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

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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.
 
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.
  
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 <cmath>2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76</cmath>, 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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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 <cmath>2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76</cmath> so the probability is <math>\frac{76}{132} = \frac{19}{33}</math>. The answer is <math>19 + 33 = \boxed{052}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:57, 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 \[2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76\] so the probability is $\frac{76}{132} = \frac{19}{33}$. The answer is $19 + 33 = \boxed{052}$.

See also

2010 AIME II (ProblemsAnswer KeyResources)
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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