Difference between revisions of "1966 AHSME Problems/Problem 34"
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== Solution == | == Solution == | ||
| − | <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>. |
== See also == | == See also == | ||
{{AHSME box|year=1966|num-b=33|num-a=35}} | {{AHSME box|year=1966|num-b=33|num-a=35}} | ||
| − | [[Category: | + | [[Category:Introductory Algebra Problems]] |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:59, 29 May 2024
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
The circumference of the wheel is 
miles. Let the time for the rotation in seconds be 
. So 
. We also know reducing the time by of a second makes increase by . So 
. Solving for we get 
. So our answer is 
.
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. 
