counting

[demoskopico]

In a village of 100 households, 75 have at least 1 DVD player, 80 have at least one cell phone, and 55 have at least 1 MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all 3 of these devices, x-y=

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

[kevin]

I assume you are having trouble finding y.

Why don't you start as follows: what is the minimum number of households with both DVD and MP3 players?

[demoskopico]

In a village of 100 households, 75 have at least 1 DVD player, 80 have at least one cell phone, and 55 have at least 1 MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all 3 of these devices, x-y=

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

d>=75
c>=80
m>=55

max= 100 clearly

100= min d + min c + min m - min (d+c) - min (d+m) - min (c+m) - 2 min (d+c+m) + max N, thus =====

min (d+c+m)= 75+80+55 - 75 - 55 - 55 +20=45

is OA C????

[kevin]

How do you justify saying that d is greater than or equal to 75. Doesn't the question say that d is equal to 75?

[demoskopico]

How do you justify saying that d is greater than or equal to 75. Doesn't the question say that d is equal to 75?


don't catch.... could you provide me with clear passages?

[kevin]

The question says that 75 households have at least one DVD player , right?