AMC 8 · 2014 · #17

Grade 6 rate-ratio

Archetype Meeting Time with Piecewise Rates └ Sub-archetype Piecewise-Leg Time Balance

ratefraction-arithmeticunit-conversion identify-subproblemsdimensional-analysis ↑ Prerequisites: ratefraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2014 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: George walks $1$ mile to school each day at a steady $3$ mph, arriving right when school begins. Today he walked the first $\tfrac{1}{2}$ mile at only $2$ mph. At what speed (in mph) must he run the remaining $\tfrac{1}{2}$ mile to still arrive on time?

Givens: Total distance to school = $1$ mile; Usual steady speed = $3$ mph (always arrives on time); Today: first $\tfrac{1}{2}$ mile walked at $2$ mph; Remaining distance = $\tfrac{1}{2}$ mile; Answer choices: (A) $4$, (B) $6$, (C) $8$, (D) $10$, (E) $12$ (mph)

Unknowns: The speed in mph George must run over the last $\tfrac{1}{2}$ mile to arrive at his usual time

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #8 Analyze the Units

This is a rate problem built on $\text{time} = \text{distance} / \text{speed}$. Tool #7 (Identify Subproblems) is the key move: split the trip into the usual full trip (to get the time budget), the slow first half (to get the time already used), and the fast second half (the unknown). Once each piece is solved, the answer follows by simple subtraction and one more $\text{speed} = \text{distance} / \text{time}$ step. Tool #8 (Analyze the Units) keeps everything in miles and hours so the final number is automatically in mph.

Execute — Answer: B

#7 Identify Subproblems 6.RP.A.3 Step 1

Find the time budget.
The usual trip is $1$ mile at $3$ mph, so the full schedule allows $\tfrac{1}{3}$ hour (which is $20$ minutes).

$$T_{\text{total}} = \dfrac{1 \text{ mi}}{3 \text{ mph}} = \tfrac{1}{3} \text{ hr} = 20 \text{ min}$$

💡 Distance divided by speed gives time — a Grade 6 unit-rate move that turns the schedule into a fixed number of minutes.

#7 Identify Subproblems 6.RP.A.3 Step 2

Find the time already used on the slow first half.
He walked $\tfrac{1}{2}$ mile at $2$ mph, which takes $\tfrac{1/2}{2} = \tfrac{1}{4}$ hour, or $15$ minutes.

$$T_{\text{first}} = \dfrac{1/2 \text{ mi}}{2 \text{ mph}} = \tfrac{1}{4} \text{ hr} = 15 \text{ min}$$

💡 Same distance-over-speed pattern, just applied to the first subproblem.

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

Subtract to get the time left for the second half.
Out of the $\tfrac{1}{3}$-hour budget, $\tfrac{1}{4}$ hour is gone, leaving $\tfrac{1}{3} - \tfrac{1}{4}$ hour.
Use a common denominator of $12$.

$$T_{\text{remain}} = \tfrac{1}{3} - \tfrac{1}{4} = \tfrac{4}{12} - \tfrac{3}{12} = \tfrac{1}{12} \text{ hr} = 5 \text{ min}$$

💡 Subtracting fractions with unlike denominators ($\tfrac{1}{3} - \tfrac{1}{4}$) is exactly the Grade 5 fraction skill.

#8 Analyze the Units 6.RP.A.3 Step 4

Apply $\text{speed} = \text{distance} / \text{time}$ to the remaining half-mile run.
The distance is $\tfrac{1}{2}$ mile and the time is $\tfrac{1}{12}$ hour, so dividing by $\tfrac{1}{12}$ is the same as multiplying by $12$.

$$\text{speed} = \dfrac{1/2 \text{ mi}}{1/12 \text{ hr}} = \tfrac{1}{2} \times 12 = 6 \text{ mph} \;\Rightarrow\; \textbf{(B)}$$

💡 Computing the unit rate "miles per hour" from a distance and a time is Grade 6 ratio reasoning.

[1] #7 6.RP.A.3 Find the time budget. The usual trip is $1$ mile at $3$ mph, so the full schedul

[2] #7 6.RP.A.3 Find the time already used on the slow first half. He walked $\tfrac{1}{2}$ mile

[3] #7 5.NF.A.1 Subtract to get the time left for the second half. Out of the $\tfrac{1}{3}$-hou

[4] #8 6.RP.A.3 Apply $\text{speed} = \text{distance} / \text{time}$ to the remaining half-mile

Review

Reasonableness: Sanity check by averaging speeds. George spent equal distance ($\tfrac{1}{2}$ mile) at $2$ mph and at $6$ mph, so the average speed is the harmonic mean $\dfrac{2 \cdot 2 \cdot 6}{2 + 6} = \dfrac{24}{8} = 3$ mph — exactly his usual pace. That confirms today's total time matches the normal time, so (B) $6$ mph is correct. (A) $4$ would be too slow — only $\tfrac{1/2}{4} = \tfrac{1}{8}$ hr $= 7.5$ min, overshooting the $5$-min budget.

Alternative: Tool #6 (Guess and Check) on the choices: each candidate $v$ mph covers $\tfrac{1}{2}$ mile in $\tfrac{1/2}{v} = \tfrac{1}{2v}$ hr. We need that to equal $\tfrac{1}{12}$ hr, so $2v = 12 \Rightarrow v = 6$. Only (B) works; the others give $\tfrac{1}{8}, \tfrac{1}{16}, \tfrac{1}{20}, \tfrac{1}{24}$ hr, none of which match $\tfrac{1}{12}$.

CCSS standards used (min grade 6)

5.NF.A.1 Add and subtract fractions with unlike denominators (Computing the remaining time $\tfrac{1}{3} - \tfrac{1}{4} = \tfrac{1}{12}$ hour by finding a common denominator of $12$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Applying $\text{time} = \text{distance}/\text{speed}$ for the usual trip ($\tfrac{1}{3}$ hr) and the slow first half ($\tfrac{1}{4}$ hr), and finally $\text{speed} = \text{distance}/\text{time} = (1/2) / (1/12) = 6$ mph for the answer.)

⭐ This AMC 8 problem only needs Grade 6 rate reasoning — distance, time, and speed — plus a single Grade 5 fraction subtraction!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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