A container initially holds 50 liters of pure disinfectant solution. If x liters of the solution is withdrawn and replaced with distilled water, and this exact process is repeated two additional times, the final quantity of pure disinfectant remaining in the container becomes 25.6 liters. What is the value of x?

The correct option is 10 liters.

Understanding the Problem:
We are given a container that initially holds 50 liters of pure disinfectant solution. A quantity of $x$ liters is drawn out and replaced with distilled water. This exact operation is repeated two additional times (making a total of 3 replacement operations). After these 3 operations, the remaining quantity of pure disinfectant in the container is 25.6 liters. We need to determine the value of $x$.

Formula for Replacement Operations:
When a container originally contains $V$ units of liquid, and $x$ units are withdrawn and replaced by water $n$ times successively, the remaining quantity of original liquid is given by:

$\text{Final Quantity}=V{(1-\frac{x}{V})}^n$

Given Values:
Initial volume of disinfectant solution, $V=50\text{liters}$
Total number of operations, $n=3$
Final quantity of pure disinfectant remaining = 25.6 liters

Step 1: Substitute the given values into the formula

$25.6=50{(1-\frac{x}{50})}^3$

Step 2: Simplify the equation
Divide both sides of the equation by 50:

${(1-\frac{x}{50})}^3=\frac{25.6}{50}$

Convert the fraction on the right-hand side to a simpler decimal:

$\frac{25.6}{50}=\frac{256}{500}=\frac{512}{1000}=0.512$

Substituting this back into the equation:

${(1-\frac{x}{50})}^3=0.512$

Step 3: Take the cube root of both sides

$1-\frac{x}{50}=\sqrt[3]{0.512}$

Since ${0.8}^3=0.512$, we have:

$1-\frac{x}{50}=0.8$

Step 4: Solve for x

$\frac{x}{50}=1-0.8$

$\frac{x}{50}=0.2$

$x=50\times 0.2$

$x=10$

Therefore, the value of $x$ is 10 liters.