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.
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