Question 491716: how many liters of 60% alcohol solution and 40% alcohol solution should be mixed to obtain twenty liters of a 45% alcohol solution?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=amount of 60% solution needed
Then 20-x=amount of 40% solution needed
Now we know that the amount of pure alcohol that exists before the mixture takes place is equal to the amount of pure alcohol after the mixture takes place, so:
0.60x+0.40(20-x)=amount of pure alcohol before the mixture takes place, and
0.45*20=amount of pure alcohol after the mixture takes place,so
0.60x+0.40(20-x)=0.45*20 simplify
0.60x+8-0.40x=9 subtract 8 from each side
0.20x=1
x=5 liters-------------------------amount of 60% solution needed
20-x=20-5=15 liters----------------amount of 40% solution needed
CK
0.60*5+0.40*15=0.45*20
3+6=9
9=9
Hope this helps---ptaylor
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