Let
x -----> ml of 20% acid
y ----> ml of pure acid
so
we have that
[tex]\begin{gathered} x+y=20 \\ y=20-x\text{ ----> equation 1} \end{gathered}[/tex][tex]0.20x+y=0.50(20)\text{ ----> equation 2}[/tex]substitute equation 1 in equation 2
[tex]\begin{gathered} 0.20x+(20-x)=0.50(20) \\ solve\text{ for x} \\ 0.20x+20-x=10 \\ x-0.20x=20-10 \\ 0.80x=10 \\ x=12.5\text{ ml} \end{gathered}[/tex]