2003 AMC 8 Problems/Problem 20

Problem

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

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

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 $4$ and $5$ is $30$ degrees (since it is 1/12 of a full circle, 360). By $4:20$ , the hour hand would have moved $\frac{1}{3}$ way from 4 to 5 since $\frac{20}{60}$ is reducible to $\frac{1}{3}$ . 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 $\boxed{10, D}$ , and we are done.

Solution 2

You may also use the formula $30h-5.5m$ to yield a result in only ~10 seconds

-sub_math

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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2003 AMC 8 Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

See Also