AMC 10 · 2009 · #23
Grade 7 probabilityRachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?
Pick an answer.
AMC 10 2009 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: Two runners circle a track from the same start line at the same moment: Rachel counterclockwise with a lap every $90$ s, Robert clockwise with a lap every $80$ s. At a uniformly random instant between the $600$-second and $660$-second marks, a photo captures the quarter of the track centered on the start line. Find the probability that both runners are inside that photo.
Givens: Rachel: counterclockwise, one lap every $90$ seconds; Robert: clockwise, one lap every $80$ seconds; Both leave the start line together at time $0$; The photo shows $\tfrac14$ of the track, centered on the start line, so it reaches $\tfrac18$ of a lap on each side of the start; The photo time is uniformly random on the interval from $10$ minutes ($600$ s) to $11$ minutes ($660$ s); Answer choices: (A) $\tfrac1{16}$, (B) $\tfrac18$, (C) $\tfrac{3}{16}$, (D) $\tfrac14$, (E) $\tfrac{5}{16}$
Unknowns: The probability that Rachel and Robert are both within the photographed arc at the chosen instant
Understand
Restated: Two runners circle a track from the same start line at the same moment: Rachel counterclockwise with a lap every $90$ s, Robert clockwise with a lap every $80$ s. At a uniformly random instant between the $600$-second and $660$-second marks, a photo captures the quarter of the track centered on the start line. Find the probability that both runners are inside that photo.
Givens: Rachel: counterclockwise, one lap every $90$ seconds; Robert: clockwise, one lap every $80$ seconds; Both leave the start line together at time $0$; The photo shows $\tfrac14$ of the track, centered on the start line, so it reaches $\tfrac18$ of a lap on each side of the start; The photo time is uniformly random on the interval from $10$ minutes ($600$ s) to $11$ minutes ($660$ s); Answer choices: (A) $\tfrac1{16}$, (B) $\tfrac18$, (C) $\tfrac{3}{16}$, (D) $\tfrac14$, (E) $\tfrac{5}{16}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #8 Analyze the Units, #5 Look for a Pattern, #7 Identify Subproblems
Tool #1 (Draw a Diagram) is the spine: model the track so 'in the photo' becomes 'within $\tfrac18$ 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 ($\tfrac18$ 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 $60$.
Execute — Answer: C
5.NF.B.4 Step 1 Model the photo as a distance
- Measure position as a fraction of a lap.
- The photo shows $\tfrac14$ of the track centered on the start line, so it stretches equally to both sides: $\tfrac14 \times \tfrac12 = \tfrac18$ of a lap on each side.
- A runner is in the photo exactly when they are within $\tfrac18$ of a lap of the start line, no matter which way they face.
💡 Centering a quarter on the line means half of it, an eighth of a lap, spills out each way.
6.RP.A.3 Step 2 Turn the arc into seconds
- Each runner covers a full lap in a fixed time, so $\tfrac18$ of a lap takes $\tfrac18$ of their lap time.
- Rachel: $\tfrac18 \times 90 = 11.25$ s.
- Robert: $\tfrac18 \times 80 = 10$ s.
- So Rachel is in the photo for the $11.25$ s before and after she crosses the start line, and Robert for the $10$ s before and after he crosses it.
💡 A steady runner spends the same slice of time on the same slice of track.
4.OA.B.4 Step 3 Find each runner at the line
- A runner sits on the start line whenever the elapsed time is a whole number of their laps.
- Rachel is at the line at multiples of $90$ s: $\ldots, 540, 630, 720, \ldots$; only $630$ s lands in $[600,660]$ (her $7$th lap).
- Robert is at the line at multiples of $80$ s: $\ldots, 560, 640, 720, \ldots$; only $640$ s lands in $[600,660]$ (his $8$th lap).
💡 You are back on the start line exactly when you have run a whole number of laps.
5.NBT.B.7 Step 4 Shade each photo window
- Attach each runner's half-width to their crossing time.
- Rachel is in the photo from $630 - 11.25 = 618.75$ s to $630 + 11.25 = 641.25$ s.
- Robert is in the photo from $640 - 10 = 630$ s to $640 + 10 = 650$ s.
- Draw these as two shaded bars on the timeline from $600$ to $660$.
💡 Each runner's window is just their crossing time plus-or-minus their own half-width.
7.NS.A.1 Step 5 Overlap the two windows
- Both runners are in the photo only where the bars overlap: from the later start $\max(618.75, 630) = 630$ s to the earlier end $\min(641.25, 650) = 641.25$ s.
- That overlap runs from $630$ s to $641.25$ s, a length of $641.25 - 630 = 11.25$ s.
💡 Two shaded bars agree only between the latest start and the earliest finish.
7.SP.C.7 Step 6 Divide to get the probability
- The photo time is uniform over the $60$-second minute, so the probability is the favorable length over the total length: $\tfrac{11.25}{60} = \tfrac{3}{16}$.
- The probability that both Rachel and Robert are in the picture is $\tfrac{3}{16}$, choice (C).
💡 For a uniformly random time, probability is just the good slice divided by the whole slice.
5.NF.B.4 Measure position as a fraction of a lap. The photo shows $\tfrac14$ of the track 6.RP.A.3 Each runner covers a full lap in a fixed time, so $\tfrac18$ of a lap takes $\tf 4.OA.B.4 A runner sits on the start line whenever the elapsed time is a whole number of t 5.NBT.B.7 Attach each runner's half-width to their crossing time. Rachel is in the photo f 7.NS.A.1 Both runners are in the photo only where the bars overlap: from the later start 7.SP.C.7 The photo time is uniform over the $60$-second minute, so the probability is the Review
Reasonableness: Every step matches the clock. Rachel truly crosses at $630$ s ($7 \times 90$) and Robert at $640$ s ($8 \times 80$), both inside the target minute. Their windows $\pm 11.25$ s and $\pm 10$ 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 — not a coincidence, since $\tfrac{11.25}{60} = \tfrac{11.25}{60}$ reduces neatly. 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 $\tfrac{3}{16}$ is choice (C).
Alternative: Tool #4 (Introduce a Variable): let $t$ be the seconds after the $600$-s mark, $0 \le t \le 60$. Rachel's distance-from-line condition is $|t - 30| \le 11.25$ (her crossing is at $t = 30$), giving $18.75 \le t \le 41.25$; Robert's is $|t - 40| \le 10$, giving $30 \le t \le 50$. Intersect the two inequalities to get $30 \le t \le 41.25$, length $11.25$, then divide by $60$. Same $\tfrac{3}{16}$, reached with absolute-value inequalities instead of a drawn timeline.
CCSS standards used (min grade 7)
4.OA.B.4Find factor pairs and recognize multiples of a whole number (Locating when each runner is on the start line — multiples of $90$ (Rachel) and $80$ (Robert) that fall in $[600,660]$.)5.NF.B.4Multiply a fraction by a fraction or whole number (Splitting the quarter-track photo into $\tfrac14 \times \tfrac12 = \tfrac18$ of a lap on each side of the start line.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Building the photo windows $630 \pm 11.25$ and $640 \pm 10$ to get the endpoints $618.75$, $641.25$, $630$, $650$.)6.RP.A.3Use ratio and rate reasoning to solve real-world problems (Converting $\tfrac18$ of a lap into seconds using each runner's lap time: $\tfrac{90}{8} = 11.25$ s and $\tfrac{80}{8} = 10$ s.)7.NS.A.1Add and subtract rational numbers on a number line (Intersecting the two time windows on the timeline and measuring the overlap length $641.25 - 630 = 11.25$ s.)7.SP.C.7Develop a uniform probability model and use it to find probabilities (Turning the uniform random photo time into probability $= \tfrac{11.25}{60} = \tfrac{3}{16}$.)
⭐ 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) $\tfrac{3}{16}$.
⭐ 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) $\tfrac{3}{16}$.
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