AMC 8 · 2022 · #22

Grade 5 rate-ratiologic

Archetype Meeting Time with Piecewise Rates └ Sub-archetype Staggered Start or Schedule Table

ratepattern-recognitionlogical-deduction physical-representationcasework ↑ Prerequisites: ratelogical-deduction

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A bus starts at Stop 0 and a girl named Zia starts at Stop 3, both at time t=0, both heading toward the library. The bus needs 2 minutes to drive between consecutive stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between consecutive stops. Every time Zia arrives at a new stop, she looks back: if the bus is already at (or has already left) the stop immediately behind her, she waits there; otherwise she keeps walking. After how many minutes from t=0 does Zia first board the bus?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

How to solve

Strategy Make a Systematic List

The motion of the bus is perfectly regular: it leaves Stop k at time t = 3k minutes (for k ≥ 1, and at t=0 for k=0), and arrives at Stop k at t = 3k - 1. So Tool #2 (Systematic List) — a small time-and-position table — lets us track both travelers in lock-step instead of trying to picture everything in our heads. Tool #1 (Diagram) helps externalize the stops as a row of dots so the "previous stop" check is concrete. Tool #5 (Pattern) is what produces the bus schedule formula t = 3k in the first place. We deliberately avoid Tool #13 (Algebra): for an AMC 8 simulation problem, a four-row table is faster and far less error-prone than setting up inequalities.

1STEP 1

Spot the pattern: each drive-2 + wait-1 cycle is 3 min, so the bus leaves Stop k at t = 3k and arrives one minute earlier at 3k - 1.

arrive Stop k: t = 3k - 1 ; leave Stop k: t = 3k

2STEP 2

Draw the stops as a row of dots 0,1,2,3,…, bus on Stop 0 and Zia on Stop 3, so 'the previous stop' is just the dot to her left.

●_bus, 0 ●₁ ●₂ ●_Zia, 3 ●₄ ●₅ ●₆ …

3STEP 3

One row per stop: Zia reaches 3,4,5,6 at t = 0,5,10,15; the bus clears the stop behind her at 6,9,12,15, so she first waits at Stop 6.

Zia at stop k & arrival t & bus leaves stop k-1 at 3(k-1) & decision ; 3 & 0 & 6 & 0 ＜ 6 → walk ; 4 & 5 & 9 & 5 ＜ 9 → walk ; 5 & 10 & 12 & 10 ＜ 12 → walk ; 6 & 15 & 15 & 15 ≥ 15 → wait

4STEP 4

Zia waits at Stop 6 from t = 15; the bus leaves Stop 5 at 15, drives 2 min, and reaches her at 15 + 2 = 17 — the moment she boards.

15 + 2 = 17 min → (A)

Answer

Walking speed vs. average bus speed: Zia covers 1 stop in 5 min, the bus covers 1 stop per 3-min cycle on average — so on average the bus is faster by a factor of

\frac{5}{3}

. Starting 3 stops behind, the bus needs about

\frac{3 × 5}{5-3}

= 7.5 Zia-stop-times ≈ 37 min if Zia kept walking forever, but because she eventually stops and waits the catch-up happens sooner. The answer 17 min is in the lowest cluster of choices (17, 19, 20, 21, 23), which fits a problem where the smart move is to stop walking rather than to keep going.

💡Key takeaway

This AMC 8 problem only needs Grade 5 "compare two number patterns side by side" thinking that you already know!

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