Water is pumped from a tank at constant rate, and no more water enters the tank. If the tank contains 19140 L at 4:47 PM and 8097 L at 5:05 PM the same day, how many liters will the tank contain at 5:11 PM that day?

Firstly, you must identify the outlet flow of the water pumped from the tank, for this, you must subtract the last volume given from the first volume:

19,140 L - 8,097 L = 11,043 L

And the minutes that passed from the first volume until the last volume given (18 minutes from 4:47 PM to 5:05 PM), so, you must divide that two values to obtain the outlet flow:

Outlet flow = [tex]\frac{Volume}{Time}[/tex]
Outlet flow = [tex]\frac{11,043 L}{18 min}[/tex]
Outlet flow = 613.5 [tex]\frac{L}{min}[/tex]

Now, you must see the next hour given (5:11 PM), if you see, from 5:05 PM to 5:11 PM has passed 6 minutes, taking into account this, you replace the equation of outlet flow to clear the volume:

Outlet flow = [tex]\frac{Volume}{Time}[/tex]

Volume = Outlet flow * time

And replace the values to obtain the new volume pumped:

Volume = 613.5 [tex]\frac{L}{min}[/tex] * 6 min
Volume = 3681 L.

At last, you must subtract these liters from the last volume identified in the tank:

New Volume in the tank = 8097 L - 3681 L
New Volume in the tank = 4416 L

The volume in the tank at 5:11 PM is 4416 Liters.

ibiscollingwood ibiscollingwood · Answer 2 · 2022-02-25T13:50:12+01:00

The tank will contain 4416 L of water at 5:11 PM .

Rate of discharge of water:

Given that, the tank contains 19140 L at 4:47 PM and 8097 L at 5:05 PM

It means that, in 18 minutes the amount of water discharge is,

[tex]=19140-8097=11043L[/tex]

The rate of flow is,

[tex]=\frac{11043}{18}=613.5L/min[/tex]

From 5:05 PM to 5:11 PM , six minutes are passes.

Water discharge in that six minute is ,

[tex]=613.5*6=3681L[/tex]

The tank will contain at 5:11 PM is,

=[tex]8097-3681=4416L[/tex]

Learn more about rate of flow here:

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