AMC 8 · 2022 · #10

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

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

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Ling drives from home at 45 mph from 8 AM to 10 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.

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

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

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 × hr cancels to mi, and mi ÷ 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.

1STEP 1

Split the trip into three graph pieces — drive out, hike, drive home — each giving one fact to test against the graphs.

trip = drive out_rising line + hike_flat line + drive home_falling line

2STEP 2

The drive out is 10 - 8 = 2 hours at 45 mph, and mph × hours cancels to miles, so the peak distance is 90 miles before the hike.

45 mi/hr × 2 hr = 90 mi

3STEP 3

The y-axis is in 30-mile units; B and C plateau at 45 mi and D at 120 mi, so only A and E plateau at the right 90 miles.

Eliminate: B, C, D — wrong peak. Survivors: A, E.

4STEP 4

The 90-mile return at 60 mph takes 90 ÷ 60 = 1.5 hr; after the hike ends at 1 PM she gets home at 2:30 PM.

(90 mi)/(60 mi/hr) = 1.5 hr → arrival at 1 PM + 1.5 hr = 2{:}30 PM

5STEP 5

A hits zero at 3 PM, but Ling arrives at 2:30 PM; E hits zero halfway between 2 and 3 PM, so E is the answer.

A ends at 3 PM — wrong. E ends at 2{:}30 PM — correct. → (E)

6STEP 6

In E the rise is 90 mi in 2 hr (slope 45) and the fall 90 mi in 1.5 hr (slope 60), so the return is steeper — E passes every test.

rising slope =

\frac{90}{2}

= 45, falling slope =

\frac{90}{1.5}

= 60 → falling is steeper ✓

Answer

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

All three landmark numbers check out: peak = 45 × 2 = 90 mi, return time = 90 ÷ 60 = 1.5 hr, total trip = 2 + 3 + 1.5 = 6.5 hr from 8 AM, which lands at 2{:}30 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.

💡Key takeaway

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

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