1999 AMC 8 Problems/Problem 2

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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]

$\text{(A)}\ 30 \qquad \text{(B)}\ 45 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90$

Solution

At $10:00$, the hour hand will be on the $10$ while the minute hand on the $12$.

This makes them $\frac{1}{6}$th of a circle apart, and $\frac{1}{6}\cdot360^{\circ}=\boxed{60\text{  (C)}}$.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
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

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