Difference between revisions of "2016 AMC 12A Problems/Problem 4"
(Created page with "==Solution== <math>x</math> must be in the middle of the list. Order the known list first: <math>\{40,50,60,90,100,200\}</math>. <math>x</math> occurs the most times, and it i...") |
(→Solution) |
||
| Line 2: | Line 2: | ||
<math>x</math> must be in the middle of the list. Order the known list first: <math>\{40,50,60,90,100,200\}</math>. | <math>x</math> must be in the middle of the list. Order the known list first: <math>\{40,50,60,90,100,200\}</math>. | ||
<math>x</math> occurs the most times, and it is the average of the list. | <math>x</math> occurs the most times, and it is the average of the list. | ||
| + | |||
<math>\#1:</math> | <math>\#1:</math> | ||
<math>x</math> is either <math>60</math> or <math>90</math> because it is the median and the mode. | <math>x</math> is either <math>60</math> or <math>90</math> because it is the median and the mode. | ||
Revision as of 23:22, 3 February 2016
Solution
must be in the middle of the list. Order the known list first: 
.
occurs the most times, and it is the average of the list.
is either 
or 
because it is the median and the mode.
is the average of the set values.
![\[x=\frac{40+50+60+90+100+200+x}{7}\]](http://latex.artofproblemsolving.com/e/3/a/e3aaaa3c3d3bbc3b6c0e069af4175e70364be48a.png)
![\[x=\frac{540+x}{7}\]](http://latex.artofproblemsolving.com/6/f/e/6fea05ca3b0890dad230d1efe0e7fd712c6337fd.png)
![\[7x=540+x\]](http://latex.artofproblemsolving.com/2/5/0/25079bdfd04955e060ee8d0acc51fb4e891038b0.png)
![\[6x=540\]](http://latex.artofproblemsolving.com/6/b/7/6b7e58a412c03874549049365a1e237aeda0de52.png)
![\[x=\boxed{\textbf{(D)}90}\]](http://latex.artofproblemsolving.com/b/b/2/bb21dc4309f65e14af76ebd324544a7bafaf6d91.png)
