Difference between revisions of "2016 AMC 10A Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
| − | Note that <math>x</math> must be the median so it must equal either <math>60</math> or <math>90</math>. You can see that the mean is also <math>x</math>, and by intuition <math>x</math> should be the greater one. <math>x=\ | + | Note that <math>x</math> must be the median so it must equal either <math>60</math> or <math>90</math>. You can see that the mean is also <math>x</math>, and by intuition <math>x</math> should be the greater one. <math>x=\boxed{\textbf{(D) }90}.</math> |
==Check== | ==Check== | ||
Revision as of 13:54, 29 November 2020
Problem
The mean, median, and mode of the 
data values 
are all equal to 
. What is the value of ?
Solution 1
Since is the mean,
Therefore, 
, so 
Solution 2
Note that must be the median so it must equal either 
or 
. You can see that the mean is also , and by intuition should be the greater one.
Check
Order the list: 
. must be either or because it is both the median and the mode of the set. Thus is correct.
Video Solution
https://youtu.be/XXX4_oBHuGk?t=163
~IceMatrix
~savannahsolver
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 AMC 10 Problems and Solutions | ||
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
