1. Stating the problem: A pump fills a tank in 2 hours, but due to a leak, it takes 2.5 hours to fill the tank. We need to find how long the leak alone would take to empty the full tank. 2. Let the pump's filling rate be $\frac{1}{2}$ tank/hour (since it fills 1 tank in 2 hours). 3. When the leak is present, it takes 2.5 hours to fill the tank, so the combined filling rate is $\frac{1}{2.5} = \frac{2}{5}$ tank/hour. 4. Let the leak's emptying rate be $x$ tank/hour. 5. The combined rate is the pump's rate minus the leak's rate: $$\frac{1}{2} - x = \frac{2}{5}$$ 6. Solve for $x$: $$x = \frac{1}{2} - \frac{2}{5} = \frac{5}{10} - \frac{4}{10} = \frac{1}{10}$$ 7. So, the leak empties $\frac{1}{10}$ of the tank per hour. 8. Therefore, the leak alone would empty the full tank in: $$\frac{1}{x} = \frac{1}{\frac{1}{10}} = 10$$ hours. Final answer: The leak can empty the tank in 10 hours.