Given:
A large hose can fill a swimming pool In 45 hours.
A small hose can fill the same pool In 50 hours.
To find:
The required time to complete the task if both hoses are used together.
Step-by-step explanation:
One hour's work of large hose is,
![(1)/(45)unit](https://img.qammunity.org/2023/formulas/mathematics/college/93lrhvln93ge3avuwj65anodwo1u3k921x.png)
One hour's work of small hose is,
![(1)/(50)unit](https://img.qammunity.org/2023/formulas/mathematics/college/ry1afitfyht8r26tx4feuercg5jsn0zbv5.png)
When both work together, the one hour's work
![\begin{gathered} (1)/(45)+(1)/(50)=(50+45)/(45(50)) \\ =(95)/(2250) \\ =(19)/(450)unit \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5f3mitpgx57y9s2kue16zc08ysxcmhi7xh.png)
We know that time is inversely proportional to the work done.
Therefore,
![\begin{gathered} T=(450)/(19) \\ =23.6842 \\ \approx23.684hours \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lhy5kdz2etsbf2ntg15xcfmbb98igsx53z.png)
Thus, if both hoses are used to fill the pool, then the time required to complete the task is 23.684 hours.
Final answer:
The time required to complete the task when both hoses work together is 23.684 hours.