Question
Pipes A, B, and C can fill a tank in 10 hours, 5 hours, and 20 hours respectively. If the efficiencies of pipes A and B are each reduced by 50% of their original efficiency and all three pipes are opened together, find the total time taken by them to fill the tank.
Solution
Let the capacity of the tank (LCM of 10, 5, and 20) be 20 liters. Efficiency of pipe A = 20/10 = 2 liters/hour Efficiency of pipe B = 20/5 = 4 liters/hour Efficiency of pipe C = 20/20 = 1 liter/hour New efficiency of pipe A = 2 × (50/100) = 1 liter/hour New efficiency of pipe B = 4 × (50/100) = 2 liters/hour Total efficiency = 1 + 2 + 1 = 4 liters/hour Required time = 20 / 4 = 5 hours
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