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