2012 AMC 8 Problems/Problem 22: Difference between revisions
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==Solution 2== | ==Solution 2== | ||
Let the values of the missing integers be <math>x, y, z</math>. We will find the bound of the possible medians. | Let the values of the missing integers be <math>x, y, z</math>. We will find the bound of the possible medians. | ||
The smallest possible median will happen when we order the set as <math>\{x, y, z, 2, 3, 4, 6, 9, 14\}</math>. The median is <math>3</math>. | The smallest possible median will happen when we order the set as <math>\{x, y, z, 2, 3, 4, 6, 9, 14\}</math>. The median is <math>3</math>. | ||
The largest possible median will happen when we order the set as <math>\{2, 3, 4, 6, 9, 14, x, y, z\}</math>. The median is <math>9</math> | The largest possible median will happen when we order the set as <math>\{2, 3, 4, 6, 9, 14, x, y, z\}</math>. The median is <math>9</math> | ||
Therefore, the median must be between <math>3</math> and <math>9</math> inclusive, yielding <math>7</math> possible medians, <math>\textbf{(D)}</math>. | Therefore, the median must be between <math>3</math> and <math>9</math> inclusive, yielding <math>7</math> possible medians, <math>\textbf{(D)}</math>. | ||
~superagh | ~superagh | ||
Revision as of 17:37, 3 October 2020
Problem
Let 
be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of ?
Solution 1
First, we find that the minimum value of the median of will be 
.
We then experiment with sequences of numbers to determine other possible medians.
Median:
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Any number greater than also cannot be a median of set .
Therefore, the answer is 
Solution 2
Let the values of the missing integers be 
. We will find the bound of the possible medians.
The smallest possible median will happen when we order the set as 
. The median is .
The largest possible median will happen when we order the set as 
. The median is
Therefore, the median must be between and inclusive, yielding possible medians, 
.
~superagh
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America. 
