SOLUTION: An oil tanker can be emptied by a main pump in 4hours an auxiliary pump can empty it in 9 hrs if the main pump is started at 9 am when should the auxiliary pump be started so the t

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Question 672642: An oil tanker can be emptied by a main pump in 4hours an auxiliary pump can empty it in 9 hrs if the main pump is started at 9 am when should the auxiliary pump be started so the tanker will be empty by noon?
Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
From 9 AM to noon is 3 hrs
Let = number of hours the main pump
is used pumping by itself
= number of hours that both pumps are pumping
---------------
The main pump's rate of pumping is ( 1 tanker / 4 hrs )
the auxilliary pump's rate is ( 1 tanker / 9 hrs )
----------------
( fraction of tanker pumped in x hrs ) +
( fraction pumped by both pumps in 3-x hrs ) = 1 whole tanker pumped

Multiply both sides by

The auxiliary pump should be started at 9:45
check answer:
In 3/4 of an hour the main pump empties
of the tanker
In 2.25 hrs, both together empty

and
OK