1953 AMC 12 Problem 10

Attempt Problem 10 of the 1953 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1953 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

The number of revolutions of a wheel, with fixed center and with an outside diameter of 66 feet, required to cause a point on the rim to go one mile is:
880880
440π\dfrac{440}{\pi}
880π\dfrac{880}{\pi}
440π440\pi
none of these
Answer: C
Concepts:circle circumferenceunit conversionrevolutions
Small hint:
In one revolution the rim point travels one circumference
Big hint:
Divide 52805280 feet by the circumference 6π6\pi feet
Solution:
The wheel’s circumference is 6π6\pi feet, so the required number of revolutions is 52806π=880π. \frac{5280}{6\pi}=\frac{880}{\pi}.
Thus, the correct answer is C.
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