IM_G05_MA_TM_Text_AY2025-26 by Uolo

Master Mathematical Thinking MATHEMATICS

Teacher Manual

MATHEMATICS

Master Mathematical Thinking

Acknowledgements

Academic Authors: Animesh Mittal, Muskan Panjwani, Anjana AR, Anuj Gupta, Gitanjali Lal, 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.

First impression 2024

Second impression 2025

FPO

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Book Title: Imagine Mathematics 5

ISBN: 978-81-979482-3-7

Published by Uolo EdTech Private Limited

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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. Numbers up to 8 Digits 6

Break-up of Topics

Place Value, Face Value and Expanded Form

Indian and International Number Systems

Comparing and Ordering Numbers

Rounding-off Numbers

Revision

Adding Numbers up to 6 Digits

Subtracting Numbers up to 6 Digits

Multiplying Numbers with 10s, 100s and 1000s

Multiplying Numbers up to 4 Digits

2. Operations on Large Numbers 10

Dividing Numbers by 10s, 100s and 1000s

Dividing 5-Digit Numbers by 3-Digit Numbers

Using DMAS

Multi-step Word Problems

Revision

Finding Factors

Prime and Composite Numbers

Divisibility by 2, 5 and 10

Divisibility by 3 and 9

3. Factors and Highest Common Factor 10

4. Multiples and Least Common Multiples 7

Prime Factorisation

Common Factors; Factor Method

Prime Factorisation Method

Long Division Method

Revision

Finding Multiples

Common Multiples

LCM by Prime Factorisation Method

LCM by Short Division Method

Word Problems

Revision

5. Fractions 7

6. Operations on Fractions 11

Reviewing Fractions

Equivalent Fractions Using Multiplication

Equivalent Fractions Using Division

Comparing Fractions

Ordering Fractions

Revision

Adding Unlike Fractions

Adding Mixed Numbers

Subtracting Unlike Fractions

Subtracting Mixed Numbers

Multiplying Fractions and Whole Numbers

Multiplying Two Fractions

Dividing a Whole Number by a Fraction

Dividing a Fraction by a Whole Number

Dividing a Fraction by a Fraction

Revision

Tenths

Hundredths

Thousandths

Conversion Between Fractions and Decimals

Percentages

7. Introduction to Decimals 10

8. Operations with Decimals 9

Like and Unlike Decimals; Converting Unlike Decimals to Like Decimals

Comparing Like Decimals; Comparing Unlike Decimals; Ordering Decimals

Rounding off Decimals to the Nearest Whole Number

Revision

Addition of Decimals

Subtraction of Decimals

Multiplying Decimals by 10, 100, 1000, ...

Multiplying Whole Numbers and Decimals; Multiplying Two Decimal Numbers

Currencies from Different Countries

Dividing Decimals by 10, 100, 1000, ...

Dividing Decimals by Whole Numbers

Revision

Types of Lines

Types of Angles

Measuring Angles

9. Lines and Angles 7

Drawing Angles

Triangles, Quadrilaterals and Polygons

Revision

Estimating Length; Measuring Lengths

Converting Between Units of Length

Word Problems on Length

Estimating Weights

Converting Between Units of Weight

Word Problems on Weights

Revision

Perimeter of Squares and Rectangles

Area of a Square; Area of a Rectangle

Area of a Triangle

Estimating Capacity; Measuring Capacity

Units of Capacity

Volume of Solids Using Unit Cubes

Volume of Solids Using the Formula

Word Problems on Volume and Capacity

Nets of 3-D Shapes

Views of Cube Structures

Reading Maps

Floor Plans and Deep Drawings Revision

Converting

Measuring

Interpreting Bar Graphs

Drawing Bar Graphs

Interpreting Pie Charts

Representing Data on a Pie Chart

Reading Line Graphs

Numbers up to 8 Digits

Learning Outcomes

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

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

Teaching Aids

Chapter: Numbers up to 8 Digits

Place Value, Face Value and Expanded Form

Learning Outcomes

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

Teaching Aids

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

Activity

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.

Learning Outcomes

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

Teaching Aids

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

Activity

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.

Learning Outcomes

Students will be able to compare numbers up to 8 digits and arrange them in ascending and descending order.

Teaching Aids

Bowls with 5 number cards with a 7-digit or 8-digit number written on them

Activity

Instruct the students to work in pairs. Distribute the bowls with number cards to each pair.

Ask the pairs to draw place value charts in their notebooks. Instruct them to pick number cards one by one and write the numbers in the place value charts.

Ask them to observe and identify the smallest number among the given numbers. Ask them the reasons for choosing the smallest number and note down the smallest number under the place value charts drawn in their notebooks. Then, from the rest of the numbers, ask them to identify the next smallest number, give the reason for their choice, and note it down under the place value charts. Instruct them to repeat this until all the numbers are arranged in an order. Ask them about the order in which it is arranged. Further, repeat the process for descending order.

Extension Idea

Ask: Interchange the digits in the tens and thousands places in 5,87,92,401 and the digits in the lakhs and hundreds places in 6,43,12,510. Order the 4 numbers in ascending order.

Say: The four numbers are 5,87,92,401, 5,87,90,421, 6,43,12,510 and 6,45,12,310. The answer is 5,87,90,421 < 5,87,92,401 < 6,43,12,510 < 6,45,12,310.

Rounding–Off

Numbers Imagine Maths Page 12

Learning Outcomes

Students will be able to round off numbers up to 8 digits to the nearest 10, 100 and 1000.

Teaching Aids

Two bowls with number cards having 8-digit numbers in one bowl and rounded-off places in another bowl

Activity

Instruct the students to work in groups of 4. Two separate bowls are kept with the 8-digit numbers and rounded–off place respectively.

A student from each group picks a card from each bowl. Ask the groups to follow the instructions written on the cards. Let them write the answers in their notebooks.

Verify their answers and give 1 point for the correct answer and 0 for the incorrect answer. Repeat this process until all the students in the group have had a chance. The group with the maximum points wins the game.

Answers

1. Place Value, Face Value and Expanded Form

Do It Together

Number name = six crore fifty-seven lakh ninety thousand two hundred eighty-four

Expanded form = 6,00,00,000 + 50,00,000 + 7,00,000 + 90,000 + 200 + 80 + 4

2. Indian and International Number System

Do It Together

1. Ten million five hundred twenty-nine thousand six hundred five = 10,529,605

Expanded form: 10,000,000 + 500,000 + 20,000 + 9000 + 600 + 5

2. 65,780,245 = sixty-five million seven hundred eighty thousand two hundred forty-five

Expanded form: 60,000,000 + 5,000,000 + 700,000 + 80,000 + 200 + 40 + 5

3. Comparing Numbers

Do It Together

1 = 1 6 = 6 4 > 1 8 = 8 0 = 0

So, 1,86,04,766 > 1,86,01,769

4. Ordering Numbers

Think and Tell

Yes, the descending order will be: 3,09,13,200 > 2,80,82,800 > 98,02,000 > 94,99,000 Do It Together

5. Forming Numbers

Think and Tell

Yes, we can form more such numbers. For example, 97,65,310; 96,37,510

Think and Tell

We chose to repeat the greatest digit to form the greatest 8-digit number using the given numbers. If we do not repeat, then it will be smaller than the greatest number and not an 8-digit number.

Do It Together

The greatest number using the given digits = 98764310. We will get the second greatest digit by interchanging place of 8 with 7.

The second greatest number using the given digits = 97864310.

6. Rounding–Off Numbers

Think and Tell

The number of vaccine doses donated by India to Bangladesh is 2,80,82,800. It lies between 2,80,82,799 and 2,80,82,849. So, it is may be rounded off to the nearest 100s.

Do It Together

To the nearest 10s: 7,39,81,506 is rounded off to 7,39,81,510.

To the nearest 100s: 7,39,81,506 is rounded off to 7,39,81,500.

To the nearest 1000s: 7,39,81,506 is rounded off to 7,39,82,000

The greatest number is 1,20,64,980.

Operations on Large Numbers 2

Learning Outcomes

Students will be able to: add numbers up to 6 digits using the column method. subtract numbers up to 6 digits using the column method. multiply numbers up to 6 digits with 10s, 100s and 1000s. multiply numbers up to 4 digits using the column method. divide numbers up to 6 digits by 10s, 100s and 1000s. divide numbers up to 5 digits by numbers up to 3 digits using the column method. solve expressions using the DMAS rule. solve multi-step word problems on using 2 or more operations.

Alignment to NCF

C-1.3: Understands and visualises arithmetic operations and the relationships among them, knows addition and multiplication tables at least up to 10  10 (pahade) and applies the four basic operations on whole numbers to solve daily life problems

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

Recap to check if students know how to add, subtract, multiply and divide numbers up to 4 digits. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

addends: numbers to be added minuend: number to be subtracted from subtrahend: number to be subtracted

Teaching Aids

Number cards (with the population of some countries written); Crossword puzzle sheets; Cards with names of grocery items; Cards with quantities (3-digit or 4-digit numbers); Word problem cards

Chapter: Operations on Large Numbers

Adding Numbers up to 6 Digits

Learning Outcomes

Students will be able to add numbers up to 6 digits using the column method.

Teaching Aids

Number cards (with the population of some countries written)

Activity

Imagine Maths Page 18

Demonstrate to the students the addition of two 6-digit numbers, 2,50,678 and 1,56,240, using the column method. Emphasise the importance of carrying over when the sum is 10 or greater in a column.

Prepare cards as shown. Malta: 5,35,064

Maldives: 5,21,021

Iceland: 3,75,318

Instruct the students to work in pairs. Each student should pick a card.

Saint Lucia: 1,80,251

Instruct them to add the total population of the two countries written on the 2 cards held by each pair using the column method in their notebooks. Discuss the results obtained and the answers in the class. Shuffle the cards and distribute them again to repeat the activity.

Extension Idea

Ask: What is the sum of the greatest 6-digit number and the smallest 6-digit number?

Say: The sum of the greatest 6-digit number (9,99,999) and the smallest 6-digit number (1,00,000) is 10,99,999.

Subtracting Numbers up to 6 Digits

Learning Outcomes

Students will be able to subtract numbers up to 6 digits using the column method.

Teaching Aids

Number cards (with the population of some countries written)

Activity

Imagine Maths Page 19

Demonstrate to the student the subtraction of two 6-digit numbers, 2,50,678 and 1,56,240, using the column method. Emphasise the importance of borrowing when the minuend is less than the subtrahend. Prepare cards with the population of some countries written on them. For instance, Malta: 5,35,064; Maldives: 5,21,021; Iceland: 3,75,318; and Saint Lucia: 1,80,251.

Instruct the students to work in pairs. Each student should pick a card.

Instruct them to find the difference between the populations mentioned on 2 cards with each pair, and write which country has a larger population.

Ask them to subtract the numbers using the column method in their notebooks. Discuss the results obtained and the answers in the class. Shuffle the cards and distribute them again to repeat the activity.

Multiplying Numbers with 10s, 100s and 1000s

Learning Outcomes

Students will be able to multiply numbers up to 6 digits with 10s, 100s and 1000s.

Teaching Aids

Crossword puzzle sheets with multiplication problems with missing numbers

Activity

Demonstrate a multiplication problem on the board by multiplying 3,21,225 by 10, 100 and then 1000. Discuss how we will first multiply the non-zero digits and then put the remaining zeroes at the end.

Write on the board:

3,21,225 × 10 = 32,12,250; 3,21,225 × 100 = 3,21,22,500; 3,21,225 × 1000 = 32,12,25,000

Distribute a copy of the crossword puzzle sheet to each student. Instruct them to solve these problems and write the answers in the puzzle. Ask them to write down the answers for multiplying the numbers by 200 and 2000 in their notebooks.

Explain to the students that they do not need to perform the multiplication. They can use their understanding to find the answers without going through all the calculations.

Multiplying Numbers up to 4 Digits

Learning Outcomes

Students will be able to multiply numbers up to 4 digits using the column method.

Teaching Aids

Cards with names of grocery items; Cards with quantities (3-digit or 4-digit numbers)

Activity

Divide the class into groups and draw the table on the board.

Instruct the groups to make 2 stacks of cards—one with the grocery items and the other with the quantities. Ask one student from each group to pick one card from each stack. The remaining students should then calculate the total cost by multiplying the rate with the quantity. For example, if the student picks a card that says ‘rice’ and 1200 kg, the total cost would be ₹143 × 1200. Instruct the students to take turns to pick more cards.

Extension Idea

Instruct: Find the product when the difference of 52,050 and 45,127 is multiplied by 250.

Say: The difference is 52,050 – 45,127 = 6923. So, the product will be 6923 × 250 = 17,30,750.

Dividing Numbers by 10s, 100s and 1000s

Learning Outcomes

Students will be able to divide numbers up to 6 digits by 10s, 100s and 1000s.

Teaching Aids

Crossword puzzle sheet with division problems with missing numbers

Activity

Demonstrate a division problem on the board by dividing 3,21,200 by 10, 100 and then 1000.

Write on the board:

3,21,200 ÷ 10 = 32,120; 3,21,200 ÷ 100 = 3212; 3,21,200 ÷ 1000 → quotient = 321 and remainder = 200

Distribute a copy of the crossword puzzle sheet to each student.

Imagine Maths Page 24

Instruct the students to complete the crossword puzzle, and write down the answers in their notebooks.

Dividing

5-Digit Numbers by

Learning Outcomes

3-Digit

Numbers Imagine Maths Page 25

Students will be able to divide numbers up to 5 digits by numbers up to 3 digits using the column method.

Teaching Aids

Cards with names of grocery items; Cards with quantities (3-digit or 4-digit numbers)

Activity

Draw a table on the board as shown. Tell the students that a grocery store orders the following list of items.

Now, write the names of items on cards and place them in one box. Write their quantities in 3-digit numbers on another set of cards and put them in another box.

Instruct the students to pick one card from each box and divide. For example, if a student gets cards for “oil” and 456 L, they have to find the rate of one unit of oil, by dividing ₹1,51,848 ÷ 456.

Extension Idea

Ask: A number when divided by 250 gives the quotient as 101 and remainder as 155. What is the number?

Say: Number = 250  101 + 155 = 25,405.

Using

Learning Outcomes

Students will be able to solve expressions using the DMAS rule.

Teaching Aids

Crossword puzzle sheet with different number problems

Activity

Begin by showing a problem from the content book and discuss it.

Distribute the crossword sheets to the students.

Instruct them to fill in the missing values in the crossword, which are numbers between 1 and 9.

Remind the students that they should keep in mind the order of operations when solving the problems. Emphasize that multiplication and division should be performed before addition and subtraction.

Extension Idea

Ask: Find the answer obtained when 225 is subtracted from 25 times 150.

Say: 25  150 – 225 = 3750 – 225 = 3525. So, the answer is 3525.

Multi-step Word Problems

Learning Outcomes

Students will be able to solve multi-step word problems on using 2 or more operations.

Teaching Aids

Word problem cards

Activity

Write the word problem on cards.

Imagine Maths Page 28

Sarah has ₹200 to spend at a bookstore. She wants to buy three books that cost ₹25 each, a notebook for ₹8 and a pen for ₹5. How much money will she have left after making these purchases?

Distribute one card to each student.

Instruct the students to circle the numbers, underline the question, box the important words, evaluate and solve. Explain to them that this is called the ‘CUBES strategy’ which we use to solve word problems.

Instruct the students to form the equation (200 – 3 × 25 – 8 – 5)

Ask the students to solve using DMAS and write the answers in their notebooks.

Extension Idea

Instruct: Create your own word problem by dividing 84 by 2, multiplying the result with 3 and subtracting 53 from it. Say: Here is an example “Lisa has 84 candies. She decides to share them equally with her sibling. Her mother bought a candy jar that has three times the number of candies that Lisa has left. Finally, she decides to give 53 candies to her friend as a gift. How many candies does Lisa have left?”

Answers

1. Adding Numbers up to 6 Digits

Do It Together

Number of poetry books = 1,26,716

Number of non-fiction books = 2,43,120

Number of fiction books = 80,133

Total number of books published by the company are 4,49,969. L TTh Th H T O

4. Multiplying Numbers up to 4 Digits

Do It Together

To find the total number of chairs, we will have to multiply 5982 and 1313.

L TTh Th H T O

2. Subtracting Numbers up to 6 Digits

Do It Together

Shirts sold in 1 year (or 12 months) = 2,85,586

Shirts sold in the first month = 19,678

Shirts sold in the remaining months = 2,65,908

Therefore, the company sold 2,65,908 shirts in 11 months.

Therefore, the total number of chairs is 5982 × 1313 = 78,54,366

5. Dividing Numbers by 10s, 100s and 1000s

Do It Together

1. 4590 ÷ 100 Q = 45 R = 90 2. 67,688 ÷ 1000 Q = 67 R = 688

6. Dividing 5-Digit Numbers by 3-Digit Numbers

Do It Together

Cost of 235 books = ₹56,745

Cost of one book = ₹56,745 ÷ 235

Therefore, the cost of one book is ₹241.

3. Multiplying Numbers with 10s, 100s and 1000s

Do It Together

Cost of 1 scooter = ₹75,250

Cost of 100 scooters = ₹75,250 × 100 = ₹75,25,000

7. Using DMAS

Do It Together

100 − 72 � 8 + 4 × 3 = 100 − 9 + 4 × 3 = 100 9 + 12 = 112 9 = 103

8. Multi-step Word Problems

Think and Tell

Yes, multiplication and division are related.

Think and Tell

No, the answer can be different.

Do It Together

Initial number of flowers = 50 × 2

Flowers given by mother = 8

Flowers given to Siya = 8 ÷ 2

Number of flowers left = 50 × 2 + 8 – 8 ÷ 2 = 108 – 4 = 104

Therefore, Liza is left with 104 flowers.

Factors and Highest Common Factor 3

Learning Outcomes

Students will be able to: find factors of a number using multiplication or division. find prime and composite numbers. identify and apply the divisibility rule for 2, 5 and 10. identify and apply the divisibility rule for 3 and 9. find the prime factorisation of a number using the factor tree. find the common factors and HCF of 2 or more numbers using the factor method. find the HCF of 2 or more numbers using the prime factorisation method. find the HCF of 2 or more numbers using the long division method.

Alignment to NCF

C-4.2: Learns to systematically count and list all possible permutations and combination given a constraint, in simple situations (e.g., how to make a committee of two people from a group of five people)

Let’s Recall

Recap to check if students know how to represent the factors of a number using different-sized rectangles. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

factor: any one of two or more numbers that are multiplied together to give a product prime number: a number that can be evenly divided only by itself or by one composite number: a number which has more than two factors

Teaching Aids

Grid paper; Crayons; Index cards or small pieces of paper with numbers divisible and not divisible by 2, 5 or 10; Index cards with numbers divisible and not divisible by 3 or 9; Small paper squares or cards; Pieces of string or yarn; Chart paper; Glue sticks; Small square cutouts made of paper; Circle cutouts made of paper; Colourful paper strips say green and blue; Ruler; Scissors

Chapter: Factors and Highest Common Factor

Finding Factors

Learning

Outcomes

Students will be able to find factors of a number using multiplication or division.

Teaching Aids

Grid paper; Crayons

Activity

Briefly revise the concept of the area of a rectangle and how it is calculated.

Give each student a sheet of grid paper. Ask them to shade squares to make as many different rectangles as possible, each with an area of 24 square units. After drawing the rectangles, instruct students to label the side lengths of each rectangle. Have students pair up and compare their rectangles. Discuss similarities and differences. Guide a class discussion by asking questions such as, “How are your rectangles similar? How are they different?”

Discuss the different side lengths and how they relate to the number 24.

Summarise the activity by emphasising that the side lengths of the rectangles are factors of 24. Reinforce the idea that drawing rectangles with the same area is a visual way to understand and identify factors.

Finally, ask the students to verify their answers by recalling multiplication facts for the number 24. Try the same activity with a rectangle that has an area of 18 square units.

Extension Idea

Ask: Veena is arranging square paper tiles to create a rectangular piece. The area of the rectangular piece is 45 cm2. Each tile is exactly 1 cm × 1 cm. How many different ways can she arrange the tiles to form a rectangular piece with an area of 45 cm2?

Say: The different ways Veena can arrange the paper tiles are: 1 cm × 45 cm, 3 cm × 15 cm and 5 cm × 9 cm. So, Veena has 3 different ways to arrange the tiles.

Prime and Composite Numbers

Learning Outcomes

Students will be able to find prime and composite numbers.

Teaching Aids

Grid paper; Crayons

Activity

Imagine Maths Page 37

Begin by revising the definition of factors with the class. Explain that factors are numbers that can multiply together to give a specific product.

Distribute a sheet of grid paper to each student.

Instruct the students to shade squares to make as many different rectangles as possible with an area of 32 square units and 13 square units on the provided grid paper. Students should label the side lengths of each rectangle directly on the grid.

Ask the students to identify the factors of 32 and 13 based on the rectangles they drew. Further, have them compare the number of factors between 32 and 13.

Ask questions such as, “Can a whole number have fewer than two factors, exactly two factors or more than two factors?”

Emphasise the connection between the number of factors and how it helps identify if a number is prime or composite.

Extension Idea

Ask: How many composite numbers are there between 30 and 50? List them. Say: The composite numbers between 30 and 50 are 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48 and 49. So, there are 14 composite numbers between 30 and 50.

Divisibility by 2, 5 and 10

Learning Outcomes

Students will be able to identify and apply the divisibility rule for 2, 5 and 10.

Teaching Aids

Imagine Maths Page 39

Index cards or small pieces of paper with a mix of numbers divisible and not divisible by 2, 5 or 10

Activity

Prepare index cards or small pieces of paper with a mix of numbers that are divisible by 2, 5 or 10, as well as numbers that are not divisible by these numbers (You can use numbers up to three digits.)

Divide the class into small groups. Distribute the prepared index cards to each group.

Instruct the students to examine the numbers on their cards and sort them into three categories: Divisible only by 2, divisible only by 5 and divisible by 2, 5 and 10.

After the sorting is complete, have each group share their categories and explain their reasoning for placing numbers in each category. Encourage discussion on the divisibility rules for 2, 5 and 10. Ask questions such as, “Are there any numbers which are divisible by only 10 and not by 2 and 5? Are there any numbers that are divisible only by 5 and 10 and not by 2?”

Engage the students in a review session where they discuss the divisibility rules as a class.

Extension Idea

Ask: The key to open a number lock must satisfy the following conditions: The number is divisible by 2; the sum of its digits is 9; and the number is greater than 100 but less than 500. Which of the following numbers is the key? a. 216 b. 333 c. 246 d. 98

Say: The only number that satisfies all the conditions is 216. Hence, 216 is the key.

Divisibility by 3 and 9

Learning

Outcomes

Students will be able to identify and apply the divisibility rule for 3 and 9.

Teaching

Aids

Index cards with numbers divisible and not divisible by 3 or 9

Activity

Start by revising the concept of multiples. Ask the students to list a few multiples of 3 and 9. Ask questions like, “How can you use the sum of the digits to determine if a number is a multiple of 3 or 9?”

Divide the class into groups.

Distribute index cards with various numbers to each group. Instruct the students to sort the numbers into 3 categories: “Divisible only by 3”, “Divisible only by 9” and “Divisible by both 3 and 9”.

Ask questions like, “Was there any strategy that helped you determine the divisibility by 3 or 9? Could you list a few numbers which are divisible only by 9?

Summarise the key learnings from the activity.

Extension Idea

Ask: What is the secret code to unlock the number lock considering the following clues: The number is divisible by 3; the sum of its digits is divisible by 9; and the number has 0 as the last digit. a. 4320 b. 6670 c. 8920 d. 210

Say: The number that satisfies all the given conditions is 4320. Hence, the secret code is 4320.

Prime Factorisation

Learning Outcomes

Students will be able to find the prime factorisation of a number using the factor tree.

Teaching Aids

Small paper squares or cards; Pieces of string or yarn; Chart paper; Glue sticks

Activity

Explain the concept of prime factorisation using factor trees.

Imagine Maths Page 40

Instruct the students to work in pairs. Distribute paper squares, pieces of string and chart paper to each pair.

Assign each pair a random number (e.g., 24, 36, 48) that they will factorise using the factor tree method. Instruct participants to write their assigned number on the paper square and stick it on the chart paper. Ask them to identify and write the first pair of factors on the squares, and connect them using a glue stick with a piece of string or yarn to represent a branch. Ask them to repeat this process until only prime numbers are left at the ends of the branches.

Ask the pairs to exchange the factor trees they created with other pairs. Discuss the similarities and differences between the factorisations.

Common Factors; Factor Method

Learning Outcomes

Imagine Maths Page 43

Students will be able to find the common factors and HCF of 2 or more numbers using the factor method.

Teaching Aids

Chart paper; Small square cutouts made of paper

Activity

Begin by introducing the concept of common factors. Write two numbers (e.g., 24 and 36) on the board.

Instruct the students to draw two overlapping circles on chart paper, representing the two numbers. Explain to them how the overlapping region in the centre will represent the common factors.

Then, in their groups, have students discuss and list the factors of each number separately. Instruct the students to write the factor of 24 and 36 on the square cutouts. Ask them to paste the factors of each number in the appropriate circles. Explain to them how the

overlapping factors should be placed in the centre region. Explain how these are the common factors of the two numbers.

Finally, instruct the students to compare the common factors within the overlapping region and select the largest one as the HCF.

Prime Factorisation Method

Learning Outcomes

Imagine Maths Page 45

Students will be able to find the HCF of 2 or more numbers using the prime factorisation method.

Teaching Aids

Circle cutouts made of paper

Activity

Select any two numbers for the purpose of our activity (say 12 and 20).

Divide the class into groups. Distribute small circle cutouts among the groups.

Instruct the students to draw a factor tree and paste the circle cutouts to represent the prime factors of each number. Then ask them to write the prime factorisation at the bottom of each tree.

Then, ask them to choose the common prime factors between the two numbers and multiply them.

Discuss how this number is the highest common factor of the given numbers.

Extension Idea

Ask: You have 48 red flowers, 60 yellow flowers and 84 white flowers. You want to make flower arrangements that have the same number of each colour. What is the greatest number of arrangements that you can make if every flower is used?

Say: 48 = 2 × 2 × 2 × 2 × 3, 60 = 2 × 2 × 3 × 5 and 84 = 2 × 2 × 3 × 7. Hence, HCF = 2 × 2 × 3 = 12. So, you can make 12 arrangements.

Long Division Method

Learning Outcomes

Imagine Maths Page 46

Students will be able to find the HCF of 2 or more numbers using the long division method.

Teaching Aids

Colourful paper strips say green and blue; Ruler; Scissors

Activity

Instruct the students to work in groups. Distribute paper strips, ruler and scissors among the groups.

Instruct the groups to cut out a strip of length 24 cm using a green strip and another strip of length 16 cm using a blue strip as shown. Ask the student to place the blue strip along the green strip as shown and cut out the remaining part of the green strip.

Have them observe that the remaining part of the green strip cut out is 8 cm long, which is smaller than the blue strip.

Instruct them to now place the smaller strip (green) along the larger strip (blue) and cut out the extra part (blue strip) as shown. Ask them to repeat this process until both strips are equal. Explain to the students how the HCF of 24 and 16 is 8 cm (the length of the last part cut out). Emphasise that a strip of length 8 cm can completely cover both strips of length 24 cm and 16 cm a complete number of times. Finally, ask the students to verify the answer using the long division method.

1. Finding Factors

Think and Tell

Yes, there can be two more ways: 12 rows of 1 dancer and 6 rows of 2 dancers. Do It Together

3. Divisibility by 2, 5 and 10 Do It Together

2 = 12 0 2 and 12 are factors of 24.

÷ 3 = 8 0 3 and 8 are factors of 24.

are factors of 24.

So, the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24

2. Prime and Composite Numbers Do It Together

1, 2, 3, 4, 6, 8, 12, 16, 24,

6. Common Factors

Do It Together

Factors of 10: 1, 2, 5 and 10.

Factors of 14: 1, 2, 7 and 14.

The common factors of 10 and 14 are 1 and 2.

7. Factor Method

Do It Together

Factors of 18 are 1, 2, 3, 6, 9 and 18.

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24

The common factors of 18 and 24 are 1, 2, 3 and 6.

The highest common factor (HCF) of 18 and 24 is 6.

8. Prime Factorisation Method

Do It Together

Prime factorisation of 16 = 2 × 2 × 2 × 2

Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3.

The common prime factorisation of 16 and 48 are 2 × 2 × 2 × 2.

Hence, HCF = 2 × 2 × 2 × 2 = 16.

9. Long Division Method

Do It Together

Since, the last divisor is 4, the HCF of 52 and 68 is 4

Multiples and Least Common Multiples

Learning Outcomes

Students will be able to: find multiples of a number using multiplication. find the common multiples and LCM of 2 or more numbers. find the LCM of 2 or more numbers using the prime factorisation method. find the LCM of 2 or more numbers using the short division method. solve word problems on finding the LCM and HCF.

Alignment to NCF

C-1.3: Understands and visualises arithmetic operations and the relationships among them, knows addition and multiplication tables at least up to 10 × 10 (pahade) and applies the four basic operations on whole numbers to solve daily life problems

Let’s Recall

Recap to check if students know how to skip count by any number to get the multiplication table for that number.

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

Vocabulary

repeated addition: adding the same number again and again least common multiple: the smallest multiple that two or more numbers have in common prime factorisation: a way of expressing a number as a product of its prime factors

Teaching Aids

A hundreds chart; A crayon box; Circle cutouts made of paper; Chart paper; Number cards; Word problems written on sheets of paper

Chapter: Multiples and Least Common Multiples

Finding Multiples

Learning Outcomes

Students will be able to find multiples of a number using multiplication.

Teaching Aids

A hundreds chart; A crayon box

Activity

Distribute a hundreds chart to each group.

Instruct the students to start from 7 and shade it. Guide them to skip count by 7 on the chart and continue shading as they land on each number. Once they have shaded all the numbers, they should write the numbers in their notebooks.

Ask questions like, “What do you notice about the numbers you shaded?”

Explain to them that the numbers shaded are part of the multiplication table of 7 and are called the multiples of 7. Repeat the activity to shade the multiples of 8 on the same chart.

Extension Idea

Ask: Which multiples of 7 lie between 57 and 93?

Say: The multiples of 7 that lie between 57 and 93 are 63, 70, 77, 84 and 91.

Common Multiples

Learning Outcomes

Students will be able to find the common multiples and LCM of 2 or more numbers.

Teaching

Aids

A hundreds chart; A crayon box

Activity

Instruct the students to work in pairs and distribute a hundreds chart to each pair. Instruct one student to start at 4 and shade it in blue. The student should then skip count by 4 and shade the top part of each cell they land on. Both students will write the shaded numbers in their notebooks. The second student will then start at 12 and shade it in red. The student should then skip count by 12 and shade the bottom part of each cell they land on. Both the students will write the shaded numbers in their notebooks.

Ask questions like, “What do you notice about the numbers you shaded? Did you find numbers which are multiples of both 4 and 12?”

Explain to them that the lowest among these common multiples is the LCM (lowest common multiple) of the two numbers.

Learning Outcomes

Students will be able to find the LCM of 2 or more numbers using the prime factorisation method.

Teaching Aids

Circle cutouts made of paper

Activity

Select any number (say 18) and walk students through the steps involved in creating a factor tree.

Now, select any two numbers for the purpose of our activity (say, 12 and 20).

Divide the class into groups. Distribute the small circle cutouts among the groups.

Instruct the students to draw a factor tree and paste the circle cutouts on the tree to represent the prime factors of each number. Ask them to write the prime factorisation at the bottom of each tree. Finally, ask them to multiply each factor by the maximum number of times it occurs in the prime factorisation of any of the given numbers.

Discuss with the student how the product is the least common multiple of the given numbers.

LCM by Short Division Method

Learning Outcomes

Students will be able to find the LCM of 2 or more numbers using the short division method.

Teaching Aids

Chart paper; Number cards

Activity

Begin by demonstrating the concept of LCM by using the short division method on the board. Instruct the students to perform the activity using the number cards.

Make groups of students and distribute the number cards to each group.

Draw a division house on the board (for the numbers 12 and 15). Provide each group with chart paper on which the division house is made.

Ask the students which prime number divides both 12 and 15. Collect their responses. Then, write 3 on the board, and let the students place the number card for 3 at the place shown on the board. Keep dividing the given numbers by the common prime numbers until the quotient is 1. Allow the students to arrange their number cards to complete their division houses. Finally, ask the students to multiply all the prime factors to find the LCM.

Ask the students to find the LCM of two other numbers by repeating the activity.

Extension Idea

Ask: What is the least common multiple of 5, 8, 10 and 15?

Say: The least common multiple of the given numbers is 120.

Learning Outcomes

Students will be able to solve word problems on finding the LCM and HCF.

Teaching Aids

Word problems written on sheets of paper

Activity

Begin the lesson by reviewing the concepts of LCM and HCF. Provide clue words and discuss when it is appropriate to use each.

Greatest Common Factor Least Common Multiple

biggest both divide first both something is repeated largest identical groups smallest identical every days most same split up next time same every hours maximum equal minimum equal

Distribute the word problem sheets with the given word problems written on them:

A factory produces 3 types of toys: type A, type B, and type C. It produces the same total number of each type of toy. The factory can package 18 type A toys, 24 type B toys, and 30 type C toys in each box. What is the smallest total number of toys of each type such that all can be packaged in boxes?

A gardener has to buy pieces of wire to build a fence around 3 rectangular flower beds of lengths 36 metres, 48 metres and 60 metres. What is the longest length of wire he can buy so that he can cover the lengths of all three flower beds without having any pieces of wire left over?

Instruct the students to read the word problem on the sheet given to them and determine whether it requires finding the LCM or HCF. They should discuss and write down the clue words/reasoning in their notebooks.

Once the correct method is identified, ask the students to solve the word problems in their notebooks using that method (LCM or HCF).

Extension Idea

Instruct: Create a word problem on finding the LCM using the numbers 4, 6 and 8.

Say: You can frame multiple word problems using the given numbers. One such word problem could be, “Arvind, Bhanu and Chitra are members of a music club. They practise playing their instruments regularly, each with a different schedule. Alex practises every 4 days. Bob practises every 6 days. Carol practises every 8 days. If they all practised today, after how many days will they need to practise on the same day again?”

Answers

1. Finding Multiples

Think and Tell

Yes, all the numbers when Ajay attends the music classes appear in the multiplication table of 5. We call such numbers multiples of 5.

Do It Together

3 × 1 = 3 3 × 2 = 6 3 × 3 = 9 3 × 4 =

So, 3, 6, 9, 12 and 15 are the first five multiples of 3.

2. Common Multiples

Do It Together

Multiples of 12: 12

Multiples of 16: 16

Common multiples of 12 and 16 are 48, 96 and so on.

The lowest common multiple of 12 and 16 is 48.

3. LCM by Prime Factorisation Method

Do It Together

4. LCM by Short Division Method

LCM = 3 × 2 × 2 × 2 × 5 = 120

So, the LCM of 24 and 15 is 120.

5. Word Problems

It Together

LCM of 3, 5 and 10 = 5 × 2 × 3 = 30

So, the smallest possible length that can be measured by each of the given scales is 30 cm.

Learning Outcomes

Students will be able to: identify types of fractions, and convert improper fractions to mixed numbers and vice versa. find equivalent fractions using multiplication. find equivalent fractions using division. compare two or more fractions. arrange fractions in ascending or descending order.

Alignment to NCF

C-1.2: Represents and compares commonly used fractions in daily life (such as ½, ¼) as parts of unit wholes, as locations on number lines and as divisions of whole numbers

Let’s Recall

Recap to check if students know the basics of fractions, the whole, numerator and denominator of a fraction. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

equivalent fraction: a fraction obtained by multiplying or dividing the numerator and the denominator by the same non-zero number proper fractions: a fraction whose numerator is less than the denominator improper fractions: a fraction whose numerator is equal to or greater than the denominator mixed fractions: fractions which are a combination of a whole number part and fractional part unit fractions: fractions with 1 as the numerator

Teaching Aids

Unshaded fraction circles divided into 3, 4 or 6 parts; Fraction strips showing 246 ,, 368 and 3 4; Fraction strips or circles showing 8 16, 4 8 and 1 2 and some additional strips that are not equivalent to 8 16; Rectangular strips of paper; Crayon box; Fraction strips showing some fractions (e.g., 123 ,, 438 and 5 6 )

Chapter: Fractions

Reviewing Fractions

Learning Objective

Students will be able to identify types of fractions, and convert improper fractions to mixed numbers and vice versa.

Teaching Aids

Unshaded fraction circles divided into 3, 4 or 6 parts

Activity

Pair up students and provide each pair with fraction circles divided into 4 or 6 parts.

Student A should shade circles to show a mixed number. For example, Student A might shade circles to show 5

1 6 . Student B’s task is to write the fraction as the number of shaded parts over the total number of parts, converting it into an improper fraction, e.g., 11 6 . Next, ask them to switch roles. Student B will now shade circles to show an improper fraction while Student A will write the number of wholes and parts shaded, converting it into a mixed number. Discuss how the mixed number and its equivalent improper fraction are related. Rotate pairs and repeat the activity to reinforce the concept of identifying and converting fractions. Ask questions like, “Will the fraction change if we change the number of total parts to 8 instead of 4 in the fraction circle showing 3 1 ? 4 ”

Equivalent Fractions Using Multiplication

Learning Objective

Students will be able to find equivalent fractions using multiplication.

Teaching Aids

Fraction strips showing 246 ,, 368 and 3 4

Activity

Distribute two sets of fraction strips to each pair. One set should have fraction strips showing 2 3 and 3 4 , while the other should contain strips showing 4 6 and 6 8 . Each student has to take one fraction strip from the first set and keep the second set unopened. Instruct the students to multiply both the numerator and the denominator by 2 for each fraction, then show the new fractions on their fraction strips. Ask them to write the new fractions next to the original fraction in their notebooks. Finally, have the students look at the fraction strips in the second set: 4 6 and 6 8 .

Ask them to compare and discuss what they notice.

Equivalent Fractions Using Division

Learning Objective

Students will be able to find equivalent fractions using division.

Teaching Aids

Fraction strips or circles showing 8 16 , 4 8 and 1 2 and some additional strips that are not equivalent to 8 16 .

Activity

Give each group of students a strip that shows 8 , 16 and some more strips. Some of these extra strips should look like 8 16 (such as 4 8 and 1 2 ), while others should be different.

Instruct the students to examine the given fraction and identify other fraction strips or circles that have fewer total parts but the same shaded area. For example, students can find fractions such as 4 8 and 1 2 , which have the same shaded area but fewer total parts.

Ask the students to write down the equivalent fractions they discover in their notebooks and observe any patterns or relationships. Encourage the students to share their observations and insights about finding equivalent fractions through division.

Ask questions like, “What is the HCF of the numerator and the denominator when the fraction is in the simplest form?”

Comparing Fractions

Learning Objective

Students will be able to compare two or more fractions.

Teaching Aids

Rectangular strips of paper; Crayon box

Activity

Imagine Maths Page 75

Divide the class into groups and distribute the rectangular paper strips to each group. Instruct each group to choose any two fractions and visually represent them by shading their respective numerators on the paper strips. After this representation, instruct the students to find the Least Common Multiple (LCM) of the chosen fractions’ denominators. Once the LCM is determined, guide the students to adjust their representations to ensure both the fractions share the same denominator.

Instruct the students to visually compare the fractions and determine which one is larger. Encourage groups to validate their visual comparison using the cross-multiplication method. Direct each group to write the larger fraction on their rectangular strips. Finally, ask the students to write their answers in their notebooks.

Extension Idea

Ask: If there are two fractions with the same numerator but different denominators, which fraction will be smaller? Say: The fraction with the larger denominator will be smaller, because a larger denominator means each part is smaller.

Learning Objective

Students will be able to arrange fractions in ascending or descending order.

Teaching Aids

Fraction strips showing some fractions (e.g., 1 4 , 2 3 , 3 8 and 5 6 )

Activity

Divide the class into groups. Provide each group with fraction strips of different fractions (e.g., 1 4 , 2 3 , 3 8 and 5 6 ). Instruct them to place the fraction strips one below the other to see which strip has the greater shaded area. Also, instruct them to arrange the fraction strips in ascending order of their shaded areas. Now, ask the students to find the Least Common Multiple (LCM) of the denominators of all four fractions. Instruct them to make equivalent fractions according to the LCM and compare the numerators to arrange the fractions in ascending order in their notebooks. Compare the results obtained in both the cases.

Extension Idea

Ask: Compare 3 16 and 16 3 . Which is smaller?

Say: On finding the LCM of the denominator, we get the fractions as 9 48 and 256 48 . 256 48 > 9 48 . Hence, 316 163 < . So, 3 16 is smaller.

1. Reviewing Fractions

5. Comparing Fractions

2. Converting Between Improper Fractions and Mixed Numbers

Do It Together

1. 8 7

into an improper fraction.

6. Ordering Fractions Do It Together

of 2, 3, 4, 6 = 12

3. Equivalent Fractions Using Multiplication

4

. Equivalent Fractions Using Division

Think and Tell

Yes, equivalent fractions give like fractions when simplified. Do It Together

Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Factors of 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.

HCF of 48 and 96 = 48

Thus,

Operations on Fractions 6

Learning Outcomes

Students will be able to: add unlike fractions. add mixed numbers. subtract unlike fractions. subtract mixed numbers. multiply a fraction by a whole number. multiply two fractions. divide a whole number by a fraction. divide a fraction by a whole number. divide a fraction by a fraction.

Alignment to NCF

C-1.2: Represents and compares commonly used fractions in daily life (such as ½, ¼) as parts of unit wholes, as locations on number lines and as divisions of whole numbers

Let’s Recall

Recap to check if students know how to add like fractions. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

unlike fraction: fractions which have different denominators mixed number: number which have a whole number part and a proper fraction reciprocal: the inverse of a number

Teaching Aids

Rectangular strips; Coloured pencils; Circular cutouts

Chapter: Operations on Fractions

Adding Unlike Fractions

Learning Outcomes

Students will be able to add unlike fractions.

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Instruct the students to work in pairs. Distribute the rectangular strips and coloured pencils among the pairs. Instruct the students to add the two fractions 1 2 and 1 3 by first showing the two fractions on the rectangular strips. Then discuss how to divide the strips so that both represent the same total number of equal parts.

Explain to them that this can be done either by finding the LCM of the denominators or by multiplying the denominators. Discuss that each strip should be divided into the same number of parts as the LCM of the denominators. Then, ask them to add both the fractions.

Finally, ask the students to convert both fractions to like fractions in their notebooks and then add the numerators.

Extension Idea

Ask: Barney has 5 6 m of cloth, and Christene has 1 2 m of cloth. What is the total length of the cloth that they both have?

Say: Total length of the

Learning Outcomes

Students will be able to add mixed numbers.

Teaching Aids

Circular cutouts; Coloured pencils

Activity

Instruct the students to work in pairs. Distribute the circular cutouts and coloured pencils among the pairs. Ask the students to show 1 1 2 and 1 1 4 on the circular cutouts. Instruct them to add the two mixed numbers. Ask the students to first put the shaded wholes together. They will then add the fractional parts the same way as they did in the previous lesson. They will finally add the wholes and parts shaded to find the answer. Instruct the students to add the two mixed numbers in their notebooks by first adding the whole number parts and then the fractional parts by converting them to like fractions and adding the numerators.

Subtracting Unlike Fractions

Learning Outcomes

Students will be able to subtract unlike fractions.

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Instruct the students to work in pairs. Distribute the rectangular strips and coloured pencils among the pairs. Ask the students to show 1 3 and 1 4 on each of the rectangular strips.

Instruct the students to subtract 1 4 from 1 3 using the rectangular strips. Discuss how to divide the strips so that both show the same total number of equal parts.

Explain to the students that this can be done either by finding the LCM of the denominators or by multiplying the denominators. Discuss that each strip should be divided into the same number of parts as the LCM of the denominators. Then from the bigger fraction cross out the same number of shaded parts as the smaller fraction to find the answer.

Ask the students to then subtract by converting both fractions to like fractions and subtracting the numerators in their notebooks.

Extension Idea

Ask: Rashmi has 7 8 m of ribbon, and she uses 1 3 m of ribbon for her project. What length of ribbon is left with her?

Say: Length of the ribbon left =

Subtracting Mixed Numbers

Learning Outcomes

Students will be able to subtract mixed numbers.

Teaching Aids

Circular cutouts; Coloured pencils

Activity

m of ribbon is left with Rashmi.

Instruct the students to work in pairs. Distribute the circular cutouts and coloured pencils among the pairs.

Ask the students to show 21 4 and 11 2 on the circular cutouts and shade them.

Instruct the students to subtract the mixed numbers by first subtracting the wholes by removing as many wholes from the minuend as the subtrahend. They will then subtract the fractional parts the same way as they did in the previous lesson.

Instruct the students to subtract the two mixed numbers in their notebooks by first subtracting the whole number parts and then the fractional parts by converting them to like fractions and subtracting the numerators.

Multiplying Fractions and Whole Numbers

Learning Outcomes

Students will be able to multiply a fraction by a whole number.

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Instruct the students to work in pairs. Distribute 3 rectangular strips and coloured pencils to each pair.

Instruct the students to multiply 2 by 1 4 using the strips. Instruct them to show 2 groups of 1 4 using the strips. Ask them how many quarters there are in 2 strips of paper. Ask them to show the quarters in two strips, using the third strip. Have a discussion around their findings regarding the number of quarters represented. Now, ask them to multiply the whole number by the fraction in their notebooks. Instruct them to multiply the whole number by the numerator and keep the denominator unchanged. Finally, have them compare their answers with those obtained using the strips.

Multiplying

Learning Outcomes

Fractions

Students will be able to multiply two fractions.

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Instruct the students to work in groups. Distribute the rectangular strips and coloured pencils among the groups.

Instruct them to multiply 1 2 by 1 3 using the strips.

Bring out the fact that they need to shade 1 3 in the rectangular strip, and then divide each part into 2 equal parts.

Once the students have shown this on their strips, ask them to draw crosses in the horizontal row to show 1 2 Discuss with the students that the part that is both coloured and has a cross is the product of the fractions. Now, ask them to solve the problem in their notebooks by multiplying the numerators and denominators. They can then compare the answers they get using both methods.

Extension Idea

Ask: How can you multiply a mixed fraction with a proper fraction?

Say: A mixed fraction can be multiplied by a proper fraction by first converting it to an improper fraction and then multiplying the numerators and denominators.

Dividing a Whole Number by a Fraction

Learning Outcomes

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

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Instruct the students to form groups. Distribute the rectangular strips and coloured pencils.

Tell them to find 3 ÷ 3 4 Ask the students to check how many three-quarters there are in three wholes using the strips and coloured pencils.

Instruct them to first divide the strips into 4 equal parts, then shade as many three-fourths as they can, in different colours. Once the students have found the answer using the strips, introduce the reciprocal method. Now, ask the students to solve the question in their notebooks by first finding the reciprocal of the fraction and then multiplying the fractions. Ask them to compare the answers obtained using both methods. Repeat the activity with another set of fractions.

Extension Idea

Ask: How will we divide a whole number by a fraction when we do not get a whole as an answer? Divide 5 ÷ 2 3 .

Say: We will take 5 strips which are divided into 3 equal parts each. Now, we will make pairs of 2 and count the total number of pairs that we get. We get 7 pairs of 2, and 1 equal part is left. Thus, 5 ÷ 2 3

Dividing

a Fraction by a Whole Number

Learning Outcomes

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

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Instruct the students to form groups and distribute the teaching aids. Tell them to divide 6 9 by 2.

Instruct the students to show 6 9 on the rectangular strip. Ask them to make 2 groups from the strips since 6 9 is to be divided by 2. Ask the students to count the number of equal parts in each group, which will form the numerator of the quotient, while keeping the denominator the same.

Now, ask the students to solve the question in their notebooks by first finding the reciprocal of the whole number and then multiplying the fractions.

Ask them to compare the answers that they get using both methods. Repeat the activity with another set of fractions.

Extension Idea

Ask: How do we divide a fraction by a whole number when we cannot directly divide the strip into equal parts?

Divide 1 2 by 3.

Say: 1 2 is the same as 3 6 . So, we will divide 3 6 by 3. Now, 3 6 ÷ 3 = 1 6 , thus, 1 2 ÷ 3 = 1 6 .

Dividing a Fraction by a Fraction Imagine Maths Page 95

Learning Outcomes

Students will be able to divide a fraction by a fraction.

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Instruct the students to form groups. Distribute the rectangular strips and coloured pencils.

Tell the students that they have to divide 3 4 by 1 3

X X

Instruct the students to show the fraction one–third on their sheet by dividing the sheet horizontally and colouring the appropriate section. Then ask them to use the same sheet to show the fraction three–fourths by dividing the sheet vertically and drawing crosses in the appropriate section. Now, ask them to count the number of equal parts that have crosses, which will represent the numerator of their answer (9). The number of equal parts that are shaded will represent the denominator of the answer (4).

Now instruct the students to solve the question in their notebooks by first finding the reciprocal of the divisor and then multiplying the fractions.

Ask them to compare the answer they get using both methods. Repeat the activity with another set of fractions.

1. Adding Unlike Fractions

6. Multiplying Two Fractions Do It

2. Adding Mixed Numbers

Do It Together

7. Dividing a Whole Number by a Fraction Think and Tell The reciprocal of 1 is 1. Do It Together

3. Subtracting Unlike Fractions

Do It Together LCM of 16 and 14 = 112

8. Dividing a Fraction by a Whole Number Do It Together

4. Subtracting Mixed Numbers Do It Together

5. Multiplying Fractions and Whole Numbers

9. Dividing a Fraction by a Fraction

Introduction to Decimals 7

Imagine Mathematics Headings

Like and Unlike Decimals; Converting Unlike Decimals to Like Decimals

Comparing Like Decimals; Comparing Unlike Decimals; Ordering Decimals

Rounding off Decimals to the Nearest Whole Number

Learning Outcomes

Students will be able to:

read and write decimal tenths in figures, words and expanded form. read and write decimal hundredths in figures, words and expanded form. read and write decimal thousandths in figures, words and expanded form. convert fractions to decimals and vice versa. convert between percentages, decimals and fractions. convert unlike decimals to like decimals. compare and order decimals. round off decimals to the nearest whole number.

Alignment to NCF

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 find fractions for the given part of a whole. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

decimals: numbers between whole numbers percentage: the number expressed as if it is part of a total which is a hundred

Teaching Aids

Decimal grids for tenths; Crayons; Number cards with various decimal numbers; Decimal grids for thousandths; Puzzle interlocking cards with a grid on one card, a fraction on the second and a decimal on the third card; Puzzle interlocking cards with a fraction on one card and a decimal on the second card and a percentage on the third card; Digit cards; Zero cards; Circular paper cut-outs (for representing dots); Cards with decimals numbers written on them; Ice-cream sticks; Paper cups

Chapter: Introduction to Decimals

Learning Outcomes

Students will be able to read and write decimal tenths in figures, words and expanded form.

Teaching Aids

Decimal grids for tenths; Crayons

Activity

Begin by discussing decimal numbers in real life and their presentation on a decimal grid. Take 0.4 as an example. Ask the students what fraction of the grid is shaded. Explain that 4 10 is called four-tenths and is written as 0.4 in decimals. Help them write this in their notebooks, in figures as 0.4, in words as “four-tenths”, and in expanded form as 0 + 4 10.

Instruct the students to work in groups. Distribute the decimal grids and crayons among each group. Ask the students to represent 0.7 and then 2.5 on the decimal grid and write each in words and in expanded form in their notebooks.

Extension Idea

Ask: Can one-tenth be further divided into 10 equal parts? What fraction of the grid would be shaded? Say: Yes, one-tenth can be further divided into 10 equal parts. 10/100 of the grid would be shaded.

Learning Outcomes

Students will be able to read and write decimal hundredths in figures, words and expanded form.

Teaching Aids

Decimal grids for tenths; Crayons

Activity

Instruct the students to work in groups. Distribute the tenth grids among the groups. Help them to convert the tenths grid to the hundredths grid by drawing horizontal lines so they can see the change. Ask them to show 35 100 on the grid using crayons. Guide them to write 0.35 as 0.35 in numbers, “thirty-five hundredths” in words, and in expanded form as, 0 + 3 10 + 5 100 in their notebooks. Next, ask them to show 0.75 on the grid and write it in words and expanded form in their notebooks.

Ask questions like, “How would you write 5.27 in words? How would you pronounce the decimal point?”

Learning Outcomes

Students will be able to read and write decimal thousandths in figures, words and expanded form.

Teaching Aids

Number cards with various decimal numbers; Decimal grids for thousandths; Crayons

Activity

Begin by asking a question such as, “What is the difference between 0.1, 0.01 and 0.001 in terms of the place value of 1?”. Discuss with the students their initial thoughts on understanding the place value of decimals. Show the thousandths grid to show 0.001.

Instruct the students to work in groups. Distribute the teaching aids among the groups. Ask them to write 5 wholes, 5 tenths, 3 hundredths and 2 thousandths in numeral, word and expanded form. Instruct the students to use grids to represent the given decimal numbers. Ask them to write the decimal numbers in words and figures in their notebooks.

Extension Idea

Ask: “What are equal decimals?” Explain with example.

Say: Decimals which have the same value are also called equivalent decimals. For example, six tenths and sixty hundreds, that is 0.6 and 0.60.

Learning Outcomes

Students will be able to convert fractions to decimals and vice versa.

Teaching Aids

Puzzle interlocking cards with a grid on one card, a fraction on the second and a decimal on the third card Activity

Demonstrate the conversion of fractions to decimals and vice versa. Recall the concept of the simplest form of fractions. Instruct the students to work in four groups.

Distribute the puzzle sets among the groups. Each set will have a fraction written on one piece, its decimal form on another and its representation on a decimal grid on the third. Ask the students to look at either the fractions or the decimal parts, do the conversion and join them with their corresponding parts. The group that completes the puzzle first wins.

Learning Outcomes

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

Teaching Aids

Puzzle interlocking cards with a fraction on one card and a decimal on the second card and percentage on the third card

Activity

Demonstrate the conversion of fractions to percentages and vice versa, as well as the conversion of percentages to decimal and vice versa.

Instruct the students to work in four groups.

Distribute the puzzle sets among the groups. Each set will have a fraction written on one piece, its decimal form on another and its percentage on another.

Ask the students to look at either the fractions or the decimal parts or percentage part, do the conversion and join them with their corresponding parts. The group that completes the puzzle first wins.

Like

and Unlike Decimals; Converting Imagine Maths Page 111 Unlike Decimals to Like Decimals

Learning Outcomes

Students will be able to convert unlike decimals to like decimals.

Teaching Aids

Digit cards; Zero cards; Circular paper cut-outs (for representing dots)

Activity

Begin by explaining the concepts of like and unlike decimals, using a visual representation such as grids.

Instruct the students to work in groups. Distribute the teaching aids among the groups. Announce both like and unlike decimals randomly.

Instruct the students to show these decimals using the digit cards and circular dots. Have them note the number of cards after the decimal points in their representations. Lead a discussion on distinguishing like and unlike decimals based on the number of cards after the decimal point. Instruct the students to convert unlike decimals to like decimals by inserting zero cards at the corresponding positions and write the converted decimals in their notebooks.

Learning Outcomes

Students will be able to compare and order decimals.

Teaching Aids

Cards with decimals numbers written on them

Activity

Demonstrate the comparison and ordering of decimals to the class. Take the students out to the playground. Instruct the students to work in groups of 5.

Draw a large number line on the floor with decimal values up to the hundredths. Place some decimal cards on a large table. Explain the rules of the relay race to the students. Ask one student from each group to run to the table, pick a card, and place themselves at the correct spot on the number line. Instruct the students to take turns in the relay race format, ensuring each member of a group gets a chance to participate. Use a stopwatch or timer to track the duration of the game. Conclude the activity with a class discussion on the learning gained from physically moving on the number line. Declare the group that finishes first, with the decimals correctly arranged on the number line, as the winning team. Back in the classroom, ask the students to write the correct order of their decimals in their notebooks.

Extension Idea

Instruct: Consider the decimal number 0.256. Rearrange the digits after the decimal point to form three new numbers. Compare these numbers and write your answers.

Say: When we compare these numbers, we can see that 0.625 is the biggest number, and 0.256 is the smallest number. We can also see that 0.562 is bigger than 0.526, hence, the correct order will be 0.625 > 0.562 > 0.526

Rounding off Decimals to the

Nearest Whole Number

Learning Outcomes

Students will be able to round off decimals to the nearest whole number.

Teaching Aids

Ice cream sticks; Paper cups

Activity

Imagine Maths Page 116

Head outdoors with the students to an area with trees. Label two trees as ‘Taller’ and ‘Shorter’. Prepare about 40 ice cream sticks and write decimal numbers up to the tenths place on each. Give each student a stick and ask them to check their number.

Explain to the student that if the number in the tenths place is 5 or more, they should run to the ‘Taller’ tree. If it’s 4 or less, they should run to the ‘Shorter’ tree. Now, those at the ‘Taller’ tree should round their decimals up to the next number, and the others should round to the current number. Ask them to write down their answers in their notebooks.

Extension Idea

Ask: How can you round off 11.47 to the nearest whole number?

Say: We can round off 11.47 to 11 as the tenths place has 4 which is less than 5.

Answers

1. Tenths

Do It Together

3.4 = 3 + 4 10 or 3 + 0.4

2. Hundredths

Do It Together Tens Ones Decimal point Tenths Hundredths 1.34 0 1 3 4

Expanded form – 1 + + 3 10 4 100 or 1 + 0.3 + 0.04

3. Thousandths

Do It Together

1. 7 tens + 5 ones + 9 hundredths + 2 thousandths = 75.092

2. 600 + 20 + 2 + 7 1000 = 622.007

3. 5 tens + 2 ones + 5 hundredths + 1 thousandths = 50 + 2 + 5 100 + 1 1000 = 52.051

4. Conversion Between Fractions and Decimals

Think and Tell

We do so because it makes changing them into decimals easier. It is like using groups of 10, 100 or 1000, which are simple and help us understand decimals better.

Do It Together 8

5. Percentages

Do It Together

As a fraction: 12.3% = 12.3 100 = 1233 1000

As a decimal: 12.3% = 12.3 100 = 0.123

6. Like and Unlike Decimals

Do It Together

Number of digits after the decimal point in 51.54 = 2

Number of digits after the decimal point in 187.37 = 2

So, the numbers of digits after the decimal point (are/are not) equal.

Thus, 51.54 and 187.37 (are/are not) like decimals.

7. Converting Unlike Decimals to Like Decimals

Do It Together

189.474 = 189.474

1.2 = 1.200

Thus, 189.474 and 1.200 are a group of like decimals.

8. Comparing Like Decimals

Do It Together

The digit in the tenths place: 5 > 0. So, 174.54 > 174.05. 174.05 is smaller.

9. Comparing Unlike Decimals

Do It Together

11. Rounding off Decimals to the Nearest Whole Number Do It Together

The digit in the tenths place: 4 > 1.

Thus, 145.40 < 145.40. So, 145.40 is greater.

10. Ordering Decimals

Do It Together

15.47 15.470 15.4

Operations with Decimals

Learning Outcomes

Students will be able to: add decimals. subtract decimals. multiply decimals by 10, 100 and 1000. multiply decimals by whole numbers and decimals. convert currencies to currencies of different countries and vice versa. divide decimals by 10, 100 and 1000. divide decimals by whole numbers.

Alignment to NCF

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 the basics of decimals and how to compare them. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

currency: medium of exchange for goods and services

Teaching Aids

Decimal grids for 10s and 100s; Crayons; Crossword puzzle cards; 10s and 100s grid with multiplication problems written on them; Play money; Items with price tags in 4 different currencies; Puzzle cards with division problems; 10s and 100s grid with division problems written on them

Chapter: Operations with Decimals

Addition of Decimals

Learning Outcomes

Students will be able to add decimals.

Teaching Aids

Decimal grids for 10s and 100s; Crayons

Activity

Instruct the students to form groups. Distribute the 10s and 100s decimal grids to each group.

Provide them with two questions around adding two tenths and two hundredths (such as 0.3 + 0.6, and 0.35 + 0.25), ask the students to colour the 10s grid in 2 different colours to show the first set of decimals and 100s grids in 2 different colours to show the second set of decimals. They will then add the shaded grids in find the answer for each problem.

Then, ask them to write the decimals using the column method in their notebooks, and add the decimals. They can then compare the answers that they got using both methods.

Ask them a question, “What is one thing that you should keep in mind while lining up the decimals?”

Extension Idea

Ask: How will you write the tenths place when you add the decimals 0.6 and 0.8?

Say: When adding 0.8 and 0.6, we see that the number in the tenths place increases to more than 9. Converting 10 tenths to 1 in the ones place, the final answer is written as 1.4.

Subtraction of Decimals

Learning Outcomes

Students will be able to subtract decimals.

Teaching Aids

Decimal grids for 10s and 100s; Crayons Activity

Instruct the students to form groups. Distribute the decimal grids to each group. Provide each group with two questions around subtracting two tenths and two hundredths (such as 0.7 – 0.3 and 0.45 – 0.25), ask them to colour the grids representing the first decimal number.

Ask questions like, “How will you show subtraction on the grid?”

Instruct them to cross out the shaded cells to show the subtraction of the second number. Discuss the answers.

Ask them to perform the subtraction using the column method in their notebooks, and compare the answers that they got using both methods.

Imagine Maths Page 124

Multiplying Decimals by 10, 100, 1000, ...

Learning Outcomes

Students will be able to multiply decimals by 10, 100 and 1000.

Teaching Aids

Crossword puzzle cards

Activity

Write 3 problems on the board one below the other: 5.142 × 10 = 51.42; 5.142 × 100 = 514.2; 5.142 × 1000 = 5142. Instruct the students to form groups.

Ask the students to look at the multiplication and discuss in their groups what they notice about the product in each case. After the discussion, facilitate a class conversation where students can share their responses and observations. Distribute the crossword puzzle card to each group. Instruct them to solve these problems and write the answers in the respective blanks.

Extension Idea

Ask: Can you use your understanding of multiplying by 10, 100 and 1000 to multiply a decimal by 10,000 or 1,00,000? What is 5.142 × 1,00,000?

Say: Yes, the same rule applies when multiplying by 10,000 or 1,00,000. For example, 5.142 × 1,00,000 = 5,14,200. Multiplying Whole Numbers and Decimals;

Learning Outcomes

Students will be able to multiply decimals by whole numbers and decimals.

Teaching Aids

10s and 100s grid with multiplication problems written on them

Activity

Instruct the students to form groups for the activity. Provide each group with decimal grids that have multiplication problems written at the top.

Instruct them to colour the grid to show one decimal vertically and draw lines to show the other decimal horizontally. Ask questions about the overlapping area that is shaded and has lines. Direct students to count the number of such cells which forms the product of the two decimals. Then, ask them to write the two decimals and the answer in their notebooks. Have students discuss among their groups how they should multiply decimals using the column method without the grid. Give them one more problem to solve.

Extension Idea

Ask: Aarav got 1.45 points in each of the 5 tests. How many points did he get? Say: To find the answer, we need to multiply 1.45 by 5. 1.45 × 5 = 7.25.

Learning Outcomes

Students will be able to convert currencies to currencies of different countries and vice versa.

Teaching Aids

Play money; Items with price tags in 4 different currencies

Activity

Begin the lesson with a small discussion about using Indian currency to buy goods in other countries. Explain to the students that different countries have their own currencies and that it is necessary to convert money when travelling internationally.

Engage the students in a small role play. For this, divide the class into groups and distribute play money representing Indian rupees to each group. Provide each group with items with price tags in 4 different currencies (such as a notebook for $6, £3 etc). Each student in the group will try to find the cost of one item in Indian currency in their notebooks. Ask them to refer to the conversion table in their Imagine Mathematics books. Ask questions like, “Can you use currency from other countries to buy things in India?”

After the activity, initiate a class discussion to allow students to share their learning.

Extension Idea

Ask: Susan bought a pencil for 65 Nepalese rupees. How much money did she spend in Indian rupees?

Say: As per the table, we know that ₹1 = NPR 0.62. So, 65 NPR = 0.62 × 65 = ₹40.3.

After the activity, initiate a class discussion to allow students to share their learning. Dividing Decimals by 10, 100, 1000, ...

Learning Outcomes

Students will be able to divide decimals by 10, 100 and 1000.

Teaching Aids

Puzzle cards with division problems

Activity

Write the 3 problems on the board one below the other: 185.7 ÷ 10 = 18.57; 185.7 ÷ 100 = 1.857; 185.7 ÷ 1000 = 0.1857. Instruct the students to form groups.

Ask the students to look at the division and discuss in their groups what they notice about the quotient in each case.

Discuss with the class that when we divide any decimal number by 10, 100 or 1000, we move as many decimal points to the left as there are 0s in the divisor.

Distribute the puzzle cards with different problems to each group. Explain that they will find problems in the puzzle where they have to divide by 10, 100 or 1000. They will keep the answer card in the middle and the question cards that give that answer all around it. Now, ask them to write the answers in their notebooks.

Ask questions like, “Did we need to actually perform division? Why did we shift the decimal point to the left, and not towards the right?”

Learning Outcomes

Students will be able to divide decimals by whole numbers.

Teaching Aids

10s and 100s grid with division problems written on them

Activity

Instruct the students to form groups for the activity. Provide each group with decimal grids of 10s and 100s with 1 problem each. (For example: 0.8 ÷ 4, 0.88 ÷ 8)

Instruct them to colour the grid to show the dividend (0.8) on the grid. Ask them to divide the shaded grid into the same number of parts as the divisor (4). The can draw lines or dots to show the 4 different parts. They will then count the decimal shown by each parts (0.2). Ask them to write the division problem and the answer in their notebooks (0.8 ÷ 4 = 0.2). They will then examine the grid and discuss among their groups how they should divide decimals using the column method. Ask them to solve the second problem using the 100s grid and the column method.

Answers

1. Addition of Decimals

2. Subtraction of Decimals Do It Together 1.

3. Multiplying Decimals by 10, 100, 1000, ... Do It Together

6. Currencies from Different Countries Do It

4. Multiplying Whole Numbers and Decimals

7. Dividing Decimals by 10, 100, 1000, ... Do It

8. Dividing Decimals by Whole numbers Do It

5. Multiplying Two Decimal Numbers

Think

Lines and Angles

Learning Outcomes

Students will be able to: identify a point, ray, line and line segment and classify the different types of lines. identify and classify the different types of angles. measure angles using a protractor. draw angles using a protractor. identify triangles, quadrilaterals and polygons and classify the types of quadrilaterals and polygons.

Alignment to NCF

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

C-3.4: 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 identify sleeping, standing and slanting lines. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

line segment: the shortest distance between 2 points, having a definite length ray: a part of a line that has one endpoint and extends indefinitely in the other direction right angle: an angle measuring 90° protractor: a device in the shape of a ‘D’ that helps measure angles of different degrees polygon: a closed and flat shape made of straight lines quadrilateral: a polygon of four sides

Teaching Aids

Circle and triangle coloured paper cutouts; Toothpicks; Sheets of paper; Glue stick; Clock with movable hands; Colourful craft paper; Ice cream sticks; Protractor; Sketch pen; Ruler; Chart paper

Chapter: Lines and Angles

Types of Lines

Learning Outcomes

Students will be able to identify a point, ray, line and line segment and classify the different types of lines.

Teaching Aids

Circle and triangle coloured paper cutouts; Toothpicks; Sheets of paper; Glue stick Activity

Begin by discussing real-life examples of a line, line segment, ray and point. Distribute toothpicks, circle and triangle coloured paper cutouts and a sheet of paper to each student.

Instruct the students to paste the toothpicks and cutouts on to the sheet of paper to show a point, a line, a line segment and a ray. Ask them to use the circle cutouts as points and the triangle cutouts as the extended ends of the line and ray.

Discuss the concepts of parallel, intersecting and perpendicular lines, and ask the students to create them on the sheet of paper.

Ask questions like, “Which pairs of lines will never meet when extended?”

Draw a figure on the board with points, lines, rays and line segments. Ask the students to identify and name different points, lines, and line segments in the figure and write them in their notebooks.

Types of Angles

Learning Outcomes

Students will be able to identify and classify the different types of angles.

Teaching Aids

Clock with movable hands; Colourful craft paper; Glue stick; Ice cream sticks Activity

Show the hands of a clock to discuss the different types of angles.

Distribute craft paper and glue sticks to each student. Ask the students to fold the paper to make pleats, then fold it in half and paste the two middle edges to make a fan. Paste sticks on the outer edges.

Once the fan is made, ask the students to fold it into different types of angles, then place and draw an outline of it in their notebooks each time.

Measuring Angles

Learning Outcomes

Students will be able to measure angles using a protractor.

Teaching Aids

Protractor

Activity

Show a protractor and the different markings on it to the class. Demonstrate how to use the protractor to measure an angle.

Instruct the students to work in pairs. Distribute the protractors to the students.

Ask them to measure the angles they outlined in the previous lesson and write the measures in their notebooks. Ask them to discuss the angle measurements with their partners.

Ask questions like, ”What is the range of angles that form acute angles? Obtuse angles?”

Extension Idea

Ask: Sunita walks to the east and then takes two turns at right angles. What angle does she make?

Say: Two turns at right angles from facing east will make a 180° angle.

Drawing Angles Imagine Maths Page 149

Learning Outcomes

Students will be able to draw angles using a protractor.

Teaching Aids

Sheets of paper; Sketch pen; Ruler; Protractor

Activity

Distribute a sheet of paper, a sketch pen, a ruler and a protractor to the students. Ask the students to draw a line segment and label it. Ask them to draw angles of any measure on both endpoints and write the measure of the angles. Instruct them to use the outer and inner scales of the protractor to draw angles at different points on the line segment. Repeat the activity for another point O.

Ask questions like, “Which scale of the protractor did you use to draw the angle at the point on the left? At the point on the right?”

Extension Idea

Ask: An angle measures 120° on the inner scale. What will it measure on the outer scale?

Say: The angle will measure 60° on the outer scale.

Learning Outcomes

Students will be able to identify triangles, quadrilaterals and polygons and classify the types of quadrilaterals and polygons.

Teaching Aids

Ice cream sticks; Chart paper

Activity

Instruct the students to form groups.

Distribute the ice cream sticks and chart paper.

Instruct the students to create a collage of polygons on the chart paper using the ice cream sticks as the side of the polygon.

Write the name of the polygon and its features, like the number of sides, the number of angles and the number of vertices under each polygon.

Ask the students to draw 2 polygons in their notebooks and label them.

Ask questions like, “How many sides and angles does a pentagon and hexagon have? Which shapes have the same number of sides and angles in them? Is there any polygon with less than 3 sides?”

Extension Idea

Ask: How do you know that a shape is not a polygon?

Say: When a shape does not have straight sides or one of the sides is open from 1 or more sides, it is not a polygon.

1. Types of Lines Do It Together

1. Number of points outside the square = 6

2. Name 2 line segments = WX, XY

3. Length of line segment ZY is 2.5 cm.

4. The line parallel to WZ is XY.

5. Draw a line passing through point E.

6. Draw a ray from point B.

2. Types of Angles

Figures may vary. Sample figures:

5. Triangles, Quadrilaterals and Polygons

Think and Tell

Yes, triangles and quadrilaterals are polygons too because they are closed figures that have 3 or more straight sides. Do It Together

Number of sides – 8

Number of angles – 8

3. Measuring Angles

Do It Together

Number of sides – 4

Number of angles – 4

Number of sides – 4

Number of angles – 4

Number of sides – 4

Number of angles – 4

4. Drawing Angles

Do It Together

Patterns and Symmetry

Learning Outcomes

Students will be able to:

extend and create repeating and rotating patterns. extend and create growing and tiling patterns. identify the rule of a number pattern or puzzle, and extend and solve it. identify symmetry and reflection in shapes and figures.

Alignment to NCF

C-1.4: Recognises, describes, and extends simple number patterns such as odd numbers, even numbers, square numbers, cubes, powers of 2, powers of 10, and Virahanka–Fibonacci numbers

C-2.3: Recognises and creates symmetry (reflection, rotation) in familiar 2D and 3D shapes

C-2.4: Discovers, recognises, describes, and extends patterns in 2D and 3D shapes

Let’s Recall

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

Vocabulary

extend (pattern): show the terms that would come next in the sequence rotate: to circle around a point overlap: to cover something partly by going over its edge reflection: an image we see in a mirror

Teaching Aids

Flowers and leaves; Watercolours; Paint brushes; Rectangle, triangle and circle paper cutouts; Slips of paper for writing a secret message; Sheets of paper; Small mirror

Chapter: Patterns and Symmetry

Extending and Creating Patterns

Learning Outcomes

Students will be able to extend and create repeating and rotating patterns.

Teaching Aids

Flowers and leaves; Watercolours; Paint brushes

Activity

Imagine Maths Page 159

Begin the class by discussing what repeating and rotating patterns are and how to extend them. Instruct the students to form groups of four. Take the students to the playground.

Instruct them to pick 2–3 flowers and some leaves from the ground, being mindful of the environment and ensuring they do not cause harm to plants. Remind them that they cannot pluck the flowers or leaves.

Distribute the watercolours and the paint brushes to the groups.

Instruct them to use the flowers and leaves they collected to create repeating or rotating patterns in their notebooks by painting one side of the leaves, flowers or petals and making impressions on paper in different ways to form the pattern.

Extension Idea

Ask: How can you create a pattern that is both repeating and rotating?

Say: We can create a pattern by showing a triangle that is rotating and a rectangle one after the other.

Growing and Tiling Patterns

Learning Outcomes

Students will be able to extend and create growing and tiling patterns.

Teaching Aids

Rectangle, triangle and circle paper cutouts

Activity

Begin by discussing growing and tiling patterns.

Distribute the cutouts of shapes to pairs, giving two kinds of shapes to each pair.

Imagine Maths Page 163

Instruct the students to work in pairs where one student makes a growing pattern with one of the shapes and the other makes a tiling pattern with the other shape.

Ask: Which shape could not be used in making a tiling pattern? Why?

Extension Idea

Ask: How can we use triangles and squares to form a tiling pattern?

Say: We can use triangles and squares to form a tiling pattern in various ways.

Learning Outcomes

Students will be able to identify the rule of a number pattern or puzzle, and extend and solve it.

Teaching Aids

Slips of paper for writing a secret message

Activity

Tell the students that a code ‘10 21 9 3 5’ hides a secret message, following the pattern of alphabets in numerical order, and ask them to decode the code. Discuss that the code hides the word ‘JUICE’.

Instruct the students to form groups of four.

Write the given code on the board.

Instruct the students to decode the message using the letters for the given numbers, such as A for 1, B for 2, C for 3 and so on. Discuss the message that the students received.

Now, ask the students to form a new secret message using the same letter-number rule on a slip of paper and pass it on to the next group. They will then decode the messages made by the other groups. The group that decodes the message first wins.

Symmetry and Reflection Imagine Maths Page 170

Learning Outcomes

Students will be able to identify symmetry and reflection in shapes and figures.

Teaching Aids

Sheets of paper; Watercolours; Paint brushes; Small mirror

Activity

Begin the activity by discussing how a figure can be divided into two identical halves, we say that the figure is symmetrical. Explain to the students that the line that divides them into identical halves is called the line of symmetry.

Distribute the sheets of paper to the students.

Instruct them to fold their sheets in half and then unfold them to make a crease in the middle of the sheet. Using water colours, they will paint a design or pattern on one side. Ask the students to fold the sheet precisely along the crease created earlier, ensuring that the painted side is facing inward. Instruct them to press down on the folded paper to transfer the colours to the other side, then unfold the sheet. For instance, the sheet may look like the given example. Next, instruct the students to use a small mirror and place it along the dotted line to see if the reflected half matches the other side. Finally, allow them to draw or cut their own shapes and test for symmetry using the mirror.

Extension Idea

Ask: Imagine cutting the shape given. Would the two halves be symmetrical?

Say: Yes, cutting the given shape in half vertically, straight down through the centre to the base, would be symmetrical.

1. Repeating Patterns

Do It Together

Answers

5. Coding and Decoding Patterns Do It Together

2. Rotating Patterns

Do It Together

3. Growing and Reducing Patterns

Do It Together Pattern What will Come Next? Type of Pattern

Growing pattern

Reducing pattern

Decode the word: CNGGREA.

Therefore, the given word is PATTERN

6. Patterns in Numbers Do It Together

4. Tiling Patterns

Think and Tell

Yes, a tangram also has a tiling pattern. Do It Together

7. Symmetry and Reflection

Think and Tell

Symmetrical objects: bottle, blackboard, etc.

Asymmetrical objects: tree, shoe, etc. Do It Together

Length and Weight 11

Learning Outcomes

Students will be able to: estimate the length of an object and measure it using a ruler. convert between the different units of length. solve word problems on adding, subtracting, multiplying and dividing lengths. estimate the weight of an object. convert between the different units of weight. solve word problems on adding, subtracting, multiplying and dividing weights.

Alignment to NCF

C-3.2: Uses an appropriate unit and tool for the attribute (like length, perimeter, time, weight, volume) being measured

C-3.3: Carries out simple unit conversions, such as from centimetres to metres, within a system of measurement

C-3.5: Devises strategies for estimating the distance, length, time, perimeter (for regular and irregular shapes), area (for regular and irregular shapes), weight, and volume and verifies the same using standard units

C-3.7: Evaluates the conservation of attributes like length and volume, and solves daily-life problems related to them

Let’s Recall

Recap to check if students know about the use of length and weight and the units used to measure them. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

estimation: the process of guessing an answer that is close to the actual answer conversion: changing the value from one form to another

Teaching Aids

A ruler; Pencils; Sketch pens; Crayons; Erasers; Paint brushes; Paper clips; Problem cards with different length conversion problems; Solution cards with corresponding converted lengths; Word problem cards with different length problems; Weight blocks of 50 mg, 10 g, 500 g and 1 kg; Classroom supplies, like pencils, water bottles, pencil boxes, erasers etc; Problem cards with different weight conversion problems; Solution cards with corresponding converted weights; Word problem cards with different weight problems

Chapter: Length and Weight

Estimating Length; Measuring Lengths

Learning

Outcomes

Students will be able to estimate the length of an object and measure it using a ruler.

Teaching Aids

A ruler; Pencils; Sketch pens; Crayons; Erasers; Paint brushes; Paper clips

Activity

Imagine Maths Page 180

Begin by asking, “How tall do you think the door of the classroom is?” or “Can you tell how long a grain of rice can be? Will it be measured in centimetres or metres?”

Show the millilitre and centimetre scales on a ruler and a pencil/pen.

Ask them questions like, “How long do you think this pencil/pen is? Will it be measured in mm or cm?”

Demonstrate to the student how to correctly use a ruler to measure the pencil in cm and mm.

Divide the class into pairs. Distribute crayons, sketch pens, erasers, paint brushes and paper clips among the pairs. Instruct each pair to take turns and guess the length of each object and then measure it using a ruler. Ask them to write down the length of each object in their notebooks.

Extension Idea

Ask: Can we measure longer objects, such as a pole or the side of a bed, with a ruler? What scale should we use? Say: Yes, we can measure longer things too, but we have to use the ruler many times for the full length. It is easier to measure longer objects using a metre scale, such as a metre rod or a measuring tape.

Converting Between Units of Length

Learning Outcomes

Students will be able to convert between the different units of length.

Teaching Aids

Problem cards with different length conversion problems; Solution cards with corresponding converted lengths Activity

Create problem cards and solution cards. Distribute 2–3 problem cards to each student and place the solution cards face down on a big table. On the floor, draw a length unit conversion line with markings for:

Ask the students to form a circle around the drawing on the floor. Ask questions like, “Which units are smaller? Which one is the bigger unit of length?”

Explain how to convert between units by multiplying or dividing by 10.

Instruct the students to look at the problem cards and read their conversion problem, such as “Convert 4.5 m into cm.” Ask them to do the conversion and collect the corresponding solution cards from the table. Instruct the students to place their problem and solution cards one below the other. Check the cards collected by all the students and ask them to do the conversion in their notebooks as well.

Extension Idea

Ask: If we have to convert kilometre(km) to millimtre(mm), should we go step by step or is a direct conversion possible?

Say: We can directly multiply kilometre(km) with 10,00,000 to convert it to millimetre(mm) rather than going step by step means from km to hectometres, than to decametres and so on.

Word Problems on Length

Learning Outcomes

Imagine Maths Page 183

Students will be able to solve word problems on adding, subtracting, multiplying and dividing lengths.

Teaching Aids

Word problem cards with different length problems

Activity

Divide the class into pairs. Distribute cards with word problems such as, “Priya bought 3 rolls of ribbon with the following lengths—Red: 15 m; Blue: 7.5 m; Golden: 13 m.”

1. What is the total length of all the ribbons?

2. How much longer is the golden ribbon than the blue?

3. What will be the length of the red ribbons in 7 rolls?

4. How many pieces of 3 m can be cut out from the red ribbon roll?

Instruct the partners to read and discuss their word problem and solve it in their notebooks using the strategy: What do we know? What do we need to know? and Solve to find the answer. Explain the strategy if needed.

Estimating Weights Imagine Maths Page 186

Learning Outcomes

Students will be able to estimate the weight of an object.

Teaching Aids

Weight blocks of 50 mg, 10 g, 500 g and 1 kg; Classroom supplies, like pencils, water bottles, pencil boxes, erasers etc.

Activity

Distribute the weight blocks among the students and ask them to feel the weight of each block in their hands. Discuss what each block weighs.

Ask questions like, “Which block feels heavier?”, “Which one is the lightest?” or “How much do they weigh?”

Divide the class into pairs. Ask the students to take out their maths textbooks, pencils, water bottles, pencil boxes, etc. Ask them to hold the objects one by one and guess the weight in milligrams, grams or kilograms. Ask the partners to discuss and write the estimated weight in their notebooks, such as:

Paper clip = about 2 mg.

Extension Idea

Ask: How does estimating the weight of different objects help us in real life?

Say: In real life, it may not be possible to weigh all objects on a balance scale and a weighing machine may not be available everywhere. Therefore, we estimate the weight of different things to make our work easier.

Converting Between Units of Weight

Learning Outcomes

Students will be able to convert between the different units of weight.

Teaching Aids

Imagine Maths Page 186

Problem cards with different weight conversion problems; Solution cards with corresponding converted weights

Activity

Create problem cards and solution cards. Distribute 2–3 problem cards to each student and place the solution cards face down on a big table. On the floor, draw a weight unit conversion line with markings for:

Ask the students to form a circle around the drawing on the floor. Ask questions like, “Which units of weight are smaller?” or “Which one is the bigger unit?”

Explain to them how to convert between units by multiplying or dividing by 10.

Instruct the students to look at the problem cards and read their conversion problem, such as Convert 7 kg into mg. Ask them to do the conversion and collect the corresponding solution cards from the table. Instruct them to place the problem and solution cards one below the other. Check the cards collected by all the students. Ask them to do the conversion in their notebooks as well.

Word Problems on Weights

Learning

Outcomes

Students will be able to solve word problems on adding, subtracting, multiplying and dividing weights.

Teaching Aids

Word problem cards with different weight problems

Activity

Divide the class into pairs. Distribute cards with word problems, such as, “Madhu bought 1 kg 500 g of apples and 6 kg of oranges.”

1. How much fruit does she have in total?

2. Madhu’s family ate 2 kg of oranges in a day. How many oranges are left now?

3. Madhu eats 500 g of apples each day. How much does she need for a week?

4. Madhu’s family consumes 2 kg of oranges each day. In how many days will they consume 6 kg of oranges?

Instruct the pairs to read and discuss their word problem and solve it in their notebooks using the strategy: What do we know? What do we need to know? and Solve to find the answer.

Extension Idea

Ask: Can you think of other word problems with weight in your day-to-day lives? Give some examples. Say: There are many problems with weight in our real lives. For example, “I buy 12 kg of rice, 5 kg of wheat flour and 2 kg of sugar. What is the total weight of the food items I bought?”

Answers

1. Estimating Length

Do It Together

Students will circle the following lengths:

2. Length of a pen: 14 cm

3. Height of a car: 160 cm

4. Length of your shoe: 12 cm

2. Measuring Lengths

Do It Together

The toffee is longer than 3 cm but shorter than 4 cm.

There are 7 mm lines after 3 cm.

Hence, the toffee is 3 cm 7 mm = 3.7 cm long.

3. Converting Between Units of Length

Do It Together

1000 mm = 1 m

1 mm = 1 1000 m

5 m 230 mm = 5 m + 230 × 1 1000 m

= 5 m + 0.230 m = 5.230 m

4. Word Problems on Length

Think and Tell

1 m = 100 cm OR 1 cm = 1 100 m

So, 1 m 55 cm = 1 m + 55 100 m

= 1 m + 0.55 m = 1.55 m

Do It Together

Length of cloth needed to make a saree = 5 m 80 cm = 5.80 m

Number of sarees required = 5

Total length of cloth needed for making 5 sarees

= 5.80 m × 5

So, 29 m of cloth is required for making 5 sarees.

5. Estimating Weights

Do It Together

Students will circle the following estimates:

2. A 2-rupee coin: 14 g

3. A photo album: 600 g

4. A tube of toothpaste: 120 g

6. Converting Between Units of Weight

Do It Together

1 g = 100 cg

5 g 230 cg = 5 × 100 cg + 230 cg = 730 cg

Think and Tell

1 kg = 1000 g OR 1 g = 1 1000 kg

So, 2 kg 106 g = 2 kg + 106 1000 kg

= 2 kg + 0.106 kg = 2.106 kg

7. Word Problems on Weights

Do It Together

Total weight of books removed from the suitcase

= 840 g + 840 g = 1680 g OR 1.680 kg

Weight of the suitcase after removing the books = 9.200 kg – 1.680 kg = 7.520 kg OR 7 kg 520 g

Perimeter and Area

Learning Outcomes

Students will be able to: find the perimeter of a square and a rectangle. find the area of a square and a rectangle and use it to find the area of compound shapes. find the area of a triangle on a square grid.

Alignment to NCF

C-3.1: Measures in non-standard and standard units and evaluates the need for standard units

C-3.2: Uses an appropriate unit and tool for the attribute (like length, perimeter, time, weight, volume) being measured

C-3.4: Understands the definition and formula for the area of a square or rectangle as length times breadth

C-3.6: Deduces that shapes having equal areas can have different perimeters and shapes having equal perimeters can have different areas

Let’s Recall

Recap to check if students know how to find the area of a figure by counting squares on a square grid. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

boundary: a line that shows the limits or edges of a certain area or space enclose: to surround or close something within a defined area compound shape: shapes formed with more than one shape

Teaching Aids

Squared paper; Crayon box

Chapter: Perimeter and Area

Perimeter of Squares and Rectangles

Learning Outcomes

Students will be able to find the perimeter of a square and a rectangle.

Teaching Aids

Squared paper

Activity

Imagine Maths Page 195

Start by discussing the perimeter of a square and a rectangle. Demonstrate the formula for the perimeter of both the shapes.

Instruct the students to work in groups. Distribute the squared paper to the groups.

Instruct the students to draw squares and rectangles on the squared paper with a perimeter of 24 cm. Each student should draw 1 square and 1 rectangle. They will then write the side lengths on the sides of the shapes drawn and the perimeter inside the shape. In their notebooks, they will calculate the perimeter of these shapes using the formula.

Discuss how, for each perimeter, only one square could be drawn, whereas multiple rectangles could be drawn.

Extension Idea

Ask: The perimeter of a rectangle is the same as that of a square. If the length of the rectangle is 7 cm and the breadth is 3 cm, find the length of the side of the square.

Say: The perimeter of the rectangle is 2 × 7 + 2 × 3 = 20 cm. Thus, the side of the square will be 20 ÷ 4 = 5 cm.

Area of a Square; Area of a Rectangle

Learning Outcomes

Imagine Maths Page 198

Students will be able to find the area of a square and a rectangle and use it to find the area of compound shapes.

Teaching Aids

Squared paper; Crayon box

Activity

Start by discussing the area of a square and a rectangle. Demonstrate the formulas used to find the area each shape.

Instruct the students to work in pairs. Distribute a sheet of squared paper and crayons to each student.

Instruct each student to shade certain areas on the squared paper using crayons of different colours to create a combined shape.

Ask them to exchange their sheets with their partners. Their partner will then find the area of the shape by splitting the shape into smaller parts. Instruct them to write their answers in their notebooks.

Learning Outcomes

Students will be able to find the area of a triangle on a square grid.

Teaching Aids

Squared paper

Activity

Start by discussing the area of a triangle. Explain to the students that the area of a triangle is half the area of a rectangle.

Instruct the students to work in pairs. Distribute a sheet of squared paper and crayons to each student.

Ask the students to draw a triangle on the squared paper such that it covers both halves and full squares. Instruct them to exchange their squared paper with their partners. Their partner will then find the area of the shape by counting the number of half and full squares covered.

Instruct them to write the answers in their notebooks.

Extension Idea

Instruct: Find the area of the given figure.

Say: The area of the first triangle is 10 ÷ 2 = 5 sq. units. Similarly, the area of the second triangle will be 6 ÷ 2 = 3 sq. units. Thus, the total area will be 5 + 3 = 8 sq. units.

Answers

1. Perimeter of a Rectangle

Do It Together

Here, length (l) = 60 cm and breadth (b) = 20 cm

Perimeter of a rectangle = 2(l + b)

The perimeter of the pillow

= 2 × (60 + 20)

= 2 × 80 = 160 cm

2. Perimeter of a Square

Do It Together

Here, side (s) = 45 cm.

Perimeter of a square = 4 × s = 4 × 45 cm

So, the perimeter of the chessboard is 180 cm.

3. Area of a Square

Think and Tell

There are 3 rows. Each row has 3 small squares. The required multiplication sentence is 3 × 3 = 9. There are 9 small boxes in total. We noticed that the number of rows and number of squares in each row are equal.

Do It Together

The area of the photo frame = 8 cm × 8 cm

The area of the photo frame = 64 sq. cm.

4. Area of a Rectangle

Do It Together

The area of the big rectangle = 12 × 6 = 72 sq. cm

The area of rectangle A + area of rectangle B = (5 × 7) + (4 × 7) = 72 + 35 + 28 = 135 sq. cm

The area of the whole shape = 135 sq. cm

5. Area of a Triangle

Do It Together

Area of rectangle 1 = 18 sq. units; Area of triangle 1 = 9 sq. units.

Area of rectangle 2 = 18 sq. units; Area of triangle 2 = 9 sq. units

Area of whole triangle = 18 sq. units

Capacity and Volume

Learning Outcomes

Students will be able to: estimate and measure the capacity of different containers. convert between units of capacity. find the volume of solids using unit cubes. find the volume of solids using the formula. solve word problems on finding capacity and volume.

Alignment to NCF

C-3.1: Measures in non-standard and standard units and evaluates the need for standard units

C-3.2: Uses an appropriate unit and tool for the attribute (like length, perimeter, time, weight, volume) being measured

C-3.3: Carries out simple unit conversions, such as from centimetres to metres, within a system of measurement

Let’s Recall

Recap to check if students know about capacity and can write express it in mL and L. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

capacity: the amount that can be held in a particular space volume: the amount of space that an object occupies unit cubes: small, equal-sized cubes used as a standard measure

Teaching Aids

Measuring jars of capacity 1 L; Containers like a cup, glass and bottle; Water for filling up containers; Conversion cards; Different–sized rectangular containers; Unit cubes; Question cards with word problems written

Chapter: Capacity and Volume

Estimating Capacity; Measuring Capacity

Learning Outcomes

Students will be able to estimate and measure the capacity of different containers.

Teaching Aids

Measuring jars of capacity 1 L; Containers like a cup, glass and bottle; Water for filling up containers

Activity

Begin by asking a few question related to capacity: How much water (in mL or L) does your water bottle hold? How much water can a syringe, the cap of a bottle or a bucket hold?

Instruct the students to form groups and distribute the containers among them.

Ask the students to guess the amount of water each containers can hold and write their estimates in their notebooks. Distribute the measuring jars and water.

Ask the students to measure the capacity of each container by filling it with water and then pouring the water into the measuring jar. Instruct them to note down the measures in their notebooks. Have a discussion around the estimated capacity and the actual capacity of the containers. Ask questions like, “Does a container with a greater height always have more capacity than a shorter container? Why?”

Units of Capacity

Learning Outcomes

Students will be able to convert between units of capacity.

Teaching Aids

Conversion cards

Activity

Start by introducing different units of capacity beyond millilitres (mL) and litres (L), such as kilolitres, decalitres, hectolitres, decilitres and centilitres. Divide the class into groups. Distribute a pack of 10 conversion cards to each group. Instruct the students to convert the given capacity into the different units of measurement as mentioned in the conversion cards, and then match the cards with measures that have the same capacity. For example, if 653 L is 6530 dL, the card with 6530 should be placed on the side reading dL.

Ask the students to write the answers in their notebooks.

Shuffle the conversion cards among the groups and repeat the activity.

Volume of Solids Using Unit Cubes

Learning Outcomes

Students will be able to find the volume of solids using unit cubes.

Teaching Aids

Different–sized rectangular containers; Unit cubes

Activity

Start by introducing the idea of volume, explaining it as the space that an object occupies. Divide the students into groups and distribute unit cubes and one container to each group. Instruct the students to fill their containers by arranging the cubes in layers until the container is full. Ask them to count the total number of cubes that could fit inside the container and write the number in their notebook. Repeat the activity by asking groups to empty their containers and then exchange them with other groups.

Discuss how the total number of units cubes tell us the volume of the shape.

Teacher tip: If containers are unavailable, old shoe boxes can be used.

Extension Idea

Ask: Can there be two containers of different sizes that fit the same number of unit cubes and have the same volume?

Say: Yes, two or more containers can have the same volume.

Volume of Solids Using the Formula

Learning Outcomes

Students will be able to find the volume of solids using the formula.

Teaching Aids

Different–sized rectangular containers; Unit cubes

Activity

Instruct the students to form groups. Distribute unit cubes and one container to each group.

Instruct them to fill the bottom of the container with a layer of cubes, count the cubes along the length and along the width and write these measures in their notebooks. Then, ask them to stack cubes along one of the edges to find the number of cubes stacked along the height and write this measure in their notebooks. They will then fill the container completely with the unit cubes.

Ask them to write the total number of unit cubes used in their notebooks. Then, instruct them to multiply the number of cubes along the length, along the width and along the height.

Discuss whether the volumes are the same and deduce the formula V = L × B × H. Finally, give them two problems on finding the volume of a container with given length, width and height to solve in their notebooks.

Word Problems on Volume and Capacity

Learning Outcomes

Students will be able to solve word problems on finding capacity and volume.

Teaching Aids

Question cards with word problems written; Unit cubes

Activity

Instruct the students to work in groups. Provide question cards with this word problem written, “A rectangular container has dimensions of 5 cm by 15 cm by 10 cm. If the container is entirely filled with unit cubes, what is the total volume of the cubes used to completely fill the container?”

Discuss what is being asked, what they need to find and how they will find the answer. Ask the students to solve the problem in their notebooks. They need to put unit cubes along the length, width and height by stacking them to find out if the total number of cubes used is the same as the volume of the container calculated using the formula.

Add rigour to the learning by providing questions that involve conversions.

Extension Idea

Instruct: Create a word problem on finding the volume of a container.

Say: There can be multiple word problems on finding the volume of a container. One such question can be — “A room is completely filled with 3125 cubical cartons measuring 25 cm. What is the volume of the room?”

Answers

1. Estimating Capacity

Do It Together

2 Capacity of a watering can 3 Capacity of a bathtub

2. Measuring Capacity

Think and Tell

We look at the markings on the cup to know how much water each cup holds.

Do It Together

The jug contains 750 mL of liquid. Its capacity is 1000 mL

3. Units of Capacity

Do It Together

1 L = 100 cL

5 L 250 cL = 5 × 100 cL + 250 cL = 5250 cL

4. Volume of Solids Using Unit Cubes

Do It Together

Layer 1 (blue) has 12 unit cubes.

Layer 2 (yellow) has 12 unit cubes.

Total number of unit cubes = 24

Volume of the given solid = 24 cu. units

5. Volume of Solids Using the Formula

Do It Together

Number of unit cubes in one layer = 5 × 2 = 10

So, volume of the box = l × b × h = 5 × 2 × 3 = 30 cu. cm

6. Word Problems on Volume and Capacity

Think and Tell

To convert 20 L 126 mL into L, we write the litres and millilitres together and place a decimal point between them. In this case, it’s 20.126 L.

Similarly, for 13 L 679 mL, the conversion is achieved by writing the litres and millilitres together and placing a decimal point between them to get 13.679 L.

Do It Together

Number of cartons that are placed along the length = 20

Number of cartons that are placed along the breadth = 12

Number of cartons in 1 layer of the room = 20 × 12 = 240

Number of cartons in 10 such layers = 240 × 10 = 2400

So, the volume of the room = 2400 cu. m

3-D Shapes on Flat Surfaces

Learning Outcomes

Students will be able to: list the features of 3-D shapes. draw 3-D shapes on a square grid. identify and draw the net of 3-D shapes. draw views of cube structures. read a map and answer questions based on it. identify the deep drawing of the given floor plan and vice versa and read them.

Alignment to NCF

C-2.1: Identifies, compares, and analyses attributes of two- and three-dimensional shapes and develops vocabulary to describe their attributes/properties

C-2.2: Describes location and movement using both common language and mathematical vocabulary; understands the notion of map (najri naksha)

Let’s Recall

Recap to check if students know how to identify basic 3-D shapes around us. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

edges: the line segments that form the shape vertex: a point where three or more edges meet in a three-dimensional object net: a 2-D figure that can be folded to form a 3-D shape map: a picture of a place printed on a flat surface

Teaching Aids

3-D wooden shapes; Slips of paper with the names of solid shapes written on each; Bowl; Squared paper; Boxes shaped like a pyramid and a prism; Scissors; Glue stick; Sellotape; Thick chart paper; Unit cube blocks; Chart paper; Word cards labelled “school gate”, “admin office”, “principal’s office”, classrooms, playground, etc.; Cardboard sheets; Sketch pens

Chapter: 3-D Shapes on Flat Surfaces

Features of 3-D Shapes

Learning Outcomes

Students will be able to list the features of 3-D shapes.

Teaching Aids

3-D wooden shapes

Activity

Begin the class by asking the students to look around the classroom and identify different 3-D shapes they see. Have a discussion bout their observations.

Instruct the students to work in groups and distribute the 3-D wooden shapes among them.

Imagine Maths Page 222

Instruct the students to move their fingers around the sides to count the number of edges, move their hand around the surface to count the number of faces and point out the corners of the solid shapes. Help them understand that some shapes do not have flat faces but curved faces. Ask them to list the features of these shapes in their notebooks. Discuss the names of the different 3-D shapes that they have.

Extension Idea

Ask: Which two 3-D shapes have 5 faces?

Say: The two 3-D shapes that have 5 faces are a square pyramid and a triangular prism

Drawing 3-D Shapes

Learning Outcomes

Students will be able to draw 3-D shapes on a square grid.

Teaching Aids

Slips of paper with the names of solid shapes written on each; Bowl; Squared paper

Activity

Imagine Maths Page 224

Demonstrate to the students how 3-D shapes can be represented on squared paper. Make slips of paper naming different solid shapes (cone, cube, prism, sphere, cuboid, etc.) and put them inside a bowl. Distribute squared paper among the students. Call the students one by one to pick a slip from the bowl. Instruct them to open page 224 of the Imagine Mathematics book to see how to draw a 3-D shape on a square grid. Ask the students to draw their chosen object on the grid, adding details and colours to make it visually appealing. Give them a chance to show their drawing to the class

Nets of 3-D Shapes

Learning Outcomes

Students will be able to identify and draw the net of 3-D shapes.

Teaching Aids

Boxes shaped like a pyramid and a prism; Scissors; Sellotape; Thick chart paper

Activity

Show the students a pyramid-shaped paper box and ask them to identify the 2-D shapes they see in its faces and how many there are. Open the box to show them the faces in the net of the box. Ask the students how 3-D objects are different from their 2-D representations.

Instruct the students to work in pairs. Distribute boxes shaped like a pyramid and prism, a scissor, sellotape and thick chart paper among the pairs.

Ask the pairs to visualise what the nets of the prism and pyramid would look like and to draw these nets on the thick chart paper. Finally, instruct them to cut out their nets and fold them to see if the shapes match the given boxes.

Extension Idea

Ask: How many types of nets does a triangular prism have?

Say: A triangular prism has nine distinct nets.

Views of Cube Structures

Learning Outcomes

Students will be able to draw views of cube structures.

Teaching Aids

Unit cube blocks

Activity

Instruct the students to work in groups of 3.

Distribute the unit cube blocks among the groups.

Instruct the groups to work together and form cube structures using the unit cube blocks. Once they have formed their structure, ask one student to look at the structure from the front and draw its front view. Ask the second student to look at the structure from the top and draw the top view and the third student to look at it from the side and draw the side view. Ask them to review the drawing they made together. Encourage them to discuss whether the views are correct or not.

Instruct them to repeat the activity by forming two more structures.

Extension Idea

Instruct: The top view, front view and side view of a shape are given. Identify the shape.

Say: The given views are of a triangular pyramid.

Learning Outcomes

Students will be able to read a map and answer questions based on it.

Teaching Aids

Chart paper; Word cards labelled “school gate”, “admin office”, “principal’s office”, classrooms, playground, etc.

Activity

Divide the students into groups of 4. Distribute the teaching aids among the groups. Ask the groups to arrange the word cards on the chart paper as per the given instructions to create a map of the school. Be specific in the instructions, such as placing the school gate, admin office and school building adjacent to each other, with the admin office positioned at the centre. Specify distances between locations; for instance, the admin office and school gate are 10 cm apart, and the school building and the admin office are 5 cm apart. Let the groups display the maps and have a discussion with the rest of the class around that. Instruct each group to write 2 questions, and exchange them with other group. The receiving group should answer these questions by referring to the maps. Ask the students to note down their answers in their notebooks.

Extension Idea

Ask: How can we calculate the distance between two places using a map?

Say: We can calculate distances between two places on a map using a piece of thread and a ruler.

Floor Plans and Deep Drawings

Learning Outcomes

Maths Page 232

Students will be able to identify the deep drawing of the given floor plan and vice versa and read them.

Teaching Aids

Cardboard sheets; Glue stick; Sellotape; Scissors; Sketch pens

Activity

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

Draw a deep drawing, as shown, on the board. Instruct the groups to cut the cardboard and make a 3-D model by pasting the cardboard sections as per the deep drawing displayed on the board. Ask them to use sketch pens to draw the windows and doors on all sides of the model according to their preferences. Draw symbols for doors and windows that they need to use on the board. Instruct them to draw the floor plan of the model drawn.

Extension Idea

Ask: Can a floor plan be associated with multiple different deep drawings?

Say: A floor plan itself is not intended to contain multiple deep drawings in the sense of representing threedimensional information. However, it is possible to have multiple detailed drawings or views for various aspects of a building.

1. Features of 3-D Shapes

Do It Together

2. Drawing 3-D Shapes Do It Together

Answers

3. Nets of 3-D Shapes Do It Together

5. Reading Maps Do It Together

1. The city park is west of the highway.

2. The hotel is east of the highway.

3. The area of the city park is 12 cm2 on the grid.

4. The area of City Lake is 20 cm2 on the grid.

5. The area of the highway is 66 cm2 on the grid.

6. Floor Plans and Deep Drawings Do It Together

4. Views of Cube Structures Do It Together

There are four windows and one door in the floor map.

Yes, the deep drawing matches the floor plan exactly since the number of doors and windows are the same in both.

Time and Temperature

Learning Outcomes

Students will be able to: convert between the different units of time (hours, minutes, seconds). calculate the elapsed time between different units of time and solve word problems on time. solve problems on measuring and calculating temperature.

Alignment to NCF

C-3.2: Uses an appropriate unit and tool for the attribute (like length, perimeter, time, weight or volume) being measured

C-3.5: Devises strategies for estimating the distance, length, time, perimeter (for regular and irregular shapes), area (for regular and irregular shapes), weight and volume and verifies the same using standard units

Let’s Recall

Recap to check if students know how to estimate the time taken for various activities. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

thermometer: a tool to measure temperature mercury: a silver-white metal

Teaching Aids

Time conversion puzzle with square pieces that have units of measurement written on all 4 sides; Year calendar; School monthly calendar; Classroom thermometer; Analog thermometer; Separate bowls with hot water, ice and normal water; Sheets with thermometers drawn on them

Chapter: Time and Temperature

Converting Between Units of Time

Learning Outcomes

Students will be able to convert between the different units of time (hours, minutes, seconds).

Teaching Aids

Time conversion puzzle with square pieces that have units of measurement written on all 4 sides

Activity

Begin the class with a short discussion about the units of time (seconds, minutes and hours) and their relationships. Display the conversions on the board for help. (e.g., 1 minute = 60 seconds, 1 hour = 60 minutes).

Instruct the students to work in groups.

Distribute the time conversion puzzles among the groups.

Ask the students to place the card with the letter A in the centre. Instruct them to look at the times mentioned on the four sides of the card. Ask them to convert each time to both larger and smaller units and look for card that match either of the answers. Ask them to place the conversion cards next to each other to make a square puzzle as shown.

Extension Idea

Ask: How many hours are there in 7200 seconds?

Say: 7200 seconds = 7200  60 = 120 minutes and 120 minutes = 120  60 = 2 hours. Thus, there are 2 hours in 7200 seconds.

Calculating Time and Duration

Learning Outcomes

Students will be able to calculate the elapsed time between different units of time and solve word problems on time.

Teaching Aids

Year calendar; School monthly calendar

Activity

Instruct the students to look at the school calendar on the wall or in their school diaries. Discuss how to read the calendar.

Instruct them to work in groups.

Distribute the school’s monthly calendar to the groups. Instruct the students to now look at the calendar and say the duration of the summer and winter vacations in days and weeks.

Ask questions like, “How many days are there between Holi and Good Friday?” or “Raksha goes out to play Holi at 9:30 a.m. and comes home at 12:45 p.m. How much time does she spend playing Holi?”

Extension Idea

Imagine Maths Page 241

Month

Event

January 3rd—Winter vacation ends 15th—Makar Sankranti 26th—Republic Day

February 14th—Saraswati Puja

March 25th—Holi

April 7th—Good Friday 14th—Ambedkar Jayanti

May 22nd—Summer Vacation starts

June 17th—Summer Vacation ends

July 29th—Muharram

August 15th—Independence Day 30th—Raksha Bandhan

September 5th—Teacher’s Day

October 2nd Gandhi Jayanti 21st—Durga Puja starts 27th—Durga Puja ends

November 14th—Children’s Day

December 23rd—Winter Vacation begins 25th—Christmas

Instruct: Create your own word problem where the start time is 1:30 p.m. and the end time is 4:20 p.m.

Say: There can be multiple word problems. One such problem could be, “Richa started driving her car at 1:30 p.m. and reached her destination at 4:20 p.m. How long did she drive her car? ”

Measuring Temperature; Converting Between Imagine Maths Page 245 Units of Temperature

Learning Outcomes

Students will be able to solve problems on measuring and calculating temperature.

Teaching Aids

Classroom thermometer; Analog thermometer; Separate bowls with hot water, ice and normal water; Sheets with thermometers drawn on them

Activity

Begin with a brief discussion about temperature. Use real-life examples, like hot coffee versus cold ice cream, to illustrate the concept. Show the classroom thermometer to the students. Discuss with them how it works, explaining the temperature scale, the metal mercury inside it and the concept of degrees.

Divide the class into groups of three and provide each group with a small thermometer.

Place three bowls of water on a desk with hot water in the first, normal water in the second and cold water in the third bowl.

Ask one student from each group to come forward and measure the temperature of the hot water. Ask them to say the reading and then return to their groups to mark the reading of the hot water on the provided sheets using a pen or pencil.

Repeat the activity with the other two students in each group, who will measure the temperatures of the normal water and the cold water. Ask them to convert the temperature measured into Fahrenheit.

Ask questions like, “Did all the groups get the same temperature for all the 3 bowls? Why? ” or “ What is the relation between Celsius and Fahrenheit?”

Extension

Idea

Ask: Which month’s temperature would be higher—December or May? Will these months be hotter or colder? Say: May is the hotter month and December is the colder month.

Answers

1. Converting from a Bigger Unit into a Smaller Unit

Think and Tell

Number of hours in a day = 24 hours

Number of seconds in an hour = 60 × 60 = 3600 seconds

Number of seconds in a day or 24 hours = 24 × 3600 = 86,400 seconds

So, the number of seconds in a day = 86,400 seconds.

Do It Together

Time remaining before the eclipse = 4.5 hours.

1 hour = 60 minutes

4.5 hours = 4.5 × 60 = 270 minutes

So, there are 270 minutes remaining before the solar eclipse.

2. Converting from a Smaller Unit into a Bigger Unit

Do It Together

Number of seconds for which Shreya juggles = 250 seconds

We know that 60 seconds = 1 minute

250 seconds = 1 250 60 ´ = 250 60

Thus, Shreya juggled the tennis balls for 4 minutes and 10 seconds.

3. Calculating Time and Duration

Do It Together

Starting time: 2 hours 30 minutes

Time for which the experiment lasted = 3 hours 45 minutes

We need to find the time at which the experiment ended, hence, we will add the data.

4. Measuring Temperature

Think and Tell

Answers may vary.

Do It Together

45°C, 22°C, 18°C, 30°C

5. Converting Between Units of Temperature

Do It Together

Initial temperature = 12°C

Increase in temperature = 8°C

New temperature = 12°C + 8°C = 20°C

Converting 20°C into °F:

× 9 5

20 × 9 5

+ 32 = °F.

+ 32 = 36°F.

So, 20°C is 68°F.

Money

Learning Outcomes

Students will be able to: solve problems on unitary method. solve problems on profit and loss.

Alignment to NCF

C-8.11: Performs simple transactions using money up to INR 100

C-8.13: Formulates and solves simple mathematical problems related to quantities, shapes, space, and measurements

Let’s Recall

Recap to check if students know the units and conversion of units of money. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

amount: the total number or quantity earn: to receive money as payment spend: to pay out money

Teaching Aids

Books; Play money; Number cards (1–9); Items to set up a stall (books, crayons, pencil boxes, etc.)

Chapter: Money

Unitary Method

Learning Outcomes

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

Teaching Aids

Books; Play money; Number cards (1–9)

Activity

Divide the students into groups of four. Distribute the play money to each group. Ask one student from each group to act as the shopkeeper and the other three to act as customers. Tell the students that the shopkeeper has 10 books which cost ₹2000.

Distribute number cards within the groups.

Instruct the students acting as customers to pick any number card which tells them how many books they will purchase from the shopkeeper. Ask them to first find the cost of 1 book, then calculate the total amount of money to be paid to buy the books and write their answers in their notebooks.

Discuss with the students how we divide to find the cost of 1 and multiply to find the cost of many.

Extension Idea

Ask: If the cost of 2 1 2 L of milk is ₹145, what will be the cost of

or Loss

Learning Outcomes

Students will be able to solve problems on profit and loss.

Teaching Aids

Play money; Items to set up a stall (books, crayons, pencil boxes, etc.)

Activity

Discuss how when a book bought for ₹50 is sold for ₹70, a profit is made. However, if the book had been sold for less than ₹50, a loss would have been incurred.

Set up a play stall in the classroom. Distribute play money to each student. Instruct the students to work in groups. One student from each group will manage the stall as the shopkeeper and will keep the crayons, books and pencil box. The student will set a cost price for each item. The other students in the group will shop at their group’s stall for 10 minutes. Encourage them to negotiate prices to get a better deal. For example, a student would want to buy an item for ₹10 that has a cost price of ₹15. Once the purchase is done, each group will figure out if they made a profit or incurred a loss based on their initial cost prices and final selling prices.

Extension Idea

Ask: A sandwich toaster that cost ₹1200 is sold for ₹2000. What is the selling price of a second toaster that cost ₹1600 if there is no profit or loss on both toasters?

Say: The profit made on the first toaster is ₹2000 – ₹1200 = ₹800. Since there was no overall profit or loss on both toasters, the selling price of the second toaster will be ₹1600 – ₹800 = ₹800.

Answers

1. Unitary Method

Think and Tell

Coins: ₹1, ₹2, ₹5, ₹10, ₹20

Banknotes: ₹10, ₹20, ₹50, ₹100, ₹200, ₹500

Do It Together

Cost of Multiple Items Cost of One Item Cost of Given Items

Cost of 3 pens = ₹60

Cost of 8 oranges = ₹80

Cost of 4 notebooks = ₹200

2. Profit or Loss

Do It Together

Cost of one pen = ₹20

Cost of one orange = ₹10

Cost of one notebook = ₹50

Cost of 5 pens = ₹100

Cost of 12 oranges = ₹120

Cost of 3 notebooks = ₹150

3. Word Problems on Money

Do It Together

1. One chocolate

CP of each chocolate = Total cost ÷ Number of units = ₹200 ÷ 20 = ₹10

SP of each chocolate = ₹20

Since SP > CP, Vansh gained a profit.

Profit = SP – CP = ₹20 – ₹10 = ₹10

Profit gained on one chocolate is ₹10.

2. All chocolates

Total CP = ₹200

Total SP = 20 × ₹20 = ₹400

Since SP > CP, Vansh gained a profit.

Total Profit = Total SP – Total CP = ₹400 – ₹200

Thus, the total profit gained is ₹200

Data Handling

Learning Outcomes

Students will be able to:

read and interpret single and double bar graphs. represent given data on a horizontal or vertical bar graph. read and interpret a pie chart. represent given data on a pie chart. read a line graph and answer questions based on it.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to read a tally marks table. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

bar graph: a graph that shows information in the form of bars of different lengths pie chart: a type of graph in which a circle is divided into sectors, each representing a part of the whole line graphs: a graph that shows the relation between two quantities in the form of a line

Teaching Aids

Sheets with a double bar graph showing data on the favourite activities of boys and girls; Chart paper; Glue sticks; Square cutouts of different colours; Pie chart sector cutouts showing 1 6 as grapes, 1 12 as kiwi, 1 12 as pomegranate, 7 24 as apple and 9 24 as banana and some extra cutouts showing different fractions; Circular chart paper with a radius of 12 cm drawn; Protractor; Coloured pencils; Sheet with a line graph showing the temperature for 6 months in a city

Chapter: Data Handling

Interpreting Bar Graphs Imagine Maths Page 265

Learning Outcomes

Students will be able to read and interpret single and double bar graphs.

Teaching Aids

Sheets with a double bar graph showing data on the favourite activities of boys and girls

Activity

Begin with a brief discussion on what bar graphs are and how they represent data. Explain that a double bar graph compares two sets of data.

Instruct the students to work in groups. Distribute sheets with double bar graphs among the groups. Instruct them to look at the bar graph. Ask questions like, “How many boys participated in tennis?”; “How many more boys participated in cricket than girls? “Ask the groups to first answer orally, then write their answers in their notebooks.

Drawing Bar Graphs

Learning Outcomes

Students will be able to represent given data on a horizontal or vertical bar graph.

Teaching Aids

Chart paper; Glue sticks; Square cutouts of different colours

Activity

Imagine Maths Page 267

Begin with a brief discussion on vertical and horizontal bar graphs. Draw the table as shown on the board.

Instruct the students to form groups of four.

Distribute 2 sheets of chart paper, square cutouts of different colours and a glue stick to each group.

Ask two students from each group to draw a vertical bar graph, and the other two to draw a horizontal bar graph using a scale of 1 unit = 5 families. Instruct them to first draw the axis and show the scale. They will then stick the coloured cutouts to show the data, where each bar can be shown as a tower of square cutouts of the same colour.

Extension Idea

Ask: How many cutouts would you stick to show the families who own a cat if the scale changes from 5 to 10?

Say: As there are 40 families; and scale = 1 unit = 10 families, we will use 4 cutouts.

Interpreting Pie Charts

Learning Outcomes

Students will be able to read and interpret a pie chart.

Teaching Aids

Pie chart sector cutouts showing 1 6 as grapes, 1 12 as kiwi, 1 12 as pomegranate, 7 24 as apple and 9 24 as banana and some extra cutouts showing different fractions

Activity

Begin with a brief discussion on what a pie chart is and how it represents data. Explain the concept of sectors or slices in a pie chart and how they represent parts of a whole. Recall the multiplication of fractions.

Instruct the students to work in groups.

Cut the pie chart into sectors and then distribute the sectors to each group.

Instruct the students that they need to join the sectors to form a pie chart. Discuss that the given pie chart shows the preferred fruit of some students.

Ask them a few questions like, “Which is the most preferred fruit?”, “How many students would prefer bananas if 120 students were surveyed?” or “How many students prefer kiwi and pomegranate?”

Ask them to solve and write the answers in their notebooks.

Extension Idea

Ask: Can you frame one question around the same pie chart that has not been discussed in class?

Say: There can be many such questions. One possible question could be, “What fraction of the students did not choose apples?”

Representing Data on a Pie Chart

Learning Outcomes

Students will be able to represent given data on a pie chart.

Teaching Aids

Circular chart paper with a radius of 12 cm drawn; Protractor; Coloured pencils

Activity

Imagine Maths Page 270

Begin with a brief review of what a pie chart is. Discuss the fact that the total angle made at the centre of the circle is 360°.

Instruct the students to work in groups. Distribute the teaching aids among the groups. Draw a table on the board showing data on the books issued by a library: Fantasy—15, Fiction—30, Non-fiction—20, Adventure—25.

Instruct the students to count the total number of books issued and find the fraction of books for each category. Then, have them calculate the angle of each category by multiplying each fraction by 360°.

Then, instruct the students to draw the sectors for each category on the circular chart paper to show the pie chart. Once they have drawn all the angles, ask them to colour and label the sectors.

Learning Outcomes

Students will be able to read a line graph and answer questions based on it.

Teaching Aids

Sheet with a line graph showing the temperature for 6 months in a city

Activity

Begin by discussing what a line graph is and how it represents data.

Show the students how to read a line graph by looking at the data on the x-axis and reading its value on the y-axis.

Instruct the students to work in groups and distribute the line graphs among them.

Ask them to read the graph carefully.

Ask them a few questions like, “Which is the hottest month?” or “What is the difference in the temperature of May and February?”

Discuss the answers in the class.

Answers

1. Interpreting Bar Graphs

Think and Tell

No, drawing bars horizontally does not make any difference to the data.

Do It Together

2. 50 + 45 = 95 runs

3. 40 + 35 + 30 + 70 + 60 = 235 runs

2. Drawing Bar Graphs

Do It Together

4. Representing Data on a Pie Chart

Do It Together

3. Interpreting Pie Charts

Do It Together

1. What is the largest expense of the family?

On comparing the fractions, 231 51010 >>

Hence, rent is the largest expense in the family.

2. If the family decides to cut their entertainment expenses by half, what fraction will they be spending on entertainment?

11111 of× 21021020 ==

Art and Crafts, 40

5. Reading Line Graphs

Do It Together

2. The highest production of food grain was in the year 1997

3. The difference in production between the year 1995 and the year 1993 was 70 − 40 = 30 tonnes

4. The total production across the years = 20 + 40 + 50 + 70 + 60 + 100 = 340 tonnes

Answers Solutions

Chapter 1 Let’s Warm–up

1. The place value of 8 in 8,60,765 is 8,00,000.

2. 4,36,536 can be written in words as: four lakh thirty-six thousand five hundred thirty-six

3. The place value of the digit 4 in 4,15,124 and 4,67,890 is the same.

4. 8,76,504 has 6 in the thousands place.

Do It Yourself 1A

1. a. 7409230: Place value: 9000 Face value: 9

b. 8656023: Place value: 50,000 Face value: 5

c. 93260075: Place value: 9,00,00,000 Face value: 9

d. 29543002: Place value: 2,00,00,000 Face value: 2

e. 61752812: Place value: 6,00,00,000 Face value: 6

f. 10081824: Place value: 800 Face value: 8

g. 43889385: Place value: 30,00,000 Face value: 3

h. 57131060: Place value: 1,00,000 Face value: 1

2. a. Indian number system: 2,16,43,332

Number name: Two crore sixteen lakh forty-three thousand three hundred thirty-two International number system: 21,643,332

Number name: Twenty-one million six hundred forty-three thousand three hundred thirty-two

b. Indian number system: 12,00,621

Number name: Twelve lakh six hundred twenty-one International number system: 1,200,621

Number name: One million two hundred thousand six hundred twenty-one

c. Indian number system: 4,62,07,219

Number name: Four crore sixty-two lakh seven thousand two hundred nineteen

International number system: 46,207,219

Number name: Forty-six million two hundred seven thousand two hundred nineteen

d. Indian number system: 9,59,10,158

Number name: Nine crore fifty-nine lakh ten thousand one hundred fifty-eight

International number system: 95,910,158

Number name: Ninety-five million nine hundred ten thousand one hundred fifty-eight

3. a. 4,41,90,887

Number name: Four crore forty-one lakh ninety thousand eight hundred eighty-seven

Expanded form: 4,00,00,000 + 40,00,000 + 1,00,000 + 90,000 + 800 + 80 + 7

b. 1,90,81,702

Number name: One crore ninety lakh eighty-one thousand seven hundred two

Expanded form: 1,00,00,000 + 90,00,000 + 80,000 + 1000 + 700 + 2

c. 81,085,432

Number name: Eighty-one million eighty-five thousand four hundred thirty-two

Expanded form: 80,000,000 + 1,000,000 + 80,000 + 5000 + 400 + 30 + 2

d. 19,854,004

Number name: Nineteen million eight hundred fifty-four thousand four

Expanded form: 10,000,000 + 9,000,000 + 800,000 + 50,000 + 4000 + 4

4. a. Sixty lakh eight thousand ninety-eight: 60,08,098

b. Twenty million five hundred sixty-nine: 20,000,569

c. Four million ninety thousand: 4,090,000

d. Eight crore one thousand two: 8,00,01,002

5. a. 10 million = 1 crore

b. 1 million = 10 lakh

c. 1 crore= 10,000 thousands

d. There are 7 zeros in 20 million

6. Area covered by the Amazon rainforest = 2722000 square miles

International number system = 2,722,000; Indian number system = 27,22,000

7. a. We need to find the greatest 7-digit number that has the smallest odd digit at its hundreds, ten thousands and lakhs place.

Greatest 7-digit number: 91,19,199

b. We need to find the smallest 8-digit number that has the digit 7 at all its odd positions, starting from the ones place. Smallest 8-digit number: 170,70,707

8. Population of India at the time of its independence = 353 million 353 million as a number can be written as 353,000,000

Challenge

1. The number of people who visited the Kumbh Mela in 1980 = 20,356,817

The number of people who visited the Kumbh Mela in 1989 = 29,304,871

To find the year when less than 25 crore= 250,000,000 people visit the Kumbh Mela we will compare the number of visitors in 1980 and 1989 with 250,000,000

20,356,817 has 8 digits and 250,000,000 has 9 digits, 20,356,817 < 250,000,000

29,304,871 has 8 digits and 250,000,000 has 9 digits, 29,304,871 < 250,000,000

Both numbers are less than 250,000,000

Hence, option c: Both years

Do It Yourself 1B

1. To compare the two numbers, we need to start from the leftmost digit and compare the digits at each place value.

76,24,578

87,90,213

In this case, the leftmost digit is 8 in the second number and 7 in the first number. Since 7 is less than 8, the number 76,24,578 is smaller than 87,90,213.

2. a. 35,72,123 < 35,78,123

b. 63,45,789 = 63,45,789

c. 2,86,73,451 > 57,81,290

d. 6,24,58,110 < 6,24,59,211

e. 82,60,154 < 89,12,620

f. 84,63,758 < 7,65,38,453

3. a. 99,00,000 10,00,000

b. 2,89,52,468 3,00,52,468

c. 9,00,000 1,00,00,000

d. 2,99,52,468 2,90,52,468

4. a. Ascending order: 93,12,820 < 1,00,36,782 < 5,00,00,367 < 8,87,21,460

Descending order: 8,87,21,460 > 5,00,00,367 > 1,00,36,782 > 93,12,820

b. Ascending order: 36,81,910 < 92,56,890 < 6,92,10,350 < 8,26,00,031

Descending order: > 8,26,00,031 > 6,92,10,350 > 92,56,890 > 36,81,910

c. Ascending order: 5,00,21,138 < 6,04,50,821 < 6,50,24,567 < 9,45,21,823

Descending order: 9,45,21,823 > 6,50,24,567 > 6,04,50,821 > 5,00,21,138

5. a. Greatest 7-digit number: 98,54,310

Smallest 7-digit number: 10,34,589

b. Greatest 7-digit number: 87,65,321

Smallest 7-digit number: 12,35,678

c. Greatest 7-digit number: 65,43,210

Smallest 7-digit number: 10,23,456

6. a. Greatest 8-digit number: 8,87,64,210

Smallest 8-digit number: 1,00,24,678

b. Greatest 8-digit number: 9,98,76,431

Smallest 8-digit number: 1,13,46,789

c. Greatest 8-digit number: 9,97,54,320

Smallest 8-digit number: 2,00,34,579

7. 5,48,79,802:

Greatest 8-digit number: 9,88,75,420

Smallest 8-digit number: 2,04,57,889

8. a. By using two different digits

Greatest 8-digit number: 9,99,99,998

Smallest 7-digit number: 10,00,000

b. By using four different digits

Greatest 8- digit number: 9,99,99,876

Smallest 7-digit number: 10,00,023

c. By using five different digits

Greatest 8-digit number: 9,99,98,765

Smallest 7-digit number: 10,00,234

9. Descending order: Russia-17,098,242 > China-9,706,961 > Australia-7,692,024 > India-3,287,590

10. Answer may vary. Sample answer: The government of a country has allocated its national budget for the upcoming fiscal year. The budget for the education sector was ₹45,678,912 and for healthcare was ₹45,345,678. Which sector had a greater budget?

Challenge

1. Given digits—5,0,1,3,7,9,6,2 and no digit was repeated.

Greatest 8-digit number: 9,76,53,210

Smallest 8-digit number: 1,02,35,679

Greatest 7-digit number: 97,65,321

Smallest 7-digit number: 10,23,567

Descending order: 9,76,53,210 > 1,02,35,679 > 97,65,321 > 10,23,567

2. By using digit 0,1,2,3,4,5,6,7,8,9 only once, the possible numbers which are closest to 40,00,00 is 40,12,356 and 39,87,654.

So, 40,12,356 – 40,00,00 = 12,356 and 40,00,00 – 39,87,654 = 12,346. Hence 39,87,654 is closest to 40,00,000.

Do It Yourself 1C

1. a. To round off 85,48,749 to the nearest tens, look at the ones digit.

Here, 9 > 5, so, 85,48,749 is rounded to 85,48,750.

b. To round off 89,05,462 to the nearest tens, look at the ones digit.

Here, 2 < 5, so, 89,05,462 is rounded to 89,05,460.

c. To round off 6,07,85,888 to the nearest tens, look at the ones digit.

Here, 8 > 5, so, 6,07,85,888 is rounded off to 6,07,85,890.

d. To rounded off to 1,56,48,950 to the nearest tens, look at the ones digit.

Here, 0 < 5, so, 1,56,48,950 is rounded to 1,56,48,950.

2. a. To round off to the nearest hundreds, look at the tens digit.

Here, 8 > 5, so, 1,25,89,183 rounded off to 1,25,89,200.

b. To round off to the nearest hundreds, look at the tens digit.

Here 6 > 5 so, 87,52,368 rounded off to 87,52,400.

c. To round off to the nearest hundreds, look at the tens digit.

Here 9 > 5 so, 68,67,790 rounded off to 68,67,800.

d. To round off to the nearest hundreds, look at the tens digit.

Here 1 < 5 so, 77,59,910 rounded off to 77,59,900.

3. a. To round off to the nearest thousands, look at the hundreds digit.

Here, 1 < 5, so, 8,97,00,110 is rounded off to 8,97,00,000.

b. To round off to the nearest thousands, look at the hundreds digit.

Here, 0 < 5, so, 53,12,069 is rounded of to 53,12,000.

c. To round off to the nearest thousands, look at the hundreds digit.

Here, 7 > 5, so, 8,21,58,701 is rounded off to 8,21,59,000.

d. To round off to the nearest thousands, look at the hundreds digit.

Here, 9 > 5, so, 5,89,89,929 is rounded off to 5,89,90,000.

4. Distance between the Earth and the moon = 238,855 miles.

To round off 238,855 miles to the nearest thousands, look at the hundreds digit.

Here, 8 > 5, so, 238,855 miles rounded off to 239,000 miles.

5. Money spent by the municipal corporation on the project = ₹65,94,830

To round off ₹65,94,830 to the nearest thousands, look at the hundreds digit.

Here, 8 > 5, so, ₹65,94,830 is rounding off to ₹65,95,000.

6. a. Greatest 7-digit number 99,63,210

Rounding off 96,63,210 to the nearest thousand gives, 99,63,000.

b. Greatest 7-digit number 96,32,100 Rounding off 99,63,210 to the nearest thousand gives 99,32,000.

Challenge

1. Smallest possible number—3,23,45,500 Biggest possible number—3,23,46,499

Chapter Checkup

1. a. Indian number system: 35,07,681—Thirty-five lakh seven thousand six hundred eighty-one; 30,00,000 + 5,00,000 + 7000 + 600 + 80 + 1

International number system: 3,507,681—Three million five hundred seven thousand six hundred eighty-one; 3,000,000 + 500,000 + 7000 + 600 + 80 + 1

b. Indian number system: 4,20,87,550—Four crore twenty lakh eighty-seven thousand nine hundred fifty; 4,00,00,000 + 20,00,000 + 80,000 + 7000 + 500 + 50

International number system: 42,087,550—Forty-two million eighty-seven thousand nine hundred fifty; 40,000,000 + 2,000,000 + 80,000 + 7000 + 500 + 50

c. Indian number system: 6,35,65,842—Six crore thirtyfive lakh sixty-five thousand eight hundred forty-two; 6,00,00,000 + 30,00,000 + 5,00,000 + 60,000 + 5000 + 800 + 40 + 2

International number system: 63,565,842—Sixty-three million five hundred sixty-five thousand eight hundred forty-two; 60,000,000 + 3,000,000 + 500,000 + 60,000 + 5000 + 800 + 40 + 2

d. Indian number system: 5,15,00,084—Nine crore fifteen lakh eighty-four; 5,00,00,000 + 10,00,000 + 5,00,000 + 80 + 4

International number system: 51,500,084—Ninety one million five hundred thousand eighty-four; 50,000,000 + 1,000,000 + 500,000 + 80 + 4

2. a. Sixty million seven hundred fifteen thousand two hundred thirty-nine Indian number system: 60,715,239

International number system: 6,07,15,239

b. Eight crore nine lakh fifty thousand two Indian number system: 8,09,50,002

International number system: 80,950,002

c. One million one hundred thousand thirty-nine Indian number system: 11,00,039

International number system: 1,100,039

3. a. Rounding off to the nearest tens: 6,45,87,120

Rounding off to the nearest hundreds: 6,45, 87,100

Rounding off to the nearest thousands: 6,45,87,000

b. Rounding off to the nearest tens: 89,09,010

Rounding off to the nearest hundreds: 89,09,000

Rounding off to the nearest thousands: 89,09,000

4. a. The digit in the crores place in 6,56,52,567 and 6,48,90,650 are the same, so, check the digit at the ten-lakh place. So, 5 > 4

6,56,52,567 > 6,48,90,650

b. The digit in the ten lakhs and the lakhs place in 90,00,518 and 90,76,757 are the same, so, check the digit at the ten thousand place. So, 0 < 7 90,00,518 < 90,76,757

c. All the digits are same. So, 34,57,879 = 34,57,879

d. 7-digit numbers are smaller than 8-digit numbers. So, 13,05,885 < 6,74,38,989

5. a. 7-digit numbers are smaller than 8-digit numbers. Check the digit at the ten lakhs place in the 7-digit numbers.

2 < 9

So, 23,56,475 < 90,87,687

Check the digit at the crores place in the 8-digit numbers.

8 < 9

So, 8,91,63,896 < 9,08,04,365

Thus, 23,56,475 < 90,87,687 < 8,91,63,896 < 9,08,04,365

b. Check the digits at the crores place.

3 < 4 < 6

The digit in the ten lakhs and the lakhs place in 6,76,12,895 and 6,76,87,980 are the same, so, check the digit at the ten thousand place.

1 < 8

So, 6,76,12,895 < 6,76,87,980

Thus, 3,24,35,678 < 4,35,46,576 < 6,76,12,895 < 6,76,87,980

6. a. Check the digits at the crores place 5 < 4 < 2

8-digit numbers are smaller than 7-digit numbers

Thus, 5,36,45,787 > 4,56,45,768 > 2,40,85,167 > 43,56,787

b. Check the digits at the crores place 9 > 4

8-digit numbers are smaller than 7-digit numbers

The digit in the ten lakhs and the lakhs place in 80,88,428 and 80,68,964 are the same, so, check the digit at the ten thousand places. 8 > 6

So, 80,88,428 > 80,68,964

Thus, 9,09,87,897 > 4,90,76,837 > 80,88,428 > 80,68,964

7. List of numbers that can be rounded off to the nearest tens as 16,48,240 are:

16,48,235; 16,48,236; 16,48,237; 16,48,238; 16,48,239; 16,48,240; 16,48,241; 16,48,242; 16,48,243; 16,48,244

8. Number of COVID-19 vaccines supplied by India to UK under the Vaccine Maitri initiative = 3,151,324

Expanded form of 31,51,324 in the Indian number system: 30,00,000 + 1,00,000 + 50,000 + 1,000 + 300 + 20 + 4

Expanded form of 3,151,324 in international number system: 3,000,000 + 100,000 + 50,000 + 1,000 + 300 + 20 + 4

9. 8-digit number which has only fives in ones period, only sevens in the thousands period, only nines in the lakhs period and only ones in the crores period is 19977555

a. Indian number system—1,99,77,555

One crore ninety-nine lakh seventy-seven thousand five hundred fifty-five International number system—19,977,555

Nineteen million nine hundred seventy-seven thousand five hundred fifty-five

b. Rounding off to the nearest tens: 1,99,77,560

Rounding off to the nearest hundreds: 1,99,77,600

Rounding off to the nearest thousands: 1,99,78,000

Challenge

1. a. The digit in the hundreds and ones place is 6: 6 6

b. The digit in the lakhs place is 4 less than the digit in the ones place: 2, , 6 6

c. The digit in the ten lakhs and ten thousands place is the smallest odd number: 12, 1 , 6 6

d. The face value of the digit in the thousands place is 5: 12, 15, 6 6

e. The digit in the tens place is the biggest 1-digit number: 12,15,696 Hence, the secret code is 12,15,696

2. The greatest 8-digit odd number by using only 5 digits not more than twice can be 99887765.

Case Study

1. Let us arrange the population of the different countries in ascending order to find which country has the least population.

Ascending order: Poland 41,026,067 < Italy 58,870,762 < France 64,756,584 < UK 67,736,802 < Germany 83,294,633 So, Poland has the least population. Hence, option c) Poland.

2. Let us arrange the population of the different countries in descending order to find which country has the greatest population.

Descending order: 83,294,633 > 67,736,802 > 64,756,584 > 58,870,762 > 41,026,067 So, Germany has the greatest population. Hence, option d) Germany.

3. The country that has approximately double the population than that of Poland = 41,026,067. Here, the country Germany population = 83,294,633 Thus, Germany

4. Ascending order: Poland 41,026,067 < Italy 58,870,762 < France 64,756,584 < UK 67,736,802 < Germany 83,294,633

5. Greatest number by rearranging Germany population is—986,43,332

Greatest number by rearranging France population—87,665,544

Greatest number by rearranging United Kingdom population—87,766,320

Greatest number by rearranging Poland population—76,642,100

Greatest number by rearranging Poland population—88,776,520

Therefore, Germany will have the greatest population.

Chapter 2

Let’s Warm-up

1. 180 ÷ 2

1. a. When the number is subtracted from itself, the difference is zero, hence, the statement is true

b. When 0 is subtracted from a number, the difference is not zero, hence, the statement is false

c. When the order of the addends is changed, the sum remains the same, hence, the statement is true.

d. The order of numbers used in subtraction cannot be changed, hence, the statement is false

2. a. TTh Th H T O 1

b. L TTh Th H T O

1 2 1 2 1

3 4 4 5 6 7

7 8 4 5 6 + 3 9 8 9 4

4 6 2 9 1 7

c. L TTh Th H T O

1 2 2 2 2

5 8 9 5 6 9

1 2 4 8 8 7 + 5 6 7 5 8

7 7 1 2 1 4

d. L TTh Th H T O

2 2 2 2 2

3 8 6 5 6 5

2 3 4 5 6 7 + 5 6 4 6 8 4 6 5 6 8 7 2 4 1 6 8

5. a. L TTh Th H T O 15 17 6 5 7 7 8 7 6 8 7 5 7 8 9 8 1 8 9 8

b. L TTh Th H T O 9 8 7 6 0 9 5 6 0 0 0 9 3 1 6 0 9

c. L TTh Th H T O 8 17

5 6 8 9 7 8

3 2 1 0 9 8

2 4 7 8 8 0

d. L TTh Th H T O 8 10

7 9 9 0 9 8

2 6 7 5 4 8

5 3 1 5 5 0

6. The greater number would be the sum of 505090 and 398460. L TTh Th H T O 1 1 1

5 0 5 0 9 0 + 3 9 8 4 6 0

9 0 3 5 5 0

Hence, the greater number will be 9,03,550.

7. Number of blazers manufactured in the year 2020 = 59,899

Number of blazers manufactured in the year 2021 = 78,906

Number of blazers manufactured in the year 2022 = 1,34,145

Total number of blazers manufactured in three years = 1,34,145 + 78,906 + 59,899

L TTh Th H T O 1 2 1 1 2 1 3 4 1 4 5 7 8 9 0 6 + 5 9 8 9 9

2 7 2 9 5 0

Hence, 2,72,950 blazers were manufactured in three years.

8. Jupiter’s diameter = 142,984 kilometres

Saturn’s diameter = 120,536 kilometres

Difference in diameter = 142,984 120,536 = 22,448 kilometres

L TTh Th H T O

9. Amount collected by NGO in the first year = ₹2,89,230

Amount collected by NGO in another year = ₹3,97,500

Total amount collected = ₹2,89,230 + ₹3,97,500 = ₹6,86,730

Amount spent by NGO = ₹3,05,700

Amount left with NGO = ₹6,86,730 ₹3,05,700

NGO is left with ₹3,81,030.

L TTh Th H T O

Challenge

1. As the numbers are giving a 6-digit number, the minimum sum will be the smallest 6-digit number = 1,00,000

Given that one of the numbers is 52,135.

The least value of the second number will be:

1,00,000 52,135 = 47,865

As the second number is also a 5-digit number, the maximum value of the number will be the greatest 5-digit number = 99,999

Hence, the range of the second number = 47,865 to 99,999

Do It Yourself 2B

1. a. The divisor and remainder in a sum can never be the same, hence, the statement is false

b. The result of multiplication is called the product; hence, the statement is false.

c. When a number is divided by another number, the result is called the quotient, hence, the statement is false

d. Anything multiplied by zero is zero, hence, the statement is true

2. a. 56,567 × 10 = 5,65,670; 56,567 × 100 = 56,56,700; 56,567 × 1000 = 5,65,67,000

b. 47,852 × 10 = 4,78,520; 47,852 × 100 = 47,85,200; 47,852 × 1000 = 4,78,52,000

c. 82,587 × 10 = 8,25,870; 82,587 × 100 = 82,58,700; 82,587 × 1000 = 8,25,87,000

d. 1,98,454 × 10 = 19,84,540; 1,98,454 × 100 = 1,98,45,400; 1,98,454 × 1000 = 19,84,54,000

3. a. 25,000 ÷ 10 = 2500; 25,000 ÷ 100 = 250; 25,000 ÷ 1000 = 25

b. 3,54,000÷ 10 = 35,400; 3,54,000 ÷ 100 = 3540; 3,54,000 ÷ 1000 = 354

c. 8,95,000 ÷ 10 = 89,500; 8,95,000 ÷ 100 = 8950; 8,95,000 ÷ 1000 = 895

d. 9,87,000 ÷ 10 = 98,700; 9,87,000 ÷ 100 = 9870; 9,87,000 ÷ 1000 = 987

TTh

L TTh Th H T O

8 4 6 × 5 6 7 8

2 7 6 8 5 4 9 2 2 0

7 2 8 6 2 8 8

L TTh Th H T O

7 8 4 × 7 8 9 0

0 0 0 0

1 0 5 6 0 5 4 2 7 2 0 0 + 4 7 4 8 8 0 0 0 5 3 5 2 5 7 6 0

5 4 3 6 2 8 4 1 7 0 3 1 4 2 0 2 8 3 6

6. Number of days in a leap year = 366

Number of hours in one day = 24 hours

Number of hours in a leap year = 366 × 24 hours = 8784 hours

7. Largest 3-digit number = 999

Smallest 2-digit number = 10

Product of largest 3-digit number and smallest 2-digit number = 999 × 10 = 9990

8. Largest 5-digit number = 99,999

Smallest 3-digit number = 100

Largest 5-digit number divided by smallest 3-digit number = 99,999 ÷ 100

Quotient = 999 and Remainder = 99

9. Per month earning of Mrs Gupta = ₹78,562

Time for which she earned = 3 year = 3 × 12 months = 36 months

Total amount earned by Mrs Gupta in three years = ₹78,562 × 36

TTh Th H T O

Hence, Mrs Gupta earned ₹28,28,232 in three years.

10. Number of words in the book = 24,645

Number of words on each page = 155

Number of pages in the book = 24,645 ÷ 155 = 159

Thus, there are 159 pages in the book.

11. Profit of the owner of the shop = ₹98,000

Money kept for himself = ₹30,000

Money left for stationary kits = ₹98,000 ₹30,000 = ₹68,000

Number of children = 100

Cost of each stationary kit = ₹68,000 ÷ 100 = ₹680

Challenge

1. Number of chairs in each row = 240

Number of rows = 380

Total number of chairs for these rows = 240 × 380 = 91,200

+ 0 0 0 0

+ 1 9 2 0

+ 7 2 0

= 9 1 2 0 0

Number of additional rows = 150

Number of chairs in each additional row = 320

Total number of chairs for additional rows = 320 × 150 = 48,000

Total number of chairs required = 91,200 + 48,000 = 1,39,200

L TTh Th H T O 3 2 0 × 1 5 0

Do It Yourself 2C

1. a. The order of DMAS is ÷, ×, +, Hence, the correct option is (ii).

b. 100 × 10 100 + 2000 ÷ 100 = 1000 100 + 20 = 1020 100 = 920

Hence, the correct option is (ii).

c. 100 ÷ 10 + 10 × 10 = 10 + 100 = 110

Hence, the correct option is (ii).

d. 63 ÷ 9 + 12 × 4 5 = 7 + 48 5 = 55 5 = 50

Hence, the correct option is (ii).

2. a. In DMAS, we first perform division/multiplication and then addition/subtraction, hence, the statement is false

b. In DMAS, the last step is subtraction, hence, the statement is true

c. 5 × 4 + 12 ÷ 3 = 24

⇒ 20 + 4 = 24

Hence, the statement is true.

d. 36 ÷ 6 3 × 2 = 2

⇒ 6 6 = 0 ≠ 2

Hence, the statement is false

3. a. 82 × 3 + 20

= 246 + 20 = 266

b. 28 ÷ 7 + 8 × 5

= 4 + 40 = 44

c. 55 ÷ 11 + 7

= 5 + 7 = 12

d. 20 + 50 ÷ 2 5

= 20 + 25 5

= 45 5 = 40

e. 16 + 8 ÷ 2 1 × 5

= 16 + 4 5

= 20 5 = 15

f. 25 4 × 12 ÷ 4 + 3

= 25 4 × 3 + 3 = 25 12 + 3

= 28 12 = 16

4. a. 120 12 + 36 ÷ 3 × 5

= 120 12 + 12 × 5

= 120 12 + 60

= 180 12 = 168

b. 279 + 321 ÷ 3 57 × 2

= 279 + 107 114

= 386 114 = 272

c. 676 + 835 × 45 ÷ 15 10

= 676 + 835 × 3 10

= 676 + 2505 10 = 3181 10 = 3171

5. Money saved by Alex = ₹120

Cost of a toy = ₹28

Cost of a book = ₹28 ÷ 2

Money left with Alex = ₹120 28 28 ÷ 2

= ₹120 28 14 = ₹92 14 = ₹78

6. Number of stickers with Anish = 50

Number of stickers given to each friend = 3

Number of friends = 4

More stickers purchased = 10

Total number of stickers = 50 3 × 4 + 10

= 50 12 + 10 = 60 12 = 48 stickers

7. Number of pencils with Sahil = 15

Number of Sahil’s friends = 3

Number of pencils given to each friend by Sahil = 15 ÷ 3 = 5

Number of pencils already with each friend = 2

Total number of pencils with each friend = 2 + 5 = 7

8. Total number of students in the school = 350

Number of students absent on Monday = 25

Total number of students present on Monday = 350 25 = 325

Money brought by each student = ₹230

Total money collected = 325 × ₹230 = ₹74,750

Challenge

1. Difference of 81 and 3 = 81 3 = 78

Now, product = 81 × 3 = 243 and quotient = 81 ÷ 3 = 27

Sum of the product and quotient = 243 + 27 = 270

So, 78 + 270 = 348

Hence, the resultant number is 348.

1. a.

L TTh Th H T O

1 1 1 1

5 6 7 8 9

+ 2 3 4 5 6

8 0 2 4 5

b. L TTh Th H T O

1 1 1 1

9 8 7 6 5 4

+ 4 5 7 7 4

1 0 3 3 4 2 8

c. L TTh Th H T O

1 1 1 1 1

6 7 3 7 7 8

+ 5 6 7 4 3 3

1 2 4 1 2 1 1

d. L TTh Th H T O

1 1 1 1 1

2 5 3 6 2 1 + 1 2 3 6 5 3 6 5 8 6

3 0 2 5 7 2

2. a. L TTh Th H T O

5 4 6 7 6 3 4 5 7 5 2 0 1 0 1

b. L TTh Th H T O 15 11

7 5 14 7 1 16

8 6 4 8 2 6

9 6 5 3 7

7 6 8 2 8 9

c. L TTh Th H T O 5 13

8 6 8 6 3 6

6 5 3 6 5

8 0 3 2 7 1

d. L TTh Th H T O 4 14

9 5 4 8 6 3

8 4 5 6 2 2 1 0 9 2 4 1

3. a. 896 × 10 = 8960; 896 × 100 = 89,600; 896 × 1000 = 8,96,000

b. 4546 × 10 = 45,460; 4546 × 100 = 4,54,600; 4546 × 1000 = 45,46,000

c. 6457 × 10 = 64,570; 6457 × 100 = 6,45,700; 6457 × 1000 = 64,57,000

d. 9876 × 10 = 98,760; 9876 × 100 = 9,87,600; 9876 × 1000 = 98,76,000

4. a. 2100 ÷ 10 = 210

2100 ÷ 100 = 21

2100 ÷ 1000 = 2 with remainder 100

b. 4086 ÷ 10 = 408 with remainder 6

4086 ÷ 100 = 40 with remainder 86

4086 ÷ 1000 = 4 with remainder 86

c. 51,200 ÷ 10 = 5120

51,200 ÷ 100 = 512

51,200 ÷ 1000 = 51 with remainder 200

d. 74,562 ÷ 10 = 7456 with remainder 2

74,562 ÷ 100 = 745 with remainder 62

74,562 ÷ 1000 = 74 with remainder 562

5. a.

6. a.

2 3 7

231 5 4 7 6 7 4 6 2 8 5 6 6 9 3 1 6 3 7 1 6 1 7 2 0

b. 1 8 0

314 5 6 7 8 5

3 1 4 2 5 3 8 2 5 1 2 2 6 5

c. 1 1 5

768 8 8 7 5 7 7 6 8 1 1 9 5 7 6 8 4 2 7 7 3 8 4 0 4 3 7

d. 1 0 1 982 9 9 8 6 6 9 8 2 1 6 6 6 9 8 2 6 8 4

7. a. 18 ÷ 2 × 3

= 9 × 3 = 27

b. 12 × 6 2 + 18 ÷ 3

= 12 × 6 2 + 6 = 72 2 + 6 = 78 2 = 76

c. 34 + 2 × 5 9 ÷ 9

= 34 + 2 × 5 1

= 34 + 10 1 = 44 1 = 43

d. 63 4 + 8 ÷ 2 × 30

= 63 4 + 4 × 30

= 63 4 + 120 = 183 4 = 179

8. 4 5

As the quotient is 45; the other number will also be 45.

9. Largest 6-digit number = 9,99,999

Smallest 6-digit number = 1,00,000

Difference = 9,99,999 − 1,00,000 = 8,99,999

10. Number of men = 3,81,53,154

Number of women = 3,15,31,873

Number of children = 6,81,231

Total number of people:

11. Total distance to be covered by Isha = 7500 km

Distance covered by Isha each day = 150 km

Number of days required to cover the distance = 7500 ÷ 150 km = 50 days

12. Number of sections in the garden = 60

Number of sections replanted = 30

Number of sections with flowers = 60 30 = 30

Number of flowers in each section = 1280

Total number of flowers in the garden = 1280 × 30 = 38,400

13. Capacity of the auditorium = 64,070

Number of people seated in one row = 430

Number of rows in the auditorium = 64070 ÷ 430 = 149

14. Number of marbles with Maya = 80

Number of friends = 6

Marbles given to each friend = 5

Number of more marbles bought = 15

Total marbles with Maya = 80 6 × 5 + 15 = 80 30 + 15 = 95 30 = 65

15. Annual income of Neha = ₹98,780

Amount spent by Neha each year = ₹50,000

Annual savings of Neha = ₹ 98,780 ₹50,000 = ₹48,780

Money saved by Neha in ten years = ₹48,780 × 10 = ₹4,87,800

16. Sum of 5,00,000 and 3,00,000 = 5,00,000 + 3,00,000 = 8,00,000

Difference of 5,00,000 and 3,00,000 = 5,00,000 3,00,000 = 2,00,000

The required number to be subtracted from the sum of 5,00,000 and 3,00,000 = 8,00,000 2,00,000 = 6,00,000

17. Number of mangoes = 5565

Number of oranges = 2 × 5565 = 11,130

Number of oranges in each of the five boxes = 11130 ÷ 5 = 2226

18. Amount in Shreya’s account = ₹45,000

Amount in her account after receiving salary = ₹45,000 × 3 = ₹1,35,000

Amount spent on rent and grocery = ₹14,500

Money left in Shreya’s account = ₹135,000 ₹14,500 = ₹1,20,500

19. Answers may vary. Sample answer:

A company manufactures 1,86,320 pencils and 98,346 erasers. The company packs 2 pencils and 1 eraser in a packet. How many pencils and erasers will be left unpacked?

Challenge

1. Population of Hassan = 1,88,000

Population of Mysore = 12,61,000

According to Statement 1, Belgaum’s population = 1,88,000 + 5,95,000 = 7,83,000

So, Statement 1 is true.

According to Statement 2:

Population of Hassan + Population of Belgaum = 1,88,000 + 7,83,000 = 9,71,000

Population of Mysore = 12,61,000

Since, 9,71,000 is less than 12,61,000, Statement 2 is also true.

Therefore, both Statement 1 and Statement 2 are true.

2. Combined balance in Account A and B = ₹8,75,632

Balance in Account A = ₹3,56,289

So, balance in Account B = ₹8,75,632 ₹3,56,289 = ₹5,19,343

Thus, with Statement 1 alone, we can find the balance in Account B.

Statement 2 doesn’t give us specific amounts. So, Statement 2 alone is not enough to find the balance in Account B. Option a: Statement 1 alone is sufficient to answer is the correct option.

Case Study

1. 5,04,975 + 7,05,136 = 12,10,111 kg

The total weight of the rice produced by Karnataka and Bihar is 12,10,111 kg.

2. a. Tamil Nadu produces 3,26,151 kg more rice than Madhya Pradesh. Hence, the statement is false b. Karnataka produces 3,15,990 kg less rice than Tamil Nadu. Hence, the statement is true.

3. Weight of the rice to be delivered = 78,432 kg

Number of bags = 456

Weight of rice in each bag = 78,432 kg ÷ 456 = 172 kg

4. Total weight of the rice mixed = 24,430 + 20,510 = 44,940 kg

Number of bags = 214

Weight of the rice in each container = 44,940 ÷ 214 = 210 kg

Chapter 3

Let’s Warm-up

1 12 1, 5, 11

2. 27 1, 3, 5

3. 30 1, 23

4 46 1, 2, 4

5. 55 1, 3, 9

Do It Yourself 3A

1. a. 1 × 14 = 14

2 × 7 = 14

Factors of 14 are 1, 2, 7 and 14.

b. 1 × 34 = 34

2 × 17 = 34

Factors of 34 are 1, 2, 17 and 34.

c. 1 × 37 = 37

Factors of 37 are 1 and 37.

d. 1 × 48 = 48

2 × 24 = 48

3 × 16 = 48

4 × 12 = 48

6 × 8 = 48

Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

e. 1 × 55 = 55

5 × 11 = 55

Factors of 55 are 1, 5, 11 and 55.

2. a. 15 ÷ 1 = 15, Remainder = 0

15 ÷ 3 = 5, Remainder = 0

15 ÷ 5 = 3, Remainder = 0

Factors of 15 are 1, 3, 5 and 15.

b. 41 ÷ 1 = 41, Remainder = 0

Factors of 41 are 1 and 41.

c. 63 ÷ 1 = 63, Remainder = 0

63 ÷ 3 = 21, Remainder = 0

63 ÷ 7 = 9, Remainder = 0

Factors of 63 are 1, 3, 7, 9, 21 and 63.

d. 72 ÷ 1 = 72, Remainder = 0

72 ÷ 2 = 36, Remainder = 0

72 ÷ 3 = 24, Remainder = 0

72 ÷ 4 = 18, Remainder = 0

72 ÷ 6 = 12, Remainder = 0

72 ÷ 8 = 9, Remainder = 0

Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.

e. 81 ÷ 1 = 81, Remainder = 0

81 ÷ 3 = 27, Remainder = 0

81 ÷ 9 = 9, Remainder = 0

Factors of 81 are 1, 3, 9, 27 and 81.

3. a. Factors of 11 are 1 and 11.

Hence, 11 is a prime number.

b. Factors of 27 are 1, 3, 9 and 27. Hence, 27 is a composite number.

c. Factors of 29 are 1 and 29.

Hence, 29 is a prime number.

d. Factors of 55 are 1, 5, 11 and 55.

Hence, 55 is a composite number.

e. Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.

Hence, 56 is a composite number.

f. Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.

Hence, 72 is a composite number.

g. Factors of 73 are 1 and 73. Hence, 73 is a prime number.

h. Factors of 81 are 1, 3, 9, 27 and 81. Hence, 81 is a composite number.

i. Factors of 83 are 1 and 83. Hence, 83 is a prime number.

j. Factors of 94 are 1, 2, 47 and 94. Hence, 94 is a composite number.

4. 11, 13, 17 and 19 are the prime numbers between 10 and 20. So, there are 4 prime numbers between 10 and 20.

5. Time period in which Mycobacterium tuberculosis divides = 17 minutes

Factors of 17 are 1 and 17. Hence there are two factors of 17.

6. 1 × 169 = 169, 13 × 13 = 169 Factors of 169 are 1, 13 and 169. Hence to put the same number of chairs in rows as the total number of rows, 13 chairs will be there in each row.

7. Answer may vary. Sample answer: Alika has a ribbon of length 27m. In how many ways she can cut the ribbon equally?

Challenge

1. True.

If 10 is a factor of a number then 2 also its factor as 2 is a factor of 10.

For example, 10 is a factor of 20 and 2 is also the factor of 20.

Do It Yourself 3B

1. Numbers divisible by 2 have 0, 2, 4, 6 or 8 as its last digit. Hence, 24, 38 and 160 will be divisible by 2.

a. 11 b. 24 c. 38 d. 49 e. 160

2. Numbers divisible by 10 will be divisible by 5. Also, numbers divisible by 10 will have 0 as their last digit. Hence, 190 and 300 will be divisible by both 5 and 10.

a. 51 b. 75 c. 37 d. 190 e. 300

3. If the sum of the digits in a number is divisible by 3, then we know that the number is divisible by 3.

a. 72: 7 + 2 = 9 (divisible by 3)

b. 63: 6 + 3 = 9 (divisible by 3)

c. 92: 9 + 2 = 11 (not divisible by 3)

d. 60: 6 + 0 = 6 (divisible by 3)

e. 130: 1 + 3 + 0 = 4 (not divisible by 3) Hence, 72, 63 and 60 are divisible by 3.

a. 72 b. 63 c. 92 d. 60 e. 130

4. If the sum is divisible by 9, we know that the number itself is divisible by 9.

a. 36: 3 + 6 = 9 (divisible by 9)

b. 45: 4 + 5 = 9 (divisible by 9)

c. 56: 5 + 6 = 11 (not divisible by 9)

d. 118: 1 + 1 + 8 = 10 (not divisible by 9)

e. 919: 9 + 1 + 9 = 19 (not divisible by 9)

Hence, 36 and 45 are divisible by 9.

a. 36 b. 45 c. 56 d. 118 e. 919

5. The number should be divisible by 5: 660, 530, 130, 370

The number has 0 as the last digit: 660, 530, 130, 370

On adding the digits of the number, the sum is divisible by 3

660: 6 + 6 + 0 = 12 (divisible by 3)

530: 5 + 3 + 0 = 8 (not divisible by 3)

130: 1 + 3 + 0 = 4 (not divisible by 3)

370: 3 + 7 + 0 = 10 (not divisible by 3)

Hence, the correct code to open the lock is 660.

6. For distributing 40 pencils, 20 erasers, 55 candies and 20 notebooks equally among the 10 kids, the numbers should be divisible by 10. The numbers which have 0 at the ones place will be divisible by 10.

40 ends with zero, 20 ends with zero. So, 40 and 20 are divisible by 10.

So, pencils, erasers and notebooks can be divided equally among the students.

Challenge

1. To put 324 blueberries and 135 chocolates in 9 cups without any left over, the numbers should be divisible by 9.

324: 3 + 2 + 4 = 9 (divisible by 9)

135: 1 + 3 + 5 = 9 (divisible by 9)

Hence, 324 blueberries and 135 chocolates can be put in 9 cups without any left over.

Do It Yourself 3C

1. Prime factorisation of 80 = 2 × 2 × 2 × 2 × 5

Hence, option b is correct.

2. 2 × 2 × 3 × 3 × 5 × 7 = 1260

Hence, option c is correct.

5. The prime factors of 365 by the division method will be 1, 5, 73, 365.

Do It Yourself 3D

1. a. Factors of 13 are 1 and 13.

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Common factor is 1.

b. Factors of 21 are 1, 3, 7 and 21.

Factors of 16 are 1, 2, 4, 8 and 16.

Common factor is 1.

c. Factors of 28 are 1, 2, 4, 7, 14 and 28.

Factors of 35 are 1, 5, 7 and 35.

Common factors are 1 and 7.

d. Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

Factors of 27 are 1, 3, 9 and 27.

Common factors = 1 and 3

e. Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.

Factors of 45 are 1, 3, 5, 9, 15 and 45.

Common factors are 1, 3, 5 and 15.

f. Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.

Factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54.

Common factors are 1, 2, 3 and 6.

g. Factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.

Factors of 22 are 1, 2, 11 and 22.

Common factors are 1 and 2.

h. Factors of 50 are 1, 2, 5, 10, 25 and 50.

Factors of 45 are 1, 3, 5, 9, 15 and 45.

Common factors are 1 and 5.

2. a. Factors of 16 are 1, 2, 4, 8 and 16.

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

Common factors = 1, 2, 4, 8

Highest common factor is 8.

b. Factors of 15 are 1, 3, 5 and 15.

Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90.

Common factors = 1, 3, 5, 15

Highest common factor is 15.

c. Factors of 25 are 1, 5 and 25.

Factors of 45 are 1, 3, 5, 9, 15 and 45.

Common factors = 1, 5

Highest common factor is 5.

d. Factors of 27 are 1, 3, 9 and 27.

Factors of 47 are 1 and 47.

Common factors = 1

Highest common factor is 1.

e. Factors of 27 are 1, 3, 9 and 27.

Factors of 81 are 1, 3, 9, 27 and 81.

Common factors = 1, 3, 9, 27

Highest common factor is 27.

f. Factors of 28 are 1, 2, 4, 7, 14 and 28.

Factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.

Common factors = 1, 2, 7, 14

Challenge

1. Number of flowers with Samantha = 132 132 = 2 × 2 × 3 × 11

Number of vases used by Samantha = Largest prime factor of 132 = 11

Highest common factor is 14.

g. Factors of 40 are 1, 2, 4, 5, 8, 10, 20 and 40.

Factors of 50 are 1, 2, 5, 10, 25 and 50.

Common factors = 1, 2, 5, 10

Highest common factor is 10.

h. Factors of 33 are 1, 3, 11 and 33.

Factors of 55 are 1, 5, 11 and 55.

Common factors = 1, 11

Highest common factor is 11. 3. a. 12

of 12 and 15 = 3

of 26 and 65 = 13

of 24, 40, 56 = 2 × 2 × 2 = 8

5. Length of box = 75 cm

Breadth of box = 85 cm

Height of box = 95 cm

Length of the longest tape that can measure the three dimensions of the box are = HCF of 75 cm, 85

and

Hence, the length of the longest tape that can measure the three dimensions of the box is 5 cm.

6. Number of days rice takes to mature = 120

Number of days wheat takes to mature = 150

Number of days will it take for the required cycle to repeat = LCM of 120 days and 150 days

Hence, the number of days it will take for the required cycle to repeat is 600 days.

Challenge

1. Capacity of three drums = 36 litres, 45 litres and 72 litres

Capacity of the largest container to measure the contents = HCF of 36 litres, 45 litres and 72 litres

Hence, the capacity of largest container to measure the contents = 9 litres

Chapter Checkup

1. a. Composite number: Factors of 49 are 1, 7 and 49

b. Prime number: Factors of 61 are 1 and 61

c. Prime number: Factors of 73 are 1 and 73

d. Composite number: Factors of 99 are 1, 3, 9, 11, 33 and 99

2. a. 1 × 16 = 16

2 × 8 = 16

4 × 4 = 16

The factors of 16 are 1, 2, 4, 8 and 16.

b. 1 × 20 = 20

2 × 10 = 20

4 × 5 = 20

The factors of 20 are 1, 2, 4, 5, 10 and 20.

c. 1 × 24 = 24

2 × 12 = 24

3 × 8 = 24

4 × 6 = 24

The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

d. 1 × 42 = 42

2 × 21 = 42

3 × 14 = 42

6 × 7 = 42

The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.

3. a. 28 ÷ 1 = 28 Remainder = 0

28 ÷ 2 = 14 Remainder = 0

28 ÷ 4 = 7 Remainder = 0

The factors of 28 are 1, 2, 4, 7, 14 and 28.

Divisible by 2 (last digit is 0, 2, 4, 6, 8)

Divisible by 3 (sum of digits is divisible by 3)

b. 36 ÷ 1 = 36 Remainder = 0

36 ÷ 2 = 18 Remainder = 0

36 ÷ 3 = 12 Remainder = 0

36 ÷ 4 = 9 Remainder = 0

36 ÷ 6 = 6 Remainder = 0

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.

c. 56 ÷ 1 = 56 Remainder = 0

56 ÷ 2 = 28 Remainder = 0

56 ÷ 4 = 14 Remainder = 0

56 ÷ 7 = 8 Remainder = 0

The factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.

d. 80 ÷ 1 = 80 Remainder = 0

80 ÷ 2 = 40 Remainder = 0

80 ÷ 4 = 20 Remainder = 0

80 ÷ 5 = 16 Remainder = 0

80 ÷ 8 = 10 Remainder = 0

The factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80.

Divisible by 5 (last digit is 0 or 5)

Divisible by 9 (sum of digits is divisible by 9)

Divisible by 10 (last digit is 10)

7. a. Factors of 25: 1, 5, 25

Factors of 45: 1, 3, 5, 9, 15, 45 Common factors: 1, 5

b. Factors of 75: 1, 3, 5, 15, 25, 75

Factors of 125: 1, 5, 25, 125 Common factors: 1, 5, 25

c. Factors of 33: 1, 3, 11, 33

Factors of 55: 1, 5, 11, 55

Common factors: 1, 11

d. Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

Factors of 156: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156

Common factors: 1, 2, 3, 4, 6, 12

8. a. Factors of 16: 1, 2, 4, 8, 16

Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Common factors: 1, 2, 4 HCF = 4

b. Factors of 25: 1, 5, 25 Factors of 65: 1, 5, 13, 65

Common factors: 1, 5 HCF = 5

c. Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

Common factors: 1, 2, 3, 4, 6, 8, 12, 24 HCF = 24

d. Factors of 150: 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150

Factors of 225: 1, 3, 5, 9, 15, 25, 45, 75, 225

Common factors: 1, 3, 5, 15, 25, 75 HCF = 75

9. a. 34 = 2 × 17 b. 34 = 2 × 17

38 = 2 × 19 51 = 3 × 17

HCF = 2 HCF = 17

c. 60 = 2 × 2 × 3 × 5 d. 105 = 3 × 5 × 7

225 = 3 × 3 × 5 × 5 180 = 2 × 2 × 3 × 3 × 5 HCF = 3 × 5 = 15 HCF = 3 × 5 = 15

10. a.

HCF (36, 63) = 9

b. 1 1 9 1 8 7 1 1 1 9 6 8 1 1 9 1 6 8 5 1 6 8 1 5 1 17 5 1 3 5 1 0

HCF (119, 187) = 17

c. 4 5 8 9 1 4 5 4 4 4 5 1 4 4 1 4 4 4 4 4 4 0

HCF (45, 89) = 1

d. 1 3 6 1 7 0 1 1 3

Remainder

HCF (136, 170) = 34

11. Answers may vary. Sample answer: For a number to have 2 as a factor it must have 0, 2, 4, 6 or 8 as its last digit. Hence, a three-digit number which has 2 as a common factor with 420 is 120.

12. Capacity of tanks = 250 litres and 425 litres

Maximum capacity of the bucket that can measure the water in tanks = HCF of 250 litres and 425 litres

2 5 0 4 2 5 1 2 5 0 1 7 5 2 5 0 1

Remainder

Hence, the maximum capacity of the bucket that can measure the water in tanks is 25 litres.

13. Number of white balloons = 24

Number of orange balloons = 16

Greatest number of balloons in each arrangement = HCF of 24 and 16

6 0 Remainder

Hence, the greatest number of balloons in each arrangement is 8.

14. Dimension of the game board = 66 inches × 24 inches

Largest size of square tile to fit the board = HCF of 66 inches and 24 inches = 6 inches

Area of the game board = 66 × 24 = 1584 square inches

HCF of 66 and 24 = 6

So, length of side of each square tile = 6 inches

Area of one square tile = 6 × 6 = 36 square inches

Number of tiles = Area of the game board

Area of one tile

=1584 ÷ 36 = 44

Hence the number of tiles he needs is 44.

15. Number of cupcakes = 240

Number of sandwiches = 160

Maximum number of uniform packets that can be made = HCF of 240 and 160 = 80

Number of cupcakes in each packet = 240 ÷ 80 = 3

Number of sandwiches in each packet = 160 ÷ 80 = 2

1 6 0 2 4 0 1

1 6 0

80 1 6 0 2 1 6 0 0

Remainder

Therefore, the maximum number of packets she can make is 80; the contents of each will be 3 cupcakes and 2 sandwiches.

16. Number of oatmeal cookies = 30

Number of chocolate chip cookies = 48

Number of plastic containers needed = HCF of 30 and 48

3 0 4 8 1

8 3 0 1

Remainder

Hence, the number of plastic containers needed = 6

Challenge

1. The greatest number that divides 178 and 128, leaving a remainder of 8 is the HCF of 178 – 8 =170 and 128 – 8 = 120

120 = 2 × 2 × 2 × 3 × 5

170 = 2 × 5 × 17

Common factors: 2, 5

HCF: 10

Thus, the greatest number that divides 178 and 128, leaving a remainder of 8 is 10.

2. Half a century = 100 ÷ 2 = 50

Number which is 2 more than half a century = 50 + 2 = 52

Number which is 16 more than a century = 100 + 16 = 116

Number that divides 52 and 116, leaving a remainder of 0 is the HCF of 52 and 116

52 = 2 × 2 × 13

116 = 2 × 2 × 29

HCF: 2 × 2 = 4

Twice the HCF: 4 × 2 = 8

Required number = 8

Case Study

1. The number of strings is an odd prime number.

Factors of 8 are 1, 2, 4 and 8. So, 8 is not a prime number. Factors of 9 are 1, 2, 3 and 9. So, 9 is not a prime number.

Factors of 7 are 1 and 7. So, 7 is a prime number.

Factors of 3 are 1 and 3. So, 3 is a prime number.

The total number of pearls is 72. The number of pearls per string must be an integer, so we check which of the prime numbers (7 or 3) can divide 72 evenly.

Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72

So, the number of strings = 3

Hence, option d is correct.

2. The maximum number of stones he can use in each, such that the number of rubies and emeralds are equal in the bangles is HCF of 18 and 24.

18 = 2 × 3 × 3

24 = 2 × 2 × 2 × 3

Common factors: 2, 3

HCF: 2 × 3 = 6

Hence, option b is correct.

3. Number of pairs of earrings a girl orders = 32

Factor tree for 32:

4. Number of jades used to make necklace = 24

Number of emeralds used to make necklace = 36

The greatest number of necklaces that the jeweler can make is HCF of 24 and 36 = 12 necklaces

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

Common factors: 2, 2, 3

HCF: 2 × 2 × 3 = 12

Chapter 4

Let’s Warm–up

1. Multiplication table of 8 60, 96, 108

2. Multiplication table of 9 40, 70, 80

3. Multiplication table of 10 24, 48, 64

4. Multiplication table of 12 36, 54, 63

Do It Yourself 4A

1. a. 8 × 1 = 8 8 × 2 = 16 8 × 3 = 24 8 × 4 = 32 8 × 5 = 40

b. 11 × 1 = 11 11 × 2 = 22 11 × 3 = 33

11 × 4 = 44 11 × 5 = 55

c. 21 × 1 = 21 21 × 2 = 42 21 × 3 = 63

21 × 4 = 84 21 × 5 = 105

d. 25 × 1 = 25 25 × 2 = 50 25 × 3 = 75

25 × 4 = 100 25 × 5 = 125

e. 50 × 1 = 50 50 × 2 = 100 50 × 3 = 150

50 × 4 = 200 50 × 5 = 250

2. On dividing 7209 by 9, we get no remainder. So, 7209 is a multiple of 9.

3. The year in which the last Purna Kumbh Mela was organised = 2013

The Purna Kumbh Mela is organised every 12 years. So, 12 × 1 = 12

12 × 24 = 24

12 × 3 = 36

Therefore, the next three Purna Kumbh Melas will be organised in 2013 + 12 = 2025, 2013 + 24 = 2037, 2013 + 36 = 2049

4. Answers may vary. Sample answer: Let us take 10 as an example.

The first ten multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90 and 100.

Since all these numbers end with a zero in the ones place, all multiples of 10 are even.

5. Both species of cicadas emerged in 2021. The first species of cicadas emerges from the ground every 13 years.

The second species of cicadas emerges every 17 years. To find when will both species will emerge together over the next century, we need to find the LCM of 13 and 17. 13 13 1 17 17 1

LCM of 13 and 17 = 221

So, the next time both species will emerge together in = 2021 + 221 = 2242

Since the next century after 2021 is from 2100 to 2199, it is clear that the two species will not emerge together over the next century.

Challenge

1. Yes, A can also be a multiple of B. For example, if B = 8, the factors are 1, 2, 4, 8

If A = 1, 2 or 4, it cannot be B’s multiple, but if A = 8, it becomes a multiple of B which is also 8.

Do It Yourself 4B

1. Multiples of

Multiples of

Multiples of 50

… So, the common multiples of 25, 75 and 50 is 150. Thus, option d is correct.

2. a. Multiples

Common multiples of 12 and 15 = 60, 120, ... 60 is the lowest among all the common multiples, so it is the LCM of 12 and 15. So, the LCM of 12 and 15 = 60

Common multiples of 9 and 63 = 63, 126, ... 63 is the lowest among all the common multiples, so it is the LCM of 9 and 63.

LCM of 9 and 63 = 63

c. Multiples of 9 5 10 15 20

Common multiples of 5 and 55 = 55, 110, ... 55 is the lowest among all the common multiples, so it is the LCM of 5 and 55.

LCM of 5 and 55 = 55

d. Multiples of 8 8 16

Multiples of 10 10 20

Common multiples of 8 and 10 = 40, 80, ... 40 is the lowest among all the common multiples, so it is the LCM of 8 and 10.

So, the LCM of 8 and 10 = 40

Common multiples of 6, 12 and 18 = 36, 72, ... 36 is the lowest among all the common multiples, so it is the LCM of 6, 12 and 18.

LCM of 6, 12 and 18 = 36

f. Multiples of 6 6 12 18 24

Multiples of 10 10 20

Multiples of 15 15 30

Common multiples of 6, 10, 15 = 30, 60, ... 30 is the lowest among all the common multiples, so it is the LCM of 6, 10 and 15.

LCM of 6, 10 and 15 = 30 g.

Multiples of 12 12 24 36 48 60 72 84

of 15 15 30

Multiples of 20 20 40 60

Common multiples of 6, 10, 15 = 60, 12, ... 60 is the lowest among all the common multiples, so it is the LCM of 12, 15 and 20.

LCM of 12, 15 and 20 = 60 h.

Multiples of 15 15 30 45 60 75 90 105 120 135 150

Multiples of 25 25 50 75 100 125 150 175 200 225 250 … Multiples of 30 30 60 90 120 150 180 210 240 270 300

Common multiples of 15, 25, 30 = 150, ... 150 is the lowest among all the common multiples, so it is the LCM of 15, 25 and 30.

LCM of 15, 25 and 30 = 150

3. Let us find the LCM of 4 and 15:

So, the LCM of 4 and 15 = 60

a. Let us find the LCM of 4 and 20:

So, the

of 4 and 20 = 20

b. Let us find the

Multiples of 25 25

of 6 and 25:

So, the LCM of 6 and 25 = 150

c. Let us find the LCM of 5 and 12:

So, the LCM of 5 and 12 = 60

d. Let us find the LCM of 8 and 30:

Multiples of 30 30 60 90

So, the LCM of 8 and 30 = 120

So, the LCM of 5 and 12 is the same as the LCM of 4 and 15 which is 60.

Thus, option c is correct.

4. The U.S. presidential elections are held every 4 years. The Senate elections are held every 6 years. To find at what year both elections held together, we need to find the LCM of 4 and 6.

Multiples of 6 6 12 18 24 30 36 42 48 54 60 …

Common multiples of 4 and 6 = 12, 24, 36, ... 12 is the lowest among all the common multiples, so it is the LCM of 4 and 6.

Lowest Common Multiple or LCM of 4 and 6 = 12 Hence, the year in which both elections held together = 2024 + 12 = 2036

5. First machine requires maintenance every 12 days. Second machine requires maintenance every 18 days. To find after how many days both machines need maintenance on the same day again, we need to find LCM of 12 and 18.

Multiples of 12 12 24 36 48 60 72 84 96

Multiples of 18 18 36 54 72 90 108 126 144

Common multiples of 12 and 18 = 36, 72, ...

36 is the lowest among all the common multiples, so it is the LCM of 12 and 18.

Lowest Common Multiple or LCM of 12 and 18 = 12 Hence, after 36 days both machines need maintenance on the same day again.

6. The first bell rings after 30 minutes in primary school. The first bell rings after 40 minutes in secondary school. To find at what time both bells ring together, we need to find LCM of 30 and 40.

So, the common multiples of 30 and 40 = 120, 240, ... Thus, after 120 minutes, both school’s bells ring together. Thus, the both bell will ring together again at 8:30 a.m. + 120 minutes = 8:30 a.m. + 60 minutes + 60 minutes = 8:30

7. Answers may vary. Sample answer:

A shopkeeper sells candles in packet of 12 and candle stands in packets of 8. what is the least number of candles and candle stands Rita should buy so that there will be one candle for each candle stand?

Challenge

1. The third multiple of 24 = 72

The third multiple of required number is half the third multiple of 24

Therefore, third multiple of required number = 72 ÷ 2 = 36 36 is a third multiple of number 12 Thus, required number = 12

b. 2 44

2 22 11 11 1 2 88 2 44 2 22 11 11 1

44 = 2 × 2 × 11

88 = 2 × 2 × 2 × 11

2 48

2 24

c. 2 12 2 6 3 3 1

2 12

2 6 3 3 1

48 = 2 × 2 × 2 × 2 × 3

12 = 2 × 2 × 3

d. 5 115 23 23 1

5 25 5 5 1

25 = 5 × 5

115 = 5 × 23

2. a. LCM of 8 and 10 = 2 × 2 × 2 × 5 = 40

b. LCM of 8 and 16 = 2 × 2 × 2 × 2 = 16

c. LCM of 16 and 10 = 2 × 2 × 2 × 2 × 5 = 80

d. LCM of 25 and 10 = 5 × 5 × 2 = 50

3. a. LCM of 16 and 24

2 16 2 24

2 8 2 12

2 4 2 6

2 2 3 3 1 1

16 = 2 × 2 × 2 × 2

24 = 2 × 2 × 2 × 3

Now, we will take 4 times two and 1 time three. So, the LCM of 16 and 24 = 2 × 2 × 2 × 2 × 3 = 48

b. LCM of 25 and 35

5 25 5 35

5 5 7 7 1 1

25 = 5 × 5

35 = 5 × 7

Now, we will take 2 times five and 1 time seven.

So, the LCM of 25 and 35 = 5 × 5 × 7 = 175

c. LCM of 36 and 45

2 36 3 45

2 18 3 15

3 9 5 5

3 3 1 1

36 = 2 × 2 × 3 × 3

45 = 3 × 3 × 5

Now, we will take 2 times two, 2 times three and 1 time five.

So, the LCM of 36 and 45 = 2 × 2 × 3 × 3 × 5 = 180

d. LCM of 63 and 105 3 63 3 105 3 21 5 35

7 7 7 7 1 1

63 = 3 × 3 × 7

105 = 3 × 5 × 7

Now, we take 2 times three, 1 time five and 1 time seven.

So, the LCM of 63 and 105 = 3 × 3 × 5 × 7 = 315

e. LCM of 18, 40 and 45

2 18 2 40 3 45

3 9 2 20 3 15

3 3 2 10 5 5 1 5 5 1 1

18 = 2 × 3 × 3

40 = 2 × 2 × 2 × 5

45 = 3 × 3 × 5

Now, we take 3 times two, 2 times three and 1 time five.

So, LCM of 18, 40 and 45 = 2 × 2 × 2 × 3 × 3 × 5 = 360

f. LCM of 72, 96 and 108 2 72 2 96 2 108 2 36 2 48 2 54 2 18 2 24 3 27 3 9 2 12 3 9 3 3 2 6 3 3 1 3 3 1 1

72 = 2 × 2 × 2 × 3 × 3

96 = 2 × 2 × 2 × 2 × 2 × 3

108 = 2 × 2 × 3 × 3 × 3

Now, we take 5 times two and 3 time three.

So, the LCM of 72, 96 and 108 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 864

g. 48, 56 and 70 2 48 2 56 2 70 2 24 2 28 5 35 2 12 2 14 7 7 2 6 7 7 1 3 3 1 1

48 = 2 × 2 × 2 × 2 × 3

56 = 2 × 2 × 2 × 7

70 = 2 × 5 × 7

Now, we take 4 times two and 1 time three, five, seven each.

So, LCM of 48, 56 and 70 = 2 × 2 × 2 × 2 × 3 × 5 × 7 = 1680

30, 60 and 90

30 = 2 × 3 × 5

60 = 2 × 2 × 3 × 5

90 = 2 × 3 × 3 × 5

Now, we take 2 times two, 2 times three and 1 time five. So, LCM of 30, 60 and 90 = 2 × 2 × 3 × 3 × 5 = 180

4. To find the LCM of 12, 24 and 56, we will take 2 three times, 3 one time and 7 one time.

LCM of 12, 24 and 56 = 2 × 2 × 2 × 3 × 7 = 168 Thus, Rohan is not right.

5. T20 cricket world cup is held every 2 years. FIFA world cup is held every 4 years. Both the world cups were last held together in 2022. To find how many times will both world cups be held together till 2040 after 2022, we need to find LCM of 2 and 4.

2 = 2 × 1

4 = 2 × 2

LCM of 2 and 4 = 4

So, the next world cups be held together till 2040 after 2022 in 2022 + 4 = 2026, 2026 + 4 = 2030, 2030 + 4 = 2034 and 2034 + 4 = 2038

So, 4 times will both world cups be held together till 2040 after 2022.

6. Answers may vary. Sample answer: Find the least length of a rope which can be cut into whole number of pieces of length 45 cm and 75 cm.

Challenge

1. Present age = Multiple of 7 7 14 21 28 35 42 49

Next year age = Present age + 1 = Multiple of 8

Also, the present age is greater than 20 but less than 80. So, from above table, I am 63 years old.

Do It Yourself 4D

1. a. 21 and 24

3 21 24

7 7 8

2 1 8 2 1 4

2 1 2 1 1

So, the LCM of 21 and 24 = 3 × 7 × 2 × 2 × 2 = 168

b. 25 and 30 2 25 30 3 25 15 5 25 5 5 5 1 1 1

So,

c. 32 and 48

and

So, the LCM of 32 and 48 = 2 × 2 × 2 × 2 × 2 × 3 = 96

d. 60 and 75 2 60 75 2 30 75 3 15 75 5 5 25 5 1 5 1 1

So, the LCM of 60 and 75 = 2 × 2 × 3 × 5 × 5 = 300

28, 42, 56

So, the LCM of 28, 42 and 56 = 2 × 2 × 2 × 3 × 7 = 168

f. 75, 100, 150 2 75 100 150 2 75 50 75 3 75 25 75 5 25 25 25 5 5 5 5 1 1 1

So, the LCM of 75, 100 and 150 = 2 × 2 × 3 × 5 × 5 = 300

g. 21, 63, 105 3 21 63 105 3 7 21 35 5 7 7 35 7 7 7 7 1 1 1

So, the LCM of 21, 63 and 105 = 3 × 3 × 5 × 7 = 315

h. 90, 135, 180

2 90 135 180

2 45 135 90

3 45 135 45

3 15 45 15

3 5 15 5

5 5 5 5 1 1 1

So, the LCM of 90, 135 and 190 = 2 × 2 × 3 × 3 × 3 × 5 = 540

2. Rohit’s method—

2 42 56

2 21 28

7 3 4

To find the LCM of 42 and 56, follow the method given below,

2 42 56

2 21 28

2 21 14

3 21 7

7 7 7 1 1

LCM of 42 and 56 = 2 × 2 × 2 × 3 × 7 = 168 Rohit’s conclusion is incorrect because he didn’t consider the highest power of 2 in the prime factorisation when calculating the LCM. Thus, Rohit is not correct and LCM of 42 and 56 is 168.

3. a. LCM of 42 and 30 b. LCM of 63, 105

2 42 70

3 21 35

5 7 35 7 7 7 1 1 3 63 105 3 21 35 5 7 5 7 7 1 1 1

LCM of 42 and 30 LCM of 63 and 105

= 2 × 3 × 5 × 7 = 210 = 3 × 3 × 5 × 7 = 315

c. LCM of 30 and 45 d. LCM of 9 and 15

2 30 45

3 15 45

3 5 15

5 5 5 1 1 3 9 15 3 3 5 5 1 5 1 1

LCM of 30 and 45 LCM of 9 and 15

= 2 × 3 × 3 × 5 = 90 = 3 × 3 × 5 = 45

e. LCM of 12 and 48

2 12 48

2 6 24

2 3 12

2 3 6

3 3 3 1 1

LCM of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48

LCM of 63, 105

LCM of 30, 45

LCM of 9, 15

LCM of 12, 48

LCM of 42, 70 90 45 48 210 315

4. Jupiter takes around 12 years to revolve around the Sun. Saturn takes around 30 years to revolve around the Sun. Jupiter, Saturn and the Sun are in a straight line at the beginning of their revolution cycle.

To find after how many years will they be in their original positions again, we need to find the LCM of 12 and 30.

2 12 30

2 6 15

3 3 15

5 1 5 1 1

So, after 60 years they will be in their original positions again.

Challenge

1. Assertion (A): Anna takes 8 minutes to complete one round, and Sylvia takes 12 minutes. If both start cycling at the same time and move in the same direction, they will meet at the starting point after 24 minutes.

Reason (R): The least common multiple (LCM) of 8 and 12 is 24. Anna takes 8 minutes to complete one round. Sylvia takes 12 minutes to complete one round.

Both Anna and Sylvia start cycling at the same time and move in the same direction.

To find after how minutes they will meet at the starting point, we need to find the LCM of 8 and 12 = 24

2 8 12 2 4 6 2 2 3 3

Therefore, after 24 minutes they will meet at the starting point.

The assertion is true and the reason explains why it is true. Thus, option (a) is correct.

Do It Yourself 4E

1. Narendra marks first date as August 7. The classes are scheduled once in every 7 days. So, the dates are, 7 × 1 = 7

7 × 2 = 14

7 × 3 = 21

7 × 4 = 28

Therefore, the dates he will mark for the classes in the month of August are 7, 14, 21 and 28.

2. To find the greatest number of groups Mrs Mehra can make, we need to find the highest common factor (HCF) of 12 and 18.

Factor of 12 1 2 3 4 6

Factor of 18 1 2 3 6 9

The HCF of 12 and 18 is 6.

This means that the greatest number of groups Mrs Mehra can make is 6.

This is because each group must have the same number of boys and girls, and 6 is the largest number that divides both 12 and 18 evenly.

3. The maximum capacity of a container that can measure the milk of either container an exact number of times is the highest common factor (HCF) of 75 and 45.

To find the HCF of 75 and 45,

Factor of 75 1 3 5 15 25 75

Factor of 45 1 3 5 9 15 45

So, the highest common factor of 75 and 45 = 15

Therefore, the maximum capacity of a container that can measure the milk of either container an exact number of times is 15 litres.

4. The first device beeps at every 60 sec and the second device beeps at every 62 sec.

To find out at what time they beep together, we need to find out the LCM of 60 and 62.

2 60 62

2 30 31

3 15 31

5 5 31 31 1 31 1 1

LCM = 2 × 2 × 3 × 5 × 31 = 1860 1 minute = 60 seconds

1 second = 1 60 minutes

1860 seconds = 1860 60 = 31 minutes

So, after 31 minutes both devices will beep together. And the time will be 10:00 a.m. + 31 minutes = 10:31 a.m.

5. Number of soccer balls = 72

Number of basketballs = 96

To find the largest possible equal groups, such that each group has the same number of soccer balls and basketballs, we need to find the HCF of 72 and 96.

Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.

Factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96

Common factors = 1, 2, 3, 4, 6, 8, 12 and 24

HCF of 72 and 96 is 24.

Therefore, 24 is the largest possible equal groups, such that each group has the same number of soccer balls and basketballs.

6. We need to find the least possible number of students in the class.

Now, the number of students will be greater than 6, 8, 12 and 16

So, we need to find the LCM in this case.

LCM = 2 × 2 × 2 × 2 × 3 = 48

Therefore, the least possible number of students in the class = 48 students

7. Sunflowers bloom every 10 weeks. Roses bloom every 5 weeks. Tulips bloom every 3 weeks.

To find after how many weeks they will all bloom together again, we need to find the LCM of 10, 5 and 3.

LCM of 2, 3 and 5 = 2 × 3 × 5 = 30

So, after 30 weeks they will all bloom together again.

8. To find the total number of stacks, we need to find the least common multiple (LCM) of 336, 240 and 96.

The LCM of 336, 240 and 96 = 2 × 2 × 2 × 2 × 2 × 3 × 5 × 7 = 3360

This means that we can stack the books in 3360 stacks, such that each stack contains the same number of books of each subject, and the height of each stack is the same. Therefore, the total number of stacks is 3360.

9. Answers may vary. Sample answer: Three pieces of ribbons 42 cm, 49 cm and 63 cm long, need to be divided into the same length. What is the greatest possible length of each ribbon?

Challenge

1. Free beverage is received after every 5th visit. Free appetizer is received after every 10th visit.

Multiples of 5 5 10 15 20 25 30 35 40 45 50

Multiples of 10 10 20 30 40 50 60 70 80 90 100

a. For both a free beverage and a free appetizer to occur on the same visit, that visit must be a multiple of both 5 and 10.

So, LCM of 5 and 10 = 2 x 5 = 10

2 5 10

5 5 5 1 1

Therefore, you will receive both a free beverage and a free appetizer on every 10th visit.

The multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90 and 100.

There are 10 multiples of 10 between 1 and 100. Thus, I will receive both a free beverage and a free appetiser on 10 visits.

b. To find the first visit where you can get both free items, we need to find the LCM of 5 and 10. So, the LCM of 5 and 10 = 2 × 5 = 10

2 5 10

5 5 5 1 1

So, on the 10th visit I will get both free items.

Chapter Checkup

1. Multiples of 9 are coloured in the number chart: 1

2. 105, 847 and 77 can be divided by 7, while 103 cannot be divided by 7. Thus, option a is correct.

3. The 9th multiple of a. 7 × 9 = 63 b. 9 × 9 = 81

c. 11 × 9 = 99 d. 13 × 9 = 117

4. 90 and 180 are common multiples of 30 and 45. Thus, option b is correct.

5. a. 2 × 3 × 3 2 × 5 × 7 2 × 3 × 3 × 5 × 7 = 630

b. 2 × 3 × 5 3 × 3 × 5 2 × 3 × 3 × 5 = 90

c. 2 × 3 × 3 × 7 2 × 3 × 11 2 × 3 × 3 × 7 × 11 = 1386

d. 3 × 3 × 5 × 7 2 × 3 × 5 × 11 2 × 3 × 3 × 5 × 7 × 11 = 6930

e. 2 × 3 × 3 × 5 2 × 2 × 3 × 5 2 × 2 × 3 × 3 × 5 = 180

6. a. 18, 27 2 18 3 27 3 9 3 9 3 3 3 3 1 1

Prime factor of 18 = 2 × 3 × 3

Prime factors of 27 = 3 × 3 × 3

LCM of 15 and 12 = 2 × 3 × 3 × 3 = 54

b. 35, 14

5 35 2 14

7 7 7 7 1 1

Prime factor of 35 = 5 × 7

Prime factors of 14 = 2 × 7

LCM of 15 and 12 = 2 × 5 × 7 = 70

c. 5, 9, 15

5 5 3 9 3 15 1 3 3 5 5 1 1

Prime factor of 5 = 5

Prime factors of 9 = 3 × 3

Prime factors of 15 = 3 × 5

LCM of 5, 9 and 15 = 3 × 3 × 5 = 45

d. 25, 40, 60

5 25 2 40 2 60

5 5 2 20 2 30 1 2 10 3 15 5 5 5 5 1 1

Prime factor of 25 = 5 × 5

Prime factors of 40 = 2 × 2 × 2 × 5

Prime factors of 60 = 2 × 2 × 3 × 5

LCM of 25, 40 and 60 = 2 × 2 × 2 × 5 × 5 × 3 = 600

7. 2 9 12 15 2 9 6 15 3 9 3 15 3 3 1 5 5 1 1 5 1 1 1

LCM of 9, 12 and 15 = 2 × 2 × 3 × 3 × 5 = 180

180 is the smallest number that is divisible by 9, 12 and 15 which also the LCM of these numbers.

Thus, option d is correct.

8. a. LCM of 12, 20 and 32

2 12 20 32

2 6 10 16

2 3 5 8

2 3 5 4

2 3 5 2

3 3 5 1

5 1 5 1 1 1 1

LCM of 12, 20 and 32 = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 480

b. LCM of 93, 62 and 120

2 93 62 120

2 93 31 60

2 93 31 30

3 93 31 15

5 31 31 5

31 31 31 1 1 1 1

So, LCM of 93, 62 and 120 = 2 × 2 × 2 × 3 × 5 × 31 = 3720

c. LCM of 15, 36 and 40

2 15 36 40

2 15 18 20

2 15 9 10

3 15 9 5

3 5 3 5

5 5 1 5

1 1 1

LCM of 15, 36 and 40 = 2 × 2 × 2 × 3 × 3 × 5 = 360

d. LCM of 45, 18 and 63

2 45 18 63

3 45 9 63

3 15 3 21

5 5 1 7

7 1 1 7 1 1 1

LCM of 45, 18 and 63 = 2 × 3 × 3 × 5 × 7 = 630

9. To find the smallest possible number of questions that could have been given for holiday homework will be the least common multiple of 3 and 13.

LCM of 3 and 13 = 3 × 13 = 39

Therefore, the smallest possible number of questions that could have been given for holiday homework is 39.

10. Three bells ring at intervals of 20 minutes, 30 minutes and 45 minutes respectively.

To find the time after which all three bells ring together, we need to find the LCM of 20, 30 and 45

20 = 2 × 2 × 5

30 = 2 × 3 × 5

45 = 3 × 3 × 5

So, LCM of 20, 30 and 45 = 2 × 2 × 3 × 3 × 5 = 180

The least common multiple (LCM) of 20, 30 and 45 is 180. So, the bells will ring together after 180 minutes, which is equal to 3 hours.

Therefore, the three bells will ring together after 3 hours.

11. El Niño occurs every 3 years.

La Niña occurs every 7 years.

To find the year in which both phenomena occur together again starting from 2020, we need to the find the LCM of 3 and 7.

3 3 7 7 1 7 1 1

So, LCM of 3 and 7 = 3 × 7 = 21

Therefore, the year in which both phenomena occur together again starting from 2020 is 2020 + 21 = 2041

Challenge

1. A multiple of a number is a number that is divisible by that number. Therefore, a multiple of 75 is a number that can be divided by 75 without leaving a remainder.

75 ÷ 15 = 5

75 ÷ 25 = 3

75 ÷ 12 = Not possible 15 and 25 are both multiples of 75, because they can be divided evenly by 75. However, 12 is not a multiple of 75, because it cannot be divided evenly by 75. Therefore, 12 is the the wrong answer. Thus, Ajay gives the wrong answer.

Thus, option (c) is correct.

2. Assertion (A): The LCM of 3 and 9 is 9.

3 3 9 3 1 3 1 1

LCM of 3 and 9 = 3 × 3 = 9

Reason (R): The LCM of any two numbers is always greater than their HCF.

Multiples are always greater than or equal to a number and factors are always less than or equal to a number.

So, we can say that the LCM of any two numbers is always greater than their HCF.

Hence, assertion and reason both are true, but the reason does not explain the assertion correctly.

Thus, the correct option is (b).

Case Study

1. a. Riya’s bus arrives every 15 minutes. Samir’s bus arrives every 20 minutes.

To find at what time the buses arrive at the same time, we need to find the LCM of 15, 20.

2 20 15 2 10 15

3 5 15 5 5 5 1 1

LCM of 20 and 15 = 2 × 2 × 3 × 5 = 60

So, the buses arrive at the same time after every 60 minutes.

Thus, the correct option is (b).

2. To find at what time Riya and Samir take the next bus, so as to reach school at the same time, we need to find the LCM of 15 and 20.

2 20 15

2 10 15

3 5 15 5 5 5 1 1

LCM of 20 and 15 = 2 × 2 × 3 × 5 = 60

Riya and Samir take the next bus so as to reach school at the same time is 8:00 a.m. + 60 minutes = 9:00 a.m.

3. Number of trees Riya planted is a multiple of 2. Number of trees Samir planted is a multiple of 5. The lowest multiple of trees planted that is common to both is the LCM of 2 and 5 = 2 × 5 = 10

2 2 5

5 1 5 1 1

Thus, Riya and Ravi planted at least 10 trees.

4. Riya’s bus arrives every 10 minutes. Samir’s bus arrives every 15 minutes. To find at what time the buses arrive at the same time when both the buses arrived together was at 3:30 p.m., we need to find the LCM of 10, 15 = 30

2 10 15

3 5 15

5 1 5 1 1

The time the buses arrive at the same time when both buses arrived together was at 3:30 p.m. is 3:30 p.m. + 30 minutes = 4:00 p.m.

5. Answers may vary.

Chapter 5

Let’s Warm-up

1. 1 3 a half

2. 1 2 two-fourths

3. 1 4 one-fourth

4. 1 8 one-third

Do It Yourself 5A

1. a. As the denominators of 4 7 and 1 4 are different, they are unlike fractions.

b. As the denominators of 5 8 and 4 8 are the same, they are like fractions.

c. As the denominators of 1 4 and 1 5 are different, they are unlike fractions.

d. As the denominators of 3 7 and 5 7 are the same, they are like fractions.

e. As the denominators of 6 18 and 7 18 are the same, they are like fractions.

2. Proper fractions have denominator greater than numerator, hence, 41212 ,, 51529 are proper fractions. Improper fractions have numerator equal to or greater than the denominator, hence, 658179 ,,, 25113 are improper fractions.

Mixed numbers are fractions which are a combination of a whole number and a proper fraction, hence, 157 4 , 5 , 6 789 are mixed fractions.

5. Total number of rice packets = 87

Number of families = 5

Number of rice packets each family get = 87 5 17 5 87 5 37 35 2

So, 87 5 = 17 2 5

Thus, each family get 17 2 5 rice packets.

6. Mixed Number 2 3 6 Improper Fraction 15 6

Challenge

1. 7 2

6 = 7 × 6 + 2 6 = 44 6 = 22 3 Prime denominator Improper fraction

Thus, the fraction that Sanchita is talking about is 22 3

Do It Yourself 5B

1. Answers may vary. Sample answers:

a. 6263 12 18 and 72147321 ×× = = ××

b. 6263 12 18 and 1523015345 ×× = = ××

c. 112 113 22 33 and 1723417351 ×× = = ××

d. 122 123 24 36 and 2024020360 ×× = = ××

2. Answers may vary. Sample answers:

a. 122 124 63 and 202102045 ÷÷ = = ÷÷

b. 142 147 72 and 562285678 ÷÷ = = ÷÷

c. 302 303 15 10 and 4222142314 ÷÷ = = ÷÷

d. 322 324 16 8 and 4822448412 ÷÷ = = ÷÷

3. a. HCF of 24 and 60 is 12. 2412 2 60125 ÷ = ÷

b. HCF of 27 and 63 is 9. 279 3 6397 ÷ = ÷

c. HCF of 24 and 72 is 24. 2424 1 72243 ÷ = ÷

d. HCF of 44 and 55 is 11. 4411 4 55115 ÷ = ÷

4. a. 228214 and 4221025 ÷÷ = = ÷÷

As, 14 , 25 ≠ the fractions 2 4 and 8 10 are not equivalent.

b. 62 33 and 51025 ÷ = ÷

As, 33 , 55 = the fractions 3 5 and 6 10 are equivalent.

c. 147221122 and 21733333 ÷÷ = = ÷÷

As 22 , 33 = the fractions 14 21 and 22 33 are equivalent.

d. 55252511 and 105250252 ÷÷ == ÷÷

As, 11 , 22 = the fractions 5 10 and 25 50 are equivalent.

5. a. 5 × 7 = 35; hence, 4 × 7 = 28 428 535 =

b. 3 × 7 = 21; hence, 8 × 7 = 56 321 856 =

c. 36 ÷ 6 = 6; hence, 12 ÷ 6 = 2 212 636 =

d. 1 × 14 = 14; hence, 5 × 14 = 70 141 705 =

6. Fraction of the Earth’s surface that is covered with water = 2 3

212

312 × × = 24 36

So, 2 3 is equivalent to 24 36

So, option d is correct.

7. Answer may vary. Sample answer: = =

Challenge

1. Green—16, Pink— 3 18 or 1 6 , Blue—1 4 or 3 12,

6, Yellow—1 4 or 3 12 ; Green, pink and purple are equivalent. Yellow and blue are equivalent.

Do It Yourself 5C

1. a. As the fractions 14 23 and 17 23 are like fractions and 3 thus, 1417,<1417 23 2 <

b. 3 4 and 7 9 can be converted into like fractions as: 39742728 ; 49369436 ×× == ××

Since, 27 < 28, 2728 3636 < Thus, 37 49 <

c. 4 5 and 6 9 can be converted to like fractions as: 49653630 ; 59459545 ×× = = ××

Since, 36 > 30 3630 , 4545 > Thus, 46 59 >

d. 1 6 and 4 20 can be converted into like fractions as: 110431012 ; 6106020360 ×× = = ××

Since, 10 < 12, 1012 6060 < Thus, 14 620 <

2. a. 432512210 ; 531533515 ×× === ××

Since 12 > 10, 1210 1515 > or 42 53 > Hence, the given statement is correct.

b. 665511 ; 12621052 ÷÷ == ÷÷

Since, 11 22 = Hence, the given statement is correct.

c. 717158119120 ; 817136178136 ×× = = ××

Since, 119120 136136 < Hence, the given statement is correct.

d. 14111212 154 144 ; 12111321112132 ×× = = ××

Since, 154144 132132 >

Hence, the given statement is correct.

3. a. LCM of 4, 6 and 7 is 84. 221114312 ;; 421614712 ××× = = = ××× 42 1436 84 8484

So, 143642 848484 << Thus, 132 674 <<

b. LCM of 2, 3, 5 and 6 is 30. 2101655115 ;;; 3105665215 ×××× === = ×××× 2062515 30303030

So, 6152025 30303030 <<<

Thus, 1125 5236 <<<

c. LCM of 4, 7, 12 and 13 is 1092. 3845156591 252 780 455 ; ; ; 138410927156109212911092 1273791 273 637 ; 4273109212911092 ××× = = = ××× ×× = = ××

So, 252273455637780 10921092109210921092 <<<<

Thus, 31575 13412127 <<<<

d. LCM of 8, 12 and 15 is 120. 71588415 ;;; 815158815 510 1210 ××× = = = ××× × = × 105 64 60 120120120 50 120

So, 506064105 120120120120 <<<

Thus, 5487 128158 <<<

4. a. LCM of 4, 5, 6 and 8 is 120. 315130 ;; 815430 324120 ; 524620 ×× = = ×× ×× = = ×× 45 30 120120 72 20 120120

So, 72453020 120120120120 >>>

Thus, 3311 5846 >>>

b. LCM of 2, 3, 8 and 9 is 72. 3988 ;; 8998 124136 ; 324236 ×× == ×× ×× = = ×× 2764 7272 2436 7272

So, 64362724 72727272 >>>

Thus, 8131 9283 >>>

c. LCM of 6, 9, 11 and 14 is 1386. 1126 199 ; ; 111261499 11541231 ; 91546231

×× = = ×× ×× = = ×× 126 99 1386 1386 154 231 1386 1386

So, 23115412699 1386138613861386 >>> Thus, 1111 691114 >>>

d. LCM of 3, 5, 6, 11 and 15 is 330. 466322430 ;;; 56615221130 2110155 ; 3110655 ××× = = = ××× ×× = = ×× 264 66 120 330 330 330 220 55 330330

So, 2642201206655 330330330330330 >>>> Thus, 42431 5311156 >>>>

5. Answer may vary. Sample answer:

6. Hours Shashank studied for = 6 13

Hours Reshma studied for = 11 23

LCM of 13 and 23 is 299.

6 × 23 13 × 23 = 138 299; 11 × 13 23 × 13 = 143 299

Since, 138 299 < 143 299

i.e. 6 13 < 11 23

Thus, Reshma studied for a longer duration.

7. Answer may vary. Sample answer: Shalini studied for 14 9 hours, Priya studied for 8 3 hours. Who studied for more hours?

Challenge

1. Statement I: Rakesh and Prerna took the same amount of time.

Statement II: The time taken by Roshan is less than that of Prerna but more than that of Swati.

Statement III: Swati took the most amount of time.

Statement IV: The time taken by Roshan is less than that of Swati but more than that of Rakesh or Preeti.

Rakeshʼs time = 4 6 = 2 3 hour

Roshanʼs time = 7 8 hour

Swatiʼs time = 1 hour

Prernaʼs time = 2 3 hour

into like fractions:

Time taken by Rakesh and Prerna = 16 24

Arranging the time taken in ascending order, 16 24 = 16 24 < 21 24 < 24 24

Thus, Rakesh < Prerna < Roshan < Swati. Hence, Statement II is false.

Chapter Checkup

3. Answers may vary. Sample answers:

a. 222324 ;; 626364 ××× === ××× 468 121824

b. 121314234 ;; 929394 ××× === ×××182736

c. 525354 ; ; 828384 ××× === ××× 101520 162432

d. 112113114 ;; 142143144 ××× = = = ××× 223344 284256

4. Answers may vary. Sample answers:

a. 162164168 ;; 242244248 ÷÷÷ = == ÷÷÷ 842 1263

b. 242243246 ;; 422423426 ÷÷÷ = = = ÷÷÷ 12 8 4 21147

c. 7537557515 ;; 1953195519515 ÷÷÷ = = = ÷÷÷ 25 15 5 65 39 13

d. 108210831086 ;; 132213231326 ÷÷÷ = = = ÷÷÷ 54 36 18 66 44 22

5. a. HCF of 42 and 56 is 14. 4214 3 56144 ÷ = ÷

b. HCF of 66 and 84 is 6. 666 11 84614 ÷ = ÷

c. HCF of 75 and 125 is 25. 7525 3 125255 ÷ = ÷

d. HCF of 72 and 136 is 8. 728 9 136817 ÷ = ÷

6. a. 221 15 2 222 ×+ ==

LCM of 2 and 5 is 10. 5251428 and 210510 = =

Since, 2528 < Then, 2528 1010 < Thus, 114 2 25 <

b. 654 4 34 6 555 ×+ ==

LCM of 5 and 4 is 20. 3413634170 and 520420 = =

Since, 136170 <

Then, 136170 2020 < Thus, 434 6 54 <

c. 243 8 33 ÷ = ÷ Thus, 24 8 3 =

d. 193 119 57 6 33339 × === ×

Thus, 157 6 39 =

7. a. LCM of 2, 5, 6 and 12 is 60. 55130 ;; 125230 412110 ; 512610 ×× = = ×× ×× = = ×× 2530 6060 4810 6060

Since, 10253048 <<< 10253048 60606060 ⇒<<< 1514 61225 ⇒<<<

b. LCM of 2, 5, 8 is 40. 1838 ;; 5858 35120 ; 85220 ×× == ×× ×× = = ×× 824 4040 1520 4040

Since, 8152024 <<< 8152024 40404040 ⇒<<< 1313 5825 ⇒<<<

c. LCM of 2, 3, 5 and 7 is 210. 1105270 ;; 2105370 630242 ; 730542 ×× = = ×× ×× = = ×× 105140 210210 180 84 210210

Since, 84105140180 <<< 84105140180 210210210210 ⇒<<< 2126 5237 ⇒<<<

d. LCM of 2, 5, 7 and 12 is 420 11351210 ;; 12352240 384460 ; 584760 ×× = = ×× ×× = = ×× 385 210 420 420 252 240 420 420

Since, 210240252385 <<< 210240252385 420420420420 ⇒<<< 14311 27512 ⇒<<<

8. Fraction of books Alex took = 1 4

Fraction of books Bailey took = 3 8

Fraction of books Casey took = 3 12

LCM of 4, 8 and 12 is 24. 163332 ;; 4683122 133 4128 ××× == = ×××

∴=< 696 242424

Thus, Alex and Casey took the least fraction of books and Bailey took the greatest fraction.

9. Fraction of population of Asia = 3 5

Fraction of population of Australia = 1 80

Fraction of population of Europe = 7 80

Fraction of population of Africa = 7 40

Fraction of population of South America = 1 20

Fraction of population of North America = 3 40

LCM of 5, 80, 40 and 20 = 80

316

516 × × = 48 80 ; 11 801 × × = 1 80 ;

801 × × = 7 80 ; 72 402 × × = 14 80 ;

204 × × = 4 80 ; 32 402 × × = 6 80

Since, 1 < 4 < 6 < 7 < 14 < 48 146714

8080808080 <<<< < 48 80

⇒ 113

802040 << < 77 8040 < < 3 5

Australia < South America < North America < Europe < Africa < Asia

Challenge

1. LCM of 2, 8, 4 and 6 = 24

112

212 × × = 12 24 ; 33 83 × × = 9 24 ;

56 46 × × = 30 24 ; 54 64 × × = 20 24

Since 9 < 12 < 20 < 30

912

2424 < < 20 24 < 30 24

3155

8264 <<<

So, 3 8 is on extreme left.

2. Assertion (A): 1 2 > 1 3 > 1 4

Reason (A): In unit fractions, the denominators are compared. The smaller the denominator, the smaller the fractions. Since the numerator never changes with a unit fraction, we have to look at the denominator to compare two fractions. The smaller the bigger the denominator, the smaller the fraction.

So, 1 2 > 1 3 > 1 4

Thus, the assertion is true, but the reason is false. Hence, the correct option is (c).

Case Study

1. HCF of 3 and 6 = 3

63 ÷ ÷ = 1 2

So, the fraction of the paper bin = 3 6 is equivalent to 1 2 Thus, the correct option is (b).

2. LCM of 6, 4 and 5 = 60

310 6 10 × × = 30 60 ; 31545 41560 × = × ; 412 512 × × = 48 60

So, 30 < 45 < 48

3045

6060 < < 48 60

3 6 < 3 4 < 4 5

Paper Bin < Plastic Bin < Glass Bin

Hence, option (a) is correct.

3. The fraction of the glass bin = 4 5

42 52 × × = 8 10

The fraction of the glass bin is equivalent to 8 10

4. Plastic bin = 3 4

Since Improper fractions have a numerator equal to or greater than the denominator, the fraction of the plastic bin is not an improper fraction because 3 < 4.

5. Answers may vary.

Chapter 6

Let’s Warm-up

1. 2 halves 1 2 together make a whole.

2. A quarter 1 4 divides a whole into four equal parts.

3. 1 3 + 1 3 + 1 3 = 1

4 Adding 1 7 seven times gives one whole.

5. 1 6 shows 1 part out of six equal parts from a whole.

Do It Yourself 6A

1. a. Among the given fractions 134 1 444 +==

So, 1 4 , 2 4 , 3 4 , 5 4

b. Among the given fractions 12112 1 24222 +=+== So, 1 8 , 5 8 , 1 2 , 2 4

c. Among the given fractions 312347 1 721777 +=+== .

So, 3 7 , 8 21 , 12 21 , 4 14

d. Among the given fractions 215235 1 525555 +=+==

So, 2 5 , 15 10 , 15 25 , 12 35

2. a. 5 9 4 10 54 109910 4540 90 8517 9018 ×+× += × + = == 5 9 4 10 54 109910 4540 90 8517 9018 ×+× += × + = == b. 4341 51555 5 1 5 +=+ == 4341 51555 5 1 5 +=+

32322121 9393 13 2 5 933 23 5 99 23 5 9 5555 99 +=+++

32322121 9393 13 2 5 933 23 5 99 23 5 9 5555 99

e. 15 151111 7714714 611 714 6 2 11 7214 1211 1414 23 14 141 9 14 9 9 1 1 1414 + ++=+

15 151111 7714714 611 714 6 2 11 7214 1211 1414 23 14 141 9 14 9 9 1 1 1414 +

××

= ×+ = =+= 21 544 4 214421 21 45204 5 5 420 10517621 202020 10517621 20 302 20 151 10 1510 1 10 11 1515 1010 ++ 202020

21 544 4 214421 21 45204 5 5 420 10517621 202020 10517621 20 302 20 151 10 1510 1 10 11 1515 1010 ××

g. 5 1 3 + 17 13 18 = 5 + 17 + 1 3 + 13 18 = 22 + 1 × 6 3 × 6 + 13 18 = 22 + 6 18 + 13 18 = 22 + 19 18 = 22 + 18 × 1 + 1 18 = 22 + 1 + 1 18 = 23 + 1 18 = 23 1 18

h. 121 121 345345 3918 3918 1622 1 12 369218 641 12 181818 12121111 1818 ++=+++++ ×× =+++ ×× =+++ =+= 121 121 345345 3918 3918 1622 1 12 369218 641 12 181818 12121111 1818 ++=+++++ ×× =+++ ×× =+++ =+=

3. a. 64642 5555 −==

b. 65146561412 126126 390168 72 222 72 111 36 37 12 12311 3 1212 ×−× −= × = = = = ×+ = = 65146561412 126126 390168 72 222 72 111 36 37 12 12311 3 1212 ×−× −= × = = = = ×+ = = c. 148 1411 11 3243824 11211 2424 11211 24 101 24 2445 5 4 2424 × −=− ×

= = ×+ = = 148 1411 11 3243824 11211 2424 11211 24 101 24 2445 5 4 2424 × −=− × =− = = ×+ = = d. 77 313 73 137137 4939 91 10 91 ×−× −= × = = 77 313 73 137137 4939 91 10 91 ×−× −= × = = e. 451487 5417 4848 2139 48 212 39 428 4239 3 88 ×+×+ −=− =− × =− × == 451487 5417 4848 2139 48 212 39 428 4239 3 88 ×+×+ −=− =− ×

f. 296321 2261 9393 247 93 73 24 933 2421 99 2421 9 31 93 ×+×+ −=− =− × =− × =− = == 296321 2261 9393 247 93 73 24 933 2421 99 2421 9 31 93 ×+×+ −=− =− × =− × =− = == g. 21 1715 714 177214151 7 14 121211 714 1212 211 7214 242211 1414 242211 14 313 2 1414 ×+×+ =

6. Weight of mangoes in a basket initially = 1 2 5 kg

Rajiv puts 2 2 kg 5 more mangoes.

Therefore, the total weight of the mangoes in the basket = 122222 kg12 5555 +=+++ 122222 kg12 5555 +=+++

7. Composition of nitrogen in the atmosphere = 39 50

Composition of oxygen in the atmosphere = 21 100

Total composition of nitrogen and oxygen = 39217821 50100100100 +=+ = 39217821 50100100100 +=+ = 99 100

Composition of other gases = 991 1 100100 −=

8. Total weight of wheat sent to help the flood victims = 1 50 2 kg

Weight of wheat that got spoilt = 3 10 4 kg

Amount of wheat that reached the victims

= 50113 0 24 kg = 10143 24 kg

= 1012 43 224 × × kg

= 20243

44 kg

= 159 4 kg 39 43 3 39 44 ×+ = = kg

Hence, 3 39 4 kg wheat is left in the drum.

9. Answer may vary. Sample answer:

Neeta filled 2 3 4 litres of petrol in her car. After a day, she again filled 1 1 4 litres of petrol. Next day, she went to visit her grandparents and used 3 3 4 of the petrol. How much petrol is left in her car?

Challenge

1. Statement 1: The total weight of apples and pears is 3 1 2 kg.

Statement 2: The weight of litchis bought is 2 1 3 kg.

Checking Statement 1:

Weight of apples = 1 1 2 kg

Weight of pears = 1 2 4 kg

Total weight of apple and pears = 121139 2424 +=+ 121139 2424 +=+ 6915 44

Hence, Statement 1 is false.

Checking Statement 2:

Weight of litchis bought = Total weight of fruits bought − Weight of apples, pears, and oranges bought 3112 7121 4243 31155 443

(as the total weight of apple and pear is 15 4 kg)

Hence, Statement 2 is true. Thus, option b is correct.

Do It Yourself 6B

1. a. Reciprocal of 2 3 = 3 2 b. Reciprocal of 5 9 = 9 5 c. Reciprocal of 5 4 = 4 5 d. Reciprocal of

5. a. 3 2 4

1 1 2 hours is 3 4 of 2 hours b. 1 week = 7 days 2 72 7 ×= 2 days is 2 7 of 1 week.

6. Total weight of the cake = 2 kg

Fraction of cake eaten by Sushen = 3 4

Weight of cake eaten by Sushen = 3 2 4 × kg =

Hence, Sushen ate 3 2 kg cake.

7. Total amount of rice with Ravi = 21 kg

Rice in each packet = 3 4 kg

Total no. of packets packed by Ravi = 3 21 4 ÷

Hence, Ravi packed 28 packets of rice.

8. Total amount of fuel in the tank = 50 L

Fraction of fuel used in the journey = 3 5

Amount of fuel used in the journey = 3 50 5 × L

= 3 510 5 ×× L = 30 L

Fuel left in the tank = 50 L – 30 L = 20 L

Hence, amount of fuel left in the tank is 20 L.

9. Capacity of container = 57 4 L

Capacity of each bottle = 3 4 L

Number of bottles than can be filled from the container = 19 3 573574 4 19 444343 × ÷=×=×=

Hence, 19 bottles can be filled from the container.

10. Answer may vary. Sample answer:

Roy run 4 5 of a km each day for 6 days. How many kms did he run in all?

Challenge

1. Capacity of the tank = 40 litres

Fraction of water in the tank = 4 5

Volume of water in the tank = 4 5 of 40 = 4 4032 5 ×= litres

Given that a quarter of the volume has been used for cooking, which is

= 1 4 of 32 = 1 328 4 ×= litres

Volume of water left in the tank = 32 − 8 = 24 litres

As Nisha requires 28 litres of water but the tank has 24 litres of water remaining, the water is not enough for washing all the utensils.

Chapter Checkup

1. a. Multiplicative inverse of 4 5 = 5 4

b. Multiplicative inverse of 18 17 = 17 18

c. Multiplicative inverse of 78 14 = 14 78

d. Multiplicative inverse of 1 5 = 5

2. a. The reciprocal of every fraction is a proper fraction. False

b. 0 ÷ any fraction = 0. True

c. The reciprocal of 0 is 0. False

d. The multiplicative inverse of any fraction is always greater than 1. False

52 517 17 9189218 1017 1818 39 27 1829 31 1 22

b. × +=+ × =+= 62 621 21 173417234 122133 343434 × +=+ × =+= 62 621 21 173417234 122133 343434

1141 36323

f. 4375 1717 17471735 17 17 7556131 171717 12 7 17 + ×+×+ = + =+= = g. 2524 36 32265 4 834 3636 82 34 326 252 163450 66632 251 8 33

2424 21 2412 e. 231287 3217 2822 723 28 74 23 248 28235 888

494231 4241 9393 407 93 73 40 933 4021 99 191 2 99

h. 21 11 32 312211 32 53 32 5233 32 109 1 66 ×+×+ = =− ×−× = × ==

5. a. 55 6 3 2 933 101 3 33 ×=×× × ==

b. ×=×× =×= 14 14 15 53 55 31442

c. ×=× ×× =×= 2323 983324 111 3412

d. ×=× × =×= 3434 516544 313 5420

6 a. 5525 52 251 12 22 ÷=× == b. 12134 2 43 4 4316 3 ÷=× =××= c 11111 ÷3= × 553 11 = 15

d. 551 ÷18= × 9918 5 = 162

7. a. 4×8+ 12×8 + 1 11 8–2=–4848 3317 =–48 33×2 17 =–4×28 6617 =–88 491 = =6 88

2×4+13×4+1 11 2+4= + 4343 913 =+ 43 9×3+13×4 = 3×4 ==6797 1212

Since the whole number parts are equal, we will compare the proper fraction part of our fraction 7×8 1712 1256 ,= ,=,. 8128×1212×89696

Clearly, 56 96 > 12 96 ⇒ 7 12 > 1 8 ⇒ 7 6 12 > 1 6 8

Thus, the sum of 21 4 and 41 3 difference of 21 8 and 81 4 b. 4 54 5 ×=

1×4+12×5+4 14 1+2= + 4545 514 =+ 45

5×5+14×4 = 4×5 25+56 = 20 ==4811 2020

Clearly, 4 is smaller than 4 1 20 .

Thus, the product of 5 and 4 5 is smaller than the sum of 1 1 4 and 4 2 5 .

c. 2 3 5 ×− 4 1642 5555 ×=−=

8. Fraction for children = 29 100

Fraction for adults = 13 20

Fraction for elderly people = 3 50

People who are not children = Adults + Elderly people

Fraction for people who are not children = 13365671 2050100100100 +=+=

9. Distance travelled by man on bicycle = 1 4 7 km

Distance travelled by man on foot = 3 2 4 km

Distance travelled by man on car = 1 10 2 km

Total distance covered by man = 4210131 742 ++ km = 4714232101 742 ×+×+×+ ++ km

= 291121

742 ++ km

= 212 2911 7422 × ++ × km

= 291142 744 ++ km

= 2953 74 + km

294537 74 ×+× = × km = 487 28 km = 11 17 28 km

Hence, the distance covered by man is 11 17 28 km.

10. Total length constructed in phase 1 = 65 km

Total length constructed in phase 2 = 125 km

Total length constructed in phase 3 = 167 km

The length constructed in phase 4 is 1 1 2 times the length constructed in phase 1.

So, the length of phase 4 = 1 165 2 × 31 6597 22 =×= km

11. Number of cars in the parking lot = 66

Fraction of parking lot occupied = 3 4

Now, by multiplying by 22 in the numerator and denominator, we get 322 3 66 442288 × == ×

Since 66 cars are in the parking lot already, it means a total of 88 cars can be parked.

12. Total number of cans = 14

Total capacity of 14 cans = 1 144 2 litre

Capacity of each can = 1 144 2 ÷ 14 litre = 14421 1 214 ×+ × litre = 2891 214 × litre = 289 28 litre = 9 10 28 litre

Hence, capacity of each can is 9 10 28 litre.

13. Total length of cloth purchased by Radhika = 20 m

Cloth used for curtains = 2 11 5 m

Cloth used for bed sheet = 5 3 6 m

Length of cloth used = 11325 56 + m = 5112365 56 ×+×+ + m

= 5723 56 + m = 576 235 457 5630 ×+× = × m

Length of cloth left = 457 20 m 30

= 20×30–457 14323 = m=4m 303030

Hence, the amount of cloth left is 23 4 30 m.

14. Answers may vary. Sample answer:

A glass of water holds cups of water. How many 1 4 cups will it take to fill the glass?

Challenge

1. Quantity of grate cheese required = 5 6 cup

Quantity we get on grating 6 cheese cubes = 531 ×= 652 cup

Remaining quantity needed = 5153 626 21 cup 63 −= ==

6 cheese cubes fill 1 2 cup.

12 cheese cubes will fill 1 cup.

And 12 4 3 = cheese cubes will fill 1 3 cup.

2. Assertion (A): Shefali has 2000 marbles, out of which 1 4 are red and the rest are green. 2 5 of the red marbles are defective.

The total number of non-defective red marbles is 300.

Reason (R): To add two unlike fractions, first the denominator is made the same by finding the equivalent fractions and then the numerators are added.

Total number of marbles = 2000

Fraction of red marbles = 1 4

Number of red marbles = 1 2000500 4 ×=

Number of defective red marble

= 22 of 500=×500=200 55

Number of non-defective red marble = 500 – 200 = 300 marbles

So, the assertion is true.

The reason explains that to add two unlike fractions, you first make the denominators the same by finding equivalent fractions and then add the numerators. This is a correct and standard procedure for adding fractions with different denominators.

So, the reason is correct. Hence, option b is correct.

Case Study

1. Total time spent = Time spent on Monday + Time spent on Thursday =2+1=+31113 4242 11+6 171 ===4hours 444

Hence, option c is the correct answer.

2. Time spent in the gardening and farming class on Wednesday = 1 2 4

Time spent in the class on Sunday = 1 1 2 hours

Difference in time spent =2–111 42 93 =–42 9–63 ==hours 44

Hence, option b is the correct answer.

3. Time spent by Sonia in animal care and welfare classes = 2 2 5 hours

Time spent by Misha in animal care and welfare classes = 45 minutes less than Sonia.

45 minutes = 3 4 hours

So, time spent by Misha =2–23 54 123 =–54 48–15 = 20 3313 ==1hours 2020

Hence, Misha spent 13 1 20 hours in the Animal care and Welfare classes.

4. Answers may vary.

Chapter 7

Letʹs Warm-up

1. ++= 1113 10101010

2. Each equal part in a hundred grid is equal to 1 100 or hundredths.

3. When a whole is divided into 10 equal parts then each part is equal to 1 10 or tenths.

4. ++++= 11111 1 55555 = 1

5. The fraction for 3 out of 10 = 3 10

Do It Yourself 7A

1. a. 45.5 = forty-five point five or forty-five and five tenths

b. 12.3 = twelve point three or twelve and three tenths

c. 42.14 = forty-two point one four or forty-two and fourteen hundredths

d. 178.64 = one hundred seventy-eight point six four or one hundred seventy-eight and sixty-four hundredths

e. 457.18 = four hundred fifty-seven point one eight or four hundred fifty-seven and eighteen hundredths

f. 25.127 = twenty-five point one two seven or twenty-five and one hundred twenty-seven thousandths

g. 174.201 = one hundred seventy-four point two zero one or one hundred seventy-four and two hundred one thousandths

h. 765.004 = seven hundred sixty-five point zero four or seven hundred sixty-five and four thousandths

2. a. 413.2 b. 72.33 c. 361.04 d. 52.402

3. a.

One point six or one and six tenths b.

One point seven two or one and seventy-two hundredths

4. a. 45.1 = 40 + 5

e. 976.07 = 900 + 70 + 6 + 7 100 or 900 +

f. 15.062

+ 0.001

d. 400 + 80 + 9 + 9 10 + 8 100 + 2 1000 = 400 + 80 + 9 + 0.9 + 0.08 + 0.002 = 489.982

6. a. 9 10 = 0.9 = 90%

b. 8 25 = 8 × 4 25 × 4 = 32 100 = 0.32 = 32%

c. 17 50 = 17 × 2 50 × 2 = 34 100 = 0.34 = 34%

7. a. 1.3 = 13 10 ⇒ 130%

b. 12.54 = 1254 ÷ 2 100 ÷ 2 = 627 50 ⇒ 1254%

c. 32.120 = 32120 1000 = 32120 ÷ 40 1000 ÷ 40 = 803 25 ⇒ 3212%

d. 2.302 = 2302 1000 = 2302 ÷ 2 1000 ÷ 2 = 1151 50 ⇒ 230.2%

8. Answers may vary. Sample answers:

a. 187.2 = 187.20 = 187.200

b. 87.02 = 87.020 = 87.0200

c. 963.14 = 963.140 = 963.1400 d. 189.221 = 189.2210 = 189.22100

9. Fractions Decimals Percentages a. 2 + 1 10 = 20 + 1 10 = 21 10

10. a.

b. 1 and15.2 5 12 12 0.2 55210 × === × 0.2 < 15.2 1 5 < 152

c. 1 and 0.125 8 == 1125 0.125 81000

1 8 < 0.125

d. 7 0.375 and 8 == 7875 0.875 81000 < 0.3750.875

0.375 < 7 8

11. Distance covered by Shivraj Dhananjay Thorat = 455.89 km 455.89 = 400 + 50 + 5 + 8 10 + 9 100 =

12. Work completed by the contractor in 7 days = 5 8 of the total work 5 8 = 5 × 125 8 × 125 = 625 1000 = 0.625

13. Number of tigers in India = 3600

Total number of tigers in the world = 6000

Percentage of the world’s tiger population in India = 3600 6000 × 100 = 60%

Challenge

1. Answers may vary. Sample answers:

We have to find two numbers between 0.9975 and 0.9998. We can only use the digits 9, 0, 8, 9, 7.

We need to create numbers that start with 0.99.

So, the two numbers can be: 0.9978 and 0.9997.

Do It Yourself 7B

1. Like decimals have same number of digits after the decimal.

a. 1.2, 5.4, 8.9, 6.54, 1.3

Not like decimals.

b. 2.23, 4.26, 4.89, 4.2, 6.584

Not like decimals.

c. 7.89, 7.2, 64.594, 45.2, 56.5 Not like decimals.

d. 81.564, 78.512, 453.125, 486.154, 86.15

Not like decimals.

2. a. 4.2 and 6.25  b. 17.23 and 691.56 

c. 11.3 and 17.8  d. 5.157 and 64.581 

3. a. 1.45 and 1.6 1.6 = 1.60 b. 81.566 and 12.2 12.2 = 12.200

c. 17.98 and 14.221 17.98 = 17.980 d. 11.001 and 11 11 = 11.000

4. a. 13.15, 1.2 → 1.2 = 1.20 b. 3.48, 1.2 → 1.2 = 1.20

c. 4.8, 1.526 → 4.8 = 4.800 d. 1.4, 47.584 → 1.4 = 1.400

5. a. 12 20 , 1 5 12 × 5

20 × 5 = 60 100 = 0.6 1 × 2 5 × 2 = 2 10 = 0.2

b. 145 , 4010

1425 350 0.35

40251000 × == ×

== 5 0.50.50 10

c. 53 , 84

b. 12.2 rounds up to 13 when rounded to the nearest whole number. False

5125 5 625 0.625

881251000 × === ×

4425100 × ==== × d. 45 , 520

325 3 75 0.750.750

c. 99.9 rounds up to 100 when rounded to the nearest whole number. True

3. a. 4.2 or 4.15

→ 4.2 = 4.20

Since, 2 > 1

Thus, 4.2 > 4.15

b. 15.64 or 15.67

55210 × ==== × 55 25 0.25

42 48 0.80.80

205100 × == ×

6. Japanese Yen = ₹0.54 = ₹0.540 Canadian Dollar = ₹60.9 = ₹60.900 US Dollar = ₹83.394 0.540, 60.900 and 83.394 are like decimals.

7. Answers may vary. Sample answer: Mahika bought 1.5 L of milk and 3.245 L of curd. Convert both the decimals into like decimals.

Challenge

1. 1 2 = 1 × 5 2 × 5 = 5 10 = 0.5

8 25 = 8 × 4

25 × 4 = 32 100 = 0.32

7 8 = 7 × 125

8 × 125 = 875 1000 = 0.875

29 40 = 29 × 25

40 × 25 = 725 1000 = 0.725

Let us convert them to like decimals,

0.5 = 0.500; 0.32 = 0.320; 0.875 = 0.875; 0.725 = 0.725

Arranging in descending order.

0.875 > 0.725 > 0.500 > 0.320

Thus, 7 8 > 29 40 > 1 2 > 8 25 So, statement 1 and statement 3 are correct. Thus, option (c) is correct.

Do It Yourself 7C 1. a.

2. a. If the digit in the tenths place is 5, then the number rounds up to the nearest whole number. True

Since, 64 < 67

Thus, 15.64 < 15.67

c. 87.654 or 87.65

→ 87.65 = 87.650

Since, 654 > 650

Thus, 87.654 > 87.65

d. 294.98 or 294.864

→ 294.98 = 294.980

Since, 980 > 864

Then, 294.98 > 294.864

4. a. 2.6 lie between 2 and 3.

The digit at tenths place = 6

So, 2.6 rounds up to 3.

b. 7.9 lie between 7 and 8.

The digit at the tenths place = 9 So, 7.9 rounds up to 8.

c. 48.2 lie between 48 and 49.

The digit at the tenths place = 2

So, 48.2 rounds down to 48.

d. 15.1 lie between 15 and 16.

The digit at the tenths place = 1

So, 15.1 rounds down to 15.

5. a. 14.14, 14.1, 14.01, 14.101 14.14 = 14.140, 14.1 = 14.100, 14.01 = 14.010, 14.101

Since, 010 < 100 < 101 < 140

Then, 14.01 < 14.1 < 14.101 < 14.14

b. 84.56, 84.5, 84.6, 84.55, 84.65 84.56; 84.5 = 84.50; 84.6 = 84.60; 84.55; 84.65

Since, 50 < 55 < 56 < 60 < 65

Then, 84.5 < 84.55 < 84.56 < 84.6 < 84.65

c. 184.2, 184.23, 184.1, 184.112

184.2 = 184.200; 184.23 = 184.230; 184.1 = 184.100; 184.112

Since, 100 < 112 < 200 < 230

Then, 184.1 < 184.112 < 184.2 < 184.23

d. 64.23, 54, 64.32, 64.22, 64.33

Since, 22 < 23 < 32 < 33

Then, 54 < 64.22 < 64.23 < 64.32 < 64.33

6. a. 45.2 lie between 45 and 46.

The digit at the tenths place = 2

So, 45.2 rounds down to 45.

45.6 lie between 45 and 46.

The digit at the tenths place = 6

So, 45.6 rounds up to 46.

45 < 46

b. 14.1 lie between 14 and 15.

The digit at the tenths place = 1

So, 14.1 rounds down to 14. 14.2 lie between 14 and 15.

The digit at the tenths place = 2

So, 14.2 rounds down to 14. 14 = 14

c. 87.6 lie between 87 and 88.

The digit at the tenths place = 6

So, 87.6 rounds up to 88.

88.2 lie between 88 and 89.

The digit at the tenths place = 2

So, 88.2 rounds down to 88. 88 = 88

d. 17.3 lie between 17 and 18.

The digit at the tenths place = 3

So, 17.3 rounds down to 17. 17.6 lie between 17 and 18.

The digit at the tenths place = 6

So, 17.6 rounds up to 18. 17 < 18

7. 205.8

The digit at the tenths place = 8 So, 205.8 rounds up to 206.

8. Let the three athletes be: Athlete 1, Athlete 2 and Athlete 3

The race was finished in:

Athlete 1 = 24.54 minutes

Athlete 2 = 24.68 minutes

Athlete 3 = 24.36 minutes

Since, 68 > 54 > 36

Then, 24.68 > 24.54 > 24.36

24.54, 24.68, 24.36

Since, 36 < 54 < 68

Then, 24.36 < 24.54 < 24.68

9. Let us convert the decimals into like decimals.

10.4 = 10.4000

11.23 = 11.2300

6.8548 = 6.8548

19.99 = 19.9900

3.5003 = 3.5003

Arrange in ascending order.

3.5003 < 6.8548 < 10.4000 < 11.2300 < 19.9900

Thus, the states in ascending order according to their population are Telangana, Rajasthan, Bihar, Maharashtra, Uttar Pradesh.

Challenge

1. Statement: a is greater than 14.5. If a is greater than 14.5, it could be any value above 14.5. For example, a could be 14.6, 14.9, 15.0, 15.3 or even 100. For numbers between 14.5 and 15.4, the rounded value to the nearest whole number is 15. For example:

– a could be 14.6, 14.9 or 15.3, which would round to 15.

– If a is 15.5 or higher, it would round to a number greater than 15.

Since the statement does not specify the range of a beyond saying it is greater than 14.5, the value of a could also be much larger (e.g., 100), and this would not round to 15. Therefore, the statement alone cannot answer the given question.

Chapter Checkup

1.2 = 1 + 0.2 = 1 + 2 10

1.48 = 1 + 0.4 + 0.08 = 1 + 4 10 + 8 100

2. a. 15.2 = fifteen and two-tenths or fifteen point two

b. 71.65 = seventy-one and sixty-five tenths or seventy-one point six five

c. 814.36 = eight hundred fourteen and thirty-six hundredths or eight hundred fourteen point three six

d. 176.801 = one hundred seventy-six and eight hundred one thousandths or one hundred seventy-six point eight zero one

4. a. 11.1

b. % 42 8 0.8 52 = 10 80 × == ×

c. % 125 60 0.6 20510 = 0 60 × == ×

d. % 3625 900 0.9 402510 = 00 90 × == ×

6. a. 12.6 1262 126 63 101025 ÷ === ÷ = 1260%

b. 52.2 5222 522 261 101025 ÷ === ÷ = 5220%

c. 20.8 2082 208 104 101025 ÷ === ÷ = 2080%

d. 25.315 253155 25315 5063 100010005200 ÷ == = ÷ = 2531.5%

7. a. 12.3 and 12.4

Since, 3 < 4 Then, 12.3 < 12.4

b. 14.5 and 14.55 → 14.5 = 14.50

Since, 50 < 55 Then, 14.5 < 14.55

c. 222.22 and 222.02

Since, 22 > 0.2 Then, 222.22 > 222.02

d. 3.003 and 3.033

Since, 3 < 33 Then, 3.003 < 3.033

8. a. 1.3 lie between 1 and 2.

The digit at the tenths place = 3

So, 1.3 rounds down to 1.

b. 99.6 lie between 99 and 100.

The digit at the tenths place = 6

So, 99.6 rounds up to 100.

c. 100.2 lie between 100 and 101.

The digit at the tenths place = 2

So, 100.2 rounds down to 100.

d. 999.9 lie between 999 and 1000.

The digit at the tenths place = 9

So, 999.9 rounds up to 1000.

9. a. 1.3, 1.31, 1.2, 1.33

→ 1.3 = 1.30; 1.31; 1.2 = 1.20; 1.33

Since, 33 > 31 > 30 > 20

Then, 1.33 > 1.31 > 1.3 > 1.2

b. 19.4, 19.44, 19.54, 19.501

→ 19.4 = 19.400; 19.44 = 19.440, 19.54 = 19.540, 19.501

Since, 540 > 501 > 440 > 400

Then, 19.54 > 19.501 > 19.44 > 19.4

c. 555.5, 555.55, 555.05, 555.555

→ 555.5 = 555.500; 555.55 = 555.550; 555.05

= 555.050; 555.555

Since, 555 > 550 > 500 > 050

Then, 555.555 > 555.55 > 555.5 > 555.05

d. 748.01, 748.101, 748.11, 748.1

→ 748.01 = 748.010; 748.101; 748.11 = 748.110; 748.1 = 748.100

Since, 110 > 101 > 100 > 010

Then, 748.11 > 748.101 > 748.1 > 748.01

10. Answers may vary. Sample answer: 3.2 3.7

11. Raj = 14.3 minutes = 14.30

Rekha = 15.2 minutes = 15.20

Utkarsh = 13.92 minutes = 13.92

Ali = 13.99 minutes = 13.99

Since, 15 > 14 > 13

For 13.92 and 13.99

92 < 99

So, 13.99 > 13.92

Then, 15.20 > 14.30 > 13.99 > 13.92

So, Utkarsh takes the least time to finish the race; thus, he is the winner.

12. Amount spent by:

Mr Jadeja = $12.99

Sarah = $13.01

Alice = $12.9 = 12.90

Jacob = $13.1 = 13.10

Since, 112 < 13

90 < 99 and 0.1 < 10

Then, 12.9 < 12.99 < 13.01 < 13.1

Challenge

1. The numbers are 0.9, 0.91, 0.93, 0.97, 1, 1.05

0.97 is not a part of the pattern.

The pattern should be 0.9, 0.91, 0.93, 0.96, 1, 1.05.

Thus, 0.97 does not follow the pattern.

2. Assertion (A): The decimal number 0.78 is less than the decimal number 0.7801.

This statement is true. When comparing 0.78 and 0.7801, we align the decimals and compare digit by digit from left to right. Since 0.7801 has additional digits beyond 0.78, and those digits make it larger, 0.78 is indeed less than 0.7801.

Reason (R): When comparing two decimal numbers, the number with more digits to the right of the decimal point is always larger.

This statement is false. The number of digits to the right of the decimal point does not determine the size of the number. What matters is the value of the digits in corresponding places. For example, 0.8 is greater than 0.78, even though it has fewer digits to the right of the decimal point.

Conclusion:

• Assertion (A) is true.

• Reason (R) is false. Therefore, the correct option is (C).

Case Study

1. The shortest height among the given heights is 420.5 metres. So, the shortest skyscraper is Jin Mao Tower. Hence, the correct option is (C).

2. The tallest height among the given heights is 678.9 metres. So, the tallest skyscraper is Merdeka 118. Hence, the correct option is (D).

3. The ascending order of the heights is:

420.5 < 475.6 < 527.7 < 599.1 < 632.0 < 678.9

So, the ascending order of the skyscrapers based on the heights is Jin Mao Tower, Wuhan Greenland Centre, CITIC Tower, Lotte World Tower, Shanghai Tower and Merdeka 118.

4. Heights of the towers when rounded to the nearest whole number.

Merdeka 118 = 679 metres

CITIC Tower = 528 metres

Jin Mao Tower = 421 metres

Lotte World Tower = 599 metres

Wuhan Greenland Centre = 476 metres

Shanghai Tower = 632 metres

5. The expanded form of the heights of the towers will be:

Merdeka 118, 678.9 = 600 + 70 + 8 + 9 10

CITIC Tower, 527.7 = 500 + 20 + 7 + 7 10

Jin Mao Tower, 420.5 = 400 + 20 + 5 10

Lotte World Tower, 599.1 = 500 + 90 + 9 + 1 10

Wuhan Greenland Centre, 475.6 = 400 + 70 + 5 + 6 10

Shanghai Tower, 632.0 = 600 + 30 + 2

Chapter 8

Letʼs Warm-up

1. 1.14 < 1.41

Do It Yourself 8A

5. a. 45.2 + 32.9 = 78.1 < 79.1

b. 62.1 + 81.009 = 143.109 > 143.009

c. 47.8 – 36.6 = 11.2

d. 125.6 – 84.934 = 40.666 > 35.666

6. Milk yielded by Gir cow = 13.4 litres

Milk yielded by Sahiwal cow = 8.55 litres

Extra milk yielded by Gir cow than Sahiwal cow = 13.4 – 8.55

Thus, Gir cow yields 4.85 litres of more milk than Sahiwal cow.

7. Raman’s height = 1.65 m

Difference between Raman’s height and his brother’s height = 0.18 m

The brother’s height = 1.65 − 0.18 = 1.47 O . t h 1 . 6 5 0 . 1 8 1 4 7

8. Weight of frog = 3.3 kg

Weight of tadpole = 0.005 kg

Total weight of frog and tadpole = 3.3 kg + 0.005 kg = 3.305 kg

Thus, the total weight of frog and tadpole is 3.305 kg.

3 . 3 0 0

+ 0 . 0 0 5

3 3 0 5

9. The distance covered by Rashi = 1.548 km

The distance covered by Prerna = 2.328 km

The distance covered by Navya = 1.986 km

The total distance covered by them = 1.548 + 2.328 + 1.986 = 5.862 km

O t h th

2 1 1

1 . 5 4 8

2 3 2 8

+ 1 9 8 6

5 8 6 2

10. a. The total weight of Rakesh and Ismail (in kg) = 34.53 + 36.35 = 70.88 kg

T O . t h

3 4 5 3

+ 3 6 . 3 5

7 0 8 8

b. The total weight of John and Shahid (in kg) = 43.5 + 40.235 = 83.735 kg

T O . t h th

4 3 . 5 0 0

+ 4 0 . 2 3 5

8 3 . 7 3 5

11. The distance between points A and B (in km) = 1.2

The distance between point B and C (in km) = 2.35

The distance between points A and C (in km) = 1.2 + 2.35 = 3.55

The total distance between points A and D (in km) = 7

The distance between points C and D (in km) = 7 − 3.55 = 3.45

O t h

7 0 0

3 5 5

3 . 4 5

Thus, the distance between C and D is 3.45 km.

12. The total money Rashmi has = ₹5555.5

The amount spent on the saree = ₹555.5

The amount spent on an umbrella = ₹55.5

The amount given to the auto driver = ₹5.5

The total amount spent = ₹555.5 + ₹55.5 + ₹5.5 = ₹616.5

The total amount left with her = ₹5555.5 − ₹616.5 = ₹4939

Th H T O . t

5 5 5 5 . 5

6 1 6 5

4 9 3 9 0

Challenge

1. The sum of the numbers = 10

The first number is greater than 4.

The second number is greater than 5.

It is given that both the numbers have only tenths place.

Let the first number = 4.1, then the second number = 5.9; 4.1 + 5.9 = 10

First number = 4.2, second number = 5.8; 4.2 + 5.8 = 10

Similarly, first number can be 4.3, 4.4, 4.5, 4.6, 4.7, 4.8 or 4.9, and the corresponding second number will be 5.7, 5.6, 5.5, 5.4, 5.3, 5.2 or 5.1

When the first number = 5.0, the second number = 5.0. In this case, the second number is not greater than 5.

Thus, the total number of combination that Rahul can find is 9.

Do It Yourself 8B

1. a. b.

0.2 × 4 = 0.8 0.4 × 5 = 2.0

2. a. 45.6 × 10 = 456

b. 15.478 × 100 = 1547.8

c. 63.157 × 1000 = 63,157

d. 4 × 2.5 = 10 2 5 × 4 1 0 0

e. 12 × 14.945 = 179.34 1 4 9 4 5 × 1 2 2 9 8 9 0 + 1 4 9 4 5 0 1 7 9 3 4 0

f. 12.3 × 35.69 = 438.987 3 5 6 9 × 1 2 3 1 0 7 0 7 7 1 3 8 0 + 3 5 6 9 0 0 4 3 8 9 8 7

3. a. 5.15 × 10 = 51.5 < 515

b. 83.482 × 1000 = 83482

c. 6 × 1.658 = 9.948 < 10.948

d. 715.15 × 8 = 5721.2 > 5720.12

e. 2.13 × 31.4 = 66.882 > 6.5882

f. 1.5 × 634.71 = 952.065 > 94.2065

4. a. 12.15 × 10 = 121.5

b. 15.26 × 1000 = 15260

c. 61.235 × 10 = 612.35

d. 14.1 × 1000 = 14100

e. 6 × 1.645 = 9.87

f. 1.1 × 47.503 = 52.2533

5. 1 Euro = ₹89.19

1000 Euros = ₹89.19 × 1000 = ₹89,190 8 9 1 9 × 1 0 0 0

8 9 1 9 0

6. Number thought by Matthew = 4.6 Six times 4.6 = 6 × 4.6 = 27.6 4 6 × 6

2 7 6

7. The total number of workers in the factory = 24

The amount of money given to each of 11 workers = ₹252.54

The total amount of money given to 11 workers = 252.54 × 11 = ₹2777.94

2 5 2 5 4 × 1 1

2 7 7 7 9 4

The amount of money given to each of 13 workers = ₹364.52

The total amount of money given to 13 workers = 364.52 × 13 = ₹4738.76

3 6 4 5 2 × 1 3

4 7 3 8 7 6

The total amount of money given to the workers = ₹2777.94 + ₹4738.76 = ₹7516.7

8. The salary of Suresh’s father = 800 Dirham

The salary of Saurav’s father = 2200 Sri Lankan rupees.

Since, 1 Dirham = ₹22.51 and 1 LKR = 0.25

Therefore, the salary earned by Suresh’s father (in ₹)

= 800 × ₹22.51 = 18,008.00

2 2 5 1

× 8 0 0

1 8 0 0 8

The salary earned by Saurav’s father (in ₹) = 2200 × ₹0.25 = 550

2 2 0 0 × 2 5

5 5 0 0 0

₹18008 − ₹550 = ₹17,458

Thus, Suresh’s father gets ₹17,458 more than Saurav’s father.

9. Total number of dustbins ordered = 96

Cost of each dustbin = ₹123.65

Total cost of 96 dustbins = ₹123.65 × 96

Multiply 12365 and 96 12365 × 96 = 1187040

So, 123.65 × 96 = 11870.4

Thus, the total cost of dustbins is ₹11,870.4.

10. 1 Pound = ₹104.06

2 Pounds = 2 × ₹104.06 = ₹208.12

1 Rand = ₹4.43

5 Rands = 5 × ₹4.43 = ₹22.15

1 US dollar = ₹82.67

6 US dollars = 6 × ₹82.67 = ₹496.02

1 Euro = ₹89.19

5 Euro = 5 × ₹89.19 = ₹445.95

1 Yuan = ₹11.51

15 Yuan = 15 × ₹11.51 = ₹172.65

The total money that Anjali has (in Indian Rupees) = ₹208.12 + ₹22.15 + ₹496.02 + ₹445.95 + ₹172.65

= ₹1344.89

Challenge

1. Given that the number thought by both Rashmi and Sneha is between 1 and 5, and the number thought by Sneha is a whole number, so the numbers can be 2, 3 and 4.

The quotient of the division of the numbers is 1.25

Also, the number thought by Sneha has 2 decimal places. The possible numbers can be found by multiplying the whole number and the quotient.

1.25 × 2 = 2.5 (the number has only one decimal place, so it cannot be the number)

1.25 × 3 = 3.75 (the number has two decimal places, so it cannot be the number)

1.25 × 4 = 5 (the number has no decimal place and also equal to 5, so it cannot be the number)

Hence, the only possible numbers that Rashmi and Sneha has thought of are 3 and 3.75.

Do

Yourself

a. 0.8 ÷ 4 = 0.2

1.6 ÷ 8 = 0.2

2. a. 15.6 ÷ 10 = 1.56

b. 51.23 ÷ 100 = 0.5123

c. 1.2 ÷ 100 = 0.012

d. 32.561 ÷ 1000 = 0.032561

e. 23.1 ÷ 1000 = 0.0231

f. 2 ÷ 10 = 0.2

4. a. 13.25 ÷ 5 = 2.65

b. 33.04 ÷ 4 = 8.26 > 8.24

c. 202.08 ÷ 8 = 25.26 < 25.28

d. 150.1 ÷ 5 = 30.02

e. 378.18 ÷ 9 = 41.52 < 42.04

f. 378.18 ÷ 9 = 42.02 < 42.04

5. Hariʹs trip expenses = ₹1,04,200 1 pound = ₹104.2

So, ₹1 = 1 104.2 pound = 10 1042 pound

₹1,04,200 = 1,04,200 × 10 1042 pounds = 1000 pounds

So, Hari should take 1000 pounds with him to England.

6. The weight of one carton = 37.8 kg ÷ 9 cartons 4 . 2 9 3 7 . 8 3 6

The weight of one carton = 4.2 kg Therefore, the weight of one carton is 4.2 kg.

7. The cost of a litre of oil = $37.92 ÷ 6 litres

6 . 3 2

6 3 7 . 9 2 3 6 1 0 1 9 1 8 1 2 1 2 0 0

The cost of a litre of oil = $6.32

Therefore, the cost of a litre of oil is $6.32. Yes, Sakshi should buy the bottles.

8. Answers may vary. Sample answer:

A community centre received a donation of 47.5 litres of hand sanitizer to be distributed equally among 5 local schools. How much hand sanitizer will each school receive to ensure an equal distribution?

Challenge

1. 1 pound = ₹104.06

1000 pounds = 1000 × ₹104.06 = ₹104,060

We know that,

1 Euro = ₹89.19

₹104,060 = 104060 ÷ 89.19 = 1166.72 euros

Thus, Mr Agarwal will get 1166.72 euros in exchange of 1000 pounds.

Chapter Checkup 1. a. 0.23 + 0.47 = 0.7

18.963 + 14.517 = 33.48

T O . t h th 1 8 . 9 6 3

e. 534.115 − 15.36 = 518.755

4. a. The decimal point has been shifted one place to the right, hence, we have to multiply the given number by 10.

45.12 × 10 = 451.2

b. The decimal point has been shifted three places to the right, hence, we have to multiply the given number by 1000.

184.351 × 1000 = 184351

c. The decimal point has been shifted two places to the right, hence, we have to multiply the given number by 100.

2.517 × 100 = 251.7

d. The decimal point has been shifted one place to the right, hence, we have to multiply the given number by 10.

2.002 × 10 = 20.02

e. The decimal point has been shifted three places to the right, hence, we have to multiply the given number by 1000.

81.36 × 1000 = 81360

f. The decimal point has been shifted three places to the right, hence, we have to multiply the given number by 1000.

1.001 × 1000 = 1001

5. a. 47.01 ÷ 10 = 4.701 b. 531.14 ÷ 100 = 5.3114

c. 8143.12 ÷ 1000 = 8.14312 d. 1.21 ÷ 1000 = 0.00121

e. 14.2 ÷ 1000 = 0.0142 f. 123.321 ÷ 100 = 1.23321

6. a. 12 × 1.54 =

8. a. 12.145 + 18.415 + 2.51 = 33.07 T O . t h th 1 2 . 1 4 5 0 2 . 5 1

−

c. 15.47 + 81.415 − 41.555 = 55.33 H T O t h th 8 1 4 1 5

31.23 + 17.28 − 11.111 =

e. 15.145 + 81.1 − 45.64 = 50.605 H T O . t h th 8 1 . 1 0 0 1 5 1 4

f. 71.64 + 16.87 − 63.998 = 24.512

9. a. 1.51 + 23.72

b. 27.63 − 2.78

c. 4 × 6.17

d. 3.26 × 7.02

10.

11. a. 1 Won = ₹0.062 6000 Won = ₹0.062 × 6000 = ₹372 b. 1 Euro = ₹89.19 150 Euro = ₹89.19 × €150 = ₹13,378.5 c. 1 Rand = ₹4.43

12. Change = ₹50 − ₹45.8 = ₹4.2 H T O . t h th 15 9 9 10

Therefore, Jason will get back ₹4.2 in change from the shopkeeper.

13. Total cost = $47.85 + $21.36 + $22.01

Total cost = $91.22

Money required = $91.22 − $90

Money required = $1.22

Therefore, Raj requires an additional $1.22 to buy all the items.

14. For every $20, George’s friend gives $1 tax.

So, the total money that George’s friend can get for exchanging into Indian currency = $168 × $20 $20 +$1 = $168 × $20 $21 = $160

Then we multiply the amount of remaining currency by the exchange rate to find the amount of Indian currency Georgeʹs friend has.

Indian currency = $160 × ₹89.12

Indian currency = ₹14,259.2

Therefore, Georgeʹs friend has ₹14,259.2 in Indian currency.

15. Petrol consumed in one hour = 6.5 L

Petrol consumed in 4 hours = 6.5 L × 4 = 26 L

Total petrol consumption = 130 26 L

Number of days = 5 5

Therefore, the car runs for a total of 5 days.

16. Answers may vary. Sample answer: John’s car consumes 0.08 gallons of fuel per mile. If he drives 45.7 miles, how many gallons of fuel does he use?

Challenge

1. Harry’s weight = 51.25 kg × 0.56 = 28.7 kg

5 1 2 5 × 5 6

3 0 7 5 0

+ 2 5 6 2 5 0

2 8 7 0 0 0

John’s weight = 28.7 kg × 0.89 = 25.543 kg

2 8 7 × 8 9

2 5 8 3

+ 2 2 9 6 0

2 5 5 4 3

Therefore, John’s weight is 25.543 kg.

2. When multiplying two decimal numbers, the product is not always less than both of the original numbers. If we multiply 1.2 by 0.5, the product is 0.6, which is smaller than 1.2, but bigger than 0.5.

Multiplying any two numbers between 0 and 1 results in a number that is smaller than each of the original numbers. For example, 0.5 × 0.4 = 0.2, which is smaller than both 0.5 and 0.4.

Hence, option (d). A is false, but R is true.

Case Study

1. Rainfall in Mumbai = 2286.3 mm

Rainfall in Delhi = 714.5 mm

Rainfall in Bangalore = 970.6 mm

Total rainfall in Mumbai, Delhi and Bangalore = 2286.3 + 714.5 + 970.6

= 3971.4 mm

So, the correct option is d.

2. Rainfall in Chennai = 1391.2 mm

Rainfall in Kolkata = 1647.8 mm

Difference between the rainfall in Chennai and Kolkata = 1647.8 – 1391.2

= 256.6 mm

So, the correct option is a.

3. Rainfall in Pune = 722.3 mm

Rainfall in Pune next year = Double the rainfall in 2023

= 2 × 722.3 mm

= 1444.6 mm

4. Combined rainfall of Delhi, Bangalore and Chennai = 714.5 + 970.6 + 1391.2

= 3076.3 mm

Rainfall in Mumbai = 2286.3 mm

Difference between the rainfall in Mumbai with the total rainfall of Delhi, Bangalore and Chennai combined = 3076.3 – 2286.3

= 790 mm

5. Adding 100 mm to each city’s rainfall.

City Mumbai Delhi Bangalore Chennai Kolkata Pune

TotalRainfall Hint:Average= Totalnumberofcities

Total rainfall in all the cities = 2386.3 + 714.5 + 970.6 + 1391.2 + 1647.8 + 722.3 = 7832.7 mm

Total number of cities = 6

Average rainfall = 7832.7 6 = 1305.45 mm

6. Answers may vary.

Chapter 9 Let’s Warm-up

Intersecting lines

Parallel lines

Ray

2. a. A line segment AB can be denoted as .AB False

b. A ray CD can be denoted as .CD False

c. A line PQ can be denoted as PQ

True

d. Two parallel lines are denoted by the symbol ll. True

Line segment – The bridge shown above has 2 endpoints and is a straight line. A bridge Sides of a railway track

Parallel lines – The sides of the railway track do not meet at any point.

Tip of a nail Sunlight

Point – The tip of a nail is equivalent to a point. Ray – The sunlight provides us light in the form of rays.

4. a. M b. M N

c. P Q

d. S T e. U X O V W

5. a. Points: A, B, C, D, E, F Line: ,, ACCBAB

b. Answers may vary. Sample answer: Line segment: FD

c. 1 ray; BE

6. The pair of swords intersect at a point and hence, they show the pair of intersecting lines.

Challenge

1. True, a line segment has two endpoints, and only one line segment can be drawn from the endpoints. P Q

2. a. Intersecting lines

Rani

Rishi

b. Parallel lines

Rani

Rishi

Do It Yourself 9B

1. Answers may vary. Sample answer:

3. ∠1 and ∠3 have angles greater than 90⁰. ∠1 and ∠3 are obtuse angles.

2. a. ∠PQR, ∠RQP or ∠Q

b. ∠LMN, ∠NML or ∠M

c. ∠XYZ, ∠ZYX or ∠Y

Acute angle Obtuse angle

Straight angle Obtuse angle

5. The time at which Rishi’s school starts = 7:30 a.m. The time at which Rishi’s school gets over = 1:30 p.m.

Acute angle Obtuse angle

The angle formed at the time when Rishi’s school starts—7:30 a.m., is an acute angle. The angle formed at the time when Rishi’s school ends—1:30 a.m., is an obtuse angle.

6. The water molecule formed by the snowflake is in the shape of a regular hexagon. The angles formed in the regular hexagon are acute angles and obtuse angles.

Challenge

1. a. When sticks OC, OD, OF, OG and OA are removed, the angle formed is greater than 90°. So, the angle formed is an obtuse angle.

b. The angle formed by sticks OB and OG is equal to 180°. So, the angle formed is a straight angle.

c. The angles formed by OG and OE are ∠GOF, ∠FOE and ∠GOE. All three angles are acute angles.

Do It Yourself 9C

1. a. Acute angle b. Obtuse angle

c. Right angle d. Acute angle

2. a. ∠MNO = 74° b. ∠XYZ = 115°

c. ∠KLM = 152° d. ∠PQR = 120°

3. a. ∠AOF = 53° b. ∠AOE = 90° c. ∠DOA = 123°

d. ∠COB = 32° e. ∠DOB = 57° f. ∠FOB = 127°

Challenge

1. The angle formed by the striker when it hits the coin through the centre forms a straight line. The measure of the straight angle is 180°.

Twice the angles of triangle = 2 × 3 = 6

So, the shape with 6 sides and 6 angles is a hexagon.

6. Drawings may vary.

Do It Yourself 9D

1. a. A quadrilateral or a parallelogram

b. A nonagon

c. A hexagon

d. A decagon

2. a.

3. a. A quadrilateral has 4 sides. True

b. An octagon has eight angles. True

c. A pentagon has five sides. True

d. A rhombus has four sides. False

4. The shape with 8 sides and 8 angles is an octagon.

5. Twice the sides of triangle = 2 × 3 = 6

P R

The sides of the triangle are PQ, PR and RQ.

Challenge

1. There are 6 quadrilaterals and 24 triangles in the given image.

Number of Quadrilaterals

Chapter Checkup

Number of Triangles

1. a. Since the diagram has arrows on both ends, it represents a line.

b. Since the diagram has an arrow on one end, it represents a ray.

c. Since the diagram has no arrows, it represents a line segment.

d. Since the diagram has no arrows, it represents a line segment.

2. a. AC is a line segment. True

b. PQ and RS are sets of parallel lines. False

c. CS is a line. False

An acute angle

A right angle

A straight angle An obtuse angle

4. a. ∠AOC = acute angle

b. ∠AOD = right

c. ∠AOF = straight

d. ∠BOF = obtuse

e. ∠BOC = acute

5. a.

11. The zebra crossings are a type of parallel lines.

12. The shape has 6 sides. So, the matrix is in the shape of a hexagon.

13. Answer may vary. Sample answer:

Challenge

1. Statement I: If two lines are perpendicular, then they intersect at a right angle. This is true. So, statement I is correct. Statement II: Perpendicular lines are lines that intersect at a 90° angle, forming four right angles at their intersection point. This is also true. So, statement II is also true. Thus, both the statements are true. Hence, option (a) is correct.

Case Study

1. Option c

There are 4 triangles in the figure.

2. Option b

When 2 right triangles and parallelogram are joined a rectangle is formed.

3. Option c

A circle can never be formed using tangram as it is made of straight lines and circle needs curves.

4. Answer may vary. Sample answer:

8. a. Quadrilateral; 4 sides, 4 angles

b. Heptagon; 7 sides, 7 angles

c. Quadrilateral; 4 sides, 4 angles

d. Octagon; 8 sides, 8 angles

9. a.

Warm-up

Do It Yourself 10A

a. It is a pattern. b. It is a pattern. c. It is not a pattern.

It is not a pattern.

The pattern is rotating 45° clockwise. b. The pattern is rotating 90° clockwise.

c. The given pattern is not following any rule as the first and second figure have the left upper one-fourth portion shaded as yellow, but the rest colours differ in the two figures.

5. It is given that the pattern made by Rakesh is rotating in half turns. The given pattern satisfies this condition. Hence, it is the pattern made by Rakesh.

6. Answers may vary. Sample answers:

The pattern is rotating 180° anti-clockwise or the pattern is rotating 180° clockwise.

Challenge

1. The given shape follows a 90°, 180°, 90°, 180° clockwise rotation. Hence, the next term would be—

Do It Yourself 10B

1. Option b is not a tiling pattern because there are gaps and overlaps.

3. a. The given pattern is increasing by one at each step, hence, the next in the pattern would be:

b. The given pattern is increasing by one at each step, hence, the next in the pattern would be:

c. The given pattern is increasing by two units at each step, hence, the next in the pattern would be:

d. The given pattern is decreasing by three at each step, hence, the next in the pattern would be:

4. The given tile does not belong to the tiling pattern because in the given tiling pattern all the stars are of red, yellow and green colour. Thus, option b is correct.

Figures may vary. Sample figure:

5. Option b, because it shows the same tiling pattern as bought by Ankit.

Flowers in fourth week =

Flowers in fifth week =

Challenge

1. The number of sides in the outer shape is increasing by 1 and that of the inner shape is decreasing by 1. Hence, the next unit would be—

Do It Yourself 10C

1. a. 15, 19, 23, 27, 31, 35

+ 4 + 4 + 4 + 4 + 4

Adding 4 to each term.

b. 6, 8, 10, 12, 14, 16, 18

+ 2 + 2 + 2 + 2 + 2 + 2

Adding 2 to each term.

c. 1, 4, 9, 16, 25, 36, 49 12 22 32 42 52 62 72

The pattern has squares of natural numbers. 12, 22, 32, 42, 52, 62, 72

2. a. H U E H U

5 3 4 5 3

b. I O Q C B

8 9 0 1 2

c. C E H B U

1 4 5 2 3

3. Flowers in first week =

Flowers in second week =

Flowers in third week =

4. Orders in January = 12

Orders in February = 24

Orders in March = 36

Do

Challenge

5. Answer may vary. Sample answer: 6 1 5 4 2 3 12

Therefore, orders in July = 12 × 7 = 84

Therefore, there were 5 red and 5 yellow flowers after 5 weeks.

The number pattern is: 12, 24, 36, … = 12 × 1, 12 × 2, 12 × 3, …

1. The inner value is 2 added to the sum of product of the column numbers.

(6 × 4) + (2 × 2) + 2 = 30

(3 × 2) + (4 × 3) + 2 = 20

(3 × 2) + (1 × 6) + 2 = 14

So, answer is 14.

Yourself 10D

1. a. The lines of symmetry for the figure can be given as:

Hence, the shape has 2 lines of symmetry.

b. The lines of symmetry for the figure cannot be given.

Hence, the given figure does not have any line of symmetry. It is asymmetrical.

c. The lines of symmetry for the figure can be given as:

Hence, the shape has 1 horizontal line of symmetry.

d. The lines of symmetry for the figure can be given as:

Hence, the shape has many lines of symmetry

d No line of symmetry.

Asymmetrical

4. a. When two halves look identical, they are called asymmetrical. False

b. The letters F, G, J and L have two lines of symmetry. False

c. The number 8 has 2 lines of symmetry. True

d. The number 3 has a horizontal line of symmetry. True

5. a. b.

7. Answer may vary. Sample answer: Here is the reflection of a comb and a coffee mug.

Challenge

1. If the 2 flats are identical, then they are the mirror image of each other. Chapter Checkup

L is equal to 2

E is equal to 1 M is equal to 4

O is equal to 16 N is equal to 8

4. According to the given code language:

So, the code for the word LEMON is 2 1 4 16 8.

5. a. It is a reducing pattern with one line decreasing each time. The next two figures can be given as:

b. It is a rotating pattern rotating 90° anti-clockwise. The next two figures can be given as:

c. It is an increasing pattern with a box increasing each time. The next two figures can be given as:

7. 331, 316, 301, 286, 271, , , − 15 − 15 − 15 − 15 − 15 − 15 − 15

So, the next three terms are:

271 − 15 = 256

256 − 15 = 241

241 − 15 = 226

8. a. First term = 1

Second term = 1 + 2 = 3

Third term = 3 + 6 = 9

Fourth term = 9 + 18 = 27

Fifth term = 27 + 54 = 81

Pattern: 1, 3, 9, 27, 81

b. First term = 15

Second term = 15 + 3 = 18

Third term = 18 + 5 = 23

Fourth term = 23 + 3 = 26

Fifth term = 26 + 5 = 31

Pattern: 15, 18, 23, 26, 31

9. a. The given number pattern is increasing by 2, 3, 4, 5, … each time. Hence, the missing numbers will be:

51 + 6 = 57; 57 + 7 = 64; 64 + 8 = 72; 72 + 9 = 81; 81 + 10 = 91

The pattern will be: 37, 39, 42, 46, 51, 57, 64, 72, 81, 91

b. The given number pattern is decreasing by 1, 2, 3, 4, … each time. Hence, the missing numbers will be:

80 − 5 = 75; 75 − 6 = 69; 69 − 7 = 62; 62 − 8 = 54; 54 − 9 = 45

The pattern will be: 90, 89, 87, 84, 80, 75, 69, 62, 54, 45

No, when you cut this house vertically in half, the both halves would not look the same.

11. Patterns in cells: 2, 4, 8, 16, 32, 64, … a. b.

12. The cells start with 2.

The next cells in the pattern are 4, 8, 16, 32, 64.

Every time, the previous number is multiplied by 2 to get the next number.

The next three numbers in the sequence will be:

64 × 2 = 128

128 × 2 = 256

256 × 2 = 512

Thus, the next three numbers in the pattern are 128, 256, 512.

Rule: Multiplication by 2.

13. Pocket money saved: 20, 40, 80, 160, ...

January = ₹20

February = ₹40 = ₹20 × 2

March = ₹80 = ₹40 × 2

April = ₹160 = ₹80 × 2

The pocket money saved in each month is double of the previous month. Therefore, the pocket money saved in:

May = ₹160 × 2 = ₹320

June = ₹320 × 2 = ₹640

July = ₹640 × 2 = ₹1280

August = ₹1280 × 2 = ₹2560

Therefore, Mihir saved ₹2560 in August.

14. Steps taken on the first day = 100

Steps taken on the second day = 120 = 100 + 20

Steps taken on the third day = 150 = 120 + 30

Steps taken on the fourth day = 190 = 150 + 40

Steps taken on the fifth day = 240 = 190 + 50

Steps taken on the sixth day = 300 + 240 = 60

So, Emily will take 300 steps on the sixth day.

Challenge

1. Observing the pattern—

The

2. 20 25 32 41 52 65

+5 +7 +9 +11 +13

So, Tara will take 65 steps on the sixth day. Hence, the assertion is correct.

From the given steps, the pattern of increase is not a constant 5 steps each day. The increase is changing by 2 steps more each day. Hence, the reason is false. Thus, option c is correct.

A is true, but R is false.

Case Study

1. b. Rectangle

2. False—The minarets have reflectional symmetry vertically.

3. reflectional symmetry.

4. No

5. Answers may vary. Sample answer: Dispose of trash in designated bins, and avoid bringing items that can cause litter.

Chapter 11

Let's Warm-up

1. A laptop is a small object. So, cm will be used to measure it.

2. A table is a long object. So, m will be used to measure it.

3. A coffee cup is a light object. So, g will be used to measure it.

4. A chair is a heavy object. So, kg will be used to measure it.

5. The height of a plant is small. So, cm will be used to measure it.

Do It Yourself 11A

1. Answers may vary Sample answers:

a. Pencil 15 cm 13 cm 15 13 = 2 cm

b. Eraser 3

c. Sharpener

d. Book

2. a. 5.5 cm b. 2.7 cm c. 9 cm d. 4.5 cm

3. a. 10 mm = 1 cm; 1 mm = 1 10 cm

45 mm = 1 × 10 45 = 45 =4.5 10 cm

b. 1000 mm = 1 m; 1 mm = 1 1000 m

547 mm = 1 ×547 1000 547 ==0.547 1000 m

c. 1000 m = 1 km; 1 m = 1 1000 km

1056 m = 1 ×1056 1000 1056 ==1.056 1000 km

d. 10 dm = 1 m; 1 dm = 1 10 m

6892 dm = 1 × 10 6892

= 6892 =689.2 10 m

e. 1 hm = 100 m 2.034 hm = 2.034 × 100 = 203.4 m

f. 10 hm = 1 km; 1 hm = 1 10 km

7.698 hm = 1 × 10 7.698 = 7.698 =0.7698 10 km

4. a 45 m 10 cm = 45 m + 10 100 m = 45 m + 0.1 m = 45.1 m

b. 16 m 80 mm = 16 × 1000 mm + 80 mm = 16000 mm + 80 mm = 16080 mm

c. 280.5 m = 280 m 5 dm

5. The length of the pencil = 13 cm 3 cm = 10 cm

1,00,000 cm = 1 km; 1 cm = 1 100000 km

10 cm = 10 100000 km = 0.0001 km

6. The cloth required for 1 skirt = 3 m 586 cm = 3 m + 500 cm + 86 cm = 3 m + 5 m + 86 cm = 8 m + 86 cm = 8.86 m

The cloth required for making 6 skirts = 6 ⨯ 8.86 m = 53.16 m = 53 m 16 cm

7. The distance travelled on the bicycle = 2 km 578 m = 2000 m + 578 m = 2578 m

The distance travelled on the bus = 21 km 870 m = 21,000 m + 870 m = 21870 m

The distance travelled on foot = 1 km 346 m = 1000 m + 346 m = 1346 m

The total distance travelled = 2578 m + 21870 m + 1346 m = 25794 m = 25000 + 794 m = 25 km 794 m

8. The length of the door = 2 m 1 dm = 2 ⨯ 10 dm + 1 dm = 20 + 1 = 21 dm

The length of the wall = 3 m 2 dm = 3 ⨯ 10 dm + 2 dm = 30 + 2 = 32 dm

The total length = 21 dm + 32 dm = 53 dm

9. Answer may vary. Sample answer:

The length of pipe A is 67 m 25 cm and the length of pipe B is 87 m 30 cm. How much more is the length of pipe B than length of pipe A?

Challenge

1. Number of days in a week = 7

Cardboard sheet chewed in 1 day = 2 cm

Cardboard sheet chewed in 7 days = 7 × 2 cm = 14 cm

So, 7 cm of the cardboard sheet was eaten in 1 week.

1. a.

2. a. 1000 g = 1 kg; 1g = 1 1000 kg

79 g = 1 × 1000 79 = 79 1000 = 0.079 kg

b. 1 g = 1000 mg

975 g = 975 ⨯ 1000 = 9,75,000 mg

c. 1 kg = 1000 g

4677 kg = 4677 ⨯ 1000 = 46,77,000 g

d. 10 dg = 1 g; 1 1 dg= 10 g

1655 dg = 1 × 10 1655

1655 = 10 = 165.5 g

e. 100 cg = 1 g; 1 cg = 1 100 g

6876 cg = 1 × 100 6876

6876 = 100 = 68.76 g

f. 1 g = 1000 mg; 1 dag = 10,000 mg

390 g 45 dag = 390 ⨯ 1000 + 45 × 10,000 = 3,90,000 + 4,50,000 = 8,40,000 mg

3. a. The bigger unit: 6 kg + 10 1000 kg = 6 kg + 0.01 kg = 6.01 kg; The smaller unit: 6 × 1000 g + 10 g = 6000 g + 10 g = 6010 g

b. The bigger unit: 16 g + 80 1000 g = 16 g + 0.08 g = 16.08 g;

The smaller unit: 16 × 1000 mg + 80 mg = 16,000 mg + 80 mg = 16,080 mg

c. 547 kg 6 g

The bigger unit: 547 kg + 6 1000 g = 547 kg + 0.006 = 547.006 kg; The smaller unit: 547.006 × 1000 = 5,47,006 mg

d. 3 g 8 cg

The bigger unit: 3 g + 8 100 g = 3 g + 0.08 g = 3.08 g;

The smaller unit: 3 × 100 cg + 8 cg = 300 cg + 8 cg = 308 cg

e. 87 kg 6 dag

The bigger unit: 87 kg + 6 100 kg = 87 kg + 0.06 kg = 87.06 kg;

The smaller unit: 87 × 100 dag + 6 dag = 8700 dag + 6 dag = 8706 dag

f. 12 hg 42 g

The bigger unit: 12 hg + 42 100 hg = 12 hg + 0.42 hg = 12.42 hg;

The smaller unit: 12 × 100 g + 42 g = 1200 g + 42 g = 1242 g

4. The weight of a four-month-old kitten = 1688 g = 1688 ÷ 1000 kg = 1.688 kg

5. The weight of an egg = 40 g

The number of eggs in 1 kg = 1000 g 1000 = 40 = 25 eggs

6. The weight of 8 cartons = 32 kg 448 g = 32,448 g

The weight of 1 carton = 32448 8 = 4056 g = 4 kg 56 g

7. Weight of apples = 2 kg 450 g

Weight of guavas = 1 kg 547 g

Weight of pears = 2 kg 136 g

Total weight of the fruits = 2 kg 450 g + 1 kg 547 g + 2 kg 136 g = 5 kg 1133 g = 5 kg + 1000 g + 133 g = 6 kg 133 g

8. The weight of two chairs = 16 kg 400 g

The weight of one chair = 10 kg 300 g

The weight of the other chair = 16 kg 400 g 10 kg 300 g

= (16 10 kg) + (400 300 g) = 6 kg 100 g

9. The weight of one conical flask = 260 g

The weight of 4 conical flasks = 4 × 260 g = 1040 g

The weight of one beaker = 150 g

The weight of 3 beakers = 3 × 150 g = 450 g

a. The total weight carried = 1040 g + 450 = 1490 g = 1 kg 490 g

b. The weight of the remaining conical flasks = 3 × 260 g = 780 g

The weight of the remaining beakers = 2 × 150 g = 300 g

The total weight = 780 g + 300 g = 1080 g = 1 kg 80 g

10. The amount of baking soda in one kg cake = 3 dg

The amount of baking soda in 17 one kg cakes = 17 × 3 dg = 51 dg = 0.0051 kg

11. Answers may vary. Sample answer:

The weight of a watermelon is 11 kg and the weight of a jackfruit is 5500 g. What is the weight of both the fruits together?

Challenge

1. 1 quintal = 1,00,000 g = 1,00,000 1000 = 100 kg

1 tonne = 1000 kg = 10 × 100 kg = 10 × 1 quintal = 10 quintals

So, 1000 kg is 10 quintals.

Chapter Checkup

1. a. m; kg b. cm; g c. m; kg

2. a. 12.5 cm b. 17.9 cm c. 22.2 cm

3. James’ cotton candy = 11.2 cm + 3 cm = 14.2 cm long

4. The length of the red straw = 12.5 cm

The length of the blue straw = 8 cm

The difference between their lengths = 12.5 8 = 4.5 cm

5. a. 1000 m = 1 km; 1 m = 1 1000 km

8 m = 1 × 1000 8 = 8 1000 = 0.008 km

b. 100 m = 1 hm; 1 m = 1 100 hm

4 hm 35 m = 4 + 35 × 1 100 = 4.35 hm

c. 1 m = 1000 mm

1232 m = 1232 × 1000 = 12,32,000 mm

d. 10 m = 1 dam; 1 m = 1 10 dam

897 m = 897 × 1 10 = 89.7 dam

6. a. 1 kg = 1000 g

5 kg = 5 × 1000 = 5000 g

b. 100 cg = 1 g; 1 cg = 1 100 g

4 g 64 cg = 4 + 64 × 1 100 = 4.64 g

c. 1 g = 1000 mg

5487 g = 5487 × 1000 = 54,87,000 mg

d. 100 dag = 1 kg; 1 dag = 1 100 kg

43 kg 7 dag = 43 + 7 × 1 100 = 43.07 kg

7. The weight of a young joey = 0.38 kg = 0.38 × 1000 g = 380 g

8. The weight of a candle = 125 g = 125 × 1000 mg = 1,25,000 mg

9. The weight that the lift is allowed to carry = 260 kg

The total weight of all the people = 85 kg + 70 kg + 58 kg + 80 kg = 293 kg No, they cannot use the lift together.

10. Answers may vary. Sample answer:

a. One 4 cm and one 8 cm

b. One 4 cm and two 8 cm

c. Three 8 cm

11. The length of each stamp = 3 cm

a. The total length of the stamps = 12 × 3 cm = 36 cm

b. No, because the total length of 12 stamps is greater than the length of the stamp album.

12. To bake one cake, the following items are required:

200 g of flour, 3 eggs, 75 g of butter, 100 g of sugar, some milk So, to bake 100 cakes the following quantities of ingredients will be needed:

The weight of the flour = 200 g × 100 = 20,000 g

= 20000 1000 kg = 20 kg

The weight of the butter = 75 g × 100 = 7500 g

= 7500 1000 kg = 7.5 kg

The weight of the sugar = 100 g × 100 = 10,000 g

= 10000 1000 kg = 10 kg

13. Answer may vary. Sample answer:

Ana buys 23.5 kg of rice and 18.25 kg of flour. How much rice and flour did she buy in all?

Challenge

1. The smallest 2-digit number = 10

The smallest 3-digit number = 100

The grandmother’s weight = (100 28) + 10 = 82 kg

So, Meenakshi’s weight = 82 =41 2 kg = 41,000 g

2. Statement 1: The father’s weight is three times the difference of Nita’s weight and Riya’s weight.

The information is alone not sufficient as it doesn’t have any data.

Statement 2: Nita’s weight = 36 kg

Riya’s weight = 1 3 × 36 kg = 12 kg

The statement alone is not sufficient as it doesn’t talk about father’s weight.

Combining the data of statements 1 and 2, we get

Difference in Nita’s weight and Riya’s weight

= 36 kg 12 kg = 24 kg

Father’s weight = 24 kg × 3 = 72 kg

So, father’s weight is 72 kg.

So, both the statements together are sufficient to find the weight of Riyaʼs father.

Hence, option c is correct.

Case Study

1. a. cm b. mm

2. Weight of the male leopard = 72 kg = 72,000 g

Weight of the striped hyena = 55,000 g

Difference in weight = 72,000 55,000 = 17,000 g

Hence, option d is correct.

3. Length of enclosure 1 = 8,894 m

Length of enclosure 2 = 8,894 m × 2 = 17,788 m

Total length = 8,894 m + 17,788 m = 26,682 m

Length of the play area = 26,682 20,000 = 6,682 m

4. Weight of 1 tiger = 72 kg

Weight of 1 lion = 176 kg

The van can shift 4 tigers, or 1 tiger and 1 lion in a trip.

5. Answers may vary.

Chapter 12

Let’s Warm-up

1. Perimeter = 10 cm

Area = 6 sq. cm

2. Perimeter = 12 cm

Area = 6 sq. cm

3. Perimeter = 12 cm

Area = 5 sq. cm

4. Perimeter = 14 cm

Area = 6 sq. cm

Do It Yourself 12A

1. a. Perimeter = 4 × 20 cm = 80 cm

b. Perimeter = 2 × 100 + 2 × 50 = 200 + 100 = 300 cm

c. Perimeter = 2 × 20 + 2 × 10 = 40 + 20 = 60 cm

d. Perimeter = 2 × 30 + 2 × 20 = 60 + 40 = 100 cm

2. Perimeter = 4s

a. s = 32 m

Perimeter = 4 × 32 = 128 m

b. s = 53 cm

Perimeter = 4 × 53 = 212 cm

c. s = 27 m

Perimeter = 4 × 27 = 108 m

d. s = 78 cm

Perimeter = 4 × 78 = 312 cm

3. Perimeter = 2l + 2b

a. Perimeter = 2 × 12 + 2 × 15 = 24 + 30 = 54 cm

b. Perimeter = 2 × 55 + 2 × 85 = 110 + 170 = 280 m

c. Perimeter = 2 × 18 + 2 × 20 = 36 + 40 = 76 m

d. Perimeter = 2 × (l + b) = 2 × (25 + 15) = 80 cm

4. a. Perimeter = 112 mm = 2 × 32 mm + 2 × b

112 = 64 + 2b

112 − 64 = 2b

48 2 b = b = 24 mm

Therefore, breadth = 24 mm

b. Perimeter = 148 mm = 4 × s = 148

4 s = s

s = 37 mm

Therefore, each side of the square = 37 mm

5. The length of each side of square-shaped garden = 45 m

The perimeter of the garden = 4 × 45 m = 180 m

He would require 180 m of wire to fence the entire garden.

6. Length of the table mat = 50 cm

Breadth of the table mat = 20 cm

Perimeter of the table mat = 2 × (l + b) = 2 × (50 + 20) = 2 × 70 cm = 140 cm

Cost of putting 1 cm lace = ₹10

Cost of putting 140 cm lace = 140 × ₹10 = ₹1400

7. Length = 60 m

Breadth = 30 m

Area = l × b = 60 × 30 = 1800 sq. m

So, the area covered by the mosque is 1800 sq. m.

8. The length of the square formed = 10 cm

The perimeter of the square = 4 × 10 cm = 40 cm

The perimeter of the square = The perimeter of the rectangle

Length = 12 cm, Perimeter = 40 cm

Breadth = ?

Perimeter = 2l + 2b

40 = 2 × 12 + 2 × b

40 = 24 + 2b

40 – 24 = 2b

16 = 2b

b = 8 cm

Thus, the breadth of the rectangle will be 8 cm.

9. Answers may vary. Sample answer:

Challenge

1. Length = 3 × Breadth ⇒ l = 3b

Perimeter = 32 m

So, 32 = 2 × (l + b)

32 = 2 × (3b + b)

32 = 2 × (4b)

32 2 = 4b

16 = 4b

b = 16 4

b = 4 m

⇒

⇒ l = 3b = 3 × 4 = 12 m

Thus, the length of the rectangle is 12 m and the breadth is 4 m.

Do It Yourself 12B

1. a. sq. cm

b. sq. cm

c. sq. m

d. sq. m

e. sq. km

f. sq. m

g. sq. cm

h. sq. m

2. Area of rectangle = Length × Breadth

a. Area = 15 cm × 22 cm = 330 sq. cm

b. Area = 58 m × 70 m = 4060 sq. m

3. Area of square = Side × Side

a. Area = 56 m × 56 m = 3136 sq. m

b. Area = 67 cm × 67 cm = 4489 sq. cm

4. The measurement of the room = 12 m × 12 m The carpet required to cover the floor = The area of Advita’s Room = 12 m × 12 m = 144 sq. m

5. a. 1 2

Area of triangle 1 = Half the area of rectangle 1 = 25 2 sq. units

Area of triangle 2 = Half the area of rectangle 2 = 10 2 sq. units

Area of whole triangle 25 2 + 10 2 = 35 2 = 17.5 sq. units.

b. 1

Area of triangle 1 = Half the area of rectangle 1 = 12 2 sq. units = 6 sq. units

Area of rectangle A = 6 cm × 8 cm = 48 sq. cm

Area of rectangle B = 8 cm × 3 cm = 24 sq. cm

Area of rectangle C = 12 cm × 2 cm = 24 sq. cm

Area of the whole shape = 48 sq. cm + 24 sq. cm + 24 sq. cm = 96 sq. cm

Area of rectangle A = 3 cm × 3 cm = 9 sq. cm

Area of rectangle B = 8 cm × 3 cm = 24 sq. cm

Area of the whole shape = 9 sq. cm + 24 sq. cm = 33 sq. cm

7. Answers may vary. Sample answers:

a. Perimeter = 36 cm

Area = 80 sq. cm

b. Area = 24 sq. cm

Perimeter = 22 cm

8. Answers may vary.

9. Length of the rectangle = 8 m

Breadth of the rectangle = 1 4 × length = 1 4 × 8 = 2 m

Area of the rectangle = 8 m × 2 m = 16 sq. m

Side of the square = 2 × breadth of the rectangle = 2 × 2 m = 4 m

Area of the square = 4 m × 4 m = 16 sq. m

So, both the square and rectangle have equal areas.

Challenge

1. Perimeter of each square = 36 cm

Side of each square = 36 4 = 9 cm

Breadth of the rectangle = Side of the square = 9 cm

Length of the rectangle = 2 × Side of the square = 2 × 9 cm = 18 cm

Perimeter of the rectangle = 2 × 18 cm + 2 × 9 cm = 36 cm + 18 cm = 54 cm

Chapter Checkup

1. Perimeter = 2l + 2b; Area = l × b

a. l = 14 cm, b = 18 cm

Perimeter = 2 × 14 + 2 × 18 = 28 + 36 = 64 cm

Area = 14 × 18 = 252 sq. m

b. l = 27 m, b = 21 m

Perimeter = 2 × 27 + 2 × 21 = 54 + 42 = 96 m

Area = 27 × 21 = 567 sq.m

c. l = 49 m, b = 33 m

Perimeter = 2 × 49 + 2 × 33 = 98 + 66 = 164 m

Area = 49 × 33 = 1617 sq. m

2. Perimeter = 4 × Side; Area = Side × Side

a. s = 80 cm

Perimeter = 4 × 80 = 320 cm

Area = 80 × 80 = 6400 sq. cm

b. s = 32 m

Perimeter = 4 × 32 = 128 m

Area = 32 × 32 = 1024 sq. m

c. s = 45 m

Perimeter = 4 × 45 = 180 m

Area = 45 × 45 = 2025 sq. m

3. a. Breadth = 24 cm; Area = 960 sq. cm

Length = Area Breadth = 960 40 cm 24 =

Perimeter = 2 × 40 + 2 × 24 = 80 + 48 = 128 cm

b. Perimeter = 420 m

Side = 420 105 4 == 105 m

Area = 105 m × 105 m = 11,025 sq. m

c. Length = 34 m; Perimeter = 124 m

2 × Breadth = 124 − 2 × 34 = 124 − 68 = 56

Breadth = 56 28 2 = = 28 m

Area = 34 × 28 = 952 sq. m

4. Area of triangle = Area of rectangle 2

Area of rectangle 1 = 2 sq. units

Area of triangle 1 = 2 2 sq. units = 1 sq. units

Area of rectangle 2 = 6 sq. units

Area of triangle 2 = 6 2 sq. units = 3 sq. units

Area of whole triangle = 1 sq. units + 3 sq. units = 4 sq. units

Area of rectangle = 24 sq. units

Area of triangle = 24 12 sq. units 2 =

Area of rectangle 1 = 15 sq. units

Area of triangle 1 = 15 sq. units 2

Area of rectangle 2 = 5 sq. units

Area of triangle 2 = 5 sq. units 2

Area of the whole triangle = 15520 10 sq. units 222 +==

Area of rectangle 1 = 12 × 3 sq. cm = 36 sq. cm

Area of rectangle 2 = 3 × 1 sq. cm = 3 sq. cm

Area of the whole figure = 36 + 3 sq. cm = 39 sq. cm

Area of rectangle 1

= 8 × 6 sq. cm

= 48 sq. cm

Area of rectangle 2

= 5 × 4 sq. cm

= 20 sq. cm

Area of rectangle 3

= 5 × 2 sq. cm

= 10 sq. cm

Area of the whole figure

= 48 + 20 + 10 sq. cm

= 78 sq. cm c.

Area of rectangle 1 = 6 × 5 = 30 sq. cm

Area of rectangle 2 = 7 × 4 = 28 sq. cm

Area of the whole figure = 30 + 28 = 58 sq. cm

6. Area of the figure = 80 sq. cm

Area of one square 80 5 = sq. cm = 16 sq. cm

So, side = 4 cm

Perimeter of the figure

= 12 × 4 cm = 48 cm

So, option c is the correct option.

7. Length of each side = 300 feet

Area = 300 × 300 feet = 90,000 sq. feet

So, the area of the platform is about 90,000 sq. feet.

8. The side of one square wooden panel = 25 cm

The area of one square wooden panel = 25 × 25 = 625 sq. cm

The total area of 8 such panels = 8 × 625 sq. cm = 5000 sq. cm

9. The area of the floor = 12 m × 15 m = 180 sq. m

The area of the carpet = 13 m × 13 m = 169 sq. m

The area of the floor that is not carpeted = 180 − 169 = 11 sq. m

10. Rectangle:

length = 4 straws; breadth = 2 straws

Perimeter = 2 × (l + b) = 2 × (4 + 2) = 12 straws

Area of the rectangle = 4 × 2 = 8

Square:

Perimeter = 12 straws

Side = 12 4 = 3 straws

Area of the square = 3 × 3 = 9

So, both the shapes have the same perimeter but the square has more area than the rectangle.

Challenge

1. Perimeter of a rectangle = 2 × length + 2 × breadth = 22 cm

Perimeter of the shape formed:

= (2 × length + 2 × breadth) × 3 – (2 × length + 2 × breadth)

= (22) × 3 – (22) = 66 – 22 = 44 cm length length breadth breadth

2. New length = 2 × Length New Width = 2 × Width

∴New Area = New Length × New Width = (2 × Length) × (2 × Width) = 4 × (Length × Width) = 4 × Area

No, the statement is not true.

Case Study

1. Area of 1 classroom = 8 m × 7 m = 56 sq. m

Area of 4 classrooms = 4 × 56 sq. m = 224 sq. m

2. Area of staff room = 14 m × 7.2 m = 100.8 sq. m

Area of art room = 12 m × 7.6 m = 91.2 sq. m

So, the staff room has more area.

3. Length of the side of the school = 7 m + 7 m + 10 m + 7 m + 7 m = 38 m

Area of the school = 38 m × 38 m = 1444 sq. m

So, option b is correct.

4. Area of the school = 1444 sq. m

Area of the ground = 18 m × 18 m = 324 sq. m

Remaining area = 1444 − 324 = 1120 sq. m

5. Answers may vary.

Chapter 13

Let’s Warm-up

1. Between a jug and a glass, a jug has more capacity.

2. If 8 glasses of juice fill a jug, then 8 glasses is the capacity of the jug.

3. Four tin cans of paint fill a bucket of paint. If the capacity of 1 tin can is 2 L, then 8 L is the capacity of the bucket.

4. Half of a container fills the bucket completely. The capacity of the container is 2 full buckets.

Do It Yourself 13A

1. A teaspoon

2. a. 1000 mL = 1 L

1 mL = 1 1000 = L

658 mL = 1 1000 = × 658 L = 0.658 L

b. 1 L = 1000 mL

8437 L = 8437 × 1000 mL = 84,37,000 mL

c. 10 dL = 1 L

1 dL = 1 10 LL

2567 dL = 1 2567 257.6 10 LL×= × 2567 L = 257.6 L

d. 100 cL = 1 L

1 cL = 1 100 L

5054 cL = 5054 × 1 100 L = 50.54 L

e. 10 L = 1 daL

1 L = 1 10 LdaL

821 L = 821 × 1 10 daLdaL = 82.1 daL

f. 136 L 80 dL = 144 L

1 L = 1000 mL

144 L = 144 × 1000 = 1,44,000 mL

3. Complete the given table.

S. No. Full form In bigger units In smaller units

a. 8 L 60 mL 8.060 L 8060 mL

b. 32 L 89 cL 32.89 L 3289 cL

c. 4012 L 3 dL 4012.3 L 40,123 dL

d. 90 kL 6 daL 90.06 L 9006 daL

4. Capacity of the bucket that Aman had = 20 L

Capacity of the bucket in mL = 20 × 1000 = 20,000 mL

So, the capacity of the bucket is 20,000 mL.

5. No, C has enough space to pour more liquid, so it has more capacity than the other two pots. Now converting mL to dL

1 mL = 1 100 dL

1050 × 1 100 dL = 10.50 dL

6. Answer may vary. Sample answer: Radhika’s home had water tank with a capacity of 500 L. What is the capacity of the tank in mL?

Challenge

1. Yes.

As there are 3 bowls of 600 mL and 8 bowls of 200 mL each.

3 × 600 mL = 1800 mL

8 × 200 mL = 1600 mL

1800 mL + 1600 mL = 3400 mL

So, in total we can easily fill up to 3400 mL capacity.

Do It Yourself 13B

1. a.

From one face we can count there are 4 cubes.

So, layer 1 has 8-unit cubes.

Number of layers = 2

So, the total number of unit cubes = 2 × 8 = 16

Volume of the given solid = 16 cu. units

This is a two layers solid. In the 1st layer, we can see 2-unit cubes on the right and left side.

And, 2-unit cubes are placed in middle. There are also two cubes in the 2nd layer.

So, the total number of unit cubes = 8

So, the volume of given solid = 8 cu. units

The number of cubes in the 1st layer = 4 × 5 = 20 cubes

So, the number cubes in four layers = 20 × 4 = 80 cubes

So, the volume of given solid = 80 cu. units

There are two layers with 4 × 3 cubes = 24 cubes

The bottom two layers are with 6 cubes in each.

So total cubes = 12 cubes

So, the volume of given solid = 36 cu. units

2. To find the number of cartons, we need to find the volume of the almirah.

So, the volume of the almirah = Volume = Length × Width × Height

Volume = 3 × 2 × 1 = 6 cu. m

3. Length = 10 cm, width = 15 cm and height = 6 cm

Volume of the container = l × w × h = 10 × 15 × 6 = 900

So, the volume of the container = 900 cu. cm

4. Length = 30 cm, width = 22 cm and height = 10 cm

Volume of the fish tank = l × w × h = 30 × 22 × 10 = 6600

So, the volume of the fish tank = 6600 cu. cm

5. a. Volume of milk = 20 mL

Increased volume = 20 mL + 10 mL = 30 mL

Volume of each biscuit = 30 mL – 20 mL = 10 mL

b. Volume of 1 biscuit = 10 mL

Volume of 3 biscuits = 3 × 10 mL = 30 mL

6. To find whose lunch box is bigger in size, we must compare the volumes of the two lunch boxes.

Rinku’s lunch box volume:

Volume = Length × Width × Height

Volume = 16 cm × 8 cm × 3 cm = 384 cu. cm

Rita’s lunch box volume:

Volume = Length × Width × Height

Volume = 10 cm × 7 cm × 3 cm = 210 cu. cm

Comparing the two volumes, 384 cu. cm > 210 cu. cm

Thus, Rinku’s lunch box is bigger in size than Rita’s lunch box.

7. Amount of sugar water prepared by Rekha = 16 L

Amount of sugar water prepared by Rekha in mL = 16 L × 1000 = 16,000 mL

Capacity of each beaker = 2000 mL

Number of beakers needed = Total volume of the sugar water Capacity of each beaker = 16000 2000 = 8

Hence, 8 beakers are needed to pour the sugar water equally.

8. The jug initially contained 1.5 litres of juice = 1500 mL (1 L = 1000 mL).

Total juice poured into the glasses = 350 mL + 890 mL = 1240 mL

Juice left in the jug = Initial amount − Total amount poured Juice left in the jug = 1500 mL − 1240 mL = 260 mL

So, there is 260 mL of juice left in the jug.

Challenge

1. Height of the blocks = 6

Volume of the building = volume of block 1 + volume of block 2

Volume of block 1 = 4 × 5 × 6 = 120 cu. units

Volume of block 2 = 2 × 2 × 6 = 24 cu. units

Total volume of the building = 120 cu. units + 24 cu. units = 144 cu. units

Chapter Checkup

1. a. Glue in a small bottle

2 L 50 L 2000

b. Paint in a bucket

c. Ketchup in a packet

d. Shampoo in a bottle

e. Water in a pool

2. The initial level of water = 500 mL

The final level of water = 800 mL

Number of match boxes = 6

Volume of 6 match box = 800 mL – 500 mL = 300 mL

Volume of 1 match box = 300 mL ÷ 6 = 50 mL

Length of solid = 4 unit

Breadth of solid = 4 unit

Height of solid = 4 unit b.

Length of solid = 8 unit

Breadth of solid = 2 unit

Height of solid = 3 unit

e. 10 L = 1 daL

1 L = 1 daL 10

243 L = 1 10 × 243 daL = 24.3 daL

f. 907 L 56 dL = 912.6 L

1 L = 1000 mL

912.6 L = 912.6 × 1000 = 9,12,600 mL

7. a. On counting, there are 12-unit cubes in the given solid, hence, the volume of the solid is 12 cu. units.

b. On counting, there are 44-unit cubes in the given solid, hence, the volume of the solid is 44 cu. units.

c. On counting, there are 15-unit cubes in the given solid, hence, the volume of the solid is 15 cu. units.

d. On counting, there are 22-unit cubes in the given solid, hence, the volume of the solid is 22 cu. units.

8. The pack originally contains 1 litre of juice = 1000 mL

Juice left in the pack = Total amount − Amount drunk by Juhi Juice left in the pack = 1000 mL − 235 mL = 765 mL

So, 765 mL of juice is left in the packet.

9. Capacity of the first flask = 850 mL

Capacity of the second flask = 1 L 250 mL = 1000 mL + 250 mL = 1250 mL

Total water the flasks can hold together = Capacity of the first flask + Capacity of the second flask = 850 mL + 1250 mL = 2100 mL

1000 mL = 1 L; 1 mL = 1 1000 L

2100 mL = 1 2100 1000 ×= × 2100 = 2.1 L or 2 L 100 mL

So, the two flasks can hold a total of 2 L 100 mL of water.

10. a. Number of cubes across length = 7, number of cubes across width = 4, number of cubes across height = 4

Volume = l × w × h = 7 × 4 × 4 = 112 cu. units

So, the volume of the container = 112 cu. units

b. Number of cubes across length = 7, number of cubes across width = 3, number of cubes across height = 3

Volume = l × w × h = 7 × 3 × 3 = 63 cu. units

So, the volume of the container = 63 cu. units

c. Number of cubes across length = 4, number of cubes across width = 3, number of cubes across height = 3

Volume = l × w × h = 4 × 3 × 3 = 36 cu. units

So, the volume of the container = 36 cu. units

11. Length = 50 cm, width = 30 cm and height = 25 cm

Volume of carton = l × w × h = 50 × 30 × 25 = 1500 × 25 = 37,500 cu. cm

12. Capacity of Seema’s bucket = 6.5 L

Tank capacity = Seema’s bucket capacity × 4

Tank capacity = 6.5 L × 4

Tank capacity = 26 L

So, the water tank can hold 26 L of water.

13. Capacity of bottle = 5 cups of water

Capacity of bowl = 5 cups of water

Capacity of jug = 5 bottles of water = 5 × 5 cups of water = 25 cups of water

Capacity of pail = 3 jugs + 7 bowls of water

= 3 × 25 cups + 7 × 5 cups = 75 cups + 35 cups = 110 cups

Capacity of a cup = 90 mL

So the capacity of pail = 110 × 90 mL = 9900 mL

1000 mL = 1 L; 1 mL = 1 1000 L

9900 mL = 1 9900

1000 ×= × 9900 = 9.9 L or 9 L 900 mL

So, the capacity of the pail is 9 L 900 mL.

Challenge

1. Length of the cube = 6 units

Width of the cube = 5 units

Height of the cube = 4 units

Volume of the block = l × b × h = 6 × 5 × 4 = 120 units

To maximise the hollow space, we need to leave a layer of cubes on each side.

l = 6 – 2 = 4 units

b = 5 – 2 = 3 units

h = 4 – 2 = 2 units

Inner volume = 4 × 3 × 2 = 24 units

Outer volume = total volume – inner volume

= 120 units – 24 units = 96 units

Since, each block Jassi uses is made up of 8 small cubes:

Number of blocks = 96 8 = 12

So, Jassi requires 12 blocks to create hollow blocks.

2. The volume of a cube will not always be greater than the volume of a cuboid. It will depend on the dimensions.

The reason given is the formula used to find the volume of a cube and a cuboid.

So, the assertion is not true, and only the reason is true.

Thus, option d is correct.

Case Study

1. a. false

The buttermilk packet comes is 100 mL.

b. true

The milk packets usually come in 500 mL or 1 L.

2. Volume of the flavoured milk = 0.45 L

Volume of the flavoured milk = 0.45 × 1000 = 450 mL

So, the volume of the flavoured milk is 450 mL.

3. Volume of milk required to make 15,000 kg of paneer = 45,000 L

Volume of milk required to make 30,000 kg of paneer

= 45,000 × 30,000 15,000 = 90,000 litres

4. Dimension of the cheese pack = 7 cm × 3 cm × 4 cm

Dimension of 10 cheese packs = (7 cm × 3 cm × 4 cm) × 10 = 7 cm × 3 cm × 40 cm

The box should have a minimum height of 40 cm to pack 10 boxes.

5. Answers may vary.

Let’s Warm-up

1. cube 2. cylindrical 3. cuboid 4. sphere

Do It Yourself 14A

1. a. b.

2. a. Cone: Vertices = 1, Edges = 1, Faces = 2

b. Sphere: Vertices = 0, Edges = 0, Faces = 1

c. Square prism: Vertices = 8, Edges = 12, Faces = 6

d. Cylinder: Vertices = 2, Edges = 0, Faces = 3

3. The Skylon Tower is cylindrical in shape.

4. a. A cube and a cuboid have 8 vertices, 6 flat faces and 12 edges.

b. A sphere has only one curved face.

c. A cylinder has 2 edges but 0 vertex.

d. A cone has one curved face and a vertex.

5. Four cubes, when placed one on top of the other form a cuboid. A cuboid has 6 flat faces, 8 vertices and 12 edges.

7. Rohan’s tent is a prism, and Simi’s tent is a pyramid.

Similarity: They both have 5 faces.

Difference: Rohan’s tent has a triangular faces, and Simi’s tent has a rectangular faces.

Challenge

1. A prism has two identical surfaces, both of which are polygonal with straight sides. A cylinder also has two identical parallel surfaces like a prism, but since they are curved and not polygonal, it is not considered a prism.

Do It Yourself 14B

1. a. The given net is of a square pyramid.

b. The given net is of a cylinder.

c. The given net is of a cone.

d. The given net is of a hexagonal pyramid.

2. The given net is of a triangular prism. The given image represents a triangular prism.

4. When the net made by Rishi is folded, the faces overlap each other.

But when the net made by Megha is folded, the triangles do not overlap, so Megha made the net correctly.

5. 6.

Rectangular Prism Rectangular Pyramid

Challenge

1. When R is on the top, P will be at the bottom.

Do It Yourself 14C

Option a is correct.

Side view:

Top view:

4. Mehar used 15 building blocks to make the shape. Front view: Side view: Top view:

Challenge

1. If the front view shows 4 dots, then the back view would show 3 dots.

If the left side view shows 5 dots, then the right-side view would show 2 dots.

Do It Yourself 14D

1. a. The hospital is to the south of the airport.

b. The hotel is to the east of the bank.

c. The airport is to the west of the school.

2. 1 cm = 25 km

5 cm = 5 × 25 km = 125 km

3. Given that the scale is 1 cm = 12 km

Distance between the college and the complex on the map = 2 cm

Actual distance = 2 × 12 km = 24 km

4. Length of the cupboard = 1 square = 2 m;

Width of the cupboard = 4 squares = 4 × 2 m = 8 m

Length of the table = 2 squares = 2 × 2 m = 4 m;

Width of the table = 3 squares = 3 × 2 m = 6 m

Length of the desks = 3 squares = 3 × 2 m = 6 m;

Width of the desks = 1 square = 2 m

5. Given that 1 cm = 2.5 km 5 cm = 5 × 2.5 km = 12.5 km

6. Myanmar, Sri Lanka and Pakistan

Challenge

1. Distance between his house and the school on the map = 2 cm

Distance between the school and the stadium on the map = 2 cm

Distance between his house and the stadium on the map = 2 cm + 2 cm = 4 cm

Scale: 1 cm = 3 km

The actual distance from his house to the stadium = 4 × 3 km = 12 km

Do It Yourself 14E

The front of the house has only one window.

4. If we enter from the back door, we will be in the kitchen.

5. We cannot show the windows on the backside, as well as either the left or right side, in the deep drawing.

6. Answers may vary. Sample answer:

3. Front view:

1. As the house has 4 windows on each lateral side, Rakesh won’t be able to show four backside windows and four windows on either the left or right side. Hence, he won’t be able to show 8 windows in the deep drawing.

8. Answers may vary. Sample answers:

2. A globe is sphere in shape. It has one curved face and no edges or vertices.

3. a. A pattern that can be cut and folded to make a model of a 3-D shape is called a net.

b. A special way of drawing a house that is deep to show its length, width and height is called a deep drawing.

c. The 2-D representation of the map of a house is called the floor plan.

4. a. Cylinder Faces = 3 Edges = 2 Vertices = 0

b. Cube Faces = 6 Edges = 12 Vertices = 8

c. Cone Faces = 2 Edges = 1 Vertices = 1

Top View Front View Side View

9. There are 4 doors in the given floor plan.

10. Closet 1 = 4 squares

Bathroom = 12 squares

Since, 3 × 4 = 12

So, the bathroom is 3 times bigger than closet 1. 11. Answer may vary. Sample answer:

12. 1 cm = 3 km

Distance from Ben’s house to May’s house = 3 cm = 3 × 3 km = 9 km

Distance from May’s house to Jule’s house = 4 cm = 3 × 4 km = 12 km

Total distance covered by Ben = 9 + 12 km = 21

Challenge

1. The minimum number of flat faces a solid can have is 4. The solid is called a tetrahedron.

3. Person 1: Top view

Person 2: Side view

Person 3: Front view

4. Answers may vary. Sample answer: Person 1 should stay away from the edge and hold onto something sturdy.

Chapter 15

Let’s Warm-up

Do It Yourself 15A

1. 1 hour = 60 minutes

a. 3 hours = 3 × 60 = 180 minutes

b. 4 hours = 4 × 60 = 240 minutes

c. 2 hours 20 minutes = 2 × 60 + 20

= 120 + 20 = 140 minutes

d. 3 hours 10 minutes = 3 × 60 + 10 = 180 + 10 = 190 minutes

e. 3 hours 50 minutes = 3 × 60 + 50 = 180 + 50 = 230 minutes

f. 4 hours 30 minutes = 4× 60 + 30 = 240 + 30 = 270 minutes

2. 1 minute = 60 seconds

a. 5 minutes = 5 × 60 = 300 seconds

b. 8 minutes = 8 × 60 = 480 seconds

c. 9 minutes 10 seconds = 9 × 60 + 10 = 540 + 10 = 550 seconds

d. 10 minutes 20 seconds = 10 × 60 + 20 = 600 + 20 = 620 seconds

e. 12 minutes 40 seconds = 12 × 60 + 40 = 720 + 40 = 760 seconds

f. 15 minutes 50 seconds = 15 × 60 + 50 = 900 + 50 = 950 seconds

3. a. 240 seconds 5 minutes 50 seconds

b. 480 seconds 8 minutes 40 seconds

c. 350 seconds 4 minutes

d. 440 seconds 9 minutes 10 seconds

e. 520 seconds 7 minutes 20 seconds

f. 550 seconds 8 minutes

4. Time taken by train to reach London = 8 hours 30 minutes = 8 × 60 minutes + 30 minutes = 480 minutes + 30 minutes = 510 minutes

Therefore, the journey is 510 minutes long.

5. Time Maya takes to read the book in the morning = 50 minutes

Time Maya takes to read the book in the afternoon = 45 minutes

Total time taken by Maya to read the book = 50 minutes + 45 minutes = 95 minutes = 1 hour 35 minutes

6. Answers may vary. Sample answer: A day has 24 hours. How many seconds does a day has?

Challenge

1. 55 minutes in the morning + 50 minutes in the evening = 105 minutes

Since her teacher asked her to practice for 1 hour and 50 minutes

1 hour and 50 minutes = 60 minutes + 50 minutes = 110 minutes

Since 105 minutes < 110 minutes, Sarah didn’t practice for the duration asked by the teacher. Hence, statement 1 is false. Difference between duration = 110 minutes – 105 minutes = 5 minutes

Therefore, she should practice for an extra 5 minutes to meet her teacherʼs requirement. Hence, statement 2 is true.

1. Taking a shower

2. Putting on a jacket

3. Going for a trip

Swimming lesson

Rainy season

1. Time shown on clock A = 1:00

Time shown on clock B = 2:00

Duration between the times shown on the two clocks:

2 hours – 1 hour = 1 hour

Thus, the duration between the time shown on the two clocks is 1 hour.

2. a. 4 hours 30 min

+ 6 hours 20 min

10 hours 50 min

b. 7 hours 5 min = 6 hours + 1 hour + 5 min

= 6 hours + 60 min + 5 min = 6 hours 65 min

6 hours 65 min

– 5 hours 30 min

1 hour 35 min

c. 3 weeks 10 days

+ 6 weeks 5 days

9 weeks 15 days

= 9 weeks + 14 days + 1 day

= 9 weeks + 2 weeks + 1 day

= 11 weeks 1 day

d. 2 years 6 months + 1 year 5 months

3 years 11 months

e. 7 weeks 20 days

– 2 weeks 5 days

5 weeks 15 days

= 5 weeks + 14 days + 1 day

= 5 weeks + 2 weeks + 1 day

= 7 weeks 1 day

f. 3 years 1 month = (2 years + 1 year) + 1 month =

2 years + 12 months + 1 month = 2 years 13 months

2 years 13 months – 7 months

2 years 6 months

3. Time: 2:40 p.m. = 2 hours 40 minutes

Time after 35 minutes = 2 hours 40 minutes + 35 minutes

2 hours 75 minutes

= 2 hours + 60 minutes + 15 minutes

= 2 hours + 1 hour + 15 minutes

= 3 hours 15 minutes

So, the time will be 3:15 p.m.

4. Meeting starts at = 9:30 a.m. = 9 hours 30 minutes

Meeting ends at = 12:45 p.m. = 12 hours 45 minutes

Meeting lasted for = 12 hours 45 minutes – 9 hours 30 minutes 3 hours 15 minutes

Therefore, the meeting lasted for 3 hours 15 minutes.

6. Summer vacation = 1 month 25 days

Winter vacation = 20 days

Total duration 1 month 25 days + 20 days 1 month 45 days

= 1 month + 30 days + 15 days

= 1 month + 1 month + 15 days

= 2 months 15 days

So, the school was closed for 2 months 15 days.

25 October 2020 25 October 2021 25 January 2021 25 January 2022 15 February 2022

1 year 3 months 1 year 21 days

So, the museum was open for 2 years 3 months and 21 days.

10 June 2019 10 June 2023 10 November 2023 5 December 2023

7. 4 years 5 months 25 days

It took 4 years 5 months 25 days for Sam to become an advanced violinist.

8. Departure time from New Delhi = 6:00 a.m.

Arrival time at Katra = 2:00 p.m.

Total time taken for the journey = 8 hours

We know that, 1 hour = 60 minutes

8 hours = 480 minutes

Total stoppage time = 2 minutes + 2 minutes + 2 minutes = 6 minutes

Total run time of the train = 480 minutes – 6 minutes = 474 minutes

9. Answers may vary. Sample answer: Yohaan cycles from 6:45 p.m. to 8:25 p.m. For how long does he cycles?

Challenge

1. Meera needs to leave her house at least 5 minutes before the bus arrives.

Time taken to walk to the bus stop = 10 minutes

So, she needs to start walking to the bus stop 10 minutes before she needs to leave the house.

Adding the time to walk to the bus stop (10 minutes) and the time she needs to leave before the bus arrives (5 minutes), we get a total of 15 minutes.

Also, time taken to get ready = 15 minutes

Total time required = 15 + 15 = 30 minutes

So, Meera needs to wake up at least 30 minutes before 8:10 a.m. to ensure she catches the bus on time.

8:10 a.m. – 30 minutes = 7:40 a.m.

Therefore, Meera should wake up at 7:40 a.m. to ensure she catches the bus on time.

Do It Yourself 15C

1. a. The temperature of a hot cup of coffee is around 85°C. Thus, option (ii) is correct.

b. The possible temperature of a snowman is 0°C. Thus, option (i) is correct.

2. a. 26°C b. 35°C c. 42°C d. 12°C

3. a 35°C to °F 9 ° 5 C

+ 32 = °F

+ 32 = 95°F

b. 80°F to °C

 5 °F32°C 9   5 8035 9  26.6°C

c. 45°C to °F 9 ° 5 C

+ 32 = °F

4532 5

113°C °F

d. 149°F to °C  5 °F32 9  = °C

 5 14932 9  = 65°C

4. a. As the temperature of Srinagar is the lowest, it is the coolest place.

b. As the temperature of Chennai is the highest, it is the hottest place.

c. Difference between the temperature of Delhi and Shimla = 41.2°C − 11.5°C = 29.7°C

d. Degree the temperature needs to rise to reach 25°C in Bangalore = 25°C – 21.2°C = 3.8°C

5. Highest temperature = 51°C

Lowest temperature = –60°C

Difference between the highest and the lowest temperature = 51 – (–60) = 111°C

Let us convert °C into °F.

111°C = 9 11132 5

= 231.8°F

Challenge

1. It is given that °C = °F

Let °C = °F = x 9 °F°C32° 5

32 5 x  x = –40°

Thus, the correct option is (d).

Chapter Checkup

1. a. 2 hours = (2 × 60 minutes) = 120 minutes = (120 × 60 seconds) = 7200 seconds

b. 1 hour 30 minutes = (1 × 60 minutes) + 30 minutes = 90 minutes = (90 × 60 seconds) = 5400 seconds

c. 2 hours 20 minutes

= (2 × 60 minutes) + 20 minutes = 140 minutes = (140 × 60 seconds) = 8400 seconds

d. 6 minutes 40 seconds

= (6 × 60 seconds) + 40 seconds

= 360 seconds + 40 seconds

= 400 seconds

2. a. 30°C to °F 9 ° 5 C

+ 32 = °F

+ 32 = 86°F

b. 175°F to °C

5 17532 9  = 79.44°C

3. Time taken to reach the destination = 6 hours 30 minutes

= 6 × 60 minutes + 30 minutes

= 360 minutes + 30 minutes

= 390 minutes

Thus, the flight takes 390 minutes to reach its destination.

4. Time for which Shreya sleeps = 6 hours 15 minutes

= 6 × 60 minutes + 15 minutes = 360 minutes + 15 minutes

= 375 minutes

Shreya sleeps for 375 minutes.

5. Match started at: 3:45 p.m. = 3 hours 45 minutes

Duration of match = 1 hour 15 minutes

Match ended at =

3 hours 45 minutes

+ 1 hour 15 minutes

4 hours 60 minutes

= 4 hours + 1 hour

= 5 hours = 5 p.m.

Thus, the match ended at 5 p.m.

6. Age at which Prashant went to school = 5 years 4 months

Prashant’s present age = 12 years 3 months

Total duration = 12 years 3 months – 5 years 4 months

= 11 years + 12 months + 3 months – 5 years 4 months

= 11 years 15 months – 5 years 4 months

11 years 15 months

– 5 years 4 months

6 years 11 months

Hence, Prashant has been at school for 6 years 11 months.

7. Sanya started drawing the picture at = 1:32 p.m. = 1 hour 32 minutes

Sanya completed the picture at = 5:15 p.m. = 5 hours

15 minutes = 4 hours 60 minutes + 15 minutes = 4 hours

75 minutes

Time duration = 4 hours 75 minutes

– 1 hour 32 minutes

3 hours 43 minutes

Hence, Sanya took 3 hours 43 minutes to draw the picture.

8. Starting date = 6 July

Number of days in July = 31

Number of days from 6 July to 31 July = 25 days

Duration = 40 days

Days left after the month of July = 40 – 25 = 15 days

So, the finishing date is 15 August.

9. 12 hours = 720 minutes

Total minutes = 720 minutes

Dividing 720 by 75, Quotient = 9

Remainder = 45

So, Kunal takes the medicine 9 times in 12 hours.

10. Train arrived Jaipur at = 10:45 a.m. = 10 hours 45 minutes

Train was late by = 1 hour 15 minutes.

The scheduled time of arrival of the train at Jaipur, 10 hours 45 minutes

– 1 hour 15 minutes

9 hours 30 minutes

Hence, the train was scheduled to arrive at Jaipur at 9:30 a.m.

11. Temperature = 32°C

Temperature decreased every minute = 2°C

Temperature reaches 16°C: 32°C – 2°C – 2°C – 2°C – 2°C – 2°C – 2°C – 2°C – 2°C

Time taken to reach 16°C = 8 minutes

Challenge

1. Holidays start on = Saturday, 25 June

Number of days in June = 30

Number of days from 25 June to 30 June = 5 days

Number of leaves = 15

Days left after the month of June = 15 days – 5 days = 10 days

So, the holidays end on Sunday, 10 July. Hence, the assertion is true.

July has 31 days; hence, the reason is also true, but it doesn’t explain assertion.

Thus, option b is correct as both the Assertion and the Reason are true, but the Reason is not the correct explanation of Assertion.

2. Time of departure from New Delhi = 6:00 a.m.

Time taken by the flight to reach London = 9 hours

35 minutes

Time at which the flight reaches London according to Indian Standard Time = 6:00 a.m. + 9 hours 35 minutes = 3:35 p.m.

Time by which London is behind Indian standard time = 5 hours 30 minutes

Time at which the flight reaches London according to London time = 3:35 p.m. – 5 hours 30 minutes = 10:05 a.m.

Thus, the flight reaches to London at 10:05 a.m.

Case Study

1. In 1990, the average daily temperature was 20°C.

a. 20°C

2. Average daily temperature in 1990 = 20°C

Average daily temperature in 2020 = 24°C

Temperature increase = 24°C – 20°C = 4°C

b. 4°C

3. The average daily temperature in 2020 was 24°C.

4. Temperature in 1990 = 20°C

Temperature in 2020 (1990 + 30) = 24°C

Temperature increase = 24°C – 20°C = 4°C

Temperature in 2050 if the same trend follows ⇒ 2050

= 2020 + 30

⇒ 24°C + 4°C = 28°C

5. The average daily temperature in the coastal town has increased over the past 30 years. So, the statement is false

6. Answers my vary. Sample answer:

Understanding temperature changes helps us know how to protect the Earth.

Chapter 16

Let’s Warm-up

1. ₹3.70 = 3.70 × 100 = 370 p

2. 412 p = ₹ 412 100 = ₹4.12

3. ₹6.10 = 6.10 × 100 = 610 p

4. 305 p = ₹ 305 100 = ₹3.05

Do It Yourself 16A

1. a. Cost of 1 candy = ₹5

Cost of 7 candies = ₹5 × 7 = ₹35

If the cost of one candy is ₹5, then the cost of 7 candies will be ₹35

b. Pocket money received for 4 weeks = ₹240

Pocket money received for 1 week = ₹240 ÷ 4 = ₹60

If Lisa received ₹240 as pocket money for 4 weeks, the amount of money she received every week was ₹60

c. Money saved in 3 months = ₹600

Money saved in 1 month = ₹600 ÷ 3 = ₹200

Riya saved ₹600 in 3 months. She saved ₹200 every month.

2. Cost of 1 dozen candles = ₹15

Cost of 1 candle = ₹15 12

Cost of 20 candles = ₹ 15 12 × 20 = ₹25

3. Cost of 35 envelopes = ₹630

Cost of 1 envelope = ₹ 630 35 = ₹18

Cost of 57 envelopes = ₹18 ⨯ 57 = ₹1026

4. Cost of 3 4 m of cloth = ₹54

Cost of 1 m of cloth = ₹54 4 3 × = ₹ 216 3 = ₹72

Cost of 1 2 m of cloth = ₹ 1216216 236 ×= ⨯ ₹72 = ₹36

5. Price of 10 grams of gold = ₹88.62

Price of 1 gram of gold = ₹88.62 10 = ₹8.862

Price of 100 grams of gold = 100 ⨯ ₹88.62 = ₹886.2

So, the price of 100 grams of gold in 1947 was ₹886.20.

6. 1 month = 4 weeks

So, 2 months = 2 × 4 weeks = 8 weeks

Number of sessions in 1 week = 3 sessions

Number of sessions in 8 weeks = 8 × 3 = 24 sessions

Cost of each session = ₹550

Cost of 24 sessions = 24 × ₹550 = ₹13,200

So, Dinesh spent ₹13,200 on physiotherapy in 2 months.

Challenge

1. The cost for 2 pencils (since the 3rd is free) is 2 × ₹2 = ₹4

Each set of 3 pencils costs ₹4.

No. of sets we can buy = ₹50 ₹4 = 12.5

12 complete sets and ₹2 remaining.

Hence, 12 sets = 12 × 3 = 36 pencils

And 1 pencil for ₹2. Thus, a total of 37 pencils can be bought for ₹50.

Do It Yourself 16B

a. The amount of money that a seller spends to pay for an item is called the cost price

b. When the cost price is higher than the selling price, then there is a loss

c. When the selling price is greater than the cost price, then the seller is said to have gained a profit.

d. A flower vase is sold at ₹725 for a profit of ₹225. Its CP is ₹500

2. Profit = ₹30

SP = ₹540

CP = SP – P = ₹540 – ₹30 = ₹510

Since ₹540 > ₹510, your cost price was less than the selling price.

3. a. CP = ₹525, SP = ₹575

Profit = SP – CP = ₹575 – ₹525 = ₹50

b. CP = ₹200, SP = ₹155

Loss = CP – SP = ₹200 – ₹155 = ₹45

c. Loss = ₹550, SP = ₹7850

CP = SP + Loss = ₹7850 + ₹550 = ₹8400

d. CP = ₹1200, Profit = ₹85

SP = CP + Profit = ₹1200 + ₹85 = ₹1285

4. CP = ₹1100

a. SP = ₹1100

Since CP = SP, there is no profit or loss. Hence profit/loss = ₹0

b. Profit = ₹240

SP = CP + Profit = ₹1100 + ₹240 = ₹1340

c. SP = ₹940

Loss = CP – SP = ₹1100 – ₹940 = ₹160

5. The price of a barrel of oil in July 1990 = ₹262.50

The price of a barrel of oil in October 1990 = ₹612.50

The increase in price = ₹612.50 – ₹262.50 = ₹350

So, the person earned a profit of ₹350 per barrel by buying in July and selling in October.

6. Answers may vary. Sample answer:

The cost price of a toy is ₹1600. What will be the selling price if it is sold at a loss of ₹400?

Challenge

1. CP of 3 kg of oranges = ₹510

CP of 1 kg of oranges = ₹510 ÷ 3 = ₹170

CP of 4 kg of apples = ₹456

SP of 1 kg of oranges = ₹32

SP of 3 kg of oranges = ₹32 × 3 = ₹96

Total CP of oranges and apples = ₹510 + ₹456 = ₹966

Total SP of oranges and apples = Total CP of oranges and apples

SP of apples if there is no profit or loss = Total CP – SP of oranges = ₹966 – ₹96 = ₹870

SP of 4 kg of apples = ₹870

SP of 1 kg of apples = ₹870 ÷ 4 = ₹217.5

Thus, the shopkeeper should sell apples at ₹217.5 per kg so that he is able to recover his loss and break even.

Do It Yourself 16C

1. SP of the TV = ₹30,000

Profit = ₹1563

Profit = SP – CP ⇒ CP = SP – Profit

CP = ₹30000 – ₹1563 = ₹28,437

2. CP of 20 pencil boxes = ₹480

CP of 1 pencil box = ₹480 ÷ 20 = ₹24

SP of 1 pencil box = ₹50

Since, SP > CP

So, it is a profit.

Profit = ₹50 − ₹24 = ₹26

3. CP of 15 kg spices = ₹3000

CP of 1 kg of spices = ₹3000 ÷ 15 = ₹200

SP of 1 kg of spices = ₹ 250

Since, SP > CP

So, it is a profit.

Profit = ₹250 − ₹200 = ₹50

4. Total CP of the old smartphone = ₹5620 + ₹530 = ₹6150

SP of the smartphone = ₹6150

Since, CP = SP

So, neither profit and nor loss.

5. CP of the bicycle = ₹7830 + ₹270 = ₹8100

SP of the bicycle = ₹8000

Since, CP > SP

So, there is loss.

Loss = CP – SP = ₹8100 – ₹8000 = ₹100

7. Loan amount = ₹40,000

Monthly payment = ₹3500

Duration of pay back = 1 year (12 months)

Total amount = monthly payment × number of months = ₹3500 × 12 = ₹42,000

Therefore, Rina paid back a total of ₹42,000 to the bank.

Interest amount = Total amount paid – Loan amount

= ₹42,000 – ₹40,000 = ₹2000

6. Bill Date: 05/02/2023

1. Loan amount = 2000 SGD

Monthly payment = ₹5000

Exchange rate: 1 SGD = ₹62.5

Loan amount in ₹ = 2000 × ₹62.5 = ₹1,25,000

Number of months = 125000 5000 = 25

Therefore, Tanya will pay off the loan in 25 months.

Chapter Checkup

1. a. Selling price is the amount of money a seller pays to buy an item for his store. False

b. When CP < SP, the seller makes a profit. True

c. Since, CP < SP. It is a profit. True

d. Cost of 5 pencils = ₹30

Cost of 1 pencil = ₹30 ÷ 5 = ₹6

If the cost of 5 pencils is ₹30, then the cost of one pencil will be ₹8. False

2. a. CP = ₹167 and SP = ₹185

Since, SP > CP, it is a profit.

So, Profit = SP – CP = ₹185 – ₹167 = ₹18

b. CP = ₹36 and SP = ₹29

Since, CP > SP, it is a loss.

So, Loss = CP – SP = ₹36 – ₹29 = ₹7

c. CP = ₹147 and SP = ₹125

Since, CP > SP, it is a loss.

So, Loss = CP – SP = ₹147 – ₹125 = ₹22

3. SP of 1 bulb = ₹15

SP of 20 bulbs = ₹15 × 20 = ₹300

Profit = ₹5

CP of 20 bulbs = SP – Profit = ₹300 – ₹5 = ₹295

CP of 1 bulb = ₹295 ÷ 20 = ₹14.75

Hence, the cost price of each bulb was ₹14.75.

4. CP = ₹780 + ₹675 = ₹1455

SP = ₹1450

CP > SP, so it is a loss.

Loss = ₹1455 − ₹1450 = ₹5

5. Total CP = (₹200 × 30) + (₹100 × 20) = ₹8000

Total number of shirts = 30 + 20 = 50

Total SP = ₹300 × 50 = ₹15,000

SP > CP, so it is a profit.

Profit = ₹15000 − ₹8000 = ₹7000

6. 4 Pencils for ₹24

1 pencil = ₹24 ÷ 4 = ₹6

10 Pencils for ₹50

1 pencil = ₹50 ÷ 10 = ₹5

So, 10 pencils for ₹50 is a better buy.

7. 36 chairs at ₹550 each

1 chair = ₹550

22 chairs at ₹14300

1 chair = ₹14300 ÷ 22 = ₹650

So, 22 chairs at ₹14300 will cost more.

8. CP of 7 phones = ₹2100 × 7 = ₹14,700

Profit = ₹700

Total SP = ₹14,700 + ₹700 = ₹15,400

SP of one phone = ₹15400 ÷ 7 = 2200

10. Sum borrowed = ₹5000

Number of months = 10

Amount to be paid each month = ₹5000 10 = ₹500

11. Answers may vary. Sample answer:

A school owner takes a ₹30,000 loan from the bank to repair his school. He pays back the loan in 20 months. He paid an interest of ₹2000. How much does he pay each month if the payments are equal?

Challenge

1. Statement 1: Each worker gets paid ₹500 every day.

Statement 2: In 1 day each worker will get 1 3 of the total amount.

Number of workers = 3

Total amount of money got by 3 workers = ₹9000

Amount of money received by each worker = ₹9000 ÷ 3 = ₹3000

Number of days each worker works = 6

Money received by each worker each day = ₹3000 ÷ 6 = ₹500

So, statement 1 is true.

In 1 day, each worker gets ₹500 which is equal to 1 18 of the total amount. Thus, statement 2 is not correct.

2. Let the salary be ₹x.

Rent = ₹1 5 × x

Remaining salary after rent =

Food = ₹1

Remaining salary after food = ₹4

Clothes and transport = ₹ 1 3 × ₹3

Remaining salary after clothes and transport = ₹3 5 × x − ₹1 5 × x = ₹2 5 × x = ₹5000

x = ₹5000 × 5 2 = ₹12,500

Case Study

1. Number of stamps Shreya earned in first year = 2

Number of stamps Shreya earned in second year = 2 × 2 = 4 stamps.

Thus, Shreya collected 4 stamps in her second year.

2. Value of 1 stamp = ₹500

Value of 6 stamps = 6 × ₹500 = ₹3000 (option c)

3. Cost of 1 stamp = ₹450

Loss of ₹25 on each stamp = ₹450 � ₹25 = ₹425 (option d)

4. Answer may vary. Sample answer: It is important to work hard and try to improve because it helps us achieve more and become better at what we do.

Chapter 17

Let’s Warm-up

1. Kavya

2. Rehan

3. Total stamps = 50 + 25 + 25 = 100; half of 100 = 50 Hence, Kavya has half of total stamps.

4. 50 + 25 + 25 = 100

5. 50 – 25 = 25

Do It Yourself 17A

1. Least: V-A = 30 students Most: V-C = 50 students

2. 0246 810121416182022242628

Do It Yourself 17B

4. a. Number of children tickets sold on Sunday = 75; Number of children tickets sold on Friday = 40 Difference between the number tickets sold = 75 – 40 = 35

b. Total number of tickets sold the four days = 100 + 30 + 110 + 40 + 140 + 60 + 160 + 75 = 715

c. Total number of tickets sold on Sunday = 160 + 75 = 235

Total number of tickets sold on Saturday = 140 + 60 = 200 Difference in the tickets sold = 235 – 200 = 35

5. Answers may vary. Sample answers:

Q1. What is the number of adult tickets sold in the 4 days?

Q2. What is the difference in the children ticket sold on Thursday to that on Sunday?

Challenge

1. The total number of tickets sold to adults on Saturday and Sunday = 140 + 160 = 300

The total number of tickets sold to adults on Monday, Tuesday and Wednesday = 2 3 of 300 = 2 3 × 300 = 200

The total number of tickets sold to the children on Friday and Saturday = 40 + 60 = 100

The total number of tickets sold to the children on Monday, Tuesday and Wednesday = 4 5 of 100 = 4 100 5  = 80

Total tickets sold on Monday, Tuesday and

4. a. Fraction of students interested in Maths = 1 4

Total number of students = 200

Number of students interested in Maths = 200 × 1 20050 4 ×= students

b. Number of students interested in Science = 40

Total number of students = 200

Fraction of students interested in Science = 401 2005 =

c. Number of students interested in EVS = 10

Total number of students = 200

Fraction of students interested in EVS = 101 20020 =

5. Answers may vary. Sample answer: How many more students are interested in sports than Maths?

Challenge

1. Total global energy consumption = 13,000 million tonnes

We know that the pie chart represents a whole.

So, Energy consumption by oil + coal + natural gas + hydropower + nuclear + Renewables = 1 +++++ 3127631

100100255025 Renewables = 1 +++++ 31272464

100100100100100 Renewables = 100 100

100 + Renewables = 100 100

Renewables = 10092 100100 = 8 100

Energy consumed by Nuclear = 1 13,000 25  = 520 million tonnes

Energy consumed by Renewables = 8 13,000 100 × = 1040 million tonnes

Total energy consumed by Nuclear and Renewables combined = 520 + 1040 = 1560 million tonnes

Thus, option (c) is correct.

Do It Yourself 17C

1. Kavya spent the highest number of hours on the internet on Wednesday and Thursday.

2. The city visited by the most number of the people is Pune. The city visited by the least number of people is Goa.

3. a. Temperature recorded during the second week = 8°C

b. The highest temperature was recorded in Week 1.

c. Temperature in the first week = 10°C; Temperature in the last week = 2°C

Difference in temperature = 10°C – 2°C = 8°C

4. a. 700 guitars were sold in the year 2017.

b. 800 guitars were sold in the year 2021.

c. The least number of guitars were sold in the year 2018.

5. Answers may vary. Sample answer:

Q1. What is the total numbers of guitars sold during the given years?

Q2. In which years did Rahul sell more than 600 guitars? Challenge

1. a. The sale of cloth in May = 4000

We need to find the month in which the sale was 8000.

In April, the sale was 8000. Thus, in April the sale of cloth was double of that of May.

Total sales

Average Number of months

Total sales = 8000 + 4000 + 7000 + 6000 + 9000 = 34,000

Average sales = 34000 5 = 6800

The average sales over the period of 5 months were 6800.

Chapter Checkup

1. a. 15 families like cat.

b. 10 + 9 = 19

19 families prefer birds and rabbits as their pets.

c. Number of families who like dogs = 14; Number of families who like fish = 12

Difference = 14 – 12 = 2

The number of families who like dogs as their pet are 2 more than the number of families who like fish as their pet.

2. a. The boys and girls read an equal number of pages on Thursday.

b. Number of pages read by boys on Wednesday = 18; Number of pages read by girls on Wednesday = 15 Hence, boys read more pages than girls.

c. Total number of pages read by girls = (12 + 11 + 15 + 16 + 18) = 72 pages

Total number of pages read by boys = (10 + 14 + 18 + 16 + 17) = 75 pages

Difference in the number of pages read = 75 – 72 = 3 pages

December

November

October

September

4. a. Fraction of the area of the Pacific Ocean 251 502 ==

Total area of the oceans = 362 million sq. km

Area of the Pacific Ocean = × 1 362 2 = 181 million sq. km

b. Fraction of the area of the Indian Ocean 101 505 ==

Fraction of the area of the Atlantic Ocean 101 505 ==

Fraction of the Indian Ocean and the Atlantic Ocean = += 112 555

Area of the Indian Ocean and the Atlantic Ocean

= × 2 362 5 = 144.8 million sq. km

c. Fraction of the area of the Antarctic Ocean = 3 50

Fraction of the area of the Arctic Ocean = 1 25 = 2 50

On comparing the fractions of the area, we get

= >> 3231 or 50505025

so, the area of the Antarctic Ocean is more.

The area by the Antarctic Ocean more than the Arctic Ocean

= −= 321

505050

× 1 362

50 = 7.24 million sq. km

5. Activity

6. a. On Friday, the fruit seller sold 15 kg fruit.

b. Fruits sold on Saturday = 30 kg; Fruits sold on Sunday = 35 kg

Total fruits sold = 30 kg + 35 kg = 65 kg

c. Sale in the initial 4 days = (20 + 25 + 10 + 20) = 75 kg

Sale in the last 3 days = (15 + 30 + 35) = 80 kg

Difference in sale = 80 kg – 75 kg = 5 kg

7. a. As the temperature of Jaipur is highest, it is the hottest city.

b. As the temperature of Ooty is lowest, it is the coldest city.

c. The temperature of Chennai is 15°C.

d. Bangalore has a temperature of 10°C.

9. a. Chocolate ice cream is the most popular. b. Blueberry is the least favourite ice cream.

c. Children who like chocolate ice cream = 7 5014 25 ×= × 50 = 14 children

d. Children who like Vanilla = 5 25 × 50 = 10 children

Children who like Strawberry = 6 5012 25 ×= × 50 = 12 children

Children who like either vanilla or strawberry = 10 + 12 = 22 children

10. Answers will vary. Sample answer: What is the difference in the number of students who like Mango flavour to the one who like blueberry flavour?

Challenge

1. Production of fertilisers in 1998 = 45 thousand tonnes

Production of fertilisers in 2003 = Double of that in 1998

= 2 × 45,000 = 90,000 tonnes

Production of fertilisers in 1997 = 60 thousand tonnes

Production of fertilisers in 2004 = 8 6 of that in 1997

= × 8 60,000 6 = 80,000 tonnes

Total production over the span of 10 years = 25,000 + 40,000 + 60,000 + 45,000 + 65,000 + 50,000 + 75,000 + 80,000 + 90,000 + 80,000

= 6,10,000 tonnes

2. Total number of students who were surveyed = 600

Fraction of students whose favourite food is Indian = 5 8

Number of students whose favourite food is Indian

= 5 8 × 600 = 375

Fraction of students whose favourite food is Italian = 3 20

Number of students whose favourite food is Italian

= 3 20 × 600 = 90

Fraction of students whose favourite food is Mexican = 1 10

Number of students whose favourite food is Mexican = 1 10 × 600 = 60

Fraction of students whose favourite food is Chinese = 1 8

Number of students whose favourite food is Chinese = 1 8 × 600 = 75

Thus, option d is incorrect. Hence, the correct option is (d).

Case Study

1. Total number of tigers = Number of tigers in Rajasthan + Number of tigers in Andhra Pradesh + Number of tigers in Bihar + Number of tigers in Telangana = 88 + 62 + 31 + 21 = 202 tigers

Thus, option b is correct.

2. Total number of tigers = 202

Total number of elephants = 84 + 74 + 25 + 57 = 240

Difference between the number of tigers and elephants = 240 – 202 = 38

Thus, option c is correct.

3. Number of Tigers and Elephants in Indian States

Telangana Rajasthan Andhra Pradesh

Number of Elephants/Tigers Elephants

4. Andhra Pradesh has 12 more elephants than tigers.

5. Answers may vary.

About the Book

Imagine Mathematics seamlessly bridges the gap between abstract mathematics and real-world relevance, offering engaging narratives, examples and illustrations that inspire young minds to explore the beauty and power of mathematical thinking. Aligned with the NEP 2020, this book is tailored to make mathematics anxiety-free, encouraging learners to envision mathematical concepts rather than memorize them. The ultimate objective is to cultivate in learners a lifelong appreciation for this vital discipline.

Key Features

• Let’s Recall: Introductory page with a quick recall of concepts learnt in previous grades

• Real Life Connect: Introduction to a new concept related to real-life situations

• Examples: Solved problems showing the correct method and complete solution

• Do It Together: Guided practice for learners with clues and hints to help solve problems

• Think and Tell: Probing questions to stimulate Higher Order Thinking Skills (HOTS)

• Error Alert: A simple tip-off to help avoid misconceptions and common mistakes

• Remember: Key points for easy recollection

• Did You Know? Interesting facts related to the application of concept

• Math Lab: Fun cross-curricular activities

• Challenge: Critical thinking questions to enhance problem-solving and analytical thinking skills

• Case Study: Scenario-based questions to apply theory to real-life situations

• QR Codes: Digital integration through the app to promote self-learning and practice

About Uolo

Uolo partners with K-12 schools to provide technology-enabled learning programs. We believe that pedagogy and technology must come together to deliver scalable learning experiences that generate measurable outcomes. Uolo is trusted by over 15,000+ schools across India, Southeast Asia and the Middle East.

ISBN 978-81-979482-3-7

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