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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...

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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myCBSEguide is a prominent streaming resource for students studying for CBSE examinations. The site offers extensive study material, practice papers, case study questions, online examinations, and other resources at the push of a click. myCBSEguide also has a mobile app for learning on the move. There’s no excuse not to try myCBSEguide with so much convenience and simplicity of access! Hence Proved!

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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.

CBSE Class 11 Maths – Chapter 1 Sets- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sets : Notes and Study Materials -pdf

• Concepts of  Sets
• Sets Master File
• Sets Revision Notes
• R D Sharma Solution : Sets
• NCERT Solution  Sets
• NCERT  Exemplar Solution Sets
• Sets : Solved Example 1

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

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)

Sets Class 11 MCQs Questions with Answers

Question 1. If A, B and C are any three sets, then A – (B ∪ C) is equal to (a) (A – B) ∪ (A – C) (b) (A – B) ∪ C (c) (A – B) ∩ C (d) (A – B) ∩ (A – C)

Answer: (d) (A – B) ∩ (A – C) Hint: Given A, B and C are any three sets. Now, A – (B ∪ C) = (A – B) ∩ (A – C)

Question 2. (A’)’ = ? (a) ∪ – A (b) A’ (c) ∪ (d) A

Answer: (d) A Hint: (A’)’ = A

Question 3. A – B is read as? (a) Difference of A and B of B and A (b) None of the above (c) Difference of B and A (d) Both a and b

Answer: (a) Difference of A and B of B and A Hint: A – B will read as difference of A and B of B and A Ex: Let A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7} Now, A – B = {2, 4}

Question 4. If A, B and C are any three sets, then A × (B ∪ C) is equal to (a) (A × B) ∪ (A × C) (b) (A ∪ B) × (A ∪ C) (c) None of these (d) (A × B) ∩ (A × C)

Answer: (a) (A × B) ∪ (A × C) Hint: Given A, B and C are any three sets. Now, A × (B ∪ C) = (A × B) ∪ (A × C)

Question 5. IF A = [5, 6, 7] and B = [7, 8, 9] then A ∪ B is equal to (a) [5, 6, 7, 8, 9] (b) [5, 6, 7] (c) [7, 8, 9] (d) None of these

Answer: (a) [5, 6, 7, 8, 9] Hint: Given A = [5, 6, 7] and B = [7, 8, 9] then A ∪ B = [5, 6, 7, 8, 9]

Question 6. Which of the following sets are null sets (a) {x: |x |< -4, x ?N} (b) 2 and 3 (c) Set of all prime numbers between 15 and 19 (d) {x: x < 5, x > 6}

Answer: (b) 2 and 3 Hint: 2 and 3 is the null set.

Question 7. IF R = {(2, 1),(4, 3),(4, 5)}, then range of the function is? (a) Range R = {2, 4} (b) Range R = {1, 3, 5} (c) Range R = {2, 3, 4, 5} (d) Range R {1, 1, 4, 5}

Answer: (b) Range R = {1, 3, 5} Hint: Given R = {(2, 1),(4, 3),(4, 5)} then Range(R) = {1, 3, 5}

Question 8. The members of the set S = {x | x is the square of an integer and x < 100} is (a) {0, 2, 4, 5, 9, 58, 49, 56, 99, 12} (b) {0, 1, 4, 9, 16, 25, 36, 49, 64, 81} (c) {1, 4, 9, 16, 25, 36, 64, 81, 85, 99} (d) {0, 1, 4, 9, 16, 25, 36, 49, 64, 121}

Answer: (b) {0, 1, 4, 9, 16, 25, 36, 49, 64, 81} Hint: The set S consists of the square of an integer less than 100 So, S = {0, 1, 4, 9, 16, 25, 36, 49, 64, 81}

Question 9. In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects? (a) 19 (b) 41 (c) 21 (d) 57

Answer: (b) 41 Hint: The number of students who took at least one of the three subjects can be found by finding out A ∪ B ∪ C, where A is the set of those who took Physics, B the set of those who took Chemistry and C the set of those who opted for Math. Now, A ∪ B ∪ C = A + B + C – (A ∩ B + B ∩ C + C ∩ A) + (A ∩ B ∩ C) A is the set of those who opted for Physics = 120/2 = 60 students B is the set of those who opted for Chemistry = 120/5 = 24 C is the set of those who opted for Math = 120/7 = 17 The 10th, 20th, 30th….. numbered students would have opted for both Physics and Chemistry. Therefore, A ∩ B = 120/10 = 12 The 14th, 28th, 42nd….. Numbered students would have opted for Physics and Math. Therefore, C ∩ A = 120/14 = 8 The 35th, 70th…. numbered students would have opted for Chemistry and Math. Therefore, B ∩ C = 120/35 = 3 And the 70th numbered student would have opted for all three subjects. Therefore, A ∪ B ∪ C = 60 + 24 + 17 – (12 + 8 + 3) + 1 = 79 Number of students who opted for none of the three subjects = 120 – 79 = 41

Question 10. { (A, B) : A² +B² = 1} on the sets has the following relation (a) reflexive (b) symmetric (c) none (d) reflexive and transitive

Answer: (b) symmetric Hint: Given {(a, b) : a² + b² = 1} on the set S. Now a² +b² = b² + a² = 1 So, the given relation is symmetric.

Question 11. Two finite sets have N and M elements. The number of elements in the power set of first set is 48 more than the total number of elements in power set of the second test. Then the value of M and N are (a) 7, 6 (b) 6, 4 (c) 7, 4 (d) 6, 3

Answer: (b) 6, 4 Hint: Let A and B be two sets having m and n numbers of elements respectively Number of subsets of A = 2m Number of subsets of B = 2n Now, according to question 2m – 2n = 48 ⇒ 2n(2m – n – 1) = 24(22 – 1) So, n = 4 and m – n = 2 ⇒ m – 4 = 2 ⇒ m = 2 + 4 ⇒ m = 6

Question 12. The range of the function f(x) = 3x – 2‚ is (a) (- ∞, ∞) (b) R – {3} (c) (- ∞, 0) (d) (0, – ∞)

Answer: (a) (- ∞, ∞) Hint: Let the given function is y = 3x – 2 ⇒ y + 2 = 3x ⇒ x = (y + 2)/3 Now x is saisfied by all values. So, Range{f(x)} = R = (-∞, ∞)

Question 13. If A, B, C be three sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C, then, (a) B = C (b) A = C (c) A = B = C (d) A = B

Answer: (a) B = C Hint: Given A, B, C be three sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C then B = C

Question 14. In 2nd quadrant? (a) X < 0, Y < 0 (b) X < 0, Y > 0 (c) X > 0, Y > 0 (d) X > 0, Y < 0

Question 15. How many rational and irrational numbers are possible between 0 and 1? (a) 0 (b) Finite (c) Infinite (d) 1

Answer: (c) Infinite Hint: There are infinite many rational and irrational numbers are possible between 0 and 1 This is because between any two numbers, there are infinite numbers.

Question 16. Empty set is a? (a) Finite Set (b) Invalid Set (c) None of the above (d) Infinite Set

Answer: (a) Finite Set Hint: In mathematics, and more specifically set theory, the empty set is the unique set having no elements and its size or cardinality (count of elements in a set) is zero. So, an empty set is a finite set.

Question 17. If A = [5, 6, 7] and B = [7, 8, 9] then A U B is equal to (a) [5, 6, 7, 8, 9] (b) [5, 6, 7] (c) [7, 8, 9] (d) None of these

Answer: (a) [5, 6, 7, 8, 9] Hint: Given A = [5, 6, 7] and B = [7, 8, 9] then A U B = [5, 6, 7, 8, 9]

Question 18. Which of the following two sets are equal? (a) A = {1, 2} and B = {1} (b) A = {1, 2} and B = {1, 2, 3} (c) A = {1, 2, 3} and B = {2, 1, 3} (d) A = {1, 2, 4} and B = {1, 2, 3}

Answer: (c) A = {1, 2, 3} and B = {2, 1, 3} Hint: Two sets are equal if and only if they have the same elements. So, A = {1, 2, 3} and B = {2, 1, 3} are equal sets.

Question 19. In a class of 50 students, 10 did not opt for math, 15 did not opt for science and 2 did not opt for either. How many students of the class opted for both math and science. (a) 24 (b) 25 (c) 26 (d) 27

Answer: (d) 27 Hint: Total students = 50 Students who did not opt for math = 10 Students who did not opt for Science = 15 Students who did not opt for either maths or science = 2 Total of 40 students in math and 13 did not opt for science but did for math = 40 – 13 = 27 So, students of the class opted for both math and science is 27

Question 20. In last quadrant? (a) X < 0, Y > 0 (b) X < 0, Y < 0 (c) X > 0, Y < 0 (d) X > 0, Y > 0

Important Questions for CBSE Class 11 Maths Chapter 1 – Sets

Important questions for cbse class 11 maths chapter 1 - sets, cbse class 11 maths chapter-1 important questions - free pdf download.

Free PDF download of Important Questions with solutions for CBSE Class 11 Maths Chapter 1 - Sets prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. CoolGyan.Org to score more marks in your Examination.

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. Not a set

Ans. A = {–1, 0, 1, 2, 3}

6. List all the elements of the set,

Ans. A = { 0,1,2,3,4,5}

7. Write the set B = {3,9,27,81} in set-builder form. Which of the following are empty sets? Justify.

8. A = { x : x ϵ N and 3 <x <4}

Ans. Empty set

Ans. Non-empty set

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

Ans. Infinite set

11. B = { x : x ϵ N and x is an even prime number}

Ans. Finite set

13. Write (–5,9] in set-builder form

Ans. { x : x ϵ R, –5 < x ≤ 9}

14. Write { x : –3 ≤ x <7} as interval.

Ans. [ –3,7)

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

Ans. 2 3 = 8

16. Write all the possible subsets of A = {5,6}. If A = {2,3,4,5}, B = { 3,5,6,7}

Ans. ∅, { 5}, {6}, {5,6}

Ans. A ∪ B = {2,3,4,5,6,7}

Ans. A ∩ B = {3, 5}

19. If A = {1,2,3,6} , B = {1, 2, 4, 8} find B – A

Ans. B – A = {4,8}

20. If A = {p, q}, B = {p, q, r}, is B a superset of A? Why?

Ans. Yes, because A is a subset of B

21. Are sets A = {1,2,3,4}, B = { x : x ϵ N and 5 ≤ x ≤ 7} disjoint? Why?

Ans. es, because A ∩ B = ∅

22. If X and Y are two sets such that n(X) = 19, n(Y) = 37 and n(X ∩ Y) = 12, find n(X ∪ Y).

Ans. n(X ∪ Y) = 44

23.Describe the set in Roster form

Ans. { 17, 26, 35, 44, 53, 62, 71, 80 }

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

A = { x:x is a letter in the word FOLLOW}

B = { y:y is a letter in the word WOLF}

Ans. A= {F, O, L, W}

B = {W, O, L, F }

25.Write down all the subsets of the set {1,2,3}

28.Write the set in roster form A = The set of all letters in the word T R I G N O M E T R Y

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

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

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

Ans. A = {A, L, O, Y}

B = {L, O, Y, A}

Hence A = B

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

Ans. {3, 4} is an element of set A.

32.Fill in the blanks.

34.Is set C = { x : x – 5 = 0} and E = {x : x is an integral positive root of the equation x 2 – 2x – 15 = 0} are equal?

Ans. C = {5}

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

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

(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.

(ii) {2, 6, 10} and {3, 7, 11} are disjoint sets

Hence false

37.Fill in the blanks

(ii) (A ∩ B)ʹ = ---------

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

Ans. A = {b, c, d, f, g, h, j}

39.Is pair of sets equal? Give reasons.

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

Ans. A = {2, 3}

B = {- 2, - 3}

[∴ x 2 + 5x + 6 = 0

x 2 + 3x + 2x + 6 = 0

x = - 2 – 3

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

Ans. (- 3, 0) = {x : x ϵ R, - 3 < x < 0}

[6, 12] = {x : x ϵ R, 6 ≤ x ≤ 12}

41.If X = {a, b, c, d}

Y = {f, b, d, g}

Find X – Y and Y – X

Ans. X – Y = {a, b, c, d} – {f, b, d, g}

Y – X = {f, b, d, g} – {a, b, c, d}

43.List all the element of the set A = { x : x is an integer x 2 ≤ 4}

Ans. {-2, -1, 0, 1, 2}

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

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

45.Write the following as interval

(i) {x : x ϵ R, - 4 < x ≤ 6}

(ii) {x : x ϵ R, 3 ≤ x ≤ 4}

Ans. (i) (-4, 6]

(ii) [3, 4]

46.If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} Find (A ∩ B) ∩ ( B ∪ C)

Ans. A ∩ B = {7, 9, 11}

B ∪ C = {7, 9, 11, 13, 15}

(A ∩ B) ∩ (B ∪ C) = {7, 9, 11}

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?

(ii) How many can speak Hindi only?

Ans. (i) 20 people can speak both Hindi and English

(ii) 480 people can speak Hindi only

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

Ans. 29% of the Indians like both grapes and pineapple.

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 play neither cricket nor tennis?

Ans. Hint : ∪ – set of people surveyed

A – set of people who play cricket

B – set of people who play tennis

Number of people who play neither cricket nor tennis

= n[(A ∪ B)"] = n(U) – n(A ∪ B)

= 450 – 200

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?

Ans. There are 280 students in the group.

5. If A = [–3, 5), B = (0, 6] then find (i) A – B, (ii) A ∪ B

Ans. (i) [–3, 0]; (ii) [–3, 6]

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.

Ans. Let A denote the set of students taking apple juice and B denote the set of students taking orange juice

=400-100-150+75=225

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

Ans. Let A=set of Indian who like apples

B= set of Indian who like oranges

n (A)= 73, n (B) = 65

38% like both

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. n(M∪P) = 20, n(M)=12

9. Let U = {1, 2, 3, 4, 5, 6} A = {2, 3} and B = {3, 4, 5}

= { 1, 4, 5, 6}

Hence proved.

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

= n (U) – [ n(A) + n (B) – n (A ∩B)]

= 1000 – [300+300-200]

= 1000 – 400

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 drink coffee, How many drink coffee, but not tea.

Ans. n (T) = 100

n (T – C) = 65

n (T ∪ C) = 210

n (T – C) = n (T) – n (T ∩ C)

65 = 100 – n (T ∩ C)

n (T ∩ C) = 35

n (T ∪ C) = n (T) + n (C) – n (T ∩ C)

210 = 100 + n (C) – 35

n (C) = 145.

13. If P (A) = P (B), Show that A = B

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.

Ans. n (M) = 12, n (M – B) = 8

Ans. (i) n (A) = n (A – B) + n (A ∩ B)

= 14 + x + x

N (B) = n (B – A) + n (A ∩ B)

but n (A) = n (B) (Given)

14 + 2x = 4x

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

= 14 + x+ 3x + x

16. If A and B are two sets such that A ∪ B= A ∩ B, then prove that A = B

Ans. Let a ϵ A, then a ϵ A ∪ B

Since A ∪ B = A ∩ B

a ϵ A ∩ B . So a ϵ B

Similarly if b ϵ B,

Then b ϵ A ∪ B. Since

A ∪ B = A ∩ B, b ϵ A ∩ B

Ans. C – B = A

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 C = the set of people who like cricket and

T = the set of people who like tennis.

n ( C ∪ T) = 56, n (C) = 40

n (C ∩ T) = 10

n (C ∪ T) = n (C) + n (T) – n (C ∩T)

65 = 40 + n (T) – 10

19. Let A, B and C be three sets A ∪ B = A ∪ C and A ∩ B = A ∩ C show that B = C

20. If ∪ = {a, e, i. o. u}

A = {a, e, i}

And B = {e, o, u}

C = {a, i, u}

Then verify that A ∩ (B – C) = (A ∩ B) – (A ∩ C)

Ans. B – C = {e, o}

A ∩ (B – C) = e

A ∩ B = {e}

A ∩ C = {a}

(A ∩ B) – (A ∩ C) = e

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 B 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?

(ii) How many like product C only?

Ans. Hint : Let A, B, C denote respectively the set of people who like product

a, b, c, d, e, f, g – Number of elements in bounded region

(i) Total number of Surveyed people = a + b + c + d + e + f + g = 43

(ii) Number of people who like product C only = g = 10

2. A college awarded 38 medals in football, 15 in basket ball 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. people got medals in exactly two of the three sports.

a + b + f + e = 38

b + c + d + f = 15

e + d + f + g = 20

a + b + c + d + e + f + g = 50

we have to find b + d + e

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

(1) chemical C 1 but not chemical C 2

(2) chemical C 2 but not chemical C 1

(4) chemical C 1 or chemical C 2

Ans. A denote the set of individuals exposed to the chemical C 1 and B denote the set of individuals exposed to the chemical C 2

=50-30 = 20

4.In a survey it was found that 21 peoples 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. a + b + c + d = 21

b + c + e + f = 26

c + d + f + g = 29

b + c = 14,c + f =15,c + d = 12

d = 4,c = 8,f = 7,b = 6,g = 10,e = 5,a = 3

like product c only = 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 medal in all the three sports, how many received medals in exactly two of the three sports?

Ans. Let A, B and C denotes the set of men who received medals in football, basketball and cricket respectively.

n (A) = 38, n (B) = 15, n (C) = 20

58 = 38 + 15 + 20 – (a + d ) – (d + c) – (b + d) + 3

18 = a + d + c + b + d

18 = a + b + c + 3d

9 = a + b + c

6.In a survey of 60 people, it was found that 25 people read news paper 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

newspaper. Find

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

(ii) The no. of people who read exactly one news paper.

Ans. a + b + c + d = 25

c + d + f +g = 26

f = 5, b = 8, d = 6, c = 3, g = 12

e = 10, a = 8

(i) a + b + c + d + e + f + g = 52

(ii) 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

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

Ans. Let C be the set of students in chemistry class and P be the set of students in physics class.

n (C) = 20, n (P) = 30

8.In a survey of 25 students, it was found that 15 had taken mathematics, 12had 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 (ii) only mathematics (iii) only physics

(iv) physics and chemistry but mathematics (v) mathematics and physics but not chemistry (vi) only one of the subjects (vii) at least one of three subjects

(viii) None of three subjects.

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?

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

Ans. ∪ = 100, a = 18

a + e = 23, e + g = 8

a + e + g + d = 26

e + g + f + c = 48

so, e = 5, g = 3, d = 0, f = 5, c = 35

(i) d + g + f + b = 0 + 3 + 5 + 10 = 18

(ii) d + g = 0 + 3 = 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

(ii) Neither Hindi nor science.

Ans. Let ∪ = all the students of the class ,H = students who like Hindi

S = Students who like Science

(i) n ( H ∪ S) = n (H) + n (S) – n (H ∩ S)

= 30 + 25 – 16

= ∪ - n (H ∪ S)

11.In a town of 10,000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C. 5% 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 (ii) B only (iii) none of A, B, and C.

Ans. x + a + c + d = 4000

y + a + d + b = 2000

z + b + c + d = 1000

a + d = 500, b + d = 300, C + d = 400 d = 200

On Solving a = 300, b = 100, c = 200

(i) x = 4000 – 300 – 200 – 200 = 3300

(ii) y = 2000 – 300 – 200 – 100 = 1400

(iii) z = 1000 – 100 – 200 – 200 = 500

None of these = 10,000 – (3300 + 1400 + 500 + 300+ 100 + 200 + 200)

= 10,000 – 6000

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 second set. Find the value of m and n.

Ans. Let A and B be two sets having m and n elements respectively

no of subsets of A = 2 m

no of subsets of B = 2 n

According to question

2 m = 56 + 2 n

2 m - 2 n = 56

2 n (2 m-n – 1) = 56

2 n (2 m-n – 1) = 2 3 (2 3 – 1)

Class 11 Maths Chapter 1 Sets MCQs

Class 11 Maths Chapter 1 Sets MCQs (Multiple-choice questions) are available here with answers. All the MCQs are prepared as per the latest exam pattern and CBSE Syllabus (2022-2023). The objective questions are prepared by our subject experts, with proper explanations. MCQs of Class 11 Maths Chapter 1 will include all the concepts of the NCERT curriculum. Practising these multiple-choice questions will help Class 11 students to score good marks in the examination.

Also check: Class 11 Maths chapter-wise MCQs

MCQs for Class 11 Maths Chapter 1 Sets

MCQs of Chapter 1 Sets Class 12 Maths are provided here with the correct option. These multiple-choice questions for sets are given here with proper explanations. The questions are based on the Sets concept and related formulas. Students are suggested to solve the questions by themselves and then verify with the provided answers.

Class 11 Maths Chapter 1 Sets MCQs – Download PDF

Q.1: How many elements are there in the complement of set A?

C. All the elements of A

D. None of these

Answer: A. 0

Explanation: The complement of a set A will contain the elements that are not present in set A.

Q.2: Empty set is a _______.

A. Infinite set

B. Finite set

C. Unknown set

D. Universal set

Answer: B. Finite set

Explanation: The cardinality of the empty set is zero, since it has no elements. Hence, the size of the empty set is zero.

Q.3: The number of elements in the Power set P(S) of the set S = {1, 2, 3} is:

Answer: B. 8

Explanation: Number of elements in the set S = 3

Number of elements in the power set of set S = {1,2,3} = 2 3

Q.4: Order of the power set P(A) of a set A of order n is equal to:

Answer: C. 2 n

Explanation: The cardinality of the power set is equal to 2 n , where n is the number of elements in a given set.

Q.5: Which of the following two sets are equal?

A. A = {1, 2} and B = {1}

B. A = {1, 2} and B = {1, 2, 3}

C. A = {1, 2, 3} and B = {2, 1, 3}

D. A = {1, 2, 4} and B = {1, 2, 3}

Answer: C. A = {1, 2, 3} and B = {2, 1, 3}

Explanation: Two sets are said to be equal if they both have the same elements.

Q.6: Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, P = {1, 2, 5}, Q = {6, 7}. Then P ∩ Q’ is :

Answer: A. P

Explanation: Given,

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

P = {1, 2, 5}

Q’ = {1, 2, 3, 4, 5, 8, 9, 10}

P ∩ Q’ = {1, 2, 5} = P

Q.7: The cardinality of the power set of {x: x∈N, x≤10} is ______.

Answer: A. 1024

Set X = {x: x∈N, x≤10}

X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Number of elements of power set of X, P(X) = 2 10 = 1024

Q.8: Write X = {1, 4, 9, 16, 25,…} in set builder form.

A. X = {x: x is a set of prime numbers}

B. X = {x: x is a set of whole numbers}

C. X = {x: x is a set of natural numbers}

D. X = {x: x is a set of square numbers}

Answer: D. X = {x: x is a set of square numbers}

X = {1, 4, 9, 16, 25,…}

X = {1 2 , 2 2 , 3 2 , 4 2 , 5 2 , …}

X = {x: x is a set of square numbers}

Q.9: If A, B and C are any three sets, then A × (B ∪ C) is equal to:

A. (A × B) ∪ (A × C)

B. (A ∪ B) × (A ∪ C)

C. (A × B) ∩ (A × C)

D. None of the above

Answer: A. (A × B) ∪ (A × C)

A, B and C are any three sets.

Now, A × (B ∪ C) = (A × B) ∪ (A × C)

Q.10: The range of the function f(x) = 3x – 2‚ is:

A. (- ∞, ∞)

C. (- ∞, 0)

D. (0, – ∞)

Answer: A. (- ∞, ∞)

Let the given function be

⇒ y + 2 = 3x

⇒ x = (y + 2)/3

Since, for all values of y, x has different values. Thus, values of x and y can range from -∞ to ∞.

So, Range{f(x)} = R = (-∞, ∞).

Video Lesson on What are Sets

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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.

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Chapter 1 Class 11 Sets

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Learn Chapter 1 Sets of Class 11 free with solutions of all NCERT Questions for CBSE Maths. All examples, formulas and exercise questions explained in an easy way. Important questions are also marked for your reference

Topics in the chapter include

• Definition - What is set and how is it used
• Depiction of set - How to represent it, Roster form, Set-builder form
• Intervals - Interval notation and how it is related to sets.
• Types of sets - Null set, Finite set, Infinite Set, Equal Set, Subset, Power set, Universal set
• Union, Intersection and Difference of Sets - Explanation and its questions
• Venn diagram - What is Venn diagram and representing union, intersection, difference of sets in the Venn Diagram
• Complement of set - How to find complement and its representation in venn diagram
• Number of elements formula - Using number of elements formula for 2 and 3 sets and solving questions
• Proof questions - Solving proof questions using properties of sets and its application of properties of sets

Explanation of individual topics with questions to practice are given in Concept wise. Check out Concept wise, or the NCERT Exercise wise. 

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Class 11 Maths solutions of important questions of chapter 1-Sets

Class 11 Maths solutions of important questions of ‘Sets’

Class 11 Maths solutions of important questions of chapter 1-Sets are going to help every maths student of class 11 in boosting their preparation for the class 11 exam of CBSE board. The solutions of important questions of chapter 1-Sets will clear your basics of maths which is required to study for gaining higher maths study. All questions are solved by an expert teacher of maths as per the CBSE norms. All important questions of chapter 1- sets of class 11 CBSE are solved in a step by step method.

Study the basics of the Sets, see the video

Q1.If A = { 1,2,3,4,5,6}, B = {2,4,6, 8} then find A – B.

Ans.We are given the sets A ={ 1,2,3,4,5,6}, B = {2,4,6, 8}

A – B = { 1,2,3,4,5,6}- {2,4,6, 8} = {1, 3, 5}

Q2. Let A and B be two sets containing 3 and 6 elements respectively. Find the maximum and the minimum number of elements in A ∪ B.

Ans.There may be the case when atleast 3 elements are common between both sets

Let a set A = {a, b, c} and B = {a, b, c, d, e, f}

∴ A ∪ B = {a, b, c, d, e, f} implies that the minimum number of elements in A ∪ B are = 6

There may be the case when there are no any elements are common between both sets

lf A = {a, b, c}, B = { d, e, f, g, h, i}

A ∪ B = {a, b, c, d, e, f,g,h,i} implies that the maximum number of elements in A ∪ B are = 9

Q3.If A = {(x,y) : x² + y²= 25  where x, y ∈ W } write a set of all possible ordered pair .

Ans.  We are given the set A = {(x,y) : x² + y²= 25  where x, y ∈ W }

All possible ordered pair of set A are following

For x = 0,y =5, x=3,y=4,for x =4, y =3,for x=5,y =0

A = {(0,5),(3,4),(4,3),(5,0)}

Q4.If A = {1,2,3}, B = {4, 5, 6} and C ={5} verify that A ∪ ( B ∩ C) = (A ∪ B) ∩ A ∪ C.

We are given the sets A = {1,2,3}, B = {4, 5, 6} ,C = {5}

B ∩ C = {5}

A ∪ ( B ∩ C) = {1,2,3,5}

(A ∪ B) and A ∪ C

A ∪ B = {1,2,3,4,5,6} and A ∪ C = {1,2,3,5}

(A ∪ B) ∩ A ∪ C = {1,2,3,5}

A ∪ ( B ∩ C) = (A ∪ B) ∩ A ∪ C, Hence proved

Q5.From the adjoining Venn diagram, write the value of the following.

(c) (A ∩ B)’

Ans. From the venn diagram ,we have

U = {1,2,3,4………15}

A = {7,9.11}. B ={11,12,13,14}

A’ = U – A = {1,2,3,4………15} – {7,9.11} = {1,2,3,4,5,6,8,10,12,13,14,15}

B’ = U – B = {1,2,3,4………15} – {11,12,13,14}= {1,2,3,4,5,6,7,8,9,10,,15}

We have,(A ∩ B) = 11

(A ∩ B)’ = U – (A ∩ B) = {1,2,3,4………15} – {11} = (1,2,3,4,5,6,7,8,9,10,12,13,14,15}

Q6. If P(A) = P(B) show that A = B.

Ans. P(A} and P(B) implies that both are power sets of A and B respectively

Every set is an element of its power set , so A ∈ P(A)

Since, we are given that

P(A) = P(B)

Therefore, A ∈ P(B)

Indicates that every element of A belongs to the set B

So, A ⊂ B……(i)

Similarly B ∈ P(B)

Indicates that every element of B belongs to the set A

So, B ⊂ A……(ii)

From (i) and (ii), we get

A = B, Hence proved

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Q7. Let A  and B be sets ; if A∩X = B∩X = ∅ and A∪X = B∪X for some set X.Show that A=B.

Ans. We are given that A∩X = B∩X = ∅ and A∪X = B∪X

To prove   A=B

Proof.  A∪X = B∪X (given)

Multiplying both sides by A∩

A∩ (A∪X) =A∩ (B∪X)

Using distributive property

(A∩ A) ∪ (A ∩ X) = (A∩ B )∪ (A∩ X)

A∩Φ = (A∩ B) ∪ Φ

A = (A∩ B) …………(i)

Multiplying both sides by B∩

B∩(A∪X) = B∩(B∪X)

(B∩ A) ∪ (B ∩ X) = (B∩ B )∪ (B∩ X)

(B∩ A) ∪φ = B ∪ φ

B = (A∩ B) …………(ii)

From (i) and (ii)

Q8.If A ={1,2,3,4,5},then write the proper subsets of A.

Ans. The number of elements in the given sets A ={1,2,3,4,5} are =5

The number of proper subsets of any set are = 2 n – 1

   Where n = number of elements = 5

  The number of proper subsets of any set are =  2 5 – 1 =32 – 1 = 31

Q9.Write the following sets in the Roster form

(i) A={x : x ∈ R, 2x+11 =15}

(ii)B={x |x² =x, x ∈ R}

(iii)C={x = x is a positive factor of the prime number p}

Ans.(i)We have, A={x : x ∈ R, 2x+11 =15}

2x+1= 15 ⇒x= 2

(ii)We have,B={x |x² =x, x ∈ R}

∴ x² = x ⇒ x²-x= 0 ⇒x(x-1)= 0 ⇒x=0,1

(iii)We have, C={x = x is a positive factor of the prime number p}

Sice positive factors of a prime u∪mer are 1 ad the number itself,we have

Q10.For all sets A,B and C show that (A – B) ∩ (A – C) = A – (B ∪ C).

Ans. Considering that

x ∈ (A – B)∩(A – C)

⇒ x ∈ (A – B) and x ∈ (A – C)

⇒ (x ∈ A and x∉ B) and (x ∈ A and x∉ C)

⇒ (x ∈ A ) and (x ∉B and  x∉ C )

⇒(x ∈ A ) x∉ (B ∪C)∈

⇒x ∈ A – (B ∪C)

⇒(A – B) ∩(A – C)⊂A – (B ∪ C)…….(i)

Now,Considering that

y ∈A – (B ∪ C)

⇒ y ∈A and y ∉  (B ∪ C)

⇒y ∈A and (y ∉B and y ∉ C)

⇒(y ∈A and y ∉B) and (y ∈A and y ∉ C)

⇒y ∈ (A – B) and y ∈ (A – C)

⇒y ∈ (A – B) ∩ y ∈ (A – C)

⇒A – (B ∪ C) ⊂(A – B) ∩(A – C)………(ii)

(A – B) ∩(A – C)= A – (B ∪ C), Hence proved

Q11.Let A,B and C be the sets such that A∪B = A∪C and A ∩B = A ∩C,show that B = C.

Ans. According to question, A ∪ B = A ∪ C and A ∩ B = A ∩ C

To show, B = C

Let us assume, x ∈ B So, x ∈ A ∪ B

Hence, x ∈ A or x ∈ C

when x ∈ A, then x ∈ B

∴ 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

Q12.Show that for any sets A and B

A = (A∩B)∪(A-B) 

Ans.We have to prove

A = (A∩B)∪(A-B)

Taking RHS and solving it

(A∩B)∪(A-B)

Using the property

A -B = A -(A∩B)

=  (A∩B) ∪ (A∩B’)

Applying distributive property

A∩(B∪C) = (A∩B) ∪(A∩C)

Replacing C = B’ in LHS

A-B = A∩(B∪B’)

= A∩ (U) [since B∪B’ = U]

= A [since A∩ U = A]

= RHS, Hence proved

Q13. Write the following sets in the roster form:

(i) A = {x : x ∈ R, 2x + 15 = 15}

(ii) B = {x : x² = x, x ∈ R}

Ans(i) It is given to us that set

A = {x : x ∈ R, 2x + 11 = 15}

2x + 11 = 15

2x = 15 – 11

Therefore in roster form it is written as  A = {2}

(ii) It is given to us that

B = {x : x² =x, x ∈ R}

x(x – 1) = 0

x = 0, x = 1

Therefore in roster form it is written as

Q14. If  A and B are subsets of the universal set U, then show that 

(i) A ⊂ A ∪ B

Ans. Let’s prove that

Let  x ∈ A or x ∈ B

If x ∈ A then x ∈ A ∪ B

Hence A ⊂ A ∪ B

Q15. A,B and C are subsets of universal set if A = {2,4,6,8,12,20}, B ={3,6.9.12.15},C ={5,10,15,20} and U is the set of all whole numbers,draw a venn diagram showing the relation of  U,A,B and C.

Q16.If A and B are two sets such that A ⊂B then find:

(i) A ∩ B    (ii) A ∪ B

Ans .(i) A ∩ B implies all the elements which are available in A and B

Let A = {1,2,3}, B = {1,2,3,4,5}⇒ A ⊂B

A ∩ B = {1,2.3} = A

A ⊂B implies that all the elements of A are in B

(ii) A ∪ B implies all the elements which are available in A or B

A ∪ B = {1,2,3,4,5} = B

∴ A ∪ B = B

Q17.Let A ={{2,3,4},{5,6},{6,7,8}}.Determine which of the following is true or false.

(i) 4 ∈ A (ii) {2,3,4} ⊂ A (iii) {6,7,8}∈ A (iv) {5,6} ∈ A (v) Φ∈ A (vi)  Φ ⊂ A

Ans. (i) False , since 4 is not an element of the set A, it is the part of the element {2,3,4}

(ii) False, {2,3,4} is an element of A,therefore it is written as {2,3,4}∈ A

(iii) True, since {6,7,8} is an element of A

(iv) True, since {5,6} is an element of A

(v) False,Φ is not an element of any set,it is a subset of every set

(vi) True, since Φ is a subset of every set

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CBSE Question Bank for Class 11 Maths Chapter 1 Sets Free PDF

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

NCERT Solutions for Class 11 Maths Chapter 1 Sets

In NCERT Solutions for class 11 Maths Chapter 1 Sets, you will learn about the description of a Set, types of sets, subsets, concepts of subsets, Venn Diagrams, Operations on Sets using Venn Diagrams, Laws of Algebra of sets, some important results on number of elements in sets.

Free download NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1, Ex 1.2, Ex 1.3, Ex 1.4, Ex 1.5, Ex 1.6 and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

Topics and Subtopics in Class 11 Maths NCERT Solutions Chapter 1 Sets

• Class 11 Sets Ex 1.1
• Class 11 Sets Ex 1.1 in Hindi Medium
• Class 11 Sets Ex 1.2
• Class 11 Sets Ex 1.2 Hindi Medium
• Class 11 Sets Ex 1.3
• Class 11 Sets Ex 1.3 Hindi Medium
• Class 11 Sets Ex 1.4
• Class 11 Sets Ex 1.4 Hindi Medium
• Class 11 Sets Ex 1.5
• Class 11 Sets Ex 1.5 Hindi Medium
• Class 11 Sets Ex 1.6
• Class 11 Sets Ex 1.6 Hindi Medium
• Class 11 Sets Miscellaneous Exercise
• Class 11 Sets Miscellaneous Exercise in Hindi Medium
• Sets Class 11 Notes Maths Chapter 1
• NCERT Exemplar Class 11 Maths Chapter 1 Sets
• RD Sharma Class 11 Solutions Chapter 1 Sets
• Chapter 1:  Sets Class 11 NCERT Book
• अध्याय 1:  समुच्चय Class 11 NCERT Book

Class 11 Maths NCERT Solutions Chapter 1 Sets Ex 1.1

Ex 1.1 Class 11 Maths Q.1: Which of the following are sets. Explain your answer. (a).  The collection of all day in a week which have the first letter S. (b).  The collection of ten most famous singers of India. (c).  A group of best football strikers of the world. (d).  The collection of all girls in your school. (e).  The collection of all odd numbers below 50. (f).  A collection of poems written by the poet Shakespeare. (g).  The collection of all prime numbers. (h).  The collection of questions in science book. (i).  A collection of most dangerous reptiles in India.

Solution: (i) The collection of all months of a year beginning with the letter J is a well-defined collection of objects because one can definitely identity a month that belongs to this collection. Hence, this collection is a set. (ii) The collection of ten most talented writer of India is not a well-defined collection because the criteria for determining a writer’s talent vary from person to person. Hence, this collection is not a set. (iii) A team of eleven best cricket batsmen of the world is not a well-defined collection because the criteria for determining a batsman’s talent may vary from person to person. Hence, this collection is not a set. (iv) The collection of all boys in your class is a well-defined collection because you can definitely identify a boy who belongs to this collection. Hence, this collection is a set. (v) The collection of all natural numbers less than 100 is a well-defined collection because one can definitely identify a number that belongs to this collection. Hence, this collection is a set. (vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined collection because one can definitely identify a book that belongs to this collection. Hence, this collection is a set. (vii) The collection of all even integers is a well-defined collection because one can definitely identify an even integer that belongs to this collection. Hence, this collection is a set. (viii) The collection of questions in this chapter is a well-defined collection because one can definitely identify a question that belongs to this chapter. Hence, this collection is a set. (ix) The collection of most dangerous animals of the world is not a well-defined collection because the criteria for determining the dangerousness of an animal can vary from person to person. Hence, this collection is not a set.

Ex 1.1 Class 11 Maths Q.2: Let P = {2, 3, 4, 5, 6, 7}. Insert the correct symbol  ∈ o r ∉  inside the given blank spaces below: (a).  2 . . . . . . . . . . P (b).  9 . . . . . . . . . P (c).  11 . . . . . . . . P (d).  4 . . . . . . . . P (e).  0 . . . . . . . . P (f).  7 . . . . . . . . P

Solution: (i) 5 ∈ A (ii) 8 ∉ A (iii) 0 ∉ A (iv) 4 ∈ A (v) 2 ∈ A (vi) 10 ∉ A

Ex 1.1 Class 11 Maths Q.3: Write the given sets in roster form: (a).  P = {y: y is an integer and -4 < y < 6}. (b).  Q = {y: y is a natural number which is <8} (c).  R = {y: y is a 2 digit natural number in which the sum of its digits is 9} (d).  S = {y: y is a prime number which is a divisor of 70} (e).  T = The set of all letters in the word ELEPHANT (f).  U = The set of all letters in the word DIVISION

Solution: (i) A = { x :  x  is an integer and –3 <  x  < 7} The elements of this set are –2, –1, 0, 1, 2, 3, 4, 5, and 6 only. Therefore, 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} The elements of this set are 1, 2, 3, 4, and 5 only. Therefore, the given set can be written in roster form as B = {1, 2, 3, 4, 5}

(iii) C = { x :  x  is a two-digit natural number such that the sum of its digits is 8} The elements of this set are 17, 26, 35, 44, 53, 62, 71, and 80 only. Therefore, this set can be written in roster form as C = {17, 26, 35, 44, 53, 62, 71, 80}

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

∴60 = 2 × 2 × 3 × 5 The elements of this set are 2, 3, and 5 only. Therefore, this set can be written in roster form as D = {2, 3, 5}.

Ex 1.1 Class 11 Maths Q.4: Write the given sets in set-builder form: (a).  {4, 8, 12, 16, 20} (b).  {3, 9, 27, 81} (c).  {4, 16, 64, 256, 1024} (d).  {1, 3, 5, 7…} (e).  {1, 8, 27, 64….1000}

Ex 1.1 Class 11 Maths Q.6: Match the following:

NCERT Solutions for Class 11 Maths Chapter 1 Sets (समुच्चय) Ex 1.1 in Hindi

Class 11 Maths NCERT Solutions Chapter 1 Sets Ex 1.2

Ex 1.2 Class 11 Maths Q.1: Which of the following given below is null set? (i).  Set of odd natural numbers which is divisible by 2. (ii).  Set of even numbers which are prime (iii).  {x: x is a natural number,  x < 5  and  x > 7 } (iv).  {y: y is a point common to any two parallel lines}

Ex 1.2 Class 11 Maths Q.2:  State whether the following sets are infinite or finite: (i).  A set of months of a year. (ii).  {1, 2, 3 ….} (iii).  {1, 2, 3…99, 100} (iv).  The set of positive integers which are greater than 100. (v).  The set of prime numbers which are less than 99

Ex 1.2 Class 11 Maths Q.3: State whether the following sets are infinite or finite: (i).  The set of lines parallel to the x – axis. (ii).  The set of letters in the vowels. (iii).  The set of numbers multiple of 10. (iv).  The set of humans living on Earth. (v).  The set of circles passing through the origin (0, 0).

Ex 1.2 Class 11 Maths Q.4: In the following set given below, state whether A = B or not: (i).  A = {w, x, y, z} B = {z, y, x, w} (ii).  A = {5, 9, 13, 17} B = {9, 5, 17, 19} (iii).  A = {4, 2, 6, 10, 8} 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 …}

Ex 1.2 Class 11 Maths Q.5 In the following set given below, is the pair of sets equal? (i). A = {3, 4} B = {y: y is solution of  y² + 5 y + 6 = 0 } (ii).  A = {a: a is a letter in the word FOLLOW} B = {b: b is a letter in the word WOLF}

Ex 1.2 Class 11 Maths Q.6: From the following sets, 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}

Class 11 Maths NCERT Solutions Chapter 1 Sets Ex 1.3

Ex 1.3 Class 11 Maths Q.1: Fill in the blanks properly using ⊂  and  ⊄ . (i).  {3, 4, 5} ____ {2, 3, 4, 5, 6} (ii).  {a, b, c} ____ {d, c, d} (iii).  {y: y is a pupil of Class 11 of the school} ____ {y: y is students of the school} (iv).  {y: y is a circle in the plane} ____ {y: y is a circle in the same plane with radius 2 unit} (v).  {y: y is an equilateral triangle in a plane} ____ {y: y is a rectangle in the same plane} (vi).  {y: y is an equilateral triangle in a plane} ____ {y: y is a triangle in the plane} (vii).  {y: y is an odd natural number} ____ {x: x is an integers}

Ex 1.3 Class 11 Maths Q.2: State whether the given statements are true or false: (i).  {b, c}  ⊄  {c, d, e} (ii).  {a, e, i}  ⊂  {x: x is a vowel in the English alphabets} (iii).  {1, 2, 3}  ⊂ {1, 2, 4, 5} (iv).  {c}  ⊂  {b, c, d} (v).  {b}  ∈  {a, b, c, d} (vi).  {y: y is an even natural no. less than 6}  ⊂  {y: y is a natural no. which can divide 36}

Ex 1.3 Class 11 Maths Q.3: Let X = {11, 12, {13, 14}, 15}. According to the given set which of the given statements are false? Explain why. (i).  {13, 14}  ⊂  X (ii).  {13, 14}  ∈  X (iii).  {{13, 14}}  ⊂  X (iv).  11  ∈  X (v).  11  ⊂  X (vi).  {11, 12, 15}  ⊂  X (vii).  {11, 12, 15}  ∈  X (viii).  {11, 12, 13}  ⊂  X (ix).  Ø  ∈  X (x).  Ø  ⊂  X (xi).  {Ø}  ⊂  X

Ex 1.3 Class 11 Maths Q.4: Write all the subsets of the given sets: (i).  {b} (ii).  {b, c} (iii).  {2, 3, 4} (iv).  Ø

Ex 1.3 Class 11 Maths Q.5: How many elements has P(X), if X = Ø ?

Ex 1.3 Class 11 Maths Q.6: Write the given in the form of intervals: (i).  {y: y  ∈  R, –5 < y ≤ 7} (ii).  {y: y  ∈  R, –13 < y < –11} (iii).  {y: y  ∈  R, 1 ≤ y < 8} (iv).  {y: y  ∈  R, 4 ≤ y ≤ 5}

Ex 1.3 Class 11 Maths Q.7:  Write the given intervals in the form of set – builder: (i).  (–4, 1) (ii).  [7, 13] (iii).  (7, 13] (iv).  [–24, 6)

Ex 1.3 Class 11 Maths Q.8: What universal set/ sets would you propose for the given sets?

(i).  The set of right triangles (ii).  The set of isosceles triangles

Ex 1.3 Class 11 Maths Q.9: X = {1, 3, 5}, Y = {2, 4, 6} and Z = {0, 2, 4, 6, 8}

Which of the given sets can be considered as the universal set for the given sets X, Y and Z? (i).  {0, 1, 2, 3, 4, 5, 6} (ii).  Ø (iii).  {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (iv).  {1, 2, 3, 4, 5, 6, 7, 8}

Class 11 Maths NCERT Solutions Chapter 1 Sets Ex 1.4

Ex 1.4 Class 11 Maths Q.1: Find the union of each of the following pairs of sets: (i). P = {1, 4, 6} and Q = {1, 3, 4} (ii). X = {a, e, i, o, u} and Y = {x, y, z} (iii). X = {x: x is a natural number and multiple of 3} and Y = {x: x is a natural number less than 6} (iv). X = {x: x is a natural number and 1 < x ≤ 5} and Y = {x: x is a natural number and 5 < x < 10} (v). X = {4, 5, 6} and Y = Φ

Ex 1.4 Class 11 Maths Q.2:  Let A = {x, y} and B = {x, y, z} Is A ⊂ B? What is A ∪ B?

Ex 1.4 Class 11 Maths Q.3: If X and Y are two sets such that X  ⊂  Y, then what is X  ∪  Y?

Ex 1.4 Class 11 Maths Q.4: If P = {1, 2, 3, 4}, Q = {3, 4, 5, 6}, R = {5, 6, 7, 8} and S = {7, 8, 9, 10} Find the following: (i).  P  ∪  Q (ii).  P  ∪  R (iii).  Q  ∪  R (iv).  Q  ∪  S (v).  P  ∪  Q  ∪  R (vi).  P  ∪  Q  ∪  S (vii).  Q  ∪  R  ∪  S

Ex 1.4 Class 11 Maths Q.5:  Find the intersection of each of the following pairs of sets: (i).  P = {1, 4, 6} and Q = {1, 3, 4} (ii).  X = {a, e, i, o, u} and Y = {x, y, z} (iii).  X = {x: x is a natural number and multiple of 3} and Y = {x: x is a natural number less than 6} (iv).  X = {x: x is a natural number and 1 < x ≤ 5} and Y = {x: x is a natural number and 5 < x < 10} (v).  X = {4, 5, 6} and Y = Φ

Ex 1.4 Class 11 Maths Q.6: If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17} Find the following: (i).  A  ∩  B (ii).  B  ∩   C (iii).  A  ∩   C  ∩   D (iv).  A  ∩   C (v).   B  ∩   D (vi).  A  ∩   (B  ∪  C) (vii).  A  ∩   D (viii).  A  ∩   (B  ∪  D) (ix).  (A  ∩   B)  ∩   (B  ∪  C) (x).  (A ∪   D)  ∩   (B  ∪  C)

Ex 1.4 Class 11 Maths Q.7: If A = {y: y is a natural number}, B ={y: y is an even natural number}, C = {y: y is an odd natural number} and D = {y: y is a prime number} Find the following: (i). A  ∩  B (ii). A  ∩  C (iii). A  ∩  D (iv). B  ∩  C (v). B  ∩  D (vi). C  ∩  D

Ex 1.4 Class 11 Maths Q.8: Which of the given pairs of sets are disjoint? (i).  A = {1, 2, 3, 4} and B = {x: x is a natural number and 4 ≤ x ≤ 6} (ii).  A = {a, e, i, o, u} and B = {c, d, e, f} (iii).  A = {x: x is an even integer} and B = {x: x is an odd integer}

Ex 1.4 Class 11 Maths Q.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} and D = {5, 10, 15, 20} Find the following: (i).  A – B (ii).  A – C (iii).  A – D (iv).  B – A (v).  C – A (vi).  D – A (vii).  B – C (viii).  B – D (ix).  C – B (x).  D – B (xi).  C – D (xii).  D – C

Ex 1.4 Class 11 Maths Q.10: If X = {a, b, c, d} and Y = {f, b, d, g} Find the following: (i).  X – Y (ii).  Y – X (iii).  X  ∩  Y

Ex 1.4 Class 11 Maths Q.11: What is R – Q, if R is the set of real numbers and Q is the set of rational?

Ex 1.4 Class 11 Maths Q.12: State whether the following statements are true or false. Give reason. (i).  A = {2, 3, 4, 5} and B = {3, 6} are disjoint sets. (ii).  A = {a, e, i, o, u } and B = {a, b, c, d} are disjoint sets. (iii).  A = {2, 6, 10, 14} and B = {3, 7, 11, 15} are disjoint sets. (iv).  A = {2, 6, 10} and B = {3, 7, 11} are disjoint sets.

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5

Ex 1.5 Class 11 Maths Q.1: Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9} X = {1, 2, 3, 4, 5} Y = {2, 4, 6} and Z = {2, 4, 5, 6}. Find the following sets: (i).   X’ (ii).  Y’ (iii).   ( X ∪ Y ) ′ (iv).   ( X ∪ Z ) ′ (v).   ( X ′ ) ‘ (vi).   ( Y – Z ) ′

Ex 1.5 Class 11 Maths Q.2: If U = {a, b, c, d, e, f, g, h}, find the complements of the given sets: (i).  W = {a, b, c} (ii).  X = {d, e, f, g} (iii).  Y = {a, c, e, g} (iv).  Z = {f, g, h, a}

Ex 1.5 Class 11 Maths Q.3: Take natural numbers as the universal set. Write the complements of the given sets: (i).  A = {y: y is an even natural number} (ii).  B = {y: y is an odd natural number} (iii).  C = {y: y is a positive multiple of 3} (iv).  D = {y: y is a prime number} (v).  E = {y: y is a natural number divisible by 3 and 5} (vi).  F = {y: y is a perfect square} (vii).  G = {y: y is perfect cube} (viii).  H = {y: y + 5 = 8} (ix).  I = {y: 2y + 5 = 9} (x).  J = {y: y  ≥  7} (xi).  K = {y: y  ∈  N and 2y + 1 > 10}

Ex 1.5 Class 11 Maths Q.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 ∪ B ) ′  =  A ′ ∩ B ′ (ii).   ( A ∩ B ) ′  =  A ′ ∪ B ′

Ex 1.5 Class 11 Maths Q.5: Draw the Venn diagrams for the following: (i).   ( A ∪ B ) ′ (ii).   A ′ ∩ B ′ (iii).   ( A ∩ B ) ′ (iv).   A ′ ∪ B ′

Ex 1.5 Class 11 Maths Q.6: Let U be the universal set that is the set of all triangles in a plane. If X is the set of all triangles with at least one angle different from  60 ∘ , what is X’ ?

Ex 1.5 Class 11 Maths Q.7: Complete the given statements using proper symbols: (i).   A ∪ A ′ (ii).  Ø  ∩ A (iii).   A ∩ A ′ (iv).   U ′ ∩ A

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6

Ex 1.5 Class 11 Maths Q.1: If A and B are two sets such that n(A) = 16, n(B) = 24 and n ( A ∪ B )  = 39. Find  n ( A ∩ B ) .

Ex 1.6 Class 11 Maths Q.2: If A and B are two sets such that  A ∪ B  has 17 elements, A has 9 elements and B has 14 elements. How many elements does  A ∩ B  have?

Ex 1.6 Class 11 Maths Q.3: In a group there are 450 people, 200 speaks Hindi and 270 can speak English. How many people can speak Ex both English and Hindi?

Ex 1.6 Class 11 Maths Q.4: If X and Y are two sets such that X has 22 elements, Y has 34 elements, and  n ( X ∩ Y )  has 10 elements, how many elements does  n ( X ∪ Y )  have?

Ex 1.6 Class 11 Maths Q.5: If A and B are two sets such that A has 45 elements, n (A ∩ B) has 15 elements and n (A ∪ B) has 70 elements, how many elements does B have?

Ex 1.6 Class 11 Maths Q.6:There are 70 people, out of which 35 like tea, 55 like coffee, and each person likes at least one of the two beverages. How many people like both tea and coffee?

Ex 1.6 Class 11 Maths Q.7: There are 70 students in a group, 35 like cricket, 15 like both tennis and cricket. How many like tennis only and not cricket? How many like tennis?

Ex 1.6 Class 11 Maths Q.8: In a committee, 60 people speak French, 30 speak Spanish and 20 speak both Spanish and French. How many speak at least one of these two languages?

NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1 in Hindi

Ex 1.1 Class 11 Maths प्रश्न 1. निम्नलिखित में कौन से समुच्चय हैं? अपने उत्तर का औचित्य बताइए। (i) j अक्षर से प्रारम्भ होने वाले वर्ष के सभी महीनों का संग्रह। (ii) भारत के दस सबसे अधिक प्रतिभाशाली लेखकों का संग्रह। (iii) विश्व के सर्वश्रेष्ठ ग्यारह बल्लबाजों का संग्रह। (iv) आप की कक्षा के सभी बालकों का संग्रह। (v) 100 से कम सभी प्राकृत संख्याओं का संग्रह। (vi) लेखक प्रेमचन्द द्वारा लिखित उपन्यासों का संग्रह। (vii) सर्भीसम पूर्णाकों का संग्रह। (viii) इस अध्याय में आने वाले प्रश्नों का संग्रह। (ix) विश्व में सबसे अधिक खतरनाक जानवरों का संग्रह। हल: (i) j से शुरु होने वाले महीनों के नाम : जनवरी, जून व जुलाई। अतः यह एक समुच्चय है। (ii) प्रतिभाशाली लेखक को परिभाषित नहीं किया जा सकता। इसीलिए यह एक समुच्चय नहीं है। (iii) सर्वश्रेष्ठ बल्लेबाज को परिभाषित नहीं कर सकते। अतः यह एक समुच्चय नहीं है। (iv) कक्षा के सभी विद्यार्थियों की संख्या निश्चित होती है। अतः यह एक समुच्चय है। (v) 100 से कम प्राकृत संख्याएँ 1, 2, 3, …….. 99 हैं। अतः यह एक समुच्चय है। (vi) लेखक प्रेमचन्द्र द्वारा लिखित उपन्यासों का संग्रह गबन, गोदान आदि द्वारा परिभाषित हैं। अतः यह एक समुच्चय है। (vii) समपूर्णांक {….. -6, -4. 2, 4, 6, ….} हैं। इसलिए यह एक समुच्चय है। (viii) इस अध्याय के प्रश्न परिभाषित हैं। अतः यह एक समुच्चय है। (ix) संसार के सबसे अधिक खतरनाक पशुओं के संग्रह को परिभाषित नही किया जा सकता। इसलिए यह एक समुच्चय नहीं है।

Ex 1.1 Class 11 Maths प्रश्न 2. मान लीजिए A = {1, 2, 3, 4, 5, 6}. रिक्त स्थानों में उपयुक्त प्रतीक ∈अथवा ∉ भरिए। (i) 5 …… A (ii) 8 …….. A (iii) 0 ……… A (iv) 4 ……. A (v) 2 …….. A (vi) 10 ……. A हल: (i) 5 ∈ A (ii) 8 ∉ A (iii) 0 ∉ 4 (iv) 4 ∈ A (v) 2 ∈ A (vi) 10 ∉ A

Ex 1.1 Class 11 Maths प्रश्न 3. निम्नलिखित समुच्चयों को रोस्टर रूप में लिखिए: (i) A = {x : x एक पूर्णाक है और -3 < x < 7} (ii) B = {x : x संख्या 6 से कम एक प्राकृत संख्या है।} (iii) C = {x : x दो अंको की ऐसी प्राकृत संख्या है जिसके अंकों का योगफल 8 है।} (iv) D = {x : x एक अभाज्य संख्या है जो 60 की भाजक है। (v) E = TRIGONOMETRY शब्द के सभी अक्षरों का समुच्चय (vi) F = BETTER शब्द के सभी अक्षरों क़ा समुच्च्य हल: (i) A = {-2, -1, 0, 1, 2, 3, 4, 5, 6} (ii) B = {1, 2, 3, 4, 5} (iii) C = {17, 26, 35, 44, 53, 62, 71, 80} (iv) D = {2, 3, 5} (v) E = {T, R, I, G, O, N, M, E, Y} (vi) F = {B, E, T, R}

Ex 1.1 Class 11 Maths प्रश्न 4. निम्नलिखित समुच्चयों को समुच्चय निर्माण रूप में व्यक्त कीजिए: (i) {3, 6, 9, 12} (i) {2, 4, 8, 16, 32} (iii) {5, 25, 125, 625} (iv) {2, 4, 6, ….} (v) {1, 4, 9, ……100} हल: (i) {x : x = 3n और 1 ≤ n ≤ 4} (ii) {x : x = 2n और 1 ≤ n ≤ 5} (iii) {x : x = 5 और 1 ≤ n ≤ 4} (iv) {x : x एक सम प्राकृत संख्या है।} (v) {x : x = n², 1 ≤ n ≤ 10}

Ex 1.1 Class 11 Maths प्रश्न 6. बाई ओर रोस्टर रूप में लिखित और दाईं ओर समुच्चय निर्माण रूप में वर्णित समुच्चयों का सही मिलान कीजिए। (i) {1, 2, 3, 6} (a) {x : x एक अभाज्य संख्या है और 6 की भाजक है। (ii) {2, 3} (b) {x : x संख्या 10 से कम एक विषम प्राकृत संख्या है। (iii) {M, A, T, H, E, I, C, S} (c) {x : x एक प्राकृत संख्या है और 6 की भाजक है। (iv) {1, 3, 5, 7, 9} (d) {x : x MATHEMATICS शब्द का एक अक्षर है: हल: (i) → (c) (ii) → (a) (iii) → (d) (iv) → (b)

NCERT Solutions for Class 11 Maths All Chapters

• Chapter 1 Sets
• 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 Permutation 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

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Download Case Study Questions for Class 11 Maths

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[PDF] Download Case Study Questions for Class 11 Maths

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

Click on the chapter to view.

Class 11 Maths Chapters

Chapter 1 Sets 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 Permutation 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

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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CBSE Class 11 Term-1 Maths 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

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Trump Picks His Running Mate, and Political Heir

Former president donald j. trump chose the 39-year-old senator j.d. vance of ohio as his vice-presidential nominee..

This transcript was created using speech recognition software. While it has been reviewed by human transcribers, it may contain errors. Please review the episode audio before quoting from this transcript and email [email protected] with any questions.

From “The New York Times,” I’m Michael Barbaro. This is “The Daily.” Today, on the first day of the Republican National Convention, Donald Trump makes his choice of a running mate. We watched it unfold in real time from Milwaukee.

[MUSIC PLAYING]

It’s Tuesday, July 16.

Are you OK?

I’m getting phone interference.

Oh no, maybe that’s from your phone. There’s like 16 phones in this car.

I’ll start it.

OK. It is 9:35 AM on Monday morning, and we are driving into downtown Milwaukee to the headquarters of the Republican National Convention. And it’s day 1, the opening hours of what will be the Republican Party’s crowning of their nominee, former President Donald Trump. And the singular event that hangs over this entire convention is the one that none of us expected, which is the attempted assassination of the presumptive Republican nominee.

And the news is just flying at us this morning. We just got word that he’s going to be announcing his running mate today, his choice of a VP. So this is a pretty extraordinary day already before 10:00 AM, and we’re going to try to make sense of it all.

We just went through the first round of secret service asking to see our badges. We are now stopped at a giant metal pop-up wall. And in front of us, cars are being slowly and very thoroughly inspected by bomb-sniffing dogs and security teams.

Hi. You guys are good. You just have to get screened through the checkpoint.

Perfect. OK, thank you. Thanks.

Secret Service? Should I open the trunk?

— turn the engine off and pop the hood for me, please?

Oh, the front hood? How do I do that? We’re in a rental car. Here we go.

She’s got it. Look at her.

And also, don’t run into my barricade. People keep doing that even though it says stop on it, you know? Keep your eyes open, all righty? Let’s be safe, everybody.

Great. Thank you so much.

So we’re headed into deeper security.

— take off your sunglasses. Can you back up for me real quick?

Good. Good to go. Carlos?

Good to go. And —

Good to go. Thank you.

There wasn’t a single package inside that bag that they didn’t search. Hi.

Check in over here. Thank you guys for coming.

Hi, I can take whoever’s next.

Once we made it through all that security, we walked into the first official event of the day, a briefing for reporters from the Trump campaign about what this first day of the convention would look like.

So we’re inside a very big ballroom that has a podium and two flags set up next to it. Wisconsin, United States, there’s about 200 journalists in here, filling a quarter of the room and a lot of red-shirted RNC staffers and volunteers who are kind of corralling us. And we’re going to wait and see what they have to say.

But just as this briefing was about to begin, we were told that it was for planning purposes only and that we couldn’t use any of the audio from it.

I just want to summarize what we heard in this news conference. Essentially, what senior advisors to the Trump campaign just told us is that this day is going to be very action-packed. It is going to begin in the early afternoon with the official technical nomination by delegates of Trump as the Republican nominee. And then in the early afternoon, around 3 o’clock or so, the nomination of a vice presidential candidate will occur.

What that means is that sometime between now, 10:30 AM or so, and 3:00 PM, we will know the identity of who Trump’s running mate will be. And finally, we are going to lay eyes on Donald Trump himself at around 9:00 PM in the evening. And the crowd is going to go wild because it’s the first time many of them, many of us will have seen him since the assassination attempt. So it’s a really packed schedule filled with a giant piece of news in the middle, which is who Trump picked as his running mate.

Mike Bender.

To get inside Trump’s decision making, we turn to our colleague Mike Bender, a politics reporter at “The Times,” who’s been on the VP beat for the past few months.

OK, we just want to bug you for a minute about what is the status of the VP choice. Last time we talked to you, you told us that there were three top contenders — Doug Burgum, governor of North Dakota, Florida Senator Marco Rubio, and Ohio Senator JD Vance. So where do things sit here at 11:00 at noon on Monday?

Just hours before he’s actually supposed to make the announcement and nominate the running mate? I mean, I literally ran into a couple of sources who are with other VP contenders who are telling me that they haven’t received word one way or the other yet. I don’t think he’s told the person yet. I mean, we’re three hours from a nomination. I seriously do not think that Trump has actually made the formal offer yet.

We have to go — I think we have to just, like, explain that to people. We’re three hours from the necessary public disclosure of this information, and it’s not clear the decision’s been made.

Yeah, no, it’s amazing. I mean, Trump is known to vacillate over big decisions. He did almost the same thing back in 2016 when he picked Mike Pence. He decided to pick Mike Pence and then spent the night before their first news conference together, complaining to aides that he had made the wrong choice, wondering if he could pick someone differently. My reporting over the last week has been that he’s been going through similar iterations on this decision.

So we’re going to lightly stalk you until we get the news. And hopefully, when we get the news, you can break it to us. Or it will be broken over our heads, and then we’ll talk about it. And we’ll make sense of it.

Yeah, definitely. I am also hoping to be the first to know. And if I’m the first to know, you guys will be the second to know.

OK thank you, Michael.

Yeah. [MUSIC PLAYING]

Just as we finished talking to Mike, he made a phone call.

Hey. You don’t think — why don’t you think it’s Marco? He called? Like, just now? OK, I just saw some folks outside for another contender who haven’t heard anything yet. So I mean, and that was 30 minutes ago. So I’m wondering if it’s happening now. No, I didn’t see that. OK. OK. OK. Thanks, bye.

So can you just tell us about that phone call you had?

Yeah, yeah, definitely. This was someone who is very close with one of the VP contenders, one of the final three. This contender just received a call from Trump’s team saying he’s not the pick.

Soon enough, Mike confirmed that Marco Rubio had been crossed off the VP list, and that Rubio wasn’t the only VP contender to receive that call.

So it’s 1:30 PM, and this is a little orthodox, but I’m peeking over the shoulder of my colleague Mike Bender. And I can see that he’s writing the following. Both Doug Burgum and Marco Rubio have been told that they are not Trump’s VP choice. So in real time, assuming that this is right, we’re left thinking that the choice is either JD Vance or some wildcard person who we haven’t even thought of.

Ladies and gentlemen, we are about to begin a very important part of our program. I would like to ask that the aisles be cleared and the delegates please take their seats. Thank you.

In the meantime, the official business of this convention began. The nomination of Trump as the Republican nominee.

Michael, tell me what’s happening.

OK. So at this point, 2:00 PM, the roll call of delegates is underway on the floor of the Convention Hall.

West Virginia, 32 delegates.

So what’s happening right now is that the delegates from each state are publicly pledging their support to Donald Trump. And the delegates from each state represent the outcome of the primary vote that happened many, many months ago. So like right now, West Virginia’s designated speaker with a red hard hat on, is saying, from the great state of West Virginia —

And to cast those 32 votes for our former and future president, Donald J. Trump!

— we are pledging our 32 delegates to Donald Trump, and on and on and on it will go. And so this is a formality, adding up all the delegates from all the states. But it’s actually a technically required component of Donald Trump becoming and accepting his party’s nomination. And Trump will become the official nominee when 1,215 delegates in this room have pledged their support to him, which will probably be in about 30 minutes.

I stand before you today on behalf of the great state of New Hampshire.

And after Trump is nominated, this entire exercise happens all over again for his running mate whose identity we still don’t know. So that’s where a lot of the suspense of this moment is, not in the obvious fact that Donald Trump is about to be nominated, but that everyone in this room is about to nominate a running mate that they don’t know the identity of, that we don’t know the identity of. And we have to know the identity of it before it happens.

And all of a sudden, after not knowing how or when we would learn Trump’s choice, we finally do. And it’s delivered in classic Trump fashion.

So I’ve just seen on Truth Social that Trump has announced his VP pick, and it is JD Vance of Ohio.

OK. “Daily” editor Rachel Quester breaking the news that it’s Vance, which we kind of thought it would be, but now we know.

After the break, Vance’s nomination and why Trump picked him.

We’ll be right back.

OK, so we’re walking to the Convention Hall. It’s a bit of a trek. It’s very hot outside and blazingly bright. And we are now entering the media doorway.

Hi. How are you? Thank you.

Where are you all trying to get?

To the floor.

To the floor? To the floor?

OK. Just right out there and then somebody on the direction.

Around 3:30 or so, we walked out onto the convention floor just as Vance was being nominated as Trump’s running mate.

USA! USA! USA!

All right, time for a little convention business here. The question is on the motion that Senator JD Vance be nominated by acclamation —

This room is about to complete the nomination of Vance as VP.

All those opposed signify by saying no. In the opinion of the chair, the “ayes” have it, and the motion is adopted. Without objection, the motion to reconsider is laid upon the table.

And just to give you a little bit of a sense of the feel of the room, JD Vance is smiling really widely. He’s leaning back. He’s laughing. Crowd is chanting JD, JD. And he seems a little in awe of the moment.

I am proud to announce that Senator JD Vance has the overwhelming support of this convention to be the next vice president of the United states.

And that’s that. JD Vance, the vice presidential nominee. We got to get out of here.

The chair is pleased —

When that was over, we headed back to where Mike Bender was working to talk to him about Trump’s choice.

So welcome to “The Daily” studio here in Milwaukee.

Can you close this door?

You can shut that, yeah. [MUSIC PLAYING]

Mike, you’re ready?

So if you believe that Donald Trump made this decision, as we think he did kind of the last minute, even if he’d been thinking about it for a long time, what’s your understanding of why he chose JD Vance?

I think Trump is making this pick on who he thinks gives him the best chance to win in November. He announced this decision in a Truth Social post and mentioned a couple of key states in that statement, essentially saying that he thinks Vance can help him win in Pennsylvania, in Michigan, Wisconsin. These are essential states for victory in November, both for Trump and President Biden. Trump won them in 2016 but lost to Biden in those same states in 2020.

Hm! He’s very openly saying, I have made a strategic choice. What’s interesting about that, Mike, is that in our conversation a couple of weeks ago when we talked about who Trump was looking at closely, you had said that he wasn’t thinking about these traditional questions of, Does my VP win me X state, Y state? You know, he was engaging in questions of personal rapport. And he got along really well with Doug Burgum.

So clearly, the strategic question became front of mind for Trump. And I just want to understand, Was there something in the race that changed that made him suddenly think about this or what?

Well, well, well, look at “The Daily” fact-checking its reporters here on air. Well, you’re not wrong either, Michael. There has been a shift in the last few weeks with Trump. He’s gone from talking about finding a running mate who could help him govern, to finding a running mate who can help him govern and someone who can help him win.

And when he ran the analysis here, when he looked at the different set of candidates he has, his ultimate decision was that JD Vance is the one who can do that. Vance will take him deeper into the states where he needs to win in the Midwest and appeal directly to the working-class, blue-collar voters he needs to capture these battlegrounds.

Explain exactly why he thinks JD Vance helps him win those three or so key states? What is it about Vance’s story that maps on to that strategy?

The answer is both biographical and rooted in policy. Vance is a child of Appalachia. He grew up in a poor, working-class neighborhood in Ohio. He served in the military. And when he decided to run for his first elected office in 2022, he was very far right on a lot of issues. He is one of the most staunch anti-abortion voices in the Senate right now. He’s an economic populist, the sort of anti-trade isolationist.

He was one of the first voices urging the Senate to vote against aid for Ukraine at a time where his party leaders were supporting that. So for Trump, when it comes to selling the idea of Trumpism and MAGA-ism, JD Vance is a very effective communicator for him, particularly in a crucial area of the country, that if Trump wins some of these Midwestern states could mean the end of Biden as president.

Got it. So we should see this decision as Trump picking someone who is ideologically extremely aligned with him, perhaps even a little further to the right than Trump, an issue like abortion, and from and of the place in the country that Trump needs to win, the Midwest, Appalachia. And that combination means, for Trump, Vance.

Yeah, exactly. I think it’s also helpful to put this in the context of the other contenders.

Vance is not bringing a new piece of the puzzle to Trumpism. Marco Rubio, the senator from Florida, would have been a Spanish-speaking advocate who may appeal more to Latinos. Governor Doug Burgum, from North Dakota, could have helped settle down that pro-business Republicans who are nervous about Trump’s unpredictability. Trump is sort of signaling here that he’s not interested in adding to the party. He doesn’t want to make the tent bigger. He’s doubling down on this sort of white working-class, pro-MAGA piece of the party that he sees not just as a way to win in November, but very clearly the path for the party in the future.

Of all the candidates Trump was thinking about for VP — and we’ve talked about this with you — they went on a journey, right, from Trump skeptic to Trump supporter. But JD Vance’s journey stood out because he was so unstinting in his criticism of Trump. When Trump first entered the political scene back in 2016, I remember interviewing Vance and talking to him about this stuff. I mean, he compared Trump to Hitler. He called Trump the opioid of the masses. He suggested Trump was a con man. That’s a lot to overcome, but he did.

But he did. I mean, this is the most stunning 180 degree political flip-flop of our time. To go from saying the sort of things that you just brought up to now being chosen as his most trusted advisor, a running mate, a number 2, who will serve as president if something should happen to him, is extraordinary. I mean, you may have to go back to post-revolutionary times when we used to pick vice presidents by who came in second place to find a vice president who has said such searing criticisms of a president.

Fascinating. And ultimately, what’s your understanding of why Trump could get past that, somebody who is not very good at accepting criticism?

No, that’s true. But as much as Trump hates criticism for his own actions and deeds, he loves the redemption narrative. And he loves being asked for forgiveness. And JD Vance has spent several years seeking Trump’s approval by going on television, making nice with all of the right-wing websites and media in order to show how much he has changed his mind on Trump, and maybe most effectively, blame the media.

That point has been proven by Vance himself, who has explained his new thinking, his evolution on Trump by saying he was lied to by a media narrative about Trump. And now that he’s gotten to know Trump and now that he’s seen him, Trump in action as president, he has changed his mind.

Right. And it sounds like he never really needed to change his mind about some of the fundamental ideas of Trumpism. He had to change his mind about Trump. He seems always to have been fundamentally aligned with the ideas that Trump embraces — economic populism, some pretty far-right social positions. He needed to change his mind on the man, not the message.

Yeah, it was very personal, I think, for both Trump and Vance in this instance. Again, Vance is someone who grew up in rural Ohio, whose family is from Appalachia, saw some of the things that Trump has railed against when it comes to manufacturing jobs that are being shipped overseas, you know, these sort of pillars and institutions of society that have failed to uphold their end of the bargain when it comes to working-class, blue-collar, small-town Americans like Vance is.

If you are the Democrats right now, Mike, and you’re absorbing this news, where do you see the greatest vulnerabilities are going to lie for Trump now picking Vance?

Democrats are definitely going to use Vance’s old words against him, this sort of library of video clips and audio interviews of Vance going after Trump. But Democrats will also seize on Vance as an extremist, whether that’s his ardent abortion view and support for a national ban and his willingness to do what his predecessor, Mike Pence, wouldn’t. Vance has been on record already saying that he would have blocked the certification of the 2020 result, and that would have helped overturn that election.

Right. So you’re saying one simple way that the Biden ticket can go after Vance is by saying that you will enable Trump to break the law, overturn the election. We should expect that.

Yeah, I think so. I mean, looking back on what happened after 2020, the system worked because there was a lot of people around Trump who maybe they weren’t guardrails, but maybe more speed bumps. And there’s no indication that Vance has any willingness to play that role in the next Trump administration.

In that vein, it was pretty widely noted that in the hours after the attempted assassination of Donald Trump, we saw JD Vance come out with a statement. It was the most strongly worded of anyone seeking to be his VP. And it had some factual problems. Here’s what he said. He said, “Today, the attempted assassination is not just some isolated incident. The central premise of the Biden campaign is that President Donald Trump is an authoritarian fascist who must be stopped at all costs. That rhetoric led directly to President Trump’s attempted assassination.” We should say. There’s no evidence that that’s true. We don’t know the motivations of the shooter. We don’t know that he consumed any of that rhetoric or that Vance is even characterizing it correctly. So what should we make of the fact that Trump chose that running mate who made that statement in this moment?

I think Vance in a lot of ways kind of embodies the id of Trump and that instinct to fight. And even though these sort of manufactured statements from the campaign are calling for unity and calling for peace, what Trump really wants —

Since the attempted assassination, right.

That’s right. What Trump really wants is someone who is going to keep fighting. You know, factual or not, I think this shows the passion and the energy Trump was looking in at running mate, valuing that interest in fighting more than —

Interest in unity and peace.

Yeah, or facts on the ground.

Right. I want to end, Mike, with something you hinted at earlier, which is when you said that Trump is looking to Vance to set a path for the future of the Republican Party. What is that path with Vance as number 2?

Vance is only 39 years old. He’s barely old enough to be president.

Right, 35 is the requirement.

Exactly. So he’s obviously going to be viewed very much as the heir apparent for Trumpism. Trump knows this. And the signal it’s sending to everyone, not just in the party but the rest of the country, is that any remnants of a debate about whether this party snaps back to its sort of pro-business establishment culture —

Pre-Trump era.

— the pre-Trump era is exactly that. It’s a pre-Trump era. It’s over for Republicans. And when it’s not Trump, it’s going to be JD Vance or someone exactly like him.

Right. In other words, Trumpism is here to stay. It is the Republican Party now that I’ve chosen JD Vance.

There’s no going back anymore, Michael.

Thank you, Mike. This is really helpful. I really appreciate it.

Thank you for having me. [MUSIC PLAYING]

Are you guys from — are you guys from Michigan?

We make “The Daily” podcast.

Back at the Convention Hall.

We wanted to ask folks from all the big swing states about the selection of JD Vance about an hour ago and what you make of that decision, and if you think it’s going to help Trump win this state.

Absolutely. Look, I think the key thing in the platform is that it is dedicated to the forgotten men and women, and that is the blue-collar workers in the flyover states. JD Vance gets that.

It was clear that Republican delegates saw JD Vance as helping Trump win those key Midwestern states that will be essential to Trump winning in November.

So if Donald Trump wins Pennsylvania, which we’re going to make damn sure he does, we’re going to work our asses off.

You think that Vance helps them do that?

And just to make sure — I want to understand why.

Because JD Vance is like the common man. He’s like the common guy.

And that like Trump, they do see Vance as the future of Trumpism.

And the other nice thing is he’s young. He’s 39 years old

Why’s that matter?

It’s good to have somebody young with somebody that’s old in case, God forbid, something ever happens to Trump.

In other words, you already see him as the successor, the inheritor of Trump’s message and the party in MAGA?

Well, yeah, he’s going to have to carry the mantle. That’s probably what’s going to end up happening. Trump is only there for four years. You need somebody afterwards for the next eight. You need somebody for the next eight after that.

OK, can I — [MUSIC PLAYING]

And you’re from Wisconsin. You’re a delegate from Wisconsin. This is important. Trump mentioned Wisconsin in announcing Vance. Why was Vance your number one choice?

I think he brings youth to the field, to the vice president. And I looked at the upcoming years in ‘28, what’ll happen. And I think he was the man that can do it in ‘28 for the Republican Party.

You’re already looking forward to the next race?

Yes, very much so.

Here’s what else you need to know today. In a stunning decision, the judge overseeing Trump’s classified documents case threw out all the charges against him. In the process, she rejected what was widely seen as the strongest federal charges against the former president. Judge Aileen Cannon, a Trump appointee, ruled that the special counsel who filed the charges had been given his job in violation of the Constitution.

That finding flew in the face of previous court decisions reaching back decades. In response, the Department of Justice said that it plans to appeal Cannon’s ruling.

Today’s episode was produced by Carlos Prieto, Clare Toeniskoetter, Jessica Cheung, Mooj Zadie, Eric Krupke, and Rikki Novetsky. It was edited by Brendan Klinkenberg and Rachel Quester, with help from Paige Cowan.

Contains original music by Dan Powell, Elisheba Ittoop and Marion Lozano, and was engineered by Alyssa Moxley. Our theme music is by Jim Brunberg and Ben Landsverk of Wonderly.

That’s it for “The Daily.” I’m Michael Barbaro. See you tomorrow.

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Hosted by Michael Barbaro

Featuring Michael C. Bender

Produced by Carlos Prieto ,  Clare Toeniskoetter ,  Jessica Cheung ,  Mooj Zadie ,  Eric Krupke and Rikki Novetsky

Edited by Brendan Klinkenberg and Rachel Quester

With Paige Cowett

Original music by Dan Powell ,  Elisheba Ittoop ,  Marion Lozano and Corey Schreppel

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On the first day of the Republican National Convention, Donald J. Trump chose his running mate: Senator J.D. Vance of Ohio.

We watched the process unfold in real time in Milwaukee.

Michael C. Bender, who covers Mr. Trump and his movement for The Times, takes us through the day.

On today’s episode

Michael C. Bender , a political correspondent covering Donald J. Trump and his Make America Great Again movement for The New York Times.

Background reading

What to know about J.D. Vance , Mr. Trump’s running mate.

Mr. Trump’s decision to pick Mr. Vance signals concern for the future of his MAGA movement.

There are a lot of ways to listen to The Daily. Here’s how.

We aim to make transcripts available the next workday after an episode’s publication. You can find them at the top of the page.

The Daily is made by Rachel Quester, Lynsea Garrison, Clare Toeniskoetter, Paige Cowett, Michael Simon Johnson, Brad Fisher, Chris Wood, Jessica Cheung, Stella Tan, Alexandra Leigh Young, Lisa Chow, Eric Krupke, Marc Georges, Luke Vander Ploeg, M.J. Davis Lin, Dan Powell, Sydney Harper, Michael Benoist, Liz O. Baylen, Asthaa Chaturvedi, Rachelle Bonja, Diana Nguyen, Marion Lozano, Corey Schreppel, Rob Szypko, Elisheba Ittoop, Mooj Zadie, Patricia Willens, Rowan Niemisto, Jody Becker, Rikki Novetsky, Nina Feldman, Will Reid, Carlos Prieto, Ben Calhoun, Susan Lee, Lexie Diao, Mary Wilson, Alex Stern, Sophia Lanman, Shannon Lin, Diane Wong, Devon Taylor, Alyssa Moxley, Olivia Natt, Daniel Ramirez and Brendan Klinkenberg.

Our theme music is by Jim Brunberg and Ben Landsverk of Wonderly. Special thanks to Sam Dolnick, Paula Szuchman, Lisa Tobin, Larissa Anderson, Julia Simon, Sofia Milan, Mahima Chablani, Elizabeth Davis-Moorer, Jeffrey Miranda, Maddy Masiello, Isabella Anderson, Nina Lassam and Nick Pitman.

Michael C. Bender is a Times political correspondent covering Donald J. Trump, the Make America Great Again movement and other federal and state elections. More about Michael C. Bender

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