Question

How many liters of a 25% solution must be mixed with a 12% solution to get 13L of a 15% SOLUTION?

Answer 1

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:

1. The total volume of the mixture:
\[
x + y = 13
\]

2. 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:
1. \( x + y = 13 \)
2. \( 0.25x + 0.12y = 1.95 \)

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**.