AMC 10 · 2009 · #23

Grade 7 probability

Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region

geometric-probabilityrateinterval-arithmetic physical-representation ↑ Prerequisites: geometric-probability

📏 Long solution 💡 4 insights

Problem

Rachel and Robert run on a circular track, both starting from the same line at the same time. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. At a random time between 10 minutes and 11 minutes after they begin, a photographer standing inside the track takes a picture showing one-fourth of the track, centered on the starting line. Find the probability that both Rachel and Robert are in the picture.

Pick an answer.

(A)

$\frac {1}{16}$

(B)

$\frac 18$

(C)

$\frac {3}{16}$

(D)

$\frac 14$

(E)

$\frac {5}{16}$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is the spine: model the track so 'in the photo' becomes 'within 1/8 lap of the start,' then draw a timeline of the minute from 600 to 660 s and shade each runner's photo windows. Tool #8 (Analyze the Units) converts an arc (1/8 lap) into seconds using each runner's lap time, since each lap fraction maps to a fixed number of seconds. Tool #5 (Look for a Pattern) finds when each runner sits on the start line inside the target minute — every 90 s for Rachel, every 80 s for Robert. Tool #7 (Identify Subproblems) splits the job into 'Rachel's window,' 'Robert's window,' then 'overlap,' so the final probability is one length divided by 60.

1STEP 1

Model the photo as a distance

Measure in laps. The quarter-lap photo is centered on the line, so it reaches 1/8 lap each way — that arc is the whole test.

photo half-width = 1/4 × 1/2 = 1/8 lap on each side

2STEP 2

Turn the arc into seconds

A steady runner needs 1/8 of a lap time to cover 1/8 lap: Rachel 11.25 s, Robert 10 s on each side of a crossing.

Rachel: 90/8 = 11.25 s; Robert: 80/8 = 10 s

3STEP 3

Find each runner at the line

A runner sits on the line at whole multiples of the lap time; inside 600–660 s that is Rachel at 630 s and Robert at 640 s.

630 = 7 × 90 ∈ [600,660]; 640 = 8 × 80 ∈ [600,660]

4STEP 4

Shade each photo window

Add each half-width to the crossing time: Rachel is in the photo on [618.75, 641.25] s, Robert on [630, 650] s.

Rachel: [618.75, 641.25]; Robert: [630, 650]

5STEP 5

Overlap the two windows

Both appear only where the bars overlap — [630, 641.25], from latest start to earliest end, a window of 11.25 s.

[630, 650]∩[618.75, 641.25] = [630, 641.25], length 11.25 s

6STEP 6

Divide to get the probability

The instant is uniform over the 60-second minute, so the probability is 11.25/60 = 3/16, choice (C).

P = 11.25/60 = 3/16 → (C)

Answer

3/16

Every step matches the clock. Rachel truly crosses at 630 s (7 × 90) and Robert at 640 s (8 × 80), both inside the target minute. Their windows ± 11.25 s and ± 10 s are exactly the eighth-lap arcs. The overlap [630, 641.25] has length 11.25 s, and 11.25/60 = 0.1875 = 3/16, a clean answer choice. The overlap (11.25 s) is naturally smaller than either single window (22.5 s and 20 s), so the joint probability being below each individual one is sensible. Answer 3/16 is choice (C).

💡Key takeaway

Turn 'in the picture' into 'within an eighth of a lap of the start,' change that arc into seconds for each runner, shade both time windows on the timeline, and the overlap (11.25 out of 60 seconds) is the probability: (C) 3/16.

Model the photo as a distance
Turn the arc into seconds
Find each runner at the line
Shade each photo window
Overlap the two windows
Divide to get the probability