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In a village of 100 households, 75 have at least one DVD player, 80

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Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 07 Nov 2016, 09:21
navigator123 wrote:
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)

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Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 07 Nov 2016, 19:50
vedantkabra wrote:
navigator123 wrote:
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)


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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Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 11 Dec 2016, 03:52
[quote="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.
hi karishma

i do not understand. how did you derive at most 55 households can have all 3 by Van diagram

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Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 10 May 2017, 18:05
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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Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 11 May 2017, 07:26
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hoopsgators wrote:
How would one solve this problem if a portion of the 100 households did not own ANY of the electronics mentioned in the question?



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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Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 24 May 2017, 05:29
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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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Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 16 Aug 2017, 15:30
I often just don't get the complexity of such kind of questions
max overlap is 55 (as this is the max number of 3 types of households that can have all three devices)
min overlap for 3 devices is 75+55+80 - 100*2 (max number for households that can have 2 devices) = 210 - 200 = 10

so 55 - 10 = 45
Am i missing something here?

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Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 25 Oct 2017, 03:54
Here,the maximum households having three will be 55
Now,if we want to get the minimum number of households having three,we need to find at least how many have two of things.
So,At least DVD and cell phone=(75+80)-100=55
At least DVD and MP3=(75+55)-100=30
At least cell and MP3 =(80+55)-100=35
now,minimum number=(75+80+55-30-35-55)-100=10
so,the difference is (55-10)=10
Hope it's clear now :-)

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In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 23 Nov 2017, 10:27
Bunuel wrote:
Hussain15 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

For me the best way to solve this problem is not use Venn diagram or formulas but to draw simple bars (note: each dash is 5):

Min overlap is 10:
-------------------- 80 phones;
-------------------- 75 DVD's;
-------------------- 55 MP3.

Max overlap is 55:
-------------------- 80 phones;
-------------------- 75 DVD's;
-------------------- 55 MP3.

55-10=45.

Answer: C.


hi Bunuel

maybe it is very obvious, but I am facing bit trouble
please guide me through my explanation

at least 55 people own MP3 players, so it is very natural to assume that maximum number of people who can have all three is 55
okay

now, to have the lowest possible overlapping

100 - 80 = 20 people have no phones (but may have DVDs)
100- 75 = 25 people have no DVDs (but may have phones)

so, you have distributed 45 mp3 players to these 45 villagers, causing no overlapping. so far so good

but we are left with 10 more mp3 players to distribute, but one cannot provide these players to the inhabitants who already have the mp3 players because doing so will add up the number of people who own mp3s to 45, a condition that is unwarranted by the question. So, these 10 people must be coming from the groups of DVDs or??? / and ???? phones holders. Here I am stumped, because

these 10 people can have phones earlier and are now getting mp3s
OR
these 10 people can have DVDs earlier and are now getting mp3s
in this case there is no overlapping ..... :sad:

OR

these people can have phones + DVDs earlier and are now getting mp3s
in this case there is overlapping.... :sad:

thanks in advance, man

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Re: In a village of 100 households, 75 have at least one DVD player, 80 [#permalink]

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New post 15 Jan 2018, 09:36
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: 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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Re: In a village of 100 households, 75 have at least one DVD player, 80   [#permalink] 15 Jan 2018, 09:36

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