SOLUTION: an oil tanker can be emptied by the main pump in four hours. an auxilary pump can empty the tanker in 9 hours. If the main pump started at 9a.m., when should the auxiliary pump be
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Question 339203: an oil tanker can be emptied by the main pump in four hours. an auxilary pump can empty the tanker in 9 hours. If the main pump started at 9a.m., when should the auxiliary pump be started so the tanker is emptied by noon? Found 2 solutions by scott8148, Edwin McCravy:Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website! in the 3 hours between 9 and noon , the main pump will empty 3/4 of the tanker (3 hrs / 4 hrs)
this means that the auxiliary pump must empty the remaining 1/4 , it will take the pump 2 ¼ hrs to do this (9 hrs * 1/4)
so the auxiliary must start at 9:45 (2 ¼ hrs before noon)
You can put this solution on YOUR website! an oil tanker can be emptied by the main pump in four hours. an auxilary pump can empty the tanker in 9 hours. If the main pump started at 9a.m., when should the auxiliary pump be started so the tanker is emptied by noon?
The main pump's emptying rate is 1/4 tanker per hour. It will be open
the entire 3 hours, so it will empty 3/4 of the tank.
The other 1/4 of tank must be emptied by the auxiliary pump.
We need to know how long it will take the auxiliary pump to empty
the remaining 1/4 tank which the main pump cannot empty in the
3 hours.
The auxiliary pump's rate is 1/9 tanker per hour. Let x be the
number of hours it takes the auxiliary pump to empty the 1/4 tank
which the main pump will not be able to empty in the 3 hours allotted.
(rate in tankers per hour)(time required) = 1/4 tank
(1/9)x = 1/4
Multiply both sides by 36
4x = 9
x = 9/4
x = 2 1/4 hours
x = 2 hours 15 minutes
So the auxiliary tank must be started 2 hours 15 minutes before 12 noon,
So it must started at 9:45 AM
Edwin