Difference between revisions of "2014 AMC 8 Problems/Problem 25"
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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> | <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> | ||
==Solution== | ==Solution== | ||
− | 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/ | + | ===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. | ||
+ | |||
+ | ===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== | ||
+ | {{AMC8 box|year=2014|num-b=24|after=Last Problem}} | ||
+ | {{MAA Notice}} |
Latest revision as of 03:39, 16 January 2021
Problem
A straight one-mile stretch of highway, feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at miles per hour, how many hours will it take to cover the one-mile stretch?
Note: mile = feet
Solution
Solution 1
There are two possible interpretations of the problem: that the road as a whole is feet wide, or that each lane is 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 feet wide, so Robert must be riding his bike in semicircles with radius feet and diameter feet. Since the road is feet long, over the whole mile, Robert rides semicircles in total. Were the semicircles full circles, their circumference would be feet; as it is, the circumference of each is half that, or feet. Therefore, over the stretch of highway, Robert rides a total of feet, equivalent to miles. Robert rides at 5 miles per hour, so divide the miles by mph (because and time = distance/rate) to arrive at hours.
Solution 2
If Robert rides in a straight line, it will take him hours. When riding in semicircles, let the radius of the semicircle , then the circumference of a semicircle is . The ratio of the circumference of the semicircle to its diameter is , so the time Robert takes is , which equals to 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.