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

This solution hasn't considered 'None'. Even if we assume that they saw that None = 0 works for both cases and hence None is immaterial, notice that when z = 100, you get a as 5. That is not correct.

The maximum value of z is 90 (the number of people with exactly 2 devices). This gives the minimum number of people with 3 devices as 10.
The minimum value of z is 0 so max value of a is 55.
Either way, to find the max/min value of z you will need to use some logic. You might as well use it for max/min value of a.

This looks the simplest approach of the lot. I wonder if it will be true for all scenarios. Bunuel any comments ?

Hi wrightvijay,

It doesn't matter whether they have given the highlighted fact above or not because it's the question of maximizing and minimizing the overlapping portion which will be maximized when everything else is minimized and vice versa.

The most (x) number of households with all three is the least of the given, so 55. That one's straightforward.

The least (y) is when you add up all of the single households and assume as many households with at least two of the three items. 75 + 80 + 55 = 210. There are 100 households and if we are to maximize the number of households with at least two of the three items, that's 200 of the 210 items. Then, 10 are still remaining which tells us that there must be at least 10 households which have three items once every household doubles up. y = 10

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.

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

Dear Moderators chetan2u, VeritasKarishma, Bunuel, Gladiator59, generis

Is the above approach correct; though the answer is correct I am a little skeptical with this approach, please suggest.

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.

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.

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

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!!

Its funny how complicated your are thinking. Here was my approach:

All the figure say they have AT LEAST ... (so maybe more)

all of them could have 100. which is the maximum.

the lowest possible number is 55(55 have at least one mp3 player)

100-55=45

This is not correct. The question does not say "at least 75 people have one DVD player". It says "75 people have at least one DVD player" which means 75 people have one or more DVDs. It doesn't mean that number of people having a DVD is 75 or more.
The maximum is 55 and minimum is 10. Check the solutions given above.

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Where does it say that every household in the village have at least one of these three devices?
There could be none of these three devices in some households.

Hi Karishma

Can you please help me to solve the question by applying the below formula ?
Total-Neither = A+B+C - (common in two) - 2(common in three)

Hey Radhika,

In this question, you cannot just plug in the numbers in a formula and get the answer. You will need to do a bit of analysis for the possible overlap since you need the maximum and minimum value of overlap of all 3.

Common in three = A + B + C - (common in two) - Total + Neither
Common in three = (75 + 80 + 55 - (common in two) - 100 + Neither)/2
Common in three = (110 - (common in two) + Neither)/2
You need to maximise "common in three". For that, imagine the 75 circle inside the 80 circle and the 55 circle is inside the 75 circle. In that case, neither = 100 - 80 = 20 and common in two = 75 - 55 = 20.
So Common in three = 55

Similarly, we will minimise Common in three.

Effectively, we have used the venn diagram only to answer the question.

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)