IM_TM_G07_MA_MB_Text_AY25-26_eBook by Uolo

Master Mathematical Thinking MATHEMATICS

Teacher Manual

7

MATHEMATICS

Master Mathematical Thinking

Acknowledgements

Academic Authors: Muskan Panjwani, Anjana AR, Anuj Gupta, Simran Singh

Creative Directors: Bhavna Tripathi, Mangal Singh Rana, Satish

Book Production: Sanjay Kumar Goel, Vishesh Agarwal

Project Lead: Neena Aul

VP, Learning: Abhishek Bhatnagar

All products and brand names used in this book are trademarks, registered trademarks or trade names of their respective owners.

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Book Title: Imagine Mathematics Teacher Manual 7

ISBN: 978-81-984519-5-8

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Preface

Mathematics is an essential tool for understanding the world around us. It is not just another subject, but an integral part of our lives. It shapes the very foundation of our understanding, personality and interaction with the world around us. Studies from across the globe have shown that proficiency in mathematics significantly influences career prospects and lifelong learning.

According to the NEP 2020, mathematics and mathematical thinking are crucial for empowering individuals in their everyday interactions and affairs. It focuses on competencies-based education, which essentially means actively and effectively applying mathematical concepts in real life. It also encourages innovative approaches for teaching maths, including regular use of puzzles, games and relatable real-world examples to make the subject engaging and enjoyable.

It is in this spirit that Uolo has introduced the Imagine Mathematics product for elementary grades (1 to 8). This product’s key objective is to eliminate the fear of mathematics by making math exciting, relatable and meaningful for children.

Imagine Mathematics positions itself on the curricular and pedagogical approach of the Gradual Release of Responsibility (GRR), which has been highly recommended by the NEP 2020, the NCF 2023 and other literature in learning and educational pedagogies. Subsequent pages elaborate further on this approach and its actualisation in this book.

This book incorporates highly acclaimed, learner-friendly teaching strategies. Each chapter introduces concepts through real-life situations and storytelling, connecting to children’s experiences and transitioning smoothly from concrete to abstract. These teacher manuals are designed to be indispensable companions for educators, providing well-structured guidance to make teaching mathematics both effective and enjoyable. With a focus on interactive and hands-on learning, the manuals include a variety of activities, games, and quizzes tailored to enhance conceptual understanding. By integrating these engaging strategies into the classroom, teachers can foster critical thinking and problem-solving skills among students. Moreover, the resources emphasise creating an enriched and enjoyable learning environment, ensuring that students not only grasp mathematical concepts but also develop a genuine interest in the subject.

In addition, the book is technologically empowered and works in sync with a parallel digital world, which contains immersive gamified experiences, video solutions and practice worksheets, among other things. Interactive exercises on the digital platform make learning experiential and help in concrete visualisation of abstract mathematical concepts. We invite educators, parents and students to embrace Imagine Mathematics and join us in nurturing the next generation of thinkers, innovators and problem-solvers. Embark on this exciting journey with us and let Imagine Mathematics be a valuable resource in your educational adventure.

Numbers up to 8 Digits 1

Imagine Mathematics Headings: Clear and concise lessons, aligned with the topics in the Imagine Mathematics book, designed for a seamless implementation.

Alignment

C-1.1:

C-4.3:

Numbers up to 8 Digits 1

Imagine Mathematics Headings

Place Value, Face Value and Expanded Form

Indian and International Number Systems

Comparing and Ordering Numbers

Numbers up to 8 Digits 1

Learning Outcomes: Clear, specific and measurable learning outcomes that show what students should know, understand, or do by the end of the lesson.

Learning Outcomes

Students will be able to:

Rounding–off Numbers

Learning Outcomes

Students will be able to: write the place value, face value, expanded form and number names for numbers up to write numbers up to 8 digits in the Indian and International number system. compare numbers up to 8 digits and arrange them in ascending and descending order. round off numbers up to 8 digits to the nearest 10, 100 and 1000.

Alignment to NCF

Numbers up to 8 Digits 1

C-1.1: Represents numbers using the place value structure of the Indian number system, numbers, and knows and can read the names of very large numbers

C-4.3: Selects appropriate methods and tools for computing with whole numbers, such as computation, estimation, or paper pencil calculation, in accordance with the context

C-5.1: Understands the development of zero in India and the Indian place value system for the history of its transmission to the world, and its modern impact on our lives and in all technology

Let’s Recall

write the place value, face value, expanded form and number names for numbers up to 8 digits. write numbers up to 8 digits in the Indian and International number system. compare numbers up to 8 digits and arrange them in ascending and descending order. round off numbers up to 8 digits to the nearest 10, 100 and 1000.

Alignment to NCF

Place Value, Face Value and Expanded Form

Recap to check if students know how to write the place value, expanded form and number 6-digit numbers.

Indian and International Number Systems

Ask students to solve the questions given in the Let’s Warm-up section.

Comparing and Ordering Numbers

C-1.1: Represents numbers using the place value structure of the Indian number system, compares whole numbers, and knows and can read the names of very large numbers

Alignment to NCF: Learning Outcomes as recommended by NCF 2023.

Vocabulary

Rounding–off Numbers

C-4.3: Selects appropriate methods and tools for computing with whole numbers, such as mental computation, estimation, or paper pencil calculation, in accordance with the context

Learning Outcomes

C-5.1: Understands the development of zero in India and the Indian place value system for writing numerals, the history of its transmission to the world, and its modern impact on our lives and in all technology

Let’s Recall

expanded form: writing a number as the sum of the values of all its digits order: the way numbers are arranged estimating: guessing an answer that is close to the actual answer rounding off: approximating a number to a certain place value for easier calculation

Teaching Aids

Recap to check if students know how to write the place value, expanded form and number names for 6-digit numbers.

Students will be able to: write the place value, face value, expanded form and number names for numbers up to 8 digits. write numbers up to 8 digits in the Indian and International number system. compare numbers up to 8 digits and arrange them in ascending and descending order. round off numbers up to 8 digits to the nearest 10, 100 and 1000.

Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

Chart papers with empty place value chart drawn; Buttons; Beads; Bowls; Digit

Alignment to NCF

5 number cards with a 7-digit or 8-digit number written on them; Two bowls with number 8-digit numbers in one bowl and rounded-off places in another bowl

Teaching Aids

Let’s Recall: Recap exercises to check the understanding of prerequisite concepts before starting a topic.

C-1.1: Represents numbers using the place value structure of the Indian number system, compares whole numbers, and knows and can read the names of very large numbers

C-4.3: Selects appropriate methods and tools for computing with whole numbers, such as mental computation, estimation, or paper pencil calculation, in accordance with the context

Let’s Recall

Chart papers with empty place value chart drawn; Buttons; Beads; Bowls; Digit cards; Bowls with 5 number cards with a 7-digit or 8-digit number written on them; Two bowls with number cards having 8-digit numbers in one bowl and rounded-off places in another bowl

Recap to check if students know how to write the place value, expanded form and number names for 6-digit numbers. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

Teaching Aids

Chart papers with empty place value chart drawn; Buttons; Beads; Bowls;

cards; Bowls

Learning Outcomes

Numbers up to 8 Digits 1

Numbers up to

Alignment to NCF

C-1.1: Represents numbers using the place value structure of the Indian number system, compares whole numbers, and knows and can read the names of very large numbers

C-4.3: Selects appropriate methods and tools for computing with whole numbers, such as mental computation, estimation, or paper pencil calculation, in accordance with the context

QR Code: Provides access to digital solutions and other interactive resources.

Learning Outcomes

Let’s Recall

Recap to check if students know how to write the place value, expanded form and number names for 6-digit numbers.

Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

Alignment to NCF

Teaching Aids

Vocabulary: Helps to know the important terms that are introduced, defined or emphasised in the chapter.

C-1.1: Represents numbers using the place value structure of the Indian number system, compares whole numbers, and knows and can read the names of very large numbers

C-4.3: Selects appropriate methods and tools for computing with whole numbers, such as mental computation, estimation, or paper pencil calculation, in accordance with the context

Let’s Recall

expanded form and number names for numbers up to 8 digits. Indian and International number system. arrange them in ascending and descending order. nearest 10, 100 and 1000.

place value structure of the Indian number system, compares whole names of very large numbers and tools for computing with whole numbers, such as mental pencil calculation, in accordance with the context zero in India and the Indian place value system for writing numerals, world, and its modern impact on our lives and in all technology

Teaching Aids: Aids and resources that the teachers can use to significantly improve the teaching and learning process for the students.

Chapter: Numbers up to 8 Digits

Recap to check if students know how to write the place value, expanded form and number names for 6-digit numbers.

Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

Teaching Aids

Place Value, Face Value and Expanded Form Imagine Maths Page 2 Learning Outcomes

write the place value, expanded form and number names for given in the Let’s Warm-up section.

the sum of the values of all its digits close to the actual answer to a certain place value for easier calculation

Place Value, Face Value and Expanded Form

Learning Outcomes

Teaching Aids

Imagine Maths Page 2

Students will be able to write the place value, face value, expanded form and number names for numbers up to 8 digits.

Chart papers with empty place value chart drawn; Buttons; Bowls; Digit cards

Activity

Teaching Aids

Chart papers with empty place value chart drawn; Buttons; Bowls; Digit cards

Activity

Instruct the students to work in small groups. Distribute the teaching aids among the groups. Keep number cards (0–9) in a bowl. Pick a card from the bowl and say the digit aloud along with a place, for example “3 in the thousands place.” Instruct the groups to place as many buttons as the digit in that place on the place value chart. Repeat this to form a 7-digit or 8-digit number. You can repeat digits in more than 1 place. Once the 7-digit or 8-digit number is formed, have each group record the expanded form and number names in their notebooks. Discuss the face value and place value of a few digits in the class. Ask the students to say the number names aloud.

Extension Idea

Activity: A concise and organised lesson plan that outlines the activities and extension ideas that are to be used to facilitate learning.

Ask: In a number 1,67,48,950, if we interchange the digit in the ten thousands place with the digit in the crores place, then what is the difference in the place values of the digits in the ten thousands place?

Extension Idea

Say: On interchanging the digits, the new number is 4,67,18,950. Difference in the place values = 40,000 – 10,000 = 30,000.

Extension Idea: A quick mathematical-thinking question to enhance the critical thinking skill.

Indian and International Number Systems Imagine Maths Page 5

Say: On interchanging the digits, the new number is 4,67,18,950. Difference in the place values = 40,000 – 10,000 = 30,000.

chart drawn; Buttons; Beads; Bowls; Digit cards; Bowls with number written on them; Two bowls with number cards having rounded-off places in another bowl

Learning Outcomes

Students will be able to write numbers up to 8 digits in the Indian and International number system.

Indian and International Number Systems Imagine Maths Page 5

Teaching Aids

Learning Outcomes

Chart papers with empty place value chart drawn; Buttons; Beads

Students will be able to write numbers up to 8 digits in the Indian and International number system.

Activity

Teaching Aids

Chart papers with empty place value chart drawn; Buttons; Beads

Activity

Arrange the class in groups of 4, with each group split into teams of 2 students. Distribute the teaching aids among the groups. Guide the students to form 7-digit or 8-digit numbers on the chart using buttons and beads as in the previous lesson. Ask them to place buttons for the Indian number system on one chart and beads for the International number system on the other. The two teams should record the expanded form of the numbers using commas in their notebooks and number names. Ask the groups to find the similarities and differences between the two systems. Discuss the answers with the class.

Extension Idea

Ask: How many lakhs are there in 10 million?

Answers: Answers, provided at the end of each chapter, for the questions given in Do It Together and Think and Tell sections of the Imagine Mathematics book.

Answers

Say: To find out, we divide 10,000,000 by 100,000. So, there are 100 lakhs in ten million.

Extension Idea

Ask: How many lakhs are there in 10 million?

Say: To find out, we divide 10,000,000 by 100,000. So, there are 100 lakhs in ten million.

Period Plan

The teacher manuals corresponding to Imagine Mathematics books for Grades 1 to 8 align with the recently updated syllabus outlined by the National Curriculum Framework for School Education, 2023. These manuals have been carefully designed to support teachers in various ways. They provide recommendations for hands-on and interactive activities, games, and quizzes that aim to effectively teach diverse concepts, fostering an enriched learning experience for students. Additionally, these resources aim to reinforce critical thinking and problem-solving skills while ensuring that the learning process remains enjoyable.

In a typical school setting, there are approximately 180 school days encompassing teaching sessions, exams, tests, events, and more. Consequently, there is an average of around 120 teaching periods throughout the academic year.

The breakdown of topics and the suggested period plan for each chapter is detailed below.

Chapters No. of Periods

1. Integers

2. Fractions and Decimals

Addition of Integers

Break-up of Topics

Subtraction of Integers

Multiplication of Integers

Division of Integers

Revision

Adding and Subtracting Fractions

Multi-step Problems on Adding and Subtracting Fractions

Multiplication of Fractions

Dividing Fractions

Multi-step Problems on Operations on Fractions

Adding and Subtracting Decimals

Multi-step Word Problems on Adding and Subtracting Decimals

Multiplying and Dividing Decimals

Multi-step Word Problems on Operations on Decimals

Revision

3. Data Handling 9 Arithmetic Mean or Mean

Mode

Median Range

Making Bar Graphs

Interpreting Bar Graphs

Revision

Chance

4. Introduction to Probability 5

Theoretical Probability

Experimental Probability

Revision

5. Simple Equations 6

6. Angle Pairs and Parallel Lines 5

Forming Linear Equations

Solving Equations: Same Operation on Both Sides

Solving Equations: Transposition

Applications of Simple Equations

Revision

Pairs of Angles

Parallel Lines and Transversals

Constructing Parallel Lines

Revision

Medians and Altitudes

Properties of Angles of a Triangle

Properties of the Sides of a Triangle

The Pythagoras Theorem

7. Triangles and Their Properties 13

8. Ratio and Proportion 6

Congruence in Shapes and Figures

SSS Congruence Criteria

SAS Congruence Criteria

ASA Congruence Criteria

RHS Congruence Criteria

Revision

Equivalent Ratios

Proportional Terms

Unitary Method

Speed, Distance and Time

Revision

Percentages, Decimals and Fractions

Percentage and Ratios

Finding Percent and Percent Change

9. Percentage, Profit and Loss, Simple Interest 12

Word Problems on Percentage

Profit and Loss Percentages

Marked Price and Discount

Word Problems on Profit and Loss

Simple Interest

Revision

10. Rational Numbers 12

Representing Rational Numbers

Rational Numbers in Standard Form; Equivalent Rational Numbers

Comparing and Ordering Rational Numbers

Rational Numbers Between Two Rational Numbers

Addition of Rational Numbers; Additive Inverse

Subtraction of Rational Numbers

Multiplication of Rational Numbers; Reciprocal or Multiplicative Inverse of a Non-zero Rational Number

Division of Rational Numbers

Revision

Constructing Triangles: SSS Criterion

Constructing Triangles: SAS Criterion

11. Construction of Triangles 6

12. Perimeter and Area 11

Constructing Triangles: ASA Criterion

Constructing Triangles: RHS Criterion

Revision

Perimeter and Area of Squares and Rectangles

Area of Parallelograms

Area of Triangles

Circumference of Circles

Area of Circles, Semicircles and Quadrants

Combined Shapes

Word Problems on Perimeter and Area

Revision

Generating Rules in Formulas and Patterns

Algebraic Expressions and Terms

Addition of Algebraic Expressions

13. Algebraic Expressions 9

14. Exponents and Powers 6

Subtraction of Algebraic Expressions

Simplifying Algebraic Expressions

Finding the Value of an Expression

Revision

Exponents and Prime Factors; Comparing Numbers in Exponential Form

Powers of Rational Numbers; Reciprocal of a Rational Number

Laws of Exponents

Expressing Large Numbers in the Standard Form

Revision

15. Symmetry 3

16. Visualising Solid Shapes 8

Reflection Symmetry

Rotational Symmetry

Revision

Features of 3-D Shapes

Nets of 3-D Shapes

Oblique and Isometric Sketches

Cross Section and Shadows

Views of 3-D Shapes

Revision

Total Number of Periods

Learning Outcomes

Students will be able to: add 2 or more integers. subtract an integer from another. multiply 2 or more integers. divide an integer by another integer.

Alignment to NCF

C-1.3: Learns about the inclusion of zero and negative quantities as numbers, and the arithmetic operations on them, as given by Brahmagupta

C-1.4: Explores and understands sets of numbers, such as whole numbers, fractions, integers, rational numbers and real numbers, and their properties, and visualises them on the number line

Let’s Recall

Recap to check if students know how to compare and order integers using a number line. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

integer: a positive or negative whole number

Teaching Aids

Integer tiles as red and yellow square cut-outs to show negative and positive integers, respectively

Chapter: Integers

Addition of Integers

Learning Outcomes

Students will be able to add 2 or more integers.

Teaching Aids

Integer tiles as red and yellow square cut-outs to show negative and positive integers, respectively

Activity

Begin the class by giving 1–2 examples where negative numbers are seen. Ask questions like: How will you write the number if the lift goes below the ground floor? Introduce integer tiles to students. Explain that the red tiles represent negative integers, and the yellow tiles represent positive integers. Bring out the fact that a pair of red and yellow tiles represents 0. Instruct students to work in pairs. Distribute integer tiles among the pairs. Instruct students to add 5 and −6 using the integer tiles. Ask them to place the integer tiles (yellow tiles) that represent 5. Ask them to now place the integer tiles (red tiles) that represent −6 below the previous set. Instruct them to remove the pairs of integer tiles that make zero. Explain that the remaining number of integer tiles (−1) would be the answer. So, 5 + (−6) = −1. Instruct students to add the integers −8 and −3, 7 and −1 using the integer tiles. Ask students to note down the results in their notebooks using the addition sentence and discuss among their groups what they notice about the answers. Discuss the rules for the addition of integers with students. Ask questions like: What is the sign of the sum of 2 negative integers? Give them practice by running a quiz around adding integers.

Subtraction of Integers

Learning

Outcomes

Students will be able to subtract an integer from another.

Teaching Aids

Imagine Maths Page 4

Integer tiles as red and yellow square cut-outs to show negative and positive integers, respectively

Activity

Begin the class by asking students the difference between the temperatures in 2 cities of −14 and −13. Discuss the answers.

Instruct students to work in pairs. Distribute integer tiles among the pairs.

Instruct students to subtract 5 from 2. Ask them to use tiles to show 2. Ask questions like: Can you take away –5 or 5 negative tiles from the 2 positive tiles?

Tell students that 2 positive tiles can be represented as 2 positive tiles and 3 neutral group of tiles. Let students figure out how to subtract and find the answer. (Help them if needed).

Instruct students to now subtract −4 from +3 using the integer tiles. Ask students to note down the result in their notebooks using the subtraction sentence. Discuss the rules for subtraction of integers with students.

Ask questions like: What is the sign of the difference of 2 negative integers? One negative and 1 positive integer? Give them practice by running a quiz around subtracting integers.

Extension Idea

Instruct: Give two numbers which give −25 as a result after subtraction.

Say: Let the first number be 15 and the second number be a. 15 − a = −25 a = 15 + 25 = 40. Hence, when 40 is subtracted from 15, it gives −25 as a result.

Multiplication of Integers Imagine Maths Page 8

Learning Outcomes

Students will be able to multiply 2 or more integers.

Teaching Aids

Integer tiles as red and yellow square cut-outs to show negative and positive integers, respectively

Activity

Discuss how, in integers, multiplication is repeated addition or subtraction with 2 or more numbers. The first number tells us how many groups we are adding or removing. The second number tells us how many are in each group and whether they are positive (+) or negative (−).

Instruct students to work in pairs. Distribute integer tiles among the pairs.

Instruct students that they need to multiply 3 and −4 using the tiles. Once students show 3 groups of −4; ask them to note down the result in their notebooks.

Ask the pairs to now multiply −3 and −4 using the tiles. Discuss that since they need to multiply −3 by −4, they need to remove 3 groups of −4, but to take away 3 groups of 4 negative tiles, they first need to place zero tiles. Discuss that on taking away 4 negative tiles from each of the 3 neutral groups, they get 12 positive tiles. Ask them to write the result in their notebooks using the division sentence.

Discuss the rules for multiplication of integers with students and run a quiz around the multiplication of integers.

Extension Idea

Ask: What will be the sign of the final product if 90 negative integers are multiplied?

Say: 90 is an even number. The final product when 90 integers are multiplied will be positive.

Division of Integers Imagine Maths

Learning Outcomes

Students will be able to divide an integer by another integer.

Teaching Aids

Integer tiles as red and yellow square cut-outs to show negative and positive integers, respectively

Activity

Instruct students to work in pairs. Distribute integer tiles among the pairs.

Instruct the pairs to show −15 ÷ 5 using the tiles by first showing −15 using red tiles (negative tiles) and then find the number of groups by dividing these tiles into 5 equal groups with 3 tiles in each group. We get −3 in each group. Ask them to note down the result in their notebooks.

Ask students to now show −15 ÷ −5 using the tiles. Explain that they will have to first place 15 negative (red) tiles. Since negative groups cannot be made, to divide −5, we keep 5 negative tiles in each group, and find the total number of groups formed. Ask them to write the result in their notebooks using the division sentence.

of tiles ÷ Number of tiles in each group = Number of groups

Discuss the rules for the division of integers with students.

Ask questions like: What is the sign of the division of 2 negative integers? One negative and 1 positive integer? Once students are thorough with the concept, run a quiz around the division of integers.

Extension Idea

Instruct: Find −2 + 3 × 15 ÷ (−5) using DMAS.

Say: −2 + 3 × 15 ÷

Number

Answers

1. Addition of Integers

Do It Together

Temperature of water in the bowl = 85°C

Temperature drops in 20 minutes = 33°C

Temperature drops in the next 10 minutes = –13°C

Total temperature drop = (–33) + (–13) = −46°C

Temperature after 30 minutes = 85°C + (−46°C) = 39°C

2. Properties of Addition of Integers

Do It Together

1. 32 + (−17) = (−17) + 32; Property used = Commutative property

2. −36 + 0 = −36; Property used = Existence of additive identity

3. (12 + 3) + (−5) = 12 + (3 + (−5)); Property used = Associative property

4. 18 + (−18) = 0; Property used = Existence of additive inverse

3. Subtraction of Integers

Do It Together

Height of Mt Everest above sea level: 29,029 feet

Depth of Assal Lake below sea level = 510 feet

Difference in depth = 29,029 − (–510) = 29,539 feet

4. Properties of Subtraction of Integers

Do It Together

1. 45 − (−27) = 72; where 72 is also an integer; Property used = Closure property

2. −100 − 0 = −100; Property used = Property of zero

3. 36 − 0 = 36; Property used = Property of zero

4. (25 – 12) − 15 ≠ 25 – (12 − 15); Property used = Associative property

5. Multiplying Integers

Think and Tell

99 is an odd number. The final product when 99 negative integers are multiplied will be negative.

Do It Together

Money lost by the store every day = ₹125

Money lost in 6 days = (–125) × 6 = −₹750

So, the store loses ₹750 in 6 days.

6. Properties of Multiplication of Integers

Think and Tell

The concept of multiplicative inverse does not apply to zero as any number divided by 0 is not defined.

Do It Together

1. (–25) × (–8) = (–8) × (–25)

Property used: Commutative property

2. 52 × 1 = 52 Property used: Existence of multiplicative identity

3. 23 × (13 + 14) = (23 × 13) + (23 × 14)

Property used: Distributive property of multiplication of integers over addition

4. (17 × 5) × (–9) = 17 × (5 × (–9))

Property used: Associative property

7. Dividing Integers

Do It Together

Number of quarters played = 4

Team A’s total score for the match = −48

Average score per quarter = (–48) ÷ 4 = −12

8. Properties of Division of Integers

Think and Tell

For the commutative property to hold true under division, a ÷ b = b ÷ a

This is only possible when a = b and a, b ≠ 0.

Do It Together

1. (–125) ÷ (–25) = 5

2. 0 ÷ 45 = 0

3. 23 ÷ (−1) = –23

4. 136 ÷ 136 = 1

Learning Outcomes

Students will be able to:

add and subtract fractions. solve multi-step word problems on adding and subtracting fractions. multiply 2 or more fractions. divide a fraction by a fraction or a whole number. solve multi-step word problems on operations on fractions. add and subtract decimals. solve multi-step word problems on adding and subtracting decimals. multiply and divide decimals.

solve multi-step word problems on operations on decimals.

Alignment to NCF

C.1.4: Explores and understands sets of numbers, such as whole numbers, fractions, integers rational numbers, and real numbers, and their properties and visualises them on the number line

C-1.6: Explores and applies fractions (both as ratios and in decimal form) in daily–life situations

Let’s Recall

Recap to check if students know how to write fractions and classify them as proper, improper, and mixed. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

fraction: number of the form a b , where a and b are whole numbers and b ≠ 0

reciprocal: the number formed after interchanging the numerator and denominator of a fraction

decimals: numbers that occur between whole numbers

decimal point: the dot present between the whole number part and the fractional part of a number

Teaching Aids

Bingo cards; Question cards; Situation cards; Question slips; Rectangular strips; Colour pencils; Puzzle cards with multiplication problems and their corresponding answers; Small bells; Stopwatch; Story cards; Question slips; Problem cards; Shopping list with 7 items; Word problem cards; Grocery list of 3 items; Flash cards

Chapter: Fractions and Decimals

Adding and Subtracting Fractions

Learning Outcomes

Students will be able to add or subtract fractions.

Teaching Aids

Bingo cards; Question cards

Activity

Explain that fractions need to be converted to like fractions to be able to add or subtract them. Make problem cards (example, 2 6 + 8 5 ; 7 6 − 1 3 ; 3 15 + 1 4 , etc.) and bingo cards (showing answers to the corresponding question cards).

Divide the class into groups. Distribute the bingo cards to each group. Shuffle the problem cards and place them face down on the table. Pick a problem card and write the problem on the board or read it out loud. Instruct the students to solve the problem, and check whether the answer lies on the bingo card. If it does, they need to cross out the number on their bingo card. The group that crosses out a row and a column first, wins. Ask the students to write the answers in their notebooks.

Extension Idea

Ask: How much will you add to the sum of 5 8 and 1 8 to get 5 4 ?

Say: So, 5 8 + 1 8 = 6 8 , and 5 4 = 10 8 . You will need to add 4 8 or 1 2 more to 6 8 to get 10 8 which is equal to 5 4 .

Multi-step Problems on Adding and Imagine Maths Page 23

Subtracting Fractions

Learning Outcomes

Students will be able to solve multi-step word problems on adding and subtracting fractions.

Teaching Aids

Situation cards; Question slips

Activity

Instruct students to form groups for the activity. Distribute the situation cards to each group.

Preeti, a baker, was baking a cake for the town’s Baking Fair. Her original recipe required 3 4 cup of sugar and 1 2 cup of butter. A friend suggested she enhance the flavour by adding 1 3 cup of chocolate chips. Preeti realised she only had 1 4 cup of chocolate chips in her cupboard.

Prepare question slips such as the ones given below and distribute among the students.

How much more of the chocolate chips does Preeti need for her recipe?

How many cups of ingredients did Preeti use in total?

Instruct students to solve these word problems in their notebooks. Discuss their approach in class. Give more word problems to the students for practice after discussion.

Extension Idea

Ask: How much sugar will Preeti need to bake double the size of the cake she was baking?

Say: Preeti will need to double the amount of sugar, i.e., 3 4 + 3 4 = 3 2 or 1 1 2 cups.

Multiplying

Fractions

Learning Outcomes

Students will be able to multiply 2 or more fractions.

Teaching

Aids

Maths Page 25

Rectangular strips; Colour pencils; Puzzle cards with multiplication problems and their corresponding answers

Activity

Instruct students to work in groups. Distribute rectangular strips and colour pencils among the groups. Instruct them to multiply 3 5 with 1 2 using the strips. Discuss how they can show 3 5 groups of 1 2 . Bring out the fact that they need to shade the strip to show 3 5 vertically and then shade the strip to show 1 2 horizontally by making crosses. Finally, ask them to write the part that is both shaded and crossed out, in their notebooks. Next, ask them to multiply the numerators of the 2 fractions and multiply the 2 denominators. They will then compare the answers to identify the rule.

Repeat the activity with one more set of fractions.

Give the students a mix of puzzle cards with their corresponding answers, where they need to multiply fractions and match with the answers. The group that finishes first is the winner.

Extension Idea

Ask: How many hours is 2 8 of a day?

Say: 2 8 of a day is equal to 2 8 of 24 hours.

2 8

× 24 = 6 hours.

Dividing

Learning Outcomes

Students will be able to divide a fraction by a fraction or a whole number.

Teaching Aids

Small bells; Stopwatch

Activity

Start the class by briefly discussing the division of fractions. Instruct students to form groups. Distribute the bells and stopwatches to each group. Inform students that they will be playing a quiz which will have fractional division problems. Explain the rules and format of the quiz.

Introduce the bell system. Tell them that the first team to ring the bell gets the opportunity to answer. Display fractional division questions on the board, such as, What is 3 5 ÷ 1 2 ?, Calculate 2 3 ÷ 4, etc. Start the timer for each question (e.g., 30 seconds). Inform teams that they can only ring the bell once they’ve heard the question. Tell them that if a team answers correctly, they earn points. Incorrect answers may result in point deductions. Keep track of each team’s score on the board. Award 1 point for each correct answer and deduct 0.25 for each incorrect one. Include double-point questions such as, solve for x: x 5 = 2 3 ÷ 1 6 . Repeat the activity with enough questions. Then, award recognition to the winning team. Discuss their answers in class and encourage them to write the solutions in their notebooks.

Multi-step Problems on Operations on Fractions

Learning Outcomes

Students will be able to solve multi-step word problems on operations on fractions.

Teaching Aids

Story cards; Question slips

Activity

Instruct students to form groups for the activity. Distribute the story cards to each group.

Imagine Maths Page 30

Sophia loves to stitch and is planning to make scrunchies for a school fundraiser. She needs to calculate the total length of cloth required. She needs 3 4 m of red cloth and 2 5 m of blue cloth for each batch of 10 scrunchies. She plans to make 5 batches for the fundraiser and wants to distribute the scrunchies equally among 4 schools.

Prepare question slips such as the ones given below and distribute them among the students.

How many metres of cloth does Sophia need for the entire fundraiser?

How many batches of scrunchies will each school receive?

What is the total length of cloth needed for the 5 batches?

Instruct students to solve these word problems in their notebooks. Discuss their approach in class. Give them more questions for practice.

Adding and Subtracting Decimals

Learning Outcomes

Students will be able to add and subtract decimals.

Teaching Aids

Problem cards; Bingo cards

Activity

Make problem cards (example: 156.154 + 51.264, 458.564 – 147.117) and bingo cards (showing answers to the corresponding problem cards).

Divide the class into groups and distribute the bingo cards to each group.

Put the problem cards in a box and shuffle them. Then, pick a card and read out the problem. Instruct students to solve it using the column method. Ask them to check whether the answer is on the bingo card. If it is, ask them to cross it out.

The group that crosses out a row or a column first wins. Ask students to write the answers in their notebooks.

Extension Idea

Ask: Nishita covered a distance of 5.32 km by bicycle, 3.645 km by bus and walked 1.2 km. What is the total distance that she covered?

Say: Total distance covered = 5.32 + 3.645 + 1.2 = 10.165 km. Multi-step

Problems on Adding and

Subtracting Decimals

Learning Outcomes

Students will be able to solve multi-step word problems on adding and subtracting decimals.

Teaching Aids

Shopping list with 7 items; Word problem cards

Activity

Create a shopping list of various items and write their prices. Divide the class into groups. Distribute the shopping lists and assign a budget of $30 to each group.

Create word problem cards like ‘How much money will you have left if you purchase a toy, book and a pen? ’

Ask 1 student from each group to come up and pick 2 word problem cards. Instruct students to solve the problems based on the shopping list and find the answer.

Ask them to write the answers in their notebooks.

Discuss the results in class.

Ask questions like: Which operations did you perform first?

BINGO CARD

Extension Idea

Instruct: Create a multi-step word problem based on depositing and withdrawing money in/from a bank account in decimals.

Say: Sample word problem: ‘Ravi deposited ₹375.5, ₹600.75 and ₹1237.5 in his bank account on three different days. If he withdrew ₹1110.25, how much money is left in his bank account?’

Multiplying and Dividing Decimals

Learning Outcomes

Students will be able to multiply and divide decimals.

Teaching Aids

Grocery list of 3 items

Activity

Divide the class into groups of 4 and distribute a grocery list of 3 items to each group.

Imagine Maths Page 38

Ask one student in each group to act as the shopkeeper and the other 3 students as customers. Instruct one student to find the cost of 5 kg of pulses, the second student to find the cost of 6 kg of rice and the third student to find the cost of 5 litres of milk. Encourage them to solve the problem using the column method of multiplication and long division. The shopkeeper should check the calculations, and each group should calculate the total amount required to purchase the given weight of the items. Ask students to write the answers in their notebooks.

Multi-step Problems on Operations on Decimals

Learning

Outcomes

Students will be able to solve multi-step word problems on operations on decimals.

Teaching

Aids

Flash cards with word problems

Activity

Instruct students to work in groups. Distribute word problem cards among the groups. Ask them to read the problem, such as:

‘Ramesh earns ₹1755.25 for working 7 hours in a day. How much money will he earn if he works 10 hours in a day? ’

Imagine Maths Page 40

Ramesh earns ₹1755.25 for working 7 hours in a day. How much money will he earn if he works 10 hours in a day?

Ask students to write down what we know and what we need to find out. Discuss how to solve the problem. Instruct them to solve the problem and discuss the answer.

Extension Idea

Ask: Create a multi-step word problem including all four operations on decimal numbers.

Say: Sample word problem: ‘Naman donated ₹1552.5 and ₹3624.75 to two charities. Sunaina donated half of the total amount that Naman donated, and Sunil donated 3 times the difference of the amounts that Naman donated. What amount did each of them donate? ’

Answers

1. Adding and Subtracting Fractions

Do It Together

1. 7 + 10 20 + 4 20 + 5 20 = 7 + 19 20 = 7 19 20

2. 329 56 –136 56 = 193 56 = 3 25 56

2. Multi-step Problems on Adding and Subtracting Fractions

Do It Together

Total juice used to make fruit punch and flavoured jelly = 4 3 4 + 6 1 3 = 10 +

3 4 + 1 3

 = 10 + 9 12 + 4 12 = 10 + 13 12 = 11 + 1 12

= 11 1 12 cups

Juice left for drinking = 17 − 11 1 12 = 5 11 12 cups

3. Multiplication of a Fraction by a Whole Number

Do It Together

Total money spent = ₹5000 × 3 8 = 15000 8 = ₹1875

Thus, Rajeev spent ₹1875

4. Multiplication of a Fraction by a Fraction

Do It Together

1. 3 1 2 × 5 = 7 2 × 5; 3 1 2 = 7 2 3 1 2 × 5 = 35 2 = 17 1 2 2. 2 2 3 × 5 1 2 = 8 3 × 11 2 2 2 3 = 8 3 5 1 2 = 11 2 2

5. Division of a Fraction by a Whole Number

Do It Together 7 1 2 = 15 2 ; 15 2 ÷ 5 = 15 2 ÷ 5 1 = 15 2 × 1 5 = 3 2

6. Division of a Whole Number by a Fraction

Do It Together

2 3 = 5 3

7. Division of a Fraction by Another Fraction

Do It Together

8. Multi-step Problems on Operations on Fractions

Do It Together

Total cake shared = 6 16 + 5 12 = 18 + 20 48 = 38 48 = 19 24 Fraction of cake remaining after sharing = 1 – 19 24 = 5 24

Fraction of cake given to her sister = 2 5 × 5 24 = 1 12

Fraction of cake remaining after sharing with the sister

9. Adding and Subtracting Decimals

Do It Together

Sum = 367.25 + 125.2 + 512 = 1004.45 … (1)

Difference = 563.24 − 321 = 242.24 … (2)

Adding (1) and (2) gives 1004.45 + 242.24 = 1246.69 H T O . t h

3 6 7 2 5 1

10. Multi-step Problems on Adding and Subtracting Decimals

Do It Together

Total cups of sugar = 18 2 5 cups = 18.4 cups

Sugar used = 2.25 + 6 3 4 + 1.45 = 2.25 + 6.75 + 1.45 = 10.45 cups

Sugar left = 18.4 – 10.45 = 7.95 cups

11. Multiplying and Dividing Decimals

Think and Tell

It is not necessary to get a natural number when we divide 2 decimal numbers.

Do It Together

Area of rectangle = 143.75 cm2; Breadth of rectangle = 11.5 cm

Length = Area Breadth = 143.75 11.5 = 12.5

Hence, the length of the rectangle is 12.5 cm.

12. Multi-step Problems on Operations on Decimals

Do It Together

Price of 12 adult tickets at ₹125.5 per ticket = 12 × 125.5 = ₹1506

Price of 5 child tickets at ₹75.5 per ticket = 5 × 75.5 = ₹377.5

Difference in the cost = ₹1506 − ₹377.5 = ₹1128.5

Data Handling 3

Learning Outcomes

Students will be able to: find the arithmetic mean or the mean of a given set of data. find the mode of a given set of data. find the median of a given set of data. find the range of a given set of data. draw single or double bar graphs for a given set of data. interpret single or double bar graphs for a given set of data.

Alignment to NCF

C-5.1: Collects, organises, and interprets the data using measures of central tendencies such as average/ mean, mode, and median

C-5.2: Selects, creates, and uses appropriate graphical representations (e.g., pictographs, bar graphs, histograms, line graphs, and pie charts) of data to make interpretations

C-8.2: Learns systematic counting and listing, systematic reasoning about counts and iterative patterns, and multiple data representations; learns to devise and follow algorithms, with an eye towards understanding correctness, effectiveness, and efficiency of algorithms

Let’s Recall

Recap to check if students know how to collect, organise and interpret data. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

observation: a piece of information we gather during a study or experiment summation: addition of a sequence of numbers frequency: the number of times a value occurs in a set of data frequency distribution: description of frequency for each possible value of a variable central tendency: a single value that represents the centre of the data set tally mark: a vertical mark used to keep count bar graph: graph that shows information in the form of bars of different lengths

Teaching Aids

Marbles; Paper cups; Cards (with data sets, e.g., 3, 5, 7, 9, 11); Number cards (1–10); Building blocks (different colours); Number cards (0 to 20); Graph paper; Coloured paper; Glue stick; Chart paper with a double bar graph; Strips of paper

Chapter: Data Handling

Arithmetic Mean or Mean

Learning Outcomes

Students will be able to find the arithmetic mean or the mean of a given set of data.

Teaching Aids

Marbles; Paper cups; Cards (with data sets, e.g., 3, 5, 7, 9, 11)

Activity

Begin by explaining what the arithmetic mean is and how to calculate it. Divide the class into groups of 4.

Distribute the marbles and paper cups to each group. Also, distribute a set of cards to each group with data sets written on each card (e.g., 3, 5, 7, 9, 11).

Instruct students to put the marbles according to the data set in different paper cups. Ask them the total number of marbles used.

Ask questions like: Do all the cups have an equal number of marbles? How can we put an equal number of marbles in each cup?

Instruct students to divide the total number of marbles by the number of cups used and to write the result in their notebooks. Explain to them how the resulting number will give them the average number of marbles per cup, which represents the arithmetic mean.

If time permits, provide additional data sets to the groups, and ask them to find the arithmetic mean.

Extension Idea

Ask: The mean of 4, 22, 12, 7, x and 13 is 11. What is the value of x?

Say: Mean = 4 + 22 + 12 + 7 + x + 13 6 = 58 + x 6 . 11 = 58 + x 6 . So, x = 66 − 58 = 8.

Mode Imagine Maths Page 50

Learning Outcomes

Students will be able to find the mode of a given set of data.

Teaching Aids

Number cards (1–10)

Activity

Begin the class by explaining the concept of mode and its use.

Lead students to the playground and mark spots on the ground from 1 to 10, spaced far apart. Create number cards from 1 to 10 with repeating numbers (ensuring there are more cards than the total number of students). Place these cards in a box and have students pick a card each. Then, instruct them to stand on the spot matching the number on their card.

Explain that the spot with the most students is the mode.

Collect the cards and return them to the box. Shuffle the cards and repeat the activity a few more times. Have students record the results in their notebooks each time.

Engage students in a discussion about their findings. Ask them to articulate why they think a particular number card appears more frequently and how this relates to finding the mode.

Extension Idea

Ask: The runs scored by a cricketer are: 120, 24, 48, 64, x, 29, 71, 3, 64, 29. If the mode is 64, then find the value of x.

Say: Since 29 and 64 both occur twice and the mode is 64, then x must be equal to 64, to make it 1 more than 29.

Median Imagine Maths Page 51

Learning Outcomes

Students will be able to find the median of a given set of data.

Teaching Aids

Building blocks (different colours)

Activity

Begin the class by explaining the concept of a median and its use.

Instruct students to form groups. Provide each group with a set of building blocks in different colours. Some colours may have more blocks or repeats than other colours.

Instruct students to use the building blocks to build towers. Ask them to note down the number of blocks used for each colour in their notebooks. Also, ask them to arrange the colours in ascending order based on the number of blocks they used for each. For example, if they used 1 blue, 2 orange, 3 yellow, 4 green and 5 red blocks, order them as: blue, orange, orange, yellow, yellow, yellow, green, green, green, green, red, red, red, red, red.

Ask students to count the total number of building blocks they used to make the tower. Guide each group to find the colour that represents the middle value of the data set. Explain how this colour represents the median of the data set.

Ask questions like: Did your group use an odd or even total number of blocks? Why do you think we have different ways to find the median for odd and even counts? How did using coloured blocks help you understand the concept of the median?

Extension Idea

Ask: The numbers 17, p, 24, p + 7, 35, 36, 46 are written in ascending order. If the median of the data is 25, what is the value of p?

Say: There are 7 observations. So, the median is the 7 + 1 2 th = 4th observation, i.e., p + 7. Hence, p + 7 = 25, and p = 18. The numbers are 18, 18, 24, 25, 35, 36 and 46.

Learning Outcomes

Students will be able to find the range of a given set of data.

Teaching Aids

Number cards (0 to 20)

Activity

Discuss what a range is for the given set of data and how to find the range for the given set of data. Divide the class into groups. Distribute the number cards to each group. Ask students to make a data of 10 random numbers using the number cards.

Ask them to arrange the data in ascending order. Ask each group to find the range of the data that they had formed.

Ask students to repeat the process with different numbers.

Ask question such as: Was the range same for all the data set? On what does the range of the data depend?

Making Bar Graphs

Learning Outcomes

Students will be able to draw single or double bar graphs for a given set of data.

Teaching Aids

Graph paper; Coloured paper; Glue stick

Activity

Begin by explaining the importance of the monsoon season in India. Explain that the purpose of the activity is to create a visual representation of the average annual rainfall during the southwest and northeast monsoon seasons across different regions of India. Draw a table on the board as shown.

Divide the class into groups and ask them to look at the table drawn. Distribute graph paper, coloured paper and a glue stick to each group. Instruct students to draw a set of axes on their graph paper, with the x-axis representing the regions and the y-axis representing the rainfall in millimetres. Ask them to label the x-axis with the names of the regions and the y-axis with appropriate intervals to accommodate the rainfall data. Using the data provided in the table, ask students to plot two bars for each region—one representing the average rainfall during the southwest monsoon and the other representing the average rainfall during the northeast monsoon by cutting and pasting the rectangular strips of coloured paper (remind students that the widths of the rectangular strips should be the same).

Encourage students to scale their axes to represent the rainfall data accurately. Instruct them that each square on the graph paper can represent a certain number of millimetres of rainfall. Ensure that students provide a title for their bar graph.

Encourage a discussion on the bar graph formed by each group.

Extension Idea

Instruct: Create questions to ask other students in the group about the bar graph. Say: You can ask a lot of questions about the bar graph. One such question could be: Which two regions receive the highest average annual rainfall during the southwest monsoon season?

Interpreting Bar Graphs

Learning Outcomes

Students will be able to interpret single or double bar graphs for a given set of data.

Teaching Aids

Chart paper with a double bar graph; Strips of paper

Activity

Begin by introducing the topic of biomes and explaining what they are. Explain that temperature is one of the factors that distinguishes different biomes (grasslands, deserts, rainforests etc.).

Prepare the double bar graph on chart paper where each biome is represented by two bars: one for the average high temperature and one for the average low temperature.

Instruct students to work in groups. Distribute the chart paper with the double bar graph, to each group.

Imagine Maths Page 60

Temperature Comparison in Different Biomes

Ask students to examine the graph closely and note any patterns or differences they observe between the biomes.

Pose questions such as: Which biome has the highest average high temperature? Which biome has the lowest average low temperature? How does the temperature range vary between different biomes? Are there any biomes where the high and low temperatures are very close together?

Provide each group with 2 strips of paper. Instruct each group to create 2 questions related to the bar graph.

Have each group pass their 2 question strips to another group. Instruct each group to answer the questions they received from another group.

Desert

Rainforest

Grassland

Biomes

Answers

1. Arithmetic Mean or Mean

Think and Tell

No, the mean of a data set cannot be more than the highest value in the data set as it represents the average value of the data set.

Do It Together

Step 1

Let us find ∑f and ∑( f × x).

Step 2

Mean = ∑( f × x) ∑f = 170 30 = 5.67

Thus, the arithmetic mean or the average marks of the students is 5.67.

2. Mode

Do It Together

Can the missing number be 5? No, because then the mode(s) will be 8 and 6

Can the missing number be 6? No, because then the mode(s) will be 6

Can the missing number be 7? No, because then the mode(s) will be 8 and 6

Can the missing number be 8? Yes, because then the mode will be 8

3. Median

Think and Tell

No, it is not necessary that the median will always be present in the data because if the number of observations is even, then the median will be the average of the middle two values, which is not necessarily the value present in the data.

Do It Together

Arrange the heights (in cm) in ascending order.

148, 154, 160, 162, 164, 169, 171

The total number of observations = 7, which is an odd (even/odd) number.

The middle observation is 4th

The median height is 162 cm.

4. Range; Choosing the Right Measure of Central Tendency

Do It Together

Using Mean and Median:

As the retailer is dealing with categorical data, he can’t use the mean or median to plan his purchase for the upcoming week.

Using Mode:

As the mode is used to recognise the most commonly occurring observation, the retailer can use the mode to plan his upcoming purchase.

The mode for the given data is medium. Therefore, only the mode can be used to plan the purchase for the upcoming week.

Range = 70 − 45 = 25

5. Making Bar Graphs

Do It Together

6. Interpreting Bar Graphs

Think and Tell

A bar graph is typically used to display categorical or qualitative data, where the categories are distinct and separate from one another. This type of data is often non-numeric or can be grouped into categories.

Think and Tell

In a standard bar graph, the width of the bars typically does not have any significance in terms of conveying data. However, the width of the bars should be consistent for a bar graph.

Do It Together

2. What percentage of students prefer indoor games in Grade 6? 75%

3. What percentage of students prefer outdoor games in Grade 7? 55%

4. 70% of students of which grade prefer outdoor games? Grade 5

Introduction to Probability 4

Learning Outcomes

Students will be able to: find the chance of an event happening using words like sure, impossible, likely, unlikely or equally likely. find the theoretical probability of an event and write it as a fraction. find the experimental probability of an event and write it as a fraction.

Alignment to NCF

C-6.2: Applies concepts from probability to solve problems on the likelihood of everyday events

Let’s Recall

Recap to check if students know how to find the ratio of two quantities and write it in its simplest form. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

chance: the likelihood of an event taking place probability: a chance that an event will take place experiment: an action that produces clearly defined results outcome: each possible result of an experiment event: a collection of one or more outcomes

Teaching Aids

Flash cards with some real-life situations written on them such as The sun rises in the west; Carry an umbrella to school on a rainy day etc,; Bag of coloured buttons: blue (12), red (15), green (6), yellow (5), white (10); Dice; Sheet of paper; Pencil; Piece of cardboard; Sketch pens; Push pins; Hook; Bowls

Chapter: Introduction to Probability

Learning Outcomes

Imagine Maths Page 70

Students will be able to find the chance of an event happening using words like sure, impossible, likely, unlikely or equally likely.

Teaching Aids

Activity

Instruct students to form groups. Distribute flash cards with some real-life situations written on them such as The sun rises in the west; Carry an umbrella to school on a rainy day. Discuss which words, such as sure, impossible, likely or unlikely, can be used to describe the event. Distribute the teaching aids to each group. Draw the given table on the board. Instruct students to copy the table into their notebooks. Ask them to count the number of different coloured buttons and write it in the table.

Ask: What is the chance of picking a blue button from the bag?

Discuss the answer where the chance of picking a blue button depends on the total outcomes, which is the total number of buttons (48). The chance of picking a blue button is 12 out of 48, which is less than half. So, the chance of this event happening is ‘unlikely ’. Encourage students to complete the table using words like likely, unlikely or equally likely

Extension Idea

Ask: What is the chance of a black button being picked from the bag?

Say: Since there are no black buttons in the bag, the chance of this event happening is impossible.

Theoretical Probability

Learning Outcomes

Students will be able to find the theoretical probability of an event and write it as a fraction.

Teaching Aids

Dice; Bag of coloured buttons: blue (12), red (15), green (6), yellow (5), white (10)

Maths Page 73

Activity

Show students a dice and ask them if they rolled the dice, what would they get. Explain that the total outcomes are 6, and the favourable outcomes in rolling a dice and getting a 6 is 1. Also, discuss how to write the theoretical probability

1 6

of the event as a fraction.

Instruct students to form groups. Distribute the resources to each group.

Ask students to count the buttons and find the theoretical probability of picking a button at random of each colour. Discuss one example with them and guide them to identify that the theoretical probability of picking a green button is 6 48

Ask questions like: What is the theoretical probability of getting a pink button? A blue button?

Experimental Probability

Learning Outcomes

Imagine Maths Page 77

Students will be able to find the experimental probability of an event and write it as a fraction.

Teaching Aids

Sheet of paper; Pencil; Piece of cardboard; Sketch pens; Push pins; Hook; Dice; Bowls

Activity

Ask students if a dice is rolled 30 times, how often the number 5 might show up. Then, discuss how, if the number 5 appears 10 times, the experimental probability of getting 5 is 10 30 .

Instruct students to form groups. Distribute the teaching aids.

Inform them that they will create a spinner for the school fair with 4 options. Draw the table on the board. Instruct students to copy the table into their notebooks.

Outcome Jackpot

Number of Times

Instruct students to take turns to spin the spinner and then list the colour that the spinner stops at in the table using tally marks. Each group must spin it 25 times. Once the table is filled, ask them to write the answer to these questions in their notebooks: What is the experimental probability of spinning a jackpot? What is the experimental probability of winning a turn? Then, encourage students to discuss their findings within their groups and compare their results with other groups.

Extension Idea

Ask: During a basketball shooting practice session, Sarah takes 30 free throws and records the number of successful shots she makes as 18. Based on the experimental probabilities, what percentage of the free throws did Sarah make successfully?

Say: Based on the experimental probabilities, 18 30 × 100 = 60. So, Sarah made 60% of her free throws successfully. This means that 60% of her attempts resulted in successful shots.

Jackpot Win a turn

Get a free juice Win a prize

Answers

1. Understanding Probability

Do It Together

1. Green balloon – Equally likely

2. White balloon – Unlikely

3. Green or orange balloon – Likely

4. Red balloon – Impossible

2. Theoretical Probability

Do It Together

Cards in the box are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

The number of cards that have prime numbers is 4

Total number of cards = 10

Therefore, the probability of drawing a prime number card = 4 10 or 2 5

3. Experimental Probability

Think and Tell

As mentioned in the story on page 74, the bag has 20 marbles out of which 4 are red. So, the theoretical probability of picking a red marble is 4 20 or 1 5 .

Experimental probability of picking a red marble = 7 20

Do It Together

P(6) = 4 15 P(7) = 9 15 or 3 5

Probability of the spinner stopping at 6 = 4 15

Probability of the spinner stopping at 7

Simple Equations

Learning Outcomes

Students will be able to: form linear equations for given statements. solve equations by using the same operation on both sides of the equation. solve equations by using the transposing method. apply simple equations to solve word problems.

Alignment to NCF

C-2.3: Forms algebraic expressions using variables, coefficients, and constants and manipulates them through basic operations

C-2.4: Poses and solves linear equations to find the value of an unknown, including to solve puzzles and word problems

C-2.5: Develops own methods to solve puzzles and problems using algebraic thinking

Let’s Recall

Recap to check if students know how to frame algebraic expressions. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

linear: straight equation: a statement in arithmetic that uses an equal sign to show the equality of two quantities variable: something that can change or that has no fixed value

Teaching Aids

Index cards or small slips of paper; Envelopes for each set of index cards; Balance scale or a makeshift scale (e.g., a ruler or a wooden beam balanced on a fulcrum); Unit cubes; Two bags with 2 unit cubes in each; One bag with 5 unit cubes; Equation cards with linear equations in one variable; Equation mazes; Crayons; Chart paper

Chapter: Simple Equations

Forming Linear Equations

Learning Outcomes

Students will be able to form linear equations for given statements.

Teaching Aids

Index cards or small slips of paper; Envelopes for each set of index cards

Activity

Prepare a set of index cards with different statements written on them. These statements should involve situations that can be modelled using linear equations in one variable, such as: If you subtract 8 from 4 times a number, the result is 20. Place each set of index cards in separate envelopes.

Divide the class into small groups. Distribute 1 envelope of index cards to each group.

In each round, have students draw a statement from the envelope and read it aloud to their group. The group members should discuss and formulate the linear equation in one variable that represents the given statement. Finally, they should write the equation in their notebooks.

If you subtract 8 from 4 times a number, the result is 20.

After each round, have a class discussion. Ask each group to share their formulated equation and reasoning.

Extension Idea

Instruct: A shipping company charges a flat fee of ₹20 plus ₹5 for each package shipped. If the total cost for shipping is ₹45, write a linear equation to represent the total cost.

Say: Let x be the number of packages shipped. The equation formed will be 45 = 20 + 5x.

Solving Equations: Same Operation on Both Sides

Learning

Outcomes

Imagine Maths Page 88

Students will be able to solve equations by using the same operation on both sides of the equation.

Teaching Aids

Balance scale or a makeshift scale (e.g., a ruler or a wooden beam balanced on a fulcrum); Unit cubes; Two bags with 2 unit cubes in each; One bag with 5 unit cubes; Equation cards with linear equations in one variable

Activity

Begin with a brief review of the concept of solving linear equations in one variable by maintaining equality on both sides of the equation.

Divide the class into small groups.

Distribute 2 equation cards, unit cubes, 2 bags with 2 unit cubes in each, 1 bag with 5 unit cubes and a balance scale to each group.

Ask the groups to read the equation, and then represent it on their balance scale using the bags and unit cubes. Instruct students to manipulate the weights on both sides of the scale to find the solution to the equation, which is the number of cubes inside each bag. Then, ask them to write the equation in their notebooks and solve it by performing the same operations on both sides to maintain the balance. Give them 2 more equations to solve without using the balance scale. Discuss the answers.

Extension Idea

Ask: Sarah is saving money for a new purse. She currently has ₹80 and plans to save ₹15 each week. How many weeks will it take for her to save a total of ₹200?

Say: It will take Sarah 8 weeks to save a total of ₹200 for her new purse.

Solving Equations: Transposition

Learning Outcomes

Students will be able to solve equations by using the transposing method.

Teaching Aids

Equation mazes; Crayons

Activity

Create or find equation mazes that include linear equations in one variable.

Imagine Maths Page 90

Begin the activity with a brief review of the transposition method for solving linear equations in one variable. Discuss 1–2 examples with students.

Instruct students to form pairs. Distribute an equation maze and crayons to each pair.

Instruct students to start at the beginning of the maze and solve each equation to find the correct solution by transposing from one side of the equation to the other. Ask them to shade the equation and its solution. The correct solution will guide them to the next part of the maze. Once students reach the exit, have them review their answers.

Applications of Simple Equations

Learning Outcomes

Students will be able to apply simple equations to solve word problems.

Teaching Aids

Index cards or small slips of paper; Chart paper

Activity

Begin by discussing the concept of linear equations in one variable and their significance in solving real-world problems.

Share a captivating story with students that involves solving a linear equation. For example: Once upon a time in Mathland, there was a magical garden where flowers bloomed in a peculiar pattern. The number of flowers in the garden increased by 3 each day, and there were already 8 flowers in the garden. As the magical gardener, your task is to find out how many days it will take for the garden to have a total of 23 flowers.

After telling the story, ask students to work in small groups. Provide each group with slips of paper and ask them to form a linear equation that shows the problem in the story. In this case, the equation would be 3x + 8 = 23, where x represents the number of days.

Collect the slips from each group and paste them on the chart paper. Discuss each group’s equation, emphasising the different ways students approached the problem. Make sure to highlight the correct equation.

Ask questions like: What strategies did you use to form the linear equation? How did you approach solving the equation?

Extension Idea

Instruct: Create a story problem that involves solving the equation 300 + 25x = 650.

Say: You can create multiple story problems for the given equation like: Emily rented a car service that charges a flat fee of ₹300 per day plus an additional ₹25 per mile driven. If Emily paid ₹650 for a one-day rental, find the number of miles she drove.

Answers

1. Forming Linear Equations

Do It Together

Let the runs scored by Aakash be X Sahil scored twice as many runs as Aakash = 2x Together, their runs fell four short of 100 = 100 – 4 = 96

This situation can be represented in the form of an equation as: x + 2x = 96

On simplifying the equation, we get, 3x = 96.

2. Solving Equations: Same Operation on Both Sides

Do It Together

1. We shall multiply both sides of the equation with 12 13

13c 11 × 12 13 = 9 × 12 13 ⇒ 13 12 × 12 13 × c = 9 × 12 13

On simplifying both sides, we get, c = 108 13

2. We shall subtract 20 from both sides of the equation.

3b + 20 – 20 = 32 – 20

On simplifying both sides, we get,

3b = 12

We should now divide both sides of the equation by 3

3 = 12 3 ⇒ b = 4

Verification: Substituting b = 4 in the given equation, we get

3 × 4 + 20 = 12 + 20 = 32 = RHS

Hence, the solution is verified as correct.

3. Solving Equations: Transposition

Do It Together

1. 13c = 26 × 12 [Transposing 12 from LHS to RHS]

13c = 312

c = 312 13 [Transposing 13 from LHS to RHS]

So, c = 24

Verification: Substituting c = 13 in the given equation, we get, 13 × 24 12 = 312 13 = 26 = RHS

Hence, the solution is verified as correct.

2. 3b = 32 – 20 [Transposing 20 from LHS to RHS]

3b = 12

b = 12 3 [Transposing 3 from LHS to RHS]

So, b = 4

Verification: Substituting b = 4 in the given equation, we get,

3 × 4 + 20 = 12 + 20 = 32 = RHS

Hence, the solution is verified as correct.

3. 1.6x – 2.4 – 0.5x = 6 [Opening the brackets of LHS]

On simplifying the RHS, we get, x = 4

4. Applications of Simple Equations

Do It Together

On simplifying the LHS, we get,

2(2x + 5) = 142 ⇒ 10x + 10 = 142

On transposing 10 from the LHS to the RHS, the operation changes from addition to subtraction.

4x = 142 − 10

On simplifying, we get,

x = 132 4 ; x = 33 metres

Breadth of the rectangular plot is 33 metres.

Length of the rectangular plot is 38 metres.

Angle Pairs and Parallel Lines

Learning Outcomes

Students will be able to: identify complementary and supplementary pairs of angles, adjacent angles, linear pair of angles and vertically opposite angles. apply their understanding of angles formed when a pair of parallel lines are cut by a transversal. construct a pair of parallel lines using a compass and ruler.

Alignment to NCF

C-3.4: Draws and constructs geometric shapes, such as lines, parallel lines, perpendicular lines, angles, and simple triangles, with specified properties using a compass and straightedge

Let’s Recall

Recap to check if students know how to measure angles using a protractor. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

angle: space between two rays starting from the same point parallel lines: lines that move in the same direction and are the same distance apart at every point

Teaching Aids

Cut-outs of angles that form complementary (30°, 60°, 25°, 65°), supplementary (135°, 45°, 110°, 70°) and vertically opposite pairs of angles (100°, 100°, 80°, 80°); Chart paper; Blank sheets; Ruler; Protractor; Compass; Colour pencils; Cut-outs of cars; Glue

Chapter: Angle Pairs and Parallel Lines

Pairs of Angles

Learning Outcomes

Students will be able to identify complementary and supplementary pairs of angles, adjacent angles, linear pair of angles and vertically opposite angles.

Teaching Aids

Cut-outs of angles that form complementary (30°, 60°, 25°, 65°), supplementary (135°, 45°, 110°, 70°) and vertically opposite pairs of angles (100°, 100°, 80°, 80°); Chart paper

Activity

Begin the class by discussing the different types of angles.

Instruct students to work in groups. Distribute cut-outs of angles and chart paper among the groups.

Ask the groups to paste any 2 angles to form a complementary angle and any 2 angles to form a supplementary angle, on the chart paper. They will then pick any 4 angles and paste them together to show 2 pairs of vertically opposite angles. Ask them to label the vertices and angles. Finally, they will list the adjacent and linear pairs of angles that they see in the figures.

Ask questions like: If the measure of an angle is 15°, what would be the measure of its complement?

Give students more questions on finding the measure of the missing angle when one angle is given based on complementary and supplementary angles.

Extension Idea

Ask: If two angles measure 45° and 135°, are they supplementary? Give a reason for your answer. Say: Yes, if two angles measure 45° and 135°, they are supplementary as they will add up to 180°.

Parallel Lines and Transversals

Learning Outcomes

Imagine Maths Page 107

Students will be able to apply their understanding of angles formed when a pair of parallel lines are cut by a transversal.

Teaching Aids

Blank sheets; Ruler; Protractor

Activity

Begin the class by discussing where students see parallel lines in their real life. Discuss that the lines made for effective parking in parking lots are parallel and the line joining them is their transversal.

Instruct students to work in groups. Distribute blank sheets, rulers and protractors among the groups. Ask the groups to draw parallel lines using the edges of the ruler. Ask them to then draw a line intersecting both the parallel lines at any angle other than a right angle. Instruct the groups to measure all the angles formed by the transversal on the parallel lines using the protractor and write the measures.

Instruct them to have a discussion around what they notice about the angles so formed. Discuss the angles that are of the same measure when a pair of parallel lines are intersected by a transversal.

Give them a problem on parallel lines and transversals on the board where 1–2 angles measures are given and students need to find the remaining angles.

Constructing Parallel Lines

Learning Outcomes

Students will be able to construct a pair of parallel lines using a compass and ruler.

Teaching

Aids

Chart paper; Compass; Ruler; Colour pencils; Cut-outs of cars; Glue Activity

Recall the lesson in which students discussed real-life examples where they see parallel lines and transversals. Show them pictures if needed.

Instruct students to work in groups. Distribute chart paper, compasses, rulers, colour pencils and cut-outs of cars among the groups.

Imagine Maths Page 110

Ask students to create a parallel parking space on the chart paper by constructing parallel lines for at least 6 cars, making sure that the distance between the lines is enough to fit each car cut-out. Instruct them to open page 110 of their Imagine Mathematics books and look at the steps of construction of parallel lines. Ask them to use the steps and create parallel lines on the chart paper given. Ensure that each student gets to construct at least one line parallel to another line. Ask them to use colour pencils to make the parking space colourful. Ask them to stick the car cut-outs on the parking spaces created.

Encourage the groups to show their creativity to the class.

Extension Idea

Ask: If the angle formed by the lines in the parking lot is 90°, what would you notice about the cars parked?

Say: If the angle formed by the lines in the parking lot is 90°, then the cars parked in the parking lot will be at 90° to the transversal.

Answers

1. Complementary

and Supplementary Angles

Think and Tell

Yes. ∠AOC and ∠COB together making a straight angle = 180°, hence the angles are supplementary.

Do It Together

1. ∠A = 60° ∠B = 30°

Complementary angles

2. C = 45° ∠D = 135° Supplementary angles

2. Adjacent Angles

Think and Tell

Yes, one of the adjacent pairs of angles are ∠XAC and ∠CAB

Do It Together

Answers may vary. Sample answers.

2. Adjacent angles of ∠ACB are ∠ACF and ∠ACE

3. Adjacent angles of ∠ECD are ∠FCE and ∠DCB

4. Adjacent angles of ∠FCD are ∠DCB and ∠ACF.

3. Linear Pair of Angles

Do It Together

1. 3x = 180°

So, x = 180°

3 = 60°

2. (3x + 7)° + (2x + 18)° = 180°

5x + 25° = 180°

So, x = 155° 5 = 31°

4. Vertically Opposite Angles

Do It Together

1. Hence, 3x – 15° = 75°

3x = 75° + 15° = 90°

3x = 90° ⇒ x = 30°

2. (2x – 10)° = 60°

2x = 70°

x = 35°

5. Parallel Lines and Transversals

Think and Tell

Yes, if line a is parallel to line b and line b is parallel to line c, then line a will be parallel to line c.

Do It Together

1. In the figure, ∠PQR = ∠XRY = 60° (corresponding angles)

∠QPR = ∠PRX = 65°

∠PRY = ∠PRX + ∠XRY = 65° + 60° = 125°

2. ∠PRQ + ∠PRY = 180° (Linear pair)

∠PRQ + 125° = 180°

∠PRQ = 180° – 125° = 55°

6. Constructing Parallel Lines

Do It Together

2.5 cm

Triangles and Their Properties

Imagine Mathematics Headings CB Page

Medians and Altitudes 118

Properties of Angles of a Triangle 121

Properties of the Sides of a Triangle 125

The Pythagoras Theorem 127

Congruence in Shapes and Figures 130

Learning Outcomes

Imagine

Congruence Criteria 135

Students will be able to: identify and draw the median and altitude in a triangle. identify and apply the interior and exterior sum property of a triangle. identify and apply the side sum property of a triangle. apply and prove Pythagoras’ theorem. find congruency in shapes and figures. use the SSS congruency criteria to solve problems based on it. use the SAS congruency criteria to solve problems based on it. use the ASA congruency criteria to solve problems based on it. use the RHS congruency criteria to solve problems based on it.

Alignment to NCF

C-3.2: Outlines the properties of lines, angles, triangles, quadrilaterals and polygons and applies them to solve related problems

C-3.5: Understands congruence and similarity as it applies to geometric shapes and identifies similar and congruent triangles

Let’s Recall

Recap to check if students know how to classify triangles on the basis of their sides and angles. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

median: a line segment that connects the vertex of a triangle to the midpoint of the opposite side altitude: a perpendicular line segment drawn from the vertex of a triangle to the opposite side centroid: the point where the medians of a triangle intersect

orthocentre: the point where the altitudes of a triangle intersect superposition: placing one thing on top of another line segments: a straight connection between two points or a portion of a line corresponding sides/angles: a pair of matching sides or angles that are in the same position in two different figures

diagonal: a straight line connecting two opposite vertices in a figure

Teaching Aids

Paper cut-outs of different types of triangles; Notebooks; Sheets of paper; Coloured straws; Squared paper; Glue stick; Sheets of coloured paper; Scissors; Cut-outs of pairs of triangles with sides of equal lengths and some distractor triangles; Straws; Strips of paper measuring 5 cm and 8 cm; Protractor; Two sets of cards with congruent triangles drawn on them; Ruler; Cut-outs of congruent right-angled triangles and some distractor triangles

Chapter: Triangles and Their Properties

Medians and Altitudes

Learning Outcomes

Students will be able to identify and draw the median and altitude in a triangle.

Teaching Aids

Paper cut-outs of different types of triangles

Activity

Instruct students to form groups. Distribute cut-outs of equilateral, isosceles, scalene, acute, right and obtuse triangles, giving 1 of each type to each group. Instruct students to fold the triangle such that the crease joins the vertex to the centre point of the opposite side. Discuss what a median is. Then, instruct them to trace the triangles in their notebooks and draw perpendiculars from each vertex of the triangle to its opposite side. Now, discuss what the altitude is.

Then, ask students about the similarities and differences between the creases obtained on the two sets of triangles and in each triangle separately. Discuss their observations and conclusions about each triangle.

Ask questions like: Did you get all the altitudes inside the triangles? Give a reason for your answer. In which triangle is the median and altitude the same?

Extension Idea

Ask: Can a triangle have all three medians of equal length? Why?

Say: No, a triangle cannot have all three medians of equal length unless it is an equilateral triangle. The medians of an equilateral triangle are equal in length, but in other triangles, they are generally of different lengths.

Yes, an equilateral triangle can have all three medians of equal length. But in other triangles, they are generally of different lengths.

Properties of Angles of a Triangle

Learning

Outcomes

Students will be able to identify and apply the interior and exterior sum property of a triangle.

Teaching

Aids

Paper cut-outs of different types of triangles; Notebooks; Sheets of paper

Activity

Instruct students to form groups.

Distribute the cut-outs, giving 1 of each type to each student of the group. Instruct students to mark the angles in the triangles, cut out the angles and then put them together to form 1 angle. Then, encourage them to discuss what they notice about the new angle formed in their groups.

Ask them about their observations and discuss the conclusions they derive. Let them verify their conclusions using the measures of the angles in their notebooks.

Ask questions like: What do you notice about the new angle formed?

Then, instruct them to trace each triangle on a sheet of paper and extend one side of the triangle to form an angle outside it. Next, instruct them to mark all the interior and exterior angles of the triangle, cut out the 2 interior opposite angles and paste them on the exterior angle such that they completely overlap with the exterior angle. Discuss the exterior angle sum property. Then, ask them about their observations and discuss the conclusions they derived. Let them verify their conclusions using the measures of the angles in their notebooks. Finally, give them questions on finding the missing angle of a triangle when the measures of 2 angles are given.

Extension Idea

Ask: Can a triangle have an exterior angle measuring 90° at two different vertices? Explain.

Say: No, a triangle cannot have an exterior angle measuring 90° at two different vertices because the sum of all exterior angles of any polygon is 360°. If the measure of an exterior angle is 90° at two different vertices, then the third exterior angle will be 180°, making the interior angle 0°, which is not possible.

Properties of the Sides of a Triangle Imagine Maths Page 125

Learning

Outcomes

Students will be able to identify and apply the side sum property of a triangle.

Teaching Aids

Coloured straws

Activity

Instruct students to form groups.

Distribute 3 straws of different colours to each group. Instruct them to cut the straws into the measures written on the board and try putting them together to form a triangle. Ask them if they can form a pattern between the measures of two sides of the triangle and its third side. Then, ask them about their observations and discuss the conclusions they derived. Discuss the triangle inequality property. Let them verify their conclusions using the measures of the sides in their notebooks.

Repeat this activity using different lengths of straws.

Give students questions in which they need to find the possible length of the third side of a triangle.

The Pythagoras Theorem Imagine Maths Page 127

Learning

Outcomes

Students will be able to apply and prove Pythagoras’ theorem.

Teaching

Aids

Squared paper; Glue stick

Activity

Instruct students to form groups. Write the measures of 3 sides of a right triangle on the board. Distribute squared paper to each group.

Instruct the students to cut 3 squares with sides that are the same as the side lengths of the right triangle. Ask the students to place the 2 smaller squares inside the largest square by cutting and pasting without leaving any gaps.

Point out that the 2 squares cover exactly the same area as the largest square.

Ask them to try deducing the relationship between the lengths of the sides of the triangle and its area and discuss the Pythagoras theorem.

Ask questions like: What is the rule of the lengths of the sides of a right-angled triangle?

Give students questions in which they need to find the length of the third side of a right-angled triangle when the lengths of 2 sides are given.

Extension Idea

Ask: A pilot is flying at an altitude of 6000 feet. If the pilot needs to descend to land on the runway, and the runway is 8000 feet away horizontally, what is the length of the descent path?

Say: (Length of the descent path)2 = 60002 + 80002 = 3,60,00,000 + 6,40,00,000 = 10,00,00,000 = 10,000 × 10,000. So, the length of the descent path = 10,000 feet.

Congruence in Shapes and Figures

Learning Outcomes

Students will be able to find congruency in shapes and figures.

Teaching Aids

Sheets of coloured paper; Scissors

Activity

Imagine Maths Page 130

Discuss real-life examples where students can see congruency (for example, biscuit packets, the leaves of a book, a bundle of currency notes, etc.) and explain why the objects are congruent. Inform students that they will be making congruent shapes using sheets of paper. Instruct students to form pairs.

Provide each pair with 2 sheets of coloured paper and a pair of scissors.

Instruct the pairs to form shapes congruent to each other. They must make at least 4 copies of each shape. Ask them to discuss and decide which shape each partner will make. Then, ask them to use different methods of making 4 copies of a shape that will be congruent. One way would be to fold the sheet multiple times and then cut out a shape, and another would be to draw shapes with sides of the same length. Once done, they will check their partner’s shapes to see if all of them are congruent to each other.

Discuss with students that there are different ways to make congruent shapes and that congruent shapes are formed when the cut-outs overlap completely.

Ask: Can you identify any objects in the classroom that are congruent to each other? If time allows, ask the students to try cutting some more shapes for further practice on the concept of congruency.

SSS Congruence Criteria

Learning Outcomes

Students will be able to use the SSS congruency criteria to solve problems based on it.

Teaching Aids

Imagine Maths Page 132

Glue stick; Sheets of paper; Cut-outs of pairs of triangles with sides of equal lengths and some distractor triangles; Straws

Activity

Begin by introducing the SSS congruence criteria.

Ask students to consider two triangles ABC, where in one triangle, AB = 3 cm, BC = 4 cm and CA = 5 cm, and in the other triangle ABC, AB = 4 cm, BC = 5 cm and CA = 3 cm. Discuss how the corresponding three sides of the triangles are equal in length, and so they are congruent according to the Side-Side-Side criteria.

Congruent Triangles Using the SSS Rule

Following the SSS rule, make cut-outs of triangles with pairs of triangles of equal side lengths and some distractors. Instruct students to work in pairs. Distribute a sheet of paper, triangle cut-outs and a glue stick to each pair.

Instruct them to measure the length of the sides of the triangles, write the measures inside the triangle and then make pairs of triangles that overlap exactly or are congruent. They should paste the pairs of triangles on the sheet of paper and label them as ‘Congruent Triangles Using the SSS Rule’.

Distribute the straws to each pair. Direct each student in the pair to cut straws of 3 lengths (2 pieces of 4 cm, 7 cm and 5 cm each). Direct each student to construct a triangle using straws of differing lengths. Ask them to match the triangle they made with their partner’s.

Ask questions like: How do you know that the 2 triangles are congruent?

Extension Idea

Ask: A triangle has a side measuring 15 cm. If this triangle is congruent to 2 other triangles, one with a side measuring 10 cm and another with a side measuring 18 cm, what are the lengths of all the sides of the first triangle?

Say: Because all 3 triangles are congruent, the lengths of all the sides of all the triangles are 15 cm, 10 cm and 18 cm.

SAS Congruence Criteria

Learning Outcomes

Students will be able to use the SAS congruency criteria to solve problems based on it.

Teaching Aids

Strips of paper measuring 5 cm and 8 cm; Glue stick; Protractor; Sheets of paper

Imagine Maths Page 133

Activity

Begin by introducing the SAS congruence criteria.

Ask students to consider 2 triangles ABC, where in one triangle AB = 3 cm and CA = 5 cm, and in the other triangle PQR, PQ = 5 cm and RP = 3 cm. The angle formed between these 2 sides measures 60°. Discuss that as the corresponding sides of both triangles, along with the angle between them, are equal, they satisfy the SAS criteria for congruence.

Instruct students to work in pairs.

Distribute a glue stick, a protractor, a sheet of paper and strips of paper measuring 5 cm and 8 cm, ensuring that each student in the pair gets one of each strip.

Instruct each student to paste 1 strip on the sheet of paper and then paste the other strip at one end of the previous strip such that they form an angle of 60°. Let them use a protractor to measure the angle. Then, guide them to draw a line on the sheet to complete the triangle. Ask them to write the measures inside the triangle. They will then cut out their triangles and superimpose them on their partner’s triangle to check if they overlap completely. Then, they will measure the length of each side and prove the congruence in their notebooks.

If time allows, give them a few more practice questions to solve in their notebooks.

ASA Congruence Criteria

Learning Outcomes

Students will be able to use the ASA congruency criteria to solve problems based on it.

Teaching Aids

Two sets of cards with congruent triangles drawn on them; Ruler; Protractor Activity

Imagine Maths Page 134

Begin by introducing the ASA congruence criteria. Ask students to consider two triangles with angles of 45° and 60° where the side between these angles measures 4 cm. Discuss that if 2 angles and the side between them are the same in both triangles, they are congruent according to the ASA criteria.

Prepare 2 identical sets of cards, one with triangle ABC and the other with triangle XYZ drawn on them, where both triangles are congruent by the ASA congruency rule.

Instruct students to form 2 groups.

Distribute the sets of cards, giving 1 set to each student in Group 1 and the second set to each student in Group 2.

Instruct students in Group 1 to measure the length of side AB and ∠A and ∠B in ΔABC, and the students in Group 2 to measure the length of side XY and ∠X and ∠Y in ΔXYZ. They will then move around the classroom to find the student whose triangle has the exact same measurements. Once they have found their partners, they will work together to prove the congruence in their notebooks.

Prompt them to consider whether this method also satisfies the criteria for SSS and SAS congruence.

Extension Idea

Ask: The measures of 2 angles of a triangle are 45° and 60°. The angles of another triangle measure 45° and 75°. If these 2 triangles have 1 side measuring 13 cm, are they both congruent using the ASA rule?

Say: 180° – 45° + 60° = 75°. Although the 3 angles in both triangles are equal, we do not know if the included side in both triangles has the same measurement. So, the triangles may not be congruent using the ASA rule.

RHS Congruence Criteria

Learning Outcomes

Students will be able to use the RHS congruency criteria to solve problems based on it.

Teaching Aids

Imagine Maths Page 135

Glue stick; Sheets of paper; Cut-outs of congruent right-angled triangles and some distractor triangles

Activity

Begin the class by introducing the RHS congruence criteria. Make cut-outs of pairs of right-angled triangles with equal lengths of the hypotenuse, one leg and the angle opposite to the hypotenuse measuring 90° (following the RHS rule) and some distractor triangles. Instruct students to work in pairs.

Distribute a sheet of paper, triangle cut-outs and a glue stick to each pair.

Congruent Triangles Using the RHS Rule

Instruct them to measure the length of the longest side, the angle opposite the longest side, which should be 90°, and one of the arms of the 90° angle. They will then write the measures inside the triangle and make pairs of triangles that overlap exactly or are congruent. They should paste the pairs of triangles on the sheet of paper and label them as ‘Congruent Triangles Using the RHS Rule’.

Extension Idea

Ask: One triangle has a right angle and a hypotenuse measuring 10 cm. Another triangle has a right angle and a side measuring 6 cm. If these 2 triangles are congruent, what is the length of the third side of the first triangle?

Say: Since the triangles are congruent, the first triangle has a hypotenuse measuring 10 cm and a side measuring 6 cm. We can find the length of the third side using Pythagoras’ theorem where 102 = a2 + 62. On solving this, we get a2 = 102 – 62 and a = 8. So, the length of the third side is 8 cm.

Answers

1. Medians and Altitudes

Think and Tell

We can show the medians of a triangle by drawing a line from the vertex to the centre of the opposite side.

Do It Together

SR = QR (Since PR is the median to the side SQ and R is the midpoint of SQ.)

SQ = SR + RQ. So, SQ = 7 + 7 = 14 cm

PG:GR = 2 : 1 (Property of a centroid)

PR = PG + GR (G lies between P and R)

So, PR = 8 + (8 ÷ 2) = 8 + 4 = 12 cm

2. Angle Sum Property of a Triangle

Do It Together

Using the angle sum property of a triangle,

BAC + ABC + ACB = 180°

(x + 5)° + (x + 10)° + (2x + 5)° = 180°

4x + 20° = 180°

So, x = (180° – 20°) ÷ 4 = 160° ÷ 4 = 40°

Using the value of x,

(x + 5)° = 45°

(x + 10)° = 50°

(2x + 5)° = 85°

The measures of the angles are 45°, 50° and 85°

3. Exterior Angle Sum Property of a Triangle

Do It Together

Step 1

PQR + QPR = PRS (Using the exterior angle sum property)

Substituting the values of the respective angles,

PQR + 80° = 140°

So, PQR = 140° − 80° = 60°

Step 2

PQR + QPR + PRQ = 180° (angle sum property)

Putting the values of the respective angles,

60° + 80° + PRQ = 180°

PRQ = 180° − (60° + 80°) = 180° − 140° = 40°

The angles of the triangle PQR are 60°, 80° and 40°

4. Properties of the Sides of a Triangle

Do It Together

We know that Side 1 + Side 2 > Side 3. (The triangle inequality property)

Finding the sum of the two sides, Side 1 + Side 2 = 4 cm + 12 cm = 16 cm

We know that Side 1 – Side 2 < Side 3. (The triangle inequality property)

Finding the difference of the two sides, Side 1 – Side 2 = 12 cm – 4 cm = 8 cm

Therefore, the length of the third side of the triangle has to be more than 8 cm but less than 16 cm

5. The Pythagoras Theorem

Think and Tell

In a right-angled triangle ABC, if AB2 = AC2 + BC2, then the right angle is ACB.

Do It Together

AB = 8 m, CD = 16 m and BD = 6 m

Draw AE || BD meeting CD at E. Then, ABDE is a rectangle.

Using the properties of a rectangle, ED = 8 m and AE = 6 m

In a right-angled triangle AEC, we have, AC2 = CE2 + AE2

Putting the values, AC2 = 82 + 62 = 64 + 36 = 100 = 10 × 10 AC = 10 m

The distance between the tops of the two poles is 10 m.

6. Congruence in Shapes and Figures

Do It Together

We know that ABCD = PQRS.

The correspondence is ABCD PQRS.

A P

D S

BC QR

PQR ABC

The corresponding sides are: AB and PQ; BC and QR; CD and RS; DA and SP.

7. SSS Congruence Criteria

Do It Together

In ΔABC and ΔDEC,

AB = DE (side)

BC = CD (side)

AC = CE (side)

Therefore, by the SSS congruence criteria, ΔABC = ΔDEC.

8. SAS Congruence Criteria

Do It Together

We know that ΔPQR = ΔXYZ.

So, PQ = XY = z

QR = YZ = 6 cm

PR = XZ = x and R = Z = y

z = 5 cm

x = 8 cm y = 50°

9. ASA Congruence Criteria

Do It Together

In ΔAOC and ΔBOD, CAO = DBO (given)

AO = BO (given)

AOC = BOD (vertically opposite angles)

Therefore, by the ASA congruence criteria, ΔAOC = ΔBOD

10. RHS Congruence Criteria

Do It Together

In ΔABD and ΔACD, ∠ABD = ACD = 90°

AD = AD (common side)

AB = AC = 4.9 cm

Therefore, by the RHS congruence criteria, ΔABD = ΔACD.

Ratio and Proportion

Learning Outcomes

Students will be able to: find equivalent ratios and write them in simplified form. check if four terms are in proportion. solve problems using the unitary method. solve problems on speed, distance and time.

Alignment to NCF

C-1.6: Explores and applies fractions (both as ratios and in decimal form) in daily-life situations

C-2.1: Understands equality between numerical expressions and learns to check arithmetical equations

Let’s Recall

Recap to check if students know the basics of ratios. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

equivalent ratios: ratios with the same relationship but different numbers proportionality condition: two ratios being equal means they are proportional means: middle terms in a proportion extremes: first and last terms in a proportion

Teaching Aids

Red and blue building blocks of the same size; Geo boards; Strings; Sheet of paper with a table showing the quantity of items to be bought and the cost of each item; Play money; Toy cars; 1-m-long road track made of cardboard; Stopwatch

Chapter: Ratio and Proportion

Equivalent Ratios

Learning Outcomes

Students will be able to find equivalent ratios and write them in simplified form.

Teaching Aids

Red and blue building blocks of the same size

Activity

Begin by revising the concept of the simplest form of a ratio. Instruct students to work in groups. Distribute the red and blue building blocks among the groups.

Ask students to build a simple structure, using both coloured building blocks. After the structures are built, instruct students to write down the ratio of the red and blue blocks used in its simplest form in their notebooks.

Ask the groups to build another structure by using double the red and blue blocks used earlier. Example: If 6 red blocks and 4 blue blocks were used earlier, ask them to use 12 red and 8 blue blocks. Ask them to write down the ratio of red and blue blocks used in its simplest form in their notebooks.

Ask them to discuss in their groups what they notice about the simplest form of both ratios. Bring out the fact that two ratios are equivalent when their simplest form is the same.

Help them understand that equivalent ratios can be found if the numerator and denominator are multiplied or divided by the same natural number. Give them questions for practice in their notebooks.

Extension Idea

Ask: What is the ratio of length to breadth of a rectangle with a perimeter of 180 cm, where the length is 10 cm more than the breadth?

Say: Let b = x cm and l = x + 10. So, 2 (x + x + 10) = 180 ⇒ x = b = 40 cm and l = 50 cm. Therefore, the ratio is 5:4.

Proportional Terms Imagine Maths Page 145

Learning Outcomes

Students will be able to check if four terms are in proportion.

Teaching Aids

Geo boards; Strings

Activity

Instruct students to work in groups. Distribute the teaching aids among the groups.

Instruct students to use the strings to make two non-concentric rectangles of different lengths and widths on the geometric board (geo board). Once the groups are done forming the rectangles, ask them to look closely at the rectangles they created and note down the ratio of length to width of both rectangles in their simplest form.

Ask questions like: Is the simplest ratio of length to width of both the rectangles equal?

If not, ask them to change the length and width of the second rectangle so that it has the same ratio of length to width as the first rectangle. Once students are done, review the rectangles formed and conclude that if the simplest ratios are equal, the ratios are said to be in proportion.

Teacher tip: If the geo board is unavailable, use a paper sheet with dots drawn at equal distances.

Unitary Method Imagine Maths Page 148

Learning Outcomes

Students will be able to solve problems using the unitary method.

Teaching Aids

Sheet of paper with a table showing the quantity of items to be bought and the cost of each item; Play money

Activity

Begin the lesson by asking questions like: How do you calculate the amount to be paid when buying a certain item from the market?

Discuss student responses. Bring out the fact that whenever we purchase multiple units of an item, we need the price of one item.

Instruct students to work in groups of 4.

Distribute the sheet of paper with a table showing the quantity of items to be bought with their costs and play money to each group.

Instruct one student to act as a shopkeeper and the other three as buyers. Ask the buyers to pick one item each and calculate the amount of money they need to pay for the quantity of items to be bought using the unitary method. Ask them to use the play money for their purchases. Ask students acting as the shopkeeper to verify the amount paid by each buyer. Instruct them to note down the answers in their notebooks.

Repeat the activity by asking students to change their roles and modify the quantities to be bought.

Extension Idea

Ask: Calculate the cost of 3.5 litres of milk using the unitary method for the given price.

Say: It is given that the cost of 2 litres of milk is ₹120. So, the cost of 1 litres of milk is 120 ÷ 2 = ₹60 and the cost of 3.5 litres of milk is 3.5 × 60 = ₹210.

Speed, Distance and Time

Learning Outcomes

Students will be able to solve problems on speed, distance and time.

Teaching Aids

Marbles; 1-m-long road track made of cardboard; Stopwatch Activity

Imagine Maths Page 150

Begin by asking questions like: How much time does it take you to reach school? How far are your homes from the school? Discuss their answers in the class.

Instruct students to form groups. Distribute the marbles, tracks and stopwatches to each group. Instruct the groups to imagine that the marbles are cars and run them on the tracks and find out how much time it takes to make 1, 2 or 5 rounds using the stopwatch. Discuss the time recorded by each group in the class.

Bring out the fact that time, distance and speed are dependent on each other. Help students derive the formula for time, speed and distance and ask them to calculate the speed of the car using the formula. Ask them to note their answers in their notebooks. Give students questions to practise finding speed, distance or time when any of the 2 are given.

Extension Idea

Ask: Sarah and Ben are planning a bike ride to the park. Sarah can bike at a speed of 10 miles per hour, while Ben can bike at 8 miles per hour. The park is 12 miles away. They decide to race. If both maintain their individual speeds, who will reach the park first?

Say: Sarah’s speed is 10 miles/hr; Ben’s speed is 8 miles/hr.

The time taken by Sarah to reach the park = 12 10 = 1.2 hours.

The time taken by Ben to reach the park = 12 8 = 1.5 hours. Hence, Sarah will reach the park first.

Answers

1. Equivalent Ratios

Think and Tell

2:3 has infinite equivalent ratios. Som examples are 4:6, 6:9, 8:12 and 10:15.

Do It Together

LCM of 3, 7, 15 and 5 = 105

1 3 = 35 105; 5 7 = 75 105; 4 15 = 28 105; 2 5 = 42 105;

Hence, the descending order is

5 7 > 2 5 > 1 3 > 4 15

2. Proportionality Condition

Do It Together

Let the fourth term be x; 4.2:1.2 :: 0.7:x

4.2 × x = 1.2 × 0.7 ⇒ x = 1.2 × 0.7 4.2 = 0.2

Hence, the fourth term is 0.2

3. Continued Proportion

Do It Together

1 3 : 2 9 :: 2 9 : x ⇒ 1 3 × x = 2 9 × 2 9 ⇒ x = 2 9 × 2 9 ÷ 1 3 ⇒

4. Unitary Method

Do It Together

Number of toys produced in 6.5 hours = 208

Number of toys produced in 1 hour = 208 ÷ 6.5 = 32

Number of toys produced in 20 hours = 32 × 20 = 640

5. Speed, Distance and Time

Do It Together

Distance covered by Kunal = 4.5 km; Kunal’s speed = 1.5 km/hr Time taken by Kunal = Distance ÷ Speed = 4.5 1.5 = 3 hr

Distance covered by Soham = 8 km; Soham’s speed = 2 km/hr; Time taken by Soham = Distance ÷ Speed = 8 2 = 4 hr

Ratio of the time taken by both of them = 3:4

Percentage, Profit and Loss, Simple Interest

Learning Outcomes

Students will be able to: convert between percentages, decimals and fractions.

convert between percentages and ratios. find the percent of a quantity and percent change.

solve word problems on percentages.

Alignment to NCF

find profit or loss percentage, cost price or selling price when any two are given.

find the marked price or discount for a given item. solve word problems on finding the profit or loss percent.

solve word problems on finding the simple interest on a given amount.

C-1.5: Explores the idea of percentage and applies it to solve problems

Let’s Recall

Recap to check if students know how to read fractions from pie charts. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

percent: a part out of 100

Teaching Aids

interest: extra money on a loan or savings

Paper slips containing decimal numbers and fractions; Decimal grids; Crayons; Scenario cards; Question slips; Puzzle cards with 4 problems, and the answer to one of them written in the centre; Question cards; Puzzle interlocking cards with the cost price written on the first piece and the selling price on the second; Puzzle interlocking pieces with profit and loss written; Pencil box; Tiffin box; water bottle; Marked price tags; Play money; Table cards containing information on the types of chocolates, their CP, SP or MP

Chapter: Percentage, Profit and Loss, Simple Interest

Percentages, Decimals and Fractions

Learning Outcomes

Students will be able to convert between percentages, decimals and fractions.

Teaching Aids

Paper slips containing decimal numbers and fractions; Decimal grids; Crayons

Activity

Begin by discussing with students how stock market shares are represented. Choose a simple fraction or decimal, like 1 3 or 0.4, and represent it on the decimal grid. Discuss the relationship between the coloured grids and the whole (100%) and how this can be expressed as a percentage.

Now, form small groups and distribute the decimal grids. On the table, place paper slips containing decimal numbers and fractions such as 0.57 or 3 5.

Instruct each group to pick a slip and ask them to colour the decimal grids accordingly.

Now, have the groups observe the coloured grids as a part of the whole (100%) and then represent the same in their notebooks as a percentage. Rotate the slips among the groups and repeat the activity.

Extension Idea

Ask: How can you convert 35% into fractions?

Say: To convert 35% into fractions, we simply divide 35 by 100 to get 35 100, which can be simplified as 7 20.

Percentage and Ratios

Learning Outcomes

Students will be able to convert between percentages and ratios.

Teaching Aids

Scenario cards; Question slips

Activity

Begin by introducing how ratio and percentage are related, and explain how we can convert between ratios and percentages.

Instruct students to form groups for the activity. Distribute the scenario cards to each group.

“In the inter-state math Olympiad of Madhya Pradesh and Delhi, 500 students participated. The ratio of students who participated from Madhya Pradesh to those from Delhi is 3:2.”

Prepare question slips such as the ones given below and distribute among the students.

What percent of the total students participated from Madhya Pradesh?

How many fewer students participated from Delhi?

Instruct students to solve these word problems in their notebooks. Discuss their approach in class. Make the students practise more questions on conversion between ratios and percentages.

Extension Idea

Ask: How many students from Delhi participated for the Olympiad if 20% did not participate?

Say: Number of students from Delhi = 200, 80% of them participated as 20% did not participate, hence, 80

100 × 200 = 160 students participated.

Finding Percent and Percent Change

Learning Outcomes

Students will be able to find the percent of a quantity and percent change.

Teaching Aids

Puzzle cards with 4 problems, and the answer to one of them written in the centre

Activity

Begin by discussing some real-life examples of percent change, such as the population change of a city in 2 years. Make puzzle cards with 4 problems and the answer to one of them written in the centre.

Instruct students to form groups for the activity. Distribute the puzzle cards to each group. Tell them that each card consists of 4 problems to be solved. Students have to solve the expressions, and match the answers with the one written on the centre card. Instruct them to encircle the problem that matches with the answer on the centre card. After the activity, discuss their answers in the class. If time permits, rotate the puzzle cards among groups and repeat the activity.

Extension Idea

Instruct: Frame a word problem where you use percent in your daily life.

Say: One such problem can be, “What percent of the day do you spend in school?”.

Word Problems on Percentage

Learning Outcomes

Students will be able to solve word problems on percentages.

Teaching Aids

Scenario cards; Question cards

Imagine Maths Page 160

Imagine Maths Page 162

Activity

Instruct students to form groups for the activity. Distribute the scenario cards to each group.

“The prices of three sweaters were ₹400, ₹500 and ₹750 in 2022. A year later, the price of each sweater became ₹600”.

Prepare question cards such as the ones given below and distribute among the students:

“What is the percentage change in the price of each sweater?”

“What should be the new price of each sweater if there is a 20% increase in price in 2024?”

Instruct students to solve these word problems in their notebooks. Discuss their approach in class. Give the students practice questions based on more such scenarios.

Profit and Loss Percentages

Learning Outcomes

Imagine Maths Page 165

Students will be able to find profit or loss percentage, cost price or selling price when any 2 are given.

Teaching Aids

Puzzle interlocking cards with the cost price written on the first piece and the selling price on the second; Puzzle interlocking pieces with profit and loss written

Activity

Start by briefly discussing the concepts of cost price, selling price, profit and loss.

Divide the class into groups and make puzzle interlocking cards with the cost price written on one piece and the selling price on the second. Also, make puzzle interlocking pieces with profit and loss written on it.

Distribute the puzzle pieces and puzzle cards among the groups. Instruct students to find whether there is a profit or a loss and join it with the appropriate puzzle card. Ask students to calculate the profit or loss percentage and write the answers in their notebooks.

Encourage students to shuffle the puzzle pieces among the groups and solve.

Marked Price and Discount

Learning Outcomes

Students will able to find the marked price or discount for a given item.

Teaching

Aids

Pencil box; Tiffin box; Water bottle; Marked price tags; Play money

Imagine Maths Page 166

Activity

Discuss about marked price and discount and how they can be found when either the marked price or the discount is given.

Divide the class into groups. Keep different discount cards ready. Display the items with price tags along with the discount on the table.

Ask each group to come forward with the play money. Ask the group members to calculate the discount price and pay for the items on the table.

For example, if the cost of a water bottle is ₹800 and the discount is 10%, students will find the discount as ₹80. They will subtract 80 from 800 and pay ₹720.

Ask question such as: On what does the discount percentage depend?

Extension Idea

Ask: What is the marked price of a winter jacket, if it is sold at ₹1800 after a discount of ₹200?

Say: MP = SP + Discount, hence, the MP of the sweater was ₹2000. Word Problems on Profit and Loss Imagine Maths Page 167

Learning Outcomes

Students will be able to solve word problems on finding the profit or loss percent.

Teaching Aids

Play money; Table cards containing information on the types of chocolates, their CP, SP or MP; Question cards

Activity

Prepare table cards containing information on the types of chocolates, their CP, SP or MP.

Set up chocolate-selling corners in the classroom.

Divide the class into groups and assign roles within each group: one student will act as the “shopkeeper”, while the others will be the “customers”. Provide each group with play money, table cards and question cards such as:

The shopkeeper sold 15 Dark Delight chocolates. Calculate the total profit made.

Due to the discount offer, a customer bought 8 Raspberry Bliss chocolates. Calculate the total discount percentage given.

Instruct students to solve the problems given. Each group should discuss and solve the problems together. The customers will use their play money to buy chocolates from the shopkeeper to verify and experience the real-life scenario. The shopkeeper must ensure they calculate the correct prices based on the given information. Rotate the roles within the groups, allowing each student to experience being the shopkeeper.

Extension Idea

Ask: Ruchika bought 6 bananas for ₹30. She mistakenly dropped 3 bananas somewhere while returning home. For how much should she sell each of the remaining bananas to earn a profit of 20%?

Say: After losing 3, Ruchika will be left with 3 bananas. CP of 1 banana = 30 3 = ₹10. SP of one banana to sell it at a profit of 20% = 100 + 20% 100 × 10 = 120 100 × 10 = ₹12. Therefore, she needs to sell each of the remaining bananas for ₹12 to make a profit of 20%.

Simple Interest

Learning Outcomes

Imagine Maths Page 171

Students will be able to solve word problems on finding the simple interest on a given amount.

Teaching Aids

Scenario cards; Question cards

Activity

Instruct students to form groups for the activity. Distribute the scenario cards to each group. Make question cards and distribute them among groups.

Discuss the scenario and instruct students to solve these word problems in their notebooks. Discuss their answers and approach in the classroom. Give students practice questions based on more such scenarios, if time permits.

Ramesh and Suresh deposited ₹2000 each in different banks at 5% and 6% interest per annum. Ramesh withdraws his money at the end of 1 year, whereas Suresh withdraws it after 2 years.

Who will get more interest at the end of 1 year and by how much?

What amount of money will Suresh get after 2 years?

What is the total interest earned by both of them?

Answers

1. Percentages, Decimals and Fractions

Do It Together

1. 56%

56% = 56 100 = 14 25

56% = 56 ÷ 100 = 0.56

2. 120% = 120 ÷ 100 = 6 5

120% = 120 ÷ 100 = 1.2

3. 550% = 550 ÷ 100 = 11 2

550% = 550 ÷ 100 = 5.5

2. Percentage and Ratios

Do It Together

The ratio of monkeys to total animals = 3:8

The percentage of monkeys at the zoo = 3 8 × 100 = 37.5%

Think and Tell

Percentage of elephants at the zoo = 5 8 × 100 = 62.5%

3. Finding Percent and Percent Change

Do It Together

1. 910

2. 35% of 520 = 520 × 35 100 = 182

Number 35% more than 520 = 520 + 182 = 702

4. Word Problems on Percentage

Do It Together

Number of people who went to the exhibition on Sunday = 845

Number of people who went to the exhibition on Monday = 169

Change in the number of people = 845 − 169 = 676

Decrease in percent =

Change in the no. of people

Number of people who went on Sunday × 100% = 676

845 × 100% = 80%

Thus, there was an 80% decrease in the number of people visiting the exhibition on Monday.

5. Profit and Loss Percentages

Do It Together

1. 100 80 × ₹1600 = ₹2000

2. 112% 100 × 2500 = ₹2800

6. Marked Price and Discount

Do It Together

80 1600 × 100% = 5%

7. Word Problems on Profit and Loss

Do It Together

Total CP of scooter = ₹47,000 + ₹8000 = ₹55,000

SP of scooter = ₹58,000

Profit% = Profit CP × 100% = ₹3000 ₹55,000 × 100% = 5.45%

8. Using the Formula

Do It Together

₹1000 × 100 10 × 5 = 100000 50 = ₹2000

9. Word Problems on Simple Interest

Do It Together

₹375 = ₹1500 × R × R 100 ; R × R = ₹375 × 100 ₹1500 = 25

R = 5%

Rational Numbers 10

Learning Outcomes

Students will be able to: identify rational numbers and represent them on a number line. find equivalent rational numbers and write them in their simplest form. compare and order rational numbers. find rational numbers between 2 rational numbers.

Alignment to NCF

add rational numbers and find the additive inverse of a rational number. subtract rational numbers. multiply rational numbers. divide rational numbers by finding the reciprocal of the divisor.

C-1.4: Explores and understands sets of numbers, such as whole numbers, fractions, integers, rational numbers and real numbers, and their properties, and visualises them on the number line

C-9.1: Recognises how concepts (like counting numbers, whole numbers, negative numbers, rational numbers, zero, concepts of algebra and geometry) evolved over a period of time in different civilisations

Let’s Recall

Recap to check if students know the concept of fractions and types of fractions. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

subunit: a small part of a unit

HCF: highest common factor/the greatest common number that divides 2 or more numbers

LCM: lowest common multiple/the least common number that is exactly divisible by 2 or more numbers

Teaching Aids

Cards with rational numbers written on them; Worksheets; Cards with 4 rational numbers written on them; Cards with sums of rational numbers written on them; Cards with subtraction problems of rational numbers written on them; Bingo cards; Dice

Chapter: Rational Numbers

Representing Rational Numbers

Learning Outcomes

Imagine Maths Page 179

Students will be able to identify rational numbers and represent them on a number line.

Teaching Aids

Cards with rational numbers written on them

Activity

Begin the lesson by introducing the concept of rational numbers and their representation. Instruct students to form groups of 4 each and name the teams: Team A, Team B, Team C and so on.

Make a number line using chalk on the floor at the front of the classroom where all the students can see it. Shuffle the cards with rational numbers (e.g., 1 2 , –3 4 etc.) and place them face down in a pile on the teacher’s desk. Each team will take turns drawing a card from the pile and discuss and decide where the rational number on the card should be plotted on the number line. Once they make their decision, they will send one member to the front to plot the number on the number line. For instance, Team A draws a card with the rational number –3 4 . They discuss and decide it should be plotted between –1 and 0 on the number line. If the number is plotted correctly, the team earns a point. If not, they lose a point. The next team takes its turn, and play continues until all rational numbers have been plotted.

After all the rational numbers have been plotted, tally the points earned by each team. The team with the most points wins the game.

Ask students to explain why certain numbers were placed where they were on the number line. Address any misconceptions or questions that arise.

Rational Numbers in Standard Form;

Imagine Maths Page

181 Equivalent Rational Numbers

Learning

Outcomes

Students will be able to find equivalent rational numbers and write them in their simplest form.

Teaching Aids

Worksheets

Activity

Begin by reviewing what rational numbers are and how they can be reduced to standard form. Explain the concepts of equivalent rational numbers to the students.

Create worksheets in a “match the following” format, containing 5 pairs of rational numbers that are equivalent to each other.

Ensure that the numbers are shuffled on both sides.

Distribute 1 worksheet to each student.

Instruct them to simplify all the rational numbers to their simplest form. Then, they will match the equivalent rational numbers with their corresponding pairs.

Ask students to discuss why simplifying fractions is important and how it helps in understanding the relationship between different rational numbers. Afterwards, discuss their responses and provide additional questions if necessary.

Comparing and Ordering Rational Numbers

Learning

Outcomes

Students will be able to compare and order rational numbers.

Teaching Aids

Cards with 4 rational numbers written on them

Activity

Begin by saying that when comparing 2 rational numbers, we need to ensure that the denominators of both numbers are the same.

Instruct students to form groups of 4.

Distribute a card containing 4 rational numbers to each group.

Guide students to determine the least common multiple (LCM) of all 4 denominators.

Ask students to form equivalent rational numbers such that all the rational numbers have the same denominators. Instruct students to arrange the rational numbers in ascending order.

Ask questions like: Why is it important to have the same denominators when comparing rational numbers?

Provide each group with another card containing 4 rational numbers and repeat the process to reinforce the concept.

Rational Numbers Between Two Rational Numbers

Learning Outcomes

Students will be able to find rational numbers between 2 rational numbers.

Teaching Aids

Cards with 4 rational numbers written on them

Imagine Maths Page 185

Activity

Begin by reviewing the concept of finding a rational number between any 2 rational numbers.

Write pairs of rational numbers on cards, for example: 1 3 and 1 2, 2 5 and 1 2 etc.

Shuffle the cards and place them face down in a stack. Divide the students into teams of 10.

Take the students to the playground.

Make a start line and arrange the teams in lines, with each line standing behind a designated starting point. Explain that the game involves a relay race to find a rational number between 2 given rational numbers.

When the game starts, the first player from each team picks up the top card from the stack, reads the pair of rational numbers and finds a rational number between them. Once the player finds a rational number, they run to a designated endpoint (e.g., a wall) and back to their team to hand over the card and write the answer in their notebooks. The next player in line repeats the process, finding a rational number between the given pair on the card and running the relay.

The game continues until all players on a team have completed the relay race. The first team to complete the relay race with all players finding rational numbers between the given pairs wins the game.

Extension Idea

Instruct: If we have 2 rational numbers, say a and b, then give any 1 rational number that lies between a and b.

Say: One rational number that lies between a and b, is a + b 2

Addition of Rational Numbers; Additive Inverse

Learning Outcomes

Imagine Maths Page 189

Students will be able to add rational numbers and find the additive inverse of a rational number.

Teaching Aids

Cards with sums of rational numbers written on them

Activity

Begin the class by reviewing the concept of addition of 2 rational numbers. Make cards with sums of rational numbers written on them.

Divide the class into 2 teams and set up the scoreboard to keep track of each team’s points. Place the cards face down on a table.

Each round consists of a member from each team competing against each other. Flip a card to reveal an addition problem.

Now, the first player from each team will solve the same problem in their notebooks. Instruct students to find the additive inverse of the sum and write it next to the sum. The player who correctly completes the task first earns a point for their team. Rotate players for each round.

The game can consist of multiple rounds, and the team with the most points at the end wins.

Extension Idea

Ask: Find the sum of the shaded portions.

The

Subtraction of Rational Numbers Imagine Maths Page 191

Learning Outcomes

Students will be able to subtract rational numbers.

Teaching Aids

Cards with subtraction problems of rational numbers written on them

Activity

Begin the class by reviewing the concept of subtraction of two rational numbers. Make cards with subtraction problems of rational numbers written on them.

Divide the class into 2 teams and set up the scoreboard to keep track of each team’s points. Place the cards face down on a table.

Each round consists of a member from each team competing against each other.

Flip a card to reveal the subtraction problem.

Now, the first player from each team will solve the same problem in their notebooks. The player who correctly completes the task first earns a point for their team. Rotate players for each round. The game can consist of multiple rounds, and the team with the most points at the end wins.

Extension Idea

Ask: To which number should

Say: Let the

Multiplication of Rational Numbers; Reciprocal or Imagine Maths Page 194

Multiplicative Inverse of a Non-zero Rational Number

Learning Outcomes

Students will be able to multiply rational numbers.

Teaching Aids

Bingo cards

Activity

Begin the class by reviewing the concept of multiplication of 2 rational numbers.

Prepare a list of 25 multiplication problems. Create bingo cards with a 5 × 5 grid. Each square on the grid should contain a rational number that is the solution of the multiplication problems in the list.

For instance:

List:

–1

2 × 3 4 ; 2 3 × –1 5 ; –2 3 × –1 4 etc.

Make sure the numbers on each bingo card are shuffled and uniquely arranged. Distribute the bingo cards and markers to each student.

Explain the rules of bingo to the students. The objective is to be the first to cover 5 spaces in a row, either horizontally, vertically or diagonally.

Call out the rational number multiplication problems one by one from the list, for instance, –5 7 × 14 –15 . Students will solve the multiplication problem and look for the answer on their bingo cards, i.e., 2 3. If they have the answer, they will cross the corresponding space with a marker.

The game continues until a student shouts “Bingo!” after successfully covering 5 spaces in a row. Verify the student’s answers to confirm the win.

Extension Idea

Ask: Look at the following: is the LHS equal to the RHS?

2 3 × –6 7 + 4 5 = –12 21 + 4 5

Say: Here, LHS = –4 105 and RHS = 24 105. So, the LHS is not equal to the RHS.

Division of Rational Numbers

Learning Outcomes

Students will be able to divide rational numbers by finding the reciprocal of the divisor.

Teaching Aids

Cards with rational numbers written on them; Dice Activity

Begin the class by reviewing the concept of division of 2 rational numbers. Divide the class into small groups of 3–4 students each.

Place a stack of cards with rational numbers written on them face down on the table.

Imagine Maths Page 195

Start with the first group. Instruct students to roll the dice. Say that the reciprocal of the number obtained is the divisor.

Now ask them to pick a card with a rational number written on it.

Instruct them to divide the rational number by the reciprocal of the number rolled. Allow the group to work together to solve the division problem and provide their answer. If correct, they earn points. Continue playing rounds with each group taking turns and determine a winner. At the end of the game, review the division problems and solutions, as a class, to reinforce learning.

Extension Idea

Ask: If the product of 2 3 with another rational number is 8

Say: Let the other rational number be

, what is the other rational number?

Answers

1. Representing Rational Numbers

Think and Tell

Yes, all fractions are rational numbers but not all rational numbers are fractions.

Do It Together

–0.5 = –3 6

2. Rational Numbers in Standard Form

Do It Together

The denominator cannot be negative; so,

The HCF of 12 and –18 is 6

Hence, the standard form of 12 –18 is –2 3 .

3. Equivalent Rational Numbers

Do It Together

1. 3 2 = 9 6 =

4. Comparing and Ordering Rational Numbers

Think and Tell

Fractions are numbers written in the p q form where p and q are whole numbers and q > 0. Rational numbers are also numbers written in p q form but here p and q are integers and q ≠ 0.

It Together

The LCM of 3, 9 and 27 is 27.

Ascending order: –13 9 < –2 9 < –3 27 < 11 3

5. Rational Numbers Between Two Rational Numbers

Do It Together

Let us multiply and divide 1 and 3 by 4 to expand the range. 1 × 4 4 = 4 4 3 × 4 4 = 12 4

We know that, 12 4 > 4 4. So, the rational numbers between 1 and 3 are: 5 4, 6 4, 7 4 , 8 4 , 9 4 , 10 4 and 11 4

6. Properties of Rational Numbers

Do It Together

We will divide the numerator and denominator by 7. 35 63 = 35 ÷ 7 63 ÷ 7 = 5 9

7. Addition of Rational Numbers

Do It Together

The LCM of the denominators 5, 4 and 12 is 60.

Rewriting the rational numbers:

2 5 = 2 × 12 5 × 12 = 24 60 3 4 = 3 × 15 4 × 15 = 45 60 5 12 = 5 × 5 12 × 5 = 25 60

So, 2 5 + 3 4 + 5 12 = 24 60 + 45 60 + 25 60 = 94 60

8. Additive Inverse

Do It Together

1. 5 14 = –5 14 2. 19 –21 = 19 21 3. –12 –17 = –12 17

9. Subtraction of Rational Numbers

Do It Together

Let the required number be x

Then, –3 8 + x = 6 7 ⇒ x = 6 7 –(–3) 8 = 6 7 + 3 8

LCM of 8 and 7 = 56. Hence, x = 48 56 + 21 56 ⇒ x = 69 56 ∴ x = 69 56

10. Multiplication of Rational Numbers

Do It Together

Thickness of each sheet = 5 7 inches

Thickness of two sheets = 2 × 5 7 inches

Therefore, thickness of the stack of two metal sheets will be 10 7 inches.

11. Reciprocal or Multiplicative Inverse of a Non-zero Rational Number

Do It Together

The given rational number is: p q = 2m × r n × 2r

So, the reciprocal of p q is: n × 2r 2m × r

12. Division of Rational Numbers

Do It Together ab = a × b –2

Construction of Triangles 11

Imagine Mathematics Headings

Constructing Triangles: SSS Criterion 203

Constructing Triangles: SAS Criterion 205

Constructing Triangles: ASA Criterion 208

Constructing Triangles: RHS Criterion 210

Learning Outcomes

Students will be able to: construct triangles using the SSS criterion. construct triangles using the SAS criterion. construct triangles using the ASA criterion. construct triangles using the RHS criterion.

Alignment to NCF

C-3.4: Draws and constructs geometric shapes, such as lines, parallel lines, perpendicular lines, angles, and simple triangles, with specified properties using a compass and straightedge

C-3.5: Understands congruence and similarity as it applies to geometric shapes and identifies similar and congruent triangles

Let’s Recall

Recap to check if students know how to construct angles and angle bisectors. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

SSS criterion: the Side−Side−Side criterion, a rule by which it is necessary to know the measures of three sides of a triangle to construct it

SAS criterion: the Side−Angle−Side criterion, a rule by which it is necessary to know the measures of two sides and one included angle of a triangle to construct it

ASA criterion: the Angle–Side–Angle criterion, a rule by which it is necessary to know the measures of two angles and one included side of a triangle to construct it

RHS criterion: the Right Angle–Hypotenuse–Side criterion, a rule by which it is necessary to know the measures of the hypotenuse and one other side of a triangle to construct it

Teaching Aids

2 sets of straws of 3 different lengths; Ruler; Compass; Sets of 3 paper strips forming the sides of a triangle; Protractor; Cardboard; Cardboard pins; Coloured string; Paper strips of 3 different lengths

Chapter: Construction of Triangles

Constructing Triangles: SSS Criterion

Learning Outcomes

Students will be able to construct triangles using the SSS criterion.

Teaching Aids

2 sets of straws of 3 different lengths; Ruler; Compass Activity

Instruct students to form pairs for the activity. Provide students with 2 sets of straws of 3 different lengths, for example, 3, 4, 5 and 7, 9, 1 cm. Ask them to form 2 triangles by connecting the straws. Now, ask them questions like: Were you able to make 2 triangles? What did you notice? Can you make another triangle using the given lengths? Guide them to understand the uniqueness of the triangle made using the SSS criterion. Encourage students to explore different combinations of straw lengths and ensure that the sum of any 2 sides is greater than the third side.

Now, instruct them to use a compass and construct a triangle using these lengths in their notebooks. Ask them to compare their triangles with their partner to see that both triangles are exactly the same.

Extension Idea

Ask: How will you construct an equilateral triangle with one side measuring 5 cm?

Say: First, draw a 5-cm base. Then, extend lines, also measuring 5 cm, from each end to complete an equilateral triangle. All the sides will be equal.

Constructing

Triangles: SAS Criterion

Learning Outcomes

Students will be able to construct triangles using the SAS criterion.

Teaching Aids

Sets of 3 paper strips forming the sides of a triangle; Protractor Activity

Imagine Maths Page 205

Instruct students to form groups for this activity. Give them paper strips, 2 of which measure 5 cm and 6 cm. Instruct them to arrange the 2 strips at a 60° angle using a protractor. They should cut the third strip and place it to form the third side of the triangle. Ask them whether the angle in their triangle is included between the sides, and if not, how they can include it.

Help them understand that if the included angle is not 60°, the students will not obtain the same triangle. Then, ask them to use the compass and ruler to construct the triangle with specified measures using the SAS criterion.

Constructing Triangles: ASA Criterion

Learning Outcomes

Students will be able to construct triangles using the ASA criterion.

Teaching Aids

Cardboard; Cardboard pins; Coloured string; Protractor

Activity

Instruct students to form groups for the activity. Give a large piece of cardboard, pins and coloured strings to each group.

Give each group a set of 2 angles and 1 included side and instruct them to create triangles by pushing the cardboard pins into the cardboard and wrapping string around these to form the side and angles. Guide them to use their protractors when necessary.

Ask questions like: ‘Would you be able to create a unique triangle if the given side is placed between any 2 of the 3 angles?’ This artistic approach reinforces the idea of creating the side and then the angles to create triangles. Then, encourage them to construct the triangle in their notebooks using the ASA criterion and a compass.

Extension Idea

Instruct: Try creating a word problem on constructing a triangle using the ASA criterion and involving a real-life situation.

Say: One such word problem can be, ‘Alex wants to build a triangular treehouse. Angle A is 40°, angle B is 60° and the base plank is 8 feet in length. Can he construct the triangle? Which congruence criteria will be used?’

Constructing Triangles: RHS Criterion

Learning Outcomes

Students will be able to construct triangles using the RHS criterion.

Teaching Aids

Paper strips of 3 different lengths; Ruler; Compass

Activity

Maths Page 210

Instruct students to form groups. Distribute 3 paper strips measuring 8 cm, 6 cm and 10 cm in length to each group.

Instruct them to try arranging the strips as the sides of a triangle. Ask them which type of triangle it is and where the longest strip is positioned. Repeat the activity with more Pythagorean triplets.

Now, provide each group with a ruler and compass and ask them to individually construct a triangle of the given measures in their notebooks. Encourage them to compare their triangles within their groups to verify that the triangles constructed are exactly the same.

Extension Idea

Ask: In a right-angled triangle, can any angle be greater than a right angle? Why?

Say: No, in a right-angled triangle, no angle can be greater than the right angle. This is because the sum of the angles in any triangle is always 180°. In a right-angled triangle, 1 angle is fixed at 90°, leaving the other 2 angles as acute angles, which are always less than 90°.

Answers

1. Constructing Triangles: SSS Criterion

Think and Tell

Yes, since the sum of any 2 sides is greater than the third side, the triangle can have sides of length 5 cm, 6 cm and 10 cm.

Do It Together

ABC = Scalene triangle

2. Constructing Triangles: SAS Criterion Do It Together

3. Constructing Triangles: ASA Criterion

4. Constructing Triangles: RHS Criterion

Do It Together

Perimeter and Area 12

Perimeter and Area of Squares and Rectangles

Area of Parallelograms

Area of Triangles

Circumference of Circles

Learning Outcomes

Area of Circles, Semicircles and Quadrants

Problems on Perimeter and Area

Students will be able to: find the perimeter and area of squares and rectangles using the formula. find the area of parallelograms. find the area of triangles. find the circumference of circles. find the area of circles, semicircles and quadrants. find the area of combined shapes made with squares, rectangles, parallelograms, triangles, circles, semicircles and quadrants. solve word problems on finding the area and perimeter.

Alignment to NCF

C-4.1: Discovers, understands, and uses formulae to determine the area of a square, triangle, parallelogram, and trapezium and develops strategies to find the areas of composite 2D shapes

Let’s Recall

Recap to check if students know how to find the perimeter and area of squares and rectangles. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

· perimeter: the total length of the boundary of a figure

· area: the amount of space occupied by an object/shape

· circumference: the perimeter of a circle

· quadrant: one-fourth of a circle

Teaching Aids

A4 sheets of paper; Square and rectangle cut-outs of different areas with names of places written (like swimming pool, park, apartments and office complex); Ruler; Glue stick; Squared paper; Pencils; Scissors; Rectangular cut-outs; Pieces of string; Compass; Cut-outs of circles, semicircles and quadrants; Cut-outs of different 2-D shapes; Word problem cards

Chapter: Perimeter and Area

Perimeter and Area of Squares and Rectangles

Learning Outcomes

Students will be able to find the perimeter and area of squares and rectangles using the formula.

Teaching Aids

A4 sheets of paper; Square and rectangle cut-outs of different areas with names of places written (like swimming pool, park, apartments and office complex); Ruler; Glue stick

Activity

Divide the class into groups of 4.

Tell students that they will calculate the area of a residential complex.

Distribute an A4 sheet of paper and the square and rectangle cutouts of different areas with names of places written (like swimming pool, park, apartments and office complex) among the groups.

Instruct students to place and stick the square and rectangular cut-outs on the A4 sheets of paper to show a layout of a residential complex. Make sure that the sizes of the sheets and the cut-outs given to the groups are the same.

Instruct students to measure the lengths of the sides of the shapes and find the area occupied by the park, the swimming pool, the office complex and the apartments, keeping in mind that 1 cm on paper will be equal to 5 m on the actual ground. They will also find the area of the remaining land by subtracting the area of the buildings from the area of the entire residential complex, which is the A4 sheet of paper.

Discuss the answers in class.

Extension Idea

Ask: A rectangular field has a perimeter of 40 m, where the width is one-fourth the length. If on expanding the field, the perimeter remains constant and its width is the same as its length, then how will the area before and after expansion differ from each other?

Say: Perimeter = 2 (l + b) = 40 and b = l 4 . So, 2 (l + l 4 ) = 40 5l 2 = 40 l = 40 × 2 5 = 16 m. Therefore, b = 16 4 = 4 m.

Also, area before expansion = 16 × 4 = 64 sq. m. After expansion, Perimeter = 2 (l + b) = 40 and b = l. So, 2 (l + l) = 40 4l = 40 l = 40 4 = 10 m. Therefore, b = 10 m. Also, area after expansion = l × b = 10 × 10 = 100 sq. m. Hence, if the breadth is the same as its length on expanding the field, the area increases by 100 – 64 = 36 sq. m.

Area of Parallelograms

Learning Outcomes

Students will be able to find the area of parallelograms.

Imagine Maths Page 219

Swimming Pool Park

Apartments

Office Complex

Teaching Aids

Squared paper; Ruler; Pencils; A4 sheets of paper; Scissors

Activity

Begin by reviewing the definition of a parallelogram, emphasising its various properties.

Instruct students to form groups of equal size.

Distribute the resources to each group. Introduce the concept of area and instruct students to draw a parallelogram of any length and breadth on squared paper. Ask them to draw a perpendicular from any of the vertices of the parallelogram to its opposite side, then cut off the triangle formed by the perpendicular and paste it on the other side of the parallelogram. Ask them to identify the shape so formed after cutting and pasting and to find its area.

Ask questions like: What is the new shape that is formed? What is its area? How is it related to the area of the parallelogram?

Discuss their observations and derive the formula for the area of a parallelogram.

Instruct them to find the area of the new shape and the parallelogram if each square on the squared paper is 1 cm long.

Give them some questions on finding the missing length of a parallelogram when the area is given.

Extension Idea

Ask: Your room has a rectangular floor with dimensions of 4 metres by 3 metres. You want to lay down a large rug in the shape of a parallelogram with a base of 2 metres. What is the maximum height the rug can have, if its area is not equal to or greater than half of the area of the room?

Say: Area of the rectangle = 4 × 3 = 12 sq. m. Area of a parallelogram = b × h or 2 × h. The area of the rug is less than half of the area of the rectangular floor. So, area of the rug < 6 or 2h < 6. That is, h < 3. So, the height of the rug can be a maximum of 3 m.

Area

Triangles

Learning Outcomes

Students will be able to find the area of triangles.

Teaching Aids

Rectangular cut-outs; A4 sheets of paper; Scissors

Activity

Begin by discussing the area of a rectangle.

Divide the class into groups.

Distribute the rectangle cut-outs (L = 10 cm, B = 6 cm) to the students.

Maths Page 221

Instruct them to cut the rectangle in a way to form triangles. Ask them to use the area of the rectangle to find the area of the triangles, then deduce the formula that they can use to find the area of a triangle with given lengths.

Ask questions like: How did you find the formula for the area of a triangle?

Discuss that when a rectangle is cut into 2 identical triangles, they can be superimposed over each other and that their area is the same.

Further, give a target area of, say, 36 sq. cm to each group and distribute squared paper and a scissors. Instruct each student to create and cut triangles of different sizes and types with the same area. Ask them to tessellate the triangular pieces and form a shape. Have them find the area of the tessellated shape.

Ask questions like: How many triangles could you form? What is the relationship between the tessellated shape and its area?

Give them some questions on finding the missing length of a triangle when its area is given.

Circumference

Learning Outcomes

Circles

Students will be able to find the circumference of circles.

Teaching Aids

Pieces of string; Ruler; A4 sheets of paper; Compass

Activity

Distribute a piece of string to each student. Ask them to wrap the string around a pen, cut its ends and measure the length of the string using the ruler. Discuss the circumference of the pen. Tell them that circumference is nothing but the perimeter of a circle.

Instruct students to form groups.

Distribute a sheet of paper, a ruler and a piece of string to each group. Draw a table on the board.

Instruct students to first copy the table into their notebooks, then draw circles of different radii using a compass on the sheet of paper. They will then cut out the circles and measure the circumference of these circles using the piece of string and a ruler. Let them write the length of the diameter and circumference in the table in their notebooks. Ask them to find the ratio of the circumference to the diameter to derive the approximate values of pi. Using this, instruct them to derive the formula for the circumference of the circle.

Ask questions like: What is the relationship between the circumference and the diameter of the circle?

Give them some questions on finding the radius when the circumference is given.

Extension Idea

Ask: You are designing a circular flower bed with a radius of 2 metres. You want to plant flowers around the edge every 30 cm. How many flowers will you need?

Say: To determine the number of flowers needed for the circular flower bed, we’ll first find the circumference of the circle and divide it by the given length of the spacing. So, number of flowers

or 42, roughly (because of the overlapping of the first and last flowers).

Radii

Area of Circles, Semicircles and Quadrants

Learning Outcomes

Students will be able to find the area of circles, semicircles and quadrants.

Teaching Aids

Cut-outs of circles, semicircles and quadrants; Scissors

Activity

Discuss the area of a rectangle and the circumference of a circle. Divide the class into groups.

Distribute cut-outs of circles with a radius of 7 cm.

Instruct students to fold the circle to find the diameter, and then draw the diameter of the circle. Then, ask them to draw as many diameters as they can in the circle by folding the circle. Now, they should cut the circle into smaller sectors, rearrange them in the form of a rectangle by placing the sectors alternately so that they get a perfect looking rectangle. Then ask them to use the formula for the area of a rectangle to find the area where the length of the rectangle is the same as the circumference of the circle and the breadth of the rectangle is the radius of the circle. Now, ask them to deduce the formula for the area of a circle.

Ask questions like: What is the formula for the area of a circle? How did you find the formula?

Distribute cut-outs of a circle, a semicircle and a quadrant to the students.

Instruct them to use their understanding of the area of a circle to find the area of the given shapes by measuring the radii of the shapes. Discuss the answers with the class.

Combined Shapes

Learning Outcomes

Students will be able to find the area of combined shapes made with squares, rectangles, parallelograms, triangles, circles, semicircles and quadrants.

Teaching Aids

Cut-outs of different 2-D shapes

Activity

Instruct students to form groups.

Distribute a few cut-outs of different shapes to each group.

Instruct students to join the different shapes to form a combined shape.

Ask each student in the group to measure the length of the sides of each shape, find its area and then add the areas together to find the total area. Ask students to write the answers in their notebook. Ask them to interchange the shapes with other groups who will then rearrange them to make a new combined shape and find the total area of the combined shape formed, in their notebooks.

Ask questions like: Is the area of the combined shape the same for all the groups? How?

Extension Idea

Ask: A circular pizza with a diameter of 24 cm is cut into 16 equal slices. You rearrange the slices to form a larger rectangle. What is the length of the rectangle? (Use π ≈ 3.14)

Say: When a circular pizza is cut into 16 equal slices and rearranged to form a rectangle, each slice becomes a triangular piece. The rectangle formed will have a length equal to half the circumference of the original circle (pizza). So, the length of the rectangle = (2πr) 2 = 1 2 × 3.14 × 24 cm = 37.68 = 38 cm (roughly).

Word Problems on Perimeter and Area

Learning Outcomes

Students will be able to solve word problems on finding the area and perimeter.

Teaching Aids

Word problem cards

Activity

Begin the class with real-life examples where finding the area of a shape is needed. Instruct students to work in groups. Distribute word problem cards to each group.

A goat is tied to one corner of a square field of side 14 m. If the length of the rope is 8 m, what is the area of the field that the goat can move around in?

A circular swimming pool is to be built in the centre of a rectangular piece of land with a circular pavement of width 1 m all around the pool. The length and width of the rectangular land are 50 m and 30 m, respectively. What is the area of the pavement if it touches the edges along the length of the land?

Instruct students to first draw a picture to show the problems in their notebooks, and then solve the problems. They will then discuss the answers within their groups. Finally, discuss the solutions in class and the challenges faced while solving the problems.

1. Perimeter and Area of Squares and Rectangles

Do It Together

Size of 1 tile to be cut = 12.5 cm × 10 cm

Area of 1 tile to be cut = 125 sq. cm

Size of tile to be used for cutting = 125 cm × 65 cm

Area of tile to be used for cutting = 8125 sq. cm

Number of tiles = 8125 ÷ 125 = 65

So, 65 tiles can be cut from the tile of size 125 cm × 65 cm.

2. Area of Parallelograms

Think and Tell

Yes. By counting the number of squares, the area of the parallelogram = area of full squares + area of half squares =

Number of full squares + 1 2 × number of half squares =

24 + 1 2 × 8 = 24 + 4 = 28 sq. units

Do It Together

1. Find the area of the parallelogram.

Area of the parallelogram = b × h = 18 × 9

= 162 sq. cm

2. Find GI, if EF = 12 cm.

Area of the parallelogram = EF × GI

GI = 162 12

GI = 13.5 cm

3. Area of Triangles

Think and Tell

Such triangles are called congruent triangles.

Do It Together

So, the area of the square = 12 × 12 = 144 sq. m

Therefore, the area of the triangle = 1 2 × 144 = 72 sq. m

4. Circumference of Circles

Do It Together

So, the perimeter of the square = 4s = 4 × 66 cm = 264 cm

Circumference of the circle = 264 cm

264 = 2 × 3.14 × r

r = 264 2 × 3.14 = 264 6.28 cm

r = 42.04 cm

Cost of 1 cm of wire = ₹4

Cost of 264 cm of wire = ₹4 × 264 = ₹1056

5. Area of Circles, Semicircles and Quadrants

Do It Together

So, the perimeter of the square = 4 × 15 cm = 60 cm

Perimeter of the circle = 60 cm

2 × 22 7 × r = 60 cm

r = 60 × 7 2 × 22 = 9.55 cm

Area of the circle = 22 7 × 9.55 × 9.55 = 286.64 sq. cm

6. Combined Shapes

Think and Tell

Circles with the same centre and different radii are called concentric circles. They are found at the cross-section of pipes, wires and rings, etc.

Do It Together

Area of the remaining sheet of paper =

= π × (24 × 24) – (2 × π × (1.6)2 + 1 4 × π × (1.6)2 + 6 × 6)

= 1810.29 – (16.09 + 2.01 + 36)

= 1810.29 – 54.0864

= 1756.19 sq. cm

7. Word Problems on Perimeter and Area

Think and Tell

The plot is in the shape of a rectangle. So, its area without the pathways, crossroads and fountain can be calculated using the formula for the area of a rectangle.

Do It Together

Area of road 1 = 150 × 2 = 300 sq. m

Area of road 2 = 150 × 2 = 300 sq. m

Area of the common square = 2 × 2 = 4 sq. m

Area of the roads = Area of road 1 + Area of road 2 – Area of the common square

= 300 + 300 – 4 = 596 sq. m

Cost of constructing 1 sq. m of road = ₹520

Cost of constructing 596 sq. m = ₹520 × 596 = ₹3,09,920

Algebraic Expressions 13

Learning Outcomes

Students will be able to: generate rules in patterns and use it to extend the pattern or create a formula using variables. identify parts of an algebraic expression and types of expressions. add algebraic expressions. subtract algebraic expressions. simplify algebraic expressions. find the value of an expression when the values of the variables are given.

Alignment to NCF

C-2.2: Extends the representation of a number in the form of a variable or an algebraic expression using a variable

C-2.3: Forms algebraic expressions using variables, coefficients, and constants and manipulates them through basic operations

Let’s Recall

Recap to check if students know how to identify and extend a number pattern. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

algebraic expression: expression consisting of numbers, constants, variables and operation symbols

Teaching Aids

Cut-outs of regular polygons like: equilateral triangle, square, regular pentagon and hexagon, octagon, etc; Expression cards with 1 algebraic expression on each card; Chart paper; Rectangle and circle cut-outs; Piece of string; Glue stick; Algebra tiles as blue, green, red and yellow paper cut-outs; Set of puzzle interlocking cards with an algebraic expression having brackets on one card, the simplified form of the same expression on the second card and its value on the third card; Expression mazes; Crayons

Chapter: Algebraic Expressions

Generating Rules in Formulas and Patterns

Learning Outcomes

Imagine Maths Page 239

Students will be able to generate rules in patterns and use them to extend the pattern or create a formula using variables.

Teaching Aids

Cut-outs of regular polygons like: equilateral triangle, square, regular pentagon and hexagon, octagon, etc.

Activity

Begin the class by revisiting the terms ‘variable’ and ‘constants’. Explain what it means when we say that the area of a rectangle is 2L + 2B and that these are called formulas.

Instruct students to work in groups.

Distribute cut-outs of regular polygons among the groups.

Instruct students to find the perimeter of the shapes by taking the length of the sides as a variable. This will help them deduce the perimeter formulas. Once they deduce the formulas, ask them to find the perimeter of the shapes by measuring the length of one of the sides using the ruler and then applying the formula to find the perimeter. They will then measure the length of all the sides to see if they get the same perimeter. Let them solve in their notebooks. Ask them to discuss in their groups to see if they got the same perimeter. Encourage them to deduce the formula for the area of a square, rectangle and triangle as well.

Discuss how to find the area of a regular polygon with ‘n’ sides which has a side length of 4 units (or any given length).

Extension Idea

Ask: What is the perimeter of a regular shape with each side measuring n units and having m sides?

Say: The perimeter of a regular polygon with each side measuring n units and having m sides can be calculated using the formula: Perimeter = n × m.

Algebraic Expressions and Terms

Learning Outcomes

Imagine Maths Page 242

Students will be able to identify parts of an algebraic expression and types of expressions.

Teaching Aids

Expression cards with 1 algebraic expression on each card; Chart paper; Rectangle and circle cut-outs; Piece of string; Glue stick

Activity

Begin with reviewing the parts of an expression. Discuss like and unlike terms and types of algebraic expressions such as monomials, binomials, etc. Create expression cards (e.g., 10pq + 4p2 – q3) with 1 expression on each card. (Make sure to cover all types of expressions in the cards).

Instruct students to work in groups of 3. Distribute 3 expression cards, and other teaching aids to each group. Instruct the groups to paste the expression card on chart paper on the left. Ask the groups to work together to form factor trees for all the expressions given using rectangular cut-outs to show the terms; and circle cut-outs to show the factors. They can use the glue stick to paste the card and the cut-outs, and connect the expression, terms and factors using the strings. Ask them to also label the expression (monomial, binomial, etc.), terms and the factors by writing on the side.

(10pq + 4p2 – q3) Algebraic Expression

Once the students are done forming the factor tree, ask them to combine the terms in all the expression cards they have and note down the like and unlike terms on the chart paper. Encourage students to present the factor trees formed in front of the class.

Ask them to draw the factor trees for the remaining 2 expressions in their notebooks.

Addition of Algebraic Expressions

Learning Outcomes

Students will be able to add algebraic expressions.

Teaching Aids

Algebra tiles as blue, green, red and yellow paper cut-outs

Activity

Introduce algebra tiles to students. Discuss the values of the various cut-outs and that these will be called algebra tiles.

Bring out the fact that a pair of 1 negative and 1 positive tile represents 0.

Instruct students to work in pairs. Distribute algebra tiles among the pairs.

Instruct students to add 3x2 – 2x + 1 and x2 + 4x – 2 using the algebra tiles. Ask them to place algebra tiles that represent 3x2 – 2x + 1. Ask them to now place algebra tiles that represent x2 + 4x – 2 below the previous set. Instruct them to remove the pairs of algebra tiles that make zero. Discuss that the remaining number of algebra tiles 4x2 + 2x – 1 would be the answer.

Instruct students to write the expressions one below the other and the answer below them. They will then try and understand the rule for adding expressions using the column method. Discuss the rules of addition with students. Then, let students add the expressions (5x2 – 7x + 5) and (–7x2 + 3x – 8) using the column method in their notebooks. Discuss the answers.

Give them more practice questions on adding algebraic expressions.

Extension Idea

Instruct: Add

Say: On adding the 2 expressions, we get

Subtraction of Algebraic Expressions

Learning Outcomes

Students will be able to subtract algebraic expressions.

Teaching Aids

Algebra tiles as blue, green, red and yellow paper cut-outs

Activity

Instruct students to work in pairs. Distribute algebra tiles among the pairs. Instruct students to subtract (2x2 – 2x + 1) from (x2 – 3x + 3). Ask them to use tiles to show (x2 – 3x + 3). Ask them questions like: Can you take away 2x2 positive tiles from the x2 positive tiles? Explain that since there is only 1 positive x2 tile, we cannot take away 2 positive x2 tiles. Tell students that to add 1 more x2 positive tile, 1 set of zero tiles needs to be added as 1 x2 positive tile and 1 –x2 neutral tile. Let students figure out how to subtract and find the answer. (Help them if needed.)

Instruct students to now subtract (4x2 – 5x + 3) from (5x2 + 2x – 4) using the algebra tiles. Discuss the horizontal and column methods of subtraction with students. Instruct students to subtract both the expressions discussed using the horizontal method and column method in their notebooks. Give them practice by running a quiz around subtracting algebraic expressions.

Extension Idea

Ask: What should be added to (5x2y – 3xy + 5x2 – 9) to get (3y2x + 8yx – x2y – 2x2)?

Say: Let A be added to (5x2y – 3xy + 5x2 – 9) to get (3y2

(3y2x + 8

Simplifying Algebraic Expressions

Learning Outcomes

Students will be able to simplify algebraic expressions.

Teaching Aids

Imagine Maths Page 252

Set of puzzle interlocking cards with an algebraic expression having brackets on one card, the simplified form of the same expression on the second card and its value on the third card

Activity

Begin by showing students an algebraic expression and then simplifying it to find the value.

Instruct students to work in groups.

Distribute sets of puzzle interlocking cards to each group.

Instruct the groups to simplify and solve the algebraic expression in their notebooks and interlock them with its solution. Record the time taken by each group to interlock all the puzzles. The group which takes the least time to interlock all the puzzles wins. Check the puzzles created and announce the winning group.

Finding the Value of an Expression

Learning Outcomes

Imagine Maths Page 253

Students will be able to find the value of an expression when the values of the variables are given.

Teaching Aids

Expression mazes; Crayons

Activity

Create or find expression mazes that include algebraic expressions.

Begin the activity with a brief review to find the value of an expression when the value of the variable is given. Discuss 1–2 examples.

Instruct students to form pairs. Distribute an expression maze and crayons to each pair.

Instruct students to start at the beginning of the maze and solve each expression by putting the values of the variables. Ask them to shade the expression and its value. The correct value will guide them to the next part of the maze.

Once students reach the exit, have them review their answers.

Extension Idea

Ask: What should be the value of c if (3a2 – 5ab + c – 2b2) equals 4 for a = 1 and b = −1?

Say: Substituting the values of a and b we get, 3(1)2 – 5 × 1 × (–1) + c – 2(–1)2 =

= –2.

Answers

1. Finding Formulas

Think and Tell

For the perimeter of a square to be equal to the area

4a = a2

4a a = a2 a

4 = a

[Dividing both sides by a]

Thus, the side of the square is 4 units.

The unit should not be cm or m.

Do It Together

1. Length of the side of the given triangle a = 15.7 cm

Perimeter of triangle = 3a = 3 × 15.7 cm = 47.1 cm

2. Length of the side of the given pentagon a = 7.5 cm

Perimeter of pentagon = 5a = 5 × 7.5 cm = 37.5 cm

3. Length of the side of the given hexagon a = 6.5 cm

Perimeter of hexagon = 6a = 6 × 6.5 = 39 cm

On comparing, the perimeter of the triangle is the greatest.

2. Number Patterns

Do It Together

The sum of the first 15 natural numbers = 15(15 + 1) 2 = 120

The sum of the first 10 natural numbers = 10(10 + 1) 2 = 55

Difference = 120 – 55 = 65

3. Parts of an Expression

Think and Tell

The coefficient of xy in the term −xy is −1.

Do It Together

Terms

Factors

The factors of the term −5a2b are −5, a,

4. Like and Unlike Terms

Do It Together

, and b

5. Types of Algebraic Expressions

Do It Together

6. Addition of Algebraic Expressions

Do It Together

4pqr + 9pr2 − 3pr and 10pr2 − 12pqr

4pqr + 9pr2 − 3pr + (− 12pqr) + 10pr2

= (4 + (− 12pqr))pqr + (9 + 10)pr2 − 3pr

= − 8pqr + 19pr2 − 3pr

7. Subtraction of Algebraic Expressions

Do It Together

Horizontal Method

(24x2z 13xz + 3yz) − (14x2z + 12xz) – 13yz)

= 24x2z 13xz + 3yz − 14x2z 12xz + 13yz

= 24x2z 14x2z – 13xz − 12xz + 3yz + 13yz

= (24 14)x2z + (–13 − 12)xz + (3 + 13)yz

= (10x2z 25xz + 16yz)

Column Method

24x2z 13xz + 3yz

14x2z + 12xz 13yz − − + (24 14)x2z + (–13 − 12)xz + (3 + 13)yz = (10x2z 25xz + 16yz)

8. Finding the Value of an Expression

Do It Together

Subtracting 5xy + 7x2 from 9x2 − 3xy + 5, we get

(9x2 − 3xy + 5) − (5xy + 7x2)

= 9x2 − 3xy + 5 − 5xy − 7x2

= (9 − 7)x2 + (−3 − 5)xy + 5 = 2x2 − 8xy + 5

Putting x = 1 and y = −1, we get

2(1)2 – 8 × 1 × (−1) + 5

= 2 + 8 + 5 = 15

Exponents and Powers

Learning Outcomes

Students will be able to: express a number in its exponential form and as a product of its prime factors. simplify an expression with rational numbers in exponent form. use the correct law of exponents to simplify exponential expressions. express large numbers in standard form.

Alignment to NCF

C-1.1: Develops a sense for and an ability to manipulate (e.g., read, write, form, compare, estimate, and apply operations) and name (in words) large whole numbers of up to 20 digits, and express them in scientific notation using exponents and powers

Let’s Recall

Recap to check if students know how to encounter and use large numbers in daily conversations. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

exponent: how many times a number is multiplied by itself prime factors: the smallest prime numbers that multiply to create a given number standard form: number in standard form will be K × 10n, where K is a number between 1 and 10 and n is an integer

Teaching Aids

Question cards to represent numbers in the exponential form; Crossword puzzle cards on powers of rational numbers; Set of puzzle pieces (question, answer and the law of exponents in each set); Puzzle cards with a large number in standard form written in the centre and options on 4 sides

Chapter: Exponents and Powers

Exponents and Prime Factors;

Imagine

Maths Page 260 & 262 Comparing Numbers in Exponential Form

Learning Outcomes

Students will be able to express a number in its exponential form and as a product of its prime factors.

Teaching Aids

Question cards to represent numbers in the exponential form

Activity

Begin with a brief discussion on prime factors of a number. Make question cards to represent numbers in the exponential form.

Instruct students to work in groups. Distribute question cards with some numbers such as 384, 70, 120, etc. among the groups.

Ask them to write the prime factors of the first number in the prime factorisation column. Instruct the groups to count and write the prime numbers that are multiplied and the number of times each prime number is multiplied, in the respective columns. Tell them that in exponential form, 384 is written as 27 × 31. Ask students what rule they noticed in the exponential representation. Bring out the fact that the exponent indicates how many times the base (the prime factor) is multiplied by itself.

Ask them to solve the remaining questions using the same approach. Run a quiz around the class to give more practice on the concept with a variety of questions.

Extension Idea

Ask: Which is greater, 23 × 5 or 22 × 52?

Say: 23 × 5 = 40 and 22 × 52 = 100. 100 > 40, therefore, 22 × 52 is greater.

Powers of Rational Numbers;

Reciprocal of a Rational Number

Learning Outcomes

Imagine Maths Page 262 & 264

Students will be able to simplify an expression with rational numbers in exponent form.

Teaching Aids

Crossword puzzle cards on powers of rational numbers

Activity

Begin by discussing and demonstrating on the board how rational numbers can be written in exponential form. Prepare crossword puzzle cards on powers of rational numbers. Instruct students to form groups for the activity. Distribute a crossword puzzle card to each group. Explain that each crossword puzzle contains expressions involving rational numbers in exponent form. Instruct students to solve the crossword puzzle by simplifying the expressions and writing down the results in exponential form and vice versa. After the activity, ask each group to share 1 expression they simplified and how they arrived at the solution. Encourage them to write the solutions in their notebooks.

Extension Idea

Ask: What will be the reciprocal of (–5)2?

Say: The reciprocal of (–5)2 is

–1

Laws of Exponents

Learning

Outcomes

Students will be able to use the correct law of exponents to simplify exponential expressions.

Teaching Aids

Set of puzzle pieces (question, answer and the law of exponents in each set)

Activity

Maths Page 265

Begin by explaining the laws of exponents to the class. Divide the class into 3 groups. Distribute the puzzle cards (question card, answer card and laws of exponents card) to each group.

Ask students to solve the question and match it with the answer. Ask them to match the question and answer with the correct law of exponent. The team to match the set first will get a point. Discuss their approaches in class and encourage them to write the solutions in their notebooks.

Extension Idea

Ask: What is the error in the RHS of this equation: (56 × 52) = 258?

Say: The base in the RHS shouldn’t be 25. Instead, the answer should be 58.

Expressing Large Numbers in the Standard Form

Learning

Outcomes

Students will be able to express large numbers in standard form.

Teaching Aids

Puzzle cards with a large number in standard form written in the centre and options on 4 sides

Begin by introducing how large numbers are expressed in standard form. Instruct students to form groups for the activity. Distribute the puzzle cards with a large number in standard form written in the centre and options on 4 sides to each group. Tell them that each card consists of 4 numbers to be expressed in the standard form. Students have to find the standard form of each number and match the answers with the one written on the centre card. Instruct them to encircle the number that matches with the answer on the centre card. After the activity, discuss their answers in class. If time permits, rotate the puzzles among groups and repeat the activity.

Answers

1. Exponential Form

Think and Tell

Exponent = n = 1

2. Exponents and Prime Factors

Do It Together

Prime factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Prime factors in exponential form = 28 × 3

3. Comparing Numbers in Exponential Form

Do It Together

1. 83 < 74

83 = 512

74 = 2401 2. 105 < 97 105 = 100,000 97 = 4,782,694

3. 184 > 48 184 = 1,04,976 48 = 65,536 4. 213 < 321 213 = 9261 321 = 10,460,353,203

4. Powers of Rational Numbers

Do It Together

5. Reciprocal of a Rational Number Do It Together

6. Multiplying Powers Do

7. Dividing Powers Do

8. Power of a Power

9. Simplifying Exponential Expressions

Do It Together

Using

10. Expressing Large Numbers in the Standard Form

Do It Together

Symmetry 15

Learning Outcomes

Students will be able to: identify and draw the reflection of a figure/shape on squared paper. find the angle of rotational symmetry in a shape or figure.

Alignment to NCF

C-2.3: Recognises and creates symmetry (reflection, rotation) in familiar 2D and 3D shapes

Let’s Recall

Recap to check if students know how to determine the lines of symmetry of different figures. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

reflection symmetry: the symmetry that a shape has if it looks the same after being flipped or reflected over a line

rotational symmetry: the symmetry that a shape has if it looks the same after being rotated by a certain angle around a central point order of rotational symmetry: the number of times a shape looks the same after being rotated 360° around its centre angle of rotation: the measure of the smallest angle through which a shape can be rotated to coincide with itself

Teaching Aids

Paper cut-outs of different shapes; Pencils; Squared paper; Small mirrors; Circular spinners made from cardboard or paper; Cut-outs of different geometric shapes; 360° protractor; Sheets of paper; Scissors

Chapter: Symmetry

Reflection Symmetry

Learning Outcomes

Imagine Maths Page 275

Students will be able to identify and draw the reflection of a figure/shape on squared paper.

Teaching Aids

Paper cut-outs of different shapes; Pencils; Squared paper; Small mirrors

Activity

Form groups for the activity. Distribute cut-outs of various shapes to each group. Instruct students to experiment with these by drawing lines of symmetry on them. Encourage them to draw lines vertically, horizontally or diagonally, aiming for identical sides on either side of the line. Ask questions around what these lines represent. Discuss their responses to bring out the meaning of reflectional symmetry.

Now, distribute the mirrors and squared paper and ask each student to draw a line through the centre and a figure on one side of the line. Next, they will pass the sheet to the student next to them, who will then complete the figures by drawing the reflected parts. Finally, they should use the mirrors to verify the symmetry of the reflections drawn on the paper.

Conclude with a brief class discussion to share observations and reinforce the concept of reflection symmetry.

Extension Idea

Ask: Can you think of 1 word that looks the same as its reflection?

Say: There are many words that look the same as their reflection. One such word is WOW.

Rotational Symmetry

Learning Outcomes

Students will be able to find the angle of rotational symmetry in a shape or figure.

Teaching Aids

Imagine Maths Page 280

Circular spinners made from cardboard or paper; Cut-outs of different geometric shapes; 360° protractor; Sheets of paper; Scissors

Activity

Begin by asking a question like: Have you ever noticed how certain snacks, like triangular chips, look identical from all corners? Discuss your answers. Form groups for the activity. Distribute spinners and various shape cut-outs. Instruct students to place the shapes on the spinner. Keep 1 shape still and rotate the other shape.

Ask students to check how the shape aligns after certain rotations. Guide students to measure the angle between the original and rotated positions for each shape using protractors. Say that this is the angle of rotation for their shape.

Now, ask them to count how many times a shape coincides with itself in 1 rotation, defining it as the order of rotational symmetry. Repeat with various shapes, prompting students to write the results in their notebooks.

Teacher tip: If the 360° protractor is not available, make one using a paper plate and markers.

Extension Idea

Ask: Create a mandala design with a rotational symmetry of order 4.

Say: You can make many designs with a rotational symmetry of order 4.

Answers

1. One Line of Symmetry

Think and Tell

The Ferris wheel has one line of symmetry.

3. Rotational Symmetry

Think and Tell

The order of rotational symmetry of the big (Ferris) wheel is 2, and the angle of rotational symmetry is 180°.

Do It Together

Rotation through 120° Rotation through 240° Rotation through 360°

Do It Together

1. a. Yes

2. a.

Yes

3. Figure (a) is symmetrical about the vertical line of symmetry, but Figure (b) is symmetrical about the horizontal line of symmetry

More Than One Line of Symmetry

Do It Together

1. a. Yes b. 4

2. a. Yes b. 3

This figure fits onto itself times in a complete revolution. So, the order of rotational symmetry is . Now, the angle of rotational symmetry = 360° order of rotational symmetry = 360° = .

4. Both Reflection and Rotational Symmetry

Do It Together

Number of lines of symmetry = 8

Order of rotational symmetry = number of lines of symmetry = 8.

Visualising Solid Shapes

Learning Outcomes

Students will be able to: identify and list the features of 3-D shapes. identify and draw nets of 3-D shapes. identify and draw oblique and isometric sketches of 3-D shapes. visualise 3-D shapes and draw their cross-section and shadows. draw different views of structures made of cubes on a square grid.

Alignment to NCF

C-3.1: Describes, classifies, and understands relationships among different types of two- and three-dimensional shapes using their defining properties/attributes

C-3.3: Identifies attributes of three-dimensional shapes (cubes, parallelepipeds, cylinders, cones), works hands-on with material to construct these shapes, and also uses two-dimensional representations of three-dimensional objects to visualise and solve problems

Let’s Recall

Recap to check if students know the properties of 2-D and 3-D shapes. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

lateral face: a face that is on the side of a 3-D shape instead of the top or bottom cross-section: the shape we get on cutting straight through a 3-D shape or an object

Teaching Aids

Cards (with shape names, number of faces, edges and vertices); Cards (with the name of a 3-D shape); Cut-outs of nets; Glue; Flash cards (with 3-D shapes and their dimensions written on them); Oblique drawing sheets; Isometric drawing sheets; Sheets of paper; Playdough; Plastic knife; Flashlight; Scissors; Book; Small cubes; Square grid paper

Chapter: Visualising Solid Shapes

Features of 3-D Shapes

Learning Outcomes

Students will be able to identify and list the features of 3-D shapes.

Teaching Aids

Cards (with shape names, number of faces, edges and vertices)

Activity

Imagine Maths Page 293

Divide the class into 4 groups. Create 4 sets of cards: one set with names of shapes, one with the number of faces, one with the number of vertices and one with the number of edges of the shapes, as in the example shown. Make the groups stand in 4 separate lines. Provide the groups with a set of flash cards each. Instruct students to move around and find the other students to match the shapes with their corresponding properties. Guide them to work together to match each shape with the correct number of faces, edges and vertices based on the information provided on their flash cards.

Nets

of 3-D Shapes Imagine Maths Page 296

Learning Outcomes

Students will be able to identify and draw nets of 3-D shapes.

Teaching Aids

Cards (with the name of a 3-D shape); Sheets of paper; Glue; Scissors

Activity

Begin the class by using one 3-D shape (e.g., a cube) and demonstrate how to draw a cut-out of its net.

Distribute cards with the name of a 3-D shape written on each (pyramids and prisms) and sheets of paper. Instruct students to draw the net of the shape on their card on the paper. When they are done, instruct them to cut the net out carefully along the lines. Once they have cut it out, ask them to fold the cut-outs along the indicated lines. Using glue, instruct them to carefully paste the edges together, ensuring a complete connection and making a solid shape. Encourage students to compare their folded shapes to the actual 3-D shapes. Ask: Can you tell me how a prism is different from a pyramid?

Extension Idea

Ask: Which cube is formed by the given net?

Say: Option B is the cube formed by the given net when connected edge to edge.

Oblique and Isometric Sketches

Learning Outcomes

Students will be able to identify and draw oblique and isometric sketches of 3-D shapes.

Teaching Aids

Flash cards (with 3-D shapes and their dimensions written on them); Oblique drawing sheets; Isometric drawing sheets

Activity

Begin by introducing the concept of oblique and isometric sketches. Ask students to look at the examples in their books of oblique and isometric sketches and discuss the characteristics of each type of sketch and how they differ. Create flash cards with different shapes drawn on them and their dimensions marked, as shown in the example. Instruct students to work in pairs. Give each pair a flash card and provide each student in the pair with an isometric sheet and an oblique sheet, respectively. Guide them to carefully examine the image on the card, and then have each student draw an isometric sketch and an oblique sketch accordingly. Once they are done, have them swap sheets with their partners and review each other’s sketches. Ask: Do both the sketches look the same? Why or why not?

Extension Idea

Ask: What makes an oblique sketch different from an isometric sketch?

Say: In an oblique sketch, there is more focus on the front side of an object, whereas in an isometric sketch, the focus is on the edge of an object.

Cross Section and Shadows Imagine Maths Page 302

Learning Outcomes

Students will be able to visualise 3-D shapes and draw their cross-section and shadows.

Teaching Aids

Playdough; Plastic knife; Sheets of paper; Flashlight

Activity

Create a 3-D shape with playdough, like a ball, and talk about its features. Discuss cross-sections of shapes by cutting one ball vertically and one ball horizontally, showing that both result in circles.

Now, instruct students to work in groups of 4.

Distribute playdough to each group and ask them to make their own 3-D shapes. Have them cut their shapes vertically and horizontally. Instruct them to then draw the obtained cross-sections in their notebooks. Encourage them to think about how the shape of the cross-section changes with different angles and orientations. Now, set up a flashlight to cast shadows on a wall. Show using the playdough ball how the shadow of the ball is circular. 5 units

Then, instruct each group to hold their shapes in front of the light to observe the shadows. Ask them to make sure they keep rotating the shapes to see the top, side and front views of the shadows.

Ask questions like: Do you think that the cross-section of an object is related to its shadow?

Extension Idea

Ask: Can you obtain a square shadow of a cuboidal box?

Say: Yes, if the light source is at an angle to one of its faces, a square shadow can be obtained.

Views of 3-D Shapes

Learning Outcomes

draw different views of structures made of cubes on a square grid.

Teaching Aids

Book; Small cubes; Square grid paper

Activity

Show the students a book from different sides and discuss the different views that they see. Draw the front, side and top view of the book, on the board. Instruct students to work in groups of 3. Distribute the square grids and unit cubes to each group.

Instruct the students to put the unit cubes together to make a cube structure. Now, each student in the group will look at the structure from a different view—front, side or top view. They will then shade the grid to show each view. Discuss in groups and check each view.

Extension Idea

Ask: Is it possible to draw a 3-D structure made of cubes if only one view is provided?

Imagine Maths Page 303

Say: No, it is not possible to draw the exact structure with just one view. A single view can correspond to multiple different 3D structures, so it does not provide sufficient information to determine the exact arrangement.

Answers

1. Features of 3-D Shapes

Think and Tell

No, a cylinder is not a prism, and a cone is not a pyramid since both prism and pyramid have a polygonal base but a cylinder and a cone does not.

Do It Together

4. Cross-section and Shadows

Do It Together

Answers may vary. Sample answer:

2. Nets of 3-D Shapes

Do It Together

Number of faces =

2 faces rectangular faces

Net of a heptagonal prism

3. Oblique and Isometric Sketches

Do It Together

5. Views of 3-D Shapes

Do It Together

Rectangle

Solutions

Chapter 1

Let’s Warm−up

1. 25 < 89

2. −32 < −16

3. −12 < 5

4. 0 > −12

5. 14 > −9

Do It Yourself 1A

1. a. 5 + (−9) = −4

b. 15 + 14 = 29

c. −28 + 59 = 31

d. (−23) + 12 = −11

e. 32 + (−122) = −90

f. −136 + (−25) = −161

g. −152 + 365 = 213

h. 158 + (−125) = 33

2. a. Additive inverse of 48 = −48

b. Additive inverse of −72 = 72

c. Additive inverse of 98 = −98

d. Additive inverse of −124 = 124

3. a. −9 − (8) = −9 + (−8) = −17

b. 17 − (−12) = 17 + (12) = 29

c. −32 − 25 = −32 + (−25) = −57

d. −20 − 38 = −20 + (−38) = −58

e. −98 − (−36) = −98 + (36) = −62

f. 223 − 54 = 223 + (−54) = 169

g. −182 − 136 = −182 + (−136) = −318

h. 156 − (−214) = 156 + (214) = 370

4. To find the missing values, we will use the rule of transposition.

a. 18 − 12 = 6

b. 10 − 17 = −7

c. 956 − = 422

Moving 422 to the LHS, we get 956 − 422. 956 − 422 = 534

So, 956 − 534 = 422

d. 215 − 136 = 79

e. −548 + = −267

Moving −548 to the RHS, we get −267 + 548 −267 + 548 = 281

So, −548 + 281 = −267

f. − 192 = 564

Moving −192 to the RHS, we get 564 + 192 564 + 192 = 756

So, 756 − 192 = 564

g. 705 + = 1202

Moving 705 to the RHS, we get 1202 − 705 1202 − 705 = 497.

So, 705 + 497 = 1202.

h. 815 − (−125) = 940

5. a. The sum of −5 and 8 is 3.

b. Subtracting 15 from −20 gives −35

c. The absolute value of −10 is 10

d. If you have 2000 in your bank account and you spend 3000, your account balance becomes ₹−1000.

e. If I have a gain of ₹310 and a loss of ₹125, my net result is ₹185

f. Adding the additive inverse of −8 to −8 gives 0

6. −3652 + 5864 = 2212

2212 − 2369 = −157

7. a. The commutative property of addition holds true for integers. True

b. The associative property of subtraction holds true for integers. False

c. The commutative property of subtraction holds true for integers. False

d. The associative property of addition holds true for integers. True

8. a. (21) + (13) = 13 + (21); Property used—Commutative property

b. −3 + 0 = −3; Property used—Existence of additive identity

c. (−35 + 13) + (−16) = −35 + (13 + (−16)); Property used— Associative property

d. (−23) − 0 = −23; Property used—Subtraction property of zero

9. a. p − (−q) = 32 − (−29) = 32 + (additive inverse of −29)

= 32 + 29 = 6

And p + q = 32 + 29 = 61

Hence, p − (−q) = p + q is verified.

b. p − q = 32 − 29 = 3 and q − p = 29 − 32 = −3

As 3 ≠ −3 the given statement is verified.

c. (p + q) + r = (32 + 29) + (−17) = 61 + (−17) = 61 − 17 = 44

and p + (q + r) = 32 + (29 + (−17)) = 32 + 12 = 44

As 44 = 44, the given statement is verified.

10. Let the other integer be x.

Given that x + (−89) = 142

⇒ x = 142 + 89 = 231

Hence, the other integer is 231.

11. a. [128 + (−78)] + [−145 − (−92)]

= [50] + [−53] = −3

b. [−214 − (124)] + [−136 + (45)]

= [−338] + [−91]

= −338 − 91 = −429

c. [262 − (−132)] − [165 − (−87)]

= [262 + 132)] − [165 + 87)]

= 394 − 252 = 142

d. [−324 + (121)] − [−145 − (213)]

= [−203] − [−358]

= −203 + 358 = 155

12. Altitude climbed by the hiker = 1200 m

Altitude descended = 500 m = −500 m

Change in altitude experience = 1200 + (−500) = 700 m

Difference in altitude = 2037 − 500 = 1537 m

Thus, the hiker is 1537 meters lower than the altitude of Mount Mitchell.

13. Points scored in the first quarter = 80

Points scored in the second quarter = −25

Points scored in the third quarter = 15

Points scored in the fourth quarter = 25 + 10 = 35

Total points scored = 80 + (−25) + 15 + 35 = 105

14. Hottest temperature = 51℃

Coldest temperature = −45℃

Difference = (51) − (−45℃) = 51℃ + 45 ℃ = 96 ℃.

15. Answer will vary. Sample answer:

The temperature was −5℃ at night and rose by 10℃ by morning. What was the temperature in the morning?

Challenge 1. Answers may vary. Sample answer.

6 + (−11) + 5 + (−5) + 2 = 6 − 11 + 5 − 5 + 2 = −3

Do It Yourself 1B

1. a. 12 × 15 = 180

b. −11 × 6 = −66

c. −21 × 13 = −273

d. −14 × (−6) = 84

e. 25 × 0 = 0

f. 56 × (−1) = −56

2. a. Odd number of negative integers = negative sign

b. Even number of negative integers = positive sign

c. Odd number of negative integers = negative sign

3. a. Multiplicative inverse of 21 = 1 21

b. Multiplicative inverse of −32 = − 1 32

c. Multiplicative inverse of −54 = 1 . 54

4. a. −11 × 5 × (−2) = −55 × −2 = 110

b. 56 × (−12) × 0 = −672 × 0 = 0

c. 12 × (−1) × (−8) = −12 × −8 = 96

d. 25 × (−12) × 3 = −300 × 3 = −900

e. −23 × 15 × (−2) × 5 = −345 × −10 = 3450

f. −6 × (−12) × −3 × (−10) = 72 × 30 = 2160

5. a. 36 × 0 = 0

b. (−123) × 1 = −123

c. (−12) × (56) = 56 × (−12)

d. 56 × (−12 + 23) = 56 × (−12) + 56 × 23

e. −2198 × (−1) = 2198

f. (123 × 143) × (−36) = 123 × (143 × (−36))

6. a. m × n = 8 × −3 = −24; n × m = −3 × 8 = −24

As −24 = −24; hence verified.

b. (m × n) × o = (8 × −3) × −5 = −24 × −5 = 120

m × (n × o) = 8 × (−3 × −5) = 8 × 15 = 120

As 120 = 120; hence verified.

c. m × (n + o) = 8 × (−3 + −5)

= 8 × −8 = −64

m × n + m × o = 8 × −3 + 8 × −5 = −24 − 40 = −64

As −64 = −64, hence verified.

7. a. [45 × (−25)] − [42 × (−35)]

= [−1125] − [−1470] = −1125 + 1470

= 345

b. [−100 × (47)] + [−27 × (−130)]

= [−4700] + [3510] = −4700 + 3510 = −1190

c. [−115 × (−80)] − [−25 × (−120)]

= [9200] − [3000]

= 9200 − 3000

= 6200

8. Rate of descending = 55 feet per minute = −55

Position of the submarine after 8 minutes

= −55 × 8 = −440

= 440 feet below sea level.

9. Scores gained on hitting the target = 10 × 20 = 200 points

Scores lost on missing the target = −6 × 25 = −150

Overall score = 200 + (−150) = 200 − 150 = 50 points

10. The lowest point of the Hawaiian Trench = 5500 m

The Lowest point of the Puerto Rico Trench = 5500 × 1.5 = 8250

So, the lowest point of the Puerto Rico Trench is 8250 m or −8250 m.

11. Answer may vary. Sample answer:

The temperature of a deep freezer decreases by −12℃ every hour. How much will be the be temperature after 1 day?

Challenge

1. Points scored per correct answer = 5

Points taken away per incorrect answer = −2

Total number of questions in the test = 25

Questions answered incorrectly = 7; questions answered correctly = 25 − 7 = 18

Points scored for 18 correct answers = 5 × 18 = 90

Points scored for 7 incorrect answers = −2 × 7 = −14

Total points scored by the student = 90 + (−14) = 76 points

Do It Yourself 1C

1. a. 72 ÷ (−4) = −18

b. (−56) ÷ 7 = −8

c. −176 ÷ (−11) = 16

d. 192 ÷ (−12) = −16

e. 344 ÷ 43 = 8

f. (−984) ÷ 12 = −82

g. (−676) ÷ (−26) = 26

h. −585 ÷ (−13) = 45

2. a. (−88 ÷ 4) ÷ −1 = −22 ÷ −1 = 22

b. 125 ÷ (125 ÷ 5) = 125 ÷ 25 = 5

c. (192 ÷ (−16)) ÷ 4 = −12 ÷ 4 = −3

d. 324 ÷ (−18 ÷ 2) = 324 ÷ −9 = −36

e. (900 ÷ (18)) ÷ 6 = 50 ÷ 6 = 8.33

f. (−3060 ÷ (−36)) ÷ 5 = 85 ÷ 5 = 17

g. 0 ÷ (−1508 ÷ 29) = 0 ÷ −52 = 0

h. (1000 ÷ (20)) ÷ 100 = 50 ÷ 100 = 0.5

3. −145 divided by −1 gives 145.

4. a. (−250) ÷ 1 = −250 True

b. 0 ÷ (−378) = −378 False since 0 ÷ (−378) = 0.

c. (−39) ÷ (−39) = 1 True

d. (−90) ÷ 3 = 30 False since the answer is (−90) ÷ 3 = −30.

5. Let the other integer be x

Given that x × (−25) = 800

⇒ x = 800 ÷ (−25) = −32

Hence, the other integer is −32.

6. a. m ÷ n = 12 ÷ 3 = 4

n ÷ m = 3 ÷ 12 = 1 4

As, 4 ≠ 1 , 4 hence verified.

b. (m ÷ n) ÷ o = (12 ÷ 3) ÷ 2 = 4 ÷ 2 = 2

m ÷ (n ÷ o) = 12 ÷ (3 ÷ 2) = 12 ÷ 1.5 = 8

As, 2 ≠ 8, hence verified.

7. Answers may vary. Sample answers:

First pair: a = −270, b = 18 Second pair: a = 30, b = −2

8. Height of each floor = 5 m

Distance travelled above basement = 20 × 5 = 100 m

Distance travelled below basement = 2 × 5 = 10 m

Distance to be covered = 100 + 10 = 110 m

Speed of the lift = 2 m/s

Time required to reach the second floor of the basement

= Distance to be covered

Speed of the lift

= 110 =55s 2 s

9. The initial position of the parachute = 1800 m above ground level (+1800)

The final position of the parachute = 300 m above the ground level (+300)

Therefore, the distance between the final and the initial position = 300 − 1800 = −1500 m

The distance covered in 1 minute = 20 m

As the parachute is coming down (descends), the distance covered in 1 minute will be (−20 m)

Therefore, time taken to cover (−1500) m = (−1500) ÷ (−20) = 75 minutes.

Therefore, the parachute will take 75 minutes to reach 300 m above the ground level.

10. Number of points lost in the first level for each fall = 2

Total Number of points lost in the first level = 10

Number of times Rahul’s character falls in the first level = 10 2 = 5 times

Number of points lost in second level for each fall = 4

Total Number of points lost in second level = 16

Number of times Rahul’s character falls in the second level = 16 4 = 4 times

Hence, Rahul’s character falls more on the first level.

11. Average = Total temperature across 7 days No. of days ()()()()()()() −+−+−+−+−+−+− = 6456322 7 ℃℃℃℃℃℃℃

Hence the average temperature of the week is −4ºC.

Challenge

1. Temperature recorded at 12 noon = 18℃

Rate of fall in temperature = 3℃ per hour

Difference in temperature at noon and 6℃ below zero = 18℃

− (−6℃) = 18℃ + 6℃ = 24℃

Number of hours it took to drop by 24℃ = 24℃ 3℃/hour = 8 hours

Hence, the temperature at 8 p.m. would be 6℃ below zero

Drop in temperature from 12 noon to midnight = 12 × 3℃ = 36℃

Temperature at midnight = 18℃ − 36℃ = −18℃.

Chapter Checkup

1. a. (−256) + 362 = 106

b. 652 + (−129) = 523

c. (−659) + (−1352) = −2011

d. 214 − (−126) = 214 + 126 = 340

e. 248 − (−369) = 248 + 369 = 617

f. (−2698) − (−1236) = −2698 + 1236 = −1462

2. a. 25 + (−63) = −38; additive inverse = 38; multiplicative inverse = 1 −38

b. −125 + 47 = −78; additive inverse = 78; multiplicative inverse = 1 −78

c. −148 + (−157) = −305; additive inverse = 305; multiplicative inverse = 1 −305

3. a. 25 × (−89) = −2225

b. (−125) × (−26) = 3250

c. (−36) × 48 = −1728

d. (245) × (−57) = −13,965

e. (1245) × (−142) = −1,76,790

f. (−2365) × (123) = −2,90,895

4. a. 363 ÷ (−11) = −33

b. (−384) ÷ 12 = −32

c. 450 ÷ (−25) = −18

d. (−5472) ÷ 36 = −152

e. 2852 ÷ 23 = 124

f. (−9594) ÷ (−41) = 234

5. a. (−156) + 389 = 389 + −156

b. 524 + −1 = 523

c. 1885 × −1= (−1885)

d. 23 × (−17) = (−17) × 23

e. −45 × (33 × 2) = (−45 × 33) (−45× 2)

f. 64 ÷ (−1) = −64

6. a. 42 ÷ (−6 + 5) = 42 ÷ (−1) = −42

b. −64 ÷ 4 × (2 − 6) = −16 × −4 = 64

c. (9 ÷ 3) + 7 × (4 ÷ 2) = 3 + 7 × 2 = 3 + 14 = 17

d. 4(−12 + 6) ÷ 3 = 4 × (−6) ÷ 3 = 4 × (−2) = −8

e. 7 × (5 + 3) ÷ 4 × (9 − 2) = 7 × 8 ÷ 4 × 7 = 7 × 2 × 7 = 98

f. (6 + 2) − 15 ÷ (5 × 2) = 8 − 15 ÷ 10 = 8 − 1.5 = 6.5

7. 135 + (−325) = −190; 253 − (−528) = 781; −190 + 781 = 591

8. Let the number to be multiplied be x. x × (−165) = 9240

x = 9240 −165 = −56

Hence, −56 should be multiplied to (−165) to get a product of 9240.

9. 25 × (−36) = −900; 2380 ÷ (−68) = −35; −900 + (−35) = −935

10. Let the unknown number be x. 23,072 ÷ x = −412

x = 23,072 ÷ (−412)

x = −56

Hence, −56 should be divided from 23072 to get (−412).

11. Anish score in:

First round = 55

Second round = −80

Third round = 65

Since Ravi’s score is double of Anish’s score So, Ravi score in:

First round = ×= 55 2110

Second round = −×=− 80 2 160

Third round = ×= 65 2130

Total score of Anish = 55 + (−80) + 65 = 40

Total score of Ravi = 110 + (−160) + 130 = 80

12. Let the amount to be deposited be x.

x + −₹2000 = ₹10,000

x = ₹10,000 + ₹2000 = ₹12,000

Hence, Suhani should deposit ₹12,000. The value that we learn from Suhani’s mother is managing personal finances.

13. Initial position of the plane = 12,000 feet = +12,000 feet; Distance climbed to reach the cruising altitude = 15,000 feet = +15,000 feet

New altitude after climbing 15,000 feet = (12,000 + 15,000) feet = +27,000 feet

After hitting the turbulence, the plane descends 23,945 feet = −23,945 feet

New altitude of the plane = 27,000 + (−23,945) = +3055 feet

14. Temperature records at 11 a.m = 6°F

Temperature drops by every hour = 3°F

To find the temperature at 4 p.m., we first determine the time from 11 a.m. to 4 p.m.

From 11 a.m. to 4 p.m. is 5 hours.

Temperature drops over 5 hours = °×=°3F 515F

Temperature records at 4 p.m. of the same day = 6°F −15 ºF = − 9°F

Challenge

1. Floor at which Kajal got into the elevator = 4th

Floor of the food court = 4 + 10 = 14th

She went down 5 floors to purchase books = 14 − 5 = 9th floor

Hence, Kajal is on the 9th floor

2. Assertion (A): The value of ×−×− −× 121620 24 8 is −20.

The value of ×−×− −× 121620 24 8 = =− 3840 20 192 = −20

Reason (R): As there are 2 negative numbers in the numerator and 1 in the denominator, the answer will be negative.

Even number of negative integers in numerator = positive sign

Odd number of negative integers in denominator = negative sign

Thus, making the final answer negative.

Both A and R are correct, and R is the correct explanation for A

Hence, option a is correct.

Case Study

1. The temperature increase when magnesium is heated from its melting point to its boiling point = 1090 ℃ − 650 ℃ = 440℃

Thus, option b is correct.

2. Magnesium has the highest boiling point.

To compare how much higher is boiling point of magnesium as compared to chlorine = Boiling point of magnesium − Boiling point of chlorine = 1090℃ − (−34℃) = 1124℃

3. The temperature range (difference of the melting and boiling points) of Mercury = 357℃ − (−38℃) = 395℃

The temperature range (difference of the melting and boiling points) of Oxygen = (−183℃) − (−227℃) = 44℃

The temperature range (difference of the melting and boiling points) of Magnesium = (1090℃) − 650℃ = 440℃

The temperature range (difference of the melting and boiling points) of Chlorine = (−34℃) − (−101℃) = 67℃

The temperature range (difference of the melting and boiling points) of Nitrogen = (−196℃) − (−210℃) = 14℃

Nitrogen element has the narrowest range.

4. Consider the boiling points of each element to know which would change state from liquid to gas when the temperature is increased from −250℃ to 0℃.

Mercury: Boiling point = 357℃ (would remain solid)

Oxygen: Boiling point = −183℃ (would be gas)

Magnesium: Boiling point = 1090℃ (would remain solid)

Chlorine: Boiling point = −34℃ (would be gas)

Nitrogen: Boiling point = −196℃ (would be gas)

The elements that would change from liquid to gas when the temperature is increased from −250℃ to 0℃ are

Oxygen, Chlorine and Nitrogen

5 × ºC

9 5 + 32

Melting point of nitrogen = −210 Melting point of nitrogen in ℉ = ()

9 210 32378 32346

℃ = −378 + 32 = –346℉

Chapter 2

Let’s Warm–up

1. The numerator is always smaller than the denominator for proper fractions.

2. A mixed fraction has a whole number part and a fractional part.

3. 8 is the denominator in 5 8.

4. Unlike decimals have different number of digits after the decimal point.

5. The fraction 25 50 can be written as 0.5 in its decimal form.

Do It Yourself 2A

1. a. 25 + 39

6+5 11 = 99 (LCM of 3 and 9 = 9)

b. 92 + 2211

9+413 = 2222 (LCM of 11 and 22 = 22)

c. 35 + 406

9+100 109 = 120120 (LCM of 40 and 6 = 120)

d. 2+1=+351514 6969

45+28 ==4731 181818 (LCM of 6 and 9 = 18)

e. 3+1=+21177 5656

102+35 ==413717 303030 (LCM of 5 and 6 = 30)

f. 7+3=+253023 4646

= 15 2 + 23 6

45+23 ===1168341 6633 (LCM of 2 and 6 = 6)

2. a. 113 124 –119 21 == 12126 –(LCM of 12 and 4 = 12)

b. 72 39 –

212 19 = 99 –(LCM of 3 and 9 = 9)

c. 2=30230 73173

14630 ==111643 737373 –(LCM of 73 and 1 = 73)

d. 53=211719 3636 3419 ===21551 6622 –3419 ===21551 6622 –(LCM of 3 and 6 = 6)

e. 74=313121 4545 15584 ==37111 202020 –(LCM of 4 and 5 = 20)

f. 63=215610 9393 5630 ==2268 999 –(LCM of 9 and 3 = 9)

3. a. 2 2 3 + 3 1 5 = (2 + 3) + 2 3 + 1 5

= 5 + 10 15 + 3 15

= 5 13 15 7 2 4 − 2 1 3 = 30 4 − 7 3 = 90 12 − 28 12 = 62 12 = 5 2 12 = 5 1 6

Now, let us compare the fractional part, since the whole number part is the same. LCM of 15 and 6 is 30 13 15 = 26 30 and 1 6 = 5 30 13 15 > 1 6

So, 5 13 15 > 5 1 6

Thus, 2 2 3 + 3 1 5 > 7 2 4 − 2 1 3

b. 1 5 8 − 2 3 = 13 8 − 2 3

= 39 24 − 16 24 = 23 24 1 2 + 3 4 = 2 4 + 3 4 = 5 4 = 1 1 4 23 24 < 1 1 4

Thus, 1 5 8 2 3 < 1 2 + 3 4

c. 8 5 7 − 4 1 4 = 61 7 − 17 4

= 244 28 − 117 28

= 127 28 = 4 15 28

4 15 28 > 3 7 8

Thus, 8 5 7 4 1 4 > 3 7 8

4. a. 2 17 + 1 17 + 3 17 = 2 + 1 + 3 17 = 6 17

b. 3 4 9 + 1 2 9 + 2 1 9 = (3 + 1 + 2) + 4 9 + 2 9 + 1 9

= 6 + 7 9 = 6 7 9

c. 2 17 + 1 17 − 3 17 = 2 + 1 − 3 17

= 0 17 = 0

d. 5 12 + 1 12 − 3 12 = 5 + 1 − 3 12

= 3 12 = 1 4

e. 15+4=15+7537293 86128612 36021+1166 = 24 ==1844917 2424

f. 12+235514 626 17734 =12+ 626

= = 72+172134 ===534172 6633 = 72+172134 ===534172 6633

72+172134 ===534172 6633

5. a. 7 7 8 − 5 2 3 = 63 8 − 17 3 = 189 24 − 136 24

= 53 24 = 2 5 24

Thus, 5 2 3 is 2 5 24 less than 7 7 8

b. 4 1 4 − 2 2 5 = 17 4 − 12 5

= 85 20 − 48 20 = 37 20

= 1 17 20

Thus, 1 17 20 must be added to 2 2 5 to get 4 1 4

c. 1 4 + 1 8 9 + 4 5 = 1 4 + 17 9 + 4 5

= 45 180 + 340 180 + 144 180

= 529 180 = 2 169 180

Thus, the perimeter of the triangle is 2 169 180 cm.

d. 7 11 + 2 9 + 7 11 + 2 9 = 14 11 + 4 9

= 126 99 + 44 99

= 170 99 = 1 71 99

Thus, the perimeter of the rectangle is 1 71 99 cm.

e. 9 2 5 − 7 1 4 = 47 5 − 29 4

= 188 20 –145 20 = 43 20 2 3 4 = 4 5 6

= 11 4 = 29 6

= 66 24 + 116 24

= 182 24 = 91 12

7 7 12 – 43 20

= 91 12 –43 20

= 455 60 –129 60

= 326 60 = 163 30 = 5 13 30

The difference in the sum of 3 2 4 and 5 4 6 and the difference in the sum of 2 9 5 and 1 7 4 is 13 5 30

6. Fraction of the wall painted by Aisha = 3 5

Fraction of the wall painted by Aisha’s sister = 1 6

Fraction of the wall painted by both of them = 3 5 + 1 6

= 18 30 + 5 30 = 23 30

Thus, Aisha and her sister painted 23 30 of the wall together.

7. Mango juice drank by Ram = 3 4 bottle

Orange juice drank by Lisa = 2 9 bottle

Total juice drank by Ram and Lisa = 3 4 + 2 9

= 27 36 + 8 36

= 35 36

Thus, Ram and Lisa drank 35 36 of a bottle of juice.

8. Length of the rectangular piece of cloth = 3 8 m

Width of the rectangular piece of cloth = 2 5 m

Amount of lace needed for the cloth = Perimeter of the rectangular piece of cloth

= +++ 3232 8585

= + 64 85 = + 3032 4040 = == 622211 11 404020 = == 622211 11 404020

Challenge

1. Shade 5 fractions that add up to one whole. One is shaded ( 1 6 ). 4+6+2+9+3 11131 24 ++++= ==1 6412882424

4+6+2+9+3 11131 24 ++++= ==1 6412882424 1 6 1 4 1 2 1 12

Thus, Amount of lace needed for the cloth = 11 1 20 m.

9. Total time spent on studies = 5 2 3 hours

Time spent studying Maths and Science = 2 4 5 hours

Time spent studying other subjects = 5 2 3 − 2 4 5 = 17 3 − 14 5

= 85 15 − 42 15 = 43 15 = 2 13 15 hours.

Thus, Shikha devotes 2 13 15 hours to other subjects.

10. Total water in the jug = 2 1 4 gallons

Water drank by Nishant each day = 4 7 gallons

Water drank by Nishant in 3 days = 4 7 + 4 7 + 4 7

= 12 7 gallons

Water left in the jug = 2 1 4 − 12 7

= 9 4 − 12 7

= 63 28 − 48 28

= 15 28

Thus, 15 28 gallons of water is left in the jug.

11. Total length of the string = 30 yards

Length of the string used to tie a parcel = 11 2 5 yards

Length of the string used to tie a box = 6 1 4 yards

Total string used by Carlton = 11 2 5 + 6 1 4

= (11 + 6) + 2 5 + 1 4

= 17 + 8 20 + 5 20

= 17 + 13 20

= 17 13 20

String left with Carlton = 30 − 17 13 20

= 30 1 − 353 20

= 600 20 − 353 20

= 247 20 = 12 7 20

Thus, 12 7 20 yards of string is left with Carlton.

Do It Yourself 2B

1. a. 9 × 6 5 = 9 × 6 5 = 54 5 = 10 4 5 b. 14 × 5 7 = 10 c. 5 3 4 × 20 = 23 4 × 20 1 = 115 1 = 115 d. 2 3 × 9 3 = 2 × 9 3 × 3 = 18 9 = 2 e. 25 9 × 27 15 = 5 f. 2 1 9 × 1 3 = 19 9 × 1 3

19 × 1 9 × 3 = 19 27

1 5 6 × 2 6 7 = 11 6 × 20 7 = 220 42 = 110 21 = 5 5 21 h. 3 5 9 × 1 8 28 = 32 9 × 36 28 = 32 7 = 4 4 7 2. a. 1 2 × 3 4 × 4 5 = 1 × 3 × 4 2 × 4 × 5 = 12 40 = 3 10 b. 1 4 9 × 3 2 × 6 = 13 9 × 3 2 × 6 1 = 13 × 3 × 6 9 × 2 × 1

c. 1 4 × 5 6 × 3 1 5 × 2 = 1 4 × 5 6 × 16 5 × 2 1 = 1 × 5 × 16 × 2 4 × 6 × 5 × 1 = 160 120 = 4 3 = 1 1 3

d. 2 1 4 × 1 1 3 × 1 1 2 = 9 4 × 4 3 × 3 2 = 9 × 4 × 3 4 × 3 × 2 = 108 24 = 9 2 = 4 1 2

3. a. 2 2 5 ×    1 7 9 − 1 2    = 12 5 ×    16 9 − 1 2 

= 12 5 × 

32 18 − 9 18    = 12 5 × 23 18 = 12 × 23 5 × 18 = 276 90 = 46 15 = 3 1 15

b. 1 4 + 

 1 2 × 1 5 6    = 1 4 +  

1 2 × 11 6 

= 1 4 + 11 12 = 3 12 + 11 12 = 14 12 = 7 6 = 1 1 6

c    2 3 4 × 1 2 × 8 9    + 2 5 3 =    11 4 × 1 2 × 8 9    + 11 3 = 11 9 + 11 3 = 11 + 33 9 = 44 9 = 4 8 9

d. 1 2 3 ×    2 5 6 + 3 1 3    = 5 3 ×    17 6 + 10 3 

= 5 3 × 

17 6 + 20 6 

= 5 3 × 37 6 = 185 18 = 10 5 18

4. a. 1 5 of 1000 grams = 1 5 × 1000 grams = 200 grams

b. 5 6 of a metre = 5 6 of 100 cm = 5 6 × 100 cm = 500 6 cm = 250 3 cm = 83 1 3 cm

c. 3 4 of 2 hours = 3 4 × 2 1

= 3 2 = 1 1 2 hours

d. 2 7 of 7 weeks = 2 7 × 49 days (Since, 1 week = 7 days; 7 weeks = 7 × 7 = 49)

= 14 days

e. 7 8 of 5 dozen = 7 8 × 60 (Since 1 dozen = 12; 5 dozens = 5 × 12 = 60)

= 420 8 = 105 2 × 1 12 = 4 3 8 dozen

f. 2 3 of a leap year = 2 3 of 366 days (Since 1 leap year = 366 days)

= 732 3 days

= 244 days

5. Human bones make up about 1 7 of the total body weight

Weight of a person = 63 kg

Bone weight of a person = 1 7 × Weight of a person

1 =×63=9 kg 7

6. Total milk with Ram = 2 1 5 litres

Fraction of milk used by Ram = 1 3

a. Milk used by Ram = 1 3 of 2 1 5 litres

= 1 3 × 11 5

= 11 15 litres

b. Milk left with Ram = 2 1 5 − 11 15

= 11 5 − 11 15 = 33 15 − 11 15

= 22 15

= 1 7 15 litres

7. Total fraction of pizza eaten by Simran and her friends = 1 3 4

Total weight of a pizza = 800 g

Total weight of pizza eaten by Simran and her friends = 1 3 4 ×

800 g = 7 4 × 800 g = 1400 g

8. Total number of cookies with Pihu = 30 cookies

Fraction of cookies given to her sister = 2 3

Fraction of cookies given to neighbours = 1 5

Total fraction of cookies given = 2 3 + 1 5

= 10 15 + 3 15 = 13 15

Number of cookies given = 13 15 × 30 = 26 cookies

Thus, 26 cookies were distributed by Pihu. 9. The ratio of the Science books, Novels, and English books in a library is 2:3:5.

Total number of books = 1500

Number of Science books

= 2 2 + 3 + 5 × 1500 = 2 10 × 1500 = 300

Number of Novels 33 =×1500=×1500=450 2+3+510

33 =×1500=×1500=450 2+3+510

Number of English books

= ×=×= ++ 5515001500750 23510

The difference in the number of science books and English books = 750 − 300 = 450 books

Challenge

1. Assertion (A): When you multiply two fractions, the product is always less than each of the individual fractions.

This is false. For example, ×= 34 2 23 which is greater than both 3 2 and 4 3

Reasoning (R): Multiplying fractions involves multiplying the numerators together and the denominators together. This is true. This is the correct method for multiplying fractions.

Hence, the correct option is option d. A is false, but R is true.

Do It Yourself 2C

1. a. Reciprocal = 11 7

b. 2 5 9 = 23 9 ; Reciprocal = 9 23

c. Reciprocal = 1 3

d. 2 1 4 = 9 4 Reciprocal = 4 9

2. a. 2 3 ÷ 6 = 2 3 × 1 6

= 2 18 = 1 9

b. 2 11 ÷ 66 = 2 11 × 1 66

= 2 726 = 1 363

c. 8 1 2 ÷ 4 = 17 2 × 1 4 = 17 8 = 2 1 8

d. 2 10 13 ÷ 1 = 2 10 13

e. 18 ÷ 3 5 = 18 1 × 5 3

= 90 3 = 30

f. 14 ÷ 7 8 = 14 1 × 8 7

= 112 7 = 16

g. 2 ÷ 5 1 6 = 2 ÷ 31 6

= 2 1 × 6 31 = 12 31

h. 3 2 5 ÷ 1 4 = 17 5 × 4

= 68 5 = 13 3 5

3. a. Perimeter of the equilateral triangle = 3 7 m

Length of each side = 3 7 m ÷ 3

= 3 7 × 1 3 = 1 7 m

Thus, the length of each side of the equilateral triangle is 1 7 m.

b. 4 5 ÷ 2 3 = 4 5 × 3 2 = 12 10 = 6 5

Thus, the product of 2 3 and 6 5 is 4 5

c. Sum of 12 8 and 8 12 = 12 8 + 8 12

= 36 24 + 16 24 = 52 24

= 13 6

96 ÷ 13 6 = 96 1 × 6 13

= 576 13 = 44 4 13

Thus, 96 divided by the sum of 12 8 and 8 12 is 44 4 13.

4. Length of the piece of the cloth = 12 cm

Length of the small piece of cloth = 2 3 cm

Number of small pieces of cloths = 12 ÷ 2 3 = 12 × 3 2 = 18 pieces

Thus, there will be 18 small pieces of cloth.

5. Area of the rectangular field = 43 3 4 cm2

Breadth of the field = 12 1 2 cm

Length of the field = area ÷ breadth = 43 3 4 ÷ 12 1 2

= 175 4 ÷ 25 2

= 175 4 × 2 25 = 7 2

= 3 1 2 cm

Thus, the length of the rectangular field is 7 2 cm.

6. Total juice in the large container = 18 1 3 litres

Capacity of each small container = 1 5 6 litres

Total number of small containers that can be filled from a large container = 18 1 3 ÷ 1 5 6

= 55 3 ÷ 11 6

= 55 3 × 6 11 = 10

Thus, 10 small containers are required.

7. Total weight of the fruits = 5 9 kg

Number of packets = 15

a. Weight of each packet = 5 9 ÷ 15

= 5 9 × 1 15 = 1 27 kg

b. Number of packets sold = 5

Weight of the packets sold = 5 × 1 27 kg

= 5 27 kg

8. Let us say there are x number of people at the restaurant.

Number of males at the restaurant = 3 7 x

Number of females at the restaurant = x − 3 7 x = 4 7 x

Difference between the number of females and the number of males = 28

4 7 x –3 7 x = 1 7 x = 28

x = 28 × 7 = 196

Number of females = 4 7 × 196 = 112

Thus, there are 112 females at the restaurant.

9. Let the total number of shirts be x

Number of printed shirts = 1 3 x

Number of remaining shirts = x –1 3 x = 2 3 x

Number of striped shirts = 1 4 × 2 3 x = 1 6 x

Number of plain shirts = x –1 3 x –1 6 x

= 6 – 2 – 1 6 x

= 3 6 x = 1 2 x

1 2 x = 96

x = 192

Thus, there are 192 shirts in total.

10. Total number of players = 72

Number of players who play football = 3 8 of 72 = 27

Players who are not football players = 72 – 27 = 45

Thus, there are 45 non-football players.

11. Fraction of money spent on house rent = 1 5

Fraction of money spent on food = 1 4

Fraction of money spent on other expenses = 2 5

Total fraction of money spent = 1 5 + 1 4 + 2 5

= 4 5 + 5 20 + 8 20

= 17 20

Fraction of money left = 1 – 17 20

= 20 20 –17 20 = 3 20

Money left with Rakul = ₹3000

3 20 of total money = ₹3000

Total money with Rakul = 20 3 × ₹3000 = ₹20,000

Thus, Rakul had ₹20,000 initially.

Challenge

1. Rate of leakage for the first hole = 1 4 gallons per half an hour

Leakage by first hole per hour = 1 4 ÷ 1 2

= 1 4 × 2 = 1 2 gallons

Rate of leakage for the second hole = 1 5 gallons per quarter of an hour

Leakage by second hole per hour = 1 5 ÷ 1 4

= 1 5 × 4 = 4 5 gallons

Total leakage by 2 holes per hour = 1 2 + 4 5

= 5 10 + 8 10 = 13 10 gallons

Total capacity of the cistern = 500 gallons

Time required to empty the complete cistern = 500 ÷ 13 10

= 500 × 10 13

= 5000 13 = 384 8 13 hours

Do It Yourself 2D

and 12.9 b. 315.36 and 218.234 H T O t h 1

c. 118.5 from 732.55 d. 438.236 from 752.23

H T O t h

2 12

7 3 2 5 5

1 1 8 . 5 0 6 1 4 0 5 H T O t h th 4 11 11 12 10

e. 856.18 from 998.856 f. 1065.235 from 1189.15

H T O t h th

7 15

9 9 8 8 5 6

8 5 6 . 1 8 0

1 4 2 6 7 6

3. Let the unknown terms be x

a. 236.45 + x = 568.2

x = 568.2 – 236.45 = 331.75

b. 514.7 – x = 241.23

x = 514.7 – 241.23 = 273.47

c. x + 321 = 512.47

x = 512.47 – 321 = 191.47

d. x − 475 = 251.487

x = 251.487 + 475 =726.487

e. 785.5 + 324.21 = 1109.71 f. 987 – 562.26 = 424.74

4. Let x be added to 389.265 to get 625.56

389.265 + x = 625.56

x = 625.56 − 389.265 = 236.295

5. Let x be subtracted from 235.125 to get 126.24

235.125 – x = 126.24

x = 235.125 – 126.24 = 108.885

6. Sum of 356.21 and 585.236 = 356.21 + 585.236 = 941.446

Difference of 829.23 – 469.251 = 829.23 – 469.251 = 359.979 941.446 – 359.979 = 581.467

7. 126.256 + 325.46 – 195.264 + 326.152 – 256.369 = 126.256 + 325.46 + (–195.264) + 326.152 + (–256.369)

Adding the positive numbers = 126.256 + 325.46 + 326.152 = 777.868

8 1 4

H T O . t h th 7 7 7 . 8 6 8 4 5 1 6 3 3 3 2 6 . 2 3 5

Adding the negative numbers = (−195.264) + (−256.369) = −451.633 777.868 – 451.633 = 326.235

8. Total ingredients = Flour + Sugar + Butter

Total ingredients = 3.75 cups + 2.5 cups + 1.25 cups = 7.5 cups

Therefore, 7.5 cups of ingredients are needed in total for the recipe.

9. Sweets bought by Ravi = 10.5 Kg

Weight of the empty box = 1 Kg 350 g = 1.350 Kg

Weight of sweets + Weight of empty box = Total weight

10.5 kg + 1.35 kg = 11.85 kg

Therefore, the total weight of the sweets and the box is 11.85 kg.

10. Ronita's current monthly expense for cable and internet = ₹635.80.

The new plan she is considering offers internet for ₹452.75 and cable for ₹126.55.

New plan expense = ₹452.75 + ₹126.55 = ₹579.30.

Monthly savings = Current expense – New plan expense

Monthly savings = ₹635.80 – ₹579.30 = ₹56.50

Therefore, Ronita would save ₹56.50 every month by switching to the new plan.

11. Let x be the size of the third application.

We know that the total size of the three applications is 986.48 KB, and the size of the first two applications is 256.36 KB and 485.84 KB, respectively.

x + 256.36 + 485.84 = 986.48

x = 986.48 – 256.36 – 485.84

x = 244.28 KB

Therefore, the third application is 244.28 KB.

12. Price of the coffee = $1.46

Price of the waffle = $0.85

Total amount spent = $2.31

Change received if paid $10 = $10.00 − $2.31 = $7.69

Hence, Sam will receive a change of $7.69.

13. Weight of sugar donated = 35.65 kg

Weight of flour donated = 78 kg 540 g = 78.54 kg

Weight of rice donated = 17 kg 320 g = 17.32 kg

Let weight of pulses donated be x kg

Total weight of the items donated =142.63 kg

35.65 + 78.54 + 17.32 + x = 142.63

131.51 + x = 142.63

x = 142.63 – 131.51

x = 11.12

Hence, weight of the pulses donated = 11.12 kg or 11 kg 120 g.

14. Amount saved in the months of June and July = ₹3652.45 and ₹4578.58

Total savings in June and July = ₹3652.45 + ₹4578.58 = ₹8231.03.

Amount spent on purchasing bicycle and helmet = ₹2468.75 and ₹689.48

Total amount spent on bicycle and helmet = ₹2468.75 + ₹689.48 = ₹3158.23.

Total amount left with Sam at the end of July = ₹8231.03 − ₹3158.23 = ₹5072.8.

15. The difference in the amount deposited and withdrawn = Total amount deposited in four months − Total amount withdrawn in four months

= (₹10,512.25 + ₹8523 + ₹9684.5 + ₹10,365) – (₹5423.58 + ₹6842.75 + ₹3754.36 + ₹4365.45)

= 390,84.75 – 20,386.14 = ₹18,698.61

Challenge

1. Distance travelled on the last three days = 79.75 + 84.5 + 76.55 = 240.8 km

Distance travelled in the initial 4 days = 82.75 + 80.5 + 78.25 + 81.5 = 323 km H

The ratio of the distance travelled in the last 3 days to the initial 4 days = 240.8 323 == 2408 3230 1204:1615

2. Increase in temperature from February to March = 9.7 – 5.5 = 4.2 ℃

Increase in temperature from February to June = 20.2 – 5.5 = 14.7

One-fourth of 14.7 = ×= 1 14.7 4 3.675

Hence, the statement is not correct.

Do It Yourself 2E

1. a. 1 2 × 1 5

6 0

1 2 0

1 8 0 0.12 × 15 = 1.8 b. 1 1 2 2 × 6 5 5 6 1 0 6 7 3 2 0 7 2 9 3 0 112.2 × 6.5 = 729.3

c. 3 2 1 × 1 3 3 9 6 3

9 6 3 0

3 2 1 0 0

4 2 6 9 3 32.1 × 1.33 = 42.693 d. 5 0 1 4 × 2 5 2 5 0 7 0 1 0 0 2 8 0 1 2 5 3 5 0 5.014 × 25 = 125.35

e. 5 2 2 5 × 5 2 3 1 5 6 7 5

1 0 4 5 0 2 6 1 2 5

2 7 3 2 6 7 5 52.25 × 5.23 = 273.2675 f. 1 2 2 5 6 × 5 2 1 1 2 2 5 6 2 4 5 1 2 6 1 2 8 0

3 8 5 3 7 6 122.56 × 5.21 = 638.5376

2. Given that, 365 × 124 = 45,260

a. 36.5 × 124 = 4526 b. 3.65 × 12.4 = 45.26

c. 36.5 × 1.24 = 45.26

3. a. 0.165 ÷ 1.5 = 1.65 ÷ 15 = 0.11 0.11 15 1.65 – 0 16 – 15 15 – 15 0 b. 42.7 ÷ 14 = 3.05 3.05 14 42.70 – 42 07 – 0 70 – 70 0

d. 23.85 ÷ 1.8 = 238.5 ÷ 18 = 13.25 13.25 18 238.50 – 18 58 – 54 45 – 36 90 – 90 0 e. 487.36 ÷ 3.2 = 4873.6 ÷ 32 = 152.3 152.3

4. Let x be the unknown value,

a. 68.527 × x = 23.63

x = 23.63 68.527 = 0.345

b. x × 1.6 = 52

x = 52 1.6 = 32.5

c. 2177.35 ÷ 35 = 62.21

d. 74.2 × 12.3 = 912.66

e. 984.576 ÷ x = 25.6

x = 984.576 25.6 = 38.46

f. 616.54 ÷ 14.5 = 42.52

5. The product of two decimal numbers = 3538.29

One number = 41.5

Hence, 41.5 × the second number = 3538.29

Second number = 3538.29 41.5 = 85.26

6. Length of the side of the given square = 124.75 cm

Area of square = side2 = 124.752

= 15,562.5625 cm2

Perimeter of square = 4 × side = 4 × 124.75 = 499 cm

7. Length of the cloth required to make 1 shirt = 2.35 m

Length of cloth required to make 48 shirts = 2.35 × 48 = 112.8 m

8. The value of 1 euro = ₹90.61

The value of 8 euro = ₹90.61 × 8 = ₹724.88

The value of 1 dollar = ₹83.73

The value of 15 dollars = ₹83.73 × 15 = ₹1255.95

Total money Mrunal has = ₹724.88 + ₹1255.95 = ₹1980.83

9. Distance covered by the car in 8 hours = 452.8 km

c. 49.08 ÷ 0.02 = 4908 ÷ 2 = 2454 2454 2 4908 – 4 09 – 8 10 – 10 08 – 8 0

Distance covered by the car in 1 hour = 452.8 8 = 56.6 km

Distance covered by the car in 2 4 5 hours or 14 5 hours = 56.6 × 14 5 = 158.48 km

10. Amount paid for sugar = 7.65 kg × ₹38/kg = ₹290.7

Amount paid for rice = 2.485kg × ₹62/kg = ₹154.07

Amount paid for flour = 5.725 kg × ₹46/kg = ₹263.35

Total amount = amount for sugar + amount for rice + amount for flour = ₹290.7+ ₹154.07+ ₹263.35 = ₹708.12

Hence, the total amount to be paid by Suhani = ₹708.12

11. Total amount of milk left = 1 1 2 litres = 1.5 litres

Milk in 8 bottles = 25.5 litres – 1.5 litres = 24 litres

8 bottles contains = 24 litres milk

1 bottle = 24 8 = 3 litres

Hence, the volume in each bottle = 3 litres

12. To find the cost of fencing, we first need to find the perimeter of the garden.

Perimeter of the garden = 2 [length + breadth] = 2 [7.25 + 5.8] = 2 [13.05] = 26.10 m

of fencing 26.10 m at ₹58.5 per metre = 26.10 × ₹58.5 = ₹1526.85

13. Total cost of the music system = ₹25,526.75

Amount paid as down payment = ₹10,250

Rest of the amount to be paid = ₹25,526.75 − ₹10,250 = ₹15,276.75

Number of installments = 15

Amount paid in each installment = 15276.75 15 = ₹1018.45

14. Cost of 2 pairs of shoes = 2 × ₹525.75 = ₹1051.50

Cost of a pair of 3 t-shirts = 3 × ₹263.50 = ₹790.50

Cost of 5 pairs of socks = 5 × ₹50.25 = ₹251.25

Cost of a bag = ₹755

Total cost of the items purchased = ₹1051.50 + ₹790.50 + ₹251.25 + ₹755 = ₹2848.25

15. Length of the cloth used for 1 banner = 2 m 75 cm = 2.75 m

Length of the cloth used for 8 banners = 8 × 2.75 m = 22 m

Length of the cloth used for 1 tablecloth = 1.55 m

Length of the cloth used for 3 tablecloths = 3 × 1.55 m = 4.65 m

Total length of the cloth used = 22 m + 4.65 m = 26.65 m

Cost of 1 m of cloth = ₹75.5

Cost of 26.65 m of cloth = 26.65 × 75.5 = ₹2012.075

The total cost of the cloth used = 26.65 m × ₹75.5 = ₹2012.075

16. Length of 25 red ribbons = 25 × 45.65 = 1141.25 m

Length of 3 blue ribbons = 3 × 520.75 = 1562.25 m

Length of 2 green ribbons = 2 × 368.15 = 736.3 m.

Total length of the ribbons = 1141.25 + 1562.25 + 736.3 = 3439.8 m

Length of ribbon left = 560.2 m

Length of ribbon left = 3439.8 – 560.2 = 2879.6 m

Challenge

1. To centre the figure on the page, Saumya needs to leave an equal margin on both ends.

Margin on each end = Page width – Figure width 2 = 22.63 cm – 12.75 cm 2 = 4.94 cm

Therefore, Saumya should leave a margin of 4.94 cm on each end so that the figure is centred on the page.

Chapter Checkup

1. a. 5 13 + 2 39

= 15 39 + 2 39 = 17 39

b. 1 9 + 5 6 = 2 18 + 15 18 = 17 18

c. 14 6 + 25 4

= 28 12 + 75 12 = 103 12 = 8 7 12

d. 1 5 7 + 4 8 14 = 12 7 + 64 14

= 24 + 64 14 = 88 14 = 44 7 = 62 7

e. 7 3 − 5 9

= 21 – 5 9 = 16 9 = 17 9

f. 15 6 –3 2 = 15 – 9 6 = 6 6 = 1

g. 8 3 5 – 3 2 15 = 43 5 –47 15 = 129 – 47 15 = 82 15 = 5 7 15

h. 12 2 7 – 5 3 8 = 86 7 –43 8 = 688 – 301 56 = 387 56 = 651 56

2. a. 1 9 of two hours = 1 9 of 120 minutes = 1 9 × 120 minutes = 13 1 3 minutes

b. 2 3 of a day = 2 3 of 24 hours

= 2 3 × 24 hours = 16 hours

c. 5 1 5 of a year = 26 5 of 365 days = 26 5 × 365 days = 1898 days

d. 3 4 of a metre = 3 4 of 100 cm = 3 4 × 100 cm = 75 cm

3. a. 122.56 + 36.2 H T O . t h 1 2 2 . 5 6 + 3 6 2 0 1 5 8 . 7 6 b. 321.4 + 12.63 H T O t h 1 3 2 1 4 0 + 1 2 . 6 3 3 3 4 0 3

c. 536.3 + 311.236 H T O t h th 5 3 6 3 0 0 + 3 1 1 . 2 3 6 8 4 7 5 3 6 d. 756.59 + 413.5 + 623.12 Th H T O t h 1 1 1

5 6 5 9 + 4 1 3 . 5 0 + 6 2 3 1 2 1 7 9 3 . 2 1

e. 336.8 – 285.123 H T O . t h th 2 13 7 9 10 3 3 6 . 8 0 0 2 8 5 1 2 3 0 5 1 . 6 7 7 f. 586.4 – 387.25 H T O . t h 4 17 16 3 10

6 5 8 4 1 2

× 12.29 = 1923.6308

e. 498.8 ÷ 4 = 124.7 124.7 4 498.8 – 4 09 – 8 18 – 16 28 – 28 0

g. 90.15 ÷ 1.5 = 901.5 ÷ 15 = 60.1 60.1 15 901.5 – 90 01 – 0 15 – 15 0 h. 23.94 ÷ 6.3 = 239.4 ÷ 63 = 3.8 3.8 63 239.4 – 189 504 – 504 0

5. a. 3 1 5 × 2 1 4 = 16 5 × 9 4 = 144 20 = 36 5 = 7 1 5 2 1 3 ÷ 4 1 6 = 7 3 ÷ 25 6 = 7 3 × 6 25 = 14 25

Thus, the product of 3 1 5 and 2 1 4 is more.

b. 25 × 4 75 = 4 3 = 1 1 3 3 1 4 + 5 4 5 = 8 + 1 4 + 4 5

= 8 + 5 20 + 16 20 = 9 1 20

Thus, the product of 25 and 4 75 is less.

c. 4 ÷ 2 7 = 4 × 7 2 = 14 2 5 × 15 = 6 14 – 6 = 8

Thus, the difference is 8.

d. 9 × 1 1 2 = 9 × 3 2 = 27 2 = 13 1 2 9 1 2 + 7 1 3 = 16 + 1 2 + 1 3 = 16 5 6 13 1 2 + 16 5 6 = 29 + 1 2 + 5 6

= 29 + 8 6 = 30 1 3

Thus, the sum is 30 1 3

6. a. 1 2 3 + 3 4 5 + 4 1 2 = (1 + 3 + 4) + 2 3 + 4 5 + 1 2

= 8 + 20 30 + 24 30 + 15 30

= 8 + 59 30

= 8 + 1 29 30 = 9 29 30

b. 2 15 17 –2 34 + 1 = 49 17 –2 34 + 1

= 98 34 –2 34 + 34 34 = 98 – 2 + 34 34 = 130 34 = 65 17 = 3 14 17

c. 2 6 7 × 1 3 × 5 7 = 20 7 × 1 3 × 5 7

= 20 × 1 × 5 7 × 3 × 7

= 100 147

d. 5 3 4 ÷ 4 2 3 ÷ 1 2 3 = 23 4 ÷ 14 3 ÷ 5 3 =    23 4 × 3 14    ÷ 5 3 = 69 56 ÷ 5 3 = 69 56 × 3 5 = 207 280

a. Distance covered in 1 1 4 hours = 48 3 4 miles

Distance covered in 1 hour = 48 3 4 ÷ 1 1 4 = 195 4 ÷ 5 4 = 195 4 × 4 5 = 39 miles

Distance covered in 2 hours = 39 miles × 2 = 78 miles

Thus, the train covers 78 miles in 2 hours.

b. 2 13 of x = 26

x = 26 × 13 2 = 169

c. 22 2 9 × x = 1800

200 9 x = 1800

x = 1800 × 9 200

x = 81

d. Let the unknown number be x

x × 2 3 = 3 5

x = 3 5 ÷ 2 3 = 3 5 × 3 2 = 9 10

e. Let the number be x.

7 × x = 180

x = 180 × 7 6 = 210

f. Let the number be x.

x × 7 47 55 = 21 3 5

x × 432 55 = 108 5

x = 108 5 × 55 432 = 11 4 = 2 3 4

8. Fraction of cake with Miya = 3 5

a. Fraction of cake eaten by Miya = 2 3 of 3 5 = 2 3 × 3 5 = 6 15 = 2 5

b. Fraction of cake left with Miya = 3 5 –2 5 = 1 5

9. Total marks = 600; marks scored by Aarav = 452.65 Marks lost = 600 – 452.65 = 147.35

10. Mass of powdered milk in each tin = 0.625 kg

Total number of tins in a carton = 12

Total mass of powdered milk in a carton = 0.625 kg/tin × 12 tins = 7.5 kg

11. Given: Average growth rate of hair = 1cm 8 mm per month = 1.8 cm/month

Growth of hair strand in 3 years or 36 months = 1.8 × 36 months = 64.8 cm

So, after three years, the length of the strand will be 64.8 cm.

12. Let the total capacity of the tank be x litres.

Fraction of water inside the tank initially = 4 5

Quantity of water drawn from the tank = 65 litres

Fraction of water left inside the tank = 7 12

4 5 x –7 12 x = 65

(48 – 35)x 60 = 65 x = 300 litres

Water left inside the tank = 7 12 x = 7 12 × 300 = 175 litres

Thus, 175 litres of water is left inside the tank.

13. Speed for the first 5 hours = 4 1 5 km per hour

Distance travelled in the first 5 hours = 5 × 4 1 5 = 5 × 21 5 = 21 km

Speed for the next 3 hours = 3 1 4 km per hour

Distance travelled in the next 3 hours = 3 × 3 1 4 = 3 × 13 4 = 39 4 hour

Total distance travelled in 8 hours =

4 = 30 3 4 km

Thus, the total distance covered in 8 hours is 30 3 4 km.

14. Weight of Sack A = 125.35 kg

Weight of Sack B = Weight in Sack A + 25.65 kg = 125.35 kg + 25.65 kg = 151 kg

Weight of Sack C = Weight in Sack B − 13.49 kg = 151 kg − 13.49 kg = 137.51 kg

Total weight of 3 sacks = 125.35 kg + 151 kg + 137.51 kg = 413.86 kg

Therefore, the total weight of rice in all three sacks is 413.86 kg.

15. a. Cost of 3 vegetable bowls at ₹155.26 per bowl = 3 × ₹155.26 = ₹465.78

Cost of 2 apple pies at ₹62.37 per apple pie = 2 × ₹62.37 = ₹124.74

Cost of 4 milk shakes at ₹ 89.49 per milk shake = 4 × ₹89.49 = ₹357.96

Total amount spent = ₹465.78 + ₹124.74 + ₹357.96 = ₹948.48

b. If the family paid ₹1000 and their total cost was ₹948.48, their change would be: ₹1000 − ₹948.48 = ₹51.52

16. Turns completed by the planet about its axis = 1 3 5 turns

a. Turns completed in a week = 1 3 5 × 7 = 8 5 × 7

= 56 5 = 11 1 5 turns

b. Number of days in a fortnight = 14 days

Number of turns completed by the planet = 1 3 5 × 14

= 8 5 × 14 = 112 5 = 22 2 5 turns

17. Fraction of distance covered by the train = 1 3

Fraction of distance covered by the bus = 2 5

Total distance covered by train and bus = 1 3 + 2 5 = 5 15 + 6 15 = 11 15

Fraction of distance covered walking = 1 –11 15 = 4 15

Distance covered by walking = 16 km 4 15 x = 16

x = 16 × 15 4 = 60 km

a. Distance covered by the bus = 2 5 × 60 = 24 km

b. Distance covered by the train = 1 3 × 60 = 20 km

c. Total distance covered = 60 km

18. Amount of sugar bought = 8.6 kg

Amount of sugar left after pouring equally in 5 bottles = 0.35 kg

Amount of sugar poured in 5 bottles = 8.6 – 0.35 = 8.25 kg

Amount of sugar poured in each bottle = 8.25 5 = 1.65 kg

Therefore, the mass of sugar in 1 bottle is 1.65 kg.

19. Distance travelled on day 1 = 150.25 km

Distance travelled on day 2 = 175.50 km

Total distance the family needs to cover = 450.75 km

Distance the family needs to drive to reach the destination = 450.75 km – 150.25 km + 175.50 km = 125 km

Amount of fuel required to travel 11.5 km = 1 litre

Amount of fuel required to travel 450.75 km = 450.75 11.5 = 39.2 litres

Challenge

1. a. Amount saved in Week 2 = ₹1254

Amount saved in Week 6 = 2.25 times of amount saved in Week 2 = ₹1254 × 2.25 = ₹2821.5

b. Amount saved in the initial 6 months = ₹1058.25 + ₹1254 + ₹1365.78 + ₹1563.75 + ₹1026.65 + ₹2821.5 = ₹9089.93

Total amount Monika willing to save by end of seventh month = ₹12,000

Amount Monika should save in the seventh month = ₹12,000 − ₹9089.93 = ₹2910.07

2. Statement 1: Julia divided the total number of cookies into 8 boxes.

Statement 1 doesn’t give any information about the total number of cookies. Hence, it is not sufficient alone.

Statement 2: Each box contains 5 cookies

Statement 2 gives information on the number of cookies in 1 box but not the total number of cookies. Hence, it is not sufficient alone.

Combining statements 1 and 2 we get that there are 8 boxes and each box has 5 cookies.

Total number of cookies = 8 × 5 = 40 cookies.

Hence, both statements together are sufficient, but neither statement alone is sufficient

So, the correct answer is option (c).

Case Study

1. Number of type A bracelets sold = 35

The cost of each type A bracelet = ₹75.5

Money earned = ₹75.5 × 35 = ₹2642.5 So, (c) is the correct option.

2. Number of type B bracelets sold = 21

The cost of each type A bracelet = ₹90.5

Money earned = ₹90.5 × 21 = ₹1900.5

Fraction of donation = 3 7 = ×= 3 1900.5 7 ₹814.5

So, (b) is the correct option.

3. Donation made on type A bracelet = ₹2642.5 × = 2 5 ₹1057 Donation made on type B bracelet

Donation made on type C bracelet = 120.5 × 28

× = 5 8 ₹2108.75

So, type C bracelet made them donate the highest amount.

4. Total amount donated = ₹1057 + ₹814.5 + ₹2108.75 = ₹3980.25

5. Answers may vary

Chapter 3

Let’s Warm-up

1. School B (26) won the maximum number of medals

2. Total number of medals won by all schools = 11 + 26 + 19 + 21 + 15 = 92 medals

3. Number of medals won by School D = 21

Number of medals won by School E = 15

Difference in the number of medals won by School D to the number of medals won by School E = 21 − 15 = 6 medals

4. Number of medals won by School C = 19

Total number of medals won by schools A, B, D and E = 11 + 26 + 21 + 15 = 73 medals

Ratio of the number of medals won by School C and the number of medals won by the rest of the schools = 19:73

Do It Yourself 3A

1. The first 10 odd numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.

Sum of the first 10 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

Sum of all observations

So, Mean =

Total number of number of observations = 100 10 = 10

2. Arranging in ascending order: 1, 1, 2, 2, 3, 4, 4, 4, 5, 5, 6, 7, 7, 7, 7, 10, 10, 12

7 occurs most frequently in the given data, so the mode of the data is 7.

3. No. of illustrations (per page) (x) No. of pages

Mean = ∑ f × x ∑ f = 467 112 = 4.17

4. Fiction, Poetry, Drama, Drama, Fiction, Fantasy, Poetry, Drama, Fantasy, Drama.

Genre Fiction Poetry Drama Fantasy

Number of Times

Chosen by Children 2 2 4 2

Since, the drama genre is chosen the most, so the modal genre is drama.

5. Mean = 92

Sum of all the observations =

+ (x + 7) = 5x + 17

Sum of all the observations = Mean × Total number of observations = 92 × 5 = 460

So, 5x + 17 = 460

5x = 460 − 17 = 443

x = 443 5 = 88.6

6. Enrollments: 1555, 1670, 1850, 1780, 1400, 2540

Arranging in ascending order: 1400, 1555, 1670, 1780, 1850, 2540

Number of observations = 6

Median = 1 2

6 2 + 1

n 2

th + 

n 2 + 1

observation = 1 2

6 2

th +

= 1 2 [3rd + 4th] observation = 1 2 [1670 + 1780]

= 1 2 [3450] = 1725

So, the median score of the team is 1725.

7. The heights of trees collected by an ecologist are as follows (in metres): 5, 7, 10, 12, 15, 18, 22, 30.

Range = Highest observation − Lowest observation = 30 − 5 = 25

8. Mean of 5 numbers = 26

So, sum of all the observations = mean × total number of observations = 5 × 26 = 130

Sum when 6 is removed from the observations = 130 − 6 = 124

Thus, the new mean = 124 4 = 31

The new mean is greater than the original mean.

9. Given that the mean of 5 numbers is 38, so the sum of the numbers will be:

Mean =

Sum of all observations 38= No. of observations 6 ⇒

Sum of 5 numbers of all observations 38= of observations 6 ⇒

Sum of 5 numbers = 38 × 5 = 190

Sum of 5 numbers

Given that when a number is added, the new mean reduces by 2 = 38 − 2 = 36

Sum of 6 numbers = 36 × 6 = 216

The value of added number = Sum of 6 numbers − Sum of 5 numbers = 216 − 190 = 26.

Hence, the added number is 26.

10. 2, 4, 6, 6, 2x + 1, 9, 9, 10, 11 are written in ascending order.

Number of observations = 9

Median = 7

Also, median = 1 2 n

th observation = 91 2

th observation = 5th observation = 2x + 1

Thus, 2x + 1 = 7

⇒ 2x = 7 − 1

⇒ 2x = 6

⇒ x = 6 2

⇒ x = 3

11. Sum of the lengths of all the rivers = 724 + 1376 + 800 + 2525 = 5425 km

Mean length = 5425 4 = 1356.25 km

So, the mean length of the rivers visited by the troop is 1356.25 km.

12. 72, 73, 72, 76, 72, 74, 73, 75, 76, 74, 72 are the scores of the team members of a golf team.

Mean would be the most appropriate tendency of measure to find the average score of the team members because the scores are relatively close and there are no extreme values that would change the average.

Minimum score = 72

Maximum score = 76

Range = Maximum score − Minimum score = 76 − 72 = 4

Challenge

1. Mean attendance from Monday to Saturday = 33

Total attendance from Monday to Saturday = 33 × 6 = 198

Mean attendance from Monday to Thursday = 32

Total attendance from Monday to Thursday = 32 × 4 = 128

Mean attendance from Thursday to Saturday = 34

Total attendance from Thursday to Saturday = 34 × 3 = 102

Total attendance from Monday to Thursday + Total attendance from Thursday to Saturday = 128 + 102 = 230

So, the attendance on Thursday = 230 − 198 = 32.

2. Since 47 values are between 1 and 5, it is likely that one or more of these numbers may occur frequently. If any of these values is the mode, it reflects the most typical or popular choice among the observations.

The mode is not affected by the 3 values more than 50. Therefore, central tendency mode will better represent the typical value of the data set.

Do It Yourself 3B

1. a. The bar graph consists of rectangular bars of equal width

b. The space between two consecutive bars are equal.

c. In a bar graph, the bars can be represented either horizontally or vertically.

d. The height of a bar represents the frequency of the corresponding data.

1. Yes, it is possible to draw two different bar graphs. We can do so by taking different scales. Answer may vary. One can be drawn using 1 division = 5000 toys and other can be drawn using 1 division = 10,000 toys as shown below.

Do It Yourself 3C

1. 1 cm = 5 ft.

5 cm = 5 × 5 = 25 ft.

2. a. The subject in which the studentʼs performance in the 2nd term was poorer than in the 1st term is English

b. The subjects in which the student perform the best in 1st term are Maths and Science.

Marks scored = 80

c. Marks scored in Hindi = 60

Marks scored in English = 70

Marks scored in Maths = 80

Marks scored in science = 80

Marks scored in Drawings = 70

Total marks scored in term 1 =

d. Term 1:

Total marks scored in term 1 = 60 + 70 + 80 + 80 + 70 = 360

Term 2:

Marks scored in Hindi = 62

Marks scored in English = 66

Marks scored in Maths = 96

Marks scored in science = 86

Marks scored in Drawings = 76

Total marks scored in Term

Difference between the total marks in term 1 and term 2 = 386 − 360 = 26

b. 80 > 50 > 30 > 20. So, merchandise S gives the most profit on selling.

4. a. The family ’s expenditure in March was ₹2200

b. The family spent the most in January and August.

Monthly expenditure in January = ₹2400 = Monthly expenditure in August

c. Monthly expenditure in January = ₹2400

Monthly expenditure in February = ₹1800

Monthly expenditure in March = ₹2200

Monthly expenditure in April = ₹2200

Monthly expenditure in May = ₹1400

Monthly expenditure in June = ₹2000

Monthly expenditure in July = ₹2000

Monthly expenditure in August = ₹2400

Monthly expenditure in September = ₹2000

Monthly expenditure in October = ₹2200

Monthly expenditure in November = ₹2000

Monthly expenditure in December = ₹1600

Total expenditure in a year = ₹2400 + ₹1800 + ₹2200 + ₹2200 + ₹1400 + ₹2000 + ₹2000 + ₹2400 + ₹2000 + ₹2200 + ₹2000 + ₹1600 = ₹24,200

d. Total expenditure in the first six months = ₹2400 + ₹1800 + ₹2200 + ₹2200 + ₹1400 + ₹2000 = ₹12,000

Total expenditure in last six months = ₹2000 + ₹2400 + ₹2000 + ₹2200 + ₹2000 + ₹1600 = ₹12,200

5. a.

Ratio = 12,000:12,200 = 60:61

Rainfall on different days of the

b. The highest rainfall was on Friday.

c. Rainfall on the first two days = 10 + 9 = 19 mm

Rainfall on the last two days = 12 + 18 = 30 mm

Ratio = 19:30

6. a. 400 people wear medium-sized shirts.

b. 300 people wear extra-large sized shirts.

c. Number of people who wear large-sized shirts = 350

Number of people who wear small-sized shirts = 360

Ratio: 350:360 = 35:36

d. Total number of people = 300 + 350 + 400 + 360 = 1410

7. a. Total number of cars sold by showroom A in the first three months = 60 + 75 + 75 = 210

b. Total number of cars sold by showroom B in the last three months = 60 + 75 + 75 = 210

c. Total number of cars sold by showroom A in the first four months = 60 + 75 + 75 + 90 = 300

Total number of cars sold by showroom B in the first four months = 45 + 60 + 90 + 75 = 270

300 > 270, so showroom A sold more cars.

d. Total number of cars sold by showroom A = 60 + 75 + 75 + 90 + 45 + 60 + 105 = 510

Total number of cars sold by showroom B = 45 + 60 + 90 + 75 + 60 + 75 + 75 = 480 Ratio = 510:480 = 17:16

8. Answer will vary. Sample answers.

a. How many cars in total were sold by Showroom B in the first three months?

b. What is the ratio of the number of cars sold by both the showrooms in the last 3 months?

Challenge

1. a. Income earned by company A in 2011 = ₹1,42,500

Profit earned in 2011 by company A = 35%

Profit% = Income − Expenditure (E) Expenditure (E) × 100

35 = 1,42,500 − E E × 100

⇒ 35 = 1,42,50,000 − 100E E

⇒ 35E = 1,42,50,000 − 100E

⇒ 35E + 100E = 1,42,50,000

⇒ 135E = 1,42,50,000

⇒ E = 1,42,50,000 135 = ₹1,05,555.56

b. Expenditure of company A in 2010 = ₹70,00,000

Income of company A in 2010 = ₹1,05,555.56

Income of company A in 2011 = ₹1,42,500

Total income = ₹1,05,555.56 + ₹1,42,500 = ₹2,48,055.56

Chapter Checkup

1. a. 8, 12, 16, 13, 9, 10, 15, 12

Mean = Sum of all the observations

Total number of observation

= 8 + 12 + 16 + 13 + 9 + 10 + 15 + 12 8 = 95 8 = 11.875

Ascending order = 8, 9, 10, 12, 12, 13, 15, 16

Number of terms = 8

Median = 1 2

[4th + 5th]

= 12 + 12 2 = 12

Mode = 12

n 2 +

b. 12, −15, −32, −11, −15, 14, −30, −18

Mean = Sum of all the observations

Total number of observation

= 12 + (−15) + (−32) + (−11) +

= −11.875

Ascending order = −32, −30, −18, −15, −15, −11, 12, 14

Number of terms = 8

Median = 1 2

[4th + 5th]

= 1 2 [−15 −15] = −15

Mode = − 15

2. a. The first 10 natural numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Mean = 1 + 2 + 3 + 4 + 5 + 6 +

Number of terms = 10

Median = 1 2

[5th + 6th]

= 5 + 6 2 = 11 2 = 5.5

b. The first 8 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19

Mean = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 8 = 77 8 = 9.625

Number of terms = 8

Median = 1 2

= 1 2 [4th + 5th] = 7 + 11 2 = 9

c. The factors of 36: 1, 2, 3, 4, 6, 9, 12, 18 and 36

Mean = 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 9 = 91 9 = 10.11

Number of terms = 9

Median = 

n + 1 2

th observation =

9 + 1 2

th observation = 5th observation = 6

3. a. In 2018, the production of food grains was the highest with 90 million tonnes.

b. Production of food grains in 2016 = 69 million tonnes

c. Production of food grains in 2015 = 60 million tonnes

Total production of food grains = 51 + 60 + 69 + 80 + 90 = 350 million tonnes

Ratio = 60:350 = 6:35

4. Run Scored 6 8 10 15 20 50 80 100 120

Frequency 1 1 1 3 1 1 1 1 1 15 occurs most frequently in the given data, so 15 is the mode.

Range = Highest value − Lowest value = 120 − 6 = 114

5. Arranging is ascending order: 1.5, 2.7, 13.6, 35.6, 52.2, 82.1, 96.5, 97, 108.5, 125.5

Median would be most suitable to represent the rainfall as the given data has sizable number of observations.

Range = Highest value − Lowest value = 125.5 − 1.5 = 124 mm

7. Ascending order: 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9

Marks x Number of

Mean = ∑ f × x ∑ f = 100 20 = 5

Median = 1 2

= 1 2 [10th + 11th] =

8. a. Number of people who use TV = 8 lakhs

Number of people who use Telephone = 10 lakhs

Number of people who use TV compared to telephones = 10 − 8 = 2 lakhs

b. Number of people who use computers = 9 lakhs

Number of people who use iPods = 8 lakhs

9 lakhs > 8 lakhs by 1 lakh, so only 1 lakh more people use computers than iPods

c. Number of people who use computers = 9 lakhs

Number of people who use TV = 8 lakhs

Number of people who use telephone = 10 lakhs

Number of people who use iPods = 8 lakhs

Number of people who use webcam = 10 lakhs

Total number of users = 9 lakhs + 8 lakhs + 10 lakhs + 8 lakhs + 10 lakhs = 45 lakhs

d. Answer may vary. Sample answer.

A suitable title can be ‘Number of Users of Different Electronic Appliances’. Answer may vary

9. a. Friday has the highest sale of bulbs, with 225 bulbs

b. Total number of bulbs sold = 175 + 200 + 125 + 150 + 225 = 875

875 bulbs were sold in the whole week.

c. Total number of bulbs sold = 175 + 200 + 125 + 150 + 225 = 875

Number of bulbs sold on Tuesday = 200

The fraction of bulbs sold on Tuesday = 2008 = 87535

d. Total number of bulbs sold = 175 + 200 + 125 + 150 + 225 = 875

Number of days in a week = 5

The average number of bulbs sold during the week

Name

of employees

Raj Sahil Nivira Yuvan Kabir

Salaries (in thousands) 26 24 25 22 27

Salaries of 5 employees

11.

= Total number of bulbs sold875

Total number of bulbs sold875

Number of days in a week 5 = =

Number of days in a week 5 = = 175 bulbs

10. a. NH 10 is the shortest of all.

b. 811 km is the length of the National Highway 9

c. Length of the National Highway 10 = 175 km

3 times the length of National Highway 10 will be 175 × 3 = 525 km, which is about the length of National Highway 3.

Questions may vary. Sample questions.

1. What is the total marks obtained by Kunal in all the subjects?

2. In which subject did Kunal score the highest marks?

3. What are the average marks obtained by Kunal?

12. 80, 80, 80, 80, 85, 85, 85, 90, 90, 95

So, the modal size is 80.

If the size of one shirt is misread as 80 instead of 85

80, 80, 80, 85, 85, 85, 85, 90, 90, 95

The mode will be 85.

13. Mean = 45

Number of students = 20

Sum of weights = 45 × 20 = 900 kg

Sum after Ankit and Suhani joined the class = 900 + 55 + 52 = 1007

So, the new mean weight of the class = 1007 20 + 2 = 1007 22 = 45.77 kg

14. Marks scored in 4 tests = 75, 82, 79, 76

Sum of marks scored in 4 tests = 75 + 82 + 79 + 76 = 312

Number of tests = 4

Sum of marks to get an average of 80 in 5 tests = 80 × 5 = 400

Minimum marks need to be secured to maintain the average of 80 = 400 − 312 = 88

So, Mohan needs to score at least 88 marks in his next test.

15. 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 41, 42, 43, 44 25, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 42, 43, 44, 45

Median weight =

n + 1 2

observation = 8th observation = 34 kg

Thus, the new median is 34 kg.

Challenge

1. The graph is misleading. The numbers on the y-axis starts at 19. It should always start at 0. The tabular form and correct graph would be:

2. Assertion (A): The ratio of the total sales of branch B2 for both years to the total sales of branch B4 for both years is 210:170

Sales of books of branch B2 in 2010 = 100 thousand

Sales of books of branch B2 in 2011 = 110 thousand

Total sales of branch B2 in two years = 100 + 110 = 210 thousand

Sales of books of branch B4 in 2010 = 90 thousand

Sales of books of branch B4 in 2011 = 80 thousand

Total sales of branch B2 in two years = 90 + 80 = 170 thousand

The ratio of total sales = 210:170

So, the assertion is true.

Reason (R): Ratio is the relation between two numbers which shows how much bigger one quantity is than another. So, the reason is true and it explains the assertion correctly. Thus, the correct answer is option a.

Case Study

1. Height of Nanga Parbat = 8300 m

Height of Annapurna = 8100 m

Height of Kanchenjunga = 8500 m

Height of Nanda Devi = 7900 m

Height of Mount Everest = 8800 m

Height of Aconcagua = 6900 m

8800 m > 8500 m > 8300 m > 8100 m > 7900 m > 6900 m

So, Mt Everest > Kanchenjunga > Nanga Parbat > Annapurna > Nanda Devi > Aconcagua is the required descending order. Thus, option c is correct

2. Highest peak = Mt Everest

Height of the highest peak = 8800 m

Lowest peak = Aconcagua

Height of the lowest peak = 6900 m

Ratio = 8800 m : 6900 m = 88 : 69

3. Average height of other five mountains

= Total height of the other five mountains

Number of other five mountains

= 8300 m 8000 m 8500 m 7900 m 6900 m 5 ++++

= 39600 =7920 5 m

Height of Mount Everest = 8800 m

Difference in height of Mt Everest to the average height of other mountains = 8800 − 7920 = 880 m

Sahil Raj Nivira Yuvan Kabir

4. Arrange the height of the mountains in ascending order

6900, 7900, 8000, 8300, 8500, 8800

Total observations = 6 (Even number)

Median = 3rd term 4th term 2 + 80008300 = =8150 2 +

Hence, median height = 8150 m

If a mountain of height 8600 is added to the list the new ascending order will be:

6900, 7900, 8000, 8300, 8500, 8600, 8800

Median = 4th term = 8300 m

5. Answer may vary

Chapter 4

6:12 1:5 5:25 4:5 3:9 1:2 8:10 1:4 4:16 1:3

Do It Yourself 4A

1. a. Getting a number greater than 5 on rolling a dice—Unlikely.

b. Getting heads on tossing a coin—Equally likely

c. A lion flying in the sky—Impossible

d. A child going to school on Monday—Likely

e. Picking a pencil from a box of pencils—Sure

2. a. Getting an even number—Equally Likely

b. Getting a number from 1 to 10—Sure

c. Getting the number 5 or 6—Unlikely

d. Getting the number 20—Impossible

3. a. Spinner stop at yellow—Impossible

b. Spinner stop at red—Equally Likely

c. Spinner stop at green—Unlikely

4. Likelihood of securing a goal in the 11th attempt if a team secures 5 goals out of 10 attempt is equally likely.

5. a. Factors of 40 = 1, 2, 4, 5, 8, 10, 20, 40 (8 numbers)

Number of cards = 40

Chance of drawing the cards is unlikely.

b. Even numbers up to 40 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40 (20 numbers)

Number of cards = 40

Chance of drawing even number card is equally likely.

6. Figrues may vary. Sample figure.

Challenge

1. Answers may vary. Sample answers:

a. Wearing a mask due to increasing pollution.

b. Travelling to the moon.

c. Christmas falls in the month of January.

d. The sun sets in the west.

2. a. Spinner B, as it has two sections out of six that represent “Down”, while spinner A only has one section out of six that represents “Down”.

b. If I have to move “Forward”, I would choose any of the two spinners, as each of them has 2 sections out of six that represent “Forward”.

Do It Yourself 4B

1. a. Numbers on the spinner = 1, 2, 3, 4, 5, 6, 7, 8

So, the number of possible outcomes = 8

b. Numbers greater than 3 = 4, 5, 6, 7, 8

So, favourable outcomes of spinning a number greater than 3 = 4, 5, 6, 7, 8

c. Even numbers = 2, 4, 6, 8

So, the number of times the arrow can land on even number = 4

d. Prime numbers = 2, 3, 5, 7

So, the number of times the arrow can land on a prime number = 4

2. Out of 5 choices, only 1 can be the correct answer.

Total number of outcomes = 5

Number of possible outcomes = 1

Probability = 1 5

3. Number of pieces of fruits = 12

Number of apples = 7

Number of peaches = 2

Number of fruits that are not peaches or apples = 12 − (7 + 2)

= 12 − 9 = 3

Probability of getting neither a peach nor an apple = 3 12 = 1 4

4. Number of letters of the English alphabet (Total number of outcomes) = 26

Getting a letter except Z = 25

Probability of getting a letter except Z = 25 26

5. Number of lettered tiles = 98

Number of blank tiles = 2

Total number of tiles = 100

a. Probability of getting a blank tile = 2 100 = 1 50

b. Probability of getting a lettered tile = 98 100 = 49 50

6. Total number of outcomes = 6

a. Number of reds = 2

Probability of getting reds = 2 6 = 1 3

b. Number of 1s = 1

Probability of getting a 1 = 1 6

c. Number of odd numbers = 3 (1, 3, 5)

Probability of getting an odd number = 3 6 = 1 2

d. Number of multiples of 2 = 3 (2, 4, 6)

Probability of getting an odd number = 3 6 = 1 2

e. Numbers less than 7 = 6

Probability of getting a number less than 7 = 6 6 = 1

f. Since there is no number 9 on the spinner, the probability of getting a 9 is 0 6, that is, 0.

7. Numbers on the dodecahedron = 12

a. Since there is one face with number 1 on the dodecahedron, the probability of getting a 1 = 1 12

b. Numbers less than 9 on the dodecahedron = 8 (1, 2, 3, 4, 5, 6, 7, 8)

Probability of getting a number less than 9 = 8 12 = 2 3

c. Numbers that are multiples of 3 on the dodecahedron = 4 (3, 6, 9, 12)

Probability of getting a multiple of 3 = 4 12 = 1 3

d. Numbers that are greater than 6 = 6 (7, 8, 9, 10, 11, 12)

Probability of getting a number greater than 6 = 6 12 = 1 2

8. Number of black crayons = 6

Number of blue crayons = 4

Number of red crayons = 5

Number of yellow crayons = 3

Number of white crayons = 2

Total number of crayons (Total number of outcomes) = 20

a. Probability of getting a black crayon = 6 20 = 3 10

b. Probability of getting a blue crayon = 4 20 = 1 5

c. Number of crayons except white = 6 + 4 + 5 + 3 = 18

Probability of not getting a white crayon = 18 20 = 9 10

d. There is no pink crayon, so the probability of getting a pink crayon = 0 20 = 0

e. Total number of black and blue crayons = 6 + 4 = 10

Probability of getting a black or blue crayons = 10 20 = 1 2

f. Total number of blue, red and yellow crayons = 4 + 5 + 3 = 12

Probability of getting a blue, red and yellow crayons = 12 20 = 3 5

9. a. The given word has 13 letters, of which 5 are letters E. The probability of picking the letter E is 5 . 13 So the statement is false.

b. The given word has 13 letters, of which 2 are letters F, and 2 are letters C. Hence, the probability of picking the letters F and C = 2 13 So, the statement is true.

c. The letters R, V and N have appeared one, thus, the probability of picking these letters = 1 . 13 Hence, the statement is false.

d. The given word has 13 letters, of which 1 is letter R. The probability of picking the letter R is 1 . 13 So, the statement is true.

Hence, statements b and d are true.

10. Numbers on the slips of paper = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25

Total numbers written on the slips (Total number of outcomes) = 25

a. Odd numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25

Number of odd numbers = 13

Probability of getting an odd number = 13 25

b. Since there are no 3-digit numbers, probability of getting a 3-digit number = 0 25 = 0

c. Numbers other than 4 = 1, 2, 3, 5, 6, 7,…, 25

Total numbers other than 4 = 24

Probability of getting a number that is not 4 = 24 25

d. Since all are positive numbers, so the probability of getting a positive number = 25 25 = 1

11. Answers may vary. Sample answer:

a. P(prime number). Theoretical Probability = 9 25

b. P(number greater than 19). Theoretical probability = 6 25

Challenge

1. Integers from 1 to 10 = 10

Total number of integers from 1 to 10 = 10

Number of integers less than 3 = 2 (1, 2)

Probability of getting a number less than 3 = 2 10 = 1 5

Amit’s error: He included number 3 also.

Do It Yourself 4C

1. Total number of spins = 14 + 7 + 11 + 8 = 40

Number of times spinning a letter A = 14

Probability of spinning a letter A = 14 40 = 7 20

Number of times spinning a letter B = 7

Probability of spinning a letter B = 7 40

Number of times spinning a letter C = 11

Probability of spinning a letter C = 11 40

Number of times spinning a letter D = 8

Probability of spinning a letter D = 8 40 = 1 5

2. Total number of times Malati records a colour = 12 + 10 + 15 + 13 = 50

Number of times she gets a red = 12

Probability of getting a red = 12 50 = 6 25

Number of times she gets a blue = 10

Probability of getting a blue = 10 50 = 1 5

Number of times she gets a green = 15

Probability of getting a green = 15 50 = 3 10

Number of times she gets a yellow = 13

Probability of getting a yellow = 13 50

3. Total number of times a coin is tossed = 40 + 60 = 100

Number of heads = 40

Number of tails = 60

Probability of getting a head = 40 100 = 2 5

Mala’s error: Mala has found the probability of getting heads over the number of times she gets a tail whereas she must have found the probability over the total number of times a coin is tossed.

4. Number of defective pieces = 5

Total number of devices = 100

Number of selected devices = 1

Probability of selected devices that are not defective = 95 100 = 19 20

5. Total number of spins = 8 + 6 + 9 + 11 + 9 + 7 = 50

a. Number of times spinning a 6 = 7

Probability of spinning a 6 = 7 50

b. Number of times spinning an even number (2, 4, 6) = 6 + 11 + 7 = 24

Probability of spinning an even number = 24 50 = 12 25

c. Number of times spinning an odd number (1, 3, 5) = 8 + 9 + 9 = 26

Probability of spinning an odd number = 26 50 = 13 25

d. Number of times spinning a number less than 3 (1, 2) = 8 + 6 = 14

Probability of spinning a number less than 3 = 14 50 = 7 25

e. Since there is no number 7 on the spinner, so the probability of spinning a 7 is 0 50 = 0.

f. Number of times not spinning a 1 (spinning 2, 3, 4, 5, 6)

= 6 + 9 + 11 + 9 + 7 = 42

Probability of not spinning a 1 = 42 50 = 21 25

6. Total number of students selected their sports = 30 + 22 + 34

+ 20 + 14 = 120

a. Number of students selected cricket = 30

Probability of student selecting cricket = 30 120 = 1 4

b. Number of students selected hockey = 20

Probability of student selecting hockey = 20 120 = 1 6

c. Number of students selected baseball = 22

Probability of student selecting baseball = 22 120 = 11 60

d. Number of students selected football = 34

Probability of student selecting football = 34 120 = 17 60

7. Total number of times a spinner is spun = 41 + 43 + 39 + 37 + 40 = 200

a. Theoretical probability of spinning 4 = 1 5 = 40 200

Number of times 4 is spun = 37

Experimental probability of spinning a 4 = 37 200

As, 40 200 > 37 200,

Theoretical probability > Experimental probability

b. Theoretical probability of spinning 3 = 1 5 = 40 200

Number of times 3 is spun = 39

Experimental probability of spinning a 3 = 39 200

As, 40 200 > 39 200,

Theoretical probability > Experimental probability

c. Only number 5 is greater than number 4 on the spinner

Theoretical probability of spinning a number greater than 4 = 1 5 = 40 200

Number of times a number greater than 4 (5) is spun = 40

Experimental probability of spinning a number greater than 4 = 40 200

As, 40 200 = 40 200,

Theoretical probability = Experimental probability

d. Number of odd numbers (1, 3, 5) on the spinner = 3

Theoretical probability of spinning an odd number = 3 5 = 120 200

Number of times an odd number (1, 3, 5) is spun = 41 + 39 + 40 = 120

Probability of spinning an odd number = 120 200

As, 120 200 = 120 200,

Theoretical probability = Experimental probability

8. Answer may vary. Sample answer:

A coin was tossed 20 times. Heads turned up 10 times on the coin. What is the theoretical and experimental probability of the coin turning up heads?

Challenge

1. Number of students randomly selected = 75

Number of students plan to go to college after completing high school = 60

Experimental probability of selecting a student who plans to go to college after completing high school = 60 75

In 700 students, the number of students who plan to go to college after completing high school would be, 700 × 60 75 = 560 students

So, 560 students are likely to go to college after completing high school.

Chapter Checkup

1. a. Impossible b. Sure

2. a. January comes before February is a sure event.

b. The sun setting in east is an impossible event.

c. A child going to school on a Wednesday is a likely event.

3. A die has numbers 1, 2, 3, 4, 5 and 6.

a. An odd number: 3 out of 6 numbers on the dice are odd numbers. So, it is equally likely to pick an odd number on rolling the dice.

b. 6: 1 out of 6 numbers is 6. So, it is unlikely to pick 6 on rolling the dice.

c. A number greater than 6: There are no numbers greater than 6. So, it is impossible to pick a number greater than 6 on rolling the dice.

d. Zero: The number zero does not exist on the dice. So, it is impossible to pick a zero on rolling the dice.

e. A number less than 10: All the numbers on the dice are less than 10. So, it is sure to pick a number less than 10 on rolling the dice.

4. Number of fiction books = 3000

Number of non-fiction books = 4000

Total number of books = 3000 + 4000 = 7000

Probability of selecting a fiction book = 3000 7000 = 3 7

5. Possible outcomes when a dice is rolled = 1, 2, 3, 4, 5, 6

Number of possible outcomes = 6

a. Possible outcomes when a number less than 3 is rolled = 1, 2

b. Possible outcomes when a number less than 3 is not rolled = 3, 4, 5, 6

c. Possible outcomes when an even number is rolled = 2, 4, 6

d. Possible outcomes when an even number is not rolled = 1, 3, 5

e. Possible outcomes when a 5 is not rolled = 1, 2, 3, 4, 6

f. Possible outcomes when a 7 is not rolled = 1, 2, 3, 4, 5, 6

a. The favourable outcomes for the event of spinning a 1 = 1, 1.

So, the number of ways the event of spinning a 1 = 2.

b. The favourable outcomes for the event of spinning a 3 = 3, 3, 3

So, the number of ways the event of spinning a 3 = 3.

c. The favourable outcomes for the event of spinning an odd number = 1, 1, 3, 3, 3.

So, the number of ways the event of spinning an odd number = 5.

d. The favourable outcomes for the event of spinning an even number = 2, 2, 2.

So, the number of ways the event of spinning an even number = 3.

e. The favourable outcomes for the event of spinning a number greater than 0 = 1, 1, 2, 2, 2, 3, 3, 3.

So, the number of ways the event of spinning a number greater than 0 = 8

f. The favourable outcomes for the event of spinning a number less than 3 = 1, 1, 2, 2, 2.

So, the number of ways the event of spinning a number less than 3 = 5.

7. Total number of outcomes = 8

a. Number of blues = 2

The probability of getting blues = 2 8 = 1 4

b. Number of 1s = 3

The probability of getting a 1 = 3 8

c. Number of even numbers (2, 4, 6, 8, 10, 2) = 5 8

d. Number of 4s = 1

The probability of getting 4 = 1 8

8. Number of discs that are red = 5

Number of discs that are yellow = 3

Number of discs that are black = 2

Total number of discs = 10

a. Probability of drawing red disc = 5 10 = 1 2

b. Probability of drawing black disc = 2 10 = 1 5

c. Probability of drawing red or black disc = 5 + 2 10 = 7 10

d. Number of drawing discs that are not black = Number of drawing discs that are red or yellow = 5 + 3 10 = 8 10 = 4 5

e. Since there are no blue discs, the probability of drawing a blue disc = 0 10 = 0

f. Since it can be any of the given card red, yellow or black, the probability of drawing the card = 10 10 = 1

9. Number of the letters of the English alphabet = 26

a. Probability of picking a letter X = 1 26

b. Number of vowels = 5 (a, e, i, o, u)

Probability of picking a vowel = 5 26

c. Probability of picking a letter M or N = 2 26 = 1 13

d. Number of unique letters in the word MATHEMATICS

= (M, A, T, H, E, I, C, S) = 8

Probability of getting a letter in the word MATHEMATICS = 8 26 = 4 13

10. Total number of students = 6 + 8 + 10 + 6 = 30

Number of students planting trees on Wednesday = 6

The probability that a randomly picked student planted a tree on Wednesday = 61 305 =

11. Total numbers on the cards = 25

a. Odd numbers from 1 to 25 = 1,3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25

Number of odd numbers = 13

Probability of getting an odd number = 13 25

b. Multiples of 5 from 1 to 25 = 5, 10, 15, 20, 25

Number of multiples of 5 = 5

Probability of getting a multiple of 5 = 5 25 = 1 5

c. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Number of factors of 24 = 8

Probability of getting a factor of 24 = 8 25

d. Numbers that contain digit 2 = 2, 12, 20, 21, 22, 23, 24, 25

Number of numbers that contain digit 2 = 8

Probability of getting a number that contain digit 2 = 8 25

12. Total number of times a tile is picked from the bag = 100

a. Number of times 2 is picked = 18

Probability of getting a 2 = 18 100 = 9 50

b. Number of times, an even number is picked (2, 4) = 18 + 21 = 39

Probability of getting an even number = 39 100

c. Number of times, 5 is not picked (1, 2, 3, 4) = 20 + 18 + 22 + 21 = 81

Probability of not getting a number 5 = 81 100

d. Number of times, a number less than 3 is picked (1, 2) = 20 + 18 = 38

Probability of getting a number less than 3 = 38 100 = 19 50

13. Probability of an event = 1 5

Number of trials = 20

Number of times the probability is expected to occur in 20 trials = 1 204 5 ×=

Challenge

1. Assertion (A): If a dice is thrown, the probability of getting a number greater than 4 and less than 5 is zero.

Reason (R): Probability of an impossible event is zero

The possible outcomes of throwing a dice are {1, 2, 3, 4, 5, 6}. In the given sample space, no number is greater than 4 and less than 5 at the same time. Hence, the probability of getting the desired outcome is impossible

Also, the probability of an impossible event = 0 a. Both A and R are true, and R is the correct explanation of A.

2. Statement 1: There are 8 red cards.

Statement 2: The probability that a randomly selected student has a green card is 2 5

Statement (1):

This statement provides the total number of red cards. Since the total number of blue cards is not given, we cannot determine the probability. This statement alone is not sufficient.

Statement (2):

This statement tells us that the probability that a randomly selected student has a green card is 2 5 As we know that the total number of students in the class is 30, we can find the number of green cards by finding the equivalent of 2 5 with base 30.

4. Number of green beads = 7

Number of black beads = 5

Number of green and black beads = 7 + 5 = 12

Total number of beads = 5 + 7 + 8 + 7 = 27

Number of beads neither green nor black = 27 − 12 = 15

P(neither green nor black beads) = 15 27 = 5 9

Chapter 5 Let’s Warm-up

Statement Expression

1. 3 more than twice a number 2n − 5

2. 2 less than five times a number

3. 2 more than thrice a number

4. 5 less than twice a number

5. 5 less than thrice a number

Do It Yourself 5A

1. An equation is a mathematical statement that shows the equality of two expressions, using the “=” sign. Hence, a and c are equations.

a. 5x + 5 = 35 is an equation.

b. 3a – 7 + 24 is not an equation.

c. 5x 2 + 3 2 = 18 5 is an equation.

2. Linear equation in one variable contains only one variable with the highest exponent as 1.

a. x + 9 = 14  b. x + 12 = 4  c. 4x + 5y = 8 

d. 7x + 3 = 10  e. x2 + 7 = 29  f. 6x + 8y + 3z = 38 

3. The students will draw 3 red balls in the RHS.

Therefore, 22612

55630 × == ×

Hence, there are 12 green cards. But with this we cannot find the asked probability, and hence this statement is also not sufficient.

Combining statements 1 and 2, number of blue cards = 30 − (Number of red cards + number of green cards) = 30 − (8 + 12) = 10

So, P(selecting student with blue card) = 101 303 =

Hence, option c. Statement 1 and 2 together are sufficient to answer.

Case Study

1. Number of red beads = 8

Total number of beads = 5 + 7 + 8 + 7 = 27

P(red bead) = 8 27

Hence option c is correct.

2. As there are no blue beads left, the impossible event is drawing a blue bead.

Hence, option c is correct.

3. Number of red beads = 8

Number of black beads = 5

Number of red or black beads = 8 + 5 = 13

Total number of beads = 5 + 7 + 8 + 7 = 27

P(red or black beads) = 13 27

2 = 5

4. a. Let the number be x Half of the number = x 2

So, the required equation is x + x 2 = 9.

b. Let the number be x Thrice the number = 3x

So, the required equation is 3x + 5 = 17.

c. Let the number be x

Half of the number = x 2

So, the required equation is x 2 = 15.

d. Let the number be x. 32 times the number = 32x

Sum of 32 times the number and 5 = 32x + 5

Half of the sum of 32 times the number and 5 = 32x + 5 2

So, the required equation is 32x + 5 2 = 27.

e. Let the number of frogs be x.

Two more than the number of frogs = x + 2

So, the required equation is x + 2 = 8.

f. Let the number of students be x

The number of students increased by 10 = x + 10

So, the required equation is x + 10 = 38.

30 = 30

Hence, the given value is the solution of the given equation.

f. 7x − 21 = 1 (x = 3)

7 × 3 – 21 = 21 – 21 = 0

0 ≠ 1

Hence, the given value is not the solution of the given equation.

8. Statements may vary. Sample statements.

a. 9 subtracted from y gives 81.

b. 5 subtracted from 4 times q gives 15.

5. a. One-tenth of a number gives 7. 10 – x = 7

b. 10 subtracted from a number gives 7. x + 27 = 77

c. A number subtracted from 10 gives 7. 1 10 x = 7

d. The sum of a number and 27 is 77. x – 10 = 7 Statement Expression

6. a. Candies with me = m

Candies with Lina = m – 3

Candies with me and Lina = 18

So, the required equation is m + m – 3 = 18 or 2m – 3 = 18.

b. Age of Tina = c years

Age of Tina’s sister = (c + 3) years

But Tina’s sister is 12 years old.

So, the required equation that represents the given situation is c + 3 = 12.

c. Let the age of Rakesh be x years.

Age of Rajat = (2x) years

Sum of their ages = x + 2x or 3x

But the sum of their ages is 18 years.

So, the required equation that represents the given condition is x + 2x = 18 or 3x = 18.

d. Let the breadth of the rectangle be x cm.

The length of the rectangle = (x + 3) cm

Perimeter of the rectangle = 2(x + x + 3) = 2(2x + 3)

But the perimeter of the rectangle = 18 cm

So, the required equation that represents the given condition is 2(2x + 3) = 18.

7. a. x − 5 = 11 (x = 7)

7 – 5 = 2

2 ≠ 11

Hence, the given value is not the solution of the given equation.

b. 2x + 4 = 10 (x = 6)

2 × 6 + 4 = 12 + 4 = 16

16 ≠ 10

Hence, the given value is not the solution of the given equation.

c. 4x − 3 = 17 (x = 5)

4 × 5 – 3 = 20 – 3 = 17

17 = 17

Hence, the given value is the solution of the given equation.

d. 8 − 2x = 6 (x = 2)

8 − 2 × 2 = 8 − 4 = 4

4 ≠ 6

Hence, the given value is not the solution of the given equation.

e. 5x + 5 = 30 (x = 5)

5 × 5 + 5 = 25 + 5 = 30

c. 30 times z equals 1650.

d. The sum of 9 times x and 5 gives 50.

e. 12 added to 4 times x is 28.

f. 2 added to one-fourth of y equals 8.

Challenge

1. Flat charges of Company A = $50

Per mile charge of Company A = $2

Amount need to pay to travel 10 miles = $2 × 10 + $50 = $70

Flat charges of Company B = $30

Per mile charge of Company B = $3

Amount need to pay to travel 10 miles = $3 × 10 + $30 = $60

Hence, Company B rental car will be cheaper to travel a distance of 10 miles.

Do It Yourself 5B

1. a. We shall first add 7 to both sides of the equation.

x – 7 + 7 = 0 + 7

On simplifying both sides of the equation, we get, x = 7

b. We shall first subtract 9 from both sides of the equation.

x + 9 − 9 = 0 – 9

On simplifying both sides of the equation, we get, x = −9

c. We shall first add 2 to both sides of the equation.

x – 2 + 2 = 6 + 2

On simplifying both sides of the equation, we get, x = 8

d. We shall first subtract 5 from both sides of the equation.

x + 5 − 5 = 3 – 5

On simplifying both sides of the equation, we get, x = −2

e. We shall first add 8 to both sides of the equation.

y – 8 + 8 = −4 + 8

On simplifying both sides of the equation, we get, y = 4

f. We shall divide both sides of the equation by 4.

4a 4 = 60 4

On simplifying both sides of the equation, we get, a = 15

g. We shall multiply both sides of the equation by 3.

b 3 × 3 = 9 × 3

On simplifying both sides of the equation, we get, b = 27

h. z 7 = 6 5

We shall multiply both sides of the equation by 7.

z 7 × 7 = 6 5 × 7

On simplifying both sides of the equation, we get,

z = 42 5 = 8 2 5

2. a. –x + 14 = 21 –x = 21 – 14 ⇒ x = –7

Verification: Substituting x = –7 in the given equation

x + 14 = 21, we get

−(−7) + 14 = 7 + 14 = 21

21 = 21

Hence, the solution is verified for its correctness.

b. 2x + 12 = 64

2x = 64 − 12 ⇒ 2x = 52

⇒ x = 26

Verification: Substituting x = 26 in the given equation

2x + 12 = 64, we get

2 × 26 + 12 = 52 + 12 = 64 64 = 64

Hence, the solution is verified for its correctness.

c. 4z − 6 = 3

4z = 3 + 6 ⇒ 4z = 9

z = 9 4

Verification: Substituting x = 9 4 in the given equation

4z − 6 = 3, we get

4 × = 9 4 − 6 = 9 – 6 = 3

3 = 3

Hence, the solution is verified for its correctness.

d. 30p 7 = 60

30p = 420 ⇒ p = 420 30 = 14

Verification: Substituting p = 14 in the given equation

30p

7 = 60, we get

30 × 14

7 = 60

60 = 60

Hence, the solution is verified for its correctness.

e. 5 + x 2 = 6

10 + x 2 = 6

10 + x = 12

x = 2

Verification: Substituting x = 2 in the given equation

56 2 x += = 6, we get

2 5+=5+1=6 2

6 = 6

Hence, the solution is verified for its correctness.

f. 1 1=4

2 x –⇒ 2 =42=8

2 x x ––

x = 8 + 2 = 10

Verification: Substituting x = 10 in the given equation

1 1=4

2 x – , we get

1 2 × 10 1 = 5 1 = 4

4 = 4

Hence, the solution is verified for its correctness.

g. 13 +12=25 4 y 13+48 =25 4 y

13y = 48 = 100

13y = 48

y = 4

Verification: Substituting y = 4 in the given equation 13 12=25 4 y + , we get

13×4 +12=13+12=25 4

25 = 25

Hence, the solution is verified for its correctness.

h. 15 10=30 3 a –1530 =30 3 a –

15a − 30 = 90 15a = 120 120 ==8 15 a

Verification: Substituting a = 8 in the given equation 15 10=30 3 a – , we get

15×8 120 10=10=4010=30 33

30 = 30

Hence, the solution is verified for its correctness

3. The answer is incorrect.

The given equation is 12 + x = 20.

We shall subtract 12 from both sides of the equation.

12 + x – 12 = 20 – 12

On simplifying both sides of the equation, we get, x = 8

So, the correct answer is 8.

4. Answers may vary. Sample answers:

a. Veena had saved some money in her money bank. She spent ₹75 on buying stationery items. If she is now left with ₹150, how much money did she have to begin with?

x – 75 = 150

Add 75 on both sides of the equation, x – 75 + 75 = 150 + 75

x = 225

b. Ankur has 249 postcards. He added new postcards to his collection. If he has 350 postcards in all, how many new postcards did he add to his collection?

249 + x = 350

Subtract 249 from both the sides of the equation, 249 + x – 249 = 350 – 249

x = 101

c. The age of Rima’s grandmother is 9 times Rima’s age. If the age of Rima’s grandmother is 72, what is the age of Rima?

9x = 72

Divide both sides with 9, 9x 9 = 72 9

x = 8

Challenge

1. We shall either multiply both sides of equation 1 by 6 or divide both sides of equation 2 by 6 to make both the equations equal.

Equation 1: 1 –5 26 x x  =+  1 665 36 x x 

6x − 2 = x + 30

Equation 2: 6x – 2 = x + 30

62+30 = 66 xx –

6230 ++ 6666 xx –1 =+5 36 x x –

Do It Yourself 5C

1. a. x + 4 = 10

On transposing 4 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 10 – 4 x = 6

b. a + 9 = 13

On transposing 9 from the LHS to the RHS, the operation changes from addition to subtraction.

a = 13 – 9 a = 4

c. x + 2 7 = 5 7

On transposing 2 7 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 5 7 –2 7 x = 3 7

d. n + 2.6 = 4

On transposing 2.6 from the LHS to the RHS, the operation changes from addition to subtraction.

n = 4 – 2.6 n = 1.4

e. x − 9 = 3

On transposing −9 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 3 + 9 x = 12

f. x − 5.4 = 12.8

On transposing −5.4 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 12.8 + 5.4 x = 18.2

g. −14 + a = 27

On transposing −14 from the LHS to the RHS, the operation changes from subtraction to addition.

a = 27 + 14 a = 41

h. p − 5 9 = 7 9

On transposing −5 9 from the LHS to the RHS, the operation changes from subtraction to addition.

p = 7 9 + 5 9 p = 12 9 = 4 3

i. x − 6.5 = 15

On transposing −6.5 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 15 + 6.5 x = 21.5

2. a. 4y = 20

On transposing 4 from the LHS to the RHS, the operation changes from multiplication to division.

y = 20 4 y = 5

b. 90 = 15y

On transposing 15 from the RHS to the LHS, the operation changes from multiplication to division.

90 15 = y 6 = y or y = 6

c. −9a = 108

On transposing −9 from the LHS to the RHS, the operation changes from multiplication to division.

a = 108 9 108 9 a = −12

d. −7m = −21

On transposing −7 from the LHS to the RHS, the operation changes from multiplication to division.

m = 21 7 m = 3

e. j 8 = 6

On transposing 8 from the LHS to the RHS, the operation changes from division to multiplication.

j = 6 × 8 j = 48

f. x 3 = −11

On transposing 3 from the LHS to the RHS, the operation changes from division to multiplication.

x = −11 × 3 x = −33

g. w −9 = −5

On transposing −9 from the LHS to the RHS, the operation changes from division to multiplication.

h. g

3.2 = 20

w = −5 × (−9) w = 45

On transposing 3.2 from the LHS to the RHS, the operation changes from division to multiplication.

g = 20 × 3.2 g = 64

i. 2x 4.5 = 30

On transposing 4.5 from the LHS to the RHS, the operation changes from division to multiplication.

2x = 30 × 4.5 = 135

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 135 2 = 67.5

3. a. 6x + 14 = 16

On transposing 14 from the LHS to the RHS, the operation changes from addition to subtraction.

6x = 16 – 14 6x = 2

On transposing 6 from the LHS to the RHS, the operation changes from multiplication to division.

x = 2 6 = 1 3

b. 7 + 5x – 3x = 21

On simplifying the LHS, we get, 7 + 2x = 21

On transposing 7 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 21 – 7 2x = 14

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 14 2 x = 7

c. 1.2x + 4.8x = 18

On simplifying the LHS, we get, 6x = 18

On transposing 6 from the LHS to the RHS, the operation changes from multiplication to division.

x = 18 6 x = 3

d. 6(5x – 2) + 2 = 30

On opening the bracket, we get,

30x – 12 + 2 = 30 30x – 10 = 30

On transposing 10 from the LHS to the RHS, the operation changes from subtraction to addition.

30x = 30 + 10 30x = 40

On transposing 30 from the LHS to the RHS, the operation changes from multiplication to division.

x = 40 30 x = 4 3 or x = 1 1 3

e. 7(x − 4) + 13 = 6

On opening the bracket, we get,

7x − 28 + 13 = 6 7x − 15 = 6

On transposing −15 from the LHS to the RHS, the operation changes from subtraction to addition.

7x = 6 + 15 7x = 21

On transposing 7 from the LHS to the RHS, the operation changes from multiplication to division.

x = 21 7 x = 3

f. 10y − 0.2(7 − y) = 19

On opening the bracket, we get,

10y − 1.4 + 0.2y = 19 10.2y − 1.4 = 19

On transposing −1.4 from the LHS to the RHS, the operation changes from subtraction to addition.

10.2y = 19 + 1.4 10.2y = 20.4

On transposing 10.2 from the LHS to the RHS, the operation changes from multiplication to division.

y = 20.4 10.2 y = 2

g. () 532+5=5+15 xx –

1510+5=5+15 xx –

155=5+15 xx –

155=15+5 xx –⇒ 10=20=2 xx

h. (10)(53) = 34 xx

4(x – 10) = 3(5 – 3x)

4x − 40 = 15 − 9x

4x − 9x = 15 + 40

13x = 55 x = 55 13

i. 5x + 12 = 0.2(10x – 15)

5x + 12 = 2x − 3

5x − 2x = −3 −12

3x = −15 x = −5

j. 3(x – 6) + 4(x + 4) = 5(x – 2)

3x – 18 + 4x + 16 = 5x – 10

7x – 2 = 5x – 10

7x – 5x = –10 + 2

2x = –8 x = –4

k. ()() 123 31–25= 232 xx++

123 6×(31)–6×(25)=6× 232 xx++

3(3x + 1) −4(2x + 5) = 9

9x + 3 − 8x − 20 = 9

x – 17 = 19 x = 26

l. 13(5–2)+4=(–9) 5 10 xx

13(5–2)–(–9)–4 510 xx = 13

10×(5–2)–10×(–9)=–4×10 5 10 xx

2(5x − 2) − 3(x − 9) = −40

10x − 4 − 3x + 27 = −40

7x + 23 = −40

7x = –40 – 23 = –63

x = –9

4. Answer may vary. Sample answer.

x + 6 = 9; x − 2 = −5; 2x = −6

Challenge

1. a. Let the number be x

The equation that represents the given situation is x − 60 = 4.

On transposing 60 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 4 + 60 x = 64

b. Let the unknown number be x

The equation that represents the given condition is 13 + x = 46.

On transposing 13 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 46 − 13 x = 33

c. Let the number be x

The equation that represents the given condition is x − 21 = 81.

On transposing 21 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 81 + 21 x = 102

d. Let the number be x.

The equation that represents the given condition is 7 × x = 56.

On transposing 7 from the LHS to the RHS, the operation changes from multiplication to division.

x = 56 7 x = 8

e. Let the number be x

The equation that represents the given condition is x 7 = 5.

On transposing 7 from the LHS to the RHS, the operation changes from division to multiplication.

x = 5 × 7 x = 35

f. Let the number be x

The equation that represents the given condition is 3x − 10 = 35.

On transposing −10 from the LHS to the RHS, the operation changes from subtraction to addition.

3x = 35 + 10 3x = 45

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 45 3 x = 15

g. Let the number be x

The equation that represents the given condition is 3x + 5 = 20.

On transposing 5 from the LHS to the RHS, the operation changes from addition to subtraction.

3x = 20 − 5 3x = 15

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 15 3 x = 5

Do It Yourself 5D

1. Let the number be x; twice the number = 2x

Twice the number decreased by 22 = 2x − 22

According to the given condition, 2x − 22 = 48

On transposing 22 from the LHS to the RHS, the operation changes from subtraction to addition.

2x = 48 + 22 2x = 70

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 70 2 x = 35

So, the required number is 35.

2. Let the number be x; half of the number = x 2

According to the given condition, x + x 2 = 72

3x 2 = 72

On transposing 2 from the LHS to the RHS, the operation changes from division to multiplication.

3x = 72 × 2

3x = 144

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 144 3 x = 48

So, the required number is 48.

3. Let one of the numbers be x; the other number will be (x + 3)

As per the given condition, x + x + 3 = 95

2x + 3 = 95

On transposing 3 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 95 − 3

2x = 92

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 92 2

x = 46

So, one of the numbers is 46.

The other number = x + 3 = 46 + 3 = 49

4. Let one of the parts be x; the other part = x + 8

According to the given condition, x + x + 8 = 20

2x + 8 = 20

On transposing 8 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 20 − 8

2x = 12

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 12 2

x = 6

So, one of the parts is 6.

The other part = x + 8 = 6 + 8 = 14

5. Let the three consecutive numbers be (x − 1), x, and (x + 1).

As per the given condition, x + (x + 1) – (x – 1) = 24

x + x + 1 – x + 1 = 24

x + 2 = 24

On transposing 2 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 24 – 2 x = 22

One of the numbers is 22.

(x − 1) = 22 – 1 = 21

(x + 1) = 22 + 1 = 23

So, the three consecutive numbers are 21, 22, and 23.

6. The sum of the measures of the angles of a triangle is 180°. Hence, x + (x + 40)° + (x – 10)° = 180°

3x + 30° = 180°

On transposing 30 from the LHS to the RHS, the operation changes from addition to subtraction.

3x = 180° – 30°

3x = 150°

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 150° 3 x = 50°

(x + 40)° = 50° + 40° = 90°

(x – 10)° = 50° – 10° = 40°

So, the angles of a triangle are 50°, 90° and 40°.

7. Perimeter of a triangle = Sum of the measures of the three sides

Hence, 2x + 3 + 2x + 5 + 3x = 148

7x + 8 = 148

On transposing 8 from the LHS to the RHS, the operation changes from addition to subtraction.

7x = 148 – 8

7x = 140

On transposing 7 from the LHS to the RHS, the operation changes from multiplication to division.

x = 140 7

x = 20 units

2x + 3 = 2 × 20 + 3 = 43 units; 2x + 5 = 2 × 20 + 5 = 45 units and 3x = 3 × 20 = 60 units

So, the measures of the three sides of the triangle are 43 units, 45 units and 60 units, respectively.

8. Let the age of Olivia be x; Age of Olivia’s mother = 3x + 5

As per the given situation, 3x + 5 = 44

On transposing 5 from the LHS to the RHS, the operation changes from addition to subtraction.

3x = 44 – 5

3x = 39

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 39 3 x = 13

So, the age of Olivia is 13 years.

9. Let the number of magazines be x

The number of books = 3x

As per the given situation, x + 3x = 320

4x = 320

On transposing 4 from the LHS to the RHS, the operation changes from multiplication to division.

x = 320 4 x = 80

Number of books = 3x = 3 × 80 = 240

10. Let the cost of a pen be x; cost of a book = 50 + x

As per the given situation, 5 × (50 + x) + 3x = 730

250 + 5x + 3x = 730 250 + 8x = 730

On transposing 250 from the LHS to the RHS, the operation changes from addition to subtraction.

8x = 730 − 250 8x = 480

On transposing 8 from the LHS to the RHS, the operation changes from multiplication to division.

x = 480 8 x = 60

So, the cost of the pen is ₹60.

Cost of the book = 50 + x = 50 + 60 = ₹110

11. Let the number of stamps with Rakesh be x

Number of stamps with Sameer = 3x + 20

Total number of stamps with Rakesh and Sameer = x + 3x + 20

But the total number of stamps with Rakesh and Sameer is 140.

So, the required equation is x + 3x + 20 = 140

On simplifying the LHS of the equation, we get, 4x + 20 = 140

We shall subtract 20 from both sides of the equation.

4x + 20 – 20 = 140 – 20

On simplifying both sides of the equation, we get, 4x = 120

We shall now divide both sides of the equation by 4.

4x 4 = 120 4 x = 30

Hence, Rakesh has 30 stamps.

12. Let the weight of one of the bags full of onions be x

Weight of the other bag = 5 + x

As per the given situation, x + 5 + x = 31

2x + 5 = 31

On transposing 5 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 31 − 5

2x = 26

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division. x = 26 2 x = 13

So, the weight of one of the bags full of onions is 13 kg.

Weight of the other bag = 5 + x = 5 + 13 = 18 kg

13. Let the present age of Satish be x. Five years from now, the age of Satish will be (x + 5).

As per the given situation, (x + 5) = 1 1 2 x (x + 5) = 3 2 x

On transposing 2 from the RHS to the LHS, the operation changes from division to multiplication.

2 × (x + 5) = 3x

2x + 10 = 3x

On transposing 2x from the LHS to the RHS, the operation changes from addition to subtraction.

10 = 3x – 2x 10 = x

Hence, the present age of Satish is 10 years.

14. Let the cost of the bat be x.

Cost of the ball = 1 5 x

As per the given situation, x + 1 5 x = 1200

6x 5 = 1200

On transposing 5 from the LHS to the RHS, the operation changes from division to multiplication.

6x = 1200 × 5

6x = 6000

On transposing 6 from the LHS to the RHS, the operation changes from multiplication to division.

x = 6000 6 x = 1000

So, the cost of the bat is ₹1000.

Cost of the ball = 1 5 x = 1 5 × 1000 = 200

So, the cost of the ball is ₹200.

15. Let the present age of Heena be x years; present age of her mother = 4x 20 years hence,

Age of Heena will be (x + 20) years.

Age of Heena’s mother will be (4x + 20) years.

As per the given situation, 4x + 20 = 2(x + 20)

4x + 20 = 2x + 40

On transposing 2x from the RHS to the LHS, the operation changes from addition to subtraction.

4x − 2x + 20 = 40

2x + 20 = 40

On transposing 20 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 40 − 20

2x = 20

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 20 2

x = 10

Hence, the present age of Heena is 10 years.

The present age of Heena's mother = 4x = 4 × 10 = 40 years

16. Let the number of cups of water held by a small jug be x

The number of cups of water held by a large jug = x + 2

As per the given condition, 2x + x + 2 = 8

On simplifying the LHS of the equation, we get, 3x + 2 = 8

We shall first subtract 2 from both sides of the equation.

3x + 2 – 2 = 8 – 2

On simplifying both sides of the equation, we get, 3x = 6

We shall now divide both sides of the equation by 3.

3x 3 = 6 3

So, x = 2.

Hence, the number of cups of water held by a small jug = 2 cups

The number of cups of water held by a large jug = 2 + 2 = 4 cups.

17. Let the number of ₹20 notes be x

Number of ₹10 notes = 3x

Given that the total value of money = ₹250

20 × x + 10 × 3x = 250

20x + 30x = 250

50x = 250

x = 250 50 = 5

Hence, Nisha has five rupees 20 notes, and 5 × 3 = 15 rupees 10 notes.

18. Answers may vary. Sample answer.

A bicycle rental shop charges a flat fee of ₹100 to rent a bike plus ₹30 per hour for the duration of the rental. Write a linear equation for the total cost of renting a bike for x hours. Find the total cost given by a person to rent the bicycle for 4 hours.

Challenge

1. Let the age of Tina be x years.

Since the difference between the ages of Savita and Tina is 4 years.

Therefore, age of Savita = x + 4

As per the given situation, x + x + 4 = 18

2x + 4 = 18

On transposing 4 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 18 − 4

2x = 14

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 14 2

x = 7

Hence, the age of Tina is 7 years.

Age of Savita = x + 4 = 7 + 4 = 11 years

Chapter Checkup

1. a. 11

2 x = 150

b. 5x = 65

c. 5p − 3 = 22

d. 2x + 36 = 48

e. 4p + 9 = 31

2. a. x – 3 = –20

We shall add 3 to both sides of the equation.

x – 3 + 3 = –20 + 3

x = –17

b. x + 5 = 8

We shall subtract 5 from both sides of the equation.

x + 5 − 5 = 8 − 5

x = 3

c. x – 3 = 7

We shall add 3 to both sides of the equation.

x – 3 + 3 = 7 + 3

x = 10

d. x + 5 19 = 3 19

We shall subtract 5 19 from both sides of the equation.

x + 5 19 − 5 19 = 3 19 − 5 19

x = –2 19

e. x + 3 8 = 1 3 8

⇒ x + 3 8 = 11 8

We shall subtract 3 8 from both sides of the equation.

x + 3 8 − 3 8 = 11 8 − 3 8

⇒ x + 8 8

⇒ x = 1

f. x − 1 3 = 2 3

We shall add 1 3 to both sides of the equation.

x − 1 3 + 1 3 = 2 3 + 1 3

x = 3 3

x = 1

3. a. 9x = 18

On transposing 9 from the LHS to the RHS, the operation changes from multiplication to division.

x = 18 9 x = 2

b. 4a = 12

On transposing 4 from the LHS to the RHS, the operation changes from multiplication to division.

a = 12 4

a = 3

c. –3y = –27

On transposing (–3) from the LHS to the RHS, the operation changes from multiplication to division.

y = –27

–3

y = 9

d. –18 = –6x

On transposing (–6) from the RHS to the LHS, the operation changes from multiplication to division.

–18

–6 = x 3 = x or x = 3

e. +10=10 x –

=1010=20 x [Transposing 10 from LHS to RHS]

f. 17=20 x =20+17=3 x [Transposing −17 from LHS to RHS]

g. +7=4 y =47=3 y [Transposing 7 from LHS to RHS]

h. q 6 = 7

q = 7 × 6 = 42 [Transposing 6 from LHS to RHS]

i. x –29 = 2

x = 2 × −29 = −58 [Transposing −29 from LHS to RHS]

4. a. 12y = 7y – 15

On transposing 7y from the RHS to the LHS, the operation changes from addition to subtraction.

12y – 7y = –15 5y = –15

On transposing 5 from the LHS to the RHS, the operation changes from multiplication to division.

y = –15 5 y = –3

b. 3x – 2 = 5x – 12

On transposing (–12) from the RHS to the LHS, the operation changes from subtraction to addition.

3x – 2 + 12 = 5x

On transposing 3x from the LHS to the RHS, the operation changes from addition to subtraction.

–2 + 12 = 5x – 3x 10 = 2x

On transposing 2 from the RHS to the LHS, the operation changes from multiplication to division.

10 2 = x 5 = x or x = 5

c. 9 2 x = 54

On transposing 2 from the LHS to the RHS, the operation changes from division to multiplication.

9x = 54 × 2

9x = 108

On transposing 9 from the LHS to the RHS, the operation changes from multiplication to division.

x = 108 9 x = 12

d. –13

2 x = 26

On transposing 2 from the LHS to the RHS, the operation changes from division to multiplication.

–13x = 26 × 2

–13x = 52

On transposing (–13) from the LHS to the RHS, the operation changes from multiplication to division.

x = 52 –13

x = –4

e. x 3 – 5 = 6

On transposing (–5) from the LHS to the RHS, the operation changes from subtraction to addition.

x 3 = 6 + 5

x 3 = 11

On transposing 3 from the LHS to the RHS, the operation changes from division to multiplication.

x = 11 × 3

x = 33

f. 9(x + 14) = 27

On transposing 9 from the LHS to the RHS, the operation changes from multiplication to division.

(x + 14) = 27 9

(x + 14) = 3

On transposing 14 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 3 – 14

x = –11

g. 3 – x 5 = 6

On transposing 5 from the LHS to the RHS, the operation changes from division to multiplication.

3 – x = 6 × 5

3 – x = 30

On transposing 3 from the LHS to the RHS, the operation changes from addition to subtraction.

–x = 30 – 3

–x = 27

On transposing (–1) from the LHS to the RHS, the operation changes from multiplication to division.

x = 27 –1

x = –27

h. 2.5(10x – 2) –21x = 15

25x – 5 – 21x = 15

4x = 20

x = 20 4 = 5

i. 2(4m – 25) = 7(25 – m)

8m – 50 = 175 – 7m

8m + 7m = 175 + 50

15m = 225

m = 225 15 = 15

j. 64 1= 4610 ll l () 543 64 = 630 ll l

105203 = 630 lll

6. Let the whole number be x

The equation that represents the given condition is 3x + 9 = 45.

() 30 ×10=5203

6 lll

() 510=220 ll

550=220 ll

52=20+50 ll

3l = 30

⇒ l = 10

k. −= 1 32 432 a aa

()42 3×3 1 = 432 a aa –––

()942 1 = 122 aa a –+ = 9481 122 aaa + = 581 122 aa –

()() 258121 aa+=− 10161212 aa+=− 12101612 aa−=+ 228 a = 28 14 2 a ==

l. ()22 35 5 43 y y + =+

352415 43 yy +−+ =

()() 3354211 yy +=+

915844 yy +=+ 158449 yy−=− 35 735 5 7 yy=⇒==

5. a. 5328 m += 52835255 mmm =−⇒=⇒= 52835255 mmm =−⇒=⇒=

b. 37 2 n +=

6 7 2 n + = 614 n += 8 n =

c. 2 610 3 o += 218 10 3 o + = 21830 o += 2126 oo=⇒= 2126 oo=⇒=

The code for moon is: 5668

On transposing 9 from the LHS to the RHS, the operation changes from addition to subtraction.

3x = 45 − 9 3x = 36

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 36 3 x = 12

7. Let the number be x.

The equation that represents the given condition is 2 3 x

On simplifying the LHS, we get, x 3 = 3

On transposing 3 from the LHS to the RHS, the operation changes from division to multiplication. x = 3 × 3 x = 9

8. Let the other part be x.

As per the given condition, x + x + 18 = 38

2x + 18 = 38

On transposing 18 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 38 − 18 2x = 20

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 20 2 x = 10 and x + 18 = 10 + 18 = 28

So, one part is 28 and the other part is 10.

9. Let the breadth of the rectangle be b.

Length of the rectangle = 5 + b

According to the given situation, 2(5 + b + b) = 250

2(5 + 2b) = 250

10 + 4b = 250

On transposing 10 from the LHS to the RHS, the operation changes from addition to subtraction.

4b = 250 – 10

4b = 240

On transposing 4 from the LHS to the RHS, the operation changes from multiplication to division.

b = 240 4 b = 60

So, the breadth of the rectangle is 60 m.

Length of the rectangle = 5 + b = 5 + 60 = 65 m.

10. Let Rohan went with ₹x to the market.

Money spent on medicines = 1 3 x

Money donated = ₹50

Money, he has left with = ₹350

The equation can be given as: 50=350 3 x x –

3150 =350 3 xx = 21501050 x –

21200600 xx=⇒=

11. Let the number be x

As per the given condition, x = 13 × (–18) + 7

x = –234 + 7

x = –227

So, the required number is –227.

12. Let the three consecutive even numbers be (x – 2), x, (x + 2).

As per the given condition, (x – 2) + x + (x + 2) = 96

3x = 96

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 96 3

x = 32

(x – 2) = 32 – 2 = 30

(x + 2) = 32 + 2 = 34

So, the three consecutive even numbers are 30, 32 and 34.

13. Let the three consecutive numbers be (x – 1), x, (x + 1).

As per the given condition, x + x + 1 – (x – 1) = 14.

2x + 1 – x + 1 = 14 x + 2 = 14

On transposing 2 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 14 – 2

x = 12

x – 1 = 12 – 1 = 11

x + 1 = 12 + 1 = 13

So, the three consecutive numbers are 11, 12, and 13.

14. Since all sides of a square are equal, hence 4x – 7 = 3x + 5.

On transposing 3x from the RHS to the LHS, the operation changes from addition to subtraction.

4x – 3x – 7 = 5

x – 7 = 5

On transposing (–7) from the LHS to the RHS, the operation changes from subtraction to addition.

x = 5 + 7

x = 12

So, the value of x is 12.

15. Let the third side be x.

As per the given condition, x + 3x + 3x = 28 7x = 28

On transposing 7 from the LHS to the RHS, the operation changes from multiplication to division.

x = 28 7 x = 4

So, the third side is 4 cm long.

3x = 3 × 4 = 12

So, the length of the sides of the isosceles triangle is 4cm; 12 cm; 12 cm.

16. Let the weight of packet q be q.

Weight of packet p = 1 1 2 + q

Weight of packet r = 2 1 2 + q

As per the given condition, q + 1 1 2 + q + 2 1 2 + q = 32

3q + 4 = 32

On transposing 4 from the LHS to the RHS, the operation changes from addition to subtraction

3q = 32 – 4 3q = 28

The weights of the packets p, q and r are 10 5 6 kg, 9 1 3 kg and 11 5 6 kg respectively.

17. Let the price of a textbook be x.

Price of a notebook = x + 10

As per the given situation, 2 × (x + 10) + 3x = 120. 2x + 20 + 3x = 120

5x + 20 = 120

On transposing 20 from the LHS to the RHS, the operation changes from addition to subtraction.

5x = 120 – 20 5x = 100

On transposing 5 from the LHS to the RHS, the operation changes from multiplication to division. x = 100 5 x = 20

So, the price of a textbook is ₹20.

Price of a notebook = x + 10 = 20 + 10 = ₹30.

18. Let the number of toffees with Ramesh be x.

Number of toffees with his sister = 2x

As per the given situation, x + 2x = 54

3x = 54

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 54 3 x = 18

So, the number of toffees with Ramesh is 18.

19. Let the present age of the son be x; present age of the father = 30 + x 12 years hence,

Age of the son = x + 12

Age of the father = 30 + x + 12

= 42 + x

As per the given situation, 42 + x = 3(x + 12)

42 + x = 3x + 36

On transposing 36 from the RHS to the LHS, the operation changes from addition to subtraction.

42 + x – 36 = 3x

x + 6 = 3x

On transposing x from the LHS to the RHS, the operation changes from addition to subtraction.

6 = 3x – x 6 = 2x

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division q = 28 3 = 9 1 3 p = 1 1 2 + 9 1 3 = 3 2 + 28 3 = 65 6 = 10 5 6 kg r = 2 1 2 + 9 1 3 = 5 2 + 28 3 = 71 6 = 11 5 6 kg

On transposing 2 from the RHS to the LHS, the operation changes from multiplication to division.

6 2 = x 3 = x or x = 3

So, the present age of the son is 3 years.

Present age of the father = 30 + x = 30 + 3 = 33 years.

Challenge

1. Assertion (A): The solution of the equation (4x − 7) = (2x − 5) is −1.

Reason (R): Adding or subtracting the same number from both sides of an equation keeps the equation balanced. (4x − 7) = (2x − 5)

4x − 2x = −5 + 7

2x = 2

2 1 2 x ==

Hence, the assertion is false.

Adding or subtracting the same number from both sides of an equation keeps the equation balanced, hence the reason is true.

So, the correct answer is option d. A is false, but R is true

2. Statement 1: Sushant scored a total of 120 marks.

Statement 2: Sushant attempted 45 questions out of the given 50 questions.

Marks are given for right answer = 5

Marks deducted for wrong answer = 2

Statement 1: Sushant scored a total of 120 marks. It does not give any information about the number of right or wrong answers attempted, and hence is alone not sufficient.

Statement 2: Sushant attempted 45 questions out of the given 50 questions.

From the above statement we can infer that Sushant attempted x correct answer and 45 – x wrong answers but still, the statement alone is not sufficient to answer the question.

Combining statements 1 and 2 and the information in the question gives

5 × x – 2 × (45 – x) = 120

5902120 xx−+=

7210 x =

30 x =

Hence, statements 1 and 2 together are sufficient to answer. Both statements together are sufficient. Thus, option c is the correct answer.

Case Study

1. The overall charge of an atom is determined by adding the charges of its electrons and protons.

The expression for an overall charge of chlorate anion can be given as: b + 17. Thus, option a is correct.

2. Given that the overall charge of carbonate anion is −2.

Charge from electron + charge from proton = Overall charge −32 + f = −2

f = −2 + 32

Hence, d is the answer

3. Charge from electron + charge from proton = Overall charge −10 + a = 1

a = 1 + 10 = 11

Hence, the charge from the proton in sodium cation is 11. Thus, the correct option is c.

4. Charge from electron + Charge from proton = Overall charge

c + 12 = 2

c = 2 − 12 = −10

Hence, the charge from electron in magnesium cation is −10.

5. Let the charge from electron in barium cation = x

Charge from proton = x + 110

Given that overall charge in barium cation = +2

We know that charge from electron + charge from proton = overall charge

Substituting the values, we get

x + (x + 110) = 2

2x + 110 = 2

2x = 2 – 110

2x = −108

x = −54

Hence charge from electron = −54

Charge from proton = −54 + 110 = 56

Chapter 6

Let’s Warm-up 1. 27° 2. 142°

Do It Yourself 6A

1. a. 80° + 100° = 180°, so the given angles are supplementary angles.

b. 45° + 135° = 180°, so the given angles are supplementary angles.

c. 60° + 120° = 180°, so the given angles are supplementary angles.

d. 35° + 125° = 160°, so the given angles are neither supplementary nor complementary.

e. 27° + 63° = 90°, so the given angles are complementary angles.

f. 40° + 53° = 93°, so the given angles are neither supplementary nor complementary.

g. 37° + 37° = 74°, so the given angles are neither supplementary nor complementary.

h. 78° + 33° = 111°, so the given angles are neither supplementary nor complementary angles.

2. a. Supplement of 33° = 180° − 33° = 147°

Complement of 33° = 90° − 33° = 57°

b. Supplement of 72° = 180° − 72° = 108°

Compliment of 72° = 90° − 72° = 18°

c. Supplement of 41° = 180° − 41° = 139°

Complement of 41° = 90° − 41° = 49°

d. Supplement of 89° = 180° − 89° = 91°

Complement of 89° = 90° − 89° = 1°

3. a. The two angles are 25° and 108°

Angles are neither supplementary nor complementary, since 25° + 108° = 133°.

b. The two angles are 52° and 38°

Angles are complementary, since 52° + 38° = 90°.

c. The two angles are 122° and 58°

Angles are supplementary, since 122° + 58° = 180°.

A D C B In rectangle ABCD,

∠A = ∠B = ∠C = ∠D = 90°.

Pair of supplementary angles:

∠A and ∠B, ∠B and ∠C, ∠C and ∠D, ∠D and ∠A, ∠D and ∠B, ∠C and ∠A.

5. a. (4x − 45)° + (3x + 50)° = 180° (Linear pair)

4x + 3x = 180° − (50° − 45°)

7x = 180° − 5° = 175° x = 175° 7 = 25°

x = 360° 3 = 120°

So, the angle is 120°.

Supplement: 180° − x = 180° − 120° = 60°

So, the supplement angle is 60°.

12. Let the angles be 5x and 4x

Given, 5x + 4x = 90°.

9x = 90°. So, x = 90° 9 = 10°

Angles are 5x = 5 × 10° = 50°

4x = 4 × 10° = 40°

b. 2 x

x 22 x

° + 2 x

x 23 x

° = 180° (Linear pair)

3x + 2x 3 × 2 = 5x 6 = 180°.

So, x = 6 5 × 180° = 216°

c. (2x + 15)° + (2x + 25)° + (x − 20)° + 90° = 360°

5x + 110° = 360°

5x = 360° − 110° = 250°.

So, x = 250° 5 = 50°

6. (3x + 10)° + (7x − 20)° = 90°

10x − 10° = 90°

10x = 90° + 10° = 100°.

So, x = 100° 5 = 10°

7. (4y + 13)° + (3y − 8)° = 180°

7y + 5° = 180°

7y = 180° − 5° = 175°. So, y = 175° 7 = 25°

Measure of angles = (4y + 13)° = (4 × 25 + 13)° = 113°

(3y − 8)° = (3 × 25 − 8)° = 67°

So, the measure of the angles are 113° and 67°.

8. a. y° + y° + y° = 135° (vertically opposite angle)

3y° = 135°. So, y = 135° 3 = 45°

b. n° + 90° + 30° = 180°

n + 120° = 180°.

So, n = 180° − 120° = 60°

m° + n° = 180° (Linear pair)

m + 60° = 180°.

So, m = 180° − 60° = 120°

c. Vertically opposite angles are equal.

So, a° + 2a° + 3a° + a° + 2a° + 3a° = 360° (sum of all angles in a circle is 360°)

12a = 360°

a = 360° 12 = 30°

9. Let the angle be x and its complement be x − 30.

x + (x − 30°) = 90°

2x = 90° + 30° = 120°. So, x = 120° 2 = 60°.

10. Vertically opposite angles are equal.

So, the measure of angle a is 30°.

11. Let the angle be x. Its supplement will be x 2

x + = 180° = > 2x + x 2 180° or 3x = 180° × 2 = 360°.

13. Let the angles be 13x and 5x Given, 13x + 5x = 180°

18x = 180°

x = 180° 18 = 10°

Angles are 13x = 13 × 10° = 130°

5x = 5 × 10° = 50°

14. Let the angles be x and y.

Supplement of x: 180° − x

Complement of y: 90° − x

(180° − x) − x = 46°

180° − x − x = 46°

2x = 180° − 46° = 134°. So, x = 134° 2 = 67°

Also, y − (90° − y) = 20° y − 90° + y = 20°

2y = 90° + 20° = 110°. So, y = 110° 2 = 55°

Challenge

1. 90° − x

180°− x = 4 9

9(90° − x) = 4(180° − x)

810° − 9x = 720° − 4x

810° − 720° = 9x − 4x

90° = 5x

x = 18°

So, the measure of the angle is 18°.

Do It Yourself 6B

1. a.

parallel

Not parallel

2. Pairs of corresponding angles: 2 and 6, 2 and 10, 6 and 10, 1 and 5, 1 and 9, 5 and 9, 3 and 7, 7 and 11, 3 and 11, 4 and 8, 8 and 12, 12 and 4.

Pairs of alternate angles: 2 and 8, 1 and 7, 3 and 5, 4 and 6, 1 and 11, 2 and 12, 3 and 9, 4 and 10, 6 and 12, 5 and 11, 7 and 9, 8 and 10.

3. a. The 110° angle should be equal to the supplement of the 72° angle since they are corresponding pair of angles.

Supplement of 72° = 180° − 72° = 108°. Since 110° ≠ 108°, the lines are not parallel.

b. Clearly, alternate exterior angles are not equal (89° ≠ 99°), so the lines are not parallel.

c. The 102° angle should be equal to the supplement of the 89° angle since they are corresponding pair of angles.

Supplement of 89° = 180°− 89° = 91°. Since 102° ≠ 91°, the lines are not parallel.

d. Clearly, alternate exterior angles are equal (109° = 109°), so the lines are parallel.

4. D A B P Q R C S l

5.9 cm 5.9 cm E F

5. P q B A

6. Angle a should be equal to the supplement of the 62° angle since they form corresponding pair of angles.

Supplement of 62° = 180° − 62° = 118°. So, a° = 118°.

Angle b should be equal to the supplement of the 72° angle since they form corresponding pair of angles.

Supplement of 72° = 180° − 72° = 108°. So, b° = 108°.

7. ∠EAC = ∠ACB = 60° (alternate angles)

∠DAB = ∠ABC = 90° (alternate angles)

∠DAB + ∠BAC + ∠CAC = 180° (Linear sum property)

∠BAC + 60° + 90° = 180°

∠BAC = 180° − 150° = 30°

8. Q n m 5.5 cm

9. XD = 5.6 cm; YD = 9 cm; ZD = 4.2 cm X Y Z D

10. ∠QPS + ∠PSR = 180° (interior angles on the same side of the transversal)

120° + ∠PSR = 180°

∠PSR = 180° − 120° = 60°

Similarly, ∠QRS + ∠PQR = 180° (interior angles on the same side of the transversal)

∠QRS + 150° = 180°

∠QRS = 180° − 150° = 30°

11. a. x° = 80° (alternate angles); y° = 32° (alternate angles)

b. m° + 110° = 180° (interior angles on the same side of the transversal)

m° = 180° − 110° = 70°

m° = n° = 70° (alternate angles)

c. p° = 60° (alternate exterior angles)

q° = 105° (alternate interior angles)

d. h° = 45° (corresponding angles)

g° = 72° (alternate angles)

12. The rain would hit the ground at an angle of 60° since they form alternate angles.

13. a. y° = 35° (alternate angles); x° = 50° (alternate angles)

b. Draw a line parallel to the 2 lines and label the angles, as shown.

116° + a° = 180° (interior angles of the same side of the transversal)

a° = 180° − 116° = 64°

136° + b° = 180° (interior angles of the same side of the transversal)

b° = 180° − 136° = 44°

x° = a° + b° = 64° + 44° = 108°

14. Rough figure.

8 cm 4 cm 6 cm

10 cm

Steps: i. Draw a rectangle ABCD of length 10 cm × 8 cm.

ii. Mark any two points A and B on any edge of the outer rectangle.

iii. Construct right angles at points A and B, respectively.

iv. Draw arcs with centres at A and B, each with a 2 cm radius, intersecting the perpendiculars at A and B at points E and F.

v. Connect points EF and extend this line on both sides to achieve the desired parallel line.

vi. Repeat the process for the other three sides.

vii. Connect the corresponding points on these lines to form the inner rectangle.

Challenge

1. w = 72°

w = x + y (alternate interior angles)

x + y = 72°

y = w 4 = 18°

x = w − y = 72° − 18° = 54°

x + z = 180° (sum of interior angles on the same side of transversal)

54° + z = 180°

z = 180° − 54° = 126°

So, z = 126°

Chapter Checkup

1. a. Neither, since 76° + 109° = 185°

b. Neither, since 24° + 56° = 80°

c. Complimentary, since 44° + 46° = 90°

d. Supplementary, since 14° + 166° = 180°

e. Neither, since 36° + 135° = 171°

f. Neither, since 50° + 150° = 200°

g. Complementary, since 36° + 54° = 90°

h. Complementary, since 78° + 12° = 90°

2. a. Supplement of 34° = 180° − 34° = 146°

b. Supplement of 119° = 180° − 119° = 61°

c. Supplement of 126° = 180° − 126° = 54°

d. Supplement of 166° = 180° − 166° = 14°

3. a. Complement of 26° = 90° − 26° = 64°

b. Complement of 40° = 90° − 40° = 50°

c. Complement of 47° = 90° − 47° = 43°

d. Complement of 86° = 90° − 86° = 4°

4. a. 2x° + (3x − 10)° = 90°

2x° + 3x° − 10° = 90°

5x° = 90° + 10° = 100°.

So, x = 100° 5 = 20°

b. (2x + 6)° + 3x° + x° = 180°

2x° + 6° + 3x° + x° = 180°

6x° = 180° − 6° = 174°.

So, x = 174° 6 = 29°

c. 3x° + 4x° + x° + (2x − 10)° + 90° = 360°

10x° + 80° = 360°

10x° = 360° − 80° = 280°.

So, x = 280° 10 = 28°

d. x° + 3x° + 4x° = 180°

8x° = 180°

x = 180° 8 = 22.5° (approx)

5. ∠CAB = 180° − 142° = 38° (Linear pair)

∠ACB = 180° − 113° = 67° (Linear pair)

∠CAB + ∠ACB + ∠ABC = 180° (Angle sum property of the triangle)

38° + 67° + ∠ABC = 180°

∠ABC = 180° − 105° = 75°

6. Since the yellow lines in the parking space are parallel,

∠m = 115° (corresponding angle)

7. a. The 120° angle should be equal to the supplement of the 70° angle since they are corresponding pair of angles. Supplement of 70° = 180° − 70° = 110°. Since 110° ≠ 120°, the lines are not parallel.

b. The 75° angle should be equal to the supplement of the 105° angle since they are corresponding pair of angles.

Supplement of 105° = 180° − 105° = 75°. Since they are equal, the lines are parallel.

c. The 110° at the bottom should be equal to the supplement of the 110° angle at the top alternate exterior angles. Supplement of 110° = 180° − 110° = 70°. Since they are not equal (110° ≠ 70°) the lines are not parallel.

8. a. y° should be equal to the supplement of the 65° angle (corresponding angles).

y° = Supplement of 65° = 180° − 65° = 115°.

b. z° = 130° (alternate exterior angles)

c. x° should be equal to the supplement of the 85° angle (alternate exterior angles).

x° = Supplement of 85° = 180° − 85° = 95°.

9. m° = 137° (alternate exterior angles); g° = 137° (corresponding angles)

° = Supplement of the 113° angle = 180° − 113° = 67° (alternate interior angles)

° = Supplement of the 113° angle = 180° − 113° = 67° (corresponding angles)

10. x° + 75° = 180° (interior angles on the same side of transversal)

° = 180° − 75° = 105°

° + (75° − 15°) = 180° (interior angles on the same side of transversal)

° = 180° − 60° = 120°

° + 75° = 180° (interior angles on the same side of transversal)

° = 180° − 75° = 105°

(w° + 15°) + z° = 180° (interior angles on the same side of transversal)

(105° + 15°) + z° = 180°

° = 180° − 120° = 60°

11. Since PQRS is a parallelogram, PS ꠱ QR and PQ ꠱ RS.

∠PQR + ∠QPS = 180° (interior angles on the same side of transversal)

∠PQR + 115° = 180°

∠PQR = 180° − 115° = 65°

∠RSP + ∠QPS = 180° (interior angles on the same side of transversal)

∠RSP + 115° = 180°

∠RSP = 180° − 115° = 65°

∠RSP + ∠QRS = 180° (interior angles on the same side of transversal)

∠QRS + 65° = 180°

∠QRS = 180° − 65° = 115°

12. ∠QDC = ∠QBA = 63° (alternate angles)

∠DCQ = ∠BAQ = 47° (alternate angles)

Challenge

1. Assertion (A): If a transversal intersects two parallel lines, then the interior angles on the same side of the transversal are supplementary angles.

Reason (R): The sum of the measures of two supplementary angles is 180°.

The assertion is correct because when a transversal intersects two parallel lines, the interior angles on the same side of the transversal are supplementary.

The reason is also correct because the sum of two supplementary angles is 180°.

The reason does not explain the assertion.

So, both assertion and reason are true, but the reason does not explain the assertion.

So, option b is true.

Since l and m are parallel lines and PQ, RS and TU are transversal.

Then, ∠4 = ∠QPS (alternate interior angles)

∠4 = 84°

∠1 = ∠QOR = 40° (vertically opposite angles)

PQ and TU are parallel and m and l are transversal.

∠2 + ∠QPT = 180° (consecutive interior angles)

∠2 = 180° − 84° = 96°

∠2 + ∠3 = 180°

∠3 = 180° − 96° = 84°

a. ∠1 = 40° b. ∠2 = 96° c. ∠3 = 84° d. ∠4 = 84°

Case Study

1. 0.2x + 60° + x + 80° = 180°

x = 180° − 140°

x = 40 ÷ 1.2 = 33.3°

x = 33.3°

2. 0.2x + 60° + 90° + angle between block A and block B = 180°

(0.2 × 33.3°) + 150° + angle between block A and block B = 180°

6.66° + 150° + angle between block A and block B = 180° angle between block A and block B = 180° − 156.66° = 23.34°

3. The headquarters and block A are parallel and the extended line of inner area is the transversal.

∠C = ∠A (corresponding angles are equal)

∠A = 0.2x + 60°

= (0.2 × 33.3°) + 60° = 66.66°

Thus, measure of ∠A = 66.66° ≈ 66.7° So, option d is correct

4. Use the information obtained in Q2 and identify that these lines are not parallel. Since the angle obtained between headquarters and training centre is 30° and that from Q2 is 23.34°.

Which helps us conclude that these lines are not parallel. Thus, option b is correct.

Chapter 7 Let’s Warm-up

1. 60°, 70°, 50° Isosceles triangle

2. 90°, 45°, 45° Obtuse-angled triangled

3. AB = BC = CA Acute-angled triangle

4. 100°, 30°, 50° Equilateral triangle

5. AB = BC Isosceles-right triangle

Do It Yourself 7A

1. a. Median is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.

b. Altitude is a line segment perpendicular to a side of the triangle.

c. The point where the altitudes intersect is called the orthocenter of the triangle.

d. An altitude may or may not lie inside the triangle.

2. A median is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.

Each triangle has three medians, one from each vertex. The point where the medians intersect is called the centroid of the triangle.

In triangle ABC, AR is the median to side BC, C is the median to side AB and BQ is the median to the side AC.

3. If BD is the median of triangle ABC, AD is given as 5 cm.

AC = AD + DC

As BD is the median, AD = DC

AC = AD + AD = 2 × AD

AC = 2 × 5 cm = 10 cm

The length of AC is 10 cm.

4. The centre of mass in a triangle will be at the centroid. The centroid is always located inside the triangle, regardless of the type of triangle (acute, obtuse or right).

Thus, option a is correct.

5. AD and CE are the two medians in a triangle ABC on the opposite sides BC and AB, respectively.

G is the centroid, therefore, AG:GD = 2:1 (property of a centroid)

AG = 2 cm

GD = 1 2 × AG = 2 2 = 1 cm

AD = 2 + 1 = 3 cm

A median divide the opposite side in two equal parts, therefore,

BD = CD = 4 cm

BC = BD + CD = 4 + 4 = 8 cm

6. Medians of the given triangle are CF, EB and AD.

a. G is the centroid, therefore, GE:BG = 2:1 (property of a centroid)

BG = 5 cm,

GE = (2 × 5)cm = 10 cm

BE = BG + GE

= 5 cm + 10 cm = 15 cm

b. CG:GF = 2:1 (property of centroid)

CG = 16 cm,

GF = 16 2 = 8 cm

CF = CG + GF = 16 cm + 8 cm = 24 cm

c. AD = 30 cm

AG:GD = 2:1 (property of centroid)

AG = 2 3 × 30 = 20 cm

GD = 1 3 × 30 = 10 cm

d. GC:GF = 2:1 (property of centroid)

GF = x

GC = 2 × x = 2x

CF = GF + GC = x + 2x = 3x

e. AG:GD = 2:1 (property of centroid)

AG = 2 × GD

9x = 2 × (5x − 1)

9x = 10x − 2

10x − 9x = 2

x = 2

AD = AG + GD

= 9x + 5x −1 = 14x − 1

= 14 × 2 − 1 = 27

7. PS is the median of a triangle PQR

Therefore, QS = RS

10x − 7 = 5x + 3

10x − 5x = 3 + 7

5x = 10

x = 2

∠PSR = (15x + 42)°

= (15 × 2 + 42)° = (30 + 42)° = 72°

∠PSR = 72°

Challenge

1. In an equilateral triangle, all sides and angles are equal, which creates a unique symmetry.

This symmetry ensures that the orthocentre and the centroid coincide at same point.

Therefore, when the orthocentre and the centroid of a triangle are the same, it is an equilateral triangle.

Do It Yourself 7B

1. a. ∠x + 35° + 40° = 180° (Angle sum property of a triangle)

∠x = 180° − 75° = 105°

b. (x + 2)° + (3x + 3)° + (2x + 10)° = 180° (Angle sum property of a triangle)

6x + 15 = 180°

6x = 180° − 15° = 165°

x = 165° 6 = 27.5°

c. x + y + y = 180° (Angle sum property of a triangle)

x + 2y = 180°

Also, 2y = 110° (Sum of two interior opposite angles is equal to the exterior angle)

y = 110 2 = 55°

x + 55° + 55° = 180°

x = 180° − 110°

x = 70°

d. 30° + 120° + x = 180° (Angle sum property of a triangle)

150° + x = 180°

x = 180° − 150°

x = 30°

y + 90° = 120° (Exterior angle property of a triangle)

y = 120° − 90°

y = 30°

2. Given: The angles of the triangle are in the ratio 5:6:7.

Let the angles be 5x, 6x and 7x. Since the sum of the angles in a triangle is 180°,

5x + 6x + 7x = 180°

Solving for x:

18x = 180°

x = 10°

Therefore, the angles are 5x = 50°, 6x = 60° and 7x = 70°.

The measures of the angles are 50°, 60° and 70°.

3. One of the exterior angles of a triangle = 130°

Ratio of interior opposite angles = 2:3

By the exterior angle property,

2x + 3x = 130°

5x = 130° ⇒ x = 26°

The angles of the triangle are—

2x = 2 × 26° = 52°;

3x = 3 × 26° = 78°, and 130° + ∠y = 180°

∠y = 180° − 130° = 50°

Therefore, the angles of the triangle are 52°, 78° and 50°.

4. a + b + c = 180°

45° + 60° + c = 180°

c = 180° − 105°

c = 75°

5. Given: In triangle ABC, angle A is 50° more than angle B and 20° less than angle C

Let angle B be x. Then, angle A = x + 50° and angle C = x + 50°

+ 20° = x + 70°.

Since the sum of the angles in a triangle is 180°.

x + (x + 50°) + (x + 70°) = 180°

Solving for x : 3x + 120° = 180°, 3x = 60°, x = 20°

Therefore, angle A = 70°, angle B = 20°, and angle C = 90°.

The angles of the triangle are 70°, 20° and 90°.

6. Given:

In triangle ABC, ∠A − ∠B = 15° and ∠B − ∠C = 30°.

Let angle B be x

Then, angle A = x + 15° and angle C = x − 30°.

Since the sum of the angles in a triangle is 180°.

(x + 15°) + x + (x − 30°) = 180°

Solving for x:

3x − 15° = 180°

3x = 195°

x = 65°

Therefore, angle A = 80°, angle B = 65°, and angle C = 35°.

7. a. To show

∠ACD = ∠A + ∠B

BA parallel to CE (given)

AC is a transversal that cuts AB and CE.

∠1 = ∠3 (alternate interior angles)

BC is a transversal that cuts lines AB and CE

∠2 = ∠4 (corresponding angles)

∠1 + ∠2 + ∠BCA = 180°

∠BCA + ∠3 + ∠4 = 180° (linear pair)

∠BCA = 180° − (∠3 + ∠4)

∠1 + ∠2 + 180° − (∠3 + ∠4) = 180°

∠1 + ∠2 = ∠3 + ∠4

Thus, ∠A + ∠B = ∠ACD

Hence proved

b. To show: ∠A + ∠B + ∠C = 180°

∠BCD = ∠ACB + ∠3 + ∠4 = 180° (Straight angle) — eq 1

Since ∠3 + ∠4 = ∠ACD, ∠3 + ∠4 = ∠1 + ∠2 (From (a))

Substituting in—eq 1,

∠BCD = ∠ACB + ∠1 + ∠2

∠BCD = ∠C+ ∠A + ∠B

∠C+ ∠A + ∠B = 180°

Hence, ∠A + ∠B + ∠C = 180°

8. PQT = TQR (given)

Let us take the angles PQT as x.

So, PQT = TQR = x

∠RQS + ∠PQU = 80° (from figure)

Therefore, ∠PQU = ∠TQU − ∠TQP

∠PQU = 90° − x° (∠TQU = 90°, since QT is perpendicular to US)

Similarly, ∠RQS = 90° − x°

Therefore, ∠PQU = ∠RQS

Since ∠RQS + ∠PQU = 80°

∠RQS + ∠RQS = 80°

2∠RQS = 80°

∠RQS = 40°

∠UQS = ∠PQU + ∠PQT + ∠TQR + ∠RQS

180° = 40° + x + x + 40°

180° = 80° + 2x 100° = 2x; x = 50°

The measure of ∠PQT is 50°.

Challenge

1. An equilateral triangle has all its angles measuring 60°. So each exterior angle measures 120°. Therefore, an equilateral triangle has measure of all its exterior angle twice than the measure of its interior angle.

So, the possible values for the third side x are 5 cm < x < 23 cm

c. For sides of lengths 34 cm and 7 cm:

34 + 7 > x

41 > x

Similarly, 7 + 34 > x

41 > x

Also, 34 − 7 < x

27 < x

So, the possible values for the third side x are 27 cm < x < 41 cm

4. Given, PS is the median, hence, QS = SR

By triangle inequality property,

PQ + QS > PS

Similarly, PR + SR > PS

Adding both inequalities, PQ + QS + PR + SR > PS + PS

Since QS + SR = QR

PQ + QR + PR > 2PS

As we have shown that the triangle inequality property holds true for this statement, hence, PQ + QR + PR > 2 PS.

5. The triangle can be constructed if the sum of any two sides is greater than the third side.

1450 + 1350 > 2200

1450 + 2200 > 1350

1350 + 2200 > 1450

Thus, the triangle can be constructed.

Do It Yourself 7C

1. a. 2 cm, 3 cm, and 4 cm

2 + 3 > 4 (True)

2 + 4 > 3 (True)

3 + 4 > 2 (True)

This set forms a triangle.

b. 1 cm, 5 cm, and 10 cm

1 + 5 < 10

This set does not form a triangle.

c. 3.2 cm, 5.6 cm, and 8.3 cm

3.2 + 5.6 > 8.3 (True)

3.2 + 8.3 > 5.6 (True)

5.6 + 8.3 > 3.2 (True)

This set forms a triangle

2. a. In triangle PSR, PS < PR + SR (By the triangle inequality property)

b. In triangle PQS, PS > PQ − QS (By the triangle inequality property)

3. a. For sides of lengths 3 cm and 6 cm:

3 + 6 > x

9 > x

Similarly, 6 + 3 > x 9 > x

And 6 − 3 < x

3 < x

So, the possible values for the third side x are 3 cm < x < 9 cm.

b. For sides of lengths 14 cm and 9 cm: 14 + 9 > x

23 > x

Similarly, 9 + 14 > x 23 > x

And, 14 − 9 < x 5 < x

Challenge

1. Given: A triangle ABC with AB = 5 cm, BC = 6 cm and AC = 7 cm. O is a point in the triangle.

There is no information about point O being joined to anything. We can assume that O is any arbitrary point in the triangle.

To prove: AO + BO + CO > 9 cm

By the Triangle Inequality Theorem,

AO + BO > AB

BO + CO > BC

CO + AO > AC

Adding these inequalities:

(AO + BO) + (BO + CO) + (CO + AO) > AB + BC + AC

2(AO + BO + CO) > AB + BC + AC

AO + BO + CO > (AB + BC + AC) 2

Since AB + BC + AC = 5 + 6 + 7 = 18 cm:

AO + BO + CO > 18 2

AO + BO + CO > 9 cm

Therefore, AO + BO + CO is greater than 9 cm.

Do It Yourself 7D

1. a. 2 cm, 3 cm and 6 cm

62 = 36

22 + 32 = 13

36 ≠ 13

This is not true, so the given lengths do not form a rightangled triangle.

b. 6 m, 8 m and 10 m

102 = 62 + 82

100 = 36 + 64

100 = 100

This is true, so the given lengths form a right-angled triangle.

c. 15 cm, 20 cm and 25 cm

252 = 152 + 202

625 = 225 + 400

625 = 625

This is true, so the given lengths form a right-angled triangle.

2. a. h = 20 m

b = 12 m

Using Pythagoras theorem,

h2 = b2 + p2

202 = 122 + p2

400 = 144 + p2

400 − 144 = p2

p2 = 256 = 16 × 16

So, p = 16 m

b. h = 25 cm

b = 7 cm

Using Pythagoras theorem,

h2 = b2 + p2

252 = b2 + 72

625 = b2 + 49

625 − 49 = b2

b2 = 576 = 24 × 24

So, b = 24 cm

c. h = 30 m

b = 24 m

Using Pythagoras theorem,

h2 = b2 + p2

302 = 242 + p2

900 = 576 + p2

900 − 576 = p2

p2 = 324 = 18 × 18

So, p = 18 m

3. Given—Pythagorean triplet (k, 24, 25)

By Pythagorean theorem,

k2 + 242 = 252

k2 = 625 − 576

k2 = 49

k2 = 7 × 7

k = 7

4. Given—The sides of a rectangle are 12 cm and 16 cm.

The diagonal of the given rectangle is the hypotenuse of the right triangle.

Using Pythagoras theorem,

h2 = b2 + p2

h2 = 162 + 122

h2 = 256 + 144

h2 = 400

h2 = 20 × 20

h = 20 cm

5. Hypotenuse = 26 cm

Base = 10 cm

Base2 + Altittude2 = Hypotenuse2

102 + Altittude2 = 262

Altittude2 = 676 − 100 = 576

Altitude = 24 cm

So, the length of the altitude is 24 cm.

6. Triangle ABC and Triangle ADC are right angled triangle

In triangle ABC, AC2 = AB2 + BC2 (by Pythagoras theorem)

x2 = 32 + 42

x2 = 9 + 16 = 25 = 5 × 5

x = 5 cm

Similarly, in triangle ADC,

x2 + y2 = 132

52 + y2 = 169

y2 = 169 − 25 = 144

y2 = 12 × 12

y = 12 cm

7. According to the question, 20 m is the hypotenuse and 12 m is the height.

To find the base.

202 = 122 + Base2

400 = 144 + Base2

Base2 = 256

Base = 16

Thus, the horizontal distance between the ball and the person is 16 m.

8. Answers may vary. Sample answer:

A ladder 13 m long is placed on the ground in such a way that it touches the top of a vertical wall 12 m high.

Find the distance of the foot of the ladder from the bottom of the wall.

Challenge

1. Let us denote the lengths of the legs of the triangle as a and b, with (a > b).

Given that the hypotenuse is 10 units and the difference between the legs is 2 units,

a2 + b2 = 102 (from the Pythagorean theorem)

a − b = 2

a = b + 2

(b + 2)2 + b2 = 102 = 100

2b2 + 4b + 4 = 100

2b2 + 4b = 100 − 4 = 96

2b2 + 4b − 96 = 0

b2 + 2b − 48 = 0

Solving the equation

(b + 8)(b − 6) = 0

The value of b cannot be negative, so b = 6 cm

a = b + 2 = 6 + 2 = 8 cm

So, the lengths of the legs are 6 cm and 8 cm.

Do It Yourself 7E

1. AB = 10 cm

AB ≅ CD

If two lines are congruent, then, their lengths are the same.

Then, AB = CD

So, CD = 10 cm

2. Figures may vary. Sample figures.

4.2cm x SU ≅ OP

SU = 4.2 cm

When the two line segments are congruent, they have the same lengths.

SU = OP ⇒ OP = x = 4.2 cm

b. L N M x X Z

55° Y

∠LMN ≅∠XYZ

∠XYZ = 55°

∠LMN = x

When two angles are congruent, the measures of the angles are equal.

Thus, ∠LMN = 55°

c. A x B D C

Thus, (A) is false, but (R) is true.

Hence, option d is correct.

2. In rectangle PQRS,

Rectangle PABD ≅ Rectangle

DBCS ≅ Rectangle ACRQ

Length of rectangle PABD

= Length of rectangle DBSC

= Length of rectangle ACRQ

Breadth of rectangle PABD

= Breadth of rectangle DBSC

= Breadth of rectangle ACRQ

Given: In rectangle ACRQ,

8 cm 4 cm

P S Q R Y

Rectangle ABCD ≅ Rectangle PSRQ

If two rectangles are congruent, then the length and the breadth of one rectangle are exactly equal to the respective length and the breadth of the other rectangle.

AB = PS ⇒ x = 4 cm

S T U 7.2 cm V x

Circle with radius ST = CST

Circle with radius VU = CVU circle C1 ≅ circle C2

If two circles are congruent, then, the radius of both the circles are of equal lengths.

VU = ST ⇒ x = 7.2 cm

4. Both the targets are circular in shape and have the same radii. If the radius of one circle is equal to the radius of the other circle, then, the circles are congruent. Hence, yes, the two circular targets are congruent. So, circle C1 ≅ circle C2

5. There will be 5 spots since the butterflies have the same wing pattern.

6. D

c b a C B O A

In the given figure, a = b = c a = ∠AOB

b = ∠BOC

c = ∠COD

∠AOC = a + b = c + b (since, a = b = c) then, ∠AOC = ∠DOC + ∠BOC = ∠DOB So, ∠AOC ≅∠DOB (or ∠BOD)

Challenge

1. Assertion (A): If two angles are congruent, their corresponding sides are also congruent.

Reason (R): Two angles are congruent if they have the same measure.

Assertion is false because congruent angles alone do not guarantee that the corresponding sides are congruent. For the sides to be congruent, the triangles must be congruent, which requires all corresponding angles and sides to be congruent.

Reason (R) is true because two angles are congruent if they have the same measure, regardless of the length of their corresponding sides.

Breadth = AQ = 5 cm

Length = AC = AB + BC = 5 + 5 cm = 10 cm

Area = Length = Breadth = 10 cm × 5 cm = 50 cm2

Since, Area of rectangle PABD = Area of rectangle DBSC = Area of rectangle ACRQ

Then,

Area of rectangle PQRS = Area of rectangle PABD + Area of rectangle DBCS + Area of rectangle ACRQ

= 50 cm2 + 50 cm2 + 50 cm2

= 150 cm2

So, the area of rectangle PQRS is 150 cm2

Do It Yourself 7F

a. In ΔPRS and ΔSQP

PR = SQ (given)

RS = QP (given)

PS = SP (common side)

b. Since the lengths of the sides of triangle PRS and SQP are equal, by SSS congruence

ΔPRS ≅ ΔSQP.

2. In ∆PRS and ∆PRQ, RS = PQ (equal side)

∠SRP = ∠QPR (equal angle)

RP = PR (common side)

So, by SAS congruence criterion, ∆PRS ≅∆PRQ

3. Given that ∠CPD = ∠BPD

∠CPD + ∠APC = 180° →∠APC = 180° − ∠CPD

∠BPD + ∠APB = 180° →∠APB = 180° − ∠BPD

∠APC = ∠APB

In ΔCAP and ΔBAP

∠APC = ∠APB (proved)

AP = AP (common side)

∠CAP = ∠BAP (Given: AD bisects ∠BAC)

So, by ASA congruence criteria

ΔCAP ≅ ΔBAP.

4. A B C D

In ΔABD and ΔCBD,

∠BAD = ∠BCD (AB = BC)

AD = DC (BD bisects AC)

∠BDA = ∠BDC (90°)

So, by ASA congruence criteria

ΔABD ≅ ΔCBD.

45º 45º 30º30º A B D C

∠DAB = 45° + 30° = 75°

∠CBA = 45° + 30° = 75°

In ΔADB and ΔBCA,

∠DAB = ∠CBA (75°)

AB = AB (common side)

∠DBA = ∠CAB (30°)

So, by ASA congruence criteria

ΔADB ≅ ΔBCA.

6. Q S R

Challenge

1. In ∆AOC and ΔBOD,

OC = OD (O is the mid-point)

∠AOC = ∠BOD (vertically opposite angles)

PR + RS = QS + RS

PS = QR

AB P PR = QS

In ∆APS and ∆BQR

∠APS = ∠BQR (90°)

PS = QR (proved)

AS = BR (given)

So, by RHS congruence criteria

∆APS ≅∆BQR.

7. A BE D C

Given: ABCDE is a regular pentagon

Hence, AB = BC = CD = ED = AE

In ∆ABC and ∆AED

AB = AE

BC = ED

AC = AD (diagonals of a regular pentagon are equal)

So, by SSS congruence criteria,

∆ABC ≅∆AED

8. A C B

a. In ∆ABD and ∆CBD

∠ADB = ∠CDB (90°)

BD = BD (common side)

AB = BC (given)

So, by RHS congruence criteria

∆ABD ≅∆CBD.

b. Since, ∆ABD ≅∆CBD.

Then, ∠A = ∠C (Corresponding parts of congruent triangles).

c. Since, ∆ABD ≅∆CBD.

Then, AD = CD (Corresponding parts of congruent triangles).

9. A B C M L P X

10. AB C DE

BC = AC (Given)

In ΔPBM and ΔPBL

∠PBM = ∠PBL (Given)

BM = BL (Given)

∠PMB = ∠PLB (90°) by ASA congruence criteria, ΔPBM ≅ ΔPBL

a. ∠DCA + ∠DCE = ∠ACE

∠ECB + ∠DCE = ∠BCD

So, ∠ACE = ∠BCD

In ∆DBC and ∆EAC

∠B = ∠A (Given)

∠ACE = ∠BCD (Proved)

So, by ASA congruence criteria

ΔDBC ≅ ΔEAC.

b. If ΔDBC ≅ ΔEAC

Then, DC = EC (Corresponding parts of congruent triangles).

11. A B R Q P C

Given that: AQ = CR

So, AQ +CQ = CR + CQ

AC = RQ

In ΔABC and ΔPRQ

∠ACB = ∠RQP (alternate interior angles)

AC = RQ (proved)

∠BAC=∠PRQ (alternate interior angles)

So, by ASA congruence criteria

ΔABC ≅ ΔPRQ.

12. The triangles necessarily need not be congruent since it does not tell anything about the lengths of the sides of the triangle.

OA = OB (O is the mid-point)

So, by SAS congruence condition

∆AOC ≅∆BOD.

OC = OD; ∠AOC = ∠BOD and OB = OA are the three equality relations of matching parts used.

Chapter Checkup

a. There can be only two acute angles in a triangle. False

b. The sum of the length of any two sides of a triangle is greater than the third side. True

c. Two angles are congruent if the sum of their measures is 180°. False

d. Two lines are congruent if they are parallel. False

2. For a set of angles to form a triangle, then by the angle sum property, the sum of the angles must be 180°.

a. 30° + 40° + 110° = 180°

The given set will form a triangle.

b. 34° + 29° + 120° = 183°

The given set will not form a triangle.

c. 50° + 120° + 10° = 180°

The given set will form a triangle.

d. 34° + 28° + 100° = 162°

The given set will not form a triangle.

3. For a set of side lengths to form a right−angled triangle, the Pythagorean theorem must be satisfied.

By the Pythagoras theorem, a2 + b2 = c2, where c is the hypotenuse.

a. 62 + 82 = 36 + 64 = 100 and 92 = 81

The given set will not form a right−angled triangle.

b. 92 + 122 = 81 + 144 = 225 and 152 = 225

The given set will form a right−angled triangle.

c. 242 + 102 = 576 + 100 = 676 and 262 = 676

The given set will form a right−angled triangle.

4. Using the angle sum property of a triangle,

a. (x − 10)° + (4x − 10)° + 70° = 180°

x − 10° + 4x − 10° + 70° = 180°

5x − 20° + 70° = 180°

5x + 50° = 180°

5x = 180° − 50° = 130°

x = 26°

∠P = (x − 10)° = (26 − 10)° = 16°

∠Q = (4x − 10) ° = (4 × 26 − 10)° = 94°

b. Using the angle sum property in triangle FGH,

x° + 80° + 50° = 180°

x + 130° = 180°

x = 50°

c. Let the angle opposite to 54° be x

Since AB = AC, triangle ABC is an isosceles triangle. In an isosceles triangle base angles are equal.

So, ∠B = 54°.

x = 54° (vertically opposite angles)

Using the angle sum property in triangle ABC,

54° + 54° + y = 180°

108° + y = 180°

y = 180° − 108°

y = 72°

5. Given : An isosceles triangle ABC with AC = BC; and AB2 = 2AC2

To prove: ABC is a right−angled triangle.

Proof:

In triangle ACB,

AC = BC

The angles corresponding to these sides are equal hence, these two angles must be less than 90 degrees.

Since AB2 = 2AC2

AB2 = AC2 + AC2

AB2 = AC2 + BC2 (since AC = BC)

By the Pythagoras theorem,

Triangle ABC is a right−angled triangle. (Hence proved)

6. In a television set, the diagonal (d), width (w) and height (h) can be represented using the Pythagoras theorem.

d2 = w2 + h2

Given: the diagonal (d) is 52 inches and the width (w) is 48 inches:

522 = 482 + h2

h2 = 522 − 482

h2 = 2704 − 2304

h2 = 400 = 20 × 20

h = 20 inches

Therefore, the height of the television screen is 20 inches.

7. In ΔABC,

AB = 6 cm

AC = 8 cm

BC = AB2 + BC2 = 62 + 82

= 36 + 64 = 100 = 10 cm

In ΔPQR,

PQ = 6 cm

PR = 10 cm

QR = PQ2 + PR2 = 62 + 102 = 36 + 100 = 136 = 2 34 cm

In ΔABC and ΔPQR,

AB = PQ

BC = PR

AC ≠ QR

So, by SSS congruence criteria ΔABC ≠ ΔPQR

Area of the rectangular sheet = 50 cm × 30 cm = 1500 cm2

If all 4 triangles are congruent, then the area of 4 isosceles triangles = area of 2 squares = 2 × 10 × 10 = 200 cm2 (If the triangles are congruent then their areas are also the same)

Area of remaining portion = 1500 − 200 = 1300 cm2

9. The situation given in the question forms right-angled triangles. Let w be the width of the street when the ladder is placed against the wall on one side of the street.

Using the Pythagorean theorem,

52 + w2 = 132 = 169

w2 = 169 − 25 = 144

w2 = 12 × 12

w = 12 m

Similarly, let z be the width of the street when the ladder is placed against the wall on the other side of the street.

122 + z2 = 132 = 169

z2 = 169 − 144 = 25 = 5 × 5

z = 5 m

Width of the street = w + z = 12 + 5 = 17 m

10.

In ΔQRT and ΔRQU,

∠QTR = ∠RUQ (90°)

QR = RQ (common hypotenuse);

QT = RU (given).

So, by the RHS congruence criteria

ΔQRT ≅ ΔRQU.

Then, ∠R = ∠Q

⇒ PQ = PR

Similarly,

PS = QT

⇒∠Q = ∠P

⇒ PR = QR

Together, we get PQ = QR = RP or ∆PQR is equilateral.

Since, all the angles of an equilateral triangle are equal then,

∠P = ∠Q = ∠R = x

So, x + x + x = 180°

3x = 180°

⇒ x = 180 3 = 60°

11.

So, the measure of each angle in ∆PQR is 60°.

50° A F E

BD C

ΔBED ≅ ΔCFD.

In ΔBED and ΔCFD,

ED = FD (given)

∠BED = ∠CFD (90°)

BD = CD (given)

So, by the RHS congruence criteria

Then, ∠B = ∠C = 50°. (Corresponding parts of congruent triangles)

12. The triangles are congruent but scaled up by a factor of 2. One section of the lattice measures 50 metres on each side, So, the new dimensions of the scaled-up section will be 50 m × 2 = 100 m

13. ∠A + ∠B + ∠C = 180° (Angle sum property)

60° + 80° + ∠C = 180°

∠C = 180° − 140° = 40°

Since ∠B and ∠C are bisected, In ∆BOC, ∠B = 40° and ∠C = 20°

∠B + ∠C + ∠O = 180°

∠O = 180° − 60° = 120°

Hence, ∠C = 40° and ∠BOC = 120°

14. Answers may vary. Sample answer: The heights of two buildings are 34 m and 29 m, respectively. If the distance between the two buildings is 12 m, what is the distance between the tops of the buildings?

Challenge

1. Assertion (A): (AC)2 = (AD)2 + 2(BD)2 + (CD)2

Reason (R): According to the Pythoras Theorem, (Hypotenuse)2 = (Perpendicular)2 + (Base)2

Assertion:

Consider ΔABC, Apply the Pythagoras theorem, (AC)2 = (AB)2 + (BC)2 (i)

Consider ΔABD, Apply the Pythagoras theorem, (AB)2 = (AD)2 + (BD)2 (ii)

Consider ΔBDC, Apply the Pythagoras theorem, (BC)2 = (CD)2 + (BD)2 (iii)

Add (ii) and (iii), (AB)2 + (BC)2 = (AD)2 + (BD)2 + (CD)2 + (BD)2

Substituting from (i), (AC)2 = (AD)2 + 2(BD)2 + (CD)2

So, the assertion is true.

Reason: The reason is true since (Hypotenus)2 = (Perpendicular)2 + (Base)2 according to the Pythagoras theorem.

The reason explains the assertion. So, both A and R are true, and R is the correct explanation of A. Thus, the correct option is a.

2. Statement 1: ∠A = ∠D, ∠B = ∠E and AB = DE

Statement 2: BC = EF.

Statement 1 provides two pairs of equal angles (∠A = ∠D and ∠C = ∠D) and one pair of equal sides (AB = DE). This is sufficient for the ASA congruence criteria because the third pair of angles will be equal due to the angle sum property of triangles.

Therefore, Statement 1 alone is sufficient to conclude that ∆ABC ≅∆DEF.

Statement 2 provides only one pair of equal sides (BC = EF). This alone is not sufficient to conclude that the triangles are congruent since it does not provide enough information to satisfy any of the congruence criteria. Hence, Statement 2 is not sufficient. Thus, option a is correct.

Case Study

1. ACB = 30°

∠B = 90°

We know that the sum of interior angles in a triangle is 180°.

So,

∠ACB + ∠ABC + ∠CAB = 180°

30° + 90° + ∠CAB = 180°

∠CAB = 180° − 120°

∠CAB = 60°

Thus, the correct option is (C).

2. AB = 24 cm

AD = BD = 24 cm ÷ 2 = 12 cm

AC = 20 cm

ΔADC forms a right-angled triangle. Apply the Pythagoras theorem, AC2 = AD2 + CD2

202 = 122 + CD2

CD2 = 202 − 122

CD2 = 400 − 144

CD2 = 256 = 16 × 16

CD = 16 cm

Thus, the correct option is (C).

3. ACB = 70°

It is given that the blue triangle is an isosceles triangle. Apply the angle sum property in a triangle.

∠ABC + ∠BAC + ∠ACB = 180°

∠ABC + ∠ABC + 70° = 180°

2∠ABC = 110°

∠ABC = ∠BAC = 55°

Now,

∠ABD = ∠ABC + x°

90° = 55° + x°

x°=35°

Let us find the value of y°.

∠ACE = ∠BCE

∠ACB + ∠ACE + ∠BCE = 360°

70° + y° + y° = 360°

2y° = 290°

y° = 145°

Thus, x° + y° = 145° + 35° = 180°

4. Consider quadrilateral AFCG.

∠FCG = 90°

Now, 15° + 110° + 90° + ∠AFC = 360°

∠AFC = 360° − 215°

∠AFC = 145°

Now, ∠AFC + ∠AFE = 180°

∠AFE = 180° − ∠AFC = 180° − 145°

∠AFE = 35°

Consider ΔAEF.

∠EAF + ∠AEF + ∠AFE = 180°

∠EAF + 110° + 35° = 180°

∠EAF = 180° − 145°

∠EAF = 35°

Thus, ∠AFC = 145° and ∠EAF = 35°.

Chapter 8

Let’s Warm-up

1. Pooja’s income = 22,000 Arushi’s income = 26,500

Pooja’s income:Arushi’s income = 22,000:26,500 = 44:53

The ratio of Pooja’s income to Arushiʼs income is 44:53

2. Highest income = 30,750

Lowest income = 18,500

Highest income:Lowest income = 30,750:18,500 = 123:74

The ratio of highest income to lowest income is 123:74.

3. Difference between Rachna’s and Amit’s income = 30,750 – 23,250 = 7500

Difference between Rohan’s and Divya’s income = 29,000 – 18,500 = 10,500

(Rachna’s income – Amit’s income):(Rohan’s income – Divya’s income)

= 7500:10,500 = 5:7

The ratio of the difference in the income of Rachna and Amit to that of Rohan and Divya is 5:7.

4. Pooja’s income + Divya’s income = 22,000 + 18,500 = 40,500

Rachna’s income + Arushi’s income = 30,750 + 26,500 = 57,250

(Pooja’s income + Divya’s income):(Rachna’s income + Arushi’s income)

= 40,500:57,250 = 162:229

The ratio of the sum of income of Pooja and Divya to that of Arushi and Rachna is 162:229.

Do It Yourself 8A

1. Total number of roses = 45

Number of red roses = 32

Number of white roses = 45 – 32 = 13

Ratio of red roses to white roses = 32:13

2. a. 3 m = 3 × 100 cm = 300 cm

120 cm:3m = 120 cm:300 cm

= 120 300 = 60 × 2 60 × 5 = 2 5 = 2:5

b. 6 L = 6 × 1000 mL = 6000 mL

6000 mL:240 mL = 6000 240 = 25 × 240 240 = 25 1 = 25:1

c. 5 km = 5 × 1000 m = 5000 m

5000 m:800 m = 5000 800 = 25 × 200 4 × 200 = 25 4 = 25:4

d. 1 hour = 3600 seconds

240 seconds:3600 seconds = 240 3600 = 240 15 × 240 = 1 15 = 1:15

3. a. 6:1 = 6 1 = 6 × 5 1 × 5 = 30 5 = 30:5

6:1 = 30:5

b. 8:5 = 8 5 = 8 × 5 5 × 5 = 40 25 = 40:25

40:25 = 8:5

c. 20:16 = 20 16 = 4 × 5 4 × 4 = 5 4 = 5:4

5:4 = 20:16

d. 55:60 = 55 60 = 11 × 5 12 × 5 = 11 12 = 11:12

55:60 = 11:12

4. a. The two ratios are 1:7 and 5:21.

Writing them in fraction form gives 1 7 and 5 21

LCM of 7 and 21 = 21.

1 7 = 3 21

On comparing 3 21 and 5 21, 3 < 5.

So, 3 21 < 5 21 or 1:7 < 5:21.

b. The two ratios are 10:5 and 14:9.

Writing them in fraction form gives 10 5 and 14 9

LCM of 5 and 9 = 45.

10 5 = 90 45; 14 9 = 70 45

On comparing 90 45 and 70 45, 90 > 70.

So, 90 45 > 70 45 or 10:5 > 14:9.

c. The two ratios are 7:15 and 6:13.

Writing them in fraction form gives 7 15 and 6 13

LCM of 15 and 13 = 195.

7 15 = 91 195; 6 13 = 90 195

On comparing 91 195 and 90 195, 91 > 90.

So, 91 195 > 90 195 or 7:15 > 6:13.

d. The two ratios are 8:15 and 9:20.

Writing them in fraction form gives 8 15 and 9 20 .

LCM of 15 and 20 = 60.

8 15 = 32 60; 9 20 = 27 60

On comparing 32 60 and 27 60, 32 > 27.

So, 32 60 > 27 60 or 8:15 > 9:20.

5. Ratios are 2:5, 7:12, 3:7, 9:2 and 11:14.

Ratios in fraction form are 2 5 , 7 12 , 3 7 , 9 2 and 11 14

LCM of 5, 12, 7, 2, 14 = 420

2 5 = 168 420; 7 12 = 245 420; 3 7 = 180 420; 9 2 = 1890 420 ; 11 14 = 330 420

As 168 < 180 < 245 < 330 < 1890;

Hence, 2:5 < 3:7 < 7:12 < 11:14 < 9:2

6. A:B = 5:11 and B:C = 7:13

LCM of 11 and 7 = 77

A B = 5 11 = 35 77 ; B C = 7 13 = 77 143

A:B:C = 35:77:143

Thus, A:C = 35:143

7. Answer may vary. Sample answer: 12 cm 9 cm 8 cm 9 cm

8. Cups of sugar in recipe = 1 2 3 = 5 3

Cups of flour in the recipe = 21 6 = 13 6

Ratio of cups of sugar to cups of flour = 5 3 : 13 6 = 10 6 : 13 6 = 10:13

9. Let the angle of triangle be 2x, 3x and 4x

2x + 3x + 4x = 180°

9x = 180°

x = 20°

Smallest angle: 2x = 2 × 20° = 40°

Largest angle: 4x = 4 × 20° = 80°

Sum of smallest and largest angles = 40° + 80° = 120°

10. Let lengths of pieces of ribbon 3x, 4x and 6x

Length of shortest piece: 3x = 12 inches

⇒ x = 4 inches

Therefore, the total length of ribbon = 3x + 4x + 6x = 13x = 13 × 4 inches = 52 inches

11. Total amount of money = ₹3700

Money received by Rehan = 2 2 + 3 + 5 × ₹3700 = 2 10 × ₹3700 = ₹740

Money received by Bobby = 3 2 + 3 + 5 × ₹3700 = 3 10 × ₹3700 = ₹1110

Money received by Heera = 5 2 + 3 + 5 × ₹3700 = 5 10 × ₹3700 = ₹1850

12. Let the initial height of the tomato plant be 7x cm, and the initial height of the pepper plant be 5x cm.

After applying the nitrogen-based fertiliser, each plant’s height increases by 9 cm. So, the new heights will be:

Tomato plant: 7x + 9

Pepper plant: 5x + 9

7x + 9

5x + 9 = 5 4

4(7x + 9) = 5(5x + 9)

28x + 36 = 25x + 45

3x = 9

x = 3

Initial heights are:

Tomato plant: 7x = 7 × 3 = 21 cm

Pepper plant: 5x = 5 × 3 = 15 cm

Thus, the height of tomato plant and pepper plant is 21 cm and 15 cm, respectively.

Challenge

1. Let the ages of Priya and Surbhi be 7x and 8x

7x + 6

8x + 6 = 10 11

77x + 66 = 80x + 60

80x – 77x = 66 – 60

3x = 6 ⇒ x = 2

Age of Priya = 7x = 7 × 2 = 14

Age of Surbhi = 8x = 8 × 2 = 16

Thus, assertion is true.

Since, Priya is 14 years and Surbhi is 16 years old. Reason is also correct, but it doesn’t explain the assertion.

Hence, both assertion and reason are true, but the reason is not the correct explanation of the assertion. So, option b is correct.

Do It Yourself 8B

1. Using, Product of extremes = Product of means

a. 5 × 18 = 6 × 15

90 = 90 = Proportional

b. 4 × 14 ≠ 3 × 16

56 ≠ 48 = Not proportional

c. 1 5 × 1 4 ≠ 2 7 × 14 5

1 20 ≠ 4 5 = Not proportional

d. 3 8 × 8 9 ≠ 7 9 × 3 21

1 3 ≠ 1 9 = Not proportional

2. a. 40:24 :: 20:x

40x = 24 × 20

40x = 480 x = 12

b. 6:8 :: x:56

8x = 56 × 6

8x = 336 x = 42

c. 2 3 : 5 7 : 4 5 :x 2 3 × x = 4 5 × 5 7 x = 3 2 × 4 7 x = 6 7

d. 3 5 : 1 2 :: x: 4 9 1 2 × x = 3 5 × 4 9 x 2 = 12 45 x = 24 45 = 8 15

3. a. 4:6 :: 6:x 4x = 36 x = 36 4 = 9

b. 12.5:2.5 :: 2.5:x 12.5x = 2.5 × 2.5 = 6.25

x = 6.25 12.5 = 1 2

c. 5 7 : 2 5 :: 2 5 :x 5 7 x = 2 5 × 2 5 = 4 25 5 7 x = 4 25

⇒ x = 28 125

d. 3 5 :4.2 :: 4.2:x 3 5 x = 4.2 × 4.2

0.6x = 17.64

⇒ x = 17.64 0.6 = 29.4

4. Amount of gasoline required to travel 126 miles = 18 gallons

Amount of gasoline required to travel 175 miles = x gallons (let)

126:18 :: 175:x Using,

Product of extremes = Product of means

126 × x = 175 × 18

x = 175 × 18 126 = 25 gallons

5. Cost of 4 TV sets = ₹1,12,000

Cost of 23 TV sets = ₹x (let)

4:1,12,000 :: 23:x 4x = 1,12,000 × 23

x = 1,12,000 × 23 4 = ₹6,44,000

Thus, the cost of 23 such TV sets is ₹6,44,000.

6. Number of days for which 270 kg grass will feed 42 cows = 21 days

Number of days for which 360 kg grass will feed 42 cows

= x days (let)

270:21 :: 360:x

270x = 21 × 360

x = 21 × 360

270 = 28 days

7. Let the number to be subtracted be x

23 – x:30 – x :: 57 – x:78 – x

23 – x

30 – x = 57 – x 78 – x 1 –23 – x 30 – x = 1 –57 – x 78 – x

30 – x – 23 + x

30 – x = 78 – x – 57 + x 78 – x 7

30 – x = 21 78 – x 546 – 7x = 630 – 21x 21x – 7x = 630 – 546

14x = 84 ⇒ x = 6

Challenge

1. Rohan, Kunal, and Vicky's ages form a continued proportion, and Kunal is twice the age of Rohan [Given].

Let Rohan’s age = r and Kunal’s age = k

Since Kunal is twice Rohan's age k = 2r Vicky’s age = v

Since their ages form a continued proportion, we have: r k = k v r 2r = 2r v [Replacing k with 2r]

1 2 = 2r v v = 4r

So, Vicky 's age is four times Rohan's age, v must be a multiple of 4.

a. 15 years: 15 is not divisible by 4, so this cannot be Vicky's age.

b. 16 years: 16 is divisible by 4, so this can be Vicky's age.

c. 32 years: 32 is divisible by 4, so this can be Vicky's age.

d. 36 years: 36 is divisible by 4, so this can be Vicky's age.

Hence the correct option is a, 15 years.

Do It Yourself 8C

1. Weight of potatoes bought in ₹16 = 2 kg

Weight of potatoes bought in ₹1 = 2 kg 16 = 1 8 kg

Weight of potatoes bought in ₹120 = 1 8 kg × 120 = 15 kg

Therefore, 15 kg potatoes can be bought for ₹120.

2. Distance travelled in 3 hours = 150 km

Distance travelled in 1 hour = 150 3 km = 50 km

Distance travelled in 9 hour = 9 × 50 km = 450 km

Therefore, the car travelled 450 km in 9 hours.

3. Cost of 15 pens = ₹300

Cost of 1 pen = ₹300 15 = ₹20

Therefore, the cost of 5 pens = 5 × ₹20 = ₹100

4. Number of times the human heart beats in 5 minutes = 375

Number of times the human heart beats in one minute

= ₹375

5 = 75

Therefore, the number of times the human heart beats in 1 hour or 60 minutes = 75 × 60 = 4500 times

5. Number of days required by 12 men to complete the work = 18 days

Number of days required by 1 man to complete the work = 12 × 18 days = 216 days

Number of days required by 8 men to complete the work = 216 8 = 27 days

Therefore, 8 men needs 27 days to complete the work.

6. Burning time of 20 cm long candle = 50 minutes

Burning time of 1 cm long candle = 50 20 = 2.5 minutes

Total length of the other candle = 20 cm + 10 cm = 30 cm

Burning time of 30 cm long candle = 30 × 2.5 cm = 75 minutes

7. Weight of 19 bags = 332.5 kg

Weight of 1 bag = 332.5 19 kg = 17.5 kg

Weight of 35 bags = 35 × 17.5 kg = 612.5 kg

Therefore, 35 such bags will weigh 612.5kg.

8. Number of shirts made by 33 m cloth = 15

Number of shirts made by 1 m cloth = 15 33 = 5 11

Number of shirts made by 110 m cloth = 5 11 × 110 = 50

9. Number of books in 1840 grams = 8 books

Number of books in 1 gram = 8 1840 = 1 230

Number of books in 4.37 kg or 4370 g = 4370 × 1 230 = 19

Therefore, 19 such books will weigh 4.37 kg.

10. Number of workers required to build the wall in 5 days = 12

Number of workers required to build the wall in 1 day = 5 × 12 = 60

Therefore, the number of workers required to build the wall in 3 days = 60 3 = 20

11. Answer will vary. Sample answer: A car travels 240 kilometres using 16 litres of petrol. How much petrol will be needed to travel 450 kilometres?

Challenge

1. Number of sweets in one box = 35 sweets

Number of sweets in five boxes = 35 × 5 sweets = 175 sweets

Number of teaspoons of flour needed for 5 sweets = 11

Number of teaspoons of flour needed for 1 sweet = 11 5

Number of teaspoons of flour need for 175 sweets = 11 5 × 175 = 385

Hence, 385 teaspoons of flour required to make sweets to fill five boxes.

So, Conclusion I follows.

Number of sweets in 5 boxes = 35 × 5 = 175 sweets

So, Conclusion II does not follow.

Thus, the correct answer is option (a). Only Conclusion I follows.

1. Distance = 400 m, Time = 25 seconds

Speed = Distance Time = 400 m 25 seconds = 16 m/s

2. Distance = 500 m

Time = 6 min = 6 × 60 seconds = 360 seconds

Speed = Distance Time = 500 360 m/s = 1.39 m/s

3. Speed = 8.5 m/s, Time = 1 minute = 60 seconds

Distance travelled = Speed × Time = 8.5 m/s × 60 s = 510 m

Thus, a whale travels 510 m in 60s.

4. Distance travelled = 3.2 km = 3200 m, Speed = 20 m/s

Time taken to finish the journey = Distance Speed = 3200 m

20 m/s = 160 seconds = 160 60 minutes = 2.67 minutes (approx.)

5. Distance between Delhi and Uttarakhand = 372 km

Speed of the car = 48 km/hr

Time taken to reach Uttarakhand = 372 48 hour = 7.75 hour = 7.75 × 60 minutes = 465 minutes

6. Distance travelled by earth = 94,00,00,000 km

Time taken = 365 days = 365 × 24 hours = 8760 hours

a. Speed of earth in km/day = 94,00,00,000 km 365 days = 25,75,342.47 km/day

b. Speed of earth in km/hour = 94,00,00,000 km 8760 hours = 1,07,305.94 km/hour

7. Distance travelled by Kirti = 36,000 m = 36000 1000 km = 36 km

Speed = 8 km/h

Time taken by Kirti to finish the journey = Distance Speed = 36 8 = 4.5 hours

Distance travelled by Mihir = 45 km

Speed = 10 km/h

Time taken by Mihir to finish the journey = Distance Speed = 45 10 = 4.5 hours

a. Both Kirti and Mihir took the same amount of time to finish the journey.

b. Since Kirti and Mihir took the same amount of time, the difference would be zero minutes.

8. Speed of jogging = 5.5 km/h

a. Rahul jogs the farthest on Monday as he spends more time jogging on Monday (17 min).

b. Time spent jogging on Monday = 17 min = 17 60 hour = 0.2833 hour

Distance covered on Monday = Speed × Time = 5.5 × 0.2833 km = 1.558 km

Time spent jogging on Tuesday = 12 min = 12 60 hour = 0.2 hour

Distance covered on Tuesday = Speed × Time = 0.2 × 5.5 km = 1.1 km

Time spent jogging on Wednesday = 15 min = 15 60 hour = 0.25 hour

Distance covered on Wednesday = Speed × Time = 0.25 × 5.5 km = 1.375 km

Time spent jogging on Thursday = 8 min = 8 60 hour = 2 15 hour

Distance covered on Thursday = Speed × Time = 2 15 × 5.5 km = 0.733 km

Time spent jogging on Friday = 0 min

Distance covered on Friday = 0 km

Challenge

1. Statement 1: The distance from P to Q is 300 km. This statement tells about the total distance and nothing about how the towns are split among the three points.

Statement 2: If the distance from P to R is half the distance from R to Q.

Given: PR = 1 2 RQ

PR + 2PR = 300 km

3PR = 300 km

PR = 100 km

So, both statements together gives the enough information to solve for the distance from P to R.

So, the answer is (c). Both statements 1 and 2 together are sufficient.

Chapter Checkup

1. 51:85 :: x:5

Using,

Product of means = Product of extremes

85x = 51 × 5

x = 51 × 5 85 = 3

2. Let the number be 7x and 8x.

7x – 24

8x – 24 = 5 6

42x – 144 = 40x – 120

42x – 40x = 144 – 120

2x = 24

x = 12

Numbers are 7x = 7 × 12 = 84

8x = 8 × 12 = 96

3. Number of widgets produced in 5 minutes = 32

Number of widgets produced in 1 minute = 32 5

Number of widgets produced in 1 hour i.e. 60 minutes

= 32 5 × 60 minutes = 384 widgets

4. Distance travelled by the plane = 2,90,000 m

Time taken = 9000

Speed of the plane = Distance Time

= 2,90,000

9000 = 32.22 m/s

5. Speed of Suman = 4 m/s

Time = 5 1 2 minutes = 11 2 minutes

= 11 2 × 60 seconds = 330 seconds

Distance travelled = Speed × Time = 4 m/s × 330 seconds = 1320 metres

6. Time taken by 25 men to construct the building = 150 days

Time taken by 1 man to construct the building = 150 × 25 days

Time taken by 30 men to construct the building = 150 × 25 30 = 125 days

7. Distance covered by the mouse = 5.5 m

Time taken = 30 seconds

Speed of the mouse = Distance Time = 5.5 30 m/s = 0.18 m/s

8. Amount of baking powder required = 1 2 tablespoon

Amount of sugar required = 13 4 tablespoon = 7 4 tablespoon

Amount of flour required = 25 8 tablespoon = 21 8 tablespoon

LCM of 2, 4 and 8 = 8

Ratio of baking powder to sugar to flour required

= 1 2 : 7 4 : 3 12

= 1 2 × 8: 7 4 × 8: 21 8 × 8

= 4:14:21

9. Distance travelled by beetle = 1.09 m Speed of beetle = 0.2 m/s = Time of flight of beetle before getting caught = 1.09 0.2 = 5.45 seconds

10. Let the length of electric pole be x Length of stick : Length of stick’s shadow :: Length of electric pole : Length of electric pole shadow

4 ft:3 ft :: x:90 ft

Using,

Product of means = Product of extremes

3x = 90 × 4

x = 90 × 4 3 = 120 feet

11. Speed of Vivek = 8.0 mph

Distance to be covered = 28 miles

Time taken to complete the race = Distance Time = 28 8 = 3.5 hours

12. Let the present ages of Suresh and Kartik be 11x and 5x Their ages five years ago are (11x – 5) years and (5x – 5) years.

11x – 5

5x – 5 = 5 2

22x – 10 = 25x – 25

25x – 22x = 25 – 10

3x = 15

x = 5

Present ages of Suresh: 11x = 11 × 5 years = 55 years

Present ages of Kartik: 5x = 5 × 5 years = 25 years

13. Total income = ₹18,000

Sunil’s expenditure = 4 4 + 1 × ₹18,000 = 4 5 × ₹18,000 = ₹14,400

Sunil’s savings = 1 4 + 1 × ₹18,000 = 1 5 × ₹18,000 = ₹3,600

Therefore, Sunil saved ₹3600 and spent ₹14,400.

14. Let the number to be added be x.

10 + x:18 + x :: 22 + x:38 + x 10 + x 18 + x = 22 + x 38 + x 1–10 + x 18 + x = 1–22 + x 38 + x 18 + x – 10 – x 18 + x = 38 + x – 22 – x 18 + x 8 18 + x = 16 18 + x 304 + 8x = 288 + 16x 16x – 8x = 304 – 288 8x = 16 x = 16 8 = 2

Thus, 2 should be added to 10, 18, 22 and 38 to make them in proportion.

Challenge

1. Amount received by Avinash = ₹28,800 LCM of 3,4,5 = 60

1 3 : 1 4 : 1 5 = 1 3 × 60: 1 4 × 60: 1 5 × 60

= 20:15:12

Let the amounts received by Kirti, Vikas and Avinash be 20x, 15x and 12x 12x = ₹28,800 x = ₹28,800 12 = ₹2400

Amount received by Kirti = 20x = 20 × ₹2400 = ₹48,000

Amount received by Vikas = 15x = 15 × ₹2400 = ₹36,000

Total amount to be distributed = ₹48,000 + ₹36,000 + ₹28,800 = ₹1,12,800

2. Statement 1: The number of marbles that can be bought for ₹30 is 300.

Statement 2: The cost of 45 marbles will be ₹3.50.

Statement 3: The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.

The cost of 1 marble: ₹1.50 15 = ₹0.10 per marble.

The number of marbles that can be bought for ₹30 = ₹30 ₹0.10 = 300 marbles.

Hence, Statement 1 is true.

The cost of 45 marbles = ₹0.10 times 45 = ₹4.50.

Hence, Statement 2 is false. The correct cost would be ₹4.50, not ₹3.50.

Statement 3 provides a definition of the unitary method, which is accurate.

Hence, Statement 3 is true.

Thus, Option b. Statements 1 and 3 are correct, but Statement 2 is incorrect.

Case Study

1. Area A: Nitrogen to Phosphorus ratio = 0.08 0.03 = 2.67

Area B: Nitrogen to Phosphorus ratio = 0.08 0.03 ≈ 0.71

Area C: Nitrogen to Phosphorus ratio = 0.02 0.01 = 2

Therefore, Area A has the highest proportion of nitrogen to phosphorus. Hence, option a is correct.

2. Nitrogen content in Area A = 0.08%

Phosphorous content in Area A = 0.03%

Ratio = 0.08 0.03 = 8 3

The ratio of nitrogen to phosphorus in Area A is 8:3.

3. Potassium content in Area B = 0.04%

Nitrogen content in Area B = 0.05%

Ratio of potassium to nitrogen content in Area B = 0.04 0.05 = 4 5

The ratio of potassium to nitrogen in Area B is 4:5.

4. True—Nitrogen content in Area C is 0.02 which is less than Area A and Area B.

5. False—Phosphorous content is highest in Area B.

Chapter 9 Let’s Warm−up

1. The most popular bakery item is coffee

2. The least popular bakery item is juice

3. The fraction of people (out of 100) who like coffee is 19 50 × 2 2 = 38 100

4. The fraction of people (out of 100) who like pastries is 1 10 × 10 10 = 10 100

5. The fraction of people (out of 100) who like juice is 1 20 × 5 5 = 5 100 .

Do It Yourself 9A

1. a. 55% = 5511 10020 = = 5511 10020 =

b. 0.36%===0.36369 100100002500

0.36%===0.36369 100100002500

c. 2.25% = 2.252259 == 10010000400 2.252259 == 10010000400 2.252259 == 10010000400

d. 56.4% = 56.4564141 == 1001000250 56.4564141 == 1001000250

56.4564141 == 1001000250

e. 1 16%=16.5%===16.516533 2 1001000200

2. a. Number of shaded squares = 62

Total number of squares = 100

Fraction of shaded squares = 62 100

Percentage of shaded portion = 62 100 × 100 = 62%

b. Number of shaded triangles = 5

Total number of triangles = 8

Fraction of the shaded triangles = 5 8

Percentage of the shaded triangles = 5 8 × 100% = 62.5%

c. Number of shaded squares = 8

Total number of squares = 16

Fraction of the shaded squares = 81 162 =

Percentage of the shaded squares = 1 2 × 100% = 50%

3. a. 48% = 48 ÷ 100 = 0.48

b. 1.25% = 1.25 ÷ 100 = 0.0125

c. 0.005% = 0.005 ÷ 100 = 0.00005

d. 1 11% 5 = 11.2% = 11.2 ÷ 100 = 0.112

e. 4 12% 10 = 12.4% = 12.4 ÷ 100 = 0.124

4. a. 0.2 = 0.2 × 100% = 20%

b. 1.2 = 1.2 × 100% = 120%

c. 2.34 = 2.34 × 100% = 234%

d. 12.1 = 12.1 × 100% = 1210%

e. 0.0048 = 0.0048 × 100% = 0.48%

5. Percentage of earth covered by land = 29%

Percentage of earth covered by water = 100% − 29% = 71%

71 ÷ 100 = 0.71

0.71 of earth’s surface is covered by water.

6. Total number of students in the class = 30 students

Percentage of students who enrolled in the art club = 20%

Ratio of the percentage of art club members to the total number of students = 20% : 100% = 1:5.

Thus, the ratio of art club members to the total number of students is 1:5

7. Percentage of blue candies = 25%

Percentage of red candies = 100% − 25% = 75%

Ratio of blue to red candies = 25%:75% = 1:3

Thus, the ratio of blue to red candies is 1:3.

8. Percentage of attendees who are children = 30%

Percentage of attendees who are adults = 100% − 30% = 70%

Ratio of children to adults = 30%:70% = 3:7.

Thus, the ratio of children to adults is 3:7.

9. Ratio of roses to tulips = 3:2

Percentage of roses = 33×100%=×100%=60% 3+25

33×100%=×100%=60% 3+25

33×100%=×100%=60%

3+25

Thus, there are 60% roses in the garden.

10. Percentage of people who prefer apple pie = 75%

Percentage of people who prefer donuts = 100% − 75% = 25%

Ratio of apple pie lovers to donut lovers = 75%:25% = 75:25 = 3:1

Thus, the ratio of apple pie lovers to donut lovers is 3:1

800 5000 × 100 5cm 5cm 3 ×100%=×100%=1% 4m25 cm 425cm 17

% = 16%

d. Let a% of 5 kg be 800 g 5 kg = 5000 g a% = 5cm 5cm 3 ×100%=×100%=1% 4m25 cm 425cm 17

11. Total number of parts = 24

Number of parts of pure gold = 22

Number of parts of other metal = 2

The ratio of pure gold to other metal = 2211 21 = 2211 21 =

So, the ratio of pure gold to other metal is 11: 1.

12. Number of cups of sugar = 3

Total number of cups = 3 + 2 = 5

Fraction of sugar to the dry ingredients = 3 5

Percentage of sugar to the dry ingredients = 3 10060% 5 ×= × 100 = 60%

So, 60% of dry ingredient is sugar.

Challenge

1. Assertion (A): 11.11% of the total people are managers.

Reason (R): The ratio of managers to employees is 1:9.

1 out of every 10 employee is a manager, which means 10% of employees are managers.

So, the assertion that 11.11% of the total people are managers is false.

If there is 1 manager for every 9 non-managers, the ratio is 1:9.

So, the reason is true.

So, the assertion is false and the reason is true. Option d is the correct answer.

Do It Yourself 9B

1. a. 25% of 250 = 25 25062.5

100 ×=

b. 30% of 750 = 30 750225

100 ×=

c. 125% of 45 = 125 4556.25

100 ×=

d. 14% of 4800 = 14 4800672

100 ×=

2. a. Let a% of 4 m 25 cm be 5 cm.

a% = 5cm 5cm 3 ×100%=×100%=1% 4m25 cm 425cm 17

5cm 5cm 3 ×100%=×100%=1% 4m25 cm 425cm 17

5cm 3

b. Let a% of 3 L be 450 mL a% = 450 mL

c. Let a% of 2 minutes 30 seconds be 75 seconds

2 minutes 30 seconds = 120 seconds + 30 seconds = 150 seconds a% = 5cm 5cm 3 ×100%=×100%=1% 4m25 cm 425cm 17

% = 50%

3. a. Let the number be a 12% of a = 48 1 12××=48 100 a 1 =48×100× 12 a

a = 400

b. Let the number be a 2 66% 3 of a = 567

2001××=567 3100 a 1

=567×100×3× 200 a a = 850.5

c. Let the number be a 125% of a = 5 hours 20 minutes = 320 minutes 1 125××=320 100 a 1 =320×100× 125 a

a = 256 minutes

a = 4 hours 16 minutes

d. Let the number be a. 20% of a = 3 m 40 cm = 340 cm 1 20××=340 100 a 1 =340×100× 20 a

a = 1700 cm

a = 17 m

4. a. Percentage decrease

Amount of decrease ×100%= Original Amount

48 =×100%=15%

b. Percentage increase

c. Percentage increase

d. Percentage decrease

= 400360 100 400 × %

= 40 400 × 100% = 10%

Percentage of money that Raju gets

= 3 2 + 3 + 5 × 100%

= 33500500150

23510 ×=×= ++ × 100%

= 30%

Amount of money that Roy gets = 5 2 + 3 + 5 × ₹500 = 55500500250 23510 ×=×= ++ ``` × ₹500 = ₹250

5. Initial concentration of acetic acid in a vinegar solution = 40%

Concentration of acetic acid in a vinegar solution after some dilution=50%

Percentage change

= 5040 40 × 100%

= 10 40 × 100% = 25%

Percentage of money that Roy gets

= 5 2 + 3 + 5 × 100%

= 55500500250

23510 ×=×= ++ ``` × 100%

= 50%

10. Percentage of copper in the alloy = 26%

Total zinc in the alloy = 1480 g

Percentage of zinc in the alloy = 100% − 26% = 74%

So, there is a 25% increase in the concentration of acetic acid.

6. Marks scored by Yash is 360 out of 500. Marks scored by Jaya is 450 out of 600.

Score percentage = Marks scored Total marks × 100

Percentage scored by Yash = 360 500 × 100 = 72%

Percentage scored by Jaya = 450 600 × 100 = 75%

So, Jaya has scored better percentage than Yash.

7. Total number of people at the event = 1440

Percentage of women in the gathering = 40%

Number of women who attended the gathering = 40 100 × 1440 = 576

Therefore, there are 576 women in the gathering.

8. Number of women employees in a factory = 450

Percentage of male employees = 55%

Percentage of female employees = 100% − 55% = 45%

Let the total number of employees be a 45% of a = 450 1

45××=450 100 a 1 =450×100× 45 a a = 1000

Thus, the total number of employees in the factory are 1000.

9. Total amount of money = ₹500

Ratio of the money that Ravi, Raju and Roy gets = 2:3:5

Amount of money that Ravi gets

= 2 2 + 3 + 5 × ₹500 = 22500500100 23510 ×=×= ++ × ₹500 = ₹100

Percentage of money that Ravi gets

= 2 2 + 3 + 5 × 100%

= 22500500100

23510 ×=×= ++ × 100% = 20%

Amount of money that Raju gets

= 3 2 + 3 + 5 × ₹500

= 33500500150

23510 ×=×= ++ × ₹500 = ₹150

Let the total quantity of the alloy be a 74% of a = 1480 g 74 1480 100 a ×= × a = 1480

a = 1480 × 100 14802000 g 74 a =×=

Thus, the total quantity of the alloy is 2000 g or 2 kg.

11. Percentage decrease in the population every year = 8%

Present population of the city = 1,69,280

Let the population of the city last year be 'a'. 169280 =×100=1,84,000 1008 a

Let the population of the city 2 years back be 'b'. 184000 =×100=2,00,000 100–8 b

Thus, the population of the city before 2 years was 2,00,000.

12. Marks scored by B = 20% more than A

Marks scored by A = 10% less than C

Marks scored by D = 20% more than C

Let the marks scored by C be 100.

Marks scored by A = 10% less than 100 = 10 100–×100=90 100

Marks scored by D = 20% more than 100 = 20 100100120 100 +×=

Percentage by which marks obtained by D is more than those obtained by A = 12090 100% 90 × = 30 1 100%33% 90 3 ×=

Thus, D obtained 1 33% 3 more marks than A.

13. Percentage decrease in the value of car every year = 20%

Value of the car after 2 years = ₹5,65,000

Value of the car after 1 year = ₹565000 100 − 20 × 100 = ₹7,06,250

Original value of the car = ₹706250 100 − 20 × 100 = ₹8,82,812.5

Thus, the original cost of the car was ₹8,82,812.5

14. Mark scored by Rahul in Mathematics = 80

Mark scored by Rahul in Science = 85

Mark scored by Rahul in Hindi = 90

Mark scored by Rahul in English = 80

Mark scored by Rahul in Social Science = 85

Average mark scored = 8085908085 5 ++++ = 420 5 = 84

Average percentage of marks scored by Rahul is 84%.

Challenge

d. Loss% = Loss2000-1800200 ×100%= ×100%=×100%=10% CP 2000 2000

Loss2000-1800200 ×100%= ×100%=×100%=10% CP 2000 2000

Loss2000-1800200 ×100%= ×100%=×100%=10% CP 2000 2000 = 10%

1. Percentage increase from 80 to 100 = ×=×= increase2010010025% initial value80 × 100

= 100 − 80

80 × 100 = ×=×= increase2010010025% initial value80 × 100 = 25%

Percentage decrease from 100 to 80 = ×= decrease 10020% initial value × 100 = 100 − 80

100 × 100 =20%

The percentage increase is 25%, while the percentage decrease is 20%.

Hence, the percentage increase from 80 to 100 is not the same as the percentage decrease from 100 to 80.

2. Percentage hike in the price of petrol = 4%

Initial price of the petrol = ₹95

Distance travelled by the man = 3456 km every month

Mileage of the car = 16 km per litre

Petrol consumed = 3456 ÷ 16 = 216 litres

Increase in the price of the petrol = 4 953.8 100 ×= × ₹95 = ₹3.8

Increase in the expenditure of the man due to petrol = 216 × ₹3.8 = ₹820.8

Thus, option (i) is correct.

3. a. Profit = SP − CP = ₹500 − ₹450 = ₹50

b. Loss = CP − SP SP = CP − Loss = ₹2364 − ₹364 = ₹2000

c. Discount = MP − SP = ₹585 − ₹450 = ₹135

Discount% = Discount135 9 100%100%23% MP 585 117 ×=×= = Discount135 9 100%100%23% MP 585 117 ×=×= = Discount135 9 100%100%23% MP 585 117 ×=×= (approx.)

Thus, the expenditure of the man increases by ₹820.8 due to the increase in the price of petrol.

Do It Yourself 9C

1. a. SP > CP, so there is gain.

Gain = SP − CP = ₹745 − ₹620 = ₹125

b. SP > CP, so there is gain.

Gain = SP − CP = ₹6000 − ₹5230 = ₹770

c. CP > SP, so there is loss.

Loss = CP − SP = ₹50,000 − ₹49,812 = ₹188

2. a. Gain = SP − CP = ₹510 − ₹405 = ₹105

Thus, option (ii) is correct.

b. Loss = CP − SP

= ₹10,235 − ₹9400 = ₹835

Thus, option (ii) is correct.

d. Discount = MP − SP = ₹3060 − ₹2550 = ₹510

Discount% = Discount510 2 100%100%16% MP 3060 3 ×=×= = Discount510 2 100%100%16% MP 3060 3 ×=×= = Discount510 2 100%100%16% MP 3060 3 ×=×=

4. Marked price of a toy car = ₹540.

Discount = 20%

Discount% = Discount amount Marked Price × 100%

Discount amount = 20 100 × 540 = ₹108

The selling price = MP − discount price = ₹540 − ₹108 = ₹432

So, the selling price of the toy is ₹432.

5. The selling price for Mia will be the cost price for her friend. Thus, the cost price of the ring for Mia’s friend will be ₹7600.

6. Cost of the scooter = ₹49,000

c. Profit% = Profit2500-2000500 ×100%= ×100%=×100%=25% CP 2000 2000

Profit2500-2000500 ×100%= ×100%=×100%=25% CP 2000 2000

Profit2500-2000500 ×100%= ×100%=×100%=25% 2000 2000 = 25%

Thus, option (i) is correct.

Cost of transportation = ₹6000

Money spent on repairs = ₹5000

Total price of the scooter to Alex’s father = ₹49,000 + ₹6000 + ₹5000 = ₹60,000

Selling price of the scooter = ₹58,000

CP > SP, so there is loss.

Loss = CP − SP = ₹60,000 − ₹58,000 = ₹2000

Thus, there is a loss of ₹2000.

7. Marked price of the water cooler = ₹8350

Discount offered = 18%

Discount% = Discount amount 100% MP ×

18% = Discount amount 100% 8350 × Discount amount = 1503

Selling price = Marked price − Discount amount = ₹8350 − ₹1503 = ₹6847

8. SP of the watches is the same.

Let the SP of each watch be x.

CP = 100 100 + Profit% × SP

CP of the first watch = 10010 1001011 xx×= + = 10010 1001011 xx×= +

CP = 100 100 − Loss% × SP

CP of the second watch = 1010 100109 xx×= = 1010 100109 xx×=

Total CP of the watch = CP of the first watch + CP of the second watch = 90x+110x 1010 200 x+x= =x 1199999

90x+110x 1010 200 x+x= =x 1199999

CP > SP, so there is loss. 200 2 –2

×100%=×100%=1%

99 xx x = 1%

×100%=×100%=1%

Thus, there is 1% loss.

9. Seling price of the article = ₹1805 Loss% = 5%

CP = 100 100 − Loss% × SP

CP = 100 100 ×1805=×1805 100–595 = 100 100 ×1805=×1805 100–595 = ₹1900

Profit = 5%

SP = 100 + Profit% 100 × CP

SP = 100+5105×1900=×1900=`1995 100100

100+5105×1900=×1900=`1995 100100 = ₹1995

10. Marked price = 35% above CP

Discount offered = 20% on MP

Let the CP be 100.

MP = 35% above 100 = 135

Discount = MPDiscount%13520 27 100%100 ×× == = MPDiscount%13520 27 100%100 ×× == = 27

SP = MP − Discount = 135 − 27 = 108

SP > CP, so there will be gain.

Profit% = Profit108100×100%=×100%=8% CP 100 –

Profit108100×100%=×100%=8% CP 100 –= 8%

Thus, the shopkeeper earns a profit/gain of 8%.

11. Selling price of 17 balls = ₹720

According to question,

Loss incurred after selling 17 balls = Cost price of 5 balls

CP of 17 balls − SP of 17 balls = CP of 5 balls

CP of 12 balls = SP of 17 balls

CP of 12 balls = ₹720

CP of 1 ball = ₹720 12 = ₹60

Thus, the cost of each ball is ₹60.

12. Selling price of the article when there is profit = ₹1920

Selling price of the article when there is loss = ₹1280

Cost price (CP) is the same in both the cases.

If the profit% and the loss% is the same, then the profit and the loss on a same article will be the same.

Profit = Loss = ₹1920 − ₹1280 2 = ₹640 2 = ₹320

CP = ₹1280 + ₹320 = ₹1600

Given: There must be a profit of 25%

SP = 100Profit%10025 CP 16002000 100 100 ++ ×=×= = 100Profit%10025 CP 16002000 100 100 ++ ×=×= × ₹1600 = ₹2000

Thus, the article should be sold at ₹2000 to make a profit of 25%.

13. Total cost price of 56 kg rice = 26 × ₹20 + 30 × ₹36 = ₹520 + ₹1080 = ₹1600

Total selling price of 56 kg rice = 56 × ₹30 = ₹1680

Profit = ₹1680 − ₹1600 = ₹80

Thus, the trader’s profit is ₹80.

14. Cost of each bulb = ₹10

Cost of 200 such bulbs = 200 × ₹10 = ₹2000

Thus, dealer must sell the article at ₹1995 to gain 5% on that article.

Number of bulbs that were fused = 5

Bulbs left = 200 − 5 = 195

Selling price of each bulb = ₹12

Total selling price of 195 bulbs = 195 × ₹12 = ₹2340

SP > CP, so there is profit.

Profit% = 2340–2000 Profit 340 ×100%= ×100=×100=17% CP 2000 2000 2340–2000 Profit 340 ×100%= ×100=×100=17% CP 2000 2000

2340–2000 340 ×100%= ×100=×100=17% 2000 2000 = 17%

Thus, the shopkeeper earns a profit of 17%.

Challenge

1. Assertion (A): The vendor buys oranges at ₹26 per dozen and sells them at 5 for ₹13. The vendor's gain percent is 20%.

Reason (R): Gain percent = Gain Cost Price × 100%

CP = ₹26 per dozen = 26 12 = ₹13 6 per orange

SP: ₹13 for 5 oranges = 13 5 = ₹2.6 per orange

Gain = SP − CP = ₹2.6 − ₹13 6 = ₹13 30

Gain percent = gain ×100% CP = 13 30 13 6 × 100% = 20%

So, the gain percent is 20%.

So, the assertion is true.

The reason explains how to calculate the gain percentage. So, the reason is true and it explains the assertion. Thus, both A and R are true, and R is the correct explanation of A.

Hence, option a is correct.

Do It Yourself 9D

1. a. 5000×5×3 PRT

SI== 100100 = ₹750

A = P + SI = 5000 + 750 = ₹5750

b. 6000×12×5 PRT

SI== 100100 = ₹3600

A = P + SI = 6000 + 3600 = ₹9600

c. 2000×10×3 PRT SI== 100100 ₹600

A = P + SI = 2000 + 600 = ₹2600

d. 2250×2×4 PRT SI== 100100 ₹180

A = P + SI = 2250 + 180 = ₹2430

2. a. P = ₹7500, R = 3% per annum and Sl = ₹2250

SI×1002250×100 T== =10 P×R7500×3 years

b. Here, Sl = 1 4 of P = 1 4 of ₹2000 = ₹500

SI×100500×100 T== =6.25 P×R2000×4 years

c. P = ₹3375, R = 15% per annum and Sl = ₹1518.75

SI×1001518.75×100 T== =3years P×R3375×15

d. P = ₹12,000, R = 4 1 2% per annum and Sl = ₹3510

SI×1003510×100 T== =6.5years P×R 1 12000×4 2

3. a.

SI×10023520×100 P== R×T6×8 ₹49,000

b. SI×1002800×100 P== R×T 1 5×9 3 ₹6000

c. SI×10023100×100 P== R×T11 5×3 22 ₹1,20,000

d. SI×10031500×100 P== R×T31 12×1 54 ₹2,00,000

4. a.

SI×1001995×100 R== =3% P×T9500×7 per annum

b. SI×1003412.5×100 R== =6.5%perannum P×T 1 12500×4 5

c. SI×10018000×100 R== =7.5%perannum P×T20000×12

d. SI×1001111.5×100 R== =0.0325%perannum P×T 1 360000×9 2

5. P = ₹40,000

R = 10% per annum

T = 2 years

SI==PRT40000×10×2 =8000 100100

A = P + SI = ₹40,000 + ₹8000 = ₹48,000

Thus, the total amount paid after 2 years is ₹48,000.

6. Total interest paid = ₹450

Rate of interest = 4.5% per annum

Time = 1 year

P = SI×100450×100 = R×T4.5×1 ₹10,000

Thus, Akila borrowed an amount of ͅ₹10,000 from her friend.

7. P = ₹2,50,000

R = 8% per annum

T = 4 years

250000×8×4 PRT

SI== =80,000 100100

Thus, Preeti has to pay ₹80,000 after 4 years.

8. SI = ₹720

R = 18%

T = 2 years

P = SI×100 R×T = 720×100 18×2 = ₹2000

He has earned 12%, so total interest = ₹720 + ₹720 × 12 100 = ₹720 + ₹86.4 = ₹806.4

New rate = SI ×100 P ×T = 806.4×100 2000×2 = 20.16%

So, the new rate is 20.16%.

9. Let the amount Abella borrowed be P. PRT

SI= 100

Total SI = SI for first 3 years + SI for next 4 years + SI for next 3 years

×5×3×8×4×12×3 17,845=++=++=15323683 100100100100100100100 PPP PPPP

P = 17845100 21,500 83 × = = ₹21,500.

Thus, she borrowed an amount of ₹21,500.

Challenge

1. Since the interest rate is constant, we can use a simple proportion to solve this.

If the money triples in 15 years, it means the interest earned is twice the original principal (3P − P = 2P) in 15 years. To quadruple the money, we need to earn thrice the original principal (4P − P = 3P).

So, if it takes 15 years to earn 2P, it will take: 15 years 3 2 P P

× 3P = 22.5 years to earn 3P.

Therefore, it will take 22.5 years for the money to quadruple itself.

Chapter Checkup

1. a. 44=×100% 55 = 80%

b. 18.75% = 18.7518753 1001000016 == = 18.7518753 1001000016 == = 18.7518753 1001000016 ==

c. 11111111

2%=%=×= 555100500 11111111

2%=%=×= 555100500

11111111

2%=%=×= 555100500

d. 25757 11==×100%=1140% 555 25757 11==×100%=1140% 555 = 1140%

e. 0.48 = 0.48 × 100% = 48%

f. 19919 4%=%=×==0.045 222100200

19919 4%=%=×==0.045 222100200

g. 5.64 = 5.64 × 100% = 564%

h. 165651 8%=%=×=0.08125 888100 165651

8%=%=×=0.08125 888100 = 0.08125

2. a. 25% of ₹460 = 251×460=×460= 115 1004 ``

b. 1 12% 2 of 20 L = 25 % 2 of 20 L = 251××20 =2.5 2100 L L

c. 1 50% 2 of 1000 = 101 % 2 of 1000 = 1011××1000=505 2100

d. 5% of 3 kg 200 g = 5% of 3200 g = 5 ×3200=160 100 g

3. a. Profit = SP − CP = ₹200 − ₹150 = ₹50

b. Loss = CP − SP = ₹1580 − ₹1235 = ₹345

c. Profit = SP − CP = ₹5235 − ₹5100 = ₹135

d. Loss = CP − SP = ₹458 − ₹369 = ₹89

4. a. 20 mL 20 mL 1 ×100%=×100%=% 6 L 6000 mL3 20 mL 20 mL 1 ×100%=×100%=% 6 L

b. 1000 m1 km ×100%=×100%=5% 20 km20 km = 5%

c. ₹2 and 10p `15 × `2 and 10 p210 p ×100%=×100%=14% `15 1500 p = 14%

d. 0.9 ×100%=20% 4.5

5. a. Percentage increase =  

Amount of increase ×100%= Original Amount =     18–12 6 ×100%=×100%=50% 12 12 = 50%

b. Percentage increase = 

Amount of increase 15 ×100%=×100%=100% Original Amount 15 = 100%

Amount of decrease

c. Percentage decrease = 

Original Amount 200 =

6. a. SP > CP, so there will be a profit. Profit% = 6550 Profit 15 100% 100%100%30% CP 50 50 ×=×=×= = ₹65 − ₹50 `50 × 100% = 6550 Profit 15 100% 100%100%30% CP 50 50

b. CP > SP, so there will be a loss.

100% 100%100%16%

Loss% = 150125 Loss 25 2 100% 100%100%16% CP 150 150 3 ×=×=×= = ₹150 − ₹125 `150 × 100% = 150125 25 2

150 150 3 ×=×=×= ` = 150125 25 2 100%100%16% 150 3 ×=×=×=

c. CP > SP, so there will be a loss.

Loss% = `1000-`896 Loss 104 ×100%= ×100%=×100%=10.4% CP `1000 1000 = ₹1000 − ₹896 `1000 × `1000-`896 Loss 104 ×100%= ×100%=×100%=10.4% CP `1000 1000 `1000-`896 Loss 104 ×100%= ×100%=×100%=10.4% CP `1000 1000 = 10.4%

7. a. Discount% = Discount amount ×100% MP = 80 100 640 × = 12.5%

b. Discount% = Discount amount ×100% MP = 289 100 1156 × = 25%

c. Discount% = Discount amount ×100% MP = 155 100 775 × = 20%

8. a. SI==PRT500×2×3 =`30 100100 ₹30

A = P + SI = ₹500 + ₹30 = ₹530

b. 550014611 PRT 22 SI 5183.75 100100 ×× === ₹5183.75

A = P + SI = ₹5500 + ₹5183.75 = ₹10,683.75

9. CP of the pen = ₹100

SP of the pen = ₹125

Since, SP > CP, so there is gain.

Gain = ₹125 − ₹100 = ₹25

Increase in expenditure

= 72–5418 1 ×100%=×100%=33% 54 54 3

Thus, the stock price increased by 1 33% 3 .

12. Total number of students in the school = 480 Ratio of boys to girls = 7:8

a. Percentage of boys in the school = 772 ×100%=×100%=46% 7+815 3 772 ×100%=×100%=46% 7+815 3

Thus, there are 2 46% 3 of the boys in the school.

b. Percentage of girls in the school = 81×100%=53% 7+8 3

Thus, there are 1 53% 3 of the girls in the school.

13. Let us denote the third number as x The first number is 20% more than the third number: 1.2x The second number is 50% more than the third number: 1.5x

The ratio of the first two numbers is: 1.2 1.5 x x = 12 15 = 4 5

So, the ratio of first two numbers is 4:5.

14. Percentage of the expenditure by Kapil out of his total income = 20%

Percentage increase in the salary = 20% Let the initial income of Kapil be 100.

Savings by Kapil initially = Income − Expenditure = 100 − 20% of 100 = 100 − 20 = 80

Increased income of Kapil = 100 + 20% of 100 = 100 + 20 = 120

Since the savings remain the same, so, savings for increased income = 80

Expenditure in increased salary = 120 − 80 = 40

Percentage increased in the expenditure =

Increase in expenditure 40-20 ×100%=×100%= Initial expenditure 20

40-20 ×100%=×100%= Initial expenditure 20

Gain% = Gain25×100%=×100%=25% CP100

Gain25×100%=×100%=25% CP100 = 25%

Thus, the gain percent is 25%.

10. Total amount spent by cobbler = ₹80 + ₹10 + ₹10 = ₹100

CP for the cobbler = ₹100 SP of each pair of shoes = ₹180 SP > CP, so there is profit.

Profit = ₹180 − ₹100 = ₹80

Profit% = Profit80100%100%80% CP 100 ×=×=

Thus, the cobbler makes a profit of 80% on each pair of shoes.

11. Initial price of the share = ₹54

Final price of the share = ₹72

Percentage increase

= 20 100%100% 20 ×= = 100%

Thus, the expenditure increased by 100%.

15. Discount% = 12%

SP of the item = ₹880

SP = MP − Discount

₹880 = MP 12 100 −× MP

₹880 = 88 100 MP

MP = ₹1000

Thus, the marked price on the item is ₹1000.

16. Discount% on the watch = 5%

Difference in the amount if the watch was sold at 6% discount = ₹27

Discount difference = Profit difference 65 MP×–MP×=27 100100

1 MP=27

100

MP = ₹2700

Thus, the marked price on the watch is ₹2700.

17. Amount borrowed by Sunita = ₹5000

Time = 2 years

Rate at which Sunita borrows money = 4% per annum

Rate at which Sunita lends money to Arya = 1 6 4 % per annum

Interest given by Sunita = 5000×4×2 =`400 100 ₹400

Interest received by Sunita from Arya = 1 5000×6×2 4 =`625 100 ₹625

Profit earned by Sunita = ₹625 − ₹400 = ₹225

Thus, Sunita gained ₹225.

Challenge

1. The ratio of seats for Mathematics, Physics and Biology in the school = 6:8:12

Let the seats for Mathematics, Physics and Biology be 6x, 8x and 12x, respectively.

Percentage increase in the number of seats for Mathematics = 40%

Increased seats for Mathematics = 6x + 40% × 6x = 6x + 2.4x = 8.4x

Percentage increase in the number of seats for Physics = 50%

Increased seats for Physics = 8x + 50% × 8x = 8x + 4x = 12x

Percentage increase in the number of seats for Biology = 75%

Increased seats for Biology = 12x + 75% × 12x = 12x + 9x = 21x

Ratio of increased seats for Mathematics, Physics and Biology in the school = 8.4x:12x:21x = 14:20:35

Thus, the ratio of the increased seats is 14:20:35.

2. Let the amount lent at 5% be x, and the amount lent at 8% be 1000 − x

The simple interest for one year at 5% is:

×5×1

SI==0.05 100 x x

2. Discount = 10% of ₹80,000 = ₹8,000

Selling Price = Marked Price − Discount = ₹80,000 − ₹8,000

So, option d is correct.

3. Percentage Profit = Profit Initial Investment × 100 = ₹0

₹50,000 × 100 = 0%

The percentage profit made by the greenhouse in the first year was 0%.

4. Simple Interest = ₹30,000 × 8 × 3 100 = ₹7,200

Total Amount = ₹30,000 + ₹7,200 = ₹37,200

The statement is false since the total amount is ₹37,200 and not ₹36,000.

5. Answer may vary

Chapter 10

Let’s Warm-up

1. Fractions with the same denominator are called like fractions

2. A fraction with the denominator greater than the numerator is called a proper fraction

3. A fraction with the numerator greater than the denominator is called an improper fraction

4. 7 14 8 is a mixed fraction.

5. 17 2 is an improper fraction.

Do It Yourself 10A

1. a. 1 out of 4 parts = 1 4 b. 3 out of 8 parts = 3 8

c. 2 out of 4 parts = 2 4 d. 4 out of 8 parts = 4 8

2. Answers may vary. Sample answers:

a. 7 1

b. 147 1.4

105 ==

c. Zero divided by anything is zero. For example, 0 5

SI= =0.081000–100 x x

The simple interest for one year at 8% is: ()() 1000–×8×1

According to the assertion, the total interest is ₹65: ()

0.05x+0.081000–=65 x Solving this,

0.05+80–0.08=65 xx

−0.03x = −15

x = 500

So, the amount lent at 5% is ₹500, and the amount lent at 8% is ₹500.

Hence, the reason is True.

Both A and R are true, but R is the correct explanation of A. So, option a is correct.

Case Study

1. Profit = Revenue − Expenses = ₹70,000 − ₹70,000 = ₹0

So, option a is correct.

d. 51 0.5

102 −==

3. a. 10 and 11 are both positive numbers, hence, a positive rational number.

b. 1 and 5 are both positive numbers, hence, a positive rational number.

c. −9 and −13 are both negative numbers, hence, a positive rational number.

d. 7 and 8 are both positive numbers, hence, a positive rational number.

4. Numerator = (3 × (–2)) = –6; Denominator = (7 × (– 4)) = –28

Fraction = = Numerator6 Denominator28

5. a.

6. a. –24÷3 –8 = 15÷35

b. –7÷7 –1 = 21÷73

c. 0.6 = 6 10 = 3 5

d. The denominator cannot be negative, hence, 70×–1 70 –70 == –98–98×–198 –7014–5 98147 ÷ = ÷

7. a. −4 11 = x 33

(−4) × 33 = 11x

x = (−4) × 33 11 = −12

b. –714 22 x =

(−7) × x = 14 × 22

x = 14 × 22 −7 = −44

c. 7 1339 x =

7 × 39 = 13x x = 7 × 39 13 = 21

d. 38 −22 = x −11

(−11) × 38 = −22x

x = (−11) × 38 −22 = 19

8. Answers may vary. Sample answer:

a. 7 13 7214

13226 × = × 7321

13339 × = × × = × 7428 13452

b. 12 20 12336 20360 × = × 12560 205100 × = × 12784 207140 × = ×

9. Fraction of pudding that Rayan ate = 1 3

Fraction of pudding that friend ate = 0.5 = 2221 4422 ÷ == ÷ Since, 11 32 ≠

They did not eat the same amount of pudding.

Challenge

1. Conclusion 1: The pizza was cut into 3 equal parts.

Conclusion 2: Maya got more pizza than Riya.

Conclusion 3: If Tia joins them, each of them will get 1 5 th of pizza.

Number of people = 3

Number of parts each ate = 1

Number of parts left = 0

Number of parts the pizza was cut in = 1 + 1 + 1 = 3

So, they cut the pizza in 3 equal parts, hence, Conclusion 1 follows.

Each ate equal amount of pizza, hence, Conclusion 2 does not follow.

If Tia joins them, number of people = 4, and the number of parts the pizza should be cut in = 4.

Hence, each will get 1 4 parts, hence, Conclusion 3 does not follow.

Do It Yourself 10B

1. a. LCM of 7 and 9 is 63. 4936 7963 × = × 5 735 9763 × = × Since, 36 > 35

Then, 3635 6363 > 45 79 ⇒>

b. Since, 2 < 7 Then, 27 33 < c. LCM of 2 and 5 is 10. 155 2510 −×− = × 3 26 5210 −×− = × Since, −5 > –6

Then, 56 1010 > 13 25 ⇒>

d. 1313 2424 = and 99 44 = LCM of 24 and 4 is 24. 13 24 9654 4624 −×− = × Since, –13 > –54

Then, 1354 2424 > 139 244 ⇒>

2. VT = TU = UW and PR = RS = SQ 0 –1 –2 V 1 P R S T U W –3 –4 –5 Q 2 3 WV is divided into three equal parts.

So, =×==−×= 3936 –3 and2 3333 WV 78 33 TU== PQ is divided into three equal parts.

So, =×==×= 1and23336 3333 PQ 45 = = 33 RS

3. Answer may vary. Sample answer.

a. Multiply –1 and 0 by 6 6 –1×=6–66;0×=0 666

So, –1–2–3–4–5 ,,,, 66666

b. LCM of 5 and 3 is 15. –43–12 5315 × = × and –2510 3515 ×− = ×

Multiply by 3 3

–123–36–103–30 ×=; ×= 1534515345

So, –31–32–33–34–35 ,,,,. 4545454545

c. Multiply by 6 6 1662612 ; 66366636 ×=×=

So, 7891011 ,,,, 3636363636

d. LCM of 10, 12, 9 and 15 is 180.

5 1890 10 18180 −×− = × , 4 1560 12 15180 −×− = × , 3 2060 9 20180 × = × , 5 1260 15 12180 × = ×−

Since 60 > −60 = −60 > −90

Then 3 9 > 5 15 > 4 12 > 5 10

So, the greatest rational number is 3 9 .

5. a. LCM of 9, 12 and 18 is 36. 1044024853157214 , ,, 943694361233618236 −×−××× ==== ××××

Since, −40 < 8 < 14 < 15

Then, <<< 10275 991812

b. LCM of 3, 6, 16 and 20 is 240.

14×1216810×151504×401601×8080 =,=,=,= 20×1224016×152406×402403×80240 14×1216810×151504×401601×8080 =,=,=,= 20×1224016×152406×402403×80240

Since, –168 < –160 < –150 < –80

Then, 144101 206163 <<<

d. LCM of 3 and 5 is 15. –2510 3515 ×− = × and 339 5315 −×− = ×

Multiply by 6 6

–106–60–96–54 ; 1569015690 ×=×=

c. LCM of 3, 5 and 7 is 105. −×−−×−−×−−×−

×××× 170703421264301201(160)105 , , ,0.5 3702105422107302102(2105)210

So, 5556575859 ,,,,. 9090909090

4. a. LCM of 3 and 5 is 15

8 × 5 3 × 5 = 40 15 , −4 × 5 3 × 5 = −20 15 , 15 15, −2 × 3 5 × 3 = −6 15

Since 40 > 15 > −6 > −20

Then 8 3 > 1 > 2 5 > 4 3

So, the greatest rational number is 8 3

b. LCM of 2, 5, 9 and 10 is 90. 1 4545 2 4590 × = × , 3 1854 5 1890 −×− = × , 4 1040 9 1090 × = × , 9 981 10 990 × = ×

Since 81 > 45 > 40 > −54

Then 9 10 > 1 2 > 4 9 > 3 5

So, the greatest rational number is 9 10 .

c. LCM of 8, 9, 10 and 11 is 3960. 1495495 84953960 × = × ; 2440880 94403960 × = × ; 33961188 103963960 × = × ; 43601584 113603960 × = ×

Since, 1584 > 1188 > 880 > 495

Then, 4321 111098 >>>

So, the least rational number is 4 11 .

Since, –126 < –120 < –105 < –70

Then, <<−< 341 0.5 573

d. LCM of 3, 5 and 6 is 30. 3618–21020–46245525 , , , 56303103056306530 ××−×−× ==== ××××

Since, –24 < –20 < 18 < 25

Then, 4235 5356 <<<

6. Warmest to coldest means higher to lower, i.e. in descending order.

a. 9280252044562464–3105315 ,, , 328084015568408105840 ×××− === ××× –141201680 7120840 ×− = ×

Since, 2520 > 2464 > −315 > −1680

Then, 944314 31587 >>>

b. 31030–3103075359436 , , , 4104041040854010440 ××−×× ==== ××××

Since, 36 > 35 > 30 > –30

Then, 9733 10844 >>>

c. –5.0,50–3944328 –3.9, 4.4, 32.8 10101010 =−===

Since, 328 > 44 > –39 > –50 Then, 32.8 > 4.4 > –3.9 > –5.0

d. –463 872576–356168504504 252 , , –1 , 772504956504504504863504 × ××−−− ==×== ×××

Since, 576 > –168 > –252 > –504

Then, 834 1 798 >>>−

7. Number of glasses filled by Raman = 11 5

Number of glasses filled by Farhan = 13 7

LCM of 5 and 7 is 35. 11777 5735 × = × and 13565 7535 × = ×

Since, 77 > 65

Then, 1113 57 >

So, Raman’s bottle can hold more water.

8. For a possible value of wheat production by another state that falls between these two figures, we can choose any rational number between 16.1 and 32.4 million tonnes. For example, a state could have produced 24.0 million tonnes, which can be expressed as the rational number 240 10 = 24.

Challenge

1. a. The largest two−digit number = 99

f. LCM of 7 and 3 is 21. 15345 7321 −×− = × and 177119 3721 × = × 451194511974 21212121 ()−−+ +==

2. a. 55 0 77 −= b. 1111 0 88 −=

c. 11 0 22 −= d. 99 0 1111 −=

3. a. 13133330 0 2623666 ×− −=−=== × 13133330 0 2623666 ×− −=−=== × 13133330 0 2623666 ×− −=−=== × b. 2×97×5 1835 27 1835 17 ==== 595×99×545454545 ––2×97×5 1835 27 1835 17 ==== 595×99×545454545 ––2×97×5 1835 27 1835 17 ==== 595×99×545454545 ––2×97×5 1835 27 1835 17 ==== 595×99×545454545 ––

The smallest 3−digit number = 100 99 100 ⇒

b. The smallest three−digit number = 100

c. 3537582140214019 87877856565656 ×× −=−=−== ×× 3537582140214019 87877856565656 ×× −=−=−== ×× 3537582140214019 87877856565656 ×× −=−=−== ××

The largest three−digit number = 999 100 999 ⇒

⇒ LCM of 100 and 999 is 99,900 9999998,901 10099999,900 × = × 10010010,000 99910099,900 × = × Since, 98,901 > 10,000 Then, 99100 . 100999 > The correct answer is option (a).

Do It Yourself 10C

1. a. LCM of 3 and 5 is 15. 12560 3515 × = × and 14342 5315 × = × 6042(6042)10234 151515155 + +===

b. 0.8 = 8 10 = 4 5

4 5 + 2 5 = 6 5

c. 22 5 + 5 5 = 27 5

d. 2 5 + −9 5 = −7 5

e. LCM of 3 and 5 is 15. 8540 3515 × = × and 8324 5315 −×− = × 4024402416 15151515 ()

+==

d. 717(1)718 1 88888 −−−+ −==== 717(1)718 1 88888 −−−+ −====

4. a. 983458350 5 99999 −+− ×++==   983458350 5 99999 −+− ×++==

b. LCM of 8, 12 and 16 is 48. 76549342202742202735 861241634848484848 −×−××−−−−+− ++=++== ××× 76549342202742202735 861241634848484848 −×−××−−−−+− ++=++== ×××

c. LCM of 50, 5 and 6 is 150. 932305252760125 503530625150150150 −××−×−− ++=++= ××× = 27+60125 9246 == 15015075 = 27+60125 9246 == 15015075 = 27+60125 9246 == 15015075

d. LCM of 33, 22 and 44 is 132. 2436138183818313 334226443132132132132132 ××−× −+=−+== ××× 2436138183818313 334226443132132132132132 ××−×

5. Answer may vary. Sample answer:

6. Let the other number be x. 14 2 5 x +=−

x = −2 + 14 5 = 1014 5 −+ = 4 5

So, the other number is 4 5 .

7. Let the other number be x. 5 7 + x = −1 x = −1 + 5 7 = −+75 7 = 2 7

So, the other number is 2 7 .

8. Length of wood = 17 4 m

Remaining length of the wood

××

17171751748568856817 = 45455420202020 ×× ==−= ×× 17171751748568856817 = 45455420202020 ×× ==−= ×× 17171751748568856817 = 45455420202020 ×× ==−= ×× 17171751748568856817 = 45455420202020 m

9. Distance from Ghaziabad to Delhi is 57 2 km

Distance from Delhi to Gurgaon is 32.7

Total distance from Ghaziabad to Gurgaon via Delhi

= 57 2 + 32.7

= 57327285327 21010 + +=

= 612 10 = 306 5 km

So, the total distance from Ghaziabad to Gurgaon via Delhi is 306 5 km.

10. Answer may vary. Sample answer: Raj bought 14 9 m of cloth. He again bought 13 8 m of cloth. How much cloth does he have in total?

Challenge

1. Statement 1: He spent ₹630 9 on tea and snacks, ₹508 2 on food, ₹44 on repairs of the taxi and saved the rest.

Statement 2: The taxi driver spent 46% of his earnings.

To find out how much he saved, we need to subtract his total expenses from his earnings.

Earnings = ₹800

Total expenses = ₹ 630508220 925  ++

₹ 630 9 

 = ₹70; ₹ 508 2 

= ₹254; ₹ 220 5     = ₹44

Total expenses = ₹(70 + 254 + 44) = ₹368

Savings = Earnings − Total expenses = ₹800 − ₹368 = ₹432

So, according to Statement 1, the taxi driver saved ₹432. Statement 2:

The taxi driver spent 46% of his earnings on various expenses.

To find out how much he saved, we can calculate the remaining percentage of his earnings, which represents his savings.

Remaining percentage = 100% − 46% = 54%

Savings = 54% of ₹800 = 220 5     54 100 220 5  

 × ₹800 = ₹432

So, according to Statement 2, the taxi driver saved ₹432. Hence, (IV) Either Statement 1 and Statement 2 alone are sufficient.

Do It Yourself 10D

c. 611611666 117117777 × ×=== × 611611666 117117777 × ×=== ×

d. 3838242 20920918015 −−×−− ×=== × 3838242 20920918015 −−×−− ×=== ×

3. a. 6263189 – –535–2105 ÷=×== 6263189 – –535–2105 ÷=×==

b. 4414 – –3–55–315 ÷=×=

c. 131441 8483246 ÷=×== 131441 8483246 ÷=×==

d. ÷=−×== 164163484 213214847 ÷=−×== 164163484 213214847

4. a. 354532545 489842898  ×× ÷×=÷

6. a. + +=== 8560856014529 12512512512525 + +=== 8560856014529 12512512512525 + +=== 8560856014529 12512512512525

⇒ multiplicative inverse 25 29 = b. + +=== 5555105 44442 + +=== 5555105 44442 ⇒ multiplicative inverse 2 5 = 7. Required number = 4849363 30930824020 −÷=−×=−=− 4849363 30930824020 −÷=−×=−=− 4849363 30930824020 −÷=−×=−=− 4849363 30930824020 −÷=−×=−=− 8. 6546957605435 15 799799796363

6520172729 =–÷–=×–=–=–8872 82016020

b. 4–51841218–4818–1618 912199–51945191519

4–51841218–4818–1618 912199–51945191519

–166–96 =×= 51995

211422614262672 137261314131413

–6089–60–5340–1780 ×=×== 9639567189

9. (m + n) ÷ (m – n) a. 131342428 2 222222224  +÷−=÷=×==−   b. 35353×35×53×35×5 +÷–=+÷–53535×33×55×33×5

    92592534–341534 =+÷–=÷=×= 15151515151515–16–16 (16)     –34–17 168 ==

10. Growth of sunflower in a day = 5 2 cm

Growth of sunflower in 2.5 days = 5 2 × 25 10 = 25 4 cm

So, the sunflower will grow 25 4 cm in 2.5 days.

11. Distance Sambhav ran in 5 days = 1 11 4 km = 114145 44 ×+ = km

Distance Sambhav ran in 1 day = 454519 5 4454 ÷=×= km

12. Length of 1 ribbon = 153 60 m.

Length of 20 such ribbons = 153 60 × 20 = 153 3 m.

So, the length of 20 such ribbons will be 153 3 m.

Challenge

1. Statement 1: The area of the garden is more than 7 m2

b. 4134213813813213 7147214141414142 ×+ +=+=+=== × 4134213813813213 7147214141414142 ×+ +=+=+=== × 4134213813813213 7147214141414142 ×+ +=+=+=== × 4134213813813213 7147214141414142 ×+ +=+=+=== × 4134213813813213 7147214141414142 ×+ +=+=+=== × 33 0 22 −=− = Additive inverse

Statement 2: The perimeter of the garden is 12.16 m.

Length of the garden = 1 4 3 m

Width of the garden = 1.75 m = 7 4 m

the

Hence, both statements are true.

Chapter Checkup

1. Numerator = (65 – 32) = 33

Denominator is 3 × 9 = 27

Rational number = 33 27 2.

3. a. HCF of 8 and 72 is 8. 881 7289 ÷ = ÷ b. HCF of 12 and 60 is 12. 12121 60125 ÷ −=− ÷ c. HCF of 3 and 45 is 3. 331 45315 ÷ −=− ÷ d. HCF of 36 and 54 is 18. 36182 54183 ÷ = ÷

4. a. 464765283058 57573535 ×+×+ +=== × 464765283058 57573535 ×+×+ +=== × 464765283058 57573535 ×+×+ +=== × 5858 0 3535 −=− = Additive inverse

5. Answers may vary. Sample answer:

a. 122 6212 × = × 133 6318 × = × 144 6424 × = ×

b. 9218 12224 × −=− × 9327 12336 × −=− × 9436 12448 × −=− ×

c. 428 7214 × = × 4312 7321 × = × 4416 7428 × = × d. 122 11222 −×− = × 133 11333 −×− = × 144 11444 −×− = ×

6. a. 48–32 7856 −× = × Hence, x = 56

b. 8216 40280 × = × Hence, x = 80

c. 4205 5255 ÷ = ÷ Hence, x = 4

d. –5420 7428 ×− = × Hence, x = −20

7. Answer may vary. Sample answer:

a. Multiply 3 5 and 4 5 , by 6 6 3618 5630 ×= and 4624 5630 ×= So 1920212223 ,,,,, 3030303030

b. Multiply 2 9 and 3 9 , by 6 6 2612 9654 ×= and 3618 9654 ×= So, 1314151617 ,,,, 5454545454

c. Multiply –1 2 and –3 2 , by 3 3 133 236 ×= and –339 236 ×= So, 45678 ,,,, 66666

d. Multiply –1 and 1, by 3 3 33 1 33 −×= and 33 1 33 ×= So, 2112 ,,0,, 3333

111161 ×+×6=+=+3 242828 1+2425 == 88

h. 13–1413–7–1334–182–91–442 ×+×+×=+ + 115115115555555 –182–91–442–715 = ==–13 5555

13–1413–7–1334–182–91–442 ×+×+×=+ + 115115115555555 –182–91–442–715 = ==–13 5555 

8. a. 797159101059019513 1015101515015010 ×+×+ +==== ×

212215451 –52521010 ×−×−− === × 212215451 –52521010 ×−×−−

c. 15 42 42 × 153 505010 == d. 428480 14801428 ÷=× 7 1808040 1479849 =×==

÷=× 7 1808040 1479849 =×==

14801428 ÷=× 7 1808040 1479849 =×==

13–1413–7–1334–182–91–442 ×+×+×=+ + 115115115555555 –182–91–442–715 = ==–13 5555

= −13 9. –5 12 × 4 15 5 ––3

Multiplicative inverse = 3 10

10. Multiply 4 9 by 4 4 4416 9436 −×− = × As 4 –9 and 16 –36 are equivalent rational numbers, they represent the same number.

11. 65865846532653297 123123412121212 ×+ +=+=+== × 65865846532653297 123123412121212 ×+ +=+=+== × = 65 + 32 12

65865846532653297 123123412121212 ×+ +=+=+== ×

65865846532653233 123123412121212 ×− −=−=−== × 65865846532653233 123123412121212 ×− −=−=−== × 65865846532653233 123123412121212 ×−

× 65865846532653233 123123412121212 ×−

8–1458–(–1)45 –×=× 9950950

8+145945 =×=× 950950 =1×=459

8–1458–(–1)45 –×=× 9950950

8+145945 =×=× 950950 =1×=459

So, 9733971297 1212123333 ÷=×= 9733971297 1212123333 ÷=×=

8–1458–(–1)45 –×=× 9950950

5010

8+145945 =×=× 950950 =1×=459

5010

8+145945 =×=× 950950 =1×=459

5010

g. 111161 ×+×6=+=+3 242828 1+2425 == 88

12. LCM of 8, 4 and 27 is 216. 427108154549872 , , 827216454216278216 ××× === ××× Since, 108 > 72 > 54 Then, 491 8274 >> difference between the greatest and the least number = 41412424221 8484288884 ×− −=−=−=== ×

13. Number of books in a library = 18,000

Mystery novels books = 1 6

Science fiction books = 2 6

Number of historical fiction books

= 18,000 –3 18,000 6

= 18,000 − 3 18,000 6

= 18,000 – 9000 = 9000

There are 9000 historical fiction books in the library.

14. Speed of the car = 212 25 km/minute

Time = 1 minute

Distance travelled in a minute = 212 25 × 1 = 212 25 km

Distance travelled in 3.75 minutes or 375 100 = 212 25 × 375 100 = 53 13159 255 × = km

So, the car travelled 159 5 km in 3.75 minutes.

15. Distance crawled in one minute = 1 3 6 = 3 6 1 6 ×+ = 19 6 inches

Distance crawled in the next minute = 2.75 inches = 11 4 inches

Total distance = 19 6 + 11 4 = ()() 19 2 11 3 12 ×+× = 38 33 12 + = 71 12 inches

Therefore, the insect crawled 71 12 inches.

16. Original price = ₹12,000

Sale price = 5 12,00010,000 6 ×= ` = ₹10,000

Tanmay paid ₹10,000 for the computer.

17. Quantity of the aquarium = 40 gallons

Aquarium filled = 20 4032 25 ×= gallons

Quantity needed = 40 – 32 = 8 gallons. Therefore, he needs 8 gallon

18. Answer may vary. Sample answer:

A gardener has 7 8 litres of liquid fertiliser. He needs to divide it equally among 1 2 litres of water to dilute it properly for his plants. How many litres of water can he mix with the fertiliser?

Challenge

1. To find the original number, we need to multiply the simplified fraction 1 7

by the removed factor (7).

Original number = 17 77

= 7 49

So, the original rational number was 7 49

2. Original price = ₹30,000

Sale price = 5 6 ×× 25,000 = 25,000

Samay paid ₹25,000 for the computer.

So, the assertion is true.

Reason: There is a sale of ₹5000 on the AC. This is true but it does not explain the assertion.

Hence, option (b). Both (A) and (R) are true, and (R) is not the correct explanation of (A).

Case Study

1. Combined Fraction = 75123 2020205 +==

So, option b is correct

2. 0.30 × 40 billion metric tons = 12 billion metric tonnes

So, option b is correct

3. Decimal: 0.25

Fraction: 1 4

4. Fraction of total emissions represented by agriculture: 6.817 or 0.17 40100 = or 0.17

5. Answer may vary.

Chapter 11

Let’s Warm-up

1. You construct an angle of 90° by bisecting a 120° angle. False

2. An angle bisector bisects the angle into 3 equal parts. False

3. When an angle bisector of a 60° angle is constructed, we get two angles of 30° each. True

4. While bisecting an angle, the width of compass can be less than half of the width of the arc. False

5. You can construct an angle of 15° by bisecting an angle of 95° False

Do It Yourself 11A

1. a. A triangle can be constructed with the length of three sides given.

b. A triangle cannot be constructed if the sum of lengths of any two sides is less than the length of third side.

c. We need the length of one side/sides to construct an equilateral triangle.

2. Scale may vary. a.

3. Scalene triangle; since none of the sides are equal.

4. Yes, a triangle can be constructed with side lengths as 5 cm, 12 cm and 13 cm since the sum of the lengths of 2 smaller sides is greater than the length of the largest side.

4. XYZ = 91°, YZX = 44° Scalene triangle since none of the sides are equal; Obtuseangled triangle since one angle is greater than 90°.

5. Equilateral triangle since all the sides are equal.

5. ΔPQR is a scalene triangle since none of the sides are equal. and acute-angled triangle since all three angles are less than 90°. QR is the longest side in the triangle.

6. BAC = 37°, BCA = 23°

Challenge

1. Perimeter of the triangle = 18 cm

Length of sides of the triangle are in the ratio 2:3:4.

2x + 3x + 4x = 18 cm

9x = 18 cm

x = 18 9 = 2 cm

Length of each side of the triangle is 2 × 2 = 4 cm, 3 × 2 = 6 cm and 4 × 2 = 8cm

Do It Yourself 11B

1. To construct a triangle with SAS criterion, we need the measure of two sides and an included angle. Hence, option b and c are the correct answers.

2. a.

1. The length of the third side is 5.3 cm.

Do It Yourself 11C

2. XZ = 6.2 cm, YZ = 3.2 cm

R = 70°

4. DF = 6.3 cm, EF = 5.1

5. Scalene triangle since none of the sides are equal; Acute angled triangle since all three angles are less than 90°. LN = 7.1 cm; MN = 8.7 cm

Unknown side = 7 cm (Unknown side)

3. BC = 6.8 cm; BAC = 46°, BCA = 44°

Unknown angles are 32° and 58°.

Challenge

1. If 3 m on ground is equal to 1 cm on paper then, 15 m = 15 3 cm = 5 cm & 6 m = 6 3 cm = 2 cm

From the construction, we can see that the unknown side measures 4.6 cm. Since, 1 cm on paper is equal to 3 m on the ground. So, the approximate length of the unknown side of Sarah's garden on ground = 4.6 × 3 m = 13.8 m

Chapter Checkup

1. a. Length of 3 sides - SSS

b. Length of 2 sides and included angle - SAS

c. Length of the hypotenuse of a right-angled triangle and one of its sides - RHS

d. Measures of 2 angles and included side - ASA

2. A = 82°, B = 60° and C = 38°. Since all three angles are less than 90°. So, the triangle is acute-angled triangle

3. Scalene triangle since none of the sides are equal.; Obtuseangled triangle since has exactly one angle that is greater than 90°, while the other two angles are less than 90°. (111°, 41°, 28°)

10. Sum of length of sides UW and VW = 6 cm + 4.9 cm = 10.9 cm

4. SAS criterion: Two sides and one included angle.

5. AC = 6 cm; BC = 4.1 cm

6. XZ =

7. Remaining angles are 55° and 80°.

8. Length of unknown side = 3.6 cm approx.

11. If 10 cm on model is equal to 1 cm on paper then, 45 cm = 45 10 cm = 4.5 cm and 30 cm = 30 10 cm = 3 cm

From the construction, we can see that the unknown side measures 3.4 cm.

The approximate length of the unknown side of the model bridge = 3.4 × 10 cm = 34 cm

Challenge

1. No, it is not possible to construct the unique triangle, since we cannot determine the position of point C uniquely.

We need the length of the other side or measure of another angle to construct the triangle uniquely.

We will use the SSS rule or SAS rule to construct this triangle.

2. Assertion(A): A triangle can be constructed with the base measuring 6 cm and the base angles measuring 50° and 95°. Reason(R): We can construct a triangle using the ASA construction rule when the measures of two angles and the length of the included side are given.

A is true since a triangle can be constructed with the base measuring 6 cm and the base angles measuring 50° and 95° using the ASA rule.

R is also true since the ASO construction rule is stated correctly and it explains the assertion correctly.

So, option c is the correct answer.

Case Study

1. Option b

All three sides of the triangle have equal lengths hence equilateral triangle.

2. Option c

We need the length of one side of the equilateral triangle as three sides of the triangle are equal.

Chapter 12

Let’s Warm-up

1. a. 60 mm = 6 cm, 80 mm = 8 cm

Perimeter = 2 cm + 14 cm

+ 12 cm + 14 cm + 6 cm

+ 8 cm + 2 cm + 6 cm

+ 2 cm + 10 cm + 8 cm + 12 cm = 96 cm

b. Area of part I = 14 × 2 = 28 sq. cm

Area of part II = 8 × 2 = 16 sq. cm

Area of part III = 14 × 2 = 28 sq. cm

Area of part IV = 2 × 2 = 4 sq. cm

Area of part V = 8 × 2 = 16 sq. cm

Total area of the figure = 28 + 16 + 28 + 4 + 16 = 92 sq. cm

2. a. Perimeter = 168.4 cm + 208 cm + 84.2 cm + 208 cm

+ 82.2 cm + 84.2 cm + 168.4 cm + 250.6 cm = 1254 sq. cm

b. Area of part I = 376.4 × 84.2 = 31,692.88 sq. cm

Area of part II = 168.4 × 82.2 = 13,842.48 sq.cm

Area of part III = 84.2 × 84.2 = 7089.64 sq. cm

Total area of the figure = 31,692.88 + 13,842.48 + 7089.64 = 52,625 sq. cm

Do It Yourself 12A

1. a. Perimeter of the rectangle = 2 (l + b) = 2 (6.8 + 3.7) = 2 × 10.5 = 21 cm

Area of the rectangle = l × b = 6.8 cm × 3.7 cm = 25.16 sq. cm

b. Perimeter of square = 4 × side = 4 × 4.5 cm = 18 cm

Area of square = side × side = 4.5 × 4.5 = 20.25 sq. cm

2. a. Perimeter of the square = 4 × side = 16 cm side = 16 ÷ 4 = 4 cm

Area of the square = side × side

= 4 × 4 = 16 sq. cm

b. Perimeter of the square = 4 × side = 32 cm side = 32 ÷ 4 = 8 cm

Area of the square = side × side

= 8 × 8 = 64 sq. cm

c. Perimeter of the square = 4 × side = 52 m side = 52 ÷ 4 = 13 m

Area of the square = side × side

= 13 × 13 = 169 sq. m

d. Perimeter of the square = 4 × side = 108 m side = 108 ÷ 4 = 27 m

Area of the square = side × side

= 27 × 27 = 729 sq. m

3. a. P = 10 cm, L = 2 cm

B = P 2 – L

= 10 2 – 2

= 5 – 2 = 3 cm

Area = L × B

= 2 × 3 = 6 sq. cm

b. P = 18 cm, L = 5 cm

B = P 2 – L

= 18 2 – 5

= 9 – 5 = 4 cm

Area = L × B

= 5 × 4 = 20 sq. cm

c. P = 48 cm, L = 21 cm

B = P 2 – L

= 48 2 – 21

= 24 – 21 = 3 cm

Area = L × B

= 21 × 3 = 63 sq. cm

d. P = 94 m, L = 38 m

B = P 2 – L

= 94 2 – 38 = 47 – 38 = 9 m

Area = L × B

= 38 × 9 = 342 sq. m

4. Perimeter of a square = 4 × side

= 4 × 15 m = 60 m

Perimeter of the rectangle = perimeter of the square = 60 m

Length of the rectangle = 18 m

Therefore, breadth of the rectangle = Perimeter 2 – Length

= 60 2 – 18 = 30 – 18

= 12 m

5. Cost of fencing a square field = perimeter of square × ₹12/m = ₹24,000

Perimeter of the square = ₹24,000 ÷ ₹12 = 2000 m

4 × side = 2000 m

So, side = 2000 m ÷ 4 = 500 m

Area = side × side

500 m × 500 m = 2,50,000 sq. m

6. B = 3 × L, A = 1083 sq. m

L × B = 1083 sq. m

L × 3 × L = 1083 sq. m

L × L = 1083 ÷ 3 = 361 = 19 × 19

So, L = 19 m

B = 3 × L

= 3 × 19 = 57 m

P = 2 (L + B)

= 2 (19 + 57)

= 2 × 76 = 152 m

Cost of fencing = ₹25 × 152 = ₹3800

7. Area of the sheet of paper = 105 cm × 45 cm = 4725 sq. cm

Area of the rectangle cut out = 20 cm × 6 cm = 120 sq. cm

Number of pieces of paper = 4725 120 = 39.375

Number of equal-sized pieces of paper, from a sheet of paper = 39

8. Perimeter of rectangle: area of rectangle = 5:3

Perimeter of the rectangle = 2 (l + b)

Area of the rectangle = length × breadth

So, 2 (l + b)

l × b = 5 3

6(l + b) = 5(l × b)

Let the length of the rectangle be 2x, and the breadth of the rectangle be 3x

Therefore, 6(2x + 3x) = 5(2x × 3x)

⇒ 6 × 5x = 5 × 6x2

⇒

30x = 30x2

30x2

30x = 1

⇒ x = 1

Length of the rectangle = 2 units,

Breadth of the rectangle = 3 units

Perimeter of the rectangle = 2 (l + b) = 2 (2 + 3) = 2 × 5 = 10 units

Area of the rectangle = length × breadth = 2 × 3 = 6 sq. units

9. The length of the Indian 10-rupee note = 123 mm

The width of the Indian 10-rupee note = 63 mm

Area of the Indian 10-rupee note = 123 mm × 63 mm = 7749 sq. mm

Perimeter of the Indian 10-rupee note = 2(123 mm + 63 mm) = 372 mm

10. Answers may vary. Sample answer: Sana is planning to tile her kitchen floor

The kitchen is rectangular in shape, with a length of 8 metres and a width of 6 metres. How much area does she need to cover with tiles?

Challenge

1. Area of the rectangle = l × b

Length of the new rectangle = 3l

Breadth of the new rectangle = b 6

Area of the new rectangle = 3l × b 6 = l × b 2

So, the ratio of new rectangle to the old rectangle = lb 2 :lb = 1:2

2. Area of rectangle = area of square = 49 sq. cm

If the length is of maximum possible measure, then the length of rectangle must be at least 49 cm.

So, breadth of rectangle will be 1 cm such that its area is 49 sq. cm.

Perimeter of the rectangle = 2 (l + b)

= 2 (49 + 1) = 2 × 50 = 100 cm

If the side of the square is 7 cm, then the area of the square will be 49 sq. cm.

So, the side of the square = 7 cm

Perimeter of the square = 4 × side = 4 × 7 = 28 cm

Since, 100 > 28, the perimeter of the rectangle is greater than the perimeter of the square.

Do It Yourself 12B

1. a. A = b × h

= 9 × 7 = 63 sq. cm

b. A = b × h

= 6 × 8 = 48 sq. cm

c. A = b × h

= 7 × 3 = 21 sq. cm

2. a. A = b × h = 11 × h

88 = 11h

So, h = 88 11 = 8 cm

b. A = b × h = 9.2 × h

152.4 = 9.2h

So, h = 152.4 9.2 = 16.56 cm

c. A = b × h = 16 × b

252.8 = 16 × b

So, b = 252.8 16 = 15.8 cm

d. A = b × h = 7.6 × b

70.68 = 7.6 × b

So, b = 70.68 7.6 = 9.3 cm

3. Base of a parallelogram = 2 × height of the parallelogram

Area of the parallelogram = 648 sq. cm

Area of a parallelogram = b × h

So, b × h = 648 or 2h × h = 648

2h2 = 648

h2 = 648 2 = 324 = (18)2

So, h = 18 cm

b = 2 × 18 = 36 cm

4. Perimeter of the parallelogram = 2 (l + b) = 64 cm

l + b = 64 2 = 32 cm

l – b = 8 cm

2l = 32 + 8 = 40 cm

l = 40

2 = 20 cm = base of the parallelogram

Height of parallelogram = 8 cm

Area of the parallelogram = b × h

= 20 × 8 = 160 sq. cm

5. Let the adjacent sides of the parallelogram be 4x and 3x

Length of the parallelogram = 8.5 cm = 4x

x = 8.5

4 = 2.125 cm

So, the other side of the parallelogram is

3 × 2.125 cm

= 6.375 cm

Perimeter = 2 (l + b) = 2 (8.5 + 6.375) = 2 × 14.875 = 29.75 cm

6. Area of parallelogram = b × h = 25 × 10 = 250 sq. cm

Area of parallelogram = b × h

250 = 15h

So, h = PM = 250 15 = 16.7 cm (approx.) 25 cm O Q M R S L P 15 cm

9 cm

7. Base length of each bunting = 12 cm

Height of each bunting = 5 cm

Area of each bunting = b × h = 12 × 5 = 60 sq. cm

The area of each bunting is 60 sq. cm.

8. a. Area of parallelogram = b × h = 21 × 10 = 210 sq. cm

Area = b × h

= LO × MF

210 = 15 × h

h = MF = 210 15 = 14 cm

b. Area of parallelogram = b × h = 30 × 12 = 360 sq. cm

Area = b × h = LO × MF

360 = b × 20

b = LO = 360 20 = 18 cm

Challenge

1. Let the length of smaller side be a

Let the length of the other side be b

a = 3 4 b

Length of the perpendicular of the larger side h1 is 18 m.

Area = a × h1 = b × h2

3 4 b × h1 = b × 18

h1 = b × 18 × 4

3 × b = 24 cm

So, the length of the perpendicular on the smaller side is 24 cm.

Do It Yourself 12C

1. a. A = 1 2 × b × h

= 1 2 × 16 × 13 = 104 sq. cm

b. A = 1 2 × b × h

= 1 2 × 10 × 8 = 40 sq. in

c. A = 1 2 × b × h

= 1 2 × 9 × 5 = 22.5 sq. cm

2. a. A = 1 2 × b × h

= 1 2 × 11 × 8

= 44 sq. cm

b. A = 1 2 × b × h

49.77 = 1 2 × 15.8 × h

h = 49.77 × 2 15.8 = 6.3 cm

c. A = 1 2 × b × h

39.90 = 1 2 × b × 7.6 cm

b = 39.90 × 2 7.6 = 10.5 cm

3. Area of triangle = 1 2 × b × h

= 1 2 × 8 × 6 = 24 sq. cm

24 = 1 2 × 10 × h

h = 24 × 2 10 = 4.8 cm

4. a. Area of triangle = 1 2 × 8 × 7.5 = 30 sq. cm

b. Area of triangle = 1 2 × 15 × CE

30 = 1 2 × 15 × CE

CE = 30 × 2 15 = 4 cm

5. Let the base and the height of the triangle be 3x and 4x, respectively.

Area of triangle = 1 2 × 3x × 4x = 96 sq. cm

⇒ 6x2 = 96 sq. cm

⇒ x2 = 96 6 = 16 = (4)2

⇒ x = 4

So, base of the triangle = 3 × 4 = 12 cm

Height of the triangle = 4 × 4 = 16 cm

6. Dimensions of the rectangle = 30 m × 25 m

Area of the rectangle = l × b = 30 × 25 = 750 sq. m

Since, the diagonals of a rectangle bisect each other at O, O is the mid-point.

So, height of the triangle = OF = 1 2 × 25 = 12.5 cm

Area of triangle AOB = 1 2 × base (length of rectangle) × height of triangle = 1 2 × 30 × 12.5 = 187.5 sq. cm

Hence, the area of the shaded region = 750 − 187.5 = 562.5 sq. cm

7. a. Area of the parallelogram = b × h = 50 × 35 = 1750 sq. cm

Area of triangle = 1 2 × b × h = 1 2 × 1750 = 875 sq. cm

Area of the shaded region = 1750 – 875 = 875 sq. cm

b. Area of rectangle = 27 × 30 = 810 sq. cm

Area of one triangle = 1 2 × 10 × 15 = 75 sq. cm

Area of the other triangle = 1 2 × 12 × 20 = 120 sq. cm

Area of the shaded region = 810 – 75 – 120 = 615 sq. cm

8. Height of the Great Pyramid of Giza = 147 m

Base of the Great Pyramid of Giza = 230 m

Area = 1 2 × b × h

= 1 2 × 230 × 147

= 16,905 sq. m

So, area of each face of the Great Pyramid of Giza is 16,905 sq. m.

Challenge

1. ABCD is a square of side 14 cm .

So, AB = AD = DC = BC = 14 cm. The ratio of side DF to FC is 1:4

So, the length of DF = 1 14 + × 14 cm

= 2.8 cm

FC = DC - DF

= 14 cm – 2.8 cm = 11.2 cm

Area of the triangle CEF = 1 2 × b × h = 1 2 × 11.2 cm × 14 cm = 78.4 sq.cm

Area of the square ABCD = 14 cm × 14 cm = 196 sq. cm

Triangle GFC is common to both square ABCD and triangle CEF. We will subtract the area of the square ABCD and triangle CEF, which will eliminate the common triangle GFC and give only the area of the shaded region.

Difference between the shaded region = area of square ABCD – area of triangle CEF

= 196 sq. cm – 78.4 sq. cm

= 117.6 sq. cm

So, the difference between the shaded region is 117.6 sq. cm.

Do It Yourself 12D

1. a. R = 14 cm, C = 2πr

C = 2 × 22 7 × 14 = 88 cm

b. R = 28 cm, C = 2πr

C = 2 × 22 7 × 28 = 176 cm

c. R = 42.7 cm, C = 2πr

C = 2 × 22 7 × 42.7 = 268.4 cm

2. a. C = 75.42 cm

C = πD

So, 75.42 = 22 7 × D

So, D = 75.42 × 7 22 = 24 cm

b. C = 352 cm

C = πD

So, 352 = 22 7 × D

So, D = 352 × 7 22 = 112 cm

c. C = 502.85 cm

C = πD

So, 502.85 = 22 7 × D = 160 cm

3. Circumference of a circular playground = 286 m

2πr = 286 m

22 7 × D = 286 m

D = 286 × 7 22 = 91 cm

So, the diameter of the playground is 91 cm.

4. Circumference of a circle – diameter = 90 cm

2πr – D = 90 cm

2πr – 2r = 90 cm

22 7 – 1

22 – 7 7

= 90 cm

2r × 15 7 = 90 cm

30r 7 = 90 cm

r = 90 × 7 30 = 630 30 = 21 cm

5. D = 2r = 50 cm r = 25 cm

Perimeter of semicircle =    1 2 × 2πr

+ D

= (3.14 × 25) + 50

= 78.5 + 50 = 128.5 cm

6. Let the radii of the two circles C1 and C2, be 5r and 6r, respectively.

Circumference of circle 1

Circumference of circle 2 = 2πr1 2πr2

= 2 × 22 7 × 5r 2 × 22 7 × 6r

= 5 6 = 5:6

7. Distance covered by a car tyre in 1 revolution = Circumference of a circle = 2πr

Distance covered in 1 revolution = 2πr

= 2 × 22 7 × 40

Distance covered in 140 revolutions = 2 × 22 7 × 40 × 140 = 3520 m = 35,200 cm

So, the distance covered by tyre in 140 revolutions is 35,200 cm.

8. Cost of fencing a circular garden = circumference × ₹40

₹36,500 = 2πr × ₹40

₹36,500 = 2 × 22 7 × r × ₹40

r = 36,500 × 7 2 × 22 × 40 = 145.17 m

9. Diameter of the wheel of a truck = 57.15 cm

Radius of the wheel of the truck = 28.575 cm

Circumference of one revolution by the truck = 2πr

= 2 × 22 7 × 28.575 cm = 179.62 cm

Total distance to be covered by a truck = Total circumference

1 km 30 m = 1030 m = 1,03,000 cm

Number of revolutions = 103000 179.62 = 573.43

Hence, the wheel of the truck will make 574 revolutions.

10. Radius of Ain Dubai = 125 m

C = 2πr

= 2 × 3.14 × 125

= 785 m

Distance between two adjacent cabins = 785 48 = 16.36 m

So, the distance between two adjacent cabins is 16.36 m.

11. Side of the square = 88 cm

Perimeter of the square = 4 × side = 4 × 88 = 352 cm

Circumference of the circle = perimeter of square = 352 cm

2πr = 352 cm

2 × 22 7 × r = 352 cm

r = 352 × 7

2 × 22 = 56 cm

Challenge

1. Diameter of the circular field = 96 m

C = πD = 3.14 × 96 = 301.44 m

Initial speed of the man is 8 m/minute

Time to complete one round = 301.44 8 = 37.68 minutes

After the break, his speed decreases by 20%

New speed = 8 × (1 – 0.2) = 6.4 m/minute

For the second round, he takes a path that reduces the distance by 10%:

301.44 × (1 − 0.1) = 301.44 × 0.9 = 271.29 m

Time to complete this reduced distance at his new speed:

271.29

6.4 = 42.38 minutes

The total time taken for both rounds including the break is:

37.68 + 5 + 42.39 = 85.07 minutes

So, the man takes 85.07 minutes totally

Do It Yourself 12E

1. a. r = 7 cm

Area of circle = πr2

= 22 7 × 7 × 7

= 154 sq. cm

b. r = 21 cm

Area of circle = πr2

= 22 7 × 21 × 21

= 1386 sq. cm

c. r = 3.5 cm

Area of circle = πr2

= 22 7 × 3.5 × 3.5

= 38.5 sq. cm

d. r = 10.5 cm

Area of circle = πr2 = 22 7 × 10.5 × 10.5

= 346.5 sq. cm

2. a. Area of the circle = 154 sq. cm

πr2 = 22 7 × r × r

= 154 sq. cm

r × r = 154 × 7 22 = 49

r2 = (7)2

r = 7 cm

d = 2r = 2 × 7 = 14 cm

b. Area of the circle = 616 sq. cm

πr2 = 22 7 × r × r = 616 sq. cm

r × r = 616 × 7 22 = 196

r2 = (14)2

r = 14 cm

d = 2r = 2 × 14 = 28 cm

c. Area of the circle = 38.5 sq. cm

πr2 = 22 7 × r × r

= 38.5 sq. cm

r2 = 38.5 × 7 22 = 12.25

r = 3.5 cm

d = 2r = 2 × 3.5 = 7 cm

d. Area of the circle = 346.5 sq. cm

πr2 = 22 7 × r × r = 346.5 sq. cm

r2 = 346.5 × 7 22 = 110.25

r = 10.5

d = 2r = 2 × 10.5 = 21 cm

3. a. Diameter of semicircle = 14 cm

Radius of semicircle = 14 2 = 7 cm

Area of semicircle = 1 2 πr2

= 22 2 × 7 × 7 × 7 = 77 sq. cm

b. Radius of a quarter = 21 cm

Area of a quarter = 1 4 πr2

= 22 4 × 7 × 21 × 21 = 346.5 sq. cm

c. Diameter of a bigger semicircle = 42 mm, Radius of a bigger semicircle = 21 mm

Area of a bigger semicircle = 1 2 πr2

= 1 2 × 22 7 × 21 × 21 = 693 sq. mm

Radius of a bigger semicircle = Diameter of a small semicircle = 21 mm

Radius of a small semicircle = 21 2 = 10.5 mm

Area of two small semicircles

= 2 × 1 2 πr2

= 2 × 1 2 × 22 7 × 10.5 × 10.5 = 346.5 sq. mm

Total area = 693 – 346.5 = 346.5 sq. mm

d. Diameter of the semicircle = 4.9 cm

Radius of the semicircle = radius of quadrant = 4.9 cm ÷ 2 = 2.45 cm

Area of semicircle = 1 2 πr2

= 22 2 7 × × 2.45 × 2.45 = 9.4325 sq. cm

Area of 2 quadrants = 2 × 1 4 πr2

= 1 2 × 22 7 × 2.45 × 2.45 = 9.4325 sq. cm

Total area = area of semicircle + area of 2 quadrants = 9.4325 + 9.4325 = 18.865 sq. cm

4. Let the radii of the two circles C1 and C2, be 4r and 7r, respectively.

Area of circle 1

Area of circle 2 = πr12 πr22 = 22 7 × 16r2 22 7 × 49r2 = 16 49 = 16:49

5. Let the radius of circle P be r.

Area of circle P

Area of circle Q = πr12 πr22 = 22 7 × r2

22 7 × 36 × 36 =

r 36

2 = 25 36 =

5 6

2 r 36 = 5 6

r = 5 6 × 36 = 30 cm

Area of circle P = πr2

= 22 7 × 30 × 30

= 2828.57 sq. cm

6. Area of the washer (including the hole) = πr2 = 22 7 × 8 × 8

= 201.14 sq. mm

Area of the hole = πr2

= 22 7 × 2.5 × 2.5 = 19.64 sq. mm

Area of the washer (excluding the hole) = 201.14 sq. mm –19.64 sq. mm = 181.5 sq. mm

7. Area of circle = 1386 sq. cm

πr2 = 22 7 × r × r = 1386

r2 = 1386 × 7 22 = 441

r2 = (21)2

r = 21 cm

Circumference = 2πr

= 2 × 22 7 × 21 = 132 cm

Perimeter of the square = circumference of the circle = 132 cm 4 × side = 132 cm

Side of the square = 132 ÷ 4 = 33 cm

Area of the square = 33 × 33 = 1089 sq. cm

Challenge

1. Radius of the cone = 24 cm

Area of the sheet of paper used to make the birthday caps

= 3 4 × πr2

= 3 4 × 22 7 × 24 × 24 = 1357.71 sq. cm

Area of 20 sheets of paper used to make birthday caps

= 20 × 1357.71 sq. cm = 27,154.2 sq. cm

Do It Yourself 12F

1. a. Total area of the figure

= Area of rectangle + area of semicircle

= (28 × 30) + 122 27 × × 14 × 14

= 840 + 308

= 1148 sq. cm

b. Total area of the figure

= Area of the rectangle – area of semicircle

= (5.6 × 8.2) − 122 27 × × 2.8 × 2.8

= 45.92 – 12.32

= 33.6 sq. cm

c. Total area of the figure

= Area of rectangle + area of semicircle

= (18 × 14) + 122 27 × × 7 × 7

= 252 + 77

= 329 sq. cm

a. Area of the shaded region

= area of square PQRS – area of a quarter circle

= (14 × 14) − 1 4 × 22 7 × 14 × 14 = 196 – 154 = 42 sq. cm

b. Area of the shaded region = area of rectangle HIJK – area of circle

= (4 × 6) − 22 7 × 2 × 2 = 24 – 12.57 = 11.43 sq. cm

c. Area of the shaded region = area of the circle – area of the triangle

22 7 × 14 × 14 − 1 2 × 28 × 14 = 616 – 196 = 420 sq. cm

3. Area of circular sheet of metal = 3.14 × 8 × 8 = 200.96 sq. cm

Area of a quarter circle = 1 4 × 3.14 × 5 × 5 = 19.625 sq. cm

Area of the remaining sheet = 200.96 – 19.625 = 181.335 sq. cm

4. Area of rectangular piece of metal sheet = 150 × 90 = 13500 sq. cm

Area of three circular sheets = 3 × 3.14 × 25 × 25 = 5887.50 sq. cm

Area of the metal sheet left = 13500 – 5887.50 = 7612.50 sq. cm

5. Area of square = 32 × 32 = 1024 sq. cm

Area of circle = 22 7 × 16 × 16 = 804.57 sq. cm

Area of the remaining portion of the cardboard = 1024 – 804.57 = 219.43 sq. cm

6. Area of rectangular sheet of paper = 50 × 38 = 1900 sq. cm

Area of 27 buttons = 27 × 22 7 × 2.5 × 2.5 = 530.36 sq. cm

Area of the remaining sheet = 1900 – 530.36 = 1369.64 sq. cm

7. Area of the circular card = 22 7 × 44 × 44 = 6084.57 sq. cm

Area of 2 circles = 2 × 22 7 × 6 × 6 = 226.29 sq. cm

Area of a rectangle = 5 × 2 = 10 sq. cm

Area of the remaining mask = 6084.57 – (226.29 + 10) = 5848.28 sq. cm

Challenge

1. Area of the big circle = 22 7 × 14 × 14 = 616 sq. cm

Area of circle 1 = 22 7 × 3.5 × 3.5 = 38.5 sq. cm

Area of circle 2 = 22 7 × 10.5 × 10.5 = 346.5 sq. cm

Area of the unshaded part of the circle = Area of circle 1 +

area of circle 2 = 38.5 + 346.5 = 385 sq. cm

Area of the shaded part of the circle = 616 – 385 = 231 sq. cm

Ratio of unshaded part to the shaded part of the circle = 231:385 = 3:5

Do It Yourself 12G

1. Length of the floor = 57.8 m

Width of the floor = 36.1 m

Floor area = l × b = 57.8 × 36.1 = 2086.58 sq. m

So, the floor area of memorial is 2086.58 sq. m.

2. Area of the garden = 3.14 × 80 × 80 = 20,096 sq. m

Area of the pond = 3.14 × 28 × 28 = 2461.76 sq. m

Area of the remaining garden = 20,096 – 2461.76 = 17,634.24 sq. m

So, the area of remaining garden = 17,634.24 sq. m

3. Area of the field = 100 × 50 = 5000 sq. m

a. Area of the crossroads = (100 × 2) + (50 × 2) – (2 × 2) sq. m = 200 + 100 – 4 sq. m = 296 sq. m

b. Area of the remaining field = (100 × 50) – 296 sq. m = 4704 sq. m

Cost of grassing per 10 sq. m = ₹110

Cost of grassing 4704 sq. m = ₹110 10 × 4704 = ₹51,744

4. Area of 1 wall = 18.4 × 10.2 = 187.68 sq. m

So, area of 4 walls = 187.68 × 4 = 750.72 sq. m

Area of 1 door = 2 m × 1 m = 2 sq. m

So, area of 2 doors = 2 × 2 = 4 sq. m

Area of 1 window = 1 m × 0.6 m = 0.6 sq. m

So, area of 3 windows = 0.6 × 3 = 1.8 sq. m

Area of 4 walls to be whitewashed

= 750.72 – (4 + 1.8)

= 750.72 – 5.8

= 744.92 sq. m

The cost of whitewashing the walls with an area of 744. 92 sq. m = ₹80 × 744.92 = ₹59,593.60

5. Circumference of a circular track = 176 m

2 × 22 7 × r = 176

r = 176 × 7 22 × 1 2 = 28 m = inner radius

Width of the track = 8 m

Outer radius = 8 m + 28 m = 36 m

Area of the track = 22 7 × 36 × 36 − 22 7 × 28 × 28

= 22 7 × (1296 – 784)

= 22 7 × 512

= 1609.15 sq. m

New length of the pool = 12 m – (2 × 1.5) m = 9 m

New breadth of the pool = 5 m – (2 × 1.5 m) m = 2 m

Area of coping = (12 × 5) – (9 × 2)

= 60 – 18

= 42 sq. m

Cost of coping 1 sq. m = ₹50

Cost of coping 42 sq. m = ₹50 × 42 = ₹2100

7. Cost of carpeting per sq. m = ₹150

Total cost of carpeting = ₹45,000

Area of the hallway = Total cost of carpeting Cost of carpeting per sq. m = 45000 150 = 300 sq. m

Length of the hallway = 300 ÷ 15 = 20 m

8. Length of the square cut out = 10 cm

New length = 60 – (2 × 10)

= 60 – 20 = 40 cm

New width = 45 – (2 × 10)

= 45 – 20 = 25 cm

Area of the base = 40 × 25 = 1000 sq. cm

Area of 4 sides:

Area of 2 sides with sides of length 10 cm and 40 cm

= 2 × 40 × 10

= 800 sq. cm

Area of 2 sides with sides of length 10 cm and 25 cm = 2 × 25 × 10 = 500 sq. cm

Total surface area to be painted = area of base + area of 4 sides

= 1000 + 800 + 500 = 2300 sq. cm

9. Circumference of circular field = 2πr = 2 × 3.14 × 80 = 502.4 m

Speed = 12 km/hr = 12,000 m/hr = 200 m/minute

Time taken = distance speed

= 502.4

200 = 2.51 minutes

Time taken = 2.51 minutes × 60 = 150.72 seconds

10. Area = 628 sq. cm

πr2 = 22 7 × r × r = 628 sq. cm

r2 = 628 × 7 22 = 199.81 = 14.14 cm

The camera can monitor an area with a radius of 15 metres, which is greater than the 14.14 metres radius of the parking lot. So yes, the security camera will cover the entire parking lot.

11. Area of rectangular lawn = 30 cm × 20 cm = 600 sq. cm

Area of the lawn that the cow can graze = Area of a quarter circle

= 1 4 × πr2 = 1 4 × 22 7 × 8 × 8 = 50.29 sq. cm

Area of the remaining lawn = 600 – 50.29 sq. cm = 549.71 sq. cm

12. Answer may vary. Sample answer:

The length of the side of a square picnic mat is 62 cm. Find the area and the perimeter of the square picnic mat.

Challenge

1. a. Radius of the tent = 25.5 + 3.5 = 29 m

Circumference formed by the tent = 2πr

= 2 × 3.14 × 29

= 182.12 m

Number of sticks required to tie the rope = circumference

÷ distance between the sticks

= 182.12 m ÷ 7 m

= 26.01 sticks

So, 26 sticks will be required.

b. Area of stage = πr2

= 3.14 × (8.5)2 = 226.865 sq. m

Area of tent = πr2 = 3.14 × (25.5)2

= 2041.785 sq. m

Area left for audience = area of tent – area of stage = 2041.785 sq. m − 226.865 sq. m

= 1814.92 sq. m

Chapter Checkup

1. a. Area of triangle = 1 2 × 6 × 6 = 18 sq. cm

b. A = b × h = 6 × 4 = 24 sq. cm

c. Area of inner circle = πr2 = 22 7 × 7 × 7 = 154 sq. cm

Area of outer circle = πr2 = 22 7 × 14 × 14 = 616 sq. cm

Area of shaded region = 616 sq. cm – 154 sq. cm = 462 sq. cm

2. Area of a square = 81 sq. cm (side)2 = (9)2 sq. cm

Side = 9 cm

Area of 4 triangles = 4 × 1 2 × b × h = 4 × 1 2 × 4.5 × 4.5 = 40.5 sq. cm

Area of the square formed by joining the midpoints of the sides of the given square = 81 – 40.5 = 40.5 sq. cm

3. Let the length of the rectangle be l and the breadth be b

Length of the rectangle = 2l, breadth of the rectangle = 1 3 b

Area of the new rectangle = 2l × 1 3 b = 2 3 lb

Area of the original rectangle = l × b

Ratio of the new rectangle to the original rectangle

= 2 3 lb : lb

= 2 3 lb lb = 2 3 1

= 2 3 = 2:3

4. Side of the square = 108 m

Area of the square = 108 × 108

= 11,664 sq. m

Area of the rectangle = area of square = 11,664 sq. m

Length × breadth = 11,664 sq. m

160 × breadth = 11,664 sq. m

b = 11,664

160 = 72.9 m

So, the width of the rectangular plot is 72.9 m.

5. Length of the floor = 6.68 m = 668 cm

Width of the floor = 8.2 m = 820 cm

Area of the floor = 668 × 820 = 5,47,760 sq. cm

Dimensions of a tile = 64 cm by 32 cm

Area of 1 tile = 64 cm × 32 cm = 2048 sq. cm

Number of tiles = 547760 2048 = 267 (approx.)

So, the number of tiles that are required will be 268 full tiles.

Cost of 1 tile = ₹20.50

Cost of 268 tiles = ₹20.50 × 267 = ₹5474

6. Area of figure 1 = 1 2 × b × h

= 1 2 × 25 × 6 = 75 sq. cm

Area of figure 2 = 1 2 × b × h

= 1 2 × 24 × 5 = 60 sq. cm

Total area of the parallelogram park = 75 sq. cm + 60 sq. cm = 135 sq. cm

7. Area of parallelogram = 1520 sq. m

b × h = 1520 sq. m

28 × h = 1520 sq. m

h = 1520 28 = 54.29 m = DP

Area of parallelogram

= 1520 sq. m

b × h = 1520 sq. m

28 × h = 1520 sq. m

h = 1520 28 = 47.5 m = BQ

8. Area of a scalene right-angled triangle = 1 2 × b × h

= 1 2 × 15 × 8 = 60 sq. cm

9. Circumference + radius of a circle = 153 cm

2πr + r = 153 cm

2 × 22 7 + 1

= 153

44 7 + 1

= 153

r × 51 7 = 153

r = 153 × 7 51 = 21 cm

The diameter = 2 × 21 = 42 cm

Circumference = 2πr = 2 × 22 7 × 21 = 132 cm

Area of the circle = 22 7 × 21 × 21 = 1386 sq. cm

10. Area of the shaded region = area of the rectangle – area of 2 quarters = (15 × 30) – area of a semicircle

= 450 − 1 2 × 22 7 × 15 × 15

= 450 – 353.57 = 96.43 sq. cm

11. Length of the rectangular lawn = 22 m

Breadth of the rectangular lawn = 15 m

Perimeter of the lawn = 2 (l + b) = 2 (22 + 15) = 2 × 37 = 74 m

Number of shrubs that can be planted in 1 m of hedge = 5

Number of shrubs that can be planted in 74 m of hedge = 74 × 5 = 370 shrubs

12. Radius of the ground = 70 m

Area of the ground = πr2 = 3.14 × 70 × 70 = 15,386 sq. m

Cost of levelling and planting is ₹120 per sq. m

Cost of levelling and painting 15,386 sq. m = 15,386 × 120

= ₹1,846,320

So, the total cost of levelling and planting grass at Wankhede Stadium is ₹1,846,320.

13. Length of the Bagh = 587 m

Width of the Bagh = 251 m

Area of the Bagh = 587 × 251 = 147,331 sq. m

So, the area of the Shalimar Bagh is 147,331 sq. m

14. Area of the circular sheet = 22 7 × 27 × 27 = 2291.14 sq. cm

Area of 4 circles with radius 1.6 cm cut = 4 × 22 7 × 1.6 × 1.6 = 32.18 sq. cm

Area of a rectangle cut = 5 × 3 = 15 sq. cm

Area of a square cut = 4 × 4

= 16 sq. cm

Area of the remaining sheet = 2291.14 – (32.18 + 15 + 16)

= 2291.15 – 63.18 = 2227.96 sq. cm

15. Outer circumference = 352 m, inner circumference = 308 m

2 × 22 7 × r1 = 352 m, 2 × 22 7 × r2 = 308 m

r1 = 352 × 7 22 × 2 , r2 = 308 × 7 22 × 2

r1 = 56 m, r2 = 49 m

Width of the track = outer radius – inner radius = 56 – 49 = 7 m

16. Let the adjacent sides of a parallelogram be 7x and 3x.

Perimeter of a parallelogram = 2 (l + b) = 120 cm

= 2 (7x + 3x) = 120

= 2 × 10x = 120

x = 120 20 = 6

So, two adjacent sides of a parallelogram are 7 × 6

= 42 cm, 3 × 6 = 18 cm.

Area of the parallelogram = b × h = 42 × 14 = 588 sq. cm

17. Circumference of Monaco Grand Prix Circuit is 3.337 km

C = 2πr = 2 × 22 7 × r

r = 3.337 7 2 22 × × = 0.53 km

Area = πr2 = 22 7 × 0.53 × 0.53 = 0.882 sq. km

Circumference of the Indianapolis Motor Speedway is around 4.023 km.

C = 2πr = 2 × 22 7 × r

r = 4.023 7 2 22 × × = 0.64 km

Area = πr2 = 22 7 × 0.64 × 0.64 = 1.287 sq. km

Difference in areas = 1.287 – 0.882 = 0.405 sq. km

18. Area of the rectangular garden = 25 m × 15 m = 375 sq. m

New length of the garden after margin inside = 25 – (1.5 + 1.5) = 22 m

New breadth of the garden after margin inside = 15 – (1.5 + 1.5) = 12 m

Area of the margin for planting lilies = 375 – (22 × 12) = 111 sq. m

Cost of 1 lily stem = ₹35

Cost of 4 lily stems = ₹35 × 4 = ₹140

Cost of 1 sq. m = ₹140

Cost of 111 sq. m = ₹140 × 111 = ₹15,540

19. Diameter = 18 cm, radius = 9 cm

Circumference = 2πr = 2 × 22 7 × 9 = 56.58 cm = 0.5658 m

Cost of putting 100 cm (= 1 m) lace around a circular cushion = ₹22

Cost of putting 56.58 cm lace around a circular cushion = ₹ 22 ÷ (100 × 56.58) = ₹12.45/m

Challenge

1. Perimeter = 80 m

2(l + b) = 80 m

l + b = 40 m

We should choose the length and breadth, such that the sum of length and breadth is 40 m and the area is the maximum.

a. l = 20 m, b = 20 m, area = 400 sq. m

b. l = 30 m, b = 10 m, area = 300 sq. m

c. l = 35 m, b = 5 m, area = 175 sq. m

For length = 20 m and breadth = 20 m, the area is the maximum.

2. Assertion (A): If the radius of a circle is 14 m then the area of the circle is 88 sq. cm and its circumference is 616 m.

Reason (R): The area of a circle is given by the formula A = πr2, while the circumference is given by C = 2πr, where r is the radius.

C = 2πr = 2 × 22 7 × 14 = 88 m

Area = πr2 = 22 7 × 14 × 14

= 616 sq. cm

So, the assertion is false.

The reason correctly states the formula to find the area and circumference.

So, the reason is true.

So, the assertion is incorrect and reason is correct.

So, option d is correct.

Case Study

1. Total length = 39 m

Length of bedroom, bathroom and living area

= 12 m + 6 m + 16 m = 34 m

Length of garden area = 39 m – 34 m = 5 m

Width of living area = width of the living area = 22 m

So, option c is correct.

2. Area of the bathroom = l × b = 6 m × 8 m = 48 sq. m

So, option d is the correct answer.

3. Perimeter of living area = 2(16 m + 22 m)

= 2 × 38 m = 76 m

So, the perimeter of the living area is 76 m.

Thus, option c is correct.

4. Perimeter of the kitchen = 2(9 + 10) = 2 × 19 = 38 m

Perimeter of the dining area = 2(9 + 8) = 2 × 17 = 34 m

Total perimeter = 38 m + 34 m = 72 m

5. Area of the bathroom = l × b = 6 m × 8 m = 48 sq. m

Area of 1 tile = 4 sq. m

Number of tiles required = 48sq. m 4sq. m = 12

So, 12 tiles are required for bathroom.

Chapter 13

Let’s Warm-up

1. 105, 100, 95, 90, 85

2. 9, 19, 29, 39, 49, 59

3. 15, 28, 41, 54, 67, 80

4. 90, 75, 60, 45, 30, 15

5. 2, 4, 8, 16, 32, 64, 128

Do It Yourself 13A

1. If 89 is the successor of a natural number n, then n = 89 − 1 = 88

2. If an even number is denoted by n; the next even number can be given as n + 2

3. Height of the triangle = 20 cm

Area of the triangle = 1 2 × base × height = 175 cm2

Length of base = 175 × 2 height = 350 20 = 17.5 cm

4. Perimeter of the square = 4 × side = 62 cm

Length of the side of the square = 62 4 = 15.5 cm

Area of the square = side2 = 15.5 × 15.5 = 240.25 cm2

5. Sum of first n natural number = n(n + 1) 2

Sum of first 12 natural number = 12(12 + 1) 2 = 12 × 13 2 = 78

Sum of the first n natural number = n (n + 1) 2

Sum of first 18 natural number = 18(18 + 1) 2 = 18 × 19 2 = 171

Sum of the first 12 natural number and the first 18 natural number = 78 + 171 = 249.

6. The rule for the given pattern: 1, 3, 5, 7, 9 is (2n − 1) where n is a natural number.

The 100th term can be given as: 2 × 100 − 1 = 200 − 1 = 199

7. Length of the base of the mirror = 8 feet

Height of the mirror = 5 feet

Area of the mirror = 1 2 × base × height

= 1 2 × 8 × 5

= 40 2 = 20 sq. feet

8. Perimeter of the regular pentagon = 30 cm

Length of the wire = Perimeter of regular hexagon

30 = 6 × length of the side of hexagon

Length of each side of hexagon = 30 cm ÷ 6 = 5 cm

9. Length of side of the square = 35 cm;

Length of rectangle = 45 cm

Area of the square = side2 = 352 = 1225 cm2

Given, that the area of the square = area of a rectangle

Area of the rectangle = length × breadth = 1225 cm2

50 × breadth = 1225 cm2

Breadth = 1225 50 = 24.5 cm

10. To find the number of bacteria after 5 hours, we can use the pattern of doubling.

The sequence goes as follows:

In 0 hours: 1 bacterium

In 1 hour: 1 × 2 = 2 bacteria

In 2 hours: 2 × 2 = 4 bacteria

In 3 hours: 4 × 2 = 8 bacteria

In 4 hours: 8 × 2 = 16 bacteria

In 5 hours: 16 × 2 = 32 bacteria

The formula for the number of bacteria after n hours is: N = 2n where, N is the number of bacteria, and n is the number of hours.

So, after 5 hours, there will be 25 = 32 bacteria.

Challenge

1. Length of Nikhil’s plot = 25 m

Perimeter of Nikhil’s plot = 75 m

Perimeter = 2 × (Length + Breadth)

75 = 2 × (25 + Breadth)

75 2 = 25 + Breadth

37.5 = 25 + Breadth

37.5 − 25 = Breadth

Breadth = 12.5 m

According to question,

Area of Kunal’s plot = 2 × Area of Nikhil’s plot

Area of Nikhil’s plot = 25 × 12.5 = 312.5 m2

Area of Kunal’s plot = 2 × 312.5 m2 = 625 m2

= 25 m × 25 m

Side of Kunal plot = 25 m

Perimeter of Kunal’s plot = 4 × 25 m = 100 m

Thus, the perimeter of Kunal’s plot is 100 m.

Do It Yourself 13B

1. a. Addition of x and y = x + y

b. One-fourth of a number a added to thrice the number b = 1 4 × a + 3b = a 4 + 3b

c. 3 less than z added to the product of x and y = z − 3 + xy

d. The product of m and n subtracted from twice the product of p and q = 2pq − mn

2. Expression Coefficient of a Coefficient of b

a. 6a2b + 12bc 6ab 6a2, 12c

b. 3ac − 10ba 3c, −10b −10a

c. 15ba2 + 12ac + 7b2c 15ba, 12c 15a2, 7bc

d. 5ca2 + 3ab − 6ca 5ca, 3b, −6c 3a

3. Expression Terms

a. 5m2 + mn 5m2 , mn

b. 6ab + 8a2bc – 2bc2 – ca 6ab, 8a2bc, –2bc2 , –ca

c. 0.5xy + 1.2x2y – 3.3y 0.5xy, 1.2x2y, –3.3y

d. 2 5 yz + 5 6 x2y + 0.6xy 2 5 yz, 5 6 x2y, 0.6xy

4. Like terms have the same algebraic factors whereas unlike terms have different algebraic factors.

a. Among the given terms, 8mn, 5mn, nm are like terms whereas 6mn2 is the odd one out.

b. Among the given terms, 3p2q2, 8q2p2 , q2p2 are like terms whereas 5p2q is the odd one out.

c. Among the given terms, 0.2xy, 2yx, 0.8xy are like terms whereas 0.5zx is the odd one out.

d. Among the given terms, 5a2b, 7ba2 , a2b are like terms whereas ab2 is the odd one out.

5. Like Terms Unlike Terms

a. (m2n, 5m2n, 8nm2), (3mn, 6nm) 5m, 7n

b. (2a2c, 3ca2), (3ac2 , c2a) 5b2a, 8b2c

c. (pqr, 9qrp, 6rpq), (8rp, 5pr) pq2

d. (5mln, 5lmn), (6lk2m, k2lm) 2klm

6. a. Expression 5xy + 8yz

5xy 8yz

5 x y 8 y z Terms Factors

b. Expression 3pq2 – 5pr + 2 3pq2 − 5pr 2

Terms Factors

3 p q q −5 p r 2

c. Expression 1.2x2y − 3xyz

1.2x2y –3xyz

1.2 x x y −3 x y z Terms Factors

Expression m + mn2 − 6l + 8nl m mn2 − 6l 8nl

Terms

Factors

m m n n − 6 l 8 n l

7. Number of paintings made by John = x

Number of paintings made by Maria = Twice as much as John = 2x

Total number of paintings with John and Maria = x + 2x = 3x

8. Width of the rectangle = w cm

Length of the rectangle = w + 5 cm

Perimeter of the rectangle = 2 × (Length + Width) = 2 × ((w + 5) + w) = (4w + 10) cm

9. Answers may vary. Sample answer:

An eraser costs ₹3 each and a pencil cost ₹1 each. The total cost of x erasers and y pencils is ₹(3x + y).

Challenge

1. i. 6mn + 2mn2 + 3 is a binomial.

ii. 3 × m is a monomial.

iii. 9pq + 2qr +5s has 9. 2. 5 as constant terms.

iv 4a2 = 5ab + 3c + 8 is terinomial with no constant term.

Statement (i) is not true since 6mn + 2mn2 + 3 is trinomial (has 3 terms) and not a binomial.

Statement (ii) is true, since 3 × m is a monomial as it has only one term 3m

Statement (iii) is false, since 9, 2, 5 are coefficients in the expression 9pq + 2qr + 5s and not constants.

Statement (iv) is not true, since 4a2 + 5ab + 3c + 8 is quadrinomial (has 4 terms) and not trinomial. So, only statement (ii) is correct. Thus, the correct answer is option d.

Do It Yourself 13C

1. a. 5xy + x = 5xy + x

b. 3p2 + 5p2 = 8p2

c. 4ab + (–3ab) = ab

d. 10x2y and 7x2y = 17x2y

2. a. 2x + 3y

+ x + y

(2 + 1)x + (3 + 1)y

= 3x + 4y

b. 2x + 3y + z

+ 2x + y z

(2 + 2)x + (3 + 1)y + (1 1)z

= 4x + 4y

c. a + 2b – 3

+ –3a + 5b + 5

(1 3)a + (2 + 5)b + ( 3 + 5)

= –2a + 7b + 2 d.

–5a + 4b 8c

4a + 7b 9c

+ 2a + 4b + 2c

( 5 + 4 + 2)a + (4 + 7 + 4)b + ( 8 9 + 2)c

= a + 15b 15c

3. a. x − 2y + x + y + y − 2x + x + 3y

= x + x − 2x + x − 2y + y + y + 3y

(1 + 1 − 2 + 1)x + (−2 + 1 + 1 + 3)y

= x + 3y

b. 3p2 + q2 − 2r2 + 4q2 + 5r2

= 3p2 + q2 + 4q2 − 2r2 + 5r2

= 3p2 + (1 + 4)q2 + (−2 + 5)r2

= 3p2 + 5q2 + 3r2

c. 4a2 + 3b2 + −2a2 + 13ab2 + 4a2 + 3ab2

= 4a2 − 2a2 + 4a2 + 3b2 + 13ab2 + 3ab2

= (4 − 2 + 4)a2 + 3b2 + (13 + 3)ab2

= 6a2 + 3b2 + 16ab2

d. 19a + 15b − 11c + 12a − 5b + 6c + 6b + 4c

= 19a + 12a + 15b − 5b + 6b − 11c + 6c + 4c

= (19 + 12)a + (15 − 5 + 6)b + (−11 + 6 + 4)c

= 31a + 16b − c

4. a. 2x + 5xy − 5x + 6xy 7ab2 − 4b + 6ac

b. 7x2 − 3y2 + 4x + 3y2 − 4x2 7pq + 4pr

c. 5ab2 − 3b + ac + 2ab2 + 5ac − b 11xy − 3x

d. 12pq + 9pr − 5rp + 5qp − 10pq −2pq + 10pr

e. 7pq2 + 3pq − 5qp − 7q2p + 10pr 4a2b − 9ab + 3c

f. 9a2b − 15ab + 3c − 5ba2 + 6ba 3x2 + 4x

5. a. 12ab2 − 3a + 5bc + 5a + 3a2b + 3ab2

= 12ab2 + 3ab2 − 3a + 5a + 5bc + 3a2b

= 15ab2 + 2a + 5bc + 3a2b

Adding 4ab2 − 7a2b + 12bc to the given sum will give:

15ab2 + 2a + 5bc + 3a2b + 4ab2 − 7a2b + 12bc

= 15ab2 + 4ab2 + 2a + 5bc + 12bc + 3a2b − 7a2b

= 19ab2 + 2a + 17bc − 4a2b

b. ab2 − 2a2b + 6bc + (−11a2b + 9bc)

= ab2 − 2a2b − 11a2b + 6bc + 9bc

= ab2 − 13a2b + 15bc

Adding 4ab2 − 7a2b + 12bc to the given sum will give:

ab2 − 13a2b + 15bc + 4ab2 − 7a2b + 12bc

= ab2 + 4ab2 − 13a2b − 7a2b + 15bc + 12bc

= 5ab2 − 20a2b + 27bc

8. Money spent to buy a shirt = ₹(3x2 + 2xy + 9y2)

Money spent on the meal = ₹(5x2 − xy + 3y2)

Total money spent = ₹(3x2 + 2xy + 9y2) + ₹(5x2 −

So, Kunal spent ₹(8x2 + xy + 12y2) in all.

Challenge

1. Assertion: The perimeter of the given triangle is 6a + 3.

Perimeter = Sum of the length of all the sides

= (a + 7) + (2a + 5) + (3a − 9) = 6a + 3

So, the assertion is correct.

Reason: The area of triangle is 1 2 × base × height

The reason is also true but it does not explain the assertion. Thus, option b is correct.

Do It Yourself 13D

1. a. ab − 9

3ab 5

– +

(1 − 3)ab + (−9 + 5) = −2ab − 4

b. 9y2 − 5xy

−3y2 + 12xy

+ –

(9 + 3)y2 + (−5 − 12)xy = 12y2 − 17xy

c. 4y2 + 5y 7

2y2 + 9y 4

+ – +

(4 + 2)y2 + (5 − 9)y + (−7 + 4)

= 6y2 − 4y − 3

d. 5b2 − 3ab + 8

b2 + 7ab

+ – +

(5 + 1)b2 + (−3 − 7)ab + 8

= 6b2 − 10ab + 8

= −10x + 7x + 8x2 − 10x2

= −3x − 2x2

b. 5mn − 6mn2 + 2n2− (mn2 + 2mn − 8n2)

= 5mn − 6mn2 + 2n2 − mn2− 2mn + 8n2

= 5mn − 2mn − 6mn2 − mn2 + 2n2 + 8n2

= 3mn − 7mn2 + 10n2

c. p2q + 5q2 − 8 − (3p2q + 2q2 + 5)

= p2q + 5q2 − 8 − 3p2q − 2q2− 5

= p2q − 3p2q + 5q2 − 2q2 − 8 − 5

= −2p2q + 3q2 − 13

d. 5ab + 6ac + 3bc + 8 − (5 − ab − ac)

= 5ab + 6ac + 3bc + 8 − 5 + ab + ac

= 5ab + ab + 6ac + ac + 3bc + 8 − 5

= 6ab + 7ac + 3bc + 3

3. 8a2 − 12a + 6b − (3a2 + 15a + 8)

= 8a2 − 12a + 6b − 3a2− 15a − 8

= 5a2 − 27a + 6b − 8

Hence, 5a2− 27a + 6b − 8 must be added to 3a2 + 15a + 8 to get 8a2− 12a + 6b

4. 5xz2 + 2xy − 7yz − (7xy − 4x2z + 12yz + 3)

= 5xz2 + 2xy − 7yz − 7xy + 4x2z − 12yz − 3

= 5xz2− 5xy − 19yz + 4x2z − 3

Hence, 5xz2− 5xy − 19yz + 4x2z − 3 must be subtracted from 5xz2 + 2xy − 7yz to get 7xy − 4x2z + 12yz + 3.

5. Sum of 6x2y + 9xy − 3y2 and x2y − 3xy + 5y2

= 6x2y + 9xy − 3y2 + x2y − 3xy + 5y2

= 7x2y + 6xy + 2y2

Sum of 7xy2 + 12xy − 10x2y and 2xy2 + 7y2 + 2x2y

= 7xy2 + 12xy − 10x2y + 2xy2 + 7y2 + 2x2y

= 9xy2 + 12xy − 8x2y + 7y2

9xy2 + 12xy − 8x2y + 7y2 − (7x2y + 6xy + 2y2)

= 9xy2 + 12xy − 8x2y + 7y2 − 7x2y − 6xy − 2y2

= 9xy2 + 6xy − 15x2y + 5y2

6. Length of ribbon with Mohan = (3m + 3) m

Length of ribbon he used = (3m − 5) m

Length of ribbon left = 3m + 3 − (3m − 5)

= 3m + 3 − 3m + 5 = 8 metres

As there is a minus sign in between, we need to flip the sign of the subtrahend, and the above algebraic tiles can be solved as: a + 7 2a + 5 3a − 9

8. Answer will vary. Sample answer:

An orchard grows two types of apple trees. The yield from the first type of apple tree can be represented by the expression 5x2 + 4x + 3 kilograms of apples, while the yield from the second type of tree is represented by 3x2 + 2x 5 kilograms. What is the total apple yield from both types of trees?

Challenge

1. Statement: P = 6a2 + 12ab + 3ab2; Q = 5ab – 8ab2; R = 15ab – 4a2 + 9

Conclusion I: 3Q = 15ab – 24ab2

Conclusion II: The value of P – R + 3Q = 10a2 + 12ab – 21ab2 – 9

According to question,

P = 6a2 + 12ab + 3ab2; Q = 5ab − 8ab2; R = 15ab − 4a2 + 9

Conclusion I:

3Q = 3 × (5ab − 8ab2 )

= 15ab − 24ab2 = RHS

So, conclusion 1 follows.

Conclusion II:

P − R + 3Q = (6a2 + 12ab + 3ab2) − (5ab − 8ab2) + (15ab − 24ab2)

P − R + 3Q = (6a2 + 12ab + 3ab2) − (15ab − 4a2 + 9) + (15ab − 24ab2)

= (6a2 + 4a2) + (12ab − 15ab + 15ab) + (3ab2 − 24ab2) − 9

= 10a2 + 12ab − 21ab2 − 9

So, Conclusion II follows.

Thus, both the conclusions follows.

Hence, option c is the correct answer.

Do It Yourself 13E

1. At a = 5

a. (a + 6) = 5 + 6 = 11

b. (2a − 3) = (2 × 5 − 3) = 10 − 3 = 7

c. (a2 − 3a + 5) = (52 − 3 × 5 + 5)

= 25 − 15 + 5 = 15

d. 3a 5 = 3 × 5 5 = 3

2. a. 5x − (y − 2x + 8y)

= 5x − y + 2x − 8y

= 7x − 9y

b. (10m − 3mn + 4m) + 5mn

= 10m − 3mn + 4m + 5mn

= 14m + 2mn

c. 3a − [5ab − (2a − 3ab)]

= 3a − [5ab − 2a + 3ab]

= 3a − 5ab + 2a − 3ab = 5a − 8ab

d. {4x2 − (3x2 + 2x + 5) + 6x}

= {4x2 − 3x2 – 2x − 5 + 6x}

= x2 + 4x − 5

3. At m = −2

a. (5m − 3m2 − 6m) − 2m2

= 5m − 3m2 − 6m − 2m2

= −5m2 − m

Adding the values of the variables to the expressions, we get −5(−2)2 − (−2)

= −20 + 2 = −18

b. 2(m2 − m) + 3(m2 + m) = 2m2 − 2m + 3m2 + 3m

= 5m2 + m

Adding the values of the variables to the expressions, we get 5(−2)2 + (−2) = 20 − 2 = 18

c. {2m − (5m2 + 5m) + 3m2} = {2m − 5m2 − 5m + 3m2}

= −2m2 − 3m

Adding the values of the variables to the expressions, we get −2(−2)2 − 3(−2) = −8 + 6

= −2

d. [6m2 − {2m + 5 − (2m2 − 8)} + 3]

= [6m2 − {2m + 5 − 2m2 + 8} + 3]

= [6m2 − 2m + 2m2 − 13 + 3]

= 8m2 − 2m − 10

8(−2)2 − 2(−2) − 10 = 32 + 4 − 10

= 26

4. At x = 1, y = −1

a. x2 + y2 = (1)2 + (−1)2 = 1 + 1 = 2

c. 3x2 − 4xy + 8 = 3(1)2 − 4(1)(−1) + 8 = 3 + 4 + 8 = 15

b. 5x2 + 2y2 − 1 = 5(1)2 + 2(−1)2 − 1 = 5 + 2 − 1 = 6

d. 2x3y + 3x2 − 5x = 2(1)3(−1) + 3(1)2 − 5 × 1 = −2 + 3 − 5 = −4

5. a. 5x + [7x2 + 2 − (2x2 + y2) + (4x − 3y)]

= 5x + [7x2 + 2 − 2x2 − y2 + 4x − 3y]

= 5x + 5x2 − y2 + 4x − 3y + 2

= 5x2 + 9x − y2 − 3y + 2

b. [2m2 + {5mn − (8n + 2nm) + 6m} − 5m2]

= [2m2 + {5mn − 8n − 2nm + 6m} − 5m2]

= [2m2 + 3mn − 8n + 6m − 5m2]

= −3m2 + 3mn − 8n + 6m

c. 2pq + [4pq2 − {3pq + 4p − (2pq2 − 2p)} − 3pq]

= 2pq + [4pq2 – {3pq + 4p – 2pq2 + 2p} – 3pq]

= 2pq + [4pq2 – 3pq – 4p + 2pq2 – 2p – 3pq]

= 6pq2 – 4pq – 6p

d. 2a + [7ab – (2a2 + 5ab – a) – {3 + (2a + a2)}]

= 2a + [7ab – 2a2 – 5ab + a – {3 + 2a + a2}]

= 2a + [7ab – 2a2 – 5ab + a – 3 – 2a – a2]

= 2a + 2ab – 3a2 – a – 3

= a + 2ab – 3a2 – 3

6. At x = −2 and y = 1

a. 4x − 3y + 5y + 2x = 6x + 2y = 6 × (−2) + 2 × 1 = −12 + 2 = −10

b. x2y + 2xy + 3xy − 4 = x2y + 5xy − 4

= (−2)2 × 1 + 5(−2)(1) − 4

= 4 − 10 − 4 = −10

7. A = 3x + 2y; B = x2 − 5x and C = y2 − 6y at x = 2 and y = 2

A − 2B + C = 3x + 2y − 2(x2 − 5x ) + y2 − 6y

= 3x + 2y − 2x2 + 10x + y2 − 6y

= 13x − 4y − 2x2 + y2

13 × 2 − 4 × 2 − 2(2)2 + (2)2

= 26 − 8 − 8 + 4 = 14

8. Amount of water in the tank = (250a2 + 100a) litres

Amount of water used by Pooja = 50a litres

Amount of water left = 250a2 + 100a − 50a

= (250a2 + 50a)

At a = 3; the amount of water left = 250 × (3)2 + 50 × 3 = 2250 + 150 = 2400 litres

9. Expression for Rohan’s age = 2x + [3x − {4x + (4 − x)}]

= 2x + [3x − {4x + 4 − x}]

= 2x + [3x − 3x − 4] = 2x − 4

At x = 5, Rohan’s age = 2 × 5 − 4 = 10 − 4 = 6 years

10. Time spent by Rahul on internet = x minutes

Time spent by Vivek on internet = 2x minutes

Time spent by Ankit on internet = (2x + 10) minutes

Total time spent = x + 2x + 2x + 10 = 5x + 10

At x = 22; time spend by Rahul = 22 minutes

Time spent by Vivek = 2 × 22 = 44 minutes

Time spent by Ankit = 2 × 22 + 10 = 54 minutes

Challenge

1. Statement: Manisha has a triangular plot with sides

(3x2 + 3y + 15) m, (2x2 – y + 30) m and (7x2 + 2x – 10) m.

Conclusion I: The perimeter of her plot is (12x2 + 2y + 2x + 35) m

Conclusion II: If x = 2 and y = 1, the perimeter of her plot is between 90 m and 100 m.

The length of the sides are (3x2 + 3y + 15), (2x2 − 7 + 30) and (7x2 + 2x − 10).

Conclusion I:

Perimeter = Sum of length of all sides

= 3x2 + 3y + 15 + 2x2 − y + 30 + 7x2 + 2x − 10

= (12x2 + 2y + 2x + 35) m

So, conclusion 1 follows.

Conclusion II:

At, x = 2 and y = 1

Perimeter = 12(2)2 + 2 × 1 + 2 × 2 + 35

= 48 + 2 + 4 + 35 = 89 metres

So, the perimeter is not between 90 m and 100 m

Conclusion II does not follow.

Hence, option a is correct.

Chapter Checkup

1. Expression Type of Expression

ab + 4a − 5ac Trinomial

3yx2 + 5xy − 3yx − 8 Quadrinomial

m × n = mn Monomial

2x2 + y Binomial

2. a. 5m2n, 6n2m − Unlike

b. 2ab, −3ba − Like

c. x2y, −5yx − Unlike

d. 2p2q2, 8q2p2 − Like

e. 3mn, −5m − Unlike

f. 6ab2, −2ab2 − Like

3. a. Expression x2y + yx x2y yzx x x y y zx

Terms

Factors

b. Expression 2cb2 + 6ab + 8c 2cb2 6ab 8c

Terms

Factors

2 c b b 6 a b 8 c

c. Expression pq – 5pq2 + 9 pq –5pq2 9 p q –5 p q q 9

Terms

Factors

4. 14 = 13 × 1 + 1

27 = 13 × 2 + 1

40 = 13 × 3 + 1

53 = 13 × 4 + 1

Hence, the rule for the given pattern is 13n + 1

The 15th term of the pattern can be given as: 13 × 15 + 1

= 196

5. Length of the base of the given triangle = 32 cm

Area of triangle = 1 2 × b × ℎ = 1 2 × 32 × ℎ = 184 cm2

Height = 184 × 2 32 = 11.5 cm

Hence, the height of the triangle is 11.5 cm.

6. a. 5a2b + (−3ab2) + (−7ab2) + 14a2b

= 5a2b − 3ab2 − 7ab2 + 14a2b

= 19a2b − 10ab2

b. 10x + 12y − 12xy − 12 + 18 − 5x − 10y + 5xy + 9xy

= 5x + 2y + 2xy + 6

7. Length of the given rectangle = (3a + 2) metres

Breadth of the given rectangle = (6a − 7) metres

Perimeter = 2 × (L + B)

= 2 × (3a + 2 + 6a − 7)

= 2 × (9a − 5)

= (18a − 10) metres

8. Perimeter = Sum of all sides

Given, that the length of the equal sides of the isosceles triangle is a and the length of the unequal side is b

Perimeter of isosceles triangle = a + a + b = 2a + b

9. At y = −3

y3 + 4y2 − 7y − 2

= (−3)3 + 4(−3)2 − 7 × (−3) − 2

= −27 + 36 + 21 − 2 = 28

10. At x = 2, y = 3

4x3 − 5y2 + 7 = 4(2)3 − 5(3)2 + 7 = 32 − 45 + 7 = −6

11. Let A must be added to x2 + 2xy + y2 to get 5x2 + 7xy

x2 + 2xy + y2 + A = 5x2 + 7xy

A = 5x2 + 7xy − (x2 + 2xy + y2)

= 5x2 + 7xy − x2 − 2xy − y2

= 4x2 + 5xy − y2

Hence, 4x2 + 5xy − y2 must be added to x2 + 2xy + y2 to get

5x2 + 7xy

12. Let A be subtracted from 5x + 8y + 12 to get −3x + 9y + 17

5x + 8y + 12 − A = −3x + 9y + 17

A = 5x + 8y + 12 − (−3x + 9y + 17)

= 5x + 8y + 12 + 3x − 9y − 17

= 8x − y − 5

Hence, 8x − y − 5 should be subtracted from 5x + 8y + 12 to get −3x + 9y + 17

13. At m = −5, n = 3

3 + 5m − 3 − 5n

= (–5)3

3 + 5 × (−5) − 3 − 5 × 3

= − 125 3 − 25 − 3 − 15

= −84.67

14. At m = −1, n = −1

a. [5 − {(4mn + 3m) − (5mn2 − 2m)} + 3mn2]

= [5 − {4mn + 3m − 5mn2 + 2m} + 3mn2]

= [5 − 4mn − 3m + 5mn2 − 2m + 3mn2]

= 5 − 4mn − 5m + 8mn2

Adding the value of the variables to the expressions, we get

5 − 4(−1)(−1) − 5(−1) + 8(−1)(−1)2

= 5 − 4 + 5 − 8

= −2

b. 5m − [8m2 − {4n2 − (4m + 5mn + 6m2) − 2n2}]

= 5m − [8m2 − {4n2 − 4m − 5mn − 6m2 − 2n2}]

= 5m − [8m2 − 2n2 + 4m + 5mn + 6m2]

= 5m − 8m2 + 2n2 − 4m − 5mn − 6m2

= m − 14m2 + 2n2 − 5mn

Adding the values of the variables to the expressions, we get (−1) − 14(−1)2 + 2(−1)2 − 5(−1)(−1)

= −1 − 14 + 2 − 5

= −18

15. Amount with Sunita = ₹(15a2 + 3ab + 3b2)

Amount spent on purchasing books = ₹(10a2 − ab + b2)

Amount left = ₹(15a2 + 3ab + 3b2 − 10a2 + ab − b2) = ₹(5a2 + 4ab + 2b2)

16. Length of rope with Sunil = (5x − 2y + 2) metres

Length of rope attached = (3x − xy + 5) metres

Total length = (5x − 2y + 2 + 3x − xy + 5) metres

= (8x − 2y − xy + 7) metres

At x = 5 and y = 2

8x − 2y − xy + 7 = 8 × 5 − 2 × 2 − 5 × 2 + 7

= 40 − 4 − 10 + 7

= 33 metres

17. Perimeter of triangular-shaped wooden frame = 96 cm

Given, that the length of each side of the wooden frame is same, hence, the length of each side of the triangle = 96 3 = 32 cm

Length of each side of square frame required = 32 cm

Perimeter of the square frame = 4 × 32 = 128 cm

Challenge

1. The given equation is

(3x2 + 14xy − ay2) − (bx2 − 12xy + 3y2 )

= 8x2 + 26xy − 12y2

(3 − b) x2 + (14 + 12)xy − (a + 3) y2

= 8x2 + 26xy − 12y2

Let us compare the LHS and RHS to find the values of a and b

3 − b = 8

b = −5

Now, a + 3 = 12

a = 9

Thus, a = 9 and b = −5.

2. Statement 1: x = 2 3 y

Simplifying the statement,

3x = 2y

3x − 2y = 0

Thus, statement 1 alone is sufficient to answer the given question.

Statement 2: y = 6

This does not tells us whether 3x − 2y = 0 or since there is no expression of x here.

Thus, statement 2 alone is not sufficient to answer the question.

Hence, option a is the correct answer.

Case Study

1. Area of the garden = plot area − area of construction of house = (4x2 + 3x − 2) − (x2 + 2x + 1) = (4x2 + 3x − 2) − x2 − 2x − 1

= 3x2 + x − 3

Hence, option c is correct.

2. If x = 2,

(x2 + 2x + 1) = (2)2 + 2(2) + 1 = 9 m2

Hence, option c is correct.

3. Cost of landscaping the garden = ₹500/m2

Area of the garden = 500 × (3x2 + x − 3) = 500 (3(2)2 + 2 − 3)

= 500(11 m2) = ₹5500

4. Area of neighbour's garden = (x2 + 4) m2

Total garden area = (x2 + 4) + 3x2 + x − 3

= (4x2 + x + 1)m2

Chapter 14

Let’s Warm-up

1. 100 hundreds 10,00,00,000 (Ten Crore)

2. 1000 hundreds 10,00,000 (Ten Lakh)

3. 100 ten thousands 1,00,00,000 (One Crore)

4. 100 lakhs 10,000 (Ten Thousand)

5. 1000 lakhs 1,00,000 (One Lakh)

Do It Yourself 14A

1. a. Base = 9; Power = 11

b. Base = −17; Power = 9

c. Base = n; Power = 10

d. Base = 3 8 ; Power = 5

e. Base = 7 11 ; Power = 8

2. a. 5 × 5 × 5 × 5 = 54

b. (–9) × (–9) × (–9) × (–9) × (–9) × (–9) × (–9) = (–9)7

c. 7 × 7 × 7 × 8 × 8 × 8 × 8 × 8 = 73 × 85

d. n × n × n × n × n × p × p × p × p × p = n5 × p5 e. 5 3

f. 4 11

× 4 11

3. a. 82 = 64; 83 = 512

b. 112 = 121; 113 = 1331

c. 182 = 324; 183 = 5832

d. 212 = 441; 213 = 9261

e. 302 = 900; 303 = 27,000

4. a. 173 = 17 × 17 × 17 = 4913

× 4 11

b. (−5)5 = (−5) × (−5) × (−5) × (−5) × (−5) = –3125

c. 43 × 34 = 4 × 4 × 4 × 3 × 3 × 3 × 3 = 64 × 81 = 5184

d. 3 2 9

= 2 9 × 2 9 ×

b. The reciprocal of (−3)4 is =

c. The reciprocal of 1 4

is = 4

d. The reciprocal of 5 1 2

7. a. 34 = 3 × 3 × 3 × 3 = 81 and 43 = 4 × 4 × 4 = 64; 81 > 64.

Hence, 34 is greater.

b. (−5)4 or 54 = (−5) × (−5) × (−5) × (−5) = 625 5 × 5 × 5 × 5 = 625; 625 = 625.

Both are equal.

c. 3

is greater.

5 7 2 = − 5 7 × − 5 7

= 25 49 and 7 5 2 = − 7 5 × 7 5 = 25 49

25 49 < 49 25

7 5 2 is greater

8. a. 113 × 122 = 11 × 11 × 11 × 12 × 12 = 1,91,664

b. 103 × 151 = 10 × 10 × 10 × 15 = 15,000

c. 73 × (−3)4 = 7 × 7 × 7 × (−3) × (−3) × (−3) × (−3) = 27,783

d. 3 4 5

× 5 × 2 3 8

= 4 5 × 4 5 × 4 5 × 5 × 3 8 × 3 8 = 9 25

9. a. 11x = 1331 11x = 11 × 11 × 11 = 113

Hence, the value of x = 3

b. 4256 61296 x  = 

4 6 x 

×××

Hence, the value of x = 4

c. 2 5 x 

= 32 3125

2 5 x 

= 4 4 4 6

= 2×2×2×2×2 5×5×5×5×5 = 5 2 5

Hence, the value of x = 5

10. The age of the banyan tree = 500 years

2 500

2 250

5 125

5 25 5 5 1

500 = 2 × 2 × 5 × 5 × 5 = 22 × 53

11. Marbles sold by Vivek = 78,125 Expressing 78125 as power of 5:

5 78125

5 15625

5 3125

5 625

5 125

5 25

5 5 1

78,125 = 5 × 5 × 5 × 5 × 5 × 5 × 5 = 57

Challenge

1. Points scored by Ramu = 3125 Expressing 3125 as powers of 5. 3125 = 5 × 5 × 5 × 5 × 5 = 55

It is given that Ramu expressed his points as 5x + 1

According to this, 5x + 1 = 55

So, x + 1 = 5 x = 4

Thus, the value of x is 4.

Do It Yourself 14B

1. a. (–9)3 × (–1)3 = ((−9) × (−1))3 [Since am × bm = (ab)m] = 93

b. (–5)2 × (–5)4 × (–5)6 = (−5)2 + 4 + 6 [Since am × an = a(m + n)] = (−5)12

[Since am × an = a(m + n)] = 7

3. a. ((9)2)2 = 92 × 2 [Since, (am)n = am × n] = 94

b. ((−12)1)3 = (−121 × 3) [Since, (am)n = am × n] = (−12)3

4. a. 40 = 1; Since a0 = 1

b. ((7)2)0 = 72×0 = 70 = 1; [Since a0 = 1]

c. (−5)0 ×

= 1 ÷ 1 = 1 [Since a0 = 1]

5. (20 + 30) × (60 + 71)

Since a0 = 1

(20 + 30) × (60 + 71) = (1 + 1) × (1 + 7) = 2 × 8 = 16.

6. a. 24 × 23 −27

b. (−3)6 ÷ (−3)3 128

c. (−5)4 × (−4)4 −17

d. 50 × (−17)1 1,60,000

7. 6p × 63 = 362

6p × 63 = 362 = (62)2 = 64 (Since, (am)n = am × n)

6p × 63 = 64

Since, am × an = a(m + n)

6p × 63 = 6p + 3

6p+3 = 64

On comparing, p + 3 = 4

p = 4 − 3 = 1

The value of p is 1

8. a. 128 6 2× (2) m m = 128 66 2× 2 m m = 212–6 × m8 – 6 [Since am ÷ an = am – n] = 26 × m2

b. 23 45 6×24 4×3 = 23 45 (2×3)×(2×2×2×3) (2×2)×3 = 22333 4 25 2×3×(2) ×3 2×3

= 2293 85 2×3×2×3 2×3 [Since am × bm = (ab)m] = 2+92+3 85 2×3 2×3 = 115 85 2×3 2×3 [Since am × an = a(m + n)]

= 211 – 8 × 35 – 5 [Since am ÷ an = am – n]

= 23

c. 583 454 25×49×121 5×7×11 = 583 222 454 5×7×11 5×7×11

= 10166 454 5×7×11 5×7×11 [Since am × bm = (ab)m] = 510 − 4 × 716 − 5 × 116 − 4 [Since am ÷ an = am – n]

= 56 × 711 × 112

d. 4 22 (5) 25×25 ab abab = 444 2222 5 5×5 ab abab = 444 433 5 5 ab ab [Since am × bm = (ab)m]

= 54 − 4 a4 − 3 b4 − 3 [Since am ÷ an = am – n]

= ab

9. a. 9 × 3n + 2 = 243 = 32 × 3n + 2 = 35 = 32 + n + 2 = 35 [Since am × an = a(m + n)]

= 3n + 4 = 35

On comparing, n + 4 = 5 n = 5 4

n = 1

b. 6 2 7 13

= 5 7 13 n

= 2×6 7 13

= +5 7 13 n

[Since am × bm = (ab)m] = 12 7 13

On comparing, 12 = n + 5 n = 12 − 5 n = 7

10. Pressure = 43 Pa

Area = 45 m2

Force = pressure × area = 43 × 45 = 43 + 5 = 48 = 65,536 N

11. Weight of the adult dolphin = 23 × 5 × 11 kg

Weight of the adult shark = 24 × 53 kg

Product of weight = 23 × 5 × 11 × 24 × 53

= 23 + 4 × 51 + 3 × 11

= 27 × 54 × 11 = 128 × 625 × 11 = 8,80,000

Challenge

1. Assertion: 46 × 56 = 206

Consider the LHS, 46 × 56 = (4 × 5)6 (Using am × bm = (ab)m) = 206 = RHS

So, the assertion is true.

Reason: am × bm = (ab)m

The reason is also true. It explains the assertion. Thus, both A and R are true and R is the correct explanation of A.

Hence, option c is the correct answer.

1. a. 864000 = 8.64 × 100000 = 8.64 × 105

b. 5500000 = 5.5 × 1000000 = 5.5 × 106

c. 4152000000 = 4.152 × 1000000000 = 4.152 × 109

d. 614000000000 = 6.14 × 100000000000 = 6.14 × 1011

2. a. 3 × 104 = 3 × 10000 = 30000

b. 4.55 × 105 = 4.55 × 100000 = 455000

c. 7.11 × 107 = 7.11 × 10000000 = 71100000

d. 9.8 × 109 = 9.8 × 1000000000 = 9800000000

3. Approximate age of earth = 4.54 × 109 years

4.54 × 109 = 4.54 × 1000000000 = 4540000000 years

4. 8,32,79,000 = 8.3279 × 10000000 = 8.3279 × 107

8.3279 × 10n = 8.3279 × 107

Comparing both sides, n = 7

5. 7 × 106 + 5 × 105 + 4 × 103 + 6 × 102 + 8 × 101 + 2 × 100 = 7000000 + 500000 + 4000 + 600 + 80 + 2 = 7504682 = 7.504682 × 106

6. a. 13568000 = 1.3568 × 10000000 = 1.3568 × 107

b. 1410000000 = 1.41 × 1000000000 = 1.41 × 109

c. 2760000000 = 2.76 × 1000000000 = 2.76 × 109

d. 140000000 = 1.4 × 100000000 = 1.4 × 108

Challenge

1. Let us simplify the expression and solve. 4343 2 55512 57571 57245 35575 7 ×× ===×= ××

So, the simplification of the given expression is 245 and not 5 49

Thus, the powers of the numbers in the numerator and denominator is negative. So, the final answer will be 245.

Chapter Checkup

1. a. 8 × 8 × 8 × 8 × 8 × 8 = 86

b. (–15) × (–15) × (–15) × (–15) × (–15) × (–15) × (–15) × (–15) = (−15)8

c. 3 7 × 3 7 × 3 7 × 3 7 × 3 7 × 3 7 × 3 7 × 3 7 × 3 7 = 9 3 7 

d. 2 11 × 2 11 × 2 11 × 2 11 × 2 11 × 2

× 4 15 × 4 15 = 6 2 11     × 7 4 15    

2. a. (p + q)4 = (3 + 4)4 = 74 74 = 7 × 7 × 7 × 7 = 2401

b. (p − q)2 = (3 − 4)2 = (−1)2 = 1

c. (2p − q)3 = (2 × 3 − 4)3 = (6 − 4)3 = 23 = 8

d. (p − 2q)2 = (3 − 2 × 4)2 = (3 − 8)2 = (−5)2 = (−5) × (−5) = 25

3. a. (7 × 2)4 = 144 = 14 × 14 × 14 × 14 = 38,416

b. (–1)9 = (–1) × (–1) × (–1) × (–1) × (–1) × (–1) × (–1) × (–1) × (–1) = −1

c. 103 × 103 = 10 × 10 × 10 × 10 × 10 × 10 = 10,00,000

d. 0 13 20

= 1

4. a. 256 =

c. 1090 = 2 × 5 × 109 2 1090 5 545 109 109 1

5. a. 15×(2)34 20 = 34 2 (3×5)×(2) 2×5 = 334 2 3×5 × (2) 2×5 = 31 34 2 3×5×(2) 2 = 32 3×5×16 4 = 27 × 25 × 4 = 2700

b. 23221 20 ×(1) 100× n n = n23 – 20 × (−1) × (10)−2 = 3 2 ×(1) 10 n = 3 102 n

c. (170 – 230) × (170 + 230) = (1 − 1) × (1 + 1) = 0 × 2 = 0 d. 4 24 23  

× 4 23 12 

= 4 2423 × 2312  

= (2)4 = 2 × 2 × 2 × 2 = 16

6. a. 7.2 × 102 < 3.8 × 103 720 < 3800

b. 5.8 × 105 < 3.6 × 108 580000 < 360000000

c. 6.9 × 108 < 4.5 × 109 690000000 < 4500000000

7. a b = 21 22 33

a b = 21 2 3

= 3 2 3

[Since am × an = a(m + n)] = 2228 33327 ×× = ××

= 88 2727

= 64

8. a. 9.1 × 106 9.1 × 107

b. 90100 91100000000

c. 9.11 × 1010 9100000

d. 91000000 9.01 × 104

9. a. (–1)3 × (–1)9 = (–1)n

= (−1)3+9 = (−1)n [Since am × an = a(m + n)] = (–1)12 = (–1)n

n = 12

b. 3n + 1 × 81 = (3)8

= 3n + 1 × 34 = (3)8

= 3n + 1 + 4 = (3)8 [Since am × an = a(m + n)]

3n + 5 = (3)8

On comparing both sides, n + 5 = 8

n = 8 − 5

n = 3

c. 52n + 1 ÷ 25 = (5)3

52n + 1 ÷ 52 = (5)3

52n + 1 − 2 = (5)3 [Since am ÷ an = am – n]

52n – 1 = (5)3

On comparing both sides,

2n − 1 = 3

2n = 4

n = 2

10. Weight of the blue whale = 202193.335 kg 202193.335 kg can be expressed in scientific notation as 2.02193335 × 105 kg.

11. Speed of the aircraft = 690 km/h

Total time taken to cover the distance = 3 h 20 mins = 3 1 3 h = 10 3 h

Total distance covered = Speed × Time = 690 × 10 3 = 2300 km

In centimetres:

1 km = 100000 centimetres

2300 km = 230000000 centimetres = 2.3 × 108 centimetres

12. The male population of a country in 2011 = 623700000 623700000 in standard form = 6.237 × 100000000 = 6.237 × 108

13. Distance between Earth and Venus = 188200000 km

Distance in standard form = 1.882 × 108

Challenge

1. Population of bacteria in the colony = 1000

Population in hour 1 = 4,000

Population in hour 2 = 16,000

Population in hour 3 = 64,000

Population in hour 4 = 2,56,000

Population in hour 5 = 10,24,000

The population between hour 2 and hour 3 = 64,000 – 16,000 = 48,000 = 4.80 × 104

The population between hour 4 and hour 5 = 10,24,000 –2,56,000 = 768,000 = 7.68 × 105

So, Statement 1 is not true while Statement 2 is true.

Thus, option b is correct.

2. Statement 1: (−48)−1 when divided by an even number leaves an even remainder.

Since (−48) is an even number, we are given that dividing (−48)−1 by an even number leaves an even remainder. This statement alone does not specify the value of b, only that b is an even number.

Statement 2: (−48)−1 when divided by b results in (−6)−1

According to the given statement, (−48)−1 ÷ b = (−6)−1, this gives us the value of b.

So, this statement alone is sufficient to find the value of b

Thus, option b is correct.

Case Study

1. Capacity of the CD = 700 MB

700 = 2 × 2 × 5 × 5 × 7 = 22 × 52 × 7

Thus, option c is correct.

2. Capacity of the hard disk = 512 GB

1 GB = 1000 MB

Capacity of hard disk = 512 × 1000 MB = 512000 MB = 5.12 × 105 MB

Thus, option b is correct.

3. Maximum capacity of the hard disk that the store sells = 32 TB

1 TB = 1000 GB

1 GB = 1000 MB

1 TB = 1000 × 1000 MB

= 1000000 MB

Maximum capacity of the hard disk that the store sells (in MB)

= 32 × 1000000 MB

= 3.2 × 107 MB

4. Capacity of the first flash drive = 64 MB

Capacity of the second flash drive = 8 GB

= 8 × 1000 MB = 8000 MB

Difference between the capacity of both the flash drives

= 8000 MB − 64 MB

= 7936 MB

= 7.936 × 103 MB

Chapter 15

Let’s Warm-up

1. Butterfly

2. Letter H

3. Plus sign

4. Sun

Do It Yourself 15A

Two lines of symmetry

One line of symmetry

Infinite lines of symmetry

Four lines of symmetry

8. Figures may vary. Sample figure.

f. No, a parallelogram does not have any line of symmetry. g. Yes

h. No, the shape is a scalene triangle, so it does not have any line of symmetry.

4. In the word "DICTIONARY," the letters that have no line of symmetry are: N and R. 5. a. b.

Challenge 1. a. b. Yes, It has 6 lines of symmetry

6. Answers may vary. Sample answers. a.

Do It Yourself 15B

More than 1

Equal to 1

More than 1 Equal to 1

More than 1 More than 1

We would shade the squares along the diagonal line at an equal distance on both sides. Answers may vary. Sample answers.

7. a. b. c. d.

Equal to 1

2. a. Angle of rotational symmetry of the image = 72°. So, the order of rotational symmetry = 360° 72° = 5.

b. Angle of rotational symmetry of the image = 120°. So, the order of rotational symmetry = 360° 120° = 3.

c. Angle of rotational symmetry of the image = 90°. So, the order of rotational symmetry = 360° 90° = 4.

d. Angle of rotational symmetry of the image = 180°. So, the order of rotational symmetry = 360° 180° = 2.

3. Figures may vary. Sample figures.

a. Letters H, I, O, X Rotation through 90° Rotation through 180° Rotation through 270° Rotation through 360°

b. Letters N, S, Z

1800

c. Letters A, B, C, D, E, K, M, T, U, V, W, Y

d. Letters F, G, J, L, P, Q, R 4. i. a.

c. Answer may vary. Sample answer.

Happy Diwali

8. a. Answers may vary. Sample answers. Red Fort, Qutub Minar, Lotus Temple

9. a. Rectangle: Reflection symmetry with 2 lines of symmetry

Rotational symmetry of order 2 x x x x x

b. Rotational symmetry of order 4. No reflective symmetry.

c. 1 line of reflection symmetry, but no rotational symmetry

b. c. d.

ii. a. 4 times b. 4 times c. 4 times d. 6 times

iii. a. 360 4 = 90° b. 360 4 = 90° c. 360 4 = 90° d. 360 6 = 60°

5. a. Anticlockwise

b. The figure has been rotated by 90°

c. No, the figure did not look the same after rotation.

d. Order of rotation = 1

6. a. Figure 1: Order of rotational symmetry = 4, angle of rotation = 90°

d. Parallelogram with rotational symmetry of order 2 and no reflective symmetry.

Challenge

1. Colour of the parts may vary. Sample figures.

a. Order 2:

b. Order 3:

c. Order 4:

Figure 2: Order of rotational symmetry = 2, angle of rotation = 180°

b. No, the figures do not show reflection symmetry. Figure 1: No, Figure 2: Yes,

7. a. 4 lines of symmetry in each figure.

b. For both figures, order = 4, angle of rotation = 360 4 = 90°

Chapter Checkup

1. a. Yes b. Yes

c. No, d. Yes e. No, f. No

Lines of symmetry may vary. Sample answer:

2. a. 1 line of symmetry

b. No line of symmetry

c. 4 lines of symmetry

d. 2 lines of symmetry

1 line of symmetry

f. 4 lines of symmetry

3. Error: The arrow head in the mirror image should be pointing the mirror.

Object Mirror image

Mirror image Correct image:

5. S. No.

A, T, U, V, W

a. Order = 1, angle of rotation = 360°

b. Order = 2, angle of rotation = 180°

c. Order = 4, angle of rotation = 90°

d. Order = 4, angle of rotation = 90°

e. Order = 2, angle of rotation = 180°

6. a. b.

7. a. Scalene triangle 1

b. Parallelogram

c. None

d. Nonagon, 140° 40° e. Isosceles trapezium

8. Reflective symmetry —

Rotational symmetry — Numbers 0 and 8 have rotational symmetry of order 2. So, the numbers that have both reflective and rotational symmetry are 0 and 8.

9. 7:15

10. Each of the given faces have 2 lines of symmetry.

11. The new position of C will be found as: The new position of C will replace H in the original figure after rotating the figure clockwise by 270° and 45° anticlockwise.

12. Answer may vary. Sample answers. a. b.

13. Answer may vary. Sample answers. a. b.

14. Figure may vary. Sample answers.

15. Minimum of 6 squares should be added to get a rotational symmetry of order 4.

16. a. Shape Q

b. P - 1, Q - 1, R - 1, S - 1, T - 1, U – 0 c. Rotational symmetry of order 1

4. a. B, C, D, E, K

c. X, O

Object

Challenge

1. a. 32

b. 32

2. The number of lines of symmetry for two regular figures of different sizes but the same number of sides are same. This is true, so the assertion is correct.

For two regular figures, if the number of lines of symmetry is the same, the figures need not be identical. This is also correct. So, the reason is also correct.

The reason does not correctly explain the assertion so the correct answer is option a.

Case Study

1. The order of rotation in a 4-cup anemometer is 4. Option a

2. a. The angle of rotation in a 3-cup anemometer is 180°. False The angle of rotation of 3 cup anemometer is 120° and not 180°.

b. The angle of rotation in a 4-cup anemometer is 90°. True

4. Angle of rotation when the wind is 1.2 mph = 90°

Angle of rotation when the wind speed is 4.8 mph = 90° × 4.8 mph 1.2 mph = 360°

A 3-cup anemometer is more efficient since it has an order of symmetry of 3, which makes it more efficient and accurate.

Chapter 16

Let’s Warm-up

c. A tetrahedron is a triangular pyramid. Edges = 6; Vertices = 4

d. Edges = 9; Vertices = 6

6. A ₹2 coin is circular/cylindrical in shape. The new shape will also be a cylinder.

7. The cube and the cuboid are two three-dimensional shapes that have the same number of faces, vertices and edges. Both the shapes have 6 faces, 8 vertices and 12 edges.

8. Difference: An octagonal prism has two identical and parallel octagonal bases and 8 rectangular lateral faces. An octagonal pyramid has one octagonal base and 8 triangular lateral faces.

Similarity: Both the shapes are 3-D shapes with octagonal bases.

Challenge

1. The 3-D shape formed with 4 congruent triangles meeting at a point will have a square base. So, the 3-D shape is a square pyramid.

2. Option c

The top of the glass greenhouse is made of a triangular prism, and the base is made of a rectangular prism.

Do It Yourself 16B

1. a. An octagonal prism has 10 faces. It has 2 octagonal identical faces and 8 rectangular faces.

b. A nonagonal pyramid has 10 faces. It has 1 nonagonal base and 9 triangular lateral faces.

1. Pentagonal prism 2. Cone

3. Triangular prism 4. Cylinder Do It Yourself 16A

1. A solid has 3 dimensions. The three dimensions are length, width and height.

2. a. The solid that has only one vertex is cone

b. If three cubes are joined side by side, the shape obtained is a cuboid

c. The number of edges in a football is 0

d. A square prism is also called a cuboid cube.

e. The polygon region of a solid shape is called a face.

3. The shape of the Ashoka Pillar is called a cylinder.

4. A prism has two identical and parallel polygonal bases and rectangular lateral faces.

A pyramid has a polygonal base and triangular lateral faces.

a. Pentagonal pyramid, since it has 1 pentagonal base

b. Square prism or cuboid, since it has 2 square bases

c. Rectangular pyramid, since it has one rectangular base

d. Hexagonal prism, since it has 2 hexagonal bases

e. Octagonal pyramid, since it has one octagonal base

5. a. Edges = 0; Vertices = 0

b. Edges = 1; Vertices = 1

c. A rectangular prism has 6 faces. It can have all rectangular faces or both square and rectangular faces.

d. A cone has 2 faces. It has 1 circular base and 1 curved lateral face.

2. A triangular prism has 2 triangular bases and 3 square/ rectangular lateral surfaces.

The different shapes needed to draw the net of a triangular prism are triangle and square/rectangle.

3. No. A square prism is not necessarily a cube. A square prism has a square bases, but the dimensions of the lateral surfaces need not be the same as that of the bases.

5. a. The net has 2 triangular congruent bases and 3 rectangular lateral faces. It is the net of a triangular prism.

b. The net has 2 congruent rectangular bases and 4 rectangular lateral faces. It is the net of a cuboid/ rectangular prism.

c. The net has a circular base and a triangular curved face. It is the net of a cone.

d. The net has 1 square base and 4 triangular lateral surfaces. It is the net of a square pyramid.

e. The net has 2 congruent circular bases and a curved face. It is the net of a cylinder.

f. The net has 1 pentagonal base and 5 triangular lateral surfaces. It is a pentagonal pyramid.

A cone has 2 faces: 1 circular base and 1 curved lateral face.

A triangular pyramid has 4 faces: 1 triangular base and 3 triangular lateral faces.

7. a. b.

8. Diameter = 14 cm. Radius = 14 2 cm = 7 cm

Circumference of the circle = Length of the rectangle = 2πr = 2 × 22 7 × 7 = 44 cm

Length of the cylinder = 15 cm

Since the length of the rectangle in the net is the same as the circumference of the circles drawn, the correct net is:

2. Do It Yourself 16C

10. A triangular prism has 2 triangular bases so the triangles are the bases. 11.

5 cm 5 cm

Challenge

5 cm

1. The 2 bangles will form the 2 circular bases. The rectangular sheet of paper will form the lateral surface. So, if Seema rolls the paper and attaches the bangles to each end edge-to-edge, she will get a cylinder.

5. a. The cuboid is of the dimensions: 5 units × 4 units × 1 unit

b. The cuboid is of the dimensions: 3 units × 5 units × 2 units

c. The cuboid is of the dimensions: 6 units × 3 units × 3 units

6. a. The cube is of the dimensions: 4 units × 4 units × 4 units

b. The cuboid is of the dimensions: 3 units × 5 units × 2 units

c. The cuboid is of the dimensions: 5 units × 2 units × 2 units

7. Reema placed three cubes of dimensions 3 cm × 3 cm × 3 cm end to end. 3 cm 3 cm 3 cm 3

The shape formed is a cuboid. Dimensions of the shape formed is 9 cm × 3 cm × 3 cm.

Challenge

1. a. b.

Do It Yourself 16D

1. a. A birthday hat is in the shape of a cone. The vertical cross section gives a triangle and the horizontal cross section gives a circle.

b. A dice is in the shape of a cube. The vertical cross section gives a square and the horizontal cross section gives a square.

c. A brick is in the shape of a cuboid. The vertical cross section gives a rectangle and the horizontal cross section gives a rectangle.

d. A ball is spherical in shape. The vertical cross section gives a circle and the horizontal cross section gives a circle.

3. Figures may vary. Sample figures. a. b. c.

6. A cube can cast a shadow in the shape of a rectangle. This occurs when the light source creating the shadow, falls at an angle on the cube.

7. Answers may vary. Sample answers.

a. Sphere b. Cone c. Cube/cuboid d. Cuboid

Challenge

1. If a cylindrical pipe is cut along the base horizontally, we get a rectangle.

Then, the length of the rectangle = length of the cylindrical pipe = 3 m = 3 × 100 cm = 300 cm

Breadth of the rectangle = diameter of the cylinder = 2 × radius = 2 × 7 cm = 14 cm

Area of the rectangle = length × breadth = 300 cm × 14 cm

= 4200 square cm2

So, the cross-sectional are of the cylinder will be 4200 cm2

Chapter Checkup

1. a. Cone

b. Cube

c. Cylinder

d. Triangular prism

2. a. Cuboid

b. Square pyramid

c. Cube

d. Cylinder

3. a. Solid that have 8 triangular faces-Octahedron

b. Solid that have 6 triangular and 2 hexagonal facesHexagonal prism

c. Solid that have 5 faces and 5 vertices-Square/rectangular pyramid

d. Solid that have a rectangular face and 2 equal circular faces-Cylinder

4. a. The net has 2 circular faces and a rectangular lateral surface. The net is of a cylinder.

b. The net has a square base and four triangular lateral faces. The net is of a square pyramid.

c. The net has 6 square faces. The net is of a cube.

d. The net has a triangular base and three triangular lateral faces. The net is of a triangular pyramid.

5. a. b. c. d.

6. Breadth = Height

Length = 2 × Breadth

2 opposite faces in the cuboid will be square.

Dimensions: 2 units × 2 units × 4 units

Figures may vary. Sample figure.

8. a. Rectangle

b. Rectangle

c. Rectangle

9. a. Horizontal: Vertical:

b. Horizontal: Vertical:

c. Horizontal: Vertical:

10. Top view Front view

11. A heptagonal prism has 2 heptagonal faces and 7 rectangular faces. So, the shapes needed are 2 heptagons and 7 rectangles. A heptagonal pyramid has 1 heptagonal face and 7 triangular faces. So, the shapes needed are 1 heptagon and 7 triangles.

12. a. b.

13. Answer may vary. Sample answer:

D C B A

When C is on the top, letter A will be at the bottom.

Challenge

1. There can be 3 possible cross section areas when Vansh cuts the cake in half.

2. Figures may vary. Sample figure.

Total number of cubes Ram placed is 15.

view Top View Side View

3. a. Number of vertices a regular octahedron have = 6

b. Option iii is the net of a dodecahedron.

4. Assertion (A): The side view of a square pyramid appears to be a triangle.

Reason (R): A square pyramid is a 3-D geometric shape with 1 square base and 4 triangular faces.

The assertion is true since the side view of a square pyramid appears to be a triangle.

The reason is also true since a square pyramid is a 3-D geometric shape with 1 square base and 4 triangular faces.

The reason explains the assertion correctly.

Thus, both A and R are true, and R is the correct explanation of A. Hence, option a is correct.

Case Study

1. The figure shown is the artificial satellite that Sanjay wants to make. The 3-D shape he will use to make it is a cuboid.

2. Number of vertices in the octahedron are 6.

Number of faces in the octahedron are 8.

Number of edges in the octahedron are 12.

3. The vertical cross section has four sides, so it is a quadrilateral.

The horizontal cross section has six sides, so it is a hexagon.

The vertical and horizontal cross sections of the octahedron are a quadrilateral and a hexagon. So, the correct option is b.

4. Answers may vary. Sample answer:

5. Answers may vary. Sample answer: Top view Front view

About the Book

The Imagine Mathematics teacher manuals bridge the gap between abstract mathematics and real-world relevance, offering engaging activities, games and quizzes that inspire young minds to explore the beauty and power of mathematical thinking. These teacher manuals are designed to be indispensable companions for educators, providing well-structured guidance to make teaching mathematics both effective and enjoyable. With a focus on interactive and hands-on learning, the lessons in the manuals include teaching strategies that will ensure engaging lessons and foster critical thinking and problem-solving skills. The teaching aids and resources emphasise creating an enriched and enjoyable learning environment, ensuring that students not only grasp mathematical concepts but also develop a genuine interest in the subject.

Key Features

• Alignment with Imagine Mathematics Content Book: Lesson plans and the topics in the learners’ books are in sync

• Learning Outcomes: Lessons designed as per clear, specific and measurable learning outcomes

• Alignment to NCF 2022-23: Lessons designed in accordance with NCF recommendations

• Built-in Recaps: Quick recall of pre-requisite concepts covered in each lesson

• Supporting Vocabulary: Systematic development of mathematical vocabulary and terminology

• Teaching Aids: Resources that the teachers may need to facilitate the lesson

• Activity: Concise and organised lesson plans that outline each activity

• Extension Ideas: Analytical opportunities upon delivery of each lesson

• Detailed Solutions: Solutions to all types of questions in the Imagine Mathematics Content Book

• Digital Assets: Access to supplementary interactive resources

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ISBN 978-81-984519-5-8