2010 AMC 12B Problem 10

The numbers $1$ through $99$ sum to $\dfrac{99\cdot100}{2}=4950.$ The average condition is $$\frac{4950+x}{100}=100x,$$ so $4950+x=10000x$ and $9999x=4950.$ Thus $x=\dfrac{4950}{9999}=\dfrac{50}{101}.$ Thus, the correct answer is B.