Question

You need a 55% alcohol solution. On hand, you have a 330 mL of a 80% alcohol mixture. How much pure water will you need to add to obtain the desired solution?

helpppppppp
You will need.............
mL of pure water
to obtain............
mL of the desired 55% solution.

Answer 1

y = amount needed at 55% alcohol.

x = amount of pure water, or 0% alcohol

so, you have on hand 330 mL of 80% alcohol, so how much alcohol only is there on that mixture? well, is 330 mL, so 80% of 330 is 330 * (80/100) = 330 * 0.8 = 264 mL.

likewise, let's convert the other perecentage values to deciaml, bearing in mind that pure water is, well just pure water, it has no alcohol, or we can say that it has 0% alcohol.

`\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{mL of }}{amount}\\ \cline{2-4}&\\ \textit{on hand}&330&0.8&264\\ \textit{pure water}&x&0&0x\\ \cline{2-4}&\\ mixture&y&0.55&0.55y \end{array}~\hfill \begin{cases} 330+x=\boxed{y}\\\\ 264+0x = 0.55y \end{cases}`

`\bf \stackrel{\textit{substituting on the 2nd equation}}{264+0x=0.55\left( \boxed{330+x} \right)}\implies 264=181.5+0.55x\implies 82.5=0.555x \\\\\\ \cfrac{82.5}{0.55}=x\implies 150=x`