Difference between revisions of "1955 AHSME Problems/Problem 2"

(Solution 2)
 
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==Solution 2==
 
==Solution 2==
  
Using the formula <math>\frac{|(60h-11m)|}{2}</math>, and plugging in 12 for h and 25 for m, we get the angle between to be 222.. Since the problem wants the smaller angle, we do <math>360-222.5 = 137.5 \Rightarrow 137°30'</math>.
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Using the formula <math>\frac{|(60h-11m)|}{2}</math>, and plugging in 12 for h and 25 for m, we get the angle between to be 222.5<math>^{\circ}</math>. Since the problem wants the smaller angle, we do <math>360-222.5 = 137.5 \Rightarrow 137^{\circ}30'\Rightarrow \fbox{B}</math>
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==See Also==
 
==See Also==
  

Latest revision as of 12:37, 4 May 2020

Problem

The smaller angle between the hands of a clock at $12:25$ p.m. is:

$\textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ$

Solution

At $12:25$, the minute hand is at $\frac{25}{60}\cdot 360^\circ$, or $150^\circ$. The hour hand moves $5$ 'minutes' every hour, or $\frac{5}{60}\cdot 360 = 30^\circ$ every hour. At $30^\circ$ every hour, the hour hand moves $\frac{1}{2}$ minutes on the clock every minute. At $12:25$, the hour hand is at $\frac{25}{2}^\circ$. Therefore, the angle between the hands is $150^\circ - \frac{25}{2}^\circ$, $137.5^\circ$, or $137^{\circ} 30^{'} \implies$ $\boxed{\mathrm{(B) 137^\circ 30'}}$.

Solution 2

Using the formula $\frac{|(60h-11m)|}{2}$, and plugging in 12 for h and 25 for m, we get the angle between to be 222.5$^{\circ}$. Since the problem wants the smaller angle, we do $360-222.5 = 137.5 \Rightarrow 137^{\circ}30'\Rightarrow \fbox{B}$

See Also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AHSME Problems and Solutions


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