Question 264824
main pump equals 4 hours.
aux pump equals 9 hours.
start at 9:00 am and want to end at 12:00 pm.
total of 3 hours.


rate * time = units


the units are 1 tank.


main pump can empty the tank in 4 hours so x*4 = 1 which means that the main pump can empty 1/4 of the tank in 1 hour.

the aux pump can empty the tank in 9 hours so x*9 = 1 which means that the aux pump can empty 1/9 of the tank in 1 hour.


the main pump starts at 900 am.


the total time allowed is 3 hours.


the main pump alone would therefore empty 3 * 1/4 = 3/4 of the tank in 3 hours.


that means that 1/4 of the tank still needs to be emptied.


the aux pump can empty 1/4 of the tank in how many hours?


the formula is rate * time = units of work.


we get 1/9 * x = 1/4 where x is the amount of time it will take the aux tank to empty 1/4 of the tank.


formula is:


1/9 * x = 1/4


multiply both sides of this equation by 9 to get:


x = 9/4 hours.


9/4 hours is the same as 2 + 1/4 hours.


subtract that from 12 pm and you get 9 + 3/4 hours is when the aux pump should be started.


90 + 3/4 hours is the same as 9:45 am.


here's what happens.


at 9:00 am the main pump starts pumping.


at 9:45 am the aux pump starts pumping.


by 12:00 pm the main pump has pumped 1/4 * 3 = 3/4 of the tank.
by 12:00 pm the aux pump has pumped 1/9 * 9/4 = 1/4 of the tank.


3/4 + 1/4 = 1 which means the tank is empty at 12:00 pm.