Problem Solving

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.

For similar problems, check here:
http://www.beatthegmat.com/group-of-students-t263753.html
http://www.beatthegmat.com/sets-t148362.html

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Student Review #1
Student Review #2
Student Review #3

The greatest possible number of households that have all three devices is 55 when all the households that have a MP3 player also have a DVD player and a cell phone. So x = 55.

Since 75 + 55 = 130 and 130 - 100 = 30, then 30 households must have both a DVD player and an MP3 player. Some of these 30 households might have a cell phone and some might not. Since 30 + 80 = 110 and 110 - 100 = 10, then 10 households must have a cell phone in addition to a DVD player and a MP3 player. In other words, a minimum of 10 households must have all three devices. So y = 10.

Therefore, x - y = 55 - 10 = 45.

Answer: C

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