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Pipes A and B can fill a tank in 7 hours and 10 hours, respectively, and pipe C can empty the full tank in 14 hours. All three pipes are opened together, but pipe A is closed after 4.5 hours. In how many hours will the remaining part of the tank be filled?
16
24
12
8
8
Use the LCM of 7, 10, 14 which is 70. Find the efficiency of each pipe (A=10, B=7, C=-5), calculate work done in 4.5 hours, and solve for the remaining time.
Pipes A and B fill the tank in 7 and 10 hours. Pipe C empties it in 14 hours. All three are open for 4.5 hours, then A is closed.
Total┬аCapacity=LCM(7,10,14)=70units
Use the LCM of 7, 10, 14 which is 70. Find the efficiency of each pipe (A=10, B=7, C=-5), calculate work done in 4.5 hours, and solve for the remaining time.
Subtracting the efficiency of pipe C (emptying pipe) instead of adding it, or forgetting to subtract the work done in the first 4.5 hours from the total capacity.
Calculate Efficiencies
Assuming total tank capacity is the LCM of 7, 10, and 14, which is 70 units. Efficiency of A = 70/7=10, B = 70/10=7, C = тИТ70/14=тИТ5.
EfficiencyA+B+CтАЛ=10+7тИТ5=12┬аunits/hr
Calculate Work for first 4.5 hours
The three pipes work together for 4.5 hours. Total work done = Rate├ЧTime.
Work=12├Ч4.5=54┬аunits
Calculate remaining work
Subtract the completed work from the total capacity to find the remaining volume.
Remaining=70тИТ54=16┬аunits
Calculate remaining time
After pipe A is closed, only pipes B and C are open. Their combined efficiency is 7тИТ5=2 units/hr.
Time=216тАЛ=8┬аhours
D is correct because the remaining 16 units of the tank are filled by pipes B and C at a net rate of 2 units/hr, taking 8 hours.
This concept is identical to standard work problems where workers join or leave a project, just replace 'workers' with 'pipes' and 'negative work' for drainage pipes.