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>. | ||
− | ==Solution== | + | ==Solutions== |
+ | ===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 < | + | 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>. |
+ | |||
+ | ===Solution 2=== | ||
+ | As before, derive that there are <math>132</math> possibilities for Dave's original and replacement gates. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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 17:03, 9 August 2018
Problem
Dave arrives at an airport which has twelve gates arranged in a straight line with exactly 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
feet or less to the new gate be a fraction
, where
and
are relatively prime positive integers. Find
.
Solutions
Solution 1
There are possible situations (
choices for the initially assigned gate, and
choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most
feet apart.
If we number the gates through
, then gates
and
have four other gates within
feet, gates
and
have five, gates
and
have six, gates
and
have have seven, and gates
,
,
,
have eight. Therefore, the number of valid gate assignments is
so the probability is
. The answer is
.
Solution 2
As before, derive that there are possibilities for Dave's original and replacement gates.
Now suppose that Dave has to walk feet to get to his new gate. This means that Dave's old and new gates must be
gates apart. (For example, a
foot walk would consist of the two gates being adjacent to each other.) There are
ways to pick two gates which are
gates apart, and
possibilities for gate assignments, for a total of
possible assignments for each
.
As a result, the total number of valid gate arrangements is and so the requested probability is
for a final answer of
.
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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