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Get access to the detailed solutions to the previous years questions asked in CMAT exam.
Let the capacity of the cistern be 600 units.
From the given data, the efficiencies of pipes A and B are 5 units/ min and 4 units/min respectively.
Let the efficiency of outlet pipe C be 'k' units/min.
Given the time taken to fill the cistern when all the three pipes are open = 100 minutes
⇒ Efficiency of pipes × time taken = Capacity of cistern
⇒ (5 + 4 − k) × 100 = 600
⇒ 9 − k = 6
⇒ k = 3
Therefore the time taken (t) by pipe C to empty the cistern
= Capacity of the cistern ÷ efficiency of pipe C
=> t = 600 ÷ 3 = 200 minutes = 3 hrs. 20 min.