Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2011 AMC 10B Problems/Problem 12: Difference between revisions

← Older edit
Putu2003 (talk | contribs)
160 edits
Trivia (talk | contribs)
editor
3 edits
mNo edit summary
 
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
== Problem==
==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 <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?
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 <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?
Line 5: Line 5:
<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>


==Solution==
==Solution 1==
===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,
Line 17: Line 16:
\end{align*} </cmath>
\end{align*} </cmath>


===Solution 2===
==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, think of it has just a circular track <math>6</math> meters in width. If the diameter of the smaller circle were <math>r,</math> then the length of the smaller circle would be <math>r \pi</math> and the length of the larger circle would be <math>(r+12) \pi.</math> Since it still takes <math>36</math> seconds longer,
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 <math>r,</math> then the length of the smaller circle would be <math>r \pi</math> and the length of the larger circle would be <math>(r+12) \pi.</math> Since it still takes <math>36</math> seconds longer,


<cmath> \begin{align*}
<cmath> \begin{align*}

Latest revision as of 08:43, 1 November 2025

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_12&oldid=264570"