Difference between revisions of "2003 AMC 8 Problems/Problem 20"
Evanwang2014 (talk | contribs) m (→Problem) |
(→Solution 2) |
||
(11 intermediate revisions by 7 users not shown) | |||
Line 5: | Line 5: | ||
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12</math> | <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | 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> | + | 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{(D) 10}</math>. |
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=6FdbHSooYSA | ||
+ | |||
+ | ~David | ||
==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}} |
Latest revision as of 09:37, 14 June 2024
Contents
[hide]Problem
What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?
Solution 1
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 .
Video Solution
https://www.youtube.com/watch?v=6FdbHSooYSA
~David
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.