2006 AMC 10A Problem 15

Attempt Problem 15 of the 2006 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

Odell and Kershaw run for 3030 minutes on a circular track. Odell runs clockwise at 250250 m/min and uses the inner lane with a radius of 5050 meters. Kershaw runs counterclockwise at 300300 m/min and uses the outer lane with a radius of 6060 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
2929
4242
4545
4747
5050
Answer: D
Concepts:relative speeddistance rate and timecircumference
Small hint:
Compute each runner’s time for one lap; they turn out equal
Big hint:
Running in opposite directions, they meet twice per lap; count how many laps fit in 3030 minutes
Solution:
Odell’s lap is 2π(50)=100π2\pi(50) = 100\pi m at 250250 m/min, taking 100π250=0.4π\frac{100\pi}{250} = 0.4\pi min. Kershaw’s lap is 2π(60)=120π2\pi(60) = 120\pi m at 300300 m/min, also 120π300=0.4π\frac{120\pi}{300} = 0.4\pi min.
Their periods are equal. Running in opposite directions, they meet at times t=k2(0.4π)t = \frac{k}{2}(0.4\pi) for k=1,2,…k = 1, 2, \ldots Requiring t≤30t \le 30 gives k≤600.4π=150π≈47.7,k \le \frac{60}{0.4\pi} = \frac{150}{\pi} \approx 47.7, so they pass 4747 times.
Thus, the correct answer is D.
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