Three interviewers, A, B, and C are interviewing 40

Answer is A = 0

If A admitted 15 are overlapping with B admission of 17 But C doesnot overlap with anybody.
Then no student will get nod from all the 3.
Hence 0 student will get admission.

i thought it was straight 12.
applicant interview by A + applicant interview by B + applicant interview by C - total.
15+17+20-40=52-40=12

Did i miss something?

Isnt the ques saying that applicant has to get admission from all the three to get admission.
"Only with three interviewers' admission can an applicant be admitted."

Can anyone please post the graphical explanation as a solution for this problem ?

'A' is the answer.

Question asks the LEAST number of applicants admitted. And, an applicant only be admitted when all three interviewers gave admission.

Interviewer C admitted- 20
Interviewer B admitted- 17
Interviewer A admitted- 15

Suppose 'B' admitted 17 candidates other than 'C' admitted. And then 'A' admitted remaining '3' (40-20-17=3) and any '12' from either 'C' or 'B'. In this case not even a single candidate is selected by all three, making the answer to be 'A'

VeritasKarishma Bunuel Could you please tell me what's wrong with my approach? I do understand Karishma's explanation here: https://gmatclub.com/forum/three-interv ... l#p1456069. I noticed that this Q is similar to 2 other Q (one of them is quoted below). Even though they are similar, approach isn't the same for both since my derived formula works for one but not the other. Could you please help me understand the difference and how to tackle such problems? Is there any common strategy/formula? Thanks!

Min (A & B & C) = (A + B + C) % Total + None
No constraint on None => Take None = 0
(15+17+20) % 40 = 52 % 40 = 12

Do you think my approach (notes posted below) make sense?

My approach

• To minimize the all item overlapping group, distribute the items as fairly as possible. Think of distributing chocolates to kids. You want to give first to everyone and then the second to everyone and so on. Take None=0 unless there's a min constraint on it.
• (Divisor: Interviewers; Dividend: Students) https://gmatclub.com/forum/three-interv ... 18918.html
○ We don't have L1 sets so can't use the formula mentioned below.
• (Divisor: Households; Dividend: Electronics) https://gmatclub.com/forum/in-a-village ... 57-40.html
○ Let "at least 1" => L1
○ 3 Overlapping Sets: Min (L1A & L1B & L1C) = (L1A + L1B + L1C) % Total Divisor + None
○ 3 Overlapping Sets: Min (A & B & C) = (A + B + C) % Total + None
• 2 Overlapping Sets: Min (A & B) = (A + B) % Total + None https://gmatclub.com/forum/a-group-of-p ... 04442.html

Similar Q with 3 Sets:

https://gmatclub.com/forum/in-a-village ... l#p2312717

Similar Q with 2 Sets:

https://gmatclub.com/forum/a-group-of-p ... l#p2365267

To get admission you must part of all three sets: A, B, and C

“At least how many applicants get admission” is essentially asking what is the MINIMUM Number of people, out of the 40 unique people, who MUST be part of all 3 sets.


40 unique people


There are: 15 + 17 + 20 = 52 admissions from interviewers


(1st) let all 40 of the unique people get one admission

52 - 40 = 12 admissions left to apportion

And

right now at this point: All 40 people have only 1 admission from exactly one of the interviewers ——> A or B or C


(2nd) we can pick 12 more people to get another “admission” such that:


12 people = have exactly 2 admissions

(Takes care of 24 admissions)

And

28 people = have exactly 1 admission

(Takes care of 28 admissions)

24 + 28 = all 52 admissions that were given by the interviewers A, B, C

Thus:
No one MUST be part of all 3 sets

0




Posted from my mobile device

Let's solve the question by completing a diagram with three overlapping circles.

There are 40 applicants in total
15 + 17 + 20 = 52
52 - 40 = 12
So, to minimize the number of applicants approved by all three interviewers, we need to place 12 applicants in areas where two circles overlap.

Here's one possible scenario:


Since it's possible to have a scenario in which 0 applicants gain admission (i.e., 0 applicants are in the intersection of all three circles), the correct answer is A

Let a be sum of ones

Let b be sum of twos

And c sum of threes.


a + 2b + 3c = 52
a + b + c = 40

Eliminating a gives

b + 2c = 12

To minimize c, we maximize b

Gives c to be 0

Posted from my mobile device

GIVEN:

• Total number of applicants = 40
• Interviewers A, B, and C selected 15, 17, and 20 applicants, respectively.
• An applicant gets admitted only when he is selected by ALL three interviewers.

TO FIND:

• Minimum number of applicants to get admission.
  
  • That is, the minimum number of applicants who are selected by all three interviewers.

SOLUTION:
We will solve this question in some neatly laid out steps. Just follow each section and make sure you understand every bit of the solution. Let’s go!

Define variables:
Let’s assign some variables first that will denote the various possibilities of selection:

• Selected by ALL three:
  
  • Say w = number of applicants selected by all three interviewers
• Selected by EXACTLY two:
  
  • x = the number of applicants selected by only A and B.
  • Y = the number of applicants selected by only A and C.
  • z = the number of applicants selected by only B and C.
• Selected by NONE:
  
  • N = the number of applicants selected by none of the three interviewers.
  
  Note: Since nothing has been said about each applicant being selected by at least one interviewer, there is a possibility that some of the applicants are selected by none of the interviewers.

Represent on Venn diagram:
Now, we’ll draw a Venn diagram that will clearly show the scenario at play here. We will use the variables as we defined above. Here goes:


Despite populating all we know, there still are some empty regions in our Venn diagram. But it’s no big deal. Using what the diagram already has, we can easily find the values of the empty regions as well. It’s always a good idea to understand all possible regions!
So, here’s what we can infer:

• Number of applicants selected by only A = 15 – x – y – w. (This is the part of circle A that does not meet any other circle.)
• Number of applicants selected by only B = 17 – x – z – w. (This is the part of circle A that does not meet any other circle.)
• Number of applicants selected by only C = 20 – y – z – w. (This is the part of circle A that does not meet any other circle.)

And done! Now, it’s time for the main question asked – the minimum possible value of ‘w’.

Finding Minimum possible ‘w’:
Looking at the Venn diagram, we can say that:
Total applicants = Selected by only A + Selected by only B + Selected by only C + Selected by only A and B + Selected by only A and C + Selected by only B and C + Selected by all A, B and C + Selected by None ----(I)

Rewriting the equation using all our variables, we get:

• 40 = (15 – x – y – w) + (17 – x – z – w) + (20 – y – z – w) + x + y + z + w + N
  ⇒ 40 = 52 – 2w – x – y – z + N
  ⇒ 2w = 12 - (x + y + z) + N ----(II)

Observe that to minimize ‘w’, we need to maximize (x + y + z) and minimize (N). Why?

- Because the larger a value we subtract from 12, the smaller the remaining difference will be, leading to a smaller ‘w’.
- And the smaller a value we add to 12, the smaller the resulting sum will be, leading to a smaller ‘w’.

PART 1: Minimize N
At the least, N can take a minimum value of ZERO. This is when each of the 40 applicants were selected by at least one of the three interviewers.
In this case, (II) becomes:

- 2w = 12 - (x + y + z) ----(III)

PART 2: Maximize (x + y + z)
Note that ‘w’, being a certain number of applicants, cannot be negative. Hence, the maximum that we can subtract from 12 is 12 itself. That is maximum (x + y + z) = 12.
Hence, from (III), we get 2w = 12 – 12 = 0.


This implies that the minimum possible value for w is ZERO.


Correct Answer: Option A


Hope this helps!


Best,
Aditi Gupta
Quant expert, e-GMAT

Hi chetan2u sir,

Can you please share your approach to this question?

Thanks in advance.

ChandlerBong

We are looking for the worst scenario where none or the least of the person are selected by all three.

More than maths, it is logic based.
For maths, you can make three circles with overlap and fit the numbers to ensure minimum in the area overlapped by all three circles.

Pure logic
Let the applicants be named 1 to 40.
A: Selects 1 to 15
B: Doesn’t select 1 to 15, but 16 to 32.
Now, there is no overlap between A and B, so surely irrespective of whom C selects, there will be none who would have been selected by all three.

So 0 is the answer.

That's a great approach, it didn't come to me initially. Thanks a lot, chetan2u for the clarity!

­I think your analogy is wrong 

region of overlap = 52-40 =12 

this 12 include 2 intersections and three intersections since it is asking min =0

40=52-II+III
III=12-II

III_Min=0 (possible0