A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how…

Concept: In work-rate problems, each person's one-day work equals 1 divided by the number of days they need alone, and the whole task equals 1 unit of work. When a helper joins in only on specific days of a repeating cycle, find the work finished in one full cycle by adding up each day's contribution, then find how many such cycles are needed to complete the whole unit of work.

1. A's one-day work = 1/20; B's one-day work = 1/30; C's one-day work = 1/60.

2. A works on every day of the schedule; B and C join him only on the third day of each 3-day block, as the question states.

3. Work finished in one 3-day cycle = 3 x (A's one-day work) + B's one-day work + C's one-day work = 3(1/20) + 1/30 + 1/60.

4. Using a common denominator of 60: 3(1/20) = 9/60, 1/30 = 2/60, 1/60 = 1/60, so the cycle's work = (9 + 2 + 1)/60 = 12/60 = 1/5.

5. Since 1/5 of the work finishes every 3 days, the full unit of work (5 x 1/5) needs 5 such cycles, so the total time = 5 x 3 = 15 days.

Cross-check: 5 complete cycles at 1/5 of the work each add up to exactly 1 full unit of work (5 x 1/5 = 1), confirming the count is consistent; B and C's help lands on days 3, 6, 9, 12 and 15 - five times in total - matching the five cycles used above.