A vessel contains 139.96 litres of a mixture of milk and water, where 80% is milk. If 24.96% of the mixture is removed and 19 litres of water is added, what is the percentage of water in the resultant mixture (approximate)?

To find the approximate percentage of water in the final mixture, we can break down the process step-by-step:

Step 1: Understand the initial composition of the mixture
The vessel initially contains 139.96 litres of a mixture of milk and water.
Since milk constitutes 80% of the mixture, water constitutes the remaining portion:
Water percentage = 100% - 80% = 20%

Step 2: Calculate the volume of the mixture left after removal
It is given that 24.96% of the mixture is removed. Therefore, the percentage of mixture remaining is:

$100\%-24.96\%=75.04\%$

Now, let's find the volume of this remaining mixture:

$\text{Remaining Volume}\approx 139.96\times 75.04\%\approx 139.96\times 0.7504��105\text{litres}$

Step 3: Find the quantity of water in the remaining mixture
Removing a portion of a homogeneous mixture does not alter the ratio of its components. Thus, the remaining 105 litres of mixture still contains 20% water.

$\text{Volume of water in remaining mixture}=20\%\text{of}105=0.20\times 105=21\text{litres}$

Step 4: Account for the added water
Next, 19 litres of pure water is added to the vessel.
This increases both the quantity of water and the total volume of the mixture.

$\text{New volume of water}=21+19=40\text{litres}$

$\text{New total volume of mixture}=105+19=124\text{litres}$

Step 5: Calculate the final percentage of water
The percentage of water in the resultant mixture is calculated as:

$\text{Percentage of water}=\left(\frac{40}{124}\right)\times 100\%\approx 32.26\%$

Approximating to the nearest integer gives 32%.

Therefore, the correct option is 32.