Solve the problem.
-How many liters of a 30% alcohol solution must be mixed with 50 liters of a 70% solution to get a 40% solution?
A) 20 L
B)15 L
C) 150 L
D) 200 L

Question

Quizplus · Accepted Answer

To solve this, we can set up an equation based on the amount of alcohol in each solution. Let x be the amount of 30% alcohol solution needed. The total amount of alcohol from the 30% solution is 0.30x liters, and from the 70% solution, it's 0.70*50 liters. The final mixture will have (x + 50) liters of solution at 40% alcohol. Setting up the equation: 0.30x + 0.70*50 = 0.40(x + 50). Solving for x gives x = 150 liters.

Solve the Problem. -How Many Liters of a 30% Alcohol Solution Must Be