Difference between revisions of "1999 AMC 8 Problems/Problem 2"
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</asy> | </asy> | ||
| − | <math>\ | + | <math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 90</math> |
==Solution 1== | ==Solution 1== | ||
Latest revision as of 20:45, 15 January 2024
Contents
Problem
What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?
![[asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); } label("$3$",2*dir(0),W); label("$2$",2*dir(30),WSW); label("$1$",2*dir(60),SSW); label("$12$",2*dir(90),S); label("$11$",2*dir(120),SSE); label("$10$",2*dir(150),ESE); label("$9$",2*dir(180),E); label("$8$",2*dir(210),ENE); label("$7$",2*dir(240),NNE); label("$6$",2*dir(270),N); label("$5$",2*dir(300),NNW); label("$4$",2*dir(330),WNW); [/asy]](http://latex.artofproblemsolving.com/4/5/0/4500534f275d08449e027ed4322baa4c284ea69e.png)
Solution 1
At 
, the hour hand will be on the 
while the minute hand on the 
.
This makes them 
th of a circle apart, and 
.
Solution 2
We know that the full clock is a circle, and therefore has 360 degrees. Considering that there are numbers, the distance between one number will be 
.
If the time is , then the hour hand will be on , and the minute hand will be on, , making them 
numbers apart, so they will be 
degrees apart, or 
See Also
| 1999 AMC 8 (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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
