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Section 2.3 : Applications of Linear Equations
7. How much of a 20% acid solution should we add to 20 gallons of a 42% acid solution to get a 35% acid solution?
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Start SolutionWe’ll start by letting \(x\) be the amount of the 20% solution we’ll need. This in turn means that we’ll have \(x + 20\)gallons of the 35% solution once we’re done mixing.
The basic word equation is then,
\[\left( \begin{array}{c}{\mbox{Amount of acid}}\\ {\mbox{in 20% solution}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Amount of acid}}\\ {\mbox{in 42% solution}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Amount of acid}}\\ {\mbox{in 35% solution}}\end{array} \right)\]We know that Amount of Acid in Solution = Percentage of Solution X Volume of Solution. This gives the following word equation.
\[\left( {0.20} \right)\left( \begin{array}{c}\,\,{\mbox{Volume of}}\\ {\mbox{20% solution}}\end{array} \right) + \left( {0.42} \right)\left( \begin{array}{c}\,\,{\mbox{Volume of}}\\ {\mbox{42% solution}}\end{array} \right) = \left( {0.35} \right)\left( \begin{array}{c}\,\,{\mbox{Volume of}}\\ {\mbox{35% solution}}\end{array} \right)\] Show Step 2So, plugging all the known information in gives the following equation that we can solve for \(x\).
\[\begin{align*}0.2x + \left( {0.42} \right)\left( {20} \right) & = 0.35\left( {x + 20} \right)\\ 0.2x + 8.4 & = 0.35x + 7\\ 0.15x & = 1.4\\ x & = 9.33\end{align*}\]So, we’ll need 9.33 gallons of the 20% acid solution.