Three workers A, B and C can complete a job in 12, 15 and 20 days respectively, working alone. All three start working together. After some time, C leaves, and A and B complete the remaining work in 2 more days. If the total time taken to finish the work is 5.5 days, for how many days did C work?

ATQ, Let total work = LCM(12,15,20) = 60 units. Rates: A = 60/12 = 5 units/day B = 60/15 = 4 units/day C = 60/20 = 3 units/day Total time = 5.5 days. A and B work the whole time; C works for t days. After C leaves, A and B take 2 more days ⇒ C worked for t = 5.5 − 2 = 3.5 days. Work done in first 3.5 days by A + B + C = (5 + 4 + 3) × 3.5 = 12 × 3.5 = 42 units. Remaining work = 60 − 42 = 18 units, done by A + B in 2 days: (5 + 4) × 2 = 9 × 2 = 18 units (consistent). So C worked for 3.5 days.