SOLUTION: tank has three fill pipes and one drain pipe that are all controlled by levers located on a control board. The lenghts of time necessary to fill or empty the tank while working alo

Question 186724: tank has three fill pipes and one drain pipe that are all controlled by levers located on a control board. The lenghts of time necessary to fill or empty the tank while working alone are listed below.P1 is started at noon and allowed to run for 10 minutes at which time p2 and p4 are opened. After some time has passed the operator realizes that he opened the drain pipe by mistake.
In an attempt to rectify the situation, he accidently closes p2 and opens p3.
The tank then fills in 15 minutes. At what time of the day the tank become full?
FIll Drain
Pipes p1 P2 P3 P4
Time(min) 20 60 90 30
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
tank has three fill pipes and one drain pipe that are all controlled by levers
located on a control board. The lenghts of time necessary to fill or empty the
tank while working alone are listed below.P1 is started at noon and allowed to
run for 10 minutes at which time p2 and p4 are opened. After some time has
passed the operator realizes that he opened the drain pipe by mistake.
In an attempt to rectify the situation, he accidently closes p2 and opens p3.
The tank then fills in 15 minutes. At what time of the day the tank become full?
FIll Drain
Pipes p1 P2 P3 P4
Time(min) 20 60 90 30
:
Let t = "some time passed"
:
P1 time = 10 + t + 15
P1 time = (t+25)
:
P2 time = t
:
P3 time = 15
:
P4 time = t + 15
:
Let the full tank = 1
:
+ + - = 1
reduce the fractions that we can
+ + - = 1
Multiply equation by 60, results:
3(t+25) + t + 10 - 2(t+15) = 60
3t + 75 + t + 10 - 2t - 30 = 60
3t + t - 2t + 75 + 10 - 30 = 60
2t + 55 = 60
2t = 60 - 55
2t = 5
t = 2.5 min is "some time has passed"
therefore total time since noon:
:
10 + 2.5 + 15 = 27.5 min, tank is full at 12:27:30
:
:
check solution on a calc; enter:
+ + - = 1

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