Difference between revisions of "2003 AMC 8 Problems/Problem 20"

Revision as of 23:47, 4 July 2013

Problem

What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}$

Solution

Imagine the clock as a circle. By $4:20$ , the hour hand would have moved $\frac{1}{3}$ way towards the five since $\frac{20}{60}$ is reducible to $\frac{1}{3}$ . The central angle formed between $4$ and $5$ would be $30$ degrees since the angle between $12$ and $6$ is $180$ degrees and $\frac{180}{6} = 30$ Since the central angle formed from $4:20$ is a third of the central angle formed between $4$ and $5$ , $30$ degrees, the answer is $\boxed{\textbf{(D)}\ 10}$ .

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.

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 ==See Also==
 {{AMC8 box|year=2003|num-b=19|num-a=21}}
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Difference between revisions of "2003 AMC 8 Problems/Problem 20"

Revision as of 23:47, 4 July 2013

Problem

Solution

See Also