SOLUTION: An oil tanker can be emptied by the main pump in 5 hours. An auxilary pump can empty the tanker in 14 hours. If the main pump is started at 7pm, when should the auxilary pump be st

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Question 211038: An oil tanker can be emptied by the main pump in 5 hours. An auxilary pump can empty the tanker in 14 hours. If the main pump is started at 7pm, when should the auxilary pump be started so that the tanker is emptied by 11pm?
Found 2 solutions by ankor@dixie-net.com, Alan3354:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
An oil tanker can be emptied by the main pump in 5 hours.
An auxilary pump can empty the tanker in 14 hours.
If the main pump is started at 7pm, when should the auxilary pump be started
so that the tanker is emptied by 11pm?
:
Let t = no. of hrs to run the Aux pump
:
The main pump will run 4 hr (7 - 11 pm)
:
Let the completed job = 1; (an empty Tanker)
:
The shared work equation
+ = 1
Multiply equation by 70 to get rid of the equation, results:
14(4) + 5t = 70
56 + 5t = 70
5t = 70 - 56
5t = 14
t =
t = 2.8 hrs to run the aux pump
:
2.8 hr = 2 + .8(60) = 2 hrs 48 min
:
Subtract 2:48 from 11:00 = 8:12 PM start the aux pump
;
;
Check solution
4/5 + 2.8/14 =
.8 + .2 = 1

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
To empty is "by 1100", just run them both.
They'll do the job in 70/19 hours, = ~3 hrs 41 minutes.
They'll be done at 10:41, "by 1100."
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If you meant "at 1100", that's a different problem.