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Difference between revisions of "2011 AMC 12B Problems/Problem 8"

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==Problem==
 
==Problem==
  
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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 width <math>6</math> meters, and it takes her <math>36</math> 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?
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Keiko walks once around a track at the same constant speed every day.  The sides of the track are straight, and the ends are semicircles.  The track has a width of <math>6</math> meters, and it takes her <math>36</math> 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?
  
 
<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==
 
==Solution==
  
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To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in time it takes her for each distance.  We are given the difference in time, so all we need to find is the difference between the distances.
+
To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in the time it takes her for each distance.  We are given the difference in time, so all we need to find is the difference between the distances.
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The track is divided into lengths and curves.  The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves.
 
The track is divided into lengths and curves.  The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves.
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The curves of the track are semicircles, but since there are two of them, we can consider both of the at the same time by treating them as a single circle.  We need to find the difference in the circumferences of the inside and outside edges of the circle.
 
The curves of the track are semicircles, but since there are two of them, we can consider both of the at the same time by treating them as a single circle.  We need to find the difference in the circumferences of the inside and outside edges of the circle.
−
 
  
 
The formula for the circumference of a circle is <math>C = 2 * \pi * r</math> where <math>r</math> is the radius of the circle.
 
The formula for the circumference of a circle is <math>C = 2 * \pi * r</math> where <math>r</math> is the radius of the circle.
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Let's define the circumference of the inside circle as <math>C_1</math> and the circumference of the outside circle as <math>C_2</math>.
 
Let's define the circumference of the inside circle as <math>C_1</math> and the circumference of the outside circle as <math>C_2</math>.
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If the radius of the inside circle (<math>r_1</math>) is <math>n</math>, then given the thickness of the track is 6 meters, the radius of the outside circle (<math>r_2</math>) is <math>n + 6</math>.
 
If the radius of the inside circle (<math>r_1</math>) is <math>n</math>, then given the thickness of the track is 6 meters, the radius of the outside circle (<math>r_2</math>) is <math>n + 6</math>.
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Using this, the difference in the circumferences is:
 
Using this, the difference in the circumferences is:
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<math> C_2 - C_1 = 2 * \pi * (r_2 - r_1) = 2 * \pi * (n + 6 - n) = 2 * \pi * 6 = 12\pi </math>
 
<math> C_2 - C_1 = 2 * \pi * (r_2 - r_1) = 2 * \pi * (n + 6 - n) = 2 * \pi * 6 = 12\pi </math>
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<math> 12\pi </math> is the difference between the inside and outside lengths of the track.  Divided by the time differential, we get:
 
<math> 12\pi </math> is the difference between the inside and outside lengths of the track.  Divided by the time differential, we get:
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<math> 12\pi \div 36 = \boxed {\textbf{(A)}\ \frac{\pi}{3}} </math>
 
<math> 12\pi \div 36 = \boxed {\textbf{(A)}\ \frac{\pi}{3}} </math>
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== See also ==
 
== See also ==
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{{AMC12 box|year=2011|num-b=7|num-a=9|ab=B}}
 
{{AMC12 box|year=2011|num-b=7|num-a=9|ab=B}}
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{{MAA Notice}}

Latest revision as of 18:35, 7 August 2023

Problem

Keiko walks once around a track at 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

To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in the time it takes her for each distance. We are given the difference in time, so all we need to find is the difference between the distances.

The track is divided into lengths and curves. The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves.

The curves of the track are semicircles, but since there are two of them, we can consider both of the at the same time by treating them as a single circle. We need to find the difference in the circumferences of the inside and outside edges of the circle.

The formula for the circumference of a circle is $C = 2 * \pi * r$ where $r$ is the radius of the circle.

Let's define the circumference of the inside circle as $C_1$ and the circumference of the outside circle as $C_2$.

If the radius of the inside circle ($r_1$) is $n$, then given the thickness of the track is 6 meters, the radius of the outside circle ($r_2$) is $n + 6$.

Using this, the difference in the circumferences is:

$C_2 - C_1 = 2 * \pi * (r_2 - r_1) = 2 * \pi * (n + 6 - n) = 2 * \pi * 6 = 12\pi$

$12\pi$ is the difference between the inside and outside lengths of the track. Divided by the time differential, we get:

$12\pi \div 36 = \boxed {\textbf{(A)}\ \frac{\pi}{3}}$

See also

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 12 Problems and Solutions

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