Solve the problem.
-How many liters of a 50% alcohol solution must be mixed with 40 liters of a 70% solution to get a 60% solution?
A) 4 L
B) 40 L
C) 8 L
D) 80 L

Question

Quizplus · Accepted Answer

To solve this problem, let's denote the amount of 50% alcohol solution needed as x liters. The total amount of alcohol in the 50% solution is 0.5x liters, and in the 40 liters of 70% solution, it is 0.7*40 liters. The final mixture will have (x + 40) liters of solution, and we want this to be a 60% alcohol solution, so the total alcohol in the final mixture should be 0.6(x + 40) liters. Setting up the equation: 0.5x + 0.7*40 = 0.6(x + 40). Solving for x gives x = 40 liters.

Solve the Problem. -How Many Liters of a 50% Alcohol Solution Must Be