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KarishmaB
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In a village of 100 households, 75 have at least one DVD player, 80 05 Jun 2020, 03:43
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sayan640
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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!!!

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.

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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There are 100 households. If I want to minimise the overlap, I will try to keep the three circles as far away from each other as possible.
So once I give cell phone to 80 households, I would like to give DVD players to the other 20. But the leftover 75 - 20 = 55 DVD players I will still need to give to the households that already have a cell phone. Though we will still have 80 - 55 = 25 households with only DVD players. Now when I distribute 65 MP3 players, I will first give 45 of them to those who have only 1 device (20 + 25) and the rest 20 will need to be given to households that will end up having all three.
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In a village of 100 households, 75 have at least one DVD player, 80 09 Jun 2020, 00:34
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VeritasKarishma
sayan640
VeritasKarishma
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!!!

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.

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.


There are 100 households. If I want to minimise the overlap, I will try to keep the three circles as far away from each other as possible.
So once I give cell phone to 80 households, I would like to give DVD players to the other 20. But the leftover 75 - 20 = 55 DVD players I will still need to give to the households that already have a cell phone. Though we will still have 80 - 55 = 25 households with only DVD players. Now when I distribute 65 MP3 players, I will first give 45 of them to those who have only 1 device (20 + 25) and the rest 20 will need to be given to households that will end up having all three.
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Please check the highlighted portion in your explanation above.
I think 25 households will have only cell-phones.
Initially 80 households have at least one cell-phones and 75 households have at least one dvd players.
55 households have both dvd and cell phones.
25 households have only cell phones and 20 households have only DVDs.

VeritasKarishma Please correct me if I said anything wrong.
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In a village of 100 households, 75 have at least one DVD player, 80 11 Jun 2020, 01:40
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sayan640

Please check the highlighted portion in your explanation above.
I think 25 households will have only cell-phones.
Initially 80 households have at least one cell-phones and 75 households have at least one dvd players.
55 households have both dvd and cell phones.
25 households have only cell phones and 20 households have only DVDs.

VeritasKarishma Please correct me if I said anything wrong.
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Yes, 25 have only cell phones and 20 only DVD players (I switched the two). Together, 45 have only 1 device and 55 have both. So 20 of these 55 will end up having all 3 devices. I am still not sure what your query is.
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In a village of 100 households, 75 have at least one DVD player, 80 23 Oct 2020, 03:04
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we need to find MAX - MIN = ?

in order to MAX we need to have as much overlap as possible between the three

in order to MIN we need the least amount of overlap between the three...
start with 80 as it is the biggest value
then going down analyze the gaps that are in the outside and max the amount that you would put in the outside region so that you can minimize the amount that overlaps
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In a village of 100 households, 75 have at least one DVD player, 80 29 Oct 2020, 22:31
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Hi yashikaaggarwal
Just having a little trouble with this one. Need your help.

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:

Now,
We are given that we need to 'maximize' the number of households that have all three of these devices.
It will be 100, right? It is given that 55 have at least one MP3 player.
It is very much possible that 100 folks have all the 3 devices.

As for minimum, I am little stuck. It should be 55.
So basically, 100-55= 45.

Is my thinking correct? I didn't use Venn diagrams, equations.
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In a village of 100 households, 75 have at least one DVD player, 80 30 Oct 2020, 04:00
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Hi yashikaaggarwal
Just having a little trouble with this one. Need your help.

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:

Now,
We are given that we need to 'maximize' the number of households that have all three of these devices.
It will be 100, right? It is given that 55 have at least one MP3 player.
It is very much possible that 100 folks have all the 3 devices.

As for minimum, I am little stuck. It should be 55.
So basically, 100-55= 45.

Is my thinking correct? I didn't use Venn diagrams, equations.
Show more
Yeah you are correct with both part, but let's understand the logic
the total is 100
and if we total all DVD users(Only 1, 2 or all 3) + Cell phone users(Only 1, 2 or all 3) + MP3 users(Only 1, 2 or all 3)
that equals to 75+80+55 = 210 which is 110 more than 100
so the largest interaction zone can be 110
BUT,
we can never go more than the original total, and
If we want to maximize those who like all 3, we have to maximize the value in the intersection. So, we have to minimize the value of the union. that can hold upto 100 value.
So, Maximum is 100

Now, Minimum.
If we want to minimize those who like all 3, we have to minimize the value in the intersection. So, we have to maximize the value of the union. the minimum value among all DVD users, Cell phone users, and MP3 users (75,80,55 is 55)
which can be a part of all 3 venn Interaction period at the lowest level
so the minimum value is 55

Difference between maximum and minimum is 100-55 = 45
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In a village of 100 households, 75 have at least one DVD player, 80 28 Feb 2021, 02:42
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Please find below an explanation that will not confuse you but create an easy path to answer such questions -:

Clearly maximum overlap can be 55 (as explained above)
For minimum overlap, we should maximize the 2 sets of people here -:
In total there are - 75 + 80 + 55 = 210
Maximize 2 sets i.e. 100*2 = 200 (we are maximizing 100 households having either 2 of the 3 equipments)
Minimum Overlap = 10

Same logic can be applied in another official question -:
https://gmatclub.com/forum/in-a-certain ... 60789.html
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In a village of 100 households, 75 have at least one DVD player, 80 20 Jan 2022, 12:18
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Okay! Some nice explanations there. Here's my take...

Total Houses: 100.

Houses with DVD player: 75
Houses w/ Cell Phones: 80
Houses w/ MP3s: 55

That's all the data we have. Now we need to find the max and min values of 'g' (remember, that represents the 'all-the-three' block in the Venn Diagram).

Lets focus on the max value of 'g':
If we go on distributing a set of those three gadgets to each house, we'll run out of MP3s after House No. 55. So, we'll stop right there. That's the max number of houses which can have all the three gadgets. (g max = 55)

Lets now focus on the min value of 'g':
We don't have to begin from the start. Let's continue from where we left in our previous step.

55 houses have all the three gadgets. Let's call it Block A. 45 houses are still empty, and this is Block B. We have 80-55 = 25 CD players, and 75-55=20 DVD players left for distribution.

Total gadgets left: 45. Total houses left: 45. Voila! Give each one away to each house left.

Now, let's minimise the no. of houses having all the three gadgets. How can we do that?

None of the houses in Block B have an MP3 player. Okay, lets remove 45 MP3s from the Block A and re-distribute it to Block B.

55-45=10. Just 10 houses now have all the three gadgets.

Focus on the second step, since that is the trickiest part of this solution. Remember, you can't remove any more DVDs or CDs from Block A, since all the houses in Block B have them already. So, you can only remove the MP3s from Block A.

There you go: 55-10 = 45!

Cheers!
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In a village of 100 households, 75 have at least one DVD player, 80 26 May 2023, 01:55
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ScottTargetTestPrep JeffTargetTestPrep Please help how to solve and approach this
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In a village of 100 households, 75 have at least one DVD player, 80 26 May 2023, 02:56
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Was just going through the question again.

Isn't the following understanding consistent with the information given in the question?

75 households have at least 1 dvd player, 15 have at least two dvd players and 5 have at least 3 dvd players and so on

Similarly for the other two items.
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In a village of 100 households, 75 have at least one DVD player, 80 12 Mar 2024, 23:06
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Thank you. This was the most helpful explanation imo.
Let me try to iterate on it.
We have a total of 100 ppl, CD: 80, DVD: 75, MP3: 55.

Lets start with CD and DVD.
- What is the maximum number of people with both CD and DVD? 75. Answer is limited by the smallest number. Example: we have 100 boys and 10 girls. Assuming a couple = 1boy+1girl. Max number of couples is 10.
- I take CD as DVD as a new group X. What is the maximum number of people with X and MP3? 55.
- Therefore max is 55.

- What is the minimum number of people with both CD and DVD? The minimum happens when number of people with just DVD is maximised.
- # of ppl with DVD = 75 = # of ppl with DVD and CP + # of ppl with just DVD
- Since there are only 20 ppl with no CP, this means that the max # of ppl with just DVD is 20
- Hence solving the above equation, min # of ppl with DVD and CP is 75 - 20 =55 = x
- taking DVD and CP as a new group X, the min # of ppl with X is 55.

- What is the min number of ppl with X and MP3? As above, the min happens when number of ppl with only MP3 is maximised
- # of people with MP3 = 55 = # of ppl with MP3 and X + # of ppl with MP3 only
- Since there are only 45 ppl with no X, this means that the max # of ppl with MP3 only is 10
- Hence solving the above equation, min # of ppl with X and MP3 is 55 - 45 = 10 = y

therefore: x-y - 55-10 = 45
rednblack89
For mainhoon and others who still didn't understand, you have to create the situation of minimum possible overlap. This is done in the way Bunuel explains it:

After the first step of distributing phones and DVDs among the 100, you get 55 as the minimum possible overlap for these 2.
Now you bring in the MP3s, of which there are 55. Remember, you are trying to make sure as few people as possible get all 3 - if possible, none. So, what do you do when you have 55 MP3s, and 45 people left who don't have an DVD and a phone? Distribute it among them first, naturally!

After doing that, you still have 10 left over. You are left with no choice but to distribute these among those who already have a DVD and phone. Hence the minimum overlap you can possibly achieve is 10.

This gives you the answer of 55 - 10 = 45.
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In a village of 100 households, 75 have at least one DVD player, 80 12 Jul 2024, 19:38
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Try to understand Bunuel explanation. Its brilliant and a time saver.
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In a village of 100 households, 75 have at least one DVD player, 80 25 Sep 2024, 23:51
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Bunuel
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
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.
Show more


Finding this the most sense to understand, however once you know, the trick that you can take the sum of the minimums, and then subtract that sum from the end total that that actually becomes the easiest way to find the minimum. So using an understanding these two methods and conjunction, makes this question a lot easier to solve or rather just understand.

Posted from my mobile device
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In a village of 100 households, 75 have at least one DVD player, 80 22 Nov 2024, 08:27
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The most they could overlap is 55 of course. And for the least, I think it's best to draw lines or venn diagrams where you make the minimum amount of overlap of all 3:

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In a village of 100 households, 75 have at least one DVD player, 80 28 Nov 2024, 07:31
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Hi Karishma,
I find the solution in this way, let me know where i m wrong.....:

Minimum value of households with all three devices --> 55
Max value of households --> 100 ( because the problem can not say that it can not happen.... each household have all devices).
So 100-55 = 45
KarishmaB
suchoudh
VeritasPrepKarishma

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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In a village of 100 households, 75 have at least one DVD player, 80 26 Dec 2024, 12:13
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Ma'am, why is none 0?
KarishmaB
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

Check out this video discussion on how to handle max-min in Sets: https://youtu.be/oLKbIyb1ZrI
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In a village of 100 households, 75 have at least one DVD player, 80 27 Dec 2024, 13:15
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I can chime in, the question mentions at least 1 for all the devices: 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. So, there no one who has none of the devices

nayanq2001
Ma'am, why is none 0?
KarishmaB
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

Check out this video discussion on how to handle max-min in Sets: https://youtu.be/oLKbIyb1ZrI
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In a village of 100 households, 75 have at least one DVD player, 80 23 Apr 2025, 07:24
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Hi, Can we solve this qn using table and Countings method as shown in Qn.3 of this video.?

GMAT Ninja Quant Ep 14: Overlapping Sets

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 28 Jun 2025, 06:05
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Same question but with only a few replacements of the words to make it a different sceanrio: https://gmatclub.com/forum/in-a-company-of-100-employees-75-have-at-laptop-80-have-at-least-one-312092.html
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In a village of 100 households, 75 have at least one DVD player, 80 17 Oct 2025, 13:44
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Same questions too

Elsapaul111
Hi, Can we solve this qn using table and Countings method as shown in Qn.3 of this video.?

GMAT Ninja Quant Ep 14: Overlapping Sets


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