I need 100 liters of 50% alcohol solution. If I only have 30% and 80% solutions in stock, how many liters of each should I mix to get the 100 liters of 50% alcohol?

Expert Answers
justaguide eNotes educator| Certified Educator

30% and 80% solutions of alcohol have to be mixed to obtain a 50% solution. The volume of the 50% solution required is 100 liters.

Let the number of liters of 30% solution to be used be x, the number of solutions of 80% solution is 100 - x.

0.3*x + 0.8*(100 - x) = 0.5*100

=> 0.3*x + 80 - 0.8x = 50

=> 0.5x = 30

=> x = 60

To obtain the 100 l of 50% solution 60 liters of 30% solution and 40 liters of 80% solution are required.