Four pipes P, Q, R and S can fill a cistern in 20 hours, 25 hours, 40 hours and 50 hours respectively. Pipe P is opened at 6:00 am, Q at 8:00 am, R at 9:00 am and S at 10:00 am. At what time will the cistern be completely full?

Introduction / Context:
This is a cistern and pipe problem where multiple inlets open at different times. Such questions test a candidate's ability to handle varying rates over different time intervals and to sum fractional contributions correctly. Each pipe has its own filling time for the whole cistern, giving a specific hourly rate. The pipes do not all operate from the start, so we must treat the process in blocks of time where the set of open pipes is constant.

Given Data / Assumptions:

• Pipe P fills the cistern alone in 20 hours.
• Pipe Q fills it alone in 25 hours.
• Pipe R fills it alone in 40 hours.
• Pipe S fills it alone in 50 hours.
• P opens at 6:00 am, Q at 8:00 am, R at 9:00 am, and S at 10:00 am.
• All pipes are inlets, and all work at constant rates.

Concept / Approach:
We convert each pipe's time to fill into a rate of filling per hour. Then we consider time intervals: from 6 to 8 only P is open, from 8 to 9 P and Q are open, from 9 to 10 P, Q, and R are open, and after 10 all four pipes are open. For each interval we compute the fraction of the cistern filled. We keep a running total until the cumulative filled fraction reaches 1. For the final interval, if needed, we may compute a partial time within the block when the cistern becomes exactly full.

Step-by-Step Solution:
Step 1: Hourly rates are: P = 1/20, Q = 1/25, R = 1/40, S = 1/50 of the cistern per hour. Step 2: From 6:00 to 8:00 am, only P works for 2 hours, filling 2 times 1/20 = 1/10 of the cistern. Step 3: From 8:00 to 9:00 am, P and Q work together for 1 hour, filling 1/20 + 1/25 = 9/100. Total filled by 9:00 am is 1/10 + 9/100 = 19/100. Step 4: From 9:00 to 10:00 am, P, Q and R work together, filling 1/20 + 1/25 + 1/40 = 23/200 in one hour. Total filled by 10:00 am is 19/100 + 23/200 = 61/200. Step 5: From 10:00 am onward, all four pipes work together with rate 1/20 + 1/25 + 1/40 + 1/50 = 27/200 of the cistern per hour. Step 6: Remaining fraction at 10:00 am is 1 - 61/200 = 139/200. Time needed at the combined rate is (139/200) divided by (27/200) = 139/27 hours which is approximately 5.148 hours, that is 5 hours and about 9 minutes. Step 7: Adding 5 hours and 9 minutes to 10:00 am gives 3:09 pm as the time when the cistern becomes full.

Verification / Alternative check:
To verify, convert 0.148 hours to minutes: 0.148 times 60 equals about 8.88 minutes, which rounds to about 9 minutes. So the final time is indeed very close to 3:09 pm. The cumulative fractions at each stage sum exactly to 1 when expressed symbolically using exact fractions. This confirms that the reasoning with sequential intervals and rates is correct and that no interval or pipe contribution has been double counted or omitted.

Why Other Options Are Wrong:

• 4:18 pm would require a longer duration after 10:00 am than the computed 5.148 hours.
• 12:15 pm and 11:09 am are too early, because by 10:00 am only 61/200 of the tank has been filled, leaving a substantial fraction still empty.
• Only 3:09 pm aligns with the exact fractional calculations of the remaining work and the combined rate of all four pipes.

Common Pitfalls:
Learners often forget to treat each interval separately and instead try to average the number of pipes, which gives incorrect results. Another common error is miscalculating fractional additions or failing to simplify rates properly. It is also easy to misinterpret the moment when each pipe starts working. Carefully tracking start times and cumulative work after each interval is essential in solving such problems accurately.

Final Answer:
The cistern will be completely full at approximately 3:09 pm.