To find out how many liters of a 25% solution and a 12% solution are needed to create 13 liters of a 15% solution, we can set up a system of equations.
Let:
- \( x \) be the volume of the 25% solution (in liters).
- \( y \) be the volume of the 12% solution (in liters).
We have two equations based on the problem statement:
-
The total volume of the mixture: \[ x + y = 13 \]
-
The total amount of solute (the percentage of solution times the volume) must equal the amount of solute in the final solution: \[ 0.25x + 0.12y = 0.15 \times 13 \]
Calculating the right side of the second equation: \[ 0.15 \times 13 = 1.95 \] So the second equation becomes: \[ 0.25x + 0.12y = 1.95 \]
Now we have the system of equations:
- \( x + y = 13 \)
- \( 0.25x + 0.12y = 1.95 \)
From the first equation, we can express \( y \) in terms of \( x \): \[ y = 13 - x \]
Now we substitute \( y \) into the second equation: \[ 0.25x + 0.12(13 - x) = 1.95 \]
Expanding the second equation: \[ 0.25x + 1.56 - 0.12x = 1.95 \]
Combining like terms: \[ (0.25 - 0.12)x + 1.56 = 1.95 \] \[ 0.13x + 1.56 = 1.95 \]
Now we isolate \( x \): \[ 0.13x = 1.95 - 1.56 \] \[ 0.13x = 0.39 \] \[ x = \frac{0.39}{0.13} = 3 \]
Now that we have \( x \), we can find \( y \): \[ y = 13 - x = 13 - 3 = 10 \]
Thus, the amounts needed are:
- 3 liters of the 25% solution
- 10 liters of the 12% solution
Therefore, you need to mix 3 liters of the 25% solution and 10 liters of the 12% solution to get 13 liters of a 15% solution.