Question 944404
a chemist has two alcohol-in-water: a 20% alcohol solution and a 50% alcohol solution. He needs 12 L of a solution that is 45% alcohol. How many liters of the two starting solutions should he mix
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let x=amt of 20% solution to mix
12-x=amt of 50% solution to mix
20%x+50%(12-x)=45%*12
.20x+6-.50x=5.4
.30x=.60
x=2
12-x=10
amt of 20% solution to mix=2 liters
amt of 50% solution to mix=10 liters