A cistern has two pipes. One can fill it with water in $8$ hours and other can empty it in $5$ hours. In how many hours will the cistern be emptied if both the pipes are opened together when $\frac{3}{4}$ of the cistern is already full of water?

### Concept & Net Emptying Rate When an emptying pipe acts faster than a filling pipe, the tank will experience a net loss of volume over time. The net emptying rate is the emptying rate minus the filling rate. $$ \text{Net Emptying Rate} = \frac{1}{\text{Empty Time}} - \frac{1}{\text{Fill Time}} $$ ### Step-by-Step Solution * **Identify Rates:** Part emptied by the outlet pipe in $1$ hour = $\frac{1}{5}$ Part filled by the inlet pipe in $1$ hour = $\frac{1}{8}$ * **Calculate Net Rate:** Since $\frac{1}{5} > \frac{1}{8}$, the cistern is being emptied. Net part emptied in $1$ hour = $\frac{1}{5} - \frac{1}{8} = \frac{8 - 5}{40} = \frac{3}{40}$ * **Determine Remaining Work:** The cistern is not full; it is only $\frac{3}{4}$ full. We only need to empty this fraction. * **Calculate Total Time:** Time required to empty $\frac{3}{4}$ of the cistern = $\frac{\text{Volume to empty}}{\text{Net hourly rate}}$ Time $= \frac{\frac{3}{4}}{\frac{3}{40}} = \frac{3}{4} \times \frac{40}{3} = \frac{40}{4} = 10$ hours. ### Exam Strategy & Shortcut Use the LCM total work method. Let the total capacity be $40$ units (LCM of $8$ and $5$). Fill rate $= +5$ units/hr. Empty rate $= -8$ units/hr. Net rate $= -3$ units/hr (emptying). Volume to empty $= \frac{3}{4} \text{ of } 40 = 30$ units. Time to empty $= \frac{30}{3} = 10$ hours. ### Common Pitfall A common mistake is calculating the time to empty a *full* cistern ($\frac{40}{3} = 13\frac{1}{3}$ hours) and selecting option (d), completely ignoring the critical detail that the cistern is only $\frac{3}{4}$ full. ### Final Answer Therefore, the correct answer is **$10\text{ hours}$**.