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 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?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Let $d$ = distance from trailhead to halfway point; total trail length $= 2d$; Chantal: $4$ mph for first $d$, $2$ mph for next $d$ (up steep), $3$ mph for last $d$ (down steep); Jean: covers distance $d$ at constant unknown speed; They start at the same moment and meet at the halfway point; Choices: (A) $\frac{12}{13}$, (B) $1$, (C) $\frac{13}{12}$, (D) $\frac{24}{13}$, (E) $2$

Unknowns: Jean's average speed in mph until they meet

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #8 Analyze the Units, #1 Draw a Diagram

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 $\text{time} = \text{distance} / \text{speed}$ 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.

Execute — Answer: A

#1 Draw a Diagram 6.RP.A.3 Step 1

Mark the trailhead, the halfway point, and the tower on a number line, distance $d$ apart.
Chantal walks $d$ at $4$ mph (flat), $d$ at $2$ mph (steep up), $d$ at $3$ mph (steep down), landing at the halfway point.
Jean walks $d$ at an unknown pace.
Same time elapsed.

$$\text{trailhead} \xrightarrow{d,\;4} \text{half} \xrightarrow{d,\;2} \text{tower} \xrightarrow{d,\;3} \text{half}$$

💡 A picture turns the wordy trip into three labeled segments and one segment for Jean.

#8 Analyze the Units 6.RP.A.2 Step 2

For each leg use $\text{time} = \dfrac{\text{distance}}{\text{speed}}$.
Leg 1: $\dfrac{d}{4}$ h.
Leg 2: $\dfrac{d}{2}$ h.
Leg 3: $\dfrac{d}{3}$ h.
Units check: miles $\div$ (miles/hour) $=$ hours.

$$t_1 = \dfrac{d}{4},\;\; t_2 = \dfrac{d}{2},\;\; t_3 = \dfrac{d}{3}$$

💡 Unit rate $r = d/t$ inverts cleanly to $t = d/r$ — Grade 6 rate reasoning.

#7 Identify Subproblems 5.NF.A.1 Step 3

Add Chantal's three leg times to get the total time elapsed when they meet.
Common denominator $12$: $\dfrac{d}{4} = \dfrac{3d}{12},\;\dfrac{d}{2} = \dfrac{6d}{12},\;\dfrac{d}{3} = \dfrac{4d}{12}$.
Sum $= \dfrac{13d}{12}$ hours.

$$T = \dfrac{d}{4} + \dfrac{d}{2} + \dfrac{d}{3} = \dfrac{3d+6d+4d}{12} = \dfrac{13d}{12}$$

💡 Same denominator first, then add — Grade 5 fraction addition.

#7 Identify Subproblems 6.RP.A.3 Step 4

Jean covered distance $d$ in the same time $T = \dfrac{13d}{12}$.
So Jean's average speed is $\dfrac{d}{T} = d \cdot \dfrac{12}{13d} = \dfrac{12}{13}$ mph.
The $d$ cancels — the answer does not depend on the actual length of the trail.

$$s_J = \dfrac{d}{\;13d/12\;} = \dfrac{12}{13}\;\text{mph}$$

💡 Dividing by a fraction = multiplying by its reciprocal; the unknown distance cancels.

#7 Identify Subproblems 4.NF.A.2 Step 5

Match to answer choices: $\dfrac{12}{13}$ is choice (A).

$$\dfrac{12}{13}\;\Rightarrow\;\textbf{(A)}$$

💡 Read the simplified fraction off the list of options.

[1] #1 6.RP.A.3 Mark the trailhead, the halfway point, and the tower on a number line, distance

[2] #8 6.RP.A.2 For each leg use $\text{time} = \dfrac{\text{distance}}{\text{speed}}$. Leg 1: $

[3] #7 5.NF.A.1 Add Chantal's three leg times to get the total time elapsed when they meet. Comm

[4] #7 6.RP.A.3 Jean covered distance $d$ in the same time $T = \dfrac{13d}{12}$. So Jean's aver

[5] #7 4.NF.A.2 Match to answer choices: $\dfrac{12}{13}$ is choice (A).

Review

Reasonableness: 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 $\dfrac{3d}{13d/12} = \dfrac{36}{13}$ mph, so Jean's is $\dfrac{12}{13}$ mph — matches. The answer is less than $1$ mph, which fits the story (heavy backpack, slower than every one of Chantal's legs).

Alternative: Tool #6 (Guess and Check) on the answer choices: pick a convenient $d$ such as $d = 12$ miles. Then Chantal's legs take $3,\,6,\,4$ hours, total $13$ hours. Jean covers $12$ miles in $13$ hours, average $\dfrac{12}{13}$ mph — directly (A). Plugging a number is faster than algebra and avoids fraction errors.

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Treating each hiking leg as a rate problem and computing Jean's overall mph.)
6.RP.A.2 Understand the concept of a unit rate and use rate language (Converting speed into time per mile via $t = d / r$ for each leg.)
5.NF.A.1 Add and subtract fractions with unlike denominators (Adding $\frac{d}{4} + \frac{d}{2} + \frac{d}{3}$ via the common denominator $12$.)
4.NF.A.2 Compare two fractions with different numerators and different denominators (Selecting the matching choice $\frac{12}{13}$ from the five options.)

⭐ 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 $\frac{d}{13d/12} = \frac{12}{13}$ mph.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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