Difference between revisions of "2012 AMC 8 Problems/Problem 11"
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− | The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, | + | ==Problem== |
+ | The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and <math>x</math> are all equal. What is the value of <math>x</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> | <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 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> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 2: Algebra== | ||
+ | Notice that the mean of this set of numbers, in terms of <math>x</math>, is: | ||
+ | |||
+ | <math>\frac{3+4+5+6+6+7+x}{7} = \frac{31+x}{7}</math> | ||
+ | |||
+ | 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: | ||
+ | |||
+ | <cmath>\dfrac{31+x}{7}=6</cmath> | ||
+ | <cmath>31+x=42</cmath> | ||
+ | <cmath>x=\boxed{\text{(D) } 11}</cmath> | ||
+ | |||
+ | ==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>. 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 | ||
+ | |||
+ | == 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}} |
Latest revision as of 12:25, 3 January 2024
Contents
Problem
The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and are all equal. What is the value of
?
Solution 1: Guess & Check
We can eliminate answer choices and
, because of the above statement. Now we need to test the remaining answer choices.
Case 1:
Mode:
Median:
Mean:
Since the mean does not equal the median or mode, can also be eliminated.
Case 2:
Mode:
Median:
Mean:
We are done with this problem, because we have found when , the condition is satisfied. Therefore, the answer is
.
Solution 2: Algebra
Notice that the mean of this set of numbers, in terms of , is:
Because we know that the mode must be (it can't be any of the numbers already listed, as shown above, and no matter what
is, either
or a new number, it will not affect
being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and
equal:
Solution 3: Balance scale
We know the unique mode must be , so the mean must be the same number
. Let's imagine a scale.
exactly stands the mid-point of the scale. Numbers of
represent the left side "weights" of the scale. Numbers of
represent the right side "weights" of the scale. On the left side, the difference of the three "weights" between
are
, respectively. It gives us the total difference is
. In order to allow the scale to keep balance, on the right side, the total difference must be
. Because we have already known the difference of the right side "weights" between
is
, partially, so the difference between
and unknown
must be
. It exactly gives us the answer:
. ---LarryFlora
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.