IM_TM_G08_MA_MB_Text_AY25-26_eBook by Uolo

8 Master Mathematical Thinking MATHEMATICS

Teacher Manual

MATHEMATICS

Master Mathematical Thinking

Acknowledgements

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

Creative Directors: Bhavna Tripathi, Mangal Singh Rana, Satish

Book Production: Sanjay Kumar Goel, Vishesh Agarwal

Project Lead: Neena Aul

VP, Learning: Abhishek Bhatnagar

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

First impression 2024

Second impression 2025

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Book Title: Imagine Mathematics Teacher Manual 8

ISBN: 978-81-984519-6-5

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Preface

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

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

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

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

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

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

Numbers up to 8 Digits 1

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

Alignment

C-1.1:

C-4.3:

Numbers up to 8 Digits 1

Imagine Mathematics Headings

Place Value, Face Value and Expanded Form

Indian and International Number Systems

Comparing and Ordering Numbers

Numbers up to 8 Digits 1

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

Learning Outcomes

Students will be able to:

Rounding–off Numbers

Learning Outcomes

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

Alignment to NCF

Numbers up to 8 Digits 1

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

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

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

Let’s Recall

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

Alignment to NCF

Place Value, Face Value and Expanded Form

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

Indian and International Number Systems

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

Comparing and Ordering Numbers

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

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

Vocabulary

Rounding–off Numbers

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

Learning Outcomes

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

Let’s Recall

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

Teaching Aids

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

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

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

Vocabulary

Alignment to NCF

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

Teaching Aids

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

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

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

Let’s Recall

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

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

Vocabulary

Teaching Aids

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

Learning Outcomes

Numbers up to 8 Digits 1

Numbers up to

Alignment to NCF

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

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

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

Learning Outcomes

Let’s Recall

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

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

Vocabulary

Alignment to NCF

Teaching Aids

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

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

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

Let’s Recall

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

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

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

Chapter: Numbers up to 8 Digits

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

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

Vocabulary

Teaching Aids

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

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

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

Place Value, Face Value and Expanded Form

Learning Outcomes

Teaching Aids

Imagine Maths Page 2

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

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

Activity

Teaching Aids

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

Activity

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

Extension Idea

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

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

Extension Idea

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

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

Indian and International Number Systems Imagine Maths Page 5

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

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

Learning Outcomes

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

Indian and International Number Systems Imagine Maths Page 5

Teaching Aids

Learning Outcomes

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

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

Activity

Teaching Aids

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

Activity

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

Extension Idea

Ask: How many lakhs are there in 10 million?

Answers: Answers, provided at the end of each chapter, for the questions given in DoItTogether and ThinkandTell sections of the ImagineMathematics book.

Answers

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

Extension Idea

Ask: How many lakhs are there in 10 million?

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

Period Plan

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

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

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

Chapters No. of Periods

1. Rational Numbers

2. Solving Equations in One Variable

3. Polygons and Quadrilaterals

4. Bar Graphs and Histograms

Break-up of Topics

Addition and Subtraction of Rational Numbers

Properties of Addition and Subtraction of Rational Numbers

Multiplication of Rational Numbers

Division of Rational Numbers

Properties of Multiplication and Division of Rational Numbers

Revision

Transposing Method

Simplifying Equations to Linear Form

Application of Linear Equations

Revision

Classifying Polygons

Properties of Quadrilaterals

Trapezium and Kite

Parallelograms

Rectangle, Square and Rhombus

Interior Angle Sum Property

Exterior Angle Sum Property

Revision

Frequency Distribution of Ungrouped Data

Frequency Distribution of Grouped Data

Drawing and Reading Bar Graphs

Drawing Double Bar Graphs

Reading Double Bar Graphs

Drawing Histograms

Reading Histograms

Revision

Sectors and Central Angles

5. Pie Charts

6. Probability

7. Squares and Square Roots

8. Cubes and Cube Roots

Theoretical Probability Experimental Probability

Square of a Number

Patterns of Square Numbers

Finding Square Roots Using Prime Factorisation

Finding Square Roots Using Long Division

Cube Cube Root Through Prime Factorisation Method

Percentages

9. Percentage and Its Applications

10. Compound Interest

11. Algebraic Expressions

12. Area of Polygons

Finding Simple Interest

Finding Compound Interest Word Problems

Terms and Factors of an Algebraic Expression

Addition of Algebraic Expressions; Subtraction of Algebraic Expressions

Multiplying a Monomial by a Monomial

Multiplying a Monomial by a Polynomial

Multiplying a Polynomial by a Polynomial Revision

Area of Figures Made with Parallelograms, Triangles and Circles

Area of a Trapezium and Rhombus

Area of General Quadrilaterals

Area of Combined Shapes Revision

13. Surface Area and Volume of Solids 9

14. Exponents and Powers 7

15. Direct and Inverse Proportions 7

16. Factorisation and Division of Algebraic Expressions 10

Surface Area of Cuboids

Surface Area of Cubes

Surface Area of Cylinders

Volume of Cuboids

Volume of Cubes

Volume of a Cylinder

Revision

Positive Integral Exponent of a Rational Number

Negative Integral Exponent of a Rational Number

Applying Laws of Exponents

Use of Exponents

Revision

Direct Proportion

Problems on Inverse Proportion

Time and Work

Pipes and Cisterns Problems

Revision

Using Common Factors

Using Grouping and Regrouping Terms

Factorisation Using Identities

Factorisation of Quadratic Trinomials

Dividing a Monomial by a Monomial

Dividing a Polynomial by a Monomial

Dividing a Polynomial by a Polynomial

Revision

Reading Double Line Graphs

Drawing Double Line Graphs

Cartesian Coordinate of a Point

17. Linear Graphs 8

Drawing a Linear Graph

Reading Linear Graphs

Revision

Total Number of Periods

Rational Numbers 1

Learning Outcomes

Students will be able to: add and subtract rational numbers given in different forms. identify and apply the properties of rational numbers for addition and subtraction. multiply rational numbers given in different form. divide rational numbers given in different forms. identify and apply the properties of rational numbers for multiplication and division.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to represent rational numbers on number line and compare and order rational numbers.

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

Vocabulary

rational number: a number expressed as P q , where p and q are integers, and the denominator ≠ 0

Teaching Aids

Bingo cards; Problem cards; Question cards with different addition and subtraction problems based on their properties; Flower cards; Crayons; Maze sheet with division problems; Table cards

Chapter: Rational Numbers

Addition and Subtraction of Rational Numbers

Learning Outcomes

Students will be able to add and subtract rational numbers given in different forms.

Teaching Aids

Bingo cards; Problem cards

Activity

Briefly talk about adding and subtracting rational numbers. Make problem cards (Example, 2 5 + 5 8 , –1 3 + 7 6 , 0.625 – 5 –3 , etc.) and bingo cards (showing answers to the corresponding problem cards).

Bingo cards Problem cards

Divide the class into groups. Distribute a bingo card to each group.

Shuffle the problem cards and put them into a box. Pick a card and read out the problem. Instruct the students to solve it (encourage them to use the LCM method) and check whether the answer is on their bingo card. If it does, they should cross out that number. The first group to cross out a row and a column on their bingo card wins!

Ask the students to write the answers in their notebooks.

Extension Idea

Ask: Which decimal number should be subtracted from 12 5 to get – 3 8 ? Say: 12 5 –

Properties of Addition and Subtraction of Imagine Maths Page 6 Rational Numbers

Learning Outcomes

Students will be able to identify and apply the properties of rational numbers for addition and subtraction.

Teaching Aids

Question cards with different addition and subtraction problems based on their properties 2 5 + 5

Activity

Briefly talk about adding and subtracting rational numbers.

Ask the students to work in pairs. Give each pair question cards with different addition and subtraction problems based on their properties, such as 2 5 + 3 4 and 3 4 + 2 5 .

Instruct each student in the pair to solve one question and then compare their answers to see if they got the same result. Help them deduce that if their answers match, it shows that rational numbers follow the commutative property of addition.

Repeat the activity for other properties of addition and subtraction by giving them different types of questions.

Ask questions like, “What should you add to or subtract from –3 2 to get –3 2 ?” Discuss student responses. Instruct the students to note down their observations and learnings in their notebooks.

Extension Idea

Ask: What is the additive inverse of 3

Say:

Multiplication of Rational Numbers Imagine Maths Page 8

Learning Outcomes

Students will be able to multiply rational numbers given in different forms.

Teaching Aids

Flower cards; Crayons

Activity

Briefly revisit the concepts of multiplying and dividing rational numbers.

Prepare flower cards with questions written in the centre and blank petals for students to fill in the answers.

Ask the students to work in groups. Distribute one flower card and crayons to each group.

Instruct them to solve the questions on the cards in their notebooks and write the corresponding answers on the petals. Ask them to shade the petals as they get the answer to each question.

After the activity, encourage the groups to exchange cards to verify their solutions. The group with all the correct answers gets to showcase their flower card on the teacher’s table for the rest of the day.

Division of Rational Numbers

Learning Outcomes

Students will be able to divide rational numbers given in different forms.

Imagine Maths Page 9

Teaching Aids

Maze sheet with division problems; Crayons

Activity

Ask the students to work in pairs. Distribute the maze sheet and crayons to each pair. Ask the students to pick one question and solve it in their notebooks. Colour the question and the answer pair in the same colour. Ask them to repeat the process for all other questions.

Ask question such as, “Did they find LCM to solve, or did they use multiplicative inverse to solve.”

Extension Idea

Ask:

Properties of Multiplication and Division of Imagine Maths Page 11

Rational Numbers

Learning Outcomes

Students will be able to identify and apply the properties of rational numbers for multiplication and division.

Teaching Aids

Table cards

Activity

Ask the students to work in groups. Distribute a card with a table drawn on it to each group. Write one problem for each property of rational numbers related to multiplication and division on the board. Instruct the students to solve the problems in their notebooks, deduce the concepts and fill in the table with their learnings. Once they are done, discuss their findings. Have them check whether their result is the same as other groups. Discuss any challenges they faced and help them bridge the gaps, if any.

Associative

Distributive

Commutative

Ask questions like, “What should be multiplied by – 1 7 to get 0?”

Give the students some practice questions and encourage them to write the solutions in their notebooks.

Answers

1.

Addition of Rational Numbers

Think and Tell

No (pi) is not a rational number, as it cannot be expressed in the form of p q

Do It Together

1. 3.6 can also be written as 36 10

The LCM of 4, 5 and 10 is 20.

Write the rational numbers with the denominator as 20. = 75 20 ; 92 20 ; 72 20

Hence, 75 + 92 + 72 = 239 = 239 20 = 11 19 20

2. On converting 6 ‒13 into a rational number with a positive denominator, we get 6 ‒13 × ‒1 ‒1 = ‒6 13

‒2.8 can also be written as = ‒28 10

The LCM of 3, 13 and 10 is 390.

Write the rational numbers with the denominator as 390 = 650 390 ; ‒180 390 ; ‒1092 390

Hence, 650 + (−180) + (‒1092) = 622 = ‒622 390 = ‒311 195

2. Subtraction of Rational Numbers

Do It Together

1. Convert both rational numbers to the same form. On converting 12 5 to decimal form, we get 12 5 = 2.4

6.25 ‒ 2.4 = 3.85 = 385 100 = 77 20 = 3 17 20 in fraction form

2. Convert 5 ‒2 to a positive denominator 5 ‒2 × ‒1 ‒1 = ‒5 2 3 1 2 = 7 2 7 2 ‒‒5 2 = 12 2 = 6

3. Properties of Addition and Subtraction of Rational Numbers

Do It Together

1. 5 2 ‒ 0 = 5 2

Property used = Subtraction property of zero

2. 5 2 + (‒3) 9 = ‒3 9 + 5 2

Property used = Commutative property

3. 12 25 + ‒12 25 = 0

Property used = Additive Inverse

4. Multiplication of Rational Numbers

Do It Together

1. Converting 8 ‒10 into positive denominator, we get, 8 ‒10 × ‒1 ‒1 = ‒8 10 5 9 × ‒8 10 = ‒40 90 = ‒4 9

2. Converting 2.5 to the same form as the other number we get, 2.5 = 25 10 25 10 × ‒20 35 = ‒10 7

5. Division of Rational Numbers Do It Together 1. ‒15

3.5 can also be written as 35 10

6. Properties of Multiplication and Division of Rational Numbers Do It Together

Property used: Division over subtraction

Property used: Multiplicative property of 0.

13 18 × 1 = 13 18

Property used: Multiplicative Identity

Solving Equations in One Variable 2

Learning Outcomes

Students will be able to: solve linear equations using the transposing method. simplify equations to the linear form and solve them. apply linear equations in solving real-life problems.

Alignment to NCF

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

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

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

Let’s Recall

Recap to check if students know about the balancing and transposing methods used to solve linear equations. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

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

Teaching Aids

Puzzle cutouts with linear equations and solutions written; Interlocking puzzles with linear equations and their solutions; Word problem cards

Chapter: Solving Equations in One Variable

Transposing Method

Learning

Outcomes

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

Teaching Aids

Puzzle cutouts with linear equations and solutions written Activity

Discuss the transposing method to solve linear equations.

Ask the students to work in groups. Distribute puzzle cutouts with linear equations and solutions to each group.

Instruct the groups to arrange the cutouts so that each linear equation is placed next to its correct solution. Ask them to solve the equations in their notebooks and then put the pieces together to complete the puzzle. Record the time taken by each group to complete the puzzle. Check the puzzles created by each group and announce the fastest or winning group.

Extension Idea

Imagine Maths Page 19

Ask: Sarah is three years older than twice her brother’s age. If Sarah is 13 years old, what is her brother’s age? Say: Let Sarah’s brother’s age be x years. Then, 2x + 3 = 13 ⇒ x = 5. So, her brother is 5 years old.

Simplifying Equations to Linear Form

Learning Outcomes

Students will be able to simplify equations to the linear form and solve them.

Teaching Aids

Interlocking puzzles with linear equations and their solutions Activity

Show the students some equations and have them simplify them to form linear equations.

Ask the students to work in groups. Distribute puzzle interlocking cards with linear equations and their solutions to each group.

Imagine Maths Page 22

Instruct the groups to simplify and solve the linear equations in their notebooks and join them to the interlocking pieces with their solutions. Record the time taken by each group to complete all the puzzles. Check the puzzles created by each group and announce the fastest or winning group.

Application

Learning Outcomes

Students will be able to apply linear equations in solving real-life problems.

Teaching Aids

Word problem cards

Activity

Ask the students to work in groups. Distribute cards to each group containing different word problems, such as:

(1) Thelengthofarectangleis5.5metresmorethan1.9timesitswidth.Iftheperimeteroftherectangleis40metres, finditslengthandwidth.

(2) Simranthoughtoftwoconsecutiveoddnumbersthathaveasumof44.Whatarethetwonumbersshethoughtof?

(3) Acabcharges₹650perdayplus₹8perkm.IfRohitpaid₹1750foraday,howmanykilometresdidhetravel?

Instruct the students to the problem carefully, create a linear equation based on the problem on their card and solve it in their notebooks. Once they have solved the problem, ask them to pass their cards to the other members of their group so that everyone can solve different problems. Ask them to discuss their answers within their group. Finally, bring the class together to discuss the answers and solutions.

Extension Idea

Instruct: Create your own word problem that involves a variable.

Say: There can be many such problems. One could be: Tomhas$20morethantwicetheamountofmoneyJanehas inherwallet.IfTomhas$50,howmuchmoneydoesJanehave?

Answers

1. Transposing Method

2. Simplifying Equations to Linear Form

Do It Together

3. Application of Linear Equations

Do It Together

Let the distance travelled be x km.

Fixed charges for a day = ₹500

Rate of the car for each km = ₹12

Total bill amount = ₹3320

Polygons and Quadrilaterals 3

Learning Outcomes

Students will be able to: classify polygons based on their properties. classify quadrilaterals based on their properties. identify the properties of trapezium and a kite and apply them to solve problems. identify properties of parallelogram and apply them to solve problems. identify properties of rhombus, a square and a rectangle and apply them to solve problems. deduce the interior angle sum property of polygons and quadrilaterals and apply it. deduce the exterior angle sum property of polygons and quadrilaterals and apply it.

Alignment to NCF

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

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

C-7.3: Proves theorems using Euclid’s axioms and postulates – for angles, triangles, quadrilaterals, circles, area-related theorems for triangles and parallelograms

Let’s Recall

Recap to check if students know how to identify open curves, closed curves, polygons and non-polygons. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

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

convex polygon: a polygon in which the measure of each angle is less than 180°

concave polygon: a polygon in which the measure of one of the angles is more than 180° supplementary angles: a pair of angles whose sum is 180° bisect: to divide a line, angle or shape into two equal parts

congruent: figures having exactly the same shape and size transversal: a line that intersects two or more lines at distinct points

Teaching Aids

Cutouts of convex and concave polygons (pentagon, hexagon, heptagon, octagon, nonagon and decagon); Cutouts of concave and convex quadrilaterals (parallelogram, rectangle, trapezium, kite, rhombus); Protractor; Ruler; Straws; Scissors; Glue; Cutouts of polygon (heptagon, octagon, nonagon, decagon); Cutouts of quadrilaterals (parallelogram, rectangle, trapezium, kite, rhombus); Sheet of paper with 4 irregular convex polygons drawn on it

Chapter: Polygons and Quadrilaterals

Classifying Polygons

Learning Outcomes

Students will be able to classify polygons based on their properties.

Teaching

Aids

Cutouts of convex and concave polygons (pentagon, hexagon, heptagon, octagon, nonagon and decagon)

Activity

Show the students some cutouts of polygons and non-polygons. Discuss the definition of polygons, emphasising that they are simple closed shapes made up of 3 or more straight sides.

Ask the students to work in groups of 5. Distribute five shape cutouts to each group.

Draw the following table on the board and ask the students to copy this table into their notebooks.

Polygon

Number of Sides (n) Diagonals Convex/Concave

Instruct each student in the group to outline one polygon in their notebook and then draw its diagonals. Ask them to complete the table and then discuss in their groups to deduce the formula for finding the number of diagonals in a polygon with n sides.

Ask questions like, “How can you find the number of diagonals in a polygon with n sides? What do you notice about the diagonals of a concave polygon?”

Ask the students to trace the other polygons from their cutouts in their notebooks and draw their diagonals. They can then check the number of diagonals that they get against what they have written in the table.

Properties of

Quadrilaterals

Learning Outcomes

Students will be able to classify quadrilaterals based on their properties.

Teaching Aids

Imagine Maths Page 40

Cutouts of concave and convex quadrilaterals (parallelogram, rectangle, trapezium, kite, rhombus); Protractor; Ruler

Activity

Begin the class by showing the students a convex and a concave quadrilateral.

Ask them to identify the similarities and the differences between the two and have a discussion on convex and concave quadrilaterals.

Divide the class into small groups or pairs. Distribute a cutout of a quadrilateral to each group.

Instruct the groups to measure each interior angle of their quadrilateral cutout using protractors. Ask them to measure the sides and the perpendicular distance between opposite sides using a ruler to check if they are parallel and equal.

Ask them what additional features they noticed that define the shape.

Draw a table on the board as shown. Ask the students to copy the table into their notebooks and record their results in the table.

Name of Quadrilateral:

Guide them to identify the type of quadrilateral based on the measures recorded.

Extension Idea

Ask: The measures of four angles of a quadrilateral are 30°, 70°, 50° and x. If the sum of these four angles is 360°, identify the type of quadrilateral.

Say: As 30° + 70° + 50° + x = 360°, then x = 360° – 150° = 210°. So, it is a concave quadrilateral.

Trapezium and Kite

Learning Outcomes

Students will be able to identify the properties of a trapezium and a kite and apply them to solve problems.

Teaching Aids

Straws; Scissors; Glue

Activity

Draw a trapezium and a kite on the board.

Explain that a quadrilateral with only one set of opposite sides parallel is a trapezium, while a quadrilateral where both pairs of adjacent sides are of equal length is a kite.

Ask the students to work in pairs. Distribute a set of straws to each pair. Instruct them to cut the straws and paste them in their notebooks to create both a trapezium and a kite using the properties of each shape. Following this, present two problems (one for each shape) involving missing angles to be solved by applying the respective properties. For example:

1. Giventhattwoadjacentanglesofatrapeziummeasure70°and110°,determinethemeasuresoftheothertwoangles.

2. Iftheanglebetweentwounequalsidesofakiteis110°,andtheanglebetweenthetwolongersidesis50°,calculate themeasuresoftheremainingtwoangles.

Ask the students to work individually on these problems and then engage in group discussions to compare and analyse their solutions.

Learning Outcomes

Students will be able to identify the properties of a parallelogram and apply them to solve problems.

Teaching Aids

Straws; Scissors; Glue

Activity

Begin the class by drawing a parallelogram on the board.

Explain that a parallelogram is a quadrilateral in which both pairs of opposite sides are equal and parallel and discuss its properties.

Ask the students to work in pairs. Distribute a set of straws to each pair. Ask them to cut the straws and paste it in their notebooks to create a parallelogram using the properties of the shape.

Following this, present a problem involving missing angles to be solved by applying the properties of a parallelogram. For example:

1. Giventhattwoadjacentanglesofaparallelogrammeasure60°and120°,determinethemeasuresoftheremaining twoangles.

Ask the students to work individually on these problems and then engage in group discussions to compare and analyse their solutions.

Extension Idea

Ask: If the opposite sides of a parallelogram measure 15 cm and 20 cm, are the two triangles formed by the longer diagonal congruent?

Say: Yes, the triangles are congruent using the SAS congruence criterion, as the opposite sides and opposite angles are equal. Additionally, the SSS congruence criterion applies since the diagonal serves as the common side.

Rectangle, Square and Rhombus

Learning

Outcomes

Imagine Maths Page 45

Students will be able to identify the properties of a rectangle, a square and a rhombus and apply them to solve problems.

Teaching

Aids

Straws; Scissors; Glue

Activity

Begin the class by drawing a rhombus, a square and a rectangle on the board.

Explain that a rectangle is a parallelogram with only right angles and that it shares this property with a square. Explain that a square which has equal sides and angles, differs from a rhombus where all the sides are equal but only the opposite angles match and one diagonal is longer than the other. Discuss all the properties of these shapes. List the properties on the board.

Ask the students to work in pairs. Distribute a set of straws to each pair. Ask them to cut the straws to create a rhombus, a square and a rectangle, in their notebooks.

Following this, present a problem for each shape that requires applying the respective properties. For example:

1. Thelengthandthebreadthofarectangleare6cmand8cm,findthelengthofitsdiagonal.

2. Findthelengthofthediagonalofasquareifitsareais100sq.cm.

3. Thelengthsofthediagonalsofarhombusare16cmand12cm.Determinetheperimeteroftherhombus.

Ask the students to work individually on these problems and then engage in group discussions to compare and analyse their solutions.

Ask the students to compare the shapes based on their properties. Also ask them what common properties these shapes share.

Interior Angle Sum Property

Learning Outcomes

Imagine Maths Page 49

Students will be able to deduce the interior angle sum property of polygons and quadrilaterals and apply it.

Teaching Aids

Cutouts of polygon (heptagon, octagon, nonagon, decagon); Cutouts of quadrilaterals (parallelogram, rectangle, trapezium, kite, rhombus); Protractor

Activity

Begin the class with a discussion on how the sum of the angles in a triangle is 180°.

Ask the students to work in groups of 5. Distribute 5 cutouts of a polygon, two of which will be quadrilaterals, to each group and ask each student in a group to take one polygon.

Draw the table on the board and ask the students to copy it into their notebooks.

Instruct the students to mark one of the vertices in the polygon, draw diagonals from that vertex to form triangles, count the number of triangles and multiply it by 180°. They will also measure all the angles of each polygon and complete the table. Ask them to then discuss within their groups how to deduce the formula to find the angle sum of a polygon with n sides. Give the students some problems where they need to find the missing angle in a given polygon. Next, instruct the students to take the quadrilateral cutout and cut the angles, turn them around to form a complete angle to deduce that the sum of all the interior angles of a quadrilateral is 360°.

Ask questions like, “What do you notice about the sum of interior angles in a quadrilateral and other polygons with 5 or more sides?”

Extension Idea

Ask: If the interior angle sum of a regular polygon is 900°, what is the measure of each angle and the total number of sides in the polygon?

Say: We know that the interior angle sum of a polygon can be found using the formula (n – 2) × 180°. Here, the interior angle sum is given as 900°. So, (n – 2) × 180° = 900° or n = 7. A polygon with 7 sides is a heptagon.

For a regular heptagon, we can find the approximate measure of each angle as 900 7 = 128.6°.

Exterior Angle Sum Property

Learning Outcomes

Imagine Maths Page 50

Students will be able to deduce the exterior angle sum property of polygons and quadrilaterals and apply it.

Teaching Aids

Sheet of paper with 4 irregular convex polygons drawn on it; Scissors; Cutouts of quadrilaterals (parallelogram, rectangle, trapezium, kite, rhombus); Protractor

Activity

Begin the class with a discussion on how the measure of a complete angle is 360°.

Instruct the students to work in groups of 4. Distribute the sheet with shapes drawn on it to each group.

Instruct each student to pick any one shape (quadrilateral or polygon with 5 or more sides) and extend all the arms of its angles to show the exterior angles. Then, they should cut out the exterior angles and turn them around to make an angle. Ask them to discuss within their groups and deduce the exterior angle sum property of a polygon and a quadrilateral.

Explain that the sum of the exterior angles in a polygon is 360°.

Give the students some problems where they need to find the missing angle in a polygon using the exterior angle sum property.

Instruct them to outline any one on a sheet of paper and extend the angles. Ask them to measure each exterior angle of the shape using a protractor and add the measures to find the sum. Help them deduce that the sum of all the exterior angles of a quadrilateral is 360°.

Ask questions like, “Is the sum of all four exterior angles the same for each quadrilateral? Discuss the exterior angle sum property of quadrilaterals.”

Extension Idea

Ask: In a quadrilateral, the exterior angles are in the ratio 1:2:3:4. What is the measure of each exterior angle of the quadrilateral?

Say: As the sum of the exterior angles is 360°, x + 2x + 3x + 4x = 360° → 10x = 360 → x = 36°. So, the measures of the exterior angles are 36°, 72°, 108° and 144°.

1. Classifying Polygons

Think and Tell

Triangles are polygons that are most commonly seen in our everyday life.

Think and Tell

No, it is not possible to have a concave polygon with three sides. A concave polygon has at least four sides.

Do It Together

Polygon

Number of Side (n) 5 6 7

Diagonals 5 9 14

Convex/Concave Concave Convex Convex

2. Properties of Quadrilaterals

Do It Together

Angles

150°, 100°, 40°, 70° 75°, 105°, 101°, 79° 216°, 36°, 40°, 68°

Type of Quadrilateral Convex Convex Concave

3. Trapezium and Kite

Think and Tell

⇒ 10x° = 180°

⇒ x° = 18°; 3x° = 54°; 6x° = 108°

⇒ a° = x° = 18°

⇒ b° = 3x° = 54°

⇒ c° = 6x° = 108°

5. Rectangle, Square and Rhombus

Do It Together

PR bisects ∠P, so, ∠RPS = 90° 2 = 45°

110° + 110° + ∠PUT + ∠RUQ = 360°

(Sum of angles around a point is 360°)

∠PUT = 70° (∠PUT + ∠PUQ form a straight angle)

Consider ΔPUT:

∠PUT + x + ∠UPT(∠RPS) = 180°

70° + x + 45° = 180°

x = 180° − (45° + 70°) = 65°

6. Interior Angle Sum Property

Do It Together

The shape has 7 sides so it is a heptagon.

Sum of angles of the heptagon = (n – 2) × 180°

Measure of the missing angle is

900° – (67° + 138° + 143° + 152° + 108° + 134°)

= 900° – 742°

= 158°.

7. Exterior Angle Sum Property

∠DAC = ∠ACD ( AD = CD )

∠DAC + ∠ACD + 54° = 180°

∠DAC = ∠ACD = 63°

Only one pair of opposite angles are equal in a kite. Do It Together 54°

∠DAC + ∠ACD + ∠CDA = 180° (angle sum property)

∠ABC + ∠BCA = ∠CAB (angles of equilateral triangle)

So, ∠ABC = ∠BCA = ∠CAB = 60°

∠A = ∠DAC + ∠BAC = 63° + 60° = 123°

∠C = ∠ACD + ∠BCA = 63° + 60° = 123°

4. Parallelograms

Do It Together

Since, in ΔABC:

⇒ x° + 6x° + 3x° = 180°

Do It Together

∠a + 122° = 180°.

So, ∠a = 180° − 122° = 58°

∠a = 58°

∠b = 180° − 81° = 99°,

∠c = 180° − 71° = 109°,

∠d = 360° − (58° + 81° + 71° + 34° + 34°) = 82°

Bar Graphs and Histograms 4

Learning Outcomes

Students will be able to:

organise and read data in frequency distribution tables for ungrouped data. organise and read data in frequency distribution tables for grouped data. draw and read single bar graphs. draw double bar graphs for given data. read double bar graphs. draw histograms for given data. read and interpret histograms.

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 data in a table. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

grouped data: data arranged in categories or ranges for easier analysis ungrouped data: individual data points without categorisation or grouping frequency distribution: a tabular representation showing the number of occurrences of each value or range in a dataset data: facts, figures, or other pieces of information that can be used to learn about something frequency: number of times a particular observation occurs in the data histogram: bar graphs which are used to represent grouped data

Teaching Aids

Chart paper; Sticky notes; Matchsticks; Glue; Strips of paper; Graph paper; Coloured paper; Slips of paper; Scissors; Chart paper with a double bar graph; Chart paper with a histogram

Chapter: Bar Graphs and Histograms

Frequency Distribution of Ungrouped Data

Learning Outcomes

Students will be able to organise and read data in frequency distribution tables for ungrouped data.

Teaching Aids

Chart paper; Sticky notes

Activity

Begin by explaining the objective of the day’s activity: to determine the most common shoe size among the students in the class. Distribute a sticky note to each student and instruct them to write down their shoe size.

Create a blank frequency distribution table on chart paper with columns labelled ‘Shoe Size’ and ‘Tally Marks’.

Ask the students to work in groups. Provide each group with the frequency distribution table drawn on chart paper. Collect the sticky notes containing the shoe sizes. Review the collected data by calling the shoe sizes from the sticky notes one by one, and have each group draw tally marks corresponding to each size on their chart paper.

Once all the data is tallied, count the tally marks for each shoe size to determine its frequency, representing the number of students wearing shoes of that size.

Instruct the students to identify the shoe size with the highest frequency to determine the most common shoe size among the Grade 8 students.

Extension Idea

Ask: What is the range of a data set which has 10 observations and all the observations are 10?

Say: Range = 10 – 10 = 0

Frequency Distribution of Grouped Data

Learning Outcomes

Imagine Maths Page 61

Students will be able to organise and read data in frequency distribution tables for grouped data.

Teaching Aids

Chart paper; Matchsticks; Glue stick

Activity

Begin by introducing the concept of grouping data. Ask the students the number of hours they study in a week and record the data in a table on the board. Discuss how to group the number of hours into ranges, like 0–2, 3–5, 6–8, 9–11, and so on.

Ask the students to form groups. Distribute chart paper to each group.

Instruct the students to convert the discrete data into grouped data, form the frequency distribution table on chart paper using the matchsticks as tally marks and writing the frequency for each group. Encourage them to showcase their work in the class.

Total

Using their tables, define related concepts such as class intervals, class limits, class marks, class size, etc. Discuss how grouping the data reduces the task of managing large sizes of ungrouped data. Help the students compare and observe the differences and similarities between the two tables. Discuss their responses and encourage them to write their learnings in their notebooks.

Extension Idea

Ask: If you group the data into intervals of 0–5 instead of 0–2 and 3–5, what differences will arise in their frequencies? Say: The interval 0–5 will include a wider range of values compared to the separate intervals of 0–2 and 3–5. Therefore, you may observe a higher frequency in the 0–5 interval compared to the frequencies observed in the 0–2 and 3–5 intervals.

Drawing and Reading Bar Graphs

Learning Outcomes

Students will be able to draw and read single bar graphs.

Teaching Aids

Graph paper; Coloured paper; Glue stick; Slips of paper; Scissors

Activity

Begin the lesson by briefly revising the concept of a bar graph. Draw a table on the board listing the activities and the number of hours spent.

Imagine Maths Page 64

Divide the class into groups and give each group graph paper, coloured paper, and a glue stick.

Instruct the students to make a visual representation of the average number of hours spent by them on extra-curricular activities per week. Instruct them to draw two axes on graph paper, with the x-axis representing the activities and the y-axis representing the hours. Ask them to label the x-axis with the names of the activities and the y-axis with the appropriate intervals to accommodate the weekly hours. Instruct them to use the data in the table to make bars for each activity. They can cut and paste coloured paper strips of appropriate lengths to create the bars. Remind them to make the strips of the same width.

Ask questions like, “How many hours a week are spent on volunteer work? What is the average number of hours per week spent on all these activities? On which activity is the greatest number of hours spent?”

Encourage everyone to answer the questions.

Discuss their understanding and ask them to write the answers in their notebooks. Give each group 2 slips of paper. Ask them to discuss and create 2 questions based on the data, write them on the slips, pass the slips to another group, and answer the questions they receive in their notebooks.

Extension Idea

Ask: What changes are required if you want to make a horizontal bar graph?

Say: To make a horizontal bar graph, the x-axis and y-axis would switch roles, with the activities labelled on the y-axis and the hours on the x-axis. The bars would then extend horizontally from the y-axis, representing the number of hours spent on each activity.

Drawing Double Bar Graphs

Learning Outcomes

Students will be able to draw double bar graphs for given data.

Teaching Aids

Graph paper; Coloured paper; Glue stick; Scissors

Activity

Imagine Maths Page 68

Begin the lesson by explaining the importance of sports in physical and mental development. Explain that the purpose of the day’s activity is to create a visual representation of the favourite sports of 130 students. Divide the class into groups. Distribute graph paper, coloured paper and glue to each group.

Draw a table on the board as shown.

Instruct the students to draw two axes on their graph paper, with the x-axis representing the sports and the y-axis representing the number of students. Ask them to label the x-axis with the name of each sport and the y-axis with appropriate intervals to accommodate the data. Using the data in the table, ask the students to plot 2 bars in two different colours for each sport—one representing the number of girls and the other representing the number of boys—by cutting and pasting rectangular strips of coloured paper of appropriate lengths. Remind them that the widths of the strips should be the same.

Encourage the students to scale their axes to represent the data accurately. Explain that each square on the graph paper can represent a certain number of students. Ensure that they give the bar graph an appropriate title.

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

Extension Idea

Instruct: Create questions to ask other students about the bar graph.

Say: You can ask a lot of questions about the bar graph. One such question could be: Whichsportwaschosenas thefavouritebythemoststudents?

Reading

Double Bar Graphs Imagine Maths Page 69

Learning Outcomes

Students will be able to read double bar graphs.

Teaching Aids

Chart paper with a double bar graph; Slips of paper; Scissors

Activity

Begin the lesson by talking about the values of compassion and social responsibility. Explain how even the smallest contributions make big impacts on the lives of people in need. Create a double bar graph on chart paper that displays the monetary contributions from a donation drive in a school over 4 weeks in July and in August. Ask the students to work in groups. Distribute the chart paper with the double bar graph to each group. Instruct them to examine the graph closely and note down any patterns or differences they observe in the contributions.

Ask questions like, “Which week had the highest collection in both months? Which week had the lowest collection? What is the range of the collections for July? What are the total and average weekly collections in both months?”

Provide each group with 2 slips of paper. Instruct each group to create 2 questions related to the bar graph. Have each group pass their slips to another group. Instruct each group to answer the questions they received.

Drawing

Histograms Imagine Maths Page 74

Learning Outcomes

Students will be able to draw histograms for given data.

Teaching Aids

Graph paper; Coloured paper; Glue stick; Scissors

Activity

Introduce the concept of using histograms to represent grouped data. Explain that the students will be creating a histogram on temperature ranges throughout a year for a city.

Draw a table on the board as shown.

Divide the class into groups. Distribute graph paper, coloured paper and a glue stick to each group. Instruct them to draw two axes on the graph paper, as they would do for a bar graph. Emphasise marking the class intervals on the x-axis. Instruct them to use the data in the table to make bars of equal width for each temperature range. They can cut and paste coloured paper strips to create the bars. Ask them to make sure that there are no gaps between the bars.

Ask the students to check whether the graph accurately shows the temperature ranges for the months by adjusting the height of the bars. Each coloured paper strip can stand for a certain number of months. Ask each group to give their graph an appropriate title.

Ask questions like, “What are the similarities and differences between a bar graph and a histogram?” Encourage everyone to share their learnings.

Extension Idea

Ask: How will the height of the bar change if I reduce the scale to half?

Say: The height of each bar will increase. For example, if 1 division on the scale initially represented 2 months, reducing the scale to 1 month means the bar’s height must be increased from 2 divisions to 4 divisions to accurately represent 4 months.

Reading Histograms

Learning Outcomes

Students will be able to read and interpret histograms.

Teaching Aids

Chart paper with a histogram

Activity

Prepare a histogram on chart paper showing the heights (in cm) of 25 students of Grade 8. Ask the students to work in groups. Distribute sheets of chart paper with the histogram to each group.

Ask the students to examine the histogram closely and answer the questions.

Ask questions like, “How many students have a height between 140 cm and 160 cm?”, “How many students are more than 150 cm tall?”, “How many students are less than 145 cm tall?” or “What is the average height of the students?”

Maths Page 75

Instruct the students to answer these questions. Discuss their responses and encourage them to write the answers in their notebooks.

Extension Idea

Ask: What percentage of the students are taller than 150 cm?

Say: 12 25 × 100% = 48%. So, 48% of the students are taller than 150 cm.

Answers

1. Frequency Distribution of Ungrouped Data

Think and Tell

In daily life, you need to organise data in activities like making to–do lists, managing schedules and tracking expenses.

Do It Together

Range = Highest value in the data set − smallest value in the data set

Therefore, the range of the given data = 4 − 0 = 4

2. Frequency Distribution of Grouped Data

Do It Together

1. What is the lower limit of the first class interval? 45

2. What is the upper limit of the last class interval? 80

3. What is the size of each class? 5

4. Which class interval has the highest frequency? (60–65)

5. Which class interval has the lowest frequency? (45–50)

6. What is the class mark of the fifth class interval? 65 + 70 2 = 135 2 = 67.5

3.

Drawing and Reading Bar Graphs Do It Together

1. Which is the longest river? Brahmaputra

2. Which is the shortest river? Kaveri

3. Length of river Mahanadi = 900 km

Length of river Kaveri = 800 km

Ratio of the length of river Mahanadi to that of river Kaveri = 900:800 = 9:8

4. Length of river Ganga = 2500 km

Length of river Narmada = 1300 km

Difference in length = 2500 – 1300 = 1200 km

4.

Drawing Double Bar Graphs Do It Together

Narmada GangaBrahmaputra Kaveri Mahanadi

5. Reading Double Bar Graphs

Think and Tell

Other products have the highest percentage increase from the previous year since the gap between the exports of 2021 and 2022 is the highest. Do It Together

2. Number of people who visited India Gate in 2019 = 700

3. Number of people who visited Taj Mahal in 2018 = 900

Number of people who visited Taj Mahal in 2019 = 1000

Percentage increase = 100 900 × 100% = 11.11%

4. Average number of visitors in 2018 = 700 + 900 + 600

+ 500 = 2700 4 = 675

Average number of visitors in 2019 = 760 + 1000 + 700

+ 500 = 2960 4 = 740

Difference = 740 − 675 = 65

6. Drawing Histograms

7. Reading Histograms

Do It Together

2. Number of employees older than 40 years = 20 + 20 + 5 + 5 = 50

5 + 5 = 195

Percentage of total employees who are older than 40 years = 50 195 × 100% = 25.64%

3. Number of employees older than 35 years = 30 + 20 + 20 + 5 + 5 = 80

Number of employees younger than 35 years = 30 + 45 + 40 = 115

Required ratio = 80:115 = 16:23

4. Number of employees who are younger than 40 years = 30 + 45 + 40 + 30 = 145

Total number of employees = 195

Percentage of total number of employees who are younger than 40 years = 145 195 × 100% = 74.36%

Pie Charts 5

Learning Outcomes

Students will be able to: find the measure of the angle for each given sector (fraction) in a circle. draw a pie chart for the given data. read 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 pie charts where the sectors are divided into halves, fourths or eighths.

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

Vocabulary

pie chart: graph representing data in a circular form sectors: divisions of a circle displaying the size of some related piece of information central angle: angle formed at the centre of a circle

Teaching Aids

Sheet with a table showing 4 genres, columns for the number of students and measures of the central angle; Sector cutouts with different central angles; Glue stick; Chart paper; Pie chart drawn on a sheet of paper; Question cards

Chapter: Pie Charts

Sectors and Central Angles

Learning Outcomes

Students will be able to find the measure of the angle for each given sector (fraction) in a circle.

Teaching Aids

Sheet with a table showing 4 genres, columns for the number of students and measures of the central angle

Activity

Begin by reminding students about pie chart representation of data. Discuss how the whole circle represents data in the form of sectors with each sector having a central angle. Discuss the central angles formed by different sectors inside the circle, like one-fourth forms a 90° angle and one-eighth forms a 45° angle.

Give them categories, such as, book genres (fiction, thriller, documentary, mystery) and ask them to pick their favourite by raising their hands when you call out each genre one by one. Note down the data in a table on the board:

Genre

Fiction

Thriller

Documentary Mystery

Number of Students Measure of the Central Angle

Ask questions like, “If 10 out of 40 students in the class like fiction, how many students out of 360 would like fiction?”

Bring out the formula for finding the central angle for this category on the board. Write the measure of the central angle for fiction on the board.

Ask the students to work in groups and copy the table drawn on the board, in their notebooks.

Instruct them to find the central angle for each category using the formula. Ask them to compare the angle measures with the other group members to check if their answers are correct.

Ask questions like, “How do you know that the measures of the angles are correct?”

Give the students some more data and ask them to find the central angle.

Extension Idea

Ask: What will be the percentage for the data category that has a central angle of 90°?

Say:

Learning Outcomes

Students will be able to draw a pie chart for the given data.

Teaching Aids

Sector cutouts with different central angles; Glue stick; Chart paper

Activity

Begin the class with a discussion on how to find the central angles for given data. Draw a table on the board with data showing favourite activity of 100 students.

Ask the students to work in groups. Distribute the chart paper, glue stick and sector cutouts with different central angles (e.g., 10°, 20°, 30°, 45°, etc.) to each group. Make sure that the cutouts with central angles needed for each category are also included. Instruct the students to read the data and calculate the central angle for each category in their notebooks. They can then discuss within their groups if all the members got the same answer.

Ask them to draw a circle on the chart paper with the same radius as in the sector cutouts. Guide the students to use the cutouts to find the angles for their respective categories and paste them onto their circles. Guide them in starting with the first sector cutout, using the horizontal radius as the base, and recording both the angle and its value. Instruct them to continue pasting the remaining sector cutouts, using the previous one as the base. The students will then label the sectors in their circles with the angles and corresponding values.

Invite the groups to showcase their work. Initiate a class discussion to review what the students have learnt. Encourage them to recreate the same pie chart in their notebooks using protractors and rulers.

Extension Idea

Ask: What change should the number of sectors undergo if you remove one category of data from the pie chart?

Say: The number of sectors will decrease by 1 with the decrease in the number of categories. The measures of all the angles will also change thereby changing the size of each remaining sector.

Reading Pie Charts

Learning Outcomes

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

Teaching

Aids

Pie chart drawn on a sheet of paper; Question cards Activity

Begin the class by discussing real-life examples where pie charts are used.

Imagine Maths Page 89

Present this scenario to the class:

Rohitrecordeddataonthedifferentfruitssoldbyafruitsellerinaday.Hesawthatthetotalsalesforthatday amountedto₹12,000.

Prepare the pie chart (as shown) on a sheet of paper and some question cards.

Whichwasthemostsoldproductoftheday?

Whatwasthesalesvalueofbananas?

Whatwasthedifferenceofthesalesofapplesandkiwis?

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

Ask the students to read the pie chart and answer the questions given on the question cards in their notebooks.

Discuss the responses with the whole class and encourage the students to write the answers in their notebooks.

Extension Idea

Ask: Can the total percentage in the pie chart be more than 100?

Say: No, a pie chart represents 100% of a whole, and each sector or slice corresponds to a portion of that whole. Therefore, the sum of all the percentages in a pie chart should always equal 100%.

Answers

1. Sectors and Central Angles

Do It Together

2. Drawing a Pie Chart

Do It Together

3. Reading Pie Charts

Think and Tell

No, the sum of all the values of all the components cannot be less than the total value.

Do It Together

1. Total Production = 27,000 × 360°

2. Production of

Production

Difference = 22,500 − 13,500 = 9000 tonnes.

Probability 6

Learning Outcomes

Students will be able to: find the theoretical probability of an event. find the experimental probability of an event.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to find the probability of a given event happening. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

probability: the chances that an event will happen event: a possible outcome or a set of possible outcomes of an activity or test experiment: an activity or test that is repeated, in which all possible outcomes are known outcome: a possible result of an activity or test

Teaching Aids

Coin; Small bags with 3 red, 4 blue, 5 yellow and 8 green marbles; Spinner 1 with numbers 2, 4, 6, 8, 10; Spinner 2 with numbers 1, 3, 5, 7, 9

Chapter: Probability

Theoretical Probability

Learning Outcomes

Students will be able to find the theoretical probability of an event.

Teaching Aids

Coin; Small bags with 3 red, 4 blue, 5 yellow and 8 green marbles; Spinner 1 with numbers 2, 4, 6, 8, 10; Spinner 2 with numbers 1, 3, 5, 7, 9

Activity

Show the students a coin and ask them what we will get if we flip a coin. Discuss the total possible outcomes and the favourable outcomes of flipping a coin, and how to write the probability of the event.

Ask the students to work in groups. Distribute a bag of marbles to each group. Ask them to count the marbles and find the theoretical probability of picking a marble of each colour at random from the bag. Discuss one example and guide them if necessary. For example: The total number of marbles (20) represents the total outcomes, and the number of red marbles (3) represents the favourable outcomes. So, the theoretical probability of picking a red marble is 3 10

Ask questions like, “What is the theoretical probability of getting a black marble? A green marble?”

Extension Idea

Show the students the two spinners.

Ask: Spinner 1 has even numbers less than 11, and Spinner 2 has odd numbers less than 10. What is the theoretical probability of getting a number less than 5 when both spinners are spun together?

Say: Total outcomes = 10, Favourable outcomes or numbers less than 5 = 4.

So, theoretical probability = 4 10 or 2 5

Experimental Probability Imagine Maths Page 105

Learning Outcomes

Students will be able to find the experimental probability of an event.

Teaching Aids

Small bags with 3 red, 4 blue, 5 yellow and 8 green marbles

Activity

Discuss the concept of experimental probability using the example of a coin. Explain that if we toss a coin 50 times, we can count how often we get heads to find the experimental probability. Discuss how if heads appears 20 times out of 50 coin tosses, the experimental probability of getting heads is 20 50 or 2 5 .

Ask the students to work in groups. Distribute a bag of marbles to each group. Draw a table on the board as shown and ask the students to copy the table into their notebooks.

Colour Red Blue Yellow Green

Number of Times Picked

Instruct the students to pick a marble from the bag and record the colour they get in the table using tally marks. Ask them to put the marble back into the bag and repeat the activity 25 times. Once the table is filled, instruct them to write the answers to these questions in their notebooks: Whatistheexperimentalprobabilityofpickinga redmarble?Whatistheexperimentalprobabilityofpickingablueorgreenmarble?

Extension Idea

Ask: On flipping a coin 20 times, we get heads 12 times and tails 8 times. What is the experimental probability of getting tails as a percentage?

Say: Total outcomes = 20, Favourable outcomes = 8. So, experimental probability = 8 20 × 100 = 40%.

Answers

1. Theoretical Probability

Think and Tell

Yes, the probability of the wheel stopping at an odd number is the same as the probability of the wheel stopping at an even number.

Do It Together

a. shows a picture of a bird?

E = Event of getting a card that shows a picture of a bird

P(E)= Number of cards that show a bird picture Total number of outcomes

 4 11

Therefore, the possibility of getting a card that shows a picture of a bird is

4 11

b. does not show a picture of a bird?

E = Event of getting a card that does not show a picture of a bird

P(E) =   7 11  

Therefore, the possibility of getting a card that does not show a picture of a bird is   7 11  

c. CD, CT, CG, CR GD, GT, GG, GR DD, DT, DG, DR HD, HT, HG, HR SD, ST, SG, SR BD, BT, BG, BR RD, RT, RG, RR

2. Experimental Probability

Think and Tell

The approximate experimental probability of getting heads if we toss a coin 1000 times is 50%.

Do It Together

We first find the experimental probability of rolling an odd number.

The bar graph shows 48 ones, 50 threes and 49 fives.

So, an odd number was rolled 48 + 50 + 49 = 147 times in a total of 300 rolls.

P(odd) = Number of times an odd number was rolled 147 = Total number of rolls 300 = 49%

Next, we find the theoretical probabiliity of rolling an odd number.

P(odd) = Number of favourable outcomes31 = = Number of possible outcomes62 = 50%

The experimental probability of rolling an odd number is 49%, which is close to the theoretical probability of 50%

Squares and Square Roots

Learning Outcomes

Students will be able to: find the square of a number. identify patterns in square numbers to help find the square of a number. find the square root of a number using prime factorisation. find the square root of a number using long division.

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

Let’s Recall

Recap to check if students know how to express a number as a product of prime factors. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

square root: a value that, when multiplied by itself, gives the original number perfect square: a number obtained by multiplying an integer by itself

Teaching Aids

Connecting cubes; Problem cards; Rule cards; Puzzle cards

Chapter: Squares and Square Roots

Square of a Number

Learning

Outcomes

Students will be able to find the square of a number.

Teaching Aids

Connecting cubes

Activity

Imagine Maths Page 114

Introduce the concept of a square of a number. Ask the students to work in groups. Distribute building blocks or connecting cubes to each group. Each member of the group takes 2, 4, 5 or 6 cubes. They will then make another row and ask them to continue making as many rows as the number given. Ask them to note the total number of cubes used. For example, if the number of cubes placed are 5, we will place 5 rows of 5 cubes each to get 25 cubes in all.

Ask question like, “How will you find the square of a larger number like 78?” For larger number, we can split the number into tens and ones, and multiply to find the answer. Instruct them to find the squares of different numbers and provide them with more practice.

Extension Idea

Ask: What is the smallest number by which 288 must be multiplied so that we get a perfect square?

Say: 288 = 2 × 2 × 2 × 2 × 2 × 3 × 3. There is one 2 that is not in a pair. So, if the number is multiplied by 2, we get a perfect square.

Patterns of Square Numbers

Learning Outcomes

Imagine Maths Page 116

Students will be able to identify patterns in square numbers to help find the square of a number.

Teaching Aids

Problem cards; Rule cards

Activity

Begin by discussing various patterns of square numbers using examples for each rule, referring to page 116 of the Imagine Mathematics book.

Prepare problem cards and respective rule cards. (Make sure problems related to all types of rules are covered.)

Ask the students to work in groups. Distribute the problem cards to each group and place the rule cards on the table. Instruct the groups to read the problem and pick the rule card that will be used to solve the problem from the table. Ask them to solve the problem using the rule in their notebooks. Then, have each group showcase their pattern to the group sitting next to them and ask them to verify it. Rotate the problem cards among the groups for more practice.

Finding Square Roots Using Prime Factorisation

Learning Outcomes

Students will be able to find the square root of a number using prime factorisation.

Teaching Aids

Puzzle cards

Activity

Begin the class by introducing the method of prime factorisation to find the square root of a number. Provide examples and guide the students with a few practice problems.

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

Ask them to match the edges of their cards until all the edges match with equivalent expressions. The final solution will be a 3 × 3 grid as shown.

Ask questions like, “With which card did you begin matching the cards? Which letter card did you place in the middle of the square? Top right corner? Bottom left corner?”

Extension Idea

Ask: If there is a square garden of area 242 sq. m, can all its sides be 22 m each?

Say: 242 = 2 × 11 × 11. As 242 isn’t the square of 22, the sides of the garden can’t be 22 m each.

Finding Square Roots Using Long Division

Learning

Outcomes

Students will be able to find the square root of a number using long division.

Teaching Aids

Puzzle cards

Activity

Begin the class by explaining the methods of finding the square root of large numbers using the long division method. Prepare the puzzle cards with square numbers and their square root.

Ask the students to form groups. Distribute cards with square numbers and corresponding square root cards to each group.

Instruct the groups to find the square root of the given numbers using the long division method and join them with the corresponding square root cards. The first group to join all the puzzle cards correctly will win.

Extension Idea

Ask: If the area of a square photo frame is 2079.36 sq. cm, what is the length of each side?

Say: Area = side × side = side2; Length of each side = 2079.36 = 45.6 cm.

Answers

1. Square of a Number

Think and Tell

No, such a number won’t be a perfect square.

Think and Tell

No, such a number won’t be a perfect square.

Think and Tell

Shobhit can buy a table with dimensions 2 × 2 = 4 sq. m.

4 sq. m. is less than 5 sq. m., so the table will fit inside the room.

Think and Tell

The square of an n-digit number will either have (n × n) − 1 or n × n digits.

Do It Together

1362 = (100 + 36)2

= (100 + 36) (100 + 36)

= 100(100 + 36) + 36(100 + 36)

= 1002 + 100 × 36 + 36 × 100 + 362

= 10,000 + 3600 + 3600 + 1296 = 18,496

2. Patterns of Square Numbers

Think and Tell

There are 2n non-perfect square numbers between the squares of n and (n+ 1).

Think and Tell

No, if a number cannot be expressed as a sum of successive odd numbers, then it is not a perfect square.

Do It Together

1. 452 = [4 (4 + 1)] × 100 + 25 = (4 × 5) × 100 + 25

= 20 × 100 + 25

= 2000 + 25 = 2025

2. 1152 = [11 (11 + 1)] × 100 + 25

= (11 × 12) × 100 + 25

= 132 × 100 + 25

= 13200 + 25 = 13,225

3. Finding Square Roots Using Prime Factorisation

Do It Together

14,580 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 5

Clearly, we do not have pairs of all prime factors, and we are left with a factor 5 which cannot be paired.

So, 5 is the least number by which 14,580 should be divided to get a perfect square.

4. Finding Square Roots Using Long Division

Think and Tell

Yes, we can find the square root of 5,53,536 using the long division method.

So, the square root of 5,53,536 is 744.

Do It Together

Cubes and Cube Roots

Learning Outcomes

Students will be able to: find the cube of a number and identify perfect cubes. find the cube root of a number using prime factorisation.

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

Let’s Recall

Recap to check if students know how to find the volume of a solid by counting unit cubes. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

prime factorisation: a way of expressing a number as a product of its prime factors cube root: any number multiplied by itself three times to get a cube perfect cube: a number obtained by multiplying an integer by itself three times

Teaching Aids

Connecting cubes; Puzzle cards

Chapter: Cubes and Cube Roots

Perfect Cube

Learning Outcomes

Students will be able to find the cube of a number and identify perfect cubes.

Teaching Aids

Connecting cubes

Activity

Introduce the concept of perfect cubes. Explain that a perfect cube is a number that can be formed by multiplying an integer by itself three times.

Show an example of a perfect cube using connecting cubes. For instance, build a 2 × 2 × 2 cube with 8 connecting cubes.

Divide the class into groups. Distribute building blocks or connecting cubes to each group.

Instruct the students to join connecting cubes to form a cubic structure. Ask them to note the number of connecting cubes forming one side and the total number of connecting cubes used to form the cubic structure.

Discuss how the total number of connecting cubes gives us the cube of the number.

Ask questions like: How can you tell if a model represents a perfect cube or not just by looking at it?

For larger numbers, like 2744, introduce the prime factorisation method for determining if the number is a perfect cube. Instruct the students to find the cubes of different numbers and give them more practice questions.

Extension Idea

Ask: What is the smallest number by which 3087 must be multiplied so that we get a perfect cube?

Say: 3087 = 7 × 7 × 7 × 3 × 3. If the number is multiplied by 3, we get a perfect cube.

Cube Root Through Prime Factorisation Method

Learning Outcomes

Students will be able to find the cube root of a number using prime factorisation.

Teaching Aids

Puzzle cards

Activity

Begin the class by introducing the method of prime factorisation to find the cube root of a number. Provide examples and guide the students through a few practice problems.

Divide the class into groups. Distribute the puzzle cards to each group.

Instruct them to match the edges of their cards until all the edges match with equivalent expressions. The final solution will be a 3 × 3 grid as shown.

Ask questions like: Which letter card did you place in the middle of the square? Top right corner? Bottom left corner? Bottom right corner?

Extension Idea

Instruct: 2*97 is a 4-digit number. If the number is a perfect cube, find the missing digit and the cube root of the number.

Say: Using the trial and error method, we can start by using 1 as the missing digit to check whether 2197 is a perfect cube. The prime factorisation of 2197 is: 13 × 13 × 13 × 1

So, 2197 is a perfect cube and the cube root of 2197 is 13. So, the missing digit is 1.

Answers

Think and Tell

Yes, if the cube of a number is a multiple of 3, it is also a multiple of 27.

The units digit of the cubes of even numbers are always even. The units digit of the cube of some numbers is the same as the units digit of the number. The units digit of the cubes of odd numbers are always odd.

1. To Check If a Number Is a Perfect Cube or Not

Do It Together

6561 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3

Clearly, not all prime factors form a triplet and we are left with a factor 3 which cannot form a triplet.

So, 3 is the lowest number by which 6561 should be divided to get a perfect cube.

2. Patterns in Cubes

Do It Together

We know that if n and n + 1 are two consecutive numbers, then (n + 1)3 − n3 = 3n (n + 1) + 1 So, 873 – 863 = 3 × 86 × (86 + 1) + 1 = 258 × 87 + 1

3. Cube Root Through Prime Factorisation Method

Percentage and Its Applications

Learning Outcomes

Students will be able to:

solve word problems on ratios and percentages. solve word problems on percentage change. solve problems on finding the profit/loss percentage, cost price or selling price if any two are given. solve word problems on finding the discount amount. solve word problems on finding the tax amount.

Alignment to NCF

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

Let’s Recall

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

Vocabulary

percentage: part of a whole represented as a fraction of 100 ratio: a relation or comparison between numbers or things based on amount or degree profit: the money that you make when you sell something at a price higher than its cost price loss: the money that you lose when you sell something at a price lower than its cost price tax: money collected by the government of a country for public use

Teaching Aids

Situation cards; Question cards; Bags of small items (like candies, pencils or buttons) with tags containing statements (like CP = ₹150, Sell at a profit of ₹15); Blank tags; Items with price tags showing the marked price and % discount offered; Play money

Chapter: Percentage and Its Applications

Ratio and Percentages

Learning Outcomes

Students will be able to solve word problems on ratios and percentages.

Teaching Aids

Situation cards; Question cards

Activity

Ask the students to work in groups. Distribute cards to each group with this situation: Inasportseventorganisedatschool,atotalof400studentsparticipated.20%ofthestudentsparticipated inathletics,15%ofthestudentsparticipatedinbasketball,30%ofthestudentsparticipatedinfootballand35%of thestudentsparticipatedincricket.

Make some question cards (as many as possible) and distribute to each group.

Whatistheratioofthenumberofstudentswhoparticipatedinathleticstothosewhoparticipatedinbasketball?

Howmanymorestudentsparticipatedincricketthaninfootball?

Instruct each student in the group to pick one question card, solve the problem in their notebook and write the answer. The group who solves the maximum number of problems correctly wins. If time permits, shuffle the question cards among the groups. Discuss the answers with the whole class.

Extension Idea

Ask: If 60% of the football participants were male, then how many female football participants were there? Say: Number of football participants = 30% of 400 = 120 participants. 60% of 120 = 72 participants were male. So, the number of female football participants = 120 – 72 = 48 participants.

Percentage Change

Learning Outcomes

Students will be able to solve word problems on percentage change.

Teaching Aids

Situation cards; Question cards

Activity

Ask the students to work in groups.

Distribute cards to each group with this situation: Aaronhasabakery.Readtheratecardthatisdisplayedinhisshop.

Imagine Maths Page 153

Afteraweek,thereweresomechangesinthepricesoftheproducts.

Make some question cards (as many as possible) and distribute to each group. Ifthereisanincreaseof10%inthepriceofbread,whatisthenewprice?

Whatisthepercentagechangeinthepriceofpizza,ifthenewpriceis₹480?

Extension Idea

Ask: What is the new price of cookies if it was first increased by 20% and then decreased by 10%?

Say: Initial increased price = ₹120 + 20% of ₹120 = ₹120 + ₹24 = ₹144. Final price of the cookies = ₹144 – 10% of ₹144 = ₹144 – ₹14.4 = ₹129.6.

Profit and Loss

Learning Outcomes

Students will be able to solve problems on finding the profit/loss percentage, cost price or selling price if any two are given.

Teaching Aids

Bags of small items (like candies, pencils or buttons) with tags containing statements inside (like CP = ₹150; Sell at a profit of ₹15); Blank tags

Activity

Revise the concepts of cost price (CP), selling price (SP), profit and loss. Discuss the formulas for finding the profit/loss percentage, cost price or selling price if any two are given. Prepare bags of small items (like candies, pencils, or buttons) with tags containing statements inside. (Make sure that the statement tags inside each bag are different.)

CP = ₹150; sell at a profit of ₹15

CP = ₹125; sell at a profit of ₹25

Ask the students to work in groups of 3. Distribute 2 blank tags and 2 bags of items (one with a profit tag and one with a loss tag) to each group.

Instruct the students to open the bags and look at the tags inside. Ask them to find the selling price (SP) and profit/loss percentage for each bag. They will then use the blank tags to prepare price tags showing the selling price (SP) and profit or loss%.

SP = ₹165; Profit percentage = 10% SP = ₹100; Loss percentage = 20%

Once they are done, instruct the groups to pass their bags to the group sitting next to them. Ensure these do not contain the price tags provided by the teacher. Each group will then look at the new tags and find the cost price of each item. Ask them to also find the total CP of the items, the total SP and the profit/loss percentage.

Extension Idea

Ask: Raj bought a vintage watch and sold it at a profit of 12%. If he had sold it for $180 more, he would have made a profit of 15%. What was the cost price of the watch?

Say: When Raj sold the watch at a 12% profit, SP was 112% of CP: SP1= CP + 0.12 × CP = 1.12CP. If he had sold it for $180 more at a 15% profit, SP would have been 115% of CP: SP2 = CP + 0.15 × CP = 1.15CP. As the difference of the selling prices is $180, we can write the equation: SP2 − SP1 = 180 ⇒ 1.15CP − 1.12CP = 180 ⇒ CP = $6000. So, the cost price of the watch was $6000.

Finding Discount

Learning Outcomes

Students will be able to solve word problems on finding the discount amount.

Teaching Aids

Items with price tags showing the marked price and % discount offered; Play money

Activity

Imagine Maths Page 161

Explain the concept of discount. Discuss the important formulas related to discount. Convert the classroom into a shop by putting some items on the table with price tags showing the marked price and % discount offered. Ask the students to work in groups. Distribute play money and the price tags for the items to each group. Instruct the students to discuss in their groups and calculate the amount that they need to pay for each item after deducting the discount amount. Ask the students to choose 2 items (with the least cost) that they want to buy and then use the play money to pay the teacher for those items. Once the students have a thorough understanding of the concept, introduce the concept of successive discounts. Ask questions like, “Does the order in which successive discounts are applied affect the final price of an item?” Give them various word problems based on discount for practice.

Extension Idea

Ask: A supermarket offers a 15% discount on the value of purchases above $50. If a customer’s total bill amount comes to $70, how much will the final bill amount be with the discount?

Say: The discount is applicable only on the amount exceeding $50.

Amount eligible for discount = Total bill amount − Threshold amount = $70 − $50 = $20; Discount = 15% of $20 = 15 100 × $20 = $3. Therefore, the customer will pay $70 − $3 = $67 after the discount.

Tax Imagine Maths Page 166

Learning Outcomes

Students will be able to solve word problems on finding the tax amount.

Teaching Aids

Situation cards; Question cards

Activity

Instruct the students to refer to page 166 and 167 of the ImagineMathematics book to understand the types of taxes applied in India.

Ask the students to work in groups. Distribute cards (with various situations) to each group.

Amitawenttopurchasesomeitemsforhernewhouse.Shepurchasedafanfor₹3500,andlightsandfittingsfor ₹25,350fromonestore;andkitchenappliancesfor₹1,28,600fromanotherstore.

Make question cards and distribute them among the groups.

Whatamountdidshepayforthefanandlightsandfittingsif18%GSTwasappliedonthefinalbill?

WhatistheSPofthekitchenappliancesiftheamountgivenisinclusiveof18%GST?

Instruct each student in the group to pick one question card and solve the problem in their notebook. Instruct the groups to exchange the situation and question cards with other groups to solve a variety of problems. Discuss the answers with the whole class.

Answers

1. Ratio and Percentages

Think and Tell

100% of any number is the number itself.

Do It Together

Let the total number of students be x.

Percentage of students who like football = 35%

Percentage of students who like basketball = (100% − 35) = 65%

Number of students who like basketball = 1300

Total number of students = 1300 65 × 100 = 2000

Think and Tell

Percentage of marks scored = 60 80 × 100% = 75%

2.

Percentage Change

Think and Tell

Original price = ₹2500

Desired SP = 100+Profit% ×= 100 æö ÷ ç ÷ ç ÷ ç ÷ ÷ ç èø CP 100+4%104×675=×675=702 100 100 æö ÷ ç ÷ ç ÷ ç ÷ ÷ ç èø

Hence, the shopkeeper should sell the T-shirt for ₹702 to gain 4%.

5. Finding Discount

Do It Together

SP = 50,000 × 100 – 15 100 æö ÷ ç

èø

Therefore, the selling price of the chain = ₹42,500

Also, given that, the jeweller makes a profit of 6.25%.

CP ꞊ SP × 100 100+Profit% æö ÷ ç ÷ ç ÷ ç ÷ ÷ ç èø = 42,500 × 100 100+6.25% æö

CP ꞊ ₹40,000

Therefore, the cost price of the chain = ₹40,000.

New price after the price goes down by 10% = 2500 −

10 100 × 2500

= 2500 − 250 = ₹2250

The price decreases to ₹2250.

Do It Together

Original length of the rope = 150 cm

Final length after reduction = 125 cm

Reduction in length = 25 cm

Reduction percentage =

Reduction in length Original length × 100% = 25 150 × 100% = 16 2 3 %.

3. Finding Profit or Loss Percentage

Do It Together

Profit = ₹(10x − 8x) = 2x

Profit% ꞊ Profit CP × 100% = 2 ×100% 8 x x = 25%

Therefore, the percent profit is 25%

4. Finding Selling Price or Cost Price

Do It Together

100 100 = ×648=×648=675 100–4%96

Now, CP = 675, desired profit% = 4%

6. Finding Successive Discounts

Do It Together

Selling price after the second discount = 90 7.2 = 82.8

Total discount allowed = 10 + 7.2 = 17.2

Therefore, a single discount equivalent to two successive discounts of 10% and 8% is 17.2%.

Therefore, 18% is a better offer.

7. Tax

Do It Together

Marked price = ₹6500

Discount = 8% of marked price = 6500 8 ×= 100 ₹520

Selling price = Marked Price – Discount = ₹(6500 520) = ₹5980

Sales tax = 8% of Selling price = 5980 8 ×= 100 ₹478.4

Therefore, Renuka pays ₹5980 + ₹478.4 = ₹6458.4

Compound Interest

Learning Outcomes

Students will be able to: find the simple interest, if the principal, rate of interest and time period are given. deduce the formula for compound interest and apply it. solve word problems on simple and compound interest.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to find simple interest. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

principal: the initial amount invested or borrowed interest rate: the price you pay to borrow money

Teaching Aids

Cards with questions related to simple interest; Cards with questions related to compound interest; Situation cards; Question cards

Chapter: Compound Interest

Finding Simple Interest

Learning Outcomes

Students will be able to find the simple interest, if the principal, rate of interest and time period are given.

Teaching Aids

Cards with questions related to simple interest

Activity

Begin the class by revising the terms associated with simple interest, such as principal, time, rate, etc. Discuss real-life scenarios where students may come across the terms. Discuss the simple interest formula. Ask the students to note down the formula in their notebooks.

Principal = `80,000

Time = 5 years

Simple interest = `50,000

Rate = ?

Prepare cards with questions on simple interest. Ask the students to work in pairs. Distribute question cards among the pairs. Ask them to solve the problem in their notebooks. Then, have them exchange cards with other groups and solve as many problems as possible.

Discuss the answers with the whole class.

Extension Idea

Ask: Sam invested a certain amount of money in a fixed deposit account. In 2 years, the total amount in the account grew to ₹5500, and in 3 years, it grew to ₹6000. What was the principal amount he invested and the annual interest rate?

Say: We know: Amount in 2 years = ₹5500; Amount in 3 years = ₹6000; Interest earned in 1 year = ₹6000 − ₹5500 = ₹500. So, the interest earned in 2 years = ₹500 × 2 = ₹1000. Principal = Amount – Interest = ₹5500 − ₹1000 = ₹4500. Simple Interest = 1000 = 4500 × × 2

Hence, the principal is ₹4500, and the rate of interest is about 11.11% p.a.

Finding Compound Interest

Learning

Outcomes

Students will be able to deduce the formula of compound interest and apply it.

Teaching Aids

Cards with questions related to compound interest

Activity

Prepare cards with questions related to compound interest. Ask the students to work in pairs. Instruct them to open page 177–178 of their ImagineMathematicsbooks, go through the page and note down their observations. Have a discussion around what they understand. Explain what compound interest is and how it is different from simple interest.

Once the students understand the application of compound interest, distribute the question cards among the pairs.

Saradepositedacertainamountofmoneyintoafixeddepositaccountwhereinterestiscompoundedannually. After2years,shereceived₹8500asthematurityamountatanannualinterestrateof8%.Whatwastheinitial amountsheinvested?

Ask the pairs to read the problem and solve it in their notebooks.

Once the students have thoroughly understood the concept, introduce compounding half-yearly. Ask the pairs to solve the same problem by taking the rate of interest as compounded half-yearly. Ask them to observe and note down the impact on the rate of interest when compounded yearly, half-yearly in their notebooks.

Extension Idea

Ask: Suman invested ₹10,000 for 2 years in two plans. Plan A earns 15% interest compounded annually. Plan B earns 14% interest compounded half-yearly. What is the total interest amount received from both plans?

Say: Amount on ₹10,000 at 15% compounded annually:

A = 10,000 2 15 ×1 =10, 000 ×1.15× 1.15 =13, 225 100 `

Amount on ₹10,000 at 14% compounded half–yearly:

A = 10,000 4 14 2 ×1 =10, 000 ×1.07× 1.07

Thus, the total interest received = ₹(13,225 – 10,000) + ₹(13,107.96 – 10,000) = ₹6,332.96.

Word Problems

Learning Outcomes

Students will be able to solve word problems on simple and compound interest.

Teaching Aids

Situation cards; Question cards

Activity

Ask the students to work in groups. Distribute different situation cards to the groups.

Sunishborrowedanamountof₹1,00,000fromafriendattherateof10%for3years. Make question cards and distribute these to the groups.

WhatamountwillSunishpayifhisfriendchargedhimsimpleinterest?

Howmuchinterestwillhepayiftheinterestiscompoundedannually?

Imagine Maths Page 184

Instruct each student in the group to pick a question card and solve the problem in their notebooks. Ask the groups to exchange their situation and question cards with other groups to solve a variety of problems. Discuss the answers with the whole class.

Answers

1. Finding Simple Interest

Do It Together

1. Interest = ₹4500 − ₹4000 = ₹500

2. The formula for simple interest is SI = P × R × T 100

We have just calculated that SI = ₹500, and we know that P = ₹4000 and T = 10 years.

Substituting these values into the formula, we get 500 = 4000××10 100 R

Rearranging and solving this, we get, R = 50,000 40,000 .

The interest rate is therefore, R = 1.25%.

2. Compound Interest: Annually

Think and Tell

Simple interest on ₹8000 at 22% per year for 3 years = 8000×22×3 =5280 100 `

Amount after 3 years = ₹8000 + ₹5280 = ₹13,280

Amount on ₹8000 at 20% compounded annually

A = 8000 3 20 ×1 = 8000 ×1.2 ×1.2 ×1.2 =13, 824 100 `

As the amount received in Plan B is more, he must choose Plan B.

Do It Together

We know that A = P 1 + R 100 

We have A = ₹6050, R = 10% and n = 2

Therefore, 6050 = P 2 10 ×1 + 100

On simplifying the RHS, we get

6050 = P 2 100 +10 × 100

6050 = P 2 110 × 100

6050 = P 2 11 × 10

6050 = P 121 × 100

P = =5000 1 6,05, 21 000

Hence, P = ₹5000

Therefore, Amit had invested a sum of ₹5000

3. Compound Interest: Half-yearly

Think and Tell

The compound interest will change as the formula given below.

A = P 1 + r n nt

Do It Together

On simplifying the RHS, we get A = ₹18,000 ×

A = ₹18,000 × 200 × 16 200 4

A = ₹18,000 × 216 200 4

A = ₹18,000 × 1.08 × 1.08 × 1.08 × 1.08

A = ₹24,488.80

So, the amount received is ₹24,488.80.

4. Word

Problems

Do It Together

Population at 1 a.m. = 16,500

Number of hours between 1 a.m. and 5 a.m. = 4

Rate of growth (R) = 2.5% per hour

Population after 4 hours = 16,500 4 2.5 ×1 + 100 = 16,500 4 100+2.5 × 100 = 16,500 4 102.5 × 100 = 16,500 () 4 ×1.025 = 16,500 × 1.025 × 1.025 × 1.025 × 1.025

= 18,212.91 or 18,213.

So, the bacteria count at 5 a.m. is 18,213. 4 16 200 1+

Algebraic Expressions 11

Learning Outcomes

Students will be able to: draw a factor tree for an expression and list the terms and factors. add and subtract algebraic expressions using the horizontal and vertical method. multiply a monomial by a monomial. multiply a monomial by a polynomial. multiply a polynomial by a polynomial.

Alignment to NCF

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

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

Let’s Recall

Recap to check if students know how to add and subtract two terms. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

variable: something that can change or that has no fixed value factor: any one of two or more numbers that are multiplied together to give a product coefficient: the constant part of an algebraic term involving multiplication algebraic expression: mathematical phrase with numbers and variables polynomial: an algebraic expression with multiple terms monomial: an algebraic expression with one term trinomial: an algebraic expression with three terms

Teaching Aids

Cards (each with a unique algebraic expression); Chart paper; Coloured pencils; Algebra tiles made using blue, green, red and yellow paper cutouts; Sheet of paper; Paper cutouts for representing the algebraic terms 1, x, y, x2 , xy; Glue

Chapter: Algebraic Expressions

Terms and Factors of an Algebraic Expression

Learning Outcomes

Imagine Maths Page 192

Students will be able to draw a factor tree for an expression and list the terms and factors.

Teaching Aids

Cards (each with a unique algebraic expression); Chart paper; Coloured pencils

Activity

Start by introducing the concept of algebraic expressions. Explain how factor trees help in finding the terms and factors of a given algebraic expression.

Ask the students to work in groups.

Distribute two cards each containing an algebraic expression (like 2t3 + 9t2 + t – 12), chart paper and coloured pencils to each group.

Instruct the students to draw a three-layered tree diagram on the chart paper for each expression on the cards. Ask them to put the expression in the first layer, the different terms of the expression in the second layer, and the factors of each term in the third layer.

Guide the students through the process of factorising the terms. Demonstrate how to factorise the first term step by step. Encourage students to collaborate within their groups to factorise the remaining terms.

Instruct them to use different colours to represent the different layers for the expression, terms, and factors, and label each layer clearly to indicate what it represents.

Give the students more practice questions on drawing factor trees for given expressions and listing the terms and factors.

Extension Idea

Instruct: Create an expression that has 5x3, –7x2, –3x and 2 as its terms.

Say: We can create many expressions using the given terms. One such expression

Addition of Algebraic Expressions;

Subtraction

of Algebraic Expressions

Learning Outcomes

Imagine Maths Page 194

Students will be able to add and subtract algebraic expressions using the horizontal and vertical method.

Teaching Aids

Algebra tiles made using blue, green, red and yellow paper cutouts

Activity

Introduce algebra tiles to the students. Explain the values of the various cutouts and that these will be called algebra tiles.

+1 = –1 = x = –x = x2 = –x2 =

Bring out the fact that a pair of negative and positive tiles together represent 0. Ask the students to work in pairs. Distribute algebra tiles among the pairs.

Instruct the students to add (2x2 + 3x + 5) and (x2 – 2x – 3) using the algebra tiles. Ask them to arrange algebra tiles to represent (2x2 + 3x + 5). Then, ask them to place algebra tiles representing (x2 – 2x – 3) below the previous set. Instruct them to remove any pair of algebra tiles that make zero. Explain that the remaining number of algebra tiles (3x2 + x + 2) would show the answer.

Answer: 3x2 + x + 2

Instruct the students to write the expressions one below the other and the answer below them. Ask them to observe the expressions and the answer and try to understand the rules for adding expressions using the vertical method.

Have a discussion on the rules of addition.

Then, ask the students to add the expressions (4x2 – 6x + 3) and (–6x2 + 2x – 7) using the vertical method in their notebooks.

Discuss the answers.

Give the students more practice questions on adding and subtracting algebraic expressions using the horizontal and vertical methods.

Extension Idea

Instruct: Subtract: 31722 –+–564 nmmnn from 24522 +–732 mnmnn

Say: On subtracting the expressions, we get 22 . 5293 +–42154 mnmnn

Multiplying a Monomial by a Monomial

Learning Outcomes

Students will be able to multiply a monomial by a monomial.

Teaching Aids

Paper cutouts for representing the algebraic terms 1, x, y and xy

Activity

Start the class by discussing the box method of multiplying two numbers. Ask the students to work in groups. Distribute the cutouts to each group.

Instruct them to use the box method to multiply two monomials, 2x and 2y, by representing the monomials using the paper cutouts. Ask them to place the two x cutouts along a row and the two y cutouts along a column, multiply them one by one and place the xy cutouts to show the products. Next, ask them to add the products to get the final answer (4xy).

Once they are done, ask them to write the answers in their notebooks and then discuss the answers within their groups. Discuss their results.

Discuss the rule of multiplying monomials.

Repeat the activity by giving the students two more multiplication problems to solve using the cutouts as well as the rule.

Extension Idea

Ask: What is the value of xy multiplied by the product of 6y and –4?

Say: 6y × (–4) = –24y. On multiplying –24y by xy, we get –24xy2 .

Multiplying a Monomial by a Polynomial

Learning Outcomes

Students will be able to multiply a monomial by a polynomial.

Teaching Aids

Paper cutouts for representing the algebraic terms x, 1, and x2

Activity

Start the class by briefly revisiting the box method of multiplying two numbers. Ask the students to work in groups. Distribute the cutouts to each group.

Instruct them to use the box method to multiply a monomial x and a polynomial (x + 1) by representing them using the paper cutouts. Ask them to place one x cutout along a row and one x and one 1 cutout along a column, multiply them one by one and place the x2 and x cutouts to show the respective products. Next, ask them to add the products to get the final answer (x2 + x).

Once they are done, ask them to write the answers in their notebooks. Discuss their results.

Discuss the rule of multiplying a monomial by a polynomial. Ask questions like: How would you multiply a trinomial by a monomial?

Repeat the activity by giving the students two more multiplication problems to solve using the cutouts as well as the rule.

Extension Idea

Ask: Verify the product of 5x2(x + 5) for x = 1

Say: 5x2(x + 5) = 5x3 + 25x2. LHS = 5x2(x + 5) and RHS = 5x3 + 25x2 . For x = 1, LHS = 5(1)2 (1 + 5) = 30, RHS = 5(1)3 + 25(1)2 = 5 + 25 = 30. Hence, LHS = RHS = 30.

Learning Outcomes

Students will be able to multiply a polynomial by a polynomial.

Teaching Aids

Paper cutouts for representing the algebraic terms x, y, 1, and xy; Sheet of paper; Glue

Activity

Ask the students to work in groups. Distribute cutouts to represent a polynomial multiplication, such as (x + 2)(y + 2) = xy + 2x + 2y + 4, a sheet of paper and glue to each group.

Instruct them to arrange and paste the terms of the two polynomials in the corresponding rows and columns. Ask questions like: How would you find the product? Instruct them to find the product and paste the respective cutout at each intersection of a row and column. Then, ask them to add the individual products and simplify to find the answer in their notebooks.

Repeat the activity with a trinomial and binomial multiplication. Demonstrate how to solve such multiplication problems on the board.

Give the students some practice questions to solve in their notebooks. Give them some problems on going from products to factors, this will enhance their learning by enabling a strong grasp of the concept.

Answers

1. Terms and Factors of an Algebraic Expression

2. Addition of Algebraic Expressions

So, the perimeter of the rectangle is

17 1 –10+ 32 æö

3. Subtraction of Algebraic Expressions

Think and Tell

If you do not evaluate the minus sign between two algebraic expressions, it means you are not performing the operation of subtraction. Instead, you are performing the operation of addition.

Do It Together

=–7+11–7+5–8+6 pqqpqq

=–7+5+11–8–7+6 pqpqqq

=–2+3–1 pqq

So, –2+3–1 pqq will be added to –5+8–6 pqq to get –7+11–7. pqq

4. Simplifying Algebraic Expressions

Do It Together

Perimeter of the triangle = (4a + 10b − 6) cm

Let the third side be X

Perimeter of a triangle = 9a − 5b + 7 + (−5a + 2b – 3) + X (4+10–6)=9–5+7+(–5+2–3)+ abababX =4+10–6–(9–5–7)–(–5+2–3) Xababab =4+10–6–9+5–7+5–2+3 Xababab =4–9+5+10+5–2–6–7+3 Xaaabbb =13–10Xb

So, the third side of the triangle is (13b 10) cm

5. Multiplying a Monomial by a Monomial

Do It Together

2a2b3c × (−a2bc2) × 5a3b3c

223332 (2×(1)×5)×(××)×(××)×(××) aaabbbccc =–774 10abc =–

6. Multiplying a Monomial by a Polynomial Do It Together

4x2y (9x2y2z + 7) 2222 (4×9)(4×7) xyxyzxy=+ 3628432 xyzxy=+

7. Multiplying a Polynomial by a Polynomial Think and Tell

Yes. While the FOIL method is a specific way to perform multiplication for binomials, the BOX method is a more general approach that can be extended to larger algebraic expressions by creating a grid for each term.

Do It Together (3x2 − 5x + 1) (x − 1) − 2x (2x + 3) = 3x2 (x − 1) − 5x (

= 22 2 (3)–(31)–(5)(51)–1–4–6 xxxxxxxxx =×××+×+

= 322 2 =3–3–55–1–4–6 ++ xxxxxxx

= 32 3–12–1 xx =

Area of Polygons

Learning Outcomes

Students will be able to: find the area of figures made with parallelograms, triangles and circles. find the area of a trapezium and a rhombus. find the area of general quadrilaterals. find the area of combined shapes.

Alignment to NCF

CG-4: Develops understanding of perimeter and area for 2D shapes and uses them to solve day-to-day life problems

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-4.3: Constructs various designs (using tiling) on a plane surface using different 2D shapes and appreciates their appearances in art in India and around the world

Let’s

Recall

Recap to check if students know how to find the perimeter and area of a triangle. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

parallelogram: a flat closed 2-D figure with opposite sides that are equal and parallel trapezium: a quadrilateral in which one pair of opposite sides is parallel rhombus: a flat, closed figure in which all four sides are equal in length, and both pairs of opposite sides are parallel

Teaching Aids

Cutout of a figure made with a parallelogram, a triangle and two semi-circles; Cutouts of a trapezium and rhombus; Chart paper; Glue stick; Cutouts of general quadrilaterals; Cutouts of a rhombus, trapezium, parallelogram and general quadrilateral; Ruler; Sheets of paper

Chapter: Area of Polygons

Area of Figures Made with Parallelograms,

Imagine Maths Page 216 Triangles and Circles

Learning Outcomes

Students will be able to find the area of figures made with parallelograms, triangles and circles.

Teaching Aids

Cutout of a figure made with a parallelogram, a triangle and two semi-circles

Activity

Revisit the formulas for finding the area of a parallelogram, triangle and circle. Ask the students to note the formulas in their notebooks.

Ask the students to work in small groups. Distribute the teaching aids to each group. Make sure that all the groups get cutouts of the same size.

Instruct the groups to measure and label the lengths of the sides of the cutouts and stick them on the chart paper using glue to create a new shape. Ask them to be careful not to overlap any of the shapes. Instruct them to find the area of the new shape created, in their notebooks.

Ask questions like: How did you find the area of the shape formed? What other shape can you form using the same cutouts?

Discuss how the answer for all the groups was the same even though the shapes were arranged differently by different groups to make the combined shapes. Give them more problems to solve if time permits.

Extension Idea

Ask: A kid’s circular swimming pool of diameter 2.8 m was made (inscribed) within a square piece of land. What is the area of the remaining piece of land?

Say: The total area of the square piece of land = 2.8 × 2.8 = 7.84 sq. m. and the area of the circular pool = 22 7 × 1.4 × 1.4 = 6.16 sq. m. The area of the remaining piece of land = 7.84 – 6.16 = 1.68 sq. m.

Area of a Trapezium and Rhombus

Learning Outcomes

Students will be able to find the area of a trapezium and a rhombus.

Teaching Aids

Cutouts of a trapezium and rhombus; Chart paper; Glue stick

Activity

Imagine Maths Page 220

Begin the class by showing the students cutouts of a trapezium and a rhombus and discussing the properties of these two shapes.

Ask the students to work in pairs. Distribute cutouts of a trapezium and rhombus to each pair. Instruct the pairs to stick the shapes on the chart paper and label the vertices. Ask them to derive the formula for finding the area of the given shapes by dividing them into smaller shapes. Let the students try to derive the formula. Give them hints like: Thetrapeziumcanbedividedinto2trianglesandonerectangletoderivetheformula. Theoppositeverticesoftherhombuscanbejoinedtoderivetheformula.

When they are done, let the students share their findings with the rest of the class. Ask them to open page 220 of their ImagineMathematics books and check whether they have derived the formula correctly or not.

Once all the students understand the formula, give them practice problems on finding the area and other measures of a trapezium and rhombus.

Extension Idea

Ask: Is it possible for a rhombus and a trapezium to have the same area? Give one example. Say: Yes, it is possible for a rhombus and a trapezium to have the same area. For example, consider a rhombus with diagonals of length 40 cm and 10 cm and a trapezium with parallel sides of lengths 8 cm and 12 cm and a height of 20 cm. Both figures would have an area of 200 cm2.

Area of rhombus = 1 2 × product of diagonals = 1 2 × 10 × 40 = 200 cm2

Area of trapezium = 1 2 × (sum of parallel sides) × height = 1 2 × (12 + 8) × 20 = 1 2 × 20 × 20 = 200 cm2

Area of General Quadrilaterals

Learning Outcomes

Students will be able to find the area of general quadrilaterals.

Teaching Aids

Cutouts of general quadrilaterals; Chart paper; Glue stick

Activity

Imagine Maths Page 224

Begin by discussing the features of a quadrilateral. Discuss various types of quadrilaterals. Ask the students to work in pairs. Distribute a cutout of a general quadrilateral and chart paper to each pair.

Instruct the pairs to stick the shape on the chart paper and label the vertices. Ask them to derive the formula for finding the area of the shape. Let the students try to derive the formula. Give them hints like: Join one vertex to the opposite vertex to form a diagonal and divide the quadrilateral into 2 triangles.

Once they are done, let the students share their findings with the rest of the class. Ask them to open page 224 of their ImagineMathematics books and check whether they have derived the formula correctly or not. Once all students understand the formula, give them practice problems on finding the area of general quadrilaterals.

Extension Idea

Ask: Rhea has a wall shaped liked a quadrilateral where h1 = 5 feet, h2 = 6 feet and BD = 21 feet. What is the area of the wall?

Say: The wall is shaped like a quadrilateral and its area can be measured using the formula 1 2 × 21 × (5 + 6) = 115.5 sq. feet.

Area of Combined Shapes Imagine Maths Page 226

Learning Outcomes

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

Teaching Aids

Cutouts of a rhombus, trapezium, parallelogram and general quadrilateral; Glue stick; Ruler; Sheets of paper

Activity

Recap the area formulas for a rhombus, trapezium and general quadrilateral.

Ask the students whether we can derive a formula for all shapes in real life. Give real-life examples of shapes or buildings that are a combination of various shapes like The Pentagon in USA.

Ask the students to work in groups. Distribute the shape cutouts, a sheet of paper and a glue stick to each group.

Instruct the groups to paste the shapes on the sheet of paper in any way they like without overlapping to make a combined shape like the given example.

Ask them to measure the lengths needed to find the area of each shape and label the figure. Each student of the group should find the area of one of the shapes and then the group should add the areas to get the area of the complete figure.

Discuss the answers.

Ask questions like: Will the area of the combined shape be the same if the quadrilaterals are arranged differently?

Give the students more questions for practice.

Answers

1. Area of Figures Made with Parallelograms, Triangles and Circles

Do It Together

1. Perimeter = 86 cm

9 + 15 + x + 12 + 12 + 12 + 15 = 86 cm

75 + x = 86

x = 11 cm

2. Length of the rectangle = 15 cm; Breadth of the rectangle = 9 cm

Area of the rectangle = length × breadth = 15 × 9 = 135 sq. cm

Side of the square = 12 cm

Area of the square = side × side = 12 × 12 = 144 sq. cm

Base of the triangle = 12 cm; Height of the triangle = 9 cm

Area of the triangle = 1 2 × base × height = 1 2 × 12 × 9 = 54 sq. cm

Total area of the figure = Area of the rectangle + Area of the square + Area of the triangle = 135 + 144 + 54 = 333 sq. cm

Fraction of the area of the triangle of the total area = 54 333 = 6 37

2. Area of a Trapezium and Rhombus

Think and Tell

To find the area of a trapezium (trapezoid) using the area of only a triangle, we need to divide the trapezium into two triangles and use the formula for the area of a triangle. We can do so by drawing a diagonal from one vertex to the opposite vertex, splitting the trapezium into two triangles.

Think and Tell

Since the diagonals bisect each other at right angles, they split up the rhombus into four congruent right triangles. We can use the Pythagorean theorem to find the length of one-half of a side of the rhombus.

If d1and d2 are the diagonals of the rhombus, the side of the rhombus can be given as:

12 + 22

Do It Together

One diagonal = 14 cm

So, 14 × d2 = 672 ⇒ d2 = 672 14 = 48 cm

To find the perimeter of the rhombus, we need to find the side of the rhombus.

Side of the rhombus = ()() 22 22 12 22 1448 + =+ 2222 =7+24 =625 =25 cm

Perimeter = 4 × side = 4 × 25 = 100 cm

3. Area of General Quadrilaterals

Do It Together

Area of quadrilateral = 1 2 × diagonal × sum of length of perpendiculars on the diagonal from the opposite vertices

= 1 2 × 25 × 12 = 150 sq. cm

Lengths of perpendiculars from opposite vertices are 4 cm and 8 cm.

4. Area of Combined Shapes

Do It Together

Area of the polygon = 22 × 6 + 22 × 9 + 22 × 9 + 1 2 × 20 × 9

= 22 × (6 + 9 + 9) + 90

= 22 × 24 + 90 = 528 + 90 = 618 sq. cm

Surface Area and Volume of Solids

Learning Outcomes

Students will be able to:

deduce and apply the formula for the surface area of a cuboid. deduce and apply the formula for the surface area of a cube. deduce and apply the formula for the surface area of a cylinder. deduce and apply the formula for the volume of a cuboid. deduce and apply the formula for the volume of a cube. deduce and apply the formula for the volume of a cylinder.

Alignment to NCF

C-5.2: Visualises and uses mathematical thinking to discover formulae to calculate surface areas and volumes of solid objects (cubes, cuboids, spheres, hemispheres, right circular cylinders/cones, and their combinations)

Let’s Recall

Recap to check if students know about perimeter and area of 2-D shapes. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

surface area: the measurement of the total area of the outer surface of a three-dimensional object volume: the measurement of the space occupied by the entire three-dimensional object

Teaching Aids

Sheets of paper; Ruler; Scissors; Cuboidal box; Unit cubes; Cubical box; Cylindrical jar; Cardboard circular discs with the same radius as the jar

Chapter: Surface Area and Volume of Solids

Surface Area of Cuboids

Learning Outcomes

Imagine Maths Page 235

Students will be able to deduce and apply the formula for the surface area of a cuboid.

Teaching Aids

Sheets of paper; Ruler; Scissors

Activity

Discuss how to find the area of rectangles using the formula by giving the measures of length and width. Discuss the nets of a cuboid and the number of rectangles that can be seen in the net of a cuboid.

Ask the students to form equal groups. Distribute the teaching aids to each group.

Instruct the groups to draw the nets of 2 cuboids of different dimensions on the sheet. Ask them to cut out the nets, measure the lengths of the sides and write the measures inside each rectangle. Instruct them to find the area of each rectangle and add the areas. Ask them what this area represents. Discuss that the sum of areas of all the rectangles form the total surface area of the cuboid.

Ask the students to use this relationship to deduce the formula for the surface area of a cuboid.

Give the students questions on finding the surface area of a cuboid given the lengths of the edges or on finding the missing length, width, or height of a cuboid when two of these dimensions and the surface area are given.

Surface Area of Cubes

Learning Outcomes

Students will be able to deduce and apply the formula for the surface area of a cube.

Teaching Aids

Sheets of paper; Ruler; Scissors

Activity

Imagine Maths Page 237

Discuss how to find the area of a square using the formula by giving the measure of its side. Discuss the nets of a cube and the number of squares that can be seen in the net of a cube.

Ask the students to form groups of 4. Distribute the teaching aids to each group.

Instruct the groups to draw the nets of 2 cubes of different dimensions on the sheet. Ask them to cut out the nets, measure the lengths of the sides and write the measures inside each square. Instruct them to find the area of each square and add the areas. Ask them what this area represents. Discuss that the sum of areas of all the squares form the total surface area of the cube.

Ask the students to use this relationship to deduce the formula for the surface area of a cube.

Give the students questions on finding the surface area of a cube given the lengths of the edges or on finding the missing side length of a cube when its surface area is given.

Extension Idea

Ask: A painted box of dimensions 8 cm × 5 cm × 3 cm has a 72 sq. cm painted surface including all its faces. What is the length (as a whole number) of the side of the largest cube that uses the same amount of paint as the box? What is the surface area of such a cube? Explain.

Say: The painted area is 72 sq. cm. The maximum area of the cube can be 72 sq. cm. So, area of cube < 72 sq. cm  6a2  72  a2  12. So, the maximum value of a can be 3 cm. Hence, the largest possible cube that can be painted has a side of length 3 cm. Surface area of a cube with side length 3 cm = 6 × (3 × 3) = 6 × 9 = 54 sq. cm.

Surface Area of Cylinders

Learning Outcomes

Imagine Maths Page 238

Students will be able to deduce and apply the formula for the surface area of a cylinder.

Teaching Aids

Sheets of paper; Ruler; Scissors

Activity

Discuss how to find the area of a rectangle and circle using the respective formulas by giving the measures of the sides and radius. Discuss the nets of a cylinder and the number of circles and rectangles that can be seen in the net of a cylinder.

Ask the students to form groups of 4. Distribute the teaching aids to each group. Instruct the groups to draw the nets of 2 cylinders of different dimensions on the sheet. Ask them to cut out the nets, measure the lengths of the sides and radii and write the measures inside the rectangle and each circle. Ask them to find the area of the rectangle and circles and add the areas. Ask them what this area represents. Discuss that the sum of areas of all the shapes form the total surface area of the cylinder.

Ask the students to use this relationship to deduce the formula for the surface area of a cylinder. Give the students questions on finding the surface area of a cylinder given the height and radius or on finding the missing height or radius of a cylinder when one of the two dimensions and the surface area are given.

Extension Idea

Ask: A cylindrical block has a diameter of 2 metres and a height of 10 metres. If both the height and diameter are doubled, how does this affect the surface area?

Say: Original total surface area = 2πr (r + h) = 2π × 1(1 + 10) = 22π sq. m, New total surface area = 2πr (r + h) = 2π × 2 (2 + 20) = 88π sq. m. So, the new surface area is 4 times the original surface area.

Volume

of Cuboids

Learning

Outcomes

Students will be able to deduce and apply the formula for the volume of a cuboid.

Teaching Aids

Cuboidal box; Unit cubes

Imagine Maths Page 241

Activity

Discuss the concept of volume and how it applies to cuboids. Emphasise that it represents the amount of space inside the cuboid.

Ask the students to form groups of 4. Distribute a cuboidal box and some unit cubes to each group.

Instruct them to find the number of unit cubes that fit inside the cuboidal box. Then, ask them to count the number of unit cubes along the length, breadth and height of the cuboidal box. Using this, ask them to establish a relationship between the total number of cubes and the number of cubes along the edges of the cuboidal box to deduce the formula for the volume of a cuboid.

Ask questions like: How is the total number of cubes inside the box related to the product of the number of cubes along the length, breadth and height?

Give the students some questions on finding the volume of a cuboid given the lengths of its edges or on finding a missing dimension given the other two dimensions and the volume of the cuboid.

Volume of Cubes Imagine Maths Page 243

Learning Outcomes

Students will be able to deduce and apply the formula for the volume of a cube.

Teaching Aids

Cubical box; Unit cubes

Activity

Discuss the concept of volume and how it applies to cubes. Emphasise that it represents the amount of space inside the cube.

Ask the students to form groups of 4. Distribute a cubical box and some unit cubes to each group.

Instruct them to find the number of unit cubes that fit inside the cubical box. Then, ask them to count the number of unit cubes along each side of the cubical box. Using this, ask them to establish a relationship between the total number of cubes and the number of cubes along each side of the cubical box to deduce the formula for the volume of a cube.

Ask questions like: How is the total number of cubes inside the box related to the product of the number of cubes along each side?

Give the students some questions on finding the volume of a cube given the length of one side or on finding the missing length given the volume of the cube.

Extension Idea

Ask: A large cubical block made of solid wood has a side length of 2 metres. You are given 64 smaller solid wooden cubes, each with a side length of 5 centimetres. If you have to carve out these cubes from the large cubical block, what is the volume of the large cubical box left, if these small cubes are carved out of the box?

Say: Volume of large cube = 2 m × 2 m × 2 m = 8 cu. m = 80,00,000 cu. cm. Volume of each tiny cube = 5 cm × 5 cm × 5 cm = 125 cu. cm. Volume of tiny cubes = 64 × 125 = 8,000 cu. cm. Volume left = 80,00,000 – 8000 = 79,92,000 cu. cm.

Volume of a Cylinder

Learning Outcomes

Students will be able to deduce and apply the formula for the volume of a cylinder.

Teaching Aids

Cylindrical jar; Cardboard circular discs with the same radius as the jar

Activity

Discuss the concept of volume and how it relates to cylinders. Emphasise that it represents the amount of space inside the cylinder.

Ask the students to form groups of 4. Distribute the teaching aids to each group.

Instruct them to find the number of discs that fit inside the cylindrical jar. Then, ask them to find the radius of each discs. Using this, ask them to establish a relationship between the total number of discs and the radius of the discs to deduce the formula for the volume of a cylinder.

Ask questions like: How is the total number of discs inside the jar related to the height of the discs?

Give the students some questions on finding the volume of a cylinder given the height and radius or on finding the missing height or radius given one of the two dimensions and the volume of the cylinder.

Extension Idea

Ask: You want to bake a cylindrical birthday cake with a fixed height of 28 cm. You have enough batter to make a cake with a total volume of 5632 cu. cm. What is the maximum diameter you can create for the cake while using all the batter?

Volume = π × D 2 2 × 28 = 7πD2. So, 7πD2 = 5632  D2 = 5632 7π  D = 5632 7π = 16 cm.

Therefore, the maximum diameter of the cake you can create with the available batter is approximately 16 cm.

Answers

1. Surface Area of Cuboids

Do It Together

Total surface area = 2 × (lb + lh + bh)

190 = 2 × (5l + 3l +15)

190 = 10l + 6l + 30

160 = 16l

l = 10 m

2. Surface Area of Cubes

Do It Together

New length of the side of the cube = ɑ + 50% of ɑ = ɑ + 1 2

= 3 2 a

Total surface area of the new cube = 6 × 3 2

Increase in the total surface area of the cube = 27 2 a

= 15 2 a2

Percentage increase in the total surface area of the cube =

= 5 4 × 100% = 125%

3. Surface Area of Cylinders

Do It Together

r = 330×7 2×22×15 m = 7 2 m

4. Volume of Cuboids

Think and Tell

Capacity is the maximum amount of a substance that an object can hold whereas volume is the total amount of space occupied by an object.

Do It Together

Breadth = 800 cm = 8 m

Capacity of the tank = 4,80,000 L

1000 L = 1 m3

4,80,000 L = 480 m3

Volume = lbh = 12 × 8 × h

h = 480 12 × 8 m = 5 m

5. Volume of Cubes

Do It Together

Side of the cube = 12 cm

Volume = Side3 = 12 × 12 × 12 cm3 = 1728 cm3

We know that, 1 cm3 = 1 mL

Total capacity of the cube = 1728 mL

6. Volume of a Cylinder

Do It Together

r2 = 79.2×7 22×70 m2 = 0.36 m2

r = 0.6 m

Exponents and Powers

Learning Outcomes

Students will be able to:

simplify expressions involving positive exponents of a rational number. simplify expressions involving negative exponents of a rational number. apply laws of exponents to simplify exponential expressions. solve word problems on expressing large numbers in the standard form.

Alignment to NCF

CG-1: Understands numbers and sets of numbers (whole numbers, fractions, integers, rational numbers, and real numbers), looks for patterns, and appreciates relationships between numbers

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

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

C-2.1: Extends the understanding of powers (radical powers) and exponents

Let’s Recall

Recap to check if students know the concepts of squares and cubes. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

binary fission: a process in which a single bacterial cell splits into two cells to multiply exponent: the number of times a number is multiplied by itself

Teaching Aids

Cards (each with rational number written on it); Cards (with exponents); Cards with operations (+, –, × and ÷); Word problem cards

Chapter: Exponents and Powers

Positive Integral Exponent of a Rational Number

Learning Outcomes

Imagine Maths Page 253

Students will be able to simplify expressions involving positive exponents of a rational number.

Teaching Aids

Cards (each with a rational number written on it)

Activity

Begin the class by explaining that when any number, such as 2, is raised to the power n (i.e., 2n ) it means that 2 is being multiplied by itself n times.

Divide the class into groups of 10. Distribute a card containing a base number, such as –2 3 , to each group. Assign a unique number (from 1 to 10) to each student within the group.

In their notebooks, instruct the first student to raise the base number to the power of 1, i.e., 1 –2–2 = 33

; the second student to raise it to the power of 2, i.e.,

; and so on up to the tenth student.

Ask the students how they can speed up calculations while ensuring accuracy. Discuss that they do not need to calculate their result from scratch; instead, they can use the result obtained by the preceding student.

Extension Idea

Ask: If 5125 = 7343 m

, what is the value of m?

Say: 3 5×5×5=5=125 and 3 7×7×7=7=343;

Therefore, m = 3.

Negative Integral Exponent of a Rational Number Imagine Maths Page 254

Learning Outcomes

Students will be able to simplify expressions involving negative exponents of a rational number.

Teaching Aids

Cards (each a with rational number written on it)

Activity

Begin the class by explaining that when any number, such as 2, is raised to the power –n(i.e., 2–n), the result is the reciprocal of the exponential power, i.e., 1 . 2n

Divide the class into groups of 10. Distribute a card containing a base number, such as 1 2 , to each group. Assign a unique integer (from –1 to –10) to each student within the group.

In their notebooks, instruct the first student to raise the base number to the power of –1, i.e.,

; the second student to raise it to the power of –2, i.e.,

; and so on up to the tenth student.

Ask how they can speed up calculations while staying accurate. Discuss how they are not required to calculate their result from scratch; instead, they can use the result obtained by the preceding student.

Extension Idea

Ask: How do we express this in decimals: 5×10–5 ?

Say: We can write it as: –5 5 5×10=5×=5×1115=5×==0.00005. 10×10×10×10×10100000100000

Applying Laws of Exponents

Learning Outcomes

Students will be able to apply laws of exponents to simplify exponential expressions.

Teaching Aids

Cards (with exponents); Cards with operations (+, –, × and ÷)

Activity

Imagine Maths Page 258

Begin the class by discussing the basic concepts of exponents and their properties. Briefly discuss the importance of simplifying exponential expressions.

Prepare cards showing rational numbers raised to some exponential power, for instance,

Prepare cards showing mathematical operators (+, –, × and ÷).

and 55

Ask the students to work in pairs. Distribute one exponent card to each student and one operator card to each pair. Instruct the pairs to combine the two exponents using the operator on their card to form a mathematical expression.

For example: –38 11 × 55

Ask them to solve the resultant expression and write the answers in their notebooks. Instruct the students to exchange cards with other pairs and solve the expressions.

Extension Idea

Ask: 2 5×125=25. xx What is the value of x?

Say: +3 334 5×125=5×(5)=5×5=5=5 xxxxxxxxx and 2224 25(5)5 == . So, 44 5=54=4=1. →→ xxx

Use of Exponents

Learning Outcomes

Imagine Maths Page 259

Students will be able to solve word problems on expressing large numbers in the standard form.

Teaching Aids

Word problem cards

Activity

Begin the class by explaining what standard form is and its importance in dealing with large numbers. Give some examples of large numbers written in standard form.

Distribute a word problem card to each student.

Thevalueofanantiqueiteminthemarket50yearsagowas₹4.Howmuchwoulditcostnowifeveryfiveyearsits valueincreasedtothepreviousvalueraisedtothepowerof2?

Instruct the students to use the CUBES strategy to solve the problem. Circulate around the classroom to provide assistance and guidance as needed.

Instruct the students to exchange their cards with their partner. Have them solve the new word problem on their card using the same CUBES strategy.

Once they have completed the word problems, review the answers together as a class.

Answers

1. Positive Integral Exponent of a Rational Number Do It Together

2. Negative Integral Exponent of a Rational Number

Think and Tell

At each step, the value is decreased to half the previous value. We can extend this pattern further as:

3. Applying Laws of Exponents Do It Together

4. Use of Exponents Do It Together

Direct and Inverse Proportions

Learning Outcomes

Students will be able to: solve problems on direct proportion. solve problems on inverse proportion. solve problems on time and work. solve problems on pipes and cisterns.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to find the simplest form and equivalent ratios of given ratios. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

variation: a change or difference in something direct proportion: when two quantities increase or decrease at the same rate inverse proportion: when one quantity increases while the other decreases, and vice versa

Teaching Aids

Counters; Paper plates; Water; Building blocks; Stopwatch; Bowls; Bottles of water; 2 funnels of different widths

Chapter: Direct and Inverse Proportions

Direct Proportion

Learning Outcomes

Students will be able to solve problems on direct proportion.

Teaching Aids

Counters; Paper plates

Activity

Ask the students to form groups. Provide each group with a set of counters to represent cookies. Instruct the students to imagine they are organising a party and that the counters are cookies. Ask them to group 4 cookies (counters) per plate. Then, have them calculate the number of cookies that can be put on 2, 3 or 10 plates.

Ask questions like, “What is the ratio of the number of plates to the number of cookies?”

Instruct the students to write the ratio of plates to cookies in its simplest form in their notebooks. Discuss their answers and guide them to deduce the concept of direct proportion. Draw a table as shown on the board and ask the students to copy it into their notebooks.

Help students notice the variation in quantities. Encourage them to write their learnings in their notebooks. Give the students some practice questions based on direct proportion.

Extension Idea

Ask: If the ratio of friends to cookies is 1:5, for how many friends should 45 cookies be made? Say: If the ratio of friends to cookies is 1:5, it means that for every group of 1 friend, there are 5 cookies. Therefore, 45 cookies should be made for 9 friends.

Problems on Inverse Proportion Imagine Maths Page 272

Learning Outcomes

Students will be able to solve problems on inverse proportion.

Teaching Aids

Counters; Water

Activity

Ask the students to form groups and give each group a set number of counters (e.g.,12 counters). Guide the students to divide the counters into groups. First into groups of 2, then into groups of 3 and finally into groups of 6. Encourage the students to observe that as the number of groups increases, the number of items in each group decreases.

Ask questions like, “How does the ratio of counters to groups affect the number of counters in each group?” Encourage the students to discuss their observations within their groups and write their learnings in their notebooks.

Ask the students to repeat the activity using 18 counters.

Extension Idea

Ask: If it takes 5 students 4 hours to clean a classroom, how many hours will it take for 10 students to clean the same classroom, assuming they work at the same rate?

Say: As the number of students is inversely proportional to the time taken, it will take 10 students 2 hours to clean the same classroom.

Time and Work

Learning Outcomes

Students will be able to solve problems on time and work.

Teaching Aids

Building blocks; Stopwatch

Activity

Ask the students to work in pairs. Provide each student an identical set of building blocks. Explain that their challenge is to individually construct 2 towers of 24 building blocks within a specified time frame. Start the timer as soon as they begin building. Once they are done, ask questions like, “What did you notice about the amount of work you could complete within the given time?” Record the time each student took to complete the construction.

Imagine Maths Page 274

Ask the students to pair up to construct 2 towers by each student constructing one tower and note how it takes less time to complete the task when more people work together. Demonstrate how to solve a problem related to time and work. Then, give the students some practice questions on time and work. Discuss their responses and encourage them to write their learnings in their notebooks.

Pipes and Cisterns Problems

Learning Outcomes

Students will be able to solve problems on pipes and cisterns.

Teaching Aids

Bowls; Bottles of water; 2 funnels of different widths; Stopwatch; Water

Imagine Maths Page 276

Activity

Take the students to an open ground. Divide the students into groups. Distribute the teaching aids to each group.

Instruct the students to pour water from the bottle into the bowl using the first funnel and record the time taken to fill the bowl completely.

Ask the students to empty the bowl. Now, instruct them to use the second funnel to fill the bowl completely and note the time taken.

Ask the students to empty the bowl again, use both funnels simultaneously to fill the bowl completely and note the time taken.

Encourage a discussion around what the students noticed about the time taken to fill the bowl in all three cases. Give them some practice problems based on pipes and cisterns.

Extension Idea

Ask: What will happen if there is one outflow pipe in the bowl? Does it affect the flow of water?

Say: Yes, the outflow pipe affects the flow of water, and the time taken to fill the bucket increases.

Answers

1. Direct Proportion

Do It Together

Length of Cloth 2 m 5 m 8 m 15 m 20 m

Cost of Cloth `150 `375 `600 `1125 `1500

2. Problems on Inverse Proportion

Do It Together

Let the total number of cows after he buys some cows be x

Number of cows 45 x

Time 60 54

More cows will consume more food, and hence it will last for fewer days.

Using the inverse proportion formula, 45 × 60 = x × 54

⇒ x = 45×60 54 = 50

Hence, the farmer bought 50 − 45 = 5 more cows.

3. Time and Work

Do It Together

Time taken by A and B to finish the work if they worked together = 30 days

(A+B)’s 1 day of work = 1 30

(A+B)’s 10 days of work = 10 30

Remaining work = 1 − 10 30 = 20 30

Now 20 30 work is done by A in 40 days.

Therefore, the whole piece of work will be done by A in 40 × 30 20 = 60 days

4. Pipes and Cisterns Problems

Think and Tell

If the outlet pipe empties the tank faster (i.e., takes less time), then it is not possible to fill the tank while both pipes are open because water will be removed at a faster rate than it is added.

Do It Together

Time taken by the pump to fill the tank without leakage = 2 hours

Work done by the pump in 1 hour without leakage = 1 2

Time taken by the pump to fill the tank with leakage = 1 2 3 = 7 3 hours

Work done by the pump in 1 hour with leakage = 3 7

Leakage per hour = 137–61 –== 271414

Time taken by leakage to empty the tank = 14 hours.

Factorisation and Division of Algebraic Expressions 16

Learning Outcomes

Students will be able to:

factorise algebraic expressions using common factors. factorise algebraic expressions by grouping and regrouping terms. factorise an expression using identities when the expression is a perfect square or difference of two squares. factorise quadratic trinomials. divide a monomial by a monomial. divide a polynomial by a monomial. divide a polynomial by a polynomial.

Alignment to NCF

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

Let’s Recall

Recap to check if students can identify monomials, binomials and polynomials, and know how to add or subtract algebraic expressions. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

factorisation: process of finding expressions that when multiplied, create a given expression

Teaching Aids

Chart paper; Pair of expression cards with polynomials having common factors; Glue stick; Paper cutouts of a rectangle and circle; Problem cards with terms of an algebraic expression written on them; Puzzle cards with 3 expressions; Cards with factorised answer for the given expression; Semi-circular problem cards with an expression, quarters showing factors of the expression in the semi circle; Expression cards (e.g., 10p2qr, 5pq); Factor cards for each expression on the expression cards and extra cards (such as, 2, 5, p, q, q, r, r, 5, p, q); Puzzle cards

Chapter: Factorisation and Division of Algebraic Expressions

Using Common Factors

Learning Outcomes

Students will be able to factorise algebraic expressions using common factors.

Teaching Aids

Chart paper; Pair of expression cards with polynomials having common factors; Glue stick; Paper cutouts of a rectangle and circle

Activity

Begin by revising the concept of identifying factors of natural numbers. Relate this to identifying factors of monomials. Create expressions cards (e.g., 6+12+4–24222324 xyxyxyxy ) with one expression on each card. Ensure that the card cover all types of expressions. Divide the class into groups of 3. Distribute 2 expression cards and the other teaching aids to each group.

Polynomial:

Terms:

Factors:

Instruct the groups to paste the expression cards on the chart paper and write the type of expression (monomial, binomial, etc.) on the left. Ask the group members to work together to form factor trees for the two expressions by pasting rectangle cutouts to show the terms and circle cutouts to show the factors. Ask them to also label the expression, terms and the factors by writing on the side.

Ask questions like, “Do you see some common factors? How can you factor them out?”

Discuss their responses to bring out that the expression can be factorised as 2

3). Discuss how the expression can be factorised without drawing a factor tree. Give the students more practice questions and ask them to write the factors and common factors in their notebooks.

Extension Idea

Ask: Can the expression 7xy + 15z be factorised?

Say: The common factor is 1, hence the expression cannot be factorised further.

Using Grouping and Regrouping Terms

Learning Outcomes

Students will be able to factorise algebraic expressions by grouping and regrouping terms.

Teaching Aids

Problem cards with terms of an algebraic expression written on them

Activity

Show the students four numbers 3, 7, 9 and 21 and ask them to identify the pairs of numbers that have common factors. Discuss how to factorise algebraic expressions by grouping and regrouping terms by identifying the

common factors. Demonstrate one or two examples on the board. Explain that this process makes factorisation simpler. Prepare sets of 4 cards each having one term of an expression, such as 3a2b – 2ac2 – 6ab + 4c2 .

Ask the students to work in groups and distribute a set of 4 cards to each group. Instruct them to arrange the cards to group the terms to be able to find the maximum number of common factors. Ask them to write the arranged expression in their notebooks and factor out the common factors to write the factored expression.

Ask questions like, “Was regrouping needed for all the expressions? How did you identify the terms to be regrouped?” Discuss their approach and rotate the problem cards among the groups for more practice.

Extension Idea

Ask: If the area of a rectangular plot is 25x4y3 – 5x2y + 15x4y2 – 3x2, how will you find the length and breadth?

Say: The length and breadth can be found by factorising the expression by regrouping terms.

25x4y3 – 5x2y + 15x4y2 – 3x2 = 25x4y3 + 15x4y2 – 5x2y – 3x2 = 5x4y2(5y + 3) – x2(5y + 3) = (5x4y2 – x2)(5y + 3); L = (5x4y2 – x2), B = (5y + 3).

Factorisation Using Identities

Learning Outcomes

Students will be able to factorise an expression using identities when the expression is a perfect square or difference of two squares.

Teaching Aids

Puzzle cards with 3 expressions; cards with factorised answer for the given expression

Activity

Prepare puzzle cards with algebraic expressions (some perfect squares, and some in the form a2 – b2), where one card has the expression and the other has the factorised answer.

Ask the students to work in groups. Distribute the puzzle cards and the factorised answer cards.

Instruct the students to factorise the expressions and match the puzzle card with factorised answer cards correctly.

Ask questions like, “Are you able to factorise the expressions using common factors or regrouping? How are these expressions different from those you have dealt with previously?” Help them deduce that not all expressions can be factorised using common factors or regrouping.

On the board, draw shapes like a square with side length (a + b) and a rectangle with length (a + b) and breadth (a – b). Ask them to multiply the dimensions to find the area. Ask questions like, “How can you use this learning to factorise?” Discuss their answers and demonstrate how identities can be used to factorise expressions.

Instruct the students to look at their expressions again to identify which of the three identities will be used for factorisation. Let them solve to write the factors and common factors in their notebooks.

Learning Outcomes

Students will be able to factorise quadratic trinomials.

Teaching Aids

Semi-circular problem cards with an expression, quarters showing factors of the expression in the semi circle; Glue stick

Activity

Define what a quadratic trinomial is and explain the steps to factorise a quadratic polynomial. Demonstrate a few examples on the board.

Prepare problem cards with quadratic trinomials written on them, and factor cards, some with correct factors and some with random factors written on them.

Divide the class into groups and distribute 1 problem card and 4 factor cards (2 correct and 2 random) to each group. Instruct the students to find the factors of the trinomial in their notebooks. Then, ask them to identify the correct factor cards as its solution and paste them with the problem card to complete the circle. Ask them to check with other groups whether they got the same factors.

Discuss the answers and let the groups exchange cards and solve.

Extension Idea

Ask: If you paid x2 + 5x + 6 rupees at a vegetable shop and bought x + 3 lemons, what is the price of each lemon?

Say: When we factorise the given polynomial, we get (x + 3) and (x + 2) as factors. So, if (x + 3) is the number of lemons, (x + 2) is the price of each lemon.

Dividing a Monomial by a Monomial

Learning Outcomes

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

Teaching Aids

Imagine Maths Page 290

Expression cards (e.g., 10p2qr, 5pq); Factor cards for each expression on the expression cards and extra cards (e.g., 2, 5, p, q, q, r, r, 5,p,q)

Activity

Begin by briefly revising division of an expression by numbers. Prepare expression cards (e.g. 10p2qr) and factor cards for each expression on the cards and some extra cards (e.g., 2, 5, p, q, q, r).

Ask the students to work in groups. Distribute the expression and factor cards to each group.

Give the students a division problem, such as 10p2qr by 5pq. Instruct them to represent the numerator and denominator as a product of prime factors by placing factor cards in the numerator and denominator. Ask them to remove the common factor cards from the numerator and denominator and group the rest. Tell them that the expression formed by the remaining cards is the answer.

Give more practice questions to the students and ask them to solve the problems first by using the factor cards and then in their notebooks.

Discuss the answers and ask them to share their learnings with the whole class.

Extension Idea

Instruct: Form a monomial division problem with the variables a,b,c where the quotient is a2 .

Say: There can be many such problems. One of them could be a3bc ÷ abc

Dividing a Polynomial by a Monomial

Learning Outcomes

Students will be able to divide a polynomial by a monomial.

Teaching Aids

Puzzle cards

Activity

Briefly explain how each term of the polynomial is divided by the monomial in polynomial by monomial division. Prepare puzzle cards where one has the dividend (e.g., 5x3 + 10x2 – 15x), another has the divisor (5x) and the third has the answer (x2 + 2x – 3) written on them.

Ask the students to work in groups of 3. Distribute 2 sets of puzzle cards to each group.

Imagine Maths Page 291

Instruct them to solve the division problem in each set to match the puzzle cards for the dividend and divisor with the card showing the correct solution. Once they are done, ask them to discuss their results in their groups and write the solution in their notebooks as well. Give the students more division problems for practice.

Dividing a Polynomial by a Polynomial

Learning Outcomes

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

Teaching Aids

Puzzle cards

Activity

Imagine Maths Page 291

Begin the class by explaining the methods of dividing a polynomial by a polynomial: by factorising and eliminating common factors and by long division. Prepare puzzle cards with polynomial division problems and their solutions.

Ask the students to work in pairs. Distribute a division problem card and some solution cards to each pair.

Tell each pair to solve the problems individually such that one partner solves by factorising and cancelling the common factors, and the other solves by the long division method. When they are done, ask them to compare to check whether they got the same answer by both methods. Finally, ask them to match the problem card with the correct solution card and showcase their completed puzzle to the class. Discuss their approaches and encourage them to write their answers in their notebooks.

Dividend

1. Using Common Factors

Think and Tell

No, multiplying a polynomial by a monomial does not change the number of terms in the original polynomial. When you multiply a polynomial by a monomial, you distribute the monomial to each term in the polynomial, but you don’t add or remove any terms.

Do It Together

1. 4a3 – 9a2b2 –9a2b2 4a3

4 × a × a × a –1 × 9 × a × a × b × b

= a2(4a – 9b2)

2. a(a – 3) – 7xy(3 – a)3

= a × (a + 7xy(a – 3)3

= (a – 3)((a + 7xy(a – 3)2)

= (a – 3)((a + 7xy(a2 – 6a + 9))

= (a – 3)(a + 7xya2 – 42xya + 63xy)

2. Using Grouping and Regrouping Terms

Do It Together

1. 3mn + 2pn + 3mq + 2pq

= 3mn + 3mq + 2pn + 2pq

= 3m(n+q) + 2p(n + q)

= (3m + 2p)(n + q)

2. x2 − ax − bx + ab x(x – a) – b(x–a) = (x–b)(x – a)

3. Factorisation When the Expression is a Perfect Square

Do It Together

1. b2 – 6ab + 9a2

= b2 + 2 × (–3a) × b + (–3a)2

= (b – 3a)2

2. 9x4 + 24x2 + 16 = (3x2)2 + 2 × 3x2 × 4 + (4)2

= (3x2 + 4)2

4. Factorisation When the Expression is a Difference of Two Squares

Do It Together

m4 − n4

= (m2)2 – (n2)2

= (m2 – n2)(m2 + n2)

= (m – n)(m+n)(m2 + n2)

5. Factorisation of Quadratic Trinomials

Do It Together

1. 2x2 + x – 6

= 2x2 + 4x – 3x – 6 = 2x(x + 2) – 3(x + 2)

= (2x – 3)(x + 2)

6. Dividing a Monomial by a Monomial

Do It Together

1. 21k9e4y

� e2 � y5 6

2. 64z9d7c6 12d2z2c3 = 64 � z9 � d7 � c6 12 � z2 � d2 � c

7. Dividing a Polynomial by a Monomial

Do It Together

2. (12a5b4c7 − 26a2b3c3 + 6a4b4c6) � 6b3a2c3 = 12a5b4c7 − 26a2b3c3 + 6a4b4c6 4b3a2

8. Dividing a Polynomial by a Polynomial Do It Together

Linear Graphs

Learning Outcomes

Students will be able to: read and interpret a double line graph. draw a double line graph using the given data. identify, plot and label points and axes on a coordinate plane. draw a linear graph using the given data. read and interpret a linear graph.

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

CG-8: Develops basic skills and capacities of computational thinking, namely, decomposition, pattern recognition, data representation, generalisation, abstraction, and algorithms in order to solve problems where such techniques of computational thinking are effective

Let’s Recall

Recap to check if students know how to read and draw bar graphs and pie charts. Ask students to solve the questions given in the Let’sWarm-up section.

Vocabulary

axis: a fixed line on a graph used for marking measurements coordinate: a unit that locates a specific point on a graph quadrant: a region that is defined by two axes

Teaching Aids

Sheet with a double line graph; Graph paper; Cards showing the population of 2 cities over 5 different years; Coloured pencils; Cards showing 4 pairs of coordinates; Sheet with a linear graph

Chapter: Linear Graphs

Reading Double Line Graphs

Learning

Outcomes

Students will be able to read and interpret a double line graph.

Teaching Aids

Sheet with a double line graph

Activity

Recall what a line graph is and what its purpose is. Ask the students to open page 298 in their Imagine Mathematics books and discuss the difference between a line graph and a double line graph.

Ask the students to work in pairs. Distribute a sheet with a double line graph to each pair. Ensure that the graph has clear labels, titles and axes. For example, it could represent a double line graph representing a comparison of the week’s sales for Company A and Company B.

Ask questions like, “On which day did Company A have the highest sales?” or “By how much did the sales of Company B exceed that of Company A on Wednesday?”

Instruct the students to analyse the graph and write the answers in their notebooks. Discuss the answers with the whole class. Give the students more questions for practice.

Extension Idea

Ask: The graph represents the time two friends spent gardening. Who spent more time gardening? Find the percentage difference in the number of hours spent.

Say: Friend A spent 28 hours gardening in total, whereas Friend B spent 16.5 hours. Friend A spent more time gardening. So, the percentage difference is: 28 − 16.5 28 × 100 = 11.5 28 × 100 = 41.07%.

Drawing Double Line Graphs

Learning

Outcomes

Students will be able to draw a double line graph using the given data.

Teaching Aids

Imagine Maths Page 300

Graph paper; Cards showing the population of 2 cities over 5 different years; Coloured pencils

Activity

Recall how to read and interpret a double line graph. Ask the students to work in pairs. Distribute graph paper and a card showing the population of 2 cities in different years to each pair. Year

A (in lakhs)

B (in lakhs)

Instruct the students to use this data to draw a double line graph on the graph paper. Provide guidance on labelling the axes and choosing an appropriate scale. Ask the students to use different colours to represent the two different data sets on the graph.

Ask them to point out the similarities and differences between single line graphs and double line graphs. If time permits, provide real-world examples of situations where double line graphs are used to compare two related sets of data for further analysis.

Extension Idea

Instruct: If the population of City C is given for different years, draw a line graph to represent the data on the same graph paper.

Say: This data can be represented on the same graph using a pencil in a third colour.

Cartesian Coordinate of a Point

Learning Outcomes

Students will be able to identify, plot and label points and axes on a coordinate plane.

Teaching Aids

Graph paper; Cards showing 4 pairs of coordinates Activity

Imagine Maths Page 303

Discuss what a cartesian plane is and how it works. Show the students some examples. Ask them to look at the graphs on page 304 of their ImagineMathematics book. Review the concepts of x-axis, y-axis, quadrants and coordinates, and demonstrate how to plot points on the coordinate plane.

Divide the class into groups of 4. Distribute graph paper and a card containing 4 pairs of coordinates to each group. For example:

(−3, 2), (−5, 4), (4, 1), (5, 0)

Instruct the students to plot each coordinate on the graph paper, labelling the points as A, B, C and D.

Ask questions like, “What quadrants do your coordinates lie in? If you join all the coordinates, does this form any shape?”

Extension Idea

Ask: A point lies equidistant from both the x-axis and y-axis in the third quadrant. What are the coordinates of the point if the area of the region is 4 sq. units?

Say: If the area is 4 sq. units and the point is equidistant from the x-axis and y-axis, then the point is at a distance of 2 units from each axis. Since it falls within the third quadrant, its coordinates are (−2, −2).

Drawing a Linear Graph

Learning Outcomes

Students will be able to draw a linear graph using the given data.

Teaching Aids

Graph paper

Activity

Discuss the concept of linear graphs. Ask the students to look at the graphs on page 306 of their Imagine Mathematics book. Explain how to draw a linear graph using given data points. Discuss the importance of labelling the axes with appropriate units and the significance of connecting the plotted points with a straight line.

Ask the students to work in pairs. Distribute graph paper to each pair. Provide different sets of data representing linear relationships, for example, distance versus time for a vehicle.

Instruct each student to plot the coordinates on the graph paper. Provide guidance on labelling the axes and selecting appropriate scales for the graph. Guide them to connect the plotted points with a straight line to represent the linear relationship.

Ask the students in each pair to review each other’s work.

Ask questions like, “How can we use the graph to guess what might happen next or understand the relationship better?”

Extension Idea

Ask: How is a graph showing simple interest earned over time different from a graph showing compound interest?

Say: Compound interest curves are non-linear due to exponential growth, showing increasing rates over time. Simple interest graphs are linear as they maintain a constant growth rate, resulting in straight lines.

Reading Linear Graphs

Learning Outcomes

Students will be able to read and interpret a linear graph.

Teaching Aids

Sheet with a linear graph

Activity

Begin by revising what linear graphs are and how they represent relationships between 2 variables. Divide the class into groups. Distribute a sheet with a linear graph on it to each group. Ensure that the graph has clear labels, titles and axes. For example, it could be a linear graph representing the perimeter and side length of squares.

Ask questions like, “What is the perimeter of a square with a side length of 4 units?” or “What is the length of the side of a square with a perimeter of 18 units?”

Instruct the students to analyse the linear graphs and answer the questions in their notebooks.

Discuss the students’ interpretations of the graphs. Summarise the key learnings from the activity and emphasise the importance of being able to read and interpret linear graphs.

Answers

1. Reading Double Line Graphs

Do It Together

1. The profit gained in the year 2012 in total was 17,000

2. The difference between the highest profit gained by Company A and the lowest profit gained by Company B is 9000.

3. The percentage decrease in the profit of Company A from 2011 to 2014 is 83.33%.

4. The percentage increase in the profit of Company B from 2011 to 2014 is 0%.

5. The company that earned more profit overall is Company B.

2. Drawing Double Line Graphs

Do It

Think and Tell

In this graph, all the points are collinear, i.e., lie on a line, so it is called a linear graph.

3. Cartesian Coordinate of a Point

4. Drawing a Linear Graph

Do It Together

5. Reading Linear Graphs

Do It Together

1. The x-axis represents time (in minutes).

2. The y-axis represents distance (in km)

3. The distance covered in 4 minutes is 2.5 km.

4. The distance of 5 km was covered in 8 minutes

Answers Solutions

Chapter 1

Let’s Warm-up

1. 1 4 < 3 5

4. 2 9 = 16 72 5. 14 5 = 2.8

Do It Yourself 1A

1. a. −2 + 2.5 = 0.5

b. 8 + (−24) = −16

c. 17 + 18.44 = 35.44

d. 4+6 4610 +== 7777

e. 57 + 911

LCM of 9 and 11 is 99. 555763 =; = 9991199

5563 8 = 9999

f. (12)(13) + 215

LCM of 2 and 15 is 30. 12)180)13)26) = and= 2301530 (((( 18026206 += 3030 30 = 103 15

2. a. 1212+=0 77

Additive inverse of − 1212+=0 77 is 1212+=0 77

b. 1313 =0 22

Additive inverse of 13 2 is 13 2

c. 55+=0 99

Additive inverse of 5 −9 is 5 9

3. a. 8+2 8210 == 11111111

b. 2738 938 11 == 13393939

c. 57252853 == 1215606060

d. = ((13)6) 815 195147 48 += 120120120

e. 9 7 − 1.3 = 9 7 − 13 10 = 90 − 91 70 = 1 70

f. 13 −4 − 2.1 = 13 4 − 21 10 = 130 − 84 40 = −214 40 = − 107 20

4. a. Closure property b. Associative property

c. Subtraction property of zero

5. a. 4 9 + 0 = 4 9;

Property used: Additive identity

b.   3 8 + 9 16   + 5 8 = 3 8 +   9 16 + 5 8 ;

Property used: Associative property

c. 3 5 − 0 = 3 5;

Property used: Subtraction property of zero

6. a. 2 5 + 3 10 = 3 10 + 2 5

Solving the LHS, LCM of 5 and 10 is 10.

2 5 + 3 10 = 4 + 3 10 = 7 10

Solving the RHS, LCM of 10 and 5 is 10.

3 10 + 2 5 = 3 + 4 10 = 7 10

As LHS = RHS, hence 2 5 + 3 10 = 3 10 + 2 5

b. 3 2 − 1 4 ≠ 1 4 − 3 2

Solving the LHS, LCM of 2 and 4 is 4.

3 2 − 1 4 = 6 −1 4 = 5 4

Solving the RHS, LCM of 2 and 4 is 4. 1 4 − 3 2 = 1 − 6 4 = − 5 4 As LHS ≠ RHS, hence 3 2 − 1 4 ≠ 1 4 − 3 2

c.   6 7 + 11 14 

Solving the LHS, LCM of 7 and 14 is 14.

6 7 + 11 14

+ 2 7 =

12 + 11 14

 + 2 7 = 23 14 + 2 7 = 23 + 4 14 = 27 14

Solving the RHS, LCM of 7 and 14 is 14. 6 7 +

= 6 7 + 15 14 = 12 + 15 14 = 27 14 As LHS = RHS, hence

6 7 + 11 14

7. 2127 += 56 x 2721 = 65 135126261 = = = 8.7 303030 x

8. 8.5 − x = −12.25 8.5 + 12.25 = x

x = 20.75 = 207583 = 1004 207583 = 1004 or 3 20 4

9. 5 6 + (−3) 8 6 12 + (−3) 24 2 8 8 (−16)

= 5 6 3 8 − 6 12 − 3 24 − 2 8 + 8 16

(LCM of 3, 8, 12, 16, 24 = 48)

= 40 − 18 − 24 − 6 − 12 + 24 48

= 4 48 = 1 12

10. The budget allocated for urban development = 1 5

The budget allocated for rural development = 11 200

The total budget allocated for both urban and rural development = 1 5 + 11 200

= 40 + 11 200 = 51 200

11. Length of the wood = 122 3 feet = 38 3 feet

Length of the wood cut from both ends = 5 8 + 5 8 = 10 8 feet

Length of the wood left = 38 3 − 10 8

= 38 × 8 − 10 × 3 24

= 304 − 30 24 = 274 24

= 137 12 = 11 5 12 feet

12. Apples = 15 kg; 4 Bananas = 2.6 kg

Total fruits purchased = 9 kg

Oranges purchased = 9 − 15 2.6 4 9 − 3.75 − 2.6 = 2.65 kg

Oranges purchased = 2.65 kg

Challenge

1. 32+6042 857 += 93636 ===150257 361818

7 18 should be subtracted to make 1 7 18 equal to the nearest whole number, such that − 77 1=1 1818 11 18 should be added to make 1 7 18 equal to the nearest whole number, such that − 711 1+=2 1818

2. Total juice prepared = 25.5 litres Juice consumed = 

8+8+2223 558 litres

= 424219 ++ 558 = 672+95 8419 += 5840 672+95

= 767 =19.175 litres 40 Juice remaining = 25.5 − 19.175 = 6.325 L

Do It Yourself 1B

1. a. 21836 ×= 5735 b. 163860832 ×== 1971337

c. 5525 ×= 818 d. 132532513 ×== 15101506

e. 124250436 ×= = 710705 f. 2332736368 ×== 5105025

2. a. 14514342 ÷= ×= 535525

b. 1241225300 ÷=×== 15 5255420 1241225300 ÷=×== 15 5255420

c. 2872830840÷=×==8 1530157105 2872830840÷=×==8 1530157105

d. 31731131131 ÷=×=×= 14114798198 31731131131 ÷=×=×= 14114798198

e. 1336131013065 ÷=×== 6101036216108

f. 364836103603 ÷=×== 101010484804

3. a. 317217 ××== 28914448

b. 8714 ×÷ 1135 875 =×× 113 14 28020 == 46233

c. −9 −7 × −6 5 ÷ −12 5 = −9 −7 × −6 5 ÷ −5 12 = 9 14

d. 54 4.5 ÷ × 79 45741260 =× ×= 1059450 144 = =2 55

e. 251615 ×÷ 4510 2516104000 =××= 4515300 ==13401 33

f. 34834344 ÷×=×× 10391089 40817 == 72030

4. a. LHS =  

537 ×× 4812 52110535 =×== 496384128

RHS = 

537 ×× 4812 15710535 =×== 3212384128

Hence, LHS = RHS

b. 537 LHS= ÷= 4812

1037 =÷ 8812

71284 =×= 8756 5737 RHS= ÷ ÷ 412812 512312 =×× 4787 6036 = 2856 1203684 == 565656

Hence, LHS = RHS

c. LHS = 537103777 = ==0 48888888 

RHS = 5375410414 ==+= 48848888

Hence, LHS ≠ RHS

5. a. 5 9 × 1 = 5 9; Property used: Multiplicative identity

b. 6 11 × 11 6 = 1; Property used: Multiplicative inverse

c. 4 7 ÷ 1 = 4 7; Property used: Division by 1

6. a. 5 2 ×   4 5 + 6 10   = 5 2 × 4 5 + 5 2 × 6 10

= 2 + 3 2 = 4 + 3 2 = 7 2

b. 7 8 ×   24 35 + 16 21   = 7 8 × 24 35 + 7 8 × 16 21

= 3 5 + 2 3

= 9 + 10 15 = 19 15

c. 8 11 ×   33 40 + 44 16   = 8 11 × 33 40 + 8 11 × 44 16

= 3 5 + 2 = 3 + 10 5 = 13 5

7. a. 5 8 ÷ 2 8 × 9 18 + 30 15 17 21 ÷ 34 7 × 24 8

= 5 8 × 8 2 × 9 18 + 30 15 17 21 × 7 34 × 24 8

= 5 2 × 1 2 + 2 1 1 2

= 5 4 + 2 1 2

= 5 + 8 − 2 4 (LCM of 1, 2 and 4 = 4) = 11 4

Multiplicative inverse = 4 11

b. 4 7 × 11 8 ÷ 33 28 + 7 4 ÷ 9 8 × 3 14

= 2 3 + 1 3 = 3 3 = 1 (LCM of 1 and 3 = 3)

Multiplicative inverse of = 1

8. Length of one rational number be x

According to the question, − 1728 ×= 63 x

x= 28 × 6 3 × (−17)

x= 168 −51 = 56 −17

x= − 56 17

Multiplicative inverse of − 56 17 is − 17 56

9. Length of an Independence Hall = 163 5

Area of the base of an Independence Hall = side × side

= 163 5 × 163 5

= 26569 25 or 1062 19 25

10. Length of each piece = 357 ÷24 8 3571 =× 824 ==35711955m=1m 1926464

11. Let the decimal number be x

According to the question, × 3 5 x = 3 2

x = ÷=×= 33355 25232

Converting 5 2 to decimal, we get 2.5

Hence, 3 5 should be multiplied by 2.5 to get 3 2 as product.

12. Let the fraction be x

According to the question, 4.2 ÷ x = 3

4.2 × = 1 x 3

⇒ == 4.2 1.4 3 x

Converting 1.4 to fraction form we get, = 147 105

c. 98271643 =+= 23666

98271643 =+= 23666 . The multiplicative inverse is 6 43

d. − 24 13 − (− 29) 39 = 72 + 29 39 = − 43 39 . The multiplicative inverse is − 39 43

e. 3515 ×= 7856 . The multiplicative inverse is 56 15

Hence, 4.2 should be divided by 7 5 to get 3 as the quotient.

13. Let the total pages be x.

Fraction read = 5 8

Fraction left = 3 8

According to the question, 3 8 of x = 120

Total pages in the book = x = 120 × 8 3 = 320

14. The cost of a zoo ticket for adults = ₹62.50

The cost of a zoo ticket for one kid = 2 5 × 62.50 = ₹25

The cost of a zoo ticket for five kids = 25 × 5 = ₹125

Challenge

1. The product of 2 15 2151 and 585183 −=−×=−

The reciprocal of the product of 215and3 58 −=−

According to the question, 1 3 3 −÷−

= − 3 x − 3 = 9

≠ 1

f. 1221252×==9 7428 . The multiplicative inverse is 1 9

g. 17 6 ÷ 51 24 = 17 6 � 24 51 = 4 3. The multiplicative inverse is 3 4

h. 13251326338 ÷= ×= 526525125 13251326338 ÷= ×= 526525125 . The multiplicative inverse is 125 –. 338

3. The base of a triangle = 156 5 m

The height of a triangle = 191 2 m

Area of a triangle = 1 2 × base × height = 1 2 × 156 5 × 191 2 = 14898 10 m = 1489.8 m

4. a. 318 LHS= ÷÷ 6510  

Hence, the quotient that we get on dividing the reciprocal of the product of 215 and 58 by the product is not the multiplicative identity.

Chapter Checkup

1. a. The sum of a rational number and its additive inverse is zero

b. The reciprocal of a negative rational number is negative.

c. The rational number 0 does not have any reciprocal.

d. Adding zero to a rational number gives the rational number itself.

2. a. 27123523 += +=; 56303030

27123523 += +=; 56303030 . The multiplicative inverse is 30 23

b. 1914571443 +== 1751515151 1914571443 +== 1751515151 . The multiplicative inverse is 51 43 .

311031 =÷×= ÷ 65864 3 =×4= 2 6 RHS=÷÷318 6510

11051025 ×5×= ×= 28288

Hence, LHS ≠ RHS

b. LHS = 318 ×+ 6510

= 11051025 ×5×= ×= 28288

= 114151 ×+= ×= 255252

d. LHS=++318 6510

1568924 =++= + 3030103030 151 == 302

RHS=++318 6510

328 =++ 61010 151015301 =+= += 301030302

Hence, LHS = RHS …

5. Let the decimal number be x.

According to the question, 9 17 + x = − 8 5 x = − 8 5 − 9 17 = −136 − 45 85 = −181 45 or −211 85

6. Let the decimal number be x

According

6969

45484548 5454

393 5454

354 5493

31 9331

8. a.

1328411143 852147222

13 42 + == 167 44

9. Length of one roll = 2 15m 3 m = 47 m 3 m

Length of 5rolls=5114721473291 ×= ×= m=82m 4434344 5rolls=5114721473291 ×= ×= m=82m 4434344 m

10. Total rice = 1209 52kg=kg 44

Number of families = 11 209 ÷11 4 2091 =× 411 2093 = =4 kgor4.75kg

Each famil= 44 ge 4 yts

11. Cricket practised in last week = 12 5 hours 6 hours

Cricket practised this week = 2 1 hours 8 hours Total hours =12+251 68 7717 =+ 68 30851359 =+= 242424 23 =14 hours 24

12. India’s GDP = 677 200 trillion dollars

Japan’s GDP = 21 5 trillion dollars

Japan’s GDP is greater than India’s GDP by = 21 5 − 677 200 = 840 − 677 200 = 163 200 trillion dollars

13. Pizza eaten by Richa = 1=37 44

Pizza eaten by Alex = 75211031 +=+= 46121212 = 75211031 +=+= 46121212

Pizza eaten by Prince = 3721 ×= 248

Total pizza eaten 73121 =++ 4128 426263 =++ 242424 ==616723 2424

14. Total money Shilpa had = ₹750

Money spent on book = 2 5 × 750 = ₹300

Money spent on stationary () 4 = ×750–300 9 4 =×450=200 9 () 4 = ×750–300 9 4 =×450=200 9 = ₹200

Money left = ₹750 − (₹300 + ₹200) = ₹750 − ₹500 = ₹250

15. Answersmayvary.Sampleanswer:

Nitin is filling small containers with water. He has 7 8 of a gallon of water and wants to fill each container with 1 4 of a gallon. How many containers can Nitin fill?

Challenge

1. Let the number be x

According to the question, 53–=80 35 xx

25–9 =80 15 xx

16 80; 15 x = x = 75

The number is 75.

2. Assertion (A): The sum of − 3 5 and − 4 4 will be positive.

Reason (R): The product of two negative rational numbers is positive.

It is true, that the product of two negative rational numbers is positive.

For example: −×−= 3412 5735

So, option b is correct. A is false, but R is true.

Case Study

1. Weight of one wheat bag = 25.5 kg

Number of wheat bags bought = 10

Total weight of wheat bought = 25.5 × 10 = 255 kg

So, option b is correct.

2. Weight of one rice bag = 12 2 3 kg

Number of rice bags bought = 24

Total weight of rice bought = 12 2 3 × 24 = 38 3 × 24 = 304 kg

Total weight of wheat bought = 25.5 × 10 = 255 kg

Difference in wheat and rice bought = 304 − 255 = 49 kg

So, option c is correct.

3. Weight of one corn bag = 37 2 kg

Number of corn bags bought = 4

Total weight of corn bought = 37 2 × 4 = 74 kg

Weight of corn purchased by one family = 1 2 kg

Number of families who purchased 74 kg corn

= 74 ÷ 1 2 = 74 × 2 = 148 families

So, option d is correct.

4. The cost price of one corn bag = ₹1850

Cost price of four corn bags = 1850 × 4 = ₹7400

Weight of one corn bag = 37 2 kg

Weight of four corn bags = 37 2 × 4 = 74 kg

The selling price of one corn bag = ₹135

The selling price of four corn bags = 135 × 74 = ₹9990

Profit earned = Selling price of four corn bags Cost price of four bags = ₹9990 ₹7400 = ₹2590

5. Weight of one rice bag = 12 2 3 kg = 38 3 kg

Weight of 15 rice bags = 15 × 38 3 kg = 190 kg

Donation of rice for one kg = 100 g

= 100 1000 = 0.1 kg

Donation of rice for 190 kg = 190 × 0.1 = 19 kg

Chapter 2

Let’s Warm-up

1. x − 4x

False, as x − 4x is not an equation.

2. 3x = 45

True, as 3x = 45 is an equation in one variable x

3. 4x – 3x + 9

False, as 4x − 3x + 9 is not an equation.

4. 5x + 2 = 3x − 5

True, as 5x + 2 = 3x − 5 is an equation in one variable x

5. 9x + 2y = 2x − 1

False, as 9x + 2y = 2x − 1 is an equation in two variables x and y

Do It Yourself 2A

1. a. 3x + 1 = 10 3x + 1 − 1 = 10 − 1 3x = 9 3x 3 = 9 3 x = 3

b. 2x + 3 = 5

2x + 3 − 3 = 5 − 3

2x = 2

2x 2 = 2 2

x = 1

c. 3x + 1 = 7 + x

3x+ 1 x= 7 +xx

2x+ 1 = 7

2x + 1 − 1 = 7 − 1

2x = 6

2x 2 = 6 2

x = 3

2. 3x + 5 = 9

3. a. 3x − 5 = 2x + 97

3x − 2x − 5 = 97

x − 5 = 97 ⇒ x = 97 + 5

x = 102

b. 5(x − 7) − 2x = 13 − x

5x − 35 − 2x = 13 − x

3x − 35 = 13 − x

3x + x − 35 = 13

4x − 35 = 13

4x = 13 + 35 ⇒ 4x = 48 ⇒ x = 12

c. 15x − 2(18 + x) = 29

15x − 36 − 2x = 29

13x − 36 = 29

13x = 29 + 36 ⇒ 13x = 65 ⇒ x = 5

d. x + 2(3 + x) = 5(x − 6)

x + 6 + 2x = 5x − 30

3x + 6 = 5x − 30

6 = 2x − 30 ⇒ 6 + 30 = 2x

36 = 2x ⇒ 18 = x

e. 4x + 7 = 1.5(x + 12)

4x + 7 = 1.5x + 18

2.5x = 11 ⇒ x = 4.4

f. 1.3(x − 3) + 2.6 = x

1.3x − 3.9 + 2.6 = x

1.3x − 1.3 = x ⇒ 1.3–=1.3 xx

⇒ 0.3x = 1.3 ⇒ x = 13 3 (or approximately 4.33)

4. Let the smaller even integer be x, and the larger even integer be x + 2.

According to the question:

22–( + 2) = 4 511 xx

22–10(+2) =4 55 xx

(22–10(+2))=220 xx

22– 10–20=220xx

12–20=220 x

12x = 240

240 = 12 x

x = 20

x + 2 = 22

So, the smaller even integer is x = 20, and the consecutive even integer is x + 2 = 22.

5. Let the smaller number be x and the larger number be 3x.

Given: If 15 is added to both numbers, we get x + 15 and 3x + 15.

According to the question:

3x + 15 = 2(x + 15)

⇒ 3x + 15 = 2x + 30

⇒ x = 15

Therefore, the smaller number is 15, and the larger number is 3 × 15 = 45.

6. Let the breadth of the rectangular field be x and the length be (2x − 7).

The perimeter of the rectangle is given: 2[(2x − 7) + x] = 196

= 2(2x − 7) + 2x = 196

= 4x − 14 + 2x = 196

= 6x = 210

= x = 35

The breadth is 35 m, and the length is 2 × 35 − 7 = 63 m.

The area is 35 × 63 = 2205 square metres.

7. a. 18 cm 15 cm

Perimeter = 54 cm 12 cm

x + 18 + 12 + 15 = 54

x + 45 = 54

x = 9 cm

Perimeter = 56 cm x 4 cm

x + 3 3x x − 1

3x + 5

x + 3 + x + 4 + 3x + x − 1 + 3x + 5 = 56 cm

9x + 11 = 56 cm

9x = 45 cm

x = 5 cm

c. 2x + 3

2(x + 1) x + 4 x + 3

Perimeter = 35 cm x x

2x + 3 + x + x + 4 + x + 3 + x + 2(x + 1) = 35 cm

2x + 3 + x + x + 4 + x + 3 + x + 2x + 2 = 35 cm

8x + 12 = 35 cm

8x = 23 cm

x = 23 8 cm or x = 2.875 cm

8. Given that the rectangles are identical. So, the measures of the sides will also be equal.

2x + 3 = x + 5 and 4y − 1 = 3y + 4

Using the transposition method,

2x − x = 5 − 3 and 4y − 3y = 4 + 1

x = 2 and y = 5

Length of the rectangles = 2x + 3

= 2 × 2 + 3 = 7 units

Breadth of the rectangles = 4 × 5 − 1 = 19 units. Area of each rectangle = 7 × 19 = 133 units.

Combined area = 133 + 133 = 266 square units.

9. Answermayvary.Sampleanswer:

a. x + 2 = 7 b. 2x + 3 = x + 8 c. 3x 5 = x + 5

Challenge

1. The numbers greater than 20 but less than 30 are 21, 22, 23, 24, 25, 26, 27, 28 and 29.

The number that satisfies the condition that the sum of the digits is the same as their product is 22 as 2 + 2 = 4 and 2 × 2 = 4.

Hence, 22 is the answer.

Do It Yourself 2B

1. a. 5 =3 2–1 x x for x = 2:

5(2)10=3 2(2)–13 ≠

No, the solution given is not correct.

b. 75 = +1–1xx

For x = 6: 75 = 6 +16–1

1 = 1

Yes, the solution given is correct.

c. 7+15+6 = 34 xx

For x = 3; 22 3 21 4 ≠ 21 4

No, the solution given is not correct.

2. a. 3–8 =63–8=30 5 x xx x ⇒

−8 = 27x ⇒ x = 8 –27

b. 57 = +2 xx

5(x + 2) = 7x ⇒ 5x + 10 = 7x 10 = 2x ⇒ x = 5

c. 0.4+5 8 x = 2.5+4 15 x

15(0.4x + 5) = 8(2.5x + 4)

6x + 75 = 20x + 32

75 − 32 = 20x − 6x

43 = 14x ⇒ x = 43 14

d. 5(2–)–4(1+) 2–xx x = 5 8

10–5–4–4 2–xx x = ⇒ 6–9 5 82–x x

⇒ 6–9 5 82–x x = 5 8

8 (6 − 9x) = 5 (2 − x) 48 − 72x = 10 − 5x

38 = 67x ⇒ x = 38 67

e. 69 = 3+15+3 xx

6(5x + 3) = 9(3x + 1)

30x + 18 = 27x + 9

3x = −9 x = −3

f. 4+12–13–7 +–=6 325 xxx

10(4+1)+15(2–1)–6(3–7) =6 30 xxx

40x + 10 + 30x − 15 − 18x + 42 = 180

70x − 18x + 37 = 180

52x = 143

x = 143 52

3. a.

3–14+6 + 23 xx = 43 6

3(3–1)+2(4+6) 43 = 66 xx

9x − 3 + 8x + 12 = 43

17x + 9 = 43

17x = 34

x = 2

Putting x = 2; 23 + 2 y = 2

3 1 + y = 2

3 =1 y y = 3

b. 4–53+5 = 1110 xx

10(4x − 5) = 11(3x + 5)

40x − 50 = 33x + 55

7x = 105

x = 15

Putting x = 15; 152+=11

3 y

5 + 2 =11 y 2 =6 y y = 1 3

4. Let the denominator of the fraction be x. Then, the numerator is x − 3.

The fraction is –3 x x .

Given: When both the numerator and denominator are decreased by 2, the new fraction becomes –5 –2 x x

According to the question, the simplest form of this new fraction is 3 4

The equation becomes:

–5

–2 x x = 3 4

Solving for x:

4(x − 5) = 3(x − 2)

4x − 20 = 3x − 6

x = 14

So, the fraction is –3 x x = 14–3 14 = 11 . 14

5. Let the denominator be x, then the original fraction will be = x+ 6 x

If both the numerator and denominator are decreased by 3, the new fraction is (x+ 6) − 3 x − 3 = 5 3

By simplifying the equation, we get: x+ 3 x − 3 = 5 3

Cross − multiplying: 3(x + 3) = 5(x − 3).

Solving for x: 3x + 9 = 5x − 15

x = 12

The original fraction is x+ 6 x = 12 + 6 12 = 18 12

6. Let o be the calories in an orange. Let p be the calories in a pear.

Let w be the calories in a watermelon. An orange has 10 less calories than a pear and 17 more calories than a watermelon.

o = p − 10 and o = w + 17

⇒ p = o + 10 and w = o − 17

3 oranges have 6 less calories than 1 pear and 3 watermelons

3o = p + 3w − 6

From the above equation, put the value of p and w

3o = o + 10 + 3(o − 17) − 6

⇒ 3o = o + 10 + 3o − 51 − 6

⇒ 3o = 4o − 47

⇒ 3o − 4o = − 47

⇒ o = −47 or o = 47

Thus, an orange has 47 calories.

Challenge

1. Statement 1: The third digit of the postal code is one more than the second digit.

Statement 2: All the digits of the postal code are prime numbers.

Statement 3: The last digit of the postal code is twice the first digit.

Statement 4: The difference of the fourth and sixth digit is the fifth digit of the postal code.

a. –1 1 = 7–273 a a

3(a − 1) = 7a − 27

3a − 3 = 7a − 27

24 = 4a

a = 6

b. 2+67–34 = 43 bb

3(2b + 6) = 4(7b − 34)

6b + 18 = 28b − 136

154 = 22b

b = 7

c. 11–443+17 = 54 cc

4(11c − 44) = 5(3c + 17)

44c − 176 = 15c + 85

29c = 261

c = 9

d. 9–7 29 = 7+3)1 (3 d d

31(9d − 7) = 29(7d + 3)

279d − 217 = 203d + 87

76d = 304

d = 4

Given, the postal code is 2abc5d. Substituting the values of a, b, c, and d, we get, the postal code as 267954. Hence, Statements 1, 3 and 4 are correct.

Do It Yourself 2C

1. a. 3x + 5 = 1 2 x − 13

b. 4y + 11 = 2y + 56

c. Let the three consecutive even numbers be n, n + 2, and n + 4.

Equation: n + (n + 2) + (n + 4) = 144

d. Let the original speed of the train be s and the time taken for the initial trip be t

Equation for the first trip: st = 240

On the return trip, the speed is increased to s + 20, and the time taken is t − 2.

Equation for the return trip: (s + 20)(t − 2) = 240

2. Let the four consecutive odd numbers be x, x + 2, x + 4 and x + 6.

x + (x + 2) + (x + 4) + (x + 6) = 352

4x + 12 = 352

4x = 340

x = 85

So, the four consecutive odd numbers are 85, 87, 89 and 91.

3. Let the three consecutive multiples of 11 be 11x, 11x + 11 and 11x + 22.

11x + (11x + 11) + (11x + 22) = 429

33x + 33 = 429

33x = 396

x = 12

So, the three consecutive multiples of 11 are 132, 143 and 154.

4. Let the angles be 3x, 4x and 5x The sum of angles in a triangle is 180°

3x + 4x + 5x = 180°

12x = 180°

x = 15°

Smallest angle = 3x = 3 times 15 = 45°

Largest angle = 5x = 5 times 15 = 75°

The difference between the largest and smallest angles is 75° − 45° = 30°

5. Let the daughter’s age be x. Puneet’s age is then 3x

The sum of their ages is given by:

x + 3x = 52

4x = 52

x = 13

Puneet’s age is 3 times 13 = 39 years.

The difference in their ages is 39 − 13 = 26 years.

6. Let the runs scored by Rohit be x. Virat’s runs are then (2x + 20). The total runs are given as:

x + (2x + 20) = 300 − 4

3x + 20 = 296

3x = 276

x = 92

So, Rohit’s score is 92, and Virat’s score is 2 × 92 + 20 = 204.

7. Let x be the total estate. Animesh left the property to be divided among his daughter, son and for donation as follows:

Daughter: 1 2 x

Son: 1 3

Given that the donation was ₹5,00,000:

So, x = ₹30,00,000 Daughter’s share =

Son’s share = 1 3

= ₹15,00,000

So, Animesh gave ₹15,00,000 to his daughter and ₹10,00,000 to his son.

8. Let the original number be x

The new number, after a 10% decrease is 0.9x, and when increased by 25%, it becomes 1.25 times 0.9x.

1.25 × 0.9x = 81

1.125x = 81 x = 81 1.125

x = 72

So, the original number is 72.

9. Let one part be x and the other part be 6000 − x

The equation is:

0.15x = 0.35(6000 − x)

0.15x = 2100 − 0.35x 0.5x = 2100

x = 4200

So, one part is 4200, and the other part is 6000 − 4200 = 1800.

10. Let the smaller angle be x and the bigger angle be 2x

Given: When 15 is added to the smaller angle and 15 is subtracted from the bigger angle, the new ratio becomes 1:1.

According to the question:

+15

2–15 x x = 1 1

x + 15 = 2x − 15

x = 30

So, the smaller angle is 30°, and the bigger angle is 2 × 30 = 60°.

11. Let Naveen’s age be 2x and his father’s age be 5x After 10 years, their ages will be (2x + 10) and (5x + 10).

2 + 101 = 5 + 102 x x

4x + 20 = 5x + 10

x = 10

Father’s present age = 5 × 10 = 50 years, Naveen’s present age = 2 × 10 = 20 years.

Father’s age at the time of Naveen’s birth = (50 − 20) years = 30 years.

12. Total marks of Prelims in IBPS PO exam = 100

Marks for correct answer = x

Marks for incorrect answer = 1 4 x

Marks for not attempted questions = 0

Number of questions attempted by Rashmi = 78

The number of questions attempted by Rashmi and that are correct = 56

Number of questions incorrect or not attempted by Rashmi = 78 – 56 = 22

Score = Marks from correct answers − Deductions

50.5 = 56x – 22

50.5 = 224x – 22x 4

202 = 202 x

202

202 = x x = 1

Hence, the marks for correct answer = 1.

13. Let the two digits be x and y, where x is the digit in the tens place, and y is the digit in the one’s place.

The number is 10x + y

According to the question, one’s digit is 0.4 times the ten’s digit:

y = 0.4x

Also,

(10y + x) + (10x + y) = 77

11x + 11y = 77

x + y = 7 …. (1)

Substituting the value of y in (1)

x + 0.4x = 7

1.4x = 7

x = 5

And, y = 0.4 × 5 = 2

So, the original number is 10x + y = 10(5) + 2 = 52.

14. Let the speed of the boat in still water be B, and the speed of the current be C.

Let the distance be x km.

The speed upstream is BC, and the speed downstream is B + C

Given that the speed of the flow of the river is 3 km per hour. Time downstream is 5 h and time upstream is 3 h.

Since Distance = Speed × Time

x = (B − C ) × 3 ……. (1)

x = (B + C ) × 5 ……... (2)

Equating (1) and (2) and substituting the values:

(B − 3) × 3 = (B + 3) × 5

3B − 9 = 5B + 15

2B = 24

B = 12 km/h

Hence, the speed of the boat in still water is 12 km/h.

Challenge

1. Total number of notes = 92

Let the number of ₹5 notes be x; Number of ₹10 notes = (92 – x)

Given that the sum made by the denominations = ₹870

So, 5x + 10(92 – x) = 870

5x + 920 – 10x = 870

⇒ 5x = 50

⇒ x = 10

Hence, the number of ₹5 notes = 10 and the number of ₹10 notes = (92 – 10) = 82.

2. Given that the distance between two places A and B is 300 km. Two cars start simultaneously from A and B in opposite directions.

The distance between them after 4 hours is 60 km.

Distance travelled by both of them = 300 − 60 = 240 km/h.

The speed of one car = s km/h

Speed of the second car = (s + 10) km/h

We know, Distance = Speed × Time

According to the question,

Distance travelled by car A in 4 hours = 4s km/h

Distance travelled by car B in 4 hours = 4(s + 10) km/h

Distance travelled by both of them = 4s + 4(s + 10) = 240

= 4s + 4s + 40 = 240

8s = 200

s = 25 km/h

and s + 10 = 25 + 10 = 35 km/h

The speed of one car is 25 km/h and that of the second car is 35 km/h.

Chapter Checkup

1. a. 2x + 1 = 5

2x = 5 − 1

2x = 4

x = 4 2 = 2

b. 2x + 2 = 6 + x

2x − x = 6 − 2

x = 4

c. 3x + 4 = 10 + x

3x − x = 10 − 4

2x = 6

x = 6 2 = 3

2. 2x + 5 = 11

3. a. 5(x + 3) = 3(1.5x + 18)

5x + 15 = 4.5x + 54

0.5x + 15 = 54

0.5x = 39 ⇒ x = 78

b. 6(2x + 11) = 8(2x − 1)

12x + 66 = 16x − 8

66 = 4x − 8

74 = 4x ⇒ x = 18.5

c. 8x − 7 − 3x = 6x − 2x − 3

5x − 7 = 4x − 3

x − 7 = −3 ⇒ x = 4

d. 10x − 5 − 7x = 5x + 15

3x − 5 = 5x + 15

−2x − 5 = 15

−2x = 20 ⇒ x = −10

e. 5(x − 1) = 2(x + 8)

5x − 5 = 2x + 16

3x − 5 = 16

3x = 21 ⇒ x = 7

f. 4x − 3 = (3x + 1) + (5x − 4)

4x − 3 = 3x + 1 + 5x − 4

4x − 3 = 8x − 3

−3 = 4x − 3 ⇒ x = 0

4. a. 39 = 2(+2)10 x x

30x = 18x + 36

12x = 36 ⇒ x = 3

b. 4(+2)+3 27 = 2+3(–1)17 x xx

4+8+34+11 27 27 ==

(2+3–3)175–317 xx xxx ⇒

17 (4x + 11) = 27 (5x − 3)

68x + 187 = 135x − 81

268 = 67x ⇒ x = 4

c. 5+11)4(+6) = 2326 ( xx

26 (5x + 11) = 92 (x + 6)

130x + 286 = 92x + 552

38x = 266 ⇒ x = 7

d. 3+1 5 x = 4(+3)–4 8 x

8 (3x + 1) = 5 (4(x + 3) − 4)

24x + 8 = 20x + 60 − 20

4x = 32 ⇒ x = 8

e. 7(x − 3) + 2 = 11(8–)–3 7 x

7x − 21 + 2 = 88–11–3 7 x

7x − 19 = 85–11 7 x

49x − 133 = 85 − 11x

60x = 85 + 133

60x = 218

x = 218 60 ⇒ x = 109 30

f. 11(2+1) 11 x = 6x − 4 − 3x

22+11 11 x = 3x − 4

22x + 11 = 11(3x − 4)

22x + 11 = 33x − 44

11 + 44 = 11 x

55 = 11x ⇒ x = 5

5. a. F = 9 +32 5 C

When F = −40

−40 = 9 + 32 5 C

−40 − 32 = 9 5 C ⇒⇒ −72 = 9 5 C

⇒ 5 –72×= 9 C ⇒ −40 = C

b. 9+44(2–1) = 107 xx

7(9x + 4) = 10 × 4(2x − 1)

63x + 28 = 80x − 40

68 = 17x x = 4

Then + 3 = 8 xy xy

Putting x = 4 ⇒

4+ 3 = 48 y y ⇒ 8(4 + y) = 12y

32 + 8y = 12y ⇒ 32 = 4y ⇒ y = 8

c. 2+1–23+2 –= 10615 xxx

3(2+1)–5(–2)3+2 = 30 15 xxx

6+3–5+103+2 = 3015 xxx

+133+2 = 3015 xx

15 (x + 13) = 30 (3 + 2x)

15x + 195 = 90 + 60x

105 = 45x ⇒ x = 105 45 ⇒⇒ x = 7 3

Then, 25+=4 xm

2 7 3 + 5 m = 4 ⇒ 6 7 + 5 m = 4

5 m = 4 − 6 7 ⇒ 5 m = 22 7 ⇒ m = 35 22

6. a.

2x + 3 = 7 for x = 2

2 × 2 + 3 = 7

4 + 3 = 7

7 = 7

b. 3 (x − 1) + 4x = 46 for x = 7

3(7−1) + 4 × 7 = 46

3 × 6 + 28 = 46

18 + 28 = 46

46 = 46

c. 2x + 3(x + 1)

15x − 2(2x + 4) = 18 25 for x = 3

2 × 3 + 3 (3 + 1)

15 × 3 − 2(2 × 3 + 4) = 18 25

6 + 3(4)

45 − 2(10) = 18 25

6 + 12

45 − 20 = 18 25

18 25 = 18 25

d. 3x + 4(x + 3) 20 = x + 2(3x − 7) 7 for x = 4

3 × 4 + 4 (4 + 3) 20 = 4 + 2 (3 × 4 − 7) 7

12 + 28 20 = 4 + 10 7

40 20 = 14 7

2 = 2

7. Let the number be represented by x.

The expression can be written as

Multiplying both sides by 3 to clear the fraction: x − 1 2 = 1 5

Adding 1 2 to both sides: x = 1 5 + 1 2 = 7 10

8. Let Shagun’s present age be x. Her mother’s age = 5x.

After twenty-five years, Shagun’s age will be x + 25 and her mother’s age will be 5x + 25.

According to question,

x + 25 = 1 2 (5x + 25) − 4

x + 25 + 4 = 1 2 (5x + 25)

x + 29 = 1 2 (5x + 25)

2x + 58 = 5x + 25

3x = 33

x = 11

Shagun’s present age is 11 years, and her mother’s present age is 5 × 11 = 55 years.

The difference in their present ages is 55 − 11 = 44 years.

9. Let the breadth of the pool be x. So, the length is x + 4.

The perimeter of the rectangle is given by 2 (L + B)

Given: Perimeter = 170 m

So, 2 × (x + (x + 4)) = 170

Therefore, 4x + 8 = 170

4x = 162

x = 40.5

The length of the pool is x + 4 = 40.5 + 4 = 44.5 m, and the breadth is x = 40.5 m

Area = 44.5 × 40.5 = 1802.25 sq. m.

10. Let the length of the plot be 15x and the breadth be 7x

Given: The cost of fencing is ₹150 per metre.

The perimeter of the rectangle is 2 × (L + B)

So, 2 × (15x + 7x) × 150 = 1,32,000

44x × 150 = 1,32,000

44x = 880

x = 20

Length = 15 × 20 = 300 m, breadth = 7 × 20 = 140 m.

Therefore, the dimensions of the plot are 300 m × 140 m.

11. Let the present ages of Kushagra and Kush be 5x and 6x, respectively.

Given: After five years, their ages will be (5x + 5) and (6x + 5) years respectively, and the sum = 43 years.

Hence, (5x + 5) + (6x + 5) = 43

Combining like terms, 11x + 10 = 43 hence, x = 3.

Present age of Kushagra = 5x = 5 × 3 = 15 years.

Present age of Kush = 6x = 6 × 3 = 18 years.

The difference between their ages is 18 − 15 = 3 years.

12. Let the angles be 7x, 17x, 19x and 29x in degrees.

The sum of angles in a quadrilateral is 360°.

7x + 17x + 19x + 29x = 360°

72x = 360°

x = 5

Therefore, the angles are 35°, 85°, 95° and 145°.

13. Let G be the total number of goats.

Given:

Grazing goats: 1 2 × G

Playing goats: 3 4 × 1 2 × G

Drinking goats: 15

G − 1 2 × G − 3 4 × 1 2 × G = 15

G − 1 2 × G − 3 8 × G = 15

G − 4 8 × G − 3 8 × G = 15

G − 7 8 × G = 15

1 8 G = 15

G = 120

Therefore, there were 120 goats in total.

14. Let x be the total money.

The share of A is 2 5 of the total money, so 2 5 × x.

The share of B is 2 3 of the remaining money, so 2 3 × 3 5 x C’s share is given as ₹600.

2 5 × x + 2 3 × 3 5 x + 600 = x

2 5 x + 2 5 x + 600 = x

4 5 x + 600 = x

1 5 x = 600 x = 3000

Therefore, the total amount x is ₹3000.

15. Let the speed of the boat be 7x and the speed of the stream be 2x

Hence, speed downstream = 7x + 2x = 9x speed upstream = 7x − 2x = 5x

Given: The boat takes 6 hours more travelling upstream than downstream.

Let t hours be the time taken downstream.

Since, Distance = Speed × Time

Hence, Distance downstream = 9x × (t) …… (1)

Distance upstream = 5x × (t + 6) …… (2)

Equating (1) and (2):

9x × (t) = 5x × (t + 6)

9t = 5t + 30

4t = 30

t = 7.5 h

The total time for the entire journey is t + t + 6 = 7.5 + 7.5 + 6 = 21 hours

16. Answersmayvary.Sampleanswer:

Sides of triangle: 2x, x, x + 4

Side of square: x + 1

Challenge

1. Let x be the time taken by Raj to write 100 pages. Raju takes twice the time taken by Raj to write the remaining pages, so 2x.

Vivek takes 2 hours more than half the time taken by Raju,

+ 2.

The total time taken to write the book is

and

x = 47 6 = 7 5 6 hours

Time taken by Vivek to write the remaining pages

= (x + 2) hours

= 7 5 6 + 2

= 9 5 6 hours

Thus, Vivek takes 9 5 6 hours to write the remaining pages.

2. Assertion (A): The difference of two numbers is 25. The larger number is x. The smaller number is x – 25.

Reason (R): Numbers which follow each other in order, without gaps, from the smallest to largest like 12, 13, 14 and 15 are consecutive numbers.

If the difference of the two numbers is 25. The larger number can be given as x and the smaller number can be given as x − 25. Hence, the assertion is True.

Numbers that follow each other in order, without gaps, from the smallest to largest are called consecutive numbers. Hence, this statement is also true but not the correct explanation of the first statement.

Both A and R are true, but R is not the correct explanation of A. Thus, option b is correct.

Case Study

1. As the customer took the cab service and travelled a distance of 12 km in city A, the equation can be given as:

12 × ₹10 + ₹d = x 10 × 12 + d = x

Thus, option c is correct.

2. Neha took the cab service and travelled a distance of 11 km in city A, and paid ₹150. The fixed charge would be:

11 × ₹10 + ₹d = 150

₹110 + ₹d = ₹150

d = 150 – 110 = ₹40

The fixed charge will be ₹40. Thus, option b is correct.

3. As Suhani is travelling in city A, the per km charge can be given as:

a × 5 + ₹15 = ₹45

5a = ₹45 − ₹15

5a = ₹30

a = ₹6/km

Thus, option c is correct.

4. The equation for the amount paid by Megha can be given as:

b × 6 + ₹25 = ₹76

6b = ₹76 − ₹25

6b = ₹51

b = 51 6 = ₹8.5/km

The equation for the amount paid by Rahul can be given as:

c × 6 + ₹15 = ₹57

6c = ₹57 − ₹15

6b = ₹42

c = 42 6 = ₹7/km

The difference in per km charge of bike and auto = ₹8.5 − ₹7 = ₹1.5/km

5. We already find the fixed charge for cab A in Q2 as ₹40. For travelling a distance of 7km in city B, Shalini will pay:

7 × ₹11 + ₹40 = ₹77 + ₹40 = ₹117

Hence, the given statement is false.

6. Answersmayvary.

Chapter 3

Let’s Warm-up

1. Open curve 2. Closed curve

3. Polygon 4. Non-simple curve

Do It Yourself 3A

1. a. A polygon with 7 sides is called a heptagon.

b. Number of sides in a hexagon (n) = 6

The number of a diagonals in

The number of diagonals in a hexagon is 9

c. A polygon with all sides and angles equal is called a regular polygon.

2. a. A triangle is a convex polygon. True

b. The number of diagonals in an octagon are 7. False

c. A circle is an example of a 2-D geometrical shape that is not a polygon. True

3. a. Since all the angles (65°, 115°, 111°, 70°, and 179°) in the polygon are less than 180°, it is a convex polygon.

b. Since one angle (195°) in the polygon is more than 180°, it is a concave polygon.

4. a. Since all the angles in the given polygon are less than 180°, it is a convex polygon.

b. Since all the angles in the given polygon are less than 180°, it is a convex polygon.

c. Since all the angles in the given polygon are not less than 180°, it is a concave polygon.

d. Since all the angles in the given polygon are not less than 180°, it is a concave polygon.

5. Since all the angles in the 6-sided polygon are 120°, which is less than 180°, it is a convex polygon.

7. Number of diagonals in a polygon = n(n − 3) 2

a. 4(4 − 3) 2 = 2 diagonals; Quadrilateral

b. 8(8 − 3) 2 = 20 diagonals; Octagon

c. 9(9 − 3) 2 = 27 diagonals, Nonagon

d. 10(10 − 3) 2 = 35 diagonals; Decagon

In a regular octagon, 5 diagonals can be drawn from one vertex.

So, 3 non-overlapping triangles can be made in a pentagon by joining the vertices.

10. Figuresmayvary.Samplefigure:

Challenge

1. The new polygon formed is a concave hexagon which is a 6-sided polygon.

Number of diagonals = n(n − 3) 2

= 6(6 − 3) 2 = 6 × 3 2 = 18 2 = 9

So the number of diagonals will be 9.

Do It Yourself 3B

1. Figuresmayvary.Samplefigures:

2. a. 190°, 60°, 30°, 80°

Since one angle measure is more than 180°, it is a concave quadrilateral.

b. 65°, 115°, 111°, 69°

Since the angles measure are less than 180°, it is a convex quadrilateral.

c. 26°, 226°, 30°, 78°

Since one angle measure is more than 180°, it is a concave quadrilateral.

3. (2y + 7)° + 119° = 180° (Interior angles on the same side of the transversal)

2y° + 126° = 180°

2y° = 180° − 126° = 54°

y = 54°

2 = 27°

(5x −2)° + 133° = 180° (Interior angles on the same side of the transversal)

5x° + 131° = 180°

5x° = 180° − 131°

x = 49° 5 = 9.8°

4. We know that the opposite sides in a parallelogram are equal. Let the breadth of the parallelogram be x Therefore, the length will be x + 15 cm.

Perimeter = x + (x + 15) + x + (x + 15) = 130 cm

4x + 30 cm = 130 cm

4x = 130 cm − 30 cm = 100 cm

x = 100 4 cm = 25 cm

Breadth = 25 cm, Length = 25 + 15 cm = 40 cm

5. Let the adjacent angles be 2x and 4x Since, the adjacent angles in a parallelogram are supplementary,

2x + 4x = 180°

6x = 180°

x = 180° 6 = 30°

Therefore,

2x = 2 × 30° = 60°

4x = 4 × 30° = 120°

Therefore, the angles in the parallelogram are 60°, 120°, 60° and 120°.

6. The diagonals of a parallelogram bisect each other. Therefore,

3b − 6 = 1 + 2b ⇒ 3b − 2b = 1 + 6 ⇒ b = 7

3a = 12 ⇒ a = 12 3 = 4

7. ∠JTK = ∠MTL = 140° (Vertically opposite angles)

We know that the diagonals in a rectangle are equal and bisect each other. Therefore,

TJ = TK

In Δ TJK,

∠TJK = ∠TKJ (Angles opposite to equal sides)

Also,

∠TJK + ∠TKJ + ∠JTK = 180° (Angle sum property of the triangle)

∠TJK + ∠TJK + 140° = 180°

2∠TJK = 180° − 140° = 40°

∠TJK = 40° 2 = 20°

Since, each angle in a rectangle is right angle,

∠TJK + ∠MJL = 90°

20° + ∠MJL = 90°

∠MJL = 90° − 20° = 70°

Hence, ∠MJL = 70° and ∠TJK = 20°.

8. In ∆QPS, Q T R S P 50° 61°

∠SQP = ∠PSQ = 50° (Angles opposite to equal sides)

∠SQP + ∠PSQ + ∠SPQ = 180° (Angle sum property of the triangle)

50° + 50° + ∠SPQ = 180°

∠SPQ = 180° − 100° = 80°

∠QPT = ∠SPQ 2 = 80° 2 = 40° (Longer diagonal bisects the angles)

Hence, ∠QPT = 40°.

In ∆ QRS,

∠SQR = ∠QSR = 61° (Angles opposite to equal sides)

∠SQR + ∠QSR + ∠SRQ = 180° (Angle sum property of the triangle)

61° + 61° + ∠SRQ = 180°

∠SRQ = 180° − 122° = 58°

∠TRS = ∠SRQ 2 = 58° 2 = 29° (Longer diagonal bisects the angles)

Hence, ∠QPT = 40° and ∠TRS = 29°

Challenge

1. Answermayvary.Sampleanswer:

S P Q R

a. Four pair of equal line segments are: PZ & ZR, ZQ & ZS, PQ & PS, RS & QR

b. Four pair of equal angles are:

∠QPS & ∠QRS, ∠PQR & ∠PSR, ∠QSR & ∠PSQ, ∠PQS & ∠RQS

given polygons. b

1 Fill in the blanks.

c. Congruent triangles: ∆QPR ≅ ∆SPR, ∆QRS ≅ ∆QPS, ∆QZR ≅ ∆SZP

a The sum of all exterior angles of a polygon is always 360°.

b The number of triangles that can be formed within a polygon with n sides

c The sum of the interior angles of a triangle is 270°.

d It is possible to have a regular polygon where each exterior angle is the right

a The sum of the interior angles in a quadrilateral is .

b Each interior angle of a regular hexagon is

d. Yes. The diagonals of a rhombus bisect the angles through which they pass.

e. In ∆QPZ and ∆SPZ,

3 Find the measure of each exterior angle of a regular polygon with: a 3 sides b 8 sides c

c The polygon in which the sum of all exterior angles is equal to the sum of interior angles is called a

d The measure of each exterior angle of a regular polygon of 18 sides is

SZ = ZQ (Diagonals bisect each other)

SP = PQ (Sides of the rhombus)

2 Write True or False.

PZ = PZ (Common arm)

Do It Yourself 3C

Therefore, ∆QPZ ≅ ∆SPZ

Do

a The sum of all exterior angles of a polygon is always 360°.

b The number of triangles that can be formed within a polygon with n sides is n

180° × 10 = 1800°

It Yourself 3C

angle in each of the following.

1 Fill in the blanks.

a. The sum of the interior angles in a quadrilateral is 360°

Do It Yourself 3C

c The sum of the interior angles of a triangle is 270°.

a The sum of the interior angles in a quadrilateral is

b. Each interior angle of a regular hexagon is 120°

b 155°

4 Find the interior angle sum of the given polygons. a b c Find the value of the unknown angle in each of the following.

So, the sum of interior angles for the given polygon is 1800°. a.

b Each interior angle of a regular hexagon is .

c. Sum of exterior angles of any polygon = 360° If sum of interior angles = 360°

So, the polygon is a quadrilateral.

1 Fill in the blanks.

Do It Yourself 3C

d It is possible to have a regular polygon where each exterior angle is the right angle.

3 Find the measure of each exterior angle of a regular polygon with: a 3 sides b 8 sides c 12 sides

1 Fill in the blanks.

c The polygon in which the sum of all exterior angles is equal to the sum of interior angles is called a

regular hexagon is .

The polygon in which the sum of all exterior angles is equal to the sum of interior angles is called a quadrilateral

4 Find the interior angle sum of the given polygons.

d The measure of each exterior angle of a regular polygon of 18 sides is

2 Write True or False.

d. Number of sides (n) = 18

a The sum of all exterior angles of a polygon is always 360°.

b Each interior angle of a regular hexagon is

a The sum of the interior angles in a quadrilateral is Each interior angle of a regular hexagon is

a The sum of the interior angles in a quadrilateral is .

The measure of each exterior angle of a regular polygon = 360 n °

6 What could be the value of the unknown exterior angle in each case?

c The polygon in which the sum of all exterior angles is equal to the sum of

9)°

sum of all exterior angles is equal to the sum of interior angles is called a .

d The measure of each exterior angle of a regular polygon of 18 sides is

b The number of triangles that can be formed within a polygon with n sides is n.

360 20 18 ° =°

c The sum of the interior angles of a triangle is 270°.

5 Find the value of the unknown angle in each of the following.

2 Write True or False.

The given polygon is a pentagon. The sum of all the interior angles in a pentagon is 540°. So, 40° + 89° + 106° + 44° + y° = 540° 279° + y° = 540° y° = 540° – 279° y° = 261°

c The polygon in which the sum of all exterior angles is equal to the sum of interior angles is called a .

d It is possible to have a regular polygon where each exterior angle is the right angle.

The measure of each exterior angle of a regular polygon of 18 sides is 20°.

unknown exterior angle in each case?

2 Write True or False.

exterior angle of a regular polygon of 18 sides is . angles of a polygon is always 360°. that can be formed within a polygon with n sides is n. angles of a triangle is 270°. regular polygon where each exterior angle is the right angle.

d The measure of each exterior angle of a regular polygon of 18 sides is .

a The sum of all exterior angles of a polygon is always 360°.

b The number of triangles that can be formed within a polygon with n sides

3 Find the measure of each exterior angle of a regular polygon with: a 3 sides b 8 sides c 12 sides

2. a. The sum of all exterior angles of a polygon is always 360°. True

b. The number of triangles that can be formed within a polygon with n sides is n. False

7 Find the number of sides of a regular polygon if each of its exterior angles measures: a 40° b 45° c

c The sum of the interior angles of a triangle is 270°.

a The sum of all exterior angles of a polygon is always 360°.

d It is possible to have a regular polygon where each exterior angle is the right

4 Find the interior angle sum of the given polygons. a b c

c. The sum of the interior angles of a triangle is 270°. False

6 What could be the value of the unknown exterior angle in each case?

b The number of triangles that can be formed within a polygon with n sides is n

d. It is possible to have a regular polygon where each exterior angle is the right angle. True

3. Each exterior angle in a regular polygon = 360 n ° , where n is the number of sides.

8 In a parallelogram, one angle measures 60° less than its adjacent angle. Determine angles in the parallelogram.

c The sum of the interior angles of a triangle is 270°.

5 Find the value of the unknown angle in each of the following.

a. 360 120 3 ° =° = 120°

3 Find the measure of each exterior angle of a regular polygon with: a 3 sides b 8 sides c

d It is possible to have a regular polygon where each exterior angle is the right angle.

9 The measures of the exterior angles of a hexagon are (3x – 4)°, (x + 4)°, (7x – 3)°, Find the measure of each angle.

4 Find the interior angle sum of the given polygons. a b c

3 Find the measure of each exterior angle of a regular polygon with:

b. 360 45 8 ° =° = 45°

The given polygon is a hexagon. The sum of all the interior angles in a hexagon is 720°. So, 112° + 130° + 114° + 113° + 155° + x° = 720° 624° + x° = 720° x° = 720° – 624° x° = 96° c.

a 3 sides b 8 sides c 12 sides

c. 360 30 12 ° =° = 30°

4. a.

4 Find the interior angle sum of the given polygons.

7 Find the number of sides of a regular polygon if each of its exterior angles measures: a 40° b 45° c 60°

5 Find the value of the unknown angle in each of the following.

Math_G8_Book.indb 52

6 What could be the value of the unknown exterior angle in each case?

180° × 2 = 360° So, the sum of interior angles for the given polygon is 360°. b. angles in a quadrilateral is .

8 In a parallelogram, one angle measures 60° less than its adjacent angle. Determine the measures of all four angles in the parallelogram.

regular polygon if each of its exterior angles measures:

Find the measure of each angle.

exterior angle of a regular polygon with: b 8 sides c 12 sides the given polygons. b c unknown angle in each of the following. b 155°

The given polygon is a quadrilateral.

9 The measures of the exterior angles of a hexagon are (3x – 4)°,

5 Find the value of the unknown angle in each of the following.

and

6 What could be the value of the unknown exterior angle in each case?

The sum of all the interior angles in a quadrilateral is 360°. (7x – 3)° + (2x + 10)° + (5x + 9)° + (3x + 4)° = 360° (17x + 20)° = 360° 17x° = 340° x° = 20° 6. a.

b 45° c 60° measures 60° less than its adjacent angle. Determine the measures of all four angles of a hexagon are (3x – 4)°, (x + 4)°, (7x – 3)°, (8x – 1)°, (2x + 3)° and (9x + 1)°.

7 Find the number of sides of a regular polygon if each of its exterior angles measures: a 40°

Math_G8_Book.indb 52 04-10-2024

8 In a parallelogram, one angle measures 60° less than its adjacent angle. Determine the measures of all four angles in the parallelogram.

6 What could be the value of the unknown exterior angle in each case?

180° × 5 = 900°

So, the sum of interior angles for the given polygon is 900°.

9 The measures of the exterior angles of a hexagon are (3x – 4)°, (x + 4)°, (7x – 3)°, (8x – 1)°, (2x + 3)° and (9x + 1)°. Find the measure of each angle.

The sum of the exterior angles in any polygon is 360°. 111° + 80° + 46° + z° = 360° 237° + z° = 360° z° = 123°

7 Find the number of sides of a regular polygon if each of its exterior angles measures: a 40° b 45° c

8 In a parallelogram, one angle measures 60° less than its adjacent angle. Determine angles in the parallelogram.

9 The measures of the exterior angles of a hexagon are (3 x + 4)°, (7x – 3)°, Find the measure of each angle.

value of the unknown exterior angle in each case? 80° 111°

10. The base of a Vijaya Stambha is a regular decagon. So, number of sides (n) = 10

We know the sum of the interior angles of a polygon = (n − 2) × 180°

So, the measure of one of the interior angles of a polygon = (–2)×180° n n

of sides of a regular polygon if each of its exterior angles measures: b 45° c 60°

c (7x – 3)° (2x + 10)° (3x + 4)° (5x + 9)° in each case?

The sum of the exterior angles in any polygon is 360°.

46° + 81° + 34° + 82° + 47° + a° = 360°

290° + a° = 360° a° = 70°

(10–2)×180° = 10

one angle measures 60° less than its adjacent angle. Determine the measures of all four parallelogram.

8×180° = 10

Hence, the measure of one of the angles of the base of the building = 144°

c m° 38° 52° 16° 57° 42° 57°

38° its exterior angles measures: c 60° adjacent angle. Determine the measures of all four – 4)°, (x + 4)°, (7x – 3)°, (8x – 1)°, (2x + 3)° and (9x + 1)°.

the exterior angles of a hexagon are (3x – 4)°, (x + 4)°, (7x – 3)°, (8x – 1)°, (2x + 3)° and (9x + 1)°. of each angle.

Challenge

1. Let the angles of the quadrilateral be x, 2x, 3x, and 4x.

x + 2x + 3x + 4x = 360°

The sum of exterior angles in any polygon is 360°.

38° + 52° + 16° + 57° + 42° + 57° + 38° + m° = 360°

300° + m° = 360°

m° = 60°

7. The measure of each exterior angle of a regular polygon = 360 n °

⇒ 360° = Measure of each exterior angle of a regular polygon

a. 360° ==9 40° n

b. 360° ==8 45° n

c. 360° ==6 60° n

04-10-2024 16:29:05

8. Let the angles in a parallelogram be x and x − 60°. Since, the adjacent angles in a parallelogram are supplementary,

x + x − 60° = 180°

2x − 60° = 180°

2x = 180° + 60° = 240°

x = 240° 2 = 120°

The other adjacent angle = x − 60° = 120° − 60° = 60° As the opposite angles in a parallelogram are equal, the angles of the parallelogram are 120°, 60°, 120° and 60°.

9. (3–4)°+(+4)°+(7–3)°+(8–1)°+(2+3)° +(9+1)°=360° xxxxx x

(3–4++4+7–3+8–1+2+3+9+1)°=360° xxxxxx

(3++7+8+2+9)+(–4+4–3–1+3+1)°=360° xxxxxx   = 360° 360° 30+0°=360°==12° 30 xx ⇒ 

So, =12° x

Then,

(3x – 4)° = (3 × 12 – 4)° = 32°

(x + 4)° = (12 + 4)° = 16°

(7x – 3)° = (7 × 12 – 3)° = 81°

(8x – 1)° = (8 × 12 – 1)° = 95°

(2x + 3)° = (2 × 12 + 3)° = 27°

(9x + 1)° = (9 × 12 + 1)° = 109°

10x = 360°

x = 360° 10 = 36°

2x = 2 × 36° = 72°

3x = 3 × 36° = 108°

04-10-2024 16:29:05

4x = 4 × 36° = 144°

Therefore, the angles of the quadrilateral are 36°, 72°, 108° and 144°.

Since, all the angles are less than 180°, it’s a convex quadrilateral.

The measure of the largest angle = 144°.

Chapter Checkup

1. a. A polygon with unequal sides and unequal angles is called an irregular polygon.

b. The sum of the interior angles of a hexagon is 720°

c. Number of diagonals in a polygon = n(n − 3) 2

Number of diagonals in a pentagon = 5(5 − 3) 2 = 5

The number of diagonals in a pentagon is 5

d. The sum of the exterior angles of any polygon is always 360°

2. a. The diagonals of a kite bisect each other at right angles. True

b. The diagonals of an isosceles trapezium are equal. True

c. The diagonals of a rectangle are perpendicular to each other. False

d. All quadrilaterals are parallelograms. False

3. The interior angles in a convex quadrilateral are less than 180°. Hence, 222° can’t be an interior angle in a convex quadrilateral.

4. a. Since all the angles in the given polygon are not less than 180°, it is a concave polygon.

b. Since all the angles in the given polygon are less than 180°, it is a convex polygon.

c. Since all the angles in the given polygon are not less than 180°, it is a concave polygon.

d. Since all the angles in the given polygon are less than 180°, it is a convex polygon.

5. The number of diagonals in any polygon is given as (–3) 3 nn , where n is the number of sides.

a. A convex quadrilateral has 4 sides. 4(4–3) 4 ==2 22

b. A nonagon has 9 sides. 9(93) 54 =9==27 22 –n ⇒

c. n ⇒ 17(17–3) 238 n=17 ==119 22

11 Find the value of the unknown interior angle in each case.

6. Sum of the interior angles of a quadrilateral is 360°.

11 Find the value of the unknown interior angle in each case.

11. a. Sum of the interior angles in a quadrilateral is 360°.

41° + 58° + 21° + l° = 360° 120° + l° = 360° l

= 360° − 120° l

= 240°

b. Sum of the interior angles in a quadrilateral is 360°.

a. 47°+ 58°+ 98° + 156° = 359°

b. 54°+ 59°+ 72°+ 108°= 293°

c. 83°+ 68°+ 82°+ 127° = 360°

Hence, option c is a set of interior angle of a quadrilateral.

21° 41° 58°

7. The measure of exterior angle of a regular polygon = 360 n ° , where n is the number of sides in the polygon.

a. 360° =90° 4

b. 360° =24° 15

c. 360° =18° 20

11 Find the value of the unknown interior angle in each case.

8. Number of sides in a polygon = 360 Measure of each exterior angle of a regular polygon °

a. 360° ==20 18° n sides

b. 360° = 36° n = 10 sides

c. 360° = 90° n = 4 sides

9. Each interior angle has a corresponding exterior angle and both of them are supplementary.

a. Supplement of 70° = 180° − 70° = 110°

11 Find the value of the unknown interior angle in each case.

21° 41° 58°

88° + 58° + 64° + m° = 360° 210° + m° = 360° m° = 360° − 210° m° = 150° c. Sum of the interior angles in a quadrilateral is 360°.

12 In a parallelogram, the adjacent angles are (2z 25)º and (3z + 10)º. Determine the parallelogram.

Supplement of 90° = 180° − 90° = 90°

Supplement of 110°= 180° − 110° = 70°

Supplement of 90° = 180° − 90° = 90°

b. Supplement of 45° = 180° − 45° = 135°

74° + 66° + 97° + n° = 360° 237° + n° = 360° n° = 360° − 237° n° = 123° d. Sum of the interior angles in a pentagon is 540°. Let the unknown angle be x°.

13 The angles of a pentagon measure x°, (x − 5)°, (x + 15)°, (3x − 44)° and (x

14 In a rhombus, one of the diagonals is of the same length as one of its sides. angles in the rhombus.

15 The angles of a hexagon are in the ratio 1:2:3:4:6:8. Find the measure of the

64°

16 Find the values of wº, xº, yº and zº.

11 Find the value of the unknown interior angle in each case. a 21° 41° 58°

12 In a parallelogram, the adjacent angles are (2z 25)º and (3z + 10)º. Determine the measures of all four angles in the parallelogram.

13 The angles of a pentagon measure x°, (x − 5)°, (x + 15)°, (3x − 44)° and (x 70)°. Find x

12 In a parallelogram, the adjacent angles are (2z 25)º and (3z + 10)º. Determine the parallelogram.

14 In a rhombus, one of the diagonals is of the same length as one of its sides. Determine the measures of the angles in the rhombus.

Supplement of 60° = 180° − 60° = 120°

Supplement of 100°= 180° − 100° = 80°

Supplement of 155° = 180° − 155° = 25°

16 Find the values of wº, xº, yº and zº.

c. Supplement of 80° = 180° − 80° = 100°

Supplement of 85° = 180° − 85° = 95°

Supplement of 95° = 180° − 95° = 85°

Supplement of 100° = 180° − 100° = 80°

10. Sum of the exterior angles of a quadrilateral is 360°.

a. p° + 48° + 92° + 117° = 360°

13 The angles of a pentagon measure x°, (x − 5)°, (x + 15)°, (3x − 44)° and (x

15 The angles of a hexagon are in the ratio 1:2:3:4:6:8. Find the measure of the smallest and the biggest angles.

p° + 257° = 360° ⇒ p° = 360° − 257° = 103°

b. q° + 89° + 93° + 51° = 360°

q° + 233° = 360° ⇒ q° = 360° − 233° = 127°

c. r° + 90° + 67° + 93° = 360°

14 In a rhombus, one of the diagonals is of the same length as one of its sides. angles in the rhombus.

e. The sum of the interior angles in a hexagon is 720°. Let the unknown angle be x°.

17 Sohan is designing a logo for his company which is in the shape of a regular diagonals connect the interior angles of the polygon? How many triangles interior angles?

16 Find the values of wº, xº, yº and zº.

12 In a parallelogram, the adjacent angles are (2z 25)º and (3z + 10)º. Determine the measures of all four angles in the parallelogram.

13 The angles of a pentagon measure x°, (x − 5)°, (x + 15)°, (3x − 44)° and (x 70)°. Find x

r° + 250° = 360° ⇒ r° = 360° − 250° = 110°

524° + x = 720°

Chapter 3 • Polygons and Quadrilaterals

15 The angles of a hexagon are in the ratio 1:2:3:4:6:8. Find the measure of the

18 In a city with several towns connected by bridges, there are 10 towns. If each town by a direct bridge, how many bridges are needed in total? How many bridges have?

f 51°

12 In a parallelogram, the adjacent angles are (2z 25)º and (3z + 10)º. Determine the parallelogram.

x° = 720° − 524° x° = 196°

14 In a rhombus, one of the diagonals is of the same length as one of its sides. Determine the measures of the angles in the rhombus.

17 Sohan is designing a logo for his company which is in the shape of a regular 12-sided polygon. How many diagonals connect the interior angles of the polygon? How many triangles can be formed by connecting these interior angles?

13 The angles of a pentagon measure x°, (x − 5)°, (x + 15)°, (3x − 44)° and (x

15 The angles of a hexagon are in the ratio 1:2:3:4:6:8. Find the measure of the smallest and the biggest angles.

16 Find the values of wº, xº, yº and zº.

18 In a city with several towns connected by bridges, there are 10 towns. If each town is connected to its adjacent town by a direct bridge, how many bridges are needed in total? How many diagonals does this network of bridges have?

17 Sohan is designing a logo for his company which is in the shape of a regular diagonals connect the interior angles of the polygon? How many triangles interior angles?

14 In a rhombus, one of the diagonals is of the same length as one of its sides. angles in the rhombus.

15 The angles of a hexagon are in the ratio 1:2:3:4:6:8. Find the measure of the 95° + 156° + 91° + 142° + 40° + x° = 720°

(3y + 5)° 96° + 133° + 105° + 95° + x° = 540° 429° + x° = 540° x° = 540° − 429° x° = 111°

18 In a city with several towns connected by bridges, there are 10 towns. If each

f. The sum of the interior angles in a heptagon is 900°. Let the unknown angle be x°.

87° 235° 254°

87° + 254° + 64° + 51° + 235° + 61° + x° = 900°

y° = 180° − (125° + 45°) = 10°

b. 3x° + 135° = 180° (interior angles on the same side of the transversal)

3x° = 180° − 135° = 45°

x° = 45 3 ° = 15°

5y° + 125° = 180° (interior angles on the same side of the transversal)

5y° = 180° − 125° = 55°

y° = 55 5 ° = 11°

25)º and (3z + 10)º. Determine the measures of all four angles in

752° + x° = 900° x° = 900° – 752° x° = 148°

x + 15)°, (3x − 44)° and (x 70)°. Find x

12. Since, the adjacent angles of a parallelogram are supplementary,

same length as one of its sides. Determine the measures of the

(2z − 25)° + (3z + 10)° = 180°

5z° − 15° = 180°

5z° = 180° + 15° = 195°

1:2:3:4:6:8. Find the measure of the smallest and the biggest angles.

z = 195° 5 = 39°

Therefore, the adjacent angles are (2z − 25)° = (2 × 39 − 25)° = 53°

5x = 180°

5y° 3x° c 79° x° D A B C (6y + 4)° (3y + 5)°

c. x° + 79° = 180° (interior angles on the same side of the transversal)

x° = 180° − 79° = 101°

6y + 4° + 3y + 5° = 180° (interior angles on the same side of the transversal)

9y + 9° = 180°

9y = 171°

y = 171 9 ° = 19°

17. Number of sides of the logo = n = 12

So, the number of diagonals = 12(12–3) =54 2

Similarly, in ∆PQS and ∆RQS,

(3z + 10)° = (3 × 39 + 10)° = 127°

Hence, the angles of the parallelogram are 53°, 127°, 53° and 127°.

x = 180° 5 = 36° Therefore,

∠RQS = ∠PQS & ∠RSQ = ∠PSQ

Hence, the diagonals of a rhombus bisect its angles.

13. Interior angle sum of a pentagon = (5 2) × 180° = 540°

14.

2x = 2 × 36° = 72°

16. Since, the diagonals bisect the angles in rhombus,

3x = 3 × 36°= 108°

x° + (x 5)° + (x + 15)° + (3x 44)° + (x − 70)° = 540°

7x − 104° = 540°

∠RSQ = ∠PSQ= 52°

∠PSR = 2 ∠PSQ= 2 × 52° = 104°

x = 92°

Hence, the angles in the parallelogram are 72°, 108°, 72° and 108°.

10 non-overlapping triangles can be formed by connecting these interior angles.

∠PSR = ∠PQR = 104° (opposite angles are equal in parallelogram)

which is in the shape of a regular 12-sided polygon. How many polygon? How many triangles can be formed by connecting these

13. Q P R S 60° 60° 60° 60° 60° 60°

bridges, there are 10 towns. If each town is connected to its adjacent needed in total? How many diagonals does this network of 04-10-2024 16:29:05

18. Let each of the 10 vertices of the decagon represent 10 towns. As each town is connected to its nearest neighbour, the sides of the decagon represent the bridges. So, there are 10 bridges in total.

∠PSR + ∠QPS = 180°

104° + ∠QPS = 180° (Adjacent angles are supplementary)

∠QPS = 180° − 104° = 76°

∠QPS = ∠QRS = 76°

Word Problems

Consider the PQRS, the diagonal is equal to the sides of the rhombus.

Consider the rhombus PQRS, where the diagonal PR is equal to the sides of the rhombus.

1. Let the angles be 2x, 3x, 4x and 5x

2x + 3x + 4x + 5x = 360°

ΔPSR and ΔPQR are equilateral triangles.

∆PSR and ∆PQR are equilateral triangles.

Thus, all the angles in these two triangles are equal to 60°. Therefore, the angles of the rhombus are 60°, 120°, 60° and 120°.

14x = 360°

And, the number of diagonals = 10(10–3) 10(7)70 ===35 222

x = 360° 14 = 25.71°

14. a. x° = 125° (Vertically opposite angles) w° = 50° (alternate angles)

15. Given the ratio of angles of hexagon= 1 : 2 : 3 : 4 : 6 : 8

Let the angles of a quadrilateral be x, 2x, 3x, 4x, 6x and 8x respectively.

z° + 135° = 180° (Linear pair)

Challenge

Angles are

2x = 2 × 25.71° = 51.42°

3x = 3 × 25.71° = 77.13°

4x = 4 × 25.71° = 102.84°

° = 180° − 135° = 45°

Interior angle sum = (6 − 2) × 180° = 720°

1. Assertion (A): The sum of the interior angles of any polygon is given by the formula (n – 2) × 180°, where n is the number of sides.

5x = 5 × 25.71° = 128.55°

x + 2x + 3x + 4x + 6x + 8x = 720°

° + y ° + z° = 180° (Angle sum property of triangle)

⇒ 24x = 720°

125° + y° + 45° = 180°

⇒ x = 720° 24 = 30°

y° = 180° − (125° + 45°) = 10°

x = 30°

b. 3x° + 135° = 180° (interior angles on the same side of the transversal)

8x = 8 × 30° = 240°

3x° = 180° − 135° = 45°

Since, all the angles are different, we can conclude that climbing structure is not a parallelogram.

The formula given to calculate the sum of the total interior angles of any polygon is correct. Thus, the assertion is correct.

2. In a parallelogram, opposite sides are equal so the length of opposite side is 24 m.

Reason (R): A polygon can be divided into n – 2 triangles. Any polygon can be divided into n – 2 non-overlapping triangles. So, this statement is also correct.

So, the smallest angle = 30°and the greatest angle = 240°.

16. a. x° = 125° (Vertically opposite angles)

x° = 45° 3 = 15°

w° = 50° (alternate angles)

5y° + 125° = 180° (interior angles on the same side of the transversal)

z° + 135° = 180° (Linear pair)

In a parallelogram, the opposite angles are equal and the adjacent angles are supplementary, so the angles are 120°, 60°, 120° and 60°.

Chapter 5

We know that the sum of interior angles of a triangle is 180°. So, the sum of interior angles inside n – 2 triangles will be (n – 2) × 180°.

z° = 180° − 135° = 45°

5y° = 180° − 125° = 55°

Let's Warm-up

Thus, the reason helps us explain the assertion. Thus, the correct answer is option a.

y° = 55°

5 = 11°

x° + y° + z° = 180° (Angle sum property of triangle)

125° + y° + 45° = 180°

c. x° + 79° = 180° (interior angles on the same side of the transversal)

° = 180°− 79° = 101°

6y + 4° + 3y + 5° = 180° (interior angles on the same side of the transversal)

9y + 9° = 180°

9y = 171° 171° = 19°

1. Exactly two pairs of adjacent sides are equal. Kite

2. Exactly one pair of opposite sides are parallel. Trapezium

a Both (A) and (R) are true, and (R) is the correct explanation of (A).

3. All angles in a quadrilateral are right angles, but only the opposite sides are equal. Rectangle

4. All sides are equal, but all angles may not necessarily be equal. Rhombus

2. Statement 1: ∠PQR = ∠SPQ = 90°

According to Statement 1, ∠PQR = ∠SPQ = 90°.

It is given that PQ is parallel to RS, and PS is parallel to QR. So, it states that all the angles inside the quadrilateral are equal to 90°. This does not tell us that the quadrilateral is a square since it does not tell us anything about the length of the sides of the quadrilateral. So, Statement 1 alone is insufficient.

Statement 2: PQ = QR

According to Statement 2, PQ = QR.

It is given that PQ is parallel to RS, and PS is parallel to QR. So, it states that all the sides of the quadrilateral are equal. But this does not tell us anything about the angles inside the quadrilateral so this statement is also insufficient alone to tell us whether the given quadrilateral is a square or not. If we put both the statements together, we can note that all the sides of the quadrilateral are equal and all the angles are equal to 90°. Thus, the shape is a square.

Hence, option d is correct.

d Both statements together are sufficient, but neither statement alone is sufficient.

Case Study

1. Each interior angle of a hexagon is 120°. So, the shape of the tile is hexagon. Thus, option c is correct.

2. Length of the park = 160 m

Breadth of the park = 120 m

Diagonal of a rectangle = (Length)+(22 Breadth)

Length of the diagonal of the park = 160+12022

= 25600 + 14400

= 40000

= 200 m

The distance at which each sprinkler is planted = 5 m

Total number of sprinklers = 200 m ÷ 5 m = 40 sprinklers

Thus, option b is correct.

3. Let the angles be 2x, 3x, 4x and 5x

2x + 3x + 4x + 5x = 360°

14x = 360°

x = 360°

x = 25.71°

Therefore, the angles are:

2x = 2 × 25.71° = 51.42°

3x = 3 × 25.71° = 77.13°

4x = 4 × 25.71° = 102.84°

5x = 5 × 25.71° = 128.55°

Since, all the angles are different, we can conclude that climbing structure is not a parallelogram.

4. The swimming pool has opposite side parallel and equal in length. Thus, the shape of the swimming pool is parallelogram.

5. Answersmayvary.

Chapter 4

Let's Warm-up

1. Ascending order:

20, 20, 20, 20, 20, 20, 20, 25, 25, 25, 25, 25, 25, 25, 25, 25, 30, 30, 30, 30, 30, 30, 30, 30, 30, 35, 35, 35, 35, 35

Temperature (in C°) Tally

Do It Yourself 4A

1. Number of students = 50

Number of students who like cricket

= 40% of 50 = 40 100 × 50 = 20

Number of students who like badminton

= 24% of 50 = 24 100 × 50 = 12

Number of students who like tennis

= 10% of 50 = 10 100 × 50 = 5

Percentage of students who don’t like any sport

= 100% (40 + 24 + 10)%

= 100% − 74% = 26%

Number of students who don’t like any sport

= 26% of 50

= 26 100 × 50 = 13

Sport

No favourite sport

Number of Students (Frequency)

Tennis is liked by the minimum number of students.

2. Ascending order:

1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5

Range = 5 − 1 = 4

3. a. The highest monthly income is ₹85 thousands. 14 individuals have the highest monthly income.

b. The lowest monthly income is ₹25 thousands. 14 individuals have the lowest monthly income.

c. Range = Maximum value of the data − Minimum value of the data = 85,000 − 25,000 = 60,000

4. Ascending order:

75, 76, 77, 78, 79, 79, 80, 81, 82, 83, 84, 85, 85, 86, 86, 87, 87, 88, 88, 88, 89, 90, 90, 91, 92, 92, 93, 94, 94, 95

Scores Number of Students (Frequency) Tally Marks 75 1 |

Do It Yourself 4B

1. a. The smaller value of a class interval is called the lowerclass limit. So, the lower-class limit of 50−60 is 50.

b. The class mark is the midpoint of each class interval.

40+50 90 ==45 22 and 50+60 110 ==55 22

c. The difference between the upper limit and the lower limit of a class interval is called the class size.

= 10

So, class size = 10

2. Ascending order:

3, 5, 7, 9, 10, 12, 13, 15, 17, 18, 19, 19, 20, 23, 24, 27, 29, 34, 34, 38, 39, 40, 40, 42, 47

Class Intervals (Ages of People)

Range = 95 − 75 = 20

5. Digits: 3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3, 2, 7, 9, 5, 0, 2, 8, 8, 4, 1, 9

Ascending order: 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9

Digits

3. 120, 120, 125, 150, 153, 203, 207, 215, 235, 263, 340, 350, 350, 400, 450, 450, 452, 489, 500, 562, 600, 645, 658, 800

Class Intervals (Electricity Bill in ₹)

6. Answersmayvary.

Challenge

1. Yes, the range will be affected as the highest value of the incorrect data set has been replaced with some other value. Range = Highest value in the data set − Lowest value in the data set.

The highest and lowest values in the incorrect data set = 45 and 22

Range of data noted by Rohan = 45 − 22 = 23

The highest and lowest values in the correct data set = 43 and 22

The range of the correct data set = 43 − 22 = 21

5. a. The number of students who scored marks less than or equal to 15 = 7 + 6 + 7 = 20

b. The number of students who scored more than or equal to 15 marks = 8

c. The class mark of the class interval with the frequency

6 = 5 + 10 15 = = 7.5 22 = 5 + 10 15 = = 7.5 22 7.5

d. Number of students who wrote the test = 7 + 6 + 7 + 8 = 28

e. Percentage of students who scored less than 15 marks

b. Number of books sold on Saturday = 1090

Required percentage = 1090 5100 × 100% = 21.37%

c. Number of books sold on Thursday: Number of books sold on Friday = 970:670 = 97:67

Number ofstudents who scored less than 15 marks = Total number of students

7 + 6 + 7 20

Number ofstudents who scored less than 15 marks = Total number of students

d. Average of books sold per day = 5100 850 6 = = 850

= × 100 = × 100 = 71.41%

7 + 6 + 7 + 828

7 + 6 + 7 20

= × 100 = × 100 = 71.41%

7 + 6 + 7 + 828

6. Answersmayvary.Sampleanswer:

What percentage of students scored more than 10 marks?

Challenge

1. Statement 1: The frequency distribution table summarising the height of 100 adults is known.

It is not possible to get the original list of the 100 individual heights from a frequency distribution table, as a frequency distribution table summarises the data by grouping the individual heights into intervals (bins or classes) and then counting the number of heights that fall into each interval. This process involves aggregation, where detailed individual data is combined into broader categories.

Statement 2: The height of the shortest person and the tallest person is known.

The height of the shortest and tallest person will also not be sufficient to identify the original list.

Combining statements 1 and 2 will also not help in identifying the original list of heights. Hence, option e is the correct answer. Both statements together are not sufficient.

Do It Yourself 4C

1. Total wheat exported to these 5 countries = 94,000 + 86,000 + 56,000 + 55,000 + 63,000 = 354,000 tonnes of wheat

a. Wheat exported to Sri Lanka = 94,000 tonnes

Percentage of wheat exported to Sri Lanka

= 94,000

3,54,000 × 100% = 26.55%

b. Wheat exported to Afghanistan = 55,000 tonnes

Percentage of wheat exported to Afghanistan = 55,000

3,54,000 × 100% = 15.5%

c. Wheat exported to Qatar = 63,000 tonnes

Wheat exported to the Yemen Republic = 86,000 tonnes

Wheat exported to Qatar: Wheat exported to the Yemen Republic = 63,000 tonnes: 86,000 tonnes = 63:86

d. Average wheat exported = Total wheatexported No. of nations

3,54,000 5 = 70,800 tonnes of wheat

2. Total number of books sold = 800 + 700 + 870 + 970 + 670 + 1090 = 5100

a. Number of books sold on Friday = 670

Required percentage = 670 5100 × 100% = 13.14%

Required ratio = (800 + 700 + 670) : (870 + 970 + 1090) = 2170:2930 = 217:293

3. a. Foreign exchange reserve in 2013 = $3500 million

b. Foreign exchange reserve in 2011 = $2500 million

Foreign exchange reserve in 2012 = $3100 million

Difference between the foreign exchange reserve for 2011 and 2012 = $3100 million − $2500 million = $600 million

c. Foreign exchange reserve in 2013 = $3500 million

Foreign exchange reserve in 2012 = $3100 million

Increase in foreign reserve = $3500 million $3100 million = $400 million

Percentage increase = $400 million $3100 million × 100% = 12.9%

d. Sum of foreign exchange reserves in 2010 and 2011 = $2800 million + $2500 million = $5300 million

Sum of foreign exchange reserves in 2012 and 2013 = $3100 million + $3500 million = $6600 million

Required ratio = $5300 million:$6600 million = 53:66 4.

a. Total export in all years = 15.6 + 14.8 + 16.7 + 17.0 + 19.4 + 19.8 = 103.3 crores

b. Average export in all years = 103.3 =17.217 6 crores

c. Export of pears in 2015 = 15.6 crores

Export of pears in 2018 = 17.0 crores

Increase in export = 17.0 − 15.6 crores = 1.4 crores

Percentage increase = 1.4 15.6 × 100% = 8.97%

d. Exports in 2016 = 14.8 crores

Exports in 2019 = 19.4 crores

Average exports in 2016 and 2019 = 14.8+19.4 2 crores = 17.1 crores

So, 2016 and 2019 had an average export of 17.1 crores.

5. Total number of students in the school = 450

a. Total number of students who passed in 2021 = 400

b. Total number of students who passed in 2018 = 300

Percentage of students who passed in 2018 = 300 ×100%=66.67%

450

c. Total number of students who passed in 2020 = 350

Percentage of students who passed in 2020 = 350 ×100%=77.78%

450

d. Total number of students who passed in 2019 = 350

Total number of students who passed in 2022 = 400

Increase in number of students = 400 − 350 = 50

Percentage increase = 50 ×100%=14.28%

350

e. Answermayvary.Sampleanswer:

What is the ratio of students who passed in 2018 to those who passed in 2020?

6. a. Total population of Delhi and Mumbai = 28 + 20 million = 48 million

b. Difference between the population of Shanghai and Cairo = 25 million − 20 million = 5 million

c. Total population of all the cities = 28 + 25 + 18 + 20 + 20 + 12 = 123 million

Average population = 123 6 million = 20.5 million

d. Difference between the population of Moscow and New York = 18 − 12 = 6 million

Challenge

1. One reason a bar graph is not suitable for displaying this data is that a bar graph is good for showing single values for each category, but the life spans are given as ranges, not single values. A bar graph can’t show the full range from the lowest to the highest number for each animal. Yes, we can plot a bar graph if we calculate the average ages of the life spans of the animals. The data will be then given as:

Crocodile 50–60 years

Tortoise 80–150 years

2 = 55 years

Do It Yourself 4D

1. a. Basmati exported in 2021 = $1660 million

Basmati exported in 2022 = $2280 million

Percentage increase in exports = 2280–1660 ×100%=37.34% 1660

b. Fruit have the least exports

c. Total exports for the year 2021 (in million $) = 301 + 1632 + 1903 + 1660 + 2969 + 2591 = $11,056 million

d. Total exports for the year 2022 = 313 + 2111 + 2099 + 2280 + 3207 + 3761 = $13,771 million

a. Total tea production by Company C in three years = 80 + 70 + 60 = 210 tonnes

b. Total tea production by Company B in 2015 and 2016 = 70 + 70 = 140 tonnes

Total tea production by Company D in three years = 50 + 40 + 50 = 140 tonnes

Required ratio = 140 tonnes : 140 tonnes = 1:1

c. Production by Company D in 2016 = 50 tonnes

Production by Company D in 2017 = 40 tonnes

Percentage increase in production by Company D from 2016 to 2017 = 50–40 ×100%=25% 40

d. Production by Company B in 2015 = 70 tonnes

Production by Company B in 2017 = 90 tonnes

Percentage increase in production by Company B from 2015 to 2017 = 90–70 ×100%=28.57% 70

Comparison of Ayesha and Ravi's Test Scores

Scale: 1 division = 10 Marks

4. a. Total number of fruit sold by vendor A = 60 + 80 + 100 + 40 + 20 = 300

b. Total number of fruit sold by vendor B = 80 + 70 + 90 + 60 + 40 = 340

c. Oranges and apples sold by vendor A = 60 + 80 = 140 Banana, pear and melon sold by vendor A = 100 + 40 + 20 = 160

Required ratio = 140:160 = 7:8

d. Banana was the most sold fruit out of all fruit by the two vendors.

e. Answermayvary.Sampleanswer: What is the ratio of bananas sold by both the vendors.

5. a. Total runs scored by Virat Kohli = 1320 + 1000 + 680 + 760 + 1520 + 1400 + 1360 = 8040

b. Total runs scored by Rohit Sharma = 1200 + 480 + 800 + 500 + 1280 + 1140 + 1540 = 6940

Crocodile Blue Whale Tortoise Dolphin Shark

c. Average runs scored by Virat = 8040 7 = 1148.57

Average runs scored by Rohit Sharma = 6940 7 = 991.43

Difference between average runs scored = 157.14

d. Runs scored by Virat Kohli before 2017 = 1320 + 1000 + 680 + 760 = 3760

Runs scored by Rohit Sharma after 2015 = 500 + 1280 + 1140 + 1540 = 4460

Required ratio = 3760:4460 = 188:223

Challenge

1. Statement 1: The ratio of sci-fi genre tickets sold by Cinema Hall 1 to Cinema Hall 2 is 8:9.

Ratio of sci–fi genre tickets sold by Cinema Hall 1 to Cinema Hall 2 = = 1809 1608

Hence, Statement 1 is incorrect.

Statement 2: Cinema Hall 2 sold 25% more horror film tickets than Cinema Hall 1.

Number of horror film tickets sold by Cinema Hall 1 = 80

Number of horror film tickets sold by Cinema Hall 2 = 100

Number of more tickets sold by Cinema Hall 2 as compared to Cinema Hall 1 = 100 – 80 = 20

Let us find 25% of the number of horror film tickets sold by Cinema Hall 1. 25% of 80 = 20

Hence, Statement 2 is correct.

Statement 3: Cinema Hall 2 sold 60% less drama film tickets than Cinema Hall 1.

Number of drama movie tickets sold by Cinema Hall 1 = 160

Number of drama movie tickets sold by Cinema Hall 2 = 100

Number of lesser tickets sold by Cinema Hall 2 as compared to Cinema Hall 1 = 160 – 100 = 60

60% of 160 = 96

Hence, Statement 3 is incorrect.

Thus, option a is correct.

Only Statement 2 is correct.

Do It Yourself 4E

a. Total rainfall over the years = 700 + 600 + 550 + 650 + 800 = 3300 mm

b. Highest rainfall occurred in 2016–2017.

c. Rainfall during 2014–2015 = 550 mm

Rainfall during 2013–2014 = 600 mm

Percentage decrease in rainfall = 600 mm–550 mm ×100%=8.33%

600 mm

d. On observing the graph, the highest difference in rain can be seen in the years 2015–2016 and 2016–2017. Hence, the highest percentage difference in rainfall over the previous year will be in the years 2015–2016 and 2016–2017.

Scale: 1 division = 2 Presidents Age Group 6 11 8 3 1 17

a. Number of presidents aged more than 66 at the time of inauguration = 3 + 1 = 4

b. Number of presidents aged more than 60 at the time of inauguration = 8 + 3 + 1 = 12

c. Number of presidents aged more than 54 at the time of inauguration = 17 + 8 + 3 + 1 = 29

a. Total number of people surveyed = 50 + 90 + 75 + 60 + 50 + 40 = 365

b. Number of people who exercised for more than 30 mins = 90 + 75 + 60 + 50 + 40 = 315

Required percentage = 315 ×100%=86.30% 365

c. Number of people who exercised for less than 120 mins = 50 + 90 + 75 + 60 = 275

Required percentage = 275 ×100%=75.34% 365

d. Number of people who exercised for less than 90 mins = 50 + 90 + 75 = 215

Number of people who exercised for more than 120 mins = 50 + 40 = 90

Required ratio = 215:90 = 43:18

4. a. Number of workers earning more than ₹300 = 100 + 150 + 180 + 120 = 550

b. Number of workers earning less than ₹500 = 50 + 80 + 100 + 150 = 380

c. Total number of workers = 50 + 80 + 100 + 150 + 180 + 120 = 680

Number of workers earning less than ₹600 = 50 + 80 + 100 + 150 + 180 = 560

Required percentage = 560 ×100%=82.35% 680

d. Number of workers earning less than ₹400 = 50 + 80 + 100 = 230

Number of workers earning more than ₹400 = 150 + 180 + 120 = 450

Required ratio = 230:450 = 23:45

e. Number of workers earning between ₹400 − ₹600 = 150 + 180 = 330

5. a. Number of students who scored more than 40 marks

= 9 + 6 + 7 + 7 + 8 + 7 = 44

Total number of students = 5 + 5 + 9 + 6 + 7 + 7 + 8 + 7 = 54

Required percentage = 44 ×100%=81.48% 54

b. Number of students who scored less than 30 marks = 5

Required percentage = 5 ×100%= 54 = 9.25%

c. Number of students who scored more than 50 marks

= 6 + 7 + 7 + 8 + 7 = 35

Number of students who scored less than 50 marks

= 5 + 5 + 9 = 19

Required ratio = 35:19

d. Number of students who scored more than 40 marks

= 9 + 6 + 7 + 7 + 8 + 7 = 44

Number of students who scored less than 40 marks

= 5 + 5 = 10

Required difference = 44 − 10 = 34

6. a. Number of employees older than 50 years = 5 + 5 = 10

b. Number of employees younger than 35 years = 30 + 45 + 40 = 115

Number of employees older than 35 years = 30 + 20 + 20

+ 5 + 5 = 80

Required ratio = 115:80 = 23:16

c. Number of employees older than 25 years = 45 + 40 + 30

+ 20 + 20 + 5 + 5 = 165

Total number of employees = 30 + 45 + 40 + 30 + 20 + 20

+ 5 + 5 = 195

Required percentage = 165 ×100%=84.6% 195

d. Answersmayvary.Sampleanswer: What is the ratio of the number of employees who are younger than 40 years to those who are older than 40 years?

Challenge

1. The data set only contains five ages (65, 73, 70, 74, 69).

Histograms are more effective and meaningful when used with larger data sets where the distribution of data can be more clearly seen. Also the ages provided are relatively close to each other, and with such a small range, a histogram won’t show a clear pattern or trend.

A bar graph will be an appropriate graphical representation for the given data set and can be given as:

Chapter Checkup

1. Ascending order: 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6

2. Ascending order: 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 16, 16, 17

3. a. Blood group AB has the highest frequency of 15. So, AB has the most donors.

b. Number of people that participated in the blood donation campaign = 14 + 10 + 15 + 9 = 48

4. a. 3 crores + 2 crores = 5 crores

b. Maharashtra and Uttarakhand are the most populated states.

c. Nagaland is the least populated state.

5. Ascending order:

59, 61, 61, 62, 62, 65, 67, 67, 70, 70, 71, 71, 71, 71, 72, 72, 72, 72, 74, 74, 75, 77, 77, 77, 78, 79, 80, 80, 80

Class Interval Frequency Tally Marks

56–60 1 |

61–65 5 ||||

66–70 4 ||||

71–75 11 |||| |||| |

76–80 8 |||| |||

Converting this data again into the exclusive form.

Class Interval Frequency Tally Marks

55–60 1 |

60–65 4 ||||

65–70 3 |||

70–75 12 |||| |||| ||

75–80 9 |||| ||||

6. a. Units of Product A sold = 20% of 5,20,000

= 20 ×5,20,000=1,04,000

100 = 1,04,000

b. Units of Product C sold

= 16% of 5,20,000

= 16 ×5,20,000=83,200

100 = 83,200

c. Units of Product B sold

= 12% of 5,20,000

= 16 ×5,20,000=83,200 100 = 62,400

Units of Product D sold

= 28% of 5,20,000

= 28 ×5,20,000=1,45,600 100 = 1,45,600

Units of Product D − Units of Product B

= 1,45,600 − 62,400 = 83,200

d. Units of Product A : Units of Product C = 104000 : 83200 = 5:4

7. a. Number of students passing from college B = 360

b. If the pass percentage of college A is the same as the pass percentage of college E, then the ratio of the pass percentage of college A and the pass percentage of college E to the total number of students in college A and the total number of students in college E is the same.Percentage of the total strength of E that is greater than the total strength of A

= 400–320 80 ×100%=×100%=25%

320 320

400–320 80 ×100%=×100%=25%

320 320 = 25%

c. Let the number of students in college B be x 90% of x = 360

x = 360 ×100=400 90 = 400

Total number of students who failed = 400 360 = 40

d. Total number of students who passed from all the colleges

= 320 + 360 + 280 + 360 + 400 = 1720

Let the number of students who failed be x 1720 : x = 43 : 9

x = 1720 ×9 =360 43 = 360

9. a. Cars sold by Dealer A = 600 + 500 + 800 + 800 + 700 = 3400

b. Cars sold by Dealer B = 700 + 600 + 700 + 900 + 700 = 3600

c. Number of cars sold in 2019 = 500 + 600 = 1100

Number of cars sold in 2022 = 700 +700 = 1400

Required ratio = 1100:1400 = 11:14

d. Cars sold by Dealer A in 2018 = 600

Total cars sold by Dealer A = 600 + 500 + 800 + 800 + 700 = 3400

Percentage = 600 ×100%=17.65% 3400

10. a. Total candidates appearing for the test in 2021 = 30 + 25 + 25 + 26 + 31 + 25 = 1,62,000

b. Total candidates appearing for the test in 2022 = 31 + 30 + 29 + 30 + 34 + 29 = 1,83,000

c. Number of candidates appearing for the test in Mumbai in 2021 = 30,000

Number of candidates appearing for the test in Bangalore in 2021 = 26,000

Required ratio = 30,000:26,000 = 15:13

d. Average number of candidates appearing in each city in 2021

= 1,62,000 =27,000 6

e. Answersmayvary.Sampleanswers:

i. What is the average number of candidates appearing for the entrance exam for the year 2022?

ii. What is the ratio of the number of candidates appearing for the entrance exam from Delhi and Bhopal in the year 2022?

Scale: 1 division = 5 units

Class Interval

12. a. Number of students taller than 140 cm

= 20 + 25 + 20 + 15 + 15 + 10 = 105 students

b. Number of students shorter than 165 cm

= 30 + 25 + 30 + 35 + 20 + 25 + 20 + 15 + 15 = 215 students

c. Number of students taller than 155 cm

= 15 + 15 + 10 = 40 students

d. Number of students shorter than 150 cm

= 30 + 25 + 30 + 35 + 20 + 25 = 165 students

Number of students taller than 150 cm

= 20 + 15 + 15 + 10 = 60 students

Required ratio = 165:60 = 33:12

a. Total number of goals scored by Messi

= 8 + 17 + 16 + 38 + 47 + 53 + 73 + 60 + 41 = 353

Total number of goals scored by Ronaldo

= 12 + 23 + 42 + 26 + 33 + 53 + 60 + 55 + 51 = 355

b. Ronaldo scored more goals over the years

c. Number of goals scored by Messi before 2010

= 8 + 17 + 16 + 38 + 47 = 126

Number of goals scored by Ronaldo after 2007

= 42 + 26 + 33 + 53 + 60 + 55 + 51 = 320

Required ratio = 126:320 = 63:160

Challenge

1. Assertion: 70.72% players scored more than 25 runs.

Reason: 127 out of 222 players scored more than 30 runs

Number of players with an average of more than 25 = 30 + 45

+ 27 + 40 + 15 = 157

Required percentage = × 157 100% 222 × 100% = 70.72%

The number of players who scored more than 30 runs = 45 + 27 + 40 + 15

Total number of players

= 20 + 15 + 30 + 30 + 45 + 27 + 40 + 15 = 222

Hence, both Assertion and Reason are true.

2. Statement 1: The average sales in the initial 4 months = $3.5 million

It is given in statement 1 that the average sales in the initial 4 months = $3.5 million

Total sales in the initial 4 months = 4 × $3.5 million = $14 million

This doesn’t help us in finding the sale of the 4th month. Statement 2: The average sales in the last 3 months is $6 million.

It is given in Statement 1 that the average sales in the last months = $6 million

Total sales in the initial 3 months = 3 × $6 million = $18 million

This doesn’t help us in finding the sale of the 4th month.

From both statements, we can conclude that, total sales in initial 4 months + total sales in last 3 months = total sales in 6 months + sales in the 4th month (as 4th month is common in both the statements)

Therefore, $14 million + $18 million = $27 million + sales in the 4th month

⇒ $32 million = $27 million + sales in the 4th month

⇒ sales in 4th month = $32 million − $27 million = $5 million

Hence, both statements 1 and 2 are sufficient to answer. Thus, option c is correct.

Case Study

1. Cost of sending 150 g parcel locally = ₹30

Cost of sending 800 g parcel to 180 km = ₹77

Cost of sending 1600 g parcel to 500 km = ₹177

Total cost paid = ₹284

Thus, option b is correct.

2. Cost of sending 350 g parcel locally = ₹35

Cost of sending 350 g parcel upto 200 km = ₹59

Percentage increase = 59352410010068.6%

Thus, option c is correct.

3. Cost of sending one 250 g parcel upto 200 km = ₹59

Cost of sending 10 such parcels = 10 × ₹59 = ₹590

The cost of sending one 250 g parcel between 201 to 1000 km = ₹71

Cost of sending 10 such parcels = 10 × ₹71 = ₹710

The total increases by ₹710 − ₹590 = ₹120.

4. Charges of sending 40 grams parcel locally = ₹18

Charges of sending 130 grams parcel locally = ₹30

Charges of sending both parcels separately = ₹18 + ₹30 = ₹48

Combined weight of the parcel = 40 + 130 = 210 grams

Charges of sending 210 grams parcel locally = ₹35

Hence, Mayra should send both items as one parcel.

Chapter 5 Let’s Warm-up

1. The fraction of students who like yellow colour is 3 20

2. The most liked colour is blue

3. The fraction of students who do not like red colour is 3 4 .

4. The least liked colour is lavender

5. The difference in the fraction of students who like pink colour to the ones who like blue colour is 1 5

Do It Yourself 5A

1. Central angle = People who know more than three languages Total no. of people × 360° = 375 1200 × 360° = 112.5°

2. Central angle = 45 100 × 360° = 162°

The central angle representing 45% of people liking biryani is 162°.

3. The pie chart for the table can be given as:

4. The pie chart for the table can be given as:

States No. of Workers

Karnataka 440

Tamil Nadu

7. The pie chart for the table can be given as:

No. of Workers from Various States Working in a Garment Factory

Andhra Pradesh, 340

Tamil Nadu, 620 Kerala, 300 Others, 100

Karnataka, 440

5. The pie chart for the table can be given as:

(in billions)

Import Items

Plastics, ₹180

Fuels and oil, ₹240

6. The pie chart for the table can be given as: Clothing Brand % of Buyers

Number of Lions

8. Answermayvary.Sampleanswer:

Below are the details of tigers in 5 national parks of India:

Jim Corbett National Park – 260 Tigers

Bandipur National Park – 150 Tigers

Nagarhole National Park – 141 Tigers

Bandhavgarh National Park – 135 Tigers

Dudhwa National Park – 135 Tigers

National Park Number of Tigers Central Angle

Jim Corbett National Park

Bandipur National Park

Nagarhole

Bandhavgarh

Jim Corbett National Park

1. Ratios of the sports given is

Football : Cricket = 2 : 3

Cricket : Badminton = 3 : 1

Football : Volleyball = 4 : 5

Volleyball : Basketball = 5 : 3

Ratios of all the sports together is:

Football : Cricket : Badminton : Volleyball : Basketball = 4 : 6 : 2 : 5 : 3

Sum of all the ratios = 4 + 6 + 2 + 5 + 3 = 20 Sports Budget Allocated

Do It Yourself 5B

1. Students admitted to Education = 8 ×5,00,000= 100 40,000

Students admitted to Arts = 20 ×5,00,000= 100 1,00,000

Students admitted to Law = 14 ×5,00,000= 100 70,000

Students admitted to Commerce = 33 ×5,00,000= 100 1,65,000

Students admitted to Science = 25 ×5,00,000= 100 1,25,000

2. a. Shared cabs show the second highest central angle in the pie chart.

Thus, shared cabs are the second most popular mode of transport.

b. Central angle associated with bus = 45°

Fraction of people preferring the bus = 45 1251 0.125

360 1000 8 ° === °

c. Since a quarter of 360° is 90°, shared cab is preferred by a quarter of the people.

3. a. Given that, the number of Type C cars sold were 500, that is 25%.

Therefore, the total number of cars sold = 500 ×100 25 = 2000.

b. Percentage of Type D cars = 18

Number of Type D cars = 18 100 × 2000 = 360.

c. Percentage of Type A and Type B cars together = 35 + 22 = 57

Number of Type A and Type B cars together

= 57 100 × 2000 = 1140

4. Total students = 2500

a. Central angle associated with orange juice = 86.4°

Number of students who like orange juice = 86.4° 360° × 2500 = 600

b. Apple juice is the least popular juice.

Number of students who like it = 36° 360° × 2500 = 250

c. Central angle for students not liking apple juice

= (360° − 36°) = 324°

Fraction = 324° 360° × 0.9 = 9 10

5. a. Given that the number of elephants is 11% = 77

Total number of animals in the zoo = 77×100 =700 11

b. Tigers are the lowest in number.

Number of tigers = 5 100 × 700 = 35

c. The difference in the percent of deer and reptiles = (28% − 25%) = 3%

Difference in number = 3 100 × 700 = 21

6. Given that teenagers who like classical music is 17% = 255

a. Number of teenagers surveyed = 255 × 100 17 = 1500

b. Number of teenagers who like rock music = 30% × 30 100

Number of teenagers who like hip hop = 34% = 34 100

Ratio of teenagers liking rock music to hip hop

= 30 100 : 34 100 = 30 : 34 = 15 : 17

7. Total plants = 1200

a. Number of sunflower plants = 82.5° 360° × 1200 = 275

Number of tulip plants = 60° 360° × 1200 = 200

Difference between number of sunflower plants and tulip plants = 75.

b. Central angle for rose plants sown = 112.5°

Faction = 112.5° =0.3125==31255 360° 10,00016

c. Number of marigold plants = 67.5° 360° × 1200 = 225

Number of jasmine plants = 37.5° 360° × 1200 = 125

Ratio of marigold to jasmine = 225 : 125 = 9 : 5

8. a. The most consumed form of energy is coal = 52%

b. The ratio of consumption of Nuclear to the consumption of Renewable energy is 2 100 : 22 100 = 2 : 22 = 1 : 11

c. The fraction of the energy consumed by Hydropower and Natural gas is 13 100 + 6 100 = 19 100

9. Answersmayvary.Sampleanswer:

a. What is the ratio of consumption of Renewable source of energy to the consumption of coal?

b. Which is the least consumed form of energy?

Challenge

1. a. Central angle for glass used in July = 26 100 × 360° = 93.6°

The difference in the value of the central angle made by the glass in both months = 93.6° 72° = 21.6°

b. Paper collected in June = 93.6° 360° × 60,000 = 15,600 tonnes

Paper collected in July = 22 100 × 60,000 = 13,200 tonnes

Total paper collected = 15,600 + 13,200 = 28,800 tonnes

c. Plastic collected in June = 126° 360° × 60,000 = 21,000 tonnes

Plastic collected in July = 32 100 × 60,000 = 19,200 tonnes

In June 21,000 − 19,200 = 1800 tonnes of more plastic was collected.

d. Percentage of cardboard collection in June = 43.2° 360° × 100% = 12%

Given, percentage of cardboard collection in July = 15%

Percentage change (increase) = 15 12 12 × 100% = 25%

Chapter Checkup

1. Given that the percentage of people preferring fresh juice = 23% Let the central angle be x; 23 = x 360° × 100 23 × 360°

100 = x x = 82.8°

2. a. Total items sold = 360

The table for the pie chart can be given as:

Items Sold Others, 36

b. Total items sold = 360 The table for the pie chart can be given as:

Number of Items Sold Central Angle

Sold

Sandwiches, 80 Burgers, 72

3. Total number of families = 80

The pie chart for the table can be given as:

No. of Families

No. of Families Owning Pets

4. Total number of students = 1800

The pie chart for the table can be given as:

Subject No. of Students Central Angle

Maths 595 595 ×360°=119° 1800

English 420

French 225

Science 385

420 ×360°= 1800

225 ×360°= 1800

385 ×360°= 1800

Others 175 175 ×360°=35° 1800 84° 45° 77° 35° 119° Others, 175

French, 225 English, 420 Maths, 595 Science, 385 No. of Students

5. Total number of people in the town = 2400

The pie chart for the table can be given as:

Type of Books No. of People Central Angle

Fantasy 280 280 ×360°=42° 2400

Adventure 600

600 ×360°= 2400

Biography 440 66° 440 ×360°= 2400

Novels 320 320 ×360°=48° 2400

Comedy 760 114° 760 ×360°= 2400

Adventure, 600 Comedy, 760 Books Read Biography, 440 Fantasy, 280 Novels, 320

6. a. The notebooks show the highest central angle in the pie chart.

Thus, notebooks are the most popular item in the store.

b. Central angle for the notebooks = 90°

Fraction = 90° 360° = 1 4

c. Central angle for party supplies = 54.3°

Percentage of party supplies sold

7. Let the number of ice creams sold in 5 days = x

Ice creams sold on Wednesday = 22% = 198

According to the question, 22% of x =198

x = 198 22 × 100 = 900

The total number of ice creams sold in 5 days = 900

8. a. The highest food producing sector in India is agriculture.

b. The central angle for livestock = 25 100 × 360° = 90°

c. The ratio of the food produced by the dairy sector and the food produced by fisheries = 10 100 : 5 100 = 10 : 5 = 2 : 1

9. a. Percentage of sales of household items = 72° ×100%=20% 360°

b. The most popular product is produce.

Sale = 108° ×1,00,000=30,000 360° `

c. The two least popular products are dairy and beverages.

Total central angle = 36° + 54° = 90°

Percentage sales = 90° 360° × 100 = 25%

10. a. Given that the number of people who prefer aerobics = 150

Let the total number of people be x; according to the question,

36° 360° of x 150

Therefore, x = 1500

Given, central angle for running = 64.8°

Number of people who prefer running

= 64.8°

360° × 1500 = 270

Percentage = 270 1500 × 100% = 18%

b. Number of people who prefer cycling = 43.2° 360° × 1500 = 180

Number of people who do not prefer to gym

= 360° − 57.6° 360° × 1500 = 1260

Ratio = 180 : 1260 = 1 : 7

c. Central angle of people who do not prefer yoga and meditation = (360° − 115.2°) = 244.8°

Percentage = 244.8° 360° × 100% = 68%

11. Answermayvary.Sampleanswer:

What was the total percentage of the people who prefer gym and cycling?

What is the value of most preferred workout?

Challenge

1. Statement 1 states the number of favourite sports. This does not helps us create a pie chart.

Statement 2 states the percentage breakdown of different sports.

When we add all the percentages, 25%, 25%, 20%, 15% and 15%, we get 100%.

So, this also states that there are five games as it completes 100%.

From this data, we can make a pie chart. So, Statement 2 alone is sufficient to answer the question. Thus, option b is correct.

2. Assertion (A) says that a country spent ₹3,60,00,000 on sports, and the money spent on Basketball is ₹5,00,000 more than on Tennis.

This is based on information from a pie chart showing how much money was spent on each sport. Reason (R) explains how to calculate the money spent on each sport using a pie chart.

The pie chart divides the total spending into sectors, based on central angles.

One can calculate the money for each sport by using this formula:

Amount spent on a sport= Central angle of the sport 360 × Total amount spent

This formula tells us how much money was spent on each sport by using the size of the angles in the pie chart. So, both A and R are true. Thus, option a is correct.

Case Study

1. Patients were aged between 18 years and 60 years = (40 + 30) 100 × 2,00,000 = 70 100 × 2,00,000 = 1,40,000 Option c is correct.

2. Number of patients above 60 years = 20 100 × 2,00,000 = 40,000

Number of patients above 60 years having moderate symptoms = 25 100 × 40,000 =10,000

3. Number of patients aged less than 18 years = 10 100 × 2,00,000 = 20,000

Number of patients who had severe symptoms and were aged less than 18 years = 15 100 × 20,000 = 3000

4. Number of COVID-19 patients between 18 years and 60 years = 1,40,000

Number of COVID-19 patients between 18 years and 60 years with mid symptoms = 60 100 × 1,40,000 = 84,000

Chapter 6 Let’s Warm-up

1. A dog flying. Impossible

2. The cycle of day and night. Sure

3. Getting heads when a coin is tossed. Equally likely

4. Drawing a blue marble from a bag containing 6 blue and 2 green marbles. Likely

5. Drawing a green marble from a bag containing 10 yellow and 3 green marbles. Unlikely

Do It Yourself 6A

1. a. The probability of an event having no chance of occurring is 0.

b. The probability of a sure event is 1

c. The possible outcomes when you toss a coin are heads and tails

d. On rolling a dice, the event of getting a prime number has 3 possible outcomes.

2. Total number of marbles in the glass jar = 40

P(drawing a single green marble) = 1 5

So, the number of green marbles in the glass jar = 4 5 × 40 = 8

Hence, option b is correct.

3. There are 26 letters in the English alphabet.

Total number of possible outcomes = 26

Number of favourable outcomes for the event of choosing any letter except Z = 25

Therefore, the theoretical probability of randomly choosing any letter except Z = 25 26

4. a. The possible outcomes when a coin is tossed are heads and tails.

Total possible outcomes of the experiment = 2 (either heads or tails)

Number of tails1

()= = Total number of outcomes2 PE

Therefore, the theoretical probability of getting a tail is 1 2

b. Number of contestants who play the game = 250

Number of contestants expected to win = probability of getting tails × number of contestants who play the game

= 1 2 × 250 = 125

So, about 125 contestants are expected to win.

5. Possible outcomes on rolling a six-sided dice are 1, 2, 3, 4, 5 and 6.

So, the total number of possible outcomes = 6

a. The event of rolling a 4 has only one favourable outcome.

P(rolling a 4) = 1 6

b. The event of rolling an odd number has 3 favourable outcomes (1, 3 and 5).

P(rolling an odd number) = 31 = 62

c. The event of rolling a number greater than 4 has 2 favourable outcomes (5 and 6).

P(rolling a number greater than 4) = 2 6 = 1 3

Number of times the dice is rolled by Amit = 12

Number of times, Amit would expect a number greater than 4 to be rolled = 12 × 1 3 = 4

6. P(pointer landing in the green section) = 4 10

P(pointer landing in the blue section) = 3 10

P(pointer landing in the yellow section) = 2 10

P(pointer landing in the black section) = 1 10

Sum of P(pointer landing in the green, blue, yellow and black sections) = 4 10 + 3 10 + 2 10 + 1 10 = 4 + 3 + 2 + 1 10 = 10 10 = 1

7. Possible outcomes = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

So, the total number of possible outcomes = 12

a. The event of rolling a number greater than 10 has two favourable outcomes (11 and 12).

P(rolling a number greater than 10) = 21 = 126

b. The event of rolling a number less than 5 has four favourable outcomes (1, 2, 3 and 4).

P(rolling a number less than 5) = 41 = 123

c. The event of rolling either a 4, 6 or 9 has three favourable outcomes.

P(rolling either a 4, 6 or 9) = 31 = 124

If the 12-sided solid is rolled 200 times, the number of times a 4, 6 or 9 can be expected = 1 × 200 = 50 4

So, we can expect either a 4, 6 or 9 to be rolled about 50 times in 200 trials.

8. Answersmayvary.Sampleanswers:

a. What is the probability of getting a prime number?

The event of rolling a prime number has five favourable outcomes (2, 3, 5, 7 and 11).

P(rolling a prime number) = 5 12

b. What is the probability of getting a number less than 7?

The event of rolling a number less than 7 has five favourable outcomes (1,2, 3, 4, 5 and 6).

P(rolling a prime number) = 6 12 = 1 2

9. There are 36 possible outcomes.

a. The event of rolling a sum of 6 on two dice has five favourable outcomes. (1 + 5, 2 + 4, 3 + 3, 4 + 2, 5 + 1).

P(rolling a sum of 6 on two dice) = 5 36

b. The event of rolling a sum of 10 on two dice has three favourable outcomes. (4 + 6, 5 + 5, 6 + 4).

P(rolling a sum of 10 on two dice) = 31 = 3612

10.

Total numbers given = 7 Number of 4’s in the spinner = 2 P(getting an 4) = 2 8 = 1

Challenge

1 a. We can make an organised list to find the sample space.

collared shirt − khakis − sneakers; collared shirt − khakis − flip-flops

collared shirt − khakis − sandals; collared shirt − jeans − sneakers

collared shirt − jeans − flip-flops; collared shirt − jeans − sandals

collared shirt − capris − sneakers; collared shirt − capris − flip-flops

collared shirt − capris − sandals;

T-shirt − khakis − sneakers

T-shirt − khakis − flip-flops; T-shirt − khakis − sandals

T-shirt − jeans − sneakers; T-shirt − jeans − flip-flops

T-shirt − jeans − sandals; T-shirt − capris − sneakers

T-shirt − capris − flip-flops; T-shirt − capris − sandals

So, 18 different outfits are possible.

b. The event of Navin wearing his jeans but not his sneakers has four favourable outcomes: collared shirt − jeans − flip-flops, collared shirt − jeans − sandals, T-shirt − jeans − flip-flops, T-shirt − jeans − sandals

Hence, the probability that Navin will wear his jeans but not his sneakers = 42 = 189

Do It Yourself 6B

1. a. Tossing a coin is an example of a random experiment.

b. Drawing a black card from a deck of playing cards is an example of a random experiment.

c. Throwing a ball is not an example of a random experiment.

d. Rolling two dice together is an example of a random experiment.

2. Experimental probability,

P(E) = Number of times Leena wins

Total number of trials = 8 48 = 1 6

Therefore, the experimental probability of winning this game is 1 6 or 0.16 or 16.67%.

3. a. The total number of songs played in a 24-hour period = 60 + 75 = 135

P(E) = Number of times a rap song is played

Total number of song played = 75 135 = 5 9 = 0.55 or 55.55%

b. In 135 trials, a pop song was played 60 times.

In the next 90 trials, a pop song is expected to be played, 60 90 × = 40 135 times

4. The total number of trials in the experiment = 50

P(picking a black ball from Basket B) > P(picking a black ball from Basket A)

Hence, I should choose Basket B for a better chance of winning.

303 () = = 505 Pvowel

The number of vowels in the bag full of 105 lettered tiles is expected to be = 3 105 × 5 = 63

5. a. This is the part where the circles overlap. 10 students play both tennis and squash.

P(bothtennisandsquash) = 10 90 or 11.11%

b. This is the part outside of both circles.

53 students play neither tennis nor squash.

P(neithertennisnorsquash) = 53 90 or 58.88%

c. These are both parts of the tennis circle.

25 + 10 = 35

35 students play tennis. 35 () = 90 Ptennis or 38.88%

d. This is the part inside either circle but not in both.

25 + 2 = 27

27 students play tennis or squash but not both.

P(tennisorsquashbutnotboth) = 27 90 or 30%

6. a. 61 students own a mobile phone.

Probability that a student owns a mobile phone

= 61 125 or 48.8%

b. 64 students do not own a mobile phone.

Probability that a student does not own a mobile phone

= 64 125 or 51.2%

c. There are 29 boys who own a mobile phone.

Probability of a boy who owns a mobile phone

= 29 125 or 23.2%

d. There are 25 girls who do not own a mobile phone.

Probability of a girl who does not own a mobile phone = 25 125 or 20%

7. a. Experimental probability,

P(E) = Number of defective pairs

Total number of pairs of jeans = 5 200 = 1 40

In 5000 trials, the number of jeans expected to be defective would be, 5000 × 1 40 = 125

So, 125 jeans are expected to be defective in a shipment of 5000.

Challenge

1. Total number of outcomes = 100

Number of outcomes of getting two tails is = 22

Experimental probability of an event

= Number of times the event occurs

Total number of trial

The experimental probability of getting two tails is = 22 100 = 11 50

So, the Assertion is true.

Experimental probability of an event

= Number of times the event occurs

Total number of trial

So, the reason is the correct explanation of Assertion. Therefore, both A and R are true, and R is the correct explanation of A.

The correct option is a.

Chapter Checkup

1. a. The probability of an event that is sure to happen is 1

b. If an event cannot occur, then its probability is 0

c. The probability of selecting P from the word ʽSPECIALʹ is 1 7

d. The probability of an event can’t be more than 1.

e. If a dice is rolled once, the probability of getting an even prime number is 1 6

2. a. The possible outcomes of flipping a fair coin are heads and tails.

b. The possible outcomes of rolling a standard six-sided dice are 1, 2, 3, 4, 5 and 6.

c. The possible outcomes of selecting a single letter at random from the English alphabet are 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 and z.

d. The possible outcomes of flipping two fair coins simultaneously are (Heads, Heads), (Heads, Tails), (Tails, Heads), (Tails, Tails).

e. The possible outcomes of rolling two standard six-sided dice simultaneously are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

3. a.

Number of cards that shows a triangle1 () = = Total number of outcomes8 PE

Therefore, the probability of getting a card that shows a triangle is 1 8 or 0.125 or 12.5%.

b. The cards that show a shape with four sides include a square, rectangle and rhombus.

Number of cards that shows a quadrilateral3 () = = Total number of outcomes 8 PE

Therefore, the probability of getting a card that shows a quadrilateral is 3 8 or 0.375 or 37.5%.

c. The cards that show a shape with more than four sides include a pentagon, an octagon and a hexagon.

Number of cards that show a shape with more than four sides 3 () = = Total number of outcomes8 PE

Therefore, the probability of getting a card that shows a shape with more than four sides is 3 8 or 0.375 or 37.5%.

d. The card that shows a shape with no sides is that of a circle.

Number of cards that shows a circle1 () = = Total number of outcomes8 PE

Therefore, the probability of getting a card that shows a circle is 1 8 or 0.125 or 12.5%.

4. Number of letters in the given word ʽAPPRECIATIONʹ = 12

Number of A’s in the given word = 2

P(getting an A) = 2 12 = 1 6

5. a. Total number of bolts = 100

Number of rusted bolts = 2 5 × 100 = 200 5 = 40

P(Rusted bolt) = 40 100 = 2 5

b. Total number of bolts = 100

Number of non-rusted bolts = 100 – 40 = 60

P(Rusted bolt) = 60 100 = 3 5

6. a. Number of apples packed in a carton = 200;

Number of rotten apples = 30

Number of good apples = 200 − 30 = 170;

P(good apple) = 170 200 = 17 20 or 85%

b. P(rotten apple) = 30 200 = 3 20 or 15%

7. Number of favourable outcomes for the event = 2

Total number of streams = 4

P(Mathematics as a subject)

= Number of favourable outcomes

Total number of outcomes = 2 4 = 1 2

8. a. Theoretical probability

P(green) = Number of favourable outcomes

Number of possible outcomes = 1 5

Experimental probability

P(green) = Number of times green colour was spun

Total number of spins

= 15

150 = 1 10

1 5 > 1 10, theoretical probability is greater in this case.

b. Theoretical probability

P(red) = Number of favourable outcomes

Number of possible outcomes = 1 5

Experimental probability

P(red) = Number of times a red colour was spun

Total number of spins

= 35

150 = 7 30

7 30 > 1 5, experimental probability is greater in this case.

c. Theoretical probability

P(blueofpurple) = Number of favourable outcomes

Number of possible outcomes = 2 5

Experimental probability

P(blueofpurple) =

Number of times a blue or purple colour was spun

Total number of spins

= 25 + 42

150 = 67 150

150 > 2 5, experimental probability is greater in this case.

d. All the five sections of the spinner are coloured.

Theoretical probability

P(any colour) = Number of favourable outcomes

Number of possible outcomes = 5 5 = 1

Experimental probability

P(colour) = Number of times any colour was spun

Total number of spins

= 25 + 15 + 33 + 35 + 42 150 = 150 150 = 1

So, the theoretical and experimental probability of the wheel stopping at a coloured section is the same that is 1.

9. There are 9 prime numbers between 1 and 25. They are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

So, the total number of possible outcomes is 9.

a. Favourable outcomes for the event of getting a one-digit number are 2, 3, 5 and 7.

So, the number of favourable outcomes = 4

P(one-digit number) 4 9 or 44.44%

b. 2 is the only even prime number.

So, the number of favourable outcomes = 1

P(even number) = 1 9 or 11.11%

c. All prime numbers between 1 and 25 are odd except 2 which is an even prime number.

So, the number of favourable outcomes = 8

P(odd number) = 8 9 or 88.88%

d. Prime numbers between 1 and 25, that are greater than 11 are 13, 17, 19, and 23.

So, the number of favourable outcomes is 4.

P(number greater than 11) = 4 9 or 44.44%

10. Possible outcomes of randomly typing a digit in the keypad of a mobile phone are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

So, the total number of possible outcomes are 10.

a. There are five even numbers from 0 to 9. They are 0, 2, 4, 6 and 8.

So, the number of favourable outcomes is 5.

P(typing an even number) = 5 10 = 1 2 or 50%

b. There is only 1 multiple of 5 from 0 to 9, that is 5.

So, the number of favourable outcomes is 1.

P(typing a multiple of 5) = 1 10 or 10%

c. Factors of 9 are 1, 3 and 9.

So, the number of favourable outcomes is 3.

P(typing a factor of 9) = 3 10 or 30%

11. Total number of bangles on display = 50

P(Customer choosing pair of pearl bangles) = 6 25

Number of pearl bangles on display = 50 × 6 25 = 12.

Challenge

1. Amita can pay the cost of the pencil using either ₹5 coin or ₹10 coin.

P(using a ₹5 coin) = 1 2

Percentage of using a ₹5 coin is = 1 2 × 100 = 50%.

So, Ranita is correct.

2. Number of blue marbles = 50

Total number of marbles = 300

The probability of drawing a blue marble is

= Number of blue marbles

Total number of marbles = 50 300 = 1 6

Number of non-blue marbles = Total number of marbles

– Number of blue marbles

= 300 – 50 = 250

The probability of drawing a non-blue marble is

= Number of non - blue marbles

Total number of marbles = 250 300 = 5 6

Percentage of non-blue marbles = 5 6 × 100 = 83.33% So, the correct option is a. Only conclusion I follows.

Case Study

1. Number of birds observed = 100

Number of sparrows spotted = 40

Probability of randomly spotting a sparrow in the park = P(sparrow) = Number of sparrows spotted

Total number of birds observed = 40 100 = 0.4 So, option b is correct.

2. Number Rashmi seeing either a pigeon or a crow = Number of pigeons spotted + Number of crows spotted = 30 + 20 = 50

Probability that Rashmi sees either a pigeon or a crow = Number of pigeons spotted + Number of crows spotted

Total number of birds observed = 50 100 = 0.5

3. Rashmi not seeing a sparrow = Number of birds other than sparrow = Total number of birds observed – Number of sparrows spotted = 100 – 40 = 60

The probability of not seeing a sparrow is = Number of birds other than sparrow

Total number of birds observed = 60 100 = 0.6

4. Number of pigeons spotted = 30

Probability of pigeon = Number of pigeons spotted

Total number of birds observed = 30 100 = 0.3

Number of pigeons spotted = 20

Probability of crow = Number of crows spotted

Total number of birds observed = 20 100 = 0.2

Probability of pigeon is more than the probability of crow. So, the statement is true.

5. Answersmayvary.

Chapter 7

Let's Warm-up

1. Among 3, 7, 8, 9, 11 and 12, the numbers 8, 11 and 12 are the only composite numbers. False

2. The prime factors of 48 are 2 and 3 only. True

3. The prime factorisation of 120 is 2 × 2 × 2 × 2 × 3 × 3 × 5. False

4. The prime factorisation of 132 is 2 × 2 × 3 × 11. True

5. Prime factorising a number using the factor tree method gives a different result than using the division method of prime factorisation. False

Do It Yourself 7A

1. a. 961

i. 1 because the unit digit of the number is 1. ii. Odd square because the number is odd.

b. 4624

i. 6 because the unit digit of the number is 4.

ii. Even square because the number is even.

c. 5329

i. 1 because the unit digit of the number is 9.

ii. Odd square because the number is odd.

d. 7688

i. 4 because the unit digit of the number is 8. ii. Even square because the number is even.

e. 19,753

i. 9 because the unit digit of the number is 3.

ii. Odd square because the number is odd.

2. a. 52 = 5 × 5 = 25

b. 72 = 7 × 7 = 49

3. a. (17)2 = 17 × 17 = 289

b. (41)2 = 41 × 41 = 1681

c. (−43)2 = (−43) × (−43) = 1849

d.    2 7 7749 =×= 8864 8

e. 2 9 –121212144 =×= 13131 –36 –1

4. a. 492 = (50 − 1)2

= (50 − 1)(50 − 1)

= 50(50 − 1) − 1(50 − 1) = 502 − 50 × 1 − 1 × 50 + 12

= 2500 − 50 − 50 + 1 = 2401

b. 592 = (50 + 9)2

= (50 + 9)(50 + 9)

= 50(50 + 9) + 9(50 + 9) = 502 + 50 × 9 + 9 × 50 + 92

= 2500 + 450 + 450 + 81 = 3481

c. 982 = (100 − 2)2

= (100 − 2)(100 − 2)

= 100(100 − 2) − 2(100 − 2) = 1002 − 100 × 2 − 2 × 100 + 22

= 10000 − 200 − 200 + 4 = 9604

d. 1052 = (100 + 5)2

= (100 + 5)(100 + 5)

= 100(100 + 5) + 5(100 + 5)

= 1002 + 100 × 5 + 5 × 100 + 52

= 10000 + 500 + 500 + 25 = 11,025

e. 1962 = (200 − 4)2

= (200 − 4)(200 − 4) = 200(200 − 4) − (200 − 4) = 2002 − 200 × 4 − 4 × 200 + 42 = 40000 − 800 − 800 + 16 = 38,416

5. a. 676 = 262 = sum of first 26 odd numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 +

25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 +

47 + 49 + 51

b. 729 = 272 = sum of first 27 odd numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 +

25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 +

47 + 49 + 51 + 53

c. 289 = 172 = sum of first 17 odd numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 +

25 + 27 + 29 + 31 + 33

d. 900 = 302 = sum of first 30 odd numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 +

27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 +

51 + 53 + 55 + 57 + 59

e. 1225 = 352 = sum of first 35 odd numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 +

25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55 + 57 + 59 + 61 + 63 + 65 + 67 + 69

6. The formula to find the nth triangular number, Tn is n n(n+1) T = 2

1 1(1+1) T = 2 = 1

2 2(2+1) T = 2 = 3

3 3(3+1) T6 == 2 4 4(4+1) T 10 == 2

5 5(5+1) T = 15 = 2 6 6(6+1) T 21 == 2 7 7(7+1) T 28 == 2 8 8(8+1) T 36 == 2 9 9(9+1) T 45 == 2 10 10(10+1) T 55 == 2

Triangular numbers from 20 to 50 = 21, 28, 36, 45

The sum of two consecutive triangular numbers is a square number such as:

21 + 28 = 49 = 72, 28 + 36 = 64 = 82, 36 + 45 = 81 = 92

7. a. 132 − 122 = 2 � 12 + 1 = 25

There are 2 n non-perfect square numbers between the squares of two consecutive numbers.

Here, n = 12 and n + 1 = 13

So, there are 2 × 12 = 24 non-perfect squares.

b. 212 − 202 = 2 � 20 + 1 = 41

There are 2 n non-perfect square numbers between the squares of two consecutive numbers.

Here, n = 20 and n + 1 = 21

So, there are 2 × 20 = 40 non-perfect squares.

c. 182 − 172 = 2 � 17 + 1 = 35

There are 2 n non-perfect square numbers between the squares of two consecutive numbers.

Here, n = 17 and n + 1 = 18

So, there are 2 × 17 = 34 non-perfect squares.

d. 262 − 252 = 2 � 25 + 1 = 51

There are 2 n non-perfect square numbers between the squares of two consecutive numbers.

Here, n = 25 and n + 1 = 26

So, there are 2 × 25 = 50 non-perfect squares.

e. 322 − 312 = 2 � 31 + 1 = 63

There are 2 n non-perfect square numbers between the squares of two consecutive numbers.

Here, n = 31 and n + 1 = 32

So, there are 2 × 31 = 62 non-perfect squares.

8. We know that if n is an odd number, then n2 = 22 (–1)(+1) + 22 nn

( ) ( ) 22 2 57157+1 5 + 2 –7 = 2

324913249+1 =+ 2 –2

32483250 =++=16241625 22

b. ( ) ( ) 22 2 63163+1 6 + 2 –3 = 2

396913961+1 =+ 2 –2

39683970 =+=+19841985 22

c. ( ) ( ) 22 2 77177+1 7 + 2 –7 = 2

592915929+1 =+ 2 –2 59285930 =+=+29642965 22

d. ( ) ( ) 22 2 39139+1 3 + 2 –9 = 2 152111521+1 =+ 2 –2 15201522 =+=+760761 22

e. ( ) ( ) 22 2 85185+1 8 + 2 –5 = 2

722517225+1 =+ 2 –2

72247226 =+=+36123613 22

9. a. 752 = [7 (7 + 1)] × 100 + 25

= 7 × 8 × 100 + 25 = 5625

b. 512 − 502 = (51 + 50) (51 − 50) = 101 × 1 = 101

c. 952 = [9 (9 + 1)] × 100 + 25

= 9 × 10 × 100 + 25 = 9025

d. 782 − 772 = (78 + 77) (78 − 77) = 155 × 1 = 155

e. 1052 = [10 (10 + 1)] × 100 + 25

= 10 × 11 × 100 + 25 = 11,025

10. a. 75 × 77 = (76 − 1) (76 + 1) = 762 − 12 = 5776 − 1 = 5775

b. 64 × 68 = (66 − 2) (66 + 2) = 662 − 22 = 4356 − 4 = 4352

c. 50 × 52 = (51 − 1) (51 + 1) = 512 − 12 = 2601 − 1 = 2600

d. 91 × 95 = (93 − 2) (93 + 2) = 932 − 22 = 8649 − 4 = 8645

e. 36 × 40 = (38 − 2) (38 + 2) = 382 − 22 = 1444 − 4 = 1440

11. The members of a Pythagorean triplet are given by 2m, m2 − 1 and m2 + 1.

a. If 2m = 12, then m = 6.

So, m2 − 1 = 62 − 1 = 36 − 1 = 35 and m2 + 1 = 62 + 1 = 36 + 1 = 37

Thus, the Pythagorean triplet in this case is given by (12, 35, 37)

If m2 − 1 = 12, then m2 = 13 ⇒ m is not an integer. If m2 + 1 = 12, then m2 = 11 ⇒ m is not an integer.

In either case, values of m are not possible.

So, the only Pythagorean triplet here is (12, 35, 37) with 12 as its smallest member.

b. If 2m = 16, then m = 8.

So, m2 − 1 = 82 − 1 = 64 − 1 = 63 and m2 + 1 = 82 + 1 = 64 + 1 = 65

Thus, the Pythagorean triplet in this case is given by (16, 63, 65)

If m2 − 1 = 16, then m2 = 17 ⇒ m is not an integer. If m2 + 1 = 16, then m2 = 15 ⇒ m is not an integer. In either case, values of m are not possible.

So, the only Pythagorean triplet here is (16, 63, 65) with 16 as its smallest member.

c. If 2m = 28, then m = 14.

So, m2 − 1 = 142 − 1 = 196 − 1 = 195 and m2 + 1 = 142 + 1 = 196 + 1 = 197

Thus, the Pythagorean triplet in this case is given by (28, 195, 197)

If m2 − 1 = 28, then m2 = 29 ⇒ m is not an integer. If m2 + 1 = 28, then m2 = 27 ⇒ m is not an integer.

In either case, values of m are not possible.

So, the only Pythagorean triplet here is (28, 195, 197) with 28 as its smallest member.

d. If 2m = 32, then m = 16.

So, m2 − 1 = 162 − 1 = 256 − 1

= 255 and m2 + 1 = 162 + 1 = 256 + 1 = 257

Thus, the Pythagorean triplet in this case is given by (16, 255, 257)

If m2 − 1 = 32, then m2 = 33 ⇒ m is not an integer. If m2 + 1 = 32, then m2 = 31 ⇒ m is not an integer.

In either case, values of m are not possible.

So, the only Pythagorean triplet here is (32, 255, 257) with 32 as its smallest member.

e. If 2m = 36, then m = 18.

So, m2 − 1 = 182 − 1 = 324 − 1 = 323 and m2 + 1 = 182 + 1 = 324 + 1 = 325

Thus, the Pythagorean triplet in this case is given by (36, 323, 325)

If m2 − 1 = 36, then m2 = 37 ⇒ m is not an integer. If m2 + 1 = 36, then m2 = 35 ⇒ m is not an integer.

In either case, values of m are not possible.

So, the only Pythagorean triplet here is (36, 323, 325) with 36 as its smallest member.

12. 42 = 16

342 = 1156

3342 = 111556

33342 = 11115556

333342 = 1111155556

3333342 = 111111555556

13. 92 = 81

992 = 9801

9992 = 998001

99992 = 99980001

999992 = 9999800001

9999992 = 999998000001

14. The sum of the first n odd numbers is n2

a. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = sum of first 8 odd numbers = 82 = 64

b. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 = sum of first 17 odd numbers = 172 = 289

c. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 = sum of first 21 odd numbers = 212 = 441

15. a. Pattern 1 = 22 = 4

Pattern 2 = 22 + (2 + 2 + 1) = 4 + 5 = 9 = 32

Pattern 3 = 32 + (3 + 3 + 1) = 9 + 7 = 16 = 42

Pattern n = n2 + (n + n + 1) = (n + 1)2

So, the rule is n2 + (2n + 1) = (n + 1)2

b. Pattern 1 = 22 = 4

Pattern 2 = 22 + (2 + 2 + 1) = 4 + 5 = 9 = 32

Pattern 3 = 32 + (3 + 3 + 1) = 9 + 7 = 16 = 42

Pattern 4 = 42 + (4 + 4 + 1) = 16 + 9 = 25 = 52

Pattern 5 = 52 + (5 + 5 + 1) = 25 + 11 = 36 = 62

Pattern 6 = 62 + (6 + 6 + 1) = 36 + 13 = 49 = 72

c. 562 = (55 + 1)2 = 552 + (2 × 55 + 1) = 3025 + (110 + 1) = 3025 + 111 = 3136

16. Velocity (v) = 12 m per second

Mass (m) = 8000 kg

Then, Kinetic Energy = 1122 =××800012 22 mv 22 = 4000×144=5,76,000 kgms

Challenge

1. Answersmayvary.Sampleanswers:

Pythagorean triples satisfy a2 + b2 = c2

a. 32 + 42 = 9 + 16 = 25 = 52 (3, 4, 5)

b. 52 + 122 = 25 + 144 = 169 = 132 (5, 12, 13)

c. 82 + 152 = 64 + 225 = 289 = 172 (8, 15, 17)

d. 22 + 242 = 4 + 576 = 625 = 252 (7, 24, 25)

Do It Yourself 7B

1. a. 1369 has 9 in its units place, so its square root will have 3 or 7 in its units place.

b. 3136 has 6 in its units place, so its square root will have 4 or 6 in its units place.

c. 5184 has 4 in its units place, so its square root will have 2 or 8 in its units place.

d. 8281 has 1 in its units place, so its square root will have 1 or 9 in its units place.

e. 9604 has 4 in its units place, so its square root will have 2 or 8 in its units place.

2. 4 4

Area = 16 square units

Area of a square = side2

Side = area= 16=4 units

3. a. 144 − 1 = 143 (1 time), 143 − 3 = 140 (2 times), 140 − 5 = 135 (3 times), 135 − 7 = 128 (4 times), 128 − 9 = 119 (5 times), 119 − 11 = 108 (6 times), 108 − 13 = 95 (7 times), 95 − 15 = 80 (8 times), 80 − 17 = 63 (9 times), 63 − 19 = 44 (10 times), 44 − 21 = 23 (11 times), 23 − 23 = 0 (12 times)

So, 144 = 12.

b. 36 − 1 = 35 (1 time), 35 − 3 = 32 (2 times), 32 − 5 = 27 (3 times), 27 − 7 = 20 (4 times), 20 − 9 = 11 (5 times), 11 − 11 = 0 (6 times)

So, 36 = 6.

c. 441 − 1 = 440 (1 time), 440 − 3 = 437 (2 times), 437 − 5 = 432 (3 times), 432 − 7 = 425 (4 times), 425 − 9 = 416 (5 times), 416 − 11 = 405 (6 times), 405 − 13 = 392 (7 times), 392 − 15 = 377 (8 times), 377 − 17 = 360 (9 times), 360 − 19 = 341 (10 times), 341 − 21 = 320 (11 times), 320 − 23 = 297 (12 times), 297 − 25 = 272 (13 times), 272 − 27 = 245 (14 times),

245 − 29 = 216 (15 times), 216 − 31 = 185 (16 times), 185 − 33 = 152 (17 times), 152 − 35 = 117 (18 times), 117 − 37 = 80 (19 times), 80 − 39 = 41 (20 times), 41 − 41 = 0 (21 times)

So, 441 = 21.

d. 900 − 1 = 899 (1 time), 899 − 3 = 896 (2 times), 896 − 5 = 891 (3 times), 891 − 7 = 884 (4 times), 884 − 9 = 875 (5 times), 875 − 11 = 864 (6 times), 864 − 13 = 851 (7 times), 851 − 15 = 836 (8 times), 836 − 17 = 819 (9 times), 819 − 19 = 800 (10 times), 800 − 21 = 779 (11 times), 779 − 23 = 756 (12 times) 756 − 25 = 731 (13 times), 731 − 27 = 704 (14 times), 704 − 29 = 675 (15 times), 675 − 31 = 644 (16 times), 644 − 33 = 611 (17 times), 611 − 35 = 576 (18 times), 576 − 37 = 539 (19 times), 539 − 39 = 500 (20 times), 500 − 41 = 459 (21 times), 459 − 43 = 416 (22 times), 416 − 45 = 371 (23 times), 371 − 47 = 324 (24 times), 324 − 49 = 275 (25 times), 275 − 51 = 224 (26 times), 224 − 53 = 171 (27 times), 171 − 55 = 116 (28 times), 116 − 57 = 59 (29 times), 59 − 59 = 0 (30 times)

So, 900 = 30.

e. 169 − 1 = 168 (1 time), 168 − 3 = 165 (2 times), 165 − 5 = 160 (3 times), 160 − 7 = 153 (4 times), 153 − 9 = 144 (5 times), 144 − 11 = 133 (6 times), 133 − 13 = 120 (7 times), 120 − 15 = 105 (8 times), 105 − 17 = 88 (9 times), 88 − 19 = 69 (10 times), 69 − 21 = 48 (11 times), 48 − 23 = 25 (12 times), 25 − 25 = 0 (13 times)

So, 169 = 13. 4. a. 5 6545 7 1309 11 187 17 17 1

6545 = 5 × 7 × 11 × 17, not a perfect square

b. 2 9054 3 4527 3 1509 503 503 1 9054 = 2 × 3 × 3 × 503, not a perfect square

c. 2 7396 2 3698 43 1849 43 43 1

d. 2 5786 11 2893 263 263 1

7396 = 2 × 2 × 43 × 43, a perfect square = (2 × 43)2 = (86)2

5786 = 2 × 11 × 263, not a perfect square

5. a. 3 6561

2187

729

243

81 3 27 3 9 3 3 1 6561 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3

6561

(3 × 3)×(3 × 3)×(3 × 3)×(3 × 3)

6. a. 2 8748

2187

729

243

20,164 = (2 × 2)×(71 × 71) = 2 × 71 = 142

c. 2 3100 2 1550 5 775 5 155 31 31 1

d. 2 15600 2 7800 2 3900 2 1950 3 975 5 325 5 65 13 13 1

2 4000 2 2000

1000

500

250

3100 = 2 × 2 × 5 × 5 × 31

31 is the least number with which when 3100 is multiplied, we get a perfect square.

3100 × 31 = 2 × 2 × 5 × 5 × 31 × 31 96,100 = (2 × 5 × 31)2 = 3102

15,600 = 2 × 2 × 2 × 2 × 3 × 5 × 5 × 13 (3 × 13 =) 39 is the least number with which when 15,600 is multiplied, we get a perfect square.

15,600 × 39 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 × 13 × 13

6,08,400 = (2 × 2 × 3 × 5 × 13)2 = 7802

4000 × 10 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

125 5 25 5 5 1 4000 = 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 (2 × 5 =) 10 is the least number with which when 4000 is multiplied, we get a perfect square.

9248 = 2 × 2 × 2 × 2 × 2 × 17 × 17

2 is the least number by which when 9248 is divided, we get a perfect square.

9248 ÷ 2 = 2 × 2 × 2 × 2 × 17 × 17

8748 × 3 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3

26,244 = (2 × 3 × 3 × 3 × 3)2 = 1622

9 3 3 1 8748 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3 3 is the least number with which when 8748 is multiplied, we get a perfect square.

b. 2 5202 3 2601 3 867 17 289 17 17 1 5202 = 2 × 3 × 3 × 17 × 17 2 is the least number with which when 5202 is multiplied, we get a perfect square.

5202 × 2 = 2 × 2 × 3 × 3 × 17 × 17 10,404 = (2 × 3 × 17)2 = 1022

40,000 = (2 × 2 × 2 × 5 × 5)2 = 2002 7. a. 2 9248 2 4624 2 2312 2 1156 2 578 17 289 17 17 1

4624 = (2 × 2 × 17)2 = 682 b. 2 5292 2 2646 3 1323 3 441 3 147

7 1 5292 = 2 × 2 × 3 × 3 × 3 × 7 × 7

3 is the least number by which when 5292 is divided, we get a perfect square.

5292 ÷ 3 = 2 × 2 × 3 × 3 × 7 × 7

1764 = (2 × 3 × 7)2 = 422

14580

Challenge

1. Let the number Sheetal could have thought of be x

So, according to the question,

x + 2 × x + 3 × 2x + 6 × 3 × 2x = perfect square

135

5 is the least number by which when 14,580 is divided, we get a perfect square.

14,580 ÷ 5 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

2916 = (2 × 3 × 3 × 3)2 = 542 d. 2 29988

45 3 15 5 5 1 14,580 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 5

14994

7497

e. 2 56628 2 28314 3 14157 3 4719

29,988 = 2 × 2 × 3 × 3 × 7 × 7 × 17 17 is the least number by which when 29,988 is divided, we get a perfect square.

29,988 ÷ 17 = 2 × 2 × 3 × 3 × 7 × 7 1764 = (2 × 3 × 7)2 = 422

x + 2x + 6x + 36x = perfect square

45x = perfect square

5 × 3 × 3 × x = perfect square

For the number to be a perfect square, it should be multiplied by 5.

So, the number that is a perfect square is 5 × 5 × 3 × 3 = 225. Therefore, the number that Sheetal could think of is 5.

Do It Yourself 7C

1. a. 81 is a 2-digit number (even).

So, 2 n = 2 2 = 1, the square root will have 1 digit.

b. 289 is a 3-digit number (odd).

So, +1 2 n = 3+1 2 = 4 2 = 2, the square root will have 2 digits.

c. 8464 is a 4-digit number (even).

So, 2 n = 4 2 = 2, the square root will have 2 digits.

d. 2304 is a 4-digit even number (even).

So, 2 n = 4 2 = 2, the square root will have 2 digits.

e. 18,496 is a 5-digit even number (odd).

So, +1 2 n = 5+1 2 = 6 2 = 3, the square root will have 3 digits.

56,628 = 2 × 2 × 3 × 3 × 11 × 11 × 13

13 is the least number by which when 56,628 is divided, we get a perfect square.

56,628 ÷ 13 = 2 × 2 × 3 × 3 × 11 × 11 4356 = (2 × 3 × 11)2 = 662

8. LCM of 6, 9, 12, 16, 18

= 2 × 2 × 2 × 2 × 3 × 3

= 16 × 9

= 144 So, the smallest square number which is divisible by 6, 9, 12, 16 and 18 is 144.

2 6, 9, 12, 16, 18

2 3, 9, 6, 8, 9

2 3, 9, 3, 4, 9

2 3, 9, 3, 2, 9

3 3, 9, 3, 1, 9

3 1, 3, 1, 1, 3 1, 1, 1, 1, 1

9. Amount collected for relief fund = ₹1296

Number of students in the class = contribution of each student in the class

= 1296 = 2×2×2×2×3×3×3×3

= 2 × 2 × 3 × 3

= 36

10. Length of the square table cloth = (27)2 = 27 × 27 = 729 sq. cm

Since the piece of cloth has an area of 721 sq. cm, we can say that this piece of cloth is not sufficient.

= 381

382 is less than 1456 by 12.

So, the least number to be subtracted from 1456 to get a perfect square is 12. Thus, the required perfect square number is 1456 − 12 = 1444.

Therefore, 1444 = 38

722 is less than 5243 by 59.

So, the least number to be subtracted from 5243 to get a perfect square is 59.

Thus, the required perfect square number is 5243 − 59 = 5184. Therefore, 5184 = 72

942 is less than 8912 by 76. So, the least number to be subtracted from 8912 to get a perfect square is 76. Thus, the required perfect square number is 8912 − 76 = 8836. Therefore, 8836 = 94

1662 is less than 27,804 by 248.

So, the least number to be subtracted from 27,804 to get a perfect square is 248.

Thus, the required perfect square number is 27,804 − 248 = 27,556. Therefore, 27556 = 166

3582 is less than 1,28,643 by 479.

So, the least number to be subtracted from 1,28,643 to get a perfect square is 479.

Thus, the required perfect square number is 1,28,643 − 479 = 1,28,164. Therefore, 128164 = 358

5608 is greater than 742 by 132. The next perfect square is 752 = 5625

So, the least number to be added to 5608 to get a perfect square is 5625 − 5608 = 17.

Thus, the required perfect square number is 5625 and 5625 = 75.

4620 is greater than 672 by 131. The next perfect square is 682 = 4624

So, the least number to be added to 4620 to get a perfect square is 4624 − 4620 = 4.

Thus, the required perfect square number is 4624 and 4624 = 68.

1,74,700 is greater than 4172 by 811. The next perfect square is 4182 = 1,74,724

So, the least number to be added to 1,74,700 to get a perfect square is 1,74,724 − 1,74,700 = 24.

Thus, the required perfect square number is 1,74,724 and 174724 = 418.

77,800 is greater than 2782 by 516. The next perfect square is 2792 = 77,841

So, the least number to be added to 77,800 to get a perfect square is 77,841 − 77,800 = 41. Thus, the required perfect square number is 77,841 and 77841 = 279.

2,83,000 is greater than 5312 by 1039. The next perfect square is 5322 = 2,83,024

So, the least number to be added to 2,83,000 to get a perfect square is 2,83,024 − 2,83,000 = 24.

Thus, the required perfect square number is 2,83,024 and 283024 = 532.

c. 8 .8 8 8 78 .85 44 64

e. 1032.9796=32.14 3 2 .1 4 3 10 32 .97 96 9 62 1 32 1 24 641 8 97 6 41 6424 2 56 96 2 56 96 0 6. a.

0.8=0.894 0 .8 9 4

256×361=2×2×2×2×2×2×2×2×19×19 =2×2×2×2×19=304

b. 17×17 289 1717 = == 2562×2×2×2×2×2×2×22×2×2×216

c. 8×32=2×2×2×2×2×2×2×2 =2×2×2×2×2×2×2×2=2×2×2×2=16

d. 92.16+0.9216=+92169216 10010000

2×2×2×2×2×2×2×2×2×2×3×3 =+ 10×10

2×2×2×2×2×2×2×2×2×2×3×3 10×10×10×10

2×2×2×2×2×32×2×2×2×2×3 9696 = + =+ 10 10010100 =9.6+0.96=10.56

8. Longest side = 22 15+20=225+400

=625=5×5×5×5 =5×5=25 m

9. 360 m 150 m

Distance she walks diagonally = 150+36022

= 22500+129600 = 152100 = 2×2×3×3×5×5×13×13

= 2 × 3 × 5 × 13 = 390 m

10. Greatest 4-digit number = 9999

9 9

9 99 99

189 18 99 17 01 1 98

Since remainder ≠ 0, 9999 is not a perfect square. So, 992 is less than 9999 by 198. So, the least number to be subtracted from 9999 to get a perfect square is 198.

Thus, the required perfect square number is 9999 − 198 = 9801.

9 9

9 98 01 81

189 17 01 17 01 0

11. Answersmayvary.Sampleanswer: 25 = 52

Challenge

1. Total number of students in the school = 2217 − x

In order to find the value of x, we need to find the least value of the number of students that has to be subtracted from 2217 to make it a perfect square such that the number of students in each row is equal to the total number of rows.

So, we can see that 2217 is greater than 472 by 8. So, we need to subtract 8 from 2217 to make it a perfect square. Thus, the least value of x is 8.

Chapter Checkup

1. a. 382 = 38 × 38 = 1444 b. (−47)2 = (−47) × (−47) = 2209

 d. 2 9 –13

–3–3 39 – = 7779 = 4 

=

= –9 13 × –9 13 = 81 169

2. a. 452 − 162 = (45 + 16) (45 − 16) = 61 × 29 = 1769

b. 65 × 67 = (66 − 1) (66 + 1) = 662 − 12 = 4356 − 1 = 4355

c. 1042 = (100 + 4)2 = (100 + 4)(100 + 4) = 100(100 + 4) + 4(100 + 4) = 1002 + 100 × 4 + 4 × 100 + 42 = 10000 + 400 + 400 + 16 = 10816

d. 120 × 124 = (122 − 2) (122 + 2) = 1222 − 22 = 14,884 − 4 = 14,880

3. 324 = 182 = sum of 18 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35

4. If 2m = 96, then m = 48.

So, m2 − 1 = 482 − 1 = 2304 − 1 = 2303 and m2 + 1 = 482 + 1 = 2304 + 1 = 2305

Thus, the Pythagorean triplet in this case is given by (96, 2303, 2305).

If m2 − 1 = 96, then m2 = 97 ⇒ m is not an integer.

If m2 + 1 = 96, then m2 = 95 ⇒ m is not an integer.

In either case, values of m are not possible.

So, the only Pythagorean triplet here is (96, 2303, 2305) with 96 as its smallest member.

If 2m = 24, then m = 12.

So, m2 − 1 = 122 − 1 = 144 − 1 = 143 and m2 + 1 = 122 + 1

= 144 + 1 = 145

Thus, the Pythagorean triplet in this case is given by (12, 143, 145)

If m2 − 1 = 24, then m2 = 25 ⇒ m = 5, but the smallest member will be 2m = 10 in this case.

If m2 + 1 = 24, then m2 = 23 ⇒ m is not an integer.

In either case, values of m are not possible.

So, the only Pythagorean triplet here is (24, 143, 145) with 24 as its smallest member.

5. a. 1 6 1 2 65 1 26 1 65 1 56 9

Since remainder ≠ 0, 9999 is not a perfect square. b. 4 6 4 21 16 16 86 5 16 5 16 0

Since remainder = 0, 2116 is a perfect square. So, 2116 = 462 Verification:

2116

1058

2116 = 2 × 2 × 23 × 23 2116 = 2×2×23×23 = 2 × 23 = 46

Since remainder = 0, 3136 is a perfect square. So, 3136 = 562

6,57,700 is greater than 8102 by 1600. The next perfect square is 8112 = 6,57,721.

So, the least number to be added to 6,57,700 to get a perfect square is 6,57,721 − 6,57,700 = 21.

Thus, the required perfect square number is 657721 and 657721=811.

8. 2 8, 9, 10 2 4, 9, 5 2 2, 9, 5

3 1, 9, 5

3 1, 3, 5 5 1, 1, 5 1, 1, 1

LCM of 8, 9, 10 = 2 × 2 × 2 × 3 × 3 × 5 = 8 × 9 × 5 = 360

To make the number a perfect square, multiply 360 by 10, then 3600 × 10 = 3600 is the required number.

So, the least number which is a perfect square and is divisible by 8, 9 and 10 is 3600.

9. The longest straight line that can be drawn on the floor of the room is the diagonal of the rectangular room.

So, length of the diagonal = 26+1822 = 676+324

= 1000 = 1010 = 31.62 m

10. Area of each square room = 256 square feet

Area of a square = Side × Side = 256 square feet

Side of the square room = 256 = 16 feet

Side of the new square room = Double the side length of original square room = 2 × 16 feet = 32 feet

Area of the new square room = 32 × 32 = 1024 square feet

Combined area of the two rooms = 256 square feet + 256 square feet = 512 square feet

To compare the area of the new room to the combined area of the two rooms, we take the ratio, 1024:512 = 2:1

Thus, the area of the new room is twice the combined area of the both the original rooms.

11. No, the square root of a number is not always smaller than the original number.

Case 1: When the square root is smaller than the original number.

In this case, the original number is bigger than 1. 4=2, Original number is 4 while the square is 2.

Case 2: When the square root is equal to the original number.

In this case, the original number is equal to 1 or 0. 1=1 , Original number is 1 while the square is 1.

Case 3: When the square root is smaller than the original number.

In this case, the original number is less than 1. (Excluding the case for 0)

0.25=0.5 , Original number is 0.25 while the square is 0.5.

12. =9.8 m L

2 =9.8 m/s g

π =3.14

π=2 TL g

9.8

=2×3.14× 9.8

=6.28×1=6.28

13. Total money collected for charity = ₹7396

Let the number of residents = x

So, the money contributed by each resident = ₹x

→ ₹ ×=7396xx ` ₹7396

2 =7396 x 2 ==7396=86xx

Therefore, there were 86 residents in the society.

Challenge

1. Statement 1 says that the x is always greater than 100. We know that 100=10

So, any number greater than 100 will have its square root greater than 10.

Thus, this statement alone is sufficient to answer the question. Statement 2 says that x is smaller than 150.

So, let us assume the number to be 81. Then 81=9, which is less than 10.

Thus, this statement is not sufficient to answer the question. Hence, option a is correct.

2. Assertion (A): The square of any positive integer is always greater than the integer itself.

Reason (R): When a number greater than 1 is multiplied

by itself, the product is always greater than the original number.

Assertion is not true since 1 is a positive integer and 1 × 1 = 1 which is not greater than 1. So, square of any positive integer is always greater than the integer itself is not true. Reason is true that when a number greater than 1 is multiplied by itself, the product is always greater than the original number.

Thus, Assertion is false while the Reason is true.

Thus, option (d) is correct.

Case Study

Area of the rectangular park = 9600 sq. m

Length of the park = 160 m

Area = Length × Breadth

9600 = 160 × Breadth

Breadth = 9600 ÷ 160 = 60 m

It is given that the square recreational zone is approximately one-fourth of the total area of the park, so, the area of the square recreational zone = 1 ×9600 4 = 2400 sq. m

Let us find the square root of 2400 to find the approximate side length of the square recreational zone.

240049 ≈ m

1. Thus, the side length of the square recreational zone is 49 m. So, option (b) is correct.

2. Length of the path = ()() 22 Length+Breadth = 160+6022

= 256003600 + = 29200 = 170.88 m

Thus, option (b) is correct.

3. The approximate area left inside the rectangular park after the square recreational zone = 9600 sq. m − 2400 sq. m = 7200 sq. m

4. Radius of the fountain = 5.6 m

Area of the fountain region = πr2 = 22 7 × 5.6 × 5.6 = 98.56 sq. m

Cost of tiling = ₹9856 per sq. m

Total cost of tiling = ₹9856 × 98.56 = ₹9,71,407.36

5. Answerwillvary.

Chapter 8

Let’s Warm-up

1. The size of a pineapple is bigger than an apple. So, pineapple occupies more space. There are an apple and a pineapple on a table. Between both fruits, pineapple occupies more space.

2. Length = 5 unit cubes, Breadth = 3 unit cubes, Height = Number of layers = 2

Volume of the solid = 5 × 3 × 2 = 30 cu. units

If a solid with 2 layers has 5 unit cubes along its length, and 3 unit cubes along its breadth, its volume is 30 cu. units.

3. Number of unit cubes in one layer = 16, Number of layers = 3

Volume of the solid = 16 × 3 = 48 cu. units

If a solid has 3 layers with 16 unit cubes in each layer, then its volume is 48 cu. units.

4. Length = 3 cm, Breadth = 2 cm, Height = 2 cm

Volume = 3 × 2 × 2 = 12 cu. cm.

So, 12 unit cubes will be in the container

If a container is 3 cm long, 2 cm wide and 2 cm high, then 12 unit cubes will be in the container.

5. Side of each cube = 2 cm, Volume of a cube = 2 × 2 × 2 = 8 cu. cm

Number of 2 cm cubes = 14, Volume = 14 × 8 = 112 cu. cm

A carton is filled with fourteen 2 cm cubes. The volume of the carton is 112 cu. cm.

Do It Yourself 8A

1. The number of cubes in each side of the cube should be 3 or 4.

2. a. Cube of 15 = 15 × 15 × 15 = 3375

b. Cube of 26 = 26 × 26 × 26 = 17,576

c. Cube of 39 = 39 × 39 × 39 = 59,319

d. Cube of 121 = 121 × 121 × 121 = 17,71,561

3. a. i. Cube of 1 = 1, digit at unit’s place of its cube = 1

ii. Cube of 0 = 0, digit at unit’s place of its cube = 0

iii. Cube of 7 = 343, digit at unit’s place of its cube = 3

iv. Cube of 0 = 0, digit at unit’s place of its cube = 0

b. Odd cubes—1331, 24,387 Even cubes—5800, 1,25,000 1331 and 24,387 have odd cubes because cubes of odd numbers are always odd. 5800 and 1,25,000 have even cubes because cubes of even numbers are always even.

c. The numbers 5800, and 24,387 are not a perfect cube because, 5800 = 2 × 2 × 2 × 5 × 5 × 29 and 24,387 = 3 × 11 × 739

4. a. 3 4

33 = –18×–18×–18 33×33×33 = 5832 35,937 –

5. a. (2.8)3 = 2.8 × 2.8 × 2.8 = 21.952 b. (1.2)3 = 1.2 × 1.2 × 1.2 = 1.728

c. (3.04)3 = 3.04 × 3.04 × 3.04 = 28.094464

d. (0.08)3 = 0.08 × 0.08 × 0.08 = 0.000512 6. a. 5 2195 439 439 1 2195 = 5 × 439, hence, it is not a perfect cube. b. 2 13824

9 3 3 1 13,824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3, a perfect cube = 23 × 23 × 23 × 33 = (2 × 2 × 2 × 3)3 = 243

So, 13,824 is a perfect cube of 24. c. 47 103823

1 1,03,823 = 47 × 47 × 47, a perfect cube = 473 So, 1,03,823 is a perfect cube of 47. d. 7 117649 7 16807 7 2401

343 7 49 7 7 1 1,17,649 = 7 × 7 × 7 × 7 × 7 × 7, a perfect cube = 73 × 73 = (7 × 7)3 = 493 So, 1,17,649 is a perfect cube of 49.

We require two more 2s to form a triplet of 2.

So, (2 × 2 =) 4 is the smallest number with which when 432 is multiplied, we get a perfect cube.

Therefore, 432 × 4 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

1728 = 23 × 23 × 33 = (2 × 2 × 3)3 = 123

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

We require two more 2s to form a triplet of 2.

So, (2 × 2 =) 4 is the smallest number with which when 128 is multiplied, we get a perfect cube.

Therefore, 128 × 4 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 512 = 23 × 23 × 23 = (2 × 2 × 2)3 = 83

2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

We require one more 2 to form a triplet of 2.

So, 2 is the smallest number with which when 2916 is multiplied, we get a perfect cube.

Therefore, 2916 × 2 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

We require two more 2s to form a triplet of 2.

So, (2 × 2 =) 4 is the smallest number with which when 3456 is multiplied, we get a perfect cube.

Therefore, 3456 × 4 = 2 × 2 × 2 × 2 × 2 × 2

× 2 × 2 × 2 × 3 × 3 × 3

13,824 = 23 × 23 × 23 × 33 = (2 × 2 × 2 × 3)3 = 243

64 = 23 × 23 = (2 × 2)3 = 43 2 576 2 288 2 144 2 72 2 36 2 18 3 9 3 3 1

b. 3 5145 5 1715 7 343 7 49 7 7 1

576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 (3 × 3 =) 9 is the smallest number by which when 576 is divided, we get a perfect cube.

Therefore, 576 ÷ 9 = 2 × 2 × 2 × 2 × 2 × 2

5145 = 3 × 5 × 7 × 7 × 7

(3 × 5 =) 15 is the smallest number by which when 5145 is divided, we get a perfect cube.

Therefore, 5145 ÷ 15 = 7 × 7 × 7 343 = 73

1728 = 23 × 23 × 33 = (2 × 2 × 3)3 = 123 2 36288 2 18144 2 9072 2 4536 2 2268 2 1134 3 567 3 189 3 63 3 21 7 7 1

55566 3 27783 3 9261 3 3087 3 1029 7 343

36,288 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

× 3 × 7 (3 × 7 =) 21 is the smallest number by which when 36,288 is divided, we get a perfect cube.

Therefore, 36,288 ÷ 21 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

55,566 = 2 × 3 × 3 × 3 × 3 × 7 × 7 × 7

(2 × 3 =) 6 is the smallest number by which when 55,566 is divided, we get a perfect cube.

Therefore, 55,566 ÷ 6 = 3 × 3 × 3 × 7 × 7 × 7

9261 = 33 × 73 = (3 × 7)3 = 213

9. 113 = sum of 11 odd numbers, where the first odd number

= 2 × (11 − 1)th triangular number + 1

= 2 × 10th triangular number + 1.

Also, 10th triangular number = sum of 10 consecutive numbers

= 2 × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 +

= 2 × 55 + 1 = 111 113 = 111 +

129 + 131 = 1331

203 = sum of 20 odd numbers, where the first odd number

= 2 × (20 − 1)th triangular number + 1 = 2 × 19th triangular number + 1

Also, 19th triangular number = sum of 19 consecutive numbers

= 2 × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14

+ 15 + 16 + 17 + 18 + 19) + 1

= 2 × 190 + 1 = 381

203 = 381 + 383 + 385 +

399 + 401 + 403 + 405 +

419 = 8000

10. a. 13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93 = (1 + 2 + 3 +

+ 6 + 7 + 8 + 9)2 = 452 = 2025

b. 13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 +

= (1 + 2 + 3 + 4 +

= 6084

11. a. 413 − 403 = 3 × 40 × (40 + 1) + 1

= 3 × 40 × 41 + 1 = 4921

b. 1003 − 993 = 3 × 99 × (99 + 1) + 1

= 3 × 99 × 100 + 1 = 29,701

12. Volume of the chocolates = 8 × 10 × 2 = 160 cu. cm

Volume of the cubical carton = 20 × 20 × 20 = 8000 cu. cm

Number of packs of chocolates = 8000 160 = 50

There are 15 cubes. Side of each cube = 2 cm

Volume of the solid = 2 × 2 × 2 × 15 = 120 cu. cm

No, even after rearrangement, the solid will not be a perfect cube.

For this, we need a total of 27 cubes. So, 27 − 15 = 12 more cubes are needed to form a perfect cube. Do It Yourself 8B 1.

46656

2 × 2 × 3 × 3 = 36

a. 27 27 − 1 = 26 (1 time); 26 − 7 = 19 (2 times); 19 − 19 = 0 (3 times)

So, 3 27 = 3.

b. 216

216 − 1 = 215 (1 time); 215 − 7 = 208 (2 times); 208 − 19 = 189 (3 times); 189 − 37 = 152 (4 times); 152 − 61 = 91 (5 times); 91 − 91 = 0 (6 times)

So, 3 216 = 6.

c. 1728

1728 − 1 = 1727 (1 time); 1727 − 7 = 1720 (2 times); 1720 − 19 = 1701 (3 times); 1701 − 37 = 1664 (4 times); 1664 − 61 = 1603 (5 times); 1603 − 91 = 1512 (6 times) 1512 − 127 = 1385 (7 times); 1385 − 169 = 1216 (8 times); 1216 − 217 = 999 (9 times); 999 − 271 = 728 (10 times); 728 − 331 = 397 (11 times); 397 − 397 = 0 (12 times)

So, 3 1728 = 12.

d. 1000

1000 − 1 = 999 (1 time); 999 − 7 = 992 (2 times); 992 − 19 = 973 (3 times); 973 − 37 = 936 (4 times); 936 − 61 = 875 (5 times); 875 − 91 = 784 (6 times) 784 − 127 = 657 (7 times); 657 − 169 = 488 (8 times); 488 − 217 = 271 (9 times); 271 − 271 = 0 (10 times)

So, 3 1000 = 10.

4. a. 3 32768 97336 = 3 2×2×2×2×2×2×2×2×2×2×2×2×2×2×2 2×2×2×23×23×23 = 32 23×2 = 32 46

b. 3 –21952 132651 = 3 –2×–2×–2×–2×–2×–2×–7×–7×–7 3×3×3×17×17×17

= –28 51

c. () 3 125×–2744 = 3 3 –343000=–5×–5×–5×–2×–2×–2×–7×–7×–7 = −70

d. ()() 3 –1728×–6859 = 3 11852352

3 2×2×2×2×2×2×3×3×3×19×19×19 = 228

5. a. Volume = 5,92,704 cu. cm

Edge of the cube = 3 volume = 3 592704

= 3 2×2×2×2×2×2×3×3×3×7×7×7

= 2 × 2 × 3 × 7 = 84 cm

b. Volume = 85,184 cu. cm

Edge of the cube = 3 volume = 3 85184

= 3 2×2×2×2×2×2×11×11×11

= 2 × 2 × 11 = 44 cm

6. a. To find the smallest number, we subtract the odd numbers successively from 408.

408 − 1 = 407 (1 time); 407 − 7 = 400 (2 times); 400 − 19 = 381 (3 times); 381 − 37 = 344 (4 times); 344 − 61 = 283 (5 times); 283 − 91 = 192 (6 times)

192 − 127 = 65 (7 times)

The next number to be subtracted from 65 is 169. But 169 > 65, so 65 must be subtracted from 408 to get a perfect cube.

So, 65 is the smallest number that must be subtracted from 408 to get a perfect cube.

Now, 408 − 65 = 343.

3 343 = ×× 3 777 = 7

b. To find the smallest number, we subtract the odd numbers successively from 2215.

2215 − 1 = 2214 (1 time); 2214 − 7 = 2207 (2 times); 2207 − 19 = 2188 (3 times); 2188 − 37 = 2151 (4 times); 2151 − 61 = 2090 (5 times); 2090 − 91 = 1999 (6 times)

1999 − 127 = 1872 (7 times); 1872 − 169 = 1703 (8 times); 1703 − 217 = 1486 (9 times); 1486 − 271 = 1215 (10 times); 1215 − 331 = 884 (11 times); 884 − 397 = 487 (12 times)

487 − 469 = 18 (13 times)

The next number to be subtracted from 18 is 547. But 547 > 18.

So, 18 must be subtracted from 2215 to get a perfect cube. So, 18 is the smallest number that must be subtracted from 2215 to get a perfect cube.

Now, 2215 − 18 = 2197.

3 2197 = 3 13×13×13 = 13

7. a. 720 − 1 = 719 (1 time); 719 − 7 = 712 (2 times); 712 − 19 = 693 (3 times); 693 − 37 = 656 (4 times); 656 − 61 = 595 (5 times); 595 − 91 = 504 (6 times) 504 − 127 = 377 (7 times); 377 − 169 = 208 (8 times)

The next number to be subtracted from 208 is 217.

But 217 > 208, so 217 − 208 = 9 must be added to 720 to get a perfect cube.

So, 9 is the smallest number that must be added to 720 to get a perfect cube.

Now, 720 + 9 = 729.

3 729 = 3 9×9×9 = 9

b. 2700 − 1 = 2699 (1 time); 2699 − 7 = 2692 (2 times); 2692 − 19 = 2673 (3 times); 2673 − 37 = 2636 (4 times); 2636 − 61 = 2575 (5 times); 2575 − 91 = 2484 (6 times); 2484 − 127 = 2357 (7 times); 2357 − 169 = 2188 (8 times); 2188 − 217 = 1971 (9 times); 1971 − 271 = 1700 (10 times); 1700 − 331 = 1369 (11 times); 1369 − 397 = 972 (12 times); 972 − 469 = 503 (13 times);

The next number to be subtracted from 503 is 547.

But 547 > 503, so 547 − 503 = 44 must be added to 2700 to get a perfect cube.

So, 44 is the smallest number that must be added to 2700 to get a perfect cube.

Now, 2700 + 44 = 2744.

3 2744 = 3 2×2×2×7×7×7 = 14

8. Let the three numbers be x, 3x and 6x.

Sum of their cubes = x3 + (3x)3 + (6x)3 = 15,616

x3 + 27x3 + 216x3 = 15,616

244x3 = 15,616

x3 = 15616 244 = 64

x = 3 64 = 3 4×4×4 = 4

3x = 3 × 4 = 12

6x = 6 × 4 = 24

So, the numbers are 4, 12 and 24.

9. Volume = 74,088 cu. cm

Length of each side of the cube = 3 volume = 3 74088

= 3 222333777 ××××××××

= 2 × 3 × 7 = 42 cm

Challenge

1. 1*28 is a perfect cube and a 4-digit number

Using the hit and trial method, putting digits 0 to 9 in reverse order in place of ‘*’ and checking if it is a perfect cube.

So, for * = 7, we get 1728 as the 4-digit number.

= 23 × 23 × 33 = (2 × 2 × 3)3 = 123

So, the missing digit is 7 and the cube root of 1728 = 12.

2 1728 2 864 2 432 2 216 2 108 2 54 3 27 3 9 3 3 1 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3, a perfect cube

Chapter Checkup

1. a. (41)3 = 41 × 41 × 41 = 68,921

b. 33 1=613 77

= 13×13×13 7×7×7 = 2197 343 c. 3 2 –9

d. (3.6)3 = 3.6 × 3.6 × 3.6 = 46.656

2. The shaded cube shows the cube of number 3.

3. a. 2 830584

2 415292

2 207646 47 103823 47 2209 47 47 1

2 210000

2 105000 2 52500

2 26250

3 13125 5 4375 5 875 5 175 5 35 7 7 1

c. 2 328500 2 164250 3 82125 3 27375 5 9125 5 1825

5 365 73 73 1

4. a. 133 = sum of 13 odd numbers, where the first odd number = 2 × (13 − 1)th triangular number + 1 = 2 × 12th triangular number + 1 Also, the 12th triangular number = sum of 12 consecutive numbers = 2 × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12) + 1 = 2 × 78 + 1 = 157 133 = 157 + 159 + 161 + 163 + 165 + 167 + 169 + 171 + 173 + 175 + 177 + 179 + 181 = 2197

3 132651 3 44217 3 14739 17 4913

8,30,584 = 2 × 2 × 2 × 47 × 47 × 47, is a perfect cube = 23 × 473 = (2 × 47)3 = 943 So, 8,30,584 is a cube of 94.

2,10,000 = 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5 × 5 × 7, is not a perfect cube.

b. 13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93 + 103 + 113 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11)2 = 662 = 4356

5. a. 3 512 1000 = 3 3 512 1000 = 8 10 = 0.8

b. 3 –2744 8000 = 3 3 –2744 8000 = 3 3 –2×–2×–2×7×7×7 2×2×2×2×2×2×5×5×5 = –2×7 2×2×5 = –14 20

c. 3 40961000000 ×

= 3 2×2×2×2×2×2×2×2×2×2×2×2 ×2×2×2×2×2×2×5×5×5×5×5×5

= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 1600

d. 3 10.648 = 3 10648 1000 = 3 3 10648 1000 = 3 3 2×2×2×11×11×11 10×10×10 = 2×11 10 = 22 10 = 2.2

6. Volume = 2,26,981 cu. m

Length of the edge of the cube

= 3 volume

= 3 226981

3,28,500 = 2 × 2 × 3 × 3 × 5 × 5 × 5 × 73, is not a perfect cube.

1,32,651 = 3 × 3 × 3 × 17 × 17 × 17, a perfect cube = 33 × 173 = (3 × 17)3 = 513 So, 1,32,651 is a cube of 51.

= 3 616161 ×× = 61 m

7. Side of the cube = 5 cm

Volume of 1 cubical bar = 5 × 5 × 5 = 125 cu. cm

Volume of 25 cubical bars = 125 × 25 = 3125 cu. cm

8. To find the smallest number, we subtract the odd numbers successively from 2789.

2789 − 1 = 2788 (1 time)

2788 − 7 = 2781 (2 times)

2781 − 19 = 2762 (3 times)

2762 − 37 = 2725 (4 times)

2725 − 61 = 2664 (5 times)

2664 − 91 = 2573 (6 times)

2573 − 127 = 2446 (7 times)

2446 − 169 = 2277 (8 times)

2277 − 217 = 2060 (9 times)

2060 − 271 = 1789 (10 times)

1789 − 331 = 1458 (11 times)

1458 − 397 = 1061 (12 times)

1061 − 469 = 592 (13 times)

592 − 547 = 45 (14 times)

The next number to be subtracted from 45 is 631. But 631 > 45, so 45 must be subtracted from 2789 to get a perfect cube.

So, 45 is the smallest number that must be subtracted from 2789 to get a perfect cube.

Now, 2789 − 45 = 2744.

3 3 4274422277=×××××7=2×7=1

9. To find the smallest number, we need to subtract the odd numbers successively from 4900.

4900 − 1 = 4899 (1 time)

4899 − 7 = 4892 (2 times)

4892 − 19 = 4873 (3 times)

4873 − 37 = 4836 (4 times)

4836 − 61 = 4775 (5 times)

4775 − 91 = 4684 (6 times)

4684 − 127 = 4557 (7 times)

4557 − 169 = 4388 (8 times)

4388 − 217 = 4171 (9 times)

4171 − 271 = 3900 (10 times)

3900 − 331 = 3569 (11 times)

3569 − 397 = 3172 (12 times)

3172 − 469 = 2703 (13 times)

2703 − 547 = 2156 (14 times)

2156 − 631 = 1525 (15 times)

1525 − 721 = 804 (16 times)

The next number to be subtracted from 804 is 817. But 817 > 804, so 817 − 804 = 13 must be added to 4900 to get a perfect cube.

So, 13 is the smallest number that must be added to 4900 to get a perfect cube.

Now, 4900 + 13 = 4913.

3 3 4913=17×17×17=17 = 17

10. 2 24500

2 12250 5 6125 5 1225

5 245 7 49

7 7 1

24,500 = 2 × 2 × 5 × 5 × 5 × 7 × 7

We require one more 2 and 7 to form a triplet of 2 and 7.

So, (2 × 7 =) 14 is the smallest number with which when 24,500 is multiplied, we get a perfect cube.

Therefore, 24,500 × 14 = 2 × 2 × 2

× 5 × 5 × 5 × 7 × 7 × 7

3,43,000 = 23 × 53 × 73 = (2 × 5 × 7)3 = 703

3,14,432 = 23 × 23 × 173 = (2 × 2 × 17)3 = 683 2 5345344

12. Area = 147 sq. cm

3 × area of square = 147 sq. cm

Area of square = 147 3 = 49 sq. cm = (7)2

Side of square = 7 cm = side of the cube

Volume of 1 cube = 7 × 7 × 7 = 343 cu. cm

Number of cubes = 6

Volume of 6 cubes = 343 × 6 = 2058 cu. cm

Challenge

1. Let the number of students getting prizes be 2x, 3x, 5x.

Product = 2x × 3x × 5x = 30x3 = 2,05,770

x3 = 205770 30 = 6859

x = 3 6859 = 3 19×19×19 = 19

Number of students getting the prize for punctuality = 19 × 2 = 38

Number of students getting the prize for respect = 19 × 3 = 57

Number of students getting the prize for discipline = 19 × 5 = 95

2. a. To find the dimensions of the cube, we need to determine the length of each side. Since the volume V of a cube is given by s3, where s is the length of a side, we can solve for s using the given volume:

V = s3

Given V = 8000 cubic metres, s3 = 8000

s = 3 8000 = 20

So, each side of the cube should be 20 metres.

b. Each tank has a side length of 20 metres. So, when eight tanks are placed adjacent to each other, the total length of the side of the resulting shape is: 20 metres × 8 = 160 metres.

Case Study

1. Volume of the cube = 1.5 × 1.5 × 1.5 = 3.375 cubic metres. Hence, 3.375 cubic metres of concrete are needed to construct a cube with an edge length of 1.5 metres.

So, the correct answer is option d.

2. To find the edge length, you need to take the cube root of the volume.

3 64 = 4 m

So, the correct answer is option a.

3. Volume of the cube = 2.5 × 2.5 × 2.5 = 15.625 cubic metres.

4. Volume of each smaller cube = 1.8 × 1.8 × 1.8 = 5.832 cubic metres

668168

334084

167042

53,45,344 = 2 × 2 × 2 × 2 × 2 × 2 × 17 × 17 × 17 × 17

Therefore, 17 is the smallest number by which when 53,45,344 is divided, we get a perfect cube.

Therefore, 53,45,344 ÷ 17 = 2 × 2 × 2

× 2 × 2 × 2 × 17 × 17 × 17

Number of cubes used to make one cubical design = 4 Volume of each cubical design = 4 × 5.832 = 23.328 cubic metres.

Chapter 9

Let’s Warm-up

1. The least preferred colour is red

2. The most preferred colour is blue

3. The percentage of people who prefer yellow is 17%

4. The percentage of people who like green is 31%

5. The number of people who prefer blue over red are 68

Do It Yourself 9A

1. a. 22 0.2= ×100%=20% 1010  ⇒ 

b. 5656 0.56= ×100%=56% 100100

c. 45 45%==0.45 100

d. 13.5 13.5%==0.135 100

e. 120.5 120.5%==1.205 100

2. a. 0.4 0.4%= 100 41 == 1000250 0.4 0.4%= 100 41 == 1000250

b. 0.125 0.125%= 100 1251 == 100000800 0.125 0.125%= 100 1251 == 100000800

c. 62.5 62.5% = 100 6255 == 10008 62.5 62.5% = 100 6255 == 10008

d. 11.5 11.5%= 100 11523 == 1000200 11.5 11.5%= 100 11523 == 1000200

e. 120%==1206 1005

3. a. 210 210% == 21:10 100

b. 0.55 0.55%= 100 5511 == 10,0002000 =11:2000

c. 65.5 65.5%= 100 655131 == 1000200 =131:200 65.5 65.5%= 100 655131 == 1000200 =131:200

d. 2×5+1 11 2%=× 55100 11 ==11:500 500 2×5+1 11 2%=× 55100 11 ==11:500 500

e. 15×5+2 15%=×21 55100 77 ==77:500 500 15×5+2 15%=×21 55100 77 ==77:500 500

4. a. 60 ×100%=2×100%=200%

b. 1 L = 1000 mL

So, 3L = 3000 mL 900 ×100=30%

c. 991÷3×100=××100=60% 553

d. 1 kg = 1000 g So, 5 kg =

c. `98 `88 98–8810100%100%

6. a. ₹190 + 20% of ₹190 = ₹190 + 20 100 � ₹190 = ₹190 + ₹38 = ₹228

b. 150 kg − 30% of 150 kg = 150 kg − 30 100 � 150 kg = 150 kg − 45 kg = 105 kg

c. 1000 L − 12.5% = 1000 L − 12.5 100 � 1000 L = 1000 L − 125 L = 875 L

d. 750 grams + 0.5% = 750 grams + 0.5 100 � 750 grams = 750 grams + 3.75 grams = 753.75 grams

7. Original price = ₹250 per litre, Increase in price = 10% New price = Original price + Increased price = ₹250 + 10 100 � 250 = ₹250 + ₹25 = ₹275

8. Let the number be x. Increased by 30%: 30 +× 100 10+3 3 =+= 1010 13 = 10 xx xxx x x Then reduced by 20%: 13×10 132013 –×= 101001010×10 2×13 13026104 –=–= 10×10100100100 xxx xxxx

Percentage change = Increased number Original number � 100% 104 –100 × 100%

13. Let the amount of money Mr Sharma saved be x

Percentage of money he left for his:

Mother = 20% of x

Sister = 30% of x

Wife = 25% of x

104–100 = ×100% 100

100 25 x x x xx x x x = Thus, there is 4% increase in the number.

41 =×100%=×100%=4%

9. Rajat’s score out of 500 In half yearly exams = 400 In final exams = 430

Percentage increase = Score in final exams Score in half yearly exams

Score in half yearly exams � 100%

430–400 = ×100%

400

30 =×100%

400 =7.5%

10. Population = 50,000

Percentage of registered voters = 35%

a. Number of registered voters = 35 ×50,000=17,500 100

b. Number of people who voted in the recent election = 60% of 17,500 = 60 ×17,500=10,500 100

c. The number of voters between 18 and 30 years old = 40% of 10,500 = 40 ×10,500=4200 100

11. Present value = ₹3,50,000, Depreciation value of a car = 10%

Value of the car after one year = Value of car − Depreciation value of a car after one year

= ₹3,50,000 − 10 100 � ₹3,50,000

= ₹3,50,000 − ₹35,000 = ₹3,15,000

Value of the car after second year = Value of car − Depreciation value of a car after two year

= ₹3,15,000 − 10 100 � ₹3,15,000

= ₹3,15,000 − ₹31,500

= ₹2,83,500

So, the value of the car is ₹2,83,500 after two years.

12. Let the total marks be x, Passing percentage = 30%

So, 30303 30% ofmarks=×== 10010010 xx x

Passing marks = 45 + 15 = 60

We can write, 3 =60 10 x

3x = 600

600 = 3 x

x = 200

So, the highest number of marks that a student can score in the examination is 200.

Brother = [100 − (20 + 30 + 25)]% = [100 − 75]% = 25% of x

100 � x= ₹4500

x = ₹18,000

So, he saved ₹18,000.

Thus, Mr Sharma saved ₹18,000 that month.

Challenge

1. Total number of smartphones = 10,000

a. 15 15% of10,000=×10,000=1500 100

So, 1500 smartphones are defective.

b. 40 40%of1500=×1500=600 100

So, 600 smartphones can be repaired.

c. 60 60%of1500=×1500=900 100

So, 900 smartphones cannot be repaired and are scrapped.

Do It Yourself 9B

1. Given that, CP = ₹39,500

a. SP = ₹35,000

CP > SP, so there is loss.

Loss = CP − SP

= ₹39,500 − ₹35,000 = ₹4500

So, there is a loss of ₹4500.

b. SP = ₹42,000

CP < SP, so there is profit.

Profit = SP − CP = ₹42,000 − ₹39,500 = ₹2500

So, there is a profit of ₹2500.

c. SP = ₹38,000

CP > SP, so there is loss.

Loss = CP − SP = ₹39,500 − ₹38,000 = ₹1500

So, there is a loss of ₹1500.

d. SP = ₹45,000

CP < SP, so there is profit.

Profit = SP − CP = ₹45,000 − ₹39,500 = ₹5500

So, there is a profit of ₹5500.

2. a. CP = ₹500, SP = ₹572 SP > CP, there is profit.

Profit = SP − CP = ₹572 − ₹500 = ₹72

Profit % = Profit CP � 100%

= 72 100%

500 ×� 100% = 14.4%

b. CP = ₹7500, SP = ₹5525

CP > SP, there is loss.

Loss = CP − SP = ₹7500 − ₹5550 = ₹1950

Loss % = Loss 100% CP ×� 100% = 1950 100% 7500 ×� 100% = 26%

c. CP = ₹6724, Loss % = 12.5%

Here, CP > SP since there is loss.

Loss % = CP–SP ×100% CP � 100%

12.5% = 6724–SP 100% 6724 ×� 100%

84050 = (6724 − SP) × 100

840.5 = 6724 − SP

SP = ₹5883.5

d. SP = 35,100, Profit % = 8%

Since, there is profit so, CP < SP

Profit % = SP–CP 100% CP × � 100%

8% = 35100–CP 100% CP ×� 100%

8CP = 3510000 − 100CP

108CP = 3510000

CP = ₹32,500

3. CP of the car = ₹4,47,200

SP = ₹5,09,808

Since, SP > CP

Profit = ₹5,09,808 − ₹4,47,200 = ₹62,608

Profit % = Profit 100% CP ×� 100% = 62,608 100% 447200 ×� 100% = 14%

4. CP of the bluetooth speaker = ₹14,500; SP = ₹14,000

Since CP > SP, loss is made; L = ₹14,500 − ₹14,000 = ₹500

Loss % = Loss500×100%=×100%=3.45% CP14,500 � 100% = Loss500×100%=×100%=3.45% CP14,500 � 100% = 3.45%

5. The purchase price of the article = ₹5000

Cost of the cartage = ₹500

Total cost price of the article = ₹5000 + ₹500 = ₹5500

Selling price of the article = ₹6600

SP > CP, there is profit.

Profit % = 6600–5500 100% 5500 ×� 100%

1100 =100%20% 5500 ×= � 100% = 20%

Thus, the shopkeeper earns 20% profit.

6. Cost price of the mixer = ₹4500

Loss % = 7%

Loss % = CP–SP 100% CP ×� 100%

7% = 4500–SP 100% 4500 ×� 100%

31500 = (4500 − SP) × 100

315 = 4500 − SP

SP = 4500 − 315

SP = ₹4185

Thus, the retailer sold the mixer for ₹4185.

7. SP of the toy car = ₹819

Loss = 9%

CP = 100 100–9 × ₹819 = ₹900

SP for a profit of 5% = 100+5 900 100 ×× 900 = ₹945

8. Total cost of the cookies = ₹10,000

Fraction of cookies sold at 20% loss = 1 4 of the total cookies

Cost of 1 4 of the cookies = 1 4 of ₹10000 = ₹2500

Cost price of the remaining cookies = ₹10,000 − ₹2500 = ₹7500

Loss on 1 4 of the cookies = 20% of ₹2500 = 20 100 × 2500 = ₹500

Selling price of 1 4 of the cookies = ₹2500 − ₹500 = ₹2000

Total profit gained = 40%

Profit earned = 40% of ₹10000

= 40 10000 100 ×� 10000 = ₹4000

Total selling price of the cookies = ₹10,000 + ₹4000 = ₹14,000

Total selling price of the remaining cookies = ₹14,000 − ₹2000 = ₹12,000

Profit earned on the remaining cookies = ₹12,000 − ₹7500 = ₹4500

Profit % for the remaining cookies = 4500 100% 7500 ×� 100% = 60%

Thus, the remaining cookies should be sold at 60% profit to gain 40% profit on the cookies.

9. Amount paid by Raj = ₹1,08,000

Profit % earned by Ramu = 20%

Loss % for Shyam when he sold the land to Raj = 10%

Let the cost of the land initially be 100.

The cost at which the land is sold to Shyam = 100 + 20% of 100 = 120

The cost at which the land is sold to Raj = 120 − 10% of 120 = 120 − 12 = 108

So, if the final cost of the land is 108 then initially it was 100.

If the cost of the land was ₹1,08,000 then the initial cost of the land will be ₹1,00,000.

Thus, Ramu paid ₹1,00,000 for the land.

10. CP of 100 books = ₹1500

Books discarded = 5

SP of 95 books = ₹1710

Since SP > CP, profit is earned.

P = ₹1710 − ₹1500 = ₹210

Profit % = 210 ×100%=14% 1500 � 100% = 14%

11. Watch 1: SP = ₹1200; P% = 20

Watch 1: CP = 100 ×1200=1000 100+20 ` � 1200 = ₹1000

Watch 2: SP = ₹1200; L% = 20

Watch 2: CP = 100 ×1200=1500 100–20 ` � 1200 = ₹1500

Total CP = ₹1000 + ₹1500 = ₹2500

Total SP = ₹1200 + ₹1200 = ₹2400

Since CP > SP, loss is incurred; L = ₹2500 − ₹2400 = ₹100

Loss % = Loss100×100%=×100%=4% CP 2500 � 100% = Loss100×100%=×100%=4% CP 2500 × 100% = 4%

12. Answersmayvary.Sampleanswer:

Maya bought a set of 5 books for ₹2200. She sold each book for ₹600. Calculate the total profit% or loss% incurred by Maya.

Challenge

1. Let the CP of the necklace be x. Profit earned by Madhuri = 7%

Profit percent earned had she sold the jewellery for ₹2940 more = 10%

⇒ Difference in profit percent = 10% − 7% = 3%

Since profit is earned on the CP, 3% of x = ₹2940

x = ₹2940 × 100 3 = ₹98,000

Do It Yourself 9C

1. Given, MP = ₹4000 Discount%

Discount=MarketPrice× 100

a. Discount = 25%; discount = 4000 × 25 100 = ₹1000

Selling Price = Marked Price − Discount = 4000 − 1000 = ₹3000

b. Discount = 20%; discount = 4000 × 20 100 = ₹800

SP = MP − Discount = 4000 − 800 = ₹3200

c. Discount = 12 1 2 % = 25 % 2 ;

discount = 4000 × 25 2×100 = ₹500

SP = MP − Discount = 4000 − 500 = ₹3500

d. Discount = 10%; Discount = 10 4000× 100 = ₹400

SP = MP − Discount = 4000 − 400 = ₹3600

2. MP = ₹900, Discount = ₹175

SP = MP − Discount

SP = ₹900 − ₹175 = ₹725

Thus, the selling price of the article is ₹725.

3. MP of the washing machine = ₹14,500; SP = ₹13,775

SP = MP − Discount

Discount = MP − SP

Discount = ₹14,500 − ₹13,775 = ₹725 Discount

Discount%=×100% MP

Discount % = 725 ×100%=5% 14,500 ` `

Thus, the discount allowed is ₹725 and the discount percentage is 5%.

4. SP of the dining table = ₹16,000; discount = 20% (100–Discount%)

WeknowthatSP=MP× 100

MP = 100 16,000×= 100–20 ` ₹20,000

Thus, the marked price of the table is ₹20,000.

5. SP of the trousers = ₹7500; discount offered = ₹500

MP = ₹7500 + ₹500 = ₹8000

Discount

Discount%=×100% MP

Discount % = 500 ×100%= 8000 ` ` 6.25%

Thus, 6.25% discount was offered on the pair of trousers.

6. Let the MP of the article = 100; first discount = 25% of ₹100 = ₹25;

SP after the first successive discount = ₹100 − ₹25 = ₹75

Second discount = 12% of reduced price = 12% of ₹75 = ₹9

SP after the second successive discount = ₹75 − ₹9 = ₹66

Total discount (x) = ₹100 − ₹66 = ₹34

Discount Discount%=×100% MP

Discount % = 34 ×100%=34% 100

7. Let the MP of the goods be ₹100

First discount = 15% of ₹100 = 15 ×100=15 100 ``

SP after the first discount = ₹100 − ₹15 = ₹85

Second discount = 10% of reduced price = 10 ×85=8.5 100 ``

SP after the second discount = ₹85 − ₹8.5 = ₹76.5

Total discount = ₹100 − ₹76.5 = ₹23.5

Discount

Discount%=×100% MP

Discount % = 23.5 ×100%=23.5% 100

8. Marked price on the article = ₹702

Discount % on the article = 15%

Price after the discount = 702 − 15% of 702 = 702 − 105.3 = ₹596.7

Additional discount offered = 10%

Final price of the article after discount = ₹596.7 − 10% of ₹596.7 = 596.7 − 59.67 = ₹537.03

Thus, the selling price of the article is ₹537.03.

9. Given, CP of the dining table = ₹15,000; MP = ₹18,000

Loss incurred = 8% = 8 ×15,000=1200 100 `

SP = CP − Loss = ₹15,000 − ₹1200 = ₹13,800

Discount MP − SP = ₹18,000 − ₹13,800 = ₹4200

Discount

Discount%=×100% MP

Discount % = 4200 ×100=23.33% 18,000

10. Let the CP be ₹100.

MP = 32% of ₹100 = ₹32

Profit to be gained = 10% of ₹100 = ₹10

SP = CP + Profit = ₹100 + ₹10 = ₹110

Discount = ₹132 − ₹110 = ₹22

% discount = 22 ×100%=16.66% 132

11. CP of the article = ₹850; P = 12% = 12 ×850=102 100 `

SP = ₹850 + ₹102 = ₹952

Discount offered = ₹200; MP = ₹952 + ₹200 = ₹1152

12. MP of the article = ₹9000; discount offered = 20%

Discount%

Discount= ×MP 100%

= 20 ×9000=1800 100 `

SP = ₹9000 − ₹1800 = ₹7200

Profit earned = 20%

CP = 100 ×7200=6000 100 +20 `

13. Answermayvary.Sampleanswer:

Maya bought a laptop at a store that was having a scale. The store offered a 15% discount on the marked price of the laptop. After applying the discount, Maya paid ₹34,000 for the laptop. If the store sold the laptop at its marked price, the shopkeeper would have made a profit of 20% on the cost price.

Challenge

1. Let the cost price of each ticket be ‘x’.

Initial cost of the tickets = Half the original value = 1 2 of x = 0.5x

Additional discount % = 10% on the reduced price

= 10% of 0.5x = 0.05x

Final price of the ticket = –0.5–0.05 xxx = 0.45 x

Total discount offered = –0.45 xx = 0.55 x

Total discount % offered = 0.55 ×100% x x = 55%

Thus, Amit gets a discount of 55%.

Do It Yourself 9D

1. SP of the vacuum cleaner = ₹8000

GST = 18%

Amount Manohar should give to the shopkeeper = 100+18 8000×=9440 100 `

Thus, Manohar should give ₹9440 to the shopkeeper.

2. SP of the computer = ₹38,000

SP of the printer = ₹8000

Total SP = ₹38,000 + ₹8000 = ₹46,000

Rate of tax = 7%

Tax = 7 100 × ₹46,000 = ₹3220

Total amount paid = ₹46,000 + ₹3220 = ₹49,220

Thus, Akash must pay ₹49,220 to buy these two items.

3. Bill amount for Renu = ₹16,240

GST = 12%

SP before GST = 100 16,240×=14,500 100+12 `

Thus, the price of the phone before the GST was added was ₹14,500.

4. Let the SP of the refrigerator be x

According to the question, 9 ×=1170 100 x

x = ₹13,000

Thus, the actual selling price of the refrigerator is ₹13,000.

5. Price of the harmonium including tax = ₹17,280

Rate of tax = 8% of SP

Let the SP be ₹100

Tax = 8 100 × ₹100 = ₹8

SP including tax = ₹100 + ₹8 = ₹108

SP of the harmonium = ₹17,280 108 × 100 = ₹16,000

Thus, the selling price of the harmonium is ₹16,000.

6. MP = ₹5000; Discount = 10% of ₹5000 = ₹500

SP = ₹5000 − ₹500 = ₹4500

Tax applied = 10% of SP = 10 ×4500=450 100 `

Amount paid = ₹4500 + ₹450 = ₹4950

7. SP of the soap including tax = ₹6.36

Tax = 6% of SP

Let the SP be ₹100

Tax = 6 100 × ₹100 = ₹6

SP including tax = ₹106

Actual SP (original price) = ₹6.36 106 × 100 = ₹6

Profit earned = 20% of CP

CP = 100 10020 + × ₹6 = ₹5

Thus, the manufacturing cost per soap is ₹5.

8. MP of the book = ₹750

Discount applied = 10%

SP of the book after discount is applied = ₹750 − 10% of ₹750 = 750 − 75 = ₹675

Tax levied = 5%

Price of the book after applying the tax = ₹675 + 5% of ₹675 = 675 + 33.75 = ₹708.75

Thus, the customer has to pay ₹708.75 to buy the book.

Challenge

1. Marked price of the shirt = ₹2500

Initial discount = 20%

Price of the shirt after discount = ₹2500 − 20% of ₹2500 = ₹2500 − ₹500 = ₹2000

Additional discount given = 5%

Price of the shirt after additional discount = ₹2000 − 5% of ₹2000 = ₹2000 − ₹100 = ₹1900

Price at which the shirt is bought = ₹2128

Total tax amount = ₹2128 − ₹1900 = ₹228

Tax % = Tax ×100% Price after discount

= 228 ×100% 1900 = 12%

Thus, the tax percentage is 12%.

Chapter Checkup

1. a. 5:2 = 5 ×100% 2 = 250%

b. 15:120 = 15 ×100% 120 = 12.5%

c. 80:125 = 80 ×100% 125 = 64%

d. 44=×100% 2525 = 16%

e. 4419 3=3×100%=×100% 555 4419

3=3×100%=×100% 555 = 380%

f. 1414=×100% 164164 = 8.54%

g. 0.003 = 0.003 × 100% = 0.3%

h. 2.789 = 2.789 × 100% = 278.9%

2. a. 46% = 46:100 = 23:50

b. 0.008% = 0.008 100 = 8 100000 = 1 12500

c. 5.8% = 5.8 100 = 0.058

d. 2 16% 3 = 50 % 3 = 50 3×100 = 50 300 = 1 6

e. 2 1% 5 = 7 % 5 = 7 500 = 0.014

f. 2 33% 3 = 101 % 3 = 101 300

3. a. 5 % 4 of ₹5000 = 5 ×5000 400 ` ₹5000 = ₹62.5

b. 1 % 8 of 1600 L = 1 ×1600 800 L = 2 L

c. 25% of 1250 cm = 25 100 × 1250 cm = 312.5 cm

d. 15.5% of 310 kg = 15.5 100 × 310 kg = 48.05 kg

4. Original Value New Value Percentage Change Increase or Decrease

a. 220 286 286–220 ×100% 220 = 30% Increase

b. 45 81 81–45 ×100% 45 = 80% Increase

c. 5400 4500 540045 – 00 5400 × 100 = 2 16 3 % Decrease

d. 12000 9000 ( ) 12000–9000 ×100% 12000 = 25% Decrease

5. a. CP = SP − Profit = 8064 − 864 = ₹7200

Profit % = 864 ×100% 7200 = 12%

b. SP < CP, there is loss. Loss = ₹460 − ₹437 = ₹23

Loss % = 23 ×100% 460 = 5%

c. SP = CP + Profit = 6800 + 476 = ₹7276

Profit % = 476 ×100% 6800 = 7%

d. SP = CP − Loss = 6400 − 192 = ₹6208

Loss % = 192 ×100% 6400 = 3%

6. a. MP = ₹625, SP = ₹500

Discount = 625 − 500 = ₹125

Discount % = Discount ×100% MP = 125 ×100% 625 = 20%

b. MP = ₹1780, Discount = 15%

Discount % = () MP–SP ×100% MP

15% = 1780–SP ×100% 1780

26700 = 178000 − 100SP

100SP = 151300

SP = ₹1513

7. Average weight of the tigers in a forest last year = 75 kg

Average weight of the tigers in the forest this year = 60 kg

Percentage change in the average weight of tigers = 75–60 75 × 100% = 20%

Thus, the percentage change in the average weight of the tigers is 20%.

8. Let the total number of students be x Percentage of boys = 65%

Number of boys = 6513 ×= 10020 x x

Number of girls = 700 13 =+700 20 x x 13+14,000 = 20 x x

20=13+14,000 xx

20–13=14,000 xx

7x = 14,000

x = 2000

So, there are 2000 students.

9. Given, SP of the refrigerator = ₹27,500; GST = 18%

GST = 18 ×27,500=4950 100 ``

Bill amount = SP + GST = ₹27,500 + ₹4950 = ₹32,450

10. Purchase price of the car for Ajay = ₹5,20,350

Overhead expenses on car = ₹2500 + ₹15,000 = ₹17,500

Total cost to Ajay = ₹5,20,350 + ₹17,500 = ₹5,37,850

Profit made = 14% on CP 14 =×5,37,850 100 =75,299 `

SP of the car = ₹5,37,850 + 75,299=` = ₹6,13,149

11. Price of the watch including tax = ₹1980 Tax = 10%

Let the SP be ₹100, then SP including VAT will be ₹110. SP of the watch (original price) = ₹1980 × 100 110 = ₹1800

Thus, the original price of the watch is ₹1800.

12. Let the CP of 1 pencil be ₹x CP of 10 pencils = ₹10x

Given, SP of 10 pencils = CP of 15 pencils = ₹15x Profit = ₹15x − ₹10x = ₹5x

Profit % = 5 ×100%=50% 10 x x

13. Total number of children = 600

Number of children who chose vegan meal = 280

Number of children who chose non-vegetarian meal = 120

Number of children who chose vegetarian meal = 600 − (280 + 120) = 600 − 400 = 200

a. Percentage of children who chose vegetarian meal = 200 ×100%=33.33%

600

b. Fraction of children who preferred a vegan meal = 2807 60015 =

c. Percentage of children who chose non-vegetarian meal = 120 ×100%=20%

600

14. Let the man’s monthly salary be x.

Salary spent on the house rent = 20 20% of =× 100 20 == 1005 xx xx

Remaining salary = 5– 4 –== 555 xx xx x

40% of the remainder 44048 =40% of =×= 5100525 xxx 44048 =40% of =×= 5100525 xxx

Salary left = ₹9600 So, 48=+9600 525 xx 48–=9600 525 xx

12 =9600 25 x

x= ₹20,000

So, the man’s monthly salary is ₹20,000.

15. Let the price of diesel be ₹x and its consumption be y. Then, expenditure = Price × Consumption ⇒ Expenditure = ₹xy

Since, the price of the diesel increases by 60%, Then, increased price = 60 +60%of =+ 100 160 ==1.6 100

Expenditure remains the same = ₹xy

Reduced consumption = Expenditure == Price1.6x1.6 105 == 168 xyy yy ` ` Expenditure == Price1.6x1.6 105 == 168 xyy yy ` `

16. Given, CP of notebooks per dozen = ₹72

CP of one notebook = ₹6

SP of a packet of 5 notebooks = ₹25

SP of one notebook = ₹5

Since, CP > SP; loss is incurred.

Loss per notebook = ₹6 − ₹5 = ₹1

17. Let the marked price of the book be ₹100

Initial discount = 20% of MP = ₹20

So, selling price after the first discount = ₹100 − ₹20 = ₹80

Second discount = 10% of reduced price = 10% of ₹80 = 10 100 × 80 = ₹8

Selling price after the second discount = 80 − 8 = ₹72

For the selling price to be ₹72, the MP must be ₹100. For the selling price to be ₹1008, the MP must be = 1008×100 7 

= ₹1400

Therefore, the marked price of the book was ₹1400.

18. MP of the TV set = ₹45,000; discount offered = 15% of MP

Discount = 15 ×45,000=6750 100 `

SP = ₹45,000 − ₹6750 = ₹38,250

Sales tax applied = 12% on SP = 12 ×38,250=4590 100 `

Total bill amount = ₹38,250 + ₹4590 = ₹42,840

Thus, Amit pays ₹42,840 for the television.

19. CP of 200 litres of oil = ₹12,000

Amount spent on packaging and transport = ₹2000 + ₹6000 = ₹8000

Total cost to the businessman = ₹12,000 + ₹8000 = ₹20,000

SP of 1 litre oil = ₹120

SP for 200 litres = ₹24,000

Since SP > CP, profit is earned.

Profit = ₹24,000 − ₹20,000 = ₹4000

Profit % = 4000 ×100%=20% 20,000

Thus, the profit percentage is 20%.

20. Let the original price of the apples be ₹x.

Reduced price of the apples after 25% drop in the price

= ₹x − 25% of ₹x

= 0.75x

It is given that, The person can buy 2 kg more apples for ₹300 with the reduced price.

Original quantity of apples bought for ₹300 at ₹x per kg = 300 x

Quantity bought at the reduced price = 300 0.75 x = 400 x

a. According to the question, 400300–2 xx =

100 2 x =

50 x =

Thus, the original price of the apples is ₹50 per kg.

b. Reduced price of the apples = 0.75x = 0.75 × ₹50 = ₹37.5

Thus, the reduced price of the apples is ₹37.5 per kg.

Challenge

1. Statement I: She neither gained nor lost.

Statement II: She lost ₹16.

Statement III: Her loss is 6.25%.

Selling price of each book = ₹120

Profit on the first book = 25%

Let us find the CP of the first book.

25% = 120–CP ×100%

25CP = 12000 − 100CP

125CP = 12000

CP = ₹96

Loss on the second book = 25%

Let us find the CP of the second book.

25% = CP–120 ×100%

25CP = 100CP − 12000

75CP = 12000

CP = ₹160

Total CP of the books = ₹96 + ₹160 = ₹256

Total SP of the book = ₹120 + ₹120 = ₹240

Since, CP > SP, there is loss.

Loss % = 256–240 ×100% 256 = 16 ×100% 256 = 6.25%

Statement I is false as there is loss.

Statement II is true as the loss is ₹16.

Statement III is true as the loss is 6.25%.

Thus, option (a) is correct.

2. Number of students in the school who are eligible to vote = 600

Statement 1: 70% of the eligible students cast their votes. So, the number of eligible students who cast their vote = 70% of 600 = 70 ×600 100 = 420

This statement tells us the number of students who voted but it does not tell us the number of students who voted for Alex.

So, Statement 1 alone is not sufficient to answer the question.

Statement 2: 60% of the eligible votes were received by Sarah. So, the number of eligible votes received by Sarah = 60% of 600 = 60 ×600 100 = 360

This statement alone does not tell us about the number of votes that were casted.

So, Statement 2 alone is not sufficient to answer the question. Combining both the statements.

Number of eligible votes received by Alex = Total votes − Number of votes received by Sarah = 420 − 360 = 60

Percentage of votes received by Alex

Number of votes received by Alex × Total number of students 100%

60 ×100 600 = 10%

Statement 1 helps us tell the total number of votes that were casted while Statement 2 tells us the number of students who voted for Sarah out of the number of voted that were casted.

From this, we can also find the number of votes that Alex got as there were only two candidates.

Thus, both statements together are sufficient to answer the question.

Hence, option (c) is correct.

Case Study

1. Total number of people reported with COVID-19 = 800,000,000

Number of people in China who were reported with COVID-19 = 100,000,000

Percentage of total people who had COVID-19 in China = 100000000 ×100% 800000000 = 12.5%

Thus, option b is correct.

2. Total number of people reported with COVID-19 = 800,000,000

Percentage of Americans who were diagnosed with COVID = 25%

Number of Americans who were diagnosed with COVID = 25 100 × 800,000,000 = 200,000,000

Thus, option c is correct.

3. Total covid cases in India = 45 million = 4,50,00,000

A = P + SI = ₹800 + ₹112 = ₹912

b. R = 7% p.a.; T = 5 years; SI = ₹904

SI = 75 904 100 100 PTRP ⇒ ×××× = 75 904

100 100 PTRP ⇒ ×××× = 75 904

100 100 PTRP ⇒ ×××× = 904100 75 P × = ×

⇒ P = ₹2582.86

A = P + SI = ₹2582.86 + ₹904 = ₹3486.84

c. P = ₹4800; R = 7.5% p.a.; SI = ₹900

SI = 100 PTR ××

48007.5900100 900 10048007.5 T T ⇒ ××× == ×

Total cases in South-Eastern Asia region = 6,00,00,000

Percentage of South-Asia region who were Indians and had covid = 45000000 60000000 × 100% = 75%

4. Total number of people in India = 1,40,00,00,000

Percentage of people who got vaccinated with primary series of COVID-19 vaccine = 70%

Number of people who got vaccinated with primary series of COVID-19 vaccine = 70% of 1,40,00,00,000 = 70 ×1,40,00,00,000 100 = 98,00,00,000

Percentage of the people who at least one booster dose who were already vaccinated = 25%

Number of the people who at least one booster dose who were already vaccinated = 25% of 98,00,00,000 = 98,00,00,000 25 × 100 = 24,50,00,000

Thus, 24,50,00,000 took at least one booster dose.

5. Answersmayvary.

Chapter 10

Let’s Warm-up

Column A

Column B

1. 10% of `1000 `140

2. 5% of `4000 `240

3. 2% of `7000 `360

4. 4% of `6000 `100

5. 12% of `3000 `200

Do It Yourself 10A

1. a. P = ₹800; R = 3.5% p.a.; T = 4 years

SI = 80043.5 100100 PTR×××× = = ₹112

⇒ T = 2.5 years. So, the time is 2.5 years

A = P + SI = 4800 + 900 = ₹5700

d. P = ₹9600; R = 8% p.a.; T = 3 months

SI = 3 96008 12 100100 PTR ×× ×× = = ₹192

A = P + SI = ₹9600 + ₹192 = ₹9792

e. P = ₹20,000; T = 2 years; A= ₹21,800

A = P + SI

₹21,800 = ₹20,000 + SI

₹21,800 − ₹20,000 = SI

₹1800 = SI

So, the simple interest is ₹1800.

SI = 20,0002 1800 100 100 PTRR ⇒ ×××× = 20,0002 1800 100 100 PTRR ⇒ ×××× = 1800100 4.5% 20,0002 R ⇒ × == ×

Thus, the rate of interest is 4.5% p.a.

2. P = $20,000 R = 4.25% T = 3 years

SI = 100 PRT = 20,000 4.25 3 100 ×× = $2550

Total amount = P + SI = $20,000 + $2550 = $22550

So, the interest earned after 3 years is $2550, and the total amount will be $22,550.

3. We know that the amount in 2 years = ₹4400

Amount in 3 years = `4600

Interest earned in 1 year

= ₹4600 − ₹4400 = ₹200

So, the interest earned in 2 years will be = ₹200 × 2 = ₹400

Since Principle = Amount − Interest = ₹4400 − ₹400 = ₹4000

Since Simple Interest = P × R × T 100

200 = 4000 × R × 1 100 200 × 100 4000 = R R = 5%

Hence, the principle is ₹4000, and the rate of interset is 5% p.a.

4. P = ₹3650, R = 10% p.a., T = 3 Jan 2016 to 17 Mar 2016 = 29 days of January + 29 days of February (since 2016 is a leap year) + 17 days of March = 75 days

SI = P × R × T 100 = 3650 × 10 × 75 100 × 366 = ₹74.79

5. Answersmayvary.Sampleanswer:

Ravi deposits ₹12,000 in a savings account at a local bank. The bank offers a simple interest rate of 6% per annum. Ravi plans to keep his money in the account for 4 years. How much interest will Ravi earn after 4 years, and what will be the total amount in his account at the end of 4 years?

Challenge

1. Total SI earned from all deposits = SI123 SI SI ++ = ₹360

SI1 = P1RT1 100 = 600 × R × 3 100 = 18R

SI2 = P2RT2 100 = 900 × R × 2 100 = 18R

SI3 = P3RT3 100 = 1500 × R × 4 100 = 60R 18R + 18R + 60R = ₹360 96R = ₹360

R = ₹360 ÷ 96 = 3.75 So, the rate of interest is 3.75%

Do It Yourself 10B

1. P = ₹1600, T = 5 years, R = 6% p.a.

A = 1+ 100 n R P æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø = 5 6 16001+ 100 æö

= 5 106 1600 100 æö ÷

èø = ₹2141.16

CI = A − P = ₹2,141.16 − ₹1600 = ₹541.16

2. A = ₹4410, T = 2 years, R = 5% p.a., Principle = ?

A = 1+ 100 n R P æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø

₹4410 = 2 5 1+ 100 P æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø

₹4410 = 2 105 100 P 

P = ₹4000

3. P = ₹60,000, A = ₹69,457.50, R = 5% p.a., T = ? =1+ 100 n R AP     5

69,457.50=60,0001+ 100 næö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø

69,457.50105 = 60,000100 næö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø

1.157625 = (1.05)n (1.05)3 = (1.05)n Therefore, n = 3 years

4. a. P = ₹50,000; R = 12%; n = 2 years

A = 2 12 1+=50,0001+ 100 100 n R P 

  2 112 =50,000 100

= 50,000 × 1.12 × 1.12 = ₹62,720

b The principle (P) at the beginning of 3rd year = Amount at the end of 2nd year So, P = `62,720; R = 12% and n = 1 year

A = 1 12 1+=62,7201+ 100 100 n R P 

1 6272×112 112 62,720= 10010 

= `70,246.40

5. When the interest is compounded half-yearly, P = ₹2,00,000; R = 8 =4% 2 and n = 18 months = 1.5 years = 3 half years ( ) 3 4

CI=×1+ =2,00,0001+2,00,000 100 100

=2,00,000×1.04×1.04×1.042,00,000 =2,24,972.82,00,000=24,972.8 n R PP éùé ù êúêæöæö ú çç÷÷ êúê÷÷ ú çç÷÷ êúêçç÷÷ ú çç÷÷ çç êúêèøèø ú ëûë û```

6. P = ₹2500, T = 2 years, R = 10% p.a.

SI = 2500×10×2 = 100100 PRT = ₹500

A = 1+ 100 n R P æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø = 2 10 25001+ 100 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø = 2 110 2500 100 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø = ₹3025

CI = A − P = ₹3025 − ₹2500 = ₹525

Difference = CI − SI = ₹525 − ₹500 = ₹25

7. The principle amount and rate for the first year will be:

P = $30,000; R = 4.75%; n = 2 year,

A = 2 2×1 4.75 1+=30,0001+ 100 100 n R P 

2 2 104.75 =30,000 =300×(104.75) 100

$32,91768.75 = So, Alison will have $32,91768.75 in the account after 2 years.

8. SI = ₹9000, T = 2 years, R = 6% p.a., CI = ?

SI = 100 PRT ×6×2

9000= 100 P

P = ₹75,000

A = 1+ 100 n R P

= 2 6 75,0001+ 100

= ₹84,270

CI = A − P = ₹84,270 − ₹75,000 = ₹9270

9. a. When the interest is compounded annually,

P = ₹15,000; R = 10% and n = 1 year

CI = ×1+ 100 n R PP 

= (15,000 × 1.1) − 15,000 = ₹16,500 − ₹15,000 = ₹1500

b. When the interest is compounded semi-annually, P = ₹15,000; R = 10 =5% 2 and n = 1 × 2 = 2 half years

CI = ×1+ 100 n R PP 

2121 =15,000××15,000 2020  

= ₹16,537 − ₹15,000 = ₹1537.5

₹15,000 lent at 10% per annum compounded semi-annually for 1 year will earn more interest. The difference is ₹1537.5 − ₹1500 = ₹37.5.

10. We have A = ₹27,783; R = 10% per annum = 10 =5% 2 per half year; n = 1 1 2 years = 3 2 years = 3 ×2=3 2 half years

We know that A = 1+ 100 n PR

Putting the value in the above formula, we get

27,783 =

33 5 21 1+ 27,783= 100 20 PP

27,783×20×20×20 == 21×21×21 P ₹24,000

11. Answersmayvary.Sampleanswer:

Anita deposits ₹10,000 in a bank that offers a 7% interest rate per annum. Calculate the difference between the simple interest and the compound interest after 3 years, if the compound interest is compounded annually.

Challenge

1. We know that A = 

1+ 100 n R P

Given that P = P; n = 3 years; R = R and A = 216 P 125

Substituting the values in the formula gives: 216 125 P =  ⇒

33 216 1+ =1+ 100125100 RR P

Taking cube root on both side we get,  ⇒

3 3 3 216 6 =1+ =1+ 1251005100 RR 61 =1==20%. 10055 RR −⇒

Hence, the rate of interest is 20%

2. SI = ×××2×10 1 = ==0.20 1001005 PTRP PP

CI = 

n R PPPPPPP

2 10 ×1+ =1+ =1.21=0.21 100

Given that 0.21P – 0.20P = 0.01P = ₹50

P = 50 =5000 0.01

Hence, the sum borrowed is ₹5000.

Do It Yourself 10C

1. P = ₹8000, R = 12% p.a., T = 9 12 years

SI = ××8000×12×9 = 100100×12 PRT

××8000×12×9 = 100100×12 PRT = ₹720

So, the interest received by Savita is ₹720.

A = P + SI = ₹8000 + ₹720 = ₹8720

So, Savita collected ₹8720 from the bank.

2. R = 8% p.a., T = 4 years, A = ₹9900, P = ?

A = P + SI, =+ 100 PRT AP , =1+ 100 RT AP æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø 8×4

9900=1+ 100 P æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø ; 100+32

9900= 100 P æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø 9900×100 = 132 P

Principle = ₹7500

So, the amount borrowed by Kanchan originally was ₹7500.

3. P = ₹3200, R = 6% p.a., A = ₹3680, T = ?

SI = A − P = ₹3680 − ₹3200 = ₹480

SI = 100 PRT

3200×6× 480= 100 T

480×100 = 3200×6 T T = 2.5 years. So, Suresh used Narendra’s money for 2.5 years.

4. A = ₹6720, P = ₹6000, T = 1 year and 6 months = 1.5 years, R = ?

SI = A − P = ₹6720 − ₹6000 = ₹720

SI = 100 PRT

6000××1.5 720= 100 R

720×100 = 6000×1.5 R R = 8% p.a.

So, the rate of interest charged by the bank is 8% p.a.

5. P = ₹25,000, R = 8.5% p.a., T = 3 years, A = ?

=1+ 100 n R AP

3 8.5 =25,0001+ 100 A

3 108.5 =25,000 100 = ₹31,932.23

6. Principle borrowed by Yogesh = ₹30,000, R = 12% p.a., T = 4 years

Amount to be paid by Yogesh to Navin after 4 years = 1+ 100 RT P æö

èø = 148 30,000 100 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷

Amount to be received by Yogesh from Akash after 4 years

1+ 100 n R P

Amount gained by Yogesh = ₹47,205.58 − ₹44,400 = ₹2805.58

7. P = 1,80,000, R for the first year = 8.5%, R for the second year = 7.5%, R for the third year = 7%, T = 3 years, amount after three years = ? Amount after

Amount after the third year =

8. A = ₹79,860, T = 3 years, R = 10% p.a. compounded annually, P = ? =1+ 100 n R AP

3 110 79,860= 100 P P = ₹60,000

9. P = ₹6400, T = 3 years, R = 1 7% 2 p.a. compounded half yearly, A = ? A

2 1+ 200 n R P

10. Present population (P) = 12,50,000, rate of increase (R) = 10%, time (T) = 4 years

after four years =

11. Present price of the plot = ₹2,00,000, rate of appreciation = 5%, T = 3 years

Expected price after 3 years =

12. Cost of the computer = ₹60,000, rate of depreciation = 40%, T = 2 years

Price of the computer after two years 60,000

13. P = ₹50,00,000 R = 6.5% T = 2 years

Compounded annually, n = 2

Total interest received by RBI = A – P = ₹56,71,125 − ₹50,00,000 = 6,71,125

So, the total interest received by RBI would be ₹6,71,125.

14. Answermayvary.Sampleanswer:

Sunita borrowed a sum of ₹10,000 from the bank at the interest rate of 10% per annum, compounded annually, and paid back an amount of ₹12,100. For how much time did she borrow the sum?

Challenge

1. Assertion (A): Raj borrows ₹80,000 for 4 years at 10% per annum, compounded annually. Then she has to pay ₹1,17,128 after 4 years.

P = ₹80,000

R = 10%

T = 4 years

A =     1+ 100 n R P

= 80,000 

4 10 1+ 100

= ₹1,17,128

So, Assertion (A) is true.

Reason (R): Interest is the ratio of amount to principle. This is incorrect. Interest is actually calculated as the difference between the amount and the principle.

A is true and R is false.

So, option c is correct.

Chapter Checkup

1. a. P = ₹3500; T = 3 years; R = 5%

SI = ×× 100 PTR

SI = 3500×3×5 =525 100 `

Amount = P + SI = ₹3500 + ₹525 = ₹4025

b. SI = ₹392, T = 3.5 years, R = 3.5% p.a., P = ?

SI = ×× 100 PTR

392= 100 P

×3.5×3.5

392×100

3.5×3.5 = P

P = ₹3200

Amount = P + SI = ₹3200 + ₹392 = ₹3592

c. P = ₹2700, R = 9% p.a., T = 146 days = 146 365 years

SI = ×× 100 PTR

= 2700× 9×146 =97.2 100×365 `

Amount = P + SI = ₹2700 + ₹97.2 = ₹2797.2

d. SI = A − P = ₹11,800 − ₹10,000 = ₹1800

SI = ×× 100 PTR

1800 = 10,000 × × 2 100 R

1800×100 ==9% 10,000×2 R

2. a. P = ₹5000, R = 9% p.a., n = 2 years, A = ?, CI = ?

A =     1+ 100 n R P =     2 9 50001+ 100 =     2 109 5000 100 = ₹5940.5

CI = A − P = ₹5940.5 − ₹5000 = ₹940.5

b. P = ₹72,000, R = 6% p.a., n = 3 years, A = ?, CI = ? A =  

3 6 72,0001+ 100 =

3 106 72,000 100 = ₹85,753.152 CI = A − P = ₹85,753.152 − ₹72,000 = ₹13,753.152

3. A = ₹16,335, R = 10% p.a. compounded annually, n = 2 years, P = ? A =

1+ 100 n R P

2 10

16,335=1+ 100 P

2 110

16,335= 100 P

16,335×100×100 = 110×110 P

P = `13,500

4. P = ₹8800, A = ₹10,648, R = 10% p.a. compounded annually, n = ?

A = 

1+ 100 n R P

10,648=88001+ 100 n

10,648110 = 8800100 n

1.21=(1.1)n 2 (1.1)=(1.1)n Hence, n = 2 years

5. P = ₹4000, A = ₹5324, n = 3 years, R = ?

A =

1+ 100 n R P

3 5324=40001+ 100 R

3 1.331=1+ 100 R

3 3 (1.1)=1+ 100 R 

Taking cube roots on both the sides, we get,

1.1=1+ 100 R æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø

1.11= 100 R -

0.1= 100 R

R = 10% per year

6. P = ₹20,000, R = 10% p.a. compounded semi-annually, n = 1 1 2 years 2 =1+ 200 n R AP   

3 2× 2 10 = 20,000×1+ 200 

CI = A − P = ₹23,152.5 − ₹20,000 = ₹3152.5

7. Let principle (P) = ₹a, R = 8% p.a. simple interest, amount (A) = ₹2a, T = ?

Using formula, A = P + SI, we get, 2a = a + SI

2a − a = SI

SI = a We know that SI = ×× 100 PRT

×8× = 100 aT a ×100=×8× aaT

100 = 8 T T = 12.5 years

8. Amount after two years = ₹780, amount after seven years = ₹1230, P = ?, rate of simple interest = ?

SI for five years = ₹1230 − ₹780 = ₹450

SI for one year = 450 =90 5 `

SI for two years = 2 × ₹90 = ₹180

Now, P = ₹780 − ₹180 = ₹600

SI = ×× 100 PRT

180= 100 R R = 15%

600××2

9. Number of people who donated blood initially = 10,000

Percentage increase in the number of people who donated the blood every year = 15%

Number of people who donated the blood after 2 years = 2 15 100001+ 100

= 10000(1.15)2 = 13,225 people

Thus, 13,225 people donated blood after 2 years.

10. P = ₹25,000, R = 8% p.a. compounded semi-annually, A = ₹28,121.60, n = ? 2 =1+ 200 n R AP     2 8 28,121.60=25,0001+ 200 n  

2 28,121.60208 = 25,000200 n 

1.124864 = (1.04)2n Since, 1.124864 = (1.04)3

Therefore, 2=3 n 3 ==1.5 2 n years

11. SI = ₹4500, T = 3 years, R = 10% p.a. SI = ×× ×3×10 4500= =15,000

For finding compound interest, we have the formula as:

CI = ×1+ 100 n R PP    

CI = 3 10 15,000×1+–15,000 100  

CI = 15,000 × (1.1)3 −15,000

CI = 19,965 − 15,000 = ₹4965

12. P = ₹5000, R = 8% p.a., Time = 3 years, CI − SI = ?

A = 1+ 100 n R P  

= 3 8 50001+ 100 

= 3 108 5000 100

= ₹6298.56

CI = A − P = ₹6298.56 − ₹5000 = ₹1298.56

SI = 5000×8×3 = =1200 100100 PRT = ₹1200

So, CI − SI = ₹1298.56 − ₹1200 = ₹98.56

13. P = ₹12,000, n = 2 years, R = 9% p.a. compounded annually, A = ?

A = 1+ 100 n R P

= 2 9 120001+ 100

= 2

14. Present population (A) = 54,080, rate of increase (R) = 4% annually, time (t) = 2 years, population two years ago (P) = ? 2 4 54,080=1+ 100 P

54,080×100×100 = =50,000 104×104 P

15. Present value of a car (A) = ₹4,73,850, rate of depreciation (R) = 10%, time (T) = 3 years, price of the car three years ago (P) = ? 3 10

4,73,850=1 100 P

4,73,850×100×100100 = =6,50,000 90×90×90 P ´ `

16. Answermayvary.Sampleanswer:

Rahul is planning to invest ₹40,000 in a high-yield savings account that offers an annual interest rate of 8%, compounded semi-annually. He wants to know how much his investment will grow to in 3 years. Calculate the compound interest earned and the total amount after 3 years.

Challenge

1. R = 10% p.a., Time = 2 years, CI − SI = ₹1, P = ? As per the given condition, CI − SI = ₹1

2. Statement I: A sum of ₹20,000 amounts to ₹24,200 at the rate of compound interest.

Statement II: The amount was invested for a period of 2 years under compound interest.

Statement I gives the information about the principle and amount, but there is no information about time so we cannot find the rate of compound interest. Hence, Statement I alone is not sufficient.

Statement II gives information about time only and hence Statement II alone is not sufficient to find the answer.

On combining statements I and II we get the information about the unknown as:

Principle = ₹20,000; Amount = ₹24,200; and Time = 2 years

On placing the known values in the formula, A = 1+ 100 n R P 

, we can find the value of R. Hence, statements I and II together are sufficient to answer.

Thus option 3 is correct.

Case Study

1. Principle invested by investor B = ₹18,000 Interest earned = ₹6480; Time = 3 years

SI = ×× 100 PTR

6480 = 18000×3×6480×100 = =12%. 100 18000×3 R R ⇒

Thus, option c is correct.

2. CI for investor C = ₹3660.8

Principle = ₹22,000; Amount = 22,000 + 3660.8 = ₹25,660.8

Rate = 8%

Putting the values in the formula, we get:

A = 1+ 100 n R P 

25,660.8 = 8 220001+ 100 n 

25660.8108 = 22000100 n

256608108 = 220000100 n

11664108 = 10000100 n

[on dividing by 22] 2 108108 = 100100 n

On equating LHS and RHS, we get n = 2 years

so, option c is correct.

3. Amount invested in 2022 = ₹12,000

Rate of simple interest = 10%

SI = ×× 100 PTR = 12000×1×10 =1200 100 `

Amount = ₹12,000 + ₹1200 = ₹13,200

Amount invested in 2023 = ₹13,200

Rate of compound interest = 12% compounded semi annually

Time = 1 year

A = 2 1+ 200 n R P 

A = 2×1 12 132001+ 200

A = 13,200 × 1.06 × 1.06 = ₹14,831.52

So, option d is correct.

4. Principle for investor D = ₹20,000

Time = 2 years

CI = ₹4642

Amount = ₹20,000 + ₹4642 = ₹24,642

A = 1+ 100 n R P 

24642 = 2 200001+ 100 R

2 24642 =1+ 20000100 R

2 12321 =1+ 10000100 R

2 2 100+ 111 = 100100 R

On equating, we get R = 11% If the rate of interest is 2% more = 11 + 2 = 13%, the amount can be given as:

A = 2 13 200001+ 100

A = 20,000 × 1.13 × 1.13 = ₹25,538

Chapter 11

Let’s Warm-up

1. 8p − 3p −5b

2. 7 + 8 5b

3. 2p + 5p

Do It Yourself 11A

1. a. –x2 + 2x + 3

The terms containing variables are: −x2, 2x b. –x2 + 5x + 6

The terms containing variables are: −x2, 5x c. –x2 + 4x + 2

The terms containing variables are: −x2, 4x d. 2x + 7

The term containing a variable is 2x.

2. a. 3 4 a2bc 3 4 aabc

The expression 2 3 –4 abc has only one term that is 2 3 –4 abc

Factors = 3 –,,,, 4 aabc

The coefficient of ‘a’ is 3 –4 abc

6a + 5 6a 5 6 a

The expression 6a + 5 has two terms 6a and 5.

Factors = 6, a; 5.

The coefficient of ‘a’ is 6. c.

The expression 8a2 − 5a+ 1 3 has three terms 8a2, −5a and 1 3

Factors = 8, a, a; −5, a; 1 3 .

The coefficient of ‘a’ in 8a2 is 8a. The coefficient of ‘a’ in −5a is −5.

7 12 7 12 a3 + 5abc 4 9 a2 10 7 12 a3 4 9 a2 4 9 aaa 5 abcaa 5abc 10

The expression 7432+5––10 12 9 aabca has four terms 3 7 12 a , 5abc, 2 4 –9 a and −10.

The factors of 7432+5––10 12 9 aabca are 74,,,;5,,,;–,,;–10 12 9 aaaabcaa

The coefficient of ‘a’ in 3 7 12 a is 2 7 12 a ;

The coefficient of ‘a’ in 5abc is 5bc;

The coefficient of ‘a’ in 2 4 –9 a is 4 –9 a

3. Pairs of like terms are circled and pairs of unlike terms are crossed out. a. 5a, 2 3 a b. 6xy, 8x2y

c. 2a2 b2 c, 9ab2c2 d. 11 7 abc, 8cba

4. a. + 8 xy , literal factors of the expression + 8 xy are x and y

In 8 x the literal coefficient of 1 8 is x.

In 8 y the literal coefficient of 1 8 is y.

b. 1 2–6 xy , literal factors of the expression 1 2–6 xy are

x and y. In 2x the literal coefficient of 2 is x and the literal coefficient of 1 –6 is y.

c. In (ab) − ab, the literal factors of the expression (ab) − ab are a,b,a,b. In ab, the literal coefficient of 1 is a and literal coefficient of −1 is b; in −ab, the literal coefficient of −1 is ab

5. A B

Length of a each blue straws = x cm

Length of a each green straws = y cm

The algebraic expression for Figure A is x + x + y = 2x + y, and the algebraic expression for Figure B is y + y + x = 2y + x

6. a. Show the 2 expressions using algebra tiles.

Place all the tiles together

Show the final answer in the form of an algebraic expression. 3x2 + 5x + 6

b. Show the minuend, 2 4–2+1 xx using algebra tiles.

To subtract, cancel out as many tiles as given in the subtrahend, 2 +3–2xx

To subtract 2 x , cancel out one 2 x tile.

To subtract +3x, we need at least three +x tiles. Since there are not enough +x tiles, we bring in 0 (3 pairs of +1x and

−1x tiles). Then cancel out the three +x tiles.

To subtract −2, we need two −1 tiles. Since there are not enough −1 tiles, we again bring in 0 (2 pairs of +1 and −1 tiles).

The remaining tiles show the final answer.

x2 x2 x2 xxxxx 1 1 1

So, the resulting algebraic expression is 3x2 − 5x + 3.

7. 3x + y + 1 and 2x − 5y − 4

3++1+2–5–4

3+2+–5+1–4=5–4–3 xyxy xxyyxy =

Verification:

3x + y + 1

+ 2x 5y 4 5x 4y 3

b. 3xy2 + 5xy − 6 and 7y2x − 2xy + 3 3xy2 + 5xy − 6 + 7y2x − 2xy + 3

= 3xy2 + 7y2x + 5xy − 2xy − 6 + 3

= 10xy2 + 3xy − 3

Verification:

2 3 xy + 5 xy 6

+ 2 7 yx 2 xy + 3 2 10 xy + 3 xy 3

c. 2 4 –5–+2 5 pp and 2 3 3+5–4 pp 22 22 2 43

–5–+2+3+5–54 –5+3–+5+2–43 54 –2++215 54 pppp pppp pp = =

Verifcation:

–2 5 p –4 5 p + 2

+ 2 3 p + 5 p 3 4

2 2 p + 21 5 p + 5 4 –

d. 76+–1 87pqp and 52++8 87pqp

7652128+–1+++8=++7 878787 pqppqppqp

Verification: 7

8 pq + 6 7 p 1

+ 5 8 pq + 2 7 p + 8 12

8 pq + 8 7 p + 7

8. a. 2 5–3–1 xx and 2 3–4+7 xx 22 22 22 2

5–3–1–(3–4+7) 5–3–1–3+47

5–3–3+4–1–72+–8 xxxx xxxx xxxxxx = = =

Verification:

5x2 3x 1 3x2 4x + 7 +

2x2 + x 8

b. 3 4 8+–5 mnn and 3 2–+11 25 mnn

33 33 3 8+––2–+411 525 =8+––2+–411 525 3 =6+–1 2 mnnmnn mnnmnn mnn

Verification:

8mn + 3n 4 5

2mn 3 1 2 n + 1 5 +

6mn + 3 3 2 n 1

c. 2 2+3+8 yy and 2 +2–4yy

22 222 2+3+8–(+2–4) =2+3+8––2+4=2+12 yyyy yyyyy

Verification:

2 y + 3y2 + 8

2 y + y2 4 +

0 + 2y2 + 12

d. 7–13–11pqq and 2pq + 8 − 4q 7–13–11–(2+8–4) =7–13–11–2–8+4=–1–15–7 pqqpqq pqqpqqpqq

Verification:

7 13 pq 11q

8 + 2 pq 4q + 1 15 pq 7q

9. a. 3(5–4)–2(7+3)+4(8–2) xxx = 15–12–14–6+32–8 xxx = –18+32–8xx = 33–26 x

b. 3232 ) 13 (2+4–( ) 5–2 24 mmmm = 3232 +2–+153 42 mmmm = –+11732 42 mm

c. 23 2 (–3+6–4–)+(2+1)–(8–5+1) xxxxxx

23 2 = –3+6–4–+2+1–8+5–1 xxxxxx

223 = –3–8+6–4+1–1–+2+5 xxxxxx

23 =–11+6–4+6 xxx = 32 6–11+6–4 xxx

d. 2 (–2)–(–+3)+(4–5) xxxx = 2 –2–+–3+4–5 xxxx

= 2 6–10–xx = 2 –+6–10 xx

10. a. Let the missing expression be x 443 ) (6–4+8)+=(+5+8+8 aaxaaa

4 34 =+5+8+8–6+4–8 xaaaaa =–5+9+843 xaa

b. Let the missing expression be a 22 –(6+6+11)=(10+4–1) axxxx

22 =10+4–1+6+6+11 axxxx

2 =16+10+10axx

c. Let the missing expression be x 222 (–19–6+6)+(–8–4–9)–=–15+8+7 nnnnxnn

222 –19–6+6–8–4–9+15–8–7= nnnnnnx

222 –19–8+15–6–4–8+6–9–7= nnnnnnx

2 –12–18–10=nnx

11. We will first find the sum of 2 (–18+–32) pp and 2 (132012) pp+− and then subtract 2 ) 40–1( 3 p from it.

Sum of 2 (–18+–32) pp and 2 (13+20–12) pp

= 22 (–18+–32)+(13+20–12) pppp

= 2 2+14–44 pp

Subtracting 2 ) 40–1( 3 p from 2 (2+14–44) pp , we get, 22 ) (2+14–44)–(40–13 ppp

22 =2+14–44–40+13 ppp

2 =15+14–84 pp

12. Let the algebraic expression to be subtracted from 3 () 7–3+7hhth to get 3 () –6+3–9hhth be x

33 ) (7–3+7–=(–6+ ) 3–9 hhthxhhth

(7–3+7–(–333 ) 6+– ) 9= hhthhhthx

33 7–3+7+6–3+9= hhthhhthx

3 =13–6+16 xhhth

13. The difference in the lengths of the ropes of Rohan and Harsh

= 22 (8+30–4)–(13–18) + mmmm

= 8m + 30m2 4 m2 13m + 18

22 =30–+8–13–4+18 mmmm

2 =(29–5+14) mm metres

14. The length of the fence required for the field

2 222 ) =(9+17) +(34+7–18+(9+2 +(1 ) 7–25)ssssss

= 2 222 9+17+34+7–18+9+2+17–25 ssssss

= 2222 9–18+2+17+34+9+17+7–25 ssssss

= 2 (10+43–1) ss metres

15. Answermayvery.Sampleanswer: Rohit has (5x + 7) fiction books, and (x + 3) not-fiction books in his library. How many books does he have in all?

Challenge

1. The wrong coefficients in the first step are circled in the solution below.

The correct solution is:

22222222 21122112 –––=––+ 35353535 ppqppqppqppq

= 21122222 ––+ 3355 pppqpq

= 1122 + 35ppq

Do It Yourself 11B

y y y y x xy xy xy xy

x xy xy xy xy

Product = −20xy b. x x x 1 1 1 1 y xy xy xy y y y y

xy xy xy y y y y

x x x 1 1 1 1

x x x 1 1 1 1 1 x x x 1 1 1 1

Product = 6xy + 8y − 9x − 12

3. a. (−5x2y2z3) (12x2yz)(3yz) = (−5 × 12 × 3) (x2y2z3 × x2yz × yz) = −180 × x2+2 × y2+1+1 × z3 = −180 × x4 × y4 × z5 = −180 x4y4z5

b. ab (−6abc + 5c − 7) = −6a2b2c + 5abc − 7ab

c. (3x − 1) (8y + 5) = 3x (8y + 5) − 1 (8y + 5) = 24xy + 15x − 8y − 5

d. (7y + 9) (3z2 + 6z − 5) = 7y (3z2 + 6z − 5) + 9 (3z2 + 6z − 5)

= 21yz2 + 42yz − 35y + 27z2 + 54z − 45

e. (2a2 − 3a + 7) (4a2 + 6a + 11)

= 2a2 (4a2 + 6a + 11) − 3a (4a2 + 6a + 11) + 7 (4a2 + 6a + 11)

= 8a4 + 12a3 + 22a2 − 12a3 − 18a2 − 33a + 28a2 + 42a + 77

= 8a4 + 32a2 + 9a + 77

f. (8p5 − p3 + 7p2 + 10p − 5) (3p2 + 9p + 11)

= 8p5 (3p2 + 9p + 11) − p3 (3p2 + 9p + 11) + 7p2 (3p2 + 9p + 11) + 10p (3p2 + 9p + 11) − 5 (3p2 + 9p + 11)

= 24p7 + 72p6 + 88p5 − 3p5 − 9p4 − 11p3 + 21p4 + 63p3 +

77p2 + 30p3 + 90p2 + 110p − 15p2 − 45p − 55

= 24p7 + 72p6 + 85p5 + 12p4 + 82p3 + 152p2 + 65p − 55

4. a. (5h + 9) (3h − 7) = 5h (3h − 7) + 9 (3h − 7)

= 15h2 − 35h + 27h − 63

= 15h2 − 8h − 63

For h = 1, 15h2 − 8h − 63 = 15 (1)2 − 8 × 1 − 63 = −56

b. (9q − 1) (6q2 − q + 5) = 9q (6q2 − q + 5) − 1 (6q2 − q + 5)

= 54q3 − 9q2 + 45q − 6q2 + q − 5 = 54q3 − 15q2 + 46q − 5

For q = −4,

54(−4)3 − 15(−4)2 + 46(−4) − 5

= 54 × (−64) − 15 × 16 + 46 × (−4) − 5

= −3456 − 240 − 184 − 5 = −3885

c. (7a2 − 4a − 2) (5a2 − 2a + 9)

= 7a2 (5a2 − 2a + 9) − 4a (5a2 − 2a + 9) − 2 (5a2 − 2a + 9)

= 35a4 − 14a3 + 63a2 − 20a3 + 8a2 −36a − 10a2 + 4a − 18

= 35a4 − 34a3 + 61a2 − 32a − 18

For a = 0, 35a4 − 34a3 + 61a2 − 32a − 18

= 35(0)4 − 34(0)3 + 61(0)2 − 32 × 0 − 18 = −18

d. (2t3 + 3t2 + 5t − 4) (3t2 − 5t + 11)

= 2t3 (3t2 − 5t + 11) + 3t2 (3t2 − 5t + 11) + 5t (3t2 − 5t + 11) − 4 (3t2 − 5t + 11)

= 6t5 − 10t4 + 22t3 + 9t4 − 15t3 + 33t2 + 15t3 − 25t2 + 55t −

12t2 + 20t − 44

= 6t5 − t4 + 22t3 − 4t2 + 75t − 44

For t = −2, 6(−2)5 − (−2)4 + 22(−2)3 − 4(−2)2 + 75(−2) − 44

= −192 − 16 − 176 − 16 − 150 − 44 = −594

5. a. 2 3 a (3a + 4b − 5c) − 4 5 a (2a − 5b + 7c) = −2a2 − 8 3 ab +

10 3 ac − 8 5 a2 + 4ab − 28 5 ac

= 22 –30–40+50–24+60–84 15 aabacaabac

= 2 –54+20–34 15 aabac

For a = −1, b = 2, c = 10, 2 –54+20–34 15 aabac

= 2 –54(–1)+20 (–1)(2)–34(–1)(10) 15

= –54–40+34024682 == 15155

b. a2b (a3 − 3b + a + 2) − ab (b4 − 2b2 − 3b) − c (a3 + b2 + 1)

= −a5b + 3a2b2 − a3b − 2a2b − ab5 + 2ab3 + 3ab2 − a3c − b2c − c

For a = −1, b = 2, c = 10, −(−1)5 (2) + 3 × (−1)2 (2)2 − (−1)3(2) − 2(−1)2(2) − (−1) (2)5 + 2(−1) (2)3 + 3(−1) (2)2 − (−1)3(10) − (2)2(10) − 10

= 1 × 2 + 3 × 1 × 4 + 1 × 2 − 2 × 1 × 2 + 1 × 32 − 2 × 8 − 3 × 4 + 1 × 10 − 4 × 10 − 10 = 2 + 12 + 2 − 4 + 32 − 16 − 12 + 10 − 40 − 10 = −24

c. (a − 5) (a4 −2b3 + 3c − 4) + 3 7 bc (7a2 − 11)

= a (a4 −2b3 + 3c − 4) − 5 (a4 −2b3 + 3c − 4) + 3 7 bc (7a2 − 11) = a5 − 2ab3 + 3ac − 4

− 5a4 + 10b3 − 15

+ 3a2

− 33 7 bc For a = −1, b = 2, c = 10; (−1)5 − 2 (−1) (2)3 + 3(−1) (10) − 4(−1) − 5(−1)4 + 10(2)3 − 15(10) + 20 + 3(−1)2(2)(10) − 33 7 (2) (10)

= −1 + 16 − 30 + 4 − 5 + 80 − 150 + 20 + 60 − 660 7 = −42 − 660 7 = −702 7

6. a. Area of rectangle = 6x2y × −5x3y2z = −30x2+3 y1+2 z = −30x

y3z sq. units

b. Area of rectangle = 117 8–+9 55 æöæö çç÷÷ çç÷÷ çç÷÷ çç÷÷ çç÷÷ èøèø xx = 8x 7 +9 5

− 11 5 7 +9 5 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø x = 56 5 x2 + 72x − 77 25 x − 99 5 = 56 5 x2 + 77 72 –25 æö

99 5 = 2 2 1800–77 56 99 +–5255 56172399 =+–sq.units 5255 xx xx

÷ ç ÷

ç ÷ ç èø x

c. Area of rectangle = 232 96554 += 117677 abcabcabc 

b1+1 c1+3 + 45 66 a1+1 b1+2 c1+1

= 5445324232 + sq.units 7766 abcabc 

d. Area of the rectangle = (3z − 1) (2z2 + 7z − 10) = 3z (2z2 + 7z − 10) − 1 (2z2 + 7z − 10)

= 6z3 + 21z2 − 30z − 2z2 − 7z + 10

= (6z3 + 19z2 − 37z + 10) sq. units

7. We know that, Volume of a cuboid = (length × breadth × height) cubic units

a. 6pqr, 3p2 qr2, 4q2r Volume of the cuboid = 6pqr × 3p2qr2 × 4q2r

= (6 × 3 × 4) × p1+2q1+1+2r1+2+1

= 72p3q4r4 cu. cm

b. Volume of the cuboids = 7 2 p2q3r4 × 1 2 pqr × 2p (q + r)

= 7 2 p2+1+1 q3+1+1 r4+1 + 7 2 p2+1+1 q3+1 r4+1+1

= 77455446+cu.cm 22pqrpqr    

c. Volume of the cuboid = (p + q) × (p − q + 12) × 8r

= 8r (p + q) (p − q) + 96r (p + q) = 8r (p2 − q2) + 96rp + 96rq

= (8rp2 − 8rq2 + 96rp + 96rq) cu. cm

d. Volume of the cuboid = − 1 2 pq(r + 1) × 7pq × 11qr

= − 77 2 p1+1 q1+1+1 r (r + 1)

   

= 777723223 ––cu.cm 22pqrpqr

8. a. Volume of the cube = 3s (stu + 1) × 3s (stu + 1) × 3s (stu + 1)

= 27s3 (stu + 1)3 cu. cm

b. Volume of the cube = −3s2t3u5 × (−3s2t3u5) × (−3s2t3u5)

= −27s2+2+2 t3+3+3 u5+5+5 = −27s6t9u15 cu. cm

c. Volume of the cube = (−6ab + 2a2 + 3) × (−6ab + 2a2 + 3)

× (−6ab + 2a2 + 3)

= (−6ab + 2a2 + 3)3 cu. cm

d Volume of the cube = 4p3q4 (9r + 2q − 5) × 4p3q4 (9r + 2q − 5) × 4p3q4 (9r + 2q − 5)

= 64p3+3+3 q4+4+4 (9r + 2q − 5)3 = 64p9q12 (9r + 2q − 5)3 cu. cm

9. (5x3 − 4x2 + 2x − 1) (x2 − 3x + 5)

= 5x3 (x2 − 3x + 5) − 4x2 (x2 − 3x + 5) + 2x (x2 − 3x + 5) − 1 (x2 − 3x + 5)

= 5x3+2 − 15x3+1 + 25x3 − 4x2+2 + 12x2+1 − 20x2 + 2x1+2

− 6x1+1 + 10x − x2 + 3x − 5

= 5x5 − 15x4 + 25x3 − 4x4 + 12x3 − 20x2 + 2x3 − 6x2 + 10x − x2 + 3x − 5

= 5x5 − 19x4 + 39x3 − 27x2 + 13x − 5

So, the coefficient of x4 in the product is −19.

10. Answersmayvary.Sampleanswer: (2x + 5) units ( x 3) units

Length = (2x + 5) units; Breadth = (x − 3) units

Area = (2x2 – x – 15) sq. units.

11. Length of the rectangular garden = (3x + 2) m, breadth of the garden = (4x − 1) m

Area = (3x + 2) (4x − 1) = 3x(4x − 1) + 2(4x − 1)

= 12x2 − 3x + 8x − 2

= (12x2 + 5x − 2) sq. m

Cost of planting the grass = ₹(2x − 3) per metre

Total cost of planting the grass = ₹(2x − 3) × (12x2 + 5x − 2)

= (2x × (12x2 + 5x − 2)) – 3(12x2 + 5x − 2)

= 24x3 + 10x2 – 4x – 36x2 – 15x + 6

= 24x3 – 26x2 – 19x + 6

Thus, the total cost of planting the grass is ₹(24x3 – 26x2 –19x + 6).

Challenge

1. Assertion: (2x + 3)(x − 1) = 2x(x − 1) + 3(x − 1)

= 2x2 − 2x + 3x − 3

= 2x2 + x − 3

The assertion is true. The reason is also true and it explains the assertion.

Thus, both A and R are true, and R is the correct explanation of A.

Hence, option a is correct.

Chapter Checkup

Expression

1. a. 6pq

Terms

Factors

Factors inside the rectangular box are the variables. Factors inside the circle are the numerical coefficients.

Expression

b. mn + 2n2 − 3m2 mn 2n2 −3m2 2 −3 –mm nnnm

Terms Factors

Factors inside the rectangular box are the variables. Factors inside the circle are the numerical coefficients.

2. a. 4x + y, 2x − 1 quadrinomial

b. 8a2b, 9pq, −6x trinomial

c. 3xy + x − 1, 8z2 − 3z + 5 monomial

d. −6x2y + 2xy 6x + 4, p2 − 7pq + 4q − 1 binomial

3. Like terms are circled and unlike terms are crossed out. a. 2a, − 3 5 a

4. a. 22 (3+2–6)+(––2+1) xxxx

Show the two expressions using algebra tiles. x2 x2 x

Place all the tiles together. Cross out each pair of one negative and one positive tile of the same factor.

The remaining tiles show the final answer.

The resulting expression is 2 2–5 x . b. 22 (–+4+2)–(3–2–3) xxxx

Show the minuend, 2 (–+4+2) xx using algebra tiles.

To subtract, cancel out as many tiles as given in the subtrahend, 3x2 − 2x − 3.

To subtract 3x2, we need three x2 tiles. Since there are not enough x2 tiles, we bring in 0 (3 pairs of x2 and −x2 tiles).

To subtract −2x, we need at least two −x tiles. Since there are not enough −x tiles, we again bring in 0 (2 pairs of x and x tiles).

To subtract −3, we need three −1 tiles. Since there are not enough −1 tiles, we again bring in 0 (3 pairs of 1 and −1 tiles).

The remaining tiles show the final answer.

The resulting expression is –4x2 + 6x + 5.

5. a. –3m2 + m – 2 and 4m2 + 6m + 7

–3m2 + m – 2 + 4m2 + 6m + 7

= –3m2 + 4m2 + m + 6m–2 + 7

= 2 +7+5mm

b. 32 7+4+–1 zzz and 32 2–6–2+2 zzz

3232 7+4+–1+2–6–2+2 zzzzzz

= 3322 7+2+4–6+–2–1+2 zzzzzz

= 32 9–2–+1 zzz

c. 23232 51 ––7+3+4–+9–6–2 77 qqqqqq

= 22233 51 +3+9–––6–7+4–2 77 qqqqqq

= 23 6 13––6–5 7 qqq

6. a 3655263 11 11+10–26+–8–9+++17 22 zzzzzzzzz

= 3655263 11 11+10–26+–8+9–––17 22 zzzzzzzzz

= 3366 552 11 11–17+10––26–+–8+9 22 zzzzzzzzz

36 52 6532 5315 =–6+9––+9 22 15 53 = 9––6+9–22 zzzzz zzzzz

b. 4224 8–5–(2–3+6) xxxxx = 4224 8–5–2+3–6 xxxxx = 42 2–7+3 xxx

c. 312424–6+8––+5–5 42 yyyy    

= 312424–6+8+–5+5 42yyyy

= 322244+–6–5+8+5 44yyyy

= 42 5 –11++13 4 yy

7. a. 2424234 7–4+3+5+6–9+8–2 ssssssss 2224443 =7+5–9–4+6–2+3+8 ssssssss =3+3+823 sss

b. 422 24 (8–3+2)–(4–5)+(7–9+6) tttttttt

422 =8–3+2–4+5+7–9+624 tttttttt

44222 =8+6–3–4–9+2+5+7 tttttttt

42 =14–16+14 ttt

8. Difference of 3 (–2+) xx and 3 (7–3+7) xx

= 33 (7–3+7)–(–2+) xxxx

= 33 7–3+7+2–xxxx

= 33 7+2+7––3 xxxx

= 3 9+6–3 xx

Adding 2 (3–2+1) xx to 3 (9+6–3) xx , we get, 3232 (9+6–3)+(3–2+1)=9+6–3+3–2+1 xxxxxxxx

= 32 9+3+4–2 xxx

9. a. 6xy2z4 × (−11x2yz2) × 3y3z = −198x1+2 y2+1+3 z4+2+1

= −198x3y6z7

b. (4y + 7) (2z2 + 5z − 6) = 4y (2z2 + 5z − 6) + 7 (2z2 + 5z − 6)

= 8yz2 + 20yz − 24y + 14z2 + 35z − 42

c. (−2p3 + 8p2 + 10p − 5) (p2 + 3p + 9)

= −2p3 (p2 + 3p + 9) + 8p2 (p2 + 3p + 9) + 10p (p2 + 3p + 9) − 5 (p2 + 3p + 9)

= −2p5 − 6p4 − 18p3 + 8p4 + 24p3 + 72p2 + 10p3 + 30p2 + 90p − 5p2 − 15p − 45

= −2p5 + 2p4 + 16p3 + 97p2 + 75p − 45

d. st(−5stu + 3u − 9)

= −5s2t2u + 3stu − 9st

e. (7a2 − 2a + 8) (3a2 + 5a + 12)

= 7a2 (3a2 + 5a + 12) − 2a (3a2+ 5a + 12) + 8 (3a2 + 5a + 12)

= 21a4 + 35a3 + 84a2 − 6a3 − 10a2 − 24a + 24a2 + 40a + 96

= 21a4 + 29a3 + 98a2 + 16a + 96

f. (6p5 − 4p3 + 2p2 + 13p − 1) (4p2 + 7p + 12)

= 6p5 (4p2 + 7p + 12) − 4p3 (4p2 + 7p + 12) + 2p2 (4p2 + 7p +

12) + 13p (4p2 + 7p + 12) − 1 (4p2 + 7p + 12)

= 24p7 + 42p6 + 72p5 − 16p5 − 28p4 − 48p3 + 8p4 + 14p3

+ 24p2 + 52p3 + 91p2 + 156p − 4p2 − 7p − 12

= 24p7 + 42p6 + 56p5 − 20p4 + 18p3 + 111p2 + 149p − 12

10. a. (9h + 4) (2h − 11)

= 9h (2h − 11) + 4 (2h − 11)

= 18h2 − 99h + 8h − 44

= 18h2 − 91h − 44

For h = 2, 18h2 − 91h − 44

= 18 (2)2 − 91 × 2 − 44

= 72 − 182 − 44 = −154

b. (2a2 − 5a − 8) (4a2 − a + 3)

= 2a2 (4a2 − a + 3) − 5a (4a2 − a + 3) − 8 (4a2 − a + 3)

= 8a4 − 2a3 + 6a2 − 20a3 + 5a2 − 15a − 32a2 + 8a − 24

= 8a4 − 22a3 − 21a2 − 7a − 24

For a = −5, 8(−5)4 − 22(−5)3 − 21(−5)2 − 7(−5) − 24

= 8 × 625 − 22 × (−125) − 21 × 25 + 35 − 24

= 5000 + 2750 − 525 + 35 − 24 = 7236

c. (7q − 8) (3q2 − 4q + 7)

= 7q (3q2 − 4q + 7) − 8 (3q2 − 4q + 7)

= 21q3 − 28q2 + 49q − 24q2 + 32q − 56

= 21q3 − 52q2 + 81q − 56

For q = 0, 21(0)3 − 52(0)2 + 81(0) − 56 = −56

d. (2t3 + 5t2 + 7t − 10) (3t2 − 10t + 15)

= 2t3 (3t2 − 10t + 15) + 5t2 (3t2 − 10t + 15) + 7t (3t2 − 10t + 15) − 10 (3t2 − 10t + 15)

= 6t5 − 20t4 + 30t3 + 15t4 −50t3 + 75t2 + 21t3 − 70t2 + 105t − 30t2 + 100t − 150

= 6t5 − 5t4 + t3 − 25t2 + 205t − 150

For t = −2, 6(−2)5 − 5(−2)4 + (−2)3 − 25 (−2)2 + 205 (−2) − 150

= 6 × (−32) − 5 × 16 − 8 − 25 × 4 + 205 × (−2) − 150

= −192 − 80 − 8 − 100 − 410 − 150 = −940

11. a. 1 5 a (8a + 5b − 11c) − 5 6 a (3a − 5b + 12c)

= − 8 5 a2 − ab + 11 5 ac − 15 6 a2 + 25 6 ab − 10ac

= 8 5 − 15 6 a2 + 25 6 − 1 ab + 11 5 − 10 ac

= –48–75 30 a2 + 25–6 6 ab + 11–50 5 ac

= − 41 10 a2 + 19 6 ab − 39 5 ac

For a = −1, b = 2, c = 10, 41 10 (−1)2 + 19 6 (−1) (2) − 39 5 (−1) (10)

= 41 –10 19 –3 + 390 5

= –123–190+2340 30 = 2027

b. −2a2b(5a3 − 7b + a + 11) − ab(9b4 −3b2 − 4b) − c

= −10a5b + 14a2b2 −2

− 2cb2 − c

−10a5

− 2cb2 − c

For a = −1, b = 2, c = 10, −10 (−1)5 (2) − 9 (−1) (2)5 + 14 (−1)2 (2)2 −2 (−1)3 (2) − 22 (−1)2 (2) + 3 (−1) (2)3 + 4 (−1) (2)2 − 2 (−1)3 (10) −2 (10) (2)2 − 10 = 20 + 288 + 56 + 4 − 44 − 24 − 16 + 20 − 80 − 10 = 214

c. (a − 7) (a4 − 6b3 + 4c − 5) + 3 7 bc (9a2 − 14) = a (a4 − 6b3 + 4c − 5) − 7 (

= a5 − 6ab3 + 4ac − 5a − 7a4 + 42b3 − 28c + 35 + 27 7 a2bc − 6bc For a = −1, b = 2, c = 10, (−1)5 − 6 (−1) (2)3 + 4 (−1) (10) − 5(−1) − 7

6 (2) (10) = −1 + 48 − 40 + 5 − 7 + 336 − 280 + 35 + 540 7 − 120 = 540 7 − 24 = 540–168 7 = 372 7

12. a. Area = 7x2y × 8x3y2z = 56x5y3z sq. units

b. Area = 109 9109 2–+11=2+11–+11 74 474 æöæöæöæö çççç÷÷÷÷ çççç÷÷÷÷ çççç÷÷÷÷ çççç÷÷÷÷ çççç÷÷÷÷ èøèøèøèø xxxxx = 9 2 x2 + 22x − 45 14 x − 110 7 = 2 308–45 9 110 +–2147 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ÷ ç èø xx 2 308–45 9 110 +–2147 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ÷ ç èø xx 9 2 x2 + 263 14 x 110 7 2 308–45 9 110 +–2147 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ÷ ç èø xx sq. unit

c. Area = 9 13 abc 232324232 41363 +=+ 769126 æ öæ ö çç÷÷ çç÷÷ çç÷÷ çç÷÷ çç÷÷ è øè ø abcabcabcabc sq. units

d. Area = (5z − 3) × (2z2 + 9z − 11) = 5z (2z2 + 9z − 11) − 3 (2z2 + 9z − 11)

= (10z3 + 45z2 − 55z − 6z2 − 27z + 33 = 10z3 + 39z2 − 82z + 33) sq. units

13. a. Volume = 7 11 p2q3r4 × 2 7 pqr × 2p (q + r) = 4 11 p4q4r5 (q + r) 44455446 = + cu.cm 1111pqrpqr æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø

b. Volume = (p + q) × (2p − 3q + 14) × 7r

= [p (2p − 3q + 14) + q (2p − 3q + 14)] × 7r

= (2p2 − 3pq + 14p + 2pq − 3q2 + 14q)7r

= 14p2r − 21pqr + 98pr + 14pqr − 21q2r + 98qr

= (14p2r − 7pqr + 98pr − 21q2r + 98qr) cu. cm

c. Volume = − 1 2 pq (r + 5) × 9pq × 10qr

= (−pqr −5pq) 45pq2r

= (−45p2q3r2 − 225p2q3r) cu. cm

14. a. Volume = 5s (stu + 2) × 5s (stu + 2) × 5s (stu + 2)

= 125s3 (stu + 2)3 cu. cm

b. Volume = 4s2t3u5 × 4s2t3u5 × 4s2t3u5

= 64s6t9u15 cu. cm

c. Volume = (−7ab + 11a2 + 2) × (−7ab + 11a2 + 2) × (−7ab + 11a2 + 2)

= (−7ab + 11a2 + 2)3 cu. cm

d. Volume = 3p3q4 (2r + 5q − 6) × 3p3q4 (2r + 5q − 6) × 3p3q4 (2r + 5q − 6) = 27p9q12 (2r + 5q − 6)3 cu. cm

15. Perimeter of the field = 2(l + b)

2 =2(9+7–4+3) xxx

2 =2(9+10–4) xx

2 =(18+20–8 xx ) cm

Perimeter of the footpath = 2(+) lb

= 2 2(4–3+7+4) xx

= 2 2(4–3+11) xx

= ( 2 8–6+22 xx ) cm

Difference between the perimeter of the field and the footpath = 22 (18+20–8)–(8–6+22) xxxx

= 22 18+20–8–8+6–22 xxxx

= 2 m ( c 10+26–30) xx

16. Area of the rectangular structure = (5x + 3) × (2x − 1)

= 5x (2x − 1) + 3 (2x − 1) = 10x2 − 5x + 6x − 3

= (10x2 + x − 3) sq. m

Cost of a sq. m of material = ₹100

Total cost of the structure = ₹100 (10x2 + x − 3)

17. Price of bananas = ₹(15x + 6)

Price of grapes = ₹(13x2 − 9x + 2)

Price of apples = 22 (13–9+2)–(–8+4–18) xxxx

= 22 13–9+2+8–4+18 xxxx

= 2 (21–13+20) xx `

Total amount spent to buy fruits

= 22 (15+6)+(13–9+2)+(21–13+20) xxxxx

= 22 13+21+15–9–13+6+2+20 xxxxx

= 2 (34–7+28) xx `

18. Population after two years = (2x3 − 3x2 + 5x) (x + 1)2

= (2x3 − 3x2 + 5x) (x2 + 2x + 1)

= 2x3 × x2 + 2x3 × 2x + 2x3 × 1 − 3x2 × x2 − 3x2

× x2 + 5x × 2x + 5x × 1

= 2x5 + x4 + x3 + 7x2 + 5x

19. Answermayvary.sampleanswer:

Sana has (x2 + 2x + 3) pencils and the cost of each pencil is (x3 + 4x + 1). What is the price of all the pencils?

Challenge

1. Length of the chart paper = (3x2 − 5 + 6x) metres

Breadth of the chart paper = (4x + 5) metres

Cutting squares from the corners of the chart paper does not change the perimeter of the chart paper. Hence, the perimeter of the chart paper = 2(l + b)

= 2(3x2 5 + 6x + 4x + 5)

= 2(3x2 + 10x)

= (6x2 + 20x) metres

2. Statement 1: Length of the rectangle = (x + 2)

This statement does not help us find the area of the rectangle.

Statement 2: The breadth of the rectangle is (3x − 4) and the value of x is 7 cm.

Breadth of the rectangle = (3x − 4)

This statement does not help us find the area of the rectangle. Combining both the statements.

Area of the rectangle = (x + 2)(3x − 4) = 3x2 + 2x − 8

= 3(7)2 + 2(7) − 8

= 149 + 14 – 8

= 153 sq. cm

Thus, both statements together are sufficient, but neither statement alone is sufficient.

Hence, option c is correct.

Case Study

1. The shopping complex is hexagon is shape. A regular hexagon can be divided into 6 triangles.

Area of each triangle = 1 2 × Base × Height

= 1 ×(214)×(38) 2 xx + –

= (x + 7)(3x − 8)

= x(3x − 8) + 7(3x − 8)

= 3x2 − 8x + 21x − 56

= 3x2 + 13x − 56

Area of the shopping complex = 6 × Area of each triangle = 6 × (3x2 + 13x − 56)

Thus, the area of the shopping complex is 6(3x2 + 13x – 56) m2. Hence, option a is correct.

2. Area of the circular fountain = Area of a circle = πr2

Radius of the circular fountain = 7(x − 13)

Area of the circular fountain = ()2 22 7(–13) 7 x ×

= 22 ×7(–13)×7(–13) 7 xx

= 154(x – 13)(3 –13)

= 154(x2 – 13x – 13x + 169)

= 154(x2 – 26x + 169) m2

Thus, the area of the circular fountain is 154(x2 + 26x – 169) m2. Hence, option b is correct.

3. Area of the floor that has to be tiled = Area of the shopping complex – Area of the fountain

= 6(3x2 + 13x − 56) − 154(x2 − 26x + 169)

= 18x2 + 78x − 336 − (154x2 − 4004x+ 26026)

= (−136x2 + 4802x − 26362) m2

Cost of tiling = ₹80 per sq. m = ₹80

Total cost of tiling = 80 × (−136x2 + 4802x − 26362)

= ₹80 (−136x2 + 4802x − 26362)

4. Length of each rectangular plot = (x − 3) m

Breadth of each rectangular plot = (x − 9) m

Area of each rectangular plot = Length × Breadth

= (x − 3)(x − 9)

= (x2 − 3x − 9x + 27)

= (x2 − 12x + 27) m2

Area of 4 rectangular plots = 4(x2 − 12x + 27) m2

Area of the remaining shopping complex = Area of shopping complex – Area of the fountain – Area of the shopping complexes

= (−136x2 + 4082x − 26362) − 4(x2 − 12x + 27)

= (−136x2 + 4082x − 26362) − (4x2 − 48x + 108)

= (−140x2 + 4130x − 26470) m2

Thus, the area of the remaining shopping complex is (−140x2 + 4130x − 26470) m2.

5. Answersmayvary.

Chapter 12

Let’s Warm-up

1. Perimeter of an equilateral triangle of length 5 cm 40 cm

2. Area of a triangle of length 4 m and height 8 m

3. Perimeter of a triangle of sides 3 cm, 4 cm and 5 cm

4. Area of a right triangle of lengths 6 m, 8 m and 10 m

5. Perimeter of a triangle of sides 20 cm, 10 cm and 10 cm

Do It Yourself

12A

1. Radius of the circular track = 120 metres

Area of circle = πr2

= 3.14 × 120 × 120 = 45,216 sq. m

2. Length of the rectangle = 14 cm

Area = 84 sq. cm

Since, Area = Length × Breadth

84 = 14 × Breadth

Breadth = 84 6 14 = cm

24 sq. m

16 sq. m

15 cm

12 cm

3. Area of a tile = 100 cm × 100 cm = 10,000 sq. cm = 1 sq. m

So, the perimeter of the rectangle = 2 × (Length + Breadth)

= 2×(14+6)=2×20=40 cm

Given that, perimeter of the square = perimeter of the rectangle

Perimeter of the square = 40 cm

4 × Side = 40 cm

Side = 40 10 4 = cm

So, the side of the square is 10 cm.

Length = 40 m + 1.5 m + 1.5 m = 43 m

Breadth = 32 m + 1.5 m + 1.5 m = 35 m

Area of the entire region = 43 m × 35 m = 1505 sq. m

Area of the park = 40 m × 32 m = 1280 sq. m

Area of the path = area of the entire region − area of the park

= 1505 – 1280 = 225 sq. m

Number of the tiles required

= Area of the path225==225

Area of one tile1

So, 225 square-shaped tiles are required.

4. Radius of the circle BC = Height of the triangle

ΔABC is a right-angled triangle.

So,

AC2=AB2 +BC2

212 = 142 + BC2

441 – 196 = BC2 = 245

BC = 245 = 7 5

Area of a quarter circle = 1 4 × Area of a circle

= 1 4 × πr2 = × 122 47 ×× 122 47 × 7 5 × 7 5 = 192.5 sq. cm

So, the area of the quarter circle is 192.5 sq. cm.

5. Diameter of the circle = 28 cm

Radius = 28 cm = 14 cm 2 28 cm = 14 cm 2

a. 25 cm 14 cm

Area of the square = 25 cm × 25 cm = 625 sq. cm

Area of a quarter circle = 1 × 4 Area of a circle

= 1 4 × πr2

= 1 4 × 22 7 × 142 = 154 sq. cm

Area of the shaded region = 625 – 154 = 471 sq. cm

AB2 + BC2 = AC2

82 + x2 = 102

28 cm 14 cm 14 cm

Side of the square = 14 cm + 14 cm = 28 cm

Area of the square = 28 cm × 28 cm = 784 sq. cm

Area of 4 quarter circles = 1 4 × × 4 Area of a circle = 4 × 1 4 × πr2

2 22 = × 14= 616 sq. cm 7

Area of the shaded region = 784 – 616 = 168 sq. cm

64 + x2 = 100 x2 = 36 x = 6

BC = 6

Distance from A to C, if walked along the length and breadth = 8 m + 6 m = 14 m.

Therefore, she must have walked = 14 m − 10 m = 4 m So, the required distance = 4 m.

8. Since, the two sides of the parallelogram are in the ratio = 4:3

Let the two sides be 4x and 3x.

Then the perimeter = 4x + 3x + 4x + 3x = 56 cm = 14x 56 ==4 14 x

Therefore, the sides of the parallelogram are 4 × 4 = 16 cm and 3 × 4 = 12 cm.

Side of the square = 14 cm + 14 cm = 28 cm

Area of the square = 28 cm × 28 cm = 784 sq. cm

Area of 2 quarter circles = 2 × 1 4 × Area of a circle

= 2 × 1 4 × πr2

= 1 2 × 22 7 × 142 = 308 sq. cm

Area of two right triangles = 2 × 1 2 × 14 × 14 = 196 sq. cm

Area of the shaded region

= 784 − 308 − 196 = 280 sq. cm

6. Area of a parallelogram = base × height

Height = 1 × 16=2 8 cm

Area = 16 × 2 = 32 sq. cm.

9. Area of the wooden frame = 162 cm × 36 cm = 5832 sq. cm

a. Area of each triangle = 1 × 6 × 8=24 sq. cm 2

b. Number of triangles = 5832 =243 24

c. Total perimeter of all the triangles used = Number of triangles × Perimeter of each triangle = 243 × (6 + 8 + 10) = 5832 cm.

10. Area of the lawn = area of the square + 3 × area of the semi-circle =

2 1 (28×28)+3×× 2 122 =784+3×××14 × 14 27 =1708sq.m r π

Fertiliser for 1 sq. m = 20 g

Fertiliser for 1708 sq. m = 20 × 1708 = 34160 = 34.16 kg

Cost of 10 kg fertiliser = ₹16

Cost of 34.16 kg fertiliser = ₹16 10 × 34.16 = ₹54.656

Y and Z are the horizontal and vertical heights of the given parallelogram, respectively.

Considering Y, area = AB × Y or CD × Y

Considering Z, area = AD × Z or BC × Z

John’s answer = CD × Y

David’s answer = BC × Z

So, both John and David are correct.

7. Let us say Aana walked from point A to C.

Distance she moved along the length and breadth

= distance from A to C = AB + BC 8 m

x 10 m

Let BC = x In right triangle ABC, using the Pythagoras rule:

Let the length of a rectangle = l and breadth = b Perimeter of each rectangle

= 2×(+)=48 cm=(+)=24 cm lblb

AB = BC = CD = DA = +=24 cm lb

Area of the square = 24 cm × 24 cm = 576 sq. cm

Area of the square = 5 × area of each rectangle

Then the area of each rectangle = 576 5 = 115.2 sq. cm

12. Area of the sign = 17 sq. m

Area of the 2 parallelograms = 2 × (7.5 × 1) = 15 sq. m

Area of the triangle = 1715=2 sq. m –

11. A B C

13. Area of the room = 40 m × 12 m=480 sq. m

Area of one tile = 0.5 m × 0.5 m=0.25 sq. m

Total number of tiles = 480 =1920 0.25

There are =80 0.5

40 brown tiles along the length, and 12 =80 0.5

24 brown tiles along with the breadth.

So, the number of brown tiles = 80 + 24 + 80 + 24 – 4 = 204

Let the number of white tiles = x

So, the number of pink tiles = twice the number of white tiles = 2x

Then, 204 + x + 2x = 1920

204 + 3x = 1920

3x = 1716

x = 572

2x = 1144

So, there are 204 brown tiles, 1144 pink tiles and 572 white tiles.

Challenge

1. A A A B B C C C (B) E E D (D)

Let the area of the parallelogram ABCD be A.

Fraction of the shaded region = 3 8

Fraction of the newly obtained shaded region = 3 8 × 3 8 = 9 64

So, the fraction of the shaded area we obtained in the end compared to the original parallelogram is 9 64

Do It Yourself 12B 1.

IE = IH + HE

32 cm = IH + 25 cm

IH = 7 cm

And GC = FD = 16 cm

Area of the trapezium ABCG = 1 2 × (6 + 16) × 7 = 77 sq. cm

Area of the rectangle CDFG = 25 × 16 = 400 sq. cm

Total area = 400 + 77 = 477 sq. cm

b. 27 cm 18 cm 25 cm 12 cm 10 cm D C B A

Area of the trapezium ABCD = 1 2 × (27 + 18) × 12

= 270 sq. cm

Area of the trapezium CDEF = 1 2 × (18 + 25) × 10

= 215 sq. cm

Total area = 270 + 245 = 485 sq. cm

16 12 2 5 F E D A B C

Height = 16 − 5 = 11 units

Area of the rectangle ABCD = 2 × 5 sq. units

Area of the trapezium CDEF = 1 2 × (2 + 12) × 11 = 77 sq. units

Total area = 10 + 77 = 87 sq. units

2. Length of the rope = 80 cm

l of the rectangle = 14 cm

b of the rectangle = 8 cm

P of the rectangle = 2 × (14 + 8) = 44 cm

Remaining length of the rope = 80 − 44 cm = 36 cm

P of the rhombus = 36 cm

Each side of the rhombus = 36 =9cm 4

3. Area = 1 2 × d1 × d2 = 1 2 × 2.69 × 2.69 = 3.62 sq. m 4. D C B F E A 12 cm 32 cm 15 cm

Area = 1 2 × 32 × (12 + 15) = 1 2 × 32 × 27 = 432 sq. cm

5. Area = 170 sq. cm

Length of the diagonal = 18 cm

Length of Perpendicular1 = 15 cm

Let length of Perpendicular2 = x cm

Area = 1 2 × 17 × (15 + x)

170 = 17 2 × (15 + x)

x = 5 cm

So, the length of the other perpendicular is 5 cm.

6. Since, the parallel sides are in the ratio 4:6

Let the sides be 4x and 6x

Height = 15 cm

Area = 1 2 × (4x + 6x) × 15 = 450 sq. cm

75x = 450

x = 450

75 = 6

So, the sides are: 4x = 4 × 6 = 24 cm and 6x = 6 × 6 = 36 cm.

7. Perimeter = 200 cm

Length of each side = 200 4 = 50 cm

Area = 672 sq. cm = 1 2 × d1 × d2

672 = 1 2 × (96) × d2

d2 = 14 cm

8. B A

D E F C

Area = 1 ×(32+25)×AE=185 2

AE = 6.5 m

DE = FC and AB = EF

DC = DE + EF + FC

32m = 2DE + 25

DE = 3.5 m

2222 ( AD= AE+DE=6.5)+(3.5)

AD = 7.38 m

So, lengths of the non-parallel sides is 7.38 m.

9. If the ratio of the sides of a trapezium is 6:13:9:11 and P = 78 cm

Then, let the sides be 6x, 13x, 9x, 11x 6+13+9+11=78=39 xxxxx

78 ==2

36 x 6=6×2=12 x 13=13×2=26 x 9=9×2=18 x 11=11×2=22 x

Area = 1 ×(22+26)×10=240 sq. cm 2

Challenge

1. Assertion (A): The area of a trapezium is equal to the area of a rectangle if the non-parallel sides are equal. This statement is not true. The area of a trapezium is generally calculated as 1 (1622)9171 2 ×+×= × (sum of parallel sides) × height, not based on the non-parallel sides being equal. Reason (R): The height of the trapezium is the perpendicular distance between the parallel sides. This statement is correct.

So, the assertion is false and the reason is true. So, option d is correct.

Area of trapezium BCDE = 1 (1622)9171 2 ×+×= sq. cm

Area of rectangle ABEF = 15×22=330 sq.sq.cmcm

area = 171 + 330 = 501 sq. cm

Area of the triangle = 1 × 60 × 55=1650 sq. cm 2

Area of the rectangle = 60 130×60=100×60=6000 sq. cm 2

Area of the semicircle =

Total area = 1650 + 6000 + 1414.28 = 9064.28

c. Area of the semicircle = 2 1229××=31.82

Area of the triangle = 191×9×10=×9×5.5=4.5×5.5=24.75 sq. cm

Total area = 31.82 + 24.75 = 56.57 sq. cm 2. a. Area of ABCH = 1 × (9+ 16) × 5=62.5 2 sq. cm

Area of CDHG = 16 × 9 = 144 sq. cm

Area of EFGD = 1 × (9 + 16) × 5=62.5 sq. cm 2

Total area = 62.5 + 144 + 62.5 = 269 sq. cm

Area of EDJF = 8 × 35=280 sq. m

Area of BCDJ = 1 × (22 + 60) × 8=328 2 sq. m

Area of ABFG =   1 × (60 + 35) + 50 × 18=1305 2 sq. m

Total area = 280 + 328 + 1305 = 1913 sq. m

Area of ABC = 1 × 14 × 5=35 sq. cm 2 sq. cm

Area of ACHG = 14 × (354)=14 × 31=434 sq. cm –

Area of DFIH = 4 × (14 – 4) = 4 × 10 = 40 sq. cm

Area of DEF = 1 ×2×10=10 sq. cm 2

Total area = 35 + 434 + 40 + 10 = 519 sq. cm

3. a.

Since, GH = HB GH = HB = 6 cm

So, area of AHB = 1 ×3 cm×6 cm=9 sq. cm 2 b.

of DEFG = 1 ×(6+4.2)×2=10.2 sq. cm 2

4. Length = 69.5 cm

Width = 30.9 cm

Area = Length × Width = 69.5 × 30.9 = 2,147.55 sq. m

So, the total area of the Parthenon is approximately 2147.55 square metres.

5. A x 60° B C O D 15 cm E F Each triangle in the hexagon has all angles of 60°, so all 6 triangles are equilateral triangles.

So, in triangle AOX, 222 =+ AOOXAX

168.75 = OX2 OX OX= 168.75=12.9913 ≈ cm

Area of triangle AOB = 1 ×15×13=97.5 2 sq. cm

So, the area of all 6 triangles = area of the hexagon = 97.5 × 6 = 585 sq. cm

Other way,

From the above figure, Diagonal = 15 + 15 cm = 30 cm (all triangles are equilateral)

Distance between the diagonal and the base = altitude of the triangle = 13 cm So,

C D E

x 15 cm 30 cm 13 cm 15 cm

Area of one trapezium = 1 ×(15+30)×13=292.5 2 sq. cm

Area of the hexagon = area of 2 trapeziums = 2 × 292.5 sq. cm = 585 sq. cm 6. 6 cm 10 cm 2 cm 2 cm 18 cm 18 cm 9 cm

Area = 22 12261226 (45×10)××+×× 272272

Area of ABEF = 2 × (8 – 3.4) = 2 × 4.6 = 9.2 sq. cm

Area of BCDE = 1 ×(2+8)×3.4=17 sq.cm 2

Total area = 9.2 + 17 = 26.2 sq. cm

8. XR = 9 m, TR = 15 m and TY = 10 m

TR = TX + XR

15 m = TX + 9 m

TX = 15 – 9

TX = 6 m

And, TX + XY = TY

6 m + XY= 10 m

XY = 4 m

XR = XY + RY

9 m = 4 m + RY

RY = 5 m

Area of ∆PXT = 1 ×6×4.4=13.2sq. m 2

Area of ∆PQYX = 1 ×(4.4+2)×4=12.8 2 sq. m

Area of ∆QRY = 1 ×5×2=5 2 sq. m

Area of ∆RST = 1 ×15×3.6=27 2 sq. m

Total area = 13.2 + 12.8 + 5 + 27 = 58 sq. m

9. Area of a tile = 25 × 20 = 500 sq. cm A F P Q R S E B C D 2.5 m 2 m 2 m 2 m 2 m 3 m

2.5 m 4 m 1 m

Area of ABS = 1 ×2.5×2=2.5 . 2 sqm sq. m

Area of BCQS = 1 ×(2.5+4)×(2+3)=16.25 . 2 sqm sq. m

Area of CQD = 1 ×4×(2+2)=8 . 2 sqm sq. m

Area of PED = 1 ×2×1=1 . 2 sqm sq. m

Area of EFRP = 1 ×(2.5+1)×(3+2)=8.75 . 2 sqm sq. m

Area of FAR = 1 ×(2+2)×2.5=5 . 2 sqm sq. m

Total area = 2.5+16.25+8+1+8.75+5=41.5 . 41.5 ×10,000 . =4,15,000 . sqm sqcmsqcm = sq. m = 41.5 × 10,000 sq. m = 4,15,000 sq. m

Number of tiles = Total area Area of a tile = 4,15,000 =830 500

So, 830 tiles are required.

10. Answersmayvary.Sampleanswer: Area = 310 sq. cm 10 cm

Challenge

1. Area of the triangular tiles = 1 ×5×12=30 sq. cm 2 sq. m

Area of bigger square tile = 12×12=144sq.sq.cmcm

Area of smaller square tile = 5×5=25 sq.sq.cmcm

Total area of the shape =4×30+144×2×25 =120+144+50 =314 sq. cm

Chapter Checkup

Area of the rhombus = 12 11××=×9.2×6.4=29.44 sq. cm 22 dd

11××=×9.2×6.4=29.44 sq. cm 22 dd

Area of shaded region = 29.44 sq. cm

2. Let the diagonals be d and 5d.

Area = 12 1 ×× 2 dd

1000 sq. cm 1 1000 . =××5 2 sqcmdd

2000 = 5d2

400 = d2

d = 20 cm and 5d = 5 × 20 = 100 cm

So, the diagonals are 20 cm and 100 cm.

3. Let the parallel sides be and 3 xx

So, 1 350=×(+3)×14 2 xx

700 =4 14 x

50 = 4x or 50 4 = x

x = 12.5 cm

And 3 = 3 × 12.5=37.5 cm x

So, the lengths of the parallel sides are 12.5 cm and 37.5 cm.

4. y x x y x x y

Since, the area of a rectangle = × lengthbreadth

But the area of parallelogram = baseheight × If the length and breadth of a rectangle are the same as the base and height of the parallelogram, then both the shapes will always have the same area.

So, Nishant’s statement is always true.

5. a.

Area of GFEH = 1 × (22+58) × 15=600 sq. cm 2

Area of ADEH = 15 × 58=870 sq. cm

Area of ABCD = 1 × (8 + 58) × 12=396 sq. cm 2

Total area = 600 + 870 + 396 = 1866 sq. cm

b. A

Area of AEP = 1 ×95×90=4275 sq. cm 2

Area of EDRP

= 1 ×(90+115)×(95+30)=12,812.5 sq. cm 2

Area of CDR = 1 ×100×115=5750 sq. cm 2

Area of ABC = 1 ×(95+95+30+100)×76=12,160 sq. cm 2

Total area =4275+12812.5+5750+12,160 = 34,997.5 sq. cm 6. A

In triangle PES, 222 ES=PE+PS

ES=6+8=36+64=100

ES=10 cm

So, EF = 10 cm + 2 cm = 12 cm

Area of DEFG = 9 × 12 = 108 sq. cm

Area of PES = 1 ×6×8=24 2 sq. cm

Area of PQRS = 8 × 8 = 64 sq. cm

Area of A = Area of DEFG − Area of PES = 108 sq. cm − 24 sq. cm

= 84 sq. cm

Area of C = Area of PQRS − Area of PES

= 64 sq. cm − 24 sq. cm = 40 sq. cm

So, the difference between A and C = 84 − 40 = 44 sq. cm 7. a.

Area of 2 quarter circles = 2 122 2×××(6)=56.57 sq. cm 47

Area of square = (6+6)×(6+6)=12×12=144 sq. cm éù êú ëû

Area of 2 circles = 2 22 2××(6)=226.28 7 sq. cm

Area of the shape = 226.28 + 144 − 56.57 = 313.71 sq. cm

b. 95 cm 24 cm 63 cm

Area of trapezium = 1 ×24+(9563)×63=1764 sq. cm 2 éù êú ëû –

Area of quarter circle = 2 122××(63)=3118.5 sq. cm 47

Total area = 1764+3118.5=4882.5 sq. cm

BC = BZ + ZY + YC

25 = 2BZ + 15 (Since, BZ = YC and AD = ZY)

BZ = 5 cm

In triangle ABZ, AB2 = AZ2 + BZ2

132 = AZ2 + 52 ⇒ AZ2 = 144 AZ = 12 cm

Area = 1 ×(15+25)×12=240 sq. cm 2 sq. cm

9. Area of 1 tile = 1 × 0.5=0.5 sq. m

Area of 15 tiles = 15×0.5=7.5 sq. m

Area of the entire floor =9×(0.5×5) =9×2.5 =22.5 sq. m

Area of the uncovered floor = 22.57.5=15 sq. m –10. 12 m 8 m 15 m

Area of the floor = 1 ×(12+15)×8=108 sq. m 2

Paint required for 10 sq. m = 1 L

Paint required for 108 sq. m = 1 ×108=10.8 10 LL

Cost of 5 L paint = 2899`

Cost of 10.8 L paint = 2899 ×10.8=6261.84 5 ` `

Amount of money David has = ₹5400 Since, ₹5400 < ₹6261.84

So, David does not have enough money to buy all the paint he needs.

11. 160 m

150 m

240 m

Area = 1 ×(160+240)×150=30,000 sq. cm 2

Amount of wheat in 1 hectare = 6500 kg

Amount of wheat in 10,000 sq. m = 6500 kg

Amount of wheat in 30,000 sq. m = 6500 ×30,000 10,000 = 19,500 kg

So, the field produces 19,500 kg of wheat.

12. Area of one triangle = 1 ×5×5=12.5 2 sq. m

Area of 4 triangles = 4 × 12.5 sq. m = 50 sq. m

Area of the swimming pool = 22 ×3×3=28.29 7 sq. m

Area of the square region = 40 m × 40 m = 1600 sq. m

Area of the grass region = 1600 sq. m − 50 sq. m −

28.29 sq. m = 1521.71 sq. m

Cost of planting per sq. m of the grass = ₹15

Cost of planting 1521.71 sq. m of grass = ₹15 × 1521.72

= ₹22.825.65

13. Answersmayvary.Sampleanswer:

A community park is designed in the shape of a rectangle with a semicircle attached to one of its shorter sides.

The rectangular part of the park is 40 metres long and 20 metres wide. Find the total area of the park.

Challenge

1. Length of ST = 4 cm

Length of the side of square = 4 cm

Area of ΔPST = Area of ΔPOT – Area of ΔPOS

Length of the diagonal of the square = 4422 + = 32 = 42 cm

Length of PO = 1 2 × Diagonal of the square = 22 cm

Length of OS = Length of PO = 22 cm

Length of OT = + 422 cm

Area of ∆POT = × 1 2 PO × OT

= 1 2 × 22 × () + 422

= () + 424 cm2

Area of ∆POS = 1 4 × Area of the square = 1 4 × 42 = 4 cm2

Area of ∆PST = ()+− 4244 = 42 cm2

2. Area of any quadrilateral = 1 2 × diagonal × sum of length of perpendiculars on the diagonal from the opposite vertices

Statement 1: The sum of the perpendiculars on the diagonal BD from the opposite vertices is given but the length of the other diagonal is unknown. So, we cannot find the area of the quadrilateral using this statement alone.

Thus, it is not sufficient to find the area of the quadrilateral.

Statement 2: The length of the BD is one-third the perimeter of the quadrilateral ABCD and the perimeter of the quadrilateral is 96 cm. With the help of this statement, we can find the length of the diagonal but this statement tells us nothing about the sum of the length of the perpendiculars on the diagonal from the opposite vertices.

Thus, this statement is also not sufficient to find the area of the quadrilateral alone.

By considering both the statements together, we know the length of the diagonal and the sum of the perpendiculars on the diagonal BD from the opposite vertices, and hence, we can find the area of the quadrilateral.

Thus, both statements together are sufficient to find the area of the quadrilateral.

Hence, option c is correct.

Case Study

1. Length of the bases of the trapezium = 70 m and 50 m

Height of the trapezium = 40 m

Area of a trapezium = 1 2 × (Sum of the bases) × Height

= 1 2 × (50 + 70) × 40 = 2400 m2

So, option c is correct.

2. Length of the first diagonal = 80 m

Length of the second diagonal = 100 m

Area of the rhombus = 1 2 × d1 × d2

= 1 2 × 80 × 100 = 4000 m2

So, option d is correct.

3. Area of the trapezium field = 1 2 × 2400 m2 = 1200 m2

Area of the rhombus field = 1 2 × 4000 m2 = 2000 m2

4. Area of the square field = 3 × 3 = 9 m2

Area of the remaining trapezium field = 2400 m2 – 9 m2 = 2391 m2

Area of the remaining rhombus field = 4000 m2 – 9 m2 = 3991 m2

Total area of the remaining fields = 2391 m2 + 3991 m2 = 6382 m2 Chapter 13

= 100 cm2 Perimeter = 45 cm Perimeter = 64 cm

= 154 cm2

cm2

Do It Yourself 13A

1. a. Lateral Surface Area = 4a2

= 4 × 17 × 17 mm2 = 1156 mm2

Total Surface Area = 6a2

= 6 × 17 × 17 mm2 = 1734 mm2

b. Lateral Surface Area = 4a2

= 4 × 21 × 21 m2 = 1764 m2

Total Surface Area = 6a2

= 6 × 21 × 21 m2 = 2646 m2

c. Lateral Surface Area = 4a2

= 4 × 28 × 28 m2 = 3136 m2

Total Surface Area = 6a2

= 6 × 28 × 28 m2 = 4704 m2

2. a. Lateral Surface Area = 2(l + b)h

= 2 × (12 + 8) × 15 cm2 = 600 cm2

Total Surface Area = 2(lb + lh + bh)

= 2(12×8+12×15+8×15) cm2

= 792 cm2

b. Lateral Surface Area = 2(l + b)h

= 2 × (60 + 7) × 18 cm2 = 2412 cm2

Total Surface Area = 2(lb + lh + bh)

= 2(60×7+ 60×18+7×18) cm2 = 3252 cm2

c. Lateral Surface Area = 2(l + b)h

= 2 × (55.5 + 11.5) × 9.25 m2 =1239.5 m2

Total Surface Area = 2(lb + lh + bh)

= 2(55.5×11.5+55.5×9.25+11.5×9.25) m2 = 2516 m2

3. a. Lateral Surface Area = 2πrh

= 2 × 2 22 ×12×14cm 7 = 1056 cm2

Total Surface Area = 2πr(h + r)

= 2 × 2 22 ×12×(14+12)cm 7

= 2 × 22 7 × 12 × 26 cm2 = 1961.14 cm2

b. Lateral Surface Area = 2πrh

= 2 × 2227××22m 72 = 484 m2

Total Surface Area = 2πr(h + r)

= 2 × 22277××+22m 722

= 22 × 51 2 = 561 m2

c. Lateral Surface Area = 2πrh

= 2 × 2227××0.40m 74

= 4.4 m2

Total Surface Area = 2πr(h + r)

= 2 × 22277××+0.40m 744

= 2 × 2227××2.15m 74

= 23.65 m2

4. a. Lateral Surface Area = 2πh(R+r)

= 2 × 2 22 ×100×(14+8)cm 7

= 13,828.57 cm2

Total Surface Area = 2π(Rh + rh + (R2 –r2))

= 2 × ( ) 222 22 ×14×100+8×100+(14–8)cm 7

= 2 × 2 22 ×(1400+800+132)cm 7

= 14,658.29 cm2

b. Lateral Surface Area = 2πh(R+r)

= 2 × 2 22 ×40×(110+100)cm 7 = 52,800 cm2

Total Surface Area = 2π ( ) 22 ) ++(–RhrhRr

= 2 × ( ) 222 22 × 40×110+100×40+(110–100) cm 7

= 2 × 22 22 ×( 4400+4000+2100)cm= 66,000 cm 7

c. Lateral Surface Area = 2πh(R+r)

= 2 × 2 22 ×120×(100+80)cm 7

= 1,35,771.43 cm2

Total Surface Area = 2π ( ) 22 ) ++ (–hRrhRr

= 2 × ( ) 222 22 ×120×100+80×120×(10080)cm 7 –

= 2 × 22 ×(12000+9600+3600) 7 = 1,58,400 cm2

5. a. Lateral Surface Area of a cube = 4a2 = 225 cm2

2 =56.25 a ⇒ a = 7.5 cm

b. Total Surface Area of a cuboid = 2(++) lblhbh

2(30×15+30 +15)hh ⇒ m2 = 1800 m2

900 + 90h = 1800 ⇒ h = 10 m

c. Total Surface Area = 2πr(h + r)

⇒ 2 221111 2×××+ dm 722 h  

= 605 dm2

1331242 + 77 h ⇒ =605 ⇒ h = 12 dm

6. Perimeter of the face of the cube = 4a = 32 cm ⇒ a = 8 cm

Total Surface Area of the cube = 6a2

6=6×8=384222 acmcm2

7. Curved Surface Area of cylinder = 2πrh = 1200 sq. cm

22 2× ×10×=1200=19.09 cm 7 hh ⇒

Total Surface Area of the cylinder = 2πr(h + r)

= 2 × 22 7 × 21 × (90 + 21) cm2 = 14,652 cm2

12. Area of four walls = 2(l + b)h

= 2(15 + 12) × 6 = 324 m2

Cost of whitewashing = ₹19 × 324 = ₹6156

Surface area to be whitewashed including ceiling

= Area of four walls + Area of ceiling

= 2(+)+ =2(15+12)×6+15×12= lbhlb

2(+)+ =2(15+12)×6+15×12= lbhlb 504 m2

Total cost of whitewashing = ₹19 × 504 = ₹9576

13. Let the breadth of the tank be ‘b’ cm.

Therefore, the length of the tank will be ‘8b’ cm

Area of the base of the tank = 8b × b = 12,800 cm2

8=12,80022cm=40 cm bb ⇒

Length of the base = 8b = 8 × 40 cm = 320 cm

Let the depth of the tank be ‘h’ cm

Lateral Surface Area = 2(l + b)h = 1,08,000 cm2

2(320+40)h = 1,08,000 ⇒ h = 150 cm

14. Radius of boiler = 2.8 m=1.4m 2 ; height of boiler = 6.3 m

Area of boiler = Lateral Surface Area + Area of base of the boiler

8. Given R = 50 2 cm = 25 cm, r = 25 − 2 cm = 23 cm, h = 140 cm

Lateral Surface Area of the cylindrical tube = 2πh(R + r) = 2 × 2 22 ×140×(25+23)cm= 7 42,240 cm2

Let the length of edges of each cube be ‘a’ units

Sum of the Surface Area of three cubes = 3 × 6a2 = 18a2

New shape is a cuboid where l = 3a, b = a, h = a

Total Surface Area of new shape = 2(++ )lblhbh

= 2(3×+3×+×)= aaaaaa 14a2

Required ratio = 14a2 : 18a2 = 7 : 9

10. Area of each face of a cube = a2

The swimming pool is shaped like a cube. Since, the swimming pool is open at the top, only five faces have to be painted.

Area of the swimming pool to be painted

= 5 × a2 = 5 × (3.5)2 m2 − 61.25 m2

Cost of painting the swimming pool = ₹25 × 61.25

= ₹1531.25

11. Radius of the base of the cylinder = 21 cm,

Height of the cylinder = 45 × thickness of each plate = 45 × 2 cm = 90 cm

= 2πrh + πr2

= 2 × 22 7 × 1.4 × 6.3 + 22 7 × (1.4)2 m2

= 61.6 m2

No. of days taken by boiler to get completely corroded = Area of boiler

Area of boiler corroding per day = 61.6 1.25 = 49.28 days

15. Answermayvary.Sampleanswer:

Lena is designing a cuboid shaped storage box for her craft supplies with dimensions of 4 feet in length, 3 feet in width, and 2 feet in height. She plans to paint only the sides of the box, to give it a fresh look. What is the area of the box she will paint?

Challenge

1. Assertion (A): The Total Surface Area of the cylinder with a radius of the base 14 cm and height of 30 cm is 3872 cm2

Reason (R): If r is the radius and h is the height of the cylinder, then the Total Surface Area = (2πrh + 2πr2)

Total Surface Area of cylinder = 2πr(r + h)

= () 22 2××141430 7 +

= 2 × 22 × 2 × 44 = 3872 cm2

So, the assertion is true.

Total Surface Area of the cylinder = 2πr(r + h) = 2πrh + 2πr2

Hence, the reason is also true and explains on how to find the total surface area of the cylinder.

Thus, option A is correct. Both A and R are true and R is the correct explanation of A.

Do It Yourself 13B

1. a. i. Volume of a cube 3 3 3 = = 16 =4096 cm a

ii. Volume of a cube = a3 = 243 = 13,824 mm3

b. i. Volume of a cuboid = lbh = 15 × 10 × 8 cm3 = 1200 cm3

ii. Volume of a cuboid = lbh = 75 × 5 × 60 cm3 = 22,500 cm3

c. i. Volume of a cylinder = πr2h = 23 22 ×3.5×10cm 7 = 385 cm3

ii. Volume of a cylinder = πr2h = 22 7 × 422 × 12 m3 = 665.28 m3

2. a. Volume of a cylinder = π(R2 − r2)h = 22 7 × (152 − 82) × 3.5 m3 = 1771 m3

b. Volume of a cylinder = π(R2 − r2)h = 223 2 7) 2 21 ×(18×cm 75 – = 3630 cm3

c. Volume of a cylinder = π(R2 − r2)h = 22 7 × (72 − 22) × 14 9 mm3 = 220 mm3

3. a. Volume of a cuboid = lbh = 17 × 5 × h = 3060 cm3 ⇒ h = 36 cm

b. Volume of a cylinder = πr2h = 23 22 ×5.5× mm 7 h = 3993 mm3 ⇒ h = 42 mm

c. Volume of a cylinder = π(R2 − r2)h = 22 7 × (212 − 112) × h m3 = 352 m3 ⇒ h = 0.35 m

d. Volume of a cube = a3 = 1331 dm3 ⇒ a = 11 dm

4. Volume of the tank = 5.2 m3 = 5.2 × 106 cm3

Also, volume of the tank = Area of the base of tank × Depth of the tank

Depth of tank = Volume of the tank Area of the base of tank = 5.2 × 106 cm3 1300 cm2 = 4000 cm

5. Longest perch that can be placed in the hall would be equal to the diagonal of the hall.

Diagonal of the hall = 222222 ++= 25+16+6 lbh 222222 ++= 25+16+6 lbh = 30.28 m

6. Let the edges of the cube be ‘a’ cm.

Diagonal of cube = a 3 = 8 3 cm ⇒ a = 8 cm

Volume of cube = a3 = 83 cm3 = 512 cm3

7. Volume of the cubical crates = a3 = 503 cm3 = 1,25,000 cm3 = 0.125 m3

Volume of the warehouse = lbh = 50m×35m×20m=35,000m3

No. of crates that can fit in the warehouse 3 3 Volumeofthewarehouse = Volumeofthecubical crates

35,000m = =2,80,000 0.125m

8. 1 litre = 1000 cm3; 60 L = 60,000 cm3

Volume of the ice cubes = a3 = 103 cm3 = 1000 cm3

No. of ice cubes that can be formed from the water in the container = 60,000 cm3 1000 cm3 = 60

9. Let the edges of the cube be ‘a’ units.

Original volume of the cube = 3a

Increased edges of the cube = 1+=503 1002 aa

New volume = 3 3 327 = 28 aa

Percentage increase in the volume of the cube 33 3

Newvolumeoriginalvolume

10. Let the edges of the cubes be x and y, respectively.

Ratio of the volumes of the cubes = 33:=1:64xy

⇒ Ratio of sides =:=1:4 xy

Ratio of the surface area of cubes 6:6=222222 := 1:4 =1 :16 xyxy = 6:6=222222 := 1:4 =1 :16 xyxy =

11. The largest cube that can be cut has the edge length of the smallest dimension of the cuboid.

Hence, the length of the edges of the cube will measure 12 cm.

Volume of this cube = a3

= 333 =12=1728 cm a

12. Let the edges of the cube be ‘a’ units.

Volume of a cube = 3a

Given, Diameter of base of cylinder = a

⇒ Radius of base of cylinder = 2 a

Height of cylinder = a

Volume of cylinder = πr2h

= π × a2 4 × a = π 4 a3

Empty space left in the cube = Volume of cube − Volume of cylinder = a3 –π 4 a3 = π 3 1–4 a

13. Length of the aluminium rod: h = 3.5 m, radius of each rod: r = 4 cm 2 = 2 cm = 0.02 m

Volume of each aluminium rod = πr2h

= 23 22 ×0.02×3.5 m= 7 0.0044 m3

Volume of aluminium = 1.76 m3

Number of rods that can be made

= Volume of aluminium

Volume of each aluminium rod = 1.76 0.0044 = 400

14. Let the length, breadth and height of the cuboid be l, b and h, respectively.

Volume of the cuboid = lbh

Increased volume of cuboid = 255040 1+ ×1×1+ 100100100 lbh

5177 =××= 4258 lbhlbh

Ratio of volumes of original cuboid and new cuboid

= 7 : =8:78:7 8 lbhlbhlbhlbh =

15. Internal radius of pipe = 4cm = =2cm=0.02m 2 r

Volume of water flowing out of pipe per second

= πr2h = 233 22 ×0.02×3 m= 0.00377 m 7

Volume of water flowing out of pipe in 45 minutes

= Volume of water flowing out of pipe per second × 45 × 60 m3

= 0.00377×45×60=10.179m3

Given, the radius of base of the cylindrical tank = 90 cm = 0.9 m

Let the level to which the water rises in the cylindrical tank be h

Volume of water flowing out of pipe in 45 minutes

= Volume of water stored in cylindrical tank

= πr2h = 233 22 ×0.9× m= 10.179 m 7 h

=3.998 400 cm hm ⇒≈

16. We have V = a × b × c and S = 2 (ab + bc + ca)

R.H.S. = 21112 bcacab SabcSabc  ++ ++=

= 22 2 abbcca Sabc  ++ ×

= 21 2 S SV

1 V = = L.H.S. Hence proved.

17. Answermayvary.Sampleanswer:

A cylindrical water tank has a diameter of 2 metres and a height of 3 metres. The tank is to be filled with water. How much water (in cubic metres) can the tank hold when it is full?

Challenge

1. Assertion (A): The surface area of a cylinder is always greater than its volume.

Reason (R): The surface area of a cylinder depends on both the radius and height.

The surface area of a cylinder is not always greater than its volume. The relationship between surface area and volume varies depending on the radius and height of the cylinder.

The reason is true: The surface area does depend on both the radius and height (2πr(h + r)).

A is false, but R is true.

Hence, option d is correct.

Chapter Checkup

1. a. Volume of a cuboid = lbh = 12 × 10 × 5 m3 = 600 m3

No. of unit cubes that can fit = 600

b. Volume of a cube = a3 = 33 = 27 m3

No. of unit cubes that can fit = 27

c. Volume of a cuboid = lbh = 10 × 5 × 8 m3 = 400 m3

No. of unit cubes that can fit = 400

d. Volume of a cube = a3 = 63 m3 = 216 m3

No. of unit cubes that can fit = 216

2. 1 litre = 1000 cm3

a. Volume of a cube 333 3 =a=150cm=33,75,000 cm=3375litres

b. Volume of a cuboid = lbh 200×150×50cm=33 1500000 cm = 200×150×50cm=33 1500000 cm = = 1500 litres

c. Volume of cylinder = πr2h = 23 22 ×800×300cm 7 = 6,03,42,8571 cm3 = 6,03,428.6 litres

d. Volume of hollow cylinder = π(R2 – r2)h = 223 22 ×(300–120×14 ) cm= 7 = 33,26,400 cm3 = 3326.4 litres

3. Let the height of the cylinder be h

Total Surface Area of cylinder = 2πr(h+ r) = 660 m2 22 2××5(5+)=660 7 h h + 5 = 21

⇒ h = 16 m

4. Let the radius of the cylinder be ‘r’ and height of the cylinder be ‘h’.

Total Surface Area of cylinder = 2πr(h + r) = 2640 m2 22 2× ×7×(+7)=2640 7 h

So, h = 53 mm

5. a. Total Surface Area = 6a2 = 9126 m2

⇒ a = 39 m

b. Lateral Surface Area = 2(l + b)h= 8100 cm2

2(60+)×45 = b 8100

60+ b = 90 ⇒ b = 30 cm

c. Lateral Surface Area = 2πrh = 88 m2 2211 2× ××=88 72 r ⇒ r = 2.55 m

d. Volume of cube = a3 = 1728 cm3

⇒ a = 3 1728 = 12 cm

e. Volume of cuboid = lbh = 63 m3

= 5 × 3 × h = 63

⇒ h = 63 15 = 4.2 m

f. Volume of cylinder = πr2h = 475.904 mm2

22 7 × 2.62 × h = 475.904

⇒ h = 22.40 mm

6. Let the edges of the original cube be ‘a’ units.

Original surface area = 6a2

New surface area after doubling the edges

= 6(2)=2422 aa

Increase in the surface area = 24 222 –6=18 aaa

% increase in the surface area

Increaseinsurfacearea = ×100%

Originalsurfacearea

= 2 × 22 7 × r = 22 cm

⇒ r = 3.5 cm

Volume of the cylinder = πr2h

= 22 7 × 3.52 × 4 = 154 cm3

13. Lateral Surface Area of the cylinder = Area of square sheet = 30 × 30 cm2 = 900 cm2

Cost of packaging the box26,250=105000cm

14. Surface area of the box = 3 Cost of packaging the box26,250=105000cm Rate of packaging the box0.25 = 3

Rate of packaging the box0.25 = 1,05,000 cm3

Given, length of the wooden box = 50 cm; Breadth of the wooden box = 30 cm

= 2 2 18 ×100%=300%

6 a a

7. Let the radius of base of cylinder be ‘r’ and its height be ‘h’.

Original curved surface area =2πrh

New curved surface area after radius is doubled and height is halved.

New curved surface area =2π(2r)

2 h = 2πrh

Change in surface area = 2πrh − 2πrh= 0

8. Total Surface Area of the cube = 6=6×42222 m=96m a

Area of each square hole = 1 × 1 m2 = 1 m2

Total area of the square holes = No. of faces of cube × Area of each square hole = 6 × 1 m2 = 6 m2

Total area of the faces of the cube after removing square holes = 96 − 6 m2 = 90 m2

9. Volume of the cube = a3

Total volume of the cubes = 13 + 63 + 83 cm3 = 729 cm3

Let the edges of the new cube be ‘a’ units.

Volume of the new cube = Volume of cubes melted together a3 = 729 cm3 ⇒ 3 =729=9cm a

Total Surface Area of the new cube = 22 6=6(9)= 486 a cm2

10. Let the outer radius and inner radius of the hollow cylinder be R and r, respectively.

Height of hollow cylinder = 180 cm

Outer curved surface area = 2πRh = 31,680 cm2 22 2×××180 7 R = 31,680

⇒ R = 28 cm

Inner volume of cylinder = Amount of air held

πr2h = 2,49,480 cm3

⇒ 2 22 ××180 7 r = 2,49,480

⇒ r2 = 441 ⇒ r = 21 cm

Thickness of the cylinder = 28 cm − 21 cm = 7 cm

11. We know that the volume of the cylinder = πr2h

Let the volumes of the cylinders be V1 and V2, respectively.

V1 = πX2Y, V2 = πY2X

V1 V2 = πX2Y

πY2X = X Y = X:Y

12. Let the radius of the base be r and its height be h. h = 4 cm

Perimeter of the base = 2πr

Using,

Surface area of the box = 2(++) lblhbh =1,05,000 cm3

2(50×30+50×+30×) hh = 1,05,000

1500 + 80h = 52500 ⇒ h = 637.5 cm

15. Total Surface Area of a cuboidal box = 2(++) lblhbh 22

=2(60×40+60×30+30×40) cm=10,800cm

Area of one coloured square sheet = 50 × 50 = 2500 cm2

No. of coloured square sheet required = 2 2 10800 cm

Total Surface Area of cuboidal box = =4.32

Area of one coloured square sheet 2500 cm = 5 (approx)

16. Volume of the brick = lbh

= 15 cm × 5 cm × 5 cm

= 375 cm3

Volume of the wall = LBH = 5 m × 1.5 m × 1.5 m = 11.25 m3

= 1,12,50,000 cm3

No. of bricks required 3 3 11250000cm

Volume of wall = = =30,000

Volume of brick 375cm

17. Volume of water flowing out of pipe per hour

= Area of cross-section of pipe × Speed of water

= 1.5 × 1.5 × 25000 = 56,250 m3

Volume of water to be filled in the tank

= Area of the tank × Level of water = 250 × 100 × 3 m3 = 75000 m3

Time required = e ta V nk olume of water to be filled in the Volume of water flowing out of pip

= 75,000 56,250 = 1 1 3 hour = 80 minutes

18. The length of the pipe = 48 cm

The external diameter of the pipe = 35 cm

The external radius of the pipe = 35 17.5 2 = cm

External volume = πr2h

2 22 (17.5)4846,200 7 =××=

= 46,200 cm3

The internal diameter of the pipe = 28 cm

The internal radius of the pipe = 28 14 2 cm = cm

Internal volume = πr2h

2 22 (14)4829,568 7 =××= = 29,568 cm3

Volume of the wood = (External volume) – (Internal volume) = (46,200 – 29,568) cm3 = 16,632 cm3

Weight of the pipe = 16,632 × 0.8 = 13,305.6 g

19. Answerwillvary.Sampleanswer:

Maria has a cylindrical flower pot with a height of 80 cm and a base that has a diameter of 40 cm. What is the volume of soil she needs to fill the pot completely?

Challenge

1. Dimensions of the rectangular sheet of paper: 44 cm by 14 cm

The cylinder is formed by rolling out a sheet of paper about its longer side.

So, the height of the cylinder h = 14 cm

And the circumference of the cylinder = 44 cm

2πr = 44

2 × 22 × r 7 = 44 r = 7 cm

The volume of the cylinder = πr2h

22 × 7 × 7 × 14 7 = 2156 cm3

2. Assertion (A): The volume of two cuboids with the same length and breadth will always be the same, regardless of their heights.

Reason (R): The volume of a cuboid is calculated by multiplying the length, breadth, and height.

Assertion: Volume of a cuboid is given as length × breadth × height. So the volume of a cuboid depends on lengths, breadth and height. If the length and breadth of two cuboids are same but the height is different then their volumes will also be different. So, the assertion is false.

Reason: Volume of a cuboid = Length × Breadth × Height

So, the reason is true.

Hence, assertion is false while the reason is true.

Thus, option c is correct.

Case Study

1. Length of the stage = 5 m

Breadth of the stage = 4 m

Height of the stage = 80 cm = 0.8 m

Let us find the Total Surface Area of the stage first and then subtract the area of the bottom from it.

Total Surface Area = 2 × (lb + bh + hl)

= 2 × ((5 × 4) + (4 × 0.8) + (0.8 × 5))

= 2 × (20 + 3.2 + 4)

= 2 × 27.2

= 54.4 m2

Area of the bottom face = lb = 5 × 4 = 20 m2

Total area that is to be painted = 54.4 – 20 = 34.4 m2

Thus, 34.4 m2 area is to be painted.

Hence, option b is correct.

2. Volume of the cubical stool = 2,16,000 cm3

Volume of a cube = (Side)3

(Side)3 = 216000 (Side)3 = 603

Side of the cubical stool = 60 cm

Total Surface Area of a cube = 6 × (Side)2

The Total Surface Area of the cubical stool = 6 × 602 = 6 × 3600 = 21,600 cm2

Thus, the Total Surface Area of the cubical stool is 21,600 cm2

Hence, option c is correct.

3. Height of the cylindrical dustbin = 30 cm

Diameter of the cylindrical dustbin = 14 cm

Radius of the cylindrical dustbin = 14 ÷ 2 = 7 cm

The coloured paper will only be covered on the curved side of the dustbin.

Curved surface area of a cylinder = 2πrh

Curved surface area of the dustbin = 2 × 22 7 × 7 × 30 = 1320 cm2

There are 8 such dustbins.

Total coloured paper required = 8 × 1320 cm2 = 10,560 cm2

Thus, 10,560 cm2 of paper is required for 8 dustbins.

4. Length of the aquarium = 120 cm

Breadth of the aquarium = 50 cm

Height of the aquarium = 1 m = 100 cm

Volume of the aquarium = lbh = 120 cm × 50 cm × 100 cm = 6,00,000 cm3

It is given that, 1 cm3 = 1 mL

Volume of the aquarium = 6,00,000 mL

We need to find the capacity of the aquarium in litres.

We know that, 1 L = 1000 mL

Volume of the aquarium in litres = 6,00,000 ÷ 1000 = 600 L

Thus, the aquarium can store 600 L of water.

5. Answersmayvary.

Chapter 14

Letʹs Warm-up

1. The square of 11 is 121 and its cube is 1331

2. If the units digit of a number is 3, then the units digit of its square is 9 and that of its cube is 7

3. The prime factorisation of 48 is 24× 3

a. Is it a perfect square? No

b. The smallest number that should the given number be divided by to get a perfect square is 3.

4. Reeti says, “125 is a perfect square”, whereas Meeti says, “125 is a perfect cube.” Meeti is correct.

Do It Yourself 14A

1. a. 46 = 4 × 4 × 4 × 4 × 4 × 4 = 4096

b. (−7)−3 = 111 ×× 777 æöæöæö ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ èøèøèø × 111 ×× 777 æöæöæö ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ èøèøèø × 111 ×× 777 æöæöæö ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ èøèøèø = 1 343 -

 =××= 

c. 3 444464 5555125

d. 4 2 3 

= 3333 ××× 2222 æöæöæöæö çççç÷÷÷÷ çççç÷÷÷÷ çççç÷÷÷÷ çççç÷÷÷÷ çççç÷÷÷÷ èøèøèøèø = 81 16

2. a. 16 81 = 2 3 × 2 3 × 2 3 × 2 3 = 4 2 3

b. 81 243 = 3 × 3 × 3 × 3 3 × 3 × 3 × 3 × 3 = 4 5 3 3 = 3−1

c. 125 64 = 5 4 × 5 4 × 5 4 = 3

4. a. 45.123 = 4 × 10 + 5 × 1 + 1 × 111 +2× + 3× 101001000 = 4 × 101 + 5 × 100 + 1 × 10−1 + 2 × 10−2 + 3 × 10−3

6. a. 33333366 4344443 3433334 +  ×=×===

b. 245 45 2 4 5 2 2 145 × 652 =(6)××52 45 =(2×3)××52 5 2 × æöæöæö ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ ççç÷÷÷ èøèøèø æöæö çç÷÷ çç÷÷ çç÷÷ çç÷÷ çç÷÷ èøèø æö æö ÷

ç ÷ ç èø èø = 45 22 85×(3)××52 2 2) 5 ( =

c. (4)343232 +× 

= [16 + 27] × 43−2 = 43 × 43−2 = 43 × 2 1 43 = 1 43 = (43)−1

d. 23 733 6714

= 33 4714 ÷ × = 33 2814 ÷ = 1 3141 2 2832 ×==

7. Let the number be x.

b. 541.056 = 5 × 100 + 4 × 10 + 1 × 1 + 0 × 111 +5×+6× 101001000 + 5 × 111 +5×+6× 101001000 + 6 × 111 +5×+6× 101001000

= 5 × 102 + 4 × 101 + 1 × 100 + 0 × 10−1 + 5 × 10−2 + 6 × 10−3

According to the question, –3 3 ÷=16 8 x æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø 33 8181×=16;×= 3 316 x x æöæö çç÷÷ çç÷÷ çç÷÷ çç÷÷ çç÷÷ èøèø 5121 ×=; 2716 x 32 = 27 x

c. 107.02356 = 1 × 100 + 0 × 10 + 7 × 1 + 0 × 11111 +2×+3×+5×+6× 10100100010000100000 11111 +2×+3×+5×+6× 10100100010000100000

= 1 × 102 + 0 × 101 + 7 × 100 + 0 × 10−1 + 2 × 10−2 + 3 × 10−3 + 5 × 10−4 + 6 × 10−5

d. 5003.671 = 5 × 1000 + 3 × 1 + 6 × 111 +7×+1× 101001000

= 5 × 103 + 3 × 100 + 6 × 10−1 + 7 × 10−2 + 1 × 10−3

5. a. 25 = 2 × 2 × 2 × 2 × 2 = 32

52 = 5 × 5 = 25

So, 25 is greater than 52

b. 27 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128

72 = 7 × 7 = 49

So, 27 is greater than 72

The number is 32 27 8. 34 4 54 ×(0.8) 65 x= æö æö çç÷÷ çç÷÷ = çç÷÷ çç÷÷ çç÷÷ èø èø –34 64 55 ×x= æöæö çç÷÷ çç÷÷ çç÷÷ çç÷÷ çç÷÷èøèø 43 835 4333 =×45(1)×2×5(1)2 = 56 5×2×35×3 × x= æöæö çç÷÷ çç÷÷çç÷÷ çç÷÷ çç÷÷ èøèø 43 835 4333 =×45(1)×2×5(1)2 = 56 5×2×35×3 × x= æöæö çç÷÷ çç÷÷çç÷÷ çç÷÷ çç÷÷ èøèø 43 835 4333 =×45(1)×2×5(1)2 = 56 5×2×35×3 × x= æöæö çç÷÷ çç÷÷çç÷÷ çç÷÷ çç÷÷ èøèø = –32 135

The number is –32 . 135

9. Weight of the insect = 5−3

Food it eats on the first day = 26 × 5−3 g = (22)3 × 5−3 g = (4 × 5−1)3 g = 3 4 5

g = (0.8)3 g = 0.512 g

Challenge

1. a. Number of views the video got in branch 1 in week 0 is 5.

Number of views the video got in branch 2 in week 0 is 4. So, the ratio is 5:4

b. Percentage increase = Increased viewsinitial views Initial views    × 100 = 255 5 × 100 = 400%

c. Number of views in branch 2 in week 2 is 64.

Number of views in branch 2 in week 6 = 256 × 4 × 4 × 4 = 16,834

Ratio of week 2 to week 6 is 64 16834 = 1 256 = 1:256

Do It Yourself 14B

1. a. 10980000 = 1098 × 104 = 1.098 × 103 × 104 = 1.098 × 107

b. 2300000000000000 = 23 × 100000000000000 = 2.3 × 101 × 1014 = 2.3 × 1015

c. 10.446678 = 1.0446678 × 101

d. 453.98785 = 4.5398785 × 102

2. a. 42 44 99

(–= 2 4 9

= 42) 4 9 +

= 2 2 4 9 = 16 81

b. 4242 777 131313

3. a. 4.21 × 108 = 8 421 10 100 × 28 82 421101042110421106 =××=×=× = 421000000 b. 1.03 × 108 = 8 103 10 100 × 28 82 103101010310103106 =××=×=× = 103000000

c. 40.857 × 10−2 = 2 408571 1000 10 × 40857 0.40857 100000 ==

d. 3.467 × 10−5 = 5 34671 1000 10 × = 8 3467 0.00003467 10 =

4. a. 2.11 × 102 = 222 211 102111010 100 ×=×× 222 211 102111010 100 ×=×× 22 0 2111021110211 =×=×= 1.67 × 10−2 = 1671167 0.0167 10010010000 ×==

1671167 0.0167 10010010000 ×==

So, 0.0167 < 211 Thus, 2.11 × 102 > 1.67 × 10−2

b. 7.3 × 10−5 = 56 73173 10 1010 ×= 73 0.000073 1000000 == 0.7 × 10−8 = 89 717 10 1010 ×= 7 0.000000007 1000000000 = = So, 0.000073 > 0.000000007 Thus, 7.3 × 10−5 > 0.7 × 10−8

5. a. 6.757869 × 105 + 980.436 × 103 = 675786.9 + 980436 = 1656222.9 = 1.6562229 × 106

b. (3.87 × 10−7 ) + (1.02 × 10−7 ) = (3.87 + 1.02) × 10−7 = 4.89 × 10−7

c. 3.6548 × 10⁸ + 154.142 × 10⁶ = 3.6548 × 10⁸ + 1.54142 × 108 (3.6548 + 1.54142) × 108 = 5.19622 × 108

d. 46157.22 × 10² – 23.1572 × 10⁵ = 4615722 – 2315720 = 2.300002 × 106

6. a. (5−9 × 53) × 155 = 5−9+3 × (3 × 5)5 = 5−6 × 35 × 55 = 5(−6+5) × 35 = 5−1 × 35 = 5 3243 55 =

b. 444444 233583 585235

(5)55 (8)22 (5)5 (8)2 (5)5(5)2 5 (8)2(8) (5)5125 2128 (8) 7. a. –62 –2 72×3×4 81×16×27 = 23–64 44–6 3×2×3×2 3×2×3 = 3 2 2 3 = 8 9 b. –30 –5 20×3×4 36×5×45 = –30 –5 4×5×3×4 4×9×5×5×9 = 2–30 22–52 2×5×3×2 2×3×5×5×3 = 2–3 24–4 2×5×3×1 2×3×5 = 55 × 3−7 = 55 × 7 1 3 = 5 5 3

× 1 9 = 3125 2187

c. 9× 18×3× t t –3 –5–5 = 9× 9×2×3× t t –3 –5–5 = 2 2 3× 2×3×3× t t –3 –5–5 = 2522 –5 3243 = 22 23 ttt × = ×

d. +1 +1 +2 16×24×2 16×22×2 mm mm= 52 53 2×22×2 2×22×2 mm mm= 52 53 ) () 2 (22 222 m m= 3 2 2 2 = 23 − 2 = 2 8. Let the missing number be x

b. 2x = 32

We know that, 2 × 2 × 2 × 2 × 2 = 32

25 = 32

This gives, x = 5

10. Amount of food carried by one ant = 2.78 × 10−3 inches

Amount of food carried by another ant = 4.56 × 10−5 inches

Total amount of food carried by both the ants

= 2.78 × 10−3 + 4.56 × 10−5

= 278 100 × 1 1000 + 456 100 × 1 100000

= 278 100000 + 456 10000000

= 27800456 10000000 + = 28256 10000000 = 0.0028256 inches

11. Number of people spoke French = 7.7 × 103

Number of people spoke German = 3.2 × 104

7.7 × 103 = 77 10 × 103

= 77 × 103−1 = 77 × 102 = 7700,

3.2 × 104 = 32 10 × 104 = 32 × 104−1 = 32 × 103 = 32000; Since 32000 > 7700,

The German language had more speakers by 32000 − 7700 = 24300

= 2.43 × 104

12. Total area of the land = 1500 sq. mile

Number of mosquitoes in per square mile of land = 80 million

Number of mosquitoes in 1500 sq. mile of land = 80 million × 1500 sq. miles = 120000 million sq. miles

13. Size of an Antarctic Ocean = 20,330,000 sq. km

Size of an Arctic Ocean = 3 4 × 20,330,000 sq. km

= 3 × 5082500 sq. km = 15247500 sq. km

= 1.5247500 × 107 sq. km

Challenge

1. Statement 1 says, 2x = 8 ⇒ 2x = 23

On comparing, x = 3

With the help of Statement 1, we cannot find the value of 2x+y, since we do not know the value of y So, Statement 1 alone is not sufficient.

Statement 2 says, 2y = 16 ⇒ 2y = 24

On comparing, y = 4

With the help of Statement 2, we cannot find the value of 2x+y, since we do not know the value of x

But if we consider both the statements together, then we know the values of both x and y

Thus, both the statements together are sufficient to find the value of 2x+y

Hence, option c is correct.

Chapter Checkup

1. a. 65.879 = 6 × 10 + 5 × 1 + 8 × 111 +7× + 9× 101001000 = 6 × 101 + 5 × 100 + 8 × 10−1

+ 5 × 10−4

c. 4876.9023 = 4 × 1000 + 8 × 100 + 7 × 10 + 6 × 1 + 9 × 111 +0× + 2× 101001000 +

2. a. 1.275 × 10−5 m is in standard form.

1.275 × 10−5 m = 1275 1000 × 10−5 = 1275 100000000 = 0.00001275

b. 1.386 × 109 cu. km is in standard form. 1.386 × 109 = 1386 1000 × 109 = 1386 × 106 = 1386000000

c. 0.0000000000000000000000000000911 kg is in usual form. 0.0000000000000000000000000000911 = 911 10000000000000000000000000000000

= 9.11 × 102 × 10−31 = 9.11 × 10−29 kg

3. 0.000000654879 is in usual form because it has not been expressed as k × 10n

4. a. 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 256 = 28

b. 59,049 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3

59,049 = 310

c. 78,125 = 5 × 5 × 5 × 5 × 5 × 5 × 5 78,125 = 57

77777 ×××= 99999 79 Reciprocalof 97is.

5. a. 4 44 8

38 Reciprocalofis. 83 53123 ÷=÷ 1225525 121212 =××× 555

7. 710 = 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 = 49 × 49 × 49 × 49 × 49 = 495

8. a. 210, 323 = (25)3 = 215, 642 = (26)2 = 212, 128−2 = (27)−2 = 2−14, 83 = (23)3 = 29

Ascending order: 2−14, 29, 210, 212, 215

⇒ 128−2, 83, 210, 642, 323

Descending order: 215, 212, 210, 29, 2−14

⇒ 323, 642, 210, 83, 128−2 b. 310, 273 = (33)3 = 39, 7292 = (36)2 = 312, 18−2 = (2 × 32)−2 = 2−2 × 3−4, 93 = (32)3 = 36

Ascending order: 2−2 × 3−4, 36, 39, 312, 310

⇒ 18−2, 93, 273, 310, 7292

Descending order: 310, 312, 39, 36, 2−2 × 3−4

⇒ 7292, 310, 273, 93, 18−2

9. x−2 = 2 1 x , Multiplicative inverse = x2

Neither Tanya nor Sayani is correct because Tanya’s answer is negative x2, and Sayani’s answer has power as 2 less than −1, and the product of each of the two gives with x−2 is x−2 × (−x)−2 = x−4 and x−2 × x−3 = x−5, which are incorrect because a number multiplied by its multiplicative inverse equals 1.

10. a. (8−2+ 5−3 + (−3) −4)0 = 1 as the power is 0. b. 3 30 2 (1.7) (0.2) 3.4

××

11. a. 62762 15231 12834234 243362736(3 ) ×××× = ×××× 764 7426523 5223 232 23 3233 + ×× = =× ××× ––12 232918=×=×=

b. 31332 215221 40342532 2435453535 ××××× = ××××× = 23−2 × 51+2+1 × 33−5+2 40 25326251250 =××=×=

c. +1 +2 +1 15×5 + 20×5×5 3×5 + 20 ×5 mm mm = +1 +1 +2 +1 5×3×5+ 20×5

3×5+ 20 ×5 mm mm = +2 +1 +2 +1

3×5+ 20×5

3×5+ 20 ×5 mm mm = +1 +1

515+20)

515+20) ( ( m m = 1

12. a. 210 ÷ 26 = 2−4 × 23x − 1

⇒ 24 = 2−4 + 3x −1

⇒ 24 = 23x −5

⇒ 4 = 3x − 5 ⇒ 9 = 3x ⇒ x = 3

b. 3 5 8 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø × –8 5 8 x æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø = –10 5 8 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø ⇒ –38 5 8 x æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø = –10 5 8 æö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç èø

⇒ 3 − 8x = −10

⇒ 13 = 8x ⇒ x = 13 8

c. 72x+1 ÷ 343 = 2401

⇒ 72x+1 ÷ 73 = 2401

⇒ 72x+1 − 3 = 74

⇒ 2x − 2 = 4

⇒ 2x = 6 ⇒ x = 3

13. + + + 2 × × (× ) × plqplq pql xxx xxx = + ++++ ()++2 plqplq pql x x = 2 + 2+2

2(++) plq pql x x = 2( + + ) 2(++) pql pql x x = 1

14. Total area of the land = 6 × 105 m;

Quantity of cotton produced = 2.4 × 106 kg

Quantity of cotton planted per metre = 6 5 × 10 2 ×1 60 .4 = 2.4 × 106 − 5 6 = 1 2.4 ×10 6 = 24 6 = 6 24 = 1 4 = 4 kg

15. Weight of 1 crate = 2.2 × 103 kg; Number of crates = 4 × 102 = 400

Weight of 400 crates = 2.2 × 103 × 400 = 2200 × 400 = 8,80,000 kg

16. Diameter of Jupiter = 13.9822 × 104 km = 1,39,822 km

Diameter of Saturn = 1.165 × 105 km

= 1165

1000 × 105 km

= 1,16,500 km

1,39,822 km > 1,16,500 km, so diameter of Jupiter is greater than the diameter of Saturn.

17. Size of a plant cell = 0.00002155 m, Size of a red blood cell = 0.0000000015 m

According to the question,

Size of a plant cell = 2 × size of red blood cell

= 2 × 0.0000000015 m

= 0.000000003 m ≠ 0.00002155 m

No, the size of a plant cell is not a double.

18. Area of the Pacific Ocean = 1.55 × 108 sq. km, Area of the Atlantic Ocean = 1.0646 × 108 sq. km

The area of the Pacific Ocean is greater by 8 8 1.55 ×10 1.0646×10 = 1.46 (approx.) times

Challenge

1. Assertion:

(a³ × b³) + 2000 = –1000, where a = 5 and b = –2

(5³ × (−2)³) + 1 = −1000 + 1 = −9991001

So, assertion is incorrect.

Reason (R): (a³ × b³) = (ab)³ and a0 = 1.

The reason is true.

Assertion is false and reason is true.

So, option d is correct.

m+3 2m–3 3m–9 ()yy y = 1024

m+32m–2

3m–9 × yy y = 1024

m+3+2m–2

3m–9 y y = 1024

3m+1

3m–9 y y = 1024

y3m+1−3m+9 = 1024

y10 = 1024, y10 = (2)10 y = 2

Case Study

1. The mass of hydrogen is 1.6735 × 10−27 kg

The mass of oxygen is 2.656 × 10−26 kg

The mass of gold is 3.27 × 10−25 kg

The mass of iron is 9.273 × 10−26 kg

Since, hydrogen is the lightest.

So, option a is the correct answer.

2. a. Gold is the heaviest element of all, so option a is incorrect.

b. Mass of oxygen = 2.656 × 10−26 kg

Mass of carbon = 1.994 × 10−26 kg

c. The mass of all the elements is written in standard form. so, option c is correct.

d. Mass of gold is 3.27 × 10−25 kg

Mass of iron is 9.273 × 10−26 = 0.9273 × 10−25 kg

So, option d is incorrect.

3. Mass of carbon = 1.994 × 10−26 kg

5 times the mass = 1.994 × 10−26 × 5 = 9.97 × 10−26 kg

The element closer to 5 times is iron whose mass is 9.273 × 10−26 kg.

4. Weight of the heaviest element = 3.27 × 10−25 kg

Weight of the lightest element = 1.6735 × 10−27 kg 25 27 3.27 10 1.6735 10 × × ––= 200:1

Chapter 15

Let’s Warm−up

1. If there are 45 boys and 39 girls in a hall. The ratio of boys to girls can be given as 39 45 False

2. The simplest form of ratio 27:6 = 9:2. True

3. 11 2 is the simplest form of 88 12 False

4. A ratio can have infinite equivalent ratios. True

5. One of the equivalent ratios of 81:30 is 27:10. True

Do It Yourself 15A

1. a. The speed of a car and the time taken by the car to cover a particular distance. No

b. The number of men required to construct a wall and the time taken by the men. No

2. a. No

b. Yes

Constant of proportion = 7 12

3. X 11 11 × 152 19 = 88

4. Cost of 9 kg of rice = ₹585 Cost of 52 kg of rice = ₹x ⇒ 9585 == 52 x x 585 ×52= 9 × 52 = ₹3380

5. No. of chocolates made in 6 hours = 450 No. of chocolates made in 4 hours = x ⇒ 6450450 == × 4= 300 46 x x chocolates

1.994 10 × × ––= 1.331

26 26 2.656 10

So, option b is incorrect.

6. 11 2 hours 60 + 30 minutes = 90 minutes No. of words Kamal can type in 15 minutes = 1020

No. of words Kamal can type in 90 minutes = x 15 90 = 1020 x ⇒ x = 1020 15 × 90 = 6120 words

7. Number of workers required for digging 6 m in a day = 15

Number of workers required for digging 42 m in a day = x 61542==×15 42 6 x x Þ 61542==×15 42 6 x x Þ = 105 workers

8. Distance covered by Shalini in 25 minutes = 1.5 km

Distance covered by Shalini in 40 minutes = x km

9. Number of minutes in a day = 24 × 60 min = 1440 min

Number of bottles manufactured in 40 min = 75

Number of bottles manufactured in 1440 min = x

4075==×75

144 40 0 4 40 x x = 2700 bottles

10. Space required for 5 boxes = 1500 cc

Space required for 150 boxes = x cc 51500150 == ×1500=45,000 150 5 x x ⇒ 51500150 == ×1500=45,000 150 5 x x ⇒ cc

11. Cab charges for 8 km = ₹204 Cab charges for 22 km = ₹x

12. No. of boxes in the shelf length of 17.6 m = 72 No. of boxes in the shelf length of 19.8 m = x 17.67219.8 == × 19.8 17.6 x x Þ 17.67219.8 == × 19.8 17.6 x x Þ 72 = 81 boxes

13. The number of bags weighs 90 kg = 5 The number of bags weighs 576 kg = x = 905 576 x × == 5 576 32 bags 90 x

14. Money earned by 20 men = Money earned by 6 women

Money earned by 1 man = Money earned by 6 20 women

Money earned by 40 men = Money earned by

6 40 20 women = Money earned by 12 women

Money earned by (40 men + 12 women) = Money earned by

(12 women + 12 women) = Money earned by 24 women

Let the amount earned by 40 men and 24 women = ₹x per day.

Therefore, = 624 2400 x =×=240049600 x `

Thus, 40 men and 12 women earn ₹9600.

Challenge

1. Number of days required to pick up 2.8 × 1010 kg dust = 14 days

Number of days required to pick up 5600 × 109 g dust = 9 10 560010 142.8 2.8101000 days × ×= ×× × 14 = 2.8 days

Hence, Rani is correct.

Do It Yourself 15B

1. a. Distance travelled by the car and time taken by the car. No b. Number of men required to paint a house and time taken by the men. Yes

2. 18 × 4 = 36 × a 18×4 ==2 36 a

3. Amount of time required by Mohit to finish the task = 6 days

Amount of work done by him in one day = 1 6

4. No. of days taken by 42 women to complete the work = 15 No. of days required for 18 women to complete the work

= 42 × 15 18 = 35

5. No. of chocolates received by each of the 28 students = 6 Since the number of students is reduced by 7, No. of chocolates received by each of the 21 students

= 28 ×6 21 = 8

Number of more chocolates received = 8 6 = 2 Thus, 2 more chocolates are received by each student.

6. Amount of time taken by A to complete the work = 14 days

Work done by A in one day = 1 14

Amount of time taken by B to complete the work = 21 days

Work done by B in one day = 1 21

Work done in one day by both A and B = 1 14 + 1 21 = 3 + 2 42

= 5 42

Number of days required by A and B to finish the work

= 42 5 days = 2 8 5 days

7. Time taken by three people to fit the door = 2 days

Time taken by 2 people to fit the door = 2 × 3 2 days = 3 days

8. No. of days wheat lasts for 4 friends = 30 days

No. of people for which wheat would last 20 days

= 30 × 4 20 = 6

Number of people came to the group in search of shelter

= 6 − 4 = 2

9. Speed of the car to cover the distance in 12 hours = 60 km/h

Speed of the car to cover the distance in 9 hours = 12 9 × 60

= 80 km/h

Increase in the speed of car = 80 km/h − 60 km/h

= 20 km/h

10. Portion of the tank emptied by P in one hour = 1 25

Portion of the tank emptied by Q in one hour = 1 30

Portion of the tank emptied by P and Q in one hour

= 11 + 2530 = 6+511 = 150150

Time required to empty the full tank = 150 11 hour = 13 7 11 hours

11. Amount of time taken by Manish to finish the piece of work

= 40 minutes

Amount of work done by Manish per minute = 1 40

Portion of work completed by Manish in 8 minutes = 8 40 = 1 5

Portion of work left = 1 − 1 5 = 4 5

Priya finishes 4 5 work in 16 minutes.

Time required by Priya to finish the whole work alone

= 5 4 × 16 minutes = 20 minutes

Amount of work done by Priya per minute = 1 20

Work done in one minute by both Manish and Priya

= 1 20 + 1 40 = 2 + 1 40 = 3 40

Number of minutes required by Manish and Priya to finish the work = 40 3 = 13 1 3 minutes

12. Portion of the tank filled each hour by Pipe A = 1 36

Portion of the tank filled each hour by Pipe B = 1 18

Portion of the tank emptied each hour by Pipe C = 1 133

Portion of the tank filled each hour = 111 + 3618 –133 121 = +–3636133 31399–36363 = –= = 3613347884788

Time required to fill the tank = 4788 363 = 13 23 121 hours

13. Portion of work done by A and C together in one day = 1 2

Portion of work done by A in one day = 1 4

Portion of work done by C in one day

= Portion of work done by A and C together in one day

− Portion of work done by A in one day

= 1 2 − 1 4 = 2 − 1 4 = 1 4

Portion of work done by B and C together in one day = 1 3

Portion of work done by B in one day

= Portion of work done by B and C together in one day − Portion of work done by C in one day 4–3 111 =–== 341212

Time taken by B to finish the work alone = 12 days

14. No. of days food will last for 1800 sailors = 27

No. of sailors on the ship for the food to last for 13 more

days or (13 + 27) = 40 days = 27 × 1800 40 = 1215

No. of sailors transferred to the boat = 1800 − 1215 = 585

15. Portion of recycling done by X in 1 hour = 1 8

Portion of recycling done by Y in 1 hour = 1 10

Portion of recycling done by Z in 1 hour = 1 12

Portion of recycling done in 1 hour by all 3 machines 111 =++ 81012

15+12+1037 = = 120120

Portion of recycling done between 11 a.m. and 1 p.m. =2×=3737 12060

Portion of recycling remaining = 3723 1 –= 6060

Portion of recycling done by X and Y in 1 hour

= 10+8 11 189 +=== 810808040

Time required to finish the remaining portion of work

= 23 60 9 40 = 234046 ×= 60927 hours = 46 ×60 27 minutes

= 102.22 minutes

Time of finishing recycling 500 bottles = 1 p.m. + 102. 22 minutes ≈ 2:43 p.m.

Challenge

1. Statement 1. Suman and Kavya, together, can complete the project in 30 days.

Statement 2. Pulkit and Suman, together, can complete the project in 20 days.

Given: Kavya and Pulkit together can complete the project in 24 days, so

One day ’s work of Kavya and Pulkit, += 1 . 24 KP

In 10 days, they can complete = 105 2412 of the project.

Therefore, the remaining work after 10 days is −= 57 1 1212 of the project.

From Statement 1, one day’s work of Suman and Kavya, += 1 30 SK .

We need Kavya’s work rate to determine how many more

days of Suman needs to complete the project; hence, Statement 1 alone is not sufficient.

From Statement 2, one day’s work for Suman and Pulkit, += 1 20 SP

We need the individual work rate of Pulkit as well; hence, Statement 2 alone is not sufficient.

Adding the equations from Statement 1 and Statement 2; +++=+ 11 3020 SKSP + +=== 13251 2 24606012 S

Solving for S, S = 1 48 . Thus, Suman’s work rate is 1 48 of the project per day.

Time required for Suman to complete the remaining work = 7 12 1 48 = 28 days.

Using both statements together, we could determine that Suman needs 28 more days to complete the remaining work.

Hence, option d is correct.

Chapter Checkup

1. a. Direct Proportion x 13 m = 13 ×15065 30 = 104 y 30 150 140 ratio 13:30 13:30 13:30

b. Inverse Proportion; 15 × 32 = 10 × 48 = 480 x 15

2. Rice flour required for 5 persons = 880 g Rice flour required for 12 persons = x g 588012 = =×880g=2112g 12 5 x x Þ

3. Amount paid for 11 days of work = ₹4950

Amount paid for 30 days of work = ₹x 11495030 = = × 30 11 x x Þ ₹4950 = ₹13,500

4. No. of days in a month = 30 No. of days Rohan works = 6 Fraction of work Rohan can do in 6 days = 6 30 = 1 5

5. Cost of each of the 35 boxes = ₹120

Cost of each box after discount of ₹20 = ₹120 − ₹20 = ₹100 No. of boxes that can be bought at ₹100 each = 35×120 = 100 42

No. of more boxes that can be bought = 42 − 35 = 7

6. No. of men required to plough the field in 7 days = 12

No. of men required to plough the field in 2 days = 12 × 7 2 = 42

7. No. of tankers required to fill the society tank in 4 hours = 8

No. of tankers required to fill the society tank in 2 2 3 hours = 4×8 8 3 = 12 tankers

8. No. of cows grazing the field for 32 days = 20 No. of cows grazing the field for 128 days = 3220 128 × = 5

9. No. of people carried in 6 trips = 90

No. of people carried in 35 trips = x 69035==90=525

35 6 x x Þ ×

10. Weight of 15 identical tiles = 127.5 kg

Weight of 50 identical tiles = x kg

15127.550 = =×127.5Kg=425kg

50 15 x x Þ

11. Let the amount of work done per hour by A, B, and C be 111 , and. ABC 111 + = 12 BC ……….(i) 111 + = 12 BC ……….(ii)

1111 + += 6 ABC ………(iii)

Putting values from (i) in (iii)

1 8 + 1 C = 1 6 8–6 11121 =–=== 68484824 C

Putting values from (ii) in (iii) 111 + = 126 A

1 A = 1 6 − 1 12 = 2 1 12 = 1 12

11112+131 += + = == 122424248 AC

Time required for A and C to finish the job together = 8 hours

12. Fraction of the tank filled by P in one hour = 1 25

Fraction of the tank filled by Q in one hour = 1 35

Fraction of the tank filled in one hour if both pipes operate simultaneously

= 117+512 + == 2535175175

Time taken by pipes to fill the tank = 175 12 = 14 7 12 hours

13. Time taken to reach school at the speed of 12 km/hr = 20 minutes

Reduced speed = 12 km/hr − 2 km/hr = 10 km/hr

Time taken to reach school at reduced speed = 12 ×20 10 minutes = 24 minutes

Time at which Krishvi will reach school = 8:00 a.m. + 24 minutes = 8:24 a.m.

14. Total amount distributed to 36 children = ₹18,900

Increased number of children = 36 + 6 = 42

Amount received by each child = ₹18,900 42 = ₹450

15. Let the time taken by the leak to empty the tub be x minutes.

Fraction of the tub emptied in one minute = 1 x Therefore, 111 =–3624 x 1113–21 = –= = 24367272 x minutes x = 72 minutes

16. No. of grains in 3 kg = 1.5 × 105 grains No. of grains in 5.5 kg = x 5 5

31.5 × 105.5 = = ×1.5 ×10 5.5 3 x x Þ grains = 2.75 × 105 grains

17. Fraction of the tank filled by Pipe M in one hour = 1 50

Fraction of the tank filled by Pipe N in one hour = 1 40

Fraction of the tank emptied by Pipe O in one hour = 1 80

Fraction of the tank filled in every hour = 111 +–

504080 = 8 +10 –5 400 = 13 400

Time taken by the tank to be filled completely = 400 13 hours = 30 10 13 hours

18. Fraction of the farm ploughed by Sonu in one day = 1 16

Fraction of the farm ploughed by Kamal in one day = 1 12

Fraction of the farm ploughed by Sonu and Kamal together in one day

= 1 16 + 1 12 = 3 + 4 48 = 7 48

Amount of work left after three days = 1–3×=1–77 4816

= 9 16

19. Bacteria’s length when enlarged 2,00,000 times = 2 cm

Bacteria’s length when enlarged 2,00,000 − 50,000 = 1,50,000 times = x cm

2,00,00021,50,000==×2cm=1.5cm 1,50,000 2,00,000 x x

Bacteria’s enlarged length = 2 – 1.5 = 0.5 cm

20. Fraction of work completed by Darsh, Yash and Pooja together each day

= 1114+2+17 + + = = 246122424

Time taken by Darsh, Yash and Pooja to finish the work = 24 7 hours = 3 3 7 hours

Challenge

1. Statement I: X and Y together can complete the work in 30 hours.

Statement II: X works for 16 hours and Y finishes the remaining work alone in 44 hours.

Let the fraction of work done by X and Y in one hour be 1 x and 1 y , respectively.

Fraction of work finished by X and Y together in one hour

= 1 x + 1 y = 1 30

Hence, 1 y = 1 30 − 1 x

Amount of work finished by X in 16 hours = 16 x

Amount of work left for Y to finish = 1 − 16 x Time taken by Y to finish the remaining work =

= 16 1–11 –30 x y = 44 hours

Solving for X, 1–=–164444 30 xx

28 x = 14 30 x = 60 hours

1111 =–= 306060 y or y = 60 hours

Hence, time required for Y to finish the work alone is 60 hours.

Since X and Y takes the same amount of time to complete the work alone, hence, Y is equally efficient as X.

2. Assertion (A): If 6 men can complete a job in 15 days, then adding 4 more men will reduce the time required to 8 days. Reason (R): Time taken to complete a task is inversely proportional to the number of men working on it.

Work = Number of men × Time taken.

For 6 men completing the job in 15 days: Work = 6 men × 15 days = 90 man-days.

If we add 4 more men then number of men = 6 + 4 = 10

Calculate the new time (T): 90 man-days = 10 men × T days

== 90 9 days 10 T

So, with 10 men, the job can be completed in 9 days, not 8 days. Thus, the assertion is false.

As the number of men increases, the time taken to complete a task decreases.

Hence, Assertion is false, and Reason is true. Thus, option b is correct.

Cash Study

1. Water required to irrigate one square metre of land by traditional flooding = 50 litres

Water required to irrigate 100 square metre land by traditional flooding = 100 × 50 = 5000 litres.

Water required to irrigate one square metre of land by drip irrigation = 20 litres

Water required to irrigate 100 square metre of land by drip irrigation = 100 × 20 = 2000 litres

Amount of water saved = 5000 – 2000 = 3000 litres. Thus, option c is correct.

2. Number of farmers = 10

Length of land each farmer has = 50 square metres

Saving per farmer when switching from traditional flooding to drip irrigation = 50 × (50 − 20) = 1500 litres.

Total savings of 10 farmers = 1500 × 10 = 15,000 litres. Thus, option b is correct.

3. Length of land irrigated by 50 litres of water by traditional flooding = 1 square metre

Length of land irrigated by 1 litre of water by traditional flooding = 1 sq. m

Length of land irrigated by 50,000 litres of water by traditional flooding = 1 50 × 50,000 = 1000 sq. m

Length of land irrigated by 20 litres of water by drip irrigation = 1 square metre

Length of land irrigated by 1 litre of water by drip irrigation = 1 sq. m 20

Length of land irrigated by 50,000 litres of water by traditional flooding = 1 20 × 50,000 = 2500 sq. m

Difference in area = 2500 sq. m − 1000 sq. m = 1500 sq. m

4. Answermayvary.

Chapter 16

Let’s Warm-up

1. (6+5)+(12+3)=17+9 xyyxyx

2. (8+5)+(12–2)=17+6 xyxxyxxyx

3. 22 (6 + 5)– 12=–6+5 zyzzy

4. (7–5)–(12+3)=4–17 xyyxxy

Do It Yourself 16A

1. a. 2 3+9xx

2 3 x 9 x 3××xx 3×3× x 2 3+9=3(+3) xxxx

b. 4–232 yy

322 4 – 2=2(2–1)yyyy

c. 22 5 +10 abab

2×2×××

2 5ab 2 10ab

5×××aab 2×5×××abb

22 5+10=5(+2) abababab

d. 6–1832 xyxy 3 6 xy 2 18 xy

2×3××××xxxy 2×3×3×××xyy

322 6–18=6(–3) xyxyxyxy

4 8c 2 12c 8–42 12 cc

2×2×2××××cccc 2×2×3××cc

8–422212=4(2–3) cccc

323 9 – 6 pqpq

32 9pq 3 6 pq

3×3××××ppqq 23 pqqq×××××

32322 9–6=3(3–2) pqpqpqpq

2. a. 2 7–217(–3) xxyxxy =

b. 3222 10–15=5(2–3) abababba

c. 4323 16–12=4(4–3) pqpqpqpq

d. 342323 18–6=6(3–1) mnmnmnmn

e. 524342 25–5=5(5–) xyxyxyxy

f. 342323 30–10=10(3–1) abababab

g. 322 11+22=11(+2) xxxx

h. 433 14–7=7(2–1) yyyy

i. 2322 17–34=17(–2) abababab

3 a. (3y + 2)(y –1 ) –2(3y + 2)

= (3y + 2)(y – 1 – 2)

= (3y + 2)(y – 3)

b. (3 + 2) (2 –1)–(3 + 2) xxx

=(3+2)(2–1–1) =(3+2)(2–2) =2(–1)(3+2) xx xx xx

c. (5–1)(3+2)–4(3+2) yyy

= {} (3+2)(5–1–4)=(3+2)(5–5) yyyy

= 5(32)(1) yy+−

d. (2+5)(–3)–(2 + 5) aaa

= (2+5)(–3–1)=(2+5)(–4) aaaa

e. (2 – 4)(3–6)–2(3–6)yyy =(3 – 6)(2–4–2)=(3 – 6)(2–6)yyyy =(3 – 6)(2–4–2)=(3 – 6)(2–6)yyyy

f. (+3)(+5)–(+3)2 xxx =(+3)(+5–(+3)) =(+3)(+5––3) =2(+3) xxx xxx x

g. 2 (4 + 7) ( + 2)–(4 + 7) =(4+7)((4+7)(+2)–1) aaa aaa

h. 3 2 (4 – 3)( + 2)–5(4 – 3) =(4–3)(+2–5(4–3)) xxx xxx

i. 3 2 } (–2)(+4)–3(+4) =(+4){(–2)–3(+4) xxx xxx

4. a. 32 2 2 2–3+4–6 = (2–3)+2(2–3) =(+2)(2–3) xxxyy xxyx xyx

b. 3+6–4–2 =3(+2)–2(+2) =(3–2)(+2) abaccdbd abcdbc adbc

c. 2 –7–2+14 =(–7)–2(–7) =(–2)(–7) xxxyy xxyx xyx

d. 2+5–4–10 =(2+5)–2(2+5) =(–2)(2+5) abacbdcd abcdbc adbc

e. 2 – 3 – 2 +6 =(–3)–2(–3) =(–3)(–2) xxx xxx xx

f. 2 3–2–15+10 =(3–2)–5(3–2) =(–5)(3–2) xxyxy xxyxy xxy

g. 2 6–3–10+5 =3(2–)–5(2–) =(3–5)(2–) xyxyx xyxyx xyx

h. 2 4–12+2–6 =4(–3)+2(–3) =(4+2)(–3) xxyxy xxyxy xxy

i. 6xy −9x2 − 4y + 6x = 3x (2y − 3x) − 2(2y − 3x) = (3x − 2) (2y − 3x)

5. Answersmayvary.Sampleanswer: (4pq + 12qr − 2ps − 6rs)

Challenge

1. Money spent on buying plants = ₹(4x2 + 8xy + x + 2y)

2 4+8 + + 2 =4(+2)+1(+2) =(4+1)(+2) xxyxy xxyxy xxy

No. of plants bought = x + 2y

Cost of each plant = Money spent on buying plants

No. of plants bought = (4x + 1)(x + 2y) x + 2y = ₹(4x + 1)

Do It Yourself 16B

1. a. 2 22 2 +8+16 =+2×4×+4 =(+4) aa aa a

b. 2 22 2 +10+25 =+2×5 ×+5 =(+5) xx xx x

c. ×× 2 22 2 +4+4 =+22 +2 =(+2) bb bb b

d. 22 22 2 81+18+ =9+2 ×9×+ =(+9) aa aa a

2. a. = = 2 22 –4 – 2 (+2)(–2) a a aa

b. 2 22 25–= 5–=(5+)(5–) x x xx

c. 2 22 9–16 =(3)–4 =(3+4)(3–4) b b bb

d. 2 22 49–16 =(7)–4 =(7+4)(7–4) p p pp

e. 22 –=(+)(–) ab abab

f. 222 22 –=()–=(+)(–) axy axy axyaxy

g. 2 22 100–81 =(10)–9 =(10+9)(10–9) b b bb

h. 22 22 –4 =()–2 =(+2)(–2) xy xy xyxy

3. a. 22 2 2 1–8+16 =1–2×4×1+(4) =(4–1) axax axax ax

b. 33 22 22 3–243 =3(–81) =3(–(9)) =3(+9)(–9) xyxy xyxy xyxy xyxyxy

c. 442222 2222 22 –=()–() =(+)(–) =(+)(+)(–) abab abab ababab

d. 4222 22 2 –81=()–(9) =(+9)(–9) =(+9) (–9)(+9) bb bb bbb

e. 2 22 2 64–176+121 =8–2×11×8+(11) =(11–8) xx xx x

f. 4224 222222 222 49–168+144 =(7)–2×7×12+(12) =(12–7) xxyy xxyy yx

g. = 2 22 2 49–56+16 =(7)–2×7×4+4 (7–4) yy yy y

h. 64 − 16y + y2 = 82 − 2 × 8 × y + y2 = (y 8)2

4. a. Using, (–)=–2+222 abaabb

22 –14+ ?=–2×7×+?bbbb

Missing term = 72 = 49

b. Using, (–)=–2+222 abaabb

222 –?+9=–? +3 xx

Missing term = 2× ×3 6 xx =

c. Using, (–)=–2+222 abaabb

?–2+=?–2××22 + cddcdd

Missing term = c2

d. Using, (–)=–2+222 abaabb

36–?+=6–?+222 rr

Missing term = 2 ×6×=12rr

5. a. 22 1 +=+2×× 2 aaaa

Constant term = 2 11 = 24

b. –8+=–2×4×+22 yyyy

Constant term = 2 4=16

c. 22 9–6=(3)–2 ×3×1dddd

Constant term = 12 = 1

d. 22 5 +5=+2 × × 2 xxxx

Constant term = 2 525 = 24

e. 22 –40+ 25=(5)–2×4×5 bbbb

Constant term = (4)2 = 16

f. 24+=2×12×+22 xxxx

Constant term = 122 = 144

g. 22–14=–2×7× xxxx

Constant term = 72 = 49

h. 22–16=–2×8× xxxx

Constant term = 82 = 64

6. a. 44222222 222222 22 –16=()–(4)=(+4) (–4)=(+4) (–(2)) =(+4)(–2)(+2) mnmnmn mnmnmn mnmnmn

b. 22 22 22 ––2–1=–(+2+1)=–(+1) abbabbab =(–(+1))(+(+1)) abab =(––1)(++1) abab

c. 222222 +4––4+4–4–abcabababc ⇒

= a2 2 × a × 2b + (2b)2 c2

= (a 2b)2 c2

= (a 2b + c)(a 2bc)

d. 2 22 22 22 +2+=+2 × ×+=+ xyxxyyxy yxyyxxyx

7. Answersmayvary.Sampleanswer:

Given the expression 9x2 − 24x + 16, factorise it using the appropriate algebraic identity.

Challenge

1. Area of the rectangular wall space = (9y2 – 4x2) sq. cm 22 (9–4)sq.cm=(3)–(2)22 =(3+2)(3–2)sq.cm yxyx yxyx

Putting the values x = 2 and y = 2

Length = 3y + 2x = 3 × 3 + 2 × 2 = 13 cm

Breadth = 3y – 2x = 3 × 3 − 2 × 2 = 5 cm

Do It Yourself 16C

1. a. Middle term = −5x = −4x − x

b. Middle term = −9x = −8x − x

c. Middle term = −11x = −8x − 3x

d. Middle term = −17x = −16x − x

2. a. x2 − 13x + 42 = x2 − 6x − 7x + 42

= x(x − 6) − 7(x − 6) = (x − 6) (x − 7)

b. x2 − 12x + 35 = x2 − 7x − 5x + 35

= x(x − 7) − 5(x − 7) = (x − 5) (x − 7)

c. x2 + 11x + 10 = x2 + 10x + x + 10

= x(x + 10) + 1(x + 10) = (x + 1) (x + 10)

d. x2 − 9x + 14 = x2 − 7x − 2x + 14

= x(x − 7) − 2(x − 7) = (x − 2) (x − 7)

Hence, option c is the correct answer.

3. a. 6x2 − 11x + 3 = 6x2 − 9x − 2x + 3

= 3x(2x − 3) − 1(2x − 3)

= (3x − 1) (2x − 3)

b. 2x2 − 7x + 6 = 2x2 − 4x − 3x + 6

= 2x(x − 2) − 3(x − 2)

= (2x − 3) (x − 2)

c. 2x2 − 5x + 2 = 2x2 − 4x − x + 2

= 2x(x − 2) − 1(x − 2)

= (2x − 1) (x − 2)

d. 3x2 − 8x + 4 = 3x2 − 6x − 2x + 4

= 3x(x − 2) − 2(x − 2)

= (3x − 2) (x − 2)

e. 9x2 − 9x + 2 = 9x2 − 6x − 3x + 2

= 3x(3x − 2) − 1(3x − 2)

= (3x − 1) (3x − 2)

f. 9x2 − 15x + 6 = 9x2 − 9x − 6x + 6

= 9x(x − 1) − 6(x − 1)

= (9x − 6) (x − 1)

g. 3x2 − 11x + 10 = 3x2 − 6x − 5x + 10

= 3x(x − 2) − 5(x − 2)

= (3x − 5) (x − 2)

h. 15x2 − 11x + 2 = 15x2 − 6x − 5x + 2

= 3x(5x − 2) − 1(5x − 2)

= (3x − 1) (5x − 2)

4. a. 3x2 − 23x + 14

= 3x2 − 21x − 2x + 14 = 3x (x − 7) − 2(x − 7) = (3x − 2) (x − 7)

The lengths and breadths could be (3x − 2) and (x − 7).

b. 5x2 − 37x + 14

= 5x2 − 35x − 2x + 14

= 5x (x − 7) − 2 (x − 7)

= (5x − 2) (x − 7)

The lengths and breadths could be (5x − 2) and (x − 7).

c. 4x2 − 18x + 14

= 4x2 − 4x − 14x + 14

= 4x (x − 1) − 14 (x − 1) = (4x − 14) (x − 1)

The lengths and breadths could be (4x − 14) and (x − 1).

d. 4x2 − 8x + 3

= 4x2 − 6x − 2x + 3

= 2x (2x − 3) − 1 (2x − 3)

= (2x − 1) (2x − 3)

The lengths and breadths could be (2x − 1) and (2x − 3).

5. Answersmayvary.Sampleanswer:

The area of a rectangular field can be given by the expression x2 + 7x + 10. What is the length and width of the garden?

Challenge

1. a. Yes

b2 c3 + 8bc4 + 12c5

= b2c3 + 6bc4 + 2bc4 + 12c5

= bc3 (b + 6c) + 2c4 (b + 6c) = (b + 6

= c3 (b + 6c) (b + 2c)

b. Yes

3c5 − 18c4 − 48c3 = 3c5 − 24c4 + 6c4 − 48c3

= 3c4 (c − 8) + 6c3 (c − 8) = (c − 8) (3c4 + 6c3 )

= 3c3 (c + 2) (c − 8)

2. The expression for the profit of the store = (3x2 – 12x – 15)

Total profit = No. of products sold × Profit per product sold

3x2 − 12x − 15

= 3 (x2 − 4x − 5)

= 3 (x2 − 5x + x − 5)

= 3 (x (x − 5) + 1 (x − 5))

= 3 (x + 1) (x − 5)

Since (x − 5) is the number of products sold, therefore the profit per product is 3(x + 1).

Do It Yourself 16D

1. a. 15x3 3x = 5x3 − 1 = 5x2

b. 15a2b4 5ab2 = 3a2 − 1 b4 − 2 = 3ab2

c. 24y4z3 6yz = 4y4 − 1 z3 − 1 = 4y3z2

d. 18x5y2z 9xyz2 = 25 − 1y2 − 1 z2 − 1 = 2x4y z

e. 36p4q3r2 12p2qr2 = 3p4−2 q3−1r2−2 = 3p2q2

f. 84j5k7r9 7jk3r2 = 12j5−1k7−3r9−2 = 12j4k4r7

2. a. 2x3 + 4x2 6x 2x = 2x3 2x 4x2 2x 6x 2x +

= x2 + 2x 3

b. 3y4 − 9y3 + 6y2 3y

= 3y4 3y 9y3 3y 6y2 3y + = y3 3y2 + 2y

= 4a5 4a 12a4 4a 8a3 4a + +

= a4 3a3 + 2a2 4a5 − 12a4 + 8a3 4a

d. 32 32 2 6–18+12 6 61812 =–+ 666 =–3+2 mmm m mmm mmm mm

e. 654 2 654 222 432 5+10–15 5 51015 =+–55 5 =+2–3 xxx x xxx xxx xxx

f. 5a3x − 15a2y2 + 3az3 3a

= 5a5x 3a 15a2y2 3a 3az3 3a +

= a2x 5ay2 + z3 5 3

3. a. 225abc (4a − 8) (5b − 15) (125 (a − 2) (b − 3)

225 × 4 × 5 × abc (a − 2) (b − 3) = 125 (a − 2) (b − 3) = 36abc

b. 144ab (a + 5) (b − 4) 36a (b − 4) = 4b (a + 5)

c. 28 (a + 3) (a2 + 3a + 70) 7(a + 3) = 4(a2 + 3a + 70)

d. 117abc (a + b) (b + c) (c + a)

78ab (b + c) (c + a) c(a + b) 3 2 =

4. a. 4x −6

x2 − 4 4x3 −6x2 + 3x −9

4x3 +0x2 −16x −6x2 + 19x −9

−6x2 + 0x +24 19x −33

b. 3y2 9y + 18

y2 − 4 3y2 −9y3 +6y2 −12y +0

3y4 +0y3 −12y2 −9y3 +18y2 −12y +0

−9y3 +0y2 +36y 18y2 −48y +0

18y2 +0y +72 −48y +72

c. x3 +5x2

x2 + 4 x5 +5x4 +4x3 +20x2 +16x +0

x5 +0x4 +4x3

5x4 +0x3 +20x2 +16x +0

5x4 +0x3 +20x3 16x +0

d. 6a3 15a2 +39a −84

a + 2 6a4 −3a3 +9a2 −6a +0

6a4 +12a3

15a3 +9a2 −6a +0

−15a3 +30a2 39a2 −6a +0

39a2 +78a −84a +0

−84a −168 168

e. 9b4 + 12b3 21b2 −28b + 42

b2 + 2 9b6 +12b5 −3b4 −4b3 +0b2 +0b +0

9b6 +0b5 −18b4 12b5 −21b4 −4b3 +0b2 +0b +0

12b5 +0b4 +24b3 −21b4 −28b3 +0b2 +0b +0

−28b3 +42b2 +0b +0 −21b4 +0b3 −42b2 −28b3 +0b2 −56b 42b2 +56b +0

42b2 +0b +84 56b –84

f. 2c4 −12c3 +30c2 −60c +108

c + 2 2c5 –8c4 +6c3 –0c2 –12c + 0

2c5 +4c4 −12c4 +6c3 +0c2 –12c + 0

–12c4 –24c3

30c3 +0c2 –12c + 0

30c3 +60c2 –60c2 –120c 108c + 0 ––––108c +216 ––216

−60c2 –12c + 0

5. a. 3x +7

x2 + 4x + 4 3x3 –5x2 +2x –1

3x3 –12x2 +12x 7x2 –10x –1

7x2 –28x +28 18x −29 ––

b. 2y2 +6y +14

2y2 – 3y – 6 4y4 +6y3 –2y2 +8y –12

4y4 –6y3 –12y2 12y3 +10y2 +8y –12 12y3 +18y2 –36y 28y2 +44y –12 28y2 –42y –84 86y +72 –––

c. 3a2 + a 2 +13 4

2a2 – a + 1 6a4 –2a3 +9a2 –5a +3

6a4 –3a3 +3a2 a3 +6a2 –5a +3 a3 –a2 2 + a 2 13a2 2 –11a 2 +3 13a2 2 –13a 4 + 13 4 − 9a 4 1 4 ––

d. x2 x +4

x3 – 2x2 – x + 2 x5 –3x4 +5x3 –2x2 –4x +8 x5 –2x4 –x3 +2x2 –x4 +6x3 –4x2 –4x +8 –x4 +2x3 +x2 –2x 4x3 –5x2 –2x +8 4x3 –8x2 –4x +8 3x2 +2x +0 –––

e. 7x +23

x2 – 3x – 2 7x3 +2x2 –9x –5

9y5 –18y4 –9y3 +18y2 24x

6. To find what number must be multiplied by (x – 7) to get (x3 – 12x2 + 38x – 21), we need to divide (x3 – 12x2 + 38x – 21) by (x – 7)

x2 −5x + 3

x – 7 x3 – 12x2 + 38x –21

x3 –7x2 −5x2 +38x –21 –5x2 +35x 3x –21 3x –21 0 –

Hence, x – 7 must be multiplied by x2 – 5x + 3.

7. Amount of water in the bucket = (x2 + 5x + 6) litres

Number of days took for drying up = (x + 3) days

Rate of evaporation per day = (x3 + 5x + 6) (x + 3)

x + 2 x + 3 x2 + 5x + 6

x2 + 2x 3x + 6 3x + 6 0 – –– –

So, the rate of evaporation per day is (x + 2) litres per day.

8. The expression used to represent the total number of chocolates = (x3 + 6x2 + 12 + 13x)

The expression used to represent the total number of students = (x + 3)

x2 + 3x + 4

x + 3 x3 +6x2 +13x +12

x3 +3x2

3x2 +13x +12

23x2 –69x –46 74x +41 ––

7x3 –21x2 –14x 23x2 +5x –5

3x2 +9x 4x +12 4x +12 0

Each kid will get (x2 + 3x + 4) chocolates.

Challenge

1. 3x2 +3x +5

x 1 3x3 +0x2 −8x +0

3x3 −3x2

3x2 −8x +0

3x2 −3x +5x +0 −5x +5 −5

Hence, 5 must be added to 3x3 8x to make it exactly divisible by x 1.

2. a3 +1

a2 + 2a – 3 a4 +2a3 –2a2 +a –1

c. 3x2 − 2xy + 6x − 4y

= x (3x − 2y) + 2(3x − 2y)

= (x + 2) (3x − 2y)

d. 9x2 − 6x − 12xy + 8y

= 3x (3x − 2) − 4y(3x − 2)

= (3x − 4y) (3x − 2)

e. 4x3 − 2x2 − 6xy + 3y

= 2x2(2x − 1) − 3y (2x − 1)

= (2x2 − 3y)(2x − 1)

f. 6x2 + 7xy − 18x − 21y

= x (6x + 7y) − 3 (6x + 7y)

= (x − 3) (6x + 7y)

4. a. a2 + 6ab + 9b2

= a2 + 2 × a × 3b + (3b)2

= (a + 3b)2

b. x2 − 10xy + 25y2

= x2 − 2 × x × 5y + (5y)2

= (x − 5y)2

a4 +2a3 –3a3 a2 +a –1

a2 +2a –3 a +2 ––

Hence, −a + 2 must be subtracted from a4 + 2a3 − 2a2 + a − 1 to make it exactly divisible by a2 + 2a − 3.

Chapter Checkup

1. a. 2x2 + 6x = 2x (x + 3)

b. 3ab − 9a = 3a (b − 3)

c. 4y3 + 8y2 = 4y2 (y + 2)

d. 7xy − 14x = 7x (y − 2)

2. a. (x + y) − (x + y) (2x − 11)

= (x + y) (1 − 2x + 11)

= − 2 (x + y) (6 − x)

b. x (x − 2z) + y (x − 2z) + (2z − x)

= (x − 2z) (x + y − 1)

c. (n − 10)2 + (10 − n)

= (10 − n) (10 − n + 1)

= (10 − n) (11 − n)

d. (3a − 1)2 − 6a + 2

= (3a − 1)2 − 2 (3a − 1)

= (3a − 1) (3a − 1 − 2)

= 3 (3a − 1) (a − 1)

e. 5x + 10y − 7 (x + 2y)2

= 5 (x + 2y) − 7 (x + 2y)2

= (x + 2y) (5 − 7x − 14y)

f. (5x + y) − (5x + y)3

= (5x + y) (1 − (5x + y)2)

3. a. 4a2 b − 6ab2 − 2ab + 3b2

= 2ab(2a − 3b) − b(2a − 3b)

= (2ab − b) (2a − 3b)

= b(2a − 1) (2a − 3b)

b. 5x2 − 10xy + 3xy − 6y2

= 5x (x − 2y) + 3y (x − 2y)

= (5x + 3y) (x − 2y)

c. 4x2 + 4xy + y2

= (2x)2 + 2 × 2x × y + y2

= (2x + y)2

d. 25x2 + 40xy + 16y2

= (5x)2 + 2 × 5x × 4y + (4y)2

= (5x + 4y)2

e. x2 − 14xy + 49y2

= x2 − 2 × x × 7y + (7y)2

= (x − 7y)2

f. 16a2 − 32ab + 16b2

= 16 (a2 − 2ab + b2)

= 16 (a − b)2

5. a. p2 − 9q2

= p2 − (3q)2

= (p + 3q) (p − 3q)

b. 4x2 − 49y2

= (2x)2 − (7y)2

= (2x + 7y) (2x − 7y)

c. a2 − 16b2

= a2 − (4b)2

= (a + 4b) (a − 4b)

d. 9m2 − 25n2

= (3m)2 − (5n)2

= (3m + 5n) (3m − 5n)

e. x2 − 36y2

= x2 − (6y)2

= (x + 6y) (x − 6y)

f. 25u2 − 4v2

= (5u)2 − (2v)2

= (5u − 2v) (5u + 2v)

6. a. 6x2 + 11x − 2

= 6x2 + 12x − x − 2

= 6x (x + 2) − 1 (x + 2)

= (6x − 1) (x + 2)

b. x2 − 5x + 6

= x2 − 3x − 2x + 6

= x (x − 3) − 2 (x − 3)

= (x − 2) (x − 3)

c. 3x2 + 12x − 15

= 3x2 + 15x − 3x − 15

= 3x (x + 5) − 3(x + 5)

= 3(x − 1) (x + 5)

d. 3x2 − 10x + 7

= 3x2 − 7x − 3x + 7

= x (3x − 7) − 1 (3x − 7)

= (x − 1) (3x − 7)

e. x2 + 14x + 48

= x2 + 8x + 6x + 42

= x (x + 8) + 6 (x + 8)

= (x + 6) (x + 8)

f. 2x2 − 3x − 5

= 2x2 − 5x + 2x − 5

= x (2x − 5) + 1(2x − 5)

= (x + 1) (2x − 5)

7. a. 6x3 + 9x2 − 12x 3x 6x3 3x 3x 3x = + 9x3 12x

= 2x2 + 3x − 4

b. 4a2 b3 c2 − 8ab2 c2 d 2abc

4a2 b3 c2 8ab2 c2 d 2abc 2abc

= = 2ab2 c − 4bcd

c. 12a3 b2 − 6a2 b3 + 18ab4 6ab2

12a3 b2

6ab2 = + 6a2 b3 18ab4 6ab2 6ab2

= 2a2 − ab + 3b2

d. 9x5 − 3x4 + 6x3 3x2

9x5

3x2 3x2 3x2 = 3x4 6x3 = 3x3 − x2 − 2x

e. 7a4 b3 c2 − 14a3 b2 c3 + 21a2 bc4

7a4 b3 c2

7a2bc2

7a2bc2 = + 14a3 b2 c3 21a2 bc4 7a2bc2 7a2bc2

= a2b2 − 2abc + 3c2

f. 15x6 − 10x5 + 5x4 5x3

15x6

5x3 5x3 5x3 = + 10x5 5x4

= 3x3 − 2x2 + x

8. a. 2x2 +8x +10

x − 2 2x3 +4x2 −6x +0

2x3 −4x2 8x2 −6x +0 8x2 −16x 10x +0 10x –20 20

b. 3x3 −3x2 −6x +12 x + 1 3x4 +0x3 −9x2 +6x +0

3x4 +3x3 −3x3 −9x2 +6x +0

−3x3 +3x2 −6x2 +6x +0

−6x2 −6x 12x +0 12x +12 −12

c. 2x2 +7x +2

2x − 1 4x3 +12x2 −3x +0 4x3 −2x2 14x2 −3x +0 14x2 7x 4x +0 4x −2 2

d. x3 −3x2 +3x 6 5

5x + 2 5x4 −13x3 +9x2 +0x +0

5x4 +2x3

−15x3 +9x2 +0x +0

−15x3 −6x2

15x2 +0x +0

15x2 +6x −6x +0

−6x 12 5 12 5

e. 3x3 +9x2 2 +3x 4 + 9 8

2x − 3 6x4 +0x3 −12x2 +0x +6

6x4 −9x3

9x3 −12x2 +0x +6

9a3 − 27x2 2 3x2 2 +0x +6

3x2 2 − 9x 4 9x 4 − 27 8 9x 8 +6 75 8

f. x2 − 3x 7 + 86 49

7x − 4 7x3 −7x2 +14x +0

7x3 −4x2 −3x2 +14x +0 3x3 + 12x 7 86x 7 +0 86x 7 − 344 49 344 49

9. a. 3x +13

x2 − 3x + 2 3x3 +4x2 −2x +1

3x3 −9x2 +6x 13x2 −8x +1 13x2 −39x +26 31x −25

b. 2x2 +2x +1

x2 − x 2 2x4 +0x3 −5x2 +0x +3

2x4 −2x3 −4x2 2x3 −x2 +0x +3 x2 +4x +3 x2 x −2 5x +5

2x3 −2x2 −4x

c. 3x + 1 2

2x2 − 3x 2 6x3 −8x2 +2x −4

6x3 −9x2 −6x x2 +8x −4 x2 − 3x 2 −1

19x 2 −3

d. 5x –16

x2 + 2x – 1 5x3 –6x2 +3x –2

5x3 +10x2 –5x −16x2 +8x –2 –16x2 –32x +16 40x −18 ––

e. x2 −4x +15 x2 + 4x + 4 x4 +0x3 +3x2 +0x –4 x4 +4x3 +4x2 −4x3 –x2 +0x –4

15x2 –60x +60 −44x −64 ––––4x3 –16x2 –16x

15x2 +16x –4

f. 2x 4x2 − 6x 3 8x3 −12x2 −6x +9 8x3 −12x2 −6x 0 9

10. Area of the garden = 64x3 − 25xy2 = x(64x2 − 25y2) = x ((8x)2 − (5y)2 ) = x (8x + 5y) (8x − 5y)

Length

Breadth

x (8x + 5y) (8x − 5y)

(8x + 5y) x (8x − 5y)

(8x − 5y) x (8x + 5y)

x (8x + 5y) (8x − 5y)

x (8x − 5y) (8x + 5y)

(8x + 5y) (8x − 5y) x

Challenge

1. x2 + 3x + 5

x2 + 4x − 2 x4 + 7x3 + 15x2 + 7x − 0 x4 + 4x3 − 2x2

3x3 + 17x2 + 7x 3x3 + 12x2 + 6x

5x2 + 13x − 5 5x2 + 20x − 10 −7x + 5 + +

Hence, (5 − 7x) must be subtracted from x4 + 7x3 + 15x2 + 7x – 5 so that the resulting polynomial is exactly divisible by x2 + 4x – 2.

2. Assertion: The factorisation of x2 + xy + 2x + 2y is (x + 2)(x − y).

x2 + xy + 2x + 2y = xy(x + y) +2(x + y)

= (x + 2)(x + y)1(x + 2)(x – y)

So, the assertion is false.

Reason: Factorisation is defined as breaking or decomposing an expression into a product of two or more expressions which when multiplied together, give the original expression.

The definition of factorisation given is correct. So, the reason is correct.

Hence, A is false, but R is true. Thus, option d is correct.

Case Study

1. We know that the area is the product of length and breadth.

Area occupied by the door

=(2ac + 5b2 + 6ac + 15bc) square units

On factorising 2ab + 5b2 + 6ac + 15bc by taking common terms out, we get

b(2a + 5b) + 3c (2a + 5b) = (2a + 5b)(b + 3c)

Hence, the possible length and breadth are (2a + 5b) units and (b + 3c) units

So, option d is correct answer.

2. The perimeter of one square-shaped window

= (12a + 20b) = 4(3a + 5b)

We know that the perimeter of a square = 4 × side On comparing, we get the length of the side of the window as: (3a + 5b) units

Area of a square = side2 = (3a + 5b)2 square units So, option b is correct.

3. Length of the wall = (–32 – 21 – 54) Area of the wall Breadth of the wall–6) ( xxx x =

x2 + 5x + 9

x − 6 (x3 − x2 − 21x − 54 x4 + 6x2

5x2 − 21x 5x2 − 30x 9x − 54 9x − 54 0 + + +

Hence, the length of the wall is (x2 + 5x + 9) units.

4. Cost of fitting one tile =

Hence, the cost of fitting one square tile is ₹(2x2 + 5x − 1).

5. Answermayvary.Sampleanswer:

Neha planted a variety of plants in her garden.

The area of the garden can be represented by polynomial (3x3 + 5x2 + 6x + 8) sq. cm. She plants each variety in a plot of area (x + 2) sq. cm.

How many flowers of each type will each plot have?

Chapter 17

Let’s Warm-up

1. The most favourite sport among the students is badminton.

2. The least favourite sport among the students is chess

3. Students like badminton more than hockey.

4. Students like chess less than cricket.

Do It Yourself 17A

1. a. A line graph is used to study how data changes over time. True

b. When interpreting a double-line graph, if one line is consistently higher than the other, it means that the data represented by that line is always greater. True

c. Line graphs can also be called histograms. False

d. In a double-line graph, the two lines must always intersect at some point on the graph. False

2. a. The most number of shirts was sold on Thursday

b. The least number of shirts was sold on Wednesday.

c. Percentage increase in the number of shirts sold from Wednesday to Thursday = Number of shirts sold on Thursday − Number of shirts sold on Wednesday

Number of shirts sold on Wednesday

d. Average number of shirts sold on 5 days

= Total number of shirts sold

Number of days

= 30 + 40 + 20 + 45 + 35 5 170 5 = = 34

Ratio of the number of shirts sold on Tuesday to the average number of shirts sold on all 5 days = 40 : 34 = 20 : 17

Challenge

1. Assertion: City A was most populated in 2004 and least populated in 2001.

Reason: In 2001, the population of City A was 60% of the population of City B.

From the given graph, city A was most populated in 2004 and least populated in 2001.

Population of city A in 2001 = 6000

Population of city B in 2001= 10,000

× 100% = 60% 10,000 6000

So, there were 60% of people in City A in comparison to City B in the year 2001.

Both the assertion and the reason are true, but the reason does not explain the assertion. Thus, option b is correct.

Do It Yourself 17B

1. a. Every point on the y-axis is of the form (x, 0). False

b. The graph of every linear equation in two variables need not be a line. False

c. A point whose y-coordinate is zero and x-coordinate is 5 will lie on the y-axis. False

d. The coordinates of origin are (0, 0). True

e. A point whose x-coordinate is zero and y-coordinate is non-zero will lie on the y-axis. True

2. a. Abscissa = 9, Ordinate = 4

b. Abscissa = 4, Ordinate = −5

c. Abscissa = 0, Ordinate = 4

d. Abscissa = −2, Ordinate = −8

e. Abscissa = 0, Ordinate = −7

3. a. Since, the x-coordinate is 0, then the point lies on the y-axis.

b. Since, the y-coordinate is 0, then the point lies on the x-axis.

c. Since, the y-coordinate is 0, then the point lies on the x-axis.

d. Since, the x-coordinate is 0, then the point lies on the y-axis.

e. Since, the x-coordinate and the y-coordinate are both 0, then the point lies on the origin.

4. a. (−8, 8) lies in the second quadrant.

b. (−9, −2) lies in the third quadrant.

c. (1, −8) lies in the fourth quadrant.

d. (8, 5) lies in the first quadrant.

e. (−5, 0) lies in the second quadrant.

5. Scale: x-axis = 1 division = 1 unit

-axis = 1 division = 20 units

5. Answersmayvary.Sampleanswer:

The number of books read by two students Ronak and Priya in 6 months is given below. Draw double line graph representing the given data.

The points do not lie in a straight line.

Scale: x-axis = 1 division = 1 unit y-axis = 1 division = 1

The line passing through (4, 4) and (−4, −4) meets the x-axis and y-axis at the origin, whose coordinates are (0, 0).

7. Scale: x-axis = 1 division = 1 unit y-axis = 1 division = 50 units

Cost of a notebook = ₹50

1 1 × ₹50 = ₹50 2 2 × ₹50 = ₹100 3 3 × ₹50 = ₹150 4 4 × ₹50 = ₹200 5 5 × ₹50 = ₹250

6 × ₹50 = ₹300

7 × ₹50 = ₹350

8 × ₹50 = ₹400

9 × ₹50 = ₹450

10 × ₹50 = ₹500

a. The cost of 7 books is ₹350.

b. The number of books purchased with ₹50 = 1

The number of books purchased with ₹550 = 1 50 × 550 = 11

8. Scale: x-axis: 1 division = 200 km y-axis: 1 division = 2 hours

9. Answersmayvary.Sampleanswer:

The dataset represents the number of hours studied and the corresponding scores on a test. Number of hours studied 1 2 3

Scale: x-axis = 2 divisions = 1 hour y-axis = 1 division = 10 marks

The graph is not linear.

Challenge

1. a. Let us denote the total fare as F (in ₹) and the distance travelled as d (in km).

Given: The initial fee (fixed charge) = ₹50

Fare for 5 km = ₹150 Fare for 10 km = ₹250

The equation for the fare is: F = 50 + 20d

Data points:

b. Graph:

Scale: x-axis: 1 division = 5 km

y-axis: 1 division = ₹100

Chapter Checkup

1. a. Abscissa = 5, Ordinate = 4; first quadrant.

b. Abscissa = −8, Ordinate = −2; third quadrant.

c. Abscissa = −9, Ordinate = −8; third quadrant.

d. Abscissa = −5, Ordinate = −10; third quadrant.

2. Scale: x-axis: 1 division = 1 unit y-axis: 1 division = 1 unit

b. The points does not lie on a line.

Scale: x-axis: 1 division = 1 unit y-axis: 1 division = 1 unit

4. Scale: x-axis: 1 division = 1 unit y-axis: 1 division = 1 unit

When we extend the line joining the points A (1, 3) and B (0, 4), it meets the x-axis at C (4,0) and y-axis at B (0, 4).

5. Scale: x-axis: 1 division = 1 year y-axis: 1 division = 2 thousands

No, from the graph we can see that A (5, 4) and B (4, 5) do not represent the same point.

3. a. The points lie on a line.

Scale: x-axis: 1 division = 1 unit y-axis: 1

= 1

6. Point Abscissa Ordinate Coordinates

A −2 1.5 (−2, 1.5)

B 3 1 (3, 1)

C 1 −3 (1, −3)

D −3 −2 (−3, −2)

Scale: x-axis: 1 division = 1

8. Scale: x-axis: 1 division = 5 minutes y-axis: 1 division = 5 m

9. Scale: x-axis: 1 division = ₹100 y-axis: 1 division = ₹500

Deposit

Simple Interest

Yes, the graph passes through the origin.

Challenge

1. The sales in 2022 = Twice the sales in 2017 = 3 × 5 crores = 15 crores

The sales in 2023 = Thrice the sales in 2016 = 2 × 6 crores = 12 crores

The sales in 2024 = All-time high. We do not know the exact sales in 2024 so we cannot plot the graph for the sales.

2. Statement 1: The coordinates of point P are (7,8).

Statement 2: The coordinates of points Q and R are (7,0) and (3,0).

Statement 1 gives information only about coordinates of point P.

Statement 2 gives information only about coordinates of point Q and R.

Hence, Statement 1 alone and Statement 2 alone are not sufficient to answer the question.

The base QR lie along the x-axis, so the base is 7 – 3 = 4 units

The height is the vertical distance from point P to line QR: 8 – 0 = 8 units

Area of a triangle = ×× 1 2 bh

= 1 2 × 8 × 4 = 16 sq. units.

Hence, c) Both statements together are sufficient to answer the question, but neither statement alone is sufficient. So, option c is correct.

Ansh

Raman

Case Study

1. The speed at which Mariam is driving is 20 km/hr.

So, the correct answer is option b.

2. The speed at which Mariam is driving = 20 km/hr.

Total time = 8 hours

Total distance covered = 20 × 8 = 160 km

3. Total distance covered in 2.5 hours = 2.5 × 20 = 50 km

So, option d is correct.

4. Increase in speed = 10 km/hr

New speed = 20 + 10 = 30 km/hr

Total distance = 60 km

Time taken = 60 ÷ 30 = 2 hours

Time taken earlier to travel 60 km = 60 ÷ 20 = 3 hours

Difference between time = 3 hours − 2 hours = 1 hour

Thus, Mariam will take 1 less hour to travel 60 km.

5. Answersmayvary.

About the Book

The ImagineMathematics teacher manuals bridge the gap between abstract mathematics and real-world relevance, offering engaging activities, games and quizzes that inspire young minds to explore the beauty and power of mathematical thinking. These teacher manuals are designed to be indispensable companions for educators, providing well-structured guidance to make teaching mathematics both effective and enjoyable. With a focus on interactive and hands-on learning, the lessons in the manuals include teaching strategies that will ensure engaging lessons and foster critical thinking and problem-solving skills. The teaching aids and resources emphasise creating an enriched and enjoyable learning environment, ensuring that students not only grasp mathematical concepts but also develop a genuine interest in the subject.

Key Features

• Alignment with Imagine Mathematics Content Book: Lesson plans and the topics in the learners’ books are in sync

• Learning Outcomes: Lessons designed as per clear, specific and measurable learning outcomes

• Alignment to NCF 2022-23: Lessons designed in accordance with NCF recommendations

• Built-in Recaps: Quick recall of pre-requisite concepts covered in each lesson

• Supporting Vocabulary: Systematic development of mathematical vocabulary and terminology

• Teaching Aids: Resources that the teachers may need to facilitate the lesson

• Activity: Concise and organised lesson plans that outline each activity

• Extension Ideas: Analytical opportunities upon delivery of each lesson

• Detailed Solutions: Solutions to all types of questions in the ImagineMathematics Content Book

• Digital Assets: Access to supplementary interactive resources

About Uolo

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ISBN 978-81-984519-6-5