8 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. 406/- Print quantity : (5,000 pcs.) Contributors Laxmi Gautam Rajendra Sapkota Prakash Ghimire Layout Oasis Desktop Ramesh Bhattarai Printed in Nepal 8 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. 406/- Print quantity : (5,000 pcs.) Contributors Laxmi Gautam Rajendra Sapkota Prakash Ghimire Layout Oasis Desktop Ramesh Bhattarai Printed in Nepal 8 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. 406/- Print quantity : (5,000 pcs.) Contributors Laxmi Gautam Rajendra Sapkota Prakash Ghimire Layout Oasis Desktop Ramesh Bhattarai Printed in Nepal
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 2022 Preface
Contents Unit-1 Sets 1. Sets...................................................................................... 2 Unit-2 Arithmetic 2. Binary and Quinary Number.......................................... 17 3. Rational and Irrational Number..................................... 26 4. Ratio and Proportion....................................................... 35 5. Unitary Method................................................................ 45 6. Profit and Loss.................................................................. 51 7. Simple Interest.................................................................. 62 Unit-3 Mensuration 8. Perimeter and Area .......................................................... 74 Unit-4 Algebra 9. Indices................................................................................ 91 10. Factorization...................................................................... 95 11. H.C.F. and L.C.M.............................................................. 106 12. Rational Expressions........................................................ 113 13. Equation and Graph..........................................................123 Unit-5 Geometry 14. Lines and Angle............................................................... 139 15. Triangle, Quadrilateral and Polygon............................ 155 16. Congruency and Similarity............................................ 187 17. Solid Shapes...................................................................... 202 18. Co-ordinates...................................................................... 211 19. Symmetry and Tessellation............................................ 227 20. Transformation................................................................. 233 21. Bearing and Scale Drawing............................................. 245 Unit-6 Statistics 22. Statistics............................................................................. 254 Specification Grid.................................................................... 272
Oasis School Mathematics-8 1 Estimated Teaching Hours 10 Contents: • Disjoint and overlapping sets • Proper and Improper subsets of given set Expected Learning outcomes: At the end of this unit, Students will be able to develop the following competencies: • Identify whether given two sets are disjoint or overlapping. • To from the subsets of given set. • To identify whether the given subset is proper subset or improper subset of given set. Material Required Model of Venn diagram, A4 size paper, scissors, etc. Sets
2 Oasis School Mathematics-8 Unit 1 Sets Flash Back Let's discuss the following questions in your classroom. • If A be the set of even number less than 10, how to represent this set by different methods? • How to identify whether the given set is singleton (unit) set or null set? • What is the meaning of finite set and infinite set? • How to form the subsets from given set? • Let's consider a set of even numbers less than 10. Then A = a set of even number less than 10. (It is the representation of set by description method)2 Again, A = {2, 4, 6, 8} It is the representation of set by listing method. Again, A = { x : x is an even number less then 10} It is the representation of set by set builder method. • A set having no element is a null set. A set having only one element is a unit set. Let's observe the following examples: • A set of even number which is prime • A set of satellite of the earth • A set of female prime minister of Nepal You know! • Representation of set by different methods • Null set and Unit set • Finite set and Infinite set • Equal and Equivalent sets • Subsets and Universal set
Oasis School Mathematics-8 3 • A set of prime number from 24 to 28. There is only one element in first and second example. So first two examples are the example of unit set. Similarly, there is no element in the last two examples. So last two examples are the example of null set. • Let's take some examples A set of star in the sky A set of prime numbers from 1 to 20. In first example, there are infinite number of elements. So it is infinite set. In second example there are finite number of elements. So it is finite set. Hence, a set having infinite number of element is infinite set and a set having finite number of elements is finite set. • Let's take two sets A = { 1, 3, 5, 7, 9} and B = { a, e, i, o, u}. The number of elements in both sets A and B is equal; but the elements are different. So A and B are equivalent sets. Again, lets take two more sets. A = { 1, 3, 5, 7, 9} and B = a set of odd numbers less than 10. The number of elements in both A and B is same. The elements in both sets are same. Hence A and B are equal sets. Hence, two sets A and B are said to be equivalent sets if they have equal number of elements. Two sets A and B are said to be equal sets if they have same elements. • Let's take any two sets. A = {1, 2} and B = { 1, 2,3}. Here, every element of set A is also in set B. So, A is the subset of B. • Let's take some sets like A = {a, b}, B = {a, e, i, o, u} C = {w, x, y, z}, D = {p, q, r, s} then, we can assume a set U = a set of English alphabet. Then every set A, B, C and D are subset of U. So, U is the universal set. Review Exercise 1.1 1. Answer the following questions. (a) Let A = a set of vowel alphabets, then write the set A by listing method and set builder method. (b) Let A = {x : x is a prime number less than 10}. Express A by listing method and description method.
4 Oasis School Mathematics-8 2. State whether the following sets are unit or null set. (a) A set of highest peak of the world. (b) A set of even number which is prime (c) A set of people having height 12 ft. (d) A set of mountain whose height is more than 9000m. 3. Identify whether the given sets are finite or infinite. (a) A set of integers. (b) A set of even numbers from 1 to 10. (c) A set of stars in the sky. (d) A set of natural numbers. (e) A set of multiples of 5 from 1 to 30. 4. Identify whether the given sets are equal or equivalent. (a) A = a set of the letters of the word ''TEA''. B = a set of the letters of the word " EAT". (b) P = {2}, Q = a set of the highest peak of the world. (c) M = {1, 3, 5, 7, 9} N = a set of odd numbers less than 10. (d) X = a set vowel alphabet Y = {a, e, i, o, u}. 5. Suggest the universal set for the given set (a) A = a set of boys of class VIII B = a set of girls of class VIII (b) M = {x : x is an odd number} N = {x : x is an even number} (c) P ={ 1}, Q = {2, 3, 5, 7}, R = {4, 6, 8, 9} (d) A = {Sunday, Saturday} B = {Monday, Tuesday, Wednesday} C = {Thursday, Friday} 6. (a) If A = {2, 4, 6, 8}, B = {2, 4}. Identify which set A or B is the subset of another set? (b) A = {1, 2, 3, 4, 5, 6}, B = {2, 3}, C = {4, 5, 6}, D = {1, 2, 3, 4, 5}, (i) Which sets are the subset of set A? (ii) Is B the subset of A? (iii) Is B the subset of D? (iv) Is D, the subset of A? Answer Consult your teacher.
Oasis School Mathematics-8 5 1.2 Overlapping and Disjoint sets Let's take any two sets A = {a, b, c, d, e} and B = {a, e, i, o, u} In the above two sets. (i) Is there any common element on both set? (ii) What are its common elements? (iii) How to show this relation in diagram? (iv) What is this diagram called? (v) What type of sets are these? • Yes, there are common elements on set A and B. • Its common elements are a and e. • We can represent this relation as shown in the diagram. • This diagram is called a Venn digram. • Sets 'A' and 'B' are overlapping sets. Hence, two sets A and B are said to be overlapping sets if they contain at least one common element. Again, Let's take any two sets. A = {2, 4, 6, 8} and B = {1, 3, 5, 7, 9}. (i) Is there any common element on both set? (ii) How to show this relation in diagram? (iii) What type of sets are these? • No, there is no common element on set A and set B. • We can represent this relation as shown in the diagram • These sets are disjoint set Hence, two sets A and B are said to be disjoint sets if they contain no common element. A B •a •a •i •o •d •u •c •b A B •6 •1 •3 •5 •7 •9 •8 •4 •2
6 Oasis School Mathematics-8 1.3 Subsets Let's take any three sets like A = {1, 2,3}, B = {1, 2} and C = {1, 3}. Observe the above three sets and discuss the answer of given questions. • Is every element of set B in set A? • Is every element of set C in set A? • Is every element of set B in set C ? • Discuss their relation. • Every element of set B is also in set A. • Every element of set C is also in set A. • Every element of set B is not in set C. • Set B is the subset of set A. • Set C is the subset of set A. • Set B is not the subset of set C. Hence, a set B is said to be the subset of set A, if every element of set B is in the set A. If B is the subset of A, it is written as B ⊆ A. If C is the subset of A, it is written as C ⊆ A. • Every set is the subset of itself • Null set is the subset of every set Remember ! How to show the relation of a set with its subset in a diagram? Observe the above figure and answer the questions given below. • What are the elements of set A? • What are the elements of set B? • Are all the elements of set B, the elements of set A also? • Elements of set A are {p, q, r, s, t, u, v} • Elements of set B are {p, q, r} • Every elements of set B are also the elements of set A. • B is the subset of set A. So, while representing the relation of a set and its subset in diagram, keep subset inside the diagram of given set. A B • v • p • q •r • u • s • t
Oasis School Mathematics-8 7 Remember ! A Q B P B ⊂ A P ⊂ Q Number of subsets of given set Formation of subsets of given set Let, A = {1, 2, 3}. Let's form the subsets of set A. φ is null set. So, it is the subset of A. {1}, {2}, {3} are the subsets of A. {1, 2}, {1, 3}, {2, 3} are the subsets of A. {1, 2, 3} is also the subsets of A. Hence, the subsets of set A are φ, {1}, {2}, {3}, {1,2}, {1, 3}, {2, 3} and {1, 2, 3}. Let, A = {a}, let's find its subsets. φ and {a} are the subsets of A. If there is 1 element, number of subsets is 2. Again, let A = {a, b} Then its subsets are φ, {a}, {b} and {a, b}. If there are 2 elements, the number of subsets is 4. Again, let A = {a, b, c} Then its subsets are φ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}. If there are 3 elements, the number of subsets is 8. From the above it is clear that, if there are n elements, then the number of subsets of the given set = 2n. Proper subset and Improper subset Let A = {1, 2}, then what are the possible subset of A? The possible subsets of A are φ, {1}, {2}, and {1, 2}. Among these subsets, {1, 2} is equal to A. So, subset {1, 2} is the improper subset of A. φ, {1} and {2} are the proper subset of A. Hence, among the subsets, the set itself is the improper subset and other subsets are proper subsets. From this it is clear that number of proper subsets is equal to 2n–1. φ is the proper subset of A. It is written as φ⊂A. {1} is the proper subset of A. It is written as {1} ⊂ A.
8 Oasis School Mathematics-8 • A set itself is the subset of given set. • A set itself is the improper subset of given set. • The number of subsets of given set is 2n. • The number of proper subsets of given set is 2n–1. Remember ! Summary Set Subsets Number of elements Number of subsets φ {1} {1, 2} {1, 2, 3} φ φ, {1} φ, {1}, {2}, {1, 2} φ, {1}, {2}, {3}, {1, 2} {2, 3}, {1, 3} {1, 2, 3} 0 1 2 3 1= 20 2 = 21 4 = 2² 8 = 2³ Say Write B is the proper subset of A B is the improper subset of A B ⊂ A B ⊆ A Exercise 1.2 1. State whether the given pairs of sets are overlapping or disjoint. (a) (b) (c) A = {a, b, c, d, e} and B = {a, e, i, o, u}. (d) P = {1, 2, ,3, 4, 5} and Q = {8, 9, 10}. (e) X = {x : x is the multiple of 3 less than 20}. Y = {x : x is the multiple of 4 less than 20}. 2. Study the given figure and identify whether the given pair of sets are disjoint or overlapping. A B A B
Oasis School Mathematics-8 9 (a) (i) A and B (ii) A and C (iii) B and C. (b) (i) P and Q (ii) Q and R (iii) P and R. (c) (i) A and C (ii) B and C (iii) A and B 3. Study the given figure and answer the questions given below. (a) Identify whether the sets A and C are disjoint or overlapping sets. (b) Are B and C disjoint sets? Give reason. (c) Write the relation of the sets A and B. 4. (a) Let A = {3, 6, 9, 12, 15, 20), B = {5, 10, 15, 20}. (i) Is there any common elements in two sets? (ii) What are these common element? (iii) What are these two sets called? (iv) Show these sets in diagram? P Q R A B C A B C A B C
10 Oasis School Mathematics-8 (b) Let P ={a, e, i, o, u}, Q = {b, c, d, f, g}. (i) Is there any common elements in two sets? (ii) What are these common elements? (iii) What are these two sets called? (iv) Show these set in diagram. 5. State whether the following statements are true or false. (a) If there are n elements in a set, the number of its subsets in 2n. (b) A set is subset of itself. (c) Null set is the subset of every set. (d) If A = B, then A is the proper subset of B. (e) If A = {p, q, r} and B = {r} then B is the improper subset of A. (f) If A is the proper subset of B, then it is written as A⊂B. 6. If A = {a, b, c, d, e, f}, B = {a, e} C = {d, e, f}, D = {a, b, p, q}. Identify whether the following statements are true or false. (a) B is the proper subset of A. (b) D is the subset of A. (c) C is the subset of A. (d) B is the subset of C. 7. If A = {0, 1, 2, 3, 4, 5}. State whether the following statements are true or false. (a) φ ⊂ A (b) {1, 2} ⊂A (c) {5} ⊆ A (d) {0, 1, 2,3, 4, 5} ⊂ A (e) O ⊆ A. 8. If A = {0, 1, 2, 3}, which of the following are subsets and improper subset of A? (a) {0, 1} (b) {0, 1, 2} (c) {2, 3} (d) {3} (e) {0, 1, 2, } 9. (a) If A = {0, 1, 2}, find the number of subsets of set A? (b) If B = {a, b, c, d}, find the number of subsets of B. (c) If P = {a, b, c, d, e, f, g}, find the number of subsets, number of proper subsets and the number of improper subsets of p. 10. Write the all possible subsets of the following sets. (a) {1} (b) {p, q} (c) {x, y, z} (d) {a, b} 11. If A = {x, y,z}, write all the possible subsets of A. Identify which one of these is not a proper subset of A.
Oasis School Mathematics-8 11 12. Copy the given table in your copy and complete the table.. Sets Subsets Proper subsets Improper subsets {p} {0,1} {x, y, z} 13. Study the given figure and answer the questions given below. (a) X and Y are disjoint sets. Is this statement true? (b) "Z is the proper subset of X". Is this statement true? (c) What are the common elements both lie on the sets X and Z? Answer Consult your teacher. Project Work 1. Make 3 pairs of set which are disjoint and similarly make 3 pairs of set which are overlapping. Show these sets in diagram and present it in your classroom. 2. List out the five favourite fruits of one friend. List out the five favourite fruits of another friend. Keep these elements in the diagram. Present the following result in the classroom. Make two sets and determine whether they are disjoint or overlapping . 3. A = Students of your class who like Football. B = Students of your class who like Cricket. C = Students of your class who like Table tennis. D = Students of your class who like basketball. E = Students of your class who like Hockey. E = Students of your class who like Baseball List some pair of sets which are disjoint and some pair of sets which are overlapping. X Y • v • p Z • q • z • s • t • r • u
12 Oasis School Mathematics-8 Miscellaneous Exercise 1. If A = letters of the word "member" and B = letters of the word 'november". (a) Are A and B overlapping sets? Why? (b) Write common elements of both A and B. (c) How many elements are common in both A and B? 2. If A = {a, e, i, o, u}, B = {a, o, u}, C = {e, i, o, p}. (a) Among the sets B and C, which one is the proper subset of A? (b) Are B and C, overlapping sets or disjoint sets? Give reason. (c) What are the possible subsets of B? 3. Observe the given figure and answer the questions given below. (a) Identify whether the sets A and B are disjoint or overlapping sets? Give reason. (b) If C represents the sets which contains the common element of A and B, write the elements of set C. (c) Write the possible subsets of set C? If one more element is there, what is the changes in the number of subsets? 4. If the subsets of a set are φ, { 0 }, {0, 1}, {0, 1, 2}, {0, 2} {1, 2}, Then, (a) Write the set whose subsets are above sets. (b) Is {1, 2} a proper subset of given set? Give reason. (c) If one more element is there in the set, how many more subsets will be there? 5. If A = {1, 2, 3, 4, 5} (a) Which one of the following is not true? (i) {1, 2} ⊂ A (ii) φ ⊂ A (iii) {1 2, 3,} ⊆ A (b) If a set B = a set of vowel alphabet. Identify whether sets Aand B are disjoint in overlapping? Give reason. (c) How many proper subsets of set A can be formed? 6. (a) What types of sets are overlapping sets and what type of sets are disjont sets? (b) If there are 4 elements in a set, how many subsets of A are there? How many of them are proper subsets and how many of them are improper subsets? A B • m • p • q • n • s • a • b • c
Oasis School Mathematics-8 13 (c) If two sets are equal, what type of sets are they? Overlapping or disjoint? 7. If P = {p, q, r, s}, Q = a set of the letter of the word "people". (a) Is there any common element in sets P and Q? (b) What type of sets are P and Q overlapping or disjoint? Give reason. (c) If there is another set R = letter of the word 'apple', what is the relation of 'R' with P and Q? 8. Study the given figure and answer the questions given below. (a) Identify whether the sets A and C, B and C, A and B are disjoint or overlapping? (b) If the set A= {a, b, c, d, e}, B = {b, c, p, q, r} and C = {x, y, z}, is this information fit for the given figure? Justify with reason. (c) What is the number of subsets of set C? Explain with reason. 9. Study the given figure and answer the questions given below. (a) Identify two pair of disjoint sets and one pair of overlapping sets. (b) P = {x, y, z, w}, Q = {p, q, r, x} and R = {a, b, c}. Substitute these elements on the given figure. (c) Is the information given here, suitable in the given figure. Give reason. 10. If φ, {a}, {b}, {a, b} are the subsets of given set, then: (a) Write the set A whose subsets are mentioned above. (b) If B = {a, b, c}, write the relation of set A and B. (c) Write the number of proper subsets of set B. C A B P Q R
14 Oasis School Mathematics-8 Assessment Test Paper Attempt all the questions. Full marks – 9 1. Study the given figure and answer the questions given below. (a) What type of sets are A and B disjoint or overlapping? Give reason. (1) (b) If A = a set of prime number less than 10. (1) B = a set of even number less than 10, then write the elements in the above diagram. (1) 2. (a) What type of sets are disjoint sets and what type of sets are overlapping sets? (1) (b) φ, {1}, {2}, {1,2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}, then find the set whose subsets are given. (1) (c) If there are 16 subsets of a set, how many elements are there in the given set. (1) 3. If A = {a, b, c}. (a) Find the subsets of set A. (1) (b) Which one among them is not a proper subset? (1) (c) A set 'B' is disjoint to set A, is it possible that B is the subset of A? (1) A B
Oasis School Mathematics-8 15 Objective Questions Choose the correct alternatives. 1. If a set has 3 elements then the number of subsets of A is (i) 8 (ii) 7 (iii) 3 2. If A = a set of the letters of the word "CAN". B =9 set of the letters of the word "CAT" then A and B are (i) disjoint sets (ii) overlapping sets (iii) equal sets 3. Which of the following statement is not true? (i) Every set is a subset of given set (ii) φ is the subset of every set. (iii) The number of proper subsets of given set is 2n. 4. Every elements of set A are the elements of set B as well, the number of elements in set B is more than that in A then (i) B is the subset of A (ii) A is the subset of B (iii) A and B are equal sets 5. In the given diagram, which of the following statement is not true? (i) A and B are overlapping sets (ii) A and C are disjoint sets (iii) A is the subsets of B A B C
16 Oasis School Mathematics-8 Estimated Teaching Hours 45 Contents • Binary and quinary numbers • Rational and Irrational numbers • Ratio and Proportion • Discount, profit and loss • Unitary Method • Simple Interest Expected Learning Outcomes At the end of this unit, students will be able to develop the following competencies: • To convert decimal number into binary and quinary number and vice-versa. • To identify rational and irrational numbers. • To write the given decimal number in scientific notation and vice versa. • To solve the problems related to ratio and proportion. • To solve the problem related to discount, profit and loss. • To identify whether the given variables are direct variables or indirect variables and solve the problem using these relations. • To develop the concept of simple interest. • To solve the problem of simple interest. Teaching Materials • Chart paper, A4 size paper, flash cards, etc. Arithmetic
Oasis School Mathematics-8 17 2.1 Flash back • Ten basic symbols are used to represent the large number. These symbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. • These numbers can be expressed as the power of 10. 52 = 5 × 101 + 2 × 10° 356 = 3 × 10² + 5 × 101 + 6 × 10° 1432 = 1 × 10³ + 4 × 10² + 3 × 101 + 2 × 10° These numbers are called Decimal number. 2.2 Binary and Quinary Numbers Binary number system This is another system of numeration. This system uses only two digits 0 and 1. In this system, numbers are expressed in terms of powers of 2. So it is base-two system. For example, 10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 8 + 0 + 0 + 1 = 9 11102 = 1 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 = 16 + 8 + 4 + 2 + 0 = 30. ∴ 10012 = 9 and 11102 = 30 binary number decimal number binary number decimal number Unit 2 Binary and Quinary Numbers You know! • Factorization of a number. • Numbers as the product of multiples.
18 Oasis School Mathematics-8 Place value chart Decimal numeration Base two grouping Binary numeration 27 26 25 24 23 22 21 20 0 0 0 1 1 12 2 1 0 102 3 1 1 112 4 1 0 0 1002 5 1 0 1 1012 6 1 1 0 1102 7 1 1 1 1112 8 1 0 0 0 10002 9 1 0 0 1 10012 10 1 0 1 0 10102 In the binary system, we see from the above table that the even numbers have zero (0) in ones digit and the odd numbers have one (1) in ones digit. I. Conversion of Binary System (2 base system) into decimal number system (10 base system): To convert a binary number system into decimal system, expand the given binary number by giving the place value of each digit in power of 2. For example, 10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 1 × 8 + 0 × 4 + 0 × 2 + 1 × 1 = 8 + 0 + 0 + 1 = 9 1000012 = 1 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 1 × 32 + 0 × 16 + 0 × 8 + 0 × 4 + 0 × 2 + 1 × 1 = 32 + 0 + 0 + 0 + 0 + 1 = 33 II. Conversion of decimal numbers into binary numbers: To convert a decimal number into a binary number system, follow the given steps. For example: Convert 30 into a binary number. 2 2 2 2 2 30 7 Remainder 0 1 1 1 1 15 3 1 0 Writing all the remainders from the bottom to the top 11110. ∴ 30 = 111102 • Divide the given number successively by 2 until the quotient is zero. • List the remainder obtained in each successive division in a column. • Write all the remainders from the bottom to the top. Which is the required binary number.
Oasis School Mathematics-8 19 Quinary Number System The number system which uses the number 5 as the base is called quinary number system. Under the quinary number system, the numbers are expressed in powers of 5. The five basic symbols used in the quinary number system are 0, 1, 2, 3 and 4. For example, 135 = 1 × 51 + 3 × 50 = 1 × 5 + 3 × 1 = 8 ∴ 135 = 810 2335 = 2 × 5² + 3 × 5¹ + 3 × 5º = 2 × 25 + 3 × 5 + 3 × 1 = 50 + 15 + 3 = 68 ∴ 2335 = 6810 Let's see the given table Decimal numbers Quinary numbers 0 0 1 15 2 25 3 35 4 45 5 105 6 115 7 125 8 135 9 145 10 205 Place value chart Base 5 system 54 53 52 51 50 Base-10 system 235 2 3 13 1425 1 4 2 47 12305 1 2 3 0 190 Conversion of quinary number system (base-5 system) into decimal number system (base - 10 system) To convert a quinary number system into decimal number system, we should expand the given number by giving the place value of each digit in powers of 5. Find the product and then the sum of all numbers.
20 Oasis School Mathematics-8 For example, 1235 = 1 × 5² + 2 × 5¹ + 3 × 5° = 1 × 25 + 2 × 5 + 3 × 1 = 25 + 10 + 3 = 38 ∴ 1235 = 3810 and 11435 = 1 × 5³ + 1 × 5² + 4 × 5¹ + 3 × 5° = 1 × 125 + 1 × 25 + 4 × 5 + 3 × 1 = 125 + 25 + 20 + 3 = 173 ∴ 11435 = 17310. Conversion of decimal numbers into quinary numbers To convert decimal numbers into quinary number system • Divide the given number successively by 5 until the quotient is zero. • List the remainder obtained in each successive division in a column. • Write all the remainders from the bottom to the top. Which is the required quinary number. For example: Convert 258 into quinary number system. Here, Writing the remainder from the bottom to the top 2013. ∴ 258 = 20135 Worked Out Examples Example 1 Display the numbers 10012 , 111012 , 1000112 and 10000012 in place value chart in binary system. Solution: Place value chart in binary system 26 25 24 23 22 21 20 10012 1 0 0 1 5 5 5 5 258 10 Remainder 3 1 0 2 51 2 0
Oasis School Mathematics-8 21 111012 1 1 1 0 1 1000112 1 0 0 0 1 1 10000012 1 0 0 0 0 0 1 Example 2 Convert the following binary numbers into decimal system. (i) 10012 (ii) 10000012 Solution: (i) 10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 1 × 8 + 0 × 4 + 0 × 2 + 1 × 1 = 8 + 0 + 0 + 1 = 9 (ii) 10000012 = 1 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 1 × 64 + 0 × 32 + 0 × 16 + 0 × 8 + 0 × 4 + 0 × 2 + 1 × 1 = 64 + 0 + 0 + 0 + 0 + 0 + 1 = 65 Example 3 Convert 59 into binary (base two) number system. Solution: ∴ 59 = 1110112 Example 4 Display the numbers 23015 , 123145 and 2304105 in place value chart of quinary number system. Solution: 55 54 53 52 51 50 23015 2 3 0 1 12345 1 2 3 1 4 2304105 2 3 0 4 1 0 2 2 2 2 2 2 59 14 Remainder 1 1 0 1 1 1 29 7 3 1 0
22 Oasis School Mathematics-8 Example 5 Convert the following quinary numbers into decimal numbers (i) 1005 (ii) 23405 (iii) 11345 Solution: (i) 1005 = 1 × 5² + 0 × 5¹ + 0 × 50 = 1 × 25 + 0 × 5 + 0 × 1 = 25 + 0 + 0 = 25 (ii) 23405 = 2 × 53 + 3 × 52 + 4 × 51 + 0 × 50 = 2 × 125 + 3 × 25 + 4 × 5 + 0 × 1 = 250 + 75 + 20 + 0 = 345 iii) 11345 = 1 × 53 + 1 × 52 + 3 × 51 + 4 × 50 = 1 × 125 + 1 × 25 + 3 × 5 + 4 × 1 = 125 + 25 + 15 + 4 = 169 Example 6 Convert the following decimal numbers into quinary numbers. (i) 145 (ii) 9999 Solution: (i) Successive remainders are taken from bottom to top 1040. ∴ 145 = 10405 (ii) ∴ 9999 = 3044445 5 5 5 5 145 5 1 Remainder 0 4 20 1 29 0 Checking 10405 = 1 × 5³ + 0 × 5² + 4 × 5¹ + 0 × 50 = 125 + 0 + 20 + 0 = 145 5 5 5 5 5 5 9999 1999 399 Remainder 4 4 4 4 0 3 79 15 3 0
Oasis School Mathematics-8 23 Example 7 Convert 21345 into binary number system. Solution: Let's convert 21345 into decimal system. Now, 21345 = 2 × 53 + 1 × 52 + 3 × 51 + 4 × 50 = 250 + 25 + 15 + 4 = 294 ∴ 21345 = 29410 .....................(i) Again, let's convert 29410 into binary number. ∴ 29410 = 1001001102 .................(ii) From (i) and (ii) ∴ 21345 = 1001001102 .....................(i) Example 8 Convert 110102 into quinary number system. Solution: Let's convert 110102 into decimal number system. Now, 110102 = 1 × 24 + 1 × 2³ + 0 × 2² + 1 × 2¹ + 0× 20 = 1 × 16 + 1 × 8 + 0 × 4 + 1 × 2 + 0 × 1 = 16 + 8 + 0 + 2 + 0 = 26 ∴ 110102 = 2610 ..........................(i) Again, let's convert 2610 into quinary number system. Remainder 0 1 1 0 0 1 0 0 1 2 2 2 2 2 2 2 2 2 294 147 73 36 18 9 4 2 1 0 5 5 5 26 1 Remainder 1 0 1 5 0
24 Oasis School Mathematics-8 ∴ 2610 = 1015 ...........................(ii) From (i) and (ii) 110102 = 1015 . Exercise 2.1 1. Display the following binary numbers in place value chart. (a) 1101002 (b) 1000100112 (c) 11110012 2. Convert each of the binary number (i.e. base 2 system) into decimal number system. (a) 102 (b) 1002 (c) 11112 (d) 10012 (e) 101002 (f) 101012 (g) 111102 (h) 11110102 (i) 10000012 (j) 11111112 3. Convert the following decimal numbers into binary numbers. (a) 45 (b) 105 (c) 355 (d) 400 (e) 529 (f) 512 (g) 190 (h) 101 (i) 723 4. Display the following quinary numbers in place value chart. (a) 423015 (b) 1230015 (c) 23101235 5. Convert the following quinary numbers into decimal number (i.e. base 10 number). (a) 1005 (b) 2105 (c) 4235 (d) 200015 (e) 32345 6. Convert the following decimal numbers into quinary numbers. (a) 150 (b) 500 (c) 1024 (d) 5054 (e) 8085 7. Express the following base two numbers into base five numbers. (a) 111102 (b) 1111112 (c) 1011012 8. Convert the following base five numbers into base two numbers. (a) 10125 (b) 44445 (c) 31045 (d) 400025 9. Answer the following questions. (a) What type of numbers are binary numbers? Which 2-digits are used in binary number?
Oasis School Mathematics-8 25 (b) Convert the binary number 110010102 into decimal number. (c) Convert the decimal number thus obtained into base-5 number. Show the relation of base-2 number and base-5 number. 10. Solve the following questions. (a) What type of numbers are quinary number? What digits are used in quinary numbers? (b) Convert a quinary number 245 into decimal number. (c) Convert the decimal number thus obtained into binary number. Project Work 1. Find out some fields how the binary numbers and quinary numbers are applicable in the technology. Present the report in your classroom. Answer 1. Consult your teacher 2. (a) 2 (b) 4 (c) 15 (d) 9 (e) 20 ( f) 21 (g) 30 (h) 122 (i) 65 (j) 127 3. (a) 1011012 (b) 11010012 (c) 1011000112 (d) 1100100002 (e) 10000100012 (f) 10000000002 (g) 101111102 (h) 11001012 (i) 10110100112 4. (a) Consult your teacher 5. (a) 25 (b) 55 (c) 113 (d) 1251 (e) 444 6. (a) 11005 (b) 40005 (c) 130445 (d) 1302045 (e) 2243205 7. (a) 1105 (b) 2235 (c) 1405 8. (a) 100001002 (b) 10011100002 (c) 1100101002 (d) 1001110001102 9. (a) Consult your teacher (b) 302 (c) 22025 10. (a) Consult your teacher (b) 89 (c) 10110012
26 Oasis School Mathematics-8 3.1 Flash Back • 1, 2, 3, 4 ..............are natural numbers. • 0, 1, 2, 3, 4 ................. are whole numbers • ................. –3, –2, –1, 0, 1, 2, 3, ................ are Integers. • Sum of two whole number is whole number but the difference of two whole number may not be a whole number, so the term Integer is introduced to define it. • Sum, difference and product of two integer but the quotient of two integers may not be an integer. N = The set of natural numbers = {1, 2, 3, 4, .......................} W = The set of whole numbers = {0, 1, 2, 3, ....................} Z = The set of integers = {............, -3, -2, -1, 0, 1, 2, 3, ............} Let's take any two integers, 2 and 5 2 + 5 = 7 is an integer. 2 – 5 = –3 is an integer. 2 × 5 = 10 is an integer. 2 ÷ 5 = 2 5 is not an integer. To make the division of integers meaningful rational number is introduced. 2 5 is a rational number in which 2 and 5 both are integers. Hence, any number in the form of p q where p ∈ Z, q ∈ Z and q ≠ 0, is a rational number. Unit 3 Rational and Irrational Numbers You know! • Natural number • Whole number • Integers
Oasis School Mathematics-8 27 • Every integer is a rational number. • Rational number is either a terminating or non terminating recurring decimals. Terminating and non-terminating recurring decimals Let's convert some rational numbers into decimal, 1 2 = 0.5 (terminating decimals) 1 4 = 0.25 (terminating decimals) 1 3 = 0.3333 ............ (non-terminating recurring decimals) 3 11 = 0.272727 ........ (non terminating recurring block of digits) 3.2 Rational Number and Irrational number A rational number is either a terminating or non-terminating recurring decimals. Thus, all integers and fractions, whether positive or negative, including zero are called rational numbers. Irrational Number A set of numbers which have non-terminating and non-recurring decimals, i.e. they cannot be expressed in the form p q where p and q both are integers and q ≠ 0, such numbers are irrational numbers. 2 = 1.4142135 ........ (non-terminating, non recurring decimals) 3 = 1.7320507… (non-terminating, non recurring decimals) ∴ π = 3.14159 … (non-terminating, non recurring decimals) Hence, an irrational number can be expressed as a non-terminating, non-recurring (non-repeating) decimal. Irrational number in number line Draw a number line. Take a point P on the number line, where OP = 1 unit. Draw OPQ = 900 . Cut OP = OQ = 1 unit. Join OQ. Now, OQ2 = OP2 + PQ2 = 1 + 1 = 2. Taking 'O' as the centre and OQ as radius, draw an arc which meets numberline at R, which represents 2 as OQ = OR. -3 -2 -1 0 Q P 1 R 2 3 2
28 Oasis School Mathematics-8 Real Numbers The set of all rational and irrational numbers taken together is called the set of real numbers. The set of real numbers is denoted by R and is written as R = Q ∪ Q' = {x : x ∈ Q or x ∈ Q' } .…(i) where Q = the set of rational numbers Q' = the set of irrational numbers N W Q R z Q This shows that every rational number is a real number and every irrational number is a real number. Thus, natural numbers constitute a proper subset of integers and the integers constitute a proper subset of rational numbers and then latter constitute a proper subset of real numbers. i.e., N ⊂ W ⊂ Z ⊂ Q ⊂ R. The relation between above set of numbers is shown in the above diagram. Here, Z+ = N = the set of positive integers or natural numbers Z– = the set of negative integers W = the set of whole numbers The classification of the set of real numbers is shown in the following diagram. Real Numbers (R) Rational numbers (Q) Integers Fractions Negative integers (Z Positive fractions - ) Zero Whole numbers (W) Negative fractions Positive integers (Z+ ) or, natural number (N) Irrational numbers (Q)
Oasis School Mathematics-8 29 Worked Out Examples Example 1 Examine whether the following numbers are rational or irrational. (i) 2 (ii) 0.5 (iii) 2 3 (iv) 49 Solution: (i) Here, 2 = 1.4142134…. This is a non-terminating, non-recurring decimal. So, 2 is an irrational number. (ii) This is a terminating decimal. So, 0.5 is a rational number. (iii) 2 3 = 0.6666............... This is a non-terminating recurring decimal. So, it is a rational number. (iv) 49 = 7. As the square root of 49 gives perfect whole number, it is a rational number. Example 2 Find rational number p q, if, (i) p = 4, q = 10 (ii) p = 0.3, q = 0.6 Solution: (i) Rational number (ii) Rational number = p q = 4 10 = 2 5 = p q = 0.3 0.6 = 1 2 Exercise 3.1 1. Identify the nature of following decimal numbers. (terminating decimals, non terminating recurring decimals, non terminating recurring block of digits and not terminating and non recurring decimals) (a) 1.25 (b) 1.3333....... (c) 3.1416 ....... (d) 1.3176521 (e) 0.272727...... (f) 0.4 2. State whether the following numbers are rational or irrational? (a) 2 (b) 3 5 (c) 5 (d) 16 (e) 1 4 (f) 1.3333... (g) 22 7 (h) – 25 (i) 3.272727...
30 Oasis School Mathematics-8 3. Convert the following numbers into decimal and state whether they are terminating or non-terminating decimal. (a) 1 5 (b) 4 5 (c) 5 3 (d) 2 3 (e) 31 3 (f) 2 (g) 3 4. Write the rational number in the form p q if, (a) p = 3, q = 5 (b) p = 0.5, q = 0.6 (c) p = 1, q = 3 (d) p = 1.5, q = 2.5 5. State whether the following statements are true or false. (a) Every integer is a rational number. (b) Every rational number is an integer. (c) 1 is the smallest natural number. (d) An rational number is a non-terminating and recurring decimal. (e) A rational number is a non-terminating and non-recurring decimal. 6. From the adjoining figure, write the elements of the following sets: (a) The set of natural numbers (N) (b) The set of whole numbers (W) (c) The set of negative integers (Z– ) (d) The set of integers (Z) (e) The set of rational numbers (Q) (f) The set of irrational numbers (g) The set of real numbers (R) 7. Answer the following questions. (a) Is 0 a rational number? Why? (b) Why 1.3333... is a rational number? (c) Why 2.4527314... is an irrational number? (d) Is π a rational number? Why? (e) Is 3 an irrational number? Why? 8. Solve the following questions. (a) In the fraction 2 5 , what type of numbers are 2 and 5? (b) What type of number is 2 5 ? Why? (c) What is the nature of the decimal 0.33333.....? What type of number is this rational or irrational? Give reason. N W •1 •2 •4 •4 •10 •0 •–4 •–2 •–1 •–3 •-10 z Q • 13 • 2 • 3 • 125 • 4 3 • 4 5 • 2 3 • –2 3 Q R
Oasis School Mathematics-8 31 9. Answer the following questions. (a) In the fraction -3 5 , what type of numbers are –3 and 5? What type of number is this? (b) What is the numerical value of π? Is it a rational or an irrational number? Give reason. (c) Why the number 1.66666......is a rational number? Justify the answer. 3.3 Scientific Notation of Numbers Large numbers which are expressed as a × 10n where, (1 ≤ a < 10) and n=0, 1, 2, … are known as the numbers written in scientific notation or in standard form. Answer 1. Consult your teacher. 2. Consult your teacher. 3. (a) 0.5, terminating (b) 0.8, terminating (c) 1.6666..., non-terminating (d) 0.6666..., non-terminating, recurring decimal (e) 3.3333..., non-terminating, recurring decimal (f) 1.41 ..........., non terminating, non recurring (g) 1.732 ........., non terminating, non recurring. 4. (a) 3 5 (b) 5 6 (c) 1 3 (d) 3 5 5. (a) true (b) false (c) true (d) false (e) false 6. (a) N = {1,2,3,4,10} (b) W = {0, 1, 2, 3, 4, 10} (c) Z = {-1, -2, -3, -4, -10} (d) Z = {-10, -4, -3, -2, -1, 0, 1, 2, 3, 4, 10} (e) Q = {-10, -4, -3, -2, -1, 0, 1, 2, 3, 4, 10, 4/5, -1/3 2/3, 4/3} (f) = { 3, 2, 13, 125 } (g) R = {-10, -4, -3, -2, -1, 0, 1, 2, 3, 4, 10, 4/5, -1/3, 2/3, 4/3, 3, 2, 13, 125 } 7. Consult your teacher. 8. Consult your teacher. 9. Consult your teacher. Project Work In a chart paper, draw a diagram to represent the real number. Show the following numbers in the suitable positions. 2 , 3 , π, 1 2 , –2 3 , 5 6 , –9 7 , 3 5 , 9 , –2, –3, –6, –9, 0, 1, 2, 6, 5, 10. In a chart paper show the diagram to show the relation among Natural number, Whole number, Integer, Rational number, Irrational number, etc.
32 Oasis School Mathematics-8 Worked Out Examples Example 1 Express in scientific notation. (i) 35,00,00,000 (ii) 53,000,000,000 Solution: (i) Here, 350,000,000 = 35 × 10000000 = 3.5 × 100000000 = 3.5 × 108 (ii) Here, 53,000,000,000 = 53 × 1000,000,000 = 5.3 × 10000000000 = 5.3 × 1010 Example 2 Express the standard form into usual forms 9.3 × 107 Solution: Here, 9.3 × 10000000 = 93 × 1000000 = 93000000. Example 3 Hair of a man grows by 0.000004 cm per second. Write it in scientific notation. Solution: Here, 0.000004 cm = 4 1000000 cm = 4 106 cm = 4 × 10–6 cm Example 4 Simplify: (i) (6.8 × 105 ) + (3.7 × 108 ) (ii) (4 × 10–5) (9 × 104 ) 6 × 107 Solution: (i) (6.8 × 105 ) + (3.7 × 108 ) = (6.8 × 100000) + (3.7 × 100000000) = (68 × 10000) + (37 × 10000000) = 680000 + 370000000 = 370680000 = 3.7068 × 108 Example 5 Convert the following scientific notation in usual form. (i) 2.5 × 10–4 (ii) 3.75 × 10–5 (iii) 3.567 × 10–7 (iv) 7.003 × 10–10 Solution: (i) 2.5 × 10–4 = 2.5 104 = 25 105 = 25 100000 = 0.00025 (ii) 3.75 × 10–5 = 3.75 105 = 375 107 = 375 10000000 = 0.0000375 (ii) (4 × 10–5) (9 × 104 ) 6 × 107 4×9 6 [ 10-5×104 107 [ = 6[10-5 × 104 × 10-7] = 6 × 10-8 =
Oasis School Mathematics-8 33 (iii) 3.567 × 10–7 = 3.567 107 = 3567 1010 = 0.0000003567 (iv) 7.003 × 10–10 = 7.003 1010 = 7003 1013 = 0.0000000007003 Exercise 3.3 1. Express the following in scientific notation. (a) 3,700 (b) 39,000 (c) 3,000000 (d) 99000000 2. Express the following in scientific notation. (a) 3.5 (b) 0.0057 (c) 0.000098 (d) 0.00307 3. Express the following standard form in usual forms. (a) 2.5 × 105 (b) 3.6 × 107 (c) 0.5 × 103 (d) 2 × 107 4. Express the numbers in the statements below in scientific notation. (a) There are 2250000 children in Nepal. (b) A cold drink industry made a profit of Rs. 76200000 in a year. (c) Approximately 257800000 litre of water is contained in Ranipokhari. (d) The average distance of the sun from the earth is 149760000 km. approximately. 5. Convert the numberin scientific notation in the statements below into the general form. (a) The velocity of light is approximately 2.98 × 108 m/sec. (b) The diameter of the sun has been approximated as 1.39 × 1012 metre. (c) The volume of the moon is approximated to be 3.7 × 1013 cubic metre. 6. Express the numbers used in the following statements in scientific notation. (a) The diameter of the hole of needle is 0.0007 cm. (b) The height of a man grows by 0.000000035 cm. per second. (c) Themassofanelectronusedtobeapproximated 0.0000000000000000000000000091 gm. 7. Simplify and express in scientific notation. (a) (7 × 10–7) (2.5 × 1011) 5 × 1010 (b) (3.5×10–6) (6.3×10-7) 4.9 × 10-15 (c) 0.275×0.005 0.00000625×0.00125 (d) 85000 × 0.00002 0.0000034
34 Oasis School Mathematics-8 8. Simplify and express in general form. (a) (3.75 × 10 – 7 ) + (4.25 × 10 – 3 ) (b) (8 × 10-7)+(1.6 × 10-10) 3.2 × 10-7 (c) (5.25 × 10-6) – (2.25 × 10-7) 3 × 10-9 9. (a) Write the fraction 75 10000 in Scientific Notation. (b) What is the product of 75 10000 with 0.005? Write it in Scientific Notation. (c) Using the result of (b), simplify: 10. (a) Write the numbers 0.003 and 150000000000 in Scientific Notation. (b) Find their product. (c) Simplify: × 0.005 2.5 × 10–6 75 10000 0.003 × 150000000000 5 × 1020 Answer 1. a. 3.7 × 103 b. 3.9 × 104 c. 3 × 106 d. 9.9×107 2. a. 3.5×100 b. 5.7×10-3 c. 9.8×10-5 d. 3.07×10-3 3. a. 250000 b. 36000000 c. 500 d. 20000000 4. a 2.25×106 b. 7.6×107 c. 2.578 × 108 litres d. 1.4976 × 108 km. 5. a. 298000000 m/second b. 1.390000000000 meter c. 37000000000000 m 6. a. 7 × 10-4 cm. b. 3.5 × 10-8cm/second c. 3.75 × 10-7gm 7. a. 3.5 × 10-6 b. 4.5 × 102 c. 1.76 × 10-5 d. 5 × 105 8. a. 0.004250375 b. 2.5005 c. 1825 9. (a) 7.5×10-3 (b) 3.75 ×10-5 (c) 15 10. (a) 3 × 10-3, Rs. 1.5 × 1011 (b) 4.5×108 , (c) 9×10-11 Project Work • Collect some Scientific facts like : size of virus, diameter of sun, distance between the sun and earth, etc. Write these in scientific notation. • From the google collect the following informations: – Distance between the sun and the earth – Distance between Kathmandu to Newyork. – Size of a virus. – Average weight of an elephant. – Size of Bacteria. Show these informations in scientific notation and normal form.
Oasis School Mathematics-8 35 Flash back • Is it possible to compare 30 kg with 50m? why? • Is it possible to compare 50cm with 60cm by difference and by division? • It is not possible to compare 30kg with 50m, as they have different units. • It is possible to compare 50cm with 60cm by difference and division since they have same units. 4.1 Ratio Let us suppose that marks obtained by Lakpa in Mathematics is 80 and the marks obtained by Dorje is 60. Let's compare their marks, Lakpa got (80-60) = 20 marks more than Dorje. Let's compare their marks by division. Marks obtained by Lakpa Marks obtained by Dorje = 80 60 = 80 60 = 4 3 = 4:3 Hence the ratio is the comparison of two quantities having same unit by division. We use the symbol : to express ratio. Write Say a : b 'a' is to 'b' or 'a' to 'b' Terms of ratio: In the ratio 4 : 5 Unit 4 Ratio and Proportion You know! • Conversion of a fraction into its lowest term • Comparison of numbers by difference and by division.
36 Oasis School Mathematics-8 4 is antecedent and 5 is consequent The first term of the ratio is antecedent and the second term is consequent. I. Properties of ratio Take a ratio a : b i.e. a b . Let's multiply both terms by 3, a × 3 : b × 3 = 3a : 3b = a : b Hence the ratio remains same if both of its terms are multiplied by a non zero constant number. Again, let's divide both terms of a ratio a : b by 2 a÷2 : b ÷ 2 = b/2 a/2 = a 2 × 2 b = a : b. Hence the ratio remains same if each term of a ratio is divided by a non zero constant number. II. Compound Ratio A new ratio formed by taking the product of two or more ratios is called compound ratio. Thus, if there are any two ratios a b and c d , then the compound ratio = a b × c d = ac bd = ac : bd. For example, if 2:3 and 4:5 are any two ratios, their compound ratio = 2 3 × 4 5 = 8 15 = 8 : 15 Worked Out Examples Example 1 Find the ratio of 75 cm and 1 m. Solution: Here 1 m = 100 cm. ∴ Ratio of 75 cm and 1 m = 75cm 100cm = 3 4 • In a ratio, both the quantities should be of the same kind and should have the same unit. • A ratio is a pure number and has no unit. Remember !
Oasis School Mathematics-8 37 Example 2 Convert the ratio 1 2 : 2 3 in the form of a : b. Solution: 1 2 : 2 3 = 1 2 × 6 : 2 3 × 6 = 3 : 4 Example 3 Find the compound ratio of 3:4, 4:5 and 5:6 Solution: The compound ratio of 3:4, 4:5 and 5:6 = 3 4 × 4 5 × 5 6 = 3 6 = 1 2 = 1:2 Example 4 If a : b = 4 : 5, b : c = 3 : 4, find a : c. Solution: Given, a : b = 4 : 5 b : c = 3 : 4 Taking compound ratio a b × b c = 4 5 × 3 4 or, a c = 3 5 ∴ a : c = 3:5. Example 5 The ratio of the number of boys and girls in a school is 3 : 4. If there are 120 boys, find the number of girls. Solution: Let, the number of girls be x, Number of boys Number of girls = 3 4 120 x = 3 4 3x = 120 × 4 x = 120 × 4 3 x = 160 ∴ Number of girls = 160. Multiply both terms by the L.C.M of denominators Alternative method Let no. of boys be 3x and no. of girls be 4x. Then, we have or, 3x = 120 or, x = 120/3 = 40 ∴ Number of girls = 4x = 4 × 40 = 160
38 Oasis School Mathematics-8 Example 6 Divide Rs. 7,200 in the ratio 3 : 4 : 5 : 6. Solution: Let four parts of the sum be 3x, 4x, 5x and 6x Here, 3x + 4x + 5x + 6x = 7200 or, 18x = 7200 or, x = 400 ∴ 3x = 3 × 400 = Rs. 1200 ∴ 4x = 4 × 400 = Rs. 1600 or, 5x = 5×400 = Rs. 2000 ∴ 6x = 6×400 = Rs. 2400 Hence four parts of money are Rs. 1200, Rs. 1600, Rs. 2000 and Rs. 2400. Example 7 Two numbers are in the ratio 3:4. If 4 is added to each of them, their ratio becomes 4:5, find the numbers. Solution: Let, the required two numbers be 3x and 4x. From the given condition, 3x + 4 4x + 4 = 4 5 or, 5(3x + 4) = 4(4x + 4) or, 15x + 20 = 16x + 16 or, 16x – 15x = 20 – 16 ∴ x = 4 ∴ Required numbers are 3x = 3 × 4 = 12 and 4x = 4 × 4 = 16 Example 8 The ratio of the present ages of a father and his son is 5:2. Five years ago, the ratio of their ages was 3 : 1. Find their present ages. Solution: Let the present ages of the father and his son be 5x years and 2x years respectively. Then, 5 years ago, age of the father = (5x – 5) years. 5 years ago, age of the son = (2x – 5) years. By the question, 5x – 5 2x – 5 = 3 1
Oasis School Mathematics-8 39 or, 1(5x – 5) = 3(2x – 5) or 5x – 5 = 6x – 15 or, 6x – 5x = 15 – 5 ∴ x = 10 Hence, the present age of the father = 5x = 5 × 10 years = 50 years and the present age of his son = 2x = 2 × 10 years = 20 years Exercise 4.1 1. Convert the given ratios into the lowest terms. (a) 15 25 (b) 32 48 (c) 24 64 2. Find the ratios of the following and reduce them to their lowest terms. (a) 75 cm and 125 cm (b) Rs. 4 and 50 paisa (c) 4 months and 2 years (d) 200 gm and 2 kg. 3. Convert the given ratios in terms of a : b. (a) 1 2 : 1 3 (b) 2 5 : 1 4 (c) 2 3 : 3 5 (d) 1 2 : 2 3 4. Find the compound ratio of the following ratios. (a) 2:3 and 3:5 (b) 4:5 and 9:16 (c) 4:7 and 3:8 (d) 3:4 and 7:12 5. Find the value of x in each of the following. (a) x 4 = 2 3 (b) 4 x = 2 3 (c) 4 7 = x 21 (d) 2 3 = 8 x 6. (a) If a : b = 5 : 8 and b : c = 4 : 3, find: a : c (b) If a : b = 2: 3 and b : c = 4 : 5, find a : c. (c) If a : b = 2:3, b:c = 4:5 and c : d = 1:2, find a : d. 7. (a) A ratio is equal to 4:5. If antecedent is 36, find the consequent. (b) If a ratio is 2:3 and its consequent is 45, find its antecedent. 8. (a) Divide Rs. 300 between A and B in the ratio 2 : 3. (b) Divide Rs. 1,500 in the ratio 1:2. (c) Divide Rs. 9,000 between A, B and C in the ratio 2 : 3 : 4. (d) An alloy contains copper, zinc and tin in the ratio of 2 : 3 : 5. Find the mass of each metal in 120 gm of alloy.
40 Oasis School Mathematics-8 9. (a) If the angles of a triangle are in the ratio 2 : 3 : 5, find the size of each angle. (b) If the ratios of the angles of a quadrilateral are 1:3:5:6, find the size of each angle. 10. (a) The ratio of the number of boys and girls in a class is 5 : 3. If there are 35 boys, find the number of girls. (b) The ratio of the ages of two sisters is 3:5. If the elder sister is 25 years old, find the age of the younger sister. (c) The ratio of monthly income of Imran and Salman is 4:5. If the monthly income of Salman is Rs. 60,000, find the monthly income of Imran. 11. (a) Two numbers are in the ratio 4:5. When 5 is added to each of them, their ratio becomes 5:6. Find the numbers. (b) Two numbers are in the ratio 3:5. When 5 is subtracted from each of term, their ratio becomes 1:2, find the numbers. (c) Two numbers are in the ratio 3:4. If 10 is added to both numerator and denominator, the new ratio becomes 4:5. Find the numbers. 12. (a) The ratio of the present ages of a father and his son is 5:2. Ten years hence, the ratio of the ages will be 2:1. Find their present ages. (b) The ratio of the present ages of a mother and her daughter is 3:1. Six years ago, the ratio of their ages was 6:1. Find their present ages. (c) Ages of two brothers are in the ratio 3:4. 5 years ago, the ratio of their ages was 2:3. Find the their present ages. 13. The ratio of number of boys to the number of girls in a school is 3 : 4. (a) Identify who are greater in number, boys or girls? (b) If the number of girls is 240, what is the number of boys? (c) If 20 boys and 40 girls remained absent, what is the ratio of present boy to the present girl. 14. There are 240 boys and 300 girls in a school. (a) What is the ratio of boys to the girls in the school? (b) In the next session 30 boys left the school and 40 new boys admitted in the school. Similarly, 20 girls left the school and 40 new girls joined, find the new ratio of boy and girl. (c) If equal number of boys remain absent in a day, then the ratio of present boys to the girls is 4:3. Find how many boys and girls are absent?
Oasis School Mathematics-8 41 Answer 1. (a) 3 5 (b) 2 3 (c) 3 8 2. (a) 3:5 (b) 8:1 (c) 1:6 (d) 1:10 3. (a) 3:2 (b) 8:5 (c) 10:9 (d) 3:4 4. (a) 2:5 (b) 9:20 (c) 3:14 (d) 7:16 5. (a) 8/3 (b) 6 (c) 12 (d) 12 6. (a) 5:6 (b) 8:15 (c) 4:15 7. (a) 45 (b) 30 8. (a) Rs. 120, Rs. 180 (b) Rs. 500, Rs. 1000 (c) Rs. 2000, Rs. 3000, Rs. 4000 (d) 24gm, 36gm, 60 gm 9. (a) 360 , 540 , 900 (b) 240 , 720 , 1200 , 1440 10. (a) 21 (b) 15yrs (c) Rs. 48,000 11. (a) 20, 25 (b) 15, 25 (c) 30, 40 12. (a) 50 yrs, 20 yrs, (b) 30 yrs, 10 yrs, (c) 15 yrs, 20 yrs. 13. (a) Consult your teacher (b) 180 (c) 4 : 5 14. (a) 4 : 5 (b) 5 : 4 (c) 50 4.2 Proportion Let's take two ratios 8:12 and 6:9. Here, 8:12 = 8 12 = 2 3 and 6:9 = 6 9 = 2 3 . i.e. 8:12 = 6:9. Such equality of two ratios is called a proportion. Hence, four quantities a, b, c and d are in proportion, if a b = c d i.e. a : b = c : d. A proportion is also written as a : b : : c : d In a proportion, first and fourth terms are called extremes whereas second and third terms are called means. If a, b, c and d are in proportion, then : a b = c d, a × d = b × c. The product of extremes = the product of means. Continued proportion Let's take any two ratios 3:6 and 6:12. Here, 3 6 = 1 2 = 1 : 2 and 6 12 = 1 2 = 1 : 2 ∴ 3 : 6 = 6 : 12 Hence, 3, 6, 6 and 12 are in proportion. i.e. 3, 6 and 12 are in proportion. This type of proportion is continued proportion. Three quantities are said to be in continued proportion if the ratio of the first to the second is equal to the ratio of second to the third. i.e. three quantities a, b and c are said to be in continued proportion if a : b = b : c, where b is called as mean proportional.
42 Oasis School Mathematics-8 • a, b, c, d are said to be proportional if a:b :: c:d. • 'd' is said to be fourth proportional to a, b, c if a b = c d . • If a, b, c, d are in proportion, first and last are extremes and second and third are means. • In proportion, product of extremes = product of means. Remember ! Worked Out Examples Example 1 Test whether the numbers 2, 3, 8 and 12 are in proportion or not. Solution: Here, Ratio of first two = 2 : 3 ratio of last two = 8:12 = 8 12 = 2 3 = 2:3 Since the ratio of the first two and the last two numbers is equal given from terms are in proportion. Example 2 The first, second and third terms of proportion are 4, 22 and 6 respectively. Find the fourth proportional. Solution: Let fourth proportional be x. Then, 4, 22, 6 and x are in proportion. Thus, we have, 4 : 22 = 6 : x or, 4 22 = 6 x or, 4x = 22 × 6 or, x = 22×6 4 = 33 Hence, the required fourth proportional is 33. Example 3 Find the mean proportional between 2 and 18. Solution: Let the mean proportional between 2 and 18 be x. Then, 2, x, 18 are in continued proportion. ∴ 2 x = x 18
Oasis School Mathematics-8 43 or, x² = 36 or, x = 6 ∴ Mean proportional is 6. Example 4 What number must be added to each of the terms 4, 9, 17 and 35 to be in proportion? Solution: Let, the required number be x. then, (4 + x), (9 + x) (17 + x) and (35 + x) are in proportion. i.e. 4 + x 9 + x = 17 + x 35 + x (4 + x) (35 + x) = (17 + x) (9 + x) or, 140 + 4x + 35 x + x² = 153 + 17x + 9 x + x² or, 140 + 39x = 153 + 26x or, 39x – 26 x = 153 – 140 or, 13x = 13 or, x = 1 ∴ The required number is 1. Exercise 4.2 1. Identify whether the given numbers are in proportion or not? (a) 1, 2, 6, 8 (b) 6, 8, 10, 20 (c) 25, 35, 50, 70 (d) 4 cm, 6cm, 12, cm, 18 cm (e) 5kg, 10kg, 20kg, 40 kg 2. Find the values of x in the following proportions. (a) 2 : 5 = 6 : x (b) 7 : 5 = x : 35 (c) 4 : x = 16 : 20 (d)x : 7 = 16 : 28 3. Find the fourth proportional in the following: (a) 1, 3, 5 (b) 2, 4, 8 (c) 15, 18, 30 (d) 10, 8, 5 4. Find the mean proportional between the following: (a) 4 and 9 (b) 4 and 16 (c) 9 and 16 (d) 5 and 20
44 Oasis School Mathematics-8 5. (a) If 3, x, 12 are in continued proportion, find the value of x. (b) If 4, 12, x, 15 are in proportion, find the value of x. (c) Two extremes of a proportion are 10 and 60. If the first mean is 15, find the second mean. (d) Two means of a proportion are 12, 8. If the first extreme is 6, find the last extreme. 6. If the numbers 10, 20, 32 and 64 are in proportion, (a) write their relation. (b) If the second number is reduced by 4 and the third is increased by 8, identify whether the numbers are in proportion or not. (c) If 5 is added to the first number and 1 is added to the last, what must be added to the third number such that the numbers are in proportion. Answer 1. Consult your teacher. 2. (a) 15 (b) 49 (c) 5 (d) 4 3. (a) 15 (b) 16 (c) 36 (d) 4 4. (a) 6 (b) 8 (c) 12 (d) 10 5. (a) 6 (b) 5 (c) 40 (d) 16 6. (a) Consult your teacher (b) Yes (c) 20 Project Work 1. Take 3 litre and 5 litre milk. Find their respective price. Calculate the ratio of volume and calculate the ratio of their price. Draw out the conclusion. 2. Visit the shop nearby your home. Find the cost of 2 kg rice and 5 kg rice. Find the ratio of weight and the ratio of their price. Draw the conclusion.
Oasis School Mathematics-8 45 Flash back Study the following questions properly and discuss the answer • Cost of a copy is Rs. 45, identify whether the cost of two copies is more or less? How to find it? • Cost of 15 pens is Rs. 450, identify whether the cost of a pen is more or less? How to calculate it? • 30 men can finish a work is 30 days. Identify whether 40 men can finish it in less or more time? why? • What is the relation of cost and number of article? whether the cost increases decrease as the number of article increases? • The cost of a copy is Rs. 45, cost of two copies will be more. It can be obtained it by multiplication. • Cost of 15 pens is Rs. 450, then the cost of a pen is less. We can obtain it by division. • As the number of article increases the cost also increases. Direct Variation Look at the following examples • The cost of 1kg rice is Rs. 90, then what is the of cost 5 kg rice? More or less? It is clear that cost of 5 kg rice is more than the cost of 1 kg rice. As the quantity increases the cost also increases. This variation is called Direct Variation. Weight(kg) Cost (Rs) 1 5 90 x (suppose) If weight and cost are direction variations then Observe the direction of arrows Unit 5 Unitary Method You know! • Calculation by multiplication and calculation by division.
46 Oasis School Mathematics-8 5 1 = x 90 or, x = 5 × 90 = Rs. 450 Hence, the cost of 5 kg rice is Rs. 450. Indirect Variation Look at the following example. 20 men can do a work in 15 days. In how many days 5 men can do the same work? If 20 men can do a work in 15 days, it is clear that for 5 men it takes more men to complete the work. As the number of men decreases, the number of days to complete the work increases, such relation is indirect variations. Men Days 20 5 15 x (suppose) If the men and days are indirect variations then, 20 5 = x 15 or, x = 20 × 15 5 = 60 days Worked Out Examples Example 1 The cost of 8 articles is Rs. 640. What is the cost of 20 articles? Solution: Cost of 8 articles is Rs. 640 Cost of 1 article is Rs.640 8 Cost of 20 articles is Rs.640 8 × 20 ∴ Required cost = Rs. 640 8 × 20 = Rs. 1600 ∴ Cost of 20 articles is Rs. 1600. Observe the direction of arrows Alternative method Articles Cost 8 20 640 x Since the number of articles and cost are direct variations 20 8 = x 640 or, 8x = 20×640 or, x = 20×640 8 = Rs. 1600 Cost of 1 article is less than the cost of 8 articles, So divide Cost of 20 articles is more than the cost of 1 article, So multiply