When determining X (the greatest number of villagers with all three), we know that X cannot exceed the device with the fewest total villagers owning it. Why? We know that if only 55 have an MP3 player, it is impossible for 56 villagers to have all three. Thus, our X is set at 55.

When determining Y (the least amount of villagers with all three), we know that Y cannot be less than the difference between the total number of villagers (100) and the device with most villagers owning it. Why? We know that it's impossible for less households to own all three than own one individually. We want to pick the device with the MOST owners in order to find the LEAST amount of possible owners of all three. The difference between the total and most owned device (cell phones) is 100 - 80 = 20. Thus, our Y is set at 20.

Plugging in both values gives us:
X - Y =
55 - 20 = 35

theCodeToGMAT wrote:In a village of hundred households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. Every village has at least one of these three devices. If X and Y are respectively the greatest and the lowest possible number of households that have all the three of these devices, X-Y is :
A) 65
B) 55
C) 45
D) 35
E) 25
45

In a village of hundred households, 75 have at least one DVD player, 80 have at least one cellphone and 55 have at least one mp3 player. Every village has a tleast one of these 3 devices. If X and Y are respectively the greatest and lowest possible number of households that have all 3 devices then X-Y is?

A) 65
B) 55
C) 45
D) 35
E) 25

Let D = DVD owners, C = cellphone owners, and M = MP3 owners.

T = D + C + M - (DC + DM + CM) - 2(DCM).

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in D, everyone in C, and everyone in M:
Those in exactly 2 of the groups (DC + DM + CM) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (DCM) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

In the problem above:
T = 100
D = 75
C = 80
M = 55.
Thus:
100 = 75 + 80 + 55 - (DC + DM + CM) - 2(DCM)
(DC + DM + CM) + 2(DCM) = 110.

MAXIMUM:
To maximize the value of DCM, we must MINIMIZE the value of DC + DM + CM.
If DC + DM + CM = 0, we get:
0 + 2(DCM) = 110
DCM = 55.

MINIMUM:
To MINIMIZE the value of DCM, we must MAXIMIZE the value of DC + DM + CM.
Since D=75, the maximum possible value of CM = 100-75 = 25.
Since C=80, the maximum possible value of DM = 100-80 = 20.
Since M=55, the maximum possible value of DC = 100-55 = 45.
Since the maximum value of DC + DM + CM = 45+20+25 = 90, we get:
90+ 2(DCM) = 110.
DCM = 10.

Thus:
x-y = 55-10 = 45.

The correct answer is C.

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