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# In a group of 68 students, each student is registered for at least one

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Intern
Joined: 02 Nov 2009
Posts: 18

Kudos [?]: 20 [0], given: 5

In a group of 68 students, each student is registered for at least one [#permalink]

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05 Dec 2009, 07:14
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Difficulty:

25% (medium)

Question Stats:

70% (00:57) correct 30% (00:58) wrong based on 152 sessions

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In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 13
B. 10
C. 9
D. 8
E. 7

[Reveal] Spoiler:
I just want to understand why this question cannot be solved by using equation

AUBUC = A + B +C - AintB - BintC - CintA + AintBintC

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-group-of-68-students-each-student-is-registered-for-at-106500.html
[Reveal] Spoiler: OA

Kudos [?]: 20 [0], given: 5

VP
Joined: 05 Mar 2008
Posts: 1468

Kudos [?]: 299 [0], given: 31

Re: In a group of 68 students, each student is registered for at least one [#permalink]

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05 Dec 2009, 07:41
sinharavi wrote:
This is a question from one of MGMAT tests.

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 13

B. 10

C. 9

D. 8

E. 7

I just want to understand why this question cannot be solved by using equation

AUBUC = A + B +C - AintB - BintC - CintA + AintBintC

no need to repeat this post

formulae-for-3-overlapping-sets-69014.html

Kudos [?]: 299 [0], given: 31

Intern
Joined: 17 Nov 2009
Posts: 37

Kudos [?]: 112 [0], given: 9

Schools: University of Toronto, Mcgill, Queens
Re: In a group of 68 students, each student is registered for at least one [#permalink]

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05 Dec 2009, 11:09
sinharavi wrote:
This is a question from one of MGMAT tests.

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 13

B. 10

C. 9

D. 8

E. 7

I just want to understand why this question cannot be solved by using equation

AUBUC = A + B +C - AintB - BintC - CintA + AintBintC

The above equation will result is 22 however the question asks for exactly two classes.

P(AnB) + P(AnC) + P(BnC) - 3P(AnBnC)
= 22- 3(3) = 22 - 9 = 13

What is the OA?
_________________

--Action is the foundational key to all success.

Kudos [?]: 112 [0], given: 9

VP
Joined: 05 Mar 2008
Posts: 1468

Kudos [?]: 299 [0], given: 31

Re: In a group of 68 students, each student is registered for at least one [#permalink]

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05 Dec 2009, 14:53
Bullet wrote:
sinharavi wrote:
This is a question from one of MGMAT tests.

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 13

B. 10

C. 9

D. 8

E. 7

I just want to understand why this question cannot be solved by using equation

AUBUC = A + B +C - AintB - BintC - CintA + AintBintC

The above equation will result is 22 however the question asks for exactly two classes.

P(AnB) + P(AnC) + P(BnC) - 3P(AnBnC)
= 22- 3(3) = 22 - 9 = 13

What is the OA?

12 History
22 Math
21 English
3 in all three
10 in both

Kudos [?]: 299 [0], given: 31

Intern
Joined: 17 Nov 2009
Posts: 37

Kudos [?]: 112 [0], given: 9

Schools: University of Toronto, Mcgill, Queens
Re: In a group of 68 students, each student is registered for at least one [#permalink]

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06 Dec 2009, 00:53
Bullet wrote:
sinharavi wrote:
This is a question from one of MGMAT tests.

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 13

B. 10

C. 9

D. 8

E. 7

I just want to understand why this question cannot be solved by using equation

AUBUC = A + B +C - AintB - BintC - CintA + AintBintC

The above equation will result is 22 however the question asks for exactly two classes.

P(AnB) + P(AnC) + P(BnC) - 3P(AnBnC)
= 22- 3(3) = 22 - 9 = 13

What is the OA?

My Apologies, that my upper post has tyoe. I don't know how it happen

P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
68 = 25+ 25+ 34 - P(A n B) – P(A n C) – P(B n C) + 3

P(A n B) + P(A n C) + P(B n C) = 19

for exactly two persons we need to find out.

P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)
= 19 - 3(3) = 10
_________________

--Action is the foundational key to all success.

Kudos [?]: 112 [0], given: 9

SVP
Joined: 16 Nov 2010
Posts: 1598

Kudos [?]: 592 [0], given: 36

Location: United States (IN)
Concentration: Strategy, Technology
Re: In a group of 68 students, each student is registered for at least one [#permalink]

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16 May 2011, 05:27
mgmat-cat-3-overlapping-sets-106500.html
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GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 592 [0], given: 36

Current Student
Joined: 03 Jun 2013
Posts: 22

Kudos [?]: 3 [0], given: 71

Concentration: Strategy, General Management
GMAT 1: 520 Q38 V32
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WE: Information Technology (Consulting)
Re: In a group of 68 students, each student is registered for at least one [#permalink]

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14 Dec 2014, 11:10
sinharavi wrote:
This is a question from one of MGMAT tests.

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 13

B. 10

C. 9

D. 8

E. 7

I just want to understand why this question cannot be solved by using equation

AUBUC = A + B +C - AintB - BintC - CintA + AintBintC

use this
Total = Group 1 + Group 2 + Group 3 - (people in 2 groups) - 2(people in all 3 groups) + none.

Kudos [?]: 3 [0], given: 71

Math Expert
Joined: 02 Sep 2009
Posts: 41886

Kudos [?]: 128688 [1], given: 12182

Re: In a group of 68 students, each student is registered for at least one [#permalink]

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15 Dec 2014, 07:29
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
sinharavi wrote:
In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 13
B. 10
C. 9
D. 8
E. 7

[Reveal] Spoiler:
I just want to understand why this question cannot be solved by using equation

AUBUC = A + B +C - AintB - BintC - CintA + AintBintC

"Each student is registered for at least one of three classes" means that there are no students who are registered for none of the classes.

Total = {people in group A} + {people in group B} + {people in group C} - {people in exactly 2 groups} - 2*{people in exactly 3 groups} + {people in none of the groups}:

68 = 25 + 25 + 34 - {people in exactly 2 groups} - 2*3 + 0 --> {people in exactly 2 groups}=10

Look at the diagram:

We need to find {people in exactly 2 groups}, so yellow section. Now, when we sum {people in group A} + {people in group B} + {people in group C} we count students who are in exactly 2 groups (yellow section) twice, so to get rid of double counting we are subtracting {people in exactly 2 groups} once.

Similarly when we sum {people in group A} + {people in group B} + {people in group C} we count students who are in exactly 3 groups (blue section) thrice (as it is the portion of all three groups), so to count this group only once we are subtracting 2*{people in exactly 3 groups}.

For more on this check: formulae-for-3-overlapping-sets-69014.html#p729340

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-group-of-68-students-each-student-is-registered-for-at-106500.html
_________________

Kudos [?]: 128688 [1], given: 12182

Re: In a group of 68 students, each student is registered for at least one   [#permalink] 15 Dec 2014, 07:29
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