SOLUTION: To get 100 gallons of mixture that is 56% acid solution, it was necessary to mix 20% acid solution with some 80% solution. How much of each was required?
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Question 356507: To get 100 gallons of mixture that is 56% acid solution, it was necessary to mix 20% acid solution with some 80% solution. How much of each was required?
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
.80X+.20(100-X)=.56*100
.80X+20-.20X=56
.60X=56-20
.60X=36
X=36/.60
X=60 GALLONS OF 80% IS USED.
100-60=40 GALLONS OF 20% IS USED.
PROOF:
.80*60+.20*40=.56*100
48+8=56
56=56
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