1964 AHSME Problems/Problem 40

Problem

A watch loses $2\frac{1}{2}$ minutes per day. It is set right at $1$ P.M. on March 15. Let $n$ be the positive correction, in minutes, to be added to the time shown by the watch at a given time. When the watch shows $9$ A.M. on March 21, $n$ equals:

$\textbf{(A) }14\frac{14}{23}\qquad\textbf{(B) }14\frac{1}{14}\qquad\textbf{(C) }13\frac{101}{115}\qquad\textbf{(D) }13\frac{83}{115}\qquad \textbf{(E) }13\frac{13}{23}$

Solution

From March 15 $1$ P.M. on the watch to March 21 $9$ A.M. on the watch, the watch passed $20 + 5 \times 24 = 140$ hours.

Since $1$ watch hour equals $\frac{24}{23 + \frac{57.5}{60}} = \frac{576}{575}$ real hour, the difference between the watch time and the actual time passed is $140 \times \left( \frac{576}{575} - 1 \right) = \frac{28}{115}$ hour $=14\frac{14}{23}$ minutes.

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1964 AHSME Problems/Problem 40

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