Question

A 20% acid solution is mixed with a 70% acid solution to get 50 liters of a 40% solution.

How much of the 70% solution is used?
A)30 liters
B)20 liters
C)25 liters

Answer 1

x = liters of solution at 20%

y = liters of solution at 70%

`\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{Sol'n at 20\%}&x&0.2&0.2x\\ \textit{Sol'n at 70\%}&y&0.7&0.7y\\ \cline{2-4}&\\ mixture&50&0.4&20 \end{array}~\hfill \begin{cases} x + y = 50\\\\ 0.2x+0.7y=20 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{x+y=50\implies x=50-y}\hspace{5em}\stackrel{\textit{substituting on the 2nd equation}}{0.2(50-y) + 0.7y=20} \\\\\\ 10-0.2y+0.7y=20\implies -0.2y+0.7y=10\implies 0.5y=10 \\\\\\ y=\cfrac{10}{0.5}\implies \boxed{y=20}`