Question 53078
I will solve this problem using two approaches to demonstrate, once again, the flexibility that is frequently available in solving mixture problems.

Let x=the amt of initial mixture that is drained and refilled with 100% antifreeze. Note: the initial mixture contains 35% antifreeze and 65%(water?).
First, we'll examine the water:

Amt of water in initial mixture (16)(.65)minus amt of water that was drained out (x)(.65) equals the amt of water in the final mixture (16)(.50). Our equation to solve is: 
(16)(.65)-.65x=(16)(.50) simplifying, we have
10.4-.65x=8 and
-.65x=-2.4
x=3.69 liters

Now, lets examine the antifreeze:
Amt of antifreeze in the initial mixture(16)(.35)minus amt of antifreeze that was drained out (x)(.35) plus the pure antifreeze that was added back (x)(1) equals amt of antifreeze in the final mixture (16)(.50).  Our equation to solve is:
(16)(.35)-.35x+x=(16)(.50) simplifying, we have
5.6+.65x=8 and
.65x=2.4
x=3.69 liters

Hope this helps -----ptaylor