Three taps can fill a cistern in 18 min, 15 min and 10 min respectively. The cistern being empty, all the three taps are kept open. After 3 min the first tap is closed. Counting time from that moment the cistern will be full in

Let total capacity of cistern is L.C.M. (18,15,10) = 90 units

Let the three taps be A, B and C respectively.

A alone can fill in 18 minutes, => A's efficiency = $$\frac{90}{18}=5$$ units/min

Similarly B's efficiency = $$\frac{90}{15}=6$$ units/min

and C's efficiency = $$\frac{90}{10}=9$$ units/min

Now, (A+B+C)'s 3 minute's work = $$(5+6+9)\times3=20\times3=60$$ units

Capacity of cistern left to be filled = $$90-60=30$$ units

$$\therefore$$ Now remaining time taken by taps B and C to fill the cistern = $$\frac{30}{(6+9)}=2$$ minutes

=> Ans - (D)