Two pipes A and B can fill a tank in 18 minutes and 24 minutes respectively. If both the pipes are opened simultaneously, then after how much time should pipe B be closed so that the tank is full in 12 minutes?

tank filled by A in one min=$$\dfrac{1}{18}$$

tank filled by B in one min=$$\dfrac{1}{24}$$

let they are opened together for t minutes.

because total time is 12min

so time taken by A alone when B is closed =(12-t)min

tank filled by A and B in t min=$$\dfrac{t}{18}+\dfrac{t}{24}$$

tank filled by A and B in t min + by A in (12-t) minutes =full tank

$$\dfrac{t}{18}+\dfrac{t}{24} + \dfrac{12-t}{18}=1$$

$$\dfrac{t}{18}+\dfrac{t}{24} + \dfrac{12}{18}-\dfrac{t}{18}=1$$

$$\dfrac{t}{24} = 1 -\dfrac{12}{18}$$

$$t(\dfrac{t}{24}+= \dfrac{6}{18}$$

$$\dfrac{t}{24} = \dfrac{1}{3}$$

$$t = \dfrac{1}{3}\times24$$

$$t=8$$min