Pipes A and B can fill a tank in one hour and two hours respectively while pipe C can empty the filled up tank in one hour and fifteen minutes. A and C are turned on together at 9 a.m. After 2 hours, only A is closed and B is turned on. When will the tank be emptied?

As per the question,

A can fill the tank in 1 hour

B can fill the tank in 2 hour

C can empty the tank in 1 hour 15 min or $$\dfrac{5}{4}$$ hour

If A and C opened together, then tank will be filled in 1 hour $$1-\dfrac{1}{\dfrac{5}{4}}=1-\dfrac{4}{5}=\dfrac{1}{5}$$

As per the question, both tap is opened for 2 hours, so the tank will be filled in 2 hour $$=\dfrac{2}{5}$$

Now, tap B and C are opened and A is closed, so the tap will empty in 1 hour $$\dfrac{4}{5}-\dfrac{1}{2}=\dfrac{3}{10}$$

So tank will be completely empty in $$=\dfrac{10}{3}$$

Hence, $$\dfrac{2}{5}$$ will empty in $$\dfrac{2}{5}\times\dfrac{10}{3}=\dfrac{4}{3}$$

Hence tank will completely empty at $$9:00+ 2 hour +1 hour 20min=12:20PM$$