Question

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?

Answer 1

Answer:

The liters that the tank will contain at 5:11 PM that day are:

• 4416 Liters.

Step-by-step explanation:

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 = $\frac{Volume}{Time}$
• Outlet flow = $\frac{11,043 L}{18 min}$
• Outlet flow = 613.5 $\frac{L}{min}$

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 = $\frac{Volume}{Time}$

• Volume = Outlet flow * time

And replace the values to obtain the new volume pumped:

• Volume = 613.5 $\frac{L}{min}$ * 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.