2010 AMC 10B Problem 14

Recall that the sum of the first $n$ integers is $\dfrac{n(n + 1)}{2}.$ Then, we have that $$\dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x,$$ which simplifies to $$99 \cdot 50 = (100^2 - 1)x$$$$= 101 \cdot 99x,$$ by difference of squares. Dividing gives us $x = \dfrac{50}{101}.$ Thus, B is the correct answer.