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MGMAT CAT5 - Group

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TaTum
 Post subject: MGMAT CAT5 - Group
 Post Posted: Thu Sep 04, 2008 1:23 am 
How many attendees are at a convention if 150 of the attendees are neither female nor students, one-sixth of the attendees are female students, two-thirds of the attendees are female, and one-third of the attendees are students?

180
300
450
600
900



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OA is [color=white]E[/color]. I got this question wrong because I first thought it was an Overlapping Group question. Later, I realized it was another type of Group question (four different groups). How can I notice this from at the beginning?
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Orchid
 Post subject: Re: MGMAT CAT5 - Group
 Post Posted: Thu Sep 04, 2008 5:37 am 
TaTum wrote:
How many attendees are at a convention if 150 of the attendees are neither female nor students, one-sixth of the attendees are female students, two-thirds of the attendees are female, and one-third of the attendees are students?

180
300
450
600
900



--
OA is [color=white]E[/color]. I got this question wrong because I first thought it was an Overlapping Group question. Later, I realized it was another type of Group question (four different groups). How can I notice this from at the beginning?


My answer is E :)
Using Venn diagram makes this easier. Let all the attendees be x, so we have the following equation:

x - 150 = 2/3 x + 1/3 x - 1/6 x ~~~>>> so x = 900 :D [/img]
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jwinawer
 Post subject:
 Post Posted: Sat Sep 20, 2008 4:04 am 
Offline
ManhattanGMAT Staff


Posts: 76
Well, just to be clear, it is an overlapping group question. The set of people can be divided into F / M and student / non-student. This can definitely be nicely solved with Venn diagrams, as the post above notes. It can also be solved by setting up a double set matrix (a 2 x 2 matrix divided into M/F, student / non-student, plus totals). Either method should work.
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