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.
_________________

Am I missing something here??? it seems straightforward......

The obvious maximum that have all 3 is 55, because you are limited by the SMALLEST number.

The minimum is simply the sum of the max of each people who DONT have the product, so:

100-80 = 20 don't have Cell
100-75 = 25 don't have DVD
and 100-55 = 45 don't have MP3

SO a total of 20+25+45 = 90 combined who might NOT have SOME combination of the 3 products. So subtract that from 100, to give you the minimum of the people who COULD have all 3 and you get 100-90 = 10.

55-10 = 45

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.
_________________

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?
_________________

Fantastic explanation. Thanks Bunuel for your explanation and Hussain15 for posting this question.
_________________

I have done it in 1:41 doing this:

x= (75+55+80-100-z)/2 for z=0 ->55
y=the same for z=100 ->10

Therefore x-y=45

PS: z is obviously the people who have 2 devices.
_________________

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
_________________

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE


_________________

I have given the solution to a similar problem. Here's the solution for this problem.

1. Let C, D and M be who have cell only, dvd only and mp3 only and CD be who have both cell and dvd only and so on.
2 CDM is minimum when C+D+M=0 and maximum when CD+CM+DM=0.
3. C+D+M+CD+DM+CM+CDM=100 -- (1) because the total of all the mutually exclusive has to sum up to 100 which is the number of households.
4. (C+CD+CM+CDM) + (D+CD+DM+CDM) +(M+CM+DM+CDM) = 80+75+55 or ,
C+D+M+2CD+2CM+2DM+3CDM =210 ----(2)
5. (2) - (1) = CD+CM+DM+2CDM=110 --- (3)
6. (3) - (1) => CDM-(C+D+M) =10 ---(4)
We can find the minimum value when we put (C+D+M) = 0 above giving minimum CDM=10
7. To find maximun value equation (1) can be changed as
C+D+M+CDM=100 -- (5) by putting CD+CM+DM=0
From (4) and (5) we have CDM=55 which is maximun CDM
8.The answer is max CDM - min CDM i.e., 55-10=45

As a shortcut for this type of problems where there are 3 items , the minimum possible for all the 3 can be found out by adding the percentages of the 3 i.e., 55+75+80=210 . and subtracting 200 i.e, 210-200=10 . the maximum can be found out by adding the minimum to 100 and dividing by 2 i.e., (100+10)/2 = 55. Both the answers are percentages.
_________________

Hi Karishma,
is this correct? I am not able to get this approach.

100=80+75+55-z-2a

where z - no. of people with exactly 2 devices
a – no of people with 3 devices
a(max) = x, a(min) = y, (x-y) ?
100 = 210-z-2a
2a = 110-z
a = (110-z)/2

Now, how can we ever take z = 0, as even in case of maximum overlap i.e. when maximum no. of people have 3 devices,
z = 20, which is the overlap b/w C(mobile device) and D(DVD).

Also how is z = 100, all 100 cannot have exactly 2 devices.