SOLUTION: An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 7 hours. If the main pump is started at 8PM, when should the auxiliary pump be s
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-> SOLUTION: An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 7 hours. If the main pump is started at 8PM, when should the auxiliary pump be s
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Question 885922: An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 7 hours. If the main pump is started at 8PM, when should the auxiliary pump be started so that the tanker is emptied by 11PM? Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! The rate of emptying for main pump is: which is ( 1 tanker / 4 hrs )
The rate of emptying for auxiliary pump is: which is ( 1 tanker / 7 hrs )
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8 PM to 11 PM is hrs
Let = time in hrs during which main pump is on = time in hrs during which both pumps are on
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For main pump alone:
fraction of tanker emptied = = fraction left to be emptied
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With both pumps on:
Multiply both sides by
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The auxiliary pump should be started at:
8 PM + 1.25 hrs = 9:15 PM
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check:
OK
