10 Author Shyam Datta Adhikari Phone : 0977-01-4313205 Approved by the Government of Nepal, Ministry of Education, Science and Technology, Curriculum Development Centre (CDC), Sanothimi, Bhaktapur.
Name : ............................................................ Class : ....................... Roll No. : ................ Section : .......................................................... School : ............................................ Publisher Oasis Publication Pvt. Ltd. Copyright The Publisher Edition First : 2067 B.S. Second : 2068 B.S. Third : 2072 B.S. Fourth : 2074 B.S. Fifth : 2076 B.S. Sixth : 2080 B.S. (New Curriculum) Price : Rs. 481/- Print quantity : (5,000 pcs.) Contributors Laxmi Gautam Rajendra Sapkota Prakash Ghimire Layout Oasis Desktop Ramesh Bhattarai Printed in Nepal 10
Preface Oasis School Mathematics has been designed in compliance with the latest curriculum of the Curriculum Development Center (CDC), the Government of Nepal with a focus on child psychology of acquiring mathematical knowledge and skill. The major thrust is on creating an enjoyable experience in learning mathematics through the inclusion of a variety of problems which are closely related to our daily life. This book is expected to foster a positive attitude among children and encourage them to enjoy mathematics. A conscious attempt has been made to present mathematical concepts with ample illustrations, assignments, activities, exercises and project work to the students in a friendly manner to encourage them to participate actively in the process of learning. I have endeavored to present this book in a very simple and interesting form. Exercises have been carefully planned. Enough exercises have been presented to provide adequate practice. I have tried to include the methods and ideas as suggested by the teachers and subject experts who participated in the seminars, and workshops conducted at different venues. I express my sincere gratitude to my friends and well wishers for their valuable suggestions. I am extremely grateful to Man Bahadur Tamang, Laxmi Gautam, Sunil Kumar Chaudhary, Ram Prasad Sapkota, Saroj Neupane for their invaluable suggestions and contributions. Sincere gratitude to Managing Director Oasis Publication for his invaluable support and cooperation in getting this series published in this shape. In the end, constructive and practical suggestions of all kinds for further improvement of the book will be appreciated and incorporated in the course of revision. Shyam Datta Adhikari Author March 2023
Contents Unit - 1 Set 1-25 1 Set...................................................................................... 2 Unit - 2 ArithMetic 26-84 2 Compound Interest......................................................... 27 3 Growth & Compound Depreciation............................ 49 4 Money Exchange............................................................. 66 Unit - 3 MenSurAtiOn 85-135 5 Pyramid............................................................................ 86 6 Combined Solid.............................................................. 105 7 Cost Estimation............................................................... 120 unit - 4 AlgebrA 136-202 8. Sequence and Series....................................................... 137 9. Quadratic Equation........................................................ 165 10 Algebraic Fraction.......................................................... 185 11 Indices............................................................................... 193 Unit - 5 geOMetry 203-266 12 Area of Triangles and Quadrilaterals.......................... 204 13 Construction.................................................................... 220 14 Circle................................................................................. 232 unit - 6 StAtiSticS AnD PrObAbility 267-311 15 Statistics............................................................................ 268 16 Probability........................................................................ 291 unit - 7 trigOnOMetry 312-323 17 Trigonometry................................................................... 313 Specification Grid................................................................... 324
Oasis School Mathematics-10 1 Set Contents • Word problems on set using a Venn diagram of 2 sets • Word problems on set using a Venn diagram of 3 sets expected Learning Outcomes At the end of this unit, students will be able to develop the following competencies: • To solve the word problems on two sets using a Venn diagram • To solve the word problems on three sets using a Venn diagram Materials Required • flash cards, chart paper, A4 size paper, sketch pen Estimated Teaching Hours 12
2 Oasis School Mathematics-10 1.1 Warm-up Activities Observe the following Venn diagram and answer the questions given below. Write the set of elements of (i) A (ii) B (iii) A and B both (iv) A or B (v) neither A nor B (vi) Number of elements in each set. 1.2 Set Operations Review Union of Sets Let A and B be any two sets. The union of the two sets A and B is denoted by A ∪ B and is defined as a set of elements which either belong to A or B or both. A ∪ B is read as 'A union B' or 'A cup B'. Symbolically, A ∪ B = { x : x ∈ A or x ∈ B}. Representation in Venn diagram: The shaded region in the Venn diagram represents A ∪ B. Intersection of Sets: Let A and B be any two sets. The intersection of the two sets A and B is denoted by A ∩ B and is defined as a set of elements which belong to A and B both. A ∩ B is read as 'A intersection B' or 'A cap B'. Symbolically, A ∩ B = {x : x ∈ A and x ∈ B} A B U •e •x •a •p •y •b •q •z •c •d •f A B U U A B U A∪B when A and B are overlapping A∪B when A and B are disjoint A∪B where B is the subset of A A B Unit 1 Set
Oasis School Mathematics-10 3 Representation in Venn diagram: The shaded region in the Venn diagram represents A ∩ B. Difference of Sets: Let 'A' and 'B' be two sets. The difference of the two sets A and B is denoted by A – B and is defined as a set of elements of A which do not belong to B. A – B is read as 'A difference B' or 'A minus B'. Symbolically, A – B = {x : x ∈ A and x ∉ B}. Similarly, B – A = { x : x ∈ B and x ∉ A}. Complement of Set Let A be a subset of the universal set U. The complement of set A is denoted by A or Ac , or A' and is defined as a set of elements of U which do not belong to A. A is read as 'A bar' or 'a complement of set A in U'. Symbolically, A = {x : x ∈ U and x ∉ A}. Symmetric Difference of Sets Let A and B be two sets. The symmetric difference of the two sets A and B is denoted by A∆B and is defined as the union of the differences A–B and B–A. A∆B is read as 'A delta B' Symbolically, A∆B = (A – B) ∪ (B – A) = {x : x ∈ A – B or x∈B – A} A B U A B U U A∩B when A and B are overlapping A∩B when A and B are disjoint A∩B where B is the subset of A A B A B U A B U The shaded region in the Venn diagram represents A – B. The shaded region in the Venn diagram represents B – A. A U A B U The shaded region in the Venn diagram represents A ∆ B. A∆B = (A ∪ B) – (A ∩ B) Note
4 Oasis School Mathematics-10 1.3 Cardinality of Finite Sets The number of distinct elements in a finite set is called the cardinal number of a set. Hence, the cardinal number of a set is the number of elements present in that sets. Observe the given Venn diagram. Find the number of elements in set A. A = {2, 3, 5, 7}. Set A has 4 elements. The cardinal number of set A is 4. Symbolically n(A) = 4 Similarly, find n(B), n(A∩B), n(A∪B), and n(A∪B). Cardinal number of union of two sets Let A and B be any two overlapping sets such that n(A) = x n(B) = y and n(A∩B) = z then n(A–B) = x–z and n(B–A) = y–z From the Venn diagram, n(A∪B) = x–z+z+y–z = x+y–z = n(A) + n(B) – n(A∩B) Hence, n(A∪B) = n(A) + n(B) – n(A∩B) If A and B are disjoint sets, n(A∩B) = 0 Then, n(A∪B) = n(A) + n(B) Some important results of cardinality of two sets: (i) n(A∪B) = n(A) + n(B) – n(A∩B) (ii) n(A∪B) = n(A) + n(B) [If A and B are disjoint sets] (iii) n(A–B) = n0 (A) = n(A) – n(A∩B) (iv) n(B–A) = n0 (B) = n(B) – n(B∩A) (v) n(A–B) = n0 (A) = n(A∪B) – n(B) (vi) n(B–A) = n0 (B) = n(A∪B) – n(A) (vii) n(A∪B) = n0 (A) + n0 (B) + n(A∩B) A 3 4 6 8 10 5 7 2 B U 1 A x–z z y–z B U A B n(A∩B) n(B–A) or n0 n(A–B) or n (B) 0 (A) n(A∪B) n(A∪B) U
Oasis School Mathematics-10 5 (viii) n(A∪B) = n(U) – n(A∪ B) (ix) n(A∆B) = n(A – B) + n (B–A) (x) n(A∆B) = n(A ∪ B) – n (B ∩ A) Worked Out examples example 1 A and B are the subsets of the universal set 'U' in which n(U) = 50, n(A) = 32, n(B) = 25 and n(A∩B) = 15. (a) Illustrate the above information in a Venn diagram. (b) Find the value of n(A∪B). Solution: Given, n(U) = 50, n(A) = 32, n(B) = 25, n(A∩B) = 15 Let n(A B) ∪ = x Remember ! Verbal form Set notations * number of elements in either A or B or both n(A∪B) * number of elements in at least one of A or B n(A∪B) * number of elements in both sets A and B n(A∩B) * number of elements only in set A n0 (A) or n(A–B) * number of elements in A but not in B n0 (A) or n(A–B) * number of elements only in set B n0 (B) or n(B–A) * number of elements in B but not in A n0 (B) or n(B–A) * number of elements neither in A nor in B n(A∪B) * number of elements in exactly one set n0 (A) + n0 (B) Note • If two sets 'A' and 'B' are disjoint, n(A∪B) = n(A) + n(B) and n (A∩B) = 0 ∴ n(A∪B) has maximum value and n(A∩B) has minimum value. • If B⊂A, n(A∪B) = n(A) and n(A∩B) = n(B) ∴ n(A∪B) has minimum value. n(A∩B) has maximum value. In this case, n(A∪B) = n(A) and n(A∩B) = n(B). A A B U U B
6 Oasis School Mathematics-10 Illustration in a Venn diagram: From the Venn diagram n(U) = 17 + 15 +10 + x or, 50 = 42 + x or, x = 50 – 42 or, x = 8 ∴ n(A B) ∪ = 8 example 2 If n(A) = 45, n(B) = 65, n(A∪B) = 85, (a) write the relation among n(B), n(B) and n(A∩B) (b) using this relation, find n(A∩B) (c) find n(A∩B), n0 (A) and n0 (B) (d) Illustrate the above information in a Venn-diagram. Solution: Here, (a) n(A) = 45, n(B) = 65, n(A∪B) = 85. n(A∪B) = n(A) + n(B) – n(A∩B) (b) 85 = 45 + 65 – n(A∩B) or, n(A∩B) = 45 + 65 – 85 or, n(A∩B) = 110 – 85 ∴ n(A∩B) = 25 (c) n0 (A) = n(A) – n(A∩B) = 45 – 25 = 20 n0 (B) = n(B) – n(A∩B) = 65 – 25 = 40 (d) Illustration in a Venn diagram A 32–15 = 17 15 x 25–15 =10 B U A 20 25 40 B U Here, n(A) = 45, n(B) = 65, n(A∪B) = 85 (a) n(A∪B) = n(A) + n(B) – (A∩B) (b) Let, n(A∩B) = x From the Venn diagram, n(A∪B) = 45 – x + x + 65 – x or, 85 = 45 + 65 – x or, 85 = 110 – x or, x = 110 – 85 Alternative method A 45–x x 65–x B
Oasis School Mathematics-10 7 A 20 25 40 B or, x = 25 ∴ n(A∩B) = 25 (c) From the Venn diagram, n0 (A) = 45 – x = 45 – 25 = 20 n0 (B) = 65–x = 65–25 = 40 (d) Illustration in the Venn diagram. example 3 If n(A) = 20, n(B) = 25, (a) on which condition n(A∪B) has maximum value? (b) what is the maximum value of n(A∪B)? (c) on which condition n(A∪B) has minimum value? (d) what is the minimum value of n(A∪B)? (e) what happens to n(A∩B) if n(A∪B) has maximum value and minimum value? Justify your answer. Solution: (a) n(A∪B) has maximum value if 'A' and 'B' are disjoint set. (b) The maximum value of n(A∪B) = n(A) + n(B) = 20 + 25 = 45 (c) n(A∪B) has minimum value if one set is the subset of another. Since, n(A) < n(B), n(A∪B) has minimum value if A ⊂ B. (d) The minimum value of n(A∪B) = n(B) [∴ A ⊂ B] = 25 (e) n(A∩B) has minimum value when n(A∪B) has maximum value. From the first Venn-diagram, ∴ The minimum value of n(A∩B) = 0. Again, n(A∩B) has maximum value when n(A∪B) has minimum value. So, the minimum value of n(A∩B) = n(A) = 20. example 4 In a class of 55 students, 15 students liked Maths but not english, and 18 students liked english but not Maths. If 5 students did not like both, (a) write the information from the questions, (b) how many students liked both subjects? (c) Represent the above information in a Venn diagram. A 20 25 B U A B U
8 Oasis School Mathematics-10 Solution: (a) Let, M and E be the sets of students who liked Maths and English respectively. Here, n0 (M) = n(M–E) = 15 n0 (E) = n(E–M) = 18 n(M E) ∪ = 5 n (U) = 55 n(M∩E) = ? We have, (b) n(U) = n0 (M) + n0 (E) + n(M∩E) + n(M E) ∪ or, 55 = 15 + 18 + 5 + n(M∩E) or, 55 = 38 + n(M∩E) or, n(M∩E) = 55 – 38 = 17 So, 17 students liked both the subjects. (c) Representation in a Venn diagram is as shown in the figure. Let, M and E be the sets of students who liked Maths and English respectively. (a) Here, n0 (M) = 15, n0 (E) = 18 n(M∪E) = 5, n(U) = 55 Let, n(M∩E) = x (b) From the Venn diagram, 15 + x + 18 + 5 = 55 or, x = 55 – 38 = 17 (c) Representation in a Venn diagram is as shown in the figure. M 15 5 17 18 E U Alternative method M 15 5 x 18 E U example 5 In an examination, 80% of the students passed in Mathematics, 70% passed in Science and 10% failed in both subjects. (a) Assuming n(U) = 100, write the cardinality of the sets mentioned above. (b) Using a Venn diagram, find the percentage of students who passed both subjects. (c) If there are 250 students, find how many students passed in both subjects. M 15 5 17 18 E U
Oasis School Mathematics-10 9 Solution: (a) Let 'M' and 'S' be the sets of students who passed in Mathematics and Science respectively. Let, n(U) = 100 then, n(M) = 80, n(S) = 70, n(M S) ∪ = 10 n(M∩S) = ? Let, n(M∩S) = x (b) From the Venn diagram, 80 – x + x + 70 – x + 10 = n(U) or, 160 – x = 100 or, x = 160 – 100 or, x = 60 ∴ n(M∩S) = 60 Hence, 60% students passed in both subjects. (c) If there are 250 students number of students who passed in both subject = 60% of 250 = 60 100 × 250 = 150. example 6 In a group of people, 50% like tea, 70% like coffee, 10% don't like both and 120 like both. (a) Assuming n(U) = x, write the cardinality of sets in terms of x. (b) Present these vale in Venn diagram and find the value of x. Solution: (a) Let 'T' and 'C' be the sets of people who like tea and coffee respectively. Let, n(U) = x = = ×= = = ×= ∩ = ∪ = = ×= 50 50x Given, n(T) 50% of x x 100 100 70 70x n(C) 70% of x x 100 100 n(T C) 120 10 10x n(T C) 10% of x x 100 100 (b) From the Venn diagram, 120 T C U –120 70x 100 10x 100 50x 100 –120 = ++ + =+ + = − − = = 50x 70x 10x üüüüüüü 100 100 100 50x 70x 10x or, x – 120 120 100 100 130x üüü 100 130x 130x or, x 120 or, 120 100 100 3
10 Oasis School Mathematics-10 = ++ + =+ + = − − = = = = × × = = 50x 70x 10x üüüüüüü 100 100 100 50x 70x 10x or, x – 120 120 100 100 130x üüü 100 130x 130x or, x 120 or, 120 100 100 3x or, 120 10 or, 3x 120 10 120 10 or, x 3 üüü ∴ Total number of people in the survey = 400. example 7 In a group of 70 people the ratio of people who like documentaries to the people who like movies is 5:4. If 10 people like both and 8 people do not like both, using a Venn diagram, find the number of people who like (a) documentaries only (b) movies only (c) show the final result in the Venn-diagram. Solution: (a) Let 'D' and 'M' be the sets of people who like documentaries and movies respectively. Given, n(U) = 70 Let n(D) = 5x and n(M) = 4x n(D∩M) = 10, n(D M) ∪ = 8 Now, D 5x–10 10 4x–10 M 8 U From the Venn diagram, n U( ) x x or, x or, x or, x or, x = − + + − + = − = + = = = 5 10 10 4 10 8 70 9 2 9 70 2 9 72 72 9 8 5 10 5 8 10 30 4 10 4810 0 0 ( ) ( ) ( ) ( ) i n D x ii and n M x = − = × − = = − = × −
Oasis School Mathematics-10 11 (b) and n0 (M) = 4x – 10 = 4 × 8 – 10 = 22 Hence, 30 people like documentaries only and 22 like movies only. (c) Illustration in the Venn-diagram. exercise 1.1 1. From the given figure, verify that: (a) Find n(A), n(B), n0 (A), n0 (B), n(A∩B) and n(A∪B) (b) n(A∪B) = n(A) + n(B) – n(A∩B) (c) n (A – B) = n (A) – n(A∩B) (d) n (B – A) = n(B) – n(A∩B). (e) n(A∪B) = n0 (A) + n0 (B) + n(A∩B) 2. (a) In the given Venn diagram, A and B are the subsets of the Universal set U. Given numbers represent their cardinal number. Find; (i) n(A∪B) (ii) n(A∩B) (iii) n(A) (iv) n(B) (v) n(U) (vi) n( A B ∩ ) (b) In the given Venn diagram, if n(U) = 150, find (i) n(A∩B) (ii) n0 (A) (iii) n0 (B) (iv) n(A∪B) (c) In the given Venn diagram, if n(U) = 60, find the value of n(A), n(B), n(A∪B) and n(A – B). 3. If n(A) = 12, n(B) = 14, (a) on which condition n(A∪B) has maximum value? and what is the maximum value of n(A∪B)? (b) on which condition n(A∪B) has minimum value? and what is the minimum value of n(A∪B)? (c) on what condition n(A∩B) has maximum and minimum value (d) find the maximum and minimum value of n(A∩B). 4. If n(A) = 40, n(B) = 60 and n(A∩B) = 10, (a) write their relation with n(A∪B) (b) find the value of n(A∪B). D 30 8 10 22 M U A a b c d e f g h i B k j U A 25 15 30 B 20 U A 80–x x 60–x B 40 U A x 20 2x B 25 U
12 Oasis School Mathematics-10 5. If n0 (A) = 35, n0 (B) = 45, n(A∩B) = 20, (a) using formula find n(A) and n(B). (b) write their relation with n(A∪B). (c) find n(A∪B) 6. (a) 'A' and 'B' are the subsets of a universal set U, in which n(U) = 43, n(A) = 25, n(B) = 18 and n(A∩B) = 7. (i) Draw a Venn diagram to illustrate the above information. (ii) Using Venn diagram, find the value of n( A B ∪ ). (b) If n(A) = 75, n(B) = 85, n ( A B ∪ ) = 15 and n(U) = 130, using a Venn diagram, find the value of (i) n(A∩B) (ii) n0 (A) (iii) n0 (B). 7. (a) If n(A) = 40, n(B) = 60 and n(A∪B) = 80, then, (i) find the value of n(A∩B). (ii) find the value of n0 (A). (iii) show this in a Venn diagram. (b) If n(U) = 200, n(M) = 160, n(N) = 150, n( M N ∪ ) = 30, find the value of (i) n(M∩N) (ii) n0 (M) (iii) n0 (N) (iv) Show the above information in a Venn diagram. 8. (a) In a survey of a group of 100 students, 68 passed in Mathematics, 60 passed in Science, 12 failed in both subjects. (i) write the cardinality of the sets mentioned above.. (ii) show these informations in the Venn diagram. (iii) find the number of students who passed in both subjects. (iv) find the number of students who passed only one subject. (b) In a survey among 90 people, it was found that 51 people liked to eat oranges and 59 liked to eat apples. If 3 people did not like to eat both the fruits, (i) how many liked to eat both fruits? (ii) how many liked to eat only oranges? (iii) how many liked to eat only apples? (iv) how many liked to eat only one fruit? (v) show the above information in a Venn diagram. 9. (a) In a survey of 100 people, it was found that 65 liked folk songs, 55 liked modern songs and 35 liked folk as well as modern songs. (i) Draw a Venn diagram to illustrate the fact. (ii) How many people did not like both songs? (b) In a survey among 350 people in a community, 200 people are influenced by materialism and 120 people are influenced by spiritualism. If 120 people are
Oasis School Mathematics-10 13 influenced by both, (i) Show the above information in the Venn diagram. (ii) Find the number of people who are influenced by neither of them. 10. (a) In a group of 70 people, 37 like tea, 52 like milk and each person likes at least one of the two drinks. (i) How many people like both tea and milk ? (ii) How many people like tea only ? (iii) How many people like milk only ? (iv) Show the above information in a Venn diagram. (b) In a school, all students play either football or volleyball or both. 300 play football, 250 play volleyball and 110 play both the games. Draw a Venn diagram to find: (i) the number of students playing football only. (ii) the number of students playing volleyball only. (iii) the total number of students. 11. (a) In a group of 200 people, 60 like milk only, 50 like curd only and 10 like none of the two. Using a Venn diagram find the number of people who like both. (b) In a class of 55 students, 15 liked Maths but not English. 18 students liked English but not Maths and 5 students did not like both. Find (i) how many liked both subjects (ii) how many liked Maths (iii) how many liked English (iv) Show this information in Venn diagram. 12. In an election, two candidates P and Q stood for the post and 40,000 people were in voters list. Voters are supposed to cast the vote for single candidate. 20,000 people cast vote for P and 15,000 cast for Q and 3,000 people cast vote even for both. (i) Show these information in Venn diagram (ii) How many people did not cast their votes? (iii) How many votes were valid? Find it. 13. (a) In a survey of a group of people, it was found that 39% of the people like tea only, 27% like coffee only and 15% do not like both. (i) What percentage of the people like both of them? (ii) What percentage of the people like tea? (iii) What percentage of the people like coffee? (iv) Represent the above information in a Venn diagram. (b) In an examination, 40% of the students passed in Mathematics only and 30% passed in Science only. If 10% of students failed in both subjects,
14 Oasis School Mathematics-10 (i) what percentage of students passed in both subjects? (ii) what percentage of students passed in Mathematics? (iii) represent all the results in a Venn diagram. 14. (a) In a survey of a group of people, it was found that 70% of the people liked Coke, 60% liked Fanta, 2,000 people liked both of them and 20% liked none of them. (i) Draw a Venn diagram to illustrate the above information (ii) Find the total number of people in the survey. (b) In a survey of some students, it was found that 60% of the students studied Commerce and 40% studied Science. If 40 students studied both the subjects and 10% didn't study any of the subjects, by drawing a Venn diagram, (i) find the total number of students. (ii) find the number of students who studied Science only. 15. (a) 32 teachers in a school like either milk or curd or both. The ratio of the number of teachers who like milk to the number of teachers who like curd is 3:2 and 8 teachers like both milk and curd. (i) Find the number of teachers who like milk. (ii) Find the number of teachers who like curd only. (iii) Represent the above information in a Venn diagram. (b) In a group of 95 students, the ratio of students who like Mathematics and Science is 4:5. If 10 of them like both the subjects and 15 of them like none of these subjects, then by drawing a Venn diagram, find how many of them (i) like only Mathematics. (ii) like only Science. (c) Out of the 120 students who 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, (i) find the number of students who passed in Mathematics. (ii) find the number of students who passed in Science. (iii) show the result in a Venn diagram. 16. (a) Out of 1500 students who appeared in an examination, 20% students failed in the examination, 40% of the failures failed in Mathematics only and 20% failures failed in Science only and 10% failures failed in other subjects. Find (i) the number of students who failed in both Mathematics and Science. (ii) show this information in a Venn diagram. (b) Out of 1000 students who appeared in an examination, 60% passed the examination. 60% of the failing students failed in Mathematics and 50% of the failing students failed in English. If the students failed in English and Mathematics only, find the number of students who failed in both subjects.
Oasis School Mathematics-10 15 17. (a) In a survey, one third children like only mango and 22 do not like mango at all. 2 5 children like orange but 12 like none of them. (i) show the above data in a Venn-diagram. (ii) how many children like both type of fruits. (b) In a survey 3 14 children like only Math and 70 do not like Math at all. Also, 9 14 children like Science but 20 like none of them. (i) show the above data in the Venn-diagram. (ii) how many students like both subjects. Answers 1. Consult your teacher 2. (a) (i) 70 (ii) 15 (iii) 40 (iv) 45 (v) 90 (vi) 75 (b) (i) 30 (ii) 50 (iii) 30 iv) 110 (c) 25, 30, 35, 5 3. Consult your teacher. 4. (b) 90 5. (a) 55, 65, (c) 100 6. (a) (ii) 7 (b) (i) 45 (ii) 30 (iii) 40 7. (a) (i) 20 (ii) 20 (b) (i) 140 (ii) 20 (iii) 10 8. (a) (iii) 40 (iv) 48 (b) (i) 23 (ii) 28 (iii) 36 (iv) 64 9. (a) (ii) 15 (b) (ii) 150 10. (a) (i) 19 (ii) 18 (iii) 33 (b) (i) 190 (ii) 140 (iii) 440 11. (a) 80 (b) (i) 17 (ii) 32 (iii) 35 12. (ii) 8000 (iii) 29,000 13. (a) (i) 19% (ii) 58% (iii) 46% (b) (i) 20% (ii) 60% 14. (a) (ii) 4,000 (b) (i) 400 (ii) 120 15. (a) (i) 24 (ii) 8 (b) (i) 30 (ii) 40 (c) (i) 70 (ii) 60 16. (a) (i) 90 (b) 40 17. (a) (ii) 8 (b) (ii) 40 Project Work • Take a class of your school and ask the students whether they like sports, music, both or none. Collect the information and present it in a Venn diagram. 1.4 Cardinality of three Sets Let A, B and C be three overlapping subsets of the universal set U. Hence, cardinal number of A = n(A) Cardinal number of B = n(B) Cardinal number of C = n(C) We have n(A∪B∪C) = n[A∪ (B∪C)] ………… (i) n[A∩(B∪C)] = n[(A∩B) ∪ (A∩C)] ………… (ii) n[(A∩B) ∩ (A∩C)] = n(A∩B∩C) ………… (iii) Now, n(A∪B∪C) = n[A∪(B∪C)] [From (i)] 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)] [From (ii)] A B C U
16 Oasis School Mathematics-10 = 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(B∩C) – n(A∩B) – n(A∩C) + n(A∩B∩C) [From (iii)] ∴ n(A∪B∪C) = n(A) + n(B) + n(C) – n(B∩C) – n(A∩B) – n(A∩C) + n(A∩B∩C) If A, B and C are disjoint sets, then n(A∪B∪C) = n(A) + n(B) + n(C) Some important results on cardinality of three sets (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∩B) = n0 (A∩B) + n(A∩B∩C) (iii) n(B∩C) = n0 (B∩C) + n(A∩B∩C) (iv) n(A∩C) = n0 (A∩C) + n(A∩B∩C) (v) n(A) = n0 (A) + n0 (A∩B) + n0 (A∩C) + n(A∩B∩C) (vi) n(B) = n0 (B) + n0 (A∩B) + n0 (B∩C) + n(A∩B∩C) (vii) n(C) = n0 (C) + n0 (A∩C) + n0 (B∩C) + n(A∩B∩C) (viii) n(A∪B∪C) = n0 (A)+n0 (B)+n0 (C)+n0 (A∩B)+n0 (B∩C)+n0 (A∩C)+n(A∩B∩C) (ix) n A( ) ∪ ∪B C = n(U) – n(A∪B∪C) A n0 (C) n(A∩B∩C) n0 (A∩C) n0 (A) n0 (A∩B) n0 (B) n0 (B∩C) n(A∪B∪C) B C U Remember ! Verbal form Set notations * number of elements in either A or B or C n(A∪B∪C) * number of elements in at least one of A or B or C n(A∪B∪C) * number of elements in all three sets n(A∩B∩C) * number of elements in exactly one set n0 (A) + n0 (B) + n0 (C) * number of elements in exactly two sets n0 (A∩B)+n0 (B∩C)+n0 (A∩C) * number of elements in none of the sets n A( ) ∪ ∪B C
Oasis School Mathematics-10 17 Worked Out examples example 1 Sets A, B and C are the subsets of the universal set U. If n(A) = 100, n(B) = 120, n(C) = 110, n(A∩B) = 60, n(B∩C) = 50, n(C∩A) = 55 and n(A∪B∪C) = 200, find n(A∩B∩C) and illustrate the above information in a Venn diagram. Solution: Here, n(A) = 100, n(B) = 120, n(C) = 110, n(A∩B) = 60, n(B∩C) = 50, n(C∩A) = 55 and n(A∪B∪C) = 200. We have, n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C) 200 = 100 + 120 + 110 – 60 – 50 – 55 + n(A∩B∩C) or, 200 = 330 – 165 + n(A∩B∩C) or, 200 = 165 + n(A∩B∩C) or, n(A∩B∩C) = 200 –165 or, n(A∩B∩C) = 35 Illustration in Venn diagram example 2 In a survey of college students about their interest in movies, the following information was obtained, 48% like Nepali movies, 40% like english movies, 31% like Hindi movies, 24% like Nepali and english movies, 19% like Nepali and Hindi movies, 13% like Hindi and english movies, 6% like all three types of movies. Find: (i) what percentage of people like none of these movies? (ii) what percentage of people like only two types of movies? Solution: Let N, E and H be the set of students who like Nepali, English and Hindi movies respectively. Let n(U) = 100 then n(N) = 48, n(E) = 40, n(H) = 31, n(N∩E)= 24 n(N∩H) = 19, n(H∩E) = 13, n(N∩E∩H) = 6, n(N∩E∩H) = ? Insert n(A∩B∩C) = 35 Find n0 (A∩B) = n(A∩B) – n(A∩B∩C) = 60–35 = 25 Similarly, find n0 (B∩C) and n0 (A∩C) n0 (A) = 100–25–35–20 = 20 Similarly, find n0 (B) and n0 (C) Steps A 35 B C U (55–35)=20 100 – (25+35+20)=20 110–(35+15+20)=40 120–(25+35+15)=45 (60 – 35) = 25 (50–35)=15
18 Oasis School Mathematics-10 We have, n(N∪E∪H) = n(N) + n(E) + n(H) – n(N∩E)–n(N∩H)–n(E∩H) + n(N∩H∩E) n(N∪E∪H) = 48 + 40 +31 – 24 – 19 – 13 + 6 = 69 Again, n N( ) ∪ ∪E H = n(U) – n(N∪E∪H) = 100 – 69 = 31 Hence, 31% people like none of these three types of movies. Again, n0 (N∩E) = n(N∩E) – n(N∩E∩H) = 24 – 6 = 18 n0 (E∩H) = n(E∩H) – n(N∩E∩H) = 13 – 6 =7 n0 (N∩H) = n(N∩H) – n(N∩E∩H) = 19 – 6 = 13 ∴ n(only two movies) = 18 + 7 +13 = 38 Hence 38% students like only two types of movies. Let, n(U) = 100, n(N) = 48, n(E) = 40, n(H) = 31, n(N∩E) = 24, n(N∩H) = 19, n(H∩E) = 13, n(N∩E∩H) = 6, n(N∪E∪H) = ? Representing above information in Venn-digram. From the Venn-diagram, n0 (N) = n(N) – 18 – 6 – 13 = 48 – 37 = 11 Again, n0 (E) = n(E) – 18 – 6 – 7 = 40 – 31 = 9 Again, n0 (H) = n(H) – 13 – 6 – 7 = 31 – 25 = 5 Now, substituting these value in the Venn-diagram. N E H 18 6 13 7 U Alternative method N E H 18 6 13 7 9 x 11 9 U Let, n(N∪E∪H) = x
Oasis School Mathematics-10 19 exercise 1.2 1. (a) From the given Venn diagram, find the following. (i) n0 (A) (ii) n0 (B) (iii) n0 (C) (iv) n0 (A∩B) (v) n0 (B∩C) (vi) n0 (A∩C) (vii) n(A∩B∩C) (viii) n(A∪B∪C) (ix) n(U) (x) n(A) (xi) n(B) (xii) n(C) (xiii) n(A∩B) (xiv) n(B∩C) (xv) n(A∩C). (b) In the given Venn diagram, if n(U) = 200, n(P∩Q) = 60, n(Q∩R) = 52, n(P∩R) = 50, find the values of (i) n(P∩Q∩R) (ii) n0 (P∩Q) (iii) n0 (Q∩R) (iv) n0 (P∩R) (c) Insert the cardinality of the following information in the Venn diagram and hence find n(U). n(A) = 14, n(B) = 10, n(C) = 22, n(A∩B∩C) = 6, n(A∩B) = 7, n(B∩C) = 9, n(A∩ C) = 11 and n A( ) ∪ ∪B C = 4. 2. (a) If n(A) = 20, n(B) = 40, n(C) = 30, (A∩B) = 10, n(B∩C) = 12, n(A∩C) = 8 and n(A∩B∩C) = 5, find n(A∪B∪C). (b) If n(A) = 65, n(B) = 50, n(C) = 35, n(A∩B) = 25, n(B∩C) = 20, n(C∩A) = 15, n(A∩B∩C) = 5 and n(U)=100, find the value of n A( ) ∪ ∪B C . (c) If n0 (A) = 25, n0 (B) = 22, n0 (C) = 32, n0 (A∩B) = 5, n0 (B∩C) = 12, n0 (A∩C) = 13, n(A∩B∩C) = 6 and n A( ) ∪ ∪B C = 15, find n(U). 3. (a) The survey of a group of people shows that 60 like tea, 45 like coffee, 30 like milk, 25 like coffee as well as tea, 20 like tea as well as milk, 15 like coffee as well as milk and 10 like all three. How many people were asked this question? Solve by drawing a Venn diagram. (b) In a class of 175 students, 100 students are studying Maths, 70 Physics, 46 Chemistry, 30 Maths and Physics, 28 Maths and Chemistry, 23 Physics and Chemistry and 18 all the three subjects. (i) What percentage of students are studying only one subject? (ii) What percentage of the students are studying none of these subjects? (c) In a group of 300 people 50 like oranges only, 60 like apples only, 70 like mangoes only, 25 like oranges and mangoes but not apple, 30 like oranges and apples but not mangoes, 35 like mangoes and apple but not oranges. If 15 people do not like all three, find the number of people who like all three fruits. A B C 8 7 6 3 7 9 15 12 U P Q R 20 x 42 25 35 U A B C U
20 Oasis School Mathematics-10 4. Out of the total candidates in an examination, 40% students passed in Maths, 45% in Science and 55% in Health. 10% students passed in Maths and Science, 20% in Science and Health, 15% in Health and Maths, no one failed in all subjects. (a) Write the cardinality of sets mentioned above. (b) Draw a Venn diagram to show the above information. (c) Calculate the percentage of students that passed in all Maths, Science and Health. 5. In a group of students, 30 study English, 35 study Science, 25 study Mathematics, 12 study English only, 15 study Science only, 10 study English and Science only and 6 students study Science and Mathematics only. (a) Write the cardinality of sets mentioned in the question. (b) Draw a Venn diagram to illustrate the above information. (c) Find how many students study all the subjects. (d) How many students are there altogether? Answers 1. (a) (i) 15 (ii) 12 (iii) 9 (iv) 8 (v) 3 (vi) 6 (vii) 7 (viii) 60 (ix) 67 (x) 36 (xi) 30 (xii) 25 (xiii) 15 (xiv) 10 (xv) 13 (b) (i) 42 (ii) 18 (iii) 10 (iv) 8 (c) 29 2. (a) 65 (b) 5 (c) 130 3. (a) 85 (b) (i) 61.71% (ii) 12.57% (c) 15 4. (c) 5% 5. (c) 4, (d) 62 Project Work Take a survey in your community among the students studying of the secondary level in different schools. Ask them about their interest in different sports. Total number of students in the survey ............................ . Number of students who like football ............................ . Number of students who like cricket ............................ . Number of students who like basketball ............................ . Number of students who like football and cricket ............................ . Number of students who like football and basketball ............................ . Number of students who like cricket and basketball ............................ . Number of students who like all three games ............................ . Number of students who like none of these ............................ . Show these informations in a Venn diagram and present in your classroom.
Oasis School Mathematics-10 21 Miscellaneous exercise 1. If n(A) = 80, n(B) = 70, n(A∩B) = 40, then: (a) Find n0 (A) and n0 (B) (b) Show the above informations in the Venn diagram (c) If n(U) = 120, find n(A∪B). (d) If B ⊂ A, what would be the value of n(A∩B) and n(A∪B)? Ans: (c) 10 (d) 70, 80] 2. Given, n(P) = 80, n(Q) = 60, n(P∪Q) = 25 (a) If n(P∩Q) = x, find the value of n0 (P) and n0(Q) in terms of x. (b) Substitute these informations in the Venn diagram. (c) If n(U) = 145, find the value of x, n0 (P) and n0 (Q) (d) If Q⊂P, what would be the changes in the value of n(P∪Q)? Ans: [(c) 20, 60, 40 (d) Increases by 40] 3. There are 40 students in a class, A be the set of students who like Mathematics, B be the set of students who like English. If 18 like English, 30 like Mathematics and 3 like neither, (a) Write the cardinality of A, B, A∪B and U (b) If, n(A∩B) = x, show these informations in the Venn diagram. (c) Find the value of n(A∩B). (d) What would be the value of n(A∩B) and n(A∪B) if all the students who like English also like Mathematics. Ans: [(c) 11, (d) 18 and 10] 4. In a survey conducted among 100 students regarding the medium of communication in an English Boarding School of Kathmandu, it was found that 80 use the English language and 25 use both English and Nepali languages. (a) Write the cardinality of sets. (b) If the school allows using English and Nepali language only, Show these information in Venn diagram. (c) Find the number of students using the Nepali language. Ans: [(c) 45)] 5. In a survey of a community, it was found that 85% people like the winter season and 65% the summer season. If there were no people who like neither season, (a) Present the above information in a Venn diagram. (b) What is the percentage of people who like both the seasons? (c) What percentage of the people like the winter season only ? (d) If there are 400 students, find how many people like only one season. Ans: [(b) 50%, (c) 35%, (d) 200] 6. In a group of 90 people, 70 like mango, 50 like both mango and orange. (a) Write the cardinality of different sets. (b) Each person like at least one of two fruits, show it in Venn diagram.
22 Oasis School Mathematics-10 (c) Find the number of people who like oranges. (d) Find the number of people who like only one fruit. Ans: [(c) 70, (d) 40] 7. In a group of 150 people, 35 like tea only, 52 like coffee only and 28 like both. (a) Write the cardinality of sets from the above information. (b) Show these informations in Venn diagram. (c) How many like none of these? (d) Show the result in a Venn diagram. Ans: [(c) 35] 8. In an examination, it was found that 55% students failed in Mathematics and 34% failed in English. If 35% students passed in both subjects, (a) Write the cardinality of different set from the above information. (b) Show these informations in the Venn diagram. (c) What percentage of students failed in Mathematics only? (d) If there are 500 students, find how many students passed in at most one subject. Ans: [(c) 41%, (d) 325] 9. In a school, 40% students play football, 30% play volleyball and 20% play both game. (a) If there are x students altogether, write the cardinality of U and other sets interns of x. (b) Substitute these informations in the Venn diagram. (c) If 90 students play neither games, find the total number of students and the number of students who play only football. (d) Show the result in the Venn diagram. Ans: [(c) 18] 10. Shankar took a survey among the 500 students in a school about their interest on taking part in Coding and Robotics, the result in shown in the given table. Robotics - 350 Coding - 400 Both Robotics and Coding - 300 (a) Write the cardinality of above sets. (b) Show these informations in the Venn diagram. (c) Find the number of students who like neither of them. (d) If 200 students do not take part in the survey, what percent of student want to take part in at most one activity. Ans: [(c) 50 (d) 57.14%] 11. In a survey among some people, one-fourth of the people like Android mobile set only, and one-third of the people like Apple mobile set. (a) Assuming n(U) = x, write the cardinality of the sets in terms of x. (b) If there are 50 people who do not like both mobile set, show these informations in the Venn diagram. (c) Calculate the value of x and the number of people who like at least one set. (d) If the people who like Android do not like Apple sets. Is there any changes on the result? Justify your answer. [Ans: (c) 120, 70 (d) No, there is no change in the result]
Oasis School Mathematics-10 23 12. Out of 60 students in a hostel, 30 like apple, 25 like orange and the ratio of students who apple only and orange only is 5:4, (a) Write the cardinality of sets mentioned in the question. (b) Present the data in the Venn diagram. (c) Using Venn diagram, find how many of them like both of them and how many of them like none of them. (d) If all the students who like orange also like apple, then what would be the number of students who like apple only. Ans:[(c) 5, 10 (d) 5] 13. In a group of 100 students, the ratio students who wants to take part in Quiz competition to the number of students who wants to take part in Spelling Competition in 3:5. (a) Write the cardinality of the sets if 30 want to take part in both competition and 10 do not like to take part in any competition. (b) Present these information in the Venn diagram. (c) Using Venn diagram, find the number of students who like to take part in only one competition. (d) If the third activity conducted by the school is speech competition and every student has to take part in at least one among three, find what will be the number of student take part in speech competition. Ans: [(c) 60, (d)10] 14. In a survey, one-third children like only mangoes, 2 5 children like oranges and 12 like none of these. (a) Write the cardinality of sets mentioned above of n(U) = x. (b) Show these informations in the Venn diagram. (c) Using Venn diagram, find the total number of students. (d) If 22 children do not like mangoes, how many like only orange? Also find how many like both fruits? [Ans: (c) 45 (d) 10, 8] 15. In survey among some students about their like on milk and tea it is found that, 40% students like milk and 180 students do not like milk. (a) Find the total number of students in the survey. (b) If one-third students like only milk, find how many students like both milk and tea. Ans: [(a) 300 (b) 20] 16. In a survey, 3 14 students like only Maths, 50% students do not like Maths and 40 students like both Maths and Science, (a) Show these informations in the Venn-diagram. (b) Find the total number of students. (c) If 20 student do not like both subject, find the number of students who like only science. (d) Show the final result in the Venn-diagram. Ans: [(b) 140, (c) 50]
24 Oasis School Mathematics-10 17. In an election of municipality, two candidates A and B stood for the post of mayor and 25000 people were in the voter list. Voters were supposed to cast the vote for a single candidate. 12000 people cast vote for A, 10000 people cast for B and 1000 people cast vote even for both. (a) Show these informations in Venn-diagram. (b) How many people didn't cast vote? (c) How many votes were valid? Find it. (d) By how much vote A won the election? Ans: [(b) 4000, (c)20,000, (d) 2000] 18. If n(A) = 48, n(B) = 51, n(C) = 40, n(A∩B) =11, n(B∩C) = 10, n(C∩A) = 9, n(A∩B∩C) = 4 and n(U) = 120, (a) show there informations in the Venn diagram. (b) using Venn diagram, find the value of n(A∪B∪C) and n A( ) ∪ ∪B C . Ans: [(b) 113, 7] 19. In a survey of food taste of a group of people, the following information obtained. 39 like Pizza, 25 like Burger, 22 like Mo : Mo, 20 like Pizza and Burger, 17 like Pizza and Mo : Mo, 9 like Burger and Mo : Mo, 4 like all three and 6 like none of these. (a) Write the cardinality of the sets mentioned above. (b) Show these informations in the Venn diagram. (c) Using Venn diagram, find the number of people who like at most two foods. (d) Find the percentage of people who like only one food. Ans: [(c) 46 (d) 12%] 20. In a survey of students, it was found that 50% of them liked English, 60% liked Mathematics and 40% liked Science. If 20% liked Maths and English, 15% liked English and Science, 25% liked Maths and Science and 5% liked all these three subjects, with the help of a Venn diagram, find (a) assuming n(U) = 100, write the cardinality of sets mentioned in the question. (b) show these informations in the Venn diagram. (c) what percentage of students like only two subjects? and what percentage of students don't like any subject? (d) if 240 students like mathematics, find how many like English. Ans: [(c) 45%, (d) 100] 21. Among the examinees of a school in Kathmandu, 40% students in Science, 45% in Maths and 55% in English obtained A+ grades. Similarly, 10% in Maths and Science, 20% in Science and English and 15% in English and Maths obtained A+ grade. If every student obtained an A+ grade in at least one subject, (a) show the above information in a Venn diagram. (b) find the percentage of students who obtained an A+ grade in all three subjects. (c) if 300 students were surveyed, how many students obtained an A+ in only one subject? Ans: [(b) 65%, (c) 195]
Oasis School Mathematics-10 25 Full Marks : 18 1. In a group of 50 people, it is found that 35 likes to watch movie. 40 like to listen music and 15 like neither. If 'A' and 'B' be the set of people who like to watch movie and to listen the music respectively. Then, (a) Write value of n(U), n(A), n(B) (1) (b) What is notation of number of people who like neither and what is its value? (1) (c) Plot these informations in the Venn-diagram and find the number of people who like both. (3) (d) Show the final result in the Venn diagram. (1) 2. If n(U) = 100, n(A) = 70, n(B) = 60, then (a) find the values of n(A-B) and n(B-A) in terms of x. (1) (b) find the value of n(A∪B). (1) (c) using this Venn-diagram, find the value of x. (3) (d) show the final result in the Venn-diagram. (1) 3. If n(A∩B) = 20, n(B∩C) = 24 and n(A∩C) = 18, then, (a) find the values of no (A∩B), no (B∩C) and no (A∩C) in terms of x. (1) (b) If n(A∩B) = 20, n(B∩C) = 16 and n(A∩C) = 24, find the value of no (A∩B), no (B∩C) and no (A∩C) in terms of x. (1) (c) If no (A) = 80, no (B) = 70 and no (C) = 75, n(A∪B∪C) = 15, find the value of n(A), n(B), n(C) in terms of x. (3) (d) Show these result in Venn-diagram. (1) A B x 20 U A B x C U
26 Oasis School Mathematics-10 ARItHMetIC Contents • Compound Interest • Growth and Depreciation • Money Exchange expected Learning Outcomes At the end of this unit, students will be able to develop the following competencies: • To solve the questions related to the CI compounded annually upto 3 years. • To calculate the CI compounded semi-annually upto 2 years. • To calculate the CI compounded quarterly upto 1 year. • To solve the problems of population growth and deprecation • To solve the problem related to C.I compounded quarterly. • To convert foreign currency into Nepali currency and Nepali currency into foreign currency from the given rate of exchange. Materials Required • Chart of exchange rate of currency issued by the Nepal Rastra Bank, Interest rate issued by Rastra Bank, etc. Estimated Teaching Hours 28
Oasis School Mathematics-10 27 2.1 Warm-up Activities Discuss the following in your class and draw a conclusion. A man kept Rs. 5000 in a bank for a year. What is the simple interest at the rate of 10% p.a.? In this question, • What is the value of principal (P)? • What is the value of rate (R)? • If the time is 1 year, how to calculate simple interest (I)? • What is the relation among P, T, R and I? • How to calculate the amount? • What does rate of interest mean? • If the rate of interest is 10% p.a., what does it mean? If the interest of Rs. 100 in 1 year is Rs. 12, rate of interest is 12% p.a. With the help of the above information, find the rate of interest if, • Interest per rupee per year is 9 paisa. • Interest per rupee per month is 1 paisa. 2.2 Compound Interest Let's compare two different examples Suntali took a loan of Rs. 5,000 at the rate of 8% p.a. for 2 years. How much interest does she have to pay? Chameli took a loan of Rs. 5,000 at the rate of 5% p.a. for 2 years. After one year, the interest of the first year is added to the principal. How much interest does she have to pay in 2 years? In the first example, the interest that Suntali has to pay is simple interest. In the second example, since the interest of the first year is added to the principal, Chameli has to pay the interest of the interest of the first year also. So it is a compound interest. Remember ! Interest of Rs. 100 in 1 year is the rate of interest. Unit 2 Compound Interest
28 Oasis School Mathematics-10 If the principal remains the same for the entire period, it is called the simple interest. Simple interest (I) = PTR 100 Where, P = Principal R = Rate of interest per annum T = Time (in year) If the principal does not remain the same for the entire period, due to the addition of interest after a certain period of time, then the money is said to be lent at compound interest. If the interest is compounded annually, the interest is added to the principal after the interval of one year. P = Principal T = Time (in year) R = Rate percent per annum Interest in one year = PTR 100 = P R × ×1 100 = PR 100 Amount in one year = P + I = P + PR 100 = P R 1 100 + ∴ Principal in the second year = P R 1 100 + Interest in the second year = P R 1 T R 100 100 + × × = Amount in the second year = PR 100 P R 1 100 + = P R 1 100 + × P R 1 100 + = P R 1 100 2 + P R R PR R 1 100 100 100 1 100 + = + P R 1 100 + +
Oasis School Mathematics-10 29 Similarly, Amount in the third year = P R 1 100 3 + Amount in the fourth year = P R 1 100 4 + ∴ Amount in Tth year = P R T 1 100 + ∴ Compound Amount (C.A.) = P R T 1 100 + Compound Interest (C.I.) = CA – P = P R T 1 100 + – P = P R T 1 100 + 1 – ∴ Compound Interest (C.I.) = P R T 1 100 + 1 – If the interest is compounded half yearly Rate = R% p.a. = R 2 % per half year Time = T years = 2T half years In this case, Compound Amount (C.A.) = P R T 1 200 2 + Compound Interest (C.I.) = P R T 1 200 1 2 + – If the interest is compounded quarterly, In this case, interest is added to the principal after every three months Then, Rate = R% p.a = R 4 % quarter year Time = T years = T 4 quarter year In this case, Compound amount (C.A) = P(1 + R 100 )4T = C. A – P = P(1 + R 100 )4T = P[(1 + R 100 )4T– 1]
30 Oasis School Mathematics-10 Remember ! * If the interest is compounded half yearly, the interest is compounded twice in a year. * If the interest is compounded quarterly, the interest is compounded 4 times in a year (compounded in every 3 months). * Compound interest compounded yearly means compounded annually. * Compound interest compounded half yearly means compounded semi-annually. Note If R1 , R2 and R3 be the rate percent per annum for the first, second and third year respectively, then compound amount for three years is given by C.A. = P R R R 1 100 1 100 1 100 1 2 3 + + + • In T years and M months, Compound Amount (C.A.) = P 1 + 1 + R 100 MR 1200 Worked Out examples example 1 Without using the formula, find the amount and compound interest on Rs. 1,200 for 3 years at the rate of 8% p.a. when the interest is compounded annually. Solution: Here, Principal (P) = Rs. 1200 Rate (R) = 8% p.a. Using, I = PTR 100 Interest in the first year = Rs. 1200 1 8 100 × × = Rs. 96 ∴ Principal in the second year = Rs. 1,200 + Rs. 96 = Rs. 1,296 Interest in the second year = Rs. 1296 1 8 100 × × = Rs. 103.68 Principal in the 3rd year = Rs. 1,296 + Rs. 103.68 = Rs. 1,399.68 Interest in the 3rd year = 1399.68 × 1 × 8100 100
Oasis School Mathematics-10 31 = Rs. 111.97 Amount in the 3rd year = Rs. 1,399.68 + Rs. 111.97 = Rs. 1,511.65 ∴ Compound interest = A – P = Rs. 1,511.65 – Rs. 1,200 = Rs. 311.65 example 2 Find the amount and compound interest of Rs. 3000 for 2 years at the rate of 8% p.a.. then: (a) If the interest is compounded annually. (b) If the interest is compounded semi-annually. (c) Compare the result of (i) and (ii) draw out the conclusion. Solution: (a) Here, Principal (P) = Rs. 3,000 Time (T) = 2 years Rate (R) = 8% p.a. We have, Annual compound amount (C.A.) = P R T 1 100 + . = Rs. 3,000 1 8 100 2 + = Rs. 3,000 1 2 25 2 + = Rs. 3,000 27 25 2 = Rs. 3,000 × 729 625 = Rs. 3499.20 Thus, Compound Amount (C.A.) = Rs. 3499.20 Also, Compound Interest = C.A. – P C.I. (A) = Rs. (3,499.20 – 3,000) = Rs. 499.20 (b) Here, Principal (P) = Rs. 3,000 Time (T) = 2 years Rate (R) = 8% p.a. We have, Semi annual compound amount (C.A.) = P R T 1 200 2 +
32 Oasis School Mathematics-10 = Rs. 3,000 1 8 200 4 + = Rs. 3,000 1 1 25 4 + = Rs. 3,000 26 25 4 = Rs. 3,000 × 456976 390625 = Rs. 3509.58 Thus, Compound Amount (C.A.) = Rs. 3509.58 Also Compound Interest (C.I.) = C.A. – P C.I. (S) = Rs. (3509.58 – 3,000) = Rs. 509.58 (c) Comparing the result of (i) and (ii). It is clear that compound interest. Compounded semi annually is more than the compound interest compounded annually. C. I. (S) > C. I. (A) From the above result, we conclude that, there will be more interest if the interest is compounded quickly. example 3 Calculate the C.I. of Rs. 5,000 for 6 months at the rate of 4% per annum compounded quarterly. Solution: Here, Principal (P) = Rs. 5,000 Time (T) = 6 months = 6 12 = 1 2 year Rate (R) = 4% p.a. We have, Compound amount (C.A.) = P(1 + R 400 ) 4T = 5000 (1 + 4 400 ) 4×1/2 = 5000 (1 + 1 100 ) 2 = 5000 ( 101 100 ) 2 = 5000 (1.01)2 = 5000 × 1.0201
Oasis School Mathematics-10 33 = Rs. 5,100.50 Compound interest (C.I.) = C.A. – P. = (5100.50 – 5000) – Rs.100.50 example 4 Calculate the compound interest of Rs. 700 lent for 3 years at 6% p.a. for the first year, 6 1 2 % p.a. for the second year and 7% p.a. for the third year, interest being compounded yearly. Solution: Here, Principal (P) = Rs. 700 Rate for the first year (R1 ) = 6% p.a. Rate for the second year (R2 ) = 6 1 2 % p.a. = 13 2 % p.a. Rate for the third year (R3 ) = 7% p.a. We have, Compound Amount (C.A.) = P R R R 1 100 1 200 1 100 1 2 3 + + + = Rs. 700 1 6 100 1 13 200 1 7 100 + + + = Rs. 700 106 100 213 200 107 100 = Rs. 855.55 ∴ Compound Interest (C.I.) = C.A. – P = Rs. 855.55 – Rs. 700 = Rs. 155.55 example 5 Find the difference between simple interest and the annual compound interest on Rs. 12,000 for 2 years at the rate of 15% p.a. Solution: Here, Principal (P) = Rs. 12000 Time (T) = 2 years Rate (R) = 15% p.a. We have, Simple interest (S.I.) = PTR 100 = 12000 2 15 100 × × = Rs. 3600
34 Oasis School Mathematics-10 Again, Compound interest (C.I.) = P R T 1 100 + 1 – = 12,000 1 15 100 1 2 + – = 12,000 115 100 1 2 – = 12,000 529 400 – 1 = 12,000 × 129 400 = 30 × 129 = Rs. 3,870 ∴ C.I. – S.I. = Rs. (3,870 – 3,600) = Rs. 270 example 6 A compound interest on a sum of money in 2 years at 8% per annum will be Rs. 320 more than the simple interest. Find the sum. Solution: Here, Principal (P) = ? Time (T) = 2 years Rate (R) = 8% p.a. We have, Simple Interest (S.I.) = PTR 100 = P× ×2 8 100 = 4 25 P Compound Interest (C.I.) = P R T 1 100 + 1 – = P 1 8 100 1 2 + –
Oasis School Mathematics-10 35 = P 108 100 1 2 – = P 729 625 – 1 = P 104 625 = 104P 625 From the given condition, C.I. – S.I. = Rs. 320 or, 104 625 4 25 P P – = 320 or, 104 100 625 P P – = 320 or, 4 625 P = 320 or, 4P = Rs. 320 × 625 or, P = Rs. 320 625 4 × or, P = Rs. 50,000 ∴ The sum is Rs. 50,000. example 7 In what time will Rs. 16,000 invested at 10% p.a. compounded semi-annually amounts to Rs. 18,522? Solution: Here, Principal (P) = Rs. 16,000 Rate (R) = 10% p.a. Compound amount (C.A.) = Rs. 18,522 We have, C.A. = P R T 1 200 2 + or, or, or, 18522 16000 = 21 20 2 T or, 9261 8000 = 21 20 2 T 18522 16000 1 10 200 18522 16000 210 200 2 2 = + ( ) = ( ) T T
36 Oasis School Mathematics-10 or, 21 20 3 = 21 20 2 T or, 2T = 3 or, T = 3 2 years ∴ Time period = 1 1 2 years. example 8 the compound amount of a sum of money in 2 years and 3 years are Rs. 5,832 and Rs. 6,298.56 respectively. Find the rate of interest compounded yearly and the sum. Solution: Here, Compound amount in 2 years = Rs. 5,832 Compound amount in 3 years = Rs. 6,298.56 We have, Compound amount = P P R T 1 100 + When, T = 2 yrs. 5,832 = P R 1 100 2 + ................... (i) When, T = 3yrs 6,298.56 = P R 1 100 3 + ................... (ii) Dividing equation (ii) by (i), 6298 56 5832 . = P R P R 1 100 1 100 3 2 + + or, 629856 583200 = P R 1 100 + 1.08 = P R 1 100 + or, P R 1 100 + = 0.08 or, R = 8 % From equation (i) 5,832 = P 1 8 100 2 + or, 5,832 = P 108 100 2
Oasis School Mathematics-10 37 or, 5,832 = P 27 25 2 or, P = 5832×(25)2 (27)2 or, P = Rs. 5,000 ∴ P = Rs. 5,000 Hence, Rate (R) = 8 % p.a. and the sum is Rs. 5,000. example 9 Compound interest on a certain sum of money for 2 years at 5% p.a. compounded annually is Rs. 820. Find the simple interest of the same sum at the same rate and for the same period of time. Solution: Here, Time (T) = 2 years Rate (R) = 5% p.a. Compound interest (C.I.) = Rs. 820 We have, C.I = P R T 1 100 + 1 – 820 = P 1 5 100 1 2 + – or, 820 = P 105 100 1 2 – or, 820 = P 21 20 1 2 – or, 820 = P 441 400 – 1 or, 820 = P × 41 400 or, P = 820 400 41 × or, P = Rs. 8000 Again, Principal (P) = Rs. 8000 Time (T) = 2yrs.
38 Oasis School Mathematics-10 Rate (R) = 5% P.a. Simple interest (I) = ? We have, I = PTR 100 = 8000 2 5 100 × × = Rs. 800 Hence, simple interest is Rs. 800. example 10 Rs. 16,000 invested at 10% p.a. compounded annually amounts to Rs. 19,360 in a certain period of time. What would be the amount of the same sum at the same rate and for the same period of time if the interest is being compounded semi-annually? Solution: Here, Principal (P) = Rs. 16,000 Rate (R) = 10% p.a. Compound amount (C.A.) = Rs. 19360 Time (T) = ? We have, Compound amount (C.A.) = P P R T 1 100 + or, 19360 = 16000 1 10 100 + T or, 19360 = 16000 11 10 T or, 19360 16000 = 11 10 T or, 121 100 = 11 10 T or, 11 10 2 = 11 10 T ∴ T = 2 years. Again, if the interest is being compounded semi-annually, We have, Compound Amount (C.A.) = P R T 1 200 2 + = Rs. 16,000 1 10 200 2 2 + ×
Oasis School Mathematics-10 39 = Rs. 16,000 21 20 4 = Rs. 16,000 21 20 4 = Rs. 16,000 × 194481 160000 = Rs. 19,448.10 example 11 the compound interest on a certain sum of money for 2 years is Rs. 330 and the simple interest on the same sum for 3 years at the same rate is Rs. 480. Find the rate percent and the sum. Solution: Here, S.I. = Rs. 480, T = 3 yrs. We have, S.I. = PTR 100 or, 480 = 3 100 P R or, 160 = PR 100 or, PR = 16000 .................. (i) Again, C.I. = Rs. 330, T = 2 yrs. We have, C.I = P R T 1 100 + 1 – or, 330 = P R 1 100 1 2 + – or, 330 = P R R 1 2 100 10000 + + 1 ² – or, 330 = PR 2 R 100 10000 + or, 330 = 16000PR 2 R 100 10000 + [from (i)] or, 330 = 320 + 8 5 R or, 330 – 320 = 8 5 R or, 10 = 8 5 R
40 Oasis School Mathematics-10 or, 50 = 8R or, R = 50 8 % = 25 4 % = 6 1 4 % From (i) P = 16000 R = 16000 4 25 × = Rs. 2,560 example 12 A man took a loan of Rs. 4,50,000 from the bank at the rate of 10% p.a. compounded annually. (a) How much loan is left if he paid Rs. 1,50,000 at the end of first year? (b) If the bank changed the interest rate 8% p.a. compounded semi-annually, after paying first installment, how much loan is left to pay if he paid the next installment of Rs. 1,50,000 at the end of second year. (c) If he wants to clear the loan at the end of third year, how much does he have to pay? Solution: (a) For the first year, Principal (P) = Rs. 4,50,000 Rate (R) = 10% p.a. Time (T) = 1 year We have, compound amount (C.A.) = P (1+ R 100) T = 4,50,000 (1+ 10 100) 1 = 4,50,000 ( 110 100)= 4,95,000. Since, he paid an installment of Rs. 1,50,000. So, the amount left to pay. = Rs. 4,95,000 – Rs. 1,50,000 = Rs. 3,45,000 (b) For second year, Principal (P) = Rs. 3,45,000 Rate (R) = 8% p.a. Time (T) = 1 year Since the interest is compound semi-annually, We have, compound amount (C.A.) = P (1+ R 200) 2T = 3,45,000 (1+ 8 200) 2×1
Oasis School Mathematics-10 41 = 3,45,000 ( 208 200) 2 = 3,73,152 Since, he paid second instalment of Rs. 1,50,000 Loan left to pay, = Rs. 3,73,152 – Rs. 1,50,000 = Rs. 2,23,152 (c) For third year, Principal (P) = 2,23,152 Rate (R) = 8% p.a. Time (T) = 1 year We have, compound amount (C.A.) = P (1+ R 200) 2T = 2,23,152 (1+ 8 200) 2×1 = 2,23,152 ( 208 200) 2 = 2,41,361.20 Hence, he has to pay Rs. 241361.20 at the end of third year to clear the loan. example 13 A man deposited Rs. 1,00,000 in a bank at the interest rate of 10% p.a. compounded annually. (a) How much interest does he get in 2 years? (b) He withdraw Rs. 11,000 from the bank and the bank reduced the interest rate 6% p.a. compounded semi annually, how much interest did he get in next 2 years. (c) If he deposited Rs. 20,000 at the end of 4 years, find the interest in the fifth year? (d) Find the total interest received by him in 5 years. Solution: Here, (a) For the two year, Principal (P) = Rs. 1,00,000 Rate (R) = 10% p.a. Time (T) = 2 year We have, compound amount (C.A.) = P (1+ R 100) T = 1,00,000 (1+ 10 100) 2 = 1,00,000 ( 110 100) = Rs. 1,21,000. The interest in the two years = Rs. (1,21,000 – 1,00,000) = Rs. 21,000
42 Oasis School Mathematics-10 (b) For next 2 years, Principal (P) = Rs. 1,10,000 Rate (R) = 6% p.a. Time (T) = 2 years We have, compound amount (C.A.) = P (1+ R 200) 2T = 1,10,000 (1 + 6 200) 2 × 2 = 1,10,000 × (1.03)2 = Rs. 1,23,805.97 Interest in third year = Rs. (1,23,805.97 – 1,10,000) = Rs. 13805.97 Again, (c) For fifth years, Principal (P) = Rs. 1,23,805.97 + Rs. 20,000 = Rs. 1,43,805.97 Rate (R) = 6% p.a. Time (T) = 1 year Since the interest is compounded half yearly. We have, compound amount (C.A.) = P (1+ R 200) 2T = Rs. 1,43,805.97 (1+ 6 200) 2×1 = Rs.1,43,805.97 ( 206 200) 2 = Rs. 1,52,563.75 Interest on 5th year = Rs. (1,52,563.67 – 1,43,805.97) = Rs. 8,757.78 (d) Hence, total interest received in 5 years = Rs. 13805.97 + Rs. 8,757.78 = Rs. 43,563.75 exercise 2.1 1. Answer the following questions. (a) According to the annual compound interest, write the relation among compound amount (C.A.), rate percent (R), time in year (T) and principal (P). (b) According to the annual compound interest, write the relation among compound interest (C.I.), rate percent (R), time in year (T) and principal (P). (c) According to the semi annual compound interest, write the relation among compound interest (C.I.), rate percent per annum (R), time in year (T) and principal (P). (d) According to the semi annual compound interest, write the relation among compound amount (A), rate percent per annum (R), time in year (T) and principal (P).
Oasis School Mathematics-10 43 (e) If the interest is compounded yearly, till what period of time are the simple interest and compound interest equal? (f) If the interest is compounded half yearly, till what time are the simple interest and compound interest are equal? (g) If the rate of interest for years consecutive 3 is R1 , R2 and R3 respectively, write the relation among C.A., T, P, R1 , R2 and R3 . (h) If the interest per rupee per year is 6 paisa, what is the rate of interest? (i) If the interest per rupee per month is 1 paisa, find the rate of interest. 2. Without using the formula, find the compound amount and the compound interest. (a) Rs. 2500 for 2 years at the rate of 10% p.a. compounded annually. (b) Rs. 3500 for 1 year at the rate of 8% p.a. compounded semi annually. (c) Rs. 60,000 for 9 months at the rate of 8% p.a. compounded quarterly. 3. Using the formula, find the compound amount and the compound interest compounded yearly of the following: (a) Rs. 1,200 for 2 years at the rate of 8% p.a. (b) Rs. 3,000 for 3 years at the rate of 10% p.a. 4. Find the compound amount and compound interest of the following case. (a) Rs. 1,000 for 1 year 6 months at the rate of 6% p.a. compounded half yearly. (b) Rs. 5,000 for 6 months at the rate of 8% p.a. compounded quarterly. 5. (a) Ram borrowed Rs. 4,800 from Shyam at the rate of 10% per annum. At the end of 2 years, (i) how much simple interest will Ram have to pay? (ii) how much compound interest will Ram have to pay? (iii) which interest is more by what percent? Justify your answer. (b) A borrowed Rs. 24,000 from B for 1 1 2 years at the rate of 10% p.a. Calculate, (i) simple interest that A has to pay to B. (ii) compound interest that A has to pay to B if the interest is compounded half-yearly. (iii) by what percent C.I. is more than S.I.? (iv) Justify your answer. (c) A man deposited Rs. 60,00,000 in a book for 1 year at the rate of 4% p.a. (i) how much C.I. will he get if the interest is compounded half yearly. (ii) how much C.I. will he get if the interest is compounded yearly. (iii) which interest is more and by how much percent?
44 Oasis School Mathematics-10 6. (a) Find the difference between the compound interest and simple interest of Rs. 2,500 for 3 years at the rate of 10% p.a. (b) A borrowed Rs. 18,000 from bank at the rate of 15% p.a. simple interest. If he lent the same amount of money to B at the same rate of compound interest compounded annually, how much would he gain after 2 years? 7. (a) Find the difference between compound interest payable annually and semi annually for an amount of Rs. 14,000 at the rate of 12% per annum for 2 years. (b) Ajit borrowed Rs. 10,000 from a bank at the rate of 6% per annum compounded annually. Immediately he lent it to Sujit at the same rate of compound interest payable semi-annually. How much did Ajit gain in 2 years ? 8. Which scheme do you prefer out of the given two rates with the given principal and time? (a) Rs. 12,000 for 2 years, C.I. at the rate of 10% p.a. or S.I. at the rate of 12% p.a. (b) Rs. 5,000 for 2 years, C.I. compounded annually at the rate of 12% p.a. or C.I. compounded semi-annually at the rate of 10% p.a. (c) A bank has fixed the rate of interest 10% p.a. semi-annually compound interest in account M and 12% per annum annually compound interest in account N. If you are going to deposit Rs. 40,000 for 2 years, in which account will you deposit and why? 9. (a) What sum invested for 3 years amounts to Rs. 9261 at 5% p.a. compound interest? (b) What sum of money produces an interest of Rs. 524.95 in 2 years at the rate of 12% per annum compounded semi-annually? 10. (a) Find the compound interest of Rs. 6,000 for 2 years if the rate of interest for the first year is 5% p.a. and for the second year 10% p.a. (b) A man borrowed Rs. 4,000 for 3 years at compound interest. How much should he pay at the end of 3 years if the rate for the first 2 years is 8% p.a. and that of the third year is 10% p.a.? 11. (a) In how many years will Rs. 7500 amount to Rs. 9408 at the rate of 12% p.a. compounded annually? (b) A sum of Rs. 25,000 invested at 8% p.a. compounded semi-annually amounts to Rs. 28,121. 60. Calculate the time period. (c) In how many years will the compound interest payable yearly on Rs. 10,000 at 10% per year be Rs. 3,310? (d) A man took a loan of Rs. 46,875. If the rate of compound interest is 4 paisa per rupee per year, in how many years will the compound interest become Rs. 5,853 ?
Oasis School Mathematics-10 45 12. (a) A man invested Rs. 3,000 in a finance company and received Rs. 3,993 after 3 years. Find the rate of interest, if the interest is payable annually. (b) At what rate of compound interest payable semi-annually on Rs. 20,000 is Rs. 2,050 in 1 year ? (c) Compound interest on Rs. 10,000 for 2 years compounded annually is Rs. 1,025. Find the interest per rupee per year. 13. (a) The difference between the compound interest and simple interest on a sum of money for 2 years at the rate 10% p.a. is Rs. 240. Find the sum. (b) A man borrowed a sum of money for 2 years at 5% p.a. simple interest, and he immediately lent the sum at compound interest at the same rate for the same period of time. In the transaction, if he gained Rs. 11, find the sum borrowed. 14. (a) The difference between the annual compound interest and the semi annually compound interest on a sum of money for 2 years at the rate of 20% p. a. is Rs. 482. Find the sum. (b) If the compound interest on a sum of money compounded semi annually in one year at 10% per annum is Rs. 40 more than the compound interest on the same sum compounded annually in the same time and at the same rate, find the sum. 15. (a) According to the system of compound interest, a sum of money in 2 years amounts to Rs. 7,260 and in 3 years it amounts to Rs. 7,986. Find the rate of interest and the sum. (b) A certain sum of money amounts to Rs. 3,380 in 2 years and Rs. 3,515.20 in 3 years at compound interest. Find the rate of interest and the sum. (c) The semi annual compound amount of a sum of money in 1 year is Rs. 400 and in 2 years is Rs. 441. Find the principal and the rate of interest. 16. (a) The compound interest of a certain sum for 2 years at 10% p.a. is Rs. 420. (i) Find the principal, (ii) What would be the simple interest on the same sum at the same rate of interest for the same period of time? (b) A man borrowed a certain sum of money at the rate of 12% p.a. simple interest. If he has paid Rs. 510 as S.I. at the end of a year. (i) Find the sum (ii) Find the compound interest is compounded semi annually at the same rate and for the same period of time? (iii) How much extra money should be pay as C.I. than S.I.? (c) A man borrowed a certain sum of money at the rate of 10% p.a. for 2 years and paid Rs. 2100 as compound interest compounded yearly.
46 Oasis School Mathematics-10 (i) Find the sum (ii) What would be the interest it is compounded half yearly? (iii) How much extra money should be paid as C.I. compounded half yearly than C.I. compounded yearly. 17. (a) What will be the compound interest on Rs. 5,000 for 2 years at the rate of 20% compounded per annum ? At what time, will the same interest be produced by the same sum at the same rate of simple interest ? (b) What will be the compound interest on Rs. 2,000 for 2 years at the rate of 10% per annum? At what rate of simple interest will the same sum produce the same interest in the same period of time? 18. (a) Mohan lends altogether Rs. 6,600 to Ram and Shyam for 2 years. Ram agrees to pay a simple interest at 15% p.a. and Shyam agrees to pay a compound interest at the same rate. If Ram paid Rs. 112.50 more than Shyam in interest, find how much he lent to each ? (b) Divide Rs. 4,500 into two parts such that the sum of the simple interest on the first at the rate of 10% p. a. for 2 years and the compound interest on the second at the same rate for the same period of time is Rs. 925. 19. (a) The compound interest on a certain sum of money for 2 years is Rs. 137.50 and the simple interest on the same sum for 3 years at the same rate is Rs. 198. Find the rate percent and the sum. (b) Compound interest on a sum of money for 2 years compounded annually is Rs. 1680. Simple interest on the same sum for the same period and at the same rate is Rs. 1600. Find the sum and the rate of interest. 20. A man has invested a certain sum to get compound interest. Compound interests for the first and second year are Rs. 200 and Rs. 210 respectively. (a) What is the C.I. in 2 years? Form the equations from above condition. (b) Find the rate and the sum. (c) Why the interest of second year is Rs. 10 more than interest in the first year? 21. The compound interest on a certain sum of money for the 1 year and 2 years are Rs. 500 and Rs. 1,050 respectively. (a) Obtained the equation from above two conditions. (b) Calculate the value of 'P' and 'R'. (c) Calculate the interest in the second year. Why it is more than the interest in the first year?