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Difference between revisions of "2014 AMC 8 Problems/Problem 25"

(I fixed the links in the See Also box that go to previous and next problems.)
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==Solution 1: ==
 
==Solution 1: ==
 
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.
 
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.
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[[User:Bsu1|Bsu1]] ([[User talk:Bsu1|talk]])Hoped this helped!
  
 
==Solution 2: ==
 
==Solution 2: ==

Revision as of 20:08, 7 September 2020

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.


Bsu1 (talk)Hoped this helped!

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)
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Problem 24 		Followed by
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All AJHSME/AMC 8 Problems and Solutions

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