From a container of milk, 5 litres of milk is replaced with 5 litres of water. This process is repeated again.Thus in two attempts the rati…

Solution
Remaining milk = Initial concentration (1 - $\frac{\textrm{quantity taken out}} {\textrm{Total amount}}$)
Concentration of milk in the final mixture = $\frac{81}{81 + 19} = \frac{81}{100}$
Let initially, milk in the container = $x$ litres
=> $81 = 100 (1 - \frac{5}{x})^2$
=> $\frac{81}{100} = (1 - \frac{5}{x})^2$
=> $(1 - \frac{5}{x})^2 = (\frac{9}{10})^2$
=> $1 - \frac{5}{x} = \frac{9}{10}$
=> $\frac{5}{x} = 1 - \frac{9}{10}$
=> $\frac{5}{x} = \frac{10 - 9}{10} = \frac{1}{10}$
=> $x = 5 \times 10 = 50$ litres