A pump can fill a tank with water in 2 hours. Because of a leak, it took $2 \frac{1}{3}$ hours to fill the tank. The leak can drain all the water of the tank in

### Concept & Net Work Rate The net rate at which a tank is filled when there is a leak is the difference between the filling rate of the pump and the emptying rate of the leak. $$ \text{Net Work Rate} = \text{Rate of Pump} - \text{Rate of Leak} $$ ### Step-by-Step Solution 1. **Determine Individual Rates:** * Time taken by the pump alone = 2 hours. * Work done by the pump in 1 hour = $\frac{1}{2}$ of the tank. 2. **Determine Combined Rate:** * Time taken by the pump and leak together = $2 \frac{1}{3} = \frac{7}{3}$ hours. * Work done by the (pump + leak) in 1 hour = $\frac{1}{7/3} = \frac{3}{7}$ of the tank. 3. **Calculate Leak Rate:** Let the leak empty the tank in $x$ hours. Its work in 1 hour is $\frac{1}{x}$. * $\text{Pump Rate} - \text{Leak Rate} = \text{Net Rate}$ * $\frac{1}{2} - \frac{1}{x} = \frac{3}{7}$ * $\frac{1}{x} = \frac{1}{2} - \frac{3}{7}$ * $\frac{1}{x} = \frac{7 - 6}{14} = \frac{1}{14}$ 4. **Find Total Time for Leak:** * Since the leak does $\frac{1}{14}$ of the work in 1 hour, it will empty the full tank in 14 hours. ### Exam Strategy & Shortcut Using the LCM method: Let total capacity = LCM(2, 7/3). A multiple of 14 works well. Let Capacity = 14 units. Efficiency of Pump = $14 / 2 = 7$ units/hr. Efficiency of Pump + Leak = $14 / (7/3) = 6$ units/hr. Leak Efficiency = $(Pump + Leak) - Pump = 6 - 7 = -1$ unit/hr. Time for leak to empty tank = $14 / 1 = 14$ hours. ### Common Pitfall A frequent error is subtracting the times directly (e.g., $7/3 - 2 = 1/3$) rather than subtracting their reciprocal rates (work done per hour). Always operate on rates, never on total times directly. ### Final Answer Therefore, the correct answer is **14 hours**.