Let $$a$$ be the original number of ounces of acid and $$w$$ be the original number of ounces of water. We can write two equations $$\frac{a}{a+w+1}=\frac{1}{5}$$ and $$\frac{a+1}{a+w+2}=\frac{1}{3}$$. Then cross-multiply to get rid of the fractions. $$5a=a+w+1$$, $$3a+3=a+w+1$$ Solve the system to get $$a=1$$ and $$w=3$$. The percentage of acid in the original mixture is $$\frac{1}{1+3}=25 \textbackslash\% $$. 
When one ounce of water is added to a mixture of acid and water, the new mixture is $$20\textbackslash\%$$ acid. When one ounce of acid is added to the new mixture, the result is $$33\frac{1}{3}\textbackslash\%$$ acid. The percentage of acid in the original mixture is ($$1973$$ AHSME Problem, Question \#$$33$$) 
Options:
1. $$22\textbackslash\%$$ 
2. $$24\textbackslash\%$$ 
3. $$25\textbackslash\%$$ 
4. $$30\textbackslash\%$$ 
5. $$33\frac{1}{3}\textbackslash\%$$ 