1957 AHSME Problems/Problem 11

Problem

The angle formed by the hands of a clock at $2:15$ is:

$\textbf{(A)}\ 30^\circ \qquad \textbf{(B)}\ 27\frac{1}{2}^\circ\qquad \textbf{(C)}\ 157\frac{1}{2}^\circ\qquad \textbf{(D)}\ 172\frac{1}{2}^\circ\qquad \textbf{(E)}\ \text{none of these}$

Solution

To find the angle of the clock, we first have to determine where the hands are. The time is $2.25$ hours, so the hour hand would be $62.5$ degrees clockwise. As the clock is a quarter of the way through the hour, the minute hand is $90$ degrees clockwise.

Thus, we can say that the angle formed by the hands is $90 - 62.5 = \boxed{\textbf{(B) } 27\frac{1}{2}}$ degrees.

1957 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1957 AHSME Problems/Problem 11

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