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 \boxed{\mathrm{(B) 137^\circ 30'}}</math>. |
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==See Also== | ==See Also== | ||
Revision as of 04:53, 8 July 2018
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
.
See Also
1955 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.