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How many liters each of a 55% acid solution and a 80% acid solution must be used to produce 50 liters of a 60% acid solution? (Round to two decimal places if necessary.)

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Given:

Let x denote the number of liter solutions.


\begin{gathered} \text{acid}+\text{acid}=\text{acid} \\ 55\text{ \%x+80 \%( 50-x)=}60\text{ \%}*50 \end{gathered}

Solve the equation for x,


\begin{gathered} 55\text{ \%x+80 \%( 50-x)=}60\text{ \%}*50 \\ (55)/(100)x+(80)/(100)(50-x)=(60)/(100)*50 \\ 0.55x+0.8(50-x)=30 \\ 0.55x+40-0.8x=30 \\ -0.25x=30-40 \\ x=-(10)/(-0.25) \\ x=40 \end{gathered}

It gives,


\begin{gathered} 40\text{ liters of 55 \% acid will be n}eeded \\ (50-x)=(50-40)=10\text{ liters of 80 \% acid will be n}eeded \end{gathered}

Answer: 40 liters of a 55% acid solution and 10 liters of 80% acid solution must be used to produce 50 liters of a 60% acid solution

User Badal Shah
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