Competition · AMC preparation · step 4 of 4
AMC 8 · 2022 · #22
Grade 5 rate-ratiologic
Pick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Find the bus timetable pattern
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.
The repeating 2+1=3 minute cycle is exactly the kind of rule-based number pattern Grade 4 students learn to generate.
4.OA.C.5Look For A PatternDraw the stops in a row
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.
Even a kindergarten-level row of objects (left/right positions) is enough to make "previous stop" unambiguous.
K.G.A.1Draw A DiagramList each walking choice
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.
Tracking two number patterns (Zia's 0,5,10,15 and the bus's 6,9,12,15) in parallel and spotting where they meet is exactly the Grade 5 "two numerical patterns" skill.
Zia keeps walking past Stops 3, 4, and 5 and first stops to wait at Stop 6, the stop she reaches exactly when the bus is leaving the stop just behind her.
▸ Why?
Laying her arrival times 0, 5, 10, 15 next to the bus's behind-stop departure times 6, 9, 12, 15 row by row, she arrives earlier than the bus at Stops 3, 4, and 5 so she walks on, but at Stop 6 the two times are equal for the first time, so that is where the wait rule is first met.
▸ Why?
Zia's arrival times at Stops 3, 4, 5, 6 are 0, 5, 10, 15 because each further stop of walking adds the same fixed 5 minutes, so n stops of walking take 5n minutes.
▸ Why?
The bus leaves the stop just behind her at 6, 9, 12, 15 because leaving each earlier stop is the same fixed 3-minute drive-and-wait cycle repeated, so leaving the stop that is k-1 ahead of the start takes 3(k-1) minutes.
▸ Why?
One cycle is 2 minutes of driving joined to 1 minute of waiting with no gap between them, and those two pieces make one whole cycle of 3 minutes.
▸ Why?
Leaving the stop that is k-1 ahead means repeating that same 3-minute cycle k-1 times, which is k-1 equal groups of 3 minutes, giving 3(k-1) minutes.
Add the final wait
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.
Adding the 2-minute drive to the 15-minute waiting moment is a one-step addition word problem within 100, a Grade 2 standard.
2.OA.A.1Make A Systematic ListThis AMC 8 problem only needs Grade 5 "compare two number patterns side by side" thinking that you already know!
- Find the bus timetable pattern
- Draw the stops in a row
- List each walking choice
- Add the final wait
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