The Common Admission Test (CAT) 2012 had a little surprise for some students in the form of a question based on four sets. As far as I can remember, questions from set theory in CAT have always been based on two or three sets and CAT 2012 was the first time a question based on four sets was asked. As a concept, this had been neglected by many coaching institutes as it had never been asked and can be time consuming to solve in the exam.

Let us first try and understand what a Venn – Diagram for four sets would look like. If there are 4 sets (A, B, C & D), then our Venn – Diagram will need 15 separate sections. Every object can either be in a particular set or not. This would give us a total of 2×2×2×2 = 16 choices. 1 of these choices would be when the object will not be any of the sets. Hence, the Venn – Diagram will need 16 – 1 = 15 separate sections.

Couple of representations that I could find on Googling Venn Diagram 4 sets are given below:

Both of them are valid, but that does not mean that you can use them. May be you could, but it would be extremely difficult to draw and manage these. Given below is the one that I have been using for solving questions based on four sets and I do not even remember where I first picked this up. It also indicates the 15 regions separately which need to be covered in a Venn – Diagram of four sets. Over and above that, it is really easy to draw which is a big plus for someone like me who once in class 6^{th} tried to draw a deer for an assignment to draw an animal and got pass marks because the teacher thought it was a buffalo. TrueStory!

I think it can be best understood if we solve couple of questions based upon it.

**Example**: In a college library, four different business newspapers - Economic Times, Business Standard, Business Line and Financial Express - are available. All students visit the library regularly but 20% of them do not read any business newspaper. The four newspapers given in the above order are read by 230,180,180 and 220 students respectively. The number of students reading exactly 2 newspapers for any two newspapers is 20. There are 30 students who read all the four news papers but there is nobody who reads exactly three out of four newspapers.

Questions based upon the above information are given below.

Let us first draw the Venn – Diagram with the information that is given to us. It will look something like what I have drawn below:

We should now try to fill the missing areas such as people reading only one newspapers.

People reading only Economic Times: 230 – (20+20+20+30) = 230 – 90 = 140

People reading only Business Standard: 180 – (20+20+20+30) = 180 – 90 = 90

People reading only Business Line: 180 – (20+20+20+30) = 180 – 90 = 90

People reading only Financial Express: 220 – (20+20+20+30) = 220 – 90 = 130

If we put the above information in the Venn – Diagram, it should look something like this:

Let us try to attempt the questions now.

**1. How many students do not read any newspaper at all?**

Answer: We know that 20% of the students do not read any newspaper. This means that 80% of the students read at least one newspaper. These 80% people are represented in the 15 regions in the above Venn – Diagram. So,

80% of total = (140 + 90 + 90 + 130) + 6×20 + 30 = 600

Total = 600/0.8 = 750

Students who do not read any newspapers = 20% of 750 = **150 students.**

**2) What percentage of the people reading Business Standard also read at least one other newspaper?**

Answer: Out of the 180 students who read Business Standard, 90 read only Business Standard. This means that the other 90 students also read at least one other newspaper.

Required percentage = (90/180) ×100 = **50%**

**3) If all the students in the college including those who do not read any newspaper read at least one newspaper, (Out of the four newspapers above) which he is not reading at present, then what is the least number of students reading all the four newspapers?**

Answer: If the above mentioned condition happens, all those who are reading no newspaper will start reading one; all those who are reading one newspaper will start reading two; all those who are reading two newspapers will start reading three and all those who are reading three newspapers will start reading all four. The ones who were reading all four newspapers earlier will continue to read all four.

Least number of students reading all four = No. of students reading three newspapers earlier + No. of students four newspapers earlier = 0 + 30 = **30.**

The Common Admission Test (CAT) 2012 had a little surprise for some students in the form of a question based on four sets. As far as I can remember, questions from set theory in CAT have always been based on two or three sets and CAT 2012 was the first time a question based on four sets was asked. As a concept, this had been neglected by many coaching institutes as it had never been asked and can be time consuming to solve in the exam.

Let us first try and understand what a Venn – Diagram for four sets would look like. If there are 4 sets (A, B, C & D), then our Venn – Diagram will need 15 separate sections. Every object can either be in a particular set or not. This would give us a total of 2×2×2×2 = 16 choices. 1 of these choices would be when the object will not be any of the sets. Hence, the Venn – Diagram will need 16 – 1 = 15 separate sections.

Couple of representations that I could find on Googling Venn Diagram 4 sets are given below:

Both of them are valid, but that does not mean that you can use them. May be you could, but it would be extremely difficult to draw and manage these. Given below is the one that I have been using for solving questions based on four sets and I do not even remember where I first picked this up. It also indicates the 15 regions separately which need to be covered in a Venn – Diagram of four sets. Over and above that, it is really easy to draw which is a big plus for someone like me who once in class 6^{th} tried to draw a deer for an assignment to draw an animal and got pass marks because the teacher thought it was a buffalo. TrueStory!

I think it can be best understood if we solve couple of questions based upon it.

**Example**: In a college library, four different business newspapers - Economic Times, Business Standard, Business Line and Financial Express - are available. All students visit the library regularly but 20% of them do not read any business newspaper. The four newspapers given in the above order are read by 230,180,180 and 220 students respectively. The number of students reading exactly 2 newspapers for any two newspapers is 20. There are 30 students who read all the four news papers but there is nobody who reads exactly three out of four newspapers.

Questions based upon the above information are given below.

Let us first draw the Venn – Diagram with the information that is given to us. It will look something like what I have drawn below:

We should now try to fill the missing areas such as people reading only one newspapers.

People reading only Economic Times: 230 – (20+20+20+30) = 230 – 90 = 140

People reading only Business Standard: 180 – (20+20+20+30) = 180 – 90 = 90

People reading only Business Line: 180 – (20+20+20+30) = 180 – 90 = 90

People reading only Financial Express: 220 – (20+20+20+30) = 220 – 90 = 130

If we put the above information in the Venn – Diagram, it should look something like this:

Let us try to attempt the questions now.

**1. How many students do not read any newspaper at all?**

Answer: We know that 20% of the students do not read any newspaper. This means that 80% of the students read at least one newspaper. These 80% people are represented in the 15 regions in the above Venn – Diagram. So,

80% of total = (140 + 90 + 90 + 130) + 6×20 + 30 = 600

Total = 600/0.8 = 750

Students who do not read any newspapers = 20% of 750 = **150 students.**

**2) What percentage of the people reading Business Standard also read at least one other newspaper?**

Answer: Out of the 180 students who read Business Standard, 90 read only Business Standard. This means that the other 90 students also read at least one other newspaper.

Required percentage = (90/180) ×100 = **50%**

**3) If all the students in the college including those who do not read any newspaper read at least one newspaper, (Out of the four newspapers above) which he is not reading at present, then what is the least number of students reading all the four newspapers?**

Answer: If the above mentioned condition happens, all those who are reading no newspaper will start reading one; all those who are reading one newspaper will start reading two; all those who are reading two newspapers will start reading three and all those who are reading three newspapers will start reading all four. The ones who were reading all four newspapers earlier will continue to read all four.

Least number of students reading all four = No. of students reading three newspapers earlier + No. of students four newspapers earlier = 0 + 30 = **30.**