Ganit Prabha Class VI Mathematics Textbook PDF

)#0+624#$*# &ODVV9,  GANITPRABHA (MATHEMATICS TEXT BOOK ) CLASS - VI WEST BENGAL BOARD OF SECONDARY EDUCATION First Edition : Dec, 2013 Second Edition : Dec, 2014 Third Edition : Dec, 2015 Forth Edition : Dec, 2016 Fifth Edition : Dec, 2017 Sixth Edition : Dec, 2018 Seventh E...

)#0+624#$*# &ODVV9,  GANITPRABHA (MATHEMATICS TEXT BOOK ) CLASS - VI WEST BENGAL BOARD OF SECONDARY EDUCATION First Edition : Dec, 2013 Second Edition : Dec, 2014 Third Edition : Dec, 2015 Forth Edition : Dec, 2016 Fifth Edition : Dec, 2017 Sixth Edition : Dec, 2018 Seventh Edition : Dec, 2019 Eighth Edition : December, 2020 COPYRIGHT WEST BENGAL BOARD OF SECONDARY EDUCATION Published by Smt. Paromita Roy Secretary-in-charge West Bengal Board of Secondary Education 77/2, Park Street, Kolkata - 700016 Printed at West Bengal Text Book Corporation Limited (Government of West Bengal Enterprise) Kolkata- 700 056  ¶˛yÓ˚ˆÏì˛Ó˚ ¢Ç!Óïyò ≤Ãhfl˛yÓòy xyõÓ˚yñ ¶˛yÓ˚ˆÏì˛Ó˚ ãòÜíñ ¶˛yÓ˚ì˛ˆÏÑ˛ ~Ñ˛!›˛ ¢yÓ≈ˆÏ¶˛Ôõ ¢õyãì˛y!s˛fÑ˛ ïõ≈!òÓ˚ˆÏ˛ô«˛ Üíì˛y!s˛fÑ˛ ¢yïyÓ˚íì˛s˛f Ó˚)ˆÏ˛ô ܈Ïv˛¸ ì%˛úˆÏì˛ ¢ì˛ƒ!ò¤˛yÓ˚ ¢ˆÏAÜ ü˛ôÌ @˘Ã£í Ñ˛Ó˚!äÈ ~ÓÇ ì˛yÓ˚ ¢Ñ˛ú òyÜ!Ó˚Ñ˛ ÎyˆÏì˛ı ¢yõy!ãÑ˛ñ xÌ≈˜Ïò!ì˛Ñ˛ Á Ó˚yã˜Ïò!ì˛Ñ˛ òƒyÎ˚!Óâ˛yÓ˚ó !â˛hs˛yñ õì˛≤ÃÑ˛yüñ !ÓŸªy¢ñ ïõ≈ ~ÓÇ v˛z˛ôy¢òyÓ˚ fl˛∫yï#òì˛yó ¢yõy!ãÑ˛ ≤Ã!ì˛¤˛y xã≈ò Á ¢%ˆÏÎyˆÏÜÓ˚ ¢õì˛y ≤Ã!ì˛¤˛y Ñ˛Ó˚ˆÏì˛ ˛ôyˆÏÓ˚ ~ÓÇ ì˛yˆÏîÓ˚ ¢Ñ˛ˆÏúÓ˚ õˆÏïƒ Óƒ!È=˛ÈÙÈ¢æõ Á ãyì˛#Î˚ ˙Ñ˛ƒ ~ÓÇ ¢Ç£!ì˛ ¢%!ò!ÿ˛ì˛ Ñ˛ˆÏÓ˚ ˆ¢Ôºyì,˛c ܈Ïv˛¸ ì%˛úˆÏì˛ó xyõyˆÏîÓ˚ Üí˛ô!Ó˚°ÏˆÏîñ xyãñ 1949 ¢yˆÏúÓ˚ 26 òˆÏ¶˛¡∫Ó˚ñ ~ì˛myÓ˚y ~£z ¢Ç!Óïyò @˘Ã£í Ñ˛Ó˚!äÈñ !Ó!ïÓÂï Ñ˛Ó˚!äÈ ~ÓÇ !òˆÏãˆÏîÓ˚ x˛ô≈í Ñ˛Ó˚!äÈ– THE CONSTITUTION OF INDIA PREAMBLE WE, THE PEOPLE OF INDIA, having solemnly resolved to constitute India into a SOVEREIGN SOCIALIST SECULAR DEMOCRATIC REPUBLIC and to secure to all its citizens : JUSTICE, social, economic and political; LIBERTY of thought, expression, belief, faith and worship; EQUALITY of status and of opportunity and to promote among them all – FRATERNITY assuring the dignity of the individual and the unity and integrity of the Nation; IN OUR CONSTITUENT ASSEMBLY this twenty- sixth day of November 1949, do HEREBY ADOPT, ENACT AND GIVE TO OURSELVES THIS CONSTITUTION.  Fundamental Rights and Fundamental Duties of Citizens of India Fundamental Rights (Article 14-35 of Indian Constitution) 1. Right to Equality z No religious instruction shall be provided in State- z The law of the country considers all citizens equally run educational institutions and no religious wor- and protects everybody equally. ship cannot be imparted against the consent of the z The State shall not discriminate against any citizen on student in such institutions. grounds only of religion, race, caste, sex, place of birth 5. Cultural and Educational Rights or any of them. z All citizens of our country have the right to protect, z There shall be equality of opportunity for all citizens in preserve, and propagate their language, script and matters relating to employment or appointment to any culture. office under the State. z The State shall not deny admission into educational z Abolition of untouchability is proclaimed and prohibited. institutes maintained by it or those that receive aids z Prohibition of acceptance and use of all titles except from it to any person based on race, religion, caste or military and academic. language. 2. Right to Freedom z All religious and linguistic minorities have the right z Freedom of speech and expression. to establish and administer educational institutions of their choice. z Freedom to assemble peacefully without arms. 6. Right to Constitutional Remedies z Freedom to form associations/unions/cooperative societies. z A person can move to Supreme Court and High Court if z Freedom to move freely throughout the territory of he/she wants to get his/her fundamental rights pro- India. tected. z Freedom to reside in any part of the country. Fundamental Duties in India z Right to carry on any trade or profession/occupa- (Article 51A of Indian Constitution) tion. 1. Abide by the Indian Constitution and respect its ideals z The convicted person will be punished in the existing and institutions, the National Flag and the National An- law. them. z A person cannot be convicted for the same offence 2. Cherish and follow the noble ideals that inspired the more than once. national struggle for freedom. z No person accused of an offence shall be compelled 3. Uphold and protect the sovereignty, unity and integrity by the State to bear witness against himself. of India. z Right to life and personal liberty. 4. Defend the country and render national service when called upon to do so. z No person can be arrested without any valid reason 5. Promote harmony and the spirit of common brotherhood and the arrested person should be given the scope to defend oneself. amongst all the people of India transcending religious, linguistic and regional or sectional diversities and to 3. Right against Exploitation renounce practices derogatory to the dignity of women. z Prohibition of labour without payment or buying and 6. Value and preserve the rich heritage of the country’s com- selling of any person. posite culture. z Prohibition of employment of children below the age 7. Protect and improve the natural environment including of 14 in any hazardous industry or factories or mines. forests, lakes, rivers and wildlife and to have compas- 4. Right to Freedom of Religion sion for living creatures. z Freedom of conscience, the freedom to profess, prac- 8. Develop scientific temper, humanism and the spirit of tice, and propagate religion to all citizens. inquiry and reform. z Right to form and maintain institutions for religious 9. Safeguard public property and to abjure violence. and charitable intents and acquire the immovable and 10. Strive towards excellence in all spheres of individual movable property. and collective activity so that the nation constantly rises z There can be no taxes, the proceeds of which are di- to higher levels of endeavour and achievement. rectly used for the promotion and/or maintenance of 11. Provide opportunities for education to his child or ward any particular religion/religious denomination. between the age of six and fourteen years.  PREFACE In the year 2011, the Government of West Bengal had constituted an 'Expert Committee' in the field of school Education to renew and reconsider the school curriculum, syllabus and textbooks following the guidelines of National curriculum Framework 2005 and Right to Education Act 2009. This book is a result of honest endeavour and tireless effort by the experts of the concerned subject in the committee. This book has been designed according to the syllabus of class VI and has been named 'Ganit Prabha'. This book follows the principles of studying Mathematics as a language so that students when exposed to problems on the language of Mathematics, can easily understand which Mathematical process, formula or method needed to be applied in solving the problem. Arithmatic, Algebra and Geometry sections have been explained in such a simple and vivid language so that all pupils can grasp the subject of Mathematics easily. The scope of Mathematics as an effective instrument in successfully solving a pupil's personal, family and different kinds of social problems has been emphasized. On behalf of the Board, I sincerely express my gratitude and thank all eminent teachers, educationists, subject experts and well known illustrators who have put in sincere effort to prepare this important book. With the help of 'Paschimbango Sarbosiksha Mission', the book will be distributed free of cost to all students. It will be a grave mistake on our part if we do not acknowledge the contribution of Dept. of Education, West Bengal Goverment, West Bengal Right to Education and Paschimbango Sarbosiksha Mission in making this project a success. The West Bengal Board of Secondary Education remains obliged. Hope this book 'Ganitprabha' published by West Bengal Goverment will play an important role in making science related topics attractive and interesting. This will also help in raising the standard of the study of Mathematics at Madhyamik level and provide a great impetus to the students. In this way, the Board intends to fulfil its responsibility towards society. I humbly request all educationists, teachers and others interested in education to bring to our notice, with an open mind, any mistake they find in the book so that the same can be rectified in the next edition. This will improve the book and help the student community at large. There is a saying in English, "even the best can be bettered." We will gladly accept all concrete criticism and feedback for the betterment of the quality of the book from all interested in and attached to the field of education. December – 2013 77/2Park Street President Kolkata - 700 016 West Bengal Board of Secondary Education  FOREWORD The honourable Chief Minister of West Bengal, Smt Mamata Bandyopadhyay constituted an 'Expert Committee' in the field of school education in 2011. This committee was given the responsibility to review, reconsider and revise the syllabus and textbooks. Based on the recommendations of the committee, curriculum, syllabus and textbooks have been prepared.From the very onset, we have followed the guidelines and vision of the National Curriculum Frame Work 2005 and Right to Education Act-2009 (RTE Act, 2009). At the same time, we have accepted Rabindranath Tagore's ideals of education as the basis of our approach. The Mathematics book at the upper Primary level is called 'Ganitprabha'. The students are guided how to solve the Mathematical problems step by step. For the benefits of the students in every sphere, the basic concepts have been presented in a lucid language and supplemented with Hands on Trial method. All care and effort had been taken to make the book interesting and engaging. We have kept in mind the student's ability of application while compiling the book. Hope this book will be whole heartedly accepted by the student community. This book 'Ganitprabha' will be distributed free of cost to the students of our state in the year 2014 with the help of 'Paschimbango Sarbosiksha Mission'. Selected educationists, teachers and subject experts have prepared the book within a very short time. The sole controller of Madhyamik Examination, West Bengal Board of Secondary Education has obliged us by approving the book.The West Bengal Board of Secondary Education, Education Department of West Bengal Government, 'Paschimbango Sarsbosiksha Misson', West Bengal Right to Education have provided us with great help on several instances. We sincerely thank them. Respected Minister of Education, Dr. Partha Chatterjee has obliged us with his valuable suggestions and comments. We express our gratitude to him. For the improvement of the quality of the book, the advice and suggestions of all who love education, are welcome. December – 2013 Nibedita Bhavan Chairman 5th Floor 'Experts Committee' Bidgannagar, Kolkata - 700091 School Education Dept W.B. Goverment ˛TEXT BOOK DEVELOPMENT COMMITTEE UNDER EXPERT COMMITTEE Planning and Design Aveek Mazumdar (Chairman Expert - Committee) Rathindranath Dey (Member Secertary, Expert Committee) Sankarnath Bhattacharya Sumona Som Tapasundar Bandyopadhyay Malay Krishna Mazumder Partha Das Pradyut Pal Cover and Illustrations Pronobesh Maiti Execution Biplab Mondal Assisted by : Anupam Dutta, Pinaki Dey Translated in English Mitali Guha Sarkar Translation Supervisors Prof. Amlanjyoti Das Deblina Mitra  CONTENTS Chapters Topics Pages 1. Revision of Previous Lessons...................................................................... 1 2. Concepts of Seven and Eight Digit Numbers.............................................. 43 3. Logical Approximation of Numbers........................................................... 53 4. Roman Numbers upto Hundered................................................................. 59 5. Concepts of Algebraic Variables................................................................. 62 6. Multiplication and Divison of a Fraction by a Whole Number and by a Fraction............................................................... 73 7. Multiplaction and Divison of Decimal Fractions With Whole Numbers and Decimal Fractions............................................90 8. Metric System............................................................................................97 9. Percentage...............................................................................................106 10. Recurring Decimal Numbers.................................................................... 114 11. Geometrical Concepts related to Regular Solides....................................122 12. L.C.M and H.C.F of 3 numbers.................................................................130 13. Data Handling and Analysis.......................................................................138 14. Concept of Line, Line Segment, Ray and Point.......................................148 15. Determination of Area and Perimeter......................................................155 16. Concept of Directed Numbers and Number Line......................................159 17. Geometrical Concepts based on Different Instruments of Geometry Box.....................................................................................176 18. Square Root..............................................................................................202 19. Measurement of Time..............................................................................218 20. Geometrical Concept of Circle................................................................230 21. Fundamental concept of Ratio and Proportion.........................................234 22. Drawing of Different Geometrical Figures..............................................249 23. Symmetry.................................................................................................269 24. Solids from Different Sides (Perspective)...............................................276 25. Fun with Numbers....................................................................................278 26. Open shapes of Regular Solids (Net)........................................................283 27. Equivalence of Fractions, Decimal Fractions, Percentage and Ratio................................................................................285 28. Let's Match...............................................................................................286 Revision of Previous Lessons Chapter : 1.1 1. Revision of Previous Lessons 1.1 Simplifications Today in our physical education class, all the 36 students of our class went to the playground for some warm up exercises. Our teacher asked us to stand in 3 rows, each having equal number of students. Therefore in each row, there would be S ÷ V students = students Later, students of class five joined us. They too arranged themselves in rows so that each row has the same number of students as ours. After students stood in 2 rows, there were 10 students extra. 1 Let's find, how many students of class five joined us. × 2 + 10 = + = 4 students of class five left for some other work. In mathematical language, the problem can be written as 36 ÷ 3 × 2 + 10 - 4 = × + - = + - = - = students of class five are there in the play ground. 1 Chapter : 1.1 Ganit Prava – Class VI Let us see what are the steps in a simplification Division Æ Multiplication Æ Addition Æ Subtraction D (Division) Æ M (Multiplication) Æ A (Addition) Æ S (Sustraction) In short DMAS If there were brackets, those must be removed first. Let us look at the order in which brackets are removed. '_____' ( ) { } [ ] Line bracket or First bracket Second bracket Third bracket Vir culum Hence, the order followed is BODMAS Now, when is 'of' operated? In simplification 'Of' is operated after 's So, we follow BODMAS rule for simplification. B Æ Bracket O ÆOf D Æ Division M Æ Multiplication A Æ Addition S Æ Subtraction 2 Revision of Previous Lessons Chapter : 1.1 Let us verify using buttons 36 ÷ 3 × 2 + 10 - 4 36 Æ 36 ÷ 3 Æ 36 ÷ 3 × 2 Æ 36 ÷ 3 × 2 + 10 Æ 36 ÷ 3 × 2 + 10 - 4 Æ Æ Æ 30 Let do ourself - 1.1 Let us try to find the values of 2. For (40 ÷ 5) × (3 × 8 - 3 + 6) 1. (30 - 24 - 6) - 8 First let us operate and then First step is to operate (40 ÷ 5) × 4 (3 × 8 - 3 + 6) (30 - 24 - 6) - 8 = (40 ÷ 5) × 4 (3 × + 6) =( - 18) - = × 4 (15 + 6) = - = 8 × 4 of 21 = = 8 × 84 = Therefore (30 - 24 - 6) - 8 = Therefore (40 ÷ 5) × 4 (3 × 8 - 3 + 6) = 3 Chapter : 1.1 Ganit Prava – Class VI 3. Let us find the value of (40 ÷ 5) ÷ 4(5-3) If we try to find the value of If there is no 'sign' after a (40 ÷ 5) ÷ 4 × (5-3), let's see number, it is what we get. We first operate and then (40 ÷ 5) ÷ 4 × (5-3) =8÷4×2 (40 ÷ 5) ÷ 4(5-3) =2×2 = 8 ÷ 4 of 2 =4 =8÷8 =1 4. Let me find the values of {25 - (4 + 9)} ÷ 3 and 25 - 4 + 9 ÷ 3. 5.Shall we get same values for (16 - 4)(5 - 3) and 16 - 4(5 - 3) 6. Let me find the values for 10 - 3 -5 and 20 ÷ 5 ÷ 2 In simplification if signs of subtraction or division are repeated and if there are no brackets, then subtraction or division is operated step by step from left side. Let us work out - 1.1 1.(A) Verify, if the values are same for all these cases (a) 20 + 8 ÷ (4 - 2) (b) (20 + 8) ÷ (4 - 2) (c) (20-8)(4-2) (d) 20 - 8 (4 - 2) (e) (20 + 8) ÷ 4 - 2 1.(B) Let us form similar simplification sums with numbers 12, 6, 3 and 1 and then find their values. 2. Let us find the value of simplification sums: (a) 256 ÷ 16 ÷ 2 ÷ 18 ÷ 9 × 2 (b) (72 ÷ 8 × 9) - (72 ÷ 8 of 9) (c) 76 - 4 - [ 6 + {19 - ( 48 - 57 - 17 )}] (d) {25 × 16 ÷ (60 ÷ 15) - 4 × ( 77 - 62)} ÷ ( 20 × 6 ÷ 3) (e) [ 16 ÷ {42 - 38 + 2}] 12 ÷ (24 ÷ 6) × 2 + 4 (f) 4 × [ 24 - { (110 - 11 + 3 × 4 ) ÷ 9}] ÷ 2 of 9 (g) 200 ÷ [88 - {(12 × 13) - 3 × (40 - 9)}] (h) (987 - 43 + 25) - 10 [5 + {(999 ÷ 9 × 3) + (8 × 9 ÷ 6) 4}] 3. Let us form a story for following simplification and then solve. (a) (12 - 2) ÷ 2 (b) {90 - (48-21)} ÷ 7 4. Let us express in mathematical language and solve. Rajdeep's father sold 125 guavas from their guava orchard for Rs 2 each at Baruipur market. With the money he bought 2 pens for Rs 5 each and 2 exercise books for Rs 20 each and divided the remaining money among two brothers and sisters to buy sweets. Let us find how much money Rajdeep got for buying sweets. 4  Revision of Previous Lessons Chapter : 1.2 1.2 H.C.F. and L.C.M. Let's study the relations between Factors and Multiples Let us draw squares as given below and try to find the relations between the numbers 1 to 15 × 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 1 2 3 4 5 6 11 12 2 2 4 6 8 10 12 14 3 3 6 9 12 4 4 8 12 5 5 10 15 20 6 6 12 18 24 7 8 9 10 11 12 12 13 14 15 Study the table and fill all the blank squares. From the table we find The multiples of 2 except zero are , The multiples of 3 except zero are , , ,........ , ,.......... Hence 2, 4, 6,.... each has a factor „ 2 Hence 3, 6, 9,... each has a factor „ 3 Leaving '0', the multiples of a number are (finite/infinite). 5 Chapter : 1.2 Ganit Prava – Class VI Let us take another number 12. Let us find out from table, whcih numbers have 12 as their multiple and circle all 12's with red. 12 is multiple of , , , , and Hence factors of 12 are , , , , and We have, 12 = 1× 2 × 2 × 3 Thus prime factors of 12 are and (Since 1 is neither prime nor composit) Let us find which numbers have 2 as one of thier multiples. 2 is multiples of &. 2 has factors & , number of factors of 2 being Hence 2 is a (prime/composit) number. But 1 has a factor and its number of factor is. Hence 1 is neither nor composit. Let us find which numbers have 6 as their multiples and mark it red. 6 is multiple of , , , and. is Thus, factor of 6 are , , and. Again 6 = × Number of factor of 6 are , hence 6 is (prime/composit). 2 is even (prime/composit) number. Every even number is a multiple of. Hence every even number greater than 2 is a (prime/composit) number. Hence we can say, that is the only even prime number. From the table note down 15 odd numbers. None of the odd numbers are divisible by (2 or 3) Let's do ourselves 1.2 From the table let us note down 20 composite numbers, 5 prime numbers and 15 even numbers. 6 Revision of Previous Lessons Chapter : 1.2 We shall find prime numbers between 1 and 100. Let us write numbers from 1 to 100 in the small squares of a square array as given below. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 (1) Firstly, we cross out 1 with a ' ' mark. Scince 1 is neither nor (2) Next we circle 2 as ÚÛ and cross out all other multiples of 2 like 4, 6, 8,... with a ' ' mark. (3) The number next to 2 is 3 which is not crossed. So we circle 3 as Ú Û and cross all other multiples of 3 like 6, 9, 12... with ' ' mark. (4) The number next to 3 which is not crossed is '5'. So we circle 5 Ú Û, and cross all the mul- tiples of 5 like 10, 15, 20,.... with a ' ' mark. (5) Continuing this way, till we will find all the numbers are either circled Ú Û or are crossed (' '). Thus, from the table we observe that, except '1' the numbers that are circled are all (prime/ composite) numbers and numbers which are crossed (' ') are all (prime/composite) numbers. In third century B.C., Greek Mathematician Eratosthenes formulated a method by which prime numbers between 1 to 100 could be identified without finding factors or multiples. This method is termed as Sieve of Eratosthenes. Let's do ourselves 1.3 Let us find the prime numbers between 101 to 200 and note down. 7 Chapter : 1.2 Ganit Prava – Class VI Arranging flowers in flower vase 2 Today we have a special occasion in our house. Several flower vases have to be arranged which are on a table. Father bought 12 sticks of Rajanigandha, 18 pieces of yellow roses and 30 pieces of red roses. Let's find in how many vases equal numbers of each flower type can be arranged. Firstly, either all 12 sticks of Rajnigandha or equal numbers of them can be put 2 12 in , , , , and vases. Since, the factors of 12 are , , , , and. Again, all 18 or equal numbers of yellow roses can be put in , , , , and vases. 18 Since, the factors of 18 are , , , , and. Similarly, all 30 or equal number of red roses can be put in , , , , , , and vases. Since factors of 30 are , , , , , , and. 30 Hence, all 12 Rajanigandha sticks, or equal numbers of it can be put in , , , , and vases. All 18 yellow roses or equal numbers of it can be put in , , , , and vases. And all 30 red roses, or equal numbers of it can be put in , , , , , , and vases. Thus 12 Rajanigandha sticks, 18 yellow roses and 30 red roses can be put in equal numbers in or or or vases. 8  Revision of Previous Lessons Chapter : 1.2 To arrange equal number of Rajanigandha sticks, yellow and red roses in maximum numbers of vases. Hence it is found, the maximum number of vases in which three types of flowers together can be arranged is. Thus the greatest common factor of 12, 18 and 30 is Or the Highest Common Factor is ? H.C.F. of 12, 18 and 30 is Let us try by Prime Factor Method 12 18 30 2 6 2 9 2 15 2 3 3 3 3 5 12 = 2 × 2 × 3 Hence highest common factor of 12, 18 and 30 is = × =6 18 = 2 × 3 × 3 ? H.C.F. of 12, 18 & 30 is. 30 = 2 × 3 ×5 By Division Method By Shorter Method 1 5 2 12,18, 30 12 18 6 30 3 6, 9, 15 -12 2 - 2, 3, 5 6 12 -12 0 Hence H.C.F. of 12, 18 & 30 is 2 × 3 = 6 0 H.C.F. of 12, 18, 30 is 6. It is found 6 is the H.C.F. of 12, 18 & 30 Let's try and find two other numbers whose H.C.F. is 6 [ Hints: 6 × 4 = &6×7= , since 4 and 7 are mutually , so 6 is the H.C.F. of & ] 9 Chapter : 1.2 Ganit Prava – Class VI Hands on trial Let us take paper strips with squares of same size Red strip a has 12 squares Blue strip b has 18 squares Yellow strip c has 30 squares (i) Let us put the red strip 'a' on blue strip 'b' and the extra portion of blue strip is cut off. 'a' 'b'..... 'a' ¬ 'b' (ii) The extra portion of blue strip is again put on red strip 'a' and once again extra portion of red strip is cut off. 'a'...... ¨ (iii) The extra portion of red strip when put on the blue piece of strip, those were found equal. (iv) Let's see from the yellow strip , how many maximum number of pieces can be cut off from the remaining portion of the blue strip. c ª ª ª ª We now observe that on the yellow strip, times of the remaining blue strip can be placed. No portion is left extra. The number of squares in the blue strip is 6. Thus, it is derived from hands on trial the H.C.F. of 12, 18 & 30 is 6. ? Highest common factor of 12, 18 & 30 is 6. 10  Revision of Previous Lessons Chapter : 1.2 3 An educational trip to Gadiyara has been arranged by my school for collecting environment related data. 32 students of class six, 36 students of class seven and 28 students of class eight are going for the trip. Some small school buses have been arranged so that each bus must accomodate equal number of students from the three classes. Let us find the total number of buses required and the number of students each bus can accomodate. 32 students of class six, 36 students of class seven and 28 students of class eight are going for the trip. 32, 36, 28 The highest common factor or H.C.F. of 32, 36 and 28 is = × =. 16,18, 14 The maximum number of buses required = × =. 8, 9, 7 In each bus, number of students of class six ÷ = Number of students of class seven = ÷ = & Number of students of class eight = ÷ = ?Total number of students in each bus = + + = 4 Let us find H.C.F. of 65, 25 and 55. 5 Let us find H.C.F. of 48, 80 and 70. 5 65,25, 55 48, 80, 72 13, 5, 11 Hence, H.C.F. of 65, 25 and 55 is 6 Let us find H.C.F. of 15 and 16. Hence, H.C.F. of 48, 80 and 70 is 15 = 1 × 3 × 5 3 15 2 16 16 = 1 × 2×2×2 Alternative method 5 2 8 Hence, H.C.F. of 15 & 16 is = 1 1 15, 16 2 So we say 15 and 16 are prime numbers. 15,16 ? Two numbers are called mutually prime if their H. C. F. is 1. Hence H.C.F. of 15 & 16 = 1 7 Let us find H.C.F. of 25, 35 and 60 by the method of Division: 12 25 35 5 60 -25 -5 Hence, H.C.F. of 25, 35 and 60 10 25 10 is -20 -10 5 10 0 -10 0 11 Chapter : 1.2 Ganit Prava – Class VI 27 28 29 30 31 32 33 34 35 36 37 38 39 40 26 41 25 71 72 73 74 75 76 77 78 79 80 42 24 70 81 43 23 69 99 100 --- --- --- --- 82 44 22 68 98 83 45 21 67 97 84 46 20 66 96 85 47 19 65 95 86 48 18 64 94 93 92 91 90 89 88 87 49 17 63 50 16 62 61 60 59 58 57 56 55 54 53 52 51 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Chandra, Mohit and Tirtha started putting fruits in the boxes in the above chart. Starting from 0, Chandra puts mangoes in the boxes at a difference of every 12 boxes, Mohit puts apple in the boxes which are at a difference of every 18 boxes, and Tirtha puts bananas at a difference of every 24 boxes. That is, Chandra puts mangoes in 12 , 24 , 36 ,.... boxes. These numbers are multiple of. Mohit puts apples in , , ,... boxes. These numbers are multiple of. & Tirtha puts bananas in 24 , 48 , 72 ,... boxes. These numbers are multiple of. It is found that Chandra, Mohit and Tirtha have kept mangoes, apples and bananas in , , ,...... boxes. These numbers are common multiples of 12, 18 and 24. The 'least numbered' cell, which has all the three fruits is. Hences, the least common multiple of 12, 18 & 24 is 72. ?L.C.M. of 12, 18 and 24 is 72. 12 Revision of Previous Lessons Chapter : 1.2 8 Let us find L.C.M. of 12, 18 and 24 by finding their prime factors. 12 18 24 Hence 12= 2 × 2 × 3 18= 2 × 3 × 3 24= 2 × 2 × 2 × 3 ?L.C.M. of 12, 18 and 24 = 2×3×2×2×3 = 72 By Shorter Method 12,18, 24 6, 9, 12 3, 9, 6 1, 3, 2 1, 1, 2 ?L.C.M. of 12, 18 & 24 = 2×2×3×3×2 = 72 1, 1, 1 9 Let us find the least number, which is divisible by 8 and 9. Let's try to find L.C.M. of 8 and 9. 8 8, 9 L.C.M. of 8 and 9 = × = 9 1, 9 1, 1 8 9 8=2×2×2 9=3×3... L.C.M. of 8 and 9 = × × × × = 13 Chapter : 1.2 Ganit Prava – Class VI Hands on Trial 'a' A strip of paper 'a' is taken, which has 5 equal square divi- sions and another strip of paper b with 6 equal square divi- 'b' sions as before. 1 strip of 'b'€ has 6 squares 1 strip of 'a'€ has 5 squares 2 strips of 'b'€ has 12 squares 2 strips of 'a'€ has 10 squares 3 strips of 'a'€ has 15 squares 3 strips of 'b'€ has 18 squares 4 strips of 'a'€ has 20 squares 5 strips of 'a'€ has 25 squares 4 strips of 'b'€ has 24 squares has 30 6 strips of 'a'€ has 30 squares 5 strips of 'b'€ squares It is observed that if 6 strips of 'a' are put on 5 strips of 'b', they will coincide. Then we can say L.C.M. of 5 and 6 is. 14  Revision of Previous Lessons Chapter : 1.2 Hands on Trial Let's draw three equal squares with equal number of square divisions on them as given below. In the first square 1-100 are written inside the small squares. In the second square, multiples of '8' are circled and then circles are cut out. From the third square, multiples of 12 are circled and then cut out. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Square paper for multiple of 8. (1) The second square paper, is placed in the first square paper (2) Over the two square papers, the third square paper is placed. It is observed that the squares on which 24, 48, 72 and were written, could only be seen. 24, 48, 72 and 96 are of 24. The least multiple of them. Square paper for multiples of 12 Hence, the least common multiple of 8 and 12 or L.C.M. =. H.C.F. of 8 and 12 is and L.C.M. of 8 and 12 is. Let's take product of H.C.F. and L.C.M. of 8 and 12. It is observed, H.C.F. × L.C.M. = × = 4 × 2× = × = product of two numbers... Product of two numbers = Product of their H.C.F. & L.C.M. 15 Chapter : 1.2 Ganit Prava – Class VI 10 At 5 O' clock in the morning, 3 cocks crowed together. The first cock crows after every 36 secs, the second cock after every 30 secs and the third cock crows after every 48 secs. When will the cocks crow together again? Let us find the prime factors of 36, 30, 48 and then find their L.C.M. 36 30 48 18 15 24 36 = 2× 2 × 3 ×3 9 12 30 = 2 ×3 ×5 6 48 = 2× 2 × 2 × 2 × 3 3 The L.C.M. of 36, 30 and 48 by finding their prime factors = 2×3×2×3×5×2×2 = 36, 30, 48 The L.C.M. of 36, 30 and 48 18, 15, 24 = × × × × × = 9, 15, 12 The three cocks will crow together again 3, 5, 4 after 720 secs = (720 ÷ 60) min. = min. By Simple Method That is at 5:12 in the morning, the cocks 1, 5, 4 will crow together. 1, 1, 4 1, 1, 1 Let's work out 1.2 1. Let us find mentally (a) 6 multiples of 5 which do not contain '0'. (b) Let us find 3 multiples of 7 which are greater than 50. (c) Let us think and find two 2 digit numbers which are multiples of 4. (d) Let's find three numbers whose factor is 4. (e) Let's find two numbers whose L.C.M. is 12 and their sum is 10. 2. (a) What are the prime factors of 14. (b) What is the least prime number? (c) Which number is neither prime nor composite? 3. (A) 42 is multiple of which of the following numbers — (a) 7 (b) 13 (c) 5 (d) 6 (B) Which of the following numbers has 11 as a factor — (a) 101 (b) 111 (c) 121 (d) 112 4. Which of the following pairs of numbers are mutually prime — (a) 5, 7 (b) 10, 21 (c) 10, 15 (d) 16, 15 5. Let us find two composite numbers which are mutually prime. 16  Revision of Previous Lessons Chapter : 1.2 6. (a) Let's write H.C.F. of mutually prime numbers. (b) Let's write L.C.M. of mutually prime numbers. 7. Finding prime factors and 1, let's find H.C.F. of the following — (a) 22, 44; (b) 54, 72 (c) 27, 64 (d) 36, 30 (e) 28, 35, 49. (f) 30, 72, 96 (g) 20, , [ taken numbers apart from '0'.] 8. Let's find H.C.F. of the following by division method — (a) 28, 35 (b) 54, 72 (c) 27, 63 (d) 25, 35, 45 (e) 48, 72, 96 9. Let's find L.C.M. of the following by finding their prime factors — (a) 25, 80 (b) 36, 39 (c) 32, 56 (d) 36, 48 & 72 (e) 25, 35 & 45 (f) 32, 40 & 84 10. Let's find , which pair of numbers are mutually prime — (a) 47, 23 (b) 25, 9 (c) 49, 35 (d) 36, 54 11. Let's find H.C.F. and L.C.M. of the following numbers.by short division method — (a) 33 and 132 (b) 90 and 144 (c) 32, 40 and 72 (d) 28, 49, 70 12. Let's find the least number which is divisible by 18, 24 & 42. 13. Let's find the greatest number which when divides 45 and 60 do not keep any remainders. 14. Let's find the product of two numbers whose L.C.M. and H.C.F are 252 and 6 respectively. 15. If the H.C.F. and L.C.M. of two numbers are 8 and 280 and one of the numbers is 56, let's find the other number. 16. Let's write two numbers whose H.C.F. is 1. 17. Let's find the maximum number of people among whom 48 rasogollas and 64 sandeshes are to be distributed in equal numbers. 18. Bibhash and his friends decided to form a drama group of either 8 or 10 people. Let's find the minimum number of people who will be able to form both types of group. 19. Panchayat has sent flower saplings to the stundent's of class VI of Jodunath Vidya Mandir, for their school garden. It was found that if the saplings can be put in rows of 20, 24 and 30, then the number of saplings in each row will be equal. Let's find the minimum number of saplings that Panchayat sent to school. 20. The perimeter of a front wheel and rear wheel of an engine are 14 dcm. and 35 dcm respec- tively. Let's find the minimum distance the engine can cover before both wheels take com- plete revolutions together. 21. For each of the following cases, let's write two numbers whose (a) H.C.F. is 7 (b) L.C.M. is 12 (c) H.C.F. is. (Put 1digit number) (d) L.C.M. is (Put 1 digit number.) 17 Chapter : 1.3 Ganit Prava – Class VI 1.3 Fraction I took a white rectangular piece of paper. I folded it horizontally into 3 equal parts and coloured each part. Portion, coloured safron = part Portion, coloured white = part Portion, coloured green = part Seema took a paper, same as mine, but coloured it different. Let's find how and what portion she coloured. Portion of safron colour = part 1 part is coloured Let's colour part 4 Portion of green colour = part Let's colour 3 part Let's 3 colour part Let's 2 colour part 5 4 5 Let's colour part Let's colour part Let's colour part Let's colour part 6 2 18  Revision of Previous Lessons Chapter : 1.3 11 Mangoes fell from the trees due to yesterday's storm. I collected them in a basket, but found few mangoes were rotten. I disposed off the rotten mangoes. Let me find what part of mangoes I disposed off 1 part of the number of mangoes collected are rotten. 3 1 In the picture part of mangoes are circled with red o 3 1 It is found, rotten mangoes are part = mangoes. 3 From picture it is found part of mangoes are fresh. ? 2 part of total number3of mangoes = mangoes. 3 Hence, from picture it is found, mangoes are rotten and, mangoes are fresh. Let's find out from my picture in which parts the different fruits are present In the picture, out of the 4 equal parts of the fruits, the apples are part. That is, in the 1 picture number of apples are part of the total number of fruits. 4 In the picture, mango is part of equal parts. That is, in the picture mango is part of the total number of fruits. 4 In the picture aside, red flowers are part of equal parts of flowers. Thus part of flowers are red flowers. White flowers are part of equal parts of flowers. Thus part of flowers are white flowers. 2 Æ˛ parts = 2 parts out of 5 equal parts = balls. 5 1 Æ˛ parts = 1 part out of 4 equal parts = coins. 4 (Let's do ourselves) 19 Chapter : 1.3 Ganit Prava – Class VI 12 I painted one of the window panes and a part of the other pane. Let me find, how much has been painted. 1 1 1× 4 + 1 5 Coloured part Æ ( 1 + 4 ) = 1 4 = 4 = 4 × + Coloured Æ (2 + )= = = × + Coloured Æ (2 + )= = = Now let me try to colour after knowing the portions to colour. 2 12 5 12 2 Æ˛ Æ 25 Æ˛ 5 10 2 20 Æ˛ Æ˛ Æ (Let's do myself) 6 11 Æ˛ (Let's do myself) Æ˛ Æ 4 20 Revision of Previous Lessons Chapter : 1.3 Let's colour the parts of an equal sized paper, 1 2 4 part part part part 6 12 1 1× 1× 1×4 1× = 3×2 = 3× = 3× = = (putting as we like ) 3 3× 4 4÷ 1 3 3÷ 1 2 2÷ = 12 12 ÷ = = =3 = = 3 9 9÷ 6 6÷ or 4 1 = 1 or 3 1 = 1 or 2 1 = 1 12 3 3 93 3 63 3 1 2 4 part is coloured 4 part = 8 part = 16 part Divided in equal Divided in equal colour the part parts and colour parts and colour 1 2 5 8 (are equivalent fractions) Hence = = = = = = 2 8 12 1 part is coloured " Let's colour 2 part 6 1 2 5 8 Hence = = = = = = (putting as we like ) 3 9 12 21 Chapter : 1.3 Ganit Prava – Class VI Let's find interesting relations between Equivalent Fractions Equivalent Numerator of first fraction Denominator of first fraction Are the products Fractions × Denominator of second fraction × Numerator of second fraction equal 2=6 2 × 15 = 5×6= yes 5 15 2=8 2 × 28 = 7×8= 7 28 1= 1 × 36 = 9× = 36 9 2 2× = 3× = = 3 15 Let's do ourselves 1.4 Let's find four equivalent fractions for each of the fractions 2 and 5. 3 9 Hands on trial In a fraction, if the numerator is increased keeping denominator same, or the denominator is increased keeping numerator same, let us find the change in the value of the fraction. Let us take 3 equal rectangular pieces of paper with 6 equal divisions in each. Let's also colour the parts as given below: 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 123456789012345678901234567890 123456789012345678901234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 12345678901234567890 12345678901234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 123456789012345678901234567890 1234567890 1234567890 1234567890 12345678901234567890 12345678901234567890 1234567890 1 2 part 3 part Part 6 6 6 It is found, for positive fractions, if numerator 1 2 3 [Let's put >/< ] Æ is increased keeping denominator same, the 6 6 6 value of fraction. Let us take another 3 equal rectangular pieces of paper with equal divisions in each and let us colour the parts as given below: 12345678901234 12345678901234 1234567890 1234567890 1234567 1234567 12345678901234 12345678901234 1234567890 1234567890 1234567 1234567 12345678901234 1234567890 1234567 12345678901234 12345678901234 1234567890 1234567890 1234567 1234567 12345678901234 1234567890 1234567 12345678901234 12345678901234 1234567890 1234567890 1234567 1234567 12345678901234 1234567890 1234567 12345678901234 1234567890 1234567 1 1 1 2 3 4 It's found, for positive fractions,if 1 1 1 Æ denominator is increased keeping [Let's put >/< ] numerator same, the value of fraction 2 3 4 22  Revision of Previous Lessons Chapter : 1.3 Let's plant trees 13 Ramita and Rahman together were planting flower trees in their garden. Ramita planted trees in 12 part and 30 Rahman planted trees in 15 part of this 25 garden. Let's find who worked more 2 3 15 Romita did 12 part = part and Rahman did part = part 30 25 5 5 2 3 [ Let's put >//< ) 0.7 0.07 (Let me put >/< ) Let me put myself any value 70 parts out of 100 Coloured Fraction Decimal = part = part square number 10 small squares 10 100 1 1 small squares 100 0.01 7 small squares 70 small squares do myself 28 Revision of Previous Lessons Chapter : 1.4 I have a 30cm. long scale. I shall measure the four sides of our window with this scale. I found that the breadth of the window is 154 cm. and its height is 186cm. Let me try to change cm. to metre. 54 Breadth of window = 154 cm = (154÷100) m = 154 m = 1m + m 100 100 = 1m + 0.54 m = 1.54m Height of window = 186 cm = ( ÷100) m = m= m 1 1 Decimal number tens units tenths ( ) hundreths ( ) 10 100 1.54 1 5 4 1.86 1 8 6 15.1 1 5 1 27.39 The length of my pen = 15cm = m = 0.15m 100 The height of my eraser = 9mm = m = 0.009m 1000 The length of my pencil = cm = m= m Decimal tenth hundreths Thousandths numbers 0.15 1 5 0.009 0 0 9 Let me put myself the values 29 Chapter : 1.4 Ganit Prava – Class VI Convert decimals to fractions 0.7 = parts of equal parts. 0.07 = part of equal parts. K K K 0.007 = parts of equal parts. 57 7 5.7 = 10 = 5 10 5 whole and 7 parts out of 10 equal parts of 1 whole. K 507 7 5.07 = 100 = 5 100 5 whole and 7 parts out of 100 equal parts of 1 whole. K 5007 7 5.007 = 1000 = 5 1000 5 whole and 7 parts out of 1000 equal parts of 1 whole. K Hence, are find, 0.7 0.07 0.007 (put myself >///// = or < in the boxes given below. (a) 5.0 0.5 (b) 72.1 72.10 (c) 68.5 68.52 (d) 72.93 729.3 (e) 42.6 42.600 (f) 2.33 3.22 (g) 924 924.00 (h) (Let us put decimal numbers on your own) 10. Let's express the following in decimal numbers : (a) six tenths, (b) nine hundredths‚ (c) two thousandths (d) Two hundred three point four five (e) Four thousands two units five thousandths 3 8 (f) Six hundred twentynine point zero zero five (g) 2 +10 (h)10 + 7 + 1000 (i) 400 + 50 + 9 + 1 100 1000 11. I had ` 5. I bought a pen for ` 3.50. Let's find how much money is left. 12. What must be added to 2.75 to get 3? 13. From a string of length 12.5 cm, Mira cuts off 8.5 cm. Let's find, what portion of string is left? 14. The length of my exercise book is cm. and its breadth is cm. Let me find the perimeter of my exercise book. 15. For an occasion at our house, father bought rice for ` 200, pulses for ` 125.50 and vegetables for ` 242.50. Let's calculate the total amount of money spent, by father. 16. In a Long-jump competition, Sahil jumped 182.88 cm and Munna jumped 179.25 cm of length. Let's find how much more did Sahil jump. 17. What must be subtarcted from 5 to get 2.172? 18. 2.647 is subtracted from 4.15. How much is to be added to the result to get 10? 19. Let's find the values of the following — (a) 0.07 + 0.09 (b) 4.11+ 1.6 (c) 312.61 + 276.72 (d) 5 - 0.555 (e) 27.56 +14.69 (f) 4.3 + 3 - 6.4 (g) 3.36 - 4.62 +2.18 (h) 2.67 -3.727 + 4.2 35 Chapter : 1.5 Ganit Prava – Class VI 1.5 Geometric Measures Mahidur, Rana, Rini and Nasir were sitting in the garden and were drawing pictures of things familiar to them. Mahidur drew the scene of a village. Mahidur's picture We found in Mahidur's picture, the two land dividers are [ parallel/intersecting] The front side of the top of the hut is [triangular/circular.] The land between two dividers is [triangular/ rectangular] In Mahidur's picture, let us mark the sides of triangular and rectangular shapes in red and their angular points with green circles. Let us also try to find acute angles, obtuse angles, right angles and straight angless in Mahidur's picture and mark them with violet colour. We also find triangular shape has angles. Rana's Picture But in Rana's picture , we see many [circular/ triangular] shapes. Let us try to draw a picture like Rana using circular shapes. Rini's Picture In Rini's figure, let us identify angular points, line segment and different angles and mark them. Let's try to find acute angle, right angle and obtuse angle and straight angle from the picture and note them separately. 36  Revision of Previous Lessons Chapter : 1.5 Nasir's picture In Nasir's picture, the railway tracks are mutually [parallel/intersecting] Similarly let's mark point, line segment, and angles with different colours. I separately coloured the different parallel line segments. I shall draw a picture too, but in a different manner. I shall take many points on a sheet of paper and will name the points. I shall join the points to draw different figures and shall try to find the number of points needed to form thses figures. Let's find — Ay y y yC y End point of ray AB is , again the points C, D and E were joined by line segments to form a triangle. y y y y y B D E Let's measure the sides of the triangle CDE and let's name the triangles h according to sides. Xy yY y y y The points , , , were joined to form a rectangle [drawn and y y y y y named] Let's draw another line segment parallel to the line segment XY y y y y y formed by joining the points and. [drawn and named] Let me join the points , and by line segment to form an obtuse angled triangle (drawn and named). Let's measure the sides of the obtuse angled triangle. According to sides, the triangle is a triangle. Let's measure the three angles of the triangle (with a protractor) and the angles are , and. Also the sum of three angles ; + + =. Points , , and are joined in order to form a quadrilateral. Let us measure the sides of the quadrilateral with a scale. 37 Chapter : 1.5 Ganit Prava – Class VI I shall now draw only triangular figures of different sort. Equilateral triangle to be coloured green. Isosceles triangle to be coloured red. Scalene triangle to be coloured brown. Equilateral triangle Three sides equal f Isosceles triangle Acute angled trian- gular figure is Two sides equal f drawn Scalene triangle Let's try to draw All three sides unequal f if it is possible Three sides equal f Is o Right angled trian- tri sce Two sides equal f an le gular figure is gl s Scalene triangle e drawn All three sides unequal f Let's try to Three sides equal f draw if it is Obtuse angled triangular figure is possible Two sides equal f iso sceles triangle drawn All three sides unequal f Scalene triangle Hence, Number of green triangular figures , therefore Equilateral triangle is always an triangle. Each angle of an equilateral triangle is measured with a protractor and is found to be equal to degree. Again, there are red triangle. Hence Isosceles triangles can be , right angled and. Number of brown coloured triangles are Hence, scalene triangle can be , and also obtuse angled triangles. I draw similar triangles of other measures. I paste these triangles on a card board and try to derive relation between types of triangle according to their angles and types from the triangle according to their sides. 38  Revision of Previous Lessons Chapter : 1.5 Today let me find circular things and draw them Button 10 Rupee Round shaped Let me draw coin plate If a line is drawn with a pencil around these things, we get circles. The length of circle is called its. We got few [quadrilaterals/circles]. It is found that the circumferences of these circles are [equal/unequal]. Now, let us find, which circle has a smaller radius and which one has a bigger radius.(do myself) I draw circles with compass and try to compare their circumferences with the help of pin and string. The length of the diameter of a circle = 2 × length of the of that circle. Let us find the length of the radius of circle — the length of the diameter of the first circle is 2 cm. Therefore its radius is cm. The length of the diameter of the second circle is 3 cm, the length of its radius is3 cm = 1.5 cm. 2 The length of the diameter of the third circle is 4 cm, then its radius is 2 cm. Radius of the smallest circle is [less/more] than the radius of the bigger circle. Hence, circumferences of the circle depends on the length of of a circle. Let's draw — I draw circles with diameters of lengths it 3 cm, 5 cm and 6 cm and try to find their circumferences are gradually increasing or not. Let's work out - 1.5 1. Let us try to place the 120,22.50,1800,1790, Acute Obtuse Right Straight following angles under 1000,390,900, 690, 910 Angle Angle Angle Angle right heading. 2. Straight angle = 2 × , hence straight angle is twice a right angle. 3. Let's find, with which set of the lengths of line segments, triangle can be drawn : SaV 2 cm, 3 cm, 4 cm, (b) 4 cm, 3 cm, 7 cm, (c) 1 cm, 3 cm, 2 cm, (d) cm, cm, and cm. 4. Let us define acute angled and obtuse angled triangles. Let's draw the triangles and measure its angles with a protractor. 5. Let's measure the perimeter of the square in the figure. 6. The diameter of circle in the figure = cm. 4 cm centre Radius = cm = cm. 39  Chapter : 1.6 Ganit Prava – Class VI 1.6 Unitary Method 20 One wall of a room of Nilajul's house has collapsed. To repair the wall properly, they appointed masons. 5 masons took `.1000 daily for the work. They worked for 6 days and completed half of the work. After 6 days 2 masons did not turn up. Hence, = masons will finish the remaining work. If more masons work, more money has to be paid. Hence, number of masons and the wage they get are in relation. In Mathematical language, Number of masons wage (`) 5 1000 3 ? 5 masons get ` 1 mason will get ` ÷ =` 3 masons will get ` × = 21 In 6 days, 5 masons finished only half of the work. But 3 masons took more days to finish the remaining the work. Why so? For the same amount of work, if number of masons are more , they would take less number of days to finish the work. Again, if the same work is done by less mumber of masons, they would take more time. Hence the relation is indirect. We can say, for a particular work, the number of masons and number of days are in relation. There is another term used for indirect relation. This indirect relation is called Inverse relation. 40 ˛1. Revision of Previous Lessons Chapter : 1.6 In Mathematical language No. of masons Number of days 5 6 3 ? Half of the work is done by 5 masons in 6 days 1 mason will take 6×5 days (more than 6 days) 2 6×5 3 masons will take (less than 6×5 days) 6×5 day ÷ 3 = = 10days 3 Hence, if number of masons is 3, they will take 10 days to finish the remaining half of the work. ? Number of extra days required = - =4 22 If number of masons be 10, let's find how many days are needed to complete the work. In Mathematical language No. of masons Time (days) 5 6 10 ? (Let us find the relation between number of masons and number of days, and work it out myself) 23 10 of us decided to visit our farmhouse for a week. We arranged food accordingly. But at the last moment 4 more friends joined us. Let's find how long will that food last. Let's find relation first If number of friends increase, the same food will last for number of days. Again, if the number of friends are less the same food will last for number of days. Hence, amount of food being fixed, number of friends are related to time. In mathematical language Number of Time (days) friends 10 1 week = days + = ? For 10 friends, the food will last for 7 days for 1 friend, the food will last for = days Hence for friends, the food will last for = days Thus, our food would last for days. 41  Chapter : 1.6 Ganit Prava – Class VI 24 A computer company manufactures 180 laptops with the help of 40 engineers in 30 days. Let us find how many engineers would be needed to manufacture 2700 laptops in 200 days. In mathematical language, Number of` Number of Number of laptops days Engineers 180 30 40 2700 200 ? To manufacture 180 laptops in 30 days, 40 engineers are needed. To manufacture 180 laptops in 1 day, 40×30 engineers are needed. 40×30 To manufacture 1 laptop in 1 day, engineers are needed 180 40×30×2700 To manufacture 2700 laptops in 1 day, engineers are needed 180 450 90 To manufacture 2700 laptops in 200 days, 40×30×2700 engineers are needed. 6 180×200 5 ? 90 engineers are needed to manufacture 2700 laptops in 200 days. Let's work out - 1.6 1. To construct an embankment on a certain part of the river Ichhamati, 40 labours are required to work for 35 days. If the work has to be finished in 28 days, let us find how many labours will be needed? 2. Rajeeb, Debangana, Masum and Tajmira can do 150 sums in 6 days. Let us find, in how many days will Rajeeb and Tajmira finish 250 sums if each of them does equal number of sums everyday. 1 3. Two labourers can polish 3 of a door in 1 day. Let us find how many labourers are needed to polish 2 of a door in 2 days. 3 4. For mid-day meal of 500 students, 175 kg of rice is required for a week. After 75 kg of rice has been used, Let us find how long will remaining rice last for 400 students. 5. 4 tractors are required to cultivate 360 bighas of land in 20 days. Let us find, how many tr actors are needed to cultivate 1800 bighas of land in 10 days. 6. In a fair, if 12 generators work for 6 hours a day, then deposited oil is finished in 7 days If generators run for 4 hours a day for 9 days, then let us find how many generators can run with that remaining amount of oil. 7. 15 vans can carry 75 quintals of vegetables in 40 mins. Let us find how long will 20 vans take to carry 100 quintals of vegetables. 8. 150 kg of wheat is stored for 20 hostel boys for 30 days. But 30 kg of wheat was wasted and 5 boys went home. With the remaining amount of wheat, Let us find for how long will the remaining boys be fed? 42 Concept of seven and eight-digit numbers Chapter : 2 2. Concept of Seven and Eight digit Numbers My uncle has a book shop. When sales are high my sister goes to the shop along with my uncle and helps him in accounting. In this January the demand for books is high, so there is heavy rush of customers at the shop. I too go to my uncle's shop with my sister in the evening. In the first week books are sold for a total amount of ` 5,08,610. But in the second week the sale amounted to ` 4,92,070. Let me try to represent with sticks and colour balls, the amount of money earned in two weeks by selling books. Amount of money earned by Amount of money earned by selling books in first week. selling book in the second week ten units ten lac thousands thousands hundreds tens lac thousands thousands hundreds tens units 5 0 8 6 1 0 4 9 2 0 7 0 Five lac eight thousand six hundred ten Four lac ninty-two thousand seventy Æ ten ten lac thousands thousands hundreds tens units lac thousands thousands hundreds tens units Total amount of Æ money ¯ in two weeks. Stick of thousand can not hold more than The stick of ten thousand can not hold more 9 balls, so the amout could not be kept on that stick than I balls, so could not be kept on that stick Æ ten ten ten lac thousands thousands hundreds tens units lac lac thousands thousands hundreds tens units ¯ The stick of lac can not hold more than The place for the new stick is named ten 9 balls, so it can not be put on that stick lac. Since extra ball in lac's place could not be Instead of 10 violet balls of lac we placed placed, another stick is needed. 1 coloured ball in place 43  Chapter : 2 Ganit Prava – Class VI Let us express the population count of year 2011 for few districts of West Bengal by sticks and balls. Population of Howrah District 48,41,638 Population of Purulia District 29,27,975 Population of Kolkata District 44,86,679 Population of Jalpaiguri Dist 38,69,675 Population of Howrah District Population of Purulia District Tlac˛ lac ten Thou H ten U Tlac˛ lac ten Thou H ten U thou thou 4 8 4 1 6 3 8 Let us Population of Kolkata District Population of Jalpaiguri District place the Tlac˛ lac ten Thou H ten U Tlac˛ lac ten Thou H ten U thou balls on thou the sticks Expanded form of a number Number Place Value In Words 7000000 7543932 Seven ten lacs five lacs Seventy five lac four ten thousands forty three thousand +500000 three thousands nine nine hundred and +40000 hundreds three tens and thirty two. +3000 two units. +900 +30 +2 2318600 Eighty five lacs five 44 Concept of seven and eight-digit numbers Chapter : 2 Let's write a seven digit number and expand it digits are seven digit number Expanded form. 1,2,3,4,6,7, 8 2 3 4 6 7 8 1 2000000 +300000+40000+ 6000 + 700 + 80+1 1,0,3,4,5,9,7 2,9,4,6,7,8,3 Using each digit only once, let us write four numbers of seven digits and then write the numbers in words. Digits are First number of Second number Third number Fourth number seven digits of seven digits of seven digits of seven digits 2539167 Twenty five lac thirty 2,5,3,9,7,1,6 nine thousand one hundred sixty seven Place value of 5 is Place value of 5 is Place value of 5 is Place value of 5 500000 50000 5000 is 500 6,7,2,3,1,5,0 Place value of Place value of Place value of Place value of 6 is 600000 6 is 60000 6 is 6000 6 is 600 9,8,2,3,7,5,4 Place value of 7 Place value of 7 Place value of 7 Place value of 7 is 700000 is 70000 is 7000 is 700 5,7,2,3,1,8,0 Place value of 8 Place value of 8 Place value of 8 Place value of 8 is 800000 is 80000 is 8000 is 800 45  Chapter : 2 Ganit Prava – Class VI Today is Sunday. It is a holiday. I found my aunt very busy since morning. She will take my cousin sister to the Polio Booth. There she will be given polio vaccine. I too went to the booth with my sister. All through the day many children were given Polio vaccine. Let's see how many children of south 24 parganas were given polio vaccine. I learnt from my father that in our district of south 24-Parganas, around 26,10,500 children were given polio vaccine. I found out that the number of such children who took polio vaccine, in my district and in three other districts also. 1 With the help of sticks and coloured balls, number of children who took Polio Vaccine in four districts are shown below. 28,22,000 (approx) children in North 24-Parganas 26,10,500 (approx.) children in South 24-Parganas. Ten lac˛ lac ten thou Thou H ten U Ten lac˛ lac Ten thou Thou H ten U 2 8 2 2 0 0 0 2 6 1 0 5 0 0 24,12,300 (approx.) children in Kolkata. 22,34,200 (approx.) children in Howrah Ten lac˛ lac Ten thou Thou H ten U Ten lac˛ lac Ten thou Thou H ten U 2 4 1 2 3 0 0 2 2 3 4 2 0 0 46 Concept of seven and eight-digit numbers Chapter : 2 Total number of children from North and South Total number of children from Kolkata and 24-Parganas who were given Polio Vaccine. Howrah who were given Polio Vaccine. Ten lac˛ lac Ten thou Thou H ten U Ten lac˛ lac Ten thou Thou H ten U 5 4 3 2 5 0 0 4 6 4 6 5 0 0 Total number of children TL l Tt T H t U from 24-Parganas North and C TL l t T H t U South, Kolkata & Howrah, who all got Polio Vaccine. In Ten Lac stick, more than I balls can not be placed. Hence another strick or a 'place' is needed. The name of this stick or 'Place' is "Crore"(c). 1 0 0 7 9 0 0 0 Instead of 10 orange balls I take 1 ball. Let us put balls in respective sticks and write the number C TL l Ten thou T H Ten U C TL l Ten thou T H Ten U 5 0 2 0 8 0 6 9 one crore eleven lac six hundred five 47  Chapter : 2 Ganit Prava – Class VI Let's count the books in the Library We have a small library in our locality. I often go there with my elder sister and read interesting books. There are 21635 books in this library. Do other libraries too have so many books? There are many libraries in Kolkata which have much more books. Like, Asiatic Society library, National Library at Alipore. etc. I came to know from my elder sister that the number of books in Asiatic Society Library is around 1,49,000 and the number of books in Nat