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

Revision as of 13:20, 31 August 2020

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: $3$

Sequence: $-2, -1, 0, 2, 3, 4, 6, 9, 14 , 15$

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 $7\implies \textbf{(D)}.$

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 11: / Line 11: @@
 Median: <math> 3 </math>
-Sequence: <math> -2, -1, 0, 2, 3, 4, 6, 9, 14 </math>
+Sequence: <math> -2, -1, 0, 2, 3, 4, 6, 9, 14 , 15</math>
 Median: <math> 4 </math>

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

Revision as of 13:20, 31 August 2020

Problem

Solution

See Also