Three pipes A, B and C can fill a tank in 10, 15 and 20 hours respectively. Pipe A was opened at 6 AM, pipe B at 7 AM and pipe C at 8 AM. At what time was the tank completely filled?
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When pipes are opened at different times, calculate the work done in each interval until all pipes are working together. Then, calculate the time to finish the remaining work with the combined rate.
Step 1: Determine the tank's capacity and individual filling rates using the LCM method.
The tank's capacity is LCM(10, 15, 20) = 60 units.
Pipe A's rate: 60/10 = 6 units/hour.
Pipe B's rate: 60/15 = 4 units/hour.
Pipe C's rate: 60/20 = 3 units/hour.
Step 2: Calculate the volume filled before all pipes are operational (i.e., by 8 AM).
From 6 AM to 7 AM (1 hour): Pipe A fills \(1 \times 6 = 6\) units.
From 7 AM to 8 AM (1 hour): Pipes A and B fill \(1 \times (6+4) = 10\) units.
Total filled by 8 AM: 6 + 10 = 16 units.
Step 3: Calculate the remaining work and the time needed to complete it.
Remaining capacity: 60 - 16 = 44 units.
From 8 AM onwards, all pipes are open. Combined rate: \(6 + 4 + 3 = 13\) units/hour.
Time to fill the remaining part: \( \frac{44}{13} \) hours.
Step 4: Convert the time and find the final time.
\( \frac{44}{13} \text{ hours} = 3\frac{5}{13} \) hours.
This is 3 hours and \( \frac{5}{13} \times 60 \) minutes = 3 hours and \( \approx 23 \) minutes.
Time to fill the tank after 8 AM is approximately 3 hours and 23 minutes.
Final Time: 8:00 AM + 3 hours 23 minutes = 11:23 AM.
The closest option is 11:20 AM.