AMC 8 · 2010 · #8
Grade 6 rate-ratioProblem
As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction 1/2 mile in front of her. After she passes him, she can see him in her rear mirror until he is 1/2 mile behind her. Emily rides at a constant rate of 12 miles per hour, and Emerson skates at a constant rate of 8 miles per hour. For how many minutes can Emily see Emerson?
Pick an answer.
AMC 8 2010 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: Emily bikes at $12$ mph; Emerson skates in the same direction at $8$ mph. Emily first spots Emerson when he is $1/2$ mile ahead of her, and last sees him in her rear mirror when he is $1/2$ mile behind her. For how many minutes is Emerson within Emily's view?
Givens: Emily's speed = $12$ mph (constant); Emerson's speed = $8$ mph (constant), same direction; Emily sees Emerson starting when he is $1/2$ mile ahead of her; Emily stops seeing him when he is $1/2$ mile behind her; Answer choices: (A) $6$, (B) $8$, (C) $12$, (D) $15$, (E) $16$ (minutes)
Unknowns: The total time, in minutes, that Emerson is visible to Emily
Understand
Restated: Emily bikes at $12$ mph; Emerson skates in the same direction at $8$ mph. Emily first spots Emerson when he is $1/2$ mile ahead of her, and last sees him in her rear mirror when he is $1/2$ mile behind her. For how many minutes is Emerson within Emily's view?
Givens: Emily's speed = $12$ mph (constant); Emerson's speed = $8$ mph (constant), same direction; Emily sees Emerson starting when he is $1/2$ mile ahead of her; Emily stops seeing him when he is $1/2$ mile behind her; Answer choices: (A) $6$, (B) $8$, (C) $12$, (D) $15$, (E) $16$ (minutes)
Plan
Primary tool: #8 Analyze the Units
Secondary: #15 Organize Information in More Ways
This is a rate problem, so Tool #8 (Analyze the Units) keeps the bookkeeping honest: miles divided by mph gives hours, which we then convert to minutes. The trick that simplifies the whole thing is Tool #15 (Reorganize Information): instead of tracking two moving people, switch to Emerson's frame of reference — pretend Emerson is standing still and Emily approaches him at the relative speed $12 - 8 = 4$ mph. Now there is one moving object covering a fixed total distance of $1$ mile (close the $1/2$-mile gap, then open another $1/2$-mile gap).
Execute — Answer: D
6.RP.A.3 Step 1 - Reorganize the problem into Emerson's frame.
- Both bike and skate move forward, but only the gap between them matters for visibility.
- Emily gains on Emerson at the relative speed $12 - 8 = 4$ mph, so we can treat Emerson as stationary and Emily as moving at $4$ mph.
💡 Switching to a single-mover frame is the Tool #15 move: same data, simpler picture.
4.MD.A.2 Step 2 - Find the total distance Emily must cover relative to Emerson during the visibility window.
- She starts $1/2$ mile behind him (he is ahead), passes him, and ends $1/2$ mile ahead of him.
- In Emerson's frame, that is a straight $1/2 + 1/2 = 1$ mile of relative travel.
💡 Adding the "closing" leg and the "opening" leg captures the whole window in one number.
6.RP.A.3 Step 3 Apply $\text{time} = \dfrac{\text{distance}}{\text{speed}}$ with consistent units (miles and mph give hours).
💡 Dividing miles by miles-per-hour cancels "miles" and leaves "hours" — Tool #8 unit check at work.
5.MD.A.1 Step 4 Convert hours to minutes because the answer choices are in minutes.
💡 Same time, just expressed in the unit the question wants — a Grade 5 unit-conversion step.
6.RP.A.3 Reorganize the problem into Emerson's frame. Both bike and skate move forward, b 4.MD.A.2 Find the total distance Emily must cover relative to Emerson during the visibili 6.RP.A.3 Apply $\text{time} = \dfrac{\text{distance}}{\text{speed}}$ with consistent unit 5.MD.A.1 Convert hours to minutes because the answer choices are in minutes. Review
Reasonableness: Emily gains $4$ miles every hour on Emerson, so closing a $1/2$-mile gap takes $\tfrac{1/2}{4} = \tfrac{1}{8}$ hour, and then opening a $1/2$-mile gap takes another $\tfrac{1}{8}$ hour. Total $= \tfrac{2}{8} = \tfrac{1}{4}$ hour $= 15$ minutes. That matches answer (D). It is also a believable amount of time — a few city blocks of mismatched cyclist-vs-skater pace, not seconds and not an hour.
Alternative: Tool #7 (Identify Subproblems) splits the window into two halves: closing the gap ($1/2$ mile at relative speed $4$ mph $= 7.5$ min) and opening the gap (another $7.5$ min). Sum: $15$ minutes — same answer (D). This is the same idea as the reference solution, just framed as two subproblems instead of one consolidated trip.
CCSS standards used (min grade 6)
4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Adding the closing leg ($1/2$ mile) and the opening leg ($1/2$ mile) into a single $1$-mile visibility window.)5.MD.A.1Convert among different-sized standard measurement units within a given system (Converting $\tfrac{1}{4}$ hour into $15$ minutes so the answer matches the choices.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Subtracting speeds to get a relative rate ($12 - 8 = 4$ mph) and then computing time $= $ distance $/$ rate $= 1/4$ hour.)
⭐ Same-direction chase problems become easy when you pretend the slower mover is standing still — then it's just one distance over one speed, the Grade 6 rate idea you already know.
⭐ Same-direction chase problems become easy when you pretend the slower mover is standing still — then it's just one distance over one speed, the Grade 6 rate idea you already know.
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