<PRE><B>Question 165519</B>
Let x=amount that needs to be drained off and replaced with pure antifreeze
Now we know that the amount of pure antifreeze left after x amount is drained off (0.4(50-x)) plus the amount of pure antifreeze added(x) has to equal the amount of pure antifreeze in the final mixture (0.5(50)). So, our equation to solve is: 
0.4(50-x) + x=0.50*50 get rid of parenthesis(distributive)
20-.4x+x=25 subtract 20 from each side
and combine like terms 
.6x=5 divide each side by 0.6
x=8.33 qts-----------------------------------ans
CK
.4(50-8.33)+8.33=25
  (50/3)+25/3=25
       75/3=25
        25=25
Hope this helps</PRE>