Excel in Mathematics Book 10 Final_CTP (2080)_compressed (1)

Book 10 MATHEMATICS MATHEMATICS vedanta Excel in Approved by the Government of Nepal, Ministry of Education, Science and Technology, Curriculum Development Centre, Sanothimi, Bhaktapur as an Additional Learning Material Vedanta Publication (P) Ltd. j]bfGt klAns];g k|f= ln= Vanasthali, Kathmandu, Nepal +977-01-4982404, 01-4962082 [email protected] www.vedantapublication.com.np vedanta

Book 10 Published by: Vedanta Publication (P) Ltd. j]bfGt klAns];g k|f= ln= Vanasthali, Kathmandu, Nepal +977-01-4982404, 01-4962082 [email protected] www.vedantapublication.com.np All rights reserved. No part of this publication may be reproduced, copied or transmitted in any way, without the prior written permission of the publisher. MATHEMATICS MATHEMATICS vedanta Excel in First Edition: B.S. 2077 (2020 A.D.) Second and Updated Edition: B. S. 2078 (2021 A. D.) Third Edition: B. S. 2079 (2022 A. D.) Fourth Revised and Updated Edition: B. S. 2080 (2023 A. D.) Price: Rs 574.00 Printed Copies: 35,000.00

Publisher's words Vedanta, the emergent publication house, is committed to cater quality school-level textbooks and reference materials that are well-researched and thoughtfully prepared. We make the utmost endeavours to ensure the readers that our products are error-free. We believe, the processes of updating, upgrading and modification of textbooks are the most significant aspects of quality education. Therefore, our textbooks which are completely based on the latest curriculum developed by Curriculum Development Centre (CDC), Nepal are also updated, upgraded and modified as per time. The textbooks of Vedanta Excel in Mathematics series are originally written by Mr. Hukum Pd. Dahal, who has been writing mathematics text books since B.S. 2052 (1995 A.D.) to till the date. Besides the Excel in Mathematics series, Mr. Dahal has written three more series: Elementary Mathematics, Maths in Action and Speedy Maths. Recently, according to the new curriculum developed by CDC, Nepal, Vedanta Excel in Mathematics Book 10 is updated, upgraded and modified by Mr. Tara Bdr. Magar, who is also involving as the editor of all the textbooks of the series. -Vedanta Publication (P) Ltd.

Forewords The series of 'Excel in Mathematics' is completely based on the contemporary pedagogical teaching learning activities and methodologies extracted from Teachers' training, workshops, seminars, and symposia. It is an innovative and unique series in the sense that the contents of each textbooks of the series are written and designed to fulfill the need of integrated teaching learning approaches. The series has focused on promoting the fundamental mathematical skills: Social skill, Logical skill, Computational skill and Problem solving skill of the learners. Excel in Mathematics is an absolutely modified and revised edition of my three previous series: 'Elementary mathematics' (B.S. 2053), 'Maths in Action (B. S. 2059)', and 'Speedy Maths' (B. S. 2066). Excel in Mathematics has incorporated applied constructivism. Every lesson of the whole series is written and designed in such a manner, that makes the classes automatically constructive and the learners actively participate in the learning process to construct knowledge themselves, rather than just receiving ready made information from their instructors. Even the teachers will be able to get enough opportunities to play the role of facilitators and guides shifting themselves from the traditional methods of imposing instructions. Each unit of Excel in Mathematics series is provided with many more worked out examples, which are arranged in the order of the learning objectives and they are reflective to the corresponding exercises. Therefore, each textbook of the series itself plays the role of a ‘Text Tutor’. There is a proper balance between the verities of problems and their numbers in each exercise of the textbooks in the series. Clear and effective visualization of diagrammatic illustrations in the contents of each and every unit in grades 1 to 5, and most of the units in the higher grades as per need, will be able to integrate mathematics lab and activities with the regular processes of teaching learning mathematics connecting to real life situations. The learner friendly instructions given in each and every learning content and activity during regular learning processes, will promote collaborative learning and help to develop learner-centred classroom atmosphere. In grades 6 to 10, the provision of ‘General section’, ‘Creative section - A’, and ‘Creative section - B’ fulfill the coverage of overall learning objectives. For example, the problems in ‘General section’ are based on the knowledge, understanding, and skill (as per the need of the respective unit) whereas the ‘Creative sections’ include the Application and Higher ability problems.

Furthermore, the evaluations of all levels of learning achievements (knowledge, understanding, application and higher ability) based on a single problem are incorporated in every exercise, according to the revised and updated specification grid developed by CDC, Government of Nepal. The provision of ‘Classwork’ from grades 1 to 5 promotes learners in constructing knowledge, understanding and skill themselves with the help of the effective roles of teacher as a facilitator and a guide. Besides, the teacher will have enough opportunities to judge the learning progress and learning difficulties of the learners immediately inside the classroom. These classworks prepare learners to achieve higher abilities in problem solving. Of course, the commencement of every unit with 'Classwork-Exercise' plays a significant role as a 'Textual-Instructor'. The 'project works' given at the end of most of the exercises in the textbooks of each grade provide some ideas to connect the learning of mathematics to the real life situations. The provision of ‘Section A’ and ‘Section B’ in grades 4 and 5 provides significant opportunities to integrate mental maths and manual maths simultaneously. Moreover, the problems in ‘Section A’ judge the level of achievement of knowledge and understanding, and diagnose the learning difficulties of the learners. The provision of ‘Looking back’ at the beginning of each unit in grades 1 to 8 plays an important role of ‘placement evaluation’ which is in fact used by a teacher to judge the level of prior knowledge and understanding of every learner to select their teaching learning strategies. The socially communicative approach by language and literature in every textbook, especially in primary level of the series, plays a vital role as a ‘textual-parents’ to the young learners and helps them overcome maths anxiety. Furthermore, as per the need of ICT in teaching and learning mathematics, we have included 'Vedanta ICT Corner' to fulfil this requirement in all classes. The Excel in Mathematics series is completely based on the latest curriculum of mathematics, designed and developed by the Curriculum Development Centre (CDC), the Government of Nepal. We do hope the students, teachers, and even the parents will be highly benefited from the ‘Excel in Mathematics’ series. Constructive comments and suggestions for the further improvements of the series from the concerned are highly appreciated.

Acknowledgment In making effective modification and revision in the Excel in Mathematics series from the previous series, We are highly grateful to the Principals, HODs, Mathematics teachers and experts, PABSON, NPABSAN, PETSAN, ISAN, EMBOCS, NISAN, UNIPS and independent clusters of many other Schools of Nepal, for providing us with opportunities to participate in workshops, Seminars, Teachers’ training, Interaction programme, and symposia as the resource person. Such programmes helped us a lot to investigate the teaching-learning problems and to research the possible remedies and reflect to the series. We are extremely grateful to Dr. Ruth Green, a retired professor from Leeds University, United Kingdom who provided us with very valuable suggestions about the effective methods of teaching-learning mathematics and many reference materials. We would like to express our deepest appreciation to Mr. Bir Bahadur Chalaune (Veer) for his praiseworthy efforts in the preparation of ICT Materials, which are presented in the Vedanta ICT Corner. We are thankful to Dr. Komal Phuyal for editing the language of the series. Moreover, we gratefully acknowledge all Mathematics Teachers throughout the country who encouraged us and provided us with the necessary feedback during the workshops/ interactions and teachers’ training programmes in order to prepare the series in this shape. We are profoundly grateful to the Vedanta Publication (P) Ltd. for publishing this series. We would like to thank Chairperson Mr. Suresh Kumar Regmi, Managing Director Mr. Jiwan Shrestha, and Marketing Director Mr. Manoj Kumar Regmi for their invaluable suggestions and support during the preparation of the series. Also, we are heartily thankful to Mr. Pradeep Kandel, the Computer and Designing Senior Officer of the publication house for his skill in designing the series in such an attractive form. -Authors

Contents S.N Chapter Page No. 1. Sets 13 1.1 Set –Looking back, 1.2 Cardinality relations of sets 2. Compound Interest 40 2.1 Simple interest – Looking back, 2.2 Principal and Interest - review, 2.3 Compound interest, 2.4 Simple Interest vs. Compound interest, 2.5 Derivation of formula of Compound Amount and Compound Interest, 2.6 Interest compounded Half-yearly and Quarter-yearly 3. Population Growth and Depreciation 61 3.1 Population growth - introduction, 3.2 Depreciation 4. Money and Money Exchange 75 4.1 Currency exchange - introduction, 4.2 Revaluation and Devaluation of currency, 4.3 Money exchange by using chain rule 5. Mensuration (I): Pyramid 89 5.1 Pyramid –introduction , 5.2 Surface area of square based pyramid, 5.3 Volume of square based pyramid, 5.4 Cone - introduction, 5.5 Surface area of cone, 5.6 Volume of cone 6. Mensuration (II): Combined Solids 107 6.1 Combined solids –introduction, 6.2 Surface area and volume of some basic 3-D shapes – Review , 6.3 Combination of square-based prism and pyramid, 6.4 Combination of square-based pyramids, 6.5 Combination of a cylinder and a hemisphere, 6.6 Combination of a cylinder and a cone, 6.7 Combination of cone and hemisphere, 6.8 Combination of two cones, 6.9 Mensuration in household activities 7 Sequence and Series 131 7.1 Sequence and Series – Looking back, 7.2 Arithmetic sequence (A.S.), 7.3 An arithmetic mean between two numbers, 7.4 Arithmetic means between two numbers, 7.5 Sum of first n terms of an Arithmetic Series, 7.6 Geometric sequence (G.S.), 7.7 A geometric mean between two numbers, 7.8 Geometric means between two numbers, 7.9 Relation between arithmetic mean and geometric mean, 7.10 Sum of first n terms of an Geometric Series 8. Quadratic Equations 162 8.1 Quadratic equations – Looking back, 8.2 Solution of quadratic equations, 8.3 Application of Quadratic equations 9. Rational Expressions 188 9.1 Rational expressions–Looking back, 9.2 Simplification of rational expressions

10. Exponential Equations 197 10.1 Indices –Looking back, 10.2 Indices- review, 10.3 Laws of indices, 10.4 Exponential equation, 10.5 Solving exponential quadratic equations, 10.6 Solving exponential equations with the given condition 11. Area of Triangles and Quadrilaterals 209 11.1 Area of triangles and quadrilaterals – Looking back, 11.2 Relation between the area of triangles and quadrilaterals 12. Geometry - Construction 231 12.1 Construction of a triangle equal in area to the given triangle, 12.2 Construction of a parallelogram equal in area to the given triangle, 12.3 Construction of a parallelogram equal in area to the given parallelogram, 12.4 Construction of a triangle equal in area to the given parallelogram, 12.5 Construction of a triangle equal in area to the given quadrilateral 13. Geometry - Circle 240 13.1 Circle-Looking back, 13.2 Definition of terms related to circle, 13.3 Theorems related to arcs and the angles subtended by them, 13.4 Theorems based on the angles subtended by the same arc, 13.5 Theorems based on cyclic quadrilateral 14. Statistics 272 14.1 Statistics – Looking back, 14.2 Statistics-review, 14.3 Measures of central tendency, 14.4 Mean of grouped and continuous data, 14.5 Median -review, 14.6 Median of grouped and continuous data, 14.7 Graphical method of determining the median, 14.8 Derivation of formula of median of grouped and continuous data, 14.9 Quartiles, 14.10 Quartiles of grouped and continuous series, 14.11 Mode of grouped and continuous data, 14.12 Derivation of formula of mode in grouped data, 14.13 Relationship between Mean, Median and Mode, 14.14 Range of grouped and continuous data 15. Probability 301 15.1 Probability -Looking back, 15.2 Definition of basic terms on probability -review, 15.3 Mutually exclusive and Non-mutually exclusive events, 15.4 Law of addition of mutually exclusive events, 15.5 Law of addition of none-mutually exclusive events, 15.6 Multiplication law of probability for independent events, 15.7 Probability tree diagram, 15.8 Multiplication law of probability of dependent events 16. Trigonometry 323 16.1 Trigonometric ratios of an acute angle of a right-angled triangle - review , 16.2 Values of trigonometric ratios of some standard angles, 16.3 Height and distance, 16.4 Angle of elevation and angle of depression, Revision and Practice Time 336 - 366 Answers 367 - 387 Model Questions 389 - 392

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 13 Vedanta Excel in Mathematics - Book 10 Classwork - Exercise 1.1 Set –Looking back 1. If A={x : x is a prime number, x ≤ 10} and B = {y: y is a factor of 12} are the subsets of a universal set U = {1, 2, 3, …, 12}, list out the elements of the following sets. a) A ∪ B = ……………………………, A ∪ B = ……………………………. b) A ∩ B = ……………………………, A ∩ B = ……………………………. c) A – B = ……………………………, A – B = ……………………………. d) B – A = ……………………………, B – A = ……………………………. e) A ∆ B = ……………………………, A ∆ B = ……………………………. 2. For any two sets X = {p, e, n}, Y = {c, o, p, y} and Z ={n, o, t, e} , find: a) n (X ∪ Y ∪ Z ) = ……………… b) n (X ∩ Y ∩ Z) = ……………… c) n (X ∪ (Y ∩ Z)) = ……………… d) n (X – (Y ∩ Z) ) = ……………… 1.2 Cardinality relations of sets Let’s take a set V = {a, e, i, o, u}. Since, it contains 5 elements. The cardinal number of the set V is 5. Thus, the number of elements contained by a set is called its cardinal number. For example: If A = {2, 3, 5, 7}, then the cardinal number of set A, i.e. n(A) = 4. The cardinal number of a set is called its cardinality. Certain relations between two or more sets can be generalized by taking the cardinalities of the sets. Such relations are very much useful in the application of set theory. 1. Cardinality relations of union of two sets If U is a universal set and the finite sets A and B are the subsets of U, then there arise the following three cases. Case- I: Cardinality relations of union of two sets A and B when A ⊂ B Let, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is a universal set. A = {4, 8} and B = {2, 4, 6, 8, 10} are the subsets of U. Unit 1 Sets

Vedanta Excel in Mathematics - Book 10 14 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets Now, A ∪ B = {2, 4, 6, 8, 10} and A ∩ B = {4, 8} Here, n (U) = 10, n (A) = 2, n (B) = 5, n (A ∪ B) = 5 Now, n (A ∪ B) = 5 = n (B) and n (A ∩ B) = 2 = n (A) Facts to remember If A ⊂ B then, (i) n (A ∩ B) = n (A) and n (A ∪ B) = n (B). (ii) n (A ∩ B) has the maximum value whereas n (A ∪ B) has the minimum value. Case- II: Cardinality relations of union of two disjoint sets A and B Let, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is a universal set. A = {1, 2, 4, 8} and B = {5, 10} are the subsets of U. Now, A ∪ B = {1, 2, 4, 5, 8, 10}, A ∩ B = φ and A ∪ B = {3, 6, 7, 9} Here, n (U) = 10, n (A) = 4, n (B) = 2, n (A ∪ B) = 6 Now, n (A ∪ B) = 6 = 4 + 2 = n (A) + n (B) Also, n (A ∪ B) = 4 = 10 – 6 = n (U) – n (A ∪ B) Facts to remember If A and B are disjoint sets then, (i) n (A ∩ B) = 0 and n (A ∪ B) = n (A) + n (B). (ii) n (A ∩ B) has the minimum value whereas n (A ∪ B) has the maximum value. Case- III: Cardinality relations of union of two overlapping sets Let, U = {1, 2, … , 10} is a universal set. A = {1, 3, 5, 7, 9} and B = {1, 2, 3, 6} are the subsets of U. Now, A ∪ B = {1, 2, 3, 5, 6, 7, 9}, ∴ n (A ∪ B) = 7 (A ∪ B) = {4, 8, 10} , ∴ n(A ∪ B) = 3 A ∩ B = {1, 3}, only A = A – B = {5, 7, 9} and only B = B – A = {2, 6} Here, n (U) = 10, n (A) = 5, n (B) = 4 and n (A ∩ B) = 2 Now, n (A ∪ B) = 7 = 5 + 4 – 2 = n (A) + n (B) – n (A ∩ B) Also, n (A ∪ B) = 3= 10 – 7 = n (U) – n (A ∪ B) Again, n (only A) = n (A – B) = 3 = 5 – 2 = n (A) –n (A ∩ B) n (only B) = n (B – A) = 2 = 4 – 2 = n (B) – n (A ∩ B) B A 2 6 1 3 5 7 9 4 8 10 U A B U 1 4 3 6 5 10 7 9 2 8 U A B 5 1 2 6 10 8 4 7 3 9

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 15 Vedanta Excel in Mathematics - Book 10 Sets Derivation of the cardinality relations of union of two sets If A and B are two overlapping finite subsets of a universal set U, n (A ∪ B) = n (A) + n (B) – n (A ∩ B) Proof: Let, n (A) = a, n (B) = b and n (A ∩ B) = c. Then, n (A – B) = a – c and n (B – A) = b – c Now, from Venn-diagram, n (A ∪ B) = (a – c) + c + (b – c) = a + b – c = n (A) + n (B) – n (A ∩ B) Proved Let’s learn a few more relations which are derived from the cardinality relation of union of two sets. 1. n (only A) = n (A – B) = no (A) = n (A) –n (A ∩ B) 2. n (only B) = no (B) = n (B) – n (A ∩ B) 3. n (A ∪ B) = n (A) + n (B) – n (A ∩ B) 4. n (A ∩ B) = n (A) + n (B) – n (A ∪ B) 5. n (A ∪ B) = no (A) + no (B) + n (A ∩ B) 6. n (A ∪ B) = n (A) + no (B) or no (A) + n (B) 7. n (U) = (A ∪ B) + n (A ∪ B) 8. n (A ∪ B) = n (U) when n (A ∪ B) = 0 9. n (A ∪ B) = n (U) – n (A ∪ B) 10. n (A ∩ B) = n (U) – n (A ∩ B) or, no (A) + no (B) + n (A ∪ B) 11. no (A) + no (B) = n (A ∪ B) – n (A ∩ B) Facts to remember Suppose, A and B are the sets of people who like apple and banana respectively, remember the following terminologies with their corresponding notations to solve the word-problems. S.N. Information Notation 1. Number of people who like (i) either apple or banana or both (ii) at least one of the fruits n (A ∪ B) 2. Number of people who like both the fruits (apple as well as banana) n (A ∩ B) 3. Number of people who (i) don’t like both the fruits (ii) like neither apple nor banana n (A ∪ B) 4. Number of people who like apple only no (A) 5. Number of people who like banana only no (B) 6. Number of people who like only one (exactly one) type of fruits no (A) + no (B) 7. Number of people who like at most one type of fruit n (A ∩ B) U A B a – c c b – c U A B n (A) n (B) n (only A) n (A ∩ B) n (only B) n (A ∪ B) www.geogebra.org/classroom/cdkr453t Classroom code: CDKR 453T Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 16 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets Worked-out Examples Example 1: If n(U) = 150, n(A) = 60, n(B) = 80 and n(A ∪ B) = 110, find (i) n(A ∪ B) (ii) n(A ∩ B) (iii) no (A) (iv) no (B). Draw a Venn-diagram to illustrate the above information. Solution: Here, n(U) = 150 n(A) = 60, n(B) = 80, n(A ∪ B) = 110 (i) Now, n(A ∪ B) = n(U) – n(A ∪ B) = 150 – 110 = 40 (ii) Also, n(A ∩ B) = n(A) + n(B) – n(A ∪ B) = 60 + 80 – 110 = 30 (iii) Again, no (A) = n(A) – n(A ∩ B) = 60 – 30 = 30 (iv) And, no (B) = n(B) – n(A ∩ B) = 80 – 30 = 50 Example 2: If A and B are two sets containing 40 and 50 elements respectively, find (i) the possible maximum and minimum values of n (A ∪ B). (ii) the possible maximum and minimum values of n (A ∩ B). Solution: Here, n (A) = 40 and n (B) = 50 In the given problem, there are two possible cases: Case I: When A and B are disjoint sets, n (A ∩ B) has minimum value The minimum value of n (A ∩ B) = 0 ∴ n (A ∪ B) = n (A) + n (B) = 40 + 50 = 90 Case II: When A ⊂ B, n (A ∩ B) has maximum value The maximum value of n (A ∩ B) = n (A) = 40 ∴n (A ∪ B) = n (A) + n (B) – n (A ∩ B) = 40 + 50 – 40 = 50 Hence, the maximum and minimum values of n (A ∪ B) are 90 and 50 respectively and the maximum and minimum values of n (A ∩ B) are 40 and 0 respectively. Example 3: 80 students appeared the SEE from Saraswoti Secondary School in B.S. 2079. In the result, it was found that 15 students got ‘A+’ grade in Mathematics, 25 got ‘A+’ grade in Science and 10 students got ‘A+’ grade in both Mathematics and Science. (i) Represent the above information in a Venn-diagram. (ii) Find the number of students who got ‘A+’ grade in Mathematics only. (iii) Find the number of students who got ‘A+’ grade either in Mathematics or in Science. (iv) Find the number of students who got below ‘A+’ grades in both the subjects. A 30 30 50 B U Venn-diagram 40 A B 40 50 B A 40 10

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 17 Vedanta Excel in Mathematics - Book 10 Sets Solution: Let S and M denote the sets of students who got ‘A+’ grade in Science and Mathematics respectively. Then, n (U) = 80, n (M) = 15, n (S) = 25 and n (M ∩ S) = 10 Now, (i) The given Venn-diagram represents the above information. (ii) no (M) = n (M) – n (M ∩ S) = 15 – 10 = 5 ∴ 5 students got ‘A+’ grade in Mathematics only. (iii) n (M ∪ S) = n (M) + n (S) – n (M ∩ S) = 15 + 25 – 10 = 30 ∴ 30 students got ‘A+’ grade either in Mathematics or in Science. (iv) n (M ∪ S) = n (U) – n (M ∪ S) = 80 – 30 = 50 Hence, 50 students got below ‘A+’ grades in both the subjects. Example 4: A survey conducted in a rural municipality regarding agricultural status of farmers, it was found that 75% farmers are involving in crops farming, 40% are involving in vegetables farming and each farmer is involving in at least one of these two farming. Find the percentage of farmers who are (i) involving in both crops and vegetables farming (ii) involving in only one of these farming (iii) not involving in vegetables farming (iv) Draw a Venn-diagram to show the given data Solution: Let C and V denote the sets of farmers who were involving in crops farming and vegetables farming respectively. Here, n (U) = 100 (i.e. 100%), n(C) = 75, n (V) = 40 and n (C ∪ V ) = 0 ∴ n (U) = n (C ∪ V) (i) Now, n (C ∪ V) = n (C) + n (V) – n (C ∩ V) or, n (C ∩ V) = n (C) + n (V) – n (C ∪ V) = 75 + 40 – 100 = 15 Thus, 15% farmers are involving in both the farming. (ii) no (C) = n (C) – n (C ∩ V) = 75 – 15 = 60 and no (V) = n (V) – n (C ∩ V) = 40 – 15 = 25 Now, no (C) + no (V) = 60 + 25 = 85 So, 85% farmers are involving in only one of these farming. (iii) n (V) =n (U) – n (V) = 100 – 40 = 60 So, 60% farmers are not involving in vegetables farming. (iv) The given data are shown in the adjoining Venn-diagram. Example 5: During ‘Visit Nepal 2020’, among 1100 Japanese tourists who came to visit Nepal through a Travel Company, 32% tourists preferred to visit Pokhara, 48% of them preferred to visit Lumbini and 30% preferred to visit other places. M 5 10 15 S U 50 C 60 15 25 V U

Vedanta Excel in Mathematics - Book 10 18 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets (i) How many tourists preferred at least one of these two places? (ii) How many tourists preferred at most one of these two places? (iii) Illustrate this information in a Venn-diagram. (iv) How many more tourists preferred Lumbini only than Pokhara only? Solution: Let P and L denote the sets of Japanese tourists who preferred to visit Pokhara and Lumbini respectively. Here, n (U) = 1100, n (P) = 32% of 1100 = 352, n (L) = 48% of 1100 = 528 and n (P ∪ L ) = 30% of 1100 = 330 (i) Now, n (P ∪ L) = n (U) – n (P ∪ L ) = 1100 – 330 = 770 ∴770 tourists preferred at least one of these two places. (ii) We know, n (P ∩ L) = n (P) + n (L) – n (P ∪ L) = 352 + 528 – 770 = 110 no (P) = n (P) – n (P ∩ L) = 352 – 110 = 242 no (L) = n (L) – n (P ∩ L) = 528 – 110 = 418 ∴The number of tourists who preferred at most one of the places = no (P) + no (L) + n (P ∪ L ) = 242 + 418 + 330 = 990 (iii) The required Venn-diagram to show the above information is given alongside. (iv) Also, no (L) – no (P) = 418 – 242 = 176 Hence, 176 more tourists preferred Lumbini only than Pokhara only. Example 6: In a group of 112 people, 56 like cold drinks only, 45 like hot drinks only and each person likes at least one of the two drinks. (i) Draw a Venn-diagram to illustrate the above information. (ii) How many people like both the drinks? (iii) How many people like cold drinks? (iv) How many people like hot drinks? Solution: Let, C and H denote the set of students who like cold drinks and hot drinks respectively. Here, n(U) = n(C ∪ H) = 112, no (C) = 56 and no (H) = 45 (i) Now, n(C ∪ H) = no (C) + no (H) + n(C ∩ H) or, 112 = 56 + 45 + n(C ∩ H) or, n(C ∩ H) = 11 (ii) From the Venn-diagram, 11 people like both the drinks. (iii) n(C) = no (C) + n(C ∩ H) = 56 + 11 = 67 ∴ 67 people like cold drink. Alternative process The number of tourists who preferred at most one of the places = n (P ∩ L ) = 1100 – 110 = 990 P 242 418 110 L U 330 C 56 11 45 H U

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 19 Vedanta Excel in Mathematics - Book 10 Sets (iv) n(H) = no (H) + n(C ∩ H) = 45 + 11 = 56 ∴ 56 people like hot drinks. Example 7: In the local level election held in 30th Baishakh 2079, two candidates Amar and Dolma stood for the post of the Mayor in a municipality. There were 30,000 voters in the voter list and they were allowed to cast vote for a single candidate. 13,500 people cast vote for Amar, 14,000 people cast vote for Dolma and 1,500 people were found to cast vote even for both the candidates. (i) Illustrate this information in a Venn-diagram. (ii) Find the number of people who cast vote. (iii) Find the number of people who did not cast vote. (iv) Find the number of valid votes. Solution: Let, A and B denote the sets of people who cast votes for Amar and Dolma respectively. Here, n(U) = 30,000, no (A) = 13,500, no (B) = 14,000 and n(A ∩ B) = 1,500. Now, (i) The given Venn-diagram illustrates the above information. (ii) The number of people who cast vote, n (A ∪ B) = no (A) + no (B) + n(A ∩ B) = 13,500 + 14,000 + 1,500 = 29,000 (iii) The number of people who did not cast vote, n (A ∪ B) = n(U) – n (A ∪ B) = 30,000 – 29,000 = 1,000 (iv) The number of valid votes = no (A) + no (B) = 13,500 + 14,000 = 27,500 Example 8: One day morning, a stationer kept a record about the people who came to buy the daily newspapers Gorkhapatra and Kantipur in his stationery. 85 people bought Gorkhapatra, 65 didn’t buy Gorkhapatra, 75 didn’t buy Kantipur and 40 people bought Kantipur but not Gorkhapatra. Fill the information in a Venn-diagram and answer the following questions. (i) How many people took part in the survey? (ii) How many people bought Gorkhapatra but not Kantipur? (iii) How many people didn’t buy both the newspapers? (iv) What percent of people bought only one newspaper? Solution: Let, G and K denote the sets of people who bought Gorkhapatra and Kantipur respectively. Here, n (G) = 85, n (G ) = 65, n (K ) = 75 and no (K) = 40. Now, (i) The number of people who took part in the survey, n (U) = n (G) + n (G ) = 85 + 65 = 150 A 13500 14000 1000 1500 B U G 50 40 25 35 K U

Vedanta Excel in Mathematics - Book 10 20 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets (ii) We have, n (K) = n (U) – n (K ) = 150 – 75 = 75 Also, n (G ∩ K) = n (K) – no (K) = 75 – 40 = 35 Again, no (G) = n (G) – n (G ∩ K) = 85 – 35 = 50 ∴50 people bought Gorkhapatra but not Kantipur. (iii) We have, n (G ∪ K) = n (G) + n(K) – n(G ∩ K) = 85 + 75 – 35 = 125 Also, n ( G ∪ K ) = n(U) – n (G ∪ K) = 150 – 125 = 25 (iv) The number of people who bought only one of the newspaper = no (G) + no (K) = 50 + 40 = 90 ∴The percent of people who bought only one newspaper = 90 150 ×100 % = 60% Example 9: In a survey of 95 students who graduated SEE from a school, 6 students are willing to join both Management and Science faculties and 11 of them are not willing to join any of these faculties. The ratio of number of students who are willing to join Management to those who are willing to join Science is 2: 3. (i) Represent the above information in a Venn-diagram. (ii) Find the number of students who are willing to join Management. (iii) Find the number of students who are willing to join Science only. (iv) Find the ratio of number of students who are willing to join Management only to those who are willing to join Science only. Solution: Let, M and S denote the sets of students who are willing to join Management and Science respectively. Here, n (U) = 95, n (M ∩ S) = 6 and n (M ∪ S) = 11 Let, n (M) = 2x and n (S) = 3x (i) The given Venn-diagram represents the above information. From Venn-diagram, n (U) = no (M) + no (S) + n (M ∩ S) + n (M ∪ S) or, 95 = (2x – 6) + 6 + (3x – 6) + 11 or, 95 = 5x + 5 or, 5x = 90 or, x = 18 (ii) The number of students who are willing to join Management, n (M) = 2x = 2 × 18 = 36 (iii) The number of students who are willing to join Science only, no (S) = 3x – 6 = 3 × 18 – 6 = 48 (iv) The number of students who are willing to join Management only, no (M) = 2x – 6 = 2 × 18 – 6 = 30 Hence, the ratio of number of students who are willing to join Management only to those who are willing to join Science only = no (M) no (S) = 30 48 = 5: 8 M 2x – 6 3x – 6 11 6 S U M 30 48 11 6 S U

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 21 Vedanta Excel in Mathematics - Book 10 Sets Example 10: In a survey, it was found that the ratio of the people who liked electric bike and petrol bike is 5 : 4 out of which 45 people liked both types of bikes, 35 liked petrol bike only, and 60 liked none of the bikes. (i) How many people liked petrol bike? (ii) Represent the above data in a Venn-diagram. (iii) How many people were participated in the survey? (iv) How many people liked at most one bike? Solution: Let, E and P denote the sets of people who liked electric bike and petrol bike respectively. Here, n (E ∩ P) = 45, no (P) = 35 and n ( E ∪ P ) = 60 Let, n (E) = 5x and n(P) = 4x. (i) Now, the number of people who liked petrol bike, n (P) = no (P) + n (E ∩ P) = 35 + 45 = 80 (iii) We have, n (P) = 80 (ii) Venn-diagram or, 4x = 80 or, x = 20 ∴n (E) = 5x = 5 × 20 = 100 and no (E) = 100 – 45 = 55 Also, n (U) = no (E) + no (P) + n (E ∩ P) + n ( E ∪ P ) = 55 + 35 + 45 + 60 = 195 Hence, 195 people were participated in the survey, (iv) The number of people who liked at most one type of bike = no (E) + no (P) + n( E ∪ P ) = 55 + 35 + 60 = 150 Example 11: In a survey of some farmers, it is found that 48 % of them use organic fertilizers, 64 % use chemical fertilizers and 10 % of them use none of the fertilizers. (i) Represent the above information in a Venn-diagram. (ii) If there are 44 farmers who use both types of fertilizers, find the number of farmers who participated in the survey. (iii) Find the number of farmers who use organic fertilizer only. (iv) If each farmer had used at least one of these fertilizers, what change would be observed in the percent of farmers who use both the fertilizers? Solution: Let, O and C be the sets of the farmers who use organic and chemical fertilizers respectively Here, n (U) = 100 (i.e. 100%), n (O) = 48, n(C) = 64 and n (O ∪ C) = 10 (i) We know, n(O ∪ C) = n(U) – n (O ∪ C ) = 100 – 10 = 90 Again, n (O ∩ C) = n (O) + n (C) – n (O ∪ C) = 48 + 64 – 90 = 22 Thus, 22% farmers use both types of fertilizers. E 5x – 45 4x – 45 60 45 P U O 26 42 10 22 C U Venn-diagram

Vedanta Excel in Mathematics - Book 10 22 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets (ii) Let, the number of farmers who participated in the survey be x. According to question, 22% of x = n (O ∩ C) or, 22% of x = 44 or, x = 200 Hence, 200 farmers participated in the survey. (iii) The number of farmers who use organic fertilizer only, no (O) = 26% of 200 = 52 (iv) The number of farmers who use chemical fertilizer, n (C) = 64% of 200 = 128 If each farmer uses at least one of these fertilizers, then n (O ∪ C) = 0 and n (O ∪ C) = n(U) Now, n (O ∩ C) = n (O) + n (C) – n (O ∪ C) = 48 + 64 – 100 = 12 Hence, 12% of farmers would use both the fertilizers. Alternative process of (i) and (ii) Let, n (U) = x then, n (O) = 48% of x = 0.48x, n(C) = 64% of x = 0.64x and n (O ∪ C) = 10% of x = 0.1x (i) Representing the given data in a Venn-diagram (ii) From Venn-diagram, n(U) = n (O) + n (C) – n (O ∩ C) + n (O ∪ C) or, x = 0.48x + 0.64x – 44 + 0.1x or, 44 = 0.22x or, x = 200 Thus, 200 farmers participated in the survey. Example 12: A survey was conducted in a village regarding the people who recently returned Nepal from foreign employments, it was found that 25% of them were planning for dairy farming but not poultry farming, 30 % for poultry farming but not dairy farming, 5% for both types of framing and 64 people were not interested in both of these farming. (i) Draw a Venn-diagram to illustrate the above information. (ii) Find the number of people participated in the survey. (iii) Find the number of people who were interested in both types of farming. (iv) Find the number of people who were interested in dairy farming Solution: Let the sets of people who were planning for dairy and poultry farming are D and P respectively. Here, n (U) = 100 (i.e. 100%), no (D) = 25, no (P) = 30 and n (D ∩ P) = 5 (i) Now, n (U) = no (D) + no (P) + n (D ∩ P) + n ( D ∪ P ) or, 100 = 25 + 30 + 5 + n ( D ∪ P ) or, n ( D ∪ P ) = 40 Thus, 40% people were not interested in both types of farming. O (0.48x–22) (0.64x–22) 0.1x 44 C U D 25 30 40 5 P U Venn-diagram

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 23 Vedanta Excel in Mathematics - Book 10 Sets (ii) Let, the number of people participated in the survey be x. Then, according to question, 40 % of x = 64 or, x = 160 ∴160 people participated in the survey. (iii) Again, from the Venn-diagram, n (D ∩ P) = 5 % of 160 = 8 ∴ 8 people were interested in both types of farming. (iv) n (D) = (25 % + 30%) of 160 = 88 ∴88 people were interested in dairy farming. Example 13: In a locality of hilly region of Nepal, 60% families do not have electricity facility, 55% do not have drinking water supply facility, and 36 families do not have both the facilities. If none of the families have both the facilities, by using Venn-diagram, find: (i) the percent of families who do not have both of these facilities. (ii) the total number of families participated in the survey. (iii) the number of families who have electricity facility only. Solution: Let the sets of families who do not have electricity and drinking water supply facilities are E and W respectively. Here, n(U) = 100 (i.e. 100%) Then, n (U) = n (E ∪ W) = 100, n (E) = 60 and n (W) = 55 (i) Now, n (E ∩ W) = n (E) + n (W) – n (E ∪ W) = 60 + 55 – 100 = 15 ∴ 15% families do not have both the facilities. (ii) Let the total number of families be x. Then, 15 % of x = 36 or, x = 240 Hence, 240 families participated in the survey. (iii) Again, percent of families who have electricity facility only = (55 – 15) % or (100 – 60) % = 40 % ∴ The number of families who have electricity facility only = 40 % of 240 = 96 Example 14: A marketing company found that, of 500 households surveyed, 140 used neither brand A nor B soaps, 110 used only brand A soap and for every household that used both brands of soap, 4 used only brand B soap. (i) How many households used both brands of soaps? (ii) How many households used only one brand of soap? (iii) How many households used brand B soap? (iv) Draw a Venn-diagram to show the above information. Solution: Let, A and B are sets of households surveyed. Here, n (U) = 500, n (A ∪ B) = 140 and no (A) = 110 Now, n (A ∪ B) = n (U) – n (A ∪ B) = 500 – 140 = 360 Let, n (A ∩ B) = x, then no (B) = 4x E 45% 40% 15% W U E 108 96 40% 36 W U

Vedanta Excel in Mathematics - Book 10 24 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets (i) We have, n (A ∪ B) = no (A) + no (B) + n(A ∩ B) or, 360 = 110 + 4x + x or, 5x = 250 or, x = 50 ∴ 50 households used both brands of soaps. (ii) Also, no (A) + no (B) = 110 + 4x = 110 + 4 × 50 = 310 ∴310 households used only one brand of soap. (iii) Again, n (B) = no (B) + n(A ∩ B) = 4x + x = 5 × 50 = 250 (iv) The given Venn-diagram shows the above information. EXERCISE 1.1 General section 1. a) From the adjoining Venn-diagram, find the cardinal numbers of the following sets. (i) n (U) (ii) n (A) (iii) n (B) (iv) n (A ∪ B) (v) n (A ∩ B) (vi) no (A) (vii) no (B) (viii) n (A ∪ B) b) By using the above cardinal numbers of sets, verify the following relations. (i) n (A ∪ B) = n (A) + n (B) – n (A ∩ B) (ii) no (A) = n (A) – n (A ∩ B) (iii) no (B) = n (B) – n (A ∩ B) (iv) no (A) + no (B) = n (A ∪ B) – n (A ∩ B) (v) n (A ∪ B) = n (U) – n (A ∪ B) (vi) n (A ∪ B) = no (A) + no (B) + n (A ∩ B) 2. a) In the given Venn-diagram, F and B are the sets of students who play football and basketball respectively. Let's find the number of students in these sets operations. (i) n(F ∪ B) (ii) n(F ∩ B) (iii) n(( F ∪ B) (iv) no (F) (v) n(B – A) (vi) n(U) b) In the adjoining Venn-diagram, N and H are the sets of people who like Nepali and Hindi movies respectively. If n(U) = 135 and n(N ∪ H) = 120, find: (i) n(N ∩ H) (ii) n((N ∪ H) (iii) no (N) (iv) no (H) 3. a) If n(U) = 50, n(A) = 25, n(B) = 27 and n(A ∩ B) = 10, find: (i) n(A ∪ B) (ii) n((A ∪ B) (iii) n(A – B) (iv) no (B) b) If n(U) = 60, n(A) = 30, n(B) = 40 and n(A ∪ B) = 55, find: (i) n(A ∩ B) (ii) n((A ∪ B) (iii) no (A) (iv) n(only B) c) P and Q are the subsets of a universal set U. If n(U) = 90, n(P) = 45, n(Q) = 35 and n(( P ∪ Q) = 15, illustrate this information in a Venn-diagram and find: (i) n(P ∪ Q) (ii) n(P ∩ Q) (iii) no (P) (iv) n(Q – P) 140 A 110 50 200 B U U A B a c f i h l g j b e d F 12 10 8 18 B U 85 – x x 55 – x N H U

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 25 Vedanta Excel in Mathematics - Book 10 Sets d) If n(A) = 65, n(B) = 80 and A ⊂ B, find: (i) n(A ∪ B) (ii) n(A ∩ B) (iii) n(A – B) (iv) n(B – A) e) If no (A) = 12, no (B) = 15, n((A ∪ B) = 11 and n(U) = 45, find: (i) n(A ∪ B) (ii) n(A ∩ B) (iii) n(A) (iv) n(B) 4. a) If n (A) = 40 and n (B) = 50, find the possible (i) maximum value of n (A ∩ B) and minimum value of n (A ∪ B). (ii) minimum value of n (A ∩ B) and maximum value of n (A ∪ B). b) If M and N are two sets containing 15 and 10 elements, find: (i) the greatest value of n (M ∩ N) and least value of n (M ∪ N). (iii) the least value of n (M ∩ N) and greatest value of n (M ∪ N). c) P and Q are the subsets of universal set U. If n (U) = 100, n (P) = 75 and n (Q) = 55, find: (i) the greatest value of n (P ∪ Q) (ii) the least value of n (P ∩ Q). Creative section – A 5. a) A survey was conducted among 125 students of class 10 studying in Gyan Mandir Secondary School regarding the places for education excursion and it was found that 65 students preferred to visit Lumbini, 75 preferred to visit Pokhara, and 25 preferred both the places. (i) Show the above information in a Venn-diagram. (ii) Find the number of students who preferred either Lumbini or Pokhara. (iii) Find the number of students who preferred neither of two places. (iv) Find the number of students who preferred Lumbini only. b) A survey was carried out in a village of Chandrapur municipality of Rautahat district regarding the local languages. Among 1,400 people participated in the survey, it was found that 600 people can speak Bhojpuri, 900 can speak Maithili and 350 people can speak Bhojpuri as well as Maithili languages. (i) Draw a Venn-diagram to show the above data. (ii) How many people can speak at least one of these languages? (iii) How many people cannot speak both the languages? (iv) How many people can speak Maithili only? c) During ‘Visit Nepal 2020’, among 7,500 Chinese tourists who visited Nepal, 60% of them had already visited Bhutan, 50 % had visited Sri Lanka, and 30 % have visited both the countries. (i) Illustrate this information in a Venn-diagram. (ii) How many tourists have visited either Bhutan or Srilanka? (iii) How many tourists have already visited Sri Lanka only? (iv) How many tourists have already visited only one of these countries? d) In a survey of a community, 55% of the people like Dashain festival, 60% like Tihar festival and 30% like both festivals. By drawing a Venn-diagram, find: (i) the percent of people who like Dashain or Tihar or both festivals. (ii) the percent of people who don’t like both the festivals. (iii) the percent of people who like only Dashain. (iv) the percent of people who like only one of these festivals.

Vedanta Excel in Mathematics - Book 10 26 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets 6. a) In a survey conducted in a community of Gorkha district regarding cultural dances, it was found that out of 500 people, 180 people can perform Sorathi dance, 250 can perform Ghatu dance and 120 can perform neither of these two dances. (i) How many people can perform either Sorathi or Ghatu dance or both? (ii) How many people can perform both the dances? (iii) How many people can perform only one of the dances? (iv) Represent the above information in a Venn-diagram. b) Last Wednesday, a School successfully accomplished its school’s day. For this program, every student participated in at least one of the activities, athletics or music. In a class of 45 students, 21 participated in athletics and 29 participated in music. (i) How many students participated in either athletics or music? (ii) How many students participated in both the activities? (iii) How many students participated only one activity? (iv) Draw a Venn-diagram to represent the above information. 7. a) Among 54 SEE appeared students from a school, 18 students got ‘A+’ grade in Mathematics only, 25 got ‘ A+’ grade in English only and 7 students did not get ‘ A+’ grade in these two subjects. (i) Represent the above information in a Venn-diagram. (ii) Find the number of students who got ‘ A+’ grade in both of these subjects. (iii) How many students got ‘ A+’ grade in Mathematics? (iv) How many more students got ‘A+’ grade in English than in Mathematics? b) In a survey of a group of people, 20 % are using cellular data but not Wi-Fi, 65% are using Wi-Fi but not cellular data and 5 % of them use neither cellular data nor Wi-Fi, (i) Represent the above information in a Venn-diagram. (ii) Find the percent of people who are using cellular data as well as Wi-Fi. (iii) Find the percent of people who are using Wi-Fi. (iv) Find the percent of people who are not using cellular data. 8. a) In a survey of 15,000 students of different schools, 40 % of them were found to have tuition classes before the SEE examination. Among them, 50 % studied only Mathematics, 30 % only Science and 10 % studied none of these two subjects. (i) Represent the above information in a Venn-diagram. (ii) How many students studied Mathematics as well as Science? (iii) How many students studied Science? (iv) Find the ratio of number of students who studied Science to those who studied Mathematics. b) Of 200 students who appeared the SEE from Dhaulagiri Secondary School, 25% students got ‘A+’ grade in different subjects. Out of the students who got ‘A+’ grade, 40% students got in English only, 20% got in Mathematics only and 30% in other subjects. (i) Show the above information in a Venn-diagram.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 27 Vedanta Excel in Mathematics - Book 10 Sets (ii) How many students got ‘A+’ grade in both the subjects? (iii) How many students got ‘A+’ grade in English? (iv) Find the ratio of number of students who got ‘A+’ grade in English to those students who got in Mathematics. 9. a) In a survey of a community, it was found that 65% of people liked folk songs, 55% liked modern songs, and 10% of people did not like both types of songs. By illustrating the above information in a Venn-diagram, find: (i) What percentage of people liked either folk songs or modern songs? (ii) If 360 people liked both types of songs, how many people were surveyed? (iii) How many people liked modern songs only? b) In a survey of some farmers of a community, 70% of them are found cultivating rice, 60% cultivating wheat, 20% are not cultivating both the crops, and 450 farmers are found cultivating both the crops. By drawing a Venn-diagram to illustrate the above information, (i) find the percentage of people who are cultivating both the crops. (ii) Find the total number of farmers participated in the survey. (iii) Find the number of farmers who are cultivating rice only. 10. a) In a survey of a group of people, it was found that 65% of them liked comedy movies, 63% liked action movies, 33% liked both types of movies, and 150 people did not like both types of movies. (i) Draw a Venn-diagram to illustrate the above information. (ii) Find the number of people participated in the survey. (iii) Find the number of people who liked both types of movies. (iv) Find the number of people who liked only one type of movies. b) In an examination, 80 % examinees passed in English, 70 % in Mathematics, 60 % passed in both the subjects, and 45 examinees failed in both subjects. (i) Draw a Venn-diagram to represent the above information. (ii) Find the number of number of examinees participated in the survey. (iii) Find the number of examinees who passed only one subject. (iv) Find the number of students who failed Mathematics. c) In an examination 45 % students passed in Science only, 25 % passed in English only and 5 % students failed in both subjects. If 200 students passed in English, find the total number of students by using Venn-diagram. 11. a) 75 students in a class like picnic or hiking or both. Out of them 10 like both the activities. The ratio of the number of students who like picnic to those who like hiking is 2 : 3. (i) Represent the above information in a Venn-diagram. (ii) Find the number of students who like picnic. (iii) Find the number of students who like hiking only. (iv) Find the percentage of students who like picnic only.

Vedanta Excel in Mathematics - Book 10 28 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets b) A survey is conducted in a community regarding the activities that are beneficial for healthy life. Out of 100 people participating in the survey, it is found that the ratio of number of people who are doing yoga and going jogging regularly in the early morning is 4 : 5. If 25 people are following both the activities and 35 people are not following both of these activities, answer the following questions. (i) Draw a Venn-diagram to represent the above information. (ii) Find the number of people who are going jogging. (iii) Find the number of people who are doing yoga only. (iv) Find the percent of people who are going jugging only. c) In a group of students, the ratio of the number of students who liked music and sports is 9 : 7. Out of which 25 liked both the activities, 20 liked music only, and 15 liked none of the activities. (i) Find the number of students who liked music. (ii) Find the number of students who liked sports. (iii) Find the total number of students in the group. (iv) Represent the above information in a Venn-diagram. 12. a) Out of 120 students appeared in an examination, the number of students who passed in Mathematics only is twice the number of students who passed in Science only. If 50 students passed in both subjects and 40 students failed in both subjects, by using a Venn-diagram, find the number of students who passed in: (i) Mathematics (ii) Science (iii) only one subject. b) In a group of 500 people, 200 like summer season only and 150 like winter season only. If the number of people who do not like both the seasons is twice the number of people who like both the seasons. By using a Venn-diagram, find the number of people who like: (i) summer season (ii) winter season (iii) at most one season. Creative section - B 13. a) In the local level election held in 30th Baishakh 2079, Mr. Harka and Mrs. Sunita were two candidates for the post of the mayor in a municipality and 25,000 voters were in the voter list. Voters were supposed to cast the vote for a single candidate. 12,000 people cast vote for Harka, 10,000 people cast vote for Sunita and 1,000 people cast vote even for both the candidates. (i) Show the above information in a Venn-diagram. (ii) How many people cast the vote? (iii) How many people didn’t cast the vote? (iv) How many votes were valid? b) There are 400 students in a school. They are allowed to cast vote either only for Ajay or for Ojaswi as their school prefect. 50 of them cast vote for both Ajay and Ojaswi and 24 did not cast the vote. The candidate Ojaswi won the election with the majority of 56 more votes than Ajay.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 29 Vedanta Excel in Mathematics - Book 10 Sets (i) How many students cast the vote? (ii) How many valid votes were received by Ajay? (iii) How many votes were valid? (iv) Show the result in Venn-diagram. 14. a) In a survey conducted among some people of a community, 650 people like meat, 550 people don’t like meat, 480 don’t like fish and 250 like meat but not fish. (i) How many people were surveyed? (ii) How many people like fish but not meat? (iii) How many people are vegetarians? b) In a group of 1,100 youths, each uses smart phones, either I-phone or Samsung or both or neither, 150 of them use only I-phone, 770 of them use Samsung and 900 use only one of these two smart phones. (i) Find the number of youths who use both of these smart phones. (ii) Find the number of youths who use neither of these smart phones. 15. a) A marketing company found that, of 200 households surveyed, 80 used neither brand A nor B soaps, 60 used only brand A soap and for every household that used both brands of soap, 3 used only brand B soap. (i) How many households used both brands of soaps? (ii) How many households used only one brand of soap? (iii) How many households used brand B soap? (iv) Draw a Venn-diagram to show the above information. b) 100 employees in an office were asked about their preference for tea and coffee. It was observed that for every 3 people who preferred tea, there were 2 people who preferred coffee and there was a person who preferred both the drinks. The number of people who drink neither of them is same as those who drink both. (i) How many people preferred both the drinks? (ii) How many people preferred only one drink? (iii) How many people preferred at most one drink? c) Due to the heavy rainfall during a few monsoon days, 140 households were victimized throughout the country. In the first phase, Nepal government has decided to provide the support of either food, shelter or both to a few victimized households. 80 households got food support, 70 got shelter and 50 households got the support of food and shelter both. The government has managed the budget of Rs 10,000 per household for food, Rs 25,000 per household for shelter and Rs 35,000 per household for food and shelter. (i) Calculate the amount of budget for food only. (ii) Calculate the amount of budget for shelter only. (iii) Calculate the total amount of budget allocated. (iv) How many households were remained to get support in the first phase?

Vedanta Excel in Mathematics - Book 10 30 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets 2. Cardinality relations of union of three sets Let, A = {a, b, c, d, e, f}, B = {a, c, e, g, h} and C = {d, e, g, i} are the subsets of a universal set U = {a, b, c, d, e, f, g, h, i, j}. Now, A ∪ B ∪ C = {a, b, c, d, e, f, g, h, i}, A ∩ B = {a, c, e}, B ∩ C= {e, g}, C ∩ A = {d, e}, A ∩ B ∩ C = {e} Here, n (A) = 6, n (B) = 5, n (C) = 4, n (A ∩ B) = 3, n (B ∩ C) = 2, n (C ∩ A) = 2, n (A ∩ B ∩ C) = 1 Also, n (A ∪ B ∪ C) = 9 … (i) Again, n (A) + n (B) + n (C) – n (A ∩ B) – n (B ∩ C) – n (C ∩ A) + n (A ∩ B ∩ C) = 6 + 5 + 4 – 3 – 2 – 2 + 1 = 9 … (ii) From (i) and (ii) we get, n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (B ∩ C) – n (C ∩ A) + n (A ∩ B ∩ C) Derivation of the cardinality relations of union of three sets If A , B and C are two overlapping finite subsets of a universal set U, n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (B ∩ C) – n (C ∩ A) + n (A ∩ B ∩ C) Proof: n(A ∪ B ∪ C) = n[A ∪ (B ∪ C)] = n(A) + n(B ∪ C) – n[A ∩ (B∪ C)] = n(A) + n(B) + n(C) – n(B ∩ C) – n[(A∩ B) ∪ (A ∩ C)] = n(A) + n(B) + n(C) – n(B ∩ C) – n[(A ∩ B) + n(A ∩ C) – n(A ∩ B ∩ C)] = n(A) + n(B) + n(C) – n(B ∩ C) – n(A∩ B) – n(A ∩ C) + n(A∩ B ∩C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A∩ C) + n(A ∩ B ∩C) Proved U A B C n (A) n (C) n (B) n (only A) n (only C) n (only A ∩ C) n (only A ∩ B) n (only B ∩ C) n (A ∩ B ∩ C) n (only B) Let’s learn a few more relations which are derived from the cardinality relation of union of three sets. 1. n (U) = n (A ∪ B ∪ C) + n (A ∪ B ∪ C) 2. n (A ∪ B ∪ C) = n (U) when n (A ∪ B ∪ C) = 0 3. a) no (A) = n(A) – n(A ∩ B) – n(A ∩ C) + n(A ∩ B ∩ C) b) no (B) = n(B) – n(A ∩ B) – n(B ∩ C) + n(A ∩ B ∩ C) c) no (C) = n(C) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C) A B b f a c d i j e h g C U www.geogebra.org/classroom/xqfjrwtr Classroom code: XQFJ RWTR Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 31 Vedanta Excel in Mathematics - Book 10 Sets 4. a) no (A ∩ B) = n(A ∩ B) – n(A ∩ B ∩ C) b) no (B ∩ C) = n(B ∩ C) – n(A ∩ B ∩ C) c) no (A ∩ C) = n(A ∩ C) – n(A ∩ B ∩ C) Facts to remember If the sets of items A, B and C are the subsets of a universal set U, then S.N. Information Notation 1. n (like at least one of item A or B or C) n (A ∪ B ∪ C) 2. n (like all three items) n (A ∩ B ∩ C) 3. n (don’t like all three items) n (A ∪ B ∪ C) 4. n (only one item), n (exactly one items) no (A) + no (B) + no (C) 5. n (only two item), n (exactly two items) no (A ∩ B) + no (B ∩C) + no (C ∩A) 6. n (like at most one item) no (A) + no (B) + no (C) + n (A ∪ B ∪ C) 7. n (like at most two items) n (A ∩ B ∩ C) Worked-out Examples Example 1: A, B, and C are subsets of a universal set U. If n(U) = 180, n(A) = 95, n(B) = 65, n(C) = 50, n(A ∩ B) = 25, n(B ∩ C) = 15, n(A ∩ C) = 20 and n(A ∩ B ∩ C) = 5, illustrate this information in a Venn-diagram and find: (i) n(only A) (ii) no (B) (iii) no (C) (iv) no (A ∩ B) (v) n( only B ∩ C) (vi) no (A ∩ C) (vii) n(A ∪ B ∪ C). Solution: Here, n(U) = 180, n(A) = 95, n(B) = 65, n(C) = 50, n(A ∩ B) = 25, n(B ∩ C) = 15, n(A ∩ C) = 20 and n (A ∩ B ∩ C) = 5 From the Venn-diagram, (i) n(only A) = 55 (ii) no (B) = 30 (iii) no (C) = 20 (iv) no (A ∩ B) = 20 (v) no (B ∩ C) = 10 (vi) no (A ∩ C) = 5 (vii) n(A ∪ B ∪ C) = n(U) – n(A ∪ B ∪ C) = 180 – (55 + 30 + 20 + 5 + 20 + 10 + 15) = 180 – 155 = 25 Hints: At first, insert n(A ∩ B ∩ C) = 5, Then no (A ∩ B) = n(only A ∩ B) = 25 - 5 = 20, Then no (B ∩ C) = n (only B ∩ C) = 15 - 5 = 10, Then no (A ∩ C) = n (only A ∩ C) = 20 - 5 = 15, Then no (A) = n (only A) = 95 – 20 – 5 – 15 = 55, Then no (B) = n (only B) = 65 – 20 – 5 – 10 = 30, Then no (C) = n (only C) = 50 – 15 – 5 – 10 = 20 Venn-diagram A B 25 55 20 15 20 5 30 10 C U

Vedanta Excel in Mathematics - Book 10 32 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets Alternative process n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) = 95 + 65 + 50 – 25 – 15 – 20 + 5 = 215 – 60 = 155 ∴ n(A ∪ B ∪ C) = n(U) – n(A ∪ B ∪ C) = 180 – 155 = 25 Example 2: In a survey, 600 students were asked what future careers they prefer. 235 of them replied they prefer to be a Doctor, 210 a Pilot, 175 an Educator, 100 Doctor as well as pilot, 70 Doctor as well as Educator, 80 pilot as well as Educator and 40 replied they prefer all three careers. (i) Draw a Venn-diagram to illustrate the above information. (ii) How many students prefer either of these careers? (iii) How many students do not prefer any of the three careers? (iv) How many students prefer only one of these careers? Solution: Let A, B, and C be the sets of students who prefer to be a Doctor, a Pilot and an Educator respectively. Here, n(U) = 600, n(A) = 235, n(B) = 210, n(C) = 175, n(A ∩ B) = 100, n(A ∩ C) = 70, n (B ∩ C)= 80 and n(A ∩ B ∩ C) = 40 (i) Illustration in a Venn diagram: (ii) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C) = 235 + 210 + 175 – 100 – 70 – 80 + 40 = 410 Hence, 410 students prefer either of three careers. (iii) n (A ∪ B ∪ C) = n (U) – n (A ∪ B ∪ C) = 600 – 410 = 190 Hence, 190 students do not prefer any of the three careers. (iv) From Venn-diagram, no (A) + no (B) + no (C) = 105 + 70 + 65 = 240 Hence, 240 students prefer only one of these careers. Example 3: In a survey 2,500 people were asked what types of movies they like. It was found that 50 % of them liked Nepali movies, 60 % liked Hindi movies and 40 % like Korean movies. Likewise, 20 % of them liked Nepali as well as Hindi movies, 15 % liked Hindi as well as Korean movies, 25 % liked Nepali as well as Korean movies and 5 % did not like these types of movies. (i) Find the number of people who liked all three types of movies. (ii) Draw a Venn-diagram to illustrate the above information. (iii) Find the number of people who liked exactly two types of movies. (iv) Find the number of people who liked at most one type of movies. Solution: Let, N, H and K denote the sets of people who liked Nepali, Hindi and Korean movies respectively Here, n (U) = 2500, n(N) = 50 % of 2500 = 1250, n(H) = 60 % of 2500 = 1500, A B 190 105 60 30 65 40 70 40 C U

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 33 Vedanta Excel in Mathematics - Book 10 Sets n(K) = 40 % of 2500 = 1000, n(N ∩ H) = 20 % of 2500 = 500, n(H ∩ K) = 15 % of 2500 = 375, n (N ∩ K)= 25 % 2500 = 625 and n (N ∪ H ∪ K) = 5 % of 2500 = 125 (i) Now, n (N ∪ H ∪ K) = 2500 – n (N ∪ H ∪ K) = 2500 – 125 = 2375 Also, n(N∪H∪K)= n(N) + n(H) + n(K) – n(N ∩ H) – n(H ∩ K) – n(N ∩ K) + n(N ∩ H ∩ K) or, 2375 = 1250 + 1500 + 1000 – 500 – 375 – 625 + n(N ∩ H ∩ K) or, n(N ∩ H ∩ K) = 125 ∴ 125 people liked all three types of movies (ii) Illustrating the above information in a Venn-diagram: (iii) From Venn-diagram, the number of people who liked exactly two types of movies = no (N ∩ H) + no (H ∩ K) + no (N ∩ K) = 375 + 250 + 500 = 1125 (iv) The number of people who liked at most one type of movies = no (N) + no (H) + no (K) + n (N ∪ H ∪ K) = 250 + 750 + 125 + 125 = 1250 Example 4: A mobile house keeps the records of sales of different brands of mobiles every year. Last year, it was found that 30 % customers bought I-phone sets, 41% bought Samsung sets, and 35% bought Xiaomi sets. Similarly, 5% customers bought I-Phone and Xiaomi sets, 7 % bought Xioami and Samsung sets, 6 % bought I-phone and Samsung sets and 2 % bought all three brands of mobile sets. (i) Show the above data in a Venn-diagram. (ii) If 240 customers bought none of these types of mobile sets, find the total number of customers kept in the record last year. (iii) Find the number of customers who bought at most two types of mobile sets (iv) Find the number of customers who bought at least two types of mobile sets Solution: Let I, X and S denote the sets of customers who bought I-phone, Xioami and Samsung mobile sets respectively and total number of customers be 100 (i.e. 100%). Here, n (U) = 100, n (I) = 30, n(X) = 35, n(S) = 41, n (I ∩ X) =5, n(X ∩ S) = 7, n (S ∩ I) = 6 and n (I ∩ X ∩ S) = 2 (i) Showing the above information in a Venn-diagram: (ii) n(I ∪ X ∪ S) = n (I) + n (X) + n (S) – n (I ∩ X) – n (X ∩ S) – n (S ∩ I) + n (I ∩ X ∩ S) = 30 + 35 + 41 – 5 – 7 – 6 + 2 = 90 N H 125 250 375 500 125125 750 250 K U I X 10% 21% 3% 2% 4% 30% 25% 5% S U

Vedanta Excel in Mathematics - Book 10 34 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets Also, n (I ∪ X ∪ S ) = n (U) – n (I ∪ X ∪ S) = 100 – 90 = 10 Thus, 10 % of customers bought none of the three types of mobile sets. Let, the total number of customer be x. Then, 10 % of x = 240 ∴ x = 2400 Hence, 2400 customers were recorded last year. (iii) The number of customers who bought at most two types of mobile sets = n (I ∩ X ∩ S ) = (100 – 2) % of 2400 = 98 % of 2400 = 2352 (iv) The number of customers who bought at least two types of mobile sets = no (I ∩ X) + no (X ∩ S) + no (S ∩ I) + n (I ∩ X ∩ S) = (3% + 4% + 5% + 2%) of 2400 = 14 % of 2400 = 336 Example 5: In a survey of a group of people, it was found that 60 of them have business, 45 have government jobs, 125 have farming, 27 have business only, 15 have government jobs only, 10 have business and government jobs only, 5 have government jobs and farming only. (i) Draw a Venn-diagram to illustrate the above information (ii) Find how many people have all three professions. (iii) How many people were there in the survey? Solution: Let B, G, and F be the sets of people who have business, government jobs, and farming respectively. Here, n(B) = 60, n(G) = 45, n(F) = 125, no (B) = 27, no (G) = 15, no (B ∩ G) = 10, no (G ∩ F) = 5 Let the number of people who have all three professions be x. (i) ∴ The number of people who have all three professions = n(B ∩ G ∩ F) = x = 15 (iii) Again, the number of people who have business and farming only = no (B ∩ F) = n(B)– (27 + 10 + x) = 60 – 37 – 15 = 8 Also, the number of people who have farming only = no (F) = n(F) – (5 + x + 8) = 125 – 13 – 15 = 97 ∴ The total number of people, n(U) = n(B ∪ G ∪ F) = 27 + 10 + 15 + 15 + 5 + 97 + 8 = 177 B G 27 10 x 15 5 F U (ii) From the Venn-diagram, n(G) = 10 + x + 5 + 15 or, 45 = x + 30 or, x = 15

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 35 Vedanta Excel in Mathematics - Book 10 Sets EXERCISE 1.2 General section 1. a) In the given Venn-diagram, A, B, and C represent the sets of tourists who visited Bardiya, Khaptad, and Rara respectively. If n(U) = 200, find (i) n(A) (ii) n(B) (iii) n(C) (iv) n(A ∪ B ∪ C) (v) n(A ∪ B ∪ C ) (vi) n(A ∩ B ∩ C) (vii) n(only A) (viii) no (B) (ix) no (C) (x) n(A ∩ B) (xi) n(B ∩ C) (xii) n(A ∩ C) (xiii) no (A ∩ B) (xiv) no (B ∩ C) (xv) no (A ∩ C) b) The sets of people who like tea, milk and coffee are denoted by T, M and C in the adjoining Venn-diagram. If n (U) =80, find: (i) n (like milk and tea) (ii) n (like all three drinks) (iii) n (like only one drink) (iv) n (like exactly two drinks) (v) n (like at least one drink) (vi) n (like neither of the drinks) c) A, B, and C are the subsets of a universal set U. Draw a Venn-diagram and insert the cardinality of the following information. n(U) = 50, n(A) = 18, n(B) = 21, n(C) = 20, n(A ∩ B) = 7, n(B ∩ C) = 5, n (A ∩ C) = 8 and n(A ∩ B ∩ C) = 3. 2. a) A universal set U has three subsets A, B and C. If n (U) = 100, n (A) = 40, n (B) = 55, n (C) = 30, n (A ∩ B) = 25, n (B ∩ C) = 20, n (C ∩ A) = 15 and n (A ∩ B ∩ C) = 10, find the values of n (A ∪ B ∪ C) and n(A ∪ B ∪ C ). Show the above data in a Venn-diagram. b) Given that, n(U) = 150, n(P) = 48, n(Q) = 56, n(R) = 44, no (P) = 17, no (Q) = 20, no (P ∩ Q) = 10, no (Q ∩ R) = 8. By using a Venn-diagram, find (i) n(P ∩ Q ∩ R) (ii) no (P ∩ R) (iii) no (R) (iv) n (P ∪ Q ∪ R) Creative section-A 3. a) A survey conducted in a group of students showed that 50 liked singing, 40 liked dancing, 35 liked acting, 20 liked singing as well as dancing, 12 liked dancing as well as acting, 18 liked acting as well as singing, and 7 liked all three activities. If every student liked at least one activity, answer the following questions by drawing a Venn-diagram. (i) How many students were asked this question? (ii) How many students liked dancing only? (iii) How many students liked acting only? (iv) How many students liked only one activity? A B 38 25 20 40 5 32 15 C U T M 14 5 6 15 10 16 9 C U

Vedanta Excel in Mathematics - Book 10 36 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets b) In an examination, out of 150 students, 65 succeeded in English, 75 in Mathematics, 60 in Nepali, 30 in English as well as in Mathematics, 25 in English as well as in Nepali, 40 in Nepali as well as in Mathematics, and 10 succeeded in all three subjects. (i) Draw a Venn-diagram to show the above information. (ii) Find the number of students who didn’t succeed in all three subjects (iii) Find the number of students who succeeded in English only. (iv) Find the number of students who succeeded in exactly one subject. c) A hotel provides the extra facilities: swimming, gym and fun-park facilities to its guests. Among 300 guests arrived in the hotel, it was found that 60% guests used swimming, 50% used gym and 40% used fun-park. Moreover, 30% used swimming and gym facilities, 20% used gym and fun-park, 15% used fun-park and swimming while 10% used all three facilities. By using the Venn-diagram, find the number of guests who used (i) none of the facilities (ii) only one facility (iii) only swimming and gym 4. a) In a survey of 900 tourists who arrived in Nepal during ‘Visit Nepal 2020’, 450 preferred to go trekking, 300 preferred rafting, 400 preferred forest safari, and 100 preferred none of these activities. If 200 preferred trekking and rafting, 110 preferred trekking and safari, 100 preferred rafting and safari, (i) how many of them preferred either of these activities? (ii) How many of them preferred all of these activities? (iii) Illustrate the information in a Venn-diagram. (iv) How many of them preferred only two activities? b) In a factory, there are 80 workers and 3 machines-A, B and C. Each worker can operate at least one machine. 50 workers can operate machine-A, 45 can operate machine-B and 40 can operate machine-C. Moreover, 25 workers can operate machines A and B, 20 workers can operate machines B and C, while 15 workers can operate machines A and C. By using a Venn-diagram, find the number of workers who can operate: (i) at least one of these machines (ii) all three machines (iii) exactly two machines 5. a) In an examination, 60% students passed in Nepali, 45% passed in English and 40% passed in Science. Likewise, 20% of them passed in Nepali and English, 15% passed in English and Science, 25% passed in Nepali and Science, and 5% students passed in all three subjects. (i) Represent the above information in a Venn-diagram. (ii) If 50 students failed in all three subjects, find the total number of students. (iii) How many students passed in English? b) Out of some people interviewed in a community about the preference of social media, it was recorded that 62% people use Facebook, 60% use Twitter and 53% use Instagram. Moreover, 35% people use Facebook and Twitter, 30% use Twitter and Instagram, 25% use Facebook and Instagram, and 10% use all these three social media. (i) Represent the above information in a Venn-diagram. (ii) If 30 people did not use any of these media, find the total number of people participated in the survey. (iii) Find the number of people who use only Twitter and Instagram.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 37 Vedanta Excel in Mathematics - Book 10 Sets Creative section-B 6. a) In a group of students, 20 study Economics, 18 study History, 21 study Science, 7 study Economics only, 10 study Science only, 6 study Economics and Science only and 3 study Science and History only. (i) Represent the above information in Venn-diagram (ii) How many students study all the subjects? (iii) How many students are there altogether? b) In a survey of a group of people, it was found that 30 of them have business, 35 have services, 25 have farming, 12 have business only, 15 have services only, 10 have business and services only and 6 have services and farming only. (i) Draw a Venn-diagram to illustrate the above information (ii) Find how many people have all three occupations? (iii) How many people were in the survey? 7. a) Each student in a class of 32 plays at least one game: cricket, football, or basketball. 20 play cricket, 18 play basketball and 25 play football. 9 play cricket and basketball, 13 play football and basketball, and 5 play all three. Find the number of students who play: (i) cricket and football (ii) cricket and football but not basketball (iii) Show the information in a Venn-diagram. b) In a group of 150 people, the number of people who like Coke, Pepsi and Sprite are equal. There are 35 people who like Coke and Pepsi, 40 people who like Pepsi and Sprite, 44 who like Sprite and Coke, 25 people like all three drinks and 34 people don’t like any of these drinks. (i) Show the information in a Venn-diagram. (ii) How many people like Coke? (iii) How many people like Pepsi but not Coke or Sprite? Project Work and Activity Section 8. Make the groups of your friends and conduct a survey inside your classroom: how many of your friends like Cricket, Football, and Cricket as well as Football. Then, tabulate the data as shown in the table given below: Total No. of students No. of students who like cricket No. of students who like football No. of students who like cricket as well as football a) Find the number of students in the following cases by using the cardinality relations of two sets. (i) students who like only cricket (ii) students who like only football (iii) students who like cricket or football (iv) students who like cricket and football (v) students who like neither cricket nor football. b) Compare these results with the surveyed data. c) Draw a Venn-diagram to show the collected data.

Vedanta Excel in Mathematics - Book 10 38 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sets 9. Conduct a survey inside your classroom and collect the data about how many of your friends like tea, coffee, milk, tea and coffee, tea and milk, coffee and milk and tea, coffee as well as milk. Then tabulate the data and compute the following numbers by using cardinality relation of three sets. a) Number of friends who like tea, coffee and milk. b) Number of friends who do not like any of these three drinks. c) Number of friends who like only (i) tea and coffee (ii) tea and milk (iii) coffee and milk. d) Number of friends who like only (i) tea (ii) coffee (iii) milk e) Show your data in a Venn-diagram OBJECTIVE QUESTIONS Tick the correct alternatives. 1. If n (A) = 11 and n (B) = 18, what is the possible maximum value of n (A ∩ B)? (A) 0 (B) 11 (C) 18 (D) 29 2. Out of 30 students of a class, 15 like to play volleyball, 20 like to play basketball and each student like to play at least one of the game. How many students like to play volleyball and basketball both? (A) 35 (B) 20 (C) 15 (D) 5 3. In a school, there are 10 teachers who teach Mathematics or Science. Out of them 6 teach Mathematics and 4 teach both the subjects, how many teachers teach Science? (A) 8 (B) 7 (C) 5 (D) 6 4. Out of 77 districts of Nepal, 27 districts have shared their boarder with India, 15 districts have shared their boarder with China and 37 districts have not shared their boarder with India and China both. How many districts have shared their boarder with China but not with India? (A) 15 (B) 13 (C) 25 (D) 38 5. In a group of 60 students, 20 passed in both Mathematics and Science and 30 passed in only one subject. How many students failed in both the subjects? (A) 40 (B) 30 (C) 10 (D) 50 6. In a survey it was found that 30 people enjoy Nepali movie, 10 people do not enjoy Nepali movies and 15 do not enjoy Hindi movie. How many people enjoy Hindi movie? (A) 25 (B) 20 (C) 15 (D) 40 7. In a group of people, 40 people favored tea, 24 people favored tea but not coffee and 30 people favored coffee, how many people favored coffee but not tea? (A) 16 (B) 10 (C) 6 (D) 14 8. In a survey of a community, 55% of the people like yoga, 65% like jugging and 10 % like neither yoga nor jugging. What percent of people like only one? (A) 30% (B) 25% (C) 35% (D) 60% 9. In a group of children, 25 like sweet and 15 like balloon. If the number of children who like sweet only is twice the number of children who like balloon only, how many children like sweet as well as balloon? (A) 5 (B) 10 (C) 20 (D) 35 10. Of the students who are in three athletic teams: basketball, football and tennis in a certain school, 5 are in all three teams, 48 are in only one team and 27 are in only two teams, how many students are there in all team? (A) 80 (B) 70 (C) 26 (D) 48

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 39 Vedanta Excel in Mathematics - Book 10 Assessment –I 1. The result of a terminal examination of Shivapuri Secondary School has been published. Out of 150 students from class X, 40 students have secured A+ grade in English, 20 have secured A+ grade in Mathematics and 100 students have secured below A+ grades in both subjects. According to this information, answer the following questions. (a) Represent the above information in a Venn-diagram. (b) How many students have secured A+ grade in both subjects? (c) Find the number of students who have secured A+ grade in exactly one of these two subjects. (d) Find the number of students have secured below A+ grades in at most one of these two subjects. 2. In a group of 60 people, a survey was conducted regarding the preference of tea and coffee and the following information was found. - One fourth of people like tea and coffee both. - 15% people like none of these drinks - Ratio of number of people who like tea only and coffee only is 4: 5. Based on the above information, answer the following questions. (a) Show the information in a Venn-diagram. (b) Find the number of people who like tea only. (c) Find the ratio of the number of people who like both the drinks and coffee only. (d) Find the number of people who don’t like at least one of the drinks. 3. There are only chicken and fish as non-veg items in a party. In a group of 100 non-veg people, 40 like only chicken and 30 like only fish. If the number of people who do not like any of the two non-veg items is double of the number of people who like both items, answer the following questions. (a) How many people like either chicken or fish? (b) Illustrate the above information in a Venn-diagram. (b) How many people like only one of these items? (c) Find the number of people who like at most one item (d) How many people do not like chicken at all? 4. A survey regarding mobile games was conducted among 110 youths and found that the numbers of youths who are fond of in PUBG, Free Fire and Ludo are 55, 40 and 35 respectively. 20 youths are fond of in PUBG and Free Fire, 10 are fond of in Free Fire and Ludo, 15 are fond of in Ludo and PUBG and 20 play none of these games. (a) Show the obtained information in a Venn-diagram. (b) Find the number of youths who are fond of in exactly one of these games. (c) Find the number of youths who are fond of in exactly two of these games. (d) Find the number of youths who are fond of in at most one of these games.

Vedanta Excel in Mathematics - Book 10 40 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Unit 2 Compound Interest Classwork - Exercise 2.1 Simple interest – Looking back 1. Mr. Sharma deposits Rs 20,000 for 3 years in a commercial bank at the rate of 6% p.a. a) Write the values of principal, rate and time duration. b) What will be the simple interest after 3 years? c) What will be the amount of balance at the end of 3rd year? d) What is the meaning of the rate of interest being 6% p.a.? 2. Let’s fill the blank spaces with correct answers. a) The simple interest on Rs 20,000 at 8% per annum for 3 years is ……………… b) The simple amount on Rs 50,000 at 4% per year for 6 years is………………… c) If a sum of Rs 5,600 amounts to Rs 8,400 in 5 years, the rate of interest is…… d) A sum yields an interest of Rs 1,100 at the rate of 10% p.a. in 2 years 6 months, the required sum is ………………… 2.2 Principal and Interest - review Interest is the charge for the privilege of borrowing money from a lender. When we borrow a loan from a bank, we should pay interest for the use of borrowed money under the certain condition. Similarly, if we deposit money in a bank, the bank pays us interest. The borrowed or deposited sum is the principal. There are three major variables which are used in the calculation of interest: principal, rate of interest, and duration of time. Interest is expressed as a rate percent per year. There are two types of calculation of interest. (i) Calculation of simple interest (ii) Calculation of compound interest Simple interest is calculated always from the original principal at the given rate percent for any interval of time. We calculate simple interest by using the following formula, Simple interest (I) = P × T × R 100 Where, P is the principal; T is the time; and R is the rate of interest.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 41 Vedanta Excel in Mathematics - Book 10 Compound interest 2.3 Compound interest Let’s study the following situations. Situation-I: A farmer borrows a loan of Rs 50,000 from Agricultural Development Bank to promote his fish farming for 2 years at the rate of 5% p.a. How much interest does he pay? Situation-II: A farmer borrows a loan of Rs 50,000 from Agricultural Development Bank to promote his fish farming for 2 years at the rate of 5% p.a. However, the interest of the first year is added to the principal after 1 year and new principal is obtained for the second year. How much interest does he pay? For the first situation: Simple interest (I) = P × T × R 100 = 50000 × 2 × 5 100 = Rs 5,000 Thus, the farmer pays Rs 5,000 simple interest. For the second situation: Simple interest of the 1st year (I1 ) = P × T × R 100 = 50000 × 1 × 5 100 = Rs 2,500 Principal for the second year = Rs 50,000 + Rs 2,500 = Rs 52,500 Simple interest of the 2nd year (I2 ) = 52500 × 1 × 5 100 = Rs 2,625 Total interest in 2 years (I) = I1 + I2 = Rs 2,500 + Rs 2,625 = Rs 5,125 Since, the interest of the first year is added to the principal and new principal is obtained for the second year, the farmer has to pay Rs 5,125 compound interest. Facts to remember 1. When the interest accrued during the first time period (say 1 year), it is added to the original principal and the amount so obtained is taken as the principal for the second time period (i.e., the next 1 year) and this keeps going on up to the fixed time then the interest so calculated is called compound interest. 2. In case of simple interest, the principal remains same. However, the principal is increased every year in case of compound interest compounded annually. 2.4 Simple Interest vs. Compound interest Suppose a person wishes to deposit Rs 10,000 in a bank at 10% per year for 4 years. Let’s observe calculation of simple interest and compound interest in the following table then suggest him/her about the better system of accruing interest. Time System of Simple interest System of Compound interest Principal Interest Principal Interest 1st year Rs 10,000 Rs 1,000 Rs 10,000 Rs 1,000 2nd year Rs 10,000 Rs 1,000 Rs 11,000 Rs 1,100 3rd year Rs 10,000 Rs 1,000 Rs 12,100 Rs 1,210 4th year Rs 10,000 Rs 1,000 Rs 13,310 Rs 1,331 Total Simple Interest (S.I.) = Rs 4,000 Compound Interest (C.I.) = Rs 4,641 www.geogebra.org/classroom/krxt7wcb Classroom code: KRXT 7WCB Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 42 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Compound interest In the 1st year, there is no difference between the simple interest and compound interest. For the compound interest system, the amount of 1st year is the principal for the 2nd year. Similarly, the amount of 2nd year is the principal of the 3rd year and so on. Facts to remember 1. As more principal grows more interest, the compound interest is more than the simple interest in each year after the first year when the interest is compounded annually. 2. In the above example, the compound interest of the first year is Rs 1,000 and that of second year is Rs 1,100. If we consider Rs 1,000 as principal and Rs 1,100 as the amount then the percentage of their difference is 10% which the rate of interest. It satisfies for the known interests of two successive years to find the rate of interest. 2.5 Derivation of formula of Compound Amount and Compound Interest Let, the original principal = Rs P the rate of interest = R% p.a. the time interval = T years Now, For the 1st year Interest (I1 ) = P × T × R% = P × 1 × R% = PR% ∴Amount (A1 ) = P + PR% = P (1 + R%) For the 2nd year Principal (P2 ) = Amount of the 1st year = P (1 + R%) Interest (I2 ) = P (1 + R%) × 1 × R% = P (1 + R%) R% ∴Amount (A1 ) = P (1 + R%) + P (1 + R%) R% = P (1 + R%) (1 + R%) = P(1 + R%)2 For the 3rd year Principal (P3 ) = Amount of the 2nd year = P (1 + R%)2 Interest (I3 ) = P (1 + R%)2 × 1 × R% = P (1 + R%)2 R% ∴Amount (A3 ) = P(1 + R%)2 + P(1 + R%)2 R%= P(1 + R%)2 (1 + R%) = P(1 + R%)3 Similarly, Amount in the 4th year (A4 ) = P (1 + R%)4 ∴ The amount in T year = P (1 + R%)T = P 1 + R 100 T Also, the interest in T year = P 1 + R 100 T – P = P 1 + R 100 – 1 T Thus, Compound Amount (C.A.) = P 1 + R 100 T and Compound Interest (C.I.) = P 1 + R 100 – 1 T 1. When the interest is compounded annually but the rate being different in different years, for example, R1 % in the first year, R2 % in the second year, R3 % in the third years, then the amount in 3 years is calculated as:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 43 Vedanta Excel in Mathematics - Book 10 Compound interest Compound Amount (C.A.) = P 1 + R1 100 1 + R2 100 1 + R3 100 Also, Compound Interest (C.I.) = P 1 + R1 100 1 + R2 100 1 + R3 100 – 1 2. When the interest is compounded annually but the time is given in ‘T’ years and ‘M’ months, then the amount is calculated as: Compound Amount = P 1 + R 100 T 1 + MR 1200 Also, Compound Interest = P 1 + R 100 T 1 + MR 1200 – 1 2.6 Interest compounded Half-yearly and Quarter-yearly If the compound interest is payable half yearly (every 6 months), then rate = R 2 % per half yearly and time = 2T half years. Now, the compound amount half yearly = P 1 + 2T R 2 × 100 = P 1 + 2T R 200 Also, the compound interest half yearly = P 1 + – 1 2T R 200 Similarly, if the compound interest is payable quarter-yearly (every 3 months), then rate = R 4 % per quarter-yearly and time = 4T quarter years. So, the compound amount quarter-yearly = P 1 + 4T R 4 × 100 = P 1 + 4T R 400 And the compound interest quarter-yearly = P 1 + – 1 4T R 400 Facts to remember 1. The interest compounded annually and semi-annually on a sum at the certain rate are equal up to the first 6 months then the interest compounded semi-annually is always more than the interest compounded annually in each 6 month period. 2. The interest compounded annually and quarterly on a sum at the certain rate are equal up to the first 3 months then the interest compounded quarterly is always more than the interest compounded annually in each 3 month period. Worked-out Examples Example 1: Calculate the compound interest of Rs 7,000 at 10% per annum for 3 years without using the formula.

Vedanta Excel in Mathematics - Book 10 44 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Compound interest Solution: Here, principal (P) = Rs 7,000 Rate of interest (R) = 10% p.a. Time (T) = 3 years Now, the interest in the first year (I1 ) = PTR 100 = 7,000 × 1 × 10 100 = Rs 700 ∴ The principal for the second year = Rs 7,000 + Rs 700 = Rs 7,700 Now, the interest in the second year (I2 ) = 7,700 × 1 × 10 100 = Rs 770 ∴ The principal for the third year = Rs 7,700 + Rs 770 = Rs 8,470 Again, the interest in the third year (I3 ) = 8,470 × 1 × 10 100 = Rs 847 ∴ The compound interest (C.I.) = I1 + I2 + I3 = Rs 700 + Rs 770 + Rs 847 = Rs 2,317 Example 2: Mr. Rayamajhi deposited Rs 60,000 in a bank at the rate of 10% per annum for 2 years. a) How much amount will he get if the bank pays the simple interest? b) How much amount will he get if the bank pays the interest compounded yearly? c) How much amount will he get if the bank pays the interest compounded half-yearly? Solution: Here, principal (P) = Rs. 60,000, rate of interest (R) = 10% p.a. and time duration (T) = 2 years. a) Simple interest = P×T×R 100 = 60,000 × 2 × 10 100 = Rs 12,000 ∴ Amount = P + I = Rs 60,000 + Rs 6,000 = Rs 66,000 Hence, he will get Rs 66,000 simple amount. b) Yearly compound amount = P 1 + R 100 T = Rs 60,000 1 + 10 100 2 = Rs 60,000 × 11 10 2 = Rs 60,000 × 121 100 = Rs 72,600 Hence, he will get the amount of Rs 72,600 if the interest is compounded yearly. c) Half-yearly compound amount = Rs 60,000 1 + R 200 2×2 = Rs 60,000 × 21 20 4 = Rs 60,000 × 194481 160000 = Rs 72,930.38 Hence, he will get the amount of Rs 72,930.38 if the interest is compounded half-yearly.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 45 Vedanta Excel in Mathematics - Book 10 Compound interest Example 3: Dheepshikha borrowed Rs 1,50,000 from a bank at the rate of 8% p.a. for 1 year. a) How much interest should she pay if it is compounded annually? b) How much interest should she pay if it is compounded semi-annually? c) How much interest should she pay if it is compounded quarterly? Solution: Here, principal (P) = Rs 1,50,000, Time (T) = 1 year and rate of interest (R) = 8% p.a. a) Interest compounded annually = P 1 + – 1 T R 100 = 1,50,000 1 + – 1 1 8 100 = 1,50,000 (1.08 – 1) = 1,50,000 × 0.08 = Rs 12,000 Hence, she should pay Rs 12,000 if the interest is compounded annually. b) Interest compounded semi-annually = P 1 + – 1 2T R 200 = 1,50,000 1 + – 1 2 × 1 8 200 = 1,50,000 [(1.04)2 – 1] = 1,50,000 (1.0816 – 1) = 1,50,000 × 0.0816 = Rs 12,240 Hence, she should pay Rs 12,240 if the interest is compounded semi-annually. c) Interest compounded quarterly = P 1 + – 1 4T R 400 = 1,50,000 1 + – 1 4 × 1 8 400 = 1,50,000 [(1.02)4 – 1] = 1,50,000 (1.08243216 – 1) = 1,50,000 × 0.08243216 = Rs 12,364.82 Hence, she should pay Rs 12,364.82 if the interest is compounded quarterly. Example 4: A sum of Rs 24,000 is deposited in a bank X at the rate of 5% p.a. simple interest. If the sum would have been deposited in bank Y at the same rate of compound interest, which bank would give more interest after 3 years and by how much? Solution: Here, principal (P) = Rs 24,000, Time (T) = 3 years and rate of interest (R) = 5% p.a. For bank X: simple interest (S.I.) = P×T×R 100 = Rs 24,000 × 3 × 5 100 = Rs 3,600 For bank Y: compound interest (C.I.) = P 1 + – 1 T R 100

Vedanta Excel in Mathematics - Book 10 46 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Compound interest = Rs 24,000 1 + – 1 3 5 100 = Rs 24,000 (1.157625 – 1) = Rs 3,783 ∴Difference of compound and simple interest = Rs 3,783 – Rs 3,600 = Rs 183 Hence, bank Y would give more interest by Rs 183. Example 5: Laxmi deposited Rs 16,000 at 10% p.a. interest compounded semi-annually for 11 2 years. Narayan deposited the equal sum at the same rate of interest compounded annually for the same duration of time. Who gets more interest and by how much? Solution: Here, principal (P) = Rs 16,000, time (T) = 11 2 years and rate (R) = 10 % p.a. Now, interest compounded semi annually (I1 ) = P 1 + R 2 ×100 – 1 2T = Rs 16,000 1 + 10 200 – 1 3 = Rs 16,000 21 × 21 × 21 20 × 20 × 20 – 1 = Rs 2,522 Also, interest compounded annually (I2 ) = P 1 + R 100 T 1 + 1 2 R 100 – 1 = Rs 16,000 1 + 10 100 1 1 + 10 200 – 1 = Rs 16,000 11 10 × 21 20 – 1 = Rs 16,000 231 – 200 200 = Rs 2,480 ∴ Difference between I1 and I2 = I1 – I2 = Rs 2,522 – Rs 2,480 = Rs 42 Hence, Laxmi will get more interest by Rs 42. Example 6: The compound interest on the sum of money in 2 years at the rate of 5 % per annum is Rs 364 more than simple interest. Find the sum. Solution: Here, time (T) = 2 years, rate (R) = 5 % and let the Principal be Rs P. Now, C.I. = P 1 + R 100 T – 1 = P 1 + 5 100 2 – 1 = 41P 400 Also, S.I = PTR 100 = P × 2 × 5 100 = P 10 From the question, C.I. – S.I. = Rs 364 or, 41P 400 – P 10 = Rs 364 or, P 400 = Rs 364 or, P = Rs 1,45,600 Hence, the required sum is Rs 1,45,600. Alternative process I2 = P 1 + R 100 T 1 + MR 1200 – 1 = 16,000 1 + 10 100 1 1 + 6 × 10 1,200 – 1 = 16,000 11 10 × 21 20– 1 = Rs 2,480 Answer checking: C.I. = 1,45,600 1 + 5 100 – 1 2 = Rs 14,924 S.I. = 1,45,600 × 2 × 5 100 = Rs 14,560 C.I. – S.I. = Rs 14,924 – Rs 14,560 = Rs 364 which is given in the question.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 47 Vedanta Excel in Mathematics - Book 10 Compound interest Example 7: Simple interest on a sum of money for 2 years at 4% p.a. is Rs 800. Find the compound interest on the same sum and at the same rate for 1 year if the interest is reckoned half-yearly? Solution: Here, time (T) = 2 years, rate (R) = 4% p.a. and simple interest (S. I.) = Rs 800 Now, S. I. = PTR 100 = P × 2 × 4 100 = 2P 25 or, 800 = 2P 25 or, P = Rs 10,000 Again, interest compounded half-yearly: C. I.= P 1 + R 2 ×100 – 1 2T = 10,000 1 + 4 200 2 × 1 – 1 = 10,000 51 × 51 50 × 50 – 1 = 10,000 × 101 2,500 = Rs 404 Hence, the required compound interest is Rs 404. Example 8: At what rate percent per annum compound interest will Rs 2,500 amounts to Rs 2,704 in 2 years? Solution: Here, principal (P) = Rs 2,500, time (T) = 2 years and C.A. = Rs 2,704 Now, C.A. = P 1 + R 100 T 2704 = 2500 1 + R 100 2 or, 2704 2500 = 100 + R 100 2 or, 52 50 2 = 100 + R 100 2 or, 100 + R 100 = 52 50 or, R = 5200 50 – 100 = 200 50 = 4 % Hence, the required rate of interest is 4 % p.a. Example 9: Mr. Balendra borrowed Rs 4,00,000 from a commercial bank at the rate of 12% p.a. compounded half yearly for 2 years. After one year, the bank changed its policy to pay the interest compounded quarterly at the same rate. a) How much interest should he have to pay in the first year? b) How much interest should he have to pay in the second year? c) How much less interest would he have to pay if the bank had charged interest compounded annually for 2 years? Solution: Here, principal (P) = Rs 4,00,000, Time (T) = 2 years and rate of interest (R) = 12% p.a. Answer checking: P = Rs 2,500 T = 2 years R= 4% p.a. C.A. = Rs 2,500 1 + 4 100 2 = Rs 2,704 which is given in the question.

Vedanta Excel in Mathematics - Book 10 48 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Compound interest a) For the first year, half-yearly compound interest (I1 )= P 1 + R 2 ×100 – 1 2T = 4,00,000 1 + 12 200 2 × 1 – 1 = 4,00,000 [(1.06)2 – 1] = Rs 4,00,000 × 0.1236 = Rs 49,440 b) For the second year, principal = amount of the first year = Rs 4,00,000 + Rs 49,440 =Rs 4,49,440 ∴ Interest compounded quarterly (I2 ) = P 1 + R 4 ×100 – 1 4T = 4,49,440 1 + 12 400 4 × 1 – 1 = 4,49,440 [(1.03)4 – 1] = 4,49,440 × 0.12550881 = Rs 56,408.68 So, total interest paid in 2 years (C.I.) = I1 + I2 = Rs 49,440 + Rs 56,408.68 = Rs 1,05,848.68 c) Interest compounded annually for 2 years = P 1 + R 100 T – 1 = Rs 4,00,000 1 + 12 100 2 – 1 = Rs 4,00,000 [(1.12)2 – 1] = Rs 4,00,000 × 0.2544 = Rs 1,01,760 ∴Difference in interests of 2 years = Rs 1,05,848.68 – Rs 1,01,760 = Rs 4,088.68 Hence, he would have to pay Rs 4,088.68 less interest if the bank had charged the interest compounded annually for 2 years. Example 10: A person took a loan of Rs 3,00,000 for 2 years at the rate of 10% annual compound interest. To reduce the interest and the loan partly, he/she paid Rs 1,40,000 at the end of the first year. a) How much should he/she has to pay at the end of the second year to clear his/her debt? b) Find the total interest paid by him/her in two years. c) If he/she had paid the loan only at the end of the second year, how much less or more interest would he/she pay? Solution: Here, principal (P) = Rs 3,00,000, time (T) = 2 years and rate of interest (R) = 10% p.a. a) For the first year: principal (P1 ) = Rs 3,00,000 and time (T) = 1 year Now, compound amount (C.A.1 ) = P1 1 + R 100 T

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 49 Vedanta Excel in Mathematics - Book 10 Compound interest = Rs 3,00,000 1 + 10 100 1 = Rs 3,30,000 Money paid at the end of the first year = Rs 1,40,000 For the second year, principal (P2 ) = Rs 3,30,000 – Rs 1,40,000 = Rs 1,90,000 Again, compound amount (C.A.2 ) = P2 1 + R 100 T = Rs 1,90,000 1 + 10 100 1 = Rs 2,09,000 Hence, he/she has to pay Rs. 2,09,000 at the end of second year to clear the bedt. b) Total amount paid in two years to clear the debt = Rs. 1,40,000 + Rs. 2,09,000 = Rs. 3,49,000 Hence, the total interest paid in two years = Rs. 3,49,000 – Rs. 3,00,000 = Rs. 49,000 c) Again, compound interest of 2 years (C.I.) = P 1 + R 100 T – 1 = 3,00,000 1 + 10 100 2 – 1 = Rs 3,00,000 [1.21 – 1] = Rs 3,00,000 × 0.21 = Rs. 63,000 Difference in two interests = Rs. 63,000 – Rs. 49,000 = Rs. 14,000 Hence, he/she has to pay Rs 14,000 more interest if the loan had been paid only at the end of the second year. Example 11: Mrs. Rai invests Rs 25,000 for 3 years at a certain rate of interest compounded annually. At the end of one year the sum amounts to Rs 27,500. Calculate: (i) the rate of interest (ii) the amount at the end of second year (iii) the amount at the end of third year. Solution: Here, principal (P) = Rs 25,000, time (T) = 1 year and amount (A) = Rs 27,500 (i) Now, C.A. = P 1 + R 100 T 27,500 = 25,000 1 + R 100 1 or, 1 + R 100 = 27,500 25,000 = 11 10 or, R = 10% ∴The rate of interest is 10% p.a. (ii) Amount at the end of second year: C.A.= P 1 + R 100 T = 25,000 1 + 10 100 2 = 25,000 × 11 10 × 11 10 = Rs 30,250 (iii) Amount at the end of third year: C.A.= P 1 + R 100 T = 25,000 1 + 10 100 3 = 25,000 × 11 10 × 11 10 × 11 10 = Rs 33,275

Vedanta Excel in Mathematics - Book 10 50 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Compound interest Example 12: A floriculturist takes a loan of Rs 40,000 from an Agricultural bank. If the rate of compound interest is 5 paisa per rupee per year, in how many years does he pay the compound interest of Rs 6,305? Solution: Here, the rate of interest (R) = 5 paise per Re 1 per year = 500 paise, i.e. Rs 5 per Rs 100 per year = 5 % per annum Also, principal (P) = Rs 40,000 and C.I. = Rs 6,305 Now, C.I. = P 1 + R 100 – 1 T or, 6,305 = 40,000 1 + 5 100 – 1 T or, 6,305 40,000 = 21 20 – 1 T or, 1261 8000 = 21 20 T – 1 or, 21 20 T = 1261 8000 + 1 or, 21 20 T = 9261 8000 = 21 20 3 or, T = 3 years So, the required time is 3 years. Example 13: Mr. Gupta deposited a sum of Rs 50,000 in a commercial bank for 3 years. If the bank provided 4% interest compounded annually for the first year and the rate of interest was gradually increased by 1% every year, how much interest did he get in 3 years? Solution: Here, principal (P) = Rs 50,000 Rate of interest in the first year, R1 = 4 % p.a. Rate of interest in the second year R2 = 5 % Rate of interest in the third year R3 = 6 % p.a. Now, compound amount (C.A.) = P 1 + R1 100 1 + R2 100 1 + R3 100 = 50,000 1 + 4 100 1 + 5 100 1 + 6 100 = 50,000 × 104 100 × 105 100 × 106 100 = Rs 57,876 ∴ Compound interest (C.I.) = C.A. – P = Rs 57,876 – Rs 50,000 = Rs 7,876 Hence, he got the interest of Rs 7,876 in 3 years. Alternative process (C.A.) = P + C. I = Rs 40,000 + Rs 6,305 = Rs 46,305 Now, C.A. = P 1 + R 100 T or, Rs 46,305 = 40,000 1 + 5 100 T or, 46,305 40,000 = 21 20 T or, 21 20 T = 9,261 8000 or, 21 20 T = 21 20 3 ∴ T = 3 years