1957 AHSME Problem 12

Comparing the numbers 10^{-49} and 2\cdot 10^{-50} we may say:

\begin{array}{l} \textbf{(A)}\ \text{the first exceeds the second by }{8\cdot 10^{-1}}\qquad\\ \textbf{(B)}\ \text{the first exceeds the second by }{2\cdot 10^{-1}}\qquad\\ \textbf{(C)}\ \text{the first exceeds the second by }{8\cdot 10^{-50}}\qquad\\ \textbf{(D)}\ \text{the second is five times the first}\qquad\\ \textbf{(E)}\ \text{the first exceeds the second by }{5} \end{array}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.