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

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Intern
Joined: 30 Apr 2016
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Re: In a village of 100 households, 75 have at least one DVD player, 80  [#permalink]

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07 Nov 2016, 10:21
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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07 Nov 2016, 20:50
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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11 Dec 2016, 04:52
[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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10 May 2017, 19:05
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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11 May 2017, 08:26
2
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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24 May 2017, 06:29
1
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.
Attachments

Sets Club.PNG [ 10.03 KiB | Viewed 1176 times ]

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

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16 Aug 2017, 16:30
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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25 Oct 2017, 04:54
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

Don't forget to press--> Kudos
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In a village of 100 households, 75 have at least one DVD player, 80  [#permalink]

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Updated on: 19 Jan 2018, 07:30
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.

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

OR

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

Originally posted by testcracker on 23 Nov 2017, 11:27.
Last edited by testcracker on 19 Jan 2018, 07:30, edited 1 time in total.
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Re: In a village of 100 households, 75 have at least one DVD player, 80  [#permalink]

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15 Jan 2018, 10:36
1
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]

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19 Jan 2018, 07:39
gmatcracker2017 wrote:
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.

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

OR

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

yes, I think I have got the issue

it is possible that these 10 people can have all three items, so minimum overlapping is 10
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Re: In a village of 100 households, 75 have at least one DVD player, 80  [#permalink]

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08 Feb 2018, 01:23
make it more simple. using ven diagram for 2 factors, dvd and cellphone first.

look at two factors, person with dvd is 75 and with cellphone is 80.
so, there are 55 person with at least both dvd and cellphone because there is only total 100. this overlapped area contain 55 persons. simple

now there is 55 person with mp3. to maximize , we put all of 55 persons with mp3 in the the overlapped area between two groups above. perfect. there are maximum 55 person with dvd,cellphone and mp3

to minimize, look at the overlapped group of 55 person with dvd and cell phone. among 55 person with mp3, we put the minimum number in the the overlapped area of the two groups.

LOOK AT area of 75 person with dvd. we can put maximum 25 persons with mp3 into group of 80 person with cellphone because we can not put more. if we put 26 there would be 75+26=101 persons. WRONG.
similarly,,look at area of 80 persons with cellphone, we can put maximum 20 person with mp3 into group of 75 persons with dvd. if we put 21, there would be 80+21= 101 person, impossible.

so, from 55, we put 25 and 20 into two above group, the number left is 10.

mimimum is 10

this is hard. but can be learned. if we can not do this, we can still get 49 on math. dont worry
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Re: In a village of 100 households, 75 have at least one DVD player, 80  [#permalink]

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10 Jan 2019, 12:43
PiyushK wrote:
My imagination about minimum overlapping percentage.

Attachment:
The attachment overlapp.jpg is no longer available

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

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05 Feb 2019, 13:25
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

Tough question, please let me know if I made a mistake in my reasoning:

The max number of people that can do all three is 55. We can't have more than that because there are only 55 people with an MP3 player.

The min number was a bit more tricky for me. If there are 75 people with a DVD player, that means that the remaining 25 people did not have a DVD player. 25 will be the MAXIMUM number of people that can have both a cellphone and MP3 player. The same applies for the cellphone and mp3 player. We get the values 20 and 45 respectively.

To further explain the reasoning above: The amount of people who own an item + the number of people who don't own that item should be equal to the total group. (Similar to the negation strategy in probability. I.E. The chances of getting ALL HEADS plus the chances of getting NO HEADS will equal to your overall data set).

Hence, we have 75 + 80 + 55 - 90 - 2*all three = 100

MIN(All three) = 10

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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26 Feb 2019, 08:56
1
it is tricky
first consider . 80 dvd and 75 cell phone, the overlap must be from 55 to 75.

consider 80 dvd and 55 mp3 because the overlap can be change from 55 to 35.

also consider 75 cell phone and 55 mp3 because the overlap can be changed from 30 to 55

to make the maximum. let put 55 mp3 into 75 cell phone and put 75 cell phone in to 80 dvd. we get 55 is maximum
to make minium. first we get minimum from one couple , suppose it is 75 and 55, the minimum is 30
now we put 80 into these 2 circles. we have 55-30=25 and 75-30=45, 25 and 45 has only one characteristeic. we put 80 into 25 and 45, we use up 70 out of 80. there is 10 out of 80 left, so 10 is put into minimum 30 and the minimum of 3 characteristics is 10

so, the diffference is 55-10=45

the point here is that we consider 2 circles first, making the overlap max and min from these 2 circles and put the 3rd circle into these 2 circles. this is the fastest and most simple way
Re: In a village of 100 households, 75 have at least one DVD player, 80   [#permalink] 26 Feb 2019, 08:56

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