Difference between revisions of "2014 AMC 8 Problems/Problem 25"
(→Problem) |
m (→Solution 2) |
||
(40 intermediate revisions by 29 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | A straight one-mile stretch of highway, | + | 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 | ||
− | + | <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> | |
− | + | ==Video Solution for Problems 21-25== | |
− | + | https://www.youtube.com/watch?v=6S0u_fDjSxc | |
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/zi01koyAVhM ~savannahsolver | ||
==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 to arrive at <math>\textbf{(B) }\frac{\pi}{10}</math> hours. | + | ===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>, 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}} |
Revision as of 12:16, 17 January 2024
Contents
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
Video Solution for Problems 21-25
https://www.youtube.com/watch?v=6S0u_fDjSxc
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 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
, 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.