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 "2014 AMC 8 Problems/Problem 25"

m (→‎Problem)
Line 12: Line 12:
 
==Solution 2: ==
 
==Solution 2: ==
 
If Robert rides in a straight line, it will take him <math>\frac{1}{5}</math> hours. When riding in semicircles,  let the radius of the semicircle <math>r</math>, then the circumference of a semicircle is <math>\pi r</math>. The ratio of the circumference of the semicircle to its diameter is <math>\frac{\pi}{2}</math>, so the time Robert takes is <math>\frac{1}{5} * \frac{\pi}{2}</math>, which equals to <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours.
 
If Robert rides in a straight line, it will take him <math>\frac{1}{5}</math> hours. When riding in semicircles,  let the radius of the semicircle <math>r</math>, then the circumference of a semicircle is <math>\pi r</math>. The ratio of the circumference of the semicircle to its diameter is <math>\frac{\pi}{2}</math>, so the time Robert takes is <math>\frac{1}{5} * \frac{\pi}{2}</math>, which equals to <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours.
 +
 +
==See Also==
 +
{{AMC8 box|year=2014|num-b=19|num-a=21}}
 +
{{MAA Notice}}

Revision as of 01:45, 3 November 2019

Problem

A straight one-mile stretch of highway, $40$ feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at $5$ miles per hour, how many hours will it take to cover the one-mile stretch?

Note: $1$ mile = $5280$ feet

[asy] size(10cm); pathpen=black; pointpen=black; D(arc((-2,0),1,300,360)); D(arc((0,0),1,0,180)); D(arc((2,0),1,180,360)); D(arc((4,0),1,0,180)); D(arc((6,0),1,180,240)); D((-1.5,1)--(5.5,1)); D((-1.5,0)--(5.5,0),dashed); D((-1.5,-1)--(5.5,-1)); [/asy] $\textbf{(A) }\frac{\pi}{11}\qquad\textbf{(B) }\frac{\pi}{10}\qquad\textbf{(C) }\frac{\pi}{5}\qquad\textbf{(D) }\frac{2\pi}{5}\qquad\textbf{(E) }\frac{2\pi}{3}$

Solution 1:

There are two possible interpretations of the problem: that the road as a whole is $40$ feet wide, or that each lane is $40$ feet wide. Both interpretations will arrive at the same result. However, let us stick with the first interpretation for simplicity. Each lane must then be $20$ feet wide, so Robert must be riding his bike in semicircles with radius $20$ feet and diameter $40$ feet. Since the road is $5280$ feet long, over the whole mile, Robert rides $\frac{5280}{40} =132$ semicircles in total. Were the semicircles full circles, their circumference would be $2\pi\cdot 20=40\pi$ feet; as it is, the circumference of each is half that, or $20\pi$ feet. Therefore, over the stretch of highway, Robert rides a total of $132\cdot 20\pi =2640\pi$ feet, equivalent to $\frac{\pi}{2}$ miles. Robert rides at 5 miles per hour, so divide the $\frac{\pi}{2}$ miles by $5$ mph (because t = d/r time = distance/rate) to arrive at $\boxed{\textbf{(B) }\frac{\pi}{10}}$ hours.

Solution 2:

If Robert rides in a straight line, it will take him $\frac{1}{5}$ hours. When riding in semicircles, let the radius of the semicircle $r$, then the circumference of a semicircle is $\pi r$. The ratio of the circumference of the semicircle to its diameter is $\frac{\pi}{2}$, so the time Robert takes is $\frac{1}{5} * \frac{\pi}{2}$, which equals to $\boxed{\textbf{(B) }\frac{\pi}{10}}$ hours.

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AJHSME/AMC 8 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=2014_AMC_8_Problems/Problem_25&oldid=110799"