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 PROBLEMSHi 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 5038 times ]
Now add all the students in the venn diagram = 14 + 5 + 13 + 3+ 4 + 6 + 5 = 50

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