The correct answer is **B: 7 days**. To solve this problem, we first determine the individual rates of work (efficiency) for A, B, and C. We use the Least Common Multiple (LCM) method to represent the total work units. Given: Time taken by A to complete the work = 10 days Time taken by B to complete the work = 12 days Time taken by C to complete the work = 15 days Calculate the Total Work Units: The LCM of 10, 12, and 15 is 60. So, let the total work be 60 units. Calculate Individual Daily Work Rates (Efficiency): A's 1-day work = $\frac{60 \text{ units}}{10 \text{ days}} = 6$ units/day B's 1-day work = $\frac{60 \text{ units}}{12 \text{ days}} = 5$ units/day C's 1-day work = $\frac{60 \text{ units}}{15 \text{ days}} = 4$ units/day Let $T$ be the total number of days required to complete the entire work. We can divide the work into three distinct phases based on when workers join or leave: **Phase 1: A, B, and C work together for the first 2 days.** * Combined 1-day work of A, B, and C = $6 + 5 + 4 = 15$ units/day. * Work done in the first 2 days = $15 \text{ units/day} \times 2 \text{ days} = 30$ units. **Phase 2: A leaves. B and C work together for a certain period.** * A leaves after 2 days. B leaves 3 days before the total completion of work. This means B and C work together starting from day 3 until $T-3$ days. * The duration of this phase is $(T-3) - 2 = T-5$ days. * Combined 1-day work of B and C = $5 + 4 = 9$ units/day. * Work done by B and C in this phase = $9 \text{ units/day} \times (T-5) \text{ days} = 9(T-5)$ units. (For this phase to exist, $T-5$ must be $\ge 0$, implying $T \ge 5$.) **Phase 3: B leaves. C works alone for the last 3 days.** * B leaves 3 days before the completion of work. Therefore, C works alone for the final 3 days. * C's 1-day work = 4 units/day. * Work done by C in the last 3 days = $4 \text{ units/day} \times 3 \text{ days} = 12$ units. **Set up the equation for Total Work:** The sum of work done in all three phases must equal the total work (60 units). Work done (Phase 1) + Work done (Phase 2) + Work done (Phase 3) = Total Work $30 + 9(T-5) + 12 = 60$ Combine the constant terms: $42 + 9(T-5) = 60$ Subtract 42 from both sides: $9(T-5) = 60 - 42$ $9(T-5) = 18$ Divide both sides by 9: $T-5 = \frac{18}{9}$ $T-5 = 2$ Add 5 to both sides to find $T$: $T = 2 + 5$ $T = 7$ days Thus, the total duration for which the work lasted is 7 days. **Verification:** * **Days 1-2 (A, B, C):** Work done = $(6+5+4) \times 2 = 15 \times 2 = 30$ units. * **Days 3-4 (B, C):** Duration $T-5 = 7-5 = 2$ days. Work done = $(5+4) \times 2 = 9 \times 2 = 18$ units. * **Days 5-7 (C alone):** Last 3 days. Work done = $4 \times 3 = 12$ units. * **Total Work =** $30 + 18 + 12 = 60$ units. This matches the assumed total work units. The calculation is consistent, and the total number of days for the work to last is 7 days.