The question in the image asks:

One tap can fill a cistern in 2 hours and another can empty the cistern in 3 hours. How long will they take to fill the cistern if both taps are opened?

Solution:

1. The rate at which the first tap fills the cistern is: $\frac{1}{2} \text{ cistern/hour}.$

2. The rate at which the second tap empties the cistern is: $\frac{1}{3} \text{ cistern/hour}.$

3. When both taps are opened, the net rate of filling the cistern is: $\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \text{ cistern/hour}.$

4. This means that in one hour, $\frac{1}{6}$ of the cistern will be filled. To fill the entire cistern, it will take: $6 \text{ hours}.$

Thus, the cistern will be filled in 6 hours.

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How would the time change if the second tap emptied the cistern in 4 hours instead of 3?
2. What happens if both taps have the same rate of filling and emptying?
3. How do you calculate the time if three taps are involved: one filling and two emptying at different rates?
4. How can this problem be visualized using flow rates and a tank diagram?
5. How would you solve the problem if the emptying tap is slower than the filling tap?

Tip: Always think of rates as contributions to a task; positive rates add to completion, and negative rates subtract from it.