Class 8 maths exercise 5.1 solutions

NCERT Solutions for Class 8 Maths Chapter 5 - Data Handling Exercise 5.1 The CBSE Class 8 Chapter 5 Data Handling has a total of 3 exercises. The PDF containing the NCERT Solutions of the first exercise, Exercise 5.1, is available here. The The NCERT textbook contains plenty of questions in it intended for the students to solve and practise. To score high marks in the Class 8 examination, solving and practising the Download PDF carouselExampleControls111 Previous Next  Access other exercise solutions of Class 8 Maths Chapter 5 – Data Handling Access answers of Maths NCERT Class 8 Chapter 5 – Data Handling Exercise 5.1 Page Number 76 1. For which of these would you use a histogram to show the data: (a) The number of letters for different areas in a postman’s bag. (b) The height of competitors in an athletics meet. (c) The number of cassettes produced by 5 companies. (d) The number of passengers boarding trains from 7.00 a.m. to 7.00 p.m. at a station. Give a reason for each. Solution: We know that a Histogram is a graphical representation of data if the data is represented using class interval. Since the cases mentioned in options (b) and (d) can be divided into class intervals, a histogram can be used to show the data. Similarly, since the cases mentioned in options (a) and (c) cannot be divided into class intervals, a histogram cannot be used to represent the data. 2. The shoppers who come to a departmental store are marked as man (M), woman (W), boy (B) or girl (G...

Frank Learning Maths Class 8 Solutions Chapter 5 Algebraic Expressions Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 5 Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions. Here students will find solutions for Exercise 5.1, 5.2 and 5.3. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum. Algebraic Expressions Exercise 5.1 Solution Question no – (1) Solution : (a) 6x + 4y – 3z 7x – 11y – 9z 14x + 8y – 6z —————————— 27x + y – 18z or = (6x + 4y – 3z) + (7x – 11y – 9z) + (14x + 8y – 6z) = 27x + y – 18z (b) 7p – 8q + 11r + 13 10p – 13q + 0 + 18 8p – 5q – 14r + 12 5p + 17q + 14r —————————— 30p – 9q + 11r + 43 (c) 30p – 9q + 11r + 43 8x 2 + 7x + 12 17x 2 – 15x – 21 – 9x 2 + 11x + 4 4x 2 – 18x + 19 —————————— 20x 2 – 15x + 14 (d) 6ax – 2by + 3cz 11ax – 16by – 15cz – 9ax – 3by + 10cz —————————— 8ax – 21by – 2cz (e) 15m 2n – 17mn + 8mn 2 13 m 2n – 15mn – 9mn 2 12m 2n + 21mn – 10cz ————————————- 40m 2n – 11mn – 15m 2n (f) 13x + 17y – 19z + 2 14x + 12y + 0 – 31 – 15x + 6z + 12 2x + y + 11z + 9 ————————————- 14x + 30y – 2z – 8 Question no – (2) Solution : (a) 12x + 7y – 15z 17x + 8y – 15z (-) (-) (+) —————————- – 5x – y + 0 (...

The NCERT Solutions for Class 8 Maths are a great resource for students who want to excel in their studies. The solutions are designed by experts in the field and provide step-by-step guidance to help students understand the concepts better. They also offer tips and tricks to make the learning process easier. NCERT Solutions for Class 8 Maths provides students with a complete and thorough understanding of the subject. It is a very popular resource among students and teachers. The Solutions provide step-by-step explanations of all the topics covered in the Class 8 Maths syllabus. All textbook questions in line with the CBSE Board are included in Infinity Learn NCERT Solutions of Math Class 8 Chapter-wise – Free PDF Download • Chapter 1 Rational Numbers • • • • • • • • • • • • • • • Class 8th NCERT Solutions for will benefit much from the math lessons provided here. They can use these as a reference for practising problems, improving their skills, and efficiently preparing for exams. Exercises are provided at the end of each chapter to allow students to put what they’ve learned into practice. Prepare for JEE Main and Advanced with Infinity Learn online JEE course. NCERT Solutions for Class 8 Maths Chapter 1 In this chapter, students will discover the independent, associative, and closure properties of real numbers, integers, whole numbers, rational numbers, and natural numbers. The role of zero and one, multiplication over addition, the representation of Rational Numbers on ...

NCERT Solutions Class 8 Maths Chapter 5 Exercise 5.1 Data Handling The NCERT solutions for Class 8 maths Chapter 5 exercise 5.1 Data Handling covers the concept of raw data, its depiction using frequency distribution charts as well as through various graphs such as ☛ Download NCERT Solutions Class 8 Maths Chapter 5 Exercise 5.1 Exercise 5.1 Class 8 Chapter 5 Download PDF More Exercises in Class 8 Maths Chapter 5 • • NCERT Solutions Class 8 Maths Chapter 5 Exercise 5.1 Tips Students should understand the fundamentals of data, how to gather it, and its representation techniques before proceeding to the The Download Cuemath NCERT Solutions PDF for free and start learning! NCERT Video Solutions for Class 8 Maths Chapter 5 Exercise 5.1 Video Solutions for Class 8 Maths Chapter 5 Exercise 5.1

Class 8 Maths Exercise 5.1 Solutions in Hindi Medium Class 8 Maths Exercise 5.1 Solutions in English Medium Class 8 Maths Exercise Solutions for 2022-2023 Class 8 Maths Exercise 5.1 Solutions for terminal exams NCERT Solutions for Class 8 Maths chapter 5 Exercise 5.1 8th Science Chapter 5 Exercise Solutions Class 8 Maths Exercise solutions for terminal exam chapter 5

NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.1 – Data Handling, has been designed by the NCERT to test the knowledge of the student on the following topics : • Looking for Information – Different types of graphs A Pictograph A bar graph Double Bar Graph • Organising Data • Grouping Data Bars with a difference NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.1 Data Handling NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.1 Data Handling Q.1 For which of these would you use a histogram to show the data? (a) The number of letters for different areas in a postman’s bag. (b) The height of competitors in an athletic meet. (c) The number of cassettes produced by 5 companies. (d) The number of passengers boarding trains from 7.00 a.m. to 7.00 p.m. at a station. Give reasons for each. Solution: In case of (b) and (d) histogram is used to show the data Histogram is a graphical representation of data if the data is measured in manner of class intervals. In case of (b), heights of different players in an athletic meet in intervals (range) of height In case of (d), the number of passengers boarding the trains at an interval of one hour at a station Q.2 The shoppers who come to a departmental store are marked as: man (M), woman (W), boy (B) or girl (G). The following list gives the shoppers who came during the first hour in the morning: W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W Make a fr...

Chapter 5 Solution West Bengal Board : Class 8 Mathematics West Bengal Board Class 8 Math Solution full Exercise 5.1, 5.2 and 5.3 by Experts. Here in this page WB Board Class 8 Student can find Ganit Prabha Class 8 Math Chapter 5 ঘনফল নির্ণয় Exercise 5.1, 5.2 and 5.3 Solution . Board WB Board Class 8 (Eight ) Subject Mathematics Book Name Ganit Prabha Chapter 5 Exercise 5.1 Solution Exercise 5.2 Solution (iii) (x 2y – z 2) = (x 6y 3 – 3x 4y 2z 2 + 3x 2yz 4 – z 6) (iv) (l + b – 2c) 3 = l 3 + b 3 – 8c 3 + 3l 2b + 3lb 2 – 6cl 2 – 12lbc – 6b 2c + 12c 2l + 12c 2b (v) (2.89) 3 + (2.11) 3 + 15 x 2.89 x 2.11 = 24.137 + 9.4 + 91. 46 = 125 (vi) (2m + 3n) 3 + (2m – 3n) 3 + 12n (4n 2 – 9n 2) = (2m) 3 + 3. (2m) 2.3n + 3. 2m. (3n) 2 + (3n) 3 + (2m) 3 – 3.(2m) 2.3n + 3.2m.(3n) 2 + (3n) 3 + 48mn 2 – 180mn 2 = 8m 3 + 36m 2n + 54mn 2 + 27n 3 + 8m 3 – 36m 2n + 54mn 2 + 27n 3 + 48mn 2 – 108mn 2 ∴ x 3 + y 3 = (x + y) 3 – 3.xy. (x+y) = (2) 3 – 3.1(2) = 8 – 6 = 2 (f) x = y + z =) x – y = z =) (x – y) 3 = (z) 3 =) x 3 – y 3 – 3xy (x – y) = z 3 =) x 3 – y 3 – 3xy. Z = z 3 =) x 3 – y 3 – z 3 -3xyz = 0 (k) m +n = 5 ; m n = 6 (m 2 + n 2) (m 3 + n 3) = (x 3 + y 3) (x 3 – y 3) = (x 2 + y 2) (x 4 – x 2y 2 + y 4) (x 3 + y 3 (x 3 – y 3) (xviii) m 3 + n 3 – m(m 2 – n 2) + n(m – n) 2 = (m – n) (m 2 + mn + n 2) – n (m – n) (m+ n) + n (m – n) (m- n) = (m – n) (m 2 + mn + n 2) – m (m + n) + n (m – n) = (m – n) (m 2 + mn + n 2 – m 2 – mn + mn – n 2) = (m – n) mn