Two pipes can fill a cistern in 20 minutes and 30 minutes respectively. Both pipes being opened, the time when the first pipe must be turned off so that the cistern may be filled in 10 minutes more will be

Let the capacity of the cistern be the LCM of 20 and 30, which is 60 units.
* Efficiency of Pipe 1 (${\text{E}}_1$) = $\frac{60}{20}=3$ units/minute
* Efficiency of Pipe 2 (${\text{E}}_2$) = $\frac{60}{30}=2$ units/minute

Let both pipes run together for $x$ minutes, after which Pipe 1 is turned off. Then, Pipe 2 runs alone for an additional 10 minutes to finish filling the cistern.

The total work done can be expressed as:

$$(E_1+E_2)\cdot x+E_2\cdot 10=60$$

Substitute the values of efficiencies into the equation:

$$(3+2)\cdot x+2\cdot 10=60$$

$$5x+20=60$$

$$5x=40$$

$$x=\frac{40}{5}=8\text{minutes}$$

Thus, the first pipe must be turned off after 8 minutes.