Question 119685
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how many gallons of 30% alchol solution and how many of 60% alchol solution must be mixed to produce 18 gallons of 50% solution
Let "x" be the amount of 30% alcohol; active ingredient = 0.3x gallons
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Then "18-x" is amt. of 60% alcohol ; act. ingre. =0.6(18-x)=10.8-0.6x gals.
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Amt. of mixture is 18 gals; act. ingre. = 0.5*18 = 9 gallons
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EQUATION:
active = active = active
0.3x + 10.8-0.6x = 9
-0.3x = -1
x = 10/3 = 3 1/3 gallons (Amount of 30% alcohol needed for the mixture.)
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Cheers,
Stan H.