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# One-sixth of the attendees at a certain convention are

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One-sixth of the attendees at a certain convention are [#permalink]

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01 Jun 2010, 04:14
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Question Stats:

71% (02:34) correct 29% (01:44) wrong based on 431 sessions

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One-sixth of the attendees at a certain convention are female students, two-thirds of the attendees are female, and one-third of the attendees are students. If 150 of the attendees are neither female nor students, what is the total number of attendees at the convention?

A. 300
B. 450
C. 600
D. 800
E. 900
[Reveal] Spoiler: OA
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01 Jun 2010, 04:41
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Let no. of Attendee are A.
Now As per the condition of the problem stated above .

We have following population in the convention as Attendee.
Total no. of females = [2/3]*A
Total no. of females as student : [1/6]*A
Total no. of students = [1/3]*A
Total no. of male as student = [1/6]*A
Total no. of males = A - [2/3]*A = [1/3]A
No. of males which are not student = [1/3]A - [1/6]*A = 150

Hence A = 900

Total no of males who are not student will be the answer as it states it should be neither female nor student
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01 Jun 2010, 04:45
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sm0k3rz wrote:
One-sixth of the attendees at a certain convention are female students, two-thirds of the attendees are female, and one-third of the attendees are students. If 150 of the attendees are neither female nor students, what is the total number of attendees at the convention?

300
450
600
800
900

Didnt really understand the solution provided to this question.

[Reveal] Spoiler:
E

Make a chart:
Attachment:

Chart.JPG [ 12.3 KiB | Viewed 8179 times ]

Next:
Female Non-student = (Total Female) - (Female Students) --> $$\frac{2}{3}x-\frac{1}{6}x=\frac{1}{2}x$$;

Total Non-students = (Total) - (Total students) --> $$x-\frac{1}{3}x=\frac{2}{3}x$$;

Total Non-students = (Female non-students) + (Male non-students) --> $$\frac{2}{3}x=\frac{1}{2}x+150$$ --> $$x=900$$.

Hope it helps.
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01 Jun 2010, 04:51
I agree, setting up a table really helps with these. It makes it easy to visualize and set up your equations.
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02 Jun 2010, 21:10
I used the formula :

Total = B + C - ( B n C ) + Neither

Let no. of Attendee are A
A = 2/3 * A + 1/3 * A - 1/6 * A + 150
gives A = 900.

i used this formula as there was no mention of Males in the Q anywhere...

But i dont know whether this method is correct or incorrect... plz help..
thanks
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27 Jun 2010, 02:26
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sag wrote:
I used the formula :

Total = B + C - ( B n C ) + Neither

Let no. of Attendee are A
A = 2/3 * A + 1/3 * A - 1/6 * A + 150
gives A = 900.

i used this formula as there was no mention of Males in the Q anywhere...

But i dont know whether this method is correct or incorrect... plz help..
thanks

Agree with this soln. There is no mention of males. Can be visualized in terms of Venn Diagram and then solved like above.
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Re: One-sixth of the attendees at a certain convention are [#permalink]

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22 Oct 2013, 11:16
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20 May 2014, 02:37
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Using Venn Diagram
Attachment:
File comment: Venn Diagram

gmat.jpg [ 25.25 KiB | Viewed 4938 times ]

Lets take number of total attendees as X.

Out of these, X/3 are students (represented by circle marked student), 2X/3 are female (represented by circle marked female) and X/3 are female students (represented by the overlapping area of both the circles)

From Venn Diagram, total number of attendees who are student , female or both will be given by

X/3 + 2X/3 - X/6 = 5X/6

Those who are not student or female = X - 5X/6 = X/6

Given that, number of attendees who are neither student nor female = 150

or, X/6 = 150
=> X = 900

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One-sixth of the attendees at a certain convention are [#permalink]

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20 Aug 2014, 16:08
f -f

s 1/6 1/6 2/6

3/6 1/6=150 4/6
-s
4/6 2/6 6/6=900
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03 Feb 2016, 00:54
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Re: One-sixth of the attendees at a certain convention are [#permalink]

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21 Apr 2016, 16:30
its simple, there are 1/6 female students so

1-1/6 equals 5/6x no female students

now, 5/6x(no female students) + 150(no female no students)= X (total people)

150= 1/6x which is x = 900
Re: One-sixth of the attendees at a certain convention are   [#permalink] 21 Apr 2016, 16:30
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