A can complete a work in 10 days, B in 12 days and C in 15 days. All of them began the work together, but A had to leave the work after 2 days of the start and B 3 days before the completion of the work. How long did the work last?

Total work = LCM(10,12,15) = 60 units. Rates: A = 6, B = 5, C = 4 units/day. All work together for 2 days: 15 × 2 = 30 units done, 30 left. B leaves 3 days before completion, so B works 4 days total (2 + 4 = 6 total work duration): B does 5 × 4 = 20 units. Remaining 40 units done by A and C at 10 units/day = 4 days. Total = 2 + 4 = 6 days... Wait, let me recalculate: Days 1-2: all three work (30 done). Days 3-4: B and C work (20 done, 50 total). Days 5-7: only C works (12 done, 62 total - exceeds 60). This doesn't work. Correct approach: Let total days = d. A works 2 days, B works (d-3) days, C works d days. 6×2 + 5×(d-3) + 4×d = 60. 12 + 5d - 15 + 4d = 60. 9d - 3 = 60. 9d = 63. d = 7.