2010 AMC 10B Problems/Problem 14

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:

100(100x)=99(50)+x 10000x=99(50)+x 9999x=99(50) 101x=50 x=\frac{50}{101}

This gives us our answer.