Question 263828

Let x=amount of 80% antifreeze needed

Initially the amount of pure antifreeze in the tank =0.30*20=6 gal. If we increase the pure antifreeze in the tank by 50% we will need to have have 6+3=9 gal of pure antifreeze((9-6)/6=50% increase).  In other words, we need to use enough of the 80% antifreeze to get 3 gal of pure antifreeze.  So our equation to solve is:
 0.80x=3
x=3.75 gal---needs to be added to the tank

CK
20*0.30+3.75*0.80=9
6+3=9
9=9

Hope this helps---ptaylor