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

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

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

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!

ScottTargetTestPrep JeffTargetTestPrep Please help how to solve and approach this

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.

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

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