A vessel contains 100 litres mixture of milk and water in the respective ratio of 22 : 3. 40 litres of the mixture is taken out from the vessel and 4.8 litres of pure milk and pure water each is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?

Quantity of milk in vessel = $$\frac{22}{25} \times 100 = 88$$ litres

=> Quantity of water = $$100 - 88 = 12$$ litres

40 litres of the mixture is taken out, i.e., $$\frac{40}{100} = (\frac{2}{5})^{th}$$

=> Milk left = $$88 - \frac{2}{5} \times 88 = 52.8$$ litres

Water left = $$12 - \frac{2}{5} \times 12 = 7.2$$ litres

Now, 4.8 lires of milk and water are added.

=> Quantity of milk in the vessel = 52.8 + 4.8 = 57.6 litres

Quantity of water in the vessel = 7.2 + 4.8 = 12 litres

$$\therefore$$ Required % = $$\frac{57.6 - 12}{57.6} \times 100$$

= $$\frac{475}{6} = 79 \frac{1}{6} \%$$