Imagine_Maths_TM_Grade6

Teacher Manual MATHEMATICS

MATHEMATICS

Master Mathematical Thinking

Grade 6

Mathematics is not just another subject. It is an integral part of our lives. It shapes the very foundation of our understanding, personality and interaction with the world around us.

However, due to the subject’s abstract nature, the stress of achieving high academic scores and complex teaching methods, most children develop a fear of mathematics from an early age. This fear not only hinders their mathematical thinking, logical reasoning and general problem solving abilities, but also negatively impacts their performance in other academic subjects. This creates a learning gap which widens over the years.

The NEP 2020 has distinctly recognised the value of mathematical thinking among young learners and the significance of fostering love for this subject by making its learning engaging and entertaining. Approaching maths with patience and relatable real-world examples can help nurture an inspiring relationship with the subject. 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 learning exciting, relatable and meaningful for children. This is achieved by making a clear connection between mathematical concepts and examples from daily life. This opens avenues for children to connect with and explore maths in pleasant, relatable, creative and fun ways.

This product, as recommended by the NEP 2020 and the recent NCF draft, gives paramount importance to the development of computational and mathematical thinking, logical reasoning, problem solving and mathematical communication, with the help of carefully curated content and learning activities.

Imagine Mathematics strongly 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 latest NCF Draft and other international educational policies. In this approach, while learning any new mathematical concept, learners first receive sufficient modelling, and then are supported to solve problems in a guided manner before eventually taking complete control of the learning and application of the concept on their own. 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 exercises among other things. Interactive exercises on the digital platform make learning experiential and help in concrete visualisation of abstract mathematical concepts.

In Imagine Mathematics, we are striving to make high quality maths learning available for all children across the country. The product maximises the opportunities for self-learning while minimising the need for paid external interventions, like after-school or private tutorial classes.

The book adapts some of the most-acclaimed, learner-friendly pedagogical strategies. Each concept in every chapter is introduced with the help of real-life situations and integrated with children’s experiences, making learning flow seamlessly from abstract to concrete. Clear explanations and simple steps are provided to solve problems in each concept. Interesting facts, error alerts and enjoyable activities are smartly sprinkled throughout the content to break the monotony and make learning holistic. Most importantly, concepts are not presented in a disconnected fashion, but are interlinked and interwoven in a sophisticated manner across strands and grades to make learning scaffolded, comprehensive and meaningful.

As we know, no single content book can resolve all learning challenges, and human intervention and support tools are required to ensure its success. Thus, Imagine Mathematics not only offers the content books, but also comes with teacher manuals that guide the pedagogical transactions that happen in the classroom; and a vast parallel digital world with lots of exciting materials for learning, practice and assessment. In a nutshell, Imagine Mathematics is a comprehensive and unique learning experience for children.

On this note, we welcome you to the wonderful world of Imagine Mathematics. In the pages that follow, we will embark on a thrilling journey to discover wonderful secrets of mathematics—numbers, operations, geometry and measurements, data and probability, patterns and symmetry, algebra and so on and so forth. Wishing all the learners, teachers and parents lots of fun-filled learning as you embark upon this exciting journey with Uolo.

Key Elements of a Lesson—A Quick Glance

Clear and concise lessons that can be implemented through the academic year with each perfectly aligned to the topics covered in the Imagine Mathematics learners’ content book.

Large Numbers 1

Large Numbers 1

Large Numbers 1

the questions given in the Let’s Warm-up section. Vocabulary 9-digit

Alignment to NCF

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

Teaching Aids

C-1.1: Develops a sense for and an ability to manipulate (e.g., read, write, form, compare, estimate, and apply operations) and name (in words) large whole numbers of up to 20 digits, and expresses them in scientific notation using exponents and powers

Let’s Recall

Recap to check if students know how to read and write 7-digit numbers.

Ask students to solve the questions given in the Let’s

Place value blocks of 10s, 100s and 1000s; Bingo cards; Number cards (0–9); Bowl; Indian place value chart; International place value chart; Clue cards to form a 9-digit number 1

Clear,

Large Numbers 1

specific and

learning

that show what students should know, understand, or do by the end of the lesson.

Learning Outcomes

Large Numbers 1

Students will be able to: read and write Roman numerals.

read and write 9-digit numbers in the Indian number system. read and write 9-digit numbers in the international number system. compare and order 9-digit numbers and form numbers using the given digits.

Alignment to NCF

Learning Outcomes

C-1.1: Develops a sense for and an ability to manipulate (e.g., read, write, form, compare, estimate, and apply operations) and name (in words) large whole numbers of up to 20 digits, and expresses them in scientific notation using exponents and powers

Large Numbers 1

Let’s Recall

Recap exercises to check the understanding of prerequisite concepts before starting a topic.

Recap to check if students know how to read and write 7-digit numbers. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

Alignment to NCF

Learning Outcomes

QR

9-digit numbers: numbers that have place values ranging from ones to crores, that is, from 10,00,00,000 to 99,99,99,999 roman numbers: a system where letters were used by ancient Romans to represent numbers, which is still used today

C-1.1: Develops a sense for and an ability to manipulate (e.g., read, write, form, compare, estimate, and apply operations) and name (in words) large whole numbers of up to 20 digits, and expresses them in scientific notation using exponents and powers

Let’s Recall

Teaching Aids

Recap to check if students know how to read and write 7-digit numbers.

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

Vocabulary

Students will be able to: read and write Roman numerals. read and write 9-digit numbers in the Indian number system. read and write 9-digit numbers in the international number system. compare and order 9-digit numbers and form numbers using the given digits.

Alignment to NCF

C-1.1: Develops a sense for and an ability to manipulate (e.g., read, write, form, compare, estimate, and apply operations) and name (in words) large whole numbers of up to 20 digits, and expresses them in scientific notation using exponents and powers

9-digit numbers: numbers that have place values ranging from ones to crores, that is, from 10,00,00,000 to 99,99,99,999 roman numbers: a system where letters were used by ancient Romans to represent numbers, which is still used today

Teaching Aids

Let’s Recall

Recap to check if students know how to read and write 7-digit numbers.

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

Place value blocks of 10s, 100s and 1000s; Bingo cards; Number cards (0–9); Bowl; Indian place value chart; International place value chart; Clue cards to form a 9-digit number 1

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

Vocabulary

9-digit numbers: numbers that have place values ranging from ones to crores, that is, from 10,00,00,000 to 99,99,99,999 roman numbers: a system where letters were used by ancient Romans to represent numbers, which is still used today

Teaching Aids

Place value blocks of 10s, 100s and 1000s; Bingo cards; Number cards (0–9); Bowl; Indian place value chart; International place value chart; Clue cards to form a 9-digit number 1

Place value blocks of 10s, 100s and 1000s; Bingo cards; Number cards (0–9); Bowl; Indian place value chart; International place value chart; Clue cards to form a 9-digit number 1

12:16:16 PM

Chapter: Large Numbers

Period Plan

The teacher manuals corresponding to Imagine Maths 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 problemsolving 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.

Roman Numerals

Indian Number System

International Number System

Comparing and Ordering Numbers

Revision 2.

Rounding Off Large Numbers

Estimation in Operations

Addition and Subtraction

Multiplication and Division

Parentheses and Order of Operations

Multi-step Word Problems

Revision

Whole Numbers on a Number Line

Addition of Whole Numbers

Subtraction of Whole Numbers

Multiplication of Whole Numbers

Division of Whole Numbers

Patterns in Whole Numbers

Revision

Reviewing Factors and Multiples

Types of Numbers

Divisibility Rules

4.

Prime Factorisation

Highest Common Factor

Lowest Common Multiple

Word Problems in LCM and HCF

Revision

6.

9

Basics of Angles

Measuring Angles

Types of Angles

Parts of a Triangle

Classification of Triangles on the Basis of Sides

Classification of Triangles on the Basis of Angles

Parts of a Quadrilateral

Types of Quadrilaterals

Parts of a Polygon

Types of Polygons

Positive and Negative Integers

Addition of Integers

Subtraction of Integers

Representing Fractions

Comparing and Ordering Fractions

Converting Fractions

Equivalent Fractions and Simplest Form

Addition and Subtraction of Fractions

Multiplication and Division of Fractions and Mixed Numbers Revision

Representing Decimals on Number Line

Place Values in Decimals

Conversion Between Fractions and Decimals

More on Decimals

Addition and Subtraction of Decimals

Multiplication of Decimals

Division of Decimals

Application of Decimals

Revision

14.

17.

Perimeter of Regular Shapes

Area on Graph Paper

Area of Squares and Rectangles

Area and Perimeter of Combined Shapes

Rules in Patterns

Forming Algebraic Expressions

Forming Algebraic Equations

Solving Linear Equations in One Variable

Ratio

Unitary

Constructing Line Segments and Their Copy

Constructing a Perpendicular Using a Compass

Constructing a Perpendicular Bisector

Constructing Angles of Given Measures

Constructing a Copy of an Angle

Constructing an Angle Bisector

Constructing 60°, 120° and 30° Angles

Constructing 45°, 75°, 150° and 135° Angles

Revision

Large Numbers 1

Learning Outcomes

Students will be able to:

read and write Roman numerals. read and write 9-digit numbers in the Indian number system. read and write 9-digit numbers in the international number system. compare and order 9-digit numbers and form numbers using the given digits.

Alignment to NCF

C-1.1: Develops a sense for and an ability to manipulate (e.g., read, write, form, compare, estimate, and apply operations) and name (in words) large whole numbers of up to 20 digits, and expresses them in scientific notation using exponents and powers

Let’s Recall

Recap to check if students know how to read and write 7-digit numbers. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

9-digit numbers: numbers that have place values ranging from ones to crores, that is, from 10,00,00,000 to 99,99,99,999

roman numbers: a system where letters were used by ancient Romans to represent numbers, which is still used today

Teaching Aids

Place value blocks of 10s, 100s and 1000s; Bingo cards; Number cards (0–9); Bowl; Indian place value chart; International place value chart; Clue cards to form a 9-digit number

Chapter: Large Numbers

Roman Numerals

Learning Outcomes

Students will be able to read and write Roman numerals.

Teaching Aids

Place value blocks of 10s, 100s and 1000s; Bingo cards

Activity

Imagine Maths Page 2

Begin by showing the students a clock with Roman numerals. Explain the 7 basic symbols used to form Roman numbers. Tell them about the Roman numerals that can be repeated, added or subtracted. Elaborate the rules used to form these numerals. You may point at the clock to demonstrate some examples as you explain. Ask the class to work in groups of 4. Distribute bingo cards with a 5 × 5 grid, of random Roman numbers up to 2000, to each group.

Use the place value blocks to form a 2-digit, 3-digit or 4-digit number (up to 2000). Ask one student from each group to come up and read the number and go back and tell his/her group at number. Instruct group members to work together to find its corresponding Roman number and cross it off from their bingo card. Repeat this process until all the Roman numbers in a line—horizontally, vertically or diagonally—are crossed off. The group to cross off the Roman numbers in all three directions first, wins the game.

Extension Idea

Ask: What should be subtracted from MDLXVIII to get MXLVIII?

Say: MDLXVIII = 1568, MXLVIII = 1048, 1568 – 1048 = 520 = DXX. So, DXX should be subtracted from MDLXVIII to get MXLVIII.

Indian Number System

Learning Outcomes

Students will be able to read and write 9-digit numbers in the Indian number system.

Teaching Aids

Number cards (0–9); Bowl; Indian place value chart

Activity

Prepare a bowl of number cards (0–9) and keep it on your table for the activity.

Imagine Maths Page 4

Ask the students to work in groups of 4 and distribute a blank Indian place value chart to each group. Instruct one student from each group to come forward and pick 9 numbers from the bowl, then go back and show the digits to his/her group. Ask the groups to form the largest and the smallest 9-digit numbers on the place value chart using those digits. Instruct the students to record the expanded form and number name in the Indian system in their notebooks.

Ask questions like: What are periods? What are all the places in the number? Does the face value of the digit change with its place value? Why and where do we put commas in the number?

Learning Outcomes

Students will be able to read and write 9-digit numbers in the international number system.

Teaching Aids

Number cards (0–9); Bowl; International place value chart

Activity

Ask the students to work in groups of 4 and distribute a blank international place value chart to each group. Reuse the number cards (0–9) from the previous lesson. Repeat the activity to get groups to form the largest and the smallest 9-digit numbers on the place value chart. Instruct the students to record the expanded form and number name in the international system in their notebooks.

Ask questions like: How is the placement of commas different in the two systems? How does it impact the place value of each digit and the number name in the two systems?

Extension Idea

Ask: How many millions are there in 16,89,04,231?

Say: 16,87,04,231 is written in the Indian number system. In the international number system, it will be written as 168,704,231. So, there are 168 millions in this number.

Comparing and Ordering Numbers

Learning Outcomes

Imagine Maths Page 10

Students will be able to compare and order 9-digit numbers and form numbers using the given digits.

Teaching Aids

Clue cards to form a 9-digit number

Activity

Prepare cards with clues to form a 9-digit number, such as: thenumber 5 is in the thousands place, 4 is in the lakhs place, the number starts with 8, the number in the tens place is 2 more than the number in the ones place, etc.

Ask the class to work in groups of 4. Distribute the clue cards to each group.

Instruct them to use the clues to form the number in their notebooks. Then, ask the groups to interchange the digits in the ones and thousands places to form a second number, then interchange the lakhs and hundreds places in the original number to form the third number, then interchange the crores and ten thousand places in the original number to form the fourth number. For example: If the number as per the clues is 3,54,87,590, the 3 new numbers are 3,54,80,597; 3,55,87,490; and 8,54,37,590.

Instruct the students to compare and arrange the numbers in ascending and descending order in their notebooks. Ask questions like: How are the two orders different from each other? How many places have the same digits and different digits? What points should be kept in mind while comparing and ordering numbers?

Extension Idea

Ask: Take the 9-digit number 87,06,98,543. If you add 1 to one place at a time (starting with the lowest) and form a new number each time, what numbers will you get? List the first four numbers and arrange them in descending order.

Say: The first four numbers would be 87,06,98,544 (87,06,98,543 + 1); 87,06,98,553 (87,06,98,543 + 10); 87,06,98,643 (87,06,98,543 + 100); 87,06,99,543 (87,06,98,543 + 1000). So, the descending order is 87,06,99,543 > 87,06,98,643 > 87,06,98,553 > 87,06,98,544.

Answers

1. Roman Numerals

Do It Together

1. 45  MCCCXC

2. 98  CX

3. 110  XLV

4. 185  XCVIII

5. 1390  CLXXXV

2. Face Value and Place Value

Do It Together

3 is in the crores place.

The place value of 3 is 3,00,00,000.

The face value of 3 is 3.

3. Expanded Form and Number Names

Think and Tell

Yes. Population of U.S.—33,99,96,563: thirty-three crore ninety-nine lakh ninety-six thousand five hundred sixty three.

Population of Russia—14,44,44,359: fourteen crore forty four lakh forty-four thousand three hundred fifty-nine.

Population of Japan—12,32,94,513: twelve crore thirty-two lakh ninety-four thousand five hundred thirteen.

Do It Together

56,70,34,219 = 500000000 + 60000000 + 7000000 + 30000 + 4000 + 200 + 10 + 9

Number name: fifty-six crore seventy lakh thirty-four thousand two hundred nineteen.

4

. Face Value and Place Value

Do It Together

4 is in the millions place.

So, the place value of 4 is 4,000,000. The face value of 4 is 4.

5. Expanded Form and Number Names

Do It Together

902,564,875 = 900000000 + 2000000 + 500000 + 60000 + 4000 + 800 + 70 + 5

Number name: nine hundred two million five hundred sixtyfour thousand eight hundred seventy-five.

6. Comparing and Ordering Numbers

Think and Tell

Population of Russia = 14,44,44,359

Population of Indonesia = 27,75,34,122

1 < 2, so 14,44,44,359 < 27,75,34,122. So, the population of Russia is less.

Think and Tell

Yes. Descending order: 33,99,96,563 > 27,75,43,122 > 14,44,44,359 > 12,32,94,513

7. Forming Numbers

Think and Tell

Yes, more such numbers would be 517860932, 783096521, 602983571, etc.

Think and Tell

The greatest 9-digit number can be formed by using the two greatest 1-digit numbers i.e. 8 and 9: 99,99,99,998. The smallest 9-digit number can be formed by using the two smallest 1-digit numbers i.e. 1 and 0: 10,00,00,000.

Do It Together

Ascending Order: 126,856,384 < 165,121,349 < 400,897,234 < 860,778,653

Descending Order: 860,778,653 > 400,897,234 > 165,121,349 > 126,856,384

Estimation and Operations on Large Numbers 2

Learning Outcomes

Students will be able to:

round off large numbers to the nearest 10, 100 or 1000. estimate the sum, difference, product or quotient in large numbers. add or subtract numbers up to 9 digits.

multiply or divide large numbers up to 9 digits by numbers up to 3 digits. use parentheses to show the order of operations. solve multi-step word problems on operations on numbers up to 9 digits.

Alignment to NCF

C-1.1: Develops a sense for and an ability to manipulate (e.g., read, write, form, compare, estimate, and apply operations) and name (in words) large whole numbers of up to 20 digits, and expresses them in scientific notation using exponents and powers

Let’s Recall

Recap to check if students know how to add and subtract 4-digit and 5-digit numbers. 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

DMAS: the order of operations (division, multiplication, addition, subtraction) that makes it easy to solve a mathematical problem with multiple operations

Teaching Aids

Sticky notes; Cards with sets of numbers with operations; Bowl; Puzzle cards; Bingo cards; Problem cards; Crayons; Puzzle interlocking cards; Question cards

Chapter: Estimation and Operations on Large Numbers

Rounding Off Large Numbers

Learning Outcomes

Students will be able to round off large numbers to the nearest 10, 100 or 1000.

Teaching Aids

Sticky notes

Activity

Imagine Maths Page 17

Begin the class by asking the students where they might use rounding off numbers in real life. Use these scenarios to review the concept of rounding off large numbers to the nearest 10, 100 and 1000.

Write various large numbers on sticky notes and paste them on the board.

Divide the class into small groups and call a student from each group up to the board to select a number. Instruct group members to work together to round off the selected number to the nearest 10, 100 and 1000, and write their answers on another sticky note to be pasted on the board next to the original number. If their answers are correct, then the group gets a point; otherwise no points. Repeat this process until all the members of the group are covered. The group with the maximum points wins!

Extension Idea

Ask: I am thinking of a number. When I round off the number to the nearest 100, it is 2,43,26,500. When I round off the same number to the nearest 10, it is 2,43,26,450. What could my number be?

Say: Considering the possible numbers before rounding off to the nearest 10, the number may lie between 2,43,26,445 and 2,43,26,454. But to meet the condition for rounding off to the nearest 100, the number would be between 2,43,26,450 and 2,43,26,549.

Estimation in Operations

Learning Outcomes

Imagine Maths Page 19

Students will be able to estimate the sum, difference, product or quotient in large numbers.

Teaching Aids

Cards with sets of numbers with operations; Bowl

Activity

Prepare some cards with sets of numbers and operations written on them, such as Sum to the nearest 100: 1,24,37,548 + 3,25,43,768 and Difference to the nearest 1000s: 98,78,563 45,47,658, and keep them in a bowl. Divide the board into 4 columns for sum, difference, product and quotient. Then, divide the class into teams of equal sizes.

Start the activity by calling the one student from each team to the front of the class. He/she should pick a card from the bowl, read the expression and quickly find the estimated answer and write it in the relevant column on the board. If the result is correct, award the team 1 point. Continue in this manner until all the students in each team have had a turn. The team with the highest total score wins the game.

Learning Outcomes

Students will be able to add or subtract numbers up to 9 digits.

Teaching Aids

Puzzle cards

Activity

Prepare puzzle cards involving addition and subtraction of numbers up to 9 digits.

Divide the class into groups of equal sizes. Distribute the puzzle cards (9 individual cards) to each group.

Instruct the students to solve the calculations on the puzzle cards and arrange the cards to form a 3 × 3 puzzle matrix.

The group which forms the matrix in the least time wins.

Review the matrix formed by each group and discuss the answers. Also ask the students to write the answers in their notebook.

Extension Idea

Ask: Tripti is keeping track of how far she walks each week. On Monday, she walks 1780 m and on Tuesday she walks 50 m farther than on Monday. If her target is to walk 2000 m each day, then by how much is she short of her target over the two days?

Say: Tripti walks 1780 m on Monday. She walks 1780 m + 50 m = 1830 m on Tuesday. Her target distance for two days is 4000 m. So, she is short of the target by 4000 – (1780 + 1830) m = 390 m.

Learning Outcomes

Students will be able to multiply or divide large numbers up to 9 digits by numbers up to 3 digits.

Teaching Aids

Bingo cards; Problem cards; Crayons

Activity

Prepare a set of cards or small pieces of paper with multiplication and division problems, like 56 × 21 or 7896 ÷ 9, written on them with their corresponding answers. Shuffle the cards and place them face down in a pile. Write the answers to the problems on bingo cards.

117636486,11,5478456

Ask the students to work in groups.

Distribute a bingo card to each group. Play the game by drawing cards, one by one, from the pile and reading the questions aloud. Ask the groups to calculate the answers (they may use their notebooks for rough work) and cross out/colour the corresponding square on their bingo card if the answer matches. The game will continue until a group successfully completes a row, column, or diagonal on their bingo card and shouts “BINGO!” to indicate their win. Verify all the answers on the winning bingo card.

Parentheses and Order of Operations

Learning Outcomes

Students will be able to use parentheses to show the order of operations.

Teaching Aids

Puzzle interlocking cards; Bowls

Activity

Imagine Maths Page 27

Make sets of puzzle interlocking cards where one card (red) has an expression involving DMAS and brackets and its interlocking card (green) has the answer. Put the red cards in one bowl and the green cards in another bowl.

Divide the class into groups and distribute bowls with both types of cards to each group. Instruct each group to pick a red card, solve the expression, and match it with its corresponding green card. Instruct them to match all the cards in the two bowls. The group which completes all the pairs first wins. Verify all the answers and ask the students to write them in their notebooks.

Extension Idea

Ask: What operations would you fill in these blanks to make the given statement true? (20 8) 2 9 = 5

Say: If we add 20 and 8, we get 28. Then, we divide it by 2 to get 14. Next, if we subtract 9 from 14, we get 5. So, the correct order of operations is: (20 + 8) ÷ 2 − 9 = 5.

Learning Outcomes

Students will be able to solve multi-step word problems on operations on numbers up to 9 digits.

Teaching Aids

Question cards

Activity

Ask the students to work in groups. Write a word story on the board and ask the students to copy that into their notebooks.

Richa is planning a birthday party for her younger sister. She wants to buy decorations, snacks and gifts. She has a budget of ₹8000 for the party. Decorations cost ₹1545, snacks cost ₹2645 and gifts cost ₹1850. Richa also wants to save some money for unexpected expenses.

Make question cards and distribute these to each group.

Instruct the students to read the questions on the cards and solve them in their notebooks.

Ask groups to exchange question cards and solve them for more practice.

Extension Idea

Ask: Use the given clues to form a word problem:

School, whiteboard, donate, buy, ₹1,55,300, ₹25,000, nine, local business

How much money is Richa left with?

If Richa spends ₹1500 on transportation, how much money will she be left with?

Say: Sample problem: A school plans to buy five new whiteboards, which costs ₹1,55,300 each. Nine local businesses have agreed to donate ₹25,000 each. How much more money does the school need to be able to buy the whiteboards?

Answers

1. Rounding Off Large Numbers

Think and Tell

To round off an 8-digit number to the largest place, that is, crores, we look at the digit to its right (that is the ten lakhs place). If the digit is > or = 5, the number rounds off to the next crores, else the number rounds off to the same crores.

Do It Together

So, the number 73,26,63,547 rounded off to the nearest 10 is 73,26,63,550.

So, the number 73,26,63,547 rounded off to the nearest 100 is 73,26,63,500.

So, the number 73,26,63,547 rounded off to the nearest 1000 is 73,26,64,000.

2. Estimating the Sum and Difference

Do It Together

1. 765 is rounded off to 800.

213 is rounded off to 200.

So, the estimated sum of 765 and 213 is 1000.

2. 984 is rounded off to 980.

356 is rounded off to 360.

So, the estimated difference of 984 and 356 is 620.

3. Estimating the Product

Do It Together

758 is rounded off to 800.

510 is rounded off to 500.

The estimated product of 758 and 510 is 4,00,000.

So, Muskan’s estimated product is incorrect.

4. Estimating the Quotient

Do It Together

332 is rounded off to 300. 25 is rounded off to 30.

300 ÷ 30 = 10.

So, the estimated quotient of 332 ÷ 25 is 10.

5. Addition and Subtraction

Think and Tell

30 kg 100 g and 14 kg 500 g are written as 30.100 kg and 14.500 kg because 1 kg = 1000 g, so 1 g = 1 1000 kg = 0.001 kg.

Do It Together

Number of children in the city = 37,65,392 – (18,90,898 + 16,54,976)

So, the number of children in the city was 2,19,518.

6. Multiplication and Division

Do It Together

Total fee collected by the school = 5678 × 16,000

So, the total fee collected by the school is 9,08,48,000.

7. Parentheses and Order of Operations

Think and Tell

Using brackets makes it easy to understand and solve the problems. So, the method is easier to use to find the answer.

Do It Together

8 × 105 = 8 × (100 + 5) = (8 × 100) + (8 × 5) = 800 + 40 = 840

8

. Multi-step Word Problems

Do It Together

Total number of tokens they have together = 152 + 84 = 236

Number of tokens needed to buy a large toy monkey = 300

Number of tokens Nikhil and Preeti need to buy the toy monkey = 300 – 236 = 64 tokens.

Whole Numbers 3

Learning Outcomes

Students will be able to:

show whole numbers on a number line and identify the predecessor and successor of a whole number. identify the properties of addition of whole numbers and add them. identify the properties of subtraction of whole numbers and subtract them. identify the properties of multiplication of whole numbers and multiply them. identify the properties of division of whole numbers and divide them. identify patterns in whole numbers and extend them.

Alignment to NCF

C-1.1: Develops a sense for and an ability to manipulate (e.g., read, write, form, compare, estimate, and apply operations) and name (in words) large whole numbers of up to 20 digits, and expresses them in scientific notation using exponents and powers

C-1.2: Discovers, identifies, and explores patterns in numbers and describes rules for their formation (e.g., multiples of 7, powers of 3, prime numbers), and explains relations between different patterns

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

Let’s Recall

Recap to check if students know how to recognise natural numbers and perform skip counting by various intervals on a number line.

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

Vocabulary

consecutive numbers: numbers one after another property: rules defining the relationship between numbers

Teaching Aids

Number cards per the number of students; Paper cups; Marbles; Beads; Number cards (1–10)

Chapter: Whole Numbers

Learning Outcomes

Imagine Maths Page 36

Students will be able to show whole numbers on a number line and identify the predecessor and successor of a whole number.

Teaching Aids

Number cards as per the number of students

Activity

Prepare stacks of number cards as per the number of students in the class.

Take the students out to the playground.

Shuffle the stack and distribute one card to each student.

Instruct the students to move around and find the student with the predecessor of their number and stand behind them, forming a line.

Once students have found their predecessors, they will naturally form a line based on the numerical order of their cards.

Repeat the activity by shuffling and distributing the cards again and asking the students to stand in front of the student that has the successor of their number.

Ask questions like: What do you think is the importance of understanding the predecessor and successor of a number?

Extension Idea

Ask: B is positioned 2 places behind A and 3 places in front of the number 26. What is A’s position?

Say: As B is 3 places in front of the number 26, B’s position is 23. As B is positioned 2 places behind A, A’s position is 21.

Addition of Whole Numbers

Learning Outcomes

Imagine Maths Page 39

Students will be able to identify the properties of addition of whole numbers and add them.

Teaching Aids

Paper cups; Marbles

Activity

Ask the students to work in groups. Distribute paper cups and marbles to each group. Instruct the groups to show 3 + 5 using the marbles and cups and write the result in their notebooks. Verify if the obtained number is a whole number. Explain that this is the closure property.

Ask them to now show 5 + 3 using the marbles and cups and note down the answer. Discuss the commutative property.

Instruct the groups to show 5 marbles in one cup and no marble in another cup. Ask them to write the addition sentence for this. Discuss the additive property of zero.

Instruct the students to again show 3 + 5 using two cups and then place 4 marbles in a third cup. Ask them to count the marbles in cups 1 and 2, then add that amount to the number of marbles in cup 3 and record the results in their notebooks. Then, ask them to count the marbles in cup 1, count the marbles in cups 2 and 3 together, add both the amounts and record the answers in their notebooks. Discuss how the sum remains the same even when we change how we group the numbers, and explain that this is the associative property.

Subtraction of Whole Numbers

Learning Outcomes

Imagine Maths Page 40

Students will be able to identify the properties of subtraction of whole numbers and subtract them.

Teaching Aids

Paper cups; Marbles

Activity

Ask the students to work in groups. Distribute paper cups and marbles to each group.

Instruct them to show 8 – 3 and then 3 – 8 using the cups and marbles and record the results in their notebooks. Discuss whether they get a whole number or the same answer in both cases and explain that subtraction of whole numbers does not satisfy the closure and commutative properties.

Discuss the property of zero by asking them to put some marbles in one cup and then to not remove any marble. Instruct the students to put 8 marbles in cup 1 and take away 3 marbles from it. Ask them to take away 2 more marbles from the remaining and find the result. Guide them in solving (8 – 3) – 2 in their notebooks. Then, ask them to show 8 – (3 −2) using the cups and marbles. Discuss how rearranging numbers affects the result of subtraction, which does not meet the associative property.

Discuss inverse operations by asking the students to show 7 – 4 and then 4 + 3 and note down the results in their notebooks.

Multiplication of Whole Numbers

Learning Outcomes

Imagine Maths Page 43

Students will be able to identify the properties of multiplication of whole numbers and multiply them.

Teaching Aids

Beads

Activity

Ask the students to work in groups. Distribute beads to each group.

Instruct them to show 6 × 3 and 3 × 6 using the beads and record the results in their notebooks as a multiplication sentence. Discuss the closure and commutative properties.

Instruct the groups to show 3 groups of zero beads and record the result in their notebooks. Discuss the multiplicative property of zero.

Ask the students to show 6 × 1 and 1 × 6 and record the results in their notebooks. Discuss the multiplicative identity. Instruct the groups to show 6 × 3 using beads and make 2 groups of 6 × 3. Help them understand that this can be shown as (6 × 3) × 2. Ask them to show 6 × (3 × 2) and check the result. Discuss the associative property and ask the groups to record the results in their notebooks.

Instruct the groups to show 6 × 3; 6 × 2; 6 × 5; and 6 × 1; and record the results in their notebooks. Discuss how (6 × 3) + (6 × 2) = 6 × 5 and (6 × 3) − (6 × 2) = 6 × 1. Discuss the distributive property of multiplication over addition and subtraction.

Extension Idea

Ask: Suppose you have one group of 13 red and 17 blue beads. How many beads are there in total if there are 10 such groups?

Say: 10 × (13 + 17) = 130 + 170 = 300. So, there are 300 beads in total.

Division of Whole Numbers

Learning Outcomes

Imagine Maths Page 46

Students will be able to identify the properties of division of whole numbers and divide them.

Teaching Aids

Beads

Activity

Ask the students to work in groups. Distribute the beads to each group.

Instruct the students to show 10 ÷ 5 by threading a total of 10 beads with 5 beads in each group and record the result in their notebooks. Instruct them to then show 5 ÷ 10.

Ask questions like: Can you divide 5 beads into groups of 10? Do we obtain the same number of groups in both cases? Discuss how division of whole numbers does not satisfy the closure and commutative properties.

Instruct the groups to show (10 ÷ 5) ÷ 2 and 10 ÷ (5 ÷ 2) and discuss the associative property.

Next, instruct them to show 10 ÷ 1 and 10 ÷ 10 and record the number of groups formed in their notebooks. Ask questions like: What did you observe on dividing the 10 beads into groups of 1 and 10?

Encourage critical thinking by asking: What happens if you attempt to divide the 10 beads into groups of zero?

Instruct the students to check all the results using the division algorithm.

Extension Idea

Instruct: Consider these problems: (20 ÷ 4) ÷ 2 and 20 ÷ (4 ÷ 2). Solve and check if the results are the same.

Say: (20 ÷ 4) ÷ 2 = 2 1 2 and 20 ÷ (4 ÷ 2) = 10. So, the results are not the same. Quotients change when the order of the numbers is changed.

Patterns in Whole Numbers

Learning Outcomes

Students will be able to identify patterns in whole numbers and extend them.

Teaching Aids

Number cards (1–10)

Activity

Ask the students to form groups of 5. Assign each student a number from 1 to 5.

Imagine Maths Page 49

Instruct Student 1 to select a number card (for example 5), multiply the number by their assigned number, and write the result in their notebook, i.e. 5 × 1 = 5. Ask Student 1 to whisper the product (5) to Student 2, who will

repeat the process to get 5 × 2 = 10. This pattern [10 × 3 = 30; 30 × 4 = 120] should continue until Student 4 whispers the calculated product to Student 5.

Ask Student 5 to determine the original number on the card selected by Student 1 based on the information received.

120 ÷ 4 = 30 30 ÷ 3 = 10 10 ÷ 2 = 5 5 ÷ 1 = 5

Extension Idea

Ask: What was the original number if the product obtained by Student 4 is 240?

Say: 240 ÷ 4 = 60 60 ÷ 3 = 20 20 ÷ 2 = 10 10 ÷ 1 = 10.

So, the original number was 10.

Answers

1. Whole Numbers on a Number Line

Think and Tell

Yes. All natural numbers 1, 2, 3... are whole numbers.

Do It Together

The predecessor of 87,452 = 87,452 – 1 = 87,451

The predecessor of 87,451 = 87,451 – 1 = 87,450

Therefore, the three consecutive numbers before 87,453 are 87,452, 87,451 and 87,450.

2. Representing Numbers on the Number Line

Think and Tell

The largest whole number is infinite because every whole number must have a successor.

Do It Together

1. If you are standing in section 3, the whole number that represents the section just before the one you’re in is section 2.

2. If you take a step to the right and move to section 5, the whole number that represents the section you were previously in is section 4.

3. Operations on a Number Line

Do It Together

Cost of 4 candies = ₹20

Cost of 1 candy = ₹20 ÷

Therefore, the cost of 1 candy is ₹5.

4. Properties of Addition

Do It Together

Money in Vaani’s bank account = ₹478

Money Vaani deposited = ₹22

Money in her sister’s bank account = ₹589

Money her sister deposited = ₹53

Total money they have = ₹478 + ₹22 + ₹589 + ₹53 = ₹1142.

5. Properties of Subtraction

Do It Together

Money the company had left = ₹9,85,610 – ₹5,54,000 = ₹4,31,610

6. Properties of Multiplication

Do It Together

Cost of 1 refrigerator = ₹30,000

Cost of 21 refrigerators = 21 × ₹30,000

Cost of 1 microwave = ₹20,000

Cost of 21 microwaves = 21 × ₹20,000

Hence, the total amount spent by the restaurant is 21 × (30,000 + 20,000) = 21 × 50,000 = ₹10,50,000

7. Properties of Division

Number of schools the books were distributed in = 233

Do It Together Total number of books = 56,458 242 233 56458 – 466 985 – 932 538 – 466

Number of books each school received = 56,458 ÷ 233 = 242 Therefore, each school received 242 books.

8. Patterns in Whole Numbers

Do It Together

Total number of employees = 4525

Bonus each employee got = ₹999

Total bonus handed out = 4525 × ₹999 = 4525 × (1000 – 1) = 4525 × 1000 – 4525 × 1 = 45,25,000 – 4525 = ₹45,20,475

4

Playing with Numbers: Factors and Multiples

Learning Outcomes

Students will be able to:

list the factors and multiples of a number. identify perfect, odd, even, prime and composite numbers. identify and apply the divisibility rule for 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11. use the factor tree and repeated division to express a number as a product of its prime factors.

find the HCF of 2 or more numbers using prime factorisation and repeated division.

find the LCM of 2 or more numbers by prime factorisation and the division method. solve word problems on LCM and HCF.

Alignment to NCF

C-1.2: Discovers, identifies, and explores patterns in numbers and describes rules for their formation (e.g., multiples of 7, powers of 3, prime numbers), and explains relations between different patterns

Let’s Recall

Recap to check if students know how to find the factors of any number. 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 multiples: number obtained when a number is multiplied by an integer prime factorisation: a way of expressing a number as a product of its prime factors lowest common multiple: the smallest multiple that two or more numbers have in common

Teaching Aids

Squared paper; Number cards; Slips with divisibility rules; Square and circle cutouts; Orange ribbon (24 cm); Yellow ribbon (36 cm); A pair of scissors; Glue; Ruler; Sheets of paper; Word problem cards

Chapter: Playing with Numbers: Factors and Multiples

Learning Outcomes

Students will be able to list the factors and multiples of a number.

Teaching Aids

Squared paper

Activity

Provide each student with a sheet of squared paper. Ask them to draw as many different rectangles as possible, each with an area of 8 square units. Ask them to explore all the different lengths and widths possible for such rectangles. When they are done drawing, ask them to label the side lengths of each rectangle.

Discuss the side lengths and how they are related to the number 8. Summarise by saying that the side lengths 1, 2, 4, and 8 are the factors of 8.

Next, instruct the students to shade squares to show a 1 × 12 rectangle with an area of 12 sq. units, and then show stacks of up to 4 rectangles of 1 × 12 each. Ask them to write the area of each new rectangle formed and how the areas are related to the number 12. Summarise by saying that the areas of 12, 24, 36 and so on are multiples of 12.

Repeat the activity for a rectangle with an area of 15 square units.

Ask questions like: How are the number of factors of a number different from the number of multiples of the same number?

Extension Idea

Ask: What type of numbers have an odd number of factors?

Say: Only square numbers have an odd number of factors. For example, five numbers which have an odd number of factors are 1, 4, 9, 16, and 25.

Types of Numbers

Learning Outcomes

Students will be able to identify perfect, odd, even, prime and composite numbers.

Teaching Aids

Squared paper

Activity

Imagine Maths Page 58

Begin the class by discussing how the numbers 6, 28 and 56 are different from 3, 11 and 55 to bring out odd and even numbers.

Ask the students to form groups of 6. Distribute a sheet of squared paper to each student.

Instruct each member of a group to take one number (out of 6, 28, 56, 3, 11, 55) and shade squares to form as many rectangles as possible with an area equal to that number. Ask them to list the factors of each number in their notebooks and then add all the factors except the number itself. Ask them to discuss in their groups what they notice about the factors and their sum.

Discuss the connection between the number of factors and whether the given numbers are prime or composite. Also discuss how the sum of the factors decides if a number is a perfect number or not.

Extension Idea

Instruct: Write any 3 pairs of twin prime numbers.

Say: Twin prime numbers are a pair of prime numbers with a difference of 2. Three pairs of twin prime numbers are (3, 5), (5, 7), (11, 13).

Divisibility Rules

Learning Outcomes

Imagine Maths Page 61

Students will be able to identify and apply the divisibility rule for 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11.

Teaching Aids

Number cards; Slips with divisibility rules

Activity

Prepare cards with numbers written on them, including some that are divisible by 2, 3, 4, 5, 6, 8, 9, 10 and 11 and some that are not.

Divide the class into 4 groups and assign each group with numbers. For example, Group 1 gets 2, 3 and 6, Group 2 gets 4 and 8, Group 3 gets 5, 9 and 10 and Group 4 gets 11.

Take the students out to the playground and distribute the slips with the divisibility rules of those numbers written on them.

Scatter the number cards all around the playground facing down.

Divisibility rule of 2: The digit in the ones place of the number is 0, 2, 4, 6 or 8.

Divisibility rule of 3: The sum of the digits of the number should be divisible by 3.

Instruct one student from each group to run around to find the number divisible by the number on their slip and bring it back. The group that collects the maximum number of cards wins.

If any of the cards have a number that is divisible by two numbers, then the student who finds it first should bring it back to their group.

Prime Factorisation

Learning Outcomes

Imagine Maths Page 64

Students will be able to use the factor tree and repeated division to express a number as a product of its prime factors.

Teaching Aids

Square and circle cutouts

Activity

Start by discussing what a factor pair of a number is.

Instruct the students to write any one factor pair of 24. Show the factor pairs of 24 using a factor tree as given.

Ask the students to work in pairs. Distribute square and circle cutouts to each pair. Instruct them to create a factor tree on their desks for the number 48 using circle cutouts for the prime factors and square cutouts for the composite factors. Ask them to then write all the prime factors in their notebooks with multiplication signs in between and multiply to check if they get the same number.

Next, instruct the students to find the factors using the repeated division method. Let them write 48 inside the division house and ask them which prime number divides 48. Instruct them to write the prime number outside the division house and write the quotient below 48 and then continue dividing in this way until the quotient is 1. Finally, ask them to compare the answers that they got using both methods.

Highest Common Factor

Learning Outcomes

Imagine Maths Page 67

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

Teaching Aids

Orange ribbon (24 cm); Yellow ribbon (36 cm); A pair of scissors; Glue; Ruler

Activity

Divide the class into groups and distribute ribbons of length 24 cm (orange) and 36 cm (yellow) to each group. Ask the students what is the maximum length of ribbon possible if they want to cut the ribbons into equal pieces without having any left.

Ask them to place the orange ribbon along the yellow ribbon as shown. Instruct them to cut the additional part of the yellow ribbon and place it along the orange ribbon. They should repeat this process until both ribbons are equal in length. Ask them to measure the length of the equal ribbons using a ruler. Explain that 12 cm (length of the last part cut) is the HCF of 36 cm and 24 cm. Next, instruct the students to find the HCF of 24 and 36 using prime factorisation and the long division method in their notebooks.

Extension Idea

Ask: What is the HCF of any 5 consecutive numbers?

Say: The HCF of any 5 consecutive numbers is 1.

Lowest Common Multiple

Learning Outcomes

Imagine Maths Page 69

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

Teaching Aids

Sheets of paper

Activity

Ask the students to work in pairs. Distribute sheets of paper to each pair.

Instruct them to draw the factor pairs of 16 and 24. Instruct the students to list the prime factors of both numbers and write each number as the multiplication of its prime factors. Ask them to identify the prime factors common to both 16 and 24, choose one shared prime factor from each pair of common factors and multiply with the factors which are not common to find the LCM.

Ask questions like: Is there any method to find the LCM quickly?

Instruct the students to draw a division house and write the two numbers inside it. Ask them which prime number divides both numbers. Collect the responses and instruct the students to write the prime number outside the division house and write the quotients below 16 and 24. They should follow these steps to continue dividing until the quotients are 1 and 1. Instruct the students to multiply the prime numbers to find the LCM.

Ask the students to compare the answers that they got using both methods.

Extension Idea

Instruct: Find the smallest number that is divisible by all the numbers from 1 to 10.

Say: To find the smallest number that is divisible by all the numbers from 1 to 10, we will find the LCM of all the numbers from 1 to 10. So, LCM = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520.

Word

Learning Outcomes

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

Teaching Aids

Word problem cards

Activity

Begin the class by discussing the concepts of LCM and HCF.

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

Kushal has a book stare. He has 15 comic books and 24 history books in each cupboard.

(a) He wants to arrange the books in such a way that each shelf has the same number of comic books and the same number of history books. What is the maximum number of shelves he can fill in this way?

(b) Kushal packed the books into cartons. He packed sets of 15 comic books and sets of 24 history books such that the total number of each type of book was the same. What is the least possible number of each type of book he packed in the carton?

Instruct the groups to read the word problem on their card and determine whether it requires finding the LCM or the HCF. Discuss the reasoning for the answers and ask the students to write the answers in their notebooks.

Extension Idea

Instruct: Create an LCM word problem using the numbers 10, 24 and 30.

Say: Sample answer: “A water pump fills a tank in 10 hours, a second pump in 24 hours, and a third pump in 30 hours. The pumps are switched on to fill tanks at the same time. Find the least amount of time after which all three pumps will complete filling a tank at the same time.”

Answers

1. Reviewing Factors and Multiples

Do It Together

2. 21

Factors = 1, 3, 7, 21

Multiples = 21, 42, 63

4. 39

Factors = 1, 3, 13, 39

Multiples = 39, 78, 117

2. Perfect Number

Do It Together

3. 27

Factors = 1, 3, 9, 27

Multiples = 27, 54, 81

The factors of 8128 are 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064, 8128.

Sum of factors (excluding itself) = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 127 + 254 + 508 + 1016 + 2032 + 4064 = 8128

Therefore, 8128 is a perfect number.

3. Even and Odd Numbers

Do It Together

1. 91—Odd

2. 61—Odd

3. 79—Odd

4. 7—Odd

5. 45—Odd

6. 82—Even

7. 39—Odd

8. 19—Odd

4. Prime and Composite Numbers

Think and Tell

There is only 1 even prime number which is 2.

Do It Together

1. Hence, 15 and 28 are co-prime numbers.

2. Factors of 21: 1, 3, 7, 21

Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40 Only 1 is the common factor Hence, 21 and 40 are co-prime numbers.

5. Divisibility Rules

Do It Together

1. 5628 is

by 4 and 6.

6

. Prime Factorisation

Do It Together

1. Prime factors of 76

2.

7. Common Factors and Common Multiples

Do It Together

1. 12, 18, 24: 6, 2

8. Highest Common Factor

9. Lowest Common Multiple

Do It Together 1. 60, 84:

10. Word Problems in LCM and HCF

Do It Together

HCF of 24, 36, 48 = 2 × 2 × 3 = 12

Thus, Emily should include 12 cookies in each pack.

5

Line Segments and Angles

Learning Outcomes

Students will be able to:

identify a plane, point, line segment, ray and line. measure and draw line segments.

identify the parts of an angle, interior and exterior of an angle, and adjacent angles. measure angles using a protractor.

classify the different types of angles.

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 about intersecting and parallel lines. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

line segment: a part of a line that has two definite end points angle: the space between two rays emerging from the same point

Teaching Aids

Set of cards showing points, lines, line segments, rays and planes; A piece of string; A pair of scissors; Ruler; Glue stick; Straws; Blue and orange buttons; Sheet of paper; Circular cutouts of radius 10 cm; Protractor; Rectangular origami sheets in different colours (15 cm × 4 cm); Ice cream sticks; Tape; Chart paper

Chapter: Line Segments and Angles

Learning Outcomes

Students will be able to identify a plane, point, line segment, ray and line.

Teaching Aids

Set of cards showing points, lines, line segments, rays and planes

Activity

Ask the students to work in groups of 3. Distribute a shuffled set of cards showing points, lines, line segments, rays and planes to each group.

Instruct the groups to look at the cards and sort them according to their features. Once they are done, have a discussion around the features of all the elements.

Instruct the students to draw a table in their notebooks with columns for each element. Ask them to draw and label the elements and list the features discussed so far. Also ask them to add 2 examples for each element that they see in everyday life.

Measuring and Drawing Line Segments

Learning Outcomes

Students will be able to measure and draw line segments.

Teaching Aids

A piece of string; A pair of scissors; Ruler; Glue stick

Activity

Imagine Maths Page 80

Begin by discussing the various units of length. Discuss the meaning of the big and small markings on the ruler. Ask the students to show various lengths such as 5 cm 3 mm in decimals.

Ask the students to work in pairs. Distribute a piece of string, a ruler, glue stick and a pair of scissors to each pair.

Instruct the students to measure 5.3 cm of the string, cut the string of that length and paste it in their notebooks. Ask them to draw a line of the same length using a ruler below the pasted thread and label it. Repeat the activity for various lengths of thread.

Extension Idea

Ask: What would be the total length of the line drawn below three pieces of string of lengths 2 cm 5 mm, 52 mm and 6.7 cm pasted end to end without any gaps?

Say: The length of the line will be the sum of the lengths of the three threads = 2.5 cm + 5.2 cm + 6.7 cm = 14.4 cm.

Learning Outcomes

Students will be able to identify the parts of an angle, interior and exterior of an angle, and adjacent angles.

Teaching Aids

Straws; Blue and orange buttons; Sheet of paper; Glue stick

Activity

Ask the students to work in pairs. Distribute 3 straws and some buttons in 2 colours to each pair.

Instruct them to stick one straw horizontally on a sheet with glue and another straw in a slanting position such that both start from the same point. Discuss what they notice about the figure. Encourage them to label the arms, vertex and angle formed.

Instruct them to paste blue buttons outside and orange buttons inside the figure. Explain that the blue buttons represent points on the exterior and orange buttons represent points on the interior of the angle.

Instruct them to stick one more straw on the sheet starting from the same vertex. Explain that the two angles formed with 3 straws are adjacent angles. Ask them to draw this figure in their notebooks, replacing straws with rays and buttons with points, and to label the rays and points.

Extension Idea

Ask: Can a point be on the exterior of an angle but on the interior of its adjacent angle?

Draw and label such a figure.

Say: Point P is exterior to angle CAB but on the interior of its adjacent angle DAC.

Measuring Angles

Learning Outcomes

Students will be able to measure angles using a protractor.

Teaching Aids

Circular cutouts of radius 10 cm; Ruler; A pair of scissors; Protractor

Activity

Recall the steps of measuring an angle using a protractor.

Imagine Maths Page 86

Ask the students to work in pairs. Distribute circular cutouts, a ruler, a pair of scissors and protractor to each pair.

Ask the students to fold the circular cutout in half and then open it. Ask them to cut it along the fold using scissors to get 2 semicircles. Ask each student to take a semicircle cutout and fold it in half, then into a tri-fold twice to get a fan. Ask them to draw lines along the creases obtained starting from the centre of the semicircle. Then, ask them to measure each of the angles drawn on the semicircle using a protractor and mark them.

have you noticed about

Types of Angles

Learning Outcomes

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

Teaching Aids

Rectangular origami sheets in different colours (15 cm × 4 cm); Ice cream sticks; Glue stick; Tape; Chart paper

Activity

Ask the students to work in groups of 6. Distribute origami sheets, ice cream sticks and glue sticks to each group.

Ask each student in a group to take one origami sheet, start from the smaller side and make accordion-style folds along the horizontal edge such that each fold is about 1 cm wide. Then, ask them to paste ice cream sticks at both ends of the folded sheet and join the sticks together at one end using tape. Instruct the groups to take the first folded sheet and stick the ice cream sticks on the chart paper so as to make an angle less than 90 .

Explain that angles less than 90 are called acute angles. Ask them to take the next sheet and stick it on the chart paper forming a 90 angle. Explain that a 90 angle is also known as a right angle. Instruct the groups in the same way so that they paste all six types of angles.

Extension Idea

Ask: What angle do the hour hand and minute hand of a clock make at 11:25?

Say: At 11:25, the hands of a clock make an obtuse angle.

Answers

1. Key Geometrical Concepts

Think and Tell

No, the surface of a volleyball is not the part of a plane. The volleyball court is a part of a plane.

Do It Together

Real‐life example Football field Stars in the sky Straw pipe Light emitted by the sun

GeometryPlanePoint Line segment Ray Figure

2. Measuring and Drawing Line Segments

Do It Together A B 7.5 cm C D 8.5 cm

3. Comparison of Line Segments

Do It Together

PQ = 4.2 cm; XY = 3.4 cm

4. Basics of Angles

Do It Together

1.

5. Measuring Angles

Do It Together

1. ∠P = 35 2. ∠Q = 140 3. ∠R = 30 4. ∠S = 30

The smallest angle is ∠R and ∠S

6. Types of Angles

Initial PositionAngle of TurnSense of RotationFinal Position

East1 right angleClockwiseSouth

North Two 1 2 turns ClockwiseNorth

South3 right angleAnticlockwiseWest

West Two 1 4 turns AnticlockwiseEast

Triangles, Quadrilaterals and Polygons 6

Learning Outcomes

Students will be able to: identify the parts of a triangle. classify triangles on the basis of their sides. classify triangles on the basis of their angles. identify the parts of a quadrilateral. classify the types of quadrilaterals. identify the parts of a polygon. classify polygons on the basis of their sides, angles and diagonals.

Alignment to NCF

C-3.1: Describes, classifies, and understands relationships among different types of two - and threedimensional shapes using their defining properties/attributes

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

Let’s Recall

Recap to check if students know how to distinguish between corners and sides of a shape. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

vertices: points where the sides of a polygon meet sides: line segments that form a polygon quadrilateral: apolygon with four sides

Teaching Aids

A4 size sheet of paper; Sheets of coloured paper; A pair of scissors; Tape; Markers; Thermocol sheets; Triangle cutouts; Protractor; Chart paper; Quadrilateral cutouts; Ruler; Geoboards; Rubber bands of different colours; Polygon cutouts

Chapter: Triangles, Quadrilaterals and Polygons

Learning Outcomes

Students will be able to identify the parts of a triangle.

Teaching Aids

A4 size sheet of paper; Sheets of coloured paper; A pair of scissors; Tape; Markers

Activity

Begin by introducing the parts of a triangle, such as sides, vertices, angles.

Ask the students to work in groups of 4. Distribute an A4 size sheet, sheets of coloured paper, and other craft materials to each group.

Instruct the students to cut out triangles of different sizes from the coloured paper and paste all the triangles on the A4 size sheet to create a design. Instruct them to label the vertices, sides and angles of each triangle.

Ask the groups to mark 8 points with the marker such that:

3 points are outside two of the triangles.

4 points are inside and outside one of the triangles.

5 points lie inside one of the triangles.

Classification of Triangles on the Basis of Sides

Learning Outcomes

Students will be able to classify triangles on the basis of their sides.

Teaching Aids

Sheets of coloured paper; A pair of scissors; Tape; Markers; Thermocol sheets

Activity

Begin by discussing the three types of triangles based on their sides.

Ask the students to work in groups of 3. Distribute the craft materials and a Thermocol sheet to each group. Instruct each group to create 3 triangles on the Thermocol sheet using a ruler and different coloured paper. Ask them to label each type of triangle it represents. Encourage creativity in designing the triangles. Instruct each group to create triangles of all three types in varying shapes and sizes.

Ask the students to collaborate to create aesthetic patterns using triangles they have created. Ask them to paste the pattern on a Thermocol sheet.

Extension Idea

Ask: Can all the sides of a triangle be unequal, but the angles be equal?

Say: No, this is not possible. If the sides of a triangle are equal, its angles will be equal too and vice-versa.

Classification of Triangles on the Basis of Angles

Learning Outcomes

Students will be able to classify triangles on the basis of their angles.

Teaching Aids

Triangle cutouts; Protractor; Chart paper

Activity

Ask the students to work in pairs. Distribute the chart paper, triangle cutouts (acute, obtuse and right triangles) and protractor to each pair.

Instruct the pairs to paste the triangle cutouts on the chart paper. Ask them to measure the angles of each cutout using a protractor and record their measurements on the chart paper. Have a discussion on what the students noticed about the angles.

Discuss the types of triangles based on the angles. Ask the pairs to classify and label the pasted triangles based on the discussion.

Extension Idea

Ask: A triangle has angles measuring177°, 2° and 1°. Is it an acute, an obtuse or a right-angled triangle?

Say: It is an obtuse triangle as one of the angles is obtuse (more than 90 degrees).

Parts of a Quadrilateral

Learning Outcomes

Students will be able to identify the parts of a quadrilateral.

Teaching Aids

Sheets of coloured paper; A pair of scissors; Tape; Markers; Chart paper

Activity

Imagine Maths Page 98

Start by explaining the fundamental concepts of a quadrilateral, including the presence of 4 sides, 4 vertices, and 4 angles.

Ask the students to work in groups. Distribute the chart paper, coloured paper and other craft materials to each group.

Instruct the groups to take the coloured paper and cut out quadrilaterals of various sizes. Ask them to paste the cutouts on the chart paper and label the vertices, sides, diagonals and angles of each quadrilateral.

Challenge the students to utilise their quadrilaterals creatively to construct new shapes or objects by pasting them on the chart paper. They can experiment with combining quadrilaterals in different arrangements to form larger polygons or abstract designs.

Learning Outcomes

Students will be able to classify the types of quadrilaterals.

Teaching Aids

Quadrilateral cutouts; Protractor; Ruler; Chart paper

Activity

Ask the students to work in groups. Distribute cutouts of different types of quadrilaterals (convex, concave, square, rectangle, trapezium, and kite), a protractor, a ruler and chart paper to each group.

Instruct the groups to paste these cutouts on the chart paper and measure the angles, sides and diagonals of each quadrilateral using a protractor and ruler.

Ask the groups to record their observations about the angles, sides and diagonals of the cutouts on the chart paper.

Discuss the types of quadrilaterals based on their features. Ask the groups to classify and label the pasted quadrilaterals based on the discussion.

Extension Idea

Ask: Name a quadrilateral that shares characteristics with both a rhombus and a rectangle.

Say: A square shares characteristics with a rectangle as all its angles are right angles. Also, all the sides of a square are equal, so it shares characteristics with a rhombus too.

Parts of a Polygon

Learning Outcomes

Students will be able to identify the parts of a polygon.

Teaching Aids

Geoboards; Rubber bands of different colours; Markers

Activity

Start by introducing the concept of polygons and their basic parts: vertices, sides, and angles.

Ask the students to work in groups of 6. Distribute geoboards, rubber bands and coloured markers to each group.

Instruct the groups to create two different polygons on the geoboards using the rubber bands. Ask them to use different colours of rubber bands to show the diagonals. Encourage them to experiment with varying the number of sides and vertices. As they create each polygon, ask the students to count the vertices, sides, diagonals and angles. Finally, ask them to recreate the shapes using coloured markers in their notebooks.

Teacher Tip: In case of unavailability of geoboards, you can stick notice board push pins on a Thermocol sheet as a substitute.

Extension Idea

Ask: There are two polygons: one with 10 sides and one with 8 sides. Which polygon will have more diagonals?

Say: The polygon with more sides has more vertices and therefore, more diagonals. Hence, the polygon with 10 sides has more diagonals.

Types of Polygons Imagine Maths Page 104

Learning Outcomes

Students will be able to classify polygons on the basis of their sides, angles and diagonals.

Teaching Aids

Polygon cutouts; Protractor; Ruler; Chart paper

Activity

Ask the students to work in groups. Distribute cutouts of different types of polygons, a protractor, a ruler and chart paper to each group.

Instruct the students to paste the cutouts on the chart paper and measure the angles, sides and diagonals of each polygon using a protractor and ruler. Ask the groups to record their observations about the angles, sides and diagonals of the polygons on the chart paper. Discuss the types of polygons based on their features. Ask the groups to classify and label the pasted polygons based on the discussion.

Extension Idea

Ask: Can you name a polygon that does not have any diagonals?

Say: Triangles are polygons that do not have any diagonals.

Answers

1. Parts of a Triangle

Do It Together

2. Only D, E, F and G are in the interior of the triangle.

2. Classification of Triangles on the Basis of Sides

Do It Together

Students will measure the lengths of the sides.

1. 2. 3. Equilateral Scalene Isoceles

3. Classification of Triangles on the Basis of Angles

Do It Together

Students will measure the angles of the triangles.

1. 2. 3.

Right-angled triangleObtuse-angled triangleAcute-angled triangle

4. Parts of a Quadrilateral

Do It Together

2. Only points Q, R and S are in the interior.

3. Only points P and T are in the exterior.

5. Types of Quadrilaterals

Do It Together P

R

Q

PQ = SR; PS = QR

6. Parts of a Polygon

Do It Together

Vertices:

Point in the interior:

Points in the exterior:

A, B, C, D, E, F, G X W, Y, Z

7. Types of Polygons

Think and Tell

All the sides in a right-angled triangle are not equal therefore it is not a regular polygon.

Do It Together

Vertices:

Diagonals: A, B, C, D, E, F, G 14

Curves and Circles 7

Learning Outcomes

Students will be able to: classify curves as open or closed, simple or non-simple. identify parts of a circle. construct a circle for the given radius. find the circumference of a circle using the formula.

Alignment to NCF

C-3.1: Describes, classifies, and understands relationships among different types of two- and three-dimensional shapes using their defining properties/attributes

Let’s Recall

Recap to check if students know how to recognise a circle in objects around them. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

loop: a shape formed by a curve when it crosses itself line segment: a part of a straight line with two end points equidistant: at equal distances from a common point intersect: meet at a point

Teaching Aids

Pieces of string; Sheets of paper; Glue; Circular cutouts; Drawing sheets; Crayons; Ruler; Pin; Pencil; Compass

Chapter: Curves and Circles

Types of Curves

Learning Outcomes

Students will be able to classify curves as open or closed, simple or non-simple.

Teaching Aids

Pieces of string; Sheets of paper; Glue

Activity

Ask the students to work in groups of 4. Distribute the teaching aids.

Imagine Maths Page 110

Guide the groups to create a curved pattern of their choice using their pieces of string and paste them on the sheet of paper. Allow time for the glue to dry.

Collect all the sheets and mix them up. Redistribute the sheets, one to each group.

Instruct the groups to identify the type of curve on the sheet as simple or non-simple and open or closed. Ask each student in a group to write a reason for their answer in their notebook. Discuss the reasons and types of curves.

Ask questions like: How did you know that a curve was a non-simple curve?

Extension Idea

Draw a curve that is non-simple and open on the board.

Ask: What do you notice about this figure?

Say: The shape is an open non-simple curve because the starting point and the end point of the curve are different.

Parts of a Circle

Learning Outcomes

Students will be able to identify parts of a circle.

Teaching Aids

Circular cutouts; Drawing sheets; Crayons; Ruler

Activity

Imagine Maths Page 113

Begin by drawing a circle on the board, clearly illustrating its various parts. Engage in a discussion to help the students identify the different parts of the circle, such as the centre, radius, diameter, chord, and secant.

Ask the students to work in pairs. Distribute a circular cutout and drawing sheet to each pair.

Instruct them to identify the centre, radius, diameter and chord, and to mark and label the specified parts on the cutout using crayons. Allow them to use different ways like folding the cutout in half to identify the diameter. They will then shade the major and minor arcs in the circle.

Ask questions like: How did you find the centre, radius and diameter of the circle?

Extension Idea

Ask: If on folding the circle in half, the crease measures 12 cm, what is the radius of the circle?

Say: On folding the circle in half, the crease obtained is the diameter. If the diameter is 12 cm, then the radius is 6 cm.

Constructing Circles

Learning Outcomes

Students will be able to construct a circle for the given radius.

Teaching Aids

Drawing sheet; Pin; Pieces of string; Pencil; Compass

Activity

Distribute the teaching aids to each student.

Imagine Maths Page 115

Instruct them to tie a pencil to one end of the string and a pin to the other end. Ask them to place the pin on the drawing sheet and then turn the string with the pencil end all the way around so that a circle is drawn. Then, ask them to measure the radius of the circle.

Next, ask the students to use the compass to construct a circle with the same radius as the circle drawn using the string on the drawing sheet. Explain that without moving the compass needle, they should rotate the compass around it in a way that the pencil draws a circle. Instruct them to keep turning until the pencil goes all the way around. Then, ask them to mark and label the centre, radius and diameter.

Ask questions like: How can you identify the centre and the diameter in the circle?

Circumference of a Circle

Learning Outcomes

Students will be able to find the circumference of a circle using the formula.

Teaching Aids

Compass; Ruler; Pieces of string

Activity

Imagine Maths Page 117

Ask the students to work in groups of 3. Distribute a compass, ruler and a piece of string to each group. Draw a table as shown on the board. Ask the students to copy it into their notebooks.

Instruct each student in a group to construct one of the three circles of radius 4 cm, 5 cm and 6 cm using a compass and then measure the distance around the boundary (circumference) of that circle. Ask the students to write the circumference (C), the diameter (D) and the value of C ÷ D in the table. Then, they should discuss in their groups how they can form the formula for C if the value of R or D is given. Discuss how the circumference of a circle is pi (π) or around 3.14 times the diameter of the circle and how to find the circumference when the radius or diameter is given. Ask the students to find the circumference of a circle with a radius of 7 cm.

Extension Idea

Ask: What is the diameter of a circle with a circumference of 88 cm?

Say: We know that the circumference = 2πr. So, r = 14 cm and diameter = 28 cm.

1. Types of Curves

Do It Together

Figures may vary. Sample figures.

Simple Open CurveSimple Closed Curve

Non-simple Open CurveNon-simple Closed Curve

2. Parts of a Circle

Think and Tell

No. The diameter of a circle is the longest chord.

Do It Together

Chord Radius Secant Diameter

3. Constructing Circles

Do It Together

4. Circumference of a Circle

3-D Shapes 8

Learning Outcomes

Students will be able to: list the features of 3-D shapes. identify and draw the nets of a 3-D shape.

Alignment to NCF

C-3.1: Describes, classifies, and understands relationships among different types of two- and threedimensional shapes using their defining properties/attributes

C-3.3: Identifies attributes of three-dimensional shapes (cubes, parallelepipeds, cylinders, cones), works hands-on with material to construct these shapes, and also uses two-dimensional representations of three-dimensional objects to visualise and solve problems

Let’s Recall

Recap to check if students know how to identify the faces of a 3-D shape. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

polygon: a 2-D shape made up of 3 or more line segments

prism: a 3-D shape with identical and parallel opposite bases and square or rectangular lateral faces

pyramid: a 3-D shape in which the base is a polygon and all the other faces are triangles tapered to a vertex

net: a 2-D figure that can be folded to form a 3-D shape

Teaching Aids

Plastic prisms and pyramids; Boxes shaped like a pyramid and a prism; A pair of scissors; Thick chart paper

Chapter: 3-D Shapes

Features of 3-D Shapes

Learning Outcomes

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

Teaching Aids

Plastic prisms and pyramids

Activity

Imagine Maths Page 122

Ask the students to work in groups. Distribute the plastic prisms and pyramids to each group. Instruct them to sort the shapes into 2 groups and then ask them what the similarities and differences between the two groups are. Explain that the base tells us the type of prism or pyramid. Ask the students to list the names of the prisms and pyramids.

Note: In case shapes are unavailable, ask the students to look at the picture of the shapes in their Imagine Mathematics books.

Ask questions like: What is one major difference between a prism and a pyramid?

Extension Idea

Ask: If you cut a cuboid diagonally, what other 3-D shape do you get?

Say: If a cuboid is cut diagonally, we get a triangular prism.

Nets of 3-D Shapes

Learning Outcomes

Students will be able to identify and draw the nets of a 3-D shape.

Teaching Aids

Boxes shaped like a pyramid and a prism; A pair of scissors; Thick chart paper

Activity

Imagine Maths Page 125

Show the students a pyramid-shaped paper box and ask them what 2-D shapes they see in its faces and how many. Then, open out 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.

Ask the students to work in pairs. Distribute the teaching aids. Allow the pairs to visualise what the nets of the prism and pyramid would look like and ask them to draw the nets on the thick chart paper. Finally, instruct them to cut out their nets and fold them to see if the shapes look the same as the given boxes.

Extension Idea

Ask: In how many ways can we create a net of a tetrahedron?

Say: A tetrahedron is also called a triangular pyramid. There are two different nets of a tetrahedron.

Answers

1. Features of 3-D Shapes

Think and Tell

Faces: 12, Edges: 30, Vertices: 20

Do It Together

NameCubePentagonal pyramid Square pyramid Hexagonal Prism Cone

Shape

Faces6 6 5 82

Edges12108181

Vertices8 6 5121

2. Nets of Cubes and Cuboids

Do It Together

There are 4 ways of placing a rectangle to form a net of a closed cuboid. A

3. Nets of Cones and Cylinders

Do It Together

1. 2. 3.

4. Nets of Pyramids and Prisms

Do It Together

Name of the shape: hexagonal pyramid

Integers 9

Learning Outcomes

Students will be able to: represent integers on a number line. add integers using the rules. subtract integers using the rules.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to find the predecessor and successor of a given number. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

predecessor: the number that comes before a number successor: the number that comes after a number integer: a positive or negative whole number or zero absolute value: distance of an integer from zero

Teaching Aids

Chart paper; Index cards; Integer tiles made using red and yellow square cutouts to show negative and positive integers, respectively

Chapter: Integers

Positive and Negative Integers

Learning Outcomes

Students will be able to represent integers on a number line.

Teaching Aids

Chart paper; Index cards

Activity

Imagine Maths Page 132

Begin the class with a discussion on how to write a number for the second floor and Basement 2, to bring out the uses of integers. Discuss more examples with the students. Ask the students to work in groups. Distribute a sheet of chart paper to each group and one index card to each student in the group. Each index card should describe a different situation involving integers.

Spending `5 4 kg decrease in weight Predecessor of 3 Successor of 6 Absolute value of –7

Instruct the groups to draw a number line from −10 to +10 on the chart paper, ensuring that it is clearly labelled. Ask the groups to read and discuss the situation on their index card, and represent the corresponding integer on the number line pasting the index card above or below the number line at that mark.

Addition of Integers

Learning Outcomes

Students will be able to add integers using the rules.

Teaching Aids

Imagine Maths Page 136

Integer tiles made using red and yellow square cutouts to show negative and positive integers, respectively

Activity

Ask the students to work in pairs. Distribute the integer tiles to each pair.

Instruct them to add 3 and −4 using the tiles. Ask them to first show the positive integer using the yellow tiles and the negative integer using the red tiles. Explain that one positive tile and one negative tile cancel each other out because they make 0, so they should be removed. Explain that the remaining number of integer tiles (−1 in this case) is the answer.

Give the students more problems to solve using the integer tiles. Ask questions like: What is the sign of the sum of two negative integers? Discuss the rules of adding integers. Give the students more addition problems to solve.

Extension Idea

Instruct: Find three integers whose sum is −25.

Say: There are multiple sets of three integers whose sum is −25. One such set could be −10, −8 and −7.

Learning Outcomes

Students will be able to subtract integers using the rules.

Teaching Aids

Integer tiles made using red and yellow square cutouts to show negative and positive integers, respectively

Activity

Begin the class by asking the students the difference between the two floors of a building: Floor 5 and Basement 2. Discuss the answers.

Ask the students to work in groups. Distribute the integer tiles to each group.

Instruct them to subtract –6 from 3. Ask them to use yellow tiles to show the integer 3.

Ask questions like: Can you take away 6 negative tiles from the 3 positive tiles? Explain that they can do so only if they add 0 tiles, that is, pairs of positive and negative tiles and that to take away 6 red tiles, they need to first add 6 red and 6 yellow tiles. Ask the students to add the required zero tiles, show the subtraction and write the subtraction sentence in their notebooks. Discuss the answer.

Instruct the students to next subtract −7 from 4 using the integer tiles and note down the result in their notebooks as a subtraction sentence.

Ask questions like: What is the sign of the answer when you subtract one negative integer from another? What is it when you subtract one negative from one positive integer? What does the sign depend on?

Discuss the rules of subtracting integers. Give the students more subtraction problems to practise solving using the rules.

Extension Idea

Ask: The difference of two integers is −34. If the subtrahend is 19, what is the minuend?

Say: We can find the minuend by adding 19 to −34.Therefore, the minuend is −34 + 19 = −15.

1. Positive and Negative Integers

Do It Together

Rudra’s House

Answers

3. Properties of Addition of Integers

Do It Together

Suhani’s House

0

+5 –1 –2 –3 Rudra’s car parking

–4

Suhani’s car parking

2. Rules for Addition of Integers

Do It Together

1200 feet beneath the surface = –1200

Rise of 635 feet = +635

Descend again 455 feet = ____ –455

Current location of vessel = (–1200) + (635) + –455 = –1020

Hence, the vessel is 1020 feet below the surface of the water.

1. –25 + 0 = –25 Existence of additive identity

2. –15 + 35 = 35 +(–15) Commutative property

3. 39+ (−12) = +27 Closure property

4. 26 + (–13 + 9) = (26 + (–13)) + 9 Associative property

4. Rules for Subtraction of Integers

Do It Together

1. –36 from 58 = 58 – (–36) = 58 + 36 = 94

2. –145 from –365 = –365 – (–145) = –365 + 145 = –220

3. –247 from 354 = 354 – (–247) = 354 + 247 = 601

4. 3654 from –5214 = –5214 – 3654 = –5214 – 3654 = –8868

5. Properties of Subtraction of Integers

Do It Together

1. –125 – 0 = –125

2. 523 – 365 = 158; 365 – 523 = –158

3. (125 – 523) – 363 = –761; 125 – (+523 –363) = –35

Fractions 10

Learning Outcomes

Students will be able to:

represent fractions and classify them as proper, improper and mixed. compare and order unlike fractions. convert improper fractions to mixed numbers and vice versa. find equivalent fractions and the simplest form of fractions. add and subtract unlike fractions. multiply and divide fractions and mixed numbers.

Alignment to NCF

C-1.2: Represents and compares commonly used fractions in daily life (such as 1 2 , 1 4  ) as parts of unit wholes, as locations on number lines and as divisions of whole numbers

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

Let’s Recall

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

Vocabulary

fractions: numbers of the form a b , where a and b are whole numbers and b 0 like fractions: fractions with the same denominator reciprocal: fraction obtained by interchanging the numerator and denominator

Teaching Aids

Fraction cards; Circle cutouts; Chart paper; A pair of scissors; Glue stick; Rectangular strips; Colour pencils; Fraction circles; Fraction strips

Chapter: Fractions

Representing Fractions

Learning Outcomes

Imagine Maths Page 147

Students will be able to represent fractions and classify them as proper, improper and mixed.

Teaching Aids

Fraction cards (including 2 proper, 2 improper and 2 mixed fractions); Circle cutouts; Chart paper; A pair of scissors; Glue stick

Activity

Begin with a discussion on proper, improper and mixed fractions. Give some examples for each type of fraction.

Ask the students to work in groups. Distribute the fraction cards and other teaching aids to each group. Explain that the fraction cards represent the fraction of pizza left after a party. Ask them to consider each circle cutout as a whole pizza. Instruct them to divide and cut the circular cutouts as per the fraction on the fraction cards and paste them on the chart paper. When they are done, ask them to exchange charts with other groups for peer review.

Ask questions like: How is an improper fraction different from a mixed number?

Comparing and Ordering Fractions

Learning Outcomes

Students will be able to compare and order unlike fractions.

Teaching Aids

Rectangular strips; Colour pencils

Activity

Ask the students to work in pairs. Distribute the rectangular strips and colour pencils to each pair.

Imagine Maths Page 149

Instruct them to show 1 3 and 3 4 on two rectangular strips and compare them. Discuss how to divide the strips such that both show the same total number of equal parts, either by finding the LCM of the denominators or by multiplying the denominators. Instruct the pairs to divide each strip into the same number of parts as the LCM of the denominators and then compare the shaded parts.

Then, ask them to compare the two fractions by first converting them to like fractions and then comparing the numerators in their notebooks.

Give the students some practise questions on ordering unlike fractions.

Extension Idea

Ask: Which is bigger: 6 1 2 or 5 1 4  ?

Say: We convert the mixed numbers into improper fractions as 13 2 and 21 4  . We know that 13 2 = 26 4 and as 26 > 21, 6 1 2 > 5 1 4

Learning Outcomes

Students will be able to convert improper fractions to mixed numbers and vice versa.

Teaching Aids

Fraction circles

Activity

Ask the students to work in pairs. Distribute the fraction circles showing improper fractions to each pair.

Instruct one student in each pair to look at the circles and write the fraction as an improper fraction and the other student to look at the same circles and write the fraction as a mixed number. Ask them to look at the fractions written by both partners and discuss how they are related and how to convert one form to the other without using fraction circles.

Discuss the answers with the whole class along with the method used for converting between improper fractions and mixed numbers without using fraction circles.

Extension Idea

Ask: My dog is 5 1 2 years old. My cat is 9 4 years old. Which is older?

Say: Here, we have a mixed number and an improper fraction. On converting the improper fraction 9 4 into a mixed number, we get 2 1 4  . As 5 1 2 > 2 1 4 , the dog is older than the cat.

Equivalent Fractions and Simplest Form

Learning Outcomes

Students will be able to find equivalent fractions and the simplest form of fractions.

Teaching Aids

Fraction strips

Activity

Imagine Maths Page 154

Ask the students to work in groups of 4. Distribute 4 fraction strips to each group. Instruct them to show 1 2 , 2 4 , 3 6 and 4 8 on the fraction strips. Once they have done so, ask them to place the number strips one below the other and discuss what they notice about the strips. Encourage them to share their observations with the rest of the class. Bring out the fact that the given fractions are equivalent.

Guide the discussion to help the students deduce the way to find equivalent fractions without using fraction strips. Ask them to find 3 more equivalent fractions for the given fraction and write these in their notebooks. Give them 2 more fractions to find equivalent fractions for without using the strips in their notebooks.

Extension Idea

Ask: Molly baked 14 cookies. Her mother ate 2 cookies, and her father ate 4 cookies. What fraction of the cookies was left? Write it in its simplest form.

Say: We know that Molly’s parents ate 6 14 of the cookies, so 8 14 was left. The simplest form of 8 14 is 4 7  .

Imagine Maths Page 156

Learning Outcomes

Students will be able to add and subtract unlike fractions.

Teaching Aids

Rectangular strips; Colour pencils

Activity

Ask the students to work in pairs. Distribute the rectangular strips and colour pencils to each pair.

Instruct them to show 1 2 and 1 3 on two rectangular strips and add the two fractions.

Ask them to find the LCM of the denominators of the fractions and then divide each strip into the same number of equal parts as the LCM; in this case, each strip will have a total of 6 equal parts. Instruct them to then count the total number of shaded parts.

Next, instruct the students to add the two fractions in their notebooks by first converting them to like fractions and then adding the numerators.

Repeat the activity for subtraction of unlike fractions.

Extension Idea

Ask: How do we solve 5 1 4 − 2 2 3 ?

Say: We change the mixed numbers to improper fractions, convert them into like fractions and then subtract.

So, 21 4 − 8 3 = 31 12

Multiplication and Division of Fractions and Mixed

Numbers

Learning Outcomes

Students will be able to multiply and divide fractions and mixed numbers.

Teaching Aids

Rectangular strips; Colour pencils

Activity

Ask the students to work in pairs. Distribute rectangular strips and colour pencils to each pair.

Imagine Maths Page 160

Instruct the pairs to multiply 3 8 and 4 5 using the strips. Ask them to first show 3 8 by dividing the strip horizontally into 8 parts and shading 3 parts using a colour pencil. Ask them to then show 4 5 on the same strip by dividing it vertically into 5 equal parts and crossing out 4 vertical parts. Ask them to now count the total number of equal parts and number of parts that have both shading and crosses. Explain that the fraction obtained is the answer of the multiplication. Then, ask them to solve the problem in their notebooks by multiplying the numerators and then the denominators. Explain that if there is a division sign between the two fractions, we replace it with a multiplication sign and write the reciprocal of the divisor and then follow the steps of multiplication.

Give the students a few problems on division and multiplication of fractions to solve in their notebooks.

Extension Idea

Ask: Jen is preparing a dish that calls for 3 4 of a cup of oil per serving. If Jen needs to prepare 2 2 3 servings, how many cups of oil will she need?

Say: We need to multiply the two fractions. 3 4 × 2 2 3 = 3 4 × 8 3 = 2. Hence, she needs 2 cups of oil for 2 2 3 servings.

Answers

1. Representing Fractions

2. Comparing and Ordering Fractions

Do It Together

LCM of 7 and 6 = 42

5 7 = 30 42 ; 3 6 = 21 42

The larger fraction is 30 42 = 5 7  .

3. Converting Fractions

Do It Together

1. 10 4 5 = (5 × 10) + 4 5 = 54 5

2. 31 6 = 5 1 6

4. Equivalent Fractions and Simplest Form

Think and Tell

The reciprocal of 0 is 0.

5. Addition and Subtraction of Unlike Fractions

Do It Together

Fraction of the cake eaten by Radha = 1 9  ; Fraction of the cake eaten by Riya = 2 3

LCM of 9 and 3 = 9

Fraction of cake eaten by Radha and Riya = 1 9 + 2 3 = 1 + 6 9 = 7 9

As the whole cake is 1.

So, the fraction of cake remaining = 1 –7 9 = 2 9

6. Addition and Subtraction of Mixed Numbers

Do It Together

On converting to improper fractions, we get:

7. Multiplying Fractions or Mixed Numbers

Think and Tell

The product of any number and its reciprocal is always 1. Hence, 5 8 × 8 5 = 1.

It Together

8. Dividing Fractions or Mixed Numbers

Decimals 11

Learning Outcomes

Students will be able to:

represent decimals on a number line.

identify the place value of digits in a decimal number and write the decimal number in its expanded form and as a number name.

convert fractions to decimals and vice versa.

find equivalent decimals and compare decimals.

add and subtract decimals up to 3 decimal places.

multiply decimals up to 2 decimal places by whole numbers or decimals up to 1 decimal place. divide decimals up to 2 decimal places by whole numbers or decimals up to 1 decimal place. solve word problems on operations on decimal 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 about decimal numbers, their parts, and how to rearrange them in increasing or decreasing order.

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

Vocabulary

equivalent decimals: decimal numbers that have the same value

Teaching Aids

Decimal cards; Blank decimal place value chart (up to thousandths); Bowl of decimal number cards; Coloured counters; Shaded grids; 10 × 10 grid; Coloured pencils; Two decks of cards (one with decimal numbers and the other with their decimal equivalents); Grids to represent decimal numbers up to 3 decimal places; Grids to represent decimal numbers up to 2 decimal places; Task cards with word problems

Chapter: Decimals

Learning Outcomes

Students will be able to represent decimals on a number line.

Teaching Aids

Decimal cards

Activity

Imagine Maths Page 167

Briefly explain the concept of decimals and their relation to the number line. Explain that tenths can be further divided into hundredths.

Take the students out to the playground and ask them to form groups of equal sizes. Give each student a decimal card.

Draw a large horizontal number line of range 1–5 on the ground. Ask the students to read the decimal number on the card and go and stand at the correct place on the number line with the card. The group with all its members to stand at the correct places first, wins the game.

Back in the classroom, instruct the students to draw a number line in their notebooks and write the answers. Give them a few more decimal numbers to represent on a number line in their notebooks.

Place Values in Decimals

Learning Outcomes

Imagine Maths Page 168

Students will be able to identify the place value of digits in a decimal number and write the decimal number in its expanded form and as a number name.

Teaching Aids

Blank decimal place value chart (up to thousandths); Bowl of decimal number cards; Coloured counters

Activity

Briefly explain the concept of decimals and their relation to tenths and hundredths. Prepare some decimal number cards and put them in a bowl.

Ask the students to work in groups. Distribute a sheet of paper with a blank decimal place value chart (up to thousandths) to each group. Assign a colour to each place and distribute counters of the same colours to each group. Pick a decimal number card from the bowl. Ask the groups to place as many counters as the digit in the correct place on the place value chart. Help them identify the place value of each digit of the number.

Instruct them to interchange the counters at the tenths place and the tens place. Ask them to identify the new number and the place value of each of its digits. Discuss the importance of the placement of the digits in a number. Ask questions like: How does the value of a digit change depending on its place in the number?

Extension Idea

Ask: What is the decimal number for 50 + 4 + 2 10 + 8 1000  ?

Say: 50 + 4 + 2 10 + 8 1000 = 50 + 4 + 0.2 + 0.008 = 54.208.

Conversion Between Fractions and Decimals

Learning Outcomes

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

Teaching Aids

Shaded grids

Activity

Imagine Maths Page 170

Begin the class by reviewing the concept of fractions and decimals. Give the students some examples on conversion to emphasise that the numbers represent the same value even though they are expressed in different forms.

Divide the class into groups of 4. Distribute the shaded grids to each group.

Ask 2 students of each group to write the fractions representing the given grids and 2 students to write the decimals representing the same grids. Then, have them compare their answers to see how one can be converted to the other.

Discuss the answers and the reasons for conversion. Give them a few more grids and ask them to write the corresponding fractions and decimals in their notebooks.

More on Decimals

Learning Outcomes

Students will be able to find equivalent decimals and compare decimals.

Teaching Aids

Imagine Maths Page 172

10 × 10 grid; Coloured pencils; Two decks of cards (one with decimal numbers and the other with their decimal equivalents)

Activity

Begin the class by reviewing the concept of like and unlike decimals. Give a few examples to emphasise on how they are different from each other. Explain that we can convert unlike decimals to like decimals by adding zeroes to form equivalent decimals and then compare them.

Ask the students to work in pairs. Distribute the teaching aids to each pair.

Instruct them to shuffle each deck of cards. Each student should draw two cards each from the decks and represent the numbers on the empty grids using coloured pencils.

Then, ask them to match grids with their partners and find out if their representations are equivalent decimals. If they are not equivalent, ask them to form equivalent decimals. Then, instruct them to compare the two representations. Explain that the student with the largest decimal number wins the round and keeps both cards.

Extension Idea

Ask: A chef is baking cookies and needs to add exactly 1.75 cups of flour. She only has measuring cups in sizes of 0.25 cups, 0.5 cups, and 1 cup. Find all the possible combinations she can use to achieve the measurement. Which combination requires the minimum number of cups?

Say: The possible combinations are: 3 × 0.5 cups + 1 × 0.25 cup; 1 × 1 cup + 1 × 0.5 cup + 1 × 0.25 cup, and 1 × 1 cup + 3 × 0.25 cups. The second option requires the minimum number of cups, that is, 3.

Learning Outcomes

Students will be able to add and subtract decimals up to 3 decimal places.

Teaching Aids

Grids to represent decimal numbers up to 3 decimal places; Coloured pencils

Activity

Begin the class by reviewing the concept of adding and subtracting decimals. Give examples to emphasise the placement of digits of decimal numbers one below the other in the column method.

Ask the students to work in pairs. Distribute the teaching aids and write an addition problem and a subtraction problem of decimals on the board.

Ask the pairs to represent the decimal numbers on the grid provided using coloured pencils. Explain that they may need more than one grid to represent the numbers. Ask them to add or subtract as per the operation on the board.

Repeat this for 5 rounds with different addition and subtraction problems of decimal numbers written on the board. The pair to get the maximum number of correct sums or differences wins the game.

Extension Idea

Ask: A theme park has 3 rides: a roller coaster costing ₹7.99, a water flume costing ₹6.50, and a mystery ride costing x dollars. You have ₹20.00 and want to go on two rides. What are the possible prices of the mystery ride if you spend exactly ₹20.00?

Say: Cost of the mystery ride if the roller coaster is included: ₹20 − ₹7.99 = ₹12.01. Cost of the mystery ride if the water flume is included: ₹20 − ₹6.50 = ₹13.50. So, the cost of the mystery ride is ₹13.50 or ₹12.01.

Multiplication of Decimals

Learning Outcomes

Imagine Maths Page 178

Students will be able to multiply decimals up to 2 decimal places by whole numbers or decimals up to 1 decimal place.

Teaching Aids

Grids to represent decimal numbers up to 2 decimal places; Coloured pencils

Activity

Begin the class by reviewing the concept of multiplying two decimals. Give examples to emphasise the placement of the digits of a decimal number at the appropriate places in the column method.

Ask the students to work in pairs. Distribute the teaching aids and write a multiplication problem of decimals, say 0.2 × 0.7, on the board.

Instruct the pairs to represent the decimal numbers on the grid using coloured pencils and multiply them. Ask them to shade the first number (0.2) vertically on the grid and shade the second number (0.7) horizontally on the same grid. Explain that the portion of the grid shaded with two colours (merged portion) is the product of the two numbers.

0.2 × 0.7 = 0.14

Repeat the activity for 5 rounds and instruct the students to write the answers in their notebooks. The pair to get the maximum correct products wins!

Extension Idea

Ask: Rubal hired a car costing ₹1500 plus 85 p for each km she travels. The car’s odometer showed 28361 km when she returned the car. How much will Rubal pay if the odometer showed 27612 km when she hired the car?

Say: Total distance travelled by Rubal = 28361 – 27612 = 749 km. So, she will pay 749 × ₹0.85 + ₹1500 = ₹2136.65.

Division of Decimals

Learning Outcomes

Imagine Maths Page 179

Students will be able to divide decimals up to 2 decimal places by whole numbers or decimals up to 1 decimal place.

Teaching Aids

Grids to represent decimal numbers up to 2 decimal places; Coloured pencils

Activity

Begin the class by reviewing the concept of dividing decimals. Give examples to emphasise the placement of the digits of a decimal number in the long division method.

Ask the students to work in pairs. Distribute the teaching aids and write a division problem of decimals on the board.

Instruct the pairs to represent the decimal numbers on the grid using coloured pencils and divide. Explain that they will need more than one grid for the division. To divide, ask them to take as many empty grids as the divisor and then shade the dividend equally in the empty grids. Explain that the number of squares coloured in each group/grid is the quotient of the two numbers.

0.54 ÷ 3 = 0.18

Repeat the activity for 5 rounds and instruct the students to write the answers in their notebooks. The pair to get the maximum number of correct answers wins!

Extension Idea

Ask: A log is 17.60 m long. It is to be cut into fence posts which must be 80 cm long. What is the largest number of fence posts that can possibly be cut from this log?

Say: Length of the log = 17.60 m = 1760 cm, length of each fence post = 80 cm. So, the largest number of fence posts that can possibly be cut from this log = 1760 ÷ 80 = 22.

Application of Decimals

Learning Outcomes

Students will be able to solve word problems on operations on decimal numbers.

Teaching Aids

Task cards with word problems

Activity

Imagine Maths Page 181

Prepare a set of task cards with a word problem involving one or more operations on decimals written on it.

Mr. Perez was cutting pieces of wood for a project. If his board was 93.6 inches total and he needed 52 inch pieces, how many pieces would he be able to cut?

Ask the students to work in groups. Distribute the task cards to each group. Ask them to work with their groups to solve the problems in their notebooks using the CUBES strategy. When they are done, rotate the task cards among the groups until all the groups have solved all the word problems in their notebooks. The group to solve each problem in the shortest possible time gets a reward. Discuss the solutions in the class and the challenges faced while solving the problems.

Answers

1. Representing Decimals on Number Line

Do It Together 12

2. Place Values in Decimals

Do It Together

Expanded form = 200 + 50 + 6 + 0.003

Number name = two hundred fifty-six point zero zero three.

3. Conversion Between Fractions and Decimals

Do It Together

1. 3.25 = 3.25 × 100 100 = 325 100 = 13 4 = 3 1 4

2. 4.14 = 4.14 × 100 100 = 414 100 = 207 50 = 4 7 50

4. Like and Unlike Decimals

Do It Together

Number of digits after the decimal point in 814.012 = 3

So, the number of digits after the decimal point are not equal. Thus, 15.45 and 814.012 are unlike decimals.

5. Equivalent Decimals

Do It Together

1.546 = 1.546, 53.2 = 53.200, 10.02 = 10.020, 18.3 = 18.300, 17.5 = 17.500, 67 = 67.000

6. Comparing Decimals

Do It Together

0.thth

48.023

48.000 48.100 48.020

On comparing, the descending order of the numbers is: 48.100 > 48.023 > 48.020 > 48

7. Addition and Subtraction of Decimals

Think and Tell

Weight he carries = 53.25 kg + 5.2 kg = 58.45 kg

Xavier can carry 65 kg – 58.45 kg = 6.55 kg more weight.

Do It Together O.th 1.71

–1.54 0.17

Thus, the difference of the heights of Aman and Rajneesh is 0.17 m.

8. Multiplication of Decimals

Do It Together

Total cost of 18 books = ₹156.56 × 18 = ₹2818.08

9. Division of Decimals

Do It Together

Cost of each packet of paint = ₹3547.5 ÷ 15 = ₹236.5

10. Application of Decimals

Do It Together

Total height reached = 3259.44 m

Climbed further again = 97.89 m

Total height reached = 3259.44 + 97.89 = 3357.33 m

Thus, the cyclist reaches a height of = 3357.33 m.

Data Handling 12

Learning Outcomes

Students will be able to: organise data in a table using tally marks. read and draw pictographs. read and draw bar graphs.

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 draw and interpret the data in a table. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

tally marks: a system of counting using vertical strokes, often in groups of five, to keep track of quantities pictograph: a table that shows the given data using pictures or symbols bar graph: a graph that shows information in the form of bars of different lengths

Teaching Aids

Ice cream sticks; Buttons of 5 colours; Chart paper; Smiley stickers; Building blocks

Chapter: Data Handling

Organising Data as Tally Marks

Learning Outcomes

Students will be able to organise data in a table using tally marks.

Teaching Aids

Ice cream sticks; Buttons of 5 colours; Chart paper

Activity

Demonstrate how to show tally marks to the students.

Ask the students to work in groups. Distribute the resources to each group.

Imagine Maths Page 187

Instruct the groups to make a tally marks table to show the number of each colour of the buttons. Ask them to first count the number of buttons of each colour and use the ice cream sticks to show the tally marks for the number. For example, if they have 3 yellow buttons, they should use 3 ice cream sticks. Then, instruct the students to create a frequency distribution table in their notebooks using tally marks. Ask questions like: How many more blue buttons do you have than red buttons?

Extension Idea

Ask: If each ice cream stick stands for 5 items instead of just 1 item, how would this change the tally marks drawn for each colour?

Say: Suppose there are 10 red buttons and 20 blue buttons. The red buttons will be represented by 2 ice cream sticks and the blue buttons will be represented by 4 ice cream sticks.

Pictographs

Learning Outcomes

Students will be able to read and draw pictographs.

Teaching Aids

Smiley stickers

Activity

Imagine Maths Page 190

Begin by asking what mode of transport students use to get to school. As each student responds, write the name of the mode of transport on the board as bus, car, bicycle, and so on.

Ask the students to work in groups. Distribute the smiley stickers to each group. Instruct the students to create a table like they did in the previous lesson but this time using symbols to show the data as 1 smiley = 1 student.

Ask the students to compare pictographs within their groups.

Ask questions like: How many students use a bicycle to get to school? What is the most used mode of transport? Which mode is the least used?

Learning Outcomes

Students will be able to read and draw bar graphs.

Teaching Aids

Building blocks; Chart paper

Activity

Ask each student to tell you the number of hours he/she spends playing outside every day. Draw a table on the board as given to show this data.

Hours 1 hour2 hours3 hours4 hours

No. of Students

Ask the students to work in groups.

Distribute building blocks and chart paper to each group.

Instruct the groups to draw the axes and write the title, label and scale on the chart paper. Next, ask them to create a bar graph to show the data by stacking the blocks, where each block represents 1 unit/1 hour and each stack represents a number of hours spent playing outside, that is, 1 hour, 2 hours, and so on. Explain that the height of each stack corresponds to the number of students that spend that many hours playing outside.

Instruct the students to draw the bar graph in their notebooks.

Ask questions like: How many students spend 3 hours playing outside? What is the maximum number of hours spent playing outside in a day? If student A spends x hours in a day, how many hours will he spend playing outside in a week?

Extension Idea

Ask: How will the number of blocks you stack change if you take the key as 1 block = 20 students?

Say: If you take 1 block = 20 students as the key, the number of blocks that you stack will reduce to one-fourth.

Answers

1. Organising Data as Tally Marks

Do It Together

135, 134, 136, 135, 132, 135, 136, 136, 133, 135, 134, 132, 135, 133, 133, 136, 134, 134, 135, 132, 132, 133, 136, 132, 134, 136, 135, 136, 132, 135

3. Drawing Bar Graphs

Think and Tell

Using bar graphs is an easier way to represent data here.

2. Pictographs

Do It Together

1. 480

2. Saturday, 800

3. Tuesday, 320

4. Interpreting Bar Graphs

1.

3.

Mensuration 13

Learning Outcomes

Students will be able to:

find the perimeter of plane shapes by adding the lengths of sides.

find the perimeter of regular shapes by using formulas.

find the area of regular and irregular shapes on squared paper.

find the area of squares and rectangles using the formula.

find the area and perimeter of combined shapes.

solve word problems on finding the perimeter and area of shapes and figures.

Alignment to NCF

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

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-3.6: 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 perimeter of a shape using a grid. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

plane figures: shapes that can be drawn on paper regular shape: plane figure in which all the sides are equal combined shapes: combination of more than one plane figure

Teaching Aids

Cutouts of polygons; Ribbon; String; Marker; Ruler; Shape drawn on squared paper; Sheets of squared paper; Crayons; Leaves; Squared paper cutouts; Word problem sheets

Chapter: Mensuration

Perimeter of Plane Figures

Learning Outcomes

Imagine Maths Page 206

Students will be able to find the perimeter of plane shapes by adding the lengths of sides.

Teaching Aids

Cutouts of polygons; Ribbon

Activity

Show the students an object with a border like a photo frame and ask them how to determine how much ribbon would be required to cover its border. Explain that we would have to measure the lengths of all the sides and add them. Ask the students to work in pairs. Provide each pair with some ribbon and a cutout of a polygon.

Instruct them to craft a card using the cutout by adding a ribbon border. Ask them to determine the total length of ribbon that will be required for the border by first measuring the length of each side of the polygon cutout and noting down the measures in their notebooks; and then adding the lengths to find the perimeter of their cutouts.

Extension Idea

Instruct: The perimeter of a pentagon is 84 cm. Three sides measure 12 cm, 15 cm, and 25 cm. The remaining two sides are equal. Determine the length of these two equal sides.

Say: The total length of the two remaining sides is 84 – (12 + 15 + 25), which equals 32. Therefore, each of the remaining sides measures 16 cm.

Perimeter of Regular Shapes

Learning Outcomes

Students will be able to find the perimeter of regular shapes by using formulas.

Teaching Aids

Cutouts of polygons; String; Marker; Ruler

Activity

Imagine Maths Page 207

Ask the students to work in groups. Distribute 3 cutouts of regular shapes (like a regular hexagon, regular octagon, etc.) and lengths of string to each group.

Instruct the students to select a point on the boundary of the shape and wrap the string around the shape along the boundary until it meets the starting point and mark it at that point using a marker. Ask them to measure the length of the string using a ruler and record it in their notebooks.

Next, instruct the students to measure the length of each side of the shape and then count the number of sides. Guide them to add the side length as many times as the number of sides. For example, if a regular polygon has 5 sides, each measuring 10 cm, the perimeter will be 10 + 10 + 10 + 10 + 10 = 5 × 10 = 50 cm. Ask them to calculate the perimeter in their notebooks.

Instruct the students to verify if the measured value (using string) matches the calculated value of the perimeter. Ask them to repeat the activity for the other regular polygons.

Extension Idea

Ask: If a regular shape with 17 sides has a perimeter of 357 cm, what is the measure of each side?

Say: The measure of each side is 357 ÷ 17 = 21 cm.

Learning Outcomes

Students will be able to find the area of regular and irregular shapes on squared paper.

Teaching Aids

Shape drawn on squared paper; Sheets of squared paper; Crayons; Leaves

Activity

Begin by showing the students a shape drawn on squared paper. Explain that as 5 squares are fully shaded or filled, the area of the shape is 5 sq. units.

Ask the students to work in pairs. Distribute a sheet of squared paper, a leaf and crayons to each pair.

Instruct the pairs to place a leaf on the squared paper and draw its outline. Ask them to colour their drawing using crayons, assigning different colours for fully filled, half-filled, more than half-filled, and less than half-filled squares. Instruct them to count each more than half square as 1, to count two halves as 1 square and to not count the less than half squares. Ask them to record the number of squares for each category in their notebooks. Then, instruct them to find the total area of the shape using the information obtained about the squares corresponding to each category.

Ask: Why do you think we do not consider the less than half-filled squares while finding the area?

Area of Squares and Rectangles Imagine Maths Page 211

Learning Outcomes

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

Teaching Aids

Squared paper cutouts of squares and rectangles

Activity

Distribute squared paper cutouts of squares and rectangles to the students along with some stickers.

Ask the students to work in pairs. Instruct them to fill the shapes using the stickers, one for each box. Then, ask them to count the total number of stickers used for a shape and record it in their notebooks.

Next, ask the students to count the stickers in a row for the length and in a column for the breadth of the shape and record these in their notebooks. They should then multiply these counts to find the total area.

Guide them to verify that the areas obtained by counting the total number of stickers and by multiplying the length and breadth are the same.

Ask: Why do you think we multiply the same length by itself for a square but multiply different lengths for a rectangle?

Give the students more problems to solve using the formula.

Extension Idea

Ask: The area of a square field is 100 sq. m. The owner wants to fence this field. The cost of fencing for 10 m is ₹500. How much would it cost to fence the entire field?

Say: As the area is 100 sq. m., the measure of each side is 10 m. So, the cost to fence the entire field would be ₹500 × 4 = ₹2000.

Learning Outcomes

Students will be able to find the area and perimeter of combined shapes.

Teaching Aids

Squared paper cutouts

Activity

Ask the students to work in pairs. Distribute squared paper cutouts to each student.

Instruct each pair to form a combined shape using the cutouts. Ask them to work together to find the perimeter of their combined shape by counting the number of squares along each side and adding the lengths of all the sides. Emphasise that they need to subtract the length of the common side once to avoid counting it twice.

When they are done, ask the students to count the total number of squares inside their combined shapes (representing the area) and record the counts in their notebooks.

Then, instruct them to find the areas of the individual shapes in the combined shapes and write the total area in their notebooks. Ask them to verify the value with the area obtained earlier and check whether they are the same.

Word Problems on Perimeter and Area

Learning Outcomes

Imagine Maths Page 215

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

Teaching Aids

Word problem sheet; Sheets of squared paper

Activity

Give each student a sheet of squared paper and a sheet with a word problem, such as: A playground is divided into two rectangular areas. The first section measures 25 metres by 20 metres, and the second section has dimensions of 15 metres by 10 metres. If the cost of planting grass is ₹50 per m2, what is the total cost of planting grass in the playground?

Instruct the students to use the CUBES strategy by circling the numbers, underlining the question, boxing the key words, evaluating and solving the problem.

Guide them to draw the figure on the squared paper.

Ask them to first find the area of the shape and then the final answer.

Extension Idea

Instruct: Create a problem that requires finding the combined area of a triangle and a square sharing the same base.

Say: A sample problem is: Aman is planning a garden area. He wants to have a square patch of grass with an equilateral triangular flower bed next to it. The side of the square patch is 5 metres, and the base of the equilateral triangle matches this side. Find the total area.

Answers

1. Perimeter of Plane Figures

Do It Together

1. Perimeter

= 7 cm + 5 cm + 6 cm + 2 cm = 20 cm

2. Perimeter

= 6 m + 2 m + 2 m + 6 m + 3 m = 19 m

3. Perimeter

= 9 cm + 5 cm + 5 cm + 11 cm + 6 cm = 36 cm

2. Perimeter of Regular Shapes

Think and Tell

The perimeter of a regular octagon with sides of 7 cm will be 8 × 7 cm = 56 cm.

Do It Together

Length of the side of heptagon = 8 m

Perimeter of heptagon = 7 × 8 m = 56 m

Length of rectangle = 15 m

Perimeter of rectangle = Perimeter of heptagon

2 × (L + B) = 56

B = 56 2 – L; B = 28 – 15 = 13 m

Hence the breadth of the rectangle is 13 m.

3. Area on Graph Paper

Do It Together

1. Fully filled squares = 33

Half-filled squares = 5

More than half-filled squares = 10

Less than half-filled squares = 3

Area = 33 + 5 × 1 2 + 10 = 45 1 2 sq. units

2. Fully filled squares = 30

Half-filled squares = 0

More than half-filled squares = 14

Less than half-filled squares = 10

Area = 44 sq. units

3. Fully filled squares = 5

Half-filled squares = 2

More than half-filled squares = 12

Less than half-filled squares = 8

Area = 18 sq. units

4. Area of Squares and Rectangles

Think and Tell

Yes.

Do It Together

Width of the room= 5 m

Length of the room = 6 m

Area of the room= Length × Breadth = 6 × 5 = 30 sq. m

Width of the rectangular tile= 0.2 m

Length of the rectangular tile= 0.3 m

Area of one tile = Length × Breadth = 0.3 × 0.2 = 0.06 sq. m

Number of tiles required = Area of the room Area of one tile = 500

Thus, the number of tiles required is 500.

5. Area and Perimeter of Combined Shapes

Do It Together

We split the shape into parts. Name them and find the missing dimensions.

Area of part A = L × B = 5 × 3 = 15 sq. cm

Area of part B = L × B = 2 × 6 = 12 sq. cm

Area of the figure = 15 + 12 = 27 sq. cm

Perimeter of the figure = 26 cm.

6. Word Problems on Perimeter and Area

Do It Together

Cost of fencing the rectangular park = ₹5200

Breadth = 8 m; Rate of fencing = ₹130

Total cost of fencing = Perimeter × Rate of fencing per m.

Perimeter = Total cost of fencing Rate of fencing per m = ₹5200 ₹130 = 40

Perimeter = 2 × (Length + Breadth)

Length = Perimeter 2 – Breadth = 20 – 8 = 12 m

Area = Length × Breadth = 20 × 12 = 240 sq. m.

Introduction to Algebra

Learning Outcomes

Students will be able to:

identify or form the rule of a pattern using variables and use it to find the nth term. use variables to write rules to deduce formulas for finding the perimeter or area of shapes. form algebraic expressions for given statements and solve them for the given value of variables. form algebraic equations for given statements. solve linear equations in one variable to find the value of the variable.

Alignment to NCF

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

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

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

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

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

Let’s Recall

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

Vocabulary

pattern: an arrangement of shapes, lines, letters, numbers, or colours that keep repeating polygon: a 2-D shape made up of 3 or more line segments

algebraic expression: a statement formed by meaningful combinations of variables, constants, and operators

linear equation: an equation in which the highest power of each variable is 1

Teaching Aids

Matchsticks; Sheet of paper with a table drawn on it; Toothpicks; Slips of paper with statements written on them; Two sets of index cards (one set with statements and the other set with the corresponding algebraic equations); Index cards (with a linear equation in one variable)

Chapter: Introduction to Algebra

Number Patterns

Learning Outcomes

Imagine Maths Page 221

Students will be able to identify or form the rule of a pattern using variables and use it to find the nth term.

Teaching Aids

Matchsticks; Sheet of paper with a table drawn on it

Activity

Begin with a discussion about patterns in everyday life.

Write a simple pattern on the board, for example, 5, 7, 9, 11, …

Ask the students to find the rule of the pattern. Discuss how to find the rule. Explain that the rule is 2 times the term number added to 3 and that this rule will help find any term of the pattern. So, the nth term is 2n + 3. Ask the students to work in groups. Distribute the matchsticks and sheets.

Shape number

12345

Number of sticks required 5 913

Rule

Number of sticks required =

Instruct them to use matchsticks to form the pattern as shown in the table. Let them form at least 5 terms of the pattern. Ask them to discuss within their groups to identify the rule of the pattern and write the rule on the sheet.

Allow the students to show one more pattern with matchsticks and identify the rule.

Extension Idea

Instruct: Consider (3, 6), (5, 10), (7, 14). Identify the rule that describes how to get the second number from the first number in every ordered pair.

Say: In each pair, the pattern is: [2 × 1 + 1, 2 × (2 × 1 + 1)] = (3, 6); [2 × 2 + 1, 2 × (2 × 2 + 1) = (5, 10) and [2 × 3 + 1, 2 × (2 × 3 + 1)] = (7, 14). So, the rule is [2n + 1, 2 × (2n + 1)].

Rules in Patterns

Learning Outcomes

Imagine Maths Page 224

Students will be able to use variables to write rules to deduce formulas for finding the perimeter or area of shapes.

Teaching Aids

Toothpicks

Activity

Say: Consider a square with sides measuring x. To find the perimeter, we just multiply the number of sides by the length, i.e., 4 × x, and for the area, we multiply the lengths of two sides, i.e., x × x

Ask the students to work in pairs. Distribute a set of toothpicks to each pair.

Instruct them to form polygons using toothpicks and write down the number of sides (which are all equal) for each polygon in their notebooks. Then, ask them to assign a variable, say x, as the length of one toothpick, calculate the perimeter of each polygon and write the results in their notebooks.

Guide them to use the toothpicks to create squares and rectangles, still using x as the length of one toothpick.

Instruct them to calculate the area of each square and rectangle using x.

Ask questions like: Is there a connection between the side length and the number of toothpicks?

Forming Algebraic Expressions

Learning Outcomes

Imagine Maths Page 225

Students will be able to form algebraic expressions for given statements and solve them for the given value of variables.

Teaching Aids

Slips of paper with statements written on them

Activity

Write a statement on the board, such as: 3 less than the perimeter of a square of side ‘a’. Discuss the algebraic expression for this statement, which is 4a – 3. Write the expression on the board. Explain that 4 is the coefficient of a, a is the variable, 3 is a constant and 4a and 3 are the terms of the expression.

Divide the class into groups. Distribute the slips of paper with statements written on them.

2 more than 4 times m 7 less than 5 times a number

Instruct the students to write the algebraic expressions for the statements written on their slips, for example, 4m + 2.

Ask the students to find the value of the algebraic expressions for any given values of the variables.

Extension Idea

Ask: For the statement, 3 less than 4 times a number is equal to 13, a student wrote the equation 3a – 4 = 13. Was his response accurate? If not, what is the correct equation for the statement?

Say: His response was not correct. The correct equation is 4a – 3 = 13.

Forming Algebraic Equations

Learning Outcomes

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

Teaching Aids

Imagine Maths Page 228

Two sets of index cards (one set with statements and the other set with the corresponding algebraic equations)

Activity

Write a statement on the board, such as: 4 more than 4 times ‘a’ is twice of 8. Discuss the algebraic equation for this statement, which is 4 + 4a = 16.

Explain that 4 + 4a is the LHS and 16 is the RHS and the highest power of a in the equation is 1, so this is a linear equation.

Divide the class into 2 equal groups.

Give one group the set of index cards containing written statements, such as 5 times m less than 10 is 5. Give the

other group the set of index cards containing the corresponding algebraic expressions for the statements, such as 10 – 5m = 5. Ensure that both sets of cards are thoroughly shuffled.

Instruct the first group to write the algebraic equations for their statements and the second group to write the possible statements for their equations.

Have each student move around the classroom to find the student that has the matching card.

Solving Linear Equations in One Variable

Learning Outcomes

Imagine Maths Page 230

Students will be able to solve linear equations in one variable to find the value of the variable.

Teaching Aids

Index cards (with a linear equation in one variable)

Activity

Write a statement on the board, such as: If a number is tripled, its result is 30. Discuss how if the number is n, then 3n = 30 n = 30 ÷ 3 n = 10.

Ask the students to work in pairs. Give each pair an index card that has a linear equation in one variable written on it. Instruct one student in each pair to use the trial and error method to find a solution and the other student to use the transposition method to solve the equation. Ask them to write the solutions in their notebooks.

When they are done, ask the pairs to compare and cross-verify their obtained solutions.

Ask: Which do you think is better: the trial and error method or the transposition method?

Extension Idea

Instruct: Find the number if 15 added to three times the number results in 45.

Say: Let the number be x. As 15 + 3x = 45, using the transposition method, we get 3x = 45 – 15 = 30 x = 30 3 = 10 x = 10.

Answers

1. Number Patterns

Think and Tell

There will be 10 × 3 = 30 seats in the tenth row.

Do It Together

Shape number 1234515

Number of sticks required 51015 20 25 75

Rule Number of sticks required = n × 5

2. Rules in Patterns

Do It Together

Perimeter of a regular pentagon = sum of the lengths of all its sides

So, P = a + a + a + a + a = 5 × a = 5a

Thus, we get the rule for the perimeter of a regular pentagon as 5a.

3. Forming Algebraic Expressions

Do It Together

Let Anil’s age be x years.

So, the age of Anil’s mother = 3x – 2 years.

4. Rules in Algebraic Expressions

Do It Together

The multiplicative identity formula is expressed as a × 1 = a, where a is any number.

5. Forming Algebraic Equations

Do It Together

Let the number be z

Adding 3 to 4 times this number can be written algebraically as 4z + 3. So, the required equation is 4z + 3 = 21.

6. Trial and Error Method

Do It Together

Here, LHS = 3x + 4 and RHS = 16

For x = 1, LHS = 3 × 1 + 4 = 3 + 4 = 7 RHS;

For x = 2, LHS = 3 × 2 + 4 = 6 + 4 = 10 RHS;

For x = 3, LHS = 3 × 3 + 4 = 9 + 4 = 13 RHS;

For x = 4, LHS = 3 × 4 + 4 = 12 + 4 = 16 = RHS;

Therefore, for x = 4, LHS = RHS = 16

So, x = 4 is the solution of the equation 3x + 4 = 16.

7. Transposition Method

Do It Together

Let Sonal have x number of friends.

Number of balloons distributed among friends = 5x. She has 3 balloons left. But the total number of balloons is 33.

So, the equation that satisfies the condition is 5x + 3 = 33.

5x + 3 = 33 5x = 33 – 3 5x = 30 x = 30 5 = 6

Therefore, Sonal has 6 friends.

Introduction to Ratio and Proportion 15

Learning Outcomes

Students will be able to: represent a ratio in the form a:b, a b or a to b. reduce a given ratio to its simplest form. find equivalent ratios for a given ratio. check whether 4 terms are in proportion. solve word problems on the unitary method.

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 about fractions and their types. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

ratio: a relation or comparison between numbers equivalent: equal in value

Teaching Aids

Number cards; Marbles in various colours (red, blue, yellow, green); Index cards (with ratios written on them); Pairs of index cards with equivalent ratios; Red and blue ribbons; A pair of scissors; Ruler; Bags with price tags; Word problem cards

Chapter: Introduction to Ratio and Proportion

Introducing Ratios

Learning Outcomes

Students will be able to represent a ratio in the form a:b, a b or a to b.

Teaching Aids

Number cards

Activity

Imagine Maths Page 237

Begin by explaining what a ratio is. Give an example of a ratio. Place some number cards on a table. Instruct each student to pick any two number cards from the table. Ask them to find the ratio of the smaller number to the bigger number. For example, if a student picks 4 and 7, the ratio would be 4:7. Instruct the students write down their ratios in their notebooks.

Repeat the activity by asking the students to write the ratio of the bigger number to the smaller number.

Ask: How does changing the order of the numbers affect the ratio?

Extension Idea

Ask: A wall has a width of 5 m. The ratio of the width of a window on the wall to that of the wall is 2:5. What is the width of the window?

Say: As the ratio is 2:5, the width of the window is 2 m.

Ratio in Simplest Form

Learning Outcomes

Students will be able to reduce a given ratio to its simplest form.

Teaching Aids

Index cards (with ratios written on them); Marbles in two different colours

Activity

Imagine Maths Page 239

Ask the students to work in groups. Distribute index cards that have a ratio written (example: 12:15) and marbles to each group.

Instruct them to use marbles of two different colours to represent the ratio (example: 12 red marbles and 15 blue marbles to show 12:15). Next, ask them to divide each set of marbles into the same number of groups (example: 3 groups of 4 red marbles and 3 groups of 5 blue marbles for 12:15). Ask them to count the number of marbles in each group. Explain that the number of marbles in each group for each colour gives the simplest form, 4:5, of the given ratio, which can be written as 4/5. Ask the students to write the ratio in the simplest form in their notebooks. Discuss how to divide the numerator and denominator by their HCF to find the simplest form. Ask questions like: Why do you think it is important to simplify a ratio?

Extension Idea

Ask: Can you divide 50 marbles into two cups in the ratio 3:7? How many marbles will there be in each cup?

Say: Yes, we can divide 50 marbles into two cups in the ratio 3:7. There will be 15 marbles in one cup and 35 in the other.

Learning Outcomes

Students will be able to find equivalent ratios for a given ratio.

Teaching Aids

Pairs of index cards with equivalent ratios; Marbles in two different colours

Activity

Ask the students to work in groups. Distribute index cards with equivalent ratios and marbles in two different colours to each group.

Instruct them to use the marbles to find the simplest form of each ratio on the cards, as they did in the previous lesson. Discuss what they notice about the simplest form of both ratios. Explain that when the simplest form of two ratios is the same, they are known as equivalent ratios.

Ask the students to find equivalent ratios of a given ratio by multiplying or dividing the numerator and denominator with the same natural number in their notebooks. Give them more problems for practise.

Proportion

Learning Outcomes

Students will be able to check whether 4 terms are in proportion.

Teaching Aids

Red and blue ribbons; Ruler; A pair of scissors

Activity

Imagine Maths Page 244

Ask the students to work in groups. Distribute red and blue ribbons, a ruler and a pair of scissors to each group.

Instruct the groups to cut red ribbons of lengths 2 cm and 6 cm, and then to cut a blue ribbon of length 3 cm. Then, ask them to think and cut another length of blue ribbon such that the ratio of the 2 cm red ribbon to the 3 cm blue ribbon is equivalent to the ratio of the 6 cm red ribbon to the length of blue ribbon they cut. Once they cut the blue ribbon, ask them to note down the ratios in their notebooks. Discuss the sign of proportion and the terms. Instruct the students to use the formula and find the correct length of the blue ribbon in their notebooks.

Then, ask them to find the length of the red ribbon if a third blue ribbon is 15 cm long and write the answer in their notebooks.

Extension Idea

Ask: Do the ratios 25 cm to 1 m 50 cm and 30 seconds to 2 minutes 30 seconds form a proportion?

Say: The ratio of 25 cm to 1 m 50 cm is 25:150 = 1:6. The ratio of 30 seconds to 2 minutes 30 seconds is 30:150 = 1:5. So, the ratios are not in proportion.

Unitary Method

Learning Outcomes

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

Teaching Aids

Marbles in various colours (red, blue, yellow, green); Bags with price tags; Word problem cards

Activity

Place the marbles in bags as follows:

4 red marbles in a bag with a price tag of ₹100;

6 blue marbles in a bag with a price tag of ₹120;

5 yellow marbles in a bag with a price tag of ₹75;

8 green marbles in a bag with a price tag of ₹320.

Create word problem cards with questions like: How much do 6 red marbles cost? or What is the cost of 3 blue marbles?

Instruct each student to pick a card, solve the problem and write the answer in their notebooks. Discuss how to solve the problems and explain the process of using the unitary method to solve them.

Extension Idea

Instruct: Create your own word problem based on unitary method.

Say: There can be many such problems. One could be: Rahul purchased 15 kg of oranges for ₹600. How much will 4 kg of oranges cost?

Answers

1. Introducing Ratios

Do It Together

1. Number of red balls = 4; Number of white balls = 9

The ratio of the number of red balls to the number of white balls = 4:9

2. Number of black balls = 5; Total number of balls = 20

The ratio of the number of black balls to the total number of balls = 5:20

2. Ratio in Simplest Form

Do It Together

1 year 8 months = 12 months + 8 months = 20 months

The required ratio = 20 months : 5 months = 20 5 = 4 1

The ratio of 1 year 8 months to 5 months is 4:1.

3. Equivalent Ratios

Think and Tell

A ratio can have infinite equivalent ratios.

Do It Together

Number of parts given to group A = 5 out of 9 parts = 5 9

Number of parts given to group B = 4 out of 9 parts = 4 9

Number of chocolates given to group A = 5 9 × 90 = 50 chocolates

Number of chocolates given to group B = 4 9 × 90 = 40 chocolates

4. Introducing Proportion

Think and Tell

They will cover 4 km by pedalling for 2 hours.

Do It Together

Ratio of Mansi’s weight to mother’s weight = 25:60 = 25 60 = 5 12

Ramesh’s weight = 30 kg; Father’s weight = 75 kg

Ratio of Ramesh’s weight to father’s weight = 30:75 = 30 75 = 2 5

Hence, the given weights are not proportion.

5. Proportional Terms

Do It Together

Let the number of cups of flour required be x.

Using proportion, 3:x::2:15

Therefore, x × 2 = 3 × 15 2x = 45 x = 45 2 = 22.5

Hence, 22.5 cups of flour are required to make 15 dozen cookies.

6. Continued Proportion

Do It Together

7:14::14:x

7 × x = 14 × 14

7x = 196 x = 28

So, the third term is 28.

7. Unitary Method

Do It Together

Cost of 1 litre of paint A = 750 15 = ₹50

Cost of 20 litres of paint B = ₹900

Cost of 1 litre of paint B = 900 20 = ₹45

Hence, paint B will be cheaper than paint A.

Sy etry 16

Learning Outcomes

Students will be able to:

draw lines of symmetry in given shapes or figures. draw the reflection of a shape or figure on squared paper.

Alignment to NCF

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

Let’s Recall

Recap to check if students know what symmetrical and non-symmetrical shapes are. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

line of symmetry: a line that divides a shape into equal and identical halves reflection: an image of the original formed in the mirror

Teaching Aids

Picture cards with figures having 0, 1 or more lines of symmetry; Sheets of paper; Shape cards with shapes/designs having 1 line of symmetry; Mirror; Squared paper with half shapes drawn

Chapter: Symmetry

Lines of Symmetry

Learning Outcomes

Students will be able to draw lines of symmetry in given shapes or figures.

Teaching Aids

Picture cards with figures having 0, 1 or more lines of symmetry; Sheets of paper

Activity

Ask the students to work in groups of 3. Distribute picture cards and sheets of paper to each group.

Instruct the groups to draw lines of symmetry on each card for the shapes/designs drawn. Ask them about the similarities and differences in these shapes/designs.

Ask if there can be designs that have no lines of symmetry or infinite lines of symmetry. Instruct them to verify their answers by creating designs on a sheet of paper that have 1, no, or more than 1 line of symmetry. Instruct them to exchange designs with another group and compare the lines of symmetry.

Extension Idea

Ask: How many lines of symmetry does the cross section of a lemon have?

Say: The cross section of a lemon is a circle, so it has infinite lines of symmetry.

Reflection and Symmetry

Learning Outcomes

Students will be able to draw the reflection of a shape or figure on squared paper.

Teaching Aids

Imagine Maths Page 257

Shape cards with shapes/designs having 1 line of symmetry; Mirror; Squared paper with half shapes drawn

Activity

Instruct each student the day before to bring a plane mirror to the class.

Ask the students to work in groups. Distribute 2–3 cards with different shapes/designs having 1 line of symmetry to each group.

Instruct the groups to place the shape cards on a piece of paper one by one and draw a vertical line on the right for each card. Guide them to place the mirror along the line such that the given shape is reflected in it. Ask them to observe and draw the reflected shape in their notebooks and compare the two shapes to tell if they are symmetrical.

Ask questions like: Is the image in the mirror the same shape and size as the original figure? Is it at an equal distance from the mirror line as the original?

Distribute a sheet of squared paper with half a shape drawn on it to each group. Ask the students to complete

the shape by drawing its other half on the squared paper itself. Ask them to exchange sheets with other groups for peer review. Help them generalise that in all cases of mirror symmetry, there is a line of symmetry along which the mirror symmetry is shown.

Extension Idea

Ask: What are the first 5 letters of the English alphabet that appear to be the same as their reflections in the mirror?

Say: A, H, I, M, O are the first 5 letters of the English alphabet that appear to be the same as their reflections.

Answers

1. Symmetry in Shapes

Think and Tell

We can draw a line of symmetry exactly at the centre of a line segment.

Do It Together

3. Reflection and Symmetry

Think and Tell

The word AMBULANCE is written in the form of its mirror image so that the person who is driving in their vehicle ahead can view it in their mirror and instantly identify the word as ‘AMBULANCE’, and give way to it in an emergency.

Do It Together

2. Symmetry in Designs or Figures

Do It Together 1

Construction of Line Segments and Angles 17

Learning Outcomes

Students will be able to:

construct a line segment and the copy of a given line segment using a compass. construct a perpendicular using a compass.

construct a perpendicular bisector using a compass. draw angles of given measures using a protractor. construct a copy of an angle using a compass. construct an angle bisector using a compass.

construct angles of 60°, 120° and 30° using a compass. construct angles of 45°, 75°, 150° and 135° using a compass.

Alignment to NCF

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

Let’s Recall

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

Vocabulary

perpendicular: two lines that intersect at a right angle, forming a 90° angle perpendicular bisector: a perpendicular that divides a line segment into two equal parts angle bisector: a line or ray that divides an angle into two equal angles

Teaching Aids

Straws; String; Ruler; Compass; Pencil; Protractor; Sheets of paper; Chart paper

Chapter: Construction of Line Segments and Angles

Constructing Line Segments and Their Copy

Learning Outcomes

Imagine Maths Page 266

Students will be able to construct a line segment and the copy of a given line segment using a compass.

Teaching Aids

Straws; String; Ruler; Compass; Pencil

Activity

Ask the students to work in pairs. Distribute a straw and a length of string to each pair. Instruct the pairs to cut the straw such that it has a length of 6.3 cm. Ask them to use the string to check the length. Then ask them to make a copy of the straw by placing it on paper and drawing a line along it.

Instruct the students to read the steps of construction of a line segment and its copy from the Imagine Mathematics book. Ask each student to draw a line segment of any length in their partner’s notebook and then use a compass to construct a copy of the line segment drawn by their partner below it.

Ask questions like: How do you know that the line segment and its copy have the same length?

Extension Idea

Ask: How will you construct a line segment equal to the sum of two lengths?

Say: First, you will copy the length of the first line segment. Then, from the end point of this line segment, you will draw a line segment of the second length.

Constructing a Perpendicular Using a Compass

Learning Outcomes

Students will be able to construct a perpendicular using a compass.

Teaching Aids

Straws; Ruler; Compass; Pencil; String; Sheets of paper

Activity

Introduce the concept of a perpendicular.

Imagine Maths Page 271

Ask the students to work in pairs. Provide each pair with 2 straws, some string, a pencil, and a sheet of paper. Give them a specific length, say 4.3 cm. Ask them to cut one straw to the given length and affix it to the sheet of paper horizontally. They should tie one end of a length of string to a pencil. Then, with the other end fixed to any point on the straw, they should stretch the string and move the pencil to draw arcs cutting the straw at 2 points, marking the two points. Next, ask them to take a longer piece of string, tie one end to the pencil, place the other end on one of the two points marked on the straw and draw an arc above the straw. They should repeat this by holding one end of the string on the other point marked on the straw. The two arcs will intersect at a point. Instruct the students to connect the initial arc’s centre to the intersection point and paste the second straw vertically along this line. This will help students understand the concept of constructing a perpendicular. Ask the students to read the steps of construction of a perpendicular in their Imagine Maths books and then use a compass to construct a perpendicular in their notebooks.

Constructing a Perpendicular Bisector

Learning Outcomes

Students will be able to construct a perpendicular bisector using a compass.

Teaching Aids

Ruler; Compass

Activity

Imagine Maths Page 273

Begin the class by discussing some real-life examples where perpendicular bisectors are used like the water sprinkler in the garden, wheel alignment in vehicles, etc.

Distribute compasses and rulers to the students.

Instruct the students to read the steps of construction of a perpendicular bisector in their Imagine Maths books. Ask them to start by constructing a line segment AB in their notebooks and then use a compass to construct a perpendicular bisector to AB and mark the point where the perpendicular bisector meets AB as O. Ask them to measure the lengths of OA and OB.

Ask questions like: What do you notice about the lengths of OA and OB? Have a discussion on how the distance from any point marked on the perpendicular bisector to the two end points of the line segment will always be equal.

Extension Idea

Ask: How can you accurately draw the perpendicular bisector for a diagonally inclined line segment on a sheet of paper?

Say: We can rotate the sheet to align the line segment horizontally, then draw the perpendicular bisector starting from the midpoint of the segment.

Constructing Angles of Given Measures

Learning Outcomes

Students will be able to draw angles of given measures using a protractor.

Teaching Aids

Protractor; Ruler

Activity

Begin the class by discussing angles and how to measure them using a protractor.

Ask the students to form pairs. Provide each student with a protractor and ruler.

Imagine Maths Page 276

Instruct the students to draw a line segment in their notebooks and label it as PQ. Ask them to use the protractor to draw angles of 45° and 57° from points P and Q, respectively, and extend them to meet at a point. Then, ask them to measure the third angle of the triangle formed. Finally, ask them to exchange notebooks to measure the angles drawn by their partners and check if the angle measure is correct.

Extension Idea

Ask: Using a protractor that measures up to 180°, how can you draw an angle of 200°?

Say: To begin drawing 200°, we will first subtract it from 360°. 360° − 200° = 160°. Then, we will draw a 160° angle and mark the reflex angle.

Constructing a Copy of an Angle Imagine Maths Page 277

Learning Outcomes

Students will be able to construct a copy of an angle using a compass.

Teaching Aids

Ruler; Compass; Protractor

Activity

Ask the students to work in pairs. Provide each pair with a ruler, compass and protractor.

Instruct the students to use the protractor to draw a 50° angle in their notebooks.

Ask them to read the steps of constructing a copy of an angle in their Imagine Maths books. Instruct them to use these steps to construct a copy of the 50° angle that they drew in their notebooks and measure the new angle to check and ensure the accuracy of the copied angle.

Ask questions like: When can the construction go wrong?

Extension Idea

Ask: What will be the measure of the copy of an angle which is the complement of 44°?

Say: The complement of 44° is 46°, so the copy of 46° will also measure 46°.

Constructing an Angle Bisector

Learning Outcomes

Students will be able to construct an angle bisector using a compass.

Teaching Aids

String; Sheets of paper; Compass; Ruler; Protractor

Activity

Imagine Maths Page 278

Begin the class by discussing examples of using angle bisectors in real life such as to divide a slice of pizza to share equally with a friend. Explain that the imaginary line from the edge of the slice to the centre dividing it into 2 equal parts is like an angle bisector.

Ask the students to work in pairs. Distribute the teaching aids.

Instruct the students to use the protractor to draw an angle of 80° on the sheet of paper. Ask them to tie a pencil to one end of the string and place the other end on the vertex of the angle drawn; then, stretch the string and draw an arc cutting the arms of the angle. From the points where the arcs cut the arms, instruct them to draw two more arcs cutting each other and draw a line segment joining the intersection point to the vertex of the angle. Ask them to measure the angles formed.

Instruct the students to use the same method of constructing an angle bisector using a compass.

Extension Idea

Draw a triangle and angle bisectors to the three angles in the triangle on the board.

Ask: What do you notice about the angle bisectors drawn to the three angles in the triangle?

Say: The angle bisectors meet at one point in the centre of the triangle.

Constructing 60°, 120° and 30° Angles

Learning Outcomes

Students will be able to construct angles of 60°, 120° and 30° using a compass.

Teaching Aids

Protractor; Compass; Ruler

Activity

Imagine Maths Page 279

Ask the students to form pairs. Provide each student with a ruler, compass and protractor.

Instruct the students to use the protractor to draw a line segment AB of length 8 cm in their notebooks.

Ask them to read the steps of constructing 60°, 30° and 120° angles in their Imagine Maths books and use these steps to construct angles of 60° using a compass on points A and B (ensuring that the rays do not meet). Then, ask them to construct an angle bisector to the 60° angle created on point A to get the 30° angle.

Instruct them to use the 60° angle constructed on point B as the base to draw another 60° to get a 120° angle at point B.

Ask the students to exchange notebooks with their partners to check each other’s work.

Constructing 45°, 75°, 150° and 135° Angles

Learning Outcomes

Students will be able to construct angles of 45°, 75°, 150° and 135° using a compass.

Teaching Aids

Chart paper; Protractor; Compass; Ruler

Activity

Ask the students to work in groups. Distribute the teaching aids to each group.

Imagine Maths Page 281

Instruct the students to make a collage of angles on chart paper by constructing angles of 45°, 75°, 150° and 135° using a compass. Each student of the group should construct one angle. Tell them that they need to use their understanding of the angles constructed in the previous lessons to construct these angles. Ask them to discuss in their groups the methods that they will use to construct these angles.

Ask questions like: How did you construct the 45°, 75°, 150° and 135° angles? Did you use the construction of angle bisectors to construct any of the angles?

Extension Idea

Ask: How will you construct an angle of 15° using a compass?

Say: We will construct an angle bisector to a 60° angle to first get two 30° angles, then an angle bisector to one of the 30° angles to get two 15° angles.

Answers

1. Constructing Line Segments and Their Copy

Do It Together

2. Constructing a Perpendicular Using Set Squares

Do It Together

F6.7 cm

3. Constructing a Perpendicular Using a Compass

Do It Together

6. Constructing a Copy of an Angle

Do It Together

7. Constructing an Angle Bisector

Do It Together Angle = 44°.

Complement of the 44° angle = 90 – 44 = 46°. On constructing the angle bisectors, we get two 23° angles.

8. Constructing 60°, 120° and 30° Angles

Do It Together

4. Constructing a Perpendicular Bisector

Do It Together

Yes, the bisectors intersect.

5. Constructing Angles of Given Measures

Do It Together

9. Constructing 45°, 75°, 150° and 135° Angles

Think and Tell

Yes, to draw an angle of 7 1 2 ° by bisecting an angle of 15°, first draw the 15° angle. Then, use a protractor to bisect the 15° angle, marking the point where the bisector intersects the angle. This intersection point will give you the 7 1 2 ° angle.

Do It Together

Solutions

Chapter 1

Let's Warm-up

1. The number 586595 can be rewritten using commas as 5,865,952. False 2. The place value and face value of 0 in 9,80,564 is 0 in each case. True 3. The expanded form of 3,83,452 is 300000 + 80000 + 3000 + 400 + 50 + 2. True

4. The number name of 7,03,478 is seven lakh three hundred four thousand seventy-eight. False 5. Between 40,73,548 and 40,72,187, 40,73,548 is greater. True

Do It Yourself

1A 1.a. 99 = 90 + 9 = XC + IX = XCIX b. 184 = 100 + 80 + 4 =

C + LXXX + IV = CLXXXIV c. 285 = 200 + 80 + 5 = CC + LXXX + V = CCLXXXV d. 438 = 400 + 30 + 8 = CD + XXX + VIII = CDXXXVIII

e. 1564 = 1000 + 500 + 60 + 4 = M + D + LX + IV = MDLXIV

f. 1287 = 1000 + 200 + 80 + 7 = M + CC + LXXX + VII = MCCLXXXVII

2. a. CDLXVI = CD + LX + VI = (500 – 100) + (50 + 10) + (5 + 1) = 400 + 60 + 6 = 466 b. XCVI = XC + VI = (100 – 10) + (5 + 1) = 90 + 6 = 96

c. CXV = C + XV = 100 + (10 + 5) = 100 + 15 = 115 d. CCCLXV = CCC + L + XV = (100 + 100 + 100) + 50 + (10 + 5) = 300 + 50 + 15 = 365

e. MDCCCLXIV = M + DCCC + LX + IV = 1000 + (500 + 100 + 100 + 100) + (50 + 10) + (5 – 1) = 1000 + 800 + 60 + 4 = 1864

f. CMXCIX = CM + XC + IX = (1000 – 100) + (100 – 10) + (10 – 1) = 900 + 90 + 9 = 999

3. a. CLXXXVII = C + LXXX + VII = 100 + (50 + 10 + 10 + 10) + (5 + 1 + 1) = 100 + 80 + 7 = 187

CXLVII = C + XL + VII = 100 + (50 – 10) + (5 + 1 + 1) = 100 +

40 + 7 = 147

187 > 147

So, CLXXXVII > CXLVII

b. XC = 100 – 10 = 90, CX = 100 + 10 = 110

90 < 110

So, XC < CX

c. CXCII = C + XC + II = 100 + (100 – 10) + (1 + 1) = 100 + 90 + 2 = 192

CC = 100 + 100 = 200

192 < 200

So, CXCII < CC

d. LXIX = LX + IX = (50 + 10) + (10 – 1) = 60 + 9 = 69

LXXII = LX + XII = (50 + 10) + (10 + 1 + 1) = 60 + 12 = 72

69 < 72

So, LXIX < LXXII

e. CXCVII = C + XC + VII = 100 + (100 – 10) + (5 + 1 + 1) = 100 + 90 + 7 = 197

CLXXV = C + LXX + V = 100 + (50 + 10+ 10) + 5 = 100 + 70 + 5 = 175

197 > 175

So, CXCVII > CLXXV

f. MCXLVI = M + C + XL + VI = 1000 + 100 + (50 – 10) + (5 + 1) = 1000 + 100 + 40 + 6 = 1146

MCMXXIV = M + CM + XX + IV = 1000 + (1000 – 100) + (10 + 10) + (5 – 1) = 1000 + 900 + 20 + 4 = 1924

1146 < 1924

So, MCXLVI < MCMXXIV

4.a. CDLXV = CD + LX + V = (500 – 100) + (50 + 10) + 5 = 400 + 60 + 5 = 465

Number just before 465 = 464 = 400 + 60 + 4 = (500 – 100) + (50 + 10) + 4 = CD + LX + IV

= CDLXIV

Number just after 465 = 466 = 400 + 60 + 6 = (500 – 100) +

(50 + 10) + 6 = CD + LX + VI

= CDLXVI

So, CDLXIV CDLXV CDLXVI

b. CMLXVII = CM + LX + VII = (1000 – 100) + (50 + 10) + (5 + 1 + 1) = 900 + 60 + 7 = 967

Number just before 967 = 966 = 900 + 60 + 6 = (1000 – 100) + (50 + 10) + 6 = CM + LX + VI

= CMLXVI

Number just after 967 = 968 = 900 + 60 + 8 = (1000 – 100) + (50 + 10) + 8 = CM + LX + VIII

= CMLXVIII

So, CMLXVI CMLXVII CMLXVIII

c. MCCCXLV = M + CCC + XL + V = 1000 + (100 + 100 + 100) + (50 − 10) + 5 = 1000 + 300 + 40 + 5 = 1345

Number just before 1345 = 1344 = 1000 + 300 + 40 + 4

= 1000 + 300 + (50 − 10) + 4

= M + CCC + XL + IV

= MCCCXLIV

Number just after 1345 = 1346 = 1000 + 300 + 40 + 6

= 1000 + 300 + (50 − 10) + 6

= M + CCC + XL + VI

= MCCCXLVI

So, MCCCXLIV CMLXVII MCCCXLVI

d. MXV = M + XV = 1000 + (10 + 5) = 1000 + 15 = 1015

Number just before 1015 = 1014 = 1000 + 10 + 4

= 1000 + 10 + 4

= M + X + IV

= MXIV

Number just after 1015 = 1016 = 1000 + 10 + 6 = 1000 + 10 + 6

= M + X + VI

= MXVI

So, MXIV MXV MXVI

5.a. CDXL – CCCXIV = (CD + XL) – (CCC + X + IV)

= [(500 – 100) + (50 – 10)] – [(100 + 100 + 100) + 10 + (5 – 1)]

= (400 + 40) – (300 + 10 –+4) = 440 – 314 = 126

= 100 + 20 + 6 = C + XX + VI = CXXVI

b. DXVIII + XCII = [D + X + VIII] + [XC + II] = [500 + 10 + (5 + 1 + 1 +

1)] + [(100 – 10) + 2]

= 518 + 92 = 610 = 600 + 10 = (500 + 100) + 10 = DCX

c. MDXL – MCCLXX = (M + D + XL) – (M + CC + LXX)

= [1000 + 500 + (50 – 10)] – [1000 + 200 + (50 + 10 + 10)]

= (1000 + 500 + 40) – (1000 + 200 + 70) = 1540 −1270 = 270

= 200 + 70 = CC + LXX = CCLXX

d. MCX + XC = (M + C + X) + (100 – 10) = 1000 + 100 + 10 + 90 = 1200

= 1000 + 200 = MCC

Word Problem 1. Number of people in one village of a town = CM = 1000 – 100 = 900

Number of people in another village of the town = CCCL = CCC + L = 300 + 50 = 350

Total number of people = 900 + 350 = 1250 = 1000 + 200 + 50 = M + CC + L = MCCL

1B 1.a. 58,76,03,653: 8 is in the crores place, so face value of 8 is 8 and place value of 8 is 8,00,00,000. b. 18,90,37,890: 9 is in the ten lakhs place, so face value of 9 is 9 and place value of 9 is 90,00,000. c.20,56,78,450: 2 is in the ten crores place, so face value of 2 is 2 and place value of 2 is 20,00,00,000.

d. 27,89,73,653: 7 is in the ten thousands place, so face value of 7 is 7 and place value of 7 is 70,000. e. 38,57,64,999: 7 is in the lakhs place, so face value of 7 is 7 and place value of 7 is 7,00,000. f. 81,87,50,658: 1 is in the crores place, so face value of 1 is 1 and place value of 1 is 1,00,00,000.

2.a. 93,27,59,191 = 90,00,00,000 + 3,00,00,000 + 20,00,000 + 7,00,000 + 50,000 + 9000 + 100 + 90 + 1 b. 11,07,64,787 = 10,00,00,000 + 1,00,00,000 + 7,00,000 + 60,000 + 4000 + 700 + 80 + 7 c. 56,42,86,604 = 50,00,00,000 + 6,00,00,000 + 40,00,000 + 2,00,000 + 80,000 + 6000 + 600 + 4 d. 70,21,35,436 = 70,00,00,000 + 20,00,000 + 1,00,000 + 30,000 + 5000 + 400 + 30 + 6

e. 89,13,45,468 = 80,00,00,000 + 9,00,00,000 + 10,00,000 + 3,00,000 + 40,000 + 5000 + 400 + 60 + 8 f. 24,85,67,835 = 20,00,00,000 + 4,00,00,000 + 80,00,000 + 5,00,000 + 60,000 + 7000 + 800 + 30 + 5 3.a. 23,34,26,578 = twenty-three crore, thirty-four lakh, twenty-six thousand, five hundred seventy-eight b. 74,36,54,679 = seventyfour crore thirty-six lakh fifty-four thousand six hundred seventynine c. 97,87,64,224 = ninety-seven crore, eighty-seven lakh, sixty-four thousand, two hundred twenty-four d. 11,98,03,456 = eleven crore ninety-eight lakh, three thousand four hundred fifty-six e. 46,57,68,979 = forty-six crore, fifty-seven lakh, sixtyeight thousand, nine hundred seventy-nine f. 57,52,65,389 = fifty-seven crore, fifty-two lakh, sixty-five thousand, three hundred eighty-nine

4.a. Eighty crore, ninety-nine lakh, forty thousand, two hundred one = 80,99,40,201 b. Sixteen crore, twenty-six lakh, ninety-two thousand one hundred forty-three = 16,26,92,143

c. Sixty-five crore, twenty-two lakh, forty-four thousand nine hundred = 65,22,44,900

5. On interchanging digits 5 and 4 in 76,47,56,229, we get 76,57,46,229. a. No, the face values will remain the same.

b. Place value of 5 in new number = 50,00,000, place value of 4 in new number = 40,000 c. Place value of 5 in old number = 50,000, difference of place values of 5 in old and new number = 50,00,000 –50,000 = 49,50,000

6. Place value of 5 in the crores place = 5,00,00,000 = 5 × 1,00,00,000 = 1,00,00,000 times 5, place value of 5 in the hundreds place = 500 = 5 × 100 = 100 times 5

So, the place value of 5 in the crores place is 99,99,900 times more than the place value of 5 in the hundreds place.

Word Problem 1. 27,25,45,619 = 20,00,00,000 + 7,00,00,000 + 20,00,000 + 5,00,000 + 40,000 + 5000 + 600 + 10 + 9

Number name: twenty-seven crore, twenty-five lakh, forty-five thousand, six hundred nineteen

1C 1.a. 8 is in the ten millions place, so the face value of 8 is 8 and place value of 8 is 80,000,000. b. 9 is in the millions place, so the face value of 9 is 9 and place value of 9 is 9,000,000. c. 2 is in the hundred millions place, so the face value of 2 is 2 and place value of 2 is 200,000,000. d. 7 is in the ten thousands place, so the face value of 7 is 7 and place value of 7 is 70,000. e. 8 is in the hundred thousands place, so the face value of 8 is 8 and place value of 8 is 800,000. f. 0 is in the thousands place, so the face value and place value both is 0.

2.a. 932,542,191 = 900,000,000 + 30,000,000 + 2,000,000 + 500,000 + 40,000 + 2000 + 100 + 90 + 1 b. 110,764,214 = 100,000,000 + 10,000,000 + 700,000 + 60,000 + 4000 + 200 + 10 + 4

c. 724,286,604 = 700,000,000 + 20,000,000 + 4,000,000 + 200,000 + 80,000 + 6000 + 600 + 4 d. 787,135,436 = 700,000,000 + 80,000,000 + 7,000,000 + 100,000 + 30,000 + 5000 + 400 + 30 + 6 e. 891,096,468 = 800,000,000 + 90,000,000 + 1,000,000 + 90,000 +

6000 + 400 + 60 + 8 f. 248,567,465 = 200,000,000 + 40,000,000 + 8,000,000 + 500,000 + 60,000 + 7000 + 400 + 60 + 5

3.a. 233,426,578 = two hundred thirty-three million, four hundred twenty-six thousand, five hundred seventy-eight b. 743,654,679 = seven hundred forty-three million, six hundred fifty-four thousand, six hundred seventy-nine c. 978,764,224 = nine hundred seventyeight million seven hundred sixty-four thousand two hundred twenty-four d. 119,803,456 = one hundred nineteen million, eight hundred three thousand, four hundred fifty-six

e. 345,768,979 = three hundred forty-five million, seven hundred sixty-eight thousand, nine hundred seventy-nine f. 575,123,389 = Five hundred seventy-five million, one hundred twenty-three thousand, three hundred eighty-nine

4.a. Eight hundred twenty-six million, one hundred twenty thousand, sixty-six = 826,120,066 b. One hundred ninety-two million, five hundred four thousand, sixteen = 192,504,016

c. Seven hundred eighteen million, seven hundred ten thousand, one hundred fifty-six = 718,710,156

5. On interchanging the digits 7 and 1 in 764,156,229, we get 164,756,229.

a. Yes, there will be a change in the place values of 7 and 1.

b. Place value of 1 = 100,000,000 and place value of 7 = 700,000

c. Place value of 7 in the original number = 700,000,000, difference = 700,000,000 – 700,000 = 699,300,000

1D 1.a. 100,356,782 < 500,040,367 < 887,210,460 < 931,124,820

b. 360,841,910 < 692,180,350 < 826,020,031 < 927,516,890

c. 500,216,138 < 604,503,821 < 650,241,567 < 945,241,823

2.a. 871,926,345 > 826,374,510 > 670,814,256 > 450,070,921

b. 962,115,108 > 801,210,450 > 678,203,001 > 423,516,789

c. 967,208,891 > 788,216,134 > 578,206,010 > 543,343,867

3.a. Greatest 9-digit number = 98,75,43,210, smallest 9-digit number = 10,23,45,789 b. Greatest 9-digit number = 87,65,43,210, smallest 9-digit number = 10,23,45,678

c. Greatest 9-digit number =98,76,54,321, smallest 9-digit number = 12,34,56,789  4.a. Greatest 9-digit number = 88,76,54,210, smallest 9-digit number = 10,24,56,788 b. Greatest 9-digit number = 99,87,65,431, smallest 9-digit number = 13,45,67,899

c. Greatest 9-digit number = 99,75,43,210, smallest 9-digit number = 10,23,45,799  5.a. Two different digits that will be used here are 9 and 8. So, the greatest 9-digit number = 99,99,98,888, and the smallest 9-digit number = 10,00,00,000 b. Five different digits that will be used here are 9, 8, 7, 6 and 5. So, the greatest 9-digit number = 99,99,98,765, and the smallest 9-digit number = 10,00,00,234

c. Four different digits that will be used here are 9, 8, 7 and 6. So, the greatest 9-digit number = 99,99,99,876, and the smallest 9-digit number = 10,00,00,023

Word Problem 1. The numbers that have at the most 9 digits from 0 to 9 were made using the flashcards. 547386291,764238105, 547689123, 632478591, 421358967

Chapter Checkup 1. a. 427 = 400 + 20 + 7 = CD + XX + VII = CDXXVII

b. 2087 = 2000 + 80 + 7 = MM + LXXX + VII = MMLXXXVII

c. 635 = 600 + 30 + 5 = DC + XXX + V = DCXXXV

d. 2015 = 2000 + 10 + 5 = MM + X + V = MMXV

e. 215 = 200 + 10 + 5 = CC + X + V = CCXV

f. 2750 = 2000 + 700 + 50 = MM + DCC + L = MMDCCL

2.a. CCCXVII = CCC + XVII = (100 + 100 + 100) + (10 + 7) = 300 + 17 = 317 b. MCLXXV = M + C + LXXV = 1000 + 100 + (50 + 10 + 10 + 5) = 1000 + 100 + 75 = 1175 c. MCDIX = M + CD + IX = 1000 + (500 –100) + (10 – 1) = 1000 + 400 + 9 = 1409 d. DCCXCVII = DCC + XC + VII = (500 + 100 + 100) + (100 – 10) + (5 + 1 + 1) = 700 + 90 + 7 = 797

e. CCXVIII = CC + X + VIII = (100 + 100) + 10 + (5 + 1 + 1 + 1) = 200 + 10 + 8 = 218

3.a. 350427681

Indian system: 35,04,27,681 = thirty-five crore four lakh twentyseven thousand six hundred eighty-one, expanded form = 30,00,00,000 + 5,00,00,000 + 4,00,000 + 20,000 + 7000 + 600 + 80 + 1

International system: 350,427,681 = three hundred fifty million, four hundred twenty-seven thousand, six hundred eighty-one, Expanded form = 300,000,000 + 50,000,000 + 400,000 + 20,000 + 7000 + 600 + 80 + 1

b. 420879502

Indian system: 42,08,79,502 = forty-two crore, eight lakh, seventynine thousand, five hundred two, Expanded form = 40,00,00,000 + 2,00,00,000 + 8,00,000 + 70,000 + 9000 + 500 + 2

International system: 420,879,502 = four hundred twenty million, eight hundred seventy-nine thousand, five hundred two, Expanded form = 400,000,000 + 20,000,000 + 800,000 + 70,000 + 9000 + 500 + 2

c. 635658421

Indian system: 63,56,58,421 = sixty-three crore fifty-six lakh fifty-eight thousand four hundred twenty-one, Expanded form = 60,00,00,000 + 3,00,00,000 + 50,00,000 + 6,00,000 + 50,000 + 8000 + 400 + 20 + 1

International system: 635,658,421 = six hundred thirty-five million, six hundred fifty-eight thousand four hundred twenty-one, Expanded form = 600,000,000 + 30,000,000 + 5,000,000 + 600,000 + 50,000 + 8000 + 400 + 20 + 1

d. 901500084

Indian system: 90,15,00,084 = ninety crore, fifteen lakh, eighty-four, Expanded form = 90,00,00,000 + 10,00,000 + 5,00,000 + 80 + 4

International system: 901,500,084 = nine hundred one million five hundred thousand eighty-four, Expanded form = 900,000,000 + 1,000,000 + 500,000 + 80 + 4

4.a. Four hundred sixty million, seven hundred twenty-two thousand, two hundred thirty-nine = 460,722,239 b. Eighty crore, nine lakh, fifty thousand, sixty-two = 80,09,50,062  c. One hundred million one hundred thousand thirty-nine = 100,100,039

d. Sixty-three crore, twelve lakhs, fifty-eight thousand, one hundred forty-three = 63,12,58,143

5.a. 656,502,567 > 648,900,650, because 5 in the ten millions place of 656,502,567 is greater than 4 in the same place of 648,900,650. b. 314,572,879 > 314,527,879, because 7 in the ten thousands place of 314,572,879 is greater than 2 in the same place of 314,527,879. c. 900,760,518 < 900,768,757, because 0 in the thousands place of 900,760,518 is less than 8 in the same place of 900,768,757. d. 113,005,885 = 113,005,885, because all the digits in both the numbers are the same.

6.a. 19,08,04,365 < 67,23,56,475 < 68,91,63,896 < 76,90,87,687

b. 324,335,678 < 435,406,576 < 676,162,895 < 676,817,980

c. 65,45,12,845 < 65,78,15,325 < 87,12,63,256 < 97,12,36,125

7.a. 25,36,45,787 > 14,56,45,768 > 14,43,56,787 > 12,40,85,167

b. 949,076,837 > 909,087,897 > 810,868,428 > 810,638,964

c. 745,125,364 > 745,124,698 > 654,785,125 > 456,125,789

8. Greatest 9-digit number: 98,65,43,210, smallest 9-digit number: 10,23,45,689

Greatest 9-digit number (except 0): 99,86,54,321, smallest 9-digit number (except 0): 11,23,45,689

Ascending order: 10,23,45,689 < 11,23,45,689 < 98,65,43,210 < 99,86,54,321

Descending order: 99,86,54,321 > 98,65,43,210 > 11,23,45,689 > 10,23,45,689

9. 444,777,888

International number system: four hundred forty-four million, seven hundred seventy-seven thousand, eight hundred eightyeight

Indian number system: 44,47,77,888 = forty-four crore fortyseven lakh seventy-seven thousand eight hundred eighty-eight

10.a. CXCVII = C + XC + VII = 100 + (100 – 10) + 7 = 100 + 90 + 7 = 197 CDXV = CD + XV = (500 – 100) + (10 + 5) = 400 + 15 = 415

197 + 415 = 612 = 600 + 10 + 2 = DC + X + II = DCXII

b. MXCVII – DCXVI = [M + XC + VII] – [DC + X + VI]

= [1000 + (100 – 10) + (5 + 1 + 1)] – [(500 + 100) + 10 + (5 + 1)]

= (1000 + 90 + 7) – (600 + 10 + 6)

= 1097 – 616 = 481 = 400 + 80 + 1 = CD + LXXX + I = CDLXXXI

Word Problem 1. 750635389

Indian number system: 75,06,35,389 = seventy-five crore six lakh thirty-five thousand three hundred eighty-nine

International number system: 750,635,389 = seven hundred fifty million, six hundred thirty-five thousand, three hundred eighty-nine

Chapter 2

Let's Warm-up

1. 65,700 + 4532 = 70,232  2. 51,000 + 4000 = 55,000

3. 13,000 – 5678 = 7322    4. 23,500 – 12,000 = 11,500

Do It Yourself

2A 1.a. Digit at ones place = 8 (> 5) Hence, 478 will be rounded off to 480. b. Digit at ones place = 5 (= 5) Hence, 7675 will be rounded off to 7680. c. Digit at ones place = 2 (< 5) Hence, 6352 will be rounded off to 6350. d. Digit at ones place = 0 (< 5) Hence, 5650 will be rounded off to 5650. e. Digit at ones place = 7 (>5) Hence, 94,267 will be rounded off to 94,270.

f. Digit at ones place = 6 (>5) Hence, 40,536 will be rounded off to 40,540. g. Digit at ones place = 7 (>5) Hence, 76,567 will be rounded off to 76,570. h. Digit at ones place = 5 (= 5) Hence, 2,54,765 will be rounded off to 2,54,770. i. Digit at ones place = 6 (>5) Hence, 8,76,746 will be rounded off to 8,76,750. j. Digit at ones place = 6 (>5) Hence, 20,45,546 will be rounded off to 20,45,550.

2.a. Digit at tens place = 4 (< 5) Hence, 545 will be rounded off to 500. b. Digit at tens place = 9 (> 5) Hence, 3098 will be rounded off to 3100. c. Digit at tens place = 7 (> 5) Hence, 4576 will be rounded off to 4600. d. Digit at tens place = 2 (< 5) Hence, 4723 will be rounded off to 4700. e. Digit at tens place = 5 (=5) Hence, 86,850 will be rounded off to 86,900. f. Digit at tens place = 0 (>5) Hence, 43,607 will be rounded off to 43,600. g. Digit at tens place = 5 (=5) Hence, 8,09,757 will be rounded off to 8,09,800.  h. Digit at tens place = 0 (< 5) Hence, 7,97,005 will be rounded off to 7,97,000. i. Digit at tens place = 9 (>5) Hence, 23,98,790 will be rounded off to 23,98,800. j. Digit at tens place = 2 (< 5) Hence, 49,46,826 will be rounded off to 49,46,800. k. Digit at tens place = 2 (<5) Hence, 3,80,08,023 will be rounded off to 3,80,08,000.  l. Digit at tens place = 4 (<5) Hence, 89,74,53,542 will be rounded off to 89,74,53,500.

3.a. Digit at hundreds place = 7 (> 5) Hence, 7747 will be rounded off to 8000. b. Digit at hundreds place = 6 (> 5) Hence, 4677 will be rounded off to 5000. c. Digit at hundreds place = 3 (< 5) Hence, 13,325 will be rounded off to 13,000. d. Digit at hundreds

place = 7 (> 5) Hence, 68,723 will be rounded off to 69,000.

e. Digit at hundreds place = 8 (> 5) Hence, 48,890 will be rounded off to 49,000. f. Digit at hundreds place = 0 (< 5) Hence, 3,42,098 will be rounded off to 3,42,000. g. Digit at hundreds place = 7 (> 5) Hence, 4,89,779 will be rounded off to 4,90,000. h. Digit at hundreds place = 5 (= 5) Hence, 65,43,567 will be rounded off to 65,44,000. i. Digit at hundreds place = 4 (< 5) Hence, 31,41,457 will be rounded off to 31,41,000. j. Digit at hundreds place = 7 (> 5) Hence, 4,65,45,758 will be rounded off to 4,65,46,000.

k. Digit at hundreds place = 0 (< 5) Hence, 1,03,09,095 will be rounded off to 1,03,09,000. l. Digit at hundreds place = 6 (> 5) Hence, 30,47,57,698 will be rounded off to 30,47,58,000.

4. All the numbers between 4,250 and 4,349 can be rounded off to the nearest hundred as 4300. 5. Greatest number that can be rounded off to the nearest thousand to give 7000 = 7499

Smallest number than can be rounded off to the nearest thousand to give 7000 = 6500 6. 7915, 7951, 7591 and 7519 can be rounded off to nearest thousand as 8000.

7519 can be rounded off to the nearest hundred as 7500.

Word Problem 1. Smallest number that can be rounded off to the nearest ten to give 5,67,790 = 5,67,785

Greatest number that can be rounded off to the nearest ten to give 5,67,790 = 5,67,794

2B 1.a. 56 rounded off to the nearest tens = 60 92 rounded off to the nearest tens = 90

Estimated sum: 60 + 90 = 150

b. 4356 rounded off to the nearest tens = 4360

9120 rounded off to the nearest tens = 9120

Estimated sum: 4360 + 9120 = 13,480

c. 89,174 rounded off to the nearest tens = 89,170

23,589 rounded off to the nearest tens = 23,590

Estimated sum: 89,170 + 23,590 = 1,12,760

d. 21,54,759 rounded off to the nearest tens = 21,54,760

9,78,422 rounded off to the nearest tens = 9,78,420

Estimated sum:

21,54,760 + 9,78,420 = 31,33,180

e. 89,73,215 rounded off to the nearest tens = 89,73,220

97,056 rounded off to the nearest tens = 97,060

Estimated sum:

89,73,220 + 97,060 = 9070280

f. 21,46,689 rounded off to the nearest tens = 21,46,690

54,202 rounded off to the nearest tens = 54,200

Estimated sum:

21,46,690 + 54,200 = 22,00,890

2.a. 8900 rounded off to the nearest thousands = 9000

5220 rounded off to the nearest thousands = 5000

Estimated difference: 9000 – 5000 = 4000

b. 9356 rounded off to the nearest thousands = 9000

4120 rounded off to the nearest thousands = 4000

Estimated difference: 9000 – 4000 = 5,000

c. 79,174 rounded off to the nearest thousands = 79,000 23,589 rounded off to the nearest thousands = 24,000

Estimated difference: 79,000 – 24,000 = 55,000

d. 21,87,759 rounded off to the nearest thousands = 21,88,000

9,78,524 rounded off to the nearest thousands = 9,79,000

Estimated difference:

21,88,000 – 9,79,000 = 12,09,000

e. 89,56,285 rounded off to the nearest thousands = 89,56,000 89,656 rounded off to the nearest thousands = 90,000

Estimated difference:

89,56,000 – 90,000 = 88,66,000

f. 34,46,667 rounded off to the nearest thousands = 34,47,000

32,095 rounded off to the nearest thousands = 32,000

111718

2188000 –979000

1209000

3447000 –32000 3415000

Estimated difference: 34,47,000 – 32,000 = 34,15,000

3.a. 78 rounded off to the nearest tens = 80

23 rounded off to the nearest tens = 20

Estimated product: 80 × 20 = 1600

b. 6786 rounded off to the nearest thousands = 7000

9123 rounded off to the nearest hundreds = 9000

Estimated product: 7000 × 9000 = 6,30,00,000

c. 8174 rounded off to the nearest thousand = 8000

3,589 rounded off to the nearest hundreds = 4000

Estimated product: 8000 × 4000 = 3,20,00,000

d. 759 rounded off to the nearest hundreds = 800

422 rounded off to the nearest hundreds = 400

Estimated product: 800 × 400 = 3,20,000

e. 215 rounded off to the nearest hundreds = 200

956 rounded off to the nearest hundreds = 1000

Estimated product: 200 × 1000 = 2,00,000

f. 406 rounded off to the nearest hundreds = 400

202 rounded off to the nearest hundreds = 202

Estimated product: 400 × 200 = 80,000

4. Answer may vary, sample answer.

a. 12 rounded off to the nearest tens = 10

9 rounded off to nearest tens = 10

Estimated product: 10 × 10 = 100

Original product: 12 × 9 = 108

Original product is greater than the estimated product.

b. 4784 rounded off to the nearest thousands = 5000

120 rounded off to the nearest hundreds = 100

Estimated product:

5000 × 100 = 5,00,000

Original product:

4784 × 120 = 5,74,080

Original product is greater than the estimated product.

c. 5437 rounded off to the nearest thousands = 5000

235 rounded off to the nearest hundreds = 200

Estimated product: 5000 × 200 = 10,00,000

Original product: 5437 × 235 = 12,77,695

Original product is greater than the estimated product.

d. 2154 rounded off to the nearest thousands = 2000

78 rounded off to the nearest tens = 80

Estimated product: 2000 × 80 = 1,60,000

Original product:

2154 × 78 = 1,68,012

Original product is greater than the estimated product.

e. 8973 rounded off to the nearest thousands = 9000

705 rounded off to the nearest hundreds = 700

Estimated product:

9000 × 700 = 63,00,000

Original product:

8973 × 705 = 63,25,965

Original product is greater than the estimated product.

f. 7146 rounded off to the nearest thousands = 7000

54 rounded off to the nearest tens = 50

Estimated product: 7000 × 50 = 3,50,000

Original product: 7146 × 54 = 3,85,884

Original product is greater than the estimated product.

5.a. 67 rounded off to the nearest tens = 70

13 rounded off to the nearest tens = 10

70 ÷ 10 = 70 10 = 7

Hence, the estimated quotient is 7.

b. 4356 rounded off to the nearest thousands = 4000

1025 rounded off to the nearest thousands = 1000

4000 ÷ 1000 = 4000 1000 = 4

Hence, the estimated quotient is 4.

c. 9174 rounded off to the nearest thousands = 9000

2008 rounded off to the nearest thousands = 2000

Hence, the estimated quotient is 4.

d. 1256 rounded off to the nearest thousands = 1000

1032 rounded off to the nearest thousands = 1000

1000 ÷ 1000 = 1000 1000 = 1

Hence, the estimated quotient is 1.

e. 715 rounded off to nearest hundreds = 700

115 rounded off to nearest hundreds = 100

700 ÷ 100 = 700 100 = 7

Hence, the estimated quotient is 7.

f. 668 rounded off to the nearest hundreds = 700

122 rounded off to the nearest hundreds = 100

700 ÷ 100 = 700 100 = 7

Hence, the estimated quotient is 7.

6.a. 87 rounded off to the nearest tens = 90

9 rounded off to the nearest tens = 10

90 ÷ 10 = 90 10 = 9

Hence, the estimated quotient is 9.

Both quotients are the same.

9004000

The estimated quotient and the original quotient are the same.

c. 8914 rounded off to the nearest thousands = 9000

358 rounded off to the nearest hundreds = 400

Hence, the estimated quotient is 22.

– 800

– 800

9

987 6 – 81

The estimated quotient and the original quotient are the same.

b. 4356 rounded off to the nearest thousands = 4000

920 rounded off to the nearest hundreds = 900

Hence, the estimated quotient is 4.

– 1432

1754 322

The estimated quotient is smaller than the original quotient.

d. 1547 rounded off to the nearest thousands = 2000

97 rounded off to the nearest tens = 100

2000 ÷ 100 = 2000 100 = 20

Hence, the estimated quotient is 20.

e. 715 rounded off to the nearest hundreds = 700

56 rounded off to the nearest tens = 60

60700 100 40 – 60 – 60 11 56715 155 43 – 112 – 56 12

The estimated quotient is smaller than the original quotient.

f. 468 rounded off to the nearest hundreds = 500

42 rounded off to the nearest tens = 40

40500 100 20 – 80 – 40 12 42468 48 6 – 42 – 42 11

The estimated quotient is greater than the original quotient.

7. New number = 415276

415276 rounded off to the nearest tens = 415280

465271 rounded off to the nearest hundreds = 465300

Required difference = 465300 – 415280 = 50020

Word Problems 1. Cost of suit = ₹3659

92

210

465300 –415280 =050020

Cost of suit rounded off to the nearest thousand = ₹4000

Cost of Saree = ₹6342

Cost of suit rounded off to the nearest thousand = ₹6000

Total money paid by Rita = ₹12,160

Total money paid rounded off to nearest thousand = ₹12,000

Total cost of suit and saree = ₹ 6000 + ₹ 4000 = ₹10,000

Cost of a bedsheet

89 Solutions

Thus, Reeti paid about ₹2000 for the bedsheet.

2. Cost of calculators = ₹560

₹ 560 rounded off to the nearest hundred = ₹600

Number of calculators bought = 45

45 rounded off to the nearest tens = 50

Estimated cost of calculators = ₹600 × 50 = ₹30,000

3. Number of apple trees in the orchard = 7869

7869 rounded off to the nearest thousand = 8000

Number of rows in the orchard = 345

345 rounded off to the nearest hundred = 300

Number of rows = 8000 ÷ 300

There are approximately 26 rows in the orchard.

2C 1.a.

11111

437548 2345

+765675 =1205568

c. 313

879436 –324255 =555181

e. 658769 –23143 =635626

f. 1111

98076576 + 32435 =98109011

2. a.

3008000

–

–

b. 111112

7698608

2313542 2313

+ 498

=10014961

d. 5647658 +21432 =5626226

111 635626 +34546 =670172

10110

701089011

98109011 –765878 =97343133

2 2 27502

+7 2 81 2 280

=750 3 9 7 8 2

3. 1412131514 84235417 9534657 –4698799 =4835858

Hence, 48,35,858, when added to 46,98,799, gives 95,34,657.

4. 1114 51414515810

625465790 –246547583 =378918207

5. 9 61010

5987700 –3657699 =2330001

Hence, 37,89,18,207, when subtracted from 62,54,65,790, gives 24,65,47,583.

59,87,700 is greater than 36,57,699 by 23,30,001.

6. 91412 0107104216

10980536

–5876989 =5103547

Word Problems 1. Number of bags of wheat = 6470

Number of bags of pulses = 23,890

Number of bags of rice = 60,000

Total number of bags of food grains = 6470 + 23,890 + 60,000 = 90,360

2. Number of books in English = 6758

Number of books in French = 9875

Number of books in Sanskrit = 4215

Number of books in other languages = 3659

Total number of books in the library = 3659 + 4215 +9875 + 6758 = 24,507

3. Number of spectators on Monday = 76,986

Number of spectators on Tuesday = 23,465 Clearly, more spectators showed up on Monday.

Difference in the number of spectators

111

60000

23890

+6470

90360

222

3659

4215

9875

+6758 =24507

76986

–23465 =53521 on both days = 76,986 – 23,465 = 53,521

4. Total capacity of the car = 87 l = 87,000 ml (1 l = 1000 ml)

Amount of petrol used = 800 ml

Amount of petrol left in the car = 87,000 ml – 800 ml = 86,200 ml

610 87000 –800 =86200

5. Distance between Mercury and Earth = 9,63,77,000 km

Distance between Earth and Jupiter = 6,56,96 000 km

Distance between Mercury and Jupiter

= Distance between Mercury and Earth + Distance between Earth and Jupiter

1111

96377000

65696000 =162073000 = 9,63,77,000 km + 6,56,96 000 km = 16,20,73,000 km

6. Income of company this year = ₹ 27,83,67,095

Increase in the income of company = ₹ 56,90,482

Income in the previous year

12

7216610

278367095

–5660482 =272676613

= ₹ 27,83,67,095 – ₹ 56,90,482 = ₹ 27,26,76,613

7. Total budget allocated by the government = ₹ 35,87,925 Money spent on materials and labour = ₹12,46,378

Payment received from the client = ₹7,23,456

Unexpected expenses = ₹89,723

Total expenses = ₹12,46,378 + ₹89,723 = ₹ 13,36,101

11111

1246378

+89723

=1336101

Total money available

= Total budget allocated by the government + Payment received from the client

Hence, the other number is 51,03,547.

11111

3587925 +723456 =4311381 = ₹ 35,87,925 + ₹7,23,456 = ₹ 43,11,381

Current budget balance = ₹ 43,11,381 – ₹13,36,101 = ₹ 29,75,280

Yes, the current budget balance is

1210

32011

4311381

–1336101

=2975280 sufficient to purchase the materials.

Required

is 19,58,996.

The number multiplied to 7 is 84,34,535.

5. Quotient = 46       6. Quotient = 9,04,35,460 Divisor = 7,28,778       Divisor = 20

Required number       Other number = 3,35,23,788         = 1,80,87,09,200

Word Problems 1. Weight of rice in one bag = 5 kg 750 g = 5000g + 750 g = 5750 g (1 kg = 1000 g)

Number of bags = 25

Total weight of rice = 5750 × 25 g = 1,43,750 g = 143kg 750 g

2. Amount of water in the vessel = 8 l 650 ml = 8000 + 650 ml = 8650 ml (1 l = 1000 ml)

Capacity of one glass = 500 ml

Number of glasses that can be fully filled = Quotient of 8650 ml ÷ 500 ml = 17 glasses

3. Number of emails sent by each person

in the HR department = 2150

Number of people in the HR department = 8

Total number of emails sent = 2150 × 8 = 17,200 emails

5750

4. Scholarship received per year = ₹3408

Number of years over which the scholarship was received = 5 years

Total scholarship received = ₹3408 × 5 = ₹17,040

5. Number of days of calling = 5

Total calls made by the receptionist = 385

Number of calls made by the receptionist each day = 385 ÷ 5 = 77

6. Number of sets of gloves = 78

Total number of gloves = 546

Total number of gloves in each set = 546 ÷ 78 = 7

7. Number of spectators at the event = 15,690 Cars leaving the parking lot per minute = 52

Number of people in each car = 3

Number of people leaving per minute = 52 × 3 = 156

×

Number of friends = 5

Number of people receiving pizza = 5 + 1

2E 1.

= 402 − 118 + 4 × 162 = 402 − 118 + 648

= 1050 – 118 = 932

c. 15 × 35 ÷ 3 7 of 21 + 20

= 15 × 35 × 7 3 × 21 + 20

= 15 × 35 × 7 × 7 + 20 = 25725 + 20 = 25,745

15615690 100

– 0 – 0

b. 168 ÷ 14 × 22 – 210 + 185 = 12 × 22 − 210 + 185 = 264 − 210 + 185 = 449 − 210 = 239

d. 35 ÷ 5 × 9 + 7 × 30 ÷ 3 of 6

= 35 ÷ 5 × 9 + 7 × 30 ÷ 3 × 6

= 7 × 9 + 7 × 10 × 6

= 63 + 420 = 483

2.a. 56 × 78 = 56 × (70 + 8) = 56 × 70 + 56 × 8 = 3920 + 448 = 4368

b. 89 × 112 = 89 × (100 + 12) = 89 × 100 + 89 × 12 = 8900 + 1068 = 9968

c. 63 × 142 = 63 × (100 + 42) = 63 × 100 + 63 × 42 = 6300 + 63 × (40 + 2) = 6300 + 63 × 40 + 63 × 2 = 6300 + 2520 + 126 = 8946

d. 34 × 7 = (30 + 4) × 7= 30 × 7 + 4 × 7 = 210 + 28 = 238

3. a. 250 + [24 – {4 ×3 + (8 – 5)}]

= 250 + [24 – {4 × 3 + 3}]

= 250 + [24 – {12 + 3}]

= 250 + [24 – 15]

= 250 + 9 = 259

c. [82 – 18 ÷ 3 × 2] + (18 – 6) ÷ 4 = [82 – 6 × 2] + 12 ÷ 4 = [82 – 12] + 3

= 70 + 3 = 73

4.a. 6 + (5 × 3) – 1 = 20

c. (10 + 8) – (4 + 1 + 5) = 8

5. a. 4 + (6 × 5) – 2 = 4 + 30 − 2 = 34 − 2 = 32

b. 8 + (8 × 1) + 3 – 8 = 8 + 8 + 3 – 8

= 19 − 8 = 11

4 + 6 × (5 – 2) = 4 + 6 × 3 = 4 + 18 = 22

8 + 8 × (1 + 3) – 8

= 8 + 8 × 4 – 8 = 8 + 32 – 8

= 40 – 8 = 32

b. (70 − 42) ÷ 14 + 15 × 3 – 7 × 5 = 28÷ 14 + 15 × 3 – 7 × 5 = 2 + 45 – 35 = 47 – 35 = 12

d. 2400 ÷ 10 × {(18 – 6) + (24 − 12)} = 240 × {12 + 12}

= 240 × 24 = 5760

b. 3 + {4 × (8 – 4)} = 19

(4 + 6) × 5 – 2 = 10 × 5 − 2 = 50 − 2 = 48

(8 + 8) × (1 + 3) – 8

= 16 × 4 – 8 = 64 – 8 = 56

a. Number sentence for pizza slices received by each person = (3 × 8) ÷ (1 + 5)

b. Pizza slices received by Damini = (3 × 8) ÷ (5 + 1) = 24 ÷ 6 = 4 slices

2. Total number of carnations = 8 × 5 + 12 × 5 = 40 + 60 = 100

2F 1. Total number of roses bought = 9 × 180 + 4 × 165 = 1620 + 660 = 2280

Bunches of 12 roses that can be made

= 2280 ÷ 12 = 190

2. Total number of cards

= 36 × 15 + 72 = 612

Number of flash cards on each page of folder = 12

51

– 60

12612 12 0 – 12

= Number of cards to be printed = 270 × 4 + 200 × 10 = 1080 + 2000 = 3080 cards.

4. Number of tomatoes produced this year = 1270

Number of tomatoes eaten by birds = 190

Number of tomatoes ruined by bugs = 230

(4 + 6) × (5 – 2) = 10 × 3 = 30

(8 + 8) × 1 + 3 – 8 = 16 × 1 + 3 – 8

= 16 + 3

– 8 = 19 – 8 = 11

=17040 92

0 – 180

180 00 0

Number of tomatoes picked by the farmer = 1270 – (190 + 230) = 1270 – 420 = 850

5. Cost of half a metre of fabric = ₹220

Cost of a full metre of fabric = ₹220 × 2 = ₹440

of the fabric paid for by Shreya 4403520 0 – 3520 8 = ₹3520

÷ 440 =

8

= 8 m – 0.5 m =

event = ₹17,600 + ₹81,000 + ₹30,900

Money left to distribute = ₹1,29,500 – ₹9200

= ₹1,20,300 1 17600 81000

Money given to each children group

= ₹1,20,300 ÷ 3 = ₹40,100

12120300 00 03 00

12

0

–

3

Chapter Checkup

129500

–9200 =120300

1.  Ten 7,65,38065,41057,67,8808,76,540

Hundred2,34,60043, 3001,23,35,0004,25,500

Thousand9,03,00083,00047,70,0002,56,79,000

2.a. 76,426 rounded off to the nearest tens = 76,430

23,143 rounded off to the nearest tens = 23,140

76,430 + 23,140 = 99,570

76,426 + 23,143 = 99,569

76430

+23140

=99570

76426 +23143 =99569

b. 4,36,476 rounded off to the nearest hundreds = 4,36,500

23,134 rounded off to the nearest hundreds = 23,100

4,36,500 – 23,100 = 4,13,400 4,36,476 – 23,134 = 4,13,342

436500

–23100

=413400

c. 9,05,042 rounded off to the nearest hundreds = 9,05,000

1321 rounded off to the nearest hundreds = 300

9,05,000 × 300 = 27,15,00,000

9,05,042 × 321 = 29,05,18,482

436476 –23134 =413342

d. 7653 ÷ 1043 = 7; Remainder = 352

7653 rounded off to the nearest thousands = 8000

1043 rounded off to the nearest tens = 1000

8000 ÷ 1000 = 8; Remainder = 0

3. a. [72 – 36 ÷ 6 of 2] + (30 – 22) + 5 = [72 – 6 × 2] + 8 + 5

= 60 + 8 + 5 = 68 + 5 = 73

c. 888 × [370 ÷ {65 + (18 ÷ 2)}]

= 888 × [370 ÷ {65 + 9}]

= 888 × [370 ÷ 74]

= 888 × 5 = 4440

b. 9 [8{12 – 7 + 5}] of 8 = 9 [8{17 – 7}] × 8 = 9 [8 × 10] × 8

= 9 × 80 × 8 = 5760

d. 60 + 13 – {(5 × 1/2 of 10) –

75 ÷ (17 – 2)}

= 73 – {(5 × 1/2 × 10) –

75 ÷ 15}

= 73 – {(5 × 5) – 5}

= 73 – {25 – 5}

= 73 – 20 = 53

4. a. Error: Square brackets need to be removed.

50 – 30 + {40 – (20 – 10)}

= 20 + {40 – 10}

= 20 + 30 = 50

b. [81 – 36 ÷ 6 of 2] + (30 – 22) ÷ 4

= [81 – 6 × 2] + (30 – 22) ÷ 4

= [81 – 12] + 8 ÷ 4 = 69 + 2 = 71

5. a. 6 × 92 = 6 × (90 + 2) = 6 × 90 + 6 × 2 = 540 + 12 = 552

b. 8 × 112 = 8 × (100 + 12) = 8 × 100 + 8 × 12 = 800 + 96 = 896

c. 9 × 162 = 9 × (100 + 60 + 2) = 9 × 100 + 9 × 60 + 9 × 2 = 900 + 540 + 18 = 1458

d. 3 × 87 = 3 × (80 + 7) = 3 × 80 + 3 × 7 = 240 + 21 = 261

6. a. 67,589 – 23,543 + 13,678 = 57,724

b.

×

Hence, option d is correct.

8. Answers may vary. Sample answers.

(30 + 3) × 3 – 6 ÷ 3 + 3 = 100

(20 + 5) × 3 – 25 ÷ 5 + 5 × 6 = 100

Word Problems 1. Number of books sold each month = 76,578

Number of books sold each month rounded off to the nearest thousand = 77,000

Number of months in a year = 12

Estimated number of books sold each year = 76,578 × 12 = 9,24,000

2. Toys manufactured on Monday = 7,65,875

7,65,875 rounded off to the nearest thousand = 7,66,000

Toys manufactured on Tuesday = 4,36,586

4,36,586 rounded off to the nearest thousand = 4,37,000

Estimated number of more toys manufactured on Monday = 7,66,000 – 4,37,000 = 3,29,000

3. Number of academic books = 12,536

Number of story books = 15,000

Number of children’s magazines = 12,549

a. Total number of books = 12,536 + 15,000 + 12,549 = 40,085

b. Number of shelves for academic books = 4

Academic books put in each shelf = 12,536 ÷ 4 = 3134

c. Number of children magazines loaned = 1018

Number of children magazines left = 12,549 – 1018 = 11,531

4. Total number of ships = 35

Capacity of each ship = 27,850 tons

Number of ships in fleet A = 9 ships

Number of ships in fleet B = 12 ships Number

a.

b.

c. Total cargo capacity of all the fleets = 35 × 27,850 tons = 9,74,750 tons

Chapter 3

Let's Warm-up

1. By 3

2. By 4

3. By 5

4. By 6

Do It Yourself

3A 1.a. Predecessor = 66,778 – 1 = 66,777 Successor = 66,778 + 1 = 66,779 b. Predecessor = 54,557 – 1 = 54,556 Successor = 54,557 + 1 = 54,558 c. Predecessor = 86,457 – 1 = 86,456 Successor = 86,457 + 1 = 86,458 d. Predecessor = 89,665 – 1 = 89,664 Successor = 89,665 + 1 = 89,666 e. Predecessor = 97,564 – 1 = 97,563 Successor = 97,564 + 1 = 97,565

2. a.

3. a. On the given number line: 0 + 10 = 10; 10 + 10 = 20; 20 + 10 = 30 and so on, So, 40 + 10 = 50; 60 + 10 = 70; 80 + 10 = 90

0

b. On the given number line: 0 + 25 = 25; 25 + 25 = 50; 50 + 25 = 75 and so on, So, 75 + 25 = 100; 125 + 50 = 150; 175 + 25 = 200; 200 + 25 = 225; 225 +

=

6.

13,214 + 1 = 13,215      13,215

7. The smallest whole number is 0. Whole numbers between 67 and 75 = 68, 69, 70, 71, 72, 73, 74 = 7

Hence, there are 7 whole numbers between 67 and 75.

8. Quotient = 1 remainder = 1

So, quotient = remainder

The whole number is 6.

Word Problems

So, 8 cookies were not burnt.

3B

+ 7 = 11 11 is a whole number.

c. 25 + 15 = 40 40 is a whole number.

e. 51 + 39 = 90 90 is a whole number.

b. 9 + 13 = 22 22 is a whole number.

d. 30 + 40 = 70 70 is a whole number.

The property being used is the closure property, which states that the sum of two whole numbers is also a whole number.

2. The commutative property states that two whole numbers can be added in any order.

a. 7 + 8 = 15 = 8 + 7 b. 9 + 10 = 19 = 10 + 9 c. 13 + 7 = 20 = 7 + 13

d. 21 + 18 = 39 = 18 + 21 e. 49 + 1 = 50 = 1 + 49

3.a. 736 + 280 + 356

736 + (280 + 356) = 736 + 636 = 1372

b. 837 + 208 + 363

(837 + 363) + 208 = 1200 + 208 = 1408

c. 453 + 647 + 1538

(453 + 647) + 1538 = 1100 + 1538 = 2638

d. 1 + 2 + 3 + 4 + 496 + 497 + 498 + 499

(1 + 2 + 3 + 4) + (496 + 497) + (498 + 499) = 10 + 993 + 997 = 10 + 1990 = 2000

e. 647 + 142 + 858 + 253

(647 + 253) + (142 + 858) = 900 + 1000 = 1900

4.a. 7845 + 999 = 7845 + 1000 – 1 = 8845 – 1 = 8844

b. 9878 + 999 = 9878 + 1000 – 1 = 10,878 – 1 = 10,877

c. 45,456 + 9999 = 45,456 + 10,000 – 1 = 55,456 – 1 = 55,455

d. 98,789 + 9999 = 98,789 + 10,000 – 1 = 1,08,789 – 1 = 1,08,788

e. 72,145 + 9999 = 72,145 + 10,000 – 1 = 82,145 − 1 = 82,144

5.a. 4570 + 0 = 0. FALSE

b. 8953 + 782 = 782 + 8953. TRUE

c. If x is a whole number so that x + x = x, then x = 0. TRUE

d. Whole numbers are commutative under addition. TRUE

6. The greatest 6-digit number is 7,65,420

The smallest 6-digit number is 2,04,567

Sum = 7,65,420 + 2,04,567 = 9,69,987

Word Problems 1. Amount Jyoti put in the money box = ₹154

Amount given by her mother = ₹50

Amount already saved = ₹146

Total amount with Jyoti = ₹154 + ₹50 + ₹146 = ₹ (154 + 146) + 50 = ₹300 + 50 = ₹350

As the result of the addition of the whole numbers is a whole number, the given situation satisfies the closure law. Commutative law = 300 + 50 = 350 = 50 + 300

Hence, the commutative law is satisfied.

2. Marks Seema got in:

Maths = 99 English = 69 Science = 91

Total marks Seema got = 99 + 69 + 91 = 99 + (69 + 91) = 99 + 160 = 259

Marks Rita got in:

Maths = 92 English = 33 Science = 84

Total marks Rita got = 92 + 33 + 84 = 92 + (33 + 84) = 92 + 117 = 209

3C 1. a. 452 – 0 = 452

c. Since, 3999 + 1 = 4000 4000 – 1 = 3999

e. Since, 7456 + 2544 = 10,000

10,000 – 2544 = 7456

b. 78,856 – 78,856 = 0

d. 7893 – 0 = 7893

f. 2365 – 1256 = 1109

2.a. 25 – 12 = 13 and 25 – 13 = 12 b. 500 – 350 = 150 and 500 – 150 = 350 c. 1000 – 400 = 600 and 1000 – 600 = 400

d. 800 – 786 = 14 and 800 – 14 = 786 e. 10,000 – 9989 = 11 and 10,000 – 11 = 9989 f. 6024 – 2365 = 3659 and 6024 – 3659 = 2365

3.a. i. 12,345 – 9 = 12,345 – (10 – 1) = 12,345 – 10 + 1 = 12,335 + 1 = 12,336 ii. 12,345 – 99 = 12,345 – (100 – 1) = 12,345 – 100 + 1 = 12,245 + 1 = 12,246 iii. 12,345 – 999 = 12,345 – (1000 – 1) = 12,345 – 1000 + 1 = 11,345 + 1 = 11,346 iv. 12,345 – 9999 = 12,345 – (10,000 – 1) = 12,345 – 10,000 + 1 = 2345 + 1 = 2346

b.i. 23,965 – 9 = 23,965 – (10 – 1) = 23,965 – 10 + 1 = 23,955 + 1 = 23,956 ii. 23,965 – 99 = 23,965 – (100 – 1) = 23,965 – 100 + 1 = 23,865 + 1 = 23,866 iii. 23,965 – 999 = 23,965 – (1000 – 1) = 23,965 – 1000 + 1 = 22,965 + 1 = 22,966 iv. 23,965 – 9999 = 23,965 –(10,000 – 1) = 23,965 – 10,000 + 1 = 13,965 + 1 = 13,966

c.i. 45,684 – 9 = 45,684 – (10 – 1) = 45,684 – 10 + 1 = 45,674 + 1 = 45,675 ii. 45,684 – 99 = 45,684 – (100 – 1) = 45,684 – 100 + 1 = 45,584 + 1 = 45,585 iii. 45,684 – 999 = 45,684 – (1000 – 1) = 45,684

– 1000 + 1 = 44,684 + 1 = 44,685 iv. 45,684 – 9999 = 45,684 –(10,000 – 1) = 45,684 – 10,000 + 1 = 35,684 + 1 = 35,685

d.i. 74,586 – 9 = 74,586– (10 – 1) = 74,586 – 10 + 1 = 74,576 + 1 = 74,577 ii. 74,586 – 99 =74,586– (100 – 1) = 74,586 – 100 + 1 = 74,486 + 1 = 74,487 iii. 74,586 – 999 = 74,586 – (1000 – 1) = 74,586

– 1000 + 1 = 73,586 + 1 = 73,587 iv. 74,586 – 9999 = 74,586 –(10,000 – 1) = 74,586 – 10,000 + 1 = 64,586 + 1 = 64,587

e.i. 84,126 – 9 = 84,126 – (10 – 1) = 84,126 – 10 + 1 = 84,116 + 1 = 84,117 ii. 84,126 – 99 = 84,126 – (100 – 1) = 84,126 – 100 + 1 =

84,026 + 1 = 84,027 iii. 84,126 – 999 = 84,126 – (1000 – 1) = 84,126

– 1000 + 1 = 83,126 + 1 = 83,127 iv. 84,126 – 9999 = 84,126 –(10,000 – 1) = 84,126 – 10,000 + 1 = 74,126 + 1 = 74,127

f. i. 99,999 – 9 = 99,999 – (10 – 1) = 99,999 – 10 + 1 = 99,989 + 1 =

99,990 ii. 99,999 – 99 = 99,999 – (100 – 1) = 99,999 – 100 + 1 =

99,899 + 1 = 99,900 iii. 99,999 – 999 = 99,999 – (1000 – 1) = 99,999

– 1000 + 1 = 98,999 + 1 = 99,000 iv. 99,999 – 9999 = 99,999 –(10,000 – 1) = 99,999 – 10,000 + 1 = 89,9 99 + 1 = 90,000

4. Smallest natural number = 1

Smallest whole number = 0

Difference = 1 – 0 = 1

5. Smallest 4-digit number = 1000

Largest 3-digit number = 999

Difference = 1000 – 999 = 1000 – (1000 – 1) = 1000 – 1000 + 1 = 0 + 1 = 1

Word Problems 1. Amount the person had = ₹10,00,000

Cost of the TV = ₹16,580

Cost of the motorcycle = ₹45,890

Cost of the flat = ₹8,70,000

Total amount spent = ₹ 16,580 + 45,890 + 8,70,000 = 9,32,470

Amount left = ₹10,00,000 – 9,32,470 = ₹67,530

2. Total amount that Chinmay had = ₹6,10,000

Amount given to Jyoti = ₹87,500

Amount given to Javed = ₹1,26,380

Amount give to John = ₹3,50,000

Total amount given by Chinmay = ₹87,500 + 1,26,380 + 3,50,000 = ₹5,63,880

Amount left = ₹6,10,000 – 5,63,880 = ₹46,120

3. Number of pens Vansh wants to sell = 78,956

Number of pens Vansh sold = 9999

Number of pens left = 78,956 – 9999 = 78,956 – (10,000 – 1) =

78,956 – 10,000 + 1 = 68,956 + 1 = 68,957.

So, he has to sell 68,957 more pens.

3D 1. a. 452 × 0 = 0

Since the result of multiplying a whole number by zero is zero then the property is a multiplicative property of zero.

b. 154 × 125 = 125 × 154

Since the two whole numbers can be multiplied in any order, then the property is a commutative property.

2.a. 5 × 356 × 8 = (5 × 8) × 356 = 40 × 356 = 14,240

b. 7 × 2354 × 9 = (7 × 9) × 2354 = 63 × 2345 = 1,48,302

c. 32 × 4512 × 3 = (32 × 3) × 4512 = 96 × 4512 = 4,33,152

d. 45 × 9875 × 12 = (45 × 12) × 9875 = 540 × 9875 = 53,32,500

e. 0 × 1213 × 5 × 542 = (0 × 5) × (1213 × 542) = 0 × 6,57,446 = 0

f. 123 × 51 × 21 × 321 = (123 × 321) × (51 × 21) = (39,483) × (1071) = 4,22,86,293

3.a. 132 × 15 + 132 × 12 = 132 × (15 + 12) = 132 × 27 = 3564

b. 256 × 10 + 256 × 14 = 256 × (10 + 14) = 256 × 24 = 6144

c. 437 × 13 + 437 × 17 = 437 × (13 + 17) = 437 × 30 = 13,110

4.a. 458 × 105 = 458 ×(100 + 5) = 458 = 100 + 458 × 5 = 45,800 + 2290 = 48,090 b. 125 × 1002 = 125 × (1000 + 2) = 125 × 1000 + 125

× 2 = 1,25,000 + 250 = 1,25,250 c. 137 × 1005 = 137 × (1000 + 5) = 137 × 1000 + 137 × 5 = 1,37,000 + 685 = 1,37,685 d. 0 × 99 = 0 × (100 – 1) = 0 ×100 – 0 × 1 = 0 − 0 = 0 e. 5014 × 999 = 5014 × (1000

– 1) = 5014 × 1000 – 5014 × 1 = 50,14,000 – 5014 = 50,08,986

5. The greatest 5-digit number = 99,999

The smallest 6-digit number = 1,00,000

Product = 99,999 × 1,00,000 = 9,99,99,00,000

Word Problems 1. Buildings owned by the businessman:

Large = 20

Number of floors = 18

Number of apartments on each floor = 5

Total number of apartments in 20 large buildings = 20 × 18 × 5 = 18 × (20 × 5) = 18 × 100 = 1800

Small = 17

Number of floors = 8

Number of apartments on each floor = 4

Total number of apartments in 17 small buildings = 17 × 8 × 4 = 17 × (8 × 4) = 17 × 32 = 544

Total number of apartments = 1800 + 544 = 2344

So, the total number of apartments is 2344.

2. Apples sold in one week = 33 kg

Apples sold in the next two weeks = 65 kg

Cost of 1 kg of apples = ₹40

Total money spent on apples = ₹40×(33+65) = ₹40 × 98 = ₹3920

So, the amount of money the vendor spent on apples in the three weeks is ₹3920.

3. Number of packs of green shirts = 18

Number of packs of blue shirts = 17

Number of shirts in 1 pack = 15

Total number of shirts = 15 × (18 + 17) = 15 × 35 = 525

So, AKASH bought 525 shirts in total.

3E 1. a. When a non–zero whole number is divided by one, the quotient is the whole number itself 9875 ÷ 1 = 9875

b. When zero is divided by a non–zero whole number, it gives the quotient zero 0 ÷ 789 = 0 c. 448 ÷ (224 ÷ 28) = 448 ÷ (8) = 56

d. 999 + (45 ÷ 45) = 999 + (1) = 1000 e. (14,500 ÷ 50) – (99 ÷ 3) = (290) – (33) = 257

2.

a. 2468 ÷ 617

(Divisor × Quotient) + Remainder = (617 × 4) + 0 = 2468 = Dividend

4

6172468

– 2468

c. 2765 ÷ 35

(Divisor × Quotient) + Remainder = (35 × 79) + 0 = 2765 = Dividend

352765

– 245

315 – 315

e.

b. 685 ÷ 100

(Divisor × Quotient) + Remainder = (100 × 6) + 85 = 685 = Dividend

100685 85

d. 27,380 ÷ 120

(Divisor × Quotient) + Remainder = (120 × 228) + 20 = 27,384 = Dividend

12027380 980

b.

c.

d.

e. 94,335 ÷ 820

Quotient = 115

= 35

82094335

4. Greatest 4-digit number = 9210 Smallest 3-digit number = 129

1299210

6. Dividing 367 by 3

Remainder = 1

1 must be subtracted from 367.

Word Problems 1. Water pumped in one minute = 850 litres.

Time taken to pump 1 litre of water = 1 850 minutes

5. Smallest 4–digit number = 1000

Let us divide 1000 by 35.

28

– 70

351000 300 – 280

Remainder = 20

Subtracting 20 from 35 = 35 – 20 = 15 So, 15 must be added to 1000.

7. The smallest number that can divide 24, 56, and 18 is the highest common factor of 24, 56, and 18.

Factors of 18: 1, 2, 3, 9

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

Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56

So, 2 is the smallest number that can divide 24, 56, and 18 without leaving a remainder.

8512750

– 425

Since, 12750 ÷ 85 = 150 minutes

1 minute = 1 60 hours

150 minutes = 150 60 =2.5 hours

So, it will take 2.5 hours to pump out 1,27,500 litres of water

2. Cloth required to make a shirt = 3 m

Number of shirts that can be made out of 1350 m of cloth = 1350 3

Since, 1350 ÷ 3 = 450.

So, 450 shirts can be made.

3. Number of students in a class = 25

Number of classes = 50

Total number of students = 50 × 25 = 1250

Total fees collection = ₹11,25,000

31350

– 15 –

Fee paid by each student = ₹11,25,000 ÷ 1250 = 1125000 1250 = 112500 125 = ₹900

So, the fee paid by each student in a month is ₹900.

3F 1.

a. 5 + 2 = 7

7 + 5 = 12

12 + 2 = 14

14 + 5 =19

19 + 2 = 21

21 + 5 = 26

5, 7, 12, 14, 19, 21, 26

b. 2 × 3 = 6

3 × 6 = 18

6 × 18 = 108 18 × 108 = 1944

2, 3, 6, 18, 108, 1944

2. a. 4452 + 99 = 4452 + 100 – 1 = 4552 – 1 = 4551

b. 2356 + 9999 = 2356 + 10000 – 1 = 12356 – 1 = 12,355

c. 7821 – 999 = 7821 – (1000 – 1) = 7821 – 1000 + 1 = 6821 + 1 = 6822  d. 98,658 – 9999 = 98,658 – (10,000 – 1) = 98,658 – 10,000 + 1 = 88,65 + 1 = 88,659

3. 1 × 1 = 1

11 × 11 = 121

111 × 111 = 12321

1111 × 1111 = 1234321

11111 × 11111 = 123454321

4.a. 45 × 9 = 45 × (10 – 1) = 45 × 10 – 45 × 1 = 450 – 45 = 405

b. 45 × 99 = 45 × (100 – 1) = 45 × 100 – 45 × 1 = 4500 – 45 = 4455

c. 45 × 999 = 45 × (1000 – 1) = 45 × 1000 – 45 × 1 = 45,000 – 45 = 44955 d. 45 × 999 = 45 × (10000 – 1) = 45 × 10000 – 45 × 1 =

4,50,000 – 45 = 449955

So, the pattern is: 405, 4455, 44955, 449955, 4499955, 44999955…

5.  1 × 8 + 1 =

6. a. 1125 +

1125 + 999 = 1125 + 1000 – 1 = 2125 – 1 = 2124

1125 + 9999 = 1125 + 10,000 – 1 = 11,125 – 1 = 11,124

b. 7852 + 9 = 7852 + 10 – 1 = 7862 – 1 = 7861

7852 + 99 = 7852 + 100 – 1 = 7952 – 1 = 7951

7852 + 999 = 7852 + 1000 – 1 = 8852 – 1 = 8851

7852 + 9999 = 7852 + 10,000 – 17,852 – 1 = 17,851

c. 9545 + 9 = 9545 + 10 – 1 = 9555 – 1 = 9554

9545 + 99 = 9545 + 100 – 1 = 9645 – 1 = 9644

9545 + 999 = 9545 + 1000 – 1 = 10,545 – 1 = 10,544

9545 + 9999 = 9545 + 10,000 – 1 = 19,545 – 1 = 19,544

d. 8455 + 9 = 8455 + 10 – 1 = 8455 – 1 = 8464

8455 + 99 = 8455 + 100 – 1 = 8555 – 1 = 8554

8455 +999 = 8455 + 100 – 1 = 9455 – 1 = 9454

8455 + 9999 = 8455 + 10,000 – 1 = 18,455 – 1 = 18,454

e. 78,425 + 9 = 78,425 + 10 – 1 = 78,435 – 1 = 78,434

78,425 + 99 = 78,425 + 100 – 1 = 78,525 – 1 = 78,524

78,425 + 999 = 78,425 + 1000 – 1 = 79,425 – 1 = 79,424

78,425 + 9999 = 78,425 + 10,000 – 1 = 88,425 – 1 = 88,424

f. 9,65,452 + 9 = 9,65,452 + 10 – 1 = 9,65,462 – 1 = 9,65,461

9,65,452 + 99 = 9,65,452 + 100 – 1 = 9,65,552 – 1 = 9,65,551

9,65,452 + 999 = 9,65,452 + 1000 – 1 = 9,66,452 – 1 = 9,66,451

9,65,452 + 9999 = 9,65,452 + 10,000 – 1 = 9,75,452 – 1 = 9,75,451

7.a. 245 × 9 = 245 × (10 – 1) = 245 × 10 – 245 × 1 = 2450 – 245 = 2205

245 × 99 = 245 × (100 – 1) = 245 × 100 – 245 × 1 = 24,500 – 245 = 24,255

245 × 999 = 245 × (1000 – 1) = 245 × 100 – 245 × 1 = 2,45,000 – 245 = 2,44,755

245 × 9999 = 245 × (10,000 – 1) = 245 × 10,000 – 245 × 1 =

24,50,000 – 245 = 24,49,755

b. 356 × 9 = 356 × (10 – 1) = 356 × 10 – 356 × 1 = 3560 – 356 = 3204

356 × 99 = 356 × (100 – 1) = 356 × 100 – 356 × 1 = 35,600 – 356 =

35,244

356 × 999 = 356 × (100 – 1) = 356 × 1000 – 356 × 1 = 3,56,000 – 356 = 3,55,644

356 × 9999 = 356 × (10,000 – 1) = 356 × 10,000 – 356 × 1 =

35,60,000 – 356 = 35,59,644

c. 445 × 9 = 445 × (10 – 1) = 445 × 10 – 445 × 1 = 4450 – 445 = 4005

445 × 99 = 445 × (10 – 1) = 445 × 100 – 445 × 1 = 44,500 – 445 =

44,055

445 × 999 = 445 × (10 – 1) = 445 × 100 – 445 × 1 = 4,45,000 – 445 =

4,44,555

445 × 9999 = 445 × (10 – 1) = 445 × 10,000 – 445 × 1 = 44,50,000 –

445 = 44,49,555

d. 654 × 9 = 654 × (10 – 1) = 654 × 10 – 654 × 1 = 6540 – 654 = 5886

654 × 99 = 654 × (100 – 1) = 654 × 100 – 654 × 1 = 65,400 – 654 =

64,746

654 × 999 = 654 × (1000 – 1) = 654 × 1000 – 654 × 1 = 6,54,000 – 654 = 6,53,346

654 × 9999 = 654 × (10,000 – 1) = 654 × 10,000 – 654 × 1 =

65,40,000 – 654 = 65,39,346

e. 751 × 9 =751 × (10 – 1) = 751 × 10 – 751 × 1 = 7510 – 751 = 6759

751 × 99 = 751 × (100 – 1) = 751 × 100 – 751 × 1 = 75,100 – 751 =

74,349

751 × 999 = 751 × (1000 – 1) = 751 × 100 – 751 × 1 = 7,51,000 – 751 = 7,50,249

751 × 9999 = 751 × (10,000 – 1) = 751 × 10,000 – 751 × 1 =

75,10,000 – 751 = 75,09,249

f. 987 × 9 = 987 × (10 – 1) = 987 × 10 – 987 × 1 = 9870 – 987 = 8883

987 × 99 = 987 × (100 – 1) = 987 × 100 – 987 × 1 = 98700 – 987 = 97,713

987 × 999 =987 × (1000 – 1) = 987 × 1000 – 987 × 1 = 987000 – 987 = 9,86,013

987 × 9999 = 987 × (10,000 – 1) = 987 × 10,000 – 987 × 1 =

98,70,000 – 987 = 98,69,013

8. Greatest 7-digit number using 4,6, and 9 if the digits can be repeated = 99,99,964

Word Problems 1. Number of students in a line = 7

Number of pencils with:

1st student = 1       2nd student = 2 × 1 = 2

3rd student = 2 × 2 = 4    4th student = 2 × 4 = 8

5th student = 2 × 8 = 16   6th student = 2 × 16 = 32

7th student = 2 × 32 = 64

So, the 7th student has 64 pencils.

2. Amount Aman gets for:

1 task = ₹99        3 tasks = ₹999

5 tasks = ₹9999       7 tasks = ₹99,999

9 tasks = ₹9,99,999     11 tasks = ₹99,99,999

13 tasks = ₹9,99,99,999

So, Aman gets ₹9,99,99,999 for 13 tasks.

Chapter Checkup 1. a. Predecessor: 1,54,698 – 1 = 1,54,697

Successor: 1,54,698 + 1 = 1,54,699 b. Predecessor: 2,54,895 – 1 = 2,54,894 Successor: 2,54,895 + 1= 2,54,896 c. Predecessor: 5,45,977 – 1 = 5,45,976 Successor: 5,45,977 + 1 = 5,45,978

d. Predecessor: 6,45,712 – 1 = 6,45,711 Successor: 6,45,712 + 1 = 6,45,713 e. Predecessor: 8,91,453 – 1 = 8,91,452 Successor: 8,91,453 + 1 = 8,91,454

= 99,00,000 – 99 = 98,99,901

7. The whole number that should be multiplied with 6,54,875 to get the number itself is 1. i.e., the multiplicative identity for the number 6,54,875 is 1.

Word Problems 1. Number of chairs = 96

Cost of each chair = ₹200

Total cost of the chairs = ₹200 × 96 = ₹19,200

Number of tables = 48

Cost of each table = ₹150

Total cost of the tables = ₹150 × 48 = ₹7200

Total cost = ₹19,200 + ₹7200 = ₹26,400

Amount given in advance = ₹3500

Amount to be given = ₹26,400 – ₹3500 = ₹22,900

So, the amount given is ₹22,900.

2. Budget for the event = ₹82,360

Amount spent on:

Decorations = ₹15,500

Music = ₹10,000

Refreshments = ₹6500

Total money left = ₹15,500 + ₹10,000 + ₹6500 = ₹32,000

Money left after the expenses = ₹82,360 – ₹32,000 = ₹50,360

So, ₹50,360 is left after the expenses.

3. Number of English books = 45

Cost of each English book = ₹100

Total cost of the English books = ₹100 × 45

Number of Hindi books = 45

Cost of each Hindi book = ₹75

Total cost of the books = ₹75 × 45

Total cost = ₹ [100 × 45 + 75 × 45] = ₹ (100 + 75) × 45 = ₹175 × 45 = ₹7875

So, the total amount the librarian has to pay is ₹7875.

4. Charge for dinner = ₹60

Charge for lunch = ₹50

Total money Dev pays for 30 days =₹ (60 + 50) × 30 = ₹110 × 30 = ₹3300

So, the money Dev has to pay for 30 days is ₹3300.

5. Total number of food boxes = 19,700

Number of food boxes distributed = 9999

Remaining number of food boxes

3.a. The smallest even natural number is 0. FALSE

b. Commutative law and associative property hold true only for addition and multiplication. TRUE c. 1 is referred to as multiplicative identity. TRUE d. 1 is referred to as additive identity. FALSE

4.a. 550 × 45 – 550 × 15 = 550 × (45 – 15) = 550 × 30 = 16,500

b. 865 × 12 + 865 × 45 = 865 × (12 + 45) = 865 × 57 = 49,305

c. 420 × 36 – 420 × 23 = 420 × (36 – 23) = 420 × 13 = 5460.

5. Divisor = 65; Quotient = 9; Remainder = 5 (Divisor × Quotient) + Remainder = Dividend (65 × 9) + 5 = 585 + 5 = 590

Dividend = 590

6. Largest 5-digit number = 99,999

Largest 2–digit number = 99

Product = 99,999 × 99 = (1,00,000 – 1) × 99 = 1,00,000 × 99 – 1 × 99

= 19,700 – 9999 = 19,700 – (10,000 – 1) = 19,700 – 10,000 + 1 = 9700 + 1 = 9701

So, the number of remaining food boxes is 9701.

6. Number of cartons = 15

Number of boxes in each carton = 10

Total number of boxes = 10 × 15 = 150

Number of oranges in each box = 12

Total number of oranges = 12 × 150 = 1800

So, Bhanu ordered 1800 oranges.

7. Number of people in the village = 45,900 2 out of 50 people in the village are illiterate.

Number of illiterate people

= 2 50 × 45,900 = 91800 50 = 9180 5

9180 ÷ 5 = 1836

So, 1836 illiterate people live in the village.

1836

59180

– 5 98

18 30

8. Length of the playground = 324 m

Width of the playground = 220 m

Perimeter = 2 × (length + width)

= 2 × (324 + 220)

= 2 × 324 + 2 × 220

= 648 + 440 = 1088 m

Distance Vicky covers going 4 times around it = 4 × 1088 m = 4352 m

So, Vicky will cover 4352 m.

Chapter 4

Let's Warm-up

Answers may vary. Sample answers:

1. 10: 2, 5      2. 15: 3, 5      3. 16: 2, 8

4. 18: 2, 9      5. 20: 4, 5

Do It Yourself

4A 1.

a. 20

1 × 20 = 20

2 × 10 = 20

4 × 5 = 20

5 × 4 = 20 (We stop here as 5 and 4 are already covered.)

So, the factors of 20 are 1, 2, 4, 5, 10, and 20.

c. 36

1 × 36 = 36

2 × 18 = 36

3 × 12 = 36

4 × 9 = 36

6 × 6 = 36

9 × 4 = 36 (We stop here as 9 and 4 are already covered.)

So, the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

e. 94

1 × 94 = 94

2 × 47 = 94

So, the factors of 94 are 1, 2, 47, and 94.

b. 26

1 × 26 = 26

2 × 13 = 26

So, the factors of 26 are 1, 2, 13, and 26.

c. 84 × 1 = 84

84 × 2 = 168

84 × 3 = 252

84 × 4 = 336

84 × 5 = 420

84 × 6 = 504

So, the first six multiples of 84 are 84, 168, 252, 336, 420, and 504.

d. 99

99 × 1 = 99

99 × 2 = 198

99 × 3 = 297

99 × 4 = 396

99 × 5 = 495

99 × 6 = 594

So, the first six multiples of 99 are 99, 198, 297, 396, 495, and 594.

3. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

Clearly, 2 and 12 are the required numbers as their sum is 14 and the product is 24.

4. The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. Clearly, 9 and 8 are the required numbers as their difference is 1 and the product is 72.

5. The factors of 63 are 1, 3, 7, 9, 21, and 63.

Clearly, 9 and 7 are the required numbers as their difference is 2 and the product is 63.

6.a. Since, 9 × 1111 = 9999

Therefore, 9999 is divisible by 9 such that there is no remainder.

b. Since, 9 × 1011 = 9099

Therefore, 9099 is divisible by 9 such that there is no remainder.

c. Since, 9 × 1,10,011 = 9,90,099

2.

g. 72

1 × 72 = 72

2 × 36 = 72

3 × 24 = 72

4 × 18 = 72

6 × 12 = 72

8 × 9 = 72

d. 88

1 × 88 = 88

2 × 44 = 88

4 × 22 = 88

8 × 11 = 88

11 × 8 = 88 (We stop here as 11 and 8 are already covered.)

So, the factors of 88 are 1, 2, 4, 8, 11, 22, 44, and 88.

f. 45

1 × 45 = 45

3 × 15 = 45

5 × 9 = 45

9 × 5 = 45 (We stop here as 9 and 5 are already covered.)

So, the factors of 45 are 1, 3, 5, 9, 15, and 45.

h. 22

1 × 22 = 22

2 × 11 = 22

So, the factors of 22 are 1, 2, 11, and 22.

9 × 8 = 72 (We stop here as 9 and 8 are already covered.)

So, the factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.

a. 37 × 1 = 37

37 × 2 = 74

37 × 3 = 111

37 × 4 = 148

37 × 5 = 185

37 × 6 = 222

So, the first six multiples of 37 are 37, 74, 111, 148, 185, and 222.

b. 62 × 1 = 62

62 × 2 = 124

62 × 3 = 186

62 × 4 = 248

62 × 5 = 310

62 × 6 = 372

So, the first six multiples of 62 are 62, 124, 186, 248, 310, and 372.

Therefore, 9,90,099 is divisible by 9 such that there is no remainder.

d. Since, 9 × 1,00,011 = 9,00,099

Therefore, 9,00,099 is divisible by 9 such that there is no remainder.

Word Problem 1. Factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100

So, 2 is a factor of 100.

Factors of 67 are 1 and 67

So, 5 is not a factor of 67.

Thus, Roy’s claim is wrong, as 67 is not a multiple of 5.

4B 1. a. Factors of 57 = 1, 3, 19, 57

Sum of factors (excluding itself) = 1 + 3 + 19 = 23

Since 57 does not meet the criterion (23 57), it is not considered a perfect number.

b. Factors of 63 = 1, 3, 7, 9, 21, 63

Sum of factors (excluding itself) = 1 + 3 + 7 + 9 + 21 = 41

Since 63 does not meet the criterion (41 63), it is not considered a perfect number.

c. Factors of 94 = 1, 2, 47, 94

Sum of factors (excluding itself) = 1 + 2 + 47 = 50

Since 94 does not meet the criterion (50 94), it is not considered a perfect number.

d. Factors of 83 = 1, 83

Sum of factors (excluding itself) = 1

Since 83 does not meet the criterion (1 83), it is not considered a perfect number.

2. Even numbers end with 0, 2, 4, 6 or 8. Odd numbers end with 1, 3, 5, 7, or 9.

a. 41 – E     b. 57 – O    c. 64 – E

d. 89 – O     e. 76 – E     f. 93 – O

g. 50 – E     h. 67 – O

3.a. The prime numbers between 40 and 50 are 41, 43, and 47.

b. The prime numbers between 60 and 70 are 61 and 67.

c. 11, 13, 17, 19, 23, and 29 are prime numbers between 10 and 30.

d. 2, 3, 5, and 7 are prime numbers between 0 and 10.

4.a. 25 is an even number. False b. 17 is a prime number. True

c. 6 is a composite number. True d. 13 and 17 are twin prime numbers. False e. 20 and 35 are co-prime numbers. False

5. Prime numbers between 60 and 90 are 61, 67, 71, 73, 79, 83, and 89.

From the list of prime numbers, we can say only one twin prime pair exists between 60 and 90, that is (71, 73).

6. The list of all composite numbers below 49 includes 4, 6, 8, 9,

10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, and 48.

7. The prime numbers between 40 and 60 are 41, 43, 47, 53, and 59.

8. a. 36 = 31 + 5 b. 58 = 53 + 5

36 = 29 + 7 c. 76 = 73 + 3

36 = 13 + 23 d. 42 = 31 + 11

36 = 17 + 19

42 = 37 + 5

42 = 29 + 13

9.a. 91 = 83 + 5 + 3 b. 77 = 71 + 3 + 3

c. 83 = 73 + 7 + 3 d. 65 = 41 + 17 + 7

65 = 47 + 13 + 5

Word Problem 1. Total number of students in the class = 36

Roll numbers up to 36 that are prime = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and 31

Roll numbers up to 36 that are composite = 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, and 36

Roll number 1 does not belong to either of the two groups.

Number of students having roll numbers that are prime = 11

Number of students having roll numbers that are composite = 24 4C 1. A number is divisible by 2 if the digit at the ones place of the number is 0, 2, 4, 6, or 8.

So, the numbers 990, 2484, 41,870 and 9,12,048 are divisible by 2.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Let us calculate the sums:

For 990: 9 + 9 + 0 = 18 (divisible by 3)

For 2484: 2 + 4 + 8 + 4 = 18 (divisible by 3)

For 41,873: 4 + 1 + 8 + 7 + 3 = 23 (not divisible by 3)

For 9,12,048: 9 + 1 + 2 + 0 + 4 + 8 = 24 (divisible by 3)

So, the numbers 990, 2484, and 9,12,048 are divisible by 3.

A number is divisible by 4 if its last two digits form a number that is divisible by 4.

For 990: The last two digits are 90, which is not divisible by 4.

For 2484: The last two digits are 84, which is divisible by 4.

Therefore, the number 2484 is divisible by 4.

For 41,873: The last two digits are 73, which is not divisible by 4.

For 9,12,048: The last two digits are 48, which is divisible by 4.

Therefore, the number 9,12,048 is divisible by 4.

So, the numbers 2484 and 9,12,048 are divisible by 4.

A number is divisible by 5 if the digit at the ones place of the number is 0 or 5.

So, the numbers 990 and 41,870 are divisible by 5.

A number is divisible by 6 if the number is divisible by both 2 and 3. We already found that 990, 2484, and 9,12,048 are divisible by both 2 and 3.

So, the numbers 990, 2484, and 9,12,048 are divisible by 6.

A number is divisible by 8 if its last three digits form a number that is divisible by 8.

For 990: The last three digits are 990, which is not divisible by 8.

For 2484: The last three digits are 484, which is not divisible by 8.

For 41873: The last three digits are 873, which is not divisible by 8.

For 9,12,048: The last three digits are 048, which is divisible by 8.

So, only the number 912048 is divisible by 8.

A number is divisible by 9 if the sum of its digits is divisible by 9. Let us calculate the sums:

For 990: 9 + 9 + 0 = 18 (divisible by 9)

For 2484: 2 + 4 + 8 + 4 = 18 (divisible by 9)

For 41,873: 4 + 1 + 8 + 7 + 3 = 23 (not divisible by 9)

For 9,12,048: 9 + 1 + 2 + 0 + 4 + 8 = 24 (not divisible by 9)

So, the numbers 990 and 2484 are divisible by 9.

A number is divisible by 10 if the digit at the ones place of the number is 0.

So, the numbers 990 and 41,870 are divisible by 10.

2.a. 161 is not a prime number. b. 167 is a prime number.

c. 217 is not a prime number. d. 367 is a prime number.

e. 223 is a prime number. f. 397 is a prime number.

g. 241 is a prime number. h. 128 is not a prime number.

3. Answers may vary. Sample answers:

a. 14    b. 48    c. 99    d. 9

4. Given that a number is divisible by both 5 and 9.

We have to find another number by which the given number is always divisible.

We know that factors of 5 are 1 and 5.

Factors of 9 are 1, 3, and 9.

Since 5 and 9 have no common factor other than 1, they are co-primes and we know that if a number is divisible by two co-primes, then it is also divisible by their product.

The product of 5 and 9 = 5 × 9 = 45.

Thus, the number is always divisible by 45.

Therefore, if a number is divisible by 5 and 9, it will be divisible by 45.

5.a. 66 and 42 are divisible by 6.

Their sum = 66 + 42 = 108 is also divisible by 6.

Their difference = 66 – 42 = 24 is also divisible by 6.

b. 650 and 390 are divisible by 13.

Their sum = 650 + 390 = 1,040 is also divisible by 13.

Their difference = 650 – 390 = 260 is also divisible by 13.

c. 72 and 99 are divisible by 9.

Their sum = 72 + 99 = 171 is also divisible by 9.

Their difference = 99 – 72 = 27 is also divisible by 9.

d. 442 and 306 are divisible by 17.

Their sum = 442 + 306 = 748 is also divisible by 17.

Their difference = 442 – 306 = 136 is also divisible by 17.

6.a. 24,016  b. 40,016

Word Problem 1. The number that corresponds to Alex’s birthdate = 1 + 7 + 7 + 2 + 0 + 1 + 0 = 18

The number that corresponds to Ben’s birthdate = 2 + 3 + 1 + 1 +

+ 0 + 1 + 0 = 10

The number that corresponds to Cloe’s birthdate = 2 + 9 + 1 + 2 + 2 + 0 + 1 + 1 = 18

18 and 10 are divisible by 2.

18 is divisible by 3, but 10 is not divisible by 3.

18 is divisible by both 2 and 3. Hence, 18 is divisible by 6, but 10 is not divisible by 6.

e. 1265 = 5 × 11 ×

=

The prime factors of 345 in ascending order are 3, 5, and 23.

3.a. 189 = 3 × 3 × 21 is not the correct prime factorisation of 189, as 21 can be further factorised as 3 × 7.

b. 252 = 2 × 6 × 3 × 7 is not the correct prime factorisation of 252, as 6 can be further factorised as 2 × 3.

The difference between 3 and 7 is 4.

6. 560 = 2 × 2 × 2 × 2 × 5 × 7

The prime factors of 560 in ascending order are 2, 5, and 7.

The product of the two smallest prime factors = 2 × 5 = 10

7. a. The smallest 5-digit number is 10,000.

10,000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

b. The largest 4-digit number is 9,999.

9999 = 3 × 3 × 11 × 101

Word Problem 1. Jane can put 6 roses, 4 lilies, and 3 tulips in each of the 6 bouquets. OR Jane can put 12 roses, 8 lilies, and 6 tulips in each of the 3 bouquets. OR Jane can put 18 roses, 12 lilies, and 9 tulips in each of the 2 bouquets.

4E 1. a. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.

The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.

Common factors of 48 and 84 are 1, 2, 3, 4, 6 and 12.

The largest common factor is 12.

So, the HCF of 48 and 84 is 12.

b. 293 is a prime number.

c. 385 = 5 × 7 × 11 is the correct prime factorisation of 385. 4. a. 210 2 105

b. The factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30.

The factors of 45 are 1, 3, 5, 9, 15, and 45.

Common factors of 30 and 45 are 1, 3, 5, and 15.

The largest common factor is 15.

So, the HCF of 30 and 45 is 15.

c. The factors of 18 are 1, 2, 3, 6, 9, and 18.

The factors of 27 are 1, 3, 9, and 27.

Common factors of 18 and 27 are 1, 3, and 9.

The largest common factor is 9.

So, the HCF of 18 and 27 is 9.

d. The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.

The factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108.

Common factors of 36, 72 and 108 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

The largest common factor is 36.

So, the HCF of 36, 72, and 108 is 36.

5. 252 = 2 × 2 × 3 × 3 × 7

The prime factors of 252 in ascending order are 2, 3, and 7.

The difference between 2 and 3 is 1.

3. a. 48601 –48 12484 –48 0

So, the HCF of 48 and 60 is 12.  So, the HCF of 63 and 105 is 21.

c. 961201 –96 24964 –96 0

The HCF of 96 and 120 is 24.  So, the HCF of 72, 96, and 120 is 24.

d. 651322 –130 26532 –64 122 –2

So, the HCF of 65 and 132 is 1.

4.a. Multiples of 60 are 60, 120, 180, 240, 300, 360, …

Multiples of 20 are 20, 40, 60, 80, 100, 120, …

The common multiples of 60 and 20 are 60, 120, …

The least common multiple of 60 and 20 is 60.

So, the LCM of 60 and 20 is 60.

b. Multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120, …

Multiples of 35 are 35, 70, 105, 140, 175, 210, …

The first common multiple of 15 and 35 is 105.

So, the LCM of 15 and 35 is 105.

c. Multiples of 4 are 4, 8, 12, 16, 20, 24, …

Multiples of 8 are 8, 16, 24, 32, 40, 48, …

Multiples of 12 are 12, 24, 36, 48, 60, 72, …

The common multiples of 4, 8 and 12 are 24, 48, …

The least common multiple of 4, 8, and 12 is 24.

So, the LCM of 4, 8, and 12 is 24.

d.

Multiples

The

of 12 are 12, 24, 36, 48, 60, 72, …

16 are 16, 32, 48, 64, 80, 96, …

common multiple of 12 and 16 is 48. So,

50

311, 11, 9 311, 11, 3

1111, 11, 1 1, 1, 1

Hence, LCM of 11, 22, and 36 = 2 × 2 × 3 × 3 × 11 = 396

Hence, LCM of 96, 128, and 240 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 5 = 1920

c. 29, 12, 36, 54

29, 6, 18, 27

39, 3, 9, 27

33, 1, 3, 9

31, 1, 1, 3 1, 1, 1, 1

Hence, LCM of 9, 12, 36, and 54 = 2 × 2 × 3 × 3 × 3 = 108

b. 296, 128, 240

248, 64, 120

224, 32, 60

212, 16, 30

26, 8, 15

23, 4, 15

23, 2, 15

33, 1, 15

51, 1, 5 1, 1, 1

d. 2102, 170, 136 251, 85, 68

251, 85, 34

351, 85, 17

517, 85, 17

1717, 17, 17 1, 1, 1

Hence, LCM of 102, 170, and 136 = 2 × 2 × 2 × 3 × 5 × 17 = 2040

7. Let us denote the two numbers as A and B.

Given: HCF = 18; LCM = 126

Let us say one of the numbers A = 54

We can use the relationship between HCF, LCM, and the numbers: HCF × LCM = A × B

Substitute the values: 18 × 126 = 54 × B

B = 18 × 126

54 = 42

So, the other number (B) is 42.

8. The HCF of the given numbers is always a factor of their LCM. Since 780 is not divisible by 25, it is not possible to have two numbers whose LCM and HCF are 780 and 25, respectively. Word Problem 1. Number of days taken by factory X to produce smartphones = 14

Number of days taken by factory Y to produce smartphones = 21

We shall now find the LCM of 14 and 21.

214

77 1

14 = 2 × 7

21 = 3 × 7

LCM of 14 and 21 = 2 × 3 × 7 = 42

So, the factories will ship their smartphones together again after 42 days.

4F 1. Number of marbles with John = 36

Number of marbles with Emily = 48

To find the maximum number of children among whom the marbles can be divided equally, we need to find the HCF of 36 and 48.

So, the largest number of cows that can evenly be divided into separate groups is 12.

4. Here, we need to find the number of days after which the gardener will water all the plants together.

212

26

12 = 2 × 2 × 3

18 = 2 × 3 × 3

LCM of 12 and 18 = 2 × 2 × 3 × 3 =

So, the gardener will water the plants together again after 36 days.

5. Number of millet cookies baked by Kim = 60

Number of chocolate chip cookies baked by Kim = 96 Here, we need to find the number of plastic containers needed to pack the cookies.

48

HCF

So, the maximum number of children among whom the marbles can be evenly distributed is 12.

2. Frequency of the bus A = 15 minutes

Frequency of the bus B = 20 minutes

The time it takes for both buses to arrive at the same time is called the least common multiple, or LCM, of the two arrival times.

315

15 = 3 × 5

20 = 2 × 2 × 5

1

LCM of 15 and 20 = 2 × 2 × 3 × 5 = 60 minutes

That means it will take 60 minutes or 1 hour for both buses to arrive at the same time again.

3. Number of cows with a farmer = 24

Number of cows with another farmer = 36

To find the largest number of cows that can evenly be divided into separate groups we need to find the HCF of 24 and 36.

224

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

HCF of 24 and 36 = 2 × 2 × 3 = 12

So, the number of plastic containers needed to pack the cookies is 12.

6. Time taken by machine X to produce toys = 25 minutes

Time taken by machine Y to produce toys = 35 minutes

Time taken by machine Z to produce toys = 40 minutes

To find the time taken by all three machines to complete their work simultaneously, we need to find the LCM of 25, 35, and 40.

All the three machines will complete their work simultaneously after 1400 minutes.

7. To find the largest number of products that can be produced so that each batch is fully utilised, we need to calculate the HCF of 18, 24, and 36.

So, the largest number of products that can be produced is 6.

8. Number of minutes taken by class P to line up for an assembly = 12 minutes

Number of minutes taken by class Q to line up for an assembly = 18 minutes

Number of minutes taken by class R to line up for an assembly = 24 minutes

To find the frequency of all three classes getting ready for the assembly together, we need to calculate the LCM of 12, 18, and 24.

Chapter Checkup

1.

a. 1 × 24 = 24

2 × 12 = 24

3 × 8 = 24

4 × 6 = 24

6 × 4 = 24 (We stop here as 6 and 4 are already covered)

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

c. 1 × 48 = 48

2 × 24 = 48

12 = 2 × 2 × 3

18 = 2 × 3 × 3

24 = 2 × 2 × 2 × 3

LCM of 12, 18, and 24 = 2 × 2 × 2 × 3 × 3 = 72

So, the classes will be ready for the assembly altogether in every 72 minutes.

9. To find the maximum length of the wooden planks, we need to find the HCF of 54, 72 and 90.

3 × 16 = 48

4 × 12 = 48

6 × 8 = 48

8 × 6 = 48 (We stop here as 8 and 6 are already covered)

So, the factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.

e. 1 × 63 = 63

3 × 21 = 63

7 × 9 = 63

9 × 7 = 63 (We stop here as 9 and 7 are already covered)

So, the factors of 63 are 1, 3, 7, 9, 21, and 63.

b. 1 × 30 = 30

2 × 15 = 30

3 × 10 = 30

5 × 6 = 30

6 × 5 = 30 (We stop here as 6 and 5 are already covered)

So, the factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30.

d. 1 × 76 = 76

2 × 38 = 76

4 × 19 = 76

So, the factors of 76 are 1, 2, 4, 19, 38, and 76.

54

72 = 2 × 2 × 2 × 3 × 3

90 = 2 × 3 × 3 × 5

HCF of 54, 72, and 90 = 2 × 3 × 3 = 18

So, the maximum length of the wooden planks can be 18 cm. 10. We need to find the least common multiple of 8, 12, and 16.

8

12 = 2 × 2 × 3

16 = 2 × 2 × 2 × 2

LCM of 8, 12, and 16 = 2 × 2 × 2 × 2 × 3 = 48

So, the buses depart simultaneously after every 48 minutes. The buses leave the station at 8:00 a.m.

The next time all three buses leave the station at the same time is at 8:48 a.m.

Word Problem 1. To find the greatest number of candies the three friends can distribute equally among themselves, we need to find the HCF of 30, 45, and 60.

g. 1 × 90 = 90

2 × 45 = 90

3 × 30 = 90

5 × 18 = 90

6 × 15 = 90

9 × 10 = 90

30 = 2 × 3 × 5

45 = 3 × 3 × 5

60 = 2 × 2 × 3 × 5

HCF of 30, 45, and 60 = 3 × 5 = 15

So, the greatest number of candies that they can equally distribute among themselves is 15.

f. 1 × 56 = 56

2 × 28 = 56

4 × 14 = 56

7 × 8 = 56

8 × 7 = 56 (We stop here as 8 and 7 are already covered)

So, the factors of 56 are 1, 2, 4, 7, 8, 14, 28, and 56.

10 × 9 = 90 (We stop here as 10 and 9 are already covered)

So, the factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.

2.  a. 16 × 1 = 16

16 × 2 = 32

16 × 3 = 48

16 × 4 = 64

16 × 5 = 80

16 × 6 = 96

So, the first six multiples of 16 are 16, 32, 48, 64, 80, and 96.

c. 92 × 1 = 92

92 × 2 = 184

92 × 3 = 276

92 × 4 = 368

92 × 5 = 460

92 × 6 = 552

So, the first six multiples of 92 are 92, 184, 276, 368, 460, and 552.

b. 48 × 1 = 48

48 × 2 = 96

48 × 3 = 144

48 × 4 = 192

48 × 5 = 240

48 × 6 = 288

So, the first six multiples of 48 are 48, 96, 144, 192, 240, and 288.

d. 105 × 1 = 105

105 × 2 = 210

105 × 3 = 315

105 × 4 = 420

105 × 5 = 525

105 × 6 = 630

So, the first six multiples of 105 are 105, 210, 315, 420, 525, and 630.

3. The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Clearly, 4 and 9 are the required numbers as their sum is 13 and the product is 36.

4.a. 54 and 36 are divisible by 9. Their sum = 54 + 36 = 90 is also divisible by 9 (as 9 + 0 = 9 is divisible by 9)

Their difference = 54 – 36 = 18 is also divisible by 9 (as 1 + 8 = 9 is divisible by 9)

b. 546 and 364 are divisible by 13.

Their sum = 546 + 364 = 910 is also divisible by 13 (as 910 ÷ 13 = 70)

Their difference = 546 – 364 = 182 is also divisible by 13 (as 182 ÷ 13 = 14)

c. 81 and 135 are divisible by 27.

Their sum = 81 + 135 = 216 is also divisible by 27 (as 216 ÷ 27 = 8)

Their difference = 135 – 81 = 54 is also divisible by 27 (as 54 ÷ 27 = 2) 5. a.

The prime factors of 675 arranged in the descending order are 5, 3.

of the two smallest prime factors =

42 = 2 × 3 × 7

56 = 2 × 2 × 2 ×

HCF of 42 and 56 = 2 × 7 = 14

60 = 2 × 2 × 3 × 5

90 = 2 × 3 × 3 × 5

HCF of 60 and 90 = 2 × 3 × 5 = 30

HCF

90 = 2 × 3 × 3 × 5

135 = 3 × 3 × 3 × 5

HCF of 90 and 135 = 3 × 3 × 5 = 45

105 = 3 × 5 × 7

147 = 3 × 7 × 7

245 = 5 × 7 × 7

HCF of 105, 147 and 245 = 7

64 = 2 × 2 × 2 × 2 × 2 × 2

96 = 2 × 2 × 2 × 2 × 2 × 3

HCF of 64 and 96 = 2 × 2 × 2 × 2 × 2 = 32

HCF of 112 and 140 is 28.    HCF of 84, 112, and 140 is 28.

HCF of 80 and 160 is 80.

HCF of 88 and 176 is 88.

HCF of 88, 176 and 264 is 88.

c. 218, 24, 54, 72

29, 12, 27, 36

29, 6, 27, 18

39, 3, 27, 9

33, 1, 9, 3

31, 1, 3, 1

1, 1, 1, 1

LCM of 18, 24, 54 and 72

= 2 × 2 × 2 × 3 × 3 × 3 = 216

LCM of 120, 200 and 160

d. 2120, 200, 160

260, 100, 80

230, 50, 40

215, 25, 20

215, 25, 10

315, 25, 5

55, 25, 5

51, 5, 1 1, 1, 1

= 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5 = 2400

e. 266, 78, 132

233, 39, 66

333, 39, 33

1111, 13, 11

131, 13, 1

1, 1, 1

LCM of 66, 78 and 132 = 2 × 2 × 3 × 11 × 13 = 1716

f. 272, 48, 81, 54

236, 24, 81, 27

218, 12, 81, 27

29, 6, 81, 27

39, 3, 81, 27

33, 1, 27, 9

31, 1, 9, 3

31, 1, 3, 1

LCM of 72, 48, 81 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 = 1296

g. 260, 90, 150, 200

230, 45, 75, 100

215, 45, 75, 50

315, 45, 75, 25

35, 15, 25, 25

55, 5, 25, 25

51, 1, 5, 5

1, 1, 1, 1

LCM of 60, 90, 150 and 200

1, 1, 1, 1

h.

296, 144, 168, 224

248, 72, 84, 112

224, 36, 42, 56

212, 18, 21, 28

26, 9, 21, 14

33, 9, 21, 7

31, 3, 7, 7

71, 1, 7, 7 1, 1, 1, 1

= 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

LCM of 96, 144, 168 and 224 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 = 2016

11. To find out a number that is divisible by 8, 12, 16, 20 and 24 we need to find the LCM of these numbers.

28, 12, 16, 20, 24

24, 6, 8, 10, 12

22, 3, 4, 5, 6

21, 3, 2, 5, 3

31, 3, 1, 5, 3

51, 1, 1, 5, 1

1, 1, 1, 1, 1

LCM of 8, 12, 16, 20 and 24 = 2 × 2 × 2 × 2 × 3 × 5 = 240

Also, we know that the greatest 4-digit is 9999.

Now if we divide 9999 by 240 the remainder is 159.

Therefore, the greatest 4-digit number exactly divisible by 240 = 9999 – 159 = 9840

12. First, we find the LCM of 7, 14, 21, 28, and 35.

27, 14, 21, 28, 35

27, 7, 21, 14, 35

37, 7, 21, 7, 35

57, 7, 7, 7, 35

77, 7, 7, 7, 7 1, 1, 1, 1, 1

LCM of 7, 14, 21, 28, and 35 = 2 × 2 × 3 × 5 × 7 = 420

Greatest 5-digit number = 99,999

We will divide 99,999 by 420 and find the remainder.

99,999 ÷ 420

Quotient = 238

Remainder = 39

As per the question, we need the greatest 5-digit number divisible by 420.

Therefore, the required number is 99,999 – 39 = 99,960

13. First, we find the LCM of 9, 12, 18, 21, 27 and 36.

29, 12, 18, 21, 27, 36

29, 6, 9, 21, 27, 18

39, 3, 9, 21, 27, 9

33, 1, 3, 7, 9, 3

31, 1, 1, 7, 3, 1

71, 1, 1, 7, 1, 1

1, 1, 1, 1, 1, 1

LCM of 9, 12, 18, 21, 27 and 36

= 2 × 2 × 3 × 3 × 3 × 7 = 756

To find the required number we add 15 to 756 which is equal to 771.

14. First, we find the LCM of 15, 25, 35 and 45.

315, 25, 35, 45

35, 25, 35, 15

55, 25, 35, 5

51, 5, 7, 1

71, 1, 7, 1

1, 1, 1, 1

LCM of 15, 25, 35 and 45

= 3 × 3 × 5 × 5 × 7 = 1575

Required number = 1575 – 17 = 1558

15. To determine the greatest 3-digit number exactly divisible by 8, 12, 18, 24 and 32, we need to find the LCM of the given numbers.

28, 12, 18, 24, 32

24, 6, 9, 12, 16

22, 3, 9, 6, 8

21, 3, 9, 3, 4

21, 3, 9, 3, 2

31, 3, 9, 3, 1

31, 1, 3, 1, 1

1, 1, 1, 1, 1

LCM of 8, 12, 18, 24, and 32 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

As we know that the greatest 3-digit number is 999, we will now divide 999 by the LCM obtained.

999 ÷ 28 gives 3 as the quotient and 135 as the remainder. Since the remainder is not zero, we subtract 135 from 999.

999 – 135 = 864 which is a multiple of 288 and exactly divisible by the given numbers.

Hence, the greatest 3-digit number exactly divisible by 8, 12, 18, 24, and 32 is 864.

Word Problems 1. Number of history books = 30

Number of science books = 45

Number of literature books = 60

We need to find the maximum number of books that can be placed on each shelf, such that each shelf contains an equal number of each type of book.

For this, we need to find the HCF of 30, 45, and 60.

30 = 2 × 3 × 5

45 = 3 × 3 × 5

60 = 2 × 2 × 3 × 5

HCF of 30, 45 and 60 = 3 × 5 = 15

Hence, the maximum number of books that can be placed on each shelf is 15.

2. To find the minimum amount of time one would have to wait for all three sprinklers to operate simultaneously, we need to find the LCM of 20, 25, and 30.

220, 25, 30

210, 25, 15

35, 25, 15

55, 25, 5

51, 5, 1

1, 1, 1

LCM of 20, 25 and 30 = 2 × 2 × 3 × 5 × 5 = 300 minutes

Hence, a person would have to wait for 300 minutes or 5 hours to watch all sprinklers operate at the same time.

3. Number of apples = 36

Number of oranges = 48

Number of bananas = 60

To find the greatest number of fruits that can be put in each basket, we need to find the HCF of 36, 48, and 60.

36

48 = 2 × 2 × 2 × 2 × 3

60 = 2 × 2 × 3 × 5

HCF of 36, 48 and 60 = 2 × 2 × 3 = 12

Hence, the greatest number of fruits that can be put in each basket is 12.

4. The traffic lights at three different road crossings change every 40 seconds, 60 seconds, and 80 seconds, respectively. We take the LCM of 40, 60, and 80.

240, 60, 80

220, 30, 40

210, 15, 20

25, 15, 10

35, 15, 5

55, 5, 5

1, 1, 1

LCM of 40, 60 and 80 = 2 × 2 × 2 × 2 × 3 × 5 = 240 seconds or 4 minutes

Therefore, after 240 seconds, the traffic lights will change simultaneously.

So, the traffic lights will change again at 4 minutes past 7 or 7:04 a.m.

5. Number of erasers with Sarah = 18

Number of pencils with Sarah = 24

Number of pens with Sarah = 30

To find the largest number of friends among whom Sarah can distribute these supplies, we need to find the HCF of 18, 24, and 30.

18 = 2 × 3 × 3

24 = 2 × 2 × 2 × 3

30 = 2 × 3 × 5

HCF of 18, 24 and 30 = 2 × 3 = 6

So, Sarah can distribute the supplies among 6 of her friends.

Chapter 5

Let's Warm-up

1. I  2. P  3. I  4. P  5. P

Do It Yourself

5A 1.a. X, Y, N, M  b. Rays – 1, Lines – 1  c. X

2. Rays lack thickness and extend in only one direction. Hence, option a and option c are correct.

3. OP represents a line. Hence, option c is correct.

4.a. A line extends in two directions. True b. A plane has a definite size. False c. A line segment is part of a ray. True

d. A ray has more endpoints than a line. False

5.a. 4  b. 6  c. 6  d. 4

Word Problem 1. Two roads crossing each other is an example of an Intersecting line.

5B 1.a. I J 5 cm

b. 7.7 cm

c. 5.2 cm

d. 35 mm = 3.5 cm 35 mm

e. 61 mm = 6.1 cm

61 mm

Students will draw line segment of lengths 5 cm, 7.7 cm, 5.2 cm, 35 mm and 61 mm.

2.a. 2.7 cm  b. 3.9 cm  c. 5.7 cm

3. 150 cm = 1500 mm; 100 dm = 10,000 mm; 10 m = 10,000 mm

1500 mm < 2000 mm < 10,000 mm = 10,000 m

150 cm < 2000 mm < 100 dm = 10 m

Hence, 150 cm is the shortest length. Option b is correct.

4.a. 1 km = 1000 m

Hence, 23 km = 23 × 1000 m = 23,000 m

b. 1 km = 1,00,000 cm

Hence, 23 km = 23 × 1,00,000 m = 23,00,000 cm

c. 1 km = 100 dam

Hence, 23 km = 23 × 100 dam = 2300 dam

d. 1 km = 10 hm

Hence, 23 km = 23 × 10 hm = 230 hm

5.a. Length of LM is 5.6 cm. So, a line segment 1 cm longer than LM will measure 6.6 cm.

S T 6.6 cm

b. Length of FG is 4 cm. So, a line segment 1 cm longer than FG will measure 5 cm.

I J 5 cm

Word Problem 1. 10 cm = 1 dm

Hence, 120 cm = 120 ÷ 10 dm = 12 dm

Hence, 150 cm = 150 ÷ 10 dm = 15 dm

5C 1.a. Interior region: Q, R, S Exterior region: A, B, C

2. ∠QON and ∠QOR have a common vertex and a common arm. Hence, ∠QON and ∠QOR are adjacent angles.

3. ∠BQB is the incorrect representation of the angle. This is incorrect because it suggests that an angle has the same initial and final points, which doesn't define an angle. An angle is formed by two distinct rays with a common endpoint.

4. Answer may vary, sample answer. P, Q and R are the newly marked points.

the pole on the left Pole A which makes a larger angle with the horizontal road.

5E 1.a. Acute angle (0° < 36° < 90°) b. Obtuse angle (90° < 93° < 180°) c. Reflex angle (180° < 192° < 360°) d. Acute angle (0° < 89° < 90°) e. Obtuse angle (90° < 126° < 180°) f. Acute angle (0° < 53° < 90°)

2. The stumps in a cricket match are perpendicular to the ground. Hence, option b is correct.

3. You would be facing the west direction by turning anti-clockwise by three right angles when you were initially facing the south direction.

4. a. One full rotation = 360°

Going from 5 to 8 = 1 4 of full rotation = 1 4 × 360° = 90°

b. One full rotation = 360°

Going from 3 to 7 = 1 3 of full rotation = 1 3 × 360° = 120°

c. One full rotation = 360°

Going from 11 to 5 = 1 2 of full rotation = 1 2 × 360° = 180°

d. One full rotation = 360°

Going from 9 to 6 = 3 4 of full rotation = 3 4 × 360° = 270°

5. a. South       b. West       c. North N E W S

N E W S

Word Problem 1. One full rotation = 360°

N E W S

Since Rohit has to leave 1 hour before the interview, he must leave his home at 2:00 pm. It means that the hour hand must be at 2 and the minute hand must be at 12.

Going from 12 to 2 = 1 6 of full rotation = 1 6 × 360° = 60°

Hence, the minute hand and hour hand will make an angle of 60° with each other.

Chapter Checkup 1.a. 9  b. 8  c. 5  d.7

2. Line segments have a finite length. They do not extend indefinitely. Hence, option d is the correct answer.

Word Problem 1. Sample Answer. Answers may vary.

5D 1. Angles are measured in degrees.

2. A protractor is used to measure angles.

3. ∠P = 34° ; ∠Q= 66° ; ∠R = 67°

4. Blue angleRed angleDifference

a.60° 35° 25°

b.62° 17° 45°

c.48° 29° 19°

d.129° 22° 107°

5.a. ∠2 and ∠3 (Common arm: BD, Common vertex: B)

∠4 and ∠5 (Common arm: BD, Common vertex: D)

b. ∠5 = 26°, ∠4 = 40°, ∠2 = 50°,∠3 = 28°    c. ∠C = 125°

Word Problem 1. The right pole, Pole B seemingly makes a smaller angle with the horizontal road. Therefore, it is more bent than

3.a. F G 4.6 cm

b. B C 77 mm

c. X Y 6.9 cm

d. D E 2.2 cm

4. 1 km = 10,000 dm

Hence, 4 km = 4 × 10,000 dm = 40,000 dm

5. Interior region: D, E, F, G Exterior region: A, B, C

6. Figures may vary. Sample Figure.

a. X B C O

8. a. Acute angle (0° < 26° < 90°)

b. Reflex angle (180° < 256° < 360°)

c. Acute angle (0° < 79° < 90°)

d. Obtuse angle (90° < 129° < 180°)

e. Obtuse angle (90° < 165° < 180°)

f. Reflex angle (180° < 189° < 360°)

10.a. One full rotation = 360°

9. a. acute angle

b. obtuse angle

c. obtuse angle

d. acute angle

e. straight angle

Going from 12 to 3 = 1 4 of full rotation = 1 4 × 360° = 90°

b. One full rotation = 360°

Going from 12 to 6 = 1 2 of full rotation = 1 2 × 360° = 180°

c. One full rotation = 360°

Going from 12 to 12 = one full rotation = 360°

11. a. East    b. West    c. North    d. South

12. At 6:00 pm, hour hand is at 6 and minute hand is at 12 so from 6 to 12 is half rotation. Now, since one full rotation is redefined as 300 degrees, therefore half rotation is equal to 150 degrees. Word Problem 1. You need to turn one right angle anti-clockwise to find yourself facing east from your initial south‐facing position.

Chapter 6

Let's Warm-up

1. This is a triangle. It has 3 sides.

2. This is a pentagon. It has 5 sides.

3. This is an octagon. It has 8 sides.

4. This is a heptagon. It has 7 sides.

5. This is a quadrilateral. It has 4 sides. Do It Yourself

6A 1. Vertices: T, R, S Sides: TR or RT, TS or ST, RS or SR Angles: ∠T or ∠RTS or ∠STR, ∠R or ∠TRS or ∠SRT, ∠S or ∠TSR or ∠RST 2. a. 6 cm, 7 cm, 9 cm – Length of all sides are unequal so the triangle is scalene. b. 15 cm, 15 cm, 15 cm –Length of all sides are equal so the triangle is equilateral. c. 19 cm, 23 cm, 29 cm – Length of all sides are unequal so the triangle is scalene. d. 9 cm, 11 cm, 9 cm – Length of the two sides are equal so the triangle is isosceles. e. 18 cm, 31 cm, 27 cm –Length of all sides are unequal so the triangle is scalene. f. 3 mm, 3 mm, 3 mm – Length of all sides are equal so the triangle is equilateral.

3. Scalene Triangle: DBC, ADC Equilateral Triangle: None Isosceles Triangle: EAD, ADB

4. a. 60°, 40°, 80° – Since all angles measure less than a right angle, the triangle is acute-angled. b. 65°, 85°, 30° – Since all

angles measure less than a right angle, the triangle is acuteangled. c. 130°, 20°, 20° – Since one of the angles measures more than a right angle, the triangle is obtuse-angled.

d. 90°, 40°, 50° – Since one of the angles measures equal to a right angle, the triangle is right-angled. e. 110°, 25°, 45° – Since one of the angles measures more than a right angle, the triangle is obtuse-angled. f. 60°, 60°, 60° – Since all angles measure less than a right angle, the triangle is acute-angled.

5. a. Since all angles measure less than a right angle, and the length of all sides are unequal the triangle is right-angled scalene. b. Since all angles measure less than a right angle, and the length of all sides are equal the triangle is acute-angled equilateral.

c. Since one of the angles measures more than a right angle, and the length of all sides are unequal the triangle is obtuse-angled scalene. d. Since one of the angles measures equal to a right angle, and the length of the two sides are equal the triangle is a right-angled isosceles.

6. a. In PQR, since QR = RP, ∠P = ∠Q. b. In ABC, since AB=BC, ∠A = ∠C. c. In MNO, since MN=MO, ∠N = ∠O. d. In XYZ, since XY = YZ = ZX, ∠X = ∠Y = ∠Z.

Word Problem 1. Measure of Sam’s Garden = 7 feet, 7 feet and 10 feet

Measure of Maria’s Garden = 5 feet, 6 feet and 7 feet

Since, two sides of Sam’s Garden have equal measures, an isosceles triangle is formed.

All three sides of Maria’s Garden are unequal, the triangle formed is scalene.

6B 1. Figures may vary. Sample figure.

Sides: EF or FE, FH or HF, HG or GH, GE or EG Vertices: E, F, G, H

2. a. In PQRS, the adjacent sides are (PS, SR), (RQ, QP), (SR, RQ), (QP, PS) and the opposite sides are (PS, QR), (SR, PQ).

b. In ABCD, the adjacent sides are (AB, BC), (BC, CD), (CD, AD), (AD, AB) and the opposite sides are (AB, CD), (AD, BC).

c. In JKLM, the adjacent sides are (JM, ML), (ML, LK), (LK, KJ), (KJ, JM) and the opposite sides are (JM, KL), (ML, JK).

3. Figures may vary. Sample figures.

4. a.

Diagonals: BD, AC     Diagonals: SU, VT

c.  J K L M Diagonals: MK, JL

5. Figures may vary. Sample figure. E G H F

The vertices of the given quadrilateral EFGH are E, F, G, H and the diagonals are EH and FG.

Word Problem 1. Let the four trees be T1, T2, T3, and T4.

Let the four stones be S1, S2, S3, and S4. The layout of the gardenFigures may vary. Sample figures.

6C 1.a. All sides of equal length and all angles measure 90° – Square b. Opposite sides are parallel and opposite angles are equal – Parallelogram c. Two pairs of adjacent sides are of equal length, and one pair of opposite angles is equal. – Kite

2. The diagonals in a square bisect each other at right angles so the angles formed will be 90°, 90°. Thus, option d is correct.

3. Since, diagonals of a rectangle are equal and bisects each other; line segments equal in length are -

LM = ON    LO = MN

LN = OM    LQ = QN = OQ = QM

4. a. This is a trapezium. b. Parallel sides: JK and ML

Non-parallel sides: JM and KL c. When JM and KL becomes equal, the quadrilateral is called as isosceles trapezium.

5. In a rhombus LONM , the equal sides are LM, MN, NO, OL. OQ = MQ

LQ = NQ

The equal angles are

∠OLM = ∠MNO

∠LON = ∠LMN

∠LQN = ∠LQM = ∠MQN = ∠NQO = ∠90°. Word Problem 1. a. Since all the angles of this quadrilateral measure less than 180°, it is a convex quadrilateral.

b. No, since none of the angles of the quadrilateral are equal and equal to 90.

c. No, since the opposite angles of the quadrilateral are unequal.

6D 1. Vertices: A, B, C, D, E, F, G, H

Sides: AB or BA, BC or CB, CD or DC, DE or ED, EF or FE, FG or GF, GH or HG, HA or AH

2. A square is a regular polygon since it has all its sides equal.

3. a. Heptagon – 7-line segments b. Nonagon – 9-line segments

c. Decagon – 10-line segments d. Pentagon – 5-line segments

e. Hexagon – 6-line segments

4. a. Number of sides = 5     b. Number of sides = 4

c. Number of sides = 3     d. Number of sides = 7

e. Number of sides = 6

5. a. The polygon with side lengths 6 cm, 6 cm, 6 cm, 9 cm, 6 cm is an irregular polygon since all the sides are unequal.

b. The polygon with angle measures as 90°,90°,90°, 90° may or may not be a regular polygon since the measure of the sides may differ.

c. The polygon with side lengths 2 cm, 2 cm, 2 cm, 2 cm and angle measuring 90°, 90°, 90°, 90° is a regular polygon since all the sides and angles are equal.

6. Since all the sides and angles in this polygon are equal, this is a regular polygon.

The logo has 9 diagonals; hence the shape is a hexagon.

Word Problem 1. Sarah planted two sunflower plants on each side of the hexagonal garden.

Total number of sunflower plants = 2 × 6 = 12

Alex planted one sunflower plant on each side of the octagonal garden.

Total number of sunflower plants = 1 × 8 = 8

Sarah planted 4 sunflower plants more than Alex.

Chapter Checkup 1. a. 7 cm, 7 cm, 10 cm – Length of two sides are equal so the triangle is isosceles b. 14 cm, 15 cm, 16 cm – Length of all sides are unequal so the triangle is scalene

c. 29 cm, 29 cm, 29 cm – Length of all sides are equal so the triangle is equilateral d. 10 cm, 12 cm, 10 cm – Length of two sides are equal so the triangle is isosceles e. 20 cm, 31 cm, 26 cm – Length of all sides are unequal so the triangle is scalene

f. 4 mm, 5 mm, 3 mm – Length of all sides are unequal so the triangle is scalene

2. a. 70°, 30°, 80° – Since all angles measure less than a right angle, the triangle is acute-angled. b. 45°, 85°, 50° – Since all angles measure less than a right angle, the triangle is acuteangled. c. 120°, 20°, 40° – Since one of the angles measures more than a right angle, the triangle is obtuse-angled.

d. 90°, 40°, 50° – Since one of the angles measures equal to a right angle, the triangle is right-angled. e. 60°, 60°, 60° – Since all angles measure less than a right angle, the triangle is acuteangled. f. 100°, 50°, 30° – Since one of the angles measures more than a right angle, the triangle is obtuse-angled.

3. Scalene triangles: AEF, BFC, ABF Equilateral triangle: DEC Isosceles triangle: FEC

4. a. Both ΔDEF and ΔPQR: Points 5, 6, 7 b. Both ΔABC and ΔDEF: Point 2 c. Only ΔDEF: Points 1, 4, 5 d. Only ΔPQR: Points 9, 10, 11, 12, 13.

5. a. RT and SU

b. Answers may vary. ∠RST, ∠STU, ∠TUR, ∠URS

c. Yes, UQ = TQ because the diagonals in a square are equal and bisect each other at right angles.

d. Pair of parallel sides: (RU ST), (UT RS)

6. a. 4 – quadrilateral b. 8 – octagon

c. 5 – pentagon d. 7 – heptagon

7.a. Number of sides = 3   b. Number of sides = 6

8.

c. Number of sides = 8   d. Number of sides = 10

Number of diagonals in a hexagon = 9

9. a. Yes, PQ and RS parallel to each other.

b. PS = QR, PQ = SR, SO = OQ, PO = RO

c. No, PR is not equal to SQ, because the vertex angles are not right angles.

10.a. Yes, MN = PM. Pair of adjacent sides of a kite are equal.

b. Yes, ∠MNQ is equal to ∠MPQ because in the isosceles triangle MNP, the angles opposite to equal sides are equal in measure.

c. Yes, ∠ONQ is equal to ∠OPQ because in the isosceles triangle NOP, the angles opposite to equal sides are equal in measure. Word Problem 1. a. Lisa will need 7 gemstones for her heptagon pendant's corners.

b. Mark will need 20 beads (10 × 2) for his decagon bracelet's sides.

c. Lisa should acquire 13 (20 – 7) more gemstones if she wants to use the same number of gemstones as Mark has beads.

Chapter 7

Let's Warm-up

between two points. The major arc and minor arc paths are both longer than the straight-line path.

1. No

7C 1. a. 3 cm        b. 3.8 cm

2. Yes

3. Yes

c. 5.6 cm

d. 4.7 cm

4. Yes

5. No

Do It Yourself

7A 1. a. b. c.d. e.

2.  a. b. c. d. e.

Closed curveOpen curveOpen curveOpen curveClosed curve 3.

4.a.

Simple curves

Non-Simple curves

b.  c.  d.

5.  Boundary of the curve

Interior of the curve

P

Q R

Exterior of the curve

Word Problem 1. The curvy track is not a simple closed curve, as it crosses itself three times. By definition, simple curves do not cross themselves.

7B 1. The red colour represents a minor sector, and the blue colour represents a major sector.

a.   b.   c.   d.

2. a. X

b. X  c. X  d. X

3. Secant is a line. Hence, it doesn’t have a finite length. Hence, option d is correct. 4. The region bounded by two radii and an arc is known as the sector. Hence, option c is correct. 5. Concentric circles do not intersect each other as they have the same centre. Hence, option d is correct.

Word Problem 1. Rajiv travels the minimum distance to reach his teacher, as a straight line is the shortest distance

2. a. 8 cm

c. 9 cm

b. 6 cm

d. 8.6 cm

3.  O 32 mm 4.7 cm

A B

4.  45 cm P Q

5.  6.2 cm   6.  4 cm 3 cm

Word Problem 1. If 4 m on the ground is 1 cm on the paper, then 20 m on the ground would be equal to 5 cm on the paper.

7D 1. a. Radius = 7 cm

Circumference = 2πr = 2 × 22 7 × 7 cm = 44 cm

b. Radius = 8.5 cm

Circumference = 2πr = 2 × 22 7 × 8.5 cm = 53.43 cm

c. Radius = 9 cm

Circumference = 2πr = 2 × 22 7 × 9 cm = 56.57 cm

d. Radius = 49 cm

Circumference = 2πr = 2 × 22 7 × 49 cm = 308 cm

e. Radius = 14 cm

Circumference = 2πr = 2 × 22 7 × 14 cm = 88 cm

2. a. Circumference = 154 cm

Radius = Circumference 2 = 154 2 × 22 7 = 154 × 7 2 × 22 = 24.5 cm

b. Circumference = 88 cm

c. Circumference = 110 cm

d. Circumference = 66 cm

e.

= 352 cm

3. Circumference of garden = 94 m

Hence, the radius of garden is 14.95 m.

4. Distance from one end of the circular park to other through the centre is the length of its diameter. Diameter of garden = 28 m

Radius =

Circumference

= 28 2 m = 14 m

Distance covered by Rohan = 2 × Circumference of the park = 2 × 88 m = 176 m

5. Radius of the innermost track = 49 m

Radius of the second track = 49 m + 7 m = 56 m

Radius of the third track = 49 m + 7 m + 7 m = 63 cm

= 2 × 22 7 × 49 m = 308 m

Circumference of the second track = 2πr = 2 × 22 7 × 56 m = 352 m

Circumference

× 63 m = 396 m

Difference in the length of the innermost and outermost tracks = 396 m – 308 m = 88 m

Word Problem 1. Circumference of the cake = 36 cm Radius of the cake = Circumference

Chapter Checkup

4. Sample answer. Answers may vary.

7.a. Radius = 2 cm Diameter = 2 × radius = 2 × 2 cm = 4 cm

b. Radius = 3.8 cm Diameter = 2 × radius = 2 × 3.8 cm = 7.6 cm

c. Radius = 9.5 cm Diameter = 2 × radius = 2 × 9.5 cm = 19 cm

d. Radius = 7.1 cm Diameter = 2 × radius = 2 × 7.1 cm = 14.2 cm

e. Radius = 4.6 cm Diameter = 2 × radius = 2 × 4.6 cm = 9.2 cm

8. a. Diameter = 7.2 cm

Radius = Diameter 2 = 7.2 cm 2 = 3.6 cm

b. Diameter = 3.6 cm

Radius = Diameter 2 = 3.6 cm 2 = 1.8 cm

c. Diameter = 9.8 mm

Radius = Diameter 2 = 9.8 cm 2 = 4.9 mm

d. Diameter = 10 m

Radius = Diameter 2 = 10 m 2 = 5 m

e. Diameter = 11.2 km

Radius = Diameter 2 = 11.2 cm 2 = 5.6 km

9. a. Radius = 10.5 cm

Circumference = 2πr = 2 × 22 7 × 10.5 cm = 66 cm

b. Radius = 6.2 cm

Circumference = 2πr = 2 × 22 7 × 6.2 cm = 38.97 cm

c. Radius = 15.7 cm

Circumference = 2πr = 2 × 22 7 × 15.7 cm = 98.68 = 98.69 cm

d. Radius = 28.5 cm

Circumference = 2πr = 2 × 22 7 × 28.5 cm = 179.14 cm

e. Radius = 20.1 cm

Circumference = 2πr = 2 × 22 7 × 20.1 cm = 126.34 cm

10. a. Circumference = 24.5 cm

Radius = Circumference 2 = 24.5 2 × 22 7 = 24.5 × 7 2 × 22 = 3.898 = 3.9 cm

b. Circumference = 14 cm

Radius = Circumference 2 = 14 2 × 22 7 = 14 × 7 2 × 22 = 2.227 = 2.2 cm

c. Circumference = 17.5 cm

Radius = Circumference 2 = 17.5 2 × 22 7 = 17.5 × 7 2 × 22 = 2.784 = 2.8 cm

d. Circumference = 10.5 cm

e. Circumference = 56 mm

11.

13.

O 2.5 cm 6.8 cm G F

14. Shortest path: Path M (as PQ is a chord)

Longest path: Path S (along the major arc)

Equal paths: Path L and Path N (combination of the two radii)

Word Problems 1. Diameter of the smallest flower bed = 8 m

Radius of the smallest flower bed = 8 m ÷ 2 = 4 m

Circumference of the smallest flower bed = 2πr = 2 × 22 7 × 4 m = 25.142 m

Diameter of the largest flower bed = 8 m + 4 × 4 m = 24 m

Radius of the largest flower bed = 24 m ÷ 2 = 12 m

Circumference of the largest flower bed = 2πr = 2 × 22 7 × 12 m = 75.428 m

Difference in the circumference of the largest and smallest flowerbeds = 75.428 m – 25.142 m = 50.286 m = 50.29 m

2. Diameter of the cards = 14 cm

Radius of the cards = 14 cm ÷ 2 = 7 cm

Length of ribbon required for one card = 2πr = 2 × 22 7 × 7 cm = 44 cm

Length of ribbon required for 10 cards = 10 × 44 cm = 440 cm

Chapter 8

Let's Warm-up

1.  Circle

4. a. 6 faces

Octagonal pyramid

b. 12 edges and 7 vertices Cone

c. 1 vertex and 2 faces

d. 16 edges

e. 7 faces and 10 vertices

Cuboid

Pentagonal prism

Hexagonal pyramid

5. Differences: A nonagonal pyramid has 9 sides, while a hexagonal pyramid has 6 sides. Also, a nonagonal pyramid has 10 vertices and 9 lateral faces, while a hexagonal pyramid has 7 vertices and 6 lateral faces.

Similarity: Both nonagonal and hexagonal pyramids are 3D shapes with a polygonal base and triangular lateral faces.

Word Problem 1.Differences:a. Number of bases: An octagonal pyramid has one base, while an octagonal prism has two bases.

b. Shape of the lateral faces: An octagonal pyramid has triangular lateral faces, while an octagonal prism has rectangular lateral faces. Similarities: a. Number of sides in the base: Both an octagonal pyramid and an octagonal prism have 8 sides in the base. b. Three-dimensional shape: Both an octagonal pyramid and an octagonal prism are three-dimensional shapes.

8B 1. a. This is the net of a hexagonal pyramid. b. This is the net of a triangular pyramid. c. This is the net of a cube. d. This is the net of a cylinder.

2. a. A sphere has no net. b. The net of a cylinder has 3 faces.

c. A cubical box is open from the top. The net of the box has 5 faces. d. A square pyramid has 1 square and 4 triangles in its net.

3. Answers may vary, here are some sample answers.

a.  b.  c.  d.

4. Answers may vary, here are some sample answers:

a.  b.  c.  d.

Cuboid  Triangular Pyramid  Cone  Pentagonal pyramid 5.

Word Problem 1. A tetrahedron has 4 plane faces; hence, the net of the tetrahedron would look like this:

2.  Rectangle

3.  Triangle

Chapter Checkup 1. a. A cube has 6 faces. b. A sphere has 0 faces.

c. A pentagonal prism has 7 faces. d. A hexagonal pyramid has 7 faces. 2. a. This is a pentagonal pyramid. It has 10 edges and 6 vertices. b. This is a cuboid. It has 12 edges and 8 vertices.

4.  Square

Do It Yourself

8A 1.a. A circle is not a solid figure. b. A triangular pyramid has 4 faces. c. The shape of a playing dice is a cube. d. A heptagonal pyramid has 8 vertices.

2. a. Cuboid b. Cylinder c. Cone d. Cube

3. a. A cube has 6 faces, 12 edges and 8 vertices. b. A cylinder has 2 circular edges. c. All the faces of a cube are identical.

d. A pentagonal prism has 5 more edges than vertices.

e. A hexagonal pyramid has 6 triangular faces. f. A cube and a squarepyramid have square bases.

c. This is a cone. It has 1 edge and 1 vertex. d. This is an octagonal prism. It has 24 edges and 16 vertices.

3. a. A pentagonal pyramid has 10 edges and 6 vertices.

b. The difference between the number of vertices and the number of faces of a decagonal prism is 8 c. The number of faces, edges and vertices in a heptagonal prism is 44. (9+21+14)

4.A cone is not a prism. 5. a. Square prism

Total number of faces = 6 Type – 4 rectangular, 2 square shaped

b. Heptagonal pyramid Total number of faces = 8 Type: 7 triangular, 1 heptagon shaped c. Hexagonal prism

Total number of faces = 8 Type: 6 parallelograms, 2 hexagons

6. a. Octagonal pyramid b. Cuboid/Rectangular prism

c. Pentagonal pyramid d. Cone

7. Answers may vary. Here are some sample answers: a.  b.  c.  d.

Octagonal pyramid        Cube       Cylinder  Triangular pyramid 8. a.   b.

9. a. The net of a heptagonal pyramid has 8 faces. b. The net of a cube has 6 faces. c. The net of a nonagonal pyramid has 10 faces.

10. Triangular pyramidCone CylinderSquare pyramid

number line. b. Out of the 2 integers, 10 and 7, 10 falls on the right on the number line. c. Out of the 2 integers, −12 and −7, −7 falls on the right on the number line. d. Out of the 2 integers, 12 and −1, 12 falls on the right on the number line. e. Out of the 2 integers, −15 and −25, −15 falls on the right on the number line.

7. a. |8| = 8 b. |−10| = 10 c. |25| = 25 d. |−35| = 35

e. |−65| = 65 8. Answers may vary. Sample answer: 5 integers less than −6 = −7, −8, −9, −10, −11 9. Answers may vary. Sample answer: 7 integers greater than −10 = −9, −8, −7, −6, −5, −4, −3 10. a. −26 > −33 b. −15 < −12 c. 47 > −47 d. 14 < 18

e. −51 < −15 11. a. −10 < −8 < −2 < 5 < 9 b. −11 < −7 < −6 < 16 < 20

c. −15 < −9 < 13 < 14 < 23 d. −45 < −11 < 15 < 35 < 63

e. −63 < −32 < −18 < 47 < 95 f. −69 < −45 < −36 < 25 < 63

12. a. 12 > 4 > −6 > −8 > −9 b. 6 > 0 > −1 > −5 > −11

c. 20 > 15 > 12 > −14 > −17 d. 32 > 25 > 17 > −3 > −15

e. 52 > 14 > −12 > −28 > −30 f. 63 > 59 > 36 > −25> −52

Word Problems 1. −7°C < −3°C < 5°C < 12°C < 18°C

2. Amount deposited = +₹1000; Amount withdrawn = −₹750

Word Problems 1. Figure may vary, here are some sample figures.

3. +35Parachute +10Helium Balloon

0Sea level –15Submarine

9B 1. a. 6 + 4 = 10

2. Figure may vary, here are some sample figures: The net of the rectangular prism would look like this—

–1–2–3–4–5–6–7–8–9–10

b. −3 + (−6) = −9

Chapter 9

Let's Warm-up

–1–2–3–4–5–6–7–8–9–10

c. 8 + (−5) = 3

–1–2–3–4–5–6–7–8–9–10

d. −6 + (−4) = −10

–1–2–3–4–5–6–7–8–9–10

e. −9 + 3 + 5 = −1

The helium balloon is in between the submarine and the parachute.

012345678910

012345678910

012345678910

012345678910

Do It Yourself

9A 1. a. +10 km; opposite = −10 km b. − ₹1000; opposite = + ₹1000 c. +5 apples; opposite = −5 apples d. −3 kg; opposite = +3 kg e. +3 runs; opposite = −3 runs f. −10 minutes; opposite = +10 minutes

–1–2–3–4–5–6–7–8–9–10

2. 012345678910 –1–2–3–4–5–6–7–8–9–10

012345678910

2. a. −4 + 6 = +2 b. −3 + (−4) = −7 c. 5 + (−10) = −5 d. 5 + 4 = +9

3. a. 12 + (−14) = −2 b. −16 + (−18) = −34 c. −25 + 63 = 38

d. 52 + (−45) = 7 e. 125 + (−256) = −131 f. −365 + (−145) = −510

4. a. 5 + (–2) = 3; –7 + 4 = −3, Since 3 > −3, 5 + (–2) > –7 + 4

b. 10 + 12 = 22; 20 + (–5) = 15, Since 22 > 15, 10 + 12 > 20 + (–5)

c. –12 + 25 = 13; –36 + 41 = 5, Since 13 > 5, –12 + 25 > –36 + 41

123456789 –1–2–3–4–5–6–7–8–9

3.  0

BA CFDE

A = −6, B = −9, C = −2, D = 3, E = 6, F = 1

4. a. Predecessor = 16, Successor = 18 b. Predecessor = −13, Successor = −11 c. Predecessor = 24, Successor = 26  d. Predecessor = −89, Successor = −87 e. Predecessor = 124, Successor = 126 5. a. −2, −1, 0, 1, 2, 3 b. −4, −3, −2, −1 c. 2, 3, 4, 5  d. −8, −7, −6, −5 e. −14, −13, −12, −11, −10, −9 6. a. Out of the 2 integers, −4 and 4, 4 falls on the right on the

d. 32 + (–52) = −20, –62 + 10 = −52, Since −20 > −52, 32 + (–52)

> – 62 + 10 e. –65 + (–42) = −107; –57 + (–34) = −91, Since −107 < −91, –65 + (–42) < –57 + (–34) f. 58 + (32) = 90; –87 + 41 = −46, Since 90 > −46, 58 + (32) > –87 + 41

5. a. (–9) + 12 + (–15) = –9 + 12 –15 = −12 b. 18 + (–25) + 28 = 18

–25 + 28 = 21 c. –32 + (–14) + (–35) = –32 –14 –35 = −81

d. 36 + (–78) + (–14) + 63 = 36 –78 –14 + 63 = 7

e. –175+ (–48) + 259 + 88 = –175 –48 + 259 + 88 = 124

6. Successor of –1025 = –1024; Predecessor of 923 = 922 –1024 + 922 = −102

7. a. Additive inverse of −8 = 8 b. Additive inverse of 12 = −12

c. Additive inverse of −14 = 14 d. Additive inverse of −36 = 36

e. Additive inverse of 45 = −45

8. a. 5 + a = 5 a = 5 − 5 a = 0 b. 7 + a = 0 a = 0 − 7 a = −7

c. 6 + a = 3 + 6 a = 9 − 6 a = 3

9. Answers may vary. Sample answer:

The three integers could be −59, 36 and −135.

Word Problems 1. Score after one play = −7

Score after another play = 12

Score after two plays = −7 + 12 = 5

2. Temperature of city on Monday = 42℃

Temperature falls by 15℃ on Tuesday; Temperature on Tuesday = 42℃ − 15℃ = 27℃

Temperature further falls by 7℃ on Wednesday; Temperature on Wednesday = 27℃ − 7℃ = 20℃

9C 1. a. −3 – (−4) = −3 + 4 = +1

–1–2–3–4–5–6–7–8–9–10

b. −10 – (5) = −10 + (−5) = −15

012345678910

–16–15–14–13–12–11–10–9–8–7–6–5–4–3–2–101234

c. 15 – (−8) = 15 + (+8) = +23

–1 0123456789101112131415161718192021 22 23

d. −2−(−6) = −2 + 6 = +4

–1–2–3–4–5–6–7–8–9–10

012345678910

2. a. −36 – 15 = −51 b. −42 – (−48) = −42 + 48 = 6

c. 69 – (−52) = 69 + 52 = 121

3. a. –125 – (–253)

b. 236 – (–541)

777

–790

c. –365 – (425)  –157

d. –414 – (–257)  128

4. Sum of 254 and −146 = 254 + (−146) = 108

Subtracting from 367 = 367 −108 = 259

5. Sum of −745 and −425 = −745 + (−425) = −1170

Subtracting −389 from −1170 = −1170 – (−389) = −1170 + 389 = −781 6. Sum of 623 and –584 = 623 + (–584) = 39

Sum of –795 and 475 = –795 + 475 = −320

Subtracting 39 from −320 = −320 – 39 = −359

7. a. Integers are associative under subtraction. – False

b. The difference of two integers is always an integer. – True

c. Zero subtracted from an integer results in the additive inverse of the integer. – False d. Subtraction is not commutative for integers. – True

8. 256 + (–365) – 124 + (–475) – 321 = 256 − 365 −124 – 475 − 321 = 256 – 1285 = −1029

9. Let the two integers be 89 and x Given, 89 + x = –423

x = –423 – 89 x = –512

Word Problems 1. Melting point of mercury = –38.9°C; boiling point of mercury = 356.7°C

Difference between the melting and boiling points = 356.7°C – (–38.9°C) = 356.7°C + 38.9°C = 395.6°C

2. Number of kilometres driven on Monday in the north = +125km

Number of kilometres driven on Tuesday in the south = −182km

Distance from the initial point = 125 + (–182)= –57 57 km towards the south

3. Initial balance in Pooja’s account = ₹ (–1250)

Amount she is willing to have in her account = ₹5000

Let the amount deposited be x

₹ (−1250) + x = ₹5000

x = 5000 + 1250 = 6250

Hence, the money she needs to deposit is ₹6250.

4. 10,500 feet

0 2514 feet

Distance between the submarine and the aeroplane = 10,500 – (–2514) = 10,500 + 2514 = 13,014 feet

Chapter Checkup 1.a. –2 is on the left of 5 on the number line.

b. 13 is on the left of 18 on the number line. c. –8 is on the left of 8 on the number line. d. –17 is on the left of –13 on the number line. e. 25 is on the left of 36 on the number line.

2. a. Chennai = 31° above 0° = +31°; Pune= 18° above 0° = +18°; Ladakh = 5° below 0° = –5°; Sonamarg = 8° below 0° = –8°

b. Since Sonmarg has the lowest temperature, it is the coldest place, and Chennai has the highest temperature, it is the hottest place.

3. 0 –2–1123

ABC DE

The point E will be marked as −2.

4. Absolute value does not have a sign with it. Hence the integers 9 and 45 are given in their absolute values.

5. a. Predecessor of −8 = −9; Successor of −8 = −7

b. Predecessor of 15 = 14; Successor of 15 = 16

c. Predecessor of −26 = −27; Successor of −26 = −25

d. Predecessor of −38 = −39; Successor of −38 = −37

e. Predecessor of 45 = 44; Successor of 45 = 46

6. +8 is larger than –16 is True +16 is smaller than +8 is False –8 is larger than +16 is False +16 is smaller than –8 is False

7. No positive integers lie between −5 and 0.

8. a. Ascending order of –25, –65, 28, 63, 100 = –65, –25, 28, 63, 100 b. Ascending order of –22, –18, –15, 16, 36 = –22, –18, –15, 16, 36 c. Ascending order of 29, 15, –9, –16, –32 = −32, −16, −9, 15, 29 d. Ascending order of –32, –17, –19, 42, 48 = –32, –19, –17, 42, 48 –22, –18, –15, 16, 36 is already arranged in ascending order.

9. a. –25 + 36 = 11 b. –142 + (–48) = −190 c. –157 + 96 = −61

d. 452 + –152 = 300 10. Largest integer = 361 Smallest integer = − 423 Sum = 361 + (−423) = −62

11. a. 7 – (–5) = 7 + 5 = 12; –5 – 9 = −14, so 12 > −14

b. 21 – 12 = 9; –30 – (–15) = −30 + 15 = −15, so 9 > −15

c. –36 – 17 = −53; 48 – 96 = −48, so −53 < −48

d. 132 – (–152) = 132 + 152 = 284; –262 – 204 = −466, so 284 > −204

e. –365 – (–542) = −365 + 542 = 177; –357 – (–534) = −357 + 534 = 177, so 177 = 177

f. 658 – (–132) = 658 + 132 = 790; –387 – 841 = −1228, so 790 > −1228

12. Predecessor of −526 = −527, successor of 658 = 659 659 – (−527) = 659 + 527 = 1186

13. Answers may vary. Sample answer: Three integers with different signs whose difference is –48 can be: −25, +42 and −65

14. a. The sum of two negative integers is always negative. True

b. The sum of an integer and its inverse is zero. True

c. The difference of two integers is always an integer. True

d. The sum of four different integers can never be zero. False

15. Numbers smaller than −858 but greater than −861 = −859, −860. −859 is the odd number smaller than −858 but greater than −861. Word Problem 1. Score in the first round = 325 points

Score in the second round = −130 points

Score in the third round = 78 points

Score after the third round = 325 + (−130) + 78 = 273 points

2. Calories burnt after working out on a trade mill = −480

Calories burnt after skipping = −260

Total calories burnt = −480 + (−260) = −740

Calories indulged with sweet = 500

Calories lost or gained = −740 + 500 = −240

As the answer has a negative sign, she lost 240 calories.

3. Initial temperature = 132℉

Temperature after drop = –250°F

Difference between the temperatures = 132℉ – (−250℉) = 132℉ + 250℉ = 382℉

4. Initial balance in the account = ₹−125

Balance after adding ₹1500 = ₹−125 + ₹1500 = ₹1375

Balance after purchasing a dress worth ₹1000 = ₹1375 − ₹1000 = ₹375

Chapter 10

Let's Warm-up

5.

Do It Yourself

10A

1.  a. Total parts = 4 Shaded parts = 3

= 3 4

c. Total parts = 20

Shaded parts = 10

Fraction = 10 20

b. Total parts = 16 Shaded parts = 6

= 6 16

d. Total parts = 9

Shaded parts = 5 Fraction = 5 9

2. a. b. c.   d.

5. a. 16 8 b. 5 7

Word Problem 1. To represent 2 3 m on a number line, we can draw a number line and divide the part between 0 and 1 into 3 equal parts. The second part will represent two parts out of the three.

10B 1. a. 5 7 and 3 7

3. Proper fractions:

Improper

and

2. a. 8 6 and 9 7

8 × 7 = 56 and 9 × 6 = 54

56 > 54

Thus, 8 6 > 9 7

c. 7 8 and 11 14

7 × 14 = 98 and 8 × 11 = 88 98 > 88

Thus, 7 8 > 11 14

3. a. 2 1 2 < 2 4 7

The whole part is the same, so we can directly compare the proper fraction.

1 × 7 = 7 and 2 × 4 = 8

7 < 8

Thus, 2 1 2 < 2 4 7

c. 3 4 6 = 3 2 3

The whole part is the same, so we can directly compare the proper fraction.

4 × 3 = 12 and 6 × 2 = 12

12 = 12

Thus, 3 4 6 = 3 2 3

b. 3 5 and 4 7

3 × 7 = 21 and 4 × 5 = 20

21 > 20

Thus, 3 5 > 4 7

d. 4 5 and 12 20

4 × 20 = 80 and 12 × 5 = 60

80 > 60

Thus, 4 5 < 12 20

b. 8 7 > 7 8

8 × 8 = 64 and 7 × 7 = 49

64 > 49

Thus, 8 7 > 7 8

d. 11 12 > 12 15

11 × 15 = 165 and 12 × 12 = 144

165 > 144

Thus, 11 12 > 12 15

4. a. 4 19  , 6 19  , 8 19  , 7 19

Since 4 < 6 < 7 < 8

Thus, 4 19 < 6 19 < 8 19 < 7 19

b. 15 28  , 4 14  , 3 7  , 18 28

Converting all the fractions into like fractions-

LCM of 28, 14, 7, 28 is 28.

4 14 × 2 2 = 8 28

3 7 × 4 4 = 12 28

Since denominators are the same, compare the numerators and order the fractions.

8 < 12 < 14 < 15

So, 8 28 < 12 28 < 14 28 < 15 28

Thus, 4 14 < 3 7 < 14 28 < 15 28

d. 7 8  , 1 2  , 3 4  , 21 16

Converting all the fractions into like fractions-

LCM of 8, 2, 4, 16 is 16.

7 8 × 2 2 = 14 16

1 2 × 8 8 = 8 16

3

4 × 4 4 = 12 16

Since denominators are the same, compare the numerators and order the fractions.

8 < 12 < 14 < 21

So, 8 16 < 12 16 < 14 16 < 21 16

Thus, 1 2 < 3 4 < 7 8 < 21 16

c. 1 3  , 4 18  , 6 9  , 21 27

Converting all the fractions into like fractions-

LCM of 3, 18, 9, 27 is 54

1 3 = 1 3 × 18 18 = 18 54

4 18 = 4 18 × 3 3 = 12 54

6 9 = 6 9 × 6 6 = 36 54

21 27 = 21 27 × 2 2 = 42 54

Since denominators are the same, compare the numerators and order the fractions.

12 < 18 < 36 < 42

So, 12 54 < 18 54 < 36 54 < 42 54

Thus, 4 18 < 1 3 < 6 9 < 21 27

5.a. 10 25 b. 21 25 c. 13 36 d. 24 36

LCM of 25 and 36 is 900.

Converting the fractions to like fractions-

10

25 = 10 25 × 36 36 = 360 900

21

25 = 21 25 × 36 36 = 756 900

13

36 = 13 36 × 25 25 = 325 900

24

36 = 24 36 × 25 25 = 600 900

Since, 325 < 360 < 600 < 756

Thus, 325 900 < 360 900 < 600 900 < 756 900

Therefore, 13 36 < 10 25 < 24 36 < 21 25

Word Problems 1. The number of hours Richa studied = 2 3 4

The number of hours Preeti studied = 2 1 2

Since the whole part is the same, we will compare the proper fractions.

LCM of 4 and 2 is 4.

1 2 × 2 2 = 2 4

Comparing the numerator: Since 3 > 2, 3 4 > 2 4

Thus, Richa studied for a longer time than Preeti.

2. Fraction of students passed in school A = 480 500

Fraction of students passed in school B = 550 600

LCM of 500 and 600 is 3000.

480

500 × 6 6 = 2880 3000

550 600 × 5 5 = 2750 3000

Since, 2880 3000 > 2750 3000

Thus, 480 500 > 550 600

In school A, a greater fraction of students passed.

3. Rounds covered by Jacky = 5 8

Rounds covered by Pritam = 7 12

LCM of 8 and 12 is 24.

5 8 × 3 3 = 15 24

7 12 × 2 2 = 14 24

Since 15 > 14

Thus, 5 8 > 7 12

Hence, Jacky covers a longer distance than Pritam.

4. Lengths of clothes purchased by Sana = 2 1 4  , 2 1 3  , 2 3 4  , 2 2 8

Since the whole part is the same, we will compare the proper fractions.

LCM of 4, 3, and 8 is 24.

1

1

Thus,

is

c. 319 12 > 21 5 12 d. 47

5. a. Bushels of apples harvested by the farmer = 5 and 1

5 and 1 4 as an improper fraction – (5

Bushels of apples harvested as an improper fraction = 21 4

b. Wood used to make the table = 18 5 feet

3 518 –15 3

Thus, 18 5 = 3 3 5

The length of the wood as a mixed number = 3 3 5 feet

Word Problems 1. Total number of cookies baked = 100

Number of siblings = 3

Share of each sibling = 100 3 33

3100 –9

10 –9 1

Thus, 100 3 = 33 1 3

Share of each sibling expressed as a mixed number = 33 1 3

2. Number of slices of the cake baked by Eva = 8

Number of slices given away by Eva = 3

Number of slices left with Eva = 5

Fraction of cake left with Eva = 5 8

Number of slices of the cake baked by Sarah = 12

Number of slices given away by Sarah = 5

Number of slices left with Sarah = 7

Fraction of cake left with Sarah = 7 12

LCM of 8 and 12 is 24.

5 8 × 3 3 = 15 24

7

× 2 2 = 14 24

Since 15 > 14, thus, 15 24 > 14 24

Therefore, 5 8 > 7 12

Hence, Eva had the greater fraction of her cake left.

1. a. 5 9 × 9 9 = 45 81 b. 17 19 × 5

= 85 95 c. 18 24 ÷ 6 6 = 3 4

2. Answers may vary. Sample answers. a.

3. Answers may vary. Sample answers.

a.

4. a. 4 5 and 12 15

4 × 15 = 60; 5 × 12 = 60

So, 4 × 15 = 5 × 12

Thus, 4 5 and 12 15 are equivalent.

c. 7 8 and 14 24

7 × 24 = 168; 8 × 14 = 112

7 × 24 8 × 14

Thus, 7 8 and 14 24 are not equivalent.

e. 7 11 and 35 44

7 × 44 = 308; 11 × 35 = 385

7 × 44 11 × 35

Thus, 7 11 and 35 44 are not equivalent.

g. 7 13 and 21 26

7 × 26 = 182; 13 × 21 = 273

7 × 26 13 × 21

Thus, 7 13 and 21 26 are not equivalent.

b. 1 3 and 5 21

1 × 21 = 21; 3 × 5 = 15

1 × 21 3 × 5

Thus, 1 3 and 5 21 are not equivalent.

d. 2 9 and 12 54

2 × 54 = 108; 9 × 12 = 108

2 × 54 = 9 × 12

Thus, 2 9 and 12 54 are equivalent.

f. 2 12 and 12 72

2 × 72 = 144; 12 × 12 = 144

2 × 72 = 12 × 12

Thus, 2 12 and 12 72 are equivalent.

5. a. HCF of 12 and 15 is 3. Divide the numerator and the denominator by 3. 12 15 ÷ 3 3 = 4 5

b. HCF of 6 and 28 is 2.

Divide the numerator and the denominator by 2. 6 28 ÷ 2 2 = 3 14

c. HCF of 16 and 144 is 16.

Divide the numerator and the denominator by 16. 16 144 ÷ 16 16 = 1 9

d. HCF of 14 and 28 is 14.

Divide the numerator and the denominator by 14. 14 28 ÷ 14 14 = 1 2

Word Problems 1. New fraction = 2 3

Subtracting 1 from the numerator and the denominator to get the original fraction = 2 – 1 3 – 1 = 1 2

Checking the fraction using (a)— The numerator is three less than twice the denominator.

1 = (2 × 2) – 3

1 = 4 – 3

1 = 1

2. Number of chocolates Riya has = 24

Number of chocolates she gave to her friend = 10

Number of chocolates she gave to her sister = 10

Number of chocolates left with Riya = (24 – (10 + 10)) = 4

Fraction of chocolates left with Riya = 4 24 = 1 6

Riya is left with 1 6 fraction of chocolates.

c.

3. a. 3 7 + 2 14

LCM of 7 and 14 is

3 7 × 2 2 = 6 14

6 14 + 2 14 = 8 14 =

c. 2 9 10 + 3 11

2

× 11 11 = 319 110

3

e. 9 3 5 + 6 2 5

Adding the Wholes = 9 + 6 = 15

Adding the Proper Fraction

= 3 5 + 2

15 + 1 = 16

b. 4 3 11 –11 22

4 3 11 = (4 × 11) + 3 11 = 47 11

LCM of 11 and 22 is 22.

47 11 × 2 2 =

94

d. 17 2 3 – 15 1 12

17 2 3 = (17 × 3) + 2

= 53 3

15 1 12 = (15 × 12) + 1 12 = 181 12 LCM of 3 and 12

We get 1 41

Word Problems

3

2

d. 5 2 3 + 2 1 3

4. a. 9 13 –1 2

1 2 × 13 13 = 13 26 18 26 –13 26 = 5 26

Weight of the empty box = 2 6 7

Weight of the iron pan inside the box = 5 2 3

Total weight = 8 11 21 kg

The total weight of the box is 8 11 21 kg.

2. Amount of sugar used in the first batch = 3 5 cups

Amount of sugar used in the second batch = 1 5 cups

Total sugar used for both the batches = 3 5 + 1 5 = 4 5 cups

The total sugar used for both the batches together = 4 5 cups

3. Total weight of candies = 9 1 2 kg = 19 2 kg

Weight of chocolate candies = 2 3 4 kg = 11 4 kg

Weight of gummy bears = 1 1 10 kg = 11 10 kg

Weight of jellybeans = 3 1 2 kg = 7 2 kg

Weight of chocolate candies, gummy bears, and jellybeans

together = 11 4 + 11 10 + 7 2 kg

LCM of 4, 10, 2 = 20

11 4 + 11 10 + 7 2 kg = 55 20 + 22 20 + 70 20 kg = 147 5 kg

Weight of lollipops = 190 20 –147 20 = 43 20 = 2 3 20 kg

4. Fraction of cookies given to Priya’s sister = 3 5

Fraction of cookies given to neighbors = 1 4

a. Fraction of cookies given away = 3 5 + 1

= 12 20 + 5 20 = 17 20

b. Fraction of cookies left with Priya = 1 – 17 20 = 3 20

2. Total raisins Maya has = 75 g

Fraction of raisins used by her = 1 5

The amount of raisins used by Maya = 1 5 × 75 = 15g

3. Total flour Emma has = 4 1 2 cups = 9 2 cups

Flour needed for one batch of cookies = 2 3 cups

can make 6 batches of cookies and 1/2 cup will be left over.

4. Answers may vary. Sample answers.

5. a. HCF of 32 and 56 is 8.

Divide the numerator and the denominator by 8

32

56 ÷ 8 8 = 4 7

b. HCF of 45 and 125 is 5.

Divide the numerator and the denominator by 5

45

125 ÷ 5 5 = 9 25

c. HCF of 66 and 94 is 2.

Divide the numerator and the denominator by 2

66

94 ÷ 2 2 = 33 47

d. HCF of 66 and 94 is 2.

Divide the numerator and the denominator by 2

62

134 ÷ 2 2 = 31 67

6. a. 10 8 and 40 32

10 × 32 = 320; 8 × 40 = 320

320 = 320

Thus, 10 8 and 40 32 are equivalent.

c. 2 5 and 10 25

2 × 25 = 50; 5 × 10 = 50

50 = 50

Thus, 2 5 and 10 25 are equivalent.

7. a. 3 1 2 and 13 5

3 1 2 = 7 2;

7 × 5 = 35; 2 × 13 = 26

35 26

Thus, the fractions are not equivalent.

c. 21 3 and 8 1

21 × 1 =21; 3 × 8 = 24

21 24

Thus, the fractions are not equivalent.

b. 3 7 and 28 32

7 × 28 = 196; 32 × 3 = 96

196 96

Thus, 3 7 and 28 32 are not equivalent.

d. 1 3 and 16 48

1 × 48 = 48; 3 × 16 = 48

48 = 48

Thus, 1 3 and 16 48 are equivalent.

b. 5 4 5 and 35 4

5 4 5 = 29 5

29 × 4 = 116; 35 × 5 = 175

116 175

Thus, the fractions are not equivalent.

d. 54 9 and 2 1 3

2 1 3 = 7 3

54 × 3 = 162; 9 × 7 = 63 162 63

Thus, the fractions are not equivalent.

c.

324 = 5 9

9. a. 2

b. 50

c. 1 7

10. a. Difference:

To compare, find a common denominator. LCM of 8 and 60 is 240.

41 8 × 30 30 = 1230 240

203

60 × 4 4 = 812 240

Since 1230 > 812, hence, the difference of 8 3 8 and 3 1 4 is more than the sum of 2 1 4 and 1 2 15

b. The product of 1 2 3 and 4 5 = 5 3 × 4 5 = 20 15 = 4 3

Sum of 1 4 and 2 4 5 = 1 4 + 14 5 = 5 20 + 56 20 = 61 20

LCM of 3 and 20 is 60.

4 3 × 20 20 = 80 60 ; 61 20 × 3 3 = 183 60

Since 183 > 61, sum of 1 4 and 2 4 5 is greater than the product of 1 2 3 and 4 5

c. The product of 3 and 4 5 = 12 5

The sum of 1 3 and 1 4 = 7 12

12

5 + 7 12 = 144 60 + 35 60 = 179 60

d. 1 5 of a dozen = 1 5 × 12 = 12 5

10 dozen = 10 × 12 = 120

Word Problems 1. Riya’s share of cookies = 3 5

LCM of 5 and 8 is 40.

3

5 × 8 8 = 24 40

Siya’s share of cookies = 5 8

5 8 × 5 5 = 25 40

Siya took a greater fraction of the cookies because 25 40 > 24 40

2. Amount of sugar Shweta has = 4 5 kg

Sugar used to make sweet dish = 3 4 kg

Sugar left = 4 5 –3 4 = 16 20 –15 20 = 1 20 kg

Shweta is left with 1 20 kg of sugar.

3. Number of students and teachers together = 1000.

Number of girls = 1 10 × 1000 = 100

Number of boys = 2 5 × 1000 = 400

Total number of students = 400 + 100 = 500

Fraction of teachers = 1000 – 500 = 500

The number of teachers is 500.

4. Total length of the cloth = 40m

Cloth used for curtains = 2 5 m

Cloth used for bedsheet = 3 3 4 m

Total cloth used = 2 5 + 3 3 4 = 2 5 + 15 4 = 8 20 + 75 20 = 83 20 m

Total cloth left = 40 – 83 20 = 800 20 –83 20 = 717 20 m = 35 17 20 m

Radha is left with 35 17 20 m of cloth.

5. Proceeds from 4 tote bags of type X = 1 13

Proceeds from 2 tote bags of type Y = 2 5

Number of X type tote bags sold = 24

Number of Y type tote bags sold = 18

a. Proceeds from type X for 40 dollars each = 24 4 × 40 × 1 13 = 18.46 dollars

b. Proceeds from type Y for 60 dollars each = 18 2 × 60 × 2 5 = 216 dollars

Total proceeds donated from Type X and Y = 18.46 + 216 = 234.46 dollars

Chapter 11

Let's Warm-up

1. 15.61 > 15.6  2. 14.5 = 14.500  3. 15.33 < 15.4

4. 17.32 > 17.3  5. 16.5 > 16.45

Do It Yourself

11A 1.a. 1.6 lies between 1 and 2. b. 6.3 lies between 6 and 7.

c. 4.8 lies between 4 and 5. d. 5.36 lies between 5 and 6.

e. 12.71 lies between 12 and 13. f. 65.02 lies between 65 and 66.

5. There are 5 divisions between any two whole numbers. So, each division here between two whole numbers is 0.2.

A represents 0.8     B represents 1.4

C represents 3.4     D represents 4.6

Word Problems

1. 33 33o C

Wednesday

33.2o C

C34.5o C35o C

2. There are 5 divisions between any two whole numbers. So, each division here between two whole numbers is 0.2 m.

A represents 0.2 m    B represents 1.6 m

C represents 3.4 m    D represents 4.4 m

11B 1.a. four hundred thirty and forty thousandths = 430.400 False

b. three thousand twenty-one and five thousandths = 321.005 True

c. three hundred four point three zero two = 304.302 True

d. eleven and twenty-three hundredths = 11.023 False

2. Part

c.

= 1248.16

d. 30 + 9 + 8 100 +

e. 5000 + 500 + 50 +

+ 0.05 + 0.005 = 5555.555

5. a. 15.053 has the digit 5 in common.

5 is at the ones place and the hundredths place.

Sum = 5 + 0.05 = 5.05

b. 164.483 has the digit 4 in common.

4 is at the ones place and the tenths place.

Sum = 4 + 0.4 = 4.4

c. 143.763 has the digit 3 in common.

3 is at the ones place and the thousandths place.

Sum = 3 + 0.003 = 3.003

d. 817.864 has the digit 8 in common.

8 is at the hundreds place and the tenths place.

Sum = 800 + 0.8 = 800.8

e. 3264.215 has the digit 2 in common.

2 is at the hundreds place and the tenths place.

Sum = 200 + 0.2 = 200.2

Word Problems 1. The cost of the Maths book = $26.98

Expanded form of 26.98 = 20 + 6 + 0.9 + 0.08 or 20 + 6 + 9 10 + 8 100

2. The lengths of the three pieces of cloth = 1.56 m, 21.5 m and 14.5 m

In words,

1.56 = one point five six or one and fifty-six hundredths

21.5 = twenty-one point five or twenty-one and five tenths

14.5 = fourteen point five or fourteen and five tenths

Expanded form

1.56 = 1 + 0.5 + 0.06 or 1 + 5 10 + 6 100

21.5 = 20 + 1 + 0.5 or 20 + 1 + 5 10

14.5 = 10 + 4 + 0.5 or 10 + 4 + 5 10

11C 1.a. 3 8 has 8 as the denominator.

8 × 125 = 1000

We need to multiply the denominator by 125 to convert 3 8 into a decimal number.

b. 7 4 has 4 as the denominator.

4 × 25 = 100

We need to multiply the denominator by 25 to convert 7 4 into a decimal number.

c. 13 2 has 2 as the denominator.

2 × 5 = 10

We need to multiply the denominator by 5 to convert 13 2 into a decimal number.

d. 37 25 has 25 as the denominator.

25 × 4 = 100

We need to multiply the denominator by 4 to convert 37 25 into a decimal number.

e. 12 5 has 5 as the denominator.

5 × 2 = 10

We need to multiply the denominator by 2 to convert 12 5 into a decimal number.

f. 89 40 has 40 as the denominator.

40 × 25 = 1000

We need to multiply the denominator by 25 to convert 89 40 into a decimal number.

2. a. 18 5 = 18 5 × 2 2 = 36 10 = 3.6   b. 97 20 = 97 20 × 5 5 = 485 100 = 4.85

c. 78 4 =

=

=

e.

3. a. 0.24 =

d. 0.125 =

4.a.

b.

c. 2.052 = 2 + 0.052 =

d. 4.075 = 4 + 0.075 =

e. 52.002 = 52 + 0.002 =

f. 25.06 = 25 + 0.06 =

5. a. 3 kg 68 g = 3 kg + 68 1000 kg = 3 kg + 0.068 kg = 3.068 kg

14 kg 50 g = 14 kg + 50 1000 kg = 14 kg + 0.05 kg = 14.05 kg

5 kg 5 g = 5 kg + 5 1000 kg = 5 kg + 0.005 kg = 5.005 kg

725 g = 725 1000 kg = 0.725 kg

45 g = 45 1000 kg = 0.045 kg

b. 4 km 15 m = 4 km + 15 1000 km = 4 km + 0.015 km = 4.015 km

8 km 40 m = 8 km + 40 1000 km = 8 km + 0.04 km = 8.04 km

3 km 4 m = 3 km + 4 1000 km = 3 km + 0.004 km = 3.004 km

750 m = 750 1000 km = 0.75 km

15 m = 15 1000 km = 0.015 km

Word Problem 1. The fraction of students that participated in the discussion = 2 5

The fraction of students who did not participate

3

5 as a decimal = 3 5 × 2 2 = 6 10 = 0.6

2. The amount of sugar used by Mr. Rathore for the pudding = 0.125 cup

0.125 as a fraction = 125 1000 = 1 8

Thus, Mr. Rathore used 1 8 cup of sugar in his pudding.

11D 1.a. Since both decimals 0.154 and 2.315 have the same number of digits after the decimal points, they are like decimals.

b. 31.2 has 1 digit after the decimal point and 153.145 has 3 digits after the decimal point. Since both decimals have different number of digits after the decimal points, they are unlike decimals.

c. Since both decimals 1.302 and 8.024 have the same number of digits after the decimal points, they are like decimals. d. Since both decimals 1.111 and 2.002 have the same number of digits after the decimal points, they are like decimals.

2.a. 1.54 = 1.540

So, 1.540 and 15.021 are like decimals.

b. 54 = 54.000

So, 54.000 and 48.021 are like decimals.

c. 1.1 = 1.10

So, 1.10 and 4.65 are like decimals.

d. 97 = 97.000

18.054 and 97.000 are like decimals.

3.a. 15.4, 15, 15.14, 15.41

The tens and the ones place are the same. Compare the tenths place.

4 > 1

Compare the hundredths place in 15.4 and 15.41

1 > 0

Thus, 15.41 is the greatest number.

b. 178.1, 178.2, 1.78, 178.15

The hundreds, tens and ones place are the same in three numbers. Compare the tenths place.

2 > 1

Thus, 178.2 is the greatest number.

c. 12.810, 12.82, 12.815, 12.825

The tens, ones and tenths place are the same in all the numbers. Compare the hundredths place.

2 > 1

Compare the thousandths place in 12.82 and 12.825

5 > 2

Thus, 12.825 is the greatest number.

4. a. 84.23 < 84.32      b. 15.75 > 15.7

c. 6.122 > 6.120        d. 0.97 > 0.87

5.a. 7.3, 8.37, 7.23, 8.32

Compare the ones place.

8 > 7

Compare the tenths place in 7.23, 7.3 and 8.37, 8.32

2 < 3

So, 7.23 < 7.3

3 = 3

Compare the hundredths place in 8.37 and 8.32

2 < 7

So, 8.32 < 8.37

Thus, 7.23 < 7.3 < 8.32 < 8.37

b. 58.37, 58.73, 58.45, 58.54

The tens and ones place is the same all the numbers. Compare the tenths place.

3 < 4 < 5 < 7

Thus, 58.37 < 58.45 < 58.54 < 58.73

c. 97.08, 97.18, 97.2, 97.8

The tens and ones place is the same all the numbers.

Compare the tenths place.

0 < 1 < 2 < 8

Thus, 97.08 < 97.18 < 97.2 < 97.8

Word Problems 1. 15.48 km = 15.480 km

12.3 km = 12.300 km

19.457 km = 19.457 km

17.5 km = 17.500 km

2. The weight of the sugar = 5 kg 70 g = 5 kg + 70 100 kg = 5 kg +

0.07 kg = 5.07 kg

The weight of the rice = 5.65 kg

5.07 < 5.65

Thus, Rajesh buys rice in greater quantity.

3. The length of the cloth bought from the first shop = 5 m 15 cm = 5 m + 15 100 m = 5 m + 0.15 m = 5.15 m

The length of the cloth bought from the second shop = 5.25 m

5.15 < 5.25

Thus, Reena bought less cloth from the first shop.

4. The distance swum by Kunal on day 1 = 1.58 km

The distance swum by Kunal on day 2 = 2.53 km

The distance swum by Kunal on day 3 = 2.39 km

The distance swum by Kunal on day 4 = 3.13 km

1.58 < 2.39 < 2.53 < 3.13

Thus, Kunal swam the most on day 4.

5. 2.54 L, 1.75 L, 2.45 L, 2.3 L

Compare the ones digit, 2 > 1, so, 1.75 L is the smallest.

Compare the tenths place in 2.54, 2.45, 2.3

5 > 4 > 3

So, 2.54 > 2.45 > 2.3

Thus, 2.54 L > 2.45 L > 2.3 L > 1.75 L

11E

c.

e. 3 + 17.64 + 81.989 + 106 HTO.thth 1111

208.629

2.a. 15.4 – 12.9       b. 14.51 – 12.79

c.

3. a. 147.6 – 48.8      (i) 63.604

b. 51.45 + 46.69     (ii) 98.8

c. 817.48 – 753.876    (iii) 98.013

d. 453.021 – 355.008  (iv) 98.14

4.a. 96.321 + 53.412 = 149.733 149.733 > 147.733

c. 851.02 + 25.999 = 877.019 877.019 < 879.019

5.a. 53.64 + 18.3 + 51.97

HTO.th 1111 53.64 18.30 +51.97 123.91

b. 148 – 20.35 = 127.65 127.65 = 127.65

d. 756.3 – 98.608 = 652.692 652.692 < 657.702

b. 17.9 + 97.74 + 56.47

HTO.thth 1221

17.90

97.74 +56.47 172.11

Thus, 53.64 + 18.3 + 51.97 = 123.91 Thus, 17.9 + 97.74 + 56.47 = 172.11

c. 814.64 – 15.903 + 12.17

Let us first find 814.64 – 15.903  Let us add 798.737 and 12.17

HTO.thth

1013

70316310

814.640 –15.903

798.737

HTO.thth

1 1

798.737

+12.170

810.907

Thus, 814.64 – 15.903 + 12.17 = 810.907

d. 71.023 + 41.846 – 12.998

Let us first add 71.023       Let us subtract 12.998 and 41.846           from 112.869

HTO.thth 1

71.023 +41.846

112.869

HTO.thth

101117

01716

112.869

–12.998

099.871

Thus, 71.023 + 41.846 – 12.998 = 99.871

e. 51.97 + 87.2 – 97.807

Let us add 51.97 and 87.2  Let us subtract 97.807 from 139.17

HTO.th 11

51.97 +87.20

139.17

HTO.thth

13811610

139.170

+97.807

41.363

Thus, 51.97 + 87.2 – 97.807 = 41.363

f. 35.143 – 23.5 + 64.52

Let us subtract 23.5 from 35.143  Let us add 11.643 and 64.52

HTO.thth

811

35.143 –23.500

11.643

Thus, 35.143 – 23.5 + 64.52 = 76.163

TO.thth

1

11.643 +64.520

76.163

Word Problems 1. The total length of the cloth with Rakhil = 25 m 12 cm

= 25 m + 12 100 m = 25.12 m

Length of the cloth used = 12.56 m

Length of the cloth left = 25.12 m – 12.56 m = 12.56 m

2. Weight of the sugar in the different rooms of the warehouse

= 584 kg 80 g, 888.045 kg, 748 kg

584 kg 80 g = 584 kg + 80 1000 kg = 584.080 kg

The total weight of the sugar stored in the warehouse

= 584.080 kg + 888.045 kg + 748 kg = 2220.125 kg

3. The amount of rainfall on Monday = 4 mm

The amount of rainfall on Tuesday = 3.5 mm

The amount of rainfall on Wednesday = 3.75 mm

The amount of rainfall on Thursday = 4.2 mm

The total amount of rainfall = 4 mm + 3.5 mm + 3.75 mm + 4.2 mm = 15.45 mm

4. The distance between Ruhi’s home and school = 6.98 km

The distance between Ruhi’s new home and school = 9 km

The distance between Ruhi’s old home and new home = 9 km – 6.98 km = 2.02 km

5. The weight of the gold chain

= 20 g 45 mg = 20 g + 45 1000 g = 20.045 g

The weight of the ring = 12.5 g

The total weight of the jewelry = 20.045 g + 12.5 g = 32.545 g

6. The amount of the money in Rajul’s bank account = ₹587.2

The amount of money deposited in his bank account = ₹300 and ₹950

The total money deposited = ₹300 + ₹950 = ₹1250

The total money in the bank account = ₹587.2 + ₹1250 = ₹1837.2

The amount of money withdrawn from the bank account = ₹450.5

The amount of money left in the bank account = ₹1837.2 – ₹450.5 = ₹1386.7

7. The amount of money spent on stationery = ₹148.6

The amount of money spent on clothes = ₹1500.5

The amount of money spent on transport = ₹238.6

The total money spent = ₹148.6 + ₹1500.5 + ₹238.6 = ₹1887.7

The total amount of money with Monica = ₹2000

The amount of money left with Monica = ₹2000 - ₹1887.7 = ₹112.3

236 ×124 944 4720 +23600 29264 3. a. 31.24 × 6 = 187.44      b. 32.6 × 1.2 = 39.12 c. 97.87 × 1.3 = 127.231      d. 157.41 × 8 = 1259.28 4. a. 51.3 × 12 = 615.6        b. 32.65 × 1.3 = 42.445    615.6 < 618.6          42.445 > 42.145 c. 103.4 × 14 = 1447.6 1447.6 > 1446.6 TO.th 11 400 3.50 3.75 +4.20 15.45 ThHTO.th 9 81010 9.00 – 6.98 2.02 TO.thth 20.045 +12.500 32.545 ThHTO.t 1 587.2 +1250.0 1837.2 ThHTO.t 713612 1837.2 –450.5 1386.7 ThHTO.t 11 1500.5 238.6 +148.6 1887.7 HTO.t 999 110101010 2000.0 +1887.7 0112.3 126

11F 1.a. 51.2 × 4 512 × 4 = 2048 51.2 × 4 = 204.8

× 9 = 1134

12.6

9 1134

1235

123.5

c. 81.7 × 11 817 × 11 = 8987 81.7 × 11 = 898.7

3489

348.9

1235 ×

14

b. 2.6 × 23.5

26 × 235 = 6110

2.6

12 ×53 36 +600 636 235 ×26 1410 +4700 6110

123 × 5131 = 631113

12.3 × 51.31 = 631.113

× 123 15393 102620 +513100 631113

d. 2.36 × 1.24

236 × 124 = 29264

5.a. Product of 15.3 and 81.36

= 15.3 × 81.36

153 × 8136 = 1244808

15.3 × 81.36 = 1244.808

b. Product of 87.64 and 23.6

= 87.64 × 23.6

8764 × 236 = 2068304

87.64 × 23.6 = 2068.304 8136

Word Problems 1. The length of the rectangular garden = 5.8 feet

The width of the rectangular garden = 3.6 feet

The area of the rectangular garden

= Length × Width = 5.8 feet × 3.6 feet

58 × 36 = 2088

Area = 20.88 sq. feet

2. The quantity of milk purchased by Anuj = 3.6 gallons

1 gallons = 4 quarts

The amount of milk purchased by Anuj in quarts = 3.6 × 4 = 14.4 quarts

3. The cost of 1 computer = $190.64

The cost of 6 such computers = $190.64 × 6 = $1143.84

4. The length of the square shaped field = 50.23 m

The area of the square shaped field = Side × Side = 50.23 m × 50.23 m = 2523.0529 sq. m

5023 × 5023

3.a.

c.

4.a.

73.6 = 73.6         23.69 > 23.59

c. 338.52 ÷ 7 = 48.36; 12.4 × 3.9 = 48.36  48.36 = 48.36

5.a. 111.24 ÷ 9 = 12.36      b. 466.44 ÷ 5.2 =

Word Problems 1. The total amount of money with Bijoy = ₹1622.4

Number of friends that Bijoy has = 3

The amount of money that each friend gets = ₹1622.4 ÷ 3 = ₹540.8

2. The special sauce required for 10 servings = 0.75 cups

The special sauce required for 1 serving = 0.75 ÷ 10 = 0.075

The special sauce required for 35 servings = 0.075 × 35 = 2.625 cups

Thus, 35 servings require 2.625 cups of special sauce.

11H 1. The distance travelled by Sophie on day 1 = 15 km 3 m = 15 km + 3 1000 km = 15.003 km

The distance travelled by Sophie on day 2 = 26 km 408 m = 26 km + 408 1000 km = 26.408 km

The total distance travelled by Sophie = 15.003 km + 26.408 km = 41.411 km

2. Mary’s weight = 1.78 kg Poppy’s weight = 1.59 kg

The total weight of two cats = 1.78 kg + 1.59 kg = 3.37 kg

3. The cost of pencil = ₹2.5

The cost of eraser = ₹5.6

The cost of notebook = ₹34.55

The total cost of three items = ₹2.5 + ₹5.6 + ₹34.55 = ₹42.65

4. The total length of ribbon with Vrisha = 15 m

The length of ribbon used for decoration = 1.6 m

The length of ribbon used in the project = 8.75 m

The total length of the ribbon used = 1.6 m + 8.75 m = 10.35 m

The length of the ribbon left with Vrisha = 15 m – 10.35 m = 4.65 m

TO.th

9

4 1010 15.00

−10.35

04.65

5. The cost of 1 kg of apples = ₹96.65

The total weight of apples purchased = 15 kg

The total cost of 15 kg of apples = 15 × ₹96.65 = ₹1449.75

6. The distance covered in 4 hours = 200.8 km

The distance covered each hour = 200.8 km ÷ 4 = 50.2 km

7. The length of a rectangle = 11.56 m

The breadth of the rectangle = 1 2 of length = 1 2 × 11.56 m = 5.78 m

The perimeter of the rectangle = 2 × (Length + Breadth) = 2 × (11.56 m + 5.78 m)

= 2 × 17.34 m = 34.68 m

Area of the rectangle = Length × Breadth = 11.56 m × 5.78 m = 66.8168 sq. m

8. The amount of flour used = 0.25 kg

The amount of sugar used = 0.1 kg

The amount of cocoa powder used = 0.275 kg

The total weight of these items used to make 1 kg cake = 0.25 kg + 0.1 kg

+ 0.275 kg = 0.625 kg

The amount of milk used to bake 1 kg

cake = 1 kg – 0.625 kg = 0.375 kg

The total weight of milk used to make 5 kg cake = 5 × 0.375 kg

a. The shaded part is 0.5 0

b. The shaded part is 1.8

c. The shaded part is 1.51

3. a. 40 + 3 + 1 10 = 40 + 3 + 0.1 = 43.1

b. 900 + 50 + 4 10 + 3 100 = 900 + 50 + 0.4 + 0.03 = 950.43

c. 600 + 40 + 9 + 8 10 + 9 100 + 1 1000 = 600 + 40 + 9 + 0.8 + 0.09 + 0.001 = 649.891

d. 10 + 3 + 5 1000 = 10 + 3 + 0.005 = 13.005

4. a. 5.3 = 5 + 0.3 or 5 + 3 10

b. 81.9 = 80 + 1 + 0.9 or 80 + 1 + 9 10

c. 145.84 = 100 + 40 + 5 + 0.8 + 0.04 or 100 + 40 + 5 + 8

d. 5.23 = 5 + 0.2 + 0.03 or 5 + 2 10 + 3 100

e. 12.056 = 10 + 2 + 0.05 + 0.006 or 10 + 2 + 5 100 + 6 1000

f. 507.633 = 500 + 7 + 0.6 + 0.03 + 0.003 or 500 + 7 + 6 10 + 3 100

Compare the digit at tenths place.

2 < 3

Compare 1.23 with 1.2 and 1.331 with 1.303 by comparing the digit at hundredths place.

0 < 3               0 < 3

So, 1.2 < 1.23           So, 1.303 < 1.331

Thus, 1.331 > 1.303 > 1.23 > 1.2

c. 153.32, 153.23, 152.1, 153.233

The digits at the hundreds place and tens place is the same.

Compare the digit at ones place.

2 < 3

So, 152.1 is the smallest.

Compare the tenths digit in rest of the numbers.

2 < 3

So, 153.32 is the largest number.

Compare 153.23 and 153.233 by comparing the hundredths digit.

0 < 3

So, 153.23 < 153.233

Thus, 153.32 > 153.233 > 153.23 > 152.1

8. a. 15 × 1.6

15 × 16 = 240

Thus, 15 × 1.6 = 24

b. 418.4 ÷ 8 = 52.3

c. 1.2 × 6.7

12 × 67 = 804

Thus, 1.2 × 6.7 = 8.04

6. a.

=

d. 51.26 =

f. 12.508 = 12508

7. a. 15.1, 15, 15.05, 14.95

=

The tens place in all the numbers is the same.

Compare the digit at ones place.

4 < 5

So, 14.95 is the smallest.

Compare the tenths digit in rest of the numbers.

0 < 1

15.1 is the largest.

Compare the hundredths digit in 15 and 15.05

0 < 5

Thus, 15.1 > 15.05 > 15 > 14.95

b. 1.23, 1.2, 1.331, 1.303

The ones place is the same in all the numbers.

d. 5.9 × 47.6

59 × 476 =28084

Thus, 5.9 × 47.6 = 280.84

e. 492.26 ÷ 5 = 98.452

f. 14.65 × 165.7

1465 × 1657 = 2427505

Thus, 14.65 × 165.7 = 2427.505

Word Problems 1. The distance jumped by Rachna = 3.6 m

The distance jumped by Nina = 3.65 m

The distance jumped by Rohini = 3.55 m

The digit at the ones place is the same.

Compare the tenths digit.

6 > 5

Compare the hundredths digit in 3.6 and 3.65

5 > 0

So, 3.65 > 3.6

Thus, Nina jumped the farthest.

2. The total amount of the bill = $38.75

The amount of the tip = $10.50

The total amount spent = $38.75 + $10.50 = $49.25

Thus, Adhvica spent $49.25 in total.

3. The distance covered in the morning = 15.486 km

The distance covered in the afternoon = 8.75 km

The total distance covered = 15.486 km + 8.75 km = 24.236 km

Thus, the cyclist covers a total distance of 24.236 km.

4. The cost of the laptop = $899.75

The amount of money saved by Bhoomi = $550.50

The total amount of money required by Bhoomi to buy the laptop = $899.75 – $550.50 = $349.25

Thus, Bhoomi needs $349.25 more to buy the laptop.

5. The cost of the shirt = $24.99

The cost of the pair of shoes = $58.50

The total amount of money spent on shopping = $24.99 + $58.50 = $83.49

The total amount of money paid = $100

The amount of money left with Himani = $100 – $83.49 = $16.51

HTO.th

99 9 1010 1010

100.00

−83.49

16.51

Thus, Himani receives $16.51 as change.

6. The weight of the apples initially at the store = 65.75 kg

The weight of the apples sold = 42.5 kg

The weight of the apples left at the store = 66.75 kg – 42.5 kg = 24.25 kg

The weight of apples that is re−stocked at the store = 28.325 kg

The total weight of the apples at the store = 23.25 kg + 28.325 kg = 51.575 kg

Thus, the total weight of the apples at the store is 51.575 kg.

7. The total amount of money earned each hour by Sarah = $10.50

The number of hours worked in a week = 15.75 hours

The total amount of money earned = 15.75 × $10.50 = $165.375

Thus, the total money earned by Sarah over the week is $165.375

8. The length of the rectangular garden = 45.6 m

The width of the rectangular garden = 1 3 × 45.6 m = 15.2 m

The perimeter of the garden = 2 × (Length + Width) = 2 × (45.6 + 15.2) = 121.6 m

The area of the garden = Length × Width = 45.6 × 15.2 = 693.12 sq. m

Thus, the perimeter of the garden is 121.6 m and its area is 693.12 sq. m

9. The initial cost of 100 text messages = ₹45

The cost of each message of first 100 messages = ₹0.10 per message

The total cost of 275 messages = Cost of first 100 messages + Cost of (275 – 100) messages = Cost of first 100 messages + Cost of 175 messages = ₹45 + (175 × ₹0.10)

= ₹45 + ₹17.5

= ₹62.5

Thus, the cost of 275 messages is ₹62.5

10. The cost of type A toys = ₹129.90

The cost of type B toys = ₹97.50

The total toys sold = 60

The total amount of money earned = ₹7567.20

Let us say, the number of toys sold for type A are x, then the number of toys sold for type B are (60 –x).

The total revenue = Revenue from selling type A toys + Revenue from selling type B toys

7567.20 = (129.90 × x) + (97.50 × (60 – x))

7567.20 = 129.90x + 5850 – 97.50x

7567.20 – 5850 = 129.90x – 97.50x

1717.20 = 32.4x

x = 53

Number of toys of type B sold = 60 – 53 = 7

Thus, 53 type A toys were sold and 7 type B toys were sold.

Chapter 12

Let's Warm-up

1. 7 students like to have milk for breakfast.

2. Fruit is liked by the least.

3. Juice is liked by the most.

4. The total number of students who took the survey is 31.

5. 2 more students like to have poha for breakfast than fruit.

12A

b. Arranging the numerical numbers in ascending or descending order is called an array. True

c. Five as tally marks is represented as  . False

d. Frequency is always the same for every observation in data. False

2. a. The collection of numerical facts is called data.

b. In data, the number of times a particular observation is repeated is called its frequency

c. The data collected in its original form is called raw data.

3. a. 20

b. 31

c. 42

d. 27

4.

5.

a. 12 students like cricket.

b. 8 students like basketball.

c. The number of students who like cricket = 12

The number of students who like football = 11

Difference between the students who like cricket and who like football = 12 – 11 = 1

d. The number of students who like basketball = 8

The number of students who like volleyball = 8

Total number of students who like basketball and who like volleyball = 8 + 8 = 16

e. Total number of students for whom the data is collected = 12 + 11 + 8 + 10 + 8 + 7 = 56

Word Problems

1.

a. Digit 6 appeared the most.

b. Digit 5 appeared the least.

c. The total number of chances played by Rashmi = 9 + 12 + 9 + 12 + 7 + 13 = 62

d. Yes, digit 1 and digit 3 appeared the same number of times and digit 2 and digit 4 also appeared the same number of times.

2.  Age Number of Student Tally Marks

a. 7 students are 12 years old.

b. 6 students are 14 years old.

c. Students that are less than 13 years old = 7 + 3 + 7 = 17

d. Students that are more than 11 years old but less than 14 years old = 7 + 7 = 14

12B 1.  Car Bus Bicycle Bike Auto Walking

2.  Monday Tuesday Wednesday Thursday Friday

3.  Brand A Brand B

Key: = 1000 children

Key: = 5000 people

Key: = 500 soaps

4. 2510 rounded to the nearest 100 is 2500 and 1460 rounded to the nearest 100 is 1500.

Cricket Football

Basketball

Volleyball

Tennis

5.  Key: = 300 bulbs

Key: = 500 children

a. The number for symbols on Tuesday = 5

The number of bulbs sold on Tuesday = 5 × 300 = 1500 bulbs

b. The number of symbols on Wednesday = 11

The number of symbols on Thursday = 10

Difference between the number of symbols on Wednesday and Thursday = 1

The number of bulbs that were sold more on Wednesday than that on Thursday = 1 × 300 = 300 bulbs.

c. The number of bulbs sold on Monday = 8 × 300 = 2400 bulbs

The number of bulbs sold on Tuesday = 5 × 300 = 1500 bulbs

Total number of bulbs sold on Monday and Tuesday together = 2400 + 1500 = 3900 bulbs

d. Day on which there are least number of symbols = Tuesday

The number of bulbs sold on Tuesday = 5 × 300 = 1500 bulbs.

e. Total number of bulbs sold in the week = (8 × 300) + (5 × 300) + (11 × 300) + (10 × 300) + (15 × 300) = 2400 + 1500 + 3300 + 3000 + 4500 = 14,700 bulbs

6. = 1000 fans and = 500 fans

a. Number of symbols in May = 4-and-a half

Number of fans manufactured in May = (4 × 1000) + 500 = 4000 + 500 = 4500 fans

b. Number of symbols in January = 5-and-a half

Number of fans manufactured in January = (5 × 1000) + 500 = 5000 + 500 = 5500 fans

c. The month in which there are maximum number of symbols = June

The number of symbols in June = 8

The number of fans manufactured in June = 8 × 1000 = 8000 fans

d. The number of fans manufactured in April = 7 × 1000 = 7000

The number of fans manufactured in February = (6 × 1000) + 500 = 6000 + 500 = 6500

The number of fans that were manufactured more in April than in February = 7000 – 6500 = 500 fans

7. Key: = 100 baskets

a. Merchant for whom there are the maximum number of symbols = Surender

Number of symbols for Surender = 10-and-a half

Number of baskets sold by Surender = (10 × 100) + 1 2 × 100 = 1000 + 50 = 1050 baskets

b. Merchant for whom there are the minimum number of symbols = Rajul

The number of symbols for Rajul = 5-and-a half

The number of baskets sold by Rajul = (5 × 100) + 1 2 × 100 = 500 + 50 = 550 baskets

c. The number of symbols for Mohan = 6-and-a half

The number of baskets sold by Mohan = (6 × 100) + 1 2 × 100 = 600 + 50 = 650 baskets

d. Total number of symbols = 7 + 6 1 2 + 10 1 2 + 8 + 5 1 2 = 37 1 2

Total number of baskets sold = (37 × 100) + 1 2 × 100 = 3700 + 50 = 3750 baskets

Word Problem 1. Key: = 120 blankets

a. Person for whom there are the maximum number of symbols = Benu

The number of symbols for Benu = 7

The number of blankets distributed by Benu = 7 × 120 = 840 blankets

b. The number of symbols for Garima = 4

The number of blankets distributed by Garima = 4 × 120 = 480 blankets

c. The number of blankets distributed by Mallika = 5 × 120 = 600 blankets

The number of blankets distributed by Cherry = 3 × 120 = 360 blankets

The number of blankets that were distributed more by Mallika than Cherry = 600 – 360 = 240 blankets

d. The total number of blankets distributed = (7 × 120) + (5 × 120) + (3 × 120) + (6 × 120) + (4 × 120) + (5 × 120) = 840 + 600 + 360 + 720 + 480 + 600 = 3600 blankets

3. Scale: 1 division = ₹1000

a. Expenditure on food = ₹7000

b. The family spends the least on clothes.

Money spent on clothes = ₹6000

c. The family spends the most on rent.

Money spent on rent = ₹9000

d. Total money spent by family = ₹7000 + ₹6000 + ₹7500 + ₹9000 + ₹7750 = ₹37,250

4. Scale: 1 division = 5 saplings

a. Saplings planted in block C = 30 b. Saplings planted in block D = 32

c. The maximum number of saplings are placed in block B.

d. Total number of saplings planted 35 + 40 + 30 + 32 + 36 = 173

5. a. Population of Cottage in 1975 = 4000

b. Population of Arsenal in 2015 = 10,000

c. Population of Britannia in 1975 = 5000

Population of Britannia in 2015 = 20,000

Increase in population of Britannia = 20,000 – 5000 = 15,000

d. Increase in population of Cottage = 15,000 – 4000 = 11,000

Increase in population of Arsenal cities = 10,000 – 1000 = 9000

Increase in population of Britannia = 20,000 – 5000 = 15,000

Thus, Britannia has the greatest increase in population over the years. The number of people increased by 15,000.

6. a. The power supply from the generators was at its highest during the time 09:00-10:00.

b. The power supply from the generators was at its lowest during the time 08:00-09:00.

Word Problems

2. a. Population in 2013 = 1,00,000

Population in 2014 = 1,35,000

Increase in population from year 2013 to year 2014 = 1,35,000 –1,00,000 = 35,000

b. Year 2011 has the lowest population.

Population in year 2011 = 70,000

c. Year 2014 has the highest population. Population in year 2014 = 1,35,000

d. Increase in population in 2012 from 2011 = 80,000 – 70,000 = 10,000

Increase in population in 2013 from 2012 = 1,00,000 – 80,000 = 20,000

Increase in population in 2014 from 2013 = 1,35,000 – 1,00,000 = 35,000

Thus, the maximum population was in the year 2014.

2.

3. a. Maths is the most favourite subject among the students.

b. Number of students who like English = 10 + 4 = 14

c. Science is the least favourite subject among the students.

d. The number of students who like Maths = 17

Total number of students = 10 + 17 + 13 + 14 = 54

Fraction of students who like Maths = 17 54

4.  School A

School B School C School D School E

School F

5.  Sohail

Ashraf Mahmood Raju Vivek

Key: = 1000 mangoes

Key: = 1000 students

9. a. The favourite colour of students is blue. 180 students like blue.

b. The least favourite colour of students is green. 80 students like green.

c. The number of students who like black = 160

The number of students who like red = 110

Difference between the number of students who like black and who like red = 160 – 110 = 50

d. Total number of students = 90 + 80 + 110 + 180 + 160 = 620 students

10. a. The number of books sold in the year 2019 = 30,000

b. The number of books sold in the year 2018 = 40,000

c. The number of books sold in the year 2020 = 33,000

Total number of books over the years = 35,000 + 40,000 + 30,000 + 33,000 + 36,000 = 1,74,000

Fraction of total books that were sold in the year 2020 = 33000 174000 = 33 174 = 11 58

d. The number of books sold in the year 2017 = 35,000

The number of books sold in the year 2021 = 36,000

The ratio of the number of books sold in years 2017 and 2021 = 35000:36000 = 35:36

6. a. The number of symbols for the month of March = 4-and-a half

The number of ice creams sold in the month of March = (4 × 150) + 1 2 × 150 = 600 + 75 = 675 ice creams

b. The month for which there are the maximum number of symbols = May

The number of ice creams sold in the month of May = 8 × 150 = 1200 ice creams

c. Number of ice creams sold in the month of January = 3 × 150 = 450 ice creams

Number of ice creams sold in the month of May = 8 × 150 = 1200 ice creams

Difference between the number of ice creams sold in the month of January and May = 1200 – 450 = 750 ice creams

d. Total number of ice creams sold = (3 × 150) + (4 × 150) + (4 × 150 + 1 2 × 150) + (6 × 150 + 1 2 × 150) + (8 × 150) + (2 × 150 + 1 2 × 150) = 450 + 600 + 675 + 975 + 1200 + 375 = 4275 ice creams

Word Problem 1. a. June has the highest temperature. Temperature in the month of June = 40°C

b. Yes, January and December have the same temperatures. Temperature in the months of December and January = 5°C

c. Temperature in the month of September = 27°C

Months which are warmer than the month of September are June, July and August.

Thus, 3 months are warmer than the month of September.

d. The month which is colder than July but warmer than September is August.

Chapter 13

Let's Warm-up

1. 2. 3. 4. 5.

14 units26 units16 units28 units18 units

Do It Yourself

13A 1. a. 11 cm + 2 cm + 5 cm + 4 cm + 6 cm + 6 cm = 34 cm

b. 13 cm + 3.5 cm + 3.5 cm + 13 cm + 4 cm = 37 cm

c. 6 cm + 2 cm + 4 cm + 18 cm + 4 cm + 2 cm + 6 cm + 22 cm = 64 cm

2. Length = 30 cm; Breadth = 12 cm

Perimeter = 2 × (Length + Breadth) = 2 × (30 + 12) = 2 × 42 = 84 cm

3. Perimeter of the square figure = 164 cm

Perimeter of a square = 4 × side

4 × side = 164 side = 164 4 = 41 cm

Hence the length of the side of the square is 41 cm.

4. Length = 25 m; Breadth = 17 m

Perimeter = 2 × (Length + Breadth) = 2 × (25 + 17) = 2 × 42 = 84 cm

Cost of fencing at ₹28/m = 84 × 28 = ₹2352

5. Length = 15 cm; Breadth = 10 cm

Perimeter = 2 × (Length + Breadth) = 2 × (15 + 10) = 2 × 25 = 50 cm

Length of wood required = 50 cm

6. Length of string = 42 cm

a. An equilateral triangle has 3 equal sides

Perimeter of equilateral triangle = Length of string

3 × Length of side = 42 Length of side = 42 3 = 14 cm

b. Perimeter of regular heptagon = Length of string

7 × Length of side = 42 Length of side = 42 7 = 6 cm

7. Length = 3 m 25 cm = 3.25 m; Breadth = 1 m 42 cm = 1.42 m

Perimeter = 2 × (Length + Breadth) = 2 × (3.25 + 1.42) = 2 × 4.67= 9.34 m

8. Length of each side of the given figure = 4 cm

Number of sides in the given figure = 12

Perimeter of the given figure = 4 × 12 = 48 cm

Let the length of the side of decagon = x; Perimeter of decagon = 10 × x

Perimeter of decagon = 48 cm

10 × x = 48 x = 48 10 = 4.8 cm

Hence the length of each side of the decagon = 4.8 cm

9. Length of the sides of the octagonal dining table = 75 cm

Length of frame = Perimeter of octagon = 75 × 8 = 600 cm = 6 m

Cost of frame = ₹125/m

Cost of framing the table = Length of frame × Cost of frame = 6 × 125 = ₹750

Word Problems 1. Height of the wall = 262 cm; Width of wall = 185 cm

Length of masking tape = Perimeter of wall = 2 × (262 + 185) = 2 × 447 = 894 cm

2. Length of football field = 77 m; Breadth of the field = 52 m

Distance covered in one round = Perimeter of football field = 2 × (Length + Breadth) = 2 × (77 + 52) = 2 × 129 = 258 m

Distance covered in 2 and a half round = 258 × 2.5 = 645 m.

13B 1. a. Fully filled squares = 30

Area = 30 sq. cm

b. Fully filled squares = 8

Half-filled squares = 7

Area = 8 + 1 2 × 7 = 11.5 sq. cm

c. Fully filled squares = 34

More than half-filled squares = 13

Less than half-filled squares = 5

Area = 34 + 13 = 47 sq. cm

2. Length of side of square = 7.2 cm

Area = side × side = 7.2 × 7.2 = 51.84 sq. cm

3. Area of rectangle = 371.2 sq. cm; Breadth = 25.6 cm

Area of rectangle = Length × Breadth = Length × 25.6

Length × 25.6 = 371.2 Length = 371.2 25.6 = 14.5 cm

Length of the given rectangle = 14.5 cm.

4. To find the tilling cost, we need to find the area of the square plot.

Length of side of the square = 15 cm

Area of square = 15 × 15 = 225 sq. cm

Cost of tilling at the rate of ₹15 per sq. cm = 225 × 15 = ₹3375

5. To find the number of tiny squares created, we need to find the area of both the big and the tiny square.

Length of side of bigger square = 10 cm

Area of bigger square = 10 × 10 = 100 sq. cm

Length of tiny square = 2 cm

Area of tiny square = 2 × 2 = 4 sq. cm

Number of tiny squares that can be created

= 100 4 = 25

6. Area of rectangular frame = 1125 sq. cm; Width = 25 cm,

Area of rectangle = 1125 = Length × Breadth = Length × 25

Length = 1125 45 = 45 cm

7. Length of the rectangular plot = 450 m

Breadth of the plot = 375 m

Area of plot = Length × Breadth = 450 × 375 = 1,68,750 sq. m

Cost of tilling at ₹15/sq.m = 1,68,750 × 15 = ₹25,31,250

8.a. Perimeter = 20 cm + 4 cm + 8 cm + 8 cm + 8 cm + 4 cm + 20 cm +

4 cm + 8 cm + 8 cm + 8 cm + 4 cm = 104 cm

Area = Area of part A + Area of part B

+ Area of part C

Area of part A = 20 × 4 = 80 sq. cm

Area of part B = 8 × 4 = 32 sq. cm

Area of part C = 20 × 4 = 80 sq. cm

Total Area = 80 + 32 + 80 = 192 sq. cm

b. Perimeter = 3 + 3 + 3 + 3 + 6 + 3 +

3 + 3 + 3 + 3 + 3 + 3 + 3 + 6 = 48 units

Area = Area of part A + Area of part B + Area of part C + Area of part D

Area of part A = 3 × 3 = 9 sq. unit

Area of part B = 12 × 3 = 36 sq. unit

Area of part C = 3 × 3 = 9 sq. unit

Area of part D = 3 × 3 = 9 sq. unit

Total Area = 9 + 36 + 9 + 9 = 63 sq. unit

2 cm 10 cm 8 cm 8 cm 4 cm

8) = 4 cm 8 4 cm 8 cm 20 cm ABC 3 3 + 3 3 3 3 3 + 3 3 3 3

A B CD 3 3 3 134

9. Size of rectangular plot = 140 m by 30 m

Area of rectangular plot = 140 × 30 = 4200 sq. m

Size of square plot = 6400 cm = 64 m

Area of square plot = 64 × 64 = 4096 sq. m

Word Problem 1. Length of the room = 6 m

Breadth of the room = 7 m

Area of room = Length × Breadth = 6 × 7 = 42 sq. m

Cost of carpet at ₹250/sq.m = 42 × 250 = ₹10,500

13C 1. Perimeter of square = 4 × side = 90 m

Length of side = 90 4 = 22.5 m

Area of square = side × side = 22.5 × 22.5 = 506.25 sq. m

2. Area of rectangular kitchen = length × breadth = 130 sq. m

Given that, length of kitchen = 10 m

10 × breadth = 130 sq. m breadth = 130 10 = 13 m

Perimeter of kitchen = 2 × (10 + 13) = 2 × 23 = 46 m

3. Length of sides of squares kept side by side = 10.5 cm

Length of new figure formed = (10.5 + 10.5) = 21 cm

Breadth of new figure (rectangle) formed = 10.5 cm

Perimeter of figure formed = 2 × (length + breadth) = 2 × (21 + 10.5) = 2 × 31.5 = 63 cm

Area of figure formed = length × breadth = 21 × 10.5 = 220.5 sq. cm

4. Perimeter of the rectangular floor = 2 × (length + breadth) = 84 m

Length of one side say breadth = 24 m; 2 × (24 + breadth) = 84 m breadth = 84 2 – 24 = 42 – 24 = 18 m

Length of carpet required = area of floor = length × breadth = 24 × 18 = 432 sq. m

5. Given that the area of the square and rectangle are equal. Length of the side of square = 3 m 50 cm = 3.5 m; length of the rectangle = 5 m

Using the area formula; 3.5 × 3.5 = 5 × breadth

Breadth = 3.5 × 3.5 5 = 2.45 m

Perimeter of rectangle = 2 × (length + breadth) = 2 × (5 + 2.45) = 2 × 7.45 = 14.9 m.

6. Dimension of sticker to be pasted = 8 cm and 13 cm

Perimeter of region on which the stickers need to be pasted = 552 cm; Length of the region = 120 cm

Perimeter of the region = 2 × (120 + breadth) = 552 cm

Breadth of the region = 552 120 – 120 = 276 – 120 = 156 cm

To find the number of stickers to be pasted, we first need to find the area of the region,

Area of the region = length × breadth = 120 × 156 = 18,720 sq. cm

Area of sticker = 8 × 13 = 104

Number of stickers to be pasted = Area of region Area of stickers = 18,720 104 = 180

7. Area of the rectangular park = 3150 sq. m

Length of the park = 90 m

To find the number of rounds, we first need to find the perimeter of the park.

Area = Length × Breadth = 90 × Breadth = 3150

Breadth = 3150 90 = 35 m

Perimeter of park = 2 × (length + breadth) = 2 × (90 + 35) = 2 × 125 = 250 m

Distance covered by Rahul = 6 km = 6000 m

Number of rounds taken by Rahul = Total Distance covered Distance covered in one round = 6000 250 = 24 rounds

Word Problem 1. Length of the room = 6 m; Perimeter = 18 m

To find the cost of tilling, we need to first find the area of the room

Perimeter of room = 2 × (length + breadth) = 2 × (6 + breadth) = 18

Breadth = 18 2 – 6 = 9 – 6 = 3 m

Area of room = Length × Breadth = 6 × 3 = 18 sq. m

Cost of tilling at ₹50/sq. m = 18 × 50 = ₹900.

Chapter Checkup 1.a. Perimeter = 10 cm + 3 cm + 2 cm + 1 cm +

3 cm + 4 cm + 2 cm + 6 cm + 9 cm + 6 cm + 2 cm = 48 cm

b. Perimeter = 4 cm + 2.5 cm + 2 cm + 2.5 cm + 6 cm + 5 cm = 22 cm

c. Perimeter = 5 cm + 10 cm + 2 cm + 3 cm + 1 cm + 2 cm + 2 cm +

2 cm + 3 cm + 10 cm = 40 cm

2. Perimeter of the rectangle = 2 × (length + breadth) = 110 cm

Breadth = 25 cm;

2 × (length + 25) = 110 cm length = 110 2 – 25 = 55 – 25 = 30 cm

3. Area of the rectangular garden = 400 sq. m

Length of the garden = 50 m

Area = Length × Breadth = 50 × Breadth = 400

Breadth = 400 50 = 8 m

4. Length of thread = Perimeter of square

Perimeter of square = 4 × side = 70

Side = 70 4 = 17.5 cm

Length of each side = 17.5 cm

5. Length = 47 m

Perimeter = 4 × side = 4 × 47 = 188 m

Cost of fencing at the rate of ₹10 per sq. m = 188 × 10 = ₹1880

6. Length = 21.6 cm; Breadth = 18.8 cm

Area = Length × Breadth = 21.6 × 18.8 = 406.08 sq. cm

7. To find the number of squares tiles needed, we need to find the area of both squares.

Length of side of square = 3 m = 300 cm

Area of square = 300 × 300 = 90,000 sq. cm

Length of square tile = 12 cm

Area of square tile = 12 × 12 = 144 sq. cm

3 m = 300 cm

Number of square tiles needed = Area of square Area of square tile = 90,000 144 = 625

8. To find the length of wire required for fencing, we need to find the perimeter.

Length of side of square field = 0.8 km = 800 m

Perimeter = 4 × side = 4 × 800 = 3200 m

If each side is to be fenced with three rows of wire, the length of wire required = 3 × 3200 = 9600 m.

9. To find the cost of painting, we first need to find the area of the cardboard.

Length = 80 cm; Breadth = 1 4 m = 25 cm

Area = Length × Breadth = 80 × 25 = 2000 sq. cm

Cost of painting = 25 paise/sq. cm = ₹0.25/sq. cm

Cost of painting 2000 sq. cm = 2000 × ₹0.25 = ₹500

10.a. Fully filled squares = 15 b. Fully filled squares = 24

Half-filled squares = 7 Half-filled squares = 0

Area = 15 + 1 2 × 7 = 18.5 sq. unitsMore than half-filled squares = 6

c. Fully filled squares =21

Half-filled squares = 5

More than half-filled squares = 3

Less than half-filled squares = 7

Area = 24 + 6= 30 sq. units

Less than half-filled squares = 4

Area = 21 + 1 2 × 5 + 3 = 26.5 sq. units

11. Length of the wire = Perimeter of rectangle = Perimeter of square = 42 cm

Perimeter of rectangle formed = 2 × (length + breadth) = 42 cm

Breadth = 5 cm;

2 × (length + 5) = 42 cm length = 42 2 – 5 = 21 – 5 = 16 cm

Area of rectangle = length × breadth = 16 × 5 = 80 sq. cm

Perimeter of square formed = 4 × side = 42 cm

Length of side of square = 42 4 = 10.5 cm

Area of square = side × side = 10.5 × 10.5 = 110.25 sq. cm.

Difference in the area = (110.25 – 80) = 30.25 sq. cm

The square formed has a greater area. The difference in the area is 30.25 sq. cm.

12. Perimeter of Sunita’s plot = 2 m + 10 m + 8 m + 4 m + 6 m + 6 m = 36 m

Area of Sunita’s plot = Area of part A + Area of part B

Area of part A = 6 × 2 = 12 sq. m

Area of part B = 8 × 4 = 32 sq. m

Total Area = 12 + 32 = 44 sq. m

Perimeter of Ramesh’s plot = 10

= 48 m

Area of Ramesh’s plot =

Area of part A + Area of part B

Area of part A = 10 × 7 = 70 sq. m

Area of part B = 7 × 5 = 35 sq. m

Total Area = 70 + 35 = 105 sq. m

Thus, Ramesh’s plot has greater area and perimeter.

13. To find the number of tiles required, we need to find the area of both swimming pool and tile

Measure of swimming pool = 5 m × 4 m = 500 cm × 400 cm

Area of swimming pool = 500 × 400 = 2,00,000 sq. cm

Measure of tile = 20 cm × 10 cm

Area of tile = 20 × 10 = 200 sq. cm

Number of tiles required = 200000 200 = 1000

14. Length of square plot = 25 m

Area of the plot = side × side = 25 m × 25 m = 625 sq. m

Area used for farming = 1 4

Area vacant = 1 – 1 4 = 3 4

Area vacant = 3 4 of total area = 3 4 × 625 = 468.75 sq. m

Word Problems 1. Length of square park = 88 m

Perimeter of square park = 4 × side = 4 × 88 = 352 m

Distance while running 3 times a day = 3 × 352 = 1056 m

Length of rectangular park = 30 m; Breadth of rectangular park = 70 m

Perimeter of rectangular park = 2 × (length + breadth) = 2 × (30 + 70) = 200 m

Distance covered while running 5 times a day = 5 × 200 = 1000 m

Sailesh runs the longer distance every day.

2. To find the area of shaded portion, we need to subtract the area of the smaller rectangle from the area of the bigger rectangle. Length of bigger rectangle = 6 m; Width = 10 m

Area of bigger rectangle = 6 × 10 = 60 sq. m

Length of smaller rectangle = 10 – (4+1) = 5 m;

Width = 6 – (2 + 2) = 2 m

Area of smaller rectangle = 5 × 2 = 10 sq. m

Area of shaded portion = 60 sq. m – 10 sq. m = 50 sq. m. Therefore, Kanika needs 50 sq. m of carpet for her room.

Chapter 14

Let's Warm-up

Number patternNext number

1.

5. 4, 8, 12…

Do It Yourself

14A 1. a. 2x 2 × x Numerical Coefficient = 2

b. 4ab 4 × ab Numerical Coefficient = 4

c. 10pqr 10 × pqr Numerical Coefficient = 10

d. – xyz  – 1 × xyz Numerical Coefficient = –1

e. 1 3 x  1 3 × x Numerical Coefficient = 1 3

f. –3abc – 3 × abc Numerical Coefficient = –3

2.a. 2, 4, 6, 8, 10, …

First term of the sequence = (2 × 1) = 2

Second term of the sequence = (2 × 2) = 4

nth term of the sequence = (2 × n) = 2n

So, General rule = (2n)

18th term = (2 × 18) = 36

b. 3, 6, 9, 12, 15, …

First term of the sequence = (3 × 1) = 3

Second term of the sequence = (3 × 2) = 6

nth term of the sequence = (3 × n) = 3n

So, General rule = 3n

18th term = (3 × 18) = 54

c. 5, 10, 15, 20, …

First term of the sequence = (5 × 1) = 5

Second term of the sequence = (5 × 2) = 10

nth term of the sequence = (5 × n) = 5n

So, General rule = 5n

18th term = (5 × 18) = 90

d. 3, 8, 13, 18, 23, …

First term of the sequence = (5 × 1) – 2 = 3

Second term of the sequence = (5 × 2) – 2 = 8

nth term of the sequence = (5 × n) – 2 = 5n − 2

So, General rule = 5n − 2

18th term = (5 × 18) – 2 = 90 – 2 = 88

3. 1, 5, 9, 13, …

First term of the sequence = (4 × 1) – 3 = 1

Second term of the sequence = (4 × 2) – 3 = 5

nth term of the sequence = (4 × n) – 3 = 4n − 3

So, General rule = 4n − 3

10th term = (4 × 10) – 3 = 40 – 3 = 37

4.a. ShapeNumber of matchsticks Rule

Shape 13 = 3 × 1

Shape 26 = 3 × 2

Shape 39 = 3 × 3

Shape 412 = 3 × 4

Number of sticks required = (3 × n) = 3n

Number of matchsticks used in the 50th shape = (3n) = 3 × 350 = 150

b. ShapeNumber of matchsticks Rule

Shape 17 = 7 × 1

Shape 214 = 7 × 2

Shape 321 = 7 × 3

Number of sticks required = (7 × n)=7n

Number of matchsticks used in the 50th shape = (7n) = 7 × 50 = 350

5. a. 3, 5, 7, 9, …

First term of the sequence = (2 × 1) + 1 = 3

Second term of the sequence = (2 × 2) + 1 = 5

nth term of the sequence = (2 × 3) + 1 = (2 × n) + 1 = 2n + 1

So,

General rule = 2n + 1

b. 5, 8, 11, 14, …

First term of the sequence = (3 × 1) + 2 = 5

Second term of the sequence = (3 × 2) + 2 = 8

nth term of the sequence = (3 × n) + 2 = 3n + 2

So, General rule = 3n + 2

Word Problem 1. First term of the sequence = 10 × 1 = 10

Second term of the sequence = 10 × 2 = 20

nth term of the sequence = 10 × n = 10n

So, General rule = 10n

Hours12345678

Cupcakes 10 × 1 = 10 10 × 2 = 20 10 × 3 = 30 10 × 4 = 40 10 × 5 = 50 10 × 6 = 60 10 × 7 = 70 10 × 8 = 80

Number of hours in a day = 24 hours.

Number of cupcakes made in 24 hours = 10 × 24 = 240.

14B 1. a. Length of two sides is same in an isosceles triangle, hence Perimeter = 2a + b

Where 'a' is the length of the two equal sides and 'b' is the base of the triangle.

b. All the four sides of a rhombus are equal, hence Perimeter = 4 × a

Where 'a' is the length each side.

c. The pairs of adjacent sides of a kite are equal, hence 2 × a + 2 × b = 2(a + b)

Where 'a' is the length of each of the two equal sides and 'b' is the length of each the other two equal sides.

d. The pairs of opposite sides of a parallelogram are equal, hence 2 × l + 2 × b = 2(l + b)

Where P is the perimeter, l is length of each of two equal sides and b is the length of each of the other two equal sides.

2. A = l × b

Where A is the area, l is length of the rectangle and b is the breadth of the rectangle.

3. a. As the length of sides of a regular hexagon are equal and it has 6 sides, the perimeter can be given as: 6 × a = 6a; where a is the length of each side.

b. As the length of sides of a regular heptagon are equal and it has 7 sides, the perimeter can be given as: 7 × a = 7a, where a is the length of each side.

c. As the length of sides of a regular nonagon are equal and it has 9 sides, the perimeter can be given as: 9 × a = 9a, where a is the length of each side.

d. As the length of sides of a regular octagon are equal and it has 8 sides, the perimeter can be given as: 8 × a = 8a, where a is the length of each side.

4. Diameter = 2 × Radius

D = 2 × r = 2r

5. As the length of sides of a regular decagon are equal and it has 10 sides, the perimeter can be given as: 10 × a = 10a, where a is the length of each side.

Word Problem 1. The perimeter for a regular polygon of 12 equal sides can be given as 12 × a = 12a, where a is the length of each side of the regular polygon.

14C 1. a. 7 × 2 + 3 – 2 = 14 + 1 = 15; as the expression has only numerical terms, it is an arithmetic expression. b. 7x + 5; as the expression has variables, it is an algebraic expression.

c. 12y; as the expression has variables, it is an algebraic expression. d. 10y – 6x; as the expression has variables, it is an algebraic expression.

2. a. x and y b. 2a, 3b, –c c. x, y, 2z d. x, –6z

e. 2x, 5 f. 4 3 x, 2 g. 5abc, –2ab, 7ac h. 2ab, 4ac, –6c

3. a. 3 + 6m b. n – 10 c. 15 × x = 15x

d. 2 × (x × y) = 2xy e. (9 × y) + 1 = 9y + 1 f. 3 × y + (x – 3) = 3y + x – 3

4. a. x + y = y + x b. (x + y) + z = x + (y + z) c. a × b = b × a

d. (a × b) × c = a × (b × c)

5. a. y subtracted from 18. b. 4 added to thrice of x

c. 3 subtracted from p d. One-fourth of a

6. Number of pencils shared = x

Number of pencils left = 7

So, total number of pencils = x + 7.

7. Number of apples in one small box = x

Total number of apples in 3 small boxes = 3 × x = 3x

Number of apples left = 5

So, the total number of apples in the large box = 3x + 5.

8. Total number of students = y

Number of students who are performing in the group dance = x

Number of students who are performing in solo = 6 then, y = x – 6

Word Problems 1. Let Rima’s age be x Mala is five years younger than Rima.

Malas’s age = x – 5

2. Number of pens given to each student = 4

Number of students = n

Total number of pens given = 4 × n = 4n

Number of pens left = 5

Total number of pens the teacher had = 4n + 5

3. Number of chocolates with Karan = m

Number of chocolates given to his younger brother = 5

Number of chocolates left = m – 5

Number of chocolates each his friend and him got = m – 5 2

4. Number of rooms on the first floor = x

Number of rooms on the ground floor = (2 × x) – 12 = 2x – 12

5. Amount spent daily = ₹y

Amount spent weekly = ₹7 × y = ₹7y

Weekly expenditure = ₹7y

Amount saved weekly = Weekly Savings = ₹z

Income for one week = Expenditure + Savings = ₹ (7y + z)

Income for three weeks = 3 × Income for one week = 3 × ₹(7y + z) = ₹(3 × 7y + 3 × z) = ₹(21y + 3z)

6. Cost of tomatoes = ₹29 per kg

Cost of cucumbers = ₹38 per kg

Weight of tomatoes = x kg

Weight of cucumbers = (x – 4) kg

Total purchase = ₹[29 × x + 38 × (x – 4)] = ₹29 x + 38x – 152 = ₹67x – 152

14D 1. a. 15 – 6 = 9

Since, there are no variables in the equation; hence it is not a linear equation.

b. 2x + 5 = 7

The highest power of x in the equation is 1. 7 = 2x + 5

The equation remains the same if its LHS and RHS are interchanged.

Hence, it is a linear equation.

c. x2 + 1 = 5

Since, the highest power of x in the equation is 2; hence it is not a linear equation.

d. 12x < 28

Since, the highest power of x in the equation is 1. But, 28 < 12x

The equation is not separated by an equal to symbol.

Hence it is not a linear equation.

e. 4 + y = 9

Since, the highest power of y in the equation is 1. 9 = 4 + y

The equation remains the same if its LHS and RHS are interchanged.

Hence it is a linear equation.

f. 23 + y > 45

The highest power of y in the equation is 1. 45 > 23 + y

The equation is not separated by an equal to symbol. Hence it is not a linear equation.

2.a. Let the number be x

When we add 3 to the number: x + 3

We get 12: x + 3 = 12

b. Let the number be x

When 7 is taken away from a number: x – 7

It leaves x – 7 = 2

c. Let the number be x.

Twice of the number = 2x

When 7 is taken away from twice a number, it leaves 3: 2x – 7 = 3

We get 2x – 7 = 3

d. Let the number be x

Half a number: 1 2 x

14 less than half a number is 7: 1 2 x – 14 = 7

e. Let the number be x product of a number and 4: 4 × x = 4x

2 added to the product of a number and 4 gives 26: 4x + 2 = 26

f. Let Rohan’s age be x.

Rohan’s mother’s age = 37

The sum of Rohan’s age and that of his mother is 46 years: x + 37 = 46

3. A number divided by 4: x 4

Increased by 3 is 10: x 4 + 3 = 10

4. Number of artifacts he has = 193

Number of artifacts sold everyday = x

Number of artifacts sold in a week = 7 × x = 7x

Number of artifacts left after a week 46

This can be written as: 193 – 7x = 46

5. Let the numerator of the fraction be x Then, the denominator = x + 5

Thus, the fraction: x x + 5

When 3 is added to both, the fraction becomes 3 4 x + 3 x + 5 + 3 = 3 4

x + 3 x + 8 = 3 4

6. Total number of plums = T

Number of plums each friend gets = P

Number of friends = 6

Then, 6 × P = T

6P = T

7. Answers may vary. Sample answers:

a. When we add 5 to the product of 4 and a number, we get 9.

b. When we add 1 to twice a number, we get 5.

c. When 7 is taken away from the product of 12 and a number, we get 5.

d. Three times the sum of a number and 1 gives 12.

Word Problems 1. Number of stamps with Badri = x

Number of stamps with Jai = x + 29

Total stamps = x + x + 29 = 73 2x + 29 = 73

2. Number of people that signed up for the soccer league = 72

Number of teams = 6

Number of people in each team = x 6x = 72

3. Let the number Tina thought of was x

When tripled = 3x

Subtracting 7 from the result = 3x – 7

Multiplying by 2 = 2 × (3x – 7) = 52

14E 1. a. If x = 1

LHS = 3x + 5 = 3(1) + 5 = 8

RHS = 8

LHS = RHS

Hence x = 1 is the solution of the given expression.

c. If x = 1

LHS = 32 – 5x = 32 – 5(1) =

32 – 5 = 27

RHS = 12

LHS RHS

If x = 2

LHS = 32 – 5x = 32 – 5(2) =

32 – 10 = 22

RHS = 12

LHS RHS

If x = 4

LHS = 32 – 5x = 32 – 5(4) =

32 – 20 = 12

RHS = 12

LHS = RHS

Hence x = 4 is the solution of the given expression.

b. If x = 0

LHS = 2(3x + 1) = 2(3(0) + 1) = 2(0 + 1) = 2(1) = 2

RHS = 8

LHS RHS

If x = 1

LHS = 2(3x + 1) = 2(3(1) + 1) = 2(3 + 1) = 2(4) = 8

RHS = 8

LHS = RHS

Hence x = 1 is the solution of the given expression.

d. If x = 1

LHS = 5x – 8 = 5(1) – 8 = 5 – 8 = –3

RHS = 7

LHS RHS

If x = 2

LHS = 5x – 8 = 5(2) – 8 = 10 – 8 = 2

RHS = 7

LHS RHS

If x = 3

LHS = 5x – 8 = 5(3) – 8 = 15 – 8 = 7

RHS = 7

LHS = RHS

Hence x = 3 is the solution of the given expression.

e. If y = 1

LHS = 8y = 8(1) = 8

RHS = 48

LHS RHS

If y = 3

LHS = 8y = 8(3) = 24

RHS = 48

LHS RHS

If y = 6

LHS = 8y = 8(6) = 48

RHS = 48

LHS = RHS

Hence y = 6 is the solution of the given expression.

2. a. If a = 10

LHS = 6 a = 6 × 10 = 60

RHS = 48

LHS RHS If a = 8

LHS = 6a = 6 × 8 = 48

RHS = 48

LHS = RHS

b. If x = 7

LHS = 3x + 2 = 3(7) + 2 = 21 + 2 = 23

RHS = 23

f. If x = 1

LHS = 3 4 x – 3 = 1 3 – 3 = 1 – 9 3 = –8 3

RHS = 0

LHS RHS

If x = 3

LHS = 1 3 x – 3 = 1 3 (3) – 3 = 1 – 3 = –2

RHS = 0

LHS RHS

If x = 9

LHS = 1 3 x – 3 = 1 3 (9) – 3 = 9 3 – 3 =

3 – 3 = 0

RHS = 0

LHS = RHS

Hence x = 9 is the solution of the given expression.

c. If y = 3

LHS = y 4 = 3 4

RHS =3

LHS RHS

If y = 7

LHS = y 4 = 7 4

RHS =3

LHS RHS

If y = 12

LHS = y 4 = 12 4 = 3

RHS = 3

LHS = RHS

3. a. x 12345678

So, for x = 7, x – 3 = 4.

b. x 246810121416

So, for x = 14, x 2 = 7.

4.a. If a = 5

LHS = 4a = 4 × 5 = 20

RHS = 20 Value = 5

LHS = RHS

Hence a = 5 is the solution of the given expression.

b. If x = 3

LHS = 2x + 5 = 2 × 3 + 5 =

6 + 5 = 11

RHS = 11

LHS = RHS

Hence x = 3 is the solution of the given expression.

c. If m = 9

LHS

= 80

RHS = 8

LHS RHS

Then, if m = 1

LHS = 9m – 1 =

= 8

RHS = 8

LHS = RHS

Hence m = 1 is the solution of the given expression.

5. a. If x = 2

LHS = 4x + 2 = 4 × 2 + 2 = 8 + 2 = 10

RHS = 10

LHS = RHS

Hence x = 2 is the solution of the given expression.

b. If y = 3

LHS = 9y – 8 = 9 × 3 – 8 = 27 – 8 = 19

RHS = 1

LHS RHS

9y – 8 = 1

9y = 1 + 8

9y = 9 y = 9 9

y = 1

Hence y = 1 is the solution of the given expression.

c. If x = 10

LHS = 5 2 x = 5 2 × 10 = 50 2 = 25

RHS = 25

LHS = RHS

Hence x = 10 is the solution of the given expression.

e. If z = 1

LHS = 5z – 6 = 5 × 1 – 6 = 5 – 6 = –1

RHS = 0

LHS RHS

5z – 6 = 0

5z = 6

z = 6 5

Hence z = 6 5 is the solution of the given expression.

6.a. x + 6 = 10

x = 10 – 6

x = 4

Hence x = 4 is the solution of the given expression.

c. m – 12 = 3

m = 3 + 12

m = 15

Hence m = 15 is the solution of the given expression.

e. 4 + k – 7 = 2

k + 4 – 7 = 2

k – 3 = 2

k = 2 + 3

k = 5

Hence k = 5 is the solution of the given expression.

7. Let the number be x 4 times the number = 4x 3 taken away = 4x − 3 = 9

4x − 3 = 9

4x = 9 + 3

4x = 12

x = 12

4 = 3

Hence the number is 3.

d. If n = 12

LHS = 2n 3 – 6 = 2 × 12 3 – 6 = 24 3 –

6 = 8 – 6 = 2

RHS = 2

LHS = RHS

Hence n = 12 is the solution of the given expression.

f. If x = 1

LHS = 3x + 2 = 3 × 1 + 2 = 3 + 2 = 5 RHS = 5

LHS = RHS

Hence x = 1 is the solution of the given expression.

b. 5 = a + 2

a + 2 = 5

a = 5 – 2

a = 3

Hence a = 3 is the solution of the given expression.

d. –7 = x + 4

x + 4 = –7

x = –4 –7

x = –11

Hence x = –11 is the solution of the given expression.

f. 3 – 4x = –13

3 + 13 – 4x = 0

16 = 4x

x = 16 4

x = 4

Hence x = 4 is the solution of the given expression.

8. Let one number be x then the other number will be x + 1.

Sum = 25

x + x + 1 = 25

2x = 25 − 1

2x = 24

x = 24

2

x = 12

x + 1 = 12 + 1 = 13

Hence the numbers are 12 and 13

9. Breadth of the rectangular field = x

Length of the rectangular field = 2x

Perimeter = 228 m = 2 × (2x + x)

= 2 × 3x = 6x

6x = 228

x = 228

6 x = 38 m

2x = 2 × 38

2x = 76 m

Hence, the length is 76 m and the breadth is 38 m.

Word Problems 1. Let the daughter’s age be x 3 times of daughter’s age = 3x 3 years more than 3 times = 3x + 3

Sahil’s age = 3x + 3 = 30

3x + 3 = 30

3x = 30 – 3

3x = 27

x = 27

3 x = 9

So, the age of his daughter is 9 years old.

2. Let the marks scored in English = x

Marks scored in Mathematics = 2x

Marks scored in Science = 70

Total marks = 220

Then, x + 2x + 70 = 220

x + 2x = 220 – 70

3x = 150

x = 150

3 x = 50

2x = 2 × 50

2x = 100

So, Sreeja scored 100 marks in Mathematics and 50 marks in English.

3. Let Romit’s age be x Rahul’s age = 2x

Difference = 2x – x = 9 2x – x = 9

Chapter Checkup

10. Let Ishita’s present age be Rekha’s present age = x + 24

After 8 years:

Ishita’s age = x + 8

Rekha’s age = x + 24 + 8 = x + 32

Also, 2 × (x + 8) = x + 32

2 × x + 2 × 8 = x + 32

2x + 16 = x + 32

2x − x = 32 − 16

x = 16

x + 24 = 16 +24 = 40

Thus, Ishita’s present age is 16 and Rekha’s present age is 40.

x = 9

2x = 2 × 9 = 18

So, Romit is 9 years old and Rahul is 18 years old.

4. Let the purse in Meenal’s purse is ₹ x

If she adds ₹51 to it, it becomes four times the original amount:

x + 51 = 4x

51 = 4x – x

51 = 3x

51 3 = x

x = 17

So, the original amount in her purse is ₹17.

5. Let the son’s age be x

Amit’s age = 4x

Six years ago:

Son’s age = x – 6

Amit’s age = 4x – 6

And, 4x – 6 = 10 × (x – 6)

4x – 6 = 10x – 60

60 – 6 = 10x – 4x

54 = 6x

x = 54

6

x = 9

4x = 4 × 36 = 36

So, the age of son is 9 years and the age of Amit is 36 years.

1. a. Number of matchsticks Rule

4 = 4 × 1

Number of sticks required = (4 × n) = 4n 8 = 4 × 2

12 = 4 × 3

b. Number of matchsticks Rule

6 = 4 × 1 + 2

10 = 4 × 2 + 2

14 = 4 × 3 + 2

c. Number of matchsticks Rule

5 = (4 × 1 + 1)

9 = (4 × 2 + 1)

13 = (4 × 3 + 1)

2. Let the number be x

Then, 0 + x = x

And x + 0 = x

So, 0 + x = x = x + 0

4.a. 3 times x = 3x

Adding 4 = 3x + 4

c. Quotient of x by 3 = x 3

Subtracting 4 = x 3 – 4

5. a. 2x + 5 = 9

Number of sticks required = (4 × n + 2) = 4n + 2

Number of sticks required = (4 × n + 1) = 4n + 1

3. Let the numbers be x, y and z Then (x × y) × z = x × (y × z)

b. 2 times y = 2y

Subtracting 2 = 2y – 2

d. y is divided by 5 = y 5 Quotient is added to 6 = 6 + y 5

The highest power of x in the equation is 1.

9 = 2x + 5

The equation remains the same if its LHS and RHS are interchanged.

Hence it is a linear equation.

b. 9x + 4 > 5

The highest power of x in the equation is 1.

5 < 9x + 4

The equation does not remain the same if its LHS and RHS are interchanged.

Hence it is not a linear equation.

c. 2y + 7 = 9

The highest power of y in the equation is 1.

9 = 2y + 7

The equation remains the same if its LHS and RHS are interchanged. Hence it is a linear equation.

6. a. If x = 1

LHS = 3x – 5 = 3 – 5 = –2

RHS = 4

LHS RHS

Equation is not satisfied.

3x – 5 = 4

3x = 4 + 5

3x = 9

x = 9 5

x = 3

b. If y = 2

LHS = 4y + 3 = 8 + 3 = 11

RHS =11

LHS = RHS

Equation is satisfied.

c. If x = 4

LHS = x 7 = 4 7

RHS = 28

LHS RHS

Equation is not satisfied.

7. Lengths of the sides of the triangle:

3x – 5

2x – 1 x + 1

a. Perimeter = 3x – 5 + 2x – 1 +

x + 1

= 3x + 2x + x – 5 – 1 + 1

=

x 7 = 28

x = 28 × 7

x = 196

Hence x = 196 is the solution of the given expression.

d. If n = 12

LHS = n 6 – 12 = 12 6 – 12 = 2 – 12 = –10

RHS = 0

LHS RHS

Equation is not satisfied.

n 6 – 12 = 0

n – 12 × 6

6 = 0

n – 72

6 = 0

n – 72 = 0

n = 72

Hence n = 72 is the solution of the given expression.

b. If perimeter = 31 cm, then 6x – 5 = 31

6x = 31+5

6x = 36

x = 36

6

x = 6

Hence x = 6 cm is the solution of the given expression.

8. Length of the rectangle = 2x + 3

a. Width of the rectangle = x + 5

Perimeter of the rectangle = 2 × (2x + 3 + x + 5) = 2 × (3x + 8)

9.a. When 30 is added to a number, we get 50.

x + 30 = 50

x = 50 – 30

x = 20

Hence x=20 is the solution of the given expression.

b. When 3 is taken away from 6 times a number we get 9.

6x – 3 = 9

6x = 9 + 3

6x = 12

x = 12

6

x = 2

Hence x = 2 is the solution of the given expression.

b. Perimeter = 43 cm = 2 × (3x + 8)

2 × (3x + 8) = 43

6x + 16 = 43

6x = 43 – 16

6x = 27

x = 27

6

x = 4.5 cm

Hence x = 4.5 is the solution of the given expression.

c. When 24 is divided by a number, we get 4.

24

n = 4

24 = 4 × n

24

4 = n n = 6

Hence n = 6 is the solution of the given expression.

d. The product of 4 and the sum of 3 times a number and 7 gives 40.

4(3x + 7) = 40

4 × 3x + 4 × 7 = 40

12x + 28 = 40

12x = 40 – 28

12x = 12

x = 12 12

x = 1

Hence x = 1 is the solution of the given expression.

10.  Let the number of marbles with Ajit = x

Number of marbles with Badri = 2x

Number of marbles with Charu = 2x + 5

Total number of marbles = x + 2x + 2x + 5 = 5x + 5 = 55

Then, 5x = 55 – 5

x = 50/5 = 10

x = 10

2x = 20

2x + 5 = 25

So, Charu has 25 marbles.

Word Problems 1. Present age of Rohit = z

a. Age after 8 years = z + 8 b. Age 5 years ago = z – 5

c. Grandmother’s age = 5z + 9 d. Grandfather’s age = 5z + 10

2. Number of drawing sheets with Sahil = x

Number of drawing sheets with Rahul = x – 2

Number of drawing sheets with Aryan = x + 6

Number of sheets Rahul and Aryan have in total = x – 2 + x + 6 = 2x + 4

Total number of sheets that the three of them have altogether = x + x – 2 + x + 6 = 3x + 4

3. Height of cuboidal box = x

Breadth of cuboidal box = 2x

Length of cuboidal box = 2x – 4

4. Price of a muffin = ₹y

Price of a cake = 10 × ₹y = ₹10y

Price of a cookie = ₹ (2y – 7)

5. Number of tennis balls ordered = 42

Number of tennis balls in one package = 3

Number of packages = x

Then, 3 × x = 42

3x = 42

6. Price of the bracelet = P = ₹115

2.5m + 35 = 115

2.5m = 115 – 35

2.5m = 80

m = 80 2.5 = 800 25 = 32

m = 32

so, the cost of materials is ₹32.

7. Let Akash’s age be x Lakhan’s age = 3x Mala’s age = 3x – 7

Sum of their ages = x + 3x + 3x – 7 = 7x – 7 = 126

7x – 7 = 126

7x = 126 + 7 = 133

x = 133 7 = 19

So, the age of Akash is 19 years old.

8. Number of sweets with Anand = x

Number of sweets with Bhanu = x + 7

Number of sweets with Kirti = 2 × (x + 7)

a. x + 7

b. 2 × (x + 7) = 2 × x + 2 × 7 = 2x + 14

Chapter 15

Let's Warm-up 1. Number of apple slices = 6 Slices eaten = 3

Fraction of apple eaten = 3 6 = 1 2

2. 8 12 = 8 ÷ 4 12 ÷ 4 = 2 3

8 12 in its simplest form is 2 3

3. 2 5 = 2 × 2 5 × 2 = 4 10

An equivalent for 2 5 is 4 10

4. As the denominator is the same, 4 7 and 6 7 are like fractions.

4

7 and 6 7 are a pair of like fractions.

5. Fraction of cake left = 1 – 4 16 = 12 16 = 3 4

The fraction of cake left is 3 4 .

Do It Yourself

15A 1.a. In a ratio a:b, the quantities a and b are referred to as the terms of the ratio.

b. The first term in a ratio is also known as the antecedent.

c. The second term in a ratio is also known as the consequent.

d. If the order of terms in a ratio changes, the ratio changes.

e. A ratio has no unit because it represents a comparison of quantities of the same units.

2.a. A ratio can have units, depending on the types of quantities being compared. - False

b. In 3:5, the consequent is 5. - True

c. The terms of a ratio a:b are called the antecedent and consequent respectively. - True

d. 10:2 is the same as 2:10. False

e. 10 litres of water can be compared with 10 km. – False

3. Number of red apples = 16; Number of green apples = 24 Ratio of red apples to green apples = 16:24

4. Number of cats = 20; Total number of animals = 30 + 20 + 10 = 60

Ratio of cats to total animals = 20:60

5. Number of boys = 25; Total number of students = 25 + 22 = 47

Ratio of boys to total students = 25:47

6. Number of roses = 15; Number of tulips = 10; Number of lilies = 5

a. Ratio of number of roses to tulips = 15:10

b. Ratio of number of lilies to roses = 5:15

c. Ratio of number of tulips to roses and lilies = 10:(15 + 5) = 10:20

7. Number of chocolates = 375; Total number of sweets = 255 + 375 = 630

Ratio of chocolates to total sweets = 375:630

8. Number of red marbles = 14; Marbles that are not black =

Number of red marbles + Number of green marbles = 14 + 12 = 26

Ratio of red marbles to marbles not black = 14:26

Word Problem 1. Number of tigers = 3; Number of bears and giraffe = 10 + 5 = 15

Ratio of tigers to bears and giraffe = 3:15

15B 1.

a. 9 : 39 46 : 37

b. 120 : 80 3 : 20

c. 138 : 111 18 : 5

d. 42 : 280 3 : 13

e. 378 : 105 15 : 1

f. 480 : 32 3 : 2

2. a. HCF of 48 and 72 = 24

48

72 = 48 ÷ 24

72 ÷ 24 = 2 3

Thus, 48:72 in its simplest form is 2:3

c. 21000 m = 21 km

HCF of 14 and 21 = 7

14

21 = 14 ÷ 7

21 ÷ 7 = 2 3

Thus, 14 km:21000 m in its simplest form is 2:3

e. 25 L = 25000 mL

HCF of 25000 and 750 = 250

25000

750 = 25000 ÷ 250

750 ÷ 250 = 100 3

Thus, 25 L:750 mL in its simplest form is 100:3

g. 9 hours = 9 × 60 = 540 minutes

b. 6 m = 6 × 100 = 600 cm

HCF of 400 and 600 = 200

400

600 = 400 ÷ 200

600 ÷ 200 = 2 3

Thus, 400 cm:6 m in its simplest form is 2:3

d. 1.5 kg = 1500 g

HCF of 1500 and 750 = 750

1500

750 = 1500 ÷ 750

750 ÷ 750 = 2 1

Thus, 1.5 kg:750 g in its simplest form is 2:1

f. 18 m = 1800 cm

HCF of 1800 and 25 = 25

1800 25 = 1800 ÷ 25

25 ÷ 25 = 72 1

Thus, 18 m:25 cm in its simplest form is 72:1

9 hours:540 minutes = 540 minutes:540 minutes = 540:540 = 1:1

Thus 9 hours:540 min in its simplest form is 1:1

h. 300 sec = 300 60 = 5 minutes

HCF of 15 and 5 = 5

15 5 = 15 ÷ 5 5 ÷ 5 = 3 1

Thus, 15 min:300 sec in its simplest form is 3:1

i. 10000 paise = 10000 100 = ₹100

₹100 :10000 paise = ₹100:₹100 = 1:1

Thus, ₹100 :10000 paise in its simplest form is 1:1

3. Amount of water = 18 litres; Amount of juice = 12 litres

Ratio of water to juice = 18:12 = 3:2

4. 24 hours = 1 day

Ratio of 24 hours to 2 days = Ratio of 1 day to 2 days = 1:2

5. HCF of two numbers = 5

Simplest form of the ratio = 3:4

The given numbers are, 3 × 5 = 15 and 3 × 4 = 20

6. Number of boys = 15; Number of girls = 20

Ratio of girls to boys = 20:15 = 20 15

HCF of 20, 15 = 5

20 ÷ 5

15 ÷ 5 = 4 3

Hence the ratio of girls to boys is 4:3.

7. Cost of 2 dozen (24) red apples = ₹360; cost of 1 red apple = 360 24 = ₹15

Cost of 8 green apples = ₹240; cost of 1 green apple = 240 8 = ₹30

Ratio of cost of 1 green apple to 1 red apple = 30:15 = 2:1

8. Length of rectangular field = 80 m; Width of field = 60 m

Perimeter of field = 2 × (l + b) = 2 × (80 + 60) = 280 m

Ratio of length to perimeter = 80:280 = 2:7

Word Problem 1. Starting time of school = 8:00 a.m; Finish time of school = 2:00 p.m

Total interval of school = 6 hours = 6 × 60 = 360 minutes

Lunch break time = 30 minutes

Ratio of lunch break to total hours of school = 30:360 = 1:12

15C 1. Answers may vary, sample answer.

a. 5:7 = 5 × 2 7 × 2 = 10 14 ; 5 × 3 7 × 3 = 15 21 ; 5 × 4 7 × 4 = 20 28

Three equivalent ratios to 5:7 = 10:14, 15:21, 20:28

b. 4:9 =

Three equivalent ratios to 4:9 = 8:18, 12:27, 16:36

c. 12:16 = 12 × 2 16 × 2 = 24 32 ; 12 × 3 16 × 3 = 36 48 ; 12 × 4 16 × 4 = 48 64

Three equivalent ratios to 12:16 = 24:32, 36:48, 48:64

d. 6:15 = 6 × 2 15 × 2 = 12 30 ; 6 × 3 15 × 3 = 18 45 ; 6 × 4 15 × 4 = 24 60

Three equivalent ratios to 6:15 = 12:30, 18:45, 24:60

e. 8:11 = 8 × 2 11 × 2 = 16 22 ; 8 × 3 11 × 3 = 24 33 ; 8 × 4 11 × 4 = 32 44

Three equivalent ratios to 8:11 = 16:22, 24:33, 32:44

f. 9:17 = 9 × 2 17 × 2 = 18 34 ; 9 × 3 17 × 3 = 27 51 ; 9 × 4 17 × 4 = 36 68

Three equivalent ratios to 9:17 = 18:34, 27:51, 36:68

g. 11:25 = 11 × 2 25 × 2 = 22 50 ; 11 × 3 25 × 3 = 33 75 ; 11 × 4 25 × 4 = 44 100

Three equivalent ratios to 11:25 = 22:50, 33:75, 44:100

h. 13:16 = 13 × 2 16 × 2 = 26 32 ; 13 × 3 16 × 3 = 39 48 ; 13 × 4 16 × 4 = 52 64

Three equivalent ratios to 13:16 = 26:32, 39:48, 52:64

2.a. 5:3 = 5 3 ; 3:4 = 3 4

LCM of 3 and 4 = 12

5 3 = 5 × 4 3 × 4 = 20 12 ; 3 4 = 3 × 3 4 × 3 = 9 12

As 20 12 > 9 12 5 3 > 3 4 or 5:3 > 3:4

c. 4:7 = 4 7 ; 5:8 = 5 8

LCM of 7 and 8 = 56

4 7 = 4 × 8 7 × 8 = 32 56 ; 5 8 = 5 × 7 8 × 7 = 35 56

As 32 56 < 35 56 4 7 < 5 8 or 4:7 < 5:8

e. 3:7 = 3 7 ; 7:9 = 7 9

LCM of 7 and 9 = 63

3 7 = 3 × 9 7 × 9 = 27 63 ; 7 9 = 7 × 7 9 × 7 = 49 63

As 27 63 < 49 63 3 7 < 7 9 or 3:7 < 7:9

g. 7:11 = 7 11 ; 5:9 = 5 9

LCM of 11 and 9 = 99

7 11 = 7 × 9 11 × 9 = 63 99 ; 5 9 = 5 × 11 9 × 11 = 55 99

As 63 99 > 55 99 7 11 > 5 9 or 7:11 >

3. Given ratios = 2 5 , 3 7 , 5 14 , 7 10

LCM of 5, 7, 14 and 10 = 70

b. 3:5 = 3 5 ; 2:9 = 2 9

LCM of 5 and 9 = 45

3 5 = 3 × 9

;

d. 3:5 = 3 5 ; 5:6 = 5 6

LCM of 5 and 6 = 30

3

5 = 3 × 6 5 × 6 = 18 30 ; 5 6 = 5 × 5 6 × 5 = 25 30

As 18 30 < 25 30 3 5 < 5 6 or 3:5 < 5:6

f. 2:7 = 2 7 ; 4:14 = 4 14 = 2 7

Hence, 2 7 = 4 14 or 2:7 = 4:14

h. 8:14 = 8 14 = 4 7 ; 4:7 = 4 7

Hence, 8 14 = 4 7 or 8:14 = 4:7

i. 12:15 = 12 15 = 4 5 ; 18:24 = 18 24 = 3 4

LCM of 5 and 4 = 20

4 5 = 4 × 4 5 × 4 = 16 20 ; 3 4 = 3 × 5 4 × 5 = 15 20

As 16 20 > 15 20 12 15 > 18 24 or 12:15 > 18:24

The given ratios can be arranged in ascending order as: 25:70 < 28:70 < 30:70 < 49:70

Thus, 5:14 < 2:5 < 3:7 < 7:10

4. Ratio in which the colours are mixed = 1:2

Quantity of first colour =3 litres

Quantity of second colour = 3 × 2 = 6 litres

5. Ratio of flour to sugar = 3:2

Given that the quantity of sugar is 600 g

So 2x = 600 x = 600 2 = 300

Amount of flour required = 3x = 3 × 300 = 900 grams.

6. Ratio of blue marbles to red marbles = 4:7

Total marbles = 132

Number of blue marbles = 4 11 × 132 = 48

Hence there are 48 blue marbles in the bag.

7. Ratio of story books and comic books = 5:4

Total number of books = 36

Number of story books = 5 9 × 36 = 20

Number of comic books = 4 9 × 36 =16

Hence there are 20 storybooks and 16 comic books.

8. Ratio of kids to adults = 5:3

Number of adults = 3x = 18 x = 18 3 = 6

Number of kids in the party = 5x = 5 × 6 = 30

9. Ratio of ages of Manan and Manish = 28:49 = 4:7

Amount that father wants to divide = ₹1650

Share of Manan = 4 11 × 1650 = ₹600

Share of Manish = 7 11 × 1650 = ₹1050

Hence Manan and Manish got ₹600 and ₹1050 respectively.

10. Ratio in which spending and saving is split = 2:3

Amount saved each month = 3x = ₹6000 x = 6000 3 = 2000

Amount spent = 2x = 2 × 2000 = ₹4000

Hence, Sneha spends ₹4000 each month

11. Ratio of students who like maths to science = 5:3

Number of students who like maths = 5x = 40 x = 40 5 = 8

Number of students who like science = 3x = 3 × 8 = 24

Hence 24 students like science.

Word Problem 1. Ratio of runs scored by team A and team B = 5:3

Runs scored by team B = 3x = 120 x = 120 3 = 40

Runs scored by team A = 5x = 5 × 40 = 200

Hence team A scored 200 runs.

15D 1. a. 4:5 = 4 5 ; 12:15 = 12 15 = 12 ÷ 3 15 ÷ 3 = 4 5

Hence the ratios are in proportion.

b. 8:5 = 8 5 ; 9:6 = 9 6 = 9 ÷ 3 6 ÷ 3 = 3 2

Hence the ratios are not in proportion.

c. 15:5 = 15 5 = 15 ÷ 5 5 ÷ 5 = 3:1 ; 12:4 = 12 4 = 12 ÷ 4 4 ÷ 4 = 3:1

Hence the ratios are in proportion.

d. 9:5 = 9 5 ; 18:10 = 8 10 = 18 ÷ 2 10 ÷ 2 = 9:5

Hence the ratios are in proportion.

e. 12:6 = 12 6 = 12 ÷ 6 6 ÷ 6 = 2:1 ; 8:4 = 8 4 = 8 ÷ 4 4 ÷ 4 = 2:1

Hence the ratios are in proportion.

f. 10:4 = 10 4 = 10 ÷ 2 4 ÷ 2 = 5 2 ; 20:8 = 20 8 = 20 ÷ 4 8 ÷ 4 = 5:2

Hence the ratios are in proportion.

2.a. 17:8 1:3 = False

b. 5:15 = 5 ÷ 5 15 ÷ 5 = 1:3; 3:9 = 3 ÷ 3 9 ÷ 3 = 1:3

As 5:15 = 3:9 = True

c. 21:4; 84:16 = 84 ÷ 4 16 ÷ 4 = 21 4

As 21:4 = 84:16 = True

d. 7:12; 21:36 = 21 ÷ 3 36 ÷ 3 = 7:12

As 7:12 = 21:36 = True

e. 55:60 = 55 ÷ 5 60 ÷ 5 = 11 22 ; 42:7 = 42 ÷ 7 7 ÷ 7 = 6:1

As 55:60 42:7 = False

f. 24:10 = 24 10 = 12:5; 60:25 = 60 ÷ 5 25 ÷ 5 = 12:5

As 24:10 = 60:25 = True

3. Answer may vary. Sample answer

a. 3:13::9:39   b. 7:42::3:18   c. 7:21::4:12

d. 2:14::8:56   e. 3:9::7:21    f. 6:30::2:10

4. a. 2:x :: 5:10

x × 5 = 2 × 10 x = 20 5 = 4

c. x:16 :: 12:24

x × 24 = 16 × 12 x = 192 24 = 8

e. 6:x :: 10:15

x × 10 = 6 × 15 x = 90 10 = 9

g. 2x:18 :: 6:12

2x × 12 = 18 × 6 x = 108 24 = 4.5

i. 30:25 :: 42:x

30 × x = 25 × 42 x = 1050 30 = 35

5.a. 12:4 :: 9:x

12 × x = 4 × 9 x = 36 12 = 3

c. 9:3 :: 6:x

9 × x = 3 × 6 x = 18 9 = 2

e. 28:42 :: 32:x

28 × x = 42 × 32 x = 1344 28 = 48

6.a. 9:3 :: 3:x

9 × x = 3 × 3 x = 9 9 = 1

c. 16:36 :: 36:x

16 × x = 36 × 36 x = 1296 16 = 81

e. 14:28 :: 28:x

14 × x = 28 × 28 x = 784 14 = 56

7. x:21 :: 45:63

x × 63 = 21 × 45 x = 945 63 = 15

9. Ratio of length to width = 5:3

Length = 5x = 15 x = 15 5 = 3

Width = 3 × x = 3 × 3 = 9 m

b. 3:6 :: 9:x

3

× x = 6 × 9 x = 54 3 = 18

d. 1:3 :: x:12

3 × x = 1 × 12 x = 12 3 = 4

f. 4:x :: 6:12

x × 6 = 4 × 12 x = 48 6 = 8

h. 8:16 :: 12:x

8 × x = 16 × 12 x = 192 8 = 24

b. 2:6 :: 1:x

2 × x = 6 × 1 x = 6 2 = 3

d. 32:8 :: 24:x

32 × x = 8 × 24 x = 192 32 = 6

f. 18:6 :: 54:x

18 × x = 54 × 6 x = 324 18 = 18

b. 2:4 :: 4:x

2 × x = 4 × 4 x = 16 2 = 8

d. 36:42 :: 42:x

36 × x = 42 × 42 x = 1764 36 = 49

f. 48:96 :: 96:x

48 × x = 96 × 96 x = 9216 48 = 192

8. x:9 :: 9:3

3 × x = 9 × 9 x = 81 3 = 27

10. The proportion of patient’s weight to medicine can be given as:

55:132 :: x:156

132 × x = 55 × 156 x = 8580 132 = 65

Hence 156 mg of medicine will be given to a patient weighing 65 kg.

11. The proportion of miles to gasoline require can be given as:

126:18 :: 175:x

126 × x = 18 × 175 x = 3150 126 = 25

Hence 25 gallons of gasoline is required to travel 175 miles.

12. The proportion of flour to cupcakes can be given as:

225:9 :: x:20

9 × x = 225 × 20 x = 4500 9 = 500

Hence Rashi does not have sufficient flour as she would need 500 grams of flour for 20 cupcakes

Amount of flour Rashi needs = 500 g – 475 g = 25 grams

Word Problem 1. The proportion of sugar to cookies can be given as: 2.5:12 :: x:30

12 × x = 2.5 × 30 x = 75 12 = 6.25 = 6 1 4 cups

15E 1. Cost of 12 pens = ₹108

Cost of 1 pen = ₹ 108 12 = ₹9

Cost of 20 pens = 20 × ₹9 = ₹180

Hence 20 pens will cost ₹180.

2. 2 km are covered in 1 hour

1 km will be covered in 1 2 hour

8 km will be covered in 1 2 × 8 = 4 hours

Hence, he will cover 8 km in 4 hours.

3. Cost of 5 kg rice = ₹100

Cost of 1 kg rice = ₹ 100 5 = ₹20

Cost of 2.5 kg rice = 2.5 × ₹20 = ₹50

Hence 2.5 kg rice will cost ₹50.

4. Number of post cards purchased in ₹375 = 125

Number of post cards purchased in ₹1 = 125 375

Number of postcards purchased for ₹180 = 125 375 × 180 = 60

Hence 60 postcards can be purchased for ₹180.

5. Number of km covered in 8 hours = 296 km

Number of km covered in 1 hour = 296 8 = 37 km

Number of km in 13 hours = 13 × 37 = 481 km

Hence the bus travels 481 km in 13 hours.

6. Months to earn ₹1350 = 1 month

Months to earn ₹1 = 1 1350

Months to earn ₹12,150 = 1 1350 × 12,150 = 9 months

Hence it will take 9 months to Amit to earn ₹12,150.

7. Temperature increases in 100 days = 25℃.

Temperature increases in 1 day = 25 100 ℃

Temperature increases in 16 days = 25 100 × 16 = 4℃.

Hence the temperature will increase by 4℃ in 16 days.

8. Number of school buses require to carry 120 students = 3

Number of buses require to carry 1 student = 3 120

Number of school buses require to carry 1000 students = 3 120 × 1000 = 25

Hence 25 school buses are required to carry 1000 students.

9. Interest paid in 1 year or 12 months = ₹18,000

Interest paid in 1 month = 18,000 12 = ₹1500

Interest paid in 7 months = ₹1500 × 7 = ₹10,500.

Hence the interest paid by Manya in 7 months = ₹10,500.

10. Cost of 24 oranges bought by Raju = ₹336

Cost of 1 orange bought by Raju = ₹ 336 24 = ₹14

Cost of 21 oranges bought by Kiran = ₹273

Cost of 1 orange bought by Kiran = ₹ 273 21 = ₹13

Hence, Kiran bought the oranges at cheaper price.

11. Number of days 52 men take to complete a piece of work = 35 days

Number of days 1 men take to complete a piece of work = 35 ×52 = 1820 days (less men more days)

Number of days 28 men take to complete the piece of work = 1820

28 = 65 days (more men, less days)

Hence, 28 men can do the piece of work in 65 days

Word Problem 1. Number of words typed in half an hour or 30 minutes = 660

Number of words typed in 1 minute = 660 30 = 22

Number of words type in 10 minutes = 22 × 10 = 220 Hence Shreya can type 220 words in 10 minutes.

Chapter Checkup 1. Number of packets of potato chips = 45; Number of packets of banana chips = 66; Number of packets of corn chips = 32

a. Ratio of potato to corn chips = 45:32

b. Ratio of banana to total chips = 66:143

2. a. 8 m = 800 cm

HCF of 800 and700 = 100

800 700 = 800 ÷ 100

700 ÷ 100 = 8 7

Thus, 8 m:700 cm in its simplest form is 8:7

b. 360 sec = 6 × 1360 60 = 6 min

Thus, 360 sec:7 min in its simplest form is 6:7

c. 5 km = 5 × 1000 = 5000 m

HCF of 5000 and 900 = 100

5000

900 = 5000 ÷ 100

900 ÷ 100 = 50 9

Thus, 5 km:900 m in its simplest form is 50:9

d. 5 L = 5 × 1000 = 5000 mL

HCF of 5000 and 750 = 250

5000

750 = 5000 ÷ 250

750 ÷ 250 = 20 3

Thus, 5 L:750 mL in its simplest form is 20:3

e. 7 kg = 7 × 1000 = 7000 g

HCF of 500 and 7000 = 500

500

7000 = 500 ÷ 500

7000 ÷ 500 = 1 4

Thus, 500 g:7 kg in its simplest form is 1:14

f. 4 hours = 4 × 60 = 240 minutes

HCF of 45 and 240 = 15

45 240 = 45 ÷ 15

240 ÷ 15 = 3 16

Thus, 45 minutes:4 hours in its simplest form is 3:16

3. Answers may vary, sample answer.

a. 3:8

Four

b. 5:9 = 5 × 2 9 × 2 = 10 18 ; 5 × 3

= 15

12:32, and 15:40

Four equivalent ratios to 5:9 = 10:18, 15:27, 20:36, and 25:45

c. 11:15 = 11 × 2 15 × 2 = 22 30 ; 11 × 3 15 × 3 = 33 45 ; 11 × 4 15 × 4 = 44 60 ; 11 × 5 15 × 5 = 55 75

Four equivalent ratios to 11:15 = 22:30, 33:45, 44:60, and 55:75

d. 4:13 = 4 × 2 13 × 2 = 8 26 ; 4 × 3 13 × 3 = 12 39 ; 4 × 4 13 × 4 = 16 52 ; 4 × 5 13 × 5 = 20 65

Four equivalent ratios to 4:13 = 8:26, 12:39, 16:52, and 20:65

e. 9:11 = 9 × 2 11 × 2 = 18 22 ; 9 × 3 11 × 3 = 27 33 ; 9 × 4 11 × 4 = 36 44 ; 9 × 5 11 × 5 = 45 55

Four equivalent ratios to 9:11 = 18:22, 27:33, 36:44, and 45:55

f. 10:19 = 10 × 2 19 × 2 = 20 38 ; 10 × 3 19 × 3 = 30 57 ; 10 × 4 19 × 4 = 40 76 ; 10 × 5 19 × 5 = 50 95

Four equivalent ratios to 10:19 = 20:38, 30:57, 40:76, and 50:95

g. 13:27 = 13 × 2 27 × 2 = 26 54 ; 13 × 3 27 × 3 = 39 81 ; 13 × 4 27 × 4 = 52 108 ; 13 × 5 27 × 5 = 65 135

Four equivalent ratios to 13:27 = 26:54, 39:81, 52:108, and 65:135

h. 15:17 = 15 × 2 17 × 2 = 30 34 ; 15 × 3 17 × 3 = 45 51 ; 15 × 4 17 × 4 = 60 68 ; 15 × 5 17 × 5 = 75 85

Four equivalent ratios to 15:17 = 30:34, 45:51, 60:68, and 75:85

4. a. 4:5 = 4 5 ; 5:7 = 5 7

LCM of 5 and 7 = 35

4 5 = 4 × 7 5 × 7 = 28 35 ; 5 7 = 5 × 5 7 × 5 = 25 35

As 28 35 > 25 35 4 5 > 5 7 or 4:5 > 5:7

c. 3:7 = 3 7 ; 7:8 = 7 8

LCM of 7 and 8 = 56

3 7 = 3 × 8 7 × 8 = 24 56 ; 7 8 = 7 × 7 8 × 7 = 49 56

As 24 56 < 49 56 3 7 < 7 8 or 3:7 < 7:8

e. 9:12 = 9 ÷ 3 12 ÷ 3 = 3 4 ; 10:13 = 10 13

LCM of 4 and 13 = 52

3 4 = 3 × 13 4 × 13 = 39 52 ; 10 13 = 10 × 4 13 × 4 = 40 52

As 39 52 < 40 52 9 12 < 10 13 or 9:12 < 10:13

5. Given ratios = 3 5 , 5 7 , 8 10 , 9 11

LCM of 5, 7, 10 and 11 = 770

b. 2:7 = 2 7 ; 3:9 = 3 9

LCM of 7 and 9 = 45

2 7 = 2 × 9 7 × 9 = 18 63 ; 3 9 = 3 × 7 9 × 7 = 21 63

As 18 63 < 21 63 2 7 < 3 9 or 2:7 < 3:9

d. 8:11 = 8 11 ; 7:9 = 7 9

LCM of 11 and 9 = 99

8 11 = 8 × 9 11 × 9 = 72 99 ; 7 9 = 7 × 11 9 × 11 = 77 99

As 72 99 < 77 99 8 11 < 7 9 or 8:11 < 7:9

f. 5:12 = 5 12 ; 9:14 = 9 14

LCM of 12 and 14 = 84

5 12 = 5 × 7 12 × 7 = 35 84 ; 9 14 = 9 × 6 14 × 6 = 54 84

As 35 84 < 54 84 5 12 < 9 14 or 5:12 < 9:14

3 5 = 3 × 154 5 × 154 = 462 770 ; 5 7 = 5 × 110 7 × 110 = 550 770 ; 8 10 = 8 × 77 10 × 77 = 616 770 ; 9 11 = 9 × 70 11 × 70 = 630 770

The given ratios can be arranged in descending order as: 630:770 < 616:770 < 550:770 < 462:770 = 9:11 > 8:10 > 5:7 > 3:5

6. a. 14:20 = 14 20 = 14 ÷ 2 20 ÷ 2 = 7 10 ; 28:40 = 28 40 = 28 ÷ 4 40 ÷ 4 = 7 10

Hence the ratios are in proportion.

b. 16:17 = 16 17 ; 20:25 = 20 25 = 20 ÷ 5 25 ÷ 5 = 4 5

Hence the ratios are not in proportion.

c. 21:19 = 21 19 ; 21:34 = 21 34

Hence the ratios are not in proportion.

d. 40:24 = 40 24 = 40 ÷ 8 24 ÷ 8 = 5 3 ; 20:12 = 20 12 = 20 ÷ 4 12 ÷ 4 = 5 3

Hence the ratios are in proportion.

e. 3:6 = 3

6 = 3 ÷ 3

6 ÷ 3 = 1:2 ; 22:44 = 22 44 = 22 ÷ 22 44 ÷ 22 = 1:2

Hence the ratios are in proportion.

f. 24:25 = 24 25 ; 29:32 = 29 32

Hence the ratios are not in proportion.

7. 12:x :: 6:36

x × 6 = 12 × 36 x = 432 6 = 72

8. 9:12 :: 12:x

x × 9 = 12 × 12 x = 144 9 = 16

9. Number of red stamps = 24; Number of blue stamps = 36

Ratio of red stamps to blue stamps = 24 36 = 24 ÷ 12 3612 = 2 3 = 2:3

Hence the ratio of red stamps to blue stamps is 2:3.

10. Cost of 4 dozen or 48 eggs = ₹288

Cost of 1 egg = ₹ 288 48

Cost of 15 eggs = ₹ 288 48 × 15 = ₹90

Hence, 15 eggs will cost ₹90.

11. Given that the numbers are in the ratio 3:5

So, 3x + 5x = 30

8x = 30 x = 30 8 = 3.75

Hence the numbers are 3 × 3.75 = 11.25 and 5 × 3.75 = 18.75

12. Time taken to cover 60 miles = 1 hour

Rime taken to cover 1 mile = 1 60 hour

Time taken to cover 180 miles = 1 60 × 180 = 3 hours

Hence, the time taken to cover 180 miles is 3 hours.

13. Ratio of apples to oranges = 3:2

Total number of fruits = 100

Number of apples = 3 5 × 100 = 60

Number of oranges = 2 5 × 100 = 40

14. Ratio of pencils to pens = 4:5

Number of pencils = 4x = 36; x = 36 4 = 9

Number of pens = 5x = 5 × 9 = 45

15. Number of bottles filled in 5 hours = 750

Number of bottles filled in 1 hour = 750 5 = 150

Number of bottles filled in next 3 hours = 150 × 3 = 450

Hence the machine will fill 450 bottles in the next 3 hours.

16. Ratio of amounts collected from first and second class = (3 × 1):(2 × 15) = 3:30 = 1:10

Amount collected from first class passengers = 1 11 × 26,400 = ₹2400.

17. Present age of mother = 36 years; age of mother after 12 years = 36 + 12 = 48 years

Present age of daughter = 36 – 20 = 16 years; age of daughter 10 years ago = 16 – 10 = 6 years

Required ratio = 48:6 = 8:1

18. Number of painters required for painting the building in 6 days = 18

Number of painters required to paint the building in 1 day = 18 × 6 = 108 (less days, more painters)

Number of painters required to paint the building in days = 180 4 = 27

Hence 27 painters are required to paint the building in 4 days.

Word Problems 1. Ratio of boys to girls = 4:5

Total number of students = 180

Number of girls = 5 9 × 180 = 100 Hence, there are 100 girls.

2. Ratio of John’s age to Mary’s age = 2:3

John’s age = 2x = 24 years; x = 24 2 = 12

Mary’s age = 3x = 3 × 12 = 36 years

3.  Ratio of mangoes to apples = 5:7

Number of apples = 7x = 140; x = 140 7 = 20

Number of mangoes = 5x = 5 × 20 = 100

4. Number of hours taken to cover 1200 km = 6 hours

Number of hours taken to cover 1 km = 6 1200 hours

Number of hours taken to cover 800 km = 6 1200 × 800 = 4 hours = 4 × 60 minutes = 240 minutes

Chapter 16

Let's Warm-up 1. Symmetrical  2. Non-symmetrical

3. Symmetrical   4. Symmetrical  5. Non-symmetrical

Do It Yourself

16A 1.

An isosceles triangle has one line of symmetry.

3.a. b.

2.a. Both l1 and l2 b. Line l2 c. Both lines l1 and l2 d. None

c. d.

4. Answers may vary. Sample answers. 5 Symmetrical objects in our surroundings are a star, a heart, a leaf, a plus sign, and a watermelon slice.

5.a.

b.

c.

d. Word Problem Pihu is right. The chessboard has two lines (diagonals) of symmetry. The horizontal and vertical lines will not divide the checks symmetrically.

16B 1. The figures given below are mirror images of each other. 2.a.

b.   c.

d. 3. a.  b.

c.  d.

e.  f.

4. a. MIRROR MIRROR

b. SYMMETRY SYMMETRY

c. REFLECTION REFLECTION

d. MATHEMATICS MATHEMATICS

5.

Thus, option b is correct.

Word Problem 1. The original word was “MOM”, since if we look at the word “MOM” using a horizontal mirror, we would see the word “WOW”.

Chapter Checkup 1.

Z S 3 D 8 U

2. Seven lines of symmetry can be drawn in the given figure.

3. a. There is 1 line of symmetry in the letter T. - True

b. A regular pentagon has 5 lines of symmetry. - True

c. All the quadrilaterals have 4 lines of symmetry. - False

d. A semicircle has one line of symmetry. - True 4.a. b. c. d.

a. 4 b. 2 c. 4 d. 4

5. a.  b.  c.  d.

6. Figure c will not look the same along both mirror lines. Reference images. a. b.

c. d.

7.a. b. c. d.

8. a.  b.

c.  d.

e.  f.

9. a. Yes, the line of symmetry marked on the given figure is correct. b. Yes, the line of symmetry marked on the given figure is correct.

c. No, the line of symmetry marked is incorrect. The correct line of line is drawn in the figure below. 10.

Word Problem1. Rahul is incorrect about the lines of symmetry in the rainbow. The rainbow has a semi-circular shape; hence, it has only one line of symmetry (vertical).

Chapter 17

Let's Warm-up 1. Line Segments = 5 Angles = 5 2. Line Segments = 5 Angles = 6 Do It Yourself

17A 1. a. A B 5.4 cm  b. P Q 6.7 cm

c. L M 6.1 cm  d. X Y 7.2 cm

2. a. A B 4.3 cm  b. D E 5.6 cm

c. G H3.9 cm  d. PQ2.7 cm

3.  C

b. BC = 4RS – 3 PQ = 4 × 3.6 – 3 × 2.3 cm = 14.4 – 6.9 = 7.5 cm

B C 7.5 cm

c. XY = RS + 2PQ = 3.6 + 2 × 2.3 cm =3.6 + 4.6 = 8.2 cm

X Y 8.2 cm

d. DE = 2RS + PQ = 2 × 3.6 + 2.3 cm = 7.2 + 2.3 = 9.5 cm

D E 9.5 cm

7. Students will construct line segments of the same length as AB, PQ and MN.

8. AB + BC + CA = 8.9 cm ABCA 2.4 cm3.3 cm3.2 cm

Word Problem 1. Draw line of suitable length and copy each line segment back to back on this line segment. Total length of path is 17.4 cm. P Q

17.4 cm

17B 1.  6.5 cm R P Q

2. 4.5 cm A B x 3.  5.6 cm

3 cm MY X

4.  C m 5.  F 3 cm 7.2 cm 3 cm XG

7. a. 5.8 cm A B

c. 8.2 cm P Q

6.  90º D E 6.1 cm 90º

b. 9 cm L M

d. 7.6 cm X Y

B 9. a. M E F

8. Yes, all the perpendicular bisectors intersect at the same point. A C S

EF D 2.8 cm

2.8 cm 2.8 cm2.8 cm

4.  C E D 4.2 cm 4.7 cm 8.9 cm

9.6 cm

5.  L N O M4.8 cm

6. Given PQ = 2.3 cm and RS = 3.6 cm

Length of DE = 4.7 cm.

a. AB = 3 RS – 2PQ = 3 × 3.6 – 2 × 2.3 cm = 10.8 – 4.6 = 6.2 cm

A B 6.2 cm

Z X Y

Z M E F 148

50º

d.

80º

65º

a. O A B

c. O

S R

b. O

P Q

d. O

X Y 3. a. O A B   b. O

c. O

OB

S R

d. O

P Q

X Y

A

4 cm

P 44º RQ P 67º R Q Q 69º RP b. A C B 29º B C A 131º B AC 20º

In the figure, ∠POQ = 126° and its supplement is ∠ROQ = 54°.

OS is the bisector of ∠ROQ.

7. Complement of 71° = 90° – 71° = 19°

In the figure, ∠YOX = 71° and its complement is ∠ZOY = 19°.

OP is the bisector of ∠ZOY.

Word Problem 1.  100º B O

X 145º OM L 55º O Q

A 60º Y O

P

Chapter Checkup 1.a. P3.6 cmQ

b. X6.7 cmY

c. A6.7 cm6.7 cm6.7 cm6.7 cmB

d. R3.6 cm3.6 cmS

2. FG = 6.4 cm  E G3.8cm6.4 cmF

3. M6.5 cmON2.6 cm

4.a. Given, UV = 6.2 cm and ST = 4.7 cm

126º 27º

O R

Z P Y X 71º 9.5º9.5º

R

WX = 3UV – 2ST = 3 × 6.2 – 2 × 4.7 cm = 19.6 – 9.4 = 9.2 cm

AB 9.2 cm

b. YZ = 5ST – 2UV = 5 × 4.7 – 2 × 6.2 cm = 23.5 – 12.4 = 11.1 cm

XY 11.1 cm

c. PQ = ST + 4UV = 4.7 + 4 × 6.2 cm = 4.7 + 24.8 = 29.5 cm

P Q 29.5 cm

d. AB = 3ST + 2UV = 3 × 4.7 + 2 × 6.2 cm = 14.1 + 12.4 = 26.5 cm

T V 26.5 cm

5. Figure may vary. Sample figure CD E F

6. a. Measure of the sides are 2 cm, 0.8 cm 2 cm and 0.8 cm.

2 cmA B 0.8 cmCD

b. Measure of the sides is 0.8 cm each.

0.8 cmWX 0.8 cmYZ

c. Measure of the sides is 1.1 cm and 1 cm.

1.1 cmYZ 1 cmYZ

7. a.

8 cm X Y

d. 7.5 cm L M

b. 9 cm P Q

8.  6.8 cm Y D C 9.  9.5 cm

4 cm N P M

c.

5.8 cm S T

10. K n

11. a. 27º  b. 136º  c. 53º  d. 73º

e. 89º

f. 101º

12. Supplement of 139° = 180° – 139° = 41°

In the figure, ∠GFH =126° and its supplement is ∠EFH = 41°. FS is the bisector of ∠ROQ.

H S E

20.5º

20.5º

139º

FG

13. Complement of 17° = 90° – 17° = 73°

In the figure, ∠WOX = 17° and its complement is ∠ZOW = 73°. OY is the bisector of ∠ZOW. 14. ∠AOD = 60°

OBA

15. a. Copying and bisecting indicated angles ∠BAE = 94° and ∠CDE = 78°, we have 47º 47º

b. Copying and bisecting indicated angles ∠PQR = 35° and ∠PRQ = 26°, we have

17.5º

17.5º Q R

X P 13º 13º R Q Y P

c. Copying and bisecting indicated angles ∠LMJ = 65° and ∠JKL = 83°, we have

L

32.5º

41.5º

R KJ

16. a. Firstly, let’s draw perpendicular at X and Y using ruler and set square.

1 cm2.1 cm2 cm J XY K

b. Yes, TJ will be equal to TK irrespective of where the point is chosen on the perpendicular bisector.

J X

1 cm

T Y K

2 cm

5.1 cm

17. No, all the perpendiculars didn’t intersect at a given point. A B C

Word Problem 1.

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: Helps to revisit students’ prior knowledge to facilitate learning the new chapter

• Real Life Connect: Introduces a new concept by relating it to day-to-day life

• Examples: Provides the complete solution in a step-by-step manner

• Do It Together: Guides learners to solve a problem by giving clues and hints

• 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

• QR Codes: Digital integration through the app to promote self-learning and practice

About Uolo

Uolo partners with K-12 schools to provide technology-based learning programs. We believe pedagogy and technology must come together to deliver scalable learning experiences that generate measurable outcomes. Uolo is trusted by over 10,000 schools across India, South East Asia, and the Middle East.

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