A container contains two liquids, $X$ and $Y$, in the ratio $5:3$. When $8\text{ liters}$ of the mixture is drawn off, and the container is filled with liquid $Y$, the ratio of $X$ to $Y$ becomes $2:3$. How many liters of liquid $X$ was in the container initially?

Let the initial quantities of liquids $X$ and $Y$ be $5k$ and $3k$ respectively. Total initial mixture $=8k$ liters.
When $8\text{ liters}$ of the mixture is drawn off:

* Liquid $X$ removed $=8\times \frac{5}{8}=5\text{ liters}$
* Liquid $Y$ removed $=8\times \frac{3}{8}=3\text{ liters}$

Remaining quantities after drawing off the mixture:

* Liquid $X=5k-5$
* Liquid $Y=3k-3$

Now, the container is completely filled up with $8\text{ liters}$ of pure liquid $Y$:

$$\text{New quantity of }Y=(3k-3)+8=3k+5$$

The new ratio becomes $2:3$:

$$\frac{5k-5}{3k+5}=\frac{2}{3}$$

$$3(5k-5)=2(3k+5)$$

$$15k-15=6k+10$$

$$9k=25⟹k=\frac{25}{9}$$

Calculate the initial quantity of liquid $X$:

$$\text{Initial }X=5k=5\times \frac{25}{9}=\frac{125}{9}=13\frac{8}{9}\text{ liters}$$