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.

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.


x - y = 55 - 10 = 45
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Well everything is on the diagram but I'll try to elaborate:

-----(-----------)---- 80 phones (black dashes represents # of people who don't have phone, so 20 people don't have phone);
-----(-----------)---- 75 DVD's (black dashes represent # of people who don't have DVD, so 25 people don't have DVD);

So you see that overlap of phone and DVD owners is 55 and 45 people don't have either phones or DVD's.

To have min overlap of 3 let 20 people who don't have phone to have MP3 and also 25 people who don't have DVD's to have MP3. So we distributed 45 MP3 and there is no overlap of 3 yet. But 10 DVD's are still left to distribute and only people to have them is the group which has both MP3 and phone.

-----(--)-------------;
-----(--)-------------;
-----(--)-------------.

Hope it's clear.
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(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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A more algebraic approach:

If we denote by N_2 the number of households with exactly two types of devices, and by N_3 the number of those with all three types, than we can write the following equation:

100=75+80+55-N_2-2N_3, from which we get that N_2+2N_3=110.
Households with exactly two types of devices we counted twice, so we have to subtract them once. Households with exactly three types of devices we counted three times, therefore we have to remove them twice.

Maximum for N_3 is 55, in which case N_2=0, meaning there are no households with exactly two types of devices, but only either with just one type or all three of them. Everybody who has an MP3 player, also has a DVD player and a cell phone.

In order to determine the minimum number of households with exactly three types of devices, let's take a look at the possible combinations of two devices. If N_3 is minimum, there should be households with exactly two types of devices but not three.

All types of devices, but also those having exactly two types, explicitly DVD players and Cell Phones - If N_{DC} is the number of households that have exactly these two types of devices, then N_{DC}+N_3\geq75+80-100=55.
All types of devices, but also those having exactly two types, explicitly DVD players and MP3 players - If N_{DM} is the number of households that have exactly these two types of devices, then N_{DM}+N_3\geq75+55-100=30.
All types of devices, but also those having exactly two types, explicitly Cell phone and MP3 players - If N_{CM} is the number of households that have exactly these two types of devices, then N_{CM}+N_3\geq80+55-100=35.

Adding the above three inequalities, we obtain that N_{DC}+N_{DM}+N_{CM}+3N_3\geq120. Since N_{DC}+N_{DM}+N_{CM}=N_2, we get that N_2+3N_3\geq120. Taking into account that N_2+2N_3=110, we obtain 110+N_3\geq{120} or N_3\geq{10}.

55-10=45

Answer C
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Bunuel,
Your answer is elegant. But I must admit I would have jumped into the Venn diagram first. So can you please explain how to solve this in that approach using the usually established formulae? It would be great to think out of the box, but sadly my mind is not such

Also in your explanation above, you mention overlap of "3", did you mean "2" = 2x5=10?
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Fantastic explanation. Thanks Bunuel for your explanation and Hussain15 for posting this question.
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Nice approach bunuel Thanks

Wow, nice method. Where did you learn this?

Merging similar topics. Please ask if anything remains unclear.
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Well its way too straight forward.

The max value(x) must be 100.
Of all 3 the least value will be min possible value = 55

X-Y = 100-55 = 45.