Question 158772
Let x=amount of 30% acid solution needed

Now we know that the amount of pure acid in the 30% solution (0.30x) plus the amount of pure acid in the 40 liters of  12% solution (0.12*40) has to equal the amount of pure acid in the solution after the mixture takes place (0.20(40+x)).  So our equation to solve is:

0.30x+0.12*40=0.20(40+x)  get rid of parens
0.30x+4.8=8+0.20x  subtract 4.8 and also 0.20x from both sides
0.30x-0.20x+4.8-4.8=8-4.8+0.20x-0.20x  collect like terms
0.10x=3.2  divide each side by 0.10
x=32 gal------------------------------amount of 30% solution needed

CK
0.30*32+0.12*40=0.20*72
9.6+4.8=14.4
14.4=14.4

Hope this helps----ptaylor