Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?
$\textbf{(A) } 9\%\qquad \textbf{(B) } 19\%\qquad \textbf{(C) } 22\%\qquad \textbf{(D) }  23\%\qquad \textbf{(E) } 25\%$
Let $m$ = cost of movie ticket
Let $s$  = cost of soda
We can create two equations:
\[m = \frac{1}{5}(A - s)\]
\[s  = \frac{1}{20}(A - m)\]
Substituting we get: 
\[m = \frac{1}{5}(A - \frac{1}{20}(A - m))\]
which yields:
\[m = \frac{19}{99}A\]
Now we can find s and we get:
\[s = \frac{4}{99}A\]
Since we want to find what fraction of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:
\[\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}\]