Explanation

Use the Double Matrix Method.
This technique can be used for most questions featuring a population in which each member has two characteristics associated with it (aka overlapping sets questions).
Here, we have a population of numbers, and the two characteristics are:
- In set A or NOT in set A
- In set B or NOT in set B

With the given information, we can set up our Matrix as follows:

Our goal is to determine which values can go in the bottom left corner (marked with a star)

Let's start by seeing what happens if we minimize the number of values that are in the TOP RIGHT corner (the number of values that are in set A BUT NOT in set B)
Since at least 2 values must be in this box, let's see what happens when there are 2 numbers there.
We get:

In this case, there are 5 numbers that are in set B but not in set A (bottom left box)


Now let's see what happens when we place a 3 in the box representing the number of values that are in set A BUT NOT in set B (i.e., the TOP RIGHT corner box):

In this case, there are 6 numbers that are in set B but not in set A (bottom left box)


Now let's see what happens when we place a 4 in the box representing the number of values that are in set A BUT NOT in set B (i.e., the TOP RIGHT corner box):

In this case, there are 7 numbers that are in set B but not in set A (bottom left box)


Now let's see what happens when we MAXIMIZE the box representing the number of values that are in set A BUT NOT in set B (i.e., the TOP RIGHT corner box):
The biggest value that can go here is 50, since numbers in the top row of boxes must add to 50.

In this case, there are 53 numbers that are in set B but not in set A (bottom left box)

So, the possible numbers that can be in the bottom left box range from 5 to 53 INCLUSIVE

Answer: B, C, D, E, F

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another slightly different approach with venn diagram.

clearly from the given conditions, A-B must be greater than or equal to 2. as such, their intersection must be less than or equal to 48. consequently B-A must be greater than 5 and any value satisfying this interval is correct. needless to say, the upper limit is 53, since n(B)=53.

It seems like one could also use a hybrid approach of the ones shown above, see attached (but maybe not).

We would start with A not B is >=2. Given A is 50, it would follow that A and B must be <=48.

Then, since we know A and B is <=48, and it's given that B is 53, then B not A must be be >=5.

It seems to make sense.

Had to read this like 3 times to understand the wording...my brain just refused to understand, at first.

Also, took a different approach because I kinda had to visualize this to make sense of what it was saying with the "not A" and the "not B". Here's how I did this.

If at least 2 are in ONLY A, then 48 are in BOTH, and 5 are ONLY B...I did a quick table to visualize

ONLY A BOTH ONLY B
2 48 53-48=5
3 47 53-47=6
... ... ...
50 0 53-0=53

So, B would have between 5 and 53 members not in A (or only in B).

This is a tricky question to grasp, indeed. Needs a lot of reasoning and inference than mathematical calculation.

Set A: 50 members
Set B: 53 members

At least 2 members in Set A but not in Set B. That means 2 to 50 members in A are not in B. The remaining members would be both in A and B.
So (50-2) to (50-50) are common members in A and B. (48 to 0)

If 0 to 48 members in B are in A, then (53-0) to (53 - 48) members will be in B but NOT in A. So the range is 53 to 5.

5 to 53 members can be in Set B but not in A.
Only 3 from the options can be excluded from this range.

What about the case where no numbers in A and B overlap?

For example, Set A could contain numbers 1-50 and set B could contain numbers 51-103 for all we know. It fulfills the requirement "At least 2 of the members in set A are not in set B" because in this case all 50 numbers in set A are not in B. In that case, "number of members in set B that are not in set A" are numbers 51-103 so it should be A,B,C,D,E,F.

Official answer not including A but don't understand why since it is possible for B to have no intersecting numbers from A.