Three taps A,B and C are used to fill a cistern. Tap A alone can fill the cistern in 9 minutes. Tap B can fill in 6 minutes and Tap C can fill in 3 minutes. How many minutes will it take to fill this cistern if all the three taps are used simultaneously?

Let the time taken to fill the cistern by 3 taps A, B and C be X, Y, and Z minutes respectively.
Then the short cut formula for,
Time taken to fill the tank when all the pipes are opened = XYZ/(XY + YZ + ZX) minutes
Here, X = 9 minutes, Y = 6 minutes and Z = 3 minutes.
Now the required time = {(9) × (6) × (3)}/{(9 × 6) + (6 × 3) + (3 × 9)} minutes
= (9 × 6 × 3)/(54 + 18 + 27)
= (9 × 6 × 3)/9{6 + 2 + 3}
= 6 × 3/(6 + 2 + 3)
= 18/11
= 1(7/11) minutes
Hence the answer is 1(7/11) minutes.