SOLUTION: How many liters of a 10% alcohol solution must be mixed wih 40L of a 50% solution to get a 40% solution

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: How many liters of a 10% alcohol solution must be mixed wih 40L of a 50% solution to get a 40% solution      Log On


   

Question 439571: How many liters of a 10% alcohol solution must be mixed wih 40L of a 50% solution to get a 40% solution
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
40L of a 50% solution equals 20L of pure alcohol.
.10x+.50(40)=.40(x+40)
.10x+20=.4x+16
4=.3x
4/.3=x
13.33=x
You need 13 1/3 liters of a 10% solution to mix up a 40% solution..