Given:
Let x denote the number of liter solutions.
[tex]\begin{gathered} \text{acid}+\text{acid}=\text{acid} \\ 55\text{ \%x+80 \%( 50-x)=}60\text{ \%}\times50 \end{gathered}[/tex]Solve the equation for x,
[tex]\begin{gathered} 55\text{ \%x+80 \%( 50-x)=}60\text{ \%}\times50 \\ \frac{55}{100}x+\frac{80}{100}(50-x)=\frac{60}{100}\times50 \\ 0.55x+0.8(50-x)=30 \\ 0.55x+40-0.8x=30 \\ -0.25x=30-40 \\ x=-\frac{10}{-0.25} \\ x=40 \end{gathered}[/tex]It gives,
[tex]\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}[/tex]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
