AMC12 2009 B

AMC12 2009 B · Q18

AMC12 2009 B · Q18. It mainly tests Rates (speed), Probability (basic).

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?

Rachel 和 Robert 在一个圆形跑道上跑步。Rachel 逆时针跑，每 90 秒跑完一圈；Robert 顺时针跑，每 80 秒跑完一圈。两人同时从同一起跑线出发。在他们开始跑后的 10 分钟到 11 分钟之间的某个随机时刻，站在跑道内的摄影师拍了一张照片，照片显示跑道的四分之一，且以起跑线为中心。求 Rachel 和 Robert 都在照片中的概率。

(A) $\frac{1}{16}$ $\frac{1}{16}$

(B) $\frac{1}{8}$ $\frac{1}{8}$

(D) $\frac{1}{4}$ $\frac{1}{4}$

(E) $\frac{5}{16}$ $\frac{5}{16}$

Answer

Correct choice: (C)

正确答案：（C）

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 $\mathrm{(C)}$.

10 分钟（$600$ 秒）后，Rachel 已完成 $6$ 圈，并且距离完成第七圈还差 $30$ 秒。由于 Rachel 跑四分之一圈需要 $22.5$ 秒，她会在第十分钟的 $18.75$ 秒到 $41.25$ 秒之间出现在照片中。10 分钟后，Robert 已完成 $7$ 圈，并且距离完成第八圈还差 $40$ 秒。由于 Robert 跑四分之一圈需要 $20$ 秒，他会在第十分钟的 $30$ 秒到 $50$ 秒之间出现在照片中。因此当照片在第十分钟的 $30$ 秒到 $41.25$ 秒之间拍摄时，Rachel 和 Robert 都会在照片中。故两人都在照片中的概率为 $\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}$。答案为 $\mathrm{(C)}$.

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