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. $\boxed{\mathrm{(B)}= \frac{50}{101}}$

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2010 AMC 10B Problems/Problem 14