Topics
Integers
- Natural Numbers
- Whole Numbers
- Negative and Positive Numbers
- Integers
- Representation of Integers on the Number Line
- Ordering of Integers
- Addition of Integers
- Subtraction of Integers
- Properties of Addition and Subtraction of Integers
- Multiplication of a Positive and a Negative Integers
- Multiplication of Two Negative Integers
- Product of Three Or More Negative Integers
- Closure Property of Multiplication of Integers
- Commutative Property of Multiplication of Integers
- Multiplication of Integers with Zero
- Multiplicative Identity of Integers
- Associative Property of Multiplication of Integers
- Distributive Property of Multiplication of Integers
- Making Multiplication Easier of Integers
- Division of Integers
- Properties of Division of Integers
Fractions and Decimals
- Concept of Fraction
- Types of Fractions
- Concept of Proper and Improper Fractions
- Concept of Mixed Fractions
- Concept of Equivalent Fractions
- Like and Unlike Fraction
- Comparing Fractions
- Addition of Fraction
- Subtraction of Fraction
- Multiplication of a Fraction by a Whole Number
- Using Operator 'Of' with Multiplication and Division
- Multiplication of Fraction
- Division of Fractions
- Concept of Reciprocals or Multiplicative Inverses
- Problems Based on Fraction
- The Decimal Number System
- Comparing Decimal Numbers
- Addition of Decimal Fraction
- Subtraction of Decimal Numbers
- Multiplication of Decimal Numbers
- Division of Decimal Numbers
- Problems Based on Decimal Numbers
Data Handling
Simple Equations
Lines and Angles
The Triangle and Its Properties
- Basic Concepts of Triangles
- Classification of Triangles based on Sides
- Classification of Triangles based on Angles
- Median of a Triangle
- Altitudes of a Triangle
- Exterior Angle of a Triangle and Its Property
- Some Special Types of Triangles - Equilateral and Isosceles Triangles
- Basic Properties of a Triangle
- Right-angled Triangles and Pythagoras Property
Comparing Quantities
- Ratio
- Concept of Equivalent Ratios
- Proportion
- Unitary Method
- Basic Concept of Percentage
- Estimation in Percentages
- Interpreting Percentages
- Conversion between Percentage and Fraction or Decimal
- Ratios to Percents
- Increase Or Decrease as Percent
- Basic Concepts of Profit and Loss
- Profit or Loss as a Percentage
- Calculation of Interest
Congruence of Triangles
- Similarity and Congruency of Figures
- Congruence Among Line Segments
- Congruence of Angles
- Congruence of Triangles
- Criteria for Congruence of Triangles
- Criteria for Similarity of Triangles
- SAS Congruence Criterion
- ASA Congruence Criterion
- RHS Congruence Criterion
- Exceptional Criteria for Congruence of Triangles
Rational Numbers
- Rational Numbers
- Equivalent Rational Number
- Positive and Negative Rational Numbers
- Rational Numbers on a Number Line
- Rational Numbers in Standard Form
- Comparison of Rational Numbers
- Rational Numbers Between Two Rational Numbers
- Addition of Rational Number
- Subtraction of Rational Number
- Multiplication of Rational Numbers
- Division of Rational Numbers
Perimeter and Area
- Basic Concepts in Mensuration
- Concept of Perimeter
- Perimeter of a Rectangle
- Perimeter of Squares
- Perimeter of Triangle
- Perimeter of Polygon
- Concept of Area
- Area of Square
- Area of Rectangle
- Triangles as Parts of Rectangles and Square
- Generalising for Other Congruent Parts of Rectangles
- Area of a Parallelogram
- Area of a Triangle
- Circumference of a Circle
- Area of Circle
- Conversion of Units
- Problems based on Perimeter
- Problems based on Area
Practical Geometry
- Construction of a Line Parallel to a Given Line, Through a Point Not on the Line
- Construction of Triangles
- Constructing a Triangle When the Length of Its Three Sides Are Known (SSS Criterion)
- Constructing a Triangle When the Lengths of Two Sides and the Measure of the Angle Between Them Are Known. (SAS Criterion)
- Constructing a Triangle When the Measures of Two of Its Angles and the Length of the Side Included Between Them is Given. (ASA Criterion)
- Constructing a Right-angled Triangle When the Length of One Leg and Its Hypotenuse Are Given (RHS Criterion)
Algebraic Expressions
Exponents and Powers
- Concept of Exponents
- Multiplying Powers with the Same Base
- Dividing Powers with the Same Base
- Taking Power of a Power
- Multiplying Powers with Different Base and Same Exponents
- Dividing Powers with Different Base and Same Exponents
- Numbers with Exponent Zero, One, Negative Exponents
- Miscellaneous Examples Using the Laws of Exponents
- Decimal Number System Using Exponents and Powers
- Crores
Symmetry
Visualizing Solid Shapes
- Definition: Unitary Method
- Steps for Applying the Unitary Method
- Example 1
- Example 2
- Example 3
- Example 4
- Real-Life Examples
- Key Points Summary
Definition: Unitary Method
The unitary method is a process used to find the value of a single unit from the value of multiple units and then find the value of multiple units from the value of a single unit.
Steps for Applying the Unitary Method
1. Express the problem using a mathematical statement.
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Write down the known information and what needs to be found.
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Example: You know 15 pens cost ₹360, but you need to find the cost of 8 pens.
2. Find the Cost of One Item (Using Division)
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Divide the total value by the number of items.
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Example: ₹360 ÷ 15 = ₹24 (So, 1 pen costs ₹24).
3. Find the Total Cost (Use Multiplication)
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Multiply the cost of 1 item by the number of items.
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Example: ₹24 × 8 = ₹192 (So, the cost of 8 pens is ₹192).
Example 1
A bunch of 15 bananas costs 45 rupees. How much will 8 bananas cost?
Solution:
- Express the given problem as a mathematical statement:
Cost of 15 bananas = ₹45
Find the cost of 8 bananas. - Find the value of one unit of the given item using division:
Cost of 1 banana = ₹45 ÷ 15 = ₹3 - Find the value of the required number of items using multiplication:
Cost of 8 bananas = ₹3 × 8 = ₹24
Therefore, the cost of 8 bananas is 24 rupees.
Example 2
For ₹384, a man can buy 12 articles. How many articles can he buy for ₹512?
Solution:
- First, find out how many articles can be bought for ₹1:
⇒ For ₹1, he can buy `12 / 384` articles - Then, find out how many articles can be bought for ₹512:
For ₹512, he can buy `12 / 384` × 512 articles = 16 articles. - Answer: 16 articles can be bought for ₹512.
Example 3
If 25 identical articles weigh 275 g, find how many articles will weigh 990 g.
Solution:
- First, find the weight of 1 article:
⇒ 1 g is the weight of `25 / 275` articles. - Then, find how many articles weigh 990 g:
And 990 g is the weight of `25 / 275` × 990 articles = 90 articles. - Answer: 90 articles will weigh 990 g.
Example 4
18 men can make 90 identical tables in one day. Find how many men will make 20 such tables in one day.
Solution:
- In one day, 90 tables are made by 18 men.
⇒ In one day, 1 table is made by `18 / 90` men
And, in one day, 20 tables are made by `18 / 90` × 20 men = 4 men - Answer: 4 men will make 20 tables.
Real-Life Examples
Suppose 6 apples cost ₹30.
- Step 1: One apple?
₹30 ÷ 6 = ₹5 - Step 2: What about 4 apples?
₹5 × 4 = ₹20
Key Points Summary
Unitary Method: Find the value of 1 unit, then scale up/down.
- Use division to find the value of 1 unit and multiplication to find the value of many units.
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Great for shopping, travel, sharing, and work problems.
Example Question 1
A motorbike travels 220 km in 5 litres of petrol. How much distance will it cover in 1.5 litres of petrol?
In 5 litres of petrol, motorbike can travel 220 km.
Therefore, in 1 litre of petrol, motorbike travels = `220/5` km.
Therefore, in 1.5 litres, motorbike travels
= `220/5 xx 1.5 "km" = 220/5 xx 15/10` km = 66 km.
Thus, the motorbike can travel 66 km in 1.5 litres of petrol.
Example Question 2
If the cost of a dozen soaps is Rs. 153.60, what will be the cost of 15 such soaps?
We know that 1 dozen = 12
Since, cost of 12 soaps = Rs. 153.60
Therefore, cost of 1 soap = `153.60/12` = Rs. 12.80
Therefore, cost of 15 soaps = Rs. 12.80 × 15 = Rs. 192.
Thus, cost of 15 soaps is Rs. 192.
Example Question 3
Cost of 105 envelopes is Rs. 350. How many envelopes can be purchased for Rs. 100?
In Rs. 350, the number of envelopes that can be purchased = 105.
Therefore, in Rs. 1, number of envelopes that can be purchased = `105/350`
Therefore, in Rs. 100, the number of envelopes that can be purchased
= `105/350 × 100 = 30`.
Thus, 30 envelopes can be purchased for Rs. 100.
Example Question 4
A car travels 90 km in 2 1/2hours.
(a) How much time is required to cover 30 km with the same speed?
(b) Find the distance covered in 2 hours with the same speed.
(a) In this case, time is unknown and distance is known. Therefore, we proceed as follows:
`2 1/2 "hours" = 5/2 "hours" = 5/2 × 60 "minutes" = 150 "minutes"`.
90 km is covered in 150 minutes.
Therefore, 1 km can be covered in `(150)/(90)` minutes.
Therefore, 30 km can be covered in `(150)/(90) × 30` minutes i.e., 50 minutes.
Thus, 30 km can be covered in 50 minutes.
(b) In this case, distance is unknown and time is known. Therefore, we proceed as follows:
Distance covered in `2 1/2 "hours (i.e.," 5/2` hours ) = 90 km.
Therefore, distance covered in 1 hour = 90 ÷ `5/2 "km" = 90 × 2/5` = 36 km.
Therefore, distance covered in 2 hours = 36 × 2 = 72 km.
Thus, in 2 hours, the distance covered is 72 km.
