Two pipes can fill a cistern in $20\text{ minutes}$ and $30\text{ minutes}$ respectively. If both pipes are opened simultaneously, after how many minutes must the first pipe be turned off so that the cistern gets completely filled in exactly $10\text{ minutes}$ more?

Let the total capacity of the cistern be the LCM of $20$ and $30$, which is $60\text{ units}$.
* Efficiency of Pipe 1 ($E_1$) = $\frac{60}{20}=3\text{ units/minute}$
* Efficiency of Pipe 2 ($E_2$) = $\frac{60}{30}=2\text{ units/minute}$

Let Pipe 1 be turned off after $t$ minutes.
For the first $t$ minutes, both pipes work together. After $t$ minutes, Pipe 1 is closed, and Pipe 2 runs alone for another $10\text{ minutes}$ to finish filling the cistern.

We can set up the work equation:

$$\text{Work done by both in }t\text{ mins}+\text{Work done by Pipe 2 alone in }10\text{ mins}=60$$

$$(E_1+E_2)\times t+E_2\times 10=60$$

$$(3+2)\times t+2\times 10=60$$

$$5t+20=60$$

$$5t=40⟹t=8\text{ minutes}$$

Thus, the first pipe must be turned off after $8\text{ minutes}$.