Difference between revisions of "1971 AHSME Problems/Problem 34"
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== Solution == | == Solution == | ||
− | <math>\boxed{\textbf{(B) }$2.60}</math>. | + | If the clock were functioning properly, the minute hand would have an angular velocity of <math>1\tfrac{\text{rev}}{\text{hr}}</math>, and the hour hand would have an angular velocity of <math>\tfrac1{12}\tfrac{\text{rev}}{\text{hr}}</math>. Thus, the minute hand would move at a velocity of <math>1-\tfrac1{12}=\tfrac{11}{12}\tfrac{\text{rev}}{\text{hr}}</math> relative to the hour hand. Thus, the minute hand would pass the hour hand every <math>\tfrac{12}{11}</math> hours, or <math>\tfrac{720}{11}</math> minutes (after the minute hand has made a full revolution relative to the hour hand). Because the slow clock takes <math>69</math> minutes to do this, one hour to this clock is actually <math>\tfrac{69}{(\tfrac{720}{11})}=\tfrac{253}{240}</math> hours. Thus, <math>8</math> hours on the slow clock is actually <math>8\cdot\tfrac{253}{240}=\tfrac{253}{30}</math> hours. Thus, the workers worked for an extra <math>\tfrac{253}{30}-8=\tfrac{13}{30}</math> hours. Because, from the problem, the workers are paid <math>1.5</math> times their usual wage when working overtime, they earn <math>$4\cdot1.5=$6</math> per hour in overtime. Therefore, they should receive <math>\tfrac{13}{30}\cdot6=\tfrac{13}{5}=2.6</math> extra dollars, so our answer is <math>\boxed{\textbf{(B) }$2.60}</math>. |
== See Also == | == See Also == | ||
{{AHSME 35p box|year=1971|num-b=33|num-a=35}} | {{AHSME 35p box|year=1971|num-b=33|num-a=35}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:18, 8 August 2024
Problem
An ordinary clock in a factory is running slow so that the minute hand passes the hour hand at the usual dial position( o'clock, etc.) but only every minutes. At time and one-half for overtime, the extra pay to which a per hour worker should be entitled after working a normal hour day by that slow running clock, is
Solution
If the clock were functioning properly, the minute hand would have an angular velocity of , and the hour hand would have an angular velocity of . Thus, the minute hand would move at a velocity of relative to the hour hand. Thus, the minute hand would pass the hour hand every hours, or minutes (after the minute hand has made a full revolution relative to the hour hand). Because the slow clock takes minutes to do this, one hour to this clock is actually hours. Thus, hours on the slow clock is actually hours. Thus, the workers worked for an extra hours. Because, from the problem, the workers are paid times their usual wage when working overtime, they earn per hour in overtime. Therefore, they should receive extra dollars, so our answer is .
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 33 |
Followed by Problem 35 | |
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All AHSME Problems and Solutions |
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