Difference between revisions of "2012 AMC 8 Problems/Problem 22"
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==Problem== | ==Problem== | ||
| − | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math> R </math> 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 <math> R </math> ?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>R</math> be a set of nine distinct integers. Six of the elements are <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, <math>9</math>, and <math>14</math>. What is the number of possible values of the median of <math>R</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<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>. | ||
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Therefore, the answer is <math>7\implies \textbf{(D)}.</math> | Therefore, the answer is <math>7\implies \textbf{(D)}.</math> | ||
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| + | ==Solution 2== | ||
| + | Let the values of the missing integers be <math>x, y, z</math>. We will find the bound of the possible medians. | ||
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| + | 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>. | ||
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| + | 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>. | ||
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| + | Therefore, the median must be between <math>3</math> and <math>9</math> inclusive, yielding <math>7</math> possible medians, so the answer is <math>\textbf{(D)}</math>. | ||
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| + | ~superagh | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/yBSrLxv0LbY ~savannahsolver | ||
==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}} | ||
Latest revision as of 08:36, 18 May 2022
Problem
Let 
be a set of nine distinct integers. Six of the elements are 
, 
, 
, 
, 
, and 
. 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, so the answer is 
.
~superagh
Video Solution
https://youtu.be/yBSrLxv0LbY ~savannahsolver
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. 
