Competition · AMC preparation · step 4 of 4

AMC 10 · 2021A · #6

Grade 6 rate-ratio
Archetype Meeting Time with Piecewise Rates └ Sub-archetype Piecewise-Leg Time Balance
ratefraction-arithmeticratio-proportion identify-subproblemsdimensional-analysis ↑ Prerequisites: rate
📏 Medium solution 💡 2 insights
Problem
Chantal and Jean start together at the trailhead. Chantal hikes at 4 mph to the halfway point, then 2 mph up the steep top half to the tower, then turns around and comes back down the steep half at 3 mph. She meets Jean exactly at the halfway point. During that same time, Jean has been hiking from the trailhead to the halfway point at a constant pace. What was Jean's average speed in miles per hour?

Pick an answer.

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

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Chantal's trip splits naturally into three legs at three different speeds — Tool #7 (Identify Subproblems) handles each leg separately, then sums the times. Tool #8 (Analyze the Units) keeps the rate relation time = distance / speed honest. A quick number-line sketch (Tool #1) makes it obvious that Jean has covered only d miles while Chantal has covered 3d miles in the same time.

1STEP 1

Mark the three points on a line

Sketch trailhead, halfway, and tower a distance d apart: Chantal walks three legs of d; Jean walks one leg of d in the same time.

trailhead → half → tower → half
2STEP 2

Write each leg's time

Use time = distance ÷ speed on each leg; the three times come out to d4\frac{d}{4}, d2\frac{d}{2}, and d3\frac{d}{3} hours.

t₁ = d/4, t₂ = d/2, t₃ = d/3
3STEP 3

Add the three leg times

Add the three times over common denominator 12: d4\frac{d}{4} + d2\frac{d}{2} + d3\frac{d}{3} = 13d12\frac{13d}{12} hours — the total elapsed time.

T = d/4 + d/2 + d/3 = (3d+6d+4d)/12 = 13d/12
4STEP 4

Find Jean's speed

Jean covered d in that same time, so his speed is d ÷ (13d12\frac{13d}{12}) = 1213\frac{12}{13} mph — the d cancels, so trail length never matters.

s_J = d/( 13d/12 ) = 12/13 mph
5STEP 5

Match against the choices

That simplified fraction, 1213\frac{12}{13}, is answer choice (A).

12/13 → (A)
Answer
12/13
Sanity check the size of the answer. Chantal walked 3d miles in time T while Jean walked d miles in the same time, so Jean's average speed must be one-third of Chantal's average speed. Chantal's average speed is 3d13d12\frac{3d}{\frac{13d}{12}} = 3613\frac{36}{13} mph, so Jean's is 1213\frac{12}{13} mph — matches. The answer is less than 1 mph, which fits the story (heavy backpack, slower than every one of Chantal's legs).
💡Key takeaway

This AMC 10 problem only needs Grade 6 rate reasoning you already know — time = distance / speed for each of Chantal's three legs, add the times with a common denominator 12 to get 13d12\frac{13d}{12} hours, then Jean's speed is d13d12\frac{d}{\frac{13d}{12}} = 1213\frac{12}{13} mph.

  • Mark the three points on a line
  • Write each leg's time
  • Add the three leg times
  • Find Jean's speed
  • Match against the choices

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