SOLUTION: How many liters of a 60% alcohol solution must be mixed with 90 liters of a 90% solution to get a 80% solution
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Question 136962
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How many liters of a 60% alcohol solution must be mixed with 90 liters of a 90% solution to get a 80% solution
Answer by
checkley77(12844)
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.6X+.9*90=.8(90+X)
.6X+81=72+.8X
.6X-.8X=72-81
-.2X=-9
X=-9/-.2
X=45 LITERS OF 60% ALCOHOL IS NEEDED.
PROOF:
.6*45+.9*90=.8(90+45)
27+81=.8(135)
108=108