MATH SOLVE

5 months ago

Q:
# 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?

Accepted Solution

A:

Answer:Auxiliary pump should be started at 6:36 PMStep-by-step explanation:Let the volume of oil tanker be V.An oil tanker can be emptied by the main pump in 5 hours, [tex]\texttt{Rate of main pump = }\frac{V}{5}[/tex]An auxiliary pump can empty the tanker in 12 hours, [tex]\texttt{Rate of auxiliary pump = }\frac{V}{12}[/tex]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 hoursThat is main pump works for 4 hours and let auxiliary pump works for t hours.We have [tex]\frac{V}{5}\times 4+\frac{V}{12}\times t=V\\\\t=12\times \left ( 1-\frac{4}{5}\right )=12\times \frac{1}{5}=2.4hours[/tex]So auxiliary pump works for 2.4 hours, It should be started at 9 - 2.4 = 6.6 PM = 6:36 PMAuxiliary pump should be started at 6:36 PM