For me the best way to solve this problem is not use Venn diagram or formulas but to draw simple bars (note: each dash is 5):

Min overlap is 10:
-------------------- 80 phones;
-------------------- 75 DVD's;
-------------------- 55 MP3.

Max overlap is 55:
-------------------- 80 phones;
-------------------- 75 DVD's;
-------------------- 55 MP3.

55-10=45.

Answer: C.
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a+b+c+d = 100............4
using 4 and 1 we get d = 20
using 4 and 2 we get a = 25
using the above values of a,d and equation 3
we get c = 55 -a-d = 55- 20-25 = 10

Hence minimum is 10.

For max :
75 will have 55 inside
80 will have the intersection of 75 and 55 i.e. 55

thus total becomes 55 + 20 + 25 = 100 where 55 = intersection of all the three
20 = 75 outside the intersection of 75 and 55
25 = 80 outside the intersection of 75,55,80

thus 55 is the max.

x-y = 55-10 = 45[/quote]


gurpreetsingh,

Is it possible to demonstrate your approach using 3 overlapping sets. That's the reason why I can't make that much sence out of your explanation because here we have 3 overlapping sets. I cant figure out why you are using just 2.

All the help is highly appreciated.
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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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(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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Nice approach bunuel Thanks

Wow, nice method. Where did you learn this?

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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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!, 11 New DS set. NEW!!!


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