A container contains a mixture of milk and water in the ratio $7:5$. When $9\text{litres}$ of mixture is drawn off and the container is filled with water, the ratio of milk and water becomes $7:9$. How many litres of milk was contained by the container initially?

Let the initial quantities of milk and water in the container be $7x\text{litres}$ and $5x\text{litres}$ respectively.
Total initial volume of the mixture = $7x+5x=12x\text{litres}$.

When $9\text{litres}$ of the mixture is drawn off:

* Quantity of milk removed = $9\times \frac{7}{12}=\frac{21}{4}\text{litres}$
* Quantity of water removed = $9\times \frac{5}{12}=\frac{15}{4}\text{litres}$

Now, the container is filled with $9\text{litres}$ of water.

* New quantity of milk = $7x-\frac{21}{4}$
* New quantity of water = $5x-\frac{15}{4}+9=5x+\frac{21}{4}$

According to the question, the new ratio of milk to water is $7:9$:

$$\frac{7x-\frac{21}{4}}{5x+\frac{21}{4}}=\frac{7}{9}$$

Taking $7$ common from the numerator on the left side:

$$\frac{7\left(x-\frac{3}{4}\right)}{5x+\frac{21}{4}}=\frac{7}{9}$$

Canceling $7$ from both sides:

$$\frac{x-\frac{3}{4}}{5x+\frac{21}{4}}=\frac{1}{9}$$

Cross-multiplying to solve for $x$:

$$9\left(x-\frac{3}{4}\right)=1\times \left(5x+\frac{21}{4}\right)$$

$$9x-\frac{27}{4}=5x+\frac{21}{4}$$

$$9x-5x=\frac{21}{4}+\frac{27}{4}$$

$$4x=\frac{48}{4}$$

$$4x=12$$

$$x=3$$

The initial quantity of milk was $7x$:

$$\text{Initial quantity of milk}=7\times 3=21\text{litres}$$

Therefore, the container initially contained $21\text{litres}$ of milk.