In a container, two types of liquids $P$ and $Q$ are in the ratio of $3:7$. If $10$ litres of the mixture is drawn off from the container and it is filled with $6$ litres of liquid $Q$, then the ratio of $P$ and $Q$ becomes $1:3$. What was the original quantity (in litres) of the mixture?

Let the original quantities of liquids $P$ and $Q$ be $3x$ and $7x$ respectively.Total original volume $=3x+7x=10x$ litres.When $10$ litres of mixture is drawn off:Liquid $P$ removed $=10\times \frac{3}{10}=3$ litresLiquid $Q$ removed $=10\times \frac{7}{10}=7$ litresRemaining amounts after drawing off:Liquid $P=3x-3$ Liquid $Q=7x-7$ Next, $6$ litres of liquid $Q$ is added:New amount of liquid $Q=7x-7+6=7x-1$ The new ratio of $P$ to $Q$ is given as $1:3$:
$$\frac{3x-3}{7x-1}=\frac{1}{3}$$
Cross-multiplying gives:
$$3(3x-3)=1(7x-1)$$
$$9x-9=7x-1$$
$$9x-7x=9-1$$
$$2x=8$$
$$x=4$$
The original quantity of the mixture $=10x=10\times 4=40$ litres.