1955 AHSME Problems/Problem 2

Problem

The smaller angle between the hands of a clock at $12:25$ p.m. is:

$\textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ$

Solution

At $12:25$ , the minute hand is at $\frac{25}{60}\cdot 360^\circ$ , or $150^\circ$ . The hour hand moves $5$ 'minutes' every hour, or $\frac{5}{60}\cdot 360 = 30^\circ$ every hour. At $30^\circ$ every hour, the hour hand moves $\frac{1}{2}$ minutes on the clock every minute. At $12:25$ , the hour hand is at $\frac{25}{2}^\circ$ . Therefore, the angle between the hands is $150^\circ - \frac{25}{2}^\circ$ , $137.5^\circ$ , or $137^{\circ} 30^{'} \implies$ $\boxed{\mathrm{(B) 137^\circ 30'}}$ .

Solution 2

Using the formula $\frac{|(60h-11m)|}{2}$ , and plugging in 12 for h and 25 for m, we get the angle between to be 222.5°. Since the problem wants the smaller angle, we do $360-222.5 = 137.5 \Rightarrow 137°30'$ .

1955 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1955 AHSME Problems/Problem 2

Contents

Problem

Solution

Solution 2

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