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GClub M03, Question 7/37 - 3 overlapping sets

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Senior Manager
Joined: 21 Jul 2009
Posts: 367
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
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GClub M03, Question 7/37 - 3 overlapping sets [#permalink]  02 Mar 2010, 18:45
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Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above?

A) 5%
B) 10%
C) 15%
D) 25%
E) 30%

I am not convinced with the explanation given. There was also an alternative explanation, which wasn't very clear either -

Even though this is not a typical 2-group problem with overlapping members, we can still apply the group formula: Total = G1 + G2 + N - B, where G1, G2 are group 1 and group 2, N is neither, and B is both. Because we have more than 2 groups, we need to adjust the formula to reflect customers purchasing 3 products and therefore being members of 3 groups, and being counted as 3 distinct customers. The formula needs to be modified as follows: Total = G1 + G2 + G3 + N - B - T*(3-1) , where T is members of three groups and B is members of only two groups.

There is no neither group because all customers purchase at least one product.

I suspect the formula for three overlapping sets used in the question. Can someone confirm it.
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Senior Manager
Joined: 01 Feb 2010
Posts: 268
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Kudos [?]: 35 [0], given: 2

Re: GClub M03, Question 7/37 - 3 overlapping sets [#permalink]  02 Mar 2010, 20:36
BarneyStinson wrote:
Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above?

A) 5%
B) 10%
C) 15%
D) 25%
E) 30%

I am not convinced with the explanation given. There was also an alternative explanation, which wasn't very clear either -

Even though this is not a typical 2-group problem with overlapping members, we can still apply the group formula: Total = G1 + G2 + N - B, where G1, G2 are group 1 and group 2, N is neither, and B is both. Because we have more than 2 groups, we need to adjust the formula to reflect customers purchasing 3 products and therefore being members of 3 groups, and being counted as 3 distinct customers. The formula needs to be modified as follows: Total = G1 + G2 + G3 + N - B - T*(3-1) , where T is members of three groups and B is members of only two groups.

There is no neither group because all customers purchase at least one product.

I suspect the formula for three overlapping sets used in the question. Can someone confirm it.

The formula is
n(AUBUC) = n(A)+n(B)+n(C)-B+A-N
where B is both (it will sum of (A&B), (B&C), (C&A)).
and A is All
and N is neither
so plugging in we get
100 = 60+50+35+10-B-0
B = 55.
exactly two = 55 - 3(10) = 25
subtracting 10% three times as this value is including in all the (A&B), (B&C), (C&A).
Senior Manager
Joined: 21 Jul 2009
Posts: 367
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Followers: 14

Kudos [?]: 98 [0], given: 22

Re: GClub M03, Question 7/37 - 3 overlapping sets [#permalink]  04 Mar 2010, 21:01
bangalorian2000 wrote:
The formula is
n(AUBUC) = n(A)+n(B)+n(C)-B+A-N
where B is both (it will sum of (A&B), (B&C), (C&A)).
and A is All
and N is neither
so plugging in we get
100 = 60+50+35+10-B-0
B = 55.
exactly two = 55 - 3(10) = 25
subtracting 10% three times as this value is including in all the (A&B), (B&C), (C&A).

Gotcha bro!!! You are indeed smart!!!
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Re: GClub M03, Question 7/37 - 3 overlapping sets   [#permalink] 04 Mar 2010, 21:01
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