Difference between revisions of "2012 AMC 8 Problems/Problem 11"

Revision as of 13:02, 1 February 2015

Problem

The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$ ?

$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$

Solution

Since there must be a unique mode, and $6$ is already repeated twice, $x$ cannot be any of the numbers already listed (3, 4, 5, 7). (If it were, the mode would not be unique.) So $x$ must be $6$ , or a new number.

Solution 1: Guess & Check

We can eliminate answer choices ${\textbf{(A)}\ 5}$ and ${\textbf{(C)}\ 7}$ , because of the above statement. Now we need to test the remaining answer choices.

Case 1: $x = 6$

Mode: $6$

Median: $6$

Mean: $\frac{37}{7}$

Since the mean does not equal the median or mode, ${\textbf{(B)}\ 6}$ can also be eliminated.

Case 2: $x = 11$

Mode: $6$

Median: $6$

Mean: $6$

We are done with this problem, because we have found when $x = 11$ , the condition is satisfied. Therefore, the answer is $\boxed{{\textbf{(D)}\ 11}}$ .

Solution 2: Algebra

Notice that the mean of this set of numbers, in terms of $x$ , is:

$\frac{3+4+5+6+6+7+x}{7} = \frac{31+x}{7}$

Because we know that the mode must be $6$ (it can't be any of the numbers already listed, as shown above, and no matter what $x$ is, either $6$ or a new number, it will not affect $6$ being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and $6$ equal:

$\frac{31+x}{7} = 6\\ 31+x = 42\\ x = \boxed{{\textbf{(D)}\ 11}}$

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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.

@@ Line 5: / Line 5: @@
 ==Solution==
-Since there must be an unique mode, and <math>6</math> is already repeated twice, <math>x</math> cannot be any of the numbers already listed (3, 4, 5, 7). (If it were, the mode would not be unique.) So <math>x</math> must be <math>6</math>, or a new number.
+Since there must be a unique mode, and <math>6</math> is already repeated twice, <math>x</math> cannot be any of the numbers already listed (3, 4, 5, 7). (If it were, the mode would not be unique.) So <math>x</math> must be <math>6</math>, or a new number.
 ===Solution 1: Guess & Check===

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