Three taps P, Q and R can fill a tank in 12 hours, 15 hours and 20 hours respectively. Tap P is kept open all the time, while taps Q and R are opened for one hour each alternately, starting with tap Q. In how many hours will the tank be completely filled?

Introduction / Context:
This is a pipes and cisterns problem with an alternating pattern. Tap P is always on, while taps Q and R are turned on alternately for one hour each. You must carefully consider the work done in each two hour cycle and then determine how many such cycles plus any extra time are needed to fill the tank fully.

Given Data / Assumptions:

• Tap P alone fills the tank in 12 hours.
• Tap Q alone fills the tank in 15 hours.
• Tap R alone fills the tank in 20 hours.
• P is open continuously.
• Q and R are open alternately, each for one hour, starting with Q.
• The tank is initially empty.

Concept / Approach:
We first compute the filling rates of each tap as a fraction of the tank per hour. Then we analyse the pattern over a 2 hour cycle: in the first hour, taps P and Q are open; in the second hour, taps P and R are open. This gives a fixed amount of work per 2 hour cycle. We then find how many whole cycles are needed to approach full capacity and how much additional time is required in the next hour of the sequence to complete the tank.

Step-by-Step Solution:
Let tank capacity be 1 unit. Rate of P = 1 / 12 tank per hour. Rate of Q = 1 / 15 tank per hour. Rate of R = 1 / 20 tank per hour. Hour 1 (P + Q): work done = 1 / 12 + 1 / 15. Compute: LCM of 12 and 15 is 60, so 1 / 12 = 5 / 60, 1 / 15 = 4 / 60, total = 9 / 60 = 3 / 20. Hour 2 (P + R): work done = 1 / 12 + 1 / 20. LCM of 12 and 20 is 60, so 1 / 12 = 5 / 60, 1 / 20 = 3 / 60, total = 8 / 60 = 2 / 15. Thus, in each 2 hour cycle, work done = 3 / 20 + 2 / 15. Compute with LCM 60: 3 / 20 = 9 / 60 and 2 / 15 = 8 / 60, sum = 17 / 60. After 3 full cycles (6 hours), work done = 3 * 17 / 60 = 51 / 60 of the tank. Remaining work = 1 − 51 / 60 = 9 / 60 = 3 / 20. Next hour in sequence is again P + Q, whose rate is 3 / 20 per hour. Time required to do remaining 3 / 20 with rate 3 / 20 is exactly 1 hour. Total time = 6 hours + 1 hour = 7 hours.

Verification / Alternative check:
You can list the work hour by hour: hour 1 (P+Q) adds 3 / 20, hour 2 (P+R) adds 2 / 15, and so on. Summing after 7 hours gives exactly 1 tank full, confirming that there is no overfilling or shortfall. The cycle based approach is usually simpler and less error prone.

Why Other Options Are Wrong:
9 hours and 11 hours are too long given the strong combined rates, while 13 hours is clearly excessive. Any time greater than 7 hours would exceed the amount of work required to fill one tank. On the other hand, less than 7 hours would not accumulate enough fraction of the tank when summing the contributions of each hour following the prescribed pattern.

Common Pitfalls:
A common mistake is to average the rates of Q and R instead of analysing the actual alternating pattern. Another error is to mistakenly assume that all three taps are open all the time, which would give a different time altogether. Always track the hour by hour pattern when taps or workers are operating alternately.

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