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1. Let there be W work. Jake, John, and Jordyn work at speeds \(\frac{W}{16}\), \(\frac{W}{48}\), and \(\frac{W}{96}\), respectively.

2. The key thing to notice is that for every 3 hours, each person must've worked exactly 1 hour. During that time \(\frac{W}{16} + \frac{W}{48} + \frac{W}{96} = \frac{6W + 2W + W}{96} = \frac{3W}{32}\) work was done.

3. After 30 hours \(\frac{3W}{32} * 10 = \frac{30W}{32}\) work was done and \(W - \frac{30W}{32} = \frac{2W}{32} = \frac{W}{16}\) work is left.

4. Since \(\frac{W}{16}\) is less than \(\frac{3W}{32}\), the time that will be needed to do the work depends on the order of the workers.

5. The job will take the minimum amount of time if we choose the faster workers first. It will take 1 extra hour for Jake to make a total of 31.

6. Likewise, the job will take the maximum amount of time if we choose the slower workers first. It will take \(2 + \frac{\frac{W}{16} - \frac{W}{48} * 1 - \frac{W}{96} * 1}{\frac{W}{16}} = 2 + (1 - \frac{\frac{3}{96}}{\frac{1}{16}}) = 2 + (1 - 0.5) = 2.5\) extra hours with Jordyn, then John, and then Jake working to make a total of 32.5.

7. Our answer will be: Maximum Hours - 32.5 and Minimum Hours - 31.
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