The correct answer is **A) $33.16\%$**. Let's break down the problem step-by-step to determine the total work and the work done by A. **Step 1: Define the individual efficiencies.** Let the efficiency of B be $E_B$ units/day. Given that A is 25% more efficient than B: $E_A = E_B + 0.25 E_B = 1.25 E_B$ Given that C is 10% less efficient than B: $E_C = E_B - 0.10 E_B = 0.90 E_B$ To simplify calculations, let's assume a common base for efficiencies. A convenient way to avoid decimals is to express these ratios as integers. If $E_B = 20$ units/day (a multiple of 4 and 10): $E_A = 1.25 \times 20 = 25$ units/day $E_C = 0.90 \times 20 = 18$ units/day So, the efficiencies are: $E_A = 25$ units/day $E_B = 20$ units/day $E_C = 18$ units/day **Step 2: Calculate work done in Phase 1.** In the first phase, A, B, and C work together for 4 days. Combined efficiency of A, B, and C = $E_A + E_B + E_C = 25 + 20 + 18 = 63$ units/day. Work done in Phase 1 = $63 \text{ units/day} \times 4 \text{ days} = 252$ units. Work done by A in Phase 1 = $E_A \times 4 \text{ days} = 25 \text{ units/day} \times 4 \text{ days} = 100$ units. **Step 3: Calculate work done in Phase 2.** In the second phase, A leaves, and B and C work for 6 more days. Combined efficiency of B and C = $E_B + E_C = 20 + 18 = 38$ units/day. Work done in Phase 2 = $38 \text{ units/day} \times 6 \text{ days} = 228$ units. Work done by A in Phase 2 = $0$ units (since A was not working). **Step 4: Calculate A's new efficiency and duration of Phase 3.** After Phase 2, B leaves, and A rejoins with his efficiency reduced by 20%. Initial efficiency of A = $E_A = 25$ units/day. Reduced efficiency of A ($E'_A$) = $25 \times (1 - 0.20) = 25 \times 0.80 = 20$ units/day. The total project duration is 18 days. Duration of Phase 1 = 4 days. Duration of Phase 2 = 6 days. Duration of Phase 3 = Total duration - Duration of Phase 1 - Duration of Phase 2 Duration of Phase 3 = $18 - 4 - 6 = 8$ days. **Step 5: Calculate work done in Phase 3.** In the third phase, A (with reduced efficiency) and C work for 8 days. Combined efficiency of $A'$ and C = $E'_A + E_C = 20 + 18 = 38$ units/day. Work done in Phase 3 = $38 \text{ units/day} \times 8 \text{ days} = 304$ units. Work done by A in Phase 3 = $E'_A \times 8 \text{ days} = 20 \text{ units/day} \times 8 \text{ days} = 160$ units. **Step 6: Calculate the total work and total work done by A.** Total work = Work (Phase 1) + Work (Phase 2) + Work (Phase 3) Total work = $252 + 228 + 304 = 784$ units. Total work done by A = Work by A (Phase 1) + Work by A (Phase 2) + Work by A (Phase 3) Total work done by A = $100 + 0 + 160 = 260$ units. **Step 7: Calculate the percentage of total work performed by A.** Percentage of work done by A = $\frac{\text{Total work done by A}}{\text{Total work}} \times 100$ Percentage of work done by A = $\frac{260}{784} \times 100$ Simplify the fraction: $\frac{260}{784} = \frac{130}{392} = \frac{65}{196}$ Percentage = $\frac{65}{196} \times 100 = \frac{6500}{196}$ Now perform the division: $6500 \div 196 \approx 33.1632...$ Rounding to two decimal places, the percentage of work done by A is approximately $33.16\%$. Therefore, the correct option is A.