Difference between revisions of "2012 AMC 8 Problems/Problem 11"
m (→Problem) |
|||
(5 intermediate revisions by 5 users not shown) | |||
Line 4: | Line 4: | ||
<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== | + | ==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> | Case 1: <math> x = 6 </math> | ||
Line 26: | Line 26: | ||
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>. | 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: | ||
+ | |||
+ | <math>\frac{31+x}{7} = 6\\ | ||
+ | 31+x = 42\\ | ||
+ | x = \boxed{{\textbf{(D)}\ 11}}</math> | ||
==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 18:16, 15 February 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 ?
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.