Question 133669
Let x=amount of 75% solution needed
Then 20-x=amount of 25% solution needed

Now we know that the amount of pure antifreeze in the 25% solution (0.25(20-x)) plus the amount of pure antifreeze in the 75% solution (0.75x)  has to equal the amount of pure antifreeze in the final mixture (0.50*20).  So our equation to solve is:

0.25(20-x)+0.75x=0.50*20  simplify

5-0.25x+0.75x=10  subtract 5 from both sides
5-5-0.25x+0.75x=10-5  collect like terms

0.50x=5  divide both sides by 0.50
x=10 qts------------------------------amount of 75% solution needed
20-x=20-10=10 qts------------------------------amount of 25% solution needed

CK
0.25*10+0.75*10=0.50*20
2.5+7.5=10
10=10


Hope this helps----ptaylor