Difference between revisions of "1957 AHSME Problems/Problem 11"
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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>67.5</math> degrees clockwise. As the clock is a quarter of the way through the hour, the minute hand is <math>90</math> degrees clockwise. | 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>67.5</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 - 67.5 = 22\frac{1}{2} | + | Thus, we can say that the angle formed by the hands is <math>90 - 67.5 = 22\frac{1}{2}^\circ</math>. This is answer <math>\boxed{\textbf{(E)}}</math>. |
==See Also== | ==See Also== | ||
− | {{AHSME box|year=1957|num-b=10|num-a=12}} | + | {{AHSME 50p box|year=1957|num-b=10|num-a=12}} |
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:AHSME]][[Category:AHSME Problems]] | [[Category:AHSME]][[Category:AHSME Problems]] |
Latest revision as of 08:09, 25 July 2024
Contents
Problem
The angle formed by the hands of a clock at is:
Solution
To find the angle of the clock, we first have to determine where the hands are. The time is hours, so the hour hand would be degrees clockwise. As the clock is a quarter of the way through the hour, the minute hand is degrees clockwise.
Thus, we can say that the angle formed by the hands is . This is answer .
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AHSME Problems and Solutions |
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