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m03 - #7

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m03 - #7 [#permalink] New post 28 Feb 2008, 21:12
Question:
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?

My Answer

B1=60%
B2=50%
B3=35%

B1+B2+B3 = 100%
B1 = 10%

Now how can I find the answer from this... I can't understand the answer they provide to this question.

Thanks in advance for the help :-D
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Re: m03 - #7 [#permalink] New post 28 Feb 2008, 21:35
Here is how I'm thinking:
x- number of customers
0.6x+0.5x+0.35x-0.1x=1.35x => 1.35x-x= 0.35x
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Re: m03 - #7 [#permalink] New post 28 Feb 2008, 21:47
The correct answer is 25%, but I'm not seeing how they got there. I did the question exactly how you outlined below.
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Re: m03 - #7 [#permalink] New post 28 Feb 2008, 22:05
I have no idea. :shock:
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Re: m03 - #7 [#permalink] New post 28 Feb 2008, 22:22
i got 35%

draw a 3 circle venn diagram

the triple overlap is 10%

the double overlap regions labeled as x, y, z

60% + 50% + 35% - x - y - z - 10% = 100%

x + y + z = 35%
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Re: m03 - #7 [#permalink] New post 29 Feb 2008, 01:42
Liquid wrote:
Question:
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?

My Answer

B1=60%
B2=50%
B3=35%

B1+B2+B3 = 100%
B1 = 10%

Now how can I find the answer from this... I can't understand the answer they provide to this question.

Thanks in advance for the help :-D



100 = m + c + a - (mc + ca + am) - 2 (mca)
100 = 60 + 50 + 35 - (mc + ca + am) - 2 (10)
(mc + ca + am) = 145 - 100 - 20
(mc + ca + am) = 25
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Re: m03 - #7 [#permalink] New post 29 Feb 2008, 02:31
GMAT TIGER wrote:


100 = m + c + a - (mc + ca + am) - 2 (mca)
100 = 60 + 50 + 35 - (mc + ca + am) - 2 (10)
(mc + ca + am) = 145 - 100 - 20
(mc + ca + am) = 25


can you explain why mca is timed by 2?
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Re: m03 - #7 [#permalink] New post 29 Feb 2008, 05:26
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Here is the theoritical equation:
T = G1 + G2 + G3 - G1G2 - G2G3 - G1G3 - 2B

where:
GxGy: both Gx and Gy (overlapping)
B: All G1, G2, and G3 (overlapping)
T: total

Derivation:
Total = G1only + G2only + G3only + G1G2 + G2G3 +G3G1 + B (covering all possible groups)
example: total students = only milk + only chicken + only apples + both milk and apple + both apple and chicken + both chicken and milk + all milk,chicken, and apple

G1only = G1 - G1G2 - G1G3 - B
G2only = G2 - G1G2 - G2G3 - B
G3only = G3 - G1G3 - G2G3 - B

plug in equations in the main equation above:
Total = G1 - G1G2 - G1G3 - B + G2 - G1G2 - G2G3 - B + G3 - G1G3 - G2G3 - B + G1G2 + G2G3 + G1G3 + B
Total = G1 + G2 + G3 - (G1G2 + G1G3 + G2G3) - 2B
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Re: m03 - #7 [#permalink] New post 29 Feb 2008, 08:17
Mishari wrote:
Here is the theoritical equation:
T = G1 + G2 + G3 - G1G2 - G2G3 - G1G3 - 2B

where:
GxGy: both Gx and Gy (overlapping)
B: All G1, G2, and G3 (overlapping)
T: total

Derivation:
Total = G1only + G2only + G3only + G1G2 + G2G3 +G3G1 + B (covering all possible groups)
example: total students = only milk + only chicken + only apples + both milk and apple + both apple and chicken + both chicken and milk + all milk,chicken, and apple

G1only = G1 - G1G2 - G1G3 - B
G2only = G2 - G1G2 - G2G3 - B
G3only = G3 - G1G3 - G2G3 - B

plug in equations in the main equation above:
Total = G1 - G1G2 - G1G3 - B + G2 - G1G2 - G2G3 - B + G3 - G1G3 - G2G3 - B + G1G2 + G2G3 + G1G3 + B
Total = G1 + G2 + G3 - (G1G2 + G1G3 + G2G3) - 2B


Great explanation.
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Re: m03 - #7 [#permalink] New post 29 Feb 2008, 11:43
This may be a little rudimentary but here is how I solved it.

I assume 100 people therefore you have:
60 milks
50 chickens
35 apples

If 10 percent bought all three then subtrack 10 from each one and we have:

50 milks
40 chickens
25 apples

We have 90 peple left. If they each bought one item there would still be 25 left. So 25 of them had to atleast buy 2 products.
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Re: m03 - #7 [#permalink] New post 02 Mar 2008, 14:19
GMAT TIGER wrote:
100 = m + c + a - (mc + ca + am) - 2 (mca)
100 = 60 + 50 + 35 - (mc + ca + am) - 2 (10)
(mc + ca + am) = 145 - 100 - 20
(mc + ca + am) = 25


Ok, cool. But why is MCA multiplied by 2?

Thanks.
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Re: m03 - #7 [#permalink] New post 02 Mar 2008, 14:31
Liquid wrote:
GMAT TIGER wrote:
100 = m + c + a - (mc + ca + am) - 2 (mca)
100 = 60 + 50 + 35 - (mc + ca + am) - 2 (10)
(mc + ca + am) = 145 - 100 - 20
(mc + ca + am) = 25


Ok, cool. But why is MCA multiplied by 2?

Thanks.



it is multiplied by two b/c it appears 3 times. but you need to count it only once. so you substract 2MCA.
does this help?
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Re: m03 - #7 [#permalink] New post 02 Mar 2008, 17:57
elmagnifico wrote:
it is multiplied by two b/c it appears 3 times. but you need to count it only once. so you substract 2MCA.
does this help?


Sort-of.... :oops:

can you elaborate.....
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Re: m03 - #7 [#permalink] New post 13 Mar 2008, 09:33
Mishari wrote:
Here is the theoritical equation:
T = G1 + G2 + G3 - G1G2 - G2G3 - G1G3 - 2B

where:
GxGy: both Gx and Gy (overlapping)
B: All G1, G2, and G3 (overlapping)
T: total

Derivation:
Total = G1only + G2only + G3only + G1G2 + G2G3 +G3G1 + B (covering all possible groups)
example: total students = only milk + only chicken + only apples + both milk and apple + both apple and chicken + both chicken and milk + all milk,chicken, and apple

G1only = G1 - G1G2 - G1G3 - B
G2only = G2 - G1G2 - G2G3 - B
G3only = G3 - G1G3 - G2G3 - B

plug in equations in the main equation above:
Total = G1 - G1G2 - G1G3 - B + G2 - G1G2 - G2G3 - B + G3 - G1G3 - G2G3 - B + G1G2 + G2G3 + G1G3 + B
Total = G1 + G2 + G3 - (G1G2 + G1G3 + G2G3) - 2B


Great explanation, THX!!!

But, If you're not so algebraic, like me =) You can also see this conceptually why this is true. Because just by looking at the derived equation, one might wonder, why? Or for me, why "-2B?" Why not just -B?

I enlightened myself by making an actual Vann diagram. With 3 post-its, I made three circles and overlapped them. Then it became so clear why the equation tells you to do -2B instead of -B or -3B

anyone like me, who just can't ACCEPT the equation as is, unless you actually SEE it, it might help.
Re: m03 - #7   [#permalink] 13 Mar 2008, 09:33
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