Suppose that one solution contains 20% alcohol and another solution contains 50% alcohol. How many liters of each solution should be mixed to make 12 liters of a 40%-alcohol solution?
A) 6 liters of 20% solution and 6 liters of 50% solution
B) 4 liters of 20% solution and 8 liters of 50% solution
C) 8 liters of 20% solution and 4 liters of 50% solution

Question

Quizplus · Accepted Answer

Let x be the number of liters from the 20% solution and y be the number of liters from the 50% solution. Then, we have the following system of equations:
x + y = 12  (since we need a total of 12 liters)
0.2x + 0.5y = 0.4(12)  (since the final solution should be 40% alcohol)
Solving for x and y, we get x = 4 and y = 8. Therefore, we need 4 liters of the 20% solution and 8 liters of the 50% solution. Choice B is the correct answer.

Suppose That One Solution Contains 20% Alcohol and Another Solution