2021 Fall AMC 12A Problems/Problem 5

The following problem is from both the 2021 Fall AMC 10A #6 and 2021 Fall AMC 12A #5, so both problems redirect to this page.

Problem

Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?

$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$

Solution 1

There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet.

Each of Oscar's leaps covers $132\div12=11$ feet, and each of Elmer's strides covers $132\div44=3$ feet.

Therefore, Oscar's leap is $11-3=\boxed{\textbf{(B) }8}$ feet longer than Elmer's stride.

~MRENTHUSIASM

Solution 2

There are $41-1=40$ gaps between the $41$ telephone poles, so Elmer takes $44 \cdot 40 = 1760$ strides in total, and Oscar takes $12 \cdot 40 = 480$ leaps in total. Therefore, the answer is $(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{\textbf{(B) }8}$ .

~MrThinker

Video Solution (Simple and Quick)

https://youtu.be/895dqtet8MA

~Education, the Study of Everything

Video Solution by TheBeautyofMath

for AMC 10: https://youtu.be/ycRZHCOKTVk

for AMC 12: https://youtu.be/jY-17W6dA3c?t=591

~IceMatrix

Video Solution by WhyMath

https://youtu.be/AEmEKsHe2i0

~savannahsolver

Video Solution by HS Competition Academy

https://youtu.be/0HjK-M6BJ8E

~Charles3829

Video Solution

https://youtu.be/TjWjrApq-OQ

~Lucas

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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 AMC 12 Problems and Solutions

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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