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CBSE Case Study Questions for Class 11 Maths Sets Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Sets in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 11 Maths Sets PDF

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Case Study Questions for Class 11 Maths Chapter 1 Sets

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[PDF] Download Case Study Questions for Class 11 Maths Chapter 1 Sets

Here we are providing case study questions for class 11 maths. In this article, we are sharing Class 11 Maths Chapter 1 Sets case study questions. All case study questions of class 11 maths are solved so that students can check their solutions after attempting questions.

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills. Here are some reasons why case study questions are important in Class 11 maths education:

• Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
• Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
• Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
• Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
• Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

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Sets- Case Based Type Questions | Mathematics (Maths) Class 11 - Commerce PDF Download

Correct Answer is Option (c) From the table, it is observed that there are 3 colleges whose fees for 4 year course is less than or equal to 6 lakhs. So, the required set is {Jaypee, Chandigarh, Kalinga}.

Question 2: If Sri and Ram wants to take admission in the same college. Find the set of college in which both Sri and Ram can take their admission. (a) {Chandigarh, Kalinga} (b)  {Jaypee} (c)  {SRM, LPU} (d)  Ø

Correct Answer is Option (a) The common colleges in which Sri and Ram both of them can take admission are Chandigarh and Kalinga. So, the required set is {Chandigarh, Kalinga}.

Question 3: Sri has taken an education loan of 3 lakhs to increase his overall budget. Find the set of college in which Ram can take admission but Sri cannot. (a)  {Birla, Thapar}   (b)  {Thapar} (c) {SRM} (d)  {Manipal, Thapar}

Correct Answer is Option (b) On increasing the budget of Sri by 3 lakh, the total budget will be 7 lakhs. So, the set of college in which she can take admission are {Manipal, LPU, Jaypee, Chandigarh, Kalinga}. But the set of colleges in which Ram can take admission are {Manipal, LPU, Thapar, Jaypee, Chandigarh, Kalinga}. Since there is one element that is not common in both the sets, the required set is {Thapar}.

Question 4: Sri has a total budget of 4 lakhs and she wanted  admission in the famous college. Will she able to take admission in the colleges given in table. Find the set of college in which she can take admission. (a) Yes, {SRM, Birla, LPU} (b)  No, {Jaypee, LPU, Kalinga} (c)  Yes, {Chandigarh, Kalinga} (d)  None of the above

Correct Answer is Option (c) Yes, there are two colleges whose fees are either equal to or less than 4 lakhs. So, the required set is {Chandigarh, Kalinga}.

Question 5:  Ram has taken an education loan of 2 lakhs to increase his overall budget. Find the number of college in which he can take admission after raising his budget. (a)  4 (b) 6 (c) 7 (d) 9

Correct Answer is Option (b) Earlier when the budget was 6 lakh, the set of colleges are {Jaypee, Chandigarh, Kalinga}. On increasing the budget by 2 lakhs, the total budget of Ram becomes 8 lakhs. So, the set of college in which he can take admission are {Manipal, LPU, Thapar, Jaypee, Chandigarh, Kalinga}.

Direction:  A survey is conducted by a career counsellor in a college to find career choice of students after the Intermediate. There are 100 students that goes for Engineering Courses, 50 wants to make their career in Medical, 100 students continue their further study in Arts. There are 10 students that go for both Engineering and Medical, and 3 goes for Medical and Arts. There are 3 students that do not go for any further studies. Question 6:  Find the total number of students on which the survey is conducted. (a)  250 (b)  230 (c)  240 (d)  238

Correct Answer is Option (c) Let A set denotes Engineering, B set denotes Medical and C set denotes Art. Draw the Venn’s diagram for the given situation. Total number of students is the sum of elements in all the regions. So, the total number of students are 240.

Question 7: Find the number of students that goes for both Medical and Art. (a) 5 (b) 2 (c)  3 (d) 4

Correct Answer is Option (c) Let A set denotes Engineering, B set Medical and C set Art. Draw the Venn’s diagram for the given situation. The region VI shows the student that goes for both Medical and Arts. So, n(B ∩ C) = 3.

Question 8: Find the number of students that goes for only engineering course. (a)  90 (b)  95 (c)  100 (d)  93

Correct Answer is Option (a) Let A set denotes Engineering, B set denotes Medical and C set denotes Art. Draw the Venn’s diagram for the given situation. So, the number of students that goes for only engineering course are 90.

Question 9: Find the number of students that goes for medical or engineering. (a)  150 (b)  140 (c)  160 (d)  No of the above

Correct Answer is Option (b) Let A set denotes Engineering, B set denotes Medical and C set denotes Art. Draw the Venn’s diagram for the given situation. The region I, II, III, IV, V, and VI shows the student that goes for Engineering or Medical. So, the required number of students is 140.

Question 10: Find the number of students that goes for Engineering or Art. (a) 150 (b) 200 (c) 190 (d) 197

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Class 11 Mathematics Case Study Questions

Table of Contents

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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Important Questions for CBSE Class 11 Maths Chapter 1 - Sets 2024-25

• Class 11 Important Question
• Chapter 1: Sets

Crucial Practice Problems for CBSE Class 11 Maths Chapter 1: Sets

Class 11th is quite important in making students understand the complex concepts of mathematics and preparing them for the JEE Main exams . Class 11th needs a significant amount of hard work from a student's point of view, and the same goes for the teachers as well. Vedantu is pushing the limits of education by helping students in their journey of making a strong foundation for the exams which they are going to give at the end of the year along with the entrance exams. Let's provide you with the briefing of the important questions for class 11 maths chapter 1 . 

In this chapter, students will be learning about the different types of sets and how to represent them. They will also get to know about empty sets, finite and infinite sets, equal sets, subsets, power sets, Universal sets, union and intersection of the given sets. Moreover, as the student studies the chapter and reaches its end, they will be able to solve the problems that use the formulas from the above topic and the difference of sets, a complement of sets, and properties of Complement. A lot of questions you are going to see in the competitive exams like JEE Main will use the concepts you have learned in class 11th as the competitive exams test you on your learning ability and come up with the answer constraint environment. 

Download CBSE Class 11 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 11 Maths Important Questions for other chapters:

Boost Your Performance in CBSE Class 11 Mathematics Exam Chapter 1 with Important Questions

Very short questions and answers (1 marks questions).

Which of the following are sets? Justify your answer.

1. The collection of all the months of a year beginning with letter  M

Ans: Set, because collection of certain and unique type of data is called a set.

2. The collection of difficult topics in Mathematics.

Ans: Not a set, because difficult topics differ person to person.

Let \[A=\{1,3,5,7,9\}\]. Insert the appropriate symbol in blank spaces:-(Question3,4) 

Ans: \[\in \]

4. 5-A                                                                                                                                                 

5. Write the set $A=\left\{ x:x\text{ is an integer},-1\le x \le 4 \right\}$ in roster form.

Ans: The elements in roster form is as shown

\[A=\{-1,0,1,2,3\}\]

6. List all the elements of the set, $A=\left\{ x:x \in Z,\dfrac{-1}{2}\le x\le \dfrac{11}{2} \right\}$                                                                                                 

Ans: All the elements are as shown 

\[A=\{0,1,2,3,4,5\}\]

7. Write the set $B=\left\{ 3,9,27,81 \right\}$ in set-builder form.

Ans: The above set in set builder form is as shown

\[B=\{x:x={{3}^{n}},n\in N\text{ and }1\le n\le 4\}\]

Which of the following are empty sets? Justify.

8. $A=\left\{ x:x\in N,3\le x \le 4 \right\}$

Ans: Empty set, because there is no natural number that lies between \[3\] and \[4\]

9. $B=\left\{ x:x\in N,{{x}^{2}}=x \right\}$                                                                                                            

Ans: Non-empty set, because there exist natural number which equals to square of itself. For example ${{1}^{2}}=1$ and so on.

Which of the sets are finite or infinite? Justify.

10. The set of all points on the circumference of a circle.                                     

Ans: Infinite set, because there are many points in the circumference of circle

11. $B=\left\{ x:x\in N\text{ and x is an even prime number} \right\}$

Ans: Finite set, because the only even prime number is two.

12. Are sets $A=\left\{ -2,2 \right\},B=\left\{ x:x\in R,{{x}^{2}}-4=0 \right\}$ equal? Why? 

Ans: Yes because the number of elements in A is equal to that of B 

13. Write $\left( -5,\left. 9 \right] \right.$in set-builder form

Ans: \[\left\{ x:x\in ,-5 \le x\le 9 \right\}\]

14. Write $A=\left\{ x:-3\le x \le 7 \right\}$ as interval

Ans: Clearly in interval the above set is written as

\[\left[ -3,\left. 7 \right) \right.\]

15. If $A=\left\{ 1,3,5 \right\}$ how many elements has P(A)? 

Ans: Clearly the number of elements in  \[P\left( A \right)={{2}^{3}}=8\]

16. Write all the possible subsets of $A=\left\{ 5,6 \right\}$.                                                        

Ans: Clearly the possible values of \[A=\left\{ 5,6, \right\}\]is given by

\[\left\{ \varphi ,\left\{ 5 \right\},\left\{ 6 \right\},\left\{ 5,6 \right\} \right\}\]

17. If  $A=\left\{ 2,3,4,5 \right\},B=\left\{ 3,5,6,7 \right\}$. Find $A\bigcup B$

Ans: Clearly \[A\bigcup B=\left\{ 2,3,4,5,6,7 \right\}\]

18.In above question find  $A\bigcap B$  

Ans: Clearly \[A\bigcap B=\left\{ 3,5 \right\}\]

19. If $A=\left\{ 1,2,3,6 \right\},B=\left\{ 1,2,4,8 \right\}$ find $B-A$

Ans: We are given with sets as shown

\[A=\left\{ 1,2,3,6 \right\}\]

\[B=\left\{ 1,2,4,8 \right\}\]

Hence \[B-A=\left\{ 4,8 \right\}\]

20. If $A=\left\{ p,q \right\},B=\left\{ p,q,r \right\}$, is B a superset of \[A\]? Why?

Ans: Yes, because A is a subset of B.

21. Are sets $A=\left\{ 1,2,3,4 \right\},B=\left\{ x:x\in N\text{ and 5}\le \text{x}\le \text{7} \right\}$ disjoint? Why?

Ans: The above mentioned sets are disjoint because  \[\left( A\bigcup B \right)=\varphi \].

22. If X and Y are two sets such that $n\left( X \right)=19,n\left( Y \right)=37,n\left( X\bigcap Y \right)=12$ find $n\left( X\bigcup Y \right)$.

Ans: We know that \[n\left( X\bigcup Y \right)\] is given by

\[n\left( X\bigcup Y \right)=n\left( X \right)+n\left( Y \right)-n\left( X\bigcap Y \right)\]

Hence we get \[n\left( X\bigcup Y \right)=44\]

23. Describe the set in Roster form.

$\left\{ x:\text{x is a two digit number such that the sum of its digits is 8 } \right\}$

Ans: The set in Roster form of above mentioned set is 

\[\left\{ 17,26,35,44,53,62,71,80 \right\}\]

24. Are the following pair of sets equal? Give reasons. 

$A=\left\{ x:\text{x is a letter in the word FOLLOW } \right\}$

$B=\left\{ x:\text{x is a letter in the word WOLF } \right\}$  

Ans: We can write above mentioned sets as shown

\[A=\left\{ F,O,L,W \right\}\]

\[n\left( A \right)=4\]

\[B=\left\{ W,O,L,F \right\}\]

\[n\left( B \right)=4\]

Hence  \[A=B\]

25. Write down all the subsets of the set $\left\{ 1,2,3 \right\}$

Ans: All the subsets of the set \[\{1,2,3\}\] is given below

\[\left\{ \varnothing ,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\} \right\}\]

26. Let $A=\left\{ 1,2,\left\{ 3,4 \right\},5 \right\}$ is $\left\{ \left\{ 3,4 \right\} \right\}\in A$ is incorrect. Give a reason.

Ans: Clearly \[\left\{ 3,4 \right\}\] is an element of set A, therefore \[\left\{ \left\{ 3,4 \right\} \right\}\] is a set containing element \[\left\{ 3,4 \right\}\] which belongs to A.

Hence, \[\left\{ \left\{ 3,4 \right\} \right\}\in A\] is correct.

27. Draw Venn diagram for $\left( A\bigcap B \right)'$

Ans: We know that 

\[\left( A\bigcap B \right)'=U-A\bigcap B\]

Hence the region is shown in the venn diagram below

                                           Fig: \[(A\cap B)'\]

28. Write the set in roster form A the set of letters in TRIGNOMETRY

Ans: The set of letters in TRIGNOMETRY in roster form is written as

\[A=\left\{ T,R,I,G,N,O,M,E,T,R,Y \right\}\]

29. Are the following pair of sets are equal? Give reasons    

A, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”.

Ans: The set of letters in ALLOY is written as

\[A=\left\{ A,,L,O,Y \right\}\]

Similarly, the set of letters in LOYAL is written as

\[B=\left\{ L,O,Y,A \right\}\]

Hence \[A=B\]

30. Write down the power set of A, $A=\left\{ 1,2,3 \right\}$ 

Ans: We know that power set is written as shown

\[P(A)=\left\{ \varnothing ,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\} \right\}\]

31.  $A=\left\{ 1,2,\left\{ 3,4 \right\},5 \right\}$ which is incorrect and why.

 (i) \[\{3,4\}\subset A\]

Ans: Clearly we can see that \[\left\{ 3,4 \right\}\in A\] 

Hence \[\left\{ 3,4 \right\}\subset A\] is incorrect

(ii) \[\{3,4\}\in A\]

Hence \[\left\{ 3,4 \right\}\in A\] is correct

32. Fill in the blanks:

(i) $A\bigcup A'$

Ans: We know that \[A\bigcup A'=U\] where U is the universal set

(ii) $\left( A' \right)'$

Ans: We know that \[\left( A' \right)'=A\]

(iii) $A\bigcap A'$

Ans: We know that \[A\bigcap A'=\varphi \]where \[\varphi \] is the universal set.

33. Write the set $\left\{\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6},\dfrac{6}{7} \right\}$ in the set builder form.

Ans: The set builder form of above set is given by

\[\left\{ \dfrac{n}{n+1}:n\text{ is a natural number less than or equal to 6} \right\}\]

34. Is set $C=\left\{ x:x-5=0 \right\}$ and

$E=\left\{ x:\text{x is an integral positive root of the equation }{{x}^{2}}-2x-15=0 \right\}$ are equal?

Ans: From set C we get

Hence \[C=\left\{ 5 \right\}\]

Also on solving the equation 

\[{{x}^{2}}-2x-15=0\]

We get the positive root as shown

Hence both the sets are equal

35. Write down all possible proper subsets of the set $\left\{ 1,\left\{ 2 \right\} \right\}$.                           

Ans: All possible proper subsets of the given set are

\[\varphi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 1,\left\{ 2 \right\} \right\}\]

36. State whether each of the following statements is true or false.

(i) $A=\left( 2,3,4,5 \right),B=\left\{ 3,6 \right\}$are disjoint sets.

Ans: Clearly we have

\[\left\{ 2,3,4,5 \right\}\bigcap \left\{ 3,6 \right\}=\left\{ 3 \right\}\ne \varphi \]

Hence the above statement is false

(ii) $A=\left( 2,6,10 \right),B=\left\{ 3,7,11 \right\}$are disjoint sets.

\[\left\{ 2,6,10 \right\}\bigcap \left\{ 3,7,11 \right\}=\varphi \]

Hence the above statement is true

37. Solve the followings:

(i) $\left( A\bigcup B \right)'$

Ans: By the properties we write

\[\left( A\bigcup B \right)'=A'\bigcap B'\]

(ii) $\left( A\bigcap B \right)'$

\[\left( A\bigcap B \right)'=A'\bigcup B'\]

38. Write the set of all vowels in the English alphabet which precede k in roster form.                        

Ans: The set of all vowels in the English alphabet which precede k in roster form is as shown

\[N=\left\{ a,e,i \right\}\]

39. Is pair of sets equal? Give reasons.

$A=\left( 2,3 \right),B=\left\{ x:\text{x is the solution of }{{x}^{2}}+5x+6=0 \right\}$

Ans: Given we have

\[A=\left\{ 2,3 \right\}\]

\[B=\left\{ x:x\text{ is the solution of }{{\text{x}}^{2}}+5x+6 \right\}\]

Now we can easily find the solution of  \[{{\text{x}}^{2}}+5x+6\] to be the set \[B=\left\{ -2,-3 \right\}\]

Hence  \[A\ne B\]

So the given pair of sets are not  equal

40. Write the following intervals in set builder form: $\left( -3,0 \right)$ and $\left[ \left( 6,12 \right) \right]$ 

Ans: The set builder form of above intervals is given by

\[\left\{ -3,0 \right\}\to \left\{ x:x\in R,-3\le x \le 0 \right\}\]

\[\left\{ 6,12 \right\}\to \left\{ x:x\in R,6\le x\le 12 \right\}\]  

41. If $X=\left\{ a,b,c,d \right\}$

$Y=\left\{ f,b,d,g \right\}$

Find $X-Y$ and $Y-X$

Ans: We are given with the following sets

\[X=\left\{ a,b,c,d \right\}\]

\[Y=\left\{ f,b,d,g \right\}\]

Hence \[X-Y=\left\{ a,c \right\}\]

Similarly, \[Y-X=\left\{ f,g \right\}\]

42. If A and B are two given sets, then represent the set $\left( A-B \right)'$, using the Venn diagram.

\[\left( A-B \right)'=U-\left( A-B \right)\] and hence the venn diagram is as shown

                                    Fig-\[(A-B)'\]

43. List all the element of the set $A=\left\{ x:x\text{ is an integer},{{x}^{2}}\le 4 \right\}$ 

Ans: The elements which we will get is as shown

\[\left\{ -2,-1,0,1,2 \right\}\]

44. From the sets given below pair the equivalent sets.

$A=\left\{ 1,2,3 \right\},B=\left\{ x,y,z,t \right\},C=\left\{ a,b,c \right\},D=\left\{ 0,a \right\}$

Ans: From the data given A and C are equivalent sets because the number of elements in each is same.

45. Write the following as interval                                                                              

(i) $\left\{ x:x\in R,-4 \le x\le 6 \right\}$

Ans : The interval form of above is given as shown

\[\left( -4,6 \right]\]

(ii) $\left\{ x:x\in R,3\le x\le 4 \right\}$

Ans: The interval form of above is given as shown

\[\left[ 3,4 \right]\]

46. If $A=\left\{ 3,5,7,9,11 \right\},B=\left\{ 7,9,11,13 \right\},C=\left\{ 11,13,15 \right\}$ Find $\left( A\bigcap B \right)\bigcap \left( B\bigcup C \right)$

Ans: From the data given we have

\[A=\left\{ 3,5,7,9,11 \right\}\]

\[B=\left\{ 7,9,11,13 \right\}\]

\[C=\left\{ 11,13,15 \right\}\]

Now \[A\bigcap B=\left\{ 7,9,11 \right\}\]

\[B\bigcup C=\left\{ 7,9,11,13,15 \right\}\]

Therefore \[\left( A\bigcap B \right)\bigcap \left( B\bigcup C \right)=\left\{ 7,9,11 \right\}\]

47. Write the set $\left\{ \dfrac{1}{3},\dfrac{3}{5},\dfrac{5}{7},\dfrac{7}{9},\dfrac{9}{11},\dfrac{11}{13} \right\}$in set builder form. 

Ans: \[\left\{ \dfrac{2n-1}{2n+1}:n\text{ is a natural number less than 7} \right\}\]

Long Questions and Answers (4 Marks Questions)

1. In a group of $800$ people, $500$ can speak Hindi and $320$ can speak English. Find

(i) How many can speak both Hindi and English?    

Ans: We will use following notation

H-People who can speak Hindi

E-People who can speak English

It is given in the question that

\[n\left( E\bigcup H \right)=800\]

\[n\left( E \right)=320\]

\[n\left( H \right)=500\]

Also we know that

\[n\left( E\bigcup H \right)=n\left( E \right)+n\left( H \right)-n\left( E\bigcap H \right)\]

800=320+500\[-n\left( E\bigcap H \right)\]

Hence on solving above we get \[20\] people can speak both Hindi and English

(ii) How many can speak Hindi only? 

Also we find that

\[n\left( E\bigcap H \right)=20\]

\[n\left( E'\bigcap H \right)=n\left( H \right)-n\left( E\bigcap H \right)\]

Hence on solving above we get \[480\] people can speak both Hindi and English

2. A survey shows that $84$ percent of Indians like grapes, whereas $45$ percent like pineapple. What percentage of Indians like both grapes and pineapple?                          

A-set of Indians who like grapes

O-set of Indians who like pineapple

\[n\left( A\bigcup O \right)=100\]

\[n\left( A \right)=84\]

\[n\left( O \right)=45\]

Now we know that 

\[n\left( A\bigcup O \right)=n\left( A \right)+n\left( O \right)-n\left( A\bigcap O \right)\]

Hence on solving the above we get 

\[n\left( A\bigcap O \right)=29\]

Therefore \[29\] percent of Indians like both apples and oranges 

3. In a survey of $450$ people, it was found that $110$ play cricket, $160$ play tennis and $70$ play both cricket as well as tennis. How many plays neither cricket nor tennis? 

S-set of surveyed people

A-set of people who play cricket

O- set of people who play tennis

\[n\left( A\bigcap O \right)=70\]

\[n\left( A \right)=110\]

\[n\left( O \right)=160\]

\[\Rightarrow n\left( A\bigcup O \right)=110+160-70=200\]

Therefore students who like neither cricket nor tennis is given by

\[n\left( A'\bigcap O' \right)=450-200=250\]

4. In a group of students, $225$ students know French, $100$ know Spanish and $45$ know both. Each student knows either French or Spanish. How many students are there in the group? 

A-set of students who know French

O- set of students who know Spanish

\[n\left( A\bigcap O \right)=45\]

\[n\left( O \right)=100\]

\[n\left( A \right)=225\]

 \[\Rightarrow n\left( A\bigcup O \right)=225+100-45=280\]

Hence there are 280 students in the group.

5. If $A=\left[ \left( -3,5 \right),B=\left( 0,6 \right) \right]$  then find 

(i) $A-B$, 

Ans: Given we have 

\[A=\left( -3,5 \right)\]

\[B=\left( 0,6 \right)\]

We know that \[A-B=A\bigcap B'\]

Hence \[A-B=\left[ -3,0 \right]\]

(ii) $A\bigcup B$

We know that \[A\bigcup B\] means occurrence of at least one 

Hence \[A\bigcup B=\left[ -3,6 \right]\]

6.  In a survey of $400$ students in a school, $100$ were listed as taking apple juice, $150$ as taking orange juice and $75$ were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.

A-set of students who like apple juice

O- set of students who like orange juice

\[n\left( A\bigcap O \right)=75\]

\[n\left( A \right)=100\]

\[n\left( O \right)=150\]

\[\Rightarrow n\left( A\bigcup O \right)=100+150-75=175\]

Therefore students who take neither apple nor orange juice is given by

\[n\left( A'\bigcap O' \right)=400-175=225\]

7. A survey shows that $73$ percent of Indians like apples, whereas $65$ percent like oranges. What percent of Indians like both apples and oranges?

A-set of Indians who like apples

O-set of Indians who like oranges

\[n\left( A \right)=73\]

\[n\left( O \right)=65\]

\[n\left( A\bigcap O \right)=38\]

Therefore \[38\] percent of Indians like both apples and oranges 

8. In a school there are $20$ teachers who teach mathematics or physics. Of these $12$ teach mathematics and $4$ teach both physics and mathematics. How many teach physics?                                                                                                                          

Ans: We will use following notation 

P-Number of physics teachers

M- Number of mathematics teachers

We are given 

\[n\left( P\bigcup M \right)=20\]

\[n\left( M \right)=12\]

\[n\left( P\bigcap M \right)=4\]

\[n\left( P\bigcup M \right)=n\left( P \right)+n\left( M \right)-n\left( M\bigcap P \right)\]

On putting the respected values and solving we get

\[n\left( P \right)=12\]

9. Let $U=\left\{ 1,2,3,4,5,6 \right\},A=\left\{ 2,3 \right\},B=\left\{ 3,4,5 \right\}$. Find $A'\bigcap B',A\bigcup B$and hence show that $A\bigcup B=A'\bigcap B'$.               

Ans : We know that 

$=\left\{ 1,4,5,6 \right\}$

$=\left\{ 1,2,6 \right\}$

$A\bigcup B=\left\{ 2,3,4,5 \right\}$

$\left( A'\bigcap B' \right)=\left\{ 1,6 \right\}$

Hence proved.

10. For any two sets A and B prove by using properties of sets that:

$\left( A\bigcap B \right)\bigcup \left( A-B \right)=A$

Ans: We write LHS and RHS as shown

 $LHS=\left( A\bigcap B \right)\bigcup \left( A-B \right)$

$=\left( A\bigcap B \right)\bigcup \left( A\bigcap {{B}^{'}} \right)$ (since $\left( A-B \right)=\left( A\bigcap {{B}^{'}} \right)$)

$=A\bigcap \left( B\bigcup {{B}^{'}} \right)$

$=A\bigcap \left( U \right)$

11. If A and B are two sets and $U$ is the universal set such that 

$n\left( U \right)=1000,n\left( A \right)=300,n\left( B \right)=300,n\left( A\bigcap B \right)=200$ find $n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)$.

$n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)=n{{\left( A\bigcup B \right)}^{'}}$

$\Rightarrow n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)=n\left( U \right)-n\left( A\bigcup B \right)$

$\Rightarrow n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)=n\left( U \right)-\left[ n\left( A \right)+n\left( B \right)-n\left( A\bigcap B \right) \right]$

$\Rightarrow n\left( {{A}^{'}}\bigcap {{B}^{'}} \right)=1000-\left[ 300+300-200 \right]=600$

12. There are $210$ members in a club. $100$ of them drink tea and $65$ drink tea but not coffee, each member drinks tea or coffee. Find how many drinks coffee. How many drink coffee, but not tea.             

Ans: Let us have following notation

S-total members in the club

T-members who drink tea

C- members who drink coffee

$n\left( T \right)=100$

$n\left( T-C \right)=65$

$n\left( T\bigcup C \right)=210=n\left( S \right)$(since $n\left( T\bigcap C \right)=0$

We know that

\[n\left( T-C \right)=n\left( T \right)-n\left( T\bigcap C \right)\]

\[\Rightarrow n\left( T\bigcap C \right)=35\]

$n\left( T\bigcup C \right)=n\left( T \right)+n\left( C \right)-n\left( T\bigcap C \right)$

$\Rightarrow n\left( C \right)=145$

Therefore $n\left( C-T \right)=110$

13. If $P\left( A \right)=P\left( B \right)$, Show that $A=B$

Ans: For every $a\in A$ 

$\left\{ a \right\}\subset A$

$\Rightarrow \left\{ a \right\}\in P\left( A \right)$

$\Rightarrow \left\{ a \right\}\in P\left( B \right)$ (since $P\left( A \right)=P\left( B \right)$)

$\Rightarrow \left\{ a \right\}\in B$

$\left\{ a \right\}\subset B$

$\Rightarrow A\subset B$

Similarly we can easily say $B\subset A$

Therefore $B=A$

14. In a class of $25$ students, $12$ have taken mathematics, $8$ have taken mathematics but not biology. Find the no. of students who have taken both mathematics and biology and the no. of those who have taken biology but not mathematics each student has taken either mathematics or biology or both.

T-total number of students

M- number of students who have taken mathematics

B- number of students who have taken biology

$n\left( M \right)=12$

$n\left( M-B \right)=8$

$n\left( M\bigcup B \right)=25$

$n\left( M\bigcup B \right)=n\left( M \right)+n\left( B-M \right)$

$\Rightarrow 25=12+n\left( B-M \right)$

$\Rightarrow n\left( B-M \right)=13$

\[n\left( M\bigcup B \right)=n\left( M-B \right)+n\left( B-M \right)+n\left( M\bigcap B \right)\]

Hence we get \[n\left( M\bigcap B \right)=4\]  

15. A and B are two sets such that $n\left( A-B \right)=14+x,n\left( B-A \right)=3x,n\left( A\bigcap B \right)=x$. Draw a Venn diagram to illustrate this information. If $n\left( A \right)=n\left( B \right)$, Find 

(i) the value of $x$ 

Ans: It is given in the question 

 $n\left( A-B \right)=14+x$

$n\left( B-A \right)=3x$

$n\left( A\bigcap B \right)=x$

The venn diagram is as shown

$n\left( A \right)=n\left( A-B \right)+n\left( A\bigcap B \right)$

$\Rightarrow n\left( A \right)=14+2x$

$n\left( A \right)=n\left( B-A \right)+n\left( A\bigcap B \right)$

$\Rightarrow n\left( B \right)=4x$

Also it is given that $n\left( B \right)=n\left( A \right)$

Hence $14+2x=4x$

$\Rightarrow x=7$

(ii) $n\left( A\bigcup B \right)$         

Ans: From the above data we have

$n\left( A\bigcup B \right)=n\left( A-B \right)+n\left( B-A \right)+n\left( A\bigcap B \right)$

$\Rightarrow n\left( A\bigcup B \right)=14+x+3x+x=14+5x$

Hence  $n\left( A\bigcup B \right)=49$ (since $x=7$)

16. If A and B are two sets such that $A\bigcup B=A\bigcap B$ , then prove that $A=B$. 

Ans: Let us have $a\in A\Rightarrow a\in A\bigcap B$ 

It is given that $A\bigcup B=A\bigcap B$

Since we have $a\in A\bigcap B$ 

Therefore $A\subset B$

And similarly $B\subset A$

Therefore $A=B$ proved

17. Prove that if $A\bigcup B=C$ and $A\bigcap B=\varphi $ then $A=C-B$ 

Ans: Given $\left( A\bigcup B \right)=C$and $\left( A\bigcap B \right)=\varphi $

$\left( A\bigcup B \right)-B=\left( A\bigcup B \right)\bigcap {{B}^{'}}$

$=\left( {{B}^{'}}\bigcap A \right)\bigcup \left( {{B}^{'}}\bigcap B \right)$

$=\left( {{B}^{'}}\bigcap A \right)$

$=A$(since $\left( A\bigcap B \right)=\varphi $)

Hence proved

18. In a group of $65$ people, $40$ like cricket, $10$ like both cricket and tennis. How many like tennis only and not cricket? How many like tennis? 

Ans: Let us have following denotion

C-the set of people who like cricket

T-the set of people who like tennis

$n\left( C\bigcup T \right)=65$

$n\left( C \right)=40$

$n\left( C\bigcap T \right)=10$

We know that 

$n\left( C\bigcup T \right)=n\left( C \right)+n\left( T \right)-n\left( C\bigcap T \right)$

$\Rightarrow 65=40+n\left( T \right)-10$

Hence we get people who like tennis as $n\left( T \right)=35$

Now people who like tennis only not cricket is given by

$n\left( T-C \right)=n\left( T \right)-n\left( C\bigcap T \right)$

$\Rightarrow n\left( T-C \right)=35-10=25$

19. Let A,B and C be three sets $A\bigcup B=A\bigcup C$ and $A\bigcap B=A\bigcap C$ show that $B=C$                                                                                                                                  

Ans: Let us have $b\in B\Rightarrow b\in A\bigcup B$

Also it is given $A\bigcup B=A\bigcup C$

Therefore $b\in A\bigcup C$

Hence we get $b\in A\text{ or }b\in C$

In both cases B is subset of C

Similarly in both cases C is subset of B

Therefore $B=C$

20. If $U=\left\{ a,e,i,o,u \right\}$

$A=\left\{ a,e,i \right\}$ and $B=\left\{ e,o,u \right\}$, $C=\left\{ a,e,i \right\}$

Then verify that \[A\bigcap \left( B-C \right)=A\bigcap B-A\bigcap C\]\[A\cap (B-C)=(A\cap B)-(A\cap C)\]

$B-C=\left\{ e,o \right\}$

$A\bigcap \left( B-C \right)=\left\{ e \right\}$

$A\bigcap B=\left\{ e,o \right\}$and

$A\bigcap C=\left\{ a \right\}$

Hence proved 

$A\bigcap \left( B-C \right)=\left( A\bigcap B \right)-\left( A\bigcap C \right)$

Very Long Questions and Answers (6 Marks Questions)

1. In a survey it is found that $21$ people like product A, $26$ people like product B   and $29$ like product C. If $14$ people like product A and B, $15$ people like product and C, $12$ people like product C and A, and $8$ people like all the three products. Find 

(i) How many people are surveyed in all? 

Ans : Let us have A, B, C denote respectively the set of people who like the products A, B, C.

Then we can have a venn diagram as shown

From the above diagram 

Total number of surveyed people is given by

$a+b+c+d+e+f+g$

$a=21,e=26,g=29,d=12,b=14,f=15,c=8$

Therefore total number of surveyed people is given by

$21+14+8+12+26+15+29=125$

(ii) How many like product C only?

Ans: The number of people who like product C only is $29$                                                                 

2. A college awarded $38$ medals in football, $15$ in basketball and $20$ in cricket. If these medals went to a total of $50$ men and only five men got medals in all the three sports, how many received medals in exactly two of the three sports?

Ans : Let us have a notation F, B, and C for medals in football, basketball, and cricket respectively

C is intersection of all A,B,C and a,e,g are intersections of A and not B, B and not C, A and not C respectively.

From the above venn diagram      

\[f=5\]                ……(a)

\[a+b+e+f=38\]……(b)

\[b+c+d+f=15\]……(c)

\[e+d+f+g=20\]……(d)

$a+b+c+d+e+f+g=50$ ……(e)

From equations (d), (e) we get as shown

$a+b+c=30$……(f)

Now from equation (b) and (f) we get as shown

$e-3=c$       …….(g)

put value of c in the equation € as shown

$a+e+g+b+e+d=50-5+3$

Also from equation (d) and (e) we get

Therefore the medals received in exactly 2 of three sports is given by solving above equations as shown

 \[b+e+d=13\]

3. There are 200 individuals with a skin disorder, $120$ had been exposed to the chemical ${{C}_{1}}$, 50 to chemical ${{C}_{2}}$, and 30 to both the chemicals ${{C}_{1}}$ and ${{C}_{2}}$. Find the number of individuals exposed to 

(i). Chemical ${{C}_{1}}$ but not chemical ${{C}_{2}}$

Ans: Let us have a following notation

A- Denote the set of individuals exposed to the chemical \[{{C}_{1}}\]

B- Denote the set of individuals exposed to the chemical \[{{C}_{2}}\]

Given 

\[n\left( S \right)=200\] 

\[n\left( A \right)=120\]

\[n\left( B \right)=50\]

\[n\left( A\bigcap B \right)=30\]

\[\therefore n\left( A\bigcap \overline{B} \right)=n\left( A \right)-n\left( A\bigcap B \right)\]

\[\Rightarrow n\left( A\bigcap \overline{B} \right)=120-30=90\]

Hence the number of individuals exposed to chemical \[{{C}_{1}}\] but not to \[{{C}_{2}}\] is \[90\]

(ii). Chemical ${{C}_{2}}$ but not chemical ${{C}_{1}}$

\[\therefore n\left( \overline{A}\bigcap B \right)=n\left( B \right)-n\left( A\bigcap B \right)\]

\[\Rightarrow n\left( \overline{A}\bigcap B \right)=50-30=20\]

Hence the number of individuals exposed to chemical \[{{C}_{2}}\] but not to \[{{C}_{1}}\]  is \[20\]

(iii). Chemical ${{C}_{1}}$ or chemical ${{C}_{2}}$

Ans : Let us have a following notation

\[\therefore n\left( A\bigcup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\bigcap B \right)\]

\[\Rightarrow n\left( A\bigcup B \right)=120+50-30=140\]

Hence the number of individuals exposed to chemical \[{{C}_{2}}\]or \[{{C}_{1}}\]  is \[140\]

4. In a survey it was found that $21$ people liked product A, $26$ liked product B and $29$ liked product C. If $14$ people liked products A and B, $12$ people like C and A, $15$ people like B and C and $8$ liked all the three products. Find now many liked product C only.

Ans: Let us have a venn diagram of above information as shown

The followings are given in the question

$a+b+c+d=21$

$b+c+e+f=26$

$c+d+f+g=29$

Also it is given in the question 

$\therefore d=4$

$\therefore f=7$

Hence the number of people who like product C only is $g=10$ 

5. A college awarded $38$ medals in football, $15$ in basketball and $20$ in cricket. If these medals went to a total of $58$ men and only three men got medals in all the three sports, how many received medals in exactly two of the three sports?

Ans: Let us denote A, B and C as the sets of men who received medals in football, basketball and cricket respectively.

\[n\left( A \right)=38\]

\[n\left( B \right)=15\]

\[n\left( C \right)=20\]

\[n\left( A\bigcup B\bigcup C \right)=58\]

\[n\left( A\bigcap B\bigcap C \right)=3\]

\[\left( A\bigcup B\bigcup C \right)=n\left( A \right)+n\left( B \right)+n\left( C \right)-\left[ n\left( A\bigcap B \right)+n\left( B\bigcap C \right)+n\left( C\bigcap A \right) \right]+n\left( A\bigcap B\bigcap C \right)\]

\[\Rightarrow 58=38+15+20-\left( a+d \right)-\left( d+c \right)-\left( b+d \right)+3\]

\[\Rightarrow 18=a+b+c+3d\]

Hence we get \[a+b+c=9\]

6. In a survey of $60$ people, it was found that $25$ people read newspaper H, $26$ read newspaper T, $26$ read newspaper I, $9$ read both H and I, $11$ read both H and T, $8$ read both T and I, $3$ read all three newspapers. Find 

i) The no. of people who read at least one of the newspapers. 

Ans: Let us have a venn diagram as shown

We are given with the following data

\[a+b+c+d=25\]

\[b+c+e+f=26\]

\[d+c+g+f=26\]

And also it is given 

\[\therefore f=5\]

\[\therefore b=8\]

\[\therefore d=6\]

\[\therefore g=12\]

\[\therefore e=10\]

\[\therefore a=8\]

The no. of people who read at least one of the newspapers is \[a+b+c+d+e+f+g=52\]

ii) The no. of people who read exactly one newspaper

The no. of people who read exactly one newspaper is \[a+e+g=30\]

7. These are $20$ students in a chemistry class and $30$ students in a physics class. Find the number of students which are either in physics class or chemistry class in the following cases.

(i) Two classes meet at the same hour.

Ans: Let \[C\] be the set of students in chemistry class and \[P\] be the set of students in physics class.

\[n\left( P \right)=30\]

Now it is given that two classes meet at the same hour and hence

\[n\left( C\bigcap P \right)=0\]

\[\therefore n\left( C\bigcup P \right)=n\left( C \right)+n\left( P \right)-0\]

\[\Rightarrow n\left( C\bigcup P \right)=20+30=50\]

Hence the number of students which are either in physics class or chemistry class when classes are at the same time is \[50\].

(ii) The two classes met at different hours and ten students are enrolled in both the courses.

\[n\left( C\bigcap P \right)=10\]

\[\therefore n\left( C\bigcup P \right)=n\left( C \right)+n\left( P \right)-10\]

\[\Rightarrow n\left( C\bigcup P \right)=20+30-10=40\]

The number of students which are either in physics class or chemistry class when the two classes met at different hours and ten students are enrolled in both the courses is.

8. In a survey of $25$ students, it was found that $15$ had taken mathematics, $12$ had taken physics and $11$ had taken chemistry, $5$ had taken mathematics and chemistry, $9$ had taken mathematics and physics, $4$ had taken physics and chemistry and $3$ had taken all three subjects. 

Find the no. of students that had taken 

(i). only chemistry 

\[n\left( M \right)=a+b+d+e=15\]

\[n\left( P \right)=b+c+f+e=12\]

\[n\left( C \right)=d+e+f+g=11\]

\[n\left( M\bigcap P \right)=b+e=9\]

\[n\left( M\bigcap C \right)=d+e=5\]

\[n\left( P\bigcap C \right)=f+e=4\]

Also it is given that \[e=3\]

\[\therefore b=6,\therefore d=2,\therefore f=1\]

Also \[\therefore a=4,\therefore g=5,\therefore c=2\]

Therefore the number of students who had taken only chemistry is \[g=5\]

(ii). only mathematics 

Therefore the number of students who had taken only mathematics is \[a=4\]

(iii). only physics 

Therefore the number of students who had taken only physics is \[c=2\]

(iv). physics and chemistry but not  mathematics 

Therefore the number of students who had taken physics and chemistry but not mathematics is \[f=1\]

(v). mathematics and physics but not chemistry 

Therefore the number of students who had taken physics and mathematics but not chemistry is \[b=6\]

(vi). only one of the subjects 

Ans : Let us have a venn diagram of above information as shown

Therefore the number of students who had taken only one of the subjects is \[\therefore a+g+c=11\]

(vii). at least one of three subjects 

Therefore the number of students who had taken atleast one of the subjects is \[a+b+c+d+e+f+g=23\]

(viii). None of three subjects.

Therefore the number of students who had taken none of the subjects is \[25-\left( a+b+c+d+e+f+g \right)=2\]

9. In a survey of $100$ students, the no. of students studying the various languages were found to be English only $18$, English but not Hindi $23$, English and Sanskrit $8$, English $26$, Sanskrit $48$, Sanskrit and Hindi $8$, no language $24$. Find 

(i) How many students were studying Hindi? 

Ans: Let the total number of students be 

Let us have the venn diagram as shown

\[a+e+g+d=26\]

\[g+e+f+c=48\]

So we get 

\[e=5,g=3,d=0,f=5,c=35\]

Therefore the number of students studying Hindi is \[f+b+g+d=18\]

(ii) How many students were studying English and Hindi?

Therefore the number of students studying Hindi and English is \[g+d=3\] 

10. In a class of $50$ students, $30$ students like Hindi, $25$ like science and $16$ like both. Find the no. of students who like 

(i) Either Hindi or Science

Ans: Let the total number of students be

Let us denote number of students who like Hindi with H and who like science with S

\[n\left( H\bigcup S \right)=n\left( H \right)+n\left( S \right)-n\left( H\bigcap S \right)\]

\[\Rightarrow n\left( H\bigcup S \right)=30+25-16=39\]

Therefore the number of students who like either Hindi or Science is \[39\]

(ii) Neither Hindi nor Science.

\[n\left( {{H}^{'}}\bigcap {{S}^{'}} \right)=T-n\left( H\bigcup S \right)\]

\[\Rightarrow n\left( {{H}^{'}}\bigcap {{S}^{'}} \right)=50-39=11\]

Therefore the number of students who like either Hindi or Science is \[39\]  

11. In a town of 10,000 families, it was found that 40% of families buy newspaper A, 20% families buy newspaper B, and 10% of families buy newspaper C. 5% of families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three papers. Find the no. of families which buy

(i) A only 

Ans: Let the total number of families be 

\[T=10,000\]

Let us have the venn diagram of above information as shown

It is given in the question that 

\[x+a+c+d=4000\]

\[y+a+b+d=2000\]

\[z+b+c+d=1000\]

\[a+d=500\]

\[b+d=300\]

\[c+d=400\]

Hence on solving we get

\[a=300,b=100,c=200\]

Therefore the number of families who buy newspaper A only is \[x=4000-300-200-20=3380\]

(ii) B only

Ans : Let the total number of families be 

Let us have the venn diagram of above informations as shown

Therefore the number of families who buy newspaper B only is \[y=2000-300-200-100=1400\]

(iii) none of A, B, and C. 

From the above we get

\[z=1000-100-200-200=500\]

Therefore the number of families who buy newspaper none of A, B or C is 

\[10000-\left[ 3300+1400+500+300+100+200+200 \right]=5000\]

12. Two finite sets have m and n elements. The total no. of subsets of the first set is 56 more than the total no. of subsets of the second set. Find the value of m and n.      

Ans: Assume A and B be two sets having m and n elements respectively

Hence we know that number of subsets will be given as shown

Number of subsets of A is \[{{2}^{m}}\]

Number of subsets of B is \[{{2}^{n}}\]

According to the question 

\[{{2}^{m}}=56+{{2}^{n}}\]

\[\Rightarrow {{2}^{m}}-{{2}^{n}}=56\]

\[\Rightarrow {{2}^{n}}\left( {{2}^{m-n}}-1 \right)=56\]

On comparing we get

\[n=3,m-n=3\] \[\Rightarrow m=6\]

Sets Important Question PDF For Download

There is no doubt that we need help when we are solving something for the first time. The same goes for the important questions for class 11 maths chapter 1. Vedantu has provided its students with some tips in the pdf which can make their learning of sets in class 11 extra questions a bit less complicated and fun.

The chapter first of the 11th notebook is easy and has all the essential questions which make students test their formula-solving skills for sets. You can quickly check out the step-by-step solutions of this chapter's important questions in the pdf format and download it offline, so it can be viewed anytime even when the person is offline. 

Important Concepts Class 11 Maths Chapter 1 Related to Sets

Given below, we have breakdown of important concepts you will study in class 11 maths chapter 1. These will help you get a better grip of the formulas and the theorems which you need to use to solve the questions. 

Equal Sets 

For sets in class 11 important questions, one needs to know about sets as they are defined as a collection of well-defined, distinct objects. On the other hand, items which come together to form a set are called elements. The condition of two sets to become equal can happen when each set's element is also a part of the other set. Likewise, if both the sets are subsets of each other, you can even say these two sets are equal. 

Venn Diagram 

It is a diagram which is used by students and mathematicians to represent sets and their relation from each other. By seeing a Venn diagram, you can determine which operation has been done on the given two sets such as the intersection of the sets and their difference. Likewise, one can easily show the subsets of a given set using these diagrams. 

Union & Intersection of Sets

In class 11 sets important questions students will learn about the concept of a cardinal number of a set which is several distinct elements or members in a finite set. With the cardinality's help, we can define the size of a set if you want to denote the cardinal number of a set A you need to write it down like this n(A).

There are three properties of which you need to remember for the cardinal numbers and these are:

If A ∩ B = ∅, then, n(A ∪ B) = n(A) + n(B) this is a Union of disjoint sets.

If A and B are two finite sets, then n(A ∪ B) = n(A) + n(B) – n(A ∩ B) which is said to be a union of two sets.

If A, B and C are three finite sets, then; n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) this shows the union of three sets. 

Sets Class 11 Extra Questions

Well, if you are preparing to give the exams this year or next year, one thing is sure you need to prepare for the additional questions which are a bit tricky, and you can't find them in your textbook as well. Students studying in 11th can prepare their academic and competitive exams by solving these additional questions from Sets class 11. 

Q.1 Which of the following sets. Explain your answer.

(a). A collection of all days which are present in a single week and starts with an alphabet S.

(b).  The collection of the ten most famous singers of India.

(c).  A group of best football strikers in the world.

(d).  The collection of all boys in your school.

(e).  The collection of all the possible odd numbers below 100.

(f).  A collection of poems which are penned down by the famous poet Shakespeare.

(g).  The collection of all prime numbers.

(h).  The collection of questions in a science book.

(i).  A collection of most dangerous reptiles in India.

Q.2: Let P = {2, 3, 4, 5, 6, 7}. Insert the correct symbol  inside the given blank spaces below:

(a).  2 . . . . . . . . . . P

(b).  9 . . . . . . . . . P

(c).  11 . . . . . . . . P

(d).  4 . . . . . . . . P

(e).  0 . . . . . . . . P

(f).  7 . . . . . . . . P

Q.3: List all the elements from the given set P = {y: y is even natural number}

Q.4: If A = {(x,y) : x² + y²= 25  where x, y ∈ W } write a set of all possible ordered pairs.

Q.5: If P(X) = P(Y) show that A = B.

Q.6: Let N  and M be sets ; if N∩M = M∩X = ∅ and N∪X = M∪X for some set X.Show that N=M.

Q.7: If X ={1,2,3,4,5}, then solve the question to find out the proper subsets of A.

Q.8: For this question Write a Roster form of the given set A={x: x ∈ R, 2x+10 =12}

Q.9 Let X and Y are the two sets which have 3 and 6 elements present in them respectively. Find the maximum and the minimum number of elements in X ∪ Y.

Q.10:  If X = {(a,b) : a² + b²= 25  where a, b ∈ W } write a set of all possible ordered pairs.

Practice Questions

Write the set {-2,7} in the set builder form.

If the set N = { 1,3,7), then how many elements have set P(N).

If the universal set (U) = { 1,2,3,4,5,6,7}, A = {2,4,6} , B= { 3,5} and C = { 1,2,4,7}, Find: A′ ∪ (B ∩ C′), and  (B – A) ∪ (A – C)

If X, Y, and Z are three sets, then X – (Y ∪ Z) is equal to.

If A = {x, y} and B = { x, y, z). Is Z a superset of Y? Why?

Chapterwise Links for CBSE Class 11 Maths

Chapter 2 - Relations and Functions

Chapter 3 - Trigonometric Functions

Chapter 4 - Principle of Mathematical Induction

Chapter 5 - Complex Numbers and Quadratic Equations

Chapter 6 - Linear Inequalities

Chapter 7 - Permutations and Combinations

Chapter 8 - Binomial Theorem

Chapter 9 - Sequences and Series

Chapter 10 - Straight Lines

Chapter 11 - Conic Sections

Chapter 12 - Introduction to Three Dimensional Geometry

Chapter 13 - Limits and Derivatives

Chapter 14 - Mathematical Reasoning

Chapter 15 - Statistics

Chapter 16 - Probability

Benefits of Solving Important Questions For Class 11 Maths Chapter 1

Let's try to find out why solving the important questions for class 11th maths chapters are pretty essential and need to be done as much as possible. 

Understanding the Formulae: many times, students might skip the derivation and keep on mugging the formula all along. Knowing your formulas is a good thing, but when you don't know which one to use to solve a question is when the problem comes. With our PDF of solved sets examples, you will be able to understand which formula will be suitable for solving the problem.

Makes Your Problem Solving Efficient: Once you get a good grip on how to solve the problem, you can easily find out which problem will take more time and start writing it before anything else. 

Covers Important Topics: With this Pdf designed by Vedantu students get to know about all the main concepts of Venn Diagrams and how to use them along with Complement's properties. As a result, students will understand every topic they need to learn for their upcoming exams. 

Confidence Booster: When you start solving a question, and it comes out that you managed to get the right answer, you feel uplifted as it boosts your morale. If you have an issue with the answer, you can find out the step by step solving of the union and intersection sets answers. 

Gives More Questions for Practise: A student needs to be solving different types of problems to sharpen their mind and test their knowledge of the subject, and these important questions do the same thing. 

There you have it, these are some of the basic concepts you are going to study in the class 11th maths chapter 1 based on Sets . The important questions are solved and were written down so that it will be easier for you to understand their language. You need to keep on practising even if you think you are done with the chapter and have enough understanding. Always revise the chapter by doing some questions before you finally appear in exams. For the important questions of class 11th maths chapter 1 , you need to put both your mind and heart to study its concepts and get them memorized.

Important Related Links for CBSE Class 11 

Important Questions for Class 11 Maths Chapter 1 - Sets offered by Vedantu is an excellent resource for students who want to excel in their mathematical studies. The questions cover all the important topics in the chapter, including the definition of sets, types of sets, operations on sets, and Venn diagrams, making it easier for students to understand and improve their mathematical skills. The questions are designed by subject matter experts according to the CBSE syllabus for Class 11 students and provide a comprehensive and detailed explanation of the concepts. Additionally, the questions offer practice exercises that help students test their understanding of the chapter and prepare for their exams. Vedantu also provides interactive live classes and doubt-solving sessions to help students clarify their doubts and improve their understanding of the chapter. Overall, the Important Questions for Class 11 Maths Chapter 1 - Sets offered by Vedantu are an essential resource for students who want to improve their mathematical skills and score well in their exams.

FAQs on Important Questions for CBSE Class 11 Maths Chapter 1 - Sets 2024-25

1. How to utilize Important Questions for CBSE Class 11 Maths Chapter 1 Sets to score well in exams?

Students can solve important questions for Class 11 Maths Chapter 1 Sets available online to score well in school exams as well as competitive exams. The extra questions provided by e-learning sites on the first chapter of Class 11 Maths can be utilized to understand what types of questions can be expected from exams. These questions are really helpful for practice and clearing the concepts related to the chapter. On platforms like Vedantu, these questions are solved by expert teachers. By referring to the PDF file of important questions for Class 11 Maths Chapter 1 Sets, students will be able to practice the chapter effectively. These questions will also help in revision.

2. Where can I find Important Questions for Class 11 Maths Chapter 1 Sets?

Vedantu caters to a well-prepared set of Important Questions for Class 11 Maths Chapter 1 Sets as well as other chapters. Vedantu is a well-known online learning site known for its top quality study materials. It selects questions for a chapter based on the exam pattern and most frequently asked questions. Vedantu provides a free PDF of Important Questions for Class 11 Maths Chapter 1 Sets. These solutions are also solved by subject matter experts to provide a clear cut understanding of the chapter. These are proven to be helpful in exam preparation and provide effective revision during exams.

3. What is the importance of Vedantu’s extra questions for Class 11 Maths Chapter 1 Sets?

Vedantu’s extra questions for CBSE Class 11 Maths Chapter 1 Sets is crucial during the exam preparation. The important questions for Class 11 Maths Chapter 1 Sets allow students to practice the chapter thoroughly. Working on these questions will make students familiar with all types of questions that can be asked in the exam. The important questions PDF file at Vedantu is designed to cover all the important topics of the chapters. These questions are based on the exam pattern and are added after referring to previous year question papers. The free PDF of CBSE Maths Class 11 Chapter 1 can be utilized at the time of revision. These are really helpful in scoring well in the paper and are a confidence booster during the exam time.

4. What are the important learning outcomes from Class 11 Maths Chapter 1 Sets?

From Class 11 Maths Chapter 1 Sets, students will learn what are sets and how to represent sets. Students will also learn about types of sets such as Empty Sets, Equal Sets and what are Subsets and how to identify them. Also, one will learn how to design Venn diagrams. The chapter also includes the knowledge of different operations on sets such as Union of sets, Intersection of sets, Difference of sets, Complement of a set, etc.  Students are also taught about practical problems on Union and Intersection of Two Sets .

5. What are the important topics of the Chapter-Sets of Class 11 Maths?

Chapter 1 'Sets' of Class 11 Maths is an entirely new concept that is quite significant for the Class 11 exams. The chapter 'Sets' consists of the following important topics that students must pay attention to:

What are sets?

Sets and their representations

Finite sets and infinite sets

Universal sets

Venn diagrams

Operation on sets: Union and intersection of sets

Complement of the sets and their properties

Practical problems of union and intersection of two sets

6. How many chapters are there in Class 11 Maths apart from Chapter 1-Sets?

Class 11 Maths has a total of 16 chapters. The following are the chapters prescribed in the NCERT textbook: 

Ch. 1: Sets

Ch. 2: Relations and Functions

Ch. 3: Trigonometric Functions

Ch. 4: Principle of Mathematical Induction

Ch. 5: Complex Numbers and Quadratic Equations

Ch. 6: Linear Inequalities

Ch. 7: Permutations and Combinations

Ch. 8: Binomial Theorem

Ch. 9: Sequence and Series

Ch. 10: Straight Lines

Ch. 11: Conic Sections

Ch. 12: Introduction to Three–dimensional Geometry

Ch. 13: Limits and Derivatives

Ch. 14: Mathematical Reasoning

Ch. 15: Statistics

Ch. 16: Probability

7. What should I keep in mind while solving Chapter 1 of Class 11 Maths?

It makes a lot of difference how you present your answers in your answer sheet. Especially when it's a subject like Maths, students should make sure that their answers look neat and tidy. There are marks allocated to the steps of any solution, therefore one should ensure that his answers are written step-wise. Don't forget to mention and highlight the formula or theorem you are using in the solution. Avoid making silly mistakes with operation signs and numbers. A single mistake can make your entire solution wrong.

8. Is Chapter Sets of Class 11 Maths an important chapter?

Sets is definitely an important chapter in Class 11 Maths. It holds a major part in the exam holding a pretty good enough weightage of marks. This chapter is a whole new chapter for any student who enters Class 11 and since it is a significant chapter, students must be very keen in comprehending this chapter thoroughly. This chapter explains various types of sets in detail and questions related to them. To study more about sets students can download the Important Questions free of cost from the Vedantu website or mobile app. 

9. How can I complete my Maths class test paper of Chapter 1 of Class 11 Maths on time?

Most of the students face the difficulty of not being able to finish their papers on time. The chief reason behind this is the lack of time management. If you waste too much time solving trivial questions, you may end up skipping some really important questions. Therefore, divide your time evenly on each question before you start the exam. Regular practice using NCERT solutions and Mock tests from Vedantu can help students learn to manage their time better. 

CBSE Class 11 Maths Important Questions

Cbse study materials.

NCERT Solutions Class 11 Maths Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 1 Sets explains the most fundamental math concepts based on sets, their types, and applications. A set is a collection of well-defined objects such as a set of players in a team, a pack of cards, etc. Sets are often used to represent and define relations and functions. This knowledge of sets is also crucial for studying many related math topics, including geometry , sequences , probability , etc. With the help of NCERT Solutions Class 11 Maths Chapter 1, students will get well-versed with this topic and its applications in real-world situations.

The topic of Sets is not only fundamental for maths, but it is also helpful for other subjects. Learning about elements of sets and operations performed on them is vital for every student. With the regular practice of Class 11 Maths NCERT Solutions Chapter 1, students will easily attain this knowledge to perform operations on sets. The sample problems and examples included in these solutions are proficient in promoting a step-by-step understanding of this topic. To learn and practice with NCERT Solutions Chapter 1 Sets, download the exercises provided in the links below.

• NCERT Solutions Class 11 Maths Chapter 1 Ex 1.1
• NCERT Solutions Class 11 Maths Chapter 1 Ex 1.2
• NCERT Solutions Class 11 Maths Chapter 1 Ex 1.3
• NCERT Solutions Class 11 Maths Chapter 1 Ex 1.4
• NCERT Solutions Class 11 Maths Chapter 1 Ex 1.5
• NCERT Solutions Class 11 Maths Chapter 1 Ex 1.6
• NCERT Solutions Class 11 Maths Chapter 1 Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 1 PDF

NCERT Solutions for Class 11 Maths Chapter 1 are designed by experts to aid effective math learning for higher grades. These comprehensively structured resources are beneficial to enhance the problem-solving abilities of students. Sets can be a confusing topic if not prepared well hence, it is necessary to revisit the theories under this topic. To practice with these solutions, click on the links of the pdf files given below.

☛ Download Class 11 Maths NCERT Solutions Chapter 1 Sets

NCERT Class 11 Maths Chapter 1   Download PDF

NCERT Solutions for Class 11 Maths Chapter 1 Sets

NCERT solutions Class 11 Maths Chapter 1 Sets are highly efficient to strengthen the reasoning skills in students. These well-structured resources offer a simplistic learning approach with the help of interesting illustrations. The gradually placed examples provided in these solutions are helpful for strategic learning of each and every concept. To practice the exercise-wise NCERT Solutions Class 11 Maths Sets, try the links given below.

• Class 11 Maths Chapter 1 Ex 1.1 - 6 Questions
• Class 11 Maths Chapter 1 Ex 1.2 - 6 Questions
• Class 11 Maths Chapter 1 Ex 1.3 - 9 Questions
• Class 11 Maths Chapter 1 Ex 1.4 - 12 Questions
• Class 11 Maths Chapter 1 Ex 1.5 - 7 Questions
• Class 11 Maths Chapter 1 Ex 1.6 - 8 Questions
• Class 11 Maths Chapter 1 Miscellaneous Ex - 16 Questions

☛  Download Class 11 Maths Chapter 1 NCERT Book

Topics Covered: NCERT solutions Class 11 Maths Chapter 1 briefly introduces sets, their representation, and applications. The important topics covered in these solutions are an introduction to sets , finite infinite sets , equal sets, subsets , power sets , universal sets , Venn diagram , operations on sets including union, intersection , difference, and complement of a set.

Total Questions: Class 11 Maths Chapter 1 Sets has 48 questions in 6 exercises plus 16 questions in one miscellaneous exercise. These questions are primarily based on the representation of sets and their basic operations.

List of Formulas in NCERT Solutions Class 11 Maths Chapter 1

Having a clear understanding of all the terms and operations performed on sets is vital for every student. NCERT Solutions Class 11 Maths Chapter 1 will help gain this knowledge quickly with the help of suitable examples. Each concept in these solutions is well-explained to promote this understanding. Some of the important terms, formulas , and concepts related to sets explained in these solutions are listed below:

• Union of Sets: If A and B are two finite sets, then their union is the sum of all elements present in A and B.

Union of two sets can be represented using the formulas:

A U B = {x: x ∈ A (or) x ∈ B} which implies any element present in A U B is either an element of set A or set B.

• Intersection of Sets: If A and B are two finite sets then their intersection is the collection of all elements present in both A and B.

The intersection of two sets can be represented using the formulas:

A ∩ B = {x: x ∈ A (and) x ∈ B} which implies any element present in A∩B is present in both A and B.

• Venn Diagram: A Venn diagram is used to represents the logical relationship between sets.

FAQs on NCERT Solutions Class 11 Maths Chapter 1

What is the importance of ncert solutions for class 11 maths chapter 1 sets.

NCERT Solutions for Class 11 Maths Chapter 1 Sets are written as per the CBSE guidelines to promote efficient learning of each and every concept. The questions provided in these solutions are competent to deliver an in-depth understanding of sets and their relations. The practical examples included in these solutions are apt to promote math proficiency in students. With thorough practice, students will also gain the right approach for scoring well in exams.

What are the Important Topics Covered in Class 11 Maths NCERT Solutions Chapter 1?

The important topics covered in the NCERT Solutions Class 11 Maths Chapter 1 are the introduction to sets, their representation, types of sets, including finite infinite sets, equal sets, subsets, power sets, and universal sets. Other important topics included in these solutions are based on operations on sets and their Venn diagrams.

Do I Need to Practice all Questions Provided in NCERT Solutions Class 11 Maths Sets?

Every question included in NCERT Solutions Class 11 Maths Sets is well-framed to facilitate the formation of crystal clear concepts. By practicing all questions available in these solutions, students can easily master the fundamentals of sets and their applications. These solutions help in getting well-versed with sets, their representations, and operations. Additionally, regular practice helps kids recall concepts within seconds when required.

How Many Questions are there in Class 11 Maths NCERT Solutions Chapter 1 Sets?

NCERT Solutions for Class 11 Maths Chapter 1 Sets has 48 questions in 6 exercises. These exercises succinctly provide a holistic understanding of the whole topic, including important terms, formulas, and operations based on sets and their applications. The simple format of these solutions is suitable for acquiring a step-by-step understanding of sets.

What are the Important Formulas in NCERT Solutions Class 11 Maths Chapter 1?

NCERT Solutions Class 11 Maths Chapter 1 explains the representation of sets and their operations. Some of the important concepts explained in these solutions are based on the relations and functions of sets. These topics are elaborately detailed with the help of examples and illustrations. This chapter is majorly based on intuition rather than simply on formulas thus, requiring students to make use of their cognitive abilities.

Why Should I Practice NCERT Solutions for Class 11 Maths Chapter 1 Sets?

NCERT Solutions for Class 11 Maths Chapter 1 Sets is a competent learning resource that offers complete guidance and practice of core concepts for higher-level studies. These solutions cover the whole syllabus of CBSE Class 11 Maths Chapter 1 to provide comprehensive learning of all topics. The format of these solutions is simple and easy for students to grasp all complex concepts. Regular practice of these resources will ensure the best results in exams.

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Sets Class 11 Notes Maths Chapter 1

January 19, 2024 by Sastry CBSE

CBSE Class 11 Maths Notes Chapter 1 Sets

Set A set is a well-defined collection of objects.

Representation of Sets There are two methods of representing a set

• Roster or Tabular form In the roster form, we list all the members of the set within braces { } and separate by commas.
• Set-builder form In the set-builder form, we list the property or properties satisfied by all the elements of the sets.

Types of Sets – Class 11 Maths Notes

• Empty Sets: A set which does not contain any element is called an empty set or the void set or null set and it is denoted by {} or Φ.
• Singleton Set: A set consists of a single element, is called a singleton set.
• Finite and infinite Set: A set which consists of a finite number of elements, is called a finite set, otherwise the set is called an infinite set.
• Equal Sets: Two sets A and 6 are said to be equal, if every element of A is also an element of B or vice-versa, i.e. two equal sets will have exactly the same element.
• Equivalent Sets: Two finite sets A and 6 are said to be equal if the number of elements are equal, i.e. n(A) = n(B)

Subset – Class 11 Maths Notes

A set A is said to be a subset of set B if every element of set A belongs to set B. In symbols, we write A ⊆ B, if x ∈ A ⇒ x ∈ B

• Every set is o subset of itself.
• The empty set is a subset of every set.
• The total number of subsets of a finite set containing n elements is 2 n .

Intervals as Subsets of R Let a and b be two given real numbers such that a < b, then

• an open interval denoted by (a, b) is the set of real numbers {x : a < x < b}.
• a closed interval denoted by [a, b] is the set of real numbers {x : a ≤ x ≤ b}.
• intervals closed at one end and open at the others are known as semi-open or semi-closed interval and denoted by (a, b] is the set of real numbers {x : a < x ≤ b} or [a, b) is the set of real numbers {x : a ≤ x < b}.

Power Set The collection of all subsets of a set A is called the power set of A. It is denoted by P(A). If the number of elements in A i.e. n(A) = n, then the number of elements in P(A) = 2 n .

Universal Set A set that contains all sets in a given context is called the universal set.

Venn-Diagrams Venn diagrams are the diagrams, which represent the relationship between sets. In Venn-diagrams the universal set U is represented by point within a rectangle and its subsets are represented by points in closed curves (usually circles) within the rectangle.

Operations of Sets Union of sets: The union of two sets A and B, denoted by A ∪ B is the set of all those elements which are either in A or in B or in both A and B. Thus, A ∪ B = {x : x ∈ A or x ∈ B}.

Intersection of sets: The intersection of two sets A and B, denoted by A ∩ B, is the set of all elements which are common to both A and B. Thus, A ∩ B = {x : x ∈ A and x ∈ B}

Disjoint sets: Two sets Aand Bare said to be disjoint, if A ∩ B = Φ.

Intersecting or Overlapping sets: Two sets A and B are said to be intersecting or overlapping if A ∩ B ≠ Φ

Difference of sets: For any sets A and B, their difference (A – B) is defined as a set of elements, which belong to A but not to B. Thus, A – B = {x : x ∈ A and x ∉ B} also, B – A = {x : x ∈ B and x ∉ A}

Complement of a set: Let U be the universal set and A is a subset of U. Then, the complement of A is the set of all elements of U which are not the element of A. Thus, A’ = U – A = {x : x ∈ U and x ∉ A}

Some Properties of Complement of Sets

• A ∪ A’ = ∪
• A ∩ A’ = Φ
• ∪’ = Φ
• Φ’ = ∪
• (A’)’ = A

Symmetric difference of two sets: For any set A and B, their symmetric difference (A – B) ∪ (B – A) (A – B) ∪ (B – A) defined as set of elements which do not belong to both A and B. It is denoted by A ∆ B. Thus, A ∆ B = (A – B) ∪ (B – A) = {x : x ∉ A ∩ B}.

Laws of Algebra of Sets – Class 11 Maths Notes

Idempotent Laws: For any set A, we have

Identity Laws: For any set A, we have

Commutative Laws: For any two sets A and B, we have

• A ∪ B = B ∪ A
• A ∩ B = B ∩ A

Associative Laws: For any three sets A, B and C, we have

• A ∪ (B ∪ C) = (A ∪ B) ∪ C
• A ∩ (B ∩ C) = (A ∩ B) ∩ C

Distributive Laws: If A, B and Care three sets, then

• A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
• A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

De-Morgan’s Laws: If A and B are two sets, then

• (A ∪ B)’ = A’ ∩ B’
• (A ∩ B)’ = A’ ∪ B’

Formulae to Solve Practical Problems on Union and Intersection of Two Sets Let A, B and C be any three finite sets, then

• n(A ∪ B) = n(A) + n (B) – n(A ∩ B)
• If (A ∩ B) = Φ, then n (A ∪ B) = n(A) + n(B)
• n(A – B) = n(A) – n(A ∩ B)
• n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

Class 11 Maths Notes

• Chapter 1  Sets Class 11 Notes
• Chapter 2 Relations and Functions Class 11 Notes
• Chapter 3 Trigonometric Functions Class 11 Notes
• Chapter 4 Principle of Mathematical Induction Class 11 Notes
• Chapter 5 Complex Numbers and Quadratic Equations Class 11 Notes
• Chapter 6 Linear Inequalities Class 11 Notes
• Chapter 7 Permutations and Combinations Class 11 Notes
• Chapter 8 Binomial Theorem Class 11 Notes
• Chapter 9 Sequences and Series Class 11 Notes
• Chapter 10 Straight Lines Class 11 Notes
• Chapter 11 Conic Sections Class 11 Notes
• Chapter 12 Introduction to Three Dimensional Geometry Class 11 Notes
• Chapter 13 Limits and Derivatives Class 11 Notes
• Chapter 14 Mathematical Reasoning Class 11 Notes
• Chapter 15 Statistics Class 11 Notes
• Chapter 16 Probability Class 11 Notes

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NCERT Solutions for Class 11 Maths Chapter 1 - Sets

NCERT Solutions for Class 11 Maths Chapter 1 Sets are prepared by our expert faculty at BYJU’S according to the latest update on the CBSE Syllabus for 2023-24. These NCERT Class 11 Solutions of Maths help the students in solving the problems adroitly and efficiently. Also, BYJU’S focuses on building step-by-step solutions for all NCERT problems in such a way that it is easy for the students to understand.

Sets are used to define the concepts of functions and relations, some of which are covered in Chapter 1 of the NCERT textbook. The Class 11 Maths Chapter 1 of NCERT, categorised under the CBSE syllabus, also has some basic definitions and operations involving the sets. It is necessary to get fundamental knowledge on Sets since the study of sequences, geometry, and probability requires it. However, this is an easy chapter among all the chapters of NCERT Class 11 Maths, to score maximum marks in the board examination. These NCERT Solutions of BYJU’S help the students who are looking for a quick and easy way of solving.

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Exercise 1.1 page: 4

1. Which of the following are sets? Justify your answer.

(i) The collection of all months of a year beginning with the letter J.

(ii) The collection of ten most talented writers of India.

(iii) A team of eleven best-cricket batsmen of the world.

(iv) The collection of all boys in your class.

(v) The collection of all natural numbers less than 100.

(vi) A collection of novels written by the writer Munshi Prem Chand.

(vii) The collection of all even integers.

(viii) The collection of questions in this Chapter.

(ix) A collection of most dangerous animals of the world.

(i) The collection of all months of a year beginning with the letter J is a well-defined collection of objects as one can identify a month which belongs to this collection.

Therefore, this collection is a set.

(ii) The collection of ten most talented writers of India is not a well-defined collection as the criteria to determine a writer’s talent may differ from one person to another.

Therefore, this collection is not a set.

(iii) A team of eleven best-cricket batsmen of the world is not a well-defined collection as the criteria to determine a batsman’s talent may vary from one person to another.

(iv) The collection of all boys in your class is a well-defined collection as you can identify a boy who belongs to this collection.

(v) The collection of all natural numbers less than 100 is a well-defined collection as one can find a number which belongs to this collection.

(vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined collection as one can find any book which belongs to this collection.

(vii) The collection of all even integers is a well-defined collection as one can find an integer which belongs to this collection.

(viii) The collection of questions in this chapter is a well-defined collection as one can find a question which belongs to this chapter.

(ix) A collection of most dangerous animals of the world is not a well-defined collection as the criteria to find the dangerousness of an animal can differ from one animal to another.

2. Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈or ∉ in the blank spaces:

(i) 5…A (ii) 8…A  (iii) 0…A

(iv) 4…A (v) 2…A (vi) 10…A

(iii) 0 ∉ A

(vi) 10 ∉ A

3. Write the following sets in roster form:

(i) A = { x :  x  is an integer and –3 <  x  < 7}.

(ii) B = { x :  x  is a natural number less than 6}.

(iii) C = { x :  x  is a two-digit natural number such that the sum of its digits is 8}

(iv) D = { x :  x  is a prime number which is divisor of 60}.

(v) E = The set of all letters in the word TRIGONOMETRY.

(vi) F = The set of all letters in the word BETTER.

(i) A = { x :  x  is an integer and –3 <  x  < 7}

–2, –1, 0, 1, 2, 3, 4, 5, and 6 only are the elements of this set.

Hence, the given set can be written in roster form as

A = {–2, –1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = { x :  x  is a natural number less than 6}

1, 2, 3, 4, and 5 only are the elements of this set

B = {1, 2, 3, 4, 5}

17, 26, 35, 44, 53, 62, 71, and 80 only are the elements of this set

C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv) D = { x :  x  is a prime number which is divisor of 60}

Here 60 = 2 × 2 × 3 × 5

2, 3 and 5 only are the elements of this set

D = {2, 3, 5}

(v) E = The set of all letters in the word TRIGONOMETRY

TRIGONOMETRY is a 12 letters word out of which T, R and O are repeated.

E = {T, R, I, G, O, N, M, E, Y}

(vi) F = The set of all letters in the word BETTER

BETTER is a 6 letters word out of which E and T are repeated.

F = {B, E, T, R}

4. Write the following sets in the set-builder form:

(i) (3, 6, 9, 12) 

(ii) {2, 4, 8, 16, 32}

(iii) {5, 25, 125, 625} 

(iv) {2, 4, 6 …}

(v) {1, 4, 9 … 100}

(i) {3, 6, 9, 12}

The given set can be written in the set-builder form as { x :  x  = 3 n ,  n ∈ N and 1 ≤  n  ≤ 4}

We know that 2 = 2 1 , 4 = 2 2 , 8 = 2 3 , 16 = 2 4 , and 32 = 2 5 .

Therefore, the given set {2, 4, 8, 16, 32} can be written in the set-builder form as { x :  x  = 2 n ,  n ∈ N and 1 ≤  n  ≤ 5}.

(iii) {5, 25, 125, 625}

We know that 5 = 5 1 , 25 = 5 2 , 125 = 5 3 , and 625 = 5 4 .

Therefore, the given set {5, 25, 125, 625} can be written in the set-builder form as { x :  x  = 5 n ,  n ∈N and 1 ≤  n  ≤ 4}.

{2, 4, 6 …} is a set of all even natural numbers

Therefore, the given set {2, 4, 6 …} can be written in the set-builder form as { x :  x  is an even natural number}.

We know that 1 = 1 2 , 4 = 2 2 , 9 = 3 2  …100 = 10 2 .

Therefore, the given set {1, 4, 9… 100} can be written in the set-builder form as { x :  x  =  n 2 ,  n ∈ N and 1 ≤  n  ≤ 10}.

5. List all the elements of the following sets:

(i) A = { x :  x  is an odd natural number}

(ii) B = { x :  x  is an integer, -1/2 < x < 9/2}

(iii) C = { x :  x  is an integer, x 2 ≤ 4}

(iv) D = { x :  x  is a letter in the word “LOYAL”}

(v) E = { x :  x  is a month of a year not having 31 days}

(vi) F = { x :  x  is a consonant in the English alphabet which proceeds  k }.

So the elements are A = {1, 3, 5, 7, 9 …..}

We know that – 1/2 = – 0.5 and 9/2 = 4.5

So the elements are B = {0, 1, 2, 3, 4}.

We know that

(–1) 2  = 1 ≤ 4; (–2) 2  = 4 ≤ 4; (–3) 2  = 9 > 4

0 2  = 0 ≤ 4, 1 2  = 1 ≤ 4, 2 2  = 4 ≤ 4, 3 2  = 9 > 4

C = {–2, –1, 0, 1, 2}

So the elements are D = {L, O, Y, A}

So the elements are E = {February, April, June, September, November}

(vi) F = { x :  x  is a consonant in the English alphabet which proceeds  k }

So the elements are F = {b, c, d, f, g, h, j}

6. Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

(i) {1, 2, 3, 6} (a) {x: x is a prime number and a divisor of 6}

(ii) {2, 3} (b) {x: x is an odd natural number less than 10}

(iii) {M, A, T, H, E, I, C, S} (c) {x: x is a natural number and divisor of 6}

(iv) {1, 3, 5, 7, 9} (d) {x: x is a letter of the word MATHEMATICS}

(i) Here the elements of this set are natural number as well as divisors of 6. Hence, (i) matches with (c).

(ii) 2 and 3 are prime numbers which are divisors of 6. Hence, (ii) matches with (a).

(iii) The elements are the letters of the word MATHEMATICS. Hence, (iii) matches with (d).

(iv) The elements are odd natural numbers which are less than 10. Hence, (v) matches with (b).

Exercise 1.2 page: 8

1. Which of the following are examples of the null set?

(i) Set of odd natural numbers divisible by 2

(ii) Set of even prime numbers

(iii) { x : x  is a natural numbers,  x  < 5 and  x  > 7}

(iv) { y : y  is a point common to any two parallel lines}

(i) Set of odd natural numbers divisible by 2 is a null set as odd numbers are not divisible by 2.

(ii) Set of even prime numbers is not a null set as 2 is an even prime number.

(iii) { x :  x  is a natural number,  x  < 5 and  x  > 7} is a null set as a number cannot be both less than 5 and greater than 7.

(iv) { y :  y  is a point common to any two parallel lines} is a null set as the parallel lines do not intersect. Therefore, they have no common point.

2. Which of the following sets are finite or infinite?

(i) The set of months of a year

(ii) {1, 2, 3 …}

(iii) {1, 2, 3 … 99, 100}

(iv) The set of positive integers greater than 100

(v) The set of prime numbers less than 99

(i) The set of months of a year is a finite set as it contains 12 elements.

(ii) {1, 2, 3 …} is an infinite set because it has infinite number of natural numbers.

(iii) {1, 2, 3 …99, 100} is a finite set as the numbers from 1 to 100 are finite.

(iv) The set of positive integers greater than 100 is an infinite set as the positive integers which are greater than 100 are infinite.

(v) The set of prime numbers less than 99 is a finite set as the prime numbers which are less than 99 are finite.

3. State whether each of the following set is finite or infinite:

(i) The set of lines which are parallel to the  x -axis

(ii) The set of letters in the English alphabet

(iii) The set of numbers which are multiple of 5

(iv) The set of animals living on the earth

(v) The set of circles passing through the origin (0, 0)

(i) The set of lines which are parallel to the  x -axis is an infinite set as the lines which are parallel to the  x -axis are infinite.

(ii) The set of letters in the English alphabet is a finite set as it contains 26 elements.

(iii) The set of numbers which are multiple of 5 is an infinite set as the multiples of 5 are infinite.

(iv) The set of animals living on the earth is a finite set as the number of animals living on the earth is finite.

(v) The set of circles passing through the origin (0, 0) is an infinite set as infinite number of circles can pass through the origin.

4. In the following, state whether A = B or not:

(i) A = { a ,  b ,  c ,  d }; B = { d ,  c ,  b ,  a }

(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}

(iii) A = {2, 4, 6, 8, 10}; B = { x :  x  is positive even integer and  x  ≤ 10}

(iv) A = { x :  x  is a multiple of 10}; B = {10, 15, 20, 25, 30 …}

(i) A = { a ,  b ,  c ,  d }; B = { d ,  c ,  b ,  a }

Order in which the elements of a set are listed is not significant.

Therefore, A = B.

(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}

We know that 12 ∈ A but 12 ∉ B.

Therefore, A ≠ B

(iii) A = {2, 4, 6, 8, 10};

B = { x :  x  is a positive even integer and  x  ≤ 10} = {2, 4, 6, 8, 10}

Therefore, A = B

(iv) A = { x :  x  is a multiple of 10}

B = {10, 15, 20, 25, 30 …}

We know that 15 ∈ B but 15 ∉ A.

5. Are the following pair of sets equal? Give reasons.

(i) A = {2, 3}; B = { x :  x  is solution of  x 2  + 5 x  + 6 = 0}

(ii) A = { x :  x  is a letter in the word FOLLOW}; B = { y :  y  is a letter in the word WOLF}

x 2  + 5 x  + 6 = 0 can be written as

x ( x  + 3) + 2( x  + 3) = 0

By further calculation

( x  + 2) ( x  + 3) = 0

x  = –2 or  x  = –3

A = {2, 3}; B = {–2, –3}

(ii) A = { x :  x  is a letter in the word FOLLOW} = {F, O, L, W}

B = { y :  y  is a letter in the word WOLF} = {W, O, L, F}

Order in which the elements of a set which are listed is not significant.

6. From the sets given below, select equal sets:

A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D = {3, 1, 4, 2}

E = {–1, 1}, F = {0,  a }, G = {1, –1}, H = {0, 1}

A = {2, 4, 8, 12}; B = {1, 2, 3, 4}; C = {4, 8, 12, 14}

D = {3, 1, 4, 2}; E = {–1, 1}; F = {0,  a }

G = {1, –1}; H = {0, 1}

8 ∈ A, 8 ∉ B, 8 ∉ D, 8 ∉ E, 8 ∉ F, 8 ∉ G, 8 ∉ H

A ≠ B, A ≠ D, A ≠ E, A ≠ F, A ≠ G, A ≠ H

It can be written as

2 ∈ A, 2 ∉ C

Therefore, A ≠ C

3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H

B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H

12 ∈ C, 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H

Therefore, C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H

4 ∈ D, 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H

Therefore, D ≠ E, D ≠ F, D ≠ G, D ≠ H

Here, E ≠ F, E ≠ G, E ≠ H

F ≠ G, F ≠ H, G ≠ H

B = D and E = G

Therefore, among the given sets, B = D and E = G.

Exercise 1.3 page: 12

1. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:

(i) {2, 3, 4} … {1, 2, 3, 4, 5}

(ii) { a ,  b ,  c } … { b ,  c ,  d }

(iii) { x :  x  is a student of Class XI of your school} … { x :  x  student of your school}

(iv) { x :  x  is a circle in the plane} … { x :  x  is a circle in the same plane with radius 1 unit}

(v) { x :  x  is a triangle in a plane}…{ x :  x  is a rectangle in the plane}

(vi) { x :  x  is an equilateral triangle in a plane}… { x :  x  is a triangle in the same plane}

(vii) { x :  x  is an even natural number} … { x :  x  is an integer}

(i) {2, 3, 4} ⊂ {1, 2, 3, 4, 5}

(ii) { a ,  b ,  c } ⊄ { b ,  c ,  d }

(iii) { x :  x  is a student of Class XI of your school} ⊂ { x :  x  student of your school}

(iv) { x :  x  is a circle in the plane} ⊄ { x :  x  is a circle in the same plane with radius 1 unit}

(v) { x :  x  is a triangle in a plane} ⊄ { x :  x  is a rectangle in the plane}

(vi) { x :  x  is an equilateral triangle in a plane} ⊂ { x :  x  is a triangle in the same plane}

(vii) { x :  x  is an even natural number} ⊂ { x :  x  is an integer}

2. Examine whether the following statements are true or false:

(i) { a ,  b } ⊄ { b ,  c ,  a }

(ii) { a ,  e } ⊂ { x :  x  is a vowel in the English alphabet}

(iii) {1, 2, 3} ⊂ {1, 3, 5}

(iv) { a } ⊂ { a .  b ,  c }

(v) { a } ∈ ( a ,  b ,  c )

(vi) { x :  x  is an even natural number less than 6} ⊂ { x :  x  is a natural number which divides 36}

Here each element of { a ,  b } is an element of { b ,  c ,  a }.

We know that a ,  e  are two vowels of the English alphabet.

(iii) False.

2 ∈ {1, 2, 3} where, 2∉ {1, 3, 5}

Each element of { a } is also an element of { a ,  b ,  c }.

Elements of { a ,  b ,  c } are  a ,  b ,  c . Hence, { a } ⊂ { a ,  b ,  c }

{ x : x  is an even natural number less than 6} = {2, 4}

{ x : x  is a natural number which divides 36}= {1, 2, 3, 4, 6, 9, 12, 18, 36}

3. Let A= {1, 2, {3, 4}, 5}. Which of the following statements are incorrect and why?

(i) {3, 4} ⊂ A

(ii) {3, 4}}∈ A

(iii) {{3, 4}} ⊂ A

(vi) {1, 2, 5} ⊂ A

(vii) {1, 2, 5} ∈ A

(viii) {1, 2, 3} ⊂ A

(xi) {Φ} ⊂ A

It is given that A= {1, 2, {3, 4}, 5}

(i) {3, 4} ⊂ A is incorrect

Here 3 ∈ {3, 4}; where, 3∉A.

(ii) {3, 4} ∈A is correct

{3, 4} is an element of A.

(iii) {{3, 4}} ⊂ A is correct

{3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.

(iv) 1∈A is correct

1 is an element of A.

(v) 1⊂ A is incorrect

An element of a set can never be a subset of itself.

(vi) {1, 2, 5} ⊂ A is correct

Each element of {1, 2, 5} is also an element of A.

(vii) {1, 2, 5} ∈ A is incorrect

{1, 2, 5} is not an element of A.

(viii) {1, 2, 3} ⊂ A is incorrect

3 ∈ {1, 2, 3}; where, 3 ∉ A.

(ix) Φ ∈ A is incorrect

Φ is not an element of A.

(x) Φ ⊂ A is correct

Φ is a subset of every set.

(xi) {Φ} ⊂ A is incorrect

Φ∈ {Φ}; where, Φ ∈ A.

4. Write down all the subsets of the following sets:

(ii) { a ,  b }

(iii) {1, 2, 3}

(i) Subsets of { a } are

Φ and { a }.

(ii) Subsets of { a ,  b } are

Φ, { a }, { b }, and { a ,  b }.

(iii) Subsets of {1, 2, 3} are

Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and {1, 2, 3}.

(iv) Only subset of Φ is Φ.

5. How many elements has P (A), if A = Φ?

If A is a set with  m  elements

n (A) =  m then  n [P (A)] = 2 m

If A = Φ we get n (A) = 0

n [P(A)] = 2 0  = 1

Therefore, P (A) has one element.

6. Write the following as intervals:

(i) { x :  x  ∈ R, –4 <  x  ≤ 6}

(ii) { x :  x  ∈ R, –12 <  x  < –10}

(iii) { x :  x  ∈ R, 0 ≤  x  < 7}

(iv) { x :  x  ∈ R, 3 ≤  x  ≤ 4}

(ii) { x :  x  ∈ R, –12 <  x  < –10} = (–12, –10)

(iii) { x :  x  ∈ R, 0 ≤  x  < 7} = [0, 7)

7. Write the following intervals in set-builder form:

(i) (–3, 0)

(ii) [6, 12]

(iii) (6, 12]

(iv) [–23, 5)

(i) (–3, 0) = { x :  x  ∈ R, –3 <  x  < 0}

(ii) [6, 12] = { x :  x  ∈ R, 6 ≤  x  ≤ 12}

(iii) (6, 12] ={ x :  x  ∈ R, 6 <  x  ≤ 12}

(iv) [–23, 5) = { x :  x  ∈ R, –23 ≤  x  < 5}

8. What universal set (s) would you propose for each of the following?

(i) The set of right triangles

(ii) The set of isosceles triangles

(i) Among the set of right triangles, the universal set is the set of triangles or the set of polygons.

(ii) Among the set of isosceles triangles, the universal set is the set of triangles or the set of polygons or the set of two-dimensional figures.

9. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universals set (s) for all the three sets A, B and C?

(i) {0, 1, 2, 3, 4, 5, 6}

(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(iv) {1, 2, 3, 4, 5, 6, 7, 8}

(i) We know that A ⊂ {0, 1, 2, 3, 4, 5, 6}

B ⊂ {0, 1, 2, 3, 4, 5, 6}

So C ⊄ {0, 1, 2, 3, 4, 5, 6}

Hence, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and C.

(ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ

Hence, Φ cannot be the universal set for the sets A, B, and C.

(iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Hence, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B, and C.

(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8}

B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}

So C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}

Hence, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B, and C.

Exercise 1.4 page: 17

1. Find the union of each of the following pairs of sets:

(i) X = {1, 3, 5} Y = {1, 2, 3}

(ii) A = { a ,  e ,  i ,  o ,  u } B = { a ,  b ,  c }

(iii) A = { x :  x  is a natural number and multiple of 3}

B = { x :  x  is a natural number less than 6}

(iv) A = { x :  x  is a natural number and 1 <  x  ≤ 6}

B = { x :  x  is a natural number and 6 <  x  < 10}

(v) A = {1, 2, 3}, B = Φ

So the union of the pairs of set can be written as

X ∪ Y= {1, 2, 3, 5}

A∪ B = { a ,  b ,  c ,  e ,  i ,  o ,  u }

(iii) A = { x :  x  is a natural number and multiple of 3} = {3, 6, 9 …}

B = { x :  x  is a natural number less than 6} = {1, 2, 3, 4, 5, 6}

A ∪ B = {1, 2, 4, 5, 3, 6, 9, 12 …}

Hence, A ∪ B = { x :  x  = 1, 2, 4, 5 or a multiple of 3}

(iv) A = { x :  x  is a natural number and 1 <  x  ≤ 6} = {2, 3, 4, 5, 6}

B = { x :  x  is a natural number and 6 <  x  < 10} = {7, 8, 9}

A∪ B = {2, 3, 4, 5, 6, 7, 8, 9}

Hence, A∪ B = { x :  x  ∈ N and 1 <  x  < 10}

(v) A = {1, 2, 3}, B = Φ

A∪ B = {1, 2, 3}

2. Let A = { a ,  b }, B = { a ,  b ,  c }. Is A ⊂ B? What is A ∪ B?

It is given that

A = { a ,  b } and B = { a ,  b ,  c }

A∪ B = { a ,  b ,  c } = B

3. If A and B are two sets such that A ⊂ B, then what is A ∪ B?

If A and B are two sets such that A ⊂ B, then A ∪ B = B.

4. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}; find

(iii) B ∪ C

(v) A ∪ B ∪ C

(vi) A ∪ B ∪ D

(vii) B ∪ C ∪ D

A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}

(i) A ∪ B = {1, 2, 3, 4, 5, 6}

(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(iii) B ∪ C = {3, 4, 5, 6, 7, 8}

(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

5. Find the intersection of each pair of sets:

(i) X = {1, 3, 5}, Y = {1, 2, 3}

So the intersection of the given set can be written as

X ∩ Y = {1, 3}

(ii) A = { a ,  e ,  i ,  o ,  u }, B = { a ,  b ,  c }

A ∩ B = { a }

(iii) A = { x :  x  is a natural number and multiple of 3} = (3, 6, 9 …}

B = { x :  x  is a natural number less than 6} = {1, 2, 3, 4, 5}

A ∩ B = {3}

(iv) A = { x :  x  is a natural number and 1 <  x  ≤ 6} = {2, 3, 4, 5, 6}

B = { x :  x  is a natural number and 6 <  x  < 10} = {7, 8, 9}

6. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find

(iii) A ∩ C ∩ D

(vi) A ∩ (B ∪ C)

(vii) A ∩ D

(viii) A ∩ (B ∪ D)

(ix) (A ∩ B) ∩ (B ∪ C)

(x) (A ∪ D) ∩ (B ∪ C)

(i) A ∩ B = {7, 9, 11}

(ii) B ∩ C = {11, 13}

(iii) A ∩ C ∩ D = {A ∩ C} ∩ D

= {11} ∩ {15, 17}

(iv) A ∩ C = {11}

(v) B ∩ D = Φ

(vi) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

= {7, 9, 11} ∪ {11}

= {7, 9, 11}

(vii) A ∩ D = Φ

(viii) A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)

= {7, 9, 11} ∪ Φ

(ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15}

(x) (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15}

= {7, 9, 11, 15}

7. If A = { x :  x  is a natural number}, B ={ x :  x  is an even natural number}

C = { x :  x  is an odd natural number} and D = { x :  x  is a prime number}, find

(iii) A ∩ D

A = { x :  x  is a natural number} = {1, 2, 3, 4, 5 …}

B ={ x :  x  is an even natural number} = {2, 4, 6, 8 …}

C = { x :  x  is an odd natural number} = {1, 3, 5, 7, 9 …}

D = { x :  x  is a prime number} = {2, 3, 5, 7 …}

(i) A ∩B = { x :  x  is a even natural number} = B

(ii) A ∩ C = { x :  x  is an odd natural number} = C

(iii) A ∩ D = { x :  x  is a prime number} = D

(iv) B ∩ C = Φ

(v) B ∩ D = {2}

(vi) C ∩ D = { x :  x  is odd prime number}

8. Which of the following pairs of sets are disjoint?

(i) {1, 2, 3, 4} and { x :  x  is a natural number and 4 ≤  x  ≤ 6}

(ii) { a ,  e ,  i ,  o ,  u }and { c ,  d ,  e ,  f }

(iii) { x :  x  is an even integer} and { x: x  is an odd integer}

(i) {1, 2, 3, 4}

{ x :  x  is a natural number and 4 ≤  x  ≤  6 } = {4, 5, 6}

{1, 2, 3, 4} ∩ {4, 5, 6} = {4}

Hence, this pair of sets is not disjoint.

(ii) { a ,  e ,  i ,  o ,  u } ∩ ( c ,  d ,  e ,  f } = { e }

Hence, { a ,  e ,  i ,  o ,  u } and ( c ,  d ,  e ,  f } are not disjoint.

(iii) { x :  x  is an even integer} ∩ { x :  x  is an odd integer} = Φ

Hence, this pair of sets is disjoint.

9. If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20},

C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}; find

(iii) A – D

(vii) B – C

(viii) B – D

(xii) D – C

(i) A – B = {3, 6, 9, 15, 18, 21}

(ii) A – C = {3, 9, 15, 18, 21}

(iii) A – D = {3, 6, 9, 12, 18, 21}

(iv) B – A = {4, 8, 16, 20}

(v) C – A = {2, 4, 8, 10, 14, 16}

(vi) D – A = {5, 10, 20}

(vii) B – C = {20}

(viii) B – D = {4, 8, 12, 16}

(ix) C – B = {2, 6, 10, 14}

(x) D – B = {5, 10, 15}

(xi) C – D = {2, 4, 6, 8, 12, 14, 16}

(xii) D – C = {5, 15, 20}

10. If X = { a ,  b ,  c ,  d } and Y = { f ,  b ,  d, g }, find

(iii) X ∩ Y

(i) X – Y = { a ,  c }

(ii) Y – X = { f ,  g }

(iii) X ∩ Y = { b ,  d }

11. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

R – Set of real numbers

Q – Set of rational numbers

Hence, R – Q is a set of irrational numbers.

12. State whether each of the following statement is true or false. Justify your answer.

(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.

(ii) { a ,  e ,  i ,  o ,  u  } and { a ,  b ,  c ,  d } are disjoint sets.

(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.

(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets. Solution:

If 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}

So we get {2, 3, 4, 5} ∩ {3, 6} = {3}

If a  ∈ { a ,  e ,  i ,  o ,  u },  a  ∈ { a ,  b ,  c ,  d }

So we get { a ,  e ,  i ,  o ,  u } ∩ { a ,  b ,  c ,  d } = { a }

Here {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ

Here {2, 6, 10} ∩ {3, 7, 11} = Φ

Exercise 1.5 Page: 20

1. Let U = {1, 2, 3; 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find

(iii) (A U C)’

(iv) (A U B)’

(vi) (B – C)’

U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {1, 2, 3, 4}

B = {2, 4, 6, 8}

C = {3, 4, 5, 6}

(i) A’ = {5, 6, 7, 8, 9}

(ii) B’ = {1, 3, 5, 7, 9}

(iii) A U C = {1, 2, 3, 4, 5, 6}

(A U C)’ = {7, 8, 9}

(iv) A U B = {1, 2, 3, 4, 6, 8}

(A U B)’ = {5, 7, 9}

(v) (A’)’ = A = {1, 2, 3, 4}

(vi) B – C = {2, 8}

(B – C)’ = {1, 3, 4, 5, 6, 7, 9}

2. If U = { a, b, c, d, e, f, g, h }, find the complements of the following sets:

(i) A = { a, b, c }

(ii) B = { d, e, f, g }

(iii) C = { a, c, e, g }

(iv) D = { f ,  g ,  h ,  a } Solution:

A’ = {d, e, f, g, h}

B’ = {a, b, c, h}

C’ = {b, d, f, h}

(iv) D = { f ,  g ,  h ,  a }

D’ = {b, c, d, e}

3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:

(i) { x :  x  is an even natural number}

(ii) { x :  x  is an odd natural number}

(iii) { x :  x  is a positive multiple of 3}

(iv) { x :  x  is a prime number}

(v) { x :  x  is a natural number divisible by 3 and 5}

(vi) { x :  x  is a perfect square}

(vii) { x :  x  is perfect cube}

(viii) { x :  x  + 5 = 8}

(ix) { x : 2 x  + 5 = 9}

(x) { x :  x  ≥ 7}

(xi) { x :  x  ∈ N and 2 x  + 1 > 10}

U = N: Set of natural numbers

(i) { x :  x  is an even natural number}´ = { x :  x  is an odd natural number}

(ii) { x :  x  is an odd natural number}´ = { x :  x  is an even natural number}

(iii) { x :  x  is a positive multiple of 3}´ = { x :  x  ∈ N and  x  is not a multiple of 3}

(iv) { x :  x  is a prime number}´ ={ x :  x  is a positive composite number and  x  = 1}

(v) { x :  x  is a natural number divisible by 3 and 5}´ = { x :  x  is a natural number that is not divisible by 3 or 5}

(vi) { x :  x  is a perfect square}´ = { x :  x  ∈ N and  x  is not a perfect square}

(vii) { x :  x  is a perfect cube}´ = { x :  x  ∈ N and  x  is not a perfect cube}

(viii) { x :  x  + 5 = 8}´ = { x :  x  ∈ N and  x  ≠ 3}

(ix) { x : 2 x  + 5 = 9}´ = { x :  x  ∈ N and  x  ≠ 2}

(x) { x :  x  ≥ 7}´ = { x :  x  ∈ N and  x  < 7}

(xi) { x :  x  ∈ N and 2 x  + 1 > 10}´ = { x :  x  ∈ N and  x  ≤ 9/2}

4. If U = {1, 2, 3, 4, 5,6,7,8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that

(i) (A U B)’ = A’ ∩ B’

(ii) (A ∩ B)’ = A’ U B’

U = {1, 2, 3, 4, 5,6,7,8, 9}

A = {2, 4, 6, 8}

B = {2, 3, 5, 7}

(i) (A U B)’ = {2, 3, 4, 5, 6, 7, 8}’ = {1, 9}

A’ ∩ B’ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9}

Therefore, (A U B)’ = A’ ∩ B’.

(ii) (A ∩ B)’ = {2}’ = {1, 3, 4, 5, 6, 7, 8, 9}

A’ U B’ = {1, 3, 5, 7, 9} U {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}

Therefore, (A ∩ B)’ = A’ U B’.

5. Draw appropriate Venn diagram for each of the following:

(i) (A U B)’

(ii) A’ ∩ B’

(iii) (A ∩ B)’

(iv) A’ U B’

6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’?

A’ is the set of all equilateral triangles.

7. Fill in the blanks to make each of the following a true statement:

(i) A U A’ = ……..

(ii) Φ′ ∩ A = …….

(iii) A ∩ A’ = …….

(iv) U’ ∩ A = …….

(i) A U A’ = U

(ii) Φ′ ∩ A = U ∩ A = A

(iii) A ∩ A’ = Φ

(iv) U’ ∩ A = Φ ∩ A = Φ

Exercise 1.6 page: 24

1. If X and Y are two sets such that  n (X) = 17,  n (Y) = 23 and  n (X ∪ Y) = 38, find  n (X ∩ Y).

n (X U Y) = 38

We can write it as

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values

38 = 17 + 23 – n (X ∩ Y)

n (X ∩ Y) = 40 – 38 = 2

n (X ∩ Y) = 2

2. If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have?

n (X U Y) = 18

18 = 8 + 15 – n (X ∩ Y)

n (X ∩ Y) = 23 – 18 = 5

n (X ∩ Y) = 5

3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

Consider H as the set of people who speak Hindi

E as the set of people who speak English

n (H ∪ E) = 400

n (H) = 250

n (E) = 200

n (H ∪ E) =  n (H) +  n (E) –  n (H ∩ E)

By substituting the values

400 = 250 + 200 –  n (H ∩ E)

400 = 450 –  n (H ∩ E)

n (H ∩ E) = 450 – 400

n (H ∩ E) = 50

Therefore, 50 people can speak both Hindi and English.

4. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?

n (S ∩ T) = 11

n  (S ∪ T) =  n  (S) +  n  (T) –  n  (S ∩ T)

n  (S ∪ T) = 21 + 32 – 11

n  (S ∪ T)= 42

Therefore, the set (S ∪ T) has 42 elements.

5. If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?

n (X ∪ Y) = 60

n (X ∩ Y) = 10

n (X ∪ Y) =  n (X) +  n (Y) –  n (X ∩ Y)

60 = 40 +  n (Y) – 10

On further calculation

n (Y) = 60 – (40 – 10) = 30

Therefore, the set Y has 30 elements.

6. In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?

Consider C as the set of people who like coffee

T as the set of people who like tea

n (C ∪ T) = 70

n (C ∪ T) =  n (C) +  n (T) –  n (C ∩ T)

70 = 37 + 52 –  n (C ∩ T)

70 = 89 –  n (C ∩ T)

n (C ∩ T) = 89 – 70 = 19

Therefore, 19 people like both coffee and tea.

7. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

Consider C as the set of people who like cricket

T as the set of people who like tennis

n (C ∪ T) = 65

n (C ∩ T) = 10

n (C ∪ T) =  n (C) +  n (T) –  n (C ∩ T)

65 = 40 +  n (T) – 10

65 = 30 +  n (T)

n (T) = 65 – 30 = 35

Hence, 35 people like tennis.

We know that,

(T – C) ∪ (T ∩ C) = T

(T – C) ∩ (T ∩ C) = Φ

n  (T) =  n  (T – C) +  n  (T ∩ C)

35 =  n  (T – C) + 10

n  (T – C) = 35 – 10 = 25

Therefore, 25 people like only tennis.

8. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?

Consider F as the set of people in the committee who speak French

S as the set of people in the committee who speak Spanish

n (S ∩ F) = 10

n (S ∪ F) =  n (S) +  n (F) –  n (S ∩ F)

n (S ∪ F) = 20 + 50 – 10

n (S ∪ F) = 70 – 10

n (S ∪ F) = 60

Therefore, 60 people in the committee speak at least one of the two languages.

Miscellaneous Exercise Page 26

1. Decide, among the following sets, which sets are subsets of one and another:

A= {x: x ∈ R and x satisfy x 2 – 8x + 12 = 0},

B = {2, 4, 6},

C = {2, 4, 6, 8…},

According to the question,

A = {x: x ∈ R and x satisfies x 2 – 8x + 12 =0}

2 and 6 are the only solutions of x 2 – 8x + 12 = 0.

Hence, A = {2, 6}

B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}

Hence, D ⊂ A ⊂ B ⊂ C

Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

(i) If x ∈ A and A ∈ B, then x ∈ B

(ii) If A ⊂ B and B ∈ C, then A ∈ C

(iii) If A ⊂ B and B ⊂ C, then A ⊂ C

(iv) If A ⊄ B and B ⊄ C, then A ⊄ C

(v) If x ∈ A and A ⊄ B, then x ∈ B

(vi) If A ⊂ B and x ∉ B, then x ∉ A

A = {1, 2} and B = {1, {1, 2}, {3}}

Now, we have,

2 ∈ {1, 2} and {1, 2} ∈ {1, {1, 2}, {3}}

Hence, we get,

We also know,

{2} ∉ {1, {1, 2}, {3}}

According to the question

Let us assume that,

And, C = {1, {0, 2}, 3}

From the question,

But, we know,

A ⊂ B and B ⊂ C

Then, we have,

B = {0, 6, 8}

C = {0, 1, 2, 6, 9}

A = {3, 5, 7}

B = {3, 4, 6}

We have, x ∉ B

3. Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. show that B = C.

A ∪ B = A ∪ C

A ∩ B = A ∩ C

Let us assume,

x ∈ A or x ∈ C

When x ∈ A, then,

∴ x ∈ A ∩ B

As, A ∩ B = A ∩ C

So, x ∈ A ∩ C

∴ x ∈ A or x ∈ C

Similarly, it can be shown that C ⊂ B

Hence, B = C

4. Show that the following four conditions are equivalent:

(i) A ⊂ B (ii) A – B = Φ

(iii) A ∪ B = B (iv) A ∩ B = A

To prove, (i) ⬌ (ii)

Here, (i) = A ⊂ B and (ii) = A – B ≠ ϕ

Let us assume that A ⊂ B

To prove, A – B ≠ ϕ

Let A – B ≠ ϕ

Hence, there exists X ∈ A, X ≠ B, but since A⊂ B, it is not possible

∴ A – B = ϕ

And A⊂ B ⇒ A – B ≠ ϕ

Let us assume that A – B ≠ ϕ

To prove: A ⊂ B

So, X ∈ B (if X ∉ B, then A – B ≠ ϕ)

Hence, A – B = ϕ => A ⊂ B

∴(i) ⬌ (ii)

Let us assume that A ⊂ B

To prove, A ∪ B = B

⇒ B ⊂ A ∪ B

Let us assume that, x ∈ A∪ B

⇒ X ∈ A or X ∈ B

Taking Case I: X ∈ B

Taking Case II: X ∈ A

⇒ X ∈ B (A ⊂ B)

⇒ A ∪ B ⊂ B

Let A ∪ B = B

Let us assume that X ∈ A

⇒ X ∈ A ∪ B (A ⊂ A ∪ B)

⇒ X ∈ B (A ∪ B = B)

Hence, (i) ⬌ (iii)

To prove (i) ⬌ (iv)

To prove, X ∈ A∩ B

Since, A ⊂ B and X ∈ B

Hence, X ∈ A ∩ B

⇒ A ⊂ A ∩ B

⇒ A = A ∩ B

Let us assume that A ∩ B = A

⇒ X ∈ A ∩ B

⇒ X ∈ B and X ∈ A

∴ (i) ⬌ (iv)

∴ (i) ⬌ (ii) ⬌ (iii) ⬌ (iv)

Hence, proved

5. Show that if A ⊂ B, then C – B ⊂ C – A.

C – B ⊂ C – A

Let us assume that x is any element such that X ∈ C – B

∴ x ∈ C and x ∉ B

Since, A ⊂ B, we have,

∴ x ∈ C and x ∉ A

So, x ∈ C – A

∴ C – B ⊂ C – A

Hence, Proved.

6. Assume that P (A) = P (B). Show that A = B

P (A) = P (B)

Let x be any element of set A,

Since, P (A) is the power set of set A, it has all the subsets of set A.

A ∈ P (A) = P (B)

Let C be an element of set B

For any C ∈ P (B),

We have, x ∈ C

Similarly, we have:

SO, we get,

If A ⊂ B and B ⊂ A

7. Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.

It is not true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)

Justification:

And, B = {1, 2}

∴ A ∪ B = {0, 1, 2}

P (A) = {ϕ, {0}, {1}, {0, 1}}

P (B) = {ϕ, {1}, {2}, {1, 2}}

∴ P (A ∪ B) = {ϕ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}

P (A) ∪ P (B) = {ϕ, {0}, {1}, {2}, {0, 1}, {1, 2}}

∴ P (A) ∪ P (B ≠ P (A ∪ B)

Hence, the given statement is false

8. Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)

A = (A ∩ B) ∪ (A – B)

Proof: Let x ∈ A

X ∈ (A ∩ B) ∪ (A – B)

X ∈ (A ∩ B)

⇒ X ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

In Case II,

⇒ X ∉ B or X ∉ A

⇒ X ∉ B (X ∉ A)

⇒ X ∉ A – B ⊂ (A ∪ B) ∪ (A – B)

∴A ⊂ (A ∩ B) ∪ (A – B) (i)

It can be concluded that, A ∩ B ⊂ A and (A – B) ⊂ A

Thus, (A ∩ B) ∪ (A – B) ⊂ A (ii)

Equating (i) and (ii),

We also have to show,

A ∪ (B – A) ⊂ A ∪ B

X ∈ A ∪ (B – A)

X ∈ A or X ∈ (B – A)

⇒ X ∈ A or (X ∈ B and X ∉A)

⇒ (X ∈ A or X ∈ B) and (X ∈ A and X ∉A)

⇒ X ∈ (B ∪A)

∴ A ∪ (B – A) ⊂ (A ∪ B) (iii)

(A ∪ B) ⊂ A ∪ (B – A)

Let y ∈ A∪B

Y ∈ A or y ∈ B

(y ∈ A or y ∈ B) and (X ∈ A and X ∉A)

⇒ y ∈ A or (y ∈ B and y ∉A)

⇒ y ∈ A ∪ (B – A)

Thus, A ∪ B ⊂ A ∪ (B – A) (iv)

∴From equations (iii) and (iv), we get:

A ∪ (B – A) = A ∪ B

9. Using properties of sets, show that: (i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A.

(i) To show: A ∪ (A ∩ B) = A

∴ A ∪ (A ∩ B) ⊂ A (i)

Also, according to the question,

A⊂ A ∪ (A ∩ B) (ii)

Hence, from equation (i) and (ii)

A ∪ (A ∩ B) = A

(ii) To show,

A ∩ (A ∪ B) = A

A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)

= A ∪ (A ∩ B)

10. Show that A ∩ B = A ∩ C need not imply B = C.

B = {0, 2, 3}

And, C = {0, 4, 5}

A ∩ B = {0}

A ∩ C = {0}

∴ A ∩ B = A ∩ C = {0}

2 ∈ B and 2 ∉ C

Therefore, B ≠ C

11. Let A and B be sets. If A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints A = A ∩ (A ∪ X) , B = B ∩ (B ∪ X) and use Distributive law)

Let A and B be two sets such that A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X for some set X.

To show, A = B

= A ∩ B (i)

Now, B = B ∩ (B ∪ X)

Hence, from equations (i) and (ii), we obtain A = B.

12. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ. Solution:

Let us assume, A {0, 1}

And, C = {2, 0}

A ∩ B = {1}

B ∩ C = {2}

∴ A ∩ B, B ∩ C and A ∩ C are not empty sets

A ∩ B ∩ C = Φ

13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee.

U = the set of all students who took part in the survey

T = the set of students taking tea

C = the set of the students taking coffee

Total number of students in a school, n (U) = 600

Number of students taking tea, n (T) = 150

Number of students taking coffee, n (C) = 225

Also, n (T ∩ C) = 100

Now, we have to find that number of students taking neither coffee nor tea i.e. n (T ∩ C’)

∴ According to the question,

n ( T ∩ C’ )= n( T ∩ C )’

= n (U) – n (T ∩ C)

= 600 – 275

∴ Number of students taking neither coffee nor tea = 325 students

14. In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

U = the set of all students in the group

E = the set of students who know English

H = the set of the students who know Hindi

∴ H ∪ E = U

Given that,

Number of students who know Hindi n (H) = 100

Number of students who knew English, n (E) = 50

Number of students who know both, n (H ∩ E) = 25

We have to find the total number of students in the group i.e. n (U)

n (U) = n(H) + n(E) – n(H ∩ E)

= 100 + 50 – 25

∴ Total number of students in the group = 125 students

15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:

(i) The number of people who read at least one of the newspapers. (ii) The number of people who read exactly one newspaper.

(i) Let us assume that,

A = the set of people who read newspaper H

B = the set of people who read newspaper T

C = the set of people who read newspaper I

Number of people who read newspaper H, n (A) = 25

Number of people who read newspaper T, n (B) = 26

Number of people who read the newspaper I, n (C) = 26

Number of people who read both newspaper H and I, n (A ∩ C) = 9

Number of people who read both newspaper H and T, n (A ∩ B) = 11

Number of people who read both newspaper T and I, n (B ∩ C) = 8

And, Number of people who read all three newspaper H, T and I, n (A ∩ B ∩ C) = 3

Now, we have to find the number of people who read at least one of the newspaper

= 25 + 26 + 26 – 11 – 8 – 9 + 3

∴ There are a total of 52 students who read at least one newspaper.

(ii) Let us assume that,

a = the number of people who read newspapers H and T only

b = the number of people who read newspapers I and H only

c = the number of people who read newspapers T and I only

d = the number of people who read all three newspapers

D = n(A ∩ B ∩ C) = 3

Now, we have:

n(A ∩ B) = a + d

n(B ∩ C) = c + d

n(C ∩ A) = b + d

∴ a + d + c +d + b + d = 11 + 8 + 9

a + b + c + d = 28 – 2d

∴ Number of people read exactly one newspaper = 52 – 22

= 30 people

16. In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

Let A, B and C = the set of people who like product A, product B and product C respectively.

Now, according to the question,

Number of students who like product A, n (A) = 21

Number of students who like product B, n (B) = 26

Number of students who like product C, n (C) = 29

Number of students who like both products A and B, n (A ∩ B) = 14

Number of students who like both products A and C, n(C ∩ A) = 12

Number of students who like both product C and B, n (B ∩ C) = 14

Number of students who like all three product, n (A ∩ B ∩ C) = 8

From the Venn diagram, we get,

Number of students who only like product C = {29 – (4 + 8 + 6)}

= {29 – 18}

= 11 students

BYJU’S provides all the concepts and solutions for Chapter 1 of Class 11 Maths in a creative and logical way. The PDF of Maths  NCERT Solutions for Class 11 Chapter 1 includes the following topics and sub-topics.

1.1 Introduction

This section contains the origin, basic definition, and applications of sets.

1.2 Sets and their Representations

Here, students are able to know what a set is exactly and how it can be represented in roster/ set-builder, and denoted using the English alphabets.

1.3 The Empty Set

This topic explains when a set is called an empty set.

1.4 Finite and Infinite Sets

Students can learn the definitions and examples of finite and infinite sets.

1.5 Equal Sets

This section gives an idea about equal and unequal sets along with solved examples.

1.6 Subsets

The main concepts covered under this section are:

1.6.1 Subsets of a set of real numbers

1.6.2 Intervals as subsets of R

1.7 Power Set

In this section, students are able to know the definition of power set, which has been derived from the concept of subsets.

1.8 Universal Set

This section is well explained with real-life examples such as population studies to define the universal set and representation using a letter.

1.9 Venn Diagrams

This section has the origin and definition of a Venn diagram. Also, it is explained using all the above topics and sub-topics.

1.10 Operations on Sets

We can perform some basic operations on sets similar to the addition or subtraction of numbers. These can be defined as below:

1.10.1 Union of sets

1.10.2 Intersection of sets

1.10.3 Difference of sets

1.11 Complement of a Set

After reading this section, students are able to understand the definition and properties of the complement of a set, along with the examples as well as practice problems.

1.12 Practical Problems on Union and Intersection of Two Sets

Practice problems are provided under this section to help the students understand in a better way about the union and intersection of two sets. These will help in their further studies to get thorough with the related concepts.

Exercise 1.1 Solutions 6 Questions

Exercise 1.2 Solutions 6 Questions

Exercise 1.3 Solutions 9 Questions

Exercise 1.4 Solutions 12 Questions

Exercise 1.5 Solutions 7 Questions

Exercise 1.6 Solutions 8 Questions

Miscellaneous Exercise On Chapter 1 Solutions 16 Questions

Key Features of NCERT Solutions for Class 11 Maths Chapter 1 – Sets

There are 6 exercises and a miscellaneous exercise in this chapter to help students understand the concepts related to Sets of the Class 11 Maths CBSE Syllabus (2023-24) in detail. The summary of the topics explained in Chapter 1 of NCERT Solutions for Class 11 Maths is listed below:

• A well-defined collection of objects is called a set.
• If a set does not contain any element, it is called an empty set.
• Finite set: A set that consists of a definite number of elements Infinite set:  A set that consists of an infinite number of elements
• Two sets A and B are said to be equal if they have the same elements.
• A set A is said to be a subset of a set B if every element of A is also an element of B. Intervals are subsets of R.
• A power set of a set A is a collection of all subsets of A. It is denoted by P(A).
• The union of two sets A and B is the set of all those elements which are either in A or B.
• The intersection of two sets A and B is the set of all elements which are common. The difference of two sets A and B in this order is the set of elements that belong to A but not to B.
• The complement of a subset A of universal set U is the set of all elements of U which are not the elements of A.
• For any two sets A and B, (A ∪ B)’ = A′ ∩ B′ and ( A ∩ B )′ = A′ ∪ B′.

Disclaimer – 

Dropped Topics – 

1.7 Power Set, Exercise 1.3 Ques. 5 1.12 Practical Problems on Union and Intersection of Two Sets Exercise 1.6 Examples 31–34 and Ques. 6–7 (Miscellaneous Exercise) Ques. 13–16 (Miscellaneous Exercise), Last Point in the Summary on the Page

Last Point in the Summary

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 1

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CBSE Class 11th Maths Value Based Questions Chapter 1 Sets PDF Download

CBSE Class 11th Maths Value Based Questions Chapter 1 Sets are the easiest questions which you see in your question paper and the scoring one all student who attempt it surely get they are just little bit difficult and examine your basic knowledge regarding the particular chapter. Maths Value Based Questions for Class 11th are available here at Free of cost. These questions are expected to be asked in the Class 11th board examination. These Maths Value Based Questions are from complete CBSE Syllabus.

CBSE Class 11th Maths Value Based Questions Chapter 1 Sets

Most of these Maths Value Based Questions are quite easy and students need only a basic knowledge of the chapter to answer these questions. Download CBSE Maths Value Based Questions for board examinations. These Maths Value Based Questions are prepared by Directorate of Education, Delhi.

CBSE Maths Value Based Questions Class 11th Chapter 1 Sets PDF

The purpose of the Maths Value Based Questions is to make students aware of how basic values are needed in the analysis of different situations and how students require to recognize those values in their daily lives. Some questions are subject related. But even if they are not, that one-minute awareness of what we write about value without any specific preparation is a good step indeed.

CBSE Maths Value Based Questions for Class 11th Chapter 1 Sets download here in PDF format. The most CBSE Maths Value Based Questions for annual examination are given here for free of cost. The additional questions for practice the Class 11th exam are collected from various sources. It covers questions asked in previous year examinations.

CBSE Maths Value Based Questions for Class 11th Chapter 1 Sets Free PDF

Class 11th books have many questions. These questions are regularly asked in exams in one or other way. Practising such most CBSE Maths Value Based Questions Chapter 1 Sets certainly help students to obtain good marks in the examinations. 

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Sets Important Questions: Detailed Solutions & Explanation

Muskan Shafi

Content Writer

Sets are well-defined and organized collections of elements and objects in Mathematics.  Set Theory is a branch of Mathematics dealing with the study of sets, their properties, symbols, and operations. Set Theory was developed and proposed by  Georg Cantor , a German mathematician.

• Sets are organized collections of objects whose elements are fixed and cannot be changed.
• The objects or items present in a set are known as their elements.
• The elements of a set can be any number, alphabetical letters, numbers, symbols, etc.
• A set is denoted by a capital letter and its elements are listed within curly brackets as  A = {1, 2, 3, 4, 5} .
• Sets are represented in three different forms namely,  Set-builder Form ,  Roster Form , and  Statement Form . 
• Set Operations are mathematical operations performed on sets and their types.
• Union of Sets ,  Intersection of Sets,  Difference of Sets, and   Complement of a Set are the four main set operations. 

What is a Set?

Read More:   NCERT Solutions for Class 11 Maths Sets

Very Short Answer Questions (1 Mark Questions)

Ques. State the various types of Sets.

Ans. The various types of Sets are: 

• Singleton Set
• Infinite Set
• Null/Empty Set
• Unequal Sets
• Disjoint Sets
• Superset and Subset
• Universal Set

Ques. Represent the set Y = {x : x is an integer, –1 x < 5} in roster form.

Ans.  The set Y = {x : x is an integer, –1 x < 5} will be expressed as   Y = {–1, 0, 1, 2, 3, 4} in Roster Form.

Ques . From the sets given, what is the pair of equivalent sets?

• A = {1, 2, 3}
• B = {x, y, z, t}
• C = {a, b, c}

Ans. A = {1, 2, 3} and C = {a, b, c} are equivalent sets.

Ques. Demonstrate the positive prime numbers that are less than two in form of a set.

Ans. The number 2 is the smallest positive prime number , and 1 is not a prime number. Thus, the set of positive prime numbers that are less than two is an empty or null set. It is expressed as P = { } or Φ.

Ques. If A = {2, 4, 8} and B = { 8, 4, 2}, what kind of sets are these?

Ans. Set A and Set B are equal to each other since both sets have the same elements. Therefore, it can be said that the elements of set A are equivalent to the elements of set B. Hence, A and B can be called equal sets .

Ques . What all elements can be included in Sets?

Ans. The elements of any set can be natural numbers, rational numbers , whole numbers, imaginary numbers, or complex numbers .

Set Detailed Video Explanation

Ques. What is a Universal Set?

Ans.  A universal set is defined as a collection of all the elements of a particular subject. Universal Set  is denoted by 'U'. For instance, set U = {List of Integers}. In this set, a set of natural numbers, even and odd numbers are all subsets of this universal set.

Ques. What are the main Set Operations?

Ans. The main set operations are as follows: 

• Union of Sets   (∪)
• Intersection of Sets (∩)
• Difference of Sets (–)
• Complement of a Set (A′ or A c )

Ques. What is Union of Sets?

Ans.  Union of sets is a set operation that lists the elements in set A and set B or the elements in both set A and set B. It is denoted by the symbol “A U B”.

For example, {3,5} ∪ {1, 5} = {1, 3, 5}

Short Answer Questions (2 Marks Questions)

Ques. Represent the given sets in Roster Form.

(i) U = {x : x is an integer and – 3 ≤ x < 7}

(ii) B = {x : x is a Natural Number which is less than 6}

Ans. (i) Given that, A = {x : x is an integer and –3 ≤ x < 7}

According to the condition, the integers will be -5, -4, -3, -2, -2, 0, 1, 2, 3, 4, 5, 6, 7, 8,…..

Thus, A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

(ii) Given that, B = {x : x is a Natural Number less than 6}

The natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, ……

Thus, B = {1, 2, 3, 4, 5}

Ques. Represent the Venn Diagram for (A ∩ B)’.

Ans.  The Venn Diagram for (A ∩ B)’ is as follows: 

Ques. Are the following pair of sets equal?

• A = {x:x is a letter in the word SORROW}
• B = {y:y is a letter in the word SOW}

Ans . Elements of Sets are: 

• A= {S, O, W}
• B = {S, O, W}

Hence, A = B.

Ques. If A = {3, 5, 7, 9, 10}, B = {7, 9, 10, 13}, and C = {10, 13, 15}. Find (A ∩ B) ∩ ( B ∪ C).

Ans.  Given sets are: 

• A = {3, 5, 7, 9, 10}
• B = {7, 9, 10, 13}
• C = {10, 13, 15}

A ∩ B = {7, 9, 10}

B ∪ C = {7, 9, 10, 13, 15}

(A ∩ B) ∩ (B ∪ C) = {7, 9, 10}

Ques. Given, X = {a, b, c, d} and Y = {f, b, d, g}, determine the final value of X – Y and Y – X.

Ans . Given Sets are: 

• X = {a, b, c, d}
• Y = {f, b, d, g}

X – Y = {a, b, c, d} – {f, b, d, g} = {a, c}

Y – X = {f, b, d, g} – {a, b, c, d} = {f, g}

Ques. Represent the given sets in the Roster Form.

(i) A = {x | x is a positive integer which is less than 10 and 2 x  – 1 is an odd number}

(ii) B = {x : x 2  + 7x – 8 = 0, x ∈ R}

Ans.  (i) Since 2 x  is an even number, 2 x  – 1 is an odd number for all positive integral values of x.

In simpler terms, 2 x  – 1 is an odd number for x = 1, 2, … , 9.

Thus, A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

(ii) x 2  + 7x – 8 = 0

On simplification, 

(x + 8) (x – 1) = 0

x = – 8 or x = 1

Thus, B = {– 8, 1}

Read More:  ​ Sets Important Questions

Ques. Given sets are A = {a, b}, and B = {a, b, c}. Is A ⊂ B? Find A ∪ B. 

Ans.   Given sets are

• B = {a, b, c}

Yes, A ⊂ B, Therefore, the union of the set's pairings can be expressed as

A∪ B = {a, b, c} = B

Ques. If U = {x : x ∈ N, x ≤ 9}, A = {x : x is an even number, 0 < x < 10}, and B = {2, 3, 5, 7}, what will be the Set (A U B)?

Ans.  It is given that,

• U = {x : x ∈ N, x ≤ 9}
• A = {x : x is an even number, 0 < x < 10}
• B = {2, 3, 5, 7}

Thus, in roster form, the sets will be

• U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
• A = {2, 4, 6, 8}

A U B = {2, 3, 4, 5, 6, 7, 8}

(A U B)’ = {1, 9}

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Long Answer Questions (3 Marks Questions)

Ques. Set S = {5, 10, 15, 20, 25, 30} is given in the roster form. Rewrite it in, (i) Set Builder Form (ii) Semantic Form

Ans.  The given set is S = {5, 10, 15, 20, 25, 30}.

(i) In set-builder notation, it will be written as S = {x : x is a multiple of 5 and 1 < x < 35, x ∈ N}.

(ii) in semantic form, it will be written as S = {Multiples of 5 less than 35 and greater than 0}.

Ques. Let A and B be two sets containing 3 and 5 elements respectively. What is the maximum and the minimum number of elements in A ∪ B?

Ans. There can be 3 elements that are common, at least, between both the given sets.

• A = {a, b, c}
• B = {a, b, c, d, e}

Therefore, A ∪ B = {a, b, c, d, e,} implies that the minimum number of elements in A ∪ B are 6.

It can also be that, there are no elements that are common between both sets.

For example,

• B = {d, e, g, h, i}

Thus, A ∪ B = {a, b, c, d, e, g, h, i}. It implies that the maximum number of elements in A ∪ B is 8.

Ques. If ∪ = {a, e, i, o, u} and 

• A = {a, e, i}
• B = {e, o, u}
• C = {a, i, u}

Verify whether the value is true or not: A ∩ (B - C) = (A ∩ B) - (A ∩ C).

Ans.  According to the sets given, 

• B – C = {e, o}
• A ∩ (B - C) = e
• A ∩ B = {e}
• A ∩ C = {a}

(A ∩ B) - (A ∩ C) = e

Now, A ∩ (B - C) = e and (A ∩ B) - (A ∩ C) = e.

Thus, A ∩ (B - C) = (A ∩ B) - (A ∩ C)

Hence Proved.

Ques. According to a survey, 73% of Kashmiris actually like apples, while around 65% of them like oranges. Determine the percentage of Kashmiris who like both apples and oranges.

Ans.  Let us assume that,

• A = Set of Kashmiris who like Apples
• B = Set of Kashmiris who like Oranges

n (A∪B) = 100

n(A∩B) = n(A) + n(B) - n(A∪B)

= 73 + 65 – 100

Thus, about 38% of the Kashmiris like both apples and oranges.

Ques. In a college, 20 professors teach mathematics or physics. If 12 teach maths and 4 teach both physics and maths, How many teach Physics?

Ans.  Given that, 

• Total Teachers n(M∪P) = 20
• Teachers teaching Maths n(M) = 12
• Teachers teaching both Maths and Physics n(M∩P) = 4
• Teachers teaching Physics n(P) =?

Thus, 

n(M∪P) = n(M) + n(P) - n(M∩P)

Thus, there are 12 teachers teaching Physics.

Very Long Answer Questions (5 Marks Questions)

Ques. There are 100 students in a Student Activity Club of which 35 like drawing and 45 like music. 10 students are interested in both drawing and music. Find the number of students that like either of them or neither of them.

Ans.  Given that,

• Total Number of Students = 100
• Number of Students liking Painting, n(P) = 35
• Number of Students liking Dancing, n(D) = 45
• Number of Students liking Both Painting and Dancing, n(P∩D) = 10

The number of students who like either of them (n(P ∪ D)) and the number of students who like neither (n(P∪D)) needs to be calculated. 

Using Sets Formula ,

n(P∪D) = n(P) + n(D) – n(P∩D)

45 + 35 - 10 = 70

Therefore, there are 70 students who like either music or drawing.

Number of Students Who like Neither Painting or Dancing = Total Students – n(P∪D) = 100 – 70 = 30 

Therefore, there are 30 students who like neither music nor drawing.

Ques. Let set A = {1, 2, 3, 4, 5, 6, 7, 8}, and set B = {3, 5, 7, 9, 11, 13}.  Find:  (i) A U B (ii) A ∩ B (iii) (A ∩ B)’

• A = {1, 2, 3, 4, 5, 6, 7, 8}
• B = {3, 5, 7, 9, 11, 13}

Union of sets is the unique elements of both sets as a whole without being repeated. Thus, 

A U B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13}

(ii) A ∩ B

Inverted U means an intersection that finds elements common to both sets.

A ∩ B = {3, 5, 7}

(iii) (A ∩ B)’

It is a complement of the set operation. Thus, 

(A ∩ B)’ = {1, 2, 4, 6, 8, 9, 11, 13}

Ques. There are 200 individuals with a dermatological disorder. 120 were exposed to the chemical C 1 , 50 to chemical C 2 , and 30 to both the chemicals C 1 and C 2 . What is the number of individuals exposed to  (i) Chemical C 1 but not chemical C 2 ? (ii) Chemical C 2 but not chemical C 1 ? (iii) Chemical C 1 or chemical C 2 ?

Ans . Set A signifies the set of individuals exposed to the chemical C 1 , while B denotes the set of individuals exposed to the chemical C 2 .

• n (U) = 200
• n (A) = 120
• n (A∩B) = 30

(i) n(A-B) = n(A) - n(A∩B) = 120-30 = 90

(ii) n(B-A) = n(B) - n(A∩B) = 50 - 30 = 20

(iii) n(A∪B) =n(A)+ n(B)- n(A∩B) = 120 + 50 - 30 = 140

Ques. In a Sports Academy, there are players that play either Baseball or Football or both. The number of players who play Baseball or Football​ or both is 11, 12 and 3 respectively. What will be the number of students playing either cricket or badminton?

Ans.  Consider that, 

• n(Baseball) = n(P)
• n(Football) = n(Q)
• n(Either Baseball or Football) = n(P∪Q)
• n(Baseball and Football) = n(P∩Q)

Given that,

• (P∩Q) = 3

Using the Sets Formula, 

n(P∪Q) = n(P) + n(Q) - n(P∩Q) = 11 + 12 - 3 = 20

Thus, there are 20 players that are playing either cricket or badminton.

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CBSE CLASS XII Related Questions

1. find the inverse of each of the matrices,if it exists  \(\begin{bmatrix} 2 & 1 \\ 7 & 4  \end{bmatrix}\), 2. find the maximum profit that a company can make, if the profit function is given by p(x) = 41−24x−18x 2, 3. let f: r→r be defined as f(x) = 3x. choose the correct answer..

• f is one-one onto
• f is many-one onto
• f is one-one but not onto
• f is neither one-one nor onto

4. Find the vector and the cartesian equations of the lines that pass through the origin and(5,-2,3).

5. if (i) a= \(\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{bmatrix}\) ,then verify that a'a=i (ii) a=  \(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos \alpha & \sin\alpha \end{bmatrix}\) ,then verify that a'a=i, 6. find the inverse of each of the matrices, if it exists.  \(\begin{bmatrix} 1 &  3\\ 2 & 7\end{bmatrix}\), similar mathematics concepts.

Mathematics

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