A tap fills a cistern in 7 hours. Another tap empties the full tank in 56…

Concept:

A pipe that finishes a job in t hours works at a rate of 1/t of the job per hour. When several pipes run at the same time their rates add algebraically: an inlet pipe counts as +1/t and an outlet pipe counts as -1/t. The time required for any portion of the job is then that portion divided by the net rate.

Application:

1. Filling tap: it fills the whole cistern in 7 hours, so its rate is 1/7 of the cistern per hour.

2. Emptying tap: it empties the full cistern in 56 hours, so its rate is -1/56 of the cistern per hour.

3. Both taps open together, net rate = 1/7 - 1/56 = 8/56 - 1/56 = 7/56 = 1/8 of the cistern per hour.

4. Portion that has to be filled = 1/2 of the cistern (the cistern starts empty).

5. Time = portion / net rate = (1/2) / (1/8) = (1/2) x 8 = 4 hours.

Cross-check (unit method):

• Take the capacity as LCM(7, 56) = 56 units. The filling tap then supplies 56/7 = 8 units per hour and the emptying tap removes 56/56 = 1 unit per hour.

• With both open the cistern gains 8 - 1 = 7 units per hour. Half the cistern is 28 units, so the time is 28 / 7 = 4 hours.

• Consistency check: at 7 units per hour the complete 56-unit cistern would take 8 hours, and because the net rate never changes, half of it takes exactly half that time.

Hence, with both taps open from the start, half the cistern is filled in 4 hours.