Question

Bruce wants to make 50 ml of an alcohol solution with a 12% concentration. He has a 10% alcohol solution and a 15% alcohol solution. The equation 0.10x + 0.15(50 – x) = 0.12(50) can be used to find the amount of 10% alcohol solution Bruce should use.

How much of the 10% alcohol solution should Bruce use?
__ mL

How much of the 15% alcohol solution should Bruce use?
__mL

Answer 1

x = 10% solution

y = 15% solution

let's start at the beginning, without the equation just yet, so let's start off by making a table for the substances and let's use the decimal format for the percentages to get the actual amounts.

`\begin{array}{lcccl} &\stackrel{mL}{quantity}&\stackrel{\textit{\% of mL that is}}{\textit{alcohol only}}&\stackrel{\textit{mL of}}{\textit{alcohol only}}\\ \cline{2-4}&\\ \textit{10\% sol'n}&x&0.10&0.10x\\ \textit{15\% sol'n}&y&0.15&0.15y\\ \cline{2-4}&\\ mixture&50&12&6 \end{array}~\hfill \begin{cases} x + y = 50\\\\ 0.10x+0.15y=6 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{x+y=50\implies y=50-x}\hspace{5em}\stackrel{\textit{substituting on the 2nd equation}}{0.10x+0.15(\underset{y}{50-x})=6} \\\\\\ 0.10x+7.5-0.15x=6\implies -0.05x+7.5=6\implies -0.05x=-1.5 \\\\\\ x=\cfrac{-1.5}{-0.05}\implies \boxed{x=30}\hspace{15em}\stackrel{50~~ - ~~30}{\boxed{y=20}}`

Answer 2

Answers are under yw