[tex]\bf \begin{array}{lccclll}
&amount&concentration&
\begin{array}{llll}
concentration\\
amount
\end{array}\\
&-----&-------&-------\\
\textit{20\% sol'n}&x&0.20&0.20x\\
\textit{50\% sol'n}&y&0.50&0.50y\\
-----&-----&-------&-------\\
mixture&60&0.40&(60)(0.40)
\end{array} [/tex]
notice above... we use the decimal format for the percentage, thus 20% is really just 20/100, 40% is 40/100 and so on.
so.. whatever "x" and "y" amounts are, we know they have to add up to 60 Liters, that is x+y = 60
and whatever the concentration amount of each is, it must add up to (60)(0.40), that is 0.20x+0.50y=(60)(0.40)
thus [tex]\bf \begin{cases}
x+y=60\implies \boxed{y}=60-x\\\\
0.20x+0.50y=(60)(0.40)\\
----------\\
0.20x+0.50\left( \boxed{60-x} \right)=(60)(0.40)
\end{cases}[/tex]
solve for "x", to see how much of the 20% solution will be needed
what about "y"? well, y = 60 - x
