SOLUTION: 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 9 AM, when should the auxiliary pump be s

Algebra ->  Rate-of-work-word-problems -> SOLUTION: 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 9 AM, when should the auxiliary pump be s      Log On


   



Question 1209016: 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 9 AM, when should the auxiliary pump be started so that the tanker is emptied by noon?
Answer by ikleyn(52754) About Me  (Show Source):
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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 9 AM, when should the auxiliary pump be started
so that the tanker is emptied by noon?
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The main pump is scheduled to work 3 hours, from 9 am to noon.

In 3 hours, the main pump will make  3/4  of the job;

the remaining  1/4  part of the job should be done by the auxiliary pump.


The auxiliary pump needs  %28%281%2F4%29%29%2F%28%281%2F9%29%29 = 9%2F4 = 21%2F4  hours.

Here  1%2F9  is the rate of work of the auxiliary tank.


Hence, the auxiliary pump should start  21%2F4  hours before noon, i.e. at 9:45 am.


ANSWER.  The auxiliary pump should start  at 9:45 am.

Solved.