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

Difference between revisions of "2006 AMC 10A Problems/Problem 15"

m (→‎Problem)
m (→‎Solution 3 by Alcumus)
 
(10 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Odell and Kershaw run for 30 minutes on a [[circle|circular]] track. Odell runs clockwise at 250 m/min and uses the inner lane with a [[radius]] of 50 meters. Kershaw runs counterclockwise at 300 m/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?  
+
Odell and Kershaw run for <math>30</math> minutes on a [[circle|circular]] track. Odell runs clockwise at <math>250 m/min</math> and uses the inner lane with a radius of <math>50</math> meters. Kershaw runs counterclockwise at <math>300 m/min</math> and uses the outer lane with a radius of <math>60</math> meters, starting on the same radial line as Odell. How many times after the start do they pass each other?  
  
−
 
+
<math>\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad</math>
−
<math>\mathrm{(A) \ } 29\qquad\mathrm{(B) \ } 42\qquad\mathrm{(C) \ } 45\qquad\mathrm{(D) \ } 47\qquad\mathrm{(E) \ } 50\qquad</math>
 
  
 
== Solution ==
 
== Solution ==
Line 17: Line 16:
 
</asy></center>
 
</asy></center>
  
−
Since <math>d = rt</math>, we note that Odell runs one lap in <math>\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}</math> minutes, while Kershaw also runs one lap in <math>\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}</math> minutes. They take the same amount of [[time]] to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are <math>\frac{30}{\frac{2\pi}{5}} \approx 23.8</math> laps run by both, or <math>\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 = 47</math> meeting points <math> \Longrightarrow \mathrm{(D)}</math>.
+
Since <math>d = rt</math>, we note that Odell runs one lap in <math>\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}</math> minutes, while Kershaw also runs one lap in <math>\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}</math> minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are <math>\frac{30}{\frac{2\pi}{5}} \approx 23.8</math> laps run by both, or <math>\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 =\boxed{\textbf{(D) } 47}</math> meeting points.
 +
 
 +
== Solution 2 ==
 +
We first find the amount of minutes, <math>k</math>, until Odell and Kershaw's next meeting. Let <math>a</math> be the angle in [[radians]] between their starting point and the point where they first meet, measured counterclockwise.
 +
 
 +
Since Kershaw has traveled <math>300k</math> meters at this point and the circumference of his track is <math>120\pi</math>, <math>a=\frac{300k}{120\pi}\cdot 2\pi</math>. Similarly, <math>2\pi-a=\frac{250k}{100\pi}\cdot{2\pi}</math> since Odell has traveled <math>250k</math> meters in the opposite direction and the circumference of his track is <math>100\pi</math>.
 +
 
 +
Solving for <math>a</math> in the second equation, we get <math>a=2\pi-\frac{250k}{100\pi}\cdot 2\pi</math>. Then, from the first equation, we have <math>\frac{300k}{120\pi}\cdot 2\pi=2\pi-\frac{250k}{100\pi}\cdot 2\pi</math>. Solving for <math>k</math>, we get <math>k=\frac{\pi}{5}</math>. After <math>k</math> minutes, they are back at the same position, except rotated, so they will meet again in <math>k</math> minutes. So the total amount of meetings is <math>\lfloor\frac{30}{k}\rfloor=\lfloor\frac{150}{\pi}\rfloor=\boxed{\textbf{(D) }47}</math>.
 +
 
 +
~apsid
 +
 
 +
== Solution 3 by Alcumus (ikr)==
 +
Since Odell's rate is <math>5/6</math> that of Kershaw, but Kershaw's lap distance is <math>6/5</math> that of Odell, they each run a lap in the same time. Hence they pass twice each time they circle the track. Odell runs <cmath>(30 \ \text{min})\left(250\frac{\text{m}}{\text{min}}\right)\left(\frac{1}{100\pi}\frac{\text{laps}}{\text{m}}\right)= \frac{75}{\pi}\,\text{laps}\approx 23.87\ \text{laps},</cmath> as does Kershaw. Because <math>23.5 < 23.87 < 24</math>, they pass each other <math>2(23.5)=\boxed{47}</math> times.
  
 
== See Also ==
 
== See Also ==
Line 24: Line 35:
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:24, 23 October 2024

Problem

Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ 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?

$\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad$

Solution

[asy] draw((5,0){up}..{left}(0,5),red); draw((-5,0){up}..{right}(0,5),red); draw((5,0){down}..{left}(0,-5),red); draw((-5,0){down}..{right}(0,-5),red); draw((6,0){up}..{left}(0,6),blue); draw((-6,0){up}..{right}(0,6),blue); draw((6,0){down}..{left}(0,-6),blue); draw((-6,0){down}..{right}(0,-6),blue); [/asy]

Since $d = rt$, we note that Odell runs one lap in $\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}$ minutes, while Kershaw also runs one lap in $\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}$ minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are $\frac{30}{\frac{2\pi}{5}} \approx 23.8$ laps run by both, or $\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 =\boxed{\textbf{(D) } 47}$ meeting points.

Solution 2

We first find the amount of minutes, $k$, until Odell and Kershaw's next meeting. Let $a$ be the angle in radians between their starting point and the point where they first meet, measured counterclockwise.

Since Kershaw has traveled $300k$ meters at this point and the circumference of his track is $120\pi$, $a=\frac{300k}{120\pi}\cdot 2\pi$. Similarly, $2\pi-a=\frac{250k}{100\pi}\cdot{2\pi}$ since Odell has traveled $250k$ meters in the opposite direction and the circumference of his track is $100\pi$.

Solving for $a$ in the second equation, we get $a=2\pi-\frac{250k}{100\pi}\cdot 2\pi$. Then, from the first equation, we have $\frac{300k}{120\pi}\cdot 2\pi=2\pi-\frac{250k}{100\pi}\cdot 2\pi$. Solving for $k$, we get $k=\frac{\pi}{5}$. After $k$ minutes, they are back at the same position, except rotated, so they will meet again in $k$ minutes. So the total amount of meetings is $\lfloor\frac{30}{k}\rfloor=\lfloor\frac{150}{\pi}\rfloor=\boxed{\textbf{(D) }47}$.

~apsid

Solution 3 by Alcumus (ikr)

Since Odell's rate is $5/6$ that of Kershaw, but Kershaw's lap distance is $6/5$ that of Odell, they each run a lap in the same time. Hence they pass twice each time they circle the track. Odell runs \[(30 \ \text{min})\left(250\frac{\text{m}}{\text{min}}\right)\left(\frac{1}{100\pi}\frac{\text{laps}}{\text{m}}\right)= \frac{75}{\pi}\,\text{laps}\approx 23.87\ \text{laps},\] as does Kershaw. Because $23.5 < 23.87 < 24$, they pass each other $2(23.5)=\boxed{47}$ times.

See Also

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_15&oldid=230499"