One tap can fill a cistern in 2 hours while another tap can empty the cistern in 3 hours. If both taps are opened together on an empty cistern, how long will it take to fill the cistern?

Introduction / Context:
Competing taps create a net rate equal to the difference between the filling and emptying rates. If the net is positive, the tank eventually fills; the time equals the reciprocal of the net rate.

Given Data / Assumptions:

• Fill tap: 2 hours ⇒ +1/2 tank/hour.
• Empty tap: 3 hours ⇒ −1/3 tank/hour.
• Tank starts empty and both run continuously.

Concept / Approach:
Net rate = 1/2 − 1/3 = 1/6 tank/hour. Time = 1 / (1/6) = 6 hours.

Step-by-Step Solution:
Compute difference: 1/2 − 1/3 = (3 − 2)/6 = 1/6.Time to fill = 1 ÷ (1/6) = 6 hours.

Verification / Alternative check:
In 6 hours: filled = 6*(1/2) = 3 tanks; emptied = 6*(1/3) = 2 tanks; net = 1 tank.

Why Other Options Are Wrong:
5 and 7–8 hours imply different net rates not supported by the given times.

Common Pitfalls:
Adding times rather than subtracting rates; sign errors when handling the outlet.

Final Answer:
6 hours