1971 AHSME Problems/Problem 34

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Problem

An ordinary clock in a factory is running slow so that the minute hand passes the hour hand at the usual dial position($12$ o'clock, etc.) but only every $69$ minutes. At time and one-half for overtime, the extra pay to which a $\textdollar 4.00$ per hour worker should be entitled after working a normal $8$ hour day by that slow running clock, is

$\textbf{(A) }\textdollar 2.30\qquad \textbf{(B) }\textdollar 2.60\qquad \textbf{(C) }\textdollar 2.80\qquad \textbf{(D) }\textdollar 3.00\qquad \textbf{(E) }\textdollar 3.30$

Solution

If the clock were functioning properly, the minute hand would have an angular velocity of $1\tfrac{\text{rev}}{\text{hr}}$, and the hour hand would have an angular velocity of $\tfrac1{12}\tfrac{\text{rev}}{\text{hr}}$. Thus, the minute hand would move at a velocity of $1-\tfrac1{12}=\tfrac{11}{12}\tfrac{\text{rev}}{\text{hr}}$ relative to the hour hand. Thus, the minute hand would pass the hour hand every $\tfrac{12}{11}$ hours, or $\tfrac{720}{11}$ minutes (after the minute hand has made a full revolution relative to the hour hand). Because the slow clock takes $69$ minutes to do this, one hour to this clock is actually $\tfrac{69}{(\tfrac{720}{11})}=\tfrac{253}{240}$ hours. Thus, $8$ hours on the slow clock is actually $8\cdot\tfrac{253}{240}=\tfrac{253}{30}$ hours. Thus, the workers worked for an extra $\tfrac{253}{30}-8=\tfrac{13}{30}$ hours. Because, from the problem, the workers are paid $1.5$ times their usual wage when working overtime, they earn $$4\cdot1.5=$6$ per hour in overtime. Therefore, they should receive $\tfrac{13}{30}\cdot6=\tfrac{13}{5}=2.6$ extra dollars, so our answer is $\boxed{\textbf{(B) }$2.60}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
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