Three taps A, B and C can fill a tank in 12 hours, 15 hours and 20 hours respectively. Tap A is kept open all the time, while taps B and C are opened one at a time in alternate hours. In how many hours will the tank be completely filled?

Introduction / Context:
This problem involves varying filling rates over time. Tap A runs continuously, while taps B and C take turns for one hour each. To handle this, we consider a full 2-hour cycle and compute how much of the tank is filled in each cycle. Then we see how many such cycles are needed and whether a partial cycle is required to reach exactly one full tank.

Given Data / Assumptions:

Tap A alone fills the tank in 12 hours.

Tap B alone fills the tank in 15 hours.

Tap C alone fills the tank in 20 hours.

Tap A is open continuously.

Taps B and C are opened alternately, each for one hour at a time.

Concept / Approach:
We compute the rates of A, B and C in terms of tank per hour. Then we calculate how much of the tank is filled in the first hour (A with B) and in the second hour (A with C). Adding these gives the work done in one 2-hour cycle. We then determine how many such cycles and possibly an extra hour are needed to fill the tank.

Step-by-Step Solution:
Step 1: Rate of A = 1 / 12 tank per hour.Step 2: Rate of B = 1 / 15 tank per hour.Step 3: Rate of C = 1 / 20 tank per hour.Step 4: In hour 1, A and B together work. Combined rate = 1 / 12 + 1 / 15 = (5 + 4) / 60 = 9 / 60 = 3 / 20.Step 5: In hour 2, A and C together work. Combined rate = 1 / 12 + 1 / 20 = (5 + 3) / 60 = 8 / 60 = 2 / 15.Step 6: Work done in a 2-hour cycle = 3 / 20 + 2 / 15 = (9 + 8) / 60 = 17 / 60 of the tank.Step 7: After 3 full cycles (6 hours), work done = 3 * (17 / 60) = 51 / 60 of the tank.Step 8: Remaining work = 1 - 51 / 60 = 9 / 60 = 3 / 20.Step 9: The next hour is like hour 1, with A and B working together at 3 / 20 tank per hour.Step 10: In that one extra hour, they exactly fill the remaining 3 / 20, so total time = 6 + 1 = 7 hours.

Verification / Alternative check:
We can verify by computing total work after 7 hours directly: in the first, third, fifth and seventh hours, taps A and B work; in the second, fourth and sixth hours, taps A and C work. Summing 4 times the work of A+B (4 * 3 / 20) and 3 times the work of A+C (3 * 2 / 15) gives 12 / 20 + 6 / 15 = 3 / 5 + 2 / 5 = 1 full tank.

Why Other Options Are Wrong:
6 hours would give only 51 / 60 of the tank, which is not completely full.

20/3 hours and 15/2 hours do not correspond to any integer number of cycles and do not yield exactly one full tank when we sum each hour’s contribution.

Common Pitfalls:
Some candidates try to average the rates of B and C instead of handling the alternating pattern correctly. Others forget that we must consider whole cycles and may stop at 6 hours without checking whether the tank is completely full. The safest approach is to compute work done over a full 2-hour cycle and then proceed carefully.

Final Answer:
The tank will be completely filled in 7 hours.