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3 overlapping sets

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3 overlapping sets [#permalink] New post 12 Jul 2010, 06:07
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A
B
C
D
E

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Question Stats:

50% (02:24) correct 50% (00:46) wrong based on 9 sessions
Of the 60 students of the IT class of 2003 at UC Berkeley, 30 use a Mac, 20 use a SPARC workstation, and 20 use a SGI. 13 use both Mac and SPARC, 5 uses both MAC and SGI, and 8 uses both SGI and SPARC. If 5 of the students use all three, how many don't use any of the three (not a Mac, not a SPARC, not a SGI, but maybe a crappy Intel machine)?

A 0
B 5
C10
D 11
E 12
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Re: 3 overlapping sets [#permalink] New post 12 Jul 2010, 07:11
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We should sum up the numbers that stand for the people we know to own one computer. Then we should subtract the nuber of people owning two computers and, finally, add the number of people owning all three.

30+20+20-(13+5+8)+5=49 => 60-49=11 it is the number of people not having any of these.

I would go for D.
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Re: 3 overlapping sets [#permalink] New post 12 Jul 2010, 07:16
ghettosquad wrote:
We should sum up the numbers that stand for the people we know to own one computer. Then we should subtract the nuber of people owning two computers and, finally, add the number of people owning all three.

30+20+20-(13+5+8)+5=49 => 60-49=11 it is the number of people not having any of these.

I would go for D.


why is that you add the number of people owning all the three?
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Re: 3 overlapping sets [#permalink] New post 12 Jul 2010, 07:42
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ramana wrote:
why is that you add the number of people owning all the three?


My reasoning is based on the inclusion-exclusion principle. This is the link to wikipedia in which it is better explained.

http://en.wikipedia.org/wiki/Inclusion% ... _principle

I hope it helps!
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Re: 3 overlapping sets [#permalink] New post 12 Jul 2010, 07:52
M=30
P=20
S=20
MnP=13
MnS=5
SnP=8
MnPnS=5

Total= M+P+S -( MnP + MnS+ MnP) + MnPnS + Neither

60=70 - 26 + 5 + N

N=11
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Re: 3 overlapping sets [#permalink] New post 12 Jul 2010, 08:04
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I think the easiest way to solve any problem like this would be to draw a Venn Diagram as I've done here.
Attachment:
File comment: Venn Diagram
Venn Diagram.JPG
Venn Diagram.JPG [ 17.94 KiB | Viewed 1198 times ]


Start by looking at the center most part of the diagram, the intersection of all three users. That's given to be 5.

The numbers mentioned for people using two machines also includes the number of people using all three. So to get the number of people using only two machines, subtract 5 from each of the numbers so you get 8,4 and 0 respectively as shown (intersection of two users)

So, now if you subtract the sum of number of two users and number of three users, you get the sole users of each machine, which goes into the other part of the circles.

Add all the numbers you have now and subtract it from 60, you get 11.

Mac + SPARC + SGI = 5
Mac + SPARC (Total) = 13
Mac + SPARC (Only) = 13 -5 = 8
Macs = 30
Mac Only = Macs - (Mac + SPARC) - (All three) = 30 - 8 - 5 = 17

Similarly, you get the other numbers as shown in the diagram. So your total number of users using these computers = 49

Hence nonusers = 60-49 =11.

Hope this helps.
Re: 3 overlapping sets   [#permalink] 12 Jul 2010, 08:04
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