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

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Note: 1 mile = 5280 feet
 
Note: 1 mile = 5280 feet
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<img>https://latex.artofproblemsolving.com/c/c/a/cca402c5b2dba1c7eb738c2305a07ec63fa15191.png</img>
  
 
==Video Solution for Problems 21-25==
 
==Video Solution for Problems 21-25==

Revision as of 11:59, 13 June 2022

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 <img>https://latex.artofproblemsolving.com/c/c/a/cca402c5b2dba1c7eb738c2305a07ec63fa15191.png</img>

Video Solution for Problems 21-25

https://www.youtube.com/watch?v=6S0u_fDjSxc

Video Solution

https://youtu.be/zi01koyAVhM ~savannahsolver

Solution(s)

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.

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 $\frac{1}{5} \cdot \frac{\pi}{2}$, which equals to $\boxed{\textbf{(B) }\frac{\pi}{10}}$ hours.

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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