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In a village of 100 households, 75 have at least one DVD

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VP
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Joined: 22 Nov 2007
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In a village of 100 households, 75 have at least one DVD [#permalink] New post 24 Dec 2007, 09:03
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x - y=

A.30
B.35
C.45
D.50
E.55
Manager
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Joined: 21 May 2007
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Re: sets [#permalink] New post 24 Dec 2007, 09:34
marcodonzelli wrote:
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x - y=

A.30
B.35
C.45
D.50
E.55


E
Whats the OA? will give the explanation if E is indeed the answer (my procedure requires a diagram and is time consuming to draw in computer. if it is incorrect, it will be a wasted effort)
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CEO
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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 [#permalink] New post 24 Dec 2007, 09:38
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Expert's post
C

(75 ∩ 55) ∩ 80 = {30..55} ∩ 80 = {10..55}
x-y=55-10=45
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 [#permalink] New post 24 Dec 2007, 09:51
walker wrote:
C

(75 ∩ 55) ∩ 80 = {30..55} ∩ 80 = {10..55}
x-y=55-10=45


I got y as 0 and realized that is not possible.
The answer is C indeed.
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 [#permalink] New post 24 Dec 2007, 11:30
I'm getting D over and over
y = 5 in my calculations.
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 [#permalink] New post 24 Dec 2007, 12:00
CaspAreaGuy wrote:
I'm getting D over and over
y = 5 in my calculations.


i wish that, like a cat, i too have 9 lives in GMAT!!

i now get y=5

whats the OA ?
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 [#permalink] New post 24 Dec 2007, 21:07
Sorry man. I guess it is C.
y should be equal to 10.
  [#permalink] 24 Dec 2007, 21:07
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In a village of 100 households, 75 have at least one DVD

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