Three taps A, B and C can fill a tank in 27, 36 and 54 minutes respectively. If all the three taps are opened, then how much time (in minutes) it will take to completely fill the tank?

Let total capacity of tank is L.C.M. (27,36,54) = 108 units

A alone can fill in 27 minutes, => A's efficiency = $$\frac{108}{27}=4$$ units/min

Similarly B's efficiency = $$\frac{108}{36}=3$$ units/min

and C's efficiency = $$\frac{108}{54}=2$$ units/min

Now, (A+B+C)'s 1 minute's work = $$4+3+2=9$$ units/min

$$\therefore$$ Time required by A,B and C together to fill the tank = $$\frac{108}{9}=12$$ minutes

=> Ans - (B)