https://file.aiquickdraw.com/mathbot/file/1726165992394b8np3mi9.jpg

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.

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?

Learn how to calculate the time required to fill a cistern when one tap fills it and another empties it at different rates. This problem uses work and time concepts and algebraic formulas. Suitable for students in Grades 8-10.

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