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

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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 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.

It's not given. It is something we have derived using logic.

Take a simpler case:

Say you have 3 siblings (A,B and C) and 5 chocolates which you want to distribute among them in any way you wish. Now you want to minimise the number of your siblings who get 3 chocolates. No one gets more than 3. What do you do?

Will you leave one sibling without any chocolates (even if he did rat you out to your folks!)? No. Because if one sibling gets no chocolates, the other siblings get more chocolates and then more of them will get 3 chocolates. So instead you give 1 to each and then give the leftover 2 to 2 of them. This way, no sibling gets 3 chocolates and you have successfully minimised the number of siblings who get 3 chocolates. Basically, you spread out the goodies to ensure that minimum people get too many of them.

This is the same concept. When you want to minimise the overlap, you basically want to spread the goodies around. You want minimum people to have all 3. So you give atleast one to all of them.

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

A Simple Approach

Cell -- 80 | DVD -- 75 | MP3 -- 55
Max/Min Overlapping section for all three. -- ??

MIN --> Two ways (choose the smaller. i.e. 10)





MAX --> One way (55)



Ans = Max - Min = 55 - 10 = 45

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)

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

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.

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.

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.

Yes, it is correct and the logic is this: To minimise number of households that own all three gadgets, we need to maximise the number of houses that do not have at least one of these. So the "do not own" circles need to cover maximum number of houses. If 20 people do not own a cell phone and 25 do not own a DVD, we need to reduce their overlap so that more houses do not own at least one item. 20+25+45 add up to 90 and that is the maximum number of houses that we can cover. So 10 houses must have all three of these gadgets.

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.