Correct Option: A Let's denote the efficiency of C as $E_C$. From the problem statement: 1. B is thrice as efficient as C. So, $E_B = 3 \times E_C$. 2. A is twice as efficient as B. So, $E_A = 2 \times E_B = 2 \times (3 \times E_C) = 6 \times E_C$. Let's assume $E_C = x$ units of work per day for simplicity. Then: $E_A = 6x$ units/day $E_B = 3x$ units/day $E_C = x$ units/day **Step 1: Calculate the total work done in the original scenario.** * **Phase 1: A, B, and C work together for 4 days.** Combined efficiency of A+B+C = $E_A + E_B + E_C = 6x + 3x + x = 10x$ units/day. Work done in 4 days = $10x \times 4 = 40x$ units. * **Phase 2: B and C work together for another 2 days (after A leaves).** Combined efficiency of B+C = $E_B + E_C = 3x + x = 4x$ units/day. Work done in 2 days = $4x \times 2 = 8x$ units. * **Phase 3: C completes the remaining work alone in 5 more days (after B leaves).** Work done by C in 5 days = $E_C \times 5 = x \times 5 = 5x$ units. Total Work = Work (Phase 1) + Work (Phase 2) + Work (Phase 3) Total Work = $40x + 8x + 5x = 53x$ units. Total time taken in the original scenario = $4 + 2 + 5 = 11$ days. **Step 2: Calculate the efficiency of the new worker D.** D's efficiency is 50% greater than B's efficiency. $E_D = E_B + 0.50 \times E_B = 1.5 \times E_B = 1.5 \times (3x) = 4.5x$ units/day. **Step 3: Analyze the hypothetical scenario (D joins C to complete the remaining work).** The initial parts of the work remain the same: * Work done in the first 4 days (A+B+C) = $40x$ units. * Work done in the next 2 days (B+C) = $8x$ units. Total work done before C has to complete the remaining part = $40x + 8x = 48x$ units. Remaining Work = Total Work - Work done so far = $53x - 48x = 5x$ units. In the original scenario, C completed this $5x$ units of work in 5 days ($5x / x = 5$ days). In the hypothetical scenario, D joins C to complete this remaining $5x$ units of work. Combined efficiency of C+D = $E_C + E_D = x + 4.5x = 5.5x$ units/day. Time taken by C+D to complete the remaining $5x$ units of work = $\frac{ ext{Remaining Work}}{ ext{Combined Efficiency of C+D}} = \frac{5x}{5.5x} = \frac{5}{5.5} = \frac{10}{11}$ days. **Step 4: Calculate the new total time for project completion.** New Total Time = Time (Phase 1) + Time (Phase 2) + Time (C+D for remaining work) New Total Time = $4 \text{ days} + 2 \text{ days} + \frac{10}{11} \text{ days} = 6 + \frac{10}{11} = \frac{6 \times 11 + 10}{11} = \frac{66 + 10}{11} = \frac{76}{11}$ days. **Step 5: Calculate how many days sooner the project would be completed.** Days sooner = Original Total Time - New Total Time Days sooner = $11 - \frac{76}{11} = \frac{11 \times 11 - 76}{11} = \frac{121 - 76}{11} = \frac{45}{11}$ days. The final answer is $\frac{45}{11}$ days. The correct option is A.