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2009 AMC 12B Problem 18

Problem 18 of 25IntermediateAlgebraCounting & Probability

Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 9090 seconds, and Robert runs clockwise and completes a lap every 8080 seconds. Both start from the start line at the same time. At some random time between 1010 minutes and 1111 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?

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Solution

The picture covers the arc within 18\dfrac{1}{8} lap of the start on each side. At 600600 s Rachel has run 6236\tfrac{2}{3} laps, 3030 s short of the line; a quarter lap takes her 22.522.5 s, so she is in view between 3011.25=18.7530 - 11.25 = 18.75 s and 30+11.25=41.2530 + 11.25 = 41.25 s of the 1010th minute. At 600600 s Robert is 4040 s from the line; a quarter lap takes 2020 s, so he is in view between 3030 and 5050 s. Both appear between 3030 and 41.2541.25 s, a window of 11.2511.25 s out of 60,60, giving probability 11.2560=316.\dfrac{11.25}{60} = \dfrac{3}{16}. Thus, the correct answer is C.

More practice

Concepts: geometric probability · distance rate and time

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.