2)A cylindrical water reservoir is served with three pumps,P,Q and R.Pump P can fill the empty reservoir in S hours while Pump Q can fill the same reservoir in 10 hours. When full ,pump Q can drain the reservoir in 3 hours;
a)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 from filled.
b)On another day,the reservoir was halfway filled with water. Pump P and R were then opened at the same time.How long did it take to completely drain the reservoir.
c)The reservoir dimensions are base diameter 14metres and height 12.5metres.The water in (b) above is filled in a water tanker of capacity 12000 litres,How many trips does the tanker make to drain the reservoir?
a) Pump P can fill the reservoir in S hours, and pump Q can fill it in 10 hours. Therefore, in 1 hour, pump P can fill 1/S of the reservoir, and pump Q can fill 1/10 of the reservoir. Pump Q can drain the reservoir in 3 hours, so in 1 hour it can drain 1/3 of the reservoir.
From 8:45am to 11:15am is a total of 2.5 hours. During this time, pump P and pump Q were both operational. Pump P developed an electrical fault at 11:15am, so only pump Q was operational after that.
In the 2.5 hours when both pumps were operational, we can calculate the amount of water filled:
Amount filled by pump P = (1/S) * 2.5
Amount filled by pump Q = (1/10) * 2.5
So, the total amount filled in the 2.5 hours is:
Total amount filled = (1/S) * 2.5 + (1/10) * 2.5
After 11:15am, only pump Q operated. We need to find the remaining amount of water needed to fill the tank.
Let's assume the capacity of the reservoir is 1 unit (volume is relative). The amount of water filled in 2.5 hours is (1/S) * 2.5 + (1/10) * 2.5, and the remaining amount to be filled is 1 - [(1/S) * 2.5 + (1/10) * 2.5].
Since pump Q can fill 1/10 of the reservoir in 1 hour, it will take 10 times longer to fill the remaining 1 - [(1/S) * 2.5 + (1/10) * 2.5] amount of water.
Time taken by pump Q to fill the remaining amount = 10 * [1 - (1/S) * 2.5 - (1/10) * 2.5] = 10 - (5/S) - 0.5
To find the time of the day when the reservoir was completely filled, we need to add the time taken by pump Q to the initial time of 2.5 hours (8:45am to 11:15am).
Total time = 2.5 hours + 10 - (5/S) - 0.5 hours
b) Pump P and R were opened at the same time when the reservoir was halfway filled. Pump P can fill the reservoir in S hours, and pump R's filling time is not given.
We know that in 1 hour, pump P can fill 1/S of the reservoir. Let's assume the capacity of the reservoir is 1 unit. The remaining amount to be filled when pump P and R are opened is 1/2.
Pump P and R together can fill 1/S + 1/R of the reservoir in 1 hour.
To find the time taken to completely drain the reservoir, we need to calculate the time when the remaining amount becomes 0. So, we can set up the equation:
(1/S + 1/R) * time taken = 1/2
Solving this equation will give us the time taken to completely drain the reservoir.
c) The capacity of the reservoir is given as 14m (base diameter) * 12.5m (height) = 175 cubic meters.
1 cubic meter is equal to 1000 liters.
Therefore, the capacity of the reservoir in liters = 175 * 1000 = 175,000 liters.
Given that the water tanker's capacity is 12,000 liters, we can calculate the number of trips it takes to drain the reservoir:
Number of trips = Capacity of the reservoir / Capacity of the water tanker
Number of trips = 175,000 / 12,000
Make sure to calculate the above expression to find the exact number of trips the water tanker makes to drain the reservoir.