A dinner costs $100 and will be shared by 7 employees. Fred shared $48

I think on case 2 from the values would be (5 ,6 , 7, 8, 9).

So you choose 10 as the lowest option that Louis would pay in which the other five employees will pay lower than her, but the highest amounts possible (5 ,6 , 7, 8, 9). Once you add everything up you can see option I is out of the window, since 5+6+7+8+9+ 10= $45; thus option I is not a possible answer since the highest possible numbers do not add up to at least $52.

Constraints:
1. Lois will pay less than $48 (as he cannot pay more than Fred)
2. His amount has to be $2 or more
3. His amount has to be the greatest among the other people

In Case 2, we are finding a solution where Lois follows the constraints and others pay the highest amount possible individually (one of the extremes, we have discussed the other extreme in Case 1), Lois pays the lowest amount possible.

After taking the initial set of 2, 3, 4, 5, and 6, we can simply remove the lowest integer and add one integer.
3,4,5,6,7 - $25
4,5,6,7,8 - $30
5,6,7,8,9 - $35
6,7,8,9,10 - $40
7,8,9,10,11 - $45

If we check the trend, there is a $5 increase in every set. You can just do 2 or 3 sets to see the pattern during the exam. Continuing this pattern, we can see that the amount cannot be less than $12 (As we are making others pay the highest possible amounts as well).

Hope this helps.