K, L and M can complete a piece of work in 20 days, 30 days and 60 days respectively when working alone. If K works every day and is assisted by L and M only on every third day, in how many days will the entire work be completed?

Introduction / Context:
This is a Time and Work question involving a repeating pattern of work over days. Three workers, K, L and M, have different individual times to complete a job. K works every day, but L and M assist him only on every third day. We must compute the effective work done in each 3-day cycle and then determine how many such cycles are needed to complete the entire job. This type of problem tests understanding of periodic work patterns and combined work rates.

Given Data / Assumptions:

• K alone can finish the work in 20 days.
• L alone can finish the work in 30 days.
• M alone can finish the work in 60 days.
• K works every day.
• On every third day, L and M also work with K; on the first and second day of each cycle, only K works.
• Total work is considered as 1 unit.
• All three work at constant rates.

Concept / Approach:
We first compute the individual daily work rates by taking reciprocals of the times. Then we analyze a 3-day cycle: on day 1 and day 2, only K works; on day 3, K, L and M work together. We compute the total work done in one full 3-day cycle. Knowing how much of the work is completed per cycle, we find how many such cycles are required to reach 1 full unit of work. If the work is completed exactly in an integer number of cycles, the total days is the number of cycles multiplied by 3.

Step-by-Step Solution:
Let total work W = 1 unit. K's daily rate = 1 / 20 units per day. L's daily rate = 1 / 30 units per day. M's daily rate = 1 / 60 units per day. On day 1: only K works ⇒ work done = 1 / 20. On day 2: only K works ⇒ work done = 1 / 20. On day 3: K, L and M all work ⇒ work done = 1 / 20 + 1 / 30 + 1 / 60. Compute: 1 / 20 + 1 / 30 + 1 / 60 = (3 + 2 + 1) / 60 = 6 / 60 = 1 / 10. Total work in one 3-day cycle = 1 / 20 + 1 / 20 + 1 / 10. = 1 / 20 + 1 / 20 + 2 / 20 = 4 / 20 = 1 / 5. So, every 3 days, they complete 1 / 5 of the job. To complete the full work (1 unit), number of such cycles required = 1 / (1 / 5) = 5 cycles. Total days taken = 5 cycles * 3 days per cycle = 15 days.

Verification / Alternative check:
After 5 cycles, total work done = 5 * (1 / 5) = 1 unit, so the entire job is completed exactly at the end of a cycle. There is no partial cycle needed. It is also reasonable that the total days (15) lies between K's individual time (20 days) and the combined time of all working every day, which would be faster than 15 days. Since L and M only assist on every third day, the overall time is longer than if all three had worked daily, but shorter than K alone; 15 days fits that expectation.

Why Other Options Are Wrong:
Option B (13 days), Option C (12 days) and Option D (10 days) all imply faster completion, which would require more daily work than the computed pattern of 1 / 5 per 3 days allows. If we divide 1 unit of work by the effective average rate over the 3-day pattern, we obtain exactly 15 days, not any of the smaller values. Thus, only 15 days satisfies the periodic working pattern precisely.

Common Pitfalls:
One common error is to assume that K, L and M work together every day and then simply use their combined rate, ignoring the given condition that L and M only help on every third day. Another mistake is to average the individual times (20, 30, 60) directly instead of computing work per cycle. The correct approach is to focus on the repeating pattern and calculate work done in one complete cycle before scaling up to the total work.

Final Answer:
The entire work will be completed in 15 days when K works every day and is assisted by L and M on every third day.