Two pumps are used to empty a sewage tank. one pump pumps 60 gallons per minute (gpm) and the second pumps 50 gpm. they pumped for 30 minutes. how many total gallons are pumped?

The total gallons of sewage water pumped is 3300 gallons .

Step-by-step explanation:

Given as :

Two pumps are used to empty a sewage tank.

The rate of pumping the tank by first pump = x = 60 gallons per minute

The rate of pumping the tank by second pump = y = 50 gallons per minute

Total time taken by both the pumps to empty tank = t = 30 minutes

Let The total gallons of sewage water pumped = z gallons

Now, According to question

Total rate of pumping the sewage tank with first and second pump = x + y

Or, Total rate of pumping the sewage tank with first and second pump = 60 gpm + 50 gpm

Or, Total rate of pumping the sewage tank with first and second pump = 110 gallons per min

Now, Total rate of pumping the sewage tank =

i.e total gallons of sewage water pumped = rate of pumping the sewage tank ×  total time taken

or, z = 110 gallons per min × t

Or, z =  110 gallons per min × 30 min

Or, z = 3300 gallons

So, The total gallons of sewage water pumped = z = 3300 gallons

Hence, The total gallons of sewage water pumped is 3300 gallons . Answer

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the answer is 11.2 (i know i’m late)