Difference between revisions of "2009 AMC 10B Problems/Problem 23"
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== Solution == | == Solution == | ||
| − | 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>. | + | 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>. |
==Solution 2 (Video solution)== | ==Solution 2 (Video solution)== | ||
Revision as of 12:13, 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?
Solution
After 
minutes 
seconds
Rachel will have completed 
laps and be 
seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in 
seconds, she will be in the picture between 
seconds and 
seconds of the tenth minute. After 10 minutes, Robert will have completed 
laps and will be 
seconds past the starting line. Because Robert runs one-fourth of a lap in 
seconds, he will be in the picture between and 
seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between and seconds of the tenth minute. So the probability that both runners are in the picture is 
. The answer is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
