AMC 10 · 2011 · #12
Grade 7 rate-ratioKeiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of 6 meters, and it takes her 36 seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?
Pick an answer.
AMC 10 2011 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: A running track has two straight sides and two semicircular ends, like a stadium. The track is $6$ meters wide. Walking around the outer edge takes $36$ seconds longer than walking around the inner edge, at the same steady speed. Find that speed in meters per second.
Givens: The track's two ends are semicircles; the two long sides are straight; The track is $6$ meters wide; Walking the outside edge takes $36$ seconds longer than the inside edge; The speed is the same constant value on both edges; Answer choices: (A) $\frac{\pi}{3}$, (B) $\frac{2\pi}{3}$, (C) $\pi$, (D) $\frac{4\pi}{3}$, (E) $\frac{5\pi}{3}$
Unknowns: Keiko's walking speed $v$ in meters per second
Understand
Restated: A running track has two straight sides and two semicircular ends, like a stadium. The track is $6$ meters wide. Walking around the outer edge takes $36$ seconds longer than walking around the inner edge, at the same steady speed. Find that speed in meters per second.
Givens: The track's two ends are semicircles; the two long sides are straight; The track is $6$ meters wide; Walking the outside edge takes $36$ seconds longer than the inside edge; The speed is the same constant value on both edges; Answer choices: (A) $\frac{\pi}{3}$, (B) $\frac{2\pi}{3}$, (C) $\pi$, (D) $\frac{4\pi}{3}$, (E) $\frac{5\pi}{3}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #8 Analyze the Units
The track length seems to depend on the unknown straight-side length and the unknown inner radius, so Tool #4 (Introduce a Variable) names them $L$ and $r$ and writes each edge as a formula. Tool #1 (Draw a Diagram) shows why the outer semicircles have radius $r+6$ while the straights stay the same. The magic is that both $L$ and $r$ cancel when we subtract, leaving a pure number. Tool #8 (Analyze the Units) then turns 'extra distance over speed equals extra time' into one clean equation for $v$.
Execute — Answer: A
7.G.B.4 Step 1 Name the parts and write the inner edge
- Let $v$ be the speed, $L$ the length of one straight side, and $r$ the radius of an inner semicircle.
- Walking the inside edge covers the two straights plus the two inner semicircles.
- Two semicircles of radius $r$ join into one full circle, whose length is $2\pi r$.
- So the inner edge has length $2L + 2\pi r$.
💡 Two matching semicircles are just a whole circle cut in half and taped back together, so their combined length is one circumference.
7.G.B.4 Step 2 Write the outer edge
- Picture the outer edge drawn $6$ meters outside the inner one.
- The straight sides run alongside the inner straights, so they still have length $L$ each.
- But each outer semicircle bulges $6$ meters farther out, so its radius is $r + 6$.
- The outer edge is therefore $2L + 2\pi(r+6)$.
💡 Moving a curved edge outward by the track's width adds that width to the turning radius, but a straight edge just shifts sideways without getting longer.
7.EE.A.1 Step 3 Subtract to find the extra distance
- The outer walk is longer only because of the wider curves.
- Subtract the two lengths: the $2L$ terms cancel and the $2\pi r$ terms cancel, leaving $2\pi \cdot 6 = 12\pi$.
- So the outer edge is exactly $12\pi$ meters longer, no matter what $L$ and $r$ are.
💡 The unknown pieces appear identically in both edges, so subtracting wipes them out and only the width's effect survives.
7.EE.B.4 Step 4 Turn the extra time into an equation
- Both walks use the same speed $v$, so time is distance divided by $v$.
- The extra distance is $12\pi$ meters and the extra time is $36$ seconds.
- Matching the units, meters divided by (meters per second) gives seconds, so $\dfrac{12\pi}{v} = 36$.
💡 Since speed is shared, only the leftover distance can explain the leftover time, and that link is time = distance / speed.
6.RP.A.3 Step 5 Solve for the speed
- Solve $\dfrac{12\pi}{v} = 36$ for $v$: multiply both sides by $v$ and divide by $36$ to get $v = \dfrac{12\pi}{36} = \dfrac{\pi}{3}$.
- Keiko walks at $\dfrac{\pi}{3}$ meters per second, which is choice (A).
💡 Once distance and time are pinned down, their ratio is the speed directly.
7.G.B.4 Let $v$ be the speed, $L$ the length of one straight side, and $r$ the radius of 7.G.B.4 Picture the outer edge drawn $6$ meters outside the inner one. The straight side 7.EE.A.1 The outer walk is longer only because of the wider curves. Subtract the two leng 7.EE.B.4 Both walks use the same speed $v$, so time is distance divided by $v$. The extra 6.RP.A.3 Solve $\dfrac{12\pi}{v} = 36$ for $v$: multiply both sides by $v$ and divide by Review
Reasonableness: Check the units and size: $\frac{\pi}{3} \approx 1.05$ meters per second is a normal walking pace. Multiplying back, the extra distance is $v \times 36 = \frac{\pi}{3}\times 36 = 12\pi$ meters, which matches the $12\pi$ we found from the geometry. The straight sides never mattered, which fits the fact that the answer choices carry no $L$.
Alternative: Ignore the straights from the start, since they take equal time on both edges. Focus only on the curved ends: the inner curves total $2\pi r$ and the outer curves total $2\pi(r+6)$, a difference of $12\pi$ meters. That $12\pi$ meters costs $36$ extra seconds, so $v = \frac{12\pi}{36} = \frac{\pi}{3}$ — the same answer (A) with no variable for the straight sides at all.
CCSS standards used (min grade 7)
7.G.B.4Know the formulas for area and circumference of a circle (Writing each edge's semicircular ends as parts of a circle of circumference $2\pi r$ (inner) and $2\pi(r+6)$ (outer).)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Subtracting the two edge lengths so the $2L$ and $2\pi r$ terms cancel, leaving $12\pi$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Building the equation $\frac{12\pi}{v} = 36$ from 'extra distance over speed equals extra time'.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Treating speed as distance divided by time and solving $\frac{12\pi}{v}=36$ to get $v=\frac{\pi}{3}$.)
⭐ On a stadium track only the curved ends grow when you step outward, so the outer lap is always $2\pi \times \text{width}$ longer — the straightaways don't matter.
⭐ On a stadium track only the curved ends grow when you step outward, so the outer lap is always $2\pi \times \text{width}$ longer — the straightaways don't matter.
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