Workers A, B, and C can complete a piece of work alone in 10 days, 12 days, and 15 days respectively. All three start working together at the same time. A leaves after working for 2 days, and B leaves 3 days before the work is completed, so C finishes the remaining work alone. In how many total days is the entire work completed?

Introduction / Context:
This is a more involved time and work problem where three workers A, B, and C join and leave the job at different times. We know their individual times to complete the work and the sequence of their participation on the job. The objective is to find the total number of days required to complete the work, taking into account changing combinations of workers over different intervals.

Given Data / Assumptions:

• A alone can complete the work in 10 days.
• B alone can complete the work in 12 days.
• C alone can complete the work in 15 days.
• All three start working together from day 0.
• A leaves after 2 days of work.
• B continues but leaves 3 days before the work is completed.
• C then completes the remaining work alone.
• Total work is considered as 1 unit.

Concept / Approach:
We break the total time into three intervals, each with a different set of workers: first interval with A, B, and C; second interval with B and C; and third interval with C alone. We express the work done in each interval in terms of the total time T and the known rates of the workers. Setting the sum of the work over all intervals equal to 1 and solving for T yields the total time required to finish the job.

Step-by-Step Solution:
Step 1: Let total work be 1 unit and let the total time of completion be T days. Step 2: A's rate = 1/10 units per day, B's rate = 1/12 units per day, C's rate = 1/15 units per day. Step 3: For the first 2 days, all three (A, B, and C) work together. Work done in this period = 2 * (1/10 + 1/12 + 1/15). Step 4: After 2 days, A leaves. From day 2 to day (T - 3), only B and C work together. The length of this interval is (T - 3) - 2 = T - 5 days. Work done in this period = (T - 5) * (1/12 + 1/15). Step 5: For the last 3 days, only C works. Work done in this period = 3 * (1/15). Step 6: Now compute the rates. First, 1/10 + 1/12 + 1/15. With a common denominator of 60, we get 6/60 + 5/60 + 4/60 = 15/60 = 1/4 units per day. Step 7: Next, 1/12 + 1/15. The common denominator is 60 again, giving 5/60 + 4/60 = 9/60 = 3/20 units per day. Step 8: And 1/15 = 1/15 units per day, so C alone does 3 * (1/15) = 3/15 = 1/5 of the work in the last 3 days. Step 9: Work in first 2 days = 2 * (1/4) = 1/2. Step 10: Work in middle period = (T - 5) * (3/20). Step 11: Work in last 3 days = 1/5. Step 12: Total work done must be 1, so 1/2 + (T - 5) * (3/20) + 1/5 = 1. Step 13: Combine constant terms. 1/2 + 1/5 = (5/10 + 2/10) = 7/10. Step 14: The equation becomes 7/10 + (3/20)(T - 5) = 1. Step 15: Subtract 7/10 from both sides giving (3/20)(T - 5) = 1 - 7/10 = 3/10. Step 16: Multiply both sides by 20/3 to solve for T - 5: T - 5 = (3/10) * (20/3) = 2. Step 17: Therefore T = 2 + 5 = 7 days.

Verification / Alternative check:
Check the work distribution with T = 7 days. First 2 days with A, B, and C give 2 * (1/4) = 1/2 of the work. Next T - 5 = 2 days with B and C give 2 * (3/20) = 6/20 = 3/10 of the work. Last 3 days with C give 1/5 of the work. Summing all: 1/2 + 3/10 + 1/5 = 5/10 + 3/10 + 2/10 = 10/10 = 1, confirming the correctness of T = 7 days.

Why Other Options Are Wrong:

• 5 days and 6 days: These are too short and would not allow the workers to complete the full work at the calculated rates.
• 8 days: Longer than necessary and would result in more than 1 unit of work being completed.

Common Pitfalls:
A frequent mistake is mishandling the time intervals, such as forgetting that B leaves 3 days before completion and miscomputing the length of the middle interval. Another issue is not properly adding the work contributions and mixing up the workers' rates. Setting up the equation with clear intervals and checking the arithmetic step by step helps avoid these problems.

Final Answer:
The entire work is completed in a total of 7 days.