Difference between revisions of "1951 AHSME Problems/Problem 12"

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== Solution ==  
 
== Solution ==  
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Using the formula <math>\frac{|60h-11m|}{2}</math>, where <math>h</math> is the hour time and <math>m</math> is the minute time, we get that the angle is <math>\frac{|60(2)-11(15)|}{2}=\frac{|120-165|}{2}=\frac{45}{2}=\boxed{\textbf{(C)}\ 22\frac {1}{2}^{\circ} }</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 21:05, 10 April 2013

Problem

At $2: 15$ o'clock, the hour and minute hands of a clock form an angle of:

$\textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 5^{\circ} \qquad\textbf{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textbf{(E)}\ 28^{\circ}$

Solution

Using the formula $\frac{|60h-11m|}{2}$, where $h$ is the hour time and $m$ is the minute time, we get that the angle is $\frac{|60(2)-11(15)|}{2}=\frac{|120-165|}{2}=\frac{45}{2}=\boxed{\textbf{(C)}\ 22\frac {1}{2}^{\circ} }$

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AHSME Problems and Solutions
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