SOLUTION: An oil tanker can be emptied by the main pump in 5hours. An auxillary pump can empty the tanker in 9 hours. If the main pump is started at 5PM, when should the auxillary pump be

Algebra ->  Rate-of-work-word-problems -> SOLUTION: An oil tanker can be emptied by the main pump in 5hours. An auxillary pump can empty the tanker in 9 hours. If the main pump is started at 5PM, when should the auxillary pump be       Log On


   

Question 338897: An oil tanker can be emptied by the main pump in 5hours. An auxillary pump can empty the tanker in 9 hours. If the main pump is started at 5PM, when should the auxillary pump be started so the tanker is emptied by 9PM?
How do I determine the "time"? I get to 1/9t=1/5 and not sure what to do.
Thank you.
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
An oil tanker can be emptied by the main pump in 5hours. An auxillary pump can empty the tanker in 9 hours. If the main pump is started at 5PM, when should the auxillary pump be started so the tanker is emptied by 9PM?
.
From 5pm to 9pm is 4 hours...
.
Let x = number of hours to run aux pump
then
4(1/5) + x(1/9) = 1
4/5 + x/9 = 1
multiplying both sides by 45:
36 + 5x = 45
5x = 9
x = 9/5 hours
.
9/5 hour is the same as
9/5 * 60 = 108 minutes
1 hour and 48 minutes (time needed to run aux pump)
.
Time to start aux pump:
9 pm minus 1 hour and 48 minutes:
7:12 PM