AMC 10 · 2006 · #15
Grade 7 geometry-2dOdell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
Pick an answer.
AMC 10 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: Two runners start on the same radial line of a circular track and run for $30$ minutes. Odell goes clockwise at $250$ m/min on the inner lane (radius $50$ m); Kershaw goes counterclockwise at $300$ m/min on the outer lane (radius $60$ m). We count how many times, after the start, the two line up on the same radial line — that is, how many times they pass each other.
Givens: Total running time is $30$ minutes; Odell: clockwise, $250$ m/min, radius $50$ m; Kershaw: counterclockwise, $300$ m/min, radius $60$ m; They begin on the same radial line (same starting angle); Answer choices: (A) 29, (B) 42, (C) 45, (D) 47, (E) 50
Unknowns: The number of times after the start that the two runners pass each other
Understand
Restated: Two runners start on the same radial line of a circular track and run for $30$ minutes. Odell goes clockwise at $250$ m/min on the inner lane (radius $50$ m); Kershaw goes counterclockwise at $300$ m/min on the outer lane (radius $60$ m). We count how many times, after the start, the two line up on the same radial line — that is, how many times they pass each other.
Givens: Total running time is $30$ minutes; Odell: clockwise, $250$ m/min, radius $50$ m; Kershaw: counterclockwise, $300$ m/min, radius $60$ m; They begin on the same radial line (same starting angle); Answer choices: (A) 29, (B) 42, (C) 45, (D) 47, (E) 50
Plan
Primary tool: #8 Analyze the Units
Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #3 Eliminate Possibilities
The runners are on different-sized circles, so comparing their meters is misleading — what lines them up is how far around the circle each has gone, i.e. laps. Tool #8 (Analyze the Units) converts each speed in meters-per-minute into minutes-per-lap by dividing by the circumference $2\pi r$; this is the move that exposes the hidden coincidence — both take the same time per lap. Tool #1 (Draw a Diagram) reframes 'pass each other' as 'same radial line', so lanes stop mattering. Tool #4 (Introduce a Variable) names one lap-time $T$ and turns opposite-direction motion into a fixed meeting interval. Finally, because the count comes out just below a whole number, Tool #3 (Eliminate Possibilities) settles the tempting $48$-vs-$47$ rounding trap against the answer list.
Execute — Answer: D
7.G.B.4 Step 1 Reframe passing as same angle
- Since the runners are in different lanes, they never collide; they 'pass each other' whenever they sit on the same radial line — the same fraction of the way around the circle.
- So the radii $50$ and $60$ only tell us the lap lengths; what we really track is each runner's angular position, measured in laps.
💡 Being side by side on the track means the same direction out from the center, so different lane sizes drop out and only the angle matters.
7.RP.A.1 Step 2 Find each runner's time per lap
- A lap length is the circumference $2\pi r$.
- Odell's lap is $2\pi(50) = 100\pi$ meters, run at $250$ m/min, taking $100\pi / 250 = 0.4\pi$ minutes.
- Kershaw's lap is $2\pi(60) = 120\pi$ meters at $300$ m/min, taking $120\pi / 300 = 0.4\pi$ minutes.
- Watch the units: $\text{m} \div (\text{m}/\text{min}) = \text{min}$.
- Both lap times are exactly equal: $T = 0.4\pi$ min per lap.
💡 Dividing meters-per-minute into the meters of one lap flips the rate into minutes-per-lap, and the bigger lane's longer track exactly cancels its owner's faster speed.
7.NS.A.1 Step 3 Turn opposite directions into a meeting interval
- Give the shared lap time the name $T = 0.4\pi$.
- Because they go opposite ways, after a time $t$ Odell has swept $t/T$ of a lap one way and Kershaw $t/T$ the other way, so the angular gap between them is $2\,(t/T)$ laps.
- They line up each time this gap reaches a whole number of laps: $2\,(t/T) = 1, 2, 3, \dots$ So passings happen every half lap-time, once every $T/2 = 0.2\pi$ minutes, the first at $t = 0.2\pi$ min.
💡 Running toward each other around the loop closes the gap twice as fast, so a whole-circle gap is used up every half of one runner's lap time.
7.NS.A.3 Step 4 Count the passings in 30 minutes
- The number of passings is how many meeting intervals fit into $30$ minutes: $30 \div 0.2\pi = \dfrac{150}{\pi} \approx 47.75$.
- Only whole passings count, and the fraction $0.75$ does not finish a $48$th one.
- Check the boundary: the $47$th passing is at $47 \times 0.2\pi = 9.4\pi \approx 29.5$ min (inside the $30$ min), while the $48$th would be at $9.6\pi \approx 30.2$ min (too late).
- So there are $47$ passings, which is $\textbf{(D) } 47$.
💡 You only count meetings that actually happen before time runs out, so a count of $47.75$ rounds down, not up.
7.G.B.4 Since the runners are in different lanes, they never collide; they 'pass each ot 7.RP.A.1 A lap length is the circumference $2\pi r$. Odell's lap is $2\pi(50) = 100\pi$ m 7.NS.A.1 Give the shared lap time the name $T = 0.4\pi$. Because they go opposite ways, a 7.NS.A.3 The number of passings is how many meeting intervals fit into $30$ minutes: $30 Review
Reasonableness: Sanity-check the size: about one passing every $0.2\pi \approx 0.63$ minutes over $30$ minutes gives roughly $30/0.63 \approx 48$, so an answer in the high-$40$s is right, ruling out $29$ and $42$. The exact value $150/\pi \approx 47.75$ is not a whole number, so it must round down to the last completed passing, $47$; the boundary check confirms the $48$th meeting falls at $\approx 30.2$ min, just after the whistle. The trap answer $\textbf{(E) } 50$ comes from rounding $47.75$ up or from mishandling the half-lap interval, and $\textbf{(C) } 45$/$\textbf{(B) } 42$ come from comparing meters instead of laps.
Alternative: Use angular speed directly: turning speed is $v/r$, so Odell spins at $250/50 = 5$ and Kershaw at $300/60 = 5$ radians per minute. Opposite directions add, so their angular positions separate at $5 + 5 = 10$ rad/min. They pass once per full circle of relative turn, $2\pi$ radians, so the count is $\dfrac{10 \times 30}{2\pi} = \dfrac{300}{2\pi} = \dfrac{150}{\pi} \approx 47.75$, giving $47$ passings — the same answer.
CCSS standards used (min grade 7)
7.G.B.4Know the formulas for area and circumference of a circle (Using circumference $2\pi r$ to turn each lane's radius into its lap length and to treat 'passing' as sharing an angle on the circle.)7.RP.A.1Compute unit rates associated with ratios of fractions (Dividing lap length (meters) by speed (meters per minute) to get each runner's minutes-per-lap and spotting that both equal $0.4\pi$.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Combining the two opposite-direction motions so their angular gap grows at the sum of the rates, giving a passing every half lap time.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Dividing $30$ minutes by the meeting interval $0.2\pi$ and interpreting $47.75$ as $47$ completed passings.)
⭐ Different lanes only change lap length, so turn each speed into minutes-per-lap; here both runners lap in the same time, they meet every half-lap since they run opposite ways, and $30 \div 0.2\pi \approx 47.75$ rounds down to $47$ real passings.
⭐ Different lanes only change lap length, so turn each speed into minutes-per-lap; here both runners lap in the same time, they meet every half-lap since they run opposite ways, and $30 \div 0.2\pi \approx 47.75$ rounds down to $47$ real passings.
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