Two inlet pipes A and B can fill a cistern in 20 hours and 15 hours, respectively. A third pipe C, acting as an outlet, can empty the completely filled cistern in 30 hours. All three pipes are opened simultaneously in the empty cistern. After 4 hours, pipe B is closed. After a further 2 hours, pipe A is also closed, while pipe C continues to operate by itself until the cistern is emptied. Find the duration for which pipe C worked alone.
11 hours
Take the capacity as LCM(20, 15, 30) = 60 units. Then A fills 3 units/hr, B fills 4 units/hr and C empties 2 units/hr.
First 4 hours (A + B + C): net 3 + 4 – 2 = 5 units/hr, so 5 × 4 = 20 units are filled.
Next 2 hours (A + C): net 3 – 2 = 1 unit/hr, so 2 more units, making 22 units in the cistern.
C alone empties 22 units at 2 units/hr in 22 ÷ 2 = 11 hours.
Check: water in = A 3 × 6 = 18 + B 4 × 4 = 16 = 34 units; water out = C 2 × (6 + 11) = 34 units, so the cistern is exactly empty at the end.
Hence, 11 hours.
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