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When Professor Wang looked at the rosters for this term's

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saxenarahul021
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When Professor Wang looked at the rosters for this term's [#permalink] New post   Updated on: 30 Sep 2014, 03:45
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When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E) had 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw that 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once, how many names will be on the combined roster?

A. 30
B. 34
C. 42
D. 46
E. 50
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E

Originally posted by saxenarahul021 on 27 Nov 2012, 01:20.
Last edited by Bunuel on 30 Sep 2014, 03:45, edited 2 times in total.
Renamed the topic and edited the tags.
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iaratul
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   30 Sep 2014, 03:44
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it's easier and more efficient to use the formula rather than a venn diagram in the actual exam.

For this problem, the appropriate formula should be:
Total = Group 1 + Group 2 + Group 3 - (Sum of two group overlaps) + All three + Neither
Total = 26 + 28 + 18 - (9 + 7 + 10) + 4 + 0 = 50

FYI, if a question asks for "Exactly" two groups, the formula is slightly modified:
Total = Group 1 + Group 2 + Group 3 - (Sum of Exactly 2 group overlaps) - (2 x All Three) + Neither
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   15 Jan 2014, 21:40
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prasannajeet wrote:
Bunuel wrote:
saxenarahul021 wrote:
When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E) had 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw that 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once, how many names will be on the combined roster?

A. 30
B. 34
C. 42
D. 46
E. 50


Total # of students 26+28+18-(9+7+10)+4=50.

Answer: E.

For more check ADVANCED OVERLAPPING SETS PROBLEMS


Hi Bunuel
Lemme know why it is +4 rather than -4

Also please solve through ven-diagram actually my i got wrong while solving in ven diagram....

Rgds
Prasannajeet


When making the Venn diagram here, start by putting in the number of students which are in all 3 sets i.e. 4
Next, E and M had 9 names in common so the overlap of E and M excluding the overlap of all 3 will be 9 - 4 = 5.
Similarly for the E and S overlap and M and S overlap.
Next, E has 26 people and after removing 5 + 4 + 3 = 12, we are left with 14 people who have taken only E. Similarly for M and S too.

Attachment:
Ques3.jpg
Ques3.jpg [ 15.92 KiB | Viewed 75106 times ]


Now add all the students in the venn diagram = 14 + 5 + 13 + 3+ 4 + 6 + 5 = 50
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   27 Nov 2012, 03:48
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saxenarahul021 wrote:
When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E) had 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw that 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once, how many names will be on the combined roster?

A. 30
B. 34
C. 42
D. 46
E. 50


Total # of students 26+28+18-(9+7+10)+4=50.

Answer: E.

For more check ADVANCED OVERLAPPING SETS PROBLEMS
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   15 Jan 2014, 20:11
Bunuel wrote:
saxenarahul021 wrote:
When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E) had 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw that 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once, how many names will be on the combined roster?

A. 30
B. 34
C. 42
D. 46
E. 50


Total # of students 26+28+18-(9+7+10)+4=50.

Answer: E.

For more check ADVANCED OVERLAPPING SETS PROBLEMS


Hi Bunuel
Lemme know why it is +4 rather than -4

Also please solve through ven-diagram actually my i got wrong while solving in ven diagram....

Rgds
Prasannajeet
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   01 Oct 2014, 01:13
It takes more time to read the problem, otherwise the Venn diagram is quite straightforward

Total students = 26 + (5+6) + 13 = 50

Answer = E
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   03 Aug 2019, 04:55
Can you elaborate the meaning of not more than once in reference of this question?
My understanding is that 5+3+6 will be listed twice and 4 will be listed thrice.

I understand from not more than once is that numbers of students who were listed only in one class. If some students are listed for two classes, then the name will appear twice.

Please tell me at what point my understanding is wrong.
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   15 Sep 2019, 05:11
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gvij2017 wrote:
Can you elaborate the meaning of not more than once in reference of this question?
My understanding is that 5+3+6 will be listed twice and 4 will be listed thrice.

I understand from not more than once is that numbers of students who were listed only in one class. If some students are listed for two classes, then the name will appear twice.

Please tell me at what point my understanding is wrong.


Your understanding is correct. "not more than once" here means a unique list of students. More like the total population with each student counted only once.

[26+28+18] - [5+3+6] (listed twice) - 2*4 (listed thrice) should give you the right answer, i.e., 50
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   14 May 2020, 05:38
VeritasKarishma wrote:
prasannajeet wrote:
Bunuel wrote:
saxenarahul021 wrote:
When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E) had 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw that 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once, how many names will be on the combined roster?

A. 30
B. 34
C. 42
D. 46
E. 50


Total # of students 26+28+18-(9+7+10)+4=50.

Answer: E.

For more check ADVANCED OVERLAPPING SETS PROBLEMS


Hi Bunuel
Lemme know why it is +4 rather than -4

Also please solve through ven-diagram actually my i got wrong while solving in ven diagram....

Rgds
Prasannajeet


When making the Venn diagram here, start by putting in the number of students which are in all 3 sets i.e. 4
Next, E and M had 9 names in common so the overlap of E and M excluding the overlap of all 3 will be 9 - 4 = 5.
Similarly for the E and S overlap and M and S overlap.
Next, E has 26 people and after removing 5 + 4 + 3 = 12, we are left with 14 people who have taken only E. Similarly for M and S too.

Attachment:
Ques3.jpg


Now add all the students in the venn diagram = 14 + 5 + 13 + 3+ 4 + 6 + 5 = 50


Hi Karishma,

Could you please tell me what does "If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once" means? As student names are listed more than once such as those who are common.
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When Professor Wang looked at the rosters for this term's [#permalink] New post   25 Sep 2020, 06:11
Kunni

It means that if a list of students is prepared, none of the students should appear more than once.
if the student is already listed, he/she should not be listed again against some other subject, though she might have enrolled for 2/3 classes.
so indirectly the question is asking to find the number of students in atleast 1 set .

Thanks
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   13 Jan 2021, 16:27
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saxenarahul021 wrote:
When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E) had 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw that 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once, how many names will be on the combined roster?

A. 30
B. 34
C. 42
D. 46
E. 50


The question asked what is the number of total proctors?

Here;
\(Total = E+M+S -(E ∩ M)-(E ∩ S)-(M ∩ S) + (E ∩ M ∩ S)\)

\(TotaL= 28+28+18-9-7-10+4\)

\(Total =50\)

The answer is \(E\)
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Re: When Professor Wang looked at the rosters for this term's [#permalink] New post   20 Jun 2023, 09:06
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Re: When Professor Wang looked at the rosters for this term's [#permalink]
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