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Difference between revisions of "1957 AHSME Problems/Problem 11"

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To find the angle of the clock, we first have to determine where the hands are. The time is <math>2.25</math> hours, so the hour hand would be <math>67.5</math> degrees clockwise. As the clock is a quarter of the way through the hour, the minute hand is <math>90</math> degrees clockwise.
 
To find the angle of the clock, we first have to determine where the hands are. The time is <math>2.25</math> hours, so the hour hand would be <math>67.5</math> degrees clockwise. As the clock is a quarter of the way through the hour, the minute hand is <math>90</math> degrees clockwise.
  
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Thus, we can say that the angle formed by the hands is <math>90 - 67.5 = 22\frac{1}{2}^\circ</math>.
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Thus, we can say that the angle formed by the hands is <math>90 - 67.5 = 22\frac{1}{2}^\circ</math>.  This is answer  <math>\boxed{\textbf{(E)}}</math>.
  
 
==See Also==
 
==See Also==
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{{AHSME box|year=1957|num-b=10|num-a=12}}
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{{AHSME 50p box|year=1957|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category:AHSME]][[Category:AHSME Problems]]
 
[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 08:09, 25 July 2024

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 $\boxed{\textbf{(E)}}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AHSME Problems and Solutions

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