Question

Is your class 14 L of 11% alcohol solution must be mixed with the 20% solution to get a 13% solution how many liters of the 3% solution are needed

Answer 1

To solve this problem, we can use the concept of the weighted average. Let's denote the number of liters of the 11% alcohol solution as 'x' and the number of liters of the 20% alcohol solution as 'y'. Based on the given information, we can set up the equation 0.11x + 0.20y = 0.13(x + y). Solving this equation, we find that the number of liters of the 3% solution needed is the same as the number of liters of the 11% solution.

Step-by-step explanation:

To solve this problem, we can use the concept of the weighted average. Let's denote the number of liters of the 11% alcohol solution as 'x' and the number of liters of the 20% alcohol solution as 'y'.

Based on the given information, we can set up the following equation:

0.11x + 0.20y = 0.13(x + y)

Next, we can solve this equation to find the value of x:

1. Expand the equation: 0.11x + 0.20y = 0.13x + 0.13y
2. Combine like terms on both sides: 0.02y = 0.02x
3. Divide both sides by 0.02:
4. y = x

So, in order to get the desired 13% solution, the number of liters of the 3% solution needed would be the same as the number of liters of the 11% solution, which is 'x'.