Pipes and Cisterns – Turn off one tap after a while, find the remaining minutes: Three taps A, B, and C can fill a cistern in 10, 15, and 20 minutes respectively. All are opened together, but C is turned off after 3 minutes. How many additional minutes will A and B require to fill the cistern?

Introduction / Context:
We again separate the process into phases. First, all three taps contribute for 3 minutes. Then only A and B continue to finish the remaining fraction.

Given Data / Assumptions:

• A: 10 min ⇒ 1/10 per min.
• B: 15 min ⇒ 1/15 per min.
• C: 20 min ⇒ 1/20 per min.
• Phase 1 duration: 3 minutes with all three; Phase 2: A + B only.

Concept / Approach:
Compute the fraction completed in the first 3 minutes and subtract from 1 to get the remaining fraction. Then divide by the A + B combined rate.

Step-by-Step Solution:
In 3 minutes: (1/10 + 1/15 + 1/20)*3 = (6/60 + 4/60 + 3/60)*3 = (13/60)*3 = 13/20.Remaining = 1 − 13/20 = 7/20.A + B rate = 1/10 + 1/15 = (3 + 2)/30 = 1/6 per min.Time = (7/20) / (1/6) = (7/20)*6 = 42/20 = 2.1 min = 2 min 6 sec.

Verification / Alternative check:
2.1 min × 60 sec/min = 126 sec ⇒ 2 min 6 sec; sums correctly to a full tank.

Why Other Options Are Wrong:
1:06, 2:00, 2:30, 3:08 do not match the computed remainder at 1/6 per minute.

Common Pitfalls:
Forgetting to convert the decimal 0.1 min into seconds or miscomputing the first 3 minutes’ contribution.

Final Answer:
2 min 6 sec