Difference between revisions of "1955 AHSME Problems/Problem 2"
(Created page with "== Problem == The smaller angle between the hands of a clock at <math>12:25</math> p.m. is: <math> \textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{...") |
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==Solution== | ==Solution== | ||
At <math>12:25</math>, the minute hand is at <math>\frac{25}{60}\cdot 360^\circ</math>, or <math>150^\circ</math>. | At <math>12:25</math>, the minute hand is at <math>\frac{25}{60}\cdot 360^\circ</math>, or <math>150^\circ</math>. | ||
| − | The hour hand moves <math>5</math> 'minutes' every hour, or <math>\frac{5}{60}\cdot 360 = 30^\circ</math> every hour. At <math>30^\circ</math> every hour, the hour hand moves <math>\frac{1}{2}</math> minutes on the clock every minute. At <math>12:25</math>, the hour hand is at <math>\frac{25}{2}^\circ</math>. Therefore, angle between the hands is <math>150^\circ - \frac{25}{2}^\circ</math>, <math>137.5^\circ</math>, or <math>137^{\circ} 30^{'} \implies \boxed{\mathrm{(B) 137^\circ 30'}}</math> | + | The hour hand moves <math>5</math> 'minutes' every hour, or <math>\frac{5}{60}\cdot 360 = 30^\circ</math> every hour. At <math>30^\circ</math> every hour, the hour hand moves <math>\frac{1}{2}</math> minutes on the clock every minute. At <math>12:25</math>, the hour hand is at <math>\frac{25}{2}^\circ</math>. Therefore, the angle between the hands is <math>150^\circ - \frac{25}{2}^\circ</math>, <math>137.5^\circ</math>, or <math>137^{\circ} 30^{'} \implies</math> <math>\boxed{\mathrm{(B) 137^\circ 30'}}</math>. |
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| + | ==Solution 2== | ||
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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 11:37, 4 May 2020
Contents
Problem
The smaller angle between the hands of a clock at 
p.m. is:
Solution
At , the minute hand is at 
, or 
.
The hour hand moves 
'minutes' every hour, or 
every hour. At 
every hour, the hour hand moves 
minutes on the clock every minute. At , the hour hand is at 
. Therefore, the angle between the hands is 
, 
, or 
.
Solution 2
Using the formula 
, and plugging in 12 for h and 25 for m, we get the angle between to be 222.5
. Since the problem wants the smaller angle, we do 
See Also
| 1955 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
