A cistern has one inlet pipe L and one outlet pipe M. Pipe L alone can fill the empty cistern in 5 hours, while pipe M alone can empty the full cistern in 8 hours. If the cistern is initially 2/5 full and both pipes are opened together, how many hours will it take to fill the cistern completely?

Introduction / Context:
This problem involves a cistern with both an inlet and an outlet pipe, starting from a partially filled state. It tests a student's ability to compute net inflow when filling and emptying occur simultaneously, and then to determine how long it takes to reach full capacity from an initial fractional level. Such mixed inlet outlet questions are standard in quantitative aptitude tests.

Given Data / Assumptions:

Pipe L is an inlet and can fill an empty cistern in 5 hours.

Pipe M is an outlet and can empty a full cistern in 8 hours.

The cistern is initially 2/5 full.

Both pipes L and M are opened together.

Flow rates are constant throughout the process.

Concept / Approach:
As with other work and time problems, we treat one full cistern as one unit of work. The rate of each pipe is the fraction of the tank handled per hour. Since L fills and M empties, the net rate is the difference between their rates. Starting from the initial volume, we compute the remaining volume to be filled and divide by the net positive rate to find the time needed. The key is correct identification of inlet versus outlet and careful fraction handling.

Step-by-Step Solution:
Let the capacity of the cistern be 1 unit.Rate of inlet pipe L = 1/5 tank per hour.Rate of outlet pipe M = 1/8 tank per hour.Net rate when both pipes are open = 1/5 - 1/8 tank per hour.Compute this difference using a common denominator of 40: 1/5 = 8/40 and 1/8 = 5/40.Net rate = 8/40 - 5/40 = 3/40 tank per hour (positive, so the cistern is filling).Initial volume = 2/5 of capacity.Remaining volume to be filled = 1 - 2/5 = 3/5 of the tank.Time required = remaining volume / net rate = (3/5) / (3/40).This equals (3/5) * (40/3) = 40/5 = 8 hours.Therefore, the cistern will be completely full after 8 hours.
Verification / Alternative check:
In 8 hours, net filled volume = 8 * (3/40) = 24/40 = 3/5 of the tank.Since the cistern started with 2/5 filled, final volume = 2/5 + 3/5 = 1 full tank.This matches the definition of full, so the calculation is consistent.
Why Other Options Are Wrong:
Times such as 3 hours, 4 hours, 5 hours or 6 hours would provide less net inflow than required to add 3/5 of the tank. For example, at 3 hours the net volume added would only be 3 * (3/40) = 9/40, which combined with 2/5 (16/40) is only 25/40, far less than full. Only 8 hours provides the exact additional 3/5 needed.

Common Pitfalls:
Students can easily confuse which pipe is filling and which is emptying if they read too quickly, leading them to add rates instead of subtracting. Another common error is to forget that the tank is not starting empty and to compute the time to fill from empty rather than from 2/5 full. Keeping track of the initial fraction and remembering that work done equals rate times time for the remaining volume is essential.

Final Answer:
When both pipes are opened together from a 2/5 full state, the cistern will be completely full in 8 hours.