Question

Three runners start running simultaneously from the same point on a $500$ -meter circular track. They each run clockwise around the course maintaining constant speeds of $4.4, 4.8,$ and $5.0$ meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?

$1,000$

$1,250$

$2,500$

$5,000$

$10,000$

Solution:

Let us find the amount of time that it takes for the runner running at $4.8$ meters per second to lap the second fastest person.

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

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

We now have to find the least time, $t,$ such that $t$ is a multiple of $1250$ and the fastest runner ends up at the starting line.

Every $1250$ seconds, the fastest runner runs $1250 \cdot 5 = 6250$ meters. Then in $2500$ seconds, the fastest runner runs $12500$ meters, which is a whole number of laps.

Thus, C is the correct answer.

2012 AMC 10A Problem 16

Solution:

Problem 16 in Other Years