Bottle A, containing 12L of 15% acid, is combined with bottle B, containing 3L of 25% acid. Bottle C is 26% acid. How much of each solution is neede to have 24L of a 20% acid solution?

To solve this problem, we can use the concept of mixture problems.

Let's assume that x represents the amount of solution from bottle A (15% acid) that is needed, and y represents the amount of solution from bottle B (25% acid) that is needed to create the desired 20% acid solution. We know that the total volume of the solution is 24L, so we have the equation:

x + y = 24 ----(equation 1)

Also, we know that the acid content is a percentage. The acid in bottle A is 15% of 12L, and the acid in bottle B is 25% of 3L. So, the total acid in the solution can be expressed as:

0.15x + 0.25y = 0.20 * 24 ----(equation 2)

Now we have a system of equations (equation 1 and equation 2). We can solve this system to determine the values of x and y.

First, we can rearrange equation 1 to solve for x:

x = 24 - y

Substituting this value of x into equation 2, we have:

0.15(24 - y) + 0.25y = 0.20 * 24
3.6 - 0.15y + 0.25y = 4.8
0.10y = 1.2
y = 1.2 / 0.10
y = 12L

Now, substituting the value of y back into equation 1, we have:

x + 12 = 24
x = 24 - 12
x = 12L

Therefore, to create 24L of a 20% acid solution, you need 12L of the 15% acid solution (bottle A) and 12L of the 25% acid solution (bottle B).