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

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(→‎Solution)
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==Solution==
 
==Solution==
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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.
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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.
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===Solution 1: Guess & Check===
 +
We can eliminate answer choices <math> {\textbf{(A)}\ 5} </math> and <math> {\textbf{(C)}\ 7} </math>, because of the above statement. Now we need to test the remaining answer choices.
  
 
Case 1: <math> x = 6 </math>
 
Case 1: <math> x = 6 </math>
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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>.
 
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>.
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===Solution 2: Algebra===
 +
Notice that the mean of this set of numbers, in terms of <math>x</math>, is:
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<math>\frac{3+4+5+6+6+7+x}{7} = \frac{31+x}{7}</math>
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 +
Because we know that the mode must be <math>6</math> (it can't be any of the numbers already listed, as shown above, and no matter what <math>x</math> is, either <math>6</math> or a new number,  it will not affect <math>6</math> being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and <math>6</math> equal:
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<math>\frac{31+x}{7} = 6\\
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31+x = 42\\
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x = \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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:10, 25 November 2013

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 an 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:

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

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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