# WORD PROBLEMS ON SETS AND VENN DIAGRAMS

## About the topic "Word problems on sets and venn diagrams"

Word problems on sets and venn diagrams is one of the topics in both school level math and quantitative aptitude. People who study quantitative aptitude to get prepared for competitive exams are stumbling to solve set theory word problems.

The reason for their stumbling is, they do not know the basic stuff to solve set theory word problems.

## Basic stuff needed to solve word problems on sets and venn diagrams

To understand, "How to solve word problems on sets and venn diagrams?", we have to know the following basic stuff.

U --------> union (or)

n --------> intersection (and)

n(AUB)  =  n(A) + n(B) - n(AnB)

n(AUBUC) = n(A)+n(B)+n(C)-n(AnB)-n(BnC)-n(AnC)+n(AnBnC)

Let us come to know about the following terms in details.

n(AuB) = Total number of elements related to any of the two events A & B.

n(AuBuC) = Total number of elements related to any of the three events A, B & C.

n(A) = Total number of elements related to  A.

n(B) = Total number of elements related to  B.

n(C) = Total number of elements related to  C.

For  three events A, B & C, we have

n(A) - [n(AnB) + n(AnC) - n(AnBnC)] =Total number of elements related to A only.

n(B) - [n(AnB) + n(BnC) - n(AnBnC)] =Total number of elements related to B only.

n(C) - [n(BnC) + n(AnC) + n(AnBnC)] =Total number of elements related to C only.

n(AnB) = Total number of elements related to both A & B

n(AnB) - n(AnBnC) = Total number of elements related to both                                                 (A & B) only.

n(BnC) = Total number of elements related to both B & C

n(BnC) - n(AnBnC) = Total number of elements related to both                                                 (B & C) only.

n(AnC) = Total number of elements related to both A & C

n(AnC) - n(AnBnC) = Total number of elements related to both                                                 (A & C) only.

For  two events A & B, we have

n(A) - n(AnB)  = Total number of elements related to A only.

n(B) - n(AnB)  = Total number of elements related to B only.

Let us consider the following example, to have better understanding of the above stuff explained using venn diagram.

Example:

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.

Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively.

Venn diagram related to the above situation:

From the venn diagram, we can have the following details.

No. of students who play foot ball = 65

No. of students who play foot ball only = 28

No. of students who play hockey = 45

No. of students who play hockey only = 18

No. of students who play cricket = 42

No. of students who play cricket only = 10

No. of students who play both foot ball &  hockey = 20

No. of students who play both (foot ball & hockey) only = 12

No. of students who play both hockey & cricket = 15

No. of students who play both (hockey & cricket) only = 7

No. of students who play both foot ball and cricket = 25

No. of students who play both (foot ball and cricket) only = 17

No. of students who play all the three games = 8

## How to solve word problems on sets and venn diagrams?

Let us go through the following example problems to know  "How to solve word problems on sets and venn diagrams ?"

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:

Step 2 :

From the venn diagram above, we have

No. of students who had taken only math = 24

No. of students who had taken only chemistry = 60

No. of students who had taken only physics = 22

Step 3 :

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

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 = n(FuHuC)

= n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC)

= 65 + 45 + 42 -20 - 25 - 15 + 8

= 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:

Step 2 :

Total number of students in the group

=  28 + 12 + 18 + 7 + 10 + 17 + 8

= 100

Hence, the total number of students in the group is 100

When students go through the above two example word problems on sets and venn diagrams, we hope they would have received answer for the question, "How to solve word problems on sets and venn diagrams?". Let us see some more examples to have better understanding on solving word problems on sets and venn diagrams

Problem 3 :

In a class of 60 students, 40 students like math, 36 like science, 24 like both the subjects. Find the number of students who like

(i) Math only, (ii) Science only  (iii) Either Math or Science (iv) Neither Math nor science

Solution :

Step 1 :

Let M and S represent the set of students who like math and science respectively.

Step 2 :

From the information given in the question, we have

n(M) = 40, n(S) = 36, n(MnS) = 24

Step 3 :

Answer (i) : No. of students who like math only

= n(M) - n(MnS)

= 40 - 24

= 16

Step 4 :

Answer (ii) : No. of students who like science only

= n(S) - n(MnS)

= 36 - 24

= 12

Step 5 :

Answer (iii) : No. of students who like either math or science

= n(M or S)

= n(MuS)

= n(M) + n(S) - n(MnS)

= 40 + 36 - 24

= 52

Step 6 :

Total no. students who like any of the two subjects = n(MuS) = 52

No. of students who like neither math nor science

= 60 - 52

= 8

Problem 4 :

At a certain conference of 100 people there are 29 Indian women and 23 Indian men. Out of these Indian people 4 are doctors and 24 are either men or doctors. There are no foreign doctors. Find the number of women doctors attending the conference.

Solution :

Step 1 :

Let M and D represent the set of Indian men and Doctors respectively.

Step 2 :

From the information given in the question, we have

n(M) = 23, n(D) = 4, n(MuD) = 24,

Step 3 :

From the basic stuff, we have

n(MuD) = n(M) + n(D) - n(MnD)

24 = 23 + 4 - n(MnD)

n(MnD) = 3

n(Indian Men and Doctors) = 3

Step 4 :

So, out of the 4 Indian doctors,  there are 3 men.

And the remaining 1 is Indian women doctor.

Hence, the number women doctors attending the conference is 1

The word problems on sets and venn diagrams explained above will give clear idea to students on solving word problems on sets and venn diagrams.

And also we hope that the word problems on sets and venn diagrams explained above would be much useful for the students who struggle to solve word problems on sets and venn diagrams.