Topics
Knowing Our Numbers
Comparing Number
 Comparing Multiple Digit of Numbers
 Compare Numbers in Ascending and Descending Order
 Compare Number by Forming Numbers from a Given Digits
 Compare Numbers by Shifting Digits
 Introducing a 5 Digit Number  10,000
 Revisiting Place Value of Numbers
 Expansion Form of Numbers
 Introducing the Six Digit Number  1,00,000
 Larger Number of Digits 7 and Above
 An Aid in Reading and Writing Large Numbers
 Using Commas in Indian and International Number System
Large Numbers in Practice
Whole Numbers
 Concept for Natural Numbers
 Concept for Whole Numbers
 Successor and Predecessor of Whole Number
 Operation on of Whole Number on Number Line
 Properties of Whole Numbers
 Closure Property of Whole Number
 Associativity Property of Whole Numbers
 Division by Zero
 Commutativity Property of Whole Number
 Distributivity Property of Whole Numbers
 Identity of Addition and Multiplication of Whole Numbers
 Patterns in Whole Numbers
Playing with Numbers
 Arranging the Objects in Rows and Columns
 Factors and Multiples
 Concept of Perfect Number
 Concept of Prime Numbers
 Concept of Coprime Number
 Concept of Twin Prime Numbers
 Concept of Even and Odd Number
 Concept of Composite Number
 Concept of Sieve of Eratosthenes
 Tests for Divisibility of Numbers
 Divisibility by 10
 Divisibility by 5
 Divisibility by 2
 Divisibility by 3
 Divisibility by 6
 Divisibility by 4
 Divisibility by 8
 Divisibility by 9
 Divisibility by 11
 Common Factor
 Common Multiples
 Some More Divisibility Rules
 Prime Factorisation
 Highest Common Factor
 Lowest Common Multiple
Basic Geometrical Ideas
 Concept for Basic Geometrical Ideas (2 d)
 Concept of Points
 Concept of Line
 Concept of Line Segment
 Concept of Ray
 Concept of Intersecting Lines
 Concept of Parallel Lines
 Concept of Curves
 Different Types of Curves  Closed Curve, Open Curve, Simple Curve.
 Concept of Polygons  Side, Vertex, Adjacent Sides, Adjacent Vertices and Diagonal
 Concept of Angle  Arms, Vertex, Interior and Exterior Region
 Concept of Triangles  Sides, Angles, Vertices, Interior and Exterior of Triangle
 Concept of Quadrilaterals  Sides, Adjacent Sides, Opposite Sides, Angle, Adjacent Angles and Opposite Angles
 Concept of Circle  Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
Understanding Elementary Shapes
 Introduction to Understanding Elementary Shapes
 Measuring Line Segments
 Concept of Angle  Arms, Vertex, Interior and Exterior Region
 Right, Straight, and Complete Angle by Direction and Clock
 Acute, Right, Obtuse, and Reflex angles
 Measuring Angles
 Perpendicular Line and Perpendicular Bisector
 Classification of Triangles (On the Basis of Sides, and of Angles)
 Equilateral Triangle
 Isosceles Triangles
 Scalene Triangle
 Acute Angled Triangle
 Obtuse Angled Triangle
 Right Angled Triangle
 Types of Quadrilaterals
 Properties of a Square
 Properties of Rectangle
 Properties of a Parallelogram
 Properties of Rhombus
 Properties of Trapezium
 Three Dimensional Shapes
 Concept of Prism
 Concept of Pyramid
Integers
Fractions
Decimals
 Concept of Decimal Numbers
 Place Value in the Context of Decimal Fraction.
 Concept of Tenths, Hundredths and Thousandths in Decimal
 Representing Decimals on the Number Line
 Interconversion of Fraction and Decimal
 Comparing Decimal Numbers
 Using Decimal Number as Units
 Addition of Decimal Numbers
 Subtraction of Decimals Fraction
Data Handling
Mensuration
Algebra
Ratio and Proportion
Symmetry
Practical Geometry
 Introduction to Practical Geometry
 Construction of a Circle When Its Radius is Known
 Construction of a Line Segment of a Given Length
 Constructing a Copy of a Given Line Segment
 Drawing a Perpendicular to a Line at a Point on the Line
 Drawing a Perpendicular to a Line Through a Point Not on It
 Drawing the Perpendicular Bisector of a Line Segment
 Constructing an Angle of a Given Measure
 Constructing a Copy of an Angle of Unknown Measure
 Constructing a Bisector of an Angle
 Angles of Special Measures  30°, 45°, 60°, 90°, and 120°
notes
Highest Common Factor:
The Highest Common Factor(HCF) of two or more given numbers is the highest (or greatest) of their common factors. It is also known as Greatest Common Divisor (GCD).
a) HCF By Prime Factorization Method:
i) Find the HCF of 24 and 32 by the prime factors method.
24 = 4 × 6
= 2 × 2 × 2 × 3
32 = 8 × 4
= 2 × 2 × 2 × 2 × 2
The common factor 2 occurs thrice in each number. Therefore, the HCF = 2 × 2 × 2 = 8.
ii) Find the HCF of 195, 312, 546.
195 = 5 × 39 = 5 × 3 × 13 
2 = 4 × 78 = 2 × 2 × 2 × 3 = 2 × 2 × 2 × 3 × 13 
6 = 2 × 273 = 2 × 3 × 91 = 2 × 3 × 7 × 13 
The common factors 3 and 13 each occur once in all the numbers.
∴ HCF = 3 × 13 = 39
iii) Find the HCF of 60, 12, 36
Let us work out this example in the vertical arrangement. We write all the numbers in one line and find their factors.
∴ HCF = 2 × 2 × 3 = 12
Note that 12 is a divisor of 36 and 60.
Note:

If one of the given numbers is a divisor of all the others, then it is the HCF of the given numbers.

If no prime number is a common divisor of all the given numbers, then 1 is their HCF because it is the only common divisor.
b) The Division Method for Finding the HCF:
i) Find the HCF of 144 and 252.
1. Divide the bigger number by the smaller one.
2. Divide the previous divisor by the remainder of this division.
3. Divide the divisor of step 2 by the remainder obtained in the division in step 2.
4. Continue like this till the remainder becomes zero. The divisor in the division in which the remainder is zero is the HCF of the given numbers.
∴ The HCF of 144 and 252 = 36.
ii) Reduce `(209)/(247)` to its simplest form.
To reduce the number to its simplest form, we will find the common factors of 209 and 247.
Let us find their HCF by the division method.
Here, 19 is the HCF. That is, the numerator and denominator are both divisible by 19.
∴ `(209)/(247) = (209 ÷ 19)/(247 ÷ 19) = 11/13`.
Example
Find the HCF of the following numbers: 20, 28, and 36.
The HCF of 20, 28 and 36 can also be found by prime factorisation of these numbers as follows:
Thus,
20 = 2 × 2 × 5
28 = 2 × 2 × 7
36 = 2 × 2 × 3 × 3.
The common factor of 20, 28, and 36 is 2 (occurring twice).
Thus, HCF of 20, 28 and 36 is 2 × 2 = 4.
Example
Two tankers contain 850 litres and 680 litres of kerosene oil respectively. Find the maximum capacity of a container which can measure the kerosene oil of both the tankers when used an exact number of times.
The required container has to measure both the tankers in a way that a count is an exact number of times. So its capacity must be an exact divisor of the capacities of both the tankers. Moreover, this capacity should be maximum. Thus, the maximum capacity of such a container will be the HCF of 850 and 680.
Hence,
850 = 2 × 5 × 5 × 17= 2 × 5 × 17 × 5 and
680 = 2 × 2 × 2 × 5 × 17= 2 × 5 × 17 × 2 × 2
The common factors of 850 and 680 are 2, 5 and 17.
Thus, the HCF of 850 and 680 is 2 × 5 × 17 = 170.
Therefore, the maximum capacity of the required container is 170 liters.
It will fill the first container in 5 and the second in 4 refills.
Example
There are 20 kg of jowar and 50 kg of wheat in a shop. All the grain is to be packed in bags. If all the bags are to have equal weights of grain, what is the maximum weight of grain that can be filled in each bag?
The weight of the grain in each bag must be a factor of 20 and 50.
Besides, the maximum possible weight must be filled in each bag. Hence, let us find the HCF of 20 and 50.
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 50: 1, 2, 5, 10, 25, 50
Common factors: 1, 2, 5, 10
Of the common factors of 20 and 50, 10 is the greatest, i.e. 10 is the HCF of the numbers 20 and 50.
Therefore, a maximum of 10 kg of grain can be filled in each bag.
Example
Find the HCF of 10, 15, 12.
10 = 2 × 5
15 = 3 × 5
12 = 2 × 2 × 3
No number except 1 is a common divisor.
Hence, HCF = 1.
Example
A shop sells a 450 g bottle of jam for 96 rupees and a bigger bottle of 600 g for 124 rupees. Which bottle is it more profitable to buy?
Let us use 150, the HCF of 450 and 600 to compare.
450 = 150 × 3, 600 = 150 × 4
∴ In the small bottle, 150 g of jam costs `96/3` = 32 rupees.
In the large bottle, 150 g of jam costs `(124)/4` = 31 rupees.
∴ Thus, it is more profitable to buy the 600 g bottle of jam.