‘P’, ‘Q’, and ‘R’ can complete a task in 10 days, 15 days, and 20 days, respectively. ‘P’ starts the task alone and is joined by ‘Q’ on the second day. On the third day, ‘Q’ is replaced by ‘R’, and again on the fourth day, ‘P’ works alone. The process repeats in the same manner. Find the total time taken to complete the task. (Round off to two decimal places)

Let the total work be the LCM of [10, 15, 20] = 60 units. Efficiency of ‘P’ = 60/10 = 6 units/day. Efficiency of ‘Q’ = 60/15 = 4 units/day. Efficiency of ‘R’ = 60/20 = 3 units/day. Work done on the first day = 6 × 1 = 6 units. Work done on the second day = (6 + 4) × 1 = 10 units. Work done on the third day = (6 + 3) × 1 = 9 units. Total work done in the first three days = 6 + 10 + 9 = 25 units. Remaining work = 60 – 25 = 35 units. Time taken to complete the remaining work = (6/6) + (10/10) + (9/9) + (10/17) ≈ 3 + (10/17) ≈ 3.59 days. Total time = 3 + 3.59 = 6.59 days.