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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 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 will be 40 seconds past the starting line.  Because Robert runs one-fourth of a lap in 20 seconds, he will be in the picture between 30 and 50 seconds of the tenth minute.  Hence both Rachel and Robert will be in the picture if it is taken between 30 and 41.25 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 ==
 
{{AMC10 box|year=2009|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2009|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2009|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2009|ab=B|num-b=17|num-a=19}}
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 12:23, 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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