# How to use venn diagrams to solve problems

• To understand, how to solve venn diagram word problems with 3 circles, we have to know the following basic stuff.
• u ——–> union (or)
• n ——–> intersection (and)

1. Theorem 1 :
2. n(AuB)  =  n(A) + n(B) – n(AnB)
3. Theorem 2 :
4. n(AuBuC) :
5. = n(A) + n(B) + n(C) – n(AnB) – n(BnC) – n(AnC) + n(AnBnC)
6. Explanation :
7. Let us come to know about the following terms in details.

8. n(AuB)  =  Total number of elements related to any of the two events A & B.
9. n(AuBuC)  =  Total number of elements related to any of the three events A, B & C.
10. n(A)  =  Total number of elements related to  A.
11. n(B)  =  Total number of elements related to  B.

12. n(C)  =  Total number of elements related to  C.
13. For  three events A, B & C, we have
14. n(A) – [n(AnB) + n(AnC) – n(AnBnC)] :
15. Total number of elements related to A only.
16. n(B) – [n(AnB) + n(BnC) – n(AnBnC)] :
17. Total number of elements related to B only.

18. n(C) – [n(BnC) + n(AnC) + n(AnBnC)] :
19. Total number of elements related to C only.
20. n(AnB) :
21. Total number of elements related to both A & B
22. n(AnB) – n(AnBnC) :
23. Total number of elements related to both (A & B) only.

24. n(BnC) :
25. Total number of elements related to both B & C
26. n(BnC) – n(AnBnC) :
27. Total number of elements related to both (B & C) only.
28. n(AnC) :
29. Total number of elements related to both A & C
30. n(AnC) – n(AnBnC) :
31. Total number of elements related to both (A & C) only.

32. For  two events A & B, we have
33. n(A) – n(AnB) :
34. Total number of elements related to A only.
35. n(B) – n(AnB) :
36. Total number of elements related to B only.

### Practice Problems

Problem 1 :

In a survey of university students, 64 had taken mathematics course, 94 had taken chemistry course, 58 had taken physics course, 28 had taken mathematics and physics, 26 had taken mathematics and chemistry, 22 had taken chemistry and physics course, and 14 had taken all the three courses. Find how many had taken one course only.

• Solution :
• Step 1 :
• Let M, C, P represent sets of students who had taken mathematics, chemistry and physics respectively
• Step 2 :
• From the given information, we have
• n(M)  =  64, n(C)  =  94,  n(P)  =  58,
• n(MnP)  =  28, n(MnC)  =  26, n(CnP)  =  22
• n(MnCnP)  =  14
• Step 3 :
• From the basic stuff, we have
• No. of students who had taken only Math
• =  n(M) – [n(MnP) + n(MnC) – n(MnCnP)]
• =  64 – [28 + 26 – 14]
• =  64 – 40
• =  24
• Step 4 :
• No. of students who had taken only Chemistry :
•                                 = n(C) – [n(MnC) + n(CnP) – n(MnCnP)]
•                                 = 94 – [26+22-14]
•                                 = 94 – 34
•                                 = 60
• Step 5 :
• No. of students who had taken only Physics :
• =  n(P) – [n(MnP) + n(CnP) – n(MnCnP)]
• =  58 – [28 + 22 – 14]
• =  58 – 36
• =  22
• Step 6 :
• Total no. of students who had taken only one course :
•                                   =  24 + 60 + 22
•                                  =  106
• Hence, the total number of students who had taken only one course is 106.
• Alternative Method (Using venn diagram) :
• Step 1 :
• Venn diagram related to the information given in the question: 1. Step 2 :
2. From the venn diagram above, we have
3. No. of students who had taken only math  =  24
4. No. of students who had taken only chemistry  =  60
5. No. of students who had taken only physics  =  22
6. Step 3 :
7. Total no. of students who had taken only one course :
8. =  24 + 60 + 22
9. =  106
10. Hence, the total number of students who had taken only one course is 106.
11. Problem 2 :

In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games. Find the total number of students in the group (Assume that each student in the group plays at least one game).

• Solution :
• Step 1 :
• Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively.
• Step 2 :
• From the given information, we have
• n(F)  =  65 , n(H)  =  45, n(C)  =  42,
• n(FnH)  =  20, n(FnC)  =  25, n(HnC)  =  15
• n(FnHnC)  =  8
• Step 3 :
• From the basic stuff, we have
• Total number of students in the group is n(FuHuC).
• n(FuHuC) is equal to
• =  n(F) + n(H) + n(C) – n(FnH) – n(FnC) – n(HnC) + n(FnHnC)
• Then, we have
• n(FuHuC)  =  65 + 45 + 42 -20 – 25 – 15 + 8
• n(FuHuC)  =  100
• Hence, the total number of students in the group is 100.
• Alternative Method (Using venn diagram) :
• Step 1 :
• Venn diagram related to the information given in the question : 1. Step 2 :
2. Total number of students in the group :
3. =  28 + 12 + 18 + 7 + 10 + 17 + 8
4. =  100
5. So, the total number of students in the group is 100.
6. Problem 3 :

In a college, 60 students enrolled in chemistry,40 in physics, 30 in biology, 15 in chemistry and physics,10 in physics and biology, 5 in biology and chemistry. No one enrolled in all the three. Find how many are enrolled in at least one of the subjects.

• Solution :
• Let C, P and B represents the subjects Chemistry, Physics  and Biology respectively.
• Number of students enrolled in Chemistry :
• n(C)  =  60
• Number of students enrolled in Physics :
• n(P)  =  40
• Number of students enrolled in Biology :
• n(B)  =  30
• No.of students enrolled in Chemistry and Physics :
• n(CnP)  =  15
• No.of students enrolled in Physics and Biology :
• n(PnB)  =  10
• No.of students enrolled in Biology and Chemistry :
• n(BnC)  =  5
• No one enrolled in all the three. So, we have
• n(CnPnB)  =  0
• The above information can be put in a venn diagram as shown below. 1. From, the above venn diagram, number of students enrolled in at least one of the subjects :
2. =  40 + 15 + 15 + 15 + 5 + 10 + 0
3. =  100
4. So, the number of students enrolled in at least one of the subjects is 100.
5. Problem 4 :

In a town 85% of the people speak Tamil, 40% speak English and 20% speak Hindi. Also 32% speak Tamil and English, 13% speak Tamil and Hindi and 10% speak English and Hindi, find the percentage of people who can speak all the three languages.

• Solution:
• Let T, E and H represent the people who speak Tamil, English and Hindi respectively.
• Percentage of people who speak Tamil :
• n(T)  =  85
• Percentage of people who speak English :
• n(E)  =  40
• Percentage of people who speak Hindi :
• n(H)  =  20
• Percentage of people who speak English and Tamil :
• n(TnE)  =  32
• Percentage of people who speak Tamil and Hindi :
• n(TnH)  =  13
• Percentage of people who speak English and Hindi :
• n(EnH)  =  10
• Let x be the percentage of people who speak all the three language. 1. From the above venn diagram, we can have
2. 100  =  40 + x + 32 – x + x + 13 – x + 10 – x – 2 + x – 3 + x
3. 100  =  40 + 32 + 13 + 10 – 2 – 3 + x
4. 100  =  95 – 5 + x
5. 100  =  90 + x
6. x  =  100 – 90
7. x  =  10%
8. So, the percentage of people who speak all the three languages is 10%.
9. Problem 5 :

An advertising agency finds that, of its 170 clients, 115 use Television, 110 use Radio and 130 use Magazines. Also 85 use Television and Magazines, 75 use Television and Radio, 95 use Radio and Magazines, 70 use all the three. Draw Venn diagram to represent these data. Find

• (i) how many use only Radio?
• (ii) how many use only Television?
• (iii) how many use Television and Magazine but not radio?
• Solution:
• Let T, R and M represent the people who use Television, Radio and Magazines respectively.
• Number of people who use Television :
• n(T)  =  115
• Number of people who use Radio :
• n(R)  =  110
• Number of people who use Magazine :
• n(M)  =  130
• Number of people who use Television and Magazines
• n (TnM)  =  85
• Number of people who use Television and Radio :
• n(TnR)  =  75
• Number of people who use Radio and Magazine :
• n(RnM)  =  95
• Number of people who use all the three :
• n(TnRnM)  =  70 1. From the above venn diagram, we have
2. (i) Number of people who use only Radio is 10
3. (ii) Number of people who use only Television is 25
4. (iii) Number of people who use Television and Magazine but not radio is 15.
5. Problem 6 :

## Venn Diagrams: Exercises

There are two classifications in this universe: English students and Chemistry students.

First I'll draw my universe for the forty students, with two overlapping circles labelled with the total in each: (Well, okay; they're ovals, but they're always called “circles”.)

Five students are taking both classes, so I'll put “5” in the overlap: I've now accounted for five of the 14 English students, leaving nine students taking English but not Chemistry, so I'll put “9” in the “English only” part of the “English” circle: I've also accounted for five of the 29 Chemistry students, leaving 24 students taking Chemistry but not English, so I'll put “24” in the “Chemistry only” part of the “Chemistry” circle: This tells me that a total of 9 + 5 + 24 = 38 students are in either English or Chemistry (or both). This gives me the answer to part (b) of this exercise. This also leaves two students unaccounted for, so they must be the ones taking neither class, which is the answer to part (a) of this exercise. I'll put “2” inside the box, but outside the two circles:

## Solving Problems with Venn Diagrams | Explained with Examples

Venn diagrams define all the possible relationships between collections of sets. The most basic Venn diagrams simply consist of multiple circular boundaries describing the range of sets. A simple Venn diagrams

The overlapping areas between the two boundaries describe the elements which are common between the two, while the areas that aren’t overlapping house the elements that are different. Venn diagrams are used often in math that people tend to assume they are used only to solve math problems. But as the 3 circle Venn diagram below shows it can be used to solve many other problems. Using a 3 circle Venn diagram to solve problems

Though the above diagram may look complicated, it is actually very easy to understand. Although Venn diagrams can look complex when solving business processes understanding of the meaning of the boundaries and what they stand for can simplify the process to a great extent. Let us have a look at a few examples which demonstrate how Venn diagrams can make problem solving much easier.

### Example 1: Company’s Hiring Process

The first example demonstrates a company’s employee short listing process.

The Human Resources department looks for several factors when short-listing candidates for a position, such as experience, professional skills and leadership competence.

Now, all of these qualities are different from each other, and may or may not be present in some candidates. However, the best candidates would be those that would have all of these qualities combined. How to use Venn diagrams to hire people

## Venn Diagram – Concept and Solved Questions

Venn diagram, also known as Euler-Venn diagram is a simple representation of sets by diagrams. The usual depiction makes use of a rectangle as the universal set and circles for the sets under consideration.

In CAT and other MBA entrance exams, questions asked from this topic involve 2 or 3 variable only. Therefore, in this article we are going to discuss problems related to 2 and 3 variables.

Let's take a look at some basic formulas for Venn diagrams of two and three elements.

n ( A ∪ B) = n(A ) + n ( B ) – n ( A∩ B)
n (A ∪ B ∪ C) = n(A ) + n ( B ) + n (C) – n ( A ∩ B) – n ( B ∩ C) – n ( C ∩ A) + n (A ∩ B ∩ C )

And so on, where n( A) = number of elements in set A.
Once you understand the concept of Venn diagram with the help of diagrams, you don’t have to memorize these formulas. ### Venn Diagram in case of two elements • Where;
X = number of elements that belong to set A only
Y = number of elements that belong to set B only
Z = number of elements that belong to set A and B both (AB)
W = number of elements that belong to none of the sets A or B
From the above figure, it is clear that
n(A) = x + z ;
n (B) = y + z ;
n(A ∩ B) = z;
n ( A ∪ B) = x +y+ z.
• Total number of elements = x + y + z + w

### Venn Diagram in case of three elements Where,
W = number of elements that belong to none of the sets A, B or C

Tip: Always start filling values in the Venn diagram from the innermost value.

Example 1:

## Venn Diagram Template for easy problem solving – free and editable!

### Highlights

Venn diagrams are a simple yet effective tool to visualise logical relationships between different groups of information. They clearly illustrate how items are similar or different.

They’ve found a place in the business world, helping to solve and illustrate complex problems in a clear, visual manner.

Use our editable Venn Diagram template and guide to easily create your own Venn diagrams.

### The basic Venn Diagram- 2 circles

At its most simple, there are two overlapping circles, each circle representing a set of items. The middle section shows what set A and set B have in common. ### Three circle Venn Diagram Template

Venn Diagrams often represent two sets of information, but by adding more circles and thus more data segments, you can create more complex diagrams. This can help you explain in a clear visual way the inter-relationship between many different related sets of information.

We’ve created a free Venn Diagram template with three circles that you can use to help with a range of problems, such as identifying new business directions, defining a target market, and sharing results in reports. Get started now. ### Create your own Venn Diagram with our template

Our Venn diagram template is extremely easy to use and share.

Conceptboard‘s collaborative whiteboard allows you to add sticky notes to each area in the Venn Diagram template or simply add external images for better visualization.

We’ve added some GIFs below to illustrate how easy it is to add content, create additional circles and export so you can share with your team or use in your presentation.

The first step when it comes to a creating a great Venn Diagram is the addition of sticky notes. With Conceptboard, you also have the option of adding external content to your charts and boards. Discover how here.

Sometimes your diagrams are extra complex and 2 or 3 circles are just not good enough! Fret not, we’ve got you covered. Add as many circles(or whatever shape you choose) with your amazing tool.

### Exporting and sharing

Conceptboard allows you to invite your team members to your board and get their feedback and collaborate from within the app itself. In case you want to export your board and use it as a you have that option as well. Simply export as an image or PDF and use as you please!