A cistern has three pipes A, B and C.A and B can fill it in 6 and 5 hours respectively and C can empty in 2 hours. If these pipes are opened in order at one, two and three O'clock then when will be the cistern be empty?

Pipe rates: A fills 1/6 per hr, B fills 1/5 per hr, C empties 1/2 per hr. At 1 PM: A opens, at 2 PM: B opens, at 3 PM: C opens. Track cumulative: 1-2 PM: A fills 1/6. 2-3 PM: A+B fill 1/6+1/5 = 11/30. Total by 3 PM = 1/6 + 11/30 = 5/30+11/30 = 16/30 = 8/15. At 3 PM, all three open. Net rate = 1/6+1/5-1/2 = 5/30+6/30-15/30 = -4/30 = -2/15 (emptying). To empty 8/15 at rate 2/15: time = (8/15)/(2/15) = 4 hours. So empty at 3 PM + 4 hrs = 7 PM. Option D correct.