Initial water content in mixture replacement

Let initial water = $w$ litres. Then initial milk = $(300 - w)$ litres.

Volume of solution removed = $2w$ litres. Fraction of solution remaining = $1 - \frac{2w}{300} = \frac{300 - 2w}{300}$.

Since no milk is added when refilling with water, the final volume of milk is: $\text{Final Milk} = (300 - w) \times \frac{300 - 2w}{300}$

We are given final milk content is $72\%$ of $300 = 216$ litres: $\frac{(300 - w)(300 - 2w)}{300} = 216$ $(300 - w)(300 - 2w) = 64800$ $90000 - 900w + 2w^2 = 64800$ $2w^2 - 900w + 25200 = 0 \implies w^2 - 450w + 12600 = 0$ $(w - 30)(w - 420) = 0$

Since capacity is $300$ L, $w = 30$ litres.