Two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes.

Work done by P in 1 minute = $\dfrac{1}{24}$

Work done by Q in 1 minute = $\dfrac{1}{32}$

Let the first pipe be turned off after x minutes.

Then, P's x minutes work + Q's 16 minutes work = 1

$\Rightarrow \dfrac{1}{24} \times x + \dfrac{1}{32} \times 16 = 1\\[1em] \Rightarrow \dfrac{x}{24} + \dfrac{1}{2} = 1\\[1em] \Rightarrow \dfrac{x}{24} = 1 - \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{x}{24} = \dfrac{1}{2}\\[1em] \Rightarrow x = \dfrac{1}{2} \times 24\\[1em] \Rightarrow x = 12$

Hence, pipe P must be closed after 12 minutes.