Workers M, N and O can complete a piece of work alone in 18 days, 36 days and 54 days respectively. M works alone for 8 days, then N and O join him and they all work together for some more days. M leaves exactly one day before the work is finished, while N and O continue. For how many days do M, N and O work together?

Introduction / Context:
This time and work question involves three workers with different individual completion times. The work is performed in stages, with some workers joining later and one worker leaving just before completion. The objective is to determine how long all three workers, M, N, and O, work together simultaneously before M leaves.

Given Data / Assumptions:

• M alone can complete the work in 18 days.
• N alone can complete the work in 36 days.
• O alone can complete the work in 54 days.
• M works alone for the first 8 days.
• Then M, N and O work together for some days.
• M leaves one day before completion; N and O continue that last day.
• All rates are constant and the work is of fixed size.

Concept / Approach:
We convert each time into a daily work rate. Then we break the process into three phases. Phase one is M alone for 8 days. Phase two is M, N and O together for an unknown number of days t. Phase three is N and O together for the final 1 day. We express the total work as the sum of work done in these phases and set it equal to 1. Solving this equation gives t, the number of days all three work together.

Step-by-Step Solution:
Let total work = 1 unit.Rates: M = 1 / 18 per day, N = 1 / 36 per day, O = 1 / 54 per day.Phase 1: M alone for 8 days, work done = 8 * (1 / 18) = 8 / 18 = 4 / 9.Phase 3: last day, only N and O work, rate = 1 / 36 + 1 / 54.Compute N + O rate: 1 / 36 = 3 / 108, 1 / 54 = 2 / 108, total = 5 / 108.Work in last day = 5 / 108.Let t be the number of days in Phase 2 when all three work together.M + N + O combined rate = 1 / 18 + 1 / 36 + 1 / 54.Compute: 1 / 18 = 6 / 108, 1 / 36 = 3 / 108, 1 / 54 = 2 / 108, so total = 11 / 108.Total work equation: (4 / 9) + t * (11 / 108) + (5 / 108) = 1.Convert 4 / 9 to denominator 108: 4 / 9 = 48 / 108.So 48 / 108 + 5 / 108 = 53 / 108; remaining work = 1 - 53 / 108 = 55 / 108.Thus t * (11 / 108) = 55 / 108, so t = 55 / 108 * 108 / 11 = 5.

Verification / Alternative check:
We can check by explicitly summing all work. M alone in 8 days does 4 / 9. All three in 5 days do 5 * 11 / 108 = 55 / 108. N and O in the last day do 5 / 108. Total work is 4 / 9 + 55 / 108 + 5 / 108 = 48 / 108 + 55 / 108 + 5 / 108 = 108 / 108 = 1, which confirms that all work is completed and that the value t = 5 is consistent.

Why Other Options Are Wrong:
If t were 3 days or 4 days, the combined work of all phases would be less than one complete job and the work would remain unfinished. If t were 6 days, the total work done would exceed one and would imply that some of the work was undone, which is not possible under constant positive rates. Only 5 days gives an exact match of total work to one job.

Common Pitfalls:
One common mistake is to forget that M does not work during the last day and to treat it as if all three work throughout. Another error is in converting fractions to a common denominator when combining rates or total work. Careful algebra and structured phase wise thinking are necessary to correctly handle multi stage work problems like this.

Final Answer:
M, N and O work together for 5 days before M leaves.