Total gallons of sewage water pumped is 3300 gallons
Solution:
Given that One pump pumps 60 gallons per minute
The second pumps 50 gallons per minute
To find: total gallons pumped
Total time taken by both the pumps to empty tank = 30 minutes
Rate of pumping the tank by first pump = x = 60 gallons per minute
Rate of pumping the tank by second pump = y = 50 gallons per minute
From given question,
total rate of pumping the sewage tank with first and second pump = x + y
x + y = 60 + 50 = 110 gallons per minute
They together pumped 110 gallons per minute
Given that they pumped for 30 minutes
So for 30 minutes, total gallons pumped is given as:
⇒ 110 x 30 = 3300 gallons
Therefore, total gallons of sewage water pumped is 3300 gallons
