This problem involves calculating the time taken to fill a tank using multiple pipes, each with different filling rates and starting times, and includes a specific condition about one pipe taking a break. Let's break down the calculation step by step.

Understanding Pipe Rates

First, we determine the rate at which each pipe fills the tank. The rate is the fraction of the tank filled per hour.

• Rate of Pipe A ($R_A$) = $\frac{1}{10}$ tank per hour.
• Rate of Pipe B ($R_B$) = $\frac{1}{15}$ tank per hour.
• Rate of Pipe C ($R_C$) = $\frac{1}{20}$ tank per hour.

Calculating Tank Filling Progress

We track the amount of the tank filled at different time points, considering the opening times and the break condition for Pipe C.

Filling from 6 AM to 7 AM

Only Pipe A is open during this hour.

• Amount filled by Pipe A = $1 \text{ hour} \times R_A = 1 \times \frac{1}{10} = \frac{1}{10}$ tank.
• Total filled by 7 AM = $\frac{1}{10}$ tank.

Filling from 7 AM to 8 AM

Both Pipe A and Pipe B are open during this hour.

• Amount filled by Pipe A = $1 \text{ hour} \times \frac{1}{10} = \frac{1}{10}$ tank.
• Amount filled by Pipe B = $1 \text{ hour} \times \frac{1}{15} = \frac{1}{15}$ tank.
• Combined rate of A and B = $\frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6}$ tank per hour.
• Amount filled between 7 AM and 8 AM = $\frac{1}{6}$ tank.
• Total filled by 8 AM = (Amount filled by 7 AM) + (Amount filled between 7 AM and 8 AM) = $\frac{1}{10} + \frac{1}{6} = \frac{3+5}{30} = \frac{8}{30} = \frac{4}{15}$ tank.

Filling from 8 AM to 11 AM

Pipe C opens at 8 AM. Pipe C works for 3 hours (from 8 AM to 11 AM) before its break.

• Duration = 3 hours.
• Pipes A, B, and C are working.
• Amount filled by Pipe A = $3 \text{ hours} \times \frac{1}{10} = \frac{3}{10}$ tank.
• Amount filled by Pipe B = $3 \text{ hours} \times \frac{1}{15} = \frac{3}{15} = \frac{1}{5}$ tank.
• Amount filled by Pipe C = $3 \text{ hours} \times \frac{1}{20} = \frac{3}{20}$ tank.
• Total amount filled between 8 AM and 11 AM = $\frac{3}{10} + \frac{1}{5} + \frac{3}{20} = \frac{12}{20} + \frac{4}{20} + \frac{3}{20} = \frac{19}{20}$ tank.
• Total filled by 11 AM = (Total filled by 8 AM) + (Amount filled between 8 AM and 11 AM) = $\frac{4}{15} + \frac{13}{20}$ tank.
  • Correction based on step-by-step cumulative calculation: Amount filled between 8 AM and 11 AM should be calculated correctly: $\frac{3}{10} + \frac{1}{5} + \frac{3}{20} = \frac{12+4+3}{20} = \frac{19}{20}$. Wait, the previous calculation of total by 8 AM was 4/15. Let's redo the total by 11 AM: Total filled by 11 AM = (Total filled by 8 AM) + (Amount filled between 8 AM and 11 AM) Total filled by 11 AM = $\frac{4}{15} + \frac{19}{20}$ -- This calculation was incorrect previously. Let's reconfirm the sum $\frac{3}{10} + \frac{1}{5} + \frac{3}{20}$. LCM is 20. $\frac{6}{20} + \frac{4}{20} + \frac{3}{20} = \frac{13}{20}$. The work done in 8 AM to 11 AM is 13/20. Total filled by 11 AM = $\frac{4}{15} + \frac{13}{20} = \frac{16 + 39}{60} = \frac{55}{60} = \frac{11}{12}$ tank.

So, by 11 AM, $\frac{11}{12}$ of the tank is filled.

Filling After 11 AM

Pipe C takes a 1-hour break from 11 AM to 12 PM. Only Pipes A and B are working.

• Remaining capacity to fill = $1 - \frac{11}{12} = \frac{1}{12}$ tank.
• Combined rate of Pipes A and B = $\frac{1}{6}$ tank per hour.
• Time needed to fill the remaining capacity = $\frac{\text{Remaining Capacity}}{\text{Combined Rate of A and B}} = \frac{1/12}{1/6}$ hours.
• Time needed = $\frac{1}{12} \times 6 = \frac{6}{12} = \frac{1}{2}$ hour.
• $\frac{1}{2}$ hour is equal to 30 minutes.

Final Filling Time

The remaining part of the tank is filled 30 minutes after 11 AM.

• Filling Time = 11 AM + 30 minutes = 11:30 AM.

Conclusion

The tank is completely filled at 11:30 AM.