What is a Venn Diagram?
- A Venn diagram (also called a set diagram or logic diagram) is a diagram that shows all possible logical relations between a finite collection of different sets.
Venn Diagram
- Typically overlapping shapes, usually circles, are used, and an area-proportional or scaled Venn diagram is one in which the area of the shape is proportional to the number of elements it contains.
- These diagrams represent elements as points in the plane and sets as regions inside curves. An element is in a set S just in case the corresponding point is in the region for S.
- Venn diagram is used for the classification of data belonging to the same category but different sub-categories.
- The comparison of the different data becomes easier through the Venn diagram, also the relationship between the data.
- Grouping the information and finding similarities and differences among them becomes easy. The different unknown parameters can be easily understood and found with the help of Venn diagrams.
The concept of the Venn diagram is very useful for solving a variety of problems in Mathematics and others. To understand more about it let’s learn some important terms related to it.
Universal Set
- Universal Set is a large set that contains all the sets which we are considering in a particular situation.
- For example: If you are searching for books for a research project, the universal set might be all the books in the library, and the books in the library that are relevant to your research project would be a subset of the universal set.
Subset
- Subset is actually a set of values that is contained inside another set i.e. we can say that set B is the subset of set A if all the values of set B are contained in set A.
- For example, if we take N as the set of all the natural numbers and W as the set of all whole numbers then,
N = Set of all Natural Numbers
W = set of all Whole Numbers
we can say that N is a subset of W as all the values of set N are contained in set W.
We represent the subset using the ⊆ symbol. In the above example, we write,
N ⊆ W
It is read as N is a subset of W.
- We use Venn diagrams to easily represent a subset of a set. The image discussing the subset of a set are given below,
Venn Diagram Symbols
There are more than 30 Venn diagram symbols. We will learn about the three most commonly used symbols in this section. They are listed below as:
Let us understand the concept and the usage of the three basic Venn diagram symbols using the image given below.
How to Draw a Venn Diagram?
Venn diagrams can be drawn with unlimited circles. Since more than three becomes very complicated, we will usually consider only two or three circles in a Venn diagram. Here are the 4 easy steps to draw a Venn diagram:
- Step 1: Categorize all the items into sets.
- Step 2: Draw a rectangle and label it as per the correlation between the sets.
- Step 3: Draw the circles according to the number of categories you have.
- Step 4: Place all the items in the relevant circles.
Example: Let us draw a Venn diagram to show categories of outdoor and indoor for the following pets: Parrots, Hamsters, Cats, Rabbits, Fish, Goats, Tortoises, Horses.
- Step 1: Categorize all the items into sets (Here, its pets): Indoor pets: Cats, Hamsters, and, Parrots. Outdoor pets: Horses, Tortoises, and Goats. Both categories (outdoor and indoor): Rabbits and Fish.
- Step 2: Draw a rectangle and label it as per the correlation between the two sets. Here, let's label the rectangle as Pets.
- Step 3: Draw the circles according to the number of categories you have. There are two categories in the sample question: outdoor pets and indoor pets. So, let us draw two circles and make sure the circles overlap.
- Step 4: Place all the pets in the relevant circles. If there are certain pets that fit both the categories, then place them at the intersection of sets, where the circles overlap. Rabbits and fish can be kept as indoor and outdoor pets, and hence they are placed at the intersection of both circles.
- Step 5: If there is a pet that doesn't fit either the indoor or outdoor sets, then place it within the rectangle but outside the circles.
Representation of Sets with Venn Diagrams
Case 1: When universal sets and a normal set have been given.
Let U be the universal set representing the sets of all natural numbers, and let A ⊆ U for A = 1,2,3,4,5
Then, by a Venn diagram, we can show as
Case 2: When two intersecting subsets of U are given.
To represent two intersecting subsets, A and B of U, we draw two circles, and the intersecting regions will be shown as
Case 3: When two disjoint sets are given.
To draw two disjoint sets, we normally draw two circles which will not intersect each other within a rectangle.
Case 4: When B ⊆ A ⊆ U
In this case, we draw two concentric circles within a rectangular region.
Case 5: Complement of a set.
Let U be the set of natural numbers, and let A = {1, 2, 3}. Then, the complement of set A is given by
A’ = U – A = {1 2, 3, 4, 5, 6,….} – {1, 2, 3} = {4, 5, 6,…}
Then, the complement of A can be shown as
Venn Diagram for Sets Operations
There are different operations that can be done on sets in order to find the possible unknown parameter, for example, if two sets have something in common, their intersection is possible. The basic operations performed on the set are,
- Union of Set
- Intersection of Set
- Complement of Set
- Difference of Set
Let’s look at these set operations and how they look on the Venn diagram.
Union of sets Venn Diagram
The Union of two or more two sets represents the data of the sets without repeating the same data more than once, it is shown with the symbol ⇢∪.
n(A∪ B) = {a: a∈ A OR a∈ B}
Properties of Union of a set
Various properties of the Union of a set are,
- A∪ B = B∪ A ⇢ [Commutative property]
- A∪ (B ∪ C) = (A∪ B) ∪ C ⇢ [Associative property]
- A ∪ U = U
- A ∪ A’ = U
- A ∪ A = A
Intersection of sets Venn Diagram
The intersection of two or more two sets means extracting only the amount of data that is common between/among the sets. The symbol used for the intersection⇢ ∩.
n(A∩ B)= {a: a∈ A and a∈ B}
Properties of Intersection of Set
Various properties of the Intersection of a set are,
- A ∩ B= B ∩ A ⇢ [Commutative property]
- (A ∩ B) ∩ C= A ∩ (B ∩ C) ⇢ [Associative property]
- A ∩ A= A
- A ∩ U= A
- A ∩ A’= φ
Complement of a set of Venn Diagram
Complementing a set means finding the value of the data present in the Universal set other than the data of the set.
n(A’) = U- n(A)
Difference of Set Venn Diagram
Suppose we take two sets, Set A and Set B then their difference is given as A – B. This difference represents all the values of set A which are not present in set B.
For example, if we take Set A = {1, 2, 3, 4, 5, 6} and set B = {2, 4, 6, 8} then A- B = {1, 3, 5}.
In the Venn diagram, we represent the A – B as the area of set A which is not intersecting with set B.
Venn Diagram for Three Sets
Three sets Venn diagram is made up of three overlapping circles and these three circles show how the elements of the three sets are related. When a Venn diagram is made of three sets, it is also called a 3-circle Venn diagram. In a Venn diagram, when all these three circles overlap, the overlapping parts contain elements that are either common to any two circles or they are common to all the three circles. Let us consider the below given example:
Here are some important observations from the above image:
- Elements in P and Q = elements in P and Q only plus elements in P, Q, and R.
- Elements in Q and R = elements in Q and R only plus elements in P, Q, and R.
- Elements in P and R = elements in P and R only plus elements in P, Q, and R.
Question for Venn Diagrams
Try yourself:In a class of 60, along with English as a common subject, students can opt to major in Mathematics, Physics, Biology or a combination of any two. 6 students major in both Mathematics and Physics, 15 major in both Physics and Biology, but no one majors in both Mathematics and Biology. In an English test, the average mark scored by students majoring in Mathematics is 45 and that of students majoring in Biology is 60. However, the combined average mark in English, of students of these two majors, is 50. What is the maximum possible number of students who major ONLY in Physics?
Explanation
Let Tm and Tb be total scores of the students majoring in Mathematics and Biology respectively.
According to the given conditions,
Tm = (M + 6) × 45
Tb = (B + 15) × 60
Also, (Tm + Tb) = (B + P + 21) × 50
45(M + 6) + 60(B + 15) = 50M + 50B + 1050
10B – 5M + 270 + 900 = 1050
10B – 5M = –120
M = 2B + 24
Here, we need to determine the maximum value of P.
∴ We need to minimize the value of B. Minimum value of B can be 0.
∴ M = 24
Again, we know that, M + B + P + 21 = 60
⇒ 24 + 0 + P + 21 = 60
⇒ P = 15
Hence, option (d).
Report a problem
Solved Examples
Directions for the question: Read the information given below and answer the question that follows.
Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections: Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection. For the 200 candidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET:
No one is below the 80th percentile in all 3 sections.
150 are at or above the 80th percentile in exactly two sections.
The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.
Number of candidates below 80th percentile in P: Number of candidates below 80th percentile in C: Number of candidates below 80th percentile in M = 4:2:1.
BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.
Q1: What best can be concluded about the number of candidates sitting for the separate test for BIE who were at or above the 90th percentile overall in CET?
Sol: The number of candidates sitting for separate test for BIE who were at or above 90th percentile in CET (a) is either 3 or 10.
Q2: If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, what is the number of candidates who are at or above the 90th percentile overall and at or above the 80th percentile in both P and M in CET?
Sol: From the given condition, g is a multiple of 5. Hence, g = 20. The number of candidates at or above 90th percentile overall and at or above 80th percentile in both P and M = e + g = 60.
Q3: If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, then how many candidates were shortlisted for the AET for AIE?
Sol: In this case, g = 20. Number of candidates shortlisted for AET = d + e + f + g = 10 + 40 + 100 + 20 = 170
Q4: If the number of candidates who are at or above the 90th percentile overall and also are at or above the 80th percentile in P in CET, is more than 100, how many candidates had to sit for the separate test for BIE?
Sol: From the given condition, the number of candidates at or above 90th percentile overall and at or above 80th percentile in P in CET = 104. The number of candidates who have to sit for separate test = 296 + 3 = 299.
Q5: In a college, 200 students are randomly selected. 140 like tea, 120 like coffee and 80 like both tea and coffee.
How many students like only tea?
How many students like only coffee?
How many students like neither tea nor coffee?
How many students like only one of tea or coffee?
How many students like at least one of the beverages?
Sol: The given information may be represented by the following Venn diagram, where T = tea and C = coffee.
• Number of students who like only tea = 60
• Number of students who like only coffee = 40
• Number of students who like neither tea nor coffee = 20
• Number of students who like only one of tea or coffee = 60 + 40 = 100
• Number of students who like at least one of tea or coffee = n (only Tea) + n (only coffee) + n (both Tea & coffee) = 60 + 40 + 80 = 180
Q6: In a survey of 500 students of a college, it was found that 49% liked watching football, 53% liked watching hockey and 62% liked watching basketball. Also, 27% liked watching football and hockey both, 29% liked watching basketball and hockey both and 28% liked watching football and basket ball both. 5% liked watching none of these games.
How many students like watching all the three games?
Find the ratio of number of students who like watching only football to those who like watching only hockey.
Find the number of students who like watching only one of the three given games.
Find the number of students who like watching at least two of the given games.
Sol: n(F) = percentage of students who like watching football = 49%
n(H) = percentage of students who like watching hockey = 53%
n(B)= percentage of students who like watching basketball = 62%
n ( F ∩ H) = 27% ; n (B ∩ H) = 29% ; n(F ∩ B) = 28%
Since 5% like watching none of the given games so, n (F ∪ H ∪ B) = 95%.
Now applying the basic formula,
95% = 49% + 53% + 62% -27% - 29% - 28% + n (F ∩ H ∩ B)
Solving, you get n (F ∩ H ∩ B) = 15%.
Now, make the Venn diagram as per the information given.