a resorvoir can be filled by an inlet pipe in 24 hours and emptied by an outlet pipe 28 hours. the foreman starts to fill the re

To determine the rates at which the inlet and outlet pipes fill and empty the reservoir, we remember that work done equals rate multiplied by time. Let’s denote the inlet rate as i and for the outlet pipe as 0. Therefore,
i(24) = 1
o(28) = 1
In this context, the '1' represents the total number of reservoirs, since the problem states the time needed for each pipe to either fill or empty a singular reservoir. Solving for rates yields:
i = 1/24 reservoirs/hour
o = 1/28 reservoirs/hour

Over the first six hours, the inlet pipe fills (1/24)(6) = 1/4 reservoirs and during the same period, the outlet pipe empties (1/28)(6) = 3/14 reservoirs. To calculate the net volume of the reservoir filled, we subtract the emptying total from the filling total:
1/4 - 3/14 = 1/28 reservoirs (note that if emptying exceeds filling, a negative value results. In such cases, treat that negative value as zero, indicating that the outlet rate surpasses the inlet rate, leading to an empty reservoir).
Now we need to find out how long it will take to fill up one reservoir since we’ve already partially filled 1/28 of it, after closing the outlet pipe. In simpler terms, we need to determine the time required for the inlet pipe to finish filling the remaining 27/28 of the reservoir. Fortunately, we have already established the filling rate for the inlet pipe, leading to the equation:
(1/24)t = 27/28
Solving for t gives us 23.14 hours. Remember to add the initial 6 hours to this result since the question seeks the total time. Thus, the final total is 29.14 hours.

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The rate of water flowing into the reservoir is one reservoir every 24 hours, while it empties at one reservoir every 28 hours. For the initial six hours when the outlet pipe was operational, the net filling rate of the reservoir was (1/24) - (1/28) = (7/168) - (6/168) = 1/168 reservoir per hour. When multiplied by 6 hours, this results in a total filling of 1/28 of the reservoir while the outlet pipe was active.

The remaining capacity of the reservoir will now be filled at a rate of 1/24 reservoir per hour. Therefore, we need to calculate how many sets of 1/24 portions fit into 27/28 reservoir. (27/28) = (162/168) = (23.142857/24). Consequently, the entire filling time for the reservoir totals 23.142857 plus 6 = 29.142857 hours.