Difference between revisions of "2014 AMC 8 Problems/Problem 25"
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| − | There are two possible interpretations of the problem: that the road as a whole is <math>40</math> feet wide, or that each lane is <math>40</math> 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 <math>20</math> feet wide, so Robert must be riding his bike in semicircles with radius <math>20</math> feet and diameter <math>40</math> feet. Since the road is <math>5280</math> feet long, over the whole mile, Robert rides <math>\frac{5280}{40} =132</math> semicircles in total. Were the semicircles full circles, their circumference would be <math>2\pi\cdot 20=40\pi</math> feet; as it is, the circumference of each is half that, or <math>20\pi</math> feet. Therefore, over the stretch of highway, Robert rides a total of <math>132\cdot 20\pi =2640\pi</math> feet, equivalent to <math>\frac{\pi}{2}</math> miles. Robert rides at 5 miles per hour, so divide the <math>\frac{\pi}{2}</math> miles by <math>5</math> mph (because t = d/r time = distance / rate) to arrive at <math>\textbf{(B) }\frac{\pi}{10}</math> hours. | + | There are two possible interpretations of the problem: that the road as a whole is <math>40</math> feet wide, or that each lane is <math>40</math> 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 <math>20</math> feet wide, so Robert must be riding his bike in semicircles with radius <math>20</math> feet and diameter <math>40</math> feet. Since the road is <math>5280</math> feet long, over the whole mile, Robert rides <math>\frac{5280}{40} =132</math> semicircles in total. Were the semicircles full circles, their circumference would be <math>2\pi\cdot 20=40\pi</math> feet; as it is, the circumference of each is half that, or <math>20\pi</math> feet. Therefore, over the stretch of highway, Robert rides a total of <math>132\cdot 20\pi =2640\pi</math> feet, equivalent to <math>\frac{\pi}{2}</math> miles. Robert rides at 5 miles per hour, so divide the <math>\frac{\pi}{2}</math> miles by <math>5</math> mph (because t = d/r time = distance/rate) to arrive at <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours. |
Revision as of 14:37, 18 March 2018
Problem
A straight one-mile stretch of highway, 
feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 
miles per hour, how many hours will it take to cover the one-mile stretch?
Note: 
mile = 
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]](http://latex.artofproblemsolving.com/1/7/1/1718fccaf6294944d241e105075b38eb48de0e55.png)
Solution
There are two possible interpretations of the problem: that the road as a whole is feet wide, or that each lane is 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 
feet wide, so Robert must be riding his bike in semicircles with radius feet and diameter feet. Since the road is feet long, over the whole mile, Robert rides 
semicircles in total. Were the semicircles full circles, their circumference would be 
feet; as it is, the circumference of each is half that, or 
feet. Therefore, over the stretch of highway, Robert rides a total of 
feet, equivalent to 
miles. Robert rides at 5 miles per hour, so divide the miles by mph (because t = d/r time = distance/rate) to arrive at 
hours.
