Description for Correct answer:
Part filled by A in 1 min = \( \Large \frac{1}{24} \)

Part filled by B in 1 min = \( \Large \frac{1}{32} \)

Let B is closed after x min. Then, [Part filled by (A + B) in x min] + [Part filled by A in (9-x) min] = 1

Therefore,

\( \Large x \left(\frac{1}{24}+\frac{1}{32}\right)+ \left(9-x\right) \times \frac{1}{24}=1 \)

=> \( \Large x \left(\frac{4+3}{96}\right)+\frac{ \left(9-x\right) }{24}=1 \)

=> \( \Large \frac{7x}{96}+\frac{ \left(9-x\right) }{24}=1 \)

=> \( \Large \frac{7x+4 \left(9-x\right) }{96}=1 \)

=> \( \Large 7x+4 \left(9-x\right)=96 \)

=> 7x+36-4x=96

=> 7x - 4x = 96 - 36

= 3x = 60 => x = \( \Large \frac{60}{3} \) = 20

Hence, B must be closed after 20 min.