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# In a village of 100 households. 75 have at least one dvd

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In a village of 100 households. 75 have at least one dvd [#permalink]  26 Jul 2008, 09:53
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In a village of 100 households. 75 have at least one dvd player, 80 have at least one cell phone, and 55 have at least one mp3 player. If X and Y are respectively the greatest and the lowest possible number of house holds that have all three devices X- Y=??

A. 25
b 35
c 45
D 55
E 65
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Re: Tough one: Venn Diag/Sets [#permalink]  26 Jul 2008, 10:22
Houses with DVD Players only = a
Houses with MP3 Players only = b
Houses with Cell Phone only = C
Houses with DVD Players and MP3 Players only = d
Houses with DVD Players and Cell Phones only = e
Houses with Cell Phones and MP3 Players only = f
Houses with all three gadgets = g

Houses with DVD Players = a+d+e+g = 75
Houses with MP3 Players = b+d+f+g = 55
Houses with Cell Phones = c+e+f+g = 80

Adding above 3 equations = a+b+c+2(d+e+f)+3g = 210

100 = a+b+c+d+e+f+g

Assume a+b+c = x and d+e+f=y
x+2y+3g=210
x+y+g=100

g will be max when y=0 => g = 55
minimum value of g is 0.

so answer = 55 - 0 = 55
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Re: Tough one: Venn Diag/Sets [#permalink]  26 Jul 2008, 10:31
abhijit_sen wrote:
Houses with DVD Players only = a
Houses with MP3 Players only = b
Houses with Cell Phone only = C
Houses with DVD Players and MP3 Players only = d
Houses with DVD Players and Cell Phones only = e
Houses with Cell Phones and MP3 Players only = f
Houses with all three gadgets = g

Houses with DVD Players = a+d+e+g = 75
Houses with MP3 Players = b+d+f+g = 55
Houses with Cell Phones = c+e+f+g = 80

Adding above 3 equations = a+b+c+2(d+e+f)+3g = 210

100 = a+b+c+d+e+f+g

Assume a+b+c = x and d+e+f=y
x+2y+3g=210
x+y+g=100

g will be max when y=0 => g = 55
minimum value of g is 0.

so answer = 55 - 0 = 55

Even i got this. However, this is not the correct answer. I also cannot point out the flaw in ur explanation.
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Re: Tough one: Venn Diag/Sets [#permalink]  26 Jul 2008, 10:34
abhijit_sen wrote:
Houses with DVD Players only = a
Houses with MP3 Players only = b
Houses with Cell Phone only = C
Houses with DVD Players and MP3 Players only = d
Houses with DVD Players and Cell Phones only = e
Houses with Cell Phones and MP3 Players only = f
Houses with all three gadgets = g

Houses with DVD Players = a+d+e+g = 75
Houses with MP3 Players = b+d+f+g = 55
Houses with Cell Phones = c+e+f+g = 80

Adding above 3 equations = a+b+c+2(d+e+f)+3g = 210

100 = a+b+c+d+e+f+g

Assume a+b+c = x and d+e+f=y
x+2y+3g=210
x+y+g=100

g will be max when y=0 => g = 55
minimum value of g is 0.

so answer = 55 - 0 = 55

Agree .... IMO also ans: 55 (Max when everybody has 3: 55, min when no body has 3: 0).

EDITED:

Explanation same as above, except little change during last steps:

Adding above 3 equations = a+b+c+2(d+e+f)+3g = 210

100 = a+b+c+d+e+f+g

from the above 2 eqn:
d+e+f+2g = 110
g = [110 - (d+e+f)]/2

g is max, when (d+e+f)=0, g= 55
g is min, when (d+e+f)=110, g=0

substracting the 2, we get g=55
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Re: Tough one: Venn Diag/Sets [#permalink]  26 Jul 2008, 10:51
GMBA85 wrote:
In a village of 100 households. 75 have at least one dvd player, 80 have at least one cell phone, and 55 have at least one mp3 player. If X and Y are respectively the greatest and the lowest possible number of house holds that have all three devices X- Y=??

A. 25
b 35
c 45
D 55
E 65

(A)

X = 55

For calculating Y, imagine three horizontal bar graphs between 2 parallel y axis.

The first one, lets call it CELL, represents mp3 players and extends from the left y-axis towards the right. The other one (DVD), representing dvd players, extends from the right y-axis towards the left. Their intersection is 30, representing how many people have both CELL and DVD.
The bar with cell phones is irrelevant as it encompasses either one in its direction (extend it from either bar graph, and it won't impact the solution).

Hence, X-Y = 55 - 30 = 25
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Re: Tough one: Venn Diag/Sets [#permalink]  27 Jul 2008, 19:33
incognito1 wrote:
GMBA85 wrote:
In a village of 100 households. 75 have at least one dvd player, 80 have at least one cell phone, and 55 have at least one mp3 player. If X and Y are respectively the greatest and the lowest possible number of house holds that have all three devices X- Y=??

A. 25
b 35
c 45
D 55
E 65

(A)

X = 55

For calculating Y, imagine three horizontal bar graphs between 2 parallel y axis.

The first one, lets call it CELL, represents mp3 players and extends from the left y-axis towards the right. The other one (DVD), representing dvd players, extends from the right y-axis towards the left. Their intersection is 30, representing how many people have both CELL and DVD.
The bar with cell phones is irrelevant as it encompasses either one in its direction (extend it from either bar graph, and it won't impact the solution).

Hence, X-Y = 55 - 30 = 25

OA is 45

will post the OE if we are not able to come up with it.
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Re: Tough one: Venn Diag/Sets [#permalink]  27 Jul 2008, 20:03
1
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GMBA85 wrote:
incognito1 wrote:
GMBA85 wrote:
In a village of 100 households. 75 have at least one dvd player, 80 have at least one cell phone, and 55 have at least one mp3 player. If X and Y are respectively the greatest and the lowest possible number of house holds that have all three devices X- Y=??

A. 25
b 35
c 45
D 55
E 65

(A)

X = 55

For calculating Y, imagine three horizontal bar graphs between 2 parallel y axis.

The first one, lets call it CELL, represents mp3 players and extends from the left y-axis towards the right. The other one (DVD), representing dvd players, extends from the right y-axis towards the left. Their intersection is 30, representing how many people have both CELL and DVD.
The bar with cell phones is irrelevant as it encompasses either one in its direction (extend it from either bar graph, and it won't impact the solution).

Hence, X-Y = 55 - 30 = 25

OA is 45

will post the OE if we are not able to come up with it.

Tricky +1.. GMBA.. nice question

(D) 45 makes sense. In the graph I mentioned earlier, we're given
CELL=80
DVD=75
MP3=55

Consider two parallel y-axis and these bar graphs between them. X (maximum number of people with all 3) is clearly 55. For Y. Draw a bar graph representing MP3 from left to right. Then draw one for DVD from right to left. Their intersection is 30. The mistake I made earlier is discounting CELL, but you HAVE to account for it. Upon minimizing CELL on this graph, the minimal common area or Y comes down to 10.

Hence X - Y = 55 - 10 = 45
Attachments

Graph.PNG [ 8.12 KiB | Viewed 896 times ]

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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 00:00
Very well explained incognito1. +1.
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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 00:46
2
KUDOS
GMBA85 wrote:
In a village of 100 households. 75 have at least one dvd player, 80 have at least one cell phone, and 55 have at least one mp3 player. If X and Y are respectively the greatest and the lowest possible number of house holds that have all three devices X- Y=??

A. 25
b 35
c 45
D 55
E 65

My Approach:

Let A be the number of house holds with 3 devices. (i.e. dvd player, cell phone and mp3 player)
Let B be the number of house holds with 2 devices.
Let C be the number of house holds with 1 device.

therefore, we can write the following expression for the number of devices:

3A + 2B + C = 210 ------- (1)

also, A+ B+C = 100 ------- (2)

Now, the maximum number of house holds can be 55 or less. But in order to have the maximum value for X, A should have the highest value.
lets try 55, with equation (1) and (2)

we get 2B+ C= 45
B+ C = 45

or B = 0, C will be 45.

Thus the maximum value of X is 55.

Now, the minimum number of house holds can be 0. But in order to have the minimum value for Y, A should have the lowest value. i.e 0 or slightly more than that.
lets try 0, with equation (1) and (2)

we get 2B+C= 210
B+C=100
on solving we get B= 110, C= -10. Thus not true.

Thus A should have a value which should make C: atleast positive and B atleast less than 100.

Trying A=5.

(1) and (2) become:

2B + C= 195
B+C = 95

B= 100, C= -5. Situation cant hold true.

Trying A= 10

(1) and (2) become:

2B+C= 180
B+C=90

B=90, A=10, C=10

Thus situation can hold true.

Thus minimum value for Y = 10

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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 08:26
GMBA85 wrote:
In a village of 100 households. 75 have at least one dvd player, 80 have at least one cell phone, and 55 have at least one mp3 player. If X and Y are respectively the greatest and the lowest possible number of house holds that have all three devices X- Y=??

A. 25
b 35
c 45
D 55
E 65

Consider all the households who have mp3 player also have the other two devices, then X = 55

consider the "worst case",
among the 55 owners of mp3 player, 20 of them do not have cell phone.
So number of villagers who have both mp3 player and cell phone is 55-20 = 35
Among the 35 villagers who have both mp3 player and cell phone, the worst case is 25 of them do not have dvd player.
So number of villagers who have all 3 devices is 35 - 25 = 10
Y = 10

X-Y = 55-10 = 45

C
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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 09:54
guys- is that gmatprep?
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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 11:32
How can you have the max is 55?
Thanks,

rahulgoyal1986 wrote:
GMBA85 wrote:
In a village of 100 households. 75 have at least one dvd player, 80 have at least one cell phone, and 55 have at least one mp3 player. If X and Y are respectively the greatest and the lowest possible number of house holds that have all three devices X- Y=??

A. 25
b 35
c 45
D 55
E 65

My Approach:

Let A be the number of house holds with 3 devices. (i.e. dvd player, cell phone and mp3 player)
Let B be the number of house holds with 2 devices.
Let C be the number of house holds with 1 device.

therefore, we can write the following expression for the number of devices:

3A + 2B + C = 210 ------- (1)

also, A+ B+C = 100 ------- (2)

Now, [i]the maximum number of house holds can be 55 or less[/i]. But in order to have the maximum value for X, A should have the highest value.
lets try 55, with equation (1) and (2)

we get 2B+ C= 45
B+ C = 45

or B = 0, C will be 45.

Thus the maximum value of X is 55.

Now, the minimum number of house holds can be 0. But in order to have the minimum value for Y, A should have the lowest value. i.e 0 or slightly more than that.
lets try 0, with equation (1) and (2)

we get 2B+C= 210
B+C=100
on solving we get B= 110, C= -10. Thus not true.

Thus A should have a value which should make C: atleast positive and B atleast less than 100.

Trying A=5.

(1) and (2) become:

2B + C= 195
B+C = 95

B= 100, C= -5. Situation cant hold true.

Trying A= 10

(1) and (2) become:

2B+C= 180
B+C=90

B=90, A=10, C=10

Thus situation can hold true.

Thus minimum value for Y = 10

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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 19:08
rahulgoyal1986 wrote:
GMBA85 wrote:
In a village of 100 households. 75 have at least one dvd player, 80 have at least one cell phone, and 55 have at least one mp3 player. If X and Y are respectively the greatest and the lowest possible number of house holds that have all three devices X- Y=??

A. 25
b 35
c 45
D 55
E 65

My Approach:

Let A be the number of house holds with 3 devices. (i.e. dvd player, cell phone and mp3 player)
Let B be the number of house holds with 2 devices.
Let C be the number of house holds with 1 device.

therefore, we can write the following expression for the number of devices:

3A + 2B + C = 210 ------- (1)

also, A+ B+C = 100 ------- (2)

Now, the maximum number of house holds can be 55 or less. But in order to have the maximum value for X, A should have the highest value.
lets try 55, with equation (1) and (2)

we get 2B+ C= 45
B+ C = 45

or B = 0, C will be 45.

Thus the maximum value of X is 55.

Now, the minimum number of house holds can be 0. But in order to have the minimum value for Y, A should have the lowest value. i.e 0 or slightly more than that.
lets try 0, with equation (1) and (2)

we get 2B+C= 210
B+C=100
on solving we get B= 110, C= -10. Thus not true.

Thus A should have a value which should make C: atleast positive and B atleast less than 100.

Trying A=5.

(1) and (2) become:

2B + C= 195
B+C = 95

B= 100, C= -5. Situation cant hold true.

Trying A= 10

(1) and (2) become:

2B+C= 180
B+C=90

B=90, A=10, C=10

Thus situation can hold true.

Thus minimum value for Y = 10

Good explanation. +1
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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 19:31
[quote="chan_nhu78"]How can you have the max is 55?
Thanks,

Hey

I took the max as 55 because, as per the given data, its clear that atleast 55 households have a probability of having all the three devices. i.e more than 55 households can not have all the three devices.

Hope the above makes sense.
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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 20:14
Damn, looks like I need to work on those paintbrush skills, or else those kudos will stop coming!

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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 21:20
incognito1 wrote:
Damn, looks like I need to work on those paintbrush skills, or else those kudos will stop coming!

You already have many : )
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Re: Tough one: Venn Diag/Sets [#permalink]  28 Jul 2008, 22:57
incognito1 wrote:
Damn, looks like I need to work on those paintbrush skills, or else those kudos will stop coming!

Well U've a very effecient conversion rate
Re: Tough one: Venn Diag/Sets   [#permalink] 28 Jul 2008, 22:57
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