No idea how to name this lol

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.

For similar problems, check here:
https://www.beatthegmat.com/group-of-stu ... 63753.html
https://www.beatthegmat.com/sets-t148362.html

GMATGuruNY wrote:

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

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:
https://www.beatthegmat.com/group-of-stu ... 63753.html
https://www.beatthegmat.com/sets-t148362.html

Can you eloborate on this part more ? Why is CM ( and the rest ) = Total - D

sapuna wrote:
Can you eloborate on this part more ? Why is CM ( and the rest ) = Total - D

CM = the number who own a cellphone and an MP3 player but NOT a DVD player.
Since 75 of the 100 people own a DVD player, the greatest number who could own both a cellphone and an MP3 player but NOT a DVD player = 100-75 = 25.
Thus, the greatest possible value of CM = 25.

Similar reasoning was used to determine that the greatest possible value of DM = 20 and that the greatest possible value of DC = 45.

GMATGuruNY wrote:

sapuna wrote:
Can you eloborate on this part more ? Why is CM ( and the rest ) = Total - D

CM = the number who own a cellphone and an MP3 player but NOT a DVD player.
Since 75 of the 100 people own a DVD player, the greatest number who could own both a cellphone and an MP3 player but NOT a DVD player = 100-75 = 25.
Thus, the greatest possible value of CM = 25.

Similar reasoning was used to determine that the greatest possible value of DM = 20 and that the greatest possible value of DC = 45.

Ah , yeah. Silly me. Of course that out of 3 items the people who own 2 at most will equal the total number of people minus one particular group. Thank you again. Keep up the good work ! : )

Guys ,

can we also solve this question through Venn Diagram , if yes can you pls explain how?

Thanks ,

Shreyans