Imagine_Maths_TM_Grade7

Teacher Manual MATHEMATICS

MATHEMATICS

Master Mathematical Thinking

Grade 7

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

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

The NEP 2020 has distinctly recognised the value of mathematical thinking among young learners and the significance of fostering love for this subject by making its learning engaging and entertaining. Approaching maths with patience and relatable real-world examples can help nurture an inspiring relationship with the subject. It is in this spirit that Uolo has introduced the Imagine Mathematics product for elementary grades (1 to 8). This product’s key objective is to eliminate the fear of mathematics by making learning exciting, relatable and meaningful for children. This is achieved by making a clear connection between mathematical concepts and examples from daily life. This opens avenues for children to connect with and explore maths in pleasant, relatable, creative and fun ways.

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

Imagine Mathematics strongly positions itself on the curricular and pedagogical approach of the Gradual Release of Responsibility (GRR), which has been highly recommended by the NEP 2020, the latest NCF Draft and other international educational policies. In this approach, while learning any new mathematical concept, learners first receive sufficient modelling, and then are supported to solve problems in a guided manner before eventually taking complete control of the learning and application of the concept on their own. In addition, the book is technologically empowered and works in sync with a parallel digital world which contains immersive gamified experiences, video solutions and practice exercises among other things. Interactive exercises on the digital platform make learning experiential and help in concrete visualisation of abstract mathematical concepts.

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

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

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

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

Key Elements of a Lesson—A Quick Glance

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

Operations on Fractions 1

Operations on Fractions 1

Operations on Fractions 1

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

Operations on Fractions 1

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

Learning Outcomes

Students

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

Operations on Fractions 1

Alignment to NCF

C-1.6:

situations

Let’s Recall Recap to check if students know how to write fractions and classify them as proper, improper, and mixed.

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

students to solve the questions given in the

Vocabulary

fraction: number of the form a b  , where a and b are whole numbers and b 0

reciprocal: the number formed after interchanging the numerator and denominator of a fraction

Teaching Aids

Bingo cards; Question cards; Situation cards; Question slips; Rectangular strips; Coloured pencils; Puzzle cards with multiplication problems and their corresponding answers; Small bells; Stopwatch

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

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

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

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

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

Multi-step

Multiplying

Multi-step

on Adding and Subtracting Fractions

Adding and Subtracting Decimals

Multi-step Word Problems on Adding and Subtracting Decimals

Multiplying and Dividing Decimals

Multi-step Word Problems on Operations on Decimals

SSS

ASA

RHS

Equivalent Ratios

Proportional Terms

Unitary Method

Speed, Distance and Time

12.

13.

Percentages, Decimals and Fractions

Percentage and Ratios

Finding Percent and Percent Change

Word Problems on Percentage

Profit and Loss Percentages

Word Problems on Profit and Loss

Simple Interest

Revision

Representing Rational Numbers

Rational Numbers in Standard Form; Equivalent Rational Numbers

Comparing and Ordering Rational Numbers

Rational Numbers Between Two Rational Numbers

Addition of Rational Numbers; Additive Inverse

Subtraction of Rational Numbers

Multiplication of Rational Numbers; Reciprocal or Multiplicative

Inverse of a Non-zero Rational Number

Division of Rational Numbers

Revision

Constructing Triangles: SSS Criterion

Constructing Triangles: SAS Criterion

Constructing Triangles: ASA Criterion

Constructing Triangles: RHS Criterion

Revision

Perimeter and Area of Squares and Rectangles

Area of Parallelograms

Area of Triangles

Circumference of Circles

Area of Circles, Semicircles and Quadrants

Combined Shapes

Word Problems on Perimeter and Area

Revision

Generating Rules in Formulas and Patterns

Algebraic Expressions and Terms

Addition of Algebraic Expressions

Subtraction of Algebraic Expressions

Finding the Value of an Expression

Simplifying Algebraic Expressions Word Problems Revision

Exponents and Prime Factors; Comparing Numbers in Exponential Form

Powers of Rational Numbers; Reciprocal of a Rational Number Laws of Exponents

Expressing Large Numbers in the Standard Form

Oblique and Isometric Sketches

Visualising Solid Objects Revision

Operations on Fractions 1

Learning Outcomes

Students will be able to:

add or subtract fractions.

solve multi-step word problems on adding and subtracting fractions. multiply 2 or more fractions.

divide a fraction by a fraction or a whole number.

solve multi-step word problems on operations on fractions.

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 write fractions and classify them as proper, improper, and mixed. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

fraction: number of the form a b  , where a and b are whole numbers and b 0

reciprocal: the number formed after interchanging the numerator and denominator of a fraction

Teaching Aids

Bingo cards; Question cards; Situation cards; Question slips; Rectangular strips; Coloured pencils; Puzzle cards with multiplication problems and their corresponding answers; Small bells; Stopwatch

Chapter: Operations on Fractions

Adding and Subtracting Fractions

Learning Outcomes

Students will be able to add or subtract fractions.

Teaching Aids

Bingo cards; Question cards

Activity

Imagine Maths Page 2

Explain that fractions need to be converted to like fractions to be able to add or subtract them. Make problem cards (example, 2 6 +

1

etc.) and bingo cards (showing answers to the corresponding question cards).

Divide the class into groups.

Distribute the bingo cards to each group.

Shuffle the problem cards and place them face down on the table. Pick a problem card and write the problem on the board or read it out loud. Instruct the students to solve the problem, and check whether the answer lies on the bingo card. If it does, they need to cross out the number on their bingo card. The group that crosses out a row and a column first, wins. Ask the students to write the answers in their notebooks.

Extension Idea

Ask: How much will you add to the sum of 5 8 and 1 8 to get 5 4  ?

Say: Since 5 8 + 1 8 = 6 8  , and 5 4 = 10 8  ; you will need to add

Multi-step Problems on Adding and Subtracting

Fractions

Learning Outcomes

Imagine Maths Page 5

Students will be able to solve multi-step word problems on adding and subtracting fractions.

Teaching Aids

Situation cards; Question slips

Activity

Instruct students to form groups for the activity. Distribute the situation cards to each group.

Preeti, a baker, was baking a cake for the town’s Baking Fair. Her original recipe required 3 4 cup of sugar and 1 2 cup of butter. A friend suggested she enhance the flavour by adding 1 3 cup of chocolate chips. Preeti realised she only had 1 4 cup of chocolate chips in her cupboard.

Prepare question slips such as the ones given below and distribute among the students.

How much more of the chocolate chips does Preeti need for her recipe?

How many cups of ingredients did Preeti use in total?

Instruct students to solve these word problems in their notebooks. Discuss their approach in the class. Give more word problems to the students for practice after discussion.

Extension Idea

Ask: How much sugar will Preeti need to bake double the size of the cake she was baking?

Say: Preeti will need to double the amount of sugar, i.e., 3 4 + 3 4 = 3 2 or 1  1 2 cups.

Multiplying Fractions

Learning Outcomes

Students will be able to multiply 2 or more fractions.

Teaching Aids

Imagine Maths Page 6

Rectangular strips; Coloured pencils; Puzzle cards with multiplication problems and their corresponding answers

Activity

Instruct the students to work in groups. Distribute rectangular strips and coloured pencils among the groups. Instruct them to multiply 3 5 with 1 2 using the strips. Discuss how they can show 3 5 groups of 1 2 . Bring out the fact that they need to shade the strip to show 3 5 vertically and then shade the strip to show 1 2 horizontally by making crosses. Finally, ask them to write the part that is both shaded and crossed out, in their notebooks. Now ask them to multiply the numerators of the 2 fractions and multiply the 2 denominators. They will then compare the answers to identify the rule.

Repeat the activity with one more set of fractions.

Give the students a mix of puzzle cards with their corresponding answers, where they need to multiply fractions and match with the answers. The group that finishes first, is the winner.

Extension Idea

Ask: How many hours is 2 8 of a day?

Say: 2 8 of a day is equal to 2 8 of 24 hours.    2 8    × 24 = 6 hours.

Learning Outcomes

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

Teaching Aids

Small bells; Stopwatch

Activity

Start the class by briefly discussing the division of fractions. Instruct students to form groups. Distribute the bells and stopwatches to each group. Inform students that they will be playing a quiz round which will have fractional division problems. Explain the rules and format of the quiz.

Introduce the bell system. Tell them that the first team to ring the bell gets the opportunity to answer. Display fractional division questions on the board, such as, What is 3 5 ÷ 1 2  ?, Calculate 2 3 ÷ 4, etc. Start the timer for each question (e.g., 30 seconds). Have teams know they can only ring the bell once they’ve heard the question. Tell them the rule that if a team answers correctly, they earn points. Incorrect answers may result in point deductions. Keep track of each team’s score on the board. Award 1 point for each correct answer and deduct 0.25 for each incorrect one. Include double-point questions such as, solve for x: x 5 = 2 3 ÷ 1 6 . Repeat the activity with enough questions. Then, award recognition to the winning team. Discuss their answers in the class and encourage them to write the solutions in their notebooks.

Multi-step Problems on Operations on Fractions

Learning Outcomes

Students will be able to solve multi-step word problems on operations on fractions.

Teaching Aids

Situation cards; Question slips

Activity

Instruct students to form groups for the activity. Distribute the story cards to each group.

Imagine Maths Page 12

Sophia loves to stitch and is planning to make scrunchies for a school fundraiser. She needs to calculate the total length of cloth required. She needs 3 4 m of red cloth and 2 5 m of blue cloth for each batch of 10 scrunchies. She plans to make 5 batches for the fundraiser and wants to distribute the scrunchies equally among 4 schools.

Prepare question slips such as the ones given below and distribute them among the students.

How many metres of cloth does Sophia need for the entire fundraiser?

How many batches of scrunchies will each school receive?

What is the total length of cloth needed for the 5 batches?

Instruct students to solve these word problems in their notebooks. Discuss their approach in the class. Give them more questions for practice.

Answers

1. Adding and Subtracting Fractions

Think and Tell

No, the associative and commutative properties do not hold true for the subtraction of fractions.

Do It Together

Add: LCM of 2, 5 and 4 is 20.

7 + 10 20 + 4 20 + 5 20 = 7 + 19 20 = 7  19 20

Subtract: LCM of 8 and 7 is 56

329 56 –136 56 = 193 56 = 3  25 56

2. Multi-step Problems on Adding and Subtracting Fractions

Do It Together

Total juice used to make fruit punch and flavoured jelly

= 11  1 12 cups

Juice left for drinking = 17 − 11  1 12 = 5  11 12 cups

3. Multiplication of a Fraction by a Whole Number

Do It Together

Total money spent = ₹5000 ×

Thus, Rajeev spent ₹1875.

= 15000

= ₹1875

4. Multiplication of a Fraction by a Fraction

Do It Together

5. Division of a Fraction by a Whole Number

Do It Together

6. Division of a Whole Number by a Fraction

Do It Together

7. Division of a Fraction by Another Fraction

It Together

8. Multi-step Problems on Operations on Fractions

Operations on Decimals 2

Learning Outcomes

Students will be able to:

add and subtract decimals. solve multi-step word problems on adding and subtracting decimals. multiply and divide decimals. solve multi-step word problems on operations on decimals.

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

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 represent decimal numbers in a place value chart and how to multiply and divide by 10, 100 and 1000.

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

Vocabulary

decimals: numbers that occur between whole numbers

decimal point: the dot present between the whole number part and the fractional part of a number

Teaching Aids

Problem cards; Bingo cards; Shopping list with 7 items; Word problem cards; Grocery list of 3 items; Flash cards with word problems

Chapter: Operations on Decimals

Learning Outcomes

Students will be able to add and subtract decimals.

Teaching Aids

Problem cards; Bingo cards

Activity

Make problem cards (example: 156.154 + 51.264, 458.564 – 147.117) and bingo cards (showing answers to the corresponding problem cards).

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

Put the problem cards in a box and shuffle them. Then, pick a card and read out the problem. Instruct the students to solve it using the column method. Ask them to check whether the answer is on the bingo card. If it is, ask them to cross it out. The group that crosses out a row or a column first wins. Ask the students to write the answers in their notebooks.

Extension Idea

Ask: Nishita covered a distance of 5.32 km by bicycle, 3.645 km by bus, and walked 1.2 km. How much total distance did she cover?

Say: Total distance covered = 5.32 + 3.645 + 1.2 = 10.165 km. Multi-step

Learning Outcomes

Students will be able to solve multi-step word problems on adding and subtracting decimals.

Teaching Aids

Shopping list with 7 items; Word problem cards

Activity

Create a shopping list of various items and write their prices. Divide the class into groups. Distribute the shopping lists and assign a budget of $30 to each group.

Create word problem cards like ‘How much money will you have left if you purchase a toy, book, and a pen? ’

Ask 1 student from each group to come up and pick 2 word problem cards.

Instruct the students to solve the problems based on the shopping list and find the answer.

Ask the students to write the answers in their notebooks. Discuss the results within the class.

Ask questions like: Which operations did you perform first?

Extension Idea

ItemCost

Pen$1.55

Book$3.75

Toy $6.23

Paper clips$0.45

Markers$2.25

Glue sticks$1.5

Calculator$11.68

Instruct: Create a multi-step word problem based on depositing and withdrawing money in/from a bank account in decimals.

Say: Sample word problem: ‘Ravi deposited ₹375.5, ₹600.75, and ₹1237.5 in his bank account on three different days. If he withdrew ₹1110.25, how much money is left in his bank account?’

Multiplying and Dividing Decimals

Learning Outcomes

Students will be able to multiply and divide decimals.

Teaching Aids

Grocery list of 3 items

Activity

Divide the class into groups of 4 and distribute a grocery list of 3 items to each group.

Imagine Maths Page 23

Ask one student to act as the shopkeeper and the other 3 students as customers.

Instruct one student to find the cost of 5 kg of pulses, the second student to find the cost of 6 kg of rice and the third student to find the cost of 5 litres of milk. Encourage them to solve the problem using the column method of multiplication and long division. The shopkeeper should check the calculations, and each group should calculate the total amount required to purchase the given weight of the items.

Ask the students to write the answers in their notebooks.

Multi-step Word Problems on Operations on Decimals

Learning Outcomes

Students will be able to solve multi-step word problems on operations on decimals.

Imagine Maths Page 26

Teaching Aids

Flash cards with word problems

Activity

Instruct the students to work in groups. Distribute word problem cards among the groups. Ask them to read the problem, such as:

‘A person earns ₹1755.25 for working 7 hours in a day. How much money will he earn if he works 10 hours in a day? ’

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

Extension Idea

A person earns ₹1755.25 for working 7 hours in a day. How much money will he earn if he works 10 hours in a day?

What do we know?

What do we need to find?

Solve to find the answer.

Ask: Create a multi-step word problem including all four operations on decimal numbers.

Say: Sample word problem: ‘Naman donated ₹1552.5 and ₹3624.75 to two charities. Sunaina donated half of the total amount that Naman donated, and Sunil donated 3 times the difference of the amounts that Naman donated. What amount did each of them donate? ’

Answers

1. Adding and Subtracting Decimals

Do It Together

Sum = 367.25 + 125.2 + 512 = 1004.45 … (1)

Difference = 563.24 − 321 = 242.24 … (2)

Adding (1) and (2) gives 1004.45 + 242.24 = 1246.69

HTO.th

367.25

125.20

+512.00

1004.45

HTO.th

563.24

–321.00

242.24

2. Multi-step Word Problems on Adding and Subtracting Decimals

Do It Together

Total cups of sugar = 18 2 5 cups = 18.4 cups

Sugar used = 2.25 + 6 3 4 + 1.45 = 2.25 + 6.75 + 1.45 = 10.45 cups

Sugar left = 18.4 – 10.45 = 7.95 cups

3. Multiplying and Dividing Decimals

Think and Tell

It is not necessary to get a natural number when we divide 2 decimal numbers.

Do It Together

Area of rectangle = 143.75 cm2; Breadth of rectangle = 11.5 cm

Length =   Area Breadth = 143.75 11.5 = 12.5

Hence, the length of the rectangle is 12.5 cm.

4. Multi-step Word Problems on Operations on Decimals

Do It Together

Price of 12 adult tickets at ₹125.5 per ticket = 12 × 125.5 = ₹1506

Price of 5 child tickets at ₹75.5 per ticket = 5 × 75.5 = ₹377.5

Difference in the cost = ₹1506 − ₹377.5 = ₹1128.5

Integers 3

Learning Outcomes

Students will be able to:

add two or more integers.

subtract an integer from another.

multiply 2 or more integers.

divide an integer by another integer.

Alignment to NCF

C-1.3: Learns about the inclusion of zero and negative quantities as numbers, and the arithmetic operations on them, as given by Brahmagupta

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 compare and order integers using a number line. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

integer: a positive or negative whole number

Teaching Aids

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

Chapter: Integers

Addition of Integers

Learning Outcomes

Students will be able to add two or more integers.

Teaching Aids

Imagine Maths Page 31

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

Activity

Begin the class by giving 1–2 examples where negative numbers are seen. Ask questions like: How will you write the number if the lift goes below the ground floor? Introduce integer tiles to the students. Explain that the red tiles represent negative integers, and the yellow tiles represent positive integers. Bring out the fact that a pair of red and yellow tiles represents 0.

Instruct the students to work in pairs. Distribute integer tiles among the pairs.

Instruct the students to add 5 and −6 using the integer tiles. Ask them to place integer tiles that represent 5. Ask them to now place integer tiles that represent −6 below the previous set. Instruct them to remove the pairs of integer tiles that make zero. Explain that the remaining number of integer tiles (−1) would be the answer. Instruct the students to add the integers −8 and −3, 7 and −1 using the integer tiles. Ask the students to note down the results in their notebooks using the addition sentence and discuss among their groups what they notice about the answers. Discuss the rules for the addition of integers with the students.

Ask questions like: What is the sign of the sum of 2 negative integers? Give them practice by running a quiz around adding integers.

Subtraction of Integers

Learning Outcomes

Students will be able to subtract an integer from another.

Teaching Aids

Imagine Maths Page 33

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

Activity

Begin the class by asking the students the difference between the temperatures in 2 cities of −14 and −13. Discuss the answers.

Instruct the students to work in pairs. Distribute integer tiles among the pairs.

Instruct the students to subtract 5 from 2. Ask them to use tiles to show 2. Ask questions like: Can you take away –5 or 5 negative tiles from the 2 positive tiles?

Tell the students that two positive tiles can be represented as 2 positive tiles and 3 neutral tiles. Let the students figure out how to subtract and find the answer. (Help them if needed).

Instruct the students to now subtract −4 from +3 using the integer tiles. Ask the students to note down the result in their notebooks using the subtraction sentence. Discuss the rules for subtraction of integers with the students. Ask questions like: What is the sign of the difference of 2 negative integers? One negative and one positive integer? Give them practice by running a quiz around subtracting integers.

Extension Idea

Instruct: Give two numbers which give −25 as a result after subtraction.

Say: Let the first number be 15 and the second number be a. 15 − a = −25 a = 15 + 25 = 40. Hence, when 40 is subtracted from 15, it gives −25 as a result.

Multiplication of Integers

Learning Outcomes

Students will be able to multiply 2 or more integers.

Teaching Aids

Imagine Maths Page 35

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

Activity

Discuss how in integers, multiplication is repeated addition or subtraction with two or more numbers. The first number tells us how many groups we are adding or removing. The second number tells us how many are in each group and whether they are positive (+) or negative (−).

Instruct the students to work in pairs. Distribute integer tiles among the pairs.

Instruct the students that they need to multiply 3 and −4 using the tiles. Once the students show 3 groups of −4; ask them to note down the result in their notebooks.

Ask the pairs to now multiply −3 and −4 using the tiles. Discuss that since they need to multiply −3 by −4, they need to remove 3 groups of −4, but to take away 3 groups of 4 negative tiles, they first need to place zero tiles. Discuss that on taking away 4 negative tiles from each of the 3 groups, they get 12 positive tiles. Ask them to write the result in their notebooks using the division sentence.

Discuss the rules for multiplication of integers with the students and run a quiz around multiplication of integers. 12

Extension Idea

Ask: What will be the sign of the final product if 90 negative integers are multiplied?

Say: 90 is an even number. The final product when 90 integers are multiplied will be positive.

Division of Integers

Learning Outcomes

Students will be able to divide an integer by another integer.

Teaching Aids

Imagine Maths Page 39

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

Activity

Instruct the students to work in pairs. Distribute integer tiles among the pairs.

Instruct the pairs to show −15 ÷ 5 using the tiles by first showing −15 using red tiles (negative tiles) and then find the number of groups by dividing these tiles into 5 equal groups with 3 tiles in each group. We get −3 in each group. Ask them to note down the result in their notebooks.

Ask the students to now show, −15 ÷ −5 using the tiles. Explain that they can do so by first placing 15 neutral tiles in 5 equal groups and then removing as many tiles as the divisor from each group. Ask them to write the result in their notebooks using the division sentence.

Discuss the rules for the division of integers with the students. Ask questions like: What is the sign of the division of 2 negative integers? One negative and one positive integer? Once the students are thorough with the concept, run a quiz around the division of integers.

Extension Idea

Instruct: Find −2 + 3 × 15 ÷ (−5) using DMAS.

Say: −2 + 3 × 15 ÷ (−5) = −2 + 3 × (−3) = −2 − 9 = −11

Answers

1. Addition of Integers

Think and Tell

We have not put a (+) sign before 160 because 160 and +160 mean the same.

Do It Together

Temperature of water in the bowl = 85°C

Temperature drops in 20 minutes = 33°C

Temperature drops in the next 10 minutes = –13°C

Total temperature drop = (–33) + (–13) = −46°C

Temperature after 30 minutes = 85°C + (−46°C) = 39°C

2. Subtraction of Integers

Do It Together

Height of Mt. Everest above sea level: 29,029 feet

Depth of Assal Lake below sea level = 510 feet

Difference in depth = 29,029 – (–510) = 29,539 feet

3. Multiplication of Integers

Think and Tell

99 is an odd number. The final product when 99 integers are multiplied will be negative.

Do It Together

Money lost by the store every day = ₹125

Money lost in 6 days = (–125) × 6 = −₹750

4. Properties of Multiplication of Integers

Think and Tell

The concept of multiplicative inverse does not apply to zero as any number divided by 0 is not defined.

Do It Together

1. (–25) × (–8) = (−8) × (−25)

2. 52 × 1 = 52

3. 23 × (13 + 14) = (23 × 13) + (23 × 14)

4. (17 × 5) × (−9) = 17 × (5 × (–9))

5. Division of Integers

Do It Together

Number of quarters played = 4

Team A’s total score for the match = −48

Average score per quarter = (–48) ÷ 4 = −12

6. Properties of Division of Integers

Think and Tell

For the commutative property to hold true under division, a ÷ b = b ÷ a

This is only possible when a = b and a, b 0.

Do It Together

1. (–125) ÷ (–25) = 5

2. 0 ÷ 45 = 0

3. 23 ÷ (−1) = –23

4. 136 ÷ 136 = 1

Measures of Central Tendency 4

Learning Outcomes

Students will be able to:

find the arithmetic mean or the mean of a given set of data.

find the mode of a given set of data.

find the median of a given set of data. choose the right measure of central tendency in solving word problems.

Alignment to NCF

C-5.1: Collects, organises, and interprets the data using measures of central tendencies such as average/ mean, mode, and median

Let’s Recall

Recap to check if students know how to collect, organise and interpret data. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

observation: a piece of information we gather during a study or experiment summation: addition of a sequence of numbers

frequency: the number of times a value occurs in a set of data frequency distribution: description of frequency for each possible value of a variable central tendency: a single value that represents the centre of the data set

Teaching Aids

Marbles; Paper cups; Cards (with data sets, e.g., 3, 5, 7, 9, 11); Number cards (1–10); Building blocks (different colours); Word problem cards; Table cards

Chapter: Measures of Central Tendency

Learning Outcomes

Students will be able to find the arithmetic mean or the mean of a given set of data.

Teaching Aids

Marbles; Paper cups; Cards (with data sets, e.g., 3, 5, 7, 9, 11)

Activity

Begin by explaining what the arithmetic mean is and how to calculate it.

Divide the class into groups of 4.

Imagine Maths Page 46

Distribute the marbles and paper cups to each group. Also, distribute a set of cards to each group with data sets written on each card (e.g., 3, 5, 7, 9, 11).

Instruct the students to put the marbles according to the data set in different paper cups. Ask the students the total number of marbles used.

Ask questions like: Do all the cups have an equal number of marbles? How can we put an equal number of marbles in each cup?

Instruct the students to divide the total number of marbles by the number of cups used. Write the result in their notebooks. Explain to them how the resulting number will give them the average number of marbles per cup, which represents the arithmetic mean.

If time permits, provide additional data sets to the groups, and ask them to find the arithmetic mean.

Extension Idea

Ask: The mean of 4, 22, 12, 7, x and 13 is 11. What is the value of x?

Say: Mean = 4 + 22 + 12 + 7 + x + 13 6 = 58 + x 6  . 11 = 58 + x 6 . So, x = 66 − 58 = 8.

Mode

Learning Outcomes

Students will be able to find the mode of a given set of data.

Teaching Aids

Number cards (1–10)

Activity

Begin the class by explaining the concept of mode and its use.

Imagine Maths Page 51

Lead the students to the playground and mark spots on the ground from 1 to 10, spaced far apart. Create number cards from 1 to 10 with repeating numbers (ensuring there are more cards than the total number of students). Place these cards in a box and have the students pick a card each. Then, instruct them to stand on the spot matching the number on their card.

Explain to the students that the spot with the most students is the mode.

Collect the cards and return them to the box. Shuffle the cards and repeat the activity a few more times. Have students record the results in their notebooks each time.

Engage the students in a discussion about their findings. Ask them to articulate why they think a particular number card appears more frequently and how this relates to finding the mode.

Extension Idea

Ask: The runs scored by a cricketer are: 120, 24, 48, 64, x, 29, 71, 3, 64, 29. If the mode is 64, then find the value of x.

Say: Since 29 and 64 both occur twice and the mode is 64, then x must be equal to 64.

Median Imagine Maths Page 55

Learning Outcomes

Students will be able to find the median of a given set of data.

Teaching Aids

Building blocks (different colours)

Activity

Begin the class by explaining the concept of a median and its use.

Instruct the students to form groups. Provide each group with a set of building blocks in different colours. Some colours may have more blocks or repeats than other colours.

Instruct the students to use the building blocks to build towers. Ask the students to note down the number of blocks used for each colour in their notebooks. Ask them to arrange the colours in ascending order based on the number of blocks they used for each. For example, if they used 1 blue, 2 orange, 3 yellow, 4 green and 5 red blocks, order them as: blue, orange, orange, yellow, yellow, yellow, green, green, green, green, red, red, red, red, red.

Ask the students to count the total number of building blocks they used to make the tower. Guide each group to find the colour that represents the middle value of the data set. Explain how this colour represents the median of the data set.

Ask questions like: Did your group use an odd or even total number of blocks? Why do you think we have different ways to find the median for odd and even counts? How did using coloured blocks help you understand the concept of the median?

Extension Idea

Ask: The numbers 17, p, 24, p + 7, 35, 36, 46 are written in ascending order. If the median of the data is 25, what is the value of p?

Say: There are 7 observations. So, the median is the 7 + 1 2  th = 4th observation, i.e., p + 7. Hence, p + 7 = 25, and p = 18.

Choosing the Right Measure of Central Tendency

Learning Outcomes

Students will be able to choose the right measure of central tendency in solving word problems.

Teaching Aids

Word problem cards; Table cards

Activity

Write different types of word problems on separate cards. Each problem should involve a different context and data set.

Divide the class into groups. Distribute a separate word problem card to each group.

An example of a word problem card could be: A supervisor is examining the heights of doors manufactured in the factory. The heights of the doors (in centimetres) are as follows: 215, 230, 190, 240, 200, 225, 180, 210. Which measure of central tendency should the supervisor use, and which measure should he avoid using?

Distribute the table cards to the students.

Observations:

Observations in ascending order: _______________________________

Total number of observations: __________

Instruct the students to read the problem carefully and determine which measure of central tendency will be the most appropriate.

Ask questions like: Which measure of central tendency is most affected by extreme observations at both ends of a data set?

Finally, reveal all the correct answers and discuss why a particular measure of central tendency was appropriate for the given problem.

Answers

1. Mean of Ungrouped Data

Think and Tell

No, the mean of a dataset cannot be more than the highest value in the data set as it represents the average value of the data set.

Do It Together

Step 1

Find the first five prime numbers.

2, 3, 5, 7, 11

Step 3

Find the mean.

Mean = 28 5 = 5.6

Step 2

Find their sum.

2 + 3 + 5 + 7 + 11 = 28

4. Mode of Grouped Data

Do It Together

Can the missing number be 5? No, because then the mode(s) will be 8 and 6.

Can the missing number be 6? No, because then the mode(s) will be 6.

Can the missing number be 7? No, because then the mode(s) will be 8 and 6.

Can the missing number be 8? Yes, because then the mode will be 8.

5. Median

Think and Tell

Thus, the mean of the first five prime numbers is 5.6.

2. Mean of Grouped Data

Do It Together

Step 1

Let us find f and ( f × x).

Step 2

Mean = ( f × x) f = 170 30 = 5.67

Thus, the arithmetic mean or the average marks of the students is 5.67.

3. Mode of Ungrouped Data

Do It Together

Step 1

Arrange the observations in ascending order.

₹120, ₹120, ₹120, ₹130, ₹130, ₹130, ₹135, ₹140, ₹200, ₹250

Step 2

Find the observation that occurs most frequently.

The observation(s) ₹120 and ₹130 occur(s) three times each.

Thus, the mode(s) of the data is/are ₹120 and ₹130

No, it is not necessary that the median will always be present in the data because if the number of observations is even, then the median will be the average of the middle two values, which is not necessarily the value present in the data.

Do It Together

Step 1

Arrange the heights (in cm) in ascending order.

148, 154, 160, 162, 164, 169, 171

Step 2

Find the total number of observations.

The total number of observations = 7, which is an odd (odd/even) number.

Step 3

Note the middle observation(s) as 4th using the formulae. The median height is 162 cm.

6. Choosing the Right Measure of Central Tendency

Do It Together

Using Mean and Median:

As the retailer is dealing with categorical (categorical/ numerical) data, he can’t (can/can’t) use the mean or median to plan his purchase for the upcoming week.

Using Mode:

As the mode is used to recognise the most (most/least) commonly occurring observation, the retailer can use the mode to plan his upcoming purchase.

The mode for the given data is medium Therefore, only the mode can be used to plan the purchase for the upcoming week.

Data Handling 5

Learning Outcomes

Students will be able to: organise data in a frequency distribution table using tally marks. draw single or double bar graphs for a given set of data. interpret single or double bar graphs for a given set of data.

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

C-8.2: Learns systematic counting and listing, systematic reasoning about counts and iterative patterns, and multiple data representations; learns to devise and follow algorithms, with an eye towards understanding correctness, effectiveness, and efficiency of algorithms

Let’s Recall

Recap to check if students know how to collect data through various methods and interpret it. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

tally mark: a vertical mark used to keep count bar graph: graph that shows information in the form of bars of different lengths

Teaching Aids

Chart paper; Sticky notes; Graph paper; Coloured paper; Glue stick; Chart paper with a double bar graph; Strips of paper

Chapter: Data Handling

Data Organisation

Learning Outcomes

Imagine Maths Page 65

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

Teaching Aids

Chart paper; Sticky notes

Activity

Explain that the purpose of the activity is to find out the most popular place among the students for a field trip.

Distribute sticky notes to the students and ask them to write down their favourite place to visit out of Science Museum, Historical Site, Art Gallery and Farm.

Prepare a blank frequency distribution table on the chart paper with columns for Favourite Place and Tally Marks.

Ask the students to work in groups.

Distribute one chart paper to each group and collect the sticky notes from the students.

Read out the names of the places written on the sticky notes one-by-one and ask the students to draw tally marks on the chart paper according to the data read out.

Count the tally marks for each place to determine the frequency (number of students who chose that place).

Identify the place with the highest frequency to determine the most popular choice among the students.

Extension Idea

Ask: In a survey of daily temperatures recorded over a week in a city, the highest temperature recorded was 32°C and the lowest temperature recorded was 18°C. What is the range of temperatures recorded during this week?

Say: The range of temperatures recorded during the week = 32°C − 18°C = 14°C.

Making Bar Graphs

Learning Outcomes

Students will be able to draw single or double bar graphs for a given set of data.

Teaching Aids

Graph paper; Coloured paper; Glue stick

Activity

Imagine Maths Page 68

Begin by explaining the importance of the monsoon season in India. Explain that the purpose of the activity is to create a visual representation of the average annual rainfall during the southwest and northeast monsoon seasons across different regions of India.

Draw a table on the board as shown.

Divide the class into groups and ask them to look at the table drawn.

Distribute graph paper, coloured paper and a glue stick to each group.

Instruct the students to draw a set of axes on their graph paper, with the x-axis representing the regions and the y-axis representing the rainfall in millimetres. Ask them to label the x-axis with the names of the regions and the y-axis with appropriate intervals to accommodate the rainfall data. Using the data provided in the table, ask the students to plot two bars for each region—one representing the average rainfall during the southwest monsoon and the other representing the average rainfall during the northeast monsoon by cutting and pasting the rectangular strips of coloured paper (remind the students that the widths of the rectangular strips should be the same).

Encourage students to scale their axes to represent the rainfall data accurately. Instruct the students that each square on the graph paper can represent a certain number of millimetres of rainfall. Ensure that students provide a title for their bar graph.

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

Extension Idea

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

Say: You can ask a lot of questions about the bar graph. One such question could be: Which two regions receive the highest average annual rainfall during the southwest monsoon season?

Interpreting Bar Graphs

Learning Outcomes

Students will be able to interpret single or double bar graphs for a given set of data.

Teaching Aids

Chart paper with a double bar graph; Strips of paper

Activity

Begin by introducing the topic of biomes and explaining what they are. Explain that temperature is one of the factors that distinguish different biomes (grasslands, deserts, rainforests etc).

Prepare the double bar graph on chart paper where each biome is represented by two bars: one for the average high temperature and one for the average low temperature.

Instruct students to work in groups. Distribute the chart paper with the double bar graph, to each group. Ask the students to examine the graph closely and note any patterns or differences they observe between the biomes.

Pose questions such as: Which biome has the highest average high temperature? Which biome has the lowest average low temperature? How does the temperature range vary between different biomes? Are there any biomes where the high and low temperatures are very close together?

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

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

Answers

1. Data Organisation

2. Making Bar Graphs

Do It Together

3. Interpreting Bar Graphs

Think and Tell

A bar graph is typically used to display categorical or qualitative data, where the categories are distinct and separate from one another. This type of data is often non-numeric or can be grouped into categories.

Think and Tell

In a standard bar graph, the width of the bars typically does not have any significance in terms of conveying data. However, the width of the bars should be consistent for a bar graph.

Do It Together

2. 75% of students prefer indoor games in Grade 6.

3. 55% of students prefer outdoor games in Grade 7.

4. 95% of students prefer outdoor games in Grade 8.

5. 70% of Grade 5 students prefer outdoor games.

Introduction to Probability 6

Learning Outcomes

Students will be able to:

find the chance of an event happening using words like sure, impossible, likely, unlikely or equally likely. find the theoretical probability of an event and write it as a fraction.

find the experimental probability of an event and write it as a fraction.

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 ratio of two quantities and write it in its simplest form. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

chance: the likelihood of an event taking place probability: a chance that an event will take place experiment: an action that produces clearly defined results outcome: each possible result of an experiment event: a collection of one or more outcomes

Teaching Aids

Sheet of paper; Pencil; Bag of coloured buttons: blue (12), red (15), green (6), yellow (5), white (10); Flash cards; Dice; Chart with columns; Piece of cardboard; Sketch pens; Push pins; Hook; Bowl

Chapter: Introduction to Probability

Chance

Learning Outcomes

Imagine Maths Page 90

Students will be able to find the chance of an event happening using words like sure, impossible, likely, unlikely or equally likely.

Teaching Aids

Bag of coloured buttons: blue (12), red (15), green (6), yellow (5), white (10); Flash cards

Activity

Instruct the students to form groups. Distribute flash cards with some real-life situations written on them such as The sun rises in the west; Carrying an umbrella to school on a rainy day. Discuss which words, such as sure, impossible, likely or unlikely, can be used to describe the event.

Distribute the teaching aids to each group. Draw the given table on the board. Instruct the students to copy the table into their notebooks. Ask them to count the number of blue buttons and write it in the table.

Ask: What is the chance of picking a blue button from the bag?

ColourNumberChance of event happening

Blue12

Red15

Green6

Yellow5

White10

Discuss the answer where the chance of picking a blue button depends on the total outcomes, which is the total number of buttons (48). The chance of picking a blue button is 12 out of 48, which is less than half. So, the chance of this event happening is ‘unlikely ’. Encourage students to complete the table using words like likely, unlikely or equally likely.

Extension Idea

Ask: What is the chance of a black button being picked from the bag?

Say: Since there are no black buttons in the bag, the chance of this event happening is impossible

Theoretical Probability

Learning Outcomes

Imagine Maths Page 92

Students will be able to find the theoretical probability of an event and write it as a fraction.

Teaching Aids

Bag of coloured buttons: blue (12), red (15), green (6), yellow (5), white (10)

Activity

Show the students a dice and ask them if they rolled the dice, what would they get. Explain that the total outcomes are 6, and the favourable outcomes in rolling a dice and getting a 6 is 1. Also, discuss how to write the theoretical probability    1 6    of the event as a fraction.

Instruct the students to form groups. Distribute the resources to each group.

Ask the students to count the buttons and find the theoretical probability of picking a button at random of each colour. Discuss one example with them and guide them to identify that the theoretical probability of picking a green button is 6 48

Ask questions like: What is the theoretical probability of getting a pink button? A blue button?

Experimental Probability

Learning Outcomes

Imagine Maths Page 96

Students will be able to find the experimental probability of an event and write it as a fraction.

Teaching Aids

Sheet of paper; Pencil; Piece of cardboard; Sketch pens; Push pins; Hook; Dice; Bowl

Activity

Ask the students if a dice is rolled 30 times, how often the number 5 might show up. Then, discuss how if the number 5 appears 10 times, the experimental probability of getting 5 is 10 30  .

Instruct the students to form groups. Distribute the teaching aids. Inform them that they will create a spinner for the school fair with 4 options.

Draw the table on the board. Instruct the students to copy the table into their notebooks.

Outcome Jackpot Win a turnGet a free juiceWin a prize Number of Times

Instruct the students to take turns to spin the spinner and then list the colour that the spinner stops at in the table using tally marks. Each group must spin it 25 times. Once the table is filled, ask them to write the answer to these questions in their notebooks: What is the experimental probability of spinning a jackpot? What is the experimental probability of winning a turn? Then, encourage the students to discuss their findings within their groups and compare their results with other groups.

Extension Idea

Ask: During a basketball shooting practice session, Sarah takes 30 free throws and records the number of successful shots she makes as 18. Based on the experimental probabilities, what percentage of the free throws did Sarah make successfully?

Say: Based on the experimental probabilities, Sarah made 60% of her free throws successfully. This means that 60% of her attempts resulted in successful shots.

Jackpot

Answers

1. Understanding Probability

Do It Together

1. Green balloon – Equally likely

2. White balloon – Unlikely

3. Green or orange balloon – Likely

4. Red balloon – Impossible

2. Theoretical Probability

Do It Together

The cards in the box are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

The number of cards that have prime numbers is 4.

Total number of cards = 10

Therefore, the probability of drawing a prime number card = 4 10 or 2 5

3. Experimental Probability

Think and Tell

Theoretical probability of getting a red marble = 1 4

Experimental probability of getting a red marble = 7 20

Do It Together

P(6) = 4 15 P(7) = 9 15 or 3 5

Probability of the spinner stopping at 6 = 4 15

Probability of the spinner stopping at 7 = 9 15 or 3 5

Simple Equations 7

Learning Outcomes

Students will be able to: form linear equations for given statements. solve equations by using the same operation on both sides of the equation. solve equations by using the transposing method. apply simple equations to solve word problems.

Alignment to NCF

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

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

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

Let’s Recall

Recap to check if students know how to frame algebraic expressions. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

linear: straight

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

Teaching Aids

Index cards or small slips of paper; Envelopes for each set of index cards; Balance scale or a makeshift scale (e.g., a ruler or a wooden beam balanced on a fulcrum); Unit cubes; Two bags with 2 unit cubes in each; One bag with 5 unit cubes; Equation cards with linear equations in one variable; Equation mazes; Crayons; Chart paper

Chapter: Simple Equations

Learning Outcomes

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

Teaching Aids

Index cards or small slips of paper; Envelopes for each set of index cards

Activity

Prepare a set of index cards with different statements written on them. These statements should involve situations that can be modelled using linear equations in one variable, such as: If you subtract 8 from four times a number, the result is 20. Place each set of index cards in separate envelopes.

Divide the class into small groups. Distribute one envelope of index cards to each group.

In each round, have students draw a statement from the envelope and read it aloud to their group. The group members should discuss and formulate the linear equation in one variable that represents the given statement. Finally, they should write the equation in their notebooks.

If you subtract 8 from four times a number, the result is 20.

After each round, have a class discussion. Ask each group to share their formulated equation and reasoning.

Extension Idea

Instruct: A shipping company charges a flat fee of ₹20 plus ₹5 for each package shipped. If the total cost for shipping is ₹45, write a linear equation to represent the total cost.

Say: Let x be the number of packages shipped. The equation formed will be 45 = 20 + 5x. Solving Equations: Same Operation on Both Sides Imagine Maths Page 105

Learning Outcomes

Students will be able to solve equations by using the same operation on both sides of the equation.

Teaching Aids

Balance scale or a makeshift scale (e.g., a ruler or a wooden beam balanced on a fulcrum); Unit cubes; Two bags with 2 unit cubes in each; One bag with 5 unit cubes; Equation cards with linear equations in one variable

Activity

Begin with a brief review of the concept of solving linear equations in one variable by maintaining equality on both sides of the equation. Divide the class into small groups.

Distribute 2 equation cards, unit cubes, 2 bags with 2 unit cubes in each, one bag with 5 unit cubes and a balance scale to each group.

Ask the groups to read the equation, and then represent it on their balance scale using the bags and unit cubes. Instruct the students to manipulate the weights on both sides of the scale to find the solution to the equation, which is the number of cubes inside each bag. Then ask them to write the equation in their notebooks and solve it by performing the same operations on both sides to maintain the balance.

Give them 2 more equations to solve without using the balance scale.

Discuss the answers.

Extension Idea

Ask: Sarah is saving money for a new purse. She currently has ₹80 and plans to save ₹15 each week. How many weeks will it take for her to save a total of ₹200?

Say: It will take Sarah 8 weeks to save a total of ₹200 for her new purse.

Solving Equations: Transposition

Learning Outcomes

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

Teaching Aids

Equation mazes; Crayons

Activity

Create or find equation mazes that include linear equations in one variable.

Imagine Maths Page 107

Begin the activity with a brief review of the transposition method for solving linear equations in one variable. Discuss 1–2 examples with the students.

Instruct the students to form pairs.

Distribute an equation maze and crayons to each pair.

Instruct the students to start at the beginning of the maze and solve each equation to find the correct solution by transposing from one side of the equation to the other. Ask them to shade the equation and its solution. The correct solution will guide them to the next part of the maze.

Once students reach the exit, have them review their answers.

Applications of Simple Equations

Learning Outcomes

Students will be able to apply simple equations to solve word problems.

Teaching Aids

Small slips of paper; Chart paper

Imagine Maths Page 110

Activity

Begin by discussing the concept of linear equations in one variable and their significance in solving real-world problems.

Share a captivating story with the students that involves solving a linear equation. For example: Once upon a time in Mathland, there was a magical garden where flowers bloomed in a peculiar pattern. The number of flowers in the garden increased by 3 each day, and there were already 8 flowers in the garden. As the magical gardener, your task is to find out how many days it will take for the garden to have a total of 23 flowers.

After telling the story, ask the students to work in small groups. Provide each group with slips of paper and ask them to form a linear equation that shows the problem in the story. In this case, the equation would be 3x + 8 = 23, where x represents the number of days.

Collect the slips from each group and paste them on the chart paper. Discuss each group’s equation, emphasising the different ways students approached the problem. Make sure to highlight the correct equation. Ask questions like: What strategies did you use to form the linear equation? How did you approach solving the equation?

Extension Idea

Instruct: Create a story problem that involves solving the equation 300 + 25x = 650.

Say: You can create multiple story problems for the given equation like: Emily rented a car service that charges a flat fee of ₹300 per day plus an additional ₹25 per mile driven. If Emily paid ₹650 for a one-day rental, find the number of miles she drove

Answers

1. Forming Linear Equations

Do It Together

Let the runs scored by Aakash be x

Sahil scored twice as many runs as Aakash = 2x

Together, their runs fell four short of 100 = 100 – 4 = 96

This situation can be represented in the form of an equation as: x + 2x = 96.

On simplifying both the sides of the equation, we get, 3x = 96

2. Solving Equations: Same Operation on Both Sides

Do It Together

Let the money with Jayant be ₹x

Money with Sahil = Twice as much as Jayant = ₹2x

So, the required equation is (x + 2x) = 240.

On simplifying the equation, we get, 3x = 240 3x

3 = 240 3

So, x = 80

Thus, Jayant has ₹80.

3. Solving Equations: Transposition

Do It Together

On simplifying both sides of the equation, we get,

1.1x = 8.4

On transposing 1.1 from LHS to RHS, the operation changes from multiplication to division.

x = 8.4 1.1

On simplifying the RHS, we get, x = 7.64

4. Applications of Simple Equations

Do It Together

On simplifying the LHS, we get, 2(2x + 5) = 142 OR 4x + 10 = 142

On transposing 10 from LHS to RHS, the operation changes from addition to subtraction, 4x = 142 – 10

On simplifying, we get, x = 132 4  ; x = 33

The breadth of the rectangular plot is 33 metres. The length of the rectangular plot is 38 metres.

Angle Pairs and Para elLines 8

Learning Outcomes

Students will be able to: identify complementary and supplementary pairs of angles, adjacent angles, linear pair of angles and vertically opposite angles. apply their understanding of angles formed when a pair of parallel lines are cut by a transversal. construct a pair of parallel lines using a compass and ruler.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to measure angles using a protractor. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

angle: space between two rays starting from the same point parallel lines: lines that move in the same direction and are the same distance apart at every point

Teaching Aids

Cutouts of angles that form complementary (30°, 60°, 25°, 65°), supplementary (135°, 45°, 110°, 70°) and vertically opposite pairs of angles (100°, 100°, 80°, 80°); Chart paper; Blank sheets; Ruler; Protractor; Compass; Coloured pencils; Cutouts of cars

Chapter: Angle Pairs and Parallel Lines

Pairs of Angles

Learning Outcomes

Imagine Maths Page 117

Students will be able to identify complementary and supplementary pairs of angles, adjacent angles, linear pair of angles and vertically opposite angles.

Teaching Aids

Cutouts of angles that form complementary (30°, 60°, 25°, 65°), supplementary (135°, 45°, 110°, 70°) and vertically opposite pairs of angles (100°, 100°, 80°, 80°); Chart paper

Activity

Begin the class by discussing the different types of angles.

Instruct the students to work in groups. Distribute cutouts of angles and chart paper among the groups.

Ask the groups to paste 2 angles to form complementary angles and supplementary angles and 4 angles to show vertically opposite angles on the chart paper. Ask them to label the vertices and angles. Finally, they will list the adjacent and linear pairs of angles that they see in the figures.

Ask questions like: If the measure of an angle is 15 degrees, what would be the measure of its complement?

Give the students more questions on finding the measure of the missing angle when one angle is given based on complementary and supplementary angles.

Extension Idea

Ask: If two angles measure 45° and 135°, are they supplementary? Give a reason for your answer.

Say: Yes, if two angles measure 45° and 135°, they are supplementary as they will add up to 180°.

Parallel Lines and Transversals

Learning Outcomes

Imagine Maths Page 123

Students will be able to apply their understanding of angles formed when a pair of parallel lines are cut by a transversal.

Teaching Aids

Blank sheets; Ruler; Protractor

Activity

Begin the class by discussing where the students see parallel lines in their real life. Discuss that the lines made for effective parking in parking lots are parallel and the line joining them is their transversal.

Instruct the students to work in groups. Distribute blank sheets, rulers, and protractors among the groups.

Ask the groups to draw parallel lines using the edges of the ruler. Ask them to then draw a line intersecting both the parallel lines at any angle other than a right angle. Instruct the groups to measure all the angles formed by the transversal on the parallel lines using the protractor and write the measures.

Instruct them to have a discussion around what they notice about the angles so formed. Discuss the angles that are of the same measure when a pair of parallel lines are intersected by a transversal.

Give them a problem on parallel lines and transversals on the board where 1–2 angles measures are given and students need to find the remaining angles.

Constructing Parallel Lines

Learning Outcomes

Students will be able to construct a pair of parallel lines using a compass and ruler.

Teaching Aids

Chart paper; Compass; Ruler; Coloured pencils; Cutouts of cars

Activity

Recall the lesson where students discussed real-life examples where they see parallel lines and transversals. Show them pictures if needed.

Instruct the students to work in groups. Distribute chart paper, compasses, rulers, coloured pencils and cutouts of cars among the groups.

Imagine Maths Page 127

Ask the students to create a parallel parking space on the chart paper by constructing parallel lines for at least 6 cars, making sure that the distance between the lines is enough to fit each car cutout. Instruct them to open page 127 of their Imagine Mathematics books and look at the steps of construction of parallel lines. Ask them to use the steps and create parallel lines on the chart paper given. Ensure that each student gets to construct at least one line parallel to another line. Ask them to use coloured pencils to make the parking space colourful. Ask them to stick the car cutouts on the parking spaces created.

Encourage groups to show their creativity to the class.

Extension Idea

Ask: If the angle formed by the lines in the parking lot is 90 degrees, what would you notice about the cars parked?

Say: If the altitudes of two right-angled triangles are parallel to each other and the bases form a transversal to the parallel lines, then the corresponding angles formed by the bases and the transversal will be equal.

Answers

1. Complementary and Supplementary Angles

Think and Tell

AOC and COB together making a straight angle = 180°, hence the angles are supplementary.

Do It Together

1. A = 60° B = 30°

Complementary angles

2. C = 45° D = 135°

Supplementary angles

2. Adjacent Angles

Think and Tell

Yes, one of the adjacent pairs of angles are XAC and CAB

Do It Together

Answers may vary. Sample answers.

2. Adjacent angles of ACB are ACF and ACE

3. Adjacent angles of ECD are FCE and DCB.

4. Adjacent angles of FCD are DCB and ACF.

3. Linear Pair of Angles

Do It Together

1. 3x = 180°

So, x = 180° 3 = 60°

2. (3x + 7)° + (2x + 18)° = 180°

5x + 25° = 180°

So, x = 155° 5 = 31°

4. Vertically Opposite Angles

Do It Together

1. 3x – 15° = 75°

3x = 75° + 15° = 90°

3x = 90° x = 30°

2. (2x – 10)° = 60°

2x = 70°

x = 35°

5. Parallel Lines and Transversals

Think and Tell

Yes, if line a is parallel to line b and line b is parallel to line c, then line a will be parallel to line c.

Do It Together

In the figure, PQR = XRY = 60° (corresponding angles)

PRQ = 55°

6. Constructing Parallel Lines

Do It Together

2.5 cm

P t BA

R Q

Triangles and Their Properties 9

Learning Outcomes

Students will be able to:

identify and draw the median and altitude in a triangle. identify and apply the interior and exterior sum property of a triangle. identify and apply the side sum property of a triangle. apply and prove Pythagoras’ theorem.

Alignment to NCF

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

Let’s Recall

Recap to check if students know about how to classify triangles on the basis of their sides and angles. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

median: a line segment that connects the vertex of a triangle to the midpoint of the opposite side altitude: a perpendicular line segment drawn from the vertex of a triangle to the opposite side centroid: the point where the medians of a triangle intersect

orthocentre: the point where the altitudes of a triangle intersect

Teaching Aids

Paper cutouts of different types of triangles; Notebooks; Sheets of paper; Coloured straws; Squared paper; Colours

Chapter: Triangles and Their Properties

Learning Outcomes

Students will be able to identify and draw the median and altitude in a triangle.

Teaching Aids

Paper cutouts of different types of triangles

Activity

Instruct students to form groups. Distribute cutouts of equilateral, isosceles, scalene, acute, right and obtuse triangles, giving 1 of each type to each group. Instruct the students to fold the triangle such that the crease joins the vertex to the centre point of the opposite side. Discuss what a median is. Then, instruct them to trace the triangles in their notebooks and draw perpendiculars from each vertex of the triangle to its opposite side. Now, discuss what the altitude is.

Then, ask them about the similarities and differences between the creases obtained on the two sets of triangles and in each triangle separately. Discuss their observations and conclusions about each triangle.

Ask questions like: Did you get all the altitudes inside the triangles? Give a reason for your answer. In which triangle is the median and altitude the same?

Extension Idea

Ask: Can a triangle have all three medians of equal length? Why?

Say: No, a triangle cannot have all three medians of equal length unless it is an equilateral triangle. The medians of an equilateral triangle are equal in length, but in other triangles, they are generally of different lengths.

Properties of Angles of a Triangle

Learning Outcomes

Imagine Maths Page 138

Students will be able to identify and apply the interior and exterior sum property of a triangle.

Teaching Aids

Paper cutouts of different types of triangles; Notebooks; Sheet of paper

Activity

Instruct the students to form groups.

Distribute the cutouts, giving 1 of each type to each student of the group. Instruct the students to mark the angles in the triangles, cut out the angles and then put them together to form one angle. Then, encourage them to discuss what they notice about the new angle formed in their groups. Ask them about their observations

and discuss the conclusions they derive. Let them verify their conclusions using the measures of the angles in their notebooks.

Ask questions like: What do you notice about the new angle formed?

Then, instruct them to trace each triangle on a sheet of paper and extend one side of the triangle to form an angle outside it. Next, instruct them to mark all the interior and exterior angles of the triangle, cut out the two interior opposite angles and paste them on the exterior angle such that they completely overlap with the exterior angle. Discuss the exterior angle sum property. Then, ask them about their observations and discuss the conclusions they derived. Let them verify their conclusions using the measures of the angles in their notebooks.

Finally, give them questions on finding the missing angle of a triangle when the measures of 2 angles are given.

Extension Idea

Ask: Can a triangle have an exterior angle measuring 90° at two different vertices? Explain.

Say: No, a triangle cannot have an exterior angle measuring 90° at two different vertices because the sum of all exterior angles of any polygon is 360°. If the measure of an exterior angle is 90° at two different vertices, then the third exterior angle will be 180°, making the interior angle 0°, which is not possible.

Properties of the Sides of a Triangle

Learning Outcomes

Students will be able to identify and apply the side sum property of a triangle.

Teaching Aids

Coloured straws

Activity

Instruct the students to form groups.

Imagine Maths Page 142

Distribute three straws of different colours to each group. Instruct them to cut the straws into the measures written on the board and try joining them together to form a triangle. Ask them if they can form a pattern between the measures of two sides of the triangle and its third side. Then, ask them about their observations and discuss the conclusions they derived. Discuss the triangle inequality property. Let them verify their conclusions using the measures of the sides in their notebooks.

Repeat this activity using different lengths of straws.

Give the students questions in which they need to find the possible length of the third side of a triangle.

The Pythagoras Theorem

Learning Outcomes

Students will be able to apply and prove Pythagoras’ theorem.

Teaching Aids

Squared paper; Colours

Imagine Maths Page 146

Activity

Instruct the students to form groups. Write the measures of three sides of a right triangle on the board. Distribute squared paper to each group.

Instruct the students to cut 3 squares with sides that are the same as the lengths of the right triangle. Ask the students to place the two smaller squares inside the largest square by cutting and pasting without leaving any gaps.

Point out that the two squares cover exactly the same area as the largest square.

Ask them to try deducing the relationship between the lengths of the sides of the triangle and its area and discuss the Pythagoras theorem.

Ask questions like: What is the rule of the lengths of the sides of a right-angled triangle?

Give the students questions in which they need to find the length of the third side of a right-angled triangle when the lengths of 2 sides are given.

Extension Idea

Ask: A pilot is flying at an altitude of 6000 feet. If the pilot needs to descend to land on the runway, and the runway is 8000 feet away horizontally, what is the length of the descent path?

Say: (Length of the descent path)2 = 60002 + 80002 = 3,60,00,000 + 6,40,00,000 = 10,00,00,000 = 10,000 × 10,000. So, the length of the descent path = 10,000 feet.

Answers

1. Medians and Altitudes

Think and Tell

We can show the medians of a triangle by drawing a line from the vertex to the centre of the side.

Do It Together

SR = QR (Since PR is the median to the side SQ and R is the midpoint of SQ.)

SQ = SR + RQ. So, SQ = 7 + 7 = 14 cm

PG:GR = 2 : 1 (Property of a centroid)

PR = PG + GR (G lies between P and R)

So, PR = 8 + (8 ÷ 2) = 8 + 4 = 12 cm

2. Angle Sum Property of a Triangle

Do It Together

Using the angle sum property of a triangle,

1 + 2 + 3 = 180°

(x + 5)° + (x + 10)° + (2x + 5)° = 180°

4x + 20° = 180°

So, x = (180° – 20°) ÷ 4 = 160° ÷ 4 = 40°

Using the value of x,

(x + 5)° = 45°

(x + 10)° = 50°

(2x + 5)° = 85°

The measures of the angles are 45°, 50° and 85°.

3. Exterior Angle Sum Property of a Triangle

Do It Together

Step 1

PQR + QPR = PRS (Using the exterior angle sum property)

Substituting the values of the respective angles, PQR + 80° = 140°

So, PQR = 140° − 80° = 60°

Step 2

The sum of all the angles of a triangle is 180°.

Therefore, PQR + QPR + PRQ = 180°

Substituting the values of the respective angles, 60° + 80° + PRQ = 180°

PRQ = 180° − (60° + 80°) = 180° − 140° = 40°

The angles of the triangle PQR are 60°, 80° and 40°.

4. Properties of the Sides of a Triangle Do It Together

Step 1

Using the triangle inequality property, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Side 1 + Side 2 > Side 3.

The third side will be smaller than the sum of the two sides.

Substituting the values of the sides of the triangle, finding the sum of the two sides,

Side 1 + Side 2 = 4 cm + 12 cm = 16 cm.

Step 2

We also know that the difference of the lengths of any two sides of a triangle is less than the length of the third side. Finding the difference of the two sides,

Side 1 – Side 2 = 12 cm − 4 cm = 8 cm

The length of the third side cannot be smaller than the difference of the two sides.

Therefore, the possible length of the third side of the triangle is between 8 cm and 16 cm.

5. The Pythagoras Theorem

Think and Tell

In a right-angled triangle ABC, if AB2 = AC2 + BC2, then the right angle is angle ACB.

Do It Together

Step 1

Draw the figure and label the vertices to show the rightangled triangle.

AB = 8 m, CD = 16 m and BD = 6 m

Draw AE || BD meeting CD at E. Then, ABDE is a rectangle. Using the properties of a rectangle, ED = 8 m and AE = 6 m

Step 2

In a right-angled triangle AEC, we have, AC2 = CE2 + AE2

Substituting the values, AC2 = 82 + 62 = 64 + 36 = 100 = 10 × 10. So, AC = 10 m

The distance between the tops of the two poles is 10 m.

Congruence 10

Learning Outcomes

Students will be able to:

find congruency in shapes and figures. use the SSS congruency criteria to solve problems based on it. use the SAS congruency criteria to solve problems based on it. use the ASA congruency criteria to solve problems based on it. use the RHS congruency criteria to solve problems based on it.

Alignment to NCF

C-3.5: Understands congruence and similarity as it applies to geometric shapes and identifies similar and congruent triangles

Let’s Recall

Recap to check if students know how to differentiate between the types of triangles based on the sides and angles.

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

Vocabulary

superposition: placing one thing on top of another line segments: a straight connection between two points or a portion of a line corresponding sides/angles: a pair of matching sides or angles that are in the same position in two different figures

diagonal: a straight line connecting two opposite vertices in a figure

Teaching Aids

Sheets of coloured paper; Pair of scissors; Glue stick; Cutouts of pairs of triangles with sides of equal lengths and some distractor triangles; Straws; Strips of paper measuring 5 cm and 8 cm; Protractor; Two sets of cards with congruent triangles drawn on them; Ruler; Cutouts of congruent right-angled triangles and some distractor triangles

Chapter: Congruence

Congruence in Shapes and Figures

Learning Outcomes

Students will be able to find congruency in shapes and figures.

Teaching Aids

Sheets of coloured paper; Pair of scissors

Activity

Imagine Maths Page 152

Discuss real-life examples where students can see congruency (for example, biscuit packets, the leaves of a book, a bundle of currency notes, etc.) and explain why the objects are congruent.

Inform students that they will be making congruent shapes using sheets of paper.

Instruct the students to form pairs.

Provide each pair with 2 sheets of coloured paper and a pair of scissors.

Instruct the pairs to form shapes congruent to each other. They must make at least 4 copies of each shape. Ask them to discuss and decide which shape each partner will make. Then, ask them to use different methods of making 4 copies of a shape that will be congruent. One way would be to fold the sheet multiple times and then cut out a shape, and another would be to draw shapes with sides of the same length. Once done, they will check their partner’s shapes to see if all of them are congruent to each other.

Discuss with the students that there are different ways to make congruent shapes and that congruent shapes are formed when the cutouts overlap completely.

Ask: Can you identify any objects in the classroom that are congruent to each other?

If time allows, ask the students to try cutting some more shapes for further practice on the concept of congruency.

SSS Congruence Criteria

Learning Outcomes

Students will be able to use the SSS congruency criteria to solve problems based on it.

Teaching Aids

Imagine Maths Page 157

Glue stick; Sheets of paper; Cutouts of pairs of triangles with sides of equal lengths and some distractor triangles; Straws

Activity

Begin by introducing the SSS congruence criteria.

Ask the students to consider two triangles ABC, where in one triangle, AB = 3 cm, BC = 4 cm and CA = 5 cm, and in the other triangle ABC, AB = 4 cm, BC = 5 cm and CA = 3 cm. Discuss how the corresponding three sides of the triangles are equal in length, and so they are congruent according to the Side-Side-Side criteria.

Following the SSS rule, make cutouts of triangles with pairs of triangles of equal side lengths and some distractors.

Instruct the students to work in pairs. Distribute a sheet of paper, triangle cutouts and a glue stick to each pair.

Instruct them to measure the length of the sides of the triangles, write the measures inside the triangle and then make pairs of triangles that overlap exactly or are congruent. They should paste the pairs of triangles on the sheet of paper and label them as ‘Congruent Triangles Using the SSS Rule’.

Distribute the straws to each pair. Direct each student in the pair to cut straws of 3 lengths (2 pieces of 4 cm, 7 cm and 5 cm each). Direct each student to construct a triangle using straws of differing lengths. Ask them to match the triangle they made with their partner’s.

Ask questions like: How do you know that the two triangles are congruent?

Extension Idea

Ask: A triangle has a side measuring 15 cm. If this triangle is congruent to two other triangles, one with a side measuring 10 cm and another with a side measuring 18 cm, what are the lengths of all the sides of the first triangle?

Say: Because all three triangles are congruent, the lengths of all the sides of all the triangles are 15 cm, 10 cm and 18 cm.

SAS Congruence Criteria

Learning Outcomes

Students will be able to use the SAS congruency criteria to solve problems based on it.

Teaching Aids

Strips of paper measuring 5 cm and 8 cm; Glue stick; Protractor; Sheets of paper Activity

Begin by introducing the SAS congruence criteria.

Imagine Maths Page 159

Ask the students to consider two triangles ABC, where in one triangle AB = 3 cm and CA = 5 cm, and in the other triangle ABC, AB = 4 cm and CA = 3 cm. The angle formed between these two sides measures 60°. Discuss that as the corresponding sides of both triangles, along with the angle between them, are equal, they satisfy the SAS criteria for congruence.

Instruct the students to work in pairs.

Distribute a glue stick, a protractor, a sheet of paper and strips of paper measuring 5 cm and 8 cm, ensuring that each student in the pair gets one of each strip.

Instruct each student to paste one strip on the sheet of paper and then paste the other strip at one end of the previous strip such that they form an angle of 60°. Let them use a protractor to measure the angle. Then, guide them to draw a line on the sheet to complete the triangle. Ask them to write the measures inside the triangle. They will then cut out their triangles and superimpose them on their partner’s triangle to check if they overlap completely. Then, they will measure the length of each side and prove the congruence in their notebooks.

If time allows, give them a few more practice questions to solve in their notebooks.

Learning Outcomes

Students will be able to use the ASA congruency criteria to solve problems based on it.

Teaching Aids

Two sets of cards with congruent triangles drawn on them; Ruler; Protractor

Activity

Begin by introducing the ASA congruence criteria. Ask the students to consider two triangles with angles of 45° and 60° where the side between these angles measures 4 cm. Discuss that if two angles and the side between them are the same in both triangles, they are congruent according to the ASA criteria.

Prepare two identical sets of cards, one with triangle ABC and the other with triangle XYZ drawn on them, where both triangles are congruent by the ASA congruency rule.

Instruct the students to form 2 groups.

Distribute the two sets of cards, giving one set to each student in Group 1 and the second set to each student in Group 2.

Instruct the students in Group 1 to measure the length of side AB and A and B in ΔABC, and the students in Group 2 to measure the length of side XY and X and Y in ΔXYZ. They will then move around the classroom to find the student whose triangle has the exact same measurements. Once they have found their partners, they will work together to prove the congruence in their notebooks.

Prompt them to consider whether this method also satisfies the criteria for SSS and SAS congruence.

Extension Idea

Ask: The measures of two angles of a triangle are 45° and 60°. The angles of another triangle measure 45° and 75°. If these two triangles have one side measuring 13 cm, are they both congruent using the ASA rule?

Say: 180° – 45° + 60° = 75°. Although the three angles in both triangles are equal, we do not know if the included side in both triangles has the same measurement. So, the triangles may not be congruent using the ASA rule.

RHS Congruence Criteria

Learning Outcomes

Students will be able to use the RHS congruency criteria to solve problems based on it.

Teaching Aids

Imagine Maths Page 164

Glue stick; Sheets of paper; Cutouts of congruent right-angled triangles and some distractor triangles

Activity

Begin the class by introducing the RHS congruence criteria.

Make cutouts of pairs of right-angled triangles with equal lengths of the hypotenuse, one leg and the angle opposite to the hypotenuse measuring 90° (following the RHS rule) and some distractor triangles.

Instruct the students to work in pairs.

Distribute a sheet of paper, triangle cutouts and a glue stick to each pair.

Instruct them to measure the length of the longest side, the angle opposite the longest side, which should be 90°, and one of the arms of the 90° angle. They will then write the measures inside the triangle and make pairs of triangles that overlap exactly or are congruent. They should paste the pairs of triangles on the sheet of paper and label them as ‘Congruent Triangles Using the RHS Rule’.

Extension Idea

Ask: One triangle has a right angle and a hypotenuse measuring 10 cm. Another triangle has a right angle and a side measuring 6 cm. If these two triangles are congruent, what is the length of the third side of the first triangle?

Say: Since the triangles are congruent, the first triangle has a hypotenuse measuring 10 cm and a side measuring 6 cm. We can find the length of the third side using Pythagoras’ theorem where 102 = a2 + 62. On solving this, we get a2 = 102 – 62 and a = 8. So, the length of the third side is 8 cm.

Answers

1. Congruence in Shapes and Figures

Think and Tell

No. The size of each family member’s hand is different, and these are not congruent.

Do It Together

We know that ΔPQR = ΔXYZ

The correspondence is PQR XYZ

a. P X b. Q Y c. R Z

The corresponding sides are:

d. PQ XY

QR YZ; RP ZX

The corresponding angles are:

e. P X

Q Y; R Z

2. SSS Congruence Criteria

Do It Together

In ΔABC and ΔDEC,

AB = DE (side)

BC = CD (side)

AC = CE (side)

Therefore, by the SSS congruence criteria, ΔABC = ΔDEC.

3. SAS Congruence Criteria

Do It Together

We know that ΔPQR = ΔXYZ.

So, PQ = XY = z

QR = YZ

PR = XZ = x

and R =  Z = y

z = 5 cm

x = 8 cm

y = 50°

4. ASA Congruence Criteria

Do It Together

In ΔAOC and ΔDOB, CAO = DBO (given)

AO = BO (given)

AOC = BOD (vertically opposite angles)

Therefore, by the ASA congruence criteria, ΔAOC = ΔDOB.

5. RHS Congruence Criteria

Do It Together

In ΔABD and ΔACD, ABD = ACD = 90°

AD = AD (common side)

AB = AC = 4.9 cm

Therefore, by the RHS congruence criteria, ΔABD = ΔACD.

Ratio and Proportion 11

Learning Outcomes

Students will be able to:

find equivalent ratios and write them in simplified form. check if four terms are in proportion. solve problems using the unitary method. solve problems on speed, distance and time.

Alignment to NCF

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

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

Let’s Recall

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

Vocabulary

equivalent ratios: ratios with the same relationship but different numbers proportionality condition: two ratios being equal means they are proportional means: middle terms in a proportion extremes: first and last terms in a proportion

Teaching Aids

Red and blue building blocks of the same size; Geo boards; Strings; Sheet of paper with a table showing the quantity of items to be bought and the cost of each item; Play money; Small cars; 1 m long road track made of cardboard; Stopwatch

Chapter: Ratio and Proportion

Equivalent Ratios

Learning Outcomes

Students will be able to find equivalent ratios and write them in simplified form.

Teaching Aids

Red and blue building blocks of the same size

Activity

Begin by revising the concept of the simplest form of a ratio.

Instruct the students to work in groups. Distribute the red and blue building blocks among the groups.

Ask the students to build a simple structure, using both coloured building blocks. After the structures are built, instruct the students to write down the ratio of the red and blue blocks used in its simplest form in their notebooks.

Ask the groups to build another structure by using double the red and blue blocks used earlier. Example: If 6 red blocks and 4 blue blocks were used earlier, ask them to use 12 red and 8 blue blocks. Ask them to write down the ratio of red and blue blocks used in its simplest form in their notebooks.

Imagine Maths Page 171

Ask them to discuss in their groups what they notice about the simplest form of both ratios. Bring out the fact that two ratios are equivalent when their simplest form is the same.

Help them understand that equivalent ratios can be found if the numerator and denominator are multiplied or divided by the same natural number. Give them questions for practice in their notebooks.

Extension Idea

Ask: What is the ratio of length to breadth of a rectangle with a perimeter of 180 cm, where the length is 10 cm more than the breadth?

Say: Let b = x cm and l = x + 10. So, 2 (x + x + 10) = 180 x = b = 40 cm and l = 50 cm. Therefore, the ratio is 5:4.

Proportional Terms

Learning Outcomes

Students will be able to check if four terms are in proportion.

Teaching Aids

Geo boards; Strings

Imagine Maths Page 174

Activity

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

Instruct the students to use the strings to make two non-concentric rectangles of different lengths and widths on the geometric board (geo-board). Once the groups are done forming the rectangles, ask them to look closely at the rectangles they created and note down the ratio of length to width of both rectangles in their simplest form.

Ask questions like: Is the simplest ratio of length to width of both the rectangles equal?

If not, ask them to change the length and width of the second rectangle so that it has the same ratio of length to width as the first rectangle. Once the students are done, review the rectangles formed and conclude that if the simplest ratios are equal, the ratios are said to be in proportion.

Teacher tip: If the geo board is unavailable, use a paper sheet with dots drawn at equal distances.

Unitary Method

Learning Outcomes

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

Teaching Aids

Imagine Maths Page 177

Sheet of paper with a table showing the quantity of items to be bought and the cost of each item; Play money

Activity

Begin the lesson by asking questions like: How do you calculate the amount to be paid when buying a certain item from the market?

Discuss student responses. Bring out the fact that whenever we purchase multiple units of an item, we need the price of one item.

Instruct the students to work in groups of 4.

Distribute the sheet of paper with a table showing the quantity of items to be bought with their costs and play money to each group.

Instruct one student to act as a shopkeeper and the other three as buyers. Ask the buyers to pick one item each and calculate the amount of money they need to pay for the quantity of items to be bought using the unitary method. Ask them to use the play money for their purchases. Ask the student acting as the shopkeeper to verify the amount paid by each buyer. Instruct them to note down the answers in their notebooks.

Repeat the activity by asking the students to change their roles and modify the quantities to be bought.

Extension Idea

Ask: Calculate the cost of 3.5 litres of milk using the unitary method for the given price.

Say: The cost of 3.5 litres of milk is ₹210.

Speed, Distance and Time

Learning Outcomes

Students will be able to solve problems on speed, distance and time.

Teaching Aids

Small cars; 1 m long road track made of cardboard; Stopwatch Activity

Imagine Maths Page 178

Begin by asking questions like: How much time does it take you to reach school? How far are your homes from school? Discuss their answers in the class.

Instruct students to form groups. Distribute the cars, tracks, and stopwatches to each group. Instruct the groups to run the cars on the tracks and find out how much time it takes to make 1, 2 or 5 rounds using the stopwatch. Discuss the time recorded by each group in the class.

Bring out the fact that time, distance and speed are dependent on each other. Help the students derive the formula for time, speed and distance and ask them to calculate the speed of the car using the formula. Ask them to note their answers in their notebooks. Give students questions to practise finding speed, distance or time when any of the 2 are given.

Extension Idea

Ask: Sarah and Ben are planning a bike ride to the park. Sarah can bike at a speed of 10 miles per hour, while Ben can bike at 8 miles per hour. The park is 12 miles away. They decide to race. If they both maintain their individual speeds, who will reach the park first?

Say: Sarah’s speed is 10 miles/hr; Ben’s speed is 8 miles/hr. The time taken by Sarah to reach the park = 12

10 = 1.2 hours. The time taken by Ben to reach the park = 12 8 = 1.5 hours. Hence, Sarah will reach the park first.

Answers

1. Equivalent Ratios

Think and Tell

2:3 has infinite equivalent ratios.

Do It Together

LCM of 3, 7, 15 and 5 = 105

1 3 = 35 105; 5 7 = 75 105; 4 15 = 28 105; 2 5 = 42 105; Hence, the descending order is 5

7 > 2 5 > 1 3 > 4 25

2. Proportionality Condition

Do It Together

Let the fourth term be x; 4.2:1.2 :: 0.7:x

4.2 × x = 1.2 × 0.7 x = 1.2 × 0.7 4.2 = 0.2

Hence, the fourth term is 0.2.

3. Continued Proportion

Do It Together

1

3 :

4. Unitary Method

Do It Together

Number of toys produced in 6.5 hours = 208

Number of toys produced in 1 hour = 208 ÷ 6.5 = 32

Number of toys produced in 20 hours = 32 × 20 = 640

5. Speed, Distance and Time

Think and Tell

4 hours 30 minutes is converted to 4.5 hours by expressing the minutes as a fraction of an hour. In this case, 30 minutes is half of 60 minutes (1 hour), so it is represented as 0.5 hours. Adding this to the 4 hours gives 4.5 hours in total.

Do It Together

Distance covered by Kunal = 4.5 km; Kunal’s speed = 1.5 km/hr

Time taken by Kunal = Distance ÷ Speed = 4.5 1.5 = 3 hr

Distance covered by Soham = 8 km; Soham’s speed = 2 km/hr;

Time taken by Soham = Distance ÷ Speed = 8 2 = 4 hr

Ratio of the time taken by both of them = 3:4

Percentage, Profit and Lo , Simple Interest 12

Learning Outcomes

Students will be able to: convert between percentages, decimals and fractions. convert between percentages and ratios. find the percent of a quantity and percent change. solve word problems on percentages.

Alignment to NCF

find profit or loss percentage, cost price or selling price when any two are given. solve word problems on finding the profit or loss percent. solve word problems on finding the simple interest on a given amount.

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 read fractions from pie charts. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

percent: a part out of 100 interest: extra money on a loan or savings

Teaching Aids

Paper slips containing decimal numbers and fractions; Decimal grids; Crayons; Scenario cards; Question slips; Puzzle cards with 4 problems, and the answer to one of them written in the centre; Question cards; Puzzle interlocking cards with the cost price written on the first piece and the selling price on the second; Puzzle interlocking pieces with profit and loss written; Play money; Table cards containing information on the types of chocolates, their CP, SP, or MP

Chapter: Percentage, Profit and Loss, Simple Interest

Percentages,

Decimals and Fractions

Learning Outcomes

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

Teaching Aids

Paper slips containing decimal numbers and fractions; Decimal grids; Crayons

Activity

Imagine Maths Page 184

Begin by discussing with the students how their academic performance is shown on a report card.

Choose a simple fraction or decimal, like 1 3 or 0.4, and represent it on the decimal grid. Discuss the relationship between the coloured grids and the whole (100%) and how this can be expressed as a percentage. Now, form small groups and distribute the decimal grids. On the table, place paper slips containing decimal numbers and fractions such as 0.57 or 3 5.

Instruct each group to pick a slip and ask them to colour the decimal grids accordingly. Now, have the groups observe the coloured grids as a part of the whole (100%) and then represent the same in their notebooks as a percentage. Rotate the slips among groups and repeat the activity.

Extension Idea

Ask: How can you convert 35% into fractions?

Say: To convert 35% into fractions, we simply divide 35 by 100 to get 35 100, which can be simplified as 7 20.

Percentage and Ratios

Learning Outcomes

Students will be able to convert between percentages and ratios.

Teaching Aids

Scenario cards; Question slips

Activity

Imagine Maths Page 187

Begin by introducing how ratio and percentage are related, and explain how we can convert between ratios and percentages.

Instruct students to form groups for the activity. Distribute the scenario cards to each group.

“In the inter-state math Olympiad of Madhya Pradesh and Delhi, 500 students participated. The ratio of students who participated from Madhya Pradesh to those from Delhi is 3:2.”

Prepare question slips such as the ones given below and distribute among the students.

What percent of the total students participated from Madhya Pradesh?

How many fewer students participated from Delhi?

Instruct students to solve these word problems in their notebooks. Discuss their approach in the class. Make the students practise more questions on conversion between ratios and percentages.

Extension Idea

Ask: How many students from Delhi qualified for the Olympiad if 20% did not qualify?

Say: Number of students from Delhi = 200, 80% of them qualified as 20% did not qualify, hence, 80 100 × 200 = 160 students qualified.

Finding Percent and Percent Change

Learning Outcomes

Students will be able to find the percent of a quantity and percent change.

Teaching Aids

Puzzle cards with 4 problems, and the answer to one of them written in the centre

Activity

Begin by discussing some real-life examples of percent change, such as the population change of a city in two years. Make puzzle cards with 4 problems and the answer to one of them written in the centre. Instruct students to form groups for the activity. Distribute the puzzle cards to each group. Tell them that each card consists of four problems to be solved. Students have to solve the expressions, and match the answers with the one written on the centre card. Instruct them to encircle the problem that matches with the answer on the centre card. After the activity, discuss their answers in the class. If time permits, rotate the puzzle cards among groups and repeat the activity.

Extension Idea

Instruct: Frame a word problem where you use percent in your daily life. Say: One such problem can be, “What percent of the day do you spend in school?”.

Word

Problems on Percentage

Learning Outcomes

Students will be able to solve word problems on percentages.

Teaching Aids

Scenario cards; Question cards

Imagine Maths Page 189

Imagine Maths Page 191

Activity

Instruct students to form groups for the activity. Distribute the scenario cards to each group.

“The prices of three sweaters were ₹400, ₹500, ₹750 respectively in 2022. A year later the price of each sweater became ₹600”.

Prepare question cards such as the ones given below and distribute among the students such as:

“What is the percentage change in the price of each sweater?”

“What should be the new price of each sweater if there is a 20% increase in price in 2024?”

Instruct students to solve these word problems in their notebooks. Discuss their approach in the class. Give the students practice questions based on more such scenarios.

Profit and Loss Percentages

Learning Outcomes

Imagine Maths Page 193

Students will be able to find profit or loss percentage, cost price or selling price when any two are given.

Teaching Aids

Puzzle interlocking cards with the cost price written on the first piece and the selling price on the second; Puzzle interlocking pieces with profit and loss written

Activity

Start by briefly discussing the concepts of cost price, selling price, profit and loss.

Divide the class into groups and make puzzle interlocking cards with the cost price written on the first piece and the selling price on the second. Also, make puzzle interlocking pieces with profit and loss written on it.

Distribute the puzzle pieces and puzzle cards among the groups. Instruct the students to find whether there is a profit or a loss and join it with the appropriate puzzle card.

Ask the students to calculate the profit or loss percentage and write the answers in their notebooks. Encourage the students to shuffle the puzzle pieces among groups and solve.

Extension Idea

Ask: What is the marked price of a winter jacket, if it is sold at ₹1800 after a discount of ₹200?

Say: MP = SP + Discount, hence, the MP of the sweater was ₹2000.

Word Problems on Profit and Loss

Learning Outcomes

Students will be able to solve word problems on finding the profit or loss percent.

Imagine Maths Page 196

Teaching Aids

Play money; Table cards containing information on the types of chocolates, their CP, SP, or MP; Question cards

Activity

Prepare table cards containing information on the types of chocolates, their CP, SP, or MP.

Set up chocolate-selling corners in the classroom.

Divide the class into groups and assign roles within each group: one student will act as the “shopkeeper”, while the others will be the “customers”. Provide each group with play money, table cards and question cards such as:

The shopkeeper sold 15 Dark Delight chocolates. Calculate the total profit made.

Dark Delight CP: ₹20 MP: ₹40 P: ₹10??

Caramel Crunch SP: ₹35L: ₹7–?

Raspberry Bliss MP: ₹70?₹10+20%

Due to the discount offer, a customer bought 8 Raspberry Bliss chocolates. Calculate the total discount percentage given.

Instruct students to solve the problems given. Each group should discuss and solve the problems together. The customers will use their play money to buy chocolates from the shopkeeper to verify and experience the real-life scenario. The shopkeeper must ensure they calculate the correct prices based on the given information. Rotate the roles within the groups, allowing each student to experience being the shopkeeper.

Extension Idea

Ask: Ruchika bought 6 bananas for ₹30. She mistakenly dropped 3 bananas somewhere while returning home. For how much should she sell each of the remaining bananas to earn a profit of 20%?

Say: After losing 3, Ruchika will be left with 3 bananas. CP of 1 banana = 30 3 = ₹10. SP of one banana to sell it at a profit of 20% = 100 + 20% 100 × 10 = 120 100 × 10 = ₹12. Therefore, she needs to sell each of the remaining bananas for ₹12 to make a profit of 20%.

Simple Interest

Learning Outcomes

Imagine Maths Page 198

Students will be able to solve word problems on finding the simple interest on a given amount.

Teaching Aids

Scenario cards; Question cards

Activity

Instruct students to form groups for the activity. Distribute the scenario cards to each group.

Make question cards and distribute them among groups.

Ramesh and Suresh deposited ₹2000 each in different banks at 5% and 6% interest per annum. Ramesh withdraws his money at the end of one year, whereas Suresh withdraws it after 2 years.

Discuss the scenario and instruct students to solve these word problems in their notebooks. Discuss their answers and approach in the classroom. Give the students practice questions based on more such scenarios, if time permits.

Who will get more interest at the end of one year and by how much?

What amount of money will Suresh get after 2 years?

What is the total interest earned by both of them?

Answers

1. Percentages, Decimals and Fractions

Do It Together

1. 0.56

2. 120% = 120 ÷ 100 = 6 5

120% = 120 ÷ 100 = 1.2

3. 550% = 550 ÷ 100 = 11 2

550% = 550 ÷ 100 = 5.5

2. Percentage and Ratios

Think and Tell

Percentage of elephants at the zoo = 5 8 × 100 = 62.5%

Do It Together

Ratio of monkeys to total animals = 3:8

Percentage of monkeys at the zoo = 3 8 × 100 = 37.5%

3. Finding Percent and Percent Change

Do It Together

1. 910

2. 35% of 520 = 520 × 35 100 = 182

Number 35% more than 520 = 520 + 182 = 702

4. Word Problems on Percentage

Do It Together

Number of people who went to the exhibition on Sunday = 845

Number of people who went to the exhibition on Monday = 169

Change in the number of people = 845 − 169 = 676

Decrease in percent =

Change in the no. of people

Number of people who went on Sunday × 100% =

676

845 × 100% = 80%

Thus, there was an 80% decrease in the number of people visiting the exhibition on Monday.

5. Profit and Loss Percentages

Do It Together

1. 100 80 × ₹1600 = ₹2000

2. 112% 100 × 2500 = ₹2800

6. Marked Price

Do It Together

80 1600 × 100% = 5%

7. Word Problems on Profit and Loss

Do It Together

Total CP of scooter = ₹47,000 + ₹8000 = ₹55,000

SP of scooter = ₹58,000

Profit% = Profit CP × 100% = ₹3000 ₹55,000 × 100% = 5.4545%

8. Using the Formula

Do It Together

₹1000 × 100 10 × 5 = 100000 50 = ₹2000

9. Word Problems on Simple Interest

Do It Together

₹375 = ₹1500 × R × R 100 ; R2 = ₹375 × 100 ₹1500 = 25; R = 5%

Rational Numbers 13

Learning Outcomes

Students will be able to:

identify rational numbers and represent them on a number line.

find equivalent rational numbers and write them in their simplest form.

compare and order rational numbers.

find rational numbers between 2 rational numbers.

Alignment to NCF

add rational numbers and find the additive inverse of a rational number.

subtract rational numbers.

multiply rational numbers.

divide rational numbers by finding the reciprocal of the divisor.

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

C-9.1: Recognises how concepts (like counting numbers, whole numbers, negative numbers, rational numbers, zero, concepts of algebra, and geometry) evolved over a period of time in different civilisations

Let’s Recall

Recap to check if students know the concept of fractions and types of fractions. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

subunit: a small part of a unit

HCF: highest common factor/the greatest common number that divides two or more numbers

LCM: lowest common multiple/the least common number that is exactly divisible by two or more numbers

Teaching Aids

Cards with rational numbers written on them; Worksheets; Cards with four rational numbers written on them; Cards with sums of rational numbers written on them; Cards with subtraction problems of rational numbers written on them; Bingo cards; Dice

Chapter: Rational Numbers

Representing Rational Numbers

Learning Outcomes

Imagine Maths Page 205

Students will be able to identify rational numbers and represent them on a number line.

Teaching Aids

Cards with rational numbers written on them

Activity

Begin the lesson by introducing the concept of rational numbers and their representation.

Instruct the students to form groups of 4 each and name the teams: Team A, Team B, Team C, and so on.

Make a number line using chalk on the floor at the front of the classroom where all the students can see it. Shuffle the cards with rational numbers (e.g., 1 2 , –3 4 etc) and place them face down in a pile on the teacher's desk.

Each team will take turns drawing a card from the pile and discuss and decide where the rational number on the card should be plotted on the number line. Once they make their decision, they will send one member to the front to plot the number on the number line. For instance, Team A draws a card with the rational number –3 4 . They discuss and decide it should be plotted between –1 and 0 on the number line. If the number is plotted correctly, the team earns a point. If not, they lose a point. The next team takes its turn, and play continues until all rational numbers have been plotted.

After all the rational numbers have been plotted, tally the points earned by each team. The team with the most points wins the game.

Ask students to explain why certain numbers were placed where they were on the number line. Address any misconceptions or questions that arise.

Rational Numbers in Standard Form; Equivalent

Rational Numbers

Learning Outcomes

Imagine Maths Page 206

Students will be able to find equivalent rational numbers and write them in their simplest form.

Teaching Aids

Worksheets

Activity

Begin by reviewing what rational numbers are and how they can be reduced to standard form. Explain the concepts of equivalent rational numbers to the students.

Create worksheets in a “match the following” format, containing five pairs of rational numbers that are equivalent to each other.

Ensure that the numbers are shuffled on both sides.

Distribute one worksheet to each student.

Instruct them to simplify all the rational numbers to their simplest form. Then, they will match the equivalent rational numbers with their corresponding pairs.

Ask the students to discuss why simplifying fractions is important and how it helps in understanding the relationship between different rational numbers. Afterwards, discuss their responses and provide additional questions if necessary.

Comparing and Ordering Rational Numbers

Learning Outcomes

Students will be able to compare and order rational numbers.

Teaching Aids

Cards with four rational numbers written on them

Activity

Imagine Maths Page 209

Begin by saying that when comparing two rational numbers, we need to ensure that the denominators of both numbers are the same.

Instruct the students to form groups of 4.

Distribute a card containing four rational numbers to each group.

Guide the students to determine the least common multiple (LCM) of all four denominators.

Ask the students to form equivalent rational numbers such that all the rational numbers have the same denominators.

Instruct the students to arrange the rational numbers in ascending order.

Ask questions like: Why is it important to have the same denominators when comparing rational numbers?

Provide each group with another card containing four rational numbers and repeat the process to reinforce the concept.

Rational Numbers Between Two Rational Numbers

Learning Outcomes

Students will be able to find rational numbers between 2 rational numbers.

Teaching Aids

Cards with four rational numbers written on them

Imagine Maths Page 211

Activity

Begin by reviewing the concept of finding a rational number between any two rational numbers.

Write pairs of rational numbers on cards, for example: 1 3 and 1 2, 2 5 and 1 2 etc.

Shuffle the cards and place them face down in a stack. Divide the students into teams of 10.

Take the students to the playground.

Make a start line and arrange the teams in lines, with each line standing behind a designated starting point. Explain that the game involves a relay race to find a rational number between two given rational numbers.

When the game starts, the first player from each team picks up the top card from the stack, reads the pair of rational numbers, and finds a rational number between them. Once the player finds a rational number, they run to a designated endpoint (e.g., a wall) and back to their team to hand over the card and write the answer in their notebooks. The next player in line repeats the process, finding a rational number between the given pair on the card and running the relay.

The game continues until all players on a team have completed the relay race. The first team to complete the relay race with all players finding rational numbers between the given pairs wins the game.

Extension Idea

Instruct: If we have two rational numbers, say a and b, then give any one rational number that lies between a and b.

Say: One rational number that lies between a and b, is a + b 2

Addition of Rational Numbers; Additive Inverse

Learning Outcomes

Imagine Maths Page 214

Students will be able to add rational numbers and find the additive inverse of a rational number.

Teaching Aids

Cards with sums of rational numbers written on them

Activity

Begin the class by reviewing the concept of addition of two rational numbers.

Make cards with sums of rational numbers written on them.

Divide the class into two teams and set up the scoreboard to keep track of each team’s points. Place the cards face down on a table.

Each round consists of a member from each team competing against each other.

Flip a card to reveal an addition problem.

Now, the first player from each team will solve the same problem in their notebooks.

Instruct the students to find the additive inverse of the sum and write it next to the sum. The player who correctly completes the task first earns a point for their team. Rotate players for each round.

The game can consist of multiple rounds, and the team with the most points at the end wins.

Extension Idea

Ask: Find the sum of the shaded portions. Say: The rational number is:

Subtraction of Rational Numbers

Learning Outcomes

Students will be able to subtract rational numbers.

Teaching Aids

Cards with subtraction problems of rational numbers written on them

Activity

Begin the class by reviewing the concept of subtraction of two rational numbers.

Make cards with subtraction problems of rational numbers written on them.

Imagine Maths Page 216

Divide the class into two teams and set up the scoreboard to keep track of each team’s points. Place the cards face down on a table.

Each round consists of a member from each team competing against each other.

Flip a card to reveal a rational number.

Now, the first player from each team will solve the same problem in their notebooks.

The player who correctly completes the task first earns a point for their team. Rotate players for each round. The game can consist of multiple rounds, and the team with the most points at the end wins.

Extension Idea

Multiplication of Rational Numbers; Reciprocal or Imagine Maths Page 219

Multiplicative Inverse of a Non-zero Rational Number

Learning Outcomes

Students will be able to multiply rational numbers.

Teaching Aids

Bingo cards

Activity

Begin the class by reviewing the concept of multiplication of two rational numbers.

Prepare a list of 25 multiplication problems. Create bingo cards with a 5 × 5 grid. Each square on the grid should contain a rational number that is the solution of the multiplication problems in the list.

For instance:

List:

–1

2 × 3 4 ; 2 3 × –1 5 ; –2 3 × –1 4 etc.

Make sure the numbers on each bingo card are shuffled and uniquely arranged.

Distribute the bingo cards and markers to each student.

Explain the rules of bingo to the students. The objective is to be the first to cover five spaces in a row, either horizontally, vertically, or diagonally.

Call out the rational number multiplication problems one by one from the list, for instance, –5 7 × 14 –15. The students will solve the multiplication problem and look for the answer on their bingo cards, i.e., 2 3. If they have the answer, they will cross the corresponding space with a marker.

The game continues until a student shouts “Bingo!” after successfully covering five spaces in a row. Verify the student’s answers to confirm the win.

Extension Idea

Ask: Look at the following: is the LHS equal to the RHS?

2 3 × –6 7 + 4 5 = –12 21 + 4 5

Say: Here, LHS = –4 105 and RHS = 24 105. So, the LHS is not equal to the RHS.

Division of Rational Numbers

Learning Outcomes

Students will be able to divide rational numbers by finding the reciprocal of the divisor.

Teaching Aids

Cards with rational numbers written on them; Dice Activity

Begin the class by reviewing the concept of division of two rational numbers.

Divide the class into small groups of 3 – 4 students each.

Place a stack of cards with rational numbers written on them face down on the table.

Start with the first group. Instruct the students to roll the dice. Say that the reciprocal of the number obtained is the divisor.

Now ask them to pick a card with a rational number written on it.

Instruct them to divide the rational number by the reciprocal of the number rolled. Allow the group to work together to solve the division problem and provide their answer. If correct, they earn points. Continue playing rounds with each group taking turns and determine a winner.

At the end of the game, review the division problems and solutions as a class to reinforce learning.

Extension Idea

Ask: If the product of 2 3 with another rational number is 8 27 , what is the other rational number?

Say: Let the other rational number be

Answers

1. Representing Rational Numbers

Think and Tell

Yes, all fractions are rational numbers but not all rational numbers are fractions.

2. Rational Numbers in Standard Form

Do

The

The HCF of 12 and –18 is 6. Hence, the standard form of 12 –18 is –2 3

3. Equivalent Rational Numbers

Do

4. Comparing and Ordering Rational Numbers

Think and Tell

Fractions are the ratio of two whole numbers. Rational numbers are the ratio of two integers.

Do It Together

The LCM of

Ascending order:

5. Rational Numbers Between Two Rational Numbers

Do It Together

Let us multiply and divide 1 and 3 by 4 to expand the range.

We know that,

So, the rational numbers between 1 and 3 are:

and

6. Properties of Rational Numbers

Do It Together

Let us divide the numerator and denominator by 7.

7. Addition of Rational Numbers

Do It Together

The LCM of the denominators 5, 4 and 12 is 60.

Rewriting the rational numbers:

So, 2 5 + 3

8. Additive Inverse

Do It Together

=

9. Subtraction of Rational Numbers

Do It Together

Let the required number be x

Then, –3

LCM of 8 and 7 = 56. Hence, x = 48 56 + 21 56 x = 69 56

10. Multiplication of Rational Numbers

Do It Together

Thickness of each sheet = 5 7 inches

Thickness of two sheets = 2 × 5 7 inches

Therefore, thickness of the stack of two metal sheets will be 10 7 inches.

11. Reciprocal or Multiplicative Inverse of a Non-zero Rational Number

Do It Together

The given rational number is: p q = 2m × r n × 2r

So, the reciprocal of p q is: n × 2r 2m × r

12. Division of Rational Numbers

Do It Together

Construction of Triangles 14

Learning Outcomes

Students will be able to: construct triangles using the SSS criterion. construct triangles using the SAS criterion. construct triangles using the ASA criterion. construct triangles using the RHS criterion.

Alignment to NCF

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

C-3.5: Understands congruence and similarity as it applies to geometric shapes and identifies similar and congruent triangles

Let’s Recall

Recap to check if students know how to construct angles and angle bisectors. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

SSS criterion: the Side−Side−Side criterion, a rule by which it is necessary to know the measures of three sides of a triangle to construct it

SAS criterion: the Side−Angle−Side criterion, a rule by which it is necessary to know the measures of two sides and one included angle of a triangle to construct it

ASA criterion: the Angle–Side–Angle criterion, a rule by which it is necessary to know the measures of two angles and one included side of a triangle to construct it

RHS criterion: the Right Angle–Hypotenuse–Side criterion, a rule by which it is necessary to know the measures of the hypotenuse and one other side of a triangle to construct it

Teaching Aids

Paper strips of three different lengths; 2 sets of straws of three different lengths; Ruler; Compass; Cardboard; Cardboard pins; Coloured string; Protractor

Chapter: Construction of Triangles

Constructing Triangles: SSS Criterion

Learning Outcomes

Students will be able to construct triangles using the SSS criterion.

Teaching Aids

2 sets of straws of three different lengths; Ruler; Compass

Activity

Imagine Maths Page 226

Instruct students to form pairs for the activity. Provide students with 2 sets of straws of three different lengths, for example, 3, 4, 5 and 7, 9, 1 cm. Ask them to form 2 triangles by connecting the straws. Now, ask them questions like: Were you able to make 2 triangles? What did you notice? Can you make another triangle using the given lengths? Guide them to understand the uniqueness of the triangle made using the SSS criterion. Encourage students to explore different combinations of straw lengths and ensure that the sum of any two sides is greater than the third side.

Now, instruct them to use a compass and construct a triangle using these lengths in their notebooks. Ask them to compare their triangles with their partner to see that both triangles are exactly the same.

Extension Idea

Ask: How will you construct an equilateral triangle with one side measuring 5 cm?

Say: First, draw a 5-cm base. Then, extend lines, also measuring 5 cm, from each end to complete an equilateral triangle. All the sides will be equal.

Constructing Triangles: SAS Criterion

Learning Outcomes

Students will be able to construct triangles using the SAS criterion.

Teaching Aids

Sets of 3 paper strips forming the sides of a triangle; Protractor

Activity

Imagine Maths Page 229

Instruct students to form groups for this activity. Give them paper strips, two of which measure 5 cm and 6 cm. Instruct them to arrange the two strips at a 60° angle using a protractor. They should cut the third strip and place it to form the third side of the triangle. Ask them whether the angle in their triangle is included between the sides, and if not, how they can include it.

Help them understand that if the included angle is not 60°, the students will not obtain the same triangle. Then, ask them to use the compass and ruler to construct the triangle with specified measures using the SAS criterion.

Triangles: ASA

Learning Outcomes

Students will be able to construct triangles using the ASA criterion.

Teaching Aids

Cardboard; Cardboard pins; Coloured string; Protractor

Activity

Instruct students to form groups for the activity. Give a large piece of cardboard, pins and coloured strings to each group.

Give each group a set of 2 angles and 1 included side and instruct them to create triangles by pushing the cardboard pins into the cardboard and wrapping string around these to form the side and angles. Guide them to use their protractors when necessary.

Ask questions like: ‘Would you be able to create a unique triangle if the given side is placed between any two of the three angles?’ This artistic approach reinforces the idea of creating the side and then the angles to create triangles. Then, encourage them to construct the triangle in their notebooks using the ASA criterion and a compass.

Extension Idea

Instruct: Try creating a word problem on constructing a triangle using the ASA criterion and involving a real-life situation.

Say: One such word problem can be, ‘Alex wants to build a triangular treehouse. Angle A is 40°, angle B is 60° and the base plank is 8 feet in length. Can he construct the triangle? Which congruence criteria will be used?’

Constructing Triangles: RHS Criterion

Learning Outcomes

Students will be able to construct triangles using the RHS criterion.

Teaching Aids

Paper strips of three different lengths; Ruler; Compass

Activity

Imagine Maths Page 234

Instruct students to form groups. Distribute three paper strips measuring 5 cm, 4 cm and 3 cm in length to each group.

Instruct them to try arranging the strips as the sides of a triangle. Ask them which type of triangle it is and where the longest strip is positioned. Repeat the activity with more Pythagorean triplets.

Now, provide each group with a ruler and compass and ask them to individually construct a triangle of the given measures in their notebooks. Encourage them to compare their triangles within their groups to verify that the triangles constructed are exactly the same.

Extension Idea

Ask: In a right-angled triangle, can any angle be greater than a right angle? Why?

Say: No, in a right-angled triangle, no angle can be greater than the right angle. This is because the sum of the angles in any triangle is always 180°. In a right-angled triangle, one angle is fixed at 90°, leaving the other two angles as acute angles, which are always less than 90°.

Answers

1. Constructing Triangles: SSS Criterion

Think and Tell

Yes, since the sum of any two sides is greater than the third side, the triangle can have sides of length 5 cm, 6 cm and 10 cm.

Do It Together

1.

ABC = Scalene triangle

PQR = Isosceles triangle

2. Constructing Triangles: SAS Criterion

3. Constructing Triangles: ASA Criterion Do It Together

4. Constructing Triangles: RHS Criterion

Do It Together

Perimeter and Area 15

Learning Outcomes

Students will be able to:

find the perimeter and area of squares and rectangles using the formula.

find the area of parallelograms.

find the area of triangles.

find the circumference of circles.

find the area of circles, semicircles and quadrants.

find the area of combined shapes made with squares, rectangles, parallelograms, triangles, circles, semicircles and quadrants.

solve word problems on finding the area and perimeter.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to find the perimeter and area of squares and rectangles. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

•

the total length of the boundary of a figure

• area: the amount of space occupied by an object/shape

•

•

Teaching Aids

the perimeter of a circle

A4 sheets of paper; Square and rectangle cutouts of different areas with names of places written (like swimming pool, park, apartments and office complex); Ruler; Glue stick; Squared paper; Pencils; Pairs of scissors; Pieces of string; Compasses; Cutouts of circles, semicircles and quadrants; Cutouts of different shapes; Word problem cards

Chapter: Perimeter and Area

Perimeter and Area of Squares and Rectangles

Learning Outcomes

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

Teaching Aids

A4 sheets of paper; Square and rectangle cutouts of different areas with names of places written (like swimming pool, park, apartments, office complex); Ruler; Glue stick

Activity

Divide the class into groups of 4.

Tell the students that today they will calculate the area of a residential complex.

Distribute an A4 sheet of paper and the square and rectangle cutouts of different areas with names of places written (like swimming pool, park, apartments and office complex) among the groups.

Instruct the students to place and stick the square and rectangular cutouts on the A4 sheets of paper to show a layout of a residential complex. Make sure that the sizes of the sheets and the cutouts given to the groups, are the same.

Instruct the students to measure the lengths of the sides of the shapes and find the area occupied by the park, the swimming pool, the office complex and the apartments, keeping in mind that 1 cm on paper will be equal to 5 m on the actual ground. They will also find the area of the remaining land by subtracting the area of the buildings from the area of the complete residential complex, which is the A4 sheet of paper.

Discuss the answers in class.

Extension Idea

Ask: A rectangular field has a perimeter of 40 m, where the width is one-fourth the length. If on expanding the field, the perimeter remains constant and its width is the same as its length, then how will the area before and after expansion differ from each other?

Say: Perimeter = 2 (l + b) = 40 and b = l 4 . So, 2 (l + l 4 ) = 40 5l

= 40 l = 40 × 2 5 = 16 m. Therefore, b = 16 4 = 4 m. Also, area before expansion = 16 × 4 = 64 sq. m. After expansion, Perimeter = 2 (l + b) = 40 and b = l. So, 2 (l + l) = 40 4l = 40 l = 40 4 = 10 m. Therefore, b = 10 m. Also, area after expansion = l × b = 10 × 10 = 100 sq. m. Hence, if the breadth is the same as its length on expanding the field, the area increases by 100 – 64 = 36 sq. m.

Area of Parallelograms

Learning Outcomes

Students will be able to find the area of parallelograms.

Imagine Maths Page 242

Teaching Aids

Squared paper; Ruler; Pencils; A4 sheets of paper; Pairs of scissors

Activity

Begin by reviewing the definition of a parallelogram, emphasising its various properties. Instruct the students to form groups of equal size.

Distribute the resources to each group. Introduce the concept of area and instruct the students to draw a parallelogram of any length and breadth on squared paper. Ask them to draw a perpendicular from any of the vertices of the parallelogram to its opposite side, then cut off the triangle formed by the perpendicular and paste it on the other side of the parallelogram. Ask them to identify the shape so formed after cutting and pasting and to find its area.

Ask questions like: What is the new shape that is formed? What is its area? How is it related to the area of the parallelogram?

Discuss their observations and derive the formula for the area of a parallelogram, Instruct them to find the area of the new shape and the parallelogram if each square on the squared paper is 1 cm long.

Give them some questions on finding the missing length of a parallelogram when the area is given.

Extension Idea

Ask: Your room has a rectangular floor with dimensions of 4 metres by 3 metres. You want to lay down a large rug in the shape of a parallelogram with a base of 2 metres. What is the maximum height the rug can have, if its area is not equal to or greater than half of the area of the room?

Say: Area of the rectangle = 4 × 3 = 12 sq. m. Area of a parallelogram = b × h or 2 × h. The area of the rug is less than half of the area of the rectangular floor. So, area of the rug < 6 or 2h < 6. That is, h < 3. So, the height of the rug can be a maximum of 3 m.

Area of Triangles

Learning Outcomes

Students will be able to find the area of triangles.

Teaching Aids

Rectangular cutouts; A4 sheets of paper; Pairs of scissors

Activity

Begin by discussing the area of a rectangle.

Divide the class into groups.

Distribute the rectangle cutouts (L = 10 cm, B = 6 cm) to the students.

Instruct them to cut the rectangle in a way to form triangles. Ask them to use the area of the rectangle to find the area of the triangles, then deduce the formula that they can use to find the area of a triangle with given lengths.

Ask questions like: How did you find the formula for the area of a triangle?

Discuss that when a rectangle is cut into 2 identical triangles, they can be superimposed over each other and that their area is the same.

Further, give a target area of, say, 36 sq. cm to each group and distribute squared paper and a pair of scissors. Instruct each student to create and cut triangles of different sizes and types with the same area. Ask them to tessellate the triangular pieces and form a shape. Have them find the area of the tessellated shape.

Ask questions like: How many triangles could you form? What is the relationship between the tessellated shape and its area?

Give them some questions on finding the missing length of a triangle when its area is given.

Circumference of Circles

Learning Outcomes

Students will be able to find the circumference of circles.

Teaching Aids

Pieces of string; Ruler; A4 sheets of paper; Compasses

Activity

Imagine Maths Page 248

Distribute a piece of string to each student. Ask them to wrap the string around a pen, cut its ends, and measure the length of the string using the ruler. Discuss the circumference of the pen. Tell them that circumference is nothing but the perimeter of a circle.

Instruct the students to form groups.

Distribute a sheet of paper, a ruler and a piece of string to each group.

Draw a table on the board.

Instruct the students to first copy the table into their notebooks, then draw circles of different radii using a compass on the sheet of paper. They will then cut out the circles and measure the circumference of these circles using the piece of string and a ruler. Let them write the length of the diameter and circumference in the table in their notebooks. Ask them to find the ratio of the circumference to the diameter to derive the approximate values of pi. Using this, instruct them to derive the formula for the circumference of the circle.

Radil

(D)

Ask questions like: What is the relationship between the circumference and the diameter of the circle?

Give them some questions on finding the radius when the circumference is given.

Extension Idea

Ask: You are designing a circular flower bed with a radius of 2 metres. You want to plant flowers around the edge every 30 cm. How many flowers will you need?

Area of Circles, Semicircles and Quadrants

Learning Outcomes

Students will be able to find the area of circles, semicircles and quadrants.

Teaching Aids

Cutouts of circles, semicircles and quadrants; Pairs of scissors

Activity

Discuss the area of a rectangle and the circumference of a circle.

Divide the class into groups.

Distribute cutouts of circles with a radius of 7 cm.

Imagine Maths Page 251

Instruct the students to fold the circle to find the diameter, and then draw the diameter of the circle. Then, ask them to draw as many diameters as they can in the circle by folding the circle. Now, they should cut the circle into smaller sectors, rearrange them in the form of a rectangle by placing the sectors alternately so that they get a perfect looking rectangle. Then ask them to use the formula for the area of a rectangle to find the area where the length of the rectangle is the same as the circumference of the circle and the breadth of the rectangle is the radius of the circle. Now, ask them to deduce the formula for the area of a circle.

Ask questions like: What is the formula for the area of a circle? How did you find the formula?

Distribute cutouts of a circle, a semicircle and a quadrant to the students.

Instruct them to use their understanding of the area of a circle to find the area of the given shapes by measuring the radii of the shapes.

Discuss the answers with the class.

Combined Shapes

Learning Outcomes

Imagine Maths Page 254

Students will be able to find the area of combined shapes made with squares, rectangles, parallelograms, triangles, circles, semicircles and quadrants.

Teaching Aids

Cutouts of different shapes

Activity

Instruct the students to form groups.

Distribute a few cutouts of different shapes to each group.

Instruct the students to join the different shapes to form a combined shape.

Ask each student in the group to measure the length of the sides of each shape, find its area and then add the areas together to find the total area. Ask the students to write the answers in their notebook.

Ask them to interchange the shapes with other groups who will then rearrange them to make a new combined shape and then find the total area of the combined shape formed, in their notebooks.

Ask questions like: Is the area of the combined shape the same for all the groups? How?

Extension Idea

Ask: A circular pizza with a diameter of 24 cm is cut into 16 equal slices. You rearrange the slices to form a larger rectangle. What is the length of the rectangle? (Use π 3.14)

Say: When a circular pizza is cut into 16 equal slices and rearranged to form a rectangle, each slice becomes a triangular piece. The rectangle formed will have a length equal to half the circumference of the original circle (pizza). So, the length of the rectangle = 1 2 × 3.14 × 24 cm = 37.68 = 38 cm (roughly).

Word Problems on Perimeter and Area

Learning Outcomes

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

Teaching Aids

Word problem cards

Activity

Begin the class with real-life examples where finding the area of a shape is needed.

Instruct the students to work in groups.

Distribute word problem cards to each group.

Imagine Maths Page 257

A goat is tied to one corner of a square field of side 14 m. If the length of the rope is 8 m, what is the area of the field that the goat can move around in?

A circular swimming pool is to be built in the centre on a rectangular piece of land with a circular pavement of width 1 m all around the pool. The length and width of the rectangular land are 50 m and 30 m, respectively. What is the area of the pavement if it touches the edges along the length of the land?

Instruct the students to first draw a picture to show the problems in their notebooks, and then solve the problems. They will then discuss the answers within their groups.

Finally, discuss the solutions in the class and the challenges faced while solving the problems.

Answers

1. Perimeter and Area of Squares and Rectangles

Do It Together

Size of 1 tile to be cut = 12.5 cm × 10 cm

Area of 1 tile to be cut = 125 sq. cm

Size of tile to be used for cutting = 102 cm × 65 cm

Area of tile to be used for cutting = 6630 sq. cm

Number of tiles = 6630 ÷ 125 = 53

So, 53 tiles are required to be cut from the tile of size 102 cm × 65 cm.

2. Area of Parallelograms

Think and Tell

Yes. By counting the number of squares, the area of the parallelogram = area of full squares + area of half squares =

Number of full squares + 1 2 × number of half squares = 24 + 1 2 × 8 = 24 + 4 = 28 sq. units

Do It Together

So, the area of the parallelogram = ST × VE

400 sq. cm = 40 cm × VE

VE = 400 40 cm = 10 cm

3. Area of Triangles

Think and Tell

Yes. Such triangles are called congruent triangles.

Think and Tell

If the park is in the shape of the square, then the area of the square park = 2 × area of the triangles = 2 × 1 2 × b × h = b × h = 60 × 60 = 3600 sq. m

Do It Together

So, area of square = 12 × 12 = 144 sq. m

Area of triangle = 1 2 × 144 = 72 sq. m

4. Circumference of Circles

Do It Together

So, the perimeter of the square = 4s = 4 × 66 cm = 264 cm

Circumference of the circle = 264 cm

264 = 2 × 3.14 × r

r = 264 2 × 3.14 = 264 6.28 cm

r = 42.04 cm

Cost of 1 cm of wire = ₹4

Cost of 264 cm of wire = ₹4 × 264 = ₹1056

5. Area of Circles, Semicircles and Quadrants

Do It Together

So, Perimeter of the square = 4 × 15 cm = 60 cm

Perimeter of the circle = 60 cm

2 × 22 7 × r = 60 cm

r = 60 × 7 2 × 22 = 9.55 cm

Area of the circle = 22 7 × 9.55 × 9.55 = 286.64 sq. cm

6. Combined Shapes

Think and Tell

Circles with the same centre and different radii are called concentric circles. They are found at the cross-section of pipes, wires and rings, etc.

Do It Together

Area of the remaining sheet of paper = 22 7 × 24 × 24 – (2 × 22 7

× 1.6 × 1.6 + 1 4 × 22 7 × 1.6 × 1.6 + 6 × 6) = 1810.29 – (16.09 +

2.01 + 36) = 1810.29 − 54.1 = 1756.19 sq. cm

7. Word Problems on Perimeter and Area

Think and Tell

The plot is in the shape of a rectangle. So, its area without the pathways, crossroads and fountain can be calculated using the formula for the area of a rectangle.

Do It Together

Area of road 1 = 150 × 2 = 300 sq. m

Area of road 2 = 150 × 2 = 300 sq. m

Area of the common square = 2 × 2 = 4 sq. m

Area of the roads = Area of road 1 + Area of road 2 – Area of the common square

= 300 + 300 – 4 = 596 sq. m

Cost of constructing 1 sq. m of road = ₹520

Cost of constructing 596 sq. m = ₹520 × 596 = ₹3,09,920

Algebraic Expre ions

Learning Outcomes

Students will be able to:

generate rules in patterns and use it to extend the pattern or create a formula using variables. identify parts of an algebraic expression and types of expressions. add algebraic expressions.

subtract algebraic expressions.

find the value of an expression when the values of the variables are given. simplify algebraic expressions. solve word problems on algebraic expressions.

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 identify and extend a number pattern. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

algebraic expression: expression consisting of numbers, constants, variables and operation symbols

Teaching Aids

Cutouts of regular polygons; Expression cards; Chart paper; Rectangle and circle cutouts; Piece of string; Glue stick; Algebra tiles as blue, green, red and yellow paper cutouts; Expression mazes; Crayons; Set of puzzle interlocking cards; Slips of paper with a situation; Slips of paper with questions

Chapter: Algebraic Expressions

Generating Rules in Formulas and Patterns

Learning Outcomes

Imagine Maths Page 264

Students will be able to generate rules in patterns and use them to extend the pattern or create a formula using variables.

Teaching Aids

Cutouts of regular polygons like: equilateral triangle, square, regular pentagon and hexagon, octagon, etc

Activity

Begin the class by revisiting the terms variable and constants. Explain what it means when we say that the area of a rectangle is 2L + 2B and that these are called formulas.

Instruct the students to work in groups.

Distribute cutouts of regular polygons among the groups.

Instruct the students to find the perimeter of the shapes by taking the length of the sides as a variable. This will help them deduce the perimeter formulas. Once they deduce the formulas, ask them to find the perimeter of the shapes by measuring the length of one of the sides using the ruler and then applying the formula to find the perimeter. They will then measure the length of all the sides to see if they get the same perimeter. Let them solve in their notebooks. Ask them to discuss in their groups to see if they got the same perimeter. Encourage them to deduce the formula for the area of a square, rectangle, and triangle as well. Discuss on how to find the area of a regular polygon with ‘n’ sides which has a side length of 4 units (or any given length).

Extension Idea

Ask: What is the perimeter of a regular shape with each side measuring n units and having m sides?

Say: The perimeter of a regular polygon with each side measuring n units and having m sides can be calculated using the formula: Perimeter = n × m

Algebraic Expressions and Terms

Learning Outcomes

Imagine Maths Page 267

Students will be able to identify parts of an algebraic expression and types of expressions.

Teaching Aids

Expression cards with one algebraic expression on each card; Chart paper; Rectangle and circle cutouts; Piece of string; Glue stick

Activity

Begin with reviewing the parts of an expression. Discuss like and unlike terms and types of algebraic expressions such as monomials, binomials etc. Create expression cards (e.g., 10pq + 4p2 – q3) with one expression on each card. (Make sure to cover all types of expressions in the cards).

Instruct the students to work in groups of 3. Distribute 3 expression cards, and other teaching aids to each group.

Instruct the groups to paste the expression card on chart paper and write down the type of expression (monomial, binomial etc.) on the left. Ask the groups to work together to form factor trees for all the expressions given using rectangular cutouts to show the terms; and circle cutouts to show the factors. They can use the glue stick to paste the card and the cutouts. Ask them to also label the expression, terms and the factors by writing on the side.

Algebraic Expression

Once the students are done forming the factor tree, ask them to combine the terms in all the expression cards they have and note down the like and unlike terms on the chart paper. Encourage the students to present the factor trees formed in front of the class.

Ask them to draw the factor trees for the remaining two expressions in their notebooks.

Addition of Algebraic Expressions

Learning Outcomes

Students will be able to add algebraic expressions.

Teaching Aids

Algebra tiles as blue, green, red and yellow paper cutouts

Activity

Imagine Maths Page 270

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

Bring out the fact that a pair of one negative and one positive tile represents 0.

Instruct the students to work in pairs. Distribute algebra tiles among the pairs.

Instruct the students to add 3x2 – 2x + 1 and x2 + 4x – 2 using the algebra tiles. Ask them to place algebra tiles that represent 3x2 – 2x + 1. Ask them to now place algebra tiles that represent x2 + 4x – 2 below the previous set. Instruct them to remove the pairs of algebra tiles that make zero. Discuss that the remaining number of algebra tiles 4x2 + 2x – 1 would be the answer.

Instruct the students to write the expressions one below the other and the answer below them. They will then try and understand the rule for adding expressions using the column method. Discuss the rules of addition with the students. Then, let the students add the expressions (5x2 – 7x + 5) and (–7x2 + 3x – 8) using the column method in their notebooks.

Discuss the answers.

Give them more practice questions on adding algebraic expressions.

Extension Idea

Instruct: Add  

Say: On adding the two expressions, we get

Subtraction of Algebraic Expressions

Learning Outcomes

Students will be able to subtract algebraic expressions.

Teaching Aids

Algebra tiles as blue, green, red and yellow paper cutouts

Activity

Instruct the students to work in pairs. Distribute algebra tiles among the pairs.

Imagine Maths Page 273

Instruct the students to subtract (2x2 – 2x + 1) from (x2 – 3x + 3). Ask them to use tiles to show (x2 – 3x + 3). Ask them questions like: Can you take away 2x2 positive tiles from the x2 positive tiles? Explain that since there is only 1 positive x2 tile, we cannot take away 2 positive x2 tiles. Tell the students that to add one more x2 positive tile, one set of zero tiles needs to be added as one x2 positive tile and one –x2 neutral tile.

Let the students figure out how to subtract and find the answer. (Help them if needed.)

Instruct the students to now subtract (4x2 – 5x + 3) from (5x2 + 2x – 4) using the algebra tiles. Discuss the horizontal and column method of subtraction with the students. Instruct the students to subtract both the expressions discussed using the horizontal method and column method in their notebooks. Give them practice by running a quiz around subtracting algebraic expressions.

Extension Idea

Ask: What should be

Say: Let A be added to (5x2

to

1 1 1

Finding the Value of an Expression

Learning Outcomes

Students will be able to find the value of an expression when the values of the variables are given.

Teaching Aids

Expression mazes; Crayons

Activity

Create or find expression mazes that include algebraic expressions.

Begin the activity with a brief review to find the value of an expression when the value of the variable is given. Discuss 1-2 examples.

Instruct the students to form pairs. Distribute an expression maze and crayons to each pair.

Instruct students to start at the beginning of the maze and solve each expression by putting the values of the variables. Ask them to shade the expression and its value. The correct value will guide them to the next part of the maze.

Once students reach the exit, have them review their answers.

Extension Idea

Ask: What should be the value of c if (3a2 – 5ab + c – 2b2) equals 4 for a = 1 and b = −1?

Say: Substituting the values of a and b we get, 3(1)2 – 5 × 1 × (–1) + c – 2(–1)2 = 4 3 + 5 + c – 2 = 4 6 + c = 4 c = –2.

Simplifying Algebraic Expressions

Learning Outcomes

Students will be able to simplify algebraic expressions.

Teaching Aids

Imagine Maths Page 277

Set of puzzle interlocking cards with an algebraic expression having brackets on one card, the simplified form of the same expression on the second card and its value on the third card

Activity

Begin with showing the students an algebraic expression and then simplifying it to find the value.

Instruct the students to work in groups.

Distribute sets of puzzle interlocking cards to each group.

Instruct the groups to simplify and solve the algebraic expression in their notebooks and interlock them with its solution. Record the time taken by each group to interlock all the puzzles. The group which takes the least time to interlock all the puzzles wins. Check the puzzles created and announce the winning group.

Word Problems

Learning Outcomes

Students will be able to solve word problems on algebraic expressions.

Teaching Aids

Slips of paper with a story; Slips of paper with questions; Chart paper

Activity

Imagine Maths Page 279

Begin by discussing the concept of algebraic expressions and their significance in solving real-world problems. Share a captivating story with the students that involves solving an algebraic expression.

Once upon a time in a city, there was a poor man who used to sell colourful balloons. He used to sell (x2 + 3x + 5) balloons during the weekdays. On weekends he used to stand outside a huge park and sell (3x2 – 2x + 3) balloons.

After telling the story, ask the students to work in small groups. Distribute the slips of paper with the story and the questions to each group.

How many balloons did he sell on weekdays if x = 5?

What is the total number of balloons sold in the week if x = 4?

Ask the students to solve the problems in their notebooks and discuss the answers in their groups. Discuss each group's expression, emphasising the different ways students approached the problem. Make sure to highlight the correct answer. Give them more problems, if time permits.

Extension Idea

Ask: Create your own word problem where you need to add and find the value of the expressions (5a2 – 2b – 7) and (3a2 – 5a + 7b).

Say: Sample word problem: Suhani went to the market and purchased a scarf for ₹(5a2 – 2b – 7) and a bag for ₹(3a2 – 5a + 7b). If a = 10 and b = 5, find the total amount spent by her.

Answers

1. Finding Formulae

Think and Tell

For the perimeter of a square to be equal to the area

4a = a2

4a a = a2 a

4 = a

Thus, the side of the square is 4 units.

The unit should not be cm or m.

Do It Together

Length of the side of the given triangle a = 15.7 cm

Perimeter of triangle = 3a = 3 × 15.7 cm = 47.1 cm

Length of the side of the given pentagon a = 7.5 cm

Perimeter of pentagon = 5a = 5 × 7.5 cm = 37.5 cm

Length of the side of the given hexagon a = 6.5 cm

Perimeter of hexagon = 6a = 6 × 6.5 = 39 cm

On comparing, the perimeter of the triangle is the greatest.

2. Number Patterns

Do It Together

The sum of the first 15 natural numbers = 15(15 + 1) 2 = 120

The sum of the first 10 natural numbers = 10(10 + 1) 2 = 55

Difference = 120 – 55 = 65

3. Parts of an Expression

Think and Tell

The coefficient of xy in the term −xy is −1.

Do It Together

Expression 15ab − 5a2b + 5a

Terms

Factors

15ab    −5a2b     5a

15  a b    −5 aa b     5    a

The factors of the term −5a2b are −5, a, a, and b

4. Like and Unlike Terms

Do It Together

9a2b

5. Types of Algebraic Expressions

Do It Together

ExpressionTermsType of Expression

9xvy + 7 − 9xy + 3xv 9xvy, 7, −9xy, 3xv Quadrinomial

2x2 + 5 2x2, 5 Binomial

−7a + 5 + 6b −7a, 5, 6b Trinomial

3x 3x Monomial

6. Addition of Algebraic Expressions

Do It Together

4pqr + 9pr2 − 3pr

+−12pqr + 10pr2

(4 + (–12))pqr + (9+ 10)pr2 − 3pr

=−8pqr + 19pr2 − 3pr

7. Subtraction of Algebraic Expressions

Do It Together

Horizontal Method Column Method

(24x2z − 13xz + 3yz) − (14x2z + 12xz) – 13yz)

= 24x2z − 13xz + 3yz −

14x2z + 12xz + 13yz

= 24x2z − 14x2z – 13xz − 12xz + 3yz + 13yz

= (24 − 14)x2z + (–13 − 12)xz + (3 + 13)yz

= (10x2z − 25xz + 16yz)

24x2z − 13xz + 3yz 14x2z + 12xz − 13yz

− −    +

(24 − 14)x2z + (–13 − 12)xz + (3 + 13)yz

= (10x2z − 25xz + 16yz)

8. Finding the Value of an Expression

Do It Together

Subtracting 5xy + 7x2 from 9x2 − 3xy + 5, we get

(9x2 − 3xy + 5) − (5xy + 7x2)

= 9x2 − 3xy + 5 − 5xy − 7x2

= (9 − 7)x2 + (−3 − 5)xy + 5 = 2x2 − 8xy + 5

Putting x = 1 and y = −1, we get

2(1)2 – 8 × 1 × (−1) + 5

= 2 + 8 + 5 = 15

9. Simplifying Algebraic Expressions

Do It Together

10x − [5x2 − {4x2 − (4 + 5x + 7x2) − 2x2}]

10x − [5x2 − {4x2 − 4 − 5x − 7x2 − 2x2}]

10x − [5x2 − {−5x2 − 4 − 5x}] = 10x − [5x2 + 5x2 + 4 + 5x]

10x − [10x2 + 4 + 5x] = 10x − 10x2 − 4 − 5x

Solving the expression, we get

10x − 10x2 − 4 − 5x = 5x − 10x2 − 4

Putting x = 2; y = 1, we get

5 × 2 − 10(2)2 − 4 = 10 – 40 − 4 = −34

10. Word Problems

Do It Together

= 5x + 3x2 + 2 − x2 − 3x + 5

= (5 − 3)x + (3 − 1)x2 + 2 + 5

= 2x + 2x2 + 7

Putting x = 10 in the above expression, we get 2 × 10 + 2 × (10)2 + 7 = 20 + 200 + 7 = 227 ice creams

Exponents and Powers 17

Learning Outcomes

Students will be able to:

express a number in its exponential form and as a product of its prime factors. simplify an expression with rational numbers in exponent form. use the correct law of exponents to simplify exponential expressions. express large numbers in standard form.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to encounter and use large numbers in daily conversations. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

exponent: how many times a number is multiplied by itself

prime factors: the smallest prime numbers that multiply to create a given number standard form: number in standard form will be K × 10n, where K is a number between 1 and 10 and n is an integer

Teaching Aids

Question cards to represent numbers in the exponential form; Crossword puzzle cards on powers of rational numbers; Cards with questions based on laws of exponents; Puzzle cards with a large number in standard form written in the centre and options on 4 sides

Chapter: Exponents and Powers

and

Factors;

Comparing Numbers in Exponential Form

Learning Outcomes

Students will be able to express a number in its exponential form and as a product of its prime factors.

Teaching Aids

Question cards to represent numbers in the exponential form

Activity

Begin with a brief discussion on prime factors of a number. Make question cards to represent numbers in the exponential form.

Instruct the students to work in groups. Distribute question cards with some numbers such as 384, 70, 120 etc. among the groups.

Ask them to write the prime factors of the first number in the prime factorisation column.

Instruct the groups to count and write the prime numbers that are multiplied and the number of times each prime number is multiplied, in respective columns. Tell them that in exponential form, 384 is written as 27 × 31. Ask the students what rule they noticed in the exponential representation. Bring out the fact that the exponent indicates how many times the base (the prime factor) is multiplied by itself.

Ask them to solve the remaining questions using the same approach. Run a quiz around the class to give more practice on the concept with a variety of questions.

Extension Idea

Ask: Which is greater, 23 × 5 or 22 × 52?

Say: 23 × 5 = 40 and 22 × 52 = 100. 100 > 40, therefore, 22 × 52 is greater.

Powers of Rational Numbers;

Reciprocal of a Rational Number

Learning Outcomes

Imagine Maths Page 286

Students will be able to simplify an expression with rational numbers in exponent form.

Teaching Aids

Crossword puzzle cards on powers of rational numbers

Activity

Begin by discussing and demonstrating on the board how rational numbers can be written in exponential form. Prepare crossword puzzle cards on powers of rational numbers. Instruct students to form groups for the activity. Distribute a crossword puzzle card to each group. Explain that each crossword puzzle contains expressions involving rational numbers in exponent form. Instruct the students to solve the crossword puzzle by simplifying the expressions and writing down the results in exponential form and vice versa. After the activity, ask each group to share one expression they simplified and how they arrived at the solution. Encourage them to write the solutions in their notebooks.

Extension Idea

Ask: What will be the reciprocal of (–5)2?

Say: The reciprocal of (–5)2 is    –1 5    2 .

Laws of Exponents

Learning Outcomes

Imagine Maths Page 289

Students will be able to use the correct law of exponents to simplify exponential expressions.

Teaching Aids

Cards with questions based on laws of exponents

Activity

Begin by explaining the laws of exponents to the class.

Divide the class into 3 groups. Provide each group with a set of cards with exponential expressions, with each set having problems on multiplication, division, and having powers of a power. Instruct the students that one member from each group will come up to the board to simplify an expression following and listing the correct exponent laws. Once one team member is done, he/she will go back to his/her seat, and the next member will come up with a different question card to solve it. The team members continue until all expressions are simplified or all members get a chance. The team that solves the most expressions correctly, wins. Conduct the activity such that the decorum of the class is maintained. Discuss their approaches in the class and encourage them to write the solutions in their notebooks.

Extension Idea

Ask: What is the error in the RHS of this equation: (56 × 52) = 258?

Say: The base in the RHS shouldn’t be 25. Instead the answer should be 58.

Expressing Large Numbers in the Standard Form

Learning Outcomes

Students will be able to express large numbers in standard form.

Teaching Aids

Imagine Maths Page 293

Puzzle cards with a large number in standard form written in the centre and options on 4 sides



Activity

Begin by introducing how large numbers are expressed in standard form. Instruct students to form groups for the activity. Distribute the puzzle cards with a large number in standard form written in the centre and options on 4 sides to each group. Tell them that each card consists of four numbers to be expressed in the standard form. Students have to find the standard form of each number and match the answers with the one written on the centre card. Instruct them to encircle the number that matches with the answer on the centre card. After the activity, discuss their answers on the class. If time permits, rotate the puzzles among groups and repeat the activity.

Answers

1. Exponential Form

Think and Tell

Exponent = n = 1

2. Exponents and Prime Factors

Do It Together

Prime factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Prime factors in exponential form = 28 × 3

3. Comparing Numbers in Exponential Form

Do It Together

1. 83 < 74

83 = 512

74 = 2401

3. 184 > 48

184 = 1,04,976

48 = 65,536

2. 105 < 97

105 = 100000

97 = 4,782,694

4. 213 < 321 213 = 9261

321 = 10,460,353,203

4. Powers of Rational Numbers

Do It Together 

2

5. Reciprocal of a Rational Number

6. Multiplying Powers

7. Dividing Powers

9. Simplifying Exponential Expressions

10. Expressing Large Numbers in the Standard Form

Sy etry 18

Learning Outcomes

Students will be able to:

identify and draw the reflection of a figure/shape on squared paper. find the angle of rotational symmetry in a shape or figure.

Alignment to NCF

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

Let’s Recall

Recap to check if students know how to determine the lines of symmetry of different figures. Ask students to solve the questions given in the Let’s Warm-up section.

Vocabulary

reflection symmetry: the symmetry that a shape has if it looks the same after being flipped or reflected over a line

rotational symmetry: the symmetry that a shape has if it looks the same after being rotated by a certain angle around a central point order of rotational symmetry: the number of times a shape looks the same after being rotated 360° around its centre

angle of rotation: the measure of the smallest angle through which a shape can be rotated to coincide with itself

Teaching Aids

Paper cutouts of different shapes; Pencils; Squared paper; Small mirrors; Circular spinners made from cardboard or paper; Cutouts of different geometric shapes; 360° protractor; Sheets of paper; Scissors

Chapter: Symmetry

Learning Outcomes

Imagine Maths Page 298

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

Teaching Aids

Paper cutouts of different shapes; Pencils; Squared paper; Small mirrors

Activity

Form groups for the activity. Distribute cutouts of various shapes to each group. Instruct them to experiment with these by drawing lines of symmetry on them. Encourage them to draw lines vertically, horizontally, or diagonally, aiming for identical sides on either side of the line. Ask questions around what these lines represent. Discuss their responses to bring out the meaning of reflectional symmetry. Now, distribute the mirrors and squared paper and ask each student to draw a line through the centre and a figure on one side of the line. They will then pass the sheet to the student next to them. They will then complete the figures by drawing the reflected parts. Then, they should use the mirrors to verify the symmetry of the reflections drawn on the paper.

Conclude with a brief class discussion to share observations and reinforce the concept of reflection symmetry.

Extension Idea

Ask: Can you think of one word that looks the same as its reflection?

Say: There are many words that look the same as their reflection. One such word is WOW.

Rotational Symmetry

Learning Outcomes

Students will be able to find the angle of rotational symmetry in a shape or figure.

Teaching Aids

Imagine Maths Page 304

Circular spinners made from cardboard or paper; Cutouts of different geometric shapes; 360° protractor; Sheets of paper; Scissors

Activity

Begin by asking a question like: Have ever noticed how certain snacks, like triangular chips, look identical from all corners? Discuss your answers. Form groups for the activity. Distribute spinners and various shape cutouts. Instruct students to rotate the shapes using the spinner and observe how the shapes align with themselves after specific rotations. Guide students to measure the angle between the original and rotated positions for each shape using protractors. Say that this is the angle of rotation for their shape. Now, ask them to count how many times a shape coincides with itself in one rotation,

defining it as the order of rotational symmetry. Repeat with various shapes, prompting the students to write the results in their notebooks.

Teacher tip: If the 360° protractor is not available, make one using a paper plate and markers.

Extension Idea

Ask: Create a mandala design with a rotational symmetry of order 4.

Say: You can make many designs with a rotational symmetry of order 4.

Answers

1. One Line of Symmetry

Think and Tell

The Ferris wheel has lines of symmetry.

3. Rotational Symmetry

Think and Tell

The order of rotational symmetry of the Ferris wheel is 2, and the angle of rotational symmetry is 180°.

Do It Together

Rotation through 120° Rotation through 240° Rotation through 360°

Do It Together

1.a. Yes b. Yes

2. a. b.

3. Figure (a) has one vertical line of symmetry. Figure (b) has one horizontal line of symmetry.

2. More Than One Line of Symmetry

Think and Tell

The Ferris wheel has 1 line of symmetry.

Do It Together

1.a. Yes b. Yes

2.a. 4 b. 3

3. lines of symmetry 4 lines of symmetry 3

120°

3 3 3

This figure fits onto itself times in a complete revolution. So, the order of rotational symmetry is  . Now, the angle of rotational symmetry = 360° order of rotational symmetry = 360° = .

4. Both Reflection and Rotational Symmetry

Do It Together

Number of lines of symmetry = 8

Order of rotational symmetry = number of lines of symmetry = 8.

Visualising Solid Shapes

Learning Outcomes

Students will be able to:

identify and list the features of 3-D shapes. identify and draw nets of 3-D shapes.

identify and draw oblique and isometric sketches of 3-D shapes. visualise 3-D shapes and draw their cross section and shadows.

Alignment to NCF

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

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

Let’s Recall

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

Vocabulary

lateral face: a face that is on the side of a 3-D shape instead of the top or bottom cross section: the shape we get on cutting straight through a 3-D shape or an object

Teaching Aids

Cards (with shape names, number of faces, edges and vertices); Cards (with the name of a 3-D shape); Cutouts of nets; Glue; Flash cards (with 3-D shapes and their dimensions written on them); Oblique drawing sheets; Isometric drawing sheets; Sheets of paper; Playdough; Flashlight

Chapter: Visualising Solid Shapes

Features of 3-D Shapes

Learning Outcomes

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

Teaching Aids

Cards (with shape names, number of faces, edges and vertices)

Activity

Divide the class into 4 groups. Create 4 sets of cards: one set with names of shapes, one with the number of faces, one with the number of vertices and one with the number of edges of the shapes, as in the example shown.

Make the groups stand in 4 separate lines.

Provide the groups with a set of flash cards each.

Imagine Maths Page 316

Instruct the students to move around and find the other students to match the shapes with their corresponding properties. Guide them to work together to match each shape with the correct number of faces, edges and vertices based on the information provided on their flash cards.

Nets of 3-D Shapes

Learning Outcomes

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

Teaching Aids

Cards (with the name of a 3-D shape); Sheets of paper

Activity

Imagine Maths Page 319

Begin the class by using one 3-D shape (e.g., a cube) and demonstrate how to draw a cutout of its net.

Distribute cards with the name of a 3-D shape written on each (pyramids and prisms) and sheets of paper.

Instruct the students to draw the net of the shape on their card on the paper. When they are done, instruct them to cut the net out carefully along the lines. Once they have cut it out, ask them to fold the cutouts along the indicated lines. Using glue, instruct them to carefully paste the edges together, ensuring a complete connection and making a solid shape.

Encourage the students to compare their folded shapes to the actual 3-D shapes.

Ask: Can you tell me how a prism is different from a pyramid?

Extension Idea

Ask: Which cube is formed by the given net?

Say: Option B is the cube formed by the given net when connected edge to edge.

Learning Outcomes

Students will be able to identify and draw oblique and isometric sketches of 3-D shapes.

Teaching Aids

Flash cards (with 3-D shapes and their dimensions written on them); Oblique drawing sheets; Isometric drawing sheets

Activity

5 units

1unit1unit

2 units 3 units

Begin by introducing the concept of oblique and isometric sketches. Ask the students to look at the examples in their books of oblique and isometric sketches and discuss the characteristics of each type of sketch and how they differ. Create flash cards with different shapes drawn on them and their dimensions marked as shown in the example. Instruct the students to work in pairs. Give each pair a flash card and provide each student in the pair with an isometric sheet and an oblique sheet, respectively. Guide them to carefully examine the image on the card, and then have each student draw an isometric sketch and an oblique sketch accordingly. Once they are done, have them swap sheets with their partners and review each other’s sketches.

Ask: Do both the sketches look the same? Why or why not?

5 units

Extension Idea

Ask: What makes an oblique sketch different from an isometric sketch?

Say: In an oblique sketch, there is more focus on the front side of an object, whereas in an isometric sketch, the focus is on the edge of an object.

Learning Outcomes

Students will be able to visualise 3-D shapes and draw their cross section and shadows.

Teaching Aids

Playdough; Sheets of paper; Flashlight

Activity

Create a 3-D shape with playdough, like a ball, and talk about its features.

Discuss cross sections of shapes by cutting one ball vertically and one ball horizontally, showing that both result in circles.

Now, instruct the students to work in groups of 4.

Distribute play dough to each group and ask them to make their own 3-D shapes. Have them cut their shapes vertically and horizontally. Instruct them to then draw the obtained cross sections in their notebooks. Encourage them to think about how the shape of the cross section changes with different angles and orientations.

Now, set up a flashlight to cast shadows on a wall. Show using the play dough ball how the shadow of the ball is circular.

Then, instruct each group to hold their shapes in front of the light to observe the shadows. Ask them to make sure they keep rotating the shapes to see the top, side, and front views of the shadows.

Ask questions like: Do you think that the cross section of an object is related to its shadow?

Extension Idea

Ask: Can you obtain a square shadow of a cuboidal box?

Say: Yes, if the light source is at an angle to one of its faces, a square shadow can be obtained.

Answers

1. Features of 3-D Shapes

Think and Tell

No, a cone is not a pyramid since a pyramid has a polygonal base but a cone does not.

Think and Tell

No, a cylinder is not a circular prism since a prism has polygonal bases but a cylinder does not. Do It

4. Cross Section and Shadows

Do It Together

Answers may vary. Sample answer: A ball A birthday cap Rectangle

Shadow of the cylinderShadow of the cone Top       Side Top       Side

2. Nets of 3-D Shapes

Do It Together

Number of faces =

2 faces rectangular faces

Net of a heptagonal prism

3. Oblique and Isometric Sketches

Do It Together

5. Views of 3-D Shapes

Solutions

Chapter 1

Let’s Warm-up

1. The numerator is always smaller than the denominator for proper fractions.  2. A mixed fraction has a whole number part and a fractional part.  3. 8 is the denominator in 5 8

4. The numerator in 17 19 is 17.  5. An improper fraction cannot be less than 1.

Do It Yourself

1A 1. a. 2 3 + 5 9

LCM of 3 and 9 is 9.

2 3 = 2 3 × 3 3 = 6 9

2 3 + 5 9 = 6 9 + 5 9 = 11 9 = 1 2 9

c.  7 13 –2 9

LCM of 13 and 9 is 117.

b.  9 22 + 2 11

LCM of 11 and 22 is 22.

2 11 = 2 11 × 2 2 = 4 22

9 22 + 2 11 = 9 22 + 4 22 = 13 22

7 13 = 7 13 × 9 9 = 63 117 and 2 9 = 2 9 × 13 13 = 26 117

7 13 –2 9 = 63 117 –26 117 = 37 117

d.

– 30

= 2 1 –30 73 = 2 × 73

e.  3 40 + 5 6

LCM of 40 and 6 is 120.

3 40 = 3 40 × 3 3 = 9 120 and 5 6 = 5 6 × 20 20 = 100 120

3 40 + 5 6 = 9 120 + 100 120 = 109 120

2. a.

c.

Let

13

15 = 26 30 and 1 6 = 5 30

13 15 > 1 6

So, 5 13 15 > 5 1 6

Thus, 2  2 3 + 3  1 5 > 7  2 4 − 2  1 3

b. 1 5 8 − 2 3 = 13 8 − 2 3 = 39 24 − 16 24 = 23 24

1 2 + 3 4 = 2 4 + 3 4 = 5 4 = 1 1 4

23 24 < 1 1 4

Thus, 1 5 8 2 3 < 1 2 + 3 4

c. 8 5 7 − 4 1 4 = 61 7 − 17 4 = 244 28 − 117 28 = 127 28 = 4 15 28

4 15 28 > 3 7 8

Thus, 8 5 7 − 4 1 4 > 3 7 8

d. 11 9 + 1 1 2 = 11 9 + 3 2 = 22 18 + 27 18 = 49 18 = 2 13 18

23 7 + 1 7 = 24 7 = 3 3 7

2  13 18 < 3 3

7 Thus, 11 9 + 1 1 2 < 23 7 + 1 7

5.a. 7 7 8 − 5 2 3 = 63 8 − 17 3 = 189 24 − 136 24 = 53 24 = 2  5 24

Thus, 5 2 3 is 2  5 24 less than 7 7 8

b. 4 1 4 − 2 2 5 = 17 4 − 12 5 = 85 20 − 48 20 = 37 20 = 1 17 20

Thus, 1 17 20 must be added to 2 2 5 to get 4 1 4

c. 1 4 + 1 8 9 + 4 5 = 1 4 + 17 9 + 4 5 = 45 180 + 340 180 + 144 180 = 529 180 = 2 169 180

Thus, the perimeter of the triangle is 2 169 180 cm.

d. 7 11 + 2 9 + 7 11 + 2 9 = 14 11 + 4 9 = 126 99 + 44 99 = 170 99 = 1  71 99

Thus, the perimeter of the rectangle is 1 71 99 cm.

e. 9 2 5 − 7 1 4 = 47 5 − 29 4 = 188 20 − 145 20 = 43 20

4 + 4 5 6 = 6 + 3 4 + 5 6 =

Thus, the difference of the sum of 2 3 4 and 4 5 6 and the difference of 9 2 5 and 7 1 4 is 5 13 30.

Word Problems 1. Fraction of the wall painted by Aisha = 3 5

Fraction of the wall painted by Aisha’s sister = 1 6

Fraction of the wall painted by both of them = 3 5 + 1 6 = 18 30 + 5 30 = 23 30

Thus, Aisha and her sister painted 23 30 of the wall together.

2. Mango juice drank by Ram = 3 4 bottle

Orange juice drank by Lisa = 2 9 bottle

Total juice drank by Ram and Lisa =

Thus, Ram and Lisa drank 35 36 of a bottle of juice.

3. Length of the rectangular field = 3 8 m

Width of the rectangular field = 2 5 m

Perimeter of the rectangular field =

32

40 = 62 40 = 1 22 40 = 1 11 20

Thus, the perimeter of the rectangular field is 1 11 20 m.

1B 1. Total water in the jug = 2 1 4 gallons

Water drank by Nishant each day = 4 7 gallons

Water drank by Nishant in 3 days = 4 7 + 4 7 +

7 = 12 7 gallons

Water left in the jug

Thus, 15 28 gallons of water is left in the jug.

2. Distance jogged by Vishwa = 3 1 4 km

Distance jogged by Muskan = 3 1 2 km

Distance jogged by Neeta = 2 1 6 km

Distance jogged by Avyaan = 3 km

Total

Thus, they jogged 11  11 12 km altogether.

3. Total time spent on studies = 5 2 3 hours

Time spent studying Maths and Science = 2 4 5 hours

Time spent studying other subjects

Thus, Shikha devotes 2  13 15 hours to other subjects.

4. Weight of the empty box = 20 6 7 kg

Weight of the iron ball = 7 2 3 kg

Total

+

kg

Thus, the total weight of the box is 28  11 21 kg.

5. Total length of the string = 30 yards

Length of the string used to tie a parcel = 11 2 5 yards

Length of the string used to tie a box = 6 1 4 yards

Total string used by Carlton = 11 2

String left with Carlton = 30 − 17  13 20 = 30 1 − 353 20 =

= 12  7 20

Thus, 12  7 20 yards of string is left with Carlton.

6. a. Most: Washing cars; Least: Washing utensils

Difference in the amount of water used = 25 2 5 −10 1 5 = 127 5 −

= 76 5 = 15 1 5

Thus, the difference between the amount of water for washing car and washing utensils is 15 1 5 litres.

b. Amount of water used in washing utensils = 10 1 5 litres

Amount of water used in washing cars = 25 2 5 litres

Total amount of water used = 25 2 5 + 10 1 5 = (25 + 10) + 2 5 + 1 5 = 35 3 5 litres

Thus, the total water used in washing utensils and cars is 35 3 5 litres.

c. Total amount of water used = 20 2 3 + 10 1 5 + 15 1 4 + 25 2 5 + 18 3 4 = (20

Thus, the total water used is 90  4 15 litres.

d. Water used for bathing = 20 2 3 litres

Water used for other activities = 18 3 4 litres

Amount of extra water used in bathing than in other activities

= 10  5 18

4. a. 1 5 of 1000 grams = 1 5 × 1000 grams = 200 grams

b. 1 6 of an hour = 1 6 of 60 minutes = 1 6 × 60 minutes = 10 minutes

c.  9 17 of 51 = 9 17 × 51 = 27

d. 3 1

of 2 5 = 3 1

e.  5 6 of a metre =

= 83 1

cm

c. 3 4 of 2 hours =

d. 2 7 of 7 weeks = 2 7 × 49 days = 14 days

e. 7 8 of 5 dozen = 7 8 × 60 = 420 8 = 105 2 × 1 12 = 4 3 8 dozen

f. 2 3 of a leap year = 2 3 of 366 days = 732 3 days = 244 days

Word Problems 1. Total oil with Ameena = 8 2 5 litres

Fraction of oil used by Ameena = 2 3

Total oil used by Ameena = 2 3 of 8 2 5 litres = 2 3 × 8 2 5 litres

= 2 3 × 42 5 litres = 84 15 litres = 28 5 litres = 5 3 5 litres

Thus, the total oil used by Ameena is 5 3 5 litres.

2. Total milk with Ram = 2 1 5 litres

Fraction of milk used by Ram = 1 3

a. Milk used by Ram = 1 3 of 2 1 5 litres = 1 3 × 11

= 11 15 litres

b. Milk left with Ram = 2 1

= 1  7 15 litres

3. Total fraction of pizza eaten by Simran and her friends = 1 3 4

Total weight of a pizza = 800 g

Total weight of pizza eaten by Simran and her friends = 1 3 4 × 800 g = 7 4 × 800 g = 1400 g

4. Total number of cookies with Pihu = 30 cookies

Fraction of cookies given to her sister = 2 3

Fraction of cookies given to neighbours = 1 5

Total fraction of cookies given = 2 3 + 1 5 = 10 15 + 3 15 = 13 15

Number of cookies given = 13 15 × 30 = 26 cookies

Thus, 26 cookies were distributed by Pihu.

5. Total number of books in the library = 1500

Fraction of science books = 1 5

Fraction of novels = 3 5

Fraction of English books = 1 − 1 5 − 3 5 = 1 5

a. Number of English books = 1 5 × 1500 = 300 books

b. Number of novels = 3 5 × 1500 = 900 books

c. Number of science books = 1 5 × 1500 = 300 books

1. a.

17 8 = 2 1 8 d. 2 10 13 ÷ 1 = 2 10 13

4. a. 3 2 5 ÷ 1 4 = 17 5 × 4 = 68 5 = 272 20

67 4 = 335 20

272 20 < 335 20

Thus, 3 2 5 ÷ 1 4 < 67 4

c. 1  7 10 ÷ 2 5 = 17 10 × 5 2 = 17 4

17 4 < 19 4

Thus, 1  7 10 by 2 5 < 19 4

b. 2 5 8 ÷ 1 2 3 = 21 8 ÷ 5 3 = 21 8 × 3 5 = 63 40

3  3 40 = 123 40

63 40 < 123 40

Thus, 2 5 8 ÷ 1 2 3 < 3  3 40

d. 1 1 3 ÷ 2  2 27 = 4 3 ÷ 56 27 = 4 3 × 27 56 = 9 14 5 7 = 10 14

9

14 < 10 14

Thus, 1 1 3 ÷ 2  2 27 < 5 7

5.a. Sum of 12 8 and 8 12 = 12 8 + 8 12 = 36 24 + 16 24 = 52 24 = 13 6

96 ÷ 13 6 = 96 1 × 6 13 = 576 13 = 44  4 13

Thus, 96 divided by the sum of 12 8 and 8 12 is 44  4 13

b.  4 5 ÷ 2 3 = 4 5 × 3 2 = 12 10 = 6 5

Thus, the product of 2 3 and 6 5 is 4 5 .

c. Perimeter of the equilateral triangle = 3 7 m

Length of each side = 3 7 m ÷ 3 = 3 7 × 1 3 = 1 7 m

Thus, the length of each side of equilateral triangle is 1 7 m.

Word Problems 1. Area of the rectangular field = 43 3 4 cm2

Breadth of the field = 12 1 2 cm

Length of the field = area ÷ breadth = 43 3 4 ÷ 12 1 2 = 175 4 ÷ 25 2

= 175 4 × 2 25 = 7 2 = 3 1 2 cm

Thus, the length of the rectangular field is 7 2 cm.

2. Total juice in the large container = 18 1 3 litres

Capacity of each small container = 1 5 6 litres

Total number of small containers that can be filled from a large container = 18 1 3 ÷ 1 5 6 = 55 3 ÷ 11 6 = 55 3 × 6 11 = 10

Thus, 10 small containers are required.

3. Total weight of the fruits = 5 9 kg

Number of packets = 15

a. Weight of each packet = 5 9 ÷ 15 = 5 9 × 1 15 = 1 27 kg

b. Number of packets sold = 5

Weight of the packets sold = 5 × 1 27 kg = 5 27 kg

4. Length of the piece of the cloth = 12 cm

Length of the small piece of cloth = 2 3 cm

Number of small pieces of cloths = 12 ÷ 2 3 = 12 × 3 2 = 18 pieces

Thus, there will be 18 small pieces of cloth.

1E 1. Let’s say there are x number of people at the restaurant.

Number of males at the restaurant = 3 7 x

Number of females at the restaurant = x − 3 7 x = 4 7 x

Difference between the number of females and the number of males = 28

4 7 x –3 7 x = 1 7 x = 28

x = 28 × 7 = 196

Number of females = 4 7 × 196 = 112

Thus, there are 112 females at the restaurant.

2. Let the total number of shirts be x

Number of printed shirts = 1 3 x

Number of remaining shirts = x –1 3 x = 2 3 x

Number of striped shirts = 1 4 × 2 3 x = 1 6 x

Number of plain shirts = x –1 3 x –1

1 2 x = 96 x = 192

Thus, there are 192 shirts in total.

3. Total number of players = 72

Number of players who play football = 3 8 of 72 = 27

Players who are not football players = 72 – 27 = 45

Thus, there are 45 non-football players.

4. Total length of ribbon with Kevin = 25 m

Length of ribbon given to his sister = 13 m

Length of ribbon left with Kevin = 25 m – 13 m = 12 m

Length of ribbon used for each project = 1 1 2 m

Number of projects = 12 ÷ 1 1 2 = 12 ÷ 3 2 = 12 × 2 3 = 8

Thus, Kevin completed 8 projects.

5. Fraction of money spent on house rent = 1 5

Fraction of money spent on food = 1 4

Fraction of money spent on other expenses = 2 5

Total fraction of money spent = 1 5 + 1 4 + 2 5 = 4 5 + 5 20 + 8 20 = 17 20

Fraction of money left = 1 – 17 20 = 20 20 –17 20 = 3 20

Money left with Rakul = ₹3000

3 20 of total money = ₹3000

Total money with Rakul = 20 3 × ₹3000 = ₹20,000

Thus, Rakul had ₹20,000 initially.

6. Rate of leakage for the first hole = 1 4 gallons per half an hour

Leakage by first hole per hour = 1 4 ÷ 1 2 = 1 4 × 2 = 1 2 gallons

Rate of leakage for the second hole = 1 5 gallons per quarter of an hour

Leakage by second hole per hour = 1 5 ÷ 1 4 = 1 5 × 4 = 4 5 gallons

Total leakage by 2 holes per hour = 1 2 + 4 5 = 5 10 + 8 10 = 13 10 gallons

Total capacity of the cistern = 500 gallons

Time required to empty the complete cistern = 500 ÷ 13 10

= 500 × 10 13 = 5000 13 = 384  8 13 hours

Thus, the two holes will take 384  8 13 hours to empty the cistern.

Chapter Checkup 1. a. Any fraction divided by 0 is not defined. True b. The associative law is applicable to the division of fractions. False c. The reciprocal of 1 is 1. True d. The commutative law is applicable to the subtraction of fractions. False

4. a. The product of 5 6 and its reciprocal is 1.

b. A fraction is said to be in its lowest form when the HCF of the numerator and denominator is 1.

c. Dividing by 1 2 is the same as multiplying by 2.

d. The reciprocal of proper fractions is always greater than 1. 5. a.

b. 2 3 of a day = 2 3 of 24 hours = 2 3 × 24 hours = 16 hours

c. 5 1 5 of a year = 26 5 of 365 days = 26 5 × 365 days = 1898 days

d. 3 4 of a metre = 3 4 of 100 cm = 3 4 × 100 cm = 75 cm

8.a.

=

c. 2 6

d. 5 3

9. a.

2 1 3 ÷ 4 1 6 = 7 3 ÷ 25

= 7 3 × 6 25 = 14 25

Thus, product of 3 1 5 and 2 1 4 is more.

b. 25 × 4 75 = 4 3 = 1 1 3

3 1 4 + 5 4 5 = 8 + 1 4 + 4 5 = 8 + 5 20 + 16 20 = 9  1 20

Thus, product of 25 and 4 75 is less.

c. 4 ÷ 2 7 = 4 × 7 2 = 14

2

5 × 15 = 6

14 – 6 = 8

Thus, the difference is 8.

d. 9 × 1 1

9 1

2 + 7 1

2 = 9 × 3 2 = 27 2 = 13 1 2

3 = 16 + 1 2 + 1 3 = 16 5 6

13 1 2 + 16 5

6 = 29 + 1 2 + 5 6 = 29 + 8 6 = 30 1 3

Thus, the sum is 30 13 .

10. a.

covered in 1 1 4 hours = 48 3 4 miles Distance

Distance

in 2 hours = 39 miles × 2 = 78 miles

Thus, the train covers 78 miles in 2 hours.

b.  2 13 of x = 26

x = 26 × 13 2 = 169

c. 22 2

200

9 × x = 1800

9 x = 1800

x = 1800 × 9 200

x = 81

d. Let the unknown number be x.

x × 2 3 = 3 5

f. Let the number be x

x × 7  47 55 = 21 3

e. Let the

6

7

x

x × 432 55 = 108 5

x = 108 5 × 55 432 = 11 4 = 2 3 4

Word Problems 1. Fraction of cake with Miya = 3 5

a. Fraction of cake eaten by Miya = 2 3 of 3 5 = 2 3 × 3 5 = 6 15 = 2 5

b. Fraction of cake left with Miya = 3 5 –2 5 = 1 5

2. Turns completed by the planet about its axis = 1 3 5 turns

a. Turns completed in a week = 1 3 5 × 7 = 8 5 × 7 = 56 5 = 11 1 5 turns

b. Number of days in a fortnight = 14 days

Number of turns completed by the planet = 1 3 5 × 14 = 8 5 × 14 = 112 5 = 22 2 5 turns

3. Let the total capacity of the tank be x litres.

Fraction of water inside the tank initially = 4 5

Quantity of water drawn from the tank = 65 litres

Fraction of water left inside the tank = 7 12

4 5 x –7 12 x = 65

(48 – 35)x 60 = 65 x = 300 litres

Water left inside the tank = 7 12 x = 7 12 × 300 = 175 litres

Thus, 175 litres of water is left inside the tank.

4. Speed for the first 5 hours = 4 1 5 km per hour

Distance travelled in the first 5 hours = 5 × 4 1 5 = 5 × 21 5 = 21 km

Speed for the next 3 hours = 3 1 4 km per hour

Distance travelled in the next 3 hours = 3 × 3 1

Total distance travelled in 8 hours = 21 + 39

=

= 3 × 13 4 = 39 4 hour

= 123 4 = 30 3 4 km

Thus, the total distance covered in 8 hours is 30 3 4 km.

5. Fraction of distance covered by train = 1 3

Fraction of distance covered by bus = 2 5

Total distance covered by train and bus = 1 3 + 2 5 = 5 15 + 6 15 = 11 15

Fraction of distance covered walking = 1 – 11 15 = 4 15

Distance covered by walking = 16 km

4 15 x = 16

x = 16 × 15 4 = 60 km

a. Distance covered by the bus = 2 5 × 60 = 24 km

b. Distance covered by the train = 1 3 × 60 = 20 km

c. Total distance covered = 60 km

6. Donation for every 5 bracelets of type A = 2 5 of the proceeds

Donation for every 3 bracelets of type B = 3 7 of the proceeds

Donation for every 4 bracelets of type C = 5 8 of the proceeds

Number of type A bracelets sold = 35

Cost of each type of A bracelet = ₹70

Number of type B bracelets sold = 21

Cost of each type of B bracelet = ₹90

Number of type C bracelets sold = 28

Cost of each type of C bracelet = ₹120

Donation by selling type A bracelet = 35 × 70 × 2 5 = ₹980

Donation by selling type B bracelet = 21 × 90 × 3 7 = ₹810

Donation by selling type C bracelet = 28 × 120 × 5 8 = ₹2100

Total amount of donation = ₹980 + ₹810 + ₹2100 = ₹3890

Thus, the total amount donated to the animal shelter from the proceeds of each type of bracelet sale is ₹3890.

Chapter 2

Let’s Warm-up

1.  351.254 × 100 = 35,125.4  2. 45.123 × 1000 = 45,123

3. 124.12 ÷ 10 = 12.412     4. 39.145 ÷ 100 = 0.39145

5. 587.2 ÷ 1000 = 0.5872

Do It Yourself

2A 1. a. 223.25 and 12.9 HTO.th 1

223.25 +12.9 236.15

c. 128.2 and 236.54 HTO.th

1

128.20 +236.54

364.74

e. 136.23, 556.14 and 25.125 HTO.thth

1 1

136.230 +556.140 +025.125 717.495

2. a. 322.36 from 438.2 HTO.th

c. 118.5 from 732.55 HTO.th

b. 315.36 and 218.234 HTO.thth 1

315.360 +218.234 533.594

d. 332, 127.456 and 122.26 HTO.thth

1 1

332.000

f. 152.214, 235.3 and 365.28 HTO.thth

1 1

152.214 +235.300 +365.280 752.794

b. 132.89 from 325.566 HTO.thth 2

192.676

d. 438.236 from 752.23 HTO.thth 4

e. 856.18 from 998.856 HTO.thth

7 15 998.856 ̵856.180 142.676

3. Let the unknown terms be x

f. 1065.235 from 1189.15 ThHTO.thth

8

a. 236.45 + x = 568.2 x = 568.2 – 236.45 = 331.75

b. 514.7 – x = 241.23 x = 514.7 – 241.23 = 273.47

c. x + 321 = 512.47 x = 512.47 – 321 = 191.47

d. x − 475 = 251.487 x = 251.487 + 475 =726.487

e. 785.5 + 324.21 = 1109.71

f. 987 – 562.26 = 424.74

4. Let x be added to 389.265 to get 625.56

389.265 + x = 625.56

x = 625.56 − 389.265 = 236.295

5. Let x be subtracted from 235.125 to get 126.24

235.125 – x = 126.24

x = 235.125 – 126.24 = 108.885

6. 126.256 + 325.46 – 195.264 +

326.152 – 256.369 = 126.256 + 325.46

+ (–195.264) + 326.152 + (–256.369)

Adding the positive numbers =

126.256 + 325.46 + 326.152 = 777.868

Adding the negative numbers = (−195.264) + (−256.369) = −451.633

777.868 – 451.633 = 326.235

7. Sum of 356.21 and 585.236

= 356.21 + 585.236 = 941.446

Difference of 829.23 – 469.251 =

829.23 – 469.251 = 359.979

941.446 – 359.979 = 581.467

8. Total ingredients = Flour + Sugar + Butter

Total ingredients = 3.75 cups + 2.5 cups + 1.25 cups = 7.5 cups

Therefore, 7.5 cups of ingredients are needed in total for the recipe.

9. a. Total distance travelled = 82.75 + 80.5 + 78.25 + 81.5 +

79.75 + 84.5 + 76.55 = 563.80 Km

b. Distance travelled on the last three days = 79.75 + 84.5 + 76.55 = 240.8 km

Distance travelled in the initial 4 days = 82.75 + 80.5 + 78.25 + 81.5 = 323 km

Difference in distance travelled = 323 – 240.8 = 82.2 km

Word Problems 1. Sweets bought by Ravi = 10.5 Kg

Weight of the empty box = 1 Kg 350 g = 1.350 Kg

Weight of sweets + Weight of empty box = Total weight 10.5 kg + 1.35 kg = 11.85 kg

Therefore, the total weight of the sweets and the box is 11.85 kg.

2. Initial weight of coffee = 2.5 kg

Weight of coffee used = 1.75 kg

Initial weight of coffee – Weight of coffee used = Remaining weight of coffee

2.5 kg − 1.75 kg = 0.75 kg

Therefore, Vineet has 0.75 kg of coffee left.

3. Amount of fuel in the morning = 56.75 gallons

Fuel left after the trip = 32.5 gallons

Fuel used = difference in fuel = Initial fuel – Remaining fuel 56.75 gallons – 32.5 gallons = 24.25 gallons

Therefore, the delivery truck used 24.25 gallons of fuel during the journey.

4. Weight of sugar bought = 7 kg 450 g = 7.450 kg

Weight of rice bought = 2.35 kg

Weight of flour bought = 5 kg 700 g = 5.700 kg

Total weight = 7.450 kg + 2.35 kg + 5.700 kg = 15.5 kg

Therefore, Swati bought a total of 15.5 kg of groceries.

5. Hottest month is June = 20.2 °C

Coldest month is February = 5.5 °C

Now the difference = 20.2 °C – 5.5 °C = 14.7 °C

2B 1. Height of a palm tree = 42.68 feet

Height of mango tree 15.23 feet shorter than the palm tree = 42.68 feet – 15.23 feet = 27.45 feet

Total height of both trees = 42.68 feet + 27.45 feet = 70.13 feet

2. Ronita's current monthly expense for cable and internet = ₹635.80.

The new plan she is considering offers internet for ₹452.75 and cable for ₹126.55.

New plan expense = ₹452.75 + ₹126.55 = ₹579.30.

Monthly savings = Current expense – New plan expense

Monthly savings = ₹635.80 – ₹579.30 = ₹56.50

Therefore, Ronita would save ₹56.50 every month by switching to the new plan.

3. Let x be the size of the third application.

We know that the total size of the three applications is 986.48 KB, and the size of the first two applications is 256.36 KB and 485.84 KB, respectively.

x + 256.36 + 485.84 = 986.48 x = 986.48 – 256.36 – 485.84 x = 244.28 KB

Therefore, the third application is 244.28 KB.

4. Amount of sugar donated = 35.65 kg

Amount of flour donated = 78 kg 540 g = 78.54 kg

Amount of rice donated = 17 kg 320 g rice = 17.32 kg

Total weight of the items donated was 142.63 kg.

The amount of pulses donated = x

Total weight = 35.65 kg + 78.54 kg + 17.32 kg + x = 142.63 kg

x = 142.63 – 131.51 = 11.12 kg

Therefore, the weight of the pulses donated was 11.12 kg.

5. Amount saved in the months of June and July = ₹3652.45 and ₹4578.58

Total savings in June and July = ₹3652.45 + ₹4578.58 = ₹8231.03.

Amount spent on purchasing bicycle and helmet = ₹2468.75 and ₹689.48

Total amount spent on bicycle and helmet = ₹2468.75 + ₹689.48 = ₹3158.23.

Total amount left with Sam at the end of July = ₹8231.03 – ₹3158.23 = ₹5072.8.

6. Amount of juice in jug A = 2 L + 125 mL = 2.125 L

Amount of juice in jug B = 1 L + 350 mL = 1.350 L

Total juice in jug A and jug B = 2.125 L + 1.350 L = 3.475 L

Remaining capacity = Total capacity – Total amount of juice = 4.75 L – 3.475 L = 1.275 L

Therefore, 1.275 L more juice can be poured into both jugs.

7. The difference in the amount deposited and withdrawn =

Total amount deposited in four months – total amount withdrawn in four months

= (₹10,512.25 + ₹8523 + ₹9684.5 + ₹10,365) – (₹5423.58 + ₹6842.75 + ₹3754.36 + ₹4365.45)

= 390,84.75 – 20,386.14 = ₹18,698.61

Word Problems 1. Rita’s distance: X to Y: 14.05 km, Y to Z: 12.7 km

Total distance travelled by Rita = 14.05 km + 12.7 km = 26.75 km

Aman’s distance: X to A: 29.3 km A to Z: 21.45 km

Total distance travelled by Aman = 29.3 km + 21.45 km = 50.75 km

Difference in distance: 50.75 km – 26.75 km = 24 km

Therefore, Aman travelled 24 km farther than Rita.

2. Price of the coffee = $1.46

Price of the waffle = $0.85

Total amount spent = $2.31

Change received if paid $10 = $10.00 − $2.31 = $7.69

Hence, Sam will receive a change of $7.69.

e.

×523 15675 104500

2612500

2732675

52.25 × 5.23 = 273.2675

g.  32215 ×224 128860

644300

6443000

7216160

322.15 × 2.24 = 721.616

2. Given that, 365 × 124 = 45,260

f.  12256 ×521

12256

245120

6128000

6385376

122.56 × 5.21 = 638.5376

h.  2116 ×1283

6348

169280

423200

2116000

2714828

21.16 × 12.83 = 271.4828

a. 36.5 × 124 = 4526 b. 3.65 × 12.4 = 45.26

c. 36.5 × 1.24 = 45.26 d. 3.65 × 1.24 = 4.526

d.

g. 23.85 ÷ 1.8

= 238.5 ÷ 18

= 13.25

j. 115.542 ÷ 4.9

= 1155.42 ÷ 49

= 23.58

23.58

49 1155.42 – 98 175

– 147 284

– 245 392

– 392

0

h. 487.36 ÷ 3.2 = 4873.6 ÷ 32

i. 1117.5 ÷ 12.5 = 11175 ÷ 125 = 89.4  89.4 125 11175.0

– 1000 1175

– 1125 500 – 500 0

k. 2039.96 ÷ 5.2 = 20399.6 ÷ 52 = 392.3 392.3

52 20399.6 – 156 479

– 468 119

– 104 156 – 156    0

4. Let x be the unknown value,

a. 68.527 × x = 23.63 x = 23.63 68.527 = 0.345

b. x × 1.6 = 52 x = 52 1.6 = 32.5

l. 4756.05 ÷ 58.5 = 47560.5 ÷ 585

= 81.3  81.3

585 47560.5 – 4680 760

– 585 1755

– 1755 0

c. 2177.35 ÷ 35 = 62.21 d. 74.2 × 12.3 = 912.66

e. 984.576 ÷ x = 25.6 x = 984.576 25.6 = 38.46

f. 616.54 ÷ 14.5 = 42.52

5. The product of two decimal numbers = 3538.29

One number = 41.5

Hence, 41.5 × _____ = 3538.29

Second number = 3538.29 41.5 = 85.26

6. Length of side of the given square = 124.75 cm

Area of square = side2 = 124.752 = 15,562.5625 cm2

Perimeter of square = 4 × side = 4 × 124.75 = 499 cm

7. Cost of 1 litre of milk = ₹54.5

Cost of 45 3 4 or 183 4 litre of milk = ₹54. 5 × 183 4 = ₹2493.375

8. Length of the cloth required to make 1 shirt = 2.35 m

Length of cloth required to make 48 shirts = 2.35 × 48 = 112.8 m

9. Length of the ribbon = 30 m 80 cm = 30.8 m

Length of each piece if divided in 11 equal pieces = 30.8 11 = 2.8 m

10. Distance covered by the car in 8 hours = 452.8 km

Distance covered by the car in 1 hour = 452.8 8 = 56.6 km

Distance covered by the car in 2 4 5 hours or 14 5 hours = 56.6 × 14 5 = 158.48 km

Word Problems 1. Cost of rice = ₹52.75 per kilogram.

Weight of rice bought = 12 kg 450 g = 12.450 kg

Total cost of rice bought = 12.450 × ₹52.75 = ₹656.7375

2. 2 m 25 cm or 2.25 m of cloth is required for one pair of pants.

Number of pants that can be stitched with a piece of 20.25 m long cloth = Total length of cloth Length per pair = 20.25 m 2.25 m = 9

Therefore, 9 pairs of pants can be stitched with a piece of 20.25 m long cloth.

3. Weight of each bag of rice = 97.5 kg

Total weight of all the bags = 682.5 kg

Number of bags = Total weight weight per bag = 682.5 Kg 97.5 Kg = 7

Therefore, the grocery owner bought 7 bags of rice.

4. Total area to be divided = 875.25 square metres

Number of people among whom the area needs to be divided = 4

Area of plot each per person receive = Total area 4 = 875.25 4 = 218.8125 square metres

2D 1. Amount paid for sugar = 7.65 kg × ₹38.5/kg = ₹294.525

Amount paid for rice = 2.485kg × ₹62.75/kg = ₹155. 93375

Amount paid for flour = 5.725 kg × ₹45/kg = ₹257.625

Total amount = amount for sugar + amount for rice + amount for flour = ₹294.525 + ₹155. 93375 + ₹257.625 = ₹708.08375

Hence, the total amount to be paid by Suhani = ₹708.08375

2. Total amount of milk left = 1 1 2 litres = 1.5 litres

Milk in 8 bottles = 25.5 litres – 1.5 litres = 24 litres

8 bottles contains = 24 litres milk

1 bottle = 24 8 = 3 litres

Hence, the volume in each bottle = 3 litres

3. To find the cost of fencing, we first need to find the perimeter of the garden.

Perimeter of the garden = 2 [length + breadth] = 2 [7.25 + 5.8] =

2 [13.05] = 26.10 m

Cost of fencing 26.10 m at ₹58.75 per metre = 26.10 × ₹58.75 = ₹1533.375

4. Cost of 2 pairs of shoes = 2 × ₹525.75 = ₹1051.50

Cost of a pair of 3 t-shirts = 3 × ₹263.50 = ₹790.50

Cost of 5 pairs of socks = 5 × ₹50.25 = ₹251.25

Cost of a bag = ₹755

Total cost of the items purchased = ₹1051.50 + ₹790.50 + ₹251.25 + ₹755 = ₹2848.25

5. Total cost of the music system = ₹25,526.75

Amount paid as down payment = ₹10,250

Rest of the amount to be paid = ₹25,526.75 − ₹10,250 = ₹15,276.75

Number of installments = 15

Amount paid in each installment = 15276.75 15 = ₹1018.45

6. Length of the cloth used for 1 banner = 2 m 75 cm = 2.75 m

Length of the cloth used for 8 banners = 8 × 2.75 m = 22 m

Length of the cloth used for 1 tablecloth = 1.55 m

Length of the cloth used for 3 tablecloths = 3 × 1.55 m = 4.65 m

Total length of the cloth used = 22 m + 4.65 m = 26.65 m

Cost of 1 m of cloth = ₹75.5

The total cost of the cloth used = 26.65 m × ₹75.5 = ₹2012.075

7. Total cost of 25 notebooks at ₹45.65 = ₹45.65 × 25 = ₹1141.25

Total cost of 3 boxes of pens at ₹520.75 = ₹520.75 × 3 = ₹1562.25

Total cost of 2 boxes of markers at ₹368.15 = ₹368.15 × 2 = ₹736.30

Total cost of all the items = ₹1141.25 + ₹1562.25 + ₹736.30 = ₹3439.80

Total amount given to shopkeeper = Total cost of all the items + Amount returned by shopkeeper = ₹3439.80 + ₹560.2 = ₹4000. Hence, Sharvil gave ₹4000 to the shopkeeper.

8. Cost of 7 kg of tomatoes = ₹250.25

Cost of 1 kg of tomatoes = ₹250.25 7 = ₹35.75

Cost of 4 1 2 or 4.5 kg of tomatoes = 4.5 × ₹35.75 = ₹160.875

9. To centre the figure on the page, Saumya needs to leave an equal margin on both ends.

Margin on each end = Page width – Figure width 2 = 22.63 cm – 12.75 cm 2 = 4.94 cm

Therefore, Saumya should leave a margin of 4.94 cm on each end so that the figure is centred on the page.

Word Problem 1. Cost of one pair of gloves = 234.75 3 = ₹78.25

Cost of one pair of socks = 165 2 = ₹82.5

1 pair of gloves and 1 pair of socks can be bought for ₹78.25 + ₹82.5 = ₹160.75

Number of pairs of gloves and socks that can be bought for ₹1286 = 1286

160.75 = 8 pairs

Therefore, Krishvi can buy 8 pairs of gloves and 8 pairs of socks for ₹1286.

Chapter Checkup

1. a. 122.56 + 36.2  HTO.th

122.56

c. 16.523 + 132.29 HTO.thth

321.4 + 12.63

e. 536.3 + 311.236

g. 525.236 + 226.21 HTO.thth

1 525.236 +226.210 751.446

2. a. 132.56 − 45.2

HTO.th

0 12 12 132.56 ̵̵045.20 087.36

c. 267.523 – 159.47

HTO.thth

2 5 17 4 12 267.523 ̵̵159.470 108.053

e. 336.8 – 285.123 HTO.thth

2 13 7 9 10 336.800 ̵̵285.123 051.677

g. 658.412 – 247.48

HTO.thth

7 13 11 658.412 ̵̵247.480 410.932

3. a. 212.5 and 35 2125 ×35 10625 63750 74375

212.5 × 35 = 7437.5

c.

e. 136.3 and 125.2 1363 ×1252 2726 68150 272600 1363000 1706476

136.3 × 125.2 = 17,064.76

h. 756.59 + 413.5 + 623.12 ThHTO.th

1 1 1 756.59 +413.50 +62312 1793.21

b. 225.3 – 152.35

HTO.th

1 12 4 12 10 225.30 ̵̵152.35 072.95

d. 347.14 − 178.5

HTO.th

2 13 16 11 347.14 ̵̵178.50 168.64

f. 586.4 – 387.25

HTO.th

4 17 16 3 10 586.40 ̵̵387.25 199.15

h. 745.14−398.475

HTO.thth

6 13 14 10 13 10 745.140 ̵̵398.475 346.665

b. 110.52 and 21.5 11052 ×215 55260

110520 2210400 2376180

110.52 × 21.5 = 2376.18

315.13 ×

=

f. 624.82 and 32.55 62482 ×3255 312410

3124100

12496400

187446000

203378910

624.82 ×32.55 = 20,337.891

g. 312.1 and 15.24

3121 ×1524 12484

62420

1560500

3121000

4756404

312.1 × 15.24 = 4756.404

h. 362.12 and 213.5 36212 ×2135 181060

1086360

3621200

72424000

77312620

362.12 × 213.5 = 77,312.62 4. a.

9. Length of the garden = 17.36 m

Width of the garden = 13.52 m

Area of the garden = length × width = 17.36 × 13.52 = 234.7072 m2

Perimeter of the garden = 2 × (length + width) = 2 × (17.36 + 13.52) = 61.76 metres

10. Given: Average growth rate of hair = 1cm 8 mm per month = 1.8 cm/month

Growth of hair strand in 3 years or 36 months = 1.8 × 36 months = 64.8 cm

So, after three years, the length of the strand will be 64.8 cm.

11. Amount earned per hour for the first 40 hours = ₹523.65

Amount earned in the first 40 hours = ₹523.65 × 40 = ₹20,946

Amount earned per hour beyond 40 hours = 1 1 2 or 1.5 times of the actual amount = ₹523.65 × 1.5 = ₹785.475

Number of hours worked beyond 40 hours = 52 1 2 – 40 = 12 1 2 hours = 12.5 hours

Amount earned in 12.5 hours at ₹785.475 per hour = ₹9818.4375

Amount earned in 52 1 2 hours of work = ₹20,946 + 9818.4375 = ₹30,764.4375

e. 100.32 ÷ 80 = 1.254

4 498.8

f. 42.7 ÷ 14 = 3.05

g.  3.024 ÷ 0.36 = 302.4 ÷ 36 = 8.4

36 302.4

h. 564.975 ÷ 15.5 = 5649.75 ÷

155 = 36.45  36.45

155 5649.75 – 465

5. Total marks = 600; marks scored by Aarav = 452.65

Marks lost = 600 – 452.65 = 147.35

6. Mass of powdered milk in each tin = 0.625 kg

Total number of tins in a carton = 12

Total mass of powdered milk in a carton = 0.625 kg/tin × 12 tins = 7.5 kg

7. Weight of Sack A= 125.35 kg

Weight of Sack B = Weight in Sack A + 25.65 kg = 125.35 kg + 25.65 kg = 151 kg

Weight of Sack C = Weight in Sack B − 13.49 kg = 151 kg − 13.49 kg = 137.51 kg

Total weight of 3 sacks = 125.35 kg + 151 kg + 137.51 kg = 413.86 kg

Therefore, the total weight of rice in all three sacks is 413.86 kg.

8. Amount of milk consumed daily = 3.25 litres

Number of days in a year = 365 days

Amount of milk consumed in a year = 3.25 × 365 = 1186.25

Therefore, the total milk consumption in a year will be 1186.25 litres.

12. a. Cost of 3 vegetable bowls at ₹155.26 per bowl = 3 × ₹155.26 = ₹465.78

Cost of 2 apple pies at ₹62.37 per apple pie = 2 × ₹62.37 = ₹124.74

Cost of 4 milk shakes at ₹ 89.49 per milk shake = 4 × ₹89.49 = ₹357.96

Total amount spent = ₹465.78 + ₹124.74 + ₹357.96 = ₹948.48

b. If the family paid ₹1000 and their total cost was ₹948.48, their change would be: Change = ₹1000 − ₹948.48 = ₹51.52

Word Problems

1. Weight of Manya’s papaya

= Weight of Rani’s watermelon

3

Weight of the papaya = 4.2 Kg 3 = 1.4 kg

2. Weight of the baby elephant before 2 years = 118.99 kg

Increase in weight after 2 years = 109.85 kg

Weight after 2 years = 118.99 kg + 109.85 kg = 228.84 kg

3. Amount of sugar bought = 8.6 kg

Amount of sugar left after pouring equally in 5 bottles = 0.35 kg

Amount of sugar poured in 5 bottles = 8.6 – 0.35 = 8.25 kg

Amount of sugar poured in each bottle = 8.25 5 = 1.65 kg

Therefore, the mass of sugar in 1 bottle is 1.65 kg.

4.a. Amount saved during the five weeks = ₹1058.25 + ₹1254 + ₹1365.78 + ₹1563.75 + ₹1026.65 = ₹6268.43

b. Lowest amount saved = ₹1026.65

Highest amount saved = ₹1563.75

Difference between the lowest and highest amount saved = ₹1563.75 – ₹1026.65 = ₹537.1

5. Cost of 1 dozen or 12 cupcakes = ₹519.60

Cost of 1 cupcake = ₹519.60 12 = ₹43.30

Amount paid for 5 cupcakes = 5 × ₹43.30 = ₹216.50.

6. Distance travelled on day 1 = 150.25 km

Distance travelled on day 2 = 175.50 km

Total distance the family needs to cover = 450.75 km

Distance the family needs to drive to reach the destination = 450.75 km – 150.25 km + 175.50 km = 125 km

Amount of fuel required to travel 11.5 km = 1 litre

Amount of fuel required to travel 450.75 km = 450.75 11.5 = 39.2 Litres

Chapter 3

Let’s Warm-up 1. 25 < 89  2. –32 < –16   3. –12 < 5

4. 15 > 8 > 0 > –12 > –36  5. –14 < –9 < –1 < 13 < 20

Do It Yourself

3A 1. a. 5 + (–9) = −4 b. 15 + 14 = 29 c. –28 + 59 = 31

d. (–23) + 12 = −11 e. 32 + (–122) = −90 f. –136 + (–25) = −161

g. –152 + 365 = 213 h. 158 + (–125) = 33

2.a. –7 + 3 = −4 b. 12 + (–5) = 7 c. –19 + (–14) = −33

d. 20 + (–30) = −10 e. –69 + 81 = 12 f. (–123) + (–235) = −358

g. 565 + (–125) = 440 h. (–485) + (–325) = −810

3.a. –9 – (8) = −9 + (−8) = −17 b. 17 – (–12) = 17 + (12) = 29

c. –32 – 25 = −32 + (−25) = −57 d. –20 – 38 = −20 + (−38) = −58

e. –98 – (–36) = −98 + (36) = −62 f. 223 – 54 = 223 + (−54) = 169

g. –182 − 136 = −182 + (−136) = −318

h. 156 – (–214) = 156 + (214) = 370

4. To find the missing values, we will use the rule of transposition.

a. 18 – 12 = 6 b. 10 – 17 = −7 c. 956 – ____ = 422

Moving 422 to the LHS, we get 956 – 422.

956 – 422 = 534

So, 956 – 534 = 422

d. 215 – 136 = 79 e. –548 + ____ = –267

Moving –548 to the RHS, we get –267 + 548

–267 + 548 = 281

So, –548 + 281 = –267

f. ____ – 192 = 564

Moving – 192 to the RHS, we get 564 + 192  564 + 192 = 756

So, 756 – 192 = 564

g. 705 + ____ = 1202

Moving 705 to the RHS, we get 1202 – 705 1202 – 705 = 497.

So, 705 + 497 = 1202.

h. 815 – (–125) = 940

5.a. The sum of –5 and 8 is 3.

b. Subtracting 15 from –20 gives −35

c. The absolute value of –10 is 10

d. If I owe ₹50 and I receive ₹30, my new balance is −₹20

e. If I have a gain of ₹310 and a loss of ₹125, my net result is ₹185

f. Adding the additive inverse of –8 to –8 gives 0

6. –3652 + 5864 = 2212

2212 – 2369 = −157

7. a. The commutative property of addition holds true for integers. True

b. The associative property of subtraction holds true for integers. False

c. The commutative property of subtraction holds true for integers. False

d. The associative property of addition holds true for integers. True

e. For an integer a, a – 0 = 0. False

8. a. (21) + (13) = 13 + (21); Property used – Commutative property

b. −3 + 0 = –3; Property used – Existence of additive identity

c.  (–35 + 13) + (−16) = –35 + (13 + (–16)); Property used –Associative property

d. (–23) – 0 = –23; Property used – Subtraction property of zero

Word Problems

1. Number of items available with the store = 754

Number of items received in shipment = 325

Total items = 754 + 325 = 1079

2. Altitude climbed by the hiker = 1200 m

Altitude descended = 500 m = −500 m

Change in altitude experience = 1200 + (−500) = 700 m

3. Points scored in the first quarter = 80

Points scored in the second quarter = −25

Points scored in the third quarter = 15

Points scored in the fourth quarter = 25 + 10 = 35

Total points scored = 80 + (−25) + 15 + 35 = 105

3B 1. a. 12 × 15 = 180 b. –11 × 6 = −66 c. –21 × 13 = −273

d. –14 × (–6) = 84 e. 25 × 0 = 0 f. 56 × (–1) = −56

g. –15 × (–24) = 360 h. 16 × (–8) = −128

2. a. –11 × 5 × (–2) = −55 × −2 = 110

b. –16 × 6 × (–18) = −96 × −18 = 1728

c. 51 × (–26) × 6 = −1326 × 6 = −7956

d. 12 × (–1) × (–8) = −12 × −8 = 96

e. 25 × (–12) × 3 = −300 × 3 = −900

f. 56 × (–12) × 0 = −672 × 0 = 0

g. –45 × (–4) × (–13) = 180 × −13 = −2340

h. 11 × 0 × (–28) = 0 × −28 = 0

3. a. –23 × 15 × (–2) × 5 = −345 × −10 = 3450

b. –8 × (–16) × 5 × (–11) = 128 × −55 = −7040

c. 13 × (–21) × (–26) × 6 = −273 x −156 = 42,588

d. –6 × (–12) × –3 × (–10) = 72 x 30 = 2160

e. 34 × (–12) × 20 × 17 = −408 × 340 = −1,38,720

f. –23 × (–24) × (–43) × 9 = 552 × −387 = −2,13,624

4.a. Odd number of negative integers = negative sign

b. Even number of negative integers = positive sign

c. Even number of negative integers = positive sign

d. Odd number of negative integers = negative sign

e. Odd number of negative integers = negative sign

5.a. 36 × 0 = 0 b. (–123) × 1 = –123 c. (–12) × (56) = 56 × (–12)

d. 56 × (–12 + 23) = 56 × (–12) + 56 × 23 e. −2198 × (–1) = 2198

f. (123 × 143) × (–36) = 123 × (143 × (–36))

Word Problems 1. Rate of descending = 55 feet per minute = −55

Position of the submarine after 8 minutes = −55 × 8 = −440 = 440 feet below sea level.

2. Profit on sale of potato chips = ₹8

Loss on sale of packets of biscuits = ₹2 = −2

a. Profit on selling 25 packets of potato chips = 8 × 25 = ₹200

b. Loss on selling 115 packets of biscuits = 2 × 115 = ₹230

3. Points scored per correct answer = 5

Points taken away per incorrect answer = −2

Total number of questions in the test = 25

Questions answered incorrectly = 7; questions answered correctly = 25 – 7 = 18

Points scored for 18 correct answers = 5 × 18 = 90

Points scored for 7 incorrect answers = −2 × 7 = −14

Total points scored by the student = 90 + (−14) = 76 points

3C 1. a. 72 ÷ (–4) = −18 b. (–56) ÷ 7 = −8 c. –176 ÷ (–11) = 16

d. 192 ÷ (–12) = −16 e. 344 ÷ 43 = 8 f. (–984) ÷ 12 = −82

g. (–676) ÷ (–26) = 26 h. –585 ÷ (–13) = 45

2. a. (–88 ÷ 4) ÷ –1 = −22 ÷ –1 = 22

b. 125 ÷ (125 ÷ 5) = 125 ÷ 25 = 5

c. (192 ÷ (–16)) ÷ 4 = −12 ÷ 4 = −3

d. 324 ÷ (–18 ÷ 2) = 324 ÷ −9 = −36

e. (900 ÷ (18)) ÷ 6 = 50 ÷ 6 = 8.33

f. (–3060 ÷ (–36)) ÷ 5 = 85 ÷ 5 = 17

g. 0 ÷ (–1508 ÷ 29) = 0 ÷ −52 = 0

h. (1000 ÷ (20)) ÷ 100 = 50 ÷ 100 = 0.5

3. −145 divided by −1 gives 145.

4.a. Division is commutative for integers. False

b. Division has the associative property for integers. False

c. The quotient of two negative integers is always negative. False

d. Division by zero is defined for integers. False

e. The quotient of zero divided by any non−zero integer is zero. True

5. Answers may vary. Sample answers.

a = −270, b = 18  a = 30, b = −2

6. (12 × –3) ÷ (–2) + (–4) = −36 ÷ (–2) + (–4) = 18 + (–4) = 14

Word Problems 1. Number of points lost in the first level for each fall = 2

Total Number of points lost in the first level = 10

Number of times Rahul’s character falls in the first level = 10 2 = 5 times

Number of points lost in second level for each fall = 4

Total Number of points lost in second level = 16

Number of times Rahul’s character falls in the second level = 16 4 = 4 times

Hence, Rahul’s character falls more on the first level.

2. Temperature recorded at 12 noon = 18℃

Rate of fall in temperature = 3℃ per hour

Difference in temperature at noon and 6℃ below zero = 18℃ –(–6℃) = 18℃ + 6℃ = 24℃

Number of hours it took to drop by 24℃ = 24℃ 3℃/hour = 8 hours

Hence, the temperature at 8 p.m. would be 6℃ below zero

Drop in temperature from 12 noon to midnight = 12 × 3℃ = 36℃

Temperature at midnight = 18℃ – 36℃ = –18 ℃.

Chapter Checkup 1. a. (–256) + 362 = 106

b. 214 – (–126) = 214 + 126 = 340 c. 652 + (–129) = 523

d. 248 – (–369) = 248 + 369 = 617 e. (–659) + (–1352) = −2011

f. (–2698) – (–1236) = −2698 + 1236 = −1462

2. a. 25 × (–89) = −2225 b. (–125) × (–26) = 3250

c. (–36) × 48 = −1728 d. (245) × (–57) = −13,965

e. (1245) × (–142) = −1,76,790 f. (–2365) × (123) = −2,90,895

3. a. 363 ÷ (–11) = −33 b. (–2652) ÷ (–51) = 52

c. 5525 ÷ (–25) = −221 d. (–5472) ÷ 36 = −152 e. 2852 ÷ 23 = 124

f. (–9594) ÷ (–41) = 234

4. To find the missing values, we will use the rules of transposition.

a. (–156) + 389 = 233 b. 524 + _____ = 256

Moving 524 to the RHS, we get 256 − 524 256 − 524 = −268

So, 524 + (−268) = 256

c. _____ – (–269) = –698

Moving –269 to the RHS, we get –698 –269. –698 –269 = –967.

So, −967 − (−269) = −698

d. 29 × _____ = (–1885)

Moving 29 to the RHS, we get –1885 ÷ 29 –1885 ÷ 29 = –65 So, 29 × (−65) = −1885

e. _____ ÷ (–17) = 56

Moving –17 to the RHS, we get 56 × –17. 56 × –17 = –952

−952 ÷ (−17) = 56

f. –568 – (–258) = −310 g. _____ × (–15) = –5475

Moving –15 to the RHS, we get –5475 ÷ –15. –5475 ÷ –15 = –365 So, −365 × (−15) = −5475

h. (2356) – _____ = 5886

Moving 5886 to the LHS, we get 2356 – 5886. 2356 – 5886 = −3530.

So, (2356) − (−3530) = 5886

i. −3261 + (−1265) = −4526 j. 3654 ÷ _____ = (–87)

Moving –87 to the LHS, we get 3654 ÷ (–87).

3654 ÷ (–87) = –42.

So, 3654 ÷ (−42) = −(87).

k. (–1265) × (–352) = 4,45,280 l. (–3562) ÷ (–137) = 26

5. a. 42 ÷ (–6 + 5) = 42 ÷ (−1) = −42

b. –64 ÷ 4 × (2 – 6) = −16 × −4 = 64

c. (9 ÷ 3) + 7 × (4 ÷ 2) = 3 + 7 × 2 = 3 + 14 = 17

d. 4(–12 + 6) ÷ 3 = 4 × (−6) ÷ 3 = 4 × (−2) = −8

e. 7 × (5 + 3) ÷ 4 × (9 – 2) = 7 × 8 ÷ 4 × 7 = 7 × 2 × 7 = 98

f. (6 + 2) – 15 ÷ (5 × 2) = 8 – 15 ÷ 10 = 8 – 1.5 = 6.5

6. a. 25 + (–63) = −38; additive inverse = 38;

multiplicative inverse = 1 −38

b. –125 + 47 = −78; additive inverse = 78; multiplicative inverse = 1 −78

c. –148 + (–157) = −305; additive inverse = 305; multiplicative inverse = 1 −305

d. 365 + 124 = 489; additive inverse = −489; multiplicative inverse = 1 489

7. a. The associative property of multiplication states that changing the grouping of the numbers being multiplied does not change the product. True

b. The additive inverse property states that every integer has an additive inverse such that the sum of the integer and its additive inverse is zero. True

c. The multiplicative inverse property states that every non-zero integer has a multiplicative inverse such that the product of the integer and its multiplicative inverse is one. True

d. The commutative property holds for the division of integers. False

e. The associative property holds for the subtraction of integers. False

8. 135 + (−325) = −190; 253 – (−528) = 781  −190 + 781 = 591

9. a. x × (–165) = 9240  x = 9240 −165 = −56

Hence −56 should be multiplied to (–165) to get a product of 9240.

10. 25 × (–36) = −900  2380 ÷ (–68) = −35  −900 + (−35) = −935

11. 23,072 ÷ x = –412  x = 23,072 ÷ (–412)  x = −56

Hence, −56 should be divided from 23072 to get (–412).

12. Initial position of the plane = 12,000 feet = +12,000 feet; distance climbed to reach the cruising altitude = 15,000 feet = +15,000 feet

New altitude after climbing 15,000 feet = (12,000 + 15,000) feet = +27,000 feet

After hitting the turbulence, the plane descends 23,945 feet = −23,945 feet

New altitude of the plane = 27,000 + (−23,945) = +3055 feet

Word Problems 1. Temperature at 11 a.m. = 6℉

Rate of drop = 3℉ per hour

Number of hours between 11 a.m. and 4 p.m. = 5 hours

Temperature drops in 5 hours = 3℉ × 5 = 15℉

Temperature at 4 p.m. = 6℉ − 15℉ = −9℉

2. Floor at which Kajal got into the elevator = 4th

Floor of the food court = 4 + 10 = 14th

She went down 5 floors to purchase books = 14 − 5 = 9th floor

Hence Kajal is on the 9th floor.

3. Amount off on each pair of shoes = ₹55

Total reduction = ₹7590

Number of pair of shoes = Total reduction Amount off on each pair = 7590 55 = 138

So, there are 138 pairs of shoes in the store.

Chapter 4

Let’s Warm-up 1. 2 households have 3 pets.

2. 15 + 12 = 27

27 households have less than 2 pets.

3. 0 × 15 + 1 × 12 + 2 × 5 + 3 × 2 + 4 × 1 = 32

The total number of pets in the neighborhood is 32.   4. There are 12 households with 1 pet each.

5. (12 + 5 + 2 + 1) − (15) = 20 − 15 = 5

The difference between the households which have pets and which do not have pets is 5. Do It Yourself

4A 1. The first 10 odd numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.

Sum of first 10 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

So, Mean = Sum of all observations

Total number of number of observations = 100 10 = 10

2. Temperatures in the city = 3°C, –7°C, –1°C, –5°C and 2°C

Sum of all observations = 3 + (–7) + (–1) + (–5) + 2 = –8°C

Total number of number of observations = 5

Mean = (–8) 5 = –1.6°C

3.

No. of illustrations (per page) (x)

No. of pages ( f ) f  × x

2 25 2 × 25 = 50

4 30 4 × 30 = 120

5 45 5 × 45 = 225

6 12 6 × 12 = 72

Total f = 25 + 30 + 45 + 12 = 112 f × x = 50 + 120 + 225 + 72 = 467

Mean = f × x f = 467 112 = 4.17

4. Mean = 56.2

Sum of all the observations = 40 + 50 + 55 + a + 58 = 203 + a

Sum of all the observations = Mean × Total number of observations = 56.2 × 5 = 281

So, 203 + a = 281 a = 78

5. Mean = 92

Sum of all the observations = (x) + (x + 2) + (x + 3) + (x + 5) + (x + 7) = 5x + 17

Sum of all the observations = Mean × Total number of observations = 92 × 5 = 460

So, 5x + 17 = 460

5x = 460 − 17 = 443

x = 443 5 = 88.6

6. Mean of 5 numbers = 26

So, sum of all the observations = mean × total number of observations = 5 × 26 = 130

Sum when 6 is removed from the observations = 130 − 6 = 124

Thus, the new mean = 124 4 = 31

The new mean is greater than the original mean.

7. Mean = 40

y + z = 76

Sum of all the observations = 30 + 25 + x + 50 + y + 45 + z = 150 + x + y + z

= 150 + 76 + x = 226 + x

Sum of all the observations = Mean × Total number of

observations = 40 × 7 = 280

So, 226 + x = 280

x = 280 − 226 = 54

Word Problems 1. Total bill = ₹2080

Number of friends = 8

Amount each friend has to pay = Mean = 2080 8 = ₹260

So, each friend has to pay ₹260.

2. Sum of the lengths of all the rivers = 724 + 1376 + 800 + 2525 = 5425 km

Mean length = 5425 4 = 1356.25 km

So, the mean length of the rivers visited by the troop is 1356.25 km.

3. Mean attendance from Monday to Saturday = 33

Total attendance from Monday to Saturday = 33 × 6 = 198

Mean attendance from Monday to Thursday = 32

Total attendance from Monday to Thursday = 32 × 4 = 128

Mean attendance from Thursday to Saturday = 34

Total attendance from Thursday to Saturday = 34 × 3 = 102

Total attendance from Monday to Thursday + Total attendance from Thursday to Saturday = 128 + 102 = 230

So, the attendance on Thursday = 230 – 198 = 32.

4B 1. a. The mode is the observation that occurs the least number of times in a data set. False

b. The mode is always one of the numbers in the data. True

c. The data 2, 4, 3, 4, 6, 3, 6, 7, 4 has mode 3. False

d. There is only one mode for any given set of data. False

2. 12, 5, 7, 10, 3, 2, 1 ,7, 6, 7, 5, 10, 7, 4, 2, 1, 4, 4

Arranging in ascending order = 1, 1, 2, 2, 3, 4, 4, 4, 5, 5, 6, 7, 7,

7, 7, 10, 10, 12

7 occurs most frequently in the given data, so 7 years is the mode.

3. 12, 14, 13, 16, 18, 15, 18, 19, 12, 14, 16, 15, 18, 11

Arranging in ascending order: 11, 12, 12, 13, 14, 14, 15, 15, 16, 16, 18, 18, 18, 19

18 occurs most frequently in the given data, so the mode of the data is 18 marks.

Since, only 1 student scored 19, so, 1 student got the highest marks.

4. 60, 52, 42, 40, 43, 44, 46, 45, 42, 41, 40, 46, 48, 46.

Arranging in ascending order: 40, 40, 41, 42, 42, 43, 44, 45, 46, 46, 46, 48, 52, 60

46 occurs most frequently in the given data, so the mode of the data is 46.

5. From the table, 4 books have been read by most number of students in the class. So, the mode is 4 books.

Genre FictionPoetryDramaFantasy

Number of Times Chosen by Children 2242

Word Problem 1. Fiction, Poetry, Drama, Drama, Fiction, Fantasy, Poetry, Drama, Fantasy, Drama. Since, the drama genre is chosen the most, so the modal genre is drama.

4C 1. a. To find the median, the data can be arranged in ascending or descending order.

b. If the number of observations is an odd number, then the median is the    n + 1 2    th observation.

c. First 6 odd numbers: 1, 3, 5, 7, 9, 11

Median = 1 2  

n 2 

th + 

observation = 1 2

The median of first 6 odd numbers is 6

2. Enrollments: 1555, 1670, 1850, 1780, 1400, 2540

Arranging in ascending order: 1400, 1555, 1670, 1780, 1850, 2540

Number of observations = 6Th

Median = 1 2    n 2     th +    n 2 + 1     th observation = 1 2    6 2     th +  

 6 2 + 1     th = 1 2 [3rd + 4th] observation = 1 2 [1670 + 1780]

= 1 2 [3450] = 1725

So, the median score of the team is 1725.

3. Runs: 60, 15, 120, 40, 80, 54, 15, 25, 62, 6, 14

Arranging in ascending order: 6, 14, 15, 15, 25, 40, 54, 60, 62, 80, 120

Number of observations = 11

Median =    n + 1 2     th observation = 11 + 1 2  th = 12 2  th = 6th observation = 40

So, the median score of the team is 40 runs.

4. 10, 20, x, 30, 40, 50, 70

Median = 30

Number of observations = 7

Median =    n + 1 2     th observation = 7 + 1 2  th = 8 2 th = 4th observation. So, x lies between 20 and 30.

Therefore, the maximum possible value of x is 30.

5. 60, 55, 54, 53, 45, 35, x, 25, 20, 17, 15, 10 is in descending order.

Number of observations = 12

a. Median = 30 = 1 2    n 2     th +    n 2 + 1     th observation = 1 2    12 2     th

+    12 2 + 1     th

1 2 [6th + 7th] = 35 + x 2 = 30

35 + x = 30 × 2 = 60

x = 60 – 35 = 25

b. 55, 54, 53, 45, 35, 25, 25, 20, 20, 17, 15, 10

Median = 1 2 [6th + 7th] = 1 2 (25 + 25) = 50 2 = 25

6. 7, 21, 12, 15, 18, 9

Arranging in ascending order: 7, 9, 12, 15, 18, 21

Number of observations = 6

a. Median time = 1 2    n 2     th +    n 2 + 1     th observation =

1 2    6 2     th +    6 2 + 1     th observation

1 2 [3rd + 4th] observation

1 2 [12 + 15] = 27 2 = 13.5 minutes

b. If the largest time value is replaced by 35: 7, 9, 12, 15, 18, 35

The 3rd and 4th observation do not change so the median will remain the same.

c. Since, the median depends on the number of observations, it will change if new students are added to the data.

7. 2, 4, 6, 6, 2x + 1, 9, 9, 10, 11 are written in ascending order.

Number of observations = 9

Median = 17

Also, median =    n + 1 2     th observation =    9+ 1 2     th observation = 5th observation = 2x + 1

2x + 1 = 17

2x = 17 – 1 = 16

x = 16 2 = 8

Word Problems 1. ₹15, ₹20, ₹10, ₹25, ₹30

Arranging in ascending order: ₹10, ₹15, ₹20, ₹25, ₹30

Number of observations = 5;

Median =    n + 1 2     th observation =    5 + 1 2     th observation

= 3rd observation = ₹20

2. Number of observations = 23

Median =    n + 1 2     th observation =    23 + 1 2     th observation

= 12th observation

So, Rohan booked the 12th seat.

4D 1. a. The measures of central tendency are mean, median and mode

b. Mean represents one mathematical equivalent for the data.

c. Median is used when you need a “middle” value of a data set that has a sizeable number of observations.

d.Mode cannot be defined for data with no repeated values.

2. 5, 7, 10, 12, 15, 18, 22, 30.

Mode should not be used for the given data as there are no repeating values.

3. 72, 73, 72, 76, 72, 74, 73, 75, 76, 74, 72 are the scores of the team members of a golf team.

Mean would be the most appropriate tendency of measure to find the average score of the team members.

4. The mode is the value that appears most frequently in the data set. If a value between 1 and 5 appears more frequently than any other value, it might be a good representation.

So, mode will better represent the typical value of the data set.

5. ₹10, ₹25, ₹50, ₹100, ₹250, ₹500, ₹1000, ₹2000, ₹5000 and ₹10,000

Since, the data has a sizeable number of observations, we will use median.

Word Problem 1. Mean = Sum of all the observations

total number of observation

= 87 + 75 + 64 + 89 + 0 + 78 + 93 + 95 + 90 + 84

10 = 755 10 = 75.5

0, 64, 75, 78, 84, 87, 89, 90, 93, 95

Median = 1 2    n 2     th +    n 2 + 1     th observation = 1 2    10 2     th +    10 2 + 1     th observation

1 2 [5th + 6th] observation

1 2 [84 + 87] = 171 2 = 85.5

Median is more accurate because it is the middle value in an ordered sequence.

Chapter Checkup 1. a. 8, 12, 16, 13, 9, 10, 15, 12

Mean = Sum of all the observations

total number of observation

= 8 + 12 + 16 + 13 + 9 + 10 + 15 + 12 8 = 95 8 = 11.875

Ascending order = 8, 9, 10, 12, 12, 13, 15, 16

Number of terms = 8

Median = 1 2    n 2     th +    n 2 + 1     th = 1 2    8 2     th +    8 2 + 1     th = 1 2 [4th + 5th]

= 12 + 12 2 = 12

Mode = 12

b. 12, −15, −32, −11, −15, 14, −30, −18

Mean = Sum of all the observations total number of observation

= 12 + (−15) + (−32) + (−11) + (−15) + 14 + (−30) + (−18) 8 = −95 8 = −11.875

Ascending order = –32, –30, –18, –15, –15, –11, 12, 14

Number of terms = 8

Median = 1 2    n 2     th +    n 2 + 1     th = 1 2    8 2     th +    8 2 + 1     th = 1 2 [4th + 5th]

= 1 2 [–15 –15] = –15

Mode = – 15

2. a. The first 10 natural numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 10 = 55 10 = 5.5

Number of terms = 10

Median = 1 2    n 2     th +    n 2 + 1     th = 1 2    10 2     th +    10 2 + 1     th = 1 2 [5th + 6th]

= 5 + 6 2 = 11 2 = 5.5

b. The first 8 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19

Mean = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 8 = 77 8 = 9.625

Number of terms = 8

Median = 1 2    n 2     th +    n 2 + 1     th = 1 2    8 2     th +    8 2 + 1     th

= 1 2 [4th + 5th] = 7 + 11 2 = 9

c. The factors of 36: 1, 2, 3, 4, 6, 9, 12, 18 and 36

Mean = 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 9 = 91 9 = 10.11

Number of terms = 9

Median =    n + 1 2     th observation =    9+ 1 2     th observation = 5th observation = 6

3. 38 kg, 41 kg, 36 kg, 39 kg and 41 kg

Mean = 38 + 41 + 36 + 39 + 41 5 = 195 5 = 39 kg

4. 2, 2, 2, 2, 3, 3, 4, 4, 5, 6, 7

Number of observations = 11

Median =    n + 1 2     th observation =    11 + 1 2     th observation = 6th observation = 3

5. Ascending order: 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9

Mean = f × x f = 100 20 = 5

Median = 1 2    n 2     th +    n 2 + 1     th = 1 2    8 2     th +    8 2 + 1     th =

1 2 [10th + 11th] = 5 + 5 2 = 10 2 = 5

6. 16, 20, 20, 20, 22, 22, 26, 26, 30

Since, 20 is occurring most frequently, the mode is 20.

7. 6, 8, 10, 15, 15, 15, 20, 50, 80, 90, 90, 100, 120

=

Mean weight = f × x f = 3848 70 = 54.97 kg

Number of terms = 70

Median weight = 1 2    n 2     th +    n 2 + 1     th = 1 2    70 2     th +    70 2 + 1     th

= 1 2 [35th + 36th] = 54 kg

Mode weight = 54 kg

9. Arranging is ascending order: 1.5, 2.7, 13.6, 35.6, 52.2, 82.1, 96.5, 97, 108.5, 125.5

Median would be most suitable to represent the rainfall as the given data has sizable number of observations.

10. 4, 6, 7, 8, 9, 9, 10.

As we need to find one value that represents the whole data, mean would be best measure of central tendency to do so.

11. 45, 42, 38, 35, 18, 10

Number of terms = 6

Median age = 1 2

= 1 2 [3rd + 4th] = 38 + 35

Total f = 20 f × x = 100

So, the median age of the family is 36.5 years.

12. 80, 80, 80, 80, 85, 85, 85, 90, 90, 95

So, the modal size is 80.

If the size of one shirt is misread as 80 instead of 85

80, 80, 80, 85, 85, 85, 85, 90, 90, 95

The mode will be 85.

13. 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22,

23, 24, 24, 25, 25, 25, 25

= 432 20 = 21.6 hours

Number of terms = 20 Median

= 1 2 [10th + 11th] = 21 + 21 2 = 21 hours

14. 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 41, 42, 43, 44 25, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 42, 43, 44, 45

Median weight =

th observation = 

th observation = 8th observation = 34 kg

15.  12, 14, 15, 15, 16, 17, 18, 18, 19, 21

Mean age = 12 + 14 + 15 + 15 + 16 + 17 + 18 + 18 + 19 + 21 10 = 165 10 = 16.5 years

Median age = 1 2    10 2     th +    10 2 + 1     th = 1 2 [5th + 6th] = 16 + 17 2 = 33 2 = 16.5 years

Since, mean and median are both equal, they can both be used to represent the typical age of the students.

Word Problems 1. Mean = 45

Number of students = 20

Sum of weights = 45 × 20 = 900 kg

Sum after Ankit and Suhani joined the class = 900 + 55 + 52 = 1007

So, the new mean weight of the class = 1007 20 + 2 = 1007 22 = 45.77 kg 2. 22, 28, 29, 30, 31

Median age =    n + 1 2     th observation =    5 + 1 2     th observation = 3rd observation = 29 years

If another person, aged 32, joins the group, the median will change. 22, 28, 29, 30, 31, 32

New Median age = 1 2    6 2     th +    6 2 + 1     th = 1 2 [3rd + 4th] = 29 + 30 2

= 59 2 = 29.5 years

3. Marks scored in 4 tests = 75, 82, 79, 76

Sum of marks scored in 4 tests = 75 + 82 + 79 + 76 = 312

Number of tests = 4

Sum of marks to get an average of 80 in 5 tests = 80 × 5 = 400

Minimum marks need to be secured to maintain the average of 80 = 400 – 312 = 88

So, Mohan needs to score at least 88 marks in his next test.

Chapter 5

Let’s Warm-up 1. 13,289 is 1 more than 13,288. Jacques Kallis scored 1 run more than Rahul Dravid.

2. Total runs scored by Ricky Ponting and Alastair Cook = 13,378 + 12,472 = 25,850

The total runs made by Ricky Ponting and Alastair Cook are 25,850.

3. Total runs scored by 5 batsmen = 15,921 + 13,378 + 13,289 + 13,288 + 12,472 = 68,348

The total runs scored by 5 batsmen are 68,348.

4. 12,472 < 13,378 < 13,288 < 13,289 < 15,921. Alastair Cook scored the runs of the 5 players.

5. Runs scored by Sachin Tendulkar = 15,921

Runs scored by Jacques Kallis = 13,289

Number of more runs scored by Sachin Tendulkar = 15,921 –13,289 = 2632

Sachin Tendulkar scored 2632 more runs than Jacques Kallis.

Do It Yourself

5A

1. Collecting data is the first step in any investigation. So, option b is the correct answer.

2. Since, the table has tally marks with frequency, so it is a frequency distribution table. So, option C is the correct answer.

3. a. Interviews, surveys and direct observations are examples of primary data. T

b. Information published in magazines and newspapers after an event is an example of secondary data. T

c. Autobiographies and memoirs are examples of primary data.

F, because autobiographies and memoirs are books of the past

4. Colour of

5. Range = 6 –

6. Arranging marks in ascending order, we get 33, 38, 43, 45, 46, 56, 63, 64, 65, 69, 72, 76, 79, 79, 82, 82, 87, 91, 99, 99

a. Range = 99 – 33 = 66

b. Passing Marks = 45

Marks below 45 = 33, 38, 43

Number of students who got marks below 45 = 1 + 1 + 1 = 3

c. Marks below 60 = 33, 38, 43, 45, 46, 56

Number of students who got marks below 60 = 1 + 1 + 1 + 1 + 1 = 1 = 6

Word Problem 1. Arranging the given data in ascending order, we get 59, 60, 60, 60, 62, 62, 62, 62, 67, 68, 68, 69, 70, 70, 70, 70, 70, 71, 71, 73, 73, 74

a. Range = 74 – 59 = 15

b. Pulse rate 60 or less = 59, 60

Number of students with pulse rate 60 or less = 1 + 3 = 4

c. Pulse rate more than 65 = 67, 68, 69, 70, 71, 73, 74

Number of students with pulse rate more than 65 = 1 + 2 + 1 +

5 + 2 + 2 + 1 = 14

5B 1. a. The bar graph consists of rectangular bars of equal width

b. The space between two consecutive bars are equal.

c. In a bar graph, the bars can be represented either horizontally or vertically.

d. The height of a bar represents the frequency of the corresponding data.

2.

5C 1. 1 cm = 5 ft.

5 cm = 5 × 5 = 25 ft.

2. a. English

b. Maths and Science

Marks scored = 80

c. Maths

Marks scored = 96

d. Marks scored in Hindi = 60

Marks scored in English = 70

Marks scored in Maths = 80

Marks scored in science = 80

Marks scored in Drawings = 70

Total marks scored in term 1 = 60 + 70 + 80 + 80 + 70 = 360

e. Total marks scored in Term 2 = 62 + 66 + 96 + 86 + 76 = 386

Difference between total marks in term 1 and term 2 = 386 –360 = 26

3. Height of Nanga Parbat = 8300 m

Height of Annapurna = 8100 m

Height of Kanchenjunga = 8500 m

Height of Nanda Devi = 7900 m

Height of Mount Everest = 8800 m

Height of Aconcagua = 6900 m

a. 8800 m > 8500 m > 8300 m > 8100 m > 7900 m > 6900 m

So, Mt. Everest > Kanchenjunga > Nanga Parbat > Annapurna > Nanda Devi > Aconcagua is the required descending order.

b. Mt. Everest has the highest peak. It’s height is 8800 m.

c. Highest peak = Mt. Everest

Height of the highest peak = 8800 m

Lowest peak = Aconcagua

Height of the lowest peak = 6900 m

Ratio = 8800 m:6900 m = 88:69

d. The second highest mountain peak is Kanchenjunga, because its height is 8500 m.

4.a.

b. 80 > 50 > 30 > 20. So, merchandise S gives the most profit on selling.

5. a. The family's expenditure in March was ₹2200.

b. The family spent the most in January and August. Monthly expenditure in January = ₹2400 = Monthly expenditure in August

c. Total expenditure in a year = ₹2400 + ₹1800 + ₹2200 + ₹2200 + ₹1400 + ₹2000 + ₹2000 + ₹2400 + ₹2000 + ₹2200 + ₹2000 + ₹1600 = ₹24,200

d. Total expenditure in the first six months = ₹2400 + ₹1800 + ₹2200 + ₹2200 + ₹1400 + ₹2000 = ₹12,000

Total expenditure in last six months = ₹2000 + ₹2400 + ₹2000 + ₹2200 + ₹2000 + ₹1600 = ₹12,200 Ratio = 12,000:12,200 = 60:61

b. Friday

c. Rainfall on the first two days = 10 + 9 = 19 mm

Rainfall on the last two days = 12 + 18 = 30 mm  Ratio = 19:30

7.a. 420 people wear medium-sized shirts.

b. 340 people wear extra-large-sized shirts.

c. Number of people who wear large-sized shirts = 390 Number of people who wear small-sized shirts = 400 Ratio: 390:400 = 39:40

d. Total number of people = 340 + 390 + 420 + 400 = 1550

8.a. Total number of cars sold by showroom A in the first three months = 60 + 75 + 75 = 210

b. Total number of cars sold by showroom B in the last three months = 60 + 75 + 75 = 210

c. Total number of cars sold by showroom A in the first four months = 60 + 75 + 75 + 90 = 300

Total number of cars sold by showroom B in the first four months = 45 + 60 + 90 + 75 = 270 300 > 270, so showroom A sold more cars.

d. Total number of cars sold by showroom A = 60 + 75 + 75 + 90 + 45 + 60 + 105 = 510

Total number of cars sold by showroom B = 45 + 60 + 90 + 75 + 60 + 75 + 75 = 480  Ratio = 510:480 = 17:16

Word Problems 1.

a. 300 visitors were there in January.

b. Number of visitors in May = 350

Number of visitors in February = 250

350 − 250 = 100.

So, number of visitors in May is more than February by 100.

c. Total number of visitors = 300 + 250 + 300 + 400 + 350 = 1600

Number of visitors in April = 400

Percentage of visitors in April = 400 1600 × 100 = 25%

2.a. Income earned by company A in 2011 = ₹1,42,500

Profit earned in 2011 by company A = 35%

Profit% = Income – Expenditure (E) Expenditure (E) × 100

35 = 1,42,500 – E E × 100   35 = 1,42,50,000 – 100E E

35E = 1,42,50,000 – 100E

35E + 100E = 1,42,50,000

135E = 1,42,50,000

E = 1,42,50,000 135 = ₹1,05,555.56

b. Expenditure of company A in 2010 = ₹70,00,000

Income of company A in 2010 = ₹1,05,555.56

Income of company A in 2011 = ₹1,42,500

Total income = ₹1,05,555.56 + ₹1,42,500 = ₹2,48,055.56 Chapter Checkup

1. Arranging the data in ascending order, we get 14, 14, 16, 16, 21, 21, 21, 21, 21, 25, 28, 28, 32, 32, 37, 43, 43, 43, 45, 45, 45

6.a. In 2018, the production of food grains was the highest with 90 million tonnes.

b. Production of food grains in 2016 = 69 million tonnes

c. Production of food grains in 2017 = 77 million tonnes

Difference = 77 – 69 = 8 million tonnes

d. Production of food grains in 2015 = 60 million tonnes

Total production of food grains = 51 + 60 + 69 + 77 + 90 = 347 million tonnes

Ratio = 60:347

7.a. Number of people who use radios = 8 lakhs

Number of people who use telephones = 10 lakhs

2 lakhs (= 10 lakhs – 8 lakhs) less people use TV than telephone.

b. Number of people who use computers = 9 lakhs

Number of people who use iPods = 8 lakhs

9 lakhs > 8 lakhs by 1 lakh, so only 1 lakh more people use computers than iPods.

c. Number of users of iPods = 8 lakhs

Number of users of webcam = 10 lakhs

Difference = 10 lakhs – 8 lakhs = 2 lakhs

d. Total number of users = 9 lakhs + 8 lakhs + 10 lakhs + 8 lakhs + 10 lakhs = 45 lakhs

e. A suitable title can be ‘Number of Users of Different Electronic Appliances’. (Answer may vary)

8.a. 175 bulbs were sold on Monday.

b. 4 appeared the least number of times.

c. 5 appeared the greatest number of times.

d. Digits 1 and 6 appear the same number of times, 7.

3. Number of birthdays in February = 4

Number of birthdays in July = 7

Number of birthdays in September = 14

Number of birthdays in November = 7

Number of birthdays in December = 9

a. September has 14 birthday is the greatest of all.

b. Total number of birthdays in February and July = 4 + 7 = 11

c. 14 > 9 by 5. So, 5 more birthdays fall in September than in December.

d. July and November as both have 7 birthdays.

4.

b. Friday has the highest sale of bulbs, with 225 bulbs.

c. 875 bulbs were sold in the whole week.

d. Total number of bulbs sold = 175 + 200 + 125 + 150 + 225 = 875

Number of bulbs sold on Tuesday = 200

Fraction of bulbs sold on Tuesday = 200 875 = 8 35

e. Average number of bulbs sold during the week = 875 5 = 175

9. The graph is misleading the numbers on the y-axis start at 19. It should always start at 0. The tabular form and correct graph would be:

Name of employeesRajSahilNiviraYuvanKabir Salaries (in thousands)2624252227

10.a. NH 10 is the shortest of all.

b. 900 km is the length of NH9.

c. Length of National Highway 10 = 400 km

3 times the length of national highway 10 will be 1200 km, which is the length of national highway 3.

Word Problems

1.

Marks obtained by Kunal in his annual examination

2. a. The given bar graph shows the sales of books (in thousands) from six branches of a publishing company during two consecutive years, 2010 and 2011.

b. Sales of books of branch B2 in 2010 = 100 thousand

Sales of books of branch B2 in 2011 = 117 thousand

Total sales of branch B2 in two years = 100 + 117 = 217 thousand

Sales of books of branch B4 in 2010 = 90 thousand

Sales of books of branch B4 in 2011 = 80 thousand

Total sales of branch B2 in two years = 90 + 80 = 170 thousand

Ratio of total sales = 217:170

c. Sales of books of branch B1 in 2010 = 80 thousand

Sales of books of branch B2 in 2010 = 100 thousand

Sales of books of branch B3 in 2010 = 60 thousand

Sales of books of branch B4 in 2010 = 90 thousand

Sales of books of branch B5 in 2010 = 120 thousand

Sales of books of branch B6 in 2010 = 40 thousand

Total sales = 80 + 100 + 60 + 90 + 120 + 40 = 490 thousand

Average sales = 490 6 = 81.67 (approx.) thousand

d. Sales of books of branch B1 in 2010 = 80 thousand

Sales of books of branch B3 in 2010 = 60 thousand

Sales of books of branch B5 in 2010 = 120 thousand

Total sales in 2010 = 80 + 60 + 120 = 260 thousand

Sales of books of branch B1 in 2011 = 100 thousand

Sales of books of branch B3 in 2010 = 80 thousand

Sales of books of branch B5 in 2010 = 137 thousand

Total sales in 2011 = 100 + 80 + 137 = 317 thousand

Total sales in both the years = 260 + 317 = 577 thousand

Chapter

6

Let’s Warm-up

Ratios   Simplest Form

1.

2.

3.

6:12

4. 5. Do It Yourself

6A 1. a. Unlikely b. Equally likely c. Impossible  d. Likely e. Sure

2. a. Equally Likely b. Sure c. Unlikely d. Impossible   3. a. Impossible b. Equally Likely c. Unlikely

b. If I have to move “Forward”, I would choose any of the two spinners, as each of them has 2 sections out of six that represent “Forward”.

5. Answers may vary. Sample answers.

a. Wearing mask due to increasing pollution

b. Travelling to the moon

c. Christmas falls in the month of January

d. Sun sets in the west.

Word Problem 1.a. Factors of 40 = 1, 2, 4, 5, 8, 10, 20, 40 (8 numbers)

Number of cards = 40

Chance of drawing the cards is unlikely.

b. Even numbers up to 40 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40 (20 numbers)

Number of cards = 40

Chance of drawing even number card is equally likely.

6B 1.a. Letters from A through K (possible outcomes) = A, B, C, D, E, F, G, H, I, J, K

b. Vowels from A to K (favourable outcomes) = A, E, I

2. a. On the spinner, there are 3 reds, 1 blue, 1 green and 1 purple. So, number of possible outcomes = 6.

b. Since there are 3 reds, the arrow can land on red 3 times.

c. Since there are 5 other colours on the spinner except purple, the arrow can land on purple 5 times.

3. a. Numbers on the spinner = 1, 2, 3, 4, 5, 6, 7, 8

So, the number of possible outcomes = 8

b. Numbers greater than 3 = 4, 5, 6, 7, 8

So, favourable outcomes of spinning a number greater than 3 = 4, 5, 6, 7, 8

c. Even numbers = 2, 4, 6, 8

So, number of times the arrow can land on even number = 4

d. Odd numbers = 1, 3, 5, 7

So, number of times the arrow can land on odd number = 4

e. Prime numbers = 2, 3, 5, 7

So, number of times the arrow can land on a prime number = 4

4. Total number of outcomes = 6

a. Number of reds = 2

Probability of getting reds = 2 6 = 1 3

b. Number of 1s = 1

Probability of getting a 1 = 1 6

c. Number of odd numbers = 3 (1, 3, 5)

Probability of getting an odd number = 3 6 = 1 2

d. Number of multiple of 2 = 3 (2, 4, 6)

Probability of getting an odd number = 3 6 = 1 2

e. Numbers less than 7 = 6

Probability of getting a number less than 7 = 6 6 = 1

f. Since there is no number 9 on the spinner, the probability of getting a 9 is 0 6, that is, 0.

5. Numbers on the dodecahedron = 12

a. Since there is one face with number 1 on dodecahedron, the probability of getting a 1 = 1 12

b. Numbers less than 9 on dodecahedron = 8 (1, 2, 3, 4, 5, 6, 7, 8)

Probability of getting a number less than 9 = 8 12 = 2 3

c. Numbers that are multiples of 3 on dodecahedron = 4 (3, 6, 9, 12)

114

d. Numbers that are greater than 6 = 6 (7, 8, 9, 10, 11, 12)

Probability of getting a number greater than 6 = 6 12 = 1 2

6. Number of letters of the English alphabet (Total number of outcomes) = 26

Getting a letter except Z = 25

Probability of getting a letter except Z = 25 26

7. Number of females (favourable outcomes) = 16

Number of males = 20

Total number of students (Total number of outcomes) = 36

Probability of choosing a female student = 16 36 = 4 9

8. Number of black crayons = 6

Number of blue crayons = 4

Number of red crayons = 5

Number of yellow crayons = 3

Number of white crayons = 2

Total number of crayons (Total number of outcomes) = 20

a. Probability of getting a black crayon = 6 20 = 3 10

b. Probability of getting a blue crayon = 4 20 = 1 5

c. Number of not a white crayons = 6 + 4 + 5 + 3 = 18

Probability of getting not a white crayon = 18 20 = 9 10

d. There is no pink crayon, so the probability of getting a pink crayon = 0 20 = 0

e. Total number of black and blue crayons = 6 + 4 = 10

Probability of getting a black or blue crayons = 10 20 = 1 2

f. Total number of blue, red and yellow crayons = 4 + 5 + 3 = 12

Probability of getting a blue, red and yellow crayons = 12 20 = 3 5

9. Numbers on the slips of paper = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25

Total numbers written on the slips (Total number of outcomes) = 25

a. Odd numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25

Number of odd numbers = 13

Probability of getting an odd number = 13 25

b. Since there are no 3-digit numbers, probability of getting a 3-digit number = 0 25 = 0

c. Numbers other than 4 = 1, 2, 3, 5, 6, 7, …, 25

Total numbers other than 4 = 24

Probability of getting a number that is not 4 = 24 25

d. Since all are positive numbers, so the probability of getting a positive number = 25 25 = 1

e. Prime numbers = 2, 3, 5, 7, 11, 13, 17, 19, 23

Number of prime numbers = 9

Probability of getting a prime number = 9 25

f. Numbers greater than 19 = 20, 21, 22, 23, 24, 25

Number of numbers greater than 19 = 6

Probability of getting a number greater than 19 = 6 25

10. Months’ name ending with letter y = January, February, May, July

Number of months ending in letter y = 4

Total number of months = 12

Probability of getting a month ending with letter y = 4 12 = 1 3

11. Integers from 1 to 10 = 10

Total number of integers from 1 to 10 = 10

Number of integers less than 3 = 2 (1, 2)

Probability of getting a number less than 3 = 2 10 = 1 5

Amit’s error: He included number 3 also.

12. Out of 5 choices, only 1 can be the correct answer.

Total number of outcomes = 5

Number of possible outcomes = 1

Probability = 1 5

13. In true-false question, either the answer is true or false. So, total number of outcomes = 2

Favourable outcomes = 1

Probability of getting a correct answer = 1 2

Word Problems 1. Number of lettered tiles = 98

Number of blank tiles = 2

Total number of tiles = 100

a. Probability of getting a blank tile = 2 100 = 1 50

b. Probability of getting a lettered tile = 98 100 = 49 50

2. Number of pieces of fruits = 12

Number of apples = 7

Number of peaches = 2

Number of fruits that are not peaches or apples = 12 – (7 + 2) = 12 – 9 = 3

Probability of getting neither a peach nor an apple = 3 12 = 1 4

6C 1. a. Total number of spins = 8 + 6 + 9 + 11 + 9 + 7 = 50

a. Number of times spinning a 6 = 7

Probability of spinning a 6 = 7 50

b. Number of times spinning an even number (2, 4, 6) = 6 + 11

+ 7 = 24

Probability of spinning an even number = 24 50 = 12 25

c. Number of times spinning an odd number (1, 3, 5) = 8 + 9 +

9 = 26

Probability of spinning an odd number = 26 50 = 13 25

d. Number of times spinning a number less than 3 (1, 2) = 8 +

6 = 14

Probability of spinning a number less than 3 = 14 50 = 7 25

e. Since there is no number 7 on the spinner, so the probability of spinning a 7 is 0 50 = 0.

f. Number of times not spinning a 1 (spinning 2, 3, 4, 5, 6) = 6 + 9 + 11 + 9 + 7 = 42

Probability of not spinning a 1 = 42 50 = 21 25

2. Total number of spins = 14 + 7 + 11 + 8 = 40

Number of times spinning a letter A = 14

Probability of spinning a letter A = 14 40 = 7 20

Number of times spinning a letter B = 7

Probability of spinning a letter B = 7 40

Number of times spinning a letter C = 11

Probability of spinning a letter C = 11 40

Number of times spinning a letter D = 8

Probability of spinning a letter D = 8 40 = 1 5

3. Total number of times a coin is tossed = 40 + 60 = 100

Number of heads = 40

Number of tails = 60

Probability of getting a head = 40 100 = 2 5

Mala’s error: Mala has found the probability of getting heads over the number of times she gets a tail whereas she must have found the probability over the total number of times a coin is tossed.

4. Total number of times Malati records a color = 12 + 10 + 15 + 13 = 50

Number of times she gets a red = 12

Probability of getting a red = 12 50 = 6 25

Number of times she gets a blue = 10

Probability of getting a blue = 10 50 = 1 5

Number of times she gets a green = 15

Probability of getting a green = 15 50 = 3 10

Number of times she gets a yellow = 13

Probability of getting a yellow = 13 50

5. Total number of students selected their sports = 30 + 22 + 34 + 20 + 14 = 120

a. Number of students selected cricket = 30

Probability of student selecting cricket = 30 120 = 1 4

b. Number of students selected hockey = 20

Probability of student selecting hockey = 20 120 = 1 6

c. Number of students selected baseball = 22

Probability of student selecting baseball = 22 120 = 11 60

d. Number of students selected football = 34

Probability of student selecting football = 34 120 = 17 60

6. Total number of numbers = 5 (1, 2, 3, 4, 5)

Theoretical probability = probability of choosing 1 = 1 5

Total number of people surveyed = 15 + 30 + 35 + 20 + 25 = 125

Number of people who chose 1 = 15

Experimental probability = probability of choosing 1 = 15 125 = 3 25 1 5 > 3 25 . So, theoretical probability is greater than experimental probability.

7. Total number of times a spinner is spun = 41 + 43 + 39 + 37 + 40 = 200

a. Theoretical probability of spinning 4 = 1 5 = 40 200

Number of times 4 is spun = 37

Experimental probability of spinning a 4 = 37 200

As, 40 200 > 37 200, Theoretical probability > Experimental probability

b. Theoretical probability of spinning 3 = 1 5 = 40 200

Number of times 3 is spun = 39

Experimental probability of spinning a 3 = 39 200

As, 40 200 > 39 200, Theoretical probability > Experimental probability

c. Only number 5 is greater than number 4 on the spinner

Theoretical probability of spinning a number greater than 4 = 1 5 = 40 200

Number of times a number greater than 4 (5) is spun = 40

Experimental probability of spinning a number greater than 4 = 40 200

As, 40 200 = 40 200, Theoretical probability = Experimental probability

d. Number of odd numbers (1, 3, 5) on the spinner = 3

Theoretical probability of spinning an odd number = 3 5 = 120 200

Number of times an odd number (1, 3, 5) is spun = 41 + 39 + 40 = 120

Probability of spinning an odd number = 120 200

As, 120 200 = 120 200, Theoretical probability = Experimental probability

8. Number of students randomly selected = 75

Number of students plan to go to college after completing high school = 60

Experimental probability of selecting a student who plans to go to college after completing high school = 60 75

In 700 students, the number of students who plan to go to college after completing high school would be, 700 × 60 75 = 560 students

So, 560 students are likely to go to college after completing high school.

Word Problem 1. Number of defective pieces = 5

Total number of devices = 100

Number of selected devices = 1

Probability of selected devices that are not defective = 95 100 = 19 50

Chapter Checkup 1. a. Impossible b. Sure

2. a. January comes before February is a sure event.

b. The sun setting in east is an impossible event.

c. A child going to school on a Wednesday is a likely event.

3. a. Equally likely b. Unlikely c. Impossible

d. Impossible e. Sure

4. a. Possible outcomes when a dice is rolled = 1, 2, 3, 4, 5, 6

Number of possible outcomes = 6

a. Possible outcomes when a number less than 3 is rolled = 1, 2

b. Possible outcomes when a number less than 3 is not rolled = 3, 4, 5, 6

c. Possible outcomes when an even number is rolled = 2, 4, 6

d. Possible outcomes when an even number is not rolled = 1, 3, 5

e. Possible outcomes when a 5 is not rolled = 1, 2, 3, 4, 6

f. Possible outcomes when a 7 is not rolled = 1, 2, 3, 4, 5, 6

5. a. The favourable outcomes for the event of spinning a 1 = 1, 1.

So, the number of ways the event of spinning a 1 = 2.

b. The favourable outcomes for the event of spinning a 3 = 3, 3, 3

So, the number of ways the event of spinning a 3 = 3.

c. The favourable outcomes for the event of spinning an odd number = 1, 1, 3, 3, 3.

So, the number of ways the event of spinning an odd number = 5.

d. The favourable outcomes for the event of spinning an even number = 2, 2, 2.

So, the number of ways the event of spinning an even number = 3.

e. The favourable outcomes for the event of spinning a number greater than 0 = 1, 1, 2, 2, 2, 3, 3, 3.

So, the number of ways the event of spinning a number greater than 0 = 8

f. The favourable outcomes for the event of spinning a number less than 3 = 1, 1, 2, 2, 2.

So, the number of ways the event of spinning a number less than 3 = 5.

6. Total number of times a tile is picked from the bag = 100

a. Number of times 2 is picked = 18

Probability of getting a 2 = 18 100 = 9 50

b. Number of times, an even number is picked (2, 4) = 18 + 21 = 39

Probability of getting an even number = 39 100

c. Number of times, 5 is not picked (1, 2, 3, 4) = 20 + 18 + 22 + 21 = 81

Probability of not getting a number 5 = 81 100

d. Number of times, a number less than 3 is picked (1, 2) = 20 + 18 = 38

Probability of getting a number less than 3 = 38 100 = 19 50

7. Total number of possible outcomes = 8

a. Number of favourable times of getting a blue = 2

Probability of getting a blue = 2 8 = 1 4

b. Number of favourable times of getting a number 1 = 3

Probability of getting a number 1 = 3 8

c. Number of favourable times of getting an even number (2, 4, 6, 8, 10) = 5

Probability of getting an even number = 5 8

d. Number of favourable times of getting a number 4 = 1

Probability of getting a number 4 = 1 8

8. Total number of students = 6 + 8 + 10 + 6 = 30

Number of students choose Sunday = 6

Probability of getting a student’s favourite day of the week = 6 30 = 1 5

9. Number of discs that are red = 5

Number of discs that are yellow = 3

Number of discs that are black = 2

Total number of discs = 10

a. Probability of drawing red disc = 5 10 = 1 2

b. Probability of drawing black disc = 2 10 = 1 5

c. Probability of drawing red or black disc = 5 + 2 10 = 7 10

d. Number of drawing discs that are not black = Number of drawing discs that are red or yellow = 5 + 3 10 = 8 10 = 4 5

e. Since there are no blue discs, the probability of drawing a blue disc = 0 10 = 0

f. Since it can be any of the given card red, yellow or black, the probability of drawing the card = 10 10 = 1

10. Number of the letters of the English alphabet = 26

a. Probability of picking a letter X = 1 26

b. Number of vowels = 5 (a, e, i, o, u)

Probability of picking a vowel = 5 26

c. Probability of picking a letter M or N = 2 26 = 1 13

d. Number of unique letters in the word MATHEMATICS = (M, A, T, H, E, I, C, S) = 8

Probability of getting a letter in the word MATHEMATICS = 8 26 = 4 13

11. Total number of numbers on the cards = 25

a. Odd numbers from 1 to 25 = 1,3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25

Number of odd numbers = 13

Probability of getting an odd number = 13 25

b. Multiples of 5 from 1 to 25 = 5, 10, 15, 20, 25

Number of multiples of 5 = 5

Probability of getting a multiple of 5 = 5 25 = 1 5

c. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Number of factors of 24 = 8

Probability of getting a factor of 24 = 8 25

d. Numbers that contain digit 2 = 2, 12, 20, 21, 22, 23, 24, 25

Number of numbers that contain digit 2 = 8

Probability of getting a number that contain digit 2 = 8 25

12. Probability of the event = 1 5

Total number of trials = 20

So, 1 5 = 1 5 × 4 4 = 4 20

So, in 20 trials, the number of favourable outcomes can be 4.

13. Number of red balls that are numbered = 5

Number of blue balls that are numbered 1 to 7 = 7

Total number of balls = 12

a. Probability of getting a red ball = 5 12

b. Probability of getting a ball numbered 3 = 2 12 = 1 6 (it could be either red or blue)

c. Probability of getting a ball numbered 6 = 1 12 (blue ball)

d. Probability of getting a blue ball numbered 1 = 1 12

e. Even numbered balls = 2 (two – one red and one blue), 4 (two – one red and one blue), 6 (one – only blue)

Probability of getting an even numbered ball = 5 12

Word Problem 1. Number of fiction books = 3000

Number of non-fiction books = 4000

Total number of books = 3000 + 4000 = 7000

Probability of selecting a fiction book = 3000 7000 = 3 7

Chapter 7

Let’s Warm-up

Statement Expression

1. 3 more than twice a number

2. 2 less than five times a number

3. 2 more than thrice a number

4. 5 less than twice a number

5. 5 less than thrice a number

Do It Yourself

2n − 5

3p + 2

2n + 3

3m – 5

5n − 2

7A 1. An equation has left hand side = right hand side, hence

a. 5x + 5 = 35 is an equation.

b. 3a – 7 + 24 is not an equation.

c. 5x 2 + 3 2 = 18 5 is an equation.

d. 8x –1 2 = 43 2 is an equation.

2. A linear equation has the highest exponent as 1, with LHS = RHS, hence

a. x + 8 = 13 is a linear equation.

b. x + 9 is not a linear equation.

c. 5x + 6 = 8 is a linear equation.

d. x + 8y = 6 is a linear equation.

e. x2 + 5 = 14 is not a linear equation.

f. 9x + 7y + z = 24 is a linear equation.

3. a. x + 9 = 14  b. x + 12 = 4  c. 4x + 5y = 8

d. 7x + 3 = 10  e. x2 + 7 = 29 f. 6x + 8y + 3z = 38

4. a.  3p + 7 is not an equation. True

b. 2a + 6 −11 is an equation. False

c. 2x2 + 3 = 19 is a linear equation. False

d. 8x + 3y = 9 is a linear equation in one variable. False

e. 8 2 = 4 denotes a numerical equation. True

5. a. Let the number be x

Half of the number = x 2

So, the required equation is x + x 2 = 9.

b. Let the number be x.

Thrice the number = 3x

So, the required equation is 3x + 5 = 17.

c. Let the number be x

Half of the number = x 2

So, the required equation is x 2 = 15.

d. Let the number be x

32 times the number = 32x

Sum of 32 times the number and 5 = 32x + 5

Half of the sum of 32 times the number and 5 = 32x + 5 2

So, the required equation is 32x + 5 2 = 27.

e. Let the number of frogs be x.

Two more than the number of frogs = x + 2

So, the required equation is x + 2 = 8.

f. Let the number of students be x

The number of students increased by 10 = x + 10

So, the required equation is x + 10 = 38.

6. Statements may vary, sample statements.

a. 9 subtracted from y gives 81.

b. 5 subtracted from 4 times q gives 15.

c. 30 times z equals 1650.

d. The sum of 9 times x and 5 gives 50.

e. 12 added to 4 times x is 28.

f. 2 added to one-fourth of y equals 8.

7.

Statement

Equation

a. One-tenth of a number gives 7. 10 – x = 7

b. 10 subtracted from a number gives 7. x + 27 = 77

c. A number subtracted from 10 gives 7. 1 10x = 7

d. The sum of a number and 27 is 77. x – 10 = 7

8.a. Candies with me = m

Candies with Lina = m – 3

Candies with me and Lina = 18

So, the required equation is m + m – 3 = 18 OR 2m – 3 = 18.

b. Age of Tina = c years

Age of Tina’s sister = (c + 3) years

But Tina’s sister is 12 years old.

So, the required equation that represents the given situation is c + 3 = 12.

c. Let the age of Rakesh be x years.

Age of Rajat = (2x) years

Sum of their ages = x + 2x OR 3x

But the sum of their ages is 18 years.

So, the required equation that represents the given condition is x + 2x = 18 OR 3x = 18.

d. Let the breadth of the rectangle be x cm.

The length of the rectangle = (x + 3) cm

Perimeter of the rectangle = 2(x + x + 3) = 2(2x + 3)

But the perimeter of the rectangle = 18 cm

So, the required equation that represents the given condition is 2(2x + 3) = 18.

Word Problems 1. The breadth of the rectangular hall is given as b metres.

Length of the rectangular hall = (3b – 4) metres

2. Number of rooms on each floor of a hotel = x

Number of rooms on 7 floors = 7x

But the number of rooms on 7 floors = 217.

So, the required equation that represents the given situation is 7x = 217.

7B 1. a. We shall first add 7 to both sides of the equation.

x – 7 + 7 = 0 + 7

On simplifying both sides of the equation, we get, x = 7

b. We shall first subtract 9 from both sides of the equation.

x + 9 − 9 = 0 – 9

On simplifying both sides of the equation, we get, x = −9

c. We shall first add 2 to both sides of the equation.

x – 2 + 2 = 6 + 2

On simplifying both sides of the equation, we get, x = 8

d. We shall first subtract 5 from both sides of the equation.

x + 5 − 5 = 3 – 5

On simplifying both sides of the equation, we get, x = −2

e. We shall first add 8 to both sides of the equation.

y – 8 + 8 = −4 + 8

On simplifying both sides of the equation, we get, y = 4

f. We shall first add 1 to both sides of the equation.

y – 1 + 1 = 1 + 1

y = 2

g. We shall first subtract 5 from both sides of the equation.

y + 5 − 5 = 5 – 5

On simplifying both sides of the equation, we get, y = 0

h. We shall first subtract 6 from both sides of the equation.

y + 6 − 6 = −6 – 6

On simplifying both sides of the equation, we get, y = −12

2.a. We shall divide both sides of the equation by 4.

4a 4 = 60 4

On simplifying both sides of the equation, we get, a = 15

b. We shall multiply both sides of the equation by 3.

b 3 × 3 = 9 × 3

On simplifying both sides of the equation, we get, b = 27

c. We shall multiply both sides of the equation by 8.

c 8 × 8 = 5 × 8

On simplifying both sides of the equation, we get, c = 40

d. 5x = 28

We shall divide both sides of the equation by 5.

5x 5 = 28 5 =5 3 5

On simplifying the LHS of the equation, we get, x = 28 5

e. 9y = 12

We shall divide both sides of the equation by 9. 9y 9 = 12 9

On simplifying both sides of the equation, we get, y = 4 3 = 1 1 3

f. z 7 = 6 5

We shall multiply both sides of the equation by 7.

z 7 × 7 = 6 5 × 7

On simplifying both sides of the equation, we get,

c. 4z – 6 = 3

We shall add 6 to both sides of the equation.

4z – 6 + 6 = 3 + 6

On simplifying both sides of the equation, we get, 4z = 9

We shall now divide both sides of the equation by 4.

4z

4 = 9 4

On simplifying LHS of the equation, we get, z = 9 4

d. 30p 7 = 60

We shall multiply both sides of the equation by 7.

30p

7 × 7 = 60 × 7

On simplifying both sides of the equation, we get, 30p = 420

We shall now divide both sides of the equation by 30.

30p 30 = 420 30

On simplifying both sides of the equation, we get, p = 14

e. 5 + x 2 = 6

We shall subtract 5 from both sides of the equation.

5 + x 2 – 5 = 6 – 5

On simplifying both sides of the equation, we get, x 2 = 1

We shall now multiply both sides of the equation by 2.

x

2 × 2 = 1 × 2

On simplifying both sides of the equation, we get, x = 2

f. 1 2 x – 1 = 4

We shall add 1 to both sides of the equation.

1 2 x – 1 + 1 = 4 + 1

5

z = 42 5 = 8 2

g. p 6 = 5 18

We shall multiply both sides of the equation by 6.

p 6 × 6 = 5 18 × 6

On simplifying both sides of the equation, we get, p = 5 3 OR p = 1 2 3 h. 30q = –5

We shall divide both sides of the equation by 30.

30q

30 = –5 30

On simplifying both sides of the equation, we get, q = –1 6

3. a. –x + 14 = 21

We shall subtract 14 from both sides of the equation.

–x + 14 – 14 = 21 – 14

On simplifying both sides of the equation, we get, –x = 7

We shall now multiply (−1) on both sides of the equation.

–x × (–1) = 7 × (–1)

On simplifying both sides of the equation, we get, x = –7

b. 2x + 12 = 64

We shall subtract 12 from both sides of the equation.

2x + 12 – 12 = 64 – 12

On simplifying both sides of the equation, we get, 2x = 52

We shall now divide both sides of the equation by 2.

2x 2 = 52 2

On simplifying both sides of the equation, we get, x = 26

On simplifying both sides of the equation, we get, 1 2 x = 5

We shall now multiply both sides of the equation by 2.

1 2 x × 2 = 5 × 2

On simplifying both sides of the equation, we get, x = 10

4. Let the number be x

Five times the number = 5x

Eleven added to five times the number = 5x + 11

But eleven added to five times the number = 61

So, the required equation is 5x + 11 = 61.

5x + 11 – 11 = 61 – 11

5x = 50

5x

5 = 50 5

On simplifying both sides of the equation, we get, x = 10

So, the required number is 10.

5. The answer is incorrect.

The given equation is 12 + x = 20.

We shall subtract 12 from both sides of the equation.

12 + x – 12 = 20 – 12

On simplifying both sides of the equation, we get, x = 8

So, the correct answer is 8.

6. Answers may vary. Sample answers: a. Veena had saved some money in her money bank. She spent ₹75 on buying stationery items. If she is now left with ₹150, how much money did she have to begin with?

x – 75 = 150

Add 75 on both sides of the equation,

x – 75 + 75 = 150 + 75 x = 225

b. Ankur has 249 postcards. He added new postcards to his collection. If he has 350 postcards in all, how many new postcards did he add to his collection?

249 + x = 350

Subtract 249 from both the sides of the equation,

249 + x – 249 = 350 – 249 x = 101

c. The age of Rima’s grandmother is 9 times Rima’s age. If the age of Rima’s grandmother is 72, what is the age of Rima?

Divide both sides with 9,

9x

9x = 72

9 = 72 9 x = 8

7. We shall either multiply both sides of equation 1 by 6 or divide both sides of equation 2 by 6 to make both the equations equal.

Word Problems 1. Money saved by Tina = ₹125

Cost of the jewellery set = ₹200

Let the extra money needed by Tina to buy the jewellery set be x.

So, the required equation is 125 + x = 200

We shall subtract 125 from both sides of the equation.

125 + x – 125 = 200 – 125

On simplifying both sides of the equation, we get, x = 75 Thus, Tina needs ₹75 more.

2. Let the number of stamps with Rakesh be x.

Number of stamps with Sameer = 3x + 20

Total number of stamps with Rakesh and Sameer = x + 3x + 20

But the total number of stamps with Rakesh and Sameer is 140.

So, the required equation is x + 3x + 20 = 140

On simplifying the LHS of the equation, we get, 4x + 20 = 140

We shall subtract 20 from both sides of the equation.

4x + 20 – 20 = 140 – 20

On simplifying both sides of the equation, we get, 4x = 120

We shall now divide both sides of the equation by 4.

4x 4 = 120 4 x = 30

Hence, Rakesh has 30 stamps.

3. Let the number of cups of water held by a small jug be x

The number of cups of water held by a large jug = x + 2

As per the given condition, 2x + x + 2 = 8

On simplifying the LHS of the equation, we get, 3x + 2 = 8

We shall first subtract 2 from both sides of the equation.

3x + 2 – 2 = 8 – 2

On simplifying both sides of the equation, we get, 3x = 6

We shall now divide both sides of the equation by 3.

3x 3 = 6 3

On simplifying both sides of the equation, we get, x = 2

So, the small jug can hold 2 cups of water.

x + 2 = 2 + 2 = 4

So, the large jug can hold 4 cups of water.

7C 1. a. x + 4 = 10

On transposing 4 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 10 – 4 x = 6

b. a + 9 = 13

On transposing 9 from the LHS to the RHS, the operation changes from addition to subtraction.

a = 13 – 9 a = 4

c. x + 2 7 = 5 7

On transposing 2 7 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 5 7 –2 7 x = 3 7

d. n + 2.6 = 4

On transposing 2.6 from the LHS to the RHS, the operation changes from addition to subtraction.

n = 4 – 2.6 n = 1.4

e. x − 9 = 3

On transposing −9 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 3 + 9 x = 12

f. −5.4 = 12.8

On transposing −5.4 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 12.8 + 5.4 x = 18.2

g. −14 + a = 27

On transposing −14 from the LHS to the RHS, the operation changes from subtraction to addition.

a = 27 + 14 a = 41

h. p − 5 9 = 7 9

On transposing −5 9 from the LHS to the RHS, the operation changes from subtraction to addition.

p = 7 9 + 5 9 p = 12 9 = 4 3

i. x − 6.5 = 15

On transposing −6.5 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 15 + 6.5 x = 21.5

2.a. 4y = 20

On transposing 4 from the LHS to the RHS, the operation changes from multiplication to division.

y = 20 4 y = 5

b. 90 = 15y

On transposing 15 from the RHS to the LHS, the operation changes from multiplication to division.

90 15 = y 6 = y OR y = 6

c. −9a = 108

On transposing −9 from the LHS to the RHS, the operation changes from multiplication to division.

a = 108 −9 −108 9 a = −12

d. −7m = −21

On transposing −7 from the LHS to the RHS, the operation changes from multiplication to division.

m = −21 −7 m = 3

e. j 8 = 6

On transposing 8 from the LHS to the RHS, the operation changes from division to multiplication.

j = 6 × 8 j = 48

f. x 3 = −11

On transposing 3 from the LHS to the RHS, the operation changes from division to multiplication.

x = −11 × 3 x = −33

g. w −9 = −5

On transposing −9 from the LHS to the RHS, the operation changes from division to multiplication.

w = −5 × (−9) w = 45

h. g 3.2 = 20

On transposing 3.2 from the LHS to the RHS, the operation changes from division to multiplication.

g = 20 × 3.2 g = 64

i. 2x 4.5 = 30

On transposing 4.5 from the LHS to the RHS, the operation changes from division to multiplication.

2x = 30 × 4.5 = 135

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 135 2 = 67.5

3. a. 6x + 14 = 16

On transposing 14 from the LHS to the RHS, the operation changes from addition to subtraction.

6x = 16 – 14 6x = 2

On transposing 6 from the LHS to the RHS, the operation changes from multiplication to division.

x = 2 6 = 1 3

b. 7 + 5x – 3x = 21

On simplifying the LHS, we get, 7 + 2x = 21

On transposing 7 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 21 – 7 2x = 14

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 14 2 x = 7

c. 1.2x + 4.8x = 18

On simplifying the LHS, we get, 6x = 18

On transposing 6 from the LHS to the RHS, the operation changes from multiplication to division.

x = 18 6 x = 3

d. 6(5x – 2) + 2 = 30

On opening the bracket, we get, 30x – 12 + 2 = 30 30x – 10 = 30

On transposing 10 from the LHS to the RHS, the operation changes from subtraction to addition.

30x = 30 + 10 30x = 40

On transposing 30 from the LHS to the RHS, the operation changes from multiplication to division.

x = 40 30 x = 4 3 OR x = 1 1 3

e. 7(x − 4) + 13 = 6

On opening the bracket, we get, 7x − 28 + 13 = 6 7x − 15 = 6

On transposing −15 from the LHS to the RHS, the operation changes from subtraction to addition.

7x = 6 + 15 7x = 21

On transposing 7 from the LHS to the RHS, the operation changes from multiplication to division.

x = 21 7 x = 3

f. 10y − 0.2(7 − y) = 19

On opening the bracket, we get,

10y − 1.4 + 0.2y = 19 10.2y − 1.4 = 19

On transposing −1.4 from the LHS to the RHS, the operation changes from subtraction to addition.

10.2y = 19 + 1.4 10.2y = 20.4

On transposing 10.2 from the LHS to the RHS, the operation changes from multiplication to division.

y = 20.4 10.2 y = 2

4.a. Let the number be x.

The equation that represents the given situation is x − 60 = 4.

On transposing 60 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 4 + 60 x = 64

b. Let the unknown number be x.

The equation that represents the given condition is 13 + x = 46.

On transposing 13 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 46 − 13 x = 33

c. Let the number be x.

The equation that represents the given condition is x − 21 = 81.

On transposing 21 from the LHS to the RHS, the operation changes from subtraction to addition.

x = 81 + 21 x = 102

d. Let the number be x.

The equation that represents the given condition is 7 × x = 56.

On transposing 7 from the LHS to the RHS, the operation changes from multiplication to division.

x = 56 7 x = 8

e.

Let the number be x

The equation that represents the given condition is x 7 = 5.

On transposing 7 from the LHS to the RHS, the operation changes from division to multiplication.

x = 5 × 7 x = 35

f. Let the number be x

The equation that represents the given condition is 3x − 10 = 35.

On transposing −10 from the LHS to the RHS, the operation changes from subtraction to addition.

3x = 35 + 10 3x = 45

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

g. Let the number be x.

x = 45 3 x = 15

The equation that represents the given condition is 3x + 5 = 20.

On transposing 5 from the LHS to the RHS, the operation changes from addition to subtraction.

3x = 20 − 5 3x = 15

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 15 3 x = 5

5. Let the whole number be x

The equation that represents the given condition is 3x + 9 = 45.

On transposing 9 from the LHS to the RHS, the operation changes from addition to subtraction.

3x = 45 − 9 3x = 36

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

6. Let the number be x

x = 36 3 x = 12

The equation that represents the given condition is 2 3 x − 1 3 x = 3.

On simplifying the LHS, we get, x 3 = 3

On transposing 3 from the LHS to the RHS, the operation changes from division to multiplication.

x = 3 × 3 x = 9

7. Let the other part be x

As per the given condition, x + x + 18 = 38 2x + 18 = 38

On transposing 18 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 38 − 18 2x = 20

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 20 2 x = 10 and x + 18 = 10 + 18 = 28

So, one part is 28 and the other part is 10.

8. Let the breadth of the rectangle be b.

Length of the rectangle = 5 + b

According to the given situation, 2(5 + b + b) = 250 2(5 + 2b) = 250 10 + 4b = 250

On transposing 10 from the LHS to the RHS, the operation changes from addition to subtraction.

4b = 250 – 10 4b = 240

On transposing 4 from the LHS to the RHS, the operation changes from multiplication to division.

b = 240 4 b = 60

So, the breadth of the rectangle is 60 m.

Length of the rectangle = 5 + b = 5 + 60 = 65 m.

Word Problems 1. Let the cost of the bat be x

Cost of the ball = 1 5 x

As per the given situation, x + 1 5 x = 1200 6x 5 = 1200

On transposing 5 from the LHS to the RHS, the operation changes from division to multiplication.

6x = 1200 × 5 6x = 6000

On transposing 6 from the LHS to the RHS, the operation changes from multiplication to division.

x = 6000 6 x = 1000

So, the cost of the bat is ₹1000.

Cost of the ball = 1 5 x = 1 5 × 1000 = 200

So, the cost of the ball is ₹200.

2. Let the weight of one of the bags full of onions be x

Weight of the other bag = 5 + x

As per the given situation, x + 5 + x = 31 2x + 5 = 31

On transposing 5 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 31 − 5 2x = 26

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 26 2 x = 13

So, the weight of one of the bags full of onions is 13 kg.

Weight of the other bag = 5 + x = 5 + 13 = 18 kg

7D 1. Let the number be x; twice the number = 2x

Twice the number decreased by 22 = 2x − 22

According to the given condition, 2x − 22 = 48

On transposing 22 from the LHS to the RHS, the operation changes from subtraction to addition.

2x = 48 + 22 2x = 70

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 70 2 x = 35

So, the required number is 35.

2. Let the number be x; half of the number = x 2

According to the given condition, x + x 2 = 72 3x 2 = 72

On transposing 2 from the LHS to the RHS, the operation changes from division to multiplication.

3x = 72 × 2 3x = 144

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 144 3 x = 48

So, the required number is 48.

3. Let one of the numbers be x; the other number will be (x + 3)

As per the given condition, x + x + 3 = 95 2x + 3 = 95

On transposing 3 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 95 − 3 2x = 92

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 92 2 x = 46

So, one of the numbers is 46.

The other number = x + 3 = 46 + 3 = 49

4. Let one of the parts be x; the other part = x + 8

According to the given condition, x + x + 8 = 20 2x + 8 = 20

On transposing 8 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 20 − 8 2x = 12

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 12 2 x = 6

So, one of the parts is 6.

The other part = x + 8 = 6 + 8 = 14

5. Let the three consecutive numbers be (x − 1), x, and (x + 1).

As per the given condition, x + (x + 1) – (x – 1) = 24

x + x + 1 – x + 1 = 24 x + 2 = 24

On transposing 2 from the LHS to the RHS, the operation changes from addition to subtraction.

x = 24 – 2 x = 22

One of the numbers is 22.

(

x − 1) = 22 – 1 = 21

(x + 1) = 22 + 1 = 23

So, the three consecutive numbers are 21, 22, and 23.

6. The sum of the measures of the angles of a triangle is 180°.

Hence, x + (x + 40)° + (x – 10)° = 180° 3x + 30° = 180°

On transposing 30 from the LHS to the RHS, the operation changes from addition to subtraction.

3x = 180° – 30° 3x = 150°

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 150°

3 x = 50°

(x + 40)° = 50° + 40° = 90°

(x – 10)° = 50° – 10° = 40°

So, the angles of a triangle are 50°, 90°, and 40°.

7. Perimeter of a triangle = Sum of the meas ures of the three sides

Hence, 2x + 3 + 2x + 5 + 3x = 148 7x + 8 = 148

On transposing 8 from the LHS to the RHS, the operation changes from addition to subtraction.

7x = 148 – 8 7x = 140

On transposing 7 from the LHS to the RHS, the operation changes from multiplication to division.

x = 140

7 x = 20 units

2x + 3 = 2 × 20 + 3 = 43 units; 2x + 5 = 2 × 20 + 5 = 45 units and 3x = 3 × 20 = 60 units

So, the measures of the three sides of the triangle are 43 units, 45 units, and 60 units, respectively.

Word Problems 1. Let the age of Olivia be x; Age of Olivia’s mother = 3x + 5

As per the given situation, 3x + 5 = 44

On transposing 5 from the LHS to the RHS, the operation changes from addition to subtraction.

3x = 44 – 5 3x = 39

On transposing 3 from the LHS to the RHS, the operation changes from multiplication to division.

x = 39 3 x = 13

So, the age of Olivia is 13 years.

2. Let the present age of Satish be x

Five years from now, the age of Satish will be (x + 5).

As per the given situation, (x + 5) = 1 1 2 x (x + 5) = 3 2 x

On transposing 2 from the RHS to the LHS, the operation changes from division to multiplication.

2 × (x + 5) = 3x 2x + 10 = 3x

On transposing 2x from the LHS to the RHS, the operation changes from addition to subtraction.

10 = 3x – 2x 10 = x

Hence, the present age of Satish is 10 years.

3. Let the number of magazines be x

The number of books = 3x

As per the given situation, x + 3x = 320 4x = 320

On transposing 4 from the LHS to the RHS, the operation changes from multiplication to division.

x = 320 4 x = 80

Number of books = 3x = 3 × 80 = 240

4. Let the cost of a pen be x; cost of a book = 50 + x

As per the given situation, 5 × (50 + x) + 3x = 730 250 + 5x + 3x = 730 250 + 8x = 730

On transposing 250 from the LHS to the RHS, the operation changes from addition to subtraction.

8x = 730 − 250 8x = 480

On transposing 8 from the LHS to the RHS, the operation changes from multiplication to division.

x = 480

8 x = 60

So, the cost of the pen is ₹60.

Cost of the book = 50 + x = 50 + 60 = ₹110

5. Let the present age of Heena be x years; present age of her mother = 4x

20 years hence,

Age of Heena will be (x + 20) years.

Age of Heena’s mother will be (4x + 20) years.

As per the given situation, 4x + 20 = 2(x + 20)

4x + 20 = 2x + 40

On transposing 2x from the RHS to the LHS, the operation changes from addition to subtraction.

4x − 2x + 20 = 40 2x + 20 = 40

On transposing 20 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 40 − 20 2x = 20

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 20 2 x = 10

Hence, the present age of Heena is 10 years.

The present age of Heena’s mother = 4x = 4 × 10 = 40 years

6. Let the age of Tina be x years.

Since the difference between the ages of Savita and Tina is 4 years.

Therefore, age of Savita = x + 4

As per the given situation, x + x + 4 = 18 2x + 4 = 18

On transposing 4 from the LHS to the RHS, the operation changes from addition to subtraction.

2x = 18 − 4 2x = 14

On transposing 2 from the LHS to the RHS, the operation changes from multiplication to division.

x = 14 2 x = 7

Hence, the age of Tina is 7 years.

Age of Savita = x + 4 = 7 + 4 = 11 years

Chapter Checkup 1. a. x + 3 8 = 1 3 8 x + 3 8 = 11 8

We shall subtract 3 8 from both sides of the equation.

x + 3 8 − 3 8 = 11 8 − 3 8 x = 8 8 x = 1

b. x − 1 3 = 2 3

We shall add 1 3 to both sides of the equation.

x − 1 3 + 1 3 = 2 3 + 1 3 x = 3 3 x = 1

c. x + 5 = 8

We shall subtract 5 from both sides of the equation.

x + 5 − 5 = 8 − 5 x = 3

d. x + 5 19 = 3 19

We shall subtract 5 19 from both sides of the equation.

x + 5 19 − 5 19 = 3 19 − 5 19 x = –2 19

e. x – 3 = –20

We shall add 3 to both sides of the equation.

x – 3 + 3 = –20 + 3 x = –17

f. x – 3 = 7

We shall add 3 to both sides of the equation.

x – 3 + 3 = 7 + 3 x = 10

2. a. z 2 – 2 = 26

We shall add 2 to both sides of the equation.

z 2 – 2 + 2 = 26 + 2 z 2 = 28

b. x + 1 = –6

We shall subtract 1 from both sides of the equation.

x + 1 – 1 = –6 – 1 x = –7

c. x + 1 5 – 3 = 27

We shall add 3 to both sides of the equation.

x + 1 5 – 3 + 3 = 27 + 3 x + 1 5 = 30

d. 2x + 11 = 17

We shall subtract 11 from both sides.

2x + 11 – 11 = 17 – 11 2x = 6

b. 4a = 12

On transposing 4 from the LHS to the RHS, the operation changes from multiplication to division.

a = 12 4 a = 3

c. –3y = –27

On transposing (–3) from the LHS to the RHS, the operation changes from multiplication to division.

y = –27 –3 y = 9

d. –18 = –6x

On transposing (–6) from the RHS to the LHS, the operation changes from multiplication to division.

–18

–6 = x 3 = x OR x = 3

e. –4x = 8

On transposing (–4) from the LHS to the RHS, the operation changes from multiplication to division.

x = 8 –4 x = –2

f. –7x = 3

On transposing (–7) from LHS to RHS, the operation changes from multiplication to division.

x = 3 –7 OR x = –3 7

4.a. Let the number be x

As per the given condition, 1 1 2 × x = 150 3 2 × x = 150

On transposing 2 from LHS to RHS, the operation changes from division to multiplication.

3x = 150 × 2 3x = 300

On transposing 3 from LHS to RHS, the operation changes from multiplication to division.

x = 300 3 x = 100

So, the required number is 100.

b. Let the number be x.

As per the given condition, 5x = 65.

On transposing 5 from LHS to RHS, the operation changes from multiplication to division.

x = 65 5 x = 13

So, the required number is 13.

c. Five times p is 5p

As per the given condition, 5p – 3 = 22.

e. 5(x – 3) = 25

We shall divide both sides by 5. 5(x – 3) 5 = 25 5 (x – 3) = 5

f. 3(y – 7) – 4 = 14

We shall add 4 to both sides of the equation.

3(y – 7) – 4 + 4 = 14 + 4 3(y – 7) = 18

3.a. 9x = 18

On transposing 9 from the LHS to the RHS, the operation changes from multiplication to division.

x = 18 9 x = 2

On transposing 3 from LHS to RHS, the operation changes from subtraction to addition.

5p = 22 + 3 5p = 25

On transposing 5 from LHS to RHS, the operation changes from multiplication to division.

p = 25 5 p = 5

So, the value of p is 5.

d. Let the number be x

As per the given condition, 2x + 36 = 48.

On transposing 36 from LHS to RHS, the operation changes from addition to subtraction.

2x = 48 – 36 2x = 12

On transposing 2 from LHS to RHS, the operation changes from multiplication to division.

x = 12 2 x = 6

So, the required number is 6.

e. As per the given condition, 4p + 9 = 31.

On transposing 9 from LHS to RHS, the operation changes from addition to subtraction.

4p = 31 – 9 4p = 22

On transposing 4 from LHS to RHS, the operation changes from multiplication to division.

p = 22 4 = 11 2

So, the value of p is 11 2 .

5.a. q 6 = 7

On transposing 6 from LHS to RHS, the operation changes from division to multiplication.

q = 7 × 6 q = 42

b. x –29 = 2

On transposing (–29) from LHS to RHS, the operation changes from division to multiplication.

x = 2 × (–29) x = –58

c. 9 2 x = 54

On transposing 2 from LHS to RHS, the operation changes from division to multiplication.

9x = 54 × 2 9x = 108

On transposing 9 from LHS to RHS, the operation changes from multiplication to division.

x = 108 9 x = 12

d. –13 2 x = 26

On transposing 2 from LHS to RHS, the operation changes from division to multiplication.

–13x = 26 × 2 –13x = 52

On transposing (–13) from LHS to RHS, the operation changes from multiplication to division.

x = 52 –13 x = –4

e. x + 10 = –10

On transposing 10 from LHS to RHS, the operation changes from addition to subtraction.

x = –10 – 10 x = –20

f. x – 17 = –20

On transposing (–17) from LHS to RHS, the operation changes from subtraction to addition.

x = –20 + 17 x = –3

g. y + 7 = 4

On transposing 7 from LHS to RHS, the operation changes from addition to subtraction.

y = 4 – 7 y = –3

h. x + 20 = –20

On transposing 20 from LHS to RHS, the operation changes from addition to subtraction.

6.a. x 3 – 5 = 6

On transposing (–5) from LHS to RHS, the operation changes from subtraction to addition.

x 3 = 6 + 5 x 3 = 11

On transposing 3 from LHS to RHS, the operation changes from division to multiplication.

x = 11 × 3 x = 33

b. 3x – 2 = 5x – 12

On transposing (–12) from RHS to LHS, the operation changes from subtraction to addition.

3x – 2 + 12 = 5x

On transposing 3x from LHS to RHS, the operation changes from addition to subtraction.

–2 + 12 = 5x – 3x 10 = 2x

On transposing 2 from RHS to LHS, the operation changes from multiplication to division.

10

2 = x 5 = x OR x = 5

c. 12y = 7y – 15

On transposing 7y from RHS to LHS, the operation changes from addition to subtraction.

12y – 7y = –15 5y = –15

On transposing 5 from LHS to RHS, the operation changes from multiplication to division.

y = –15 5 y = –3

d. –22 = 14 – 9x

On transposing 14 from RHS to LHS, the operation changes from addition to subtraction.

–22 – 14 = –9x –36 = –9x

On transposing (–9) from RHS to LHS, the operation changes from multiplication to division.

–36 –9 = x 4 = x OR x = 4

e. 13y = –12y + 100

On transposing (–12y) from RHS to LHS, the operation changes from subtraction to addition.

13y + 12y = 100 25y = 100

On transposing 25 from LHS to RHS, the operation changes from multiplication to division.

y = 100 25 y = 4

f. 5(x – 3) = 10

On transposing 5 from LHS to RHS, the operation changes from multiplication to division.

(x – 3) = 10 5 (x – 3) = 2

On transposing (–3) from LHS to RHS, the operation changes from subtraction to addition.

x = 2 + 3 x = 5

g. 9(x + 14) = 27

On transposing 9 from LHS to RHS, the operation changes from multiplication to division.

(x + 14) = 27 9 (x + 14) = 3

On transposing 14 from LHS to RHS, the operation changes from addition to subtraction.

x = 3 – 14 x = –11

h. 3 – x 5 = 6

x = –20 – 20 x = –40

On transposing 5 from LHS to RHS, the operation changes from division to multiplication.

3 – x = 6 × 5 3 – x = 30

On transposing 3 from LHS to RHS, the operation changes from addition to subtraction.

–x = 30 – 3 –x = 27

On transposing (–1) from LHS to RHS, the operation changes from multiplication to division.

x = 27 –1 x = –27

i. 5x + 12 = 2

On transposing 12 from LHS to RHS, the operation changes from addition to subtraction.

5x = 2 – 12 5x = –10

On transposing 5 from LHS to RHS, the operation changes from multiplication to division.

x = –10 5 x = –2

7. Let the number be x.

As per the given condition, x = 13 × (–18) + 7 x = –234 + 7 x = –227

So, the required number is –227.

8. Let the three consecutive even numbers be (x – 2), x, (x + 2).

As per the given condition, (x – 2) + x + (x + 2) = 96 3x = 96

On transposing 3 from LHS to RHS, the operation changes from multiplication to division.

x = 96 3 x = 32

(x – 2) = 32 – 2 = 30

(x + 2) = 32 + 2 = 34

So, the three consecutive even numbers are 30, 32, and 34.

9. Let the three consecutive numbers be (x – 1), x, (x + 1).

As per the given condition, x + x + 1 – (x – 1) = 14.

2x + 1 – x + 1 = 14 x + 2 = 14

On transposing 2 from LHS to RHS, the operation changes from addition to subtraction.

x = 14 – 2 x = 12

x – 1 = 12 – 1 = 11

x + 1 = 12 + 1 = 13

So, the three consecutive numbers are 11, 12, and 13.

10. Since all sides of a square are equal, hence 4x – 7 = 3x + 5.

On transposing 3x from RHS to LHS, the operation changes from addition to subtraction.

4x – 3x – 7 = 5 x – 7 = 5

On transposing (–7) from LHS to RHS, the operation changes from subtraction to addition.

x = 5 + 7 x = 12

So, the value of x is 12.

11. Let the third side be x.

As per the given condition, x + 3x + 3x = 28 7x = 28

On transposing 7 from LHS to RHS, the operation changes from multiplication to division.

x = 28 7 x = 4

So, the third side is 4 cm long.

3x = 3 × 4 = 12

So, the length of the sides of the isosceles triangle is 4cm; 12 cm; 12 cm.

Word Problems 1. Let the weight of packet q be q

Weight of packet p = 1  1 2 + q

Weight of packet r = 2 1 2 + q

As per the given condition, q + 1  1 2 + q + 2  1 2 + q = 32

3q + 4 = 32

On transposing 4 from LHS to RHS, the operation changes from addition to subtraction

3q = 32 – 4 3q = 28

On transposing 3 from LHS to RHS, the operation changes from multiplication to division

q = 28 3 = 9 1 3

p = 1  1 2 + 9  1 3 = 3 2 + 28 3 = 65 6 = 10 5 6 kg

r = 2 1 2 + 9 1 3 = 5 2 + 28 3 = 71 6 = 11  5 6 kg

The weights of the packets are 9 1 3 kg, 10 5 6 kg, and 11 5 6 kg respectively.

2. Let the price of a textbook be x

Price of a notebook = x + 10

As per the given situation, 2 × (x + 10) + 3x = 120. 2x + 20 + 3x = 120 5x + 20 = 120

On transposing 20 from LHS to RHS, the operation changes from addition to subtraction.

5x = 120 – 20 5x = 100

On transposing 5 from LHS to RHS, the operation changes from multiplication to division.

x = 100 5 x = 20

So, the price of a textbook is ₹20.

Price of a notebook = x + 10 = 20 + 10 = ₹30.

3. Let the number toffees with Ramesh be x.

Number of toffees with his sister = 2x

As per the given situation, x + 2x = 54 3x = 54

On transposing 3 from LHS to RHS, the operation changes from multiplication to division.

x = 54 3 x = 18

So, the number of toffees with Ramesh is 18.

4. Let the present age of the son be x; present age of the father = 30 + x 12 years hence,

Age of the son = x + 12

Age of the father = 30 + x + 12 = 42 + x

As per the given situation, 42 + x = 3(x + 12) 42 + x = 3x + 36

On transposing 36 from RHS to LHS, the operation changes from addition to subtraction.

42 + x – 36 = 3x x + 6 = 3x

On transposing x from LHS to RHS, the operation changes from addition to subtraction.

6 = 3x – x 6 = 2x

On transposing 2 from RHS to LHS, the operation changes from multiplication to division.

6 2 = x 3 = x OR x = 3

So, the present age of the son is 3 years.

Present age of the father = 30 + x = 30 + 3 = 33 years.

Chapter 8

Let’s Warm-up 1.   27°  2.  142°

3.  180°  4.  213°

Do It Yourself

8A

1. a. Yes, since 80° + 100° = 180°

b. Yes, since 45° + 135° = 180°

c. Yes, since 60° + 120° = 180°

d. No, since 35° + 125° = 160°

e. Yes, since 70° + 110° = 180°

2.a. Supplement of 33° = 180° – 33° = 147°

b. Supplement of 72° = 180° – 72° = 108°

c. Supplement of 123° = 180° – 123° = 57°

d. Supplement of 51° = 180° – 51° = 129°

e. Supplement of 98° = 180° – 98° = 82°

3.a. Yes, since 27° + 63° = 90°

b. No, since 40° + 53° = 93°

c. No, since 78° + 33° = 111°

d. No, since 37° + 37° = 74°

e. Yes, since 46° + 44° = 90°

4.a. Complement of 1° = 90° – 1° = 89°

b. Complement of 19° = 90° – 19° = 71°

c. Complement of 89° = 90° – 89° = 1°

d. Complement of 41° = 90° – 41° = 49°

e. Complement of 37° = 90° – 37° = 53°

5.a. Angles are neither supplementary nor complementary, since 25° + 108° = 133°.

b. Angles are neither supplementary nor complementary, since 32° + 34° = 66°.

c. Angles are complementary, since 52° + 38° = 90°.

d. Angles are supplementary, since 122° + 58° = 180°.

6.a. (4x + 50)° + (3x – 50)° = 180° (Linear pair)

7x = 180°. So, x = 180° 7 = 25.7° (approx)

b.    x 2    ° +    x 3    ° = 180° (Linear pair)

3x + 2x

3 × 2 = 5x 6 = 180°. So, x = 6 5 × 180° = 216°

c. (2x + 15)° + (2x + 25)° + (x – 20)° + 90° = 360°

5x + 110° = 360°

5x = 360° – 110° = 250°. So, x = 250° 5 = 50°

7.a. y° + y° + y° = 135° (vertically opposite angle)

3y° = 135°. So, y = 135° 3 = 45°

b. n° + 90° + 30° = 180°

n + 120° = 180°. So, n = 180° – 120° = 60°

m° + n° = 180° (Linear pair)

m + 60° = 180°. So, m = 180° – 60° = 120°

c. 2a° 3a° a° vertically opposite angle

a° + 2a° + 3a° = 180°

6a = 180°

a = 180° 6 = 30°

8. 90°

90°

90°

A D C

90°

B In rectangle ABCD, A = B = C = D = 90°.

Pair of supplementary angles: A and B, B and C, C and D, D and A, D and B, C and A.

9. (3x + 10)° + (7x – 20)° = 90°

10x – 10° = 90°

10x = 90° + 10° = 100°. So, x = 100° 10 = 10°

10. (4y + 13)° + (3y – 8)° = 180°

7y + 5° = 180°

7y = 180° – 5° = 175°. So, y = 175° 7 = 25°

Measure of angles = (4y + 13)° = (4 × 25 + 13)° = 113°

(3y – 8)° = (3 × 25 – 8)° = 67°

So, the measure of the angles are 113° and 67°.

11. Let the angle be x and its complement be x – 30.

x + (x – 30°) = 90°

2x = 90° + 30° = 120°. So, x = 120° 2 = 60°.

12. Let the angle be x

Complement of the angle = 90° – x

Supplement of the angle = 180° – x

Given,

180° – x = 3(90° – x)

180° – x = 270° – 3x

3x – x = 270° – 180°

2x = 90°

x = 90° 2 = 45°

13. Let the angle be x. Its supplement will be x 2

x + = 180° = > 2x + x 2 180° or 3x = 180° × 2 = 360°.

x = 360° 3 =120°

Supplement: 180° – x = 180° – 120° = 60°

14. Let the angles be 5x and 4x

Given, 5x + 4x = 90°.

9x = 90°. So, x = 90° 9 = 10°

Angles are 5x = 5 × 10° = 50°

4x = 4 × 10° = 40°

15. Let the angles be 13x and 5x

Given, 13x + 5x = 180°

18x = 180°

x = 180° 18 = 10°

Angles are 13x = 13 × 10° = 130°

5x = 5 × 10° = 50°

16. Let the angles be x and y

Supplement of x: 180° – x

Complement of y: 90° – x (180° – x) – x = 46°

180° – x – x = 46°

2x = 180° – 46° = 134°. So, x = 134° 2 = 67°

Also, y – (90° – y) = 20°

y – 90° + y = 20°

2y = 90° + 20° = 110°. So, y = 110° 2 = 55°

Word Problem 1. 70°, since they are vertically opposite to each other.

8B 1. Pairs of corresponding angles: 2 and 6, 2 and 10, 6 and 10, 1 and 5, 1 and 9, 5 and 9, 3 and 7, 7 and 11, 3 and 11, 4 and 8, 8 and 12, 12 and 4.

Pairs of alternate angles: 2 and 8, 1 and 7, 3 and 5, 4 and 6, 1 and 11, 2 and 12, 3 and 9, 4 and 10, 6 and 12, 5 and 11, 7 and 9, 8 and 10.

2. a. The 110° angle should be equal to the supplement of the 72° angle since they are corresponding pair of angles.

Supplement of 72° = 180° – 72° = 108°. Since 110° 72°, the lines are not parallel.

b. Clearly, alternate exterior angles are not equal (89° 99°) so lines are not parallel.

c. The 102° angle should be equal to the supplement of the 89° angle since they are corresponding pair of angles.

Supplement of 89° = 180°– 89° = 91°. Since 102° 91°, the lines are not parallel.

d. Clearly, alternate exterior angles are equal (109°= 109°) so lines are parallel.

3. a. x° = 80° (alternate angles); y° = 32° (alternate angles)

b. m° + 110° = 180° (interior angles on the same side of the transversal)

m° = 180° – 110° = 70°

m° = n° = 70° (alternate angles)

c. p° = 60° (alternate exterior angles)

q° = 105° (alternate interior angles)

d. h° = 45° (corresponding angles)

g° = 72° (alternate angles)

4. QPS + PSR = 180° (interior angles on the same side of the transversal)

120° + PSR = 180°

PSR = 180° – 120° = 60°

Similarly, QRS + PQR = 180° (interior angles on the same side of the transversal)

QRS + 150° = 180°

QRS = 180° – 150° = 30°

5. EAC = ACB = 60° (alternate angles)

DAB = ABC = 90° (alternate angles)

DAB + BAC + CAC = 180° (Linear sum property)

BAC + 60° + 90° = 180°

BAC = 180° – 150° = 30°

6. Angle a angle should be equal to the supplement of the 62° angle since they form corresponding pair of angles.

Supplement of 62° = 180° – 62° = 118°. So, a° = 118°.

Angle b angle should be equal to the supplement of the 72° angle since they form corresponding pair of angles.

Supplement of 72° = 180° – 72° = 108°. So, b° = 108°.

7.a. y° = 35° (alternate angles); x° = 50° (alternate angles)

b. Draw a line parallel to the 2 lines and label the angles, as shown.

116° + a° = 180° (interior angles of the same side of the transversal)

a° = 180° – 116° = 64°

136° + b° = 180° (interior angles of the same side of the transversal)

b° = 180° – 136° = 44°

x° = a° + b° = 64° + 44° = 108°

Word Problems 1. The rain would hit the ground at an angle of 60° since they form alternate angles.

60°

60°

8C Answers may vary. Sample answer. 1. a.

2.  D A B P

C S l

Q R

3. Figures may vary. 4. P

P m n

Sample answer: m is not parallel to n Q R

5.  Q n

5.9 cm 5.9 cm EF

q B A

m 5.5 cm   6. XD = 5.6 cm; YD = 9 cm; ZD = 4.2 cm

Word Problem 1. Rough figure. 10 cm

8 cm 4 cm 6 cm

Steps:i. Draw a rectangle ABCD of length 10 cm × 8 cm.

ii. Mark any two points A and B on any edge of the outer rectangle.

iii. Construct right angles at points A and B, respectively.

iv. Draw arcs with centres at A and B, each with a 2 cm radius, intersecting the perpendiculars at A and B at points E and F.

v. Connect points EF and extend this line on both sides to achieve the desired parallel line.

vi. Repeat the process for the other three sides.

vii. Connect the corresponding points on these lines to form the inner rectangle.

Chapter Checkup 1. a. Yes, since 14° + 166° = 180°.

b. No, since 50° + 150° = 200° c. No, since 36° + 135° = 171°

d. No, since 40° + 135° = 175° e. No, since 76° + 109° = 185°

2.a. No, since 24° + 56° = 80° b. No, since 50° + 16° = 66°

c. Yes, since 36° + 54° = 90° d. Yes, since 44° + 46° = 90°

e. Yes, since 78° + 12° = 90°

3.a. Supplement of 34° = 180° – 34° = 146°

b. Supplement of 119° = 180° – 119° = 61°

c. Supplement of 126° = 180° – 126° = 54°

d. Supplement of 24° = 180° – 24° = 156°

e. Supplement of 166° = 180° – 166° = 14°

4.a. Complement of 47° = 90° – 47° = 43°

b. Complement of 20° = 90° – 20° = 70°

c. Complement of 26° = 90° – 26° = 64°

d. Complement of 40° = 90° – 40° = 50°

e. Complement of 86° = 90° – 86° = 4°

5.a. (x + 9)° + 2x° = 180° (Linear pair)

3x° + 9° = 180°

3x° = 180° – 9° = 171°. So, x = 171° 3 = 57°

b. x° + x° + x° = 180°

3x° = 180°

x° = 180°

3 = 60°

c.  2x° + (3x – 10)° = 90°

2x° + 3x° – 10° = 90°

5x° = 90° + 10° = 100°. So, x = 100° 5 = 20°

d. (2x + 6)° + 3x° + x° = 180°

2x° + 6° + 3x° + x° = 180°

6x° = 180° – 6° = 174°. So, x = 174° 6 = 29°

e. 3x° + 4x° + x° + (2x – 10)° + 90° = 360°

10x° + 80° = 360°

10x° = 360° – 80° = 280°. So, x = 280° 10 = 28°

f. x° + 2x° + 4x° = 180°

7x° = 180°

x = 180° 7 = 25.7° (approx)

6. CAB = 180° – 142° = 38° (Linear pair)

ACB = 180° – 113° = 67° (Linear pair)

CAB + ACB + ABC = 180° (Angle sum property of the triangle)

38° + 67° + ABC = 180°

ABC = 180° – 105° = 75°

7. m° = 137° (alternate exterior angles); g° = 137° (corresponding angles)

y° = Supplement of the 113° angle = 180° – 113° = 67° (alternate interior angles)

h° = Supplement of the 113° angle = 180° – 113° = 67° (corresponding angles)

8. a. From the figure, we can see that the alternate exterior angles are not equal (120° 130°). Hence, the lines p and q are not parallel.

b. The 120° angle should be equal to the supplement of the 70° angle since they are corresponding pair of angles. Supplement of 70° = 180° – 70° = 110°. Since 110° 120°, the lines are not parallel.

c. The 75° angle should be equal to the supplement of the 105° angle since they are corresponding pair of angles.

Supplement of 105° = 180° – 105° = 75°. Since they are equal, the lines are parallel.

d. The 110° at the bottom should be equal to the supplement of the 110° angle at the top alternate exterior angles.

Supplement of 110° = 180° – 110° = 70°. Since they are not equal (110° 70°) the lines are not parallel.

9.a.

b.

l m

Supplement of 105° = 180° – 105° = 75°

Supplement of 65° = 180° – 65° = 115°

Supplement of 120° = 180° – 120° = 60°

Supplement of 55° = 180° – 55° = 125°

Supplement of 125° = 180° – 125° = 55°

Supplement of 110° = 180° – 110° = 70°

Supplement of 140° = 180° – 140° = 40°

Supplement of 130° = 180° – 130° = 50°

10. a. y° should be equal to the supplement of the 65° angle (corresponding angles).

y° = Supplement of 65° = 180° – 65° = 115°.

b. z° = 130° (alternate exterior angles)

c. m° should be equal to the supplement of the 70° angle (corresponding angles).

m° = Supplement of 70° = 180° – 70° = 110°.

d. x° should be equal to the supplement of the 85° angle (alternate exterior angles).

x° = Supplement of 85° = 180° – 85° = 95°.

11.a. g° + h° = 180° (interior angles on the same side of transversal)

b. x° + y° = 180° (interior angles on the same side of transversal)

c. g° = 153° (alternate angles)

d. g° + h° = 180° (interior angles on the same side of transversal)

153° + h° = 180°

h° = 180° – 153° = 27°

e. x° + h° + 53° = 180°

x° + 27° + 53° = 180°

x° + 80° = 180°

x° = 180° – 80° = 100°

f. x° + y° = 180° (interior angles on the same side of transversal)

y° = 180° – 100° = 80°

12. x° + 75° = 180° (interior angles on the same side of transversal)

x° = 180° – 75° = 105°

y° + (75° – 15°) = 180° (interior angles on the same side of transversal)

y° = 180° – 60° = 120°

w° + 75° = 180° (interior angles on the same side of transversal)

w° = 180° – 75° = 105°

(w° + 15°) + z° = 180° (interior angles on the same side of transversal)

(105° + 15°) + z° = 180°

z° = 180° – 120° = 60°

13. Since PQRS is a parallelogram, PS ꠱ QR and PQ ꠱ RS.

PQR + QPS = 180° (interior angles on the same side of transversal)

PQR + 115° = 180°

PQR = 180° – 115° = 65°

RSP + QPS = 180° (interior angles on the same side of transversal)

RSP + 115° = 180°

RSP = 180° – 115° = 65°

RSP + QRS = 180° (interior angles on the same side of transversal)

QRS + 65° = 180°

QRS = 180° – 65° = 115°

14. QDC = QBA = 63° (alternate angles)

DCQ = BAQ = 47° (alternate angles)

Word Problem 1. Since the yellow lines in the parking space are parallel, m = 115° (corresponding angle)

Chapter 9

Let’s Warm-up

1. 60°, 70°, 50° Isosceles triangle

2. 90°, 45°, 45° Obtuse-angled triangle

3. AB = BC = CA Acute-angled triangle

4. 100°, 30°, 50° Equilateral triangle

5. AB = BC Isosceles-right triangle

Do It Yourself

9A 1. a. Median is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.

b. Altitude is a line segment perpendicular to a side of the triangle.

c. The point where the altitudes intersect is called the orthocenter of the triangle.

d. An altitude may or may not lie inside the triangle.

2. If BD is the median of triangle ABC, AD is given as 5 cm.

AC = AD + DC

As BD is the median, AD = DC

AC = AD + AD = 2 × AD

AC = 2 × 5 cm = 10 cm

The length of AC is 10 cm.

3. AD and CE are the two medians in a triangle ABC on the opposite sides BC and AB, respectively.

G is the centroid, therefore, AG:GD = 2:1 (property of a centroid)

AG = 2 cm

GD = 1 2 × AG = 2 2 = 1 cm

AD = 2 + 1 = 3 cm

A median divide the opposite side in two equal parts, therefore,

BD = CD = 4 cm

BC = BD + CD = 4 + 4 = 8 cm

4. Medians of the given triangle are CF, EB and AD.

a. G is the centroid, therefore, GE:BG = 2:1 (property of a centroid)

BG = 5 cm,

GE = 2 × 5cm = 10 cm

BE = BG + GE = 5 cm + 10 cm = 15 cm

b. CG:GF = 2:1 (property of centroid)

CG = 16 cm,

GF = 16 2 = 8 cm

CF = CG+GF = 16 cm + 8 cm = 24 cm

c. AD = 30cm

AG:GD = 2:1 (property of centroid)

AG = 2 3 × 30 = 20 cm

GD = 1 3 × 30 = 10 cm

d. GC:GF = 2:1 (property of centroid)

GF = x

GC = 2 × x = 2x

CF = GF + GC = x + 2x = 3x

e. AG:GD = 2:1 (property of centroid)

AG = 2 × GD

9x = 2 × (5x − 1)

9x = 10x − 2

10x − 9x = 2

x = 2

AD = AG + GD

= 9x + 5x −1 = 14x – 1

= 14 × 2 – 1  = 27

5.c. Equilateral triangle

6. PS is the median of a triangle PQR

Therefore, QS = RS

10x – 7 = 5x + 3

10x – 5x = 3 + 7

5x = 10

x = 2

PSR = (15x + 42)°

= (15 × 2 + 42)° = (30 + 42)° = 72°.

PSR = 72°

7. If point M is equidistant from points X and Y, it means that M lies on the perpendicular bisector of XY. Therefore, the correct answer is (b) The perpendicular bisector of XY

Word Problem 1. Option e; Shyam would find the centroid of the triangular table.

The centroid is the point where the three medians of a triangle intersect. The centroid is a stable point that ensures even distribution of weight, which is essential for the stability of the table.

9B 1.a. A triangle can have two acute angles.

b. If two acute angles of a right-angled triangle are equal, then the measure of the acute angle is 45°.

c. The exterior angle is greater than each of the interior opposite angles.

2.a. x + 35° + 40° = 180° (Angle sum property of a triangle)

x = 180° − 75° = 105°

b. 4x + 3x + 50° = 180° (Angle sum property of a triangle)

7x = 180° − 50° = 130°

7x = 130°

x = 130°

7

c. (x + 2)° + (3x + 3)° + (2x + 10)° = 180° (Angle sum property of a triangle)

6x + 15 = 180°

6x = 180° – 15° = 165°

x = 165°

6

d. x + y + 60° = 180°; (Angle sum property of a triangle)

y + 60° = 120° (Exterior angle sum property of a triangle)

Therefore, y = 120° – 60° = 60°

y = 60°

hence, x + 120° = 180°

x = 180° – 120° = 60°

x = 60°

e. x + y + y = 180° (Angle sum property of a triangle)

x + 2y = 180°—eq 1

Also, since 2y = 110°

Therefore y = 55°

Substituting the value of y, in—eq 1

x + 110° = 180°

x = 180° − 110°

x = 70°

f. Since the given triangle is an isosceles triangle, the angles opposite to the equal sides are equal.

Hence, x = 48° + 48° (Exterior angle property of a triangle)

x = 96°

g. x + 30° = 110° (Exterior angle property of a triangle)

x = 110° – 30° = 80°

x = 80°

30º

110º

x

h. 30° + 120° + x = 180° (Angle sum property of a triangle)

150° + x = 180°

x = 180° – 150°

x = 30°

y + 90° = 120° (Exterior angle property of a triangle)

y = 120° – 90°

y = 30°

i. 2x + 3x = 150° (Exterior angle property of a triangle)

5x = 150°

x = 150°

5

x = 30°

3. One of the exterior angles of a triangle = 130°

Ratio of interior opposite angles = 2:3

By the exterior angle property,

2x + 3x = 130°

5x = 130° x = 26°

angles of the triangle are –

2x = 2 × 26° = 52°;

3x = 3 × 26° = 78°, and 130° + y = 180°

y = 180° – 130° = 50°

Therefore, the angles of the triangle are 52°, 78° and 50°.

4. One angle of the triangle is 60°, and the difference between the two angles is 20°.

Let the remaining two angles be, x and x − 20°.

60° + x + (x − 20°) = 180° (angle sum property)

Solving for x:

2x – 20° = 120°

2x = 140°

x = 70°

Therefore, the remaining two angles are 70° and 50°.

The angles of the triangle are 60°, 70° and 50°.

5. Given: The angles of the triangle are in the ratio 5:6:7.

Let the angles be 5x, 6x, and 7x. Since the sum of the angles in a triangle is 180°,

5x + 6x + 7x = 180°

Solving for x:

18x = 180°

x = 10°

Therefore, the angles are 5x = 50°, 6x = 60°, and 7x = 70°.

The measures of the angles are 50°, 60° and 70°.

6. Given: In triangle ABC, angle A is 50° more than angle B and 20° less than angle C

Let angle B be x. Then, angle A = x + 50° and angle C = x + 50° + 20° = x + 70°.

Since the sum of the angles in a triangle is 180°.

x + (x + 50°) + (x + 70°) = 180°

Solving for x: 3x + 120° = 180°, 3x = 60°, x = 20°

Therefore, angle A = 70°, angle B = 20°, and angle C = 90°.

The angles of the triangle are 70°, 20° and 90°.

7. Given:

In triangle ABC, A – B = 15° and B – C = 30°.

Let angle B be x

Then, angle A = x + 15° and angle C = x − 30°.

Since the sum of the angles in a triangle is 180°.

(x + 15°) + x + (x − 30°) = 180°

Solving for x:

3x − 15° = 180°

3x = 195°

x = 65°

Therefore, angle A = 80°, angle B = 65°, and angle C = 35°.

8. Given: The angles of a triangle are arranged in ascending order with a difference of 10° between consecutive angles.

Let the smallest angle be x

Then, the other two angles are x + 10° and x + 20°. Since the sum of the angles in a triangle is 180°.

x + (x + 10°) + (x + 20°) = 180°

Solving for x:

3x + 30° = 180°

3x = 150°, x = 50°

Therefore, the three angles are 50°, 60°, and 70°.  The angles of the triangle are 50°, 60°, and 70°.

9. Given: BC is extended to form a ray BD. And CE ‖ BA.

a. To show: ACD = A + B

ACD = 1 + 2 (Exterior angle

sum property of a triangle)

Hence, ACD = A + B

b. To show: A + B + C = 180°

BCD = ACB + 3 + 4 = 180° (Straight angle) — eq 1

Since 3 + 4 = ACD, 3 + 4 = 1 + 2 (From (a))

Substituting in—eq 1,

BCD = ACB + 1 + 2

BCD = C+ A + B

C+ A + B = 180°

Hence, A + B + C = 180°

Word Problem 1.

Given: QT is perpendicular to SU; PQT = TQR = x; RQS + PQU = 80°

Therefore, PQU = TQU − TQP

PQU = 90° – x°

Similarly, RQS = 90° – x°

Therefore, PQU = RQS

Since RQS + PQU = 80°

RQS + RQS = 80° 2 RQS = 80°

RQS = 40°

UQS = PQU + PQT + TQR + RQS

180° = 40° + x + x + 40°

180° = 80° + 2x 100° = 2x; x = 50°

The measure of PQT is 50°.

9C 1. a. 2 cm, 3 cm, and 4 cm

2 + 3 > 4 (True)

2 + 4 > 3 (True)

3 + 4 > 2 (True)

This set forms a triangle.

b. 4 cm, 9 cm, and 12 cm

4 + 9 > 12 (True)

4 + 12 > 9 (True)

9 + 12 > 4 (True)

This set forms a triangle.

c.  1 cm, 5 cm, and 10 cm

1 + 5 > 10 (False)

This set does not form a triangle.

d. 2.3 mm, 3.4 mm, and 7.8 mm

2.3 + 3.4 > 7.8 (False)

This set does not form a triangle.

e. 3.2 cm, 5.6 cm, and 8.3 cm

3.2 + 5.6 > 8.3 (True)

3.2 + 8.3 > 5.6 (True)

5.6 + 8.3 > 3.2 (True)

This set forms a triangle.

2.a. For sides of lengths 3 cm and 6 cm:

3 + 6 > x

9 > x

Similarly, 6 + 3 > x

9 > x

And 6 − 3 < x

3 < x

So, the possible values for the third side x are 3 cm < x < 9 cm.

b. For sides of lengths 14 cm and 9 cm:

14 + 9 > x

23 > x

Similarly, 9 + 14 > x

23 > x

And, 14 − 9 < x

5 < x

So, the possible values for the third side x are 5 cm < x < 23 cm

c. For sides of lengths 34 cm and 7 cm:

34 + 7 > x

41 > x

Similarly, 7 + 34 > x

41 > x

Also, 34 – 7 < x

27 < x

So, the possible values for the third side x are 27 cm < x < 41 cm

d. For sides of lengths 23 cm and 27 cm:

23 + 27 > x 50 > x

Similarly, 27 + 23 > x

50 > x

Also, 27 − 23 < x

4 < x

So, the possible values for the third side x are 4 cm < x < 50 cm

3. Given: S is any point on the side QR of triangle PQR.

a. By the triangle inequality property – PS < PR + SR

b. By the triangle inequality property – PS > PQ – QS

4. Given, PS is the median, hence, QS = SR

By triangle inequality property,

PQ + QS > PS

Similarly, PR + SR > PS

Adding both inequalities, PQ + QS + PR + SR > PS + PS

Since QS + SR = QR

PQ + QR + PR > 2PS

As we have shown that the triangle inequality property holds true for this statement, hence, PQ + QR + PR > 2 PS.

5. Given: A triangle ABC with AB = 5 cm, BC = 6 cm and AC = 7 cm. O is a point in the triangle.

There is no information about point O being joined to anything. We can assume that O is any arbitrary point in the triangle.

To prove: AO + BO + CO > 9 cm

By the Triangle Inequality Theorem,

AO + BO > AB

BO + CO > BC

CO + AO > AC

Adding these inequalities:

(AO + BO) + (BO + CO) + (CO + AO) > AB + BC + AC

2(AO + BO + CO) > AB + BC + AC

AO + BO + CO > (AB + BC + AC)/2

Since AB + BC + AC = 5 + 6 + 7 = 18 cm:

AO + BO + CO > 18/2

AO + BO + CO > 9 cm

Therefore, AO + BO + CO is greater than 9 cm.

6. Given: O is any point in a triangle ABC and AO, BO and CO are joined.

To show: 2(AO + BO + CO) > AB + BC + AC.

By the Triangle Inequality Theorem for triangles,

AO + BO > AB

BO + CO > BC

CO + AO > AC

Adding these inequalities:

(AO + BO) + (BO + CO) + (CO + AO) > AB + BC + AC

Hence, 2(AO + BO + CO) > AB + BC + AC.

Word Problem 1. Let the distance between Delhi to Najafgarh be x, that of Najafgarh to Dwarka be y and Dwarka to Delhi be z.

Applying the triangle inequality property,

x + y > z

Hence, the distance travelled from New Delhi to Dwarka is less, than the sum of other two distances.

9D 1.a. 2 cm, 3 cm, and 6 cm

62 = 22 + 32

36 = 4 + 9

36 13

This is not true, so the given lengths do not form a right-angled triangle.

b. 6m, 8m, and 10m

102 = 62 + 82

100 = 36 + 64

100 = 100

This is true, so the given lengths form a right-angled triangle.

c. 15cm, 20cm, and 25cm

252 = 152 + 202

625 = 225 + 400

625 = 625

This is true, so the given lengths form a right-angled triangle.

2.a. h = 20 m

b = 12 m

Using Pythagoras theorem,

h2 = b2 + p2

202 = 122 + p2

400 = 144 + p2

400 – 144 = p2

p2 = 256 = 16 × 16

So, p = 16 m

b. h = 25 cm

b = 7 cm

Using Pythagoras theorem,

Dwarka

h2 = b2 + p2

252 = b2 + 72

625 = b2 + 49

625 – 49 = b2

b2 = 576 = 24 × 24

So, b = 24 cm

c. h = 30 m

b = 24 m

Using Pythagoras theorem,

h2 = b2 + p2

302 = 242 + p2

900 = 576 + p2

900 – 576 = p2

p2 = 324 = 18 × 18

So, p = 18 m

3. Given – Pythagorean triplet (k, 24, 25)

By Pythagorean theorem,

k2 + 242 = 252

k2 = 625 – 576

k2 = 49

k2 = 7 × 7

k = 7

4. Given - Sides of a rectangle are 12 cm and 16 cm.

The diagonal of the given rectangle is the hypotenuse of the right triangle.

Using Pythagoras theorem,

h2 = b2 + p2

h2 = 162 + 122

h2 = 256 + 144

h2 = 400

h2 = 20 × 20

h = 20 cm

5. Given – hypotenuse = 25cm; base = 7cm

Let the altitude of the triangle be p cm.

Using Pythagoras theorem,

h2 = b2 + p2

252 = 72 + p2

625 = 49 + p2

625 – 49 = p2

p2 = 576 = 24 × 24

So, p = 24 cm

6. In triangle ABC,

AC2 = AB2 + BC2 (by Pythagoras theorem)

x2 = 32 + 42

x2 = 9 + 16 = 25 = 5 × 5

x = 5cm

Similarly, in triangle ADC,

xdone2 + y2 = 132

52 + y2 = 169

y2 = 169 – 25 = 144

y2 = 12 × 12

y = 12 cm

7. Let Rahul’s starting point be R and ending point be T.

Using Pythagoras theorem in the triangle RST,

RT2 = RS2 + TS2

X2 = 152 + 82

X2 = 225 + 64 = 289

X2 = 17 × 17

X = 17 m

Therefore, the distance between Rahul’s starting point and ending point is 17 metres.

Word Problem 1. Given, height of the tree (AA’) = 9m

Let C be the point from where the tree is broken, and B be the point where the top of the tree touches the ground.

Let AB = x m

AC = 4m and BC = 5m

Using Pythagoras theorem in the triangle ABC,

(BC)2 = (AB)2 + (AC)2

52 = x2 + 42

25 = x2 + 16

25 − 16 = x2

x2 = 9 = 3 × 3

So, x = 3

Thus, the distance of the point where the top touches the ground from the base of the tree is 3 m.

Chapter Checkup 1.a. If the three angles of a triangle are equal, then each of them is equal to 60°.

b. In a right−angled triangle, two of its sides are also its altitudes.

c. In an equilateral triangle, the median and altitudes are denoted by the same line segment.

d. The sum of the angles of a triangle is 180°.

2.a. There can be only two acute angles in a triangle. − False

b. The Pythagoras theorem is only applicable for a right− angled triangle. − True

c. The exterior angles of a triangle are equal to the difference of the opposite interior angles. − False

d. The sum of the length of any two sides of a triangle is greater than the third side. − True

3. For a set of angles to form a triangle, then by the angle sum property, the sum of the angles must be 180°.

a. 30° + 40° + 110° = 180°

The given set will form a triangle.

b. 34° + 29° + 120° = 183°

The given set will not form a triangle.

c. 50° + 120° + 10° = 180°

The given set will form a triangle.

d. 34° + 28° + 100° = 162°

The given set will not form a triangle.

4. For a set of side lengths to form a right−angled triangle, Pythagorean Theorem must be satisfied.

By Pythagoras Theorem, a2 + b2 = c2, where c is the hypotenuse.

a. 62 + 82 = 36 + 64 = 100 and 92 = 81 The given set will not form a right−angled triangle.

b. 92 + 122 = 81 + 144 = 225 and 152 = 225

The given set will form a right−angled triangle.

c. 242 + 102 = 576 + 100 = 676 and 262 = 676

The given set will form a right−angled triangle.

5. Using angle sum property of a triangle,

a. (x − 10)° + (4x − 10)° + 70° = 180°

x − 10° + 4x − 10° + 70° = 180°

5x − 20° + 70° = 180°

5x + 50° = 180°

5x = 180° − 50° = 130°

X = 26°

P = (x − 10)° = (26 − 10)° = 16°

Q = (4x – 10) ° = (4 × 26 – 10)° = 94°

b. Using angle sum property in triangle FGH,

x° + 80° + 50° = 180°

x + 130° = 180°

x = 50°

c. Let the angle opposite to 54° be x.

x = 54° (vertically opposite angles)

Using angle sum property in triangle ABC,

54° + 54° + y = 180°

180° + y = 180°

y = 180° – 104°

y = 72°

d. Using the exterior angle property of a triangle,

x + 65° = 130°

Hence, x = 65°

Using the angle sum property of a triangle,

65° + 65° + y = 180°

y = 180° − 130° = 50°

y = 50°

6. Given – BC = 6m, AB = 10m

L as the length of the ladder,  d as the distance from the base of the ladder to the wall, and

h as the height above the ground where the ladder touches the wall.

According to the Pythagorean Theorem:

l2 = d2 + h2

Substitute the values−

102 = 62 + h2

100 = 36 + h2

h2 = 64 = 8 x 8

h = 8

Therefore, the ladder touches the wall 8 metres above the ground.

7. Given : An isosceles triangle ABC with AC = BC; and AB2 = 2AC2 To prove: ABC is a right−angled triangle.

Proof:

In triangle ACB,

AC = BC

The angles corresponding to these sides are equal hence, these two angles must be less than 90 degrees.

Since AB2 = 2AC2

AB2 = AC2 + AC2

AB2 = AC2 + BC2 (since AC = BC)

By Pythagoras theorem,

Triangle ABC is a right−angled triangle. (Hence proved)

8. In a television set, the diagonal (d), width (w), and height (h) can be represented using Pythagoras theorem.

d2 = w2 + h2

Given: the diagonal (d) is 52 inches and the width (w) is 48 inches:

522 = 482 + h2

522 = 482 + h2

h2 = 522 − 482

h2 = 2704 − 2304

h2 = 400 = 20 × 20

h = 20 inches

Therefore, the height of the television screen is 20 inches.

9. The situation forms a right−angled triangle, where the floor of the delivery truck, the ramp, and the ground form the three sides.

Using the Pythagorean Theorem, we have:  h2 + d2 = L2

Given that the height (h) is 48 cm and the horizontal distance (d) is 14 cm:

482 + 142 = L2

2304 + 196 = L2

L2 = 2500 = 50 × 50

L = 50 cm

Therefore, the length of the ramp is 50 cm.

10. The situation in the question forms a right−angled triangle. The shortest length of the rope is the height of the building.

Given that the height (h) of the building is 6 m and the horizontal distance (d) is 8 m:

h2 + d2 = l2

62 + 82 = l2

36 + 64 = l2

l2 = 100

l = 10 m

Therefore, the shortest length of the rope is 10 metres.

11. Given: Triangle ABC with

A = 60° and B = 80°, and the bisectors of B and C meet at O.

By angle sum property of in triangle ABC,

A + B +

C = 180°

Substituting the values, 80° + 60° + C = 180°

C = 180° − 140° = 40°

OBC = 40° (BO is the angle bisector)

OCB = 20° (CO is the angle bisector)

In triangle BOC,

OBC + OCB + BOC = 180°

40° + 20° + BOC = 180°

BOC = 180° − 60°

BOC = 120°

12. Since, ABC and CBP form a linear pair.

ABC + CBP = 180°

ABC +2 1 = 180° [BO is the bisector of CBP]

2 1 = 180° − ABC

1 = 90° − 1 2 ABC .....(1)

Again, ACB and QCB form a linear pair.

Therefore,

ACB + QCB = 180°

ACB + 2 2 = 180°

2 2 = 180° − ACB

2 = 90° − 1 2 ACB ...(2)

In triangle BOC, we have

1+ 2 + BOC = 180°

90° − 1 2 ABC + 90° − 1 2 ACB + BOC = 180°

180° − 1 2 ( ABC + ACB) + BOC = 180°

BOC = 1 2 ( ABC+ ACB)

BOC = 1 2 (180° − BAC)

BOC = 90° − 1 2 A (Hence proved)

13. Side BC of triangle ABC is produced to D.

ACD = B + A

ACD = 2 + 1 ...(i)

Side AC of triangle ABC is produced to E.

BAE = B + C

BAC = 2 + 3 ...(ii)

4.2cm

And side AB of triangle ABC is produced to F.

CBF = C + A

CBF = 3 + 1 ...(iii)

Adding (i), (ii) and (iii), we get:

ACD + BAE + CBF = 2 ( 1 + 2 + 3)

= 2 × 180° = 360° = 4 × 90° = 4 right angles

4.2cm

Hence, the sum of the exterior angles so formed is equal to four right angles.

Word Problems 1. In this situation, the diagonal divides the rectangle into two right triangles.

Let the width of the field be w and the length be l.

Using Pythagoras theorem,

w2 + l2 = x2

92 + 122 = x2

81 + 144 = x2

x2 = 225

x2 = 15 × 15

x = 15 metres

Hence, the players ran a distance of 15 metres.

2. The situation given in the question forms right angled triangles. Let w be the width of the street when the ladder is placed against the wall on one side of the street.

Using the Pythagorean Theorem,

52 + w2 = 132 = 169

w2 = 169 – 25 = 144

w2 = 12 × 12

w = 12 m

Similarly, let z be the width of the street when the ladder is placed against the wall on the other side of the street.

122 + z2 = 132 = 169

z2 = 169 – 144 = 25 = 5 × 5

z = 5 m

Width of the street = w + z = 12 + 5 = 17 m

Chapter 10

Let’s Warm-up 1. A point where two sides of a triangle meet is known as the vertex of a triangle.

2. Each angle of an equilateral triangle is of the measure 60°

3. A right-angled triangle has one angle of 90°.

4. A triangle has 3 vertices and 3 sides.

5. In an obtuse-angled triangle, the remaining two angles are acute. Do It Yourself

10A 1. AB = 10cm

AB CD

If two lines are congruent, then, their lengths are the same.  Then, AB = CD

So, CD = 10 cm

LMN = x

When two angles are congruent, the measures of the angles are equal.

Thus, LMN = 55°

Rectangle ABCD Rectangle PSRQ

If two rectangles are congruent, then the length and the breadth of one rectangle are exactly equal to the respective length and the breadth of the other rectangle.

Circle with radius ST = CST

Circle with radius VU = CVU CST CVU

If two circles are congruent, then, the radius of both the circles are of equal lengths.  VU = ST x = 7.2 cm

3. In rectangle PQRS,

Rectangle PABD Rectangle

DBCS Rectangle ACRQ

Length of Rectangle PABD =

Length of Rectangle DBSC =

Length of Rectangle ACRQ

Breadth of Rectangle PABD =

Breadth of Rectangle DBSC =

Breadth of Rectangle ACRQ

Given: In Rectangle ACRQ

Breadth = AQ = 5 cm

Length = AC = AB + BC = 5 + 5 cm = 10 cm

Area = Length Breadth = 10 cm × 5 cm = 50 cm2

Since, Area of Rectangle PABD = Area of Rectangle DBSC =

Area of Rectangle ACRQ

Then,

Area of rectangle PQRS = Area of rectangle PABD + Area of rectangle DBCS + Area of rectangle ACRQ

= 50 cm2 + 50 cm2 + 50 cm2 = 150 cm2

So, area of rectangle PQRS is 150 cm2

4. DE C

A BF

5. Given: UVW XYZ U V W

Then, X = U

XY = UV

WU = ZX

6. Since ABC DEF B

As the sides of the square ABCD are equal to the sides of square EFCD, Square ABCD Square EFCD, as AB = EF; AD = DE; BC = CF and DC = DC.

X Y Z

7. L W

MU N V

If, LMN UVW

Then, from the figure, L = U; M = V and N = W

So, the angles corresponding to L, M, N are U, V and W respectively.

8. X

M U N

V Given that, UMV XNY

Y

Since, if two arcs of a circle are congruent, then their corresponding chords are equal, so chord UV = chord XY Hence, UV:XY = 1:1

9. Q P

T S If, PQR = TQS

R

So,

PQS = TQR.

Then, adding RQS on LHS and RHS

PQR + RQS = TQS + RQS

Since, PQR + RQS = PQS and TQS + RQS = TQR

When the measure of two angles is the same, the two angles are congruent.

∠PQS ≅∠TQR

10. O A B C D c a b

In the given figure, a = b = c

a = AOB

b = BOC

c = COD

AOC = a + b = c + b (since, a = b = c)

then, AOC = DOC + BOC = DOB

So, AOC ∠DOB (or ∠BOD)

Word Problems 1. Let the two rectangular sheets of paper be ABDC and PQSR.

In rectangle ABDC and rectangle PQSR,

AC = PQ

And, AB = PR rectangle ABDC rectangle PRSQ.

So, the two sheets of paper are congruent under the correspondence ABDC PRSQ.

2. Both the targets are circular in shape and have the same radii.

If the radius of one circle is equal to the radius of the other circle, then, the circles are congruent.

Hence, yes, the two circular targets are congruent.

3. Let the two triangles be ΔABC and ΔPQR C A B

12 13

5

Since, ΔABC ΔPQR

Then, AB = PQ = 13 cm

BC = QR = 5 cm

CA = RP = 12 cm

R Q P

So, the measurements of the other triangle are also 5 cm, 12 cm and 13 cm.

2.  L O P MN

Given: LM = LN

MP and NO are medians.

Since, median of a triangle bisects the side to which it is drawn Then, NP = PL and LO = OM

LM = LO + OM and LN = LP + PN

So, LO + OM = LP + PN LO = OM = LP = PN … (1)

a. In ΔOMN and ΔPNM

OM = PN (from (1))

MN = NM (common side)

NO = MP (medians corresponding to equal sides of an isosceles triangle are equal)

So, by SSS congruence criteria,

ΔOMN ΔPNM

b. If, ΔOMN ΔPMN, then MP = NO

c. In LPM and LON

LP = LO (from (1))

MP = NO (proved)

LM = LN (given)

So, by SSS congruence criteria

So, LPM LON

3. P S Q R

4. A BE D C

In PQS and PRS

PR = PQ ( PQS is isosceles)

SR = SQ ( SQR is isosceles)

PS = PS (common side)

So, by SSS congruence criteria,  PQS PRS

Given: ABCDE is a regular pentagon

Hence, AB = BC = CD = ED = AE

In ABC and AED

AB = AE

BC = ED

AC = AD (diagonals of a regular pentagon are equal)

So, by SSS congruence criteria,  ABC AED

5. PS Q

AB R

In APS and BQR

AP = BQ AS = BR

PS = QR

So, APS BQR.

6. BC

Given that:

PR = QS

Adding RS on both sides

PR + RS = QS + RS

PS = QR … (1)

A D In ABC

AC = 5 cm

AB = 3 cm

In DCB

BD = 6 cm

DC = 3 cm

In ABC and DCB,

AB = DC = 3 cm

BC = BC

AC DB

So, ABC DCB

In AOB and DOC

P R S

10B 1.  Q

a. In ΔPRS and ΔSQP

PR = SQ (given)

RS = QP (given)

PS = SP (common side)

b. So, by SSS congruence  ΔPRS ΔSQP.

7. A B C

O D

AO = OC (diagonals of a square bisect each other)

BO = OD (diagonals of a square bisect each other)

AB = DC (sides of a square)

So, by SSS congruence criteria,

AOB DOC.

Similarly, In AOD and BOC

AO = OC (diagonals of a square bisect each other)

BO = OD (diagonals of a square bisect each other)

AD = BC (sides of a square)

So, by SSS congruence criteria,  AOD BOC.

8. A B

D C In ΔABC and ΔCDA

AB = CD (opposite sides of a parallelogram are equal)

BC = DA (opposite sides of a parallelogram are equal)

AC = CA (common side)

So, by SSS congruence criteria,  ΔABC ΔCDA

9. N M O

L Given:

LMN is an equilateral triangle.

O is the midpoint of MN.

Joining LO.

In ΔLMO and ΔLNO

LM = LN (sides of an equilateral triangle)

LO = LO (common sides)

MO = NO (O is the midpoint of MN)

So, by SSS congruence criteria,  ΔLMO ΔLNO.

Word Problem 1. Let ΔABC represent the triangle inside the hanger.

Then, let ΔPQR be the triangle of exact same measurements drawn by Jay.

Using SSS congruence criteria ΔABC ΔPQR

Measurements of the hanger:

AB = AC = 21 cm

BC = 30 cm

So, the measurements of the triangle made by Jay should be:

PQ = PR = 21 cm

QR = 30 cm

10C 1. In PRS and PRQ, RS = PQ (equal side)

SRP = QPR (equal angle)

RP = PR (common side)

So, by SAS congruence criterion, PRS PRQ

2. In ADC and ABC

AD = AB (given)

DAC = BAC (given)

AC = AC (common side)

So, by SAS congruence criterion, ADC ABC.

3. In PQR and QPS, QR = PS (given)

PQR = QPS (given)

PQ = QP (common side)

So, by SAS congruence criterion,  PQR QPS.

Then, PR=QS (congruent parts of congruent triangles are equal in measure).

4. In AOD and BOC.

AO = BO (given)

AOD = BOC (vertically opposite angles)

DO = CO (given)

So, by SAS congruence criterion, AOD BOC.

5. In BMP and BLP,

BM = BL (given)

MBP = LBP (BX is the bisector of ABC)

BP = BP (common side)

So, by SAS congruence criterion,  BMP BLP.

6.a. In ADB and ΔADC

AB = AC (given)

BAD = CAD (AD bisects A)

AD = AD (common side)

So, by SAS congruence criterion, ADB ADC.

b. AB = AC; BAD = CAD and AD = AD.

c. Since, ADB ADC

Then, BD = DC (congruent parts of congruent triangles)

7. ABC ΔDEF, by SAS congruence criterion.  ABC is congruent to DEF only when

AB = DE (given)

BAC = EDF (given)

And AC = DF.

So, when AC = DF, the two triangles will be congruent by SAS congruence criteria.

8. In AOC and ΔBOD,

OC = OD (O is the mid-point)

AOC = BOD (vertically opposite angles)

OA = OB (O is the mid-point)

So, by SAS congruence condition  AOC BOD.

OC = OD; AOC = BOD and OB = OA are the three equality relations of matching parts used.

9. ABC is an isosceles triangle and CD is bisector of ACB.

In ACD and ΔBCD

AC = BC (given)

ACD = BCD (By construction)

CD = CD (Common side)

Thus, ACD BCD (By SAS congruence criterion)

So, CAB = CBA (corresponding parts of congruent triangles).

Word Problem 1. Let ABCD be a rectangular sheet of paper.

AC is diagonal of the rectangle.

In ABC and ADC AB = DC (opposite sides of rectangle)

BAC = DCA (alternate interior angles)

AC = CA (Common side)

So, by SAS congruence criterion,  ABC CAD.

Hence, the triangles congruent to each other.

10D 1. Given that CPD = BPD

CPD + APC = 180° APC = 180° – CPD

BPD + APB = 180° APB = 180° – BPD

APC = APB

In ΔCAP and ΔBAP

APC = APB (proved)

AP = AP (common side)

CAP = BAP (Given: AD bisects BAC)

So, by ASA congruence criteria

ΔCAP ΔBAP.

2. A

B C D

In ΔABD and ΔCBD,

BAD = BCD (AB = BC)

AD = DC (BD bisects AC)

BDA = BDC (90°)

So, by ASA congruence criteria

ΔABD ΔCBD.

40º

40º

So, by ASA congruence criteria  ΔADB ΔADC.

b. DAB = DAC; AD = AD and BDA= CDA

c. Yes.

Since, ΔADB ΔADC

Then, BD = DC (Corresponding parts of a Congruent triangles).

R Q

P C

AB C

AB C

60º 60º

60º 60º

3.5 cm 3.5 cm

3. 3.5 cm 3.5 cm

40º

D EF

ΔABC ΔFED.

D C

4. 45º 45º 30º30º A B

D EF

40º

In ΔABC and ΔFED,  CAB = DFE (40°)

AB = FE (3.5 cm)

ABC = FED (60°)

So, by ASA congruence criteria

DAB = 45° + 30° = 75°

CBA = 45° + 30° = 75°

In ΔADB and ΔBCA,

DAB = CBA (75°)

AB = AB (common side)

DBA = CAB (30°)

So, by ASA congruence criteria

ΔADB ΔBCA.

5. A

B

C M L P X

In ΔPBM and ΔPBL

PBM = PBL (Given)

BM = BL (Given)

PMB = PLB (90°) by ASA congruence criteria, ΔPBM ΔPBL

Then, PM = PL (Corresponding parts of Congruent triangles).

6. AB C

DE a. DCA + DCE = ACE

ECB + DCE = BCD

So, ACE = BCD

In  DBC and EAC

B = A (Given)

BC = AC (Given)

ACE = BCD (Proved)

So, by ASA congruence criteria  ΔDBC ΔEAC.

b. If ΔDBC ΔEAC

Then, DC = EC (Corresponding parts of Congruent triangles).

7. A

C D B

X

8. A D BC

K E

In ΔACD and ΔABD

CAD = BAD (AX bisects BAC)

AD = AD (common side)

CDA = BDA (AX bisects BDC)

So, by ASA congruence criteria  ΔACD ΔABD.

AD = AE and DB = EC

So, AD + DB = AE + EC

AB = AC

In ΔABE and ΔACD

BAE = CAD (common angle)

AB = AC (proved)

ABE = ACD (given)

by ASA congruence criteria  ΔABE ΔACD.

9. A BC D

a. In ΔADB and ΔADC

DAB = DAC (AD bisects A)

AD = AD (common side)

BDA = CDA (90°)

10. A B

Given that: AQ = CR

So, AQ +CQ = CR + CQ

AC = RQ

In ΔABC and ΔPRQ

ACB = RQP (alternate interior angles)

AC = RQ (proved)

BAC= PRQ (alternate interior angles)

So, by ASA congruence criteria

ΔABC ΔPRQ.

Word Problem 1. In ΔABC and ΔDEF, B = E (60°)

AB = DE (6 inches)

A = D (30°)

So, by ASA congruence criteria

ΔABC ΔDEF.

10E 1.  AC

B D

ABD CBD.

In ABD and CBD

BDA = BDC (90°)

AB = BC (ABC is isosceles)

BD = BD (common side)

So, by RHS congruence criteria

2. Q S R

AB P

PS = QR (proved)

AS = BR (given)

PR = QS

PR + RS = QS + RS

PS = QR

In  APS and BQR

APS = BQR (90°)

So, by RHS congruence criteria

APS BQR.

3. A C B

D

a. In ABD and CBD

ADB = CDB (90°)

BD = BD (common side)

AB = BC (given)

So, by RHS congruence criteria

ABD CBD.

b. Since, ABD CBD.

Then, A = C (Corresponding parts of Congruent triangles).

c. Since, ABD CBD.

Then, AD = CD (Corresponding parts of Congruent triangles).

4. CB

A 90° 4 cm 10cm FE

So, ABC DEF.

5. P LN R M Q

D 90° 3 cm 10cm

In ABC and DEF,

ACB =  DFE (90°)

AB = DE (10cm)

But, AC DF or CB FE

In LPM and NPM,

PLM = PNM (90°)

LM = NM (given)

PM = PM (common side)

So, by RHS congruence criteria LPM NPM.

Since, LPM NPM.

Then, LPM = NPM (Corresponding parts of Congruent triangles).

6. A C BD E

In ABC and EBD,

ACB = EDB (90°)

BC = BD (given)

AB = EB (given)

So, by RHS congruence criteria  ABC EBD.

7. In ABD and ACD,

ADB = ADC (90°)

AD = AD (common side)

AB = AC (ΔABC is isosceles)

So, by RHS congruence criteria

ABD ACD.

Then, BD = CD

And, B = C  (Corresponding parts of Congruent triangles).

8. A C F E

D B

9. C B

In ABC and EFD,

ABC = EFD (90°)

AB = EF (given)

AC = ED (given)

So, by RHS congruence criteria

ABC EFD.

A R Q

P If, ABC PQR

Since, corresponding parts of congruent triangles are equal

AB = PQ = 12 cm

BC = QR = 5 cm

Since, AC = AB2 + BC2 = 122 + 52 = 144 + 25 = 169 = 132 = 13 cm

AC = PR = 13 cm

Word Problem 1. In ABC and DEF,

ABC = DEF (90°)

AB DE or BC EF

So, by RHS congruence criteria ABC ~ DEF.

Chapter Checkup 1. a. Any two congruent figures look the same. TRUE

b. Two angles are congruent if the sum of their measures is 180°. FALSE

c. Two lines are congruent if they are parallel. FALSE

d. If two rectangles have equal areas then they are always congruent. FALSE

e. Two circles with the same radii are congruent. TRUE

2. P Q R O

3. Given, ABC PQR

A = P 75° = (2y + 5)°

U T S V

2y = 75 – 5 y = 70 2 y = 35°

B = Q  (x – 7)° = 70° x = 70 + 7 x = 77°

(Corresponding parts of Congruent triangles).

4. L M N O

b. Given,   LMN OMN

Then, LMN = OMN (Corresponding parts of Congruent triangles).

So, MN bisects LMO.

5. If PQR XYZ.

Then, P = X  (Corresponding parts of Congruent triangles).

(2x + 15)° = (5x – 60)°

15° + 60° = 5x – 2x

75° = 3x

x = 75° 3 = 25°

F O

6. D E G

In DOE and GOF,

DO = GO (given)

DOE = GOF (vertically opposite angles)

EO = FO (given)

So, by SAS congruence criteria  DOE GOF

7. C D A

CAB = DBA (115°)

AB = AB (common side)

So, by SAS congruence criteria

ΔACB ΔADB

8. 45° 55° C

D B A

9. In ΔAZX and ΔAZY,  AZX = AZY (90°)

AZ = AY (given)

AZ = AZ (common side)

So, by RHS congruence criteria

ΔAZX ΔAZY

CAB = 95° + 20° = 115°

DBA = 95° + 20° = 115°

In ΔACB and ΔADB,  AC = BD (given)

If ΔABC ΔDCB  Then, ABC = DCB = 55° (Corresponding parts of congruent triangles)

If ROP SVU and QOP TVU  Then, ROP – QO

P  SVU – TVU  ROQ SVT

a. In LMN and OMN

LM = OM (given)

MN = MN (common side)

LN = ON (given)

So, by SSS congruence criteria  LMN OMN.

Then, ZX = ZY (Corresponding parts of congruent triangles).

10. 50°

A F E BD C

ΔBED ΔCFD.

In ΔBED and ΔCFD,  ED = FD (given)

BED = CFD (90°)

BD = CD (given)

So, by RHS congruence criteria

Then, B = C = 50°. (Corresponding parts of congruent triangles)

11. ABC PQR

Then, BC = QR (corresponding parts of congruent triangles)

So, BC QR = 1

P T U

12. SQ R

ΔQRT ΔRQU.

In ΔQRT and ΔRQU,  QTR = RUQ (90°)

QR = RQ (common hypotenuse);

QT = RU (given).

So, by RHS congruence criteria

Then, R = Q PQ = PR

Similarly,

PS = QT

Q = P PR = QR

Together, we get PQ = QR = RP or PQR is equilateral.

Since, all the angles of an equilateral triangle are equal then,

P = Q = R = x

So, x + x + x = 180°

3x = 180° x = 180 3 = 60°

So, the measure of each angle in PQR is 60°.

Word Problems

1. B

10 cm 10 cm

A C

Area of the rectangular sheet = 50 cm × 30 cm = 1500 cm2

If all 4 triangles are congruent, then

Area of 4 isosceles triangles = area of 2 squares = 2 × 10 × 10 = 200 cm2 (If the triangles are congruent then their areas are also the same)

Area of remaining portion = 1500 – 200 = 1300 cm2

2. In ΔABC,

AB = 6 cm

AC = 8 cm

BC = AB2 + BC2 = 62 + 82 =

36 + 64 = 100 = 10 cm

In ΔPQR,

PQ = 6 cm

PR = 10 cm

QR = PQ2 + PR2 = 62 + 102 = 36 + 100 = 136 = 2 34 cm

In ΔABC and ΔPQR,

AB = PQ

BC = PR

AC QR

So, by SSS congruence criteria

ΔABC ΔPQR

Chapter 11

Let’s Warm-up 1. Pooja’s income = 22,000

Arushi’s income = 26,500

Pooja’s income: Arushi’s income = 22,000:26,500 = 44:53

The ratio of Pooja's income to Arushiʼs income is 44:53.

2. Highest Income = 30,750

Lowest Income = 18,500

Highest Income:Lowest Income = 30,750:18,500 = 123:74

The ratio of highest income to lowest income is 123:74

3. Difference between Rachna’s and Amit’s Income = 30,750 –23,250 = 7500

Difference between Rohan’s and Divya’s Income = 29,000 – 18,500 = 10,500

(Rachna’s Income – Amit’s Income):(Rohan’s Income – Divya’s Income)

= 7500:10,500 = 5:7

The ratio of the difference in the income of Rachna and Amit to that of Rohan and Divya is 5:7

4. Pooja’s Income + Divya’s Income = 22,000 + 18,500 = 40,500  Rachna’s Income + Arushi’s Income = 30,750 + 26,500 = 57,250

(Pooja’s Income + Divya’s Income):(Rachna’s Income + Arushi’s Income)

= 40,500:57,250 = 162:229

The ratio of the sum of income of Pooja and Divya to that of Arushi and Rachna is 162:229.

Do It Yourself

11A 1. Total number of roses = 45

Number of red roses = 32

Number of white roses = 45 – 32 = 13

Ratio of red roses to white roses = 32:13

2.a. 3 m = 3 × 100 cm = 300 cm

120 cm:3m = 120 cm:300 cm

= 120 300 = 60 × 2 60 × 5 = 2 5 = 2:5

b. 6 L = 6 × 1000 mL = 6000 mL

6000 mL:240 mL = 6000 240 = 25 × 240 240 = 25 1 = 25:1

c. 5 km = 5 × 1000 m = 5000 m

5000 m:800 m = 5000 800 = 25 × 200 4 × 200 = 25 4 = 25:4

d. 1 hour = 3600 seconds

240 seconds:3600 seconds = 240 3600 = 240 15 × 240 = 1 15 = 1:15

3.a. 6:1 = 6 1 = 6 × 5 1 × 5 = 30 5 = 30:5

6:1 = 30:5

b. 8:5 = 8 5 = 8 × 5 5 × 5 = 40 25 = 40:25

40:25 = 8:5

c. 20:16 = 20 16 = 4 × 5 4 × 4 = 5 4 = 5:4

5:4 = 20:16

d. 55:60 = 55 60 = 11 × 5 12 × 5 = 11 12 = 11:12

55:60 = 11:12

4.a. LCM of 7 and 21 = 21

1 7 = 3 21

As 3 < 5; 3 21 < 5 21 ; Hence, 1:7 < 5:21

b. LCM of 5 and 9 = 45

10

5 = 90 45 ; 14 9 = 70 45

As 90 > 70; 10 5 > 14 9 ; Hence, 10:5 > 14:9

c. LCM of 15 and 13 = 195

7 15 = 91 195; 6 13 = 90 195

As 91 > 90; 7 15 > 6 13 ; Hence, 7:15 > 6:13

d. LCM of 15 and 20 = 60

8 15 = 32 60 ; 9 20 = 27 60

As 32 > 27; 8 15 > 9 20 ; Hence, 8:15 > 9:20

5. LCM of 5, 12, 7, 2, 14 = 420

2 5 = 168 420; 7 12 = 245 420; 3 7 = 180 420; 9 2 = 1890 420 ; 11 14 = 330 420

As 168 < 180 < 245 < 330 < 1890; Hence, 2:5 < 3:7 < 7:12 < 11:14 < 9:2

6. A:B = 5:11 and B:C = 7:13

LCM of 11 and 7 = 77

A B = 5 11 = 35 77 ; B C = 7 13 = 77 143

A:B:C = 35:77:143

Thus, A:C = 35:143

7. Perimeter of a rectangle = 2(l + b) = 2 × (12 + 8) cm = 2 × 20 cm = 40 cm

Perimeter of a square = 4a = 4 × 9 cm = 36 cm

Required ratio = 40 cm 36 cm = 4 × 10 cm 4 × 9 cm = 10:9

8. Let the numbers be 7x and 5x

Given, (7x + 9):(5x + 9) = 5:4

7x + 9

5x + 9 = 5 4

28x + 36 = 25x + 45

28x – 25x = 45 – 36 = 9

3x = 9

x = 3

The number are,

7x = 7 × 3 = 21

5x = 5 × 3 = 15

9. Let the number of pencils, erasers and sharpeners be 4x, 5x and 3x respectively.

Given, Number of pencils: 4x = 20 x = 20 4 = 5

Therefore, number of erasers = 5x = 5 × 5 = 25

Therefore, number of sharpeners = 3x = 3 × 5 = 15

10. To compare the performance of hockey teams in the given years we will compare the ratio of matches won to the matches lost.

Ratio of matches won to the matches lost in 2020 = 16:3

Ratio of matches won to the matches lost in 2021 = 18:6

LCM of 3 and 6 = 6

16 3 = 32 6 ;

As 32 > 18; 32 6 > 18 6 . Hence, 16:3 > 18:6.

It can be concluded, that the hockey team performed better in 2020.

11. Cups of sugar in recipe = 1 2 3 = 5 3

Cups of flour in the recipe = 21 6 = 13 6

12. Let the angle of triangle be 2x, 3x and 4x

2x + 3x + 4x = 180°

9x = 180°

x = 20°

Smallest angle: 2x = 2 × 20° = 40°

Largest angle: 4x = 4 × 20° = 80°

Sum of smallest and largest angles = 40° + 80° = 120°

13. Let lengths of pieces of ribbon 3x, 4x and 6x.

Length of shortest piece: 3x = 12 inches x = 4 inches

Therefore, total length of ribbon = 3x + 4x + 6x = 13x = 13 × 4 inches = 52 inches

Word Problems 1. Weight of Nikita’s bat = 2.45 pounds

Weight of Mahi’s bat = 1.95 pounds

Ratio of the weight of the bats = 2.45 1.95 = 4.9 3.9 = 49 39 = 49:39

2. Let the weights of fertilisers be 4x and 3x.

Given, Weight of the first type fertiliser: 4x = 240 g x = 240 4  = 60 g

Therefore, the weight of the second type fertiliser: 3x = 3 × 60 g = 180 g

3. Total amount of money = ₹3700

Money received by Rehan = 2 2 + 3 + 5 × ₹3700 = 2 10 × ₹3700 = ₹740

Money received by Bobby = 3 2 + 3 + 5 × ₹3700 = 3 10 × ₹3700 = ₹1110

Money received by Heera = 5 2 + 3 + 5 × ₹3700 = 5 10 × ₹3700= ₹1850

4. The ratio 2 5 : 7 10 = 4 10 : 7 10 = 4:7

Let the sides of rectangle be 4x, 7x, 4x and 7x.

Sum of the sides = 4x + 7x = 55 cm

11x = 55 cm

x = 55 11 = 5 cm

Length of the sides: 4x = 4 × 5 cm = 20 cm

7x = 7 × 5 cm = 35 cm

5. Let the ages of Priya and Surbhi be 7x and 8x

7x + 6

8x + 6 = 10 11

77x + 66 = 80x + 60

80x – 77x = 66 – 60

3x = 6 x = 2

Age of Priya = 7x = 7 × 2 = 14

Age of Surbhi = 8x = 8 × 2 = 16

6. Total amount of salad = 450 g

Weight of apples = 2 2 + 3 + 1 × 450 g = 2 6 × 450 g= 150 g

Weight of grapes = 3 2 + 3 + 1 × 450 g = 3 6 × 450 g= 225 g

Weight of strawberries = 1 2 + 3 + 1 × 450 g = 1 6 × 450 g = 75 g

Difference in the weight of grapes and apples = 225 g – 150 g = 75 g

11B 1. Using, Product of extremes = Product of means

a. 5 × 18 = 6 × 15

90 = 90 = Proportional

b. 4 × 14 3 × 16

56 48 = Not proportional

c. 1 5 × 1 4 2 7 × 14 5

1

20 4 5 = Not proportional

d. 3 8 × 8 9 7 9 × 3 21

1 3 1 9 = Not proportional

2.a. 40:24 :: 20:x

40x = 24 × 20

40x = 480

x = 12

b. 6:8 :: x:56

8x = 56 × 6

8x = 336

x = 42

c. 2 3 : 5 7 : 4 5 :x

2

3 × x = 4 5 × 5 7

3

2x = 4 7 x = 6 7

d. 3 5 : 1 2 :: x: 4 9

1 2 × x = 3 5 × 4 9 x 2 = 12 45

x = 24 45

3.a. 4:6 :: 6:x

4x = 36

x = 36 4 = 9

b. 12.5:2.5 :: 2.5:x

12.5x = 2.5 × 2.5 = 6.25

x = 6.25

12.5 = 1 2

c. 5 7 : 2 5 :: 2 5 :x

5 7 x = 2 5 × 2 5 = 4 25

5 7 x = 4 25 x = 28 125

d. 3 5 :4.2 :: 4.2:x

3

5 x = 4.2 × 4.2

0.6x = 17.64 x = 17.64 0.6 = 29.4

4. Using, Product of extremes = Product of means

21d = (24 + d) × 7

21d = 7d + 168

14d = 168 d = 168 14 = 12

5. Amount of gasoline required to travel 126 miles = 18 gallons

Amount of gasoline required to travel 175 miles = x gallons (let)

126:18 :: 175:x

Using,

Product of extremes = Product of means

126 × x = 175 × 18

x = 175 × 18 126 = 25 gallons

6. Days required to earn ₹3000 = 5 days

Days required to earn ₹5000 = x days (let)

₹3000:5 days :: ₹5000:x

3000x = 5000 × 5

x = 5000 × 5 3000 = 25 3 days

7. Cost of 4 TV sets = ₹1,12,000

Cost of 23 TV sets = ₹x (let)

4:1,12,000 :: 23:x

4x = 1,12,000 × 23

x = 1,12,000 × 23 4 = ₹6,44,000

Thus, the cost of 23 such TV sets is ₹6,44,000.

8. Number of days for which 270 kg grass will feed 42 cows = 21 days

Number of days for which 360 kg grass will feed 42 cows = x days (let)

270:21 :: 360:x

270x = 21 × 360

x = 21 × 360 270 = 28 days

9. Amount of cloth ₹1518.75 would buy = 27 m

Amount of cloth ₹1068.75 would buy = x m

1518.75:27 :: 1068.75:x

1518.75x = 27 × 1068.75

x = 27 × 1068.75

1518.75 = 19

Thus, 19 metres of cloth can be purchased from ₹1068.75

10. Let the number to be subtracted be x

23 – x:30 – x :: 57 – x:78 – x

23 – x

30 – x = 57 – x 78 – x

1 –23 – x 30 – x = 1 –57 – x 78 – x

30 – x – 23 + x 30 – x = 78 – x – 57 + x 78 – x

7

30 – x = 21 78 – x

546 – 7x = 630 – 21x

21x – 7x = 630 – 546

14x = 84 x = 6

Word Problems 1. Money spent in 35.5 weeks = $2,307.50

Money spent in 15 weeks = x

35.5:2,307.50 :: 15:x

Using,

Product of extremes = Product of means

35.5x = 2,307.50 × 15

x = 2,307.50 × 15 30.5 = $975

Thus, Sam spent $975 in 15 weeks.

2. Rohan’s age:Kunal’s age :: Kunal’s age:Vicky’s age

Using, Product of extremes = Product of means

8:16 :: 16:x

8x = 16 × 16

x = 16 × 16 32 = 32 years

Thus, Vicky’s age is 32 years.

11C 1. Weight of potatoes bought in ₹16 = 2 kg

Weight of potatoes bought in ₹1 = 2 kg 16 = 1 8 kg

Weight of potatoes bought in ₹120 = 1 8 kg ×120 = 15 kg

Therefore, 15 kg potatoes can be bought for ₹120

2. Distance travelled in 3 hours = 150 km

Distance travelled in 1 hour = 150 3 km = 50 km

Distance travelled in 9 hour = 9 × 50 km = 450 km

Therefore, the car travelled 450 km in 9 hours

3. Cost of 15 pens = ₹300

Cost of 1 pen = ₹300 15 = ₹20

Therefore, the cost of 5 pens = 5 × ₹20 = ₹ 100

4. Number of times the human heart beats in 5 minutes = 375

Number of times the human heart beats in one minute = ₹375

5 = 75

Therefore, the number of times the human heart beats in 1 hour or 60 minutes = 75 × 60 = 4500 times

5. Burning time of 20 cm long candle = 50 minutes

Burning time of 1 cm long candle = 50 20 = 2.5 minutes

Total length of the other candle = 20 cm + 10 cm = 30 cm

Burning time of 30 cm long candle = 30 × 2.5 cm = 75 minutes

6. Weight of 19 bags = 332.5 kg

Weight of 1 bag = 332.5 19 kg = 17.5 kg

Weight of 35 bags = 35 × 17.5 kg = 612.5 kg

Therefore, 35 such bags will weigh 612.5kg

7. Number of shirts made by 33 m cloth = 15

Number of shirts made by 1 m cloth = 15 33 = 5 11

Number of shirts made by 110 m cloth = 5 11 × 110 = 50

11D

8. Number of books in 1840 grams = 8 books

Number of books in 1 gram = 8 1840 = 1 230

Number of books in 4.37 kg or 4370 g = 4370 × 1 230 = 19

Therefore, 19 such books will weigh 4.37 kg

9. Number of workers required to build the wall in 5 days = 12

Number of workers required to build the wall in 1 day = 5 × 12 = 60

Therefore, the number of workers required to build the wall in 3 days = 60 3 = 20

10. Number of days required by 12 men to complete the work = 18 days

Number of days required by 1 man to complete the work = 12 × 18 days = 216 days

Number of days required by 8 men to complete the work = 216 8 = 27 days

Therefore, 8 men needs 27 days to complete the work.

11. Number of sweets in one box = 35 sweets

Number of sweets in five boxes = 35 × 5 sweets = 175 sweets

Number of teaspoons of flour needed for 5 sweets = 11

Number of teaspoons of flour needed for 1 sweet = 11 5

Number of teaspoons of flour need for 175 sweets = 11 5 × 175 sweets = 11 × 35 = 385

Hence, 385 teaspoons of flour required to make sweets to fill five boxes.

Word Problems

1. Time taken to type 12 pages = 40 minutes

Time taken to type 1 page = 40 12 minutes = 10 3 minutes

Therefore, the time taken to type 90 pages = 10 3 × 90 minutes = 300 minutes = 300 60 = 5 hours

2. Amount spent by Rahul on 15 pens = ₹172.5

Cost of 1 pen for Rahul = ₹172.5 15 = ₹11.5

Amount spent by Sunita on 20 pens = ₹210

Cost of 1 pen for Sunita = ₹210 20 = ₹10.5

a. Sunita bought the pens for a lower price.

b. Cost of 1 pen bought by Keshav = ₹11.5 × 1.5 = ₹17.25

Cost of 25 pens = ₹17.25 × 25 = ₹431.25

1. Distance = 400 m, Time = 25 seconds

Speed = Distance Time = 400 m 25 seconds = 16 m/s

2. Speed = 8.5 m/s, Time = 1 minute = 60 seconds

Distance travelled = Speed × Time = 8.5 m/s × 60 s = 510 m

3. Distance travelled = 3.2 km = 3200 m, Speed = 20 m/s

Time taken to finish the journey = Distance Speed = 3200 m 20 m/s = 160 seconds = 160 60 minutes = 2.67 minutes (approx.)

4. Distance = 500 m

Time = 6 min = 6 × 60 seconds = 360 seconds

Speed = Distance Time = 500 360 m/s = 1.39 m/s

5. Distance = 270 km = 2,70,000 m

Speed = 18 m/s

Time taken = 2,70,000 18 = 15,000 seconds = 15,000 60 minutes = 250 minutes

6. Distance between Delhi and Uttarakhand = 372 km

Speed of the car = 48 km/hr

Time taken to reach Uttarakhand = 372 48 hour = 7.75 hour = 7.75 × 60 minutes = 465 minutes

7. Distance travelled by earth = 94,00,00,000 km

Time taken = 365 days = 365 × 24 hours = 8760 hours

Speed of earth in km /day = 94,00,00,000 km 365 days = 25,75,342.47 km/day

Speed of earth in km/hour = 94,00,00,000 km 8760 hours = 1,07,305.94 km /hour

Word Problems 1. Distance travelled by Kirti = 36,000 m = 36000 1000  km = 36 km

Speed = 8 km/h

Time tame taken by Kirti to finish the journey = Distance Speed = 36 8 = 4.5 hours

Distance travelled by Mihir = 45 km

Speed = 10 km/h

Time tame taken by Mihir to finish the journey = Distance Speed = 45 10 = 4.5 hours

a. Both Kirti and Mihir took the same amount of time to finish the journey.

b. Since Kirti and Mihir took the same amount of time, the difference would be zero minutes.

2. Speed of jogging = 5.5 km/h

a. Rahul jogs the farthest on Monday as he spends more time jogging on Monday (17 min).

b. Time spent jogging on Monday = 17 min = 17 60 hour = 0.2833 hour

Distance covered on Monday = Speed × time = 5.5 × 0.2833 km = 1.558 km

Time spent jogging on Tuesday = 12 min = 12 60 hour = 0.2 hour

Distance covered on Tuesday = Speed × time = 0.2 × 5.5 km = 1.1 km

Time spent jogging on Wednesday = 15 min = 15 60 hour = 0.25 hour

Distance covered on Wednesday = Speed × time = 0.25 × 5.5 km = 1.375 km

Time spent jogging on Thursday = 8 min = 8 60 hour = 2 15 hour

Distance covered on Thursday = Speed × time = 2 15 × 5.5 km = 0.733 km

Time spent jogging on Friday = 0 min

Distance covered on Friday = 0 km

3. Distance to school = 10.8 km = 10,800 m

Time taken by Drishti = 30 minutes = 30 × 60 seconds = 1800 seconds

Speed of Drishti = 10,800 1800 m/s = 6 m/s

Time taken by Som = 45 minutes = 45 × 60 seconds = 2700 seconds

Speed of Som = 10,800 2700 m/s = 4 m/s

Difference in speed of Som and Kirti = 6 m/s – 4m/s = 2m/s

Kirti is faster than Som by 2 m/s.

Chapter Checkup 1. 51:85 :: x:5

Using,

Product of Means = Product of extremes

85x = 51 × 5

x = 51 × 5 85 = 3

2. Let the number be 7x and 8x.

7x – 24

8x – 24 = 5 6

42x – 144 = 40x – 120

42x – 40x = 144 – 120

2x = 24

x = 12

Numbers are 7x = 7 × 12 = 84

8x = 8 × 12 = 96

3. Let the number added be x

3 + x

5 + x = 5 7

21 + 7x = 25 + 5x

7x – 5x = 25 – 21

2x = 4

x = 2

Thus, 2 should be added to each term of the ratio 3:5 to make it 5:7.

4. (49 – y):56 :: y:9

Using,

Product of Means = Product of extremes

56y = (49 – y) × 9

56y = 49 × 9 – 9y

56y + 9y = 49 × 9

65y = 441

y = 441 65 = 6.785 (approx.)

5. Let the length of electric pole be x.

Length of stick : Length of stick’s shadow :: Length of electric pole :

Length of electric pole shadow

4 ft:3 ft :: x:90 ft

Using,

Product of Means = Product of extremes

3x = 90 × 4

x = 90 × 4 3 = 120 feet

6. Amount of baking powder required = 1 2 tablespoon

Amount of sugar required = 13 4 tablespoon = 7 4 tablespoon

Amount of flour required = 25 8 tablespoon = 21 8 tablespoon

LCM of 2, 4 and 8 = 8

Ratio of baking powder to sugar to flour required

= 1 2 : 7 4 : 3 12

= 1 2 × 8: 7 4 × 8: 21 8 × 8

= 4:14:21

7. Weight of flour required for 12 loaves = 3 kg

Weight of flour required for 1 loaf = 2 15 kg = 1 4 kg

Weight of flour required for 20 loaves = 1 4 × 20 kg = 5 kg

8. Diesel consumed to travel 99 km = 22 litres

Diesel consumed to travel 1 km = 22 99 litre = 2 9 litre

Therefore, the amount of diesel it will consume to travel 500 km = 2 9 × 500 litre = 111.11 litre

9. Distance travelled by beetle = 1.09 m

Speed of beetle = 0.2 m/s = Time of flight of beetle before getting caught = 1.09 0.2 = 5.45 seconds

10. Number of widgets produced in 5 minutes = 32

Number of widgets produced in 1 minute = 32 5

Number of widgets produced in 1 hour i.e 60 minutes = 32 5 × 60 minutes = 384 widgets

11. Distance travelled by the plane = 2,90,000 m

Time taken = 9000

Speed of the plane = Distance Time = 2,90,000 9000 = 32.22 m/s

12. Time taken by 25 men to construct the building = 150 days

Time taken by 1 man to construct the building = 150 × 25 days

Time taken by 30 men to construct the building = 150 × 25 30 = 125 days

13. Distance covered by the mouse = 5.5 m

Time taken = 30 seconds

Speed of the mouse = Distance Time = 5.5 30 m/s = 0.18 m/s

14. Speed of Suman = 4 m/s

Time = 5  1 2 minutes = 11 2 minutes = 11 2 × 60 seconds = 330 seconds

Distance travelled = Speed × Time = 4 m/s × 330 seconds = 1320 metres

15. Speed of Vivek = 8.0 mph

Distance to be covered = 28 miles

Time taken to complete the race = Distance Time = 28 8 = 3.5 hours

16. Present age of father = 42 years

Age of father after ten years = 52 years

Age of son = 14 years

Age of son after ten years = 24 years.

Required ratio = 52:24 = 52 24 = 4 × 13 4 × 6 = 13 6 = 13:6

17. Let the number to be added be x.

10

Thus, 2 should be added to 10, 18, 22 and 38 to make them in proportion.

18. Let the present ages of Suresh and Kartik be 11x and 5x. Their ages five years ago are (11x – 5) years and (5x – 5) years.  11

Present ages of Suresh: 11x = 11 × 5 years = 55 years

Present ages of Kartik: 5x = 5 × 5 years = 25 years

Word Problems 1. Amount earned by Sam in a week = ₹1750

Amount earned by Sam in one day = ₹1750 7 = ₹250

Amount earned by Sam in month of October (31 days) = ₹ 250 × 31 = ₹7750

Therefore, Sam will earn ₹7750 in the month of October

2. Total income = ₹18,000

Sunil’s expenditure = 4 4 + 1 × ₹18,000 = 4 5 × ₹18,000 = ₹14,400

Sunil’s Savings = 1 4 + 1 × ₹18,000 = 1 5 × ₹18,000 = ₹3,600

Therefore, Sunil saved ₹3600 and spent ₹14,400

3. Distance between home and work for Mr Sharma = 62.5 km

Distance between home and work for Mrs Sharma = 81.2 km

Speed of Mr Sharma = 50 km/h

Speed of Mrs Sharma = 58 km/h

a. Time taken by Mr Sharma = Distance Speed = 62.5 km 50 km/h = 1.25 hour

Time taken by Mrs Sharma = Distance Speed = 81.2 km 58 km/h = 1.4 hour

b. Mr. Sharma will reach home first as he is taking less time.

c. Difference between the times taken by Mrs Sharma and Mr Sharma = 1.4 hour – 1.25 hour = 0.15 hour

4. Amount received by Avinash = ₹28,800

LCM of 3,4,5 = 60  1 3 : 1 4 : 1 5 = 1 3 × 60: 1 4 × 60: 1 5 × 60 = 20:15:12

Let the amounts received by Kirti, Vikas and Avinash be 20x, 15x and 12x

12x = ₹28,800

x = ₹28,800 12 = ₹2400

Amount received by Kirti = 20x = 20 × ₹2400 = ₹48,000

Amount received by Vikas = 15x = 15 × ₹2400 = ₹36,000

Total amount to be distributed = ₹48,000 + ₹36,000 + ₹28,800 = ₹1,12,800

Chapter 12

Let’s Warm-up 1. The most popular bakery item is coffee.

2. The least popular bakery item is juice.

3. The fraction of people (out of 100) who like coffee is 38 100 .

4. The fraction of people (out of 100) who like pastries is 10 100

5. The fraction of people (out of 100) who like juice is 5 100 .

Do It Yourself

12A 1. a. 2.25% = 2.252259 10010000400

b. 55% = 5511 10020

c. 180% = 1809 1005

d. 56.4% = 56.4564141 1001000250

e. 1 16.516533 16%16.5% 2 1001000200

f. 6.56513 6.5% 1001000200

g. 0.36369 0.36% 100100002500

h. 36018 360% 1005

i. 1 58.255825233 58%58.25% 4 10010000400

j. 64%%1193193 33300

2.a. 11 100%20% 55

b. 22 100%10% 2020

c. 1818 100%72% 2525

d. 551100%% 300030006

e. 30003000 100%75,000% 44

f. 2121 100%105% 2020

g. 121 5100%525% 44

h. 363 121001260% 55

i. 33 100%75% 44

j. 66 100%60% 1010

3.a. 48% = 48 ÷ 100 = 0.48

b. 1.25% = 1.25 ÷ 100 = 0.0125

c. 12.64% = 12.64 ÷ 100 = 0.1264

d. 0.005% = 0.005 ÷ 100 = 0.00005

e. 123.5% = 123.5 ÷ 100 = 1.235

f. 1.2% = 1.2 ÷ 100 = 0.012

g. 1 11% 5 = 11.2% = 11.2 ÷ 100 = 0.112

h. 4 12% 10 = 12.4% = 12.4 ÷ 100 = 0.124

i. 3 81% 50 = 81.06% = 81.06 ÷ 100 = 0.8106

j. 1 145% 8 = 145.125% = 145.125 ÷ 100 = 1.45125

4. a. 0.2 = 0.2 × 100% = 20%

b. 0.34 = 0.34 × 100% = 34%

c. 0.547 = 0.547 × 100% = 54.7%

d. 0.984 = 0.984 × 100% = 98.4%

e. 0.0048 = 0.0048 × 100% = 0.48%

f. 1.2 = 1.2 × 100% = 120%

g. 2.34 = 2.34 × 100% = 234%

h. 12.1 = 12.1 × 100% = 1210%

i. 15.23 = 15.23 × 100% = 1523%

j. 142.214 = 142.214 × 100% = 14221.4%

5. a. Number of shaded squares = 8

Total number of squares = 16

Fraction of the shaded squares = 81 162

Percentage of the shaded squares = 1 100%50% 2

b. Number of shaded triangles = 5

Total number of triangles = 8

12B

Fraction of the shaded triangles = 5 8

Percentage of the shaded triangles = 5 100%62.5% 8

c. Number of shaded squares = 50

Total number of squares = 100

Fraction of the shaded squares = 501 1002

Percentage of the shaded squares = 1 100%50% 2

Word Problems 1. Percentage of pizza eaten by Vivan = 40%

Percentage of pizza left by Vivan = 100% – 40% = 60%

Fraction of pizza left = 603 1005

Thus, 3 5 of the pizza is left.

2. Number of students who passed = 6

Total number of students = 8

Number of students who did not pass = 8 – 6 = 2

Fraction of students who did not pass = 21 84

Percentage of students who did not pass the examination = 1 100%

4 = 25%

Thus, 25% of the students did not pass the examination.

3. Percentage of inflation in a year = 4.5%

4.5% as decimal = 4.5 ÷ 100 = 0.045

4. Percentage of packages delivered without damage = 95%

Percentage of packages delivered with damage = 100% – 95% = 5%

Fraction of packages delivered with damage = 51 10020 Thus, 1 20 of the shipment was damaged.

5. Marks scored by Yash = 360 marks out of 500 marks

Percentage of marks scored by Yash = 360 100%72% 500

Marks scored by Jaya = 450 marks out of 600 marks

Percentage of marks scored by Jaya = 450 100%75% 600 75% > 72%

Thus, Jaya scored a better percentage of marks.

6. Marks scored in Mathematics = 80

Marks scored in Science = 85

Marks scored in Hindi = 90

Marks scored in English = 80

Marks scored in Social science = 85

Total number of marks scored = 80 + 85 + 90 + 80 + 85 = 420 marks

Total number of marks in the examination = 5 × 100 = 500 marks

Percentage of the marks scored = 420 100% 500 = 84%

Thus, the average percentage of the marks scored by Rahul is 84%.

1. Total number of students in the class = 30 students

Percentage of people who enrolled in the art club = 20%

Ratio of art club members to the total number of students = 20%:100% = 1:5

Thus, the ratio of art club members to the total number of students is 1:5.

2. Percentage of blue candies = 25%

Percentage of red candies = 100% − 25% = 75%

Ratio of blue to red candies = 25%:75% = 1:3

Thus, the ratio of blue to red candies is 1:3.

3. Percentage of attendees who are children = 30%

Percentage of attendees who are adults = 100% – 30% = 70%

Ratio of children to adults = 30%:70% = 3:7.

Thus, the ratio of children to adults is 3:7.

4. Cups of flour required for the recipe = 2 cups

Cups of sugar required for the recipe = 3 cups

Percentage of the ingredients that is sugar = 33100%100%60%

2 35 Thus, sugar is 60% of the total ingredients.

5. Ratio of roses to tulips = 3:2

Percentage of roses = 33100%100%60% 325

Thus, there are 60% roses in the garden.

6. Number of challenges completed (out of 10) by the player = 8

Percentage of the challenges completed = 8 100%80% 10

Thus, the player completed 80% of the challenges successfully.

7. Percentage of people who prefer apple pie = 75%

Percentage of people who prefer donuts = 100% – 75% = 25%

Ratio of apple pie lovers to donut lovers = 75%:25% = 75:25 = 3:1

Thus, the ratio of apple pie lovers to donut lovers is 3:1.

Word Problems 1. Ratio of managers to employees = 1:9

Ratio of managers to the total people = 1:10

Percentage of managers in the company = 1 100%10% 10

Thus, 10% of the total employees in the company are managers.

12C 1. a. 25% of 250 = 25 25062.5 100

b. 30% of 750 = 30 750225 100

c. 125% of 45 = 125 4556.25 100

d. 14% of 4800 = 14 4800672 100

2.a. Let a% of 4 m 25 cm be 5 cm.

a% = 5 cm 5 cm 3 100% 100%1%

m 25 cm425 cm 17

b. Let a% of 3 L be 450 mL

a% = 450 mL 450 mL 100% 100%15% 3 L 3000 mL

3.a. Let the number be a

12% of a = 48 1 12 48 100 a 1 48100 12

a = 400

b. Let the number be a 2 66% 3 of a = 567

567 3100 a 1 5671003 200

c. Let the number be a

of a = 5 hours 20 minutes = 320 minutes 1

= 256 minutes

minutes

d. Let the number be a 20% of a = 3 m 40 cm = 340 cm

1 20 340 100 a

1 340100 20 a

a = 1700 cm

a = 17 m

4. a. Percentage decrease =

48 100%15% 320

b. Percentage

c. Percentage

=

5. a. 3%

minutes = 3% of 1200 seconds = 3 120036 100 seconds

b. Let a% of 350 kg be 3 kg 500 g

a% × 350 kg = 3 kg 500 g

a

= 1

Thus, 1% of 350 kg is 3 kg 500 g.

c. Let the number be a 1

of 24 = a

d. Let us find 1 12% 2 of 2500.

1

12% 2 of 2500 = 251 2500312.5 2100

1 12% 2 more than 2500 = 312.5 + 2500 = 2812.5

Word Problems 1. Number of women employees in a factory = 450

Percentage of male employees = 55%

Percentage of female employees = 100% – 55% = 45%

Let the total number of employees be a 45% of a = 450 1 45 450 100 a

1 450100 45 a

a = 1000

Thus, the total number of employees in the factory are 1000.

2. Total amount of money = ₹500

Ratio of the money that Ravi, Raju, and Roy gets = 2:3:5

Amount of money that Ravi gets = 22500500100

23510 ```

Percentage of money that Ravi gets = 22100%100%20% 23510

Amount of money that Raju gets = 33500500150

23510 ```

Percentage of money that Raju gets = 33100%100%30% 23510

Amount of money that Roy gets = 55500500250

23510 ```

Percentage of money that Roy gets = 55100%100%50% 23510

12D

1. Percentage of copper in the alloy = 26%

Total zinc in the alloy = 1480 g

Percentage of zinc in the alloy = 100% – 26% = 74%

Let total quantity of the alloy be a

74% of a = 1480 g

74 1480

100 a

100 14802000 g 74 a

Thus, the total quantity of the alloy is 2000 g or 2 kg.

2. Total number of men at the gathering = 1440

Percentage of women = 40%

Percentage of men = 100% – 40% = 60%

Let the total number of people be a. 60% of a = 1440

60 1440

100 a

100 14402400 60 a

Thus, there are 2400 people at the gathering.

3. Percentage of total votes that candidate A gets = 12%

Percentage of people who did not cast any vote = 54%

Total number of people who voted = 1,53,680

Percentage of total votes that candidate B gets = 100% – 12% – 54% = 34%

Number of total votes that candidate B gets = 34 1,53,600 100 = 52,224

Thus, candidate B gets 52,224 votes.

4. Percentage of students who score more than 80% marks = 35%

Students who scored between 50% and 80% marks = Half of the students

Percentage of students who scored between 50% and 80% marks = 1 100%50% 2

Percentage of students who scored less than 50% = 100% –35% – 50% = 15%

Total number of students = 500

Number of students who scored less than 50% marks = 15% of 500 = 15 50075

100

Thus, 75 students scored less than 50%.

5. Percentage decrease in the population every year = 8%

Present population of the city = 1,69,280

Let the population of the city last year be 'a'. 169280 1001,84,000 1008 a

Let the population of the city 2 years back be 'b'.

b

184000 1002,00,000 1008

Thus, the population of the city before 2 years was 2,00,000.

6. Marks scored by B = 20% more than A

Marks scored by A = 10% less than C

Marks scored by D 20% more than C

Let the marks scored by C be 100.

Marks scored by A = 10% less than 100 = 10 10010090 100 −

Marks scored by D = 20% more than 100 = 20 100100120

100

Percentage by which marks obtained by D is more than those obtained by A = 12090 100% 90

= 30 1 100%33%

90 3

Thus, D obtained 1 33% 3 more marks than A.

7. Population of the town = 10,000

Percentage increase in males after 1 year = 10%

Percentage increase in females after 1 year = 12%

Total percentage increase in the population = 11%

Population after 1 year = 11

10,00010,00010,000110011,100 100

Let the initial population of the males in the town be a

Population of females initially in the town = 10,000 – a

According to the question, 10 12 1 10,000111,100 100 100 aa

1.1a + (10,000 – a) 1.12 = 11,100

1.1a + 11,200 – 1.12a = 11,100

11,200 – 11,100 = 1.12a – 1.1a

0.02a = 100

a = 5000

Thus, the male population initially was 5000.

8. Increase in salary initially = 20%

Decrease in salary afterwards = 10%

Let the initial salary be 100.

Salary after first increment = 20 10010010020120 100

Salary after decrement = 10 12012012012108 100 − −

So, the salary becomes 108 after increment of 20% and then a decrement of 10%.

Percentage change in salary = 108100 8 100%100%8% 100 100

Thus, there is 8% increase in the salary.

9. Cost of pomegranate = Twice of that of pear

Cost of pear = 25% more than cost of an orange

Let the cost of the orange be 100.

Cost of pear = 25% more than 100 = 25 10010010025125 100

Cost of pomegranate = 2 × 125 = 250

Total cost of 4 pears, 2 pomegranates and 3 oranges = (4 × 125) + (2 × 250) + (3 × 100) = 500 + 500 + 300 = 1300

Given: There is an increase of 10% in the price of each fruit.

New cost of orange = 10% more than 100 = 100 + 10 = 110

New cost of pear = 10% more than 125 = 125 + 12.5 = 137.5

New cost of pomegranate = 10% more than 250 = 250 + 25 = 275

New cost of 4 pears, 2 pomegranates and 3 oranges = (4 × 137.5) + (2 × 275) + (3 × 110) = 550 + 550 + 330 = 1430

Percentage increase in the cost of 4 pears, 2 pomegranates and 3 oranges = 14301300130100%100%10% 1300 1300

10. Initial number = 4000

Percentage increase in the number = 25%

New number = 25 40004000400010005000 100    

Percentage decrease in the new number = 30%

New number = 30 50005000500015003500 100  − −   

Percentage change in the final number = 40003500 500 100%100%12.5% 4000 4000 

  

Thus, there is a change of 12.5% in the initial value.

11. Percentage decrease in the value of car every year = 20%

Value of the car after 2 years = ₹5,65,000

Value of the car after 1 year = 565000 1007,06,250 10020 ` `

Original value of the car = 706250 1008,82,812.5 10020 ` `

Thus, the original cost of the car was ₹8,82,812.5

Word Problem 1. Percentage hike in the price of petrol = 4%

Initial price of the petrol = ₹95

Distance travelled by man = 3456 km every month

Mileage of the car = 16 km per litre

Petrol consumed = 3456 ÷ 16 = 216 litres

Increase in the price of the petrol = 4 953.8 100 ``

Increase in the expenditure of the man due to petrol = 216 × ₹3.8 = ₹820.8

Thus, the expenditure of the man increases by ₹820.8 due to the increase in the price of petrol.

12E 1.a. SP > CP, so there is gain.

Gain = SP – CP = ₹745 – ₹620 = ₹125

b. SP > CP, so there is gain.

Gain = SP – CP = ₹6000 – ₹5230 = ₹770

c. CP > SP, so there is loss.

Loss = CP – SP = ₹50,000 – ₹49,812 = ₹188

2. a. Gain = SP – CP = ₹510 – ₹405 = ₹105

Thus, option (ii) is correct.

b. Loss = CP – SP = ₹10,235 – ₹9400 = ₹835

Thus, option (ii) is correct.

c. Profit% = Profit25002000500 100% 100%100%25% CP 2000 2000

Thus, option (i) is correct.

d. Loss% = Loss20001800200 100% 100%100%10% CP 2000 2000

Thus, option (i) is correct.

3. a. Profit = SP – CP = ₹500 – ₹450 = ₹50

b. Loss = CP – SP

SP = CP – Loss = ₹2364 – ₹364 = ₹2000

c. Discount = MP – SP = ₹585 – ₹450 = ₹135 Discount% = Discount135 9 100%100%23% MP 585 117 (approx.)

d. Discount = MP – SP = ₹3060 – ₹2550 = ₹510

= Discount510 2 100%100%16%

3060 3

12F

4.a. SP > CP, so there is profit.

Profit = SP – CP = ₹2400 – ₹1800 = ₹600

Profit% = Profit600 1 100%100%33%

CP 1800 3

b. CP > SP, so there is loss.

Loss = CP – SP = ₹500 – ₹480 = ₹20

Loss% = Loss20100%100%4%

CP 500

c. SP > CP, so there is profit.

Profit = ₹2415 – ₹2300 = ₹115

Profit% = Profit115100%100%5%

CP 2300

5.

a. CP = ₹150 and SP = ₹175  Profit = 25%

b. CP = ₹250 and SP = ₹225  Loss = 1 8% 3

c. CP = ₹200 and SP = ₹250  Profit% = 2 16% 3

d. CP = ₹3000 and SP = ₹2750  Loss = 20%

e. CP = ₹4500 and SP = ₹3600  Loss = 10%

Word Problems 1. Selling price (SP) of the toy = 540

Profit = 20%

100

CP SP 100Profit% 100

CP 540 10020

100 CP540 120

CP450

Thus, the cost price of the toy is ₹450.

2. Cost of each horse = ₹4000

Total cost of selling horses = 2 × ₹4000 = ₹8000

Profit% for the first horse = 25%

Cost price of first horse = 100 100 SP 40003200

100Profit%10025

Cost price of the second horse = ₹8000 – ₹3200 = ₹4800

Thus, the man incurs a loss of 2 16 3 % on the second horse.

3. Cost of each bulb = ₹10

Cost of 200 such bulbs = 200 × ₹10 = ₹2000

Number of bulbs that were fused = 5

Bulbs left = 200 – 5 = 195

Selling price of each bulb = ₹12

Total selling price of 195 bulbs = 195 × ₹12 = ₹2340

SP > CP, so there is profit.

Profit% = Profit23402000340 100% 10010017%

CP 2000 2000

Thus, the shopkeeper earns a profit of 17%.

4. Cost of the scooter = ₹49,000

Cost of transportation = ₹6000

Money spent on repairs = ₹5000

Total price of the scooter to Alex’s father = ₹49,000 + ₹6000 + ₹5000 = ₹60,000

Selling price of the scooter = ₹58,000

CP > SP, so there is loss.

Loss = CP – SP = ₹60,000 – ₹58,000 = ₹2000

Thus, there is a loss of ₹2000.

1. The selling price for Mia will be the cost price for her friend. Thus, the cost price of the ring for Mia’s friend will be ₹7600.

2. Cost price of the sofa for Anu = ₹19,700

Transportation charges for Anu = ₹400

Total cost of the sofa for Anu = ₹19,700 + ₹400 = ₹20,100

Selling of the sofa for Anu = ₹20,000

Loss for Anu = ₹20,100 – ₹20,000 = ₹100

Cost price of the sofa for Manu = ₹21,200

Transportation charges for Manu = ₹250

Total cost of the sofa for Manu = ₹21,200 + ₹250 = ₹21,450

Selling of the sofa for Manu = ₹21,400

Loss for Manu = ₹21,450 – ₹21,400 = ₹50

Thus, Anu incurred a bigger loss.

3. Marked price of the water cooler = ₹8350

Discount offered = 18%

Discount% = Discount amount 100% MP

18% = Discount amount 100% 8350

Discount amount = 1503

Selling price = Marked price – Discount amount = ₹8350 –₹1503 = ₹6847

4. Selling price of the article when there is profit = ₹1920

Selling price of the article when there is loss = ₹1280

Cost price (CP) is the same in both the cases.

If the profit% and the loss% is the same, then the profit and the loss on a same article will be the same.

Profit = Loss = 19201280 640 320 22 `` ` `

CP = ₹1280 + ₹320 = ₹1600

Given: There must be a profit of 25%

SP = 100Profit%10025 CP 16002000 100 100 ``

Thus, the article should be sold at ₹2000 to make a profit of 25%.

5. Selling price of 17 balls = ₹720

According to question,

Loss incurred after selling 17 balls = Cost price of 5 balls

CP of 17 balls – SP of 17 balls = CP of 5 balls

CP of 12 balls = SP of 17 balls

CP of 12 balls = ₹720

CP of 1 ball = 720 60 12 ` `

Thus, the cost of each ball is ₹60.

6. Total cost price of 56 kg rice = 26 × ₹20 + 30 × ₹36 = ₹520 + ₹1080 = ₹1600

Total selling price of 56 kg rice = 56 × ₹30 = ₹1680

Profit = ₹1680 – ₹1600 = ₹80

Thus, the trader’s profit is ₹80.

7. Seling price of the article = ₹1805

Loss% = 5%

CP = 100100180518051900 100595 `

Profit = 5%

SP = 1005105190019001995 100100 `

Thus, dealer must sell the article at ₹1995 to gain 5% on that article.

8. SP of the watches is the same.

Let the SP of each watch be x

CP of first watch = 10010 1001011 xx

CP of second watch = 1010 100109 xx

Total CP of the watch = 101090110200 1199999 xx xx x

12G

CP > SP, so there is loss. 200 2 2 99 99

xx x

Loss% 100%100%1% 200 200 99 99

Thus, there is 1% loss.

9. Marked price = 35% above CP

Discount offered = 20% on MP

Let the CP be 100.

MP = 35% above 100 = 135

Discount = MPDiscount%13520 27 100%100

SP = MP – Discount = 135 – 27 = 108

SP > CP, so there will be gain.

Profit% = Profit108100 100% 100%8% CP 100

Thus, the shopkeeper earns a profit/gain of 8%.

10. CP of 20 articles = SP of a articles

Let the CP of each article be ₹1

SP of a articles = ₹1 × 20 = ₹20

Profit = ₹(20 – a)

Profit% = 25%

SPCP Profit% 100% CP

20 25%100% a a

25a = (20 – a) × 100

25a = 2000 – 100a

25a + 100a = 2000

125a = 2000

a = 16

Thus, the value of a is 16.

Word Problem 1. Let the CP be ₹100

Profit = 320% of CP

Profit = 320% of ₹100 = ₹320

SP = CP + Profit = ₹100 + ₹320 = ₹420

New CP = ₹25% more than ₹100 = ₹125

SP remains constant so, new SP = ₹420

New Profit = ₹420 - ₹125 = ₹295

Profit% in terms of SP = Profit295100%100%70% SP 420

Thus, the profit is approximately 70% of the selling price.

1.a. PRT500053 SI 750 100100 `

b. PRT6000125 SI 3600 100100 `

c. PRT2000103 SI 600 100100 `

d. PRT225024 SI 180 100100 `

2.a. 1 200042 PRT 2 SI 200 100100 `

Total amount = ₹2000 + ₹200 = ₹2200

b. 1 3000123 PRT 3 SI 1200 100100 `

Total amount = ₹3000 + ₹1200 = ₹4200

c. 100004511 PRT 25 SI 2340 100100 `

Total amount = ₹10,000 + ₹2340 = ₹12,340

d. 200005411 PRT 22 SI 4950 100100 `

Total amount = ₹20,000 + ₹4950 = ₹24,950

3.a. Here, SI = 1 4 of P = 1 4 of ₹2000 = ₹500 SI100500100 T 6.25 PR20004 years

b. SI1002250100 T 10 PR75003 years

c. SI1001518.75100 T3 PR337515 years

d. SI1003510100 T 6.5 PR 1 120004 2 years

4.a. SI1002800100 P 6000 RT1 59 3 `

b. SI10023520100 P 49,000 RT68 `

c. SI10023100100 P 1,20,000 RT11 53 22 `

d. SI10031500100 P 2,00,000 RT31 121 54 `

5.a. SI1001995100 R 3% PT95007 per annum

b. SI1003412.5100 R 6.5

PT 1 125004 5 % per annum

c. SI10018000100 R 7.5% PT2000012 per annum

d. SI1001111.5100 R 0.0325%

PT 1 3600009 2 per annum

Word Problems 1. Total interest paid = ₹450

Rate of interest = 4.5% per annum

Time = 1 year

P = SI100450100 10,000 RT4.51 `

Thus, Akila borrowed an amount of ͅ₹10,000 from her friend.

2. P = ₹2,50,000

R = 8% per annum

T = 4 years

PRT25000084 SI 80,000 100100

Thus, Preeti has to pay ₹80,000 after 4 years.

3. P = ₹40,000

R = 10% per annum

T = 2 years

PRT40000102 SI 8000 100100

A = P + SI = ₹40,000 + ₹8000 = ₹48,000

Thus, the total amount paid after 2 years is ₹48,000.

4. SI = ₹1000

R = 10% per annum

Time = 5 years

SI1001000100 P 2000 RT105

Thus, the money lent by Nitish was ₹2000.

5. Total amount = 7 4 of P

A = P + SI

7 PPSI

4

So, SI = 3 4 P

T = 3 years

3 100 SI100

4 R 25% PT3 P P per annum

Thus, the rate of interest is 25%.

6. SI = ₹720, R = 18% and T = 2 SI100720100

P 2000 RT182 `

Extra interest amount for 1 year= 12% of P = 12% of ₹2000 = ₹240

Extra interest amount for 2 years = 240 × 2 = 480

Total SI = ₹720 + ₹480 = ₹1200 SI1001200100 R 30% PT20002 per annum

Hence, the rate of interest is 30%.

7. Let the amount Abella borrowed be P. PRT SI 100

Total SI = SI for first 3 years + SI for next 4 years + SI for next 3 years

538412315323683

17,845 100100100100100100100 PPPPPPP

P = 17845100

83 `

21,500

Thus, she borrowed an amount of ₹21,500.

8. Let the money borrowed by one person be x. Money borrowed by the other will be 1000 – x.

Total SI = SI for the first person + SI for the second person

62 = 100081 51 100100 x x − 580008

62 100100 xx

6200 = 5x + 8000 – 8x

3x = 8000 – 6200

3x = 1800

x = ₹600

₹1000 – x = ₹400

Thus, ₹600 is lent at 5% per annum and ₹400 is lent at 8% per annum.

Chapter Checkup 1. a. 44 100%80% 55

b. 18.75% = 18.7518753 1001000016

c. 2%%11111111 555100500

d. 25757 11 100%1140% 555

2. a. 0.48 = 0.48 × 100% = 48%

b. 4%%19919 0.045 222100200

c. 5.64 = 5.64 × 100% = 564%

d. 8%%165651 0.08125 888100

3. a. 25% of 460 = 251460460115 1004

b. 1 12% 2 of 20 L = 25 % 2 of 20 L = 251 20 2.5 2100 L L

c. 1 50% 2 of 1000 = 101 % 2 of 1000 = 1011 1000505 2100

d. 5% of 3 kg 200 g = 5% of 3200 g = 5 3200160 100 g

e. 40% of 6 km 400 m = 40% of 6400 m = 40 64002560 100 m

= 2 km 560 m

4. a. 0.9 100%20% 4.5

b.  20 mL 20 mL 1 100% 100%% 6 L 6000 mL3

c. 2 and 10 p210 p 100% 100%14% 15 1500 p ` `

d. 1000 m1 km 100%100%5% 20 km 20 km

e. 0.25 L 1 100%7% 3.5 L 7

5. a. Let the number be a 1 4%20 8 a 331 20 8100 a 20100828 484 3333 a

b. Let the number be a 1 5%500 2 a 111 500 2100 a 500210010 9090 11 11 a

c. Let the number be a 1 12%933 3 a 371 933 3100 a 933310032 7564 37 37 a `

d. Let the number be a. 1 50%80 m 4 cm804 cm 12 a

6011 804 12100 a 80412100195 1605 cm 601 601 a

e. Let the number be a. 24% × a = 3 L 250 mL = 3250 mL 24 3250 100 a 32501002 13,541 mL 24 3 a

6. a. Percentage increase = Amount of increase 100% Original Amount    

1812 6 100%100%50% 12 12    

b. Percentage increase = Amount of increase 15 100%100%100% Original Amount 15        

7.a. Profit = SP – CP = ₹200 – ₹150 = ₹50

b. Loss = CP – SP = ₹1580 – ₹1235 = ₹345

c. Profit = SP – CP = ₹5235 – ₹5100 = ₹135

d. Loss = CP – SP = ₹458 – ₹369 = ₹89

8. a. SP > CP, so there will be profit.

b. CP > SP, so there will be loss.

c.

9. a.

c.

10.a. PRT50023 SI 30 100100 `

A = P + SI = ₹500 + ₹30 = ₹530

b. 550014611

22 SI 5183.75 100100 `

A = P + SI = ₹5500 + ₹5183.75 = ₹10,683.7

Word Problems 1. CP of the pen = ₹100

SP of the pen = ₹125

Since, SP > CP, so there is gain.

Gain = ₹125 – ₹100 = ₹25

Gain% = Gain25100%100%25%

CP 100

Thus, the gain percent is 25%.

2. Total amount spent by cobbler = ₹80 + ₹10 + ₹10 = ₹100

CP for the cobbler = ₹100

SP of each pair of shoes = ₹180

SP > CP, so there is profit.

Profit = ₹180 – ₹100 = ₹80

Profit% = Profit80100%100%80% CP 100

Thus, the cobbler makes a profit of 80% on each pair of shoes.

3. Initial price of the share = ₹54

Final price of the share = ₹72

Percentage increase = Amount of increase 100% Original Amount    

725418 1 100%100%33% 54 54 3

Thus, the stock price increased by 1 33% 3 .

4. Total number of students in the school = 480

Ratio of boys to girls = 7:8

a. Percentage of boys in the school = 772 100%100%46%

7815 3

Thus, there are 2 46% 3 of the boys in the school.

b. Percentage of girls in the school = 81100%53% 783

Thus, there are 1 53% 3 of the girls in the school.

5. Let the third number be a.

First number = 20% more than a = a + 20% × a = a + 0.2a = 1.2a

Second number = 50% more than a = a + 50% × a = a + 0.5a = 1.5a

Ratio of first two numbers = 1.2a:1.5a = 4:5

Thus, the ratio of first two numbers is 4:5.

6. The ratio of the number of boys and girls in a college = 5:6

Let the number of boys and girls be 5x and 6x respectively.

Percentage increase in the number of boys = 40%

Increased number of boys in the school = 5x + 40% × 5x = 5x

+ 2x = 7x

Percentage increase in the number of girls = 50%

Increased number of girls in the school = 6x + 50% × 6x = 6x +

3x = 9x

New ratio of the number of boys to girls = 7x:9x = 7:9

Thus, the new ratio of the number of boys to the number of girls is 7:9.

7. Percentage of the expenditure by Kapil out of his total income = 20%

Percentage increase in the salary = 20%

Let the initial income of Kapil be 100.

Savings by Kapil initially = Income – Expenditure = 100 – 20% of 100 = 100 – 20 = 80

Increased income of Kapil = 100 + 20% of 100 = 100 + 20 = 120

Since the savings remain the same, so, savings for increased income = 80

Expenditure in increased salary = 120 – 80 = 40

Percentage increased in the expenditure =

 

Increase in expenditure 4020 100% 100% Initial expenditure 20

20 100%100%

20

Thus, the expenditure increased by 100%.

8. Cost of each shirt = ₹250

Tax on each shirt = ₹1.2

Total cost for each shirt = ₹250 + ₹1.2 = ₹251.2

Cost of 150 shirts = 150 × ₹251.2 = ₹37,680

Cost of transportation = ₹1500

Labour charges = ₹500

Total cost price for the shirts = ₹37,680 + ₹1500 + ₹500 = ₹39,680

Profit% for the shirts = 30%

Total selling price for the shirts = CP + Profit = ₹39,680 + 30% of ₹39,680 = ₹39,680 + ₹11,904 = ₹51,584

Selling price of each shirt = ₹51,584 ÷ 150 = 67 343 75 `

Thus, Jon should sell each shirt at 67 343 75 ` to earn a profit of 30%.

9.  Discount% = 12%

SP of the item = ₹880

SP = MP – Discount

₹880 = MP 12 100 − MP

₹880 = 88 100 MP

MP = ₹1000

Thus, the marked price on the item is ₹1000.

10. Discount% on the watch = 5%

Difference in the amount if the watch was sold at 6% discount = ₹27

Discount difference = Profit difference 65

MPMP27 100100 −

1 MP27

100

MP = ₹2700

Thus, the marked price on the watch is ₹2700.

11. Rate of interest = 14% per annum

Total amount paid back to the bank = ₹42,000

Time = 3 years

A = P + SI

P143

42,000P 100

42,000 = P + 0.42P

42,000 = 1.42P

33

P29,577 71

Thus, the principal amount that the man took was 33 29,577 71 ` .

12. Amount borrowed by Sansa = ₹5000

Time = 2 years

Rate at which Sansa borrows money = 4% per annum

Rate at which Sansa lends money to Arya = 1 6 4 % per annum

Interest given by Sansa = 500042 400 100 `

Interest received by Sansa from Arya = 1 500062 4 625 100 `

Profit earned by Sansa = ₹625 – ₹400 = ₹225

Thus, Sansa gained ₹225.

13. The ratio of seats for Mathematics, Physics. and Biology in the school = 6:8:12

Let the seats for Mathematics, Physics. and Biology be 6x, 8x and 12x respectively.

Percentage increase in the number of seats for Mathematics = 40%

Increased seats for Mathematics = 6x + 40% × 6x = 6x + 2.4x = 8.4x

Percentage increase in the number of seats for Physics = 50%

Increased seats for Physics = 8x + 50% × 8x = 8x + 4x = 12x

Percentage increase in the number of seats for Biology = 75%

Increased seats for Biology = 12x + 75% × 12x = 12x + 9x = 21x

Ratio of increased seats for Mathematics, Physics, and Biology in the school = 8.4x:12x:21x = 14:20:35

Thus, the ratio of the increased seats is 14:20:35.

14. Percentage decrease in the value of television each year = 10%

Initial cost of the television = ₹1,60,000

Value of the television after 1 year = 1,60,000 – 10% of 160,000 = 1,60,000 – 16,000 = ₹1,44,000

Value of the television after 2 years = 1,44,000 – 10% of 1,44,000 = 1,44,000 – 14,400 = ₹1,29,600

Thus, the cost of television after 2 years is ₹1,29,600.

15. Let the MP of the product be 100.

CP for the wholesaler = 100 – 40% of 100 = 100 – 40 = 60

SP for the wholesaler = 100

Profit = 100 – 60 = 40

Profit% = Profit40 2 100%100%66% CP 60 3

Thus, the wholesaler earns a profit of 2 66% 3 on the product.

Chapter 13

Let’s Warm-up 1. Fractions with the same denominator are called like fractions.

2. A fraction with the denominator greater than the numerator is called a proper fraction.

3. A fraction with the numerator greater than the denominator is called an improper fraction.

4. 7 14 8 is a mixed fraction.

5. 17 2 is an improper fraction.

Do It Yourself

13A 1. a. 1 out of 4 parts = 1 4 b. 3 out of 8 parts = 3 8

c. 2 out of 4 parts = 2 4 d. 4 out of 8 parts = 4 8

e. 2 out of 5 parts = 2 5

2. Numerator = (3 × (–2)) = –6; Denominator = (7 × (– 4)) = –28

Fraction = 6 28 Numerator Denominator

3. a. 10 and 11 are both positive numbers, hence, a positive rational number.

b. 1 and 5 are both positive numbers, hence, a positive rational number.

c. −3 and –5 are both negative numbers, hence, a positive rational number.

d. 7 and 8 are both positive numbers, hence, a positive rational number.

e. 5 is a positive number and –9 is a negative number, hence, a negative rational number.

4.a.

7. a.

d.  1929.5 22211− − e. 45315 105335 − −

8. Answers may vary. Sample answers.

a. 7 13

7214

13226

9. 9 763

–63 00

2918

7963 − −

10. Answers may vary. Sample answers.

a. 7 1 b. 147 1.4 105

c. Zero divided by anything is zero. 0 5

d. 51 0.5 102 −

Word Problems 1. Pudding Ryan ate = 1 3

Pudding Rayna ate = 2221 4422

Since, 11 32

They did not eat the same amount of pudding.

2. Number of people = 3

Number of parts each ate = 1

Number of parts left = 0

Number of parts the pizza was cut in = 1 + 1 + 1 = 3

So, they cut the pizza in 3 equal parts.

12B 1.

a. LCM of 7 and 9 is 63.

4936

7963

5 735

9763

Since, 36 > 35

Then, 3635 6363

45 79

b. Since, 2 < 7 Then, 27 33

c. LCM of 2 and 5 is 10. 155 2510 − − 3 26

Since, −5 > –6

56

d. 1313 2424 and 99 44

LCM of 24 and 4 is 24.

13

24

9654

4624 − −

Since, –13 > –54

Then, 1354 2424

139 244

e. 4 5 and 33 55

Since, 4 > –3

Then, 43 55

2.a. We can write 1 1 1

LCM of 1, 3 and 5 is 15.

8540 3515 ; 4520 3515 − − ; 11515 11515 ; 236 5315 − − ;

Since, –20 < –6 < 15 < 40

Then, (4)281.

353

So, the greatest rational number is 8 3

b. LCM of 2, 5, 9 and 10 is 90.

14545

24590 ; 31854 51890 − − ; 41040 91090 ; 9981 10990

Since, –54 < 40 < 45 < 81

Then, 3419 59210

So, the greatest rational number is 9 10

c. LCM of 3, 4, 9 ad 24 is 72.

22448

32472 ; 21836 41872 ; 3824 9872 ; 133 24372

Since, 3 < 24 < 36 < 48

Then, 1322 24943

So, the greatest rational number is 2 . 3

d. LCM of 11, 22 and 44 is 44.

8432

11444 − − ; 21242 22244 ; 33 44 ; 5420 11444

Since, –32 < 20 < 33 < 42

Then, 853321 11114422

So, the greatest rational number is 21 . 22

3. a. LCM of 8, 9, 10 and 11 is 3960.

1495495

84953960 ; 2440880 94403960 ; 33961188 103963960 ; 43601584 113603960

Since, 1584 > 1188 > 880 > 495

Then, 4321 111098

So, the least rational number is 1 8

b. LCM of 16, 12, 9 and 15 is 720.

845360

1645720 − − ; 460240 1260720 − − ; 380240 980720 ; 548240 1548720 −− − −

Since, 240 > –240 = –240 > –360

Then, 3458 9121516

So, the least rational number is 8 16

c. LCM of 1, 4, 3 and 5 is 60.

16060

16060 ; 31545 41560 ; 22040 32060 ; 41248 51260

Since, 60 > 48 > 45 > 40

Then, 432 1 543

So, the least rational number is 2 . 3

d. As none of the given rational numbers is negative, 0 is the smallest amongst the given set of rational numbers.

4. a. 41560 715105 − − ; 7749 157105 − −

Since, –60 < –49

Then, 47 715

b. 12336 13339 ; 21326 31339

Since, 36 > 26

Then, 122 133

c. 3618 5630 ; 5525 6530

Since, 18 < 25

Then 35 56

d. 521013339 ; 62124312 − −− −

Since, –10 > –39

Then, 513 64

5. a. LCM of 9, 12 and 18 is 36.

–10440 9436 − ; 248 9436 ; 5315 12336 ; 7214 18236

Since, −40 < 8 < 14 < 15

Then, 10275 991812

b. LCM of 3, 6, 16 and 20 is 240.

–1412168

2012240 − ; 1015150 1615240 − − ; 440160 640240 − − ; –18080 380240 −

Since, –168 < –160 < –150 < –80

Then, 144101 206163

c. LCM of 3, 5,7 and 6 is 210. 17070

− −− −− −

6. a. LCM of 9, 12 and 18 is 36. 1044024853157214

− −

, , , 943694361233618236

Since, 15 > 14 > 8 > –40

Then, 57210 121899

b. LCM of 3, 6, 16 and 20 is 240.

1412168101515044016018080

− −− −− −− −

, , , 20122401615240640240380240

Since, –80 > –150 > –160 > –168

Then, 110414 316620

− − −− − −

c. LCM of 3, 5, 7 and 6 is 210. 17070–342126430120–335105 , , , 370210542210730210635210

Since, –70 > –105 > –120 > –126

Then, 1343 3675

7. Answer may vary. Sample answer.

a. Multiply both –1 and 0 by 6 6 666

1 ;00 666 −

So, 12345 ,,,, 66666

b. LCM of 5 and 3 is 15. 4312 5315 − − and –2510 3515 −

Multiply both by 3 3

1233610330 ; 1534515345

So, 3132333435 ,,,,. 4545454545

370210

− − ; 342126430120335105 , , 542210730210635210

Since, –126 < –120 < –105 < –70

Then, 3431 5763

d. LCM of 5, 13, 25 and 19 is 6175. –3123537056475285082471976 , , , 512356175134756175252476175

−

83252600

− −

193256175

Since, –3705 < –2600 < 1976 < 2850

Then, 3886 5192513

e. LCM of 3, 5 and 6 is 30. 3618–21020–46245525 , , , 56303103056306530 − −

Since, –24 < –20 < 18 < 25

Then, 4235 5356

f. LCM of 9, 7 and 15 is 315. 1353524590–3451351121231 , , , 9353157453157453151521315

−

Since, –135 < 35 < 90 < 231

Then, 31211 79715

c. Multiply both by 6 6

1662612 ; 66366636

So, 7891011 ,,,, 3636363636

d. LCM of 3 and 5 is 15. –2510 3515 − and 339 5315 − −

Multiply both by 6 6

106609654 ; 1569015690

So, 5556575859 ,,,,. 9090909090

e. Multiply both 2 and –1 by 2 2

2;12422 2222 −

So, 1123 ,0,,,. 2222

8. VT = TU = UW and PR = RS = SQ

WV is divided into three equal parts.

So, 3936 –3 2 3333 W andV −

PQ is divided into three equal parts.

So, 3336 1 2 3333 P andQ

45

33 RS

9. Warmest to coldest means higher to lower, i.e. in descending order.

a. 9280252044562464–3105315 , , , 328084015568408105840 −

–141201680

7120840 −

Since, 2520 > 2464 > −315 > −1680

Then, 944314 31587

b. 31030–3103075359436 , , , 4104041040854010440 −

Since, 36 > 35 > 30 > –30

Then, 9733 10844

c. –5.0,50–3944328 –3.9, 4.4, 32.8 10101010 −

Since, 328 > 44 > –39 > –50

Then, 32.8 > 4.4 > –3.9 > –5.0

d. –463 872576–356168504504 252 , , –1 , 772504956504504504863504

Since, 576 > –168 > –252 > –504

Then, 834 1 798 −

10. a. Largest two−digit number = 99

Smallest 3−digit number = 100

99

100

b. Smallest three−digit number =100

Largest three−digit number= 999

100

999

LCM of 100 and 999 is 99,900

9999998,901

10099999,900

10010010,000

99910099,900

Since, 98,901 > 10,000

Then, 99100 . 100999

The correct answer is option (a).

Word Problems 1. Number of glasses filled by Raman = 11 5

Number of glasses filled by Farhan = 13 7

LCM of 5 and 7 is 35.

11777

5735 and 13565 7535

Since, 77 > 65

Then, 1113

57

So, Raman’s bottle can hold more water.

2. Thickness of Radha’s packet = 1 3

Thickness of Kashish’s packet = 2 7

LCM of 3 and 7 is 21.

177

3721 and 236 7321

Since, 7 > 6

Then, 12 37

So, Radha bought the thicker paper.

13C 1. a. 292911 5555

b. LCM of 3 and 5 is 15. 8540

3515 and 8324 5315 − − 4024402416 1515151 ) 5 (    

c. LCM of 3 and 5 is 15.

12560

3515 and 14342 5315

6042(6042)10234 151515155

d. LCM of 7 and 3 is 21. 15345

7321 − − and 177119 3721 451194511974 2 () 1212121 −−

e. 1215121527

17171717

2.a. 1311311313 88888888 − −

b. 983458350 5 99999  − −   

c. 1581584 1313131313 −−

d. 3565935659118 77777

e. 12131012131015 1717171717 − −

3.a. 55 0 77 − b. 1111 0 88 − c. 11 0 22 −

d. 99 0 1111 − e. 55 0 99 −

4. a. 32321 5555 −

b. 13133330 0 2623666 − − −

c. 717(1)718 1 88888 −−− −

d. 2729751835183517 59599545454545 − − −

e. 3537582140214019 87877856565656 − − −

5. a. LCM of 8, 12 and 16 is 48. 76549342202742202735 861241634848484848 − − −−−− −

b. LCM of 50, 5 and 6 is 150.

932305252760125

−

503530625150150150

27601259246

− −−−

15015075

c. LCM of 33, 22 and 44 is 132. 2436138183818313

334226443132132132132132

6. Second rational number = sum – first rational number = 1410144

2 555 −−

7. Number = 5752 1 777 −− − −−

8.a. 6437242124213 744728282828

b. 27279 1 9999 − −

c.

d.

e.

9. Answer may vary. Sample answer.

13D

Word Problems 1. Milk Kanak was boiling = 12215 2 222 L

Milk evaporated = 1 4 L

Milk left = 515211011019 242244444 − − − − L

2. Length of rope = 1 12 5 m = 125161 55  m

Length of pieces cut = 1033103333033303363 7 393399999 m

Remaining length of the rope = 61617526 7 555 − − m

3. Money earned = ₹800

Money spent on tea and snacks = 680 9 `

Money spent on food = 510 2 `

Money spent on repairs of the taxi = 220 5 `

Total money spent = 680510220 925 ` 680105104522018 910245518 ` 680022,9503960 909090 ` 680022,950396033,710 90 90 ``

Money saved 33,71072,00033,71038,290 800 425.45 90 90 90  −    ` ``

38,290 425.45 90 ` `

4. Length of wood = 17 4  m

Length of wood cut = 3.4 m = 342 102  m 17 5  m

Remaining length of the wood = 171717517485688568 454554202020 − 1 7 2 0  m

1. a. 323263 54542010

b. 3434123 87875614 −− −−

c. 611611666 117117777

d. 3838242 20920918015 −− −−

e. 464624 757535 −−− −

2.a. 8 5 b. 11 9 c. 1

d. 9 71

3.a. 6263189 – –535–2105

b. 4414 – –3–55–315

c. 131441 8483246

d. 164163484 –213214847

e. 1119111211

1212121919

4.a. 354532545 489842898  − − − −   

 − − − − −   

6520172729

8872 82016020

2. Cost of 1 8 2 m fabric = 17541301 75 444 ` `` 1(821)17

8 222

Cost of 1 m fabric = 301173012301 4241734 `

3. Length of the ribbon = 153 3 m

Length of each piece = 1531531153 20 332060 m

b. 45184121848181618

   

91219951945191519

16696

51995



4. Distance Sambhav ran in 5 days = 1 11 4 km = 114145 44 km

Distance Sambhav ran in 1 day = 454519 5 4454 km

  

c. 211422614262672

137261314131413

d. 31592315934527 52435242108

 −−− −−−  

4583604 10272703    

5. (m + n) ÷ (m – n)

5. Area of the garden = 2 411343114313791 m 34343412

Chapter Checkup 1. Numerator = (65 – 32) = 33

Denominator is 3 × 9 = 27

Rational number = 33 27

a. 131342428 2 222222224

 − −   

b. 353533553355 535353355335

92592534341534

3417

168

c. 42424242262721 777777777663 −−− −   

6. Required number = 2424813312 39 1388 −

7. 811 1 118

8. 54555425 2524258 −

3. a.

9. 6546957605435 15 799799796363  − −    60896053401780



  

9639567189

10. Required number = 4849363 30930824020 − − − −

11. a. 8560856014529 12512512512525 multiplicative inverse 25 29

b. 5555105 44442 multiplicative inverse 2 5

Word Problems 1. Cost of 1 2 2 kg fruits = 311543463 115 444 ` `` 1(221)5 2 222

Cost of 1 kg fruits = 4635463246322463 42454524

4463 46.3 1010 ` `

6.

may vary. Sample answer.

So 1920212223 ,,,,, 3030303030

b. Multiply 2 9 and 3 9 , by 6 6

2612 9654 and 3618 9654

So, 1314151617 ,,,, 5454545454

d. Multiply –1 and 1, by 3 3 33

1 33 − and 33 1 33

So, 2112 ,,0,, 3333

c. Multiply –1 2 and –3 2 , by 3 3

133 236 and –339 236

So, 45678 ,,,, 66666

e. Multiply 0 and –1 , 2 by 6 6

6 00

6 and 166 2612

So, 12345 ,,,, 1212121212

7.a. 797159101059019513 1015101515015010

b. 212215451 –52521010 − −−

c. 15 42 42 153 505010

d. 428480 14801428 1808040 71479849

e. 1212222 1–2123303 2422444 − −

f. 523952123913(624507)1173 –131213121561564 − −

g. 12 1 13 65 1 5 60 5 1 55

h. – 451345 14714     745 21326   

8.a. 8 25 94 1 5   

b.

7147214141414142

33 0 22 − − = Additive inverse

10. 12 –5 4 15 ––3 5     2 9 5 3  −    4 1 5 − —422 33  −  

43210

Multiplicative inverse = 3 10

11. Multiply 4 9 by 4 4

4416

9436 − −

As 4 –9 and 16 –36 are equivalent rational numbers, they represent the same number.

12. 65865846532653297 123123412121212

65865846532653233 123123412121212 − − − −

So, 9733971297 1212123333

13. LCM of 7, 3, 5 and 9 is 315

4451801105105263126535175

, , , 7453153105315563315935315

Since, 105 < 126 < 175 < 180

Then, 1254 3597 .

Sum of middle two numbers = 252955182543 59594545

Average = Sum 4343143 or Sum22 2 4545290

5 1019 1 9999 

  

b. 8–1458(1)458145945459 –

99509509509505010

c. 11116112425 63 24282888

d. 131413–7–133418291–442 115115115555555

   

e. 1926 212 310 5 6419235 21 7 13–516 48 4 9 3544 643815448154483815344816 7 351631635163

     

384571683872133836,065182 5485485

9.a.

5858

14. LCM of 8, 4 and 27 is 216. 427108154549872 , , 827216454216278216

Since, 108 > 72 > 54

Then, 491 8274

difference between the greatest and the least number = 41412424221

8484288884 − − − −

15. 141514141515196225421 15141514210210

14151414151519622529

15141514210210 − −− −

So, 42129421210421 2102102102929 −

16. 43141110117 5 ––252222    

43114111017 –25524    

4316917

10524 −

436917

10104

1121711241710618309

1041044020

Word Problems 1. Part of the book read in 2 hours = 3 7

Part of the book read in 1 hour = 3313 2 77214

Now, 13217 3 222

Part of the book read in 7 2 hours 733 2144

2. Length of a thread = 1202141 20 222 cm

Length of a thread cut = 1491493142741 62623666 cm

Remaining length = 414141341123418241 262366663 − − − cm

Therefore, the length of the remaining thread is 41 3 cm.

3. Number of books in a library = 18,000.

English = 1 6

Hindi = 2 6

Number of French books = 12 18,000 18,00018,000 66  − −   

3118,00018,00018,00018,00090009000 62   − −      

There are 9000 French books in the library.

4. Average speed of the truck = 30 km/hr

Distance covered in 1 hour = 30 km

Distance covered in 11 2 hours = 11 30165 2 km

Therefore, it will cover 165 km

5. Area = length × breadth = 2 140507000 m 339

The area of the park is 2 7000 m 9

6. Quantity of the aquarium = 40 gallons

Aquarium filled = 20 4032 25 gallons

Quantity needed = 40 – 32 = 8 gallons.

Therefore, he needs 8 gallon

7. Original price = ₹12,000

Sale price = 5 12,00010,000 6 `

Tanmay paid ₹10,000 for the computer.

8. Distance crawled in one minute = 136119 3 666 inches

Distance crawled in the next minute = 342311 2 444 inches

Total distance = 191119411614271 64642412 inches

Therefore, the insect crawled 71 12 inches.

Chapter 14

Let’s Warm-up 1. You construct an angle of 90° by bisecting a 120° angle. False

2. An angle bisector bisects the angle into 3 equal parts. False

3. When an angle bisector of a 60° angle is constructed, we get two angles of 30° each. True

4. While bisecting an angle, the width of compass can be less than half of the width of the arc. False

5. You can construct an angle of 15° by bisecting an angle of 95°. False

Do It Yourself

14A 1. a. A triangle can be constructed with the length of three sides given.

b. A triangle cannot be constructed if the sum of lengths of any two sides is less than the length of third side

c. We need the length of one side/sides to construct an equilateral triangle.

2. a.

4. In the triangle, the sum of the lengths of smaller sides is greater than the length of the largest side. Here, 5 cm + 12 cm > 13 cm so the triangle is feasible.

3. Scalene Triangle

5. Equilateral Triangle 6.

8. Perimeter of triangle = 18 cm

The ratio of length of sides = 2:3:4 The length of the sides are

14B 1. To construct a triangle with SAS criterion, we need the measure of two sides and an included angle. Hence, option b and c are the correct answers.

2.a.              b.

14D 1. Unknown side 4.6 cm    2. Unknown side = 7 cm

3. BC = 6.8 cm; BAC = 46°,    4. Unknown angles are  BCA = 44°            32° and 58°.

3. XYZ = 91°, YZX = 44°  Scalene triangle; Obtuseangled triangle

4. ΔPQR is a scalene and acute-angled triangle. QR is the longest side in the triangle.

5. BAC = 37°, BCA = 23°

Word Problem 1.

XZ = 6.2 cm, YZ = 3.2

3. R = 70°          4. DF = 6.3 cm, EF =

5. ZXY = 54°

Word Problem 1. If 3 m on ground is equal to 1 cm on paper then, 15 m = 15 3 cm = 5 cm & 6 m = 6 3 cm = 2 cm

From the construction, we can see that the unknown side measures 4.6 cm.

Approximate length of Sarah’s garden on ground = 4.6 × 3 m = 13.8 m

Chapter Checkup 1. a. SSS b. SAS c. RHS d. ASA

2. A = 82°, B = 60° and C = 38°. So, the triangle is acute-angled triangle

3. Scalene triangle; Obtuseangled triangle (111°, 40°, 28°)

4. AC = 6 cm; BC = 4.1 cm  5. XZ = 3.8 cm; YZ = 4.7 cm

5. Scalene triangle; Acute angled triangle LN = 7.1 cm; MN = 8.7 cm

Word Problem 1.

6. Remaining angles     7. SAS criterion.    are 55° and 80°.

8. Length of unknown side = 3.6 cm  9.

10. Sum of length of sides UW and VW = 6 cm + 4.9 cm = 10.9 cm

Word Problem 1. If 10 cm on model is equal to 1 cm on paper then, 45 cm = 45 10 cm = 4.5 cm & 30 cm = 30 10 cm = 3 cm

From the construction, we can see that the unknown side measures 3.4 cm. Approximate length of unknown side of model bridge = 3.4 × 10 cm = 34 cm

Chapter 15

Let’s Warm-up 1. a. 60 mm = 6 cm, 80 mm = 8 cm

Perimeter = 2 cm + 14 cm + 12 cm + 14 cm + 6 cm + 8 cm + 2 cm + 6 cm + 2 cm + 10 cm + 8 cm + 12 cm = 96 cm

b. Area of part I = 14 × 2 = 28 sq. cm

Area of part II = 8 × 2 = 16 sq. cm

Area of part III = 14 × 2 = 28 sq. cm

Area of part IV = 2 × 2 = 4 sq. cm

Area of part V = 8 × 2 = 16 sq. cm

Total area of the figure = 28 + 16 + 28 + 4 + 16 = 92 sq. cm

2. a. Perimeter = 168.4 cm + 208 cm + 84.2 cm + 208 cm + 82.2 cm + 84.2 cm + 168.4 cm + 250.6 cm = 1254 sq. cm

b. Area of part I = 376.4 × 84.2 = 31,692.88 sq. cm

Area of part II = 168.4 × 82.2 = 13,842.48 sq. cm

Area of part III = 84.2 × 84.2 = 7089.64 sq. cm

Total area of the figure = 31,692.88 + 13,842.48 + 7089.64 = 52,625 sq. cm.

Do It Yourself

15A 1. a. Perimeter of rectangle = 2 (l + b) = 2 (6.8 + 3.7) = 2 × 10.5 = 21 cm

Area of rectangle = l × b = 6.8 cm × 3.7 cm = 25.16 sq. cm

b. Perimeter of square = 4 × side = 4 × 4.5 cm = 18 cm.

Area of square = side × side = 4.5 × 4.5 = 20.25 sq. cm.

c. Perimeter of rectangle = 2 (l + b) = 2 (8.2 + 5.3) = 2 × 13.5 = 27 cm.

Area of rectangle = l × b = 8.2 cm × 5.3 cm = 43.46 sq. cm.

d. Perimeter of square = 4 × side = 4 × 5.9 cm = 23.6 cm

Area of square = side × side = 5.9 × 5.9 = 34.81 sq. cm.

2. a. Perimeter of square = 4 × side = 8 cm, the side = 8 ÷ 4 = 2 cm

Area of square = side × side = 2 × 2 = 4 sq. cm

b. Perimeter of square = 4 × side = 16 cm, side = 16 ÷ 4 = 4 cm

Area of square = side × side = 4 × 4 = 16 sq. cm

c. Perimeter of square = 4 × side = 32 cm, side = 32 ÷ 4 = 8 cm

Area of square = side × side = 8 × 8 = 64 sq. cm

d. Perimeter of square = 4 × side = 52 m, side = 52 ÷ 4 = 13 m

Area of square = side × side = 13 × 13 = 169 sq. m

e. Perimeter of square = 4 × side = 108 m, side = 108 ÷ 4 = 27 m

Area of square = side × side = 27 × 27 = 729 sq. m

3. a. P = 10 cm, L = 2 cm

B = P 2 – L = 10 2 – 2 = 5 – 2 = 3 cm

Area = L × B = 2 × 3 = 6 sq. cm

b. P = 18 cm, L = 5 cm

B = P 2 – L = 18 2 – 5 = 9 – 5 = 4 cm

Area = L × B = 5 × 4 = 20 sq. cm

c. P = 48 cm, L = 21 cm

B = P 2 – L = 48 2 – 21 = 24 – 21 = 3 cm

Area = L × B = 21 × 3 = 63 sq. cm

d. P = 94 m, L = 38 m

B = P 2 – L = 94 2 – 38 = 47 – 38 = 9 m

Area = L × B = 38 × 9 = 342 sq. m

4. Perimeter of a square = 4 × side = 4 × 15 m = 60 m

Perimeter of rectangle = perimeter of square = 60 m

Length of rectangle = 18 m

Breadth of rectangle = Perimeter 2 – Length = 60 2 – 18 = 30 – 18 = 12 m

5. Perimeter of a square: area of a square = 5:8

Perimeter of square = 4 × side, Area of square = side × side

So, 4 × side side × side = 5 8 4 side = 5 8   4 × 8 = 5 × side.

Side = 4 × 8 5 = 32 5 = 6.4

Area of square = 6.4 × 6.4 = 40.96 sq. units

6. Area of rectangle = area of square = 49 sq. cm

If the length is of maximum possible measure, then the length of rectangle must be at least 49 cm. So, breadth of rectangle will be 1 cm such that its area is 49 sq. cm.

Perimeter of rectangle = 2 (l + b) = 2 (49 + 1) = 2 × 50 = 100 cm

If the side of the square is 7 cm, then the area of the square will be 49 sq. cm.

So, the side of the square = 7 cm.

Perimeter of the square = 4 × side = 4 × 7 = 28 cm

Since, 100 > 28, perimeter of rectangle is greater than the perimeter of the square.

7. Perimeter of rectangle: area of rectangle = 5:3

Perimeter of rectangle = 2 (l + b)

Area of rectangle = length × breadth

So, 2 (l + b) l × b = 5 3

6(l + b) = 5(l × b)

Let the length of the rectangle be 2x and the breadth of the rectangle be 3x

Therefore, 6(2x + 3x) = 5 (2x × 3x)

6 × 5x = 5 × 6x2

30x = 30x2

30x2

30x = 1

x = 1

Length of the rectangle = 2 units, breadth of the rectangle = 3 units

Perimeter of the rectangle = 2 (l + b) = 2 (2 + 3) = 2 × 5 = 10 units

Area of the rectangle = length × breadth = 2 × 3 = 6 sq. units

Word Problems 1. Cost of fencing a square field = perimeter of square × ₹12/m = ₹24,000

Perimeter of square = ₹24,000 ÷ ₹12 = 2000 m

4 × side = 2000 m. So, side = 2000 m ÷ 4 = 500 m

Area = side × side = 500 m × 500 m = 2,50,000 sq. m

2. Length of the playground = 750 m

Breadth of the playground = 250 m

a. Area of the playground = length × breadth = 750 × 250 = 1,87,500 sq. m

Cost of levelling per 100 sq. m = ₹16

Cost of levelling per sq. m = ₹ 16 100 × 1,87,500 = ₹30,000

b. Perimeter of playground = 2 (l + b) = 2 (750 + 250) = 2 × 1000 = 2000 m = 2 km

Total distance travelled of the playground = 6 km

Number of rounds = 6 km ÷ 2 km = 3

So, 3 rounds will be taken to run 6 km on the playground.

3. Let the length of the rectangular field be 3x and the breadth of the rectangular field be 2x

Perimeter of the rectangular field = 2 (l + b) = 2 (3x + 2x) = 2 ×

5x = 10x

Given, perimeter of the rectangular field = 320 m

So, 10x = 320 m

x = 320 ÷ 10 = 32 m

Therefore, L = 3 × 32 = 96 m, B = 2 × 32 = 64 m

Area = L × B = 96 × 64 = 6144 sq. m

Cost of reaping the field per 50 sq. m = ₹25

Cost of reaping the field per sq. m = ₹ 25 50 × 6144 = ₹3072

4. B = 3 × L, A = 1083 sq. m

L × B = 1083 sq. m

L × 3 × L = 1083 sq. m

L × L = 1083 ÷ 3 = 361 = 19 × 19. So, L = 19 m

L = 19 m, B = 3 × L = 3 × 19 = 57 m

P = 2 (L + B) = 2 (19 + 57) = 2 × 76 = 152 m

Cost of fencing = ₹25 × 152 = ₹3800

5. Side of the square = 21 cm.

Perimeter of the square = 4 × side = 4 × 21 = 84 cm.

Let the length and breadth be 3x and 4x, respectively.

According to the question,

Perimeter of the rectangle = perimeter of the square = 84 cm

2 (l + b) = 84 cm

2 (3x + 4x) = 84

2 × 7x = 84

14x = 84. So, x = 84 14 = 6

length = 3x = 3 × 6 = 18 cm

breadth = 4x = 4 × 6 = 24 cm

Area of square = side × side = 21 × 21 = 441 sq. cm

Area of rectangle = length × breadth = 18 cm × 24 cm = 432 sq. cm

Since, 441 > 432, square encloses more area.

6. Area of 1 wall = 18.4 × 10.2 = 187.68 sq. m. So, Area of 4 walls = 187.68 × 4 = 750.72 sq. m

Area of 1 door = 2 m × 1 m = 2 sq. m. So, Area of 2 doors = 2 × 2 = 4 sq. m

Area of 1 window = 1 m × 0.6 m = 0.6 sq. m. So, Area of 3 windows = 0.6 × 3 = 1.8 sq. m

Area of 4 walls to be whitewashed = 750.72 – (4 + 1.8) = 750.72 – 5.8 = 744.92 sq. m.

The cost of whitewashing the walls with an area of 744.92 sq. m = ₹80 × 744.92 = ₹59,593.60

7. Area of the sheet of paper = 105 cm × 45 cm = 4725 sq. cm

Area of the rectangle cut out = 20 cm × 6 cm = 120 sq. cm

Number of pieces of paper = 4725 120 = 39.375.

Number of equal-sized pieces of paper, from a sheet of paper = 39

8. Area of the room = L × B = 18 × 10.5 = 189 sq. m

Area of the room occupied by the square table = side × side = 10 m × 10 m = 100 sq. m.

Area of the room left unoccupied = 189 – 100 = 89 sq. m.

9. Cost of carpeting per m = ₹150

Total cost of carpeting = ₹45,000.

Area of the hallway = 45000 150 = 300 sq. m

Length of the hallway = 300 15 = 20 m

15B

1. a. A = b × h = 6 × 4 = 24 sq. cm

b. A = b × h = 9 × 7 = 63 sq. cm

c. A = b × h = 7 × 3 = 21 sq. cm

d. A = b × h = 6 × 8 = 48 sq. cm

2. a. A = b × h = 11 × h

88 = 11h. So, h = 88 11  = 8 cm

b. A = b × h = 17.2 × h

349.16 = 17.2 × h. So, h = 349.16 17.2 = 20.3 cm

c. A = b × h = 16 × b

252.8 = 16 × b. So, b = 252.8 16 = 15.8 cm

d. A = b × h = 7.6 × b

70.68 = 7.6 × b. So, b = 70.68 7.6 = 9.3 cm

3.  Perimeter of the parallelogram = 2 (l + b) = 64 cm

l + b = 64 2 = 32 cm

l – b = 8 cm

2l = 32 + 8 = 40 cm  l = 40 2 = 20 cm = base of the parallelogram

Height of parallelogram = 8 cm

Area of the parallelogram = b × h = 20 × 8 = 160 sq. cm

4. Base of a parallelogram = 2 × height of the parallelogram

Area of the parallelogram = 648 sq. cm = b × h = 2h × h = 2h2

h2 = 648 2 = 324 = (18)2.So, h = 18 cm

b = 2 × 18 = 36 cm

5. a. Area of parallelogram = b × h = 8 × 10 = 80 sq. cm

b. Area of parallelogram = b × h = 80 = 15 × h

Altitude corresponding to the base JK = 80 15 = 5.3 cm

6. Let the adjacent sides of the parallelogram be 4x and 3x

Length of the parallelogram = 8.5 cm = 4x x = 8.4 4 = 2.125 cm

So, the other side of the parallelogram is 3 × 2.125 cm, which is equal to 6.375 cm.

Perimeter = 2 (l + b) = 2 (8.5 + 6.375) = 2 × 14.875 = 29.75 cm

7. Area of parallelogram = b × h = 25 × 10 = 250 sq. cm

Area of parallelogram = b × h = 15h = 250

h = PM = 250 15 = 16.7 cm (approx.)

LS

8. Area of parallelogram = 450 sq. cm = b × h

h = 450 b = 450 30 = 15 cm

Applying Pythagoras property to ΔADE, we get

AD2 = DE2 + AE2

172 = 152 + AE2

289 = 225 + AE2

AE2 = 289 – 225 = 64

AE2 = 82

AE = 8 cm

AB = AE + EB

30 = 8 + EB

EB = 30 – 8 = 22 cm

9. a. Area of parallelogram = b × h = 21 × 10 = 210 sq. cm

Area = b × h = LO × MF

210 = 15 × h

h = MF = 210 15 = 14 cm

b. Area of parallelogram = b × h = 30 × 12 = 360 sq. cm

Area = b × h = LO × MF

360 = b × 20

B = LO = 360 20 = 18 cm

Word Problem 1. Area of the parallelogram = b × h

576 sq. m = base × 18

base = 576 18 = 32 m

BC = 3 4 × base = 3 4 × 32 m =

24 m

9 cm R M 25 cm 15 cm A E D C B 30 cm 17 cm AO D P 18 × base C B base 3

Area = CP × DA (DA = BC)

P O Q 4

Second pathway = CP = 576 24 = 24 m

Thus, the length of the second pathway is 24 m.

15C 1. a. A = 1 2 × b × h = 1 2 × 16 × 13 = 104 sq. cm

b. A = 1 2 × b × h = 1 2 × 7 × 10 = 35 sq. cm

c. A = 1 2 × b × h = 1 2 × 10 × 8 = 40 sq. in

d. A = 1 2 × b × h = 1 2 × 9 × 5 = 22.5 sq. cm

2. a. A = 1 2 × b × h

44 = 1 2 × 11 × h

h = 44 × 2 11 = 8 cm

c. A = 1 2 × b × h

128 = 1 2 × b × 16

h = 128 × 2 16 = 16 cm

49.77 = 1 2 × 15.8 × h

h = 49.77 × 2 15.8 = 6.3 cm

d. A = 1 2 × b × h

39.90 = 1 2 × b × 7.6 cm

b = 39.90 × 2 7.6 = 10.5 cm

3. A = 1 2 × b × h = 1 2 × 16 × 10 = 80 sq. cm

4. Area of triangle = 1 2 × b × h = 1 2 × 8 × 6 = 24 sq. cm

24 = 1 2 × 10 × h

h = 24 × 2 10 = 4.8 cm

5. a. Area of triangle = 1 2 × 8 × 7.5 = 30 sq. cm

b. Area of triangle = 1 2 × 15 × CE

30 = 1 2 × 15 × CE

CE = 30 × 2 15 = 4 cm

6. Let the base and the height of the triangle be 3x and 4x, respectively.

Area of triangle = 1 2 × 3x × 4x = 96 sq. cm

6x2 = 96 sq. cm

x2 = 96 6 = 16 = (4)2

x = 4

So, base of the triangle = 3 × 4 = 12 cm and height of the triangle = 4 × 4 = 16 cm.

7. Dimensions of the rectangle = 30 m × 25 m

Area of the rectangle = l × b = 30 × 25 = 750 sq. m

Since, the diagonals of a rectangle bisect each other at O, O is the mid-point. So, height of the triangle = OF = 1 2 × 25 = 12.5 cm.

Area of triangle AOB = 1 2 × base (length of rectangle) × height of triangle = 1 2 × 30 × 12.5 = 187.5 sq. cm

Hence, area of shaded region = 750 – 187.5 = 562.5 sq. cm

8. a. Area of the parallelogram = b × h = 50 × 35 = 1750 sq. cm

Area of triangle = 1 2 × b × h = 1 2 × 1750 = 875 sq. cm

Area of the shaded region = 1750 – 875 = 875 sq. cm

b. Area of rectangle = 27 × 30 = 810 sq. cm

Area of one triangle = 1 2 × 10 × 15 = 75 sq. cm

Area of the other triangle = 1 2 × 12 × 20 = 120 sq. cm

Area of the shaded region = 810 – 75 – 120 = 615 sq. cm

9. Area of rectangle = 28 × 22 = 616 sq. cm

Area of 4 white triangles = 4 × 1 2 × 11 × 14 = 4 × 77 = 308 sq. cm

Word Problem 1. Area of a rectangular lawn = 30 × 15 = 450 sq. m

Area of 4 flowerbeds = 4 × 1 2 × 2 × 2 = 8 sq. m

Area of the lawn excluding flowerbeds = 450 – 8 = 442 sq. m

15D

1. a. R = 5 cm, C = 2πr = 2 × 22 7 × 5 = 31.43 cm

b. R = 8 cm, C = 2πr = 2 × 22 7 × 8 = 50.29 cm

c. R = 10 cm, C = 2πr = 2 × 22 7 × 10 = 62.86 cm

d. R = 12 cm, C = 2πr = 2 × 22 7 × 12 = 75.43 cm

e. R = 18 cm, C = 2πr = 2 × 22 7 × 18 = 113.14 cm

2. a. D = 10 cm, R = 5 cm, C = 2πr = 2 × 22 7 × 5 = 31.43 cm

b. D = 15 cm, R = 7.5 cm, C = 2πr = 2 × 22 7 × 7.5 = 47.14 cm

c. D = 24 cm, R = 12 cm, C = 2πr = 2 × 22 7 × 12 = 75.43 cm

d. D = 30 cm, R = 15 cm, C = 2πr = 2 × 22 7 × 15 = 94.29 cm

e. D = 40 cm, R = 20 cm, C = 2πr = 2 × 22 7 × 20 = 125.71 cm

3. a. R = 16 cm, D = 32 cm

C = 2πr = 2 × 22 7 × 16 = 100.57 cm

Ratio = 100.57 32 = 3.14285…, which is close to 3.14.

b. R = 20 cm, D = 40 cm

C = 2πr = 2 × 22 7 × 20 = 125.71 cm

Ratio = 125.71 40 = 3.14285…, which is close to 3.14.

c. R = 24 cm, D = 48 cm

C = 2πr = 2 × 22 7 × 24 = 150.86 cm

Ratio = 105.86 48 = 3.14285…, which is close to 3.14.

d. R = 36 cm, D = 72 cm

C = 2πr = 2 × 22 7 × 36 = 226.29 cm

Ratio = 226.29 72 = 3.14285…, which is close to 3.14.

4. a. C = 75.42 cm

C = πD. So, 75.42 = 22 7 × D. So, D = 75.42 × 7 22 = 24 cm

b. C = 603.42 cm

C = πD. So, 603.42 = 22 7 × D = So, D = 603.42 × 7 22 = 192 cm.

c. C = 502.85 cm

C = πD. So, 502.85 = 22 7 × D = So, D = 502.85 × 7 22 = 160 cm.

d. C = 352 cm

C = πD. So, 352 = 22 7  × D = So, D = 352 × 7 22 = 112 cm

5. Let the radii of the two circles C1 and C2, be 5r and 6r, respectively.

circumference of circle 1

circumference of circle 2 = 2πr1 2πr2 = 2× 22 7 × 5r 2 × 22 7 × 6r = 5 6 = 5:6

6. Circumference of a circle – diameter = 90 cm

2πr – D = 90 cm

2πr – 2r = 90 cm

2r    22 7 – 1    = 90 cm

2r    22 – 7 7    = 90 cm

2r × 15 7 = 90 cm

30r

7 = 90 cm

r = 90 × 7 30 = 630 30 = 21 cm

7. D = 2r = 50 cm r = 25 cm

Perimeter of semicircle =    1 2 × 2πr     + D = (3.14 × 25) + 50 =

78.5 + 50 = 128.5 cm

8. Circumference of a circular playground = 296 m

2πr = 296 m

22 7 × D = 296 m

D = 296 × 7 22 = 94.18 m

9. Distance covered by a car tyre in 1 revolution = Circumference of a circle = 2πr

Distance covered in 1 revolution = 2πr = 2 × 22 7 × 25 cm = 157.14 cm

Distance covered in 120 revolutions = 157.14 × 120 = 18,857.14 cm or 188.57 m

10. Cost of fencing a circular garden = circumference × ₹40

₹36,500 = 2πr × ₹40

₹36,500 = 2 × 22 7 × r × ₹40

r = 36,500 × 7 2 × 22 × 40 = 145.17 m

11. Diameter of the wheel of a truck = 57.15 cm

Radius of the wheel of the truck = 28.575 cm

Circumference of one revolution by the truck = 2πr = 2 × 22 7 × 28.575 cm = 179.62 cm

Total distance to be covered by a truck = Total circumference = 1 km 30 m = 1030 m = 103000 cm

Number of revolutions = 103000 179.62 = 573.43

Hence, the wheel of the truck will make 574 revolutions.

12. Length of the minute hand = 20 cm = radius of the clock

Circumference of the circular clock = 2πr = 2 × 22 7 × 20 = 125.71 cm

So, minute hand covers 125.71 cm of the clock in 1 hour.

Word Problems 1. Diameter = 18 cm, radius = 9 cm

Circumference = 2πr = 2 × 22 7 × 9 = 56.58 cm

Cost of putting 100 cm (= 1 m) lace around a circular cushion = ₹22

Cost of putting 56.58 cm lace around a circular cushion = ₹ 22 100 × 56.58 = ₹12.45

2. Diameter = 96 m, radius = 48 m

Circumference of the field by walking around it twice = 2 × 2πr = 2 × 2 × 22 7 × 48

= 2 × 301.71 m = 603.43 m

Speed of the man = 8 m/min

Time taken by the man to walk around the field = circumference speed in 1 min = 603.43 8 = 75.43 minutes

3. Side of the square = 88 cm

Perimeter of the square = 4 × side = 4 × 88 = 352 cm

Circumference of the circle = perimeter of square = 352 cm

2πr = 352 cm

2 × 22 7 × r = 352 cm

r = 352 × 7 2 × 22 = 56 cm

4. Radius = 80 m

Circumference of the circular field = 2πr = 2 × 3.14 × 80 = 502.4 m

Speed of cycling = 12 km/h = 12000 3600 = 10 3 m/s

Time taken by Seema to cycle around the circular field = circumference speed in 1 s = 502.4 × 3 10 = 150.72 s

15E 1. a. Area of circle = πr2 = 22 7 × 3 × 3 = 28.29 sq. cm

b. Area of circle = πr2 = 22 7 × 6 × 6 = 113.14 sq. cm

c. Area of circle = πr2 = 22 7 × 12 × 12 = 452.57 sq. cm

d. Area of circle = πr2 = 22 7 × 15 × 15 = 707.14 sq. cm

e. Area of circle = πr2 = 22 7 × 25 × 25 = 1964.29 sq. cm

2. a. D = 10 cm, R = 5 cm

A = πr2 = 22 7 × 5 × 5 = 78.57 sq. cm

b. D = 18 cm, R = 9 cm

A = πr2 = 22 7 × 9 × 9 = 254.57 sq. cm

c. D = 26 cm, R = 13 cm

A = πr2 = 22 7 × 13 × 13 = 531.14 sq. cm

d. D = 35 cm, R = 17.5 cm

A = πr2 = 22 7 × 17.5 × 17.5 = 962.5 sq. cm

e. D = 48 cm, R = 24 cm

A = πr2 = 22 7 × 24 × 24 = 1810.29 sq. cm

3. a. Area of circle = 50.28 sq. cm

πr2 = 22 7 × r × r = 50.28 sq. cm

r × r = 50.28 × 7 22 = 16 (approx.)

r2 = (4)2

r = 4 cm

d = 2r = 2 × 4 = 8 cm

b. Area of circle = 154 sq. cm

πr2 = 22 7 × r × r = 154 sq. cm

r × r = 154 × 7 22 = 49

r2 = (7)2

r = 7 cm

d = 2r = 2 × 7 = 14 cm

c. Area of circle = 616 sq. cm

πr2 = 22 7 × r × r = 616 sq. cm

r × r = 616 × 7 22 = 196

r2 = (14)2

r = 14 cm

d = 2r = 2 × 14 = 28 cm

d. Area of circle = 1257.14 sq. cm

πr2 = 22 7 × r × r = 1257.14 sq. cm

r × r = 1257.14 × 7 22 = 400 (approx.)

r2 = (20)2

r = 20 cm

d = 2r = 2 × 20 = 40 cm

4. a. Diameter of semicircle = 14.5 cm, radius of semicircle = 14.5 2 = 7.25 cm

Area of semicircle = 1 2 πr2 = 1 2 × 22 7 × 7.25 × 7.25 = 82.60 sq. cm

b. Radius of a quarter = 18 mm

Area of a quarter = 1 4 πr2 = 1 4 × 22 7 × 18 × 18 = 254.57 sq. mm

c. Diameter of a bigger semicircle = 42 mm, radius of a bigger semicircle = 21 mm

Area of a bigger semicircle = 1 2 πr2 = 1 2 × 22 7 × 21 × 21 =

693 sq. mm

Radius of a bigger semicircle = Diameter of a small semicircle = 21 mm

Radius of a small semicircle = 21 2 = 10.5 mm

Area of two small semicircles = 2 × 1 2 πr2 = 2 × 1 2  × 22 7 × 10.5 × 10.5 = 346.5 sq. mm

Total area = 693 – 346.5 = 346.5 sq. mm

d. Diameter of a semicircle = 16 cm, radius of a semicircle = 8 cm = radius of a quarter circle

Area of semicircle = 1 2 πr2 = 1 2 × 22 7 × 8 × 8 = 100.57 sq. cm

Area of two quarters = 2 × 1 4 πr2 = 2 × 1 4 × 22 7 × 8 × 8 = 100.57 sq. cm

Total area = 100.57 + 100.57 = 201.14 sq. cm

5. Let the radii of the two circles C1 and C2, be 4r and 7r, respectively.

Area of circle 1

Area of circle 2 = πr12 πr22 = 22 7 × 16r2 22 7 × 49r2 = 16 49 = 16:49

6. Let the radius of circle P be r.

Area of circle P

Area of circle Q = πr12 πr22 = 22 7 × r2 22 7 × 36 × 36 =    r 36    2 = 25 36 =    5 6    2 r 36 = 5 6 r = 5 6 × 36 = 30 cm

Area of circle P = πr2 = 22 7 × 30 × 30 = 2828.57 sq. cm

7. Perimeter of rectangle = 2(l + b) = 2 (26 + 18) = 2 × 44 = 88 cm

Circumference of a circle = 2πr = 2 × 22 7 × r = 88 cm

r = 88 × 7 2 × 22 = 14 cm

Area of the circle = πr2 = 22 7 × 14 × 14 = 616 sq. cm

8. Area of the washer (including the hole) = πr2 = 22 7 × 8 × 8 = 201.14 sq. mm

Area of the hole = πr2 = 22 7 × 2.5 × 2.5 = 19.64 sq. mm

Area of the washer (excluding the hole) = 201.14 sq. mm –19.64 sq. mm = 181.5 sq. mm

9. Area of circle = 1386 sq. cm

πr2 = 22 7 × r × r = 1386

r2 = 1386 × 7 22 = 441

r2 = (21)2

r = 21 cm

circumference = 2πr = 2 × 22 7 × 21 = 132 cm

Perimeter of the square = circumference of the circle = 132 cm

4 × side = 132 cm

Side of the square = 132 ÷ 4 = 33 cm

Area of the square = 33 × 33 = 1089 sq. cm

10. Area of circular flower garden = 628 sq. m

Radius of the area occupied by the sprinkler = 15 m

Area of the region occupied by the sprinkler = πr2 = 22 7 × 15 ×

15 = 707.15 sq. m

Area of region covered by the sprinkler > area of the entire garden

So, yes, the sprinkler can water the entire garden.

Word Problems

1. Area of rectangular lawn = 30 cm × 20 cm = 600 sq. cm

Area of the lawn that the cow can graze = Area of a quarter

circle = 1 4 πr2 = 1 4 × 22 7 × 8 × 8

= 50.29 sq. cm

Area of the remaining lawn = 600 – 50.29 sq. cm = 549.71 sq. cm

2. Area of the sheet of paper used to make the birthday cap =

3 4 πr2 = 3 4 × 22 7 × 24 × 24 = 1357.71 sq. cm

15F 1. a. Total area of the figure = Area of the rectangle – area of the semicircle

= (9.5 × 8.2) − 1 2 × 22 7 × 4.75 × 4.75

= 77.9 – 35.46

= 42.44 sq. cm

b. Total area of the figure = Area of rectangle + area of semicircle

= (22 × 30) + 1 2 × 22 7 × 11 × 11

= 660 + 190.14

= 850.14 sq. cm

c. Total area of the figure = Area of rectangle + area of semicircle

= (18 × 8) + 1 2 × 22 7 × 4 × 4

= 144 + 25.14

= 169.14 sq. cm

d. Total area of the figure = Area of triangle + area of semicircle

= 1 2 × 30 × 25 + 1 2 × 22 7 × 15 × 15

= 375 + 353.57

= 728.57 sq. cm

2. a. Area of the shaded region = area of square PQRS – area of a quarter circle

= (15 × 15) − 1 4 × 22 7 × 15 × 15

= 225 – 176.79

= 48.21 sq. cm

b. Area of the shaded region = area of rectangle HIJK – area of circle

= (4 × 6) − 22 7 × 2 × 2

= 24 – 12.57

= 11.43 sq. cm

c. Area of the shaded region = area of the circle – area of the triangle

= 22 7 × 7.5 × 7.5 − 1 2 × 15 × 7.5

= 176.79 – 56.25

= 120.54 sq. cm

d. Area of the shaded region = area of the outer circular figure – area of the inner circular figure

= [(100 × 70) + 22 7 × 35 × 35] − [(100 × 60) + 22 7 × 30 × 30]

= [7000 + 3850] – [6000 + 2828.57]

= 10850 − 8828.57

= 2021.43 sq. cm

3. Radius = 7.5 cm, diameter = 15 cm

So, PQ = QR = RS = 5 cm

Area of semicircle with diameter PQ = 1 2 × 3.14 × 1 2 PQ × 1 2 PQ

= 1 2 × 3.14 × 2.5 × 2.5

= 9.8125 sq. cm

Area of semicircle with diameter QS = 1 2 × 3.14 × 1 2 QS × 1 2 QS

= 1 2 × 3.14 × 5 × 5

= 39.25 sq. cm

Area of semicircle = 1 2 × 3.14 × 7.5 × 7.5 = 88.3125 sq. cm

Area of the shaded region = Area of the semicircle with diameter PQ + [area of the semicircle – area of semicircle with diameter QS]

= 9.8125 sq. cm + [88.3125 – 39.25]

= 9.8125 + 49.0625

= 58.875 sq. cm

4. Area of rectangular piece of metal sheet = 150 × 90 = 13500 sq. cm

Area of three circular sheets = 3 × 3.14 × 25 × 25 = 5887.50 sq. cm

Area of the metal sheet left = 13500 – 5887.50 = 7612.50 sq. cm

5. Area of square = 32 × 32 = 1024 sq. cm

Area of circle = 22 7  × 16 × 16 = 804.57 sq. cm

Area of the remaining portion of the cardboard = 1024 –804.57 = 219.43 sq. cm

6. Area of circular sheet of metal = 3.14 × 8 × 8 = 200.96 sq. cm

Area of a quarter circle = 1 4 × 3.14 × 5 × 5 = 19.625 sq. cm

Area of the remaining sheet = 200.96 – 19.625 = 181.335 sq. cm

7. Area of rectangular sheet of paper = 50 × 38 = 1900 sq. cm

Area of 27 buttons = 27 × 22 7 × 2.5 × 2.5 = 530.36 sq. cm

Area of the remaining sheet = 1900 – 530.36 = 1369.64 sq. cm

8. Area of a rectangular sheet of paper = 8 × 12 = 96 sq. cm

Area of a semicircle = 1 2 × 3.14 × 6 × 6 = 56.52 sq. cm

Area of the leftover sheet = 96 – 56.52 = 39.48 sq. cm

9. Area of the big circle = 22 7  × 14 × 14 = 616 sq. cm

Area of circle 1 = 22 7  × 3.5 × 3.5 = 38.5 sq. cm

Area of circle 2 = 22 7 × 10.5 × 10.5 = 346.5 sq. cm

Area of the unshaded part of the circle = Area of circle 1 + area of circle 2 = 38.5 + 346.5 = 385 sq. cm

Area of the shaded part of the circle = 616 – 385 = 231 sq. cm

Ratio of unshaded part to shaded part of the circle = 231:385 = 3:5

Word Problem 1. Area of the circular card = 22 7 × 44 × 44 = 6084.57 sq. cm

Area of 2 circles = 2 × 22 7 × 6 × 6 = 226.29 sq. cm

Area of a rectangle = 5 × 2 = 10 sq. cm

Area of the remaining mask = 6084.57 – (226.29 + 10) = 5848.28 sq. cm

15G 1. New length of the path = 8 m + 2 m + 2 m = 12 m

New breadth of the path = 15 m + 2 m + 2 m = 19 m

Area of the path = (12 × 19) sq. m – (8 × 15) sq. m = 228 sq. m – 120 sq. m = 108 sq. m

2. Area of the garden = 3.14 × 80 × 80 = 20,096 sq. m

Area of the pond = 3.14 × 25 × 25 = 1962.50 sq. m

Area of the remaining garden = 20,096 – 1962.50 = 18,133.50 sq. m

3. Area of the field = 100 × 50 = 5000 sq. m.

a. Area of the crossroads = (100 × 2) + (50 × 2) – (2 × 2) sq. m = 200 + 100 – 4 sq. m = 296 sq. m

b. Area of the remaining field = (100 × 50) – 296 sq. m = 4704 sq. m

Cost of grassing per 10 sq. m = ₹110

100 cm

50 cm 2 m

Cost of grassing 4704 sq. m = ₹110 10 × 4704 = ₹51,744

4. Area of an external circle = πr2 = 22 7 × 35 × 35 = 3850 sq. cm

Area of an internal circle = πr2 = 22 7 × 27 × 27 = 2291.15 sq. cm

Therefore, the area of the circular ring = 3850 – 2291.15 sq. cm = 1558.86 sq. cm

5. Circumference of a circular track = 176 m

2 × 22 7 × r = 176

r = 176 × 7 22 × 1 2 = 28 m = inner radius

Width of the track = 8 m

Outer radius = 8 m + 28 m = 36 m

Area of the track = 22 7 × 36 × 36 − 22 7 × 28 × 28 = 22 7 × (1296 – 784) = 22 7 × 512 = 1609.15 sq. m

Word Problems 1. New length of the pool = 12 m – (2 × 1.5) m = 9 m

New breadth of the pool = 5 m – (2 × 1.5 m) m = 2 m

Area of coping = (12 × 5) – (9 × 2) = 60 – 18 = 42 sq. m

Cost of coping 1 sq. m = ₹50

Cost of coping 42 sq. m = ₹50 × 42 = ₹2100

2. Area of the garden without the pavements = (70 × 70) – [(70 × 10) + (70 × 10) – (10 × 10)]

= 4900 – [700 + 700 – 100]

= 4900 – 1300 = 3600 sq. m

3. Area of the sheet of paper = 25 × 10 = 250 sq. cm

New length of the ribbon = 25 – 2 – 2 = 21 cm

New breadth of the ribbon = 10 – 2 – 2 = 6 cm

Area of the sheet of paper not covered by the ribbon = 21 × 6 = 126 sq. cm

Area of the sheet covered by the ribbon = 250 – 126 = 124 sq. cm

4. Circumference of a circular park = 264 m

2πr = 264 m

2 × 22 7 × r = 264 m

r = 264 × 7 22 × 1 2 r = 42 m

Inner radius = 42 m

Area of the circular park = πr2 = 22 7 × 42 × 42 = 5544 sq. m

Outer radius = 42 m + 2.5 m = 44.5 m

Area of the road = 22 7 × 44.5 × 44.5 – 5544 = 6223.65 − 5544 = 679.65 sq. m

Cost of constructing the road per sq. m = ₹220

Cost of constructing the road 679.65 sq. m = ₹220 × 679.65 = ₹1,49,523

5. a. Width of the path = 10 m – 6 m = 4 m, that is 2 m on each side

b. Length of the lawn = 15 m – 2 m = 13 m

Area of the lawn = 6 m × 13 m = 78 sq. m

c. Area of the path = Area of lawn including the path – area of the lawn

= (15 × 10) – 78 sq. m = 150 – 78 = 72 sq. m

Chapter Checkup 1. a. Area of triangle = 1 2 × 6 × 6 = 18 sq. cm

b. Area of triangle = 1 2 × 8 × 8 = 32 sq. cm

c. Area of triangle = 1 2 × 4 × 9 = 18 sq. cm

2. Area of a square = 81 sq. cm (side)2 = (9)2 sq. cm  Side = 9 cm

Area of 4 triangles = 4 × 1 2 × b × h = 4 × 1 2 × 4.5 × 4.5 = 40.5 sq. cm

Area of the square formed by joining the midpoints of the sides of the given square = 81 – 40.5 = 40.5 sq. cm

3. Let the length of the rectangle be l and the breadth be b Length of the rectangle = 2l, breadth of the rectangle = 1 3 b

Area of the new rectangle = 2l × 1 3 b = 2 3 lb

Area of the original rectangle = l × b

Ratio of the new rectangle to the original rectangle = 2 3 lb : lb = 2 3 lb lb = 2 3 1 = 2 3 = 2:3

4. Length of the hall = 5 × breadth of the hall

Perimeter of the hall = 2 (l + b)

42 = 2 (5b + b)

42 = 2 × 6b

42 = 12b

b = 42 12 = 7 2 = 3.5 m

So, length of the hall = 5 × 3.5 m = 17.5 m, and breadth of the hall = 3.5 m.

5. Side of the square = 108 m

Area of the square = 108 × 108 = 11,664 sq. m

Area of the rectangle = area of square = 11,664 sq. m

Length × breadth = 11,664 sq. m

160 × breadth = 11,664 sq. m

b = 11,664 160 = 72.9 m

So, the width of the rectangular plot is 72.9 m.

6. Length of the floor = 6.68 m = 668 cm

Width of the floor = 8.2 m = 820 cm

Area of the floor = 668 × 820 = 5,47,760 sq. cm

Dimensions of a tile = 64 cm by 32 cm

Area of 1 tile = 64 cm × 32 cm = 2048 sq. cm

Number of tiles = 547760 2048 = 268 (approx.)

So, number of tiles that are required will be 268 full tiles.

Cost of 1 tile = ₹20.50

Cost of 268 tiles = ₹20.50 × 268 = ₹5494

7. Area of figure 1 = 1 2 × b × h = 1 2 × 25 × 6 = 75 sq. cm

Area of figure 2 = 1 2 × b × h = 1 2 × 24 × 5 = 60 sq. cm

Total area of the parallelogram park = 75 sq. cm + 60 sq. cm = 135 sq. cm

8. Area of a parallelogram = area of a square = side × side = 12 × 12 = 144 sq. m

Area of the parallelogram = b × h   144 = 20 m × h

h = 144 20 = 7.2 m

9. Area of parallelogram

= 1520 sq. m

b × h = 1520 sq. m

28 × h = 1520 sq. m

h = 1520

28 = 54.29 m = DP

Area of parallelogram

= 1520 sq. m

b × h = 1520 sq. m

28 × h = 1520 sq. m

h = 1520

28 = 47.5 m = BQ

10.  Area of a scalene right-angled triangle = 1 2 × b × h = 1 2 × 15 × 8 = 60 sq. cm

11. a. Base of the parallelogram = 12 cm – 5 cm = 7 cm, height of parallelogram = 8 cm

Area of parallelogram = b × h = 7 × 8 = 56 sq. cm

b. Area of each triangle = 1 2 × b × h = 1 2 × 5 × 8 = 20 sq. cm

c. Area of two triangles = 20 + 20 = 40 sq. cm

Ratio of area of shaded portion to the remaining area of the rectangle = 56:40 = 7:5

12. Circumference + radius of a circle = 153 cm

2πr + r = 153 cm

r    2 × 22 7 + 1    = 153

r

   44 7 + 1 

 = 153

r × 51 7 = 153

r = 153 × 7 51 = 21 cm

The diameter = 2 × 21 = 42 cm

Circumference = 2πr = 2 × 22 7 × 21 = 132 cm

Area of the circle = 22 7 × 21 × 21 = 1386 sq. cm

13. Area of the shaded region = area of the rectangle – area of 2 quarters = (15 × 30) – area of a semicircle

= 450 − 1 2 × 22 7 × 15 × 15

= 450 – 353.57 = 96.43 sq. cm

14. Circumference of the circular fence = 2πr = 2 × 22 7 × 10 = 62.86 m

15. Let the diameter of circle 1 be 5x and that of circle 2 be 7x

So, radius of circle 1 = 5 2 x, radius of circle 2 = 7 2 x

Ratio of area of two circles = πr12 πr22 = r12 r22 =    5 2 x    2    7 2 x    2 = 25 4 x2 49 4 x2 = 25:49

16. Perimeter of a square = 4 × side = 44 cm side of a square = 44 4 = 11 cm

Area of square = side × side = 11 × 11 = 121 sq. cm

Circumference of a circular ring = 44 cm

2πr = 2 × 22 7 × r = 44

r = 44 × 7 2 × 22 = 7 cm

Area of circular ring = 22 7 × 7 × 7 = 154 sq. cm

Since 154 > 121, area of circular ring is more than the area of square by 154 – 121 = 33 sq. cm.

17. Area of the rectangular garden = 85 m × 60 m = 5100 sq. m

Area of the circular pond = 22 7 × 24 × 24 = 1810.29 sq. m

Area of the garden left out = 5100 – 1810.29 sq. m = 3289.71 sq. m

18. Area of the circular sheet = 22 7 × 27 × 27 = 2291.14 sq. cm

Area of 4 circles with radius 1.6 cm cut = 4 × 22 7 × 1.6 × 1.6 =

32.18 sq. cm

Area of a rectangle cut = 5 × 3 = 15 sq. cm

Area of a square cut = 4 × 4 = 16 sq. cm

Area of the remaining sheet = 2291.14 – (32.18 + 15 + 16) = 2291.15 – 63.18 = 2227.96 sq. cm

19. Let R be the outer radius and r be the inner radius.

Area of the cross-section of the pipe = Area of the outer circle –

area of the inner circle

= πR2 − πr2

= 3.14 × 4.5 × 4.5 – 3.14 × 3.5 × 3.5

= 63.585 – 38.465

= 25.12 sq. cm

20. Area of the rectangular park = 24 × 15 = 360 sq. m

New length of the rectangular path inside the park = 24 – (2 + 2) = 20 m

New breadth of the rectangular path inside the park = 15 – (2 + 2) = 11 m

Area of the rectangular area inside the park = 20 × 11 = 220 sq. m

Area of the footpath = 360 – 220 = 140 sq. m

21. Let r1 be the radius of the field and r2 be the radius of the road.

Circumference of the field = 264 m, Circumference of the road = 308 m

2 × 22 7 × r1 = 264 m, 2 × 22 7 × r2 = 308 m

r1 = 264 × 7 22 × 2 , r2 = 308 × 7 22 × 2

r1 = 42 m, r2 = 49 m

Width of the road = radius of the road – radius of the field = 49 m – 42 m = 7 m

22. Let the adjacent sides of a parallelogram be 7x and 3x

Perimeter of a parallelogram = 2 (l + b) = 120 cm

= 2 (7x + 3x) = 120

= 2 × 10x = 120

x = 120 20  = 6

So, two adjacent sides of a parallelogram are 7 × 6 = 42 cm, 3 × 6 = 18 cm.

Area of the parallelogram = b × h = 42 × 14 = 588 sq. cm

23. Outer circumference = 352 m, inner circumference = 308 m

2 × 22 7 × r1 = 352 m, 2 × 22 7 × r2 = 308 m

r1 = 352 × 7 22 × 2 , r2 = 308 × 7 22 × 2

r1 = 56 m, r2 = 49 m

Width of the track = outer radius – inner radius = 56 – 49 = 7 m

Word Problems 1. Length of the rectangular lawn = 22 m

Breadth of the rectangular lawn = 15 m

Perimeter of the lawn = 2 (l + b) = 2 (22 + 15) = 2 × 37 = 74 m

Number of shrubs that can be planted in 1 m of hedge = 5

Number of shrubs that can be planted in 74 m of hedge = 74 × 5 =

370 shrubs

2. Area of the rectangular garden = 25 m × 15 m = 375 sq. m

New length of the garden after margin inside = 25 – (1.5 + 1.5) = 22 m

New breadth of the garden after margin inside = 15 – (1.5 + 1.5) = 12 m

Area of the margin for planting lilies = 375 – (22 × 12) = 111 sq. m

Cost of 1 lily stem = ₹35

Cost of 4 lily stems = ₹35 × 4 = ₹140

Cost of 1 sq. m = ₹140

Cost of 111 sq. m = ₹140 × 111 = ₹15,540

Chapter 16

Let’s Warm-up 1. 105, 100, 95, 90, 85

2. 9, 19, 29, 39, 49, 59

3. 15, 28, 41, 54, 67, 80

4. 90, 75, 60, 45, 30, 15

5. 2, 4, 8, 16, 32, 64, 128

Do It Yourself

16A

1. Perimeter of the square = 4 × side = 62 cm

Length of side of the square = 62 4 = 15.5 cm

Area of the square = side2 = 15.52 = 240.25 cm2

2. Height of the triangle = 20 cm

Area of the triangle = 1 2 × base × height = 175 cm2

Length of base = 175 × 2 height = 350 20 = 17.5 cm

3. If an even number is denoted by n; the next even number can be given as n + 2

4. Perimeter of the regular pentagon = 30 cm

Length of the wire = Perimeter of regular hexagon = 6 × length of the side of hexagon = 30 cm

Length of the side of hexagon = 30 6 = 5 cm.

5. Length of side of the square = 35 cm; Length of rectangle = 45 cm

Area of square = side2 = 352 = 1225 cm2

Given, that the area of a square = area of rectangle

Area of the rectangle = length × breadth = 1225 cm2

45 × breadth = 1225 cm2

Breadth of the rectangle = 1225 45 = 27.22 cm

6. If 89 is the successor of a natural number n, then n = 89 – 1 = 88

7. Sum of first 12 natural number = n(n + 1) 2 = 12(12 + 1) 2  =

12 × 13 2 = 78

Sum of first 18 natural number = n (n + 1) 2 = 18(18 + 1) 2 = 18 × 19 2 = 171

Sum of first 12 natural number and first 18 natural number = 78 + 171 = 249.

8. The rule for the given pattern: 1, 3, 5, 7, 9 is (2n − 1). The 100th term can be given as: 2 × 100 – 1 = 200 – 1 = 199

Word Problems 1. Length of base of the triangular shaped window = 8 feet

Length of height of the triangular shaped window = 5 feet

Area of the triangular shade = 1 2 × base × height = 1 2 × 8 × 5 = 20 feet2

2. Length of Nikhil’s plot = 25 m

Breadth of Nikhil’s plot = half of its length = 1 2 × 25 = 12.5 m

Area of Nikhil’s plot = length × breadth = 25 × 12.5 = 312.5 m2

Area of Kunal’s square plot = double of Nikhil’s rectangular plot = 2 × 312.5 m2 = 625 m2

16B

1. a. Addition of x and y = x + y

b. One-fourth of a number a added to thrice the number b =

1

4 × a + 3b = a 4 + 3b

c. 3 less than z added to the product of x and y = z – 3 + xy

d. The product of m and n subtracted from twice the product of p and q = 2pq – mn

2. Expression Coefficient of a Coefficient of b

a. 6a2b + 12bc 6ab 6a2, 12c

b. 3ac – 10ba 3c, –10b –10a

c. 15ba2 + 12ac + 7b2c 15ba, 12c 15a2, 7bc

d. 5ca2 + 3ab – 6ca 5ca, 3b, –6c 3a

e. 2.5ab – 3.2a2b 2.5b, –3.2ab 2.5a, –3.2a2

f. 1 3 b2 + 3ab 3b 1 3 b, 3a

3. Expression Terms

a. 5m2 + mn 5m2 , mn

b. 3pq + 5q2p 3pq, 5q2p

c. 7y3 + 8xy2 – 5xy 7y3, 8xy2, –5xy

d. 6ab + 8a2bc – 2bc2 – ca 6ab, 8a2bc, –2bc2 , –ca

e. 0.5xy + 1.2x2y – 3.3y 0.5xy, 1.2x2y, –3.3y

f. 2 5 yz + 5 6 x2y + 0.6xy 2 5 yz, 5 6 x2y, 0.6xy

4. a. Expression5xy + 8yz

Terms Factors

5xy    8yz

5 xy  8  y  z

c. Expression5ab – 8ba2 + 5c

Terms Factors

b. Expression3pq2 – 5pr + 2

Terms Factors

5ab  – 8ba2  5c

5 a b –8 b a a  5 c

e. Expression1.2x2y – 3xyz

Terms Factors

1.2x2y   –3xyz

1.2 x xy   –3 x y z

3pq2 – 5pr   2

3 p q q –5 p r  2

d. Expression m + mn2 – 6l + 8nl m mn2  – 6l  8nl

Terms Factors

m m n n   – 6 l   8 n l

f. Expression 1 2 ab2 – 3a2b + 7

Terms Factors

1 2 ab2  –3a2b  7

1 2 a b b  –3 a a b  7

5. a. Among the given terms, –2xy2, 7xy2, 5xy2 are like terms whereas –2x2y is unlike term.

b. Among the given terms, 8mn, 5mn, nm are like terms whereas 6mn2 is unlike term.

c. Among the given terms, 3p2q2, 8q2p2 , q2p2 are like terms whereas 5p2q is unlike term.

d. Among the given terms, 5a2b, 7ba2 , a2b are like terms whereas ab2 is unlike term.

e. Among the given terms, 0.2xy, 2yx, 0.8xy are like terms whereas 0.5zx is unlike term.

f. Among the given terms, 1 6 m2n, 0.6nm2, 0.4m2n are like terms whereas 2 5 n2m is unlike term.

6. Like terms

Unlike terms

a. (m2n, 5m2n, 8nm2), (3mn, 6nm)5m, 7n

b. (2a2c, 3ca2), (3ac2 , c2a) 5b2a, 8b2c

c. (pqr, 9qrp,6rpq), (8rp, 5pr) pq2

d. (5mln, 5lmn), (6lk2m, k2lm)2klm

7. a.  6mn + 2mn2 + 3 is a binomial. False

b. 3 × m is a monomial. True

c.  9pq + 2qr + 5s is a trinomial. True

d. 4a2 + 5ab + 3c + 8 is a trinomial. False

Word Problem 1. In the expression 3a2 – 2a + 5, 5 is the constant term.

16C 1. a. 5xy + x = 5xy + x

b. 3p2 + 5p2 = 8p2

c. 4ab + (–3ab) = ab

d. 2ac2 + 5ac2 = 7ac2

e. 3mn + (–5mn) = –2mn

f. 10x2y and 7x2y = 17x2y

2. a 2x + 3y

+ x + y

(2 + 1)x + (3 + 1)y

= 3x + 4y

b 2x + 3y + z

+ 2x + y – z

(2 + 2)x + (3 + 1)y + (1 – 1)z

= 4x + 4y

c a + 2b – 3

+ –3a + 5b + 5

(1 – 3)a + (2 + 5)b + (–3 + 5)

= –2a + 7b + 2

d –5a + 4b – 8c

4a + 7b – 9c

+ 2a + 4b + 2c

(–5 + 4 + 2)a + (4 + 7 + 4)b + (–8 – 9 + 2)c

= a + 15b – 15c

3. a. x – 2y + x + y + y – 2x + x + 3y

= x + x – 2x + x – 2y + y + y + 3y

(1 + 1 – 2 + 1)x + (–2 + 1 + 1 + 3)y = x + 3y

b. 3p2 + q2 – 2r2 + 4q2 + 5r2

= 3p2 + q2 + 4q2 – 2r2 + 5r2

= 3p2 + (1 + 4)q2 + (–2 + 5)r2 = 3p2 + 5q2 + 3r2

c. 4a2 + 3b2 + –2a2 + 13ab2 + 4a2 + 3ab2

= 4a2 – 2a2 + 4a2 + 3b2 + 13ab2 + 3ab2

= (4 – 2 + 4)a2 + 3b2 + (13 + 3)ab2 = 6a2 + 3b2 + 16ab2

d. 19a + 15b – 11c + 12a – 5b + 6c + 6b + 4c

= 19a + 12a + 15b – 5b + 6b – 11c + 6c + 4c

= (19 + 12)a + (15 – 5 + 6)b + (–11 + 6 + 4)c = 31a + 16b – c

4. a. 2x + 5xy − 5x + 6xy 7ab2 − 4b + 6ac

b. 7x2 − 3y2 + 4x + 3y2 − 4x2 7pq + 4pr

c. 5ab2 − 3b + ac + 2ab2 + 5ac − b 11xy − 3x

d. 12pq + 9pr − 5rp + 5qp − 10pq −2pq + 10pr

e. 7pq2 + 3pq − 5qp − 7q2p + 10pr 4a2b − 9ab + 3c

f. 9a2b − 15ab + 3c − 5ba2 + 6ba 3x2 + 4x

5 a. 12ab2 − 3a + 5bc + 5a + 3a2b + 3ab2

= 12ab2 + 3ab2 – 3a + 5a + 5bc + 3a2b = 15ab2 + 2a + 5bc + 3a2b

Adding 4ab2 – 7a2b + 12bc to the given sum will give:

15ab2 + 2a + 5bc +3a2b + 4ab2 – 7a2b + 12bc

= 15ab2 + 4ab2 + 2a + 5bc + 12bc + 3a2b – 7a2b

= 19ab2 + 2a + 17bc – 4a2b

b. ab2 − 2a2b + 6bc + (−11a2b + 9bc)

= ab2 – 2a2b – 11a2b + 6bc + 9bc = ab2 – 13a2b + 15bc

Adding 4ab2 – 7a2b + 12bc to the given sum will give:

ab2 – 13a2b + 15bc + 4ab2 – 7a2b + 12bc

= ab2 + 4ab2 – 13a2b – 7a2b + 15bc + 12bc

= 5ab2 – 20a2b + 27bc

6. Perimeter = Sum of all sides

Perimeter of the given triangle = a + 7 + 2a + 5 + 3a – 9 = 6a + 3 7 1 2 mn2 –2

10

= 1 10 mn2 –7 15 nm + 1 9 77 n2

Word Problem 1. Length of Sudha’s copper wire = 5x + 9 cm

Length of Rakesh’s copper wire = 3x + 4 cm

Total length of wire = 5x + 9 + 3x + 4 = (8x + 13) cm

16D 1. a.

ab – 9

3ab – 5

–   +

(1 – 3)ab + (–9 + 5)

= –2ab – 4

b 9y2 – 5xy

–3y2 + 12xy

+   –

(9 + 3)y2 + (–5 – 12)xy

= 12y2 – 17xy

c. 4y2 + 5y – 7

– 2y2 + 9y – 4

+  –  +

(4 + 2)y2 + (5 – 9)y + (–7 + 4) = 6y2 – 4y – 3

d 5b2 – 3ab + 8

– b2 + 7ab

+  –  +

(5 + 1)b2 + (–3 – 7)ab + 8

= 6b2 – 10ab + 8

2. a. –10x + 8x2 – (10x2 – 7x) = –10x + 8x2 – 10x2 + 7x

= −10x + 7x + 8x2 – 10x2 = –3x – 2x2

b. 5mn – 6mn2 + 2n2– (mn2 + 2mn – 8n2) = 5mn – 6mn2 + 2n2 –mn2– 2mn + 8n2

= 5mn – 2mn – 6mn2 – mn2 + 2n2 + 8n2= 3mn – 7mn2 + 10n2

c. p2q + 5q2 – 8 – (3p2q + 2q2 + 5) = p2q + 5q2 – 8 – 3p2q – 2q2– 5

= p2q – 3p2q + 5q2 – 2q2– 8 – 5 = –2p2q + 3q2– 13

d. 5ab + 6ac + 3bc + 8 – (5 – ab – ac) = 5ab + 6ac + 3bc + 8 – 5 + ab + ac

= 5ab + ab + 6ac + ac + 3bc + 8 – 5 = 6ab + 7ac + 3bc + 3

3. Sum of 5ab + 6ac and 8ac – 7ab + 8 = 5ab + 6ac + 8ac – 7ab + 8

= 5ab – 7ab + 6ac + 8ac + 8 = –2ab + 14ac + 8

Subtracting the result from 12ab + 5ac – 3 gives:

12ab + 5ac – 3 – (–2ab + 14ac + 8) = 12ab + 5ac – 3 + 2ab – 14ac – 8

= 14ab – 9ac – 11

4. 8a2 – 12a + 6b – (3a2 + 15a + 8) = 8a2– 12a + 6b – 3a2– 15a – 8

= 5a2 – 27a + 6b – 8

Hence, 5a2– 27a + 6b – 8 must be added to 3a2 + 15a + 8 to get 8a2– 12a + 6b

5. 5xz2 + 2xy – 7yz – (7xy – 4x2z + 12yz + 3) = 5xz2 + 2xy – 7yz –

7xy + 4x2z – 12yz – 3 = 5xz2– 5xy – 19yz + 4x2z – 3

Hence, 5xz2– 5xy – 19yz + 4x2z – 3 must be subtracted from 5xz2 + 2xy – 7yz to get 7xy – 4x2z + 12yz + 3.

6. Sum of 4x – y + 11 and – y – 12 = 4x – y + 11 – y – 12 = 4x –2y – 1

On subtracting 4x – 2y – 10 from 4x – 2y – 1, we get

4x – 2y – 1 – (4x – 2y – 10) = 4x – 2y – 1 – 4x + 2y + 10 = 9

7. Sum of 6x2y + 9xy – 3y2 and x2y – 3xy + 5y2 = 6x2y + 9xy – 3y2 + x2y – 3xy + 5y2

= 7x2y + 6xy + 2y2

Sum of 7xy2 + 12xy – 10x2y and 2xy2 + 7y2 + 2x2y = 7xy2 + 12xy –

10x2y + 2xy2 + 7y2 + 2x2y

= 9xy2 + 12xy – 8x2y + 7y2

9xy2 + 12xy – 8x2y + 7y2 – (7x2y + 6xy + 2y2) = 9xy2 + 12xy – 8x2y

+ 7y2 – 7x2y – 6xy – 2y2

= 9xy2 + 6xy – 15x2y + 5y2

8. P = 6a2 + 12ab + 3ab2; 3Q = 3(5ab – 8ab2) = 15ab – 24ab2;

R = 15ab – 4a2 + 9

P – R + 3Q = 6a2 + 12ab + 3ab2 – (15ab – 4a2 + 9) + 15ab – 24ab2

= 6a2 + 12ab + 3ab2 – 15ab + 4a2 – 9 + 15ab – 24ab2

= 6a2 + 4a2 + 12ab – 15ab + 15ab + 3ab2 – 24ab2 – 9

= 10a2 + 12ab – 21ab2 – 9

Word Problem 1. Length of ribbon with Mohan = (3m + 3) m

Length of ribbon he used = (3m – 5) m

Length of ribbon left = 3m + 3 – (3m – 5) = 3m + 3 – 3m + 5 = 8 metres

16E 1. At a = 5

a. (a + 6) = 5 + 6 = 11

b. (2a – 3) = (2 × 5 – 3) = 10 – 3 = 7

c. (a2 – 3a + 5) = (52 – 3 × 5 + 5) = 25 – 15 + 5 = 15

d. 3a 5 = 3 × 5 5 = 3

2. At m = –3

a. 2m + 7 = 2 × –3 + 7 = –6 + 7 = 1

b. 2m2 + 2m – 3 = 2(–3)2 + 2 × (–3) – 3 = 2 × 9 – 6 – 3 = 18 – 9 = 9

c. m3 + 3m2 – 4 = (–3)3 + 3(–3)2 – 4 = –27 + 27 – 4 = –4

d. 4m3 – 4m2 + 2m – 10 = 4(–3)3 – 4(–3)2 + 2(–3) – 10 = –108 –

36 – 6 – 10 = –160

3. At x = 1, y = –1

a. x2 + y2 = (1)2 + (–1)2 = 1 + 1 = 2

b. 5x2 + 2y2 – 1 = 5(1)2 + 2(–1)2 – 1 = 5 + 2 – 1 = 6

c. 3x2 – 4xy + 8 = 3(1)2 – 4(1)(–1) + 8 = 3 + 4 + 8 = 15

d. 2x3y + 3x2 – 5x = 2(1)3(–1) + 3(1)2 – 5 × 1 = –2 + 3 – 5 = –4

4. At a = 2 and b = 3

a. 5ab – 2a + 3b = 5 × 2 × 3 – 2 × 2 + 3 × 3 = 30 – 4 + 9 = 35

b. 2a2 + 3ab – 4a = 2(2)2 + 3 × 2 × 3 – 4 × 2 = 8 + 18 – 8 = 18

c. 3a3 – 2(a + 2b) = 3(2)3 – 2(2 + 2 × 3) = 24 – 16 = 8

d. 3(a3 – ab) + 3a2 – 7b + 8 = 3((2)3 – 2 × 3) + 3(2)2 – 7 × 3 + 8 = 6

+ 12 – 21 + 8 = 5

5. At x = –2 and y = 1

a. 4x – 3y + 5y + 2x = 6x + 2y = 6 × (–2) + 2 × 1 = –12 + 2 = –10

b. x2y + 2xy + 3xy – 4 = x2y + 5xy – 4 = (–2)2 × 1 + 5(–2)(1) – 4 =

4 – 10 – 4 = –10

c. y2 – 3y + 2x + 4y + 2y2 – 4x = 3y2 + y – 2x = 3(1)2 + 1 – 2(–2) =

3 + 1 + 4 = 8

6. A = 3x + 2y; B = x2 – 5x and C = y2 – 6y at x = 2 and y = 2

A – 2B + C = 3x + 2y – 2(x2 – 5x ) + y2 – 6y

= 3x + 2y – 2x2 + 10x + y2 – 6y = 13x – 4y – 2x2 + y2

13 × 2 – 4 × 2 – 2(2)2 + (2)2 = 26 – 8 – 8 + 4 = 14

7. At a = 20; a3 – 2a2 + 5a – 100 = (20)3 – 2(20)2 + 5 × 20 – 100 = 8000 – 800 + 100 – 100 = 7200

8. Given that, m2 + 2mn + o = –7 for m = –1 and n = 1. Substituting the value of m and n in the given equation, (–1)2 + 2 × –1 × 1 + o = –7 1 – 2 + o = –7 o = –7 + 1 = –6

Word Problem 1. Amount of water in the tank = (250a2 + 100a) litres

Amount of water used by Pooja = 50a litres

Amount of water left = 250a2 + 100a – 50a = (250a2 + 50a)

At a = 3; the amount of water left = 250 × (3)2 + 50 × 3 = 2250 + 150 = 2400 litres

16F 1. a. 5x – (y – 2x + 8y) = 5x – y + 2x – 8y = 7x – 9y

b. (10m – 3mn + 4m) + 5mn = 10m – 3mn + 4m + 5mn = 14m + 2mn

c. 3a – [5ab – (2a – 3ab)] = 3a – [5ab – 2a + 3ab] = 3a – 5ab + 2a – 3ab = 5a – 8ab

d. {4x2 – (3x2 – 2x + 5) + 6x} = {4x2 – 3x2 + 2x – 5 + 6x} = x2 + 8x – 5

2. At m = –2

a. (5m – 3m2 – 6m) – 2m2 = 5m – 3m2 – 6m – 2m2 = –5m2 – m –5(–2)2 – (–2) = –20 + 2 = –18

b. 2(m2 – m) + 3(m2 + m) = 2m2 – 2m + 3m2 + 3m = 5m2 + m

5(–2)2 + (–2) = 20 – 2 = 18

c. {2m – (5m2 + 5m) + 3m2} = {2m – 5m2 – 5m + 3m2} = –2m2 – 3m

–2(–2)2 – 3(–2) = –8 + 6 = –2

d. [6m2 – {2m + 5 – (2m2 – 8)} + 3] = [6m2 – {2m + 5 – 2m2 + 8} +

3] = [6m2 – 2m + 2m2 – 13 + 3] = 8m2 – 2m – 10

8(–2)2 – 2(–2) – 10 = 32 + 4 – 10 = 26

3.a. 5x + [7x2 + 2 – (2x2 + y2) + (4x – 3y)] = 5x + [7x2 + 2 – 2x2 –

y2 + 4x – 3y] = 5x + 5x2 – y2 + 4x – 3y + 2 = 5x2 + 9x – y2 – 3y + 2

b. [2m2 + {5mn – (8n + 2nm) + 6m} – 5m2] = [2m2 + {5mn – 8n

– 2nm + 6m} – 5m2] = [2m2 + 3mn – 8n + 6m – 5m2] = –3m2 + 3mn – 8n + 6m

c. 2pq + [4pq2 – {3pq + 4p – (2pq2 – 2p)} – 3pq] = 2pq + [4pq2

– {3pq + 4p – 2pq2 + 2p} – 3pq] = 2pq + [4pq2 – 3pq – 4p + 2pq2 – 2p – 3pq] = 6pq2 – 4pq – 6p

d. 2a + [7ab – (2a2 + 5ab – a) – {3 + (2a + a2)}] = 2a + [7ab –

2a2 – 5ab + a – {3 + 2a + a2}] = 2a + [7ab – 2a2 – 5ab + a – 3 –

2a – a2] = 2a + 2ab – 3a2 – a – 3 = a + 2ab – 3a2 – 3

4. At a = 1, b = –2

a. [2a2 – {(3ab + 6a) – (2b2 – 2ab)} + 3a2] = [2a2 – {3ab + 6a –

2b2 + 2ab} + 3a2] = [2a2 – {5ab + 6a – 2b2} + 3a2] = [2a2 – 5ab

– 6a + 2b2 + 3a2] = 5a2 + 2b2 – 5ab – 6a

5(1)2 + 2(–2)2 – 5(1)(–2) – 6 × 1 = 5 + 8 + 10 – 6 = 17

b. 5a – [3a2 – {2b2 – (4ab + 5a + 7a2) – 2b2}] = 5a – [3a2 – {2b2 –

4ab – 5a – 7a2 – 2b2}] = 5a – [3a2 – 2b2 + 4ab + 5a + 7a2 + 2b2] =

5a – 3a2 + 2b2 – 4ab – 5a – 7a2 – 2b2 = –10a2 – 4ab

–10(1)2 – 4(1)(–2) = –10 + 8 = –2

c. 7a – [8ab – {5b2 + (–3ab – 8a + 6ab) + 2b2} – 4ab] = 7a – [8ab

– {5b2 – 3ab – 8a + 6ab + 2b2} – 4ab] = 7a – [8ab – 7b2 – 3ab + 8a

– 4ab] = 7a – [ab – 7b2 + 8a] = 7a – ab + 7b2 – 8a = –a – ab + 7b2

–1 – (1)(–2) + 7(–2)2 = –1 + 2 + 28 = 29

d. b2 + [(2a2 + 5ab + b2) – (6ab + 2b2) – (5ab + 2a2)] = b2 + [2a2 + 5ab + b2 – 6ab – 2b2 – 5ab – 2a2] = b2 – 6ab – b2 = –6ab

–6(1)(–2) = 12

5. At x = 3 and y = 2

5(x3 + y3 – 2x2y) – [2(x2 – y2 + 2xy) – {y3 – 4(x + 2)}]

= 5x3 + 5y3 – 10x2y – [2x2 – 2y2 + 4xy – {y3 – 4x – 8}]

= 5x3 + 5y3 – 10x2y – [2x2 – 2y2 + 4xy – y3 + 4x + 8]

= 5x3 + 5y3 – 10x2y – 2x2 + 2y2 – 4xy + y3 – 4x – 8

= 5x3 + 6y3 – 10x2y – 2x2 + 2y2 – 4xy – 4x – 8

5(3)3 + 6(2)3 – 10(3)2(2) – 2(3)2 + 2(2)2 – 4 × 3 × 2 – 4 × 3 – 8 = 135 + 48 – 180 – 18 + 8 – 24 – 12 – 8 = –51

Word Problem 1. Expression for Rohan’s age = 2x + [3x – {4x + (4 – x)}] = 2x + [3x – {4x + 4 – x}] = 2x + [3x – 3x – 4] = 2x – 4

At x = 5, Rohan’s age = 2 × 5 – 4 = 10 – 4 = 6 years

16G 1. Number of apples with Rohan = x

Number of apples with Riya = 2x

Number of apples with Nikhil = 2x + 4

Total number of apples altogether = x + 2x + 2x + 4 = 5x + 4

2. Let Shobha’s age = x

Age of Shobha’s mother = 2x

Age of grandmother = 5years more than the sum of Shobha and her mother’s age = x + 2x + 5 = 3x + 5

3. Distance travelled by the car = (2x2 + 3x + 4) metres

Distance travelled by the train = (5x2 + 10x + 15) metres

Difference in distance travelled by the train to that of the car = (5x2 + 10x + 15) – (2x2 + 3x + 4) = 5x2 + 10x + 15 – 2x2 – 3x – 4 =

(3x2 + 7x + 11) metres

4. Money spent on shirt = ₹(3x2 + 2xy + 9y2)

Money spent on food = ₹(5x2 – xy + 3y2)

Total money spent = 3x2 + 2xy + 9y2 + 5x2 – xy + 3y2 = ₹(8x2 + xy + 12y2)

5. Length of sides of the triangular plot = (3x2 + 3y + 15) m, (2x2 – y + 30) m and (7x2 + 2x – 10) m

Perimeter = Sum of length of all sides = 3x2 + 3y + 15 + 2x2 –y + 30 + 7x2 + 2x – 10 = (12x2 + 2y + 2x + 35) m

At, x = 2 and y = 1

Perimeter = 12(2)2 + 2 × 1 + 2 × 2 + 35 = 48 + 2 + 4 + 35 = 89 metres

Word Problem 1. Time spent by Rahul on internet = x minutes

Time spent by Vivek on internet = 2x minutes

Time spent by Ankit on internet = (2x + 10) minutes

Total time spent = x + 2x + 2x + 10 = 5x + 10

At x = 22; time spend by Rahul = 22 minutes

Time spent by Vivek = 2 × 22 = 44 minutes

Time spent by Ankit = 2 × 22 + 10 = 54 minutes

Chapter Checkup 1. Length of base of the given triangle = 32 cm

Area of triangle = 1 2 ×b× ℎ = 1 2 × 32 × ℎ = 184 cm2

Height = 184 × 2 32 = 11.5 cm

Hence, the height of the triangle is 11.5 cm.

2. 14 = 13 × 1 + 1

27 = 13 × 2 + 1

40 = 13 × 3 + 1

53 = 13 × 4 + 1

Hence, the rule for the given pattern is 13n + 1

The 15th term of the pattern can be given as: 13 × 15 + 1 = 196

3. Perimeter = Sum of all sides

Given, that the length of the equal sides of the isosceles triangle is

a and the length of the unequal side is b

Perimeter of isosceles triangle = a + a + b = 2a + b

4.a. 5m2n, 6n2m – Unlike

b. 2ab, –3ba – Like

c. x2y, –5yx – Unlike

d. 2p2q2, 8q2p2 – Like

e. 3mn, –5m – Unlike

f. 6ab2, –2ab2 – Like

5. a. Expression x2y + yx x2y yzx

Terms Factors

x xy y  zx

b. Expression2cb2 + 6ab + 8c

Terms Factors

2cb2  6ab   8c

2 c b b 6 a b   8 c

c. Expression pq – 5pq2 + 9

Terms Factors

pq  –5pq2  9

p q –5 p q q  9

6. Expression Type of expression ab + 4a – 5ac Trinomial

3yx2 + 5xy – 3yx – 8Quadrinomial

m×n = mn Monomial

2x2 + y Binomial

7. a. 5a2b + (–3ab2) + (–7ab2) + 14a2b = 5a2b – 3ab2 – 7ab2 + 14a2b = 19a2b – 10ab2

b. 10x + 12y – 12xy – 12 + 18 – 5x – 10y + 5xy +9xy = 5x + 2y + 2xy + 6

8. Let A must be added to x2 + 2xy + y2 to get 5x2 + 7xy x2 + 2xy + y2 + A = 5x2 + 7xy

A = 5x2 + 7xy – (x2 + 2xy + y2) = 5x2 + 7xy – x2 – 2xy – y2 = 4x2 + 5xy – y2

Hence, 4x2 + 5xy – y2 must be added to x2 + 2xy + y2 to get 5x2 + 7xy

9. Let A be subtracted from 5x + 8y + 12 to get – 3x + 9y + 17

5x + 8y + 12 – A = –3x + 9y + 17

A = 5x + 8y + 12 – (–3x + 9y + 17) = 5x + 8y + 12 + 3x – 9y – 17 = 8x – y – 5

Hence, 8x – y – 5 should be subtracted from 5x + 8y + 12 to get –3x + 9y + 17

10. Length of the given rectangle = (3a + 2) metres

Breadth of the given rectangle = (6a – 7) metres

Perimeter = 2 × (L + B) = 2 × (3a + 2 + 6a – 7) = 2 × (9a – 5) = (18a –

10) metres

11. At y = –3

y3 + 4y2 – 7y – 2 = (–3)3 + 4(–3)2 – 7 × (–3) – 2 = –27 + 36 + 21 – 2 = 28

12. At x = 2, y = 3

4x3 – 5y2 + 7 = 4(2)3 – 5(3)2 + 7 = 32 – 45 + 7 = –6

13. –5(2x – 1) + 7x – 21 = –10x + 5 + 7x – 21 = –3x – 16

Substituting the value of x = –2; –3 × (–2) – 16 = 6 – 16 = –10

14. At if p = –15

p3 – 2p – 270 = (–15)3 – 2(–15) – 270 = –3375 + 30 – 270 = –3615

15. Given that 4x3 + 2x – 5a = 5 atx = –2

Substituting the value of x, we get 4(–2)3 + 2(–2) – 5a = 5 –32 – 4 – 5a = 5 5a = –41 a = –8.2

16. –3(x2 + 3xy) + 3 – xy = –3x2 – 9xy + 3 – xy = –3x2 – 10xy + 3

Atx = 5 and y = – 4;

–3x2 – 10xy + 3 = –3(5)2 – 10 × 5 × (–4) + 3 = –75 + 200 + 3 = 128

17. At m = –5, n = 3

m3 3 + 5m – 3 – 5n = (–5)3 3 + 5 × (–5) – 3 – 5 × 3 = – 125 3 – 25 – 3 –

15 = –84.67

18. 5 (x + 2) + 5x – 10 + 5xy = 5x + 10 + 5x – 10 + 5xy = 10x + 5xy

At x = –3, y = –2

10x + 5xy = 10 × (–3) + 5 × (–3) × (–2) = –30 + 30 = 0

19. At m = –1, n = –1

a. [5 – {(4mn + 3m) – (5mn2 – 2m)} + 3mn2] = [5 – {4mn + 3m – 5mn2 + 2m} + 3mn2] = [5 – 4mn – 3m + 5mn2 – 2m + 3mn2] = 5 – 4mn – 5m + 8mn2

5 – 4(–1)(–1) – 5(–1) + 8(–1)(–1)2 = 5 – 4 + 5 – 8 = –2

b. 5m – [8m2 – {4n2 – (4m + 5mn + 6m2) – 2n2}] = 5m – [8m2 – {4n2 –4m – 5mn – 6m2 – 2n2}] = 5m – [8m2 – 2n2 + 4m + 5mn + 6m2] = 5m – 8m2 + 2n2 – 4m – 5mn – 6m2 = m – 14m2 + 2n2 – 5mn (–1) – 14(–1)2 + 2(–1)2 – 5(–1)(–1) = –1 – 14 + 2 – 5 = –18 c. 8n – [5mn – {5m2 + (–3mn + 8n – 7mn) + 2m2} – 4n] = 8n – [5mn –{5m2 – 10mn + 8n + 2m2} – 4n] = 8n – [5mn – 5m2 + 10mn – 8n –

2m2 – 4n] = 8n – [15mn – 7m2 – 12n] = 8n – 15mn + 7m2 + 12n = 20n – 15mn + 7m2

20(–1) – 15(–1)(–1) + 7(–1)2 = –20 – 15 + 7 = –28

d. mn2 – [(5nm2 + 4mn – 3mn2) – (6mn – 2nm2) – (5mn – 2mn2)] = mn2 – [5nm2 + 4mn – 3mn2 – 6mn + 2nm2 – 5mn + 2mn2] = mn2 – [7nm2 –7mn – mn2] = mn2 – 7nm2 + 7mn + mn2 = 2mn2 – 7nm2 + 7mn

2(–1)(–1)2 – 7(–1)(–1)2 + 7(–1)(–1) = –2 + 7 + 7 = 12

Word Problems 1. Perimeter of triangular shaped wooden frame = 96 cm

Given, that the length of each side of the wooden frame is same, hence, the length of each side of the triangle = 96 3 = 32 cm

Length of each side of square frame required = 32 cm

Perimeter of the square frame = 4 × 32 = 128 cm

2. Amount with Sunita = ₹(15a2 + 3ab + 3b2)

Amount spent on purchasing books = ₹(10a2 – ab + b2)

Amount left = ₹(15a2 + 3ab + 3b2 – 10a2 + ab – b2) = ₹(5a2 + 4ab + 2b2)

3. Length of rope with Sunil = (5x – 2y + 2) metres

Length of rope attached = (3x – xy + 5) metres

Total length = (5x – 2y + 2 + 3x – xy + 5) metres = (8x – 2y – xy + 7)

metres

At x = 5 and y = 2

8x – 2y – xy + 7 = 8 × 5 – 2 × 2 – 5 × 2 + 7 = 40 – 4 – 10 + 7 = 33 metres

Chapter 17

Let’s Warm-up

1. 100 hundreds  10,00,00,000 (Ten Crore)

2. 1000 hundreds  10,00,000 (Ten Lakh)

3. 100 ten thousands  1,00,00,000 (One Crore)

4. 100 lakhs  10,000 (Ten Thousand)

5. 1000 lakhs  1,00,000 (One Lakh)

Do It Yourself

17A 1.a. Base = 9; Power = 11

b. Base = −17; Power = 9

c. Base = n; Power = 10

d. Base = 3 8 ; Power = 5

e. Base = 7 11 ; Power = 8

f. Base = 11 18 ; Power = 15

2. a. 5 × 5 × 5 × 5 = 54

b. (–9) × (–9) × (–9) × (–9) × (–9) × (–9) × (–9) = (–9)7

c.

d.

3. a.

1 raise to the power any natural number gives 1 as the result.)

b.

c.

d. 43 × 34 = 4 × 4 × 4 × 3 × 3 × 3 × 3 = –5184

e.  3 2 9     = 2 9 × 2 9 × 2 9 = 8 729

f.  2 11 12     = 11 12     × 11 12     = 121 144

3 3 7 × 4 2 3     = 3 7     × 3 7     × 3 7     × 2 3     × 2 3     × 2 3     × 2 3     =

1029

2480 2240

2960

2480

2240

2120

260

230

315

1

5. a. 82 = 64; 83 = 512   b. 112 = 121; 113 = 1331

c. 182 = 324; 183 = 5832   d. 212 = 441; 213 = 9261

e. 302 = 900; 303 = 27,000

6. a. 34 = 3 × 3 × 3 × 3 = 81 and 43 = 4 × 4 × 4 = 64; 81 > 64. Hence, 34 is greater.

b. (−5)4 or 54 = (−5) × (−5) × (−5) × (−5) = 625 and 5 × 5 × 5 × 5 = 625; 625 = 625. Both are equal.

   

7. a. The reciprocal of 26 is 6 1 2    

b. The reciprocal of (−3)4 is = 4 1 3    

c. The reciprocal of 1 4     is = 4

d. The reciprocal of 5 1 2     is = (−2)5

8. a. 113 × 122 = 11 × 11 × 11 × 12 × 12 = 1,91,664

b. 103 × 151 = 10 × 10 × 10 × 15 = 15,000

c. 73 × (−3)4 = 7 × 7 × 7 × (−3) × (−3) × (−3) × (−3) = 27,783

d. (−12)2 × 102 – 112 = (−12) × (−12) × 10 × 10 − 11 × 11 = 14,279

e.  3 4 5     × 5 × 2 3 8     = 4 5 × 4 5 × 4 5 × 5 × 3 8 × 3 8 = 9 25

f.  3 3 4 ×(1) 5     + 23 32 × 73     = 444 ×××(1)×(1)×(1) 555     +

33222 ×××× 77333     = 64 125 + 8 147

LCM of 125 and 147 is 18375.

64 125 × 147 147 = 9408 18375

8 147 × 125 125 = 1000 18375

64 125 + 8 147 = 94081000 + 1837518375 = 8408 18375

9.  2 5 x     = 32 3125

2 5 x   

 = 2×2×2×2×2 5×5×5×5×5 = 5 2 5    

Hence, the value of x = 5

Word Problem 1. Marbles sold by Vivek = 78,125

Expressing 78125 as power of 5:

78,125 = 5 × 5 × 5 × 5 × 5 × 5 × 5 = 57

578125

515625

53125

5625

5125

525

55 1

17B 1. a. 105 × 109 = 10(5+9) = 1014 [Since am × an = a(m+n)]

b. (–5)2 × (–5)4 × (–5)6 = (−5)2+4+6 = (−5)12 [Since am × an = a(m+n)]

c.  2 7 9     × 5 7 9     = 2+5 7 9     = 7 7 9     [Since am × an = a(m+n)]

d.  3 2 5     × 6 2 5     × 9 2 5     = 3+6+9 2 5     = 18 2 5    

[Since am × an = a(m+n)]

e. 24 × 54 = (2 × 5)4 = (10)4 [Since am × bm = (ab)m]

f. (–9)3 × (–1)3 = ((−9) × (−1))3 = 93 [Since am × bm = (ab)m]

g.  10 4 3     × 10 3 4     = 10 43 × 34     = (−1)10 = 1

[Since am × bm = (ab)m]

h. (−5)3 × 3 9 5     × 3 2 3     = 3 92 5×× 53         = (3 × 2)3 = (6)3

[Since am × bm = (ab)m]

2. a. 1315 ÷ 139 = 1315−9 = 136 [Since am ÷ an = am–n]

b. (–11)4 ÷ (–11)1 = (−11)4−1 = (−11)3 [Since am ÷ an = am–n]

c.  9 4 7     ÷ 3 4 7     = 93 4 7     = 6 4 7     [Since am ÷ an = am–n]

d. 57 ÷ 37 = 7 5 3     ÷=mm m Sinceab a b       

e.

12 2 5     ÷ 12 3 7     =

6. a. 24 × 23 −27

b. (−3)6 ÷ (−3)3 128

c. (−5)4 × (−4)4 −17

d. 50 × (−17)1 1,60,000

7. 6p × 63 = 362

6p × 63 = 362 = (62)2 = 64 (since, (am)n = am×n)

6p × 63 = 64

Since, am × an = a(m+n)

6p × 63 = 6p+3

6p+3 = 64

On comparing, p + 3 = 4

p = 4 − 3 = 1

The value of p is 1

8. a.  128

6 2× (2) m m = 128 66 2× 2 m m

= 212–6 × m8–6 [Since am ÷ an = am–n]

= 26 × m2

b.  23 45 6×24 4×3 = 23 45 (2×3)×(2×2×2×3) (2×2)×3

= 22333 4 25 2×3×(2) ×3 2×3  

= 2293 85 2×3×2×3 2×3 [Since am × bm = (ab)m]

12 2 5 3 7

      = 12 27 × 53     = 12 14 15    

÷=mm m Sinceab a b       

f. (–15)8 ÷ (5)8 = 8 15 5     = (−3)8 ÷=mm m Sinceab a b       

3. a. ((9)2)2 = 92×2 = 94 [since, (am)n = am×n]

b. ((−12)1)3 = (−121×3) = (−12)3 [since, (am)n = am×n]

c.  2 4 3 2          = 4×2 3 2     = 8 3 2     [since, (am)n = am×n]

d.  3 3 1 8          = 3×3 1 8     = 9 1 8     [since, (am)n = am×n]

4. a. 40 = 1; Since a0 = 1

b. ((7)2)0 = 72×0 = 70 = 1; Since a0 = 1

c. (−5)0 × 0 3 8     = 1 × 1 = 1 Since a0 = 1

d.  0 x y     ÷ 0 y x     = 1 ÷ 1 = 1 Since a0 = 1

5. (20 + 30) × (60 + 71)

Since a0 = 1 (20 + 30) × (60 + 71) = (1 + 1) × (1 + 7) = 2 × 8 = 16.

= 2+92+3 85 2×3 2×3 = 115 85 2×3 2×3 [Since am × an = a(m+n)]

= 211–8 × 35–5 [Since am ÷ an = am–n]

= 23

c.  583 454 25×49×121 5×7×11 =

  

583 222 454 5×7×11 5×7×11

= 10166 454 5×7×11 5×7×11 [Since am × bm = (ab)m]

= 510−4 × 716−5 × 116−4 [Since am ÷ an = am–n]

= 56 × 711 × 112

d.  4 22 (5) 25×25 ab abab = 444 2222 5 5×5 ab abab

= 444 433 5 5 ab ab [Since am × bm = (ab)m]

= 54−4 a4−3 b4−3 [Since am ÷ an = am–n]

= ab

9. a. 9 × 3n+2 = 243

= 32 × 3n+2 = 35

= 32+n+2 = 35 [Since am × an = a(m+n)]

= 3n+4 = 35

On comparing, n + 4 = 5

n = 1

b.  6 2 7 13          = 5 7 13 n   

= 2×6 7 13     = +5 7 13 n    [Since am × bm = (ab)m]

On comparing, 12 = n + 5

n = 12 − 5

n = 7

Word Problem 1. Rice bought by Ronita at ₹42 per kg = 28 kg

Amount spent = ₹42 × 28 = ₹1176

Rice bought by Suman at ₹63 per kg = 14 kg

Amount spent = ₹63 × 14 = ₹882

The product of the amounts spent by both girls = 1176 × 882 =

10,37,232 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 7 × 7

The product of prime factorisation in exponential form = 24 × 33 × 74

1. a. 864000 = 8.64 × 100000 = 8.64 × 105

b. 5500000 = 5.5 × 1000000 = 5.5 × 106

c. 4152000000 = 4.152 × 1000000000 = 4.152 × 109

d. 614000000000 = 6.14 × 100000000000 = 6.14 × 1011

2. a. 3 × 104 = 3 × 10000 = 30000

b. 4.55 × 105 = 4.55 × 100000 = 455000

c. 7.11 × 107 = 7.11 × 10000000 = 71100000

d. 9.8 × 109 = 9.8 × 1000000000 = 9800000000

3. 40,075 is expressed in standard form = 4.0075 × 10000 =

4.0075 × 104

Hence, the power of 10 is 4.

4. 8,32,79,000 = 8.3279 × 10000000 = 8.3279 × 107

8.3279 × 10n = 8.3279 × 107

Comparing both sides, n = 7

5. 7 × 106 + 5 × 105 + 4 × 103 + 6 × 102 + 8 × 101 + 2 × 100 = 7000000 + 500000 + 4000 + 600 + 80 + 2

= 7504682 = 7.504682 × 106

6. The value of 6,43,298 in expanded exponential form

6 × 105 + 4 × 104 + 3 × 103 + 2 × 102 + 9 × 101 + 8 × 100

7. a. 13568000 = 1.3568 × 10000000 = 1.3568 × 107

b. 1410000000 = 1.41 × 1000000000 = 1.41 × 109

c. 2760000000 = 2.76 × 1000000000 = 2.76 × 109

d. 140000000 = 1.4 × 100000000 = 1.4 × 108

Word Problem 1.a. Approximate age of earth = 4.54 × 109 years.

4.54 × 109 = 4.54 × 1000000000 = 4540000000 years

Chapter Checkup 1. 7,776 in exponential form with base 6 = 6 × 6 × 6 × 6 × 6 = 65

2. a. 123 = 12 × 12 × 12 = 1728

b. 103 × 103 = 10 × 10 × 10 × 10 × 10 × 10 = 10,00,000

c. (7 × 2)4 = 144 = 14 × 14 × 14 × 14 = 38,416

d. (–1)9 = (–1) × (–1) × (–1) × (–1) × (–1) × (–1) × (–1) × (–1) × (–1) = −1

e. (–22)0 = 1  f.  0 13 20     = 1

3. a. (p + q)4 = (3 + 4)4 = 74

74 = 7 × 7 × 7 × 7 = 2401

b. (p − q)2 = (3 − 4)2 = (−1)2 = 1

c. (2p − q)3 = (2 × 3 − 4)3 = (6 − 4)3 = 23 = 8

d. (p − 2q)2 = (3 − 2 × 4)2 = (3 − 8)2 = (−5)2 = (−5) × (−5) = 25

4. a. 8 × 8 × 8 × 8 × 8 × 8 = 86

b. (–15) × (–15) × (–15) × (–15) × (–15) × (–15) × (–15) × (–15) = (−15)8

c. (–m) × (–m) × (–m) × (–m) × n × n × n × n × n × n × n = (−m)4 × n7

d. 2 × 2 × 2 × 2 × a × a × a × a × a × b × b × b × b × b × b × b × b = 24 ×

a5 × b8

5. a. 256 =

2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 28

2 256 2128

b. 582 = 2 × 3 × 97

2582

3291

9797 1

c. 1090 = 2 × 5 × 109

21090

5545 109109 1

e. 3418 = 2 × 1709 23418

17091709 1

6. a. 7.2 × 102 < 3.8 × 103

720 < 3800

b. 5.8 × 105 < 3.6 × 108

580000 < 360000000

c. 6.9 × 108 < 4.5 × 109

690000000 < 4500000000

7. a.  15×(2)34 20 = 34 2 (3×5)×(2) 2×5

= 334 2 3×5 × (2) 2×5 = 3314 2 3×5×(2) 2

= 32 3×5×16

4 = 27 × 25 × 4 = 2700

d. 2050 = 2 × 5 × 5 × 41 = 2 × 52 × 41

22050

51025

5205

4141 1

b.  23221 20 ×(1) 100× n n = n23 – 20 × (−1) × (10)−2 = 3 2 ×(1) 10 n = 3 102 n

c. (170 – 230) × (170 + 230) = (1 − 1) × (1 + 1) = 0 × 2 = 0

d.  4 24 23     × 4 23 12     = 4 2423 × 2312     = (2)4 = 2 × 2 × 2 × 2 = 16 8.  a b     = 2 2 5     × 1 2 5     a b     = 2+1 2 5     = 3 2 5     [Since am × an = a(m+n)]

= 2×2×2 5×5×5     = 8 125    

Therefore 3 a b     = 8 125     × 8 125     × 8 125     = 512 1953125    

The value of 3 a b     = 512 1953125    

9. a. 4750000 = 4.75 × 1000000 = 4.75 × 106

b. 70000000 = 7 × 10000000 = 7 × 107

c. 12 million = 1.2 × 10000000 = 1.2 × 107

d. 289 crores = 2.89 × 1000000000 = 2.89 × 109

10. a. 9.1 × 106

b. 70600

c. 4.36 × 1010

d. 13000000

11. a. (–1)3 × (–1)9 = (–1)n

= (−1)3+9 = (−1)n [Since am × an = a(m+n)]

= (–1)12 = (–1)n

n = 12

b. 3n+1 × 81 = (3)8

= 3n+1 × 34 = (3)8

= 3n+1+4 = (3)8 [Since am × an = a(m+n)]

3n+5 = (3)8

On comparing both sides, n + 5 = 8 n = 8 − 5

n = 3

c. 52n+1 ÷ 25 = (5)3

52n+1 ÷ 52 = (5)3

52n+1−2 = (5)3 [Since am ÷ an = am–n]

52n–1 = (5)3

On comparing both sides,

2n − 1 = 3

2n = 4

n = 2

3+21 3 11      [Since am × an = a(m+n)] = 24 3 11    

On comparing, 24 = 3n n = 8 e.  3 4 5 3          = 2 5 3 n    

4×3 5 3     = 2 5 3 n      [since, (am)n = am×n] 12 5 3 

= 2 5 3 n 

On comparing, 12 = 2n n = 6

1.3 × 107

43600000000

9100000

7.06 × 104

On

n

Word Problems 1. The male population of a country in 2011 = 623700000

623700000 in standard form = 6.237 × 100000000 = 6.237 × 108

2. Speed of the aircraft = 690km/h

Total time taken to cover the distance = 3 h 20 mins = 3 1 3 h = 10 3 h

Total Distance covered = Speed × Time = 690 × 10 3 = 2300 km

In centimetres:

1 km = 100000 centimetres

2300 km = 230000000 centimetres = 2.3 × 108 centimetres.

Chapter 18

Let’s Warm-up

1.  Human face  Two lines of symmetry

2.  Letter H  One line of symmetry

3.  Plus sign  Infinite lines of symmetry

4.  Sun  Four lines of symmetry

Do It Yourself

18A 1. a. b. c. d.

e. f. g. h.

2.a. Yes b. No c. Yes d. Yes e. No,

f. No, a parallelogram does not have any line of symmetry.

g. Yes

h. No, the shape is a scalene triangle so does not have any line of symmetry.

3. a. e.

b. f.

4.a. b. c.

c. g.

d. h.

5. Answers may vary. Sample answers.

a.

b.

We would shade the squares along the diagonal line at an equal distance on both sides.

Answers may vary. Sample answers.

6. The letter “S” of the word ‘MATHEMATICS’ has no line of symmetry.

7.

8. a. b.   c.   d.

Word Problem

1.a. b. Yes, It has 6 lines of symmetry

18B 1. a. More than 1  b. equal to 1  c. more than 1 d. equal to 1  e. more than 1  f. more than 1 g. equal to 1  h. equal to 1

2. a. 360 72 = 5  b. 360 120 = 3  c.  360 90 = 4  d. 360 180 = 2

3. i. a.

4. Figures may vary. Sample figures are provided.

a. Letters H, I, O, X Rotation through 90° Rotation through 180° Rotation through 270° Rotation through 360°

b. Letters N, S, Z o

o 1800

c. Letters A, B, C, D, E, K, M, T, U, V, W, Y

d. Letters F, G, J, L, P, Q, R

5. a. Rectangle: Reflection symmetry with 2 lines of symmetry

Rotational symmetry of order 2 x x x x

x

b. Rotational symmetry of order 4 No reflective symmetry.

c. 1 line of reflection symmetry but no rotational symmetry

d. Parallelogram with rotational symmetry of order 2 and no reflective symmetry.

6. a. Anticlockwise

b. The figure has been rotated by 90°

c. No, the figure did not look the same after rotation.

d. Order of rotation = 1

7. a. Figure 1: Order of rotational symmetry = 4, angle of rotation = 90°

Figure 2: Order of rotational symmetry = 2, angle of rotation = 180°

b. No, the figures do not show reflection symmetry.

Figure 1: No, b.

ii. a. 4 times  b. 4 times  c. 4 times  d. 6 times

iii. a.  360 4 = 90°  b. 360 4 = 90°  c. 360 4 = 90°  d. 360 6 = 60°

c.     d.

Figure 2: Yes,

8. Colour of the parts may vary. Sample figures.

a. Order 2:

b. Order 3:

c. Order 4: Word Problem 1. a

b. For both figures, order = 4, angle of rotation = 360 4 = 90°

Chapter Checkup 1. a. Yes  b. Yes

c. No, d. Yes  e. No, f. No

Lines of symmetry may vary. Sample answer:

2.a. 1 line of symmetry

b.  No line of symmetry

c. 4 lines of symmetry

d. 2 lines of symmetry

e. 2 lines of symmetry.

f. 1 line of symmetry

g. 4 lines of symmetry

h. 4 lines of symmetry

3.a. b. c.

4.  S. No. FigureRotation by 90° Rotation by 180° Rotation by 270° Rotation by 360° Number of times a shape looks the same

a. Order = 1, angle of rotation = 360°

b. Order = 2, angle of rotation = 180°

c. Order = 4, angle of rotation = 90°

d. Order = 4, angle of rotation = 90°

e. Order = 2, angle of rotation = 180°

5. a. B, C, D, E, K

b. A, T, U, V, W c. X, O

6. a. Scalene triangle 1

b. Parallelogram

c. None

d. Nonagon, 140° 40°

e. Trapezium 1

7. Reflective symmetry -

Rotational symmetry -

Numbers 0 and 8 have rotational symmetry of order 2.

So, numbers that have both reflective and rotational symmetry are 0 and 8.

8. Error: The arrow head in the mirror image should be pointing the mirror.

ObjectMirror image

Correct image: ObjectMirror image

9. The new position of C will be found as: The new position of C will replace H in the original figure after rotating the figure clockwise by 270° and 45° anticlockwise.

10. Answer may vary. Sample answers.

a. b.

11. Answer may vary. Sample answers.

a. b.

12. 7:15

13. Each of the given faces have 2 lines of symmetry.

14. Figure may vary. Sample answers.

15. Minimum of 6 squares should be added to get a rotational symmetry of order 4.

c. The number of edges in a football is 0

d. A square prism is also called a cuboid cube

e. The polygon region of a solid shape is called a face

2. A solid is a 3-dimensional shape.

The three dimensions are length, width and height.

3. A prism has two identical and parallel polygonal bases and rectangular lateral faces.

A pyramid has a polygonal base and triangular lateral faces.

a. Pentagonal pyramid, since it has 1 pentagonal base

b. Square prism or cuboid, since it has 2 square bases

c. Rectangular pyramid, since it has one rectangular base

d. Hexagonal prism, since it has 2 hexagonal bases

e. Octagonal pyramid, since it has one octagonal base

4. a. Edges = 0; Vertices = 0

b. Edges = 1; Vertices = 1

c. a tetrahedron is a triangular pyramid. Edges = 6; Vertices = 4

d. Edges = 9; Vertices = 6

5. Cylinder

6. A ₹2 coin is circular/cylindrical in shape. The new shape will also be a cylinder.

7. The cube and the cuboid are two three-dimensional shapes that have the same number of faces, vertices, and edges. Both the shapes have 6 faces, 8 vertices and 12 edges.

8. Difference: An octagonal prism has two identical and parallel octagonal bases and 8 rectangular lateral faces. An octagonal pyramid has one octagonal base and 8 triangular lateral faces.

Similarity: Both the shapes are 3-D shapes with octagonal bases.

Word Problem 1. The 3-D shape formed with 4 congruent triangles meeting at a point will have a square base. So, the 3-D shape is a square pyramid.

19B 1. a. An octagonal prism has 10 faces. It has 2 octagonal identical faces and 8 rectangular faces.

b. A nonagonal pyramid has 10 faces. It has 1 nonagonal base and 9 triangular lateral faces.

c. A rectangular prism has 6 faces. It can have all rectangular faces or both square and rectangular faces.

d. A cone has 2 faces. It has 1 circular base and 1 curved lateral face.

2.

a. b.

Word Problem 1.a. Shape Q

b. P - 1, Q - 1, R - 1, S - 1, T - 1, U - 0

c. Rotational symmetry of order 1

Chapter 19

Let’s Warm-up 1. Pentagonal prism  2. Cone

3. Triangular prism  4. Cylinder

Do It Yourself

19A 1. a. The solid that has only one vertex is cone

b. If three cubes are joined side by side, the shape obtained is a cuboid

c. d.

3. a. The net has 2 triangular congruent bases and 3 rectangular lateral faces. It is the net of a triangular prism.

b. The net has 2 congruent rectangular bases and 4 rectangular lateral faces. It is the net of a cuboid/rectangular prism.

c. The net has a circular base and a triangular curved face. It is the net of a cone.

d. The net has 1 square base and 4 triangular lateral surfaces. It is the net of a square pyramid.

e. The net has 2 congruent circular bases and a curved face. It is the net of a cylinder.

f. The net has 1 pentagonal base and 5 triangular lateral surfaces. It is a pentagonal pyramid.

4.

A cone has 2 faces: 1 circular base and 1 curved lateral face.

5.a.

A triangular pyramid has 4 faces: 1 triangular base and 3 triangular lateral faces.

b.

6. A triangular prism has 2 triangular bases and 3 square/ rectangular lateral surfaces.

The different shapes needed to draw the net of a triangular prism are triangle and square/rectangle.

7. No. A square prism is not necessarily a cube. A square prism has a square bases but the dimensions of the lateral surfaces need not be the same as that of the bases.

8. Diameter = 14 cm. Radius = 14 2 cm = 7 cm

Circumference of the circle = Length of the rectangle = 2πr = 2 × 22 7 × 7 = 44 cm

Length of the cylinder = 15 cm

Since the length of the rectangle in the net is the same as the circumference of the circles drawn, the correct net is:

44 cm 15 cm

9. 4 cm

cm

10. A triangular prism has 2 triangular bases so the triangles are the bases.

Word Problem 1. The 2 bangles will form the 2 circular bases. The rectangular sheet of paper will form the lateral surface. So, if Seema rolls the paper and attaches the bangles to each end edgeto-edge, she will get a cylinder.

19C 1. 2. 3.

4.a.

b.

c.

5. a. The cuboid is of the dimensions: 5 units × 4 units × 1 unit

b. The cuboid is of the dimensions: 3 units × 5 units × 2 units

c. The cuboid is of the dimensions: 6 units × 3 units × 3 units

6. a. The cube is of the dimensions: 4 units × 4 units × 4 units

b. The cuboid is of the dimensions: 3 units × 5 units × 2 units

c. The cuboid is of the dimensions: 5 units × 2 units × 2 units

7. a.

b.

Word Problem 1. Reema placed three cubes of dimensions 3 cm × 3 cm × 3 cm end to end.

3 cm3 cm3 cm 3 cm 3 cm

The shape formed is a cuboid. Dimensions of the shape formed is 9 cm × 3 cm × 3 cm.

d.    Front view        Top view      Side view

19D 1. a. A birthday hat is in the shape of a cone. Vertical: triangle, Horizontal: circle

b. Dice is in the shape of a cube. Horizontal: square, Vertical: square

c. Brick is in the shape of a cuboid. Horizontal: rectangle, Vertical: rectangle

d. Ball is spherical in shape. Horizontal: circle, Vertical: circle

2. a. b. c. d. e.

5. a.   Front view       Top view    Side view

b. Front view     Top view       Side view

3. Figures may vary. Sample figures. a. b. c. d. e.

4. a. Front view      Top view       Side view

b.  Front view      Top view      Side view

c.   Front view    Top view   Side view

6. A cube can cast a shadow in the shape of a rectangle. This occurs when the light source creating the shadow, falls at an angle on the cube.

7. Answers may vary. Sample answer.

a. Sphere  b. Cone  c. Cube/cuboid  d. Cuboid

Word Problem 1. If a cylindrical pipe is cut horizontally, we get a rectangle.

Then, Length of the rectangle = length of the cylindrical pipe = 3 m = 3 × 100 cm = 300 cm

Breadth of the rectangle = Circumference of the cylindrical pipe = 2 × 22 7 × 7 cm = 44 cm

Area of the rectangle = length × breadth = 300 cm × 44 cm = 13,200 cm2

So, the cross-sectional area of the cylinder will be 13,200 cm2

Chapter Checkup 1. a. Cone  b. Triangular pyramid

c. Triangular prism  d. Cylinder  e. Cube

2. a. Cuboid  b. Square pyramid  c. Cube  d. Cylinder

3. a. Octahedron  b. Hexagonal prism

c. Square/rectangular pyramid  d. Cylinder

4. a. Triangular pyramid  b. Cube  c. Square pyramid

d. Cone  e. Cylinder

5.a. b. c. d.

c.   Front view       Top view      Side view

6. Breadth = Height

Length = 2 × Breadth

2 opposite faces in the cuboid will be square.

Dimensions: 2 units × 2 units × 4 units

Figures may vary. Sample figure.

7.i. ii.

8. a. Rectangle  b. Rectangle  c. Rectangle

9.a. Horizontal:   Vertical:

b. Horizontal:    Vertical:

c. Horizontal:      Vertical:

10. a.  Front view:      Top view:      Side view:

b. Front view:      Top view:      Side view:

c. Front view:      Top view:      Side view:

d. Front view:      Top view:      Side view:

e. Front view:      Top view:      Side view:

11.  A heptagonal prism has 2 heptagonal faces and 7 rectangular faces. So, the shapes needed are 2 heptagons and 7 rectangles. A heptagonal pyramid has 1 heptagonal face and 7 triangular faces. So, the shapes needed are 1 heptagon and 7 triangles.

12.a. b.

13. Answer may vary. Sample answer: E

DCBA F

When the net is folded, keeping C on top, A will be at the bottom.

Word Problems 1. A cube has 6 faces. The given net only has 3 faces. If Maya adds 3 more faces, she can complete this net in 11 different ways to make a cube.

2. Dimensions of the cuboid shape cake = 20 cm × 10 cm × 4 cm 4 cm 20 cm

10 cm

Vansh will get two possible cross-sections when the cake is cut in half: vertical cross-section and horizontal cross-section. Vertical cross-sectional area = 4 cm × 10 cm = 40 cm2

Horizonal cross-sectional area = 20 cm × 10 cm = 200 cm2

3. Figures may vary. Sample figure. 2 cm 2 cm

Total number of cubes Ram placed is 15.

Front view       Top View    Side View

About the Book

Imagine Mathematics seamlessly bridges the gap between abstract mathematics and real-world relevance, offering engaging narratives, examples and illustrations that inspire young minds to explore the beauty and power of mathematical thinking. Aligned with the NEP 2020, this book is tailored to make mathematics anxiety-free, encouraging learners to envision mathematical concepts rather than memorize them. The ultimate objective is to cultivate in learners a lifelong appreciation for this vital discipline.

Key Features

• Let’s Recall: Helps to revisit students’ prior knowledge to facilitate learning the new chapter

• Real Life Connect: Introduces a new concept by relating it to day-to-day life

• Examples: Provides the complete solution in a step-by-step manner

• Do It Together: Guides learners to solve a problem by giving clues and hints

• Think and Tell: Probing questions to stimulate Higher Order Thinking Skills (HOTS)

• Error Alert: A simple tip off to help avoid misconceptions and common mistakes

• Remember: Key points for easy recollection

• Did You Know? Interesting facts related to the application of concept

• Math Lab: Fun cross-curricular activities

• QR Codes: Digital integration through the app to promote self-learning and practice

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