Six taps can fill an empty cistern in $8$ hours. How much more time will be taken, if two taps go out of order? Assume all the taps supply water at the same rate.

Using the total work formula, the total work needed to fill the cistern is six taps multiplied by eight hours, equalling forty-eight tap-hours. If two taps go out of order, only four taps remain working. Dividing the total work of forty-eight tap-hours by the four working taps gives a new filling time of twelve hours. The increase in time is twelve hours minus the original eight hours, resulting in an extra four hours.

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