2014 AMC 8 Problems/Problem 25

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

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,));
D((-1.5,0)--(5.5,0),dashed);
D((-1.5,-1)--(5.5,-1));
 (Error making remote request. Unknown error_msg)

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

Solution

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 to arrive at $\textbf{(B) }\frac{\pi}{10}$ hours.

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2014 AMC 8 Problems/Problem 25

Problem

Solution