1. **State the problem:** A worker completes tasks 25% faster but must redo 10% of the work due to errors. We want to find the overall efficiency compared to before. 2. **Define variables:** Let the original time to complete a task be $T$. 3. **Calculate new time per task without rework:** Since the worker is 25% faster, the new time is $T_{new} = T \times (1 - 0.25) = 0.75T$. 4. **Account for rework:** The worker must redo 10% of the work, so the total work done is $1 + 0.10 = 1.10$ times the original work. 5. **Calculate total new time including rework:** Total time is $T_{total} = 0.75T \times 1.10 = 0.825T$. 6. **Calculate overall efficiency:** Efficiency is inversely proportional to time, so $$\text{Efficiency} = \frac{T}{T_{total}} = \frac{T}{0.825T} = \frac{1}{0.825} \approx 1.2121.$$ 7. **Interpretation:** The worker is approximately 21.21% more efficient overall compared to before. **Final answer:** Overall efficiency is about 121.21% of the original efficiency.