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by jainrahul1985 » Tue Dec 06, 2011 9:10 am
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

OA C . Easy way to solve this question ?
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by batmannavneet » Tue Dec 06, 2011 10:07 am
We have -

D = 75, no D = 25
C = 80, no C = 20
M = 55, no M = 45

So, D+C+M - (DC+CM+DM) - 2DCM = 100
or, DC+CM+DM + 2DCM = (75+80+55) - 100 = 110 ------1

To mimimize DCM, we have DC+CM+DM = 25+20+45 = 90 ---- 2
To maximize DCM, we make DC+CM+DM = 0 -----3

Therefore,
From 1 and 2, after solving equation, we have DCM = 10
From 1 and 3, after solving equation, we have DCM = 55

Difference of the 2 is = 45, pick answer C
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by GMATGuruNY » Tue Dec 06, 2011 10:12 am
jainrahul1985 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

OA C . Easy way to solve this question ?
T = G1 + G2 + G3 - (those in 2 of the groups) - 2*(those in all 3 groups)

The big idea with overlapping groups is to subtract the overlaps. When we count everyone in the 3 groups, those in 2 of the groups will be counted twice, so they need to subtracted from the total once. Those in all 3 groups will be counted 3 times, so they need to be subtracted from the total twice.

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

Minimum value of DCM:
To minimize DCM, we need to maximize DC + DM + CM.
Since there are 100 households, the maximum value of DC + DM + CM + DCM = 100.
Subtracting this equation from the first equation, we get:
DC + DM + CM + 2(DCM) - (DC + DM + CM + DCM) = 110-100
DCM = 10.

Maximum value of DCM:
To maximize DCM, we need to minimize DC + DM + CM.
Let DC + DM + CM = 0.
Substituting DC + DM + CM = 0 into DC + DM + CM + 2(DCM) = 110, we get:
0 + 2(DCM) = 110
DCM = 55.

Max - min = 55-10 = 45.

The correct answer is C.
Last edited by GMATGuruNY on Thu Dec 08, 2011 5:54 pm, edited 1 time in total.
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by GmatMathPro » Thu Dec 08, 2011 10:13 am
jainrahul1985 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

OA C . Easy way to solve this question ?
Minimum:

To find the minimum, add 75+80+55=210, and then subtract twice the total number of households: 210-2*100=10, so minimum is 10. Edit: note that if you get a negative number with this approach, the minimum is zero.

Maximum:

Add 75+80+55=210, subtract the total, and divide this number by 2, ignoring the remainder: 210-100=110. 110/2=55. This number, or the least of the three individual groups, will be the maximum, whichever is less. In this case they are both 55, so the maximum is 55.

55-10=45.

Note that the maximum in situations like this, where they specify that each household has at least one thing, will not always just be the lowest of the three values. For example, if we had had DVD-60, Cell phone-65, mp3-55, then if we put 55 as the number of households that have all 3, then we could put 10 households as having just a cell phone and 5 households as having just a dvd player, but that would only give us a total of 70 households, and we don't have any more space in any of the circles to get the total number up to 100. If they hadn't said every household has at least one thing, we could just say that there could be 30 households that don't own any of them and be done. But if every household has to have at least one of these things, we have a problem. In this case, the maximum number of households that own all three would be 40, by the method outlined above.
Last edited by GmatMathPro on Thu Dec 08, 2011 2:52 pm, edited 1 time in total.
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by chieftang » Thu Dec 08, 2011 1:40 pm
GmatMathPro wrote:
jainrahul1985 wrote: Note that the maximum in situations like this, where they specify that each household has at least one thing, will not always just be the lowest of the three values. For example, if we had had DVD-60, Cell phone-65, mp3-55, then if we put 55 as the number of households that have all 3, then we could put 10 households as having just a cell phone and 5 households as having just a dvd player, but that would only give us a total of 70 households, and we don't have any more space in any of the circles to get the total number up to 100. If they hadn't said every household has at least one thing, we could just say that there could be 30 households that don't own any of them and be done. But if every household has to have at least one of these things, we have a problem. In this case, the maximum number of households that own all three would be 40, by the method outlined above.
Another note: In the case you specify above, then the minimum calculation results in a negative number. 180-2*100 = -20.
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by GmatMathPro » Thu Dec 08, 2011 2:46 pm
chieftang wrote:
GmatMathPro wrote:
jainrahul1985 wrote: Note that the maximum in situations like this, where they specify that each household has at least one thing, will not always just be the lowest of the three values. For example, if we had had DVD-60, Cell phone-65, mp3-55, then if we put 55 as the number of households that have all 3, then we could put 10 households as having just a cell phone and 5 households as having just a dvd player, but that would only give us a total of 70 households, and we don't have any more space in any of the circles to get the total number up to 100. If they hadn't said every household has at least one thing, we could just say that there could be 30 households that don't own any of them and be done. But if every household has to have at least one of these things, we have a problem. In this case, the maximum number of households that own all three would be 40, by the method outlined above.
Another note: In the case you specify above, then the minimum calculation results in a negative number. 180-2*100 = -20.
Right, thanks. To clarify, if you get a negative number using that approach, then the minimum would be zero.
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by gunjan1208 » Fri Dec 09, 2011 3:23 am
Hi GMATPro,

Many thanks. Could you please let me know if I there is any online source I can read and know more about venn? Its little tough for me to grasp.

Regards
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