Difference between revisions of "1955 AHSME Problems/Problem 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. | + | 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. 
