SOLUTION: An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 7 hours. If the main pump is started at 7 pm, 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 7 hours. If the main pump is started at 7 pm, when should the auxiliary pump be       Log On


   

Question 413815: An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 7 hours. If the main pump is started at 7 pm, when should the auxiliary pump be started so that the tanker is emptied by 10 pm?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The tanker is to be emptied in 2 stages in 3 hrs.
Let t = hours that main pump alone is working
Amount pumped by main pump =
(1) %281%2F4%29%2At+=+t%2F4
That is: (1 tank)/(4 hrs) x (t hrs)
The auxilliary pump + main pump will then pump out:
(2)%281%2F4%29%2A%283+-+t%29+%2B+%281%2F7%29%2A%283+-+t%29
If I add (1) and (2), I should get 1 (1 tank pumped)
t%2F4+%2B+%283+-+t%29%2F4+%2B+%283+-+t%29%2F7+=+1
Multiply both sides by 4%2A7
7t+%2B+7%2A%283+-+t%29+%2B+4%2A%283+-+t%29++=+28
7t+%2B+%283+-+t%29%2A11+=+28
7t+%2B+33+-+11t+=+28
4t+=+5
t+=+5%2F4 hrs
So, the main pump runs alone for 1 hr 15 min. That means
the auxilliary pump comes on at 8:15
check answer:
t%2F4+%2B+%283+-+t%29%2F4+%2B+%283+-+t%29%2F7+=+1
%285%2F4%29%2F4+%2B+%283+-+%285%2F4%29%29%2F4+%2B+%283+-+%285%2F4%29%29%2F7+=+1
5%2F16+%2B+7%2F16+%2B+4%2F16+=+1
+1+=+1
OK