Imagine_Maths_TM_Grade4

MATHEMATICS

Teacher Manual

MATHEMATICS

Master Mathematical Thinking

Grade 4

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.

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 6 Digits 1

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

Numbers up to 6 Digits 1

Numbers up to 6 Digits 1

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

Learning Outcomes

Students will be able to:

Numbers up to 6 Digits 1

write the place value, face value and expanded form for 5-digit numbers.

represent 5-digit numbers using the correct periods.

write the number name for 5-digit numbers.

write the place value, face value and expanded form for 6-digit numbers.

represent 6-digit numbers using the correct periods.

write the number names for 6-digit numbers.

compare numbers up to 6 digits.

arrange numbers up to 6 digits in ascending or descending order.

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

form 6-digit numbers using the given digits with/without repeating digits.

round off numbers up to 6 digits to the nearest 10, 100 or 1000 using a number line.

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

Let’s Recall

Recap to check if students know how to find the place value of a digit in a number.

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

Vocabulary

round off: to write the closest 10, 100 or 1000 number instead of the actual number

Teaching Aids

Alignment to NCF

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

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

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

5-rod abacus; Place value charts; Number cards for 5-digit numbers and their corresponding number name cards; 6-rod abacus; Number cards (0–9) in yellow, blue and red; Comma cards; Number cards for 6-digit numbers and their corresponding number name cards; Numbers cards for 2-digit, 3-digit and 4-digit numbers

Let’s Recall

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

Vocabulary

round off: to write the closest 10, 100 or 1000 number instead of the actual number

Teaching Aids

5-rod abacus; Place value charts; Number cards for 5-digit numbers and their corresponding number name cards; 6-rod abacus; Number cards (0–9) in yellow, blue and red; Comma cards; Number cards for 6-digit numbers and their corresponding number name cards; Numbers cards for 2-digit, 3-digit and 4-digit numbers

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.

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

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 handson 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.

Place Value and Expanded Form in 5-digit Numbers; Face Value in 5-digit Numbers

Representing 5-digit Numbers

5-digit Number Names

Place Values and Expanded Form in 6-digit Numbers; Face Value in 6-digit Numbers

Representing 6-digit Numbers

6-digit Number Names

Comparing Numbers

Ordering Numbers

Forming Numbers

Rounding off Numbers

Revision

Simple Vertical Addition

Vertical Addition with Regrouping

Adding Numbers Horizontally

Story Sums

Simple Vertical Subtraction

Vertical Subtraction with Regrouping

Story Sums

Addition and Subtraction Together

Story Sums

Estimating the Sum

Estimating the Difference

Revision

3.

4. Division

9

5.

6.

7.

8

Horizontal Method

Multiplying by Expanding the Bigger Number

Vertical Method

Multiplication by a 2-digit Number

Multiplication by a 3-digit Number

Word Problems

Estimating the Product

Revision

Division by 1-digit Numbers

Dividing by Tens

Division of Numbers Up to 4 digits

Dividing by Multiples of 100s and 1000s

Word Problems

Estimating the Quotient

Revision

Finding Multiples

Common Multiples

Finding Factors Using Multiplication

Finding Factors Using Division

Common Factors

Revision

Halves, Quarters and Thirds

Equivalent Fractions

Ordering and Comparing Like Fractions

Ordering and Comparing Unlike Fractions with Same Numerators

Proper and Improper Fractions

Adding Fractions

Subtracting Fractions

Revision

Points, Rays and Lines

Line Segments

Open and Closed Figures, Simple and Non-simple Figures

Polygons

Circles and Its Parts

Revision

Conversion Between Rupees and Paise

Organising Data

Creating Pictographs

Interpreting a Pictograph

Creating Bar Graphs

Interpreting Bar Graphs

Pie Charts

Revision

Numbers up to 6 Digits 1

Learning Outcomes

Students will be able to:

write the place value, face value and expanded form for 5-digit numbers. represent 5-digit numbers using the correct periods.

write the number name for 5-digit numbers.

write the place value, face value and expanded form for 6-digit numbers. represent 6-digit numbers using the correct periods.

write the number names for 6-digit numbers.

compare numbers up to 6 digits.

arrange numbers up to 6 digits in ascending or descending order. form 6-digit numbers using the given digits with/without repeating digits.

round off numbers up to 6 digits to the nearest 10, 100 or 1000 using a number line.

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

Let’s Recall

Recap to check if students know how to find the place value of a digit in a number.

Warm-up section.

Chapter: Numbers up to 6 Digits

Place Value and Expanded Form in 5-digit Imagine Maths Page 2

Numbers; Face Value in 5-digit Numbers

Learning Outcomes

Students will be able to write the place value, face value and expanded form for 5-digit numbers.

Teaching Aids

5-rod abacus

Activity

Help students recall the abacus representation for each place value.

Divide the class into small groups and provide each group with an abacus (If unavailable, students can draw in their notebooks).

Assign a 5-digit number to each group (for example: 32487).

Instruct the groups to represent the given number on the abacus and write the place value and face value of each digit in their notebooks.

Next, ask them to write the expanded form of the number. For example: 32487 = 30000 + 2000 + 400 + 80 + 7.

Give the students more numbers to practise and ask them to write the place value, face value and expanded form.

Extension Idea

Ask: Suppose the digits in the tens and thousands places in the number 15,483 are interchanged. Will the place value and face value of the digits change?

Say: After interchanging the digits, the new number is 18,453. The place value of the digits would change but not the face value.

Representing 5-digit Numbers

Learning Outcomes

Students will be able to represent 5-digit numbers using the correct periods.

Teaching Aids

Place value charts

Activity

Imagine Maths Page 4

Demonstrate how 5-digit numbers are represented using commas between different periods. Ask the students to work in groups. Distribute the place value charts to each group. Instruct the students to listen to the clues and write the digits in their correct places using commas. Read out clues like: The digit 2 is in the tens place and the thousands place. The digit 5 is in the ones place and the ten thousands place. The digit in the hundreds place is one more than the digit in the ones place. Students will then write the number in their notebooks placing the commas at the correct places.

Repeat the activity with more numbers and clues.

Extension Idea

Ask: Where would you place the commas in 567891?

Say: The commas are placed after the ones period and the thousands period. Here, 891 falls under the ones period and 67 falls under the thousands period. So, 567891 will be written as 5,67,891.

Learning Outcomes

Students will be able to write the number name for 5-digit numbers.

Teaching Aids

Number cards for 5-digit numbers and their corresponding number name cards

Activity

Demonstrate how to write the number name of a 5-digit number. Create a set of number cards with the corresponding names for 5-digit numbers.

Ask the students to work in groups. Distribute 10 number cards and their corresponding number name cards to each group.

Instruct the groups to shuffle the cards and arrange them facing down on a big table. Ask one student from the group to flip over two cards at a time. The number card should match the card with its corresponding number name. If the number matches its name, the player keeps the cards and plays another turn. If not, then the next student gets the chance. Continue this till all the students in a group get a turn.

The student who gets the maximum number of matches wins.

Place Values and Expanded Form in 6-digit Imagine Maths Page 8 Numbers; Face Value in 6-digit Numbers

Learning Outcomes

Students will be able to write the place value, face value and expanded form for 6-digit numbers.

Teaching Aids

6-rod abacus

Activity

Instruct the students to form groups. Provide each group with an abacus. Explain that the different rods on the abacus represent different the place values of 1s, 10s, 100s, 1000s, 10000s and 100000s.

Give the students a 6-digit number (for example: 345678) and instruct them to represent it on the abacus. Use different colours for each place value rod for better visualisation. Then, ask them to write the place value and face value of the digits in the number in their notebooks.

Ask the students to manipulate the numbers on the abacus to create different 6-digit numbers. Encourage them to move the beads in each rod to understand the impact on the overall value.

Ask the students to draw the abacus in their notebooks, if abacus is unavailable.

Use the abacus to demonstrate the expanded form of a 6-digit number. For example: 345678 = 300000 + 40000 + 5000 + 600 + 70 + 8.

Extension Idea

Ask: What is the expanded form of a number which has six thousands less than 84,236?

Say: Six thousands less than 84,236 = 84,236 – 6000 = 78,236. So, the expanded form of 78,236 = 70000 + 8000 + 200 + 30 + 6.

Learning Outcomes

Students will be able to represent 6-digit numbers using the correct periods.

Teaching Aids

Number cards (0–9) in yellow, blue and red; Comma cards

Activity

Demonstrate how 6-digit numbers are represented using commas.

Assign a different coloured card to the digits in each period. For example, use yellow cards for the ones period, blue for the thousands period, and red for the lakhs period.

Ask the students to work in groups. Distribute number and comma cards to each group. Instruct them to create a 6-digit number, say 723560, using the number cards. They should use the assigned colours for each period, placing the cards side by side. Ask them to insert comma cards after every set of period cards to classify the position of each period.

Give the students more 6-digit numbers to practise.

6-digit Number Names

Learning Outcomes

Students will be able to write the number names for 6-digit numbers.

Teaching Aids

Number cards for 6-digit numbers and their corresponding number name cards

Activity

Imagine Maths Page 10

Demonstrate how to write the number name of a 6-digit number. Create a set of number cards with numbers and their corresponding names for 6-digit numbers.

Ask the students to work in groups. Distribute 10 number cards and their corresponding number name cards to each group.

Instruct the groups to shuffle the cards and scatter all the cards on a big table facing down. Ask one student from the group to flip over two cards at a time. The number card should match the card with its corresponding number name. If the number matches its name, the player keeps the cards and plays another turn. If not, then the next student gets the chance. Continue this till all the students in a group get a turn. The student who gets the maximum number of matches wins.

Comparing Numbers

Learning Outcomes

Students will be able to compare numbers up to 6 digits.

Teaching Aids

6-rod abacus

Activity

Imagine Maths Page 12

Ask the students to form 3 groups. Provide each group with too many resources required. We can write again that they can draw it.

Instruct the students to represent a 6-digit number on one abacus (say, 4,14,358) and another 6-digit number on the other abacus (say, 4,56,125).

Ask them to first compare the beads in the lakhs place of each abacus, then say if the number of beads is the same, and then compare the beads in the ten thousands place. Explain that if the number of beads is different, then the number with more beads in the same place is greater than the other number. Ask the students to write the comparison of the numbers in their notebooks using the appropriate signs.

Give the students more 6-digit numbers to compare to practise.

Extension Idea

Ask: If the hundreds and the ten thousands digits in the number 8,15,367 are interchanged, which will be greater: the original number or the new number?

Say: The original number is 8,15,367 and the new formed number is 8,35,167. 8,35,167 > 8,15,367. Thus, the new number is greater than the original number.

Ordering Numbers

Learning Outcomes

Imagine Maths Page 14

Students will be able to arrange numbers up to 6 digits in ascending or descending order.

Teaching Aids

Number cards with 6-digit numbers

Activity

Take the students to the playground and make a number line on the ground (make sure it covers a broad range of 6-digit numbers).

Shuffle and arrange the number cards with 6-digit numbers face down in the playground.

Ask the students to form groups. Instruct one student from each group to pick one number card and place it at the correct position on the number line. Then, the next student from each group should pick a card and place it on the number line. Continue this activity until all the students in the group get a chance.

Back in the classroom, give some numbers to the students and ask them to arrange them in ascending and descending order.

Imagine Maths Page 16

Learning Outcomes

Students will be able to form 6-digit numbers using the given digits with/without repeating digits.

Teaching Aids

Number cards (0–9) in yellow, blue and red

Activity

Ask the students to form groups. Distribute number cards (0–9) to each group.

Instruct the students to form the largest 6-digit number using the number cards when no digit is repeated. Hint that they should use the six largest digits to make the required number.

Ask the students to form the smallest 6-digit number when no digit is repeated and then form the smallest 6-digit number when one digit can be repeated once.

Ask the students to note down the numbers so formed in their notebooks.

Ask questions like: What is the smallest 6-digit number that can be formed using any of the digits from 0 to 9 when no digit is repeated?

Extension Idea

Ask: Use the digits, 7, 9, 0, and 2 to make the smallest 6-digit number. The digits 0 and 2 can be repeated once.

Say: The given digits are 7, 9, 0, 2 and the digits 0 and 2 can be repeated once. So, the smallest number will be 2,00,279.

Rounding off Numbers

Learning Outcomes

Imagine Maths Page 19

Students will be able to round off numbers up to 6 digits to the nearest 10, 100 or 1000 using a number line.

Teaching Aids

Number cards for 2-digit, 3-digit and 4-digit numbers

Activity

Take the students to the playground and draw large number lines on the ground for numbers 10 to 100, 100 to 1000 and 1000 to 10,000. Divide the class into groups. Make number cards with different numbers.

Call 1 student from each group and give a number card to each of them.

Ask them to go and stand near the number on the number line.

Instruct the remaining students to identify the nearest 10, 100 or 1000 and round off the numbers. If the number rounds up, the students standing on the number line can jump forward and if the number rounds down, they can jump backwards.

Repeat this activity for all the students in the group by assigning different number cards to each student.

Extension Idea

Ask: A number when rounded to the nearest 1000 gives 156,000 and rounded to the nearest 100 gives 156,400. How many such numbers are possible?

Say: Numbers that can be rounded to the nearest 1000 as 156,000 lie between 155,500 and 156,499 while numbers that can be rounded to the nearest 100 as 156,400 lie between 156,350 and 156,449. Of these, the numbers greater than 156,400 lie between 156,400 and 156,449. Thus, there are 49 such possible numbers.

1. Place Value and Expanded Form in 5-digit Numbers

5. Place Values and Expanded Form in 6-digit Numbers

2. Face Value in 5-digit Numbers

Think and Tell

Yes, the place value and face value of the ones digit in any number is always the same.

6. Face Value in 6-digit Numbers

The face value of the number in the Thousands place is 7.

3. Representing 5-digit Numbers

7. Representing 6-digit Numbers

8. 6-digit Number Names

4. 5-digit Number Names

Do

The

9. Comparing Numbers

Do

It Together

Step 2: We check the digit in the lakhs place. The digits in the lakhs place are the same. So, we cannot decide yet.

Step 3:

We now move to the next place, which is the ten thousands place.

Here, 5 > 4

Therefore, 7,53,278 > 7,43,271.

10. Ordering Numbers

Do It Together

3,68,109 > 1,76,902 and 75,045 > 60,438

So, the numbers can be sequenced as 60,438; 75,045; 1,76,902; 3,68,109.

The ascending order is: 60,438 < 75,045 < 1,76,902 < 3,68,109.

The descending order is 3,68,109 > 1,76,902 > 75,045 > 60,438.

11. Forming Numbers

Think and Tell

14,807; 14,870; 14,780; 87,140; 87,410

Think and Tell

7,48,801; 8,74,810; 7,41,880; 8,87,104; 8,17,404

Think and Tell

Largest number: 9,99,999

Smallest number: 1,00,000

Do It Together

Part 1: Forming 5-digit numbers first, with no repeating digits. We need the highest digit in the leftmost place to form bigger numbers.

96541

The largest 5-digit number that can be formed is 96541 We need the smallest digit in the leftmost place to form smaller numbers.

14569

The smallest 5-digit number that can be formed is 14569.

Part 2: Let us form 6-digit numbers first, with 1 repeating digit.

We need the highest digit in the leftmost place to form bigger numbers. So, 9 should repeat!

996541

The largest 6-digit number that can be formed is 996541

We need the smallest digit in the leftmost place to form smaller numbers. So, 1 should repeat! 114569

The smallest 6-digit number that can be formed is 114569.

12. Rounding off Numbers

Do It Together

Rounding off to the nearest 100.

90,135 is between 90,100 and 91,200, but is closer to 90,100

So, 90,135 can be rounded off to 90,100.

Rounding off to the nearest 1000

90,135 is between 90,000 and 91,000, but is closer to 90,000

So, 90,135 can be rounded off to 90,000.

Addition and Subtraction 2

Learning Outcomes

Students will be able to:

add two or more 4-digit or 5-digit numbers vertically (without regrouping).

add two or more 4-digit or 5-digit numbers vertically (with regrouping).

add two or more 4-digit or 5-digit numbers horizontally (without regrouping).

solve word problems on adding 4-digit or 5-digit numbers.

subtract a 4-digit or 5-digit number from a 5-digit number (without regrouping).

subtract a 4-digit or 5-digit number from a 5-digit number (with regrouping).

solve word problems on subtracting 5-digit numbers.

solve expressions that have both addition and subtraction of 5-digit numbers.

solve story problems that have both addition and subtraction of 5-digit numbers.

estimate the sum of two 5-digit numbers.

estimate the difference of two 5-digit numbers.

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 addition and subtraction of numbers up to 3 digits.

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

Vocabulary

estimate:

approximate value of a number or a sum

Teaching Aids

Place value blocks of 1000s, 100s, 10s and 1s; Place value charts; Counters; Word problem sheets; 5-rod abacus; Sets of 3 puzzle interlocking cards with expressions on 2 cards and the answer on 1 card; Number lines showing 10 divisions drawn on a sheet of paper

Chapter: Addition and Subtraction

Simple Vertical Addition

Learning Outcomes

Imagine Maths Page 26

Students will be able to add two or more 4-digit or 5-digit numbers vertically (without regrouping).

Teaching Aids

Place value blocks of 1000s, 100s, 10s and 1s

Activity

Ask the students to work in groups. Distribute the place value blocks to each group.

Instruct them to add the numbers 1317 and 3181 using the blocks.

Then, ask them to write the two numbers one below the other by writing the digits as per their place value.

Instruct them to first add the digits in the ones place, then the tens, then the hundreds and finally the thousands place. Ask them to write the sum below the addend digits.

Instruct the students to check the vertical addition result with the result they got using place value blocks. Repeat the activity with the numbers 58,124 and 1525.

Teacher Tip: If resources are unavailable, cutouts showing the different place value blocks can be distributed.

Vertical Addition with Regrouping

Learning Outcomes

Imagine Maths Page 28

Students will be able to add two or more 4-digit or 5-digit numbers vertically (with regrouping).

Teaching Aids

Place value blocks of 1000s, 100s, 10s and 1s

Activity

Ask the students to work in groups. Distribute place value blocks to each group.

Instruct them to add the numbers 1825 and 2386 using the blocks. Tell them that if the sum of the number of blocks in any place exceeds 9, they should regroup to the next higher place. Explain that ten 1s make one 10, ten 10s make one 100 and ten 100s make one 1000.

Then, instruct the students to write the numbers one below the other with the digits as per their place value. Ask them to first add the ones digits, then the tens, then the hundreds and finally the thousands. Tell them that if the sum of the digits in any place exceeds 9, they should regroup to the next higher place.

Instruct the students to check the vertical addition result with the result they got using place value blocks. Repeat the activity with the numbers 48,635 and 32,478.

Extension Idea

Ask: When do you get a 6-digit number on adding two 5-digit numbers?

Say: When the sum of the digits in the ten thousands place of the two numbers is more than 9, we get a 6-digit number.

Learning Outcomes

Students will be able to add two or more 4-digit or 5-digit numbers horizontally (without regrouping).

Teaching Aids

Place value charts; Counters

Activity

Ask the students to work in groups. Distribute 3 place value charts to each group.

Instruct them to show the numbers 53,423 and 1404 on separate place value charts by placing the same number of counters as the face value of the digits in each number.

Then, ask them to count the total number of counters in the ones column of both place value charts and write the digit for the total in the ones place of the third place value chart to show the addition. Instruct them to repeat these steps for the counters in the other places of both numbers. Explain that the result they got is the sum of the two numbers.

Repeat the activity with the numbers 45,214 and 23,675. Once the students are done, ask them to replace the counters with digits and then add.

Ask questions like: Will you get the same answer if the digits on both grids are written vertically?

Story Sums

Learning Outcomes

Students will be able to solve word problems on adding 4-digit or 5-digit numbers.

Teaching Aids

Word problem sheets

Activity

Imagine Maths Page 30

Distribute the word problem sheets to the students. Instruct them to read the problem: A construction company used 14,567 bricks to build one part of a building and 8345 bricks for another part. How many bricks did they use in total? Discuss what they understand about the question. Ask them to write down what they know, what they need to find and then solve to find the answer using the column method.

Once the students are done, discuss the answer with the whole class.

Ask questions like: How many bricks would the construction company have used if they used 256 more bricks in another part?

Extension Idea

Ask: Create your own word problem on adding 5234, 1845 and 1056.

Say: There can be multiple word problems on adding three 4-digit numbers. One such problem could be: A school purchased 5234 notebooks, 1845 pens, and 1056 markers for the classrooms. How many school supplies did they buy in total?

Learning Outcomes

Students will be able to subtract a 4-digit or 5-digit number from a 5-digit number (without regrouping).

Teaching Aids

5-rod abacus

Activity

Ask the students to work in groups. Distribute an abacus to each group.

Instruct them to subtract 4230 from 75,650 using the abacus by first showing the bigger number on it and then removing the same number of beads as the smaller number.

Then, ask them to write the numbers one below the other by writing the digits as per their place value. Instruct them to first subtract the digits in the ones place, then the tens, hundreds, thousands, and finally the ten thousands place. Ask them to write each difference below the two digits they subtract with.

Instruct the students to check the vertical subtraction result with the result they got using the abacus. Repeat the activity to subtract 12,356 from 34,466.

Teacher Tip: If abacuses are unavailable, students can draw an abacus in their notebooks and show the beads by drawing circles.

Ask questions like: Can the difference of two numbers be greater than the minuend? Why or why not?

Vertical Subtraction with Regrouping

Learning Outcomes

Imagine Maths Page 35

Students will be able to subtract a 4-digit or 5-digit number from a 5-digit number (with regrouping).

Teaching Aids

5-rod abacus

Activity

Ask the students to work in groups. Distribute an abacus to each group.

Instruct the groups to subtract 5257 from 98,512 by first showing the bigger number on the abacus and then taking away the same number of beads as the smaller number. Tell them that if the number of beads to be taken away in any place is more than the number of beads available to subtract from, then they should remove one bead from the next higher place and add ten beads to the lower place.

Then, instruct the students to write the numbers one below the other with the digits lined up as per their place value and then subtract. Tell them that if the digit in the minuend is less that the digit in the subtrahend, they should regroup from the next higher place.

Instruct the students to check the vertical subtraction result with the result they got using the abacus. Repeat the activity with the numbers 81,135 and 32,127.

Extension Idea

Ask: Can you think of any two numbers which give the difference as 13,254 on subtraction?

Say: We can find the numbers by first taking any number. If the number taken is greater than 13,254, subtract 13,254 from it to get the second number. If the number taken is less than 13,254, add it to 13,254 to get the second number. Let us take the number 24,165. 24,165 – 13,254 = 10,911. Hence 24,165 and 10,911 are two such numbers which give 13,254 on subtraction.

Learning Outcomes

Students will be able to solve word problems on subtracting 5-digit numbers.

Teaching Aids

Word problem sheets; 5-rod abacus

Activity

Distribute the word problem sheets and 5-rod abacuses.

Instruct the students to read the problem: The distance between two cities is 3456 miles. A car has travelled 2198 miles so far. How many more miles does it need to travel to reach the destination? Discuss what they understand about the question. Ask them to write down what they know and what they need to find in their notebooks and then solve to find the result using the abacus.

Once the students are done, discuss the answer with the whole class. Have them solve one more problem with 5-digit numbers.

Extension Idea

Ask: Create your own word problem where you need to subtract a 4-digit number from a 5-digit number.

Say: There can be multiple word problems on subtracting a 4-digit number from a 5-digit number. One such problem could be: Emily had ₹15,678 in her bank account. She withdrew ₹4849 to pay her house rent. How much money did she have left in her account?

Learning Outcomes

Students will be able to solve expressions that have both addition and subtraction of 5-digit numbers.

Teaching Aids

Sets of 3 puzzle interlocking cards with expressions on 2 cards and the answer on 1 card

Activity

Begin by taking an expression like 25,635 + 31,458 – 23,147 that involves both addition and subtraction. Explain that we need to first add the two numbers and then subtract the third number from the sum.

Ask the students to work in groups. Distribute the sets of puzzle interlocking cards to each group.

Explain that two expressions have the same answer and the students have to join those two expressions with their answer. Ask them to use their notebooks to find the result. The group that solves all the puzzles first, wins. Ask questions like: Does the result change if we perform the subtraction operation first?

Learning Outcomes

Students will be able to solve story problems that have both addition and subtraction of 5-digit numbers.

Teaching Aids

Word problem sheets

Activity

Distribute the word problem sheets to the students. Instruct them to read the problem: Suhani saved ₹5078 in January, spent ₹2845 in February, and then saved ₹3256 in March. What is her total saving at the end of March?

Discuss what they understand about the question. Ask them to write down what they know and what they need to find in their notebooks. Discuss what operation needs to be carried out first. Ask them to perform the operation using vertical addition or subtraction. Once they have solved the problem, discuss the answer with the whole class.

Extension Idea

Ask: Create your own word problem for 12,562 + 14,258 – 19,365.

Say: Multiple word problems can be written for this expression. One such problem could be: A clothing company produced 12,562 shirts in a month and 14,258 shirts in another month. If the company sold 19,365 shirts in the two months, how many shirts was it left with?

Estimating the Sum

Learning Outcomes

Students will be able to estimate the sum of two 5-digit numbers.

Teaching Aids

Number lines from 20,000 to 30,000 and 60,000 to 70,000 drawn on a sheet of paper

Activity

Imagine Maths Page 43

Recall how to round off 2-digit or 3-digit numbers to the nearest 10 and 100. Divide the class into small groups. Distribute the number line sheets to each group.

Instruct the students to find the estimated sum of 24,145 and 65,712 by first marking the two numbers on the number line, then finding the nearest thousand for each number, marking the rounded off number and finally, adding the rounded off numbers.

Then, ask the students to write the question and its answer in their notebooks and discuss the digit that needs to be checked while rounding off to the nearest 1000.

Ask questions like: Is the sum after rounding off to the thousands and to the ten thousands the same?

Give the students one more question to solve using both methods.

Learning Outcomes

Students will be able to estimate the difference of two 5-digit numbers.

Teaching Aids

Number lines showing 10 divisions drawn on a sheet of paper

Activity

Ask the students to work in groups. Distribute the number line sheets to each group.

Instruct the students to find the estimated difference of 24,362 and 15,236 by first marking the two numbers on the number line, then finding the nearest thousand for each number, marking the rounded off number and finally, subtracting the rounded off numbers.

Then, ask the students to find the answer in their notebooks and compare the answers that they got using both methods.

Repeat the process with more 5-digit numbers.

Answers

1. Simple Vertical Addition

Think and Tell

Ten thousand is the fifth place in a number. A 4-digit number has 4 places, hence there cannot be a ten-thousand in a 4-digit number.

Do It Together

So, 65,234 + 2345 = 67,579

2. Vertical Addition with Regrouping

5. Simple Vertical Subtraction

Do It Together

So, 84,467 + 2893 = 87,360.

3. Adding Numbers Horizontally

Do It Together

So, 12,344 + 1115 = 13,459.

4. Story Sums

Do It Together

What do we know?

Number of chocolates produced in the morning = 3456

Number of chocolates produced in the evening = 4257

What do we need to find?

Total number of chocolates produced in a day = 3456 + 4257

Solve to find the answer.

So, the factory produced 7713 chocolates during the day.

So, 75,234 – 3121 = 72,113.

6. Vertical Subtraction with Regrouping

Do It Together

So, 78,131 – 9993 = 68,138.

7. Story Sums

Do It Together

What do we know?

Stamps with Sunaina = 8455

Stamps with her sister = 6712

What do we need to find?

Stamps with Sunaina – Stamps with Sunaina’s sister

Solve to find the answer.

8455 – 6712 = 1743

So, Sunaina has 1743 more stamps than her sister. Check your answer

8455 – 1743 = 6712

So, the answer is correct.

8. Simplify

Do It Together

46,798 + 1457 – 21,020

Let us find the sum of 46,798 + 1457

46,798 + 1457 = 48,255

Let us now subtract 21,020 from the sum.

48,255 – 21,020 = 27,235

So, 46,798 + 1457 – 21,020 = 27,235.

9

. Story Sums

Do It Together

Total capacity of the stadium = 45,000

Number of men watching the match = 25,765

Number of women = 11,567

What do we need to find?

How many seats were left empty?

Solve to find the answer.

45,000 – 25,765 – 11,567 = 19,235 –11,567 = 7668

10. Estimating the Sum

Think and Tell

To round off a number to the nearest 10,000, check the digit in the thousands place to decide whether to round up or round down. If the thousands are 5000 or more, round up. If they are 4999 or less, round down.

Do It Together

13,567

28,082

Rounded off to the nearest thousand

Round off to the nearest thousand

The estimated sum = 14,000 + 28,000 = 42,000

Actual sum = 13,567 + 28,082 = 41,649

11. Estimating the Difference

Do It Together

78,111

21,991

Rounded off to the nearest thousand

Round off to the nearest thousand

14,000

28,000

78,000

22,000

The estimated difference = 78,000 – 22,000 = 56,000

The actual difference of 78,111 – 21,991 = 56,120.

Multiplication 3

Learning Outcomes

Students will be able to:

multiply a 3-digit or 4-digit number by a 1-digit number by writing them horizontally.

a 3-digit or 4-digit number by a 1-digit number by expanding the bigger number.

them vertically.

by writing them vertically.

word problems on multiplying numbers up to 4 digits by numbers up to 3 digits.

the product of two numbers.

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 x 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 use repeated addition to multiply. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

multiplicand: number to be multiplied

multiplier: number by which we multiply

product: number obtained from multiplication

Teaching Aids

Price list of 3 items; Number cards with 3-digit, 4-digit, and 1-digit numbers; Bill template showing 4 grocery items with the cost per kg and the quantity bought; Multiplication cards; Bingo cards; Number cards; Chart paper; Glue stick; Word problem cards; Puzzle cards

Chapter: Multiplication

Horizontal Method

Learning Outcomes

Imagine Maths Page 49

Students will be able to multiply a 3-digit or 4-digit number by a 1-digit number by writing them horizontally.

Teaching Aids

Price list of 3 items

Activity

Divide the class into groups of 4. Provide each group with a price list of 3 items.

Wheat ₹223 3 kg

Ask one student in each group to pose as a shop owner and the remaining students as customers.

Instruct one student in the group to find the total price of rice, the second student to find the total price of dal and the third student to find the total price of wheat. Tell them that they need to find the product using horizontal multiplication. The shop owner should check the calculations.

Ask the students to write the answers in their notebooks.

Extension Idea

Ask: What will be the total cost of all the items if along with the listed items, 2 L of oil is also purchased at ₹110 per litre?

Say: Cost of 3 kg of rice = ₹399, Cost of 2 kg of dal = ₹282, Cost of 3 kg of pulses = ₹669, Cost of 2 L of oil = ₹220. So, the total cost = ₹399 + ₹282 + ₹669 + ₹220 = ₹1570.

Multiplying by Expanding the Bigger Number

Learning Outcomes

Imagine Maths Page 50

Students will be able to multiply a 3-digit or 4-digit number by a 1-digit number by expanding the bigger number.

Teaching Aids

Number cards with 3-digit, 4-digit, and 1-digit numbers

Activity

Demonstrate the multiplication of a 3 or 4-digit number by a 1-digit by expanding the bigger number.

Ask the students to work in groups. Distribute the number cards to the each group (cards with bigger numbers like 543, 4122, 3548 and smaller numbers like 5, 6, 7, 8, 9).

Ask the students to expand the bigger number and find the product using the boxes, as shown in the example for 326 × 7.

Discuss the answers and ask the students to write the answers in their notebooks.

Imagine Maths Page 52

Learning Outcomes

Students will be able to multiply a 3-digit or 4-digit number by a 1-digit number by writing them vertically.

Teaching Aids

Bill template showing 4 grocery items with the cost per kg and the quantity bought

Activity

Demonstrate how to multiply a 3- or 4-digit number by a 1-digit number vertically. Ask the students to work in groups of 4. Distribute the bill templates (example below) to each group.

Instruct the students to complete the bill for 4 items and also find the total bill amount. Instruct each student from the group to calculate the total cost of one item and then add the amounts to get the total bill amount.

Invite some groups to come up and present their bills and discuss the answers.

Extension Idea

Ask: Which purchase will cost more: 5 kg of walnuts at ₹449 per kg or 3 kg of hazelnuts at 634 per kg?

Say: 5 kg of walnuts at ₹449 per kg will cost more.

Multiplication by a 2-digit Number

Learning Outcomes

Imagine Maths Page 53

Students will be able to multiply a 3-digit or 4-digit number by a 2-digit number by writing them vertically.

Teaching Aids

Multiplication cards; Bingo cards

Activity

Demonstrate how to multiply a 3- or 4-digit number by a 2-digit number vertically. Make multiplication cards (for example: 329 × 43, 566 × 25, 728 × 12) and Bingo cards (showing answers to the corresponding multiplication cards).

Divide the class into groups and distribute the Bingo cards with numbers in each cell, to each group. Put the multiplication cards inside a box and shuffle these. Pick a card and read out the multiplication problem. Instruct the students to solve it and check whether the answer is on their Bingo card. If so, they should cross out the number. The group that crosses out a row vertically or horizontally first wins. Ask the students to write the answers in their notebook.

Extension Idea

Instruct: Arrange the following items in increasing order of their costs: 25 handkerchiefs at ₹266 per unit, 18 packets of toffees at ₹399 per packet, 12 knives at ₹165 per piece.

Say: 12 knives at 165 per piece < 25 handkerchiefs at 266 per unit < 18 packets of toffees at 399 per packet.

Imagine Maths Page 55

Learning Outcomes

Students will be able to multiply a 3-digit or 4-digit number by a 3-digit number by writing them vertically.

Teaching Aids

Multiplication cards; Number cards; Chart paper; Glue stick

Activity

Demonstrate how to multiply a 3- or 4-digit number by a 3-digit number vertically. Make multiplication cards (for example: 311 × 1743, 534 × 825, 456 × 912) and number cards (showing the answers to the corresponding multiplication cards).

Ask the students to work in groups. Distribute the multiplication cards, number cards and chart paper to each group.

Instruct the groups to pick a multiplication card, solve the multiplication problem, match it with the correct number card and paste these on the chart paper. Ask them to do this for all the cards.

Ask the students to also write the answers in their notebooks.

Word Problems

Learning Outcomes

Imagine Maths Page 56

Students will be able to solve word problems on multiplying numbers up to 4 digits by numbers up to 3 digits.

Teaching Aids

Word problem cards

Activity

Ask the students to work in groups. Distribute the word problem cards to each group. Ask them to read the problem.

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

Extension Idea

Instruct: Create your own word problem where the number of rows and flowers to be planted in each row are known. Say: There can be multiple such problems. One such problem could be:

Seema has a garden. She has planted roses in 234 rows, and each row contains 2,587 roses. How many roses are there in total in Seema’s garden?

Maria has a flower garden. She has planted sunflowers in 387 rows, and each row contains 4,592 sunflowers. How many sunflowers are there in total in Maria’s garden?

What do we know?

What do we need to find?

Solve to find the answer.

Estimating the Product

Learning Outcomes

Students will be able to estimate the product of two numbers.

Teaching Aids

Puzzle cards

Activity

Begin by recalling the term estimation. Discuss some real-life scenarios where we use estimation.

Ask the students to work in groups. Distribute the puzzle cards to each group. (Puzzle cards can be made and printed beforehand. Ensure that the puzzle shows only one correct option.)

Tell the students that the number in the green part shows the estimated product of one of the problems in the yellow parts.

Instruct the groups to estimate the answers for all the problems and match with the answer in the green part.

Ask the students to write the answers in their notebooks.

Answers

1. Horizontal Method

Do It Together

Step 1

Multiply by ones.

4132 × 2 = 4

Step 3

Multiply by hundreds.

4132 × 2 = __ __ __ 4

The product of 4132 and 2 is .

Step 2

Multiply by tens.

4132 × 2 = 4

Step 4

Multiply by thousands.

4132

2. Multiplying by Expanding the Bigger Number

Do It Together

The expanded form of 639

The product is 5400

The

3. Vertical Method

Do It Together

4. Multiplication by a 2-digit Number

Think and Tell

Greatest 4-digit number = 9999 and greatest 2-digit number = 99. Product of 9999 and 99 = 9,89,901. Thus, the product of a 4-digit number and a 2-digit number cannot be a 7-digit number.

5. Multiplication by a 3-digit Number

Do It Together

The product of 752 and 417 is 3,13,584.

6. Word Problems

Do It Together

Number of people for whom food is provided in a day = 6315

Number of days =

Total number of people for whom food is provided = 6315 ×

The free meal service provides food for people.

7. Estimating the Product

Think and Tell

487 rounded to the nearest ten is 490 and 487 rounded to the nearest hundred is 500.

Do It Together

Step 1

Round off both the numbers.

148 150 879

Step 2

Multiply the rounded off numbers.

150 × =

The estimated product of 148 and 879 is .

The product of 1148 and 39 is 44,772

Division 4

Learning Outcomes

Students will be able to:

divide a 3-digit or 4-digit number by a 1-digit number using the long division method. divide numbers up to 4 digits by 10s.

divide a 4-digit number by a 2-digit number using the long division method.

divide numbers up to 4 digits by 100s and 1000s.

solve word problems on dividing numbers up to 4 digits by numbers up to 2 digits. estimate the quotient of 3-digit or 4-digit numbers divided by 2-digit numbers.

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 divide a 1-digit or 2-digit number by a 1-digit number. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

estimation: the process of guessing an answer close to the actual answer

Teaching Aids

Place value blocks of 100s, 10s and 1s; Place value chart; Counters; Chart paper; Word problem written on a sheet of paper with space given for each element of the CUBES strategy; Pieces of a jigsaw puzzle with division problems and estimated quotients written on them

Chapter: Division

Division by 1-digit Numbers

Learning Outcomes

Imagine Maths Page 63

Students will be able to divide a 3-digit or 4-digit number by a 1-digit number using the long division method.

Teaching Aids

Place value blocks of 100s, 10s and 1s

Activity

Begin by recalling division as the process of equally sharing items among groups. Use the problem 448 ÷ 4 to illustrate the concept. Ask the students to work in groups. Distribute place value blocks to each group.

Instruct them to use the place value blocks to represent the dividend, 448. Ask them to divide each place value into 4 groups ensuring that each group has the same number of blocks. Ask the students to count the blocks in each group to find the quotient. Show the same division using the long division method. Instruct the students to solve 452 ÷ 4 using the place value blocks and the long division method.

Extension Idea

Ask: Is there a way to check if the quotient obtained for the division problem 452 ÷ 4 is correct?

Say: Yes. When there is no remainder, the dividend should be equal to Quotient × Divisor. Here, Dividend = 452. Quotient × Divisor = 113 × 4 = 452. We get Dividend = Quotient × Divisor. So, our answer is correct.

Dividing by Tens

Learning Outcomes

Students will be able to divide numbers up to 4 digits by 10s.

Teaching Aids

Place value chart; Counters

Activity

Ask the students to work in groups. Distribute the place value charts and counters. Ask the students to represent the number 140 by placing counters on the place value chart.

Imagine Maths Page 67

Discuss how dividing by 10 means moving each counter one place to the right on the chart. Let the students physically move each counter one place to the right on the chart while discussing the division process. As they move the counters, emphasize that the one counter representing 100 moves from the 100s column to the 10s column, and the four counters representing 40 move from the 10s column to the 1s column, showing that 140 ÷ 10 = 14.

Ask questions like: What is 8200 ÷ 20? Collect student responses.

Explain how they can cancel the zeroes in the ones place of both numbers and then divide the remaining digits mentally or using the long division method.

Learning Outcomes

Students will be able to divide a 4-digit number by a 2-digit number using the long division method.

Teaching Aids

Chart paper

Activity

Demonstrate how to draw an area model on the chart paper to represent a division problem such as 1224 ÷ 12. Ask the students to break 1224 into two numbers that are divisible by 12 and that are easy to work with.

For example: 1200 + 24 = 1224. Then, let them find the side lengths of each of the smaller rectangles. Finally, explain that the missing side of the first rectangle is 100, because 12 × 100 = 1200; that the missing side of the second rectangle is 2, because 12 × 2 = 24; and that on adding the side lengths, we get the quotient.

Ask the students to work in groups. Give them one division problem, such as 5025 ÷ 25, and let the groups construct their own area models. Instruct them to also divide 5025 by 25 using the long division method and then compare the answers that they got using the two methods.

Ask them to verify if the quotient and the remainder so obtained is correct using the equation, Dividend = (Quotient × Divisor) + Remainder. Dividing

Learning Outcomes

Students will be able to divide numbers up to 4 digits by 100s and 1000s.

Teaching Aids

Place value chart; Counters

Activity

Ask the students to work in groups. Distribute the place value chart and counters. Ask the students to represent the number 2140 by placing 2 counters in the 1000s column, 1 counter in the 100s column, 4 counters in the 10s column, and 0 in the 1s column. Instruct them to divide 2140 by 100.

Ask the students to move the counters to the right as many times as the number of zeroes in the divisor. Explain how 21 is the quotient and 40 is the remainder.

Give the students one more division problem and ask them to solve it using the place value chart.

Extension Idea

Ask: What is 4884 divided by 400?

Say: The quotient is 12 (400 × 12 = 4800) and the remainder is 84 (4884 – 4800 = 84). Thus, 4884 ÷ 400 gives Quotient = 12 and Remainder = 84.

Word Problems

Learning Outcomes

Imagine Maths Page 72

Students will be able to solve word problems on dividing numbers up to 4 digits by numbers up to 2 digits.

Teaching Aids

Word problem written on a sheet of paper with space given for each element of the CUBES strategy

Activity

Distribute the sheet with the word problem to the students.

Instruct them to circle the numbers, underline the question, and box the key words. Discuss what they need to find. Ask them to evaluate the problem, solve it using the vertical method and write the answer.

Ask questions like: How will you find out if your answer is correct?

Extension Idea

There are 7,200 pencils in a stationery store. If they are bundled into packs of 45, how many complete packs will there be?

Solve and Check!

Ask: Create your own word problem where you need to divide 1372 by 14.

Say: There can be several word problems on division of 1372 by 14. One such problem could be: Mary has 1372 apples. She wants to distribute them equally in 14 baskets. How many apples will each basket contain?

Estimating the Quotient

Learning Outcomes

Imagine Maths Page 75

Students will be able to estimate the quotient of 3-digit or 4-digit numbers divided by 2-digit numbers.

Teaching Aids

Pieces of a jigsaw puzzle with division problems and estimated quotients written on them

Activity

Begin by recalling the term estimation in division. Discuss some real-life scenarios where we use estimation. Show the students how to estimate the quotient of 257 ÷ 14 using a number line drawn on the board.

Ask the students to work in groups. Distribute the jigsaw puzzle pieces (at least 5 sets) to each group. Instruct the groups to first round off the dividend and divisor in the division problems on the puzzle pieces, estimate their quotient to the highest place and join the problem pieces with the correct answer pieces.

Extension Idea

Ask: Which two problems will give the same estimated quotient as 3956 ÷ 43? A.

Say: Options B and D will give the same quotient as 3956 ÷ 43.

Answers

1. Division by 1-digit Numbers

Think and Tell

No, the division of a 2-digit number by a 1-digit number can give a 2-digit number. For example, 85 ÷ 5 = 17.

Do It Together

2 37

9 2133 – 18

9 × 3 = 27

Thus, 2133 ÷ 9 = 237

2. Dividing by Tens

1.

3. Division of Numbers Up to 4 digits

Think and Tell

No, the division of a 4-digit number by a 2-digit number can give a 3-digit number. For example, 9,855 ÷ 27 = 365.

5. Word Problems

Do It Together

Rohit has 525 marbles. He wants to make groups of 25 each. How many such groups can he make? Use the CUBES method to solve.

1. Circle the numbers.

2. Underline the question.

3. Box the keywords.

4. Evaluate. Numbers of groups formed = Total number of marbles

Thus, Rohit makes 21 such groups.

6. Estimating the Quotient

Think and Tell

We did not round 9000 because it is already in the thousands.

Do It Together

572 rounded off to the nearest 100 is 600

11 rounded off to the nearest 10 is 10.

So, the estimated quotient is 60

4.

Dividing

Do It Together

1.

by Multiples of 100s and 1000s

2. 4565 by 2000

The quotient is 2 and the remainder is 565.

Multiples and Factors 5

Learning Outcomes

Students will be able to:

find the multiples of a number using the multiplication tables. find the common multiples of two or more numbers. find the factors of 2 numbers using square tiles and multiplication. find the factors of 2 numbers using division. find the common factors of 2 numbers and then their highest common factor.

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 10x10 (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 or combination given a constraint, in simple situations (e.g., how to make a committee of two people from a group of five people)

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

Let’s Recall

Recap to check if students know that multiplication and division are closely related. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

multiple: product that we get when one number is multiplied by another number factor: number that divides another number completely common factor: number that can evenly divide a set of two or more numbers

Teaching Aids

Number grid 1–100; Counters; Crayons

Chapter: Multiples and Factors

Finding Multiples

Learning Outcomes

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

Imagine Maths Page 81

Teaching Aids

Number grid 1–100; Crayons

Activity

Ask the students to work in pairs. Distribute the number grids and crayons to each pair.

Instruct the students to take jumps of 4 and shade the numbers on the grid using a particular colour. Once they have shaded the multiples of 4, have a discussion on what they noticed. Bring out the fact that the numbers shaded come in the multiplication table of 4 and are called multiples of 4. Ask them to write the multiples of 4 in their notebooks. Then, instruct the students to find the first five multiples of 11 on the same number grid using another colour. Ask them to write the first five multiples of 11 in their notebooks.

Ask questions like: What are the first five multiples of 11? Is 57 a multiple of 11? Why?

Extension Idea

Ask: How many multiples does a number have? Is 1 a multiple of every number?

Say: Every number has an infinite number of multiples. No, 1 is not a multiple of every number.

Common Multiples

Learning Outcomes

Students will be able to find the common multiples of two or more numbers.

Teaching Aids

Number grid 1–100; Crayons

Activity

Imagine Maths Page 84

Ask the students to work in pairs. Distribute the number grids and crayons to each pair. Instruct the students to imagine that the number grid shows the numbered floors of a building where they run the canteens and the party halls. To show that every second floor has a canteen, they should write ‘C’ on every second number on the grid. To show that every fifth floor has a party hall, they should write ‘P’ on every fifth number on the grid. Then, in their notebooks, they should write down the numbers on which they wrote C and P. Finally, ask them to circle the numbers that had both C and P written on them.

Ask questions like: What numbers had both C and P written on them?

Explain that the numbers that are circled are called the common multiples of 2 and 5. Ask the students to find the multiples and common multiples of 6 and 8 using the grid by shading the numbers.

Learning Outcomes

Students will be able to find the factors of 2 numbers using square tiles and multiplication.

Teaching Aids

Counters

Activity

Discuss the different ways in which 4 counters can be arranged with equal rows and equal columns.

Ask the students to work in groups. Distribute 20 counters to each group.

Instruct them to arrange the 20 counters in different ways so that they are placed in equal rows and equal columns. Each student in the group should show one arrangement. Then, in their notebooks, they will write the multiplication sentence for each arrangement.

Ask questions like: What arrangements did you get for 20 counters and what were their multiplication sentences?

Bring out the fact that the numbers that form the multiplication sentences are called the factors of that number.

Extension Idea

Ask: Are there any numbers that have only 2 factors? Which are those numbers?

Say: Yes, some numbers have only 2 factors and are divisible only by 1 and themselves. For example, 2 = 1 × 2; 5 = 1 × 5. Such numbers are called prime numbers.

Finding Factors Using Division

Learning Outcomes

Students will be able to find the factors of 2 numbers using division.

Teaching Aids

Counters

Activity

Imagine Maths Page 91

Discuss with the students how if we divide 15 apples into 3 groups and get 5 apples in each group, then 3 and 5 are factors of 15 because on dividing 15 by 3 or 5, we get no remainder.

Ask the students to work in groups. Distribute 24 counters to each group.

Instruct the groups to arrange an equal number of counters in 2 rows, then in 3 rows, and then in 4, 6, 8 and 12 rows. Then, in their notebooks, they will write the division sentences and the factors and list the numbers that leave no remainder.

Bring out the fact that if they are able to arrange the counters in the specified number of rows, that number would be a factor of the given number of counters.

Extension Idea

Ask: What is the smallest number that has exactly 3 factors?

Say: The smallest number that has exactly 3 factors is 4. The factors of 4 are 1, 2 and 4.

Learning Outcomes

Students will be able to find the common factors of 2 numbers and then their highest common factor.

Teaching Aids

Counters

Activity

Ask the students to work in pairs. Distribute the counters to each pair.

Instruct the students to find the factors of 12 and 15 using different arrangements of counters. Then, in their notebooks, they will list the factors, colour the common factors of both numbers and circle the common factor that is the highest.

Factors of 12 = 1, 2, 3, 4, 6, 12

Factors of 15 = 1, 3, 5, 15

Highest Common Factor = 3

Instruct the students to next find the factors of 20 and 30, write the common factors and find the highest common factor.

Extension Idea

Ask: When finding common factors of 2 numbers, can one of the numbers be the common factor of itself and the other number?

Say: Yes, a number can be a common factor of itself and another number. For example, for the numbers 5 and 15, the number 5 is a common factor of 5 and 15.

Answers

1. Finding Multiples

Think and Tell

All the circled numbers are part of the multiplication table of 2.

2. Common Multiples

Think and Tell

We stopped at 20 because there were only 20 floors in the building.

Do It Together

Step 1: Find multiples of 2 to find floors with the canteen.

Step 2: Find multiples of 5 to find floors with the party hall.

Step 3: Find the common floors by finding the common

Step 4: Tell the answer in complete sentence.

We can find the canteen and party hall together on the floors

10 and 20

3. Finding Factors Using Multiplication

Think and Tell

Sheela aunty should go with 8 plates with 1 apple in each.

Think and Tell

When finding factors by multiplication we always start with multiplying by 1 and stop when any factor starts repeating. This may be time consuming. The quicker method to find factors is division.

Multiply by 1

Multiply by 2 × =

Multiply by 3 Not possible

Multiply by 4 × =

Multiply by 5 × =

Multiply by 6 not possible

Should we multiply further?

So, the factors of 20 are .

4. Finding Factors Using Division

Think and Tell

No, we do not need to go beyond 6 to find more factors of 36.

5. Common Factors

Think and Tell

Sheela aunty stopped after making 4 plates as 4 children received an equal number of each type of fruit and there were no other combinations left.

Think and Tell

The lowest common factor of 4 and 8 is 1. The highest common factor of 4 and 8 is 4.

Fractions 6

Learning Outcomes

Students will be able to:

find the fraction of a whole and write it using symbols, find the fraction of a collection, identify halves, quarters and thirds of a whole or a collection. find equivalent fractions, write a fraction in its simplest form. identify like and unlike fractions, compare and order like fractions. compare and order unlike fractions. identify proper fractions, improper fractions and mixed numbers, convert mixed numbers to improper fractions, convert improper fractions to mixed numbers. add two like fractions, solve word problems on adding two like fractions. subtract two like fractions, solve word problems on subtracting two like fractions.

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 concept of a whole, halves and quarters. Ask students to solve the questions given in the Let’s Warm-up section.

Chapter: Fractions

Halves, Quarters and Thirds

Learning Outcomes

Imagine Maths Page 100

Students will be able to find the fraction of a whole and write it using symbols, find the fraction of a collection, identify halves, quarters and thirds of a whole or a collection.

Teaching Aids

Buttons; Circle cutouts divided into 4 equal parts and 3 equal parts; Crayons; Glue stick

Activity

Ask the students to work in groups. Distribute buttons (say, 24 buttons) and circle cutouts divided into 4 equal parts to each group.

Ask the students to find 1 4 of the total number of buttons. Instruct them to divide the total number of buttons into 4 equal groups and ask the value of one group.

Instruct the students to shade 1 4 of the circle cutout and stick 1 4 of the buttons on the shaded part.

Instruct the students to write the fraction inside the circle.

Repeat the activity to find 2 3 of the buttons.

Extension Idea

Ask: If a quarter of the buttons are further divided into halves, then how many buttons will there be in each equal part?

Say: The total number of buttons is 24. 1 4 of 24 is 6 and half of 6 is 3. So, there will be 3 buttons in each equal part.

Equivalent Fractions

Learning Outcomes

Students will be able to find equivalent fractions, write a fraction in its simplest form.

Imagine Maths Page 104

Teaching Aids

Pairs of circle cutouts divided into 4 and 8 parts

Activity

Divide the class into groups of 4. Distribute the circle cutouts to each group.

Instruct the students to shade 1 4 , 2 4 , 2 8 and 4 8 of the circle cutouts and write the shaded fraction inside the circle. Ask them to compare the 1 4 shaded cutout with the 2 8 shaded cutout and the 2 4 shaded cutout with the 4 8 shaded cutout. Bring out the fact that the compared fractions are equivalent fractions as they represent the same value. Help them find more equivalent fractions for 2 8 and 4 8 by multiplying or dividing the numerator and denominator by the same whole number in their notebooks. Ask the students whether they can divide the numerator or denominator of the fraction 1 4 further by any number. Discuss the simplest form of the fractions.

Ordering and Comparing Like Fractions

Learning Outcomes

Imagine Maths Page 107

Students will be able to identify like and unlike fractions, compare and order like fractions.

Teaching Aids

Rectangular strips; Crayons

Activity

Begin by discussing like and unlike fractions. Bring out the fact that like fractions have the same denominator and unlike fractions have different denominators.

Ask the students to work in groups. Distribute rectangular strips of equal lengths to each group.

Instruct the students to represent 4 7 and 5 7 on the rectangular strips by drawing and shading the fractions.

Instruct them to place the rectangular strips one below the other and compare the shaded parts. Ask them to say which fraction is bigger by looking at the strips.

Explain that for like fractions, the greater the numerator, the greater the fraction. Instruct the students to write the answer in their notebooks.

Ordering and Comparing Unlike Fractions

Imagine Maths Page 108 with Same Numerators

Learning Outcomes

Students will be able to compare and order unlike fractions.

Teaching Aids

Rectangular strips of equal length divided into 5, 7, 9 and 11 parts; Crayons

Activity

Ask the students to work in groups. Distribute rectangular strips of equal length divided into 5, 7, 9 and 11 parts to each group.

Instruct the students to shade 2 parts in each rectangular strip and write the fractions for the shaded parts of these strips in their notebooks.

Then, ask them to arrange all the rectangular strips one below the other and look at the shaded parts.

Ask the students to compare and order the fractions and write the answers in the notebooks. Explain that for unlike fractions with the same numerators, the smaller the denominator, the greater the fraction.

Extension Idea

Ask: Riya ate 4 8 of a pizza and Swati ate 3 6 from the same size of pizza. Who ate more pizza?

Say: Simplest form of 4 8 = 1 2 and simplest form of 3 6 = 1 2 . Hence both of them ate same amount of pizza.

Proper and Improper Fractions

Learning Outcomes

Imagine Maths Page 111

Students will be able to identify proper fractions, improper fractions and mixed numbers, convert mixed numbers to improper fractions, convert improper fractions to mixed numbers.

Teaching Aids

Sheets with unshaded fraction circles divided into 6 parts

Activity

Ask the students to work in groups.

Distribute sheets with the circles divided into 6 equal parts to each group.

Ask the students to show 4 6 by shading the first circle and 10 6 by shading the second and third circles. As them to write the fractions below the circles. Discuss proper and improper fractions based on the shaded circles.

Help the students write the mixed fraction as a combination of a whole number and a fraction. Discuss the division and multiplication methods to convert an improper fraction to a mixed fraction and vice versa. Repeat the activity for some other fractions. Ask the students to make the conversions and write the answers in their notebooks.

Extension Idea

Ask: How many 2 5  s are there in 1 3 5 ?

Say: 1 3 5 = 8 5 . So, there are four 2 5  s in 1 3 5 .

Adding Fractions

Learning Outcomes

Imagine Maths Page 115

Students will be able to add two like fractions, solve word problems on adding two like fractions.

Teaching Aids

Fraction strips; Crayons

Activity

Ask the students to work in groups. Distribute fraction strips to each group. Instruct the students to divide the fraction strip into 6 equal parts. Ask them to find 1 6 + 4 6 by shading 1 6 of the fraction strip with one colour and 4 6 more in the same strip with another colour.

Ask the students what fraction of the total strip is shaded. Instruct them to write the answer in their notebooks.

Give the students a word problem. Raju has 1 8 m of cloth and he buys 5 8 m more. What is the total length of cloth with Raju?

Instruct the students to use the fraction strips and write the answers in their notebooks.

Extension Idea

Ask: What is 2 7 + 3 7 + 5 7

Say: 2 7 + 3 7 + 5 7 = 2 + 3 + 5 7 = 10 7 = 1 3 7

Learning Outcomes

Students will be able to subtract two like fractions, solve word problems on subtracting two like fractions.

Teaching Aids

Fraction strips; Crayons

Activity

Ask the students to work in groups. Distribute fraction strips to each group. Instruct the students to divide the fraction strip into 8 equal parts and then to find 5 8 –2 8 .

Tell them to shade 5 8 on the fraction strip, and then cross out 2 8 of the shaded part.

Then, ask the students what fraction of the total strip is shaded. Instruct them to write the answers in their notebooks.

Give the students a word problem. Sid had 5 6 litres of paint. He used 1 6 litres of paint. How much paint was he left with?

Instruct the students to use the fraction strips to find the amount of paint left. Ask them to write the answers in their notebooks.

Extension Idea

Ask: Find the missing fraction: 3 1 6 –

. Thus, the missing fraction is 2 2 6 .

Answers

1. Halves, Quarters and Thirds

Do

It Together

Total number of ice creams = 36

Fraction of ice creams sold = 5 6

Number of ice creams sold = 5 6 of 36

Step 1: Total number of ice creams = 36; Denominator = 6  Numerator = 5

Step 2: Hence, the ice-cream seller sold 30 ice-creams.

2. Simplest Form of a Fraction

Think and Tell

We have not divided the numerator and denominator by the common factor 1 because it does not have any effect after division as any number divided by 1 is the number itself.

Do It Together

Equivalent fraction of 5 25 = 5 × 2 25 × 2 = 10 50

The simplest form of 5 25 can be given as:

Factors of 5 = 1, 5

Factors of 25 = 1, 5, 25

Common factors of 5 and 25 = 1 and 5

Dividing the numerator and denominator with the common factor 5 we get, 5 ÷ 5 25 ÷ 5 = 1 5

3. Ordering and Comparing Like Fractions

Think and Tell

No, equivalent fractions cannot be like fractions because like fractions have the same denominator and equivalent fractions have different denominators (but the same value).

Do It Together

As the denominators of the fractions are the same, the given set of fractions are like fractions.

On comparing and arranging the numerators in ascending order, we get: 2 < 4 < 7 < 8

The given fractions can be arranged in ascending order as: 2 9 < 4 9 < 7 9 < 8 9

4. Ordering and Comparing Unlike Fractions with Same Numerators

Do It Together

As the denominators of the fractions are different, the fractions are unlike fractions.

Since the numerators are the same, we compare and arrange the denominators in ascending order, we get:

3 < 5 < 10 < 12 < 15

The fractions can be arranged in descending order as:

7 3 > 7 5 > 7 10 > 7 12 > 7 15

5. Testing for Equivalence by Cross-multiplying

Do It Together

6. Changing Improper Fractions to Mixed Numbers

7. Changing Mixed Numbers to Improper Fractions

Do It Together

Step 1: Whole number part × Denominator = 4 × 6 = 24

Step 2: Product + Numerator = 24 + 3 = 27

Step 3: Improper fraction = Product + Numerator Denominator = [27] 6

8. Whole Numbers as a Fraction

1. 2.

Do It Together 6 [ 6 ] = 1 [ 20 ] 5 = 4

9. Adding Fractions

Do It Together

Weight of bananas = 2 1 4 kg = 9 4 kg (Converting mixed number to improper fraction)

Weight of apples = 1 1 4 kg = 5 4 kg

Total weight = 9 4 + 5 4 = 14 4 kg

Converting 14 4 kg to a mixed fraction, we get 31 2 kg.

10. Subtracting Fractions

Do It Together

Length of ribbon bought = 2 1 5 m = 11 5 m (Converting mixed number to improper fraction).

Length of ribbon used = 2 5 m

Length of ribbon left = 11 5 –2 5 = 9 5 m

Converting 9 5 m to a mixed fraction, we get 1 4 5 m.

7

Lines and 2-D Shapes

Learning Outcomes

Students will be able to:

identify and draw a point, a ray and a line. measure and draw line segments in centimetres.

identify types of curves as open, closed, simple and non-simple. classify the different types of polygons based on their sides and angles. identify the different parts of a circle and construct circles with given radii.

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

Let’s Recall

Recap to check if students know how to identify 2-D shapes and find them in 3-D shapes. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

radius: distance from the centre to the edge of the circle

diameter: a straight line that joins two points on the edge of a circle and passes through the centre circumference: the line that forms the outside edge of a circle

Teaching Aids

2 sets of cards each representing points, lines, line segments and rays; Straws of lengths 5 cm and 6 cm; Ruler; Sheets with numbered points; 4 sets of cards with details about the shape, its name, the number of sides, and the number of angles, respectively; Circle cutouts; Drawing sheets; Thread; Glue sticks; Compass; Pencil

Chapter: Lines and 2-D Shapes

Points, Rays and Lines

Learning Outcomes

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

Teaching Aids

2 sets of cards each representing points, lines, line segments and rays

Activity

Make stacks of 8 cards with 2 each representing points, lines, line segments and rays.

Ask the students to work in groups of 4. Shuffle and distribute a stack of cards to each group. Instruct group members to discuss what they notice about each card. Ask them to sort the cards based on the feature they noticed. Help them list the features in their notebooks.

Imagine Maths Page 123

Then, ask the students to draw points, line segments, lines and rays in their notebooks and label them. Discuss where we may see these in real life.

Line Segments

Learning Outcomes

Students will be able to measure and draw line segments in centimetres.

Teaching Aids

Straws of lengths 5 cm and 6 cm; Ruler

Activity

Distribute straws of lengths 5 cm and 6 cm to the students.

Imagine Maths Page 126

Instruct the students to measure the lengths of the straws using a ruler and write the measurements in their notebooks. Once they are done, ask them to draw line segments of the same lengths as the straws in their notebooks and mark the initial and end points of each line segment with letters such as A and B. Ask them to measure some more objects around them such as pencils, erasers etc. and draw line segments showing those lengths.

Extension Idea

Ask: Suppose you measured the pencil’s length with one end at 3 cm on the ruler and the other end at 11 cm. What is the actual length of the pencil?

Say: The actual length of the pencil is 11 cm – 3 cm = 8 cm.

Simple and Non-simple Figures

Learning Outcomes

Students will be able to identify types of curves as open, closed, simple and non-simple.

Teaching Aids

Sheets with numbered points

Activity

Lead the students to the playground. Instruct them to form a circle by joining hands. Once the circle is established, say that the circle is complete and closed. Ask some students to step out while the rest remain in place, creating an incomplete circle. Say that the circle is now open.

Explain that if the starting point coincides with the end point, the curve is considered closed; conversely, if they differ, the curve is open.

Lead the students back to the classroom and ask them to work in pairs. Distribute the sheets with numbered points to each pair.

Explain that you will call out numbers from 1 to 12, one by one, and the students have to join the points without lifting their pencils. Call out the numbers in such a way so as to form one simple and one non-simple curve. Direct the students to identify the type of figure created through this observation and record their findings in their notebooks.

Extension Idea

Ask: Draw a non-simple open curve.

Say: This is an example of a non-simple open curve. Polygons Imagine Maths Page 133

Learning Outcomes

Students will be able to classify the different types of polygons based on their sides and angles.

Teaching Aids

4 sets of cards with details about the shape, its name, the number of sides, and the number of angles, respectively

Activity

Create 4 sets of cards for each shape, including one with a picture of the shape, a second with the name, a third with the number of sides and corners, and a fourth with one unique feature.

Divide the class into 4 groups. Distribute all the name cards to one group, picture cards to the second group, side/corner cards to the third group, and unique feature cards to the fourth group.

Instruct the students to move around and find the other three cards that match their shape. The first group of four students to correctly form a set wins. Repeat the activity by shuffling the cards among the students. Instruct the students to write the details of the respective shapes in their notebooks.

Learning Outcomes

Students will be able to identify the different parts of a circle and construct circles with given radii.

Teaching Aids

Circle cutouts; Drawing sheets; Thread; Glue sticks; Compass; Pencil

Activity

Ask the students to work in pairs. Distribute a circle cutout, thread, glue stick, drawing sheet, compass and pencil to each pair.

Instruct the students to fold the circle cutout in half and then unfold it. Ask them to rotate the cutout and fold it in half once more and then unfold it. Explain that the point where the creases intersect is identified as the centre of the circle, and that the crease itself represents the diameter. Explain that the radius of a circle is half of the diameter.

Ask the students to paste the thread at a point at one end of the diameter, encircling the cutout until it returns to that starting point. Explain that the looping thread is referred to as the circumference. Ask the students to mark the discussed parts on the cutout.

After the pairs are done with their labelling, demonstrate how to use a compass to draw a circle. Discuss why it is important to hold the needle of the compass steady at one point. Instruct the students to set the compass width to 3 cm using a ruler and draw a circle on the sheet. Ask them to label all the parts of a circle discussed so far on the circle they have drawn.

Extension Idea

Ask: The distance between the centre and the circumference of a circle is 5 cm. What is the diameter of the circle?

Say: The distance between the centre and the circumference of a circle is the radius. If the radius is 5 cm, then the diameter is 2 × 5 cm = 10 cm.

Answers

1. Points, Rays and Lines

Think and Tell

One line can pass through two given points.

Do It Together

The points are: A, B, C, D, E and F

The figure has rays: DE , AB and CF .

The lines in the figure are: Ɩ and m.

Line Ɩ is the same as FC and line m is the same as AD.

2. Line Segments

Think and Tell

Yes, a ray and a line segment can be part of the same line.

Think and Tell

Yes, a line segment can be measured using the ruler whose zero mark is missing. We just need to keep in mind the point where it starts from.

Do It Together

The length of the comb is 7 cm.

3. Open and Closed Figures

Do It Together

3. 4.

Closed figure Open figure

4. Simple and Non-simple Figures

Do It Together

1. 2. 3. 4.

Simple figure Non-simple figure Non-simple figure Simple figure

5. Polygons

Do It Together

1. 2. 3. 4.

Pentagon Quadrilateral Hexagon Triangle

6. Parts of a Circle

Think and Tell

A circle can have infinite diameters.

Think and Tell

Yes, the radius of a circle does impact its circumference. The greater the length of the radius, the greater the circumference.

Do It Together

Centre Radii Diameter OA, , and O OB OC OD BC

Representing 3-D Shapes 8

Learning Outcomes

Students will be able to: draw different views of objects. identify the nets of a cube and a cuboid and draw them. read a map and answer questions on it.

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 about basic 2-D and 3-D shapes. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

nets: 2-D representations of solid shapes map: a picture of a place printed on a flat surface

Teaching Aids

Book; Water bottle; Pencil box; Cake box shaped like a cube; Cake boxes shaped like a cuboid; Thick chart paper; Word cards labelled with names of places like Hotel, Park, Market, etc.; Chart paper; Glue stick

Chapter: Representing 3-D Shapes

Views of Objects

Learning Outcomes

Students will be able to draw different views of objects.

Teaching Aids

Book; Water bottle; Pencil box

Activity

Imagine Maths Page 143

Show the students a glass and turn it around to show how it looks from different sides. Ask questions on how the glass looks from the top, side and front. Draw its 3 views on the board.

Divide the class into groups of 3. Distribute 3 objects to each group (any other object available in class can also be used).

Instruct the groups to place the objects one by one on the table and draw different views of different objects. Each student in the group should draw the side view of one object, the top view of the second object and the front view of the third object. Allow them to turn the objects around to see the view that they are drawing. Ask the students to draw the different views in their notebooks.

Extension Idea

Ask: What is one object that has the same front, side and top view?

Say: Any object that is shaped like a sphere or a cube has the same front, side and top view. For example, a sugar cube.

Nets of 3-D Shapes

Learning Outcomes

Students will be able to identify the nets of a cube and a cuboid and draw them.

Teaching Aids

Cake box shaped like a cube; Cake boxes shaped like a cuboid; Thick chart paper

Activity

Imagine Maths Page 146

Show the students the cube-shaped cake box and open it out. Ask the students what they see. Then, close it up to form the box again.

Discuss how a 3-D shape can be represented in 2-D form and what nets are.

Ask the students to work in groups of 4. Distribute the chart paper and cuboidal boxes to the students.

Ask them to look at the box, visualise how the box would look when opened out and draw its net on thick chart paper. Each student should draw a net and then compare the nets with the other members of their group. They should then fold the net to see the box that they get. Finally, they should open out the cuboidal box to see if the net that they have drawn have the same 2-D shapes that are there in the box.

Ask questions like: Is it possible to draw different nets for the same solid shape?

Extension Idea

Ask: If you join two cubes end to end then what will be the net of the new shape formed?

Say: When two cubes are joined end to end, a cuboid is formed. So, the net will be that of a cuboid.

Learning Outcomes

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

Teaching Aids

Word cards labelled with names of places like Hotel, Park, Market, etc.; Chart paper; Glue stick

Activity

Ask the students to work in groups. Distribute word cards labelled with different places to each group. For example, distribute cards labelled Hotel, Hospital, Post Office, Park, School, etc.

Hotel Park Hospital Post Office School Market

Instruct the students to create a map by pasting the word cards on chart paper. Give them instructions on where to place word cards such that all the groups get the same map, like the given sample.

Ask questions like: How will you reach the school from the park? Which direction is the hotel from the post office? Instruct the students to write the answers in their notebooks. Discuss the importance of maps.

Extension Idea

Park Hotel Hospital Post Office Market

Ask: What would be the actual distance between the hotel and the hospital if the distance between them is 5 cm on the map? 1 cm on paper = 3 km.

Say: 1 cm = 3 km. So, the actual distance is 5 × 3 = 15 km.

1. Views of Objects

Think and Tell

No, not all objects look different when seen from different views. Some objects shaped like a cube or a sphere look the same from all the views.

2. Nets of 3-D Shapes

Do It Together

3. Maps

Think and Tell

No, Riya will not follow the same route if she is standing outside the supermarket.

Do It Together

Mohan's House

Patterns and Symmetry

Learning Outcomes

Students will be able to:

identify the rule in a repeating or rotating pattern and extend the pattern. identify the rule in a growing or reducing pattern and extend the pattern. identify the rule in a number pattern and extend the pattern. identify shapes that tile and extend a tiling pattern. decode puzzles or secret messages and solve them. identify and draw lines of symmetry in shapes and figures.

draw the reflection of a shape or figure along the mirror line.

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

Vocabulary

number pattern: sequence of numbers in a fixed repetitive way

tessellations: patterns made by repeating and fitting together shapes without any gaps or overlaps symmetry: an arrangement where a line can be drawn dividing a figure into two identical parts

Teaching Aids

2 buntings showing a pattern and a non-pattern made with Christmas decorations; Christmas decorations (baubles, stars, bells, etc.); Pieces of string; Cut vegetables such as okra, onion, mushroom, broccoli, etc.; Paint in different colours; Sheets of paper; Sudoku grids; Cardboard cutouts of different shapes; Blank slips of paper; Cards depicting everyday objects including shapes and figures; Grid paper; Small mirrors

Chapter: Patterns and Symmetry

Repeating Patterns

Learning Outcomes

Imagine Maths Page 157

Students will be able to identify the rule in a repeating or rotating pattern and extend the pattern.

Teaching Aids

2 buntings showing a pattern and a non-pattern made with Christmas decorations; Christmas decorations (baubles, stars, bells, etc.); Pieces of string

Activity

Show the two buntings—one forming a pattern and the other forming a non-pattern. Ask students to say what they notice about the two buntings.

Discuss what patterns and non-patterns are and how the pattern shows repeating and rotating decorations.

Ask the students to work in groups. Distribute Christmas decorations and pieces of string to each group.

Instruct the students to create their own festive buntings using decorations and string, focusing on creating repeating and rotating patterns using the Christmas decorations. Discuss the variety of patterns created by the groups and celebrate the creativity within the class.

Teacher Tip: If decorations are not available, give the students cardboard cutouts with holes punched.

Growing and Reducing Patterns

Learning Outcomes

Imagine Maths Page 158

Students will be able to identify the rule in a growing or reducing pattern and extend the pattern.

Teaching Aids

Cut vegetables such as okra, onion, mushroom, broccoli, etc.; Paint in different colours; Sheets of paper

Activity

Discuss growing and reducing patterns as patterns where units increase and decrease respectively at each step. Ask the students to work in groups. Distribute the vegetable cutouts to each group. Instruct them to dip the cutouts in paint and imprint the shapes onto a sheet of paper, forming beautiful growing or reducing patterns. Encourage them to try different layouts such as vertical and horizontal. Initiate a class discussion and ask them to point out the unit that is added or reduced at each step.

Ask questions like: How do you know the pattern is growing? How do you know it is reducing?

Extension Idea

Ask: Can a pattern have both growing and reducing units? If yes, show such pattern.

Say: Yes, a pattern can have both growing and reducing units. One such pattern could be:

Learning Outcomes

Students will be able to identify the rule in a number pattern and extend the pattern.

Teaching Aids

Sudoku grids

Activity

Explain the rules of Sudoku of placing numbers 1 to 9 in each row, column, and 3×3 sub-grid without repetition. Guide the students through the initial steps of solving the Sudoku puzzle, demonstrating how to identify and fill in numbers based on the rules.

Organise the students into groups. Distribute Sudoku grids to each group. Allow them to solve the puzzles independently, encouraging them to record their answers in the provided blank spaces. Offer help as needed.

Facilitate a brief discussion on the number patterns they observed during the activity.

Extension Idea

Ask: Create an increasing pattern starting from 4, where each number is double the previous number. Say: The pattern would be 4, 8, 16, 32, ... Tiling Patterns and Tessellations Imagine Maths Page 164

Learning Outcomes

Students will be able to identify shapes that tile and extend a tiling pattern.

Teaching Aids

Cardboard cutouts of different shapes

Begin with a discussion on tiling patterns and tessellations. Explain that tiling patterns with no gaps and overlaps become tessellations. Show them some examples of tiling patterns.

Ask the students to work in groups.

Distribute shape cutouts to each group, such as hexagons, squares, random shapes, jigsaw puzzle pieces, etc.

Ask the students to arrange these cutouts to create tilling patterns. Help them note whether their patterns are tessellations or non-tessellations and write their findings in their notebooks.

Encoding and Decoding Patterns

Learning Outcomes

Students will be able to decode puzzles or secret messages and solve them.

Teaching Aids

Blank slips of paper

Activity

Imagine Maths Page 167

Begin by providing guidance on the key to encode, which involves assigning numerical values to each letter, following the pattern A = 1, B = 2, and so on up to Z.

Ask the students to work in pairs. Distribute two blank slips of paper to each pair.

25

81 16 16 25

Instruct one student in each pair to write an affirmation like “BE CALM” or “BE HAPPY” on their slip. Instruct the other student in each pair to write an encoded number for a different affirmation on their slip. Ask them to exchange slips, and encode or decode the message, depending on the slip they received.

Ask the students to record their answers in their notebooks. Engage them in a class discussion to share insights and decode the affirmations among themselves.

Extension Idea

Ask: If you were to share a secret message during a mission, what method or code would you choose to encode the message “MISSION ACCOMPLISHED”?

Say: You can use a code known as Caesar cipher. This involves shifting each letter of the message by a fixed number of positions in the alphabet. If we apply a Caesar cipher with a shift of 3 to the message “MISSION ACCOMPLISHED,” the encoded message would be: “PLVVLRQ DEERPSOLVKHG”

Symmetry

Learning Outcomes

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

Teaching Aids

Cards depicting everyday objects including shapes and figures

Activity

Imagine Maths Page 169

Ask the students to work in groups. Provide them with a collection of cards featuring everyday shapes and objects such as leaves, traffic signs, cups, butterflies, flags, squares, rectangles, etc.

Instruct them to observe and identify whether each shape or figure has symmetry. Explain the concept of figures having more than one line of symmetry. Guide them to fold the cards to discover symmetry in the figures. Instruct them to draw lines of symmetry using a pencil.

Ask questions like: Which figure had symmetry? How did you find out?

Extension Idea

Ask: Can you draw a rangoli pattern with 2 lines of symmetry? If yes, draw one such pattern. Say: Yes, we can draw rangoli patterns with 2 lines of symmetry. There can be many such patterns. One could be:

Learning Outcomes

Students will be able to draw the reflection of a shape or figure along the mirror line.

Teaching Aids

Grid paper; Small mirrors

Activity

Start by drawing a simple image on the board and showing its reflection using a mirror.

Ask the students to work in pairs. Give each student a sheet of grid paper with a line drawn down the centre and a small mirror.

Ask both students in the pair to draw a shape on one side of the line drawn. Once they are done, ask them to exchange drawings with their partners and let them complete the figure by drawing its mirror image on the other side of the line drawn.

Ask them to place the mirror along the line to check if their figure matches the reflection of the original shape.

Answers

1. Repeating Patterns

4

2. Growing and Reducing Patterns

Length, Weight and Capacity 10

Learning Outcomes

Students will be able to:

estimate and measure the length of an object. convert between metres and centimetres. estimate and measure the weight of an object. convert between grams and kilograms. estimate and measure the capacity of a container. convert between litres and millilitres.

Alignment to NCF

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

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

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

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 the units and tools to measure length, weight and capacity. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

estimate: to guess an answer that is close to the actual answer length: horizontal distance from one end to the other end weight: measurement of the heaviness of an object capacity: the amount that can be held in a particular space

Teaching Aids

Ruler; Measuring tape; 1 m strips of paper; 10 cm strips of paper; Glue stick; Grains of rice; Small digital weighing scale; Weights of 100 g, 200 g, 250 g and 500 g; Measuring cups for 1 L and 100 mL; Water bottles of capacity 200 mL, 700 mL and 1 L with no labels; Bottle of water

Chapter: Length, Weight and Capacity

Common Units of Length

Learning Outcomes

Students will be able to estimate and measure the length of an object.

Teaching Aids

Ruler; Measuring tape

Activity

Imagine Maths Page 179

Start by showing the students lengths of 1 mm, 1 cm, 1 m using measuring tools such as rulers and measuring tapes. Give them some examples of long distances to introduce kilometre as a unit.

Ask the students to work in groups. Distribute a ruler and measuring tape to each group.

Instruct the students to look at various objects around the classroom. Ask them to estimate the unit that they can use to measure their lengths. Ask them to guess the lengths of their book, pencil and desk and write the object name and its estimated length in their notebooks. Then, instruct them to use the ruler to measure the lengths of the objects. Ask them to write the actual measures next to the estimated measures and see how close these were to their estimates.

Extension Idea

Ask: What would be the length of a line formed by placing 5 pencils end to end of length 12 cm each?

Say: Length of 1 pencil = 12 cm; length of 5 such pencils = 5 × 12 = 60 cm. So, the line will be 60 cm long.

Interchanging the Units

Learning Outcomes

Students will be able to convert between metres and centimetres.

Teaching Aids

1 m strips of paper; 10 cm strips of paper; Ruler; Measuring tape; Glue stick

Activity

Imagine Maths Page 181

Ask the students to work in groups of 10. Distribute one 1 m strip of paper, ten 10 cm strips of paper, a ruler and a measuring tape to each group. Ask the groups to first measure the longer strip of paper using the measuring tape and then the 10 cm strips using a ruler. Discuss what measures they got in metres. Then, ask them to use the glue stick to paste the 10 cm strips across the length of the 1 m strip without leaving any gaps in between. Ask questions like: What is the relation between metres and centimetres?

Give some conversion problems to the students to solve in their notebooks. Discuss the answers.

Extension Idea

Ask: Find the relation between millimetres and centimetres using your ruler. How many mm is 4 cm?

Say: There are 10 mm lines between two cm lines, so 10 mm = 1 cm. 4 cm = 4 × 10 = 40 mm.

Common Units of Weight

Learning Outcomes

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

Teaching Aids

Grains of rice; Small digital weighing scale

Activity

Set up a weighing station in the classroom with a digital weighing scale. Start by showing the students the weight of some objects like a few grains of rice, a pencil, and a thick book on the scale. Allow students to come up and read the measures on the weighing scale.

Ask the students to work in groups.

Instruct them to look at various objects around the classroom such as a piece of paper, an eraser, a lunch box, a water bottle, etc. Ask them to estimate the unit that they can use to measure their weights, like mg, g or kg. Ask them to guess the weight of their Imagine Mathematics book, school bag and pencils and write the object name and its estimated weight in their notebooks.

Then, instruct the students to come up and use the weighing scale to measure the weight of the objects. Ask them to write the actual measures next to the estimated measures and see how close these were to their estimates.

Ask questions like: What is lighter: 1 kg of cotton or 1 kg of iron nails?

Interchanging the Units

Learning Outcomes

Students will be able to convert between grams and kilograms.

Teaching Aids

Small digital weighing scale; Weights of 100 g, 200 g, 250 g and 500 g

Activity

Set up a weighing station in the classroom with a digital weighing scale.

Ask the students to work in groups.

Imagine Maths Page 186

Instruct the students of each group to come up to the weighing station and place the gram weights on the scale to make up an exact weight of 1 kg. Once a group gets 1 kg on the scale, ask them to write the measures of the weights used in their notebooks. For example, if a group uses two 500 g weights, they will write 500 g, 500 g = 1 kg.

Once all the groups do the activity, ask them all to add the weights.

Discuss their answers and ask them to find a relation between grams and kilograms. Also emphasise that there can be multiple combinations of grams that can be used to make 1 kg but the total weight in grams will always add up to 1000 g to make 1 kg.

Ask them to write the relation in their notebooks.

Give the students some measures in kg like 4 kg and 7 kg. Ask them to convert the weights to grams in their notebooks. Discuss the answers.

Extension Idea

Ask: What is heavier: A bag weighing 5 kg 650 g or a bag weighing 5700 g?

Say: 5 kg 650 g = 5000 g + 650 g = 5650 g. 5700 > 5650; hence, the bag weighing 5700 g is heavier.

Learning Outcomes

Students will be able to estimate and measure the capacity of a container.

Teaching Aids

Measuring cups for 1 L and 100 mL; Water bottles of capacity 200 mL, 700 mL and 1 L with no labels; Bottle of water

Activity

Show the students a measure of 10 mL of water using the millilitre cup and then 500 mL of water and 1 L of water using a litre cup.

Ask the students to work in groups. Distribute measuring cups, water bottles and empty bottles of different capacities to each group.

Instruct the students to first estimate the capacity of each empty bottle and write down their estimates in their notebooks. Then, they should measure the capacity of each bottle by filling it with water and then pouring the water into the measuring cups. Tell them to read the measure written on the measuring cup at the level up to where it is filled. Ask them to write the actual capacity next to the estimated one and see how close these were to their estimates.

Ask questions like: Which of these has a capacity of more than 500 litres: a bathtub, a bucket of water, or a large swimming pool?

Interchanging the Units

Learning Outcomes

Students will be able to convert between litres and millilitres.

Teaching Aids

Measuring cups for 1 L and 100 mL; Bottle of water

Activity

Imagine Maths Page 190

Ask the students to work in groups. Distribute the measuring cups and a bottle of water to each group. Instruct the students to use the 100 mL cup to pour water into the 1 L cup as many times as it takes to fill the 1 L cup and count the number of times the 100 mL cup was used.

Discuss how many 100 mL cups were used to fill the 1 L cup. Ask the students to multiply the capacity of one small cup by the total number of cups used to find the relation between litres and millilitres in their notebooks.

Ask questions like: You have a water bottle that holds 500 mL and your friend has a bottle that holds 1 2 L. Whose bottle holds more water?

Extension Idea

Ask: What is more: 4 L 540 mL or 5440 mL?

Say: 4 L 540 mL = 4000 mL + 540 mL = 4540 mL. 5440 > 4540; hence; 5440 mL is more.

Answers

1. Common Units of Length

Think and Tell

We would use a measuring tape to measure a road as rulers can measure the length in cm but measuring tapes are used for longer lengths in metres.

Do It Together

1.  a thread  a ruler  a measuring tape

2. The length of 1 pipe = 7 m.

The length of 5 pipes = 5 × 7 m = 35 m.

So, the total length of the pipe that the plumber needs to fit is 35 m.

2. Interchanging the Units

Do It Together

The gardener needs to measure a length of 600 cm. We know that, 100 cm = 1 m

So, 600 cm = 6 m

The garden needs to use a metre long stick 6 time to mark the lines. 600 cm

1 m 1 m 1 m 1 m 1 m 1 m

3. Common Units of Weight

Do It Together

1. 1 kg 3 kg 2 1 2 kg

2. Total weight of fruit = 1 kg + 3 kg + 2 1 2 kg

So, Raj bought a total of 6 1 2 kg fruit today.

3. 1 kg of pears = ₹220

The price of 2 kg of pears = 2 × ₹220 = ₹440

4. Interchanging the Units

Do It Together

Total weight of 4 apples = 1 kg 40 g

Remember, 1 kg = 1000 g

Total weight of 4 apples in g = 1000 g + 40 g = 1040 g

Now, the weight of 1 apple in gram would be: 1040 g ÷ 4 = 260 g.

5. Common Units of Capacity

Do It Together

1.

a. b. c. d.

2. 400 mL + 500 mL + 700 mL + 300 mL = 1900 mL

3. To get 1000 mL of coloured water in one jug, Rehaan, would need to empty jug c and d completely.

6. Interchanging the Units

Do It Together

Total juice in the bottle = 2 L

We know that 1 L = 1000 mL.

So, a bottle of 2 L capacity contains 2 × 1000 = 2000 mL.

So, if Rima poured 200 mL into a glass, the remaining juice in the bottle is:

2000 mL – 200 mL = 1800 mL

Remember, 1000 mL = 1 L.

So, 1800 mL = 1 L 800 mL. This will be the juice remaining in the bottle.

Perimeter and Area 11

Learning Outcomes

Students will be able to:

find the perimeter of different shapes on squared paper. find the length of the missing side of a polygon, when the perimeter and length of the other sides are given. find the area of irregular shapes on squared paper.

Alignment to NCF

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

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 convert length from one unit to another. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

perimeter: total distance covered along the boundary of a closed figure

area: total space covered by a closed figure

Teaching Aids

Piece of string; Ruler; Triangle and rectangle cutouts; Squared paper with squares of 1 cm; Cutout of a figure drawn with lengths written inside and one missing length; Squared paper; Coloured pencils; Leaves from the playground

Chapter: Perimeter and Area

Perimeter of Polygons

Learning Outcomes

Students will be able to find the perimeter of different shapes on squared paper.

Teaching Aids

Imagine Maths Page 197

Piece of string; Ruler; Triangle and rectangle cutouts; Squared paper with squares of 1 cm

Activity

Instruct the students to pick up their Imagine Mathematics book. Ask them to move their fingers around the edges of the front of the book. Explain that this is the boundary of the front cover and that the distance around the boundary is called its perimeter.

Instruct the students to form groups of 3. Distribute the squared paper.

Ask the students to draw the outline of a figure on squared paper and pass that figure to the group next to them. Each group will then find the perimeter of the shape on the squared paper.

Discuss the answers with the students. Then, distribute the shape cutouts.

Instruct the students to measure the perimeter of the shapes using a piece of string and a ruler. Ask them to write the measures in their notebooks.

Extension Idea

Ask: How will you draw a figure with a perimeter of 20 units on squared paper? Draw figures to show.

Say: Different figures can be drawn with a perimeter of 20 units. E.g. we can draw a rectangle by shading 7 squares along the length and 3 squares along the width.

Finding the Missing Side

Learning Outcomes

Imagine Maths Page 199

Students will be able to find the length of the missing side of a polygon, when the perimeter and length of the other sides are given.

Teaching Aids

Cutout of a figure drawn with lengths written inside and one missing length

Activity

Discuss how to find the length of a missing side when the lengths of the remaining sides and the perimeter are given.

Distribute the cutouts of a figure drawn with lengths written inside and one missing length. Instruct the students to find the length of the missing side in the cutout and write the answer in their notebooks.

Ask questions like: What is the missing length? How did you find it?

Area of Irregular Shapes

Learning Outcomes

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

Teaching Aids

Squared paper; Coloured pencils; Leaves from the playground

Activity

Distribute sheets of squared paper to the students.

Instruct the students to form groups of 3. Ask them to go out to the playground and pick 2 leaves. In the class, ask them to place the leaves on the squared paper, draw their outlines by tracing them, and shade the figures. Then, they will find the area of each figure drawn by taking the full and more than full shaded squares as 1, two halves as 1 and ignoring the less than half-shaded squares. Finally, they will write the area in their notebooks.

Extension Idea

Ask: The given figure shows a grass patch around a swimming pool. Which of these have more area?

Say: The area of the swimming pool is 6 square units and the area of the grass patch is 10 square units. So, the grass patch has more area.

Answers

1. Perimeter of Polygons

Think and Tell

Yes, we can find the perimeter of a shape with straight sides using only a ruler. We can first measure the length of the sides using the ruler and then add the lengths to find the perimeter.

Do It Together

A: Perimeter = 12 units

B: Perimeter = 16 units

C: Perimeter = 36 m

2. Finding the Missing Side

Do It Together

3 cm + 5 cm + 2 cm + 2 cm + 2 cm + 5 cm + 3 cm + the length of the missing side = 34 cm

This is the same as 22 cm + the length of the missing side = 34 cm

So, the length of the missing side = 34 cm – 22 cm = 12 cm

3. Area of Irregular Shapes

Think and Tell

Yes, we can trace any small shape on the grid and find its area by counting squares inside the shape.

Do It Together

1. 18 sq. units

2. 15 sq. units

Time 12

Learning Outcomes

Students will be able to:

read time to the nearest minute and write it in two ways.

read time on a 24-hour clock and write it as 00:00.

convert time from a 12-hour clock to a 24-hour clock and vice versa.

convert time between hours and minutes.

find elapsed time in days, weeks, months and years.

Alignment to NCF

C-8.10: Performs simple measurements of time in minutes, hours, days, weeks, and months

Let’s Recall

Recap to check if students know that our different activities can be measured in hours, minutes, and seconds.

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

Vocabulary

analogue clock: a clock that shows the time by the hour, minute, and second hands duration: the time from the start to the end of an activity

Teaching Aids

Large clock with movable hands; Paper plates; Markers; Big and small straws; Glue sticks; Analogue clock in 24-hour format with movable hands; Puzzle interlocking cards; Clock with minute markings; 2 sets of cards with the time written in minutes on one set and the corresponding time in hours on the other set; Classroom calendar; School’s event calendar

Chapter: Time

a.m. and p.m.

Learning Outcomes

Students will be able to read time to the nearest minute and write it in two ways.

Teaching Aids

Imagine Maths Page 214

Large clock with movable hands; Paper plates; Markers; Big and small straws; Glue sticks

Activity

Show the students the clock with the time set at 12 o’clock. Discuss how to read the time while moving the hour and minute hands. Show a few times, such as 5:35, on the clock.

Ask questions like: How do you know if the given time is 5:35 in the morning or evening? Discuss how to read time with a.m. and p.m. and their meanings.

Ask the students to work in groups of 3. Distribute paper plates, markers, big and small straws and glue sticks to each group.

Instruct the groups to mark the numbers as in a clock on the plate using markers. Ask them to show 5 o’ clock in the morning, 12:25 in the afternoon and 10:10 at night using the small and big straws as the hour and minute hands. Ask them to write the times in a.m. and p.m. in their notebooks.

Extension Idea

Ask: Tanya starts from home at 11:45 a.m. and reaches the market after 2 hours. Will she reach the market in a.m. time or p.m. time? What time will she reach?

Say: Two hours after 11:45 is 1:45, which is past noon. So, it will be p.m. time.

24-hour Clock

Learning Outcomes

Students will be able to read time on a 24-hour clock and write it as 00:00.

Teaching Aids

Analogue clock in 24-hour format with movable hands

Activity

Show the students the 24-hour clock.

Ask questions like: What are the two formats of time that you see on this clock?

Imagine Maths Page 217

Discuss the differences and similarities in the time formats. Explain that this clock shows the time in the 12-hour as well as the 24-hour format. Write on the board: 1:00 p.m. = 13:00 hours.

Ask the students to work in groups. Distribute the 12-hour clocks created by them in the previous lesson. Instruct them to convert the 12-hour clock to a 24-hour clock. Ask them to tell time in the 24-hour format for different times in the 12-hour format, such as 6 o’ clock in the evening, 10 p.m. at night, 1 o’ clock after midnight by looking at the clocks created. Instruct the students to note down each 12-hour time and its respective 24-hour format in their notebooks.

Changing 24-hour Clock to 12-hour

Learning Outcomes

Students will be able to convert time from a 12-hour clock to a 24-hour clock and vice versa.

Teaching Aids

Analogue clock in 24-hour format with movable hands; Puzzle interlocking cards

Activity

Show the students the 24-hour clock. Ask them if they can tell the time in the 24-hour format without looking at the clock. Explain on the board how to change 11:30 a.m. and 11:30 p.m. into the 24-hour format and vice versa.

Ask the students to work in groups of 3. Distribute the puzzle interlocking cards to each group.

Ask the students to look at the time in the 24-hour format and look for the same time in the 12-hour format and word format. Ask them to join the three interlocking cards to complete one puzzle. The group that completes all the puzzles first wins!

Time in Hours and Minutes

Learning Outcomes

Students will be able to convert time between hours and minutes.

Teaching Aids

14:30 hours

Half past two

18:45 hours

Imagine Maths Page 221

Clock with minute markings; 2 sets of cards with the time written in minutes on one set and the corresponding time in hours on the other set

Activity

Show the time on a 12-hour clock. Explain that 1 minute passes when the minute hand moves from one marking to another. The hour hand moves a full cycle from one number to the next, showing that an hour has passed.

Ask the students to count the number of minute markings on the clock.

Discuss how to change hours into minutes and vice versa.

Take the students out to an open ground.

Ask the students to work in 2 groups. Distribute 2 sets of cards with the time written in minutes on one set to group 1 and the corresponding time in hours on the other set to group 2. Ask the students to look at the time card with them and look for their partner having the same time card in another unit. Ask them to stand with their partners once they have found them.

Repeat the activity by shuffling and exchanging the cards among the groups.

Extension Idea

Ask: Rohan’s school starts at 8:00 a.m. and ends at 2:30 p.m. How many minutes does he spend in school?

Say: There are 6 hours 30 minutes between 8:00 a.m. and 2:30 p.m. So, Rohan spends 6 × 60 minutes + 30 minutes = 390 minutes in school.

Learning Outcomes

Students will be able to find elapsed time in days, weeks, months and years.

Teaching Aids

Classroom calendar; School’s event calendar

Activity

Begin by showing the students a classroom calendar. Ask questions like: How many weeks are there in a month? How many months are there in a year? How many days are there in a year? Instruct the students to write the relations between days, weeks, months and years in their notebooks. Ask the students to work in groups. Distribute the school’s event calendar to each group.

Instruct them to look at the calendar and find the duration of the events. Ask questions like: For how many days will the Science Fair last? For how many weeks will the Art & Craft Show last? How many months are there between the Science Fair and the Music and Dance Show?

Science Fair

School Events

Starting Monday, June 25th to July 2nd

Math Projects Exhibition

Starting Tuesday, August 20th to August 25th

Art & Craft Show

Starting Thursday, Nov. 26th to Dec. 8th

Music and Dance Show

Starting Monday, Jan. 29th to Feb. 4th

Instruct the students to find the elapsed time in each case and write it in their notebooks.

Extension Idea

Ask: If it was a Wednesday on January 3, 2024, what was the date, month and year a week before?

Say: Move back on the calendar by 7 days. It was December 27, 2023 a week before.

Answers

1. a.m. and p.m.

Think and Tell

We may tell time by the time of day, such as 6:15 in the morning, 12 o’ clock noon, 5:45 in the evening, and 10:30 at night.

Do It Together

Wakes up at 6:10 a.m.

Eats breakfast at 8:20 a.m.

Comes back from school at 1:15 p.m.

Rides his bicycle at 4:15 p.m.

Goes to bed at 9:07 p.m.

Every morning, Manu wakes up at 6:10 a.m. He gets ready for school and eats breakfast by 8:20 a.m. Then, he walks to school. Manu’s school starts at 9 a.m. and ends at 1:15 p.m. Every afternoon, Manu rides his bicycle at 4:15 p.m. He goes to bed by 9:07 p.m.

2. Changing 12-hour to 24-hour Clock Time

Do It Together

1. Add 12 to the hour value: 5 + 12 = 17

2. Write the minutes as they are: 50

3. Replace p.m. with hours: 17:50 hours

Nura is on time.

3. Changing 24-hour Clock to 12-hour Clock Time

Do It Together

Time in the railway station clock = 18:00 hours.

It is more (less/more) than 12. So, it is p.m. (a.m./p.m.) time. Now, we subtract 12 from the first (first/last) 2 digits: 18 − 12 = 6.

Write down the minutes as they are. Thus, the time by a 12-hour clock is = 6:00 p.m.

4. Time in Hours and Minutes

Think and Tell

The answer will depend on the school timings.

Do It Together

Start time = 10:30 p.m.

End time = 06:30 p.m.

Duration in hours = 20

Duration in minutes = 20 hours × 60 = 1200 minutes

5. Time in Days, Weeks, Months and Years

Think and Tell

Total hours in a week = 24 × 7 = 168 hours

Total minutes in a week = 168 × 60 = 10,080 minutes

Do It Together

Total camping days = 15 days

Total days in December = 31 days. So, remaining days in December = 31 – 24 = 7 days.

In the next month of January, remaining days for camping = 15 – 7 = 8 more days.

The date of return = 8 January.

Money 13

Learning Outcomes

Students will be able to:

write money amounts in words and figures. convert between rupees and paise. read a bill and answer questions based on it. make a bill when the item name, unit cost of each item and number of items bought are given. make an expense list for the expenses incurred over a given period of time. solve word problems on adding, subtracting, multiplying and dividing money amounts.

Alignment to NCF

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

Let’s Recall

Recap to check if students know about currency notes and coins and how to count them to get an amount. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

money: the coins or paper notes of a country used to buy things bill: details of how much we need to pay for items or services bought expense: money spent on buying different things or services

Teaching Aids

5 envelopes with play money of 5 different amounts; Cards with rupee and paise amounts written; A stationery bill for 3 items; Printouts of a blank bill template; 5 items such as marbles, pencils, flowers, bottles, and candies, with price tags; Slips with expenses written; Word problem sheets

Chapter: Money

Express Money in Words

Learning Outcomes

Students will be able to write money amounts in words and figures.

Teaching Aids

5 envelopes with play money of 5 different amounts

Activity

Imagine Maths Page 233

Show the students play money notes and coins adding up to ₹20.50. Discuss how to write the amount in 2 different ways—in figures as ₹20.50 and in words as twenty rupees and fifty paise.

₹20.50

Twenty rupees and fifty paise

Ask questions like: What do the numbers on the left and right of the dot show?

Ask the students to work in groups. Distribute play money envelopes to each group.

Instruct the students to count the money in each envelope one by one and write the amount in words and figures in their notebooks. Ensure that each student in a group counts the money in one envelope.

Conversion Between Rupees and Paise

Learning Outcomes

Students will be able to convert between rupees and paise.

Teaching Aids

Cards with rupee and paise amounts written

Activity

Imagine Maths Page 234

Ask: I want to buy a toffee for 100 paise. How much will it cost in rupees? Explain that 100 paise make ₹1 and 200 paise make ₹2.

Ask the students to work in groups. Distribute the rupee cards and paise cards to each group.

₹125.05

₹12.50 ₹13.87 ₹31.78

12505p 1250p 1387p 3178p

Instruct the students to pick a rupee card and write the amount in their notebooks. Then, they should convert the rupee amount to the paise amount and look for a card with a paise amount that matches the rupee amount. They need to match all the rupee amount cards to their corresponding paise amount cards showing the same amount.

The group that matches all the cards first wins.

Instruct the students to write all the rupee and paise amounts in their notebooks.

Reading Bills

Learning Outcomes

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

Teaching Aids

A stationery bill for 3 items

Activity

Show the students the stationery bill. Discuss the different elements of a bill.

Ask the students to work in groups. Distribute a bill to each group. Ask them to read the details of the bill.

Ask questions like: When was this bill created? How many items were bought from the store? What was the cost of each brush? Which item cost the least?

Imagine Maths Page 236

Instruct the students to discuss these questions in groups and find the correct detail on the bill to answer each question. Also ask them to find the total bill amount.

Teacher Tip: The students can be asked to bring a bill of a purchase made by their parents to class and then read the bill in class to answer some general questions.

Extension Idea

Ask: If the total cost of 5 pencil cases is ₹500, what is the rate of 1 pencil case?

Say: The rate of 1 pencil case = ₹500 ÷ 5 = ₹100.

Making Bills

Learning Outcomes

Imagine Maths Page 238

Students will be able to make a bill when the item name, unit cost of each item, and number of items bought are given.

Teaching Aids

Printouts of a blank bill template; 5 items such as marbles, pencils, flowers, bottles, and candies, with price tags

Activity

Display a banner for ABC Stationers. Place the items on the table. Pretend to be the shop owner. Explain that the students will be the shoppers. Distribute the bill templates.

Ask the students to work in groups of 5. Each group will send 1 student up to the table to pick 3 items that they need to buy.

Instruct each student to make a bill for the 3 items selected for their group. Ask them to discuss if they all got the same bill amount.

Discuss the bill amounts at the end of the activity.

Extension Idea

Ask: If the cost of 2 kg of potatoes is ₹60, what is the cost of half a kg of potatoes?

Say: The cost of 2 kg of potatoes is ₹60, which means the cost of 1 kg is ₹30. So, the cost of half a kg of potatoes is ₹15.

Expense List

Learning Outcomes

Imagine Maths Page 241

Students will be able to make an expense list for the expenses incurred over a given period of time.

Teaching Aids

Slips with expenses written

Activity

Instruct the students to read how to make expense lists on Page 241 of the Imagine Mathematics book, and draw a template in their notebooks.

Ask the students to work in groups. Distribute the expense slips to each group.

Instruct each group to imagine that they are going on a trip and have ₹2000. Ask them to make their own expense list using at least 3 expense slips they want to include in their list, making sure that the total amount does not exceed ₹2000.

Ask questions like: How much did you save after all the expenses?

Extension Idea

Ask: You have spent ₹1725 on a holiday. What would be your total expense if you go on 3 such trips, spending the same amount on each trip?

Say: Expense on one trip = ₹1725. So, the expense of 3 such trips would be 3 × ₹1725 = ₹5175.

Word Problems on Money

Learning Outcomes

Imagine Maths Page 243

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

Teaching Aids

Word problem sheets

Activity

Distribute sheets with the given word problem written on each: I bought a bottle for ₹125, a lunch box for ₹100 and a book for ₹225. How much did I spend in all?

Instruct the students to discuss in their groups: What is given? What is asked? and How do we solve?

Ask the students to solve the problem on the sheet. Repeat the activity with a new word problem.

I bought a bottle for ₹125, a lunch box for ₹100 and a book for ₹225. How much did I spend in all?

What is given?

Answers

1. Express Money in Words

Do It Together

2. False

Do It Together

3. True

2. Conversion Between Rupees and Paise

Do It Together

₹635.23

₹4126.24

₹5386.15

paise

paise

paise ₹8256.32

3. Reading Bills

paise

Do It Together Bill No. 9812

Dan’s Stationery Bill Date:

5. Expense List

Do It Together

4.

5.

2. Cost of 1 packet of eraser = ₹50

Cost of 2 packets of erasers = 50 × 2 = ₹100

3. Cost of 6 notebooks = ₹180.00

Cost of 1 notebook = ₹180 ÷ 6 = ₹30

4. Total amount to be paid = ₹80.00 + ₹50.00 + ₹180.00 + ₹70.00 + ₹100.00 = ₹480.00

5. Amount of change that Sudha will get = ₹500 – ₹480.00 = ₹20.00

4. Making Bills

Think and Tell

The cost of 1 kg of bananas = ₹40

So, the cost of 1 4 kg of bananas = ₹40 ÷ 4 = ₹10

Total expenditure = ₹8525

Amount saved = ₹10,000 – 8525 = ₹1475

6. Word Problems on Money

Do It Together

Amount spent on rent = ₹5000

Amount spent on electricity bill = ₹1800

Amount spent on food = ₹7000

Amount spent on miscellaneous = ₹2000

Total amount spent in a month = ₹15,800

Savings per month = Money earned – Money spent = ₹20,000 – 15,800 = 4200

Savings per year = ₹4200

Data Handling

Learning Outcomes

Students will be able to:

collect and organise data in a tally marks table. draw a pictograph to show data. read and interpret a pictograph and answer questions based on it. draw a bar graph to show data. read and interpret a bar graph and answer questions based on it. draw and interpret a pie chart.

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

Vocabulary

tally mark: a vertical mark used to keep count

pictograph: a table that shows the given data using pictures or symbols

bar graph: a graph that shows information in the form of bars of different lengths

pie chart: a graph in which a circle is divided into sectors and each represents a part of the whole

Teaching Aids

Sticky notes; Chart paper with a table drawn for collecting data; Square cutouts in different colours representing ice creams; Chart paper; Chart paper with a pictograph drawn on it; Squared paper; Coloured pencils; Sheets of paper with a bar graph drawn; Strips of paper; Pre-cut sectors of circles representing pie chart segments; Extra sections that do not fit into the pie chart; Glue

Chapter: Data Handling

Organising Data

Learning Outcomes

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

Teaching Aids

Sticky notes; chart paper with a table drawn for collecting data

Activity

Imagine Maths Page 250

Begin the activity by discussing the concept of tally marks and their use in data collection.

Instruct students to work in groups.

Distribute sticky notes to students in each group.

Ask them to write down their favourite place out of Hill station, Sea Beach, Historical places and Grandparent’s home to visit on it. Read out each place written on the sticky notes one by one and instruct each group to work together to draw tally marks on the chart paper according to the data collected.

Once all the sticky notes have been read and tallied, review the completed tally chart as a group. Discuss the results, emphasising the place with the highest and lowest tallies.

Ask questions like: Which place got the maximum votes?

Creating Pictographs

Learning Outcomes

Students will be able to draw a pictograph to show data.

Teaching Aids

Imagine Maths Page 254

Square cutouts in different colours representing ice creams (brown, pink, green, orange); Chart paper

Activity

Start by discussing what a pictograph is and how it represents data using pictures or symbols.

Distribute square cutouts of different colours representing ice creams to each student. Explain that each colour represents a different flavour: brown for chocolate, pink for strawberry, green for mint, and orange for mango. Instruct the students to count the number of each flavour and record their counts in a pictograph. Ask the students to draw the symbols for each flavour according to the counts. Ask students to use a scale of 2 to draw the symbols.

Ask questions like: How many symbols did you draw to represent the number of students who liked chocolate ice cream? Strawberry ice cream? Mint ice cream?

Extension Idea

Ask: If 40 students like chocolate ice cream, and the scale changes to 1 symbol = 10 ice creams, how many symbols will you draw to show the data?

Say: You can draw 40 ÷ 10 = 4 symbols to show the number of students who like chocolate ice cream.

Learning Outcomes

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

Teaching Aids

Chart paper with a pictograph to show app notifications of 4 people

Activity

Instruct students to work in groups. Distribute one chart paper (with a pictograph drawn on it) to each group.

Instruct each group to examine the pictograph and discuss among themselves how it represents the app notifications for Gary, Phillip, Vivian, and Shelly. Encourage them to count the symbols and make sense of the data.

Ask questions like: Which person had 24 notifications this week? How many notifications were received by Shelly and Gary? Which person had the most notifications this week?

Extension Idea

App Notifications in One Week

Key: = 4 notifications

Ask: Create a question based on the pictograph drawn on the chart paper. Say: You can create multiple questions on the given pictograph. One such question could be: How many more notifications did Shelly receive than Vivian?

Creating Bar Graphs

Learning Outcomes

Students will be able to draw a bar graph to show data.

Teaching Aids

Squared paper; Coloured pencils

Activity

Start by discussing how bar graphs are useful for displaying information. Introduce the various components of a bar graph. Draw a table on the board as shown.

Divide the students into groups, ensuring each group has squared paper and coloured pencils.

Instruct each group to create a bar graph on the squared paper based on the provided data. Explain that each square on the grid represents one mile, and students should colour in the appropriate number of squares to represent the miles ran for each month. Emphasise the importance of labelling the graph accurately, including the months and a scale for the number of miles.

Learning Outcomes

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

Teaching Aids

Sheets of paper with a bar graph drawn; Strips of paper

Activity

Review key components of a bar graph such as axes, bars, and labels. Distribute one sheet of paper (with a bar graph drawn) to each group.

Ask the students two questions based on the provided bar graph, such as: How many students chose math as their favourite subject? Which subject has the fewest number of students who chose it as their favourite?

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

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

Pie Charts

Learning Outcomes

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

Teaching Aids

Imagine Maths Page 270

Pre-cut sectors of circles representing pie chart segments; Extra sections that do not fit into the pie chart; Chart Paper; Glue

Activity

Start by discussing the purpose of pie charts in visually representing data and the importance of accurately constructing and interpreting them.

Draw a table on the board as shown.

Instruct students to work in groups. Provide each group with pre-cut sectors of circles that represent pie chart segments. Include some extra sections that do not fit into the pie chart. Instruct students to arrange the sectors to construct a pie chart to show the favourite book genre of students where 1 4 like story books, 1 2 like Science fiction, and the remaining like non-fiction.

Have students glue the sectors onto a chart paper to create their pie charts. Encourage creativity in colourcoding and labelling each section.

Ask each group to come up with two questions related to their pie chart, such as: What genre is the most popular among the surveyed individuals? What fraction of respondents prefer Non-Fiction books? Groups swap questions with another group to answer. Instruct each group to answer the questions they received.

Extension Idea

Ask: If the total number of students surveyed was 80, how many students liked story books?

Say: Number of students who liked story books = 1 4 × 80 = 20 students.

Answers

1. Organising Data

Do It Together

How many students voted for cricket? 17

How many students voted for basketball? 10

7 students voted for football. Represent this in tally marks:

How many students in total voted for their favourite sport? 36

2. Creating Pictographs

Think and Tell

A pictograph is a better way to represent data than data tables as it is visually appealing.

Do It Together

One represents 50 wall clocks.

Monday → 300 clocks = 300 ÷ 50 = 6

Tuesday → 350 clocks = 350 ÷ 50 = 7

Wednesday → 250 clocks =

Thursday → 400 clocks =

Friday → 300 clocks =

Saturday → 200 clocks =

Monday

Tuesday

Wednesday

Thursday Friday Saturday

3. Interpreting a Pictograph

Do It Together

1. Thus, the most candy boxes were sold on Friday

2. Number of boxes sold on Tuesday = 1 × 5 = 5

Difference between the number of candy boxes sold on Monday and on Tuesday = 25 – 5 = 20

Thus, 20 more candy boxes were sold on Monday than on Tuesday.

3. Total candy boxes = 23 × 5

So, the number of candy boxes sold in the entire week = 115

4. Creating Bar Graphs

Do It Together

5. Interpreting Bar Graphs

Do It Together

2. On which day were the most cars parked? Monday

3. Number of cars parked on Tuesday = 4

Number of cars parked on Thursday = 6

Difference between the number of cars parked in a lot on Tuesday and Thursday = 6 – 4 = 2

6. Pie Charts

Do It Together

1. In a circle chart, the large (L) size covers the largest part of the circle.

So, large size is sold the most.

2. In a circle chart, small and medium sizes cover equal parts of the circle.

So, small (S) and medium (M) sized T-shirts are sold in equal numbers.

3. So, the number of large-sized T-shirts = 1 2 of 600 = 300.

4. So, the number of small-sized T-shirts bought = 1 8 of 600 = 75.

5. So, the number of medium-sized T-shirts bought = 1 8 of 600 = 75.

6. So, the number of extra-large sized T-shirts bought = 1 4 of 600 = 150.

Chapter 1

4. 4563 → 60  5. 9958 → 9000

Do It Yourself

1A a. 17,372; Seventeen thousand three hundred seventy-two

b. 43,890; Forty-three thousand eight hundred ninety

c. 74,065; Seventy-four thousand sixty-five

d. 80,379; Eighty thousand three hundred seventy-nine

e. 33,510; Thirty-three thousand five hundred ten

2. a. Twelve thousand three hundred twenty-one - 12,321

b. Thirty-four thousand six hundred - 34,600 c. Seventy-eight thousand five - 78,005 d. Fifty thousand ten - 50,010

3. a. 56938

Place value of

9 = 9 × 100 = 900

Face value of 9 = 9

c. 25401

Place value of

5 = 5 × 1000 = 5,000

Face value of 5 = 5

b. 65899

Place value of

6 = 6 × 10000 = 60,000

Face value of 6 = 6

d. 89376

Place value of

6 = 6 × 1 = 6

Face value of 6 = 6

4. a. 36,789 has 8 tens b. 47,690 has 9 tens c. 32,478 has 7 tens

d. 67,698 has 9 tens

5. a. Value in thousands place in 45,687 = 5

b. Value in thousands place in 65,690 = 5

c. Value in thousands place in 78,483 = 8

d. Value in thousands place in 96,152 = 6

6. a. 40000 + 6000 + 300 + 20 + 2 = 46,322 b. 50000 + 0 + 700

+ 50 + 7 = 50,757 c. 70000 + 3000 + 0 + 60 + 1 = 73,061

d. 90000 + 6000 + 400 + 0 + 8 = 96,408

7. Thirty-eight thousand four hundred thirty-seven. 38437 = 30000 + 8000 + 400 + 30 + 7

8. The digit in tens place = 4

The digit in thousands place = 8

Digit at one’s place = (4 + 8) ÷ 2 = 6

Digit at hundreds place = 6 – 6 = 0

So, the required number = 8046.

1B

1. a. 1,97,637; One lakh ninety-seven thousand six hundred thirty-seven  b. 3,65,021; Three lakh sixty-five thousand twenty-one  c. 6,32,845; Six lakh thirty-two thousand eight hundred forty-five  d. 8,24,137; Eight lakh twenty-four thousand one hundred thirty-seven 2. a. Four lakh eighteen thousand three hundred = 4,18,300

b. Six lakh twenty thousand = 6,20,000

c. Eight lakh five thousand two hundred sixty-four = 8,05,264

d. Seven lakh twenty thousand fifty = 7,20,050

3. a. The place value of the digit 5 in the number 2,05,649 is five thousand. FALSE b. In the number 3,42,658, the place value of the digit 3 is 3 × 1,00,000. FALSE c. The place value of the digit 5 in the number 5,49,803 is 5,00,000 which is 4,90,214 more than 9786. TRUE  4. a. Place value of digit 5 is 500000, 8 is 80000, 4 is 4000, 7 is 700, 3 is 30, 6 is 6; Expanded form: 500000 + 80000 + 4000 + 700 + 30 + 6 b. Place value of digit 7 is 700000, 4 is 4000, 3 is 300, 9 is 90, 1 is 1; Expanded form: 700000 + 4000 + 300 + 90 + 1 c. Place value of digit 3 is 300000, 7 is 70000, 9 is 900, 4 is 40, 3 is 3, Expanded form: 3,70,943 = 3 × 1,00,000 + 7 × 10,000 + 0 × 1,000 + 9 × 100 + 4

× 10 + 3 × 1 d. Place value of digit 9 is 900000, 8 is 80000, 5 is 5000, 4 is 400, 1 is 1; Expanded form: 9,00,000 + 80,000 +

5000 + 400 + 1  5. a. 400000 + 10000 + 8000 + 200 + 20 + 2 = 4,18,222 b. 500000 + 40000 + 0 + 100 + 40 + 7 = 5,40,147

c. 700000 + 40000 + 9000 + 0 + 20 + 1 = 7,49,021 d. 900000 + 80000 + 2000 + 900 + 0 + 2 = 9,82,902  6. The place value of 1 = 100000, 8 = 80000, 4 = 4000, 8 = 800, 0 = 0, 0 = 0, One lakh eighty-four thousand eight hundred.

7. The digits in the Tens and the Thousands place is 3 and 9.

The digit in the Ones place = 3 + 3 = 6

The digit in the Ten Thousand place = 4 + 3 = 7

The digit in the Hundred place = 7 – 3 = 4

The number is 79,436

1C 1. a. 24,614 < 41,700 b. 50,092 < 51,320

c. 3,15,720 < 4,13,265 d. 72,184 > 72,157

e. 8,74,126 > 8,24,510 f. 4,35,071 < 4,35,261

g. 3,54,680 = 3,54,680  2. a. 14,390 < 37,935 < 40,765 < 79,430

b. 27,880 < 32,860 < 59,573 < 66,773 c. 4,67,943 < 4,88,392 < 8,33,067 < 8,64,853 d. 7,06,583 < 7,20,157 < 7,48,546 < 7,59,404

3. a. Pacific ocean; 36,161 > 27,840 > 23,810 > 23,740 > 18,264

4. Calories Supriya ate = 15,248

Calories Supriya’s brother ate = 18,396

Since, 15,248 < 18,396.

So, Supriya’s brother consumed more calories this week.

5. As, 684 > 450 > 420 > 230

So, Strawberry > Vanilla > Mango > Chocolate

Word Problem 1. Amount given by father = 11,200

Amount required by Anna to buy books = 11,700

Since, 11,200 < 11,700

So, the amount Anna has is not sufficient to buy the books.

1D 1. a. 4, 2, 7, 6, 5

Smallest number = 24,567

Greatest number = 76,542

c. 5, 0, 2, 1, 7, 4

Smallest number = 1,02,457

Greatest number = 7,54,210

e. 2, 3, 9, 8, 0, 1

Smallest number = 1,02,389

Greatest number = 9,83,210

2. a. 2, 1, 7, 4, 9

Smallest number = 1,12,479

Greatest number = 9,97,421

c. 6, 9, 1, 2, 7

Smallest number = 1,12,679

Greatest number = 9,97,621

e. 4, 9, 1, 2, 0

Smallest number = 1,00,249

Greatest number = 9,94,210

b. 6, 1, 3, 7, 8

Smallest number = 13,678

Greatest number = 87,631

d. 8, 6, 2, 5, 9

Smallest number = 25,689

Greatest number = 98,652

b. 3, 8, 5, 0, 1

Smallest number = 1,00,358

Greatest number = 8,85,310

d. 8, 1, 0, 9, 7

Smallest number = 1,00,789

Greatest number = 9,98,710

3. Answers may vary. Sample answers:

5,12,215

5,23,325

5,34,435

5,45,545

4. 6, 2, 0, 8

Smallest number = 20,068

Greatest number = 88,620

Difference = 88,620 – 20,068 = 68,552

5. Digits arranged in ascending order by Raghav = 1,35,678 Digits arranged in descending order by Ishaan = 8,76,531

As 8,76,531 > 1,35,678, so, Ishan has the larger number.

1E 1. a. 134 is between 130 and 140, but is closer to 130

b. 569 is between 560 and 570, but is closer to 570

c. 161 is between 160 and 170, but is closer to 160

d. 1468 is between 1460 and 1470, but is closer to 1470

e. 47,121 is between 47,120 and 47,130, but is closer to 47,120

2. a. 174 is between 100 and 200, but is closer to 200

b. 1653 is between 1600 and 1700, but is closer to 1700

c. 7610 is between 7600 and 7700, but is closer to 7600

d. 2447 is between 2400 and 2500, but is closer to 2400

e. 23,492 is between 23,400 and 23,500, but is closer to 23,500

3. a. 1653 is between 1000 and 2000, but is closer to 2000

b. 6573 is between 6000 and 7000, but is closer to 7000

c. 34,784 is between 34,000 and 35,000, but is closer to 35,000

d. 87,301 is between 87,000 and 88,000, but is closer to 87,000

e. 90,123 is between 90,000 and 91,000, but is closer to 90,000

4. 23,912 can be rounded off to the nearest thousand as 24,000. Ramesh should order around 24,000 saplings.

5. Ria and Sia were in the park for about 90 minutes.

Chapter Checkup 1. a. 38,237 = Thirty-eight thousand two hundred thirty-seven. b. 4,56,321 = Four lakh fifty-six thousand three hundred twenty-one. c. 9,70,540 = Nine lakh seventy thousand five hundred forty. d. 8,06,399 = Eight lakh six thousand three hundred ninety-nine.  2. a. Forty-eight thousand three hundred twenty-one = 48,321 b. One lakh thirty-four thousand six hundred = 1,34,600 c. Seventy-eight thousand six hundred ten = 78,610 d. Nine lakh ten thousand forty-five = 9,10,045

3. a. Place value of digit 4 is 40000, 8 is 8000, 3 is 300, 6 is 60 and 1 is 1. Expanded form of 48,361 = 40000 + 8000 + 300 + 60 + 1

b. Place value of digit 8 is 80000, 7 is 7000, 1 is 100, 0 is 0 and 9 is 9. Expanded form of 87,109 = 80000 + 7000 + 100 + 9

c. Place value of digit 4 is 400000, 5 is 50000, 8 is 8000, 3 is 300, 2 is 20 and 0 is 0. Expanded form = 4,58,320 = 400000 + 50000 + 8000 + 300 + 20 + 0

d. Place value of digit 6 is 600000, 9 is 90000, 2 is 2000, 0 is 0, 4 is 40 and 2 is 2. Expanded form = 6,92,042 = 600000 + 90000 + 2000 + 40 + 2

4. a. 5,69,385

Place value = 9000

Face value = 9

c. 2,54,010

Place value = 50,000

Face value = 5

b. 1,65,899

Place value = 60,000

Face value = 6

d. 9,37,676

Place value = 600

Face value = 6

5. a. The place value of the digit 5 is written incorrectly. 6,85,486 = 6 × 100000 + 8 × 10000 + 5 × 1000 + 4 × 100 + 8 × 10 + 6 × 1 b. The place value of the digits 1, 3, 5 and 3 are written incorrectly. 2,13,548 = 2 × 100000 + 1 × 10000 + 3 × 1000 + 5 × 100 + 4 × 10 + 8 × 1 6. a. 80000 + 2000 + 300 + 20 + 2 = 82,322 b. 300000 + 50000 + 0 + 700 + 50 + 7 = 3,50,757 c. 200000 + 70000 + 3000 + 0 + 60 + 1 = 2,73,061 d. 700000 + 90000 + 6000 + 400 + 0 + 8 = 7,96,408  7. a. 64,614 > 51,700 b. 85,592 >  81,320 c. 48,184 > 48,157 d. 2,18,720 < 3,14,265 e. 7,84,126  < 7,84,510 f. 4,35,893 = 4,35,893 8. a. Ascending order: 37,880 < 42,860 < 46,773 < 69,573 Descending order: 69,573 > 46,773 > 42,860 > 37,880 b. Ascending order: 23,752 < 24,431 < 25,409 < 28,540 Descending order: 28,540 > 25,409 > 24,431 > 23,752

c. Ascending order: 64,012 < 64,393 < 64,520 < 64,905

Descending order: 64,905 > 64,520 > 64,393 > 64,012

d. Ascending order: 93,854 < 3,58,801 < 3,95,701 < 8,26,750

Descending order: 8,26,750 > 3,95,701 > 3,58,801 > 93,854

e. Ascending order: 7,13,725 < 7,26,890 < 7,58,645 < 7,89,371

Descending order: 7,89,371 > 7,58,645 > 7,26,890 > 7,13,725

f. Ascending order: 5,80,723 < 5,81,945 < 5,87,206 < 5,88,205

Descending order: 5,88,205 > 5,87,206 > 5,81,945 > 5,80,723

9. a. 3429

Nearest 10 = 3430

Nearest 100 = 3400

Nearest 1000 = 3000

c. 39,887

Nearest 10 = 39,890

Nearest 100 = 39,900

Nearest 1000 = 40,000

b. 6126

Nearest 10 = 6130

Nearest 100 = 6100

Nearest 1000 = 6000

d. 53,475

Nearest 10 = 53,480

Nearest 100 = 53,500

Nearest 1000 = 53,000

10. 4,85,345 rounded off to the nearest 1000 is 4,85,000. But according to Rohan, we get 4,85,300. That’s why Rohan is not correct.  Word Problems 1. a. ₹25,879 < ₹25,907 < ₹54,768 < ₹97,463

Saree 4 < Saree 1 < Saree 3 < Saree 2

b. Approximate cost:

Saree 1 = ₹26,000

Saree 2 = ₹97,000

Saree 3 = ₹55,000

Saree 4 = ₹26,000

2. 78,342 < 85,674 < 1,56,810 < 3,67,185

a. English is the most used language. b. Spanish is the least used language. c. Spanish < Japanese < Chinese < English

Chapter 2

Let's Warm-up 1. a. 455 – 5 = 450 b. 987 + 13 = 1000

2. a. 692 – 30  436 b. 355 + 345  662

c. 556 – 120  700

Do It Yourself

4.  carry

6. The numbers should be written according to the place value for addition. Steve did not write the answers according to the place value. Jim has added the numbers correctly.

4720 5138 9858

2372 2348

7. 1782 1340 1008 1032

2790

Word Problems 1. Animals rescued last year = 1000 Animals rescued this year = 1000 + 1145 = 2145 2145 animals were rescued this year.

2. Cars produced in 2021 = 45,821

Cars produced in 2022 = 45,821 + 1208 = 47,029 47,029 cars were produced in 2022.

3. The number of wild buffaloes = 81,232

The number of calves born = 7342

Total population = 81,232 + 7342 = 88,574 2B 1. a.

5. 57,801 – 13,456 =

6. 17,890 – 1829 =

44,345 should be added to 13,456 to get 57,801.

So, 16,061 should be subtracted from 17,890 to get 1829.

7. 12,345 – 2335 =

10,010 is 2335 less than 12,345.

8. 76,512 – 13,467 =

13,467 + 63,045 = 76,512.

Word Problems 1. Amit's number, Y = 61,020 = Rahul's number + 1234

Rahul's number = Y – 1234 = 61,020 – 1234 –

the number thought by Rahul is 59,786.

2. Bulbs produced this year = 78,770 Bulbs produced last year = 78,770 – 1672 =

77,098 bulbs were produced in the previous year.

2C 1. a. 1299 + 8772 – 1001 = 10071 – 1001 = 9070

b. 1661 + 571 – 1006 = 2232 – 1006 = 1226

c. 15,679 – 1654 + 20,865

= 15,679 + 20,865 – 1654

= 36,544 – 1654 = 34,890

e. 77,241 – 17,711 – 28,978

= 59,530 – 28,978 = 30,552

2.  + – –

34,618 + 17,803 – 28,671 – 11,190

= 52,421 – 28,671 – 11,190

= 23,750 – 11,190

= 12,560

d. 9283 – 7724 + 882

= 9283 + 882 – 7724

= 10,165 – 7724 = 2441

f. 70,918 –16,621 – 27,002

= 54,297 – 27,002 = 27,295

3. The weight of the harvested wheat = 6543 kg

The weight of harvested rice = 4298 kg

The weight of wheat sold = 3785 kg

The weight of the rice sold = 1932 kg

The total weight of the food harvested = 6543 + 4298 = 10,841 kg

The total weight of the food sold = 3785 + 1932 = 5717 kg

The total weight of the food left with farmer = 10,841 kg –5717 kg

= 5124

So, 5124 kg of food is left with the farmer.

4. Total amount of money in Sara’s bank account = ₹8752

The amount of money withdrawn to buy a gift = ₹3256

The amount of money deposited in her account = ₹9823

The amount of money in her account now = ₹8752 – ₹3256 + ₹9823

= ₹8752 + ₹9823 – ₹3256

= ₹18,575 – ₹3256 = ₹15,319

Sara has ₹15,319 left in her account.

5. The stickers in Asifa’s collection = 12,345

The stickers given to her friend = 3789

The stickers received from her cousin = 6543

The stickers in her collection now = 12345 – 3789 + 6543 = 12345 + 6543 – 3789

= 18,888 – 3789

= 15,099

So, 15,099 stickers are left with Asifa.

Word Problem 1. The number of books in the library = 8236

The number of books received from donation = 1534

The number of books received from the group of citizens = 9712

The number of books damaged = 672

The number of books in the library now = 8236 + 1534 + 9712 – 672 = 9770 + 9712 – 672 = 19,482 – 672 = 18,810

2D 1. 1245 rounded off to the nearest 100 = 1200

2456 rounded off to the nearest 100 = 2500

The estimated sum = 1200 + 2500 = 3700

2. 9013 rounded off to the nearest 100 = 9000

3578 rounded off to the nearest 100 = 3600

The estimated difference = 9000 – 3600 = 5400

3. 23,456 rounded off to the nearest 1000 = 23000

56,771 rounded off to the nearest 1000 = 57000

The estimated sum = 23000 + 57000 = 80,000

4. 97,761 rounded off to the nearest thousand = 98,000

87,112 rounded off to the nearest thousand= 87,000

The estimated difference = 98,000 – 87,000 = 11, 000

5. The estimated sum of 72,374 and 16,773 to the nearest ten thousand = 70,000 + 20,000 = 90,000

The estimated sum of 67,124 and 28,974 to the nearest ten thousand = 70,000 + 30,000 = 1,00,000

The estimated sum of 67,124 and 28,974 is more.

Word Problems 1. The cupcakes produced in the morning = 1346

The cupcakes produced in the evening = 2313

The difference = 2313 – 1346 = 967

The estimated difference to the nearest thousand = 1000 So, about 1000 more cupcakes were produced in the evening than in the morning.

2. The number of steps taken by Sarah = 2347

The number of steps required to complete 10,000 steps = 10,000 – 2347 =

The estimated number of steps rounded off to the nearest thousand = 8000

So, Sarah should walk about 8000 more steps to complete 10,000 steps.

Chapter Checkup 1. a.

4. 13,567 – 12,357 = 1210

2. The largest 5-digit number = 99,999

The smallest 4-digit number = 1000 Their sum = 99,999 + 1000 = 1,00,999

5. The sum of 45,998 and 1321 = 47,319

The difference of 45,998 and 1321 = 44,677

So, the difference between 47,319 and 44,677 = 2642

6. 45,223 + 12,678 + 16,941

7. 81,654 – 53,217 – 2345

81,654 – 53,217 = 28,437 – 2345 =

8. The estimated sum of 5678 + 1665 rounded off to the nearest thousand = 6000 + 2000 = 8000

The actual sum = 7343

9. The estimated difference of 1835 – 1346 rounded off to the nearest thousand = 2000 – 1000 = 1000

The actual difference = 489

10. b. The estimated sum of 5457 and 1108 estimated to the nearest hundreds = 5500 + 1100 = 6600

11. b. The estimated difference of 7890 and 3889 estimated to the nearest thousands = 8000 – 4000 = 4000

Word Problems 1. The length of the river Ganga = 2520 km ≈ 2500 km

The length of the river Yamuna = 1376 km ≈ 1400 km

The approximate sum of their lengths = 2500 km + 1400 km = 3900 km

The approximate difference of their lengths = 2500 km – 1400 km = 1100 km

2. The number of bricks ordered for a project = 8327

The number of bricks ordered for another project = 9912

Their estimated sum = 8300 + 9900 = 18,200

3. The weight of the bananas sold (in pounds) = 17,645

The weight of apples sold (in pounds) = 24,891

Their estimated total weight = 18,000 + 25,000 = 43,000 pounds

Chapter 3

2.

7200

c.

6

d. 1002 × 2

1002 = 1000 + 0 + 0 + 2

2 2 × 1000 = 2000

2000 + 0 + 0 + 4 = 2004

e. 2908 × 4

2908 = 2000 + 900 + 0 + 8

Checking

d.

Vertical

9 × 6 = 54 tens (5 hundreds 4 tens) = 4 tens in tens place

9 × 8 = 72 hundreds + 5 hundreds = 77 hundreds.

Checking answer using the method of expanding the bigger number:

3

e. 7251 × 4

Vertical Method 1

2

Checking answer using the method of expanding the bigger number:

f.

Checking answer using the method of expanding the bigger number:

7000

3B

3D 1. Entry fee at the club = ₹423

of tourists = 9

paid = 423 × 9

2. Number of days Jupiter has in a year = 4333

Number of years = 5

Number of days in 5 Jupiter years = 4333 × 5 There will be 21,665 days in 5 years.

3. Number of calories Riti burns in a day = 427 Number of days in January = 31

Number of calories burnt in January = 427 × 31

burn 13,237 calories in January.

4. Number of members = 28

Budget of tickets = ₹1,00,000

Cost of 1 plane ticket = ₹3879

Total cost of the tickets = 3879 × 28

The total cost is ₹1,08,612.

Hence, the total cost of tickets won’t be in their budget.

5. Distance the bus covers in a day = 145 km.

Number of days in the year 2025 = 365 (there are 365 days in a year)

Distance covered in 2025 = 145 × 365

The bus will cover 52,925 km in 2025.

of seats = 755

Number of shows for Horror Show = 32

Number of people who watched Horror Show = 755 × 32

Hence 24,160 people watched the horror story show.

b. Number of shows for Fun with Mary = 45

Total number of seats = 755 Number

= 250

7. Bundle of notes with Karan = ₹200

Number of notes with Karan = 86

Number of notes Karan gave to Ramesh = 24

Notes left with Karan = 86 − 24 = 62 Amount left with Karan = 200

8.

Amount left with Kabir = ₹9000 − ₹4801 = ₹4199

₹4199 is left with Kabir.

9. Number of pens purchased in February = 350

Number of pens purchased in November = 265

Total pens in 2017 = 350 + 265 = 615

There are 615 pens in all.

Cost of 1 pen = ₹9

Amount spent on 615 pens = 615 × 9

The total cost of the pens is ₹5535.

3E 1. a. 235 × 13

Rounding off 235 and 13 to nearest 10

235 = 240

13 = 10

240 × 10 = 2400

b. 582 × 84

Rounding of 582 and 84 to nearest 10

582 = 580

84 = 80

580 × 80 = 46,400

c. 809 × 96

Rounding off 809 and 96 to nearest 10

809 = 810

96 = 100

810 × 100 = 81,000

d. 409 × 962

Rounding off 409 and 962 to nearest 10.

409 = 410

962 = 960

410 × 960 = 3,93,600

e. 849 × 167

Rounding off 849 and 167 to nearest 10.

849 = 850

167 = 170

850 × 170 = 1,44,500

f. 655 × 845

Rounding off 655 and 845 to nearest 10

655 = 660

845 = 850

660 × 850 = 5,61,000

2. a. 169 × 74

Rounding off to nearest 100:

169 = 200

74 = 100

200 × 100 = 20,000

b. 518 × 96

Rounding off to nearest 100:

518 = 500

96 = 100

500 × 100 = 50,000.

c. 222 × 668

Rounding off to nearest 100:

222 = 200

668 = 700

200 × 700 = 1,40,000

d. 874 × 228

Rounding off to nearest 100:

874 = 900

228 = 200

900 × 200 = 1,80,000

3. a. 109 × 54

Actual Product = 109 × 54 = 5886

Rounding off to nearest 10

109 = 110

54

Rounding off to nearest 100

109 = 100

54 = 100

100 × 100 =10,000

Rounding off to 10 (550) is closer to 5886.

b. 182 × 95

Actual product = 182 × 95 = 17,290

Round off to nearest 10

182 = 180

95 = 100

180 × 100 = 18,000

Round off to nearest 100

182 = 200

95 = 100

200 × 100 = 20,000

Round off to 10 (18,000) is closer to 17,290

c. 444 × 777

Actual product = 444 × 777= 3,44,988

Rounding off to nearest 10

444 = 440

777 = 780

440 × 780 = 3,43,200

Rounding off to nearest 100

444 = 400

777 = 800

400 × 800 = 3,20,000

Rounding off to 10 (3,43,200) is closer to 3,44,988

d. 976 × 862

Actual Product

976 × 862 = 8,41,312

Rounding off to nearest 10

976 = 980

862 = 860

980 × 860 = 8,42,800

Rounding off to nearest 100

976 = 1000

860 = 900

1000 × 900 = 9,00,000

Rounding off to 10 (8,42,800) is closer to 8,41,312

4. Actual Product: 616 × 717 = 4,41,672

Estimated Product

Rounding off 616 to the nearest 100 = 600

Rounding off 717 to the nearest 100 = 700

600 × 700 = 4,20,000

Difference in the products = 4,41,672 – 4,20,000 = 21,672

5. Number of stands in the stadium = 17

Approximate number of stands (rounded off to 10) = 20

Number of seats in the stadium = 238

Approximate number of seats (rounded off to 10) = 240

Approximate number of bottles required = 240 × 20 = 4800

Approximately, they will put out 4800 bottles.

Chapter Checkup

1. a.

4. a. 893 × 84 (to the nearest ten)

c. 143 × 78 (to the nearest 100)

143 = 100

78 = 100

100 × 100 = 10,000

d. 862 × 376 (to the nearest 100 )

862 = 900

376 = 400

900 × 400 = 3,60,000

5. Number of steps in Qutub Minar = 379

Total number of steps when climbed up and down = 379 × 2 = 758

Therefore, the total number of steps climbed by the worker is 758.

6. Number of steps Ruby walks in a day = 5567

Number of days in a week = 7

Number of steps in a week = 5567 × 7 Ruby walks 38,969 steps in a week.

7.

ran by Swati everyday = 750 m

9. Amount deposited in the library every month = ₹4555

Amount deposited in 12 months = 4555 × 12

Ratan will deposit ₹54,660 in 12 months.

10. Amount of waste produced every day = 314 kg

Total days from March to May = 31 + 30 + 31 = 92

Amount of waste produced from March to May = 314 × 92 = 28,888 kg They produced 28,888 kg of waste from March to May.

Word Problems 1. Number of schools = 24

Number of equipments each school gets = 3 Cost of each equipment = ₹394

Total number of equipment required = 24 × 3 = 72

Cost of all equipment = 394 × 72 = ₹23,368 Hence, all the equipment will cost ₹28,368.

2. Number of people visiting the exhibition everyday = 467

Rounding off 467 to nearest 100 = 500

Number of days in a year = 365

Rounding off 365 to nearest 100 = 400

Total people visiting in a year = 500 × 400 = 2,00,000

Around 2,00,000 people visits the exhibition every year.

Chapter 4

Let's Warm-up

Quotient = 897

Quotient = 1234

2. (Quotient × Divisor) = Dividend

1776

b. 672 ÷ 8

Quotient × Divisor = 84 × 8 = 672 = Dividend

Quotient × Divisor = (444 × 4) = 1776 = Dividend

c. 1656 ÷ 9

d. 280 ÷ 5

3

Quotient × Divisor = 184 × 9 = 1656 = Dividend

3. a.

Quotient × Divisor = 56 × 5 = 280 = Dividend

4. a. False b. True c. True d. True

Word Problems 1. Number of chapters in the book = 6

Total number of pages in the book = 1410

Number of pages in each chapter = 1410 ÷ 6

So, there are 235 pages in each chapter.

2. Number of bananas = 145 dozen

Number of bananas in 1 dozen = 12

Number of bananas in 145 dozen = 145 × 12 = 1740

Number of bananas in each box = 6

∴

So, 3 must be added to 4587 so that it can be divided by 5 with no remainders.

Number of boxes = 1740

c. 4897

d. 9876

(Quotient × Divisor) +

(Quotient × Divisor) + Remainder = (201 × 49) + 27 = 9876 = Dividend

21 7 9 6 – 6 3 1 6 6 – 1 4 7 1 9 21 × 3 = 63 21 × 7 = 147  b. 19 × 3 = 57 1 3 1 19 2 4 9 6 – 1 9 5 9 – 5 7 2 6 – 1 9 7

4. a. 486 ÷ 10

The quotient is 48 (40 × 10 = 480)

Remainder is 6 (486 – 480)

c. 986 ÷ 100

The quotient is 9 (9 × 100 = 900)

Remainder is 86 (986 – 900)

e. 7894 ÷ 1000

The quotient is 7 (7 × 1000 = 7000)

The remainder is 894 (7894 – 7000)

5. 1 hour = 60 minutes

1 minute = 1 60 hours

b. 9765 ÷ 10

The quotient is 976 (976 × 10 = 9760)

Remainder is 5 (9765 – 9760)

d. 3479 ÷ 100

The quotient is 34 (34 × 100 = 3400)

The remainder is 79 (3479 – 3400)

f. 5555 ÷ 1000

The quotient is 5 (5 × 1000 = 5000)

The remainder is 555 (5555 – 5000)

1200 minutes = 1200 × 1 60 = 1200 60 = 20 hours.

So, there are 20 hours in 1200 minutes.

Number

of sections = 25

there are 41 students in each section.

Number

of ₹100 notes = 9

notes

amount of ₹10 and ₹100 = 900 + 60 = ₹960

Number of ₹500 notes = 4500 ÷ 500 = 4500 500 = 9

So,

of bottles produced in 46 days = 644 bottles

Number of stamps on each page = 31

Number of pages used = 1240 ÷ 31 = 40

1 4 31 1 2 4 0 – 1 2 4 0 0

3. Amount of money received in the month of April = ₹1500

Number of days in the month of April = 30

Amount of money received per day = 1500 ÷ 30

= 1500 30 = 150 3 = 50

So, the person received ₹50 each day for the month of April.

4. Number of books in the library = 8255

Number of books on each shelf = 13

Number of shelves = 8255 ÷ 13

5. Number of candy packets = 45

Number of candies in each packet = 45

2 4 5 × 4 5 1 2 2 5 + 1 8 0 0 2 0 2 5

So, there are 2025 candies.

Number of packets = 2025 ÷ 15

So, 135 smaller packets are needed.

4D 1. a. 149 ÷ 9

Rounding off to the nearest 10

→ 149 = 150 and 9 = 10

So, 150 ÷ 10 = 15

9

1 6 9 1 4 9 – 9 5 9 – 5 4 5

Variation in quotients = 16 – 15 = 1

c. 434 ÷ 31

Rounding off the numbers to nearest 10

1

4

=

6 3 5 13 8 2 5 5 – 7 8 4 5 – 3 9 6 5 – 6 5 0

1 3 5 15 2 0 2 5 – 1 5 5 2 – 4 5 7 5 – 7 5 0

b. 897 ÷ 18

Rounding off to the nearest 10

→ 897 = 900 and 18 = 20

900 ÷ 20 4 5

8 0

–

0 0 – 1 0 0 0

Quotient = 45

7 2

–

Variation in quotients = 49 – 45 = 4

Rounding off to the nearest 10

2 4 40 9 6

0

Quotient = 14

Exact answer:

Quotient = 24

Exact answer:

Quotient = 14

Variation in quotients = 14 – 14 = 0

2. a. 478 ÷ 97

Rounding off to the nearest 100

→ 478 = 500

Rounding off to nearest 100

→ 97 = 100

500 ÷ 100 = 5

Quotient = 5 Actual

Quotient = 24

Variation in quotients = 24 – 24 = 0

b. 2799 ÷ 31

Rounding off to the nearest 100

→ 2799 = 2800

Rounding off to the nearest 100

→ 31 = 0

Quotient = 4 ∴

c. 879 ÷ 48

Rounding off 879 to nearest 100

→ 879 = 900

Rounding off 48 to nearest 100

→ 48 = 0 900 ÷ 0 = 900 50 = 18

1 8

5 9 0

Actual answer:

By how much did the exact answers vary: 93 – 90 = 3

d. 8744 ÷ 58

Rounding off to nearest 100

→ 8744 = 8700

Rounding off to nearest 10 → 58 =

c. Compare the estimated quotient and the actual quotient to find the greater quotient.

∴ 1002 – 1001 = 1  The actual quotient is greater. 4.

Quotient = 145

5. Number of people = 860

Estimated number of people = 900 (nearest to the 100s)

Number of people on each table = 10

Number of tables = 900 ÷ 10 = 900 10 = 90

So, approximately 90 tables are needed.

Word Problems 1. Number of children = 238

Approximate number of children = 240 (nearest to 10)

Number of children in a tent = 4

Number of tents needed for the camp = 240 ÷ 4 = 60

So, approximately 60 tents are required.

2. Amount of money distributed = ₹5734

Estimated amount to nearest 10 = ₹5730

Number of children in one group = 11

Number of children in two groups = 22

Estimated number of students to nearest 10 = 20

Estimated amount = 5730 ÷ 20 = 286

So, the estimated amount each student gets is ₹286.

Chapter Checkup

1. a. 47 ÷ 10

The quotient is 4. (4 × 10 = 40)

d. 4000 ÷ 100

b. 489 ÷ 10

The quotient is 48. (48 × 10 = 480)

The quotient is 40. e. 4789 ÷ 1000

The quotient is 4. (4 × 1000 = 4000)

c. 145 ÷ 100

The quotient is 1. (1 × 100 = 100)

f. 8500 ÷ 1000

The quotient is 8. (8 × 1000 = 8000)

5. a. A number divisible by 5, always ends with 0 or 5. Adding 1 to 154 will give us 155. On dividing 155 with 5 there will be no remainder.

b. Let’s divide 1745 by 9 8 should be subtracted from 1745 to get it completely divide by 9. Subtract 8 from 1745.

c. Let’s divide 341 by 17 2 0

17 3 4 1 – 3 4 1

1 is remaining. 17 – 1 = 16 16 should be added to 341.

d. Let’s divide 7134 by 26. 10 should be subtracted from 7134 so that it get completely divided by 26.

6. a. 448 ÷ 9

Rounding off 448 = 450

b. 1459 ÷ 4

Rounding off 1459 = 1460

c. 779 ÷ 13

Rounding off 779 = 780

d. 4577 ÷ 20

Rounding off 4577 = 4580

e. 8797 ÷ 16

Rounding off 8797 = 8800

5 0 16 8 8 0 0

8 0

8

7. a. 112 ÷ 11

Rounding off 112 to nearest 100 = 100

Rounding off 11 to nearest 10 = 10 100 ÷

b. 489 ÷ 9

Rounding off 489 to nearest 100 = 500

Rounding off 9 to nearest 10 = 10

500 ÷ 10 = 50

c. 1548 ÷ 52

Rounding off 1548 to nearest 100 = 1500

Rounding off 52 to nearest 10 = 50

1500 ÷ 50 = 30

8. Amount of rupees = ₹1540

Number of friends = 14

d. 6987 ÷ 49

Rounding off 6987 to nearest 100 = 7000

Rounding off 49 to nearest 10 = 50

7000 ÷ 50 = 140

Amount received by each friend = 1540 ÷ 14

So, each friend will get ₹110.

9. Number of boxes = 45

Number of chocolates = 19

Total number of chocolates = 45 × 19 = 855

Number of boxes = 53

Number of chocolates = 27

Total number of chocolates = 53 × 27 = 1431

Total number of chocolates in both the boxes = 1431 + 855 = 2286

Number of boxes with each box having 18 chocolates = 2286

18

So, he packed 127 boxes with 18 chocolates each.

10. Number of flowers = 1260

Half of 1260 = 1260 ÷ 2 = 630

Number of bunches with 14 flowers in each = 630 ÷ 14 = 45 bunches

Number of bunches with 15 flowers in each = 630 ÷ 15 = 42 bunches

Total number of bunches = 45 + 42 = 87 bunches.

Word Problems 1. Number of passengers = 108

Number of rows = 12

Number of seats = 108 ÷ 12

So, there are 9 seats in each row.

2. Number of cookies = 195

Number of packets = 13

Number of cookies in each packet = 195 ÷ 13 = 15 1 5

13 1 9 5 – 1

So, Mark put 15 cookies in each packet.

3. Number of students = 1025

Number of sections = 25

Number

So, there are 41 students in each section.

4. Number of oranges in boxes = 432

Rounding off 432 to nearest 100 = 400

Number of oranges = 43

Rounding off 43 to nearest 10 = 40

Estimated number of boxes packed = 400 ÷ 40 = 10

So, around 10 boxes of oranges are packed.

5. 1 hour = 60 minutes; 1 minute = 60 seconds

So, 1 hour = 60 × 60 seconds = 3600 seconds.

∴ 1 second = 1 3600 hours.

→ 7200 seconds = 7200 3600 = 72 ÷ 36

So, there are 2 hours in 7200 seconds.

6. Weight of tomatoes = 1200 kgs

Weight of tomatoes in grams = 1200 × 1000 = 12,00,000 g

Weight of 1 tomato = 100 g

Total number of tomatoes = 12,00,000 ÷ 100 = 12,000 tomatoes

As the tomatoes are divided equally, number of tomatoes in each type of box = 1200 ÷ 2 = 600.

Number of tomatoes in the first type of box = 15

Number of type 1 box needed to hold 600 tomatoes = 600 ÷ 15 = 400

Number of tomatoes in the second type of box = 25

Number of type 2 box needed to hold 600 tomatoes = 600 ÷ 25 = 240

Total number of boxes needed = 400 + 240 = 640

Chapter 5 Let's Warm-up

125 = 41

The first five multiples of 7 are: 7, 14, 21, 28, 35.

b. 8 × 1 = 8 8 × 2 = 16 8 × 3 = 24 8 × 4 = 32 8 × 5 = 40

The first five multiples of 8 are: 8, 16, 24, 32, 40.

c. 9 × 1 = 9 9 × 2 = 18 9 × 3 = 27 9 × 4 = 36 9 × 5 = 45

The first five multiples of 9 are: 9, 18, 27, 36, 45.

d. 10 × 1 = 10 10 × 2 = 20 10 × 3 = 30 10 × 4 = 40 10 × 5 = 50

The first five multiples of 10 are: 10, 20, 30, 40, 50.

e. 11 × 1 = 11 11 × 2 = 22 11 × 3 = 33 11 × 4 = 44 11 × 5 = 55

The first five multiples of 11 are: 11, 22, 33, 44, 55.

f. 12 × 1 = 12 12 × 2 = 24 12 × 3 = 36 12 × 4 = 48 12 × 5 = 60

The first five multiples of 12 are: 12, 24, 36, 48, 60.

g. 13 × 1 = 13 13 × 2 = 26 13 × 3 = 39 13 × 4 = 52 13 × 5 = 65

The first five multiples of 13 are: 13, 26, 39, 52, 65.

h. 14 × 1 = 14 14 × 2 = 28 14 × 3 = 42 14 × 4 = 56 14 × 5 = 70

The first five multiples of 14 are: 14, 28, 42, 56, 70.

i. 15 × 1 = 15 15 × 2 = 30 15 × 3 = 45 15 × 4 = 60 15 × 5 = 75

The first five multiples of 15 are: 15, 30, 45, 60, 75.

j. 16 × 1 = 16 16 × 2 = 32 16 × 3 = 48 16 × 4 = 64 16 × 5 = 80

The first five multiples of 16 are: 16, 32, 48, 64, 80.

k. 17 × 1 = 17 17 × 2 = 34 17 × 3 = 51 17 × 4 = 68 17 × 5 = 85

The first five multiples of 17 are: 17, 34, 51, 68, 85.

l. 18 × 1 = 18 18 × 2 = 36 18 × 3 = 54 18 × 4 = 72 18 × 5 = 90

The first five multiples of 18 are: 18, 36, 54, 72, 90.

m. 19 × 1 = 19 19 × 2 = 38 19 × 3 = 57 19 × 4 = 76 19 × 5 = 95

The first five multiples of 19 are: 19, 38, 57, 76, 95.

n. 20 × 1 = 20 20 × 2 = 40 20 × 3 = 60 20 × 4 = 80 20 × 5 = 100

The first five multiples of 20 are: 20, 40, 60, 80, 100.

o. 25 × 1 = 25 25 × 2 = 50 25 × 3 = 75 25 × 4 = 100 25 × 5 = 125

The first five multiples of 25 are: 25, 50, 75, 100, 125.

3. a. The 6th multiple of 10 is: 10 × 6 = 60

b. The 9th multiple of 13 is: 13 × 9 = 117

c. The 11th multiple of 9 is: 9 × 11 = 99

d. The 5th multiple of 12 is: 12 × 5 = 60

e. The 4th multiple of 15 is: 15 × 4 = 60

f. The 5th multiple of 25 is: 25 × 5 = 125

4. The multiples of 9 are:

9 × 1 = 9 9 × 2 = 18 9 × 3 = 27 9 × 4 = 36 9 × 5 = 45 9 × 6 = 54

9 × 7 = 63 9 × 8 = 72 9 × 9 = 81 9 × 10 = 90 9 × 11 = 99

9 × 12 = 108 9 × 13 = 117 9 × 14 = 126 9 × 15 = 135

The first 8 odd multiples of 9 are: 9, 27, 45, 63, 81, 99, 117, 135.

5. The multiples of 12 are:

12 × 1 = 12 12 × 2 = 24 12 × 3 = 36 12 × 4 = 48 12 × 5 = 60

12 × 6 = 72

The first 6 even multiples of 12 are: 12, 24, 36, 48, 60, 72.

6. a. 5 7 5 15 – 7 5

0 Leaves 0 as the remainder.

On dividing 75 by 5, we get 0 as the remainder. So, 75 is a multiple of 15.

b. 7 6 8 9 – 6 3

5 Leaves 5 as the remainder.

On dividing 68 by 7, we get 5 as the remainder. So, 68 is not a multiple of 7.

c. 8 6 4 8 – 6 4

0 Leaves 0 as the remainder.

On dividing 64 by 8, we get 0 as the remainder. So, 64 is multiple of 8.

d. 11 8 8 8 – 8 8

0 Leaves 0 as the remainder.

On dividing 88 by 11, we get 0 as the remainder. So, 88 is multiple of 11.

e. 12 9 6 3 – 9 6

0 Leaves 0 as the remainder.

On dividing 96 by 12, we get 0 as the remainder. So, 96 is multiple of 12.

f. 17 5 2 3 – 5 1

1 Leaves 1 as the remainder.

On dividing 52 by 17, we get 1 as the remainder. So, 52 is not a multiple of 17.

7. The multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152.

The multiples of 8 greater than 75, but less than 150 are: 80, 88, 96, 104, 112, 120, 128, 136, 144.

The multiples of 11 are: 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154.

The multiples of 11 greater than 75, but less than 150 are: 77, 88, 99, 110, 121, 132, 143.

Word Problem 1. The number of stickers in the packet = 50

Every 5th sticker is a special glittery sticker, so it is a multiple of 5. The glittery sticker will be the stickers that are on the following numbers: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.

5B 1. a. 2 and 3

The multiples of

The

9. The factors of the numbers between 5 and 15 are as follows:

6: 1, 2, 3 and 6

7: 1 and 7

8: 1, 2, 4 and 8

9: 1, 3 and 9

10: 1, 2, 5 and 10

11: 1 and 11

12: 1, 2, 3, 4, 6 and 12

13: 1 and 13

14: 1, 2, 7 and 14

So, 12 has the greatest number of factors between 5 and 15. 10. 4 is the smallest number which has exactly 3 factors. They are: 1, 2 and 4.

Word Problems 1. She can arrange the eggs in 4 ways, which are as follows:

2 × 8 = 16  8 × 2 = 16  1 × 16 = 16  16 × 1 = 16  4 × 4 = 16

Hence, Tina can arrange the eggs in five ways.

2. The total numbers of biscuits = 72

The different possible arrangements of biscuits in the packets are as follows:

h. 35, 50

The factors of 35 are: 1, 5, 7 and 35

The factors of 50 are: 1, 2, 5, 10 and 50

The common factors of 35 and 50 are: 1 and 5

i. 54, 64

The factors of 54 are: 1, 2, 3, 6, 9, 18, 27 and 54

The factors of 64 are: 1, 2, 4, 8, 16, 32 and 64

The common factors of 54 and 64 are: 1 and 2

j. 72, 81

The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72

The factors of 81 are: 1, 3, 9, 27 and 81

The common factors of 72 and 81 are: 1, 3 and 9

2. a. 6 b. 8 c. 12 d. 3 e. 4 f. 18

a. 6 7 8 13 – 7 8

0 Leaves 0 as the remainder.

On dividing 78 by 6, we get 0 as the remainder. So, 6 is a factor of 78.

b. 6 9 6 16 – 9 6

0 Leaves 0 as the remainder.

On dividing 96 by 6, we get 0 as the remainder. So, 6 is a factor of 96.

c. 3 7 8 26 – 7 8

6

4

3

5D 1. a. 8, 10

The factors of 8 are: 1, 2, 4 and 8

The factors of 10 are: 1, 2, 5 and 10

The common factors of 8 and 10 are: 1 and 2 b. 12, 15

The factors of 12 are: 1, 2, 3, 4, 6 and 12

The factors of 15 are: 1, 3, 5 and 15

The common factors of 12 and 15 are: 1 and 3

c. 13, 16

The factors of 13 are: 1 and 13

The factors of 16 are: 1, 2, 4, 8 and 16

The common factor of 13 and 16 is 1 d. 14, 20

The factors of 14 are: 1, 2, 7 and 14

The factors of 20 are: 1, 2, 3, 4, 5 and 20

The common factors of 14 and 20 are: 1 and 2

e. 16, 18

The factors of 16 are: 1, 2, 4, 8 and 16

The factors of 18 are: 1, 2, 3, 9 and 18

The common factors of 16 and 18 are: 1 and 2 f. 20, 30

The factors of 20 are: 1, 2, 4, 5, 10 and 20

The factors of 30 are: 1, 2, 3, 5, 10 and 30

The common factors of 20 and 30 are: 1, 2, 5 and 10

g. 33, 44

The factors of 33 are: 1, 3, 11 and 33

The factors of 44 are: 1, 2, 4, 11 and 44

The common factors of 33 and 44 are: 1 and 11

0 Leaves 0 as the remainder.

On dividing 78 by 3, we get 0 as the remainder. So, 3 is a factor of 78.

d. 3 9 6 21

– 9 6

0 Leaves 0 as the remainder.

On dividing 96 by 3, we get 0 as the remainder. So, 3 is a factor of 96.

3. a. 16, 24

The factors of 16 are: 1, 2, 4, 8 and 16

The factors of 24 are: 1, 2, 3, 4, 8, 12 and 24

The common factors of 16 and 24 are: 1, 2, 4 and 8

The lowest common factor of 16 and 24 is 1

The highest common factor of 16 and 24 is 8

b. 21, 42

The factors of 21 are: 1, 3, 7 and 21

The factors of 42 are: 1, 2, 3, 6, 7, 21 and 42

The common factors of 21 and 42

42 are: 1, 3, 7 and 21

The lowest common factor of 21 and 42 is 1

The highest common factor of 21 and 42 is 21

c. 63, 18

The factors of 63 are: 1, 3, 7, 9 and 63

The factors of 18 are: 1, 2, 3, 6, 9 and 18

3 7 9 63

2 3 6 9 18

The common factors of 63 and 18 are: 1, 3 and 9

The lowest common factor of 63 and 18 is 1

The highest common factor of 63 and 18 is 9

d. 55, 100

The factors of 55 are: 1, 5, 11 and 55

The factors of 100 are: 1, 2, 4, 5, 10, 25, 50 and 100

The common factors of 55 and 11 are: 1 and 5

The lowest common factor of 55 and 100 is 1

The highest common factor of 55 and 100 is 5

e. 48, 84

The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48

The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 21, 28, 42 and 84

The common factors of 48 and 84 are: 1, 2, 3, 4, 6 and 12

The lowest common factor of 48 and 84 is 1

The highest common factor of 48 and 84 is 12

4. a. The biggest common factor of 24 and 36 is 3. False

b. 11 and 13 have no common factors. False

c. 0 is the common factor of all the numbers. False

d. 15 and 25 have a total of 3 common factors. False

e. 6 is a common factor of 18, 30 and 66. True

f. The lowest common factor of 20, 34, 39 and 42 is 1. True

5. The factors of 14 are: 1, 2, 7 and 14

The factors of 45 are: 1, 3, 5, 9, 15 and 45 45 has more factors than 14.

So, Raj is wrong.

Word Problem 1. The number of apple saplings with Bhawar Lal = 36

He can distribute the saplings equally by giving the saplings in the factors of 36

The factors of 36 are: 1, 2, 3, 4, 6, 12, 18 and 36

The number of orange saplings with Bhawar Lal = 48 He can distribute the saplings equally by giving the saplings in the factors of 48

The factors of 48 are: 1, 2, 4, 6, 8, 12, 24 and 48

The common factors of 36 and 48 are: 1, 2, 4, 6 and 12

The greatest common factor of 36 and 48 is 12. He can distribute the saplings to a maximum of 12 children, where each child will get 3 apple saplings and 4 orange saplings.

Chapter Checkup  1. a. The factors of 45 are: 1, 3, 5, 9, 15 and 45

b. The factors of 66 are: 1, 2, 3, 6, 11, 22, 33 and 66

c. The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72

d. The factors of 88 are: 1, 2, 4, 8, 11, 22, 44 and 88

e. The factors of 98 are: 1, 2, 7, 14, 49 and 98

f. The factors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120

g. The factors of 156 are: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78 and 156

h. The factors of 180 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90 and 180

i. The factors of 200 are: 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100 and 200

j. The factors of 222 are: 1, 2, 3, 6, 37, 74, 111 and 222

2. a. 7, 14

The factors of 7 are: 1 and 7

The factors of 14: 1, 2, 7 and 14

The common factors are: 1 and 7

b. 24, 30

The factors of 24 are: 1, 2, 3, 4, 6, 12 and 24

The factors of 30 are: 1, 2, 3, 5, 6, 15 and 30

The common factors of 24 and 30: 1, 2, 3 and 6

c. The factors of 9 are: 1, 3 and 9

The factors of 12: 1, 2, 3, 4, 6 and 12

The common factors: 1 and 3

d. The factors of 20 are: 1, 2, 4, 5 and 20

The factors of 25 are: 1, 5 and 25

The common factors of 20 and 25 are: 1 and 5

3. a. The factors of 5 are: 1 and 5

The factors of 20 are: 1, 2, 4, 5, 10 and 20

The common factors of 5 and 20 are: 1 and 5

As we can see, 4 is not their common factor.

b. The factors of 20 are: 1, 2, 4, 5, 10 and 20

The factors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50 and 100

The common factors of 20 and 100 are: 1, 2, 4, 10 and 20

As we can see, 4 is their common factor.

c. The factors of 12 are: 1, 2, 3, 4, 6 and 12

The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18 and 36

The common factors of 12 and 36 are: 1, 2, 3, 4, 6 and 12

As we can see, 4 is their common factor.

d. The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60

The factors of 200 are: 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100 and 200

The common factors of 60 and 200 are: 1, 2, 3, 4, 5, 10 and 20

As we can see, 4 is their common factor.

4. a. 7 × 1 = 7  7 × 2 = 14  7 × 3 = 21  7 × 4 = 28   7 × 5 = 35

The first five multiples of 7 are: 7, 14, 21, 28 and 35.

b. 11 × 1 = 11 11 × 2 = 22 11 × 3 = 33 11 × 4 = 44 11 × 5 = 55

The first five multiples of 11 are: 11, 22, 33, 44 and 55.

c. 15 × 1 = 15 15 × 2 = 30 15 × 3 = 45 15 × 4 = 60 15 × 5 = 75

The first five multiples of 15 are: 15, 30, 45, 60 and 75.

d. 23 × 1 = 23 23 × 2 = 46 23 × 3 = 69 23 × 4 = 92 23 × 5 = 115

The first five multiples of 23 are: 23, 46, 69, 92 and 115.

e. 30 × 1 = 30 30 × 2 = 60 30 × 3 = 90 30 × 4 = 120 30 × 5 = 150

The first five multiples of 15 are: 30, 60, 90, 120 and 150.

5. a. 2 and 5

The

of 4 and 14 is 28.

e.

The smallest common multiple of

and

is

The

The

The

common multiple of

and

is

The smallest common multiple of 11 and 22 is 22

The smallest common multiple of 20 and 24 is 120

6. a. The multiples of 4 are as follows:

4 × 1 = 4 4 × 2 = 8 4 × 3 = 12 4 × 4 = 16 4 × 5 = 20 4 × 6 = 24

4 × 7 = 28 4 × 8 = 32

The multiples of 4 that are smaller than 30 are: 4, 8, 12, 16, 20, 24 and 28

b. The multiples of 6 are as follows:

6 × 1 = 6 6 × 2 = 12 6 × 3 = 18 6 × 4 =24 6 × 5 = 30 6 × 6 = 36

6 × 7 = 42 6 × 8 = 48

The multiples of 6 that are smaller than 50 are: 6, 12, 18, 24, 30, 36, 42 and 48

c. The multiples of 8 are as follows:

8 × 3 = 24 8 × 4 = 32 8 × 5 = 40 8 × 6 = 48 8 × 7 = 56

8 × 8 = 64 8 × 9 = 72 8 × 10 = 80

The multiples of 8 that are greater than 24, but smaller than 80 are: 32, 40, 48, 56, 64 and 72

d. The multiples of 7 are as follows:

7 × 1 = 7 7 × 2 = 14 7 × 3 = 21 7 × 4 = 28 7 × 5 = 35 7 × 6 = 42

7 × 7 = 49 7 × 8 = 56 7 × 9 = 63 7 × 10 = 70 7 × 11 = 77

7 × 12 = 84

The multiples of 7 that are divisible by 2 are: 14, 28, 42, 56, 70 and 84

e. The multiples of 10 are as follows:

10 × 1 = 10 10 × 2 = 20 10 × 3 = 30 10 × 4 = 40 10 × 5 = 50

10 × 6 = 60 10 × 7 = 70 10 × 8 = 80 10 × 9 = 90 10 × 10 = 100

10 × 11 = 110 10 × 12 = 120 10 × 13 = 130 10 × 14 = 140

10 × 15 = 150 10 × 16 = 160 10 × 17 = 170 10 × 18 = 180

10 × 19 = 190 10 × 20 = 200

The multiples of 10 that are smaller than 200, but divisible by 3 are: 30, 60, 90, 120, 150 and 180.

f. The multiples of 12 are as follows:

12 × 1 = 12 12 × 2 = 24 12 × 3 = 36 12 × 4 = 48

12 × 5 = 60 12 × 6 = 72 12 × 7 = 84 12 × 8 = 96 12 × 9 = 108

12 × 10 = 120 12 × 11 = 132 12 × 12 = 144 12 × 13 = 156

12 × 14 = 168 12 × 15 = 180 12 × 16 = 192

The multiples of 12 that are bigger than 50, but smaller than 200 are: 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180 and 192.

7. a. 4 and 8

The factors of 4 are: 1, 2 and 4

The factors of 8 are: 1, 2, 4 and 8

The common factors of 4 and 8 are: 1, 2 and 4

b. 6 and 10

The factors of 6 are: 1, 2, 3 and 6

The factors of 10 are: 1, 2, 5 and 10

The common factors of 6 and 10 are: 1 and 2

c. 9 and 15

The factors of 9 are: 1, 3 and 9

The factors of 15 are: 1, 3, 5 and 15

The common factors of 9 and 15: 1 and 3

d. 12 and 15

The factors of 12 are: 1, 2, 3, 4, 6 and 12

The factors of 15 are: 1, 3, 5 and 15

The common factor of 12 and 15 are: 1 and 3

e. 25 and 60

The factors of 25 are: 1, 5 and 25

The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 20, 30 and 60

The common factors are: 1 and 5

f. 28 and 42

The factors of 28 are: 1, 2, 7, 14 and 28

The factors of 42 are: 1, 2, 3, 6, 7, 14, 21 and 42

The common factors of 28 and 36 are: 1, 2, 7 and 14

g. 36 and 81

The factors of 36 are: 1, 2, 3, 4, 6, 9, 18 and 36

The factors of 81 are: 1, 3, 9, 27 and 81

The common factors are: 1, 3 and 9

h. 41 and 87

The factors of 41 are: 1 and 41

The factors of 87 are: 1, 3, 29 and 87

The common factors are 1

8. a. 1 and 6 are factors of 7. No

b. 1 is the smallest factor of 21. Yes

c. 1 is the smallest and only factor of 31. No

d. 2 and 4 are factors of 8. Yes

e. 6 and 9 are factors of 54. Yes

f. 246 is a multiple of 3. Yes

g. 306 is a multiple of 9. Yes

h. 16 is the highest common factor of 32 and 64. No

9. The odd numbers between 11 to 20 are: 13, 15, 17 and 19

The multiples of 13 are: 13, 26, 39, 52 and 65

The multiples of 17 are: 17, 34, 51, 68 and 85

The multiples of 15 are: 15, 30, 45, 60 and 75

The multiples of 19 are: 19, 38, 57, 76 and 95

10. The church 1 bell rings every 60 minutes, so the intervals between the ringing are in the multiples of 60.

The bell will ring after intervals of: 60, 120 and 180 minutes.

The church 2 bell rings every 45 minutes, so the intervals between the ringing are in the multiples of 45.

The bell will ring after intervals of: 45, 90, 135 and 180 minutes.

The common interval is 180 minutes.

So, after 180 minutes, both the church bells will ring together.

Word Problems 1. The numbers of cars: 18

She can arrange the cars in the factors of 18.

The factors of 18 are: 1, 2, 3, 6, 9 and 18

The number of teddy bears: 24

She can arrange the teddy bears in factors of 24.

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24

As we can see, the greatest common factor of 18 and 24 is 6

Hence, a maximum of 6 groups can be made.

2. Number of roses in a group: 5

Multiples of 5 are: 5, 10, 15, 20 and 25

Number of lilies in a group: 4

Multiples of 4 are: 4, 8, 12, 16 and 20

As we can see, their common multiple is 20.

Therefore, Megha should purchase 20 flowers of each type.

Chapter 6

Let's Warm-up 1. There are 10 fish in the aquarium.

2. The fraction of yellow fish is 2 10 of all fish.

3. The fraction of pink fish is 1 10 of all the fishes.

4. The fraction of red fish is 4 10 of all the fish.

5. The fraction of purple fish is 3 10 of all the fish.

Do It Yourself

6A 1. a.  b.  c.  d.  e.  f.

2. a. 11 18 b. 10 30 c. 15 25 d. 8 16

Number of objects in 1 group = 12

So, 1 6 of 72 = 12

e. Total number of objects = 33

Total number of groups = 3

Number of objects in 1 group = 11

Number of objects in 2 groups = 11 × 2 = 22

So, 2 3 of 33 = 22

f. Total number of objects = 90

Total number of groups = 6

Number of objects in 1 group = 15

Number of objects in 5 groups = 15 × 5 = 75

So, 5 6 of 90 = 75

g. Total number of objects = 84

Total number of groups = 7

Number of objects in 1 group = 12

Number of objects in 3 groups = 12 × 3 = 36

So, 3 7 of 84 = 36

h. Total number of objects = 96

Total number of groups = 5

Number of objects in 1 group = 12

Number of objects in 5 groups = 12 × 5 = 60

So, 5 8 of 96 = 60

i. Total number of objects = 75

Total number of groups = 5

Number of objects in 1 group = 15

Number of objects in 4 groups = 15 × 4 = 60

So, 4 5 of 75 = 60

j. Total number of objects = 49

c.

3. a.  b.

So, 1 3 of the collection = 3 butterflies.

7 9 of the collection = 7 butterflies.

4. a. Total number of objects = 32

Total number of groups = 4

Number of objects in 1 group = 8

So, 1 4 of 32 = 8

b. Total number of objects = 56

Total number of groups = 8

Number of objects in 1 group = 7

So, 1 8 of 56 = 7

c. Total number of objects = 65

Total number of groups = 5

Number of objects in 1 group = 13

So, 1 5 of 65 = 13

d. Total number of objects = 72

Total number of groups = 6

So, 4 9 of the collection = 4 butterflies.

Total number of groups = 7

Number of objects in 1 group = 7

Number of objects in 2 groups = 7 × 2 = 14

So, 2 7 of 49 = 14

k. Total number of objects = 81

Total number of groups = 9

Number of objects in 1 group = 9

Number of objects in 4 groups = 9 × 4 = 36

So, 4 9 of 81 = 36

l. Total number of objects = 88

Total number of groups = 8

Number of objects in 1 group = 11

Number of objects in 3 groups = 11 × 3 = 33

So, 3 8 of 88 = 33

5. Number of roses in the shop = 48

Fractions of roses wilted = 1 4

Number of roses wilted = 1 4 of 48 = 12

Fractions of roses sold = 3 4

Number of roses sold = 3 4 of 48 = 36

Number of roses wilted and sold = 12 + 36 = 48.

Number of roses left in the shop = 48 – 48 = 0

Word Problems 1. Total number of parts = 10

Number of shaded parts = 5

Fraction represented by the shaded part = 5 10

2. Flowers in the bunch = 48

Fraction of Tulip flowers = 1 6

Number of Tulip flowers = 1 6 of 48 = 8

3. Number of apples in the basket = 30

Fraction of green apples = 3 5

Number of green apples = 3 5 of 30 = 18

4. Number of pencils in the box = 48

Fraction of blue pencils = 3 4

Number of blue pencils = 3 4 of 48 = 36

5. Number of balloons at the party = 75

Fraction of red balloons = 2 5

Number of red balloons = 2 5 of 75 = 30

Number of white balloons = 75 – 30 = 45 6B 1. a.  = 2 3 = 4 6

b. = 2 3 = 8 12

c. = 1 3 = 3 9

d.

2. Answer may vary. Sample answer:

a. First equivalent

Second

Third equivalent

Fourth equivalent fraction of

So, 3 4 = 6 8 , 9 12 , 12 16 , 15 20

b. First equivalent fraction of

Second

Third equivalent fraction of 2 7 = 2

8 28

Fourth equivalent fraction of 2 7 = 2 × 5 7 × 5 = 10 35

So, 2 7 = 4 14 , 6 21 , 8 28 , 10 35

c. First equivalent fraction of 1 5 = 1 × 2 5 × 2 = 2 10

Second equivalent fraction of 1 5 = 1 × 3 5 × 3 = 3 15

Third equivalent fraction of 1 5 = 1 × 4 5 × 4 = 4 20

Fourth equivalent fraction of 1 5 = 1 × 5 5 × 5 = 5 25

So, 1 5 = 2 10 , 3 15 , 4 20 , 5 25

d. First equivalent fraction of 1 4 = 1 × 2 4 × 2 = 2 8

Second equivalent fraction of 1 4 = 1 × 3 4 × 3 = 3 12

Third equivalent fraction of 1 4 = 1 × 4 4 × 4 = 4 16

Fourth equivalent fraction of 1 4 = 1 × 5 4 × 5 = 5 20

So, 1 4 = 2 8 , 3 12 , 4 16 , 5 20

3. a. Factors of 30 = 1, 3, 5, 6, 10, 15, 30

Factors of 90 = 1, 3, 5, 6, 10, 15, 30, 90

Divide the numerator and denominator with the biggest common factors, that is, 30.

30 ÷ 30

90 ÷ 30 = 1 3 . This cannot be divided further.

Hence, the simplest form of 30 90 = 1 3

b. Factors of 15 = 1, 3, 5, 15

Factors of 45 = 1, 3, 5, 15, 45

Divide the numerator and denominator with the biggest common factors, that is, 15.

15 ÷ 15

45 ÷ 15 = 1 3 . This cannot be divided further.

Hence, the simplest form of 15 45 = 1 3

c. Factors of 12 = 1, 2, 4, 6, 12

Factors of 24 = 1, 2, 4, 6, 12, 24

Divide the numerator and denominator with the biggest common factors, that is, 12.

12 ÷ 12

24 ÷ 12 = 1 2 . This cannot be divided further.

Hence, the simplest form of 12 24 = 1 2

d. Factors of 10 = 1, 2, 5, 10

Factors of 50 = 1, 2, 5, 10, 50

Divide the numerator and denominator with the biggest common factors, that is, 10.

10 ÷ 10

50 ÷ 10 = 1 5 . This cannot be divided further. Hence, the simplest form of 10 50 = 1 5

e. Factors of 2 = 1, 2

Factors of 26 = 1, 2, 13, 26

Divide the numerator and denominator with the biggest common factors, that is, 2.

2 ÷ 2

26 ÷ 2 = 2 13 . This cannot be divided further.

Hence, the simplest form of 2 26 = 1 13

f. Factors of 5 = 1, 5

Factors of 65 = 1, 5, 65

Divide the numerator and denominator with the biggest common factors, that is, 5.

5 ÷ 5

65 ÷ 5 = 1 13 . This cannot be divided further. Hence, the simplest form of 5 65 = 1 13

g. Factors of 6 = 1, 2, 3, 6

Factors of 42 = 1, 2, 6, 7, 14, 42

Divide the numerator and denominator with the biggest common factors, that is, 6.

6 ÷ 6

42 ÷ 6 = 1 7 . This cannot be divided further.

Hence, the simplest form of 6 42 = 1 7

h. Factors of 8 = 1, 2, 4, 8

Factors of 26 = 1, 2, 13, 26

Divide the numerator and denominator with the biggest common factors, that is, 2.

8 ÷ 2

26 ÷ 2 = 4 13 . This cannot be divided further.

Hence, the simplest form of 8 26 = 4 13

4. a. 1 2 × 3 3 = 3 ( ) b. 1 5 × ( ) 3 = 3 ( )

1 � 3 2 � 3 = 3 6 1 � 3 5 � 3 = 3 15

c. ( ) 12 ÷ 2 ( ) = 1 6 d. 1 3 × 7 ( ) = ( ) 21

2 ÷ 2

12 ÷ 2 = 1 6 1 � 7 3 � 7 = 7 21

e. 15 25 ÷ 5 ( ) = 3 ( ) f. 2 7 × ( ) 4 = 8 ( )

15 ÷ 5 25 ÷ 5 = 3 5 2 � 4 7 � 4 = 8 28

g. 5 8 × 6 ( ) = ( ) 48 h. 27 45 ÷ ( ) 9 = 3 ( )

5 � 6 8 � 6 = 30 48 27 ÷ 9 45 ÷ 9 = 3 5

5. a. 20 35 b. 3 8 c. 12 30 d. 9 21 e. 2 41 f. 5 24 g. 8 14 h. 7 42

6C 1. a. 1 9 , 4 9 , 7 9 , 5 9 — Like Fractions.

b. 2 11 , 6 11 , 10 11 , 9 11 — Like Fractions.

c. 13 15 , 14 15 , 1 15 , 12 15 — Like Fractions.

d. 4 8 , 4 14 , 4 7 , 4 12 — Unlike Fractions.

2. a. 1 3  =

Therefore, 8 10  >  5 15 Therefore, 5 7  > 5 9

h. 9 16  >  7 15

9 7 9 45 < 63

16 7 15 135 > 112

,

5 6 , 5 7 , 5 11 , 5 9 , 5 13 , 5 10

c. 4 10 , 8 10 , 2 10 , 9 10 , 3 10 , 7 10

Same denominators, so will compare the numerators.

The ascending order of fraction is 2 10 < 3 10 < 4 10 < 7 10 < 8 10 < 9 10

d. 2 4 , 2 10 , 2 7 , 2 9 , 2 6 , 2 3

since the numerators are same, we compare and arrange the denominators in descending order we get: 10>9>7>6>4>3. So, the ascending order of fraction is 2 10 < 2 9 < 2 7 < 2 6 < 2 4 < 2 3

e. 11 16 , 9 16 , 4 16 , 15 16 , 12 16 , 7 16

Same denominators, so will compare the numerators.

The ascending order of fraction is 4 16 < 7 16 < 9 16 < 11 16 < 12 16 < 15 16

f. 7 8 , 7 10 , 7 11 , 7 12 , 7 7 , 7 9

since the numerators are same, we compare and arrange the denominators in descending order we get: 12>11>10>9>8>7. So,

6. a. 9 12 , 4 12 , 3 12 , 10 12 , 1 12 , 7 12

Same denominators, so will compare the numerators. The descending order of fraction is 10 12

b. 7 12 , 7 15 , 7 11 , 7 9 , 7 8 , 7 10

since the numerators are same, we compare and arrange the denominators in ascending order we get: 8<9 <10 <11<12<15. So, the descending order of

c. 15 17 , 8 17 , 13 17 , 9 17 , 7 17 , 10 17

Same denominators, so will compare the numerators. The

d. 3 9 , 3 10 , 3 7 , 3 4 , 3 6 , 3 5

since the numerators are same, we compare and arrange the denominators in ascending order we get: 4<5<6<7<9<10. So, the descending order of

e. 1 8 , 6 8 , 4 8 , 5 8 , 2 8 , 7 8

Same denominators, so will compare the numerators. The

since the numerators are same, we compare and arrange the denominators in ascending order we get: 6<7<8<10<11<15. So, the descending order of fraction is

6D 1. a. 2 3 – Proper Fraction b. 15 8 – Improper Fraction

c. 3 2

6 – Mixed Fraction d. 16 9 – Improper fraction

e. 2  6 10 – Mixed Fraction f. 3 7 – Proper Fraction

g. 2 1 9 – Mixed Fraction h. 13 6 – Improper fraction

c. 7 2 6 = 7 × 6

2

= 44 6 d. 4 1 9 = 4 × 9 + 1 9 = 37 9

e. 3 8 10 = 3 × 10 + 8 10 = 38 10 f. 6 5 7 = 6 × 7 + 5 7 = 47 7

g. 5 5 9 = 9 × 5 + 5 9 = 50 9 h. 9 2 6 = 9 × 6 + 2 6 = 56 6

4. a. 6 3 = 6 ÷ 3 3 ÷ 3 = 2 1 = 2 b. 24 8 = 24 ÷ 4 8 ÷ 4 = 6 2 = 3

c. 30 6 = 5 d. 36 9 = 36 ÷ 9 9 ÷ 9 = 4 1 = 4 e. 20 10 = 20 ÷ 10 10 ÷ 10 = 2 1 = 2

f. 35 7 = 35 ÷ 7 7 ÷ 7 = 5 1 = 5 g. 45 9 = 45 ÷ 9 9 ÷ 9 = 5 1 = 5

h. 18 6 = 18 ÷ 6 6 ÷ 6 = 3 1 = 3

5. Fraction of apple pie Susan had = 1 3 4

Amount of apple pie with Susan = 4 × 1 + 3 4 = 7 4

6E 1. a. 1 3 + 1 3 = 1 + 1 3 = 2 3 b. 1 8 + 5 8 = 1 + 5 8 = 6 8 = = 3 4

c. 2 6 + 3 6 = 2 6 + 3 6 = 2 + 3 6 = 5 6 d. 1 9 + 5 9 = 1 + 5 9 = 6 9 = 6 ÷ 3 9 ÷ 3 = 2 3

e. 2 6 8 + 1 2 8 = 2 × 8 + 6 8 + 1 × 8 + 2 8 = 22 8 + 10 8 = 22 + 10 8 = 32 8 = 4

+ 3 5

= 2 × 9 + 1

÷ 3 = 17 3 = 5 2 3 h. 11 6 + 1 1

1 × 6 + 1

= 11 6 + 7 6 = 11 + 7 6 = 18 6 = 3

4. Part of garden planted with Tomatoes = 3 6

Part of garden planted with cucumber = 2 6

Total parts of garden planted = 3

+ 2

John has planted 5 6 part of his garden.

5. Number of hours spent on Reading book = 3 5

Number of hours spent on Drawing = 1 5

Total time spent on both activities = 3

Suhani spent 4 5 of an hour in both activities.

Ria is left with 3 4 m of cloth.

2. Part of pages read on Monday = 1 4

Part of pages read on Tuesday = 2 4

Total parts of pages he read = 1 4 + 2 4 = 3 4

Varun has read 3 4 of the pages of his book.

3. Distance travelled by Sunil on Saturday = 1 1

km farther on Sunday than Saturday.

4. Part of packet of cookies with Manya = 1 2

Part of packet of cookies Manya ate = 7 8

Part of packet of cookies Manya left with =

Manya is left with 3 8 part of packet of cookies.

5. Length

Lenth of cycling track is 5 6

6. Part of cement bag used in one project = 4 5

Part of cement bag used in another project = 1 5

Total cement used in projects = 4 5 + 1 5 = 5 5 = 1

Cement remaining with worker = 1 – 1 = 0

There is no cement left with worker.

Chapter Checkup 1. a.  b. c.  d.

2. a. 1 6 of 18 flowers

1 6 of 18 flowers is 3 flowers (18 ÷ 6 = 3)

c. 1 4 of 36 boxes

1 4 of 36 boxes is 9 boxes.

(36 ÷ 4 = 9)

b. 1 3 of 27 cakes

1 3 of 27 cakes is 9 cakes.

(27 ÷ 3 = 9)

d. 1 5 of 50 balloons

1 5 of 50 balloons is 10 balloons.

(50 ÷ 5 = 10)

3. Answer may vary. Sample answer:

a. 5

6

First equivalent fraction of 5 6 = 5 × 2 6 × 2 = 10 12

Second equivalent fraction of 5 6 = 5 × 3 6 × 3 = 15 18

Third equivalent fraction of 5 6 = 5 × 4 6 × 4 = 20 24

Fourth equivalent fraction of 5 6 = 5 × 5 6 × 5 = 25 30

So, 5 6 = 10 12 , 15 18 , 20 24 , 25 30

b. 7

8

First equivalent fraction of 7 8 = 7 × 2 8 × 2 = 14 16

Second equivalent fraction of 7 8 = 7 × 3 8 × 3 = 21 24

Third equivalent fraction of 7 8 = 7 × 4 8 × 4 = 28 32

Fourth equivalent fraction of 7 8 = 7 × 5 8 × 5 = 35 40

So, 7 8 = 14 16 , 21 24 , 28 32 , 35 40

c. 3 9

First equivalent fraction of 3 9 = 3 × 2 9 × 2 = 6 18

Second equivalent fraction of 3 9 = 3 × 3 9 × 3 = 9 27

Third equivalent fraction of 3 9 = 3 × 4 9 × 4 = 12 36

Fourth equivalent fraction of 3 9 = 3 × 5 9 × 5 = 15 45

So, 3 9 = 6 18 , 9 27 , 12 36 , 15 45

d. 2

7

First equivalent fraction of 2 7 = 2 × 2 7 × 2 = 4 14

Second equivalent fraction of 2 7 = 2 × 3 7 × 3 = 6 21

Third equivalent fraction of 2 7 = 2 × 4 7 × 4 = 8 28

Fourth equivalent fraction of 2 7 = 2 × 5 7 × 5 = 10 35

So, 2 7 = 4 14 , 6 21 , 8 28 , 10 35 4.

7. a. 8 11 , 4 11 , 3 11 , 6 11 , 1 11 , 7 11

Same denominators, so will compare the numerators.

Ascending order: 1 11 < 3 11 <

Descending

b. 7 8 , 7 12 , 7 11 , 7 9 , 7 13 , 7 10

since the numerators are same, we compare and arrange the denominators in descending order we get: 13>12>11>10>9>8. So,

Ascending order: 7 13 < 7 12 < 7 11 < 7 10 < 7 9 < 7 8

since the numerators are same, we compare and arrange the denominators in ascending order we get: 8<9<10<11<12<13. So, Descending order:

c. 4 17 , 8 17 , 2 17 , 9 17 , 3 17 , 15 17

Same denominators, so will compare the numerators.

d. 4 12 , 4 10 , 4 7 , 4 9 , 4 6 , 4 8

since the numerators are same, we compare and arrange the denominators in descending order we get: 12>10>9>8>7>6. So,

Ascending order: 4 12 < 4 10 < 4 9 < 4 8 < 4 7 < 4 6

since the numerators are same, we compare and arrange the denominators in ascending order we get: 6<7<8<9<10<12. So,

Descending order: 4 6 > 4 7 > 4 8 > 4 9 > 4 10 > 4 12

e. 11 19 , 9 19 , 4 19 , 15 19 , 12 19 , 7 19

Same denominators, so will compare the numerators.

Ascending order: 4 19 < 7 19 < 9 19 < 11 19 < 12 19 < 15 19

Descending order: 15 19 > 12 19 > 11 19 > 9 19 > 7 19 > 4 19

f. 3 8 , 3 10 , 3 11 , 3 12 , 3 7 , 3 9

since the numerators are same, we compare and arrange the denominators in descending order we get: 12>11>10>9>8>7. So,

Ascending order: 3 12 < 3 11 < 3 10 < 3 9 < 3 8 < 3 7

since the numerators are same, we compare and arrange the denominators in ascending order we get: 7<8<9<10<11<12. So,

Descending order:

c. 5 1

4 = 4 × 3 + 1 4 = 13 4 b. 2 3 7 = 2 × 7 + 3 7 = 17 7

6 = 5 × 6 + 1 6 = 31 6 d. 3 3 5 = 5 × 3 + 3 5 = 18 5

e. 1  5 10 = 10 × 1 + 5 10 = 15 10 f. 7  5 6 = 7 × 6 + 5 6 = 47 6

g. 4 5 8 = 4 × 8 + 5 8 = 37 8 h. 6 2 5 = 5 × 6 + 2 5 = 32 5

a. 2 5 + 1 5 = 2 + 1 5 = 3 5 b. 3 8 + 7 8 = 3 + 7 8 = 10 8 = 10 ÷ 2 8 ÷ 2 = 5 4 = 1 1

c. 1 2

d. 3 1

+ 5 4 = 1 × 4 + 2 4 + 5 4 = 6 4 + 5 4 = 6 + 5

= 11 4 = 2 3

+ 5 8 = 3 × 8 + 1 8 + 5 8 = 25 8 + 5 8 = 25 + 5 8 = 30 8 = 30 ÷ 2 8 ÷ 2 = 15 4 = 3 3

e. 3 3

+ 1 2

= 3 × 6 + 3 6 + 1 × 6 + 2 6 = 21 6 + 8 6 = 21 + 8 6 = 29 6 = 4 5

f. 11 7 + 5 7 = 11 + 5 7 = 16 7 = 2 2 7

g. 4 1 9 + 3 4 9 = 4 × 9 + 1 9 + 3 × 9 + 4 9 = 37 9 + 31 9 = 37 +

Word Problems 1. Number of pieces of cake cut by mother = 20

Fraction of pieces Sunita ate = 3 5

Number of pieces Sunita ate = 3 5 of 20 = 12

Sunita ate 12 pieces of cake.

2. Length of first jump = 2 9 m

Length of second jump = 3 9 m

Total length frog covered = 2 9 + 3 9 = 5 9

Frog covered 5 9 m.

3. Quantity of rice with Shopkeeper = 28 kg

Fraction of quantity of rice he sold = 5 7

Quantity of rice he sold = 5 7 of 28 = 20 kg

Remaining quantity of rice = 28 – 20 = 8 kg

Shopkeeper is left with 8 kg of rice.

4. Full length of wire = 10 m

Fraction of wire Sunita cut = 4 5 m

Length of wire Sunita cut = 4 5 of 10 = 8 m

Length of wire Sunita is left with = 10 – 8 = 2 m

Sunita is left with 2 m

5. Quantity

Vessel contained 1 1 2 L milk.

6. Length

7. Total strength of class = 36

Fraction of Students absent in class = 5

Number

15 students were absent on Friday

8.  Number of cows with Farmer = 56

Fraction of cows grazing in the field = 3 7

Number of cows grazing in the field = 3 7 of 56 = 24

Remaining cows who are in barn = 56 – 24 = 32 cows

The remaining 32 cows, who are in barn.

9. Number of pieces Maria cut the pizza in = 8

Number of pieces Maria ate = 3

Number of pieces Maria’s brother ate = 2

Number of pieces they ate altogether = 3 + 2 = 5

Fraction of pieces they ate = 5 8

Maria and her brother ate 5 8 part of pizza.

10. Money with Sudha = 2 5 of 50 = ₹20

Money with Ravi = 1 2 of 50 = ₹25

Ravi has more money than Sudha.

Chapter 7

Let’s Warm-up

1.  Circle

2.

3.

4. Do It Yourself

Triangle Square Rectangle

7A 1. a. A ray has no end points. True

b. A line has no end points. True

c. Only one line can pass through a point. False

d. The light from a torch is an example of a ray. True

e. A line extends endlessly in both directions. True

f. A point has only length. False

2. a.

Q P

Ray PQ

b. C D Line CD

c. Q Point Q

3. a. Point A, Point B, Point C, Point D, Point O, Point P, and Point Q are points.

b. OA , OB , OC , OP and OQ are rays. c. BC and PQ are lines.

4. E and F are points.

(HK)⃗ and (EG)⃗ are rays.

a and d are lines.

7B 1. Measurements may vary depending on the size of the book.

a. b.

Edge of the book = 4 cm

Edge of the deck of cards = 2 cm

2. Measurements may vary. Sample figures are given.

a. A B 6 cm   b. A B 9 cm

c. A B 10 cm   d. A B 14 cm

3. The figure has 6 line segments. 2cm 3 cm

3 cm 3 cm 3 cm 2cm

4. Length of Tina’s pencil = 10 cm

Length of Sheena’s pencil = 4 cm

Total length of the two pencils = 10 cm + 4 cm = 14 cm

5. Length of Anu’s line segment = 6 cm

Length of Jiya’s line segment = 6 cm + 4 cm = 10 cm

Anu’s line segment:

A B 6 cm

Jiya’s line segment:

A B 10 cm

Word Problem 1. The correct length of the plank is 25 cm.

7C 1. a. Open figure

b. Open figure

c. Closed figure

d. Closed figure

e.

Open figure

f. Open figure

2. Letters C, U, and S from the English alphabet are open figures.

3. Letters B, D, and O from the English alphabet are closed figures.

4. C. The figure is continuous.

7D 1. a.  b.  c.  d.  e.  f.

Simple figures - b, c, d, f

Non-simple figures - a, e

2. (Representations may vary.)

Simple figures:

Non-simple figures:

3. b. The boundaries cross themselves.

4. Simple figures:

Non-simple figures:

5. Yes, a figure can be non-simple and open.

7E 1. a. All simple closed shapes are polygons. False

b. A shape that crosses itself is not a simple closed shape. True

c. A polygon can be formed with two lines. False

d. A hexagon has 7 sides. False

2. a.  b.  c.  d.  e.  f.

a, c, and f show non-simple closed figures.

b, d, and e show simple open figures.

3. a.  b.  c.  d.  e.  f.

b, c, e, and f show simple figures.

4. (Representations may vary.)

a. Triangle

b. Rectangle

c. Pentagon

d. Hexagon

e. Heptagon

5. a. Polygon b.  c.  d.  e.

Word Problem 1. The closed figure can have a maximum of 4 sides. 1 2 3 4

7F 1. a. Every point on a circle is at the same distance from the centre

b. All the radii of a circle are equal in length.

c. A circle can have an infinite number of diameters.

d. The length of the boundary of a circle is called its circumference.

e. A circle has only one centre.

2. Centre Radius

3. a. iii. 5 cm

b. ii. 2

c. iv. Radius = Diameter 2

4. Representations may vary.

a. 2 cm 2 cm O  b. 5 cm 5cm O

c. 6 cm 6 cm O A

d. 8 cm 8 cm O

5. Representations may vary.

a. 7 cm O

b. Radius = 8 2 = 4 cm 4 cm O

6. Distance from the centre of the ground to its boundary = radius = 16 m

Diameter = radius × 2 = 16 × 2 = 32 m

The diameter of the field is 32 m.

7. Diameter = 6 cm

So, Radius = 6 2 = 3 cm   3 cm C D

Chapter Checkup 1. a. iii b. i. ray c. iv. 12 d. ii. 2 e. iii. light from a torch f. iii. point g. iii. open figure h. iii. 8 i. iv. radius

j. iii. 10 cm

2. The radius shown by the compass = 4 cm

Diameter = 2 × radius

Diameter = 2 × 4 = 8 cm

The diameter of the circle that is drawn with the given compass is 8 cm.

3. B – closed and non-simple, C – open and simple, D – closed and simple, U – open and simple, 7 – open and simple, 0 – closed and simple, 8 – closed and non-simple, S – open and simple

None of the given letters and numbers are polygons.

4. Hexagon

5.

Polygon

Name Rectangle Triangle Pentagon Decagon

Number of sides 4 3 5 10

6. a. Diameter = 4 cm, Radius = diameter 2 = 4 2 = 2 cm

Measurements may vary. Sample figure is given.

2 cm O  b. Radius = 2 cm 2 cm O  c. Radius = 5 cm 5cm O

d. Diameter = 6 cm

Radius = diameter 2 = 6 2 = 3 cm 3 cm O

7. Mary should take the route AB.

8. AB = 12 cm

OP = radius of the small circle = 4 cm

Since OP, OB, and OD are the radii of the same circle, OP = OB = OD = 4 cm

Distance between the points A and D = AB – OB – OD = 12 cm –

4 cm – 4 cm = 12 cm – 8 cm = 4 cm

Therefore, AD = 4 cm

9. Diameter of the jar = 16 cm

Radius of the jar = 16 2 = 8 cm

Radius of the lid to be bought = 8 cm + 2 cm = 10 cm

10. Diameter of the original frisbee = 18 cm

Radius of the original frisbee = 18 2 = 9 cm

Radius of the frisbee to be made = 9 cm − 2 cm = 7 cm

7 cm

Word Problems 1. The shortest distance is 8 m + 8 m = 16 m.

2. The final shape of the area with no grass will be a circle.

Chapter 8

Let’s Warm-up a.  3-D shape

b. 2-D shape

c. 2-D shape d. 2-D shape

Do It Yourself 8A 1.

e. 3-D shape

Top

Top

Top

2. a. Suhani’s house is on the second road. - True

b. If Kavita steps out of her house on the first road, the bank will be to the left of her house. False

c. The post office is the nearest place to the factory. False

d. The restaurant is in front of the park. True

e. To reach the factory, one has to go to the second road. True

3. a. If Rani is on mall road facing Rose street, the police station will be to her right

b. Sam’s house is on MG road

c. The restaurant is in front of Rani’s house.

d. Supermarket is centre of all the roads.

e. Sam’s house in front of Brooke park.

4. Bob can use different routes to reach bank which are as follows:

i. To reach the bank, Bob will take the right turn and move straight, then he will take the first right turn and move straight. He will then take a left turn and move straight and stop in front of the bank.

ii. To reach the bank Bob will take the right turn and move straight. He will take the third right turn and move straight, then take again right turn and move straight and he will stop in front of bank.

iii. To reach the bank Bob will take the right turn and move straight. He will take a second right turn and move straight, then take a left turn and will stop in front of the bank.

iv. To reach the bank Bob first took a left turn and moved straight, then again took a left turn and moved straight. He will take a left turn and move straight and then the bank will be there after crossing 2 cuts.

5. Truck will not cross the Farm Road to reach the burning house.

3.

4. Side C will be opposite of purple side when net is folded to make a box.

D 

A B C

5. Answers may vary. Sample answers: a.

b.

8. a. The mall is between John’s house and Rohit’s house.

b. Park is next to the school.

c. The grocery store is nearest to Kavya’s house.

d. If Rohit is on First Road with the mall to his left, he needs to take a right turn to reach the Pizza house.

9. a. Rita’s house is farthest from the school.

b. Mina’s house is not opposite to Anand garden.

c. 4 roads meet at the Central Chowk.

10. Answers may vary. 2 samples are given below.

Chapter 9 Let's Warm-up 1.  , 2. 11111, 111111

3.  ,

Do It Yourself

9A 1. a.

4. EEEEE, FFFFFF

neither growing, nor reducing b. growing

c. growing

2. a. b.

c. d.

e.

3.  a. b.

c.

d.

e.

9B 1. a. 13 is added each time to the previous number.

13, 26, 39, 52, 65, 78

b. 20 is added to the previous number each time.

10, 30, 50, 70, 90, 110

c. We are adding +1, +2, +3, +4, +5, +6 each time.

1, 2, 4, 7, 11, 16, 22

d. We are subtracting 10 from the previous number each time. 84, 74, 64, 54, 44, 34

e. We are adding +5, +10, +15, +20, +25 each time. 10, 15, 25, 40, 60, 85

2. Answer may vary. Sample answer:

4. Answer may vary. Sample answer:

10 + 15 = 25

35 – 15 = 20

20 + 25 = 45

45 + 35 = 80

25 + 35 = 60

60 + 80 = 140

9C 1. a. Yes b. No c. Yes d. No 2. a.  b.

c.  d.

Yes, every time the answer increases by 2.

3.  1 3 4

2 1 is parallelogram

2, 3, 4, 5, 7 are triangles

6 is square

9D 1. a. MOODLE b. QUIZ c. ATTEMPT d. READ e. NEXT f. FINISH g. ANSWERS h. PAGE

2. a. KEEP IT UP - 11-5-5-16 9-20 21-16

b. SAVE WATER - 19-1-22-5 23-1-20-5-18

c. PLANT TREES - 16-12-1-14-20 20-18-5-5-19

d. FANTASTIC WORK - 6-1-14-20-1-19-20-9-3 23-15-18-11

e. REDUCE REUSE RECYCLE - 18-5-4-21-3-5 18-5-21-19-5 18-5-3-25-3-12-5

9E 1. a. Yes b. No c. No d. Yes e. Yes f. No g. No h. No 2. a.     b.    c. d.      e.   f. g.  h.     i.

3.  A B C D E F G

H I J K L M N

O P Q R S T U

V W X Y Z

A, B, C, D, E, H, I, K, M, O, T, U, V, W, X and Y are symmetrical.

A, M, T, U, V, W, Y has vertical line of symmetry.

B, C, D, E, K has horizontal line of symmetry.

H, I, O, X has both vertical and horizontal line of symmetry. 9F 1. a.  b.  c.

d.  e.  f.

g.  h.  i.

j.  k.  l.

Chapter Checkup 1. a.

b.  c.

d.

e.

f.

g. h.

a. BEST WISHES → 3-9-37-39 45-17-37-15-9-37

b. PLANT A TREE → 31-23-1-27-39 1 39-35-9-9

c. SAVE PAPER → 37-1-43-9 31-1-31-9-35

d. RECYCLE → 35-9-5-49-5-23-9

3. a. We are adding 20 in each term.

110, 130, 150, 170, 190, 210, 230

b. We are adding +8, +9, +10 and so on each time. 234, 242, 251, 261, 272, 284, 297

c. We are subtracting 9 from each term.

140, 131, 122, 113, 104, 95, 86

d. We are subtracting 110 each time.  890, 780, 670, 560, 450, 340, 230

e. We are adding 11 each time.  111, 122, 133, 144, 155, 166

f. We are dividing by 4 each time.  2304, 576, 144, 36, 9

4. a. Yes b. Yes c. No d. No

e.  f.  g.  h.

i.   j.   k.   l.

5. a.  b.  c.  d.

e.  f.   6. a.  b.

c.  d.  e.  f.

g.  h.  i.  j.

k.  l.  m.  n.

o. Word Problem 1.  A circle can have an infinite number of lines of symmetry.

LESS THAN 1 METRE

MORE THAN 1 METRE

Chapter

Do It Yourself

10A

1. a. We would use a ruler to measure the length of a grain of rice.  b. We would use a measuring tape to measure the height of a door. c. We would use a measuring tape to measure the length of a desk. d. We would use a ruler to measure the width of a book.  2. a. 4 cm, or 40 mm.

b. 3 cm, or 30 mm. c. 10 cm, or 100 mm. d. 7 cm, or 70 mm.

3. a. 2 cm 7 mm, or 27 mm. b. 1 cm 5 mm, or 15 mm.

c. 2 cm 5 mm, or 25 mm. d. 2 cm 4 mm, or 24 mm.

4. a. 4 cm A B b. 12 cm A B

c. 11 cm 2 mm A B d. 17 cm 5 mm A B

5. We know that 1 m = 100 cm

a. 2 m = 2 × 100 = 200 cm b. 9 m = 9 × 100 = 900 cm

c. 12 m = 12 × 100 = 1200 cm

d. 11 m 34 cm = (11 × 100) + 34 = 1100 cm + 34 cm = 1134 cm

e. 14 m 67 cm = (14 × 100) + 67 = 1400 cm + 67 cm = 1467 cm

f. 15 m 22 cm = (15 × 100) + 22 = 1500 cm + 22 cm = 1522 cm

6. We know that 100 cm = 1 m

a. 400 cm = 400 ÷ 100 = 4 m b. 900 cm = 900 ÷ 100 = 9 m

c. 1200 cm = 1200 ÷ 100 = 12 m

d. 136 cm = 100 cm + 36 cm

We know that 100 cm = 1 m

So, 100 cm + 36 cm = 1 m 36 cm

e. 475 cm = 400 cm + 75 cm

We know that 100 cm = 1 m

So, 400 cm + 75 cm = (400 ÷ 100) m + 75 cm = 4 m 75 cm

f. 925 cm = 900 cm + 25 cm

We know that 100 cm = 1 m

So, 900 cm + 25 cm = (900 ÷ 100) m + 25 cm = 9 m 25 cm

g. 1125 cm = 1100 cm + 25 cm

We know that 100 cm = 1 m

So, 1100 cm + 25 cm = (1100 ÷ 100) m + 25 cm = 11 m 25 cm

h. 1250 cm = 1200 cm + 50 cm

We know that 100 cm = 1 m

So, 1200 cm + 50 cm = (1200 ÷ 100) m + 50 cm = 12 m 50 cm

i. 1520 cm = 1500 cm + 20 cm

We know that 100 cm = 1 m

So, 1500 cm + 20 cm = (1500 ÷ 100) m + 20 cm = 15 m 20 cm

7. Remember, 1 km = 1000 m

a. 5 km = 5 × 1000 m = 5000 m b. 9 km = 9 × 1000 m = 9000 m

c. 13 km = 13 × 1000 m = 13,000 m

d. 16 km 165 m = (16 × 1000) m + 165 m = 16000 m + 165 m

= 16,165 m

e. 40 km 175 m = (40 × 1000) m + 175 m = 40000 m

+ 175 m = 40,175 m

f. 95 km 54 m = (95 × 1000) m + 54 m

= 95000 m + 54 m = 95,054 m

8. Remember, 1000 m = 1 km

a. 1400 m = 1000 m + 400 m = 1 km 400 m

b. 1600 m = 1000 m + 600 m = 1 km 600 m

c. 2200 m = 2000 m + 200 m = 2 km 200 m

d. 1336 m = 1000 m + 336 m

We know that 1000 m = 1 km

So, 1000 m + 336 m = 1 km + 336 m = 1 km 336 cm

e. 1475 m = 1000 m + 475 m

We know that 1000 m = 1 km

So, 1000 m + 475 m = 1 km + 475 m = 1 km 475 m

f. 1925 m = 1000 m + 925 m

We know that 1000 m = 1 km

So, 1000 m + 925 m = 1 km + 925 m = 1 km 925 m

g. 2125 m = 2000 m + 125 m

We know that 1000 m = 1 km

So, 2000 m + 125 m = (2000 ÷ 1000) km + 125 m = 2 km 125 m

h. 4250 m = 4000 m + 250 m

We know that 1000 m = 1 km

So, 4000 m + 250 m = (4000 ÷ 1000) km + 250 m = 4 km 250 m

i. 7520 m = 7000 m + 520 m

We know that 1000 m = 1 km

So, 7000 m + 520 m = (7000 ÷ 1000) km + 520 m = 7 km 520 m

Word Problems 1. The number of ribbons = 5

The length of each ribbon = 24 cm 5 mm

The total length of the ribbon = (5 × 24) cm + (5 × 5) mm = 120 cm 25 mm

So, the total length of the ribbon is 120 cm 25 mm.

2. The length of each cloth piece = 30 cm

The total number of cloth pieces cut = 4

a. The total length of 4 pieces of cloth = 30 × 4 = 120 cm

b. Remember, 1 m = 100 cm

120 cm = 100 cm + 20 cm = 1 m 20 cm

Thus, the total length of the piece of cloth that was cut = 1 m 20 cm

3. The car was driven for 4 km.

We know that 1 km = 1000 m

So, 4 km = 4 × 1000 = 4000 m

Thus, Raman's uncle drove the car for 4000 m.

4. We know that 1 m = 100 cm

The width of the table = 2 m 30 cm

Width of the table in centimetres = (2 × 100) cm + 30 cm = 200 cm + 30 cm = 230 cm

In the same way:

The height of the table = 1 m 12 cm

Height of the table in centimetres = 100 cm + 12 cm = 112 cm

Thus, the table is 220 cm wide and 112 cm tall.

5. The distance from Jiya’s home to park = 1 km 500 m

The distance from the park to Jiya’s home = 1 km 500 m

So, the total distance that Jiya’s father jogs in a day

= 1 km 500 m + 1 km 500 m OR

1 km + 1 km + 500 m + 500 m = 2 km + 1000 m

We know that 1 km = 1000 m

The total distance that Jiya’s father jogs in a day = 2 km + 1 km  = 3 km

The total distance that Jiya’s father jogs in 2 days = 2 × 3 km = 6 km

10B

1. a. The weight on the scale = 2 and half kg

We know that 1 kg = 1000 g

Thus, the weight in grams = 2000 + 500 = 2 kg 500 g

b. The weight on the scale = Half kg

We know that 1 kg = 1000 g

Thus, the weight in grams = 500 g

c. The weight on the scale = 1 and half kg

We know that 1 kg = 1000 g

Thus, the weight in gram = 1000 + 500 = 1 kg 500 g

2. Remember, 1 g = 1000 mg

a. 3000 mg = 3000 ÷ 1000 = 3 g

b. 7000 mg = 7000 ÷ 1000 = 7 g

c. 10000 mg = 10000 ÷ 1000 = 10 g

d. 1100 mg = 1000 mg + 100 mg

We know that 1 g = 1000 mg

So, 1000 mg + 100 mg = 1 g 100 mg

e. 2467 mg = 2000 mg + 467 mg

We know that 1 g = 1000 mg

So, 2000 mg + 467 mg = 2 g 467 mg

f. 5967 mg = 5000 mg + 967 mg

We know that 1 g = 1000 mg

So, 5000 mg + 967 mg = 5 g 967 mg

3. Remember, 1 kg = 1000 g

a. 5000 g = 5000 ÷ 1000 = 5 kg

b. 6000 g = 6000 ÷ 1000 = 6 kg

c. 10000 g = 10000 ÷ 1000 = 10 kg

d. 4500 g = 4000 g + 500 g

We know that 1000 g = 1 kg

So, 4000 g + 500 g = (4000 ÷ 1000) kg + 500 g = 4 kg 500 g

e. 6557 g = 6000 g + 557 g

We know that 1000 g = 1 kg

So, 6000 g + 557 g = (6000 ÷ 1000) kg + 557 g = 6 kg 557 g

f. 9782 g = 9000 g + 782 g

We know that 1000 g = 1 kg

So, 9000 g + 782 g = (9000 ÷ 1000) kg + 782 g = 9 kg 782 g

4. Remember, 1 g = 1000 mg

a. 5 g = 5 × 1000 mg = 5000 mg

b. 16 g = 16 × 1000 mg = 16,000 mg

c. 8 1 2 g = (8 × 1000) mg + 1 2 g = 8000 mg + 1 2 g

We know that 1 g = 1000 mg; so 1 2 g = 500 mg

Thus, 8000 mg + 1 2 g = 8000 mg + 500 mg = 8500 mg

d. 4 g 102 mg = (4 × 1000) mg + 102 mg = 4000 + 102 = 4102 mg

e. 15 g 770 mg = (15 × 1000) mg + 770 mg = 15000 + 770 = 15,770 mg

f. 9 g 802 mg = (9 × 1000) mg + 802 mg = 9000 + 802 = 9802 mg

5. Remember, 1 kg = 1000 g

a. 5 kg = 5 × 1000 g = 5000 g

b. 17 kg = 17 × 1000 g = 17,000 g

c. 10 1 2 kg = (10 × 1000) g + 1 2 kg = 10,000 g + 1 2 kg

We know that 1 kg = 1000 g; so 1 2 kg = 500 g

Thus, 10000 g + 1 2 kg = 10,000 + 500 g = 10,500 g

d. 5 kg 10 g = (5 × 1000) g + 10 g = 5000 + 10 = 5010 g

e. 15 kg 25 g = (15 × 1000) g + 25 g = 15,000 + 25 = 15,025 g

f. 16 kg 820 g = (16 × 1000) g + 820 g = 16,000 + 820 = 16,820 g

Word Problems 1. Jaya’s weight = 38 kg 300 g

We know that 1 kg = 1000 g

So, 38 kg 300 g = (38 × 1000) g + 300 g = 38,000 + 300 = 38,300 g

Thus, Jaya’s weight in grams is 38,300 g.

2. The price of 1 kg of tomatoes = ₹150

The price of 2 kg of tomatoes = ₹150 + ₹150 = ₹300

The price of 1 2 kg of tomatoes = ₹150 ÷ 2 = ₹75

Thus, the price Rama aunty will pay to buy 2 1 2 kg of tomatoes = ₹300 + ₹75 = ₹375

3. The cost of a 500 g pack of rice = ₹65

We know that 1 kg = 1000 g = 500 g + 500 g

So, the cost of 1 kg pack of rice = ₹65 + ₹65 = ₹130

The cost of a 5 kg pack of rice = ₹130 × 5 = ₹650

Thus, Sudhir has to pay ₹650 to buy a 5 kg pack of rice.

4. The total weight of the onions sold by the vegetable seller = 145 kg 500 g

The price of 1 kg of onions = ₹30

The price of 145 kg of onions = ₹30 × 145 = ₹4350

We know that 1 kg = 1000 g. So, 500 g = 1 2 kg

The price of 500 g or 1 2 kg of onions = ₹30 ÷ 2 = ₹15

Thus, the total amount of money earned by the vegetable seller for 145 kg 500 g of onions = ₹4350 + ₹15 = ₹4365

5. The amount of wheat that 1 sack contains = 9 kg 500 g

The number of sacks the farmer packed = 9

The amount of wheat the farmer used to pack 9 sacks

= 9 × (9 kg 500 g) = 9 × 9 kg and 9 × 500 g

= 81 kg 4500 g

We know that 1 kg = 1000 g

So, 4500 g = 4000 g + 500 g = (4000 ÷ 1000) kg + 500 g = 4 kg 500 g

Thus, amount of wheat in 9 sacks = 81 kg + 4 kg + 500 g = 85 kg 500 g

10C 1. a. The amount of liquid in the measuring jug = 1 4 mL

We know that 1 L = 1000 mL

1

4 of 1000 mL = 1000 ÷ 4 = 250 mL

b. The amount of liquid in the measuring jug = 1000 mL

We know that 1000 mL = 1 L

Thus, there is 1 L liquid in the measuring jug.

2. Remember, 1000 mL = 1 L

a. 3500 mL = 3000 mL + 500 mL = (3000 ÷ 1000) L + 500 mL

= 3 L 500 mL

If 1000 mL = 1 L, then 500 mL = 1 2 L

So, 3 L 500 mL = 3 1 2 L

b. 5000 mL = 5000 ÷ 1000 L = 5 L

c. 10500 mL = 10000 ml + 500 mL = (10000 ÷ 1000) L + 500 mL

= 10 L 500 mL

If 1000 mL = 1 L, then 500 mL = 1 2 L

So, 10 L 500 mL = 10 1 2 L

3. Remember, 1000 mL = 1 L

a. 1700 mL = 1000 mL + 700 = 1 L 700 mL

b. 5286 mL = 5000 mL + 286 mL = (5000 ÷ 1000) L + 286 mL

= 5 L 286 mL

c. 7650 mL = 7000 mL + 650 mL = (7000 ÷ 1000) L + 660 mL

= 7 L 650 mL

d. 8235 mL = 8000 mL + 235 mL = (8000 ÷ 1000) L + 235 mL

= 8 L 235 mL

e. 9250 mL = 9000 mL + 250 mL = (9000 ÷ 1000) L + 250 mL

= 9 L 250 mL

f. 11300 mL = 11000 mL + 300 mL = (11000 ÷ 1000) L + 300 mL

= 11 L 300 mL

4. The number of small packs with 250 mL juice each = 4

The total amount of juice in these packs = 250 mL × 4 = 1000 mL

We know that 1 L = 1000 mL

So, there is 1 L juice in 4 packs of 250 mL juices.

Similarly, the number of bottles with 1 L juice each = 2

So, the total juice in 2 bottles = 2 L

Thus, the total juice in all packs and bottles = 1 L + 2 L = 3 L

5. The amount of tea 1 cup can hold = 100 mL

The amount of tea needed to fill in 6 cups = 100 mL × 6 = 600 mL

So, 600 mL of tea is needed in the pot to fill 6 cups of tea.

Word Problems 1. The number of bottles Coco needs to fill = 3

The capacity of each bottle = 2 L

The total capacity of 3 bottles = 2 L × 3 = 6 L

So, Coco would use 6 L of water to fill all three bottles.

2. The capacity of the jug of milkshake = 2 L

We know that 1 L = 1000 mL. So, 2 L = 2000 mL

The capacity of each glass = 200 mL

The number of 200 mL glasses that can be filled with 2000 mL of milkshake in the jug = 2000 ÷ 200 = 10. Thus, 10 glasses can be filled with one full jug of milkshake.

3. a. The number of glasses of lemonade JJ and her sister have sold so far = 8

The capacity of each glass = 250 mL

The total capacity of 8 glasses sold = 250 mL × 8 = 2000 mL

We know that 1000 mL = 1 L. So, 2000 mL = 2 L

Thus, JJ and her sister have sold 2 L of lemonade so far.

b. The number of glasses of lemonade JJ and her sister have sold so far = 8

The earnings from 1 glass of lemonade sold = ₹25

The earnings from 8 glasses of lemonade sold = ₹25 × 8 = ₹200

Thus, JJ and her sister have earned ₹200 so far by selling lemonade.

Chapter Checkup 1. Drawings may vary. Check the length of the lines drawn. Sample answers:

a. 12 cm 6 mm       b. 15 cm 8 mm

12 cm 6 mm A B

c. 25 cm 5 mm

15 cm 8 mm A B

25 cm 5 mm A B

2. Remember, 1 m = 100 cm

a. 205 cm = 200 cm + 5 cm = (200 ÷ 100) m + 5 cm = 2 m 5 cm

b. 507 cm = 500 cm + 7 cm = (500 ÷ 100) m + 7 cm = 5 m 7 cm

c. 764 cm = 700 cm + 64 cm = (700 ÷ 100) m + 64 cm = 7 m 64 cm

3. Remember, 1 km = 1000 m

a. 1205 m = 1000 m + 205 m = 1 km 205 m

b. 5763 m = 5000 m + 763 m = (5000 ÷ 1000) km + 763 m = 5 km 763 m

c. 6049 m = 6000 m + 49 m = (6000 ÷ 1000) km + 49 m = 6 km 49 m

4. Remember, 1 kg = 1000 g

a. 5065 g = 5000 g + 65 g = (5000 ÷ 1000) kg + 65 g = 5 kg 65 g

b. 4600 g = 4000 g + 600 g = (4000 ÷ 1000) kg + 600 g = 4 kg 600 g

c. 7450 g = 7000 g + 450 g = (7000 ÷ 1000) kg + 450 g = 7 kg 450 g

d. 10,500 g = 10,000 g + 500 g = (10,000 ÷ 1000) kg + 500 g = 10 kg 500 g

5. Remember, 1 kg = 1000 g

a. 2 kg 500 g = (2 × 1000) g + 500 g = 2000 g + 500 g = 2500 g

b. 4 kg 600 g = (4 × 1000) g + 600 g = 4000 g + 600 g = 4600 g

c. 5 kg 750 g = (5 × 1000) g + 750 g = 5000 g + 750 g = 5750 g

d. 12 kg 500 g = (12 × 1000) g + 500 g = 12,000 g + 500 g = 12,500 g

6. Remember, 1000 mL = 1 L  a. 2000 mL = (2000 ÷ 1000) L = 2 L

b. 7000 mL = (7000 ÷ 1000) = 7 L c. 12,500 mL = 12,000 mL + 500 mL

We know that 1000 mL = 1 L and 500 mL = 1 2 L

So, 12,000 mL + 500 mL = (12,000 ÷ 1000) L + 500 mL = 12 L + 1 2 L = 12 1 2 L

7. Remember, 1 L = 1000 mL

a. 7200 mL = 7000 L + 200 mL = (7000 ÷ 1000) L + 200 mL = 7 L 200 mL

b. 8660 mL = 8000 L + 660 mL = (8000 ÷ 1000) L + 660 mL = 8 L 660 mL

c. 16,250 mL = 16,000 L + 250 mL = (16,000 ÷ 1000) L + 250 mL = 16 L 250 mL

8. We know that 1 L = 1000 mL

So, my mother has 1000 mL of milk.

The amount of milk mother gave me = 350 mL

The amount of milk mother gave to my brother = 175 mL

Total milk consumed by me and my brother = 350 + 175 = 525 mL

Thus, the amount of milk left = 1000 – 525 = 475 mL

9. The capacity of 1 jar of oil = 3 L

The capacity of 2 jars of oil = 3 + 3 = 6 L

We know that 1 L = 1000 mL

So, 6 L = 6 × 1000 = 6000 mL

The capacity of the smaller jars = 500 mL

The number of 500 ml jars that can be filled with 6000 mL of oil = 6000 ÷ 500 = 12.

Thus, Tara can fill 12 jars with 2 jars of 3 L cooking oil in each jar.

Word Problems 1. The price of 1 kg of sugar = ₹60

a. The price of 2 kg of sugar = ₹60 × 2 = ₹120

b. The price of 5 kg of sugar = ₹60 × 5 = ₹300

c. The price of 12 kg of sugar = ₹60 × 12 = ₹720

The price of 500 g of sugar = ₹60 ÷ 2 = ₹30

The total price of 12 kg and 500 g of sugar = ₹720 + ₹30 = ₹750

d. The price of 25 kg of sugar = ₹60 × 25 = ₹1500

The price of 500 g of sugar = ₹60 ÷ 2 = ₹30

The total price of 25 kg and 500 g of sugar = ₹1500 + ₹30 = ₹1530

2. The total amount of juice in the bottle = 1 L

We know that 1 L = 1000 mL

The capacity of the juice glass filled by Rima = 200 mL

The amount of the juice left in the bottle after filling a 200 mL glass = 1000 mL – 200 mL = 800 mL

Thus, 800 mL of juice is left in the bottle now.

Chapter 11

Let's Warm-up 1. 25 cm = 250 mm  2. 3 m = 300 cm

3. 6 m 30 cm = 630 cm  4. 5 km = 5000 m  5. 6 km 40 m = 6040 m

11A 1. On measuring the length of each figure, the thread over figure Y is longer than the thread over figure X. So, the perimeter of figure Y is longer. 2. a.

3. a. Perimeter = 4 cm + 4 cm + 4 cm + 4 cm = 16 cm

b. Perimeter = 3 cm + 2 cm + 2 cm + 3 cm + 3 cm + 2 cm + 2 cm + 3 cm = 20 cm

c. Perimeter = 14 m + 9 m + 6 m + 5 m + 4 m + 4 m + 8 m = 50 m

d. Perimeter = 5 cm + 4 cm + 3 cm + 5 cm + 6 cm = 23 cm

e. Perimeter = 20 m + 30 m + 20 m + 15 m + 30 m + 20 m = 135 m

f. Perimeter = 8 cm + 8 cm + 6 cm + 6 cm + 8 cm = 36 cm

4. a. Perimeter = 5 cm + 4 cm + 2 cm + 1 cm + 3 cm + 3 cm + 6 cm + Length of unknown side

32 cm = 24 cm + The length of the unknown side

The length of the unknown side = 32 cm – 24 cm = 8 cm

b. Perimeter = 30 m + 12 m + 17 m + 20 m + 17 m + The length of the unknown side

128 m = 96 m + The length of the unknown side

The length of the unknown side = 128 m – 96 m = 32 m

c. Perimeter = 11 cm + 7 cm + 11 cm + 6 cm + 6 cm + The length of the unknown side

46 cm = 41 cm + The length of the unknown side

The length of the unknown side = 46 cm – 41 cm = 5 cm

5. a. 4 cm

b. 15 mm

7 mm 8 mm

Perimeter = 4 cm + 4 cm + 7 cm + 8 cm + 3 cm + 4 cm = 30 cm

6. Perimeter = 4 cm + 4 cm + 4 cm + 10 cm = 22 cm

Perimeter = 12 mm + 15 mm + 7 mm + 8 mm + 5 mm + 23 mm = 70 mm

Word Problems

1. To find the distance around the park, we will find the perimeter of the park.

Perimeter of the park = 150 m + 100 m + 150 m + 100 m = 500 m

So, the distance covered around the park would be 500 m.

2. To find the length of sewing thread around the pillow cover, we will find the perimeter of the pillow cover.

Perimeter of the pillow cover = 30 cm + 45 cm + 30 cm + 45 cm = 150 cm

Perimeter of 3 pillow covers = 150 × 3 = 450 cm

So, the length of the sewing thread required to sew around 3 pillow covers is 450 cm.

3. The length of every side of the tiles in each pattern = 25 cm

Number of tiles in each pattern = 6

a. The perimeter of the tiles in each pattern = 25 cm + 25 cm + 25 cm + 25 cm = 100 cm

The perimeter of the pattern = 100 × 6 = 600 cm

b. The number of sides of the tiles on the outer side of the pattern = 14

The perimeter of the pattern = 25 cm × 14 = 350 cm

All these six rectangles have area equal to 12 sq. units.

b. The perimeter of the rectangle with sides 12 and 1 = 12 + 12 + 1 + 1 = 26 units

The perimeter of the rectangle with sides 6 and 2 = 6 + 6 + 2 + 2 = 16 units

The perimeter of the rectangle with sides 4 and 3 = 4 + 4 + 3 + 3 = 14 units

So, the perimeter is not the same as the area.

5. Vegetable patch

f.

a. The area of the playing field in School A = 21 sq. units

b. The area of the playing field in School B = 19 sq. units

c. Since 21 is more than 19, so School A has a bigger playing field

Word Problem 1. The number of the full squares (before eating) = 20

Full squares ( ) 12 12

Half squares ( ) 0 0

More than half squares ( ) 1 1

Less than half squares ( ) 4 0

Total area (approx.) - 13 sq. units

The area of the chocolate bar he ate = The area of the full chocolate bar – area of chocolate bar left = 20 – 13 = 7 sq. units

b. The area of the chocolate bar before eating = 20 sq. units

a. Figure Area (sq. units) Perimeter (units)

b. (i) Figure Q and figure S have the same area but different perimeters.

(ii) Figure R and figure S have the same perimeter but different areas. (iii) Figure P and figure T have the same area and perimeter. 7. a.

The perimeter of figure A is 10 cm.

The perimeter of figure B is 14 cm.

The perimeter of figure C is 14 cm.

Figure B and Figure C have the same perimeter.

3. a. Perimeter = 3 cm + 2 cm + 2 cm + 1 cm + 4 cm = 12 cm

b. Perimeter = 7 m + 9 m + 6 m + 12 m = 34 m

c. Perimeter = 3 m + 4 m + 6 m + 4 m + 3 m + 4 m = 24 m

d. Perimeter = 6 cm + 8 cm + 3 cm + 8 cm + 5 cm = 30 cm

e. Perimeter = 8 cm + 7 cm + 10 cm = 25 cm

f. Perimeter = 4 m + 3 m + 6 m + 10 m + 16 m = 39 m

4. a. Perimeter = 3 cm + 5 cm + E + 4 cm + 10 cm + 9 cm

39 cm = 31 cm + E  E = 39 cm – 31 cm = 8 cm

b. Perimeter = 4 cm + 2 cm + 7 cm + E + 7 cm + 2 cm + 4 cm

5.

29 cm = 26 cm + E  E = 29 cm – 26 cm = 3 cm

b.

c.

All these have a perimeter of 20 units. (Drawings may vary)

Word Problems 1.

Both of these shapes have an area of 8 sq. units. (drawings may vary)

Chapter 12

Let's Warm-up 1. 6 hours is the same as 360 minutes.

2. 360 seconds is equal to 6 minutes

3. There are 7200 seconds in 2 hours.

4. 40 minutes is less than (less than, greater than or equal to) 1 hour.

5. 2 hours 15 minutes is more than (less than, greater than or equal to) 130 minutes.

Do It Yourself

12A

a. 4:30 a.m.  b. 10 a.m.  c. 4:30 p.m.

d. 10:00 p.m.  e. 5:00 p.m.

2. a. 6:30 a.m.

b. 2:00 p.m.

d. 6:00 p.m.

p.m.

3. a. 11:30 a.m. b. 02:15 a.m. c. 11:59 p.m. d. 6:44 p.m.

4. a. 8:20 b. 11:47 c. 1:28 d. 4:44

5. a. 3:36 b. 4:12 c. 10:24 d. 12:14

Word Problems 1. Akhil goes for practice at 11:30 in the afternoon would be written as 11:30 a.m.

2. Isha went for the movie at 10:30 a.m.

The time after 2 hours: 1 hour 1 hour

10:30 a.m. 11:30 a.m. 12:30 p.m.

Isha comes back at 12:30 p.m.

12B 1. a. 15:28 hours b. 23:56 hours c. 00:00 hours

d. 23:59 hours  2. a. 10:40 p.m. b. 6:25 p.m.

c. 11:24 p.m. d. 1:03 p.m.

3. a. 09:00 a.m. 12:15 hours

b. 05:45 p.m. 20:20 hours

c. 02:30 a.m. 23:50 hours

d. 12:15 p.m. 17:45 hours

e. 08:20 p.m. 09:00 hours

f. 11:50 p.m. 02:30 hours

4. Starting time of the train – 23:15 hours

Starting time of the train in a 12-hour clock- 11:15 p.m.

The train reaches its destination at – 17:30 hours

The arrival time in a 12-hour clock – 5:30 p.m.

5. The arrival time of the train—20:00 hours.

The time on Rama’s wrist watch—7:45 p.m.

The time on Rama’s wrist watch in the 24-hours format—19:45 hours

Yes, Rama is on time. The train will come in 15 minutes.

Word Problems 1. The departure time of the flight—1445 hours 1445 hours in the 12-hours format = 2:45 p.m.

The boarding pass will be given 2 hours before departure.

The time before 2 hours:

1 hour 1 hour

12:45 p.m. 1:45 p.m. 2: 45 p.m.

The boarding pass will be given at 12:45 p.m.

2. The boarding time = 1645 hours = 4:45 p.m.

Niru reaches the station at 3:15 p.m.

The time remaining for the train to arrive.

1 hour 30 minutes

3:15 p.m. 4:15 p.m. 4:45 p.m.

1 hour and 30 minutes are remaining for the train to arrive.

12C 1. a. There are 60 minutes in an hour.

b. There are 24 hours in a day.

2. a. 1 hour = 60 minutes

7 hours = 7 × 60 minutes = 420 minutes

7 hours = 420 minutes

b. 1 hour = 60 minutes

3 hours = 3 × 60 = 180 minutes.

3 hours > 115 minutes

c. 1 hour = 60 minutes

6 hours = 6 × 60 minutes = 360 minutes

6 hours = 360 minutes

d. 1 hour = 60 minutes

10 hours = 10 × 60 minutes = 600 minutes

10 hours > 360 minutes

3. a. 5 340 300 –40

60 hours minutes

60 hours minutes

–30

340 minutes = 5 hours      450 minutes = 7 hours 40 minutes 30 minutes

c. 9 560 540 –20

60 hours minutes

11 675 60 ––75 60 15

60 hours minutes

560 minutes = 9 hours 20 minutes 675 minutes = 11 hours 15 minutes

4. a. 12:00 noon to 12:00 midnight = 12 hours 12:00 midnight to 12:30 = 30 minutes

Duration = 12 hours 30 minutes.

b. 05:06 p.m. to 10:06 p.m. = 5 hours

10:06 pm to 10:55 pm = 49 minutes

Duration = 5 hours 49 minutes

c. 14:25 hours to 20:25 hours = 6 hours.

20:25 to 20:45 = 20 minutes

Duration = 6 hours 20 minutes

12D

d. 10:15 hours to 23:15 hours = 13 hours

23:15 hours to 23:30 hours = 15 minutes

Duration= 13 hours 15 minutes

5. The starting time of the movie = 7:30 p.m.

Duration = 2 hours 15 minutes.

1 hour 1 hour

7:30 p.m. 8:30 p.m. 9:30 p.m.

15 minutes after 9:30 p.m. = 9:45 p.m.

The movie will end at 9:45 p.m.

Word Problems 1. The starting time of the doctor = 10:15 a.m.

The return time of the doctor = 1 p.m.

The time taken by the doctor:

b. The total days in September = 30

The days remaining in September = 30 – 5 = 25 days

The days in October = 31 days

The days in November = 2 days

The duration = 25 + 31 + 2 = 58 days

c. 12.05.20 = 12 May

The total days in May = 31

The days remaining = 31 – 12 = 19 days

The total days in June = 10 days

The duration = 19 + 10 = 29 days.

d. 07.06.23 = 7 June, 23 and

23.07.23 = 23 July, 23

The days in June = 30

The days remaining = 30 – 7 = 23 days

The days in July = 23

The duration = 23 + 23 = 46 days

10:15 a.m. 11:00 a.m. 12:00 noon 1:00 p.m.

45 minutes 1 Hour 1 Hour = 2 hours 45 minutes.

The doctor spent 2 hours and 45 minutes with his patients.

2. The time taken to do the Math homework = 30 minutes

The time taken to write English notes = 15 minutes

The time taken to revise all other subjects = 70 minutes.

The total study time = 30 minutes + 15 minutes + 70 minutes = 115 minutes

3. The starting time of reading the book = 16:30

Duration of reading the book = 45 minutes

The time at which Nihil stopped reading:

45 minutes = 30 minutes + 15 minutes

30 minutes 15 minutes

16:30 hours 17:00 hours 17:15 hours

He stopped reading the book at 17:15 hours.

4. Starting time = 20:30 hours

Time taken = 50 minutes

50 minutes = 30 minutes + 20 minutes.

30 minutes 20 minutes

20:30 hours 21:00 hours 21:20 hours

He finished his homework at 21:20 hours.

5. The time of departure from Mumbai = 2315 hours = 11:15 p.m.

The time of arrival in Trivandrum = 1300 hours = 1:00 p.m.

The duration of the journey: 11:15 p.m.

12

The duration of the journey is 13 hours and 45 minutes.

1. a. There are 366 days in a leap year.

b. If 03.03.03 is a Friday, then the next Sunday will be on 05.03.03.

c. February is a month with 28 or 29 days.

d. 2 years = 24 months.

e. 50 months = 4 years and 2 months

2. a. 19 November 1996 = 19.11.1996

b. 15 August 1947 = 15.08.1947

c. 29 July 2023 = 29.07.2023

d. 28 February 2004 = 28.02.2004

3. a. Total days in June = 30

The days remaining in June = 0

The number of days in July = 23

The duration is 23 days

4. The number of days spent in Kashmir = 8 days

The number of days spent in Delhi = 9 days

The number of days spent in Dehradun = 4 days

The total number of days = 8 + 4 + 9 = 21 days

We know that, 1 week = 7 days

21 days = 3 weeks

Ram had a 3-week vacation.

5. 2 weeks = 14 days

Joby takes the leave from 5 March

The days until Joby returns to school = 5 March + 14 days = 19 March

Joby will rejoin the school on 19 March.

Word Problems 1. New Year begins on 1 January.

10 December is a Friday.

10 + 7 = 17 December will be Friday

17 + 7 = 24 December will be Friday.

24 + 7 = 31 December will be Friday.

1st January will be a Saturday.

2. The day of 25th December—Monday.

Riya’s birthday = 3rd Monday after 25th December.

The 1st Monday after 25th December = 25 + 7 = 1st January

The 2nd Monday after 25th December = 1 + 7 = 8th January.

The 3rd Monday after 25th December = 8 + 7 = 15th January.

Riya’s birthday is on 15th January.

3. The manufacturing date of the chocolate = 12.12.2023 = 12th December 2023

The expiry date is 18 months from the date of manufacturing

18 months = 12 months + 6 months

12 months after 12th December 2023 = 12th December 2024

6 months after 12th December 2024 = 12th June 2025

The chocolate will expire on 12th June 2025 or 12.06.2025

Chapter Checkup 1. a. This morning Emily woke up at 7 a.m.

b. She took 45 minutes to get ready, then it was 7:45 a.m.

c. She had her lunch at 12:30 p.m. in the cafeteria with her friends.

2. a. 6:42 b. 7:12 c. 12:56 d. 7:29 e. 10:21 f. 5:14

3. a. 2 hours after 4:30 a.m. in the morning - 6:30 a.m.

b. 3 hours after 8:45 p.m. in the evening - 11:45 p.m.

c. 1 hour after 10:00 p.m. at night - 11:00 p.m.

d. 4 hours after 1:20 p.m. in the afternoon - 5:20 p.m.

e. 1 hour before 11:40 a.m. in the morning - 10:40 a.m.

f. 2 hours before 9:15 p.m. in the evening - 7:15 p.m.

g. 3 hours before 12:00 noon - 9:00 a.m.

h. 1 hour after 7:55 p.m. at night - 8:55 p.m.

i. 2 hours before 2:30 p.m. in the afternoon - 12:30 p.m.

j. 3 hours after 6:10 a.m. in the morning - 9:10 a.m.

4. a. 6:30 a.m. = 06:30 hours b. 7:55 a.m. = 07:55 hours

c. 9:25 a.m. = 09:25 hours d. 12:00 a.m. = 00:00 hours

e. 1:03 p.m. = 13:03hours f. 09:15 p.m. = 21:15 hours

g. 10:24 p.m. = 22:24 hours h. 11:59 p.m. = 23:59 hours

5. a. 14:20 hours - 2:20 p.m. b. 15:45 hours - 3:45 p.m.

c. 21:12 hours - 9:12 p.m. d. 04:30 hours - 4:30 a.m.

e. 10:10 hours - 10:10 a.m. f. 14:55 hours - 2:55 p.m.

g. 20:18 hours - 8:18 p.m. h. 18:12 hours - 6:12 p.m.

i. 23:06 hours - 11:06 p.m. j. 16:50 hours = 4:50 p.m.

6. a. 1 hour = 60 minutes

2 hours = 60 × 2 = 120 minutes

b. 60 minutes = 1 hour

120 minutes = 120 ÷ 60 = 2 hours

c. 1 hour = 60 minutes

3 hours = 3 × 60 = 180 minutes.

180 minutes + 30 minutes = 210 minutes

d. 60 minutes = 1 hour

450 minutes = 450 ÷ 60

450 ÷ 60 = 7 hours 30 minutes

e. 1 hour = 60 minutes

g. 11:00 a.m. to 2 p.m. = 3 hours

2 p.m. to 2:45 p.m. = 45 minutes

The duration = 3 hours 45 minutes.

h. 5:30 to 6:00 p.m. = 30 minutes

6:00 p.m. to 9:00 p.m. = 3 hours

9:00 p.m. to 9:15 p.m. = 15 minutes

Duration = 3 hours + 15 minutes + 30 minutes = 3 hours 45 minutes.

i. 8:45 a.m. to 9:00 a.m. = 15 minutes

9:00 a.m. to 11:00 a.m. = 2 hours

11:00 to 11:30 = 30 minutes

The duration = 2 hours + 15 minutes + 30 minutes = 2 hours 45 minutes

j. 12:15 p.m. to 3:15 p.m. = 3 hours

3:15 p.m. to 3:30 p.m. = 15 minutes

The duration= 3 hours 15 minutes.

8. The starting time of colouring = 17:30 hours

The time taken to complete colouring= 100 minutes

100 minutes = 60 minutes + 40 minutes

7 450 420 –30

60 hours minutes

1 hour 15 minutes = 60 minutes + 15 minutes = 75 minutes.

f. 1 90 60 –30

60 hour minutes 90 minutes = 1 hour 30 minutes

g. 1 hour = 60 minutes

4 hours = 4 × 60 = 240 minutes

240 minutes + 20 minutes = 260 minutes

h. 60 minutes = 1 hour

300 minutes = 300 ÷ 60 = 5 hours

i. 1 hour = 60 minute

5 hours = 60 × 5 = 300 minutes

300 minutes + 45 minutes = 345 minutes

j. 60 minutes = 1 hour

390 minutes = 390 ÷ 60 = 6 hours 30 minutes

7. a. 8:00 a.m. to 2:00 p.m. = 6 hours

2:00 p.m. to 2:45 p.m. = 45 minutes

Duration = 6 hours 45 minutes.

b. 9:30 a.m. to 4:30 p.m. = 7 hours

4:30 to 5:00 = 30 minutes

5:00 to 5:15 = 15 minutes

6

60 hours minutes

–30

Duration = 7 hours + 30 minutes + 15 minutes = 7 hours 45 minutes.

c. 7:00 p.m. to 11:00 p.m. = 4 hours

11:00 p.m. to 11:30 p.m. = 30 minutes

Duration = 4 hours 30 minutes

d. 2:00 p.m. to 6 p.m. = 4 hours

6:00 p.m. to 6:45 p.m. = 45 minutes

The duration = 4 hours 45 minutes.

e. 10:15 a.m. to 1:30 p.m.

10:15 a.m. to 1:15 p.m. = 3 hours

1:15 p.m. to 1:30 p.m. = 15 minutes

The duration = 3 hours and 15 minutes.

f. 3:20 p.m. to 4:00 p.m. = 40 minutes

4:00 p.m. to 6:00 p.m. = 2 hours

6:00 p.m. to 6:10 p.m. = 10 minutes

The duration = 2 hours + 40 minutes + 10 minutes = 2 hours 50 minutes.

60 minutes = 1 hour

1 hour after 17:30 = 18:30

40 minutes after 18:30 = 19:10

Siya will finish colouring at 19:10 hours

9. The duration of meditation per day = 30 minutes

The duration of meditation in 4 days = 30 × 4 = 120 minutes.

120 minutes to hours

60 minutes = 1 hour

120 minutes = 120 ÷ 60 = 2 hours

Sravan meditated for 2 hours.

10. The duration of the yoga practice per day = 45 minutes

The duration of the yoga practice every week = 45 × 7 = 315 minutes.

60 hours minutes

5 315 300 –15

She practices yoga for 5 hours and 15 minutes per week.

Word Problems 1. The starting time of the movie = 1530 hours

The ending time of the movie = 1815 hours

The duration between 1530 to 1815

1530 to 1730 = 2 hours

1730 to 1815 = 45 minutes

The duration of the movie = (2 hours 45 minutes) – 20 minutes  = 2 hours 25 minutes

2. The departure time of the bus = 8:45 p.m.

The arrival time of the bus in Bangalore = 4:45 a.m.

The duration between 8:45 p.m. to 4:45 a.m. = 8 hours

8 hours to minutes:

8 hours = 8 × 60 = 480 minutes

The total time of the journey is 480 minutes.

3. 7th December is a Sunday

The New Year is on 1st January

Next Sunday = 7 + 7 = 14th December

The Sunday after 14th December = 14 + 7 = 21st December

The Sunday after 21st December = 21 + 7 = 28th December

1st January comes 4 days after 28th December.

The New Year will be on Thursday.

4. Today’s date = 10 January

The days in January = 31

The days left in January = 31 – 10 = 21 days.

The days left for Anil’s birthday = 45 – 21 = 24

Anil’s birthday will be on 24 February.

Chapter 13

Let's Warm-up 1. We can exchange  5  ₹10 notes for a ₹50 note.

2. We can exchange  5  ₹20 notes for a ₹100 note.

3. For a book that costs ₹120 we can give  1  note of ₹100 and  2  notes of ₹10.

Do It Yourself

13A 1. a. ₹154.56 = One hundred fifty-four rupees and fifty-six paise. b. ₹217.85 = Two hundred seventeen rupees and eighty-five paise. c. ₹396.48 = Three hundred ninety-six rupees and forty-eight paise. d. ₹469.05 = Four hundred sixty-nine rupees and five paise. e. ₹679.21 = Six hundred seventy-nine rupees and twenty-one paise. f. ₹748.49 = Seven hundred forty-eight rupees and forty-nine paise.

2. a. Three hundred eighty-nine rupees and sixty-three paise = ₹389.63 b. Five hundred forty-two rupees and eighty-three paise = ₹542.83 c. Six hundred fifty-two rupees and thirtynine paise = ₹652.39 d. Seven hundred thirty-three rupees and forty-two paise = ₹733.42 e. Eight hundred sixty-three rupees and seventy-seven paise = ₹863.77 f. Nine hundred seventy-four rupees and three paise = ₹974.03

3. a. ₹578.24 = 57824 paise b. ₹647.12 = 64712 paise

c. ₹846.25 = 84625 paise d. ₹945.37 = 94537 paise

e. ₹1015.48 = 101548 paise f. ₹1247.69 = 124769 paise

4. a. ₹635.12 b. ₹746.24 c. ₹847.61

d. ₹974.56 e. ₹1125.64 f. ₹1354.89

5. a. India Yen (¥)

b. China Pound (£)

c. United States Rupees (₹)

d. Japan Euro (€)

e. United Kingdom Dollar ($)

f. Germany Yuan (元)

13B

1. a. The cost of 4 combs = ₹116

The cost of 1 comb = ₹116 ÷ 4 = ₹29

b. The cost of 1 safety pin = ₹10

So, the cost of 3 safety pins = ₹10 × 3 = ₹30

c. The total bill amount = ₹150.50 + ₹116.00 + ₹72.00 + ₹20.00 + ₹30.00 = ₹388.50

2. a. The cost of 10 party blowers = ₹150

The cost of 1 party blower = ₹150 ÷ 10 = ₹15

b. The cost of 1 gift bag = ₹100

The cost of 30 gift bags = ₹100 × 30 = ₹3000

c. The cost of 1 packet of balloons = ₹40

So, the cost of 2 packets of balloons = ₹40 × 2 = ₹80

d. The total amount spent by Suhaas on the celebration = ₹320 + ₹150 + ₹750 + ₹1500 + ₹600 = ₹3320

3. The bill for Mariya:

Supermarket

Bill No. 1456/B

Bill Date: 05/11/2023

4. The bill for Ravi:

Supermarket

Bill No. 2364/A

Bill Date: 08/11/2023

4

The total amount with Ravi = ₹1000

The total bill amount = ₹181.50

The amount left with Ravi after his purchase = ₹818.50

5. The bill for Sam:

The total amount with Sam = ₹750

The total bill amount = ₹825.50

Clearly, Sam cannot purchase all the items.

Extra money required to purchase all the items

= ₹825.50 – ₹750 = ₹75.50 Toy Store

Bill No. 3654/D

Bill Date: 12/11/2023

13C 1. a. The total expenditure = ₹4000 + ₹1500 + ₹325.75 + ₹4500 + ₹625 + ₹950 = ₹11,900.75

So, the statement is true as ₹11,900.75 is more than ₹10,000.

b. The expenditure on education = ₹4000

The expenditure on food = ₹4500

Since ₹4000 is less then ₹4500, so the statement is false as the expenditure on education is less than the expenditure on food.

c. The expenditure on transportation = ₹325.75

The expenditure on clothing = ₹625

So, the statement is true as the expenditure on transportation is less than the expenditure on clothing.

d. The total expenditure = ₹11,900.75

So, the statement is true as the total expenditure is less than ₹12,000.

2. a. The amount spent on transportation = ₹48

The amount spent on milk = ₹50.50

So, the amount spent on transportation is less than the amount spent on milk.

b. The amount spent on medicine = ₹223.15

c. The highest amount is spent on groceries.

d. The total amount spent in a day = ₹325.25 + ₹50.50 + ₹48 + ₹125 + ₹223.15 = ₹771.90

3. The total earning of Kapil = ₹18,000

The total expenditure = ₹8000 + ₹500 + ₹1200 + ₹1500 + ₹775 + ₹1000 = ₹12,975

So, the money saved at the end of the month = ₹18,000 –₹12,975 = ₹5025

4. The total expenses of Kavya = ₹5000 + ₹3500 + ₹1000 + ₹1500 + ₹75 + ₹1625 = ₹13,400

The total expenses of Minal = ₹4500 + ₹4600 + ₹1750 + ₹1800 + ₹300 + ₹1260 = ₹14,210

13D

Since the expenses of Minal are more, so Kavya saves more money.

The difference between Kavya's and Minal's savings = ₹14,210 – ₹13,400 = ₹810.

5. The total expenses of Kunal = ₹10,230

The total savings of Kunal = ₹2765

So, the total money with Kunal at the starting of the month = ₹10,230 + ₹2765 = ₹12,995

1. The rent per day = ₹99

The number of days in January = 31

The rent paid in January = ₹99 × 31 = ₹3069

2. Pocket money of Mahi per week = ₹385

Number of days in a week = 7

Pocket money of Mahi per day = ₹385 � 7 = ₹55

3. The cost of 1 toy = ₹225

The cost of 4 toys = ₹225 × 4 = ₹900

The amount given to the shopkeeper = ₹1000

The amount that Aarav will get back from the shopkeeper = ₹1000 – ₹900 = ₹100

4. The ticket price for 1 adult = ₹275

The ticket price for 5 adults = ₹275 × 5 = ₹1375

The ticket price for 1 child = ₹125

The ticket price for 3 children = ₹125 × 3 = ₹375

Therefore, the total ticket price for 5 adults and 3 children = ₹1375 + ₹375 = ₹1750.

5. The total money collected by school = ₹5000

The refund given to 1 student = ₹20

The refund given to 20 students = ₹20 × ₹25 = ₹500

The money spent on the trip = ₹5000 – ₹500 = ₹4500

Word Problems 1. The money paid for 2 chocolate bars and 1 ice cream = ₹186

The cost of 1 ice cream = ₹60

So, the cost of 2 chocolate bars = ₹186 – ₹60 = ₹126

The cost of 1 chocolate bar = ₹126 � 2 = ₹63

2. The cost of 1 vanilla pastry = ₹25

The cost of 12 vanilla pastries = ₹25 × 12 = ₹300

The cost of 1 plum pastry = ₹35

The cost of 15 plum pastries = ₹35 × 15 = ₹525

The total amount paid = ₹300 + ₹525 = ₹825

Chapter Checkup

1. In

Figures In

Words

₹235.45 Two hundred thirty-five rupees and forty-five paise

₹356.72 Three hundred fifty-six rupees and seventy-two paise

₹582.30 Five hundred eighty-two rupees and thirty paise

₹648.47 Six hundred forty-eight rupees and forty-seven paise

₹790.52 Seven hundred ninety rupees and fifty-two paise

₹978.65 Nine hundred seventy-eight rupees and sixty-five paise

2. a. ₹236.45 = 23,645 paise  b. ₹345.78 = 34,578 paise

c. ₹598.14 = 59,814 paise  d. ₹894.69 = 89,469 paise

e. ₹1054.54 = 1,05,454 paise  f. ₹1568.17 = 1,56,817 paise

g. ₹1864.58 = 1,86,458 paise  h. ₹2045.65 = 2,04,565 paise 3. a. 3651 paise = ₹36.51  b. 4865 paise = ₹48.65

c. 5631 paise = ₹56.31  d. 7856 paise = ₹78.56

e. 12,567 paise = ₹125.67  f. 36,575 paise = ₹365.74 g. 65,237 paise = ₹652.37  h. 75,685 paise = ₹756.85 4. a. The cost of 3 packets of pencils = ₹120

So, the cost of 1 packet of pencils = ₹120 ÷ 3 = ₹40

b. The cost of 1 eraser = ₹10

The cost of 10 erasers = ₹10 × 10 = ₹100

c. The cost of 1 notebook = ₹50

So, the cost of 2 notebooks = ₹50 × 2 = ₹100

d. The total bill amount = ₹120 + ₹50 + ₹100 + ₹300 + ₹160 = ₹730

e. The total money with Sunita = ₹800

The total bill amount = ₹730

The amount left with her = ₹800 – ₹730 = ₹70

5. The picnic bill:

Food Store

Bill No. 3521/A

Bill Date: 25/11/2023

6. a. The money spent on medicines = ₹1500

b. The money spent on groceries and milk = ₹5500 + ₹3000 = ₹8500

c. Total expenditure = ₹18,400

d. Savings = Total earnings – total expenditure = ₹22,000 – ₹18,400 = ₹3600

7. The total amount earned by Rohan per month = ₹22,000

The total expenditure of Rohan = ₹3500 + ₹2600 + ₹4000 + ₹2000 + ₹1500 + ₹1200 = ₹14,800

The total savings of Rohan = ₹22,000 – ₹14,800 = ₹7200

The total earning of Mohit = ₹23,000

Total expenditure of Mohit = ₹16,300

The total savings of Mohit = ₹23,000 − ₹16,300 = ₹6700

Clearly, Rohan saves ₹500 more per month than Mohit.

So, Rohan saves ₹500 × 12 = ₹6000 more per year than Mohit.

8. The money saved by Naina in 1 week = ₹75

Total money required = ₹900

So, the number of weeks required to save money = ₹900 ÷ ₹75 = 12.

9.  Food Store

No. 3581/D

10. The monthly earning of Priya = ₹15,000

The annual savings of Priya = ₹18,000

The monthly savings of Priya = ₹18,000 ÷ 12 = ₹1500

Monthly expenditure = Monthly earning – monthly savings = ₹15,000 – ₹1500 = ₹13,500

Word Problems 1. The earning of Rohan in 1 day = ₹55

The number of days in 2 weeks = 14

The money earned by Rohan in 2 weeks = ₹55 × 14 = ₹770

2. The price of a saree = ₹895

The price of a coat = ₹1263

The money left with Mahi = ₹1526

The money that Mahi had in the beginning = ₹895 + ₹1263 + ₹1526 = ₹3684

3. The money with Kunal = ₹5000

The money given to Suhani = ₹1550

The remaining money = ₹5000 – ₹1550 = ₹3450

The money divided equally in 3 cousins = ₹3450

The money received by each cousin = ₹3450 ÷ 3 = ₹1150.

4. The money paid by each friend = ₹98

The money paid by 5 friends = ₹98 × 5 = ₹490

The total money paid by 5 friends and Rahul = ₹600

So, the money paid by Rahul = ₹600 – ₹490 = ₹110

Chapter 14

Let's Warm-up 1. There are 5 lions. They are the most in number.  2. There is 1 cat. It is the least in number.  3. There are 4 giraffes. They are 1 less in number than lions 4. There are 3 elephants. They are 1 more in number than the crocodiles.

5. There are 2 crocodiles. They are more in number than the cat Do It Yourself

14A 1. D, G, G, H, D, H, G, D, H, G

The number of times G has been listed is 4.

2. 9, 8, 5, 8, 6, 9, 4, 5, 2, 9, 8, 7

The number of students who obtained 9 marks are 3.

3. |||| |||| |||| |||| |||||

4. b, Green tea

5. People from Delhi = 12 People from Mumbai = 19

Total = 12 + 19 = 31

So, there are 31 people from Delhi and Mumbai together. 6.

e.

2. Digit

a.

a. 1 has appeared least number of times.

b. 4 and 6 have appeared the most number of times.

c. The numbers that have appeared equal number of times are:

2, 3 and 9 = 3 times 4 and 6 = 7 times

3. a. Number of marbles collected in week 6 = 5 + 5 + 5 + 4 = 19

b. The maximum number of marbles were collected in week 2 and week 4.

c. Number of marbles collected in week 3 = 14

Number of marbles collected in week 6 = 19

Number of extra marbles collected in week 6 than in week 3 = 19 – 14 = 5

d. Total number of marbles collected in all the weeks = 19 + 24 + 14 + 24 + 15 + 19 = 115

4.

a. Hours spend on playing games = 28

Hours spend on watching TV = 30

Total = 28 + 30 = 58 hours

b. Studying at home

c. Hours spend on playing games = 28 hours spend on time with family = 22

Total = 28 + 22 = 50 hours

d. Family time

14B 1. a. pictograph is a way to represent data using images and symbols. True b. A pictograph makes the data representation visually interesting and easy to understand. True

c. Pictographs cannot be used to compare two quantities. False

2.  Indoor Games Number of Students

Ludo

Chess

Carrom board

Marbles

Puzzle games

Key = 4 students

3.  Village Number of Animals

A B C D E

Key = 10 animals

d. 10 capsicums

4.  Ice Cream Flavor Number of Votes

Chocolate

Vanilla

Strawberry Mango

Apricot

Key = 1 vote

5. c is the correct option

14C 1. Since 1 picture = 4 fruits

Pictures of fruits given = 26

So, 26 pictures = 26 × 4 = 104 fruits

2. c. 12 students scored grade A and 14 students scored grade B.

So, a total of 26 students scored grades higher than grade C.

3. a. City A = 7 × 25 cm = 175 cm

City D = 2 × 25 cm = 50 cm

b. City B = 10 × 25 cm = 250 cm

City E = 3 × 25 cm = 75 cm

Rainfall at City B > Rainfall at City E

4. a. The number of pictures is the minimum on Monday.

So, the milkman sells the least number of bottles on Monday.

b. Bottles sold on Wednesday = 3 × 15 = 45 bottles

Bottles sold on Thursday = 6 × 15 = 90 bottles

Total = 45 + 90 = 135 bottles

c. Bottles sold on Friday = 7 × 15 = 105 bottles

Bottles sold on Monday and Wednesday = 5 × 15 = 75 bottles

Difference = 105 – 75 = 30 bottles

d. On Friday, he sold 7 × 15 = 105 bottles and On Saturday, he sold 8 × 15 = 120 bottles

e. Thursday

f. Bottles sold on Monday and Tuesday together

= 7 × 15 = 105 bottles

Bottles sold on Wednesday and Thursday together

= 9 × 15 = 135 bottles

Difference = 135 – 105 = 30 bottles

5. a. Apples consumed in school 1 = 10 × 35 = 350 apples

Apples consumed in school 2 = 2 × 35 = 70 apples

Apples consumed in school 3 = 3 × 35 = 105 apples

Oranges consumed in school 1 = 4 × 45 = 180 oranges

Oranges consumed in school 2 = 2 × 45 = 90 oranges

Oranges consumed in school 3 = 6 × 45 = 270 oranges

In school 2 and school 3, the number of students who consumed oranges is more than those who consumed apples.

b. Number of students who consumed apples at school 1 = 10 × 35 = 350 students

Number of students who consumed apples at school 3 = 3 × 35 = 105 students

Difference = 350 − 105 = 245 students

c. In school 2, 2 × 35 = 70 students consumed apples and 2 × 45 = 90 students consumed oranges.

Total = 70 + 90 = 160 students

Word Problems 1. a. Number of Math books = 10 × 10 = 100 books

Number of Science books = 9 × 10 = 90 books

Total = 100 + 90 = 190 books

b. Number of Computer books = 8 × 10 = 80

Number of Hindi books = 6 × 10 = 60

Difference = 80 – 60 = 20 books

c. Number of Social studies books = 8 × 10 = 80

Number of English books = 7 × 10 = 70

Total = 80 + 70 = 150 books

d. Difference between number of Hindi books and Science books

= 90 – 60 = 30

e. Number of Math and Hindi books = 100 + 60 = 160

Number of Computer and Science books = 80 + 90 = 170

Number of Computer and Science books is more.

2. a. 7 complete books of Hindi + 1 half book of Hindi = 150

7 and 1 2 books = 150 = 20 + 20 + 20 + 20 + 20 + 20 + 20 + 10

So, 1 book = 20    Key: = 20 books

b. Hindi books = 150

English books = 9 × 20 = 180 books

Math books = 7 × 20 = 140 books

Science = 5 1 2 × 20 = 110 books

History = 4 × 20 = 80 books

So, there are 110 science books.

c. History has 80 books and English has 180 books.

Difference = 100 books

d. History has 80 books and Hindi has 150.

Total = 230 books

Age Group

14E

1. a. LPG is used highest in number. b. 10 houses

2. Hair bands = 12 Hair clips = 6 Rings = 2 Watches = 2

Wallets = 4 Total = 12 + 6 + 2 + 2 + 4 = 26

3. a. Blue is the favourite colour among the students.

b. Green is the least favourite colour among the students.

c. 3 students  d. 2 + 5 = 7 students

4. a. Highest number of students participated from Grade 5.

b. Lowest number of students participated from Grade 3.

c. Grade 4

d. Students participated from grade 1 = 50

Students participated from grade 3 = 40

Difference = 50 – 40 = 10 students

e. Students participated from grade 5 = 120

Students participated from grade 2 = 90

Difference = 120 – 90 = 30

5. a. Number of pencils bought = 8

b. Number of notebooks = 6

Number of erasers = 10

Difference = 10 – 6 = 4

c. Most bought item = 12

Least bought item = 6  Difference = 6

d. Total items bought = 10 + 8 + 12 + 6 = 36

Word Problem 1. Total number of books with Sanjay = 4 + 6 + 8 + 2 = 20 books

Books donated by Sanjay = 1 4 of 20 = 5

Books left with on the bookshelf = 20 – 5 = 15

14F 1. 1 2 2. a. Three-fourths b. One-fourth

c. Lawn tennis is liked by 1 4 of 60 = 15 children

Lawn tennis is not liked by 60 – 15 = 45 children

3. a. False b. False c. False d. True

4. a. Asia b. Africa

c. Total parts = 60 + 42 + 32 + 24 + 18 + 14 + 10 = 200

Fraction of total area covered by Europe = 14 200 = 7 100

d. Fraction of total area covered by Australia and South America = 10 + 24 200 = 34 200 = 17 100

e. Antarctica

5. Category A = 1 3 , category B = 1 4 and category C = 5 12

1 4 < 1 3 < 5 12

So, Category B is smallest, Category C is biggest Thus, category B, category A, category C

Word Problem 1. a. Fraction of playing = 50 100 = 1 2

b. Fraction for cycling = 25 100 = 1 4

c. 10 people voted for walking. 100 – 10 = 90 people do not like walking.

So, the fraction = 90 100 = 9 10

Chapter Checkup 1. |||| |||| |||| = 14  Option a represents 14.

2. 4 × 5 = 20 paintings  Option a

3. 5  |||||  5 is wrongly represented.  Option d

4. a. |||| ||||  b. |||| |||| |||

c. |||| |||| |||| d. |||| |||| |||| |||

5. a. 10 kids like to go to national museum.

b. 15 children would like to go to adventure island.

c. Number of children who would like to go to the zoo = 20

Number of children who would like to go to the rail museum = 5

How many more likes to go to the zoo = 20 – 5 = 15

d. 20 + 15 + 10 + 5 = 50

6. a. 22 bicycles were sold in week 1.

b. Bicycle sold in week 1 = 22

Bicycles sold in week 4 = 15

Total bicycles sold in Weeks 1 and 4 = 22 + 15 = 37

c. 22 + 29 + 16 + 15 + 25 = 107

7. b. 4 × 50 = 200. So, Madhav harvested 200 pumpkins

8. 30 students prefer dancing in grade 4 and 25 students prefer dancing in grade 3.  Difference = 30 – 25 = 5

9. 1 4 fraction of students goes to school by car.

Fraction of students who do not go to school by car = 1 – 1 4

= 4 4 –1 4 = 3 4

10. Fraction for chess = 10 40 = 1 4

Fraction for ludo = 20 40 = 1 2

Fraction for carrom = 5 40 = 1 8

Word Problem 1. Cakes sold on Day 1 = 6

So, cakes sold on day 5 = 2 × 6 = 12

Cakes sold on Day 3 = 12

So, cakes sold on Day 6 = 12 2 = 6

Total cakes sold over 6 days = 6 + 9 + 12 + 6 + 12 + 6 = 51

About the Book

Imagine Mathematics seamlessly bridges the gap between abstract mathematics and real-world relevance, offering engaging narratives, examples and illustrations that inspire young minds to explore the beauty and power of mathematical thinking. Aligned with the NEP 2020, this book is tailored to make mathematics anxiety-free, encouraging learners to envision mathematical concepts rather than memorize them. The ultimate objective is to cultivate in learners a lifelong appreciation for this vital discipline.

Key Features

• Let’s Recall: Helps to revisit students’ prior knowledge to facilitate learning the new chapter

• Real Life Connect: Introduces a new concept by relating it to day-to-day life

• Examples: Provides the complete solution in a step-by-step manner

• Do It Together: Guides learners to solve a problem by giving clues and hints

• Think and Tell: Probing questions to stimulate Higher Order Thinking Skills (HOTS)

• Error Alert: A simple tip off to help avoid misconceptions and common mistakes

• Remember: Key points for easy recollection

• Did You Know? Interesting facts related to the application of concept

• Math Lab: Fun cross-curricular activities

• QR Codes: Digital integration through the app to promote self-learning and practice

About Uolo

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