Question
Download Solution PDFPipe A can fill a tank in 12 hours. Pipe B can fill the same tank in 16 hours. Pipe C can empty the full tank in 24 hours. Pipe A, B and C are opened alternatively for 1 hour each. If B is opened first, then how many hours will they take to fill the empty tank?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Time by Pipe A = 12 hours
Time by Pipe B = 16 hours
Time by Pipe C = 24 hours (emptying)
Order: B, A, C (1 hour each)
Formula Used:
Total Work = LCM of filling times (ignoring emptying for total work)
Efficiency = Total Work / Time
Calculation:
Total Work = LCM(12, 16, 24) = 48 units
Efficiency of Pipe A = 48 / 12 = 4 units/hour
Efficiency of Pipe B = 48 / 16 = 3 units/hour
Efficiency of Pipe C = 48 / 24 = -2 units/hour (negative as it empties)
Work done in the first 3 hours (B, A, C):
Hour 1 (Pipe B): +3 units
Hour 2 (Pipe A): +4 units
Hour 3 (Pipe C): -2 units
Net work done in 3 hours = 3 + 4 - 2 = 5 units
Number of 3-hour cycles to fill most of the tank:
Number of cycles ≈ Total Work / Net work per 3 hours = 48 / 5 = 9.6 cycles
Work done in 9 cycles (27 hours) = 9 × 5 = 45 units
Remaining work = Total Work - Work done in 27 hours = 48 - 45 = 3 units
After 27 hours, the 28th hour is Pipe B's turn.
Pipe B's efficiency is 3 units/hour.
Work done by Pipe B in the 28th hour = 3 units.
Remaining work after 28 hours = 3 - 3 = 0 units.
The tank is full at the end of the 28th hour.
Total time taken = 28 hours.
They will take 28 hours to fill the empty tank.
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