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

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==Problem==
 
==Problem==
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<!-- 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>
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<!-- 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>
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==Solution 1==
 
==Solution 1==
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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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Therefore, the answer is <math>7\implies \textbf{(D)}.</math>
  
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==Solution 2==
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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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==Solution 2==
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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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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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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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if we put x, y, z in front of 2, 3 can be the median
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==Video Solution==
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if we put x, y, z after 14, 9 is the median
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https://youtu.be/yBSrLxv0LbY ~savannahsolver
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therefore, 3,4,6,9,x,y,z can all be the medians
 
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There are then <math> \boxed{\textbf{(D)}\ 7} </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}}

Latest revision as of 08:36, 18 May 2022

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 $R$.

Therefore, the answer is $7\implies \textbf{(D)}.$

Solution 2

Let the values of the missing integers be $x, y, z$. We will find the bound of the possible medians.

The smallest possible median will happen when we order the set as $\{x, y, z, 2, 3, 4, 6, 9, 14\}$. The median is $3$.

The largest possible median will happen when we order the set as $\{2, 3, 4, 6, 9, 14, x, y, z\}$. The median is $9$.

Therefore, the median must be between $3$ and $9$ inclusive, yielding $7$ possible medians, so the answer is $\textbf{(D)}$.


~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

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