This problem involves calculating the time taken to fill a tank with three pipes operating in a specific cyclic pattern. We need to determine the rate of each pipe and the net fill rate over one cycle.

Pipe Filling and Emptying Rates

First, let's establish the rate at which each pipe works:

• Pipe A (Inlet): Fills the tank in 15 minutes. Rate = \(\frac{1}{15}\) of the tank per minute.
• Pipe B (Inlet): Fills the tank in 20 minutes. Rate = \(\frac{1}{20}\) of the tank per minute.
• Pipe C (Outlet): Empties the tank in 30 minutes. Rate = \(-\frac{1}{30}\) of the tank per minute (negative sign indicates emptying).

Cycle Analysis

The pipes operate in a 3-minute cycle:

• Minute 1: Only Pipe A operates. Fill = \(+\frac{1}{15}\).
• Minute 2: Only Pipe B operates. Fill = \(+\frac{1}{20}\).
• Minute 3: Only Pipe C operates. Empty = \(-\frac{1}{30}\).

Let's calculate the net change in the tank's water level over one full 3-minute cycle:

Net Fill per Cycle = Rate A + Rate B + Rate C

Net Fill per Cycle = \(\frac{1}{15} + \frac{1}{20} - \frac{1}{30}\)

To add these fractions, find a common denominator, which is 60:

Net Fill per Cycle = \(\frac{4}{60} + \frac{3}{60} - \frac{2}{60} = \frac{4 + 3 - 2}{60} = \frac{5}{60} = \frac{1}{12}\) of the tank.

So, in every 3 minutes, \(\frac{1}{12}\) of the tank is filled.

Calculating Total Filling Time

The cycle repeats until the tank is full. We need to find how many full cycles are needed and then calculate the time for the final partial cycle.

Let \(N\) be the number of full 3-minute cycles.

After \(N\) cycles (which take \(3N\) minutes), the tank will be \(\frac{N}{12}\) full.

Consider the state of the tank just before the last few minutes. The maximum amount the tank can be filled in one minute is by Pipe A (\(\frac{1}{15}\)). If the tank reaches a level from which it can be filled completely within the next minute (or minutes of the cycle), we need to calculate that precisely.

Let's estimate the number of cycles needed. If \(\frac{1}{12}\) is filled in 3 minutes, roughly the whole tank (1) would take \(1 \div \frac{1}{12} \times 3 = 12 \times 3 = 36\) minutes. This is an estimate; the cyclical nature requires careful calculation for the end phase.

Let's consider filling the tank up to a point where the remaining volume can be filled by Pipe A or Pipe B within their respective turns.

Consider 11 cycles:

Time elapsed = \(11 \times 3 = 33\) minutes.

Volume filled = \(11 \times \frac{1}{12} = \frac{11}{12}\) of the tank.

Remaining volume = \(1 - \frac{11}{12} = \frac{1}{12}\) of the tank.

Now, the cycle starts again:

Minute 34 (Cycle 12 begins): Pipe A operates. Fill = \(\frac{1}{15}\).

New volume = \(\frac{11}{12} + \frac{1}{15} = \frac{55}{60} + \frac{4}{60} = \frac{59}{60}\) of the tank.

Remaining volume = \(1 - \frac{59}{60} = \frac{1}{60}\) of the tank.

Minute 35 (Cycle 12 continues): Pipe B operates. Pipe B's rate is \(\frac{1}{20}\) per minute.

We need to fill the remaining \(\frac{1}{60}\) of the tank using Pipe B.

Time needed = \(\frac{\text{Remaining Volume}}{\text{Rate of Pipe B}} = \frac{1/60}{1/20} = \frac{1}{60} \times 20 = \frac{20}{60} = \frac{1}{3}\) minutes.

So, the tank gets filled during the 35th minute, specifically after \(\frac{1}{3}\) of that minute has passed.

Total time = Time for 11 cycles + Time for Pipe A's turn + Time for Pipe B to finish.

Total time = \(33 \text{ minutes} + 1 \text{ minute} + \frac{1}{3} \text{ minutes} = 34\frac{1}{3}\) minutes.

Final Answer Calculation Check

The total time taken to fill the tank is the sum of the time for 11 full cycles and the time taken in the 12th cycle until the tank is full:

Total Time = \((11 \times 3) + 1 + \frac{1}{3} = 33 + 1 + \frac{1}{3} = 34\frac{1}{3}\) minutes.

This matches Option D.