One acetic acid solution is 80% water and another is 45% water. How many liters of each solution should be mixed to produce 10 liters of a solution that is 73% water?
A)5 liters of the 80% solution and 5 liters of the 45% solution
B)8 liters of the 80% solution and 2 liters of the 45% solution
C)5 liters of the 80% solution and 5 liters of the 45.0% solution
D)3 liters of the 80% solution and 7 liters of the 45% solution
E)7 liters of the 80% solution and 3 liters of the 45% solution

Question

Quizplus · Accepted Answer

This problem can be solved using the method of weighted averages or a system of equations. The goal is to find a mixture that results in a solution that is 73% water. Let $x$ be the amount of the 80% solution and $y$ be the amount of the 45% solution. We have two conditions: $x + y = 10$ (the total volume is 10 liters) and $0.80x + 0.45y = 0.73 \times 10$ (to ensure the final mixture is 73% water). Solving these equations, we find that $x = 8$ and $y = 2$, meaning 8 liters of the 80% solution and 2 liters of the 45% solution are needed.

One Acetic Acid Solution Is 80% Water and Another Is