Pipes and Cisterns – Reduced initial flow, then full flow completes in a known time: Two pipes can fill a cistern in 30 minutes and 36 minutes, respectively. Initially both are partially clogged so that only 5/6 of the normal water flows through the first pipe and 9/10 through the second. After some time, the obstructions are removed, and from that moment the cistern is filled in 15 1/2 minutes. How long was it before the full flow began?

Introduction / Context:
The process has two phases: a reduced-flow phase and a full-flow phase. The total work (1 tank) equals the sum of work done in each phase. The second phase duration is given (15.5 minutes) at full rate; we use that to infer how much was already completed, and hence how long the reduced phase lasted.

Given Data / Assumptions:

• Normal rates: 1/30 and 1/36 per min.
• Reduced rates: (5/6)*(1/30) = 1/36; (9/10)*(1/36) = 1/40.
• Full-flow (normal) total rate: 1/30 + 1/36 = 11/180 per min.
• From removal onward: time to finish = 15.5 min = 31/2 min.

Concept / Approach:
Remaining portion at the moment of removal = (full rate) * 15.5 = (11/180)*(31/2) = 341/360. Therefore, the portion completed during reduced flow = 1 − 341/360 = 19/360. Divide this by the reduced combined rate to find the reduced-phase duration.

Step-by-Step Solution:
Reduced combined rate = 1/36 + 1/40 = 19/360 per min.Work done in reduced phase = 19/360 of tank.Time in reduced phase = (19/360) / (19/360) = 1 minute.

Verification / Alternative check:
Check totals: reduced 1 min + full 15.5 min = 16.5 min overall; full-phase work equals 341/360, reduced equals 19/360; sum is 1.

Why Other Options Are Wrong:
Any duration other than 1 minute would not make the remaining fraction equal to 341/360 at the switch.

Common Pitfalls:
Treating 15.5 minutes as the time to fill from empty at full rate; it is only for the remaining portion after obstructions are removed.

Final Answer:
1 minute