1. **Problem statement:** A pump can fill a tank in 2 hours. Due to a leak, it takes 2.5 hours to fill the tank. We need to find how many hours the leak alone would take to empty the full tank. 2. **Define variables:** Let the rate of the pump be $P$ tanks per hour. Let the rate of the leak be $L$ tanks per hour. 3. **Pump’s rate:** The pump fills 1 tank in 2 hours, so $$P = \frac{1}{2} \text{ tank/hour}$$ 4. **Combined rate (pump minus leak):** Due to the leak, filling takes 2.5 hours, so $$P - L = \frac{1}{2.5} = \frac{2}{5} \text{ tank/hour}$$ 5. **Calculate leak’s rate:** Substitute pump rate $P = \frac{1}{2}$: $$\frac{1}{2} - L = \frac{2}{5}$$ 6. **Solve for $L$:** $$L = \frac{1}{2} - \frac{2}{5} = \frac{5}{10} - \frac{4}{10} = \frac{1}{10} \text{ tank/hour}$$ 7. **Leak emptying time:** The leak empties $\frac{1}{10}$ tank per hour, so time to empty one full tank is $$\text{Time} = \frac{1}{L} = \frac{1}{\frac{1}{10}} = 10 \text{ hours}$$ **Final Answer:** The leak can empty the full tank in 10 hours.