This problem involves calculating the time taken to fill a tank when one tap fills it and another simultaneously empties it. We need to find the net filling rate.
When both taps are open, the net filling rate is the difference between the filling rate and the emptying rate:
Net Rate = (Rate of Tap A) - (Rate of Tap B)
Net Rate = $ \frac{1}{40} \, \text{tank/min} - \frac{1}{60} \, \text{tank/min} $
To subtract these fractions, find a common denominator, which is 120:
Net Rate = $ \left( \frac{1 \times 3}{40 \times 3} \right) - \left( \frac{1 \times 2}{60 \times 2} \right) \, \text{tank/min} $
Net Rate = $ \frac{3}{120} - \frac{2}{120} \, \text{tank/min} $
Net Rate = $ \frac{1}{120} \, \text{tank/min} $
The time taken to fill the tank is the reciprocal of the net filling rate:
Time = $ \frac{1}{\text{Net Rate}} $
Time = $ \frac{1}{\frac{1}{120} \, \text{tank/min}} $
Time = $ 120 $ minutes
Therefore, the time taken to fill the tank when both taps are opened simultaneously is 120 minutes.
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A water tank can be emptied in 40 minutes by a pipe of d 'diameter, so how long will it take for a 2d diameter pipe to be emptied?
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Two pipes X and Y can fill an empty tank in 16 hours and 20 hours respectively. Pipe Z alone can empty the completely filled tank in 25 hours. Firstly both pipes X and Y are opened and after 6 hours pipe Z is also opened. What will be the total time (in hours) taken to completely fill the tank?