CMAT 2018 - Slot 1 Quantitative Aptitude Question 17
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Quantitative Aptitude
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.