Question

Canon needs to produce 1000 milliliters of 36% alcohol solution. At his disposal he has 40% alcohol solution and 20% alcohol solution. How much of each does he need in order to produce his desired solution?

Answer 1

x = milliliters of 40% alcohol

y = milliliters of 20% alcohol

so just alcohol alone in each quantity will just be "x" * (40/100) = 0.4x and "y" * (20/100) = 0.2y, whilst trying to get 1000 mL of 36% alcohol only.

`\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{mL of }}{amount}\\ \cline{2-4}&\\ \textit{40\% solution}&x&0.4&0.4x\\ \textit{20\% solution}&y&0.2&0.2y\\ \cline{2-4}&\\ mixture&1000&0.36&360 \end{array}~\hfill \begin{cases} x+y=1000\\\\ 0.4x+0.2y=360 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{x+y=1000\implies }x=1000-y ~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.4(1000-y)+0.2y=360} \\\\\\ 400-0.4y+0.2y=360\implies 40-0.4y+0.2y=0\implies 40-0.2y=0 \\\\\\ 40=0.2y\implies \cfrac{40}{0.2}=y\implies \boxed{\stackrel{mL}{200}=y} ~\hfill \boxed{\underset{mL}{\stackrel{1000~~ - ~~200}{x=800}}}`