Question

How many gallons each of 25% alcohol and 5% alcohol should be mixed to obtain 20 gal of 8% alcohol?

Answer 1

`\begin{array}{lcccl} &\stackrel{solution}{gallons}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{gallons of }}{pure~alcohol}\\ \cline{2-4}&\\ \textit{25\% solution}&x&0.25&0.25x\\ \textit{5\% solution}&y&0.05&0.05y\\ \cline{2-4}&\\ mixture&20&0.08&1.6 \end{array}~\hfill \begin{cases} ~\hfill x + y &= 20\\\\ 0.25x+0.05y&=1.6 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{x+y=20\implies \underline{y = 20-x}} \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{0.25x~~ + ~~0.05(\underline{20-x})~~ = ~~1.6}\implies 0.25x+1-0.05x=1.6 \\\\\\ 0.20x+1=1.6\implies 0.2x=0.6\implies x=\cfrac{0.6}{0.2}\implies \boxed{x=3}`