Let $x$ be the number of liters drained and replaced. How many liters of bleach are removed when you drain $x$ liters of a 20% solution, and how many liters of bleach are added when you pour in $x$ liters of a 40% solution?

The final mixture is still 50 liters at 28% bleach. Write the expression for the total bleach after draining and replacing in terms of $x$, and set it equal to $0.28 \cdot 50$.

In Desmos, define the function for bleach after replacement: type y = 10 + 0.2x. Then enter the final bleach amount as a horizontal line: y = 14.

Zoom so you can see where the line $y = 10 + 0.2x$ intersects $y = 14$. Click the intersection point; the x-coordinate of this point is the number of liters that must be drained and replaced.

The original solution is 50 liters and is 20% bleach.

Compute the liters of bleach:

So, there are 10 liters of pure bleach in the original solution.

Let $x$ be the number of liters drained and replaced.

• When you drain $x$ liters of a 20% solution, the bleach removed is $0.20x$ liters.
• Bleach remaining after draining:

• Then you add $x$ liters of a 40% solution, which adds $0.40x$ liters of bleach.

Total bleach after draining and replacing is:

After the procedure, the total volume is still 50 liters, and the solution is 28% bleach.

So the final amount of bleach is:

This must equal the bleach amount we found in Step 2:

Solve the equation

Subtract 10 from both sides:

Divide both sides by $0.20$ (or multiply by 5):

So, $\boxed{20}$ liters of the original solution were drained and replaced.