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Case Study Questions for Class 6 Maths Chapter 1 Knowing Our Numbers

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Here in this article, we are providing case study questions for class 6 maths. Here you will find case study questions for Class 6 Maths Chapter 1 Chapter 1 Knowing Our Numbers.

Knowing Our Numbers Case Study Questions

Case Study Question 1:

The population of Delhi in 2017 was 19072564 and it increased to 25704625 in 2021.

(i) Write the population of 2021 in words. (a) two crore fifty seven lakh four thousand six hundred twenty five (b) twenty five lakh seventy thousand six hundred twenty five (c) two hundred fifty seven thousand four hundred twenty five. (d) twenty five crore seventy lakh forty six thousand and twenty five.

(ii) The successor of 19072564 is. (a) 19072563 (b) 19072565 (c) 29072564 (d) 19072574

(iii) What is the place value of ‘ 7 ‘ in 19072564 (a) 7 (b) 70000 (c) 7000 (d) 700

(iv) Write the population of 2017 in words according to International system of numeration.

(v) Write the population of 2021 in expanded form.

Maths Class 6 Chapter List

Latest chapter list (2023-24).

Chapter 1 Knowing Our Numbers Case Study Questions Chapter 2 Whole Numbers Case Study Questions Chapter 3 Playing with Numbers Case Study Questions Chapter 4 Basic Geometrical Ideas Case Study Questions Chapter 5 Understanding Elementary Shape Case Study Questions Chapter 6 Integers Case Study Questions Chapter 7 Fractions Case Study Questions Chapter 8 Decimals Case Study Questions Chapter 9 Data Handling Case Study Questions Chapter 10 Mensuration Case Study Questions Chapter 11 Algebra Case Study Questions Chapter 12 Ratio and Proportion Case Study Questions

Old Chapter List

Chapter 1 Knowing Our Numbers Chapter 2 Whole Numbers Chapter 3 Playing with Numbers Chapter 4 Basic Geometrical Ideas Chapter 5 Understanding Elementary Shape Chapter 6 Integers Chapter 7 Fractions Chapter 8 Decimals Chapter 9 Data Handling Chapter 10 Mensuration Chapter 11 Algebra Chapter 12 Ratio and Proportion Chapter 13 Symmetry Chapter 14 Practical Geometry

Deleted Chapter:

Tips for Answering Case Study Questions for Class 6 Maths in Exam

1. Comprehensive Reading for Context: Prioritize a thorough understanding of the provided case study. Absorb the contextual details and data meticulously to establish a strong foundation for your solution.

2. Relevance Identification: Pinpoint pertinent mathematical concepts applicable to the case study. By doing so, you can streamline your thinking process and apply appropriate methods with precision.

3. Deconstruction of the Problem: Break down the complex problem into manageable components or steps. This approach enhances clarity and facilitates organized problem-solving.

4. Highlighting Key Data: Emphasize critical information and data supplied within the case study. This practice aids quick referencing during the problem-solving process.

5. Application of Formulas: Leverage pertinent mathematical formulas, theorems, and principles to solve the case study. Accuracy in formula selection and unit usage is paramount.

6. Transparent Workflow Display: Document your solution with transparency, showcasing intermediate calculations and steps taken. This not only helps track progress but also offers insight into your analytical process.

7. Variable Labeling and Definition: For introduced variables or unknowns, offer clear labels and definitions. This eliminates ambiguity and reinforces a structured solution approach.

8. Step Explanation: Accompany each step with an explanatory note. This reinforces your grasp of concepts and demonstrates effective application.

9. Realistic Application: When the case study pertains to real-world scenarios, infuse practical reasoning and logic into your solution. This ensures alignment with real-life implications.

10. Thorough Answer Review: Post-solving, meticulously review your answer for accuracy and coherence. Assess its compatibility with the case study’s context.

11. Solution Recap: Before submission, revisit your solution to guarantee comprehensive coverage of the problem and a well-organized response.

12. Previous Case Study Practice: Boost your confidence by practicing with past case study questions from exams or textbooks. This familiarity enhances your readiness for the question format.

13. Efficient Time Management: Strategically allocate time for each case study question based on its complexity and the overall exam duration.

14. Maintain Composure and Confidence: Approach questions with poise and self-assurance. Your preparation equips you to conquer the challenges presented.

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NCERT Solutions Class 6 Maths Chapter 1 Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers will help the students take their understanding of numbers a bit further and explore topics like shifting digits, expanding brackets, roman numerals, estimating sum or difference, products, outcomes of number situations, and many more. This chapter will also help the students revise all the operations on numbers like multiplication , addition , subtraction , and division covered in the previous classes. Presented below is an in-depth analysis of Class 6 Maths NCERT Solutions Chapter 1.

These solutions will help the students understand large numbers through real-life examples while also exploring their unit conversions. They will also get to learn about the International and Indian systems of numeration along with an introduction to the concept of roman numerals. Let us now take a deeper look at the different sub-topics covered in NCERT Solutions Class 6 Maths Chapter 1 and also you can find some of these in the exercises given below.

• NCERT Solutions Class 6 Maths Chapter 1 Ex 1.1
• NCERT Solutions Class 6 Maths Chapter 1 Ex 1.2
• NCERT Solutions Class 6 Maths Chapter 1 Ex 1.3

NCERT Solutions for Class 6 Maths Chapter 1 PDF

NCERT Solutions Class 6 Maths Chapter 1 Knowing Our Numbers contains all the important questions, images, explanations, and formulas covered in the chapter. These solutions are aimed at helping the students understand and solve complex concepts such as the use of commas in larger numbers. Commas are used to mark thousands, lakhs, and crores. Let us do a detailed exercise-wise overview of this chapter using the PDF links given below :

☛ Download Class 6 Maths NCERT Solutions Chapter 1 Knowing Our Numbers

NCERT Class 6 Maths Chapter 1   Download PDF

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Students must have already studied the formation of numbers in the previous classes. In this chapter, they will be introduced to the concept of arranging numbers in ascending as well as in descending order and shifting the digits to form new numbers. All these ideas are explained along with real-life examples in the NCERT Solutions Class 6 Maths Chapter 1 Knowing Our Numbers.

• Class 6 Maths Chapter 1 Ex 1.1 - 4 Questions
• Class 6 Maths Chapter 1 Ex 1.2 - 12 Questions
• Class 6 Maths Chapter 1 Ex 1.3 - 5 Questions

☛ Download Class 6 Maths Chapter 1 NCERT Book

Topics Covered: The important topics covered under Class 6 maths NCERT Solutions Chapter 1 are natural numbers , comparison of numbers, place value of a digit, estimation of the numbers, roman numerals , and importance of brackets. Questions related to all these topics must be practiced regularly to score excellent marks in the exams.

Total Questions: Class 6 Maths Chapter 1 Knowing Our Numbers has a total of 21 well-researched questions out of which 12 are straightforward, 3 are of moderate level, and 6 are complex level problems. 

List of Formulas in NCERT Solutions Class 6 Maths Chapter 1

NCERT solutions class 6 maths chapter 1 does not cover any formulas . However, there are some pointers that the students must remember while solving problems from this chapter. These pointers will help them avoid mistakes while giving a crystal clear understanding of the topics covered in this chapter. Let us now go through these pointers one by one :

• Given two numbers, if one has a large number of digits, then it is greater. In the case where the number of digits is the same, the number with the larger leftmost digit is considered to be the greater number.
• We put commas after 3 digits in the Indian system. These commas are put after 3,5,7, thus separating the thousand, lakh, and crore. However, in the International System, we put commas after every 3 and 6 digits from right, separating thousands, billions, and further billions.
• We use estimation in order to get a rough figure of the numbers involved. This gives us a quick idea of the answer. Estimation is done by rounding numbers.

Important Questions for Class 6 Maths NCERT Solutions Chapter 1

Faqs on ncert solutions for class 6 maths chapter 1, why are ncert solutions class 6 maths chapter 1 important.

NCERT Solutions Class 6 Maths Chapter 1 will help kids identify pain points and find ways to solve questions in an efficient manner. They focus on some of the fundamental concepts that act as a building block for a kid’s mathematical journey. Topics such as estimation have practical applications in our daily life as it gives an idea of the quantity. Thus these solutions are extremely important for students considering their utility in everyday life.

Do I Need to Practice all Questions Provided in NCERT Solutions for Class 6 Maths Knowing Your Numbers?

All the questions in NCERT Solutions are curated by experts covering a range of topics. NCERT Solutions for Class 6 Maths Chapter 1 will help you explore all the topics in detail by solving a range of problems. Thus practicing all the questions of this chapter is important for a strong mathematical foundation.

What are the Important Topics Covered in NCERT Solutions Class 6 Maths Chapter 1?

NCERT Solutions Class 6 Maths Chapter 1 covers a range of essential topics such as estimating numbers, use of commas, place value, expanding brackets, and roman numerals. Questions related to all these topics are explained in a step-wise manner in the NCERT solutions.

How Many Questions are there in Class 6 Maths NCERT Solutions Chapter 1 Knowing Your Numbers?

There are a total of 21 sums in the NCERT Solutions Class 6 Maths Chapter 1 Knowing Your Numbers. These questions are distributed in 3 exercises. Out of 21 questions, 10 are easy sums, 5 are moderately difficult problems, while 6 are complex sums that need brainstorming.

How can CBSE Students utilize NCERT Solutions for Class 6 Maths Chapter 1 effectively?

In order to effectively utilize the NCERT Solutions Class 6 Maths Chapter 1 students must go through the detailed solutions of all the questions, explore the chapter summary, and theory of the chapter. Students must give adequate time to all the exercises and solve questions in a step-wise manner.

Why Should I Practice NCERT Solutions Class 6 Maths Knowing Your Numbers?

NCERT Solutions Class 6 Maths Chapter 1 Knowing Your Numbers will help you revise the previous concepts from earlier classes like place values, whole numbers, number system, etc. Apart from that, this chapter will help you understand the applications of interesting concepts in your day-to-day life.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Get here the NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers and Class 6 Maths Chapter 1 Try These Solutions and Practice Tests for revision. It is given here in Hindi and English Medium prepared for academic session 2024-25. According to rationalised syllabus and new books for class 6 Mathematics for CBSE 2024-25, there are only two exercises in chapter 1 Knowing our numbers.

6th Maths Chapter 1 Solutions for CBSE Board

• Class 6 Maths Chapter 1 Try These
• Class 6 Maths Exercise 1.1 in English
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6th Maths Chapter 1 Solutions for State Boards

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Class 6 Maths Chapter 1 Practice Test 6th Maths Chapter 1 Test 1 6th Maths Chapter 1 Test 2 6th Maths Chapter 1 Test 3 6th Maths Chapter 1 Test 4 6th Maths Chapter 1 Test 5 6th Maths Chapter 1 Test 6

Get class VI Maths Exercise 1.1 and 1.2 at Tiwari Academy in simplified way. 6th Maths Solutions PDF and Video in English and Hindi Medium are prepared in such a way that student can understand it easily. We have updated it for new session based on latest textbooks from NCERT (https://ncert.nic.in/) website. Find the Solutions of Prashnavali 1.1 and 1.2 in Hindi. We are following the latest CBSE Syllabus 2024-25. We work for your help free of cost. Separate links are given to download solutions in PDF file format. In case of any hassle in finding the solutions, please inform us. We will help you at our level best.

These NCERT Solutions are based on latest CBSE – NCERT Textbooks for the CBSE exams 2024-25. Download NCERT Solutions in PDF format to use it offline or use as it is online without downloading.

In 6 Maths Chapter 1 Knowing Our Numbers, we will study about comparing the number (smaller or greater), selecting the smallest or greatest numbers, order of numbers (ascending or descending).

Ascending order: Ascending order means arrangement from the smallest to the greatest. Descending order: Descending order means arrangement from the greatest to the smallest.

Concepts of Place values, face values and questions based on the numbers as follow: Starting from the greatest 6-digit number, write the previous five numbers in descending order. Starting from the smallest 8-digit number, write the next five numbers in ascending order. The Indian System of Numeration: In our Indian System of Numeration, Commas are used to mark thousands, lakhs and crores. We use ones, tens, hundreds, thousands and then lakhs and crores. The first comma comes after hundreds place and marks thousands. The second comma comes two digits later. It comes after ten thousands place and marks lakh. The third comma comes after another two digits. It comes after ten lakh place and marks crore.

Important Questions on 6 Maths Chapter 1

Fill in the blank: 1 lakh = _______________ ten thousand.

Fill in the blank: 1 lakh = 10 ten thousand

Place commas correctly and write the numerals: Seventy-three lakh seventy-five thousand three hundred seven.

Insert commas suitable and write the names according to indian system of numeration: 87595762.

8,75,95,762 Eight crore seventy-five lakh ninety-five thousand seven hundred sixty-two.

Estimate each of the following using general rule: 730 + 998

730 round off to 700 998 round off to 1000 Estimated sum 1700

Estimate the following product using general rule: 578 x 161

578 x 161 578 round off to 600 161 round off to 200 The estimated product = 600 x 200 = 1,20,000

We are here to help you. For educational help any time you can leave a message, we will call you with in 24 hours. Our prime motive is to help the students without any delay free of cost. NCERT Books and their solutions are given in offline as well as online mode.

How many exercises, questions, and examples are there in chapter 1 of class 6th Maths?

There are 2 exercises in chapter 1 (Knowing our Numbers) of class 6th Maths. In the first exercise (Ex 1.1), there are 4 questions. Questions 1 and 2, each having five parts, and questions 3 and 4, each having four parts. In the second exercise (Ex 1.2), there are 12 word problem questions. So, there are in all 16 questions in chapter 1 (Knowing our Numbers) of class 6th Maths. There are 6 examples in chapter 1 (Knowing our Numbers), which are good for exams point of view.

What are the main topics to study in chapter 1 of class 6th Maths?

In chapter 1 of class 6th Maths, students will study:

• 1. Comparing Numbers.
• 2. How many numbers can you make?
• 3. Shifting digits.
• 4. Introducing 10,000.
• 5. Revisiting place value.
• 6. Introducing 1, 00,000.
• 7. Larger numbers.
• 8. An aid in reading and writing large numbers.
• 9. Use of commas.
• 10. Large Numbers in Practice.

Is chapter 1 of class 6th Maths difficult?

Chapter 1 of class 6th Maths is neither too easy nor too difficult. It lies in the middle of easy and difficult because some parts of this chapter are easy, and some are difficult. However, the difficulty level of any chapter varies from student to student. So, Chapter 1 of class 6th Maths is easy or not depends on students also. Some students find it complicated, some find it simple, and some find it in the middle of simple and difficult.

How much time, students need to do chapter 1 of class 6th Maths?

Students need a maximum of 5-6 days to do chapter 1 of class 6th Maths if they give at least 1-2 hours per day to this chapter. This time is an approximate time. This time can vary because no students have the same working speed, efficiency, capability, etc.

Chapter 2: Whole Numbers »

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Class 6 Maths Chapter 1 Question and Answers

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Class 6 Maths Chapter 1 Important Questions – Knowing Our Numbers

Maths is an important subject we study in school. In Class 6, students will learn the basics of the subject, which will be needed in higher classes. The first chapter is about learning numbers. Maths deals with numbers, and students must identify numbers.

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In this chapter, students will study larger numbers like thousands, lakhs, etc. They will learn how to express these numbers with the help of commas. The chapter also includes addition and subtraction of larger numbers, how to find the largest among given numbers, etc. This is an easy chapter, but students must practice questions to build their concepts.

Extramarks is a leading company that provides all the important study materials related to CBSE and NCERT. Our experts have made the Important Questions Class 6 Maths Chapter 1 to help the students in practice. They have collected the questions from the textbook exercises, CBSE sample papers, CBSE past years’ question papers, NCERT Exemplars, and important reference books. They have solved the questions, and experienced professionals have further checked the answers to ensure the best quality of the content.

We provide a wide range of study materials to students, and you can download these after registering on our official website. You will find the CBSE syllabus, CBSE sample papers, CBSE revision notes, CBSE extra questions, CBSE past years’ question papers, NCERT books, NCERT solutions, NCERT Exemplars, NCERT important questions, vital formulas, and many more.

Get Access to CBSE Class 6 Maths Important Questions with Solutions

Also, get access to CBSE Class 6 Maths Important Questions for other chapters too:

Knowing Our Numbers Class 6 Extra Questions with Solutions

Our experts have made the question series to help students. They have collected the questions from textbook exercises, CBSE sample papers, CBSE past years’ question papers, NCERT Exemplars, and important reference books. They have also solved the questions so that students can follow the answers. Experienced professionals have further checked the solutions to ensure the best quality of the content. Thus, the Important Questions Class 6 Maths Chapter 1 will help students  score better in exams. The important questions are-

Question 1.

Fill in the blanks:

(i) One lakh = ………….. ten thousand.

(ii) 1 million = ………… hundred thousand.

(iii) 1 crore = ………… ten lakh.

(iv) 1 crore = ………… million.

(v) 1 million = ………… lakh.

(i) 1 lakh = ten ten thousand.

(ii) 1 million = ten hundred thousand.

(iii) 1 crore = ten ten lakh

(iv) 1 crore = ten million

(v) 1 million = ten lakh

Question 2.

(i) 1 metre = ____millimetres.

(ii) 1 centimetre = ____ millimetres.

(iii) 1 kilometre = ____ millimetres.

(iii) 10, 00, 000

Question 3.

(i) 1 gram = ___ milligrams.

(ii) 1 litre = ___ millilitres.

(iii) 1 kilogram = ___ milligrams.

(iii) 10,00,000

Question 4.

Place the commas correctly and write the numerals :

(i) Seventy-three lakh seventy-five thousand three hundred seven.

(ii) Nine crore five lakh forty-one.

(iii) Seven crore fifty-two lakh twenty-one thousand three hundred two.

(iv) Fifty-eight million four hundred twenty- three thousand two hundred two.

(v) Twenty-three lakh thirty thousand ten.

(i) 73,75,307

(ii) 9,05,00,041

(iii) 7,52,21,302

(iv) 5,84,23,202

(v) 23,30,010.

Question 5.

Insert commas in the numbers suitably and write their names according to the Indian System of Numeration:

(i) 87595762

(ii) 8546283

(iii) 99900046

(iv) 98432701

(i) 8,75,95,762 (Eight crore seventy-five lakh ninety-five thousand seven hundred sixty- two)

(ii) 85,46,283 (Eighty-five lakh forty-six thousand two hundred eighty-three)

(iii) 9,99,00,046 (Nine crore ninety-nine lakh forty-six)

(iv) 9,84,32,701 (Nine crores eighty-four lakh thirty-two thousand seven hundred one)

Question 6.

Insert commas in the numbers suitably and write their names according to the International System of Numeration:

(i) 78921092

(ii) 7452283

(iii) 99985102

(iv) 48049831

(i) 78,921,092 (Seventy-eight million nine hundred twenty-one thousand ninety-two)

(ii) 7,452,283 (Seven million four hundred fifty- two thousand two hundred eighty-three)

(iii) 99,985,102 (Ninety-nine million nine hundred eighty-five thousand one hundred two)

(iv) 48,049,831 (Forty-eight million forty-nine thousand eight hundred thirty-one)

Question 7.

A number in which the Sum of all of its factors is equal to twice the number is called a ___ number.

Question 8.

The numbers which have more than just two factors are called ___ numbers.

Question 9.

Two is the only ___ number which is even.

Question 10.

Two numbers having only one as a common factor are called ___ numbers.

Question 11.

The Lowest Common Multiple ( LCM) of two or more given numbers is always the lowest of their common ___.

Question 12.

The Highest Common Factor  (HCF) of two or more than two given numbers is also known as the highest of their common ___.

Question 13.

The product of the place values of the two 2’s in 428721 is

(iii) 400000

(iv) 40000000

(iii): Place the values of 2’s in 428721 are 20000 and 20

∴ The required product = 20000 × 2 = 400000

Question 14.

Number 3 × 10000 + 7 × 1000 + 9 × 100 + 0 × 10 + 4 is the same as

(iii) 37904

(iv) 379409

(ii) : 3 × 10000 + 7 × 1000 + 9 × 100 + 0 × 10 + 4

= 30000 + 7000 + 900 + 4 = 37904

Question 15.

If one is added to the greatest 7-digit number, then it will be equal to

(i) 10 thousand

(ii) 1 lakh

(iii) 10 lakh

(iv) One crore

(iv) : The greatest 7-digit number = 99,99,999

Now, 99,99,999 + 1 = 1,00,00,000

Question 16.

The greatest number in which on rounding off to the nearest thousands gives 5000, is

(iv) : (1) Rounding off 5001 to nearest thousands = 5000

(2) Rounding off 5559 to nearest thousands = 6000

(3) Rounding off 5999 to nearest thousands = 6000

(4) Rounding off 5499 to nearest thousands = 5000

And 5499 > 5001

Question 17.

Keeping the place of six in the number 6350947 same, the smallest number which can be obtained by rearranging other digits is

(i) 6975430

(ii) 6043579

(iii) 6034579

(iv) 6034759

(iii) : Tire new number formed = 6034579

Question 18.

The smallest four-digit number having three different digits is

(iv): The smallest 4-digit number with three different digits is 1002.

Question 19.

The number of all the whole numbers between 38 and 68 is

(iii): There are 29 whole numbers between 38 and 68.

Question 20.

The product of the successor and the predecessor of 999 is

(ii) 998000

(iii) 989000

(ii) : Successor of the number 999 = 999 + 1 = 1000

Predecessor of the number 999 = 999 – 1 = 998

Hence, their product = 998 1000 = 998000

Question 21.

Write in expanded form :

(ii) 574021

(iii) 8907010

(i) 74836 is equal to = 7 × 10000 + 4 × 1000 + 8 × 100 + 3 × 10 + 6 × 1

(ii) 574021 is equal to = 5 × 100000 + 7 × 10000 + 4 x 1000 + 0 × 100 + 2 × 10 + 1 × 1

(iii) 8907010 is equal to = 8 × 1000000 + 9 × 100000 + 0 × 10000 + 7 × 1000 + 0 × 100 + 1 × 10 + 0 × 1

Question 22.

A book exhibition was held for 4 days in a school. The number of the tickets sold on the counter on the first, second, third, and the final day was – 1094, 1812, 2050, and 2751. Find the total number of tickets that sold on all four days.

Number of the tickets sold on the first day = 1094

Number of the tickets sold on the second day = 1812

Number of the tickets sold on the third day = 2050

Number of the tickets sold on the final day = 2751

∴Total number of the tickets sold on all of these four days = 1094 + 1812 + 2050 + 2751 = 7,707.

Question 23.

Shekhar is a famous cricket player. He has so far scored a total of 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Shekhar has so far scored a total of 6980 runs

He wishes to complete a total of 10,000 runs.

Therefore total number of the runs needed by him are = 10,000 – 6980 = 3020 runs

Question 24.

Which of the following given statements is not true?

(i) Both the addition and multiplication are associative for whole numbers.

(ii) Zero is the identity for the multiplication of whole numbers.

(iii) Addition and multiplication are commutative for whole numbers.

(iv) Multiplication is distributive over addition for whole numbers.

(ii): Zero is the identity for the addition of whole numbers.

Question 25.

(i) 0 + 0 = 0

(ii) 0 – 0 = 0

(iii) 0 × 0 = 0

(iv) 0 – 0 = 0

(iv) : 0 + 0 is not defined.

Question 26.

The predecessor of 1 lakh is

(iii) 999999

(iv) 100001

(ii) : 1 lakh = 100000

∴ Predecessor of 100000 = 100000 – 1 = 99999

Question 27.

The successor of 1 million is

(i) Two million

(ii) 1000001

(iii) 100001

(ii) : 1 million = 1000000

∴ Successor of 1000000 = 1000000 + 1 = 1000001

Question 28.

The number of all the even numbers between 58 and 80 is

(i) : Even numbers between the numbers 58 and 80 are 60, 62, 64, 66, 68, 70, 72, 74, 76, 78.

So, these are ten even numbers between 58 and 80.

Question 29.

The Sum of the number of primes numbers between 16 to 80 and between 90 to 100 is

(iii) : Prime numbers between 16 to 80 are – 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79.

So, there are total 16 prime numbers between 16 to 80.

Also, 97 is the only one prime number between 90 to 100.

So, there is only one prime number between 90 to 100.

∴ Required sum = 16 + 1 = 17

Question 30.

(i) The HCF of the two distinct prime numbers is 1

(ii) The HCF of two coprime numbers is 1

(iii) The HCF of two consecutive even numbers is 2

(iv) The HCF of an even number and an odd number is always even

(iv): The HCF of an even and an odd number is always said to be an odd number.

Question 31.

In an election, the successful candidate was registered 5,77,500 votes, and his nearest rival had secured 3,48,700 votes. By what total margin did the successful candidate win the election?

Number of the votes secured by the successful candidate = 5,77,500

Number of the votes secured by his nearest rival = 3,48,700

Therefore, margin of the votes is necessary to win the election = 5,77,500 – 3,48,700 = 2,28,800

Question 32.

Kirti bookstore sold books worth a total of ₹2,85,891 in the 1st week of June and books  worth a total of ₹4,00,768 in the second week of the month. How much was the total sale for the two weeks together? And in which week was the total sale greater and by how much?

Books sold in the first week of the month June are worth ₹2,85,891

Books sold in the second week of the month are worth ₹4,00,768

Therefore, the total sale of the books in the two weeks together is

= ₹2,85,891 + ₹4,00,768 = ₹6,86,659

In the 2nd week of the month, the sale of total books was greater.

Therefore, the difference in the sale of books

= ₹4,00,768 – ₹2,85,891 = ₹1,14,877

So, in the second week of June, the total sale of books was more than ₹1,14,877.

Question 33.

Find the difference below between the highest and the lowest numbers that is written using the digits 6, 2, 7, 4, and 3, each only once.

Given digits = 6, 2, 7, 4, 3

Greatest number is = 76432

Least number is = 23467

Therefore, difference = 76432 – 23467 = 52,965

Question 34.

A machine, on average, manufactures 2,825 screws a day. How many screws did it produce in January 2006?

The number of screws that are manufactured in a day = 2,825.

Number of screws that are manufactured in month of January = 31 x 2825 = 87,575

Question 35.

The total distance between the school and the house of a student is 1 km and 875 m. Every day she walks both the ways. Find the total distance she covered in six days.

Distance between the school and her house = 1 km 875 m = (1000 + 875) m = 1875 metre.

Total Distance travelled by the student from school to home and from home to school is = 2 x 1875 = 3750 m

Distance travelled by the student in six days is = 3750 m x 6 – 22500 m = 22 km 500 m.

Therefore, the total distance covered in 6 days = 22 km 500 m.

Question 36.

A merchant had ₹78,592 with her. She placed an order to purchase 40 radio sets at ₹1200 each. How much total money will remain with her after the purchase?

Amount of money present with the merchant = ₹78,592

Total Number of the radio sets = 40

Price of one of the radio set = ₹1200

Therefore, the cost of 40 radio sets = ₹1200 x 40 = ₹48,000

Remaining money left with the merchant = ₹78,592 – ₹48000 = ₹30,592

Hence, the amount of ₹30,592 will remain with her after purchasing the following radio sets.

Question 37.

A vessel has four litres and 500 ml of curd. How many total glasses, each of 25 mL capacity, can be filled?

Quantity of the curd in a vessel = 4 l 500 mL = (4 x 1000 + 500) mL = 4500 mL.

Capacity of 1 glass = 25 mL

Therefore the number of glasses = 4500/25

Question 38.

A student has multiplied the number 7236 by 65 instead of multiplying by 56. Calculate by how much was his answer greater than the right answer?

The student had multiplied the number 7236 by 65 instead of multiplying by 56.

The difference between the two above multiplications is = (65 – 56) x 7236 = 9 x 7236 = 65124

(We don’t have to do both the multiplication)

Hence, the answer that is greater than the correct answer is 65,124.

Question 39.

Estimate each of the following given numbers using the general rule:

(i) 730 + 998

(ii) 796 – 314

(iii) 12,904 + 2,888

(iv) 28,292 – 21,496

Rounding off 730 nearest to hundreds = 700

Rounding off 998 nearest to hundreds = 1,000

∴ 730 + 998 = 700 + 1000 = 1700

Rounding off 796 nearest to hundreds = 800

Rounding off 314 nearest to hundreds = 300

∴ 796 – 314 = 800 – 300 = 500

Rounding off 12,904 nearest to thousands = 13000

Rounding off 2888 nearest to thousands = 3000

∴ 12,904 + 2,888 = 13000 + 3000 = 16000

Rounding off 28,292 nearest to thousands = 28,000

Rounding off 21,496 nearest to thousands = 21,000

∴ 28,292 – 21,496 = 28,000 – 21,000 = 7,000

Question 40.

Estimate the following given products using the general rule:

(i) 578 x 161

(ii)5281 x 3491

(iii) 1291 x 592

(iv) 9250 x 29

(i) 578 x 161 = 600 x 200 = 1,20,000

(ii) 5281 x 3491 = 5000 x 3000 = 1,50,00,000

(iii) 1291 x 592 = 1300 x 600 = 7,80,000

(iv) 9250 x 29 = 9000 x 30 = 2,70,000

Question 41.

Which of the following is not true?

(i) (7 + 8) + 9 = 7 + (8 + 9)

(ii) (7 × 8) × 9 = 7 × (8 × 9)

(iii) 7 + 8 × 9 = (7 + 8) × (7 + 9)

(iv) 7 × (8 + 9) = (7 × 8) + (7 × 9)

(iii) : 7 + 8 × 9 = 7 + 72 = 79,

(7 + 8) × (7 + 9) = 15 × 16 = 240

and 79 ≠ 240

Question 42.

The length of the river ‘Narmada’ is 1290 km. Its length in metres is – _____.

As, 1290 km = (1290 × 1000) m = 1290000 m

Question 43.

The total distance between Srinagar and Leh is 422 km. The same distance in metres is – _____.

As, 422 km= (422 × 1000) m = 422000 m

Question 44.

Writing numbers from the greatest to the smallest is called an arrangement in ___ order.

Question 45.

By reversing the order of the digits of the greatest number made by the five different non-zero digits, we get the new number which is the number of _____ five digits.

By reversing the order of the digits of the greatest number made by the five different non-zero digits, the new number present is the smallest number of these digits.

Question 46.

By adding 1 to the greatest ___ digit number, we get the number ten lakh.

As, greatest six-digit number = 999999

By adding one to 999999, we get 1000000.

Question 47.

The number five crore twenty-three lakh seventy-eight thousand four hundred one can also be written, using the commas, in the Indian System of Numeration as.

5, 23, 78, 401

Question 48.

In the Roman Numeration, the symbol X can be subtracted from – ___, M and C only.

Question 49.

The number 66 in Roman numerals is.

LXVI : 66 = LXVI

Question 50.

The total population of Pune was 2,538,473 in 2001. Rounded off to the nearest thousands, the population was ___.

Question 51.

The smallest whole number is ___.

0 : 0 is the smallest whole number.

Question 52.

The successor of number 106159 is ___.

As, Successor of 106159 is 106159 + 1, i.e., 106160

Question 53.

400 is the predecessor of the number ___.

As, 400 is the predecessor of 400 + 1, i.e., 401

Question 54.

___ is the successor of the largest three digit number.

As, Largest three digit number = 999

And the successor of 999 is 999 + 1, i.e., 1000

Question 55.

If the number 7254*98 is to be divisible by the number 22, then the digit at * is

(iii) : 7254 * 98 is divisible by the number 22 only if it is divisible by both 2 and 11.

Given that the number is even. Therefore it is divisible by the number 2.

7254 * 98 is divisible by 11, only if

(7 + 5 + * + 8) – (2 + 4 + 9) or (20 + *) – 15 or 5 + * is also divisible by 11.

∴ The digit at * place should be filled by 6.

Question 56.

The largest number which will always divide the Sum of any pair of consecutive odd numbers is

(ii)The Sum of any pair of the consecutive odd numbers results in the form of a multiple of 4.

∴ The required largest number is 4.

Question 57.

A number is divisible by five and six. It may not be divisible by

(iv): The Least Common Multiple also known as LCM of 5 and 6 is 30.

And also 30 is divisible by the numbers 10, 15 and 30 but not by the number  60.

Question 58.

The greatest number which will always divide the product of the predecessor and successor of an odd natural number other than 1, is

(ii): As the odd natural numbers other than 1 are – 3, 5, 7, 9 and so on.

Now, we know that the predecessor and successor of 3 are – 2 and 4 respectively, and their product is two × four = 8

Similarly, we know that the predecessor and the successor of 5 are – 4 and 6, respectively, and their product is four × 6 = 24.

Thus, the above shows that the greatest number which always divides the product of the predecessor and the successor of an odd natural number other than 1 is 4.

Question 59.

A person had only ₹ 1000000 with him. He purchased a coloured-T.V. for ₹ 16580, a motorcycle for ₹ 45890 and a flat for ₹ 870000. How much money was left with him?

The total amount a person had was = ₹ 1000000

The total amount he spent on a colour T.V. was = ₹ 16580

The amount he spent on a motorcycle was = ₹ 45890

The amount he spent on a flat was = ₹ 870000

∴ Total amount he spent is = ₹ (16580 + 45890 + 870000) = ₹ 932470

Thus, the total amount left with him = ₹ 1000000 – ₹ 932470 = ₹ 67530

Question 60.

Out of 180000 tablets of Vitamin A, a total of 18734 are distributed among the students in the district. Find the total number of remaining vitamin tablets.

Total tablets of Vitamin A are = 180000

Total number of tablets distributed among the students in the district = 18734

∴ The number of total remaining vitamin tablets = 180000 – 18734 = 161266

Question 61.

Chinmay only had ₹ 610000. He gave a total of ₹ 87500 to Jyoti, ₹ 126380 to Javed and ₹ 350000 to John. How much money was left with him?

Chinmay had a total amount = of ₹ 610000

The total amount he gave to Jyoti = ₹ 87500

The total amount he gave to Javed = ₹ 126380

The total amount he gave to John = ₹ 350009

Total amount given by Chinmay is = ₹ (87500 + 126380 + 350000) = ₹ 563880

Thus, the amount left with him

= ₹ 610000 – ₹ 563880 = ₹ 46120

Benefits of Solving Class 6 Maths Chapter 1 Extra Questions 

Practise is very important, and it helps students in several ways. Our experts have made the question series to help students in practise. Thus, the questions will help students in several ways, and they will be worthy of their time. The benefits of solving the Important Questions Class 6 Maths Chapter 1 are as follows-

• The experts have collated the questions from different sources. They have taken help from the textbook exercises, CBSE sample papers, CBSE past years’ question papers, NCERT Exemplars and important reference books. Thus, students don’t have to search for questions in different sources: they will find them in a single pdf. Thus, the Class 6 Maths Chapter 1 Important Questions will help them in practise and boost their confidence.
• The experts have not only collected the questions, but they have also provided the solutions. Thus, students can follow the solution if they cannot solve the questions. Also, they can check their answers with the provided explanations. Thus, the Maths Class 6 Chapter 1 Important Questions will help students boost their confidence and improve their exam preparation.
• Many students tend to fear maths because they don’t understand the subject. Their doubts must be cleared, and practice can help boost confidence. They must build the habit of solving questions to build interest in the subject matter. The experts have done a good job of collating the questions for students. They can solve these questions regularly, which will help them improve their knowledge. Thus, the Chapter 1 Class 6 Maths Important Questions will improve their exam preparation.

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Q.1 Which of the following is the representation of number 74 according to roman numerals?

(b). XXXXXXXIV

(d). DCCXLV

Ans (a). LXXIV

Q.2 What is the greatest 7 digit number formed by using the digits 4 , 9 , 1 and 6? Note that each digit should be used at least once.

(a). 99,99,641

(c). 99,66,441

(d). 11,11,469

Given digits:

9 > 6 > 4 > 1

The greatest 7 digit number using the digits 4, 9, 1 and 6 is 99,99,641.

Q.3 Which one of the following is the estimated product of 47 and 215?

Rounding off 215 to the nearest hundreds, we get 200.

Rounding off 47 to nearest tens, we get 50.

Estimated product

Thus, 10,000 is the estimated product of 47 and 215.

Q.4 Write 645340001 using comma in International System of Numeration.

645,340,001

Q.5 a) How many thousands make a million? b) How many lakhs make a crore?

a) 1000 thousands make 1 million. (? 1 million = 1,000,000 = 1000 thousands) b) 100 lakhs make a crore. (? 1 crore = 1,00,00,000 = 100 lakhs)

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Cbse important questions for class 6 maths, chapter 2 - whole numbers.

Chapter 3 - Playing with Numbers

Chapter 4 - basic geometrical ideas, chapter 5 - understanding elementary shapes, chapter 6 - integers, chapter 7 - fractions, chapter 8 - decimals, chapter 9 - data handling, chapter 10 - mensuration, chapter 11 - algebra, chapter 12 - ratio and proportion, chapter 13 - symmetry, chapter 14 - practical geometry, faqs (frequently asked questions), 1. is class 6 maths chapter 1 easy.

The first chapter of Class 6 Maths provides a few basic ideas related to numbers. They will learn how to express bigger numbers, such as in thousands, lakhs, crores, etc. They will also learn how to use commas to write larger number, add or subtract large numbers, etc. This is an easy chapter because most students have ideas regarding lakhs, crores, or other units of numbers. Thus, students won’t have problems understanding the subject matter if they follow the textbook closely. Students can take help from the Important Questions Class 6 Maths Chapter 1 prepared by the experts of Extramarks, and they will find a wide variety of questions to solve.

2. How can the question series help students?

Practice is very important for getting better marks in exams. Sometimes, more than the textbook exercises are needed, and students should get help from other sources. The experts of Extramarks have made the question series with help from different sources. They have collated the questions from the textbook exercise, CBSE sample papers, CBSE past years’ question papers, and important reference books. They have solved the questions, and experienced professionals have further checked the answers to ensure the best quality of the content. Thus, the Important Questions Class 6 Maths Chapter 1 will help students score better in exams. It will also help boost their confidence and incline interest in the subject matter.

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NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. It helps the students to understand slowly and to get practice well to become perfect and again a good score in their examination.

These materials are prepared based on Class 6 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 6 Solutions Maths Chapter 1 Knowing Our Numbers are in accordance with the latest CBSE guidelines and marking schemes

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NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers PDF Download

The NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers provides detailed and explained Solutions for all Questions in the textbook. By referring to the Chapter 1 Knowing Our Numbers answers, students can understand each and every step; accordingly can form their own strategies to solve Questions in a proper and systematic way during further preparations. 

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers PDF

The NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers PDF prove to be worthy as it provides a lot of Questions to solve; accordingly they can score well. Students can access the Chapter 1 Knowing Our Numbers Questions of Class 6 Maths through the Selfstudys website from their comfort zone. In the process of solving Chapter 1 Knowing Our Numbers Questions, students can build a strong and solid foundation for the chapter. 

Where can Students Find the NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers?

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• Visit the Selfstudys website. 

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Characteristics of NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers PDF

The characteristics of NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers PDF is the distinctive quality which each and every student needs to know; some of the important features are discussed below: 

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• RD Sharma Solutions
• Chapter 1 Knowing Our Numbers

RD Sharma Solutions for Class 6 Maths Chapter 1: Knowing Our Numbers

The subject experts present the concepts in a clear and precise manner based on the IQ level of students. In previous years, students have learned about basic addition, subtraction, multiplication and division. By solving this Chapter, the students will understand the sequence of numbers and interesting problems based on it.

Here, the students will learn two methods of expressing numbers in words and digits, i.e., the Indian System and International System. The solutions PDF helps students to solve exercise-wise problems on a daily basis. RD Sharma Solutions for Class 6 Chapter 1 Knowing Our Numbers are provided here.

• RD Sharma Solutions Class 6 Maths Chapter 1 Knowing Our Numbers
• RD Sharma Solutions Class 6 Maths Chapter 2 Playing with Numbers
• RD Sharma Solutions Class 6 Maths Chapter 3 Whole Numbers
• RD Sharma Solutions Class 6 Maths Chapter 4 Operations on Whole Numbers
• RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers
• RD Sharma Solutions Class 6 Maths Chapter 6 Fractions
• RD Sharma Solutions Class 6 Maths Chapter 7 Decimals
• RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra
• RD Sharma Solutions Class 6 Maths Chapter 9 Ratio, Proportion and Unitary Method
• RD Sharma Solutions Class 6 Maths Chapter 10 Basic Geometrical Concepts
• RD Sharma Solutions Class 6 Maths Chapter 11 Angles
• RD Sharma Solutions Class 6 Maths Chapter 12 Triangles
• RD Sharma Solutions Class 6 Maths Chapter 13 Quadrilaterals
• RD Sharma Solutions Class 6 Maths Chapter 14 Circles
• RD Sharma Solutions Class 6 Maths Chapter 15 Pair of Lines and Transversal
• RD Sharma Solutions Class 6 Maths Chapter 16 Understanding Three Dimensional Shapes
• RD Sharma Solutions Class 6 Maths Chapter 17 Symmetry
• RD Sharma Solutions Class 6 Maths Chapter 18 Basic Geometrical Tools
• RD Sharma Solutions Class 6 Maths Chapter 19 Geometrical Constructions
• RD Sharma Solutions Class 6 Maths Chapter 20 Mensuration
• RD Sharma Solutions Class 6 Maths Chapter 21 Data Handling – I (Presentation of Data)
• RD Sharma Solutions Class 6 Maths Chapter 22 Data Handling – II (Pictographs)
• RD Sharma Solutions Class 6 Maths Chapter 23 Data Handling – III (Bar Graphs)
• Exercise 1.1 Chapter 1 Knowing Our Numbers
• Exercise 1.2 Chapter 1 Knowing Our Numbers
• Exercise 1.3 Chapter 1 Knowing Our Numbers
• Exercise 1.4 Chapter 1 Knowing Our Numbers
• Exercise 1.5 Chapter 1 Knowing Our Numbers
• Exercise 1.6 Chapter 1 Knowing Our Numbers
• Exercise 1.7 Chapter 1 Knowing Our Numbers
• Exercise 1.8 Chapter 1 Knowing Our Numbers
• Exercise 1.9 Chapter 1 Knowing Our Numbers
• Exercise 1.10 Chapter 1 Knowing Our Numbers
• Objective Type Questions Chapter 1 Knowing Our Numbers

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Access answers to Maths RD Sharma Solutions for Class 6 Chapter 1: Knowing Our Numbers

Exercise 1.1 page: 1.7.

1. Write each of the following in numeral form:

(i) Eight thousand twelve.

(ii) Seventy thousand fifty three.

(iii) Five lakh seven thousand four hundred six.

(iv) Six lakh two thousand nine.

(v) Thirty lakh eleven thousand one.

(vi) Eight crore four lakh twenty five.

(vii) Three crore three lakh three thousand three hundred three.

(viii) Seventeen crore sixty lakh thirty thousand fifty seven.

(i) The numeral form of eight thousand twelve is 8,012.

(ii) The numeral form of seventy thousand fifty three is 70,053.

(iii) The numeral form of five lakh seven thousand four hundred six is 5, 07, 406.

(iv) The numeral form of six lakh two thousand nine is 6, 02, 009.

(v) The numeral form of thirty lakh eleven thousand one is 30, 11, 001.

(vi) The numeral form of eight crore four lakh twenty five is 8, 04, 00, 025.

(vii) The numeral form of three crore three lakh three thousand three hundred three is 3, 03, 03, 303.

(viii) The numeral form of seventeen crore sixty lakh thirty thousand fifty seven is 17, 60, 30, 057.

2. Write the following numbers in words in the Indian system of numeration:

(ii) 4,05,045

(iii) 35, 42, 012

(iv) 7, 06, 04, 014

(v) 25, 05, 05,500

(vi) 5, 50, 50, 050

(vii) 5, 03, 04, 012

(i) 42,007 is written as forty two thousand seven in the Indian system of numeration.

(ii) 4,05,045 is written as four lakh five thousand forty five in the Indian system of numeration.

(iii) 35, 42, 012 is written as thirty five lakh forty two thousand twelve in the Indian system of numeration.

(iv) 7, 06, 04, 014 is written as seven crore six lakh four thousand fourteen in the Indian system of numeration.

(v) 25, 05, 05,500 is written as twenty five crore five lakh five thousand five hundred in the Indian system of numeration.

(vi) 5, 50, 50, 050 is written as five crore fifty lakh fifty thousand fifty in the Indian system of numeration.

(vii) 5, 03, 04, 012 is written as five crore three lakh four thousand twelve in the Indian system of numeration.

3. Insert commas in the correct positions to separate periods and write the following numbers in words:

(iii) 857367

(iv) 9050784

(v) 10105607

(vi) 10000007

(vii) 910107104

(i) 4375 by inserting commas is written as 4, 375.

(ii) 24798 by inserting commas is written as 24, 798.

(iii) 857367 by inserting commas is written as 8,57,367.

(iv) 9050784 by inserting commas is written as 90,50,784.

(v) 10105607 by inserting commas is written as 1,01,05,607.

(vi) 10000007 by inserting commas is written as 1,00,00,007.

(vii) 910107104 by inserting commas is written as 91,01,07,104.

4. Write each of the following in expanded notation:

(iii) 10205

(iv) 235060

(i) 3057 = 3 × 1000 + 0 × 100 + 5 × 10 + 7 × 1

(ii) 12345 = 1 × 10000 + 2 × 1000 + 3 × 100 + 4 × 10 + 5 × 1

(iii) 10205 = 1 × 10000 + 0 × 1000 + 2 × 100 + 0 × 10 + 5 × 1

(iv) 235060 = 2 × 100000 + 3 × 10000 + 5 × 1000 + 0 × 100 + 6 × 10 + 0 × 1

5. Write the corresponding numeral for each of the following:

(i) 7 × 1000 + 2 × 1000 + 5 × 100 + 9 × 10 + 6 × 1

(ii) 4 × 100000 + 5 × 1000 + 1 × 100 + 7 × 1

(iii) 8 × 1000000 + 3 × 1000 + 6 × 1

(iv) 5 × 10000000 + 7 × 1000000 + 8 × 1000 + 9 × 10 + 4

(i) 7 × 1000 + 2 × 1000 + 5 × 100 + 9 × 10 + 6 × 1 = 72, 596.

(ii) 4 × 100000 + 5 × 1000 + 1 × 100 + 7 × 1 = 4, 05, 107.

(iii) 8 × 1000000 + 3 × 1000 + 6 × 1 = 80, 03, 006.

(iv) 5 × 10000000 + 7 × 1000000 + 8 × 1000 + 9 × 10 + 4 = 5, 70, 08, 094.

6. Find the place value of the digit 4 in each of the following:

(i) 74983160

(ii) 8745836

(i) The place value of digit 4 in 74983160 is 4 × 10, 00, 000 = 40, 00, 000.

(ii) The place value of digit 4 in 8745836 is 4 × 10, 000 = 40, 000.

7. Determine the product of the place values of two fives in 450758.

The place value of first five is 5 × 10 = 50

The place value of second five is 5 × 10, 000 = 50, 000

So the required product = 50 × 50, 000 = 25, 00, 000.

Therefore, the product of the place values of two fives in 450758 is 25, 00, 000.

8. Determine the difference of the place values of two 7’s in 257839705.

The place value of first seven is 7 × 100 = 700

The place value of second seven is 7 × 10, 00, 000 = 70, 00, 000

So the required difference = 70, 00, 000 – 700 = 69, 99, 300

Therefore, the difference of the place values of two 7’s in 257839705 is 69, 99, 300.

9. Determine the difference between the place value and the face value of 5 in 78654321.

The given number is 78654321

The place value of five = 5 × 10, 000 = 50, 000

The face value of five = 5

So the required difference = 50, 000 – 5 = 49, 995

Therefore, the difference between the place value and the face value of 5 in 78654321 is 49, 995.

10. Which digits have the same face value and place value in 92078634?

We know that the face value depends on the value of the digit and the place value depends on its place of occurrence.

So the digits which have same place and face value in a number are the ones digit and zeros of the number.

The given number is 9, 20, 78, 634

We know that 4 which is the ones digit and 0 which is the lakhs digit have same face and place value.

11. How many different 3-digit numbers can be formed by using the digits 0, 2, 5 without repeating any digit in the number?

The different 3-digit numbers which can be formed by using the digits 0, 2, 5 without repeating any digit in the number are 205, 250, 502 and 520.

Therefore, four 3 digit numbers can be formed by using the digits 0, 2, 5.

12. Write all possible 3-digit numbers using the digits 6, 0, 4 when

(i) repetition of digits is not allowed

(ii) repetition of digits is allowed.

(i) The possible 3 digit numbers using the digits 6, 0, 4 when repetition of digits is not allowed are 604, 640, 406 and 460.

(ii) The possible 3 digit numbers using the digits 6, 0, 4 when repetition of digits is allowed are 400, 406, 460, 466, 444, 404, 440, 446, 464, 600, 604, 640, 644, 646, 664, 606, 660 and 666.

13. Fill in the blank:

(i) 1 lakh = ………. Ten thousand

(ii) 1 lakh = ……… Thousand

(iii) 1 lakh = ……… Hundred

(iv) 1 lakh = ………. ten

(v) 1 crore = ………. ten lakh

(vi) 1 crore = ……… lakh

(vii) 1 crore = ……… Ten thousand

(viii) 1 crore = ……… thousand

(ix) 1 crore = ……….. Hundred

(x) 1 crore = …………. Ten

(i) 1 lakh = 10 Ten thousand

(ii) 1 lakh = 100 Thousand

(iii) 1 lakh = 1000 Hundred

(iv) 1 lakh = 10000 Ten

(v) 1 crore = 10 ten lakh

(vi) 1 crore = 100 lakh

(vii) 1 crore = 1000 Ten thousand

(viii) 1 crore = 10000 thousand

(ix) 1 crore = 100000 Hundred

(x) 1 crore = 1000000 Ten

Exercise 1.2 page: 1.13

1. Write each of the following numbers in digits by using international place value chart. Also, write them in expanded form

(i) Seven million three hundred three thousand two hundred six.

(ii) Fifty five million twenty nine thousand seven.

(iii) Six billion one hundred ten million three thousand seven.

(i) Seven million three hundred three thousand two hundred six is written as 7, 303, 206 using international place value chart.

It can be written in expanded form as

7, 303, 206 = 7 × 1000000 + 3 × 100000 + 0 × 10000 + 3 × 1000 + 2 × 100 + 0 × 10 + 6 × 1

(ii) Fifty five million twenty nine thousand seven is written as 55, 029, 007 using international place value chart.

55, 029, 007 = 5 × 10000000 + 5 × 1000000 + 0 × 100000 + 2 × 10000 + 9 × 1000 + 0 × 100 + 0 × 10 + 7 × 1

(iii) Six billion one hundred ten million three thousand seven is written as 6, 110, 003, 007 using international place value chart.

6, 110, 003, 007 = 6 × 1000000000 + 1 × 100000000 + 1 × 10000000 + 0 × 1000000 + 0 × 100000 + 0 × 10000 + 3 × 1000 + 0 × 100 + 0 × 10 + 7 × 1

2. Rewrite each of the following numerals with proper commas in the international system of numeration. Also, write the number name of each in the international system of numeration

(ii) 4035672

(iii) 65954923

(iv) 70902005

(i) 513625 is written as 513, 625 which is five hundred thirteen thousand six hundred twenty five in the international system of numeration.

(ii) 4035672 is written as 4, 035, 672 which is four million thirty five thousand six hundred seventy two in the international system of numeration.

(iii) 65954923 is written as 65, 954, 923 which is sixty five million nine hundred fifty four thousand nine hundred twenty three in the international system of numeration.

(iv) 70902005 is written as 70, 902, 005 which is seventy million nine hundred two thousand five in the international system of numeration.

3. Write each of the following numbers in the International system of numeration:

(i) Forty three lakh four thousand eighty four.

(ii) Six crore thirty four lakh four thousand forty four.

(iii) Seven lakh thirty five thousand eight hundred ninety nine only.

(i) Forty three lakh four thousand eighty four can be written as 4, 304, 084 which is four million three hundred four thousand eighty four using International system of numeration.

(ii) Six crore thirty four lakh four thousand forty four can be written as 63, 404, 044 which is sixty three million four hundred four thousand forty four using International system of numeration.

(iii) Seven lakh thirty five thousand eight hundred ninety nine only can be written as 735, 899 which is seven hundred thirty five thousand eight hundred ninety nine only using International system of numeration.

4. Write the following numbers in the Indian system of numeration:

(i) Six million five hundred forty three thousand two hundred ten.

(ii) Seventy six million eighty five thousand nine hundred eighty seven.

(iii) Three hundred twenty five million four hundred seventy nine thousand eight hundred thirty eight.

(i) Six million five hundred forty three thousand two hundred ten can be written as 65, 43, 210 which is sixty five lakh forty three thousand two hundred ten using the Indian system of numeration.

(ii) Seventy six million eighty five thousand nine hundred eighty seven can be written as 7, 60, 85, 987 which is seven crore sixty lakh eight five thousand nine hundred eighty seven using the Indian system of numeration.

(iii) Three hundred twenty five million four hundred seventy nine thousand eight hundred thirty eight can be written as 32, 54, 79, 838 which is thirty two crore fifty four lakh seventy nine thousand eight hundred thirty eight using the Indian system of numeration.

5. A certain nine digit number has only ones in ones period, only twos in the thousands period and only threes in millions period. Write this number in words in the Indian system.

It is given that a nine digit number has only ones in ones period, only twos in the thousands period and only threes in millions period.

So we get the number as 333, 222, 111

We can write it as 33, 32, 22, 111 which is thirty three crore thirty two lakh twenty two thousand one hundred and eleven using the Indian system.

6. How many thousands make a million?

We know that

One thousand = 1000 and One million = 1, 000, 000

Number of thousands which make a million = One million/ One thousand

By substituting the values

Number of thousands which make a million = 1, 000, 000/ 1000

Number of thousands which make a million = 1000

Therefore, 1000 thousands make a million.

7. How many millions make a billion?

One million = 1, 000, 000 and One billion = 1, 000, 000, 000

Number of millions which make a billion = One billion/ One million

Number of millions which make a billion = 1, 000, 000, 000/ 1, 000, 000

Number of millions which make a billion = 1000

Therefore, 1000 millions make a billion.

8. (i) How many lakhs make a million?

(ii) How many lakhs make a billion?

(i) We know that

One lakh = 1, 00, 000 and One million = 1, 000, 000

Number of lakhs which make a million = One million/ One lakh

Number of lakhs which make a million = 1, 000, 000/ 1, 00, 000

Number of lakhs which make a million = 10

Therefore, 10 lakhs make a million.

(ii) We know that

One lakh = 1, 00, 000 and One billion = 1, 000, 000, 000

Number of lakhs which make a billion = One billion/ One lakh

Number of lakhs which make a billion = 1, 000, 000, 000/ 1, 00, 000

Number of lakhs which make a billion = 10, 000

Therefore, 10, 000 lakhs make a billion.

9. Write each of the following in numeral form:

(i) Eight million seven hundred eight thousand four.

(ii) Six hundred seven million twelve thousand eighty four.

(iii) Four billion twenty five million forty five thousand.

(i) Eight million seven hundred eight thousand four can be written as

8 × 1, 000, 000 + 7 × 100, 000 + 0 × 10, 000 + 8 × 1000 + 0 × 100 + 0 × 10 + 4 × 1 = 8, 708, 004

(ii) Six hundred seven million twelve thousand eighty four can be written as

6 × 100, 000, 000 + 0 × 10, 000, 000 + 7 × 1, 000, 000 + 0 × 100, 000 + 1 × 10, 000 + 2 × 1000 + 0 × 100 + 8 × 10 + 4 × 1 = 607, 012, 084

(iii) Four billion twenty five million forty five thousand can be written as

4 × 1, 000, 000, 000 + 0 × 100, 000, 000 + 2 × 10, 000, 000 + 5 × 1, 000, 000 + 0 × 100, 000 + 4 × 10, 000 + 5 × 1000 + 0 × 100 + 0 × 10 + 0 × 1 = 4, 025, 045, 000

10. Write the number names of each of the following in international system of numeration:

(i) 435, 002

(ii) 1, 047, 509

(iii) 59, 064, 523

(iv) 25, 201, 905

(i) 435, 002 can be written as four hundred thirty five thousand and two in international system of numeration.

(ii) 1, 047, 509 can be written as one million forty seven thousand five hundred and nine in international system of numeration.

(iii) 59, 064, 523 can be written as fifty nine million sixty four thousand five hundred and twenty three in international system of numeration.

(iv) 25, 201, 905 can be written as twenty five million two hundred one thousand nine hundred and five in international system of numeration.

Exercise 1.3 page: 1.16

1. How many four digit numbers are there in all?

We know that the 10 digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

0 cannot be used in thousands place so only nine digits can be used.

10 digits can be used in hundreds, tens and units place

The number of four digit numbers = 9 × 10 × 10 × 10 = 9000

Therefore, 9000 four digit numbers are there in all.

2. Write the smallest and the largest six digit numbers. How many numbers are between these two.

We know that the smallest digit is 0 which cannot be used in the highest place value.

So 1 which is the second smallest digit can be used in the highest place value

The required smallest six digit number is 100000

We know that the largest digit is 9 which can be used in any place

The required largest six digit number is 999999

So we get the difference = 999999 – 100000 = 899999

Therefore, the smallest six digit number is 100000, the largest six digit number is 999999 and 899999 numbers are between these two numbers.

3. How many 8-digit numbers are there in all?

0 cannot be used in the highest place value and 9 can be used in the highest place value

So the 10 digits can be used in the remaining places of 8 digit numbers

The total number of 8 digit numbers = 9 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 90000000

Therefore, 90000000 eight digit numbers are there in all.

4. Write 10075302 in words and rearrange the digits to get the smallest and the largest numbers.

The given number 10075302 can be written as one crore seventy five thousand three hundred and two.

To get smallest 8 digit number using 0, 1, 2, 3, 5 and 7

We use 1 which is the smallest digit in the highest place and largest digit 7 at the units place

Further we put 5 in the tens place, 3 in the hundreds place and 2 in thousands place

So the required smallest number is 10002357

To get largest 8 digit number using 0, 1, 2, 3, 5 and 7

We use 7 which is the largest digit in the highest place value, 5 in a place after highest place, 3 as the next one, 2 as the smallest digit and then 1.

So the required largest number is 75321000.

5. What is the smallest 3-digit number with unique digits?

102 is the smallest 3-digit number with unique digits.

6. What is the largest 5-digit number with unique digits?

98765 is the largest 5-digit number with unique digits.

7. Write the smallest 3-digit number which does not change if the digits are in reverse order.

101 is the smallest 3-digit number which does not change if the digits are in reverse order.

8. Find the difference between the number 279 and that obtained on reversing its digits.

The reverse of 279 is 972

Difference between both the numbers = 972 – 279 = 693

Therefore, the difference between the number 279 and that obtained on reversing its digits is 693.

9. Form the largest and smallest 4-digit numbers using each of digits 7, 1, 0, 5 only once.

The largest 4 digit number = 7510

Smallest 4 digit number = 1057

Therefore, the largest and smallest 4-digit numbers using each of digits 7, 1, 0, 5 only once is 7510 and 1057.

Exercise 1.4 page: 1.18

1. Put the appropriate symbol (<, >) in each of the following boxes:

(i) 102394 ☐ 99887

(ii) 2507324 ☐ 2517324

(iii) 3572014 ☐ 10253104

(iv) 47983505 ☐ 47894012

(i) 102394 > 99887

(ii) 2507324 < 2517324

(iii) 3572014 < 10253104

(iv) 47983505 > 47894012

2. Arrange the following numbers in ascending order:

(i) 102345694, 8354208, 6539542, 63547201, 12345678

(ii) 1808090, 1808088, 181888, 190909, 16060666.

(i) It can be written as

6539542 < 8354208 < 12345678 < 63547201 < 102345694

(ii) It can be written as

181888 < 190909 < 1808088 < 1808090 < 16060666.

3. Arrange the following numbers in descending order:

(i) 56943300, 56943201, 5695440, 56944000, 5694437

(ii) 1020216, 1020308, 1021430, 893245, 893425.

56944000 > 56943300 > 56943201 > 5695440 > 5694437

1021430 > 1020308 > 1020216 > 893425 > 893245.

Exercise 1.5 page: 1.21

1. How many milligrams make one kilogram?

One million or ten lakh milligrams make one kilogram.

2. A box of medicine tablets contains 2, 00, 000 tablets each weighing 20mg. What is the total weight of all the tablets in the box in grams? In kilograms?

It is given that

Weight of each tablet = 20mg

Weight of 2, 00, 000 tablets = 2, 00, 000 × 20

We get the weight of 2, 00, 000 tablets = 40, 00, 000mg

Total weight of tablets in the box = 40, 00, 000mg

It can be written as

1g = 1000mg

So the weight of box having tablets = 40, 00, 000/ 1000 = 4000g

We know that 1kg = 1000g

So the weight of box having tablets = 4000/ 1000 = 4kg

Therefore, the total weight of all the tablets in the box is 4000g or 4kg.

3. Population of Sundarnagar was 2, 35, 471 in the year 1991. In the year 2001 it was found to have increased by 72, 958. What was the population of the city in 2001?

Population of Sundarnagar in 1991 = 2, 35, 471

Increase in population in 2001 = 72, 958

Population of the city in 2001 = Population of Sundarnagar in 1991 + Increase in population in 2001

Population of the city in 2001 = 2, 35, 471 + 72, 958

By addition

Population of the city in 2001 = 3, 08, 429

Therefore, the population of the city in 2001 is 3, 08, 429.

4. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final days were respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

Number of tickets sold at the counter on the first = 1094

Number of tickets sold at the counter on the second = 1812

Number of tickets sold at the counter on the third = 2050

Number of tickets sold at the counter on the final day = 2751

Total number of tickets sold = 1094 + 1812 + 2050 + 2751 = 7707

Therefore, the total number of tickets sold on all the four days is 7, 707.

5. The town newspaper is published everyday. One copy has 12 pages. Everyday 11,980 copies are printed. How many pages are in all printed everyday? Every month?

1 copy of newspaper contains = 12 pages

So the number of pages in 11, 980 copies = 12 × 11, 980 = 1, 43, 760

Hence, 1, 43, 760 pages are printed everyday

Number of pages printed in a month = 1, 43, 760 × 30 = 43, 12, 800

Therefore, 1, 43, 760 pages are printed every day and 43, 12, 800 pages are printed every month.

6. A machine, on an average, manufactures 2825 screws a day. How many screws did it produce in the month of January 2006?

Number of screws produced per day = 2, 825

So the number of screws produced in the month of January = 2, 825 × 31 = 87, 575

Therefore, the machine produced 87, 575 screws in the month of January 2006.

7. A famous cricket player has so far scored 6978 runs in test matches. He wishes to complete 10, 000 runs. How many more runs does he need?

Runs scored by a famous cricket player = 6978

So the runs required to complete 10, 000 runs = 10, 000 – 6978 = 3022

Therefore, 3, 022 runs are needed by the cricket player to complete 10, 000 runs.

8. Ravish has ₹ 78, 592 with him. He placed an order for purchasing 39 radio sets at ₹ 1234 each. How much money will remain with him after the purchase?

Amount with Ravish = ₹ 78, 592

Cost of 1 radio set = ₹ 1234

Number of radio sets purchased = 39

Amount spent to purchase 39 radio sets = ₹ (1234 × 39) = ₹ 48126

Amount remaining with Ravish = Amount with Ravish – Amount spent to purchase 39 radio sets

Amount remaining with Ravish = ₹ 78, 592 – ₹ 48126 = ₹ 30466

Therefore, ₹ 30466 money will remain with Ravish after the purchase.

9. In an election, the successful candidate registered 5, 77, 570 votes and his nearest rival secured 3, 48, 685 votes. By what margin did the successful candidate win the election?

Number of votes registered by successful candidate = 5, 77, 570

Number of votes secured by nearest rival = 3, 48, 685

Margin of victory the successful candidate obtain to win the election = Number of votes registered by successful candidate – Number of votes secured by nearest rival

Margin of victory the successful candidate obtain to win the election = 5, 77, 570 – 3, 48, 685 = 2, 28, 885

Therefore, the margin of victory for the successful candidate is 2, 28, 885.

10. To stitch a shirt 2m 15cm cloth is needed. Out of 40m cloth, how many shirts can be stitched and how much cloth will remain?

Length of cloth = 40m

Length of cloth required to stitch a shirt = 2m 15cm = 200 + 15 = 215cm

Number of shirts that can be stitched using 40m cloth = 4000/ 215 = 18.60

We know that the number of shirts should be a whole number, 18 shirts can be stitched.

Length of cloth required to stitch 18 shirts = 215 × 18 = 3870cm

So the remaining cloth = 4000 – 3870 = 130cm = 1.3m = 1m 30cm

Therefore, 18 shirts can be stitched and 1m 30cm cloth will remain.

11. A vessel has 4 litre and 650ml of curd. In how many glasses, each of 25ml capacity, can it be distributed?

Total amount of curd = 4 litre 650ml

We know that 1 litre = 1000ml

Total amount of curd = 4650ml

Capacity of each glass = 25ml

Number of glasses of curd which can be distributed = Total amount of curd/ Capacity of each glass

Number of glasses of curd which can be distributed = 4650/ 25 = 186

Therefore, 186 glasses each of 25ml capacity of curd can be distributed.

12. Medicine is packed in boxes, each such box weighing 4kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800kg?

Capacity of van carrying medicine boxes = 800kg

Capacity of van carrying medicine boxes = 8, 00, 000g

Weight of each box = 4kg 500g = 4500g

Total number of boxes that can be loaded in a van = Capacity of van carrying medicine boxes/ Weight of each box

Total number of boxes that can be loaded in a van = 8, 00, 000/ 4500 = 177.77

Number of boxes should be a whole number

Weight of 177 boxes = 177 × 4500 = 7, 96, 500g which is under the permissible limit

Weight of 178 boxes = 178 × 4500 = 8, 01, 000g which is beyond the permissible limit

Therefore, 177 boxes can be loaded in a van which cannot carry beyond 800kg.

13. The distance between the school and the house of a student is 1km 875m. Everyday she walks both ways between her school and home. Find the total distance covered by her in a week.

Distance between school and house = 1km 875m = 1875m

Distance covered by a student per day = 2 × 1875 = 3750m

Total distance covered by the student in a week = 7 × 3750 = 26250m = 26km 250m

Therefore, the total distance covered by her in a week is 26km 250m.

Exercise 1.6 page: 1.26

1. Round off each of the following numbers to nearest tens:

(vi) 12, 096

(vii) 10, 908

(viii) 28, 925

(i) 84 is rounded off to the nearest tens as 80.

(ii) 98 is rounded off to the nearest tens as 100.

(iii) 984 is rounded off to the nearest tens as 980.

(iv) 808 is rounded off to the nearest tens as 810.

(v) 998 is rounded off to the nearest tens as 1000.

(vi) 12, 096 is rounded off to the nearest tens as 12, 100.

(vii) 10, 908 is rounded off to the nearest tens as 10, 910.

(viii) 28, 925 is rounded off to the nearest tens as 28, 930.

2. Round off each of the following numbers to nearest hundreds:

(ii) 7, 289

(iii) 8, 074

(iv) 14, 627

(v) 28, 826

(vi) 4, 20, 387

(vii) 43, 68, 973

(viii) 7, 42, 898

(i) 3, 985 is rounded off to the nearest hundreds as 4, 000.

(ii) 7, 289 is rounded off to the nearest hundreds as 7, 300.

(iii) 8, 074 is rounded off to the nearest hundreds as 8, 100.

(iv) 14, 627 is rounded off to the nearest hundreds as 14, 600.

(v) 28, 826 is rounded off to the nearest hundreds as 28, 800.

(vi) 4, 20, 387 is rounded off to the nearest hundreds as 4, 20, 400.

(vii) 43, 68, 973 is rounded off to the nearest hundreds as 43, 69, 000.

(viii) 7, 42, 898 is rounded off to the nearest hundreds as 7, 42, 900.

3. Round off each of the following numbers to nearest thousands:

(ii) 9, 600

(iii) 4, 278

(iv) 7, 832

(vi) 26, 019

(vii) 20, 963

(viii) 4, 36, 952

(i) 2, 401 is rounded off to the nearest thousands as 2, 000.

(ii) 9, 600 is rounded off to the nearest thousands as 10, 000.

(iii) 4, 278 is rounded off to the nearest thousands as 4, 000.

(iv) 7, 832 is rounded off to the nearest thousands as 8, 000.

(v) 9, 567 is rounded off to the nearest thousands as 10, 000.

(vi) 26, 019 is rounded off to the nearest thousands as 26, 000.

(vii) 20, 963 is rounded off to the nearest thousands as 21, 000.

(viii) 4, 36, 952 is rounded off to the nearest thousands as 4, 37, 000.

4. Round off each of the following to the nearest tens, hundreds and thousands:

(ii) 1, 049

(iii) 45, 634

(iv) 79, 085

(i) 964 is rounded off to the nearest tens as 960.

964 is rounded off to the nearest hundreds as 1, 000.

964 is rounded off to the nearest thousands as 1, 000.

(ii) 1, 049 is rounded off to the nearest tens as 1, 050.

1, 049 is rounded off to the nearest hundreds as 1, 000.

1, 049 is rounded off to the nearest thousands as 1, 000.

(iii) 45, 634 is rounded off to the nearest tens as 45, 630.

45, 634 is rounded off to the nearest hundreds as 45, 600.

45, 634 is rounded off to the nearest thousands as 46, 000.

(iv) 79, 085 is rounded off to the nearest tens as 79, 090.

79, 085 is rounded off to the nearest hundreds as 79, 100.

79, 085 is rounded off to the nearest thousands as 79, 000.

5. Round off the following measures to the nearest hundreds:

(ii) Rs 850

(iii) Rs 3, 428

(iv) Rs 9, 080

(v) 1, 265 km

(vii) 550cm

(viii) 2, 486m

(xii) 820mg

(i) Rs 666 is rounded off to the nearest hundreds as Rs 700.

(ii) Rs 850 is rounded off to the nearest hundreds as Rs 900.

(iii) Rs 3, 428 is rounded off to the nearest hundreds as Rs 3, 400.

(iv) Rs 9, 080 is rounded off to the nearest hundreds as Rs 9, 100.

(v) 1, 265 km is rounded off to the nearest hundreds as 1, 300km.

(vi) 417m is rounded off to the nearest hundreds as 400m.

(vii) 550cm is rounded off to the nearest hundreds as 600cm.

(viii) 2, 486m is rounded off to the nearest hundreds as 2, 500m.

(ix) 360gm is rounded off to the nearest hundreds as 400gm.

(x) 940kg is rounded off to the nearest hundreds as 900kg.

(xi) 273l is rounded off to the nearest hundreds as 300l.

(xii) 820mg is rounded off to the nearest hundreds as 800mg.

6. List all numbers which are rounded off to nearest ten as 370.

The list of numbers which are rounded off to nearest ten as 370 are

365, 366, 367, 368, 369, 370, 371, 372, 373, 374.

7. Find the smallest and greatest numbers which are rounded off to the nearest hundreds as 900.

The smallest number which is rounded off to the nearest hundreds as 900 is 850.

The greatest number which is rounded off to the nearest hundreds as 900 is 949.

8. Find the smallest and greatest numbers which are rounded off to the nearest thousands as 9000.

The smallest number which is rounded off to the nearest thousands as 9000 is 8, 500.

The greatest number which is rounded off to the nearest thousands as 9000 is 9, 499.

Exercise 1.7 page: 1.29

1. Estimate the following by rounding off each factor to nearest hundreds:

(i) 730 + 998

(ii) 796 – 314

(iii) 875 – 384

(i) It can be rounded off to the nearest hundreds as 700 + 1000 = 1, 700.

(ii) It can be rounded off to the nearest hundreds as 800 – 300 = 500.

(iii) It can be rounded off to the nearest hundreds as 900 – 400 = 500.

2. Estimate the following by rounding off each factor to nearest thousands:

(i) 12, 094 + 2, 888

(ii) 28, 292 – 21, 496

(i) It can be rounded off to the nearest thousands as 13, 000 + 3, 000 = 16, 000.

(ii) It can be rounded off to the nearest thousands as 28, 000 – 21, 000 = 7, 000.

3. Estimate the following by rounding off each number to its greatest place:

(i) 439 + 334 + 4, 317

(ii) 8, 325 – 491

(iii) 1, 08, 734 – 47, 599

(iv) 898 × 785

(v) 9 × 795

(vi) 87 × 317

(i) It can be rounded off to its greatest place as 400 + 300 + 4, 000 = 4, 700.

(ii) It can be rounded off to its greatest place as 8, 000 – 500 = 7, 500.

(iii) It can be rounded off to its greatest place as 1, 00, 000 – 50, 000 = 50, 000.

(iv) It can be rounded off to its greatest place as 900 × 800 = 7, 20, 000.

(v) It can be rounded off to its greatest place as 10 × 800 = 8, 000.

(vi) It can be rounded off to its greatest place as 90 × 300 = 27, 000.

4. Find the estimated quotient for each of the following by rounding off each number to its greatest place:

(i) 878 ÷ 28

(ii) 745 ÷ 24

(iii) 4489 ÷ 394

(i) It can be rounded off to its greatest place as 900 ÷ 30 = 30.

(ii) It can be rounded off to its greatest place as 700 ÷ 20 = 35.

(iii) It can be rounded off to its greatest place as 4000 ÷ 400 = 10.

Exercise 1.8 page: 1.30

1. Write the expression for each of the following statements using brackets:

(i) Four multiplied by the sum of 13 and 7.

(ii) Eight multiplied by the difference of four from nine.

(iii) Divide the difference of twenty eight and seven by 3.

(iv) The sum of 3 and 7 is multiplied by the difference of twelve and eight.

(i) The expression for Four multiplied by the sum of 13 and 7 is 4 × (13 + 7).

(ii) Eight multiplied by the difference of four from nine is 8 × (9 – 4).

(iii) Divide the difference of twenty eight and seven by 3 is (28 – 7) ÷ 3.

(iv) The sum of 3 and 7 is multiplied by the difference of twelve and eight is (3 + 7) × (12 – 8).

2. Simplify each of the following:

(i) 124 – (12 – 2) × 9

(ii) (13 + 7) × (9 – 4) – 18

(iii) 210 – (14 – 4) × (18 + 2) – 10

On further simplification we obtain

124 – (12 – 2) × 9

= 124 – 10 × 9

= 124 – 90

(13 + 7) × (9 – 4) – 18

= 20 × 5 – 18

= 100 – 18

210 – (14 – 4) × (18 + 2) – 10

= 210 – 10 × 20 – 10

= 210 – 200 – 10

Exercise 1.9 page: 1.31

Simplify each of the following:

1. 7 × 109

2. 6 × 112

3. 9 × 105

4. 17 × 109

5. 16 × 108

6. 12 × 105

7. 102 × 103

8. 101 × 105

9. 109 × 107

1. On simplification of 7 × 109 we get

7 × 109 = 763.

2. On simplification of 6 × 112 get

6 × 112 = 672.

3. On simplification of 9 × 105 we get

9 × 105 = 945.

4. On simplification of 17 × 109 we get

17 × 109 = 1853.

5. On simplification of 16 × 108 we get

16 × 108 = 1728.

6. On simplification of 12 × 105 we get

12 × 105 = 1260.

7. On simplification of 102 × 103 we get

102 × 103 = 10506.

8. On simplification of 101 × 105 we get

101 × 105 = 10605.

9. On simplification of 109 × 107 we get

109 × 107 = 11663.

Exercise 1.10 page: 1.35

1. Write the roman-numerals for each of the following:

(i) The roman-numeral of 33 is XXXIII.

(ii) The roman-numeral of 48 is XLVIII.

(iii) The roman-numeral of 76 is LXXVI.

(iv) The roman-numeral of 95 is XCV.

2. Write the following in Roman numerals:

(i) The roman-numeral of 154 is CLIV.

(ii) The roman-numeral of 173 is CLXXIII.

(iii) The roman-numeral of 248 is CCXLVIII.

(iv) The roman-numeral of 319 is CCCXIX.

3. Write the following in Roman numerals:

(i) The roman-numeral of 1008 is MVIII.

(ii) The roman-numeral of 2718 is MMDCCXVIII.

(iii) The roman-numeral of 3906 is MMMCMVI.

(iv) The roman-numeral of 3794 is MMMDCCXCIV.

4. Write the following in Roman numerals:

(iii) 44000

(i) The roman-numeral of 4201 is MVCCI.

(ii) The roman-numeral of 10009 is XIX.

5. Write the following in Hindu-Arabic numerals:

(iii) LXXII

(i) The Hindu-Arabic numerals of XXVI is 26.

(ii) The Hindu-Arabic numerals of XXIX is 29.

(iii) The Hindu-Arabic numerals of LXXII is 72.

(iv) The Hindu-Arabic numerals of XCI is 91.

6. Write the corresponding Hindu-Arabic numerals for each of the following:

(ii) CLXXII

(iii) CCLIV

(iv) CCCXXIX

(i) The Hindu-Arabic numerals of CIX is 109.

(ii) The Hindu-Arabic numerals of CLXXII is 172.

(iii) The Hindu-Arabic numerals of CCLIV is 254.

(iv) The Hindu-Arabic numerals of CCCXXIX is 329.

7. Write the corresponding Hindu-Arabic numerals for each of the following:

(iii) KKCXXIII

(iv) KKKDCXL

(i) The Hindu-Arabic numerals of KXIX is 1019.

(ii) The Hindu-Arabic numerals of KDLXV is 1565.

(iii) The Hindu-Arabic numerals of KKCXXIII is 2123.

(iv) The Hindu-Arabic numerals of KKKDCXL is 3640.

8. Write the following in Hindu-Arabic numerals:

9. Which of the following are meaningless?

(ii) KKKCCXI

In the given question we know that (i), (iii) and (iv) are meaningless.

Objective Type Questions PAGE: 1.36

Mark the correct alternative in each of the following:

1. The difference between the place value and face value of 8 in 658742 is

(a) 0 (b) 42 (c) 735 (d) 693

The given options are incorrect.

Place value of 8 in 6,58,742 = 8,000

Face value of 8 = 8

Difference between them = 8,000 − 8 = 7,992

2. The difference between the place values of 6 and 3 in 256839 is

(a) 3 (b) 9 (c) 6800 (d) 5930

Place value of 6 in 2,56,839 = 6,000

The place value of 3 in 2,56,839 = 30

Difference between them = 6,000 − 30 = 5,970

3. The difference of the smallest three digit number and the largest two digit number is

(a) 100 (b) 1 (c) 10 (d) 99

The option (b) is the correct answer.

The smallest three-digit number is 100 and the largest two-digit number is 99.

So the difference between them= 100 − 99 = 1

4. The largest three digit number formed by the digits 8, 5, 9 is

(a) 859 (b) 985 (c) 958 (d) 589

We know that the largest number is formed by writing the digits in descending order.

5. The smallest three digit number having three distinct digits is

(a) 123 (b) 101 (c) 102 (d) 201

The option (c) is the correct answer.

We know that the smallest three digit numbers are 0, 1 and 2

By arranging them in ascending order we get 012 which is a two digit number.

Hence, 102 is the smallest three digit number.

6. The largest three digit number having distinct digits is

(a) 987 (b) 789 (c) 999 (d) 900

The option (a) is the correct answer.

The largest three digit numbers are 7, 8 and 9. The largest number is obtained by arranging them in descending order.

7. The difference between the largest three digit number and the largest three digit number with distinct digits is

(a) 10 (b) 0 (c) 12 (d) 13

We know that Largest three digit number = 999

Largest three-digit number with distinct digits = 987

Difference between them = 999 − 987 = 12

8. The product of the place values of two threes in 53432 is

(a) 9000 (b) 90000 (c) 10000 (d) 99000

We know that 3 in the second place from right is at tens place.

So the place value of 3 at tens place = 30

We know that 3 in the fourth place from right is at thousands place.

So the place value of 3 at thousands place = 3,000

Product of the place values = 3,000 × 30 = 90,000

9. The smallest counting number is

(a) 0 (b) 1 (c) 10 (d) None of these

We know that the smallest digit is zero and the smallest counting number is 1.

10. The total number of 4 digit numbers is

(a) 8999 (b) 9000 (c) 8000 (d) 9999

We know that the smallest four digit number = 1,000

Largest four digit number = 9,999.

So the total number of four-digit numbers = [9,999 − 1,000 ] + 1 = 9,000

11. The number of 3 digit numbers formed by using digits 3, 5, 9, taking each digit exactly once, is

(a) 3 (b) 4 (c) 5 (d) 6

The option (d) is the correct answer.

359, 395, 539, 593, 935 and 953 are the numbers.

12. Total number of numbers which when rounded off to nearest ten give us 200 is

(a) 9 (b) 10 (c) 8 (d) 7

195, 196, 197, 198, 199, 200, 201, 202, 203, 204 are the numbers that when rounded off to nearest ten give us 200.

13. The smallest number which when rounded off to the nearest hundred as 600, is

(a) 550 (b) 595 (c) 604 (d) 599

The number from 550 to 649 are rounded off to the nearest hundred as 600. Hence, the smallest number is 550.

14. The greatest number which when rounded off to the nearest thousand 7000, is

(a) 6500 (b) 6549 (c) 7499 (d) 6499

Numbers from 6500 to 7499 are rounded off to the nearest thousand as 7000. Hence, the greatest number is 7499.

15. The difference between the greatest and smallest numbers which when rounded off a number to the nearest tens as 540, is

(a) 10 (b) 9 (c) 8 (d) 10

We know that the greatest number that when rounded off to the nearest tens will become 540 = 544

Least number that when rounded off to the nearest tens will become 540 = 535

So, the difference between them = 544 − 535 = 9

16. The difference between the greatest and smallest numbers which when rounded off a number to the nearest hundred as 6700, is

(a) 100 (b) 99 (c) 98 (d) 101

We know that the greatest number which when rounded off to the nearest hundred will become 6,700 = 6749

Least number which when rounded off to the nearest hundred will become 6,700 = 6650

So, the difference between them = 6,749 − 6,650 = 99

17. The difference between the greatest and the smallest numbers which when rounded off to the nearest thousand as 9000, is

(a) 1000 (b) 990 (c) 999 (d) 900

We know that the greatest number which when rounded off to the nearest thousand becomes 9,000 = 9499

Smallest number which when rounded off to the nearest thousand becomes 9,000 = 8500

So, the difference between them = 9,499 − 8,500 = 999

18. Which of the following numbers is equal to 1 billion?

(a) 10 lakh (b) 1 crore (c) 10 crore (d) 100 crore

1 billion = 100 crore

19. In the international place value system, we write one million for

(a) 1 lakh (b) 10 lakh (c) 100 lakh (d) 1 crore

One million = 10 lakh

20. Which of the following is not meaningful?

(a) XIV (b) XXXV (c) XXV (d) VX

In Roman numerals V is 5 and X is 10

5 cannot be written before 10

21. Which of the following is not meaningful?

(a) XXII (b) XII (c) XVV (d) XIV

5 cannot be written two times.

Chapter 1 Knowing Our Numbers has 10 exercises with problems based on methods to represent numbers in figures or in words. The concepts covered in RD Sharma Solutions Chapter 1 are as follows:

• Introduction
• Natural numbers and whole numbers
• Methods of numeration
• Indian System of Numeration
• The place value and the face value of a digit in a numeral
• The international system of numeration
• Knowing more about numbers
• Comparison of numbers
• Large numbers in practice
• Estimating sum, difference, product and quotient
• Using brackets
• Roman Numerals

Chapter Brief of RD Sharma Solutions for Class 6 Maths Chapter 1 – Knowing Our Numbers

The students can download PDFs of RD Sharma Solutions to solve exercise-wise problems effortlessly and improve their understanding of the topics. These solutions are prepared by experienced faculty after vast research conducted on each concept. The exercise-wise solutions are explained in simple language to make it easy for the students to understand them.

Numbers play a very important role in determining the quantity or amount of things around us. From calling a friend to counting the number of eggs in the fridge, numbers are very important in day-to-day life. By using PDFs of the solutions, students can solve the problems in a shorter duration to boost their exam preparation.

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Knowing Our Numbers Class 6 Notes CBSE Maths Chapter 1 (Free PDF Download)

• Revision notes
• Chapter 1 Knowing Our Numbers

Revision Notes for CBSE Class 6 Maths Chapter 1 - Free PDF Download

A number is a mathematical unit that can be used to count, calculate, or mark things. In this Class 6 Maths Chapter 1 Notes on Knowing our Numbers, we will discuss counting numbers and their comparison, understanding the larger numbers, use of commas in numbers, estimation of numbers.

The Class 6 Maths Notes Chapter 1 created by the experts in Vedantu will help students to revise all the concepts of Knowing our Numbers before their exams. These revision notes on Chapter 1 Class 6 Maths are prepared according to the NCERT curriculum so that students can revise all the concepts of Knowing our Numbers without any doubts. The Revision notes will play a pivotal role before exams as it covers all the important concepts of Knowing our Numbers which are explained in a crisp and easy way. These Class 6 Maths Chapter 1 Notes will help students to revisit all the concepts and revise so that they can score good marks in their board exams.

Students can download the free PDF of Knowing our Numbers Class 6 Notes from the Vedantu platform. Vedantu also provides free PDF notes to various chapters of Class 6. You can also register Online for NCERT Class 6 Science tuition on Vedantu.com to score more marks in CBSE board examination. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students.

Access Class 6 Mathematics Chapter 1 – Knowing our Number in 30 Minutes

When two numbers are given, the one with more digits is larger. If the number of digits in two numbers is the same, the number with the bigger leftmost digit is the larger. If this digit is the same as the previous one, we move on to the next digit, and so on.

When creating numbers from supplied digits, it's important to check to verify if the requirements for forming the numbers are met. Thus, we must employ all four digits to construct the largest four-digit number from 7,8,3,5 without repeating a single digit; the greatest number can only contain 8 as the leftmost digit.

1000 is the lowest four-digit number (one thousand). It comes after the three-digit number 999. Similarly, 10,000 is the lowest five-digit figure. It's a ten-digit number that comes after the greatest four-digit number, 9999. Furthermore, 100,000 is the lowest six-digit figure. It is one lakh and comes after 99,999, the highest five-digit figure. In a similar way, this is true for higher digit numbers.

Using commas makes it easier to understand and write big figures. In the Indian numeration system, commas appear after the first three digits, starting on the right, and every two digits beyond that. Thousand, lakh, and crore are separated by commas after 3,5 and 7 digits, respectively. Starting from the right, commas are put after every three numbers in the International system of numeration. After three and six figures, commas separate thousand and million, respectively.

In many aspects of daily life, large numbers are required. For example, to calculate the number of pupils at a school, the number of people in a hamlet or town, the amount of money spent or received in major transactions (buying and selling), and to measure vast distances, such as between cities in a country or around the world.

Remember that a kilo represents 1000 times larger, a centi represents 100 times smaller, and a milli represents 1000 times smaller, thus 1 kilometre equals 1000 metres, 1 metre equals 100 centimetres or 1000 millimetres, and so on.

There are a few instances where we don't require an exact quantity but simply a fair guess or estimate. For example, when mentioning the estimated number of people that watched a certain international hockey match, such as 51,000, we do not need to give the actual figure.

Estimation is the process of estimating a quantity to the desired level of precision. So, depending on our needs, 4117 can be estimated to 4100 or 4000, i.e. to the closest hundred or thousand.

We must estimate the result of number operations in a variety of scenarios. This is accomplished by rounding the numbers and obtaining a rapid, approximate response.

Checking solutions by estimating the outcome of numerical operations is useful.

We may prevent misunderstanding by using brackets in cases when we need to do more than one number operation.

We utilise the Hindu-Arabic number system. The Roman numeral system is another way to write numbers.

Knowing Our Numbers Class 6 Notes

Class 6 maths chapter 1 notes.

Here let us look into some of the important concepts covered in Class 6 Chapter 1 Maths Notes:

Introduction to Numbers

Comparing Numbers

Shifting digits

Introducing 10,000

Revisiting place value

Introducing 1,00,000

Larger numbers

Reading and writing large numbers

Use of commas

Large Numbers in Practice

Estimating to the nearest hundreds by rounding off

Estimating to the nearest thousands by rounding off

Estimating outcomes of number situations

To estimate sum or difference

To estimate products

Using Brackets

Expanding brackets

Roman Numerals

Now let us revise each topic of Knowing our Numbers Class 6 Notes briefly.

We compare numbers to check whether they are greater than, or less than or equal to.

Ex: When we compare 24 and 36, we can conclude that 36 is greater than 24 or 24 is less than 36.

Similarly, when we compare larger numbers such as 5005 and 5010, we can say the number 5010 is larger than the number 5005. This is because the hundreds place and thousands place numbers are the same as 5 and 0, but when we compare the tens and ones place number we can conclude that the number 5010 is greater than the number 5005.

Shifting Digits

Shifting digits means that exchanging the digits to different places to compare them.

Ex: Consider the number 787. Here if we exchange the digits to a different place like shifting the number 8 from tens place to one place we get a new number 778. If we compare the original number with the new number we can conclude that the original number 787 is greater than the new number 778.

Similarly, if we have a number 777. Here if we exchange the digits to different places we will still end up with the same number as all the digits are the same in this number.

We know that there are 99 two-digit numbers from 1 to 99. After 99 we will have three-digit numbers till 999. After 999 we will start with four-digit numbers till 9999. Now if we add 1 to 9999 we will get a five-digit number called “Ten thousand (10000)”.

Revisiting Place Value

The place gives the value of each digit in a number.

Ex: Place value of the digit 7 in the number 75 is 7 × 10 = 70 as 7 is in the tens place.

Place value of digit 6 in the number 675 is 6 × 100 = 600 as 6 is in the hundreds place.

Place value of the digits 4 in the number 4675 is 4 × 1000 = 4000 as 4 is in the thousands place.

We got to know that when we add 1 to the largest 4 digit number we will get 10,000. Similarly, when we add 1 to the largest 5 digit number we will get the smallest 6 digit number called one lakh (1,00,000).

Larger Numbers

Larger numbers are considerably larger than those we use commonly in our daily life.

Ex: When we are counting the number of students in the class which will be usually a 2 digit number. But when counting the numbers of students in the school it will either be a 3 digit or 4 digit number which is larger than the 2 digit number. Similarly, if we are counting a number of students from 10 different schools then this number will be very large, say a 5 digit number.

So understanding the larger number will allow students to apply them wherever necessary in a proper way.

Use of Commas

Using commas in the numbers will make it easy to read the large numbers. Also using commas we can differentiate between the Indian number system and the International number system.

Ex: When we write a large number 456783 without commas it will be difficult to read, but if we separate these digits with commas we can easily read. So writing the number with commas 4,56,783 which is read as four lakh fifty-six thousand seven hundred and eighty-three.

Commas can also be used to differentiate between the Indian number system and the International number system. Consider the same number 4,56,783 in the Indian system which has three commas but in the International number system the number is just separated by two commas 456,783 which is read as four thousand fifty-six and seven hundred and eighty-three.

Estimation is done to find a value that is close enough to the original value. Estimation is just rounding off the number to the nearest whole number to make it easily readable.

When estimating, the general rule is to look at the digit to the right of the one we want to estimate.

Estimation of numbers is usually done when we are dealing with larger numbers.

Ex: When we are counting the number of students in the school, it will take a lot of time to find the headcount of each student. At this time we will give an estimated number of students. Say if the total number of students in the school is 1139, we can give the count as 1140 which is a rounded-off number and also easy to read.

To Estimate Sum or Difference

In this section of Class 6 Maths Chapter 1 Notes, we will find how to estimate the numbers when they are being added or subtracted.

Ex: When we are adding two large numbers say 8705 and 5611. We will estimate these two numbers to be 8700 and 5600 whose sum is 14300. So the sum of the numbers 8705 and 5611 will be closer to this estimated answer.

Similarly when we are subtracting two large numbers 8720 and 5645. Here we will estimate the numbers to 8700 and 5600. If we subtract these two numbers we get the difference as 3100 which is the estimated difference between 8720 and 5645.

To Estimate Products

Round the numbers to some similar numbers that we can easily multiply to estimate the multiplication result (product).

Ex: If we have to find the product of two numbers 485 and 92, we will round off the numbers like 500 and 100. So the estimated product is 50000.

When we are doing multiple operations on the numbers we use brackets to avoid confusion and to read the numbers properly.

Ex: If we have to find the solution to the equation 24 + 45 × 10. It will be difficult to find the answer as we don’t know which operation we have to use first. Do we have to use addition first or multiplication? This is when the application of brackets will play a pivotal role in determining the answer. So now using brackets we can write the equation as (24 + 45) × 10. By using the BODMAS rule first we will operate the numbers inside the brackets, so we will perform the addition first and then we will multiply the numbers. So the answer will be 69 × 10 = 690.

Roman Numerals are represented by using the combinations of different letters from the Latin alphabet.

Ex: The roman numbers I, II, III, IV, V, VI, VII, VIII, IX, X denote the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively from the Hindu-Arabic numeral system.

The Class 6 Chapter 1 Maths Notes will help students to revise all the basic and important concepts of Knowing our Numbers. As these revision notes are prepared according to the NCERT curriculum students can use these revision notes to prepare for their exams. The Class 6 Maths Notes Chapter 1 is available completely free on the Vedantu platform where students can download a PDF version of these revision notes to ace their exams.

FAQs on Knowing Our Numbers Class 6 Notes CBSE Maths Chapter 1 (Free PDF Download)

1. Why do we Have to Compare the Numbers?

By comparing numbers we can find whether the numbers are greater than or smaller than or equal to and arrange them in a decreasing or increasing order.

2. What is Meant by the Estimation of Numbers?

The method of estimating or approximating or rounding off numbers in which the value is used for some other purpose in order to prevent complicated calculations is known as number estimation.

3. What are the Uses of Brackets in Representing the Numbers?

Brackets are symbols that are used in pairs to group objects. Brackets denote solutions that are greater than or equal to the number, or that are less than or equal to it.

4. What is Chapter 1- Knowing our numbers all about according to Revision Notes of Class 6?

A number is actually the mathematical value we use for counting and measuring various objects. Through numbers, we can add, multiply, subtract, and divide. In the Chapter-Knowing Our Numbers, you will learn a comparison of numbers, expansion of a number and the smallest and the largest numbers.

There are two kinds of number systems used to write numbers: Roman numerals and Indo-Arabic numerals. Roman numerals are usually seen in clocks, school timetables, page numbers, etc.

5. What are the numbers according to Revision Notes of Chapter 1 of Class 6 Maths?

Numbers are written using only these ten symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Such symbols are referred to as figures or digits.

To identify the place values of given digits of a number, you have to multiply them with the values of the places they occupy. To understand the concept of numbers in Class 6 Maths, you must download the NCERT solutions by Vedantu which will help you to clear out your concepts easily and quickly. They are free of cost and also available on Vedantu Mobile app.

6. How can I teach Chapter 1 of Class 6 Maths to students of Class 6?

To teach Chapter 1 Knowing Our Numbers to Class 6, you have to learn a variety of different topics coupled with a tiny bit of revision. Topics included in this list are :

Patterns in number.

Indian system of numbers.

A universal system of numbers.

Estimation.

Comparison of numbers, greatest and smallest number.

Knowing large numbers till 1 crore.

To teach these concepts and get help with your methods, refer to Revision Notes of Chapter 1 of Class 6 Maths. These are tailor-made to help students understand concepts.

7. What are shifting digits according to Revision Notes of Chapter 1 of Class 6 Maths?

Shifting the digits of any given number refers to nothing but exchanging the digits of the number into other places.

You can choose to gain further knowledge about the Revision Notes of Chapter 1 of Class 6 Maths. These activities are included in the solutions provided in the NCERT solutions by Vedantu.

Shifting digits is important for understanding the chapter as well as the future concepts properly. So, make sure you have a clear concept.

8. Explain the difference between Ascending and Ascending order to number according to Revision Notes of Chapter 1 of Class 6 Maths.

Ascending Order refers to the arrangement of numbers from the smallest to the largest.

Descending Order, on the other hand, refers to arranging numbers  from the greatest number to the smallest number.

Ascending and descending order of numbers is a critical concept that will help you throughout your practical life too. Thus, it is important that you take help from authentic guides such as NCERT solutions by Vedantu.

NCERT Solutions

Cbse links for class 6.

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Class 6 (Old)

Unit 1: knowing our numbers, unit 2: whole numbers, unit 3: playing with numbers, unit 4: integers, unit 5: fractions, unit 6: decimals, unit 7: algebra, unit 8: basic geometrical ideas, unit 9: understanding elementary shapes, unit 10: ratio and proportion, unit 11: mensuration, unit 12: data handling.

Chapter 1 Knowing Our Numbers Class 6 Notes Maths

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