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Difference between revisions of "1999 AMC 8 Problems/Problem 2"

(Created page with "== 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) ...")
 
 
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<math>\text{(A)}\ 30 \qquad \text{(B)}\ 45 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90</math>
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<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 90</math>
  
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==Solution==
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==Solution 1==
  
 
At <math>10:00</math>, the hour hand will be on the <math>10</math> while the minute hand on the <math>12</math>.
 
At <math>10:00</math>, the hour hand will be on the <math>10</math> while the minute hand on the <math>12</math>.
  
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This makes them <math>\frac{1}{6}</math>th of a circle apart, and <math>\frac{1}{6}\cdot360^{\circ}=\boxed{60\text{ (C)}}</math>.
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This makes them <math>\frac{1}{6}</math>th of a circle apart, and <math>\frac{1}{6}\cdot360^{\circ}=\boxed{\textbf{(C) } 60}</math>.
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==Solution 2==
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We know that the full clock is a circle, and therefore has 360 degrees. Considering that there are <math>12</math> numbers, the distance between one number will be <math>360\div 12=30</math>.
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If the time is <math>10:00</math>, then the hour hand will be on <math>10</math>, and the minute hand will be on, <math>12</math>, making them <math>2</math> numbers apart, so they will be <math>60</math> degrees apart, or <math>\boxed{\textbf{(C) } 60}</math>
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==See Also==
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{{AMC8 box|year=1999|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 20:45, 15 January 2024

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]

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

Solution 1

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 $\frac{1}{6}\cdot360^{\circ}=\boxed{\textbf{(C) } 60}$.

Solution 2

We know that the full clock is a circle, and therefore has 360 degrees. Considering that there are $12$ numbers, the distance between one number will be $360\div 12=30$. If the time is $10:00$, then the hour hand will be on $10$, and the minute hand will be on, $12$, making them $2$ numbers apart, so they will be $60$ degrees apart, or $\boxed{\textbf{(C) } 60}$

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

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