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 99(100)/2+x=99(50)+x. We must divide this by the total number of terms, which is 100. We get: (99(50)+x)/100. This is equal to 100x, as stated in the problem. We have: (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=50/101

This gives us our answer.