2012 AMC 10A Problem 16

Attempt Problem 16 of the 2012 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 2012 AMC 10A solutions, or check the answer key.

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16.

Three runners start running simultaneously from the same point on a 500500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4,4.8,4.4, 4.8, and 5.05.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?

1,0001,000

1,2501,250

2,5002,500

5,0005,000

10,00010,000

Answer: C
Concepts:relative speedleast common multipledistance rate and time
Difficulty rating: 1600
Solution:

Let us find the amount of time that it takes for the runner running at 4.84.8 meters per second to lap the slowest person.

We must have that 4.8x4.4x=500 4.8x - 4.4x = 500 x=1250, x = 1250, where xx is the amount of time it takes for the faster runner to lap the other.

Note that 4.41250=5500,4.4 \cdot 1250 = 5500, which means that these two runners always intersect at the starting line.

We now have to find the least time, t,t, such that tt is a multiple of 12501250 and the fastest runner ends up at the starting line.

Every 12501250 seconds, the fastest runner runs 12505=62501250 \cdot 5 = 6250 meters. Then in 25002500 seconds, the fastest runner runs 1250012500 meters, which is a whole number of laps.

Thus, C is the correct answer.

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