1. **Problem statement:** The water supply in a hotel is enough for all guests for 8 days. We need to find how many days the water supply will last if only 40% of the guests are present. 2. **Understanding the problem:** The amount of water is fixed. If fewer guests use the water, it will last longer. The number of days the water lasts is inversely proportional to the number of guests. 3. **Formula:** Let $D$ be the number of days the water lasts for 40% of the guests. Since the water lasts 8 days for 100% guests, we use the inverse proportionality: $$D \times \text{percentage of guests} = 8 \times 100\%$$ 4. **Substitute the values:** $$D \times 40\% = 8 \times 100\%$$ 5. **Convert percentages to decimals:** $$D \times 0.4 = 8 \times 1$$ 6. **Solve for $D$:** $$D = \frac{8 \times 1}{0.4}$$ 7. **Simplify the fraction:** $$D = \frac{8}{0.4} = \frac{8}{\cancel{0.4}} \times \frac{\cancel{10}}{10} = \frac{80}{4} = 20$$ 8. **Answer:** The water supply will last for **20 days** if only 40% of the guests are present. **Final answer: 20 days (Option B)**