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Difference between revisions of "2010 AIME II Problems/Problem 4"
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==Solution==
 
==Solution==
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There are <math>12 \times 11 = 132</math> potential gate assignments. We need to count the valid ones.
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There are <math>12 \cdot 11 = 132</math> possible situations (<math>12</math> choices for the initially assigned gate, and <math>11</math> choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most <math>400</math> feet apart.
  
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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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If we number the gates <math>1</math> through <math>12</math>, then gates <math>1</math> and <math>12</math> have four other 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, and gates <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math> have eight. Therefore, the number of valid gate assignments is <math>2\cdot(4+5+6+7)+4\cdot8 = 76</math>, so the probability is <math>\frac{76}{132} = \frac{19}{33}</math>. The answer is <math>19 + 33 = \boxed{052}</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>.
 
  
 
== See also ==
 
== See also ==

Revision as of 13:56, 26 February 2016

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 $12 \cdot 11 = 132$ possible situations ($12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart.

If we number the gates $1$ through $12$, then gates $1$ and $12$ have four other gates within $400$ feet, gates $2$ and $11$ have five, gates $3$ and $10$ have six, gates $4$ and $9$ have have seven, and gates $5$, $6$, $7$, $8$ have eight. Therefore, the number of valid gate assignments is $2\cdot(4+5+6+7)+4\cdot8 = 76$, so the probability is $\frac{76}{132} = \frac{19}{33}$. The answer is $19 + 33 = \boxed{052}$.

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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