AMC10 2006 A

AMC10 2006 A · Q15

AMC10 2006 A · Q15. It mainly tests Rates (speed), Circle theorems.

Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?

Odell 和 Kershaw 在一个圆形跑道上跑 30 分钟。Odell 顺时针以 250 米/分钟的速度跑，使用半径 50 米的内道。Kershaw 逆时针以 300 米/分钟的速度跑，使用半径 60 米的外道，从与 Odell 相同的径向线开始。起步后他们相遇多少次？

(A) 29 29

(B) 42 42

(D) 47 47

(E) 50 50

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Since Odell's rate is 5/6 that of Kershaw, but Kershaw's lap distance is 6/5 that of Odell, they each run a lap in the same time. Hence they pass twice each time they circle the track. Odell runs $(30\ \text{min})\left(250\ \frac{\text{m}}{\text{min}}\right)\left(\frac{1\ \text{laps}}{100\pi\ \text{m}}\right)=\frac{75}{\pi}\ \text{laps}\approx 23.87\ \text{laps},$ as does Kershaw. Because $23.5<23.87<24$, they pass each other $2(23.5)=47$ times.

（D）由于 Odell 的速度是 Kershaw 的 $\frac{5}{6}$，但 Kershaw 的单圈距离是 Odell 的 $\frac{6}{5}$，所以他们跑完一圈所用时间相同。因此他们每绕跑道一圈会相互超过两次。Odell 跑了 $(30\ \text{min})\left(250\ \frac{\text{m}}{\text{min}}\right)\left(\frac{1\ \text{laps}}{100\pi\ \text{m}}\right)=\frac{75}{\pi}\ \text{laps}\approx 23.87\ \text{laps},$ Kershaw 也是如此。因为 $23.5<23.87<24$，他们相互超过 $2(23.5)=47$ 次。

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