2012 AMC 10A Problem 16

Two runners on a circular track are at the same point exactly when the distance between them is a whole number of laps. After $t$ seconds the gap between two runners is (difference of speeds) $\times\ t$.

Speeds $4.4$ and $4.8$ differ by $0.4$ m/s, so they coincide when $0.4t$ is a multiple of $500$, i.e. $t$ is a multiple of $1250$.

Speeds $4.8$ and $5.0$ differ by $0.2$ m/s, so they coincide when $0.2t$ is a multiple of $500$, i.e. $t$ is a multiple of $2500$.

(The pair $4.4$ and $5.0$ differ by $0.6$, giving multiples of $\frac{2500}{3}$, but this is automatically satisfied once the other two conditions hold.)

All three are together when $t$ is a common multiple of $1250$ and $2500$; the first such time is $\operatorname{lcm}(1250, 2500) = 2500$ seconds.

The answer is $\boxed{\textbf{(C)}\ 2,500}$.

On a loop, "together" is a statement about relative position, so subtract speeds and work with gaps rather than absolute positions. Each pair produces a period, and the group meets at the least common multiple of the pairwise periods. Only two of the three pairs are needed, since the third gap is the sum of the other two.

Check the choices from the smallest: at $t = 1000$, gaps are $400$ and $200$ m (not multiples of $500$); at $t = 1250$, gaps are $500$ and $250$; at $t = 2500$, gaps are $1000$ and $500$, both multiples of $500$. So $2500$ is the first time that works.

• Finding when each runner returns to the start (multiples of 500/4.4 etc.) instead of when they are together anywhere on the track.
• Stopping at $t = 1250$ after checking only the slowest two runners.