Sometimes, to solve some tough questions, we need to make inferences. Those inferences may not be apparent at first but once you practice, they do become intuitive. Today we will discuss one such inference based high level question of an official GMAT practice test.

**Question**: In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:

*(A) 65*

*(B) 55*

*(C) 45*

*(D) 35*

*(E) 25*

**Solution**: We need to find the value of x – y

What is x? It is the greatest possible number of households that have all three devices

What is y? It is the lowest possible number of households that have all three devices

Say there are 100 households and we have three sets:

Set DVD including 75 households

Set Cell including 80 households

Set MP3 including 55 households

We need to find the values of x and y to get x – y.

We need to maximise the overlap of all three sets to get the value of x and we need to minimise the overlap of all three sets to get the value of y.

Maximum number of households that have all three devices:

We want to bring the circles to overlap as much as possible.

The smallest set is the MP3 set which has 55 households. Let’s make it overlap with both DVD set and Cell set. These 55 households are the maximum that can have all 3 things. The rest of the 45 households will definitely not have an MP3 player. Hence the value of x must be 55.

Note here that the number of households having no device may or may not be 0 (it doesn’t concern us anyway but confuses people sometimes). There are 75 – 55 = 20 households that have DVD but no MP3 player. There are 80 – 55 = 25 households that have Cell phone but no MP3 player. So they could make up the rest of the 45 households (20 + 25) such that these 45 households have exactly one device or there could be an overlap in them and hence there may be some households with no device. In the figure we show the case where none = 0.

Now, let’s focus on the value of y i.e. minimum number of households with all three devices:

How will we do that? Before we delve into it, let us consider a simpler example:

Say you have 3 siblings (A,B and C) and 5 chocolates which you want to distribute among them in any way you wish. Now you want to minimise the number of your siblings who get 3 chocolates. No one gets more than 3. What do you do?

Will you leave out one sibling without any chocolates (even if he did rat you out to your folks!)? No. Because if one sibling gets no chocolates, the other siblings get more chocolates and then more of them will get 3 chocolates. So instead you give 1 to each and then give the leftover 2 to 2 of them (one each). This way, no sibling gets 3 chocolates and you have successfully minimised the number of siblings who get 3 chocolates. Basically, you spread out the goodies to ensure that minimum people get too many of them.

This is the same concept.

When you want to minimise the overlap, you basically want to spread the goodies around. You want minimum people to have all three. So you give at least one to all of them. Here there will be no household which has no device. Every household will have at least one device.

So you have 80 households which have cell phone. The rest of the 20 households say, have a DVD player so the leftover 55 households (75 – 20) with DVD player will have both a cell phone and a DVD player. There are 55 households who already have two devices and 45 households with just one device.

Now how will you distribute the MP3 players such that the overlap between all three is minimum? Give the MP3 players to the households which have just one device so 45 MP3 player households are accounted for. But we still need to distribute 10 more MP3 players. These 10 will fall on the 55 overlap of the previous two sets. Hence there are a minimum of 10 households which will have all three devices. This means y = 10

x – y = 55 – 10 = 45

Answer (C)

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