To get a job at Company X, an applicant must be recommended by

Question

To get a job at Company X, an applicant must be recommended by three interviewers. Out of 30 applicants, 15 were recommended by the first interviewer, 17 by the second and 20 by the third interviewer. What is the least number of applicants who would have to have been recommended by all three interviewers?

A. 0
B. 2
C. 3
D. 5
E. 10

A. 0 
B. 2
C. 3
D. 5
E. 10

Bunuel · Accepted Answer

The least number of applicants that  both  the first and the second interviewer could have recommended is 15 + 17 - 30 = 2. This scenario occurs when all applicants are recommended either by the first interviewer, the second interviewer, or by both (in other words, when no applicant is recommended by neither, or None = 0). Now, these 2 applicants, who were recommended by both the first and the second interviewers, could have been overlooked by the third interviewer, who could have recommended any 20 out of the remaining 28 applicants. Therefore, the least number of applicants that must have been recommended by all three interviewers is 0.

Answer: A.

Hope it helps.

mx39 · Answer

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 Formula 
So, 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

diya7 · Answer

15+17+20 =52 -30 = 22 that could overlap , how do we know we have to allocate 2 to 15 and 17 and the rest to third category ?

Bunuel · Answer

22 is the maximum number of applicants who could have been recommended by more than one interviewer. The question, however, asks for the least number of applicants recommended by all three interviewers. For that, we first find the least number that both the first and second interviewer could have recommended, which is 15 + 17 - 30 = 2. Then, we check if the 20 applicants recommended by the third interviewer can be distributed such that they don't overlap with these 2. Since there are 30 applicants in total, this is possible: the 2 applicants recommended by both the first and second interviewers could have been overlooked by the third interviewer, who could have recommended any 20 out of the remaining 28 applicants.

KarishmaB · Answer

Max min in overlapping sets is discussed here:

Start with the largest set and put it down: 20. We have 10 remaining.
When we put down 17, we avoid overlap so 10 go to the remaining and 7 overlap with the 20 set.
When we put down 15, to avoid overlap, we put all 15 (shaded region) on the 23 where there is no overlap. Hence minimum overlap of all three is 0.

Attachment:

Screenshot 2024-06-24 at 2.35.11 PM.png

Answer (A)

Spoorthii · Answer

If the question would be - 'what is the maximum number of applicants recommended by all 3 interviewers? - Would the answer then be 15? 
Lets say we start with 20. The second number - 17 - would fall within this 20. Likewise, the third number - 15 - would fall within 17.

Does that mean 10 of them were recommended by neither?

AnkitNandwani1 · Answer

Applicants Interviewer 1 (Y/N) Interviewer 2 (Y/N) Interviewer 3 (Y/N)
1 N N Y
2 N Y N
3 N Y N
4 N Y N
5 N Y N
6 N Y Y
7 N Y Y
8 N Y Y
9 N Y Y
10 N Y Y
11 N Y Y
12 N Y Y
13 N Y Y
14 N Y Y
15 N Y Y
16 Y N N
17 Y N N
18 Y N N
19 Y N Y
20 Y N Y
21 Y N Y
22 Y N Y
23 Y N Y
24 Y N Y
25 Y Y N
26 Y Y N
27 Y Y N
28 Y Y N
29 Y Y N
30 Y Y N

slabarca · Answer

For me:
Formula: Total A + Total B + Total C - (A and B) - (B and C) - (C and A) - 2(All of three) - None of three (this does not exist) = Total interviewed

Let's replace:

Formula: 15 + 17 + 20 - 5 - 2 - 14 - 2(All of three (our "x")) - 0 = 30

If we develop, you get x = 0,5

Then, the least amount that could work for this is 0

ndwz · Answer

Bunuel , could you please link similar types of questions to practice thank you!!

Bunuel · Answer

Ghoskrp3 · Answer

I approached by pure substitution method, I saw option "0", then immediately attempted to see if I can draw these 3 sets Venn diagram like that in my rough book.

Taking First, Second & Third interviewer values as "F", "S" & "T", I approached to keep most of the value in the interceptions to minimize values and avoiding at all costs flowing into "only 1 set" category and "all 3 sets category" of the diagram. I made the below Venn diagram:

https://gmatclub.com/forum/download/file.php?id=81308

Since I was able to achieve a diagram as such, so the answer "0" couldn't be ruled out.

A. 0

siddharth_ · Answer

Say we have numbers from 1 to 30.

Minimising overlap can be imagined as - 
1st selected - 1 to 15
2nd selected - 14 to 30
3rd selected - 1 to 10 and 21 to 30

Thus there's no number selected by all 3.

Pr4n · Answer

Although while doing it during my mock I didn't get it right as I went down the approach of cases method and couldn't think of it. And the explanations offered here, I truly don't understand.

But I wanted GmatNinja overlapping sets video and that technique can help you think of cases better.

For this question,

We know that there will be no applicants that will be recommended by none.
Now each candidate can be recommended by either 1,2 or 3 of them.
Question asks us least possible number of candidates that were recommended by 3.

Let the number of candidates recommended by 1 be x, 2 be y and 3 be z

from there we can create the following table as shown by GmatNinja

Recommended by 
 Number of Candidates 
 Countings 
 
  1 
 x 
 x 
 
  2 
 y 
 2*y 
 
  3 
 z 
 3*z 
 
  Total 
 x+y+z 
 x+2y+3z

Now we know x+y+z = 30 as given in the question, and we also know the countings is 52 i.e., 15+17+20

this question now boils down to whether x+2y+3z= 52 be solved with min value of z.

From here i substituted z=0

and for x+2y = 52, and x+y= 30

I can solve as i know x and y are integers. In this case, y is 22 and x is 8. So we have our answer.

This might seem longer but considering the question asked us to minimise and we can just put values and create linear eqn with 2 variables makes it more intuitive if you can't think of cases at once.

For venn diagram method, if someone is struggling to imagine

Imagine that the interviewer who recommended 20 candidates had only done so with overlap with others. 10 - 10 with each of the other interviewers, and then you will instantly be able to see that 0 is possible.

Pratham_Patil24 · Answer

Hi Karishma,

I watched your video on maxima and minima in overlapping sets where in for minima we start with smallest number of the set, why did we start with largest number here when we needed minima?
Also how important is this topic from real test point of view?

egmat · Answer

Pratham_Patil24  Here's what I think might be happening (I'm sure  KarishmaB  will be able to elaborate further)

The confusion might be arising because there are  two different types  of "minimum" problems in overlapping sets:

Type 1: Minimum guaranteed overlap  (like finding minimum in  A \cap B )
→ Here we DO start with the smallest set
→ Example: "At least how many people must like both coffee AND tea?"

Type 2: Minimize triple overlap while satisfying all constraints  (this problem)
→ Here we start with the largest set for maximum flexibility
→ We're trying to  avoid  triple overlap, not guarantee it

Why Starting with Largest Works Here:

When we want to  minimize   |A \cap B \cap C| , we need to  maximize the spread  of recommendations. Think of it like this:

- Starting with  20  people (largest) gives us the most "canvas" to work with
- We can then strategically place the  17  to overlap with these  20  (creating double overlaps)
- Finally, we place the  15  to avoid creating any triple overlaps

If we started with the smallest ( 15 ), we'd have less flexibility to spread out the larger sets later, potentially forcing unwanted triple overlaps.

Process Diagnosis: 
Your confusion stems from applying the "minimum guaranteed intersection" strategy to a "minimize while satisfying constraints" problem. These require opposite approaches - one forces overlap, the other avoids it.

Here is a q  uick recognition pattern that might help you: 
- Question asks "must be" or "guaranteed minimum" → Start with SMALLEST set
- Question asks "least possible" with multiple constraints → Start with LARGEST set for flexibility
- The word "least" + "would have to have been" in this problem signals we want the minimum  possible , not minimum  guaranteed

Regarding GMAT Importance: 
Overlapping sets appear in approximately  5-8\%  of GMAT quant questions. While not the most frequent topic, it's definitely tested, especially at higher score levels. The good news: once you master the  2-3  main patterns (like this distinction), these become very manageable points to earn.

I do hope this clears the confusion. If there is still something not clear enough, feel free to ask!

NikunjC · Answer

Hello, I used this approach, and got the right answer, is this the correct approach though?
Using the formula I fet

30 = 15+17+20 - (Boths) - 2 (All 3) + No

This gives me

2A + B = 22, and then A = (22 - B)/2.
Since i want to minimize A I have to maximize B, and the B can be maximized to 22, which would give the A = 0, which is the smallest A can be.
Hence A = 0

Is the thinking right ??

Goldenfuture · Answer

a + 2b + 3c = 42
N-none + b + 2c= 42
b+2c - none = 12
If b = 12
then c can be 0 and so can none.

a+ 24 = 42
a = 18 
Possible

luisdicampo · Answer

Deconstructing the Question  
Total Applicants = 30.
We need to find the  minimum  intersection of three sets:
1. Recommended by I1 (15 people)
2. Recommended by I2 (17 people)
3. Recommended by I3 (20 people)

Method: The "Rejection" Approach (Fastest)  
Instead of looking at approvals, look at rejections . A candidate needs 0 rejections to get the job. To minimize the successful candidates, we maximize the spread of the rejections.

1. Rejections by Interviewer 1:  30 - 15 = 15  rejections.
2. Rejections by Interviewer 2:  30 - 17 = 13  rejections.
3. Rejections by Interviewer 3:  30 - 20 = 10  rejections.

Total Rejections available:   15 + 13 + 10 = 38 .

We have 30 candidates and 38 rejections.
Can we ensure that  every single candidate  gets at least one rejection ?
Yes, because  38 > 30 .
We can give one rejection to each of the 30 applicants, and we would still have 8 rejections left over (which would result in some applicants getting 2 rejections).

Since it is mathematically possible for every applicant to receive at least one rejection, the number of applicants receiving  three recommendations  (zero rejections) can be  0 .

Answer:  A

Adit_ · Answer

Is it also feasible to calculate 
52-30 (which is extra) = 22

2x+3y = 22 is the equation I formed since the extras are either the two overlaps or 3 overlaps

To find the lowest we need some value of y such that 2x+3y=22

Since 22 is already even we can replace y as 0 and thus thats the answer I found.

Is this a right approach?

To get a job at Company X, an applicant must be recommended by

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Recommended by	Number of Candidates	Countings
1	x	x
2	y	2*y
3	z	3*z
Total	x+y+z	x+2y+3z

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To get a job at Company X, an applicant must be recommended by

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