Question 57310
Let x=amt of pure antifreeze that is mixed in
Then 20-x=amt of 40% antifreeze solution that is mixed in
We know that the amt of pure antifreeze that is mixed in (x)(1) plus the amt of pure antifreeze in the 40% solution that is mixed in (20-x)(.40) equals the amt of pure antifreeze in the final solution (20)(.50).
Thus, our equation to solve is:

(x)(1)+(20-x)(.40)=(20)(.50) Simplifying, we get:
x+8-.4x=10 or
.6x=2; 6x=20
x=3.3333 qts of pure antifreeze
20-x=20-3.3333=16.6667 qts of the 40% antifreeze solution

Hope this helps-----ptaylor