A vessel contains a mixture of milk and water in the respective ratio of 14 : 3. 25.5 litres of the mixture is taken out from the vessel and 2.5 litres of pure water and 5 litres of pure milk is added to the mixture. If the resultant mixture contains 20% water, what was the initial quantity of mixture in the vessel before the replacement? (in litres)

Let the total quantity of mixture in the vessel initially = $$17x$$ litres

=> Quantity of milk = $$\frac{14}{17} \times 17x = 14x$$ litres

Quantity of water = $$17x - 14x = 3x$$ litres

Acc. to ques,

=> $$\frac{14x - (\frac{14}{17} \times 25.5) + 5}{3x - (\frac{3}{17} \times 25.5) + 2.5} = \frac{80}{20}$$

=> $$\frac{14x - 21 + 5}{3x - 4.5 + 2.5} = \frac{4}{1}$$

=> $$\frac{14x - 16}{3x - 2} = \frac{4}{1}$$

=> $$14x - 16 = 12x - 8$$

=> $$14x - 12x = 16 - 8$$

=> $$x = \frac{8}{2} = 4$$

$$\therefore$$ Initial quantity of mixture in the vessel before the replacement = $$17 \times 4 = 68$$ litres