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| − | ==Solution==
| + | #REDIRECT [[2016 AMC 10A Problems/Problem 7]] |
| − | <math>x</math> must be in the middle of the list. Order the known list first: <math>\{40,50,60,90,100,200\}</math>.
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| − | <math>x</math> occurs the most times, and it is the average of the list.
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| − | <math>\#1:</math>
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| − | <math>x</math> is either <math>60</math> or <math>90</math> because it is the median and the mode.
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| − | <math>\#2:</math>
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| − | <math>x</math> is the average of the set values.
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| − | <cmath>x=\frac{40+50+60+90+100+200+x}{7}</cmath>
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| − | <cmath>x=\frac{540+x}{7}</cmath>
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| − | <cmath>7x=540+x</cmath>
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| − | <cmath>6x=540</cmath>
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| − | <cmath>x=\boxed{\textbf{(D)}\text{ 90}}</cmath>
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