AMC 10 · 2021 · #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)

(C)

$~\frac{13}{12}$

(D)

$~\frac{24}{13}$

(E)

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

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

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

t₁ =

\frac{d}{4}

, t₂ =

\frac{d}{2}

, t₃ =

\frac{d}{3}

3STEP 3

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

T =

\frac{d}{4}

\frac{d}{2}

\frac{d}{3}

\frac{3d+6d+4d}{12}

\frac{13d}{12}

4STEP 4

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

s_J =

\frac{d}{\frac{13d}{12}}

\frac{12}{13}

mph

5STEP 5

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

\frac{12}{13}

→ (A)

Answer

\frac{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

\frac{3d}{\frac{13d}{12}}

\frac{36}{13}

mph, so Jean's is

\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 $\frac{13d}{12}$ hours, then Jean's speed is $\frac{d}{\frac{13d}{12}}$ = $\frac{12}{13}$ mph.

More like this