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.

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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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

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.

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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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

Can u please solve this using Venn diagram and formula?

Gagan

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!