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## RD Sharma Solutions for Class 6 Maths

RD Sharma Solutions for Class 6 PDF can be downloaded effortlessly by the students to score well in the exam. The faculty at Studentcare.com prepare solutions for the exercise wise problems after conducting wide research on each concept. Students can solve problems from the RD Sharma textbook by referring the solutions to self analyse their areas of weaknesses. It helps students to solve problems of higher difficulty level in an efficient manner.

The solutions PDF is the best reference guide for the students, which help them to gain better conceptual knowledge. Time management skills can be improved among students by solving problems using solutions designed by our experts. RD Sharma Solutions for Class 6 Maths are available here.

Students can access various study materials like worksheets, sample papers, and previous year question papers to obtain a better academic score. RD Sharma Solutions for Class 6 sample papers can be practiced regularly by the students in order to understand the types of questions that would appear in the exam.

Solutions PDF can be used as reference material by the students to obtain a better hold on the concepts which are covered under each chapter. By solving problems from the RD Sharma textbook, students using solutions mainly improve their problem-solving abilities, which are important from the exam point of view.

Solving problems on a daily basis as per the RD Sharma textbook improves knowledge about concepts covered in each chapter. Our faculty, having vast experience, prepare chapter-wise solutions which can be used to solve problems from lower to the higher difficulty level. The examples which are present before each exercise help students improve their problem solving and analytical thinking abilities.

RD Sharma Solutions for Class 6 Maths Chapter 1 Knowing our Numbers

Chapter 1 of the RD Sharma textbook helps students understand the concept of numbers which include natural numbers and whole numbers, methods of numeration, place value, face value, and comparison of numbers. The solutions contain shortcuts that can be followed to solve problems relevant to the latest CBSE guidelines.

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

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

• Introduction
• Natural numbers and Whole numbers
• Methods of Numeration
• Indian System of Numeration
• Place value and Face value of a digit in a numeral
• International system of numeration
• Comparison of Numbers
• Large numbers in practice
• Estimation
• Estimating sum, difference, product and quotient
• Using brackets
• Roman Numerals

RD Sharma Solutions for Class 6 Maths Chapter 2 Playing with numbers

Chapter 2 of the RD Sharma textbook contains problems based on important concepts like factors and multiples, common factors and multiples, prime and composite numbers, prime factorization, tests for divisibility of numbers, and general properties. These concepts are explained in an interactive manner by our faculty to help students score well in the exam.

Numbers play a very important role in our day to day life. Numbers can be used to count various things which are used on a daily basis. As it plays a vital role, it is important that the students gain a better knowledge about numbers. To know more about the concepts which are covered in this Chapter, students can download the solutions PDF for further reference.

Chapter 2 has 11 exercises solved by expert Mathematicians based on CBSE guidelines for Class 6.Some of the topics which are covered in Chapter 2 are given below:

• Introduction
• Factors and Multiples
• Common Factors and Multiples
• Prime and Composite Numbers
• The Sieve of Eratosthenes
• Prime Factorization
• Tests for Divisibility of Numbers
• Some general properties of divisibility
• Highest Common Factor
• Some applications of HCF
• Lowest Common Multiple
• Some applications of LCM
• Some properties of HCF and LCM of given numbers

RD Sharma Solutions for Class 6 Maths Chapter 3 Whole Numbers

Chapter 3 of RD Sharma Solutions has problems based on natural numbers with their properties, whole numbers, successor and predecessor of a whole number. The solutions are explained in simple language which matches the understanding abilities of students to boost their exam preparation.

Chapter 3 has 1 exercise which describes the properties of whole numbers and methods followed in solving problems. The major concepts which are explained in Chapter 3 are listed down:

• Introduction
• Natural Numbers
• Properties of Natural Numbers
• The number zero
• Whole Numbers
• Successor and Predecessor of a Whole Number
• Representation of Whole Numbers on Number Line
• Properties of Whole Numbers

The solutions of chapter wise problems are prepared by subject matter experts based on the understanding capacity of students. The PDF of chapter wise solutions can be used by the students as a vital resource to improve their problem solving abilities. Each question of the exercise is solved using various methods to make them easily understandable by the students.

We know that the numbers from 0, 1, 2, 3 are whole numbers, where a whole number is either zero or natural number. We use whole numbers in our daily life to count things and to obtain the required quantity. So, students can use the PDF to obtain further knowledge about concepts which are explained in this Chapter.

RD Sharma Solutions for Class 6 Maths Chapter 4 Operations on Whole Numbers

The exercise wise problems of chapter 4 are solved by our subject experts to help students ace the exam. The important topics which are covered in this chapter are properties of addition, subtraction, multiplication, and division based on the RD Sharma textbook of the latest CBSE syllabus.

Chapter 4, Operations on Whole Numbers, has 5 exercises based on addition, subtraction, division and multiplication operations on whole numbers. Some of the important topics in Chapter 4 are given below:

• Introduction
• Properties of Subtraction
• Properties of Multiplication
• Division
• Properties of Division
• Patterns

Addition, Multiplication, Subtraction and Division along with the properties are the concepts which are focused upon in this Chapter. Solving the exercise wise problems mainly help students gain confidence in answering questions in a shorter duration. The students can download PDF to know more about the methods and formulas which are used in solving problems.

RD Sharma Solutions for Class 6 Maths Chapter 5 Negative Numbers and Integers

Both positive and negative numbers, including zero, are called integers. This chapter explains basic concepts like the introduction to integers, representation on a number line, and various operations on integers. RD Sharma solutions designed by subject experts can be used by students to solve exercise wise problems.

Chapter 5, Negative Numbers and Integers, has 4 exercises with problems based on types of integers and operations. Chapter 5 contains some of the important concepts which are listed below:

• The Need for Integers
• Introduction of Integers
• Representation of Integers on Number Line
• Ordering of Integers
• Negative of a Negative Integer
• Absolute Value of an Integer
• Properties of Addition of Integers
• Subtraction of Integers

The solutions which are designed by subject experts are an educational aid for the students to solve problems from easy to higher difficulty level. It contains a set of questions based on problem solving abilities of students. RD Sharma solutions PDF can be referred by the students to obtain good academic score in the exam.

Some of the applications of integers in our daily lives are in hockey scores, maps and altitude levels. The solutions can be downloaded either chapter wise or exercise wise based on the requirement. Students can download PDF in order to get a better knowledge about the concepts which are explained in this chapter.

RD Sharma Solutions for Class 6 Maths Chapter 6 Fractions

A number that represents a small part of a whole either single or a group of things is called a fraction. The faculty at studentecare.com has design solutions after conducting vast research on each concept. The students can use the solutions while solving chapter-wise problems, with respect to the latest CBSE guidelines.

Chapter 6, Fractions, has 9 exercises which explains the methods used in solving problems along with diagrammatic representation. The Chapter 6 contains some of the basic concepts like:

• Fraction
• Representation of fractions on the number line
• Fraction as division
• Types of fractions
• Equivalent fractions
• Fractions in lowest terms
• Like and Unlike Fractions
• Comparing and ordering of fractions
• Addition and Subtraction of like fractions
• Addition and subtraction of unlike fractions

Concepts which are explained in Class 6 are important as they are continued in higher classes as well. This Chapter contains methods of solving types of fractions using operations. The solutions are prepared by subject matter experts in order to improve conceptual knowledge and logical thinking abilities among students.

Some of the applications of fractions in our daily life are ingredients used for cooking and their proportion, portion of painted wall and the amount of sugar in the vessel etc. The PDF can be downloaded to get further information regarding the concepts which are explained in this Chapter.

RD Sharma Solutions for Class 6 Maths Chapter 7 Decimals

The 7th chapter of the RD Sharma textbook explains important concepts like place and face value, comparison of decimals, some useful decimal notation, and problems based on them. The solutions contain explanations in simple language to improve conceptual knowledge, important from the exam point of view. Students can download the solutions PDF effortlessly to improve their speed in solving problems.

Chapter 7, Decimals, has 9 exercises which explains the methods of solving problems along with shortcut tips which are essential. Chapter 7 helps us understand some of the major concepts which are mentioned below:

• Introduction
• Decimals as an extension of place value table
• Decimals as Fractions
• Expressing Fractions as Decimals
• Expressing Decimals as Fractions
• Comparing Decimals
• Some uses of Decimal Notation
• Equivalent Decimals
• Like and Unlike Decimals
• Addition of Numbers with Decimals
• Subtraction of Numbers with Decimals

The students can solve problems of RD Sharma textbook for Class 6 using the solutions as a reference material. This mainly improves confidence among students in solving tricky and lengthy problems easily. The PDF is easily accessible by the students and can be used as a reference material which speeds up exam preparation.

We use Decimals in daily lives, when we deal with money, length and weight of materials etc. It provides accurate value of required things. Here, by downloading PDF of solutions, the students can understand the concepts which are covered in this Chapter and methods of solving problems.

RD Sharma Solutions for Class 6 Maths Chapter 8 Introduction to Algebra

Addition, subtraction, division, and multiplication using numbers are done with the help of algebraic equations. Here, the students obtain a brief idea about the concepts which are covered in this chapter based on the prescribed syllabus. This chapter covers concepts like basic operations on literals and numbers, powers of literal numbers, variables, and constants.

Chapter 8, Introduction to Algebra, has 2 exercises which explains the various operations on algebraic terms and the concept of variables and constants. The major concepts which are talked about in Chapter 8 are as listed below:

• Introduction
• Use of Letters to Denote Numbers
• Basic Operations on Literals and Numbers
• Powers of Literal Numbers
• Variables and Constants

The solutions are created by experts with the aim of helping students perform well in the exam. The well designed solutions for each chapter help students improve conceptual knowledge and problem solving abilities. By solving the exercise wise problems, students self analyse the areas of weaknesses and work on them.

Some of the major applications of Algebra in day to day lives are Finance, long term investments, the lowest cost method of Financing, Sports, Technology, Business etc. By downloading the PDF of solutions, students gain a better knowledge about Algebra and methods used in solving exercise wise problems.

RD Sharma Solutions for Class 6 Maths Chapter 9 Ratio Proportion and Unitary Method

The primary aim of creating solutions is to help students ace the exam with good scores. Chapter 9 contains important topics like ratio, comparison of ratios, proportion, and unitary method. Students can make use of solutions PDF prepared by experts as a reference guide to perform well in the exam.

Chapter 9, Ratio, Proportion and Unitary Method, has 4 exercises which helps students solve problems using various methods. The concepts which are explained in Chapter 9 are as follows:

• Ratio
• Proportion
• Unitary Method

RD Sharma Solutions created by subject matter experts are according to the current CBSE syllabus and guidelines. The PDF of solutions can be downloaded and referred while solving the Chapter wise problems of RD Sharma textbook. Students by using the PDF of solutions can obtain a better understanding of concepts covered in the Chapter.

Ratio and Proportion are used in our daily lives and has various important applications like grocery shopping, recipes and cooking, planning vacation trips with families etc. Students can download the PDF and use it to know more about the concepts which are explained in this Chapter.

RD Sharma Solutions for Class 6 Maths Chapter 10 Basic Geometrical Concepts

The various shapes and figures which we come across in geometry are known as geometrical figures. Here, we mainly learn about the construction of figures and their properties. The students get an idea about the steps followed to construct geometric shapes and some important terms.

Geometry is the branch of mathematics that deals with sizes, shapes and dimensions of figures, angles, lines, surfaces, etc. The word originated from Greek word “Geometron” where “Geo” and “metria” means earth and measurement respectively. It is one of the most basic geometrical ideas for class 6.

The basic geometrical concepts are dependent on three basic concepts. They are-

• Points, Lines, Planes and Angles.
• Proof.
• Perpendicular and parallel.
• Triangles.
• Similarity.
• Right triangles and trigonometry.
• Transformations.

RD Sharma Solutions for Class 6 Maths Chapter 11 Angles

Objects which have two arms joined together with a common initial point is called an angle. In Chapter 11, students gain knowledge about concepts like various types of angles and special angles. The solutions PDF can be downloaded for free by the students who wish to score well in the exam.

An angle is a figure formed by two rays with the same initial point. The various types of angles which are explained in this Chapter are straight angle, complete angle, zero angle, right angle and an obtuse angle. The students can download PDF of RD Sharma Solutions for Class 6 Chapter 11 Angles for free.

RD Sharma Solutions for Class 6 Maths Chapter 12 Triangles

The important concepts based on triangles are explained in simple and understandable language in this chapter. The topics which are discussed in this chapter are types of triangles based on the length of sides and angles. The students can use the PDF to solve the problems of the RD Sharma textbook effortlessly.

If in a triangle, all three sides are unequal, then it is called a Scalene Triangle. If in a triangle, any two sides are equal, then it is called an Isosceles Triangle. If in a triangle, all three sides are equal, then it is called an Equilateral Triangle.

RD Sharma Solutions for Class 6 Maths Chapter 13 Quadrilaterals

We know that a four-sided polygon is called a quadrilateral. It is divided into various types based on the number of sides. By using the solutions PDF, students obtain an overall idea about the concepts which are explained under each exercise. They can self analyse their knowledge about the entire chapter by solving problems on a daily basis.

There are mainly 6 types of quadrilaterals which are:

• Trapezium.
• Parallelogram.
• Rectangle.
• Rhombus.
• Square.
• Kite.

RD Sharma Solutions for Class 6 Maths Chapter 14 Circles

Chapter 14 of the RD Sharma textbook contains 2 exercise solutions that cover major concepts like the circle, interior, and exterior of a circle, chords and arcs, semi-circle, and semi-circular region. Methods used in solving problems and shortcut tips are also highlighted with the aim of boosting exam preparation of students.

Centre: The center point of the entire circle. Point C is the centre of the circle. Radius: The line segment that connects the centre of circle to any point on the circle. Circumference: It is the distance around the circle. Arc: An arc is the part of circumference.

RD Sharma Solutions for Class 6 Maths Chapter 15 Pair of Lines and Transversal

Students can easily access chapter wise solutions of RD Sharma textbook, as per the latest CBSE guidelines. This chapter deals with important concepts like parallel lines, transversals, and angles made by transversals. The students can self analyse their knowledge about the entire chapter by solving problems using solutions PDF.

A transversal refers to a line which passes through two lines lying in the same plane at two different points. Moreover, in the transversal, the two certain lines can be parallel or non- parallel. Thus, the angles which form when a transversal intersects two lines are corresponding angles and alternate angles.

RD Sharma Solutions for Class 6 Maths Chapter 16 Understanding Three Dimensional Shapes

In this chapter, we learn about various three-dimensional structures we come across in our daily lives. The concepts which are talked about in this chapter are cuboid, cube, cylinder, right circular cone, sphere, pyramid, and tetrahedron. The solutions contain explanations in an interactive manner to make the subject fun for the students.

The shapes which can be measured in 3 directions are called three-dimensional shapes. These shapes are also called solid shapes. Length, width, and height (or depth or thickness) are the three measurements of the three-dimensional shapes. Sphere, cone, cylinder, cuboid, cube are examples of three-dimensional shape.

RD Sharma Solutions for Class 6 Maths Chapter 17 Symmetry

Chapter 17 of the RD Sharma textbook deals with topics like symmetry and the construction of symmetrical figures as per the prescribed CBSE syllabus. Students can download the solutions PDF effortlessly either exercise or chapter wise and practice problems on a daily basis. It improves time management skills among students, which is important from the exam point of view.

Symmetry And Line Of Symmetry Definition

The symmetry of an object is defined as one half of the object is a mirror image of the other half. When an object is split into half, both the sides are exactly the same. The line which divides them is called the line of symmetry.

RD Sharma Solutions for Class 6 Maths Chapter 18 Basic Geometrical Tools

Geometrical tools are very important in the construction of various structures. It is a very important chapter as some of the concepts will be continued in higher classes as well. Chapter 18 consists of problems based on important tools like ruler, divider, compasses, protractor, and set squares along with the steps followed in construction using these tools.

The chapter teaches the students to use the basic tools of mathematics to measure various shapes and dimensions around us. Geometrical ideas are reflected in all forms of art, measurements, architecture, engineering, cloth designing etc. In order to measure and construct the geometrical figures, knowledge of geometry becomes important.

This chapter deals with the following concepts:

• Points
• Line segment
• Line
• Intersecting lines
• Ray
• Curve
• Polygon
• Circle, chord, radius, diameter, circumference, and semi-circles

RD Sharma Solutions for Class 6 Maths Chapter 19 Geometrical Constructions

Here, we learn more about the basic construction of geometrical structures which are important from the exam point of view. The students can improve their speed in solving problems more accurately using RD Sharma solutions designed by our faculty. These solutions help students to analyse the kind of questions that would appear in the exam.

RD Sharma Solutions for Class 6 Maths Chapter 20 Mensuration

When we see plane figures, the first thing that comes to our mind is their boundaries and origin. So, this chapter helps students obtain a good hold on important concepts like closed and open curves, region, the perimeter of a rectangle, square and equilateral triangle, and its area. The students can solve exercise wise problems using solutions PDF.

Mensuration is a branch of mathematics which is a topic in Geometry. It is a study of various geometrical shapes, their length, breadth, Volume, and area for 2D as well as 3D shapes. Some important terminologies included in this topic are covered below.

RD Sharma Solutions for Class 6 Maths Chapter 21 Data Handling I Presentation of Data

The method of presenting data in tabular form which makes it easy for the observer to understand is an important concept from the exam point of view. The students can easily score marks in this chapter by understanding the steps to be followed in representing the given numerical data.

Data handling means collecting the set of data and presenting in a different form. Data is a collection of numerical figures that represents a particular kind of information. The collection of observations which are gathered initially is called the raw data. Data can be in any form. It may be words, numbers, measurements, descriptions or observations. Data handling is the process of securing the research data is gathered, archived or disposed of in a protected and safe way during and after the completion of the analysis process.

Presentation of data is very important in order to make it easily understandable.

RD Sharma Solutions for Class 6 Maths Chapter 22 Data Handling II Pictographs

The method of representing given numerical data using picture symbols is called pictograph. The solutions are prepared by experts keeping in mind the grasping abilities of students. PDF format of solutions is made available in order to boost the exam preparation of students.

After collection and organization of the given set of data, the next step is to get some useful information from it. So the data should be represented pictorially. The study of numerical data through pictures or graphs is known as the pictorial representation or graph of the data. The method of representing numerical data by using picture symbols is called a pictograph.

RD Sharma Solutions for Class 6 Maths Chapter 23 Data Handling III Bar Graphs

The pictograph is very time consuming and the students can use bar graphs in order to overcome this issue. The method of representing the numerical data in the form of bars are called bar graphs. The construction of bar graphs and interpretation are the concepts that are explained in this chapter.

Representing numerical data with the help of pictures is time-consuming. We can make use of an alternative method to overcome this issue, i.e. representing the data using Bar Graphs to make it easily understandable by the students. In this Chapter, we learn about the construction and interpretation of Bar Graphs.

Representation of numerical data by bars having the same width kept horizontally or vertically with equal spacing is called the Bar Graph.

## Description Below File

### Benefits of RD Sharma Solutions for Class 6 Maths

• The chapter-wise and exercise-wise solutions can be easily downloaded by the students, based on their requirement.
• To make learning more interesting, each solution is provided with a pictorial representation to improve analysing skills among students.
• The solutions improve speed in solving problems more accurately, as per the prescribed CBSE syllabus.
• The students can self analyse their areas of weaknesses and can work on them to score good marks.

### Is RD Sharma Solutions the right book for CBSE Class 6 students?

Yes RD Sharma is the right book for CBSE Class 6 Students as it provides lots of questions to practice. The presentation of each topic in the book is described uniquely, which means the topics covered in the book are not stretched beyond their necessary length. They yet vividly describe all the variations of a particular concept, applied to different questions.

### Name the chapters present in RD Sharma Class 6 Maths Textbook.

There are 23 chapters are there in RD Sharma Class 6 Maths Textbook. Viz, Knowing our Numbers, Playing with numbers, Whole Numbers, Operations on Whole Numbers, Negative Numbers and Integers, Fractions, Decimals, Introduction to Algebra, Ratio, Proportion and Unitary Method, Basic Geometrical Concepts, Angles, Triangles, Quadrilaterals, Circles, Pair of Lines and Transversal, Understanding Three Dimensional Shapes, Symmetry, Basic Geometrical Tools, Geometrical Constructions, Mensuration, Data Handling – I (Presentation of Data), Data Handling – II (Pictographs) and Data Handling – III (Bar Graphs).

### How many exercises does the RD Sharma Solutions provide for Class 6 students?

Exercises of RD Sharma Textbook vary according to the concepts covered in the chapters. Some chapters have 6 and some have only one. It depends on the topic of the lesson.

How many four digit numbers are there in all?

Solution:

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.

Solution:

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?

Solution:

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

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.

Solution:

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?

Solution:

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

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

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

1. Find the HCF of the following numbers using prime factorization method:

(i) 144, 198

(ii) 81, 117

(iii) 84, 98

(iv) 225, 450

(v) 170, 238

(vi) 504, 980

(vii) 150, 140, 210

(viii) 84, 120, 138

(ix) 106, 159, 265

Solution:

(i) 144, 198

We know that the prime factorization of 144 = 2 × 2 × 2 × 2 × 3 × 3

The same way prime factorization of 198 = 2 × 3 × 3 × 11

Hence, HCF of 144, 198 is 2 × 3 × 3 = 18

(ii) 81, 117

We know that prime factorization of 81 = 3 × 3 × 3 × 3

The same way prime factorization of 117 = 3 × 3 × 13

Hence, HCF of 81, 117 = 3 × 3 = 9

(iii) 84, 98

We know that prime factorization of 84 = 2 × 2 × 3 × 7

The same way prime factorization of 98 = 2 × 7 × 7

Hence, HCF of 84, 98 = 2 × 7 = 14

(iv) 225, 450

We know that prime factorization of 225 = 3 × 3 × 5 × 5

The same way prime factorization of 450 = 2 × 3 × 3 × 5 × 5

Hence, HCF of 225, 450 = 3 × 3 × 5 × 5 = 225

(v) 170, 238

We know that prime factorization of 170 = 2 × 5 × 17

The same way prime factorization of 238 = 2 × 7 × 17

Hence, HCF of 170, 238 = 2 × 17 = 34

(vi) 504, 980

We know that the prime factorization of 504 = 2 × 2 × 2 × 3 × 3 × 7

The same way prime factorization of 980 = 2 × 2 × 5 × 7 × 7

Hence, HCF of 504, 980 = 2 × 2 × 7 = 28

(vii) 150, 140, 210

We know that prime factorization of 150 = 2 × 3 × 5 × 5

The same way prime factorization of 140 = 2 × 2 × 5 × 7

Prime factorization of 210 = 2 × 3 × 5 × 7

Hence, HCF of 150, 140, 210 = 2 × 5 = 10

(viii) 84, 120, 138

We know that prime factorization of 84 = 2 × 2 × 3 × 7

The same way prime factorization of 120 = 2 × 2 × 2 × 3 × 5

Prime factorization of 138 = 2 × 3 × 23

Hence, HCF of 84, 120, 138 = 2 × 3 = 6

(ix) 106, 159, 265

We know that prime factorization of 106 = 2 × 53

The same way prime factorization of 159 = 3 × 53

Prime factorization of 265 = 5 × 53

Hence, HCF of 106, 159, 265 = 53

2. What is the HCF of two consecutive

(i) Numbers

(ii) even numbers

(iii) odd numbers

Solution:

(i) We know that the common factor of two consecutive numbers is 1.

Hence, HCF of two consecutive numbers is 1.

(ii) We know that the common factors of two consecutive even numbers are 1 and 2.

Hence, HCF of two consecutive even numbers is 2.

(iii) We know that the common factors of two consecutive odd numbers is 1.

Hence, HCF of two consecutive odd numbers is 1.

3. HCF of co-prime numbers 4 and 15 was found as follows:

4 = 2 × 2 and 15 = 3 × 5

Since there is no common prime factor. So, HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Solution:

No. It is not correct.

The HCF of two co-prime numbers is 1.

We know that 4 and 15 are co-prime numbers having common factor 1.

Therefore, HCF of 4 and 15 is 1.

1. Write down the smallest natural number.

Solution:

1 is the smallest natural number.

2. Write down the smallest whole number.

Solution:

0 is the smallest whole number.

3. Write down, if possible, the largest natural number.

Solution:

The largest natural number does not exist.

4. Write down, if possible, the largest whole number.

Solution:

The largest whole number does not exist.

5. Are all natural numbers also whole numbers?

Solution:

Yes. All natural numbers are also whole numbers.

6. Are all whole numbers also natural numbers?

Solution:

No. All whole numbers are not natural numbers.

7. Give successor of each of the following whole numbers:

(i) 1000909

(ii) 2340900

(iii) 7039999

Solution:

(i) The successor of 1000909 is 1000910.

(ii) The successor of 2340900 is 2340901.

(iii) The successor of 7039999 is 7040000.

8. Write down the predecessor of each of the following whole numbers:

(i) 10000

(ii) 807000

(iii) 7005000

Solution:

(i) The predecessor of 10000 is 9999.

(ii) The predecessor of 807000 is 806999.

(iii) The predecessor of 7005000 is 7004999.

4. What is the difference between the largest number of five digits and the smallest number of six digits?

Solution:

99999 is the largest number of five digits

100000 is the largest number of six digits

Difference = 100000 – 99999 = 1

Therefore, 1 is the difference between the largest number of five digits and smallest number of six digits.

5. Find the difference between the largest number of 4 digits and the smallest number of 7 digits.

Solution:

9999 is the largest number of 4 digits

1000000 is the smallest number of 6 digits

Difference = 1000000 – 9999 = 990001

Therefore, 990001 is the difference between the largest number of 4 digits and the smallest number of 7 digits.

6. Rohit deposited Rs 125000 in his savings bank account. Later he withdrew Rs 35425 from it. How much money was left in his account?

Solution:

Money deposited in savings bank account = Rs 125000

Money withdrawn = Rs 35425

So the money which is left out in his account = 125000 – 35425 = Rs 89575

Hence, Rs 89575 is left in his account.

7. The population of a town is 96209. If the number of men is 29642 and that of women is 29167, determine the number of children.

Solution:

Population of a town = 96209

No. of men = 29642

No. of women = 29167

Total number of men and women = 29642 + 29167 = 58809

So the number of children = Population of a town – Total number of men and women

Number of children = 96209 – 58809 = 37400

Hence, there are 37400 children.

8. The digits of 6 and 9 of the number 36490 are interchanged. Find the difference between the original number and the new number.

Solution:

It is given that

Original Number = 39460

Number after interchanging 6 and 9 = 36490

Difference between them = 39460 – 36490 = 2790

Therefore, the difference between the original number and new number is 2970.

1. Write the opposite of each of the following:

(i) Increase in population

(ii) Depositing money in a bank

(iii) Earning money

(iv) Going North

(v) Gaining a weight of 4kg

(vi) A loss of Rs 1000

(vii) 25

(viii) – 15

Solution:

(i) The opposite of Increase in population is Decrease in population.

(ii) The opposite of Depositing money in a bank is Withdrawing money from a bank.

(iii) The opposite of earning money is Spending money.

(iv) The opposite of Going North is Going South.

(v) The opposite of gaining a weight of 4kg is losing a weight of 4kg.

(vi) The opposite of a loss of Rs 1000 is a gain of Rs 1000.

(vii) The opposite of 25 is – 25.

(viii) The opposite of – 15 is 15.

2. Indicate the following by using integers:

(i) 25o above zero

(ii) 5o below zero

(iii) A profit of Rs 800

(iv) A deposit of Rs 2500

(v) 3km above sea level

(vi) 2km below level

Solution:

(i) 25o above zero is + 25o.

(ii) 5o below zero is – 5o.

(iii) A profit of Rs 800 is + 800.

(iv) A deposit of Rs 2500 is + 2500.

(v) 3km above sea level is + 3.

(vi) 2km below level is – 2.

1. Write each of the following divisions as fractions:

(i) 6 ÷ 3

(ii) 25 ÷ 5

(iii) 125 ÷ 50

(iv) 55 ÷ 11

Solution:

(i) The division 6 ÷ 3 can be written as 6/3.

(ii) The division 25 ÷ 5 can be written as 25/5.

(iii) The division 125 ÷ 50 can be written as 125/50.

(iv) The division 55 ÷ 11 can be written as 55/11.

2. Write each of the following fractions as divisions:

(i) 9/7

(ii) 3/11

(iii) 90/63

(iv) 1/5

Solution:

(i) The fraction 9/7 can be written as 9 ÷ 7.

(ii) The fraction 3/11 can be written as 3 ÷ 11.

(iii) The fraction 90/63 can be written as 90 ÷ 63.

(iv) The fraction 1/5 can be written as 1 ÷ 5.

1. Choose the decimal (s) from the brackets which is (are) not equivalent to the given decimals:

(i) 0.8 (0.80, 0.85, 0.800, 0.08)

(ii) 25.1 (25.01, 25.10, 25.100, 25.001)

(iii) 45.05 (45.050, 45.005, 45.500, 45.0500)

Solution:

(i) 0.8 (0.80, 0.85, 0.800, 0.08)

We know that 0.85 and 0.08 are not equivalent to the given decimal

For 0.85, 5 is in the hundredth place and for 0.8 the hundredth value is 0.

For 0.08, 0 is in the tenth place and for 0.8 the tenth value is 8.

(ii) 25.1 (25.01, 25.10, 25.100, 25.001)

For 25.01, 0 is in the tenth place and for 25.001 the tenth value is 0.

(iii) 45.05 (45.050, 45.005, 45.500, 45.0500)

For 45.005, 0 is in the hundredth place and for 45.05 the hundredth value is 5

For 45.500, 5 is in the tenth place and for 45.05 the tenth value is 0.

2. Which of the following are like decimals:

(i) 0.34, 0.07, 5.35, 24.70

(ii) 45.05, 4.505, 20.55, 20.5

(iii) 8.80, 17.08, 8.94, 0.27

(iv) 4.50, 16.80, 0.700, 7.08

Solution:

(i) 0.34, 0.07, 5.35, 24.70

The given values are like decimals because equal number of digits are present after the decimal point.

(ii) 45.05, 4.505, 20.55, 20.5

The given values are unlike decimals because different number of digits are present after the decimal point.

(iii) 8.80, 17.08, 8.94, 0.27

The given values are like decimals because equal number of digits are present after the decimal point.

(iv) 4.50, 16.80, 0.700, 7.08

The given values are unlike decimals because different number of digits are present after the decimal point.

3. Which of the following statements are correct?

(i) 8.05 and 7.95 are like decimals.

(ii) 0.95, 0.306, 7.10 are unlike decimals.

(iii) 3.70 and 3.7 are like decimals.

(iv) 13.59, 1.359, 135.9 are like decimals.

(v) 5.60, 3.04, 0.45 are like decimals.

Solution:

(i) Correct because the two decimals have same number of digits present after decimal point.

(ii) Correct because the three decimals have different number of digits present after decimal point.

(iii) Incorrect because the two decimals have different number of digits present after decimal point.

(iv) Incorrect because the three decimals have different number of digits present after decimal point.

(v) Correct because the three decimals have same number of digits present after decimal point.

4. Convert each of the following sets of unlike decimals to like decimal:

(i) 7.8, 7.85

(ii) 2.02, 3.2

(iii) 0.6, 5.8, 12.765

(iv) 5.296, 5.2, 5.29

(v) 4.3294, 43.29, 432.94

Solution:

(i) 7.8, 7.85

In the given values we know that 7.85 contains 2 digits after decimal point so by changing 7.8 as 7.80

Hence, 7.80 and 7.85 are like decimals.

(ii) 2.02, 3.2

In the given values we know that 2.02 contains 2 digits after decimal point so by changing 3.2 as 3.20

Hence, 2.02 and 3.20 are like decimals.

(iii) 0.6, 5.8, 12.765

In the given values we know that 12.765 contains 3 digits after decimal point so by changing 0.6 as 0.600 and 5.8 as 5.800

Hence, 0.600, 5.800 and 12.765 are like decimals.

(iv) 5.296, 5.2, 5.29

In the given values we know that 5.296 contains 3 digits after decimal point so by changing 5.2 as 5.200 and 5.29 as 5.290

Hence, 5.296, 5.200 and 5.290 are like decimals.

(v) 4.3294, 43.29, 432.94

In the given values we know that 4.3294 contains 4 digits after decimal point so by changing 43.29 as 43.2900 and 432.94 as 432.9400

Hence, 4.3294, 43.2900 and 432.9400 are like decimals.

1. Write each of the following products in exponential form:

(i) a × a × a × a × …….. 15 times

(ii) 8 × b × b × b × a × a × a × a

(iii) 5 × a × a × a × b × b × c × c × c

(iv) 7 × a × a × a …….. 8 times × b × b × b × …… 5 times

(v) 4 × a × a × …… 5 times × b × b × ……. 12 times × c × c …… 15 times

Solution:

(i) a × a × a × a × …….. 15 times is written in exponential form as a15.

(ii) 8 × b × b × b × a × a × a × a is written in exponential form as 8a4b3.

(iii) 5 × a × a × a × b × b × c × c × c is written in exponential form as 5a3b2c3.

(iv) 7 × a × a × a …….. 8 times × b × b × b × …… 5 times is written in exponential form as 7a8b5.

(v) 4 × a × a × …… 5 times × b × b × ……. 12 times × c × c …… 15 times is written in exponential form as 4a5b12c15.

2. Write each of the following in the product form:

(i) a2 b5

(ii) 8x3

(iii) 7a3b4

(iv) 15 a9b8c6

(v) 30x4y4z5

(vi) 43p10q5r15

(vii) 17p12q20

Solution:

(i) a2 b5 is written in the product form as a × a × b × b × b × b × b.

(ii) 8x3 is written in the product form as 8 × x × x × x.

(iii) 7a3b4 is written in the product form as 7 × a × a × a × b × b × b × b.

(iv) 15 a9b8c6 is written in the product form as 15 × a × a …… 9 times × b × b × … 8 times × c × c × ….. 6 times.

(v) 30x4y4z5 is written in the product form as 30 × x × x × x × x × y × y × y × y × z × z × z × z × z.

(vi) 43p10q5r15 is written in the product form as 43 × p × p …. 10 times × q × q …. 5 times × r × r × …. 15 times.

(vii) 17p12q20 is written in the product form as 17 × p × p …. 12 times × q × q × ….. 20 times.

3. Write down each of the following in exponential form:

(i) 4a3 × 6ab2 × c2

(ii) 5xy × 3x2y × 7y2

(iii) a3 × 3ab2 × 2a2b2

Solution:

(i) 4a3 × 6ab2 × c2 is written in exponential form as 24a4b2c2.

(ii) 5xy × 3x2y × 7y2 is written in exponential form as 105x3y4.

(iii) a3 × 3ab2 × 2a2b2 is written in exponential form as 6a6b4.

4. The number of bacteria in a culture is x now. It becomes square of itself after one week. What will be its number after two weeks?

Solution:

Number of bacteria in a culture = x

It is given that

Number of bacteria becomes square of itself in one week = x2

So the number of bacteria after two weeks = (x2)2 = x4

Hence, the number of bacteria after two weeks is x4.

5. The area of a rectangle is given by the product of its length and breadth. The length of a rectangle is two-third of its breadth. Find its area if its breadth is x cm.

Solution:

It is given that

Area of rectangle = l × b

Length = (2/3) x cm

So the area of the rectangle = (2/3) x × x = 2/3 x2 cm2

Hence, the area of rectangle is (2/3) x2 cm2.

6. If there are x rows of chairs and each row contains x2 chairs. Determine the total number of chairs.

Solution:

Number of rows of chairs = x

Each row contains = x2 chairs

So the total number of chairs = number of rows of chairs × chairs in each row

We get

Total number of chairs = x × x2 = x3

Hence, the total number of chairs is x3.

6. You are standing in a class-room facing north. In what direction are you facing after making a quarter turn?

Solution:

By making a quarter turn (90o), I will be facing towards east if I turn to my right hand and if I turn to my left hand, I will be facing towards west.

7. A bicycle wheel makes four and a half turns. Find the number of right angles through which it turns.

Solution:

We know that the wheel of a bicycle covers 360o in one turn.

It can be written as

360/90 = 4 right angles

We know that in four and half turns the wheel turns by 4 (4.5) = 18 right angles

Hence, the number of right angles through which it turns is 18.

8. Look at your watch face. Through how many right angles does the minute-hand moves between 8: 00 O’ clock and 10: 30 O’ clock?

Solution:

We know that the time interval between 8: 00 O’ clock and 10: 30 O’ clock is two and half hours

The minute hand turns 360o in 1 hour

360/90 = 4 right angles

So in two and half hours the minute hand turns by 2.5 (4) = 10 right angles.

Hence, the minute hand turns by 10 right angles.

9. If a bicycle wheel has 48 spokes, then find the angle between a pair of adjacent spokes.

Solution:

The central angle in a bicycle is 360o which consists of 48 spokes.

So the angle between a pair of adjacent spokes = 360/48 = 7.5o

Hence, the angle between a pair of adjacent spokes is 7.5o.

Types of Angles

• Acute Angle: An angle whose measure is less than 90° is called an acute angle.
• Right Angle: An angle whose measure is 90° is called right angle.
• Obtuse Angle: An angle whose measure is greater than 90° but less than 180° is called an obtuse angle.
• Straight Angle:
• Reflex Angle:
• Zero Angle:

What is the measure of right angle Class 6?

Answer: An acute angle measures between 0° and 90°; an obtuse angle measures between 90° and 180°; a straight angle measures 180°; a right angle measures 90°; a zero angle measures 0° and a complete angle measures 360°.

What type of angle is 26 degrees?

acute angle-an angle between 0 and 90 degrees. right angle-an 90 degree angle. obtuse angle-an angle between 90 and 180 degrees.

(i) Triangle – A plane figure formed by three non-parallel line segments is called a triangle.

(ii) Parts or elements of a triangle – The three sides, three angles and three vertices of a triangles are called parts or elements of a triangle.

(iii) Scalene triangle – A triangle in which all the sides are different is called a scalene triangle.

(iv) Isosceles triangle – A triangle whose two sides are equal is called an isosceles triangle.

(v) Equilateral triangle – A triangle whose all sides are equal to one another is called an equilateral triangle.

(vi) Acute triangle – A triangle whose all the angles are acute is called acute-angled triangle or an acute triangle.

(vii) Right triangle – A triangle whose one angle is a right angle is called a right-angled triangle or a right triangle.

(viii) Obtuse triangle – A triangle whose one angle is obtuse is called an obtuse-angled triangle or an obtuse triangle.

(ix) Interior of a triangle – The region which lies inside the triangle is called the interior of a triangle.

(x) Exterior of a triangle – The region which lies outside the triangle is called the exterior of a triangle.

What is the correct name for a quadrilateral?

Parallelogram is the proper name of the quadrilateral.

A quadrilateral is a polygon. In fact it is a 4-sided polygon, just like a triangle is a 3-sided polygon, a pentagon is a 5-sided polygon, and so on.

What makes a shape a quadrilateral?

A quadrilateral is a polygon that has exactly four sides. (This also means that a quadrilateral has exactly four vertices, and exactly four angles.)

A quadrilateral is a 4-sided two-dimensional shape. There are 4 types of quadrilaterals: parallelograms, trapezoids, rectangles, and rhombuses. A rectangle has 4 right angles. A rhombus is a quadrilateral with all 4 sides of the same length.

1. Explain the following:

(i) Circle

(iii) Centre

(iv) Diameter

(v) Chord

(vi) Interior of a circle.

Solution:

(i) Circle – A circle is a set of all those points in a plane whose distance from a fixed point remains constant.

(ii) Radius – The radius of a circle is the distance between the all the points of the circle to its centre.

(iii) Centre – The centre of a circle is a fixed point which is at a constant distance from all the points.

(iv) Diameter – A line segment passing through the centre of a circle, and having its end-points on the circle is called a diameter of the circle.

(v) Chord – A line segment with its end-points lying on a circle is called the chord of the circle.

(vi) Interior of a circle – The part of a plane inside the circle consisting of all the points is called the interior of a circle.

What is diameter of a circle class 6?

Diameter of a circle is a chord of the circle, not radius. It is the longest chord of the circle. A perpendicular drawn from the centre of the circle to the chord, bisects the chord. It is a line passing through the circle that intersects the circle at two points.

State the properties of a transversal.

The properties of a transversal are that first one being over here, the vertically opposite angles are equal. Further, the corresponding angles are equal and the interior angles which form on the same side of the transversal are supplementary. Finally, the alternate angles are equal.

Name any four objects from your environment, which have the form of

(i) a cuboid (ii) a cube

Solution:

(i) The objects which have the form of a cuboid are a compass box, a lunch box, a duster and a book.

(ii) The objects which have the form of a cube are a chalk box, a disc, a tissue box and a cubical cabin.

How do you describe a three-dimensional shape?

In geometry, a three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.

Why is it important to learn about 3D shapes?

Learning shapes not only helps children identify and organize visual information, it helps them learn skills in other curriculum areas including reading, math, and science. ... Learning shapes also helps children understand other signs and symbols.

What is line of symmetry class 6?

When a figure is folded along a line such a way that the two parts exactly fit on top of each other, then the figure is said to have a line symmetry. Line of symmetry is the line which divides figure into two identical parts and these are mirror image of each others.

Why is symmetry important?

Symmetry is a fundamental part of geometry, nature, and shapes. It creates patterns that help us organize our world conceptually. We see symmetry every day but often don't realize it. Geometry software is a very important tool in developing and testing individual ideas in the classroom.

Which triangle has 2 lines of symmetry?

The division of triangles into scalene, isosceles, and equilateral can be thought of in terms of lines of symmetry. A scalene triangle is a triangle with no lines of symmetry while an isosceles triangle has at least one line of symmetry and an equilateral triangle has three lines of symmetry.

What are the steps in writing a construction class 6?

Steps for construction:

1. Draw a line BX and take a point A, such that AB is equal to 4.5 cm.
2. Draw ∠ABP = 60° with the help of protractor.
3. With A as the centre and a radius of 5 cm, draw an arc cutting PB at C.
4. Draw AC.
5. Now, draw ∠BCY = 60°.
6. Then, draw ∠ABW, such that ∠ABW is equal to∠CAX, which cut the ray CY at D

## Introduction to Mensuration

### Mensuration

• Mensuration is the branch of mathematics that deals with the measurement of length, area or volume of various geometric shapes.

### Shapes

• A shape is the form of an object.
• Examples of two-dimensional shapes are square, rectangle and triangle, and of three-dimensional shapes are cube, cuboid and sphere.

### What is the difference between mensuration and geometry?

Mensuration refers to the calculation of various parameters of shapes like the perimeter, area, volume, etc. whereas, Geometry deals with the study of properties and relations of points and lines of various shapes.

### What are 2D and 3D Mensuration?

2D mensuration deals with the calculation of various parameters like area and perimeter of 2-dimensional shapes like square, rectangle, circle, triangles, etc.

3D mensuration is concerned with the study and calculation of surface area, lateral surface area, and volume of 3-dimensional figures like a cube, sphere, cuboid, cone, cylinder, etc.

## Types of data

Data handling method can be performed based on the types of data. The data is classified into two types, such as:

• Qualitative Data
• Quantitative Data

Qualitative data gives descriptive information of something whereas quantitative data gives numerical information about something. Here, the quantitative data is further divided into two. They are discrete data and continuous data. The discrete data can take only certain values such as whole numbers. The continuous data can take a value within the provided range.

## How to Represent Data?

The data can be usually represented in any one of the following ways. They are:

• Bar Graph
• Line Graphs
• Pictographs
• Histograms
• Stem and Leaf Plot
• Dot Plots
• Frequency Distribution
• Cumulative Tables and Graphs