Workers X, Y and Z can complete a certain piece of work in 30, 40 and 50 days respectively. All three start together, but Y leaves 4 days before the work is finished. In how many days in total is the work completed?

Introduction / Context:
This time and work problem combines different individual work rates and a change in team composition shortly before completion. It requires setting up an equation involving the number of days worked at different combined rates and then solving for the total time taken to finish the work.

Given Data / Assumptions:

• X alone can complete the work in 30 days.
• Y alone can complete the work in 40 days.
• Z alone can complete the work in 50 days.
• All three start working together.
• Y leaves 4 days before the work is completed; during the last 4 days only X and Z continue.
• We assume constant work rates and no idle time.

Concept / Approach:
The approach is to express the total work as 1 unit and write the total work done as the sum of work done in two phases: an initial phase where all three work together, and a final phase where only two workers continue. If T is the total time, then during (T minus 4) days all three work together, and during the last 4 days only X and Z work. We use the sum of the two contributions to equal 1 and solve for T.

Step-by-Step Solution:
Step 1: Let total work = 1 unit.Step 2: Rate of X = 1 / 30 work per day.Step 3: Rate of Y = 1 / 40 work per day.Step 4: Rate of Z = 1 / 50 work per day.Step 5: Rate of X + Y + Z together = 1 / 30 + 1 / 40 + 1 / 50.Step 6: Compute this sum with denominator 600: 1 / 30 = 20 / 600, 1 / 40 = 15 / 600, 1 / 50 = 12 / 600, so combined rate = (20 + 15 + 12) / 600 = 47 / 600.Step 7: Rate of X + Z together = 1 / 30 + 1 / 50 = 20 / 600 + 12 / 600 = 32 / 600 = 4 / 75.Step 8: Let total time taken be T days. For (T - 4) days, all three work; for last 4 days, only X and Z work.Step 9: Total work equation: (T - 4) * (47 / 600) + 4 * (4 / 75) = 1.Step 10: Note that 4 * (4 / 75) = 16 / 75.Step 11: Rearrange: (T - 4) * (47 / 600) = 1 - 16 / 75.Step 12: Compute 1 - 16 / 75 = (75 / 75 - 16 / 75) = 59 / 75.Step 13: So (T - 4) = (59 / 75) * (600 / 47) = (59 * 8) / 47 = 472 / 47.Step 14: Therefore T = 472 / 47 + 4 = (472 + 188) / 47 = 660 / 47 days.

Verification / Alternative check:
We can verify quickly by computing approximate value. 660 / 47 is about 14.04 days. In this period, X works the entire time, Y works for about 10.04 days, and Z also works the entire time. Approximating total work done using these durations and rates will give close to 1 unit, which is consistent and confirms that the exact algebraic solution 660 / 47 days is correct.

Why Other Options Are Wrong:

• 640/47 days: This choice slightly underestimates the time, so total work done would be less than 1 unit.
• 680/47 days: This gives a slightly larger time, leading to total work more than 1 unit.
• 665/47 days: This is a close but incorrect fraction and does not satisfy the work equation when substituted.

Common Pitfalls:
Students may mistakenly assume that all three work for the same duration or forget that Y leaves exactly 4 days before completion. Another common mistake is miscalculating combined rates or mishandling fractions, especially when denominators like 30, 40 and 50 are involved. Careful use of a common denominator such as 600 helps avoid these issues.

Final Answer:
The work is completed in 660/47 days.