Given that the chemist has one solution that is 40% alcohol and another that is 55% alcohol.
We can find how much of each in liters must she used to make 21 L of a solution that is 50% alcohol below.
Since we don't have the quantity of each solution of the alcohol we would be mixing, we would assume the following.
[tex]\begin{gathered} solution1\text{ 1=x} \\ \text{solution 2 =}(\text{21-x)} \end{gathered}[/tex]Therefore, we would have
[tex]\begin{gathered} \frac{40}{100}x+\frac{55}{100}(21-x)=\frac{50}{100}\times21 \\ 0.4x+0.55(21-x)=0.5\times21 \\ 0.4x+11.55-0.55x=10.5 \\ -0.15x=10.5-11.55 \\ -0.15x=-1.05 \\ x=\frac{-1.05}{-0.15} \\ x=7 \end{gathered}[/tex]Since x =7, we would then get the quantity of each alcohol as
[tex]\begin{gathered} solution1\text{ 1=}7l \\ \text{solution 2=}(21-7) \\ =14 \end{gathered}[/tex]Therefore, the answer is
[tex]\begin{gathered} 7l\text{ of the first solution} \\ 14l\text{ of the second solution} \end{gathered}[/tex]