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

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==Solution==
 
==Solution==
  
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>, and we are done.
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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 D, which is <math>\boxed{10}</math>, and we are done.
  
 
==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}}

Revision as of 11:45, 11 October 2019

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

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 D, which is $\boxed{10}$, and we are done.

See Also

2003 AMC 8 (ProblemsAnswer KeyResources)
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

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