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

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       Log On


   

Question 348181: 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 ankor@dixie-net.com(22740) 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?
:
Let a = no. of hrs required by the Aux pump
:
let the completed job = 1 (An empty tank)
:
The Main pump will operate from 9 til 12: 3 hrs
:
A shared work equation:
3%2F4 + a%2F9 = 1
Multiply equation by 36, results:
9(3) + 4a = 36
27 + 4a = 36
4a = 36 - 27
4a = 9
a =9%2F4 hrs
Change this to Minutes
9%2F4 * 60 = 135 min which is 2 hrs, 15 min
;
Find the starting time: 11:60 - 2:15 = 9:45 AM start the Aux pump
:
:
Check solution
a = 9/4 = 2.25 hrs
3%2F4 + 2.25%2F9 =
3%2F4 + 1%2F4 = 1; confirms our solution