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

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(Solution)
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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>.  
  
There are then <math> \boxed{\textbf{(D)}\ 7} </math> possible medians of set <math> R </math>.
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There are then <math> \boxed{\textbf{(C)}\ 6} </math> possible medians of set <math> R </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 12:50, 27 October 2013

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

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: $4$

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 $\boxed{\textbf{(C)}\ 6}$ possible medians of set $R$.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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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