AMC 8 · 2006 · #13
Grade 6 rate-ratioProblem
Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time in the morning do they meet?
Pick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Cassie leaves Escanaba at $8\!:\!30$ AM biking toward Marquette at $12$ mph. Brian leaves Marquette at $9\!:\!00$ AM biking toward Escanaba at $16$ mph. The route is $62$ miles long. At what time do they meet?
Givens: Cassie: starts at $8\!:\!30$ AM from Escanaba, speed $12$ mph; Brian: starts at $9\!:\!00$ AM from Marquette, speed $16$ mph; Total route length: $62$ miles; Answer choices: (A) $10\!:\!00$, (B) $10\!:\!15$, (C) $10\!:\!30$, (D) $11\!:\!00$, (E) $11\!:\!30$
Unknowns: The clock time at which Cassie and Brian meet
Understand
Restated: Cassie leaves Escanaba at $8\!:\!30$ AM biking toward Marquette at $12$ mph. Brian leaves Marquette at $9\!:\!00$ AM biking toward Escanaba at $16$ mph. The route is $62$ miles long. At what time do they meet?
Givens: Cassie: starts at $8\!:\!30$ AM from Escanaba, speed $12$ mph; Brian: starts at $9\!:\!00$ AM from Marquette, speed $16$ mph; Total route length: $62$ miles; Answer choices: (A) $10\!:\!00$, (B) $10\!:\!15$, (C) $10\!:\!30$, (D) $11\!:\!00$, (E) $11\!:\!30$
Plan
Primary tool: #8 Analyze the Units
Secondary: #1 Draw a Diagram
This is a textbook rate/distance/time problem, so Tool #8 (Analyze Units) is the natural primary. Speeds are in mph and times are in minutes — convert minutes to hours so that speed $\times$ time gives miles cleanly. Tool #1 (Draw a Diagram) turns the word problem into a $62$-mile segment with Cassie and Brian as arrows pointing at each other; once drawn, the head-start distance and the closing speed are easy to read off. We do not reach for Tool #13 (Algebra) because the picture plus unit-tracking gives the answer in three short steps.
Execute — Answer: D
4.OA.A.3 Step 1 - Draw the route and mark the head start.
- Picture Escanaba on the left, Marquette on the right, and a $62$-mile segment between them.
- Cassie starts $30$ minutes earlier than Brian, so at $9\!:\!00$ AM (when Brian sets off) she has already been pedaling for half an hour.
💡 Drawing the two riders with arrows toward each other makes the "closing the gap" idea visible — a Grade 4 multi-step word-problem move.
5.MD.A.1 Step 2 - Convert Cassie's head start to hours, then to miles.
- Speeds are in miles per hour, so the time must be in hours: $30 \text{ min} = \tfrac{1}{2} \text{ hr}$.
- Multiplying speed by time gives the distance Cassie covers before Brian starts.
💡 Grade 5 unit conversion: minutes to hours, then the "hr" units cancel so the answer comes out in miles, exactly what the problem needs.
4.OA.A.3 Step 3 - Find the gap at $9\!:\!00$ AM and the closing speed.
- After Cassie's $6$-mile head start, only $62 - 6 = 56$ miles separate the two riders.
- Because they ride toward each other, the gap shrinks by both speeds added together each hour.
💡 Two riders heading toward each other close the gap at the sum of their speeds — like two arrows squeezing inward on the diagram.
6.RP.A.3 Step 4 Divide the gap by the closing speed to get the time after $9\!:\!00$ AM, then add it to Brian's start time.
💡 Grade 6 rate reasoning: time $=$ distance $\div$ rate. The miles cancel and hours pop out, which is exactly what the clock-time question wants.
4.OA.A.3 Draw the route and mark the head start. Picture Escanaba on the left, Marquette 5.MD.A.1 Convert Cassie's head start to hours, then to miles. Speeds are in miles per hou 4.OA.A.3 Find the gap at $9\!:\!00$ AM and the closing speed. After Cassie's $6$-mile hea 6.RP.A.3 Divide the gap by the closing speed to get the time after $9\!:\!00$ AM, then ad Review
Reasonableness: Check both riders' positions at $11\!:\!00$ AM. Cassie has been riding $2.5$ hours, so she has covered $12 \times 2.5 = 30$ miles from Escanaba. Brian has been riding $2$ hours, so he has covered $16 \times 2 = 32$ miles from Marquette. Together: $30 + 32 = 62$ miles — exactly the route length, so they are at the same point. The answer is consistent. Also, the other choices fail this check: at $10\!:\!30$ AM (choice C), Cassie has gone $24$ mi and Brian $24$ mi, totaling only $48 < 62$, so they have not met yet.
Alternative: Tool #13 (Convert to Algebra): let $t$ be the hours after $9\!:\!00$ AM until they meet. Cassie's distance from Escanaba is $6 + 12t$, Brian's distance from Marquette is $16t$, and the two must add to $62$: $6 + 12t + 16t = 62 \Rightarrow 28t = 56 \Rightarrow t = 2$. Meeting time is $9\!:\!00 + 2 = 11\!:\!00$ AM, choice (D). The algebra version arrives at the same equation Tool #8 builds piece by piece.
CCSS standards used (min grade 6)
4.OA.A.3Solve multistep word problems using the four operations, including problems with whole-number remainders (Setting up the multi-step plan — head start, remaining gap, closing speed, meeting time — and combining the arithmetic results into a single answer.)5.MD.A.1Convert among different-sized standard measurement units within a given measurement system (Converting the $30$-minute head start to $\tfrac{1}{2}$ hour so it matches the mph units before multiplying by Cassie's speed.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Treating speed as a rate ($\text{distance per hour}$) to compute $\tfrac{56\,\text{mi}}{28\,\text{mph}} = 2$ hours of riding before the meeting.)
⭐ When two travelers move toward each other, peel off any head-start distance first, then divide the remaining gap by the sum of their speeds — the answer falls out in one step.
⭐ When two travelers move toward each other, peel off any head-start distance first, then divide the remaining gap by the sum of their speeds — the answer falls out in one step.
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