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# Foodmart customers regularly buy at least one of the

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VP
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27 Jan 2008, 23:20
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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 2 of the above products?
 5%
 10%
 15%
 25%
 30%

What is the right formula to apply? I used : x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases), considering that everybody bought smtg.
Thus, substituting, we obtain that y=1/4 x
Director
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Re: Sets with >=3 elements [#permalink]

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28 Jan 2008, 04:53
Yup, that looks good

60+50+35 = 145
145-10(2) = 125
125-100 = 25
25% of customers purchase 2 of the above products
Intern
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Re: Sets with >=3 elements [#permalink]

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05 Feb 2008, 12:11
The explanation on the practice test to the problem goes...

Let bi denote the percentage of buyers who regularly buy i products, and bx the percentage of buyers who regularly purchase product x.

We can construct these equations:
b1 + 2b2 + 3b3 = bm + bc + ba = 60% + 50% + 35% = 145%
b1 + b2 + b3 = 100%
b3 = 10%

Subtract the second one from the first one:
b2 + 2b3 = 45%
b2 = 25%

How does one explain the multipliers for terms in b1 + 2b2 + 3b3 above? Does that mean that if there are 4 products that the equation would read b1 + 2b2 + 3b3 + 4b4?
Director
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Re: Sets with >=3 elements [#permalink]

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05 Feb 2008, 12:24
marcodonzelli 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 2 of the above products?
 5%
 10%
 15%
 25%
 30%

What is the right formula to apply? I used : x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases), considering that everybody bought smtg.
Thus, substituting, we obtain that y=1/4 x

P1 - one product, P2 - two poducts, P3 - three products

P1+P2+P3 = 100
P1+2*P2+3*P3 = 60+50+35 = 145
P3 = 10 -> P2 = (145-100)-2*10 = 45-20 = 25 -> D
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Re: Sets with >=3 elements [#permalink]

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05 Feb 2008, 13:22
Hi,

x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases),

can someone explain how this formula is derived.
I am not sure how we get 2z. It says 10% but all 3 , so wont it be
1/10 x.

Thanks
-Jack
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Re: Sets with >=3 elements [#permalink]

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05 Feb 2008, 19:51
marcodonzelli 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 2 of the above products?
 5%
 10%
 15%
 25%
 30%

What is the right formula to apply? I used : x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases), considering that everybody bought smtg.
Thus, substituting, we obtain that y=1/4 x

Venn Diagram is my favorite here. Slow but true.

10= All 3
CA: A-10
MA: C-10
CA: B-10

M: 60 - (A-10+10+C-10) --> 70-A-C
C: 50 - (A-10+10+B-10) --> 60-A-B
A: 35 - (B-10+10+C-10) --> 45-B-C

70-A-C
60-A-B
45-B-C
A-10
B-10
C-10
10
--> 155-A-B-C=100 ---> A+B+C=55 Now we want only 2 items so now its just A+B+C-30 =25

D
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Re: Sets with >=3 elements [#permalink]

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07 Feb 2008, 01:00
jackychamp wrote:
Hi,

x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases),

can someone explain how this formula is derived.
I am not sure how we get 2z. It says 10% but all 3 , so wont it be
1/10 x.

Thanks
-Jack

It is a derivation of the Venn diagram

you'll see that we have to get rid of the double results and the triple ones. actually in this case worked...
Manager
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Re: Sets with >=3 elements [#permalink]

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07 Feb 2008, 16:00
Can someone provide the standard formula for 3 Venn diagram and 2 Venn Diagram.

Thanks
JAck
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Re: Sets with >=3 elements [#permalink]

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11 Oct 2011, 15:12
maratikus wrote:
marcodonzelli 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 2 of the above products?
 5%
 10%
 15%
 25%
 30%

What is the right formula to apply? I used : x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases), considering that everybody bought smtg.
Thus, substituting, we obtain that y=1/4 x

P1 - one product, P2 - two poducts, P3 - three products

P1+P2+P3 = 100
P1+2*P2+3*P3 = 60+50+35 = 145
P3 = 10 -> P2 = (145-100)-2*10 = 45-20 = 25 -> D

how do we know P1+P2+P3 = 100?
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Concentration: General Management, Entrepreneurship
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Re: Sets with >=3 elements [#permalink]

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11 Oct 2011, 21:18
i'm also a bit confused by everyone else's work, this is the way I solved it

the equation is

P(A or B or C ) = P (A) + P (B) + P (C) - P(A & B) - P(A & C) - P(B & C) + P (A & B & C)
also, the probability of A or B or C is 100% since A,B,C are the only sets in this space.

when you add up P(A), P(B), P(C), you are also over adding in case they have intersections of 2 or 3 *draw a venn diagram with 3 intersection circles and it can help make sense . Therefore you subtract out the places where two of them overlap. However, as you were subtracting the 2 set overlaps, each time you're subtracting any 3 set overlap as well. So you end up subtracting 3x for the 3-set overlap there, but recall in the beginning you added 3x for the 3-set overlap for A,B,C, so net is zero for that 3 set overlap. You can't ignore the 3-set overlap so you add it back in as the last component.

Anyway, the problem goes

60+50+35 - (2 set overlaps) + (3 set overlap) = 100
145 - (2 set overlaps) + 10 = 100
155 - (2 set overlaps) = 100
(2 set overlaps) = 55

That means that 2 set overlaps add up to 55%. Meaning P(A&B) + P(A&C) + P(C&B) = 55

What's important here is you're back to the same problem of over adding because there are probabilities where all 3 sets overlap. so if you notice you're actually adding the 3 set overlap 3x, once in each term obviously. Now normally you would think ok, I have to subtract the 3-set overlap out 2x, leaving just net 1. And that would be 100% correct if you are looking for Probability of people buying 2things AND 3 things. But the question asks for only people that buy both (in other words the two set overlaps).

Therefore you would subtract 55 - 3(3 set overlaps) = 25% getting rid of the 3 set overlaps entirely.
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Re: Sets with >=3 elements [#permalink]

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11 Oct 2011, 23:07
Some good methods have been discussed above. Let me add a small discussion on Venn diagrams

You will not have any confusion if you visualize it. The total number of people is 100 (assume it since numbers are in %). These 100 people are spread around in the 3 circles. One person can be in only one area.
Attachment:

Ques6.jpg [ 17.57 KiB | Viewed 1330 times ]

60+50+35 (= 145) is more than 100 because
60 = the entire left top circle = the left top red part + x + z + 10.
50 = the entire right top circle = the right top red part + x + y + 10
35 = the bottom circle = the bottom red part + y + z + 10
so x, y and z are counted twice and 10 is counted thrice

45 = x + y + z + 2*10
x+ y + z = 25
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 09 Jun 2011 Posts: 145 Followers: 0 Kudos [?]: 29 [0], given: 1 Re: Sets with >=3 elements [#permalink] ### Show Tags 12 Oct 2011, 23:48 @Karishma: How did you get: "45 = x + y + z + 2*10 x+ y + z = 25" I lost the part here.. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6844 Location: Pune, India Followers: 1935 Kudos [?]: 12009 [0], given: 221 Re: Sets with >=3 elements [#permalink] ### Show Tags 13 Oct 2011, 08:46 OptimusPrimea1 wrote: @Karishma: How did you get: "45 = x + y + z + 2*10 x+ y + z = 25" I lost the part here.. 145 is 45 more than 100. Why is it 45 extra? Because x, y and z were counted twice (so they appear once extra) and 10 was counted thrice ( so it appears twice extra). check out the diagram. These extras make up the 45. In 100, there is no double/triple counting. It is equal to the actual number of people. That is why 45 = x + y + z + 2*10 25 = x+y+z _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Sets with >=3 elements   [#permalink] 13 Oct 2011, 08:46
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