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In a group of 68 students, each student is registered for at [#permalink]
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18 Dec 2010, 11:55
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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. Twentyfive students are registered for History, twentyfive students are registered for Math, and thirtyfour 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
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Re: MGMAT CAT 3 Overlapping Sets [#permalink]
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tonebeeze wrote: What is the quickest method to solve 3Set, Overlapping Set problems? Are they common on the GMAT (if you are scoring 46+ on Quant)?
In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twentyfive students are registered for History, twentyfive students are registered for Math, and thirtyfour 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 "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}=10Answer: B. Look at the diagram: Attachment:
untitled.PNG [ 5.66 KiB  Viewed 22292 times ]
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: formulaefor3overlappingsets69014.html#p729340As for your question: yes, questions on 3 overlapping sets are quite common for the GMAT. Hope it helps.
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Re: MGMAT CAT 3 Overlapping Sets [#permalink]
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18 Dec 2010, 12:27
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Hi!
There are two approaches you can take for 3 overlapping set questions. Which one works best for you is really a matter of preference.
Many people find Venn diagrams to be the best approach  draw 3 circles that have both double and triple overlapping sections. Venn diagrams are an excellent tool, especially for visual learners.
On the other hand, if you're an equation kind of guy (or gal), there are two different equations that come in handy:
True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3)  (# in exactly 2 groups)  2(# in all 3 groups);
and:
True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups).
There are a few other variations of those equations as well. In theory, you should add "+ total in none of the groups" to each equation, but I don't think I've ever seen a 3set question on the GMAT in which everyone wasn't a member of at least one group.
Even at high levels of the exam, 3 set questions aren't particularly common  most test takers see 0 or 1 of them, rarely 2.



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Re: MGMAT CAT 3 Overlapping Sets [#permalink]
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21 Nov 2013, 04:05
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Bunuel wrote: 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 2 groups}.
Bunuel, did you mean that we subtract 2*{people in exactly 3 groups} right? not people in exactly 2 groups....



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Re: MGMAT CAT 3 Overlapping Sets [#permalink]
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18 Dec 2010, 13:15
tonebeeze Bunuel's explanation at the link he provides was very helpful for me. I recommend sticking to the equations as the venn diagrams tend to throw me off.



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Re: MGMAT CAT 3 Overlapping Sets [#permalink]
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Re: MGMAT CAT 3 Overlapping Sets [#permalink]
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07 Sep 2012, 12:05
siddharthasingh wrote: Can this question be done with the help if matrix? If yes, then please let me know the procedure. The matrix method of 2 X 2 works well for two overlapping sets. For three overlapping sets we would need a 3D matrix of 2 X 2 X 2, which is hard (if not impossible) to visualize on a 2D paper or computer screen. So, stick with the Venn diagrams or the formulas.
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Re: In a group of 68 students, each student is registered for at [#permalink]
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Re: In a group of 68 students, each student is registered for at [#permalink]
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07 Sep 2012, 13:16
siddharthasingh wrote: Why can't a 3x3 matrix be used here? We have 3 types of classes, call them A, B, C. Each student can be or not in A  2 possibilities, can be or not in B, also 2 possibilities, can be or not in C, another two possibilities. So there is a total of 2 x 2 x 2 = 8 different subsets and not 3 x 3 = 9. If for two overlapping sets A and B you would use a 2 x 2 matrix (A nonA, B nonB), you don't have where to put the information regarding the third set C. Imagine a cube of 2 X 2 X 2, such that you add the third characteristic related to C on the vertical axis, above the base of 2 x 2 for A and B. The 3D cube of dimensions 2 X 2 x 2 for three overlapping sets is the parallel of the 2D matrix of dimensions 2 X 2 for two overlapping sets.
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Re: In a group of 68 students, each student is registered for at [#permalink]
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Re: MGMAT CAT 3 Overlapping Sets [#permalink]
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Re: In a group of 68 students, each student is registered for at [#permalink]
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30 Jan 2016, 17:57
Suppose A=all 3 B=History and english C=history and math D=math and english E=english only F=history only G=math only
we know that: A+B+C+D+E+F+G=68 A+B+C+F=25 A+C+D+G=25 A+B+D+E=34 A=3
OK, so we have B+C+F=22 C+D+G=22 B+D+E=34
add all these 3: 2B+2C+2D+E+F+G=75 we then have: B+C+D+E+F+G=65
substract from first one the second one: B+C+D=10 B,C,D  2 objects only.



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Re: In a group of 68 students, each student is registered for at [#permalink]
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19 Feb 2016, 18:40
Hi Brunel, I am also new to Venn Diagrams  can I calculate this problem as 683(253)(253)(343)=10? Even though the answer is negative, if I transform it to positive  seems to be correct Thanks!



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Re: In a group of 68 students, each student is registered for at [#permalink]
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12 Sep 2016, 13:02
The master information collation on all things Overlapping Sets. Thanks to this(and the authors ) for making life simpler overlappingsetsmadeeasy205636.html
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Re: In a group of 68 students, each student is registered for at [#permalink]
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30 Nov 2016, 02:04
tonebeeze wrote: In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twentyfive students are registered for History, twentyfive students are registered for Math, and thirtyfour 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 Solution: Let student who attended exactly 1 course= X, 2 courses =Y, 3 courses = Z (GIVEN Z=3) Total = X+Y+Z 68= X+Y+3 X+Y=65 (1) SUM of all the courses : h + m + e = X +2Y +3Z : Reason : every course considered common between 2 is repeated 2 times, while Z is repeated 3 times : each times whenever we take 2 courses together ( x is in anb , bnc, anc) 25+25+34 = X+2Y +3(3) 849 = X+2Y X+2Y = 75 X+Y=65 FROM EQUATION (1) solving this we get : Y=10




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