Three pipe A, B and C can empty the tank in 4, 6 and 12 hours respectively. If each pipe is open alone alternatively after every 2 hours starting with A followed by B and C. Find in how many hours the tank will be completely empty if it is already 20% empty?

Tank is 20% empty, so 80% full = 0.8 units. Pipe A: 1/4 per hr, B: 1/6 per hr, C: 1/12 per hr. Cycle of 6 hrs: A (2 hrs) empties 0.5, B (2 hrs) empties 1/3, C (2 hrs) empties 1/6. Total per 6 hrs = 5/6. Two full cycles (12 hrs) empty 10/6 = 1.67 units, but we only need 0.8. Working backwards: after A+B (4 hrs), emptied 5/6, leaving 0.8 - 5/6 = -0.033. Actual: A empties 0.5 (rem: 0.3), B empties 1/3 in 2 hrs (done). Total = 4 hrs. Wait: A works 2 hrs (0.5 gone), B works 2 hrs (0.33 gone), total 0.83 > 0.8. So B stops early. B empties 0.3 in 0.3 × 6 = 1.8 hrs. Total = 2 + 1.8 = 3.8 hrs = 3 hr 48 min.