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Set Theory Question

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Manager
Joined: 12 Oct 2011
Posts: 128

Kudos [?]: 258 [1], given: 23

GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40

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21 Feb 2012, 11:06
1
KUDOS
Let's say I have a question like this:

I have 3 different groups A, B and C and now I'm given the percentages of how many people are in each group. There can also be people in 2 groups or in all three groups and I have information about the number of people who are in two groups. If the question ask how many people are in all groups, I can use the following formula:

100 = A + B + C – [AB + AC + BC] – [2*ALL]

But my question is why do you have to subtract the number of people in all three groups twice? Can anybody elaborate on that please?

Kudos [?]: 258 [1], given: 23

Math Expert
Joined: 02 Sep 2009
Posts: 42618

Kudos [?]: 135758 [0], given: 12708

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21 Feb 2012, 11:20
BN1989 wrote:
Let's say I have a question like this:

I have 3 different groups A, B and C and now I'm given the percentages of how many people are in each group. There can also be people in 2 groups or in all three groups and I have information about the number of people who are in two groups. If the question ask how many people are in all groups, I can use the following formula:

100 = A + B + C – [AB + AC + BC] – [2*ALL]

But my question is why do you have to subtract the number of people in all three groups twice? Can anybody elaborate on that please?

The following post addresses exactly the question you have: formulae-for-3-overlapping-sets-69014.html#p729340

Hope it helps.
_________________

Kudos [?]: 135758 [0], given: 12708

Manager
Joined: 12 Oct 2011
Posts: 128

Kudos [?]: 258 [0], given: 23

GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40

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21 Feb 2012, 11:32
Yes it does. Thanks a lot.

Kudos [?]: 258 [0], given: 23

Intern
Joined: 13 Jan 2012
Posts: 39

Kudos [?]: 18 [1], given: 0

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22 Feb 2012, 09:05
1
KUDOS
Let's say you had two intersecting sets...A and B; and you would like to calculate A union B.

How would you do it?
..... A + B - (A intersection B). Right?

Now, ask yourself why did you subtract (A intersection B) just once?
... because when you added A and B, the common element (A intersection B) got counted twice. You needed it just once, so you subtracted it once.

Applying the same logic, to three sets:
When you calculate [A union B union C] by adding A + B + C, the (A intersection B intersection C) piece gets counted three times. You need it accounted for just once, so you subtract it twice.

Makes sense?

Kudos [?]: 18 [1], given: 0

Re: Set Theory Question   [#permalink] 22 Feb 2012, 09:05
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