AMC 8 · 2022 · #10

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition

rategraph-readingslope-intercept identify-subproblemsdimensional-analysis ↑ Prerequisites: rategraph-reading

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

One sunny day, Ling decided to take a hike in the mountains. She left her house at $8 \, \textsc{am}$ , drove at a constant speed of $45$ miles per hour, and arrived at the hiking trail at $10 \, \textsc{am}$ . After hiking for $3$ hours, Ling drove home at a constant speed of $60$ miles per hour. Which of the following graphs best illustrates the distance between Ling’s car and her house over the course of her trip?

Pick an answer.

(A)

(distance-time graph) peak 90 mi at hours 3-6, returns to 0 at hour 8 (slower return)

(B)

(distance-time graph) peak 45 mi at hours 3-6, returns to 0 at hour 8

(C)

(distance-time graph) peak 45 mi at hours 3-6, returns to 0 at hour 7.5

(D)

(distance-time graph) peak 120 mi at hours 3-6, returns to 0 at hour 9.3

(E)

(distance-time graph) peak 90 mi at hours 3-6, returns to 0 at hour 7.5

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Toolkit + CCSS Solution

Understand

Restated: Ling drives from home at $45$ mph from $8\,\text{AM}$ to $10\,\text{AM}$, hikes for $3$ hours while the car stays parked, then drives home at $60$ mph. We must pick the distance-vs-time graph (out of A–E) whose three pieces — a rising line, a flat segment, and a steeper falling line — match those facts.

Givens: Departure $8\,\text{AM}$, arrival at trail $10\,\text{AM}$ (a $2$-hour drive); Driving speed to the trail $= 45$ mph (constant); Hike length $= 3$ hours (car parked the whole time); Return driving speed $= 60$ mph (constant); Five candidate graphs A–E with peak distances and return times labelled on the axes

Unknowns: Which lettered graph (A, B, C, D, or E) correctly shows the car's distance from home over the whole trip

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #8 Analyze the Units, #7 Identify Subproblems

Five concrete graphs are given, so Tool #3 (Eliminate Possibilities) is the natural AMC multiple-choice move: compute the three landmark facts the right graph must satisfy, then knock out any graph that fails even one. Tool #8 (Analyze the Units) keeps the rate computations honest — mph $\times$ hr cancels to mi, and mi $\div$ mph cancels to hr — so the $90$-mile peak and the $1.5$-hour return time are watertight. Tool #7 (Identify Subproblems) makes us treat the trip as three separate pieces (drive out, hike, drive home) instead of staring at the whole picture at once.

Execute — Answer: E

#7 Identify Subproblems 5.G.A.2 Step 1

Split the trip into three subproblems matching the three graph segments: (1) drive to the trail, (2) hike with car parked, (3) drive home.
Each segment will give us one fact to test against the graphs.

$$\text{trip} = \underbrace{\text{drive out}}_{\text{rising line}} + \underbrace{\text{hike}}_{\text{flat line}} + \underbrace{\text{drive home}}_{\text{falling line}}$$

💡 Reading a real-world story off a coordinate graph — matching each event to a piece of the line — is Grade 5 graphing-real-world-problems work.

#8 Analyze the Units 4.MD.A.2 Step 2

Compute the peak distance.
The drive to the trail lasts $10 - 8 = 2$ hours at $45$ mph, and "mph $\times$ hours" cancels to miles.
So the car gets $90$ miles from home before the hike starts.

$$45 \,\tfrac{\text{mi}}{\text{hr}} \times 2 \,\text{hr} = 90 \,\text{mi}$$

💡 Distance = speed $\times$ time with matching units is a Grade 4 distance/time word-problem skill.

#3 Eliminate Possibilities 5.G.A.2 Step 3

Use the $90$-mile peak to eliminate graphs whose plateau height is wrong.
The y-axis is labelled in $30$-mile units.
Graphs B and C plateau at $45$ miles, and graph D plateaus at $120$ miles, so all three are ruled out.
Only A and E plateau at the correct height of $90$ miles.

Eliminate: $\text{B}, \text{C}, \text{D}$ — wrong peak. Survivors: $\text{A}, \text{E}$.

💡 Reading a coordinate value off the y-axis to compare with $90$ miles is Grade 5 coordinate-plane reasoning.

#8 Analyze the Units 4.MD.A.2 Step 4

Compute when Ling gets home, then use that to decide between A and E.
The return drive covers $90$ miles at $60$ mph, taking $90 \div 60 = 1.5$ hours.
Starting at $1\,\text{PM}$ (after the $3$-hour hike that began at $10\,\text{AM}$), she arrives home at $2{:}30\,\text{PM}$.

$$\dfrac{90 \,\text{mi}}{60 \,\tfrac{\text{mi}}{\text{hr}}} = 1.5 \,\text{hr} \;\Rightarrow\; \text{arrival at } 1\,\text{PM} + 1.5\,\text{hr} = 2{:}30\,\text{PM}$$

💡 Time = distance $\div$ speed with matching mi-and-mph units is the same Grade 4 distance/time relationship, just rearranged.

#3 Eliminate Possibilities 5.G.A.2 Step 5

Eliminate using the arrival time.
Graph A's falling line hits zero at $3\,\text{PM}$ (one full grid square past $2\,\text{PM}$), but Ling actually arrives at $2{:}30\,\text{PM}$ — half a square past $2\,\text{PM}$.
Graph E hits zero exactly halfway between $2\,\text{PM}$ and $3\,\text{PM}$.
So A is out and E is the answer.

$\text{A}$ ends at $3\,\text{PM}$ — wrong. $\text{E}$ ends at $2{:}30\,\text{PM}$ — correct. $\;\Rightarrow\; \textbf{(E)}$

💡 Reading the x-coordinate where a graph crosses zero is straightforward Grade 5 coordinate-plane work.

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

Sanity-check that E's slopes also match the speeds.
Because $60 > 45$, the return drive should be steeper than the outbound drive.
In E, the rising line covers $90$ mi in $2$ hr (slope $45$) and the falling line covers $90$ mi in $1.5$ hr (slope $60$), so the return is indeed steeper.
Graph E passes every test.

$$\text{rising slope} = \tfrac{90}{2} = 45,\; \text{falling slope} = \tfrac{90}{1.5} = 60 \;\Rightarrow\; \text{falling is steeper} \;\checkmark$$

💡 Recognising that the steeper segment corresponds to the faster unit rate (mph) is Grade 6 unit-rate reasoning.

[1] #7 5.G.A.2 Split the trip into three subproblems matching the three graph segments: (1) dri

[2] #8 4.MD.A.2 Compute the peak distance. The drive to the trail lasts $10 - 8 = 2$ hours at $4

[3] #3 5.G.A.2 Use the $90$-mile peak to eliminate graphs whose plateau height is wrong. The y-

[4] #8 4.MD.A.2 Compute when Ling gets home, then use that to decide between A and E. The return

[5] #3 5.G.A.2 Eliminate using the arrival time. Graph A's falling line hits zero at $3\,\text{

[6] #8 6.RP.A.2 Sanity-check that E's slopes also match the speeds. Because $60 > 45$, the retur

Review

Reasonableness: All three landmark numbers check out: peak $= 45 \times 2 = 90$ mi, return time $= 90 \div 60 = 1.5$ hr, total trip $= 2 + 3 + 1.5 = 6.5$ hr from $8\,\text{AM}$, which lands at $2{:}30\,\text{PM}$ — exactly where graph E's line touches zero. The return slope ($60$) being steeper than the outbound slope ($45$) is a fourth independent check, and E satisfies that too.

Alternative: Tool #1 (Draw a Diagram) gives an even cleaner path: sketch the three key points $(8\,\text{AM},\,0)$, $(10\,\text{AM},\,90)$, $(1\,\text{PM},\,90)$, $(2{:}30\,\text{PM},\,0)$ on blank axes, connect them, then compare the resulting shape to A–E by eye. The matching graph is E, with no algebra needed beyond the two rate computations.

CCSS standards used (min grade 6)

4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money (Computing the $90$-mile peak from $45 \times 2$ and the $1.5$-hour return from $90 \div 60$ using the distance-speed-time relationship.)
5.G.A.2 Represent real-world and mathematical problems by graphing points (Reading peak heights and zero-crossing times off the candidate graphs' coordinate axes and matching them to the computed landmark points.)
6.RP.A.2 Understand the concept of a unit rate and use rate language (Recognising that the faster return speed ($60$ mph) corresponds to a steeper line segment than the outbound speed ($45$ mph), confirming graph E.)

⭐ This AMC 8 problem only needs Grade 6 unit-rate reasoning — distance, speed, and time — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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