Difference between revisions of "2012 AMC 8 Problems/Problem 11"
Larryflora (talk | contribs) |
Larryflora (talk | contribs) (→Solution 3: Balance scale) |
||
Line 39: | Line 39: | ||
==Solution 3: Balance scale== | ==Solution 3: Balance scale== | ||
− | We know the mode must be <math>6</math>, so the mean must be <math>6</math>. Let's imagine a scale. <math>6</math> exactly stands in a 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 know the total difference of the right side "weights" between <math>6</math> is <math>0+1=1</math>, so the difference between <math>6</math> and <math>x</math> must be <math>+5</math>. It exactly gives us the answer:<math>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 in a 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 know the total difference of the right side "weights" between <math>6</math> is <math>0+1=1</math>, so the difference between <math>6</math> and <math>x</math> must be <math>+5</math>. It exactly gives us the answer:<math>6+5= \boxed{{\textbf{(D)}\ 11}}</math>. ---LarryFlora |
==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 21:11, 20 August 2021
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 in a 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 . We know the total difference of the right side "weights" between is , so the difference between and must be . It exactly gives us the answer:. ---LarryFlora
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.