Question 703232:
how many gallons of an 80% alcohol solution must be mixed with 30 gallons of 24% solution to obtain a solution that is 70% alcohol. Answer by checkley79(3341) (Show Source):
You can put this solution on YOUR website! .80X+.24*30=.70(30+X)
.80X+7.2=21+.70X
.80X-.70X=21-7.2
.10X=13.8
X=13.8/.10
X=138 GAL OF 80% SOLUTION IS USED.
PROOF:
.80*138+.24*30=.70(30+138)
110.4+7.2=.70*168
117.6=117.6