Difference between revisions of "2016 AMC 10A Problems/Problem 7"
(→Solution 2) |
(→Video Solution) |
||
Line 22: | Line 22: | ||
Order the list: <math>\{40,50,60,90,100,200\}</math>. <math>x</math> must be either <math>60</math> or <math>90</math> because it is both the median and the mode of the set. Thus <math>90</math> is correct. | Order the list: <math>\{40,50,60,90,100,200\}</math>. <math>x</math> must be either <math>60</math> or <math>90</math> because it is both the median and the mode of the set. Thus <math>90</math> is correct. | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/qDqe8pztaJQ | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
==Video Solution== | ==Video Solution== |
Revision as of 14:57, 25 June 2023
Contents
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.
~bjc
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 (CREATIVE THINKING)
~Education, the Study of Everything
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.