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Difference between revisions of "2012 AMC 8 Problems/Problem 22"

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==Solution==
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==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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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>.  
  
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There are then <math> \boxed{\textbf{(D)}\ 7} </math> possible medians of set <math> R </math>.
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 +
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==Solution 2==
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For those who don't like bashing problems out, use this method.
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2,3,4,6,9,14,x,y,z
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if we put x, y, z in front of 2, 3 can be the median
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if we put x, y, z after 14, 9 is the median
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therefore, 3,4,6,9,x,y,z can all be the medians
 
There are then <math> \boxed{\textbf{(D)}\ 7} </math> possible medians of set <math> R </math>.
 
There are then <math> \boxed{\textbf{(D)}\ 7} </math> possible medians of set <math> R </math>.
  

Revision as of 01:32, 3 November 2017

Problem

Let $R$ 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 $R$ ?

$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8$




Solution 1

First, we find that the minimum value of the median of $R$ will be $3$.

We then experiment with sequences of numbers to determine other possible medians.

Median: $3$

Sequence: $-2, -1, 0, 2, 3, 4, 6, 9, 14$

Median: $4$

Sequence: $-1, 0, 2, 3, 4, 6, 9, 10, 14$

Median: $5$

Sequence: $0, 2, 3, 4, 5, 6, 9, 10, 14$

Median: $6$

Sequence: $0, 2, 3, 4, 6, 9, 10, 14, 15$

Median: $7$

Sequence: $2, 3, 4, 6, 7, 8, 9, 10, 14$

Median: $8$

Sequence: $2, 3, 4, 6, 8, 9, 10, 14, 15$

Median: $9$

Sequence: $2, 3, 4, 6, 9, 14, 15, 16, 17$

Any number greater than $9$ also cannot be a median of set $R$.

There are then $\boxed{\textbf{(D)}\ 7}$ possible medians of set $R$.


Solution 2

For those who don't like bashing problems out, use this method. 2,3,4,6,9,14,x,y,z if we put x, y, z in front of 2, 3 can be the median if we put x, y, z after 14, 9 is the median therefore, 3,4,6,9,x,y,z can all be the medians There are then $\boxed{\textbf{(D)}\ 7}$ possible medians of set $R$.

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

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