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Difference between revisions of "2010 AIME II Problems/Problem 4"

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==Problem==
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== Problem ==
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Dave arrives at an airport which has twelve gates arranged in a straight line with exactly <math>100</math> feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks <math>400</math> feet or less to the new gate be a fraction <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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Dave arrives at an airport which has twelve gates arranged in a straight line with exactly <math>100</math> feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the [[probability]] that Dave walks <math>400</math> feet or less to the new gate be a fraction <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m+n</math>.
  
 
==Solution==
 
==Solution==
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There are <math>12 \times 11 = 132</math>  potential gate assignments. We need to count the valid ones.
 
There are <math>12 \times 11 = 132</math>  potential gate assignments. We need to count the valid ones.
  
 
Number the gates <math>1</math> through <math>12</math>. Gates <math>1</math> and <math>12</math> have four gates within <math>400</math> feet. Gates <math>2</math> and <math>11</math> have five. Gates <math>3</math> and <math>10</math> have six. Gates <math>4</math> and <math>9</math> have have seven. Gates <math>5</math> through <math>8</math> have eight.
 
Number the gates <math>1</math> through <math>12</math>. Gates <math>1</math> and <math>12</math> have four gates within <math>400</math> feet. Gates <math>2</math> and <math>11</math> have five. Gates <math>3</math> and <math>10</math> have six. Gates <math>4</math> and <math>9</math> have have seven. Gates <math>5</math> through <math>8</math> have eight.
  
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Therefore, the number of valid gate assignments is <math>2 \times (4+5+6+7+2\times8) = 76</math>.
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Therefore, the number of valid gate assignments is <math>2 \times (4+5+6+7+2\times8) = 76</math>, and the probability is <math>\frac{76}{132} = \frac{19}{33}</math>. Hence, the answer is given by <math>19 + 33 = \boxed{052}</math>.
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<math>\frac{76}{132} = \frac{19}{33}</math>.
 
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Answer: <math>\boxed{052}</math>.
 
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== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=3|num-a=5|n=II}}
 
{{AIME box|year=2010|num-b=3|num-a=5|n=II}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 10:28, 6 April 2010

Problem

Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $100$ feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks $400$ feet or less to the new gate be a fraction $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

There are $12 \times 11 = 132$ potential gate assignments. We need to count the valid ones.

Number the gates $1$ through $12$. Gates $1$ and $12$ have four gates within $400$ feet. Gates $2$ and $11$ have five. Gates $3$ and $10$ have six. Gates $4$ and $9$ have have seven. Gates $5$ through $8$ have eight.

Therefore, the number of valid gate assignments is $2 \times (4+5+6+7+2\times8) = 76$, and the probability is $\frac{76}{132} = \frac{19}{33}$. Hence, the answer is given by $19 + 33 = \boxed{052}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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