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

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==Problem==
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a
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
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==Solution==
 
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IF YOU READ THIS YOU ARE GAY
==Video Solution for Problems 21-25==
 
https://www.youtube.com/watch?v=6S0u_fDjSxc
 
 
 
==Solution(s)==
 
 
===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.
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GYATTTTTT
 
 
===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} \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 16:41, 28 September 2024

a

Solution

IF YOU READ THIS YOU ARE GAY

Solution 1

GYATTTTTT

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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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