A tap fills a cistern in 14 hours. Another tap empties the full tank in 18 hours. How long (in hours) will it take to fill the tank completely, if the tank is empty and both the taps are open together?
This question involves calculating the combined time taken to fill a cistern when one tap is filling it and another tap is simultaneously emptying it. We need to determine the net rate at which the cistern is filled.
First, let's determine the rate at which each tap works. The rate is the portion of the cistern that the tap can fill or empty in one hour.
When both taps are open, the filling tap adds water, and the emptying tap removes water. The net rate of filling is the difference between the filling rate and the emptying rate.
Net Filling Rate = (Rate of Filling Tap) - (Rate of Emptying Tap)
Net Filling Rate = $\frac{1}{14} - \frac{1}{18}$
To subtract these fractions, we find a common denominator. The least common multiple (LCM) of 14 and 18 is 126.
Net Filling Rate = $\frac{1 \times 9}{14 \times 9} - \frac{1 \times 7}{18 \times 7}$
Net Filling Rate = $\frac{9}{126} - \frac{7}{126}$
Net Filling Rate = $\frac{9 - 7}{126}$
Net Filling Rate = $\frac{2}{126}$
Simplifying the fraction:
Net Filling Rate = $\frac{1}{63}$
This means that when both taps are open, $\frac{1}{63}$ of the cistern is filled every hour.
To find the total time it takes to fill the entire cistern (1 whole cistern), we take the reciprocal of the net filling rate.
Time = $\frac{\text{Total Work}}{\text{Net Filling Rate}}$
Time = $\frac{1}{\frac{1}{63}}$
Time = $63$ hours
Therefore, it will take 63 hours to fill the tank completely when both taps are open.
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