The answer is 6.

The total number of members = 59 distributed into --- 22(poetry), 27(history), and 28 (writing).
Now total the number of distributed members = 77(this number includes the number of students counted twice and counted thrice.
Out of 77 deduct 6 for members who have enrolled in two clubs = 71 now deduct 59 (for actual number of members)= 12 = members counted thrice.
But wait, this number includes members who have been counted thrice (As 6 into 2 gives you 12), the answer is 6.
Hence, C.

Hi,

Let e1 be the number of students who signed up for exactly 1 club
Let e2 be the number of students who signed up for exactly 2 clubs
Let e3 be the number of students who signed up for exactly 3 clubs
Now, e1+e2+e3 = 59
e1+2e2+3e3 = 22+27+28 = 77
Given that e2 =6
Subtract one equation from the other, we get e2+2e3 = 18 =>6+2e3 = 18 =>e3 = 6

Hence, C

Answer is 2

@ Frankenstine: The answer is 2 not 6..
6 counts remain in the end, but they are 2 members counted thrice!
Correct me if I am wrong!

I made a mistake, my bad.
Edited the first post!

winniethepooh wrote:The answer is 2.

The total number of members = 59 distributed into --- 22(poetry), 27(history), and 28 (writing).
Now total the number of distributed members = 77(this number includes the number of students counted twice and counted thrice.
Out of 77 deduct 12 for members who have enrolled in two clubs = 65 now deduct 59 (for actual number of members)= 6 = members counted thrice.
But weight this number includes members who have been counted thrice (As 3 into 2 gives you 6), the answer is 2.
Hence, A.

Hi,
You are subtracting all the members instead of subtracting the ones choosing exactly one club(counting once). Think about it!

Thanks a lot Frank!

Always prepare a venn diagram for such problems. Makes it easier to understand.
Here is the venn diagram of ur problem.
Poetry, History and Writing abbreviated as P, H and W respectively

As per formula in a 3 circle venn diagram

Total number of students=P+H+W-{(P Intersection H)+(P Intersection W)+(H Intersection W)}
-2(P Intersection H Intersection W)+number of ppl in no club ....(1)


(P Intersection H) + (P Intersection w) + (H Intersection W) = 6..... (Number of ppl enrolled in exactly two groups is 6)

Also, P+H+W=77 and total number of students=59 and all students have to compulsorily join one club hence 'number of ppl in no club'=0

substituting these values in (1)

59=77-6-2(P Intersection H Intersection W)+0


2(P Intersection H Intersection W)=71-59=12

(P Intersection H Intersection W)=6

So number of students in all six groups is 6....hence C

22+27+28-6-2x = 59
gives

2x= 12 meaning x=6.

knight247 wrote: Total number of students=P+H+W-{(P Intersection H)+(P Intersection W)+(H Intersection W)}
-2(P Intersection H Intersection W)+number of ppl in no club

Hi there, Please can you explain why you have taken 2 in the last part of the above mentioned formula??

pankajks2010 wrote:

knight247 wrote: Total number of students=P+H+W-{(P Intersection H)+(P Intersection W)+(H Intersection W)}
-2(P Intersection H Intersection W)+number of ppl in no club

Hi there, Please can you explain why you have taken 2 in the last part of the above mentioned formula??

That is simply a variation of the formula. Read the following link for more info

https://www.urch.com/forums/gmat-problem ... rmula.html


I am abbreviating Intersection with I.Refer to my venn diagram while reading this explanation
From the problem, it is given that no of ppl in exactly two clubs is 6.
i.e. (P I H) + (P I W) + (H I W)=6 .....(1)
Correct? Well, not exactly... because (P I H) + (P I W) + (H I W) contains (P I H I W) thrice coz the intersection of each circle with the subsequent circle contains (P I H I W). Therefore three intersections will contain (P I H I W) thrice. So from the above statement we have to deduct
3(P I H I W). making it

(P I H) + (P I W) + (H I W)-3(P I H I W)=6
(P I H) + (P I W) + (H I W)=6 + 3(P I H I W).....(2)

Plug this into our regular statement of

Total no of items=P+H+W-{(P I H) + (P I W) + (H I W)}+(P I H I W)+Ppl in no group

We Get,

Total no of items=P+H+W-{6 + 3(P I H I W)}+(P I H I W)+Ppl in no group

Total no of items=P+H+W- 6 - 3(P I H I W)+(P I H I W)+Ppl in no group

Total no of items=P+H+W- 6 - 2(P I H I W)+Ppl in no group

Rest is just value subtitution. All clear?

i have grasped it now and am thankful

also here is a very useful post

https://www.beatthegmat.com/formulas-for ... 16479.html