SOLUTION: please help an oil tanker can be emptied by the main pump in 4 hours. an auxiliary pump can empty the tanker in 9 hours. if the main pump is started at 9am , when should the auxil

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Question 264824: please help
an oil tanker can be emptied by the main pump in 4 hours. an auxiliary pump can empty the tanker in 9 hours. if the main pump is started at 9am , when should the auxiliary pump be started so that the tanker is emptied by noon?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
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.