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

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Question 1207783: 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?

My equation:
(x/4) + (x/9) = 1

Yes?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
The pumping rates:
Main, 1%2F4 units of Tanks per Hour
Auxiliary, 1%2F9 Tanks per hour


(?)

t amount of time with only main pump
v amount of time both pumps together
t%281%2F4%29%2Bv%281%2F4%2B1%2F9%29=1

The whole amount of time for emptying the tank from 9am to noon is 3 hours.
t%2Bv=3

This would seem to give a system of equations
system%28t%2F4%2Bv%2F4%2Bv%2F9=9%2Ct%2Bv=3%29.

Simplify the first equation, take second equation and substitute for v, into the first (now simplified) equation.

9t%2B9v%2B4v=36
9t%2B13v=36
Thinking again what I said, maybe easier to use elimination method, unless I already made a mistake somewhere...

system%289t%2B13v=36%2Ct%2Bv=3%29
and question asks essentially for t.

system%289t%2B13v=36%2C13t%2B13v=39%29
E2-E1
4t=3
highlight%28t=3%2F4%29-------------------9:45 AM

Answer by ikleyn(52747) About Me  (Show Source):
You can put this solution on YOUR website!
.
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?

My equation:
(x/4) + (x/9) = 1

Yes?
~~~~~~~~~~~~~~~~~~~~~~~


        Your equation is incorrect.

        It is inapplicable to this problem.

        This problem can be solved mentally, without using any equations. 


According to the schedule, the main pump drains the oil during 3 hours (from 9 am to the noon).

During this time, it makes  3%2F4  of the job.

Hence,  1%2F4  of the job should be performed by the auxiliary pump.


The auxiliary pump has the rate of work  1%2F9  of the job per hour.


To make  1%2F4  of the job, this auxiliary pump needs  %28%281%2F4%29%29%2F%28%281%2F9%29%29 = 9%2F4 hours = 21%2F4 hours.


It means that the auxiliary pump should start working  21%2F4 hours before the noon.


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

Solved (mentally).