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In a Village...

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In a Village...

by sk8ternite » Wed Aug 26, 2009 2:09 pm
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
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by rahulmehra13 » Wed Aug 26, 2009 9:39 pm
IMO C

Greatest number of household to have all three = 55

To calculate Lowest:

Total houses = 100

Suppose house number 1 to 75 have DVD players
and house number 21 to 100 have Cell Phones

Now, house no. 1 to 20 have only DVDs
and house no. 76 to 100 have only cell phones.

so total 45 houses have only 1 thing.

Now we have 55 mp3 players, out of 55 mp3s ,45 will go to above houses (1-20 & 76-100) and rest 10 will go to houses numbered 21 to 75 (note 21-75 already have both dvd and cell phone). That means lowest no of houses to have all three = 10

X-Y = 55-10 = 45
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by heshamelaziry » Thu Aug 27, 2009 1:43 am
Could somebody please put forward a more detailed explanation for this problem ????????


Thank You
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by viju9162 » Thu Aug 27, 2009 5:05 am
The greatest and lowest possible number of households that have all three of these devices would be 100 because in total there are 100 houses and each house can have at least 100 devices...

The minimum can be 55 ( mp3 player). Hence 100-55=45
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