Question

A chemist has 20% and 50% solutions of acid available. How many liters of each solution should be mixed to obtain 600 liters of 21% acid solution?

Answer 1

`x=\textit{Liters of solution at 20\%}\\\\ ~~~~~~ 20\%~of~x\implies \cfrac{20}{100}(x)\implies 0.2 (x) \\\\\\ y=\textit{Liters of solution at 50\%}\\\\ ~~~~~~ 50\%~of~y\implies \cfrac{50}{100}(y)\implies 0.5 (y) \\\\\\ \textit{600 Liters of solution at 21\%}\\\\ ~~~~~~ 21\%~of~600\implies \cfrac{21}{100}(600)\implies 126 \\\\[-0.35em] ~\dotfill`

`\begin{array}{lcccl} &\stackrel{Liters}{quantity}&\stackrel{\textit{\% of Liters that is}}{\textit{acid only}}&\stackrel{\textit{Liters of}}{\textit{acid only}}\\ \cline{2-4}&\\ \textit{1st Sol'n}&x&0.2&0.2x\\ \textit{2nd Sol'n}&y&0.5&0.5y\\ \cline{2-4}&\\ mixture&600&0.21&126 \end{array}~\hfill \begin{cases} x + y = 600\\\\ 0.2x+0.5y=126 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{x+y=600}\implies x=600-y \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{0.2(600-y)+0.5y=126}\implies 120-0.2y+0.5y=126 \\\\\\ 120+0.3y=126\implies 0.3y=6\implies y=\cfrac{6}{0.3}\implies \boxed{y=20} \\\\\\ \stackrel{\textit{since we know that}}{x=600-y}\implies \boxed{x=580}`