Vessel contains 40 liters wine. If x liters of wine is replaced with water, this process is repeated two more times and then quantity of wine in the vessel is 20.48 liters, find the value of x?

The correct option is 8 liters.

Step-by-Step Explanation:

When a vessel contains an initial amount of pure liquid and a certain quantity is replaced by water repeatedly, the remaining quantity of the original liquid after $n$ operations can be found using the standard replacement formula:

$\text{Remaining Quantity}=\mathrm{Initial\; Quantity}\times {(1-\frac{x}{V})}^n$

Given Data:

Initial volume of wine, $V=40$ liters

Volume of wine replaced each time = $x$ liters

The process is performed once initially and then repeated two more times, which means total number of operations, $n=1+2=3$.

Final quantity of wine remaining = $20.48$ liters

Substituting the values into the formula:

$20.48=40\times {(1-\frac{x}{40})}^3$

Divide both sides by $40$:

${(1-\frac{x}{40})}^3=\frac{20.48}{40}$

${(1-\frac{x}{40})}^3=0.512$

Taking the cube root on both sides:

$1-\frac{x}{40}=\sqrt[3]{0.512}$

Since ${0.8}^3=0.512$:

$1-\frac{x}{40}=0.8$

Rearranging to solve for $x$:

$\frac{x}{40}=1-0.8$

$\frac{x}{40}=0.2$

$x=40\times 0.2=8$

Therefore, the value of $x$ is 8 liters.