Difference between revisions of "1966 AHSME Problems/Problem 34"
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== Solution == | == Solution == | ||
| − | <math>\ | + | <math>The circumference of the wheel is </math>\frac{11}{5280}<math> miles. Let the time for the rotation in seconds be </math>t<math>. So </math>rt=\frac{11}{5280}*3600<math>. We also know reducing the time by </math>\frac{1}{4}<math> of a second makes </math>r<math> increase by </math>5<math>. So </math>(r+5)(t-\frac{1}{4})=\frac{11}{5280}*3600<math>. Solving for </math>r<math> we get </math>r=10<math>. So our answer is </math>(B)<math> </math>10<math>.</math> |
== See also == | == See also == | ||
Revision as of 23:05, 12 September 2015
Problem
Let 
be the speed in miles per hour at which a wheel, 
feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by 
of a second, the speed is increased by 
miles per hour. Then is:
Solution
\frac{11}{5280}
t
rt=\frac{11}{5280}*3600
\frac{1}{4}
r
5(r+5)(t-\frac{1}{4})=\frac{11}{5280}*3600
r
r=10
(B)$$ (Error compiling LaTeX. Unknown error_msg)10
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 33 |
Followed by Problem 35 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
