Description for Correct answer:
Before the crack appeared, Pipe A could fill \( \Large \frac{1}{2} \) the tank in an hour.

Now it takes 2 hours 30 minutes to fill the tank. That is \( \Large \frac{5}{2} \) hours.

Hence, it fills \( \Large \frac{2}{5} \)th of the tank every hour.

The crack at the bottom accounts for this reduction in the amount.

The crack, therefore, drains \( \Large \frac{1}{2}-\frac{2}{5}=\frac{5-4}{10}=\frac{1}{10} \)th of the tank every hour.

As the crack drains a tenth of the tank every hour, it will drain the entire tank in 10 hours if Pipe A is kept closed.