A cylindrical water reservoir is served with three pumps:P,Q and R.Pump P can fill an empty reservoir in 5 hours while pump Q can fill the same reservoir in 3 hours.
a)One day,the reservoir was empty and pumps P and Q opened at 8:45 a.m.At 11:15 a.m,pump P developed an electric fault.Pump Q continued operating alone till the reservoir was full.Find the time of the day that the reservoir was completely 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's dimension are;base diameter 14 metres and height 12.5metres.The water in (b) above is filled in with 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 5 hours, so the amount of water it fills in 1 hour is 1/5 of the reservoir's capacity. Similarly, pump Q can fill the reservoir in 3 hours, so the amount of water it fills in 1 hour is 1/3 of the reservoir's capacity.
From 8:45 a.m. to 11:15 a.m., there is a time period of 2.5 hours.
During this time period, pump P fills 2.5/5 = 1/2 of the reservoir's capacity.
Therefore, the remaining capacity of the reservoir is 1 - 1/2 = 1/2.
From 11:15 a.m. onwards, pump Q operates alone until the reservoir is completely filled.
Since pump Q fills 1/3 of the reservoir's capacity in 1 hour, it takes 3/1 = 3 hours for pump Q to fill the remaining 1/2 of the reservoir's capacity.
Therefore, pump Q fills the reservoir completely at 11:15 a.m. + 3 hours = 2:15 p.m.
b) Pump P can fill the reservoir in 5 hours, so the amount of water it fills in 1 hour is 1/5 of the reservoir's capacity. Similarly, pump R can fill the reservoir in x hours, so the amount of water it fills in 1 hour is 1/x of the reservoir's capacity.
If pump P and pump R operate together, they can fill the reservoir in 1/5 + 1/x of its capacity.
Since the reservoir is halfway filled, the remaining capacity is 1/2.
Therefore, we have the equation: 1/5 + 1/x = 1/2.
To solve this equation, we can multiply both sides by 10x to get: 2x + 10 = 5x.
This simplifies to: 3x = 10.
Therefore, x = 10/3 = 3.33 hours.
So, it takes approximately 3.33 hours to completely drain the reservoir.
c) The total capacity of the reservoir is given by the formula for the volume of a cylinder: V = πr^2h, where r is the radius (diameter/2) and h is the height.
The base diameter of the reservoir is 14 meters, so the radius is 14/2 = 7 meters.
The height of the reservoir is 12.5 meters.
Therefore, the capacity of the reservoir is V = π(7^2)(12.5) = 1925π cubic meters.
1 cubic meter is equal to 1000 liters.
Therefore, the capacity of the reservoir is 1925π × 1000 = 1925000π liters.
The water tanker has a capacity of 12000 liters.
To find the number of trips the tanker makes to drain the reservoir, we divide the capacity of the reservoir by the capacity of the tanker: 1925000π/12000 = 160π.
Therefore, the water tanker makes approximately 160π trips to drain the reservoir.