2021 AMC 10A Problem 6

1. They start together and meet at the same moment, so they walk for exactly the same amount of time. If we find Chantal's time, we know Jean's.
2. The problem never says how long the trail is, so the answer cannot depend on it. Choose a friendly number: let the halfway point be $12$ miles from the trailhead. ($12$ divides evenly by the speeds $4$, $2$ and $3$.)
3. Chantal's walk until they meet has three parts, each $12$ miles long:
- trailhead to halfway at $4$ mph: $12 \div 4 = 3$ hours;
- halfway up to the tower at $2$ mph: $12 \div 2 = 6$ hours;
- tower back down to halfway at $3$ mph: $12 \div 3 = 4$ hours.
4. Her total time is $3 + 6 + 4 = 13$ hours.
5. In those same $13$ hours Jean walks from the trailhead to the halfway point, which is $12$ miles.
6. Jean's average speed is $12 \div 13 = \frac{12}{13}$ mph.

The answer is $\boxed{\textbf{(A)}\ \frac{12}{13}}$.

Average speed is total distance divided by total time, never the average of the speeds. The shared walking time links the two hikers, so the whole problem is finding Chantal's time. When a length is not given, choosing a convenient value (here a number every speed divides) keeps the arithmetic free of fractions.

Use a letter instead of $12$. If the halfway point is $d$ miles away, Chantal's time is
$$\frac{d}{4} + \frac{d}{2} + \frac{d}{3} = \frac{3d + 6d + 4d}{12} = \frac{13d}{12} \text{ hours},$$
and Jean covers $d$ miles in that time, so his speed is $d \div \frac{13d}{12} = \frac{12}{13}$ mph. The $d$ cancels, as expected.