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Difference between revisions of "2011 AMC 10B Problems/Problem 12"

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<math> \textbf{(A)}\ \frac{\pi}{3} \qquad\textbf{(B)}\ \frac{2\pi}{3} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \frac{5\pi}{3}</math>
 
<math> \textbf{(A)}\ \frac{\pi}{3} \qquad\textbf{(B)}\ \frac{2\pi}{3} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \frac{5\pi}{3}</math>
  
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==Solution==
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==Solution 1==
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===Solution 1===
 
  
 
Let <math>s</math> be Keiko's speed in meters per second, <math>a</math> be the length of the straight parts of the track, <math>b</math> be the radius of the smaller circles, and <math>b+6</math> be the radius of the larger circles. The length of the inner edge will be <math>2a+2b \pi </math> and the length of the outer edge will be <math>2a+2\pi (b+6).</math> Since it takes <math>36</math> seconds longer for Keiko to walk on the outer edge,
 
Let <math>s</math> be Keiko's speed in meters per second, <math>a</math> be the length of the straight parts of the track, <math>b</math> be the radius of the smaller circles, and <math>b+6</math> be the radius of the larger circles. The length of the inner edge will be <math>2a+2b \pi </math> and the length of the outer edge will be <math>2a+2\pi (b+6).</math> Since it takes <math>36</math> seconds longer for Keiko to walk on the outer edge,

Revision as of 09:03, 19 December 2024

Problem

Keiko 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?

$\textbf{(A)}\ \frac{\pi}{3} \qquad\textbf{(B)}\ \frac{2\pi}{3} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \frac{5\pi}{3}$

Solution 1

Let $s$ be Keiko's speed in meters per second, $a$ be the length of the straight parts of the track, $b$ be the radius of the smaller circles, and $b+6$ be the radius of the larger circles. The length of the inner edge will be $2a+2b \pi$ and the length of the outer edge will be $2a+2\pi (b+6).$ Since it takes $36$ seconds longer for Keiko to walk on the outer edge,

\begin{align*} \frac{2a+2b \pi}{s} + 36 &= \frac{2a+2\pi (b+6)}{s}\\ 2a+2b\pi +36s &= 2a+2b\pi +12\pi\\ 36s&=12\pi\\ s&=\boxed{\textbf{(A)} \frac{\pi}{3}} \end{align*}

Solution 2

It is basically the same as Solution 1 except you can completely disregard the straight edges of the track since it will take Keiko the same time to walk that length for both inner and outer circles. Instead, focus on the circular part. If the diameter of the smaller circle were $r,$ then the length of the smaller circle would be $r \pi$ and the length of the larger circle would be $(r+12) \pi.$ Since it still takes $36$ seconds longer,

\begin{align*} \frac{r \pi}{s} + 36 &= \frac{(r+12)\pi}{s}\\ r \pi + 36s &= r \pi + 12 \pi\\ s&=\boxed{\textbf{(A)} \frac{\pi}{3}} \end{align*}

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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