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In a village of 100 households, 75 have at least one DVD

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Re: In a village of 100 households, 75 have at least one DVD [#permalink] New post 21 Jul 2013, 00:04
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I have given the solution to a similar problem. Here's the solution for this problem.

1. Let C, D and M be who have cell only, dvd only and mp3 only and CD be who have both cell and dvd only and so on.
2 CDM is minimum when C+D+M=0 and maximum when CD+CM+DM=0.
3. C+D+M+CD+DM+CM+CDM=100 -- (1) because the total of all the mutually exclusive has to sum up to 100 which is the number of households.
4. (C+CD+CM+CDM) + (D+CD+DM+CDM) +(M+CM+DM+CDM) = 80+75+55 or ,
C+D+M+2CD+2CM+2DM+3CDM =210 ----(2)
5. (2) - (1) = CD+CM+DM+2CDM=110 --- (3)
6. (3) - (1) => CDM-(C+D+M) =10 ---(4)
We can find the minimum value when we put (C+D+M) = 0 above giving minimum CDM=10
7. To find maximun value equation (1) can be changed as
C+D+M+CDM=100 -- (5) by putting CD+CM+DM=0
From (4) and (5) we have CDM=55 which is maximun CDM
8.The answer is max CDM - min CDM i.e., 55-10=45

As a shortcut for this type of problems where there are 3 items , the minimum possible for all the 3 can be found out by adding the percentages of the 3 i.e., 55+75+80=210 . and subtracting 200 i.e, 210-200=10 . the maximum can be found out by adding the minimum to 100 and dividing by 2 i.e., (100+10)/2 = 55. Both the answers are percentages.
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In a village of 100 [#permalink] New post 13 Oct 2013, 04:57
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 lowest possible number of households that have all three of these devices, x – y is:

A. 65
B. 55
C. 45
D. 35
E. 25
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Re: In a village of 100 [#permalink] New post 13 Oct 2013, 05:11
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mitulsarwal 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 lowest possible number of households that have all three of these devices, x – y is:

A. 65
B. 55
C. 45
D. 35
E. 25


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Re: 800 Score: Overlapping Set Problem [#permalink] New post 17 Oct 2013, 21:03
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VeritasPrepKarishma wrote:
hiredhanak: I am assuming you are looking for a venn diagram solution to this question..

It is pretty simple.
First of all maximum number of households: We want to bring the circles to overlap as much as possible.
80 - Cell phone
75 - DVD
55 - MP3
Lets take Cell phone and DVD circles since they will have maximum overlap. They must overlap in 55 households so that total number of households is 100. Now put the MP3 households in a way to maximize all three overlap.
Attachment:
Ques1.jpg

So at most 55 households can have all 3.

Now, minimum number of households: We want to take the circles as far apart from each other as possible. Now put the MP3 households in a way to minimize all three overlap. So make the MP3 households occupy the shaded region i.e. region occupied by DVD players alone and cell phone alone. You will be able to adjust 45 MP3s outside the common area but you will need to put 10 of the MP3 households in the common area. So minimum overlap is 10.
Attachment:
Ques2.jpg


x - y = 55 - 10 = 45


Responding to a pm:

When discussing the maximum overlap case, none NEEDN'T be 0. It may be, it may not be.

Put the three circles within each other. The 75 circle within the 80 circle and the 55 circle within the 75 circle. The overlap will be 55 in that case and none = 20. The figure only shows one of the possible ways of obtaining the maximum.

In the case of minimum, you would want the circles to lie as far apart as possible. If none is anything other than 0, the circles would need to overlap more. Say none = 10, the circles of 80 and 75 would need to have an overlap of 65. So the 55 circle can occupy 25 but an overlap of 30 will be needed. Hence minimum overlap will increase. TO minimize the overlap, we will need None = 0.
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Re: 800 Score: Overlapping Set Problem [#permalink] New post 17 Dec 2013, 22:52
noboru wrote:
I have done it in 1:41 doing this:

x= (75+55+80-100-z)/2 for z=0 ->55
y=the same for z=100 ->10

Therefore x-y=45

PS: z is obviously the people who have 2 devices.




Hi Karishma,
is this correct? I am not able to get this approach.

100=80+75+55-z-2a

where z - no. of people with exactly 2 devices
a – no of people with 3 devices
a(max) = x, a(min) = y, (x-y) ?
100 = 210-z-2a
2a = 110-z
a = (110-z)/2

Now, how can we ever take z = 0, as even in case of maximum overlap i.e. when maximum no. of people have 3 devices,
z = 20, which is the overlap b/w C(mobile device) and D(DVD).

Also how is z = 100, all 100 cannot have exactly 2 devices.
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Re: 800 Score: Overlapping Set Problem [#permalink] New post 18 Dec 2013, 01:44
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cumulonimbus wrote:
noboru wrote:
I have done it in 1:41 doing this:

x= (75+55+80-100-z)/2 for z=0 ->55
y=the same for z=100 ->10

Therefore x-y=45

PS: z is obviously the people who have 2 devices.




Hi Karishma,
is this correct? I am not able to get this approach.

100=80+75+55-z-2a

where z - no. of people with exactly 2 devices
a – no of people with 3 devices
a(max) = x, a(min) = y, (x-y) ?
100 = 210-z-2a
2a = 110-z
a = (110-z)/2

Now, how can we ever take z = 0, as even in case of maximum overlap i.e. when maximum no. of people have 3 devices,
z = 20, which is the overlap b/w C(mobile device) and D(DVD).

Also how is z = 100, all 100 cannot have exactly 2 devices.


This solution hasn't considered 'None'. Even if we assume that they saw that None = 0 works for both cases and hence None is immaterial, notice that when z = 100, you get a as 5. That is not correct.

The maximum value of z is 90 (the number of people with exactly 2 devices). This gives the minimum number of people with 3 devices as 10.
The minimum value of z is 0 so max value of a is 55.
Either way, to find the max/min value of z you will need to use some logic. You might as well use it for max/min value of a.
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Re: In a village of 100 households, 75 have at least one DVD [#permalink] New post 04 Apr 2014, 23:31
Stay away venn diagram for problems of this kind..
The maximum number is 55, you don't need any computing for this.

For finding the minimum possible household, just find out the number of households that would not have these gadgets

Number of houses that don't have a cellphone, DVD and MP3 are, 20, 25 and 45 respectively

When there is minimum overlap the number of households that cannot have all the three gadgets together is the sum of these three numbers which is 90. So the minimum possible number of households that can have all three gadgets is 10

55-10 = 45
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Re: 800 Score: Overlapping Set Problem [#permalink] New post 11 May 2014, 21:54
mitulsarwal wrote:
shs0145 wrote:
Am I missing something here??? it seems straightforward......

The obvious maximum that have all 3 is 55, because you are limited by the SMALLEST number.

The minimum is simply the sum of the max of each people who DONT have the product, so:

100-80 = 20 don't have Cell
100-75 = 25 don't have DVD
and 100-55 = 45 don't have MP3

SO a total of 20+25+45 = 90 combined who might NOT have SOME combination of the 3 products. So subtract that from 100, to give you the minimum of the people who COULD have all 3 and you get 100-90 = 10.

55-10 = 45


i think this is the simplest way to solve this, even i did it the same way. :-D


This looks the simplest approach of the lot. I wonder if it will be true for all scenarios. Bunuel any comments ?
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Re: 800 Score: Overlapping Set Problem [#permalink] New post 19 May 2014, 05:22
himanshujovi wrote:
mitulsarwal wrote:
shs0145 wrote:
Am I missing something here??? it seems straightforward......

The obvious maximum that have all 3 is 55, because you are limited by the SMALLEST number.

The minimum is simply the sum of the max of each people who DONT have the product, so:

100-80 = 20 don't have Cell
100-75 = 25 don't have DVD
and 100-55 = 45 don't have MP3

SO a total of 20+25+45 = 90 combined who might NOT have SOME combination of the 3 products. So subtract that from 100, to give you the minimum of the people who COULD have all 3 and you get 100-90 = 10.

55-10 = 45


i think this is the simplest way to solve this, even i did it the same way. :-D


This looks the simplest approach of the lot. I wonder if it will be true for all scenarios. Bunuel any comments ?



This is a wrong approach. Imagine you have number like 80, 75, 65 instead of 55. In this case the maximum would not be 65 as suggested but would be 60.

Even for minimum case this will not work. You have to use the approached specified above.

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Re: 800 Score: Overlapping Set Problem [#permalink] New post 19 May 2014, 19:52
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mittalg wrote:
This is a wrong approach. Imagine you have number like 80, 75, 65 instead of 55. In this case the maximum would not be 65 as suggested but would be 60.

Even for minimum case this will not work. You have to use the approached specified above.

Hope it helps!!!

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Actually the method used above is fine. You will get a max of 65, not 60. Imagine the circles one inside the other. 75 inside the 80 and 65 inside the 75. All 65 will have all 3 products.
For minimum, you need to spread them as far away as possible. 80 and 75 will have an overlap of 55. So after spreading 65 on 45, you will be left with 20 which will need to overlap with the 55. Hence minimum will be 20.

In the method used above, people who do not own at least one product will be 20, 25 and 35. Spread them out as far apart as possible, you get 20+25+35 = 80
Minimum number of people who must have all 3 = 100 - 80 = 20

So you can go with people who have products or those who don't. Either way, you get the same answer.
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Re: 800 Score: Overlapping Set Problem [#permalink] New post 19 May 2014, 19:57
VeritasPrepKarishma wrote:
mittalg wrote:
This is a wrong approach. Imagine you have number like 80, 75, 65 instead of 55. In this case the maximum would not be 65 as suggested but would be 60.

Even for minimum case this will not work. You have to use the approached specified above.

Hope it helps!!!

Kudos if you like!!


Actually the method used above is fine. You will get a max of 65, not 60. Imagine the circles one inside the other. 75 inside the 80 and 65 inside the 75. All 65 will have all 3 products.
For minimum, you need to spread them as far away as possible. 80 and 75 will have an overlap of 55. So after spreading 65 on 45, you will be left with 20 which will need to overlap with the 55. Hence minimum will be 20.

In the method used above, people who do not own at least one product will be 20, 25 and 35. Spread them out as far apart as possible, you get 20+25+35 = 80
Minimum number of people who must have all 3 = 100 - 80 = 20

So you can go with people who have products or those who don't. Either way, you get the same answer.


Yes, you are right Karishma. When I did the analysis, I mis-assumed that all of the 100 people have at least one of the 3 products which is not the case. Sorry for the confusion.
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Re: In a village of 100 households, 75 have at least one DVD [#permalink] New post 19 May 2014, 22:15
Its funny how complicated your are thinking. Here was my approach:

All the figure say they have AT LEAST ... (so maybe more)

all of them could have 100. which is the maximum.

the lowest possible number is 55(55 have at least one mp3 player)

100-55=45
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Re: In a village of 100 households, 75 have at least one DVD [#permalink] New post 20 May 2014, 08:53
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dansa wrote:
Its funny how complicated your are thinking. Here was my approach:

All the figure say they have AT LEAST ... (so maybe more)

all of them could have 100. which is the maximum.

the lowest possible number is 55(55 have at least one mp3 player)

100-55=45


This is not correct. The question does not say "at least 75 people have one DVD player". It says "75 people have at least one DVD player" which means 75 people have one or more DVDs. It doesn't mean that number of people having a DVD is 75 or more.
The maximum is 55 and minimum is 10. Check the solutions given above.
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Re: In a village of 100 households, 75 have at least one DVD   [#permalink] 20 May 2014, 08:53
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