Last visit was: 07 Sep 2026, 07:03 It is currently 07 Sep 2026, 07:03
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
User avatar
cejismundo
User avatar
Joined: 17 Jul 2006
Last visit: 18 Jan 2007
Posts: 21
Own Kudos:
2
Posts: 21
Kudos: 2
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 21 Jul 2006, 16:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 belong to club Z. If 2 belong to all three organizations and 3 belong to exactly two organizations, how many belong to none of these organizations?

a. 1
b. 2
c. 4
d. 6
e. 11



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
ps_dahiya
User avatar
Joined: 20 Nov 2005
Last visit: 15 Oct 2019
Posts: 1,486
Own Kudos:
1,256
Concentration: Strategy, Entrepreneurship
Schools:Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Posts: 1,486
Kudos: 1,256
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 21 Jul 2006, 16:13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
D

A+B+C-AB-BC-CA+ABC + NONE = TOTAL
Given
A = 5, B = 7 , C = 9
ABC = 2
AB+BC+CA-3ABC = 3 i.e AB+BC+CA - 3 * 2 = 3 i.e AB+BC+CA = 9

Put given values in above equation
5+7+9-9+2+NONE = 20
NONE = 6
User avatar
cejismundo
User avatar
Joined: 17 Jul 2006
Last visit: 18 Jan 2007
Posts: 21
Own Kudos:
2
Posts: 21
Kudos: 2
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 21 Jul 2006, 22:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I dont understand how you get
AB+BC+CA-3ABC = 3 i.e AB+BC+CA - 3 * 2 = 3 i.e AB+BC+CA = 9

can you please explain?
User avatar
ps_dahiya
User avatar
Joined: 20 Nov 2005
Last visit: 15 Oct 2019
Posts: 1,486
Own Kudos:
1,256
Concentration: Strategy, Entrepreneurship
Schools:Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Posts: 1,486
Kudos: 1,256
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 22 Jul 2006, 11:30
Kudos
Add Kudos
Bookmarks
Bookmark this Post
cejismundo
I dont understand how you get
AB+BC+CA-3ABC = 3 i.e AB+BC+CA - 3 * 2 = 3 i.e AB+BC+CA = 9

can you please explain?
Show more

AB - Includes that are in A as well as B and also include ABC
BC - Includes that are in B as well as C and also include ABC
CA - Includes that are in C as well as A and also include ABC

AB + BC + CA includes ABC three times. So get that are part of exactly two we need to subtract ABC three times.
Make venn diagram and it should be clear.

Hope this helps.
User avatar
freetheking
User avatar
Joined: 22 May 2006
Last visit: 22 Jan 2007
Posts: 188
Own Kudos:
1,071
Location: Rancho Palos Verdes
Posts: 188
Kudos: 1,071
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 22 Jul 2006, 12:02
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Total = n(X) + n(Y) + n(Z) - (n(X&Y)+n(Y&Z)+n(Z&X)) + n(X&Y&Z) + None
20 = 5+7+9 - (3+3+3) + 2 + None
None = 6
User avatar
ps_dahiya
User avatar
Joined: 20 Nov 2005
Last visit: 15 Oct 2019
Posts: 1,486
Own Kudos:
1,256
Concentration: Strategy, Entrepreneurship
Schools:Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Posts: 1,486
Kudos: 1,256
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 22 Jul 2006, 13:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
freetheking
Total = n(X) + n(Y) + n(Z) - (n(X&Y)+n(Y&Z)+n(Z&X)) + n(X&Y&Z) + None
20 = 5+7+9 - (3+3+3) + 2 + None
None = 6
Show more


Where it is written that n(X&Y) = 3 and n(Y&Z) = 3 and n(Z&X) =3????
User avatar
freetheking
User avatar
Joined: 22 May 2006
Last visit: 22 Jan 2007
Posts: 188
Own Kudos:
1,071
Location: Rancho Palos Verdes
Posts: 188
Kudos: 1,071
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 22 Jul 2006, 13:46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ps_dahiya
freetheking
Total = n(X) + n(Y) + n(Z) - (n(X&Y)+n(Y&Z)+n(Z&X)) + n(X&Y&Z) + None
20 = 5+7+9 - (3+3+3) + 2 + None
None = 6

Where it is written that n(X&Y) = 3 and n(Y&Z) = 3 and n(Z&X) =3????
Show more


I was just assuming one possible way...
as long as (n(X&Y)+n(Y&Z)+n(Z&X)) = 9, it won't have any effect.
User avatar
ps_dahiya
User avatar
Joined: 20 Nov 2005
Last visit: 15 Oct 2019
Posts: 1,486
Own Kudos:
1,256
Concentration: Strategy, Entrepreneurship
Schools:Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Posts: 1,486
Kudos: 1,256
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 22 Jul 2006, 13:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
freetheking
ps_dahiya
freetheking
Total = n(X) + n(Y) + n(Z) - (n(X&Y)+n(Y&Z)+n(Z&X)) + n(X&Y&Z) + None
20 = 5+7+9 - (3+3+3) + 2 + None
None = 6

Where it is written that n(X&Y) = 3 and n(Y&Z) = 3 and n(Z&X) =3????

I was just assuming one possible way...
as long as (n(X&Y)+n(Y&Z)+n(Z&X)) = 9, it won't have any effect.
Show more


But here we have number like this. If we modify the numbers then certainly it will be wrong.
User avatar
Futuristic
User avatar
Joined: 28 Dec 2005
Last visit: 07 May 2009
Posts: 415
Own Kudos:
52
Posts: 415
Kudos: 52
of 20 adults, 5 belong to club X, 7 belong to club Y, and 9 22 Jul 2006, 14:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Some basics first.....

Let U be the universal set. Let A, B, C be member sets within U.

Place a venn diagram in front of yourself with intersecting sets A, B,C within the universal set U. Lets find an equation that must hold true.

U = a - ab - ca + abc (part of set A not intersecting with anything else)
+b - bc - ca + abc (part of set B not intersecting with anything else)
+c - ca - bc + abc ((part of set C not intersecting with anything else)
+ ab - abc (part common to A and B but not C)
+ bc - abc (part common to B and C but not A)
+ ca - abc (part common to C and A but not B)
+ abc (part common to all 3 sets)
+ C(a + b + c) (complement of union of A, B, C)

Which simplifies to

U = a+b+c-(ab+bc+ca)+abc-C(a+b+c)

This formula above holds true unconditionally. In this problem, the trick lies in plugging in the values for (ab+bc+ca). Elements that are members of 2 sets would fall into (AB+BC+CA) but there is a catch. (AB+BC+CA) includes ABC 3 times. So, number of members of exactly 2 groups
= ab - abc + bc -abc + ca - abc.

The other elements can be readily plugged in as explained above, so I won't go into that.

Note that similarly, for 2 sets, the rule is:

U = a + b + C(a+b) -ab

Hope this helps



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!