A cylindrical water reservoir is served with 3 pumps, P,Q and R.Pump P can fill the empty reservoir in 5hours while pump Q can fill the same reservoir in 10hours.when full pump R can drain the same reservoir in 3hours.One day the reservoir was empty and pumps P and Q opened at 8.45am.At 11.15am pump P developed an electrical fault.Pump Q continued operating alone till the reservoir was full .Find the time of the day that the reservoir was completely filled.

To solve this problem, we need to find the rate at which pump Q can fill the reservoir and then use that rate to determine how long it would take pump Q to fill the reservoir alone.

Let's represent the rate at which pump P fills the reservoir as 1/5 of the reservoir per hour, and the rate at which pump Q fills the reservoir as 1/10 of the reservoir per hour. Pump R drains the reservoir at a rate of 1/3 of the reservoir per hour.

When pumps P and Q operated together from 8.45 am to 11.15 am, they worked for a total of 2.5 hours. During this time, the combined rate at which the pumps filled the reservoir is (1/5 + 1/10) = 3/10 of the reservoir per hour.

Therefore, the amount of water added to the reservoir during this time is (3/10) * 2.5 = 3/4 of the reservoir.

After pump P developed an electrical fault at 11.15 am, pump Q continued operating alone. We need to find out how long it would take pump Q to fill the remaining 1/4 of the reservoir.

Since pump Q fills the reservoir at a rate of 1/10 of the reservoir per hour, the time it would take to fill 1/4 of the reservoir is (1/4) / (1/10) = 10/4 = 2.5 hours.

Therefore, the reservoir was completely filled at 11.15 am + 2.5 hours = 1.45 pm.

So, the time of the day that the reservoir was completely filled is 1.45 pm.