Question 241532: how many gallons of a mixture containing 80% alcohol should be added to 6 gallons of a 25% solution togive 30% solution? Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! .80x+.25*6=.30(6+x)
.80x+1.5=1.8+.30x
.80x-.30x=1.8-1.5
.5x=.3
x=.3/.5
x=.6 gal. of 80% mixture is used.
Proof:
.80*.6+.25*6=.30(6+.6)
.48+1.5=.30*6.6
1.98=.3*1.6
1.98=1.98