An empty reservoir has a capacity of 30 litres. An inlet pipe fills water at 5 litres per minute, and an outlet pipe empties water at 4 litres per minute. The pipes operate alternately for one-minute intervals, with the inlet pipe operating first. After how many minutes will the reservoir become completely full for the first time?

Introduction / Context:
This problem involves alternating inflow and outflow. One pipe fills the reservoir for a minute, then the other pipe empties for a minute, and this pattern repeats. We must track the volume after each minute to find the first time the reservoir reaches exactly 30 litres. This is a good example of a discrete step process rather than a constant-rate process.

Given Data / Assumptions:

Reservoir capacity = 30 litres.

Inlet pipe rate = 5 litres per minute.

Outlet pipe rate = 4 litres per minute.

Pipes operate in 1-minute intervals, starting with the inlet pipe.

The water level cannot exceed 30 litres; we stop once it hits 30 litres for the first time.

Concept / Approach:
Since the pipes act in alternate 1-minute blocks, the volume in the reservoir changes in discrete steps. In each 2-minute cycle, the net effect is an increase of 1 litre (fill 5, then empty 4). We can model the process cycle by cycle and then determine at which minute within a cycle the reservoir first reaches 30 litres.

Step-by-Step Solution:
Step 1: Start with 0 litres at time 0.Step 2: Minute 1 (inlet): volume increases by 5 litres to 5 litres.Step 3: Minute 2 (outlet): volume decreases by 4 litres to 1 litre.Step 4: Thus, every 2-minute cycle increases the volume by 1 litre overall.Step 5: After n complete 2-minute cycles, volume = n litres. Time elapsed = 2n minutes.Step 6: We also have an inlet action at the start of each new cycle. After n cycles and one more inlet minute, volume = n + 5 litres.Step 7: We need to find the smallest integer minute when the volume first reaches 30 litres.Step 8: From the stepwise pattern, just before the final filling minute, volume should be 25 litres so that a +5 litre fill takes it to exactly 30.Step 9: Volume equals 1 litre after the first 2-minute cycle, 2 litres after the second, and so on. So volume equals 25 litres after 25 cycles.Step 10: 25 cycles correspond to 25 * 2 = 50 minutes.Step 11: At minute 51, the inlet pipe runs and adds 5 litres, taking the volume from 25 litres to 30 litres for the first time.

Verification / Alternative check:
If we simulate the process, we see the volume sequence at the end of inlet minutes as 5, 6, 7, and so on, and at the end of outlet minutes as 1, 2, 3, and so on. Following this pattern shows that we reach 25 litres after 50 minutes and then hit 30 litres after the inlet runs at minute 51, which corroborates the analytical reasoning.

Why Other Options Are Wrong:
49.5 minutes is not an integer and does not match the discrete 1-minute switching pattern.

50 minutes leaves the reservoir at 25 litres, not yet full.

52 minutes would mean one extra outlet minute after the reservoir has already become full, which is beyond the first time it hits full capacity.

Common Pitfalls:
Students often treat the net rate as 1 litre per 2 minutes and divide 30 by 0.5 to get 60 minutes, ignoring that the reservoir may become full during the filling minute within a cycle. It is important to track the sequence minute by minute, especially for discrete alternating processes, rather than relying solely on an average net rate per cycle.

Final Answer:
The reservoir will become completely full for the first time after 51 minutes.