Difference between revisions of "2010 AMC 10B Problems/Problem 14"

Latest revision as of 19:42, 26 May 2020

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-We must find the average of the numbers from 1 to 99 and x in terms of x. The sum of all these terms is \frac{99(100)}{2}+x=99(50)+x. We must divide this by the total number of terms, which is 100. We get: \frac{99(50)+x}{100}. This is equal to 100x, as stated in the problem. We have: \frac{99(50)+x}{100}=100x. We can now cross multiply. This gives:
+#redirect [[2010 AMC 12B Problems/Problem 10]]
-(100x)=99(50)+x
-x=99(50)+x
-x=99(50)
-x=50
-x=\frac{50}{101}
-This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math>