Difference between revisions of "2012 AMC 8 Problems/Problem 22"
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Any number greater than <math> 9 </math> also cannot be a median of set <math> R </math>. | Any number greater than <math> 9 </math> also cannot be a median of set <math> R </math>. | ||
| − | Therefore, the | + | Therefore, the answer is <math>7\implies \textbf{(D)}.</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=21|num-a=23}} | {{AMC8 box|year=2012|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:17, 28 September 2019
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
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 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
