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Two taps can fill a cistern in 2 hours and 45 hours respectively. A third tap can empty it in 45 hours. How long (in hours) will it take to fill the empty cistern, if all of them are opened together?
2
3
4
5
2
Since Taps 2 and 3 have the same rate (45 hours) but opposite functions (filling and emptying), they cancel each other out entirely, leaving only Tap 1 to fill the cistern.
Tap 1 fills in 2 hours, Tap 2 fills in 45 hours, and Tap 3 empties it in 45 hours.
Total┬аWork=LCM(Time1тАЛ,Time2тАЛ,Time3тАЛ)
Since Taps 2 and 3 have the same rate (45 hours) but opposite functions (filling and emptying), they cancel each other out entirely, leaving only Tap 1 to fill the cistern.
Students often fail to realize the emptying tap effectively neutralizes the filling tap of the same duration, leading to complex calculations.
Calculate Work and Efficiencies
Let the total capacity of the cistern be the LCM of 2 and 45, which is 90 units. Efficiency = (Units) / (Time).
Efficiency1тАЛ=290тАЛ=45,Efficiency2тАЛ=4590тАЛ=2,Efficiency3тАЛ=тИТ4590тАЛ=тИТ2
Find Combined Efficiency
When all three taps work together, the net rate of filling is the sum of their individual efficiencies.
Net┬аEfficiency=45+2тИТ2=45
Calculate Total Time
Time taken to fill the cistern is total capacity divided by the net efficiency.
Time=4590тАЛ=2┬аhours
A is correct because the net efficiency of the two taps working for 45 hours cancels out, resulting in the cistern being filled by the first tap in 2 hours.
This logic of cancelling terms is frequently used in 'Pipe and Cistern' and 'Time and Work' problems to avoid tedious algebraic equations.
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