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

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==Problem==
 
==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?
 
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?
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<asy>size(10cm); pathpen=black; pointpen=black;
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D(arc((-2,0),1,300,360));
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D(arc((0,0),1,0,180));
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D(arc((2,0),1,180,360));
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D(arc((4,0),1,0,180));
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D(arc((6,0),1,180,240));
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D((-1.5,-1)--(5.5,-1));</asy>
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Note: 1 mile = 5280 feet
  
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Note: <math>1</math> mile = <math>5280</math> feet
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<math> \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} </math>
  
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==Video Solution for Problems 21-25==
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==Video Solution==
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https://www.youtube.com/watch?v=6S0u_fDjSxc
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https://youtu.be/zi01koyAVhM ~savannahsolver
  
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==Solution(s)==
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==Solution==
 
===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 <math>t = \frac{d}{r}</math> and 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 <math>t = \frac{d}{r}</math> and time = distance/rate) to arrive at <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours.
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===Solution 2===
 
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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} \cdot \frac{\pi}{2}</math>, which equals to <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours.
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=24|after=Last Problem}}
 
{{AMC8 box|year=2014|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:04, 3 October 2024

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? [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));[/asy] Note: 1 mile = 5280 feet

$\textbf{(A) }\frac{\pi}{11}\qquad\textbf{(B) }\frac{\pi}{10}\qquad\textbf{(C) }\frac{\pi}{5}\qquad\textbf{(D) }\frac{2\pi}{5}\qquad\textbf{(E) }\frac{2\pi}{3}$

Video Solution

https://youtu.be/zi01koyAVhM ~savannahsolver

Solution

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 $t = \frac{d}{r}$ and time = distance/rate) to arrive at $\boxed{\textbf{(B) }\frac{\pi}{10}}$ hours.

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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

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