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

(Created page with "The angle formed by the hands of a clock at <math>2:15</math> is: <math>\textbf{(A)}\ 30^\circ \qquad \textbf{(B)}\ 27\frac{1}{2}^\circ\qquad \textbf{(C)}\ 157\frac{1}{2}^\c...")
 
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==Problem==
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The angle formed by the hands of a clock at <math>2:15</math> is:
 
The angle formed by the hands of a clock at <math>2:15</math> is:
  
 
<math>\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}</math>
 
<math>\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}</math>
 
==Solution==
 
==Solution==
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In order 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>62.5</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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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>62.5</math> degrees clockwise. As the clock is a quarter of the way through the hour, the minute hand is <math>90</math> degrees clockwise.
  
 
Thus, we can say that the angle formed by the hands is <math>90 - 62.5 = \boxed{\textbf{(B) } 27\frac{1}{2}}</math> degrees.
 
Thus, we can say that the angle formed by the hands is <math>90 - 62.5 = \boxed{\textbf{(B) } 27\frac{1}{2}}</math> degrees.

Revision as of 20:03, 28 October 2023

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.

See Also

1957 AHSME (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
All AHSME Problems and Solutions

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