Difference between revisions of "1953 AHSME Problems/Problem 10"
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We know that the radius of the wheel is <math>3</math> feet, so the total circumference of the wheel is <math>6\pi</math> feet. We also know that one mile is equivalent to <math>5280</math> feet. It takes <math>\frac{5280}{6\pi}</math> revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is <math>\boxed{\textbf{(C) } \frac{880}{\pi}}</math>. | We know that the radius of the wheel is <math>3</math> feet, so the total circumference of the wheel is <math>6\pi</math> feet. We also know that one mile is equivalent to <math>5280</math> feet. It takes <math>\frac{5280}{6\pi}</math> revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is <math>\boxed{\textbf{(C) } \frac{880}{\pi}}</math>. | ||
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+ | ==See Also== | ||
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+ | {{AHSME 50p box|year=1953|num-b=9|num-a=11}} | ||
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+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:47, 1 April 2017
Problem
The number of revolutions of a wheel, with fixed center and with an outside diameter of feet, required to cause a point on the rim to go one mile is:
Solution
We know that the radius of the wheel is feet, so the total circumference of the wheel is feet. We also know that one mile is equivalent to feet. It takes revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AHSME Problems and Solutions |
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