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In a village of 100 households, 75 have at least one DVD player, 80 19 May 2014, 20:52
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mittalg

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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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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In a village of 100 households, 75 have at least one DVD player, 80 19 May 2014, 23:15
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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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In a village of 100 households, 75 have at least one DVD player, 80 20 May 2014, 09:53
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dansa
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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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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In a village of 100 households, 75 have at least one DVD player, 80 21 Jun 2015, 18:50
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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 put it in the shaded region. You will need to put 10 of the MP3 households in the common area. So minimum overlap is 10.

Ok, and how did you get to the number 10?

(25 + 20 =) 45 households have either only Cell or only DVD Player. Out of the 55 households who have MP3 Players, put 45 in these areas so that all three do not overlap. But the rest of the (55 - 45 =)10 households that have MP3 players need to be put in the common region consisting of 55 households that have both Cell and DVD Player. Hence overlap of all three will be 10.
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________________________________

Where does it say that every household in the village have at least one of these three devices?
There could be none of these three devices in some households.
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In a village of 100 households, 75 have at least one DVD player, 80 22 Jun 2015, 05:29
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________________________________

Where does it say that every household in the village have at least one of these three devices?
There could be none of these three devices in some households.
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Hi wrightvijay,

It doesn't matter whether they have given the highlighted fact above or not because it's the question of maximizing and minimizing the overlapping portion which will be maximized when everything else is minimized and vice versa.
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In a village of 100 households, 75 have at least one DVD player, 80 22 Jun 2015, 20:29
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Where does it say that every household in the village have at least one of these three devices?
There could be none of these three devices in some households.
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It's not given. It is something we have derived using logic.

Take a simpler case:

Say you have 3 siblings (A,B and C) and 5 chocolates which you want to distribute among them in any way you wish. Now you want to minimise the number of your siblings who get 3 chocolates. No one gets more than 3. What do you do?

Will you leave one sibling without any chocolates (even if he did rat you out to your folks!)? No. Because if one sibling gets no chocolates, the other siblings get more chocolates and then more of them will get 3 chocolates. So instead you give 1 to each and then give the leftover 2 to 2 of them. This way, no sibling gets 3 chocolates and you have successfully minimised the number of siblings who get 3 chocolates. Basically, you spread out the goodies to ensure that minimum people get too many of them.

This is the same concept. When you want to minimise the overlap, you basically want to spread the goodies around. You want minimum people to have all 3. So you give atleast one to all of them.
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In a village of 100 households, 75 have at least one DVD player, 80 28 Jul 2015, 23:04
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The most (x) number of households with all three is the least of the given, so 55. That one's straightforward.

The least (y) is when you add up all of the single households and assume as many households with at least two of the three items. 75 + 80 + 55 = 210. There are 100 households and if we are to maximize the number of households with at least two of the three items, that's 200 of the 210 items. Then, 10 are still remaining which tells us that there must be at least 10 households which have three items once every household doubles up. y = 10

x-y= 55-10=45
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In a village of 100 households, 75 have at least one DVD player, 80 Updated on: 27 Nov 2020, 04:46
Originally posted by XavierAlexander on 29 Aug 2015, 15:19.
Last edited by XavierAlexander on 27 Nov 2020, 04:46, edited 1 time in total.
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A Simple Approach



Cell -- 80 | DVD -- 75 | MP3 -- 55
Max/Min Overlapping section for all three. -- ??

MIN --> Two ways (choose the smaller. i.e. 10)
Attachment:
min-1.png
min-1.png [ 6.91 KiB | Viewed 5949 times ]

Attachment:
min-2.png
min-2.png [ 7.59 KiB | Viewed 5934 times ]


MAX --> One way (55)
Attachment:
max.png
max.png [ 7.54 KiB | Viewed 5901 times ]


Ans = Max - Min = 55 - 10 = 45


Hussain15
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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In a village of 100 households, 75 have at least one DVD player, 80 19 Mar 2016, 04:01
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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
Show more





Hi Karishma

Can you please help me to solve the question by applying the below formula ?
Total-Neither = A+B+C - (common in two) - 2(common in three)
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In a village of 100 households, 75 have at least one DVD player, 80 20 Mar 2016, 23:39
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VeritasPrepKarishma
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





Hi Karishma

Can you please help me to solve the question by applying the below formula ?
Total-Neither = A+B+C - (common in two) - 2(common in three)
Show more

Hey Radhika,

In this question, you cannot just plug in the numbers in a formula and get the answer. You will need to do a bit of analysis for the possible overlap since you need the maximum and minimum value of overlap of all 3.

Common in three = A + B + C - (common in two) - Total + Neither
Common in three = (75 + 80 + 55 - (common in two) - 100 + Neither)/2
Common in three = (110 - (common in two) + Neither)/2
You need to maximise "common in three". For that, imagine the 75 circle inside the 80 circle and the 55 circle is inside the 75 circle. In that case, neither = 100 - 80 = 20 and common in two = 75 - 55 = 20.
So Common in three = 55

Similarly, we will minimise Common in three.

Effectively, we have used the venn diagram only to answer the question.
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In a village of 100 households, 75 have at least one DVD player, 80 07 Nov 2016, 10:21
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Well its way too straight forward.

The max value(x) must be 100.
Of all 3 the least value will be min possible value = 55

X-Y = 100-55 = 45.
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Bunuel VeritasPrepKarishma Does this approach work? This was what I though of too because :

100 = No. of households
55 = No. of MP3s for sure (so assuming that min of 55 households have all 3)
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In a village of 100 households, 75 have at least one DVD player, 80 07 Nov 2016, 20:50
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vedantkabra
navigator123
Well its way too straight forward.

The max value(x) must be 100.
Of all 3 the least value will be min possible value = 55

X-Y = 100-55 = 45.

Bunuel VeritasPrepKarishma Does this approach work? This was what I though of too because :

100 = No. of households
55 = No. of MP3s for sure (so assuming that min of 55 households have all 3)
Show more

This is incorrect.

How can 100 households have all 3 when only 55 households have MP3 players?
55 is the MAXIMUM number of households that can have all 3, not minimum.
and 10 is the minimum number of households that must have all 3 (explained in solutions on first page)
So 55 - 10 = 45
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In a village of 100 households, 75 have at least one DVD player, 80 10 May 2017, 19:05
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How would one solve this problem if a portion of the 100 households did not own ANY of the electronics mentioned in the question?
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In a village of 100 households, 75 have at least one DVD player, 80 11 May 2017, 08:26
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hoopsgators
How would one solve this problem if a portion of the 100 households did not own ANY of the electronics mentioned in the question?
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The question does not mention whether there are some households that own no electronics. There could be some.

Look at the figure here: https://gmatclub.com/forum/in-a-village ... ml#p825632

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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In a village of 100 households, 75 have at least one DVD player, 80 24 May 2017, 06:29
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I could not load the Venn Diagram, Please try to visualize it. ( The diagram is posted below)

a, b, and c one product each.

x, y, and z two products each.

d all three products.


a + b + c + d + x + y + z = 100
If you assume there is no one who uses two products each, then x = y = z = 0.
So, a + b + c + d = 100 ----------------(1)
a + d = 80 ------------------------------(2)
b + d = 75 ------------------------------(3)
c + d = 55 ------------------------------(4)

From 2, 3, and 4,
a + b + c + 3d = 210 ----------------(5)
a + b + c + d + 2d = 210

From 1 and 5,
100 + 2d = 210
d = 55 --------------Maximum.

If you assume there is no one who uses one product each, then a = b = c = 0.
So, d + x + y + z = 100 ----------------(6)
x + y + d = 80 ------------------------------(7)
x + z + d = 75 ------------------------------(8)
y + z + d = 55 ------------------------------(9)

From 7, 8, and 9,
2x + 2y + 2z + 3d = 2(x + y + z + d ) + d = 210 ----------------(10)

From 6 and 10,
2(100) + d = 210
d = 10 --------------Minimum.

So Maximum – minimum = 55 – 10 = 45.
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In a village of 100 households, 75 have at least one DVD player, 80 15 Jan 2018, 10:36
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VeritasPrepKarishma
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
Show more

Responding to a pm: Explaining the min case.

We want minimum overlap of all three. So we need to spread the 3 around in such a way that all 3 overlap the least. Out of 100 households, 80 have a cell phone. We have 20 households leftover without a cell phone. Now 75 have a DVD so we reduce overlap by putting 20 DVDs in the leftover 20 households. Rest 75 - 20 = 55 will need to be overlapped with DVDs.

So now we have 20 households with just DVDs, 25 with just cell phones and 55 with both.
Now we also have to distribute 55 MP3s. Note that overlap of all 3 has to be reduced as much as possible. So we should give minimum to the households that already have both. The rest of the 45 households get MP3s (not these 45 have 2 things each). But we still have 10 MP3s leftover. These will need to be given to the households which have DVDs and cell phones both. So minimum overlap of all 3 is 10.
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In a village of 100 households, 75 have at least one DVD player, 80 10 Jan 2019, 12:43
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My imagination about minimum overlapping percentage.

Attachment:
The attachment overlapp.jpg is no longer available
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After spending all day obsessing over this problem your diagram inspired me to find a solution that works for me.

We know the formula for three overlapping sets is A+B+C-(exactly two overlap) -2*(exactly three overlap) = Total



So A+B+C-(exactly two overlap)-2*(exactly three overlap)=100. Now when we add up A,B,C we get 75+55+80=210, and subtracting 100, we have 110 extra households that need to be distributed among the three. We can only distribute to A,B,C until their total equals 100 as shown in the image I have attached (since there are max 100 holds. This means we can distribute 20, 25, and 45 = 90 items total. Subtracting that from our 110 extra households, and we get 20. Remember, that since from the formula we count the three overlap twice, the minimum 3 overlap area will be 20/2=10. Now, we know the max overlap is the smallest of the three groups = 55, therefore our answer is X-Y = 55-10=45 D
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In a village of 100 households, 75 have at least one DVD player, 80 13 Jul 2019, 03:14
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ajithkumar
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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Dear Moderators chetan2u, VeritasKarishma, Bunuel, Gladiator59, generis

Is the above approach correct; though the answer is correct I am a little skeptical with this approach, please suggest.
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In a village of 100 households, 75 have at least one DVD player, 80 16 Jul 2019, 03:30
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ajithkumar
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

Dear Moderators chetan2u, VeritasKarishma, Bunuel, Gladiator59, generis

Is the above approach correct; though the answer is correct I am a little skeptical with this approach, please suggest.
Show more

Yes, it is correct and the logic is this: To minimise number of households that own all three gadgets, we need to maximise the number of houses that do not have at least one of these. So the "do not own" circles need to cover maximum number of houses. If 20 people do not own a cell phone and 25 do not own a DVD, we need to reduce their overlap so that more houses do not own at least one item. 20+25+45 add up to 90 and that is the maximum number of houses that we can cover. So 10 houses must have all three of these gadgets.
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In a village of 100 households, 75 have at least one DVD player, 80 02 Jun 2020, 04:33
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mittalg

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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VeritasKarishma
Why do you think 80 and 75 will have an over lap of 55 ?
There are 20 and 25 people respectively who dont own either dvd or cell phone.
We can distribute 45 MP3 among them in order to minimise the overlap of three products.
Since total no. of MP3 as stated by mittalg is 65 , so ( 65 - 45 ) = 20 are remaining and this 20 mp3 will be in the common portion of all three products.
I agree with your answer and the second approach mentioned by you though.
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