Difference between revisions of "2021 Fall AMC 10A Problems/Problem 6"

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Revision as of 20:06, 22 November 2021

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

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 leap 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

See Also

2021 Fall AMC 10A (ProblemsAnswer KeyResources)
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

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