NCERT Solutions Class 11 Mathematics

When it comes to academics, classes 10, 11, and 12 are undoubtedly the most important years for a student. While classes 10 and 12 are critical from the point of view of board exams, class 11 is crucial as it helps prepare you for class 12 exams. Then, it becomes vital to have a solid grip on the various subjects in class 11 as it sets your basics right. The in-depth knowledge of fundamentals you gain in class 11 will only help you  ace your board examinations, which will ultimately help you pursue your chosen career from a top college. 

Mathematics is perhaps the most feared subject among many students, and yet it is also the most scored one. If you are also struggling with certain concepts in Mathematics, NCERT solutions class 11 Mathematics from Extramarks will prove to be an excellent resource in preparing you for the examinations. Extramarks offers comprehensive solutions for all classes – from NCERT solutions class 1 to NCERT solutions class 12 – helping you ace your exams with top scores!

The NCERT solutions class 11 are designed systematically, covering all the 16 chapters included in the CBSE syllabus. We aim at providing a comprehensive learning experience for the student and the overall development of the various concepts. This is why, in our NCERT class 11 Mathematics solutions, you will find detailed explanations, step-by-step solutions to Mathematics problems, practice questions, shortcut tips, quizzes, and more. Now, you can clear your doubts and become a Mathematics wizard with a strong hold on basic mathematical concepts!

Class 11 Mathematics NCERT Solutions 

The NCERT solutions class 11 Mathematics are specially solved and organised by subject matter experts with years of experience in CBSE board education. These solutions have been developed to help you clear your doubts and provide you with a step-by-step guide to solving complex mathematical questions. 

All the chapters in the NCERT class 11 textbook include unique mathematical concepts which need regular practice. The key to scoring higher marks in mathematics is being consistent. Solving as many questions as you can, given at the end of every chapter will increase your speed and develop your problem-solving skills. Hence, you must access the NCERT Solutions from Extramarks to enhance your preparation level. 

With Extramarks’ NCERT Solutions, you can get access to all the 16 chapters in the class 11 NCERT Mathematics book in a downloadable format. In addition to your textbook exercises, in-text solved examples, and guide books, these solutions will be a valuable study resource, preparing you for different types of questions asked in the examination. You can download the NCERT Mathematics book class 11 solutions by clicking on the link given, free of cost, and start studying at any time!

NCERT Solutions for Class 11 Mathematics Chapters

The class 11 NCERT Mathematics book consists of a total of 16 chapters. If you have any confusion regarding any of the mathematical concepts, you can refer to our NCERT class 11 Mathematics solutions to have an all-round and more concise understanding of the various chapters as a whole. These solutions will provide accurate guidance, written in an easy-to-understand language and systematically organised. When you download the solutions from Extramarks, you can navigate chapter-wise as well as exercise-wise Mathematics solutions as per your convenience.

NCERT Solutions for Class 11 Mathematics 

Looking for accurate study resources to help you prepare better for your class 11 Mathematics examination? If yes, then you’re at the right place! To make it even more convenient for you, we offer detailed solutions to the class 11 Mathematics NCERT book exercises. Created by top-notch and experienced faculty in a simple language, these solutions will come in handy when you sit down to study.

Once you are through with the CBSE Mathematics syllabus and have a clear understanding of the fundamentals, it is best to start solving several sample papers and revising the various concepts for a deeper understanding. With reliable study material available at your disposal, this will become extremely easy. The chapter-wise solutions ensure that you can start with the basic concepts,  move on to the more complex topics, and utilise your time judiciously. To ensure better learning, our experts have provided easy solutions to even the most complex questions in a logical manner. 

Moreover, our solutions and study materials are designed in accordance with the latest CBSE curriculum and NCERT textbook exercises. With the rigorous practice of the different concepts, you can achieve higher marks in your class 11 exams. In addition, the solutions are 100% accurate and as per the CBSE guidelines, so rest assured that you can attempt all the difficult questions in the exam with ease.

Needless to say, NCERT Mathematics book class 11 solutions is all you need to be the top scorer in class 11 Mathematics exams.

NCERT Solutions Mathematics Chapter 1 Class 11 – Sets (Term I)

This chapter introduces the concept of sets. The important topics in this chapter include equal sets, empty sets, subsets, universal sets, and power sets. Along with this, you will learn about various set operations and how to draw a Venn diagram. The fundamentals of this chapter will be useful when studying the second chapter – relations and functions. It will also help you  understand the concepts of probability, geometry, sequences, and more. The properties of different types of sets are also covered in this chapter. This chapter includes a total of 6 exercises, along with a miscellaneous exercise, covering a wide range of questions to strengthen your learning.

Mathematics NCERT Solutions Class 11 Chapter 2 – Relations and Functions (Term II)

This chapter illustrates the concept of connecting pairs of objects from two different sets and establishing relations between these objects. Along with this, you will also learn about functions, which is one of the most important topics in the class 11 Mathematics NCERT book. Various concepts in this chapter include functions as a special type of relation, domain and range of functions, pictorial representation, domain and co-domain of relations, Cartesian product of sets, functions and their graphs, and more. 

Class 11 Mathematics NCERT Solutions Chapter 3 – Trigonometric Functions (Term II)

In the third chapter, you will learn about the concept of trigonometric ratios and trigonometric functions as well as their properties. The chapter begins with the basics of trigonometric identities, trigonometric ratios of different angles, trigonometric functions of sum and difference of two angles, trigonometric equations, the domain and range of trigonometric functions and their graphs, and the application of these concepts. All these topics are covered among the 4 exercises, along with a miscellaneous exercise. Through the various numerical questions, you get to understand the dynamic world of trigonometry.

NCERT Solutions Mathematics Class 11 Chapter 4 – Principle of Mathematical Induction

If you are interested in mathematical induction, this chapter will get you familiar with the various concepts which will also be useful when you pursue higher studies. It includes different unique topics such as the steps of the principle of mathematical induction, induction, the process of proving the induction, and several examples to explain these concepts thoroughly. All the problems in this chapter are mostly application-based.

NCERT Solutions for Class 11 Mathematics Chapter 5 – Complex Numbers and Quadratic Equations (Term I)

This chapter discusses the concepts of real numbers, complex numbers, and the square root of negative numbers. The concept of a complex number is critical for higher mathematics, so it is best to put extra focus when revising this chapter. You will be introduced to topics like algebraic properties of complex numbers, the fundamental theorem of algebra, solutions of quadratic equations in the complex number system, and the square root of the complex number. With so many topics to cover, Extramarks’ NCERT class 11 Mathematics solutions will be a useful resource in preparing for the exam.

NCERT Solutions Mathematics Chapter 6 Class 11 – Linear Inequalities (Term II)

The sixth chapter can be interesting to study, once you are thorough with the various concepts. As the name suggests, it explores the concept of how two quantities can be unequal rather than being equivalent. In the NCERT Mathematics book class 11 solutions, you will find detailed solutions to different topics like algebraic solutions to linear inequalities in one variable, their representation on the number line, rules of inequality, and the graphic representation of linear equations in two variables. There are a total of 3 exercises along with a miscellaneous exercise covered in this chapter.

NCERT Class 11 Mathematics Solutions Chapter 7 – Permutation and Combination (Term II)

The concepts of permutation and combination apply to a variety of fields apart from mathematics. That is why having a strong grip on these concepts is absolutely crucial.A permutation is each of the several possible ways in which a series of objects can be arranged. Whereas, combination is the way of selecting different objects from a set where the order of selection doesn’t matter. This chapter will discuss topics like fundamental principles of counting, factorial, the derivation of the formula of permutation and combination, and their application. 

NCERT Solutions Class 11 Mathematics Chapter 8 – Binomial Theorem

The Binomial Theorem is used to solve complex numerical and difficult computations for higher powers. This chapter will discuss the history and proof of the binomial theorem for positive integral indices. You may use the simplified expression of (a + b)n, where n is a rational number. Along with this, you will study Pascal’s triangle, the middle term, the binomial theorem for any index, the general and middle terms in binomial expansion, and their simple application. 

NCERT Solutions Class 11 Mathematics Chapter 9 – Sequences and Series (Term I)

Chapter 9 of the Class 11 Mathematics NCERT book is an important chapter and carries a high weightage in the exam. The arranged list of numbers is a sequence and the sum of all terms in a sequence is a series. This chapter introduces you to the concepts of Arithmetic Progression (A.P.), Geometric Progression (G.P.), Arithmetic Mean (A.M.), terms of a G.P., the sum of the first n numbers of a G.P., formulae for the sum of special series, infinite G.P. and its sum- Geometric Mean (G.M.), the relation between A.M. and G.M., and more. To score better in this chapter, you need to understand and memorise several formulas to accurately solve the questions.

NCERT Solutions Class 11 Mathematics Chapter 10 – Straight Lines (Term I)

Learn about  straight lines and their properties in this chapter. This is an important chapter for geometry where regular practice is the key to scoring better. Chapter 10 of the NCERT Class 11 Mathematics solutions comprises topics like the distance formula, section formula, area of a triangle, slope formula, conditions of collinearity of three points, equation of a straight line, parallel and perpendicular lines, etc. The various exercises will help you understand this chapter better.

NCERT Solutions Class 11 Mathematics Chapter 11 – Conic Sections (Term II)

In mathematics, the conic section is the curve obtained from the intersection of the surface of a cone and a plane. Conics include curves like circles and ovals, as well as conic segments, hyperbolas, and parabolas. In addition, the chapter will also introduce you to the concept and application of the x, y, and z-axis. The NCERT Mathematics book class 11 solutions cover various topics like the difference between various conics, the definition of parabola, line and a parabola, tangents to a parabola, ellipse, eccentric circles, etc.

NCERT Solutions Mathematics Class 11 Chapter 12 – Introduction to Three Dimensional Geometry (Term II)

This chapter extensively deals with coordinate geometry in a three-dimensional space. A point in space has three coordinates. This chapter introduces you to the concepts of coordinate axes and coordinate planes, coordinates of a point, the distance between two points, direction ratios of a line, the relationship between direction cosines of a line, section formula, etc. This is a highly scoring chapter where regular practice can help you ace the exam.

NCERT Solutions Mathematics Class 11 Chapter 13 – Limits and Derivatives (Term II)

The chapter, Limits and Derivatives, is an introduction to Calculus. Calculus is the study of changes in the value of a function as the points in the domain change. To begin with, you will get the basic idea of limits and derivatives along with some definitions of the algebra of limits.. The chapter includes several important theorems that you will need to learn to improve your understanding of the chapter and score well. You can study using the NCERT class 11 Mathematics solutions from Extramarks.

Class 11 Mathematics NCERT Solutions Chapter 14 – Mathematical Reasoning

This is a relatively easy chapter that deals with some basic ideas of mathematical reasoning. In this chapter, you are required to arrive at solutions to problems as per the scenario provided, using deductive reasoning. This chapter is extremely engaging and also important for upcoming competitive exams. Here, you will learn the basics of mathematical reasoning, open and compound statements, logic, statements, tables for basic logical connections, quantifiers and quantified statements, and more. 

Class 11 Mathematics NCERT Solution Chapter 15 – Statistics (Term I)

When it comes to higher mathematics, statistics is a crucial chapter. In this chapter, you will learn about some measures of dispersion and methods of calculation for grouped and ungrouped data. The different concepts introduced in this chapter include measures of dispersion, mean deviation, range, variance, and standard deviation of grouped and ungrouped data, analysis of frequency distributions, and more. Once you are clear with the topics and methods of attempting the questions, you can easily score better.

NCERT Solutions Class 11 Mathematics Chapter 16 – Probability (Term II)

Probability is a much-used concept in various fields, and this chapter explores this topic in great detail using real-life examples. You should ensure a solid understanding of the core concepts and applications in this chapter to solve the maximum number of questions from this chapter. It revolves around concepts such as probability, events, types of events, random experimenting, simple and compound events, sure and impossible events, the probability of ‘and’, ‘not’, and ‘or’ events, and so on. The NCERT class 11 Mathematics solutions from Extramarks cover all the important topics and NCERT exercises from this chapter.

Class 11 Mathematics Solutions for Students

Most students fear Mathematics because it can be a tricky subject to prepare for. It can be a highly scoring subject only when you practice  similar types of questions regularly. While your class 11 Mathematics NCERT book should be your main focus, practising extra questions can also prove to be helpful. Various subject experts at Extramarks have compiled and solved all the Mathematics NCERT textbook questions in a logical manner.

Moreover, the solutions are organised chapter-wise, which makes it easier for you to navigate and find the exercises you are looking for. And the best part is that the downloadable format allows you to access these solutions conveniently, anytime, from anywhere. So rest assured, you will be learning from the best resources available with a single click.

Chapter-Wise Weightage of Marks Class 11 Mathematics

When you’re preparing for an exam, knowing the marking scheme and the chapter-wise distribution of marks can really help you prepare better. This way, you can put extra effort and focus on your weak chapters that carry a high weightage.

The chapter-wise marks weightage for class 11 Mathematics is as below-

Unit No. Unit Name Marks
I Sets and Functions 29
II Algebra 37
III Coordinate Geometry 13
IV Calculus 06
V Mathematical Reasoning 03
VI Statistics and Probability 12
Total 100

Marking Scheme for Class 11 Mathematics – First Term

CBSE has divided the curriculum into two terms along with the syllabus, with 7 chapters in Term I out of 16 chapters. The Mathematics class 11 exam is of a total of 50 marks in both terms I and  II. Here, the internal assessment carries 10 marks while the written examination has 40 marks in each term.

Here is the marking scheme for Term I of Class 11 Mathematics

S.No. Chapters Marks
1. Chapter 1- Sets 11
2. Chapter 2- Relations and Function
3. Chapter 5- Complex Numbers and Quadratic Equations 13
4. Chapter 9- Sequences and Series
5. Chapter 10- Straight Lines 6
6. Chapter 13- Limits and Derivatives (Only Limits) 4
7. Chapter 15- Statistics 6
Total 40
Internal Assessment 10
Grand Total 50

Internal Assessment Class 11 Term I

Internal Assessment Marks
Periodic/Unit tests 05 marks
Activities and Projects- Activity file + term end assessment + viva 05 marks
Total 10

Marking Scheme for Class 11 Mathematics – Second Term

For a total of 40 marks in the written exam, the second term includes 7 chapters from a total of 16 chapters in the Mathematics NCERT syllabus.

Here is the marking scheme for the second term of the class 11 Mathematics exam-

S.No. Chapters Marks
1. Chapter 3- Trigonometric Functions  8
2. Chapter 6- Linear Inequalities 11
3. Chapter 7- Permutation and Combination
4. Chapter 11- Conic Sections
5. Chapter 12- Introduction to Three-Dimensional Geometry 9
6. Chapter 13- Limits and Derivatives (Derivatives only) 6
7. Chapter 16- Probability 6
Total 40
Internal Assessment 10
Grand Total 50

Internal Assessment Class 11 Term II

Internal Assessment Marks
Periodic/Unit tests 05 marks
Activities and Projects- Activity file + term end assessment + viva 05 marks
Total 10

You can achieve better marks and improve your  academic performance with consistent practice and a smart approach to studying. The NCERT class 11 Mathematics solutions will give you a competitive edge in preparing for your exam with their smart approach.

Solutions for Future Reference

When you download NCERT solutions class 11 Mathematics, you not only get access to study material to ace your class 11 examinations, but you can also use these for your future reference. The chapters covered in the class 11 Mathematics NCERT book are an essential part of the various competitive exams like JEE Main, JEE Advanced, CAT, CET, etc. All the questions covered will act as valuable revision notes when you start preparing for these entrance exams. 

Moreover, they are created in a straightforward language, which will allow you to understand the complex topics very easily and retain them for longer. With this in-depth knowledge, you can surely appear for your competitive exam with increased confidence and concentration.

So, once you download the NCERT Solutions, you can save them for future reference. These solutions will provide guidance for an enhanced learning experience and you can practice these chapters regularly to ace your exams!

Covering All Chapters in One Solution

The NCERT Mathematics book class 11 solutions is a key study resource that covers all the chapters from the NCERT textbook in a single set. You can access 100% accurate solutions to different chapters including, sets, relations and functions, complex numbers, linear inequalities, quadratic equations, limit and derivatives, introduction to three-dimensional geometry, and more. All the NCERT exercise questions are solved by experienced faculty, providing you with step-by-step methods to solve a particular question. 

With all the solutions available in one place, you don’t have to search in different places for different chapters when you begin studying. All you have to do is download the NCERT solutions class 11 and get instant access. All the concepts are explained in great detail which will clear your doubts and increase your confidence while attempting different questions during the actual examination.

Help in Exam Preparation

As a class 11 student, you must be aware that the mathematical concepts learned in class 11 will also be useful when you reach class 12. Not only that, but class 11 Mathematics questions are also a crucial part of several competitive exams. Since our NCERT solutions are provided systematically, you can grasp the theory behind each topic quickly while at the same time improving your time management skills. This will eventually help you in the actual exam, where time management and a thorough understanding of the difficult topics are essential for achieving higher marks.Moreover, remember that being consistent should be your top priority. The more you practise, the better you get at quickly solving questions.

Key Benefits of NCERT Solutions Class 11 Mathematics

  • NCERT solutions from Extramarks are available in downloadable format, free of cost. This allows you to study at your own pace, completing each topic by devoting the proper time
  • These solutions are compatible with all devices. So you can download them on your desktop, tablet, or smartphone and prepare for exams within your comfort zone
  • The NCERT class 11 Mathematics solutions will also assist your classroom studies by providing a reference for your homework and assignments
  • Certain chapters like probability, statistics, 3D geometry, sets, linear inequalities, etc., are relatively simple topics that you can easily master with enough practice and revision. Our comprehensive NCERT solutions class 11 Mathematics will help you do the needful
  • The detailed solutions will also help you prepare faster for the exam. This is because whenever you get stuck on a question, you can simply refer to these solutions to clear your doubts or understand how the problem was solved
  • Since Mathematics is a complex subject, having accurate solutions at your disposal can ease your worries in preparing for the exam

Moreover, not just mathematics, but Extramarks also offers complete NCERT solutions for all the class 11 subjects in a simple language, ready to download. And if that’s not enough, you can also access complete study resources for all classes, NCERT solutions class 12, NCERT solutions class 10, NCERT solutions class 9, NCERT solutions class 7, and so on. Needless to say, Extramarks is your one-stop solution for valuable study material for all classes and subjects.

Extramarks Gives You Competitive Edge

Extramarks’ NCERT solutions Class 11 Mathematics, as well as other subjects, gives you the confidence, clarity, and knowledge to attempt several difficult questions based on unique mathematical concepts with ease. Various education boards, including CBSE, design their school curriculum for classes 11 and 12 based on the NCERT textbooks for students. The concepts covered in the textbook offer you comprehensive knowledge of the basics, which can help you grasp even the most critical topics effortlessly. 

And to further help you with that, NCERT class 11 Mathematics solutions by Extramarks also help you better understand the NCERT topics, improving your speed of learning. Students can use our detailed solutions to prepare notes for self-study and last-minute revisions and get ahead of their peers with their preparation.

These solutions are designed methodically which allows you to practice all the exercise questions in a logical manner and improve your mathematical skills. And not only that. With regular practice and by solving different types of questions, you get to improve your speed, concentration level as well as time management skills. 

NCERT Solutions for Class 11 Mathematics – Preparation Tips

Preparing for a subject like mathematics can be overwhelming as well as tricky. Yet it is one of the most scoring subjects. You can score well only if you have a solid understanding of the fundamentals as well as the regular practice of solving various questions. Here are some effective tips you can follow to prepare efficiently and achieve higher marks in the exam-

  • Create a study schedule to devote time to all the chapters and keep track of your progress. You can check the marking scheme to dedicate more time to your weak topics or chapters that carry a higher weightage of marks
  • Having the best study resources can only take your exam preparation to the next level. You can use the NCERT solutions from Extramarks to boost your understanding and solve various types of questions easily
  • Try to make short notes while revising as it will not only help you gain clarity but also identify any doubts you may have about any topic
  • Once you are through with the NCERT syllabus, make sure to attempt as many questions and sample papers as you can to enhance your learning process
  • While attempting the sample papers or mock tests, time yourself to evaluate how much time you are spending on different questions and if you’re able to complete the paper in the standard 3 hours. This way, you can improve your speed and time management skills by practising more for topics that take time to solve.When preparing, completing the syllabus on time is extremely important as you must take some time for revision as well
  • Regular self-assessment will help you judge your preparation level and identify the key areas where you need more practice
  • Preparing for Mathematics can be overwhelming, so it is always important to keep your cool and take regular breaks in between studies to relax and unwind your brain
  • Remember, for a subject like Mathematics, consistency is the key. Practice on a regular basis and revise several times to easily pass your class 11 exams.

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

(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = [ a, e, i, o, u} B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
B = {x: x is a natural number and 6 < x < 10}

(v) A = {1, 2, 3}, B = Φ


(i) X = {1, 3, 5} Y = {1, 2, 3}
X ∪ Y = {1, 2, 3, 5}
(ii) A = [a, e, i, o, u} B = {a, b, c}
A ∪ B = {a, b, c, e, i, o, u}
(iii )A = {x: x is a natural number and multiple of 3}
= {3, 6, 9, 12, 15, … }
B = {x: x is a natural number less than 6}
= {1, 2, 3, 4, 5}
A∪B = {1, 2, 3, 4, 5, 6, 9, 12, 15, …}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
= {2, 3, 4, 5, 6}
B = {x: x is a natural number and 6 < x < 10}
= {5, 6, 7, 8, 9}
A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
(v) A = {1, 2, 3}, B = Φ
A ∪ B = {1, 2, 3} = A

Q.2 Find the intersection of each pair of sets of

(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = [ a, e, i, o, u} B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
B = {x: x is a natural number and 6 < x < 10}
(v) A = {1, 2, 3}, B = Φ


(i) X ∩ Y = {1, 3}
(ii) A ∩ B = {a}
(iii) A = {x: x is a natural number and multiple of 3}
= {3, 6, 9, 12, …}
B = {x: x is a natural number less than 6}
= {1, 2, 3, 4, 5}
A ∩ B = {3}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
= {2, 3, 4, 5, 6}
B = {x: x is a natural number and 6 < x < 10}
= {7, 8, 9}
A ∩ B = {} = Φ
(v) A ∩ B = {} = Φ

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

(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)


(i) A ∩ B = {7, 9, 11}
(ii) B ∩ C = {11, 13}
(iii) A ∩C ∩ D = {} = Φ
(iv) A ∩ C = {11}
(v) B ∩ D = {} = Φ
(vi) A ∩ (B ∪C) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}
(vii) A ∩ D = {} = Φ
(viii) A ∩ (B ∪ D)
= {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17}
= {7, 9, 11}
(ix) (A ∩ B ) ∩ ( B ∪ C )
= {7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}
(x) (A ∪ D) ∩ (B ∪ C)
= {3, 5, 7, 9, 11, 15, 17} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11, 15}

Q.4 If A = {x: x is a natural number },
B = {x: x is an even natural number}

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

(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D

(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D


A = {x: x is a natural number}
= {1, 2, 3, 4, 5, 6, 7, …}
B = {x: x is an even natural number}
= {2, 4, 6, 8, …}
C = {x: x is an odd natural number}
= {1, 3, 5, 7, …}
D = {x: x is a prime number}
= {2, 3, 5, 7,…}

(i) A∩B = {2, 4, 6, …} = B
(ii) A∩C = {1, 3, 5, 7, …} = C
(iii) A ∩ D = {2, 3, 5, 7,…} = D
(iv) B ∩ C = {} = Φ
(v) B ∩ D = {2}
(vi) C ∩ D = {3, 5, 7, …}
= {x: x is an odd prime number}

Q.5 Which of the following pairs of sets are disjoint

(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6 }
(ii) {a, e, i, o, u} and { c, d, e, f}
(iii) {x: x is an even integer} and {x: x is an odd integer}


(i) Let A = {1, 2, 3, 4} and
B = {x: x is a natural number and 4 ≤ x ≤ 6 } = {4, 5, 6}
A ∩ B = {4}
So, sets A and B are not disjoint sets.
(ii) Let P = {a, e, i, o, u} and Q = {c, d, e, f}
P ∩ Q = {e}
So, sets P and Q are disjoint sets.
(iii) Let C = {x: x is an even integer}
= {…, – 4, – 2, 2, 4, 6, … }
D = {x: x is an odd integer}
= {…, –5, –3, –1, 1, 3, 5,…}
C ∩ D = {} = Φ
So, sets C and D are disjoint sets.

Q.6 State whether each of the following statement is true or false. Justify your answer.

(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii) {a, e, i, o, u } and { a, b, c, d }are disjoint sets.
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.


(i) False,
Since, {2, 3, 4, 5} ∩ {3, 6} = {3}
So, given sets are not disjoint sets.

(ii) False,
Since, {a, e, i, o, u} ∩{a, b, c, d}= {a}
So, given sets are not disjoint sets.

(iii) True,
Since, {2, 6, 10, 14} ∩ {3, 7, 11, 15} = {}
= Φ
So, given sets are disjoint sets.

(iv) True,
Since, {2, 6, 10} ∩ {3, 7, 11} = {}
= Φ
So, given sets are disjoint sets.

Q.7 Fill in the blanks to make each of the following a true statement:
(i) A ∪ A′ = . . .
(ii) Φ′ ∩ A = . . .
(iii) A ∩ A′ = . . .
(iv) U′ ∩ A = . . .


(i) A ∪ A′ = U
(ii) Φ′ ∩ A = A
(iii) A ∩ A′ = Φ
(iv) U′ ∩ A = Φ

Q.8 Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C


Given:      AB=AC and AB=ACTo prove: B=CProof: Let xBxAB[BAB]          xAC[AB=AC]          xA or xCCase1:If xA and also, xB          xAB          xAC[AB=AC]          xA and xCxC and xBBC...(i)Case2:If xC and also, xA          xAC          xAB[AB=AC]          xA and xBxB and xCCB...(ii)From case 1 and case 2, we haveB=C Hence proved.

Q.9 Show that the following four conditions are equivalent:

(i) A ⊂ B
(ii) A – B = Φ
(iii) A ∪ B = B
(iv) A ∩ B = A


Taking (i) to prove (ii):
We have, A−B = { x ∈ A : x ∉ B }
Since, A⊂B.
Therefore, there is no element in A which does not belong to B.
∴ A−B = ∅
Hence, (i) ⇒ (ii).
Taking (ii) to prove (iii):
We have, A−B = ∅ ⇒ A ⊂ B ⇒ A ∪ B = B
Hence, (ii) ⇒ (iii)
Taking (iii) to prove (iv):
We have, A ∪ B = B ⇒ A ⊂ B ⇒ A ∩ B = A
∴ (iii) ⇒ (iv)
Taking (iv) to prove (i):
We have, A ∩ B = A ⇒ A ⊂ B
So,  (iv) ⇒ (i)
Therefore, (i) ⇔ (ii) ⇔ (iii) ⇔ (iv)

Q.10 Show that if A ⊂ B, then C – B ⊂ C – A.


Let x ∈ C−B ⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [ ∵ A ⊂ B ]
⇒ x ∈ C − A
Hence, C−B ⊂ C−A

Q.11 Is it true that for any sets A and B, P (A )∪ P (B) = P (A ∪ B)? Justify your answer.


Let A = {1, 2, 3} and B = {3, 4}
then A∪B = {1, 2, 3, 4}
P(A) = {Φ, {1}, {2}, {3},{1,2},{2, 3},{3, 1},{1, 2, 3}}
P(B) = {Φ, {3}, {4}, {3,4}}
P(A∪B) = {Φ, {1}, {2}, {3}, {4}, {1,2}, {1, 3}, {2, 3},
{2, 4}, {3, 4},{4, 1}, {1, 2, 3}, {2, 3, 4},
{3, 4, 1}, {4, 1, 2}, {1, 2, 3, 4}}
P(A) ∪ P(B) = {Φ, {1}, {2}, {3}, {4}, {1,2},{2, 3},
{3, 1}, {3, 4}, {1, 2, 3}}
In this way we can see that
P(A) ∪ P(B) ≠ P(A∪B)

Q.12 Show that A ∩ B = A ∩ C need not imply B = C.


Let A = {1, 2, 3}, B = {1, 4} and C = {1, 5}
Then, A ∩ B = {1} and A ∩ C = {1}
Here, A ∩ B = A ∩ C = {1}
But, B ≠ C.

Q.13 Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A = B.


Given: A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X
Since,  A = A ∩ ( A ∪ X )
= A ∩ ( B ∪ X ) [ ∵ A ∪ X = B ∪ X ]
= ( A ∩ B ) ∪ ( A ∩ X )
= ( A ∩ B ) ∪ ϕ [ ∵ A ∩ X = ϕ ]
A = ( A ∩ B ) …(i)
And      B = B ∩ ( B ∪ X )
= B ∩ ( A ∪ X ) [ ∵ A ∪ X = B ∪ X ]
= ( B ∩ A ) ∪ ( B ∩ X )
= ( A ∩ B ) ∪ ϕ [ ∵ B ∩ X = ϕ ]
B = ( A ∩ B ) …(ii)
From (i) and (ii), we have A = B

Q.14 Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ.


We can show these conditions on sets A, B and C by taking an example:
Let A = {1, 2, 3}, B = {1, 2, 6}, C = {6, 3, 7}.
Then, A ∩ B = {1, 2}, B ∩ C = {6} and C ∩ A = {3}
Therefore, A ∩ B, B ∩ C and C ∩ A are non-empty.
So, A ∩ B ∩ C = Φ.
Therefore, A = {1, 2, 3}, B = {1, 2, 6}, C = {6, 3, 7}.

Q.15 Find the mean deviation about the mean for the data in Exercises 1 and 2.
Q.1: 4, 7, 8, 9, 10, 12, 13, 17
Q.2: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44



Mean=4+7+8+9+10+12+13+178        =808        =10Mean deviation={|410|+|710|+|810|+|910|+|1010|+|1210|+|1310|+|1710|8}       =6+3+2+1+0+2+3+78       =248       =3

Thus, the mean deviation is 3.


Mean=38+70+48+40+42+55+63+46+54+4410       =50010       =50Mean deviation={|3850|+|7050|+|4850|+|4050|+|4250|+|5550|+|6350|+|4650|+|5450|+|4450|10}        =12+20+2+10+8+5+13+4+4+610        =8410=8.4Thus, the mean deviation is 8.4.

Q.16 Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49


3. Ascending order of data:10,11,11,12,13,13,14,16,16,17,17,18Here, n=12(even)Median=(n2)thterm+(n2+1)thterm2        =(122)thterm+(122+1)thterm2       =6thterm+7thterm2       =13+142       =272Median=13.5Mean deviation about median      ={|1013.5|+|1113.5|+|1113.5|+|1213.5|+|1313.5|+|1313.5|+|1413.5|+|1613.5|+|1613.5|+|1713.5|+|1713.5|+|1813.5|12}      =(3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5)12      =2812      =2.33(Approx.) 4. Ascending order of data:36,42,45,46,46,49,51,53,60,72 Here, n=10(even)Median=(n2)thterm+(n2+1)thterm2      =(102)thterm+(102+1)thterm2      =5thterm+6thterm2      =46+492      =952Median=47.5Mean deviation about median      ={|3647.5|+|4247.5|+|4547.5|+|4647.5|+|4647.5|+|4947.5|+|5147.5|+|5347.5|+|6047.5|+|7247.5|10}      =(11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5)10      =7010=7

Q.17 5. Find the mean deviation about the mean for the data in Exercises 5 and 6.

Xi 5 10 15 20 25
fi 7 4 6 3 5


Xi 10 30 50 70 90
fi 4 24 28 16 8


xififixixix¯fixix¯5735514=963104401014=416156901514=16203602014=6182551252514=1155Total25350158 Meanx¯=fixifi           =35025           =14Mean deviation about mean           =fixix¯fi           =15825=6.32Thus, the mean deviation about mean is 6.32. xififixixix¯fixix¯104401050=4016030247203050=20480502814005050=00701611207050=203209087209050=40320Total8040001280 Meanx¯=fixifi          =400080          =50Mean deviation about mean          =fixix¯fi          =128080=16Thus, the mean deviation about mean is 16.

Q.18 Find the mean deviation about the median for the data in Exercises 7 and 8.

Xi 5 7 9 10 12 15
fi 8 6 2 2 2 6
Xi 15 21 27 30 35
fi 3 5 6 7 8



n = 26evenMedian = n2th term+n2+1th term2         = 262th term+262+1th term2         = 13th term+14th term2         = 7+72         = 7Mean deviation about median         = fixi-Mefi         = 8426         = 3.23Thus, Mean deviation about median is 3.23.


xifiC.f.xiMedianfixiMedian15331530=154521582130=945276142730=318307213030=00358293530=540Total29148 n = 29oddMedian= n+12th term        = 29+12th term        = 15th term        = 30Mean deviation about Median        = fixi-Mefi        = 14829        = 5.1Thus, the Mean deviation about Median is 5.1.

Q.19 Find the mean deviation about the mean for the data in Exercises 9 and 10.


Income per day 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
Number of persons 4 8 9 10 7 5 4 3
Height in cms 95-105 105-115 115-125 125-135 135-145 145-155
Number of boys 9 13 26 30 12 10



Incomefixiui=xAhfiuixix¯fixix¯010045041650358=30812321002008150324150358=20816642003009250218250358=10897230040010350110350358=8804005007450=A00450358=92644500600555015550358=192960600700465028650358=2921168700800375039750358=3921176Total50467896 Mean=A+h×fiuifi, where ui=xiAh and h=100      =450+100×4650      =4502×46      =45092      =358Mean deviation about mean        =fixix¯fi        =789650        =157.92


Classintervalfixiui=xiAhfiuixix¯fixix¯951059100327100125.3=25.3227.710511513110226110125.3=15.3196.911512526120126120125.3=5.3137.812513530130=A00130125.3=4.714113514512140112140125.3=14.7176.414515510150220150125.3=24.7247Total100471128.8 Mean=A+h×fiuifi, where ui=xiAh and h=10      =130+10×47100      =1304.7      =125.3Mean deviation about Mean      =fixix¯fi      =1128.8100      =11.288      =11.29ApproxThus, Mean deviation about Mean is 11.29.

Q.20 Find the mean deviation about median for the following data :

Marks 0-10 Oct-20 20-30 30-40 40-50 50-60
Number of Girls 6 8 14 16 4 2


Marksfixic.f.|xiMe|fi|xix¯|010656|527.85|=22.85137.1102081514|1527.85|=12.85102.82030142528|2527.85|=2.8539.93040163544|3527.85|=7.15114.4405044548|4527.85|=17.1568.6506025550|5527.85|=27.1554.3Total50517.1N=50(even)N2=502=25So, the median class is 2030.Median=l+(N2)Cf×h       =20+251414×10[C=14,f=14​  and h=10]       =20+7.85=27.85Mean deviation about Meadian    =517.150   =10.34Thus, Mean deviation about Meadian is 10.34.

Q.21 Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Age 16-20 21-25 36-30 31-35 36-40 41-45 46-50 51-55
Number 5 6 12 14 26 12 16 9


The given table can be arranged as given below:

Marksfixic.f.|xiMe|fi|xix¯|15.520.55185|1838|=2010020.525.562311|2338|=159025.530.5122823|2838|=1012030.535.5143337|3338|=57035.540.5263863|3838|=0040.545.5124375|4338|=56045.550.5164891|4838|=1016050.555.5953100|5338|=15135Total100735 N=100(even)N2=1002=50So, the median class is 35.540.5. Median=l+(N2)Cf×h       =35.5+503726×5[C=37,f=26​  and h=5]       =35.5+2.5=38Mean deviation about Meadian   =fi|xiMe|fi   =735100   =7.35Thus, Mean deviation about Meadian is 7.35.

Q.22 Find the mean and variance for each of the data in Exercies 1 to 5.

Q1. 6, 7, 10, 12, 13, 4, 8, 12
Q2. First n natural numbers.
Q3. First 10 multiples of 3.

Xi 6 10 14 18 24 28 30
fi 2 4 7 12 8 4 3
Xi 92 93 97 98 102 104 109
fi 3 2 3 2 6 3 3



Mean(x¯)=6+7+10+12+13+4+8+128 =728 =9 xi(xix¯)(xix¯)2669=39779=2410109=1112129=3913139=416449=525889=1112129=39Total74Variance(σ2)=(xix¯)2N =748 =9.25


Mean of n natural numbers=i=1nxin=(n(n+1)2)n=n+12Variance(σ2)=i=1nxi2(x¯)2n =(12+22+32+...+n2)n(n+12)2=1n×n(n+1)(2n+1)6(n2+2n+14)Variance(σ2)=2n2+3n+16(n2+2n+14)=4n2+6n+23n26n312=n2112


Mean of 10 multiples of 3          =(3+6+9+12+15+18+21+24+27+30)10          =16510=16.5Variance(σ2)=i=1nxi2(x¯)2n          =(32+62+92+122+152+182+212+242+272+302)10(16.5)2           =346510272.25          =346.5272.25         Variance(σ2)=74.25


xififixi(xix¯)(xix¯)2fi(xix¯)26212619=13169338104401019=981324147981419=52517518122161819=11122481922419=5252002841122819=981324303903019=11121363Total407601736Mean (x¯)=i=1nfixin =76040 =19 Variance (σ2)=i=1nfi(xix¯)2n Variance(σ2)=173640 =43.4


xififixi(xix¯)(xix¯)2fi(xix¯)292327692100=86419293218693100=7499897329197100=392798219698100=2481026612102100=24241043312104100=416481093327109100=981243Total222200640Mean(x¯)=i=1nfixin = 220022 =100Variance(σ2)=i=1nfi(xix¯)2n Variance(σ2)=64022 =29.09

Q.23 Find the mean and standard deviation using short-cut method.

Xi 60 61 62 63 64 65 66 67 68
fi 2 1 12 29 25 12 10 4 5


xifidi=xiA1fidixix¯xix¯2fixix¯2602484163261133399621222424486329129112964=A25000006512112111266102202440674312393668542041680Total1000286 Meanx¯=A+i=1nfidin         =64+0100=64Varianceσ2=i=1nfixix¯2n              =286100              =2.86Standard deviation              =2.86              =1.69 Approx.

Q.24 Find the mean and variance for the following frequency distributions in Exercises 7 and 8.


Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210
Frequencies 2 3 5 10 3 5 2
Classes 0-10 10-20 20-30 30-40 40-50
Frequencies 5 8 15 16 6



Classesfixiui=xiA30ui2fiuifiui203021539618306034524612609057511559012010105=A0000120150313511331501805165241020180210219539618Total30276 Mean=A+ i=1 n f i u i n ×h =105+ 2 30 ×30 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaaxMaacaWLjaGaamytaiaadwgacaWGHbGaamOBaiabg2da9iaadgeacqGHRaWkdaWcaaqaamaaqahabaGaamOzamaaBaaaleaacaWGPbaabeaakiaadwhadaWgaaWcbaGaamyAaaqabaaabaGaamyAaiabg2da9iaaigdaaeaacaWGUbaaniabggHiLdaakeaacaWGUbaaaiabgEna0kaadIgaaeaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7cqGH9aqpcaaIXaGaaGimaiaaiwdacqGHRaWkdaWcaaqaaiaaikdaaeaacaaIZaGaaGimaaaacqGHxdaTcaaIZaGaaGimaaaaaa@5F1B@ =107   Variance(σ2)=h2n2[nin(fiui2)(infiui)2]    =(30)2(30)2[30×76(2)2]   =22804=2276


Classesfixiui=xiA10ui2fiuifiui2010552410201020815118820301525=A0000304016351116164050645241224Total501068Mean=A+i=1nfiuin×h    =25+1050×10    =27   Variance(σ2)=h2n2[nin(fiui2)(infiui)2]     =(10)2(50)2[50×68(10)2] =125[3400100]   =132

Q.25 Find the mean, variance and standard deviation using short-cut method

Height in cms 70-75 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115
No. of children 3 4 7 7 15 9 6 6 3


Classes f i x i u i = x i A 5 u i 2 f i u i f i u i 2 7075 3 72.5 4 16 12 48 7580 4 77.5 3 9 12 36 8085 7 82.5 2 4 14 28 8590 7 87.5 1 1 7 7 9095 15 92.5=A 0 0 0 0 95100 9 97.5 1 1 9 9 100105 6 102.5 2 4 12 24 105110 6 107.5 3 9 18 54 110115 3 112.5 4 16 12 78 Total 60 6 254 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DD5F@ Mean=A+i=1nfiuin×h =92.5+660×5    =92.5+0.5    =93Variance(σ2)=h2n2[nin(fiui2)(infiui)2]    =(5)2(60)2[60×254(6)2]    =253600[1524036]    =253600×15204    =105.52Standard deviation   =105.52   =10.27

Q.26 The diameters of circles (in mm) drawn in a design are given below:

Diameters 33-36 37-40 41-44 45-48 49-52
No. of circles 15 17 21 22 25

Calculate the standard deviation and mean diameter of the circles.


Classe-intervafixiui=xi-A4ui2fiufiui232.5-36.51534.5-24-306036.5-40.51738.5-11-17-1740.5-44.52142.5=A000044.5-48.52246.511222248.5-52.52550.52450100Total10025199  Varianceσ2 = h2n2ninfiui2infiui2                  = 421002100×199-252                  = 162519900-625                  = 1625×19275                  = 30.84Standard deviation                  = 30.84                  = 5.55           Mean = A+i=1nfiuin×h                  = 42.5+25100×4                  = 42.5+1                  = 43.5

Q.27 From the data given below state which group is more variable, A or B?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Group A 9 17 32 33 40 10 9
Group B 10 20 30 25 43 15 7


For group A:

Marks(A)fixiui=xiA10ui2fiuifiui21020915392781203017252434683040323511323240503345=A0000506040551140406070106524204070809753927811506342  Mean(x¯)=A+i=1nfiuin×h=45+6150×10=450.4=44.6Variance(σ2)=h2n2[nin(fiui2)(infiui)2] Variance(σ2)=(10)2(150)2[150×342(6)2]          =1225(5130036)      =1225×51264          =227.84Standard deviation(σ)          =227.84          =15.09 Marks(A)fixiui=xiA10ui2fiuifiui210201015393090203020252440803040303511303040502545=A0000506043551143436070156524306070807753921631506366 Mean(x¯)=A+i=1nfiuin×h=45+6150×10=450.4=44.6 Variance(σ2)=h2n2 [nin(fiui2)(infiui)2]Variance(σ2)=(10)2(150)2 [150×366(6)2]         =1225(5490036)         =1225×54864         =243.84Standard deviation(σ)         =243.84         =15.61

Since, both groups have equal mean. Group B has greater standard deviation. So, group B is more variable than group A.

Q.28 From the prices of shares X and Y below, find out which is more stable in value:

X 35 54 52 53 56 58 52 50 51 49
Y 108 107 105 105 106 107 104 103 104 101


For prices of share X:

Mean(x¯)=35+54+52+53+56+58+52+50+51+4910         =51010         =51 xi(xix¯)(xix¯)2353551=16256545451=39525251=11535351=24565651=525585851=749525251=11505051=11515151=00494951=24350Variance(σ2)=i=1n(xix¯)2N=35010=35Standard deviation(σ)=35=5.91C.V.(share X)=σx¯×100=5.9151×100=11.59(Appox.)

For prices of share Y:

Mean(y¯)=108+107+105+105+106+107+104+103+104+10110         =105010         =105yi(yiy¯)(yiy¯)2108108105=39107107105=24105105105=00105105105=00106106105=11107107105=24104104105=11103103105=24104104105=11101101105=41640Variance(σ2)=i=1n(xix¯)2N=4010=4Standard deviation(σ)=4 =2C.V.(share Y)=σy¯×100=2105×100=1.9<11.59C.V. of prices of share Y is lesser than that of share X.Thus, the prices of share Y are more stable than that of share X.

Q.29 An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

Firm A Firm B
No. of wage earners 586 648
Mean of monthly wages Rs. 5253 Rs. 5253
Variance of distribution of wages 100 121

1. Which firm A or B pays larger amount as monthly wages?
2. Which firm, A or B, shows greater variability in individual wages?


i.         Monthly wage of firm A = Rs. 5253     Number of workers in firm A = 586     Total paid amount in wages = Rs. 5253 × 586                                       = Rs. 3078258           Monthly wage of firm B = Rs. 5253     Number of workers in firm A = 648      Total paid amount in wages = Rs. 5253 x 648                                        = Rs.3403944Thus, firm B pays the larger amount as monthly wages.ii. Variance of distribution of wages in firm A = 100               Standard deviation of wages in firm A = 100                                                      = 10       Variance of distribution of wages in firm B = 121                 Standard deviation of wages in firm B = 121                                                        = 11Since the average monthly wages in both the firms is same,i.e., Rs. 5253, therefore, the firm with greater standard deviationwill have more variability.Thus, the firm B has greater variability in the individual wages.

Q.30 The following is the record of goals scored by team A in a football session:

No. of goals scored 0 1 2 3 4
No. of matched 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?


No. of goals x i No. of matches f i x i x ¯ x i x ¯ 2 f i x i x ¯ 2 0 1 02=2 4 4 1 9 12=1 1 9 2 7 22=0 0 0 3 5 32=1 1 5 4 3 42=2 4 12 25 30 Mean = i=1nfixin       = 5025       = 2Variationσ2= i=1nfixix¯2n= 3025= 1.2Standard deviation= 1.2= 1.09Standard deviation of team B= 1.25 goalsMean of team B= 2The team with lower standard deviation is more consistent.Therefore, team A is more consistent.

Q.31 The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

i=150xi=212,  i=150xi2=902.8,   i=150yi=261,  i=150yi2=1457.6Which is more varying, the length or weight?


i=150xi=212,  i=150xi2=902.8,Mean=i=150xi50            =21250            =4.24Variance(σ2)=i=150xi250(i=150xi50)2                    =902.850(21250)2                    =18.05617.978                    =0.078Standandard deviation(σ)                   =0.078                   =0.28C.V. of length=σx¯×100                     =6.6i=150yi=261,  i=150yi2=1457.6Mean(y¯)=i=150yi50                  =26150  =5.22Variance(σ2)=i=150yi250(i=150yi50)2                    =1457.650(26150)2                    =29.1527.25                    =1.89Standandard deviation (σ)                    =1.89                    =1.37C.V. of weight=σy¯×100                    =1.375.22×100                    =26.24 Since, C.V. of weights is greater than the C.V. of lengths.Therefore, weights are more varing than lengths.

Q.32 The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.


Let remaining two observations be x and y.

Therefore, the series is 6, 7, 10, 12,12,13,x and y.Then, Mean=6+7+10+12+12+13+x+y8 9=60+x+y8        72=60+x+y  x+y=7260  x+y=12 ...(i)Variance=i=18(xix¯)28    9.25={(69)2+(79)2+(109)2+(129)2+(129)2+(139)2+(x9)2+(y9)2}8    74=9+4+1+9+9+16+x218x+81+y218y+81    74=210+x218(x+y)+y2    74=210+x218(12)+y2 [From eqation (i)]    74=210+x2216+y2    80=x2+y2     x2+y2=80 ...(ii) Squarring equation (i), we get MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaWGtbGaamyCaiaadwhacaWGHbGaamOCaiaadkhacaWGPbGaamOBaiaadEgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gadaqadaqaaiaadMgaaiaawIcacaGLPaaacaGGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaaa@52AF@     (x+y)2=(12)2x2+2xy+y2=144         2xy=144(x2+y2)         2xy=14480         2xy=64 ...(iii)Now,(xy)2=x2+y22xy       =8064(xy)2=16      xy=±4 ...(iv)From equation (i) and (iv), we get      x=8 and y=4(By taking, xy=4)orx=4 and y=8(By taking, xy=4)Thus, the remaining two observations are 4 and 8.

Q.33 The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.


Let remaining two observations be x and y.

Therefore, the series is 2, 4, 10, 12,14,x and y.Then, Mean=2+4+10+12+14+x+y7 8=42+x+y7      56=42+x+y    x+y=5642             x+y=14 ...(i)         Variance=i=18(xix¯)27        16={(28)2+(48)2+(108)2+(128)2+(148)2+(x8)2+(y8)2}7  112=36+16+4+16+36+x216x+64+y216y+64  112=236+x216(x+y)+y2  112=236+x216(14)+y2[From eqation (i)]  112=236+x2224+y2  100=x2+y2     x2+y2=100 ...(ii)Squarring equation (i), we get   (x+y)2=(14)2x2+2xy+y2=196       100+2xy=196 [From equation (i)]          2xy=96 ...(iii)Now,(xy)2=x2+y22xy       =10096(xy)2=4      xy=±2 ...(iv) From equation (i) and (iv), we get      x=8 and y=6(By taking, xy=2)or  x=6 and y=8(By taking, xy=2)Thus, the remaining two observations are 6 and 8.

Q.34 The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.


       Mean of 6 observations=8          Sum of 6 observations=6×8          =48When each observation is multiplied by 3,New mean=(3x6)=3×486          =24Standard deviation=4Variance of 6 observations =16       x26(x6)2=16              x26(8)2=16                    x26=16+64       x2=6×80           =480When each observation is multiplied by 3,  (3x)2=9x2          =9×480New variance of 6 observations         =(3x)26(3x6)2         =9×4806(24)2         =720576         =144New standard deviation=144         =12Thus, new mean and new standard deviatiion of 6 observations are 24 and 12 respectively.


Given that x¯ is the mean and σ2 is the variance of n observations x1, x2, ...,xn. Prove that the mean and variance of the observations ax1, ax2, ax3, ., axn are ax¯ and a2σ2, respectively, (a¹0).


              Mean of n observations = x¯                      x1+x2+x3+...+xnn  = x¯ ...i      Variance of n observations  = σ2 x12+x22+x32+...+xn2nx1+x2+x3+...+xnn2 = σ2x12+x22+x32+...+xn2nx¯2 = σ2  x12+x22+x32+...+xn2n = σ2+x¯2 ...iiMean of ax1, ax2, ax3, ., axn = ax1+ ax2+ ax3+ .+ axnn      = ax1+ x2+ x3+ .+ xnnMean of ax1, ax2, ax3, ., axn = ax¯Variance of ax1, ax2, ax3, ., axn= a2x12+a2x22+a2x32+...+a2xn2n - ax1+ax2+ax3+...+axnn2= a2x12+x22+x32+...+xn2n - a2x1+x2+x3+...+xnn2= a2σ2+x¯2-a2x¯2= a2σ2+a2x¯2-a2x¯2  = a2σ2Thus, the mean and the variance of ax1,ax2,ax3,.,axn are ax¯ and a2σ2.

Q.36 The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted.
(ii) If it is replaced by 12.


      Mean value of 20 observations=10         Total sum of 20 observations=200Standard deviation of 20 observations=2  Variance of 20 observations=4    x220(x20)2=4        x220(10)2=4                       x220=100+4                       x2=104×20      Incorrect observation=8(i)If wrong item is omitted, then Remaining observations=201 =19 New mean of remaining observations=200819=19219 =10.1New variance of remaining observations =x28219(x819)2 =104×208219(200819)2 =201619(10.1)2 =106.1102.01 =4.09New standard deviation of remaining observations =4.09 =2.02(ii) When 8 is replaced by 12,                New mean of 20 observations=2008+1220 =20420 =10.2New variance of 20 observations =x282+12220(x8+1220)2New variance of 20 observations=104×2082+12220(2008+1220)2 =208064+14420(10.2)2  =216020104.04  =108104.04New variance of 20 observations=3.96New standard deviation=1.9899   =1.99(Approx)

Q.37 The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
Subject Mathematics Physics Chemistry
Mean 42 32 40.9
Standard 12 15 20
Which of the three subjects shows the highest variability in marks and which shows the lowest?


C.V. of marks in Mathematics=σMean×100     =1242×100     =28.57            C.V. of marks in Physics=σMean×100      =1532×100     =46.87    C.V. of marks in Chemistry=σMean×100     =2040.9×100     =48.89Since, C.V. of Chemistry is the greatest and C. V. of Mathematics is the lowest.So, highest variability is in the marks of Chemistry and lowest variability is in the marks of Mathematics.

Q.38 The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.


Mean value of 100 observations=20Total sum of 100 observations=2000Standard deviation of 100 observations=3Variance of 100 observations=9      x2100(x100)2=9         x2100(20)2=9                        x2100=400+9                       x2=409×100   Incorrect observations=21,21,18If wrong observations are omitted, then Remaining observations=1003 =97New mean of remaining observations=200021211897 =194097 =20New variance of remaining observations =x221221218297(x21211897)2 =4090044144132497(20)2 =3969497400 =409.216400 =9.216      New standard deviation=9.216 =3.036

Thus, the new mean and new standard deviation are 20 and 3.036 respectively.

Q.39 Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) – 47°30′
(iii) 240°
(iv) 520°


Since, 1°=π180°i    25°=25°×π180°             =5π36°   radiansii47°30=47.5° 30=0.5°                  =47.5°×π180°  radians                   =19π72  radiansiii       240°=240°×π180°  radians                   =4π3  radiansiv       520°=520°×π180°  radians                   =26π9  radians


Find the degree measures corresponding to the followingradian measures  Use π=227.i1116ii4iii5π3iv7π6


Since, 1 radian=180°π(i)1116  radians=1116×180°π =114×45°π =114×45°×722 =14×45°×72 =(3938)° =39°+38×60 =39°+22+12 =39°+22+12×60 =39°2230 (ii)4  radians=4×180°π =4×180°×722 =(229111)° =(229°+111×60) =(229°+5+511×60) =(229°+5+27) =229°527 (Approx.)(iii) 5π3  radians=5π3×180°π =53×180° =300°(iv)7π6  radians=7π6×180°π =76×180° =210°

Q.41 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?


Number of revolutions made by a wheel in 1 minute=360Number of revolutions made by a wheel in 1 second=36060                                                                         =6Angle covered in 1 revolution by wheel=2π RadiansAngle covered in 6 revolutions by wheel=6×2π                                                         =12π RadiansThus, wheel turns 12π radians in 1 second.


Find the degree measure of the angle subtended at thecentre of a circle of radius 100 cm by an arc of length 22 cm(Use π=227).


Radius of circle=100 cm   Length​ of arc =22cmAngle(θ) subtended at the centre by an arc    =ArcRadiusradiansθ=22100radiansθ=22100×180°π[ 1radian=180°π]θ=22100×180°×722    =1.8°×7    =12.6°    =12°+0.6×60     =12°+36    =12°36Thus, angle subtended by an arc of 22 cm at the centre is 12°36′ .

Q.43 In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.


Diameter of a circle=40cm      Radius​ of a circle=402cm =20cm          Length of chord=20cm

Since,ΔOAB is equilateral triangle.So,​            AOB=60° =π3 radiansSince,        angle=Length of arcRadius π3 Radian=Length of arc AB20 cm Length of arc AB=20 cm × π3 =20π3 cm

Q.44 If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.


Let radii of circles be r1 and r2 respectively.Angle subscribed by arc in first circle=60°  =60°×π180°  radians  =π3  radiansAngle subscribed by arc in second circle  =75°  =75°×π180°  radians  =5π12  radiansSince, length of both arcs is same.So,             Length of arc of first circle=Length of arc of second circleπ3  radian×r1=5π12  radian×r2      r1r2=(5π12)(π3) =54 r1:r2=5:4Thus, the ratio of radii of circles is 5:4.

Q.45 Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm


Length of pendulum=75 cmiLength of arc described by pendulum=10cmAngle swept by pendulum=1075Rad. θ=arcr                                     =215radiansiiLength of arc described by pendulum=15cmAngle swept by pendulum=1575Rad. θ=arcr                                     =15radiansiiiLength of arc described by pendulum                                     =21cmAngle swept by pendulum=2175Rad. θ=arcr                                     =725radians


Find the values of other five trigonometric functions in Exercises 1 to 5.

1.cos x=12, x lies in third quadrant. 2.sin x=35, x lies in second quadrant. 3.cot x=34, x lies in third quadrant. 4.sec x=135, x lies in fourth quadrant. 5.tan x=512, x lies in second quadrant.


1.secx=1cosx       =112       =2xliesinthirdquadrant.sin2x+cos2x=1             sinx=1cos2x                   =1122            sinx=34sinx=32  x lies in third quadrant.cosecx=1sinx          =132          =23   tanx=sinxcosx          =3212          =3x lies in third quadrant.   cotx=1tanx          =13x lies in third quadrant.Thus, sinx=32, cosecx=23,secx=2,tanx=3andcotx=13. 2. sinx=3/5cosecx=1/sinx           =135cosecx=53sin2x+cos2x=1cosx=±1sin2x        =±1352cosx=±1625        =45           x lies in II quadrant.secx=1cosx        =145        =54tanx=sinxcosxtanx=3545        =34  x lies in II quadrant.cotx=1tanx        =134        =43        x lies in II quadrant.Thus, cos x=45,secx=54 , cosecx=53,tan x=34andcotx=43. 3.tanx=1cotx       =134       =43cosec2x=1+cot2x            =1+342            =1+916cosecx=±2516           =54x lies in third quadrant.    sinx=1cosecx           =154           =45   secx=1+tan2x           =1+432           =1+169   secx=±259           =53x lies in third quadrant.   cos x=35x lies in third quadrant.Thus,  sinx=45,cosecx=54,cosx=35,secx=53andtanx=43. 4.cos x=1sec x       =1135       =513 x lies in fourth quadrant.sin x=1cos2x       =±15132       =1213 x lies in fourth quadrant.cosec x=1sin x           =11213           =1312    tan x=sin xcos x =(1213)(513)=125 [x lies in fourth quadrant.]cot x=1tanx =1(125) =512Thus, sin x=1213,cos x=513,tan x=125,cot x=512and cosec x=1312. 5.cot  x=1512 cot x=1tan  x        =125sec2x=1+tan2x         =1+5122         =1+25144secx=±169144        =1312 x lies in second quadrant.cosx=1secx        =11312        =1213 x lies in second quadrant.cosec2x=1+cot2x           =1+1252           =1+14425cosecx=±16925            =135 x lies in second quadrant.     sinx=1cosecx             =1135sinx=513 x lies in second quadrant.Thus, sinx=513,cosx=1213,cotx=125,secx=1312andcosecx=135.

Q.47 Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°
7. cosec (– 1410°)

8. tan19π3 9. sin(11π3) 10. cot(15π4)


6.sin 765° = sin 2 x 360° + 45°               = sin 45° sinn×360°+θ=sinθ               =12 7.cosec 1410°=cosec1410°        cosecθ=cosecθ   =cosec4×360°30°   =cosec 30°cosec n×360°θ=cosecθ   =2 8.tan19π3=tan6π+π3 tan19π3=tanπ3[ tan (2+θ)=tanθ]       =3 9.sin11π3=sin11π3 sinx=sin x    =sin4ππ3    =sinπ3         sin 2x=sin xsin11π3=32 10.cot 15π4=cot 15π4 cot x=cot x    =cot 4ππ4    =cotπ4 cot 2x=cot x    =1


Prove that:sin2π6+cos2π3tan2π4=12


L.H.S.:sin2π6+cos2π3tan2π4=(12)2+(12)2(1)2=14+141=1+144=24=12=R.H.S.Hence proved.


Prove that:2sin2(π6)+cosec2(7π6)cos2(π3)=32


L.H.S.:2sin2π6+cosec27π6cos2π3=2(12)2+cosec2(π+π6)×(12)2      =2×14+14×cosec2(π6)       =12+14×(2)2      =12+1      =32=R.H.S.Hence proved.


Prove that:cot2π6+cosec5π3+3tan2π6=6


L.H.S:cot2π6+cosec5π6+3tan2π6=(3)2+cosec(ππ6)+3(13)2        =3+cosec(π6)+3×13[cosec(πx)=cosec x]        =3+2+1        =6=R.H.S.Hence proved.


Prove that:2sin2(3π4)+2cos2(π4)+2sec2(π3)=10


L.H.S.:2sin2(3π4)+2cos2(π4)+2sec2(π3)=2sin2(ππ4)+2(12)2+2(2)2 =2sin2(π4)+2(12)+8=2(12)2+1+8=2(12)+1+8=1+1+8=10=R.H.S.

Q.52 Find the value of: (i) sin75° (ii) tan15°


(i) sin 75°=sin(45°+30°)=sin 45° cos 30° + sin 30° cos 45°[ sin(x+y)=sin x cos y+sin y cos x]=12×32+12×12=322+122=3+122(ii) tan 15°=tan (45°30°)=tan 45° tan 30°1 + tan 45° tan 30°=1131+1×13 =313+1×3131=(31)231=323+12   tan 15°=4232=23


cos (π4x)cos(π4y)sin(π4x)sin(π4y)=sin (x+y)


Since,cosA cosB sinA sinB = cos (A + B)L.H.S. =cos (π4 x) cos (π4 y) sin (π4 x) sin (π4 y)= cos {(π4 x) + (π4 y)}= cos (π2 x y)= cos {π2 (x + y)}= sin (x + y) [ cos (π2 θ) = sin θ]= R.H.S. Hence proved.


Prove the following:tan (π4 + x)tan (π4 x) = (1 + tan x1 tan x)2


L.H.S:tan (π4 + x)tan (π4 x) = (tan π4 + tan x1 tan π4 tan x)(tan π4 tan x1 + tan π4 tan x)      [ tan (A + B) = tan A + tan B1 tan A tan Btan (A B) = tan A tan B1 + tan A tan B]= (1 + tan x1 1.tan x)(1 tan x1 + 1.tan x)= (1 + tan x1 tan x) × (1 + tan x1 tan x)= (1 + tan x1 tan x)2= R.H.S. Hence proved.


Prove the following:cos(π+x)cos(x)sin(πx)cos(π2+x)=cot2x


L.H.S.:cos(π+x)cos(x)sin(πx)cos(π2+x)=cos x cos xsin x×sin x[cos(π+x)=cosx         cos(x)=cosx      sin(πx)=sinx    cos(π2+x)=sinx]    =cot2x    =R.H.S.Hence proved.


cos(3π2 +x)cos (2π+x)[cot(3π2x)+cot(2π + x)] = 1


L.H.S. = cos (3π2+ x)cos(2π+x)[cot (3π2 x) +cot (2π+x)]= sin x.cosx [tanx +cot x]= sin x.cos x [sinxcosx+cosxsinx]= sin x.cos x[sin2x +cos2xsin x.cosx] = sin2x+ cos2x=1 =R.H.S.

Q.57 sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos x


L.H.S.:sin (n + 1) x sin (n +2)x + cos (n +1)x cos (n + 2) x           = cos (n +1)x cos (n +2)x +sin (n +1) x sin (n + 2) x           = cos {(n +1) x (n+2)x}[ cosA cosB sinA sinB=cos(A +B)]           =cos (nx + x nx2x)           =cos (x)                  [ cos (A)=cosA]           =cos x=R.H.S.


cos(3π4 + x) cos(3π4 x) =2 sin x


L.H.S.:cos(3π4+x)cos(3π4 x)=2sin((3π4 +x)+(3π4x)2) sin((3π4 +x)(3π4 x)2)=2 sin (2(3π4)2)sin (2x2) =2sin(π π4)sin x=2 sin (π π4)sinx=2 sinπ4×sin x= 2× 12× sin x= 2sinx = R.H.S.

Q.59 Prove that:
sin2 6x – sin2 4x = sin 2x sin 10x


L.H.S.=sin2 6x sin 2 4x  =(sin 6x+sin4x)(sin6x sin 4x) [a2b2=(ab)(a+b)]  = {2sin(6x+ 4x2) cos(6x4x2)} {2 cos(6x + 4x2) sin(6x 4x2)}  ={2sin5x cos x}{2cos5xsin x}  =(2 sinx cos x) (2sin5xcos5x)  = sin2x.sin 10x [ sin 2x = 2 sin x cos x]=R.H.S.Hence proved.

Q.60 Prove that:
cos2 2x – cos2 6x = sin 4x sin 8x


L.H.S.= cos22x cos26x  =(cos2x+cos 6x) (cos 2xcos 6x)   ={2 cos(2x+ 6x2) cos (2x 6x2)}{2sin (2x 6x2)sin(2x+ 6x2)}  ={2 cos4xcos (2x)}{2sin(2x)sin 4x}  =(2sin2xcos2x)(2sin4xcos 4x)  =sin4x.sin8x[sin 2x=2 sinx cos x]=R.H.S.Hence proved.

Q.61 Prove that:
sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x


sin 2x + 2 sin 4x + sin 6x =2 sin 4x +sin 6x +sin 2x  = 2 sin 4x +2sin (6x+ 2x2) cos(6x 2x2)  =2 sin 4x + 2 sin4xcos2x  = 2 sin 4x (1 + cos2x)  = 2 sin 4x (2cos2x)  =4 cos2x sin 4x  = R.H.S.

Q.62 Prove that:
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)


We have, cot 4x (sin 5x +sin 3x)=cot x (sin 5x  sin 3x)The given equation can be written as:sin 5x+sin 3xsin 5xsin 3x=cot xcot 4x L.H.S:sin 5x +sin 3xsin 5x sin 3x =2 sin(5x+ 3x2)cos(5x 3x2)2 cos 5x+3x2 sin 5x 3x2  = sin 4xcos xcos4xsin x  = (cosxsin x)(cos 4xsin 4x)  =cot xcot4x  = R.H.S.Hence proved.


Prove that:cos9xcos5xsin17xsin3x=sin2xcos10x


L.H.S.:cos9xcos5xsin17xsin3x=2sin(9x+5x2)sin(9x5x2)2cos(17x+3x2)sin(17x3x2)[cosAcosB=2sin(A+B2)sin(AB2)      sinAsinB=2cos(AB2)sin(AB2)]   =sin(7x)sin(2x)cos(10x)sin(7x) =sin2xcos10x=R.H.S.Hence​ proved.




L.H.S.:sin5x+sin3xcos5x+cos3x=2sin(5x+3x2)cos(5x3x2)2cos(5x+3x2)cos(5x3x2)    =sin4xcosxcos4xcosx    =tan4x    =R.H.S.Hence proved.




L.H.S.:sinxsinycosx+cosy=2cosx+y2sinxy22cosx+y2cosxy2                 =sinxy2cosxy2                =tanxy2                =R.H.S.Hence proved.




L.H.S.:sinx+sin3xcosx+cos3x=sin3x+sinxcos3x+cosx=2sin(3x+x2)cos(3xx2)2cos(3x+x2)cos(3xx2)=sin2xcos2x=tan2x=R.H.S.Hence proved.


Provethat:sinx sin3xsin2x cos2x = 2sinx


L.H.S.:sinx sin 3xsin2xcos2x = (sin3x sin x)(cos2x sin2x)= 2cos (3x+ x2)sin (3x x2)cos2x=2cos2xsin xcos 2x=2sin x=R.H.S.


Provethat:cos 4x+cos3x+cos 2xsin4x +sin3x+ sin2x=cot 3x


L.H.S.:cos4x+ cos 3x + cos2xsin 4x + sin3x + sin 2x= cos 4x + cos 2x+ cos 3xsin 4x+ sin 2x+ sin 3x= 2cos(4x + 2x2) cos(4x 2x2) + cos 3x2sin (4x + 2x2) cos(4x 2x2)+ sin 3x= 2cos 3x cos x +cos3x2 sin 3x cos x+ sin 3x = cos 3x (2 cosx + 1)sin3x (2 cos x+ 1)cos 4x +cos3x + cos 2xsin 4x+ sin 3x+sin2x = cot3x =R.H.S. Hence​ proved.

Q.69 Prove that:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1


cot 3x = cot (2x + x) =cot 2x cot x 1cot 2x + cot xcot 3x (cot 2x + cot x)= cot 2x cot x 1cot 3x cot 2x + cot 3x cot x = cot 2x cot x 11 = cot x cot 2x cot 2x cot 3x cot 3x cot xor cot x cot 2x cot 2x cot 3x cot 3x cot x = 1.Hence​ proved.


Prove that:tan 4x = 4 tan x (1 tan2x)1 6 tan2x + tan4x


L.H.S:tan 4x = tan 2 (2x)    = 2 tan 2x1 tan2 2x     = 2 (2 tan x1 tan2x)1 (2 tan x1 tan2x)2    = (4 tan x1 tan2x){(1 tan2x)2 4 tan2x(1 tan2x)2}    = 4 tan x1 tan2x × (1 tan2x)21 2 tan2x + tan4x 4 tan2x    = 4 tan x(1 tan2x)1 6 tan2x + tan4x = R.H.S.Hence proved.

Q.71 Prove that:
cos 4x = 1 – 8sin2 x cos2 x


L.H.S. : cos 4x = cos 2 (2x)        = 1 2sin2 (2x) [ cos 2x = 1 sin2 x]        = 1 2(2 sin x cos x)2 [ sin 2x = 2 sin x cos x]        = 1 2 (4 sin2x cos2x)        = 1 8 sin2x cos2x        = R.H.S.

Q.72 Prove that:
cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1


L.H.S.=cos 6x=cos3(2x)=4cos32x3cos2x=4(2cos2x1)33(2cos2x1)=4(8cos6x12cos4x+6cos2x1)6cos2x+3=32cos6x48cos4x+24cos2x46cos2x+3=32cos6x24cos4x+18cos2x1=R.H.S.Hence proved.




Since, tanπ3=3 and tan4π3=tanπ+π3=tanπ3=3Therefore, principal solutions are x=π3 and 4π3.General solution of tanx=3tanx=3       =tanπ3tanx=tan+π3 x=+π3,where nZ.




Since, secx=2secπ3=2 and sec5π3=sec(2ππ3)=secπ3=2Therefore, principal solutions are x=π3 and 5π3.General solution:secx=2=secπ3=sec(2±π3)x=2±π3,where​ nZ.




Since, cot x=3     cot x=cotπ6      =cot(ππ6)      =cot(5π6)and     cot x=cot(2ππ6)      =cot(11π6)Therefore, the principal solution are x=5π6 and 11π6.General solution:              cotx=3     =cot(5π6)     =cot(+5π6)x=+5π6,where​ nZ.


Find the principal and general solutions of the followingequation : cosec x = 2


Since, cosec x = cosec (π6) cosec x = 2 = cosec (π6) = cosec (π + π6) = cosec (7π6)andcosec x = cosec (2π π6) = cosec 11π6Therefore, the principal solutions are 7π6, 11π6.General solution:cosec x = cosec (7π6) = cosec ( + (1)n 7π6) x = + (1)n 7π6, n Z.


Find the general solution for each of the followingequation : cos 4x = cos 2x


We​​  have,cos 4x = cos 2x = cos (2 ± 2x) [The general solution of cos θ = cos α θ = 2 ± α] 4x = 2 ± 2xFor(+) ve sign:     4x = 2 + 2x     2x = 2 x = For () ve sign:     4x = 2 2x     6x = 2 x = 13 The general solutions are x = 13 and , where n Z.


Find the general solution for each of the followingequation : cos 3x + cos x cos 2x = 0


We have, cos 3x + cos x cos 2x = 02 cos (3x + x2) cos (3x x2) cos 2x = 02 cos 2x cos x cos 2x = 0cos 2x (2 cos x 1) = 0 cos 2x = 0 2x = (2n + 1) π2      x = (2n + 1) π4, n Z or 2 cos x 1 = 0 cos x = 12 = cos π3      x = 2 ± π3, n Z.


Find the general solution for each of the followingequation : sin 2x + cos x = 0


We have, sin 2x + cos x = 0 2 sin x cos x + cos x = 0 cos x (2 sin x + 1) = 0Either cos x = 0 x = (2n + 1) π2, n Z.or (2 sin x + 1) = 0 sin x = 12 sin x = sin 7π2 x = + (1)n 7π2, n Z.Thus, the general solution of given equation is: + (1)n 7π2 ​  or (2n + 1) π2, n Z.




We have, sec2 2x = 1 tan 2x 1 + tan2 2x = 1 tan 2x tan2 2x + tan 2x = 0 tan 2x (tan 2x + 1) = 0Either tan 2x = 0 2x =        x = 2, where n  Z.or   tan 2x + 1 = 0 tan 2x = 1       tan 2x = tan π4      = tan (π π4)       tan 2x = tan 3π4       2x = + 3π4       x = 2 + 3π8, where n Z.The general solution of the given equation is:x = 2, 2 + 3π8, where n Z.


Find the general solution for each of the followingequation : sin x + sin 3x + sin 5x = 0


We have,   sin x+sin 3x+sin 5x=0   sin 5x+sin x+sin 3x=0   2sin(5x+x2)cos(5xx2)+sin 3x=0     2sin 3x cos 2x+sin 3x=0          sin 3x (2cos 2x+1)=0Either sin3x=03x=x=3,nZ.or 2 cos 2x+1=0cos 2x=12           2x=2±2π3             x=±π3,nZ.Therefore,​​ the general solution of the given equations is:x=3​​ or±π3,nZ.

Q.82 In any triangle ABC, if a = 18, b = 24, c = 30, find

1. cos A, cos B, cos C
2. sin A, sin B, sin C


1.Since,cosA=b2+c2a22bc                  =242+30218222430                  =576+9003241440                  =11521440                  =45cosB=c2+a2b22ca        =302+18224223018        =900+3245761080        =6481080        =35cosC=a2+b2c22ab         =182+24230221824         =324+576900864         =0864         =0

2.Since,cosA=b2+c2a22bc                  =242+30218222430                  =576+9003241440                  =11521440                  =45andsinA=1cos2A             =1452             =11625             =35Since,sinAa=sinBb=sinCc        3518=sinB24=sinC30        3518=sinB24 and 3518=sinC30      3×245×18=sinB and 3×305×18=sinC            45=sinB and  1=sinCThus, sinA=35, sinB=45, sinC=1.


For any triangle ABC, prove thata + bc = cos (A B2)sin C2


L.H.S. = a + bc=k sin A + k sin Bk sin C [ asin A = bsin B = csin C = k (let)]=sin A + sin Bsin C=2 sin (A + B2) cos (A B2)2 sin C2 cos C2=2 sin (π2 C2) cos (A B2)2 sin C2 cos C2=cos (C2) cos (A B2)sin C2 cos C2 =cos (A B2)sin C2 = R.H.S.Thus, it is proved.


For any triangle ABC, prove thata bc = sin (A B2)cos C2


L.H.S. = a bc=k sin A k sin Bk sin C [ asin A = bsin B = csin C = k (let)]=sin A sin Bsin C=2 cos (A + B2) sin (A B2)2 sin C2 cos C2=cos (π2 C2) sin (A B2)sin C2 cos C2=sin (C2) sin (A B2)sin C2 cos C2 =sin (A B2)cos C2 = R.H.S.Thus, it is proved.


For any triangle ABC, prove thatsin BC2 = bca cos A2


We have,         b ca=k sin B k sin Ck sin A      [asin A=bsin B=csin C=k (let)]     =sin B sin Asin A     =2 cos (B + C2) sin (B C2)2 sin A2 cos A2     =cos (π2 A2) sin (B C2)sin A2 cos A2     =sin (A2) sin (B C2)sin A2 cos A2     =sin (B C2)cos A2 sin (B C2) = b ca cos A2Therefore, it is proved.

Q.86 For any triangle ABC, prove that a (b cos C – c cos B) = b2 – c2


We have,bcosCccosB=b×a2+b2c22abc×a2+c2b22ac                        =a2+b2c22aa2+c2b22a                        =a2+b2c2a2c2+b22a                        =2b22c22aabcosCccosB=b2c2Hence, it is proved.


For any triangle ABC, prove thata(cosCcosB)=2(bc)cos2A2


We aregiventhat:   acosCcosB=2bccos2A2  cosCcosB2cos2A2=bcaL.H.S.=cosCcosB2cos2A2         =cosA2sinBC2cos2A2sinπ2θ=sinθ         =cosB+C2cosB+C2×sinBC2cosA2  Multiply and divide by cosB+C2         =2cosB+C22cosπ2A2×sinBC2cosA2          =2cosB+C2sinBC22sinA2cosA2          =sinBsinCsinA2cosx+y2sinxy2=sinxsinysinθ=2sinθ2cosθ2          =kbkckasinAa=sinBb=sinCc=k Let          =bca=R.H.S.Thus, acos Ccos B=2bccos2A2 is proved.


For any triangle ABC, prove thatsin(BC)sin(B+C)=b2c2a2


R.H.S.=b2c2a2  =k2sin2Bk2sin2Ck2sin2A [asinA=bsinB=csinC=k (let)]  =sin2Bsin2Csin2A  =sin(B+C)sin(BC)sin2A [ sin2xsin2y=sin(x+y)sin(xy)]   =sin(πA)sin(BC)sin2A [ B+C=πA]  =sin A sin(BC)sin2A [ sin(πA)=sinA]  =sin(BC)sin{π(B+C)} [ A=π(B+C)]  =sin(BC)sin(B+C) [ sin(πθ)=sinθ]  =L.H.S.Thus, it is proved.


For any triangle ABC, prove that(b+c) cos B + C2 = a cos BC2


We are given : (b + c) cos B + C2 = a cos B C2or               (b + c)a = cos B C2cos B + C2L.H.S.=  (b + c)a  =k sin B + k sin Ck sin A [ asin A = bsin B = csin C = k (let)]  =sin B + sin Csin A  =2 sin(B + C2) cos (B C2)2 sin A2 cos A2[sin x + sin y= 2 sin (x + y2) cos (x y2)]  =sin (π2 A2) cos (B C2)sin A2 cos A2  =cos A2 cos (B C2)sin (π2 B + C2) cos (A2) [ sin (π2θ) = cosθ]   =cos (B C2)cos (B + C2) = R.H.S.Thus,  (b + c) cos B + C2 = a cos B C2 is proved.

Q.90 For any triangle ABC, prove that a cos A + b cos B + c cos C = 2 a sin B sin C


L.H.S.=a cos A+b cos B+c cos C  =k sin A cos A+k sin B cos B+k sin C cos C[ asinA=bsinB=csinC=k (let)]  =k2(2 sin A cos A+2 sin B cos B+2 sin C cos C)  =k2 (sin 2A+sin 2B+sin 2C)  =k2 (2 sin2A+2B2.cos2A2B2+sin 2C)  =k2 {2 sin (A+B).cos(AB)+sin 2C}  =k2 {2 sin (πC).cos(AB)+2 sin C cos C}[ A+B+C=π]  =k2 {2 sin C.cos (AB)+2 sin C cos(πAB)}[ sin(πθ)=sinθ]   = k sin C {cos (AB)cos (A+B)}  =k sin C[2sin(AB+A+B2)sin{A+B(AB)2}]  =2 k sin C.sin A.sin B  =2 (k sin A).sin B.sin C  =2a sin B.sin C[sin Aa=k (let)]  =R.H.S.Thus, it is proved.


For any triangle ABC, prove thatcosAa+cosBb+cosCc=a2 + b2 + c22abc


L.H.S.=cosAa+cosBb+cosCc  =1a(b2 + c2 a22bc) + 1b (c2 + a2 b22ca) +1c (a2 + b2 c22ab)  =b2 + c2 a2 + c2 + a2 b2 + a2 + b2 c22abc  =b2 + c2 + a22abcHence, it is proved.

Q.92 For any triangle ABC, prove that
(b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0


(b2c2)cot A=(b2c2)cos Asin A =(b2c2)cos Aka[ sin Aa=sin Bb=sin Cc=k]=(b2 c2)ka × b2 + c2 a22bc=b4 + b2 c2 b2 a2 b2 c2 c4 + c2 a22kabc(c2 a2) cot B=(c2 a2)cos Bsin B=(c2 a2)cos Bkb=(c2 a2)kb × c2 + a2 b22bc=c4 + c2 a2 c2 b2 a2 c2 a4 + a2 b22kabc(a2 b2) cot C=(a2 b2)cos Csin C=(a2 b2) cos Ckc=(a2 b2)kc × a2 + b2 c22ab=a4 + a2 b2 a2 c2 a2 b2 b4 + b2 c22kabcL.H.S. = (b2 c2) cot A + (c2 a2) cot B + (a2 b2) cot C= b4 + b2 c2 b2 a2 b2 c2 c4 + c2 a22kabc +c4 + c2 a2 c2 b2 a2 c2 a4 + a2 b22kabc+a4 + a2 b2 a2 c2 a2 b2 b4 + b2 c22kabc= (b4 + b2 c2 b2 a2 b2 c2 c4 + c2 a2 + c4 + c2 a2 c2 b2a2 c2 a4 + a2 b2 + a4 + a2 b2 a2 c2 a2 b2 b4 + b2 c2)2kabc=02kabc = 0= R.H.S.Thus, it is proved.


For any triangle ABC, prove thatb2c2a2sin  2A+c2a2b2sin  2B+a2b2c2sin  2C=0


b2 c2a2 sin  2A= (b2 c2a2) 2sinA cos A=(b2 c2a2) 2(ka) cosA [ sin Aa=sin Bb=sin Cc = k]=(b2 c2a2) 2(ka) b2 + c2 a22bc=k (b4 c4 a2 b2 + a2 c22abc)Similarly,c2 a2b2sin  2B = k (c4 a4 c2 b2 + a2 b22abc) and a2 b2c2 sin  2C = k(a4 b4 a2 c2 + b2 c22abc)L.H.S. = b2 c2a2 sin  2A + c2 a2b2 sin  2B + a2 b2c2 sin  2C=k (b4 c4 a2 b2 + a2 c22abc) + k (c4 a4 c2 b2 + a2 b22abc)+ k (a4 b4 a2 c2 + b2 c22abc)=k (b4 c4 a2 b2 + a2 c2 + c4 a4 c2 b2 + a2 b2+ a4 b4 a2 c2 + b2 c22abc)=k (02abc)= 0 = R.H.S.Hence proved.

Q.94 A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35 m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.


Let PQ be a tree of height h m and A be a point on ground as AQ = 35 m.

APR=90°60° =30° APQ=30°       PAQ=60°15°     =45°       PQA=180°PAQAPQ     =180°45°30°     =105°Applying sin rule in ΔPQA, we get    sin45°PQ=sin30°AQ=sin105°APsin45°PQ=sin30°35   (12)PQ=(12)35  352=PQ2  PQ=352Thus, the height of the tree is 352m.

Q.95 Two ships leave a port at the same time. One goes 24 km per hour in the direction N45°E and other travels 32 km per hour in the direction S75°E. Find the distance between the ships at the end of 3 hours.


Speed of first ship = 24 km/hr
Speed of second ship = 32 km/hr
Distance (OP) covered in 3 hours
= 24 × 3
= 72 km

Distance (OQ) covered in 3 hours
= 32 × 3
= 96 km
Angle POQ = (90° – 45°) + (90° – 75°)
= 45° + 15°
= 60°

By using cosine formula in ΔOPQ, we getPQ2=OP2+OQ22×OP×OQ cos60° =(72)2+(96)22×72×96×12 =5184+92166912  PQ=7488 =86.53  km


Two trees, A and B are on the same side of a river. From apoint C in the river the distance of the trees A and B is 250 mand 300 m, respectively. If the angle C is 45°, find thedistance between the trees (use 2=1.44).


Let A and B be the positions of two trees respectively. The distance of first tree from C is 250m and that of second tree is 300 m.
Angle C is 45°.

Applying cosine formula in ΔABC, we get                 cos C=a2 + b2 c22ab         cos 45°=(300)2 + (250)2 c22(300)(250)        12=90000 + 62500 c2150000 1500002×22=152500AB2    15000022=152500AB2150000 × 1.4142=152500AB2           106050=152500AB2                 AB=152500106050                       =46450                  AB=215.5 mThus, distance between two trees is 215.5m.

Q.97 Prove that:



L.H.S.=2cosπ13cos9π13+cos3π13+cos5π13=cos(π13+9π13)+cos(π139π13)+cos(π10π13)+cos(π8π13)=cos(10π13)+cos(8π13)cos10π13cos8π13 =cos(8π13)cos8π13=0=R.H.S.

Q.98 Prove that:
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0


L.H.S.=(sin 3x+sin x) sinx+(cos 3xcos x) cos x=sin 3x sin x+sin2x+cos 3x cos xcos2x=cos 3x cos x+sin 3x sin xcos2x+sin2x=cos (3xx)(cos2xsin2x)[cos (AB)=cos A cos B+sin A cos B]=cos 2xcos 2x [ cos2xsin2x=cos 2x]=0=R.H.S.Hence proved.

Q.99 Prove that:

(cos x+cos y)2 + (sin x sin y)2 = 4 cos2x + y2


L.H.S. : (cos x + cos y)2 + (sin x-sin y)2 = (2cos x + y2cos x – y2)2 + (2cos x + y2sin x y2)2= 4cos2 (x + y2)(cos2 x y2+sin2 x y2)= 4 cos2 (x + y2) × 1 [Q sin2x + cos2x = 1] =4cos2 (x + y2)= R.H.S. Hence proved.

Q.100 Prove that:

(cos xcos y)2+(sin xsin y)2=4 sin2 xy2


L.H.S. : (cos x – cos y)2 + (sin x sin y)2={2sin (x + y2)sin (x y2)}2 + {2cos (x + y2)sin( x y2)}2=4sin2 (x y2){sin2(x + y2)+cos2 (x + y2)}= 4 sin2 (x y2) × 1[Q sin2x + cos2x = 1] =4sin2 (x y2)=R.H.S.Hence proved.

Q.101 Prove that:
sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x


L.H.S.=sin x + sin 3x + sin 5x + sin 7x    =sin 7x + sin x + sin 5x + sin 3x     =2 sin (7x + x2) cos (7x x2)+2sin (5x + 3x2) cos (5x 3x2)    =2 sin 4xcos 3x + 2 sin 4x cos x    =2 sin 4x (cos 3x + cos x)    =2 sin 4x {2 cos(3x + x2) cos (3x x2)}    =4 sin 4x cos 2x cos x    =4 cos x cos 2x sin 4x    =R.H.S.Hence proved.

Q.102 Prove that:

(sin 7x+sin 5x)+(sin 9x+sin 3x)(cos 7x+cos 5x)+(cos 9x+cos 3x)=tan 6x


L.H.S.=(sin 7x+sin 5x)+(sin 9x+sin 3x)(cos 7x+cos 5x)+(cos 9x+cos 3x) =(2sin7x+5x2cos7x5x2)+(2sin9x+3x2cos9x3x2)(2cos7x+5x2cos7x5x2)+(2cos9x+3x2cos9x3x2)=(2 sin 6x cos x) + (2 sin 6x cos 3x)(2 cos 6x cos x) + (2 cos 6x cos 3x)=2 sin 6x(cos x + cos 3x)2 cos 6x(cos x + cos 3x)=tan 6x=R.H.S.Hence proved.


Prove that:sin3x+sin2xsinx=4sinxcosx2cos3x2


L.H.S. = sin 3x+sin 2x sin x         = sin 3x sin x + sin 2x         = 2cos3x + x2sin3x x2+2sin x cos x         = 2cos 2x sin x + 2sin x cos x         = 2sin xcos 2x + cos x         = 2sin x2cos2x + x2cos2x x2         = 4sin x cos3x2 cosx2         = 4sin x cosx2 cos3x2         = R.H.S. Hence proved.


Findsinx2, cosx2 and tanx2 in each of the following: 8. tan x=43, x in quadrant II 9. cos x=13,x in quadrant III 10.sinx=14,x in quadrant II


8.Since,tanx=43,xinquadrantII       sec2x=1+tan2x                 =1+432                 =1+169                 =259         secx=±259                 =53x is in II quadrant.      cosx=35  2cos2x21=35      2cos2x2=135                 =25       cos2x2=15        cosx2=±15cosx2=15x lies in II quadrant,So, x2 lies in I quadrant.        =15×55        =55sinx2=1cos2x2          =±1152          =±115          =25x2 lies in I quadrant.          =25×55          =255  tanx2=sinx2cosx2          =2515=2So, the values of sinx2, cosx2 and tanx2 are 55,255 and2 respectively. 9.We have,        cosx=132cos2x21=13    2cos2x2=113               =23      cos2x2=13       cosx2=±13cosx2=13x lies in III quadrant,So, x2 lies in II quadrant.        =13×33        =33   sinx2=1cos2x2            =±1132            =±113            =23  x2 lies in II quadrant.            =23×33            =63    tanx2=sinx2cosx2            =2313=2So, the values of sinx2, cosx2 and tanx2 are 63,63 and2 respectively. 10.We have,sinx=14,xinquadrantII    cosx=1sin2x      cosx=±1142              =±1116              =154xinquadrantIISince,cosx=12sin2x2        154=12sin2x2       2sin2x2=1+154         sin2x2=12+158          sinx2=±12+158                  =8+2154x2 lies in I quadrant.     andcosx=2cos2x21154=2cos2x21cos2x2=12158cosx2=±12158        =82154x2 lies in I quadrant.tanx2=sinx2cosx2        =8+215482154        =8+2158215×8+2158+215        =8+2156460        =8+2152tanx2=4+15So, the values of sinx2, cosx2 and tanx2 are 8+2154,82154 and 4+15 respectively.

Q.105 A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is
(i) an ace (ii) black card.


(a) There are 52 points in the sample space.
(b) Number of ace of spades = 1

P(ace of spades)=152Thus, the probability of an ace of spades is 152.

(c) (i) Number of aces in a deck of 52 cards = 4

P(ace)=452=113Thus, the probability of an ace card is 113.

(ii) Number of black cards in a deck of 52 cards

P(ace)=2652=12Thus, the probability of an ace card is 12.

Q.106 A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed, find the probability that the sum of numbers that turn up is (i) 3 (ii) 12


Sample events when one coin(faced 1 and 6) and a dice: S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}n(S)=12(i) Events in which sum is 3: E={(1,2)}P(E)=112 Thus, the required probability is 112.(ii) Events in which sum is 12: E={(6,6)}P(E)=112Thus, the required probability is 112.

Q.107 There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?


Number of people in a council = 10
Number of women in a council = 6

P(woman)=610=35Thus, the required probability is 35.

Q.108 A fair coin is tossed four times, and a person win ₹ 1 for each head and lose ₹1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.


Sample space when a coin is tossed four times:

A person gain on getting each head = ₹ 1
A person loses on getting each tail = ₹ 1.50
Gain on getting HHHH = ₹ 4
Gain on getting HHHT = ₹ 1.50
Gain on getting HHTH = ₹ 1.50
Gain on getting HTHH = ₹ 1.50
Gain on getting THHH = ₹ 1.50
Loss on getting HHTT = ₹ 1.00
Loss on getting HTHT = ₹ 1.00
Loss on getting THHT = ₹ 1.00
Loss on getting THTH = ₹ 1.00
Loss on getting TTHH = ₹ 1.00
Loss on getting HTTH = ₹ 1.00
Loss on getting TTTH = ₹ 3.50
Loss on getting TTHT = ₹ 3.50
Loss on getting THTT = ₹ 3.50
Loss on getting HTTT = ₹ 3.50
Loss on getting TTTT = Rs. 6

       P(Winning4)=116P(Winning1.50)=416   =14   P(Losing1.00)=616   =38   P(Losing3.50)=416=14    P(Losing6.00)=116

Q.109 Three coins are tossed once. Find the probability of getting

(i) 3 heads
(ii) 2 heads
(iii) at least 2 heads
(iv) at most 2 heads
(v) no head (vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) at most two tails


Sample space of tossing three coins is:

i P3 heads=18ii P2 heads=38iii Pat least 2 heads=48=12iv Pat most2 heads=78v PNo head=18vi P3 tails=18vii Pexactly 2 tails=38viii PNo tail=18ix Pat most2 tails=78


If ​ 211 is the probability of an event, what is the probability of the event ‘not A’.


Let probability of an event be P(A).Then,       P(A)=211P(notA)=1P(A)          =1211          =11211          =911Thus, the probability of the event ‘not A’ is 911.

Q.111 A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is
(i) a vowel (ii) a consonant


Number of letters in the word “ASSASSINATION=13Number of vowels in the word “ASSASSINATION=6Number of consonants in the word “ASSASSINATION=136 =7(i)            P(a vowel)=613 [P(E)=n(E)S(E)](ii) P(a consonant)=713

Q.112 In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?


Total numbers from 1 to 20 = 20
Numbers to be selected = 6

Number of ways to select 6 numbers from 20 numbers =20C6 =20!6!(206)! =20×19×18×17×16×15×14!6×5×4×3×2×1×14! =38760Number of ways to fix 6 numbers for winning lottery = 1   P(Winning the prize)=138760

Q.113 Check whether the following probabilities P(A) and P(B) are consistently defined

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8


(i) Here,​ P(AB)>P(A) [P(AB)=0.6 and P(A)=0.5]    So, the probabilities P(A) and P(B) are not consistently defined.(ii) Here,P(AB)>P(A) and P(AB)>P(B).    So, the probabilities P(A) and P(B) are consistently defined.


Fill in the blanks in following table:P(A)P(B)P(AB)P(AB)(i)1315115.……(ii)0.35.……0.250.6(iii)0.50.35.……0.7


i PAB=PA+PBPAB                   =13+15115                   =5+3115                   =715ii PAB=PA+PBPAB               0.6=0.35+PB0.25        0.60.1=PB               PB=0.5iii  PAB=PA+PBPAB                         0.7=0.5+0.35PAB          PAB=0.850.7                                 =0.15


Given P(A) = 35  and P(B) = 15. Find P(A or B), if A and B are mutually exclusive events.


Since, A and B are mutually exclusive events. So,AB=ϕP(AB)=0P(AB)=P(A)+P(B)P(AB)     =35+150     =45Therefore, P(AorB) is 45.


If E and F are events such that P(E) = 14, P(F) = 12 and P(E and F) = 18, find (i)P(E or F), (ii)P(not E and not F).


Given: P(E)=14, P(F)=12 and P(EF)=18(i) P(EF)=P(E)+P(F)P(EF)          =14+1218          =2+418          =58Therefore, P(EorF) is 58.(ii) P(notE and not F)=P(EF)        =P(EF)[By Demorgan’s Law]        =1P(EF)        =158        =858        =38Therefore, P(notE and not F) is equal to 38.

Q.117 Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.


P(not E or not F) =0.25                    P(EF)=0.25                    P(EF)=0.25[By Demorgarn’s law]             1P(EF)=0.25[P(E)=1P(E)]     P(EF)=10.25     P(EF)=0.75            EF0So, E and F are not mutually exclusive.

Q.118 A and B are events such that P(A) = 0.42,
P(B) = 0.48 and P(A and B) = 0.16.
Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B)


Given:P(A)=0.42, P(B)=0.48 and P(AB)=0.16(i)    P(notA)=P(A)         =1P(A)         =10.42         =0.58(ii)    P(notB)=P(B)       =1P(B)       =10.48       =0.52(iii)P(AorB)=P(AB) =P(A)+P(B)P(AB) =0.42+0.480.16 =0.74

Q.119 In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.


PMathematics​ = 40%               PM​​​ = 0.40PBiology​ = 30%         PB ​= 0.30 PMathematics and Biology=10% PMB =0.10 PMathematics or Biology=PMB     =PM+PBPMB      =0.40+0.300.10      =0.60Therefore, probability that student wiall be studying Mathematics or Biology is 0.60.

Q.120 In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?


Let probability of passing in both examinations  be  P(A) and P(B) respectively.Then, we haveP(A)=0.8, P(B)=0.7 and P(AB)=0.95 P(AB)=P(A)+P(B)P(AB)      =0.8+0.70.95      =0.55Thus, the probability of passing in both the examination is 0.55.

Q.121 The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?


Let the probability that a student will pass the final examination in both English and Hindi =P(EH) P(EH)=0.5Let the probability that a student will not pass the final examination in neither English nor Hindi =P(EH)        P(EH)=0.1         P(EH)=0.1[By Demorgan’s law]   1  P(EH)=0.1           P(EH)=10.1           P(EH)=0.9The probability of passing the English examination        =P(E)        =0.75                                   P(EH)=P(E)+P(H)P(EH)         0.9 =0.75+P(H)0.5       P(H)=0.90.75+0.5        =0.65Therefore, the probability of passing the Hindi examination is 0.65.

Q.122 In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one