Question

An oil tanker can be emptied by the main pump in 5 hours. An auxiliary pump can empty the tanker in 12 hours. If the main pump is started at 5 PM​, when should the auxiliary pump be started so that the tanker is emptied by 9 PM​?

Answer 1

Auxiliary pump should be started at 6:36 PM

Explanation:

Let the volume of oil tanker be V.

An oil tanker can be emptied by the main pump in 5 hours,

`\texttt{Rate of main pump = }(V)/(5)`

An auxiliary pump can empty the tanker in 12 hours,

`\texttt{Rate of auxiliary pump = }(V)/(12)`

We need to find when should the auxiliary pump be started so that the tanker is emptied by 9 PM if the main pump is started at 5 PM​.

Difference in times = 4 hours

That is main pump works for 4 hours and let auxiliary pump works for t hours.

We have

`(V)/(5)* 4+(V)/(12)* t=V\\\\t=12* \left ( 1-(4)/(5)\right )=12* (1)/(5)=2.4hours`

So auxiliary pump works for 2.4 hours,

It should be started at 9 - 2.4 = 6.6 PM = 6:36 PM

Auxiliary pump should be started at 6:36 PM