Hello everyone, I hope you are all having a wonderful day.

I have a question regarding Question 13 in the Data Analysis practice from ETS’s Official Quant Reasoning book. The question and answer they provide is as follows:

Currently, I do not understand how they got the answer of B, C, D, E, and F. The way I interpreted the question was, can x in this situation not simply be zero? There is not indication that the sets have to have any intersection, and therefore, set B could have as few as 3 unique values (if all 50 values of A are contained in B), or as many as 53. I’m hoping someone could help explain, conceptually, why I am missing the ball here.

Have a great rest of your day!

I think you’re conflating multiple things here. You claim that x could be 0, so then “the members in set B that are not in set A” has to be 53 (which is not wrong). It’s just one example, so it does give you one of the possible answers.

However, I don’t follow how that then leads to:

therefore, set B could have as few as 3 unique valif all 50 values of A are contained in B), or as many as 53.

Anyway, this would imply x = 50, which clearly doesn’t work because set A (alone: without any intersection with B) would then have 0 members. This contradicts the “at least 2 of the members in set A are not in set B” because again every member of set A is also a member of set B.

As for how I would do it:

We know that |A| = 50 and |B| = 53. Also, we’re given that |A| - |A \cap B| \geq 2, and so we obtain |A \cap B| \leq 48

Our task is to find the range for |B| - |A \cap B|, and we can do this with the help of the above inequality.

53 - 48 = 5 \leq |B| - |A \cap B| \leq |B| = 53