Bunuel wrote:

John has 3 solutions: a 12% saline solution, a 8% vinegar solution, and a 15% alcohol solution. He mixes 3 liters of the alcohol solution with 3 liters of the vinegar solution. What is the minimum amount of the saline solution he must add if the resulting mixture must be at least 2% saline solution?

A. 7.1 liters

B. 3 liters

C. 2.4 liters

D. 1.2 liters

E. 1.1 liters

Let X be the amount of Saline Solution, that needs to be added.

We have Total amount of Final Solution = 3 + 3 + X = 6 + X liters

For the final solution to be 2% Saline, we need 2% of (6 + X) = 12% of X

Hence 2*(6+X) = 12*X

Solving we, get X = 1.2 liters

Answer D.

Thanks,

GyM