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}} degrees. < | + | Thus, we can say that the angle formed by the hands is <math>90 - 67.5 = 22\frac{1}{2}}</math> degrees. <math>\boxed{\textbf{(E) }</math>. |
==See Also== | ==See Also== | ||
Revision as of 20:08, 3 June 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 $90 - 67.5 = 22\frac{1}{2}}$ (Error compiling LaTeX. Unknown error_msg) degrees. $\boxed{\textbf{(E) }$ (Error compiling LaTeX. Unknown error_msg).
See Also
| 1957 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
