In a village of hundred households, 75 have atleast one DVD player, 80 have atleast one cellphone and 55 have atleast one mp3 player. Every village has atleast 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
Tough Word Problem
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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
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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?
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?
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Hi nishatfarhat87,
This question was discussed in more detail here:
https://www.beatthegmat.com/a-concept-ba ... 70064.html
GMAT assassins aren't born, they're made,
Rich
This question was discussed in more detail here:
https://www.beatthegmat.com/a-concept-ba ... 70064.html
GMAT assassins aren't born, they're made,
Rich
- GMATGuruNY
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In my solution above, the following equation was derived:[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?
(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.
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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.[email protected] wrote:
Why are each of the 2 overlaps (CM,DM,DC) subtracted from 100?
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.)