Problema Solution
An oil tanker can be emptied by the main pump in 5 hours. An auxiliary pump can empty the tanker in 11 hours. If the main pump is started at 7 pm, when should the auxiliary pump be started so that the tanker is emptied by 11pm?
Answer provided by our tutors
The rate of the main pump is: v1 = 1/5 tanker per hour
The rate of the auxiliary pump is: v2 = 1/11 tanker per hour
The time of the main pump is: t1 = 11 - 7 = 4 hr.
Lets denote the time that the auxiliary pump is working by t2.
v1*t1 + v2*t2 = 1
4*(1/5) + t2*(1/11) = 1
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click here to see the equation solved for t2
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t2 = 2.2 hr
t2 = 2 hr 0.2*60 min
t2 = 2 hr 12 min
11 pm - 2 hr 12 min = 8:48 pm
The auxiliary pump should be started at 8:48 pm.