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Difference between revisions of "2009 AMC 10B Problems/Problem 23"

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== Solution ==
 
== Solution ==
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After <math>10</math> minutes <math>(600</math> seconds<math>),</math> Rachel will have completed <math>6</math> laps and be <math>30</math> seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in <math>22.5</math> seconds, she will be in the picture between <math>18.75</math> seconds and <math>41.25</math> seconds of the tenth minute.  After 10 minutes Robert will have completed <math>7</math> laps and will be <math>40</math> seconds past the starting line.  Because Robert runs one-fourth of a lap in <math>20</math> seconds, he will be in the picture between <math>30</math> and <math>50</math> seconds of the tenth minute.  Hence both Rachel and Robert will be in the picture if it is taken between <math>30</math> and <math>41.25</math> seconds of the tenth minute.  So the probability that both runners are in the picture is <math>\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}</math>.  The answer is <math>\mathrm{(C)}</math>.
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After <math>10</math> minutes <math>(600</math> seconds<math>),</math> Rachel will have completed <math>6</math> laps and be <math>30</math> seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in <math>22.5</math> seconds, she will be in the picture between <math>18.75</math> seconds and <math>41.25</math> seconds of the tenth minute.  After 10 minutes, Robert will have completed <math>7</math> laps and be <math>40</math> seconds from completing his eighth lap.  Because Robert runs one-fourth of a lap in <math>20</math> seconds, he will be in the picture between <math>30</math> seconds and <math>50</math> seconds of the tenth minute.  Hence both Rachel and Robert will be in the picture if it is taken between <math>30</math> seconds and <math>41.25</math> seconds of the tenth minute.  So the probability that both runners are in the picture is <math>\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}</math>.  The answer is <math>\mathrm{(C)}</math>.
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==Solution 2 (Video solution)==
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Video: https://youtu.be/eZjJ5MQV47o
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~DaBobWhoLikeMath
  
 
== See also ==
 
== See also ==

Latest revision as of 13:23, 2 July 2021

The following problem is from both the 2009 AMC 10B #23 and 2009 AMC 12B #18, so both problems redirect to this page.

Problem

Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?

$\mathrm{(A)}\frac {1}{16}\qquad \mathrm{(B)}\frac 18\qquad \mathrm{(C)}\frac {3}{16}\qquad \mathrm{(D)}\frac 14\qquad \mathrm{(E)}\frac {5}{16}$

Solution

After $10$ minutes $(600$ seconds$),$ Rachel will have completed $6$ laps and be $30$ seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in $22.5$ seconds, she will be in the picture between $18.75$ seconds and $41.25$ seconds of the tenth minute. After 10 minutes, Robert will have completed $7$ laps and be $40$ seconds from completing his eighth lap. Because Robert runs one-fourth of a lap in $20$ seconds, he will be in the picture between $30$ seconds and $50$ seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between $30$ seconds and $41.25$ seconds of the tenth minute. So the probability that both runners are in the picture is $\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}$. The answer is $\mathrm{(C)}$.

Solution 2 (Video solution)

Video: https://youtu.be/eZjJ5MQV47o ~DaBobWhoLikeMath

See also

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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
2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 12 Problems and Solutions

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