Difference between revisions of "2012 AMC 8 Problems/Problem 11"
Vaathsalya (talk | contribs) m (→Solution) |
|||
| Line 3: | Line 3: | ||
<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> | ||
| + | |||
| + | ==Solutions== | ||
===Solution 1: Guess & Check=== | ===Solution 1: Guess & Check=== | ||
Revision as of 15:34, 16 January 2021
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 ?
Solutions
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:
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. 
