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

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Intern
Joined: 24 Nov 2012
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When Professor Wang looked at the rosters for this term's [#permalink]

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27 Nov 2012, 01:20
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Difficulty:

15% (low)

Question Stats:

69% (02:42) correct 31% (01:48) wrong based on 92 sessions

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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
[Reveal] Spoiler: OA

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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Re: When Professor Wang looked at the rosters for this term's [#permalink]

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27 Nov 2012, 03:48
Expert's post
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.

For more check ADVANCED OVERLAPPING SETS PROBLEMS
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Re: When Professor Wang looked at the rosters for this term's [#permalink]

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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.

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]

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15 Jan 2014, 21:40
Expert's post
1
This post was
BOOKMARKED
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.

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 [ 15.92 KiB | Viewed 2381 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]

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30 Sep 2014, 03:44
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]

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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

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Re: When Professor Wang looked at the rosters for this term's   [#permalink] 01 Oct 2014, 01:13
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