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Two taps can fill a cistern in 2 hours and 54 hours respectively. A third tap can empty it in 54 hours. How long (in hours) will it take to fill the empty cistern, if all of them are opened together?
3
5
2
4
2
Since Tap 2 and Tap 3 operate at the exact same rate (54 hours) with opposite functions (filling and emptying), their net contribution is zero. Thus, only Tap 1 remains active, filling the tank in 2 hours.
Tap 1 fills cistern in 2 hours, Tap 2 fills cistern in 54 hours, Tap 3 empties cistern in 54 hours.
Net Work Rate=T11+T21−T31
Since Tap 2 and Tap 3 operate at the exact same rate (54 hours) with opposite functions (filling and emptying), their net contribution is zero. Thus, only Tap 1 remains active, filling the tank in 2 hours.
Students often blindly apply the LCM formula for all three taps without noticing that the second and third taps cancel each other out completely.
Define individual rates
Let the total capacity of the cistern be the LCM of 2, 54, and 54, which is 54 units. The work rates per hour are: Tap 1 = +27 units/hr, Tap 2 = +1 unit/hr, and Tap 3 = −1 unit/hr.
Rate1=27,Rate2=1,Rate3=−1
Calculate combined work rate
When all three taps are opened together, add the filling rates and subtract the emptying rate: Net Rate=27+1−1=27 units per hour.
Net Rate=27 units/hr
Compute total time required
Divide the total capacity of the cistern by the net combined rate to find the total time: Time=2754=2 hours.
Time=2754=2 hours
C is correct because the combined net work rate of all three taps fills the 54-unit cistern in exactly 2 hours.
This inverse rate problem is structurally identical to relative speed problems in Time, Speed and Distance.
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