Difference between revisions of "1953 AHSME Problems/Problem 10"

(Created page with "== Problem == The number of revolutions of a wheel, with fixed center and with an outside diameter of <math>6</math> feet, required to cause a point on the rim to go one mile...")
 
 
Line 8: Line 8:
  
 
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>.
 +
 +
 +
 +
==See Also==
 +
 +
{{AHSME 50p box|year=1953|num-b=9|num-a=11}}
 +
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:47, 1 April 2017

Problem

The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:

$\textbf{(A)}\ 880 \qquad\textbf{(B)}\ \frac{440}{\pi} \qquad\textbf{(C)}\ \frac{880}{\pi} \qquad\textbf{(D)}\ 440\pi\qquad\textbf{(E)}\ \text{none of these}$

Solution

We know that the radius of the wheel is $3$ feet, so the total circumference of the wheel is $6\pi$ feet. We also know that one mile is equivalent to $5280$ feet. It takes $\frac{5280}{6\pi}$ revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is $\boxed{\textbf{(C) } \frac{880}{\pi}}$.


See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS