SOLUTION: How many gallons of an 80% alcohol solution must be mixed with 30 gallons of a 22% solution to obtain a solution that is 70% alcohol?

Question 125288: How many gallons of an 80% alcohol solution must be mixed with 30 gallons of a 22% solution to obtain a solution that is 70% alcohol?
Found 2 solutions by checkley71, amalm06:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
.8X+.22*30=.7(X+30)
.8X+6.6=.7X+21
.8X-.7X=21-6.6
.1X=14.4
X=14.4/.1
X=144 GALLONS OF 80% SOLUTION IS REQUIRED.
PROOF
.8*144+.22*30=.7(144+30)
115.2+6.6=121.8
121.8=121.8

Answer by amalm06(224) (Show Source): You can put this solution on YOUR website!
(x)(0.80)+(30)(0.22)=(30+x)(0.70)
0.8x+6.6=21+0.7x
0.1x=14.4
x=144 gallons (Answer)

Alternatively, we may use the method of alligation.
0.70-0.22=0.48
0.80-0.70=0.10
Then T1/T2=0.10/0.48=0.2083333
Since T1=30, we have 30/T2=0.2083333
T2=144 gallons (Answer)

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