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2021 AMC 10A Problem 6

2021 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: October 28, 2026

Linear Equations & Word Problems · difficulty 2 of 5 · about 3 min · position P1-10

Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at $4$ miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to $2$ miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at $3$ miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?

A) $\frac{12}{13}$
B) $1$
C) $\frac{13}{12}$
D) $\frac{24}{13}$
E) $2$

Answer

Key insight

Chantal and Jean walk for the same amount of time. Pick an easy halfway distance, such as 12 miles, and add up Chantal's three walking times.

Solution

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

Why this works

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.

Alternative approach

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.

The trap

Averaging Chantal's three speeds, or forgetting that Jean only walks to the halfway point, not all the way to the tower.

Concise solution

Let the halfway point be $d$ miles from the trailhead, so the tower is $2d$ miles away.

Chantal's trip until the meeting has three legs, each of length $d$ :
- trailhead to halfway at $4$ mph: $\frac{d}{4}$ hours;
- halfway to tower at $2$ mph: $\frac{d}{2}$ hours;
- tower back to halfway at $3$ mph: $\frac{d}{3}$ hours.

Total time: $\frac{d}{4} + \frac{d}{2} + \frac{d}{3} = \frac{3d + 6d + 4d}{12} = \frac{13d}{12}$ hours.

In that same time Jean walks from the trailhead to the halfway point, a distance of $d$ . His average speed is
$\frac{d}{13d/12} = \frac{12}{13}\text{ mph}.$

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

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)
Introduction to Algebra, Chapter 7: Proportion (§7.4)
The Art of Problem Solving, Volume 1: the Basics, Chapter 6: Quadratic Equations (§6.4)

Techniques

Set up the equation/formula and compute; no special trick needed · Substitute to simplify (u = x+1/x, shifting, scaling)