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 $67.5$ 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 $90 - 67.5 = 22\frac{1}{2}^\circ$ . This is answer $\boxed{\textbf{(E)}}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1957 AHSME Problems/Problem 11

Contents

Problem

Solution

See Also