An oil tanker can be emptied by the main pump in 4 hrs. An auxiliary pump can empty the tanker in 9 hrs. If the main pump is started at 9a.m. when should the auxiliary pump be started so that the tanker is emptied by noon?

Question

anonymous · Answer

The main pump will empty 3/4 of a tanker in 3 hours, 12noon - 9a.m.
Let x be the time for the auxiliary pump to empty 1/4 of a tanker.  Solve the following for x:$$x \frac{1}{9}= \frac{1}{4}$$x=9/4 hours or 2hours 15 minutes.
12:00-2hr 15min -> 9:45 a.m.