rishianand7 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

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

Hi GMATGuru,

I don't understand this part. Can you please simplify it further for me:

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.

Why are each of the 2 overlaps (CM,DM,DC) subtracted from 100?

[email protected] wrote:Hi GMATGuru,

I don't understand this part. Can you please simplify it further for me:

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.

Why are each of the 2 overlaps (CM,DM,DC) subtracted from 100?

In my solution above, the following equation was derived:
(DC + DM + CM) + 2(DCM) = 110.

To determine the least possible value for DCM (households with all 3 devices) we must calculate the greatest possible value for DC+DM+CM (households with exactly 2 of the devices).

Of the 100 households, 75 own D.
Thus, the greatest possible value for CM -- households that own only C and M -- is 25.
Of the 100 households, 80 own C.
Thus, the greatest possible value for DM -- households that own only D and M -- is 20.
Of the 100 households, 55 own M.
Thus, the greatest possible value for DC -- households that own only D and C -- is 45.

Result:
The greatest possible value for DC+CM+CM = 45+20+25 = 90.

[email protected] wrote:
Why are each of the 2 overlaps (CM,DM,DC) subtracted from 100?

You could think of these as the overlaps between the three pairs of two groups, then treat them exactly as you would a typical two-group Venn diagram.. CM is the overlap between C and M, so it must be subtracted one from the C-M set, DM is the overlap between D and M, so it must be subtracted once, and DC is the overlap between D and C, so it must be subtracted once. All three of these belong to the whole group (100), so they must all be subtracted once.

Another way of thinking about it is that anybody who is in C and M (but not D) is counted TWICE: once in C and once in M. But you only want to count them ONCE, so you subtract them ONCE as well to correct for the overcount. (2 counts - 1 overcount = 1 actual count.)