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Two pipes A and B can fill the tank in 40 and 60 minutes, respectively. Both the pipes are opened together. After how many minutes should the pipe B be turned off, so that the tank be fill in 30 minutes?
15 min
25 min
20 min
28 min
15 min
Calculate LCM(40, 60) = 120 units. Efficiencies: A = 3 units/min, B = 2 units/min. A works all 30 mins (90 units). Remaining 30 units (120-90) are filled by B. Time for B = 30 / 2 = 15 minutes.
Pipe A fills in 40 minutes, Pipe B fills in 60 minutes. Total time to fill tank is 30 minutes. Pipe B is turned off.
40xтАЛ+60xтАЛ=1 where x is time B is on, or using work efficiency: Work=Efficiency├ЧTime
Calculate LCM(40, 60) = 120 units. Efficiencies: A = 3 units/min, B = 2 units/min. A works all 30 mins (90 units). Remaining 30 units (120-90) are filled by B. Time for B = 30 / 2 = 15 minutes.
Students often solve for the time A is open alone instead of identifying the time B was active before being shut off.
Calculate Efficiency
Assume total tank capacity is LCM of 40 and 60, which is 120 units. Efficiency of A is 120/40=3 units/min and efficiency of B is 120/60=2 units/min.
EffAтАЛ=3,EffBтАЛ=2
Calculate work done by A
Pipe A remains open for the entire duration of 30 minutes. The total work done by A is 3├Ч30=90 units.
WAтАЛ=3├Ч30=90
Calculate remaining work for B
The remaining capacity to be filled by pipe B is 120тИТ90=30 units.
WremтАЛ=30
Find time B was open
Pipe B fills at a rate of 2 units/min. The time taken to fill 30 units is 30/2=15 minutes.
tBтАЛ=230тАЛ=15
A is correct because pipe B needs to work for 15 minutes to fill the remaining portion of the tank.
This LCM approach is universal for all 'Time and Work' problems, including those with negative efficiency (leaking pipes).
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