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Two taps can fill a cistern in 2 hours and 53 hours respectively. A third tap can empty it in 53 hours. How long (in hours) will it take to fill the empty cistern, if all of them are opened together?
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5 2024/12/20-TEAMMC Q APPLICATION
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Since the rate of filling by the second tap (1/53) exactly cancels the rate of emptying by the third tap (1/53), the effective rate is just the rate of the first tap (1/2).
Tap 1 fills in 2 hours. Tap 2 fills in 53 hours. Tap 3 empties in 53 hours.
Total Time=T11+T21−T311
Since the rate of filling by the second tap (1/53) exactly cancels the rate of emptying by the third tap (1/53), the effective rate is just the rate of the first tap (1/2).
Students often get confused by the specific fraction '53' and attempt complex arithmetic, failing to notice that the second and third taps neutralize each other's work.
Define rates
Let the rates be R1=1/2, R2=1/53, and R3=−1/53 (negative because it empties).
Rnet=21+531−531
Simplify net rate
The terms 1/53 and −1/53 cancel out, leaving the net rate equal to the rate of the first tap alone.
Rnet=21
Calculate time
Since Rnet=1/2, the time taken to fill the cistern is the reciprocal of the net rate.
Time=1/21=2 hours
A is correct because the filling rate of the second tap and the emptying rate of the third tap negate each other, leaving only the first tap to fill the cistern in 2 hours.
This concept is a classic example of 'Work Cancellation' which frequently appears in speed/time/distance problems involving headwinds or water currents.
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