Solution $$1$$.  Say we add $$N$$ ounces of water to the shaving lotion. Since half of an $$9$$ ounce bottle of shaving lotion is alcohol, we know that we have $$\dfrac{9}{2}$$ ounces of alcohol. We want $$\dfrac{9}{2}=0.3(9+N)$$ (because we want the amount of alcohol, $$\dfrac{9}{2}$$, to be $$30\textbackslash\%$$, or $$0.3$$, of the total amount of shaving lotion, $$9+N$$). Solving this, we find that $$9=0.6(9+N)\Rightarrow9=5.4+0.6N\hspace{0pt}\Rightarrow3.6=0.6N\hspace{0pt}\Rightarrow6=N$$. So, the total amount of water we need to add is $$6$$.  Solution $$2$$.  The concentration of alcohol after adding $$n$$ ounces of water is $$\dfrac{4.5}{9+n}$$.  To get a solution of $$30\textbackslash\%$$ alcohol, we solve $$\dfrac{4.5}{9+n}=\dfrac{3}{10}$$  $$45=27+3n$$  $$18=3n$$  $$6=n\Rightarrow6$$. 
The number of ounces of water needed to reduce $$9$$ ounces of shaving lotion containing $$50\textbackslash\%$$ alcohol to a lotion containing $$30\textbackslash\%$$ alcohol is: ($$1953 $$ AHSME Problem, Question \#$$9$$) 
A: $$3$$ 
B: $$4$$ 
C: $$5$$ 
D: $$6$$ 
E: $$7$$ 