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2021 Fall AMC 12A Problem 5

Problem 5 of 25EasierCombinatoricsArithmetic

Elmer the emu takes 4444 equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in 1212 equal leaps. The telephone poles are evenly spaced, and the 4141st pole along this road is exactly one mile (52805280 feet) from the first pole. How much longer, in feet, is Oscar’s leap than Elmer’s stride?

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Solution

There are 4040 gaps between the first and 4141st poles, so each gap is 528040=132\dfrac{5280}{40} = 132 feet. Elmer’s stride is 13244=3\dfrac{132}{44} = 3 feet and Oscar’s leap is 13212=11\dfrac{132}{12} = 11 feet, a difference of 11−3=811 - 3 = 8 feet. Thus, the correct answer is B.
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