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all three of these devices

Problem Solving — algebra and arithmetic
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by sanju09 » Sat Aug 14, 2010 2:05 am
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 is:

A. 65

B. 55

C. 45

D. 35

E. 25

[spoiler]Source: gmatclub.com

oops, moderators, please move it to the right bin[/spoiler]
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by kvcpk » Sat Aug 14, 2010 3:05 am
sanju09 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 is:

A. 65

B. 55

C. 45

D. 35

E. 25

[spoiler]Source: gmatclub.com[/spoiler]
Among 75, 80, 55 - 55 is the least.
Hence the maximum possible value is 55.

Now, 25 do not have DVD player, 20 do not have cell phone, 45 do not have mp3 player.
Hence minimum of them is 20.

Hence difference = 55-20 = 35.
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by debmalya_dutta » Wed Aug 25, 2010 5:07 pm
sanju09 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 is:

A. 65

B. 55

C. 45

D. 35

E. 25

[spoiler]Source: gmatclub.com

oops, moderators, please move it to the right bin[/spoiler]
Is the OA 45?
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by Tani » Sat Aug 28, 2010 9:55 am
If you add the three devices you will see that there are a total of 210. That means that at least 10 must have all three.



As stated before, the most that can have all three is equal to the smallest number for any individual device or 55.

The difference then is 45
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