Question 765298
The question really should be changed.  How about this way:


A 5 gallon radiator containing a mixture of water and antifreeze solution. When tested , it was found to have only 40% antifreeze. How much must be drained out  AND REPLACED WITH PURE ANTIFREEZE so that the radiator will then contain 5 GALLONS OF THE desired 50% antifreeze solution?


Let v = the amount of solution to remove and then replaced with pure antifreeze


{{{(5*40-v*40+v*100)/5=50}}}
Note that the desired final volume must be the same as the starting volume in the five-gallon tank.


Arithmetic steps to solving:
(5*40+(100-40)v)/5=50
{{{5*40+60v=5*50}}}
{{{60v=5*50-5*40}}}
{{{60v=50}}}
{{{highlight(v=5/6)}}}, meaning, take out five-sixths of a gallon and replace with 100% antifreeze.