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DVD, Cell phones, and MP3 players in village

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DVD, Cell phones, and MP3 players in village

by shulapa » Wed Feb 25, 2009 2:11 pm
Hi all,

In a village of 100 housholds, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. Every villager has at least one of these three devices. If X and Y are respectively the greatest and lowest number of house holds that have all three of these devices, X-Y is:

a. 65
b. 55
c. 45
d. 35
e. 25

OA: C
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Re: DVD, Cell phones, and MP3 players in village

by x2suresh » Wed Feb 25, 2009 2:16 pm
shulapa wrote:Hi all,

In a village of 100 housholds, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. Every villager has at least one of these three devices. If X and Y are respectively the greatest and lowest number of house holds that have all three of these devices, X-Y is:

a. 65
b. 55
c. 45
d. 35
e. 25

OA: C
clearly greatest possible nuber = 55


Overlap of DVD AND Cell Phone with least possible numbe r= 75+80 -100= 55

Overlap of (DVD AND Cell Phone with least possible number) and MP3 PLAYER least possible nuber = 55+55-100 = 10



Answer = 55-10=45
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by awesomeusername » Wed Feb 25, 2009 4:15 pm
Look at these other explanations. For someone like myself, who needs to use visuals, there's a post by a user named "calgmatter" who makes it a bit more intuitive.

https://www.manhattangmat.com/forums/vie ... &view=next

The least number of households than can have ALL 3 is:

|----------------------100 households--------------------|

<--------------------80 cell---------------------->20 no cell

25 No DVD<-------------------75 DVD------------------->

<25MP3--><10 MP3>..................................<20MP3>

...............|overlap3|.................................................

How many households overlap for all 3? 10. So Y=10.

55-10 = 45.
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by manishpal » Sat Sep 24, 2011 3:44 am
reply by vibhusethi on manhattan site..
pls read it carefully.it is the easiest way to slove this quest,



I found this way of thinking easier, does it make sense?

max households with no dvd = 25
no cell = 20
no mp3 = 45

Total max possible that dont have all = 90

In other words, atleast 10 households have all 3.

Finding max was easier, i.e 55.

55-10 is the answer.
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by zaarathelab » Tue Oct 11, 2011 10:48 pm
This is a tough question.

It's easy to figure out the maximum no. of households with all three devices (55) but difficult to figure out the minimum number of households with all three.

Can some experts pls explain the quickest way to solve this?
Success = Max(Hardwork) + Min(Luck)
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