Pipe A is an inlet pipe that can fill an empty cistern in 69 hours, while pipe B is an outlet pipe that can empty a full cistern in 46 hours. Starting with the cistern full, the two pipes are operated one at a time for one hour each, beginning with pipe B and then alternating. After how much time will the cistern become empty?

Introduction / Context:
This problem involves alternating usage of a filling pipe and an emptying pipe on an initially full cistern. The net effect over a repeated two hour cycle must be studied to find how long it takes until the cistern becomes empty. It is a good example of combining rates and dealing with long time spans expressed in days and hours.

Given Data / Assumptions:
- Pipe A fills a completely empty cistern in 69 hours
- Pipe B empties a completely full cistern in 46 hours
- The cistern is initially full
- First hour: only pipe B is open
- Second hour: only pipe A is open
- This two hour pattern repeats (B then A, one hour each)
- No other leaks or sources of water are present

Concept / Approach:
We compute the fraction of the cistern emptied or filled by each pipe in one hour. Over a two hour cycle (B then A) we find the net change in water level. Since the pattern repeats identically, the volume after k cycles is a linear sequence. We determine the number of cycles needed until the cistern is exactly empty and then convert the total time into days and hours.

Step-by-Step Solution:
Step 1: Rate of pipe B (emptying) = 1/46 cistern per hour. Step 2: Rate of pipe A (filling) = 1/69 cistern per hour. Step 3: First hour (pipe B only): change in level = -1/46. Step 4: Second hour (pipe A only): change in level = +1/69. Step 5: Net change per 2 hour cycle = -1/46 + 1/69. Step 6: Take LCM of 46 and 69 which is 138, so net change = (-3 + 2) / 138 = -1/138 of the cistern per 2 hour cycle. Step 7: Let the cistern be full at time zero. After k cycles, volume fraction = 1 - k/138. Step 8: For the cistern to be just empty, set 1 - k/138 = 0 which gives k = 138 cycles. Step 9: Each cycle lasts 2 hours, so total time = 138 * 2 = 276 hours. Step 10: Convert 276 hours into days and hours: 1 day = 24 hours; 276 / 24 = 11 days with remainder 276 - 11 * 24 = 12 hours.

Verification / Alternative check:
We can confirm by noting that in 138 cycles the net fraction removed is 138 * (1/138) = 1 cistern. Since the net effect per cycle is always negative, the cistern will be empty exactly at the end of a full number of cycles. The conversion from hours to days is also straightforward: 11 * 24 = 264 hours, plus 12 hours equals 276 hours, which matches the computed total.

Why Other Options Are Wrong:
11 days 10 hours: Corresponds to 274 hours, which is slightly less than the required 276 hours, so the cistern would not yet be empty.
11 days 7 hours: Even smaller total time and therefore definitely insufficient to empty the cistern.
1 day 13 hours: Much too small compared with the slow net emptying rate of 1/138 per 2 hours.

Common Pitfalls:
A common mistake is to average the individual times rather than the rates, or to forget that B empties and A fills, so their effects have opposite signs. Another pitfall is to try to track hour by hour instead of grouping into cycles, which makes the problem tedious and error prone. Always compute net change over one full pattern cycle and then scale up. Careless conversion between hours and days can also lead to wrong final options.

Final Answer:
The cistern will become empty after a total of 11 days 12 hours of alternating the two pipes as described.