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

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Senior Manager
Joined: 13 May 2007
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In a group of 68 students, each student is registered for at [#permalink]  25 Aug 2007, 22:35
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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?
13
10
9
8
7

VP
Joined: 10 Jun 2007
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Kudos [?]: 142 [0], given: 0

Re: PS # Set theory [#permalink]  25 Aug 2007, 22:40
empty_spaces 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?
13
10
9
8
7

10

(25+25+34) - "exactly two" - 2*3 = 68
"exactly two" = 10
Director
Joined: 03 May 2007
Posts: 888
Schools: University of Chicago, Wharton School
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Re: PS # Set theory [#permalink]  25 Aug 2007, 22:48
empty_spaces 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?
13
10
9
8
7

68 = 25 +25 +34 - (ab + bc + ac) - 2(3)
ab + bc + ac = 1o
Senior Manager
Joined: 13 May 2007
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Re: PS # Set theory [#permalink]  25 Aug 2007, 22:49
Bkk/fistail, could you tell me why did you do "2 * 3 "

bkk145 wrote:
empty_spaces 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?
13
10
9
8
7

10

(25+25+34) - "exactly two" - 2*3 = 68
"exactly two" = 10
VP
Joined: 10 Jun 2007
Posts: 1464
Followers: 6

Kudos [?]: 142 [0], given: 0

Re: PS # Set theory [#permalink]  25 Aug 2007, 22:55
empty_spaces wrote:
Bkk, could you tell me why did you do "2 * 3 "

bkk145 wrote:
empty_spaces 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?
13
10
9
8
7

10

(25+25+34) - "exactly two" - 2*3 = 68
"exactly two" = 10

Basically, you count "all three" twice because question ask for "exactly two". If the question ask "two and more", then you do not have to multiply 2 since it is already count "all three" once.

The formula usually given for this type of problem use "Two or more" as the principle.
Senior Manager
Joined: 13 May 2007
Posts: 252
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Re: PS # Set theory [#permalink]  25 Aug 2007, 23:01
ok i am confused here,

isnt the formula like :

A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) + ( A intersection B intersection C)

bkk145 wrote:
empty_spaces wrote:
Bkk, could you tell me why did you do "2 * 3 "

bkk145 wrote:
empty_spaces 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?
13
10
9
8
7

10

(25+25+34) - "exactly two" - 2*3 = 68
"exactly two" = 10

Basically, you count "all three" twice because question ask for "exactly two". If the question ask "two and more", then you do not have to multiply 2 since it is already count "all three" once.

The formula usually given for this type of problem use "Two or more" as the principle.
VP
Joined: 10 Jun 2007
Posts: 1464
Followers: 6

Kudos [?]: 142 [0], given: 0

Re: PS # Set theory [#permalink]  25 Aug 2007, 23:19
empty_spaces wrote:
A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) + ( A intersection B intersection C)

This is exactly what I mean. Let's use class here, so we have students taking classes A, B, and C.

Take this part as an example: (A intersection B)
This part includes students who take both "A and B" AND "A and B and C"

Same thing for (B intersection C)
This part includes students who take both "B and C" AND "A and B and C"

Which is same thing for (C intersection A)

So, you can see that (X intersection Y) counts students who take both classes as well as all three classes. Therefore, when the question ask for "exactly two", you have to multiply "all three" by 2 because "exactly two" doesn't include "A and B and C". Hope this helps.

Oh , also, I think this is the correct formula:
A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) - ( A intersection B intersection C)
Senior Manager
Joined: 13 May 2007
Posts: 252
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Kudos [?]: 7 [0], given: 0

Re: PS # Set theory [#permalink]  25 Aug 2007, 23:50
thanks Bkk for your effort although am still confused, maybe my brain isnt really working at 3 in the night.

bkk145 wrote:
empty_spaces wrote:
A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) + ( A intersection B intersection C)

This is exactly what I mean. Let's use class here, so we have students taking classes A, B, and C.

Take this part as an example: (A intersection B)
This part includes students who take both "A and B" AND "A and B and C"

Same thing for (B intersection C)
This part includes students who take both "B and C" AND "A and B and C"

Which is same thing for (C intersection A)

So, you can see that (X intersection Y) counts students who take both classes as well as all three classes. Therefore, when the question ask for "exactly two", you have to multiply "all three" by 2 because "exactly two" doesn't include "A and B and C". Hope this helps.

Oh , also, I think this is the correct formula:
A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) - ( A intersection B intersection C)
Re: PS # Set theory   [#permalink] 25 Aug 2007, 23:50
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