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

$ \textbf{(A)}\  \text{the first exceeds the second by }{8\cdot 10^{\minus{}1}}\qquad\\ 
\textbf{(B)}\ \text{the first exceeds the second by }{2\cdot 10^{\minus{}1}}\qquad \\
\textbf{(C)}\ \text{the first exceeds the second by }{8\cdot 10^{\minus{}50}}\qquad \\
\textbf{(D)}\ \text{the second is five times the first}\qquad \\
\textbf{(E)}\ \text{the first exceeds the second by }{5}$
1. We are given two numbers: \(10^{-49}\) and \(2 \cdot 10^{-50}\). We need to compare these numbers and determine by how much the first number exceeds the second number.
2. To find the difference between the two numbers, we subtract the second number from the first number:
   \[
   10^{-49} - 2 \cdot 10^{-50}
   \]
3. Notice that \(10^{-49}\) can be rewritten as \(10 \cdot 10^{-50}\). This allows us to express both numbers with the same exponent:
   \[
   10^{-49} = 10 \cdot 10^{-50}
   \]
4. Substituting this into the subtraction, we get:
   \[
   10 \cdot 10^{-50} - 2 \cdot 10^{-50}
   \]
5. Factor out \(10^{-50}\) from both terms:
   \[
   10^{-50} (10 - 2)
   \]
6. Simplify the expression inside the parentheses:
   \[
   10 - 2 = 8
   \]
7. Therefore, the difference is:
   \[
   10^{-50} \cdot 8 = 8 \cdot 10^{-50}
   \]

The final answer is \(\boxed{8 \cdot 10^{-50}}\).