Question 73871: How many gallons of a 40% alcohol solution must be mixed with 70 gallons of a 21% slution to obtain a solution that is 30% alcohol Found 2 solutions by checkley75, stanbon:Answer by checkley75(3666) (Show Source):
You can put this solution on YOUR website! .40*X+70*.21=(70+X).30
.4X+14.7=21+.3X
.4X-.3X=21-14.7
.1X=6.3
X=6.3/.1
X=63 GALLONS OF 40% ARE NEEDED
PROOF
.4*63+70*.21=(70+63).3
25.2+14.7=133*.3
39.9=39.9
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YOU MUST HAVE COPIED THE PROBLEM WRONG.
A 30% MIXTURE WITH A 21% MIXTURE CANNOT MAKE A 20% MIXTURE.
You can put this solution on YOUR website! How many gallons of a 40% alcohol solution must be mixed with 70 gallons of a 21% solution to obtain a solution that is 30% alcohol.
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21% solution DATA:
Amt = 70 gals ; amt of alcohol = 0.21*70 = 14.7 gals
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40% solution DATA:
Amt = x gals ; amt of alcohol = 0.40x gals
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30% solution DATA:
Amt = 70+x gals ; amt of alcohol = 0.3(70+x) = 21 + 0.3x gals
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EQUATION:
alcohol + alcohol = alcohol
14.7 + 0.4x = 21 + 0.3x
0.1x = 6.3
x=63 gallons (amt of 40% solution that must be added)
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Cheers,
Stan H.
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