In a town of 10 million households there are 3 newspapers in circulati

A is not the right answer as the difference is 3M and if any household subscribes three papers then the number will be less than 3M

3M is the maximum limit for the households which are subscribing more than one paper.

i Think below method is easy to understand.

The total number of house holds buying newspaper = X+Y+Z+XY+XZ+YZ+XYZ = 9---(1)
Total number of newspapers sold = X+Y+Z+2XY+2XZ+2YZ+3XYZ = 12---(2)
If we need maximum number of households then no house hold should buy all three.
applying this in 1 and 2

X+Y+Z+XY+XZ+YZ = 9
X+Y+Z+2XY+2XZ+2YZ=12
So XY+XZ+YZ = 3(Maximum)

If we need the minimum number of households then no one should buy 2 newspapers.
1,2 will become
X+Y+Z+XYZ = 9
X+Y+Z+3XYZ=12
2XYZ=3
XYZ=1.5(minimum)

Applying this on options we can conclude that E is the answer

12 mil is the number of newspapers distributed.
9 mil is the number of households in which they are distributed.
Since each household gets at least one paper, say 9 mil are distributed, one each, in the 9 mil households. We are left with 3 mil newspapers. Each one of these 3 mil papers is distributed as second/third paper in some household.

Let's focus on distributing just these 3 mil papers now.

The maximum number of households in which these 3 mil papers can be distributed is 3 mil. Each paper goes to a different household. So number of households subscribing to more than 1 newspaper is \(\leq 3\)mil. But there is no such option.

Say, if I give 2 papers (a total of 3 papers) in the same household, and the rest of them in different households, how many households will get covered? (3 mil - 1)
How can I minimize the total number of households? By giving 2 papers in as many households (a total of 3 papers) as possible. If I give 2 papers in each household, I will need to cover 1.5 mil households. So number of households receiving more than 1 paper should be at least 1.5 mil.

Answer (E)

Karishma if we tried to solve this algebraically: (numbers in million)

AuBuC=A+b+C-{exactly 2}-2*{exactly 3} --->9=12-{exactly 2}-2*{exactly 3} ---> {exactly 2}+2*{exactly3}=3 (1)

Is this the 3 that you mention in your explanation? (red phrase).

Then what do we say? we want to find {exactly 2}+{exactly 3}... Based on (1) i guess I have to find the value of {exactly 3}.. then what?

Yes it is. 3 mil newspapers are second or third papers.

{exactly 2}+2*{exactly3}=3
Maximize households -> Distribute each of the 3 mil papers in two paper homes. {exactly3} = 0
{exactly2} = 3
Number of households subscribing to more than 1 paper = 3 mil

Minimize households -> Distribute each of the 3 mil papers in 3 paper homes. {exactly 2} = 0
2*{exactly 3} = 3
{exactly 3} = 1.5 mil
Number of households subscribing to more than 1 paper = 1.5 mil

Perfect explanation as always. Thanks.

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