Part of the cistern filled by pipe P in 1 min = $\dfrac{1}{12}$
Part of the cistern filled by pipe Q in 1 min = $\dfrac{1}{16}$
Suppose Q should be closed after x minutes
i.e., Pipe P and Q will be open for initial x minutes then P will be open for another (9-x) min
x$\left(\dfrac{1}{12}+\dfrac{1}{16}\right)+(9-x)\dfrac{1}{12}$=1
$\dfrac{7x}{48}+\dfrac{9}{12}-\dfrac{x}{12}$=1
7x+36−4x=48
3x=12
x=4