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2012 AMC 8 Problems/Problem 11: Difference between revisions

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<math> \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 </math>
<math> \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 </math>
==Solution==
Since there must be an unique mode, we can eliminate answer choices <math> {\textbf{(A)}\ 5} </math> and <math> {\textbf{(C)}\ 7} </math>. Now we need to test the remaining answer choices.
Case 1: <math> x = 6 </math>
Mode: <math> 6 </math>
Median: <math> 6 </math>
Mean: <math> \frac{37}{7} </math>
Since the mean does not equal the median or mode, <math> {\textbf{(B)}\ 6} </math> can also be eliminated.
Case 2: <math> x = 11 </math>
Mode: <math> 6 </math>
Median: <math> 6 </math>
Mean: <math> 6 </math>
We are done with this problem, because we have found when <math> x = 11 </math>, the condition is satisfied. Therefore, the answer is <math> \boxed{{\textbf{(D)}\ 11}} </math>.


==See Also==
==See Also==
{{AMC8 box|year=2012|num-b=10|num-a=12}}
{{AMC8 box|year=2012|num-b=10|num-a=12}}

Revision as of 10:43, 24 November 2012

The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, 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 an unique mode, we can eliminate answer choices ${\textbf{(A)}\ 5}$ and ${\textbf{(C)}\ 7}$. 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 $\boxed{{\textbf{(D)}\ 11}}$.

See Also

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
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