Difference between revisions of "2003 AMC 8 Problems/Problem 20"
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==Solution== | ==Solution== | ||
| − | Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between <math>4</math> and <math>5</math> is <math>30</math> degrees (since it is 1/12 of a full circle, 360). By <math>4:20</math>, the hour hand would have moved <math>\frac{1}{3}</math> way from 4 to 5 since <math>\frac{20}{60}</math> is reducible to <math>\frac{1}{3}</math>. One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is <math>\boxed{10}</math>, and we are done. | + | Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between <math>4</math> and <math>5</math> is <math>30</math> degrees (since it is 1/12 of a full circle, 360). By <math>4:20</math>, the hour hand would have moved <math>\frac{1}{3}</math> way from 4 to 5 since <math>\frac{20}{60}</math> is reducible to <math>\frac{1}{3}</math>. One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is <math>\boxed{10, D}</math>, and we are done. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2003|num-b=19|num-a=21}} | {{AMC8 box|year=2003|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:46, 11 October 2019
Problem
What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?
Solution
Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between 
and 
is 
degrees (since it is 1/12 of a full circle, 360). By 
, the hour hand would have moved 
way from 4 to 5 since 
is reducible to . One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is 
, and we are done.
See Also
| 2003 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
