AMC 8 · 2022 · #22

Grade 5 rate-ratiologic

Archetype Meeting Time with Piecewise Rates

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

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus $3$ stops behind. After how many minutes will Zia board the bus?

$\textbf{(A) } 17 \qquad \textbf{(B) } 19 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 23$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Bus driving time between consecutive stops $= 2$ min; Bus dwell time at each stop $= 1$ min; Zia walking time between consecutive stops $= 5$ min; At $t=0$: bus at Stop $0$, Zia at Stop $3$, both moving toward the library; Zia's decision rule: at Stop $k$, wait if (and only if) the bus has reached or left Stop $k-1$; Answer choices: (A) $17$, (B) $19$, (C) $20$, (D) $21$, (E) $23$ (minutes)

Unknowns: The total time in minutes from $t=0$ until Zia first boards the bus

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #1 Draw a Diagram, #5 Look for a Pattern

The motion of the bus is perfectly regular: it leaves Stop $k$ at time $t = 3k$ minutes (for $k \ge 1$, 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.

Execute — Answer: A

#5 Look for a Pattern 4.OA.C.5 Step 1

First build the bus schedule by spotting its pattern.
Each "drive $2$ min, wait $1$ min" cycle takes $3$ minutes, so the bus leaves Stop $0$ at $t=0$, leaves Stop $1$ at $t=3$, leaves Stop $2$ at $t=6$, and in general leaves Stop $k$ at $t = 3k$.
It arrives at Stop $k$ one minute earlier, at $t = 3k - 1$.

$$\text{arrive Stop } k: t = 3k - 1 \quad ; \quad \text{leave Stop } k: t = 3k$$

💡 The repeating $2+1=3$ minute cycle is exactly the kind of rule-based number pattern Grade 4 students learn to generate.

#1 Draw a Diagram K.G.A.1 Step 2

Now draw the stops as a row of dots and label them $0, 1, 2, 3, 4, 5, 6, \ldots$ Mark the bus at Stop $0$ and Zia at Stop $3$ at $t=0$.
This makes the phrase "the previous stop" concrete: when Zia is at Stop $k$, the previous stop is just the dot immediately to her left.

$$\underbrace{\bullet}_{\text{bus, }0} \;\; \bullet_1 \;\; \bullet_2 \;\; \underbrace{\bullet}_{\text{Zia, }3} \;\; \bullet_4 \;\; \bullet_5 \;\; \bullet_6 \;\; \cdots$$

💡 Even a kindergarten-level row of objects (left/right positions) is enough to make "previous stop" unambiguous.

#2 Make a Systematic List 5.OA.B.3 Step 3

Make a systematic list with one row per Zia decision.
Zia arrives at Stops $3, 4, 5, 6, \ldots$ at times $t = 0, 5, 10, 15, \ldots$ (she gains $5$ minutes per stop).
At each row, write the time the bus leaves the previous stop, $3(k-1)$, and compare.

$$\begin{array}{c|c|c|c} \text{Zia at stop } k & \text{arrival } t & \text{bus leaves stop } k-1 \text{ at } 3(k-1) & \text{decision} \\ \hline 3 & 0 & 6 & 0 < 6 \Rightarrow \text{walk} \\ 4 & 5 & 9 & 5 < 9 \Rightarrow \text{walk} \\ 5 & 10 & 12 & 10 < 12 \Rightarrow \text{walk} \\ 6 & 15 & 15 & 15 \ge 15 \Rightarrow \text{wait} \end{array}$$

💡 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.

#2 Make a Systematic List 2.OA.A.1 Step 4

Zia waits at Stop $6$ from $t = 15$ onward.
The bus, having left Stop $5$ at $t = 15$, takes $2$ minutes to drive to Stop $6$, arriving at $t = 15 + 2 = 17$.
That is the moment Zia boards.

$$15 + 2 = 17 \text{ min} \;\Rightarrow\; \textbf{(A)}$$

💡 Adding the $2$-minute drive to the $15$-minute waiting moment is a one-step addition word problem within $100$, a Grade 2 standard.

[1] #5 4.OA.C.5 First build the bus schedule by spotting its pattern. Each "drive $2$ min, wait

[2] #1 K.G.A.1 Now draw the stops as a row of dots and label them $0, 1, 2, 3, 4, 5, 6, \ldots$

[3] #2 5.OA.B.3 Make a systematic list with one row per Zia decision. Zia arrives at Stops $3, 4

[4] #2 2.OA.A.1 Zia waits at Stop $6$ from $t = 15$ onward. The bus, having left Stop $5$ at $t

Review

Reasonableness: 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 $5/3$. Starting $3$ stops behind, the bus needs about $3 \times 5 / (5-3) = 7.5$ Zia-stop-times $\approx 37$ 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.

Alternative: Tool #1 (Draw a Diagram) used as the *primary* tool: sketch a time-vs-position chart with two piecewise lines — the bus (sawtooth: slope $1/2$ during driving, flat during dwell) and Zia (slope $1/5$ while walking, flat while waiting). The first time the two lines meet is the boarding time. The meeting happens at position $6$ and $t = 17$, confirming the answer (A) without any inequalities.

CCSS standards used (min grade 5)

K.G.A.1 Describe positions of objects using above, below, beside, in front of (Laying out the bus stops as a left-to-right row of dots so that "the previous stop" is the dot immediately to the left.)
2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Computing the final boarding time $15 + 2 = 17$ minutes once Zia is waiting at Stop $6$.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Generating the bus's leave-time pattern $0, 3, 6, 9, 12, 15, \ldots$ from the rule "drive $2$ min, wait $1$ min".)
5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships (Comparing Zia's arrival-time pattern $0, 5, 10, 15, \ldots$ to the bus's previous-stop-departure pattern $6, 9, 12, 15, \ldots$ row by row to spot the first time the bus condition is satisfied.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 5)

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