venn

Question

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 ?

batmannavneet · Answer

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

GMATGuruNY · Answer

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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GmatMathPro · Answer

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.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

chieftang · Answer

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.

GmatMathPro · Answer

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.

gunjan1208 · Answer

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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