Pipes and Cisterns – Two inlets and one outlet acting together: An empty cistern can be filled by tap 1 in 3 hours and by tap 2 in 4 hours, while tap 3 can empty a full cistern in 5 hours. If all three taps are opened at the same time starting from empty, in how much time will the cistern become full?

Introduction / Context:
When two inlets and one outlet operate together, their rates algebraically combine. The filling taps contribute positively, while the emptying tap contributes negatively. We then invert the net rate to get time.

Given Data / Assumptions:

• Tap 1 fills in 3 h ⇒ rate = 1/3 per h.
• Tap 2 fills in 4 h ⇒ rate = 1/4 per h.
• Tap 3 empties in 5 h ⇒ rate = 1/5 per h (negative for filling).
• Start from an empty cistern.

Concept / Approach:
Net rate = 1/3 + 1/4 − 1/5. The time to fill is the reciprocal of this net rate.

Step-by-Step Solution:
Compute net rate: LCM 60 ⇒ 1/3 = 20/60, 1/4 = 15/60, 1/5 = 12/60.Net = (20 + 15 − 12)/60 = 23/60 per h.Time = 1 / (23/60) = 60/23 h ≈ 2 h 36 min 31 s.

Verification / Alternative check:
Approximation: 60/23 ≈ 2.61 h; multiplying 2.61 by 23/60 ≈ 1 confirms consistency.

Why Other Options Are Wrong:
2 h 40 min is a rounding overshoot; 1 h 56 min is too small; 3 h is too large relative to the computed net rate.

Common Pitfalls:
Adding times directly or forgetting the outlet should be subtracted.

Final Answer:
60/23 h