Imagine_Maths_TM_Grade5

MATHEMATICS

Teacher Manual

MATHEMATICS

Master Mathematical Thinking

Grade 5

Foreword

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

Numbers Up to 8 Digits 1

Numbers Up to 8 Digits 1

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.

Numbers Up to 8 Digits 1

Numbers Up to 8 Digits 1

Learning Outcomes as recommended by the NEP 2020 and the latest National Curricular Framework (NCF).

Alignment to NCF

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

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

C-5.1: Understands the development of zero in

Alignment to NCF

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

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

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

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

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

Let’s Recall

place value for easier calculation

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

Teaching Aids

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

Vocabulary expanded form: writing a number as the sum of the values of all its digits order: the way numbers are arranged

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

estimating: guessing an answer that is close to the actual answer rounding off: approximating a number to a certain place value for easier calculation

Teaching Aids

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

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

QR Code: Access to digital solutions and other interactive resources.

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

Numbers Up to 8 Digits 1

Learning Outcomes

Students will be able to: write the place value, expanded form and number names for numbers

to 8 digits. write numbers up to 8 digits in the Indian and International number system. compare numbers up to 8 digits and arrange them in ascending and descending order. round off numbers up to 8 digits to the nearest 10, 100 and 1000.

Numbers Up to 8 Digits 1

Alignment to NCF

Learning Outcomes

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

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

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

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

Alignment to NCF

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

Let’s Recall

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

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

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

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

Let’s Recall

Vocabulary expanded form: writing a number as the sum of the values of all its digits order: the way numbers are arranged estimating: guessing an answer that is close to the actual answer

Alignment to NCF

rounding off: approximating a number to a certain place value for easier calculation

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

Teaching Aids

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

Vocabulary

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

Let’s Recall

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

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

Chart paper with empty place value chart drawn; Buttons; Beads; Bowl; Number cards; Bowls with 5 number cards with a 7-digit or 8-digit number written on them; Two bowls with number cards having 8-digit numbers in one bowl and rounded-off places in another bowl 1

Teaching Aids

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

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

Vocabulary

Chart paper with empty place value chart drawn; Buttons; Beads; Bowl; Number cards; Bowls with 5 number cards with a 7-digit or 8-digit number written on them; Two bowls with number cards having 8-digit numbers in one bowl and rounded-off places in another bowl 1

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

Teaching Aids

Chart paper with empty place value chart drawn; Buttons; Beads; Bowl; Number cards; Bowls with 5 number cards with a 7-digit or 8-digit number written on them; Two bowls with number cards having 8-digit numbers in one bowl and rounded-off places in another bowl 1

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

Period Plan

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

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

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

Adding

Subtracting Numbers Up to 6 Digits

Multiplying Numbers with 10s, 100s and 1000s

Multiplying Numbers Up to 4 Digits

Dividing Numbers by 10s, 100s and 1000s

Dividing 5-digit Numbers by 3-digit Numbers

Using DMAS

Multi-step Word Problems

Revision

Finding

Prime

Divisibility

Divisibility

Prime

Prime

Long

Finding

Common Multiples

LCM

LCM by Short Division

Word Problems

Revision

11

Reviewing Fractions

Equivalent Fractions Using Multiplication

Equivalent Fractions Using Division

Comparing Fractions

Ordering Fractions

Revision

Adding Unlike Fractions

Adding Mixed Numbers

Subtracting Unlike Fractions

Subtracting Mixed Numbers

Multiplying Fractions and Whole Numbers

Multiplying Two Fractions

Dividing a Whole Number by a Fraction

Dividing a Fraction by a Whole Number

Dividing a Fraction by a Fraction

Revision

Tenths

Hundredths

Thousandths

Conversion Between Fractions and Decimals

9

Types of Decimals

Comparing and Ordering Decimals

Rounding off Decimals

Revision

Addition of Decimals

Subtraction of Decimals

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

Multiplying Whole Numbers and Decimals; Multiplying Two Decimals

Currencies from Different Countries

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

Dividing Decimals by Whole Numbers

Revision 9.

Measuring

Drawing

Symmetry

Estimating Length; Measuring Lengths

Units of Length

Word Problems on Length Estimating Weights

Units of Weight

Word Problems on Weights

Perimeter of Squares and Rectangles

Area of a Square; Area of a Rectangle

Area of a Triangle Revision

Estimating

Units of

Word

Interpreting

Drawing Bar Graphs

Interpreting Pie Charts

Representing Data on a Pie Chart

Reading Line Graphs

Revision

Numbers Up to 8 Digits 1

Learning Outcomes

Students will be able to:

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

Alignment to NCF

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

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

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

Let’s Recall

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

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

Vocabulary

expanded form: writing a number as the sum of the values of all its digits order: the way numbers are arranged

estimating: guessing an answer that is close to the actual answer rounding off: approximating a number to a certain place value for easier calculation

Teaching Aids

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

Chapter: Numbers Up to 8 Digits

Place Value, Expanded Form and Face Value

Learning Outcomes

Imagine Maths Page 2

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

Teaching Aids

Chart paper with empty place value chart drawn; Buttons; Bowl; Number cards

Activity

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

Extension Idea

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

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

Indian and International Number System

Learning Outcomes

Imagine Maths Page 6

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

Teaching Aids

Chart paper; Buttons; Beads; Number cards

Activity

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

Extension Idea

Ask: How many lakhs are there in 10 million?

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

Learning Outcomes

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

Teaching Aids

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

Activity

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

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

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

Extension Idea

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

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

Rounding off Numbers

Learning Outcomes

Imagine Maths Page 13

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

Teaching Aids

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

Activity

Instruct the students to work in groups of 4. Write a few 8-digit numbers on one set of cards and the place to which they need to be rounded off on another set of cards, say 4,76,87,693 on one card; round off to thousands on another; and keep these cards in two separate bowls.

Call a student from each group and instruct him/her to pick a card from each bowl. Ask the groups to consider two cards as 1 and follow the instructions written on the cards. Let them write the answers in their notebooks. Verify their answers and give 1 point for the correct answer and 0 for the incorrect answer. Repeat this process until all the students in the group have had a chance. The group with the maximum points wins the game.

Answers

1. Place Value, Expanded Form and Face Value

Think and Tell

Yes, these are:

2874477: 28,74,477 – twenty-eight lakh seventy-four thousand four hundred seventy-seven in Goa and 45546836: 4,55,46,836 – four crore fifty-five lakh forty-six thousand eight hundred thirty-six in Haryana.

Do It Together

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

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

2. Indian and International Number System

Do It Together

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

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

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

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

3. Comparing Numbers

Do It Together

4. Ordering Numbers

Think and Tell

Yes, the descending order will be:

3,09,13,200 > 2,80,82,800 > 98,02,000 > 94,99,000

Think and Tell

Yes, we can form more such numbers.

Think and Tell

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

Do It Together

Ascending Order: 16,52,349; 86,70,653; 1,26,56,384; 4,00,97,234

Descending Order: 4,00,97,234; 1,26,56,384; 86,70,653; 16,52,349

5. Rounding off Numbers

Think and Tell

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

Do It Together

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

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

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

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

Operations on Numbers Up to 6 Digits 2

Learning Outcomes

Students will be able to:

add numbers up to 6 digits using the column method.

subtract numbers up to 6 digits using the column method.

multiply numbers up to 6 digits with 10s, 100s and 1000s.

multiply numbers up to 4 digits using the column method.

divide numbers up to 6 digits by 10s, 100s and 1000s.

divide numbers up to 5 digits by numbers up to 3 digits using the column method. solve expressions using the DMAS rule.

solve multi-step word problems on using 2 or more operations.

Alignment to NCF

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

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

Let’s Recall

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

Vocabulary

addends: numbers to be added

minuend: number to be subtracted from subtrahend: number to be subtracted

Teaching Aids

Number cards; Crossword puzzle sheet; 2 stacks of cards; Word problem cards

Chapter: Operations on Numbers Up to 6 Digits

Adding Numbers Up to 6 Digits

Learning Outcomes

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

Teaching Aids

Number cards (with the population of some countries written)

Activity

Imagine Maths Page 19

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

Prepare cards with the population of some countries written on them. For instance, Malta: 5,35,064; Maldives: 5,21,021; Iceland: 3,75,318; and Saint Lucia: 1,80,251.

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

Instruct them to add to find the total population of the two countries written on the cards.

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

Extension Idea

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

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

Subtracting Numbers Up to 6 Digits

Learning Outcomes

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

Teaching Aids

Number cards (with the population of some countries written)

Activity

Imagine Maths Page 20

Demonstrate the subtraction of two 6-digit numbers, 2,50,678 and 1,56,240, using the column method.

Emphasize the importance of borrowing when the minuend is less than the subtrahend.

Prepare cards with the population of some countries written on them. For instance, Malta: 5,35,064; Maldives: 5,21,021; Iceland: 3,75,318; and Saint Lucia: 1,80,251.

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

Instruct them to find the difference of the populations mentioned on their cards.

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

Learning Outcomes

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

Teaching Aids

Crossword puzzle sheets with multiplication problems with missing numbers

Activity

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

Write on the board:

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

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

Explain that in this crossword puzzle, they will find problems where they will have to multiply by 10, 100 and 1000. Instruct them to solve these problems and write the answers in the puzzle.

Then, in their notebooks, ask them to write down the answers for multiplying the numbers by 200 and 2000.

Discuss how they don’t actually need to perform the multiplication and should use their understanding to find the answers without going through all the calculations.

Multiplying Numbers Up to 4 Digits

Learning Outcomes

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

Teaching Aids

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

Activity

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

Imagine Maths Page 24

Instruct the groups to make 2 stacks of cards – one with the grocery items and the other with the quantities.

Ask one student in each group to pick one card from each stack. The remaining students should then calculate the total cost by multiplying the rate with the quantity. For example, if the student gets cards for ‘rice’ and 1200, the total cost would be ₹143 × 1200. Instruct the students to take turns to pick more cards.

Extension Idea

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

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

Dividing Numbers by 10s, 100s and 1000s

Learning Outcomes

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

Teaching Aids

Crossword puzzle sheet with division problems with missing numbers

Activity

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

Write on the board:

3,21,200 ÷ 10 = 32120; 3,21,200 ÷ 100 = 3212; 3,21,200 ÷ 1000 → quotient = 321 and remainder = 200

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

Imagine Maths Page 25

Explain that in this crossword puzzle, they will find problems where they will have to divide by 10, 100, and 1000. Instruct the students to complete the crossword puzzle, and then, in their notebooks, ask them to write down the answers.

Dividing 5-digit Numbers by 3-digit Numbers

Learning Outcomes

Imagine Maths Page 26

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

Teaching Aids

Cards with names of grocery items; Cards with quantities 3-digit numbers

Activity

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

Now, write down the names of items on cards and place them in one box.

Write their quantities in 3-digit numbers on another set of cards, and put them in another box.

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

Extension Idea

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

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

Learning Outcomes

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

Teaching Aids

Crossword puzzle sheet with different number problems

Activity

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

Distribute the crossword sheets to the students.

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

Emphasize that they need to remember that multiplication and division are performed before addition and subtraction.

Extension Idea

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

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

Multi-step Word Problems

Learning Outcomes

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

Imagine Maths Page 29

Teaching Aids

Word problem cards

Activity

Write the word problem on cards.

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

Distribute one card to each student.

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

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

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

Extension Idea

Instruct: Create your own word problem by dividing 84 by 2, multiplying the result with 3 and subtracting 53 from it.

Say: Here is an example: Lisa has 84 candies. She decides to share them equally with her sibling. Her mother bought a candy jar that has three times the number of candies that Lisa has left. Finally, she decides to give 53 candies to her friend as a gift. How many candies does Lisa have left?

Answers

1. Adding Numbers Up to 6 Digits

Do It Together

Number of poetry books = 26,716

Number of non-fiction books = 43,120

Number of fiction books = 80,133

Total number of books in the library is 1,49,969

2. Subtracting Numbers Up to 6 Digits

Do It Together

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

Shirts sold in 1 month = 4640

Shirts sold in 11 months = 2,80,946

Therefore, the company sold 2,80,946 shirts in 11 months.

4

. Multiplying Numbers Up to 4-digits

Do It Together

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

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

Do It Together

Cost of 1 scooter = ₹75,250

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

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

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

Do It Together

67,688  1000

Quotient = 67

Remainder = 688

Thus, quotient = 67 and remainder = 688

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

Do It Together

Cost of 235 books = ₹56,745

of one book = ₹56,745 ÷ 235 Therefore, the cost of one book is ₹ 110 241 235 .

7. Using DMAS

Do It Together

8. Multi-step Word Problems

Think and Tell

Yes.

Think and Tell

No.

Do It Together

Initial number of flowers = 50 × 2

Flowers given by mother = 8

Flowers given to Siya = 8 ÷ 2

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

Step 1

Divide (÷) = 50 × 2 + 8 – 4

Step 2

Multiply (×) = 100 + 8 – 4

Step 3

Add (+) = 108 – 4

Therefore, Liza is left with 104 flowers.

Step 4

subtract (–) = 104

Factors and Highest Common Factor 3

Learning Outcomes

Students will be able to:

find factors of a number using multiplication or division. find prime and composite numbers.

identify and apply the divisibility rule for 2, 5 and 10.

identify and apply the divisibility rule for 3 and 9. find the prime factorisation of a number using the factor tree. find the common factors and HCF of 2 or more numbers using the factor method. find the HCF of 2 or more numbers using the prime factorisation method. find the HCF of 2 or more numbers using the long division method.

Alignment to NCF

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

Let’s Recall

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

Vocabulary

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

Teaching Aids

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

Chapter: Factors and Highest Common Factor

Finding Factors

Learning Outcomes

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

Teaching Aids

Grid paper

Activity

Briefly revise the concept of the area of a rectangle and how it’s calculated (Area = Length × Width).

Give each student a sheet of grid paper. Ask them to draw as many different rectangles as possible, each with an area of 24 square units. After drawing the rectangles, ask students to label the side lengths of each rectangle. Pair students up and have them compare their rectangles. Discuss similarities and differences. Guide a class discussion by asking questions like: How are your rectangles similar? How are they different?

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

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

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

Extension Idea

Ask: Veena is arranging square tiles to create a rectangular garden. The area of the garden is 45 m2. Each tile is exactly 1 m × 1 m. How many different ways can she arrange the tiles to form a rectangular garden with an area of 45 m2?

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

Prime and Composite Numbers

Learning Outcomes

Students will be able to find prime and composite numbers.

Teaching Aids

Grid paper

Activity

Imagine Maths Page 39

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

Distribute a sheet of grid paper to each student.

Instruct the students to draw as many different rectangles as possible with an area of 32 square units and 13 square units on the provided grid paper. Students should label the side lengths of each rectangle directly on the grid. Ask students to identify the factors of 32 and 13 based on the rectangles they drew. Further, ask them to compare the number of factors between 32 and 13.

Ask questions such as: Can a whole number have fewer than two factors? Exactly two factors? More than two factors? Emphasize the connection between the number of factors and whether a number is prime or composite.

Extension Idea

Ask: How many composite numbers are there between 30 and 50? List them.

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

Divisibility by 2, 5 and 10

Learning Outcomes

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

Teaching Aids

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

Activity

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

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

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

After the sorting is complete, have each group share their categories and reasoning for placing numbers in each category. Encourage discussion on the divisibility rules for 2, 5 and 10. Ask questions such as: Are there any numbers which are divisible by 2, 5 and 10?

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

Extension Idea

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

a. 216 b. 333 c. 495 d. 98

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

Divisibility by 3 and 9

Learning Outcomes

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

Teaching Aids

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

Activity

Imagine Maths Page 41

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

Divide the class into groups.

Distribute index cards with various numbers to each group. Instruct students to sort the numbers into 2 categories: “Divisible by 3” and “Divisible by 9”.

Ask questions like: Was there any strategy that helped to determine the divisibility by 3 or 9? Could you list a few numbers which are divisible by both 3 and 9?

Summarise the key learnings from the activity.

Extension Idea

Ask: What is the secret code to unlock the number lock considering the following clues: The number is divisible by 3; the sum of its digits is divisible by 9; and the number has 0 as the last digit?

a. 4320 b. 6670 c. 8920 d. 210

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

Learning Outcomes

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

Teaching Aids

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

Activity

Explain the concept of prime factorisation using factor trees.

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

Assign each pair a random number (e.g., 24, 36, 48) that they will factorise using the factor tree.

Instruct participants to write their assigned number on the paper square and stick it on the chart paper. Ask them to identify and write the first pair of factors on the squares, and connect them using a glue stick with a piece of string or yarn to represent a branch. Ask them to repeat this process until only prime numbers are left at the ends of the branches.

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

Factor Method

Learning Outcomes

Imagine Maths Page 47

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

Teaching Aids

Chart paper; Small square cut-outs made of paper

Activity

Begin by introducing the concept of common factors.

Write two numbers (e.g., 24 and 36) on the board.

Instruct the students to draw two overlapping circles on the chart paper, representing the two numbers. Explain how the overlapping region in the centre will represent the common factors. Then, in their groups, students will discuss and list the factors of each number separately. For 24, factors include 1, 2, 3, 4, 6, 8, 12, and 24. For 36, factors include 1, 2, 3, 4, 6, 9, 12, 18, and 36.

Ask the students to use one colour of square cut-outs to represent the factors of the first number and paste them in the corresponding circle. Then ask them to use another colour for the factors of the second number and paste them in its circle. Explain how the overlapping factors should be placed in the centre region. Explain how these are the common factors of the two numbers. Finally, instruct the students to compare the common factors within the overlapping region and pick the largest one as the HCF.

Learning Outcomes

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

Teaching Aids

Circle cut-outs made of paper

Activity

Select any two numbers for the purpose of our activity (say 12 and 20). Divide the class into groups. Distribute small circle cut-outs among the groups.

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

Finally, ask them to multiply the common prime factors between the two numbers. Discuss how this number is the highest common factor of the given numbers.

Extension Idea

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

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

Division Method

Learning Outcomes

Imagine Maths Page 49

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

Teaching Aids

Colourful paper strips say green and blue; Ruler; Pair of scissors

Activity

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

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

Ask them to observe that the remaining part of the green strip cut out is of length 8 cm and is smaller than the blue colour strip, i.e. the strip of length 16 cm.

Instruct them to now place the smaller strip (green) along the larger strip (blue) and cut out the extra part (blue strip) as shown.

Ask them to repeat the activity until both strips are equal.

Explain how the HCF of 24 and 16 is 8 cm (the length of the last part cut out). It can be seen that a strip of length 8 cm can cover both the strips of length 24 cm and 16 cm a complete number of times. Finally, ask the students to verify the answer using the long division method.

1. Finding Factors

Do It Together

4. Divisibility by 3 and 9

Think and Tell

If a number is divisible by 10, it is also divisible by 2 and 5.

2. Prime and Composite Numbers

Do It Together

3.

6. Common Factors

Do It Together

Factors of 10: 1, 2, 5 and 10

Factors of 14: 1, 2, 7 and 14

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

7. Factor Method

Do It Together

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

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

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

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

8. Prime Factorisation Method

Do It Together

Factors of 16 = 2 × 2 × 2 × 2

Factors of 48 = 2 × 2 × 2 × 2 × 3

The common factors of 16 and 48 are 2, 2, 2 and 2.

Hence, HCF = 2 × 2 × 2 × 2 = 16.

9. Long Division Method

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

Multiples and Least Common Multiples 4

Learning Outcomes

Students will be able to:

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

Alignment to NCF

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

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

Let’s Recall

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

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

Vocabulary

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

Teaching Aids

Hundreds chart; Different coloured markers or pencils; Circle cutouts made of paper; Chart paper; Number cards; Word problems written on sheets of paper

Chapter: Multiples and Least Common Multiples

Finding Multiples

Learning Outcomes

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

Teaching Aids

Hundreds chart

Activity

Distribute a hundreds chart to each student.

Imagine Maths Page 55

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

Ask questions like: What do you notice about the numbers you shaded?

Discuss that the numbers shaded come in the multiplication table of 7 and are called the multiples of 7. Repeat the activity to shade the multiples of 8 on the same chart.

Extension Idea

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

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

Common Multiples

Learning Outcomes

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

Teaching Aids

Hundreds chart; Different coloured markers or pencils

Activity

Instruct the students to work in pairs and distribute a hundreds chart to each pair.

Imagine Maths Page 58

Instruct one of the students to start at 4 and shade it in blue. The student should then skip count by 4 and shade the top part of each cell they land on. Both students will write the shaded numbers in their notebooks. The second student will then start at 12 and shade it in red. The student should then skip count by 12 and shade the bottom part of each cell they land on. Both the students will write the shaded numbers in their notebooks. Ask questions like: What do you notice about the numbers you shaded? Did you find numbers which are multiples of both 4 and 12?

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

LCM

Learning Outcomes

Imagine Maths Page 60

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

Teaching Aids

Circle cutouts made of paper

Activity

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

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

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

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

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

LCM by Short Division Method

Learning Outcomes

Imagine Maths Page 62

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

Teaching Aids

Chart paper; Number cards

Activity

Demonstrate the concept of LCM by using the short division method on the board and ask the students to perform the activity as instructed using the number cards.

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

12, 15

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

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

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

Extension Idea

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

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

Word Problems

Learning Outcomes

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

Teaching Aids

Word problems written on sheets of paper

Activity

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

Greatest Common Factor Least Common Multiple

biggest both divide first both something is repeated largest identical groups smallest identical every days

most same split up next time same every hours

maximum equal minimum equal

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

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

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

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

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

Extension Idea

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

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

Answers

1. Finding Multiples

Think and Tell

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

The first five multiples of 4 can be obtained by multiplying 4 with 1, 2, 3, 4 and 5

So, we have,

2. Common Multiples

Common multiples of 12 and 16 are 48, 96 and so on. The lowest common multiple of 12 and 16 is 48

3. LCM by Prime Factorisation Method

4. LCM by Short Division Method

5. Word Problems

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

Fractions 5

Learning Outcomes

Students will be able to:

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

Alignment to NCF

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

Let’s Recall

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

Vocabulary

equivalent fraction: a fraction obtained by multiplying or dividing the numerator and the denominator by the same non-zero number

proper fractions: a fraction whose numerator is less than the denominator

improper fractions: a fraction whose numerator is equal to or greater than the denominator mixed fractions: fractions which are a combination of a whole number part and fractional part unit fractions: fractions with 1 as the numerator

Teaching Aids

Unshaded fraction circles divided into 3, 4 or 6 parts; Fraction strips showing 246 ,, 368 , and 3 4; Fraction strips or circles showing 8 16 and some additional strips; Fraction strips showing some fractions (e.g., 123 ,, 438 , and 5 6 ); Rectangular strips of paper; Markers or coloured pencils

Chapter: Fractions

Reviewing Fractions

Learning Objective

Imagine Maths Page 70

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

Teaching Aids

Unshaded fraction circles divided into 3, 4 or 6 parts

Activity

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

Student A should shade circles to show a mixed number. For example, Student A might shade 5 1 6 parts of a circle. Student B’s task is to write the fraction as the number of shaded parts upon the total number of parts, converting it into an improper fraction, 11 6 . Next, ask them to switch roles, with student B shading circles to show an improper fraction and student A writing the number of wholes and parts shaded, converting it into a mixed number.

Rotate pairs and repeat the activity to reinforce the concept of identifying and converting fractions. Ask questions like: Will the fraction change if we change the number of total parts to 8 instead of 4 in the fraction circle showing 3 1 ? 4

Equivalent Fractions Using Multiplication

Learning Objective

Students will be able to find equivalent fractions using multiplication.

Teaching Aids

Fraction strips showing 246 ,, 368 , and 3 4

Activity

Distribute two sets of fraction strips to each pair.

Imagine Maths Page 74

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

Ask them to compare and discuss what they notice.

Equivalent Fractions Using Division

Learning Objective

Students will learn to find equivalent fractions using division.

Teaching Aids

Fraction strips or circles showing 8 16 and some additional strips

Activity

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

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

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

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

Comparing Fractions

Learning Objective

Students will be able to compare two or more fractions.

Teaching Aids

Rectangular strips of paper; Markers or coloured pencils

Activity

Imagine Maths Page 79

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

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

Extension Idea

Ask: If there are two fractions with the same numerator but different denominators, which fraction will be smaller?

Say: The fraction with the larger denominator will be smaller, because a larger denominator means each part is smaller.

Learning Objective

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

Teaching Aids

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

Activity

Divide the class into groups. Provide each group with fraction strips of different fractions (e.g., 1 4 , 2 3 , 3 8 , and 5 6 ). Instruct them to place the fraction strips one below the other to see which strip has the greater shaded area. Also, instruct them to arrange the fraction strips in ascending order.

Now, instruct the students to find the Least Common Multiple (LCM) of the denominators of all four fractions. Instruct them to make equivalent fractions according to the LCM and compare the numerators to arrange the fractions in ascending order in their notebooks.

Compare the results obtained in both the cases.

Extension Idea

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

Say: 316 163 < . So, 3 16 is smaller.

Answers

1. Reviewing Fractions

Do It Together 1. 98772779 999 ×++ ==

2. Equivalent Fractions Using Multiplication

Do It Together

4

. Comparing Fractions

Do It Together

3 2 or 11 6

Apply the cross-multiplication method.

3 × 6 =18

2 × 11 =22

18 < 22

Thus, 3 2 < 11 6

5

.

Ordering Fractions

3. Equivalent Fractions Using Division

Think and Tell

Yes, equivalent fractions give like fractions when simplified.

Do It Together

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

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

HCF = 48

Simplest form for 481 962 =

Thus, the ascending order is <<< 2353 3462

Operations on Fractions 6

Learning Outcomes

Students will be able to:

add unlike fractions.

add mixed numbers.

subtract unlike fractions.

subtract mixed numbers.

multiply a fraction by a whole number.

multiply two fractions.

divide a whole number by a fraction.

divide a fraction by a whole number.

divide a fraction by a fraction.

Alignment to NCF

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

Let’s Recall

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

Vocabulary

unlike fraction: fractions which have different denominators mixed number: fractions which have a whole number part and a proper fraction

reciprocal: the inverse of a number

Teaching Aids

Rectangular strips; Coloured pencils; Circular cutouts

Chapter: Operations on Fractions

Adding Unlike Fractions

Learning Outcomes

Students will be able to add unlike fractions.

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Imagine Maths Page 87

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

Instruct the students to add the two fractions 1 2 and 1 3 by first showing the two fractions on the rectangular strips. Then discuss how to divide the strips in a way where both show the same total number of equal parts. Explain that this can be done either by finding the LCM of the denominators or multiplying the denominators. Discuss that each strip should be divided into the same number of parts as the LCM of the denominators.

Ask the students to then add the 2 fractions by converting both fractions to like fractions and adding the numerators in their notebooks.

Extension Idea

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

Say: Total length of the cloth = 5

Adding Mixed Numbers

Learning Outcomes

Students will be able to add mixed numbers.

Teaching Aids

Circular cutouts; Coloured pencils

Activity

Imagine Maths Page 88

Instruct the students to work in pairs. Distribute the circular cutouts and coloured pencils among the pairs. Ask the students to show 1 1 2 and 1 1 4 on the circular cutouts. Instruct them to add the given two mixed numbers. Ask the students to first put the shaded wholes together. They will then add the fractional parts the same way as they did in the previous lesson.

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

Subtracting Unlike Fractions

Learning Outcomes

Students will be able to subtract unlike fractions.

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Imagine Maths Page 91

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

Ask the students to show 1 3 and 1 4 on each of the rectangular strips.

Instruct the students to subtract 1 4 from 1 3 on the rectangular strips. Discuss how to divide the strips in a way where both show the same total number of equal parts.

Explain that this can be done either by finding the LCM of the denominators or multiplying the denominators. Discuss that each strip should be divided into the same number of parts as the LCM of the denominators. Ask the students to then subtract by converting both fractions to like fractions and subtracting the numerators in their notebooks.

Extension Idea

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

Say: Length of ribbon left = 7 8 –1 3 = 21 24 –8 24 = 13 24 m

Subtracting Mixed Numbers

Learning Outcomes

Students will be able to subtract mixed numbers.

Teaching Aids

Circular cutouts; Coloured pencils

Activity

Imagine Maths Page 92

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

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

Instruct them to subtract the two given mixed numbers.

Ask the students to subtract the shaded wholes by removing as many wholes from the minuend as the subtrahend. They will then subtract the fractional parts the same way as they did in the previous lesson.

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

Multiplying Fractions and Whole Numbers

Learning Outcomes

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

Teaching Aids

Rectangular strips; Coloured pencils

Activity

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

Imagine Maths Page 95

Instruct the students to multiply 2 by 1 4 using the strips. Instruct them to show 2 groups of 1 4 using the strips. Ask them how many quarters there are in 2 strips of paper. Ask them to show the quarters in two strips, using the third strip. Have a discussion around what they found.

Now, ask them to multiply the whole number by the fraction in their notebooks. Instruct them to multiply the whole number by the numerator and keep the denominator unchanged. They will then compare the answers with the answers they got using the strips.

Multiplying Two Fractions

Learning Outcomes

Students will be able to multiply two fractions.

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Imagine Maths Page 96

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

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

Have a discussion around how they can show 1 2 groups of 1 3 .

Bring out the fact that they need to shade 1 3 in the rectangular strip, and then divide each part into 2 equal parts. Once the students show the same on the strip, ask them to draw crosses in the horizontal row to show 1 2 .

Discuss with the students that the part that is both coloured and has a cross is the product.

Now, ask them to solve the problem in their notebooks by multiplying the numerators and the denominators. They can then compare the answers that they get using both methods.

Extension Idea

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

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

Dividing a Whole Number by a Fraction

Learning Outcomes

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

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Imagine Maths Page 98

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

Tell them to find 3 ÷ 3 4 . Ask the students to check how

many three-quarters there are in three wholes using the strips and coloured pencils.

Let them explore and get to the answer. Help them if needed. Once the students get to the answer using the strips, introduce the reciprocal method.

Now, ask the students to solve the question in their notebooks by first finding the reciprocal of the fraction and then multiplying the fractions. Ask them to compare the answers that they get using both methods. Repeat the activity with one more set of fractions.

Extension Idea

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

Say: We will take 5 strips which are divided into 3 equal parts each. Now, we will make pairs of 2 and count the total number of pairs that we get. We get 7 pairs of 2 and 1 equal part is left. Thus, 5 ÷ 2 3 = 71 2 .

Dividing a Fraction by a Whole Number

Learning Outcomes

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

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Imagine Maths Page 100

Group 1 Group 2

3

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

Instruct the students to show 6 9 on the rectangular strip. Ask the students to make 2 groups out of them since 6/9 is to be divided by 2. Ask the students to count the number of equal parts in each group, that form the numerator of the quotient, with the denominator remaining the same.

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

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

Extension Idea

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

Divide 1 2 by 3.

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

Dividing a Fraction by a Fraction

Learning Outcomes

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

Teaching Aids

Rectangular strips; Coloured pencils

Activity

Instruct the students to form groups.

Distribute the rectangular strips and coloured pencils.

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

Imagine Maths Page 101

Instruct the students to show the fraction 1/3 on the sheet by dividing the sheet horizontally and drawing crosses to show it. Then ask them to use the same sheet to show the fraction ¾ by dividing the sheet vertically and colouring to show it. Then ask them to count the number of equal parts that have the cross which is the numerator of the answer (9) while the number of equal parts that are shaded shows the denominator of the answer (4).

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

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

1. Adding Unlike Fractions

Answers

6. Multiplying Two Fractions

Do It Together

2. Adding Mixed Numbers

Do

3. Subtracting Unlike Fractions Do

7. Dividing a Whole Number by a Fraction

Think and Tell

The reciprocal of 1 is 1.

Do It Together

8. Dividing a Fraction by a Whole Number Do It

4. Subtracting Mixed Numbers

Do It Together

5. Multiplying Fractions and Whole Numbers

9. Dividing a Fraction by a Fraction

7

Introduction to Decimals

Learning Outcomes

Students will be able to:

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

Alignment to NCF

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

Let’s Recall

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

Vocabulary

decimals: numbers between whole numbers

expanded form: expressing a number as the sum of the values of all the digits in a number rounding off: approximating a number to a certain place value for easier calculation

Teaching Aids

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

Chapter: Introduction to Decimals

Tenths

Learning Outcomes

Imagine Maths Page 107

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

Teaching Aids

Decimal grids for tenths; Crayons

Activity

Begin by discussing decimal numbers in real life and their presentation on a decimal grid.

Take 0.4 as an example. Ask the students what fraction of the grid is shaded. Explain that

4 10 is called four-tenths and is written as 0.4 in decimals. Help them write this in their notebooks, in figures as 0.4, in words as “four-tenths”, and in expanded form as 0 + 0.4.

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

Extension Idea

Ask: Can one-tenth be further divided into 10 equal parts? What fraction of the grid would be shaded?

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

Hundredths

Learning Outcomes

Imagine Maths Page 111

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

Teaching Aids

Decimal grids for tenths; Crayons

Activity

Instruct the students to work in groups. Distribute the tenth grids among the groups. Help them to convert the tenths grid to the hundredths grid by drawing horizontal lines so they can see the change. Ask them to show 35 100 on the grid using crayons. Guide them to write 0.35 as 0.35 in numbers, “thirty-five hundredths” in words, and in expanded form as: 0 + 3 10 + 5 100 in their notebooks. Next, ask them to show 0.75 on the grid and write it in words and expanded form in their notebooks. Ask questions like: How would you write 5.27 in words? How would you pronounce the decimal point?

Learning Outcomes

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

Teaching Aids

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

Activity

Begin by asking a question such as the difference between 0.5, 0.05 and 0.005 in terms of the place value of 5. Discuss with the students the initial thoughts on understanding the place value of decimals. Use a grid divided into 10 equal parts to represent 0.5. Demonstrate to them how to further divide it into 100 and 1000 to represent 0.05 and 0.005. Write each decimal number in words and in expanded form on the board.

Instruct the students to work in groups. Distribute the teaching aids among the groups. Instruct the students to use grids to represent the given decimal numbers. Ask them to write the decimal numbers in words and figures in their notebooks.

Extension Idea

Instruct: Write 5 wholes, 5 tenths, 3 hundredths and 2 thousandths in numeral, word and expanded form.

Say: The number in figures can be written as: 5.532; in words as: five and five hundred thirty-two thousandths; and in expanded form as: 5 + 532 101001000 ++ . Conversion

Learning Outcomes

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

Teaching Aids

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

Activity

Demonstrate the conversion of fractions to decimals and vice versa. Recall the concept of the simplest form of fractions.

Instruct the students to work in 4 groups.

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

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

Learning Outcomes

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

Teaching Aids

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

Activity

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

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

0.16 0.713 0.1

0.4

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

Comparing and Ordering Decimals

Learning Outcomes

Students will be able to compare and order decimals.

Teaching Aids

Cards with decimals numbers written on them

Activity

Demonstrate the comparison and ordering of decimals to the class.

Imagine Maths Page 122

Take the students out to the playground. Instruct the students to work in groups of 5. Draw a large number line on the floor with decimal values up to hundredths. Place some decimal cards on a large table. Explain the rules of the relay race to the students. Ask one student from each group to run to the table, pick a card, and place themselves at the correct spot on the number line. Instruct the students to take turns in the relay race format, ensuring each member of a group gets a chance to participate. Use a stopwatch or timer to track the duration of the game. Conclude the activity with a class discussion of the learning gained from physically moving on the number line. Declare the group that finishes first with the decimals correctly arranged on the number line as the winning team. Back in the classroom, ask the students to write the correct order of their decimals in their notebooks.

Extension Idea

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

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

Rounding off Decimals

Learning Outcomes

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

Teaching Aids

Ice cream sticks; Paper cups

Activity

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

Explain to the student that if the number in the tenths place is 5 or more, they should run to the ‘Taller’ tree. If it’s 4 or less, they should run to the ‘Shorter’ tree.

Now, those at the ‘Taller’ tree should round their decimals up to the next number, and the others should round to the current number. Ask them to write down their answers in their notebooks.

Extension Idea

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

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

1.

7.

2.

Whole Numbers and Tenths

1.

2.

3.

8. Equivalent Decimals

3

0.8

9. Converting Fractions to Decimals

Think and Tell

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

3.4

4

5

6

10. Converting Decimals to Fractions

11.

Like Decimals

Do It Together

Number of digits after the decimal point in 51.54 = 2

Number of digits after the decimal point in 187.37 = 2

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

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

12.

Unlike Decimals

Do It Together

Number of digits after the decimal point in 3.145 = 3

Number of digits after the decimal point in 54.14 = 2

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

Thus, 3.145 and 54.14 (are/are not) unlike decimals.

13.

Converting Unlike Decimals to Like Decimals

Do It Together

189.4745 = 189.4745

1.2 = 1.2000

Thus, 189.4745 and 1.2000 are a set of like decimals.

14. Comparing Like Decimals

Do It Together

15. Comparing Unlike Decimals

Do It Together

The

16. Ordering Fractions

17.

Rounding off to the Nearest Whole Number

Operations with Decimals 8

Learning Outcomes

Students will be able to:

add decimals.

subtract decimals.

multiply decimals by 10, 100 and 1000.

multiply decimals by whole numbers and decimals.

convert currencies to currencies of different countries and vice versa.

divide decimals by 10, 100 and 1000.

divide decimals by whole numbers.

Alignment to NCF

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

Let’s Recall

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

Vocabulary

currency: medium of exchange for goods and services

Teaching Aids

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

Chapter: Operations with Decimals

Learning Outcomes

Students will be able to add decimals.

Teaching Aids

Decimal grids for 10s and 100s; Crayons

Activity

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

Provide them with two questions around adding two tenths and two hundredths (such as 0.3 + 0.6, and 0.35 + 0.25), ask them to colour the 10s and 100s grids to first show the decimals, and then add the shaded grids in each to show the answer for each problem.

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

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

Extension Idea

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

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

Learning Outcomes

Students will be able to subtract decimals.

Teaching Aids

Decimal grids for 10s and 100s; Crayons

Activity

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

Ask questions like: How will you show subtraction on the grid?

Guide them if needed, and ask them to cross out the shaded cells to show the subtraction of the second number. Discuss the answers.

Ask them to perform subtraction using the column method in their notebooks, and compare the answers that they got using both methods. 0.7 – 0.3

Learning Outcomes

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

Teaching Aids

Crossword puzzle card

Activity

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

Ask the students to look at the multiplication and discuss in their groups what they notice about the product in each case. Discuss student responses.

Distribute the crossword puzzle card to each group. Instruct them to solve these problems and write the answers in the respective blanks.

Extension Idea

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

Say: Yes, the same rule applies to multiplying by 10,000 or 1,00,000. 5.142 × 1,00,000 = 5,14,200

Multiplying Whole Numbers and Decimals; Imagine Maths Page 137

Multiplying Two Decimals

Learning Outcomes

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

Teaching Aids

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

Activity

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

Instruct them to colour the grid to show one decimal vertically and draw lines to show the other decimal horizontally. Ask questions around the overlapping area that is shaded and has lines. Direct students to count the number of such cells which forms the product of the 2 decimals. Then, ask them to write the 2 decimals and the answer in their notebooks. They will then discuss among their groups how they should multiply decimals using the column method without the grid.

Give them one more problem to solve.

Extension Idea

Ask: Aarav got 1.45 points in each of the 5 tests. How many points did he get?

Say: To find the answer, we need to multiply 1.45 by 5. 1.45 × 5 = 7.25.

Learning Outcomes

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

Teaching Aids

Things with price tags in 4 different currencies

Activity

Begin the lesson with a small discussion on using Indian currency to buy goods in other countries. Explain the concept of different countries having their own currencies and the necessity of converting money when travelling internationally.

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

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

Extension Idea

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

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

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

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

Learning Outcomes

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

Teaching Aids

Puzzle cards with division problems

Activity

Imagine Maths Page 141

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

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

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

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

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

Learning Outcomes

Students will be able to divide decimals by whole numbers.

Teaching Aids

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

Activity

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

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

Answers

1. Addition of Decimals

2. Subtraction of Decimals

6. Currencies from Different Countries

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

2.

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

1.

3.

4. Multiplying Whole Numbers and Decimals

8. Dividing Decimals by Whole numbers

5. Multiplying Two Decimals

Think

Lines and Angles 9

Learning Outcomes

Students will be able to:

identify a point, ray, line and line segment and classify the different types of lines.

identify and classify the different types of angles.

measure angles using a protractor. draw angles using a protractor.

identify triangles, quadrilaterals and polygons and classify the types of quadrilaterals and polygons.

Alignment to NCF

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

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

Let’s Recall

Recap to check if students know how to identify sleeping, standing and slanting lines. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

line segment: the shortest distance between 2 points having a definite length ray: a part of a line that has an end point and extends indefinitely in the other direction right angle: an angle measuring 90° protractor: a device in the shape of a D that helps to measure angles of different degrees

Teaching Aids

Circle and triangle coloured paper cutouts; Toothpicks; Sheets of paper; Glue sticks; Clock with movable hands; Colourful craft paper; Ice cream sticks; Protractors; Sketch pens; Rulers; Chart paper

Chapter: Lines and Angles

Types

of Lines

Learning Outcomes

Imagine Maths Page 148

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

Teaching Aids

Circle and triangle coloured paper cutouts; Toothpicks; Sheet of paper; Glue stick

Activity

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

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

Discuss parallel, intersecting and perpendicular lines, and ask the students to create them on the sheet of paper. Ask questions like: Which pairs of lines will never meet when extended?

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

Types of Angles

Learning Outcomes

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

Teaching Aids

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

Activity

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

Imagine Maths Page 153

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

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

Measuring Angles

Learning Outcomes

Students will be able to measure angles using a protractor.

Teaching Aids

Protractor

Activity

Imagine Maths Page 157

Show a protractor and the different markings on it. Demonstrate how to use it to measure an angle. Instruct the students to work in pairs. Distribute the protractors to the students.

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

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

Extension Idea

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

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

Drawing Angles

Learning Outcomes

Students will be able to draw angles using a protractor.

Teaching Aids

Sheets of paper; Sketch pen; Ruler; Protractor

Activity

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

Imagine Maths Page 158

X Y

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

Extension Idea

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

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

Learning Outcomes

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

Teaching Aids

Ice cream sticks; Chart paper

Activity

Instruct the students to form groups.

Distribute the ice cream sticks and chart paper.

Instruct the students to create a collage of polygons on the chart paper using the ice cream sticks as sides. Write the name of the polygon and its features, like the number of sides, the number of angles and the number of vertices under each polygon.

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

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

Extension Idea

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

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

Answers

1. Types of Lines

Ray Do It Together

2. Types of Angles

Acute Angle

Straight

Do

3.

Measuring and Drawing Angles

Drawing Angles

Do It Together

2.

4. Triangles, Quadrilaterals and Polygons

Think and Tell

Yes, we can make more shapes with 4 sides and 4 angles. A square, rectangle, rhombus, parallelogram, trapezium or kite have 4 sides and 4 angles each.

Think and Tell

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

Do It Together

Number of sides – 8

Number of angles – 8

Number of sides – 4

Number of angles – 4

Number of sides – 4

Number of angles – 4

Number of sides – 4

Number of angles – 4

Patterns and Symmetry 10

Learning Outcomes

Students will be able to: extend and create repeating and rotating patterns. extend and create growing and tiling patterns. identify the rule of a number pattern or puzzle, and extend and solve it. identify symmetry in shapes and figures and draw lines of symmetry.

Alignment to NCF

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

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

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

Let’s Recall

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

Vocabulary

extend (pattern): show the terms that would come next in the sequence

rotate: circle around a point

overlap: to cover something partly by going over its edge

Teaching Aids

Flowers and leaves; Water colours; Paint brushes; Rectangle, triangle and circle paper cutouts; Slips of paper for writing a secret message; White sheets

Chapter: Patterns and Symmetry

Extending and Creating Patterns

Learning Outcomes

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

Teaching Aids

Flowers and leaves; Water colours; Paint brushes

Activity

Imagine Maths Page 168

Begin the class by discussing what repeating and rotating patterns are and how to extend them.

Instruct the students to form groups of 4. Take the students to the playground.

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

Distribute the water colours and the paint brushes to the groups.

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

Extension Idea

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

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

Growing and Tiling Patterns

Learning Outcomes

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

Teaching Aids

Rectangle, triangle and circle paper cutouts

Activity

Begin by discussing growing and tiling patterns.

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

Imagine Maths Page 172

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

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

Extension Idea

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

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

Imagine Maths Page 175

Learning Outcomes

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

Teaching Aids

Slips of paper for writing a secret message

Activity

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

Instruct the students to form groups of 4.

Write the given code on the board.

12 9 6 5 9 19

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

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

Symmetry

Learning Outcomes

Imagine Maths Page 181

Students will be able to identify symmetry in shapes and figures and draw lines of symmetry.

Teaching Aids

Sheets of paper; Water colours; Paint brushes

Activity

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

Instruct the students to fold their sheets in half and unfold them to make a crease in the middle of the sheet. Using water colours, they will paint a design or pattern on one side.

Ask the students to fold the sheet precisely along the crease created before, ensuring the painted side is facing inward. Instruct them to press down on the folded paper to transfer the colours to the other side and unfold the sheet. For instance, the sheet may look like the given example.

Ask: What do you notice? Is the pattern symmetrical?

Extension Idea

Ask: This paper has a secret message written in code. Can you figure out what it says?

Say: If we put the sheet in front of the mirror, we can see that the secret message is ‘MATH IS FUN’.

1. Repeating Patterns

Think and Tell

Zebra crossing

Do It Together

2. Rotating Patterns

Do It Together

3. Growing and Reducing Patterns

Do It Together Pattern What will come next?

Answers

5. Coding and Decoding Patterns

Do It Together

Decode the word: CNGGREA.

We can see that:

Therefore, the given word is PATTERN.

6. Patterns in Numbers

Think and Tell

Yes. The next number is twice of the number before it.

Do It Together

Growing pattern

Reducing pattern

4. Tiling Patterns

Think and Tell

Yes, a tangram also has a tiling pattern.

Do It Together

7. Drawing Lines of Symmetry

Think and Tell

Symmetrical objects: Bottle, Blackboard etc.

Asymmetrical objects: Tree, Shoe, etc.

Do It Together

Length and Weight

Learning Outcomes

Students will be able to:

estimate the length of an object and measure it using a ruler. convert between the different units of length.

solve word problems on adding, subtracting, multiplying and dividing lengths. estimate the weight of an object.

convert between the different units of weight.

solve word problems on adding, subtracting, multiplying and dividing weights.

Alignment to NCF

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

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

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

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

Let’s Recall

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

Vocabulary

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

Teaching Aids

Ruler; Pencils; Sketch pens; Crayons; Erasers; Paint brushes; Paper clips; Problem cards with different length conversion problems; Problem cards with different weight conversion problems; Solution cards with corresponding converted lengths; Solution cards with corresponding converted weights; Word problem cards with different length problems; Word problem cards with different weight problems; Weight blocks of 50 mg, 10 g, 500 g and 1 kg; Classroom supplies

Chapter: Length and Weight

Estimating Length; Measuring Lengths

Learning Outcomes

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

Imagine Maths Page 189

Teaching Aids

Ruler; Pencils; Sketch pens; Crayons; Erasers; Paint brushes; Paper clips

Activity

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

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

Ask: How long do you think this pencil/pen is? Will it be measured in mm or cm?

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

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

Extension Idea

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

Units of Length

Learning Outcomes

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

Teaching Aids

Imagine Maths Page 191

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

Activity

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

Thousands kilometre (km)

Hundreds hectometre Tens decametre Ones metre (m)

Tenths decimetre Hundredths centimetre (cm)

Thousandths millimetre (mm)

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

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

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

Extension Idea

Ask: Why do we need to convert between different units of length?

Say: It is easier to understand bigger lengths, such as the length of a road in bigger units and shorter lengths such as the length of a rice grain in smaller units. Thus, sometimes we need to convert between the closest units as per the lengths of objects.

Word Problems on Length

Learning Outcomes

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

Teaching Aids

Word problem cards with different length problems

Activity

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

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

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

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

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

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

Estimating Weights

Learning Outcomes

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

Teaching Aids

Weight blocks of 50 mg, 10 g, 500 g and 1 kg; Classroom supplies

Activity

Imagine Maths Page 195

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

Ask questions like: Which block feels heavier? Which one is the lightest? How much do they weigh?

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

Paper clip = about 2 mg.

Extension Idea

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

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

Units of Weight

Learning Outcomes

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

Teaching Aids

Imagine Maths Page 195

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

Activity

Create problem cards and solution cards. Distribute 2–3 problem cards to each student and place the solution cards face down on a big table. Draw a weight unit conversion line and markings on the floor.

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

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

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

Word Problems on Weights

Learning Outcomes

Imagine Maths Page 197

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

Teaching Aids

Word problem cards with different weight problems

Activity

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

1. How much fruit does she have in total?

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

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

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

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

Extension Idea

Ask: Can you think of other word problems with weight in your day-to-day lives? Give some examples.

Say: There are many problems with weight in our real lives. For example: I buy 12 kg of rice, 5 kg of wheat flour and 2 kg of sugar. What’s the total weight of the food items I bought?

1. Estimating Length

Do It Together

Students will circle the following lengths:

2. Length of a pen: 14 cm

3. Height of a car: 160 cm

4. Length of your shoe: 12 cm

2. Measuring Lengths

Do It Together

The toffee is longer than 3 cm but shorter than 4 cm. There are 7 mm lines after 3 cm.

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

3. Units of Length

Do It Together

1000 mm = 1 m

1 mm = 1 1000 m

5 m 230 mm = 5 m + 230 × 1 1000 m

= 5 m + 0.230 m = 5.230 m

4. Word Problems on Length

Think and Tell

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

So, 1 m 55 cm = 1 m + 55 100 m = 1 m + 0.55 m = 1.55 m

Do It Together

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

Number of sarees required = 5

Total length of cloth needed for making 5 sarees = 5.80 m × 5

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

Answers

5. Estimating Weights

Do It Together

Students will circle the following estimates:

2. A 2-rupee coin: 14 g

3. A photo album: 600 g

4. A tube of toothpaste: 120 g

6. Units of Weight

Do It Together

1 g = 100 cg

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

7. Word Problems on Weights

Think and Tell

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

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

= 2 kg + 0.106 kg = 2.106 kg

Do It Together

Total weight of books removed from the suitcase

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

Weight of the suitcase after removing the books

= 9.200 kg – 1.680 kg = 7.520 kg OR 7 kg 520 g

Perimeter and Area 12

Learning Outcomes

Students will be able to:

find the perimeter of a square and a rectangle. find the area of a square and a rectangle and use it to find the area of compound shapes. find the area of a triangle on a square grid.

Alignment to NCF

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

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

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

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

Let’s Recall

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

Vocabulary

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

Teaching Aids

Squared paper; Crayons

Chapter: Perimeter and Area

Learning Outcomes

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

Teaching Aids

Squared paper

Activity

Imagine Maths Page 203

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

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

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

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

Extension Idea

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

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

Area of a Square; Area of a Rectangle

Learning Outcomes

Imagine Maths Page 206

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

Teaching Aids

Squared paper; Crayons

Activity

Start by discussing the area of a square and a rectangle. Bring out the formula for the area of both the shapes.

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

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

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

Area of a Triangle Imagine Maths Page 210

Learning Outcomes

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

Teaching Aids

Squared paper

Activity

Start by discussing the area of a triangle. Bring out the fact that the area of a triangle is half that of a rectangle. Instruct the students to work in pairs. Distribute a sheet of squared paper and crayons to each student.

Ask the students to draw a triangle on the squared sheet such that the triangle covers halves and full squares.

Ask them to then exchange the squared paper with their partners. Their partner will then find the area of the shape by counting the number of half and full squares covered.

Instruct them to write the answers in their notebooks.

Extension Idea

Instruct: Find the area of the given figure.

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

Answers

1. Perimeter of a Rectangle

Do It Together

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

Perimeter of a rectangle = 2l + 2b

The perimeter of the pillow = 2 × 60 + 2 × 20 = 120 + 40 cm = 160 cm

2. Perimeter of a Square

Do It Together

Here, side (s) = 45 cm.

Perimeter of a square = 4 × 45 cm

So, the perimeter of the chessboard is 180 cm.

3

. Area of a Square

Think and Tell

There are 5 rows. Each row has 5 small squares. The required multiplication sentence is 5 × 5 = 25. There are 25 small boxes in total. We noticed that the number of rows and number of squares in each row are equal.

Do It Together

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

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

4. Area of a Rectangle

Do It Together

The area of rectangle A = 3 × 2 sq. cm = 6 sq. cm

The area of rectangle B = 2 × 12 sq. cm = 24 sq. cm

The area of the whole shape = 6 + 24 sq. cm = 30 sq. cm

5.

Area of a Triangle

Do It Together

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

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

Area of whole triangle = 18 sq. units

Capacity and Volume

Learning Outcomes

Students will be able to:

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

Alignment to NCF

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

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

Let’s Recall

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

Vocabulary

capacity: the amount that can be held in a particular space

volume: the amount of space that an object occupies unit cubes: small, equal-sized cubes used as a standard measure

Teaching Aids

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

Chapter: Capacity and Volume

Estimating Capacity; Measuring Capacity

Learning Outcomes

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

Teaching Aids

Imagine Maths Page 216

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

Activity

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

Instruct the students to form groups. Distribute the containers among the groups.

Instruct the students to guess the amount of water that the containers can hold, and write in their notebooks. Distribute the measuring jars and the water.

Instruct the students to measure the capacity of each container by filling it with water and pouring the water into the measuring jar, and note down the measures in their notebooks.

Have a discussion around the estimated capacity and the actual capacity of the containers. Ask questions like: Does a container with a greater height always have more capacity than a shorter container? Why?

Units of Capacity

Learning Outcomes

Students will be able to convert between units of capacity.

Teaching Aids

Conversion cards

Activity

Start by introducing different units of capacity beyond millilitres (mL) or litres (L), like kilolitres, decalitres, hectolitres, decilitres and centilitres.

Divide the class into groups. Distribute a pack of 10 conversion cards to each group. Instruct the students to convert the given capacity into different units of measurement as mentioned in the conversion cards, and then match the cards with measures having the same capacity. E.g., 653 L is 6530, so the card with 6530 should be placed on the side reading dL.

Ask the students to write the answers in their notebooks.

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

Imagine Maths Page 217

Volume of Solids Using Unit Cubes

Learning Outcomes

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

Teaching Aids

Different sized rectangular containers; Unit cubes

Activity

Start by introducing the idea of volume, explaining it as the space that an object occupies.

Divide the students into groups. Distribute unit cubes and one container to each group.

Instruct the students to fill in their containers by arranging the cubes in layers till the container gets filled. Ask them to count the total number of cubes that could fit inside the container and write in their notebook.

Repeat the activity by asking groups to empty their containers and then exchange with other groups.

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

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

Extension Idea

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

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

Volume of Solids Using the Formula

Learning Outcomes

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

Teaching Aids

Different sized rectangular containers; Unit cubes

Activity

Imagine Maths Page 222

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

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

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

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

Word Problems on Volume and Capacity

Learning Outcomes

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

Teaching Aids

Question cards with a word problem written; Unit cubes

Activity

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

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

Add rigour to the learning by giving them questions where conversions are involved.

Extension Idea

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

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

Answers

1. Estimating Capacity

Do It Together

2. Measuring Capacity

Think and Tell

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

Do It Together

The jug contains 750 mL of liquid.

3. Units of Capacity

Do It Together

1 L = 100 cL

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

4. Volume of Solids Using Unit Cubes

Do It Together

Layer 1 has 12 unit cubes.

Layer 2 has 12 unit cubes.

Total number of unit cubes = 24

Volume of the given solid = 24 cu. units

5. Volume of Solids Using the Formula

Do It Together

Number of unit cubes = 5 × 2 = 10

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

6. Word Problems on Volume and Capacity

Think and Tell

No, the volume of water in the two containers will not be the same.

Think and Tell

To convert 20 L 126 mL into L, we write the litres and millilitres together and place a decimal point between them.

In this case, it’s 20.126 L.

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

Do It Together

Number of cartons that are placed along the length = 20

Number of cartons that are placed along the breadth = 12

Number of cartons in 1 such layer = 20 × 12 = 240

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

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

3-D Shapes on Flat Surfaces 14

Learning Outcomes

Students will be able to:

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

Alignment to NCF

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

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

Let’s Recall

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

Vocabulary

edges: the line segments that form the shape

vertex: a point where three or more edges meet in a three-dimensional object

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

map: a picture of a place printed on a flat surface

Teaching Aids

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

Chapter: 3-D Shapes on Flat Surfaces

Features of 3-D Shapes

Learning Outcomes

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

Teaching Aids

3-D wooden shapes

Activity

Begin the class by asking the students, to look around and identify different 3-D shapes that they see in the classroom. Have a discussion around that.

Instruct the students to work in groups. Distribute 3-D wooden shapes among the groups.

Imagine Maths Page 231

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

Extension Idea

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

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

Drawing 3-D Shapes

Learning Outcomes

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

Teaching Aids

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

Activity

Imagine Maths Page 233

Demonstrate to the students how 3-D shapes can be represented on a grid. Make slips of paper naming different solid shapes (cone, cube, prism, sphere, cuboid, etc.) and put them inside a bowl.

Distribute squared paper among the students.

Call the students one by one to pick a slip. Instruct them to open page 233 of the Imagine Mathematics book and see how to draw a 3-D shape on a square grid. Ask them to draw the chosen object on the grid, adding details and colours to make it visually appealing. Give them a chance to show their drawing to the class.

Nets of 3-D Shapes

Learning Outcomes

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

Teaching Aids

Boxes shaped like a pyramid and prism; Pair of scissors; Sticking tape; Thick chart paper

Activity

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 them how 3-D objects are different from their 2-D representations.

Ask the students to work in pairs. Distribute boxes shaped like a pyramid and prism, a pair of scissors, cello tape and a thick chart paper among the pairs.

Ask the pairs to visualise what the nets of the prism and pyramid with them 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: How many types of nets does a triangular prism have?

Say: A triangular prism has nine distinct nets.

Views of Cube Structures

Learning Outcomes

Students will be able to draw views of cube structures.

Teaching Aids

Unit cube blocks

Activity

Instruct the students to work in groups of 3.

Distribute the unit cube blocks among the groups.

Imagine Maths Page 236

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

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

Extension Idea

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

Say: The given views are of a triangular pyramid.

Learning Outcomes

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

Teaching Aids

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

Activity

Divide the students into groups of 4. Distribute the teaching aids among the groups.

Ask the groups to arrange the word cards on the chart paper as per the given instructions to create a map of the school. Be specific in the instructions, such as placing the school gate, admin office, and school building adjacent to each other, with the admin office positioned at the centre. Specify distances between locations; for instance, the admin office and school gate are 10 cm apart, and the school building and the admin office are 5 cm apart. Let the groups display the maps and have a discussion with the rest of the class around that. Ask questions based on the map created. Ask them to note down the answers in their notebooks.

Extension Idea

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

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

Floor Plans and Deep Drawings

Learning Outcomes

Imagine Maths Page 242

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

Teaching Aids

Cardboard sheets; Glue sticks; Sticking tape; Pair of scissors; Sketch pens

Activity

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

Extension Idea

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

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

Answers

1. Features of 3-D Shapes

2. Drawing 3-D Shapes

Do It Together

3. Nets of 3-D Shapes

Do

5. Reading Maps

Think and Tell

To draw the same picture in a 1 2 cm grid, the grid to be chosen should be half the size of the squares. So, the size of the squares would change.

Do It Together

1. The city park is west of the highway.

2. The hotel is east of the highway.

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

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

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

6. Floor Plans and Deep Drawings

4. Views of Cube Structures

Do It Together

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

The deep drawing doesn’t match the floor plan, as one of the doors is not present in the floor plan.

Time and Temperature

Learning Outcomes

Students will be able to:

convert between the different units of time (hours, minutes, seconds). calculate the elapsed time between different units of time. solve word problems on conversion and calculation of time. solve problems on measuring and calculating temperature.

Alignment to NCF

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

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

Let’s Recall

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

Vocabulary

thermometer: a tool to measure temperature

mercury: a silver-white metal

Teaching Aids

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

Chapter: Time and Temperature

Converting Between Units of Time

Learning Outcomes

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

Teaching Aids

Time conversion puzzle having square pieces that have units of measurement written on the 4 sides

Activity

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

Instruct the students to work in groups.

Distribute the time conversion puzzles among the groups.

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

Extension Idea

Ask: How many hours are there in 7200 seconds?

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

Calculating Duration

Learning Outcomes

Students will be able to calculate the elapsed time between different units of time.

Teaching Aids

Year calendar; School monthly calendar

Activity

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

Instruct them to work in groups.

Distribute the school’s monthly calendar to the groups.

Instruct the students to now look at the calendar and say the duration of the summer and winter vacations in days and weeks.

Ask questions like: How many days are there between Holi and Good Friday?

Calculating Time

Learning Outcomes

Imagine Maths Page 250

Month Event

January

3rd – Winter vacation ends

15th – Makar Sankranti

26th – Republic Day

February 14th – Saraswati Puja

March 25th – Holi

April 7th – Good Friday

14th – Ambedkar Jayanti

May 22nd – Summer Vacation starts

June 17th – Summer Vacation ends

July 29th – Muharram

August 15th – Independence Day

30th - Rakshabandhan

September 5th – Teacher’s Day

October

2nd – Gandhi Jayanti

21st – Durga Puja starts 27th – Durga puja ends

November 14th – Children’s Day

December 23rd – Winter vacation begins 25th – Christmas

Imagine Maths Page 253

Students will be able to solve word problems on conversion and calculation of time.

Teaching Aids

Word problem sheets

Activity

Instruct the students to work in groups.

Distribute the word problem sheets among the groups. Ask them to read the problem: Suhani starts reading a book at 6:45 p.m. and finishes at 8:20 p.m. How much time did she spend reading the book? and note what is given in the problem, what they need to find out and then solve the problem to find the answer.

Instruct the students to solve the word problem and discuss the answer in the class.

Extension Idea

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

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

Measuring Temperature

Learning Outcomes

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

Teaching Aids

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

Activity

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

Divide the class into groups of 3 and provide each group with a small thermometer.

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

Ask one student from each group to come forward and measure the temperature of the hot water. Ask them to say the reading and then go back to their groups and mark the reading of hot water in the provided sheets with a red pen or pencil.

Repeat the activity with the other two students in each group where they will measure the temperatures of the normal water and the cold water.

Ask questions like: Did all the groups get the same temperature for all the 3 bowls? Why?

Extension Idea

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

Say: May. May is the hotter month and December is the colder month.

Answers

1. Converting from a Bigger Unit into a Smaller Unit

Think and Tell

Number of hours in a day = 24 hours

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

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

So, the number of seconds in an hour = 86,400 seconds.

Do It Together

Time remaining before the eclipse = 4.5 hours.

1 hour = 60 minutes

4.5 hours = 4.5 × 60 = 270 minutes

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

2. Converting from a Smaller Unit into a Bigger Unit

Do It Together

Number of seconds for which Shreya juggles = 250 seconds

We know that 60 seconds = 1 minute

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

3. Calculating Duration

Do It Together

Start Year: 2015; End Year: 2023

Start Month: June (6th month); End Month: August (8th month)

Difference in years: 2023 − 2015 = 8 years

Difference in months: 8 − 6 = 2 months

The tree took 8 years and 2 months to grow to its full height.

4. Calculating Time

Do It Together

Starting time: 2 hours 30 minutes

Time which the experiment lasted = 3 hours 45 minutes

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

2 hours

+ 3 hours 30 Minutes

45 Minutes

5 hours 75 Minutes

60 + 15 Minutes

1 hours

6 hours 15 Minutes

The experiment ended at 6 hours 15 minutes = 6:15 p.m.

5. Measuring Body Temperature

Think and Tell

Answers may vary.

Do It Together

45°C, 22°C, 18°C, 30°C

6. Measuring Temperature Around Us

Do It Together

Initial temperature = 12°C

Increase in temperature = 8°C

New temperature = 12°C + 8°C = 20°C

Money 16

Learning Outcomes

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

Alignment to NCF

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

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

Let’s Recall

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

Vocabulary

amount: the total number or quantity (for example, ‘the amount to be paid’) earn: to receive money as pay spend: to pay money out

Teaching Aids

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

Chapter: Money

Unitary Method

Learning Outcomes

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

Teaching Aids

Books; Play money; Number cards (1–9)

Activity

Divide the students into groups of 4. Distribute the play money to each group.

Imagine Maths Page 261

Ask one student from each group to act as a shopkeeper and the other three to act as customers. Tell the students that the shopkeeper has 10 books which cost ₹2000.

Distribute number cards within groups.

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

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

Extension Idea

Ask: If the cost of 2 1 2 L of milk is ₹145, what will be the cost of 1 4 4 L of milk?

Say: The cost of 1 L of milk will be ₹145 ÷ 2 1 2 = ₹58 and the cost of 1 4 4 L of milk is ₹58  41 4 = ₹246.50.

Profit or Loss

Learning Outcomes

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

Teaching Aids

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

Activity

Imagine Maths Page 263

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

Set up a play stall in the classroom. Distribute the play money to each student. Instruct the students to work in groups. One student from each group will manage the stall as a shopkeeper and he/she gets to keep the crayons, books and pencil box. The student will set up a cost price for each item. The other students of the group will shop at their group’s stall for 10 minutes. Encourage them to negotiate prices to get a better deal. For example, a student would want to buy an item for ₹10 that has a cost price of ₹15.

Once the purchase is done, each group will figure out if they made a profit or incurred a loss based on their initial cost prices and final selling prices.

Extension Idea

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

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

Answers

1. Unitary Method

Think and Tell

Coins: ₹1, ₹2, ₹5, ₹10, ₹20

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

Do It Together Cost

= ₹500 Cost of 3 notebooks = ₹1500

2. Profit or Loss

Think and Tell

Ram sold ₹6 pens for ₹10 to gain a profit.

3. Cost Price and Selling Price

Think and Tell

It is a profit since SP > CP.

Do It Together

4. Word Problems in Money

Do It Together

1. One chocolate

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

SP of each chocolate = ₹20

Since SP > CP, Vansh gained a profit.

Profit = SP – CP = ₹20 – ₹10 = ₹10

Thus, the profit gained by Vansh is ₹10

2. All chocolates

Total CP = ₹200

Total SP = 20 × ₹20 = ₹400

Since SP > CP, Vansh gained a profit.

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

Thus, the total profit gained by Vansh is ₹200.

Data Handling

Learning Outcomes

Students will be able to: read and interpret a single and double-bar graph. represent given data on a horizontal or vertical bar graph. read and interpret a pie chart. represent given data on a pie chart. read a line graph and answer questions based on it.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to read a single bar graph. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

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

Teaching Aids

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

Chapter: Data Handling

Learning Outcomes

Students will be able to read and interpret a single and double-bar graph.

Teaching Aids

Sheets with a double-bar graph showing data on the favourite activity for boys and girls

Activity

Begin with a brief discussion on what bar graphs are and how they represent data. Explain that a double-bar graph compares 2 sets of data.

Instruct the students to work in groups.

Distribute sheets with double-bar graphs drawn among the groups. Instruct them to look at the bar graph and answer questions like: How many boys participated in tennis?; How many more boys participated in cricket than girls? Make the groups first answer orally, then write the answers in their notebooks.

Drawing Bar Graphs Imagine Maths Page 272

Learning Outcomes

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

Teaching Aids

Chart paper; Glue sticks; Square cutouts of different colours

Activity

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

Instruct the students to form groups of 4.

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

Ask 2 students from each group to draw a vertical bar graph, and a the other 2 to draw a horizontal bar graph for a scale of 1 unit = 5 families. Ask them to first draw the axis and show the scale. They will then stick the coloured cutouts to show the data where each bar can be shown as a tower of square cutouts of the same colour.

Extension Idea

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

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

Learning Outcomes

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

Teaching Aids

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

Activity

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

Instruct the students to work in groups.

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

Instruct the students that they need to join the sectors to form a pie chart. Discuss that the given pie chart shows the preferred fruit of some students. Ask them a few questions like: Which is the most preferred fruit? How many students would prefer bananas if 120 students were surveyed? How many students prefer kiwi and pomegranate?

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

Extension Idea

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

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

Learning Outcomes

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

Teaching Aids

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

Activity

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

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

Instruct the students to count the total number of books issued and find the fraction of books for each category, and then find the angle of each category by multiplying each fraction calculated by 360°. Then, instruct the students to draw the sectors for each category on the circular chart paper to show the pie chart. Once they are done drawing all the angles, ask them to colour and label the sectors.

Learning Outcomes

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

Teaching Aids

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

Activity

Begin by discussing what a line graph is and how it represents data. Show the students how to read a line graph by looking at the data on the x-axis and reading its value on the y-axis.

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

Instruct them to read the graph carefully and answer the questions asked.

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

Discuss the answers in the class.

Answers

1. Interpreting Bar Graphs

Think and Tell

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

Do It Together

1. 70 runs

2. 50 + 45 = 95 runs

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

2. Drawing Bar Graphs

Do It Together

4. Representing Data on a Pie Chart

Do It Together Activity

Sports, 35

Cooking, 20

Gardening, 25

Art and Crafts, 40

5. Reading Line Graphs

Do It Together

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

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

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

3. Interpreting Pie Charts

Do It Together

1. What is the largest expense of the family? Rent

On comparing the fractions, 231 51010 >>

Hence rent is the largest expense in the family.

2. If the family’s total monthly income is ₹25,000, how much is spent on rent and bills combined?

Total fraction of amount spent on rent and bills

= 214151 51010102 + +===

1 of 2 ₹25,000 = ₹12,500

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

11111 of 21021020 =×=

Answers Solutions

Chapter 1

Let's Warm-up 1. The place value of 8 in 8,60,765 is 8,00,000. 2. The 4,36,536 can be written in words as: four lakh thirty-six thousand five hundred thirtysix 3. The place value of 4 in 4,15,124 and 4,67,890 is the same.  4. 8,76,504 has 6 in the thousands place.

Do It Yourself

1A 1. a. Place value: 3,000,000

Face value: 3

b. Place value: 9,00,00,000

Face value: 9

c. Place value: 100,000

Face value: 1

d. Place value: 20,000,000

Face value: 2

e. Place value: 60,000,000

Face value: 6

f. Place value: 30,000

Face value: 3

g. Place value: 0

Face value: 0

h. Place value: 800

Face value: 8

2. a. - Indian System: 2,16,43,332

- Number Name: Two crore sixteen lakh forty-three thousand three hundred thirty-two

Expanded Form:

2,00,00,000 10,00,000 6,00,000 40,000 3000 300 30 2 +++ ++++

- International System: 21,643,332

- Number Name: Twenty-One million six hundred forty-three thousand three Hundred thirty-two

- Expanded Form:

20,000,000 1,000,000 600,000 40,000 3000 300 30 2 +++ ++++

b. - Indian System: 12,00,621

- Number Name: Twelve lakh six hundred twenty-one

Expanded Form: 10,00,000 2,00,000 600 20 1 ++++

- International System: 1,200,621

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

- Expanded Form: 1,000,000 + 200,000 + 600 + 20 + 1

c. 46207219

- Indian System: 4,62,07,219

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

Expanded Form: 4,00,00,000 + 60,00,000 + 2,00,000 + 7000 + 200 + 10 + 9

- International System: 46,207,219

- Number Name: Forty-Six million two hundred seven thousand two hundred nineteen

- Expanded Form: 40,000,000 + 6,000,000 + 200,000 + 7000 + 200 + 10 + 9

d. 95910158

- Indian System: 9,59,10,158

- Number Name: Nine crore fifty-nine lakh ten thousand one hundred fifty-eight Expanded Form: 9,00,00,000 + 50,00,000 + 9,00,000 + 10,000 + 100 + 50 + 8

- International System: 95,910,158

- Number Name: Ninety-Five million nine hundred ten thousand one hundred fifty-eight

- Expanded Form: 90,000,000 + 5,000,000 + 900,000 + 10,000 + 100 + 50 + 8

e. 7409230

- Indian System: 74,09,230

- Number Name: Seventy-Four lakh nine thousand two hundred thirty

- Expanded Form: 70,00,000 + 4,00,000 + 9000 + 200 + 30

- International System: 7,409,230

- Number Name: Seven million four hundred nine thousand two hundred thirty

- Expanded Form: 7,000,000 + 400,000 + 9000 + 200 + 30

f. 8656023

- Indian System: 86,56,023

- Number Name: Eighty-Six lakh fifty-six thousand twenty-three

Expanded Form: 80,00,000 + 6,00,000 + 50,000 + 6000 + 20 + 3

- International System: 8,656,023

- Number Name: Eight million six hundred fifty-six thousand twenty-three

- Expanded Form: 8,000,000 + 600,000 + 50,000 + 6000 + 20 + 3

g. 67890240

- Indian System: 6,78,90,240

- Number Name: Six crore seventy-eight lakh ninety thousand two hundred forty

Expanded Form: 6,00,00,000 + 70,00,000 + 8,00,000 + 90,000 + 200 + 40

- International System: 67,890,240

- Number Name: Sixty-Seven million eight hundred ninety thousand two hundred forty

- Expanded Form: 60,000,000 + 7,000,000 + 800,000 + 90,000 + 200 + 40

h. 34565892

- Indian System: 3,45,65,892

- Number Name: Three crore forty-five lakh sixty-five thousand eight hundred ninety-two

Expanded Form: 3,00,00,000 + 40,00,000 + 5,00,000 + 60,000 + 5000 + 800 + 90 + 2

- International System: 34,565,892

- Number Name: Thirty-Four million five hundred sixtyfive thousand eight hundred ninety-two

- Expanded Form: 30,000,000 + 4,000,000 + 500,000 + 60,000 + 5000+ 800 + 90 + 2

3. a. 60,08,098  b. 20,000,569  c. 4,090,000   d. 8,00,01,002

4. a. 10 million = 1 crore  b. 1 million = 10 lakh   c. 1 crore = 10,000 thousands  d. There are 7 zeros in 20 million  e. 100 lakhs = 10 millions  f. 5 crores = 50 millions

5. a. 1 lakh = 100 thousands  b. 1 million = 1000 thousands

c. 1 crore = 10,000 thousands

6. a. 91,19,199  b. 1,70,70,707

Word Problem 1. International number system = 30,145,765; Indian number system = Three crore one lakh forty-five thousand seven hundred sixty-five.

1B 1. To compare the two numbers, you start from the leftmost digit and compare the digits at each place value.

76,24,578

87,90,213

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

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

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

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

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

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

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

3. a. 56,53,437 < 97,56,424

b. 47,98,560 > 40,87,895

c. 4,23,75,449 > 46,67,087

d. 48,75,268 < 8,65,43,959

e. 98,38,765 > 89,38,453

f. 1,76,54,534 > 98,54,763

4. Answer may vary. Sample answer:

5,48,79,802 can be rearranged to form 5,79,48,280.    5,79,48,280 > 5,48,79,802

5. a.

c. 9,00,000

d.

Word Problem 1. Number of words Sanket type = 25,85,120

Number of words Mahesh type = 24,32,214

Since, 25,85,120 > 24,32,214. Hence, Sanket types more words in a week than Mahesh.

1C 1. a. 93,12,820 < 1,00,36,782 < 5,00,00,367 < 8,87,21,460

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

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

2. a. 8,26,34,510 > 6,70,81,256 > 4,50,00,921 > 87,92,345

b. 8,01,20,450 > 6,78,20,001 > 92,11,108 > 42,56,789

c. 9,67,28,891 > 7,88,21,134 > 5,78,20,010 > 5,43,33,867

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

Smallest 7-digit number: 10,34,589

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

Smallest 7-digit number: 12,35,678

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

Smallest 7-digit number: 10,23,456

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

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

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

Smallest 8-digit number: 1,13,46,789

c. Greatest 8-digit number: 9,97,54,320

Smallest 8-digit number: 2,00,34,579

5. a. Greatest 8-digit number: 9,99,99,998

Smallest 7-digit number: 10,00,000

b. Greatest 8-digit number: 9,99,98,765

Smallest 7-digit number: 10,00,234

c. Greatest 8-digit number: 9,99,99,876

Smallest 7-digit number: 10,00,023

1D

1. a. 85,48,750  b. 89,05,460  c. 6,07,85,890

d. 1,56,48,950  e. 5,40,86,570  f. 3,09,24,560

2. a. 1,25,89,200  b. 87,52,400  c. 68,67,800

d. 77,59,900  e. 5,07,87,300  f. 43,43,600

3. a. 8,97,00,000  b. 53,12,000  c. 8,21,59,000

d. 5,89,90,000  e. 86,76,000  f. 23,75,000

4. Answer may vary. Sample answer:

a. 99,63,210; Rounded off to nearest thousand: 99,63,000

b. 96,32,100; Rounded off to nearest thousand: 96,32,000

5. The number would have lied between 3,23,45,500 and 3,23,46,449.

Word Problem 1. Rounding off ₹65,94,830 to nearest 1000s gives ₹65,95,000.

The municipal corporation spent ₹65,95,000 on repairing the roads.

Chapter Checkup

1. a. 3507681

- Indian Number System: 35,07,681- Thirty-Five Lakh Seven Thousand Six Hundred Eighty-One

Expanded form = 30,00,000 + 5,00,000 + 7000 + 600 + 80 + 1

- International Number System: 3,507,681- Three Million Five Hundred Seven Thousand Six Hundred Eighty-One

- Expanded Form: 3,000,000 + 500,000 + 7000 + 600 + 80 + 1

b. 42087950

- Indian Number System: 4,20,87,950 - Four Crores Twenty

Lakhs Eighty-Seven Thousand Nine Hundred Fifty

Expanded Form: 4,00,00,000 + 20,00,000 + 80,000 + 7000 +

900 + 50

- International Number System: 42,087,950 - Forty-Two

Million Eighty-Seven Thousand Nine Hundred Fifty

- Expanded Form: 40,000,000 + 2,000,000 + 80,000 + 7000 +

900 + 50

c. 63565842

- Indian Number System: 6,35,65,842- Six Crores Thirty-Five

Lakhs Sixty-Five Thousand Eight Hundred Forty-Two

Expanded Form: 6,00,00,000 + 30,00,000 + 5,00,000 + 60,000 + 5000 + 800 + 40 + 2

- International Number System: 63,565,842 - Sixty-Three

Million Five Hundred Sixty-Five Thousand Eight Hundred

Forty-Two

- Expanded Form: 60,000,000 + 3,000,000 + 500,000 + 60,000 + 5000 + 800 + 40 + 2

d. 901500084

- Indian Number System: 90,15,00,084- Ninety Crore Fifteen

Lakh Eighty-Four

Expanded Form: 90,00,00,000 + 10,00,000 + 500,000 + 80 + 4

- International Number 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

2. a. 60,715,239  b. 8,09,50,002

c. 1,100,039  d. 12,58,043

3. a. 6,56,52,567 > 6,48,90,650  b. 90,00,518 < 90,76,757

c. 34,57,879 = 34,57,879  d. 13,05,885 < 6,74,38,989

4. a. 23,56,475 < 90,87,687 < 8,91,63,896 < 9,08,04,365

b. 3,24,35,678 < 4,35,46,576 < 6,76,12,895 < 6,76,87,980

5. a. 5,36,45,787 > 4,56,45,768 >2,40,85,167 > 43,56,787

b. 9,09, 87, 897 > 4,90,76,837> 80,88,428 > 80,68,964

6. Greatest 8-digit number: 9,87,65,320

Smallest 8-digit number: 2,03,56,789

Greatest 7-digit number: 98,76,532

Smallest 7-digit number: 20,35,678

Ascending order: 20,35,678 < 98,76,532 < 2,03,56,789 < 9,87,65,320

Descending order: 9,87,65,320 > 2,03,56,789 > 98,76,532 > 20,35,678

7. a. Rounding off to nearest tens: 6,45,87,120

Rounding off to nearest hundreds: 6,45, 87,100

Rounding off to nearest thousands: 6,45,87,000

b. Rounding off to nearest tens: 89,09,010

Rounding off to nearest hundreds: 89,09,000

Rounding off to nearest thousands: 89,09,000

c. Rounding off to nearest tens: 1,08,75,760

Rounding off to nearest hundreds: 1,08,75, 800

Rounding off to nearest thousands: 1,08,76,000

d. Rounding off to nearest tens: 24,89,700

Rounding off to nearest hundreds: 24,89,700

Rounding off to nearest thousands: 24,90,000

8. List of numbers that can be rounded off to nearest tens as 16,48,240 are:

16,48,235; 16,48,236; 16,48,237; 16,48,238; 16,48,239; 16,48,240; 16,48,241; 16,48,242; 16,48,243; 16,48,244.

9. a. Indian number system: 1,99,77,555

One Crore Ninety-Nine Lakh Seventy-Seven Thousand Five Hundred Fifty-Five

International Number system: 19,977,555

Nineteen Million Nine Hundred Seventy-Seven Thousand Five Hundred Fifty-five.

b. Answer may vary. Sample answer:

Rearranging the numbers: 1,77,55,995

Hence, 1,99,77,555 > 1,77,55,995

c. Rounding off to nearest tens: 1,99,77,560

Rounding off to nearest hundreds: 1,99,77,600

Rounding off to nearest thousands: 1,99,78,000

10.

a. The digit in the hundreds and ones place is 6: 6 __ 6

b. The digit in the lakhs place is 4 less than the digit in the ones place: 2, ____ ____, 6 ____ 6

c. The digit in the ten lakhs and ten thousands place is the smallest odd number: 12, 1 ____, 6 ____ 6

d. The face value of the digit in the thousands place is 5: 12, 15, 6 ____ 6

e. The digit in the tens place is the biggest 1-digit number: 12,15,696

Hence, the

code is 12,15,696

2A 1. a. When the number is subtracted from itself, the difference is zero. Hence the statement is true

b. When O is subtracted from a number, the difference is not zero, hence the statement is false.

c. When the order of the addends is changed, the sum remains the same, hence the statement is true

d. The order of numbers used in subtraction cannot be changed; hence the statement is false

b.

6. The greater number would be the sum of 505090 and 398460.

Hence, the greater number will be 9,03,550.

Word Problems 1. Number of blazers manufactured in the year 2020 = 59,899

Number of blazers manufactured in the year 2021 = 78,906

Number of blazers manufactured in the year 2022 = 1,34,145

Total number of blazers manufactured in three years = 1,34,145 + 78,906 + 59,899

Hence, 2,72,950 blazers were manufactured in three years.

2. Amount withdrawn by Rashi = ₹1,45,000

Amount left in Rashi’s account = ₹67,000

Total amount in Rashi’s account = ₹1,45,000 + ₹67,000

Total amount in Rashi’s account = ₹2,12,000

3. Amount collected by NGO in the first year = ₹2,89,230

Amount collected by NGO in another year = ₹3,97,500

Total amount collected = ₹3,97,500 + ₹2,89,230 = ₹6,86,730

Amount spent by NGO = ₹3,05,700

Amount left with NGO = ₹6,86,730 ₹3,05,700

NGO is left with ₹3,81,030.

2B 1.

a. The divisor and remainder in a sum can never be the same, hence the statement is false.

b. The result of multiplication is called the product; hence the statement is false.

c. When a number is divided by another number, the result is called the quotient, hence the statement is false.

d. Anything multiplied by zero is zero, hence the statement is true.

2. a. 56,567 × 10 = 5,65,670; 56,567 × 100 = 56,56,700; 56,567 × 1000 = 5,65,67,000

b. 34,263 × 10 = 3,42,630; 34,263 × 100 = 34,26,300; 34,263 × 1000 = 3,42,63,000

c. 47,852 × 10 = 4,78,520; 47,852 × 100 = 47,85,200; 47,852 × 1000 = 4,78,52,000

d. 1,98,454 × 10 = 19,84,540; 1,98,454 × 100 = 1,98,45,400; 1,98,454 × 1000 = 19,84,54,000

e. 82,587 × 10 = 8,25,870; 82,587 × 100 = 82,58,700; 82,587 × 1000 = 8,25,87,000

3. a. 25,000 ÷ 10 = 2500; 25,000 ÷ 100 = 250; 25,000 ÷ 1000 = 25

b. 2,45,000 ÷ 10 = 24,500; 2,45,000 ÷ 100 = 2450; 2,45,000 ÷ 1000 = 245

c. 3,54,000÷ 10 = 35,400; 3,54,000 ÷ 100 = 3540; 3,54,000 ÷ 1000 = 354

d. 8,95,000 ÷ 10 = 89,500; 8,95,000 ÷ 100 = 8950; 8,95,000 ÷ 1000 = 895

e. 9,87,000 ÷ 10 = 98,700; 9,87,000 ÷ 100 = 9870; 9,87,000 ÷ 1000 = 987

3

3

5

4

Word Problems

1. Per month earning of Mrs Gupta = ₹78,562

Time for which she earned = 3 year = 3 × 12 months = 36 months

Total amount earned by Mrs Gupta in three years = ₹78,562 × 36

Hence, Mrs Gupta earned ₹28,28,232 in three years.

2. Number of words in the book = 99,545

Number of words on each page = 255

Number of pages in the book = 99,545 ÷ 255

There are 390 pages with 255 words each, plus 1 additional page with 95 words only.

Hence, there are a total of 391 pages.

3. Number of chairs in each row = 2400

Number of rows = 380

Total number of chairs for these rows = 2400 × 380 = 9,12,000

Number of additional rows = 150

Number of chairs in each additional row = 320

Total number of chairs for additional rows = 320 × 150 = 48,000

Total

of chairs required = 9,12,000 + 48,000 = 9,60,000

4. Profit of the owner of the shop = ₹98,000

Money kept for himself = ₹30,000

Money left for distributing = ₹98,000 ₹30,000 = ₹68,000

Number of staff members = 100

Amount received by each staff member = ₹68,000 ÷ 100 = ₹680

2C 1. a. 100 × 10 100 + 2000 ÷ 100 = 1000 100 + 20 = 1020 100 = 920

b. The order of DMAS is ÷, ×, +,

c. 8 × 8 8 = 64 8 = 56

d. 100 ÷ 10 + 10 × 10 = 10 + 100 = 110

e. 63 ÷ 9 + 12 × 4 5 = 7 + 48 5 = 55 5 = 50

2. a. 9 + 6 4 × 2 = 15 8 = 7  b. 17 + 8 × 3 = 17 + 24 = 41

c. 28 ÷ 7 + 8 × 5 = 4 + 40 = 44  d. 55 ÷ 11 + 7 = 5 + 7 = 12

e. 82 × 3 + 20 = 246 + 20 = 266

f. 20 + 50 ÷ 2 5 = 20 + 25 5 = 45 5 = 40

g. 16 + 8 ÷ 2 1 × 5 = 16 + 4 5 = 20 5 = 15

h. 25 4 × 12 ÷ 4 + 3 = 25 4 × 3 + 3 = 25 12 + 3 =

28 12 = 16

3. a. 5 × 4 + 12 ÷ 3 = 24 ⇒ 20 + 4 = 24; hence the statement is true b. In DMAS, we first perform division/ multiplication and then addition/subtraction, hence the statement is false c. In DMAS, the last step is subtraction, hence the statement is true

d. 36 ÷ 6 3 × 2 = 2 ⇒ 6 6 = 0 ≠ 2; hence the statement is false

4. a. 40 + 60 ÷ 20 × 16 4 = 40 + 3 × 16 4 = 40 + 48 4 = 88 4 = 84

b. 80 + 20 × 20 10 ÷ 5 = 80 + 400 2 = 480 2 = 478

c. 279 + 321 ÷ 3 57 × 2 = 279 + 107 114 = 386 114 = 272

d. 724 150 ÷ 50 × 4 + 58 = 724 3 × 4 + 58 = 724 12 +

58 = 782 12 = 770

e. 676 + 835 × 45 ÷ 15 10 = 676 + 835 × 3 10 = 676 + 2505 10 = 3181 10 = 3171

f. 120 12 + 36 ÷ 3 × 5 = 120 12 + 12 × 5 = 120 12 + 60 =

180 12 = 168

g. 15 × 20 ÷ 10 5 + 15 = 15 × 2 5 + 15 = 30 5 + 15 =

45 5 = 40

h. 39 4 × 12 ÷ 3 = 39 4 × 4 = 39 16 = 23

Word Problems 1. Money saved by Alex = ₹120

Cost of a toy = ₹28

Cost of a book = ₹28 ÷ 2

Money left with Alex = ₹120 28 28 ÷ 2 = ₹120 28 14 = ₹92 14 = ₹78

2. Number of stickers with Anish = 50

Number of stickers given to each friend = 3

Number of friends = 4

More stickers purchased = 10

Total number of stickers = 50 3 × 4 + 10 = 50 12 + 10 =

60 12 = 48 stickers

3. Number of pencils with Sahil = 15

Number of Sahil’s friends = 3

Number of pencils given to each friend by Sahil = 15 ÷ 3 = 5

Number of pencils already with each friend = 2

Total number of pencils with each friend = 2 + 5 = 7

4. Total number of students in the school = 350

Number of students absent on Monday = 25

Total number of students present on Monday = 350 25 = 325

Donations brought by each student = ₹230

Total money collected = 325 × ₹230 = ₹74,750

Word Problems 1. 5 cm  2. 7m  3. 9 litre

Chapter Checkup

1. a. L TTh Th H T O

6

6 9 9 5 3 7

c. L TTh Th H T O

1 1 1 1 9 8 7 6

+ 4

d. L TTh Th H T O

1

e.

TTh Th

e. L TTh Th H T O

9 7 5 8 4 3

9 7 1 1 9 7

f. L TTh Th H T O 4 14

8

3. a. 325 × 10 = 3250; 325 × 100 = 32,500; 325 × 1000 = 3,25,000  b. 896 × 10 = 8960; 896 × 100 = 89,600; 896 × 1000 = 8,96,000  c. 4546 × 10 = 45,460; 4546 × 100 = 4,54,600; 4546 × 1000 = 45,46,000  d. 6457 × 10 = 64,570; 6457 × 100 = 6,45,700; 6457 × 1000 = 64,57,000  e. 7656 × 10 = 76,560; 7656 × 100 = 7,65,600; 7656 × 1000 = 76,56,000  f. 9876 × 10 = 98,760; 9876 × 100 = 9,87,600; 9876 × 1000 = 98,76,000

2. a. L TTh Th H T O

5 4 6 7 6

3

b. L TTh Th H T O 15

c.

5. a. 2100 ÷ 10 = 210

2100 ÷ 100 = 21

2100 ÷ 1000 = 2 with remainder 100

b. 4086 ÷ 10 = 408 with remainder 6

4086 ÷ 100 = 40 with remainder 86

4086 ÷ 1000 = 4 with remainder 86

c. 3235 ÷ 10 = 323 with remainder 5

3235 ÷ 100 = 32 with remainder 35

3235 ÷ 1000 = 3 with remainder 235

d. 51,200 ÷ 10 = 5120

51,200 ÷ 100 = 512

51,200 ÷ 1000 = 51 with remainder 200

e. 65,200 ÷ 10 = 6520

65,200 ÷ 100 = 652

65,200 ÷ 1000 = 65 with remainder 200

f. 74,562 ÷ 10 = 7456 with remainder 2

74,562 ÷ 100 = 745 with remainder 62

74,562 ÷ 1000 = 74 with remainder 562

6.   a.

3 3

As the quotient is 45; the other number will also be 45.

9. Largest 6-digit number = 9,99,999

Smallest 6-digit number = 1,00,000

Difference = 9,99,999 1,00,000 = 8,99,999

10. Sum of 5,00,000 and 3,00,000 = 5,00,000 + 3,00,000 = 8,00,000

Difference of 5,00,000 and 3,00,000 = 5,00,000

3,00,000 = 2,00,000

The required number to be subtracted from the sum of 5,00,000 and 3,00,000 = 8,00,000 2,00,000 = 6,00,000

Word Problems

1. Number of men in the town = 4,10,900

Number of women in the town = 3,05,987

Number of children in the town = 1,78,678

Population of the town = 4,10,900 + 3,05,987 + 1,78,678 = 8,95,565

2. Total distance to be covered by Isha = 7500 km

Distance covered by Isha each day = 150 km

Number of days required to cover the distance = 7500 ÷ 150 days = 50 days

3. Number of sections in the garden = 60

Number of sections replanted = 30

Number of sections with flowers = 60 30 = 30

Number of flowers in each section = 1280

Total number of flowers in the garden = 1280 × 30 = 38,400

4. Annual income of Neha = ₹98,780

Amount spent by Neha each year = ₹50,000

Annual savings of Neha = ₹ 98,780 ₹50,000 = ₹ 48,780

Money saved by Neha in ten years = ₹48,780 × 10 = ₹ 4,87,800

5. Number of marbles with Maya = 80

Number of friends = 6

Marbles given to each friend = 5

Number of more marbles bought = 15

Total marbles with Maya = 80 6 × 5 + 15 = 80 30 + 15 = 95 30 = 65

6. Number of mangoes = 5565

Number of oranges = 2 × 5565 = 11,130

Number of oranges in each of the five boxes = 11130 ÷ 5 = 2226

7. Capacity of the auditorium = 64,070

Number of people seated in one row = 430

Number of rows in the auditorium = 64070 ÷ 430 = 149

8. Amount in Shreya’s account = ₹45,000

Amount in her account after receiving salary = ₹45,000 × 3 = ₹1,35,000

Amount spent on rent and grocery = ₹14,500

Money left in Shreya’s account = ₹135,000 ₹14,500 = ₹1,20,500

Chapter 3

3A

3 × 16 = 48

4 × 12 = 48

6 × 8 = 48

Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

d. 1 × 34 = 34

2 × 17 = 34

Factors of 34 are 1, 2, 17 and 34.

e. 1 × 37 = 37

Factors of 37 are 1 and 37.

2. a. 15 ÷ 1 = 15 Remainder = 0

15 ÷ 3 = 5 Remainder = 0

15 ÷ 5 = 3 Remainder = 0

Factors of 15 are 1, 3, 5 and 15

b. 72 ÷ 1 = 72 Remainder = 0

72 ÷ 2 = 36 Remainder = 0

72 ÷ 3 = 24 Remainder = 0

72 ÷ 4 = 18 Remainder = 0

72 ÷ 6 = 12 Remainder = 0

72 ÷ 8 = 9 Remainder = 0

Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.

c. 41 ÷ 1 = 41 Remainder = 0

Factors of 41 are 1 and 41.

d. 81 ÷ 1 = 81 Remainder = 0

81 ÷ 3 = 27 Remainder = 0

81 ÷ 9 = 9 Remainder = 0

Factors of 81 are 1, 3, 9, 27 and 81.

e. 63 ÷ 1 = 63 Remainder = 0

63 ÷ 3 = 21 Remainder = 0

63 ÷ 7 = 9 Remainder = 0

Factors of 63 are 1, 3, 7, 9, 21 and 63.

3. a. Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56. Hence, 56 is a composite number.

b. Factors of 73 are 1 and 73. Hence, 73 is a prime number.

c. Factors of 81 are 1, 3, 9, 27 and 81. Hence, 81 is a composite number.  d. Factors of 27 are 1, 3, 9 and 27. Hence, 27 is a composite number.  e. Factors of 55 are 1, 5, 11 and 55. Hence, 55 is a composite number.  f. Factors of 83 are 1 and 83. Hence, 83 is a prime number.  g. Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72. Hence, 72 is a composite number.  h. Factors of 11 are 1 and 11. Hence, 11 is a prime number.  i. Factors of 94 are 1, 2, 47 and 94. Hence, 94 is a composite number.  j. Factors of 29 are 1 and 29. Hence, 29 is a prime number.

4. 11, 13, 17 and 19 are the prime numbers between 10 and 20. So, there are 4 prime numbers between 10 and 20.

5. True. If 10 is a factor of a number, 2 is also its factor as 2 is a factor of that number. For example, 2 and 10 both are factors of 20, 2 and 10 both are factors of 30, etc.

Word Problem 1. Possible ways of arranging 189 chairs are Rows Columns Total chairs

3B 1. Numbers divisible by 2 have 0, 2, 4, 6 or 8 as its last digit. Hence 24, 38 and 160 will be divisible by 2.

a. 11   b. 24 c. 38 d. 49   e. 160

2. Numbers divisible by 10 will be divisible by 5 too. Also, numbers divisible by 10 have 0 as their last digit. Hence 190, 300, 90 and 200 will be divisible by both 5 and 10.

a. 51  b. 75  c. 37   d. 190 e. 300 f. 90

g. 63  h. 25   i. 200 j. 101

3. If the sum is divisible by 3, we know that the number itself is divisible by 3.

72: 7 + 2 = 9 (divisible by 3)

63: 6 + 3 = 9 (divisible by 3)

60: 6 + 0 = 6 (divisible by 3)

Hence, 72, 63 and 60 are divisible by 3.

a. 72 b. 63  c. 92   d. 60 e. 130

4. If the sum is divisible by 9, we know that the number itself is divisible by 9.

36: 3 + 6 = 9 (divisible by 9.)

45: 4 + 5 = 9 (divisible by 9.)

Hence, 36 and 45 are divisible by 9.

a. 36 b. 45 c. 56  d. 118  e. 919

5. The 3 number should be divisible by 5: 660, 530, 365

The number has 0 as the last digit: 660, 530

On adding the digits of the number, the sum is divisible by 3.

660: 6 + 6 + 0 = 12 (divisible by 3)

530: 5 + 3 + 0 = 8 (not divisible by 3)

Hence, the correct code to open the lock is 660.

Word Problem 1. For putting 324 blueberries and 135 chocolates in 9 cups without any left over, the numbers should be divisible by 9.

324: 3 + 2 + 4 = 9 (divisible by 9).

135: 1 + 3 + 5 = 9 (divisible by 9).

Hence, 324 blueberries and 135 chocolates can be put in 9 cups without any left over.

3C 1. Prime factorization of 80 = 2 × 2 × 2 × 2 × 5. Hence, option b is correct.

2. Prime factorization of 56 = 2 × 2 × 2 × 7. Hence, option b is correct.

3. 2 × 2 × 3 × 3 × 5 × 7 = 1260. Hence, option c is correct.

4.

Word

132

Number of vases used by Samantha = Largest prime factor of 132 = 11

3D 1. a. Factors of 13 are 1 and 13.

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Common factor is 1.

b. Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.

Factors of 45 are 1, 3, 5, 9, 15 and 45.       Common factors are 1, 3, 5 and 15.

c. Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.       Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.       Common factors are 1, 2, 3, 4, 6, 8, 12 and 24.

d. Factors of 14 are 1, 2, 7 and 14.       Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.       Common factors are 1 and 2.

e. Factors of 21 are 1, 3, 7 and 21.       Factors of 16 are 1, 2, 4, 8 and 16.       Common factor is 1.

f. Factors of 18 are 1, 2, 3, 6, 9 and 18.       Factors of 27 are 1, 3, 9 and 27.       Common factors are 1, 3 and 9.

g. Factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.        Factors of 22 are 1, 2, 11 and 22.       Common factors are 1 and 2.

h. Factors of 28 are 1, 2, 4, 7, 14 and 28.       Factors of 35 are 1, 5, 7 and 35.       Common factors are 1 and 7.

i. Factors of 50 are 1, 2, 5, 10, 25 and 50       Factors of 45 are 1, 3, 5, 9, 15 and 45.       Common factors are 1 and 5.

j. Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.

Factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54.

Common factors are 1, 2, 3 and 6.

2. a. Factors of 16 are 1, 2, 4, 8 and 16.

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

Common factors = 1, 2, 4, 8

Highest common factor is 8.

b. Factors of 25 are 1, 5 and 25.

Factors of 45 are 1, 3, 5, 9, 15 and 45.

Common factors = 1, 5       Highest common factor is 5.

c. Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.

Common factors = 1,2, 3, 4, 6, 9, 12, 18, 36       Highest common factor is 36.

d. Factors of 12 are 1, 2, 3, 4, 6 and 12.

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

Common factors = 1,2, 3, 4, 6       Highest common factor is 6.

e. Factors of 27 are 1, 3, 9 and 27.

Factors of 81 are 1, 3, 9, 27 and 81.

Common factors = 1, 3, 9, 27       Highest common factor is 27.

f. Factors of 40 are 1, 2, 4, 5, 8, 10, 20 and 40.

Factors of 50 are 1, 2, 5, 10, 25 and 50.

Common factors = 1, 2, 5, 10

Highest common factor is 10.

g. Factors of 15 are 1, 3, 5 and 15.

Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90.

Common factors = 1, 3, 5, 15

Highest common factor is 15.

h. Factors of 28 are 1, 2, 4, 7, 14 and 28.

Factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.

Common factors = 1, 2, 7, 14

Highest common factor is 14.

i. Factors of 27 are 1, 3, 9 and 27.

Factors of 47 are 1 and 47.

Common factors = 1       Highest common factor is 1.

j. Factors of 33 are 1, 3, 11 and 33.        Factors of 55 are 1, 5, 11 and 55.        Common factors = 1, 11       Highest common factor is 11.

3. a. 12 = 2 × 2 × 3

15 = 3 × 5

HCF = 3

b. 15 = 3 × 5

30 = 2 × 3 × 5

HCF = 3 × 5 = 15

c. 32 = 2 × 2 × 2 × 2 × 2

40 = 2 × 2 × 2 × 5

HCF = 2 × 2 × 2 = 8

d. 45 = 3 × 3 × 5

60 = 2 × 2 × 3 × 5

HCF = 3 × 5 = 15

e. 36 = 2 × 2 × 3 × 3

54 = 2 × 3 × 3 × 3

HCF = 2 × 3 × 3 = 18

f. 26 = 2 × 13

65 = 5 × 13

HCF = 13

g. 48 = 2 × 2 × 2 × 2 × 3

60 = 2 × 2 × 3 × 5

HCF = 2 × 2 × 3 = 12

h. 75 = 3 × 5 × 5

90 = 2 × 3 × 3 × 5

HCF = 3 × 5 = 15

i. 52 = 2 × 2 × 13

78 = 2 × 3 × 13

HCF = 2 × 13 = 26

j. 24 = 2 × 2 × 2 × 3

40 = 2 × 2 × 2 × 5

56 = 2 × 2 × 2 × 7

HCF = 2 × 2 × 2 = 8

4.

a. 16 6 4 4 6 4 0

Remainder

HCF (16, 64) = 16

Remainder       HCF (18, 24) = 6

5. Capacities of containers are 180 litres and 162 litres respectively.

Capacity of the container that can be used to fill both containers = HCF of 180 litres and 162 litres.

Remainder

HCF (180, 162) = 18, hence, the capacity of the largest container that can be used to fill the containers completely = 18 litres.

Word Problems 1. Length of box = 75 cm

Breadth of box = 85 cm

Height of box = 95 cm

Length of longest tape that can measure the three dimensions of the box are = HCF of 75 cm, 85 cm and 95 cm.

Remainder

HCF (150, 225) = 75

e. 9 0

Remainder

HCF (96, 120) = 24

Remainder

Remainder

HCF

Remainder

Hence, the length of the longest tape that can measure the three dimensions of the box is 5 cm.

2. Length of pieces of fabric = 42 m, 49 m and 63 m

Greatest possible length of each curtain = HCF of 42 m, 49 m and 63 m.

Hence, the greatest possible length of each curtain is 7 m.

3. Capacity of three drums = 36 litres, 45 litres and 72 litres

Capacity of largest container to measure the contents = HCF of 36 litres, 45 litres and 72 litres

Hence, the capacity of largest container to measure the contents = 9 litres

Chapter Checkup 1. a. 1 × 20 = 20

2 × 10 = 20

4 × 5 = 20

Factors of 20 are 1, 2, 4, 5, 10 and 20

b. 1 × 12 = 12

2 × 6 = 12

3 × 4 = 12

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

c. 1 × 16 = 16

2 × 8 = 16

4 × 4 = 16

Factors of 16 are 1, 2, 4, 8 and 16

d. 1 × 24 = 24

2 × 12 = 24

3 × 8 = 24

4 × 6 = 24

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

e. 1 × 42 = 42

2 × 21 = 42

3 × 14 = 42

6 × 7 = 42

Factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.

2. a. 56 ÷ 1 = 56 Remainder = 0

56 ÷ 2 = 28 Remainder = 0

56 ÷ 4 = 14 Remainder = 0

56 ÷ 7 = 8 Remainder = 0

Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.

4.

Divisible by 2 (last digit is 0, 2, 4, 6, 8)

Divisible by 3 (sum of digits is divisible by 3)

a. 35

b. 84

c. 93

450

5.

b. 28 ÷ 1 = 28 Remainder = 0

28 ÷ 2 = 14 Remainder = 0

28 ÷ 4 = 7 Remainder = 0

Factors of 28 are 1, 2, 4, 7, 14 and 28.

c. 54 ÷ 1 = 54 Remainder = 0

54 ÷ 2 = 27 Remainder = 0

54 ÷ 3 = 18 Remainder = 0

54 ÷ 6 = 9 Remainder = 0

Factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54.

d. 80 ÷ 1 = 80 Remainder = 0

80 ÷ 2 = 40 Remainder = 0

80 ÷ 4 = 20 Remainder = 0

80 ÷ 5 = 16 Remainder = 0

80 ÷ 8 = 10 Remainder = 0

Factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80.

e. 36 ÷ 1 = 36 Remainder = 0

36 ÷ 2 = 18 Remainder = 0

36 ÷ 3 = 12 Remainder = 0

36 ÷ 4 = 9 Remainder = 0

36 ÷ 6 = 6 Remainder = 0

Factors of 80 are 1, 2, 3, 4, 6, 9, 12, 18, 36

3.

a. Composite number: Factors of 12 are 1, 2, 3, 4, 6 and 12.

b. Prime number: Factors of 31 are 1 and 31.

c. Composite number: Factors of 38 are 1, 2, 19 and 38.

d. Composite number: Factors of 46 are 1, 2, 23 and 46.

e. Prime number: Factors of 61 are 1 and 61.

f. Composite number: Factors of 99 are 1, 3, 9, 11, 33 and 99

g. Composite number: Factors of 57 are 1, 3, 19 and 57.

h. Prime number: Factors of 73 are 1 and 73.

i. Prime number: Factors of 47 are 1 and 47.

j. Composite number: Factors of 49 are 1, 7 and 49.

Divisible by 5 (last digit is 0 or 5)

Divisible by 9 (sum of digits is divisible by 9)

Divisible by 10 (Last digit is 0)

91 =

7. a. Factors of 25: 1, 5, 25

Factors of 45: 1, 3, 5, 9, 15, 45

Common factors: 1, 5

b. Factors of 75: 1, 3, 5, 15, 25, 75

Factors of 125: 1, 5, 25, 125

Common factors: 1, 5, 25

c. Factors of 33: 1, 3, 11, 33

Factors of 55: 1, 5, 11, 55

Common factors: 1, 11

d. Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

Factors of 156: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156

Common factors: 1, 2, 3, 4, 6, 12

8. a. Factors of 16: 1, 2, 4, 8, 16

Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Common factors: 1, 2, 4

HCF = 4

b. Factors of 25: 1, 5, 25

Factors of 65: 1, 5, 13, 65

Common factors: 1, 5

HCF = 5

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

Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

Common factors: 1, 2, 3, 4, 6, 8, 12, 24

HCF = 24

d. Factors of 150: 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150

Factors of 225: 1, 3, 5, 9, 15, 25, 45, 75, 225

Common factors: 1, 3, 5, 15, 25, 75

HCF = 75

9. a. 34 = 2 × 17

38 = 2 × 19

HCF = 2

b. 34 = 2 × 17

51 = 3 × 17

HCF = 17

c. 60 = 2 × 2 × 3 × 5

225 = 3 × 3 × 5 × 5

HCF = 3 × 5 = 15

d. 105 = 3 × 5 × 7

180 = 2 × 2 × 3 × 3 × 5

HCF = 3 × 5 = 15

3 6 6 3 1 3 6 2 7 3 6 1

10. a. Remainder

0

HCF (36, 63) = 9

b. 1 1 9 1 8 7 1

1 9

8 1 1 9 1

8

1 6 8 1

1 17

HCF (119, 187) = 17

c. 4 5 8 9 1

HCF (45, 89) = 1

d. 1 3 6 1 7 0 1

HCF (136, 170) = 34

1 3

11. Answers may vary. Sample answer:

For a number to have 2 as a factor it must have 0, 2, 4, 6 or 8 as its last digit. Hence, a three-digit number which has 2 as a common factor with 420 is 120.

12. Capacity of tanks = 250 litres and 425 litres.

Maximum capacity of the bucket that can measure the water in tanks = HCF of 250 litres and 425 litres

Hence, the number of plastic containers needed = 6.

Chapter 4

Let's Warm-up

Hence, the maximum capacity of the bucket that can measure the water in tanks is 25 litres.

13. Number of white balloons = 24

Number of orange balloons = 16

Greatest number of balloons in each arrangement = HCF of 24 and 16

A

B

1. Multiplication table of 8 60, 96, 108

2. Multiplication table of 9 40, 70, 80

3. Multiplication table of 10 24, 48, 64

4. Multiplication table of 12

Do It Yourself

Hence, the greatest number of balloons in each arrangement is 8.

Word Problems

1. Dimension of game board = 66 inches × 24 inches

Largest size of square tile to fit the board = HCF of 66 inches and 24 inches = 6 inches

2. Number of cupcakes = 240

Number of sandwiches = 160

Maximum number of uniform packets that can be made = HCF of 240 and 160 = 80

Number of cupcakes in each packet = 240 ÷ 80 = 3

Number of sandwiches in each packet = 160 ÷ 80 = 2

Therefore, the maximum number of packets she can make is 80; the contents of each will be 3 cupcakes and 2 sandwiches.

3. Number of oatmeal cookies = 30

Number of chocolate chip cookies = 48

Number of plastic containers needed = HCF of 30 and 48

36, 54, 63

4A 1. a. 8 × 1 = 8  8 × 2 = 16  8 × 3 = 24  8 × 4 = 32

8 × 5 = 40

b. 11 × 1 = 11  11 × 2 = 22  11 × 3 = 33  11 × 4 = 44

11 × 5 = 55

c. 21 × 1 = 21  21 × 2 = 42  21 × 3 = 63  21 × 4 = 84

21 × 5 = 105

d. 25 × 1 = 25  25 × 2 = 50  25 × 3 = 75  25 × 4 = 100

25 × 5 = 125

e. 50 × 1 = 50  50 × 2 = 100  50 × 3 = 150  50 × 4 =

200 50 × 5 = 250

2. On dividing 7209 by 9, we get no remainder. So, 7209 is a multiple of 9.

8 0 1

9 7 2 0 9 7 2 2 0 0 0 0 0 9 9 0

3. The 8th multiple of 15 is equal to 15 × 8 = 120

4. First four multiples of 9 are

9 × 1 = 9 9 × 2 = 18 9 × 3 = 27 9 × 4 = 36

The first and third multiple of 9 are odd numbers, while second and fourth multiples of 9 are even numbers.

5. Answers may vary. Sample answer: Let us take 10.

Then, 1st ten multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.

As, 0 is at the ones place so, all multiples of 10 are even.

6. a. Multiples of 8 that are less than 32 are 8, 16, 24

b. Multiple of 15 that are less than 100 are 15, 30, 45, 60, 75, 90

c. Multiples of 19 that lie between 57 and 152 are 76, 95, 114, 133

d. Multiples of 21 that lie between 105 and 210 are 126, 147, 168, 189

Word Problem 1. Narendra marks first date as August 7. The classes are scheduled once in every 7 days. So, Dates are, 7 × 1 = 7

7 × 2 = 14

7 × 3 = 21

7 × 4 = 28

Therefore, the dates he will mark for the classes in the month of August are 7, 14, 21 and 28.

4B

2.

the lowest among all the common multiples,

it is the LCM of 6, 9 and 18.

Lowest Common Multiple or LCM of 6,9 and 18 = 18

5. a. Let us check the LCM of 4 and 15,

=

=

LCM of 8 and 30 = 120

So, the LCM of 5 and 12 is the same as the LCM of 4 and 15 which is 60.

Thus, option c is correct.

Word Problem 1. The first bell rings after 30 minutes in primary school.

The first bell rings after 40 minutes in secondary school. To find at what time both bells ring together we need to find LCM of 30 and 40

So, the common multiples of 30 and 40 = 120, 240 … Thus, after 120 minutes, both school’s bells ring together. 4C 1. a.

5 35

3 75

Now, we take 5 two times, 3 and 7 one time. So, LCM of 35 and 75 = 5 × 5 × 3 × 7 = 525

b. LCM of 44 and 88

2 44

2 22

Now, we take 3 times two and 1 time 11. So, LCM of 44 and 88 = 2 × 2 × 2 × 11 = 88

c. LCM of 48 and 12

2 48

2 24

2 12

2 6

3 3 1

48 = 2 × 2 × 2 × 2 × 3

12 = 2 × 2 × 3

Now, we take 4 times two and 1 time 3. So, LCM of 48 and 12 = 2 × 2 × 2 × 2 × 3 = 48

d. LCM of 25 and 115

5 25

5 5 1

25 = 5 × 5

115 = 5 × 23

5 115 23 23 1

Now, we take 2 times five and 1 time 23. So, LCM of 25 and 115 = 5 × 5 × 23 = 575

2. LCM of 12, 24, and 56 = 2 × 2 × 2 × 3 × 7 = 168

So, Rohan is not right.

3. a. LCM of 8 and 10 = 2 × 2 × 2 × 5 = 40

b. LCM of 8 and 16 = 2 × 2 × 2 × 2 = 16

c. LCM of 16 and 10 = 2 × 2 × 2 × 2 × 5 = 80

d. LCM of 25 and 10 = 5 × 5 × 2 = 50

4. a. 12 and 15

2 12

2 6 3 3 1

12 = 2 × 2 × 3

15 = 3 × 5

3 15 5 5 1

Now, we will take 2 times two, 1 time three and 1 time

five. So, LCM of 12 and 15 = 2 × 2 × 3 × 5 = 60

b. 15 and 20

3 15 2 20

5 5 2 10 1 5 5 1

15 = 3 × 5

20 = 2 × 2 × 5

Now, we will take 2 times two, 1 time three and 1 time

five. So, LCM of 15 and 20 = 2 × 2 × 3 × 5 = 60

c. 18 and 24

2 18 2 24

3 9 2 12

3 3 2 6 1 3 3 1

18 = 2 × 3 × 3

24 = 2 × 2 × 2 × 3

Now, we will take 3 times two and 2 time three. So, LCM of 18 and 24 = 2 × 2 × 2 × 3 × 3 = 72

d. 25 and 35

5 25 5 35

5 5 7 7 1 1

25 = 5 × 5

35 = 5 × 7

Now, we will take 2 times five and 1 time seven. So, LCM of 25 and 35 = 5 × 5 × 7 = 175

e. 36 and 45

2 36 3 45

2 18 3 15

3 9 5 5

3 3 1 1

36 = 2 × 2 × 3 × 3

45 = 3 × 3 × 5

Now, we will take 2 times two, 2 times three and 1 time five.

So, LCM of 36 and 45 = 2 × 2 × 3 × 3 × 5 = 180

f. 63 and 105

3 63 3 105

3 21 5 35

7 7 7 7

1 1

63 = 3 × 3 × 7

105 = 3 × 5 × 7

Now, we take 2 times three, 1 time five and 1 time seven. So, LCM of 63 and 105 = 3 × 3 × 5 × 7 = 315

g. 16 and 24

2 16 2 24

2 8 2 12

2 4 2 6

2 2 3 3

16 = 2 × 2 × 2 × 2

24 = 2 × 2 × 2 × 3

Now, we will take 4 times two and 1 time three. So, LCM of 16 and 24 = 2 × 2 × 2 × 2 × 3 = 48 h. 18 and 20

2 18 2 20

3 9 2 10

3 3 5 5 1 1

18 = 2 × 3 × 3

20 = 2 × 2 × 5

Now, we will take 2 times two, 2 times three and 1 time five.

So, LCM of 18 and 20 = 2 × 2 × 3 × 3 × 5 = 180 i. 10 and 18

2 10 2 18

2 5 3 9 1 3 3 1

10 = 2 × 5

18 = 2 × 3 × 3

Now, we will take 1 time two, 2 times three and 1 time

5. So, LCM of 10 and 18 = 2 × 3 × 3 × 5 = 90

5. a. LCM of 18, 40 and 45

2 18 2 40 3 45

3 9 2 20 3 15

3 3 2 10 5 5

1 5 5 1 1

18 = 2 × 3 × 3

40 = 2 × 2 × 2 × 5

45 = 3 × 3 × 5

Now, we take 3 times two, 2 times three and 1 time

five.

So, LCM of 18, 40 and 45 = 2 × 2 × 2 × 3 × 3 × 5 = 360

b. LCM of 72, 96 and 108

2 72 2 96 2 108

2 36 2 48 2 54

2 18 2 24 3 27

3 9 2 12 3 9

3 3 2 6 3 3

1 3 3 1 1

72 = 2 × 2 × 2 × 3 × 3

96 = 2 × 2 × 2 × 2 × 2 × 3

108 = 2 × 2 × 3 × 3 × 3

Now, we take 5 times two and 3 time three.

So, LCM of 72, 96 and 108 = 2 × 2 × 2 × 2 × 2 × 3 × 3 ×

3 = 864

c. 48, 56 and 70

2 48 2 56 2 70

2 24 2 28 5 35

2 12 2 14 7 7

2 6 7 7 1

3 3 1 1

48 = 2 × 2 × 2 × 2 × 3

56 = 2 × 2 × 2 × 7

70 = 2 × 5 × 7

Now, we take 4 times two and 1 time three, five, seven each.

So, LCM of 48, 56 and 70 = 2 × 2 × 2 × 2 × 3 × 5 × 7 =

1680

d. 30, 60 and 90

2 30 2 60 2 90

3 15 2 30 3 45

30 = 2 × 3 ×

60 = 2 × 2 ×

90 = 2 × 3 × 3 × 5

Now, we take 2 times two, 2 times three and 1 time five.

So, LCM of 30, 60 and 90 = 2 × 2 × 3 × 3 × 5 = 180

e. 28, 56 and 84

2 28 2 58 2 84

2 14 29 29 2 42

7 7 1 3 21 1 7 7 1

28 = 2 × 2 × 7

56 = 2 × 2 × 2 × 7

84 = 2 × 2 × 3 × 7

Now, we take 3 times two, 1 time three and 1 time seven.

So, LCM of 28, 56 and 84 = 2 × 2 × 2 × 3 × 7 = 168

f. 36, 54 and 81

2 36 2 54 3 81

2 18 3 27 3 27

3 9 3 9 3 9

3 3 3 3 3 3

1 1 1

36 = 2 × 2 × 3 × 3

54 = 2 × 3 × 3 × 3

81 = 3 × 3 × 3 × 3

Now, we take 2 times two, 4 time three.

So, LCM of 36, 54 and 81 = 2 × 2 × 3 × 3 × 3 × 3 = 324

Word Problem 1. Let us find the LCM of 4 and 10.

2 4 2 10

2 2 5 5 1 1

4 = 2 × 2

10 = 2 × 5

So, LCM of 4 and 10 = 2 × 2 × 5 = 20

Thus, 20th student will get both chocolate and candy.

4D 1. a. LCM of 25 and 30

5 25, 30

5 5, 6

6 1, 6

1, 1

So, LCM of 25 and 30 = 5 × 5 × 6 = 150

b. 32 and 48

2 32 48

2 16 24

2 8 12

2 4 6

2 2 3

3 1 3

1 1

So, the LCM of 32 and 48 = 2 × 2 × 2 × 2 × 2 × 3 = 96

c. 42 and 70

2 42 70

3 21 35

5 7 35

7 7 7

1 1

So, the LCM of 42 and 70 = 2 × 3 × 5 × 7 = 210

d. 12 and 48

2 12 48

2 6 24

2 3 12

2 3 6

3 3 3 1 1

So, the LCM of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48

e. 60 and 75

2 60 75

2 30 75

3 15 75

5 5 25

5 1 5

1 1

So, the LCM of 60 and 75 = 2 × 2 × 3 × 5 × 5 = 300

f. 9 and 15

3 9 15

3 3 5

5 1 5

1 1

So, the LCM of 9 and 15 = 3 × 3 × 5 = 45

g. 21 and 24

3 21 24

7 7 8

2 1 8

2 1 4

2 1 2

1 1

So the LCM of 21 and 24 = 3 × 7 × 2 × 2 × 2 = 168

h. 30 and 45

2 30 45

3 15 45

3 5 15

5 5 5

1 1

So, the LCM of 30 and 45 = 2 × 3 × 3 × 5 = 90

i. 63 and 105

3 63 105

3 21 35

5 7 35

7 7 7

1 1

So, the LCM of 63 and 105 = 3 × 3 × 5 × 7 = 315

2. a. 28, 42, 56

2 28 42 56

2 14 21 28

2 7 21 14

3 7 21 7

7 7 7 7

1 1 1

So, the LCM of 28, 42 and 56 = 2 × 2 × 2 × 3 × 7 = 168

b. 75, 100, 150

2 75 100 150

2 75 50 75

3 75 25 75

5 25 25 25

5 5 5 5

1 1 1

So, the LCM of 75, 100 and 150 = 2 × 2 × 3 × 5 × 5 = 300

c. 96, 144, 192

2 96 144 192

2 48 72 96

2 24 36 48

2 12 18 24

2 6 9 12

2 3 9 6

3 3 9 3

3 1 3 1

1 1 1

So, the LCM of 96, 144 and 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = 576

d. 90, 135, 180

2 90 135 180

2 45 135 90

3 45 135 45

3 15 45 15

3 5 15 5

5 5 5 5

1 1 1

So, the LCM of 90, 135 and 190 = 2 × 2 × 3 × 3 ×

3 × 5 = 540

e. 102, 136, 170

2 102 136 170

2 51 68 85

2 51 34 85

3 51 17 85

5 17 17 85

17 17 17 17

1 1 1

So, the LCM of 102, 136 and 170 = 2 × 2 × 2 × 3 ×

5 × 17 = 2040

f. 21, 63, 105

3 21 63 105

3 7 21 35

5 7 7 35

7 7 7 7

1 1 1

So, the LCM of 21, 63 and 105 = 3 × 3 × 5 × 7 = 315

3. Rohit’s method –

2 42, 56

2 21, 28

7 3, 4

But to find LCM of 42 and 56, we need to solve it like below,

2 42, 56

2 21, 28

2 21, 14

3 21, 7

7 7, 7 1, 1

LCM of 42 and 56 = 2 × 2 × 2 × 3 × 7 = 168

Rohit’s conclusion is incorrect because he didn’t consider the highest power of 2 in the prime factorization when calculating the LCM.

Thus, Rohit is not correct and LCM of 42 and 56 is 168.

4. 2 12 20 32

2 6 10 16

2 3 5 8 2 3 5

2 3 5 2

3 3 5 1

5 1 5 1

So, LCM of 12, 20 and 32 = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 480

So, LCM = 480

Thus, Akash is correct as he has considered all the prime factors.

5. To find the prime factors of 16, 24, 36 and 54,

So, LCM of 16, 24, 36 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432

So, 72 is not the LCM of given numbers. Thus, Ajay is not correct.

Word Problem 1. To complete a round, Anna takes 8 minutes and, Sylvia takes 12 min.

To find when they both meet again we need to find the LCM of 8 and 12.

LCM of 8 and 12 = 2 × 2 × 2 × 3 = 24

Thus, after 24 min they both meet at the same point.

4E 1. We need to find the least possible number of students in the class. 2

Now, the number of students will be greater than 6, 8, 12 and 16

So, we need to find the LCM in this case.

LCM = 2 × 2 × 2 × 2 × 3 = 48

Therefore, the least possible number of students in the class = 48 students.

2. We need to find the bells toll together after 12 noon. And the all four bells toll at intervals of 8, 14, 18 and 24 minutes

So, we need to find the LCM of these numbers.

LCM of 8, 14, 18 and 24 = 2 × 2 × 2 × 3 × 3 × 7 = 504

So, after 504 min bells toll together. It means after 8 hour and 24 min. So, all four bells will toll together at 8:24 p.m.

3. First device beeps at every 60 sec and 2nd device beeps at every 62 sec

We need to find out the LCM of 60 and 62 = 2 × 2 × 3 × 5 × 31 = 1860

So, after 31 minutes both devices beep together. And the time will be 10:31 a.m.

4. Free beverage is received after every 5th visit. Free appetizer is received after every 10th visit.

a. On 10th, 20th, 30th, 40th, 50th, 60th, 70th, 80th, 90th and 100th visit you receive a free beverage and a free appetizer.

b. To find the first visit where you can get both free items, we need to find the LCM of 5 and 10.

So, LCM of 5 and 10 = 2 × 5 = 10

So, on 10th visit I’ll get both free items.

5. To find which customer get both the good we need to find the Least common multiple of 8 and 12

8 = 2 × 2 × 2

12 = 2 × 2 × 3

So, LCM of 8 and 12 = 2 × 2 × 2 × 3 = 24

So, 24th customer will get both the items.

Word Problems 1. The maximum capacity of a container that can measure the milk of either container an exact number of times is the highest common factor (HCF) of 75 and 45. To find the HCF of 75 and 45,

Factor of 75 1 3 5 15 25 75

Factor of 45 1 3 5 9 15 45

So, the highest common factor of 75 and 45 = 15

Therefore, the maximum capacity of a container that can measure the milk of either container an exact number of times is 15 litres.

2. To find the total number of stacks, we need to find the least common multiple (LCM) of 336, 240, and 96.

2 336 240 96

2 168 120 48

2 84 60 24

2 42 30 12

2 21 15 6

3 21

The prime factorization of 336 is 2 × 2 × 2 × 2 × 3 × 7.

The prime factorization of 240 is 2 × 2 × 2 × 2 × 3 × 5.

The prime factorization of 96 is 2 × 2 × 2 × 2 × 2 × 3.

The LCM of 336, 240, and 96 = 2 × 2 × 2 × 2 × 2 × 3 × 5 × 7 = 3360. This means that we can stack the books in 3360 stacks, such that each stack contains the same number of books of each subject, and the height of each stack is the same. Therefore, the total number of stacks is 3360.

3. To find the greatest number of groups Mrs. Mehra can make, we need to find the highest common factor (HCF) of 12 and 18.

Factor of 12 1 2 3 4 6

Factor of 18 1 2 3 6 9

The HCF of 12 and 18 is 6.

This means that the greatest number of groups Mrs. Mehra can make is 6.

This is because each group must have the same number of boys and girls, and 6 is the largest number that divides both 12 and 18 evenly.

4. The smallest length of a room into which an exact number of carpets of length 12 metres and 9 metres can fit is the least common multiple (LCM) of 12 and 9.

The prime factorization of 12 = 2 × 2 × 3.

The prime factorization of 9 = 3 × 3.

The LCM of 12 and 9 = 2 × 2 × 3 × 3 = 36.

Therefore, the smallest length of a room into which an exact number of carpets of length 12 metres and 9 metres can fit is 36 metres.

Chapter Checkup

1.

2. 9th multiple of

a. 7 × 9 = 63

b. 9 × 9 = 81

c. 11 × 9 = 99

d. 13 × 9 = 127

e. 16 × 9 = 144

3. 180 is the smallest number that is divisible by 9, 12 and 15 which also the LCM of these numbers. Thus, option d is correct.

4. 105, 847 and 77 can be divided by 7 while 103 cannot be divided by 7.

Thus, option a is correct.

5. 90 and 180 are common multiples of 30 and 45. Thus, option b is correct.

6.

A B LCM of A and B

a. 2 × 3 × 3 2 × 5 × 7

b. 2 × 3 × 5 3 × 3 × 5

c. 2 × 3 × 3 × 7 2 × 3 × 11

d. 3 × 3 × 5 × 7 2 × 3 × 5 × 11

e. 2 × 3 × 3 ×

7. a. LCM of 63 and 105

3 63 105

3 21 35

5 7 35

So,

b.

2 93 62 120

2 93 31 60

2 93 31 30

3 93 31 15

5 31 31 5

31 31 31 1

1 1 1

So,

× 3 × 3 × 5 = 90

× 3 × 3 × 7 × 11 = 1386

= 6930

c. LCM of 112, 140 and 168

2 112 140 168

2 56 70 84

2 28 35 42

2 14 35 21

3 7 35 21

5 7 35 7

7 7 7 7

1 1 1

So, LCM of 112, 140 and 168 = 2 × 2 × 2 × 2 × 3 × 5 × 7 = 1680

d. LCM of 121, 132 and 330

2 121 132 330

2 121 66 165

3 121 33 165

5 121 11 55

11 121 11 11

11 11 1 1

1 1 1

LCM of 121, 132 and 330 = 2 × 2 × 3 × 5 × 11 × 11

= 7260

e. LCM of 16, 40, and 56

2 16 40 56

2 8 20 28

2 4 10 14

2 2 5 7

5 1 5 7

7 1 1 7

1 1 1

LCM of 16, 40, and 56 = 2 × 2 × 2 × 2 × 5 × 7 = 560

f. LCM of 12, 20 and 32

2 12 20 32

2 6 10 16

2 3 5 8

2 3 5 4

2 3 5 2

3 3 5 1

5 1 5 1 1 1 1

LCM of 12, 20 and 32 = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 480

g. LCM of 125, 75 and 275

3 125 75 275

5 125 25 275

5 25 5 55

5 5 1 11 11 1 1 11 1 1 1

LCM of 125, 75 and 275 = 3 × 5 × 5 × 5 × 11 = 4125

h. LCM of 45, 18 and 63

2 45 18 63

3 45 9 63

3 15 3 21

5 5 1

7 1 1

45,

i. LCM of 15, 36 and 40

2 15 36 40

2 15 18

2 15 9 10

LCM

8. a. The LCM of 3 and 9 is 9. True

b. The LCM of any two numbers is always greater than their HCF. True

c. The least number which is divisible by 45 and 135 is 135. True

d. The LCM of 5, 15, and 45 is 90. False

9. a. 15, 12

3 15 2 12

Prime

b.

5 35

c.

d.

Prime

Prime

LCM

Prime factor of 5 = 5 Prime factors of 9 = 3 × 3

Prime factors of 15 = 3 × 5

e.

LCM of 5, 9 and 15 = 3 × 3 × 5 = 45

Prime factor of 8 = 2 × 2 × 2 Prime factors of 10 = 2 × 5

Prime factors of 12 = 2 × 2 × 3

LCM of 8, 10 and 12 = 2 × 2 × 2 × 3 × 5 = 120

f. 6, 10, 16

2

3

Prime factor of 6 = 2 × 3 Prime factors of 10 = 2 × 5

Prime factors of 16 = 2 × 2 × 2 × 2

LCM of 6, 10 and 16 = 2 × 2 × 2 × 2 × 3 × 5 = 240

10. Three bells ring at intervals of 20 minutes, 30 minutes, and 45 minutes respectively.

To find the time after which all three bells ring together, we need to find the LCM of 20, 30 and 45

20 = 2 × 2 × 5

30 = 2 × 3 × 5

45 = 3 × 3 × 5

So, LCM of 20, 30 and 45 = 2 × 2 × 3 × 3 × 5 = 180

The least common multiple (LCM) of 20, 30, and 45 is 180.

So, the bells will ring together after 180 minutes, which is equal to 3 hours.

Therefore, the three bells will ring together after 3 hours. Word Problems 1. To find the largest square tile that can fit into a rectangular area, we need to find the highest common factor (HCF) of the length and width of the rectangle.

Factors of 16 = 1, 2, 4, 8

Factors of 24 = 1, 2, 3, 4, 6, 8, 12

So, the HCF of 16 and 24 is 8. This means that the largest square tile that can fit into the board game is 8 inches by 8 inches. Therefore, the dimensions of the largest square tile that Satish can use are 8 inches by 8 inches.

2. To find the smallest possible number of questions that could have been given for holiday homework will be the least common multiple of 3 and 13.

LCM of 3 and 13 = 3 × 13 = 39

Therefore, the smallest possible number of questions that could have been given for holiday homework is 39.

3. A multiple of a number is a number that is divisible by that number. Therefore, a multiple of 75 is a number that can be divided evenly by 75.

75 ÷ 15 = 5

75 ÷ 25 = 3

75 ÷ 12 = Not possible

15 and 25 are both multiples of 75, because they can be divided evenly by 75. However, 12 is not a multiple of 75, because it cannot be divided evenly by 75. Therefore, the wrong answer is 12. Thus, Ajay gives the wrong answer.

Chapter 5

Let's Warm-up 1.

Do It Yourself

5A 1. a. As the denominators of 4 7 and 1 4 are different, they are unlike fractions.

b. As the denominators of 5 8 and 4 8 are the same, they are like fractions.

c. As the denominators of 1 4 and 1 5 are different, they are unlike fractions.

d. As the denominators of 3 7 and 5 7 are the same, they are like fractions.

e. As the denominators of 6 18 and 7 18 are the same, they are like fractions.

2. Proper fractions have denominator greater than numerator, hence, 41212 ,, 51529 are proper fractions. Improper fractions have numerator equal to or greater than the denominator, hence, 65817926 ,,,, 2511317 are improper fractions.

Mixed numbers are fractions which are a combination of a whole number and a proper fraction, hence, 15784 4 , 5 , 6 , 7 , 15 78997 are mixed fractions.

3.

Thus, each friend gets 17 2

chocolates.

denominator. Improper fraction.

Thus, the fraction that Sanchita is talking about is 22 3

5B 1. Answers may vary. Sample answers:

a. 4243 8 12 and 92189327 ×× == ××

b. 112 113 22 33 and 1723417351 ×× == ××

c. 6263 12 18 and 1523015345 ×× == ××

d. 122 123 24 36 and 2024020360 ×× == ××

e. 142 143 28 42 and 2825628384 ×× == ××

f. 6263 12 18 and 72147321 ×× == ××

2. Answers may vary. Sample answers:

a. 122 124 63 and 202102045 ÷÷ == ÷÷

b. 142 147 72 and 562285678 ÷÷ == ÷÷

c. 322 324 16 8 and 4822448412 ÷÷ == ÷÷

d. 642 644 32 16 and 8024080420 ÷÷ == ÷÷

e. 302 303 15 10 and 4222142314 ÷÷ == ÷÷

f. 442 444 22 11 and 5222652413 ÷÷ == ÷÷

3. a. HCF of 24 and 60 is 12. 2412 2 60125 ÷ = ÷

b. HCF of 27 and 63 is 9. 279 3 6397 ÷ = ÷

c. HCF of 44 and 55 is 11. 4411 4 55115 ÷ = ÷

d. HCF of 91 and 119 is 7. 91713 119717 ÷ = ÷

e. HCF of 44 and 66 is 22. 4422 2 66223 ÷ = ÷

f. HCF of 24 and 72 is 24. 2424 1 72243 ÷ = ÷

4. a. 4412413 and 8421644 ÷÷ == ÷÷

As 13 , 24 ≠ the fractions 4 8 and 12 16 are not equivalent.

b. 147221122 and 21733333 ÷÷ == ÷÷

As 22 , 33 = the fractions 14 21 and 22 33 are equivalent.

c. 55(2525)11 and 105250252 ÷÷ == ÷÷

As, 11 , 22 = the fractions 5 10 and 25 50 are equivalent.

d. 6030 22 and 903033 ÷ = ÷

As, 22 , 33 = the fractions 60 90 and 2 3 are equivalent.

e. 228214 and 4221025 ÷÷ == ÷÷

As, 14 , 25 ≠ the fractions 2 4 and 8 10 are not equivalent.

f. 62 33 and 51025 ÷ = ÷

As, 33 , 55 = the fractions 3 5 and 6 10 are equivalent.

5. a. 5 × 7 = 35; hence, 4 × 7 = 28

428

535 =

b. 1 × 14 = 14; hence, 5 × 14 = 70

141

705 =

c. 36 ÷ 6 = 6; hence, 12 ÷ 6 = 2

212

636 =

d. 15 × 4 = 60; hence, 21 × 4 = 84

1560

2184 =

e. 3 × 7 = 21; hence, 8 × 7 = 56

321

856 =

f. 11 × 5 = 55; hence, 5 × 5 = 25

525

1155 =

Word Problems 1. Fraction of pizza eaten by Tarun = 6 8

6 8 = 6 ÷2 8 ÷2 = 3 4

3 4 × 5 5 = 15 20

So, 6 8 is equivalent to 15 20 .    Thus, option d is correct.

2. Total number of cookies = 12

Number of cookies Rahul ate in the morning = 3

Number of cookies Rahul ate in the evening = 5

Total number of cookies he ate = 3 + 5 = 8

Cookies left with Rahul = 12 – 8 = 4

Fraction of cookies left with him = 4 12 = 4 ÷ 4 12 ÷ 4 = 1 3

So, 1 3 of cookies are left with Rahul.

5C 1. a. As the fractions 14 23 and 17 23 are like fractions and 1417 1417, 2323 <<

b. 4 5 and 6 9 can be converted to like fractions as:

49653630 ; 59459545 ×× == ××

Since, 36 > 30 3630 , 4545 >

Thus, 46 59 >

c. 11 13 and 2 3 can be converted into like fractions as:

113 213 332 26 ; 13339331339 ×× === ××

Since, 33 > 26, 3326 3939 >

Thus, 112 133 >

d. 1 6 and 4 20 can be converted into like fractions as:

110431012 ; 6106020360 ×× == ××

Since, 10 < 12, 1012 6060 <

Thus, 14 620 <

e. 3 4 and 7 9 can be converted into like fractions as:

39742728 ; 49369436 ×× == ××

Since, 27 < 28, 2728 3636 <

Thus, 37 49 <

f. 12 15 and 14 16 can be converted into like fractions as:

12161415 192 210 ; 15162401615240 ×× == ××

Since, 192 < 210, 192210 240240 <

Thus, 1214 1516 <

2.

a. 432512210 ; 531533515 ×× === ××

Since 12 > 10, 1210 1515 > or 42 53 >

Hence, the given statement is correct.

b. 665511 ; 12621052 ÷÷ == ÷÷

Since, 11 . 22 =

Hence, the given statement is correct.

c. 717158119120 ; 817136178136 ×× == ××

Since, 119120 136136 <

Hence, the given statement is correct.

d. 14111212 154 144 ; 12111321112132 ×× == ××

Since, 154144 132132 >

Hence, the given statement is correct.

e. 41587 60 56 ; 715105157105 ×× == ××

Since, 6056 105105 ≠

Hence, the given statement is incorrect.

f. 21749 34 36 ; 917153179153 ×× == ××

Since, 34 153 is not greater than 36 153

Hence, the given statement is incorrect.

3. a. LCM of 4, 6 and 7 is 84. 221114312 ;; 421614712 ××× === ××× 42 1436 84 8484

So, 143642 848484 <<

Thus, 132 674 <<

b. LCM of 4, 7, 12 and 13 is 1092. 3845156591 252 780 455 ; ; ; 138410927156109212911092 1273791 273 637 ; 4273109212911092

××× === ××× ×× == ××

So, 252273455637780 10921092109210921092 <<<<

Thus, 31575 13412127 <<<<

c. LCM of 8, 12 and 15 is 120. 71588415 ;;; 815158815 510 1210

105 64 60 120120120

So, 506064105 120120120120 <<<

Thus, 5487 128158 <<<

d. LCM of 2, 3, 5 and 6 is 30. 2101655115 ;;; 3105665215 ×××× ==== ×××× 2062515 30303030

So, 6152025 30303030 <<<

Thus, 1125 5236 <<<

e. LCM of 2, 4, 5 and 7 is 140. 135335520170 ;;; 435435720270

So, 355670100105 150140140140140 <<<<

Thus, 12153 45274 <<<<

××××

40

=== ××××

100 75 120120

So, 40486075100 120120120120120 <<<<

Thus, 12155 35286 <<<<

45 30 72 120120120

×

=

So, 72453020 120120120120 >>>

××××

××××

27642436 72727272

Thus, 8131 9283 >>>

c. LCM of 3, 5, 6, 11 and 15 is 330.

4663224302110 ;;; 566152211303110 155 ; 655 ×××× === ×××× × == × 264 66 120 330 330 330 220 55 330330

>>>>

Thus, 42431 5311156 >>>>

d. LCM of 6, 9, 11 and 14 is 1386. 1126 1991154 ; ;

1386138613861386

Thus, 1111 691114 >>>

30364045 60606060

Solutions 117

So, 45403630 60606060 >>>

Thus, 3231 4352 >>>

f. LCM of 3, 4, 8 and 25 is 600. 31502200524 ;;; 415032002524

375 875

So, 450400225120 600600600600 >>>

Thus, 3235 43825 >>>

5. a. LCM of 8 and 13 is

13526465

So, 104104104104 <<<

Thus, 1485 88138 <<<

b. LCM of 2, 5, 8 and 12 is 120. 16012417152310

230

120

So, 2460230255 120120120120 <<<

Thus, 112317 52128 <<<

c. LCM of 5, 6 and 7 is 210. 144218351742

714150

210210

So, 150588630714 210210210210 <<<

Thus, 5141817 7565 <<<

d. LCM of 3, 6, 15 and 19 is 570.

LCM of 13 and 23 is 299.

6 × 23

13 × 23 = 138 299; 11 × 13 23 × 13 = 143 299

Since, 138 299 < 143 299

i.e. 6 13 < 11 23

Thus, Reshma studied for longer duration.

2. Time taken by:

Rakesh = 4 6 hours

Roshan = 7 8 hours

Swati = 1 hour

Prerna = 2 3 hours

LCM of 3, 6 and 8 is 24.

∴ 4 × 4 6 × 4 = 16 24 ; 7 × 3 8 × 3 = 21 24 ; 1 × 24 1 × 24 = 24 24 ; 2 × 8 3 × 8 = 16 24

As the time taken by Rakesh and Prerna same, the    order can be given as:

So, 16 24 < 21 24 < 24 24

Thus, 2 3 hours = 4 6 hours < 7 8 hours < 1 hour

453817301695 1710 510 1520 ;;; 15385701930570695570

2190 380 3190570

××× === ××× × = ×

So, 38051015201710 570570570570 <<<

Thus, 2171645 319615 <<<

Word Problems 1. Hours Shashank studied for = 6 13

Hours Reshma studied for = 11 23

3.

may vary. Sample answers:

a. 525354 ; ; 828384

b. 222324 ;; 626364

c. 112113114 ;; 142143144

d. 424344 ;; 929394 ×××

4. Answers may vary. Sample answers:

a. 162164168 ;; 242244248 ÷÷÷

b. 242243246 ;; 422423426

c. 108210831086 ;; 132213231326

d. 7537557515 ;; 1953195519515

5. a. HCF of 42 and 56 is 14.

7. a. 221 15 2 222 ×+ ==

LCM of 2 and 5 is 10. 5251428 and 210510 ==

Since, 2528 <

Then, 2528 1010 <

Thus, 114 2 25 <

b. 654 4 34 6 555 ×+ ==

LCM of 5 and 4 is 20. 3413634170 and 520420 ==

Since, 136170 <

56144 ÷ = ÷

4214 3

b. HCF of 75 and 125 is 25. 7525 3 125255 ÷ = ÷

c. HCF of 66 and 84 is 6.

666 11

84614

d. HCF of 72 and 136 is 8. 728 9 136817 ÷ = ÷

6. a. LCM of 8 and 24 is24.

53 15 8324 × = ×

Since, 515 824 =

Hence, the fractions are equivalent.    b. LCM of 7 and 42 is 42.

Then, 136170 2020 <

Thus, 434 6 54 <

c. 243 8 33 ÷ = ÷

Thus, 24 8 3 =

d. 193 119 57 6 33339 × === ×

Thus, 157 6 39 =

8. a. LCM of 2, 4, 5 and 6 is 60. 115210212130 ;;; 415610512230 ×××× ==== ×××× 15202430 60606060

46 24

7642 × = ×

Since, 424 742 ≠     Hence, the fractions are not equivalent.    c. LCM of 3 and 24 is 24. 28 1618

382424 × =≠ ×

Since, 218 324 ≠

Hence, the fractions are not equivalent.    d. LCM of 3 and 92 is 276.

19246138 and 327692276 ==

92138 276276 ≠

Since, 146 392 ≠     Hence, the fractions are not equivalent.

Since, 15202460 <<< 15202460 60606060 →<<< 1221 4652 →<<<

b. LCM of 2, 6, 8 and 11 is 264. 9241132133 ;;; 11242132833 544 644

××× === ××× × = ×

216 132 33 264 264264 220 264

Since, 33132216220 <<< 33132216220 264264264264 →<<< 1195 82116 →<<<

c. LCM of 3, 6, 8 and 9 is 72. 3951252438 ;;; 8961232498 ×××× ==== ×××× 276012024 7272 7272

Since 242760120 <<<

242760120

72727272 →<<<

3355 9863 →<<<

d. LCM of 3, 7, 8 and 10 is 840. 78421201280 588 240 280 ;;; 108484071208403280840

××× === ×××

10. a. LCM of 3, 5, 8 is 120. 12241015840 ;;;

524815340

××× === ××× × = ×

288 150320 120 120120

5105 525 8105840

× = ×

Since, 240280525588 <<<

240280525588

588840840840 →<<<

2157 73810 →<<<

9. a. LCM of 2, 5, 6 and 12 is 60. 55130412110 ;;; 125230512610 ×××× ==== ×××× 25304810 60606060

Since, 10253048 <<< 10253048 60606060 →<<<

1514 61225 →<<<

336320288150 120120120120 →>>>

1481210 5358 →>>>

××××

==== ×××× 30404548 60606060

4321 5432 →>>>

c. LCM of 3, 5, 17 is 2805. 49351756119165 ;;; 3935556117165

××× × = ×

40404040

b. LCM of 2, 5, 8 is 40. 183835120 ;;; 585885220 ×××× ==== ×××× 8241520

Since, 8152024 <<< 8152024 40404040 →<<<

1313 5825 →<<<

c. LCM of 2, 3, 5 and 7 is 210. 1105270630 ;;; 2105370730

××× === ××× ×

105140180 210210210 84 210

84105140180

210210210210 →<<<

2126 5237 →<<<

d. LCM of 2, 5, 7 and 12 is 420 11351210384

;;; 12352240584

14311

27512 →<<<

Since, 9537 > 3740 > 3135 > 1870 9537374031351870

2805280528052805 →>>>

174192

53173 →>>>

d. 44917581752 5;;4;5 993191999 ===

931

×××× ==== ××××

988 171 171171 171

988969931729

171171171171 →>>>

52174981 93919 →>>>

Thus, 71745 554 93919 >>>

Word Problems 1. Number of slices in each cake = 12

Total number of slices eaten = 47

Fraction of the cake eaten = 47 12 3 12 47 36

11

2. Fraction of stamps Ryan took = 2 5

Fraction of stamps Sarah took = 3 8

LCM of 5 and 8 is 40.

28163515 and 58408540 ×× == ××

Since, 1615 4040 >

So, 23 58 >

Thus, Ryan took a greater fraction of stamps.

3. Fraction of marbles Alex took = 1 4

Fraction of marbles Bailey took = 3 8

Fraction of marbles Casey took = 3 12

LCM of 4, 8 and 12 is 24. 163332 ;; 4683122 ××× === ××× 696 242424

133 4128 ∴=<

Thus, Alex and Casey took the least fraction of marbles and Bailey took the greatest fraction.

4. Fraction of cake eaten by:

Sara = 3 10

Mia = 1 12

Tina = 1 18

Sancy = 3 18

LCM of 10, 12 and 18 is 180. 318 113 ;;; 1018121818 × ==== × 54151030 180180180180

Since, 10153054 <<<    Then, 10153054 180180180180 →<<<

1133 18121810 →<<<

Chapter 6

Let's Warm-up

1 2 halves 1 2 together make a whole.

2 A quarter 1 4 divides a whole into four equal parts.

3 111 333 ++= 1 4 Adding 1 7 seven times gives one whole.

5. 1 6 shows 1 part out of six equal parts from a whole.

Do It Yourself

6A 1. a. Among the given fractions 134 1 444 +== . So, 1 4 , 2 4 , 3 4 , 5 4

b. Among the given fractions 12112 1 24222 +=+== . So, 1 8 , 5 8 , 1 2 , 2 4

c. Among the given fractions 312347 1 721777 +=+== . So, 3 7 , 8 21 , 12 21 , 4 14

d. Among the given fractions 215235 1 525555 +=+== . So, 2 5 , 15 10 , 15 25 , 12 35

2. a. 5 9 4 10 4540 54 8517 109910909018 ×+×+ +==== ×

b. 43415 1 515555 +=+==

c. 2 6 27 7127191 1 318361818181818 × +=+=+==

Word

Length

Total length of the new rope = 15 3 1 36

2. Weight of mangoes in a basket initially = 1 2 5 kg

Rajiv puts 2 2 kg 5 more mangoes

Therefore, total weight of the mangoes in the basket = 122222 kg12

3. Time duration of the first task = 11 7 minutes

Time duration of the second task = 5 1 2 minutes

Time duration of the third task = 74 15 minutes

Total time taken by Zaid to complete all tasks

=++ minutes

kg

5.

×+×+×−× −=−=−= ×

b. 543331 233104 31 2310 53 43434334 29

c. 1111169713 5223 34343434

 +−+=+−+     ×+××+× ==−==  ××

164 9374133 916724 2 43 43121212

d. 1111161053 5321 33223322

 −−−=−−−      =−=−=−=  

161053 62 211 3232

Word Problems 1. Navneet studied on Tuesday

173451227 12175 31 7 41 2 434312121212 ×−××+ −=−==== hours

2. Total wheat in the drum = 1 50 2 kg

Wheat used by Sanjay = 3 10 4 kg

Wheat left in the drum = 50113 0 24 kg = 10143 24 kg

Hence, 3 39 4 kg wheat is left in the drum.

3. Total weight of fruits = 3 7 4 kg

Weight of apples = 1 1 2 kg

Weight of pears = 1 2 4 kg

Weight of oranges = 2 1 3 kg

Weight of litchis = 713112 21 4243  −++  

kg.

Hence, the weight of litchis purchased = 2

c.

Word

number of pages in book = 100

Fraction of pages read by Mahesh = 2 5

Number of pages read by Mahesh

pages

Hence, Mahesh has read 40 pages.

2. Total weight of the cake = 2 kg

Fraction of cake eaten by Sushen = 3 4

Weight of cake eaten by Sushen = 3 2 4 × kg = 3 2 22 × × kg = 3 2 kg = 1 1 2 kg

Hence, Sushen ate 3 2 kg cake.

3. Total amount of fuel in the tank = 50 L

Fraction of fuel used in the journey = 3 5

Amount of fuel used in the journey = 3 50 5 × L = 3 510 5 ×× L = 30 L

Fuel left in the tank = 50 L – 30 L = 20 L

Hence, amount of fuel left in the tank is 20 L.

4. Total number of marbles with Shefali = 2000

Fraction of marbles of red colour = 1 4

Number of marbles of red colour = 1 2000 4 × = 500

Fraction of marbles of blue colour = 1 2

Number of marbles of blue colour = 1 2000 2 × = 1000

Number of marbles of green colour 2000–(1000500)=+ 20001500500=−=

Hence, Shefali has 500 green marbles.

c.

3.

b.

c.

d.

e.

f.

4.

c.

d.

e.

f.

5.

b. 171711 17 99179

c.

d.

e. 5454136324

Word Problems 1. Total amount of rice with Ravi= 21 kg Rice in each packet = 3 4 kg

Total no. of packets packed by Ravi = 3 21 4 ÷

44 21 73 28 33 ×=××=

Hence, Ravi packed 28 packets of rice.

2. Capacity of container = 57 4 L

Capacity of each bottle = 3 4 L

Number of bottles than can filled from container = 19 3 573574 4 19 444343 × ÷=×=×=

Hence, 19 bottles can be filled from the container.

Chapter Checkup

1. a. 52 39 517171017273 9189218181818292

+=+=+=== ×× = 1 1 2

b. 62 621 21122133 173417234343434

c.

e. 183 1864 645464118 5155315151515

126

b.

c.

5.

b.

c.

d.

e.

f.

g. 2323111 9833243412

h. 3434313 5165445420 ×=×=×= ×

i. 6126121 121812633 ×=×= ×

j. 93 4274 1 93693343 × ×=×= ××

6.

7. a. 5525251 12 5222 ÷=×==

b. 12134424316 433 ÷=×=××=

c. 1111111 3 55315 ÷=×=

d. 5515 18 9918162 ÷=×=

e. 43 123124416 17417317317 × ÷=×=×=

f. 53 15515663 2462456454 × ÷=×=×= ×

8. a. False  b. True  c. False  d. True  e. False

9. a. The multiplicative inverse of 1 2 2 is 215 . 2 ; 522  =   multiplicative inverse = 2 5    

b. There are 58 quarters in 1 14. 2 12929129 14 ; 458 22242  =÷=×=   

c. Dividing by 1 3 is the same as multiplying by 3

d. 1 is the reciprocal of itself.

10 a. 48 128 1 11 3317

484848

332 176617491 6 4288888

797

Since the whole number parts are equal, we will compare the proper fraction part of our fraction 78 1712 1256 , , ,. 8128121289696 × ==

Clearly, 56 96 > 12 96 ⇒

1 25 4 14 514

×+×+

×+×+ ====

c. 2 3 5 ×− 4 1642 5555 ×=−=

Word Problems 1. Distance jumped by Raghav = 4 3 9 m

Distance jumped by Utkarsh more than Raghav = 7 8 m

Distance jumped by Utkarsh = 47 3 98 + m = 934 7 98 ×+ + m

= 317 98 + m = 31879 98 ×+× × m = 24863 72 + m = 311 72 m 23 4 72 = m

Hence, the distance jumped by Utkarsh is 4 23 72 m.

2. Distance travelled by man on bicycle = 1 4 7 km

Distance travelled by man on foot = 3 2 4 km

Distance travelled by man on car = 1 10 2 km

Total distance covered by man = 4210131 742 ++ km

= 4714232101 742 ×+×+×+ ++ km

= 291121 742 ++ km

= 212 2911 7422 × ++ × km

= 291142 744 ++ km

= 2953 74 + km

294537 74 ×+× = × km = 487 28 km = 11 17 28 km

Hence, the distance covered by man is 11 17 28 km

3. Height of Shubankar = 147 1 2 cm

Shubangi is shorter by = 1 11 3 cm

Therefore, height of Shubangi = 147 11 11 23 cm

= 147 2 11131 23 ×+×+ cm

= 29534 23 cm

= 2953 342 23 ×−× × cm

= 88568 6 cm

= 817 6 cm 1 136 6 = cm

Hence, Shubangi is 1 136 6 cm tall.

4. Number of cars in the parking lot = 66

Fraction of parking lot occupied = 3 4

Now, by multiplying by 22 in th0065 numerator and denominator, we get 322 3 66 442288 × == ×

Since, 66 cars are in the parking lot already, it means total 88 cars can be parked.

5. Total number of cans = 14

Total capacity of 14 cans = 1 144 2 litre

Capacity of each can = 1 144 2 ÷ 14 litre

= 14421 1 214 ×+ × litre

= 2891 214 × litre = 289 28 litre = 9 10 28 litre

Hence, capacity of each can is 9 10 28 litre.

6. Total length of cloth purchased by Radhika = 20 m Cloth used for curtains = 2 11 5 m

Cloth used for bed sheet = 5 3 6 m

Length of cloth used = 11325 56 + m = 5112365 56 ×+×+ + m

= 5723 56 + m = 576 235 457 5630 ×+× = × m

of cloth left = 457 20 m 30 = 2030 457 143 3030 ×− = m

= 23 4 30 m

Hence, the amount of cloth left is 23 4 30 m.

Chapter 7

Let's Warm-up

1. ++= 1113 10101010 .

2. Each equal part in a hundred grid is equal to 1 100 or hundredths

3. When a whole is divided into 10 equal parts then each part is equal to 1 10 or tenths

4. ++++= 11111 1 55555 = 1

5. The fraction for 3 out of 10 = 3 10

Do It Yourself

7A 1.

1.6 2.3

One and six-tenths. Two and three-tenths.

2. a. 0.5 = zero and five tenths or zero point five

b. 4.0 = four and zero tenths or four point zero

3. a. 52.1  b. 413.2  c. 100.5  d. 805.7

4. a. 45.1 = 40 + 5 + 1 10 or 40 + 5 + 0.1

b. 17.9 = 10 + 7 + 9 10 or 10 + 7 + 0.9

c. 143.5 = 100 + 40 + 3 + 5 10 or 100 + 40 + 3 + 0.5

d. 548.1 = 500 + 40 + 8 + 1 10 or 500 + 40 + 8 + 0.1

e. 789.4 = 700 + 80 + 9 + 4 10 or 700 + 80 + 9 + 0.4

f. 985.3 = 900 + 80 + 5 + 3 10 or 900 + 80 + 5 + 0.3

5. a. 5 10 = 0.5  b. 4 + 1 10 = 4 + 0.1 = 4.1

c. 3 + 8 10 = 3 + 0.8 = 3.8   d. 9 10 = 0.9

e. 78 + 5 10 = 78 + 0.5 =78.5  f. 80 + 6 + 8 10 = 86 + 0.8 = 86.8

Word Problems 1. 1255 cm cannot be the height of a person. To write his correct height in centimeters, Raj should put the decimal point after one place from the right. So, his height in centimeters with the correct decimal placement would be 125.5 cm.

2. Rashmi ate 3 10 = 0.3 slices of pizza.

7B 1. a. 1.25  b. 3.61

2.  1.72

3. a. 42.14 = forty-two and fourteen-hundredths or forty-two point one four

b. 87.08 = eighty-seven and eight hundredths or eighty-seven point zero eight

c. 81.87 = eighty-one and eighty-seven hundredths or eightyone point eight seven

d. 178.64 = one hundred seventy-eight point six four or one hundred seventy-eight and sixty-four hundredths

e. 874.09 = eight hundred seventy-four point zero nine or eight hundred seventy-four and nine hundredths

f. 457.18 = four hundred fifty-seven point one eight or four hundred fifty-seven and eighteen hundredths

4. a. 72.33  b. 417.02  c. 26.47  d. 361.04

5. a. 40 + 5 + 1 10 + 3 100 or 40 + 5 + 0.1 + 0.03

b. 70 + 90 + 6 100 or 70 + 9 + 0.06

c. 40 + 1 + 3 10 + 7 100 or 40 + 1 + 0.3 + 0.07

d. 400 + 80 + 6 + 7 10 + 2 100 or 400 + 80 + 6 + 0.7 + 0.02

e. 800 + 70 + 9 + 3 10 + 4 100 or 800 + 70 + 9 + 0.3 + 0.04

f. 900 + 70 + 6 + 7 100 or 900 + 70 + 6 + 0.07

6. a. 4 100 = 0.04  b. 14 + 2 100 = 14 + 0.02 = 14.02

c. 43 + 4 10 + 8 100 = 43 + 0.4 + 0.08 = 43.48

d. 500 + 3 + 4 100 = 503 + 0.04 = 503.04

e. 70 + 3 100 = 70 + 0.03 = 70.03

f. 800 + 60 + 7 + 1 10 + 7 100 = 867 + 0.1 + 0.07 = 867.17

Word Problem 1. Distance travelled by the cyclist = 23.48 km 23.48 = 20 + 3 + 4 10 + 8 100 or 20 + 3 + 0.4 + 0.08

7C 1. a. 25.127 = twenty-five and one hundred twenty-seven thousandths or twenty-five point one two seven

b. 42.872 = forty-two and eight hundred seventy-two thousandths or forty-two point eight seven two

c. 16.012 = sixteen point zero one two or sixteen and twelve thousandths

d. 174.201 = one hundred seventy-four and two hundred one thousandths or one hundred seventy-four and two zero one

e. 965.045 = nine-hundred sixty-five point zero four five or nine-hundred sixty-five and forty-five thousandths

f. 765.004 = seven hundred sixty-five and four thousandths or seven hundred sixty-five point zero zero four

2. a. 31.352  b. 52.402  c. 12.041  d. 302.005

3. a. 62 15.0621050.060.002 or105 1001000

b.

c.

d. 1 111.0011001010.001 or 100101

e. 486.1454008060.10.040.005 or 145 400806 101001000

f. 154.0311005040.030.001 or 10050

4. a. = 6 0.006 1000

b. ++=++= 25 14 140.020.00514.025 1001000 c. +++=++= 16 106 160.10.00116.106

d. +=+= 8 1061060.008106.008

e. +++=++= 37 5006

6. Answers may vary. Sample answers:

a. 45.6 = 45.60 =

c. 87.02 = 87.020 = 87.0200

d. 963.14 = 963.140 = 963.1400

e. 12.701 = 12.7010 = 12.70100

f. 189.221 = 189.2210 = 189.22100

Word Problems 1. Rice required for the recipe = 1.269 kg

1.269 = one and two hundred sixty-nine thousandths or one point two six nine

2. Distance travelled by the person = 3.659 km

> 5.50.25

1 5.5> 4

f. 5 12 and 8

== 5625 0.625 81000

> 120.625

12 > 5 8

Word Problems 1. Cake eaten by Ravi = 0.375 of a cake

0.375 = 375 1000 = 375 ÷ 125 1000 ÷ 125 = 3 8

2. Work completed by the contractor in 7 days = 5 8 of the total work

5 8 = 5 × 125

8 × 125 = 625 1000 = 0.625

7E 1. Like decimals have same number of digits after the decimal.

a. 1.2, 5.4, 8.9, 6.54, 1.3    Not like decimals.

b. 2.23, 4.26, 4.89, 4.2, 6.584    Not like decimals.

c. 7.89, 7.2, 64.594, 45.2, 56.5    Not like decimals.

d. 81.564, 78.512, 453.125, 486.154, 86.15    Not like decimals.

2. a. 4.2 and 6.25 

c. 11.3 and 17.8 

e. 10.001 and 12.690 

3. a. 1.45 and 1.6

1.6 = 1.60

c. 17.98 and 14.221

17.98 = 17.980

e. 8.21 and 3.560

8.21 = 8.210

4.  a. 13.15, 1.2

→ 1.2 = 1.20

c. 4.8, 1.526

→ 4.8 = 4.800

e. 53.23, 17.164

→ 53.23 = 53.230

5. a. 12 20 , 1 5

× 5 = 60 100 = 0.6

b. 17.23 and 691.56 

d. 5.157 and 64.581 

f. 1.111 and 1.11 

b. 81.566 and 12.2

12.2 = 12.200

d. 11.001 and 11 11 = 11.000

f. 1.101 and 967.1 967.100

b. 3.48, 1.2

→ 1.2 = 1.20

d. 1.4, 47.584

→ 1.4 = 1.400

f. 1.002, 348.1

→ 348.1 = 348.100

b.

c.

d.

e.

Word Problem 1. Company’s profit in 2021 = 23.2 crores

2021: 23.2 = 23.200 Company’s profit in 2022 = 26.987 crores

2022: 26.987 Company’s profit in 2023 = 24.54 crores

2023: 24.54 = 24.540

23.200, 26.987 and 24.540 are like decimals.

a.

b.

2. a. 4.2 or 4.15

→=4.24.20

Since, 2 > 1

Thus, 4.2 > 4.15

b. 4.3 or 4.5

Since, 3 < 5

Thus, 4.3 < 4.5

c. 15.64 or 15.67

Since, 64 < 67

Thus, 15.64 < 15.67

d. 87.654 or 87.65

→=87.6587.650

Since, > 654650

Thus, > 87.65487.65

e. 186.6 or 186.5

Since, 6 > 5

Thus, 186.6 > 186.5

f. 294.98 or 294.864

→=294.98294.980

Since, > 980864

Then, > 294.98294.864

3. a. 15.64 or 15.46

Since, 64 > 45

Then, > 15.6415.46

b. 89.14 or 89.4

→=89.489.40

Since, < 1440

Then, < 89.1489.4

c. 81.174 or 81.714

Since, < 174714

Then, < 81.17481.714

d. 853.68 or 853.6

→=853.6853.60

Since, > 6860

Then, 853.68853.6 >

e. 47.654 or 47.650

Since, > 654650

Then, > 47.65447.650

4. a. 14.14, 14.1, 14.01, 14.101

14.14 14.1 14.01 === 14.140,14.100,14.010,14.101

Since, <<< 010100101140

Then, <<< 14.0114.114.10114.14

b. 84.56, 84.5, 84.6, 84.55, 84.65

84.5684.5084.6084.5584.65 ;84.5;84.6;; ==

Since, <<<< 5055566065

Then, <<<< 84.584.5584.5684.6084.65

c. 184.2, 184.23, 184.1, 184.112

184.2 184.23 184.1 184.200;184.230;184.100;184.112 ===

Since, 100 < 112 < 200 < 230

Then, 184.1 < 184.112 < 184.2 < 184.23

d. 64.23, 54, 64.32, 64.22, 64.33

Since, 22 < 23 < 32 < 33

Then, 54 < 64.22 < 64.23 < 64.32 < 64.33

5. a. 7.4, 7.5, 7.44, 7.54

→ 7.4 = 7.40 and 7.5 = 7.50

Since, 54 > 50 > 44 > 40

Then, 7.54 > 7.5 > 7.44 > 7.4

b. 48.6, 48.36, 48.63, 48.66

→ 48.6 = 48.60

Since, 66 > 63 > 60 > 36

Then, 48.66 > 48.63 > 48.6 > 48.36

c. 921.4, 921.26, 921.35, 921.41

→ 921.4 = 921.40

Since, 41 > 40 > 35 > 26

Then, 921.41 > 921.4 > 921.35 > 921.26

d. 185.5, 185.55, 185.45, 185.501

→ 185.5 = 185.500; 185.55 = 185.550; 185.45 = 185.450

Since, 550 > 501 > 500 > 450

Then, 185.55 > 185.501 > 185.5 > 185.450

Word Problems 1. Quantity of water in a bottle = 0.75 litres

Quantity of water in another bottle = 0.9 litres

0.75 and 0.9

→ 0.9 = 0.90

Since,75 < 90

Then, 0.75 < 0.9.

So, the bottle of capacity 0.9 litres has more water in it.

2. Temperature on:

Monday = 34.2 °C

Tuesday = 34.5 °C

Wednesday = 33.7 °C

Thursday = 35.2 °C

Since, 33 < 34 < 35

For 34.2 and 34.5 → 2 < 5

So, 34.2 < 34.5

i.e. 33.7 °C < 34.2 °C < 34.5 °C < 35.2 °C

3. Let the three athletes be: Athlete 1, Athlete 2 and Athlete 3

The race was finished in:

Athlete 1 = 24.54 minutes

Athlete 2 = 24.68 minutes

Athlete 3 = 24.36 minutes.

Since, 68 > 54 > 36

Then, 24.68 > 24.54 > 24.36.

7G 1. a. If the digit in the tenths place is 5, then the number rounds up to the nearest whole number. TRUE

b. If the digit in the ones place is 4 or less, the number always rounds down to the lower whole number. TRUE

c. 12.2 rounds up to 13 when rounded to the nearest whole number. FALSE

d. 99.9 rounds up to 100 when rounded to the nearest whole number. TRUE

2. a. 2.6 lie between 2 and 3.

Digit at tenths place = 6     So, 2.6 rounds up to 3.

b. 7.9 lie between 7 and 8.

Digit at tenths place = 9

So, 7.9 rounds up to 8.

c. 48.2 lie between 48 and 49.

Digit at tenths place = 2

So, 48.2 rounds up to 48.

d. 15.1 lie between 15 and 16.

Digit at tenths place = 1

So, 15.1 rounds up to 15.

e. 71.3 lie between 71 and 71.

Digit at tenths place = 3

So, 71.3 rounds up to 71.

f. 156.4 lie between 156 and 157.

Digit at tenths place = 4

So, 156.4 rounds up to 156.

3. a. 31.9 lie between 31 and 32.

Digit at tenths place = 9

So, 31.9 rounds up to 32.

b. 14.5 lie between 14 and 15.

Digit at tenths place = 5

So, 14.5 rounds up to 15.

c. 123.4 lie between 123 and 124.

Digit at tenths place = 4

So, 123.4 rounds up to 123.

d. 543.1 lie between 543 and 544.

Digit at tenths place = 1

So, 543.1 rounds up to 543.

4.  22.9 lie between 22 and 23.

Digit at tenths place = 9

So, 22.9 rounds up to 23.

21.2 lie between 21 and 22.

Digit at tenths place = 2

So, 21.2 rounds up to 21.

Thus, 22.9 when rounded off to the nearest whole number, gives 23.

5. a. 45.2 lie between 45 and 46.

Digit at tenths place = 2

So, 45.2 rounds up to 45. 45.6 lie between 45 and 46.

Digit at tenths place = 6

So, 45.6 rounds up to 46.

45 < 46.

b. 14.1 lie between 14 and 15.

Digit at tenths place = 1

So, 14.1 rounds up to 14.

14.2 lie between 14 and 15.

Digit at tenths place = 2

So, 14.2 rounds up to 14.

14 = 14

c. 87.6 lie between 87 and 88.

Digit at tenths place = 6

So, 87.6 rounds up to 88.

88.2 lie between 88 and 89.

Digit at tenths place = 2     So, 88.2 rounds up to 88.

88 = 88

d. 17.3 lie between 17 and 18.

Digit at tenths place = 3

So, 17.3 rounds up to 17.

17.6 lie between 17 and 18.

Digit at tenths place = 6

So, 17.6 rounds up to 18.

17 < 18

e. 197.1 lie between 197 and 198.

Digit at tenths place = 1

So, 197.1 rounds up to 197.

198.6 lie between 198 and 199.

Digit at tenths place = 6

So, 198.6 rounds up to 199.

197 < 199

Word Problems

1. Amount of rainfall received in a year by the city = 15.7 inches

15.7 lie between 15 and 16.

Digit at tenths place = 7

So, 15.7 rounds up to 16.

2. Marks scored in the test = 88.5

88.5 lie between 88 and 89.

Digit at tenths place = 5

So, 88.5 rounds up to 89.

Chapter Checkup 1. a.

1.2 = 1 + 0.2 = 1 + 2 10

b.

1.48 = 1 + 0.4 + 0.08 = 1 + 4 10 + 8 100

2. a. 15.2 = fifteen and two-tenths or fifteen point two

b. 71.65 = seventy-one and sixty-five tenths or Seventy-one point six five

c. 152.1 = one hundred fifty-two and one tenths or one hundred fifty-two point one

d. 814.36 = eight hundred fourteen and thirty-six hundredths or eight hundred fourteen point three six

e. 176.254 = one hundred seventy-six and two hundred fifty-four thousandths or one hundred seventy-six point two five four

f. 176.801 = one hundred seventy-six and eight hundred one thousandth or one hundred seventy-six point eight zero one

3. a. 40.101  b. 63.4  c. 103.05  d. 420.011

e. 31.005  f. 95.223

4. a. 11.1

Tens Ones Decimal Point Tenths

Hundreds Tens Ones Decimal point Tenths

Tens Ones Decimal point Tenths Hundredths

Hundreds Tens Ones Decimal point Tenths Hundredths

e.

Tens Ones

Hundreds

7.

Since, 3 < 4

Then, 12.3 < 12.4

b. 14.5 and 14.55

→ 14.5 = 14.50

Since, 50 < 55

Then, 14.5 < 14.55

c. 102.3 and 102.25 are not like decimals.

We can write them as like decimals as:

→ 102.3 = 102.30

Since, 30 > 25

Then, 102.3 > 102.25

d. 222.22 and 222.02

Since, 22 > 0.2

Then, 222.22 > 222.02

e. 187.540 and 187.504

Since, 540 > 504

Then, 187.540 > 187.504

f. 3.003 and 3.033

Since, 3 < 33

Then, 3.003 < 3.033

8. a. 14.23, 14.2, 14.3, 14.25

→ 14.23; 14.2 = 14.20; 14.3 = 14.30; 14.25

Since, 20 < 23 < 25 < 30

Then, 14.2 < 14.23 < 14.25 < 14.3

b. 87.64, 87.6, 87.5, 87.7

→ 87.64; 87.5 = 87.50; 87.6 = 87.60; 87.7 = 87.70

Since, 50 < 60 < 64 < 70

Then, 87.5 < 87.6 < 87.64 < 87.7

c. 235.64, 235.66, 235.6, 235.666

→ 235.64 = 235.640; 235.66 = 235.660; 235.6 = 235.600; 235.666

Since, 600 < 640 < 660 < 666

Then, 235.6 < 235.64 < 235.666

d. 888.88, 888.888, 888.8, 888.08

→ 888.88; = 888.880; 888.888; 888.8 = 888.800; 888.08 = 888.080

Since, 080 < 800 < 880 < 888

Then, 888.08 < 888.8 < 888.88 < 888.888

9. a. 1.3, 1.31, 1.2, 1.33

→ 1.3 = 1.30; 1.31; 1.2 = 1.20; 1.33

Since, 33 > 31 > 30 > 20

Then, 1.33 > 1.31 > 1.3 > 1.2

b. 19.4, 19.44, 19.54, 19.501

→ 19.4 = 19.400; 19.44 = 19.440, 19.54 = 19.540, 19.501

Since, 540 > 501 > 440 > 400

Then, 19.54 > 19.501 > 19.44 > 19.4

c. 555.5, 555.55, 555.05, 555.555

→ 555.5 = 555.500; 555.55 = 555.550; 555.05 = 555.050; 555.555

Since, 555 > 550 > 500 > 050

Then, 555.555 > 555.55 > 555.5 > 555.05

d. 748.01, 748.101, 748.11, 748.1

→ 748.01 = 748.010; 748.101; 748.11 = 748.110; 748.1 = 748.100

Since, 110 > 101 > 100 > 010

Then, 748.11 > 748.101 > 748.1 > 748.01

10. a. 1.3 lie between 1 and 2

Digit at tenths place = 3

So, 1.3 rounds up to 1.

b. 14.8 lie between 14 and 15

Digit at tenths place = 8

So, 14.8 rounds up to 15.

c. 99.6 lie between 99 and 100

Digit at tenths place = 8

So, 99.6 rounds up to 100.

d. 100.2 lie between 100 and 101

Digit at tenths place = 2

So, 100.2 rounds up to 100.

e. 106.7 lie between 106 and 107

Digit at tenths place = 7

So, 106.7 rounds up to 107.

f. 999.9 lie between 999 and 1000

Digit at tenths place = 9

So, 999.9 rounds up to 1000.

Word Problems 1. Raj = 14.3 minutes = 14.30

Rekha = 15.2 minutes = 15.20

Utkarsh = 13.92 minutes = 13.92

Ali = 13.99 minutes = 13.99

Since, 15 > 14 > 13

For 13.92 and 13.99

92 < 99

So, 13.99 > 13.92

Then, 15.20 > 14.30 > 13.99 > 13.92

So, Utkarsh takes the least time to finish the race, thus, he is the winner.

2. Time taken by Mr. Goyal to finish the task = 40.6 minutes 40.6 lies between 40 and 41.

Digit at tenths place = 6

So, 40.6 rounds up to 41.

3. 156. 25 = 100 + 50 + 6 + 0.2 + 0.05 or 100 + 5 + 6 + 2 10 + 5 100

4. Amount spent by:

Mr. Jadeja = $12.99

Sarah = $13.01

Alice = $12.9 = 12.90

Jacob = $13.1 = 13.10

Since, 112 < 13

90 < 99 and 0.1 < 10

Then, 12.9 < 12.99 < 13.01 < 13.1.

Chapter 8

Let's Warm-up 1. 1.14 < 1.41  2. 1.4 = 1.400

3. 3.15 < 3.5  4. 2.33 > 2.3  5. 1.54 > 1.45

Do It

4. a. 3.5 + 154.2           53.9

b. 41.56 + 12.34         88.58

c. 31.23 + 57.35         51.003

d. 45.913 + 5.09         157.7

5. a. 45.2 + 32.9 < 79.1    b. 14.5 + 31.47 = 45.97

c. 62.1 + 81.009 > 143.009  d. 61.81 + 17.568 = 79.378

Word Problems

1. The total amount of money spent on the matchbox = ₹0.50

The amount of money spent on the bar of soap = ₹8.75

The amount of money spent on the pen = ₹6.5

The total amount of money spent by Rani = ₹0.50 + ₹8.75 +

₹6.5 = ₹15.75  T O t h

2. The weight of the cookie (in grams) = 10 + 12.82 + 8.72 = 31.54

3. The distance covered by Rashi = 1.548 km

The distance covered by Prerna = 2.328 km

The distance covered by Navya = 1.986 km

The total distance covered by them = 1.548 + 2.328 + 1.986 = 5.862 km

4. a. The total weight of Rakesh and Ismail (in kg) = 34.53 + 36.35 = 70.88 kg T O . t h

b. The total weight of John and Shahid (in kg) = 43.5 + 40.235 = 83.735 kg

T O t h th 4 3 5 0

8B 1. a. 1

2.

e.

f.

3. a. 45.3 25.6          550.584

b. 51.48 11.68          678.877

c. 896.4 345.816        19.7

d. 789.987 111.11        39.8

4. a. 47.8 36.6 = 11.2     b. 96.4 51.84 > 44.46

c. 125.6 84.934 > 35.666   d. 256.71 59.357 > 196.353

e. 512.2 461.718 < 51.482   f. 827.601 739.798 > 87.802

5. a. The difference between 12.45 and 9.496 is 2.954.

H T O t h th 0

b. 77.617 is 15.84 less than 93.457. H

Word Problems 1. The quantity of the mixture = 48.6 L

The quantity of pure milk = 41.2 L

The quantity of water in the mixture = 48.6 41.2 = 7.4 L

T O . t

4 8 . 6

4 1 2

0 7 4

2. Iyer’s height (in m) = 1.65

Difference between Iyer’s height and his brother’s height = 0.18 m

The brother’s height = 1.65 0.18 = 1.47

3. The distance between points A and B (in km) = 1.2

The distance between point B and C (in km) = 2.35

The distance between points A and C (in km) = 1.2 + 2.35 = 3.55

The total distance between points A and D (in km) = 7

The distance between points C and D (in km) = 7 3.55 = 3.45

4. The total money Rashmi has = ₹5555.5

The amount spent on the saree = ₹555.5

The amount spent on an umbrella = ₹55.5

The amount given to the auto driver = ₹5.5

The total amount spent = ₹555.5 + ₹55.5 + ₹5.5 = ₹616.5

The total amount left with her = ₹5555.5 − ₹616.5 = ₹4939

d. 4 × 2.5 = 10  e. 8 × 5.45 = 43.60 2 5

f. 12 × 14.945 = 179.34

1 4 9 4 5

2 9 8 9 0

+ 1 4 9 4 5 0

1 7 9 3 4 0

g. 1.6 × 1.58 = 2.528

1 5 8 ×

6

4 8

1

2 5 2 8

h. 12.3 × 35.69 = 438.987 3 5 6 9

4 3 8 9 8 7

3. a.  5.15 × 10 < 515  b. 83.482 × 1000 = 83482

c. 6 × 1.658 < 10.948  d. 715.15 × 8 > 5720.12

e. 2.13 × 31.4 > 6.5882  f. 1.5 × 634.71 > 94.2065

4. a. 12.15 × 10 = 121.5   b. 15.26 × 1000 = 15260

c. 61.235 × 10 = 612.35  d. 14.1 × 1000 = 14100

e. 6 × 1.645 = 9.87 f. 1.1 × 47.503 = 52.2533

5. Number thought by Matthew = 4.6

Six times 4.6 = 6 × 4.6 = 27.6

2 7 6

Thus, Matthew thinks of the number 27.6

Word Problems

1. The cost of 1 kg oranges = ₹41.23

The cost of 5 kg oranges (in ₹) = 41.23 × 5 = 206.15 4

Thus, the cost of 5 kg oranges is ₹206.15

2. The total number of workers in the factory = 24

The amount of money given to each of 11 workers = ₹252.54

The total amount of money given to 11 workers = 252.54 × 11 = ₹2777.94

2

The amount of money given to each of 13 workers = ₹364.52

The total amount of money given to 13 workers = 364.52 × 13 = ₹4738.76

The total amount of money given = ₹2777.94 + ₹4738.76 = ₹7516.7

3. The cost of 1 kg rice and 1 kg pulses = ₹75.5

The cost of 1 kg rice = ₹43.75

The cost of 1 kg pulses = ₹75.5 ₹43.75 = ₹31.75

Thus, the cost of 5 kg pulses = ₹31.75 × 5 = ₹158.75

3 1 7 5

× 5

1 5 8 7 5

8D 1. 1 Euro = ₹89.19

1000 Euros = ₹89.19 × 1000 = ₹89,190

8 9 1 9

× 1 0 0 0

8 9 1 9 0

2. The cost of the gift = 55 Yuan

1 Yuan = ₹11.51

The cost of the gift in ₹ = ₹11.51 × 55 = ₹633.05

3. The money Shweshitha wanted (in pounds) = 500

1 pound = ₹104.06

500 pounds = ₹104.06 × 500 = ₹52,030

4. The salary of Suresh’s father = 800 Dirham

The salary of Saurav’s father = 2200 Sri Lankan rupees.

Since, 1 Dirham = ₹22.51 and 1 LKR = 0.25

Therefore, the salary earned by Suresh’s father (in ₹) = 800 × ₹22.51 = 18,008.00

The salary earned by Saurav’s father (in ₹) = 2200 × ₹0.25 = 550

₹18008 ₹550 = ₹17,458

Thus, Suresh’s father gets ₹17,458 more than Saurav’s father.

5.  The amount Asmitha has (in pounds) = 180

Amount Asmitha has (in rands) = 54

Since 1 pound = ₹104.06 and 1 rand = ₹4.43, Converting into Indian Rupees— 180 × ₹104.06 = 18730.8

The total amount that Asmita has (in Indian Rupees) = ₹18730.8 + ₹239.22 = ₹18,970.02

Word Problems 1. 1 Pound = ₹104.06

2 Pounds = 2 × ₹104.06 = ₹208.12

1 Rand = ₹4.43

5 Rands = 5 × ₹4.43 = ₹22.15

1 US dollar = ₹82.67

6 US dollars = 6 × ₹82.67 = ₹496.02

1 Euro = ₹89.19

5 Euro = 5 × ₹89.19 = ₹445.95

1 Yuan = ₹11.51

15 Yuan = 15 × ₹11.51 = ₹172.65

The total money that Anjali has (in Indian Rupees) = ₹208.12 + ₹22.15 + ₹496.02 + ₹445.95 + ₹172.65 = ₹1344.89

8E 1. a. 0.8 ÷ 4 = 0.2

b. 1.6 ÷ 8 = 0.2

2. a. 15.6 ÷ 10 = 1.56 b. 154.64 ÷ 10 = 15.464

c. 51.23 ÷ 100 = 0.5123  d. 513.47 ÷ 100 = 5.1347

e. 1.2 ÷ 100 = 0.012   f. 32.561 ÷ 1000 = 0.032561

g. 23.1 ÷ 1000 = 0.0231 h. 2 ÷ 10 = 0.2

i. 11.236 ÷ 1000 = 0.011236

3.

5. a. 13.25 ÷ 5 = 2.65  b. 33.04 ÷ 4 > 8.24

c. 202.08 ÷ 8 < 25.28  d. 150.1 ÷ 5 = 30.02

e. 332.16 ÷ 8 > 41.22  f. 378.18 ÷ 9 < 42.04

6. Resulting number = 1.54265

So, 154.265 should be divided by 100 to get 1.54265

154.265 ÷ 100 = 1.54265

Word Problems 1. The weight of one carton = 37.8 kg ÷ 9 cartons 4 . 2

9 3 7 . 8 3 6 1 0 1 8 1 8 0 0

The weight of one carton = 4.2 kg Therefore, the weight of one carton is 4.2 kg.

2. The total weight of the rice = 70.728 kg

The total number of baskets = 12

The weight of the rice in each basket = 70.728 kg ÷ 12 baskets

The weight of the rice in each basket = 5.894 kg

Therefore, the weight of the rice in each basket is 5.894 kg.

3. The amount of money that Hari needs to take to England (in pounds) = ₹1,04,000 ÷ ₹104

The amount of money that Hari needs to take to England = 1000 pounds

Therefore, Hari needs to take 1000 pounds with him to England.

4. The cost of a litre of oil = $37.92 ÷ 6 litres

The cost of a litre of oil = $6.32

Therefore, the cost of a litre of oil is $6.32.

Chapter Checkup

1. a. 0.23 + 0.47 = 0.7 b. 1.4 0.62 = 0.78

4.

a. The decimal point has been shifted one place to the right, hence we have to multiply the given number by 10.

45.12 × 10 = 451.2

b. The decimal point has been shifted three places to the right, hence we have to multiply the given number by 1000.

184.351 × 1000 = 184351

c. The decimal point has been shifted two places to the right, hence we have to multiply the given number by 100.

2.517 × 100 = 251.7

d. The decimal point has been shifted one place to the right, hence we have to multiply the given number by 10.

2.002 × 10 = 20.02

e. The decimal point has been shifted three places to the right, hence we have to multiply the given number by 1000.

81.36 × 1000 = 81360

f. The decimal point has been shifted three places to the right, hence we have to multiply the given number by 1000.

1.001 × 1000 = 1001

2.

c. 9 × 32.14 = 289.26

e. 41.5 × 12.45 = 516.675

6. a. 47.01 ÷ 10 = 4.701 b. 531.14 ÷ 100 =

c. 8143.12 ÷ 1000 = 8.14312  d. 1.21 ÷ 1000 = 0.00121

e. 14.2 ÷ 1000 = 0.0142 f. 123.321 ÷ 100 = 1.23321

7. a. 12.145 + 18.415 + 2.51 = 33.07 T O t h th 1 2 1 4 5

b. 41.54 + 56.81 12.379 = 85.971

H T O . t h th

c. 15.47 + 81.415 41.555 = 55.33

H T O . t h th

d. 31.23 + 17.28 11.111 = 37.399

H T O . t h th 3 1 2 1

9. a. 1.51 + 23.72          24.85

b. 27.63 2.78           22.8852

c. 4 × 6.17             24.68

d. 3.26 × 7.02           25.23

10. a. 1 1 . 3 6

11. a. 1 Won = ₹0.062

6000 Won = ₹0.062 × 6000 = ₹372

b. 1 Euro = ₹89.19

€150 = ₹89.19 × €150 = ₹13,378.5

c. 1 Rand = ₹4.43

850 Rand = ₹4.43 × 850 Rand = ₹3,765.5

d. 1 Dirham = ₹22.51

100.5 Dirham = ₹22.51 × 100.5 = ₹2,262.255

e. 1 NPR = ₹0.62

50,000 NPR = ₹0.62 × 50,000 NPR = ₹31,000

Word Problems 1. Change = ₹50 ₹45.8 = ₹4.2

H T O . t h th

4

Therefore, Jason will get back ₹4.2 in change from the shopkeeper.

2. Hours of sleep per day = 84.7 hours/7 days

Hours of sleep per day = 12.1 hours

Therefore, gorillas sleep for an average of 12.1 hours a day.

3. Total distance to be covered = 8.75 km

Distance covered = 4.79 km

Distance remaining = 8.75 km − 4.79 km

3

Distance remaining = 3.96 km

Therefore, Tim has to travel 3.96 km more to reach his destination.

4. Total cost = $47.85 + $21.36 + $22.01

Total cost = $91.22

Money required = $91.22 − $90

Money required = $1.22

Therefore, Raj requires an additional $1.22 to buy all of the items he wants to buy.

5. For every $20, George’s friend gives $1 tax.

So, total money that George’s friend can get for exchanging into Indian currency = $168 × $20 $20 +$1 = $168 × $20 $21 = $160

Then we multiply the amount of remaining currency by the exchange rate to find the amount of Indian currency George's friend has.

Indian currency = $160 × ₹89.12

Indian currency = ₹14,259.2

Therefore, George's friend has ₹14,259.2 in Indian currency.

6. Oil consumed in one hour = 6.5 L

Oil consumed in 4 hours = 6.5 L × 4 = 26 L

Total oil consumption = 133.9 L 26

Number of days = 5.15

Therefore, the car runs for a total of 5.15 days.

7. Harry’s weight = 51.25 kg × 0.56 = 28.7 kg

Chapter 9

Let's Warm-up

1.

2.

A slanting line

3.

A standing line

d. Two parallel lines are denoted by the symbol ll. True

3.

4. A sleeping line

A bridge Sides of a railway track

Tip of a nail Sunlight

Line segment Parallel lines Point Ray

4. a. M b. M N

A sleeping line Maths Grade 5 Book_Chapter 7-12.indb 147 15-12-2023

5. A sleeping line

a. Line

b. Intersecting lines

c. Parallel lines

d. Ray

2. a. A line segment AB can be denoted as .AB False

b. A ray CD can be denoted as .CD False

c. A line PQ can be denoted as PQ  True

5. a. Points: A, B, C, D, E, F

Line: ,, ACCBAB

W

c. P Q d. S T e. U X O V

b. Answers may vary. Sample answer: Line segment: FD

c. 1 ray; .BE

Word Problem 1. The pair of swords is an example of intersecting lines.

9B 1. Answers may vary. Sample answer: B C

AB

A BC

2. a. ∠ PQR, ∠ RQP or ∠ Q  b. ∠ LMN, ∠ NML or ∠ M

c. ∠ XYZ, ∠ ZYX or ∠ Y

3. ∠ 1 and ∠ 3 are obtuse angles.

4.

Acute angle Obtuse angle

Straight angle

Obtuse angle

5. 8 right angles: ∠ BGD, ∠ BGA, ∠ FGA, ∠ FGD, ∠ CHG, ∠ CHD, ∠ EHG, ∠ EHD

Word Problem 1. 7:30 a.m.—acute angle

1:30 p.m.—Obtuse angle

The time at which Rishi’s school starts = 7:30 a.m.

The time at which Rishi’s school gets over = 1:30 p.m.

Acute angle Obtuse angle

The angle formed at the time when Rishi’s school starts—7:30 a.m., is an acute angle.

The angle formed at the time when Rishi’s school ends— 1:30 a.m., is an obtuse angle.

9C 1. Answers may vary. Sample answers:

a. ∠XYZ, ∠ZYX  b. ∠POR, ∠ROP  c. ∠AOB, ∠BOA d. ∠LMN, ∠NML

2. a. 74°  b. 115°  c. 152°  d. 120°

3. a. 53°  b. 90°  c. 123°  d. 32°  e. 57° f. 127°

4.

5. 95°

R P Q

6 cm

Word Problem 1. When Priya walks straight from her house, she is moving in a straight line. Then, when she takes a right turn, she makes a 90-degree angle at the turning point.

9D 1. The sides of the ∆PQR are PQ or QP, QR or RQ and PR or RP.

2. a. A quadrilateral or a parallelogram

b. A nonagon

c. A hexagon

d. A decagon

3.

4. A quadrilateral is a 4-sided shape. Therefore, option c is not a quadrilateral. It is a pentagon. It has 5 sides.

5. a. A polygon is a closed figure that has two or more straight sides. False    b. An octagon has eight angles. True

c. A pentagon has five sides. True d. A rhombus has five sides. False

6.



  

Word Problem 1. A triangle has three sides and three angles. Since the shape we need a shape must have twice the sides and angles of a triangle, the shape must have six sides and six angles. Hence, the correct answer is a hexagon as it has six sides and six angles.

Chapter Checkup

1. a. Since the diagram has arrows on both ends, it represents a line.

b. Since the diagram has an arrow on one end, it represents a ray.

c. Since the diagram has no arrows, it represents a line segment.

d. Since the diagram has no arrows, it represents a line segment.

2. a. AC is a line segment. True

b. PQ and RS are sets of parallel lines. False

c. CS is a line. False

3.

An acute angle

A right angle

A straight angle

8. a. Quadrilateral; 4 sides, 4 angles

b. Heptagon; 7 sides, 7 angles

c. Quadrilateral; 4 sides, 4 angles

d. Octagon; 8 sides, 8 angles

9. a. b. c. d.

O

  10. N M 125° 10 cm

Word Problems 1. The zebra crossings are made up of parallel lines.

An obtuse angle

4. a. ∠AOC = acute angle  b. ∠AOD = right

c. ∠AOF = straight    d. ∠BOF = obtuse

e. ∠BOC = acute

5. a. 150°  b. 130°  c. 170°

6. a. 120°  b. 155°  c. 129°  d. 90°

7.

a.  65°

b.  115° c.  75°

d.  140°

2. A decagon has 10 sides and 10 angles. We need a shape with the half the number of sides and angles i.e., 5 sides and 5 angles. Hence, Shweta made a pentagon.

10

e.  155° f.  180°

2.

d.

It is not a pattern.

3.

The pattern is rotating 90° in clockwise direction.

4. a. The pattern is rotating 45° Clockwise.

b. The pattern is rotating 90° Clockwise.

c. The given pattern is not following any rule as the first and second figure has left upper one‑fourth portion shaded as yellow but the rest colors differ in the two figures.

d. The pattern is rotating 180° Anti‑Clockwise. [Or] The pattern is rotating 180° Clockwise.

5. a.

b.

Word Problem 1. It is given that the pattern made by Rakesh is rotating at half turns, the given pattern satisfies the condition. Hence, it is the pattern made by Rakesh.

d.

10B 1.

The number of sides of each figure in each pentagon is in the pattern: 6, 5, 4, 3. Hence the fourth pattern will have triangle.

2. a. The given pattern is increasing by one at each step, hence, the next in pattern would be:

b. The given pattern is increasing by one at each step, hence, the next in pattern would be:

c. The given pattern is increasing by two units at each step, hence, the next in pattern would be:

d. The given pattern is decreasing by three at each step, hence, the next in pattern would be:

3. Option b. is not a tiling pattern because there are gaps and overlaps.

4.

5. a. The given tile doesn’t belong to the tiling pattern because in the given tiling pattern all the stars are of red, yellow and green colour.

b.

Word Problem 1. Option b, because it shows the same tiling pattern as bought by Ankit.

10C

1. a. 15, 19, 23, 27, 31, 35

+ 4 + 4 + 4 + 4 + 4

Adding 4 to each term.

b. 6, 8, 10, 12, 14, 16, 18

+ 2 + 2 + 2 + 2 + 2 + 2

Adding 2 to each term.

c. 1, 4, 9, 16, 25, 36, 49

12 22 32 42 52 62 72

The pattern has squares of natural numbers. 12, 22, 32, 42, 52, 62, 72

d. 1, 6, 11, 16, 21, 26

+5 +5 +5 +5 +5

Adding 5 to each term

e. 1, 1, 2, 3, 5, 8, 13

+ + + + + +

Each term is the sum of the two preceding terms.

2. a. H U

b.

3. Flowers in first week =

Flowers in second week =

The number pattern is:

1, 3, 7, 15, ?

+2 +4 +8 +16

×2 ×2 ×2

Number of tablets added on Friday = 15 + 16 = 31

Therefore, 31 tablets were added on Friday.

Flowers in third week =

Flowers in fourth week =

Flowers in fifth week =

4. 60

60 − 1 = 59

59 – 2 = 57

57 – 3 = 54

54 – 4 = 50

50 – 5 = 45

45 – 6 = 39

39 – 7 = 32

32 – 8 = 24

24 – 9 = 15

10D 1. a. The lines of symmetry for the figure can be given as:

Hence, the shape has 2 lines of symmetry

b. The lines of symmetry for the figure can’t be given.

Hence, the given figure does not have any line of symmetry. It is asymmetrical

c. The lines of symmetry for the figure can be given as:

1

12

Hence, the shape has 1 horizontal line of symmetry

d. The lines of symmetry for the figure can be given as:

5. Answer may vary. Sample answer: 6

2 3

5 4

Word Problems 1. Orders in January = 12

Orders in February = 24

Orders in March = 36

Therefore, there were 5 red and 5 yellow flowers after 5 weeks.

Hence, the shape has many lines of symmetry.

2. a.     b

Therefore, the numbers upto the 10th term are: 60, 59, 57, 54, 50, 45, 39, 32, 24, 15.

The number pattern is: 12, 24, 36 … = 12 × 1, 12 × 2, 12 × 3 …

Therefore, orders in July = 12 × 7= 84.

2. Number of tablets added on Monday = 1

Number of tablets added on Tuesday = 3

Number of tablets added on Wednesday = 7

Number of tablets added on Thursday = 15

No line of symmetry. No line of symmetry.

c. d.

3. a. A   b.  H    c. T    d. G Asymmetric

4. a. When two halves look identical, they are called asymmetrical. False

b. The letters F, G, J and L have two lines of symmetry. False

c. The number 8 has 2 lines of symmetry. True

d. The number 3 has horizontal line of symmetry. True 5.

Word Problem 1. If the 2 flats are identical then they are the mirror image of each other.

3. a. It is a reducing pattern with one line decreasing each time. The next two figures can be given as:

b. It is a rotating pattern rotating 90° anti-clockwise. The next two figures can be given as:

c. It is an increasing pattern with a box increasing each time. The next two figures can be given as:

Chapter Checkup 1. a.

b.

c.

2. d. The figure is symmetrical and the line of symmetry can be given as:

f. The figure is symmetrical and the line of symmetry can be given as:

g. The figure is symmetrical and the line of symmetry can be given as:

h. The figure is symmetrical and the line of symmetry can be given as:

4. a. First term = 5

Second term = 5 + 11 = 16

Third term = 16 + 11 = 27

Fourth term = 27 + 11 = 38

Fifth term = 38 + 11 = 49

Pattern: 5, 16, 27, 38, 49

b. First term = 10

Second term = 10 × 2 + 1 = 21

Third term = 21 × 2 + 1 = 43

Fourth term = 43 × 2 + 1 = 87

Fifth term = 87 × 2 + 1 = 175

Pattern: 10, 21, 43, 87, 175

c. First term = 1

Second term = 1 + 12 = 2

Third term = 2 + 22 = 6

Fourth term = 6 + 62 = 42

Fifth term = 42 + 422 = 1806

Pattern: 1, 2, 6, 42, 1806

d. First term = 15

Second term = 15 + 3 = 18

Third term = 18 + 5 = 23

Fourth term = 23 + 3 = 26

Fifth term = 26 + 5 = 31

Pattern: 15, 18, 23, 26, 31

5. 331, 316, 301, 286, 271, ____, ____, ____

− 15 − 15 − 15 − 15 − 15 − 15 − 15

So, the next three terms are:

271 − 15 = 256

256 − 15 = 241

241 − 15 = 226

6. a. The given number pattern is increasing by 10 each time. Hence, the missing numbers will be:

35 + 10 = 45; 45 + 10 = 55; 55 + 10 = 65; 65 + 10 = 75; 75 + 10 = 85; 85 + 10 = 95

The pattern will be:

5, 15, 25, 35, 45, 55, 66, 75, 85, 95

b. The given number pattern is increasing by 3 times, each time. Hence, the missing numbers will be:

243 × 3 = 729; 729 × 3 = 2187; 2157 × 3 = 6561; 6561 × 3 = 19683; 19683 × 3 = 59049

The pattern will be:

3, 9, 27, 81, 243, 729, 2187, 6561, 19,683, 59,049

c. The given number pattern is increasing by 2, 3, 4, 5 … each time. Hence the missing numbers will be:

51 + 6 = 57; 57 + 7 = 64; 64 + 8 = 72; 72 + 9 = 81; 81 + 10 = 91

The pattern will be:

37, 39, 42, 46, 51, 57, 64, 72, 81, 91

d. The given number pattern is decreasing by 1, 2, 3, 4,… each time. Hence the missing numbers will be:

80 − 5 = 75; 75 − 6 = 69; 69 − 7 = 62; 62 − 8 = 54; 54 − 9 = 45

The pattern will be:

90, 89, 87, 84, 80, 75, 69, 62, 54, 45

7.

8.

Petals in fourth flower = 14 = 11 + 3

Therefore, the number of petals next three flowers will have

Petals in fifth flower = 14 + 3 = 17

Petals in sixth flower = 17 + 3 = 20

Petals in seventh flower = 20 + 3 = 23

3. Patterns in cells: 2, 4, 8, 16, 32, 64, …

The next pattern has twice cells as the previous pattern.

Next three terms:

64 × 2 = 128

128 × 2 = 256

256 × 2 = 512

4. Steps taken on first day = 100

Steps taken on second day = 120 = 100 + 20

Steps taken on third day = 150 = 120 + 30

Steps taken on fourth day = 190 = 150 + 40

Steps taken on fifth day = 190 + 50 = 240

Steps taken on sixth day = 240 + 60 = 300

So, Emily will take 300 steps on the sixth day.

5. Pocket money saved: 20, 40, 80, 160, ...

January = ₹20

February = ₹40 = ₹20 × 2

March = ₹80 = ₹40 × 2

April = ₹160 = ₹80 × 2

The pocket money saved in each month is double of the previous month. Therefore, the pocket money saved in:

May = ₹160 × 2 = ₹320

June = ₹320 × 2 = ₹640

July = ₹640 × 2 = ₹1280

August = ₹1280 × 2 = ₹2560.

Therefore, Mihir saved ₹2560 in August.

Chapter 11

Let's Warm-up 1. The length of your mathematics textbook is 20 cm. 2. The length of your table in the classroom is 1 m.

3. If a child weighs 5000 g, then his weight in kg will be 5 kg

4. If the weight of a 10-rupee coin is about 8 g, then the weight of this coin in mg will be 8000 mg. 5. If a newborn baby’s weight is 2 kg, then 2000 is its weight in g.

Do It Yourself

11A 1. Answers may vary Sample answers:

9. L E M O N

2 1 4 16 8

10. Similarity: Terms are being multiplied by the 2 in both the patterns.

2, 4, 8, 16

×2 ×2 ×2

3, 6, 12, 24

×2 ×2 ×2

Difference: All the numbers in the first pattern are multiples of 2 whereas in the second pattern they are all multiples of 3.

Word Problems 1. We will get the smallest number if we arrange the digits in an ascending order.

So, 125789 is the smallest number.

2. Petals in first flower = 5

Petals in second flower = 8 = 5 + 3

Petals in third flower = 11 = 8 + 3

c.

7.698 hm = 1 10 × 7.698 = 7.698 0.7698 10 = km

d. 1000 m = 1 km; 1 m = 1 1000 km

1056 m = 1 1056 10561.056 10001000 ×== km

e. 1000 mm = 1 m; 1 mm = 1 1000 m

547 mm = 1 547 5470.547 10001000 ×== m

f. 1 hm = 100 m

2.034 hm = 2.034 ⨯ 100 = 203.4 m

4. a. 45 m 10 cm = 45 m + 10 100 m = 45 m + 0.1 m = 45.1 m

b. 16 m 80 mm = 16 × 1000 mm + 80 mm = 16000 mm + 80 mm = 16080 mm  c. 280.5 m = 280 m 5 dm

5. The length of the pencil = 13 cm − 3 cm = 10 cm

1,00,000 cm = 1 km; 1 cm = 1 100000 km

10 cm = 10 100000 km = 0.0001 km

Word Problems 1. The distance travelled on the bicycle = 2 km 578 m = 2000 m + 578 m = 2578 m

The distance travelled on the bus = 21 km 870 m = 21,000 m + 870 m = 21870 m

The distance travelled on foot = 1 km 346 m = 1000 m + 346 m = 1346 m

The total distance travelled = 2578 m + 21870 m + 1346 m = 25794 m = 25000 + 794 m = 25 km 794 m

2. The initial length of the jeans = 75 cm 6 mm

The reduced length of the jeans = 72 cm 3 mm

The length of the jeans torn = 75 cm 6 mm − 72 cm 3 mm = 3 cm 3 mm cm mm

6

3. The cloth required for 1 skirt = 3 m 586 cm = 3 m + 500 cm + 86 cm = 3 m + 5 m + 86 cm = 8 m + 86 cm = 8.86 m

The cloth required for making 6 skirts = 6 ⨯ 8.86 m = 53.16 m = 53 m 16 cm

4. The length of the door = 2 m 1 dm = 2 ⨯ 10 dm + 1 dm = 20 + 1 = 21 dm

The length of the wall = 3 m 2 dm = 3 ⨯ 10 dm + 2 dm = 30 + 2 = 32 dm

The total length = 21 dm + 32 dm = 53 dm 11B 1.

a. Frog

c. Dresser

5000 kg 5000 g 30 g

350 g 70 kg 500 kg

b. Baseball Cap

d. Iron

2. a. 1000 g = 1 kg; 1g = 1 1000 kg

79 g = 1 1000 × 79 = 79 1000 = 0.079 kg

b. 1 kg = 1000 g

4677 kg = 4677 ⨯ 1000 = 46,77,000 g

c. 100 cg = 1 g; 1 cg = 1 100 g

6876 cg = 1 100 × 6876 6876 100 = = 68.76 g

d. 10 dg = 1 g; 1 1 dg 10 = g

1655 dg = 1 10 × 1655 1655 10 = = 165.5 g

e. 1 g = 1000 mg

975 g = 975 ⨯ 1000 = 9,75,000 mg

f. 1 g = 1000 mg; 1 dag = 10,000 mg

390 g 45 dag = 390 ⨯ 1000 + 45 ⨯ 10,000 = 3,90,000 + 4,50,000 = 8,40,000 mg

3. a. The bigger unit: 6 kg + 10 1000 kg = 6 kg + 0.01 kg = 6.01 kg;

The smaller unit: 6 × 1000 g + 10 g = 6000g + 10 g = 6010 g

b. The bigger unit: 16 g + 80 1000 g = 16 g + 0.08 g = 16.08 g;

The smaller unit: 16 × 1000 mg + 80 mg = 16,000 mg + 80 mg =16,080 mg

c. The bigger unit: 547.6 1000 kg = 0.5476 kg;

The smaller unit: 547.6 × 1000 = 5,47,600 mg

d. 3 g 8 cg

The bigger unit: 3 g + 8 100 g = 3 g + 0.08 g = 3.08 g;

The smaller unit: 3 × 100 cg + 8 cg = 300 cg + 8 cg = 308 cg

e. 87 kg 6 dag

The bigger unit: 87 kg + 6 100 kg = 87 kg + 0.06 kg = 87.06 kg;

The smaller unit: 87 × 100 dag + 6 dag = 8700 dag + 6 dag = 8706 dag

f. 12 hg 42 g

The bigger unit: 12 hg + 42 100 hg = 12 hg + 0.42 hg = 12.42 hg;

The smaller unit: 12 × 100 g + 42 g = 1200 g + 42 g = 1242 g

4. The weight of a four-month-old kitten = 1688 g = 1688 ÷ 1000 kg = 1.688 kg

5. The weight of an egg = 40 g

The number of eggs in 1 kg = 1000 g 1000 25 40 == eggs.

Word Problems 1. Weight of apples = 2 kg 450 g

Weight of guavas = 1 kg 547 g

Weight of pears = 2 kg 136 g

Total weight of the fruits = 2 kg 450 g + 1 kg 547 g + 2 kg 136 g = 5 kg 1133 g = 5 kg + 1000 g + 133 g = 6 kg 133 g

2. The weight of two chairs = 16 kg 400 g

The weight of one chair = 10 kg 300 g

85 g 5 kg 200 kg

100 g 15 kg 2 kg

The weight of the other chair = 16 kg 400 g – 10 kg 300 g = (16 – 10 kg) + (400 – 300 g) = 6 kg 100 g

3. The weight of an egg = 65 g

1 dozen = 12 eggs

3 dozen = 12 × 3 = 36 eggs

The weight of 3 dozen eggs = 36 × 65 = 2340 g = 2.34 kg

4. The weight of 8 cartons = 32 kg 448 g = 32,448 g

The weight of 1 carton = == 32448 4056 g 8 4 kg 56 g

5. The weight of one soap bar = 258 g

The weight of 4 soap bars = 4 × 258 g = 1032 g

The weight of one detergent cake = 352 g

The weight of 3 detergent cakes = 3 × 352 g = 1056 g

a. The total weight carried = 1032 g + 1056 g = 2088 g = 2 kg 88 g

b. The weight of the remaining soap bars = 3 × 258 g = 774 g

The weight of the remaining detergent cakes = 2 × 352 g = 704 g

The total weight = 774 g +704 g = 1478 g = 1 kg 478 g

6. The amount of baking soda in one kg cake = 3 dg

The amount of baking soda in 17 one kg cakes = 17 × 3 dg = 51 dg = 0.0051 kg

Chapter Checkup

1. a. m; kg  b. cm; g  c. m; kg

2. a. 12.5 cm  b. 8.5 cm  c. 12 cm  d. 13.5 cm

e. 17.9 cm  f. 22.2 cm

3. James’ cotton candy = 11.2 cm + 3 cm = 14.2 cm

4. The length of the red straw = 12.5 cm

The length of the blue straw = 8 cm

The difference between their lengths = 12.5 – 8 = 4.5 cm

5. a. 1000 m = 1 km; 1 m = 1 1000 km

8 m = 1 1000 × 8 = 8 1000 = 0.008 km

b. 100 m = 1 hm; 1 m = 1 100 hm

4 hm 35 m = 4 + 35 × 1 100 = 4.35 hm

c. 1 m = 1000 mm

1232 m = 1232 × 1000 = 12,32,000 mm

d. 10 m = 1 dam; 1 m = 1 10 dam

897 m = 897 × 1 10 = 89.7 dam

6. a. 1 kg = 1000 g

5 kg = 5 × 1000 = 5000 g

b. 100 cg = 1 g; 1 cg = 1 100 g

4 g 64 cg = 4 + 64 × 1 100 = 4.64 g

c. 1 g = 1000 mg

5487 g = 5487 × 1000 = 54,78,000 mg

d. 100 dag = 1 kg; 1 dag = 1 100 kg

43 kg 7 dag = 43 + 7 × 1 100 = 43.07 kg

7. The weight of a young joey = 0.38 kg = 0.38 × 1000 g = 380 g

8. The weight of a candle = 125 g = 125 × 1000 mg = 1,25,000 mg

9. The smallest 2-digit number = 10

The smallest 3-digit number = 100

The grandmother’s weight = (100 − 28) + 10 = 82 kg

So, Meenakshi’s weight = 82 41 2 = kg.

10. Answers may vary. Sample answer:

a. One 4 cm and one 8 cm

b. One 4 cm and two 8 cm

c. Three 8 cm

Word Problems 1. The length of each stamp = 3 cm

a. The total length of the stamps = 12 × 3 cm = 36 cm

b. No, because the total length of 12 stamps is greater than the length of the stamp album.

2. The weight that the lift is allowed to carry = 260 kg

The total weight of all the people = 85 kg + 70 kg + 58 kg + 80 kg = 293 kg

No, they cannot use the lift together.

Yes, we must follow the rules of the lift to avoid overloading, and ensure the safety of all passengers.

3. To bake one cake, the following items are required:

200 g of flour, 3 eggs, 75 g of butter, 100 g of sugar, some milk

So, to bake 100 cakes the following quantities of ingredients will be needed:

The weight of the flour = 200 g × 100 = 20,000 g = 20000 1000 kg = 20 kg

The weight of the butter = 75 g × 100 = 7500 g = 7500 1000 kg = 7.5 kg

The weight of the sugar = 100 g × 100 = 10,000 g = 10000 1000 kg = 10 kg

Chapter 12

Do It Yourself

12A

1. Perimeter = 22lb +

a. Perimeter = ×+×=+= 212215243054 cm

b. Perimeter = ×+×=+= 255285110170280 m

c. Perimeter = ×+×=+= 218220364076 m

2. Perimeter = 4s

a. s = 32 m

Perimeter = ×= 432128 m

b. s = 53 cm

Perimeter = ×= 453212 cm

c. s = 27 m

Perimeter = ×= 427108 m

d. s = 78 cm

Perimeter = 478312cm ×=

3. a. Perimeter = 420cm80cm ×=

b. Perimeter = ×+×=+= 2100250200100 300 cm

c. Perimeter = ×+×=+= 220210402060 cm

d. Perimeter = ×+×=+= 2302206040100 cm

4. a. 2

= 52 + 76 = 128 m

b. 4 × 21 = 84 m

4 × 42 = 168 m

4 × 55 = 220 m

4 × 74 = 296 m

4 × 96 = 384 m

5. a. Perimeter = 112 mm = 232 mm2 b ×+×

112642b =+

112642b −=

48 2 b =

b

= 24 mm

Therefore, breadth = 24 mm.

b. Perimeter = =× 148 mm4 s

148 4 s = 37 mm s =

Therefore, each side of the square = 37 mm.

Word Problems 1. The length of each side of square shaped garden = 45 m

The perimeter of the garden 445m180 m. =×=

He would require 180 m of wire to fence the entire garden.

2. The length of the square formed = 10 cm

The perimeter of the square 410cm40 cm =×=

The perimeter of the square = The perimeter of the rectangle

Length = 12 cm, Perimeter = 40 cm

Breadth = ?

Perimeter = 22lb +

402122 b =×+×

40 = 24 + 2b

40 – 24 = 2b

16 = 2b

b = 8 cm.

Thus, the breadth of the rectangle will be 8 cm.

12B

1. a. board → sq. m b. your Book → sq. cm

c. classroom → sq. m d. your school → sq. m

e. your city → sq. km f. lake → sq. m

g. table → sq. m h. a farmer’s field → sq. m

2. Area of rectangle = Length × Breadth

a. Area = 15 cm × 22 cm = 330 sq. cm

b. Area = 47 m × 36 m = 1692 sq. m

c. Area = 58 m × 70 m = 4060 sq. m

d. Area = 88 cm × 62 cm = 5456 sq. cm

3. Area of square = Side × Side

a. Area = 56 m × 56 m = 3136 sq. m

b. Area = 67 cm × 67 cm = 4489 sq. cm

c. Area = 24 m × 24 m = 576 sq. m

d. Area = 40 cm × 40 cm = 1600 sq. cm

4. a. 1 Area of triangle 1 = Half the area of rectangle 1

Area of triangle = 15 2 sq. units = 7.5 sq. units

b. 1 2

Area of triangle 1 = Half the area of rectangle 1 = 25 2 sq. units

Area of triangle 2 = Half the area of rectangle 2 = 10 2 sq. units

Area of whole triangle +== 251035 17.5 sq.units. 222

c. 1

5. a.

Area of triangle 1 = Half the area of rectangle 1 12 2 → sq. units

6 → sq. units.

Area of rectangle A = 6 cm × 8 cm = 48 sq. cm

Area of rectangle B = 8 cm × 3 cm = 24 sq. cm

Area of rectangle C = 12 cm × 2 cm = 24 sq. cm

Area of the whole shape = 48 sq. cm + 24 sq. cm + 24 sq. cm = 96 sq. cm

b. Area of rectangle A = 5 cm × 5 cm = 25 sq. cm

Area of rectangle B = 4 cm × 6 cm = 24 sq. cm

Area of the whole shape = 25 sq. cm + 24 sq. cm = 49 sq. cm

c. Area of rectangle A = 3 cm × 3 cm = 9 sq. cm

Area of rectangle B = 8 cm × 3 cm = 24 sq. cm

Area of the whole shape = 9 sq. cm + 24 sq. cm = 33 sq. cm

6. Answers may vary.

a. Perimeter = 36 cm

Area = 80 sq. cm

b. Area = 24 sq. cm

Perimeter = 22 cm

Word Problems 1. The measurement of the room = 12 m × 12 m

The carpet required to cover the floor = The area of Advita’s Room 12 m × 12 m = 144 sq. m

2. The perimeter of the rectangular field = 80 m

The breadth of the field = 20 m

Perimeter = 22lb +

802220 l =×+×

20 m l == 20 m

Area = ×=×=2020400 sq.m lb

Thus, area of the field is 400 sq. m.

Chapter Checkup 1. Perimeter = 22lb + ; Area = lb ×

a. l = 14 cm, b = 18 cm

Perimeter 2142182836 64 cm =×+×=+=

Area = 14 × 18 = 252 sq. cm

b. l = 27 m, b = 21 m

Perimeter 227221544296 m =×+×=+=

Area =×=2721567 sq.m

c. l = 49 m, b = 33 m

Perimeter =×+×=+= 2492339866 164 m

Area 49331617 sq. m =×=

d. l = 70 cm, b = 57 cm

Perimeter 270257140114254 cm =×+×=+=

Area = 70 × 57 = 3990 sq. cm

2. Perimeter = 4 × Side; Area = Side × Side

a. s = 23 cm

Perimeter 42392 cm =×= cm

Area 2323529 sq.cm =×= sq. cm

b. s = 32 m

Perimeter 432128 m =×=

Area 32321024sq.m=×= sq. m

c. s = 45 m

Perimeter 445180 m =×= m

Area 45452025 sq.m =×= sq. m

d. s = 80 cm

Perimeter =×=480320 cm

Area = 80 × 80 = 6400 sq. cm

3. a. Breadth = 24 cm; Area = 960 sq. cm

Length = Area Breadth = 960 40 cm 24 = cm

Perimeter = 2 × 40 + 2 × 24 = 80 + 48 = 128 cm

b. Perimeter = 420 m

Side = 420 105 m 4 = m

Area = 105 m × 105 m = 11025 sq. m

c. Length = 34 m; Perimeter = 124 m

2 × Breadth = 1242341246856 −×=−=

Breadth = 56 28 2 = m

Area = 34 × 28 = 952 sq. m

4. Area of rectangle A = 5 cm × 8 cm = 40 sq. cm

Perimeter of rectangle A 252826 cm =×+×= cm

Area of rectangle B =×= 4 cm 11 cm44 sq. cm cm

Perimeter of rectangle B 2421130 cm =×+×= cm

Area of rectangle C = 7 × 6 = 42 sq. cm

Perimeter of rectangle C 272626 cm =×+×= cm

Rectangle B has the largest area while rectangle A and rectangle C have the smallest perimeters.

5. a. 6 cm

1 cm 1 2

12 cm 3 cm 3 cm 1 cm

Area of rectangle 1 = 12 × 3 sq. cm = 36 sq. cm

Area of rectangle 2 = 3 × 1 sq. cm = 3 sq. cm

Area of the whole figure = 36 + 3 sq. cm = 39 sq. cm

11 m 2 1 3

b. 8 m 2 m 2 m 5 m 6 m 5 m

Area of rectangle 1 = 8 × 6 sq. cm = 48 sq. cm

Area of rectangle 2 = 5 × 4 sq. cm = 20 sq. cm

Area of rectangle 3 = 5 × 2 sq. cm = 10 sq. cm

4 m

Area of the whole figure = 48 + 20 + 10 sq. cm = 78 sq. cm

3 cm

cm 7 cm 5 cm 1 2 3 4 3 cm

cm 9 cm

4 cm 1 cm

c. 5 cm 5 cm 5 cm 6 cm

Area of rectangle 1 = 4 × 6 sq. cm = 24 sq. cm

Area of rectangle 2 = 9 × 1 sq. cm = 9 sq. cm

Area of rectangle 3 = 4 × 4 sq. cm = 16 sq. cm

Area of rectangle 4 = 3 × 5 sq. cm = 15 sq. cm

Area of the whole figure = 24 + 9 + 16 +15 sq. cm = 64 sq. cm

6. Area of triangle = Area of rectangle 2

a.

b.

c. 1 2

Area of rectangle = 24 sq. units

Area of triangle

= 24 12 sq. units 2 =

Area of rectangle = 12 sq. units

Area of triangle 12 6 sq. units 2 ==

Area of rectangle 1 = 2 sq. units

Area of triangle 1 = 2 2 sq. units = 1 sq. units

Area of rectangle 2 = 6 sq. units

Area of triangle 2 = 6 2 sq. units = 3 sq. units

Area of whole triangle = 1 sq. units + 3 sq. units = 4 sq. units

d. 2 1

Area of rectangle 1 = 15 sq. units

Area of triangle 1 = 15 sq. units 2

Area of rectangle 2 = 5 sq. units

Area of triangle 2 = 5 sq. units 2

Area of the whole triangle = 15520 10 sq. units 222 +==

7.

10 cm

10 cm

10 cm

10 cm

10 cm

10 cm

Perimeter of the new shape formed = 10 cm + 10 cm + 10 cm + 10 cm + 10 cm = 50 cm

Thus, option d is correct option.

8. Area of the figure = 80 sq. cm

Area of one square 80 .16 . 5 sqcmsqcm== sq. cm = 16 sq. cm

So, Side = 4 cm

Perimeter of the figure = 12 × 4 cm = 48 cm.

So, option c is the correct option.

9. Perimeter of a rectangle = 2length2breadth22 cm ×+×=

Perimeter of the shape formed: = ( ()2length2breadth)32 length2 breadth ×+××−×+× (2 × length + 2 × breadth)

= (22) × 3 – (22)

= 66 – 22

= 44 cm

10. New length 2 Length =×

New Width 2 Width =×

New Area New length New Width ∴=×

= (2 × Length) × (2 × Width)

= 4 × (Length × Width)

= 4 × Area

No, the statement is not true.

Word Problems 1. The side of one square wooden panel 25 cm =

The area of one square wooden panel =×=2525625 sq.cm

The total area of 8 such panels 8625 sq.cm5000 sq.cm =×=

2. The area of the floor 12m15m180sq.m=×=

The area of the carpet =×= 13 m 13 m169 sq.m

The area of the floor that is not carpeted =−=18016911 sq.m

Chapter 13

Let's Warm-up

1. Between a jug and a glass, a jug has more capacity.

2. If 8 glasses of juice fill a jug, then 8 glass is the capacity of the jug.

3. Four tin cans of paint fill a bucket of paint. If the capacity of 1 tin can is 2 L, then 8 L is the capacity of the bucket.

4. 5 L = 5000 mL

5. 8 L = 8000 mL

Do It Yourself

13A 1. A teaspoon 250 L 50 mL 5 mL 250 mL Pan 250 mL 3 L 2 mL 800 L

2. a. 1000 mL = 1 L

1 mL = 1 1000 = L

658 mL = 1 1000 = × 658 L = 0.658 L

b. 1 L = 1000 mL

8437 L = 8437 × 1000 mL = 84,37,000 mL

c. 10 dL = 1 L

1 dL = 1 10 LL

2567 dL = 1 2567 257.6 10 LL×= × 2567 L = 257.6 L

d. 100 cL = 1 L

1 cL = 1 100 L

5054 cL = 5054 × 1 100 L = 50.54 L

e. 10 L = 1 daL

1 L = 1 10 LdaL

821 L = 821 × 1 10 daLdaL = 82.1 daL

f. 136 L 80 dL = 144 L

1 L = 1000 mL

144 L = 144 × 1000 = 1,44,000 mL

3. Complete the given table.

S. No. Full form In bigger units In smaller units

a. 8 L 60 mL 8.060 L 8060 mL

b. 32 L 89 cL 32.89 L 3289 cL

c. 4012 L 3 dL 4012.3 L 40,123 dL

d. 90 kL 6 daL 90.06 L 9006 daL

4. No, C has enough space to pour more liquid so it has more capacity than the other two pots. Now converting mL to dL

1 = 1 dL

1 mL = 1 100 dLdL

3050 × 1 100 dLdL = 30.50 dL

5. Yes. As there are 3 bowls of 600 mL and 8 bowls of 200 mL. So, in total we can easily fill upto 3400 mL capacity. Word Problems 1. Rate of dripping water is 150 mL in 2 hours.

So, water dripping in one hour = 150 mL 2 = 75 mL

Water dripping in 10 hours = 75 mL × 10 = 750 mL

We know that, 1 L = 1000 mL or 1 mL = 1 1000 L

So, 750 mL = 750 1000 L = 0.75 L

2. Number of cups = Total volume of tea (in mL) Volume of each cup (in mL)

Number of cups = 1050 150 mL

Number of cups = 7

So, we can fill 7 tea cups, each with a volume of 150 mL, from the 1050 mL bottle of tea. 13B 1.

2. Find the volume of each solid. Circle the solid with the greater volume.

a. b.

c. d.

a. From one face we can count there are 4 cubes.

So, layer 1 has 8-unit cubes.

Number of layers = 2

So, total number of unit cubes = 2 × 8 = 16

Volume of the given solid = 16 cu. units

b. This is a two layers solid. In 1st layer we can see 2-unit cubes on the right and left side.

And, 2-unit cubes are placed in middle. Two cubes are also there on 2nd layer.

So, total number of unit cubes = 8

So, the volume of given solid = 8 cu. units

c. The number of cubes in 1 layer = 4 × 5 = 20 cubes

So, the number cubes in four layers = 20 × 4 = 80 cubes

So, the volume of given solid = 80 cu. units

d. There are two layers with 4 × 3 cubes = 24 cubes

Bottom two layers are with 6 cubes in each, So total cubes = 12 cubes

So, the volume of given solid = 36 cu. units

3.

3 cm

2cm

4.

5 cm

Length = 10 cm, width = 15 cm and height = 6 cm

Volume of the container = l × w × h = 10 × 15 × 6

So, the Volume of the container = 900 cubes = 900 cu. cm

3 cm

2cm

5 cm

Length = 30 cm, width = 22 cm and height = 10 cm

Volume of the fish tank = l × w × h = 30 × 22 × 10

So, the Volume of the fish tank = 6600 cubes = 6600 cu. cm

5. To find whose lunch box is bigger in size, we must compare the volumes of the two lunch boxes.

1. Rinku’s lunch box volume:

Volume = Length × Width × Height

Volume = 16 cm × 8 cm × 3 cm = 384 cubic cm

2. Rita’s lunch box volume:

Volume = Length × Width × Height

Volume = 10 cm × 7 cm × 3 cm = 210 cubic cm

Comparing the two volumes, 384 cubic cm > 210 cubic cm

Thus, Rinku’s lunch box is bigger in size than Rita’s lunch box.

Word Problems 1. To find the number of cartons we need to find the volume of the Almirah

So, the volume of almirah = Volume = Length × Width × Height

Volume = 3 × 2 × 1 = 6 cubic m

2. 1 L = 1000 mL

So, 16 litres of juice is equal to:

16 L = 16 × 1000 mL = 16,000 mL

Now, we can find out how many bottles are needed by dividing the total volume of juice by the volume each bottle can hold:

Number of bottles = Total volume of juice (in mL)/Volume of each bottle (in mL)

Number of bottles = = 16,000 mL 4 4,000 mL

So, total 4 bottles are needed to hold 16 L juice.

3. The jug initially contained 1.5 litres of juice = 1500 millilitres (1 litre = 1000 millilitres).

Total juice poured into the glasses = 350 mL + 890 mL = 1240 mL

Juice left in the jug = Initial amount − Total amount poured

Juice left in the jug = 1500 mL − 1240 mL = 260 mL

So, there is 260 mL of juice left in the jug.

Chapter Checkup

1. Tick () the best estimate.

a. Glue in a small bottle

2 L 50 L 2000 L 50 mL

b. Paint in a bucket

50 L 50 mL 20 L 1 mL

c. Ketchup in a packet

500 mL 20 mL 2 L 50 L

d. Shampoo in a bottle 250 mL 2 L 250 L 5 L

e. Water in a pool

20 L 500 mL 6 L 5000 L

2. The given container holds 750 mL, and it seems to hold three-fourth of its capacity. So, the capacity of the container is about 1 L.

3.

5.

6. a.

b.

Length of solid = 4 unit

Breadth of solid = 4 unit

Height of solid = 4 unit

b. 1 L = 1000 mL 6869 L = 1000 × 6869 mL = 68,69,000 mL

c. 10 dL = 1 L

dL = 1 L 10

dL = 1 10 × 9796 L = 979.6 L

d. 100 cL = 1 L 1 cL = 1 L 100

3456 cL = 1 100 × 3456 L = 34.56 L

e. 10 L = 1 daL

1 L = 1 daL 10

243 L = 1 10 × 243 daL = 24.3 daL

f. 907 L 56 dL = 912.6 L

1 L = 1000 mL

912.6 L = 912.6 × 1000 = 9,12,600 mL

Length of solid = 8 unit

Breadth of solid = 2 unit

Height of solid = 3 unit

7. a. On counting, there are 12-unit cubes in the given solid, hence the volume of the solid is 12 cu. units.

b. On counting, there are 44-unit cubes in the given solid, hence the volume of the solid is 44 cu. units.

c. On counting, there are 15-unit cubes in the given solid, hence the volume of the solid is 15 cu. units.

d. On counting, there are 22-unit cubes in the given solid, hence the volume of the solid is 22 cu. units.

8. Volume of solid A = 3 × 3 × 3 = 27 cu. units

Volume of solid B = 3 × 3 × 3 = 27 cu. units

Volume of solid C = 3 × 3 × 3 = 27 cu. units

As the volume of all solids are 27 cu. units, the solids are of equal volume.

9. a. Number of cubes across length = 7, number of cubes across width = 4, number of cubes across height = 4

Volume = l × w × h = 7 × 4 × 4 = 112 cu. units

So, the volume of container = 112 cu. units

b. Number of cubes across length = 7, number of cubes across width = 3, number of cubes across height = 3

Volume = l × w × h = 7× 3 × 3 = 63 cu. units

So, the volume of the container = 63 cu. units

c. Number of cubes across length = 4, number of cubes across width = 3, number of cubes across height = 3

Volume = l × w × h = 4 × 3 × 3 = 36 cu. units

So, the volume of the container = 36 cu. units

10. Length = 50 cm, width = 30 cm and height = 25 cm

Volume of carton = l × w × h = 50 × 30 × 25 = 1500 × 25 = 37,500 cu. cm

Word Problems 1. The pack originally contains 1 litre of juice = 1000 mL

Juice left in the pack = Total amount − Amount drunk by Juhi

Juice left in the pack = 1000 mL − 235 mL = 765 mL

So, 765 mL of juice is left in the packet.

2. Capacity of first flask = 850 mL

Capacity of second flask = 1 L 250 mL = 1000 mL + 250 mL = 1250 mL

Total water the flasks can hold together = Capacity of the first flask + Capacity of the second flask = 850 mL + 1250 mL = 2100 mL

1000 mL = 1 L; 1 mL = 1 1000 L

2100 mL = 1 2100 1000 ×= × 2100 = 2.1 L or 2 L 100 mL

So, the two flasks can hold a total of 2 L 100 mL of water.

3. Capacity of Seema’s bucket = 6.5 L

Tank capacity = Seema’s bucket capacity × 4

Tank capacity = 6.5 L × 4

Tank capacity = 26 L

So, the water tank can hold 26 L of water.

4. Capacity of bottle = 5 cups of water

Capacity of bowl = 5 cups of water

Capacity of jug = 5 bottles of water = 5 × 5 cups of water = 25 cups of water

Capacity of pail = 3 jugs + 7 bowls of water = 3 × 25 cups + 7 × 5 cups = 75 cups + 35 cups = 110 cups

Capacity of a cup = 90 mL

So the capacity of pail = 110 × 90 mL = 9900 mL

1000 mL = 1 L; 1 mL = 1 1000 L

9900 mL = 1 9900 1000 ×= × 9900 = 9.9 L or 9 L 900 mL

So, the capacity of the pail is 9 L 900 mL.

Chapter 14

Let's Warm-up

1. cube  2. cylindrical  3. cuboid  4. sphere

Do It

14A

2. Cone: Faces = 2; Edges = 1; Vertices = 1

Sphere: Faces = 0; Edges = 0; Vertices = 0

Cuboid: Faces = 6; Edges = 12; Vertices = 8

3. Square prism: 2 square faces and 4 rectangular faces. Cylinder: 2 circular identical faces and one curved face.

4.

5. a. Answer may vary. Sample answer: Difference: Pyramid has 1 base and prism has 2 bases. Similarity: They are solid figures with three dimensions. Word Problem 1. Rohan’s tent is a prism, and Simi’s tent is a pyramid. Similarity: They both have 5 faces.

Difference: Rohan’s tent has a triangular base, and Simi’s tent has a rectangular base.

14B 1. The given shape is a square pyramid. The net of a square pyramid is: a.

2. The given net is of a triangular prism. The given image represents a triangular prism. a.

c. d. 4. a. b. 5.

Ice cube Ice cream cone

c. d.

4. Front View:

Side View:

Word Problem 1. When the net made by Rishi is folded, the faces overlap each other.

But when the net made by Megha is folded, the triangles do not overlap, so Megha made the net correctly.

14C 1.

Top View:

5. As per the given views, the structure made by Meena is: b.

Word Problem 1. Mehar used 15 building blocks to make the shape.

Front view

Side View:

Top View:

14D

1. a. The hospital is to the south of the airport.

b. The hotel is to the east of the bank.

c. The airport is to the west of the school.

2. 1 cm = 25 km

5 cm = 5 × 25 km = 125 km

3. Given that scale = 1 cm = 12 km

Distance between the college and the complex on the map = 2 cm

Actual distance = 2 × 12 km = 24 km

4. Length of cupboard = 1 square = 2 m; width of cupboard = 4 squares = 4 × 2 m = 8 m

Length of table = 3 squares = 3 × 2 m = 6 m; width of table = 2 squares = 2 × 2 m = 4 m

Length of desks = 1 square = 2 m; width of desks = 4 squares = 3 × 2 m = 6 m

5. Given that 1 cm = 2.5 km

5 cm = 5 × 2.5 km = 12.5 km

Word Problem 1. 1 cm = 5 km

5 cm + 2 cm = 7 cm

7 cm = 7 × 5 km = 35 km

14E 1.

2. a. A pattern that can be cut and folded to make a model of a 3-D shape is called a net.

b. A special way of drawing a house that is deep to show its length, width, and height is called a deep drawing.

c. Cube and cuboid have the same number of faces, edges, and vertices.

d. The 2-D representation of the map of a house is called the floor plan.

3. a. Cylinder

Faces =3

Edges = 2 Vertices = 0

b. Cube Faces = 6 Edges = 12 Vertices = 8

c. Cone Faces = 2 Edges = 1 Vertices = 1

d. Sphere Faces = 0 Edges = 0 Vertices = 0

e. Cuboid Faces = 6 Edges = 12 Vertices = 8

f. Square Pyramid Faces =5

Edges = 8 Vertices = 5

4. a. b.

2. The front of the house has only one window.

3.

4. If we enter from the back door, we will be in the kitchen.

5. We cannot show the windows of the backside as well as either the left or right side on the deep drawing.

Word Problem 1. As the house has 4 windows on each lateral side, Rakesh won’t be able to show four backside windows and four windows on either the left or right side. Hence, he won’t be able to show 8 windows in the deep drawing.

Chapter Checkup 1.

a. b. c.

5.

c. d.

7. Answers may vary. Sample answers:

Chapter 15 Let's Warm-up

1. Taking a shower

Months

2. Putting on a jacket

Days

3. Going for a trip

Minutes

4. Swimming lesson

Seconds

b. 3

Window Door

Window Window Window

8. There are 4 doors in the given floor plan.

9. Closet 1 = 4 squares

Bathroom = 12 squares

Since, 3 × 4 = 12

5. Rainy season

Window

Hours

Do It Yourself

15A 1. 1 hour = 60 minutes

a. 3 hours 360180=×= minutes

b. 4 hours 460240=×= minutes

c. 2 hours 20 minutes 2602012020140 =×+=+=

140 minutes

d. 3 hours 10 minutes 3601018010190 =×+=+=

So, the bathroom is 3 times bigger than closet 1.

10. Answer may vary. Sample Answer:

Word Problem 1. 1 cm = 3 km

Distance from Ben’s house to May’s house = 3 cm = 3 × 3 km = 9 km

Distance from May’s house to Jule’s house = 4 cm = 3 × 4 km = 12 km

Total distance covered by Ben = 9 + 12 km = 21 km

a. Yes Apar

Floor Plan A

Bedroom 1 Bedroom 2 Bathroom Hall

190 minutes

e. 3 hours 50 minutes 3605018050230 =×+=+=

230 minutes

f. 4 hours 30 minutes 4603024030270 =×+=+=

270 minutes

2. 1 minute = 60 seconds

a. 5 minutes 560300=×= seconds

b. 8 minutes 860480=×= seconds

c. 9 minutes 10 seconds 9601054010550 =×+=+=

550 seconds

d. 10 minutes 20 seconds 10602060020620 =×+=+=

620 seconds

e. 12 minutes 40 seconds 12604072040760 =×+=+=

760 seconds

f. 15 minutes 50 seconds 15605090050950 =×+=+=

950 seconds

3. 60 minutes = 1 hour; 1 minute = 1 60 hour

a. 120 minutes = 120 2 60 = hours

b. 180 minutes = 180 3 60 = hours

c. 150 minutes = (120 + 30) minutes =

120 hours30 minutes2

60 += hours 30 minutes

d. 200 minutes = (180 + 20) minutes = 180 hours20 minutes3

60 += hours 20 minutes

4. a. 240 seconds

b. 480 seconds

c. 350 seconds

d. 440 seconds

e. 520 seconds

f. 550 seconds

5. 1 hour = 60 minutes

1 minute = 60 seconds

5 minutes 50 seconds

8 minutes 40 seconds

4 minutes

9 minutes 10 seconds

7 minutes 20 seconds

8 minutes

1 hour and 20 minutes = 1 × 60 minutes + 20 minutes = 80 minutes = 80 × 60 seconds = 4800 seconds

Word Problems 1. Time Maya takes to read the book in the morning = 50 minutes

Time Maya takes to read the book in the afternoon = 45 minutes

Total time taken by Maya to read the book = 50 minutes + 45 minutes = 95 minutes = 1 hour 35 minutes

2. Time taken by train to reach its destination = 8 hours

30 minutes = 8 × 60 minutes + 30 minutes = 480 minutes + 30 minutes = 510 minutes

Therefore, the journey is 510 minutes long.

3. Time for which Sarah practiced the piano in the morning = 55 minutes

Time for which Sarah practiced the piano in the evening = 50 minutes

Total time = 55 minutes + 50 minutes = 105 minutes = 1 hour 45 minutes.

Therefore, she practices piano for 1 hour 45 minutes in total.

15B

1. a. 3 weeks 10 days + 6 weeks 5 days

9 weeks 15 days

= 9 weeks + 14 days + 1 day

= 9 weeks + 2 weeks + 1 day

= 11 weeks 1 day

b. 5 weeks 2 days

+ 25 days

5 weeks 27 days

= 5 weeks 21 days + 6 days = 5 weeks + 3 weeks + 6 days = 8 weeks 6 days

c. 2 years 6 months + 1 year  5 months

3 years 11 months

d. 5 years + 11 months

= 5 years 11 months

2. a. 7 weeks 20 days

– 2 weeks 5 days

5 weeks 15 days

= 5 weeks + 14 days + 1 day

= 5 weeks + 2 weeks + 1 day

= 7 weeks 1 day

b. 10 weeks 10 days

– 5 weeks

5 weeks 10 days

= 5 weeks + 1 week + 3 days

= 6 weeks 3 days

c. 11 years 7 months

– 2 years 4 months

9 years 3 months

d. 3 years 1 month = (2 years + 1 year) + 1 month = 2 years + 12 months + 1 month = 2 years 13 months

2 years 13 months

7 months

2 years 6 months

3. Suhani’s age = 13 years 5 months

Kunal’s age = Suhani’s age + 3 years 10 months

13 years 5 months

+ 3 years 10 months

16 years 15 months

= 16 years + 12 months + 3 months

= 16 years + 1 year + 3 months

= 17 years 3 months

Hence, Kunal’s age is 17 years 3 months.

4.

25 October 2020 25 October 2021 25 January 2021 15 February 2021

1 year 3 months

21 days

The museum was open for 1 year 3 months 21 days which is the same as 15 months 21 days.

5. Manya’s age = 15 years 7 months

Shubhi’s age = Manya’s age – 8 months

15 years 7 months

– 8 months

14 years 11 months

Megha’s age = Shubhi’s age – 3 years 5 months

14 years 11 months

– 3 years  5 months

11 years 6 months

Thus, Megha is 11 years 6 months old.

Word Problems 1. Summer vacation = 1 month 25 days

Winter vacation = 20 days

Total duration

1 month 25 days

+ 20 days

1 month 45 days

= 1 month + 30 days + 15 days

= 1 month + 1 month + 15 days

= 2 months 15 days

So, the school was closed for 2 months 15 days.

4 years 5 months 25 days

It took 4 years 5 months 25 days for Sam to become an advanced violinist.

15C 1. a. 5 hours + 55 min

5 hours 55 min

b. 4 hours 30 min

+ 6 hours 20 min

10 hours 50 min

c. 7 hours 5 min = 6 hours + 1 hour + 5 min = 6 hour + 60 min + 5 min = 6 hours 65 min

6 hours 65 min

– 5 hours 30 min

1 hour  35 min

d. 10 hours 10 min = 9 hours + 1 hour + 10 min = 9 hours + 60 min + 10 min = 9 hours 70 min

9 hours 70 min

– 45 min

9 hours 25 min

2. Starting time: 10:30 a.m. = 10 hours 30 minutes.

10 hours 30 minutes

+ 3 hours 10 minutes

13 hours 40 minutes.

Converting into 12-hour clock = 1:40 p.m.

3. Meeting starts at = 9:30 a.m. = 9 hours 30 minutes

Meeting ends at = 12:45 p.m. = 12 hours 45 minutes

Meeting lasted for = 12 hours 45 minutes

– 9 hours 30 minutes

3 hours 15 minutes

Therefore, the meeting lasted for 3 hours 15 minutes.

4. Time: 2:40 p.m. = 2 hours 40 minutes

Time after 35 minutes = 2 hours 40 minutes

+  35 minutes

2 hours 75 minutes

= 2 hours + 60 minutes + 15 minutes

= 2 hours + 1 hour + 15 minutes

= 3 hours 15 minutes

So, the time will be 3:15 p.m.

5. Train stops in the day for = 40 minutes and 25 seconds

Train stops in the night for = 45 minutes and 40 seconds

Total duration =

40 minutes 25 seconds

+ 45 minutes 40 seconds

85 minutes 65 seconds

= 85 minutes + 60 seconds + 5 seconds

= 85 minutes + 1 minute + 5 seconds

= 86 minutes 5 seconds = 1 hour 26 minutes 5 seconds

So, the train stops for 1 hour 36 minutes 5 seconds in total.

Word Problems 1. Online class starts at: 10:00 a.m. = 10 hours

Online class ends at: 11:30 a.m. = 11 hours 30 minutes

The class lasted for =

11 hours 30 minutes

– 10 hours

1 hour 30 minutes = 1 × 60 minutes + 30 minutes = 90 minutes

Therefore, the class lasted for 90 minutes.

2. Event started at: 2:00 p.m. = 2 hours

Event ended at: 5:30 p.m. = 5 hours 30 minutes

Event lasted for = 5 hours 30 minutes

– 2 hours . 3 hours 30 minutes

So, the event lasted for 3 hours 30 minutes.

3. Film begins at: 7:15 p.m. = 7 hours 15 minutes

Film ended at: 9:30 p.m. = 9 hours 30 minutes

Total duration of the movie = 9 hours 30 minutes

– 7 hours 15 minutes

2 hours 15 minutes

So, the duration of the film is 2 hours 15 minutes

4. Raman started to drive at: 11:20 a.m. = 11 hours 20 minutes

He reached his destination at: 4:45 p.m.

Converting into 24-hour clock = 16:45 = 16 hours 45 minutes

Total duration =

16 hours 45 minutes

–11 hours 20 minutes

5 hours 25 minutes

Hence, it took Raman 5 hours 25 minutes to reach the destination.

15D 1. a. The temperature of a hot cup of coffee is around 85 °C.

b. The possible temperature of a snowman is 0 °C.

2. 26 °C, 35 °C, 42 °C, 12 °C

3. a. As the temperature of Srinagar is lowest. It is the coolest place.

b. As the temperature of Chennai is highest. It is the hottest place.

c. Difference between the temperature of Delhi and Shimla = 41.2 °C − 11.5 °C = 29.7 °C.

d. Degrees the temperature needs to rise to reach 25 °C in Bangalore = 25 °C – 21.2 °C = 3.8 °C

4. Highest temperature = 25.6 °C

Lowest temperature = 18.4 °C

Difference between the temperatures = 25.6 °C – 18.4 °C = 7.2 °C

5. 29.3 °C > 26.0 °C > 23.9 °C > 23.6 °C

City B > City D > City C > City A

Word Problem 1. Temperature of Jammu and Kashmir = 17.3 °C

Temperature of Kerala = 39.8 °C

Difference between the temperature of the two cities = 39.8 °C–17.3 °C = 22.5 °C

Chapter Checkup 1. a. 2 hours = (2 × 60 minutes) = 120 minutes = (120 × 60 seconds) = 7200 seconds

b. 1 hour 30 minutes = (1 × 60 minutes) + 30 minutes = 90 minutes = (90 × 60 seconds) = 5400 seconds

c. 2 hour 20 minutes = (2 × 60 minutes) + 20 minutes = 140 minutes = (140 × 60 seconds) = 8400 seconds

d. 5 minutes 20 seconds = (5 × 60 seconds) + 20 seconds = 300 seconds + 20 seconds = 320 seconds

e. 6 minutes 40 seconds = (6 × 60 seconds) + 40 seconds = 360 seconds + 40 seconds = 400 seconds

f. 6 minutes 55 seconds = (6 × 60 seconds) + 55 seconds = 360 seconds + 55 seconds = 415 seconds

2. a. 90 minutes = 90 60 hours = 1.5 hours

b. 115 minutes = 60 minutes + 55 minutes = 1 hour 55 minutes

c. 160 minutes = 120 minutes + 40 minutes = 120 60 hours + 40 minutes = 2 hours 40 minutes

d. 280 seconds = 240 seconds + 40 seconds = 240 60 minutes + 40 seconds = 4 minutes 40 seconds

3. Time taken to reach the destination= 6 hours 30 minutes

= 6 × 60 minutes + 30 minutes

= 360 minutes + 30 minutes

= 390 minutes

Thus, the flight takes 390 minutes to reach its destination.

4. Time for which Shreya sleeps = 6 hours 15 minutes =

6 × 60 minutes + 15 minutes

= 360 minutes + 15 minutes = 375 minutes

Shreya sleeps for 375 minutes.

5. Match started at: 3:45 p.m. = 3 hours 45 minutes

Duration of match = 1 hour 15 minutes

Match ended at = 3 hours 45 minutes

+ 1 hour  15 minutes

4 hours 60 minutes

= 4 hours + 1 hour

= 5 hours = 5 p.m.

Thus, the match ended at 5 p.m.

6. Age at which Prashant went to school = 5 years 4 months

Prashant’s present age = 12 years 3 months

Total duration

= 12 years 3 months – 5 years 4 months

= 11 years + 12 months + 3 months – 5 years 4 months

= 11 years 15 months – 5 years 4 months

11 years 15 months

– 5 years  4 months

6 years 11 months

Hence, Prashant has been at school for 6 years 11 months.

7. a. 11:00 a.m. = 11 hours

11 hours

+ 1 hour 20 minutes

12 hours 20 minutes

= 12:20 p.m.

If a meeting starts at 11:00 a.m. and lasts for 1 hour and 20 minutes, it will end at 12:20 p.m.

b. 7:45 p.m. = 7 hours 45 minutes

7 hours 45 minutes

+ 1 hour 15 minutes

8 hours 60 minutes

= 8 hours + 1 hour = 9 hours = 9 p.m.

If you start reading a book at 7:45 p.m. and read for 1 hour and 15 minutes, you will stop reading at 9:00 p.m.

c. 2:15 p.m. = 2 hours 15 minutes

2 hours 15 minutes

+ 50 minutes

2 hours 65 minutes

= 2 hours + 60 minutes + 5 minutes = 2 hours + 1 hour + 5 minutes = 3 hours 5 minutes = 3:05 p.m.

An online class begins at 2:15 p.m. and runs for 50 minutes. The class will end at 3:05 p.m.

8. Highest temperature = 37.5 °C

Lowest Temperature = 21.2 °C

Temperature Difference = 37.5 °C – 21.2 °C = 16.3 °C

9. Starting date = 6 July

Number of days in July = 31

Number of days from 6 July to 31 July = 25 days

Duration = 40 days

Days left after the month of July = 40 – 25 = 15 days

So, the finishing date is 15 August.

10. 12 hours = 720 minutes

Total minutes = 720 minutes

Dividing 720 by 75,

Quotient = 9

Remainder = 45

So, Kunal takes the medicine 9 times in 12 hours.

Word Problems 1. Holidays start on = Saturday, 25 June.

Number of days in June = 30

Number of days from 25 June to 30 June = 5 days

Number of leaves = 15

Days left after the month of June = 15 days – 5 days = 10 days

So, the holidays end on Sunday, 10 July.

2. Sanya started drawing the picture at = 1:32 p.m. = 1 hour 32 minutes

Sanya completed the picture at = 5:15 p.m. = 5 hours 15 minutes = 4 hours 60 minutes + 15 minutes = 4 hours 75 minutes

Time duration = 4 hour 75 minutes

– 1 hour 32 minutes 3 hours 43 minutes

Hence, Sanya took 3 hours 43 minutes to draw the picture.

3. Temperature = 32 °C

Temperature decreased every minute = 2 °C.

Temperature reaches 16 °C:

32 °C – 2 °C – 2 °C – 2 °C – 2 °C – 2 °C – 2 °C – 2 °C – 2 °C

Time taken to reach 16 °C = 8 minutes

4. Train arrived Jaipur at = 10:45 a.m. = 10 hours 45 minutes

Train was late by = 1 hour 15 minutes.

The scheduled time of arrival of the train at Jaipur, 10 hours 45 minutes

– 1 hour 15 minutes

9 hours  30 minutes

Hence, the train was scheduled to arrive at Jaipur at 9:30 a.m.

Chapter 16

Let's Warm-up 1. ₹3.70 = 3.70 × 100 = 370 p

2. 412 100 = ₹4.12 = 412 p  3. ₹6.10 = 6.10 × 100 = 610 p

4. 305 100 = ₹3.05 = 305 p

Do It Yourself

16A 1. a. Cost of 7 candies = ₹35

Cost of 1 candy = ₹35 ÷ 7 = ₹5

If the cost of one candy is ₹5, then the cost of 7 candies will be ₹35.

b. Pocket money received for 4 weeks = ₹240

Pocket money received for 1 week = ₹240 ÷ 4 = ₹60

If Lisa received ₹240 as pocket money for 4 weeks, the amount of money she received every week was ₹60

c. Money saved in 3 months = ₹600

Money saved in 1 month = ₹600 ÷ 3 = ₹200

Riya saved ₹600 in 3 months. She saved ₹200 every month.

d. Cost of 10 cupcakes = ₹400

Cost of 1 cupcake = ₹400 ÷ 10 = ₹40

Cost of 55 cupcakes = 55 ⨯ ₹40 = 2200

A bakery sells 10 cupcakes for ₹400. They sell one cupcake for ₹40 and 55 cupcakes for ₹2200.

e. Cost of 10 crayons = ₹50

Cost of 1 crayon = ₹50 ÷ 10 = ₹5

Cost of 3 crayons = 3 ⨯ ₹5 = ₹15

If the price of a pack of 10 crayons is ₹50, then the cost of 3 crayons will be ₹15

2. Cost of 1 dozen candles = ₹15

Cost of 1 candle = ₹15 12

Cost of 20 candles = ₹ 15 12 × 20 = ₹25

3. Cost of 35 envelopes = ₹630

Cost of 1 envelope = ₹ 630 35 = ₹18

Cost of 57 envelopes = ₹18 ⨯ 57 = ₹1026

4. Rupees earned for 10 chores = ₹200

Rupees earned for 1 chore = ₹ 200 10 = ₹20

Rupees earned for 9 chores = ₹20 ⨯ 9 = ₹180

5. Cost of 3 4 m of cloth = ₹54

₹ 216 3

Cost of 1 2 m of cloth = ₹ 1216216 236 ×=₹ 1216216 236 ×= = ₹36

Word Problems 1. First Shopkeeper

Cost of 15 kg sugar = ₹450

Cost of 1 kg sugar = ₹ 450 30 15 = = ₹30

Second Shopkeeper

Cost of 20 kg sugar = ₹500

Cost of 1 kg sugar = ₹ 500 25 20 =  = ₹25

Thus, it is better to buy from the second shopkeeper.

2. Cost of 1 chocolate = ₹25

Numbers of chocolates she can buy with ₹200 = 200 8 25 =

Thus, Anita can buy 8 chocolates.

Since 8 × ₹25 = ₹200, Anita will not have any money left.

3. Cost of 10 bags of rice = ₹1230

Cost of 1 bag = ₹ 1230 123 10 = = ₹123

Since, 1 bag = 3 kg rice

Cost of 3 kg rice = ₹123

So, the cost of 1 kg rice = ₹123 41 3 = = ₹41

Thus, the cost of 1 kg rice is ₹41.

16B 1. Profit = SP – CP

Loss = CP – SP

a. SP = ₹85 + ₹1200 = ₹1285

b. CP = ₹200 − ₹55 = ₹145

c. CP = ₹550 + ₹7850 = ₹8400

d. SP = ₹525 − ₹70 = ₹455

2. a. When the cost price is higher than the selling price, then loss has been incurred

b. When the selling price is greater than the cost price, then the seller is said to have gained a profit.

c. The amount of money that a seller spends to pay for an item is called the cost price.

d. Profit = SP – CP → CP = SP – Profit = ₹725 – ₹225. A flower vase is sold at ₹725 for a profit of ₹225. Its CP is ₹500.

3. CP of 30 buckets = ₹1200

CP of one bucket = ₹40

Profit on one bucket = ₹27

SP of one bucket = ₹40 + ₹27 = ₹67

4. CP = ₹2500 + ₹435 = ₹2935

SP = ₹3500

Since, SP > CP

So, it is a profit. Profit = ₹3500 − ₹2935 = ₹565

5.

CP of 1 book = ₹70

CP of 350 books = ₹70 × 350 = ₹24,500

Total CP = ₹24,500

SP of 160 books = ₹80 × 160 = ₹12,800

Remaining number of books

= 350 – 160 = 190

SP of 190 books = ₹25 × 190 = ₹4750

Total SP = ₹12,800 + ₹4750 = ₹17,550

Since, Total SP < Total CP.

So, it is a loss.

Total loss = Total CP – Total SP = ₹24,500 − ₹17,550 = ₹6950.

So, the total loss on the whole transaction is ₹6950.

Word Problems

1. CP of 8 sharpeners = ₹40

CP of 1 sharpener = ₹ 40 8 = ₹5

Profit of 1 sharpener = ₹2

So, SP of 1 sharpener = CP + Profit = ₹5 + ₹2 = ₹7

Total SP = ₹7 × 8 = ₹56

So, Profit = ₹56 − ₹40 = ₹16

2. CP of 3 kg of oranges = ₹510

CP of 1 kg of oranges = ₹ 510 170 3 = = ₹170

SP of 1 kg of oranges = ₹32

SP of 3 kg of oranges = ₹32 × 3 = ₹96

Total CP of oranges and apples = ₹510 + ₹456 = ₹966

SP of apples if there is no profit or loss = Total CP –

SP of oranges

= ₹966 – ₹96 = ₹870

SP of 4 kg of apples = ₹870

SP of 1 kg of apples = ₹870 ÷ 4 = ₹217.5

16C 1. CP of 9 cups = ₹270

CP of 1 cup = ₹ 270 30 9 = = ₹30

SP of 1 cup = ₹50

Since, SP > CP

So, it is a profit

Thus, Profit = ₹50 − ₹30 = ₹20

2. CP of 1 ball = ₹9

CP of 27 balls = ₹9 × 27 = ₹243

SP of 27 balls = ₹250

Since, SP > CP

So, it is a profit.

Total Profit = ₹250 − ₹243 = ₹7

3. CP of 150 books = ₹3000

Total number of books = 150

CP of each book = ₹3000 ÷ 150 = ₹20

SP of 1 book = ₹15

Since, SP < CP

So, it is a loss.

Loss on each book = ₹20 − ₹15 = ₹5

4. CP of 7 apples = ₹49

SP of 1 apple = ₹5

SP of 7 apples = 7 × ₹5 = ₹35

Since, SP < CP

So, it is a loss.

Total loss = ₹49 − ₹35 = ₹14

5. CP of 20 pencil boxes = ₹480

CP of 1 pencil box = ₹ 480 24 20 = = ₹24

SP of 1 pencil box = ₹50

Since, SP > CP

So, it is a profit.

Profit = ₹50 − ₹24 = ₹26

6. Total CP of the old smartphone = ₹5620 + ₹530 = ₹6150

SP of the smartphone = ₹6150

Since, CP = SP

So, neither profit and nor loss.

7. CP of the bicycle = ₹7830 + ₹170 = ₹8000

SP of the bicycle = ₹8000

Since, CP = SP

So, neither profit and nor loss.

8. CP of all the toys = ₹990 + ₹1080 + ₹50 = ₹2120

SP of each toy = ₹70

SP of all the toys = ₹70 × 30 = ₹2100

Since, SP < CP

So, it is a loss.

Loss = ₹2120 − ₹2100 = ₹20

9. SP of the TV = ₹30,000

Profit = ₹1563

Profit = SP – CP ⇒ CP = SP – Profit

CP = ₹30000 – ₹1563 = ₹28,437

10. SP of the refrigerator = ₹18500

Loss = ₹1750

Loss = CP – SP ⇒ CP = SP + Loss

CP = ₹18500 + ₹1750 = ₹20,250

Chapter Checkup 1. a. Unitary method is used to find the value of a single unit and then using that value to find the value of multiple units.

b. Cost of 9 apples = ₹54

Cost of 1 apple = ₹54 ÷ 9 = ₹6

If the cost of one apple is ₹6, then the cost of 9 apples will be ₹54.

c. Rama bought a ring for ₹260 and sold it for ₹500. The cost price of the ring is ₹260.

d. When the cost price is higher than the selling price, then the seller is said to have incurred a loss.

2. a. Selling price is the amount of money a seller pays to buy an item for his store. FALSE

b. When CP < SP, the seller makes a profit. TRUE

c. Since, CP < SP. It is a profit. TRUE

d. Cost of 5 pencils = ₹30

Cost of 1 pencil = ₹30 ÷ 5 = ₹6. FALSE

3. 4 Pencils for ₹24

1 pencil = ₹ 24 6 4 = = ₹6

10 Pencils for ₹50

1 pencil = ₹50 5 10 = = ₹5

So, 10 pencils for ₹50 is a better buy.

4. 36 chairs at ₹550 each

1 chair = ₹550

22 chairs at ₹14300

1 chair = ₹14300 650 22 = = ₹650

So, 22 chairs at ₹14300 will cost more.

5. a. CP = ₹167 and SP = ₹185

Since, SP > CP, it is a profit.

So, Profit = SP – CP = ₹185 – ₹167 = ₹18

b. CP = ₹36 and SP = ₹29

Since, CP > SP, it is a loss.

So, Loss = CP – SP = ₹36 – ₹29 = ₹7

c. CP = ₹147 and SP = ₹125

Since, CP > SP, it is a loss.

So, Loss = CP – SP = ₹147 – ₹125 = ₹22

6. For 9 days of work = ₹6705

For 1 day of work = ₹ 6705 745 9 = = ₹745

For 28 days = ₹745 × 28 = ₹20,860

So, Ram will get paid ₹20,860 in 28 days.

7. CP of 12 glass bowls = ₹660

Number of bowls that broke = 3

Bowls left = 12 – 3 = 9

Loss incurred = ₹63

Total SP of 9 bowls = ₹660 – ₹63 = ₹597

SP of each bowl = ₹597 ÷ 9 = ₹66.33.

So, the selling price of 1 bowl is ₹66.33.

8. Number of workers = 3

Amount paid for 6 days to three workers = ₹3000

Amount paid for 1 day to three workers= ₹3000 ÷ 6 = ₹500

Amount paid for 1 day to 1 worker = ₹500 ÷ 3 = ₹166.67

So, each worker gets paid ₹166.67 every day.

9. CP of 7 phones = ₹2100 × 7 = ₹14,700

Profit = ₹700

Total SP = ₹14,700 + ₹700 = ₹15,400

SP of one phone = ₹ 15400 2200 7 = = ₹2200

10. Stamps earned in the first year = 2

Stamps earned in the second year = 2 × 2 = 4

Total stamps earned = 2 + 4 = 6

Value of 1 stamp = ₹450

Value of 6 stamps = 6 × ₹450 = ₹2700

Profit = ₹50 = SP – CP

SP = ₹2700 + ₹50= ₹2750

So, the selling price is ₹2750.

Word Problems 1. SP of one bulb = ₹15

SP of 20 bulbs = ₹15 × ₹20 = ₹300

Profit = ₹5

CP of 20 bulbs = ₹300 − ₹5 = ₹295

CP of 1 bulb = ₹ 295 14.7525 20 = = ₹14.75

2. CP = ₹780 + ₹675 = ₹1455

SP = ₹1450

CP > SP, so it is a loss.

Loss = ₹1455 − ₹1450 = ₹5

3. CP of 50 flowers = ₹750

Number of flowers left after keeping 4 flowers = 50 – 4 = 46

SP of 46 flowers = 46 × ₹12 = ₹552

CP > SP, so loss incurred.

Loss = CP – SP = ₹750 – ₹552 = ₹198

4. Total CP = (₹200 × 30) + (₹100 × 20) = ₹8000

Total number of shirts = 30 + 20 = 50

Total SP = ₹300 × 50 = ₹15,000

SP > CP, so it is a profit.

Profit = ₹15000 − ₹8000 = ₹7000.

Chapter 17

Let's Warm-up

1. Kavya

2. Rehan

3. Total stamps = 50 + 25 +25 = 100; half of 100 = 50. Hence, Kavya has half of total stamps.

4. 50 + 25 + 25 = 100

5. 50 – 25 = 25

Do It Yourself

17A 1. Least: V-A = 30 students Most: V-C = 50 students

2. a. The number of cows in the village were 55

b. There were 20 more buffalo than horse in the village.

3.

c. The animal that had the least number was the horse

c. Number of tickets sold to adults on Friday = 110; Number of tickets sold to children on Sunday = 75

Difference between the tickets sold = 110 – 75 = 35

d. Total number of tickets sold on Sunday = 160 + 75 = 235

Total number of tickets sold on Saturday = 140 + 60 = 200

in the tickets sold = 235 – 200 = 35

5.

0246

810121416182022242628

Monday

TuesdayWednesday Thursday Friday

17B 1. a. Food consumes the largest portion of her pocket money.

b. 1 4 × ₹100 = ₹25 goes towards savings.

c. 1 10 × ₹100 = ₹10

2. a. Walking is least preferred.

b. 2 5 × 200 = 80 students travel by bicycle.

c. 1 5 × 200 = 40 students take the rickshaw to school.

3.

5.

d.

17C 1. Kavya spent the highest number of hours on the internet on Wednesday and Thursday.

2. The city visited by the most number of the people is Pune. The city visited by the least number of people is Goa.

3. a. Temperature recorded during the second week = 8 °C

b. The highest temperature was recorded in Week 1.

c. Temperature in first week = 10 °C; Temperature in last week = 2 °C

Difference in temperature = 10 °C – 2 °C = 8°C

4. a. 700 guitars were sold in the year 2017.  b. 800 guitars were sold in the year 2021.

c. The least number of guitars were sold in the year 2018.

5. a. The cricketer hit a century in Match 2 and Match 6.   b. The lowest score of the cricketer is 10 runs.

c. The cricketer scored less than a century 5 times.

d. The total runs scored by the cricketer in the initial 4 matches = 30 + 100 + 10 + 40 = 180

Word Problem 1. a. The factory sold 8000 items of clothing in the month of April.

b. The factory sold most items in August. The items sold were 9000.

c. Sales of April month = 8000; sales of July month = 2000

Difference in sales = 8000 – 6000 = 2000

Chapter Checkup

1.

a. 15 families like cat.

b. 10 + 9 = 19

19 families prefer birds and rabbits as their pets

c. Number of families who like dogs = 14;

Number of families who like fish = 12

Difference = 14 – 12 = 2

The number of families who like dogs as their pet are 2 more than the number of families who like fish as their pet.

2. a. The boys and girls read an equal number of pages on Thursday.

b. Number of pages read by boys on Wednesday = 18;

Number of pages read by girls on Wednesday = 15

Hence, boys read more pages than girls.

c. Number of pages read by boys on Tuesday = 14.

d. Total number of pages read by girls = (12 + 11 + 15 + 16 + 18) = 72 pages

Total number of pages read by boys = (10 + 14 + 18 + 16 + 17) = 75 pages

Difference in the number of pages read = 75 – 72 = 3 pages

4.

5.

6.

a. On Friday, the fruit seller sold 15 kg fruit.

b. Fruits sold on Saturday = 30 kg; Fruits sold on Sunday = 35 kg

Total fruits sold = 30 kg + 35 kg = 65 kg

c. Sale in the initial 4 days = (20 + 25 + 10 + 20) = 75 kg

Sale

7. a. The café had least customers at 5:00 p.m.

b. There were 25 customers in the cafe at 8 p.m.

8. a. As the temperature of Jaipur is highest, it is the hottest city.

b. As the temperature of Ooty is lowest, it is the coldest city.

c. The temperature of Chennai is 15 °C

d. Bangalore has a temperature of 10 °C.

9.

10.

Word Problem 1. a. Chocolate ice cream is the most popular.

b. Blueberry is the least favorite ice cream.

c. Children who like chocolate ice cream = 7 5014 25 ×= × 50 = 14 children

d. Children who like Vanilla = 1 5010 5 ×= × 50 = 10

Children who like Chocolate = 6 5012 25 ×= × 50 = 12

Children who like either vanilla or chocolate = 10 + 12 = 22

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