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

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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>.
 
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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==Solution==
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==Solutions==
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===Solution 1===
 
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>.
 
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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===Solution 2===
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As before, derive that there are <math>132</math> possibilities for Dave's original and replacement gates.
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Now suppose that Dave has to walk <math>100k</math> feet to get to his new gate.  This means that Dave's old and new gates must be <math>k</math> gates apart.  (For example, a <math>100</math> foot walk would consist of the two gates being adjacent to each other.)  There are <math>12-k</math> ways to pick two gates which are <math>k</math> gates apart, and <math>2</math> possibilities for gate assignments, for a total of <math>2(12-k)</math> possible assignments for each <math>k</math>.
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As a result, the total number of valid gate arrangements is <cmath>2\cdot 11 + 2\cdot 10 + 2\cdot 9 + 2\cdot 8 = 76</cmath> and so the requested probability is <math>\tfrac{19}{33}</math> for a final answer of <math>\boxed{052}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 16:03, 9 August 2018

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 $m+n$.

Solutions

Solution 1

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 $19 + 33 = \boxed{052}$.

Solution 2

As before, derive that there are $132$ possibilities for Dave's original and replacement gates.

Now suppose that Dave has to walk $100k$ feet to get to his new gate. This means that Dave's old and new gates must be $k$ gates apart. (For example, a $100$ foot walk would consist of the two gates being adjacent to each other.) There are $12-k$ ways to pick two gates which are $k$ gates apart, and $2$ possibilities for gate assignments, for a total of $2(12-k)$ possible assignments for each $k$.

As a result, the total number of valid gate arrangements is \[2\cdot 11 + 2\cdot 10 + 2\cdot 9 + 2\cdot 8 = 76\] and so the requested probability is $\tfrac{19}{33}$ for a final answer of $\boxed{052}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AIME Problems and Solutions

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