A pump can fill a tank with water in $3\text{hours}$. Due to leakage in the bottom, it takes $3\frac{1}{2}\text{hours}$ to fill it. In what time will the leak empty the fully filled tank?

Let the filling pump be $A$ and the leak be $B$.
Work rate of pump $A$ alone = $\frac{1}{3}$ per hour.
Work rate of pump $A$ and leak $B$ together = $\frac{1}{3\frac{1}{2}}=\frac{1}{\frac{7}{2}}=\frac{2}{7}$ per hour.

Let the work rate of leak $B$ be $-\frac{1}{t}$ (since it empties).

$$\frac{1}{3}-\frac{1}{t}=\frac{2}{7}$$

$$\frac{1}{t}=\frac{1}{3}-\frac{2}{7}$$

$$\frac{1}{t}=\frac{7-6}{21}=\frac{1}{21}$$

$$t=21\text{hours}$$

Thus, the leak will empty the tank in $21\text{hours}$.