Question 401322
Let x=amount of coolant that must be drained and replaced with water to achieve a 50% antifreeze concentration

Now we now that the amount of pure antifreeze remaining after x amount of coolant is drained out is 0.65(3.2-x) and this has to equal the amount of pure antifreeze in the final mixture after the water is added (0.50*3.2).  So our equation to solve is:
0.65(3.2-x)=0.50*3.2 simplify
2.08-0.65x=1.6 subtract 2.08 from each side
-0.65x=-0.48
x=0.738 liters ----amount of coolant that must be drained and replaced

CK
0.65*(3.2-0.738)=0.50*3.2
0.65*2.462=1.6
1.6003~~~~1.6

Hope this helps--ptaylor