2012 AMC 8 Problems/Problem 11


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 $x$?


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:

\[\dfrac{31+x}{7}=6\] \[31+x=42\] \[x=\boxed{\text{(D) } 11}\]

Solution 3: Balance scale

We know the unique mode must be $6$, so the mean must be the same number $6$. Let's imagine a scale. $6$ exactly stands the mid-point of the scale. Numbers of $3,4,5$ represent the left side "weights" of the scale. Numbers of $6,7, x$ represent the right side "weights" of the scale. On the left side, the difference of the three "weights" between $6$ are $-3, -2, -1$, respectively. It gives us the total difference is $-6$. In order to allow the scale to keep balance, on the right side, the total difference must be $+6$. Because we have already known the difference of the right side "weights" between $6$ is $0+1=1$, partially, so the difference between $6$ and unknown $x$ must be $+6-1=+5$. It exactly gives us the answer:$x=6+5= \boxed{{\textbf{(D)}\ 11}}$. ---LarryFlora

Video Solution by OmegaLearn


~ pi_is_3.14

Video Solution

https://youtu.be/dloxxgDBm88 ~savannahsolver

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AJHSME/AMC 8 Problems and Solutions

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