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| − | == Problem ==
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_5]] |
| − | Elmer the emu takes <math>44</math> equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in <math>12</math> equal leaps. The telephone poles are evenly spaced, and the <math>41</math>st pole along this road is exactly one mile (<math>5280</math> feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
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| − | <math>\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15</math>
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| − | == Solution ==
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| − | There are <math>41-1=40</math> gaps between the <math>41</math> telephone poles, so the distance of each gap is <math>5280\div40=132</math> feet.
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| − | Each of Oscar's leap covers <math>132\div12=11</math> feet, and each of Elmer's strides covers <math>132\div44=3</math> feet. Therefore, Oscar's leap is <math>11-3=\boxed{\textbf{(B) }8}</math> feet longer than Elmer's stride.
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| − | ~MRENTHUSIASM
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| − | ==See Also==
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| − | {{AMC10 box|year=2021 Fall|ab=A|num-b=5|num-a=7}}
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| − | {{MAA Notice}}
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