Difference between revisions of "2012 AMC 8 Problems/Problem 22"
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<math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8 </math> | <math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8 </math> | ||
| − | ==Solution == | + | ==Solution 1== |
First, we find that the minimum value of the median of <math> R </math> will be <math> 3 </math>. | First, we find that the minimum value of the median of <math> R </math> will be <math> 3 </math>. | ||
| Line 40: | Line 40: | ||
Therefore, the answer is <math>7\implies \textbf{(D)}.</math> | Therefore, the answer is <math>7\implies \textbf{(D)}.</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | 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 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>. | ||
| + | |||
| + | ~superagh | ||
==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 17:37, 3 October 2020
Contents
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
