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Tap S alone fills a cistern in 2 hours; while tap T alone fills the same cistern in 3 hours. A new tap U is attached to the bottom of the cistern which can empty the completely filled cistern in 6 hours. Sunny started all three taps together at 9 a.m. When will the tank be full?
10:30 a.m.
12 p.m.
11:15 a.m.
9:45 a.m.
10:30 a.m.
Use LCM of (2, 3, 6) = 6 units. S does 3 units/hr, T does 2 units/hr, U does -1 unit/hr. Net rate = 3 + 2 - 1 = 4 units/hr. Total time = 6 / 4 = 1.5 hours.
Tap S = 2 hours to fill, Tap T = 3 hours to fill, Tap U = 6 hours to empty. Taps started at 9 a.m.
Rate=Time1тАЛ, NetRate=S+TтИТU
Use LCM of (2, 3, 6) = 6 units. S does 3 units/hr, T does 2 units/hr, U does -1 unit/hr. Net rate = 3 + 2 - 1 = 4 units/hr. Total time = 6 / 4 = 1.5 hours.
Forgetting that tap U is an outlet tap (negative work) or failing to add the time to the start time correctly.
Define Rates
Convert the time taken by each tap into units of work per hour. Assume total volume = 6 units (LCM of 2, 3, 6).
S=26тАЛ=3u/h,T=36тАЛ=2u/h,U=тИТ66тАЛ=тИТ1u/h
Calculate Combined Work
Add the rates of the filling taps and subtract the rate of the emptying tap.
NetRate=3+2тИТ1=4units/hr
Determine Time Taken
Divide the total capacity by the net rate to find the total time required to fill the cistern.
Time=46тАЛ=1.5hours=1hour30minutes
Find Final Time
Add the duration to the starting time of 9:00 a.m.
9:00a.m.+1h30m=10:30a.m.
A is correct because the combined net efficiency of the three taps results in the cistern filling in exactly 1 hour and 30 minutes, starting from 9 a.m.
This is fundamentally identical to Time and Work problems where workers build, and a destructor works against them; always assign a negative sign to the outlet pipe.
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