Apologies in advance for the very long-winded answer. I think this is an interesting question worth unpacking because there's actually so much going on here: word problems, overlapping sets, even some elementary number properties. And all these concepts are generalizable to questions just like this one.
To begin, I think this question is really a three-group overlapping sets problem in disguise. While it may be worded as a min/max question, it fits very nicely into the Venn diagram and formulaic frameworks. I prefer the latter. Recall that
Total = # entities in Group 1 + # entities in Group 2 + # entities in Group 3 - (2 * # entities in all 3) - (# entities in exactly 2) + (# entities in no group). In this case, they are all assigned a group, so that's one less thing to worry about.
Step One, Model a Venn Diagram or Work with the FormulaSo, plugging these numbers in, we get the following:
\(30 = 15 + 17 + 20 - 2a - b\), where
a is the number of elements belonging to all three groups and
b is the number of elements belonging to only two.
With this in mind, we can simplify it down to \(30 = 52 - (2a + b)\).
Something should emerge here: that the number of applicants belonging to more than two groups has to equal 22. Algebraically, it means \(2a + b\) = 22. That was the overlapping sets portion.
Now comes the basic number properties and more general exam strategy portion.
Step Two, Test for Values:This question gets really interesting here because we're being asked for the
least number of individuals who are assigned to all three groups.
What should stand out to you is that b has to be a multiple of 2 (this is that knowledge of number properties coming into play), because 2a is always going to resolve to a multiple of 2. This is super important when it comes to generalizing to other min/max style overlapping sets questions like this. With this in mind, let's test for answer choices, starting with 0 because we're being asked for the least number.
Recall that \(2a + b = 22\), so if \(a\) is 0, that leaves \(b\) as 22. Is that possible? Certainly. It would just mean that nobody gets a job and 22 people are recommended by two of the interviewers. Since this works, it leaves
AC A as the correct answer.A fun thought experiment with this one is to add the caveat that at least one person is hired, or at most 20 interviewees receive two recommendations. This would mean \(2(1) + b = 20\), making the least number of applicants recommended by all three interviewers equal to 1. I mentioned above that this strategy is generalizable. Here is a very difficult Two-Part Analysis question straight from a GMAT Club mock (and very
OG-like) that tests you on this very concept:
https://gmatclub.com/forum/at-rocket-brown-elementary-school-there-are-150-students-and-three-sports-teams-hockey-tennis-and-football-rocket-brown-students-are-allowed-to-participate-in-as-many-424579.html