A cistern can be filled by two pipes in 8 minutes and 10 minutes, separately. Both the pipes are opened together for a certain time, but due to an obstruction, only 5/8 of the full quantity of water flowed through the former pipe and 3/5 through the latter pipe. However, the obstruction was suddenly removed, and the cistern was filled in 3 minutes from that moment. How long did it take before the full flow began?

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Accepted Answer

Step 1: Calculate the rate of each pipe when working at full capacity.
- Pipe 1 fills the cistern in 8 minutes, so its rate is 1/8 of the cistern per minute.
- Pipe 2 fills the cistern in 10 minutes, so its rate is 1/10 of the cistern per minute.

Step 2: Calculate the reduced rates due to obstruction.
- Pipe 1 flows at 5/8 of its full rate: (5/8) * (1/8) = 5/64 per minute.
- Pipe 2 flows at 3/5 of its full rate: (3/5) * (1/10) = 3/50 per minute.

Step 3: Let the time before obstruction removal be t minutes.
- Water filled during obstruction: (5/64 + 3/50) * t = ( (5/64) + (3/50) ) * t
- Find common denominator for rates: 64 and 50 → 3200
  (5/64) = (5*50)/(64*50) = 250/3200
  (3/50) = (3*64)/(50*64) = 192/3200
  Sum = (250 + 192)/3200 = 442/3200 = 221/1600
- Water filled during obstruction = (221/1600) * t

Step 4: After obstruction removal, both pipes work at full capacity.
- Combined rate = 1/8 + 1/10 = (5/40 + 4/40) = 9/40 per minute.
- Time after obstruction removal = 3 minutes.
- Water filled after obstruction = (9/40) * 3 = 27/40

Step 5: Total water filled = 1 (full cistern)
- So, (221/1600) * t + 27/40 = 1

Step 6: Solve for t
- Convert 27/40 to denominator 1600: 27/40 = (27*40)/(40*40) = 1080/1600
- Equation: (221/1600) * t + 1080/1600 = 1
- Multiply both sides by 1600:
  221t + 1080 = 1600
  221t = 1600 - 1080 = 520
  t = 520 / 221 ≈ 2.3525 minutes

Step 7: Convert decimal to fraction
- 0.3525 ≈ 6/17 (since 6/17 ≈ 0.3529)
- So, t ≈ 2 6/17 minutes

Answer: 2 6/17 minutes

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