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

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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:
 
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:
  
−
<math>\frac{31+x}{7} = 6\\
+
<cmath>\dfrac{31+x}{7}=6</cmath>
−
31+x = 42\\
+
<cmath>31+x=42</cmath>
−
x = \boxed{{\textbf{(D)}\ 11}}</math>
+
<cmath>x=\boxed{\text{(D) } 11}</cmath>
  
 
==Solution 3: Balance scale==
 
==Solution 3: Balance scale==
−
We know the unique mode must be <math>6</math>, so the mean must be the same number <math>6</math>. Let's imagine a scale. <math>6</math> exactly stands the mid-point of the scale. Numbers of <math>3,4,5</math> represent the left side "weights" of the scale. Numbers of <math>6,7, x</math> represent the right side "weights" of the scale. On the left side, the difference of the three "weights" between <math>6</math> are <math>-3, -2, -1</math>, respectively. It gives us the total difference is <math>-6</math>. In order to allow the scale to keep balance, on the right side, the total difference must be <math>+6</math>. We have already known the difference of the right side "weights" between <math>6</math> is <math>0+1=1</math>, so the difference between <math>6</math> and unknown <math>x</math> must be <math>+5</math>. It exactly gives us the answer:<math>x=6+5= \boxed{{\textbf{(D)}\ 11}}</math>. ---LarryFlora
+
We know the unique mode must be <math>6</math>, so the mean must be the same number <math>6</math>. Let's imagine a scale. <math>6</math> exactly stands the mid-point of the scale. Numbers of <math>3,4,5</math> represent the left side "weights" of the scale. Numbers of <math>6,7, x</math> represent the right side "weights" of the scale. On the left side, the difference of the three "weights" between <math>6</math> are <math>-3, -2, -1</math>, respectively. It gives us the total difference is <math>-6</math>. In order to allow the scale to keep balance, on the right side, the total difference must be <math>+6</math>. Because we have already known the difference of the right side "weights" between <math>6</math> is <math>0+1=1</math>, partially, so the difference between <math>6</math> and unknown <math>x</math> must be <math>+6-1=+5</math>. It exactly gives us the answer:<math>x=6+5= \boxed{{\textbf{(D)}\ 11}}</math>. ---LarryFlora
 +
 
 +
 
 +
==Solution 4: Must be an integer(Super fast)==
 +
Since all the answer choices are integers, we know that <math>\frac{3+4+5+6+6+7+x}{7}</math>, which simplifies to <math>\frac{31+x}{7}</math>, must be an integer. The only answer choice that satisfies this is <math>\boxed{{\textbf{(D)}\ 11}}</math>.
 +
 
 +
~NXC
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/TkZvMa30Juo?t=1946
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution==
 +
https://youtu.be/dloxxgDBm88 ~savannahsolver
  
 
==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}}

Latest revision as of 15:11, 16 November 2024

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 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


Solution 4: Must be an integer(Super fast)

Since all the answer choices are integers, we know that $\frac{3+4+5+6+6+7+x}{7}$, which simplifies to $\frac{31+x}{7}$, must be an integer. The only answer choice that satisfies this is $\boxed{{\textbf{(D)}\ 11}}$.

~NXC

Video Solution by OmegaLearn

https://youtu.be/TkZvMa30Juo?t=1946

~ pi_is_3.14

Video Solution

https://youtu.be/dloxxgDBm88 ~savannahsolver

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

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

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