How many liters of 50% alcohol solution and 20% alcohol solution must be mixed to obtain 9 liters of 30% alcohol solution?

Question

You need  liters of 50% solution and  liters of 20% solution.

Answer 1

Let x = the amount of 50% alcohol

9-x = the amount of 20% alcohol

Then

{ 50x + 20(9-x) }/(x + 9 - x) = 30 or { 50x + 180 -20x } = 270

30x = 90 or x = 3 liters of the 50% solution

9-x = 6 liters of the 20% solution