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

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GMAT Instructor
Joined: 04 Jul 2006
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In a village of 100 households, 75 have at least one DVD [#permalink]  23 Sep 2006, 08:48
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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^2-y^2 is a number from

(A) 1000 to 2500 (B) 2501 to 2800 (C) 2801 to 3000 (D) 3001 to 3100 (E) 3001 to 3150
Director
Joined: 13 Nov 2003
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Location: BULGARIA
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Kudos [?]: 27 [0], given: 0

The max number for X is 55.
The MIn is 100=210-( households with 2 of the items)-2( Households with all 3) the lowest value for households with all 3 items is when the households with 2 items are max or 108. Then we have only 1 household with all 3 items.
The ans is 54*56=3024 or D
Intern
Joined: 14 Jul 2005
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Location: California
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Hi Kevin
whats the answer for this one. I am getting B.
Edited:
Sorry ... I agree with A.

Last edited by 800_gal on 19 Dec 2006, 13:43, edited 1 time in total.
Manager
Joined: 19 Aug 2006
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Max number for X is 55

Now for lowest possibility

Max no common between DVD and CP =100- (75+85)= 55

Max no common between DVD and MP3 = 100 - (75+55)=30

So common among all 3 will be 55-30= 25

so X^2 - Y^2

(55)^2 - (25)^2 =3025-625 =2400

So ANS A
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