AMC 8 · 2010 · #8

Grade 6 rate-ratio

Archetype Meeting Time with Piecewise Rates └ Sub-archetype Constant-Rate Closing Speed

rateunit-conversion identify-subproblemsdimensional-analysis ↑ Prerequisites: ratefraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

Emily bikes at 12 mph; Emerson skates in the same direction at 8 mph. Emily first spots Emerson when he is

\frac{1}{2}

mile ahead of her, and last sees him in her rear mirror when he is

\frac{1}{2}

mile behind her. For how many minutes is Emerson within Emily's view?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Analyze the Units

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 $\frac{1}{2}$ -mile gap, then open another $\frac{1}{2}$ -mile gap).

1STEP 1

Both move forward, but only the gap matters — freeze Emerson and let Emily close in at the relative speed 12 - 8 = 4 mph.

relative speed = 12 - 8 = 4 mph

2STEP 2

Emily starts $\frac{1}{2}$ mile behind, passes him, and ends $\frac{1}{2}$ mile ahead — in his frame that is $\frac{1}{2}$ + $\frac{1}{2}$ = 1 mile of relative travel.

\frac{1}{2}

\frac{1}{2}

= 1 mile

3STEP 3

Units are consistent, so time = distance / speed = $\frac{1}{4}$ = $\frac{1}{4}$ hour.

time = (1 mi)/(4 mi/hr) =

\frac{1}{4}

4STEP 4

The choices are in minutes, so convert $\frac{1}{4}$ hour to 15 minutes → (D).

\frac{1}{4}

hr × (60 min)/(1 hr) = 15 min → (D)

Answer

Emily gains 4 miles every hour on Emerson, so closing a

\frac{1}{2}

-mile gap takes

\frac{\frac{1}{2}}{4}

\frac{1}{8}

hour, and then opening a

\frac{1}{2}

-mile gap takes another

\frac{1}{8}

hour. Total =

\frac{2}{8}

\frac{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.

💡Key takeaway

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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