A, B and C can complete a work in 10, 12 and 15 days respectively. They started the work together. But A left the work 5 days before its completion. B also left the work 2 days after A left. In how many days was the work completed? (C.P.O., 2007)

### Concept & Leaving Work Sequence Formulate an equation by expressing each person's actual working days relative to the total number of days taken to complete the job. $$ \frac{\text{Days by A}}{\text{A's Time}} + \frac{\text{Days by B}}{\text{B's Time}} + \frac{\text{Days by C}}{\text{C's Time}} = 1 $$ ### Step-by-Step Solution * **Given:** A = 10 days, B = 12 days, C = 15 days. Let total completion time be $T$ days. * C did not leave, so C worked for all $T$ days. * A left 5 days before completion, so A worked for $(T - 5)$ days. * B left 2 days after A left. This means B left $5 - 2 = 3$ days before completion. So B worked for $(T - 3)$ days. * Total work equation: $\frac{T - 5}{10} + \frac{T - 3}{12} + \frac{T}{15} = 1$ * The LCM of 10, 12, 15 is 60. Multiply the equation by 60: $6(T - 5) + 5(T - 3) + 4T = 60$ * Expand and solve: $6T - 30 + 5T - 15 + 4T = 60$ $15T - 45 = 60$ $15T = 105$ $T = 7$ * The total time taken is 7 days. ### Exam Strategy & Shortcut Use LCM method: Total work = LCM(10, 12, 15) = 60 units. Efficiencies: A=6, B=5, C=4. Assume A and B stayed until the end. A left 5 days early. If A stayed, extra work = $5 \times 6 = 30$ units. B left 3 days early (5 - 2). If B stayed, extra work = $3 \times 5 = 15$ units. New total work = $60 + 30 + 15 = 105$ units. Now, A, B, and C work together for the full $T$ days. Combined efficiency = $6 + 5 + 4 = 15$. Total time $T = 105 / 15 = 7$ days. ### Common Pitfall Miscalculating B's timeline. "B left 2 days after A left" means B left $5 - 2 = 3$ days before completion, not $T - 2$ days. ### Final Answer Therefore, the correct answer is **7**.