Allied Publication Pvt. Ltd. Kathmandu, Nepal Phone : 01-5388827 The Leading Allied Mathematics For Class 10 Approved by Curriculum Development Center, GON Sanothimi, Bhaktapur, Nepal Author Ashok Dangol M.Ed. Maths (TU)
Publisher Allied Publication Pvt. Ltd. Kathmandu, Nepal Ph.: 01-5388827, 5378629 Email : [email protected] Author Ashok Dangol Copyright Author Edition 1 st 2080 Language Editor Nariswor Gautam Graphics Sunil Shrestha Printing Quantity 5000 Special Thanks Nabaraj Pathak Dileep Maharjan Jit Bahadur Khanal Binod Kumar Yadav Ashim Maskey Santosh Gupta Saraswati Shakya Sarala Devkota Rasmi Waiba Tamang Dhan Bahadur K.C. Lalbabu Prasad Yadav Jibnath Sharma Chandra Bahadur Dangol Nabaraj Paudel Nishchal Lama Dilkumar Maharjan Kamal Nepal ICT Supporter Ashok Dangol
This Allied The Leading Mathematics - 10 is basically meant for making the teachers and taught active while teaching and learning mathematics. The contents and extent of the series are strictly contained and arranged in accordance with the new vision and mission of the latest New Curriculum 9-10 of Secondary Level of CDC, Nepal. This series is basically an outcome of my untiring effort and patience. The long and dedicated service in teaching and popularization of mathematics has been a great asset in preparing this series. It has been designed as a textbook for English medium private and government school students with a new approach. This book provides maximum benefit to both teachers and students because of the following unique features: Unique features of this book; Ö Arranged especially focusing on child psychology of teaching and learning mathematics which are based on the Areas of Secondary Curriculum 9-10. Ö Prepared with the firm belief that “Mathematics begins at Home, grows in the Surroundings and takes shape in School (HSS Way)”, and sincere attempts have been made to make the learner, the teacher and reader feel “Mathematics is Fun, Mathematics is Easy and Mathematics is Everywhere (FEE Concept)”. Ö Written the focus of students’ activities and easily perform teaching-learning activities for teachers. Ö Well arranged the four colours of the whole book supports to find easily Themes, Chapters, Lessons, Examples, Practices, and other topics from Content. Ö Every Theme begins with its estimated teaching hours (Theory + Practical), competency, learning outcomes, revision contents for pre-knowledge. Ö Highlighted the important terms, notes and key points. Ö Included sufficiently all types of Classwork Examples and Home Assessments from simple to complex with suitable figures and reasons. Ö Provided QR Code to present from Projector and learn from Mobile set. Ö Included Project Works and Their Formats as self-practice to the students at home for some days' activities to memories long time about the entire chapters. Ö Included Confidence Level Tests as self-evaluations for Success and Competent by themselves. Ö Available Individual Practical Evaluation Sheets at the end. It is very much hoped that with all the above features, this book will be found really fruitful by teachers and students alike. Thank Allied Publication Pvt. Ltd., Kathmandu, Nepal for taking responsibility for publishing this book, Nariswor Gautam for language editing, and Sunil Shrestha for an attractive design. I would like to extend my sincere gratitude to the persons whose ideas or creations are directly or indirectly incorporated into the text. I would like to extend my thanks to the teachers and the students who helped me to verify the answers and to check the manuscript of this book. Also, many thanks to the schools that applied this book and suggested it to me. Finally, I heartily welcome criticisms, feedbacks and suggestions from readers so that it may appear with revise from in the coming edition and will be gratefully and thankfully acknowledged and honored. Author Preface to First Edition
Units Chapters Lessons Pages I SETS 1. Sets REVISITING SETS 10 1.1 Cardinality of Operations of Two Sets 11 Classwork Examples 16 Practice - 1.1 21 1.2 Cardinality of Operations of Three Sets 26 Classwork Examples 29 Practice - 1.2 32 Confidence Level Test - I 35 Additional Practice - I 36 II ARITHMETIC 2. Compound Interest REVISITING ARITHMETIC 40 2.1 Compound Interest 42 Classwork Examples 46 Practice - 2.1 54 3. Compound Growth and Depreciation 3.1 Compound Growth 59 Classwork Examples 62 Practice - 3.1 65 3.2 Compound Depreciation 69 Classwork Examples 71 Practice - 3.2 73 4. Money Exchange 4.1 Currency and Exchange Rate 76 Classwork Examples 78 Practice - 4.1 81 Confidence Level Test - II 86 Additional Practice - II 87 III MENSURATION 5. Area and Volume REVISITING MENSURATION 92 5.1 Right Pyramid 95 Classwork Examples 101 Practice - 5.1 102 5.2 Right Cone 106 Classwork Examples 110 Practice - 5.2 112 5.3 Combined Solids 115 Classwork Examples 119 Practice - 5.3 121 5.4 Cost Estimation on Construction 124 Classwork Examples 130 Practice - 5.4 135 Confidence Level Test - III 139 Additional Practice - III 140 CONTENT
IV ALGEBRA 6. Sequence and Series REVISITING ALGEBRA 144 6.1 Mean/s of Arithmetic Sequence 146 Classwork Examples 148 Practice - 6.1 152 6.2 Sum of Arithmetic Series 155 Classwork Examples 157 Practice - 6.2 163 6.3 Mean/s of Geometric Sequence 167 Classwork Examples 170 Practice - 6.3 174 6.4 Sum of Geometric Series 177 Classwork Examples 178 Practice - 6.4 180 7. Quadratic Equation 7.1 Solution of Quadratic Equation 183 Classwork Examples 187 Practice - 7.1 191 7.2 Verbal Problems on Quadratic Equations 193 Classwork Examples 195 Practice - 7.2 203 8. Algebraic Fractions 8.1 Simplification of Algebraic Fractions 208 Classwork Examples 210 Practice - 8.1 211 9. Indices 9.1 Solving Exponential Equations 214 Classwork Examples 216 Practice - 9.1 218 Confidence Level Test - IV 220 Additional Practice - IV 221 V GEOMETRY 10. Triangle and Quadrilateral REVISITING GEOMETRY 228 10.1 Theorems on Area of Parallelograms and Triangles 230 Classwork Examples 236 Practice - 10.1 243 11. Construction 11.1 Construction of Triangles or Quadrilaterals with Equal Area 251 Classwork Examples 252 Practice - 11.1 253 11.2 Construction of Triangles or Quadrilaterals with Equal Area 255 Classwork Examples 255 Practice - 11.2 259 12. Circle 12.1 Theorems on Arc and Its Corresponding Angles of Circle 261 Classwork Examples 270 Practice - 12.2 273 12.2 Theorems on Inscribed Angle of SemiCircle 276 Classwork Examples 278 Practice - 12.2 280
12.3 Theorem on Cyclic Quadrilateral 283 Classwork Examples 286 Practice - 12.3 289 Confidence Level Test - V 293 Additional Practice - V 294 VI STATISTICS AND PROBABILITY 13. Statistics REVISITING STATISTICS 302 13.1 Mean of Grouped Data 304 Classwork Examples 307 Practice - 13.1 309 13.2 Median of Grouped Data 312 Classwork Examples 315 Practice - 13.2 317 13.3 Mode of Grouped Data 321 Classwork Examples 323 Practice - 13.3 324 13.4 Quartiles of Grouped Data 327 Classwork Examples 329 Practice - 13.4 331 14. Probability REVISITING PROBABILITY 335 14.1 Probability on Mutually Exclusive Events 337 Classwork Examples 339 Practice - 14.1 341 14.2 Probability on Independent and Dependent Events 346 Classwork Examples 347 Practice - 14.2 349 14.3 Probability by Tree Diagram 353 Classwork Examples 354 Practice - 14.3 361 Confidence Level Test - VI 366 Additional Practice - VI 367 VII TRIGONOMETRY 15. Trigonometry REVISITING TRIGONOMETRY 372 15.1 Height and Distance 373 Classwork Examples 374 Practice - 15.1 378 Confidence Level Test - VII 383 Additional Practice - VII 384 Appendix Extra Materials I . Practice Question Set 385 II. Classroom Individual Form 388 III. Project and Practical Work Evaluation Sheet 389 IV. Internal Evaluation Form 390 V. Format of Project Work 391
Sets PB Allied The Leading Mathematics-10 Sets 9 Specification Grid Unit Areas Total working hour Knowledge Understanding Application Higher ability Total number of Items Total number of questions Total Marks No. of items Marks No. of items Marks No. of item Marks No. of item Marks I Sets 12 1 1 1 1 1 3 1 1 4 1 6 Solving the behaviour problems related to operations and cardinality of set. Marks Weightage : 6 Estimated Working Hours : Competency Learning Outcomes To solve the behaviour problems related with up to three sets by using operations, Venn diagram and cardinality of sets. 1. Sets 1.1 Cardinality of Operations of Two Sets 1.2 Cardinality of Operations of Three Sets Chapters / Lessons SETS I UNIT 12 (Th. + Pr.) HSS Way: Mathematics begins at Home, grows in the Surroundings and takes shape in School.
Sets 10 Allied The Leading Mathematics-10 Sets 11 Set and Set Relations Set synonymous "aggregate", "bunch", "collection", "dump", "family", "group", etc. "A group of well-defined objects" Set Notation Set : A, B, ..., X, Y, Z Elements : a, b, c, ..., x, y, z Set membership : x ∈ X Non-membership : x ∉ A Representation (a) By statement form : 1. A set of numbers less than 2. (b) By Roster form : 1. A = {a, b, ..., y, z} (c) By set-builder form : A = {x: x satisfies a property P(x)} = {x: x ∈ P(x)} = {x | P(x)} Some Special Characteristics 1. A set is basically unordered. 2. It consists of distinct (or unequal) objects. 3. A power set may consist of sets. 4. A set may have nothing else or may be empty. 5. A set may have a fixed number of elements Universal set : U = {all elements under discussion} Empty Set : {None element}, φ = { } Singleton Set : {Only one element}, e.g., {2} Finite Set : Set having countable no. of elements, e.g., {2, 4, 6} Infinite Set : Set having uncountable no. of elements}, e.g., {2, 4, 6, 8, ...} Some Standard Sets of Numbers N = {1, 2, 3, ..., 11, 12, ...} W = {0, 1, 2, 3, ..., 11, 12, ...} Z = {..., –2, – 1, 0, 1, 2, ...} Q = {p/q : p, q in Z and q ≠ 0} Q* = {The set of irrational numbers} R = {The set of real numbers} R × R= The Cartesian plane. Set Relations Subset A ⊆ B if x ∈ A then x ∈ B. If A ⊆ B and B ⊆ A, A = B. Proper Subset A set not containing at least one element of the other set. In other words, A ⊂ B = {x : if x ∈ A, y ∈ A then x ∈ B, y ∉ B } Improper Subset A set containing every element of the other set. In other words, A ⊆ B = {x : if x ∈ A then x ∈ B} Rational numbers Natural numbers Irrational numbers Real Numbers Whole numbers Integers R E V I S I T I N G S E T S Set Operations Union of Two Set : Set of all elements, A ∪ B = {x : x ∈ A or x ∈ B} Intersection of Two Sets: Set of common elements only, A ∩ B = {x : x ∈ A and x ∈ B} Difference of Two Sets : Set of elements not containing another set, A – B = A \ B = {x : x ∈ A and x ∉ B} Complement of a Set : Set of elements not containing in the given set, A = Ac = A' = ~ A = {x : x ∉ A} = U – A, A = A Symmetric Difference of Two Sets : A ∆ B = (A – B) ∪ (B – A) = {x : [x ∈ A and x ∉ B] or [x ∈ B and x ∉ A]} Disjoint sets : No common element, i.e., A ∩ B = φ Overlapping sets : At least one common element, i.e., A ∩ B ≠ φ Power Set : P(X) = 2x = {A | if x ∈ A then x ∈ X} Equivalent Sets: Equal no. of elements in sets, A ≡ B (or A ≈ B) to mean A is equivalent to B Equal Sets: Sets formed by the same elements, A = B if A ⊆ B and B ⊆ A Cardinality or Cardinal Number: No. of elements in the given set, |A| or # (A) or n(A) i. n(A ∪ B) = no(A) + no(B) + n(A ∩ B) = n(A) + n(B) – n(A ∩ B) ii. n(U) = no(A) + no(B) + n(A ∩ B) + n(A ∪ B) c = n(A) + n(B) – n(A ∩ B) + n(A ∪ B) c Venn Diagram V e n n J o h n A B C U 2 3 4 6 7 5 8 1 Diagram 1923 Venn 1834
Sets 10 Allied The Leading Mathematics-10 Sets 11 1.1 Cardinality of Operations of Two Sets At the end of this topic, the students will be able to: ¾ solve the behaviour problems on the cardinality of operations of two sets using Venn diagram. Learning Objectives I Introduction In mathematics, the cardinality of a set is the number of elements in the given set. For example, the set P = {2, 3, 5, 7} contains 4 elements, and therefore as a cardinality of 4. It is the cardinal number of the finite set. In the 1890s, Georg Cantor (1845-1918) generalized the concept of cardinality to infinite sets, which allowed one to distinguish between the different types of infinity and to perform arithmetic on them. The cardinality of a set is also called its size. Basically, through cardinality, we define the size of a set. The cardinal number of a set A is denoted as n(A) or |A|, where A is any set and n(A) or |A| is the number of elements in the set A. A Venn diagram is used to visually represent the logical relationship on the operations of sets like as fundamental operations on numbers and that helps us to solve examples based on these sets. There are four main set operations which include set union, set intersection, set complement, and set difference. Ovals, circles and rectangles are used for drawing Venn diagram. Venn diagrams are also called logic or set diagrams. These types of Venn diagram were introduced by an English mathematician and logician John Venn (1834-1923). The operations of Venn diagrams are used in survey of any field. Venn diagrams are widely used in set theory, logic, mathematics, businesses, teaching, computer science, data science, biotechnology, engineering, probability, statistics, arts, etc. Georg Cantor Pets Mammals Cow Cat Dog Parrot Animal Four Legs Some uses of operations of sets Venn diagram in colour mixing Venn diagram in artificial intelligence data science CHAPTER 1 SETS
Sets 12 Allied The Leading Mathematics-10 Sets 13 II Cardinality of Operations of Two Sets All around us, we find many types of objects. Some of them are material and others non-material (or) ideal. Each of them is called an element or a member. Sometimes, a set without an element is also needed, sometimes only distinct from other and sometimes common only. Given two or more sets, we can put together the elements of the two sets or non of them. Two cases arise; (a) Two sets A and B may have no element in common to both (i.e., A and B are disjoint). (b) Two sets A and B have at least one element in common to both ( i.e., A and B overlapping or intersecting). If a set A has m distinct elements, its cardinality is denoted by n(A) = m. In the same way, the cardinality of a set B with n distinct elements is denoted by cardinality n(B) = n. When we take all elements of A and B together, the total number of elements may not be m + n. This is because repeated elements are counted only once. In other words, the number of elements in the union of A and B, n(A ∪ B), (a) will be m + n if A and B are disjoint, i.e. if A ∩ B = φ or empty or the number of elements in A ∩ B is equal to 0. Therefore, n(A ∪ B) = n(A) + n(B). Also, n(A – B) = m and n(B – A) = n. (b) will be m + n – t, the number of elements common to both (t), if A and B are intersecting or overlapping, i.e., m + n – t, if the number of elements in A ∩ B is not equal to 0. Therefore, n(A ∪ B) = n(A) + n(B) – n(A ∩ B). Also, n(A – B) = m – t and n(B – A) = n – t. (c) will be m, the number of A itself if B is subset of A (B ⊂ A), where A ∪ B = A and A ∩ B = B. Therefore, n(A ∪ B) = n(A) and n(A ∩ B) = n(B). Also, n(A – B) = m – n and n(B – A) = 0. In the above Venn diagrams, n(A ∪ B) = p. The number of elements in the union of two non-empty sets is known as the cardinality of union of two sets. If A and B are two non-empty sets, the cardinality of union of A and B is denoted by n(A ∪ B) and is formulated as follows : Case I : Cardinality of union of two disjoint sets A and B; Let U = {Natural Numbers less than 10} and its subsets A = {First three even numbers} and B = {Odd Number less than 9} then find; (a) n(U), n(A), n(B), n(A ∩ B), n(A ∪ B), n(A – B), n(B – A), n(A B) (b) (i) compare n(A ∪ B) and n(A) + n(B). (ii) define the relation between n(U), n(A ∪ B) and n(A B). Solution : (a) Here, U = {1, 2, ,3 ...., 10}, A = {2, 4, 6} and B = {1, ,3 5, 7} ∴ n(U) = 10, n(A) = 3 and n(B) = 4 Now, A ∩ B = φ. ∴ n(A ∩ B) = n(φ) = 0. and A ∪ B = {1, 2, 3, 4, 5, 6, 7} ∴ n(A ∪ B) = 7 A – B = {2, 4, 6} ∴ n(A – B) = 3 = n(A) B – A = {1, 3, 5, 7} ∴ n(B – A) = 4 = n(B) Also, A ∪ B = U – (A ∪ B) = {8, 9, 10}∴ n(A ∪ B) = 3 A p B U m n A B m t n p U A B p A B 2 3 4 6 9 8 10 U 5 7 1
Sets 12 Allied The Leading Mathematics-10 Sets 13 (b) (i) We have, n(A) = 3, n(B) = 4 ∴ n(A) + n(B) = 3 + 4 =7 Thus, n(A ∪ B) = n(A) + n(B) for disjoint sets A and B. (ii) Since n(U) = 10, n(A ∪ B) = 7 and n(A ∪ B) = 3 so, n(U) = 10 = 7 + 3 ∴ n(U) = n(A ∪ B) + n(A ∪ B) = n(A) + n(B) + n(A ∪ B) for disjoint sets A and B. Case II : Cardinality of union of two overlapping/intersecting sets A and B Let U = {1, 2, 3, ......., 9} and its subsets A = {1, 3, 6} and B = {1, 3, 5, 7} then find; (i) n(A ∩ B), n(A ∪ B), n(A – B), n(B – A) and n(A ∪ B). (ii) Define the relation between n(A ∪ B), n(A), n(B) and n(A ∩ B). (iii) Define the relation between n(A ∪ B), n(A – B), n(B – A) and n(A ∩ B). (iv) Write the relation between n(U), n(A ∩ B) and n(A ∪ B). Solution: (i) We have, A ∩ B = {1, 3} ∴ n(A ∩ B) = 2 A ∪ B = {1, 3, 5, 6, 7} ∴ n(A ∪ B) = 5 A – B = {6} ∴ n(A – B) = 1 B – A = {5, 7} ∴ n(B – A) = 2 A ∪ B = {2, 4, 8, 9} ∴ n(A ∪ B) = 4 (ii) Since n(A) = 3, n(B) = 4 and n(A ∩ B) = 2, n(A ∪ B) = 5 = 3 + 4 – 2. ∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B) for intersecting sets A and B. (iii) Since n(A – B) = 1, n(B – A) = 2 and n(A ∩ B) = 2, n(A ∪ B) = 5 = 1 + 2 + 2. ∴ n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B) for overlapping sets A and B. or, n(A ∪ B) = no(A) + no(B) + n(A ∩ B) for overlapping sets A and B. (iv) Since n(U) = 9, n(A ∪ B) = 5 and n(A ∪ B) = 4 then, n(U) = 9 = 5 + 4. ∴ n(U) = n(A ∪ B) + n(A ∪ B) and n(U) = n(A) + n(B) – n(A ∩ B) + n(A ∪ B) for overlapping sets A and B. or, n(U) = no(A) + no(B) + n(A ∩ B) + n(A ∪ B) for overlapping sets A and B. Case III : Cardinality of union of two overlapping/intersecting sets A and B Let U = {1, 2, 3, ......., 9} and its subsets A = {1, 3, 6, 9} and B = {1, 3, 6} then find; (i) n(A ∩ B), n(A ∪ B), n(A – B), n(B – A) and n(A ∪ B). (ii) What are equal to n(A ∪ B) and n(U)? Solution: (i) We have, A = {1, 3, 6, 9} ∴ n(A) = 4 B = {1, 3, 6} ∴ n(B) = 3 A ∩ B = {1, 3, 6} ∴ n(A ∩ B) = 3 A ∪ B = {1, 3, 6, 9} ∴ n(A ∪ B) = 4 A – B = {9} ∴ n(A – B) = 1 B – A = { } ∴ n(B – A) = 0 A ∪ B = {2, 4, 5, 7, 8} ∴ n(A ∪ B) = 5 (ii) From the above information, n(A ∪ B) = n(A) and n(U) = n(A) + n(A ∪ B). 6 1 3 5 7 A B U 4 8 9 2 6 1 3 5 7 B A U 9 8 4 2
Sets 14 Allied The Leading Mathematics-10 Sets 15 Derivation of Relations Let n(A) = m, n(B) = n and n(A ∩ B) = t. Then n(A – B) = m – t and n(B – A) = n – t. From the diagram it is clear that, n(A ∪ B) = n(A – B) + n(A ∩ B) + n(B – A) = m – t + t + n – t = m + n – t = n(A) + n(B) – n(A ∩ B) “Alternatively” n(A ∪ B) = n(A – B) + n(A ∩ B) + n(B – A) = n(A – B) + n(A ∩ B) + n(B – A) + n(A ∩ B) – n(A ∩ B) [ Add and subtract n(A ∩ B) on the same expression.] = n(A) + n(B) – n(A ∩ B) [∴n(A)= n(A – B) + n(A ∩ B) and n(B) = n(B – A) + n(A ∪ B).] Thus, we have n(A ∪ B) = n(A) + n(B) – n(A ∩ B) or, n(A ∩ B) = n(A) + n(B) – n(A ∪ B) We know that n(A – B) = no(A) and n(B – A) = no(B). So, n(A ∪ B) = no(A) + no(B) + n(A ∩ B) or, n(A ∩ B) = n(A ∪ B) – no(A) – no(B) Therefore, n(U) = n(A ∩ B) + n(A ∪ B) = no(A) + no(B) + n(A ∩ B) + n(A ∪ B) Furthermore, n(A) = n(A) + n(A ∩ B), n(B) = no(B) + n(A ∩ B), n(A ∪ B) = n(A) + no(B) or n(B) + no(A) n(A) = no(B) + n(A ∪ B) and n(B) = no(A) + n(A ∪ B) Points to be Remembered If the sets A and B are non-empty subsets of the universal set U then, 1. For disjoint sets A and B, (i) n(A ∩ Β) = n(φ) = 0, n(A – B) = n(A ∩ B) = n(A) and n(B – A) = n(A ∩ B) = n(B). (ii) n(A ∪ B) = n(A) + n(B) (iii) n(U) = n(A ∪ B) + n(A ∪ B) = n(A) + n(B) + n(A ∪ B). m–t n–tt A B U A B n(A B) n(B – A) = no(B) n(A B) n(A – B) = no(A) n(A) n(U) n(B) U A B U
Sets 14 Allied The Leading Mathematics-10 Sets 15 2. For overlapping sets A and B, (i) n(A) = no(A) + n(A ∩ B) and n(B) = no(B) + n(A ∩ B). (ii) n(A ∪ B) = n(A) + no(B) or n(B) + no(A), where no(A) = n(A – B) = n(A ∩ B) and no(B) = n(B – A) = n(A ∩ B). (iii) n(A ∪ B) = no(A) + no(B) + n(A ∩ B). (iv) n(A ∪ B) = n(A) + n(B) – n(A ∩ B). (v) n(U) = n(A ∪ B) + n(A ∪ B). (vi) n(U) = no(A) + no(B) + n(A ∩ B) + n(A ∪ B) = n(A) + no(B) + n(A ∪ B). (vii) n(U) = n(A) + no(B) + n(A ∪ B) = no(A) + n(B) + n(A ∪ B). (viii) n(U) = n(A) + n(B) – n(A ∩ B) + n(A ∪ B). (ix) n(A) = no(B) + n(A ∪ B) and n(B) = no(A) + n(A ∪ B) 3. Some Technical Terms SN Technical Terms Symbols (i) Number of elements in A or B or both, n(A or B) n(A ∪ B) (ii) Number of elements in A and B, n(A and B) n(A ∩ B) (iii) No. of elements in A as well as B or both A and B n(A ∩ B) (iv) Number of elements only in A no(A) (v) Number of elements containing in A but not in B no(A) (vi) Number of elements only in B no(B) (vii) Number of elements containing in B but not in A no(B) (viii) Number of elements in only one set or (A – B) ∪ (B – A) (symmetric difference) no(A) + no(B) = n(A ∆ B) (ix) No. of elements that lie at least one set or (either A or B) n(A ∪ B) (x) No. of elements neither A nor B or none of A and B n(A ∪ B) (xi) No. of elements that lie on at most one set n(U) – n(A ∩ B) or n(A ∩ B) no(A) n(A∩B) no(B) n(A∪B)C A B
Sets 16 Allied The Leading Mathematics-10 Sets 17 Example-1 A and B are the subsets of a universal set U where n(U) = 80, n(A) = 42, n(B) = 35 and n(A ∩ B) = 20. (a) What is cardinality of a set ? (b) Illustrate the above information in a Venn diagram. (c) Find n(A ∪ B) and n(A ∆ B). (d) Why is n(A ∆ B) not more than n(A ∪ B) ? Solution: Here, n(U) = 80, n(A) = 42, n(B) = 35, n(A ∩ B) = 20 Now, we have (a) The number of elements in the given set is called its cardinality. (b) Illustrating the above information in a Venn diagram alongside. (c) n(A ∪ B) = n(A) + n(B) – n(A ∩ B) = 42 + 35 – 20 = 57 and n(A ∆ B) = no(A) + no(B) = [n(A) – n(A ∩ B)] + [n(B) – n(A ∩ B)] = [42 – 20] + [35 – 20] = 22 + 15 = 37 (d) n(A ∆ B) is not more than n(A ∪ B) because (A ∆ B) is proper subset of (A ∪ B). Example-2 E is the set of students who like English and N, who like Nepali and U, all students of the class. If n(U) = 200, n(E) = 170, n(N) = 150, n(E ∪ N) = 7 then, (a) What is the meaning of n(N) = 150 ? (b) Represent the above information in a Venn diagram by supposing n(E ∩ N) = x. (c) Find the value of x. (d) Why is the value of x less than n(A ∪ B) ? Solution : Here, n(U) = 200, n(E) = 170, n(N) = 150, n(E ∪ N) = 7 (a) n(N) = 150 means the number of elements in the set N is 150. (b) Representing the above information in a Venn diagram by supposing n(E ∩ N) = x. (c) From the Venn diagram, or, 200 = (170 – x) + x + (150 – x) + 7 or, x = 327 – 200 ∴ x = 127. (d) The value of x is less than n(E ∪ N) because the set E ∩ N is the proper subset of E ∪ N. A B U n(U) = 80 n(A) = 42 n(A∩B) = 20 n(B) = 35 E N U 170–x x 150–x 7 n(U) = 200
Sets 16 Allied The Leading Mathematics-10 Sets 17 Example-3 60 candidates in an examination were offered Mathematics and Computer. Every student choose at least one of these subjects and some choose both, 55 take Optional Mathematics and 45 choose Computer. (a) If M and C are the sets of students who were offered Mathematics and Computer respectively, what is the meaning of n(M ∩ C)? (b) Present the above information in set notation. (c) Present the given information in a Venn diagram. (d) How many students were offered both subjects? How many students were offered just Mathematics ? (e) Why is n(M ∪ C) not less than n(M ∩ C) ? Give reason. Solution : (a) Here, M = A set of students who were offered Mathematics C = A set of students who were offered Science ∴ n(M ∩ C) is the number of candidates of examination where were offered both Optional Mathematics and Computer. (b) Every student chooses at least one of these subjects. So, n(M ∪ C) = 0 Then, n(M ∪ C) = 60, n(M) = 55 and n(C) = 45, (c) Presenting the given information in a Venn diagram. (d) Now, n(M ∪ C) = n(M) + n(C) – n(M ∩ C) or, 60 = 55 + 45 – n(M ∩ C) or, n(M ∩ C) = 100 – 60 = 40 Thus, 40 students were offered both subjects. Again, no(M) = n(M) – n(M ∩ C) = 55 – 40 = 15 Thus, 15 were offered just Mathematics. (e) n(M ∪ C) is not less than n(M ∩ C) because (M ∪ C) is the superset of (M ∩ C). Example-4 In a group of 90 students, 54 talk in English language (E), 51 talk in Newari language (N) and 9 talk in neither. (a) Present the above information in set notation. (b) Find the number of students who talk in only one language. (c) Show the above information in the same Venn diagram. (d) What is the relation of n(E ∩ N) with n(E) and n(N)? Solution : (a) Let U = A set of all students, N = A set of students who talk Newari language, and E = A set of students who talk English language, Then, n(U) = 90, n(E) = 54, n(N) = 51, n(N ∪ E) = 9 ∴ n(N ∪ E) = n(U) – n(N ∪ E) = 90 – 9 = 81 (b) Now, we know that n(N ∪ E) = n(N) + n(E) – n(N ∩ E) or, 81 = 54 + 51 – n(N ∩ E) or, n(N ∩ E) = 105 – 81 = 24. M 55 0 C n(U) = 60 45 U
Sets 18 Allied The Leading Mathematics-10 Sets 19 ∴ The number of students who talk in only one language, n(N ∆ E) = n(N – E) + n(E – N) = no(N) + no(E) = [n(N) – n(N ∩ E)] + [n(E) – n(N ∩ E)] = [54 – 24] + [51 – 24] = 30 + 27 = 57 Thus, 57 students take only one language. (c) Showing the above information in the same Venn diagram alongside, (d) n(E ∩ N) is the number of the common students who talk in both languages English (E) and Newari (N). Example-5 In a survey of newspaper reading habit, 45% like to read. “The Himalayan Times’’, 50% like to read “The Kathmandu Post’’ and 25% like both. (a) Write the above information in set notation. (b) Draw a Venn diagram to represent the above information. (c) Find the percentage of people who do not like either. (d) Why is n(H ∪ K) not written percentage? Solution : (a) Let U = A universal set of all people who like to read newspaper, H = A set of people who like to read The Himalayan Times, K = A set of people who like to read The Kathmandu Post. Then, n(U) = 100 (say), n(H) = 45, n(K) = 50, n(H ∩ K) = 25, n(H ∪ K) = ? (b) Drawing the Venn diagram to represent the above information, (c) n(U) = n(H) + n(K) – n(H ∩ K) + n(H ∪ K) or, 100 = 45 + 50 – 25 + n(H ∪ K) or, 100 – 70 = n(H ∪ K) or, n(H ∪ K) = 30 Hence, 30% of them do not like either. (d) n(H ∪ K) is not written percentage because the percentage is taken as certain parts out of 100 equal parts, but the cardinality n(H ∪ K) is the number of elements of the set (H ∪ K). E 30 2724 N U 9 H 45 25 K n(U) = 100 50 U
Sets 18 Allied The Leading Mathematics-10 Sets 19 Example-6 In an annual examination, 68% examinees passed Science, 74% examinees passed Mathematics and 8% examinees failed both subjects. If 25 examinees passed both subjects then, (a) present the above information in set notation. (b) Find the number of examinees who passed Science only. (c) Illustrate the above information in a Venn diagram. (d) Show the number of examinees who passed Science only in percentage. Solution: (a) Let, U = A universal set of total examinees in an annual examination S = A set of examinees who passed Science and M = A set of examinees who passed Mathematics. Then, n(U) = x (suppose), n(S) = 68% of x = 68x 100 n(M) = 74% of x = 74x 100 , n(S ∪ M) = 8% of x = 8x 100 , n(S ∩ M) = 25, no(S) = ? (b) Now, we have, n(U) = n(S) + n(M) – n(S ∩ M) + n(S ∪ M) or, x = 68x 100 + 74x 100 – 25 + 8x 10 or, 100x = 68x + 74x – 2500 + 8x or, 2500 = 150x – 100x or, 2500 = 50x or, x = 50. ∴ n(S) = 68 × 50 100 = 34, n(M) = 74 × 50 100 = 37, n(S ∪ M) = 8 × 50 100 = 4. ∴ no(S) = n(S) – n(S ∩ M)= 34 – 25 = 9. Hence, 9 examinees passed Science only. (c) Drawing the Venn-diagram to illustrate the above information; (d) The number of examinees who passed Science only in percentage is, no(S) n(U) × 100% = 9 50 × 100% = 18%. We can solve it by converting decimal form and percentage into whole number. 9 25 12 S 4 M U
Sets 20 Allied The Leading Mathematics-10 Sets 21 Example-7 In a group of 60 women, the ratio of women who like only Teej and only Tihar is 4:3. 11 women of them like both of them. If there are no women who don’t like both of them then, (a) present the above information in set notation. (b) Find the number of women who like Teej. (c) Show the fact in a Venn-diagram. (d) Find the ratio of the number of women who like Tihar to that of Teej. Solution : (a) Let U = A universal set of all women in a group, J = A set of women who like Teej and R = A set of women who like Tihar. Then, no(J) = 4x and no(R) = 3x (suppose) n(U) = 60, n(J ∩ R) = 11. n(J) =? Since there are no women who do not like both of the festivals and n(J ∪ R) = 0, so n(U) = n(J ∪ R) = 60. (b) Now, we have n(J ∪ R) = no(J) + no(R) + n (J ∩ R) or, 60 = 4x + 3x + 11 or, 60 – 11 = 7x or, 7x = 49 or, x = 7 ∴ no(J) = 4x = 4 × 7 = 28 no(R) = 3x = 3 × 7 = 21. ∴ n(J) = no(J) + n(J ∩ R) = 28 + 11 = 39. ∴ n(R) = no(R) + n(J ∩ R) = 21 + 11 = 32. Thus, 39 women like Teej. (c) Representing the above fact in a Venn diagram. (d) The ratio of the number of women who like Tihar to that of Teej is, n(R) n(J) = 32 39 = 32:39. 11 J R 28 11 21 U
Sets 20 Allied The Leading Mathematics-10 Sets 21 PRACTICE 1.1 Read Think Understand Do Keeping Skill Sharp 1. (a) What is cardinality of union of two non-empty sets ? Define. (b) Write the definition of cardinality of complement of union of two non-empty sets. (c) What represents the set notation n(A ∪ B) and n(A ∩ B) ? (d) What is the meaning of no(P) ? (e) Write the relation among n(U), n(M), n(N) and n(M ∪ N), where M and N are the disjoint subsets of U. 2. Circle ( ) the correct answer. (a) If P, Q and P ∪ Q are non-empty subsets of a universal set U, which relation is true for their cardinalities ? (i) n(U) = n(P) + n(Q) (ii) n(U) = n(P) + n(Q) – n(P ∪ Q) (iii) n(U) = n(P) – n(Q) + n(P ∪ Q) (iv) n(U) = n(P) + n(Q) + n(P ∪ Q) (b) Write the relation among n(U), n(A), n(B), n(A ∩ B) and n(A ∪ B), where A and B are the overlapping subsets of U. (i) n(U) = n(A) + n(B) – n(A ∪ B) (ii) n(U) = n(A) + n(B) + n(A ∪ B) (iii) n(U) = n(A) + n(B) – n(A ∩ B) + n(A ∪ B) (iv) n(U) = n(A) + n(B) + n(A ∩ B) + n(A ∪ B) (c) Write the relation among n(U), no(V), no(B), n(V ∩ B) and n(V ∪ B), where V and B are the overlapping subsets of U. (i) n(U) = no(V) + no(B) – n(V ∪ B) (ii) n(U) = no(V) + no(B) + n(V ∪ B) (iii) n(U) = no(V) + no(B) – n(V ∩ B) + n(V ∪ B) (iv) n(U) = no(V) + no(B) + n(V ∩ B) + n(V ∪ B) (d) If P and Q are non-empty overlapping subsets of a universal set U, which relation is true for their cardinalities ? (i) n(U) = n(P) + n(Q) + n(P ∪ Q) (ii) n(U) = no(P) + n(Q) + n(P ∩ Q) (iii) n(P ∪ Q) = no(P) + n(Q) (iv) n(P ∪ Q) = n(P) + no(Q) (e) It is given that n(A) = 15, n(B) = 12 and n(A ∪ B) = 20, what is the value of n(A ∩ B)? (i) 47 (ii) 7 (iii) 17 (iv) 23 (f) It is given that no(P) = 10, no(Q) = 7 and n(P ∪ Q) = 25, what is the value of n(P ∩ Q)? (i) 8 (ii) 42 (iii) 23 (iv) 22
Sets 22 Allied The Leading Mathematics-10 Sets 23 Check Your Performance Answer the following questions for each problem. 3. Suppose U = {x : x is a positive integer less than 20}, A = {x : x is a multiple of 5} and B = {x : x is an even number}. (a) What are the cardinality of U, A, and B? (b) Construct the above information in a Venn diagram. (c) Find n(A ∩ B) and n(A ∪ B). (d) Establish the relation among n(A), n(B), n(A ∩ B) and n(A ∪ B). (e) Are A ∪ B and B ∪ A equal ? Give reason. 4. Suppose U = {x/x is a natural number up to 10}, M = {x/x is prime number} and N = {x/x is an odd number}. (a) What is the cardinality of a set ? Find the values of n(U), n(M) and n(N). (b) Illustrate the above information in a Venn diagram. (c) Find n(M ∩ N), n(M ∆ N) and n(M ∪ N). (d) Establish the relation among n(U), n(M ∩ N), n(M ∆ N) and n(M ∪ N). (e) Why is n(M ∩ N) not more than n(M) or n(N) ? Give suitable reason. 5. It is given that n(A) = 5, n(B) = 6 and n(A ∩ B) = 2, where A and B are proper subsets of U. (a) Write the cardinality of the union of A and B. (b) Illustrate the above information in a Venn diagram. (c) Find n(A ∪ B) and n(A ∆ B). (d) Write the relation between n(A ∪ B), n(A ∩ B) and n(A ∆ B). 6. It is given that n(P) = 16, n(Q) = 20 and n(P ∪ Q) = 35, n(P∪ Q) = 9 and n(U) = 39, where P and Q are proper subsets of U. (a) Define the cardinality of the intersection of P and Q. (b) Illustrate the above information in a Venn diagram. (c) Find n(P ∩ Q) and n(P ∆ Q). (d) Write the relation between n(U), n(P ∩ Q), n(P ∪ Q) and n(P ∆ Q). 7. The diagram shows the results of survey of 80 families taken in a particular place to find the types of animals owned by each family. (a) Write the definition of the cardinality of the complement of B ∪ C. (b) Present the given information in the Venn diagram in set notation. (c) Find the number of families who did not take buffaloes and cows. (d) Write the relation between n(B), no(C) and n(B ∪ C). U Cow (C) 23 12 28 Buffaloes (B)
Sets 22 Allied The Leading Mathematics-10 Sets 23 8. 50 students of a class who use either blue colour or yellow colour or both. 40 use yellow colour and 20 use blue colour. Answer the following questions: (a) Write the above information in set notation. (b) What is the meaning of n(B ∩ Y) from the given Venn diagram? (c) Fill the above information in the same Venn diagram. (d) How many students use green colour? How many students use just blue colour ? (e) Why is n(B ∩ Y) not more than n(B) or n(Y) ? Give reason. 9. Out of 50 students in a class who were asked whether they had played football or volleyball, 28 had played volleyball and 4 had played both and 12 played football. (a) Present the above information in set notation. (b) Illustrate the given information in a Venn diagram. (c) How many students had played neither football nor volleyball? (d) Which is equal to the complement of either not playing football or not playing volleyball? 10. In a survey of 150 students, it was found that 25 students drink neither tea nor coffee, 85 drink tea and 75 drink coffee. (a) Write the above information in set notation. (b) Show the given information in a Venn diagram. (c) Find out the number of students who drink both tea and coffee. (d) Which is equal to the number of the students of complement of only one drink or none of both drinks? 11. In a certain school, 40% of the pupils play basketball, 25% play volleyball and 10% play both basketball and volleyball. (a) Present the above information in set notation. (b) Illustrate the above information in a Venn diagram. (c) What percent of the pupils in the school play neither basketball nor volleyball? (d) What is the percent of pupils of complement of who play basketball nor volleyball ? 12. In the survey of a community, 55% of the people like to listen to the radio, 65% like to watch the television and 35% don't like to listen to the radio as well as to watch the television. (a) Write the above information in set notation. (b) Illustrate the above information in a Venn diagram. (c) What percent of the people in the community like to listen to the radio as well as to watch the television. (d) What is the percent of pupils of complement of who like both listening and watching television ? B Y U
Sets 24 Allied The Leading Mathematics-10 Sets 25 13. Of the 30 cars passing through Motor Testing Station, 7 had defective brakes but not defective light and 18 cars had defective lights. (a) Present the above information in set notation. (b) Show the given information in a Venn diagram. (c) How many cars had not defective both lights and brakes? (d) Why is the number of cars defective lights more than both defective lights and brakes? 14. Out of 105 students, 80 passed Health, 72 in Account, 10 failed both subjects and 7 did not appear in the examination. (a) Write the above information in set notation. (b) Present the given information in a Venn diagram. (c) Find the number of students who passed at least one subject. (d) Why is the number of students who passed Account not less than Account only? 15. Out of group of 600 Japanese tourists who visited Nepal, 60% have been already to Khokana, Lalitpur and 45% to Changunarayan, Bhaktapur and 10% of them have been to both places. (a) Write the above information in set notation. (b) Illustrate the above information in a Venn diagram. (c) How many Japanese tourists have visited at most one place? (d) Why is the number of tourists not represented in percentage ? 16. In an examination, 45% of students passed Maths only, 40% passed Science only and 8% students failed both subjects. If 188 students passed Science then, (a) Represent the given information in set notation. (b) Draw the above information in a Venn diagram. (c) Find the total number of students. (d) How many total students in percentage are there in the examination ? 17. Among 90 girls, 30 people like to wear frock only, 25 people like to wear pants only and 18 people do not like to wear both clothes. (a) Write the given information in set notation. (b) Present the above information in a Venn diagram. (c) Find the ratio of the girls who like to wear frock and pants. (d) Compare the number of girls who like to wear and pants. 18. In an election of a municipality, two candidates M and N stood for the post of Mayor and 50,000 people were in the voter list. Votes were supposed to cast the vote for a single candidate. The number of votes secured by M is thrice of the number of votes secured by N. 400 people didn't cast vote and 500 votes were invalid. [Hints: The votes secured by M and N are valid votes.] (a) Present the given information in set notation.
Sets 24 Allied The Leading Mathematics-10 Sets 25 (b) Show the above information in a Venn diagram. (c) How many votes were secured by M and N separately? Find it. (d) Compare the number of votes secured by M and N. 19. In a survey of some people, it was found that the ratio of the people who liked pop songs and rap songs is 7 : 9. Out of which, 60 people liked both songs, 30 liked rap songs only and 40 liked none of the songs. (a) Represent the given information in set notation. (b) Draw the above information in a Venn diagram. (c) Find the number of people who did not like pop songs. (d) Compare the number of people who liked pop song only and rap song only. 20. In a survey among some students, it was found that 50 liked tea, 100 like coffee and all of them liked at least one drink. The number of students who liked coffee only is 3 times the number of students who liked tea only. (a) Represent the above information in the set notation. (b) Draw a Venn diagram to illustrate the above information. (c) Find the number of students who liked both drinks. (d) Find he number of students who liked at most one drink. Project Work Make a group of students at least 15 students and ask the following questions to the students about their favourite mobiles as iPhone-Apple and Samsung-Android: (i) How many students who like iPhone-Apple mobile? (ii) How many students who like Samsung-Android mobile? (iii) How many students who do not like both mobiles? Show the above information in a Venn diagram. Find the following operations of the sets with the separate Venn diagram: (a) How many students who like iPhone-Apple mobile or Samsung-Android mobile or both? (b) How many students who like both iPhone-Apple and Samsung-Android mobiles? (c) How many students who like only iPhone-Apple mobile? (d) How many students who like Samsung-Android mobile but not iPhone-Apple mobile? (e) How many students who like at least one type of mobiles ? (f) How many students who like at most one type of mobiles ? (g) How many students who like at most both types of mobiles ? (h) How many percentage of students who like iPhone-Apple mobile? (i) How many percentage of students who like only Samsung-Android mobile ? 3. (c) 1, 11 4. (c) 3, 3, 6 5. (c) 9, 7 6. (c) 1, 34 7. (c) 17 8. (d) 10, 10 9. (c) 14 10. (c) 35 11. (c) 45% 12. (c) 55% 13. (c) 5 14. (c) 88 15. (c) 510 16. (c) 400 17. (a) 47:42 18. (c) 36825, 12275 19. (c) 70 20. (c) 25 (d) 100 Answers
Sets 26 Allied The Leading Mathematics-10 Sets 27 1.2 Cardinality of Operations of Three Sets At the end of this topic, the students will be able to: ¾ solve the behaviour problems on the cardinality of operations of three sets by using Venn diagram. Learning Objectives I Introduction There are so many types of objects around us. Some of them have common characteristics and some have distinct characteristics. For example, we discuss different kinds of animals, chargeable electric appliances and colour mixing as shown in the Venn diagram below. The Venn Fig. (i), (ii) and (iii) are overlapping of three sets. Some of the objects have not all common characteristics. For example, we discuss different kinds of animals, Limpiadhuara-Kalapani of Nepal with India and China, Prallelogram-Rectanglesquare, and Kathmandu-Lalitpur-Bhaktapur as shown in the Venn diagram below. In Venn Fig (iv), the set A is intersecting with the sets R and M, but R and M are disjoint. In Venn Fig (v), Limpiyadhura-Kalapani area is claimed by Nepal (N) and India (I), but not China (C). So, N and I are as overlapping sets, but C is disjoint. In Venn Fig. (vi), S is intersection of R and M, R and M are subset of P. In Venn Fig (vii), K, L and B are distinct districts. So, they are disjoint sets. M T T U R G B S W P Y T Buffaloes U L Cats Red Green Blue Venn Fig. (i) Venn Fig. (ii) Venn Fig. (iii) I ∩ N (I) (N) (C) (K) (L) (B) Square (S) Parallelogram (P) Rectangle (R) Rhombus (M) Venn Fig. (iv) Aquatic Animals A Venn Fig. (v) Venn Fig. (vi) Venn Fig. (vii)
Sets 26 Allied The Leading Mathematics-10 Sets 27 II Cardinality of Operations of Three Sets Three non-empty sets have more operations. Let's discuss on finding the number of the operations of three sets in different cases as mentioned below: A U B a C b c A U B a C b c A U B a d C b c e A U B a C b d c e A U B a C b c d e f A U B a C c b d e A U B a C b d c A U B a C b c d A U B a C b c d f t e The number of elements in the indicated operations of three non-empty sets is the cardinality of the related operations. The cardinalities of different parts of the operations of three intersection non-empty sets are as shown alongside. In the Venn diagram, n(A) = No. of elements in A; n(B) = No. of elements in B; n(C) = No. of elements in C; n(A ∩ B) = No. of elements in both A and B; n(B ∩ C) = No. of elements in both B and C; n(A ∩ C) = No. of elements in both A and C; n(A ∩ B ∩ C) = No. of elements in A, B and C; no(A) = No. of elements in only A; no(B) = No. of elements in only B; no(C) = No. of elements in only C; no(A ∩ B) = No. of elements in only A and B; no(B ∩ C) = No. of elements in only B and C; no(A ∩ C) = No. of elements in only A and C; n(A∪B∪C) = No. of elements neither of A, B and C. n(U) n(A) n(B) n(C) no(A) no(A) no(A) no(A∩B) no(A∩B∩C) n(A∪B∪C) no(B∩C) no(A∩C)
Sets 28 Allied The Leading Mathematics-10 Sets 29 Derivation of Cardinality of Operations of Three Sets Case I: Cardinality of union of three disjoint sets A, B and C If A, B and C are three disjoint sets as shown in the Venn diagram then, n(A ∪ B ∪ C) = n{A ∪ (B ∪ C)} = n(A) + n(B ∪ C) = n(A) + n(B) + n(C) ∴ n(A ∪ B ∪ C) = n(A) + n(B) + n(C) for disjoint sets A, B and C. Case II: Cardinality of union of three disjoint sets A, B and C If A, B and C are three overlapping sets as shown in the Venn diagram then, 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) ∩ (A∩C)} = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C). ∴ n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – (A ∩ C) + n(A ∩ B ∩ C) for overlapping sets A, B and C. From the above Venn diagram, n(A) = no(A) + no(A ∩ B) + no(A ∩ C) + n(A ∩ B ∩ C) n(B) = no(B) + no(A ∩ B) + no(B ∩ C) + n(A ∩ B ∩ C) n(C) = no(C) + no(A ∩ C) + no(B ∩ C) + n(A ∩ B ∩ C) ∴ n(A ∪ B ∪ C) = no(A) + no(B) + no(C) + no(A ∩ B) + no(B ∩ C) + no(A ∩ C) + n(A∩B∩C) for overlapping sets A, B and C. Also, n(A ∪ B ∪ C) = n(A ∪ B) + no(C) and so on. n(A ∪ B ∪ C) = n(A) + no(B) + no(C) + no(B ∩ C) and so on. Again, n(U) = n(A ∪ B ∪ C) + n(A ∪ B ∪ C) n(U) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – (A∩C) + n(A∩B∩C) + n(A ∪ B ∪ C) n(U) = no(A) + no(B) + no(C) + no(A∩B) + no(B∩C) + no(A∩C) + n(A∩B∩C) + n(A∪B∪C) Also, n(U) = n(A ∪ B) + no(C) + n(A ∪ B ∪ C) and so on. n(U) = n(A) + no(B) + no(C) + no(B ∩ C) + n(A ∪ B ∪ C) and so on. A B U C n(A ∪ B ∪ C) no(A) no(B) no(C) n(ABC) no(AB) no(BC) no(AB) U A B C
Sets 28 Allied The Leading Mathematics-10 Sets 29 Points to be Remembered If the sets A, B and C are non-empty subsets of the universal set U then, 1. For disjoint sets A, B and C, (i) n(A ∪ B ∪ C) = n(A) + n(B) + n(C). (ii) n(U) = n(A ∪ B) + n(A ∪ B ∪ C) = n(A) + n(B) + n(C) + n(A ∪ B ∪ C). 2. For overlapping sets A, B and C, (i) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) (ii) n(A) = no(A) + no(A ∩ B) + no(A ∩ C) + n(A ∩ B ∩ C) n(B) = no(B) + no(A ∩ B) + no(B ∩ C) + n(A ∩ B ∩ C) n(C) = no(C) + no(A ∩ C) + no(B ∩ C) + n(A ∩ B ∩ C) (iii) n(A ∪ B ∪ C) = no(A) + no(B) + no(C) + no(A ∩ B) + no(B ∩ C) + no(A ∩ C) + n(A ∩ B ∩ C) for overlapping sets A, B and C. n(A ∪ B ∪ C) = n(A ∪ B) + no(C) and so on. n(A ∪ B ∪ C) = n(A) + no(B) + no(C) + no(B ∩ C) and so on. (iv) n(U) = n(A ∪ B ∪ C) + n(A ∪ B ∪ C) n(U) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – (A∩C) + n(A∩B∩C) + n(A ∪ B ∪ C) n(U) = no(A) + no(B) + no(C) + no(A∩B) + no(B∩C) + no(A∩C) + n(A∩B∩C) + n(A∪B∪C) n(U) = n(A ∪ B) + no(C) + n(A ∪ B ∪ C) and so on. n(U) = n(A) + no(B) + no(C) + no(B ∩ C) + n(A ∪ B ∪ C) and so on. Example-1 Suppose A = {2, 4, 6}, B = {1, 3, 5, 7} and C = {2, 3, 5, 7} are the given sets. Then, (a) What is the cardinality of A ∪ B ∪ C ? (b) Find n(A ∪ B ∪ C). (c) Illustrate the above information in a Venn diagram. (d) Why is n(A ∩ B ∩ C) equal to 0 ? Solution : (a) The cardinality of A ∪ B ∪ C is the number of the elements in A ∪ B ∪ C. (b) Here, A = {2, 4, 6} ∴ n(A) = 3, B = {1, 3, 5, 7} ∴ n(B) = 4, C = {2, 3, 5, 7} ∴ n(C) = 4. A ∩ C = {2, 4, 6} ∩ {2, 3, 5, 7} = {2} ∴ n(A ∩ C) = 1. A ∩ B = {2, 4, 6} ∩ {{1, 3, 5, 7} = {} ∴ n(A ∩ B) = 0. B ∩ C = {1, 3, 5, 7} ∩ {2, 3, 5, 7} = {3, 5, 7} ∴ n(B ∩ C) = 3. A ∩ B ∩ C = {2, 4, 6} ∩ {{1, 3, 5, 7} ∩ {2, 3, 5, 7} = {} ∴ n(A ∩ B ∩ C) = 0. n(A∪B∪C) no(A) no(B) no(C) n(ABC) no(AB) no(BC) no(AB) U A B C
Sets 30 Allied The Leading Mathematics-10 Sets 31 Now, we know that n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(A∩C) – n(B∩C) + n(A∩B∩C) = 3 + 4 + 4 + 0 – 1 – 3 + 0 = 7. (c) Drawing a Venn diagram for the above information, (d) n(A ∩ B ∩ C) is equal to 0 because there is no element in the sets A, B and C as common i.e. A ∩ B ∩ C is an empty set. Example-2 If n(A) = 44, n(B) = 49, n(C) = 40, no(A) = 15, no(B) = 17, no(A ∩ B) = 12, no(B ∩ C) = 6, (A ∪ B ∪ C) = 5 and n(A ∩ B ∩ C) = x then, (a) Define cardinality of A ∩ B ∩ C. (b) Illustrate the above information in a Venn diagram. (c) Find the value of x and n(U). (d) What is the relation between n(A ∪ B ∪ C) and n(A ∪ B ∪ C) ? Solution : Here, n(A) = 44, n(B) = 49, n(C) = 40, no(A) = 15, no(B) = 17, no(A ∩ B) = 12, no(B ∩ C) = 6, n(A ∪ B ∪ C) = 5 and n(A ∩ B ∩ C) = x = ?, n(U) = ? (a) The number of the elements in the intersection of all the three sets A, B and C is called cardinality of A ∩ B ∩ C. (b) Illustrating the above information in the same Venn diagram, (c) Now, we know that, x = 49 – 17 – 12 – 6 [ From the Venn diagram] = 14 no(A C) = 44 – 15 – 12 – 14 [ no(AB) = n(A) – no(A) – no(AB) – n(ABC)] = 3 no(C) = 40 – 6 – 3 – 14 = 17 [ no(C) = n(C) – no(B C) – no(A C) – n(A B C)] ∴ n(U) = 15 + 17 + 17 + 12 + 6 + 3 + 14 + 5 = 89 [ n(U) = no(A) + no(B) + no(C) + no(AB) + no(BC) + no(A C) + n(A B C) + n(A ∪ B ∪ C)] (d) n(A ∪ B ∪ C) and n(A ∪ B ∪ C) are equal to the number of complement of their own sets to each other and their sum is equal to the cardinality of the universal set U. i.e. n(Complement of A ∪ B ∪ C) = n(A ∪ B ∪ C) and n(A ∪ B ∪ C) = n(A∪B∪C) So, n(A ∪ B ∪ C) + n(A ∪ B ∪ C) = n(U). A B 1 C 6 4 2 3 5 7 n(B) = 49 n(A) = 44 n(C) = 40 A C 5 B U 17 12 15 x 6
Sets 30 Allied The Leading Mathematics-10 Sets 31 Example-3 In a survey of people, 60 children liked to play Pubg game, 40 like to play Pokemon game, 45 like to play Roblox game, 25 like to play Pubg game and Pokemon game, 20 like to play Pubg game and Roblox game, 27 liked to play Pokemon game and Roblox game and 12 liked to play all three games. If every child liked to play at least one game then, (a) Define cardinality of a universal set. (b) Find how many children were involved in the survey. (c) Find how many children who like to play only Pokemon game. (d) Show the above information in a Venn diagram. (e) Why did no children like to play Pokemon game or Pubg game or Roblox game ? Solution: Let U = Total children who involved in the survey, T = A set of children who liked to play Pubg game, M = A set of children who liked to play Pokemon game, C = A set of children who liked to play Roblox game. Then, n(T) = 60, n(M) = 40, n(C)= 45, n(T M) = 25 n(T C) = 20, n(M C) = 27, n(T M C) = 12 Since every children liked to play at least one game, so n(T ∪ M ∪ C) = ? (a) The number of all members involving in the survey is called cardinality of a universal set. (b) Now, we have n(T M C) = n(T) + n(M) + n(C) – n(TM) – n(TC) – n(MC) + n(TMC) = 60 + 40 + 45 – 25 – 20 – 27 + 12 = 85. Thus, 85 children were involved in the survey. (c) Again, no(T M) = n(T M) – n(T M C) = 25 – 12 = 3 no (M C) = n(M C) – n(T M C) = 27 – 12 = 15 no (M) = n(M) – no(T M) – no(MC) – n(TMC) = 40 – 13 – 15 – 12 = 0. Hence, there are no children who liked to play only Pokemon game. (d) Showing this information in the Venn-diagram, (e) No children liked to play Pokemon game or Pubg game or Roblox game because every child liked at least one game in the survey. T M n(T) = 60 n(M) = 40 n(C) = 40 C 10 27 13 0 8 12 15
Sets 32 Allied The Leading Mathematics-10 Sets 33 PRACTICE 1.2 1. The figure shows the number of elements each region in the Venn diagram. (a) Write the dentition of the cardinality of A ∪ B ∪ C. (b) Find the following: (i) n(A) (ii) n(A ∩ B) (iii) n(A ∪ B) (iv) no(A C) (v) no(C) (vi) no(A B) (vii) n(A B C) (viii) n(A ∪ B) (ix) n(A ∪ B ∪ C) (c) Write the relation between n(A ∪ B ∪ C) and n(A ∪ B ∪ C). 2. Let a universal set be the students in Biswo Bhasa campus and the number of students shown in the region of the Venn diagram. F = Set of students who learn French, G = Set of students who learn German, L = Set of students who learn Latin (a) How many students learn French? (b) How many students learn Latin and German? (c) How many students learn French and German but not Latin? (d) How many students learn Latin only? (e) How many students learn none of these three languages? 3. The Venn diagram below shows the number of pupils in a class of 35 students who study one or other of the sciences: Biology (B), Physics (P) and Chemistry (C) or none of the three. (a) How many pupils do not study Science? (b) How many pupils study just one of these three Sciences? (c) Write down in terms of x the number of pupils who are studying one or more of these Sciences. (d) Find the value of x. 4. It is found that 15% of the pupils at a school do not jog or swim or cycle, 8% do all three of these activities, 15% swim and cycle but do not jog, 15% do both jog and swim, 20% swim only, 18% cycle only and 42% do two or more activities. (a) Define cardinality of complement of set. (b) Workout the percentage of pupils who, (i) do both jog and cycle. (ii) swim or cycle. (iii) do not swim. (iv) do not jog, swim and cycle. (d) Show the above information in a Venn diagram. (e) Why is the percentage of total pupils who involve in the survey 100% ? Give reason. U A C B 4 5 6 1 3 4 3 0 U F G L 6 5 2 2 1 0 4 3 7 0 x 9 x x+2 3 2 C P B U
Sets 32 Allied The Leading Mathematics-10 Sets 33 5. Out of 138 students in a school, 60 take computer, 100 take optional math and 68 take account. If 28 take computer and optional math, 44 take optional math and accounts and 20 take computer and accounts. (a) Write the definition of the cardinality of all the three subjects. (b) How many students take all the three subjects? Find it. (c) How many students take only two subjects? Find it. (d) Draw a Venn diagram to represent the above information. (e) Why is the cardinality of all the three subjects less than the cardinality of only two subjects? 6. In a survey of 100 people, 65 read the Kantipur, 45 read the Gorkhapatra, 40 read the Rajdhani, 25 read the Kantipur as well as Gorkhapatra, 20 read the Kantipur as well as the Rajdhani, 15 read the Gorkhapatra as well as the Rajdhani and 5 read all the three news papers. (a) Define the cardinality of neither reading all the three newspapers. (b) Show the above information in a Venn diagram. (c) How many people didn’t read all the three newspapers? Find it by using Venn diagram. (d) How many people read at most one newspaper? Find it. (e) Why is the cardinality of all the three subjects less than the cardinality of any two subjects? 7. In a group of people, 23 people like to eat rice, 17 like to eat bitten rice, 18 like to eat soup rice, 6 people like to eat rice only, 9 like to eat bitten rice only, 5 like to eat bitten rice and soup rice and 3 like to eat soup rice only. (a) Write the definition of the cardinality of the union of all the three meals. (b) Illustrate the above information in a Venn diagram. (c) How many people like to eat all the meals? Find it by using Venn diagram. (d) How many people are there altogether? Find it. (e) Why is the cardinality of all the three meals not more than the cardinality of the union of all the three meals? 8. In a class of students, 20 students study Mathematics, 21 study Science, 18 study Nepali, 7 study Mathematics only, 10 study Science only, 6 study Mathematics and Science only and 3 study Science and Nepali only. (a) Write the definition of the cardinality of students who study at least one subject. (b) Present the above information in a Venn diagram. (c) How many students study all the subjects? Find it. (d) How many students are there altogether? Find it. (e) Why is the cardinality of all three subjects not more than the cardinality of the union of all the three subjects?
Sets 34 Allied The Leading Mathematics-10 Sets 35 2. (a) 18 (b) 6 (c) 5 (d) 1 (e) 2 3. (a) 3 (b) 9 (c) 3x + 20 (d) 4 4. (b) (i) 12% (ii) 80% (iii) 50% (iv) 92% 5. (b) 2 (c) 86 6. (c) 5 (d) 50 7. (c) 4 8.(c) 2 (d) 41 Answers Project Work Collect the information to like eating apple or mango or banana or any two and three types of fruits among some students. Answer the following questions: (a) List out the number of students who like apple or mango or banana or any two and three types of fruits separately. (b) List out the number of students who like only two or three types of fruits separately. (c) How many students who unlike all the three types of fruits ? Write the answers of the following questions: (i) How many students who like only apple, mango and banana separately ? (ii) How many students who like apple and mango, mango and banana, apple and banana separately? (iii) How many students who dislike apple, mango and banana separately ? (iv) How many students are there participated in the survey ? (v) What percentage of students who like mango and banana? (vi) What percentage of students who do not like apple and mango ? (vii) What percentage of students who like all the three types of fruits ? (viii)What percentage of students who like only two types of fruits?
Sets 34 Allied The Leading Mathematics-10 Sets 35 CONFIDENCE LEVEL TEST - I Unit I : Sets Class: 10, The Leading Maths Time: 45 mins. FM: 24 Attempt all questions. 1. A and B are the subsets of a universal set U, where n(U) = 85, n(A) = 42, n(B) = 45 and n(A ∪ B) = 10. (a) Define cardinality of the complement of union of two sets. [1] (b) Illustrate the above information in a Venn diagram. [1] (c) Find n(A ∩ B) and n(A ∆ B). [3] (d) Write the relation between n(A ∪ B), n(A ∩ B) and n(A ∆ B). [1] 2. In a group of 250 students, 128 like to use i-phone (I), 118 like to use Android-phone (A) and 9 like to use both. (a) Define the union of two sets. [1] (b) Show the above information in the same Venn diagram. [1] (c) Find the number of students who do not like to use both mobiles. [3] (d) Why is n(I ∪ A) not less than n(I ∩ A) ? [1] 3. In a group of 75 women, the ratio of women who use only TikTok and only Messenger is 5:3. 16 women of them use both of them and 11 women use neither of them. (a) Present the above information in set notation. [1] (b) Show the fact in a Venn-diagram. [1] (c) Find the number of women who like TikTok. [3] (d) Find the ratio of the number of women who like Messenger to that of TikTok. [1] 4. In a survey of people, 70 people like to drink tea, 50 people like to drink milk, 50 people like to drink coffee, 30 like to drink tea and milk, 25 like to drink tea and coffee, 30 like to drink milk and coffee and 15 like to drink all the three drinks. But, every people like at least one drink. (a) Define cardinality of a Universal set. [1] (b) Show the given information in Venn Diagram. [1] (c) How many people were involved in the survey and the number of people who like to drink only milk? Find. [3] (d) Why do no people like to drink milk or tea or coffee? [1] Best of Luck
Sets 36 Allied The Leading Mathematics-10 Sets 37 Additional Practice – I Cardinality of Union of Two Sets 1. (i) In a class of 25 students, 15 students like to play cricket and 12 students like to play football. Also, each student likes to play at least one of two games. (a) How many students like to play both games? (b) Illustrate the above information in a Venn diagram. (ii) In a group of 65 students 35 like tea, 50 like milk and each person likes at least one of the two drinks. (a) How many people like both tea and milk? (b) How many people like only one drink? (c) Show the above information in a Venn diagram. 2. (i) In a class of 50 students, 25 students like to play volleyball, 35 like to play football and 15 like to play both games. (a) How many students do not like to play any games? (b) Illustrate the above information by a Venn diagram. (ii) A survey conducted among 202 students of a school, the following information are found where 125 like mango, 135 like papaya and 22 dislike both the fruits; find the following; (a) find the number of students who like both fruits by using a Venn diagram. (b) find the number of students who like mango only. (iii) In a group of 55 people, 20 people like to read Kantipur but not Nagarik and 19 people like to read Nagarik but not Kantipur. (a) If 7 people did not like both, how many people like both newspapers? (b) Represent the above information in a Venn diagram. 3. (i) In an examination, 68% of the examinees passed Mathematics, 74% English and 46% both subjects. (a) Draw a Venn-diagram to illustrate the above information. (b) Calculate the total number of examinees appeared if 6 failed both the subjects. (ii) In a survey of the students, it was found that 65% of the students studied BA course, 60% of the students studied BN course, 330 students studied both courses and 8% didn’t study both courses, then (a) show the above information in a Venn diagram. (b) find the total number of students to take part in the survey.
Sets 36 Allied The Leading Mathematics-10 Sets 37 4. (i) In a group of76people,the ratio of people who like to drinkCoca-Cola and Pepsi is 5:3. If 7 of them like both the soft drinks and 11 of them like none of the soft drinks, (a) draw a Venn diagram, find how many of them like only Coca-Cola. (b) find how many of them like only Pepsi. (ii) Out of 33 students appeared in an examination, the number of students who passed Mathematics only is twice the number of students who passed English only. If 15 students passed both subjects and 9 students failed both subjects then, (a) find the number of students who passed Mathematics. (b) find the number of students who passed English. (c) show the result in a Venn diagram. 5. (i) In a survey, two-thirds of students like Mathematics and one-fourth like only Science. If 34 students like other subjects, (a) how many students like Mathematics and only Science ? Find it. (b) show the above information in a Venn diagram. (ii) In a community, two-thirds of people like folk music and 25% like only modern music. If 44 people like other musics and 100 people like both musics, how many people like (a) modern music (b) only folk music ? (c) show the above information in the same Venn diagram. (iii) In a group of people, 33 1 3 % people like only mango and 34 did not like mango at all. Also, 26 2 3% people like orange but 24 dislike none of them.(a) Show the above information in a Venn diagram. (b) How many people like both types of fruit? Find it. Cardinality of Union of Three Sets 6. (i) Out of the total candidates in an examination, 40% students passed Mathematics, 35% Science and 50% Health. If 8% passed Mathematics and Science, 18% Science and Health and 14% Health and Mathematics. (a) Draw a Venn diagram to show the above information. (b) Calculate the percentage of students who passed all the three subjects. (ii) Out of 50 students in an examination, 20 offered Physics, 18 offered Chemistry and 15 offered Biology, 10 offered none of these subjects, 5 offered Physics and Biology, 5 offered Chemistry and Biology and 4 offered Physics and Chemistry. (a) Present the above information in Venn diagram. (b) Find the number of students who offered all the three subjects. (c) What percentage do offer exactly two subjects?
Sets 38 Allied The Leading Mathematics-10 Sets PB 7. (i) In a survey of students, it was found that 50% of them like English, 55% like Mathematics and 45% like Science. If 20% like Mathematics and English, 15% like English and Science, 20% like Mathematics and Science and 5% like all the three subjects, (a) What percentage like only two subjects? (b) What percentage doesn’t like any subject? (ii) In a survey of students every students study of least one of the three subjects Maths, English or Science. The number of students who study only one subject is twice the number of students who study only two subjects. If 20 students study all the three and 80 students who don't study only two subjects. (a) Find the total number of students in the survey. (b) Show the above information in a Venn diagram. 8. (i) In a recent survey among some girls, it was found that 55% of them wanted to use Leagoo mobile 35% wanted to use Huawei mobile, 50% wanted to use Oppo mobile, 25% wanted to use Leagoo and Oppo, 20% wanted to use Oppo and Huawei, 15% want Leagoo and Huawei and 10% wanted all three types of mobile. If 58 girls did not want to use all these mobiles, find the total number of girls involved in the survey by using a Venn diagram. (ii) In a survey of students, every student studies at least one of the three subjects Population, Environment Science and History. The number of students who study only one subject is twice the number of students who study only two students. If 30 students study all three and 70 students who don't study only two subjects, what is the total number of students in the survey? Find it by using Venn diagram. (iii) In the First Term Exam, every student passed in at least one of the subjects Maths, English and Science. 30 students passed in Maths and English, 25 passed in English only and 13 passed in Science only. The students passed in Maths only and in Science are in the ratio of 4:9. If 32% of the students passed only in Maths find the total number of students participation in the exam. 1. (i) (b) 2 (ii) (a) 20 (b) 45 2. (i) (a) 5 (ii) (a) 80 (b) 45 (iii) (a) 9 3. (i) (b) 50 (ii) (b) 1000 4. (i) (a) 38 (b) 20 (ii) (a) 41 (b) 28 5. (i) (a) 272, 102 (ii) (a) 232 (b) 252 (iii) (b) 6 6. (i) (b) 15% (ii) (b) 1 7. (i) (a) 40% (b) 0% (c) 11, 22% (ii) (a) 110 8. (i) 580 (ii) 90 (iii) 100 Answers
Arithmetic PB Allied The Leading Mathematics-10 Compound Interest 39 Solving and using the arithmetic behaviour problems related to daily life. Marks Weightage : 13 Estimated Working Hours : Competency Learning Outcomes 1. To solve the problems related to compound interest using the present tax system, bank and other financial institutions of the country. 2. To solve the behaviour problems related to compound growth. 3. To solve the behaviour problems related to currency and exchange rate. 2.1Compound Interest 2.2Compound Growth and Depreciation - Compound Growth - Compound Depreciation 2.3Money Exchange - Currency and Exchange Rate Chapters / Lessons ARITHMETIC II UNIT 28 (Th. + Pr.) Specification Grid Unit Areas Total working hour Knowledge Understanding Application Higher ability Total number of Items Total number of questions Total Marks No. of items Marks No. of items Marks No. of item Marks No. of item Marks II Arithmetic 28 2 2 2 3 3 5 2 3 9 3 13 FEE Concept: "Maths is Fun, Maths is Easy and Maths is Everywhere"
Arithmetic 40 Allied The Leading Mathematics-10 Compound Interest 41 WARM-UP Tax Definition Formula Example Tax A amount to be paid to the government for the service provided to its citizens. Tax amount = Tax rate of Taxable amount Tax = 10% of rent of Rs. 150000 = 0.1 × Rs. 150000 = Rs. 15000 Tax rate The charged rate fixed by the government on the taxable income of taxable amount Tax rate = Tax amount Taxable amount×100% Tax = Rs. 120, Taxable amount = Rs. 1200 Tax rate= 120 1200×100%=10% Income tax Tax levied directly on personal income at certain percent Income Tax = Income Tax Rate × Taxable Income Income tax = 1% of Rs. 600000 = 0.01 × Rs. 600000 = Rs. 6000 Discount The reducing amount on selling or buying goods by the seller to the buyer Discount amount = D% of Marked price MP = Rs. 2300, D% = 12% D = 12% of Rs. 2300 = 0.12×Rs. 2300 = Rs. 276 Selling price (SP/ ASP) The price after reducing discount on the marked price of the goods ASP = MP – D MP = Rs. 2300, Discount amount = Rs. 276 ASP = Rs. 2300 – Rs. 260 = Rs. 2040 Value Added Tax A consumption tax on the amount by which the value of an article has been increased at each stage VAT amount = VAT% of Actual selling price of goods ASP=Rs.2040, VAT%=13% VAT = 13% of Rs. 2040 = 0.13×Rs. 2040 = Rs. 256.50 SP with VAT (VSP) The final price of goods adding VAT amount that is paid by customer VSP = ASP + VAT ASP = 2040, VAT = 256.50 VSP = 2040 + 256.50 = Rs.2305.30 Commission and Dividend Definition Formula Example Commission The amount paid to an agent or company at a fixed rate for the service provided of any task by the agent or the company Commission = Commission Rate × Total sales amount Rate = 3%, Total sales amount = Rs. 50000 Commission amount = 3% of Rs. 50000 = Rs. 1500 Net sales amount The amount after giving commission amount on the sales amount Net sales amount = Total sales amount – Commission amount Total sales amt = Rs. 1000 Com. amt = Rs. 150 Net sales amt = Rs. 1000 – Rs. 150 = Rs. 850 R E V I S I T I N G A R I T H M E T I C
Arithmetic 40 Allied The Leading Mathematics-10 Compound Interest 41 Total income The income of a person with his/her salary, commission, tips and other sources Total income = Salary + Commission amount Salary = Rs. 10000, Com. amt = Rs. 550 Total income = Rs. 10550 Bonus A sum of money added to a person's wages as a reward for good performance Bonus amount = Some percent of Salary Salary = Rs. 45689 Bonus rate = 120% Bonus amt = 120% of Rs. 45689 = Rs. 54826.80 Dividend A sum of money paid regularly (typically annually) by a company to its shareholders out of its profits Dividend amount = Rate o × Net profit Net profit = Rs. 45000000 Rate of dividend = 16% Dividend amt = 16% of Rs.45000000 =Rs.7200000 Home Arithmetic Arithmetic involving the study of numbers, especially the properties of operations such as addition, subtraction, multiplication and division for calculation household goods that is not exactly pure arithmetic Cost of 1 egg is Rs. 15. Total cost of 30 eggs of crate is Rs. 425. Uses of Home Arithmetic Household arithmetic plays some meaningful role only in those Nepalese towns and cities where facilities for water, electricity, telephone, taxi, IT, etc. and good markets are available. The cost or price is also recorded against each such item. The total of all costs are calculated and mentioned in a slip. Such a slip of paper is called a bill. An important aspect of household arithmetic is reading of a bill. In our day-to-day life, we come across various types of bills by social requirement. In all cities and some villages of Nepal, the government has provided the facilities mentioned below: 1. Electricity 2. Drinking Water 3. Telephone 4. Taxi Commercial Arithmetic Topic Description Formulae Profit and Loss When SP > CP, it becomes profit (P). P = P% of CP = SP – CP or, SP = CP + P and CP = SP – P When SP < CP, it becomes loss (L). L = L% of CP = CP – SP or, SP = CP – L and CP = SP + L Profit % (P%) = P CP × 100% = SP – CP CP × 100%. Loss % (L%) = L CP × 100% = CP – SP CP × 100%. SP = CP + P% of CP = CP(1+P%) = CP(100+P) 100 SP = CP – L% of CP = CP(1–L%) = CP(100–L) 100 Simple Interest Interest: Paid extra money regularly at a particular rate for the use of money lent. Simple Interest (SI): Interest taken at a particular rate (R) for money lent (P) in the last time (T). Amount (A): The sum of money lent and simple interest. SI = PTR 100 = PMR 1200 (M < 12) = PDR 36500 (D < 30) SI = A – P => P = A – SI A = SI + P = P +PTR 100 = P(1 + TR 100) = P(100+TR) 100 R E V I S I T I N G A R I T H M E T I C
Arithmetic 42 Allied The Leading Mathematics-10 Compound Interest 43 2.1 Compound Interest At the end of this topic, the students will be able to: ¾ solve the problems related to compound interest on the present banking system of bank and other financial institutions of the country. Learning Objectives I Introduction We have to work with money in our daily activities. When we save some money, we deposit in the financial companies. Why do we deposit meaning in the financial company? What are we profits when the extra money is deposited in the bank or other financial institutes? The banks or the financial companies use the deposited money in other incoming fields. They get some profits by account holders from these fields and some money of them is given their account holders instead of uses of their money, called interests. All commercial banks (A Level), development and merchant banks (B Level), finance companies (C Level), micro credit development banks (D Level), cooperatives and other financial institutes are controlled by Nepal Rastra Bank (NRB) (Estd: 1956); The Central Bank of Nepal under certain rules and regulations such as investment capitals, interest rates of deposits and loans, liquidity of currency, transactions, etc. Activity 1 Pramod borrowed Rs. 1000 from Binod at the rate of 10% per annual (p.a.) for 3 years from 2080 to 2082. There might be two different ways of paying interest shown in the table below. Now, discuss about interests and amounts. One way of paying interest Other way of paying interest Year Principal in Rs. (P) Interest in Rs. (I) Amount (P + I) Principal in Rs. (P) Interest in Rs. (I) Amount 10% C. I. 10 C. I. (P + I) 2080 1000 100 100 1100 1000 100 100 1100 2081 1000 100 200 1200 1100 110 210 1210 2082 1000 100 300 1300 1210 121 331 1541 C1 C2 C3 C4 C5 C6 C7 C8 C9 CHAPTER 2 COMPOUND INTEREST
Arithmetic 42 Allied The Leading Mathematics-10 Compound Interest 43 "Alternatively" Interest for 1 year 10% of Rs. 1000 = Rs. 100 ∴ Interest for 3 years = 3 × 100 = Rs. 300 By next method, Interest for 3 years = PTR 100 = 1000 × 3 × 10 100 = 300 Instead of paying each year 10% of the principal; the interest to be paid each year may be accumulated to the principal as the amount and the same rate of interest might be applicable on the total amount for the next year. At the end of every year, Pramod pays the same interest to Binod, but not added with new principal. At the end of every year, Pramod pays the same interest to Binod and adds the interest with new principal. In this way, working out interest each year on the principal is called Simple Interest (SI). Working out the interest in this way is called the Compound Interest (CI). Where C. I. = Cumulative interest, C1 = Column 1, C2 = Column 2 and son on. ⇒ Interest worked out on principal for the given time at once is called Simple Interest. ⇒ Interest worked on both principal and the accumulated interest is called the Compound Interest. III Derivation of Yearly Compound Amount and Compound Interest Activity 2 Santosh borrowed Rs. 400 from Dinesh at the rate of 5 paisa per rupee per year for 3 years compounded annually. But Sauhard borrowed Rs. P from Ramesh at the rate of Rs. R per 100 rupee per year for T years compounded annually. How much interest and total amount do Santosh and Sauhard pay to Dinesh and Ramesh respectively ? Let 'P' be the principal, 'R%', the interest rate per annum and 'T', the number of year (time). The compound interest is denoted by symbol CI. We derive CI by using the first principle (definition). Particular Case General Case Principal, P = Rs. 400 Principal = P Rate (R) = 5 paisa per rupee per year = 5% Rate = Rs. R per 100 rupees per year = R% Time (T) = 3 years Time = T yrs Then, for the 1st year; For the 1st year; Principal, P1 = Rs. 400 Principal (P1) = P Interest (I1) = Rs. 400 × 5% = Rs. 20 Amount (A1) = Rs. 400 + Rs. 20 = Rs. 420 Interest (I1) = PR% Amount (A1) = P + PR% For the 2nd year; For the 2nd year; New Principal, P2 = Amount A1 = Rs. 420 New Principal, P2 = P + I = P + PR% = P(1+R%)1 Interest (I2) = Rs. 420 × 5% = Rs. 21 Amount (A2) = Rs. 420 + Rs. 21 = Rs. 441 Interest (I2) = (P + PR%)R% = P(1 + R%) R% Amount (A2) = P(1 + R%)1 + P(1 + R%) R% For the 3rd year; For the 3rd year; Principal, P3 = Amount (A2) = Rs. 441 Principal,P3 = P(1+R%)+P(1+R%)R%=P(1+R%)2 Oh! it is the same as the SP as, SP = CP(1 + P%)
Arithmetic 44 Allied The Leading Mathematics-10 Compound Interest 45 Interest (I3) = Rs. 441 × 5% = Rs. 22.05 Amount (A3) = Rs.441+Rs.22.05 = Rs.463.05 Interest (I3) = P(1 + R%)2 R% Amount (A3) = P(1 + R%)2 + P(1 + R%)2 R% At the end of the third year, Amount A3 = Rs. 463.5 = Rs.400(1 + 5%)3 At the end of the third year, Amount A3 = P(1+R%)2 + P(1+ R%)2 R% = P(1 + R%)3 . Hence, Santosh pays Rs. 463.05 to Dinesh in total and Rs. 463.05 – Rs. 400 =Rs. 63.05 for interest only. But, in the case of payment of Sauhard, the power of (1 + R%) is increasing by 1 for successively increasing in time in years. Then, the paying amount to Ramesh by Sauhard is, Compound Amount (CA/AT) = P(1 + R%)T and Compound Interest (CA/IT) = CA – P = P(1 + R%)T – P = P[(1 + R%)T – 1]. This formula can be derived as before. We simply have to replace R% by R 100 , since R percent per annum = R 100 per rupee per annum = R% as below: CA = P(1 + R%)T = P 1 + R 100 = P 100 + R 100 T and CI = P 100 + R 100 T – 1 for complete year. Similarly, for fraction year M 12, i.e., T years and M months; CA = P(1 + R%)T 1 + MR% 12 = P 1 + R 100 T 1 + MR 1200 and CI = P (1 + R%)T 1 + MR% 12 – 1 = P 1 + R 100 T 1 + MR 1200 – 1 Note: 1. If T ≤ 1, then CI = SI. 2. When the rates of interest are different for consecutive year; A = P(1 + R1%)T1(1 + R2%)T2..... and CI = P[(1 + R1%)T1(1 + R2%)T2 + ...... – 1]. 3. The sum (ST) and difference (DT) of CI and SI; ST = P 1 + R 100 T + 1 + TR 100 and DT = P 1 + R 100 T – 1 + TR 100 4. If the compound amounts for T years and (T + 1) years are CAT and CA(T+1) respectively then, (i) Rate of interest (R) = CAT+1 CAT – 1 × 100% (ii) Principal (P) = CAT (1+R%)T 5. If the compound interests for 1 year and 2 years are CI1 and CI2 respectively then, (i) Rate of interest (R) = CI2 CI1 – 1 × 100% (ii) Principal (P) = CI1 × 100% R Oh! It is the same as the SA as, SA = P(1 + TR%) = P 1 + TR 100 = P 100 + TR 100
Arithmetic 44 Allied The Leading Mathematics-10 Compound Interest 45 IV Derivation of Half Yearly (Semi-annually) Compound Amount and Compound Interest Some financial companies provide semi-annually interest to their account holders. When we deposit some amount in the Fixed Deposit (FD) account in the bank or other financial companies, they provide the interest in our general account in each six-month. Here, the rate of interest may or may not be the same in both accounts general and fixed deposit. In this situation, there is not fully compound interest. If we want in the form of compound interest, we will deposit the interest in the next fixed deposit account at the same rate in every six months. When the compound interest is accumulated half-yearly (semi-annually), the interest is calculated at the rate of R 2 with the period of 2T half years. Hence, the formula reduces to; A = P 1 + R% 2 2T and CI = P 1 + R% 2 2T – 1 or, A = P 1 + R 200 2T and CI = P 1 + R 200 2T – 1 or, A = P 200 + R 200 2T and CI = P 200 + R 200 2T – 1 for completed half year. Similarly, for fraction year, i.e., T years and less than 6 months; A = P 1 + R% 2 2T 1 + MR% 12 and CI = P 1 + R% 2 2T 1 + MR% 12 – 1 V Derivation of Quarter Yearly Compound Amount and Compound Interest Almost financial companies provide quarter yearly (each three months) interest to their account holders. When we deposit some amount in our account of the bank, they provide the interest in our general account in each three month. In the same process as motioned in half-yearly, if we want in the form of compound interest, we will deposit the interest in the next fixed deposit account at the same rate in every three months. When the compound interest is accumulated quarter-yearly, the interest is calculated at the rate of R 4 with the period of 4T half years. Hence, the formula reduces to; A = P 1 + R% 4 4T and CI = P 1 + R% 4 4T – 1 or, A = P 1 + R 400 4T and CI = P 1 + R 400 4T – 1 or, A = P 400 + R 400 4T and CI = P 400 + R 400 4T – 1 for completed quarter year. Remarks: 1. Compound Amount for one-third year (four months), CA4 = A = P 1 + R% 3 3T and CI = P 1 + R% 3 3T – 1 2. Compound Amount for monthly, CAM = A = P 1 + R% 12 12T and CI = P 1 + R% 12 12T – 1
Arithmetic 46 Allied The Leading Mathematics-10 Compound Interest 47 3. Compound Amount for daily (called meter interest, personal meter interest is challenge for gov.), CAD = A = P 1 + R% 365 365T and CI = P 1 + R% 365 365T – 1 , where D < 30 4. Compound Amount for year, month and day, CA = A = P 1 + R% T 1 + MR% 12 1 + DR% 365 and CI = P 1 + R% T 1 + MR% 12 1 + DR% 365 – 1 , where T > 1, M < 12, D < 30 Note: Almost banks in Nepal, provide daily interest in three months (Baishakh to last of Asar) by using simple interest formula SI = PTR 36500 as shown in model balance sheet of a bank below. Date Particulars (R% = 5%) Debit (Withdraw) Credit (Deposit) Balance in Rs. Remarks (Hidden part, using software) 2080.01.05 Open balance 5000.00 5000.00 ----- 2080.02.03 By cheque 500.00 4500.00 Int. = 5000×29×5 36500 = 19.86 2080.02.32 By ATM 1000.00 3500.00 Int. = 4500×29×5 36500 = 17.88 2080.03.10 Bank transfer 4000.00 7500.00 Int. = 3500×10×5 36500 = 4.79 2080.03.31 Interest sum 64.11 7564.11 Int. = 7500×21×5 36500 = 21.58 2080.03.31 Interest Tax 3.21 7560.90 Int. = 5% of 64.11 = 3.21 Example-1 Anushka plans to deposit Rs. 75000 for 2 years in any bank. (a) How is compound interest distinct from simple interest ? (b) How much amount will she get at 13% simple interest when she deposits it in a bank N ? Calculate it. (c) How much amount will she get at 12% interest being compounded annually when she deposits it in another bank M ? Calculate it. (d) Which bank is better to deposit the sum and by how much more amount will she get from the better bank ? Solution: (a) Simple interest is based on the principal amount of a loan or deposit, but compound interest is based on the principal amount and the interest that accumulates on it in every period. (b) Here, Principal (P) = Rs. 75000, Rate of interest (R) = 13% p.a., Time (T) = 2 years, Simple amount (SA) = ?
Arithmetic 46 Allied The Leading Mathematics-10 Compound Interest 47 Now, we have, SA = P 100 + TR 100 = Rs. 75000 100 + 2 × 13 100 = Rs. 75000 × 126 100 = Rs. 75000 × 1.26 = Rs. 94500. Hence, she will get Rs, 94500 when she deposits her amount in a bank N. (c) Here, Principal (P) = Rs. 75000, Rate of compound interest (R) = 12% p.a., Time (T) = 2 years, Compound amount (CA) = ? Now, we have, CA = P 100 + R 100 T = Rs. 75000 100 + 12 100 2 = Rs. 75000 × 112 100 2 = Rs. 75000 × (1.12)2 = Rs. 75000 × 1.2544 = Rs. 94080 Hence, she will get Rs, 94080 when she deposits her amount in another bank M. (d) Since the bank N gives more interest than the bank K, so the bank N is better than the bank M. ∴ More amount to be gotten by the better bank N = Rs. 94500 – Rs. 94080 = Rs. 420. Example-2 Ramesh plans to deposit Rs. 250000 for 3 years in any bank. (a) What is compound interest? Define it. (b) How much interest will he get at 12% being compounded annually when he deposits it a bank A ? Calculate it. (c) How much interest will he get at 11% being compounded semi-annually when he deposits it in another bank B ? Calculate it. (d) In which bank will he deposit it to get more interest and by how many percent ? Solution: (a) The interest on savings calculated on both the initial principal and the accumulated interest in every period, is called compound interest. (b) Here, Principal (P) = Rs. 250000, Rate of interest (R) = 12% p.a., Time (T) = 3 years, Compound Interest annually (CIA) = ? Now, we have, CIA = P 100 + R 100 T – 1 = 250000 100 + 12 100 3 – 1 = 250000 112 100 3 – 1 = 250000 [(1.12)3 – 1] = 250000 [1.404928 – 1] = 250000 × 0.404928 = 101232 ∴ Ramesh will get Rs. 101232 as interest at 11% being compounded annually when he deposits the sum in the bank A.
Arithmetic 48 Allied The Leading Mathematics-10 Compound Interest 49 (c) Here, Principal (P) = Rs. 250000, Rate of interest (R) = 11% p.a., Time (T) = 3 years, Compound Interest semi-annually (CIH) = ? We have, CIH = P 200 + R 200 2T – 1 = 250000 200 + 11 200 2 × 3 – 1 = 250000 211 200 6 – 1 = 250000 [(1.055)6 – 1] = 250000 [1.378842868 – 1] = 250000 × 0.378842868 = 94710.70 ∴ Ramesh will get Rs. 94710.70 as interest at 11% being compounded annually when he deposits the sum in the bank B. (d) Here, CIA (Rs. 101232) > CIH (Rs. 94710.70). So, the interest in the bank A is more than that of B. Hence, Ramesh will deposit the sum to get more interest in the bank A. More percent in interest of the bank A than the bank B = CIA – CIH CIH × 100% = Rs. 101232 – Rs. 94710.70 Rs. 94710.70 × 100% = Rs. 6521.30 Rs. 94710.70 × 100% = 0.0689 × 100% = 6.89% Example-3 A man deposits the sum of Rs. 2000 for 2 years in a bank. (a) Which formula is used for the finding interest compounded half yearly. (b) Find the interest compounded half yearly at the rate of 8% per annum in a scheme A. (c) Find the interest compounded yearly at the rate of 8% per annum in another scheme B. (d) Which is better scheme A or B ? Why? Solution : (a) The formula CI = P 1 + R% 2 2T – 1 is used for finding interest compounded half yearly. (b) Given, Principal (P) = Rs. 2000, Time (T) = 2 yrs., Rate (R) = 8% per year, Compound Interest for half yearly (CI) = ? Now, we, have AH = P 1 + R% 2 2T = Rs. 2 000 1 + 8 200 2×2 = Rs. 2000 (1.04)4 = Rs. 2340. ∴ CIH = AH – P = Rs. (2340 – 2000) = Rs. 340. "Alternatively" CIH = P 1 + R% 2 2T – 1 = Rs. 2000 1 + 8 200 2×2 – 1 = Rs. 2000 [(1.04)4 – 1] = Rs. 2000 [1.17 – 1] = Rs. 2000 × 0.17 = Rs. 340
Arithmetic 48 Allied The Leading Mathematics-10 Compound Interest 49 (c) Compound Interest for yearly (CIY) = P 1 + R% T – 1 = Rs. 2000 1 + 8 100 2 – 1 = Rs. 2000 [(1.08)2 – 1] = Rs. 2000 [1.1664 – 1] = Rs. 2000 × 0.1664 = Rs. 332.80 (d) Here, CIH > CIY. So, the scheme A is better than the scheme B because the scheme A has more interest than scheme B. Example-4 The difference between the compound interest and simple interest on the sum of money in 2 years at 10% per annum is Rs. 75. (a) Let us suppose the required sum be x. Find the simple interest in term of x. (b) Find the compound interest in term of x. (c) Find the value of x. What is the sum ? Write. Solution : (a) Here, Principal = x (say), Rate of interest (R) = 10% p.a., Time (T) = 2 years ∴ Simple interest (SI) = P × T × R 100 = P × 2 × 10 100 = Rs. 0.20x (b) Compound interest (CI) = P[(1 + R%)T – 1] = x [(1 + 10%) T – 1] = x 1 + 10 100 2 – 1 = x [1.21 – 1] = 0.21x (c) By the question, CI – SI = 75 or, 0.21x – 0.20x = 75 or, 0.01x = 75 or, x = Rs. 7500. Hence, the required sum is Rs. 7500. Example-5 A bank provides the interest at 5 paisa per rupee per year. (a) What is the rate of compound interest ? (b) In how long time the compound interest for the sum of Rs. 200000 compounded yearly at the same of (a) is Rs. 31525?. (c) What is the simple interest on the same sum at the same rate for the same time ? Find it. (d) Which interest is more between compound interest and simple interest and by how many percentage ? Find. Solution: (a) Rate of interest (R) = 5 paisa per Re.1 per 1 year = 5 × 100 paisa per Rs. 100 per 1 year = 5%,
Arithmetic 50 Allied The Leading Mathematics-10 Compound Interest 51 (b) Given, Sum of money (P) = Rs. 200000, Compound Interest (CI) = Rs. 31525, Time (T) = ? Now, we know that; CI = P 1 + R 100 T – 1 or, 31525 = 200000 1 + 5 100 T – 1 or, 31525 200000 = 1.05T – 1 or, 0.157625 + 1 = 1.05T or, 1.053 = 1.05T ∴ T = 3 yrs. Hence, the required duration of time is 3 years. (c) Simple interest for the given information (SI) = PRT 100 = 200000 × 5 × 3 100 = 30000 (d) Here, Rs. 31525 > Rs. 30000. So, CI > SI. More percent in CI than SI = CI – SI SI × 100% = 31525 – 30000 30000 × 100% = 1525 30000 × 100% = 5.08% Example-6 Rs. 16000 is deposited for 2 years at 15% per annum compounded quarter annually. (a) Which formula is used for finding the compound amount as quarter yearly interest? (b) Find the compound amount for the given information. (c) Find the compound interest. Solution: (a) Compound Amount (CA) = P 1 + R 400 4T is used for finding compound amount when the time is given in years and quarter year. (b) Principal (P) = Rs.16000, Rate (R) = 15%, Time (T) = 2 years ⸫ Compound Amount (CA) = P 1 + R 400 4T = 16000 1 + 15 400 4 × 2 = 16000 × 1.03758 = 16000 × 1.34247 = Rs. 21479.53. (c) CI = Rs. (21479.53 – 16000) = Rs. 5479.53.