\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad</math> | ||
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~apsid | ~apsid | ||
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+ | == Solution 3 by Alcumus == | ||
+ | Since Odell's rate is <math>5/6</math> that of Kershaw, but Kershaw's lap distance is <math>6/5</math> that of Odell, they each run a lap in the same time. Hence they pass twice each time they circle the track. Odell runs <cmath>(30 \ \text{min})\left(250\frac{\text{m}}{\text{min}}\right)\left(\frac{1}{100\pi}\frac{\text{laps}}{\text{m}}\right)= \frac{75}{\pi}\,\text{laps}\approx 23.87\ \text{laps},</cmath> as does Kershaw. Because <math>23.5 < 23.87 < 24</math>, they pass each other <math>2(23.5)=\boxed{47}</math> times. | ||
== See Also == | == See Also == |
Latest revision as of 17:49, 13 August 2023
Problem
Odell and Kershaw run for minutes on a circular track. Odell runs clockwise at
and uses the inner lane with a radius of
meters. Kershaw runs counterclockwise at
and uses the outer lane with a radius of
meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
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]](http://latex.artofproblemsolving.com/1/e/d/1ed90d93483a84bc0ef2a6c44c0257bb56aecc19.png)
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.
Solution 2
We first find the amount of minutes, , until Odell and Kershaw's next meeting. Let
be the angle in radians between their starting point and the point where they first meet, measured counterclockwise.
Since Kershaw has traveled meters at this point and the circumference of his track is
,
. Similarly,
since Odell has traveled
meters in the opposite direction and the circumference of his track is
.
Solving for in the second equation, we get
. Then, from the first equation, we have
. Solving for
, we get
. After
minutes, they are back at the same position, except rotated, so they will meet again in
minutes. So the total amount of meetings is
.
~apsid
Solution 3 by Alcumus
Since Odell's rate is that of Kershaw, but Kershaw's lap distance is
that of Odell, they each run a lap in the same time. Hence they pass twice each time they circle the track. Odell runs
as does Kershaw. Because
, they pass each other
times.
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 |
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