Imagine_Maths_CB_Grade8 by Uolo

8 MATHEMATICS

NEP 2020 based

NCF compliant |

CBSE aligned

MATHEMATICS Master Mathematical Thinking

Grade 8

Maths Grade 8 Chapter 1-6.indb 1

08-01-2024 14:44:58

Fo re wo rd

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. ii

Maths Grade 8 Chapter 1-6.indb 2

08-01-2024 14:44:58

Rational Numbers

Let's Recall In our previous grades, we learned about natural numbers and whole numbers. We know that integers are positive and negative whole numbers. We also know how to read, write and represent fractions and decimals. We know that these are all rational numbers. Let us recap how to add and subtract integers, fractions and decimals.

K ey El emen t s o f a C h apt e r— a Q u i c k G lanc e Adding Integers

Subtracting Integers

Like Sign

Unlike Sign

Add and keep the sign.

Subtract and keep the larger number sign.

4+2=6

–3 – (–2) = –3 + 2 = –1

(–4) + 3 = –1

(–4) + (–2) = –6

Adding and Subtracting Fractions Like Fraction

• Convert to like decimals • Align the decimal points • Add or subtract 1 2 .6 3 1 .3 0 1 3 .9 3

1 2 .6 3 − 1 .3 0 1 1 .3 3

page with a

Letʼs Warm-up

quick warm-up

Unlike fraction

• Add or subtract the numerators.

• Convert to like fraction by taking the LCM.

• Keep the common denominator.

• Add or subtract as done in like fractions.

2 4 2+4 6 + = = 5 5 5 5

2 2 10 − 6 4 – = = 3 5 15 15

Concept

Understanding Rational Numbers

introduction

Fill in the blanks. 5+2 4 3

__________________

2 52.31 + 61.2

__________________

3 − 23 + 69

__________________

5−7 4 4 8

__________________

5 89 − 47.25

__________________

Shubhi and Rishabh collected data on the temperature of different cities across the world in December.

Real Life Connect

with a real-life

Rishabh: Shubhi, can you convert the temperature of Yakutsk and Dubai in fractions? Shubhi: Let me try Rishabh! We can write

example

−25.5°C as −255 °C and 22.5°C as 225°C. 10 10 Rishabh: But −255°C is not a fraction! 10 Both the kids got confused about what they should call this number. Let’s help them out!

Temperature in ℃

Adding and Subtracting Decimals

Introductory

Change the sign of subtrahend and solve.

22.5

20 10 0

–10

–20

–30

–25.5 Yakutsk

Dubai

–18.3

–14.2

Fraser

Golmud

City

Rational Numbers On converting −25.5°C and 22.5°C to fractions, Shubhi got −255°C 10 and 225°C. These numbers are called rational numbers. 10 p A rational number is a number that is expressed as , where p and q q are integers, and the denominator should never be equal to zero.

I scored _________ out of 5.

3 4

A quick-thinking question

56 12

15 For example, 12.5, −15, 25, 6 and −8 are all rational numbers. kg of apples, 2.6 kg of bananas and some oranges from the market. 7 9 4 Suppose, she purchased 9 kg of fruit. What is the weight of theand oranges she purchased? Identifying Representing Rational Numbers We know how to represent integers, 2 fractions and decimals on a litresrational in the firston a A school prepared 25.5 litres of juice for its players.number The players 8represent line. Let us drank see how to numbers 5 3 number line. half. In the second half, the players drank 2 litres more than in the first half. What is the 8 We have marked the number line in equal Rational Number −1 = − 0.5 volume of the juice left at the end? 2 intervals of 0.5 keeping 0 in the center.

–12 10

Whole

7.5

All positive integers and zero 0, 1, 2, 3, 4,....

Natural

All positive integers, excluding zero 1, 2, 3, 4,....

Fun fact, related to the concept

Did You Know? −1 2

Knowledge of algebraic expression 0 0.5 1 1.5 2 can help us plan and schedule our

2.5

−2 −1.5 −1 −0.5

Multiplication and Division of Rational Numbers days, as we are often estimating

Decimal Number

Integer

Multiplication of Rational Numbers

and solving for unknown variables.

Represent 5 and − 4 on the number line. 3 3 5 2 We can write as 1 . So, it will lie between 1 and 2 on the number line. Divide the space between 1 and 2 3 3 into 3 equal parts and mark 5. 3

Example 1

The multiplication of rational numbers is similar to that of fractions. Product of numerators Product of two rational numbers = Product of denominators For example,

Integers

All positive and negative whole numbers, and zero ....–3, –2, –1, 0, 1, 2, 3,....

Suhani purchased

Think and Tell

Is pi a rational number?

12 5 60 7 -3 -21 -7 × = and × = = 13 7 91 9 8 72 24

Remember!

The product of two rational

numbers with similar signs is What if the rational numbers are in different forms? positive and with opposite signs -4 Let us multiply and 0.56 is negative. 13 56 14 = We will first convert them into a similar form as 0.56 = . 100 25 (–4) 14 (–56) = × 7 Puneet’s age is three times the age of his daughter. If the sum of their ages is 52 years, find the difference 325 13 25 Pointing out between their ages.

An important

8 The sum of two numbers is 45 and the numbers are in the ratio 2:1. What are the numbers? 13 and –8.5. Error Alert! 25 commonly made 9 The total cost of a table and a chair is ₹15,550. The cost of the table is ₹550 more than the cost of the chair. What is the cost of the table? -85 p . Write the numbers in the form ; hence, –8.5 = p 10 Virat scores 20 more runs than twice the runs by Rohit. Together, mistakes and Convert thescored numbers intotheir runs are four runs short of a triple 10 q 7 Puneet’s age is three times the age of his daughter. If the sum of their ages is 52 years, find the difference century. What are the individual scores of Virat and Rohit, respectively? q 13 -85 -1105 The product of two rational form and then multiply. between their ages. × = 11 Amit has a certain number of pencils that cost ₹4 per pencil. He again purchased 10 more than twice the number 25 10 250 how to avoid of pencils he had. If the total cost of all the pencils is ₹290, how many pencils did Amit have initially? numbers with similar signs is 8 The sum of two numbers is 45 and the numbers are in the ratio 2:1. What are the numbers? -1105 -221 to his daughter, his property to his son and donated the rest of = HCF of 1105 and 250 = 5; hence, 12 Animesh left one-half of his property 2 8.1 one-third 2 of27 54 18 positive and withThe opposite signs and a chair is ₹15,550. cost of the table is ₹550 more than the cost of the chair. What them 250 9 The total cost of a table 50 2.7 = then how much property his property. If the donation was×₹5,00,000 did × = he give = to his daughter and son? 3 3 3 10 30 10 is the cost of the table? 145 231 One of the two digits of a two-digit number is 40% of the other digit. If you interchange the digits of this two-digit is negative. mand . add the resulting number to the original number, you get 77. What is the original number? m and13 number Example 14 The length and width of a rectangular park are 3 9 10 Virat scores 20 more runs than twice the runs scored by Rohit. Together, their runs are four runs short of a triple What is the perimeter of the park? 14 The speed of the flow of a river is 3 km per hour. A boat goes upstream between two points in 5 hours while it century. What are the individual scores of Virat and Rohit, respectively?

point to keep in

Example 13

Find the product of

Remember!

mind

10 more thanof twice the number= 2 × (L + B) = 2 11 Amit has a certain number of pencils that cost ₹4 per pencil. He again purchased Perimeter rectangle of pencils he had. If the total cost of all the pencils is ₹290, how many pencils did Amit have initially?

435 + 231

145 231 + 3 9

covers the same distance going downstream in 3 hours. What is the speed of the boat in still water?

15 A sum of ₹770 is made up of denominations of ₹5 and ₹10. If the total number of notes is 72 then find the number of notes of each denomination.

666 1332 = = 148 m 9 9

to× his son and donated=the rest 12 Animesh left one-half of his property to his daughter, one-third of his property2 × of

9 and son?1 his property. If the donation was ₹5,00,000 then how much property did he give to his daughter

16 Two cars simultaneously start from A and B in opposite directions and the distance between them after 4 hours is 60 km. What is the speed of each car?

number and add the resulting number to the original number, you get 77. What is the original number?

1 Multiply 5 and 16 14 9 -7 covers the same distance going downstream in 3 hours. What is the speed of the boat in still water? Do It The speed of the flow of a river is 3 km per hour. A boat goes upstream between two points in 5 hours while it Together

HOTS:

15 A sum of ₹770 is made up of denominations of ₹5 and ₹10. If the total number of notes is 72 then find the

Applicative

2 Multiply 2.5 and -19 3

Points to Remember

Chapter end

An equation is a mathematical statement with an equal symbol between two expressions.

•

Any operation done on one side of an equation must be done on the other side of the equation so that the equation remains true.

•

We use the transposition method to collect similar terms on one side. Change the signs of terms while using the transposition method and moving terms from one side to the other side.

's' km per h

•

While solving the word problems, identify the unknown terms, form an equation and solve the value of the unknown term.

•

The following rules can be applied while solving linear equations:

summary

60 km. What is the speed of each car?

('s' + 10) km per h A

60 km 300 km

Points to Remember

A multidisciplinary and fun

•

An equation is a mathematical statement with an equal symbol between two expressions.

•

Any operation done on one side of an equation must be done on the other side of the equation so that the equation remains true.

•

We use the transposition method to collect similar terms on one side. Change the signs of terms while using the transposition method and moving terms from one side to the other side.

•

While solving the word problems, identify the unknown terms, form an equation and solve the value of the unknown term. Math Lab The following rules can be applied while solving linear equations: Aim: Solve linear equations in one variable

•

Materials Required: Chalk, stopwatch and equation cards

The same number can be added or subtracted from both sides of the equation.

The same number can be multiplied or divided on both sides of the equation.

Mark can the starting line and finishing to line the in theother classroom or in an There A2 term be transposed side ofopen thearea. equation with its sign changed.

Setting: In groups of 4 Method: 1

Prepare equation cards (unsolved equation on cards (should have linear equations)). could be different laps.

activity

Each team lines up behind the starting line and the first person of each team is given an equation card.

The student solves the equation and runs towards the finishing line with the correct solution.

Once they reach the finishing line, they hand the card to the next teammate. The next

Use a stopwatch to calculate the total time taken by each team to complete the race.

The team which has the fastest completion time wins the relay race.

•

number of notes of each denomination.

questions

60 km 300 km

16 Two cars simultaneously start from A and B in opposite directions and the distance between them after 4 hours is

and analytical

's' km per h

('s' + 10) km per h

Hence, the perimeter of the rectangular park is 148 m. 13 One of the two digits of a two-digit number is 40% of the other digit. If you interchange the digits of this two-digit

classroom

Rational

Any number that can be written as: a fraction –5.25

The same number can be added or subtracted from both sides of the equation.

The same number can be multiplied or divided on both sides of the equation.

A term can be transposed to the other side of the equation with its sign changed.

QR Code: Access to interactive digital resources

person solves the next equation and continues the relay.

Chapter Checkup 1

Find the value of unknown value of x. a

iii

Solve using the balancing method. a 4x + 14 = 56

b 3x + 4 = x + 18

c 5x – 7 = 2x +11

d 3x – 4 =1 – 2x

e 5x + 48 = 3(4x – 5)

f 11(x + 1) = 12(x – 1)

Solve using the transposition method.

Which of these is the correct solution of the equation?

a 5(x + 3) = 3(1.5x + 18)

b 6(2x + 11) = 8(2x – 1)

d 10x – 5 – 7x = 5x + 15

e 5(x – 1) = 2(x + 8)

a 5x + 7 = 35 for x = 7

b 3(x – 2) = 4(x + 3) for x = 18

d 4(x + 1) – 3(2x – 18) = 2x – 2 for x = 15

Chapter 2 • Solving Equations in One Variable

Maths Grade 8 Chapter 1-6.indb 3

c 8x – 7 – 3x = 6x – 2x – 3 f

4x – 3 = (3x + 1) + (5x – 4)

c 4x + 3 – 2(x + 1 ) = 8x for x = e 5(x + 9) – 2x = 90 for x = 15

1 6

08-01-2024 14:45:01

G rad ual R e le ase of Re spon si bi li t y

The Gradual Release of Responsibility (GRR) is a highly effective pedagogical approach that empowers students to learn progressively by transitioning the responsibility from the teacher to the students. This method involves comprehensive scaffolding—including modelling, guided practice, and ultimately fostering independent application of concepts. GRR, endorsed and promoted by both the NEP 2020 and NCF, plays a pivotal role in equipping teachers to facilitate age-appropriate learning outcomes and enabling learners to thrive. The GRR methodology forms the foundation of the Imagine Mathematics product. Within each chapter, every unit follows a consistent framework: 1. I Do (entirely teacher-led)

2. We Do (guided practice for learners supported by the teacher) 3. You Do (independent practice for learners) GRR Steps

Unit Component

Snapshot Understanding Rational Numbers Shubhi and Rishabh collected data on the temperature of different cities across the world in December. Rishabh: Shubhi, can you convert the temperature of Yakutsk and Dubai in fractions? Shubhi: Let me try Rishabh! We can write 225°C. −25.5°C as −255 °C and 22.5°C as 10 10 Rishabh: But −255°C is not a fraction! 10 Both the kids got confused about what they should call this number. Let’s help them out!

Real Life Connect Theoretical Explanation

Temperature in ℃

Real Life Connect

22.5

20 10 0

–10

–20

–30

–25.5 Yakutsk

Dubai

–18.3

–14.2

Fraser

Golmud

City

Rational Numbers On converting −25.5°C and 22.5°C to fractions, Shubhi got −255°C 10 225°C. These numbers are called rational numbers. 10 A rational number is a number that is expressed as p, where p and q q According to the question, are integers, and the denominator should never be equal to zero. 1000x + 1200(54 − x) = 58800 ⇒ 1000x − 1200x = 58800 6 and −8 are +all64800 rational numbers. For example, 12.5, −15, 25, 7 = 6000 9 200x = 64800 − 58800 ⇒ 200x

Rational

Any number that can be written as: a fraction

and

I do

–5.25 56 12

Whole

–12 10 7.5

All positive integers and zero 0, 1, 2, 3, 4,....

Natural

x = 30 ⇒ 54 − x =and 54 − 30 = 24 Identifying Representing Rational Numbers Example 5

Integers

All positive and negative whole numbers, and zero ....–3, –2, –1, 0, 1, 2, 3,....

All positive integers, excluding zero 1, 2, 3, 4,....

Thus, Jayahow sellsto30 tickets for ₹1000 and 24 tickets for ₹1200.on a We know represent integers, fractions and decimals number Let us see toto represent numbers on a Twenty-five years later, Suman’s age will Suman’sline. present age is how equal one-fifthrational of her mother's age. number line. be 4 years less than half the age of her mother. Find their present ages.

y Let us the mother’s present to be y years; Suman’s present age = Weassume have marked the number lineage in equal Rational Number −1 = −50.5 2 intervals of 0.5 keeping 0 in the center. According to the condition, y + 25 = 5

y + 25 −4 2

−1 2

Multiply left hand side by 2 and right hand side by 5. −2 −1.5 −1 −0.5 0 2y + 250 = 5(y + 25) − 40 ⇒ 2y + 250 = 5y + 125 − 40 5y − 2y = 250 − 85Integer ⇒ 3y = 165

Examples

Example 1

Example 6

0.5

1.5

2.5

⇒ y = 55

Decimal Number

y 55 Suman’s present age = = = 11 years 5 5 Represent 5 and − 4 on the number line. 3 present 3 age is 11 years and her mother's age is 55 years old. Thus, Suman’s We can write 5 as 1 2. So, it will lie between 1 and 2 on the number line. Divide the space between 1 and 2 3 3 5. into 3 equal parts and A positive number is 5mark times 3 more another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers? Let us assume the first number to be x; second number = 5x

According to question, 5x + 21 = 2(x + 21) 5x + 21 = 2x + 42 3x = 21

⇒x=7

⇒ 5x − 2x = 42 − 21 ⇒ 5x = 35

Thus, the numbers are 7 and 35. Example 7

If the sum of three consecutive multiples of 9 is 108, then find the second multiple. Let us assume the smallest multiple of 9 to be 9x. Next multiple = 9(x + 1); Last multiple = 9(x + 2)

So, 9x + 9(x + 1) + 9(x + 2) = 108 27x + 27 = 108

⇒ 27x = 81

⇒ x = 81 ÷ 27

⇒x=3

So, we can find, 9x = 9 × 3 = 27, 9(x + 1) = 9 × 4 = 36, 9(x + 2) = 9 × 5 = 45 Thus, the three consecutive multiples of 9 which add up to 108 are 27, 36, 45. Example 8

Maths Grade 8 Chapter 1-6.indb 4

Tanya attempts a test in which there are 60 questions. For each correct answer, x marks will be awarded, and for each incorrect answer, one-fourth of correct answer marks will be deducted, and 0 marks will be awarded for un-attempted questions. If Riya attempts 45 questions out of which 29 are correct, what is the value of x if Riya is awarded 200 marks?

08-01-2024 14:45:01

71°

88°

38°

60°

111°

59°

50°

37°

Sum of exterior angles of a hexagon

60° + 71° + 53° + 88° + 38° + 50° = 360°

111° + 47° + 106° + 59° + 37° = 360°

Similarly, if a polygon has n number of sides, then, the measure of each exterior angle of a regular 360º . n Find the measure of each exterior angle in the figure. polygon = Example 5

We know that the exterior angle sum = 360°

(3x − 5)°

→ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360° → 30x = 360° → x = 12°

GRR Steps

∠A = (2x - 1)° = 23°; ∠B = (3x - 5)° = 31°;

Unit Component

∠C = (8x + 3)° = 99°; ∠D = (7x - 2)° = 82°; ∠E = (4x + 1)° = 49°; ∠F = (6x + 4)° = 76° Example 6

→ (2x - 1)° + (3x - 5)° + (8x + 3)° + (7x - 2)° + (4x + 1)° + (6x + 4)° = 360°

(2x − 1)° F (6x + 4)°

(8x + 3)°

C E (7x − 2)° D (4x + 1)°

Snapshot

The exterior angle sum of a 30-sided regular polygon is 360°, find the measure of each angle. Number of sides = n = 30

360º 360º = = 12º. n 30 Find the measure of the missing angles a, b, c and d. Solve these in your notebook and write the answers below. ∴ the measure of each exterior angle of a regular polygon =

Do It Together

We do

Do It Together

c 34°

Error Alert!

34°

122°

The sum of an interior angle and the adjacent exterior angle can never be greater than 180°.

71°

81° Do It Yourself 3C

∠a + 122° = 180°. So, ∠a = 180° - 122° = 58° 1 Fill in the blanks. ∠a = 58°; ∠b = _____, ∠c = _____, ∠d = _____

115°

90°

a The polygon in which the sum of all exterior angles is equal to the sum of interior angles is called ______.

Chapter • Polygons sum of interior angles of a polygon of n sides is ______ right angles. b 3The c The measure of each exterior angle of a regular polygon of 18 sides is ______. d In case of a regular polygon that has 8 sides, each interior angle is ______ and each exterior angle is ______.

2 Write if true or false. a The sum of all exterior angles of a polygon is always 360°. b It is possible to have a polygon whose exterior angle is 175°. c It is possible to have a regular polygon whose each exterior angle is

Do It Yourself 3C

180º . d Each exterior angle = (n - 2) × n

1 8

th of a right angle.

3 Find the measure of each exterior angle of the following regular polygons with: 1 Fill in the blanks.

a 3 sides b 5 sides c 8 sides d 12 sides e 24 sides a The polygon in which the sum of all exterior angles is equal to the sum of interior angles is called ______.

4 Find the number of sides of the regular polygons if each of their exterior angles are:

Do It Yourself

b The sum of interior angles of a polygon of n sides is ______ right angles. a 40 b 45 c 60 d 72 c The measure of each exterior angle of a regular polygon of 18 sides is ______.

e 90

Math Lab

5 The measure of the exterior angle of a hexagon is (3x – 4)°, (x + 4)°, (7x – 3)°, (8x – 1)°, (2x + 3)°, (9x + 1)°. Find the

d In case of a regular polygon that has 8 sides, each interior angle is ______ and each exterior angle is ______. measure of each angle.

Word Problems

2 Write if true or false.

Exploring Polygons Word Problem sum of Individual all exterior angles of a polygon is always 360°. a The Setting:

b It is possible to have a polygon whose exterior angle is 175°. Materials Required: Pen and paper 1 a regular heptagon. What 1 A carpenter is building a new table. The table is in the shape of c It is possible to have a regular polygon whose each exterior angle is th of a right angle. Method:is the measure of each interior angle of the tabletop. 8 180º . d Each exterior angle = (n - 2) ×

Each student will explore 1 n the classroom and write the names of 5 polygons they spot. 3 Find the of each angle the following regularorpolygons They will thenexterior identify the of polygon as concave convex.with: 2 measure a 3 sides

b 5 sides

c 8 sides

d 12 sides

to Remember The student who completes the task first, wins. 3 Points 4 Find the number of sides of the regular polygons if each of their exterior angles are:

You do

e 24 sides

45 a• 40A polygon isba simple c 60 d 72 e 90 closed curve made up of straight line segments, with no overlapping or self-intersecting parts. 5 The measure of the exterior angle of a hexagon is (3x – 4)°, (x + 4)°, (7x – 3)°, (8x – 1)°, (2x + 3)°, (9x + 1)°. Find the • Theofpolygons can be classified into different types on the basis of their number of sides and measure each angle. vertices.

Chapter Checkup •

A diagonal of a polygon is a line segment that joins two vertices of the polygon, which are not

Word Problem already joined by the adjacent vertices of the polygon.

• In a convex polygon, all the diagonals lie inside the polygon, whereas in a concave polygon, at 1 Fill in the blanks. least diagonalislies outside the table. polygon. A carpenter building a new The table is in the shape of a regular heptagon. What 1 one a• A polygon all and angles equalangle iswith called ________. is with the measure of each ofnthe tabletop. The number of sides diagonals in ainterior polygon number of sides = n(n - 3) ÷ 2

If a polygon has number sides, thenisthe sum of all the interior angles is (n - 2) × 180° b• The sum of the n interior angles of of a hexagon ________. The sumthat of exterior angles any polygon is always a constant, c• A polygon has at least one in interior angle more than 180° is called:i.e., 360°.

Chapter Checkup

•

If a polygon has n sides, the measure of each exterior angle of a regular polygon = 360° ÷ n.

Points to Remember

d The number of diagonals in a pentagon is ________.

e The sum of the exterior angles of any polygon is always ________.

40 2 •HowAmany diagonals will each of these polygons polygon is a simple closed curve madehave? up of straight line segments, with no overlapping or

self-intersecting parts.

a Convex quadrilateral

b Triangle

c Polygon with 17 sides

• The polygons can be classified into different types on the basis of their number of sides and 3 Findvertices. the number of sides of the polygon that has the given number of diagonals:

• a A27 diagonal of a polygon is a line segment b 44 c 55that joins two vertices d 62 of the polygon, which e 67 are not already joined by the adjacent vertices of the polygon. 4 Identify the polygons as concave or convex. • In a convex polygon, all the diagonals lie inside the polygon, whereas in a concave polygon, at a least one diagonal b lies outside the polygon. c d • •

The number of diagonals in a polygon with n number of sides = n(n - 3) ÷ 2

If a polygon n has number of sides, then the sum of all the interior angles is (n - 2) × 180°

sum of angle exterior in any polygon is always a constant, i.e., 360°. 5 •FindThe the interior sumangles of the polygons with sides: • a If9 a polygon hasbn sides, the measure cof each exterior angled of21 a regular polygone= 33 360° ÷ n. 13 16

Find the number of sides whose interior angle sum is: Pearson, P. D., & Gallagher, G. (1983). Contemporary Educational6 Psychology.

a 1800

b 2700

c 3240

d 3780

Fisher, D., & Frey, N. (2021). Better learning through structured teaching: A framework for the gradual release of responsibility.

e 4500

7 Find each exterior angle of the regular polygons with the number of sides.

Fisher, D., & Frey, N. (2014). Checking for understanding: Formative assessment techniques for your classroom. a 9 b 15 c 20

d 36

e 45

d 90°

e 120°

8 Find the number of sides of a regular whose exterior angle is:

Gradual Release of Responsibility

Maths Grade 8 Chapter 1-6.indb 5

a 40° Chapter 3 • Polygons

b 45°

c 72°

08-01-2024 14:45:03

C o nt e nt s

Rational Numbers �� 1 • Understanding Rational Numbers • Operations on Rational Numbers

2 7

Solving Equations in One Variable �� 19 • Solving Linear Equations

Polygons �� 32

• Understanding Polygons and their Properties

4 5

Construction of Quadrilaterals �� 58 • Constructing Quadrilaterals

6 Classification and Tabulation

of Data �� 73 • Frequency Distribution of Grouped and Ungrouped Data

Bar Graphs and Histograms �� 85 • Graphical Representation of Ungrouped Data 86 • Graphical Representation of Grouped Data 95

Pie Charts �� 106 • Drawing and Reading Pie Charts

107

9 Probability �� 119 • Understanding Probability

10 11 12

120

Squares and Square Roots �� 131 • Square of a Number • Square Roots

150 155

Percentages �� 163

• Real Life Applications on Percentages

Compound Interest �� 185

164

• Application of Percentages

172

• Simple Interest and Compound Interest 186

15 Algebraic Expressions: Addition and

Subtraction �� 201 • Algebraic Expressions 202 • Addition and Subtraction of Algebraic Expressions 206

ultiplication of Algebraic M Expressions �� 215 • Multiplying Algebraic Expressions

216

Visualising Solid Shapes �� 231 • Views, Maps and Polyhedrons

232

18 Area of Polygons �� 245 • Area of Figures Made With Polygons

246

19 Surface Area and Volume

of Solids �� 261 • Surface Area • Volume

262 267

Exponents and Powers �� 274 • Exponents

275

21 Direct and Inverse Proportions �� 288 • Proportion

132 140

Cubes and Cube Roots �� 149 • Cube of a Number • Cube Root of a Number

Profit/Loss, Discount and Tax �� 171

Quadrilaterals �� 43 • Understanding Quadrilaterals and their Properties

289

actorisation and Division of F Algebraic Expressions �� 301 • Factorisation of Algebraic Expressions • Division of Algebraic Expressions

302 308

Linear Graphs �� 315

Playing with Numbers �� 331

• Line Graphs and Linear Graphs • Numbers and Divisibility Rules

316 332

Answers �� 342

UM24CB_8_FM.indd 6

11-01-2024 14:22:40

Rational Numbers

Let's Recall In our previous grades, we learnt about natural numbers and whole numbers. We know that integers are positive and negative whole numbers. We also know how to read, write and represent fractions and decimals. We know that these are all rational numbers. Let us recap how to add and subtract integers, fractions and decimals. Adding Integers

Subtracting Integers

Like Sign

Unlike Sign

Add and keep the sign.

Subtract and keep the larger number sign.

4+2=6

Change the sign of the subtrahend and solve. –3 – (–2) = –3 + 2 = –1

(–4) + 3 = –1

(–4) + (–2) = –6 Adding and Subtracting Decimals • Convert to like decimals. • Align the decimal points. • Add or subtract. 1 2 .6 3 + 1 .3 0 1 3 .9 3

1 2 .6 3 − 1 .3 0 1 1 .3 3

Adding and Subtracting Fractions Like Fractions

Unlike Fractions

• Add or subtract the numerators.

• Convert to like fractions by taking the LCM.

• Keep the common denominator.

• Add or subtract as done in like fractions.

2 4 2+4 6 + = = 5 5 5 5

2 2 10 − 6 4 – = = 3 5 15 15

Letʼs Warm-up Fill in the blanks. 5 + 2 4 3

__________________

2 52.31 + 61.2

__________________

3 − 23 + 69

__________________

7 − 5 8 4

__________________

5 89 − 47.25

__________________

I scored _________ out of 5.

Maths Grade 8 Chapter 1-6.indb 1

08-01-2024 14:45:04

Understanding Rational Numbers Shubhi and Rishabh collected data on the temperature of different cities across the world in December. Rishabh: Shubhi, can you convert the temperature of Yakutsk and Dubai into fractions? Shubhi: Let me try Rishabh! We can write −25.5 °C as −255 °C and 22.5 °C as 225 °C. 10 10 −255 Rishabh: But °C is not a fraction! 10 Both kids were confused about what they should call this number. Let’s help them out!

Temperature in ℃

Real Life Connect

22.5

20 10 0

–10

–20

–30

–25.5 Yakutsk

Dubai

–18.3

–14.2

Fraser

Golmud

City

Rational Numbers On converting −25.5 °C and 22.5 °C to fractions, Shubhi got −255 °C 10 and 225 °C. These numbers are called rational numbers. 10 A rational number is a number that is expressed as p, where p and q q are integers, and the denominator should never be equal to zero.

Rational

Any number that can be written as a fraction –5.25 56 12

For example, 12.5, −15, 25, 6 and −8 are all rational numbers. 7 9

Whole

–12 10 7.5

All positive integers and zero 0, 1, 2, 3, 4,....

Natural

Identifying and Representing Rational Numbers

All positive integers, excluding zero 1, 2, 3, 4,....

We know how to represent integers, fractions and decimals on a number line. Let us see how to represent rational numbers on a number line. We have marked the number line in equal intervals of 0.5 with 0 in the centre.

Integers

All positive and negative whole numbers, and zero ....–3, –2, –1, 0, 1, 2, 3,....

Rational Number −1 = − 0.5 2 −1 2

−2 −1.5 −1 −0.5

Integer

Example 1

0.5

1.5

2.5

Decimal Number

Represent 5 and − 4 on the number line. 3 3 5 We can write as 1 2. So, it will lie between 1 and 2 on the number line. Divide the space between 1 and 2 3 3 into 3 equal parts and mark 5. 3

Maths Grade 8 Chapter 1-6.indb 2

08-01-2024 14:45:04

We can write −4 as −11. So, it will lie between –1 and –2 on the number line. Divide the space between –1 3 3 and –2 into 3 equal parts and mark –4. 3 −2 −1 0 1 2 −5 3

Example 2

−4 3

−3 −2 3

−1 3

Mark − 2, 1, 1.75 and –2.25 on the number line.

1 1 The number line has been marked at equal intervals of in both directions. 4 4 7 − 8 − 9 −2= , 1 = , 1.75 = and −2.25 = 4 4 4 4 −2

−1

−2 −10 −9 4 4 Do It Together

Represent

−8 4

−7 4

−6 4

−5 4

−1 −3 −2 4 4

−1 4

1 4

2 4

1 8

2 8

3 4

4 4

5 4

6 4

7 4

9 4

10 4

−2, 5, −1 and 0.5 on the number line. 8 8

–7 8

–6 8

–4 8

–1 8

4 8

Comparing and Ordering Rational Numbers Rational numbers can be compared by changing them to the same form. Let us recall how to write rational numbers in standard form and equivalent rational numbers. Standard Form of Rational Numbers A rational number is said to be in standard form if: • both the numerator and denominator contain no common factors other than 1. • the denominator has a positive integer. To convert rational numbers to standard form, divide both the numerator and denominator by their HCF. For example, the standard form of 6 = 6 ÷ 3 = 2 . 9 9÷3 3 Equivalent Rational Numbers

Think and Tell

Is pi a rational number?

Two rational numbers are equivalent if their standard forms are equal. To find the equivalent rational number of a rational number, we multiply or divide the numerator and denominator by the same natural number. For example,

1 1×2 2 1×3 3. 2 3 1 = = ; = So, and are equivalent to . 4 4 × 2 8 4 × 3 12 8 12 4

Chapter 1 • Rational Numbers

Maths Grade 8 Chapter 1-6.indb 3

08-01-2024 14:45:04

Two or more rational numbers can be compared and ordered using the steps given. Step 1: Write all the rational numbers in the same form.

p q Step 3: Find equivalent rational numbers with the denominator as the LCM.

Step 2: If the rational numbers are in form and their denominators are different, find the LCM of the denominators. Step 4: Compare the numerators of the equivalent rational numbers. The rational number with the greater numerator is the greater rational number.

2 5 and using the above steps. 3 7 Both the rational numbers are already in the same form, so we will find the LCM of 3 and 7 = 21. 2 2 × 7 14 5 5 × 3 15 = = and = = 3 3 × 7 21 7 7 × 3 21 14 15. As 14 < 15, < 21 21 2 5 Hence, < . 3 7 Now, what if the rational numbers are in different forms? How do we compare a fraction with a decimal or a decimal with an integer? Let us compare

To compare different kinds of rational numbers, we convert them into the same form. −7. 3 p −7 The comparison can be done by either converting −2.5 to form or by converting to integer form. q 3 Let us compare −2.5 and

Method 1 −25 −5 = 10 2 LCM of 3 and 2 = 6 −2.5 =

−5 −5 × 3 −15 −7 −7 × 2 −14. = = ; = = 2 2×3 6 3 3×2 6 −15 −14 As −15 < −14, < 6 6 − 7. Hence, −2.5 < 3

Let us now arrange the rational numbers LCM of 7, 3, 9 and 5 = 315

Method 2 −7 = −2.33 3 −2.5 < − 2.33 So, −2.5 <

−7. 3

−5 7 2 3 , , and in ascending order. 7 3 9 5

−5 × 45 −225 7 × 105 735 2 × 35 70 3 × 63 189 . = , = , = and = 7 × 45 315 3 × 105 315 9 × 35 315 5 × 63 315 Compare the numerators of the rational numbers: −225 < 70 < 189 < 735 The rational numbers can be written as

−225 70 189 735 . < < < 315 315 315 315 −5 2 3 7. So, the rational numbers in ascending order are < < < 7 9 5 3 −4. Give any three equivalent rational numbers for 5 We can write the rational numbers in ascending order as

Example 3

−4 −4 × 2 −8 = = 5 5 × 2 10

So, three equivalent rational numbers for

−4 − 4 × 3 −12 = = 5 5×3 15

−4 −4 × 4 −16 = = 5 5×4 20

−4 −8 −12 −16. are , and 5 10 15 20

Maths Grade 8 Chapter 1-6.indb 4

08-01-2024 14:45:05

Example 4

4 Suman and Kamal had equal amounts of fruit in their lunch box. Suman ate of the fruit while Kamal 9 8 ate of the fruit. Who ate more fruit? 11 To compare the given numbers, we will find the LCM of 9 and 11. LCM (9, 11) = 99 Write the equivalent rational numbers: Compare the numerators: 44 < 72

4 × 11 44 8 × 9 72. = and = 9 × 11 99 11 × 9 99

Think and Tell How is

4 8 . < 9 11 Therefore, Kamal ate more fruit than Suman.

−4 different from 4 ? 5 −5

Hence,

Do It Together

13, 8, 1.7 in descending order. 6 3 p To compare and order the rational numbers, first convert them to form. q Therefore, the rational numbers can be written as: _____________________ Arrange the rational numbers − 3,

Find the LCM of the denominators and write the equivalent rational numbers. The LCM of 1, 6, 3 and 10 is _____________________. The equivalent rational numbers are: ____, ____, ____, ____ Compare the numerators: _____________________ Descending order = ____>____>____>____

Rational Numbers Between Two Rational Numbers Rational numbers between two rational numbers can be found using their mean or equivalent rational numbers. By Finding the Mean

By Finding Equivalent Rational Numbers

p m p m If q and n are two rational numbers and q < n ,

• Find the LCM of the denominators and write the equivalent rational numbers.

1 p m  p m  +  is a rational number between q and n . 2  q n 

• Write the rational numbers in increasing order. For example, to find the rational numbers between

For example, a rational number between −21 +10   −3 2 −11 and = 1  −3 + 2  = 1 × =   5 7 2 5 7 2 35 70

(

Example 5

)

1 and 8, we find the LCM of 3 and 9 and write the 3 9 1 3 8 8 equivalent rational numbers as = and = . 3 9 9 9 3 8 H ence, the rational numbers between and are 9 9 4, 5, 6 and 7 . 9 9 9 9

Find two rational numbers between 2 and 3. A rational number between 2 and 3:

Chapter 1 • Rational Numbers

Maths Grade 8 Chapter 1-6.indb 5

1 (2 + 3) = 2

5 5 ;2 < < 3 2 2

08-01-2024 14:45:05

A rational number between 2 and 5:  2 + = ;2 < <  = 2  2  2  4 4 2 2 2 1

Example 6

5

19

Hence, the two rational numbers between 2 and 3 are 5 and 9. 2 4 Give four rational numbers between 1 and 1 . 3 2 LCM of 3 and 2 = 6. Hence,

1 2 1 3 = and = 3 6 2 6

Here, the new equivalent rational numbers do not have any numbers between them. So, we multiply and divide each of them by 5. 2 2 × 5 10 3 3 × 5 15 = = and = = 3 6 × 5 30 6 6 × 5 30 Four rational numbers between Do It Together

10 15 11 12 13 14 . and = , , and 30 30 30 30 30 30

7 . Give five rational numbers between − 4 and 12 9 LCM of 9 and 12 = ____; hence, − 4 = ____ and 7 = ____ 9 12 − 4 7 Five rational numbers between and = ________________________________ 9 12

Do It Yourself 1A 1 Represent the rational numbers on the number line. a

4 5

b -

2 4

7 4

d -

6 5

2 Which rational number does point Q represent on the number line? Q

-2

3 Convert the rational numbers into their standard form. a

3 9

-2 16

-1 c

8 -28

-16 -42

15 -21

-3 14

2 16 ______ 9 72

18 ______ -3.2 5

19 53 ______ 2 7

4 Give three equivalent rational numbers for the given rational numbers. a

5 8

-9 12

6 8

5 Compare and fill in the blanks with <, > or =. a

1 3 ______ 4 5

-5 7 ______ 8 12

c - 0.4 ______

-6 -5 ______ 11 12

14 ______ 2.8 5

Maths Grade 8 Chapter 1-6.indb 6

08-01-2024 14:45:06

6 Arrange the rational numbers in ascending order. 9 1 8 5 a -3.5, , 2 , , 5 4 7 2

-3 2 7 , 1.5, 4.2, 2 , 2 5 5

5 1 , -0.75, 1.25, 1 , 7 9 6 10

8 2 , -2, 1.25, 1 , 13 9 4 4

16 1 , 3 , -2.5, 3, -18 3 4 5

3 -25 36 17 , 2.1, , , 5 2 5 2

7 Circle the rational numbers that lie between -3 and -2. -13 5

-16 8

8 Give four rational numbers between

-21 6

-18 8

-14 -26 and . 9 12

Word Problem 1

Simran bought piece of cloth?

15 m of red cloth and Raj bought 4.6 m of green cloth. Who bought the larger 4

Operations on Rational Numbers Real Life Connect

Shubhi and Rishabh have now started looking at the temperature in their city over the past 5 days. They record the data in the form of rational numbers.

Day

Temperature

Monday

-3 °C 2 16 °C 5 9 °C 5 23 °C 5 -3 °C 4

Both of them performed various operations on the rational numbers and found different data. Let us see how!

Tuesday

Addition and Subtraction of Rational Numbers

Thursday

Addition of Rational Numbers

Friday

Wednesday

The addition of rational numbers is similar to the addition of fractions. However, we first have to convert all the rational numbers to be added into rational numbers with positive denominators. Let us use the data collected by Rishabh and Shubhi and look at the cases and steps of addition of rational numbers. Same Denominators

16 9 + 5 5 (16 + 9) 25 = = =5 5 5

Different Denominators

2 −3 + 9 2

The LCM of 2 and 5 is 10.

Write the equivalent rational numbers with the denominator as the LCM. −3 × 5 −15 9 × 2 18 = and = 10 2×5 5 × 2 10 −15 18 −15 + 18 3 + = = 10 10 10 10

Chapter 1 • Rational Numbers

Maths Grade 8 Chapter 1-6.indb 7

08-01-2024 14:45:06

Adding Rational Numbers with Negative Denominators Let us add 16 and 3 5 −4

Converting 3 into a rational number with a positive denominator, we get 3 × −1 = −3 −4 −4 −1 4 LCM of 5 and 4 = 20 Write the rational numbers with denominators as 20: 16 × 4 = 64 and −3 × 5 = −15 5 × 4 20 4×5 20 So, 16 + (−3) = 64 + (−15) = 64 − 15 = 49. 5 4 20 20 20 20 Example 7

Add

18 −19 , and 2.5. 4 8

2.5 can also be written as 25. 10 The LCM of 4, 8 and 10 is 40.

Write the rational numbers with the denominator as 40: 18 × 10 = 180, −19 × 5 = −95 and 25 × 4 = 100 4 × 10 40 8 × 5 40 10 × 4 40 18 (−19) 25 Hence, + + 4 8 10 180 + (−95) + 100 = 180 − 95 + 100 = 185 = 37 40 40 40 40 40 8 Example 8

A shopkeeper sold 2

1 3 2 kg, 5 kg, 1.5 kg and 4 kg of rice to four customers. Find the total amount of 4 8 3

rice sold by the shopkeeper.

The weights can be expressed as rational numbers as: 2

1 9 3 43 15 3 2 14 = , 5 = , 1.5 = = and 4 = 4 4 8 8 10 2 3 3

Total weight of rice sold = LCM of 4, 8, 2 and 3 = 24 Hence, the total rice sold = Do It Together

9 43 3 14 + + + 4 8 2 3 54 129 36 112 54 + 129 + 36 + 112 331 19 + + + = = kg = 13 kg 24 24 24 24 24 24 24

1 Add 15, 23 and 3.6. 4 5

2 Add 5, 6 and ‒ 2.8. 3 ‒13

______________________________________________________________________________________________________________________ ______________________________________________________________________________________________________________________

Subtraction of Rational Numbers We know that subtraction is the opposite or inverse of addition. To subtract two rational numbers, we simply add the additive inverse of the rational number to be subtracted to the first rational number. Also, the rational numbers should have a positive denominator. 17 7 17 (-7) 34 + (-21) 34 - 21 13 For example, - = + = = = 6 6 3 2 3 2 6

Maths Grade 8 Chapter 1-6.indb 8

08-01-2024 14:45:07

What if the rational numbers are in different forms? We will first convert the numbers into the same form. 23 Let us subtract from 7.85. 5 Method 1

Convert

23 = 4.6 5

Method 2

Convert 7.85 to a fraction.

23 to a decimal. 5

7.85 =

785 100

785 23 785 (- 23) = + 100 5 100 5

7.85 - 4.6 = 3.25

785 + (- 460) 785 - 460 325 13 = = = 100 100 100 4

What if we have more than two rational numbers with different signs between them? Let us find out! Simplify

2 5 1 4 + - + 3 9 6 3

To simplify the above statement, we will find the LCM of all the denominators and convert them to like rational numbers. LCM of (3, 9, 6) = 18 Hence,

2 5 (- 1) 4 12 + 10 - 3 + 24 43 + + + = = 3 9 6 3 18 18

Simplify:

Example 9

7 8 -3 -9 + + 6 9 12 15

Example 10

7 8 -3 -9 7 8 -3 + + = + + additive inverse of 6 9 12 15 6 9 12 -9 + 15 7 8 3 -9 ⇒ + + + (The LCM of 6, 9, 12 and 6 9 12 15 15 is 180.)

Do It Together

521 kg. Suppose 12 the empty drum weighs 12.5 kg. What is the weight of the sugar in the drum? 521 Weight of drum with sugar = kg 12 125 25 Weight of empty drum = 12.5 kg = = kg 10 2 A drum full of sugar weighs

Weight of sugar =

521 25 521 (- 25) - = + 12 2 12 2

⇒

210 160 45 - 108 + + + 180 180 180 180

⇒

210 + 160 + 45 - 108 307 = 180 180

521 (- 150) + (The LCM of 12 and 2 is 12.) 12 12

521 - 150 371 = kg 12 12

1 Subtract 12 from 6.25. 5

2 Subtract

5 1 from 3 . ‒2 2

Chapter 1 • Rational Numbers

Maths Grade 8 Chapter 1-6.indb 9

08-01-2024 14:45:07

Properties of Addition and Subtraction of Rational Numbers Closure Property If

For example:

a c a c a c and are rational numbers, then + and - are also rational numbers. b d b d b d

-5 3 -7 -1 + = or (Rational number) 7 14 14 2

Commutative Property

For Addition If

For example: - 4.5 -

a a c c a c and are rational numbers, then + = + . b b d d b d

For Subtraction

a c a c c a and are rational numbers, then - ≠ - . b d b d d b

5 3 13 3 5 13 + = and + = 7 14 14 14 7 14 5 3 3 5 So, + = + 7 14 14 7

8 7 1 7 8 -1 - = but - = 9 9 9 9 9 9 8 7 7 8 So, - ≠ 9 9 9 9

For example:

Associative Property

For Addition

For Subtraction

a c e a c e If , and are rational numbers, then + + b d f b d f a c e = + + . b d f For example:

a c e a c e , and are rational numbers, then - b d f b d f a c e ≠ - - . b d f

5 5 32 2 3 7 + + = + = 2 10 2 10 5 10

For example:

2 3 5 2 28 32 + + = + = 5 10 2 5 10 10 Additive Identity If

Example 11

a a is a rational number, then + b b -a = 0. b For example:

Verify if m – n is a rational number if m = m-n=

Example 12

-2 -2 +0= 3 3

5 4 3 1 5 –3 – = = 10 5 10 10 5 5

4 3 5 4 1 7 - - = = 5 5 10 5 10 10

Additive Inverse

a a is a rational number, then + 0 b b a a =0+ = . b b For example:

1 = -5 (Rational number) 2

5 -5 + =0 8 8

Subtraction Property of Zero If

a is a rational number, then b a a -0= . b b For example:

3 3 -0= 5 5

–9 6 and n = . 11 22

-9 6 -18 - 6 -24 -12 = = = = rational number 11 22 22 22 11

5 3 7 and are three rational numbers, then prove that they are not associative under subtraction. 4 8 12

If ,

a c e a c e - ≠ b d f b d f 5 3 7 7 7 7 5 3 7 5 - 5 35 So, - = = and - = = 4 8 12 8 12 24 4 8 12 4 24 24

The associative property for subtraction states that

7 35 ≠ ; hence, rational numbers are not associative under subtraction. 24 24

Do It Together

Fill in the blanks using the properties of rational numbers. 1 5 - _______ = 5 2 2

2 5 + (-3) = -3 + _______ 2 9 9

3 12 + _______ = 0 25

Maths Grade 8 Chapter 1-6.indb 10

08-01-2024 14:45:08

Do It Yourself 1B 1 Solve. a

4 6 + 7 7

d -2 + 4 + 2.5

8 (-2) 11 11

9 2 - 1.3 7 (-9)

e 8 + (-24) + 14.5

2 Solve. a

5 (-7) + 9 11

b f

9 38 13 39

3 What decimal should be added to

f 17 + 18.44 + (-12)

8 (-14) 7 11 15 5

-12 (-13) + 2 15

-5 7 12 15

13 9 - 2.1 -4 16

-13 (-6) 8 15

17 (- 16) 2 5 3 7

21 -27 to get ? 5 6

4 What fraction should be subtracted from 8.5 to get -12.25? 3 7

2 3

5 A rope is 5 m long. Another rope is 1 m shorter than it. What is the total length of the ropes? 6 Ravi earned ₹480 in 1 day. He spent ₹ he save?

153 253 on tea and snacks, ₹ on food and he saved the rest. How much did 2 2

7 Name the property illustrated through each of the operations. a

a c - = a rational number b d

-12 7

a c e a c e + + = + + b d f b d f

a a -0= b b

13 2

5 21 d -9 14

8 Find the additive inverse of the rational numbers. a

9 Simplify

5 (-3) 6 (-3) 2 8 + - + - 6 8 7 14 5 (-15)

8 9

5 3

7 6

10 Find the rational number that must be added and subtracted so that they will make the sum + - to the nearest whole number.

11 Era has the given weight of grains in her pantry.

Wheat 25.5 kg

a How much wheat and rice does Era have?

Rice 12

2 kg 3

Corn 37 kg 2

b What is the difference in the weight of the wheat and corn?

Word Problems 1 2

A carpenter bought a piece of wood 12 How long is the piece of wood now?

1 2 7 kg in the first week, 4 kg in the second week, 4 kg 5 3 8 in the third week and the rest in the last week. How much flour did he use in the last week? Jimmie bought 18 kg of flour. He used 3

Chapter 1 • Rational Numbers

Maths Grade 8 Chapter 1-6.indb 11

2 5 feet long. He then cut off feet from both ends. 3 8

08-01-2024 14:45:09

3 4

15 kg of apples, 2.6 kg of bananas and some oranges from the market. 4 Altogether, she purchased 9 kg of fruit. What is the weight of the oranges she purchased? 2 A school prepared 25.5 litres of juice for its players. The players drank 8 litres in the first 5 3 half. In the second half, the players drank 2 litres more than in the first half. What was the 8 volume of the juice left at the end? Suhani purchased

Multiplication and Division of Rational Numbers Multiplication of Rational Numbers The multiplication of rational numbers is similar to that of fractions. Product of numerators Product of two rational numbers = Product of denominators For example,

12 5 60 7 -3 -21 -7 × = and × = = 13 7 91 9 8 72 24

What if the rational numbers are in different forms? -4 Let us multiply and 0.56. 13 56 14 . We will first convert them into the same form as 0.56 = = 100 25 (–4) 14 (–56) × = 13 25 325 Example 13

Example 14

13 and –8.5. 25 p -85. Write the numbers in the form ; hence, –8.5 = q 10 13 -85 -1105 × = 25 10 250 -1105 -221 HCF of 1105 and 250 = 5; hence, = 250 50 145 231 The length and width of a rectangular park are m and m. 3 9 What is the perimeter of the park? Find the product of

Perimeter of the rectangle = 2 × (L + B) = 2 2×

435 + 231 2 666 1332 = × = = 148 m 9 1 9 9

Remember! The product of two rational numbers with similar signs is positive and with opposite signs is negative.

Error Alert! Convert the numbers into the same form before multiplying to avoid errors. 2 8.1 × 2.7 = 3 3

2 27 54 18 × = = 3 10 30 10

145 231 + 3 9

Hence, the perimeter of the rectangular park is 148 m. Do It Together

1 Multiply: 5 and 16 9 -7

2 Multiply: 2.5 and -19 3

Maths Grade 8 Chapter 1-6.indb 12

08-01-2024 14:45:10

Division of Rational Numbers We know that division is the opposite of multiplication. To divide a rational number by another rational number, we multiply the dividend by the multiplicative inverse of the divisor. That is

a c a d ÷ = × b d b c Divisor

Dividend

Did You Know?

15 3 15 6 90 15 ÷ = × = = 4 6 4 3 12 2 To divide rational numbers given in different forms, we first convert them to the same form and then divide. For example:

For example:

8 951. m of cloth for ₹ What is 3 4 the cost of the cloth per meter? Cost of

Example 16

12 13 (-5) 3 9 × ÷ × ÷ 5 18 8 5 8

951 8 951 3 2853 ÷ = × =₹ 4 3 4 8 32

1 Divide: –15 � 21 8 5

Simplify

1 times 11

Solve the numbers inside the brackets first:

8 951 m of cloth = ₹ 3 4

Cost of 1 m of cloth =

Do It Together

the size of Jupiter.

13 13 2 13 1 13 ÷ 2 = ÷ = × = 5 5 1 5 2 10

Meeta bought

Example 15

The size of Earth is

12 13 -8 3 8 × × × × 5 18 5 5 9

12 (-104) 24 -29952 -1664 × × = = 5 90 45 20250 1125

2 Divide: 3.5 �

(–19) 25

__________________________________________________

Properties of Multiplication and Division of Rational Numbers Closure Property For Multiplication

For Division

a a c c and are rational numbers, then × will also be b d b d a rational number. If

For example:

2 (−3) −6 × = (rational number) 5 2 10

a a c c and are rational numbers, then ÷ is not b d b d necessarily a rational number. If

For example:

6 ÷ 0 = not defined (not a rational number) 11

Commutative Property For Multiplication If

a c c a a c and are rational numbers, then × = × . b d d b b d

For example: 1 × 2 = 2 and 2 × 1 = 2 6 7 6 42 7 42

Chapter 1 • Rational Numbers

Maths Grade 8 Chapter 1-6.indb 13

For Division If

a c c a a c and are rational numbers, then ÷ ≠ ÷ . b d b d d b

For example:

5 2 25 20 2 5 = = ÷ ÷ but 10 5 5 10 20 25

08-01-2024 14:45:11

Associative Property For Multiplication If

a, c

and

For Division

are rational numbers, then

b For example:

2 2

∴

d 5 5

14 1 14 × = 10 2 20

2 14 7 × = 2 10 20

2 1 7 × × 2 2 5

a, c

and

are rational numbers, then

≠

For example: 3 ÷ 2 ÷ 1 = 21 ÷ 1 = 21 8 4 7 8 8 3

∴

8 ÷

16 21 3 ÷ = 4 7 64

≠

2 1 3 ÷ ÷ 4 7 8

Distributive Property Multiplication over addition and subtraction a, c

and

are rational numbers, then a

a, c

a e a c × + × b f b d and a c a e e a c × − × − × = b d f b f b d b

Division over addition and subtraction

and

Multiplicative Identity

× 1 = 1 × 3

×1=

÷1=

−

d c

c e a e ÷ + ÷ d f b f and c e e a e ÷ = ÷ − ÷ d f f b f

2 −1 2 −1 = ÷ = 4 4 4 2 and 1 2 3 2 4 12 −4 −1 ÷ ÷ = = = 2 4 4 8 4 4 8 2 1 3 2 4

Multiplicative Inverse

a is a rational number, then b b

2 5 2 5 = ÷ = 4 2 4 4 and 1 2 3 2 4 12 20 5 ÷ + ÷ = + = = 2 4 4 8 4 4 8 2 1 3 + 2 4

For example:

× 0 = 0 ×

a is a rational number, then b

= 0.

6 ×0=0 11

If ÷ (−1) =

= 1. =1

Division by itself and inverse

a is a rational number, then b

Division by 1 and −1 If

Multiplicative Property of 0

a is a rational number, then b a

1 3 2 1 1 1 × = × = 2 2 4 4 8 4 and 1 3 1 2 3 2 1 × × = - = 2 4 8 8 8 2 4

are rational numbers, then

For example: 1 × 3 + 2 = 1 × 5 = 5 2 2 4 4 8 4 and 1 3 1 2 3 2 5 × + × = + = 2 4 4 8 8 8 2

−a b

a is a non-zero rational number, then b

−a b

= −1

Maths Grade 8 Chapter 1-6.indb 14

08-01-2024 14:45:12

Example 17

6 8 7 8 using the distributive property of division over subtraction. – � � 5 11 12 11 a c a c e e e − − ÷ ÷ = The distributive property of division over subtraction states that ÷ b d b d f f f 6 8 8 6 8 7 7 So, ÷ − = − ÷ ÷ 5 11 11 12 12 5 11 Solve

8 = 11

⇒ 6 × 12 − 7 × 5 ÷ 60

⇒ Example 18

8 37 37 11 407 ÷ × = = 60 11 60 8 480

Give the multiplicative inverse of −2 7. = 7 −2

Multiplicative inverse of Do It Together

72 − 35 60

8 11

−2 . 7

Fill in the blanks using the properties of rational numbers. 1

1 5 3 − ÷ = 4 8 2

−

3 ÷ 2

38 16 and 7 −19

5 14 by 5 3

−28

6 × 8

= 0

−5 and 5 8

−12 and 4.2 7

23 and −3.2 5

−13 by 3.6 6

34 51 by 7 19

× 1 =

13 18

Do It Yourself 1C 1 Multiply. a e

and

18 7

2 Divide.

7 (−30)

13 and −2.5 15

−51 and −28 5

12 −4 by 5 25

31 by −7 14

−36

and

−27 44

by 4.8

65 −14

3 If the product of two rational numbers is 28 and one of them is −17 , find the other. Also find the multiplicative 3

inverse of the result.

4 Fill in the blanks. a d

−9 −7

5 If a =

5 4

8 ×

× −6 5

,b=

−7 9

÷ 3 8

−13 2

b −4.5 ÷

____ =

____

and c =

25 4

16 5

____

÷ 1.5 =

____

−8

f 3.4 ÷

−7 3

14 5

____

−4 = ____ 9

then verify

a a × (b × c) = (a × b) × c

b (a − b) ÷ c = a ÷ c − b ÷ c

c a×c=c×a

d (a − b) − c ≠ a − (b − c)

6 Sunita has a square plot of length 185 m. What is the area and perimeter of the plot? 7 A ribbon of length

357 8

Chapter 1 • Rational Numbers

Maths Grade 8 Chapter 1-6.indb 15

m has been cut into 24 equal pieces. What is the length of each piece?

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8 A zoo ticket usually costs ₹ 250 for elders. For kids, they are priced at buy 5 kids’ tickets?

5 8

of the usual cost. How much will it cost to

9 The area of a rectangular room is 389 m2. If its length is 247 m, find its breadth. 4

10 Simplify and find the multiplicative inverse of the answer. Also find the multiplicative inverse of the result.

5 2 9 31 17 34 11 ÷ × ÷ × + 8 8 7 13 15 21 7

11 During the summer holidays, Mansi read a book. After reading pages are there in the book?

5 8

of the book, 120 pages were left. How many

Word Problems 1

Kavya bought 8 rolls of paper towels. The total length of all the rolls was 88 length of 5 such rolls?

1 8

m. What is the

2 3 Kapil earns ₹18,000 per month. He spends of the total income on rent, of the remaining 9 7 1 amount on food and of the rest on shopping. How much will he be able to save in a year? 4

Points to Remember • A rational number is a number that is expressed as p , where p and q are integers, and the q denominator should never be equal to zero. A negative rational number is always less than a positive rational number. • To compare two or more rational numbers, make their denominators the same and compare the numerators. • There is an infinite number of rational numbers between two rational numbers. • The operations on rational numbers are similar to fractions. The signs need to be kept in mind, like with integers.

Math Lab Setting: In groups of 4 Materials Required: Cards with properties of operations written on them, paper and pen. Method: 1 Prepare cards with different properties of operations written on them. For example, 1 3 5 1 5 3 5 + ÷ = ÷ + ÷ 2 5 7 2 7 5 7 2

Distribute an equal number of cards among the groups.

Maths Grade 8 Chapter 1-6.indb 16

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3 Each group identifies the property and writes its name along with the solution on the paper. The group that finishes all the sums correctly first, wins!

Chapter Checkup 1 Which rational number does point P represent on the given number line? P

−2

−1

2 Give a rational number equivalent to −9 that has 87 as the denominator. 29

3 Compare and fill in the boxes with <, > or =. a

−16

a 1.2,

5 2

, −1

1 −7 5 , , 2 6 6

−12

12 32 f 7.2 5 8 −6 Arrange the given rational numbers in descending order. e

−5 7

8 3

, 1.75, −0.85, −1

5 1

−12

−8 3

6 Give four rational numbers between 7 If m =

−12 5

and

1 −3 8 ,n= and o = then verify: 6 5 10

a m ÷ (n ÷ o) ≠ (m ÷ n) ÷ o

18 4

5 Circle the rational numbers that do not lie between −2 and −1. a

−15

18 5

21 19

1.75

29 3

51 6

, −5, 3.25, 2

−10

3 21 , 6 5 −15 8

b m × (n + o) = m × n + m × o

c m−n≠n−m

d (m + n) + o = m + (n + o)

8 Fill in the blanks. a The sum of a rational number and its additive inverse is __________. b The reciprocal of a negative rational number is __________. c The rational number __________ does not have any reciprocal. d Adding __________ to a rational number gives the rational number itself.

9 The sum of two rational numbers is −

8 5

. If one of the numbers is 9 , find the other. 17

10 Which rational number should be subtracted from

12 7

to get

−15 ? 4

11 Find the area of a triangle with a base of 156 m and a height of

191 2

5 12 Solve and find the multiplicative inverse of the answer. a e

7 −2 + 5 6 −13 5

Chapter 1 • Rational Numbers

Maths Grade 8 Chapter 1-6.indb 17

b f

3 7

19 17

8 +

14 (−51)

9 2

−

17 6

(−8) 3

24 59

d h

−12 7

−24 13

× −

−21 4

(−29) 39

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13 One roll of ribbon is 15

2 3

m long. What is the length of 5 full and one-fourth of such rolls?

14 Find (m + n) ÷ (m − n), where m =

5 6

−8

and n =

15 The given graph shows the time taken by different runners to finish a 500 m race. What is the average time taken by all the runners to complete the race in minutes?

Time taken in sec

105

100.26

100 95

98.55 91.63

90.75

90 85

Satish

Itisha

Rohan

Rhea

Runner

16 Simplify

12 7

−5 9

2 8

7 5

−

Between which two whole numbers does the result lie?

(−5) 16

Word Problems 1

2 3

Rohit practised cricket for 12

5 1 hours last week. He practised 2 hours this week. For how 6 8

many hours did Rohit practice cricket in two weeks? 1 Nikita distributed 52 kg of rice equally among 15 families. How much rice did each family 4 get? 256 175 Vivek has a rectangular plot of length m and width m, while Shweta has a square 5 3 189 plot of length m. Whose plot has a larger area? Also, find the difference of the areas of 2 the plots.

3 5 1 pizza, Alex ate more pizza than Richa and Prince ate 1 times what Richa 4 6 2 ate. How much pizza did they eat together?

Richa ate 1

Instead of multiplying a number by

wrong and the correct answer is 80, then find the number. 2 4 Shilpa had ₹750. She spent of the money on purchasing a book and of the remaining 5 9 amount on stationery. How much money was left with her?

3 3 , Ronita divided it by . If the difference between the 5 5

Maths Grade 8 Chapter 1-6.indb 18

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Solving Equations in One Variable

2 Let's Recall

Neeta is learning about the ages of all the members in her family. She asks her mother about her brother’s age. Her mother says that her age is 5 years more than 5 times her brother’s age. She also tells Neeta that she is 40 years old. Neeta wants to calculate her brother’s age. We know how to solve the equations of the type ax + b = c by the balancing and transposing methods. Let us recall the two methods. Balancing Method

Transposing Method

5x + 5 = 40

Subtracting 5 from both sides of the equation, we get,

Transposing 5 from the LHS to the RHS, we get, 5= x 40 − 5

5 x + 5 − 5 = 40 − 5

Simplifying the RHS of the equation, we get,

Simplifying both sides of the equation, we get,

5 x = 35

Transposing 5 from the LHS to the RHS, we get,

Dividing by 5 on both sides of the equation, we get,

x =7

5 x 35 = 5 5 Simplifying both sides of the equation, we get, x =7

So, Neeta’s brother is 7 years old.

Letʼs Warm-up

Match the linear equations with their solutions. Linear Equation

Solution of the Equation

1 2x + 4 = 8

x=7

x = −6

3x = 9 2

3 3x − 6 = 15

x=6

4 −5x = 40

x=2

−

x = 3 2

x = −8

I scored _________ out of 5.

Maths Grade 8 Chapter 1-6.indb 19

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Mean, Median Mode Solving Linearand Equations Real Life Connect

Vimala and Jaya are raising money by selling tickets for a hockey tournament. Both Vimala and Jaya have sold the same number of tickets. Jaya has sold thirty-six tickets more than Lata. Vimala has sold three times as many as Lata has. So, how many tickets have Lata sold?

Transposing Method Let us denote the number of tickets sold by Lata as x. So, the number of tickets sold by Jaya is 36 + x. The number of tickets sold by Vimala is 3x. Now, according to the situation, Vimala and Jaya sold the same number of tickets. 3x = 36 + x

Let us now solve this equation using different methods. Balancing Method Representing the expressions on two sides of a simple balance.

For the balancing method, we perform the same mathematical operations on both sides of the equation, so that the balance is not disturbed. The equation is 3x = x + 36 Subtract x from both sides, 3x − x = x + 36 − x ⇒ 2x = 36 Divide both sides by 2,

2x 36 = ; ⇒ x = 18 2 2

Thus, Lata sold 18 tickets. Solve: 5x + 21 = 2x + 39 5x + 21 – 21 = 2x + 39 – 21 5x = 2x + 18 ⇒ 5x – 2x = 2x + 18 – 2x 3x = 18 ⇒ x=6

3 x 18 = 3 3

Maths Grade 8 Chapter 1-6.indb 20

08-01-2024 14:45:18

Let us understand some important terms: Equation: 2x + 2 = 6 + x An expression is a combination of terms (variables or numbers or both) that are connected using mathematical operations.

An equation is a mathematical statement with an equal symbol between two expressions. 2x + 2 Expression

6+x Expression

A linear equation in one variable is an equation that has linear expressions with only one variable. 2x

Variable

LHS: 2x + 2

Equal to symbol

Equation

RHS: 6x + 2

Solution of a Linear Equation The method of finding the value of the variable so that it satisfies the equation is called solving a linear equation. The value of the variable obtained from the method is called the solution of the linear equation. Let us understand the transposition method. Transposition is the process of shifting a term of an equation to the other side simply by changing its ‘sign’ or by applying an opposite operation. When any term of an equation can be taken to the other side of the equation with a change of its sign, then this method is called transposition. Solve for x: 5x + 12 = 3x + 4 Steps for the transposition method: Step 1: I dentify the variable and constants.

Step 2: Remove the brackets to

simplify the

LHS and RHS. 5x + 12 = 3x + 4

No brackets here

Variable: x Constants: 12, 4

Step 3: M ove all variable

terms to one side

and all the constant terms to the other

Step 4: S implify the LHS

and RHS to solve for the variable.

side.

5x – 3x = 4 – 12

2x = – 8 x = –4

Remember! While transposing, keep the variable terms on one side of the equation and the constants on the other side.

Chapter 2 • Solving Equations in One Variable

Maths Grade 8 Chapter 1-6.indb 21

08-01-2024 14:45:19

Example 1

Solve using the balancing method. 1

3x + 47 = 95

5x + 15 = 3x + 33

3x + 47 – 47 = 95 – 47

5x + 15 −15 = 3x + 33 − 15

3x = 48

5x = 3x + 18 ⇒ 5x − 3x = 3x + 18 − 3x

x = 16

2x = 18 ⇒ x=9

2 x 18 = 2 2

Remember! The mathematical operation is reversed when we move terms from one side to the other side of the equal symbol.

Example 2

Solve using the transposition method. 1

Do It Together

8x – 14 = 3x + 11

Error Alert!

6x – 25 = 4x – 13

8x – 3x = 14 + 11

6x – 4x = 25 – 13

5x = 25

2x = 12

x = 5

x=6

x+3=7 On solving the equation:

x=7+3 x = 10

Find the value of x. 2

x=7–3 x=4

3x + 4(x – 1 ) = 10  3x + __________________ = 10  __________________  __________________

___________________________________________________

 __________________

___________________________________________________

Do It Yourself 2A 1 Find the value of the unknown weights. a

2 Find the value of x using the balancing method. a 6x + 4 = 22

b 2x + 3 = x + 18

c 7x + 45 = 4x + 78

d 2x + 17 = 9x – 32

e 11x + 1 = 3

f 5x – 11 = 2x + 7

Maths Grade 8 Chapter 1-6.indb 22

08-01-2024 14:45:21

3 Find the value of x using the transposition method. a 3x – 5 = 2x + 97

b 5(x – 7) – 2x = 13 – x

c 15x – 2(18 + x) = 29

d x + 2(3 + x) = 5(x – 6)

e 9x + 3 = 13(2x + 8)

f 5(x –17) = 7(x –19)

4 Solve for the value of x in the following equations. a 3x + 4(x – 2) = 41

b x – 1.5(x – 3) = 2(x − 3)

c 3(1.5x + 1) = 16.5

d 2.4x + 4.8 = 1.2x + 6

e 4x + 7 = 1.5(x + 12)

f 1.3(x – 3) + 2.6 = x

5 One number is twice another number. If 30 is subtracted from both numbers, then one of the new numbers becomes four times that of the other new number. Find the numbers.

6 Find two consecutive even integers so that two-fifths of the smaller number exceeds two-elevenths of the larger number by 4.

7 One number is 3 times the other number. If 15 were to be added to both numbers, then one of the new numbers becomes twice that of the other new number. Find the numbers.

8 Solve for the value of x. 18 cm

x Perimeter = 54 cm 12 cm 15 cm

b x+3

2x + 3 Perimeter = 35 cm

c 4 cm

Perimeter = 56 cm x – 1

x+4 x+3

2(x + 1)

3x + 5

Word Problems 1

After 30 years, Monty’s age will be 6 more than thrice his present age. What is Monty’s

Harsh’s mother is 30 years older than him. In 12 years, Harsh’s mother will be three times

The length of a rectangular field is 7 less than twice its breadth. The perimeter of the

Ravi deposited ₹25,000 in his bank account. He deposited currency notes of ₹50, ₹100 and

The ratio of Naveen’s age to his father’s age is 2:5. After 10 years, the ratio of their ages will

present age?

as old as Harsh. What are the present ages of Harsh and his mother? rectangle is 196 m. Find the area of the field.

₹500 in the ratio of 6:2:1. Find the number of notes of each denomination. be 1:2. What was Naveen’s father’s age at the time of Naveen’s birth?

Simplifying Equations to Linear Form Remember, Lata, Jaya and Vimala sold hockey tickets. They sold the tickets online as well as offline. The cost of the online ticket was ₹200 more than that of the offline ticket. Five-eighths of the online ticket’s cost is the same as three-fourths of the offline ticket’s cost. Let us find the cost of each type of ticket. Let the cost of the offline ticket be ‘y’.

Chapter 2 • Solving Equations in One Variable

Maths Grade 8 Chapter 1-6.indb 23

08-01-2024 14:45:21

5 3 ( y + 200) = y 8 4

1 y = 125 8

y = 1000 y + 200 = 1000 + 200 = 1200

5 3 y + 125 = y 8 4 3 5 y− y= 125 4 8

Thus, the cost of the offline ticket is ₹1000 and the cost of the online ticket is ₹1200. Solve for x. x +8 3

x +3 2

Step 1: Multiply the numerator of the one Step 2: S implify both sides of side with the denominator of the

the equation.

other side. x+8 3

x+3 2

Step 3: A pply the transposition

method to find the value of x. 16 – 9 = 3 x – 2 x

2x + 16 = 3x + 9

x =7

2( x + 8) = 3( x + 3) Example 3

2 x + 7 5x + 4 = 3 5 5(2x + 7) = 3(5x + 4) Solve for x:

Example 4

Solve:

10x + 35 = 15x + 12

22 x − 143 − 5 x + 15 = x −9+ 5 11

35 – 12 = 15x – 10x 23 = 5x x=

Do It Together

(2x -- 13) x -- 3 x -- 9 = +1 − 5 11 5 ( x − 9) + 5 11(2 x − 13) − 5( x − 3) = 5 × 11 5

17 x − 128 = x −4 11 17x − 128 = 11x − 44 ⇒ 17x − 11x = 128 − 44

23 5

6x = 84 ⇒ x = 14

3x + 5 1 = 2x + 1 3 3(3x + 5) = 1(2x + 1) Solve for m:

_________________________________________________________________________________________________________________________ _________________________________________________________________________________________________________________________

Do It Yourself 2B 1 Write YES if the correct solutions are given for the linear equations, else write NO. a x = 2 for

5x = 3 2x − 1

2 Solve. a d

5x − 4 7

8 9

6 9 = 3n + 1 5n + 3

b x = 6 for

b e

z + 5 6

−

(4x + 1) 3

7 5 = x +1 ( x − 1)

z +1 9

c x = 3 for

z + 3 = 4

(2 x − 1) 2

−

(3 x − 7) 5

7x + 1 3

y − (4 − 3 y )

2 y − (3 − y )

5x + 6 4

1 4

6 =

Maths Grade 8 Chapter 1-6.indb 24

08-01-2024 14:45:22

3 Find the value of x using the cross-multiplication method. a d

3x − 8

= 6

5(2 − x ) − 4(1 + x ) 2 − x

5 8

4 Answer the given questions. a Solve: b Solve:

3x − 1 2

4x + 6 3

5 7 = x x + 2

6 9 = 3m + 1 5m + 3

0.4x + 5 8

2.5 x + 4 15

2 3 43 + = 2. and find the value of y if = x y 6

2 x 4x − 5 3 x + 5 + = 11 . and find the value of y if = 3 y 11 10

The numerator of a fraction is 3 less than the denominator. If the numerator and the denominator both are decreased by 2 then the simplest form of the fraction becomes

3 . What is the fraction? 4

Two angles are in the ratio 1:2. When 15 is added to the smaller angle and 15 is subtracted from the bigger angle,

Solve the values of the variables to decode the postal code: 2abc5d.

the new ratio of the angles becomes 1:1. Find the angles.

a −1

7a − 27

1 3

2b + 6 4

7b − 34 3

11c − 44 5

3c + 17 4

9d − 7

7d + 3

29 31

What is the postal code?

Word Problem 1

The present ages of Abhijeet and Sahil are in the ratio 2:1. Four years ago, the ratio of their ages was 5:2. Find the present ages of Abhijeet and Sahil.

Application of Linear Equations Remember, Lata and her friends sold hockey tickets. Jaya sold 54 tickets in total. If she earned a total of ₹58,800 and sold two types of tickets, one for ₹1000 and the other for ₹1200, then how many tickets of each type did she sell? Linear equations help us in solving real-world problems. We need to set up an equation according to the condition given and solve it to find the value of the unknown. This method consists of two steps: 1

Translating the word problem into symbolic language

Solving the equation

Let us solve a few real-life problems. Let the tickets sold for ₹1000 be x. Then the tickets sold for ₹1200 will be 54 – x.

Chapter 2 • Solving Equations in One Variable

Maths Grade 8 Chapter 1-6.indb 25

08-01-2024 14:45:23

According to the question, 1000x + 1200(54 − x) = 58800 ⇒ 1000x + 64800 − 1200x = 58800 200x = 64800 − 58800 ⇒ 200x = 6000 x = 30 ⇒ 54 − x = 54 − 30 = 24 Thus, Jaya sells 30 tickets for ₹1000 and 24 tickets for ₹1200. Example 5

Suman’s present age is equal to one-fifth of her mother's age. Twenty-five years later, Suman’s age will be 4 years less than half the age of her mother. Find their present ages. y Let us assume the mother’s present age to be y years; Suman’s present age = 5 According to the condition, y + 25 = 5

y + 25 −4 2

Multiply the left hand side by 2 and right hand side by 5. 2y + 250 = 5(y + 25) − 40 ⇒ 2y + 250 = 5y + 125 − 40 5y − 2y = 250 − 85 ⇒ 3y = 165 ⇒ y = 55 Suman’s present age =

y 55 = = 11 years 5 5

Thus, Suman’s present age is 11 years and her mother's age is 55 years. Example 6

A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers? Let us assume the first number to be x; second number = 5x According to question, 5x + 21 = 2(x + 21)

5x + 21 = 2x + 42 ⇒ 5x − 2x = 42 − 21 3x = 21 ⇒ x = 7 ⇒ 5x = 35

Thus, the numbers are 7 and 35. Example 7

If the sum of three consecutive multiples of 9 is 108, then find the second multiple. Let us assume the smallest multiple of 9 to be 9x. Next multiple = 9(x + 1); Last multiple = 9(x + 2) So, 9x + 9(x + 1) + 9(x + 2) = 108 27x + 27 = 108 ⇒ 27x = 81 ⇒ x = 81 ÷ 27 ⇒ x = 3 So, we can find, 9x = 9 × 3 = 27, 9(x + 1) = 9 × 4 = 36, 9(x + 2) = 9 × 5 = 45 Thus, the three consecutive multiples of 9 which add up to 108 are 27, 36, 45.

Example 8

Tanya takes a test in which there are 60 questions. For each correct answer, x marks will be awarded, and for each incorrect answer, one-fourth of a correct answerʼs marks will be deducted, and 0 marks will be awarded for questions not attempted. Riya attempts 45 questions out of which 29 are correct. What is the value of x if Riya is awarded 200 marks?

Maths Grade 8 Chapter 1-6.indb 26

08-01-2024 14:45:23

Marks awarded for each correct answer = x Marks deducted for each incorrect answer = Total number of questions attempted = 45

x 4

Did You Know?

Total number of correct answers = 29

Knowledge of algebraic expressions

Total number of incorrect answers = 45 – 29 = 16

days, as we are often estimating

Total marks scored = 200

can help us plan and schedule our and solving for unknown variables.

 x 200 So, (29 × x ) −  16 ×  = 4  29x − 4x = 200 ⇒ 25x = 200 x=

200 ⇒x=8 25

Thus, the value of x is 8 marks. Do It Together

A car rental company charges ₹500 per day plus ₹12 per km. If the bill is ₹3320 for a day, then how many km were driven? Let the distance travelled be x km. Fixed charges for a day = __________________ Rate of the car for each km = __________________ Total bill amount = __________________

Do It Yourself 2C 1

Form the equation for the cases given. a Thrice x added to 5 is equal to the difference of half of x and 13. b The sum of four times y and 11 is equal to the sum of two times y and 56. c The sum of three consecutive even numbers is 144. d A train travels from point A to point B, a distance of 240 miles. On the return trip, the train’s speed is increased by 20 mph and takes 2 hours less.

The sum of four consecutive odd numbers is 352. What are the numbers?

The sum of three consecutive multiples of 11 is 429. What are the numbers?

A number is decreased by 10% and the new number obtained is increased by 25%, so the result is 81. Find the

The angles of a triangle are in the ratio 3:4:5. What is the difference between the largest angle and the smallest

Divide 6000 into two parts so that 15% of one number is equal to 35% of the other number.

number.

angle?

Chapter 2 • Solving Equations in One Variable

Maths Grade 8 Chapter 1-6.indb 27

08-01-2024 14:45:23

Puneet’s age is three times the age of his daughter. If the sum of their ages is 52 years, find the difference of

The sum of two numbers is 45 and the numbers are in the ratio 2:1. What are the numbers?

The total cost of a table and a chair is ₹15,550. The cost of the table is ₹550 more than the cost of the chair. What

their ages.

is the cost of the table?

10 Virat scores 20 more runs than twice the number of runs scored by Rohit. Together, their total is four runs short of a triple century. What are the individual scores of Virat and Rohit, respectively?

11 Amit has a certain number of pencils that cost ₹4 per pencil. He again purchased 10 more than twice the number of pencils he had. If the total cost of all the pencils is ₹280, how many pencils did Amit have initially?

12 Animesh left one-half of his property to his daughter, one-third of his property to his son and donated the rest of his property. If the donation was worth ₹5,00,000 then how much did he give to his daughter and son?

13 The ones digit of a two digit number is 0.4 times the tens digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 77. What is the original number?

14 The speed of the flow of a river is 3 km per hour. A boat goes upstream between two points in 5 hours while it covers the same distance going downstream in 3 hours. What is the speed of the boat in still water?

15 A sum of ₹870 is made up of denominations of ₹5 and ₹10. If the total number of notes is 92 then find the number of notes of each denomination.

16 Two cars simultaneously start from A and B in opposite directions and the distance between them after 4 hours is 60 km. What is the speed of each car?

's' km per h

('s' + 10) km per h A

60 km

300 km

Points to Remember • An equation is a mathematical statement with an equal symbol between two expressions. • Any operation done on one side of an equation must be done on the other side of the equation so that the equation remains true.

• We use the transposition method to collect similar terms on one side. Change the signs of terms while using the transposition method and moving terms from one side to the other side. • While solving the word problems, identify the unknown terms, form an equation and solve the value of the unknown term. • The following rules can be applied while solving linear equations:

• The same number can be added or subtracted from both sides of the equation. • The same number can be multiplied or divided on both sides of the equation.

• A term can be transposed to the other side of the equation with its sign changed.

Maths Grade 8 Chapter 1-6.indb 28

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Math Lab Aim: Solve linear equations in one variable Materials Required: Chalk, stopwatch and equation cards Setting: In groups of 4 Method: 1

Prepare equation cards (unsolved equations on cards (should have linear equations)).

Mark the starting line and finishing line in the classroom or in an open area. There 2 could be different laps. Each team lines up behind the starting line and the first person of each team is given an 3 equation card. The student solves the equation and runs towards the finishing line with the correct 4 solution. Once they reach the finishing line, they hand the card to the next teammate. The next 5 person solves the next equation and continues the relay. 6

Use a stopwatch to calculate the total time taken by each team to complete the race.

The team which has the fastest completion time wins the relay race.

Chapter Checkup 1

Find the value of x. a

Solve using the balancing method. a 4x + 14 = 56

b 3x + 4 = x + 18

c 5x – 7 = 2x +11

d 3x – 4 =1 – 2x

e 5x + 48 = 3(4x – 5)

11(x + 1) = 12(x – 1)

Solve using the transposition method. a 5(x + 3) = 3(1.5x + 18)

b 6(2x + 11) = 8(2x – 1)

c 8x – 7 – 3x = 6x – 2x – 3

d 10x – 5 – 7x = 5x + 15

e 5(x – 1) = 2(x + 8)

4x – 3 = (3x + 1) + (5x – 4)

Which of these is the correct solution of the equation? a 5x + 7 = 35 for x = 7

b 3(x – 2) = 4(x + 3) for x = 18

d 4(x + 1) – 3(2x – 18) = 2x – 2 for x = 15

Chapter 2 • Solving Equations in One Variable

Maths Grade 8 Chapter 1-6.indb 29

c 4x + 3 – 2(x + 1 ) = 8x for x = e 5(x + 9) – 2x = 90 for x = 15

1 6

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Find the value of x using the cross-multiplication method. a

3x 9 = 2( x + 2) 10

4( x + 2) + 3 27 = 2 x + 3( x − 1) 17

4( x + 3) − 4 3x + 1 = 5 8

11(8 − x ) − 3 7( x − 3) + 2 = 7

3m + 2 3 = 9m + 4 5

5 z + 14 =3 2z

7 3 5w + = w − 14 2 2

3 b+7 5 4 = 2 4 b−4 5

11(2 x + 1) 11

4( x + 6) 26

= 6 x − 4 − 3x

Solve the equations. a

2 y − 1 17 = y + 3 12

2 5 1 − = 5a 3a 15

Answer the questions. a If= F

9 C + 32, then find C when F = –40. 5

b Solve

x + y 3. 9 x + 4 4(2 x − 1) = = and then find the value of y if 8 xy 10 7

c Solve

2 5 2x + 1 x − 2 3 + 2x = 4. − = and then find the value of m if + x m 10 6 15

Find the value of the unknown variables to decode the phone number. Use the letters in the subparts to add the values. 8ab00c1d2e

a 9(a + 1) − 4(a + 2) = 6 d

5 x + 11

2d + 7 1 = 3(7d − 8) 4

b 11(b − 2) = 2(b + 1) + 3 e

11e − 3(2e − 5) 3

(7e − 21)

2(c − 2) 1 = 3(3c − 5) 3

Place the correct mathematical operations (+, –, ×, ÷) so that the equations are true. a 2x __ 3 = 7 for x = 2 c

2x __ 3(x __ 1)

15x __ 2(2x __ 4)

18 for x = 3 25

b 3(x __ 1) __ 4x = 46 for x = 7 d

3x __ 4(x __ 3) x __ 2(3x __ 7) = for x = 4 20 7

10 Two numbers are in the ratio 4:9 and the difference of the numbers is 250. What is the sum of the numbers? 1 1 1 from a number and multiply the result by then you get . What is the number? 11 If you subtract 2

12 Three consecutive even numbers add up to 696. What are the numbers? 13 The numerator of a fraction is less than its denominator by 8. If both the numerator and the denominator are increased by 17 then the fraction obtained is

5 in its simplest form. Find the original fraction. 6

Word Problems 1 The ages of Kushagra and Kush are in the ratio 5:6. Five years later the sum of their ages will be 43 years. What is the age difference between Kushagra and Kush?

2 The present age of Shagun is one-fifth of her mother’s age. After twenty-five years, her age will be 4 less than half of her mother’s age. Find the difference of their present ages.

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3 The perimeter of a rectangular swimming pool is 170 m. If the length of the pool is 4 m more than the breadth of the pool, what is the area of the pool?

4 Sheetal has currency notes in denominations of ₹50, ₹100 and ₹500. The ratio of the number of notes is 12:8:7 respectively. If the total amount of money with Sheetal is ₹24,500 how many notes of each denomination does she have?

5 There is a rectangular plot for a school. The length and breadth of the plot are in the ratio of 15:7. At the rate of ₹150 per metre, it will cost ₹1,32,000 to fence the plot. What are the dimensions of the plot?

6 Vikram took goats to the field. Half of the goats were grazing. Three-fourths of the

remaining goats were playing nearby. The remaining 15 were drinking water from the pond. How many goats were there in total?

7 The ratio of the speed of a boat to the stream is 7:2. The boat takes 6 hours more travelling upstream than downstream. What is the time taken by the boat for the entire journey?

8 The angles of a quadrilateral are in the ratio 7:17:19:29. What is the measure of the angles? 9 A book has 400 pages. Raj finishes writing 100 pages in x minutes. Raju takes twice the time taken by Raj to write the next 150 pages. Vivek takes 2 hours more than half the time taken

by Raju to write the remaining pages. If the total time taken to write the book was 33 hours and 20 minutes then what is the value of x?

10 ₹x is to be divided among three friends A, B, and C. The share of A is two-fifths of the total

amount, the share of B is two-thirds of the remaining amount and C’s share is ₹600. What is the value of x?

11 A father is 20 years older than his son. 12 years ago, the age of the father was six times that of his son. Find their present ages.

12 The age of the father is three times the age of his son. After five years, the age of the father will be 2

1 times the age of his son. Find the present ages of the father and son. 2

Chapter 2 • Solving Equations in One Variable

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3 3 Polygons Let's Recall

Shapes can be made with lines, curves or both lines and curves. A curve is a line that is not straight. It can be bent and twisted in any direction. Curves can be open, closed, simple or non-simple. Open Curve

Closed Curve

Simple Curve

Non-simple Curve

Shapes made with only line segments are called polygons. Let us see some polygons and non-polygons: Polygons: Closed

Non-polygons: Open

The region inside a closed shape is called the interior. The region outside the closed shape is called the exterior. Exterior

Letʼs Warm-up

Interior

Name the shapes as open curves, closed curves, polygons, non-simple curves. 1

__________________ 3

__________________

__________________ 4

__________________ I scored _________ out of 4.

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Understanding Polygons and their Properties Real Life Connect

Mr. Singh owns a tile shop. He sells tiles in a lot of different shapes like triangles, rectangles, hexagons etc.

We know that these are called 2-D shapes. Some two-dimensional shapes are made up of straight lines and connected at their endpoints. These are called polygons. Now, imagine if all of these shapes were made of curves. Could we tile them? A polygon is a simple closed curve made up of 3 or more sides, with no overlapping or self-intersecting parts. Polygons

Remember! In a polygon, the number of sides is always equal to the number of vertices.

Non Polygons

Did You Know? The term “polygon” comes from the Greek words “poly,” meaning many, and “gonia,” meaning angles.

Think and Tell In the given polygon ABCDE, AB, BC, CD, DE and EA are the sides. A, B, C, D and E are the vertices. ∠A, ∠B, ∠C, ∠D and ∠E are the angles. Let us see and learn some more terms.

Chapter 3 • Polygons

Maths Grade 8 Chapter 1-6.indb 33

Which polygon do you see most often in your everyday life?

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Diagonals are line segments joining opposite vertices of a polygon. E.g. AC, AD C

Interior of a Polygon is the region enclosed by the sides of the polygon.

Adjacent Sides are sides that share a common vertex (corner). They are next to each other in the sequence of sides. E.g. (AB-BC, BC-CD, CD-DE, DE-EA)

Exterior of a Polygon is the region outside the sides of the polygon.

Polygons can be convex or concave

Adjacent Angles are angles that share a common side (arm). They are next to each other in the sequence of angles. E.g. (∠A-∠B, ∠C-∠D, ∠D-∠E, ∠E-∠A)

Convex Polygon: In these polygons, the measure of each angle is less than 180º. Concave Polygons: In these polygons, the measure of one of the angles is more than 180º.

Polygons can be regular or irregular

Regular Polygon: In these polygons, the lengths of all the sides are equal. Irregular Polygon: In these polygons, the lengths of all the sides are not equal.

Classifying Polygons Polygons can be classified on the basis of their number of sides and vertices. Triangle - 3 sides

Quadrilateral - 4 sides

Pentagon - 5 sides

Hexagon - 6 sides

Heptagon - 7 sides

Octagon - 8 sides

Nonagon - 9 sides

Decagon - 10 sides

We know that a diagonal of a polygon is a line segment that joins two vertices of the polygon, which are not already joined by the adjacent sides of the polygon.

Remember! A polygon has to have at least 3 sides.

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Let us draw the diagonals in some polygons.

Polygon Triangle

Pentagon

Octagon

Decagon

Number of Sides (n)

Number of Diagonals

n (n − 3) 2

3 (3 − 3) =0 2

5 (5 − 3) =5 2

8 (8 − 3) = 20 2

10 (10 − 3) = 35 2

If a polygon has n number of sides, then the number of diagonals is n (n − 3) . 2 In a convex polygon, all the diagonals lie inside the polygon, whereas in a concave polygon, at least one diagonal lies outside the polygon. Diagonals in a

concave polygon Diagonals in a

convex polygon Example 1

Find the number of diagonals in a nonagon. Number of sides in a nonagon (n) = 9

Think and Tell

Number of diagonals in a nonagon =

polygon with three sides?

n (n − 3) 9 (9 − 3) = 27 = 2 2 Therefore, a nonagon has 27 diagonals. Example 2

Identify the given polygons as concave or convex. 1

Concave Polygon

Convex Polygon Do It Together

Is it possible to have a concave

Draw the diagonals in the given polygons. Find the number of diagonals using the formula. Identify the polygon as convex or concave.

Polygon

Number of sides (n) Diagonals

5 9

Convex/Concave

Chapter 3 • Polygons

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Do It Yourself 3A 1 Fill in the blanks. a A polygon with 7 sides is called a __________________. b The number of diagonals in a hexagon is __________________. c A polygon with all sides and angles equal is called a __________________. d A polygon with 10 sides is called a __________________. e A regular polygon is a 2-D shape where the sides are all __________________ line segments of __________________.

2 Write True or False. a A triangle is a convex polygon. b The number of diagonals in an octagon is 7. c A circle is an example of a 2-D geometrical shape that is not a polygon. d A rectangle is a regular polygon.

3 Classify the following as convex or concave polygons based on the angles of the polygons. a 65º, 115º, 111º, 70º, 179º

b 176º, 105º, 110º, 66º, 195º, 165º, 83º

4 Classify the polygons as concave or convex. a

5 A polygon has 6 sides and all the angles in the polygon measure 120°. Is this a convex or concave polygon? 6 Draw a hexagon and draw all the diagonals in the polygon. 7 Use the formula to find the number of diagonals in polygons with the number of sides given. Name the polygon. a 4 sides

b 8 sides

c 10 sides

8 In a regular octagon, how many diagonals can be drawn from one vertex? 9 How many non-overlapping triangles can we make in a polygon having 5 sides by joining the vertices?

Word Problem 1

Rajan drew a regular polygon with 7 sides. Name the polygon. He then shifted one of the vertices inside the polygon. How many diagonals does this polygon have? Identify the polygon as concave or convex.

Maths Grade 8 Chapter 1-6.indb 36

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Angle Sum Property of Polygons

Exterior Angle

We know that an angle is formed when two straight lines meet at a common vertex.

• Any angle that is formed in the polygon is called an interior angle. • The angle between a side of a polygon and an extended adjacent side that is

Interior Angle

formed outside the polygon is called an exterior angle.

The sum of all the interior angles of a triangle is 180°. What about polygons with four or more sides? Since a triangle has three sides, we find the measurements of the angles accordingly. Let us find the sum of the angles of any polygon. Step 1: Mark one of the vertices in the polygon. Step 2: Draw diagonals from that vertex to form triangles. Count the number of triangles. Step 3: Since the sum of the angles in a triangle is 180°, multiply 180° by the number of triangles formed inside each polygon.

1 Pentagon

Quadrilateral

Hexagon

Heptagon

Number of sides

4 sides

5 sides

6 sides

7 sides

Number of Triangles

180° × 2 = 360°

180° × 3 = 540°

180° × 4 = 720°

180° × 5 = 900°

From this we can deduce that the sum of the interior angles of a polygon is the product of 2 less than the number of sides of a polygon and 180°. This can be written as: Sum of interior angles of a polygon = (n – 2) × 180° where n is the number of sides of the polygon. The angle sum property doesn’t depend on whether the polygon is a concave polygon or a convex polygon. Let us find the measure of the missing exterior angle in the polygon. The shape has 6 sides and is therefore a hexagon.

126°

Sum of angles in a hexagon = (6 - 2) × 180° = 720°

126°

Let the missing angle be x.

106°

720° = 126° + 126° + 106° + 129° + 147° + x

720° = 634° + x

147°

So, x = 720° - 634° = 86° Example 3

Find the sum of the interior angles in a convex octagon using the angle sum property. Number of sides in a convex octagon n = 8 Using the angle sum property, Sum of interior angles = (8 - 2) × 180° = 1080°

Chapter 3 • Polygons

Chapter 3.indd 37

Example 4

129°

Find the interior angle sum in a nonagon. Number of sides in a convex octagon n = 9 Using the angle sum property, S um of interior angles = (9 - 2) × 180° = 7 × 90° = 1260°

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Do It Together

Find the sum of the angles in the figure. Find the measure of the missing angle. The shape has 7 sides so it is a _____________. Sum of angles of the _____________ = (n - 2) × 180 Measure of the missing angle is _____________.

138° 143°

67°

152° 108°

? 134°

Do It Yourself 3B 1 Fill in the blanks. a A quadrilateral has _______ as the sum of its interior angles. b The sum of the interior angles of a polygon of n sides is _______ right angles. c Each interior angle of a regular hexagon is _______. d If the number of sides of a regular polygon is _______, then the interior angle sum is 1800°.

2 Write if true or false. a The sum of the interior angles of a triangle is 270°. b The number of triangles that can be formed within a polygon with n sides is n. c The maximum measure of the interior angle for a regular polygon is 180°. d It is possible to have a polygon whose sum of interior angles is 540°. e If each interior angle of a regular polygon measures 168°, the polygon has 30 sides.

3 Find the interior angle sum of a polygon with: a 7 sides

b 13 sides

c 17 sides

d 20 sides

e 22 sides

4 Is it possible to construct a polygon, the sum of whose interior angles is 20 right angles? If yes, find the number of sides of the polygon.

5 If the interior angle sum of a regular polygon is 2520° then find the measure of each angle and the total number of sides in the polygon.

Word Problem 1

To manufacture a tile, the worker needs to determine the measure of each interior angle of the tile. What is the measure of each interior angle if the tile is a regular octagon?

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Exterior Angle Sum Property of Polygons The angle between a side of a polygon and an extended adjacent side that is formed outside the polygon is called an exterior angle.

The sum of the exterior angles in any polygon is always a constant, i.e. 360°. Interior angle + Exterior angle = 180°.

The angle formed on a

∴ Exterior angle = 180° - Interior angle

Let us calculate the measure of exterior angles to see what the total is. We see that the sum of the exterior angles in any polygon is 360°. 53°

38°

106°

71°

88°

Remember!

60°

59°

50°

straight line is always 180°. 47°

111° 37°

Sum of exterior angles of a hexagon

60° + 71° + 53° + 88° + 38° + 50° = 360°

111° + 47° + 106° + 59° + 37° = 360°

Similarly, if a polygon has n number of sides, then the measure of each exterior angle of a regular 360º . n Find the measure of each exterior angle in the figure. polygon = Example 5

We know that the exterior angle sum = 360°

(3x − 5)°

→ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°

→ (2x - 1)° + (3x - 5)° + (8x + 3)° + (7x - 2)° + (4x + 1)° + (6x + 4)° = 360° → 30x = 360° → x = 12°

(2x − 1)° F (6x + 4)°

(8x + 3)°

∠A = (2x - 1)° = 23°; ∠B = (3x - 5)° = 31°;

∠C = (8x + 3)° = 99°; ∠D = (7x - 2)° = 82°;

(7x − 2)° D

∠E = (4x + 1)° = 49°; ∠F = (6x + 4)° = 76° Example 6

E (4x + 1)°

The exterior angle sum of a 30-sided regular polygon is 360°. Find the measure of each angle. Number of sides = n = 30

Do It Together

a 34°

122°

34°

71°

Error Alert! The sum of an interior angle and the adjacent exterior angle can never be greater than 180°.

b 81°

∠a + 122° = 180°. So, ∠a = 180° - 122° = 58° ∠a = 58°; ∠b = _____, ∠c = _____, ∠d = _____ Chapter 3 • Polygons

Maths Grade 8 Chapter 1-6.indb 39

115°

90°

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Do It Yourself 3C 1 Fill in the blanks. a The polygon in which the sum of all exterior angles is equal to the sum of interior angles is called a ______. b The measure of each exterior angle of a regular polygon is ______. c The measure of each exterior angle of a regular polygon of 18 sides is ______. d In case of a regular polygon that has 8 sides, each interior angle is ______ and each exterior angle is ______.

1 8

th of a right angle.

3 Find the measure of each exterior angle of a regular polygon with: a 3 sides

b 5 sides

c 8 sides

d 12 sides

e 24 sides

4 Find the number of sides of a regular polygon if each of its exterior angles measures: a 40°

b 45°

c 60°

d 72°

e 90°

5 The measures of the exterior angles of a hexagon are (3x – 4)°, (x + 4)°, (7x – 3)°, (8x – 1)°, (2x + 3)°, (9x + 1)°. Find the measure of each angle.

Word Problem 1

A carpenter is building a new table. The table is in the shape of a regular heptagon. What is the measure of each exterior angle of the tabletop.

Points to Remember • A polygon is a simple closed curve made up of straight line segments, with no overlapping or self-intersecting parts. • The polygons can be classified into different types on the basis of their number of sides and vertices.

• A diagonal of a polygon is a line segment that joins two vertices of the polygon, which are not already joined by the adjacent sides of the polygon.

• In a convex polygon, all the diagonals lie inside the polygon, whereas in a concave polygon, at least one diagonal lies outside the polygon. •

The number of diagonals in a polygon with n number of sides = n(n - 3) ÷ 2.

•

The sum of the exterior angles in any polygon is always a constant, i.e., 360°.

•

If a polygon n has number of sides, then the sum of all the interior angles is (n - 2) × 180°. If a polygon has n sides, the measure of each exterior angle of a regular polygon = 360° ÷ n.

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Math Lab Exploring Polygons Setting: Individual Materials Required: Pen and paper Method: 1

Each student will explore the classroom and write the names of 5 polygons they spot.

They will then identify the polygon as concave or convex.

The student who completes the task first, wins.

Chapter Checkup 1 Fill in the blanks. a A polygon with unequal sides and unequal angles is called an _________. b The sum of the interior angles of a hexagon is ________. c A polygon that has at least one interior angle more than 180° is called a _________ polygon. d The number of diagonals in a pentagon is ________. e The sum of the exterior angles of any polygon is always ________.

2 How many diagonals will each of these polygons have? a Convex quadrilateral

b Triangle

c Polygon with 17 sides

3 Identify the polygons as concave or convex. a

4 Find the interior angle sum of the polygons with the following number of sides: a 9

b 13

c 16

d 21

e 33

5 Find the number of sides of the polygons whose interior angle sum is: a 1800°

b 2700°

c 3240°

d 3780°

e 4500°

6 Find each exterior angle of the regular polygons with the given number of sides. a 9

b 15

c 20

d 36

e 45

7 Find the number of sides of a regular polygon whose exterior angle measures: a 40°

b 45°

c 72°

d 90°

e 120°

If two adjacent angles of a parallelogram are (5x - 5)° and (10x + 35)° then find the ratio of these angles.

The angles of a pentagon measure x°, (x - 5)°, (x + 15)°, (3x - 44)° and (x - 70)°. Find x.

Chapter 3 • Polygons

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10 The angles of a quadrilateral are in the ratio 1:2:3:4. Find the difference between the largest and the smallest angle.

11 The angles of a hexagon are in the ratio 1:2:3:4:6:8. Find the measure of the smallest and the biggest angles. 12 The measure of four angles of a heptagon is the same and each of the other three angles measures 120°. Find the measure of the unknown angles.

13 Is it possible to have a regular polygon of which the exterior angle is 50°? 14 Each interior angle of a polygon is 5 times the exterior angle of the polygon. Find the number of sides. 15 Find the total number of diagonals if the interior angle sum of a polygon is 1800°. 16 The sum of all the interior angles of a polygon is 4 times the sum of its exterior angles. Find the number of sides in the polygon. Also, find the measure of each exterior angle and each interior angle.

17 Find the minimum interior angles and maximum exterior angles possible in a regular polygon. 18 The ratio between an exterior angle and the interior angle of a regular polygon is 1:2. Find the: a measure of each exterior angle

b measure of each interior angle.

c number of sides in the polygon.

Word Problems 1

spider builds its web in the shape of a regular hexagon. How many diagonals does the A spider have in its web? If it builds a web with an octagon, how many diagonals would there be?

arika drew a polygon where all the sides and angles were equal. Is the polygon regular or S irregular?

ohan is designing a logo for his company which is in the shape of a regular 12-sided S polygon. How many diagonals connect the interior angles of the polygon? How many triangles can be formed by connecting these interior angles?

ama works in a factory that produces footballs. She needs to calculate the angle measure R of the regular pentagons on the soccer ball. What is the measure of each interior angle of the regular pentagons?

akshi drew a triangle in which the second angle is double the first angle and the third angle S is 40°. Find all three angles of the triangle that Sakshi made.

In a city with several towns connected by bridges, there are 10 towns. If each town is connected to every other town by a direct bridge, how many bridges are needed in total? How many diagonals does this network of bridges have?

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4 Quadrilaterals

Let's Recall

Quadrilaterals are geometric shapes with 4 sides, 4 corners or vertices and 4 angles. They can be seen in books, roofs of buildings, earrings, etc.

Quadrilaterals come in various forms, each with its unique properties. To understand quadrilaterals better, let's explore some key terms. Diagonals

Opposite Angles

Adjacent Sides

These are the line segments joining opposite vertices of a quadrilateral.

These are the angles in a quadrilateral with no common arm.

E.g. – RT, US

E.g. – (∠T, ∠R ), (∠U, ∠S)

Adjacent sides of a quadrilateral are sides that share a common vertex (corner). They are next to each other in the sequence of sides.

Opposite Sides Opposite sides of a quadrilateral are sides that do not share a common vertex. They are located on opposite sides of the quadrilateral. E.g. – (RS, UT), (ST, RU)

E.g. – (RU, UT), (SR, ST), etc. Adjacent Angles S

Adjacent angles of a quadrilateral are angles that share a common side. E.g. – (∠R, ∠U ), (∠U, ∠T), (∠T, ∠S), (∠S, ∠R)

Letʼs Warm-up Match the columns based on the quadrilateral PQRS given below. Column A

Column B

1 Vertices

PS, QR

2 Sides

∠Q, ∠S

3 Adjacent Sides

P, Q, R, S

4 Opposite sides

PS, SR

5 Opposite Angles

PS, SR, RQ, QP

P S

I scored _________ out of 5.

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Understanding Quadrilaterals and their Properties Real Life Connect

Once upon a time, in a quiet neighbourhood, Max lived with his uncle, Charlie. One day, they found a colourful kite soaring in the park. Charlie explained that the kite was shaped like a quadrilateral. The kite’s tail was a rectangle, a type of quadrilateral. Max was thrilled. He realised that shapes were all around him, even in a quiet neighbourhood. Max couldn’t wait for more adventures with Uncle Charlie, discovering quadrilaterals everywhere they went.

Classifying Quadrilaterals There are different kinds of quadrilaterals depending on their properties, like the kite being flown by Max that was a different quadrilateral than the tail of the kite, which was a rectangular strip. Let’s explore some more information about quadrilaterals.

Properties of Quadrilaterals In this section, we will see some properties of quadrilaterals and understand regions in and around quadrilaterals.

Interior of a Quadrilateral The interior of a quadrilateral refers to the region enclosed by the four sides of the quadrilateral.

Exterior of a quadrilateral Interior of a quadrilateral

Exterior of a quadrilateral

Exterior of a Quadrilateral The exterior of a quadrilateral refers to the region outside the four sides of the quadrilateral.

Quadrilaterals can be convex or concave B

Convex Quadrilaterals In these quadrilaterals, the measure of each angle is less than 180°. D

C Q

Concave Quadrilaterals R P

In these quadrilaterals, the measure of one of the angles is more than 180°. Here, ∠R is greater than 180°. S

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Let’s say, we have 2 quadrilaterals with the interior angle measures as given. Identify the type of each quadrilateral. 1 Quadrilateral ABCD with ∠A = 45°, ∠B = 105°, ∠C = 95°, ∠D = 115°. Since none of the angles measures more than 180°, ABCD is a convex quadrilateral. 2 Quadrilateral PQRS with ∠P = 34°, ∠Q = 201°, ∠R = 71°, ∠S = 54°. ∠Q = 201° which is greater than 180°. So, PQRS is a concave quadrilateral. Example 1

Among the following angles, which one cannot be an interior angle in a convex quadrilateral? 1 73°

2 85°

3 173°

4 211°

In a convex quadrilateral, no angle can exceed 180°. Therefore, the 211° angle cannot be one of the angles in a convex quadrilateral. Example 2

Which of the following angles can be one of the interior angles in a concave quadrilateral? 1 33°

2 119°

3 183°

4 Any of the 3 angles given in a, b and c

In a concave quadrilateral, one of the angles can be more than 180°. Therefore, the correct option is d since any of the 3 angles can be interior angles in a concave quadrilateral. Example 3

Place the points J, K, L, M, N and O given below in the quadrilaterals so that: Only points J, L and M are in the interior.

Only points K, L and O are in the interior.

Only points L, M and N are in the interior.

J L

N Do It Together

Classify the following as convex or concave quadrilaterals based on the angles of the quadrilaterals. Angles

150o, 100o, 40o, 70o

Type of Quadrilateral

75o, 105o, 101o, 79o

216o, 36o, 40o, 68o

Convex

Trapezium and Kite Trapezium: A quadrilateral with exactly one pair of opposite sides parallel is a trapezium. Let us see trapezium shapes in real life.

Did You Know? The word “trapezium” comes from the Greek “trapezion,” meaning “a little table.” It was originally used to describe a quadrilateral with no parallel sides.

Chapter 4 • Quadrilaterals

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Properties of a trapezium

Given below are the properties of a trapezium. • • •

One pair of opposite sides is parallel and the other pair of opposite sides is non-parallel. Here, AB ǁ DC. When the non-parallel sides in a trapezium are equal, it is known as an isosceles trapezium. PS = QR. The diagonals in an isosceles trapezium are equal. PR = QS. A

Supplementary angles in trapezium ABCD are ∠A + ∠D = 180°

∠B + ∠C = 180°

Kite A kite is a quadrilateral with two pairs of consecutive sides of equal length.

Think and Tell How many pairs of opposite angles in a kite are equal?

Example 4

•

AC ⊥ BD

•

Diagonal BD bisects the angle ∠B and ∠D

•

∠A = ∠C.

•

Only diagonal BD bisects the diagonal AC and not vice-versa.

•

Diagonal BD divides the kite into two congruent halves.

Look at the kite. Find ∠CAD, ∠ADC and ∠ABC. B

In ΔABC, ∠BAC = ∠BCA = 50° (Angles opposite to equal sides) ∠BAC + ∠BAC + ∠ABC = 180° (Angle sum property of the triangle)

50° + 50° + ∠ABC = 180° ⇒ ∠ABC = 180° - 100° = 80°

50°

Also, ∠ADB = ∠CDB = 27° (BD bisects ∠D) Hence, ∠ADC = ∠ADB + ∠CDB = 27° + 27° = 54°

27°

In Δ ADC, ∠ACD = ∠CAD (Angles opposite to equal sides) ∠ACD +∠CAD +∠ADC = 180° (Angle sum property of the triangle) 2∠CAD + 54° = 180°

2∠CAD = 180° - 54° = 126°. So, ∠CAD = 63°

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Example 5

PQRS is a trapezium where PQ ǁ SR. Find the values of x and y. P

Do It Together

107°

126°

(2y + 3)°

(4x − 2)°

From the figure,

Similarly,

∠P + ∠S = 180° (supplementary angles)

∠Q + ∠R = 180° (supplementary angles)

2y + 3 + 107 = 180

4x - 2 + 126 = 180

2y = 180 - 110 = 70 70 y= = 35 2

4x = 180 - 124 = 56 56 x= = 14 4

Find the measure of ∠A and ∠C in the figure given below. B A

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

∠DAC = ∠ACD (________________________) ∠DAC + ∠ACD + ________ = _____

∠ABC = ∠BCA = ∠CAB (angles of equilateral triangle) So, ∠ABC = ∠BCA = ∠CAB = _________

54° D

Parallelograms A parallelogram is a quadrilateral with opposite sides that are both parallel and equal in length. We see parallelograms in roofs, solar panels and many other places. Can you think of 2 other places where you see parallelograms? Properties of a parallelogram •

Opposite angles of a parallelogram are equal. Here, ∠P = ∠R;

∠S = ∠Q. • Opposite sides of a parallelogram are parallel and equal. Here, PQ = SR, PS = QR; PQ ǁ SR; PS ǁ QR. • Diagonals of a parallelogram bisect each other. Here, the diagonals PR and SQ bisect each other at point O.

Let us see some proofs of parallelograms. 1 I f a pair of opposite sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram. Let PQRS be a quadrilateral such that PQ = SR and PQ ǁ SR. Consider ΔPQR and ΔRSP, then P PQ = SR (Given) S PR = RP (Common side) ∠QPR = ∠SRP (Alternate angles as PQ ǁ RS) Hence, ΔPQR ≅ ΔRSP (SAS Congruency rule) Therefore, ∠PRQ = ∠RPS (Corresponding parts of congruent triangles) Q Since, ∠PRQ and ∠RPS are alternate angles, PS ǁ QR R As the opposite sides have been shown to be parallel, we can conclude that PQRS is a parallelogram. Chapter 4 • Quadrilaterals

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2 Any two adjacent angles of a parallelogram are supplementary. Let PQRS be a parallelogram. Consider PS ǁ QR and PQ as their transversal. Then, ∠SPQ and ∠RQP are supplementary as they are the interior angles on the same side of the transversal PQ.

Similarly, consider PS ǁ QR and SR as their transversal. ∠PSR and ∠QRS are supplementary as they are the interior angles on the same side of the transversal SR.

Extending the same reasoning it can be shown that ∠QPS & ∠RSP and ∠SRQ & ∠PQR are also supplementary considering PQ ǁ RS along with PS and QR as the transversals respectively. 3 Diagonals of a parallelogram bisect each other. Let PQRS be a parallelogram whose diagonals QS and PR intersect at T. Consider ΔPTQ and ΔRTS, then PQ = SR

(Opposite sides of a parallelogram)

∠TQP = ∠TSR

(Alternate angles as PQ ǁ RS)

∠PTQ = ∠RTS

(Vertically opposite angles)

Hence, ΔPTQ ≅ ΔRTS

(AAS criteria of congruency)

Therefore, PT = TR and QT = ST (Corresponding parts of congruent triangles) 4 Opposite sides and angles of the parallelogram are equal. Let PQRS be a parallelogram with diagonal PR shown.

Consider ΔPQR and ΔRSP, then

a PR = RP (common side) b ∠QPR = ∠SRP

(Alternate angles as PQ ǁ RS)

(Alternate angles as PS ǁ QR)

∠QRP = ∠SPR

Hence, ΔPQR ≅ ΔRSP

(ASA axiom of congruency)

Therefore, ∠Q = ∠S, QR = PS and PQ = SR (Corresponding parts of congruent triangles) From (b) and (c), ∠QPR + ∠SPR = ∠SRP + ∠QRP ⇒ ∠P = ∠Q. Therefore, all the opposite sides and angles have been proven to be equal.

Rhombus, Square and Rectangle Rectangle

A rectangle is a quadrilateral with four right angles and opposite sides of equal length.

Error Alert! Diagonals of a parallelograms may or may not bisect angles through which they pass.

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Properties of a rectangle •

All the angles in a rectangle are right angles and equal.

Here, ∠A = ∠B = ∠C = ∠D = 90°

•

Here, AB ǁ DC; AD ǁ BC and AB = DC; AD = BC •

Opposite sides of a rectangle are parallel and equal. The diagonals are equal and bisect each other.

Rhombus A rhombus is a four-sided geometric shape with all sides of equal length and opposite angles of equal measure, but the angles are not necessarily right angles.

Properties of a Rhombus: • • • •

All the sides of a rhombus are equal. Here, PQ = QR = RS = SP

Q O

Opposite sides of a rhombus are parallel. Here, PQ ǁ SR; PS ǁ QR

Diagonals bisect each other at right angles. Here, PO = RO; OQ = OS Opposite angles of a rhombus are equal. Here, ∠P = ∠R; ∠S = ∠Q

Square

A square is a quadrilateral with equal sides and four right angles.

Properties of a Square: • • • •

Example 6

All the sides of a square are equal. Here, AB = BC = CD = DA

B O

All the angles in a square are right angles. ∠A = ∠B = ∠C = ∠D = 90° Opposite sides are parallel. Here, AB ǁ DC; AD ǁ BC.

iagonals are equal and bisect each other at right angles. AO = CO; D OD = OB.

Given that ABCD is a parallelogram. Find the values of x and y. We know that opposite sides in a parallelogram are equal. Hence, 2x + 3 = 35 cm

4y + 7 = 23 cm

2x = 35 - 3 = 32 32 x= = 16 cm 2

4 y = 23 – 7 = 16 16 y= = 4 cm 4

Chapter 4 • Quadrilaterals

Maths Grade 8 Chapter 1-6.indb 49

35 cm

D 23 cm

4y + 7 B

2x + 3

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Example 7

If the diagonals of a rhombus measure 10 cm and 24 cm, determine the length of one of its sides. Let PQRS be a rhombus where diagonals PR = 24 cm and QS = 10 cm. Since, the diagonals of a rhombus bisect each other at right angles OQ = 5 cm and OP = 12 cm. P

In ΔPOQ, we have OQ2 + OP2 = PQ2 (Pythagoras' Theorem) 52 + 122 = PQ2

25 + 144 = PQ2 PQ2 = 169. So, PQ = 13 cm Example 8

The diagonals of rectangle PQRS intersect at T. If ∠QTR = 44°°, find ∠TPS.

R P

∠QTR = ∠PTS = 44° (vertically opposite angles) Since the diagonals of a rectangle are equal and they bisect each other, PT = ST ⇒ ∠TPS = ∠TSP (Angles opposite to equal sides)

44°

∠TPS + ∠TSP + ∠PTS = 180° (Angle sum property of triangles) 2∠TPS + 44° = 180° 2∠TPS = 136° ∠TPS = Example 9

136° = 68° 2

If one of the sides of a parallelogram is 15 mm and its perimeter is 80 mm, find the lengths of the other sides. Since opposite sides in a parallelogram are equal, let the sides be 15 mm, x, 15 mm and x. 15 mm + x + 15 mm + x = 80 mm (Given) 2x + 30 mm = 80 mm 2x = 80 mm – 30 mm 2x = 50 mm or x = 25 mm Hence, the lengths of the sides in the parallelogram are 15 mm, 25 mm, 15 mm and 25 mm.

Example 10

JKLM is a rhombus. Find the values of x॰, y॰ and z॰. In ΔJKM, JM = JK (Sides of a rhombus) Therefore, ∠JMK = ∠JKM = x° Also, 120° = ∠JMK + ∠JKM (Exterior angle is the sum of 2 interior angles) x° + x° = 120° or 2x° = 120° ⇒ x° = 60°

120° x°

Also, x° = y° = 60° (Alternate angles)

In ΔTLM, y° + z° + ∠LTM = 180° (Angle sum property of triangles) Substituting the values, 60°+ z°+ 90° = 180°

y°

z°

z° + 150° = 180° ⇒ z° = 180° - 150° = 30°

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Find the values of a and b if PQRS is a parallelogram.

We know that diagonals in a parallelogram bisect each other. Therefore,

a-3=5⇒a=3+5=8

a−

a + 2b = 18

⇒ 8 + 2b = 18

10 =5 2 Hence, the values of a and b are 8 and 5, respectively. ⇒ 2b = 18 - 8 = 10 ⇒ b = Do It Together

Example 11

a O +2 b R

ABCD is a parallelogram and PQRS is a square. Find the value of the unknown angles in the given figures. D

a°

c°

b°

3x°

110°

x°

6x°

x°

_________________________________________________________________________________________________________________________ _________________________________________________________________________________________________________________________

Do It Yourself 4A 1 Place the points A, B, C, D, E and F given below in the quadrilaterals, such that:

a Only points A, B and C are in the interior.

b Only points B, C and D are in the interior.

B F

c Only points B, F and E are in the interior.

2 Classify the following as convex or concave quadrilaterals based on the angles of the quadrilateral. a 190o, 60o, 30o, 80o

b 65o, 115o, 111o, 69o

3 JKLM is a rectangle. Find the value of ∠MJL and ∠TJK. J

c 26o, 226o, 30o, 78o

4 Find the value of x and y in the figure shown below.

P 119°

T 140° M

5 Use the properties of a kite to find the value of ∠QPT and ∠TRS.

(2y + 7)°

(5x − 2)°

quadrilateral a parallelogram?

)

-6

(2a

11)

(5a

)

-1

(1 + 2b)

50°

(3b

6 What value of 'a' and ‘b’ would make the given

61° T

133°

Chapter 4 • Quadrilaterals

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7 PQRS is a parallelogram in which one angle measures 80°. Determine the measures of the remaining three angles in the parallelogram.

8 The length of a parallelogram is 15 cm longer than its breadth. If the perimeter of the parallelogram is 130 cm, find the length and breadth of the parallelogram.

9 WXYZ is a parallelogram where two adjacent angles have a ratio of 2:4. Calculate the measures of all the angles in the parallelogram.

10 PQRS is a rhombus whose diagonals intersect at Z. Draw a rough diagram and answer the questions given below. a Name any four pairs of line segments that are equal in length.

b Name any four pairs of angles that are equal in measure.

c Identify three pairs of congruent triangles in this rhombus.

d Is ∠QRZ = ∠ZRS? e Is ΔQPZ ≅ ΔSPZ?

Word Problem 1

Emily is walking through a square park with sides of length 200 metres. What is the shortest distance she can travel to reach the opposite corner?

Angle Sum Property of Quadrilaterals Interior Angle Sum Property of Quadrilaterals The sum of the interior angles of a quadrilateral is 360°. Let us prove it using figures. Consider the quadrilateral ABCD. Notice the measures of the angles ∠A, ∠B, ∠C and ∠D below. If we cut out each of the interior angles of the quadrilateral and place them side by side, we will get a complete angle, i.e., 360°. This is true for any quadrilateral. The angles will always add up to 360°. Hence, we can conclude that the sum of the angles C of a quadrilateral is 360°.

100°

B 115°

75°

Exterior Angle Sum Property of Quadrilaterals

The angles that are formed between one side of a quadrilateral and another line extended from an adjacent side are called its exterior angles. Here, ∠w, ∠x, ∠y and ∠z are the interior angles. ∠1, ∠2, ∠3 and ∠4 are the exterior angles.

The sum of the measures of the exterior angles of a quadrilateral is 360°. Hence, ∠1 + ∠2 + ∠3 + ∠4 = 360°.

70°

100°

115°

75°

D P ∠1

∠4 ∠x

∠w

∠z ∠2

∠y

∠3

Linear pairs of exterior and interior angles are as follows: ∠1 and ∠w, ∠2 and ∠z, ∠3 and ∠y, ∠4 and ∠x ∠1 = 120°, ∠2 = 60°, ∠3 = 60° and ∠4 = 120° 120° + 60° + 60° + 120° = 360° So, ∠1 + ∠2 + ∠3 + ∠4 = 360°

Remember! Every interior angle in a convex quadrilateral has its corresponding exterior angle.

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Example 12

Find the value of x° in the diagram given below.

Example 13

ind the value of y° in the diagram given F below. 109°

x°

85°

y°

71°

79° 114°

y°

Using the interior angle sum property of the quadrilateral,

U sing the exterior angle sum property of the quadrilateral,

x° + 278° = 360°

2 y° + 180° = 360°

x° + 85° + 79° + 114° = 360°

y ° + y° + 71° + 109° = 360°

x° = 360° - 278° = 82°

2 y° = 360° - 180° = 180° 180° y ° = = 90° 2

If one of the angles in a parallelogram is 65°, find the measure of the other angles.

Example 14

Since opposite angles in a parallelogram are equal, let the angles be 65°, x°, 65° and x°. 65° + 65° + x° + x° = 360° (Angle sum property of quadrilaterals) 130° + 2x° = 360° 2x° = 360° - 130° = 230° ⇒ x° = 230° ÷ 2 = 115°. Hence, the angles of the parallelogram are 65°, 115°, 65° and 115°. Do It Together

Match the set of interior angles of quadrilaterals in the left column with the correct missing angle in the right column. Interior Angles of a Quadrilateral

Missing Angle

1 55°, 75°,________, 145°

80°

2 35°, ________ , 110°, 140°

85°

3 60°, 100°, 120°, ________

40°

4 ________ , 90°, 70°, 160°

75°

Do It Yourself 4B 1 Which of the sets of angles given below is NOT a set of interior angles of a quadrilateral? a 47°, 58°, 98°, 157°

b 54°, 59°,72°,108°

c 63°, 107°, 74°, 116°

d 83°, 68°, 82°, 127°

2 Which of the sets of angles given below is NOT a set of exterior angles of a quadrilateral? a 57°, 82°, 98°, 123°

b 58°, 52°, 77°, 125°

c 65°, 85°, 85°, 125°

d 49°, 78°, 103°, 130°

Chapter 4 • Quadrilaterals

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3 Find the value of the unknown angle in each of the following. a

23°

75°

20° 100°

b° 93°

71° a°

110°

22°

e°

68°

c° 81° f

d°

51°

54°

28°

138°

79°

58°

15°

f°

47°

4 What could be the value of the unknown exterior angle in each case? a

76°

y°

103°

z°

111°

x°

131°

46°

78°

80°

103° d

86°

126°

51° 100°

t°

u°

53°

u°

111°

5 Find the value of unknown angles in each of the cases given below. a

83° x°

(2x + 10)°

37°

y°

94° z°

4y°

(3y − 30)°

y°

27°

x° z°

68°

6 In a parallelogram, two adjacent angles are (3y - 10)° and (2y + 15)°. Find the measures of all four angles of the parallelogram.

7 In a parallelogram, one angle measures 60° less than its adjacent angle. Determine the measures of all four angles in the parallelogram.

8 RSTU is a parallelogram in which the sum of two opposite angles is 150°. Find the measures of all four angles in the parallelogram.

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9 Find the values of all the angles in the quadrilateral given below

10 Find the values of unknown angles such that ABCD is a parallelogram. A

(7x − 3)° (2x + 10)°

(3x + 4)° (5x + 9)°

z°

73°

x° y° 45°

Word Problem 1

Rahul pins four nails to a soft wooden board. He then proceeds to tie threads between these nails so that they form a quadrilateral. The total length of thread used is 12 cm.

The angles of this quadrilateral are in the ratio 1:2:3:4. Is the quadrilateral formed by the threads a convex or a concave quadrilateral?

Points to Remember •

A quadrilateral has four sides and four vertices.

• The sum of the interior angles of a quadrilateral is 360°. The sum of the exterior angles of a quadrilateral is 360°. • The opposite sides of a parallelogram are parallel and equal and the diagonals bisect each other.

• The opposite sides and angles of a parallelogram are equal and any two adjacent angles of a parallelogram are supplementary.

• In a trapezium, one pair of opposite sides is parallel while the other pair of sides is non-parallel. •

Adjacent pairs of sides are equal in a kite.

Math Lab Setting: In groups of 3-4

Quadrilateral Sorting Challenge

Materials Required: Quadrilateral cards (pre-made or printed), large poster paper or whiteboard, markers, scissors, tape Method: 1 Discuss the concept of quadrilaterals like rectangles, squares, rhombuses, trapeziums, parallelograms, etc. 2

Create cards with quadrilateral names and key properties.

Chapter 4 • Quadrilaterals

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Draw a table with four columns, label it with quadrilateral names, and tape down cards.

Divide the class into groups and provide cards and sorting table access.

5 Instruct groups to sort cards into the correct columns based on properties, encouraging discussions. 6

Gather the class to discuss choices and have groups explain their reasoning.

Chapter Checkup 1 State whether the following statements are true or false. a The diagonals of a kite bisect each other at right angles. b The diagonals of an isosceles trapezium are equal. c The opposite angles of a trapezium may be equal. d The diagonals of a rectangle are perpendicular to each other. e All quadrilaterals are parallelograms.

2 Which of the interior angles given below would not belong to a convex quadrilateral? a 100°

b 178°

c 222°

d 2°

3 Classify the points as lying in the interior and exterior of the quadrilateral. P R

4 Which of the sets of angles given below is a set of interior angles of a quadrilateral? a 47°, 58°, 98°, 156°

b 54°, 59°, 72°, 108°

c 63°, 100°, 74°, 116°

d 83°, 68°, 82°, 127°

5 Which of the sets of angles given below is a set of exterior angles of a quadrilateral? a 57°, 82°, 98°, 120°

b 53°, 69°, 77°, 111°

c 65°, 85°, 85°, 125°

d 49°, 75°, 103°, 130°

6 Write the set of exterior angles corresponding to the set of interior angles given below. a 70°, 90°, 110°, 90°

b 45°, 60°, 100°, 155°

c 80°, 85°, 95°, 100°

7 Find the value of the unknown interior angle in each case. 41°

88°

97°

m°

l° 21°

58°

64°

58°

66°

n° 74°

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8 Find the value of the unknown exterior angle in each case. a

p° 117°

48°

93°

r°

93°

89°

92°

67°

q°

9 In a parallelogram, the adjacent angles are (2z - 25)º and (3z + 10)º. Determine the measures of all four angles in the parallelogram.

10 The length of a parallelogram is 20 cm longer than its breadth. If the perimeter of the parallelogram is 140 cm, find the length and breadth of the parallelogram.

11 WXYZ is a parallelogram in which one angle is 140º. Find the measures of the other three angles in the parallelogram.

12 In a parallelogram, two adjacent angles are in the ratio 2:3. Determine the measures of all four angles in the parallelogram.

13 In a rhombus, one of the diagonals is of the same length as one of its sides. Determine the measures of the angles in the rhombus.

14 Find the values of wº, xº, yº and zº. a

125° w° y°

50°

x° z° 135°

135°

125°

3x°

5y°

15 Prove that the diagonals of a rhombus bisect its angles.

(3y + 5)° (6y + 4)° 79°

x°

C Q

16 Find the measure of all the interior angles of the rhombus PQRS.

T S

52°

Word Problems 1

At the local playground, there is a climbing structure made of four bars forming a

quadrilateral shape. The angles at the corners are measured and found to be in the ratio 2:3:4:5. Determine whether the climbing structure is a parallelogram.

A tennis court is designed with four sides that form a parallelogram shape. The length of

one side is 24 m, and the angle at the top corner measures 120°. What is the length of the opposite side, and what is the measure of the angles at the other corners?

Chapter 4 • Quadrilaterals

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3 Construction of 5 Quadrilaterals Let's Recall

Rhea drew a figure with different quadrilaterals. She asked her friend Gyan to guess the shape. Gyan was confused since all of them had 4 sides, 4 angles and 4 vertices. Let us identify the shapes using their properties.

Kite

Trapezium

Rhombus

Properties of some special quadrilaterals A

Trapezium

Parallelogram

Opposite sides are equal and parallel

•

• All sides are equal

• Opposite angles are equal, and diagonals bisect each other

E D

C B

•

Rhombus

Opposite sides are parallel Opposite angles are equal

Diagonals bisect each other at right angles Kite

• Two pairs of adjacent sides are equal

B C

A D D

• Diagonals intersect at right angles

Trapezium • One pair of sides is parallel, and the other pair is non-parallel

• Diagonals in an isosceles trapezium are equal A

Letʼs Warm-up

Name the quadrilateral by reading the clues. 1 Exactly two pairs of adjacent sides are equal. _________________ 2 Exactly one pair of opposite sides is parallel. _________________ 3 All angles in a quadrilateral are right angles but only opposite sides are equal. _________________ 4 All sides are equal but all angles may not necessarily be equal. _________________ I scored _________ out of 4.

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Constructing Quadrilaterals Real Life Connect

Rahul looked at building sketches and wondered how his architect aunt, Nilima, designed such amazing structures.

Nilima told him that she uses rulers and protractors to create shapes, which amazed Rahul. He was particularly eager to learn about making shapes, especially quadrilaterals. To create quadrilaterals, we need a minimum number of elements. For instance, for a unique triangle, we need three measurements (at least one being a side). A quadrilateral requires at least five of its ten elements (four sides, four angles and two diagonals) to be given for a unique construction. You can construct a quadrilateral with the following sets of at least five elements. 1

Four sides and one diagonal

Three sides and two diagonals

Three angles and two included sides

Three sides and two included angles

Four sides and one angle

Did You Know? In the 7th century, Indian mathematician, Brahmagupta, explored properties of unique quadrilaterals in his book “Brāhma-sphuṭa-siddhānta.”

Let us discuss constructing quadrilaterals in each of the cases listed above one by one.

Construction with 4 Sides and 1 Diagonal A quadrilateral can be constructed when the measures of any 4 sides and one diagonal are given.

1 D raw a line segment PR = 6.5 cm.

2 D raw two arcs: one with the centre at P (5.3 cm radius) and another with the centre at R (4.2 cm radius), intersecting at Q.

cm 3 5.

Chapter 5 • Construction of Quadrilaterals

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6.5 cm

5.3 cm

4.2

6.5 cm

3 Join PQ and QR.

4.5 cm

3.7 cm

Draw a rough sketch of the quadrilateral PQRS with the marked measurements. We may divide quadrilateral PQRS into two triangles namely RSP and PRQ.

4.2 cm

Let us construct a quadrilateral PQRS in which PQ = 5.3 cm, QR = 4.2 cm, RS = 4.5 cm, SP = 3.7 cm and PR = 6.5 cm.

6.5 cm

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4 D raw two arcs on the opposite side of PR: one with the centre at P (3.7 cm radius) and another with the centre at R (4.5 cm radius), intersecting at S.

5 Join PS and RS.

5. 3

cm 5. 3 P

S Example 1

6.5 cm

4.2

6.5 cm

m N

Construct a quadrilateral LMNO in which LM = 6.8 cm, MN = 4.5 cm, NO = 5.2 cm, OL = 3.9 cm and LN = 7.1 cm.

Draw a sketch of a quadrilateral LMNO with measurements roughly marked. We may divide the quadrilateral LMNO into two triangles, namely ∆LMN and ∆LNO.

3.9 cm

7.1 cm

6.8 cm M

1 Draw a line segment LN = 7.1 cm.

4.5

2 D raw two arcs: one with the centre at L (6.8 cm radius) and another with the centre at N (4.5 cm radius), intersecting at M. 3 Join LM and MN.

9 cm

5 Join OL and ON. LMNO is the required quadrilateral.

86° 7.1 cm

4 D raw two more arcs: one with the centre at L (3.9 cm radius) and another with the centre at N (5.2 cm radius), intersecting at O.

Do It Together

4.5 cm

c O 5.2

Construct a quadrilateral FEDC with the given measures. Complete the rough figure. CD = 4.3 cm, DE = 3.6 cm, EF = 5.6 cm, FC= 4.7 cm, CE = 6 cm

Construction with 3 Sides and 2 Diagonals

S 3.5 cm

1 Draw a line segment PQ = 5.5 cm.

2 Draw two arcs: one with the centre at P (4.9 cm radius) and another with the centre at Q (6.7 cm radius), intersecting at S.

S 4.9 cm

5.5 cm

7 6.

5.5 cm

3 Join PS and QS.

Draw a sketch of the quadrilateral PQRS with the measurements roughly marked.

1 6.

Let us construct a quadrilateral PQRS in which PQ = 5.5 cm, PS = 4.9 cm, RS = 3.5 cm, diagonal PR = 6.1 cm and diagonal QS = 6.7 cm.

4.9 c

A quadrilateral can be constructed when the measures of any three sides and two diagonals are given.

5.5 cm

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4 Draw two arcs: one with the centre at P (6.1 cm

radius) and another with the centre at S (3.5 cm radius), intersecting at R.

6.2

5.8

3 cm

Construct a quadrilateral DAVE in which DA = 3.6 cm, DE = 5 cm, EV = 3 cm, diagonal AE = 6.2 cm and diagonal DV = 5.8 cm. Find the length of side AV. Draw a sketch of the quadrilateral DAVE with the measurements roughly marked.

5 cm

1 Draw a line segment DE = 5 cm.

5.1 cm 6.2

3.6 cm

5. D Do It Together

5.5 cm

3.6 cm

4.9 cm

5.5 cm

3.5 cm

Example 2

4.9 cm

5 Join RS and QR. PQRS is the required quadrilateral.

2 D raw two arcs: one with the centre at D (5.8 cm radius) and another with the centre at E (3 cm radius), intersecting at V.

3 Join DV and EV.

3 cm

4 D raw two arcs: one with the centre at D (3.6 cm radius) and another with the centre at E (6.2 cm radius), intersecting at A.

cm 5 cm

5 J oin DA EA and AV. DAVE is the required quadrilateral. On measuring, we get the length of AV as 5.1 cm.

Construct a quadrilateral LISA in which LI = 4.2 cm, SI = 5.5 cm, SA = 3.8 cm, diagonal IA = 6.7 cm and diagonal LS = 7.6 cm. Determine the length of the side LA.

L D

Construction with 3 Angles and 2 Included Sides

120°

3 cm

A quadrilateral can be constructed when any three angles and 2 included sides are given.

Let us construct a quadrilateral ABCD in which AB = 4.9 cm, BC = 3 cm, ∠A = 60°, ∠B = 90° and ∠C = 120°. Draw a rough sketch of the quadrilateral ABCD and mark its measurements. Chapter 5 • Construction of Quadrilaterals

Maths Grade 8 Chapter 1-6.indb 61

60° A

90°

4.9 cm

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1 D raw a line segment AB = 4.9 cm.

2 Construct ∠XAB = 60° at A.

3 Construct ∠YBA = 90° at B. X

90°

60° A

4.9 cm

4 U sing B as the centre and a radius of 3 cm, draw an arc cutting BY at C. X

4.9 cm

120°

3 cm B

60°

4.9 cm

5 A t C, construct ∠ZCB = 120°, so that ZC and XA intersect at D. ABCD is the required quadrilateral.

90°

60°

C 3 cm

90° A

60°

4.9 cm

Remember! Keep the compass width consistent when constructing the 90° and 120° angles.

Construct a quadrilateral PQRS in which PQ = 3.9 cm, QR = 4.5 cm, ∠P = 75°, ∠Q = 100° and ∠R = 60°. Measure the length of PS. Draw a rough sketch of the quadrilateral PQRS with its measurements. 1 Draw a line segment PQ = 3.9 cm.

X Y

R 60°

S 3.3 cm

75°

3.9 cm

0 10

4.5 cm

60°

75° 100°

4.5 cm

Example 3

3.9 cm Q

2 Construct ∠XPQ =75° at P. 3 Draw ∠YQP= 100° at Q using a protractor. 4 U sing Q as the centre and a radius of 4.5 cm, draw an arc cutting QY at R. 5 A t R, construct ∠ZRQ = 60°, so that ZR and XP intersect at S. PQRS is the required quadrilateral. On measuring, we get the length of PS as 3.3 cm.

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Do It Together

Construct a quadrilateral PUSH in which PU = 4.9 cm, SU = 5.5 cm, ∠P = 75°, ∠U = 100° and ∠S = 60°. Find the sum of the lengths of sides PH and SH. Rough Figure H

S 60° ___ cm

____°

75°

P ________ cm U

Construction with 3 Sides and 2 Included Angles

Construct ∠XBC = 60° at B. X

30°

6.6 cm

60°

Draw a rough sketch of the quadrilateral ABCD with its measurements.

3.3

Construct a quadrilateral ABCD in which AB = 5.4 cm, BC = 6.6 cm, CD = 3.3 cm, ∠B = 60° and ∠C =30°.

1 D raw a line segment BC = 6.6 cm.

5.4 c m

A quadrilateral can be constructed when the measures of any three sides and two included angles are given.

3 U sing B as the centre and a radius of 5.4 cm, draw an arc to cut XB at A. X

5.4

60° 6.6 cm

4 Construct ∠YCB = 30° at C. X

6.6 cm

5 U sing C as the centre and a radius of 3.3 cm, draw an arc to cut YC at D.

6.6 cm

30°

60°

6.6 cm

6 J oin AD. ABCD is the required quadrilateral. X

A cm

Chapter 5 • Construction of Quadrilaterals

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60°

5.4

cm 5.4

60°

5.4

D 30°

6.6 cm

60°

30°

6.6 cm

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Construct a quadrilateral PQRS in which PQ = 4.4 cm, QR = 5.7 cm, RS = 2.5 cm, ∠Q = 60° and ∠R = 100°. Find the length of PS.

3 Using Q as the centre and a radius of 4.4 cm, draw an arc to cut XQ at P.

4.4 cm Do It Together

100°

5.7 cm

4 Construct ∠YRQ = 100° at R, using the protractor.

2.5 cm

60°

1 Draw a line segment QR = 5.7 cm. 2 Construct ∠XQR = 60° at Q.

4.1 c

60° 100° Q 5.7 cm R

2.5 c

Draw a rough sketch of the quadrilateral PQRS with its measurements. Y

4.4 c m

Example 4

5 U sing R as the centre and a radius of 2.5 cm, draw an arc to cut YR at S.

6 J oin PS. PQRS is the required quadrilateral. On measuring, we get the length of SP as 4.1 cm.

Construct a quadrilateral JKLM in which JK = 5.7 cm, KL = 4.4 cm, LM = 2.8 cm, ∠K = 60°° and ∠L = 120°°. Find the measure of ∠J. Rough Figure M 120°

____

2.7

60° P

Draw a rough sketch of the quadrilateral PQRS with its measurements. 2 Construct an angle of 60° at Q.

5.6 cm

Construct a quadrilateral PQRS in which PQ = 4.7 cm, QR = 3.6 cm, RS = 5.6 cm, SP = 2.7 cm and ∠Q = 60°.

1 D raw a line segment PQ = 4.7 cm.

60° ____

________ cm

3.6

A quadrilateral can be constructed when the measures of four sides and one angle are given.

____ cm

Construction with 4 Sides and 1 Angle

4.7 cm

3 W ith the centre Q and radius as equal to 3.6 cm, draw an arc to cut XQ at R. X

R 3.6 cm

4.7 cm

60° Q

4.7 cm

60° P

4.7 cm

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4 D raw an arc with the centre at P (2.7 cm radius) and another arc with the centre at R (5.6 cm radius), intersecting the first at S.

5 J oin SP and SR. PQRS is the required quadrilateral.

R 3.6

4.7 cm

1 Draw a line segment AB = 4.7 cm.

3.5 c

5.5 cm

75° A 4.7 cm B

3 W ith the centre at B and the radius equal to 3.5 cm, draw an arc to cut XB at C.

Do It Together

2 Construct an angle of 75° at B.

5 cm

Draw a rough sketch of the quadrilateral PQRS with its measurements.

3.5

Construct a quadrilateral ABCD in which AB = 4.7 cm, BC = 3.5 cm, CD = 5 cm, AD = 5.5 cm and ∠B = 75°. Find the measure of diagonal AC.

Example 5

5.5

4.7 cm

60°

R 3 .6

m 6c

75° 4.7 cm

4 D raw an arc with the centre at A (5.5 cm radius) and another arc with the centre at C (5 cm radius), intersecting the first at D. 5 Join AD and CD. ABCD is the required quadrilateral.

On measuring, we get the length of diagonal AC as 5.1 cm.

Construct a quadrilateral EFGH in which EF = 4.7 cm, FG = 3.8 cm, GH = 4 cm, HE = 7 cm and ∠F = 90° Find the sum of the lengths of both diagonals.

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Do It Yourself 5A 1 Construct quadrilaterals where the lengths of 4 sides and 1 diagonal are given. a Quadrilateral LMNO: LM = 5.6 cm, MN = 4.2 cm, NO = 7.8 cm, LO = 6.5 cm, MO = 8.1 cm b Quadrilateral XYZW: XY = 6.3 cm, YZ = 4.7 cm, ZW = 5.2 cm, WX = 7.4 cm, WY = 8.6 cm

2 Construct a quadrilateral GLOW with the given measures. Measure the unknown side. a GL = 5.6 cm, OW = 3.4 cm, GW = 4.8 cm, GO = 6.7 cm, LW = 7.3 cm b GL= 5.2 cm, LO = 4.3 cm, WG = 3.8 cm, GO = 6.1 cm, WL = 7.2 cm

3 Construct the following quadrilaterals when the measures of three angles and two sides are given. Find the measures of both diagonals.

a ABCD: AB = 7 cm, AD = 4 cm, ∠A = 75°, ∠B = 110° and ∠D = 95° b WXYZ: WX = 8 cm, XY = 6 cm, ∠W = 120°, ∠X = 85° and ∠Y = 75°

4 Construct quadrilateral ABCD with the given measures. a AB = 3.5 cm, BC = 5.4 cm, CD = 4.9 cm, ∠B = 125° and ∠C = 80° b AB = 4.2 cm, BC = 5.3 cm, CD = 3.6 cm, ∠B = 135° and ∠C = 60°

5 Construct the quadrilaterals given below, where the measures of 4 sides and one of the angles are known. a RSTU: RS = 2.9 cm, ST = 3.6 cm, TU = 4 cm, RU = 6 cm, ∠S = 90° b JKLM: JK = 3.9 cm, KL, 3.1 cm, LM = 3.6 cm, MJ = 3.3 cm, ∠ J = 60°

Word Problem 1

You are creating a poster for your school’s geometry fair and want to include a unique quadrilateral shape. You know that you want to use a quadrilateral with four sides

measuring 9 cm, 6 cm, 12 cm and 7 cm, and you want one of the angles between two

consecutive sides to be 60 degrees. How can you use your ruler and compass to accurately draw this quadrilateral for your poster? Draw two quadrilaterals that fit this description.

Constructing Special Quadrilaterals Some quadrilaterals are special and some of their measures are equal. Some examples of special quadrilaterals are squares, rectangles, rhombuses, trapeziums, kites and parallelograms. The properties of special quadrilaterals help in constructing them. So, we do not need 5 measures to construct special quadrilaterals. Constructing a Square We know that a square has all its sides equal, all angles measure 90°, and its diagonals are equal in length and bisect each other at right angles. A square can be constructed when 1 of its diagonals or 1 of its sides are given.

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Let us construct a square where the length of its diagonal is 4.6 cm.

It is known that the diagonals of a square are equal in length and bisect each other at right angles. This property can help us construct the square.

Let’s say, the square is PQRS and the diagonals bisect at O.

2.3 cm

Q O 4.6 cm

Given:

PR = QS = 4.6 cm (diagonals are equal)

PO = OR = QO = OS = 2.3 cm (Diagonals bisect each other)

Q 2.3 cm

1 Draw a line segment PR = 4.6 cm. 2 Draw the right bisector XY of PR, intersecting PR at O. 3 With the centre O and radius equal to 1 × 4.6, i.e., 2.3 cm, draw arcs on either side 2 of PR, cutting XY at Q and S.

O P 4.6 cm

4 Join PQ, QR, RS and SP. PQRS is the required square.

Think and Tell Is there any other measure, other than the diagonal, that can help us construct a square?

Constructing a Rhombus A rhombus has sides of equal length, equal opposite angles and the diagonals bisect each other at right angles. Let us construct a rhombus whose diagonals are of lengths 5 cm and 7 cm.

It is known that the diagonals of a rhombus bisect each other at right angles. This property can help us construct the rhombus. Let’s say, the rhombus is PQRS and the diagonals bisect at O.

Q 2.5 cm

PR = 7 cm, PO = OR = 3.5 cm (Diagonals bisect each other)

QS = 5 cm, QO = OS = 2.5 cm (Diagonals bisect each other)

cm O

3. P

S X

1 Draw a line segment PR = 7 cm. 2 Draw XY, its perpendicular bisector and let it intersect PR at O. 3 U sing O as the centre and a radius of 2.5 cm, draw two arcs on opposite sides of PR to intersect XY at Q and S.

2.5

Given:

Q O 2.5 cm R 7 cm

4 Join PQ, QR, RS and SP. PQRS is the required rhombus.

Constructing a Rectangle

We know that a rectangle has its opposite sides equal, all angles measure 90° and its diagonals are equal in length and bisect each other. Let us construct a rectangle PQRS, given that side PQ = 6.2 cm and the diagonal PR = 7.8 cm. We know that the opposite sides are equal and all the interior angles in the rectangle are right angles. This property can help us construct the rectangle. Let’s say, the rectangle is PQRS and PR is its diagonal. Chapter 5 • Construction of Quadrilaterals

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7.8 P

6.2 cm

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Given:

PR = 7.8 cm (Length of the diagonal)

PQ = RS = 6.2 cm (Opposite sides of a rectangle are equal) ∠Q = 90° (All angles are right angles)

1 Draw a line segment PQ = 6.2 cm.

2 Draw XQ ⊥ PQ at point Q. 3 Using P as the centre and a radius of 7.8 cm, draw an arc cutting QX at R.

4 D raw two arcs: one with the centre at R and a 6.2 cm radius, and another with the centre at P and a radius equal to QR, ensuring they intersect at point S.

cm 90° Q

6.2 cm

5 Join SP and SR. PQRS is the required rectangle. Constructing a Trapezium

Let us construct a trapezium EFGH in which EF ॥ GH, HE = 3.7 cm, FE = 4.5 cm, GH = 6.7 cm and ∠H = 60°. We know that one pair of opposite sides is parallel in a trapezium. This property can help us construct the trapezium.

4.5 cm

cm 3.7

We know that a trapezium has one pair of parallel sides and in an isosceles trapezium, the non-parallel sides are equal and the diagonals are equal.

60°

6.7 cm

Given: EF ॥ GH; HE = 3.7 cm, FE = 4.5 cm, GH = 6.7 cm.

1 Draw a line segment GH = 6.7 cm. 2 Construct an angle of 60° at H. E

4.5 cm

4 Construct XY parallel to HG, passing through E.

3.7

60°

3 U sing H as the centre and a radius of 3.7 cm, draw an arc to cut the ray at E.

6.7 cm

5 Using E as the centre and a radius of 4.5 cm, draw an arc to cut EY at F. 6 Join GF. EFGH is the required trapezium.

Constructing a Parallelogram

In a parallelogram, the opposite sides are equal and parallel, the opposite angles are equal and the diagonals bisect each other.

Let us construct a parallelogram one of whose sides is 6 cm and whose diagonals are 7 cm and 8 cm. D

3.5

It is known that the diagonals of a parallelogram bisect each other. This property can help us construct the parallelogram. Let’s say the parallelogram is ABCD and the diagonals bisect each other at O.

AC = 8 cm, AO = OC = 4 cm (Diagonals bisect each other)

6 cm

AB = 6 cm

3.5

Given:

BD = 7 cm, BO = OD = 3.5 cm (Diagonals bisect each other)

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1 Draw a line segment AB = 6 cm. 2 D raw two arcs: one with the centre at A and a 3.5 cm radius, and another with the centre at B and a 4 cm radius, intersecting at O.

3 Join OA and OB.

4 E xtend AO to meet C, making OC equal to AO and extend BO to meet D, making OD equal to BO. 5 Join AD, BC and CD. ABCD is the required parallelogram.

m 4c A

cm B

6 cm

Constructing a Kite

In a kite, two pairs of adjacent sides are equal and the diagonals intersect at right angles.

Let us construct a kite in which the measures of two adjacent sides are 5 cm and 3 cm and the diagonal passing through the common vertex of these sides is 4 cm.

4 cm

Y 3

Given: WY = 4 cm (Length of the diagonal); WX = XY = 5 cm (Adjacent sides are equal); WZ = YZ = 3 cm (Adjacent sides are equal)

It is known that the two pairs of adjacent sides have equal lengths in a kite. This property can help us construct the kite. Let’s say, WXYZ is a kite and WY is its diagonal.

Z X

1 Draw a line segment WY = 4 cm.

4 D raw two more arcs: one with the centre at W (5 cm radius) and another with the centre at Y (5 cm radius), intersecting at X.

3 Join WZ and YZ.

2 D raw two arcs: one with the centre at W (3 cm radius) and another with the centre at Y (3 cm radius), intersecting at Z.

4 cm

Y cm

m 3c

5 Join WX and YX. WXYZ is the required kite.

Z Example 6

Construct a parallelogram PQRS in which PQ = 3 cm, QR = 4 cm and diagonal PR = 6 cm. It is known that the opposite sides of a parallelogram are equal. This property can help us construct a parallelogram. Let’s say the parallelogram is PQRS and PR is its diagonal.

6 P

3 cm

4 cm

Given: PQ = RS = 3 cm, RQ = SP = 4 cm, PR = 6 cm.

Error Alert! You may or may not able to construct a quadrilateral with any random set of measures of elements. Do It Together

Construct a parallelogram with its adjacent sides measuring 3 cm and 4 cm, respectively and the included angle being 60°. Complete the given rough diagram of the parallelogram before starting the construction.

Chapter 5 • Construction of Quadrilaterals

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Rough Diagram

Construction

Do It Yourself 5B 1 Construct a square with sides that measure 5 cm. 2 Construct a rhombus ABCD that has a side of 5 cm and an angle of 120°. 3 Construct a rectangle whose adjacent sides measure 5.6 cm and 4.5 cm. 4 Construct a parallelogram with a side length of 4.9 cm and diagonals of length 5.4 cm and 8 cm. 5 Construct a parallelogram PQRS in which PQ = 4.6 cm, QR = 6.7 cm and ∠Q = 45°. 6 Construct a trapezium ABCD in which AB ॥ CD, AB = 5.8 cm, CD = 2.5 cm, ∠A = 67° and AD = 3.6 cm. What is the measure of the largest angle in the quadrilateral?

Word Problem 1

You are making a greeting card, and you want to create a square-shaped window in the

front cover. The diagonals of this window are 8 cm. Construct a quadrilateral using a ruler and compass.

Points to Remember • A quadrilateral can be constructed if the measures of 5 out of its 10 elements are given. • Diagonals of parallelograms, rectangles, rhombuses and squares bisect each other. • A quadrilateral can be constructed when the measures of 4 sides & 1 diagonal, 3 sides & 2 diagonals, 3 sides & 2 included angles, 3 angles & 2 included sides or 4 sides & 1 angle are given. • You need the knowledge of fewer elements to construct special quadrilaterals as their properties help you determine other unknown elements. • The sum of the interior angles of a quadrilateral is 360°.

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Math Lab Quadrilateral Pattern Creation Materials Required: Drawing paper, rulers, compasses, pencils, protractors, coloured pencils/markers (optional) Instructions: 1 Start by reviewing the basic construction techniques for drawing quadrilaterals using a ruler and a compass.

2 Ask each student to brainstorm and plan a design or pattern that they want to create using quadrilaterals. They should consider the type of quadrilaterals they want to include (e.g., squares, rectangles, parallelograms, and so on) and the measurements for each side and angle.

3 Allow students to begin constructing their chosen patterns on their drawing paper. They should use their rulers, compasses and protractors. 4 Encourage students to get creative with their patterns. They can experiment with different colours and arrangements of the quadrilaterals to make their designs visually appealing. 5 Assess each student’s pattern on the basis of creativity, precision in construction and the ability to explain their design choices.

Chapter Checkup 1 Construct quadrilaterals with the given lengths of 4 sides and 1 diagonal. a Quadrilateral UVWX: UV = 3.9 cm, VW = 5.1 cm, WX = 4.5 cm, XU = 6.8 cm, VX = 7.2 cm. b Quadrilateral STUV: ST = 4.5 cm, TU = 5.7 cm, UV = 6.2 cm, VS = 7.3 cm, SU = 8.9 cm. c Quadrilateral KLMN: KL = 6.1 cm, LM = 3.5 cm, MN = 5.8 cm, KN = 4.2 cm, LN = 7.4 cm.

2 Construct a quadrilateral PINE with the given lengths of 3 sides and 2 diagonals. Find the measure of the unknown side.

a PI = 6.9 cm, EN = 3.2 cm, IN = 5.5 cm, PN = 4.7 cm, IE = 7.6 cm. b PI = 7.4 cm, NE = 3.2 cm, PE= 5 cm, IE = 6.8 cm, PN = 4.7 cm. c PI = 6.5 cm, IN = 4.2 cm, PE = 3.9 cm, PN = 7.1 cm, IE = 8.3 cm.

3 Construct the following quadrilaterals when the measures of 3 angles and 2 included sides are given. Find the measures of both of its diagonals.

a Quadrilateral EFGH: EF = 6 cm, FG = 5 cm, ∠E = 60°, ∠F = 45°, and ∠G = 120° b Quadrilateral IJKL: IJ = 4.9 cm, IL = 7 cm, ∠I = 70°, ∠J = 90° and ∠L = 115° c Quadrilateral MNOP: MN = 6.5 cm, MP = 4.5 cm, ∠M = 100°, ∠N = 90° = and ∠P = 135°

Chapter 5 • Construction of Quadrilaterals

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4 Construct quadrilaterals WXYZ with the given measures. a WX = 4.5 cm, XY = 6.4 cm, YZ = 3.9 cm, ∠X = 150° and ∠Y = 60° b WX = 4.6 cm, XY = 5.5 cm, YZ = 6.6 cm, ∠X = 80° and ∠Y = 100° c WX = 3.2 cm, XY= 5.7 cm, YZ = 4.4 cm, ∠X = 135° and ∠Y = 60°

5 Construct the quadrilaterals given below where 4 sides and one of the angles are known. a Quadrilateral ABCD: AB = 4.6 cm, BC = 4.1, CD = 5.6 cm, AD = 5.2 cm and ∠B = 70° b Quadrilateral PQRS: PQ = 3.9 cm, QR = 3.3 cm, RS = 6 cm, PS = 5 cm and ∠Q = 80° c Quadrilateral WXYZ: WX = 7.6 cm, XY= 6.8 cm, YZ = 5.2 cm, ZW = 4.6 cm and ∠Y = 120°

6 Construct a square whose each diagonal measures 5.8 cm. 7 Construct a rhombus PQRS in which: a PQ = 4 cm and SQ = 6 cm

b QS = 6.6 cm and QR = 3.7 cm

8 Construct a rectangle ABCD when it is given that: a One side measures 4 cm and the diagonal has a length of 6.6 cm. b The diagonals measure 5.6 cm and the angle between them is 60°. c Its adjacent sides measure 3.7 cm and 2.9 cm.

9 Construct a parallelogram ABCD, given that BD = 4.6 cm, AC = 6 cm and the angle included between the diagonals is 60°.

10 Construct a trapezium EFGH in which EF ॥ GH, EF = 4.5 cm, FG = 3.7 cm, GH = 5 cm and ∠F = 40°. Can we construct more than one trapezium in this case?

11 Construct a parallelogram ABCD in which AC = 5.5 cm, AB = 3.6 cm and the altitude of AM from A to CD is 2.6 cm.

Word Problem 1

You are working on a craft project and need to create a rhombus-shaped outline with

a diagonal of length 6 cm and a side of length 4.2 cm. How can you use your ruler and compass to draw the rhombus accurately?

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Classification and Tabulation of Data

6 Let's Recall

We have learnt that data is collected and organised in tables, making it easier to understand. The given table shows the major crop-producing states in India. This collection of information is called data. From the data set, we can conclude that: 3 of the states produce rice. 2 of the states produce wheat. 2 of the states produce pulses. 1 of the states produces maize. So, we can say that rice is the most produced crop in India.

State

Crop Produced

Haryana

Wheat

Odisha

Rice

West Bengal

Rice

Uttar Pradesh

Wheat

Rajasthan

Pulses

Tamil Nadu

Rice

Madhya Pradesh

Pulses

Karnataka

Maize

Letʼs Warm-up

The given table shows the temperatures in a month every day for 30 days. Calender MON

TUE

WED

THU

FRI

SAT

SUN

20 Complete the tally chart.

20 Temperature (in °C) 20

Tally

Total

25 30 35

I scored _________ out of 4.

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Frequency Distribution of Grouped and Ungrouped Data Real Life Connect

Ahan loves reading books. Every time he finishes reading a book, he writes down its title, author’s name and how much he likes it on a scale of 10. He writes them in his diary as: Book Title

Author

Rating Out of 10

Number of Pages

Summer’s Child

Diane Chamberlain

416

Sanctuary

Caryn Lix

320

The Diabolic

S. J. Kincaid

416

Last Star Burning

Caitlin Sangster

400

Rabbit and Robot

Andrew Smith

448

We now know that the details that Ahan collects in his reading log is data. Data is a collection of numerical figures that represents a specific kind of detail called an observation. The above data is recorded in a table in an unorganised form. When numerical data is collected in its original form and in an unorganised form, the data is called ungrouped data or raw data. We use data to remember information about all sorts of things in our everyday lives, like keeping track of grades in school, the number of goals scored in sports, or even the weather forecast throughout the day. This study of numerical data is called statistics. The stages of statistical study are: 1 Collection of data

Think and Tell

2 Organisation and presentation of data

Where do you need to organise data in your day-to-day life?

3 Analysis and interpretation of data Types of Data Qualitative Data

Quantitative Data

It is descriptive, i.e., it describes the quality of data.

It refers to the information that can be counted or measured, i.e., it can be assigned a numerical value.

For example, the qualitative data in Ahan’s reading log refers to the rating he gave to each book out of 10.

For example, the quantitative data in Ahan’s reading log includes information like the number of pages in each book.

When we organise raw data by arranging them in rows or columns, it is called an array. The data can be arranged either in ascending order or descending order. An array is like a bookshelf in Ahan’s room where he keeps his books neatly arranged, with each book representing a piece of data. So, in Ahan’s reading log, an array would be a list that holds all the data about the books.

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Organising Data Organising data is like arranging your books neatly on a bookshelf or putting them in specific categories. In the reading log table, Ahan has organised data about five books he has read.

Did You Know? Ungrouped data is also called discrete data.

This format makes it easy for Ahan to track the books read and find specific information about each book. By organising data, you can quickly access the information you need and then discover trends or patterns.

Frequency Distribution of Ungrouped Data We saw how Ahan tracks book genres like mystery, horror, and science fiction in his reading log. This helps to know how many times each genre appears, which genres Ahan reads most frequently, and which are less common. The number of times a number or an observation occurs in a given set of data is called frequency. Let us create a frequency distribution table and record the number of pages in each book he has read. Book Title

Summer’s Child

Sanctuary

The Diabolic

Last Star Burning

Rabbit and Robot

Number of Pages

416

320

416

400

448

Step 1: List all the unique values (the different numbers of pages) from Ahan’s reading log as 416, 320, 416, 400 and 448.

Step 2: Count how many times each unique value appears in the data. This gives us the frequency. 416: 2 books; 320: 1 book; 400: 1 book; 448: 1 book

Step 3: Create a table that displays the unique values and their corresponding frequencies. Number of Pages

Frequency

Tally Marks

416

320

400

448

This table is a frequency distribution of the ungrouped data which shows how many times each specific number of pages appeared in Ahan’s reading log. Ahan read two books with 416 pages, one book with 320 pages, another with 400 pages and a final one with 448 pages. If Ahan records the number of pages in each book he has read, the range is the difference in the number of pages between the thinnest and thickest books. Here, the range of the number of pages in the books = 448 - 320 = 128 The range refers to the difference of the smallest and largest values in the data. For example, Sarah recorded the marks her classmates scored out of 10 on a recent test. 5, 5, 7, 8, 9, 10, 8, 9, 7, 10, 9, 5, 8, 7, 8, 9, 10, 10, 5, 8 Chapter 6 • Classification and Tabulation of Data

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Array the data and form a frequency table. On arranging the data in ascending order, we get: 5, 5, 5, 5, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10 The frequency distribution table can be drawn as: Marks

Tally Marks

Number of students (Frequency)

||||

|||

||||

Total Example 1

Studies suggest that on an average, students should spend about 5–10 hours per week on extra-curricular activities. Given below are the number of hours 30 students spend on extra-curricular activities per week. 5, 6, 7, 3, 5, 4, 7, 6, 5, 3, 7, 6, 4, 5, 6, 5, 6, 7, 3, 5, 4, 7, 6, 5, 3, 7, 6, 4, 5, 6

Array the data in a frequency table and find the range of the given data. Step 1: List the unique values in ascending order.

We get: 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7

Step 2: Count the frequency and create the Frequency Distribution Table. Hours spent on extra-curricular activities

Tally marks

Number of students (Frequency)

||||

|||||||

|||||

Total

Step 3: Find the range of the given data.

Range = Highest value in the data set – Smallest value in the data set =7–3=4

Do It Together

Suppose we have data representing the number of siblings that 30 individuals have. The data set is as follows: 2, 3, 4, 2, 1, 3, 0, 2, 4, 3, 2, 1, 1, 0, 0, 0, 1, 3, 4, 2, 2, 2, 1, 1, 3, 1, 1, 0, 2, 0

Array the data and form a frequency table. Also, find the range of the given data. Number of Siblings

Tally Marks

Frequency

|||||

|||| |||

_________

Total

_________

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Range = Highest value in the data set − Smallest value in the data set Therefore, the range of the given data = _____ − _____ = _____

Do It Yourself 6A 1 In a class, there are 50 students. Each student was asked what their favourite sport was. 40% of the students said cricket, 24% said badminton, 20% said tennis, 10% said football, and the rest did not have any favourite sport.

Prepare a frequency distribution table for the number of students who like a sport. Which sport is liked by the minimum number of students?

2 This table is the frequency distribution that shows the monthly income levels of middle-class families in India. Monthly Income (in thousands) (in ₹)

Frequency

a How many individuals have the highest monthly income? b How many individuals have the lowest monthly income? c What is the range?

3 The given data represents the ages of 20 people in a neighbourhood: 25, 30, 35, 40, 25, 28, 35, 42, 30, 32, 28, 22, 25, 28, 40, 45, 35, 30, 32, 28 Prepare a frequency distribution table for the above data.

4 The given data represents the scores (out of 100) of 30 students on a maths test. The data set is as follows: 85, 92, 78, 90, 88, 76, 95, 82, 79, 88, 94, 87, 89, 91, 75, 83, 80, 92, 84, 86, 77, 81, 93, 79, 88, 90, 86, 85, 94, 87 Array the data and form a frequency table. Also, find the range of the given data.

5 Suppose you have data representing the number of holidays taken in a year by 30 individuals. The holiday data is as follows:

2, 3, 2, 5, 3, 4, 1, 3, 5, 2, 3, 4, 2, 1, 3, 4, 2, 5, 1, 2, 3, 4, 1, 4, 5, 1, 2, 4, 1, 5 Array the data and form a frequency table. Also, find the range of the given data.

6 The value of π up to 39 decimal places is 3.141592653589793238462643383279502884197 Make a frequency distribution table for the digits 0 to 9 before and after the decimal point.

Grouping Data COVID-19, initially detected in India in early 2020, has had a profound impact on the nation’s healthcare system, economy, and daily life, prompting extensive vaccination campaigns and public health measures. The given table shows the number of COVID-19 cases in various countries around the world.

Chapter 6 • Classification and Tabulation of Data

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Country

Number of COVID-19 Cases (in crores)

France

India

Germany

Italy

United Kingdom

United States

Spain

Brazil

Japan

Turkey

While creating a frequency distribution for this data, we tracked the number of COVID cases for each country separately. This approach is useful when we need to analyse large data sets. However, it is sometimes beneficial to group data to get a broader overview. One common way to group data is by using ranges or intervals. For instance, we can group the number of COVID cases in ranges, such as 0–2 crores and 3–5 crores. Grouping data can make it easier when we need to deal with large data sets. When individual observations are arranged in groups such that a frequency distribution table of these groups helps provide a convenient way of presenting or analysing data is known as grouped data.

Drawing a Frequency Distribution Table of Grouped Data Let us group the number of COVID cases into the ranges 0–2, 3–5, 6–8 and 9–11. These ranges are called class intervals. When the whole range of variable values is classified in some groups in the form of intervals, then each such interval is known as a class interval. The range of a class interval is called the class limit. The smaller value of a class interval is called the lower class limit, and the greater value is called the upper class limit. For example, in the class interval 3–5, 3 is the lower class limit and 5 is the upper class limit. The difference between the upper limit and the lower limit of a class interval is called the class size. For the class interval of 3–5, the class size is 2. The class mark is the midpoint of each class interval. For the class interval 3-5, the class mark is 3 + 5 = 4. 2

Did You Know? Sir Ronald Fisher introduced the concept of data handling or statistics. Indian mathematicians P. C. Mahalanobis and C. R. Rao have also played a major role in the field of statistics.

Remember! class mark =

upper limit + lower limit 2

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Remember!

Error Alert!

When we take the sum of all the frequencies, it should be equal to the total number of observations.

The class size of all the class intervals must not differ.

Types of Frequency Distribution Exclusive Form (Continuous Form) Exclusive class intervals can be defined as 10–15, 15–20, 20–25 and so on. In this interval, the upper limit of one class is the lower limit of the next class.

In this form, the lower and upper limits are known as the true lower limit and the true upper limit of the class interval.

Inclusive Form (Discontinuous Form) Inclusive class intervals can be defined as 9–14, 15–20, 21–26 and so on.

In this interval, the upper limit as well as the lower limit are included in the class.

Listed below are the steps to identify the rules of choosing class intervals for a grouped frequency distribution table: 1 Identify the highest and lowest data values and find the difference between them. 2 Decide the number of class intervals needed (between 5 to 15 classes). 3 Draw the frequency distribution table to show the data. Based on the above information, we can form a frequency distribution table as follows: Number of COVID cases in each Country (in crores) (Range)

Tally Marks

Frequency

0–2

|||

3–5

||||

6–8

9–11

Reading a Frequency Distribution Table of Grouped Data In the above table, we grouped the data by ranges and counted the number of COVID cases falling into each range. This grouped frequency distribution provides a concise summary of the number of cases all around the world. Let us now read what information the table provides. •

There were 0–2 crore COVID cases in 3 countries.

•

There were 3–5 crore COVID cases in 5 countries.

•

There were 6–8 crore COVID cases in 1 country.

•

There were 9–11 crore COVID cases in 1 country.

Chapter 6 • Classification and Tabulation of Data

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Example 2

The marks of 30 students in a science test are given below. Class Intervals

Tally Marks

Frequency

0–5

5–10

|||

10–15

|||| |||

15–20

|||| |

20–25

||||

25–30

|||| ||

Total

Answer the following questions. 1 What is the lower limit of the first class interval?

2 What is the upper limit of the last class interval? 30 3 What is the size of each class?

4 Which class interval has the highest frequency? 5 Which class interval has the lowest frequency?

(10 - 15) (0 - 5)

6 What is the class mark of the second class interval? Example 3

5 + 10 15 = = 7.5 2 2

Read the frequency distribution table showing the ages of 50 people in a neighbourhood. Answer the questions that follow. Class Intervals

Tally Marks

Frequency

0–10

|||

10–20

|||| |||| |

20–30

|||| |||| |||

30–40

|||| |||| ||||

40–50

|||| |||

Total

1 What is the upper limit of the first class interval? 10 2 What is the upper limit of the last class interval? 50 3 What is the size of each class? 10 4 Which class interval has the highest frequency?

(30 - 40)

5 Which class interval has the lowest frequency? (0 - 10) 30 6 What is the class mark of the second class interval? 10 + 20 = = 15 2 2

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Do It Together

The data below shows the annual income (in thousands) (in ₹) of 40 families in a town. 71

Prepare a frequency distribution table for the data using the class interval of 5, in your notebooks. Answer the following questions. 1 What is the lower limit of the first class interval?

_________

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

_________

3 What is the size of each class?

_________

4 Which class interval has the highest frequency?

_________

5 Which class interval has the lowest frequency?

_________

6 What is the class mark of the fifth class interval?

_________

Do It Yourself 6B 1 The following is the distribution of weights (in kg) of 46 persons: i

What is the lower limit of class 50–60?

Find the class marks of the classes 40–50 and 50–60.

iii

What is the class size?

2 Look at the given observations. These figures show the percentage of crops that were harvested in various seasons.

53, 61, 48, 60, 78, 68, 55, 100, 67, 90, 75, 88, 77, 37, 84, 58, 60, 48,

Weight (in kg)

Persons

a 30–40

b 40–50

c 50–60

d 60–70

e 70–80

62, 56, 44, 58, 52, 64, 98, 59, 70, 39, 50, 60

a Arrange these observations in ascending order, say 30 to 39 as the first group, 40 to 49 as the second group, and so on.

Now, answer the following questions: b What is the highest observation?

c What is the lowest observation?

d What is the range?

e How many observations are 75 or more?

How many observations are less than 50?

3 The observations given below show the ages of people who watch a particular TV channel. Prepare a grouped frequency table for the data:

20, 39, 42, 5, 12, 19, 47, 27, 7, 13, 40, 38, 24, 34, 15, 40, 10, 9, 3, 29, 17, 34, 23, 18, 19

Chapter 6 • Classification and Tabulation of Data

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4 Look at the given frequency distribution table. Number of Siblings

0–1

2–3

4–5

6–7

8–9

Frequency

Answer the following questions: a How many individuals in the data set have 2 or 3 siblings? b What is the total number of individuals with 4 or 5 siblings? c Calculate the percentage of individuals with 0–1 sibling. d Find the range of the number of siblings.

5 The given table shows the marks obtained (out of 20) by students in a class test. Read the table and answer the following questions. Marks

0–5

5–10

10–15

15–20

Frequency

a How many students scored marks less than 15? b How many students scored more than or equal to 15 marks? c What is the class mark of the class interval with a frequency of 6? d How many students wrote the test? e What percentage of students scored less than 15 marks?

Word Problems 1 The marks obtained by 27 students of a class in an examination are given below:

23, 18, 13, 10, 21, 7, 1, 13, 21, 13, 15, 24, 16, 3, 23, 5, 6, 8, 7, 9, 12, 20, 10, 2, 23, 24, 0

Draw a grouped frequency distribution table.

2 The weight (in kilograms) of 25 students in a class was recorded. Form a grouped

frequency distribution table for the data given below and answer the following questions. 25, 24, 20, 25, 16, 15, 18, 20, 25, 16, 20, 16, 15, 18, 25, 16, 24, 18, 25, 15, 27, 20, 20, 27, 25. a Find the size of each class. b Which class intervals have the same number of students? c

How many students weigh 18 kg to 26 kg?

d What is the class mark of the class interval with the frequency 2?

3 The heights of 25 students, measured in centimetres, were recorded as follows: 155, 149, 160, 162, 154, 149, 157, 158, 163, 171, 170,152, 155, 163, 172, 162, 162, 154, 159,

161, 171, 173, 149, 157, 155

Represent the above data in a grouped frequency distribution table, using tally marks.

4 The electricity bill amounts (in ₹) for each of the 24 houses in a village are given below. Construct a frequency table.

215, 203, 120, 350, 800, 600, 350, 400, 120, 340, 150, 562, 452, 125, 658, 235, 645, 450, 207, 489, 263, 500, 153, 450

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Points to Remember • A collection of numerical figures that represent a similar kind of information is called data. • A table is called a frequency distribution table when it shows how many times each specific number or figure appears in a data set. • In grouped data, individual observations are arranged in groups such that a frequency distribution table of these groups helps provide a convenient way of presenting or analysing the data. • We follow these basic steps when drawing a frequency distribution table for a data set that contains multiple observations: • Find the lowest and highest values of the variables. • Decide the size of the class intervals. • Write the frequency of each interval.

Math Lab Setting: In pairs Materials Required: A dice, a pen and a sheet of paper Method: 1

Divide the students into pairs.

Each pair rolls the dice 15 times.

Record the observation after each roll.

Create a frequency distribution table to record the observations.

The first pair to complete the frequency table of the 15 observations wins.

Chapter Checkup 1 Construct a frequency table for each data set. a 4, 3, 6, 5, 2, 4, 3, 3, 6, 4, 2, 3, 2, 2, 3, 3, 4, 5, 6, 4, 2, 3, 4 b 6, 7, 5, 4, 5, 6, 6, 8, 7, 9, 6, 5, 6, 7, 7, 8, 9, 4, 6, 7, 6, 5

2 Represent the following data in the form of a frequency distribution. 1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2, 2.

3 A dice was thrown 25 times, and the following scores were obtained: 6, 4, 2, 5, 3, 3, 1, 2, 6, 1, 3, 1, 2, 3, 4, 5, 6, 6, 2, 4, 4, 4, 5, 6, 1 Prepare a frequency table of the scores. Chapter 6 • Classification and Tabulation of Data

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4 A survey was conducted among students, and the observations below show that the students are preparing for

the entrance examinations when they are 12–18 years old. Prepare a frequency table based on the following ages: 13, 14, 13, 12, 14, 13, 14, 15, 13, 14, 13, 14, 16, 12, 14, 13, 14, 15, 16, 13, 14, 13, 12, 17, 13, 12, 13, 13, 13, 14

5 The given table shows the number of donors of each blood group in a hospital’s blood donation campaign. Blood Group

Frequency

a Which blood group has the most donors? b How many people participated in the blood donation campaign?

6 Complete the table given below: Class Limit

Class Interval

Lower Limit

Upper Limit

Class Size

Class Mark

0–10 10–20 20–30 30–40 40–50

Word Problems 1

The populations of some Indian states are given in the table below. State Population (in crores)

Meghalaya Maharashtra Bihar Andhra West Nagaland Uttarakhand Pradesh Bengal 3

a What is the population of Meghalaya and Nagaland together? b Which is the most populated state? c Which is the least populated state?

survey was taken across 20 houses, and the residents were asked about the number of cars A registered in their households. The results were recorded as follows: 4, 1, 4, 0, 2, 1, 1, 2, 1, 0, 4, 2, 3, 2, 0, 2, 1, 0, 3, 2 Present this data in a frequency distribution table. Also, find the maximum number of cars registered by a household.

Dr. Radha recorded the pulse rate (per minute) of 29 persons: 1, 75, 71, 72, 70, 65, 77, 72, 67, 80, 77, 62, 71, 74, 79, 67, 80, 77, 62, 71, 74, 61, 70, 80, 72, 59, 6 78, 71, 72 She wants to construct a frequency table in the inclusive form, taking the class interval 61–65 of equal width. Now, if she needs to convert this data again into the exclusive form in a separate table, how can she do that?

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Bar Graphs and Histograms

7 Let’s Recall

Bar graphs are a representation of numerical data with the help of rectangular bars which have equal widths. Each bar represents a category and the area of the bar is proportional to the frequency or count of that category. y

Scale: 1 division = 10 Look at the bar graph; it shows the number of different types ofFruit fruit sold by a fruit vendor. 100

How many oranges were sold?90 60

y 90 80

60 50

What is the difference between the number of 40 pears and melons sold? Number of pears sold = 40

30 20

Number of melons sold = 30 10 Difference = 40 – 30 = 10

Let's Warm-up

Number of Fruit

How many apples were sold? 70 70

70 60 50 40 30

20 Orange Apple Banana Pear 10 Fruit

Melon

Orange Apple Banana Pear Fruit

Look at the bar graph below and fill in the blanks.

Melon

Scale: 1 division = ₹2000

10,000 8000 6000 4000 2000 A

1 Investor A made an investment of ₹_______.

Amount Invested (in ₹)

Scale: 1 division = 10 Fruit s

100

Scale: 1 division = ₹2000

10,000 8000 6000

4000 B C D Investors2000

2 The investment made by D is ₹_______.

x A

B C Investors

3 The highest investment is made by _______. 4 The difference between the investments of B and C is ₹_______. 5 The total investment made by all the investors is ₹_______. I scored _________ out of 5.

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Mean, Median and Mode of Ungrouped Data Graphical Representation Real Life Connect

India, often hailed as an agricultural powerhouse, has made significant strides in the global wheat market over the past few decades. With its diverse agro-climatic zones, extensive cultivation practices and technological advancements, India has emerged as a prominent wheat exporter on the world stage.

Bar Graphs India exports wheat to multiple countries. Let us see approximately how many tonnes of wheat India exported to different countries in the year 2021–22. Country

Quantity (in tonnes)

Sri Lanka

94,000

Yemen Republic

86,000

Indonesia

56,000

Afghanistan

55,000

Qatar

63,000

Drawing and Reading Bar Graphs Let us plot the data on a bar graph.

• All the bars are of uniform width and the space between all the bars is consistent. • Bars are drawn over one axis and the values of the variables are shown across the other axis. The height of a bar represents the value of the variable. What is the average of the wheat exported to the 5 countries? Total wheat exported to these 5 countries = 94,000 + 86,000 + 56,000 + 55,000 + 63,000 = 3,54,000 tonnes of wheat

100,0000

Scale: 1 division = 10,000 tonnes

90,000 Quantity (in tonnes)

• A bar graph is a pictorial representation of ungrouped data.

80,000 70,000 60,000 50,000 40,000 30,000 20,000 10,000

Average = 354000 = 70,800 tonnes of wheat 5 What percentage of the total wheat is exported to Qatar?

Sri Lanka

Yemen Indonesia Afghanistan Qatar Republic

Country

Wheat exported to Qatar = 63,000 tonnes Total wheat exported = 94,000 + 86,000 + 56,000 + 55,000 + 63,000 = 354,000 tonnes 63000 Percentage = × 100% = 17.8% 354000

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Example 1

The number of mathematics books sold by a publishing house on 6 different days is shown using a bar graph. Read the bar graph and answer the questions. 1 How many books were sold on Wednesday? 2 W hat was the ratio of the books sold on Monday to the books sold on Tuesday? Number of books sold on Monday = 800 Number of books sold on Tuesday = 700 Ratio of the number of books sold on Monday to those sold on Tuesday = 800:700 = 8:7 3 W hat percentage of the total books were sold on Thursday?

Number of Mathematics Books

Number of books sold on Wednesday = 870

1200

Number of books sold on Thursday = 970 Total number of books sold = 800 + 700 + 870 + 970 + 670 + 1090 = 5100

Scale: 1 division = 100 books

1100 1000 900

800 700 600 500 400 300 200 100 ay

d on

Percentage of the total books that were sold on 1 970 Thursday = × 100% = 19 % 51 5100

d es

a sd

a rid

d ur

Day

4 What is the average number of books sold per day over the 6 days?

Total number of books sold = 800 + 700 + 870 + 970 + 670 + 1090 = 5100 Average number of books sold per day over the week = 5100 ÷ 6 = 850 Example 2

The data shows the foreign exchange reserves of a country. Draw a bar graph to represent the data and answer the questions. Year

Foreign Exchange Reserve (in $million)

2010

2800

2009 2011 2012 2013

2600 2500 3100 3500

1 The foreign exchange reserve in 2013 is how many times of that in 2011? Foreign exchange reserve in 2011 = $2500 million Foreign exchange reserve in 2013 = $3500 million 3500 ÷ 2500 = 1.4 times 2 By what percentage did the foreign exchange reserve change in 2010 from the previous year? Foreign exchange reserve in 2010 = $2800 million Foreign exchange reserve in 2009 = $2600 million Change in the foreign exchange reserve

Chapter 7 • Bar Graphs and Histograms

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Percentage change =

200

2600

× 100% ꞊ 7 % 13

3 W hat is the ratio of the number of years in which the foreign exchange was above the average reserve to those in which the reserve was below the average reserve? Total foreign exchange reserve in the 5 years = 2600 + 2800 + 2500 + 3100 + 3500 = $14,500 million Average foreign exchange reserve for the 5 years = 14,500 ÷ 5 = $2900 million Years in which foreign exchange was above the average reserve, that is, in 2012 and 2013 = 2

y Foreign Exchange Reserves (in $million) Foreign Exchange Reserves (in $million)

= $2800 million – $2600 million = $200 million

Scale: 1 division = $500 million

5000 4500 4000 3500 3000 2500 2000 5000 1500

Scale: 1 division = $500 million

4500 1000 4000 500 3500

2009

3000

2010

2500

2012

2011 Year

2013

2000 reserve = 2009, 2010 and 2011 = 3 Years in which foreign exchange was the below the average

Hence, the required ratio is 2:3.

1000

The length (in km) of some major rivers in India are given 500 below. Draw a bar graph for the data and answer the questions. 1300

Ganga

2500

Brahmaputra

2900 y

Ratio of the length of river Mahanadi to that of river Kaveri = _______ : _______ = _______ : _______

an ad i

ut m ap

Ka ve

ng ah

1000

na di

ah a

ve ri

500

Length of river Kaveri = ________

River

1500

tra

Length of river Mahanadi = 900 km

2000

ap u

4 W hat is the ratio of the length of river Mahanadi to that of river Kaveri?

2500

Scale: 1 division = __________

Shortest river = ________

3000

Br ah

3 Which is the shortest river?

500

ng a

Longest river = ________

1000

2 Which is the longest river?

1500

1 division = ______ km

2000

rm ad

Scale of the graph:

2013

2500900

1 What is the scale of the graph?

3000800

Mahanadi

2012

Scale: 1 division = __________

Kaveri

Length of a River (in km)Length of a River (in km)

Narmada

2011 Year

2010

River

2009 Length (in km)

Do It Together

1500

River

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5 What is the difference of the lengths of river Ganga and Narmada? Length of river Ganga = _______ Length of river Narmada = _______ Difference in length = _______ − _______ = _______

Do It Yourself 7A 1

Refer to the bar graph for the wheat exported by India to different countries and answer the questions. a What percentage of the total export was exported to Sri Lanka? b What percentage of the total export was exported to Afghanistan? c What was the ratio of the wheat exported to Qatar to that exported to the Yemen Republic? d What is the average quantity of wheat exported to these 5 nations?

Refer to the bar graph used in example 1 and answer the questions. a What percentage of the total books were sold on Friday? b What percentage of the total books were sold on Saturday? c What is the ratio of the number of books sold on Thursday to the number of books sold on Friday? d What is the ratio of the number of books that is less than the average number of books sold and the number of books that is more than the average number of books sold?

Refer to the bar graph used in example 2 and answer the questions. a What is the foreign exchange reserve for 2013? b What is the difference between the reserves for 2011 and 2012? c What is the percentage increase in the reserve for 2013 from that for the previous year? d What is the ratio of the reserves of 2010 and 2011 together to the reserves of 2012 and 2013 together?

The data shows the export of pears in different years. Draw a bar graph for the given data and answer the questions.

Year

2015

2016

2017

2018

2019

2020

Export (in crores)

15.6

14.8

16.7

17.0

19.4

19.8

a What was the total value of exports in all the years? b What was the average export value of pears for the given period of time? c What was the percentage increase in 2018 from 2015? d Which two years have an average export value of ₹17.1 crores?

Chapter 7 • Bar Graphs and Histograms

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5 The bar graph shows the number of students who

the bar graph and answer the questions.

Number of Students

500 450 400

b What percentage of the students passed in 2018? 350 c What percentage of the students passed in 2020?

300 250

450 Scale: 1 division = 50 students 400

Number of Students

students in the school was 450 every year. Read a How many students passed in 2021?

Scale: 1 division = 50 students

500

passed in an examination. The total number of

350 300 250 200 150 100

d What was the percentage increase in the number 200

of students passing in 2022 from those passing in150 2019? 100

2018

Word Problem

2021 2022

2020 Year

2021 2022

2020 Year

w sco

iro

sco

iro

City

or wY

ai Ne

k Sh or

ai an

lhi De

ai gh

populations of Moscow and New York?

d What is the difference between the

the six cities?

lhi

c What is the average population of

populations of Shanghai and Cairo?

Population (in millions)

Scale: 1 division = 5 million 35 The bar graph shows the approximate populations of different cities in the world. Read the bar 30 y graph and answer the questions. Scale: 1 division = 5 million 25 35 a What is the total population of Delhi 20 30 and Mumbai? 15 25 b What is the difference between the 10 20 Population (in millions)

2019

Double Bar Graphs India exports fruit, cereals, dairy products and many other products. Let us see the products that India exported in April–September 2021 and April–September 2022. 2021 (April–September) (in million $)

2022 (April–September) (in million $)

Fruit

301

313

Cereal preparations & miscellaneous processed items

1632

2111

Meat, dairy & poultry products

1903

2099

Basmati rice

1660

2280

Non-basmati rice

2969

3207

Other products

2591

3761

Products

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Drawing Double Bar Graphs Let us represent the data on a double bar graph to see the increase or decrease in the exports of various products. A double bar graph shows two sets of data simultaneously. It helps us compare two sets of data. y

Scale: 1 division $ 500 million Write the = scale of the graph. Scale: = $ 500 million 40001 division y Scale: 1 division = $ 500 million

2022 2021 2022

3761 3761

2591 2591

3207 3207

2969 2969

2280 2280

1660 1660

2099 2099

2022 19033761 1903

2099

2111

Draw two rectangular bars

representing the data.

500 Fruit NonCereal Other Meat, x Basmati NonOther Fruit Cereal Meat, Basmati preparations & Dairy rice basmati products preparations & Dairy rice basmati products & rice Fruitmiscellaneous Cereal Meat, Basmati NonOther miscellaneous & rice processed & poultry preparations Dairy rice basmati products processed poultry items products miscellaneous & rice items products processed poultryProduct Product items products 313

500

313

1000

1632

1500

2111 2111 2591

2000

2021

16323207 1632

2500

313 2969

3000

3500 4000 3000 3500 2500 3000 2000 2500 1500 2000 1000 1500 500 1000

301 2280 301

3500

1660

1903 Exports (in millons $) $) Exports (in millons

4000

301

Label both the axes.

Exports (in millons $)

Product

Consider the data shown below for the sales of different brands of laptops by a company during 2020 y and 2021. Draw a bar graph toy represent the data. Number of Laptops (in thousands)

120 Number of laptops Number of laptops 110 Brand sold in 2020 100 sold in 2021 90(in thousands) (in thousands) 80 70 A 80 90 60 50 B 95 90 40 30 C 75 100 20 10 D 100 80

115

A 115

100

110

120 y 110 120 100 110 90 80 100 70 90 60 80 50 70 40 60 30 50 40 20 30 10 20 C 10 D Brand

Scale: 1 division = 10 thousand

2020

2021 2020

2021

x Brand Brand

x x

Consider the data shown for the production of cars during the years 2012–14 from different manufacturers. Complete they bar graph. y Scale: 1 division = 2 thousand

2012–13 18 2013–14 16 (in thousands) (in thousands) Number of Cars (in thousands)

Manufacturer

14 12 10 8 6 4

8 10 20 15

2 A

Chapter 7 • Bar Graphs and Histograms

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Number of Cars Number of Cars (in thousands) (in thousands)

Do It Together

Scale: 1 division = 10 thousand Scale: 1 division = 10 thousand

Number of Laptops Number of Laptops (in thousands) (in thousands)

Example 3

20 y 18 20 16 18 14 16 12 14 10 12 8 10 6 8 4 6 2 4 2

Scale: 1 division = 2 thousand

2012-13

2013-14

2013-14 2012-13 2013-14

Manufacturer

A A

x B

Manufacturer Manufacturer

x x

08-01-2024 14:48:30

Reading Double Bar Graphs

2022

3500 3000 2500 2000

3761

3207

2280

1660

2099

1903

500

2111

1000

1632

1500

Meat, Basmati Fruits Cereal Other Nonpreparations & dairy basmati products rice miscellaneous and rice processed poultry products items

hich product category had the greatest 2 W increase in exports?

The product category with the greatest increase in exports is other products.

Product

hat is the average of the percentage increase in ythe export of each 3 W Scale: 1 division = 10 tonnes 100 product category? Production of Tea (in tonnes)

90 313 – 301 × 100% = 3.98% 301 80 Percentage increase in exports of cereal preparations & 70 2111 – 1632 60 miscellaneous processed items = × 100% = 29.35% 1632 50

Percentage increase in exports of fruit =

2021

Scale: 1 division = $ 500 million

313

Amount of non-basmati rice exported in 2022 = $3207 million 3207 – 2969 Percentage increase in exports = 2969 × 100% ≈ 8%

4000

301

Amount of non-basmati rice exported in 2021 = $2969 million

Exports (in millions $)

hat was the percentage increase in 1 W the exports of non-basmati rice?

2591

Read the bar graph and answer the questions.

2969

Consider the bar graph shown below for the export of various products by India in the year 2021 and 2022.

2016

2017

Which product has the highest percentage increase from the previous year?

Percentage increase in exports of meat, dairy &40 poultry products = 30

2015

Think and Tell

2099 – 1903 × 100% = 10.29% 1903

2280 – 1660 Percentage increase in exports of basmati rice =20 × 100% = 37.34% 1660 10

Percentage increase in exports of non-basmati rice =

3207 – 2969 × 100%B = 8.01% C A 2969

Company

3761 – 2591 Percentage increase in exports of other products = × 100% = 45.16% 2591 Average percentage increase =

The bar graph shows the production of tea by different companies in different years. Read the bar graph and answer the questions. y hat is the production of tea by 1 W different companies in the year 2016? Production of tea by company A in 2016 = 60 tonnes Production of tea by company B in 2016 = 70 tonnes Production of tea by company C in 2016 = 70 tonnes

Production of Tea (in tonnes)

Example 4

3.98% + 29.35% + 10.29% + 37.34% + 8.01% + 45.16% 134.13 = % = 22.355% 6 6

100 90 80

Scale: 1 division = 10 tonnes

2015

2016

2017

70 60 50

40 30 20

B Company

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Production of tea by company D in 2016 = 40 tonnes Total tea production in 2016 = 60 + 70 + 70 + 40 = 240 tonnes he total production of the four companies in 2015 is what percentage of the production by Company 2 T B and D together in 2017? Total production of the four companies in 2015 = 50 + 70 + 60 + 50 = 230 tonnes Total production by company B and D in 2017 = 90 + 50 = 140 tonnes

230 × 100% = 164.28% 140 3 What is the ratio of the production by Company B in the three years to the production by Company C in the three years? Required percentage =

Production by company B in the three years = 70 + 70 + 90 = 230 tonnes Production by company C in the three years = 60 + 70 + 80 = 210 tonnes The required ratio = 230:210 = 23:21 Do It Together

The bar graph shows the number of people who visited four different places in two different years. Read the bar graph and answer the questions. y

umber of people who visited 1 N Qutub Minar in 2018 = 700

hat was the percentage increase 3 W in the number of people visiting the Taj Mahal in 2019 from 2018?

Number of People

umber of people who visited 2 N India Gate in 2019 = ______

Scale: 1 division = 100 people

1000

Number of people who visited Taj Mahal in 2018 = 900 Number of people who visited Taj Mahal in 2019 = ______

900 800 700 600 500

2018

400

2019

300 200 100 Qutab Minar

Percentage increase = ______

Taj Mahal

India Gate

Lotus Temple

Place

4 What is the difference between the average number of visitors in 2018 and 2019? Average number of visitors in 2018 = _______ Average number of visitors in 2019 = _______ Difference = ____ − ____ = ____

Do It Yourself 7B 1 Refer to the double bar graph shown for the export of various products in the years 2021 and 2022 and answer the given questions.

a What was the percentage increase in the exports of basmati rice? b Which product has the least exports? Chapter 7 • Bar Graphs and Histograms

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What were the total exports for the year 2021?

What were the total exports for the year 2022?

2 Refer to the bar graph in example 4 and answer the questions. a What is the total production of tea by company C in three years? b What is the ratio of the tea produced by company B in 2015 and 2016 to the total tea produced by company D in three years?

c What is the percentage increase in the tea production by company D from 2016 to 2017? d What is the percentage increase in the tea production by company B from 2015 to 2017?

3 Refer to the bar graph shown in Do it Together above and answer the questions. a What was the percentage increase in the number of people visiting Qutab Minar in 2018 from those visiting in 2019? b How many people in total visited India Gate in both years? c What is the ratio of the total number of people who visited the Lotus temple to those who visited the Taj Mahal? d Which place had the greatest percentage increase in the number of visitors?

4 The double bar graph shows the different types of fruit sold by two vendors. Read the bar graph and answer the given questions.

Scale: 1 division = 10 fruits 100 by vendor A? 90 80 b What is the total number of fruits sold 70 y Vendor A by vendor B? Scale: 1 division = 10 fruits 60 100 Vendor B 50 c What is the ratio of the oranges and 90 40 80 30 apples sold to the other three types of 70 20 fruit sold by vendor A? Vendor A 60 10 Vendor B x 50 d Which was the highest-selling fruit Orange Apple Banana Pear Melon 40 out of the 5 types of fruit by the two 30 Fruit 20 vendors together? 10 y x Scale: 1 division = 20 tickets Apple Banana The double bar graph shows the number of movie tickets sold by two different cinemaPear halls forMelon different genres of 200 Orange Fruit 180 movies. Read the bar graph and answer the questions. 160 a What is the total number of movie tickets 140 y Cinema Scale: 1 division = 20 tickets 120 sold for a comedy genre by both theatres? hall 1 200 100 180 80 b How many tickets were sold in total by Cinema 160 60 hall 2 cinema hall 1? 140 40 Cinema 120 20 c How many tickets were sold in total by hall 1 100 x Horror Sci-fi Action Comedy Drama cinema hall 2? 80 Cinema Movie Genre 60 hall 2 d What percentage of tickets for sci-fi is 40 20 sold by Cinema 2 in comparison with x Cinema 1? Horror Sci-fi Action Comedy Drama Movie Genre

Number of Tickets Sold Number of Tickets Sold

Number of Fruits Number of Fruits

a What is the total number of fruits sold

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Word Problem 1 The double bar graph shows the runs scored by Virat Kohli and Rohit Sharma in different years. Read the bar graph and answer the questions.

a How many runs were scored in total by Virat Kohli in the given years? b How many runs were scored in total by Rohit Sharma in the given years? c What is the difference of the average runs scored by Virat Kohli and Rohit Sharma? d What is the ratio of the runs scored by Virat Kohli before 2017 and runs scored by Rohit Sharma after 2015?

Virat Kohli

Scale: 1 division = 200 runs

1600

Rohit Sharma

1400

Runs

1200 1000 800 600 400 200 2013

2014

2015

2016

2017

2018

2019

Years

Graphical Representation of Grouped Data Real Life Connect

In the city of Metropolis, a team of public health organisations organised a survey to keep a check on the health of the people. They maintained the data for the number of people and the number of minutes they exercised daily.

Chapter 7 • Bar Graphs and Histograms

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Histograms A histogram is a special type of bar graph. It is a graphical representation of grouped data. The data collected for the number of people who exercise for different numbers of minutes is shown below. Number of minutes

Number of people (in thousands)

0–30

30–60

60–90

90–120

120–150

150–180

Did You Know? The histogram was first described by Karl Pearson.

Drawing Histograms Let us draw a histogram for the above data.

Take the frequency on the y-axis.

Number of minutes people spent exercising

100 90

Number of People

80 70 60 50 40 30 Draw bars of equal width.

20 10 30

120

Number of Minutes

150

180

210

Take class intervals on the x-axis.

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Error Alert! The width of the bars in a histogram must be consistent. y

20 25 30 35 40

Example 5

x 45 50 55 20 60 25 30 35 40

x x 25 60 30 35 40 45 50 55 20 60 25 30 35 40 45 50 55 60 45 5020 55

The data shows the daily wages of workers in a factory. Draw a histogram for the data given. Wages (in ₹)

100–200

200–300

300–400

400–500

500–600

600–700

Number of workers

100

150

180

120

Wages of Workers

200 180 160

Number of Workers

140 120 100 80 60 40 20

100

Chapter 7 • Bar Graphs and Histograms

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200

300

400 500 Wages (in ₹)

600

700

800

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Do It Together

The data shows the marks scored by students in an examination of 100 marks. Draw a histogram for the given data. Marks

20–30

30–40

40–50

50–60

60–70

Number of Students

70–80

80–90

Marks Scored by Students 40 10

90–100

y 40

Marks Scored by Students 30

Number of Students

30 25

20 15

100

Marks 20

100

Marks

Reading Histograms The histogram shows the number of minutes of exercise done.

y 100

How many people exercised for more than 1 hour? 1 hour = 60 minutes

Number of People

What percentage of the total number of people exercised 80 for less than 90 minutes? 70

Total number of people = 50 + 90 + 75 + 60 + 50 + 40 = 365 60

Number of people who exercised for less than 90 minutes 50 = 50 + 90 + 75 = 215

80 Number of People

So, the number of people who 100 exercised for more than 60 minutes = 75 + 60 + 50 + 40 = 225 people

40 30

10 60 90 120 150 180 210

Percentage of all the people who exercised for less than 30 90 minutes 215 = × 100% = 58.9% 365

60 90 120 150 180 210

Number of Minutes

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Example 6

The marks obtained by students in a test are given below. Draw a histogram to represent the data and answer the questions. 48, 41, 35, 85, 99, 57, 69, 87, 76, 41, 23, 35, 22, 48, 47, 86, 97, 65, 53, 52, 84, 97, 74, 73, 34, 46, 98, 99, 56, 57, 36, 74, 98, 46, 65, 66, 97, 89, 85, 84, 86, 47, 74, 73, 71, 66, 61, 67, 34, 23, 24, 24, 43, 54 Marks

20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

Mark Scored by Students

Number of Employees

Number of Students

10 Number 9 of 5 5 9 6 7 7 8 7 8 Students 7 1 How many students scored more than 30 marks? 6 Y 5 Mark Scored by Students Number of students who scored more than 30 marks 10 4 = 5 + 9 + 6 + 7 + 7 + 8 + 7 = 49 9 3 2 If the pass mark is 40, how many students did not pass the8 2 7 test? 1 6 Number of students who did not pass the test = 5 + 5 = 10 5 0 20 30 40 50 60 70 80 90 100 3 How many students appeared for the test? 4 Marks 3 Total number of students who appeared for the test = 5 + 5 + 9 + 6 + 7 + 7 + 8 + 7 = 54 2 The histogram shows the ages of employees in an office. Observe the graph and answer the 1 Y questions. Age of EmployeesX 50 0 20 30 40 50 60 70 80 90 100 1 How many employees are older than 30 years? Marks 45 Number of Students

Do It Together

Number of employees older than 30 years = 40 + 30 + 20 + 20 + 540+ 5 = 120

Percentage of total employees who are older than 40 years = ________ 3 W hat is the ratio of employees older than 35 years to those younger than 35 years? Number of employees older than 35 years = ________ Number of employees younger than 35 years = ________ Required ratio = ________ : ________

Number of Employees

2 What percentage of the total number of employees is older than3540 years? Y 30 Age of Employees Number of employees older than 40 years = ________ 50 25 Total number of employees = ________ 45 20 40

20 15

20 25 30 35 40 45 50 55 60 Marks

10 5

20 25 30 35 40 45 50 55 60

4 What percentage of the total number of employees is younger than 40 years?Marks Number of employees who are younger than 40 years = ________ Total number of employees = ________

Percentage of total number of employees who are younger than 40 years = ________

Chapter 7 • Bar Graphs and Histograms

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Do It Yourself 7C 1

Refer to the histogram for the number of minutes people exercised and answer the questions. a What is the total number of people on whom the survey was conducted? b What percentage of the total number of people exercised for more than 30 minutes? c What percentage of the total number of people exercised for less than 120 minutes? d What is the ratio of the number of people who exercised less than 90 minutes and the number of people who exercised more than 120 minutes?

Refer to the histogram in Example 5 and answer the questions. a How many workers earn more than ₹300? b How many workers earn less than ₹500? c What percentage of the total number of workers earn less than ₹600? d What is the ratio of the number of workers who earn less than 400 to the number of workers who earn more than 400?

e How many workers make between ₹400–₹600?

Refer to the histogram in Example 6 and answer the questions. a What percentage of the total number of students scored more than 40 marks? b What percentage of the total number of students scored less than 30 marks? c What is the ratio of students who scored more than 50 marks to students who scored less than 50 marks? d What is the difference between the number of students who scored more than 40 marks and the number of students who scored less than 40 marks?

Refer to the histogram in Do It Together above and answer the questions. a How many employees are older than 50 years? b What is the ratio of the number of employees who are younger than 35 years to those who are older than 35 years? c What percentage of the total number of employees is older than 25 years? d How many employees are there in total?

The data shows the monthly rainfall for a city over different years. Draw a histogram to represent the data and answer the questions. Year

2012-2013

2013-2014

2014-2015

2015-2016

2016-2017

Rainfall (in mm)

700

600

550

650

800

a What was the total rainfall over the years? b During which year did the highest rainfall occur? hat was the percentage increase/decrease in the rainfall during 2014–15 in comparison with the c W previous year? d Which two years had the highest percentage difference in rainfall over the previous year?

100

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Word Problem 1

The data shows the ages of US presidents at the time of their inauguration. Draw a histogram to represent the data and answer the questions. (Consider only the ages in years when drawing the histogram.) Age Group

42–48

48–54

54–60

60–66

66–72

72–above

Number of Presidents

a How many presidents were more than 66 years old at the time of their inauguration? b How many presidents were more than 60 years old at the time of their inauguration? c How many presidents were more than 54 years old?

Points to Remember •

The collection of numerical facts of some information is called data.

•

Ungrouped data can be represented using a bar graph.

•

The number of times a particular observation occurs in the data is called its frequency.

• Histograms are special types of bar graphs which are used to represent grouped data. In histograms, the class intervals are shown on the horizontal axis and the heights of the rectangular bars show the frequency of the class interval. There is no gap between the class intervals.

Math Lab The Shoe Size Survey! Aim: To introduce students to the concept of histograms and help them create one based on real data. Materials Required: Chart paper, markers, sticky notes, measuring tape or ruler Settings: Whole class Method: 1 2 3

Explain to the students what a histogram is.

Draw a horizontal axis on the chart paper which represents the shoe size of the students. Draw a vertical axis on the chart paper which represents the number of students.

4 Ask each student to measure their shoe size using the measuring tape or ruler and write the sizes on the sticky notes. 5 6

Use markers to draw a histogram for the data collected. Help the students to interpret the data.

Chapter 7 • Bar Graphs and Histograms

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Chapter Checkup 1 The graph shows the percentage break-up of sales of units of different products in 2020. Read the bar graph and answer the questions. The total number of units sold by the a How many units of product A were sold? b How many units of product C were sold? c What is the difference between the number of units of product B

Percentage

company was 5,20,000.

and units of product D sold?

d What is the ratio of the units of product A and units of product

40 36 32 28 24 20 16 12 8 4

Scale: 1 Division = 4%

C sold?

Product

2 The bar graph shows the number of students who passed out of different colleges in 2022. Read the bar graph and answer the questions.

b The pass percentage of college A is the same as the pass

percentage of college E. By what percent is the total strength of college E more than the total strength of college A?

c If 90% of students passed out of college B, how many students

failed in that college?

d The ratio of the total students who passed out of all the

Number of Students

a How many students passed out of college B? 400 360 320 280 240 200 160 120 80 40

Scale: 1 Division = 40 students

colleges to the total students who failed in all the colleges is

43:9. How many students failed in total from all the colleges?

College

3 The bar graph shows the number of students who attended the workshop on different days of a week. Read the bar graph and answer the questions.

b What percentage of the total number of students attended

the workshop on Monday?

c What is the ratio of the number of students who attended the

workshop on Monday to those who attended the workshop on Wednesday?

d What percentage is the number of students who attended the

Number of Students

a How many students in total attended the workshop?

200 180 160 140 120 100 86 60 40 20

Scale: 1 Division = 20 students

Monday

workshop on Tuesday of those who attended on Friday?

Tuesday Wednesday Thursday

Day

Friday

4 The data shows the number of students of two different schools who participated in different activities at a function. Draw a bar graph to represent the given information. Activity

Athletics

Dance

Arts

Music

Theatre

Number of students from school A

Number of students from school B

102

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5 The bar graph shows the cars sold by two different

dealers. Read the bar graph and answer the questions.

Scale: 1 Division = 100 cars

Number of cars Sold

1000 900 800 700 600 500 400 300 200 100

a How many cars were sold by dealer A in total? b How many cars were sold by dealer B in total? c What is the ratio of the cars sold in 2019 to those sold

in 2022?

Dealer A Dealer B

d What percent of the total number of cars sold by

2018

2020 Year

2019

dealer A were sold in 2018?

6 The bar graph shows the number of biscuits sold by

answer the questions.

a What percentage of the total sales of C2 for both

years is the total sales of C4 for both years? ?

b What are the average sales of all the companies

in the year 2019?

c What percentage of the average sales of C1, C2

and C3 in the year 2019 is the average sales of

Number of Biscuits (in thousands)

5 companies in two years. Read the bar graph and

2021

2022

Scale: 1 Division = 20 thousand biscuits

120 100 80

2019 2020

60 40 20

C3, C4 and C5 in the year 2020?

d What are the average sales of all the companies

Company

2021

2022

in the year 2020?

total in the year 2022?

c What is the ratio of the number

of candidates appearing for the

entrance exam from Mumbai and Bangalore in the year 2021?

d What is the average number of

25 20 15 10 5

candidates appearing for the

Mumbai

Bhopal

Delhi

entrance exam for the year 2021?

25 lakhs 29 lakhs

b How many candidates appeared in

31 lakhs 34 lakhs

total in the year 2021?

26 lakhs 30 lakhs

a How many candidates appeared in

Scale: 1 division = 5 Lakhs

25 lakhs 29 lakhs

graph and answer the questions.

Number of Students (in lakhs)

exam from various cities. Read the bar

25 lakhs 30 lakhs

candidates appearing for an entrance

30 lakhs 31 lakhs

7 The bar graph shows the number of

Bangalore Chandigarh Kolkata City

e What percentage is the number of candidates from Delhi in 2021 of the number of candidates from Bhopal in

the same year?

8 Draw the histogram for the given data. Class Interval

5–15

15–25

25–35

35–45

45–55

55–65

Frequency

Chapter 7 • Bar Graphs and Histograms

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9 The histogram shows the heights of students in a school. Study the histogram and answer the questions. y

Height of Students

Number of Students

35 30 25 20 15 10 5 120 125 130 135 140 145 150 155 160 165 170 Height (in cm)

a How many students are taller than 140 cm? b How many students are shorter than 165 cm? c What is the ratio of the students who are shorter than 150 cm to those taller than 150 cm? d How many students are taller than 155 cm?

10 The histogram shows the number of cars using the road from 6 a.m. to 12 p.m. Read the histogram and answer the given questions.

y 140

Number of Cars

120 100 80 60 40 20 6

7 8 9 10 11 12 Time

a How many cars are there in total? b How many cars use the road after 9 a.m.? c How many cars use the road before 11 a.m.? d What is the ratio of the number of cars that use the road between 6 a.m. and 9 a.m. to the number of cars that

used the road after 10 a.m.?

104

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Word Problems 1

The histogram shows the frequency distribution of average runs scored by players in a cricket tournament.

a How many players scored an average of less

than 30 runs?

b How many players scored an average of c What percentage of players scored an

average of less than 40 runs?

d How many players are there in total? e What percentage of players scored more

40 Number of Players

more than 20 runs?

than 25 runs?

Scale: 1 division = 5 players

35 30 25 20 15 10 5

What is the ratio of players who scored more

10 15 20 25 30 35 40 45 50 Average

than 25 runs and players who scored less than 25 runs?

2 Consider the data given for the number of goals scored by Lionel Messi and Cristiano Ronaldo

in various years for their respective clubs. Draw a double bar graph for the data and answer the questions. Year

2005-06 2006-07 2007-08 2008-09 2009-10 2010-11 2011-12 2012-13 2013-14

Messi

Ronaldo

a How many goals were scored in total by Messi and Ronaldo individually over the years? b Who scored more goals over the years? c What is the ratio of the number of goals scored by Lionel Messi before 2010 and the number

of goals scored by Cristiano Ronaldo after 2007?

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38 Pie Charts Let’s Recall We have learnt that a pie chart is a type of graph that represents data in a circular graph. The circle represents all the available data, such as favourite fruits or school subjects. Each slice of a pie represents a different category. Let us read the pie chart showing the number of books read by 4 friends. Shalini, Priya, Aman and Kuldeep read 40 books in six months. They represent the data using a pie chart. The pie chart shows the fraction of the total number of books read by each person. Fraction of Books Read

1 1 • Aman read of the books = × 40 = 20 books 2 2 • Shalini read

1 1 of the books = × 40 = 10 books 4 4

• Priya and Kuldeep read

Priya

Aman

Shalini

1 1 of the books = × 40 = 5 books 8 8 Kuldeep

Letʼs Warm-up

The given pie chart shows the favourite colour of some students of class 8. Read the pie chart and answer the questions.

Yellow, 15

Favourite Colour Lavender, 10

Pink, 15

Blue, 35

Red, 25

1 The fraction of students who like yellow is __________________. 2 The most liked colour is __________________. 3 The fraction of students who do not like red is __________________. 4 The least liked colour is __________________. 5 T he difference of the fraction of students who like pink and the fraction of students who like blue is __________________. I scored _________ out of 5.

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Drawing and Reading Pie Charts Real Life Connect

Suman earns ₹20,000 per month. She is planning her monthly budget. She noted all her expenditures and savings in the form of a table.

Items

Amount

House Rent

₹4000

Food

₹7000

Bills

₹3000

Others

₹2500

Savings

₹3500

Drawing Pie Charts We know that a pie chart uses a circle to represent the data and hence is also called a circle graph. The whole circle represents all the data we have. The circle is divided into sectors, with each sector proportional to the size of the represented observation. Let us use the above data and learn to draw a pie chart.

Sectors and Central Angles

We know that a complete circle represents 360°. The figure below shows how to find the central angle to draw the sectors. Sector

A sector displays the size of some related piece of information.

Remember! The angle made at the centre of a circle is always 360°. Central angle

Value of the component × 360° Sum of all components

Using the above formula, the central angles for Suman’s data can be given as: Items

Amount

Measure of central angle

House Rent

₹4000

4000 × 360° = 72° 20000

Food

₹7000

7000 × 360° = 126° 20000

Bills

₹3000

3000 × 360° = 54° 20000

Others

₹2500

2500 × 360° = 45° 20000

Savings

₹3500

3500 × 360° = 63° 20000

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Example 1

Find the value of the central angle corresponding to 30 comic books out of 150 books. Central angle =

Example 2

Do It Together

Value of the components 30 × 360° = × 360° = 72° Sum of all components 150

Find the central angles for the given activities. Activity

Duration

Measure of central angle

Sleep

8 × 360° = 120° 24

School

7 × 360° = 105° 24

Tuitions

3 × 360° = 45° 24

Homework

2 × 360° = 30° 24

Others

4 × 360° = 60° 24

Remember! The sum of the measures of the central angles is equal to 360°.

Find the central angles for the given modes of transport. Activity

Duration

Measure of central angle

Bike

130

130 × 360° = 78° 600

Car

Bus

200

Walk

Others

110

Drawing a Pie Chart Let us draw a pie chart using the central angles calculated in the previous section. Items

Amount

Measure of central angle

House Rent

₹4000

4000 × 360° = 72° 20000

Food

₹7000

7000 × 360° = 126° 20000

Bills

₹3000

3000 × 360° = 54° 20000

Others

₹2500

2500 × 360° = 45° 20000

Savings

₹3500

3500 × 360° = 63° 20000

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Step 1: Draw a circle of any radius. Draw the first angle, Step 2: Draw the rest of the angles, keeping the previous keeping the horizontal radius as the base. Write the angle and the value of the data. Monthly Budget

angle as the base. Colour the sectors in different colours.

Monthly Budget

House Rent, 72°

Savings, ₹3500

₹4000

Others,

₹2500 Bills,

House Rent,

₹3000

Example 3

₹4000

63° 72° 45° 54° 126°

Food,

₹7000

Draw a pie chart for the data showing the number of students participating in different school activities. Activity

Dancing

Singing

Painting

Sketching

Running

No. of Students

300

200

100

150

250

The pie chart for the table can be drawn as:

Example 4

Activity

No. of Students

Dancing

300

300 × 360° = 108° 1000

Singing

200

200 × 360° = 72° 1000

Painting

100

100 × 360° = 36° 1000

Sketching

150

150 × 360° = 54° 1000

Running

250

250 × 360° = 90° 1000

Central Angle

School Activities Sketching, 150 Painting, 100

54° 36°

Running, 250 90°

72° 108° Singing, 200

Dancing, 300

Draw a pie chart for the table showing the number of books sold by 5 shopkeepers. Shopkeeper

Amit

Vinay

Keshav

Arun

Mark

Books sold

200

180

300

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The pie chart for the table can be drawn as:

Do It Together

Shopkeeper

No. of Books

Central Angle

Amit

200

200 × 360° = 90° 800

Vinay

180

180 × 360° = 81° 800

Keshav

80 × 360° = 36° 800

Arun

300

300 × 360° = 135° 800

Mark

40 × 360° = 18° 800

Number of books sold Arun, 300

Keshav, 80

36°

135°

81°

18°

Mark, 40

90° Amit, 200

Vinay, 180

Draw a pie chart showing the types of movies liked by teenagers in a city. Type of Movies

No. of Teenagers

Central Angle

Action

335

335 × 360° = 67° 1800

Thriller

420

Science Fiction

215

Mystery

390

Comedy

440 215 × 360° = 43° 1800

Do It Yourself 8A 1 There are 1200 residents in a society, of which 375 people know more than three languages. What would be the central angle representing people who know more than three languages?

2 If 45% of people like biryani, what would be the central angle representing this data? 3 Draw a pie chart for the data of workers from various states working at a garment factory. State

Karnataka

Tamil Nadu

Andra Pradesh

Kerala

Others

No. of workers

440

620

340

300

100

4 Draw a pie chart for the following table showing the import of various products by a country. Item

Fuels and oil

Furniture

Plastics

Gems

Chemicals

Import (in billions)

₹240

₹150

₹180

₹150

₹180

5 Draw a pie chart for the table showing the sale of different brands of smartphones on an e-commerce website. Smartphone brand Sales

Brand A 1000

Brand B 1050

Brand C 550

Brand D 650

Others 350

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6 Draw a pie chart for the table showing the percentage of buyers of different brands of clothes. Clothing brand

Brand M

% of buyers

Brand N

Brand O

Brand P

Others

Word Problem 1

Ruchika scored the given marks in her annual examination. Draw a pie chart for the marks obtained by Ruchika. Subject

Marks secured

English 115

Maths

Science

140

100

Hindi 120

EVS

125

Reading Pie Charts The next day, Suman saw the annual budget of her state in the local newspaper in the form of a pie chart. Below are two pie charts representing the same data. The first pie chart shows the data in degrees, while the second pie chart shows the data in percentages. Let us learn how to read and interpret these pie charts. Annual Budget

Annual Budget

Healthcare

Agriculture 79.2° 43.2° 100.8°

Healthcare 38%

Agriculture 22%

136.8°

Infrastructure 12 %

Infrastructure Education 28%

Education

Value of Component Total value ×

Central angle 360°

Value of Component Percentage value Total value × 100

Let us find the amounts spent on different sectors using the above pie chart if the annual budget of the state government is ₹3,12,000 crore.

Error Alert! Never divide the central angle by 100 to find the value of a component. 45º × Total value 100

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Amount spent on agriculture =

79.2° 22 × 312000 or 312000 × = ₹68,640 crore 360° 100

Amount spent on healthcare

Did You Know?

show surveys, election results,

136.8° 38 × 312000 or 312000 × = ₹118,560 crore 360° 100

Amount spent on infrastructure

Pie charts can also be used to business reports, etc.

43.2° 12 × 312000 or 312000 × = ₹37,440 crore 360° 100 100.8° 28 Amount spent on education = × 312000 or 312000 × = ₹87,360 crore 360° 100 =

Think and Tell

Can the sum of the values of all components be less than the total value?

Example 5

Read the pie chart showing the different items sold by a bakery on a day and answer the given questions if the total sales in one day amounted to ₹24,000. Sales 1 Which was the most sold product of the day? – Rolls and Buns 2 What was the sales value of cupcakes? 20 24000 × 100 = ₹4800

3 What was the difference of the sales of cookies and doughnuts? Cookies = ₹24000 ×

14 = ₹3360 100

Doughnuts = ₹24000 ×

Others 12%

Rolls and Buns 44%

Cookies 14% Cupcakes 20% Doughnuts 10%

10 = ₹2400 100

Difference = ₹3360 − ₹2400 = ₹960 Example 6

The pie chart represents the expenditure on different items for constructing a flat in Delhi. If the expenditure incurred on cement is ₹1,12,500, find the total cost of constructing the flat. Expenditure on cement = ₹1,12,500 Let the total cost of constructing the flat = ₹x We know that, Value of component = ⇒ 1,12,500 =

75° ×x 360°

Central angle × Total value 360°

360° ⇒ x = 1,12,500 × = 5,40,000 75° Hence, the total cost of constructing the flat is ₹5,40,000.

Steel Brick

Expenditure Labour

45° 100° 50° 75°

Cement

90° Timber

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Do It Together

Read the pie chart showing the annual agricultural production of an Indian state and answer the given questions if the production of wheat is 27,000 tonnes. 1 What is the total production?

Agricultural Production Maize

The central angle of wheat = 120° = _______

Gram

2 What is the difference in the production of sugar and rice? 100° Production of sugar = ×_______ = _______ 360° Production of rice = ______________

30°

Total production = 27,000 ×

Wheat 50° 120° 60° 100°

Rice

Sugar

Difference = ______________

Do It Yourself 8B 1

The pie chart shows the number of students admitted to different college faculties. If the total number of

students is 5 lakhs, find the number of students admitted to each faculty.

Law 14%

College Faculties Education 8% Arts

Commerce 33%

20%

Science 25% Mode of Transport Others Metro

2 The pie chart shows data related to a survey on office-going people in a city.

Bus

a What is the second most popular mode of transport?

45° 45° 110° 90° 70°

b What fraction of of people prefer the bus? c Which mode of transport is preferred by a quarter of the people?

Shared cab

Own vehicle

3 Read the pie chart on the cars sold by a showroom in a year and answer the given questions if the number of type C cars sold was 500.

a What was the total number of cars sold? b What was the number of type D cars sold?

Type D 18%

Cars Sold

Type A 35%

c How many Type A and Type B cars were sold together? Type C 25% Type B 22% Chapter 8 • Pie Charts

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4 2500 students from a certain school were asked about their favourite juice. Read the pie chart and answer the given questions.

a How many students chose orange juice as their favourite? b Which juice is the favourite of the least number of students? How many students like it?

c What fraction of the students do not have apple juice as their favourite?

Favourite Juice

Litchi

Orange

68.4° 86.4° 36° 61.2° 108°

Apple

Pineapple

Mango

5 Read the pie chart showing the number of animals in a zoo and answer the questions if there are 77 elephants in the zoo.

Animals in the Zoo Tigers 5%

a What is the total number of animals in the zoo? b Which animal has the lowest number, and how much? c What is the difference of the numbers of deer and reptiles?

Deers 28%

Others 31%

Elephants 11% Reptiles 25%

6 Read the pie chart showing the types of flowers planted by a farmer in his field and answer the given questions if the total number of flowers planted is 1200.

Type of Flowers

a How many more sunflowers are planted than tulips?

Tulip

b What is the fraction of roses planted?

Rose

c What is the ratio of marigolds to jasmines planted?

60° 112.5° 37.5° 82.5° 67.5°

Jasmine

Sunflower

Marigold

7 The number of girls interested in different games was plotted in a pie chart. If the number of girls interested in basketball is 648:

Tennis

a What is the percentage of girls interested in volleyball?

Games Cricket

b What is the ratio of girls who like swimming to those who like cricket? c What fraction of girls like tennis and basketball?

Swimming

Volleyball

54°

72° 59.4° 45° 129.6° Basketball

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Word Problem 1

Manya surveyed some teenagers in her city about their favourite music.

255 teenagers like classical music. Read the chart and answer the questions. a How many teenagers were surveyed?

b What is the ratio of teenagers who like rock music to those who like hip-hop? Type of Music Jazz Hip-hop 19% 34%

Rock 30%

Classical 17%

Points to Remember •

A pie chart is a type of graph that represents data in a circular graph.

•

Central angle =

Value of the component × 360° Sum of all components • To draw a pie chart, we make a circle with its radius marked. The first angle is drawn, keeping the radius as the base. The rest of the angles are drawn, keeping the previous angle as the base. Percentage value Central angle • Value of component = × Total value = Total value × 100 360°

Math Lab Setting: In groups of 5

What’s Your Favourite?

Materials Required: Paper, pen, protractor, compass, ruler, pencil and eraser. Method: ollect different sets of data within the classroom, like favourite colour, fruit, vegetable 1 C and subject of the students. 2 Ask the students to tabulate the data.

3 Distribute the different data sets collected among groups.

4 T he groups show the collected data with the help of a pie chart showing the values in both degrees and percentages. 5 The group that creates the pie chart first wins!

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Chapter Checkup 1 If 23% of the people in a locality prefer fresh juice, what would be the central angle representing this data? 2 Mark the central angles in the pie charts shown below. Items Sold

Others, 8

Sandwiches, 66

Rice, 35

Spices, 10

Pulses, 15

No. of families

Ice creams, 54

Cat

Dog

Rabbit

Birds

Maths 595

English 420

French 225

Fantasy 280

Adventure 600

Science 385

Biography 440

Others 175

b What is the fraction of the notebooks sold in the

month?

Novels 320

Comedy 760

Stationery Items

The pie chart shows the sales at a stationery shop in a month.

Party supplies

a Which is the most popular item in the store?

Office supplies

c What is the percentage of the party supplies sold at the

shop?

Others

Draw a pie chart for the given data showing the types of books read by people in a town. Type of Books No. of people

Draw a pie chart for the given data showing the favourite subject of students at a school. Subject No. of students

French fries, 45

Draw a pie chart for the given data showing the types of pets owned by the families in a locality. Pet

Burgers, 60

Shakes, 75

Bread, 12

Items Sold

Art supplies 54.3°

61.2°

79.2°

90° 75.3°

Notebooks

School supplies The pie chart shows the percentages of ice creams sold by a vendor in 5 days. If the vendor sold 198 ice creams on Wednesday, then find the total number of ice creams sold in 5 days. Friday 21%

Ice Creams Sold

Monday 24%

Thursday 16%

Wednesday 22%

Tuesday 17%

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Read the pie chart showing the sale of various products at a supermarket and answer the questions if the total sales are ₹1,00,000.

Sale

Personal care

Produce 90°

Household

108° 54°

36°

72°

Dairy

Beverages a What was the percentage of the sales of household items? b What was the sales value of the most popular product? c What was the total percentage of the sales for the two least popular products?

The pie chart shows the types of workouts preferred by the people in a society. Workout Others

Running 43

64.8° 43.2° 3 115.2° 6° .2°

Gymming

57.6°

Cycling Aerobics

Yoga and meditation a If 150 people prefer aerobics, then what percentage of the people prefer running? b What is the ratio of people who prefer cycling to those who do not prefer gymming? c What percentage of the people do not prefer yoga and meditation?

10 A company collects 60,000 tonnes of recyclable material every month. The pie charts shown below give the details of the kinds of materials collected in June and July.

Collection in July

Collection in June Others

72°

Paper 93.6°

126°

43.2°

72°

° 25.2

Glass

Others 5%

Paper 22%

Glass 26%

Cardboard

Plastic

Cardboard 15% Plastic 32%

a What is the difference of the central angles made by glass in the two months?

b How many tonnes of paper were collected in both months together? c In which month was more plastic collected, and by how much?

d What is the percentage change in the cardboard collection in the two months?

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Word Problems 1

Rohit paid ₹7200 for electricity consumption. The pie chart shows the percentage of electricity consumption by different kinds of appliances in Rohit’s home. a What amount of electricity bill was paid

Bathroom 10%

for the Electronics category?

b What is the central angle for the

electricity consumption for appliances in the kitchen category?

c Appliances under which category used

Electricity Consumption Kitchen 24%

Lighting 20% Laundary 14%

the least electricity?

Electronics, 32%

Mariya collected the data on favourite ice cream flavours of students at her school. Draw a pie chart for the data collected by Mariya. Ice cream flavour Number of Students

Vanilla

Chocolate

Mango

Guava

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9 Probability Let's Recall Sanvi and Aditya are playing a game with their father. Their father rolls the dice. What are the numbers that can come up? Is it possible that he gets the number 7?

A dice has only 6 numbers, so it is not possible to get the number 7 on rolling the dice. Probability is a way of guessing how likely something is to happen. It helps us figure out if something is sure, impossible, likely, unlikely or equally likely to happen compared to other events. Let us look at the chance of the numbers coming up on a dice when rolled.

Getting a number from 1 to 6:

Getting an even number:

Getting the number 7:

Sure, since there are only 6 numbers on a dice.

Equally likely, since 3 out of 6 numbers are even on a dice.

Impossible, since the number does not appear on a dice.

Let’s Warm-up Write the chance of the given events happening using words like sure, impossible, likely, unlikely and equally likely. 1 A dog flying. 2 The cycle of day and night. 3 Getting heads when a coin is tossed. 4 Drawing a blue marble from a bag containing 6 blue and 2 green marbles. 5 Drawing a green marble from a bag containing 10 yellow and 3 green marbles. I scored _________ out of 5.

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Mean, Median and Mode Understanding Probability Real Life Connect

Naina and her brother Sam are visiting the local village fair. They go to the Wheel of Fortune stall. They are excited to play the game since they will get a prize each time the wheel stops at an even number. The wheel has numbers from 1 to 10. Naina says 8 and spins the wheel. The wheel stops at 8. Sam: Wow! You guessed right! Now, it’s my turn. I want the number 12. Naina laughs upon hearing this and says, ‘No chance! You lose a point.’ In other words, we can say that probability helps us make smart guesses, a bit like predicting what might happen when you play a game, spin a wheel, flip a coin or roll a dice. For example, the chance of the wheel stopping at a number more than 10 is an impossible event. The chance of the wheel stopping at numbers from 1 to 10 is a sure event. Let us recall some terms. Chance of an event happening is its probability.

Outcome is a result of some activity.

Event: Collection of some or all possible outcomes of an activity is called an Event. Total Outcomes: The total number of possible outcomes is called total outcomes. E.g., in a wheel showing 5 colours, the total outcomes are 5.

Favourable Outcomes: The number of favourable outcomes helps us predict the chances of an event happening.

The chance of the occurrence of an event can be measured mathematically, and it is called the probability of the event. Events which have no chance of occurring have a probability of 0.

In general, the probability of a sure event is 1 and the probability of an event that has no chance of occurring or is impossible is 0. For any other event, probability varies between 0 to 1. Total numbers on the wheel = Total Outcomes = Possible Outcomes = 10 Sure: The wheel stopping at a number from 1 to 10.

Impossible: The wheel stopping at the number 12.

Here, possible outcomes (10) = Favourable outcomes (10)

Here, the favourable outcomes = 0

Likely: The wheel stopping at a number greater than 4.

Unlikely: The wheel stopping at a number less than 3.

Favourable outcomes = 6 (Numbers 5, 6, 7, 8, 9, 10)

Favourable outcomes = 2 (Numbers 1 and 2) Favourable outcomes < 5

Here, favourable outcomes > 5 (half of the possible outcomes)

Equally Likely: The wheel stopping at a number from 1 to 5. Favourable outcomes = 5 (Numbers 1, 2, 3, 4, 5) Favourable outcomes = 5

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Theoretical Probability Since the wheel shows numbers from 1 to 10, anytime the wheel is spun, it will stop at a number from 1 to 10. In other words, the possible outcomes of the event are the wheel stopping at the numbers 1 to 10. The total outcomes will be 10. Let us see how. Probability of Occurrence of an Event The term ‘chance’ when used in real life is measured mathematically and is called the probability of the event. Let E be an event. Then probability of an event E is the ratio of the number of outcomes favourable to the event E to the number of all possible outcomes of the experiment. P(E) =

Number of outcomes favourable to the event E Number of all possible outcomes of the experiment

Where, P(E) = Probability of an Event E and P(E) lies between 0 and 1. In the above case, the probability of the wheel stopping at an even number can be calculated using the formula. Here, E = Event of the wheel stopping at an even number Total outcomes = 10 (Total numbers on the wheel) and Favourable outcomes = 5 Applying the formula, P(E) = P(E) = P(E) =

Number of outcomes favourable to the event E

Even number Even number

Even number

Number of all possible outcomes of the experiment Total number of even numbers

Even number

Total numbers on the wheel

5 1 = 10 2

So, the theoretical probability of the wheel stopping at 5 1 an even number is or . 10 2

Theoretical probability gives exact probabilities based on mathematical principles in ideal conditions. It’s like predicting the chance of something based on what we know and the rules.

Even number

Think and Tell Is the probability of the wheel stopping at an odd number the same as the probability of the wheel stopping at an even number?

For example, if we roll a fair six-sided dice, we know each number has a 1 in 6 chance of showing up because there are 6 sides. Theoretical probability is used for simple, well-defined situations like rolling dice and tossing coins. When we find the set of all possible outcomes of more than one event, it is called the sample space. You can use tables and tree diagrams to find the sample space of two or more events.

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For example, you randomly choose a crust and style of pizza.

Select Crust

Crispy Thin

Select Style

Soft Pan

Use a tree diagram to find the sample space.

Crispy Thin Hawaiian

Mexican

Pepperoni

Vegetarian

Hawaiian

Mexican

Pepperoni

Vegetarian

Soft Pan

There are 8 different outcomes in the sample space. The 8 different pizzas possible are: CH, CM, CP, CV SH, SM, SP, SV Example 1

Rahul tosses a coin before a cricket match. What is the probability that he gets: 1 heads?

2 tails?

The possible outcomes when a coin is tossed are heads or tails. Total possible outcomes of the experiment = 2 (either heads or tails) 1 He gets heads. E = Event of getting a head Numbers of heads P(E) = Total number of outcomes 1 = 2 1 Therefore, the possibility of getting a head is . 2 Example 2

2 He gets tails. E = Event of getting a tail Number of tails P(E) = Total number of outcomes =1 2 Therefore, the possibility of getting a tail is 1. 2

A fruit box contains 4 oranges and 8 apples. If a fruit is taken out randomly, what is the probability of it being an apple? E = Event of getting an apple P(E) =

Number of apples 8 2 = = Total number of fruits 12 3

2 Therefore, the possibility of getting an apple is . 3

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Do It Together

A well-shuffled deck of animal picture cards is placed face down on a table. What is the probability that an animal card is drawn at random: Animals

1 shows a picture of a bird? E = Event of getting a card that shows a picture of a bird Number of cards that show a bird picture P(E) = Total number of outcomes = Therefore, the possibility of getting an animal card that shows a bird picture is .

Birds

2 does not show a picture of a bird? E = Event of getting a card that does not show a picture of a bird = Therefore, the possibility of getting an animal card that does not show a bird picture is .

Do It Yourself 9A 1 Write the chance of the given events happening using words like sure, impossible, likely, unlikely and equally likely. a Selecting a vowel from the letters a, e, i, o, u. b If today is a Saturday, tomorrow will be a Friday. c Drawing a white ball from a bag containing 8 white and 4 red balls. d Getting the number 2, when a dice is rolled. e Getting a tail when a coin is tossed.

2 Fill in the blanks. a The probability of an event having no chance of occurring is ___ (0/1). b The probability of a sure event is ___ (0/1). c The possible outcomes when you toss a coin are d On rolling a dice, the event of getting a prime number has

and

. possible outcomes.

3 At a village fair, you have a choice of spinning Spinner A or Spinner B. You win a

prize if the spinner lands on a section with a star in it. Which spinner should you choose if you want a better chance of winning? Give a reason for your answer.

4 At a school fair, you have a choice of randomly picking a ball from Basket A or

Basket B. Basket A has 5 green balls, 3 black balls and 8 yellow balls. Basket B

has 7 green balls, 4 black balls and 9 yellow balls. You can win a prize if you pick a black ball. Which basket should you choose if you want a better chance of winning? Write the probability of choosing a black ball.

Chapter 9 • Probability

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5 A game board has a spinner with 10 equal-sized sections, of which 4 are green, 3 are blue, 2 are yellow and 1 is black. What is the sum of the probabilities of the pointer landing in the green, blue, yellow and black sections?

6 Each letter of the alphabet is printed on an index card. What is the theoretical probability of randomly choosing any letter except Z?

7 A glass jar contains 40 marbles in different colours such as black, green, blue and yellow. The probability of 1 drawing a single green marble is . What does this mean? 5 a There are 5 green marbles in the glass jar. b There are 8 green marbles in the glass jar. c There are 10 green marbles in the glass jar. d There is only one green marble in the glass jar. 1

8 In a game, the probability of correctly guessing which of the 5 boxes contains a tennis ball is . About how many 5

winners would be expected if 60 contestants played the game?

9 Neeta is playing a game using a fair coin. Contestants win the game if the coin lands tail up. a What is the theoretical probability that the coin will land tail up? b If 250 contestants play the game, about how many of them are expected to win?

10 Amit rolls a six-sided dice labelled 1 to 6. a Find the probability of rolling a 4. b Find the probability of rolling an odd number. c If Amit rolls the dice 12 times, how many times should he expect a number greater than 4 to be rolled?

11 A 12-sided solid has equal-sized faces numbered 1 to 12. a Find the probability of getting a number greater than 10. b Find the probability of getting a number less than 5. c If the 12-sided solid is rolled 200 times, how many times would you expect either a 4, 6, or 9 to be rolled?

12 A black dice and a blue dice are rolled. If all numbers are equally likely, what is the probability that the sum of the numbers that appear on their faces is equal to: a 5?

b 6?

c 10?

13 In a two-step experiment, a coin is tossed and a dice is rolled. a What are the possible outcomes? b How many possible outcomes are there? c What is the probability of: i

tossing a head and rolling a 6?

ii tossing a head?

iii tossing a tail and rolling a number less than 3?

iv rolling a 5 or 6?

vi not rolling a 6?

tossing a head and not rolling a 6?

14 Ajay designs a game for the village fair. Contestants roll two dice at the same time. If the sum of the numbers on the two dice is 7, the player wins. About how many winners should Ajay expect if 500 contestants play his game?

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Word Problems 1

On a game show, a contestant randomly draws a chip from a bag and replaces it. Each chip says either win or lose.

The theoretical probability of drawing a winning chip is a How many chips are in the bag?

3. The bag contains 9 winning chips. 10

b Out of 20 contestants, how many do you expect to draw a winning chip?

Navin is getting dressed. He considers two different shirts, three pairs of pants and three pairs of shoes. He chooses one of each of the articles at random. Shirts

Pants

Shoes

Collared shirt

Khakis

Sneakers

T-shirt

Jeans

Flip-flops

Capris

Sandals

a How many different outfits are possible? b What is the probability that he will wear his jeans but not his sneakers?

Experimental Probability The wheel of fortune that Naina and Sam spin records the number of times the wheel lands on an even number. The picture shows a digital scorecard to record the number of people who have won a prize. Sam is excited to know that 60 people from a total of 100 people who have spun the wheel got lucky! The wheel has 10 sections numbered 1 to 10. 5 sections are odd and the remaining 5 are even. Even though the theoretical probability of spinning an even number is equal to that of an odd number, the actual results of the experiment have turned out to be different.

ODD

EVEN

Thus, the experimental probability of spinning an even number in this case is 60 or 60%. This is more than the theoretical probability of 100 spinning an even number which is 50%. Such an experiment where we cannot predict the exact result is called a random experiment. Here, each outcome has an equal chance of occurring. A Random Experiment is one whose outcome cannot be predicted exactly in advance. That is, a random experiment is an experiment, which, when repeated under identical conditions, does not produce the same outcome every time.

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Experimental probability gives the probability based on observed data. We can find the experimental probability by physically doing tests or experiments. The accuracy of experimental probability depends on the number of trials conducted. That is, as the number of trials increases, we get more accurate results. The experimental probability of an event E can be calculated using the formula, P(event) ꞊ Number of times the event occurs Total number of trials For example, if we toss a coin 100 times and it comes up heads 47 times, the experimental probability of getting heads is 47 out of 100. Experimental probability is used when theoretical probability is difficult to calculate.

Think and Tell What will be the approximate experimental probability of getting heads if we toss a coin 1000 times?

Example 3

A coin is tossed 50 times to see how often tails shows up. If tails appears 20 times, what is the experimental probability of getting tails? E = Event of getting tails

Heads

Experimental Probability, P(E) = Number of times tails appears = 20 = 2 Total number of trials 50 5

Tails

30 times

20 times

Therefore, the experimental probability of getting tails is 2, 0.40, or 40%. 5 Example 4

Amit randomly pulls a coloured block from a bag. He records the colour and then puts the block back into the bag. The table shows the results of his experiment. If he does the experiment 50 times, predict the number of times he will pull a black block from the bag. Colour

Number of Pulls

Blue 3

Black 4

Yellow

Green

The total number of trials in Amit’s experiment = 3 + 4 + 6 + 7 = 20 P(black) = Number of black blocks = 4 Total number of trials 20

In 20 trials, Amit pulled a black block 4 times.

In 50 trials, Amit would pull a black block, 50 × 4 ꞊ 10 times 20 So, Amit will pull a black block about 10 times. Do It Together

The bar graph shows the results of rolling a dice 300 times. How does the experimental probability of rolling an odd number compare with the theoretical probability? We first find the experimental probability of rolling an odd number.

P(odd) = Number of times an odd number was rolled _____ � Total number of rolls

Times Rolled

The bar graph shows _____ ones, _____ threes, and _____ fives. So, an odd number was rolled _____ + _____ + _____ = _____ times in a total of 300 rolls.

Rolling a Dice

50 48

53 50 52

49 48

46 44

Number Rolled

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Next, we find the theoretical probability of rolling an odd number. Number of favourable outcomes P(odd) = _____ � Number of possible outcomes The experimental probability of rolling an odd number is _____%, which is close to the theoretical probability of _____%.

Do It Yourself 9B 1

Which of the following are examples of a random experiment? a Tossing a coin

b Drawing a black card from a deck of playing cards

c Throwing a ball

d Taking a marble out of a bag containing coloured marbles

e Rolling two dice together

f Measuring the length of a table

marked ‘WIN’. Leena played this game many times and recorded her results. She won 8 times

and lost 40 times. Use Leena’s data to find the experimental probability of winning this game.

To play a game, you spin the spinner shown. You win if the arrow lands in one of the areas

WI N

In tennis, Govind serves an ace, a ball that can’t be returned, 4 out of the 10 times he serves. a What is the experimental probability that Govind will serve an ace in the first match of the next game? b Make a prediction about how many aces Govind will have for the next 40 serves.

During a 24-hour period, the ratio of pop songs to rap songs played on a radio station is 60:75. a What is the experimental probability that the next song played will be a rap song? b Out of the next 90 songs, how many would you expect to be pop songs?

A bag contains 50 marbles. You randomly draw a marble

from the bag, record its colour and then replace it. The table shows the results after 30 draws.

Colour

Frequency

Blue 3

Green

Red

Vowel

Consonant

Yellow 6

a Find the experimental probability of drawing a red marble. b Predict the number of red marbles in the bag.

A board game uses a bag of 105 lettered tiles. You randomly choose a tile and then return it to the bag. The table shows the number of vowels and the number of

30 times

consonants after 50 draws.

20 times

Predict the number of vowels in the bag.

The Venn diagram shows two sports played by 90 Grade 8 students. What is the probability that a randomly chosen student plays:

a both tennis and squash?

Sports Played by Grade 8 Students

b neither tennis nor squash? c tennis? d tennis or squash but not both?

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Tennis 25

Squash 2

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We held a survey of all the students in our school to find out how many own a mobile phone. The results are shown in the two-way table.

Owns a mobile phone

Does not own a mobile phone

Total

Female

Male

Total

125

If we choose a student at random from our school, what is the probability that the student: a owns a mobile phone?

b does not own a mobile phone?

c is a male who owns a mobile phone?

d is a male who does not own a mobile phone?

e is a female who does not own a mobile phone?

Table 1

Table 1 shows the results of flipping two coins 12 times.

a What is the experimental probability of getting two tails? Using this probability,

how many times can you expect to get two tails in 600 trials?

b Table 2 shows the results of flipping the same two coins 100 times. What is the experimental probability of flipping two tails? Using this probability, how many times can you expect to flip two tails in 600 trials?

Table 2

c Why is it important to use a large number of trials when using experimental probability to predict results?

Word Problem 1 You randomly select 200 pairs of jeans and find 5 defective pairs. About how many pairs of jeans do you expect to be defective in a shipment of 5000?

Points to Remember • Probability is the concept which numerically measures the degree of certainty of the occurrence of an event. •

An operation which can produce some well-defined outcomes is called an experiment.

• An experiment in which all possible outcomes are known and the exact outcome cannot be predicted in advance is called a random experiment. •

The collection of all or some of the possible outcomes is called an event.

•

Probability of an event =

•

The probability of an event always lies between 0 and 1.

•

The probability of a sure event is 1 and an impossible event is 0.

•

Probability cannot be negative.

•

The sum of the probabilities of all possible outcomes of an event is 1.

Favourable outcomes Total number of outcomes

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Math Lab Tossing Coins Aim: In groups of 3 Materials Required: A coin of ₹1, paper and pencil Setting: In groups of 3 Method: 1 Toss the coin 10 times consecutively and record the result in a table. 2 Count the number of times a head appears and number of times a tail appears. alculate the probability of getting heads by dividing the number of times heads 3 C appeared by the total number of tosses. Similarly, calculate the probability of getting tails. 4 I ncrease the number of tosses to 20 and take the cumulative result of the previous tosses. Again, calculate the probabilities. 5 As the number of tosses increases, the probabilities tend to get nearer to 1. 2

Chapter Checkup 1 Fill in the blanks.

a The probability of a sure event is _____.

b If an event cannot occur, then its probability is _____. c The probability of selecting P from the word SPECIAL is _____. d The probability of an event can’t be more than _____. e If a dice is rolled once, the probability of getting an even prime number is _____.

2 Write all possible outcomes of the following events. a Flipping a fair coin b Rolling a standard six-sided dice c Selecting a single letter at random from the English alphabet d Flipping two fair coins simultaneously e Rolling two standard six-sided dice simultaneously

3 Sneha draws a card at random from the cards given. What is the probability of drawing a card that shows a shape with:

a three sides? b four sides? c more than four sides? d no sides?

Chapter 9 • Probability

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rectangle

Square

Edges: 4 Corners: 4

triangle Edges: 3 Corners: 3

octagon

pentagon

circle

Edges: 5 Corners: 5

hexagon Edges: 0 Corners: 0

Edges: 8 Corners: 8

rhombus Edges: 6 Corners: 6

Edges: 4 Corners: 4

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4 Find the probability of getting an A on random selection of a letter from the word APPRECIATION. 5 A college student has to select a stream with the given subject combinations: a Physics, Chemistry, Mathematics

b Physics, Chemistry, Biology

c Accountancy, Mathematics, Economics

d English, Hindi, Social Sciences

Find the probability of getting mathematics as a subject on a random selection of the stream. 50

Compare the theoretical and experimental probabilities of the events.

a Wheel stopping at green

b Wheel stopping at red

c Wheel stopping at blue or purple

d Wheel stopping at a colour

Times Spun

6 The bar graph shows the result of spinning a wheel 150 times.

30 20

Fortune Wheel 33

10 0 Blue Green Yellow Red Purple

7 Prime numbers between 1 and 25 are written on identical slips, put in a box and

Colour

mixed up. If a slip is drawn at random, what is the probability of getting: a a one-digit number?

b an even number?

c an odd number?

d a number greater than 11?

8 The keypad of a mobile phone has digits from 0 to 9 on it. If a person enters a digit at random, what is the probability of typing:

a an even number?

b a multiple of 5?

c a factor of 9?

d a natural number?

e a prime number?

a composite number?

9 In a store, 50 pairs of bangles are on display. The probability that a customer will choose a pair of pearl bangles is 6 . How many pearl bangles are on display? 25

10 Amita has a ₹1 coin, a ₹2 coin, a ₹5 coin and a ₹10 coin in her purse. She needs to pay for a pencil that costs ₹4. State the probability of her using the ₹5 coin.

11 Two coins are tossed together. Find the probability of getting: a 2 tails.

b at least 1 head.

12 Two dice are tossed together. Find the probability of getting: a the same number on both dice.

b even numbers on both dice.

Word Problems 1 A box contains 100 bolts, two-fifths of which are rusted. One bolt is drawn at random from the box. Find the probability that it is: a a rusted bolt.

b not a rusted bolt.

2 200 apples are packed in a carton out of which 30 apples are rotten. If one apple is taken out from the carton at random, find the probability of getting a: a good apple.

b rotten apple.

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Squares and Square Roots

Let's Recall

Any number except 0 and 1 can be expressed as a product of its factors. This is called factorisation of the number. A number that has only two factors, the number 1 and itself, is called a prime number, whereas a number that has more than two factors is called a composite number. For example, look at the factors of 6 and 7 given below.

1×7꞊7

1 × 6 ꞊ 6, 2 × 3 ꞊ 6

factors of 7

factors of 6 4 factors Composite Number

2 factors

Prime Number

Every composite number can be expressed as a product of prime factors, no matter what the order is. This is called prime factorisation of the number. Using factor trees 36 3

Choose any factor pair of 36. Circle the prime factor.

12 3

4 2

Continue to factorise every number that is not prime.

36 = 3 × 3 × 2 × 2

Using the division method Choose the smallest prime factor.

3 1

Divide by the chosen prime factor and write the quotient here.

Keep choosing prime factors and dividing until you get a quotient of 1. Stop factorising when you get 1. 36 = 2 × 2 × 3 × 3

Letʼs Warm-up Write True or False. 1

Among 3, 7, 8, 9, 11 and 12, the numbers 8, 11 and 12 are the only composite numbers. __________

The prime factors of 48 are 2 and 3 only.

__________

The prime factorisation of 120 is 2 × 2 × 2 × 2 × 3 × 3 × 5.

__________

The prime factorisation of 132 is 2 × 2 × 3 × 11.

__________

rime factorising a number using the factor tree method gives a different 5 P result than using the division method of prime factorisation.

__________

I scored _________ out of 5.

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Square of a Number Real Life Connect

Shobhit wants to buy a square table for his home. He requires a table which occupies an area of less than 5 sq. m due to the lack of space. What size table can he purchase?

Square Numbers

`5000

Look at the figures. Each number has been arranged using squares of the same size. Some form complete squares and some do not. 1

A square

Not a perfect square

A square

A number that forms a complete square is a square number or a perfect square. For example, in the above picture, the numbers 1 and 4 form a complete square. These numbers can also be written as a product of two like factors. For example, 1 can be written as 1 × 1 = 12 and 4 can be written as 2 × 2 = 22. So, the square of 1 is 1 and the square of 2 is 4. Some more square numbers are listed below in the given table.

This means 'square'. 22 ꞊ 2 × 2 ꞊ 4

Numbers 1 to 10

If a number has 4 or 6 in its ones place, then its square ends in 6.

Squares of even numbers are always even.

Number

Square

Meaning

Product

1×1

2×2

3×3

4×4

5×5

6×6

7×7

8×8

9×9

102

10 × 10

100

In square numbers, the digits always end in 0, 1, 4, 5, 6 or 9 and not 2, 3, 7 or 8.

Think and Tell If a number upon division by 3 leaves the remainder 2, is it a perfect square?

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Let us see the squares of numbers from 11 to 20. Numbers 11 to 20

If a number has 1 or 9 in its ones place, then its square ends in 1.

If a number has 3 or 7 in its ones place, then its square ends in 9.

If a number has 2 or 8 in its ones place, then its square ends in 4.

Number

Square

Meaning

Product

112

11 × 11

121

122

12 × 12

144

132

13 × 13

169

142

14 × 14

196

152

15 × 15

225

162

16 × 16

256

172

17 × 17

289

182

18 × 18

324

192

19 × 19

361

202

20 × 20

400

Think and Tell If a number upon division by 4 leaves a remainder of 2 or 3, is it a perfect square?

Squares of odd numbers are always odd.

If a number has 0 in its ones place, then its square ends in 0. Also, there are an even number of zeroes at the end of a perfect square.

Remember! An area is the amount of space an object occupies. Area of a square = s × s

Now, let us help Shobhit finding the amount of space a square table occupies. To find the amount of space that the table occupies, we must find the area of the table. Since the table is in the shape of a square, it is sufficient to find the square of one side of the table. That is, one side of the table = 1 m. Square of this side of the table = 1 × 1 = 1 sq. m So, Shobhit can buy this table since the space it will occupy will be less than 5 sq. m. Square of Rational Numbers We can find the square of rational numbers by dividing the square of its numerator by the square of its denominator. For example, to find the square of a positive rational number 3 , we write it as: 4 3 2=3×3=3×3= 9 4 4 4 4 × 4 16 −4 Similarly, to find the square of a negative rational number , we write it as: 5 2 −4 = −4 × −4 = (−4) × (−4) = 16 5 5 5 5×5 25

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Square of Large Numbers It was easy to find the square of numbers up to 20. But to find squares of numbers beyond 20 is quite difficult. To find the square of numbers beyond 20, we use the distributive property. For example, to find the square of 24, we write it as: We know that 24 = 20 + 4 or 30 – 6. 242 = (20 + 4)2

Remember! Multiplying two negative integers gives us a positive integer.

242 = (30 − 6)2

= (20 + 4) (20 + 4)

= (30 − 6) (30 − 6)

= 20 (20 + 4) + 4 (20 + 4)

= 30 (30 − 6) − 6 (30 − 6)

= 202 + 20 × 4 + 4 × 20 + 42

= 302 + 30 × (− 6) + (− 6) × 30 + (− 6)2

= 400 + 80 + 80 + 16 = 576

= 900 − 180 − 180 + 36 = 576

How do we check if a number is a perfect square? To check if a number is a perfect square, we use prime factorisation. If on prime factorising the number, we get pairs of equal prime factors and no factor is left over, then the given number is a perfect square. For example, let us check if 784 and 3400 are perfect squares. For 784

Think and Tell How many digits will there be in a square of a number?

For 3400

Find the prime factorisation of 784.

Find the prime factorisation of 3400.

784

3400

392

1700

196

850

425

784 = 2 × 2 × 2 × 2 × 7 × 7 Group the prime factors into pairs of equal factors until no factor is left over. 784 = 2 × 2 × 2 × 2 × 7 × 7 Since there are pairs of prime factors and no factor is left over, 784 is a perfect square. Example 1

3400 = 2 × 2 × 2 × 5 × 5 × 17 Group the prime factors into pairs of equal factors until no factor is left over. 3400 = 2 × 2 × 2 × 5 × 5 × 17 Since all the prime factors do not form pairs, 3400 is not a perfect square.

Find the value of: 1

6 2 7 6 2 = 6 × 6 = 6 × 6 = 36 7 7 7 7 × 7 49

−11 2 12 −11 2 = −11 × −11 = (−11) × (−11) = 121 12 12 12 12 × 12 144

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Example 2

Find the squares of the given numbers. 1

432 432 = (40 + 3)2

Example 3

1172 1172 = (100 + 17)2

= (40 + 3) (40 + 3)

= (100 + 17) (100 + 17)

= 40 (40 + 3) + 3 (40 + 3)

= 100 (100 + 17) + 17 (100 + 17)

= 402 + 40 × 3 + 3 × 40 + 32

= 1002 + 100 × 17 + 17 × 100 + 172

= 1600 + 120 + 120 + 9 = 1849

= 10000 + 1700 + 1700 + 289 = 13,689

Show that 3600 is a perfect square. Also, find the number whose square is 3600. 3600 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

3600

Since there are pairs of equal prime factors and no factor is left over, 3600 is a perfect square.

1800

900

450

225

Now, 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 22 × 22 × 32 × 52 ⇒ (2 × 2 × 3 × 5)2 = 602 So, 3600 is a perfect square of 60. Example 4

Find the least number by which 7220 should be multiplied to get a perfect square number. 7220 = 2 × 2 × 5 × 19 × 19

7220

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

2 5

3610

361

So, 5 is the least number which 7220 should be multiplied by to get a perfect square. Example 5

1805

Find the smallest square number which is divisible by each of the numbers 8, 10, 12, and 16. 2

8, 10, 12, 16

4, 5, 6, 8

2, 5, 3, 4

1, 5, 3, 2

1, 5, 3, 1

1, 5, 1, 1 1, 1, 1, 1

The least number divisible by each of the given numbers is their LCM. LCM of 8, 10, 12, 16 = 2 × 2 × 2 × 2 × 3 × 5 = 240 Now, let us prime factorise 240. 240 = 2 × 2 × 2 × 2 × 3 × 5

Clearly, we do not get pairs of all prime factors and we are left with the factors 3 and 5 which cannot be paired.

240

120

So, 240 is not a perfect square. To make it a perfect square, it must be multiplied by 15 (3 × 5). Thus, the required square number is 240 × 15 = 3600.

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Do It Together

Look at the numbers. Answer the questions. 1 61

2 224

3 235

4 719

a How many digits will be there in the squares of each of the numbers? 61 is a 2-digit number. So, it will have either (2 × 2) – 1 or 2 × 2 digits, that is, either _____ digits or _____ digits. 224, 235 and 719 are 3-digit numbers. So, it will have either (3 × 3) – 1 or ___ × ___ digits, that is, ____ digits or ____ digits. b Which of the numbers are not perfect squares? Give reasons. The numbers that end in 2, 3, 7 or 8 are never perfect squares. So, the numbers _____ and _____ are not perfect squares.

Do It Yourself 10A 1

Look at the digits in the units place of each of the given numbers. Answer the given questions. a i

5329

961

4624

7688

19,753

83,245

925

4 5

208

What will be the units digits of the squares of the numbers?

ii Which of the numbers will have odd or even squares? iii Which of the numbers are not perfect squares?

Find the square of the numbers. a

7 8

−

12 13

3 Check if the given numbers are perfect squares using prime factorisation. Also, find the number whose square is the number given. a

6545

9054

7396

5786

26,569

4 Find the smallest number by which each number should be multiplied to get a perfect square. In each case, find the number whose square is the new number. a

5780

9072

3267

6272

18,432

43,740

5 Find the smallest number by which the number given should be divided to get a perfect square. In each case, find the number whose square is the new number. a

3456

8575

5808

35,972

95,400

67,392

Find the smallest square number that is divisible by each of the numbers 16, 18, 24, 28 and 30.

7 Five friends, Sheetal, Shyna, Hitesh, Rohit and Hetal, play a number game. Sheetal thinks of a positive integer, which Shyna then doubles. Hitesh then triples Shyna’s number. Finally, Hetal multiplies Hitesh’s number by 6.

Rohit notices that sum of these four numbers is a perfect square. What is the smallest number that Sheetal could have thought of?

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Word Problems 1 A gardener grows 45 plants in each row of his square garden. How many plants are there in total in his garden?

2 In a school, 63 students stand in a row. If the total number of rows is the same as the number of students in each row, then find the total number of students in the school.

Patterns of Square Numbers We know that the side of Shobhit’s square coffee table is 1 m. What if the side of the coffee table was 2 m? What would the difference in their areas be? Let us see!

If the side of the table is 1 m, then its area is 12, that is, 1 sq. m.

If the side of the table is 2 m, then its area is 22, that is, 4 sq. m. Difference in their areas = 4 – 1 = 3. We can also solve this by using a shortcut method: 1 is an odd number and 2 is an even number but both are consecutive numbers. For such consecutive numbers, the difference of their squares is equal to either the sum of the numbers or twice the smaller number plus one. That is, if n is any number, then (n + 1)2 – n2 = n + n + 1 = 2n + 1

So, here, 22 – 12 = 2 + 1 = 3 = (2 × 1) + 1

Think and Tell How many non-perfect square numbers are there between the squares of two consecutive numbers?

Some more patterns of square numbers are given below. 1

= 1

1+3

= 4

1+3+5

= 9

1+3+5+7

= 12

32 = 9 = 4 + 5

= 32

112 = 121 = 60 + 61

= 22

72 = 49 = 24 + 25

= 16 = 42

152 = 225 = 112 + 113

1 + 3 + 5 + 7 + 9 = 25 = 52

…

…………….. Rule: Sum of the first n odd numbers is n2.

Rule: If n is an odd number, then n2 = (n – 1) + 2

1 × 3 + 1 = 4 = 22

5 × 7 + 1 = 36 = 62

2 × 4 + 1 = 9 = 32

6 × 8 + 1 = 49 = 72

3 × 5 + 1 = 16 = 42

7 × 9 + 1 = 64 = 82

4 × 6 + 1 = 25 = 52

8 × 10 + 1 = 81 = 92

(n2 + 1) . 2

Think and Tell If a number cannot be expressed as a sum of successive odd numbers, is the number then a perfect square?

Rule: If n is an odd (or even) number, then n (n + 2) + 1 = (n + 1)2.

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112 1 dot

... 1+2

3 dots

...... 1+2+3 6 dots

111

... .......... 1+2+3+4 ... 10 dots

1111

111112

1 + 3 = 4 = 22 3 + 6 = 9 = 32 6 + 10 = 16 = 42 Rule: The sum of two consecutive triangular numbers is always a square number. 152 = 225 = 200 + 25 = 2 × 100 + 25 = [1 (1 + 1)] × 100 + 25 252 = 625 = 600 + 25 = 6 × 100 + 25 = [2 (2 + 1)] × 100 + 25 352 = 1225 = 1200 + 25 = 12 × 100 + 25 = [3 (3 + 1)] × 100 + 25 452 = 2025 = 2000 + 25 = 20 × 100 + 25 = [4 (4 + 1)] × 100 + 25 Rule: If a = tens digit and ‘5’ is the units digit, then (a 5)2 = [a (a + 1)] × 100 + 25 1 × 3 = (2 – 1) (2 + 1) = 22 – 12 = 4 – 1 = 3

Pythagorean Triplets

11 × 13 = (12 – 1) (12 + 1) = 122 – 12

32 + 42 = 9 + 16 = 25 = 52 62 + 82 = 36 + 64 = 100 = 102

= 144 – 1 = 143 2 × 4 = (3 – 1) (3 + 1) = 32 – 12 = 9 – 1 = 8 10 × 12 = (11 – 1) (11 + 1) = 112 – 12 = 121 – 1 = 120 Rule: If (a + b) and (a – b) are two numbers, then (a + b) (a − b) = a2 – b2 Express 225 as the sum of 15 odd numbers.

Example 6

Rule: If m is any non-zero number and m > 1, then we have (2m)2 + (m2 – 1)2 = (m2 + 1)2 Example 7

We know that the sum of the first n odd numbers is n2.

Example 8

Express 772 as the sum of two consecutive numbers.

We know that if n is an odd number, then 2 (n2 + 1) 2 (n – 1) n = + . 2 So, 225 = 15 2 2 (772 – 1) (772 + 1) = Sum of the first 15 odd numbers So, 772 = + 2 2 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + (5929 – 1) (5929 + 1) = + 19 + 21 + 23 + 25 + 27 + 29 2 2 5928 5930 = + = 2964 + 2965 2 2 Find a Pythagorean triplet if one of its smallest members is 20. We know that if 2m, m2 – 1 and m2 + 1 are said to form a Pythagorean triplet, where m is any non-zero number and m > 1, then (2m)2 + (m2 – 1)2 = (m2 + 1)2. If 2m = 20, then m = 10. So, m2 – 1 = 102 – 1 = 100 – 1 = 99 m2 + 1 = 102 + 1 = 100 + 1 = 101

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Thus, the Pythagorean triplet in this case is (20, 99, 101). If m2 – 1 = 20, then m2 = 21 ⇒ m is not an integer. If m2 + 1 = 20, then m2 = 19 ⇒ m is not an integer. In either case, values of m are not possible. So, the only Pythagorean triplet here is (20, 99, 101) with 20 as its smallest member. Do It Together

What is the square of the numbers? 1 45 2 115 Each of the numbers ends in 5. So, we know that if a = tens digit and ‘5’ is the units digit, then (a5)2 = [a (a + 1)] × 100 + 25. 1 452 = [4 (4 + 1)] × 100 + 25 = ( = 2 1152 = [

(

× 100 + 25

+ 25 =

+ 1)] × =

) × 100 + 25 = +

=( +

)×

Do It Yourself 10B 1 Express the numbers as the sum of odd numbers. a 676

b 729

c 289

d 900

e 1225

f 1024

2 List all the triangular numbers between 20 to 50. Express them as square numbers. 3 Find the difference of the squares of the consecutive numbers. How many non-perfect squares are there among its squares? a 12, 13

b 20, 21

c 17, 18

d 25, 26

e 31, 32

f 48, 49

4 Express the squares of the given odd numbers as the sum of two consecutive numbers. a 57

b 63

c 77

d 39

e 85

f 17

b 512 – 502

c 952

d 782 – 772

e 1052

f 992 – 982

b 64 × 68

c 50 × 52

d 91 × 95

e 36 × 40

5 Evaluate. a 752

6 Find. a 75 × 77

7 Find a Pythagorean triplet whose smallest member is given below. a 12

b 16

8 Observe the pattern.

Write the missing numbers. 42 =

34 =

1156

334 =

111556

3334 =

333342 =

1___5__6 2

= 111111555556

Chapter 10 • Squares and Square Roots

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c 28

d 32

e 36

f 48

9 Observe the pattern.

Fill in the missing digits. 92 = 2

99 = 2

999 = 2

9999 =

999992 =

9999992 =

9801

998001

99980001

9___8____1

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10 Without adding, find the sum. a 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 b 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 c 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41

Word Problem 1

Shreya made the pattern using stickers of circles. Look at the pattern. Answer the questions.

a How would you express each square pattern as a square of a number? Write the rule.

b Find the rule of the pattern up to the 6th term. c Use the rule to find the value of 562.

22 + (2 + 2 + 1)

Square Roots Real Life Connect

Rakesh, a farmer, owns a mango orchard. To increase his sales in the summer season, Rakesh wants to plant 9025 mango trees so that there are as many rows as there are trees in each row. Let us see how can we find the number of rows of mango trees he wishes to plant.

Finding Square Roots Using Prime Factorisation

The square root of a number is the value that can be multiplied by itself to get the original number. It is the inverse of a square number. It is denoted using the symbol ‘√’, called the radical symbol. For example, the square of 4 is 16 but the square root of 16 is 4. We write it as: 16 = 4 × 4 = 4.

16 16

Sides

Area

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Given below is a list of square roots of different numbers. Statement

Inference

1 =1

Statement

36 = 6 × 6 = 6

6 = 36

1 =1

Inference

2 =4

4 = 2×2 =2

7 = 49

49 = 7 × 7 = 7

32 = 9

9 = 3×3 =3

82 = 64

64 = 8 × 8 = 8

4 = 16 2

5 = 25

81 = 9 × 9 = 9

100 = 10 × 10 = 10

16 = 4 × 4 = 4

9 = 81

25 = 5 × 5 = 5

10 = 100

We can find square roots using different methods: 1 Using successive subtraction

2 Using prime factorisation

3 Using long division

Let us understand how we can find the square root of a number using successive subtraction. We know that the sum of the first n odd numbers is n2. ⇒ Every square number can be expressed as a sum of successive odd numbers, starting from 1. For example, for a square number 64, we subtract the odd numbers successively from 64. 64 – 1 = 63

(1 time)

63 – 3 = 60

(2 times)

60 – 5 = 55

(3 times)

55 – 7 = 48

(4 times)

48 – 9 = 39

(5 times)

39 – 11 = 28

(6 times)

28 – 13 = 15

(7 times)

15 – 15 = 0

(8 times)

Error Alert! Taking the square root of a number NEVER halves its value.

Square root of 121 =

121 = 121 = 60.5 2

As soon as we get 0, we stop here. Here, odd numbers are subtracted 8 times. So,

Square root of 121 =

121 = 11

64 = 8.

To find the square root using prime factorisation, we follow the steps below. Step 1

Find all the prime factors of the number.

Step 2

Group the same prime factors together to form pairs.

Step 3

Select one prime factor from each pair and find the product. The product is the square root of the number given.

For example, let us find the number of rows of mango trees the farmer wishes to plant. We know that each row contains as many trees as the number of rows. ⇒ The number of rows is equal to the number of plants in each row. Let the number of plants be x. So, the number of rows is also x. So, the total number of plants = number of plants × number of rows ⇒ 9025 = x × x ⇒ 9025 = x2 ⇒ 9025 = x

9025

1805

⇒ x = 5 × 5 × 19 × 19

361

⇒ x = (5 × 5) × (19 × 19)

By prime factorisation, we get

⇒ x = 5 × 19 = 95

So, the number of rows and number of plants in each row will be 95. Chapter 10 • Squares and Square Roots

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Square Root of a Rational Number We can find the square root of rational numbers by dividing the square root of its numerator by the square 16 root of its denominator. For example, to find the square root of a positive rational number , we write it as: 25 4 16 2 × 2 16 = = 2×2×2×2= = 5 25 25 5 5×5 Example 9

Find the square root of 49, using repeated subtraction.

Example 10

Find the square root of 1764, using prime factorisation.

For a square number 49, we subtract the odd numbers successively from 49. 49 – 1 = 48

(1 time)

48 – 3 = 45

(2 times)

45 – 5 = 40

(3 times)

40 – 7 = 33

(4 times)

33 – 9 = 24

(5 times)

24 – 11 = 13

(6 times)

13 – 13 = 0

(7 times)

169 = 225

169 225

441

3 7 7

882 147 49 7 1

1764 = 2 × 2 × 3 × 3 × 7 × 7 So, 1764 = (2 × 2) × (3 × 3) × (7 × 7) =2×3×7

Here, odd numbers are subtracted 7 times. So, 49 = 7. Find the value of

1764

As soon as we get 0, we stop there.

Do It Together

= 42

169 . 225

Do It Yourself 10C 1

What will be the units digit of the square root of the numbers? a 1369

3136

c 5184

d 8281

e 9604

f 2916

d 900

e 169

f 100

d 24,649

e 29,929

2 Find the square root of the numbers using successive subtraction. a 144

b 36

441

3 Find the square root of the numbers using prime factorisation. a 6561

4 Find. a

2401 2500

b 8836

3364 3600

20,164

4761 5625

625 1156

4096 6241

5 For each of the numbers, find the smallest number by which it should be multiplied to get a perfect square. Also, find the square root of the square number. a 8748

b 5202

c 3100

d 15,600

4000

f 75,712

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6 Find the smallest number by which the numbers should be divided to get a perfect square. Also, find the square root of the square number. a 9248

b 5292

c 14,580

d 29,988

e 56,628

f 4,08,980

7 Find the smallest square number that is divisible by 6, 9, 12, 16, 18.

Word Problems 1

The students of a class collected ₹1296 for a COVID relief fund. The contribution of each

student was as much as the number of students in that class. How many students were in the class?

If there is a piece of cloth of area 721 sq. cm, can we make a square tablecloth out of it where the length of each side is 27 cm?

Finding Square Roots Using Long Division We know that Rakesh planted 9025 mango trees in his mango orchard. He has another orchard in which he wishes to plant 1089 orange trees in rows so that the number of rows is the same as the number of trees. To find the number of rows, we can also use the long-division method. Let us see how can we use the long-division method to find the answer. For very large numbers such as 15,625 and 58,081, the prime factorisation method to find their square roots is time consuming. In such cases, we use the long-division method. For example, let us find the square root of 58,081. Step 1

58081 has an odd number of digits. Starting from the ones place, make pairs of digits by placing a bar over them. If the number of digits is odd, even the single digit will have a bar over it.

Step 2

Find the largest number whose square is less than or equal to the first pair (from the left). Take this number as the divisor and also as the quotient. Now, find its product and subtract the result from the first paired number to get the remainder. 2 Here, the remainder is 1. 2 5 80 81

5 80 81

Step 3

Bring down the number under the next bar to the right of the remainder. 2

Here, the new dividend is 180. 2 5 80 81 4 4 1 80

Chapter 10 • Squares and Square Roots

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Step 4

Double the quotient and write it with a blank on its righthand side. 2

2 5 80 81 4

4 1 80

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Step 5

uess the largest possible number to fill in the blank G that also becomes the new digit in the quotient.

Step 6

Repeat steps 3 to 5, as many times as required.

Here, the new divisor is multiplied with the new quotient to get a product less than or equal to the dividend. This means, as 43 × 3 = 129 and 44 × 4 = 176, we choose the new digit as 4. Now, divide in the same way and find the remainder. 2 4

2 4 1

2 5 80 81

4 4 4 1 80 1 76

4 81

2 5 80 81

4 4 4 1 80 1 76

4 81

Step 7

As the remainder is 0 and no digits are left in the number, 58081 = 241.

4 81 4 81 0

Think and Tell Can you now find the square root of 5,53,536 using the long-division method?

Now, we can also find the number of rows required to plant orange trees, using the long-division method. 3 3

3 10 89 9 63 1 89 1 89 0

1089 has even number of digits (4). So, there will be 2 (= 4 ÷ 2) digits in the square root. Therefore, if n is the number of digits, n digits in the square 2 (n + 1) root of an even number and digits in 2

then there will be

Think and Tell How many digits are there in the square root of a number?

the square root of an odd number.

So, 33 rows of orange trees will be plant by Rakesh.

5 .6 2

Square Root of Decimal Numbers Find the square root of 31.5844. We can find the square root of decimal numbers in the same way as the square root of whole numbers. So, 31.5844 = 5.62 Approximate Square Root by the Division Method If the square root of the number is to be corrected to n decimal places, find the square root by long division up to (n + 1) decimal places first and then, round off the square root up to n decimal places. For example, let us find the square root of 5 correct to three decimal places. So, 5 = 2.2360 (up to 4 decimal places) = 2.236 (up to 3 decimal places) Thus, 5 = 2.236.

5 31 .58 44 25 106 6 58 6 36 1122 22 44 22 44 0

2.2 3 6 0

2 5 .00 00 00 00 4 42 1 00 84 443 16 00 13 29 2 71 00 4466 2 67 96 3 04 00 44720 0 3 04 00

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Example 11

Find the square root of 7,12,336.

Example 12

8 4 4

0 .3 4 6

8 71 23 36 64 164 7 23 6 56 67 36 1684 67 36

Example 13

712336 = 844.

Find the square root of For

0 0.12 00 00 00 0 3 12 9 3 00 64 2.56 686 44 00 44 00 2.84 00 6924 2.76 96 7 04

So,

Find the estimated square root of 0.12 correct to three decimal places.

So, 0.12 = 0 .3464 (up to 4 decimal places) = 0.346 (up to 3 decimal places) Thus, 0.12 = 0.346.

784 and (40 × 160). 841

784 , we find the square root of 784 and 841 first. 841

2 8 2 7 84 4 48 3 84 3 84 0

2 9 2 8 41 4 49 4 41 4 41 0

Now, 784 = 28, 841 = 29 So,

784 841

28 29

For (40 × 160), we find the square root of each number first. That is, (40 X 160) = 40 × 160 = 2 × 2 × 2 × 5 × 2 × 2 × 2 × 2 × 2 × 5

= 2×2×2×2×2×2×2×2×5×5

= 80

Example 14

=2×2×2×2×5

What lowest number must be subtracted from 5,31,606 to get a perfect square? 7292 is less than 5,31,606 by 165. So, the lowest number to be subtracted from 5,31,606 to get a perfect square is 165. Thus, the required perfect square number is 5,31,606 – 165 = 5,31,441.

Example 15

7 2 9

7 53 16 06 49 142 4 16 2.84 1 32 06 1449 1 30 41 1 65

A ladder is placed with its foot 8 m away from the bottom of a wall 15 m high. The top of the ladder reaches the top of the wall. Find the length of the ladder. Height of the wall (perpendicular) = 15 m Width of the ground from the wall to the ladder (base) = 8 m

15 m

The ladder forms a right-angled triangle with the wall. ladder

Length of the ladder (hypotenuse) squared = 152 + 82

= 225 + 64 = 289 = 172

The height of the ladder is 17 m. Chapter 10 • Squares and Square Roots

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ground 8m

145

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Find the square root of 160.0225.

Do It Together

1 . 1 1 60 .02 25 1 60 2

Did You Know? The construction of pyramids and many other structures in nature involves the concept of squares

So,

and square roots.

160.0225 = __________________________

Do It Yourself 10D 1

How many digits would be there in the square roots of the numbers (without any calculation)? a 81

18,496

14,641

77,284

c 1,04,976

d 1,26,736

e 1,45,161

5243

c 8912

d 27,804

e 1,28,643

f 3,13,684

square roots of the perfect squares so obtained. b

4620

c 1,74,700

d 77,800

e 2,83,000

f 5,43,291

Find the square roots of the decimal numbers. b

42.5104

78.8544

8101.8001

e 12

15.21 34.81

1032.9796

Find the square root of the numbers correct to three decimal places. a 0.8 g

Find the lowest number which must be added to each of the numbers to get a perfect square. Also, find the

a 7.0225

2304

square roots of the perfect squares so obtained.

a 5608

Find the least number which must be subtracted from each of the numbers to get a perfect square. Also, find the a 1456

c 8464

Find the square root by the long-division method. a 65,536

b 289

b 85

8 ×

f 9.6

5.5

Simplify. a

256 × 361

289 256

92.16 +

0.9216

Find the greatest number with four digits which is a perfect square. Also, find the square root of the number

Find the lowest number with five digits which is a perfect square. Also, find the square root of this number.

obtained.

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Word Problems 1

Find the longest side of a right triangular field with the other two sides of 15 m and 20 m.

Beginning from her house, Nancy walks 150 m north and then 360 m west to reach a

friend’s house. On her way back, she walks diagonally until she reaches her own house. What distance did she walk while coming back?

Points to Remember •

If a number has 4 or 6 in its ones place, then its square ends in 6.

•

Squares of even numbers are always even, and those of odd numbers are always odd.

•

In square numbers, the digits always end in 0, 1, 4, 5, 6 or 9 and not 2, 3, 7 or 8.

• To check if a number is a perfect square, we use prime factorisation. If on prime factorising the number, we get pairs of equal prime factors and no factor is left over, then the number is a perfect square. •

The sum of the first n odd numbers is n2.

• If m is any non-zero number and m > 1, then we have (2m)2 + (m2 – 1)2 = (m2 + 1)2. So, 2m, m2 – 1, m2 + 1 are said to form a Pythagorean triplet.

• The square root of a number is the value that can be multiplied by itself to get the original number. It is the inverse of a square number. It is denoted using the symbol ‘√’, called the radical symbol.

Math Lab Square Root Art Gallery! Setting: In groups of 5 Materials Required: Large sheets of paper, rulers, coloured pencils, markers, or paints Method: All 5 members of each group must follow these steps. tart with a square as the base shape for your artwork. Choose any size for your initial 1 S square. se the concept of square roots to create smaller squares within the original square. The 2 U size of the smaller squares can be determined by taking the square root of the area of the larger square. se coloured pencils, markers or crayons to fill in your squares and create visually appealing 3 U designs. Experiment with colours, patterns and arrangements of different-sized squares. 4 Students who can use the concept of square roots correctly win a reward. 5 Create an art gallery of these designs on your class bulletin board to showcase your artwork.

Chapter 10 • Squares and Square Roots

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Chapter Checkup 1

Find the squares of the rational numbers. a

3 4

8 11

c –

65 × 67

2 c 85

b –

3 7

d –

9 13

Find the value of: 2 2 a 45 – 16

d 120 × 124

Express 324 as the sum of odd numbers.

Find Pythagorean triplets whose smallest members are 96 and 24.

Which of the numbers are perfect squares? Find their square roots. Verify your answer. a 265

2116

c 8839

3136

72,361

1,06,928

Find the square roots of the decimal numbers. a 7.8961

7 15

b 400.4001

21.6225 9.1204

d 33.64 × 67.24

Find the square roots correct to two decimal places. a 3.7589

b 81.0943

8 Find the reciprocal of the square root of

c 6512.8007

d 467.892

121 . 196

9 Find the lowest number which must be added to 6,57,700 to get a perfect square. Also, find the square root of the perfect square so obtained.

10 Find the greatest square number divisible by each of the numbers 8, 9 and 10.

Word Problem 1 The length and breadth of a rectangular room are 26 m and 18 m, respectively. What is the length of the longest straight line that can be drawn on the floor of the room?

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Cubes and Cube Roots

Let's Recall

1 Counting unit cubes layer by layer

length

2 Counting unit cubes along the sides To find the volume of this solid by counting cubes layer by layer, we have

To find the volume of this solid by counting cubes along the sides, we have

Number of unit cubes in layer 2 = 12

Total number of unit cubes = 12 + 12 = 24

length 4 unit cubes

Layer 2

br un ea it dth cu be s

height 2 unit cubes

Layer 1

Number of unit cubes in layer 1 = 12

br ea dt

height

The volume of a solid is the amount of space it occupies. For example, take this solid made of unit cubes. Each side of the solid has a different number of unit cubes. Its volume can be found in two ways:

Volume of the given solid = 4 × 3 × 2

= 24 cu. units

The volume of the given solid is 24 cu. units.

Letʼs Warm-up Fill in the blanks.

1 T here are an apple and a pineapple on a table. Between both the fruits, ___________ occupies more space. 2 I f a solid with 2 layers has 5 unit cubes along its length and 3 unit cubes along its breadth, its volume is ___________ cu. units. 3 If a solid has 3 layers with 16 unit cubes in each layer, then its volume is ___________ cu. units.

4 I f a container is 3 cm long, 2 cm wide, and 2 cm high, then ___________ cm cubes will be in the container. 5 A carton is filled with fourteen 2 cm cubes. The volume of the carton is ___________ cu. cm.

I scored _________ out of 5.

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Cube of a Number Real Life Connect

Jiya received a Rubik’s cube as a gift on her birthday. Upon examining the cube, she noticed it was composed of 27 smaller cubes. She counted the number of small cubes along the length, width, and height of the cube and determined that there were 3 small cubes along each dimension. She noticed a relation between the number of small cubes along the length, width and height of the Rubik’s cube and the total number of small cubes. Can you think what it was?

Perfect Cube Look at the figures. Each number has been represented using unit cubes. Like squares, some form complete cubes and some do not.

11 1 1 11 11 1 1 1 1 1 1 11 1 11 11 A A1cube 11 cube

22 1

11 1

111 11 1

11 1

1 221 1 22 221 1 1 1 11 2 2 2 2 2perfectCube 2 2 1Not 2 a perfect Cube Not a perfect Cube Not a perfect Not a perfect Cube Not a perfect Cube Not a Cube 1 22 1 1 2 2unit 3 3unit 4 4unit 1Cube 1unit unitcube cube unitcubes cubes unitcubes cubes unitcubes cubes 2 A1cube Not a perfect Not 2 a perfect Cube Not Cube 2 2 a 2perfect A cube Not a perfect Cube 1 Not Cube 2 Cube 22 cube 1 a perfect 242unit 22 cube Not a perfect 1 unit cube 2 unit cubes 3 unit cubes cubes A cube Not a perfect cube Not a perfect Not a perfect 1 1 unit cube 2 unit cubes 3 unit cubes 4 unit cubes 1 A cube A cube Not a perfect Not a perfect 1 Not aCube perfect CubeNot a perfect Not aCube perfect Cube Not aCube perfect Cube

1unit unit cube A cube 1 cube cube 1 unit Acube cube 1 × 1 × 1 1=1unit 1unitcube

2 cubes unit unit Cube cubes Not acubes perfect Cube 3 unit cubes Not a3perfect 2 unit perfect Cube perfect Cube 2Not unita cubes 3Not unita cubes 3 unit cubes cubes 3 unit 2 × 212unit ×unit 1cubes 1 ×cubes 2×1

2 2

unitCube cubes Not a 4 perfect 4 unit cubes perfect Cube 4Not unita cubes 4 unit cubes 4 unit 2 cubes ×2×1

2 2

2 2 2Nota2aperfect 2 Not 2 Cube Not Not Not perfectCube Cube 2 Not perfect Cube Nota aperfect perfectCube Cube Nota aperfect perfectCube Cube 2 a aperfect 2 7 7unit 2 5 unit cubes 6 unit cubes cubes 8 8unit cubes 2 2 2 5 unit cubes 6 unit cubes unit cubes unit cubes 2 Not 2 a perfect Cube 2 Not a perfect Cube Not2a perfect Cube Not2a perfect Cube 2 2 2 2 Not a perfect Cube Not a perfect Cube Not a perfect Cube Not a perfect Cube 282 unit 25 unit 2 6 unit 2 cubes × ×2 2× ×? ? 2 2× cubes 2× ×22 2× ×?2? 22 2× ×2 2× ×?2? ×2 2× ×22 2=2=8 8 72unit2 cubes 22 2 cubes 2 5 unit cubes 6 unit2cubes 7 unit cubes 8 unit cubes 2 Not a perfect 2Cube Not a perfect Cube Not a perfect Cube Not a perfect Cube 2 × 2 ×Cube ? 2 × 2 ×Cube ?2 2 × 2 ×Cube ? 2perfect × 2 × 2Cube =8 Not a perfect Not a perfect Not a perfect Not a 2 ×Not 2×a ? perfect Cube 2 ×Not 2×a ? perfect Cube 2 ×Not 2×a ? perfect Cube 2 × 2Not × 2 =a 8 perfect Cube 5 unit cubes 7 unit cubes 8 unit cubes perfect Cube 6 unit cubes perfect Cube perfect Cube perfect Cube 5Not unita cubes 6Not unita cubes 7Not unita cubes 8Not unita cubes unit cubes 6 unit cubes 7 unit cubes unit cubes 2 × 2 × ? 255 2 × 2 × ? 2 × 2 × ? 2 × 2 × 2 =2 8×82 8× unit × unit 2 × ?cubes 26 × unit 2 × ?cubes 27 × unit 2 × ?cubes 2 =cubes 8 2×2×? 2 × 2cube ×? 2 × 2 × ? cube 2A ×2×2=8 Not a perfect cube Not a perfect Not a perfect 2×2×? 2×2×? 2×2×? 2 × cube 2×2=8 2

5 unit cubes 2×2×?

6 unit cubes 2×2×?

7 unit cubes 2×2×?

8 unit cubes 2×2×2=8

So, when we make a cube using 1 unit cube or 8 unit cubes, we get a complete cube. 1 and 8 are called cube numbers or perfect cubes.

We see that each perfect cube has the same length, breadth and height. So, to find the number of unit 443 3 of unit cubes on one side, three times. The product so cubes in a perfect cube, we multiply the number 3 43 4 obtained is called the cube of a number.

Height Height

Height Height Height Height Height Height

4 143×3 1 × 1 = 13. For example, we can write the cube of 1 as 4 3

Similarly, the cube of 2 is written as42 × 2 × 2 = 23.

64 4 64 Thus, the cube of 1 is 1 and the cube 4 4of 2 is 8. 6464 33 4h 64 LeLe 64 h 64 t tbelow 3in the table. Some more cube numbers are listed nngg d d i L i 3 tt e Le th 64 64 dW ng ngt hh idthWiW 3 Le h Volume Volume Wth 3 h h 64 3 64 ng LtehLnLe t 364 d i d Volume t 64 i id th th gentng Volume W W W id hgth 150 th W Volume Volume Volume Volume

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64 64 64

23 = 2 × 2 × 2 = 8. This means ‘cube’.

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Cubes of even numbers are always even. Number Cube

1×1×1 2×2×2

Product

3×3×3 27

4×4×4 64

125

216

343

512

729

5×5×5 6×6×6 7×7×7 8×8×8 9×9×9

103

10 × 10 × 10 1000

The units digit of the cube of some numbers is the same as the units digit of the number. Number

Cube

113

123

133

143

153

163

173

183

193

203

× 11

× 12

× 13

× 14

× 15

× 16

× 17

× 18

× 19

× 20

× 11

× 12

× 13

× 14

× 15

× 16

× 17

× 18

× 19

× 20

1331

1728

2197

2744

3375

4096

4913

5832

6859

8000

Meaning Product

Cubes of odd numbers are always odd. Cubes of Negative Integers Look at the cube numbers.

Think and Tell

(−2)3 = −2 × −2 × −2 = −8

cube numbers? Also, if a cube of a number is a

(−1)3 = −1 × −1 × −1 = −1

What can you say about the units digit of other

(−3)3 = −3 × −3 × −3 = −27

multiple of 3, is it also a multiple of 27?

…

(−10)3 = −10 × −10 × −10 = −1000

In each case, the cube of a negative number is negative. So, we can say that the cube of a negative integer is always negative. Cubes of Rational Numbers Like squares of rational numbers, we can also find the cubes of rational numbers in the same way. To find the cube of a rational number, we multiply it thrice. For example, to find the cube of a rational 5 number , we write it as: 7 3 5 5 5 5 = × × = 5 × 5 × 5 = 125 7 7 7 7 × 7 × 7 343 7

Similarly, to find the cube of a negative rational number, −8 11

−8 −8 −8 (−8) × (−8) × (−8) −512 × × = = 11 × 11 × 11 1331 11 11 11

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Did You Know?

The famous Rubik’s cube is a popular puzzle invented by Ernő Rubik in 1974. Also, the largest Rubik’s cube in the world is 2.022 m × 2.022 m × 2.022 m.

−8 we write it as: 11

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To Check If a Number Is a Perfect Cube or Not Like perfect squares, we use prime factorisation to check if a number is a perfect cube or not. If, on prime factorising the number, we get a triplet of the same prime factors leaving behind no other factor, then the number is a perfect cube. For example, let us check if 728 and 3375 are perfect cubes. For 728

Step 1: Find the prime factorisation of 728. 2 2 2 7 13

728 364 182 91 13 1

728 = 2 × 2 × 2 × 7 × 13

Step 2: Group the prime factors into triplets of the same factors until no factor is left over. 728 = 2 × 2 × 2 × 7 × 13 Since all the prime factors do not form a triplet, 728 is not a perfect cube. Example 1

For 3375

Step 1: Find the prime factorisation of 3375. 5 5 5 3 3 3

3375 = 5 × 5 × 5 × 3 × 3 × 3

Step 2: Group the prime factors into triplets of the same factors until no factor is left over.

3375 = 5 × 5 × 5 × 3 × 3 × 3 Since all prime factors form a triplet and no factor is left over, 3375 is a perfect cube.

Find the perfect cube of:

1 (−18)3 = −18 × −18 × −18 = −5832 2

6 10

6×6×6 6 6 6 × × = = 216 10 10 10 10 × 10 × 10 1000

3 (2.08)3 = 208 100 Example 2

3375 675 135 27 9 3 1

208 208 208 8998912 × × = = 8.998912 100 100 100 1000000

Remember! Every cube number is written as a product of three like factors.

What is the smallest number by which 3456 may be multiplied so that the product is a perfect cube? 3456 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 Clearly, not all prime factors form a triplet and we are left with a factor 2 which cannot form a triplet. We require two more 2s to form a triplet of 2. So, (2 × 2 = 4) is the smallest number with which 3456 is multiplied to get a perfect cube.

Do It Together

Find the least number by which 59,049 may be divided so that the product is a perfect cube. 59,049 = 3 × 3 × _________________________________

3 3

2 3456 2 1728 2 864 2 432 2 216 2 108 2 54 3 27 3 9 3 3 1 59049 19683

Clearly, not all prime factors form a triplet and we are left with a factor ___________ which cannot form a triplet. So, ___________ is the lowest number by which 59,049 should be divided to get a perfect cube.

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Patterns in Cubes Like perfect squares, we can also create certain patterns for perfect cubes. Pattern 1 •

1 = 1 = 13

•

3 + 5 = 8 = 23

•

7 + 9 + 11 = 27 = 33

•

13 + 15 + 17 + 19 = 64 = 43

•

…

Pattern 2

…

Rule: If n3 = sum of n odd numbers, then the first odd number = 2 × (n – 1)th triangular number + 1, where (n – 1)th triangular number = sum of (n – 1) consecutive numbers.

•

13 + 23 = 1 + 8 = 9 and (1 + 2)2 = 32 = 9

•

13 + 23 + 33 = 1 + 8 + 27 = 36 and (1 + 2 + 3)2 = 62 = 36

•

13 + 23 + 33 + 43 = 1 + 8 + 27 + 64 = 100 and (1 + 2 + 3 + 4)2 = 102 = 100

•

…

Rule: 13 + 23 + 33 + 43 + … + n3 = (1 + 2 + 3 + 4 + … + n)2

Pattern 3 • 23 – 13 = 1 + 2 × 1 × 3 • …

• 33 – 23 = 1 + 3 × 2 × 3

• 43 – 33 = 1 + 4 × 3 × 3

…

Rule: If n and n + 1 are two consecutive numbers, then (n + 1)3 – n3 = 3n(n + 1) + 1. Example 3

Use the first pattern above to find the value of 83. We know that n3 = the sum of n odd numbers where the first odd number = 2 × (n – 1)th triangular number + 1. Also, (n – 1)th triangular number = sum of (n – 1) consecutive numbers 83 = sum of 8 odd numbers, where the first odd number

where the 7th triangular number

= 2 × (8 – 1)th triangular number + 1

= sum of 7 consecutive numbers

= 2 × 7th triangular number + 1

= 2 × (1 + 2 + 3 + 4 + 5 + 6 + 7) + 1 = 2 × 28 + 1

So, 8 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512 Example 4

= 56 + 1 = 57

Use the second pattern given above and find the value of 13 + 23 + 33 + 43 + 53 + 63 + 73. We know that 13 + 23 + 33 + 43 + … + n3 = (1 + 2 + 3 + 4 + … + n)2 So, 13 + 23 + 33 + 43 + 53 + 63 + 73 = (1 + 2 + 3 + 4 + 5 + 6 + 7)2 = 282 = 784

Do It Together

Use the third pattern given above, to find the value of 873 – 863. We know that if n and n + 1 are two consecutive numbers, then (n + 1)3 – n3 = ___________. So, 873 – 863 = 3 × ______________________ = ______________________ = ______________________

Chapter 11 • Cubes and Cube Roots

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Do It Yourself 11A 1

Find the cubes of the numbers. a 15

f 88

121

1331

2 Look at the digits in the units place of each of the numbers. Answer the questions. a 8

f 5800

24,387

1,25,000

216

What will be the digit in the ones place of the cube of each number?

ii Which of the numbers will have odd or even cubes? iii Which of the numbers are not perfect cubes?

3 Find the cube of the rational numbers. a

4 7

8 9

6 8

18 f −

h − 2

b −

e − 5

4 Find the cube of the decimal numbers.

13 d 1

2.8

1.2

0.56

0.08

4.7

5.89

3.04

9.1

5 Check if the numbers are perfect cubes using prime factorisation. Also, find the number whose cube is the number.

a 2195

13,824

c 1,03,823

1,17,649

3,89,010

8,57,375

6 Find the smallest number by which the number should be multiplied to get the perfect cube. In each case, find the number whose cube is the new number. a

432

128

48,672

1,21,500

2916

3456

10,125

7 What is the least number by which the number should be divided to get the perfect cube? In each case, find the number whose cube is the new number. a 576

5145

36,288

55,566

1,58,760

3,97,535

8 Use pattern 1 to find the values of 113, 153, 183, and 203. 9 Use pattern 2 to evaluate the given expressions. a 13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93

b 13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93 + 103 + 113 + 123

c 13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93 + 103 + 113 + 123 + 133 + 143 + 153

10 Use pattern 3 to find the difference of: a 413 and 403

563 and 553

883 and 893

1003 and 993

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Word Problems 1

How many packs of chocolates of dimensions 8 cm × 10 cm × 2 cm can be

Mitali forms a solid as shown below. It is made up of 2 cm cubes. What is

placed in a cubical carton of length 20 cm?

the volume of this solid? Can we rearrange the cubes to form a perfect cube? If not, how many more unit cubes will be needed to form a perfect cube?

1 1

Cube Root of a Number Real Life Connect

A cube 1 unit cube

Not a perfect Cube 3 unit cubes

The cube root of a number is the value that can be multiplied by itself to get the original number. It is the inverse of a cube number. It is denoted using the symbol ‘3 ̕ .

Index

Height ng

idt

Radicand

64 3

Inference

13 = 1

3 1 =3 1×1×1 =1

23 = 8

3 8 =3 2×2×2 =2

33 = 27

3 27 = 3 3 × 3 × 3 = 3

43 = 64

3 64 = 3 4 × 4 × 4 = 4

53 = 125

3 125 = 3 5 × 5 × 5 = 5

63 = 216

3 216 = 3 6 × 6 × 6 = 6

73 = 343

3 343 = 3 7 × 7 × 7 = 7

83 = 512

3 512 = 3 8 × 8 × 8 = 8

93 = 729

3 729 = 3 9 × 9 × 9 = 9

103 = 1000

3 1000 = 3 10 × 10 × 10 = 10

Not a perfect Cube 8 unit cubes 2×2×2=8

43 4

Not a perfect Cube 7 unit cubes 2×2×?

Statement

Chapter 11 • Cubes and Cube Roots

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Not a perfect Cube 6 unit cubes 2×2×?

Cube Root Through Prime Factorisation Method

Given below is a list of cube roots of different numbers.

Not a perfect Cube 4 unit cubes

Not a perfect Cube 5 unit cubes 2×2×?

We write it as: 3 64 = 3 4 × 4 × 4 = 4.

Not a perfect Cube 2 unit cubes

Snigdha bought a cubical aquarium of volume 1,95,112 cu. cm. She wanted to put it on a square table so that it would fit the table exactly. For this, she needed the measures of the aquarium.

For any number n, 3 n is called the radical, where 3 is called the index and n is called the radicand. For example, the cube of 4 is 64 but the cube root of 64 is 4.

Volume

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We can find cube roots using different methods:

Error Alert!

1 Using prime factorisation 2 Using estimation

The cube root of a number is NEVER one-third of its value.

To find the cube root using prime factorisation, we follow the steps.

Cube root of 1331 = 3 1331 =

For example, for a cube number 10,648, we prime factorise the number.

1331 = 443.67 3

Cube root of 1331 = 3 1331 = 11

Step 2: Group the same prime factors together in the form of a triplet.

Step 1: Find all the

prime factors of the number.

10648

5324

2662

1331

121

10648 = 2 × 2 × 2 × 11 × 11 × 11 So, 3 10648 = 3 2 × 2 × 2 × 11 × 11 × 11

= 2 × 11 = 22

Step 3: Select one prime factor from each group

and find the product. The product is the cube root of the number.

Now, let us find the measure of the aquarium.

We know that the volume of the aquarium is 1,95,112 cu. cm.

⇒ One side of the aquarium is the same as the cube root of its volume.

195112

So, by prime factorisation, we get

97556

1,95,112 = 2 × 2 × 2 × 29 × 29 × 29

48778

So, 3 195112 = 3 2 × 2 × 2 × 29 × 29 × 29

24389

841

= 2 × 29 = 58

Therefore, one side of the fish aquarium would be 58 cm. Cube root of a negative perfect cube

To find the cube root of a negative perfect cube, we find the cube root of its absolute value and multiply it by −1. For example, to find the cube root of −343, we write it as: 3

−343 = 3 −7 × −7 × −7 = −7 because 3 343 = 3 7 × 7 × 7 = 7

Cube root of a rational number

We can find the cube root of a rational number by dividing the cube root of its numerator by the cube root 2197 of its denominator. For example, to find the cube root of a rational number , we write it as: 2744 3

3 13 × 13 × 13 13 2197 3 2197 13 =3 = = = 3 2 × 7 2744 14 2744 2×2×2×7×7×7

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For a negative rational number , we write it as: 3 –5832 – 3 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 −2 × −3 × −3 −18 –5832 = 3 = = = 3 3 19 19 6859 6859 19 × 19 × 19 Cube root of the product of integers

To find the cube root of the product of integers, we multiply the cube root of one integer by the cube root of the other integer. For example, to find the cube root of 27 × (−64), we write it as: 3 27 × (–64) = 3 27 × 3 –64 =3 3×3×3 ×− 3 2×2×2×2×2×2 = 3 × − 4 = – 12

Cube root of a perfect cube using a pattern

We know that 03 = 0, 13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, … and so on.

So, from the above, we get

0×1 ×6 2 23 – 13 = 8 – 1 = 7 = 1 + 1 × 6 = 1 + 1 × 2 × 6 2 13 – 03 = 1 – 0 = 1 = 1 + 0 × 6 = 1 +

2×3 3 – 2 = 27 – 8 = 19 = 1 + 1 × 6 + 2 × 6 = 1 + ×6 2 3×4 43 – 33 = 64 – 27 = 37 = 1 + 1 × 6 + 2 × 6 + 3 × 6 = 1 + ×6 2 4×5 53 – 43 = 125 – 64 = 61 = 1 + 1 × 6 + 2 × 6 + 3 × 6 + 4 × 6 = 1 + ×6 2 3

Rule: (n + 1)3 – n3 = 1 +

n (n + 1)

1 = 13 1 + 7 = 8 = 23 1 + 7 + 19 = 27 = 33 1 + 7 + 19 + 37 = 64 = 43 1 + 7 + 19 + 37 + 61 = 125 = 53

×6

…

and so on.

Any cube number can be written as the sum of 1, 7, 19, 37, 61, …,271 and each of these odd numbers can be obtained using (n + 1)3 – n3 = 1 + n (n + 1) × 6. 2 So, to find the cube root of any number, we successively subtract these odd numbers until we get 0. As soon as we get 0, we count the number of times the subtraction is carried out. This number is the cube root of the given number. For example, for a cube number 64, we subtract the odd numbers successively from 64. 64 – 1 = 63

(1 time)

56 – 19 = 37

(3 times)

63 – 7 = 56

37 – 37 = 0

(2 times) (4 times)

As soon as we get 0, we stop here.

Here, odd numbers are subtracted 4 times. So, 3 64 = 4. Example 5

Find the cube root of 4913, using prime factorisation. 4913 = 17 × 17 × 17

So, 3 4913 = 3 17 × 17 × 17 = 17

Chapter 11 • Cubes and Cube Roots

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17 4913 17 289 17 17 1

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Example 6

Find the cube root of 729, using successive subtraction.

For a cube number 729, we subtract the odd numbers successively from 729. 729 – 1 = 728

(1 time)

728 – 7 = 721

(2 times)

721 – 19 = 702

(3 times)

702 – 37 = 665

(4 times)

665 – 61 = 604

(5 times)

604 – 91 = 513

(6 times)

513 – 127 = 386 (7 times)

386 – 169 = 217

(8 times)

217 – 217 = 0

(9 times)

As soon as we get 0, we stop there.

Here, odd numbers are subtracted 9 times. So, 3 729 = 9. Example 7

Find the smallest number that must be subtracted from 624 to make it a perfect cube. Also, find the cube root of the perfect cube obtained. To find the smallest number, we subtract the odd numbers successively from 624. 624 – 1 = 623 (1 time)

616 – 19 = 597 (3 times)

560 – 61 = 499 (5 times)

408 – 127 = 281 (7 times)

623 – 7 = 616 (2 times)

597 – 37 = 560 (4 times)

499 – 91 = 408 (6 times)

281 – 169 = 112 (8 times)

The next number to be subtracted from 112 is 217. But 217 > 112, so 112 must be subtracted from 624 to get a perfect cube. So, 112 is the smallest number that must be subtracted from 624 to get a perfect cube. Now, 624 – 112 = 512

So, 3 512 = 3 (2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2) = 2 × 2 × 2 = 8. Example 8

Three numbers are in the ratio 1:2:3. If the sum of their cubes is 15,43,500, then find the numbers. Let the numbers be x, 2x and 3x.

Then, x3 + (2x)3 + (3x)3 = 15,43,500 ⇒ x3 + 8x3 + 27x3 = 15,43,500 ⇒ 36x3 = 15,43,500

15,43,500 = 42,875 36 ⇒ x = 3 42875 ⇒ x3 =

⇒ 42,875 = 5 × 5 × 5 × 7 × 7 × 7

So, x = 35

42875

8575

1715

343

7 1

So, 3 42875 = 3 5 × 5 × 5 × 7 × 7 × 7

=5×7 = 35

⇒ 2x = 2 × 35 = 70, 3x = 3 × 35 = 105

Therefore, the numbers are 35, 70 and 105. Do It Together

Find the value of 1

–1728 2744

3 1728 –1728 = – = ____________________ 3 3 2744 2744

8 × (−125)

8 × (−125) = 3 8 × 3 −125 = ____________________

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Do It Yourself 11B 1 Find the cube root using prime factorisation. a

8000

15,625

21,952

46,656

1,10,592

2 Find the value of: a

32768 97336

–21952 132651

3 125 × (−2744)

3 (−1728) × (−6859)

3 Find the cube root of the numbers using successive subtraction. a

216

512

1728

1000

2,38,328 cu. cm

1331

4 Find the length of each edge of each given cube if its volume is: a

5,92,704 cu. cm.

85,184 cu. cm

5 Find the smallest number that must be subtracted from each number to make it a perfect cube. Also, find the cube root of the perfect cube obtained. a

408

2215

4150

1392

6893

6 Find the smallest number that must be added to each number to make it a perfect cube. Also, find the cube root of the perfect cube obtained. a

720

2700

4021

9201

5811

7 Three numbers are in the ratio 1 : 3 : 6. If the sum of their cubes is 15,616, then find the numbers. 8 1*28 is a 4-digit number. If the number is a perfect cube, find the missing digit and the cube root of the number.

Word Problem 1

A sculptor is creating a structure using cubes for his next project. If the structure has a volume of 74,088 cu. cm., then what is the length of each side?

Cube Root Through Estimation We know that the volume of the cubical aquarium is 1,95,112 cu. cm. To find the measures of the sides of the aquarium, we can also use the method of estimation. For this, we follow the steps.

Chapter 11 • Cubes and Cube Roots

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Think and Tell How many digits are there in the cube root of a number?

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195 lies between 125 and 216. 53 = 125 and 63 = 216 5<6 So, the tens place of the cube root = the smaller number = 5

Step 1: Form groups of three, starting from the right.

195 112

Units digit of 112 = 2 Its cube = 8

Group 2

Units digit of the cube root of the number = 8

Group 1

Step 2: Look at the units place of

the perfect cube and determine the digit in the units place of its cube root.

Step 3: Look at the number in

group 2. Find the cubes between which this number lies. The cube root which is less than or equal to this number will be the digit in the tens place of the cube root.

So, 3 195112 = 58. Therefore, each side of the aquarium is 58 cm. Example 9

Find the cube root of 50,653 using estimation.

50 653

Group 2

Group 1

50 lies between 27 and 64. 3

Units digit = 3, its cube = 27

(3 = 27 and 4 = 64)

So, the units digit of the cube root of the number = 7.

3<4 So, the tens digit of the cube root of the given number = the smaller number = 3 Thus, 3 50653 = 37. Do It Together

Guess the cube root of 5,92,704. 592 704 Group 2

Group 1

___________________________ 83 = 512 and 93 = 729 ____________________________________________________

Units digit = 4, its cube = 64.

So, the units digit of the cube root of the number = _______________.

Thus, 3 592704 = __________.

Do It Yourself 11C 1 Find the cube root using estimation. a

1000

6859

17,576

35,937

79,507

2 Guess the cube root of each of the numbers. Check your answer by finding its cube. a

1,32,651

1,40,608

3,00,763

4,74,552

3 If the volume of a cube is 7,78,688 cu. cm, then guess its side length.

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4 Find the value of X if 3 15125 × X = 55. 5 If a3 – 41 = 1,17,608, then find the value of a.

Word Problem 1

The volume of a cubical box of cashew nuts is 29.791 cu. cm. What is its surface area? (Hint: Surface area of a cube = 6 (side)2)

Points to Remember • A number that forms a complete cube is a cube number or a perfect cube. • Cubes of even numbers are always even and those of odd numbers are always odd. • The cube of a negative integer is always negative. • We use prime factorisation to check if a number is a perfect cube. If on prime factorising the number, we get a triplet of the same prime factors leaving behind no other factor, then the number is a perfect cube. • The sum of the cubes of the numbers is equal to the square of their sum. • The difference of the cubes of two consecutive numbers is equal to three times their product more than 1. • The cube root of a number is the value that can be multiplied by itself twice to get the original number. It is the inverse of a cube number. It is denoted using the symbol ‘3 ’. • We can find the cube root of a number using prime factorisation and estimation. • To find the cube root of a negative perfect cube, we find the cube root of its absolute value and multiply it by −1.

Math Lab Setting: In groups of 5

Cube Root Art Challenge

Materials Required: Large sheets of paper, rulers, coloured pencils, markers or paints. Method: All 5 members of each group must follow these steps.

1 Start with creating multiple cubes of different sizes, using coloured paper. You may use rulers to ensure that the cubes have uniform dimensions.

2 For each cube you create, calculate its volume and write it on the cube. The volume should correspond to a perfect cube.

(The one who creates the highest number of cubes perfectly wins the first part of the challenge.)

3 Think about the relationships between the cube sizes and their volumes and present them in the class.

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Chapter Checkup 1 Find the cube of each of the numbers. a 41

16 7

2,10,000

−

2 9

9.8

3.6

2 Which of the following are perfect cubes? Find the number whose cube is the number. a 8,30,584

3,28,500

1,32,651

3 What is the least number by which 24,500 should be multiplied to get a perfect cube? Find the number of which the new number is the cube.

4 Find1 the smallest number by which 53,45,344 should be divided to get the perfect cube. Find the number of which

1 1

the cube is the new number. 1

Not a perfect Cube Not a perfectof: Cube 5 Using patterns, find the value 3 unit cubes 4 unit cubes

A cube 1 unit cube

Not a perfect Cube 2 unit cubes

a 13

1 + 22 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 1

A cube 1 unit cube

Not a perfect Cube 2 unit cubes

1 1

Not a perfect Cube 3 unit cubes

Not a perfect Cube 4 unit cubes

2 6 Find the cube root of the numbers using estimation. Check your answer by using prime factorisation. A cube 1 unit cube

a 2,50,047

Not a perfect Cube 2 unit cubes

7,29,000 2

Not a perfect Cube 3 unit cubes

Not a perfect Cube 4 unit cubes

8,84,736

1,03,823

Not a perfect Cube a perfect Cube 7 Find the length of the edgeNotof 7 unit cubes 8 unitthe cubes cube if its volume is 2,26,981 cu. m.

Not a perfect Cube 5 unit cubes 2×2×?

Not a perfect Cube 6 unit cubes 2×2×?

2×2×?

8 Evaluate:

Not a perfect Cube 5 unit cubes 2×2×?

Not a perfect Cube 6 unit cubes 2×2×?

Not a perfect Cube 7 unit cubes 2×2×?

–2744 b 3 8000

2×2×2=8

512 a 3 1000

Not a perfect Cube 5 unit cubes 2×2×?

Not a perfect Cube 6 unit cubes 2×2×?

Not a perfect Cube 8 unit cubes 2×2×2=8

4096 ×1000000

Not a perfect Cube 7 unit cubes 2×2×?

Not a perfect Cube 8 unit cubes 2×2×2=8

10.648

9 Find the smallest number that must be subtracted from 2789 to make it a perfect cube. Also, find the cube root of 43

3 4the perfect cube obtained.

Height

4 4 10 Find the smallest number that must be added to 4900 to make it a perfect cube.64Also, find the cube root of the

perfect cube obtained.

h idt 11

Height

64ook at the two solids. What is the difference of their L Le 64 h ng out without volumes? Find actual multiplication. idt Volume th W

Volume

idt

Volume 12 Look at the solid. The shaded portion of the solid

has an area of 147 sq. cm. What is its volume?

Word Problems 1

Caren is making chocolate bars in the shape of a cube of side 5 cm. What is the volume of 25 such bars?

In a competition, a school decides to award prizes for three values: punctuality, respect and discipline. The number of students getting prizes is in the ratio 2:3:5. If the product of the numbers is 2,05,770, find the number of students getting the prize for each value.

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12 Percentages Let's Recall Mr. Martin’s company has 5 different departments where employees work. The graph below shows the distribution of employees across the various departments.

Sales department

17%

keti

31%

11%

Sale

17%

14%

3 W hich department has the highest number of people working in it?

Fina n

Admin and Marketing

Adm in

2 I n which two departments is the number of employees the same?

Scale: 1 division = 5%

50 45 40 35 30 25 20 15 10 5

Acc oun ts

Percentage People PercentageofofPeople

11%

Mar

1 What percentage of the people work in the finance department?

Department

Letʼs Warm-up

Read the graph below which depicts the survey result of favourite colours and fill in the blanks. 2 The most favoured colour is _____________. 3 T he percentage of people who prefer yellow is _____________. 4 The percentage of people who like green is _____________. 5 T he number of people who prefer blue over red are _____________.

Number of Students

1 The least preferred colour is _____________. 100

Number of students

Percentage

80 60 40 20 19 10%

Red

35 17%

Yellow

64 31%

Green

87 42%

Blue

Colour

I scored _________ out of 5.

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Real Life Applications on Percentages Real Life Connect

In the Indian parliament, there are 542 members, of which 78 are females and the rest are males.

Ratio and Percentages

What is the ratio of females to total members of parliament? A ratio means comparing two quantities and this comparison can be done using fractions or percentages. Ratio of females to total members = 78:542 or 39:271 What percentage of the total members are females? A percentage is a fraction with its denominator as ‘100’. Thus, percentage means per 100 and is denoted by %. In other words, percent means how many hundredths. 39 × 100� ꞊ 14106� or 14.39% 271 271 So, 14.39% of the parliament are females.

Think and Tell

What is 100% of any number?

Example 1

Remember! A fraction can be converted into a percentage by multiplying it by 100.

1 Convert 0.25 into a percentage. 0.25 ꞊ 25 100 ꞊ 25 × 100 � ꞊ 25� 100

3 Convert 31% into a fraction. 31% ꞊ 13 ꞊ 13 × 1 4 4 4 100 4 ꞊ 13 400

2 Express 5% as a ratio. 5% ꞊ 5 ꞊ 1 ꞊ 1:20 100 20 4 C onvert 15.3% into a decimal. 15.3% ꞊ 15.3 100 ꞊ 0.153

Error Alert! We always shift the decimal point two places to the right when we multiply by 100 and two places to the left when we divide by 100. 11.23 × 100 = 0.1123 11.23 ÷ 100 = 1123 Example 2

11.23 × 100 = 1123

11.23 ÷ 100 = 0.1123

1 If 40% of a number is 120, find the number. Let the required number be x. Then, 40� of x ꞊ 120 40 × x ꞊ 120 100 x ꞊ 100 × 120 40 x ꞊ 300

Remember! To convert a percentage value into a fraction or a decimal, divide by 100, remove the percentage symbol and then simplify if required.

2 A rya has spent 45% of her salary and she has ₹5500 remaining. Find her salary. Let Arya’s salary be x.

Arya spent 45% of x ꞊ 45 × x ꞊ 9 x 100 20 9 11 Money Arya has left ꞊ x − x ꞊ x 20 20 Money Arya has left = ₹5500 Therefore, 11x ꞊ ₹5500 20 x ꞊ 5500 × 20 = ₹10,000 11 Thus, Arya’s salary is ₹10,000.

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Do It Together

In a school, 35% of the students like playing football and the remaining students like basketball. If 1300 students like basketball, find the number of students in the school. Let the total number of students be x. Percentage of students who like football = 35% Percentage of students who like basketball = (100% − 35�) = 65% Number of students who like basketball = ___________

Think and Tell If you score 60 out of 80 in your maths test, what percentage of marks did you score?

Total number of students = ___________

Do It Yourself 12A 1 Convert the decimals into percentages. a 0.56

b 0.2

c 0.14

d 0.5

e 0.42

c 120.5%

d 675%

e 0.5%

c 62.5%

d 11.5%

e 120%

c 65.5%

d 152%

e 0.55%

2 Convert each of the percentages into a decimal. a 13.5%

b 45%

3 Express each of the percentages as a fraction. a 0.4%

b 0.125%

4 Express each of the percentages as a ratio. a 210% b 21% 5 5 Answer the questions.

a What percentage of 30 is 60? c What percentage of 3 is

9 ? 5

b What percentage of 3 L is 900 mL? d What percentage of 5 kg is 1500 grams?

Word Problems 1 In a school, 35% of the students are boys and the remaining are girls. If the number of girls is 650, find the total number of students in the school.

2 Samaira scored 460 marks out of 500 in her final exam, whereas Mysha scored 370 marks out of 450. Who scored better?

3 If 45% of a number is 135, find the number. 4 Sarah drinks a 500 mL smoothie every day. If the smoothie is made of 80% fruit and 20% yogurt, how many millilitres of yogurt are in her daily smoothie?

5 If 65% of the students in a class passed the maths test and there are 40 students in total, how many students passed the test?

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6 In a town with a population of 50,000, the number of registered voters is 35% of the

total population. Of the registered voters, 60% voted in the recent election. Of those who voted, 40% were aged between 18 and 30 years old and the rest were older.

Calculate:

a The total number of registered voters.

b The number of people who voted in the recent election. c The number of voters between 18 and 30 years old.

7 In an examination, a student has to score 30% marks to pass. A student gets 45 marks

and fails by 15 marks. Find the highest number of marks that a student can score in the examination.

8 A man left 20% of his money for his mother, 30% for his sister, 25% for his wife and the remaining ₹4500 for his brother. How much money did he leave?

9 A company manufactures smartphones. During quality testing, it is found that 15% of the smartphones are defective. Out of the defective phones, 40% can be repaired and sold

at a lower price, whereas the remaining 60% are beyond repair and are scrapped. If the company manufactures 10,000 smartphones. Then find out: a How many smartphones are defective. b How many defective smartphones can be repaired. c How many defective smartphones cannot be repaired and are scrapped.

Percentage Change Remember that there were 78 women in the Indian parliament. If there were 117 women in the next parliament session, what would be the percentage increase in the number of women? Percentage change = Amount of increase/decrease × 100� Original amount 117 − 78 39 × 100� ꞊ × 100� ꞊ 50� 78 78 Thus, the number of females in the next parliament session would increase by 50%. Percentage increase ꞊

Error Alert! Always ensure that you use the original value in the denominator when calculating the percentage change. � change ꞊ 39 × 100� 117

� change ꞊

39 × 100� 78

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Example 3

The price of an article is increased by 20%. Find the original price of the article, if the increased price is ₹3000. Let the original price of the article be ₹x.

Increase in the price = 20% of ₹x = ₹ 20 × x = ₹ x 100 5 Increased price = ₹x + ₹ x ꞊ ₹6x 5 5 Increased price = ₹3000 ₹6x = 3000. So, x = ₹2500 5

Example 4

The price of sugar goes up by 10%. By what percentage must Vijay reduced his consumption of sugar so that his expenditure on sugar remains the same? Let the consumption of sugar originally be x kg and let its cost be ₹100. New cost of x kg of sugar = ₹110 Now, ₹110 can buy x kg of sugar.

₹100 can buy x × 100 kg of sugar ꞊ 10x kg of sugar 110 11 Reduction in consumption = x − 10x = x kg of sugar 11 11

(

)

Think and Tell

Reduction in percentage of consumption x Reduction in consumption 11 = × 100% = × 100� = 9 1 � Original consumption x 11

What will happen in the above question if the price goes down by 10%?

Hence Vijay must reduce his consumption by 9 1 � 11 Do It Together

The length of a piece of rope is reduced from 150 cm to 125 cm. Find the percentage decrease. Original length of the rope = ____ cm Final length after reduction = _____ cm Reduction in length = ______ cm

Reduction percentage =

Reduction in length × 100� ꞊ ______ � Original length

Do It Yourself 12B 1 Find the percentage increase: a When 20 is increased to 24.

b When 30 is increased to 40.

d When 100 is increased to 110.

e When 5 kg is increased to 15 kg.

c When 50 is increased to 90.

2 Find the percentage decrease: a When 30 is reduced to 15.

b When 45 is reduced to 20.

d When 150 is reduced to 100.

e When 200 is reduced to 180.

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c When 98 is reduced to 88.

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3 Find the number which is: a 5% more than 80.

b 45% more than 180.

c 34% less than 340.

d 25% less than 190.

e 9% less than 168.

f 20% more than 80.

a Increase ₹190 by 20%.

b Decrease 150 kg by 30%.

c Increase ₹1500 by 60%.

d Decrease 1000 L by 12.5%.

e Increase 750 grams by 0.5%.

4 Solve the given questions.

5 A number is increased by 30% and then reduced by 20%. Find the net increase or decrease in the number. 6 The price of olive oil is increased by 10%. If the original price was ₹250 per litre, find the new price.

Word Problems 1 Rajat scored 400 marks out of 500 in the half-yearly exams. He scored 430 marks out of 500 in his final exams. What was the percentage increase in the marks scored by Rajat?

2 A factory had 1200 female workers in the year 2021. If the number of female workers

increased by 12% in 2022, find the total number of female workers working in the factory in 2022.

3 The population of a city increases by 5% every year. If the current population of the city is 1,28,000, what would be the population of the city after 3 years?

4 The value of a car depreciates every year by 10%. Find its value after 2 years, if its present value is ₹3,50,000.

Points to Remember • Percentage means per hundred or out of hundred. The symbol “%” stands for percent, i.e., 1 . 100 • Any fraction can be converted into a percentage by multiplying it by 100%. • To convert any decimal into a percentage, move the decimal two places to the right and put the % symbol. • Percentage increase/decrease in quantity ꞊ Change in quantity × 100�. Original quantity • Percentage increase/decrease is the same as percentage change.

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Math Lab Budget for a Vacation Aim: To understand the concept of percentage and apply its use in budgeting. Materials Required: Paper and pencils. Setting: In groups of 3. Process: 1 Provide a hypothetical budget of ₹10,000. xplain that they need to allocate percentages of their budget to different categories 2 E such as transportation, accommodation, food, activities and souvenirs. tudents allocate their funds based on their research and find realistic costs for each 3 S category based on their destination. tudents will use percentages to break their budget down and represent the data in a pie 4 S chart.

Chapter Checkup 1 Convert the ratios into percentages. a 5:2

b 3:9

c 15:120

2 Express each of the fractions as a percentage. a 4 b 34 c 22 25 5 3

d 11:70

3 5

e 80:125

14 164

3 Convert each of the decimals into a percentage. a 0.003

b 1.35

4 Express each of the percentages as a ratio. a 46� b 162� 3

c 2.789

d 16.6

e 0.567

c 31�

d 140�

e 950�

c 652�

d 332�

e 152�

d 12�

e 41�

5 Express each of the percentages as a fraction. a 0.008�

b 2.46�

6 Convert each of the percentages into a decimal. a 1.8�

b 0.06�

c 5.8�

7 Answer the questions. a What percentage of 60 is 54? c What percentage of 70 is 8.5?

b What percentage of 125 is 25? d What percentage of 6 is 5?

8 Find the value. a

5� of ₹500 4

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1� of 1600 L 8

25� of 120 cm

15.5� of 310 kg

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9 Find the percentage change. a When 215 is reduced to 210

b When 45 is increased to 86

c When 450 is increased to 530

d When 110 is increased to 250

10 600 children are having lunch at a school picnic. Out of these, 280 choose to have a vegan meal, 120 choose to have a non-vegetarian meal and the remaining children prefer a vegetarian meal. a What percentage of the children choose a vegetarian meal? b What fraction of the children prefer a vegan meal? c What percentage of the children choose a non-vegetarian meal?

Word Problems 1 Last year Mrs. Sheela weighed 75 kg. This year she weighs 60 kg. Find the percentage change in her weight.

2 A bag consist of 450 balls. Out of these, 16% are green, 24% are blue, 10% are white, 8% are red and the remaining balls are yellow. Find the number of balls of each colour.

3 If 65% of the students in a school are boys and the number of girls is 700, find the total number of students in the school.

4 The value of a car depreciates every year by 30%. Find its value after 2 years, if its present value is ₹9,00,000.

5 An engineering student has to secure 35% marks to pass. He gets 70 and fails by 35 marks. Find the maximum number of marks.

3 1 4 4 the mixture that contains 8.7 kg of sand. Also, determine the weight of the mixture that will

6 A mixture consists of 30 % sand, 45 % gravel, and the rest is cement. Find the weight of contain 3.5 kg of cement.

7 The price of diesel increases by 60%. By what percentage should Mithun reduce his consumption so that there is no change in his expenditure?

8 In a school election, there are two candidates running for the position of student body

president: Sarah and Alex. There are 600 students in the school who are eligible to vote.

On the election day, 70% of the eligible students cast their votes. 60% of the eligible votes were received by Sarah and rest were received by Alex. a How many students voted in the election? b What percentage of the total number of students voted for Alex? c How many votes did Sarah get in total?

9 A reduction of 25% in the price of apples enables a person to buy 2 kg more apples for ₹300. Find:

a The original price per kg of apples.

b The reduced price per kg of apples

10 A man spends 20% of his salary on house rent. After spending 40% of the remainder, he is left with ₹9600. Find his monthly salary.

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Profit/Loss, Discount and Tax

13 Let's Recall

The same watch is on sale at three different stores. If the cost price of the watch is ₹3000, which store makes the most profit?

00 ₹34

00 ₹32

00 ₹35

Store A

Store B

Store C

We know that a profit is made if SP > CP. Profit of Store A = ₹3500 − ₹3000 = ₹500

Profit of Store B = ₹3200 − ₹3000 = ₹200

Profit of Store C = ₹3400 − ₹3000 = ₹400

As 500 > 400 > 200, the most profit is made by store A.

A washing machine is on sale at two stores. If the cost price of the machine is ₹45,500, what is the profit or loss incurred by the stores?

₹50,000

₹42,500

Store A

Store B

We know that a profit is made if SP > CP and a loss is made if CP > SP. Profit of Store A = ₹50,000 − ₹45,500 = ₹4500

Letʼs Warm-up

Loss of Store B = ₹45,500 − ₹42,500 = ₹3000

Fill in the blanks. Home Décor Store purchases a bed for ₹25,000 and sells it for ₹27,000. The same store purchases a mattress for ₹10,000 and sells it for ₹9000. 1 The cost price of the bed is ______. 2 The selling price of the bed is ______. 3 The selling price of the mattress is ______. 4 The profit earned on the sale of the bed is ______. 5 The loss incurred on the sale of the mattress is ______. I scored _________ out of 5.

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Application of Percentages Real Life Connect

Sam went to purchase a new television for his home at the nearby electronics shop. Sam asked for a 32-inch television. The purchase price of the television was ₹95,000. The shopkeeper usually sells the television for ₹1,05,000, but due to the festive season, he sold the television for ₹94,000.

Profit and Loss We know that when the cost price > selling price, a loss is incurred and when the selling price > cost price, a profit is made. Let us find the profit or loss incurred for the television sold at the usual price and festive price. Selling Price Price at which article is sold

₹1,05,000 − ₹95,000 ꞊ ₹10,000 If SP > CP; Profit = SP − CP

Cost Price Price at which article is purchased

₹95,000 − ₹94,000 ꞊ ₹1000

If CP > SP; Loss = CP − SP

Hence the shopkeeper would have earned a profit of ₹10,000 if he sold the television at ₹1,05,000, but he incurred a loss of ₹1000 by selling the television at ₹94,000.

Finding Profit or Loss Percentage The percentage of profit or loss is the amount of profit or loss represented as a percentage of the total. The percentage of profit and loss of the electronics shopkeeper can be given as: Profit percentage ꞊ Profit × 100% CP ꞊ 10,000 × 100 ꞊ 10.5% 95,000

Example 1

Loss percentage ꞊ Loss × 100% CP ꞊ 1000 × 100 ꞊ 1.05% 95,000

A shopkeeper buys a bicycle for ₹1200 and sells it for ₹1500. Find his profit or loss percentage. Cost price of the bicycle = ₹1200 Selling price of the bicycle = ₹1500 Since SP > CP, there is a profit. Profit = SP – CP = 1500 – 1200 = ₹300 Profit 300 × 100% = × 100% = 25% CP 1200 Therefore, the shopkeeper makes a profit of 25%. Profit percentage =

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Example 2

Ramesh purchases a T.V. for ₹5000 and pays ₹250 for its transportation. If he sells the T.V. for ₹5075, find his percentage of profit or loss. Price at which the T.V. is bought = ₹5000 Overheads in the form of transportation = ₹250 Therefore, the total cost price of the T.V. = (5000 + 250) = ₹5250 Selling price of the T.V. = ₹5075 Since SP < CP, there is a loss. Loss = CP – SP = 5250 – 5075 = ₹175

Therefore, percentage loss = Loss × 100% = 175 × 100% ꞊ 10 % = 3.33% CP 5250 3 Therefore, Ramesh incurred a loss of 3.33%. Example 3

By selling 240 pens, Kumar lost an amount equal to the CP of 30 pens. Find his loss percentage. Let us assume that the cost price of each pen is ₹1. CP of 240 pens = ₹240 Loss = CP of 30 pens = 30 × 1 = ₹30

Loss percentage = Loss × 100% = 30 × 100% = 12.5% CP 240 Therefore, Kumar’s loss percentage is 12.5%. Do It Together

If the cost price of 10 greeting cards is equal to the selling price of 8 greeting cards, find the gain or loss percent. Let the cost price of each card be ₹x. Then, CP of 8 cards = ₹8x SP of 8 cards = CP of 10 cards = ₹10x Thus, CP = ₹8x and SP = ₹10x Since SP > CP there is a profit. Profit = ₹(10x − 8x) = _____

Profit% ꞊ Profit × 100% = _______ = ______ CP Therefore, the percent profit is _____.

Finding Selling Price or Cost Price We have already studied the meaning of the terms Selling Price and Cost Price. We will now look at some of the formulas associated with these two terms. When CP and Profit/Loss % are given SP ꞊ 100 + Profit� × CP 100 or SP ꞊ 100 - Loss� × CP 100

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When SP and Profit/Loss % are given CP ꞊ CP ꞊

100 × SP 100 + Profit� or 100 × SP 100 − Loss�

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Example 4

A shopkeeper purchased an old washing machine for ₹2500 and spent ₹500 on transportation and repairs. He then sold the same machine at a profit of 25%. Find the selling price of the machine. Price at which washing machine was bought = ₹2500 Overheads in the form of transportation and repairs = ₹500 Therefore, the total cost price of the washing machine = (2500 + 500) = ₹3000 Profit percentage = 25%

Selling Price = 100 + Profit� × CP 100 = 100 + 25 × 3000 = 125 × 3000 = 3750 100 100 Hence, the selling price of the washing machine is ₹3750. Example 5

On selling a fan for ₹810, Sachin makes a profit of 8%. For how much did he purchase it? SP of the fan = ₹810 Profit% = 8%

100 × SP 100 + Profit� 100 = × 810 100 + 8 = 100 × 810 108 = ₹750

CP of the fan =

Hence, Sachin purchased the fan for ₹750. Example 6

On selling a table for ₹1974, Rakesh loses 6%. For how much did he purchase it? SP of the table = ₹1974, loss% = 6%

100 × SP 100 − Loss� = 100 × 1974 100 − 6 = 100 × 1974 94 = ₹2100

CP of the table =

Hence, Rakesh purchased the table for ₹2100. Do It Together

By selling a T-shirt for ₹648, a shopkeeper loses 4%. For how much should he sell it to gain 4%? SP of the T-shirt = ₹648; Loss% = 4% CP of the T-shirt =

100 × SP 100 − Loss�

= ________ = ______ = ______ Now, CP = ______, desired profit% = 4% Desired SP = 100 + Profit� × CP = ________ = ______ = ______ 100 Hence, the shopkeeper should sell the T-shirt for ₹_____ to gain 4%.

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Do It Yourself 13A 1 A Home Décor Store purchases a dining table for ₹40,000. What will be the profit earned or loss incurred if it is sold at the prices given? a

₹35,000

₹42,000

₹38,000

₹45,000

2 Fill in the blanks. a CP = ______; SP = ₹572; Profit = ₹72

CP = ₹7282; SP = ______; Profit = ₹208

c CP= ₹9684; SP = ______; Loss = ₹684

CP = ______; SP = ₹1973; Profit = ₹273

e CP = ₹6,76,000; SP = ______; Loss = ₹18,500

CP = ______; SP = ₹7894; Loss = ₹306

3 A retailer bought a mixer from a wholesale dealer for ₹4500 and sold it for ₹6000. Find his profit or loss percent. 4 A shopkeeper purchased an article for ₹1,25,000 and sold it at a profit of ₹5000. Find the profit percentage. 5

A bookshop owner purchased 100 notebooks for ₹15 each. However, 5 notebooks had to be thrown away as

a few pages had been torn from them. The remaining notebooks were sold at ₹18 each. Find the gain or loss percent.

6 By selling a motorcycle for ₹56,320, the shopkeeper loses 12%. Find the price he had paid to buy the bike. 7 A fruit vendor buys oranges at the rate of 5 for ₹40 and sells them at ₹9 per orange. Find his profit percent.

Word Problems 1 Akash buys a car for ₹4,46,000 and sells it for ₹5,20,000. What is his profit? 2 Manish had to sell a saree for ₹1700, for which he had paid ₹2000 when he purchased it. What is the loss?

3 Rahul purchased a bluetooth speaker from an online website at ₹14,500 and sold it to his friend for ₹14,000. Find his profit or loss percentage.

4 By selling a bicycle for ₹819, Vinit loses 9%. At what price should he sell it to make a profit of 5%?

5 Rakesh sells two watches for ₹1200 each. On one watch, he gains 20% and on the other he loses 20%. What is the cost price of each watch, and what is his total profit or loss percentage?

6 Madhuri sold her gold necklace at a profit of 7%. If she had sold it for ₹2940 more, she would have got a 10% profit. Find the cost price.

7 A coat was sold by a shopkeeper at a profit of 5%. If it had been sold for ₹1650 less, he would have incurred a loss of 5%. Find the cost price.

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Discount It is the festive season of Diwali and hence the owner of the electronics shop is offering a 10% discount on other appliances as well. The price printed on a music system was ₹40,000. Sam thought of purchasing the music system as well. He wonders how much the music system will cost him. Let us find out.

Finding Discount Reduction in the usual price of something. Discount = Marked Price – Selling Price

The price at which the article is sold. or

Price on the label of an article.

Marked Price × Discount � 100 The discount amount expressed as a percentage of the original price.

The discount received by Sam on the music system can be given as: ₹40,000 × 10 ꞊ ₹4000 100 The cost of the music system for Sam = Price of the article after discount = ₹40,000 − ₹4000 ꞊ ₹36,000 If the shopkeeper had made a profit of ₹1250 after offering the discount, then what was the cost price of the music system? Let us find out! If the marked price, profit/ loss and discount are given, then the cost price of the article can be found as: CP = MP – Discount − Profit

CP = MP – Discount + Loss

Hence, the cost price of the music system = ₹40,000 − ₹4000 − ₹1250 ꞊ ₹34,750 Example 7

A book with a marked price of ₹600 is available at a discount of 18%. Find the discount offered and the price at which the book is available for sale. Marked price of the book = ₹600; Discount given = 18% Discount = MP × Discount Percentage ꞊ 600 × 18 ꞊ ₹108 100 100 Therefore, the discount given on the book = ₹108 We know that, SP = MP – Discount SP = ₹ (600 − 108) = ₹492

Remember! An 18% discount is the same as paying 82% of the original price.

Therefore, the book is available for ₹492. Example 8

A shopkeeper sold an article for ₹1326 after allowing a discount of 15% on its marked price. Find the marked price of the article. Let the marked price of the article be ₹x. Discount allowed = 15%; Selling price of the article = ₹1326 We know that SP ꞊ MP × 100 − Discount Percentage 100 1326 ꞊ x × 100 − 15 ⇒ 1326 ꞊ x × 85 100 100 1326 × 100 x꞊ = ₹1560 85 Therefore, the marked price of the article was ₹1560.

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Example 9

The marked price of a radio is ₹12,000. Find the discount percentage allowed on the radio if it is sold for ₹10,500. Marked price of the radio = ₹12,000; Selling price of the radio = ₹10,500 Discount = MP – SP = ₹ (12,000 – 10,500) = ₹1500

Therefore, discount percentage = Discount × 100� = 1500 × 100� = 12.5� MP 12,000 Do It Together

As part of a Diwali offer, a jeweller allows a discount of 15%. Even after giving the discount, he makes a profit of 6.25%. Amit buys a gold chain which was marked at ₹50,000. Find the cost price of this chain for the jeweller. Marked price of the chain = ₹50,000; Discount allowed = 15% SP ꞊ MP × 100 − Discount Percentage 100 SP = 50,000 × 100 − 15 100

Therefore, the selling price of the chain = ₹_____. Also, given that, the jeweller makes a profit of 6.25%. CP ꞊ SP ×

100 100 + Profit Percentage

CP ꞊ ₹______

Therefore, the cost price of the chain = ₹_____.

Finding Successive Discounts When two discounts are given one after the other, such discounts are called successive discounts. Here the first discount is given on the marked price and the second discount is given on the reduced price. For example, a shopkeeper offers a 10% discount on T-shirts and a further discount of 7% to increase the sales. Here, 10% and 7% are successive discounts. If the marked price of a T-shirt is ₹2400, the selling price can be found as given below. Marked price of the T-shirt = ₹2400

First discount ꞊ 10� on MP ꞊ 10� of 2400 ꞊ 2400 × 10 ꞊ ₹240 100 Selling price after the first discount = 2400 – 240 = ₹2160

Second discount = 7% of reduced price = 7% of ₹2160 = 7 × 2160 = ₹151.2 100 Selling price after the second discount = 2160 – 151.2 = ₹2008.8 Therefore, the selling price of the T-shirt after deducting two successive discounts is ₹2008.8. Example 10

A shopkeeper was selling all his items at a 20% discount. During the festive season, he offered a 25% discount over and above the existing discount. If Ramesh bought a shirt which was marked ₹1200, how much did he pay for it? Marked price of the shirt = ₹1200

20 = ₹240 100 Selling price after the first discount = 1200 – 240 = ₹960 First discount = 20% on MP = 20% of 1200 = 1200 ×

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Second discount = 25% of reduced price = 25% of ₹960 =

25 × 960 = ₹240 100

Selling price after the second discount = 960 – 240 = ₹720 Therefore, Ramesh paid ₹720 for the shirt. Example 11

A television set was sold for ₹57,600 after giving successive discounts of 10% and 20% respectively. What was the marked price? Let the marked price of the television set be ₹100. First discount = 10% of MP = 10% of 100 = ₹10 So, selling price after the first discount = ₹100 − ₹10 = ₹90 Second discount = 20% of reduced price = 20% of ₹90 = 90 × Selling price after the second discount = 90 – 18 = ₹72 For the selling price to be ₹72, the MP must be ₹100. For the selling price to be ₹57,600, the MP must be =

20 = ₹18 100

57,600 × 100 = ₹80,000. 72

Therefore, the marked price of the television set was ₹80,000. Example 12

Find the single discount percentage equivalent to two successive discounts of 20% and 10%. Let the marked price of an article be ₹100. Then, first discount on it = 20% of ₹100 = 100 ×

20 = ₹20 100

Selling price after the first discount = ₹ (100 − 20) = ₹80

Second discount on the reduced price = 10% of ₹80 = 80 ×

10 = ₹8 100

Selling price after the second discount = ₹80 − ₹8 = ₹72; Net selling price = ₹72 Single discount equivalent to given successive discounts = (100 − 72)% = 28% Do It Together

Which is a better offer: two successive discounts of 10% and 8% or a single discount of 18%? Let the marked price be ₹100. First discount = 10% of ₹100 = ₹10 Selling price after the first discount = ₹100 − ₹10 = ₹90 Second discount = 8% of ₹90 = 90 ×

8 = ₹7.2 100

Selling price after the second discount = Total discount allowed =

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

is a better offer.

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Do It Yourself 13B 1 The marked price of a radio is ₹4000. Find the discount and the selling price if the rate of discount is as given a 25%

b 20%

1 c 12 % 2

d 10%

2 The marked price of a washing machine is ₹14,500 and its selling price is ₹13,775. Find the discount allowed and the percentage discount.

3 The selling price of a dining table is ₹16,000. Find the marked price if a 20% discount is allowed. 4 If ₹175 is the discount offered on an article whose marked price is ₹900, then find its selling price. 5 A pair of trousers was sold for ₹7500 after a discount of ₹500 was offered on it. What was the percentage of the discount?

6 An article is sold after offering successive discounts of ₹78 and ₹24. If its marked price is ₹702, then find the selling price of the article.

7 Two successive discounts of 25% and 12% on an article are equivalent to a single discount of x %. What is x? 8 An article is sold for ₹51,000 after a discount of 15% was offered on it. Find the marked price of the article. 9 A coat is sold for ₹7920 after offering successive discounts of 12% and 10%. What is the marked price of the coat? 10 A profit of 12% can be earned by selling an article after offering a discount of ₹200. If the cost price of the article is ₹850, then find its marked price.

11 An article is marked at ₹9000. After giving a discount of 20%, it is sold at a profit of 20%. Find the cost price of the article.

Word Problems 1 The marked price of an article is ₹17,800. The shopkeeper allows a discount of 25%. What is the selling price of the article?

2 The cost price of a dining table is ₹15,000, and its marked price is ₹18,000. If a shopkeeper sells it at a loss of 8%, then what is the rate of discount offered by him?

3 The marked price of an article is 32% above its cost price. What is the rate of discount a shop owner can offer so that he gains 10%?

4 A shopkeeper allows a discount of 15% on all the goods purchased from his shop. On

request, he further allows a discount of 10% on the new price of the goods. What is the overall rate of discount given to the customer?

Chapter 13 • Profit/Loss, Discount and Tax

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Tax Sam purchases a TV for ₹94,000 and a music system for ₹36,000 from an electronics store. He asks the owner for the bill. The owner hands over a bill as shown. Bill No. 1254 S.No.

Item

Quantity

Music System

Date: 07/10/2023

Rate

Amount

94,000

₹94,000

36,000

₹36,000

Subtotal

₹1,30,000

Total

₹1,53,400

Tax (18%)

₹23,400

Sam has some questions in his mind as he looks at the bill. 1 Do we need to pay tax on the purchase of electronic items? 2 Is tax charged on the marked price or selling price of the items? 3 Is the tax charged by the electronics store in his bill correct? We will look for answers to these questions by looking at the tax structure in India.

Tax is the money which is collected by the government from the citizens of the country to provide them with the best facilities and infrastructure.

The prevailing tax structure in India before the introduction of Goods and Services Tax on July 1, 2017, is as shown below. Tax Structure in India before GST

Direct Tax

Income Tax

Wealth Tax

Indirect Tax

Central Excise

Customs Duty

VAT/ Sales Tax

Service Tax

However, for the purpose of this chapter we are going to focus only on the Sales Tax and Value Added Tax (VAT). Sales Tax Sales tax is charged by the government on the sale of different commodities. It is collected by the seller from the customer and given to the government. Therefore, this tax is always calculated on the selling price of an article and is added to the value of the bill.

Error Alert! We always add the percentage value of sales tax. Selling Price = ₹1000, Sales Tax = 5%

1000 –

100 = 1000 – 50 = ₹950

× 1000

1000 +

100

× 1000

= 1000 + 50 = ₹1050

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Value Added Tax Value Added Tax is the tax that is charged on goods and services at each stage of a supply chain. In the case of sales tax, a hefty amount of tax is paid by the customer alone whereas, in the case of VAT, everyone in the supply chain (manufacturer, wholesaler, dealer, retailer, customer) pays a small amount of tax adding up to the same big amount.

Value Added Tax Manufacturer

Value Added Tax Wholesaler

Value Added Tax Retailer

Consumer

Goods and Services Tax (GST) Goods and Services Tax (GST) is an indirect tax, which has replaced many indirect taxes like VAT, Central Excise duty, Sales Tax, Service Tax etc. It is described as one tax for one nation. The new tax structure in India after the introduction of the Goods and Services Tax on July 1, 2017, is as shown below. Tax Structure in India after GST

Direct Tax

Income Tax

Indirect Tax

Wealth Tax

Goods and Services Tax (GST)

Customs Duty

GST on electronic items in India is 18%. Thus, electronic items are indeed taxable in India as per the latest tax laws. Also, tax is always calculated on the selling price of an article and is added to the value of the bill. Thus, the electronic store that Sam visited correctly included the tax amount in the bill. Example 13

Amit bought a Bluetooth speaker marked at ₹9600. He paid a sales tax of 5% for it. What is the sales tax and the bill amount to be paid? Value of the bluetooth speaker = ₹9600 Sales tax rate = 5%

5 = ₹480 100 Bill amount = ₹9600 + ₹480 = ₹10,080 Sales tax = 5% of ₹9600 = 9600 ×

Example 14

Lakhan bought a suitcase for ₹2520 including VAT. The VAT for this item is 5%. What was the price of the suitcase before VAT was added? Also state how much the VAT is. The cost price paid by Lakhan = ₹2520 If the original value was ₹100, Lakhan spends ₹105.

Did You Know?

For every ₹105 that Lakhan spends, the original cost = ₹100

France was the first country

So, for ₹2520 which Lakhan spends, the original cost 2520 = 100 × = ₹2400 105

to implement the GST in 1954; since then, an estimated 140 countries have adopted this tax system in some form or another.

VAT = ₹2520 − ₹2400 = ₹120

Chapter 13 • Profit/Loss, Discount and Tax

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Navin bought a watch for ₹5310 that included a GST of 18%. Find the amount of GST and the actual price of the watch before GST was added.

Example 15

Let the actual price of the watch be ₹x; Amount paid by Navin = ₹5310; GST = 18% According to the question, x+

18 100x + 18x × x = 5310 ⇒ = 5310 100 100

118 x 5310 × 100 = 5310; x = ; x = ₹4500 100 118 So, the amount of GST imposed on the watch is ₹ (5310 − 4500) = ₹810 Thus, the actual price of the watch is ₹4500, and the amount of GST is ₹810. Do It Together

Renuka buys a cooker marked at ₹6500. She gets a discount of 8%. If the sales tax is 8%, find the amount she pays for the cooker. Marked price = ₹6500 Discount = 8% of marked price = _____ = _____ Selling price = Marked Price – Discount = ₹(6500 – _____) = ₹_____ Sales tax = 8% of Selling price = ________________________ Therefore, Renuka pays ₹_____________________________

Do It Yourself 13C 1 Akash bought a computer for ₹38,000 and a printer for ₹8000. If the rate of sales tax is 7% for these items, find the price he must pay to buy these two items.

2 Ajay sells a harmonium for ₹17,280 including the VAT. What is the selling price he will get after depositing the tax to the government at the rate of 8%?

3 The sales tax on a refrigerator is 9%. If the tax amount is ₹1170, find the actual selling price. 4 Amit got a discount of 10% on the suit he bought. The marked price was ₹5000 for the suit. If he had to pay a sales tax of 10% on the price at which he bought it, how much did he pay?

5 The selling price including VAT on a cooking range is ₹19,610. If the VAT is 6%, what is the original price of the cooking range?

6 A soap manufacturer can sell soap at ₹6.36 per piece after adding the VAT at the rate of 6%. Still, he will make a profit of 20% from the sale. What is his manufacturing cost per soap?

7 Manohar wants to buy a vacuum cleaner that costs ₹8000. What amount should he give to the shopkeeper if the GST on the vacuum cleaner is 18%?

Word Problem 1

Renu bought a mobile phone for ₹16,240 that included GST of 12%. Find the price of the mobile phone before the GST was added.

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Points to Remember •

The percentage of profit or loss is always calculated on the cost price.

•

The price printed on an article or written on a label attached to it is called its Marked Price.

• The deduction allowed from the marked price of an article is called a discount. A discount is always calculated on the marked price. • The price of an article after deducting discount from the marked price is called the selling price. It is the price of the article to be paid by the customer.

• If two or more discounts are allowed one after the other, then such discounts are known as successive discounts. •

There are two types of tax: direct tax and indirect tax.

•

GST stands for Goods and Services Tax and is a type of indirect tax.

Math Lab Comparing Discounts! Setting: In groups of 3 Materials Required: Pen, notebook Method: 1 Draw the table on the board. all out the names of the items available for sale across three different stores: Store 1, 2 C Store 2, and Store 3. 3 Instruct students that Store 1 is offering a 40% discount on all items. 4 Instruct students that Store 2 is offering a 50% discount on all items. 5 Instruct students that Store 3 is offering a 30% discount on all items. 6 Let the students perform their calculations and choose their best buy. Which one is the best buy? Store 1

Store 2

Store 3

A pair of sneakers

Regular price: ₹4500

Regular price: ₹4900

Regular price: ₹3900

Watch

Regular price: ₹5000

Regular price: ₹5500

Regular price: ₹4500

Sunglasses

Regular price: ₹6000

Regular price: ₹6500

Regular price: ₹5500

Discount Offered

40%

50%

30%

Chapter 13 • Profit/Loss, Discount and Tax

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Chapter Checkup 1

Find the missing terms. CP a

₹8064

₹864

Profit %

₹476 ₹8250

₹750

Loss

₹460

₹437

Find the missing terms. a

₹468

₹6400

Profit

₹6800

Loss %

₹52 ₹192

Find the unknown in each case. a MP = ₹625, SP = ₹525, Discount =?

b MP = ₹1780, Discount = 15%, SP =?

The cost price of 15 pencils is the selling price of 10 pencils. What is the percentage of profit?

The marked price of a television is 25% more than the cost price. It is sold at a discount of 10%. If the marked price is ₹12,000, then find the cost price, selling price, and the profit. What will be the percentage of profit?

Find the single discount percentage which is equivalent to two successive discounts of 20% and 5%.

Find the bill amount of a refrigerator which costs ₹27,500 and GST is charged at 18%.

A man bought notebooks at ₹72 per dozen and sold them at ₹25 for a packet of 5 books. What was his profit or

loss per notebook?

A businessman buys 200 litres of vegetable oil at ₹60 per litre. He spends ₹2000 on packaging and ₹6000 on

transport. He then sells 1-litre packets for ₹120 each. Does he make a profit? What is the percentage of profit or loss?

10 Amit buys a television set whose marked price was ₹45,000. He then gets a discount of 15% and pays a sales tax of 12%. How much does he pay for the television set?

Word Problems 1 A man sells a sofa for ₹3210 making a profit of 7%. What would have been his profit per cent if he had sold it for ₹3360?

2 Ajay bought a second-hand car for ₹5,20,350. He polished the car for ₹2500, replaced old

parts for ₹15,000, and then resold the car. He managed to make a profit of 14%. How much did he sell the car for?

3 Renuka buys some winter clothes from a shop at a discount of 20%. If the marked price for a pair of gloves is ₹186, a jacket is ₹839, and a sweater is ₹345, how much did she pay in total after the discount?

4 Tulsi bought a watch for ₹1980 including VAT at 10%. Find the original price of the watch.

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14 Compound Interest Let's Recall Ajay wants to invest ₹10,000 for 10 years. Which bank should he choose? Community Bank

Society Bank

Open a Savings Account Today!

5% interest per year

6% interest per year

Ajay saw the advertisements for Community Bank and Society Bank. He chose Society Bank as it offers a greater rate of interest on deposits. Ajay wants to purchase a bike using a loan. Which bank should he choose? Community Bank

Society Bank

Apply for a loan!

9% interest per year

10% interest per year

Ajay saw the advertisements for Community Bank and Society Bank. He chose Community Bank as it charges a lower rate of interest on loans.

Letʼs Warm-up

Match the interest rate with the amount of interest. Column A

Column B

1 10% of ₹1000

₹140

2 5% of ₹4000

₹240

3 2% of ₹7000

₹360

4 4% of ₹6000

₹100

5 12% of ₹3000

₹200

I scored _________ out of 5.

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Mean, and Mode SimpleMedian Interest and Compound Interest Real Life Connect

Narendra wants to invest ₹8000 for three years.

Simple interest Compound interest

• Plan A earns 4% simple interest per year. Amount

• Plan B earns 4% interest compounded annually. Narendra wonders which plan he should choose.

Finding Simple Interest

Year

Look at the table below that shows the account balances for 3 years under Plan A. Plan A Year

Principal

4% Simple interest

Balance at the End of the Year

₹8000

₹320

₹8320

₹8000

₹320

₹8640

₹8000

₹320

₹8960

On analysing the table carefully, Narendra understands that the interest in Plan A will always be calculated only on the principal deposited by him. This type of interest is called simple interest (SI or I). Rate (R)

Principal (P) Money borrowed or deposited. Simple Interest (SI or I)

SI =

P×R×T 100

The extra amount paid/earned by the borrower/depositor.

Rate of interest on which the principal amount is given. Time (T)

Time period for which money is borrowed or deposited.

Remember! Simple interest is interest that is calculated only on the initial sum (the “principal”) borrowed or deposited.

Principal Money borrowed or deposited.

Amount The total sum which is paid at the end of the period.

A = P + SI

Simple Interest The extra amount paid/earned by the borrower/depositor.

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Example 1

A man borrowed ₹15,000 from a cooperative bank at the rate of 8% per annum for 4 years. Find the simple interest and the amount he has to pay back. Principal (P) = ₹15,000; Rate of interest (R) = 8% per annum; Time (T ) = 4 years Simple interest =

P × R × T 15,000 × 8 × 4 = = ₹4800 100 100

Amount = Principal + Simple Interest = ₹(15,000 + 4800) = ₹19,800 Example 2

Ankur borrowed a sum of ₹2000 from a bank and returned ₹2480 after 3 years. Find the simple interest and the rate of interest per annum. Amount = Principal + Interest ₹2480 = ₹2000 + Interest

Error Alert!

₹2480 − ₹2000 = Interest

Time is always considered in years while calculating interest amounts.

₹480 = Interest So, the simple interest is ₹480. Simple interest =

Finding the simple interest earned on ₹500 at 6% per annum for 18 months:

P×R×T 100

500 × 6 × 18 100

2000 × R × 3 ₹480 = 100 480 × 100 =R 2000 × 3

I = ₹540

R = 8%

P×R×T 100

500 × 6 × 18 100 × 12 I = ₹45

Thus, the rate of interest is 8%. Example 3

Sonali deposits a certain amount of money in a bank. If the interest rate of the bank decreases from 3 1 3 % to 3 % per annum, she receives ₹100 less in 2 years. Find the sum of money she deposited. 4 2 Rate of interest earlier = 3 Rate of interest later = 3

3 % 4

1 % 2

Difference in rate of interest = 3 Simple interest =

100 =

P×R×T 100

3 1 1 %–3 %= % 4 2 4

1 ×2 4 100

P×

P = 100 × 100 ×2 P = ₹20,000 Hence, Sonali deposited ₹20,000.

Chapter 14 • Compound Interest

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The graph shows an amount of ₹4000 deposited into a savings account for 10 years with simple interest. Balance in Savings Account (in ₹)

Do It Together

Scale: 1 division = ₹200

4600 4400 4200 4000 3800 3600 0

4 6 Years

1 How much interest has been earned, in total, over the last 10 years? The amount of interest earned will be the difference of the final value (the value after 10 years) and the initial amount invested. Interest = ₹_______________ 2 What is the rate of interest?

P×R×T . 100 We have just calculated that SI = ____ and we know that P = ____ and T =____. The formula for simple interest is SI =

Substituting these values into the formula, we get ____ = 4000 × R × 10 100 Rearranging and solving this, we get, R = ____. The interest rate is therefore, R = ____ %.

Do It Yourself 14A 1 Find the missing figures using the given data. Principal

Simple Interest

Amount

₹5780

______

₹6240

₹3520

₹250

______

₹850

______

₹972

₹9450

₹550

______

₹72

₹672

______

₹1233

₹7288

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2 Find the simple interest and the amount. Principal

Rate p.a.

Time

Simple Interest

Amount

₹800

3.5%

4 years

______

₹4750

2 years

______

₹160

10%

years

______

₹8500

8.5%

1 year

______

₹9600

3 months

______

1 2

3 Find the principal. Simple Interest

Time

Rate p.a.

Principal

₹60

73 days

______

₹57

3 years

______

₹904

5 years

______

4 Find the interest rate and simple interest. Principal

Amount

Simple Interest

Time

Rate p.a.

₹7500

₹8325

______

2 years

______

₹20,000

₹21,800

______

2 years

______

₹625

₹641.25

______

146 days

______

5 Find the time. Principal

Simple Interest

Rate p.a.

Time

₹4800

₹900

7.5%

______

₹3500

₹700

______

₹500

₹31.25

12.5%

______

₹600

₹12.60

3.5%

______

6 Find the amount directly using the formula A = P 1 +

100

in the given cases.

Principal

Rate (p.a.)

Time

₹450

5 years

₹5000

10%

3 years

₹3200

7.5%

2 years

₹720

8.5%

3 years

7 A borrowed amount of ₹4000 amounts to ₹5400 in 5 years. How much will ₹5600 amount to in 3 years at the same rate?

Chapter 14 • Compound Interest

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8 Find the interest on a deposit of ₹3650 from 3 January, 2016 at the rate of 10% p.a. up to 17 March, 2016. 9 A certain amount of money amounts to ₹4400 in two years and ₹4600 in three years. Find the principal and rate of interest.

10 Draw a graph for the given table of values, with suitable scales. Simple Interest on Deposits for a Year Deposit (₹)

1000

2000

3000

4000

5000

Simple Interest (₹)

160

240

320

400

a Does the graph pass through the origin? b Use the graph to find the interest on ₹2,500 for a year. c To get an interest of ₹280 per year, how much money should be deposited?

Word Problems 1 Savita deposited ₹8000 in a bank at the rate of 12% per annum. She withdrew the money

after 9 months. What is the interest she received? What was the amount she collected from the bank?

2 Kanchan borrowed a sum of money from Sonali at 8% interest per annum. After 4 years, Kanchan had to give Sonali ₹9900 to clear the debt. What was the amount Kanchan borrowed originally?

3 Narendra lent a sum of ₹3200 to Suresh at the rate of 6% interest per annum. Suresh paid Narendra ₹3680 to clear the debt. How long did Suresh use Narendra’s money?

4 Aditya paid ₹6720 to clear a debt of ₹6000 to the bank after 1 year and 6 months. What is the rate of interest charged by the bank?

5 The bar graph shows the sum invested by 4 different persons, A, B, C and D, in a bank on simple interest. Study the graph and answer the given question.

The interest received by A after 4 years at some rate of interest p.a. is ₹3,840. If the rate of interest is 2% more, what amount would A receive for the same period? Amount Invested (in ₹)

Scale: 1 division = ₹5000

25000 18000

20000 15000 10000

22000

20000

12000

5000 A

B C Name of Investors

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Finding Compound Interest Remember how Narendra planned to invest his money? Let us look at another table that shows the account balances for 3 years under Plan B. Plan B

Year

Principal

4% Compound Interest

Balance at the End of the Year

₹8320

₹332.80

₹8652.80

1 3

₹8000

₹320

₹8652.80

₹8320

₹346.11

₹8998.91

On analysing the table carefully, Narendra understands that any interest earned by him is added back to the principal amount deposited by him. This amount, in turn, is used to calculate the interest for the subsequent year. He likes Plan B and renames it as ‘Interest on interest plan’. This type of interest is called compound interest (CI). Money earned by Plan A = ₹8960 Money earned by Plan B = ₹8998.91 He is delighted to know that Plan B will earn ₹8998.91 − ₹8960 = ₹38.91 more than Plan A after three years. So, he chooses Plan B. Let us find the compound interest on P = ₹1000, R = 5% p.a. for 2 years. Interest after 1 year:

Interest after next year:

₹1000 × 0.05 × 1

₹1050 × 0.05 × 1

= ₹50

= ₹52.50

To find the CI, we subtract the principal from the amount compounded annually. So, CI = A – P = ₹1102.50 – ₹1000 = ₹102.50

Think and Tell

Total amount after two years = ₹1000 + ₹50 + ₹52.50 = ₹1102.50

Note that simple interest would produce a total amount of only ₹1000 [1 + 0.05 × 2] = ₹1100. Here, SI = ₹100.

Narendra wants to invest ₹8000 for 3 years. • Plan A earns 22% simple interest per year. • Plan B earns 20% interest compounded annually. Which plan should he choose?

The additional ₹2.50 is the interest on ₹50 at 5% for 1 year. To find a formula for compound interest, first suppose you deposit ₹1000 in a savings account that earns 6% interest compounded annually. Balance at the end of year 1 = Principal + Annual Interest

= ₹1000 + ₹1000 × 0.06

= ₹1000(1 + 0.06) = ₹1000(1.06)

Principal

Annual Interest

Balance at the End of Year (B)

₹1000

₹1000(0.06)

B = ₹1000(1.06)

₹1000(1.06)

₹1000(1.06)(0.06)

B = ₹1000(1.06)2

₹1000(1.06)2

₹1000(1.06)2(0.06)

B = ₹1000(1.06)3

₹1000(1.06)3

₹1000(1.06)3(0.06)

B = ₹1000(1.06)4

₹1000(1.06)4

₹1000(1.06)4(0.06)

B = ₹1000(1.06)5

Chapter 14 • Compound Interest

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Extending the pattern in the table, we can find the balance after 20 years. Balance at the end of year 20 = ₹1000(1.06)20 Using the pattern in the table, we can write a general formula for the balance in an account that earns interest compounded annually after t years.

A = P (1 + r)t

where P = Principal (initial deposit) r = Annual interest rate (in decimal form)

Did You Know?

The interest rate in India averaged 6.37 percent from 2000 until 2023, reaching an all-time high of 14.50 percent in August of 2000 and a record low of 4.00 percent in May of 2020.

t = Time (in years) A = Amount after t years

Compound Interest: Annually The formula deduced above can be rewritten as:

A=P 1+

where P = Principal (initial deposit)

R 100

R = Annual interest rate (in %) n = Time (in years) A = Amount after n years Example 4

Find the compound interest on ₹10,000 at 4% per annum for 3 years compounded annually.

Example 5

Find the time required for ₹15,625 to become ₹19,683 at 8% per annum compounded annually. R 100

P = ₹10,000, R = 4% per annum, n = 3 years

We know that A = P 1 +

A=P 1+

We have A = ₹19,683, P = ₹15,625 and R = 8% n 8 Therefore, 19,683 = 15,625 1 + 100

R 100

4 A = ₹10,000 1 + 100 A = ₹10,000 100 + 4 100

19,683 108 = 15,625 100

A = ₹10,000

104 100

A = ₹10,000 ×

104 104 104 × × 100 100 100

A = ₹1 × 104 × 104 × 1.04

On simplifying the RHS, we get, 19,683 = 15,625

Therefore, Compound Interest = ₹11,248.64 – ₹10,000 = ₹1248.64

27 25

27 27 × 27 × 27 = 25 25 × 25 × 25 27 25

A = ₹11,248.64

Compound Interest = Amount – Principal

27 = 25

n n

On comparing the LHS with the RHS, we get, n=3

Therefore, the time required is 3 years.

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Example 6

If a sum of ₹1200 amounts to ₹1452 at a certain rate of interest compounded annually for 2 years, find the rate of interest. n 2 R R We know that A = P 1 + . Here, A = ₹1452, P = ₹1200 and n = 2. Therefore, 1452 = 1200 1 + 100 100 R 1452 = 1+ 100 1200

On simplifying the LHS, we get 121 R = 1+ 100 100 11 10

R = 1+ 100

R R R 11 11 − 10 1 11 R =1+ ⇒ −1= ⇒ = ⇒ = 100 100 10 10 10 100 10 100 100 R= = 10 10 Therefore, the rate of interest is 10%. Do It Together

Amit receives ₹6050 after investing some money for 2 years at 10% per annum compounded annually. Find the actual amount invested by Amit. R We know that A = P 1 + 100

We have A = ₹_____, R = ___% and n = ___ 10 Therefore, 6050 = P 1 + 100

On simplifying the RHS, we get 100 + 10 6050 = P 100

___________________ 6050 = P

11 10

___________________ P = ___________________ Hence, P = ₹_____ Therefore, Amit had invested a sum of ₹_____.

Compound Interest: Half-yearly and Quarterly The interest is compounded half-yearly When the interest is added to the principal after every half year, the interest is said to be compounded half-yearly. As interest is calculated twice in one year, we will divide the rate of annual interest by 2 and double the time.

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So, the formula for the amount is modified to: R 2n R 2n 2 A=P 1+ OR A = P 1 + 200 100 The interest is compounded quarterly When the interest is added to the principal after every three months, the interest is said to be compounded quarterly. As interest is calculated four times in one year, we will divide the rate of interest by 4 and multiply the time by 4. So, the formula for the amount is modified to: A=P 1+ Example 7

R 4

OR A = P 1 +

100

R 4n 400

What will be the compound interest on ₹40,000 at 16% per annum for 1.5 years, if interest is compounded half-yearly? The interest is compounded half-yearly. So, the amount received after 1.5 years will be calculated using the formula R A=P 1+ 200

We have P = ₹40,000, R = 16% and n = 1.5 years Therefore, A = ₹40,000 1 +

16 200

2 × 1.5

200 + 16 16 A = ₹40,000 1 + = ₹40,000 200 200

216 = ₹40,000 200

On simplifying the RHS, we get A = ₹40,000

108 100

A = ₹40,000 ×

= ₹40,000

27 25

27 27 27 × × 25 25 25

A = ₹64 × 27 × 27 × 1.08 = ₹50,388.48 Compound Interest = Amount – Principal Therefore, CI = ₹50,388.48 − ₹40,000 = ₹10,388.48 Example 8

Find the amount Ankur needs to pay back after borrowing ₹2,00,000 at 4% per annum for 1 year if the interest is compounded quarterly. The interest is compounded quarterly. So, the amount paid after 1 year will be calculated using the formula: A=P 1+

R 400

We have P = ₹2,00,000, R = 4% and n = 1 year Therefore, A = ₹2,00,000 1 +

4 400

On simplifying the RHS, we get 1 100

A = ₹2,00,000

100 + 1 100

A = ₹2,00,000

101 100

A = ₹2,00,000 1 +

4×1

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A = ₹2,00,000 ×

101 101 101 101 × × × 100 100 100 100

A = ₹2 × 101 × 101 × 10.1 × 1.01

Do It Together

A = ₹2,08,120.802 So, Ankur will pay back ₹2,08,120.802 after 1 year.

Ajay deposited ₹18,000 in a bank at 16% per annum for 9 months. Find the amount received by him if the interest is compounded quarterly. The interest is compounded quarterly. So, the amount paid after 9 months will be calculated using the formula: A=P 1+

R 400

We have P = ₹18,000, R = 16% and n = 9

16 4 × 12 Therefore, A = ₹18,000 1 + 400

9 of a year 12

On simplifying the RHS, we get 4 A = ₹18,000 1 + 100

_________________________ A = ₹18,000

104 100

____________________

A = ₹ ___________________ So, the amount received by Ajay is ₹__________.

Do It Yourself 14B 1 Find the simple interest on ₹2500 for 2 years at 10% per annum. If the bank pays compound interest, 2

compounded annually, what is the difference in the interest earned in both cases?

Find the amount and the compound interest on ₹1600 for 5 years at 6% per annum, compounded annually.

3 The simple interest on a sum of money for 2 years at 6% per annum is ₹9000. Find the compound interest on that sum at the same rate compounded yearly and for the same period.

4 What sum of money will amount to ₹4410 in 2 years at 5% per annum, compounded annually? 5 At what rate per cent per annum compound interest will ₹8000 amount to ₹8820 in 6 months, if the interest is compounded quarterly?

6 In how much time will ₹60,000 amount to ₹69,457.50 at 5% per annum compounded yearly?

Word Problems 1

Rahul took out a loan of ₹25,000 from the bank to renovate his house. If the rate of interest is 8.5% per annum compounded annually, what amount will he have to pay to the bank after 3 years to discharge his debt?

2 Yogesh borrowed ₹30,000 from his friend Navin at 12% per annum simple interest. He

further lent it to his friend, Akash at the same rate but compounded annually. Find his gain after 4 years.

3 Kamal deposited a sum of ₹6400 in a post office for 3 years, compounded half yearly at 7 % 2 per annum. What amount will he get on maturity?

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Rohit borrowed ₹18,000 from a finance company at 16% per annum, compounded 1 quarterly. What amount of money will discharge his debt after 1 years? 4

Mr. Kumar took out a loan of ₹1,80,000 from a finance company to purchase machinery 1 for his factory. If the company charges compound interest at 8 % per annum during 2 1 the first year, 7 % per annum during the second year and 7% per annum during the 2 third year, how much will he have to pay after 3 years?

Lalit paid ₹79,860 after 3 years for a loan from the bank. If the bank charges interest at 10% per annum, compounded annually, find the sum of money borrowed by him.

Word Problems Narendra buys a plot of land near a city to invest his money. The present cost of land per square metre is ₹16,000. In the past two years, the land’s price has increased at the rate of 20% per annum. Narendra expects that the price will increase at the same rate for the coming two years. He is excited to know the appreciated price of the land after two years. Let us help him out! The price of land appreciates, i.e., every year the price of the land goes on increasing. If the rate of appreciation is constant (rate of R% p.a.), we can find the expected cost of the land after a certain number of years (t years) by using the same formula that we used for calculating compound interest, where P is the original price. A=P 1+

R 100

Present cost of land per square metre = ₹16,000 The rate of appreciation = 20%

R Appreciated cost after 2 years = A = P 1 + 100 2 20 A = 16,000 1 + 100 = 16,000

100 + 20 100

= 16,000

120 100

Remember! Appreciation means an increase in the value of an asset over time and depreciation means a decrease in the value of an asset over time.

16,000 × 120 × 120 = 160 × 12 × 12 = ₹23,040 100 × 100

So, the price of the land after two years would be ₹23,040 per square metre. If by chance the cost decreases, then we say that the value has depreciated. If the depreciation is at a constant rate, then we can calculate the depreciated value by the formula: A=P 1−

R 100

The formula for compound interest can be applied in various other situations. Wherever there is a constant rate of increase or decrease in the population, this formula can be used. 196

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Let us suppose that the population of a town increases at a constant rate of R% every year. Let the given population be P. At the end of one year, the population will increase by R%. So, the population at the end of 1 year = P 1 +

R 100

At the end of the second year, the population P 1 + 2

R R 1+ =P 1+ . 100 100

R R will increase by R% and will become P 1 + 100 100

At the end of n years, the population will become P 1 +

R . 100

For example, the population of a town was 1,28,000, two years ago. If the rate of increase in population is 10% per annum, then the present population of the town = P 1 + The present population of a town is 3,20,000. If it increases at the rate of 5% p.a., what will be the population after two years?

Example 9

Example 10

= 1,28,000 1 +

10 100

= 1,54,880

The cost of a new television set is ₹22,000. Its value depreciates every year at the rate of 20%. What will the depreciated price be after two years?

Present population (P) = 3,20,000

The cost of the television set = ₹22,000

Rate of increase (R) = 5%, Time (n) = 2 years

Rate of depreciation = 20%

Population after two years

Time = 2 years

= 3,20,000 1 +

Price after two years = 22,000 × 1 −

5 100

100 + 5 = 3,20,000 100 105 = 3,20,000 100

= 22,000

105 105 = 3,20,000 × × 100 100 After two years, the population will be 3,52,800.

20 100

100 − 20 100

2 2

= 22,000 ×

80 100

= 22,000 ×

80 80 × 100 100

= ₹14,080

= 3,52,800

Do It Together

R 100

So, the depreciated price of the television set after two years is ₹14,080.

Bacteria is growing at the rate of 2.5% per hour. If the bacterial count at 1 a.m. is 16,500, find the bacterial count at 5 a.m. Population at 1 a.m. = __________ Number of hours between 1 a.m. and 5 a.m. = __________ Rate of growth (R) = __________ Population after 4 hours = 16,500 1 +

Chapter 14 • Compound Interest

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2.5 100

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= 16,500

100 + 2.5 100

= ____________________ = 16,500(1.025)4 = 16,500 × ____________________ = ____________________ So, the bacteria count at 5 a.m. is __________.

Do It Yourself 14C 1

The population of a city increases every year at the rate of 10%. If the population at present is 12,50,000, what

will be the population after 4 years?

The population of a town increases at a rate of 7% every year. If the present population is 90,000, what will it be

after two years?

The price of a plot increases at a constant rate of 5% every year. Find its expected price after 3 years if the present

price is ₹2,00,000.

A car which costs ₹2,50,000 depreciates by 10% every year. What will the car be worth after three years?

A new computer costs ₹60,000. The depreciation is 40% every year. Find the price of the computer after two years.

Production in a factory is increasing at a rate of 4% per year. If the production in 2017 was 15,625 units, find the

number of units produced in 2020.

A refrigerator was purchased for ₹45,000. Its value depreciates at the rate of 3% p.a. What would be its value

after 3 years?

The cost of an antique vase increases by 6% every year. If its current price is ₹2500, what will be its price after

3 years?

The price of a mobile phone depreciates at a rate of 6% every year. If its current cost is ₹20,000, what will be its

price after 4 years?

10 11 12 13 14

The fare for a certain route of buses increases at a rate of 10% each year. If the current fare is ₹5, find the fare

after 2 years.

The value of a residential flat constructed at a cost of ₹10,00,000 is appreciating at the rate of 10% every year.

What will be its approximate value after 4 years?

The population of bacteria in a culture is increasing at the rate of 3% per hour. Find the approximate population

of bacteria at the end of 2 hours, if it was 4,50,000 initially.

The value of a constructed house increased from ₹1,60,000 to ₹2,30,400 in 2 years. What is the appreciation rate

of the value of the house?

The present value of Ajay’s car is ₹3,88,800. He purchased the car 2 years ago. If the value of the car depreciates

at a rate of 10% p.a., how much did Ajay pay for the car?

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Points to Remember Simple Interest

Compound Interest

Simple interest is always calculated on the initial principal amount.

The total of initial principal and accumulated interest is used to calculate compound interest. The amount at the end of one year is the principal for the next year.

The principal is the same every year. Simple interest is the same every year.

Compound interest is different every year.

Simple interest is less than compound interest. Simple interest =

Compound interest is greater than simple interest. Compound Interest = A − P

P×R×T

=P 1+

100

The growth remains uniform over the years.

R 100

–P

The growth has been rapid over the years.

Math Lab Simple Interest Vs Compound Interest Setting: In groups of 4 Materials Required: Paper and pen Method: 1 Divide the class into groups. Each group should have two teams (A and B). 2 Provide each group, that is, both the teams with the same value of the principal, a conversion period and a rate of interest. The data provided to each group should be different. 3 Team A needs to calculate Compound Interest and Team B needs to calculate Simple Interest. 4 Now, compare the amount of Compound Interest of Team A with the Simple Interest of Team B for each group. Find out which amount is greater.

Chapter Checkup 1

Find the simple interest and the amount. Principal

12%

3.25 years

Time

Simple Interest

Amount

₹560

73 days

_____

₹2700

146 days

_____

Time

Rate p.a.

Principal

3.5%

_____

₹10,000

b c

Rate% p.a.

_____

Find the principal. Simple Interest a

₹4.50

5 months

₹392

3.5 years

Chapter 14 • Compound Interest

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4.5%

_____

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Find the interest and the rate percent. Principal ₹10,000

Amount ₹11,800

Simple Interest

Time

_____

Rate % p.a.

2 years

_____

4 A sum of money amounts to ₹780 in two years and ₹1230 in seven years. Find the principal and rate of simple interest.

In how much time will a sum of money double itself if invested at 8% simple interest per annum?

Find the compound interest on ₹7000 at 10% p.a. compounded annually for 4 years.

Find the compound interest on ₹30,000 at 12% p.a. for 2 years compounded annually.

Find the interest earned on ₹5000 for 2 years at 12% per annum compounded annually.

Find the amount and compound interest if the interest is compounded annually. S. No.

Principal

Rate % p.a.

Time (years)

Amount

Compound Interest

₹5000

_____

₹72,000

_____

Find the principal if it amounted to ₹16,335 compounded annually, at the rate of 10% p.a. in two years.

In how many years will ₹8800 amount to ₹10,648 at 10% p.a. compounded annually?

At what rate % of compound interest will ₹4000 amount to ₹5324 in three years if compounded yearly?

Find the compound interest on a sum of ₹7000 at 8% per annum for two years, compounded annually.

14 Calculate the compound interest on ₹20,000 for 1 annually.

1 2

years at the rate of 10% per annum compounded semi-

Find the compound interest on ₹1,60,000 for 2 years at 10% per annum, compounded semi-annually.

A sum of ₹25,000 invested at 8% p.a. compounded semi-annually amounts to ₹28,121.60. Find the time.

17 The difference of simple interest and compound interest at the rate of 10% for two years is ₹1. What is the principal if the interest is compounded annually?

Word Problems 1

Kishan borrowed ₹5000 at 8% p.a. for 3 years. Find the difference of compound interest

Aakash deposited ₹12,000 in a bank for 2 years. It is compounded annually at 9% p.a.

3 4 5

and simple interest, assuming that interest is compounded annually. What amount will he receive on maturity (at the end of 2 years)?

The population of a town increases by 4% annually. If the present population is 54,080, what was the population two years ago?

The present cost of a mobile phone is ₹15,000. If its value decreases every year by 5%, find its cost 2 years ago.

The value of a car depreciates at the rate of 10% per year. A car which was bought three years ago is now worth ₹4,73,850. What was its original price?

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Algebraic Expressions: Addition and Subtraction

Let's Recall Look at the number sentences. 7 +

= 15

6−

= 4

3 × ? = 18

In each case, the left-hand side of the sentence has symbols, shapes, or objects to represent unknown numbers. We can also represent unknowns using letters of the English alphabet. Such sentences where letters are used are called either expressions or equations. There are two types of expressions: Numeric expressions and algebraic expressions. Expressions that involve only numbers with different arithmetic operations (+, −, ×, ÷) are called numeric expressions. For example, 8 + 2, 9 – 6, 2 × 3, 14 , etc. 7

Expressions that involve both numbers and letters of the English alphabet with different arithmetic operations (+, −, ×, ÷) are called algebraic expressions. Algebraic Expression

For example, x + 5, 3 – y, 4s,



Each number is a constant and each letter is a variable. p , etc. 7

3−y

Constant

Like numbers, operations can also be performed on variables. For example, we can add and subtract terms: 2x and 3x. Addition of two terms

Variable

Subtraction of two terms

2x + 3x = 5x

3x − 2x = x

Letʼs Warm-up Match the following.

1 8p – 3p –5b 2 7 + 8 5b 3 2p + 5p 5p 4 9b – 4b 15 5 –9b + 4b 7p I scored _________ out of 5.

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Mean, Median and Mode Algebraic Expressions Real Life Connect

After school, Srishti had some free time and was getting bored. To keep herself busy, she started playing with toothpicks. She made a beautiful pattern of triangles and parallelograms using toothpicks, as shown below. Step 1

Step 2

3 sides

Step 3

8 sides

13 sides

After the initial 3 steps, she thought of getting more toothpicks to create a pattern of up to 10 steps. Let us help Srishti identify the number of toothpicks required to continue the pattern up to 10 steps.

Terms and Factors of an Algebraic Expression To find the number of toothpicks for the 10th step by counting them at every step would be difficult. So, let us first identify the rule for finding the number of toothpicks at any step and then use it to find the number of toothpicks needed at the 10th step. Step 1

Step 2

3 sides

Step 3

8 sides

13 sides

Step No.

Number of Toothpicks

Number of Sides

3+5×0

3+5×1

3+5×2

We see that for step 1, 3 toothpicks are used and for all the next steps the number of toothpicks increases by 5. So, we can say that the rule for the pattern will be 3 + 5x, where x = (step number – 1). So, the number of toothpicks required for the 10th step = 3 + (5 × 9) = 3 + 45 = 48 toothpicks. Now, notice that the rule for this pattern (3 + 5x) is nothing but an algebraic expression. Variable

Constants Term 1

3 + 5x

Algebraic Expression

Term 2

Every expression has some terms and factors. Each of the constants and variables in an expression are called factors. These factors are either combined together with the help of a ‘×’ sign or they stand alone as a single constant or variable to form terms of an algebraic expression. Terms, in turn, are then connected to each 202

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other by either ‘+’ or ‘–’ signs to form an expression. Given below is a factor tree to identify the terms and factors in an expression. Expression

5p2q + 7pq2 Terms

5p2q p

7pq2 7

Numerical factors

Literal factors Factors

A constant factor is called a numerical factor. A variable factor is called a literal factor. Coefficient Factors of a term are made up of coefficients. There may be numerical or literal coefficients. The literal factor is the coefficient of the product of the rest of the factors of the term. The case for the constant coefficient is similar. Numerical Coefficient

Constant

3X2 + 5y4 – 8

Literal Coefficient

Remember! A factor of the term is always the coefficient of

Term

the product of the rest of the factors of the term.

In 3x , 3 is the coefficient of x and x is the coefficient of 3. 5 is the coefficient of y4. Like and Unlike Terms

Two or more terms with the same variables are called like terms, whereas two or more terms with different variables are called unlike terms. For example, look at the expressions given below. Same Variables

Different Variables

Think and Tell 2x, 5x; 4y2, 3y2

2x, 5y; 4zy2, 3yz2

Like Terms

Unlike Terms

Chapter 15 • Algebraic Expressions: Addition and Subtraction

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Why are 4zy2 and 7yz2 not like terms?

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Types of Algebraic Expressions An algebraic expression can have one or more than one term. Based on the number of terms, we categorise algebraic expressions as: Monomial

Binomial

Trinomial

Quadrinomial

1 term

2 terms

3 terms 1 Example: z2 – 5x + 2, a3 + 2 3ab + b

4 terms 3 2 Example: z – 2xy – 5x2 + 1, 4 a3 + b2 – a + 5

Example: 2x, 3y2,

Polynomial:

7 az 13

3 Example: z2 + 1, a3 – 2 4

An algebraic expression with one or more terms. Monomials, binomials, trinomials and quadrinomials are all polynomials.

Did You Know? Polynomials are used in coding theory to create error-correcting codes, which are vital for reliable data transmission and storage.

Example 1

Sort the expressions as monomials, binomials, trinomials, and quadrinomials. 2 3

3x + 1, 2y, 6x2 – 7y2 + 5, − yz,

5 5 3 2 1 1 7 p – 2q + r, 9p2 + 6q2 − , 9xy − pq + 2q – , p – 2 , 9abc, 11xyz − yz + 2x – 6 6 4 5 2 6 8 3

Monomials

Binomials 3x + 1, 1 2 p– 2 3

2 yz, 3 9abc

2y, −

Example 2

Trinomials 2

Quadrinomials

6x – 7y + 5,

5 2 pq + 2q – , 6 5 1 7 11xyz − yz + 2x – 6 8 9xy −

5 p – 2q + r, 6 3 9p2 + 6q2 − 4

Look at the algebraic tiles. Use these tiles to form algebraic expressions. x

–x2

−5xyz

x 2

–x + 4x + 2

Complete the table. Expressions

3x + 2 Do It Together

Terms −5xyz

Factors

9x + 1

9, x; 1

1 7 7xy − xy2 + 2z – 6 8

1 7 7, x, y; − , x, y, y; 2, z; – 6 8

3 3 − x2 – 7y2 + 4 2

Coefficient of x

3 − x 4

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Do It Yourself 15A 1

Look at the given algebraic tiles. Write algebraic expressions for the tiles. Also, identify the terms containing variables. a

–x2

1 1

1 –x2

2 Use a factor tree and write the terms, factors, and coefficient of ‘a’ in the expressions. a –

3 2 a bc 4

c 8a2 – 5a +

b 6a + 5

1 3

2 d 9ab – a – 1 7

4 7 3 a + 5abc – a2 – 10 9 12

3 Classify the expressions as monomials, binomials, trinomials or quadrinomials. 1 7 5 3 1 5 5 1 5 11 2x + 5, −3y, −4x2 + a2 + 5, − ab, g – 2h + i, 11p2 + q2 − , 9xy − pq + 2q + p2, p – , stu, 2 8 3 4 4 6 2 3 6 6 13 5 11pqr − yz + 2p – 6 8

4 Circle the pairs of like terms and cross out the pairs of unlike terms. 2 a 5a, − a 3

b 6xy, 8x2y

c −2a2b2c, 9ab2c2

d −

5 Identify the numerical factor of the expressions in Q2.

11 abc, −8cba 7

e 3x2, 4x2

6 Identify the numerical coefficient of the expressions in Q3. 7 Form algebraic expressions for the statements. Also, write the literal factors and literal coefficients of the numerical coefficient in the expressions. a The sum of x and y, divided by 8. b The difference of twice of x and one-sixth of y. c The product of a and b subtracted from the difference of a and b. d Seven-eighths of a number a multiplied by the sum of four times a and b

8 Which of these are polynomials? Explain your answer. a −y+5 e

c −5x2y +

b −3

11 3 – 2h + 6p 4k

f 9k2p +

1 2 1 kq − 8 4

g 9

3 +5 2a

7 ab 11 5 2 h − bc + 2b + c2 – 6 5a d −

Word Problem 1

Simha bought a few blue straws of length x cm each and a few green straws of length y cm each. Use the pictures to write the algebraic expressions.

Chapter 15 • Algebraic Expressions: Addition and Subtraction

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Addition and Subtraction of Algebraic Expressions Real Life Connect

Three friends, Yamini, Harsh, and Damini are playing with marbles. Yamini has some marbles. Harsh has 12 more. Damini has three times the marbles that Yamini and Harsh have together. How many marbles are there in total? Let us see!

Adding and Subtracting Expressions Addition of Algebraic Expressions To find the total number of marbles the three friends have, we need to first find the number of marbles each of them has, then add all of them to arrive at the answer. Let us understand this! To add two or more algebraic expressions, like terms are put together and added. For example, consider the two expressions: 3x2 – 7x + 8 and 2x2 + 3x – 5. Let's add the expressions 3x2 – 7x + 8 and 2x2 + 3x – 5 using algebra tiles. 1 Show the two expressions using algebra tiles. x2

-- x

–1 –1 –1

3x2 – 7x + 8

–1 –1

2x2 + 3x – 5

2 Place all the tiles together. Cross out each pair of one negative and one positive tile of the same factor.

-- x

-1

3 The remaining tiles show the final answer. x2

-- x

5x2 – 4x + 3

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Now, let us add the two expressions using the horizontal method. (3x2 – 7x + 8) + (2x2 + 3x – 5) = 3x2 + 2x2 – 7x + 3x + 8 – 5 (Place the like terms together) = 5x2 – 4x + 3 We can also add the two expressions vertically, as shown below. 3x2 − 7x + 8 + 2x2 + 3x − 5 5x2 − 4x + 3 This is called the vertical method of adding two or more expressions. Now, to find the total number of marbles the three friends, Yamini, Harsh, and Damini have, we need to add the three expressions. Let us first write the 3 expressions. Let the number of marbles with Yamini be x.

Error Alert!

Number of marbles with Harsh = x + 12

Unlike terms are never added together.

Number of marbles with Damini = 3(x + x + 12) = 3(2x + 12) = 6x + 36

3 + 4x + 7 = 14x

Total number of marbles = (x) + (x + 12) + (6x + 36) = x + x + 12 + 6x + 36 = 8x + 48 Add the expressions.

Example 3

1 5p + q – 6 and 8p − 1 q + 3 2 5p + q – 6 + 8p − = 5p + 8p + q − = 13p +

Example 4

1 q–3 2

2 − 2 x2y2 − 6xy + 3 and 2y2x2 + 3xy − 1 5 4 4

1 q + 3. 2

−

1 q–6+3 2

2 2 2 3 1 x y − 6xy + + 2y2x2 + 3xy − 5 4 4

=− =

2 2 2 3 1 x y + 2y2x2 − 6xy + 3xy + − 5 4 4

8 2 2 1 x y – 3 xy + 5 2

Srikanth walked (2x + y + 5) km to the north, then (5x – 1) km to the east and then south-east. How far did he walk? Total distance travelled by Srikanth = (2x + y + 5) + (5x – 1) +

= 2x + 5x + y +

5 = 7x + 1 y + 6 6

(

)

5 So, Srikanth travelled 7x + 1 y + 6 km. 6 Do It Together

(

The sides of a rectangle are 3a2 + 2ab −

(

Side 1 of the rectangle = 3a2 + 2ab − Perimeter of the rectangle = 2(l + b)

)

(

)

( )

5 y + 2 km to the 6

5 y+2 6

11 6+5 5 y + 5 – 1 + 2 = 7x + y + 6 = 7x + y+6 6 6 6

(

)

1 1 3 cm and − a2 – 7ab + cm. What is its perimeter? 2 6 4

1 cm 2

(

Side 2 of the rectangle = −

Chapter 15 • Algebraic Expressions: Addition and Subtraction

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3 + 4x + 7 = 3 + 7 + 4x = 10 + 4x

)

1 2 3 a – 7ab + cm 6 4

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(( (

= 2 3a2 + 2ab −

(

)(

1 1 3 + − a2 – 7ab + 2 6 4

Then, perimeter = 2 3a2 + 2ab −

1 1 2 3 − a – 7ab + 2 6 4

)

))

)

1 = 2 3a2 − ____ + 2ab – _____ − 2 + ____ = 2(________________)

= 2 × ____ – 2 × _____ + 2 × ____) = ________________

= ________________ So, the perimeter of the rectangle is ________________ cm.

Subtraction of Algebraic Expressions Like addition, we can also subtract two or more algebraic expressions in the same way. But in subtraction, before putting like terms together, we evaluate the minus sign first. For example, consider the two expressions: 5x2 – 2x + 1 and 4x2 + x – 3. Let us first represent them using algebra tiles. 1 Show the minuend, 5x2 – 2x + 1 using algebra tiles. x2

-- x

5x2 – 2x + 1

-- x 1

2 To subtract, cancel out as many tiles as given in the subtrahend, 4x2 + x – 3. a To subtract 4x2, cancel out four x2 tiles.

-- x

-- x 1

b T o subtract +x, we need at least one +x tile. Since there are not enough +x tiles, we bring in 0 (a pair of +1x and –1x tiles). Then cancel out one +x tile.

-- x

-- x 1

-- x

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o subtract –3, we need three –1 tiles. Since there are not enough –1 tiles, we bring in 0 (3 pairs of T +1 and –1 tiles) 1

-- X

-11

d The remaining tiles show the final answer. 2

-- x

x2 – 3x + 4

Now, let us understand how to subtract the two expressions without the algebra tiles. Let us first evaluate the minus sign and then put the like terms together. So, (5x2 – 2x + 1) − (4x2 + x – 3) = 5x2 – 2x + 1 − 4x2 − x + 3 2

= 5x − 4x – 2x – x + 1 + 3

Think and Tell What happens if we do not evaluate the minus sign?

= x2 – 3x + 4 This method of subtracting two or more expressions is called the horizontal method. We can also subtract the two expressions vertically, as shown below. 5x2 − 2x + 1 4x2 + x − 3 − − + 2 − x 3x + 4

This is called the vertical method of subtracting two or more expressions. Example 5

Subtract the expressions. 3 q – 5 from 8p − q + 9 4 3 8p − q + 9 – (2p + q – 5) 4 3 = 8p − q + 9 – 2p − q + 5 4 3 3 = 8p – 2p – q − q + 9 + 5 = 6p − 1 q + 14 4 4

1 2p +

Example 6

2 – 1 a2b – 3ab + 5 from − a2b – 5ab + 1 4 6 1 5 −a 2b – 5ab + 1 – (− a2b – 3ab + ) 4 6 1 5 = −a2b – 5ab + 1 + a2b + 3ab − 4 6 1 1 5 –3 2 = −a2b + a2b – 5ab + 3ab + 1 − = a b – 2ab + 6 4 6 4

Vicky bought a shirt for ₹(9x + 15) and a pair of trousers for ₹(11x – 10). If he paid ₹7000, then how much change did he get? Cost of a shirt = ₹(9x + 15) Cost of a pair of trousers = ₹(11x – 10)

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The amount of money Vicky gave to the shopkeeper = ₹7000 The amount of change he got back = 7000 – (9x + 15 + 11x – 10) = 7000 – (9x + 11x + 15 – 10) = 7000 – (20x + 5) = 7000 − 5 – 20x = 6995 – 20x So, Vicky gets back ₹(6995 −20x). Do It Together

What should be added to −5pq + 8q – 6 to get −7pq + 11q – 7? Let the expression to be added be X. So, according to the question, we get −5pq + 8q – 6 + X = −7pq + 11q – 7 X = −7pq + 11q – 7 – (−5pq + 8q – 6) = −7pq + 11q – 7 ___ 5pq ___ 8q ___ 6 = −7pq + _____ + 11q – _____ – 7 + _____ = ______________ So, _____________ will be added to −5pq + 8q – 6 to get −7pq + 11q – 7.

Simplifying Algebraic Expressions Let us understand how to simplify two or more expressions. For example, consider the expression: 2x + 5 – (5x – 1) + (7x + 2) We can write it as: 2x + 5 – (5x – 1) + (7x + 2) = 2x + 5 – 5x + 1 + 7x + 2 = 2x – 5x + 7x + 5 + 1 + 2 = 2x + 7x – 5x + 8 = 4x + 8 Example 7

Simplify: (4xy2 – 3y + 1) + (9y – 1) – (2y2x + 5y – 3) (4xy2 – 3y + 1) + (9y – 1) – (2y2x + 5y – 3) = 4xy2 – 3y + 1 + 9y – 1 – 2y2x − 5y + 3 = 4xy2 −2y2x – 3y + 9y − 5y + 1 – 1 + 3 = 2xy2 + y + 3

Example 8

Subtract p2q3 – pq + 5 from the sum of 8p2q3 + 2pq + 1 and According to the question, let us first add and then subtract.

–1 pq + 6. 2

8p2q3 + 2pq + 1 + ( –1 pq + 6) 2 = 8p2q3 + 2pq – 1 pq + 1 + 6 = 8p2q3 + 1 1 pq + 7 2 2 Now, 8p2q3 + 1 1 pq + 7 – (p2q3 – pq + 5) = 8p2q3 + 1 1 pq + 7 – p2q3 + pq − 5 2 2 = 8p2q3 − p2q3 + 1 1 pq + pq + 7 – 5 = 7p2q3 + 2 1 pq + 2 2 2 210

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Do It Together

The perimeter of a triangle is (4a + 10b – 6) cm. If two of its sides are (9a – 5b + 7) cm and (−5a + 2b – 3) cm, find the third side. Side 1 = (9a – 5b + 7) cm

Side 2 = (−5a + 2b – 3) cm

Side 3 = ?

Perimeter of the triangle = (4a + 10b – 6) cm Let the third side be X. Perimeter of a triangle = 9a – 5b + 7 + (−5a + 2b – 3) + X _________________________________________________________________________________________________________________ _________________________________________________________________________________________________________________

Do It Yourself 15B 1 Solve the given problems using algebra tiles. a (2x2 + 4x + 4) + (x2 + x + 2)

b (4x2 – 2x + 1) – (x2 + 3x – 2)

2 Add the expressions horizontally. Verify your answer by solving them vertically. b 3xy2 + 5xy – 6 and 7y2x – 2xy + 3

a 3x + y + 1 and 2x – 5y – 4 c −5p2 −

4 5

p + 2 and 3p2 + 5p −

7 8

pq +

6 7

p – 1 and

5 8

pq +

2 7

p+8

3 Find the difference of the expressions horizontally. Verify your answer by solving them vertically. 1 1 and 2mn – n3 + 5 2 5 d 7 – 13pq – 11q and 2pq + 8 – 4q

a 5x2 – 3x – 1 and 3x2 – 4x + 7

b 8mn + n3 −

c 2y + 3y2 + 8 and y2 + 2y – 4

4 Simplify the expressions.

(2m3 + 4m2) −

(5m3 – 2m2)

a 3(5x – 4) – 2(7x + 3) + 4(8x – 2)

c (−3x2 + 6x3 – 4 – x) + (2x + 1) – (8x2 – 5x + 1)

d (x – 2) – (x2 – x + 3) + (4x – 5)

5 Find the missing expression. a (6a4 – 4 + 8a) + ______________ = (a4 + 5 + 8a + 8a3) b ______________ − (6x2 + 6x + 11) = (10x2 + 4x – 1) c (−19n2 – 6n + 6) + (−8n2 – 4n – 9) – ______________ = (−15n2 + 8n + 7) d ______________ – (4q2 – 5q) + (7q – 9q2 + 6q4) = (8q4 – 3q2 + 2q)

6 Subtract (40 – 13p2) from the sum of (−18p2 + p – 32) and (13p + 20p2 – 12). 7 What should be subtracted from (7h – 3ht + 7h3) to get (−6h + 3ht – 9h3)? 8 How much is (6y5 – 2) greater than (9y5 + 5 + 3y2)? 9 Find the errors in the problem. Also, find the correct solution. 2 3

p2 –

1 5

pq2 –

1 3

p2 –

2 5

pq2 = =

2 2 1 2 1 2 2 2 p – pq – p – pq 3 4 4 5

8p2 – 3p2 12

–

5pq2 – 8pq2 20

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5p2 12

3pq2 20

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Word Problems 1

Harsh has a rope of length (m2 + 13m – 18) metres and Rohan has a rope of length

A field is in the shape of a quadrilateral whose sides are (9s2 + 17) metres, (34s + 7 – 18s2)

(8m + 30m2 – 4) metres. What is the difference of the lengths of their ropes?

metres, (9s + 2s2) metres and (17s2 – 25) metres. If it needs to be fencing on all sides, find

the length of fencing required for the field.

Points to Remember • Each of the constants and variables in an expression is called a factor. These factors are either combined together with the help of the ‘×’ sign or stand alone as a single constant or variable to form terms of an algebraic expression. • A factor of the term is always the coefficient of the product of the rest of the factors of the term. • Two or more terms with the same variables are called like terms, whereas two or more terms with different variables are called unlike terms. • Expressions with 1, 2, 3 or 4 terms are called monomials, binomials, trinomials and quadrinomials respectively. • An expression that has one or more terms is called a polynomial. So, all monomials, binomials, trinomials, and quadrinomials are polynomials. • To add or subtract two or more algebraic expressions, like terms are put together and added or subtracted. In subtraction, before putting like terms together, we evaluate the minus sign first.

Math Lab Expression Treasure Hunt! Setting: In groups of 5 Materials Required: Flash cards with questions based on simplifying algebraic expressions written on them, small prizes and rewards, small envelopes or containers to hide the treasures. Method: All 5 members of each group must follow these steps. 1 Each group must search for envelopes or containers in which treasures are kept. Each envelope also contains a flash card with a question written on it. (Treasures may be hidden in any envelope or container in the classroom. It may be any kind of a gift or reward that students may like. For example, a set of 5 chocolates, crazy bouncing ball, etc.) 2 Read the question on the flash card in the envelope and solve the question. 3 If a group solves the question correctly, the group gets to keep the treasure inside the envelope, else they leave the treasure as it is in the envelope or container. 4 Continue the activity as long as time permits. The group to get the highest number of treasures wins the game.

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Chapter Checkup 1 Look at the factor trees. Label the variables, terms, factors, and numerical coefficients in each of the expressions. 6pq2 + 8p2q

6pq2 6

p q q

− mn + 2n2 − 3m2

8p2q

− mn

p p q

–m

2 Match the following.

2n2

− 3m2

2 n n

–3 m m

a 4x + y, 2x – 1

quadrinomial

b 8a b, 9pq, −6x

trinomial

c 3xy + x – 1, 8z2 − 3z + 5

monomial

d −6x y + 2xy – 6x + 4, p – 7pq + 4q – 1

3 Which of these are polynomials? a −5

binomial

3 +7 x

c −2a2b +

7 +5 2a

–

5 2 pq + 2q + p2 – 6 5a

6pq, −4x2y

c −7a2b3c, 8ab2c3

−

5 1 abc, − cba 6 2

b −

4 Circle the like terms and cross out the unlike terms. a 2a, − 3 a

5 Solve using algebra tiles. a (3x2 + 2x – 6) + (–x2 – 2x + 1)

b (–x2 + 4x + 2) – (3x2 – 2x – 3)

a −3m2 + m – 2 and 4m2 + 6m + 7

b 7z3 + 4z2 + z – 1 and 2z3 – 6z2 – 2z + 2

6 Add the expressions. c q2 –

5 3 1 q – 7, 3q2 + 4 – q3, and 9q2 – 6q – 2 7 7

7 Subtract the expressions.

1 1 z + z6 + 17z3 from 11z3 + 10z6 – 26z + z5 2 2 1 2 3 2 4 4 c − y + 5y – 5 from y – 6y + 8 2 4

a 8z5 – 9z2 +

b 2x2 – 3x + 6x4 from 8x4 − 5x2

8 Simplify the expressions.

a (7s2 – 4s4 + 3s) + (5s2 + 6s4) − (9s2 – 8s3 + 2s4) b (8t4 – 3t2 + 2t) – (4t2 – 5t) + (7t – 9t2 + 6t4) c (7k2 + 5k3 – 4k4) – (8k4 + 6) – (9k3 + 3k2)

9 Add (3x2 − 2x + 1) to the difference of (−2x3 + x) and (7x – 3 + 7x3). 10 What should be added to (6k5 + 1) to get (10k5 + 9k)? 11 By how much is (4p + 6p4 – 5) smaller than (2p – 8p4)? 12 What should be taken away from (6c – 4c4 – 3) to get (−3 + c4 – 3 + 8c2)?

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Word Problems 1

the footpath all around it. Find the difference in the

perimeter of the field with and without the footpath. 2

9x2 + 7x − 4 4x2 − 3x + 7

This is a rectangular field, where the shaded region is

(All measures are in cm.)

3x 4

Hetal’s mother went to the market to buy some fruit. She bought bananas for ₹(15x + 6), grapes for ₹(13x2 – 9x + 2) and apples for ₹(−8x2 + 4x – 18) less than the cost of the

grapes. Find the total amount she spent to buy the fruit.

(

)

Ananya has a sheet of chart paper whose sides are (4x + 5) metres and (3x2 – 5 + 6x) metres. If she cuts squares from all the corners of the chart paper of size perimeter of the remaining chart paper.

1 2 x + x metres, find the 3

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Multiplication of Algebraic Expressions

16 Let's Recall

2 rows

Arrays in multiplication are arrangements of sets of objects, pictures, or shapes representing numbers in the form of rows and columns. For example, this array has 2 rows and 3 columns. It can be described as a 2 by 3 array. The word ‘by’ represents multiplication. We can write different number sentences for this array as: 2×3=6 3×2=6 2+2+2=6 3+3=6

3 columns

We can also use arrays to represent situations where one or both numbers can be written in expanded form. For example, look at the given arrays. 12 × 2 can be written as (10 + 2) × 2. As an array, this can be represented as shown here. 2

12 × 13 can be written as (10 + 2) × (10 + 3). Representing such complex number sentences using shapes or objects can be a tedious task. So, this can be represented as blank arrays, as shown here. 3

Letʼs Warm-up Match the given array representation with its multiplication sentence. 1

10 2

11 × 4

8×9

6×3

11 × 7

12 × 12

I scored _________ out of 5.

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Mean, MedianAlgebraic and Mode Multiplying Expressions Real Life Connect

A newly built party hall has an underground parking lot. A guide map of the parking lot has been put up by the owners of the hall at different places in the parking lot. Party Hall

Parking

The owners decide to allot the cleaning of the parking lot to a third-party vendor. To decide on the pricing of cleaning the parking lot, the vendor wants to know its area. To decide on a suitable vendor, the owners want to know the price quoted by each vendor.

Multiplying Polynomials Let us understand how we can find the area of the parking lot. For this, let us assume that one side of the blue part of the parking lot is x units, and one side of the green part is 1 unit, as shown below. x

x+4

Then, the area of the parking lot = area of the big rectangle = length × breadth

= (x + 3) × (x + 4)

x+3

= x × (x + 4) + 3 × (x + 4) 1

= x2 + 4x + 3x + 12

= x2 + 7x + 12

Here, (x + 3) and (x + 4) are binomials. All binomials are polynomials.

Therefore, the area of the parking lot is (x2 + 7x + 12) sq. units.

Multiplying a Monomial by a Monomial To multiply two monomials, we simply find the product of their numerical coefficients and the variables. For example, let us multiply the two monomials: 4xy and 2y. 4xy × 2y = (4 × 2) × x × (y × y) = 8xy2.

Remember! All monomials, binomials, trinomials and quadrinomials are polynomials.

Constants and variables are put together.

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Example 1

Find the product of 5x2yz, (−2xy), 9xy2z. 5x2yz × (−2xy) × 9xy2z = (5 × (−2) × 9) × (x2 × x × x) × (y × y × y2) × (z × z) = −90x4y4z2

Example 2

Multiply (12a2b) by (−5abc3). Verify your answer if a = 1, b = −1, c = 2.

Remember!

(12a2b) by (−5abc3) = (12a2b) × (−5abc3)

In expressions with exponents, when bases are the same, powers are added.

= (12 × (−5)) × (a2 × a) × (b × b) × c3 = −60a3b2c3

Verification: (12a2b) × (−5abc3) = (12 × (1)2 × (−1)) × (−5 × (1) × (−1) × (2)3)

= −12 × (−5 × −8) = −12 × 40 = −480

−60a3b2c3 = −60 × (1)3 × (−1)2 × (2)3 = −60 × 1 × 1 × 8 = −480

Therefore, (12a2b) × (−5abc3) = −60a3b2c3. Example 3

Find the volume of a box of length (3a2b) m, breadth (2ab) m, and height (5ab2) m.

Do It Together

Find the product of 2a2b3c, −a2bc2, and 5a3b3c. 2a2b3c × (−a2bc2) × 5a3b3c

= (2 × (−1) × 5) × (a2 × a2 × a3) × (___ × ___ × ___) × (___ × ___ × ___)

Length = (3a2b) m, Breadth = (2ab) m, Height = (5ab2) m

= −______________________________

Volume = length × breadth × height = 3a2b × 2ab × 5ab2

= (3 × 2 × 5) × (a2 × a × a) × (b × b × b2) = 30a4b4

Multiplying a Monomial by a Polynomial We multiply each term of the polynomial by the monomial using the distributive property. Let us see how we can multiply a binomial and a trinomial by a monomial. Multiplying a Monomial by a Binomial Let us consider the binomial (3y + 1) and multiply each of its terms by the monomial (3x). Step 1: Represent each of the

polynomials using algebraic tiles.

polynomials in the same way as it is done in the multiplication chart. Then, cancel the unlike tiles (or zero pairs of tiles), if any. y

3x (3y + 1)

Step 2: Multiply the two

x x x

Chapter 16 • Multiplication of Algebraic Expressions

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Step 3: Calculate the total area of the rectangle so formed.

n multiplying, we count the O number of tiles of each term to find the area. Number of xy’s = 9 Number of x’s = 3 So, the area = 9xy + 3x

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Mathematically, 3x (3y + 1) = (3x × 3y) + (3x × 1) = 9xy + 3x Multiplying a Monomial by a Trinomial Let us consider the trinomial (2x2 – x + 1) and multiply each of its terms by the monomial (3y). This can be shown using the box multiplication method, as given below. 2x2

–x

3y 6x2y

–3xy

So, 3y(2x2 – x + 1) = 6x2y + (–3xy) + 3y

Remember!

= 6x2y – 3xy + 3y

In the box method of multiplication, we expand

We can also multiply a monomial by a trinomial horizontally as shown.

each factor, multiply using the grid, and then add all the products to get the final product.

3y (2x2 – x + 1) = (3y × 2x2) – (3y × x) + (3y × 1) = 6x2y − 3xy + 3y

Example 4

Find the product of:

1 7a2bc3 (6a2b – 11)

Example 5

2 5a2b3c5 (3a2b + 2abc3 – 5ab4c)

= (7a2bc3 × 6a2b) – (7a2bc3 × 11)

= (5a2b3c5 × 3a2b) + (5a2b3c5 × 2abc3) – (5a2b3c5 × 5ab4c)

= 42a4b2c3 – 77a2bc3

= 15a4b4c5 + 10a3b4c8 – 25a3b7c6

Multiply (7p3q3r) by (6 – 11p2q2r2). Verify your answer if p = 1, q = −1, r = −2. (7p3q3r) × (6 – 11p2q2r2) = (7p3q3r × 6) − (7p3q3r × 11p2q2r2) = 42 p3q3r – 77p5q5r3 Verification: (7p3q3r) × (6 – 11p2q2r2) = (7 × (1)3 × (−1)3 × (−2)) × (6 − 11 × (1)2 × (−1)2 × (−2)2)

= (7 × 1 × (−1) × (−2)) × (6 − 11 × 1 × 1 × 4) = (7 × 2) × (6 – 44) = 14 × (−38) = −532

42p3q3r – 77p5q5r3 = 42 × (1)3 × (−1)3 × (−2) – 77 × (1)5 × (−1)5 × (−2)3 = 42 × 1 × (−1) × (−2) – 77 × 1 × (−1) × (−8) = 42 × 2 – 77 × 8 = 84 – 616 = −532 Therefore, (7p3q3r) × (6 – 11p2q2r2) = 42p3q3r – 77p5q5r3. Do It Together

What is the product of 4x2y and (9x2y2z + 7)? 4x2y (9x2y2z + 7) = (4x2y × ______) + (4x2y × ______) = ______ + ______

Multiplying a Polynomial by a Polynomial For two or more terms in a polynomial, we multiply each term of one polynomial by each term of the other polynomial and then solve the products by putting like terms together to get the final product. Let us see how can we multiply two binomials, a binomial and a trinomial, two trinomials and so on. 218

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Multiplying a Binomial by a Binomial Multiply (x + 5) and (x + 7) To multiply a binomial by a binomial, we follow the given steps. Step 1: Represent each of the

Step 2: Multiply the two

polynomials using algebraic tiles.

polynomials in the same way as in the multiplication chart. Then, cancel the unlike tiles (or zero pairs of tiles), if any.

(x + 5)(x + 7)

Step 3: Calculate the total area of the rectangle so formed. On multiplying, we count the number of tiles of each term to find the area.

(x + 5)(x + 7) x

1 1 1 1 1 1 1

x 1 1 1 1 1

1 1 1 1 1 1 1

Number of x2’s = 1 Number of x’s = 12 Number of 1’s = 35 So, the area = x2 + 12x + 35

x x x x x x x

1 1 1 1 1

x x x x x

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Mathematically, we use the FOIL method.

The FOIL method is described in this way.

Outside

F: Multiply the First terms (x × x) of each binomial.

First

(x + 5)

O: Multiply the Outer terms (x × 7) of each binomial. I: Multiply the Inner terms (5 × x) of each binomial.

Inside Last

L: Multiply the Last terms (5 × 7) of each binomial. (x + 5) (x + 7) = (x × x) + (x × 7) + (5 × x) + (5 × 7) = x2 + 7x + 5x + 35 = x2 + 12x + 35

(x + 7)

Think and Tell Is this FOIL method similar to the BOX

Multiplying a Binomial by a Trinomial

method of multiplication? How?

Let us consider the trinomial (5x2 – 2x + 3) and multiply it with the binomial (2x − 1). This can also be shown using the BOX method of multiplication, as given below. Box Method

Using Distributive Property

5x2

–2x

10x3

–4x2

(2x – 1) (5x2 – 2x + 3)

= 2x (5x2 – 2x + 3) – 1 (5x2 – 2x + 3)

–1

–5x2

–3

= (2x × 5x2) – (2x × 2x) + (2x × 3) − (1 × 5x2) + (1 × 2x) − (1 × 3)

10x3 – 4x2 + 6x – 5x2 + 2x − 3 = 10x3 – 4x2 – 5x2 + 2x + 6x – 3 3

= 10x – 9x + 8x – 3

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= 10x3 – 4x2 + 6x – 5x2 + 2x − 3 = 10x3 – 4x2 – 5x2 + 2x + 6x – 3 = 10x3 – 9x2 + 8x – 3

219

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Multiplying a Trinomial by a Trinomial Let us consider the trinomial (x2 – 5x + 6) and multiply it by the trinomial (x2 + 2x − 1). This can also be shown using the BOX method of multiplication.

Box Method

Using Distributive Property

–5x

–5x3

6x2

2x3

–10x2

12x

–1

–x2

–6

(x2 + 2x − 1) (x2 – 5x + 6) = x2 (x2 – 5x + 6) + 2x (x2 – 5x + 6) – 1 (x2 – 5x + 6) = (x2 × x2) – (x2 × 5x) + (x2 × 6) + (2x × x2) − (2x × 5x) + (2x × 6) – (1 × x2) + (1 × 5x) – (1 × 6)

x4 − 5x3 + 6x2 + 2x3 – 10x2 + 12x – x2 + 5x − 6

= x4 − 5x3 + 2x3 + 6x2 – 10x2 – x2 + 12x + 5x − 6 = x4 − 3x3 – 5x2 + 17x – 6

We can also multiply the trinomials using the distributive property of multiplication.

= x4 − 5x3 + 6x2 + 2x3 – 10x2 + 12x – x2 + 5x − 6 = x4 − 5x3 + 2x3 + 6x2 – 10x2 – x2 + 12x + 5x − 6 = x4 − 3x3 – 5x2 + 17x – 6

Multiplying a Quadrinomial by a Trinomial Let us consider the quadrinomial (x4 + 2x2 – x + 5) and multiply each of its terms by each of the terms of the trinomial (x2 − 2x − 1). This can also be shown using the BOX method of multiplication. Box Method

Using Distributive Property

-x

2x4

-x3

5x2

-2x

-2x5

-4x3

2x2

-10x

–1

-x4

-2x2

-5

= x 6 + 2x4 − x3 + 5x2 – 2x5 – 4x3 + 2x2 – 10x – x4 − 2x2 +x−5

= x 6 – 2x5 + 2x4 – x4 − x3 – 4x3 + 2x2 + 5x2 − 2x2 – 10x +x–5 = x 6 – 2x5 + x4 − 5x3 + 5x2 – 9x – 5 Example 6

(x2 − 2x − 1) (x4 + 2x2 – x + 5)

= x 2 (x4 + 2x2 – x + 5) − 2x (x4 + 2x2 – x + 5) – 1 (x4 + 2x2 – x + 5)

= (x2 × x4) + (x2 × 2x2) − (x2 × x) + (x2 × 5) − (2x × x4) – (2x × 2x2) + (2x × x) − (2x × 5) − (1 × x4) − (1 × 2x2) + (1 × x) − (1 × 5) = x 6 + 2x4 − x3 + 5x2 – 2x5 – 4x3 + 2x2 – 10x – x4 − 2x2 +x−5

= x 6 – 2x5 + 2x4 – x4 − x3 – 4x3 + 2x2 + 5x2 − 2x2 – 10x +x–5 = x 6 – 2x5 + x4 − 5x3 + 5x2 – 9x – 5

Find the product. 1 (3x – 7) and (5x2 + 2x – 1)

2 (2p2 – 5p + 4) and (5p4 + p2 – 3p + 6)

(3x – 7) (5x2 + 2x – 1)

( 2p2 – 5p + 4) (5p4 + p2 – 3p + 6)

= 3x (5x2 + 2x – 1) – 7 (5x2 + 2x – 1)

=2 p2 (5p4 + p2 – 3p + 6) – 5p (5p4 + p2 – 3p + 6) + 4 (5p4 + p2 – 3p + 6)

= (3x × 5x2) + (3x × 2x) – (3x × 1) – (7 × 5x2) – (7 × 2x) + (7 × 1) = 15x3 + 6x2 – 3x −35x2 – 14x + 7 = 15x3 −29x2 – 17x + 7

=1 0p6 + 2p4 – 6p3 + 12p2 – 25p5 – 5p3 + 15p2 – 30p + 20p4 + 4p2 – 12p + 24 =1 0p6 – 25p5 + 2p4 + 20p4 – 6p3 – 5p3 + 12p2 + 15p2 + 4p2 – 30p – 12p + 24 = 10p6 – 25p5 + 22p4 – 11p3 + 31p2 – 42p + 24

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Example 7

If a train travels at (35a2 + 17a – 11) km/hour for (5a – 4) hours, how far has the train travelled? Speed of the train = (35a2 + 17a – 11) km/hour; Time = (5a – 4) hours Distance covered = Speed × Time = (35a2 + 17a – 11) × (5a – 4) = 35a2 (5a – 4) + 17a (5a – 4) – 11 (5a – 4) = (35a2 × 5a) – (35a2 × 4) + (17a × 5a) − (17a × 4) − (11 × 5a) + (11 × 4) = 175a3 – 140a2 + 85a2 – 68a – 55a + 44 = 175a3 – 55a2 – 123a + 44 So, the train travelled (175a3 – 55a2 – 123a + 44) km.

Do It Together

Simplify: (3x2 – 5x + 1) (x – 1) – 2x (2x + 3) (3x2 – 5x + 1) (x – 1) – 2x (2x + 3) = 3x2 (x – 1) – 5x (x – 1) + 1 (x – 1) – (2x × 2x) – (2x × 3) = (3x2 × _____) – (3x2 × _____) – (5x × _____) + (5x × _____) + x – 1 – 4x2 – _____ = ______________________________________________________________________________________________________________ = ______________________________________________________________________________________________________________ = ______________________________________________________________________________________________________________

Do It Yourself 16A 1

Look at the algebraic tiles. Identify the expressions. Draw and complete the multiplication chart of expressions. Write the product you get. a

1 1 1 1

–x –x

1 1 1

–1 –1 –1 –1 –1

–x

y y

Find the value of the polynomials using algebraic tiles.

Find the products.

a −5x × 4y

a −5x2y2z3, 12x2yz, 3yz 2

b ab (−6abc + 5c – 7)

c (3x – 1) (8y + 5)

e (2a − 3a + 7) (4a + 6a + 11) 2

g (p + 5p + 11p − 6) (2p + 5p + 10)

c (2y – 3)(3x + 4)

d (7y + 9) (3z + 6z – 5) 3

b 5x(–2x + 7)

f 9z (2z – 5)

h (8p − p + 7p + 10p – 5) (3p + 9p + 11)

Multiply and verify the result for the values of the variables. a (5h + 9) (3h – 7), for h = 1

b (9q – 1) (6q2 – q + 5), for q = −4

c (7a2 − 4a − 2) (5a2 − 2a + 9), for a = 0

d (2t3 + 3t2 + 5t − 4) (3t2 − 5t + 11), for t = −2

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Simplify. a

−

2 4 a (3a + 4b – 5c) − a (2a – 5b + 7c) 3 5

b −a2b (a3 – 3b + a + 2) – ab (b4 −2b2 − 3b) – c (a3 + b2 + 1) c (a – 5) (a4 – 2b3 + 3c – 4) +

3 bc (7a2 − 11) 7

Find the value of each of the results for a = −1, b = 2, c = 10.

Find the area of the rectangles using the lengths and breadths. a 6x2y, −5x3y2z

7 2 34 1 p q r , pqr, 2p (q + r) 2 2

7x+9 5

9 abc, 11

6 2 3 5 2 a bc + ab c 7 6

d (3z – 1), (2z2 + 7z – 10)

b (p + q), (p – q + 12), 8r

c –

1 pq (r + 1), 7pq, 11qr 2

Find the volume of the cubes using the edges (all measures are in centimetres). a 3s (stu + 1)

8x – 11 , 5

Find the volume of the cuboids using the lengths, breadths and heights (all measures are in centimetres). a

b −3s2t3u5

c (−6ab + 2a2 + 3)

d 4p3q4 (9r + 2q – 5)

Find the coefficient of the x4 term in the product of (5x3 − 4x2 + 2x − 1) and (x2 − 3x + 5).

Word Problem 1

Sara wants to build a rectangular garden of length (3x + 2) m and breadth (4x – 1) m. What is its area?

Applying Identities to Multiply We know that one side of the blue part of the parking lot is x units and one side of the green part is 1 unit, as shown.

To find the area of the parking lot, we can also apply identities.

x x+3

For this, let us first understand the meaning of identity and different kinds of identities that can be used.

x+4

We all know that an equation is a statement of equality that is true for only certain values of the variable in it. It is not true for all the values of the variable. But an identity is a statement of equality that is true for all the values of the variable. Let us understand the difference between the two using an example. Look at the given equalities.

1 1 1

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x2 + 5x + 6 = 0

x2 + 5x + 6 = (x + 2) (x + 3)

Let us evaluate this for some values of x, say x = 0, Let us evaluate this for some values of x, say 1, −1, 2, −2, … x = 0, 1, −1, 2, −2, … So, for x = 0, x2 + 5x + 6 = (0)2 + 5 × 0 + 6 = 6 For x = 1, x2 + 5x + 6 = (1)2 + 5 × 1 + 6 = 12 For x = −1, x2 + 5x + 6 = (−1)2 + 5 × (−1) + 6 = 2 For x = 2, x2 + 5x + 6 = (2)2 + 5 × 2 + 6 = 20 2

For x = −2, x + 5x + 6 = (−2) + 5 × (−2) + 6 = 0 and so on. From the above, it is clear that x2 + 5x + 6 = 0 is not true for all values of the variable in it. It is true for only one value of the variable, that is, x = −2. So, this is an equation. x+4 x

So, for x = 0, LHS = x2 + 5x + 6

= (0)2 + 5 × 0 + 6 = 6

RHS = (x + 2) (x + 3) = (0 + 2) (0 + 3) = 2 × 3 = 6 For x = 1, LHS = x2 + 5x + 6

= (1)2 + 5 × 1 + 6 = 12

RHS = (x + 2) (x + 3) = (1 + 2) (1 + 3) = 3 × 4 = 12

For x = 2, LHS = x2 + 5x + 6 = (2)2 + 5 × 2 + 6 = 20 RHS = (x + 2) (x + 3) = (2 + 2) (2 + 3) = 4 × 5 = 20 and so on.

Here, in each case, LHS = RHS.

So, it is clear that x2 + 5x + 6 = (x + 2) (x + 3) is true for all values of the variable in it. Thus, x2 + 5x + 6 = (x + 2) (x + 3) is an identity.

One of the most useful identities is (x + a) (x + b) = x (x + b) + a (x + b) = x2 + xb P+ ax + ab (x + a) (x + b) = x2 + x (a + b) + ab x+4

x a square 1 PQRS 1 1 of 1 length (a + b) units, as shown in the figure. 11 Consider 1

Here, PE = b, ES = a, PH = b, HQ = a

x+3

x Area of big square of side (a + b) = area of square PHFE + area of rectangle HQGF + area of square FGRI + area of rectangle FISE 1

(a + b)

Some more standard identities are derived below.

(side)2 = (side)2 + length × breadth + (side)2 + length × breadth

E P a S (a + b)

x+3

2 2 2 1 (a + b) = b + b × a + a + b × a

1 = b2 + ab + a2 + ab

(a + b)

b b2

(a + b) F ab b

H G Q a2 a

= a2 + ab + ab + b2 = a2 + 2ab + b2 Therefore, (a + b)2 = a2 + 2ab + b2.

Similarly, from the side PS, cut off a rectangle of length (a – b) units and breadth b units. So, here, PE = a – b, ES = b, PI = a – b, IQ = b Area of square PIFE = area of big square − area of rectangle IQRH − area of rectangle FHSE

2 Consider a square PQRS of length a units, as shown in the figure. From the side PQ, cut off a rectangle of length a units and breadth b units, as shown in the figure so that PI = a – b.

(a – b)2 (a – b)

b(a – b) a

E P b

b(a(a – b) – b)

I G Q b2 b R

(a – b)2 H (a – b)

a E

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(a – b)

b(a – b)

G b2223 R

H a–b

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(a – b)2 (a – b)

a E

b(a – b) F

b(a – b)

(side)2 = (side)2 – (area of rectangle IQGF + area of square GRHF) − length × breadth 2

(a − b) = a – [b × (a – b) + b ] − b × (a − b)

G R

= a2 – [ab – b2 + b2] – ab + b2 = a2– ab – ab + b2 = a2 − 2ab + b2 Therefore, (a − b)2 = a2 − 2ab + b2. 3 Consider a square PQRS of length a units, as shown in the figure. From this square, cut off a square of length b units such that PE = EF = (a – b) units.

a–b

Area of square PQRS – area of FGRH = area of rectangle PEHS + area of rectangle EQGF 2

b(a – b)

a(a – b) F

(side) − (side) = length × breadth + length × breadth

a2 − b2 = [a × (a – b)] + [b × (a − b)]

b2 b

Q a–b

So, here, PS = a, EQ = b, QG = a – b, HR = b

G R

= (a – b) (a + b) Therefore, a2 − b2 = (a – b) (a + b). Now, let us find the area of the parking lot using one of these identities. Area of the parking lot = area of rectangle = length × breadth = (x + 4)(x + 3) Let us consider a = 4 and b = 3. So, we apply identity: (x + a)(x + b) = x2 +x (a + b) + ab 2

(x + 4) (x + 3) = x + x (4 + 3) + (4 × 3) = x + 7x + 12

Think and Tell Can you use the distributive property to derive these identities? How?

Therefore, the area of the parking lot is (x2 + 7x + 12) sq. units.

Using Identities Now, let us learn how to apply these identities. For example, let us find the product of (x + 6) and (x + 6).

Did You Know?

(x + 6) × (x + 6) = (x + 6)2

Algebraic identities play a

Here, a = x and b = 6. So, we apply (a + b)2 = a2 + 2ab + b2.

cryptography, where they are used

crucial role in coding theory and

So, (x + 6) = x + 2 × x × 6 + 6 = x + 12x + 36

to design error-correcting codes and encryption algorithms.

Therefore, (x + 6)2 = x2 + 12x + 36.

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Example 8

Find the product of the given binomials. 2 (2p + 3q) (2p – 3q)

Here, a = 6, b = 8.

Here, a = 2p, b = 3q.

So, we apply (x + a) (x + b) = x2 + x (a + b) + ab

So, we apply a2 − b2 = (a – b) (a + b).

(x + 6) (x + 8) = x 2 + x (6 + 8) +6×8

(2p + 3q) (2p – 3q) = (2p)2 – (3q)2

= x + 14x + 48

1 1 y, b = z. 2 3 So, we apply (a − b)2 = a2 − 2ab + b2 Here, a =

1 1  1 1   y − z  y − z 3  2 3  2

= 4p2 – 9q2

2 1  1  1  1  =  y  − 2 ×  y  ×  z +  z 2  3  3  2 

= Example 9

1 1  1 1   y − z  y − z 3  2 3  2

1 (x + 6) (x + 8)

Using identities, evaluate.

1 2 1 1 y – yz + z2 4 3 9

1 (59)2

2 99 × 102

(59)2 = (60 – 1)2

99 × 102 = (100 – 1) (100 + 2) = (100 + (−1)) (100 + 2)

Here, a = 60, b = 1. So, we apply (a − b)2 = a2 − 2ab + b2 Here, x = 100, a = −1, b = 2. So, we apply

Example 10

(60 – 1)2 = (60)2 − 2 × 60 × 1 + (1)2

(x + a) (x + b) = x2 + x (a + b) + ab

= 3600 – 120 + 1

(100 + (−1)) (100 + 2) = (100)2 + 100 (−1 + 2) + (−1 × 2)

= 3481

= 10,000 + 100 − 2 = 10,000 + 100 – 2 = 10,098

If x +

1 = 10, find the value of: x 1

1 x2 + x+

2 x4 +

1 = 10 x 2

 1 2  x +  = (10) x  

x2 + x2 +

x2 1

So, x2 +

+2×x×

1 = 100 x

 2 1  2  x + 2  = (98) x   Using the identity: (a + b)2 = a2 + 2ab + b2, we get x4 + x4 +

= 100 – 2

x4 +

= 98

Chapter 16 • Multiplication of Algebraic Expressions

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= 98 (from a.) x2 Squaring both the sides, we get

+ 2 = 100

Using the identity: (a + b)2 = a2 + 2ab + b2, we get 1

Now, as x2 +

Squaring both sides, we get

x2 +

x4 1

So, x4 +

+ 2 × x2 ×

= 9604

+ 2 = 9604 = 9604 – 2 1

= 9602

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Do It Together

Find the value of the expression 16y2 + 48xy + 36x2, when x = 1, y = 2. 16y2 + 48xy + 36x2 = (4y)2 + 2 × 4y × 6x + (6x)2 Here, a = 4y, b = 6x So, (____)2 + 2 × ____ × ____ + (____)2 = (____ + ____)2

Error alert! NOT all algebraic expressions can be solved using identities. For example,

For x = 1, y = 2, (____ + ____)2 = (____ × 2 + ____ × 1)2 = (____ + ____)2 = ____2 = ____

(5x + 2) (3x – 1)

Therefore, the value of 16y2 + 48xy + 36x2 is ____.

Can be solved using distributive property.

Simplifying Expressions Using identities, we can simplify complex expressions. For example, let us simplify the expression (3a + 5)2 – (3a – 5)2 using identities.

(3x – 1) (3x – 1) (3x – 1) (3x – 1) = (3x – 1)2

Here, a = 3a and b = 5 in each pair of brackets. So, we apply two identities: (a + b)2 = a2 + 2ab + b2, (a − b)2 = a2 − 2ab + b2 (3a + 5)2 – (3a – 5)2 = (3a)2 + 2 × 3a × 5 + 52 – ((3a)2− 2 × 3a × 5 + 52) = 9a2 + 30a + 25 – (9a2 − 30a + 25) = 9a2 + 30a + 25 – 9a2 + 30a − 25 = 60a Example 11

Simplify: (5x2 + 3y2)2 – 6x2y2 Here, a = 5x2, b = 3y2. So, we apply (a + b)2 = a2 + 2ab + b2 (5x2 + 3y2)2 – 6x2y2 = (5x2)2 + 2 × (5x2) × (3y2) + (3y2)2 = 25x4 + 30x2y2 + 9y4 Now, (5x2 + 3y2)2 – 6x2y2 = 25x4 + 30x2y2 + 9y4 – 6x2y2 = 25x4 + 30x2y2 – 6x2y2 + 9y4 = 25x4 + 24x2y2 + 9y4

Example 12

Find the continued product of (3x + y) (3x – y) (9x2 + y2) (81x4 + y4). (3x + y) (3x – y) (9x2 + y2) (81x4 + y4) = [(3x + y) (3x – y)] (9x2 + y2) (81x4 + y4) Here, a = 3x, b = y. So, we apply a2 − b2 = (a – b) (a + b). [(3x + y) (3x – y)] (9x2 + y2) (81x4 + y4) = [(3x)2 – y2] (9x2 + y2) (81x4 + y4) = (9x2 – y2) (9x2 + y2) (81x4 + y4) Similarly, (9x2 – y2) (9x2 + y2) (81x4 + y4) = [(9x2 – y2) (9x2 + y2)] (81x4 + y4) = [(9x2)2 – (y2)2] (81x4 + y4) = (81x4 – y4) (81x4 + y4) Similarly, (81x4 – y4) (81x4 + y4) = [(81x4)2 – (y4)2] = 6561x8 – y8 So, (3x + y) (3x – y) (9x2 + y2) (81x4 + y4) = 6561x8 – y8

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Do It Together

Simplify: (p2 + 2p + 2) (p2 − 2p + 2) (p2 + 2p + 2) (p2 − 2p + 2) = p2 (p2 − 2p + 2) + 2p (p2 − 2p + 2) + 2 (p2 − 2p + 2) = (___ × ___) – (___× 2p) + (___× ___) + (___ × ___) – (___ × 2p) + (___ × ___) + (2p × ___) – (2 × 2p) + (2 × 2) = ______________________________________________________________________________________________________________ = ______________________________________________________________________________________________________________

Do It Yourself 16B 1

Find the products of the polynomials. a (6a + 7b) (6a + 7b)

3 2 7 2 a + b 5 6

x4 +

1 2 1 2 p − q 6 5

1 2 1 2 p − q 5 6

p8 − 1 p8

x4 +

c (4.7x + 3.4y) (4.7x + 3.4y) 2 11

3p –

p8 − 1 p8

2x− 1y 3 5

3p –

2 11

2x− 1y 5 3

Use a suitable identity to evaluate the expressions. a (85)2

b (192)2 c (108)2 d 88 × 114

f 5.2 × 6.8

5 1 a + b 6 2

Find the value of the expressions when x =

7 2 4 x − y 8 5

4 5 ,y= − . 5 6

b 9x2 + 30xy + 25y2

6 11

5p –

c 36x2 + 60xy + 75y2

e (4.5x + 1.7y)2

d x2 + 4xy + y2

Simplify: 3 2 5 2 a + b 4 6

3 2 5 2 a − b 7 8

a (5a – 4b)2 – (7a + 2b)2 + 4a2 – 6ab

c (p2 + q2) (p2 – q2) + (q2 + r2) (q2 − r2) + (r2 + p2) (r2 − p2)

d (a + b)(a – b)(a2 + b2)

– 3a2b2

Find the continued product of: 3 4

x2 +

9 16

If x +

1 = 7, then find the value of: x

If x −

1 = x

a x2 +

x−

3 4

a x2 +

a 9x2 + 42xy + 49y2

b (9a – 5b) (9a – 5b)

b (x + 7) (x – 7) (x2 + 49)

b x4 +

 1  2 1   x + 2   x + x  x   

x8 +

11 , then find the value of:

b x4 +

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If x2 + y2 = 15 and xy = 5, then find the value of: a x+y

b x–y

c x4 – y4

1 1 xy (9x – 5y)2 – xy (9x + 5y)2 2 2

e x8 – y8

(x – y)2 (x + y)2

Verify the equalities using identities. a (3p + 8)2 – 96p = (3p – 8)2

b (5ab + 7b)2 – (5ab – 7)2 = 140ab2

c (2x + 5y)2 – (2x – 5y)2 – 40xy = 0

10 Find the value of x if 4x = 982 – 382. 11 Using identities, find the value of: 2

(34.12) – (22.90)

21.56 −10.34

( 4.57 + 7.81) c p2 + a2, if p + a = 12 and pa = −2 32.58 −20.20

Word Problem 1

Jack went to a bookshop and bought a book. While at the store, he found another interesting book for ₹12 less than four times the price of the first book. If the number of second books he bought is

as many as the price of the first book, then what is the total price he must pay for the books? Write your answer as an expression.

Points to Remember • To multiply two monomials, we simply find the product of their numerical coefficients and the variables. • To multiply a monomial by a polynomial, we multiply each term of the polynomial by the monomial using the distributive property. • To multiply two polynomials, we multiply each term of one polynomial by each term of the other polynomial and then solve the products by putting like terms together to get the final product. • An identity is a statement of equality that is true for all the values of the variable.

Math Lab Algebraic Expressions Storytelling Setting: In groups of 3 Materials Required: Blank A4-sized paper for each group, markers or coloured pens, a list of algebraic expressions and identities Method: All 3 members of each group must follow these steps.

1 Each group collects a story prompt related to algebraic expressions from the teacher. For example, in a magical forest, there were 3x trees, and each tree had 2y birds living in it. Write a story about what happens in the forest. (Each story must have the expressions and identities including their solutions.

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Complete the story within the time limit set by the teacher.

3 After completing the story, present your story to the class one by one within the time limit set by the teacher. (This time limit should be less than the time limit for writing the story.) 4 The group who completes the story and narrates it as well in the least time, wins the storytelling competition.

Chapter Checkup 1

Find the products. 2 4 2 2 3 a 6xy z , −11x yz , 3y z

2 b (4y + 7)(2z + 5z – 6)

3 2 2 c (−2p + 8p + 10p − 5)(p + 3p + 9)

d st(−5stu + 3u – 9)

2 2 e (7a − 2a + 8)(3a + 5a + 12)

5 3 2 2 f (6p − 4p + 2p + 13p − 1)(4p + 7p + 12)

Multiply and verify the result for the values of the variables. a (9h + 4)(2h – 11), for h = 2

2 2 b (2a − 5a − 8)(4a − a + 3), for a = −5

2 c (7q – 8)(3q – 4q + 7), for q = 0

3 2 2 d (2t + 5t + 7t − 10)(3t − 10t + 15), for t = −2

Simplify. Find the value of each of the results for a = −1, b = 2, c = 10. 1 5 a(3a – 5b + 12c) a − a(8a + 5b – 11c) − 5 6 2 3 4 2 3 2 b −2a b(5a – 7b + a + 11) – ab(9b −3b − 4b) – c(2a + 2b + 1) 4 3 c (a – 7)(a – 6b + 4c – 5) +

3 bc (9a2 − 14) 7

Find the area of the rectangles using the lengths and breadths. 2 3 2 a 7x y, 8x y z

2x –

2 d (5z – 3), (2z + 9z – 11)

7 p2q3r4, 2 pqr, 2p (q + r) 11 7

9 abc, 13

4 2 3 1 a bc + ab2c 7 6

b (p + q), (2p – 3q + 14), 7r

c −

1 pq (r + 5), 9pq, 10qr 2

2 3 5 b 4s t u

2 c (−7ab + 11a + 2)

3 4 d 3p q (2r + 5q – 6)

Find the products of the polynomials. a (8a + 11b) (8a + 11b)

2 2 7 2 a + b 5 6

Chapter 16 • Multiplication of Algebraic Expressions

UM24CB0817.indd 229

Find the volume of the cubes using the edges (all measures are in cm). a 5s (stu + 2)

9 x + 11 4

Find the volume of the cuboids using the lengths, breadths, and heights (all measures are in cm). a

10 , 7

2 2 7 2 a + b 6 5

3 2 1 2 p − q 6 5

3 2 1 2 p − q 5 6

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7 11

2p –

4 1 x− y 5 3

7 11

1 4 x− y 3 5

x4 +

x4 −

h (11a – 7b) (11a – 7b)

p8 −

(3.5x + 7.4y) (3.5x + 7.4y)

p8 +

Use suitable identities to evaluate the given expressions. 2 a (88)

6.4 × 5.8

2 b (212)

2 c (53)

2 g (197)

2 e (2.2x + 3.3y)

d 45 × 124

1 8 a+ b 2 11

Find the value of the expressions when x =

−4 2 , y= . 9 3

2 2 a 16x + 64xy + 64y

2 2 b 4x + 20xy + 25y

3 2 5 x − y 8 11

3p –

5 11

2 2 c 49x + 84xy + 36y

10 Simplify. 2 2 2 a (2a – 3b) – (5a + 6b) + 6a – 11ab

1 2 1 2 a + b 4 6

11 Find the continued product of: a

4 5

x−

4 5

(x2 +

16 25

2 2 3 2 a − b 8 7

12 If x +

1 1 1 = −5, find the value of: a x2 + 2 b x4 + 4 x x x

13 If x −

1 = x

– 5a2b2

2 b (x + 8) (x – 8) (x + 64)

2 15 , then the value of: a x +

4 b x +

14 If x2 + y2 = 17 and xy = −4, find the value of: a x+y

b x–y

4 4 c x –y

1 xy (7x – 8y)2 – 1 xy (7x + 8y)2 2 2

8 8 e x –y

Word Problems 1

A construction company is working on a project. They need to find the total cost of materials for building a rectangular structure with a length of (5x + 3) m and a width of (2x − 1) m. If each square metre of material costs ₹100, what is the total cost?

A scientist is studying the growth of a population of rabbits over time. The population of rabbits at the start is represented by the polynomial (2x3 − 3x2 + 5x) and the population

growth rate per year is represented by the polynomial (x + 1). Find the population after two years.

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Visualising Solid Shapes

17 Let's Recall

Everywhere we go, we see a number of different objects in different shapes. We have already learnt about 2-D shapes that have only length and breadth. These 2-D shapes further make 3-D shapes that have length, breadth and height. The faces of all the solid shapes are 2-D shapes. Let’s recap some properties of some solids: If the two end faces of a solid figure are parallel and congruent, then the solid is called a prism. A solid with a polygonal base and flat triangular faces that join at a common vertex is called a pyramid.

A triangular prism has 5 faces, 9 edges, and 6 vertices.

A triangular-based pyramid has 4 faces, 4 vertices including the apex, and 6 edges.

A square/rectangular prism has 6 faces, 12 edges, and 8 vertices.

A square-based pyramid has 5 faces, 5 vertices including the apex, and 8 edges.

Letʼs Warm-up

Identify the type of prism or pyramid based on the clues. 1 I have a triangular base, with 3 triangular faces meeting at a vertex. _____________________________ 2 I have two parallel and congruent square faces. _____________________________ 3 I have a square base with 4 flat triangular faces. _____________________________ 4 I have two triangular bases and 3 rectangular faces. _____________________________ I scored _________ out of 4.

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Mean, andPolyhedrons Mode Views,Median Maps and Real Life Connect

Aradhya’s mother took her to a toy shop to buy her a birthday gift. On the racks were toys of different solid shapes. Mother: Do you also want this cone-shaped birthday cap? Aradhya: Yes mom. But why did it look like a triangle when it was on the rack? It now looks like a circle when it is put in the shopping bag. Mother: This is because you are now looking at it from a different angle!

Views of 3-D Shapes 3-D Shapes 3-D shapes are solids with three measurements, i.e. length, breadth and height. For example, cubes, cuboids, cones, cylinders, and spheres are all solid shapes. Sometimes 2 or more solid shapes are combined to form a new shape. Let us see some of them.

A flower vase has a sphere surmounted by a cylinder.

A hut has a cylinder surmounted by a cone.

An ice cream has a cone surmounted by a hemisphere.

A house has a cuboid surmounted by a triangular prism.

The 2-D shapes in 3-D shapes help us draw the 3-D shapes from different views. These solids look different when we look at them from different sides. So, every solid can be drawn from different perspectives. For example, here are the side, front and top views of a tractor.

Side view

Front view

Top view

Now, let us have a look at this cup. What do you think is the front view, side view and top view of the cup? Top view

Front View

Side View

Top View

Side view Front view

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Let us now see the front view, the side view and the top view of some 3-D shapes. Top View

Square Pyramid

Cube

Front View

Side View

Front View

Top View

Can you draw the views of other 3-D shapes? We can draw the image of these blocks from different perspectives. Top View

Front View

Top View

Side View

Side View Front View Example 1

Identify the combination of solids. Draw its front, side and top view. Solid

Front View

Side View

Top View

It has a cone mounted on a cylinder. Example 2

Identify the top view, front view and side view of the given object.

Top View Example 3

Front View

Side View

Draw the front view, side view and the top view of the cube structure. Solid

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Front View

Side View

Top View

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Do It Together

Draw the front view, side view and top view of the solid. Solid

Front View

Side View

Top View

Do It Yourself 17A 1 Identify the 3-D shapes of which these figures are made. a

2 For each object, one of the views is given. Identify the view. b

___________ view

3 Draw the front, side and top views of the 3-D shapes. a

4 Draw the 3 views of the solid figures made with two or three 3-D shapes. a

Draw the front, side and top views of each of the solid figures. a

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Word Problem 1

Sara has a little brother who is playing with blocks. She arranges

6 cubes in a shaped block like that shown in the figure. If she places 2 more blocks following the same pattern, what will be the front, side and top view of the solid obtained? Draw the views to show.

Mapping Space Around Us Aradhya and her mother, after finishing their shopping at the supermarket, decide to drop by the neighbourhood children’s park before going home. But, they do not know how to get there. Aradhya: Mom, what should we do now? Mother: Don’t worry Aradhya! We can use a map. Let us understand what a map is. A map is a symbolic representation that emphasises connections between objects or places. We have been using maps ever since we were little. Maps can be of different types, like route maps, city maps, country maps etc. A map has no reference or perspective, meaning objects closer to the observer are depicted as the same size as those farther away. So, we use a ‘scale’, which is the ratio of the distance on the paper and the actual distance on the ground.

Remember! Maps are not always very accurate.

Think and Tell

Will it be useful to have the same scale for a route map and a city/country map?

But, how do we read a map and what can we understand while reading a map? Aradhya’s mother took a route map out of her bag. Let us find out from the map how can they reach the children’s park from the supermarket. On the map, the distance between the supermarket and the children’s park is 1.5 cm + 1.2 cm = 2.7 cm. On the scale we have 1 cm = 5 km. So, 2.7 cm = 2.7 × 5 = 13.5 km. So, the children’s park is 13.5 km from the supermarket.

Chapter 17 • Visualising Solid Shapes

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4 cm

Scale: 1 cm = 5 km

1.5 cm

Supermarket

1 cm

Hospital

1.2 cm 2 cm

1.2 cm

1.5 cm

Home

School

1 cm

Children Park

2.2 cm

5 cm

1.3 cm

4 cm

1.6 cm 2.3 cm

Cafe

1.1 cm

3 cm

Police Station

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Example 4

Read the above map and answer the given questions.

Think and Tell

1 What is the shortest route from home to school?

How is a map different from a

From the map, we can see that, the school is 2.2 cm from home. So the shortest route from home to the school is 2.2 × 5 = 11 km.

picture?

2 Which is the shortest route from the cafe to the hospital? How much is the distance? The shortest route from the cafe to the hospital is via home.

Distance on the map = 1.3 + 2 = 3.3 cm. Actual distance = 3.3 × 5 = 16.5 km. 3 Which place is the farthest from home?

The police station is 4 cm from home. So, the farthest place is the police station which is 4 × 5 = 20 km from home. Example 5

The scale on a map is 0.1 cm = 10 m. Find the distance on the map for an actual distance of 1 km. Distance on the map for an actual distance of 10 m = 0.1 cm. Distance on the map for an actual distance of 1000 m =

0.1 10

× 1000 =

Thus, the distance on map for an actual distance of 1 km is 10 cm. Do It Together

100

× 1000 = 10 cm.

Read the map given above. Find the actual distance between the places using the scale. 1 School to Café

2 C hildren’s Park to the School via the Supermarket and Hospital.

On the map, the distance from the school to the café = 1.6 + 2.3 = _________

O n the map, the distance from the children's park to the school = ______________________

Scale = 1 cm = 5 km Actual distance = _________ × 5 = _________

Actual distance = _________.

Do It Yourself 17B

Sport’s Ground

1 cm

14 cm

Library

Garden

Book Store

2 cm

the main building?

16 cm

2 cm

d Is the library or the sport’s ground nearer to

0.6 cm

2.5 cm

c Is the library or the garden nearer to the main

2 cm

b Which place is nearest to the main building? building?

Reception

5 cm

a Which place is the farthest from the main building?

Scale: 1 cm = 10 m

Gate

3.6 cm

Read the map and answer the following questions.

1 cm

1.3 cm

2 cm

Main Building

Canteen Restroom

Look at the map given in Q1. Find the actual distance between: a the garden and the main building.

b the main building and the reception.

c the garden and the restroom.

d the gate and the sports ground.

The scale of a map is 1 cm = 5.5 km. Find the actual distance for the distance shown on the map in cm as given below. a 12.5 cm

b 4.7 cm

c 7.3 cm

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The scale on a map is 1 mm = 6 m. Find the distance on the map for an actual distance of 72 m.

Draw a map for the following: a Your classroom using symbols for different areas and objects. b The way from your house to the nearest market.

A running trail from Sector 60 Gurgaon to the Damdama lake is 16.5 km. What would be the distance on the map in cm, if the scale is 1 mm = 0.5 km?

Word Problem 1

Sudha invited her friends to her birthday party. She decided to hand out the invitation cards with

a map giving instructions on to how to reach her house on the day of the party. Imagine that you are Sudha and draw the required map.

Polyhedrons We have learnt in previous grades that all two-dimensional figures with a countable number of sides are called polygons. The polygons in which all the interior angles are strictly less than 180° are called convex polygons.

Convex Polygon

The polygons in which all the sides are of the same measurement are called regular polygons.

Regular Polygon

Concave Polygon

Irregular Polygon Face

We also know that The flat surface of a 3-D shape is called a face.

Vertex

The corner of a 3-D shape is called a vertex.

Edge

The side of a 3-D shape is called an edge.

Think and Tell

The 3-D shapes with flat polygonal faces, straight edges

A regular polygon is always convex in

and vertices are called polyhedrons.

nature.

Let us look at the figures and identify their faces, vertices and edges. E

Faces: Squares ABDC, BHFD, HGEF, GACE, ABHG and CDFE.

C G A

H B

ABDCEGHF is a cube. Edges: AB, BD, DC, CA, AG, GH, HB, HF, FD, FE, EC, EG Vertices: A, B, C, D, E, F, G and H. E

Chapter 17 • Visualising Solid Shapes

B D

237 C

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G A

ABCDFE is a triangular prism. Faces: Rectangles ABCD, AEFD, CFEB, and triangles CDF, ABE. Edges: AB, BC, CD, DA, EF, DF, FC, AE and EB.

B A

Vertices: A, B, C, D, E and D F. D

B D

C C

Faces: ∆ABC, ∆ABD, ∆BCD and ∆ACD.

C A

ABCD is a triangular pyramid.

C A

Edges: AB, BC, CA, BD, CD and AD. B Vertices: A, B, C and D.

• If all the faces in a 3-D shape, are convex polygons, then the polyhedron is said to be a convex polyhedron.

• If all the faces in a 3-D shape, are regular polygons, then the polyhedron is said to be a regular polyhedron.

Figure 1

Figure 2

Figure 3

Figure 4

Figure 1, 2, 3 and 4 are examples of convex polyhedrons. Figure 2 and 4 are examples of regular polyhedrons. Figure 1 and 3 are examples of irregular polyhedrons.

Remember! If even one face is not a convex polygon, then the polyhedron is not convex.

Think and Tell

Is it possible to have a polyhedron with 7 edges?

Prisms A prism is a three-dimensional polyhedron, in which the end faces or the bases are identical and parallel and the lateral faces are parallelograms.

When the base of a prism is a

When the lateral edges in a prism are

prism.

is said to be a right prism.

said to be an oblique prism.

regular polygon, it is called a regular perpendicular to each other, the prism not perpendicular, then the prism is

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Rectangular Lateral Faces Triangular Bases

Square Bases 4 rectangular

3 rectangular

lateral sides

Triangular Prism

Square Prism

5 faces, 9 edges and 6 vertices

6 faces, 12 edges and 8 vertices

Rectangular Bases

Pentagonal Bases

5 rectangular

4 rectangular

lateral sides

Rectangular Prism

Pentagonal Prism

6 faces, 12 edges and 8 vertices

7 faces, 15 edges and 10 vertices

Pyramids A pyramid is a polyhedron with a polygonal base and triangular lateral faces with common vertices which join at a common point called the apex.

Think and Tell

What do all prisms have in common?

Triangular Lateral Faces Triangular Base 3 triangular

lateral faces

Square Base 4 triangular

lateral faces

Triangular Pyramid

Square Pyramid

4 faces, 6 edges and 4 vertices

5 faces, 8 edges and 5 vertices

Rectangular Base

Pentagonal Base

4 triangular

lateral faces Rectangular Pyramid

5 faces, 8 edges and 5 vertices

5 triangular

lateral faces Pentagonal Pyramid

6 faces, 10 edges and 6 vertices vertex vertex vertex vertex face face edgeedge face face edge edge

Chapter 17 • Visualising Solid Shapes

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vertex

Regular Octahedron A polyhedron with 8 lateral triangular faces that have equal dimensions is called a regular octahedron.

edge

face

We can also say that it is a combination of two square pyramids. vertex

It has 8 faces, 12 edges and 6 vertices. face edge

Example 6

Non¯polyhedrons When a solid has at least one curved face, it is said to be a non-polyhedron.

Think and Tell

Cylinders, spheres and cones are such examples.

called?

What are prisms with 7, 8, 9 and 10 edges

A polyhedron has an octagon as the base and all the other lateral faces are triangular in shape. Name the shape. Write the number of faces, vertices and edges in the shape. Since the base is an octagon (8 sides) and the lateral faces are triangular, the given polyhedron is an octagonal pyramid. It has 9 faces, 9 vertices and 16 edges.

Example 7

Name and write the differences between the two figures. The figure is a hexagonal prism.

It has 2 identical hexagonal bases and 6 rectangular lateral faces. There are 8 faces, 18 edges, and 12 vertices. Do It Together

The figure is a hexagonal pyramid. It has 1 hexagonal base and 6 triangular lateral faces. There are 7 faces, 12 edges, and 7 vertices.

A polyhedron has two octagonal bases that are parallel and the rest of the lateral faces are rectangular in shape. What is the name of the polyhedron? How many faces, vertices and edges does it have? The polyhedron is an ____________________________________________. Vertices = __________

Faces = 10

Edges = __________

Do It Yourself 17C 1

Fill in the blanks.

a If all the faces of a polyhedron are regular, then it is a ____________ polyhedron.

b A rectangular prism is also called a ____________.

c The polygonal regions forming a polyhedron are called ____________.

d A triangular pyramid is also called a ____________.

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Can a polyhedron have these as their faces?

What is the minimum and the maximum number of faces that these shapes can have?

If we have four congruent equilateral triangles, what more do we need to make a pyramid?

If ABCDEFGHIJKL is a hexagonal prism, write the names of its faces, edges and vertices.

A pyramid has the base as a convex heptagon. Label the vertices and name the faces, edges and vertices.

Is a cylinder a polyhedron? Why or why not?

a Three triangles

b Four triangles

a Prism

c A square and four triangles

b Pyramid

Word Problem 1

Sam built a solid with a decagon and some triangles and made a pyramid-like solid. How many faces, edges and vertices does this solid have?

Euler’s Formula We have learnt about the features of different polyhedrons. Euler's Formula helps us identify if a solid is a polyhedron or not. Let us see the number of faces, vertices and edges in some solids. Faces (F)

Vertices (V)

Edges (E)

F + V – Edges

Triangular Prism

5+6–9=2

Cuboid

6 + 8 – 12 = 2

Pentagonal Prism

7 + 10 – 15 = 2

This proves that for any solid shape, the sum of its faces and vertices is always 2 more than the number of edges. This is called Euler’s Formula Euler’s Formula: Faces + Vertices – Edges = 2 F+V-E=2

Example 8

A polyhedron has 14 faces and 24 vertices. How many edges does the polyhedron have? F = 14; V = 24 Euler’s formula: F+V-E=2

14 + 24 - E = 2 → 38 - E = 2 → E = 38 - 2 = 36 Therefore, the polyhedron has 36 edges.

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Did You Know? The famous Swiss mathematician Leonard Euler was the one who discovered Euler’s Formula.

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Find the number of edges in a polyhedron that has 2 hexagonal faces and 6 rectangular lateral faces.

Example 9

Example 10

We know that in a hexagonal pyramid:

We know that a hexagonal prism has 2 hexagonal faces and 6 rectangular lateral faces.

Faces (F) = 7, Vertices (V) = 7, Edges (E) = 12

Euler’s formula: F + V - E = 2

F+V-E

Euler’s formula: F + V - E = 2

So, Vertices (V) = 12, Faces (F) = 8, Edges (E) = ? 8 + 12 - E = 2 ⇒ 20 - E = 2

= 7 + 7 - 12 = 14 - 12 = 2

20 - 2 = E ⇒ E = 18

S ince F + V – E = 2, Euler’s formula is verified.

Therefore, the hexagonal prism has 18 edges. Do It Together

V erify Euler’s formula for a hexagonal pyramid.

Using Euler’s Formula, fill in the blanks. Faces

Vertices

Edges

Name of the Polyhedron

__________

Do It Yourself 17D 1 How is a prism different from a pyramid? 2 Verify Euler’s formula for a: a Triangular prism

b Nonagonal pyramid

c Regular octahedron

3 Find the number of faces, vertices or edges for polyhedrons with certain features. a 5 faces and 9 edges

b 8 faces and 12 vertices

c 9 faces and 16 edges

d 20 faces and 12 vertices

If a prism has a base polygon of n sides, find the number of faces, vertices and edges.

Verify Euler’s formula for the given figures. a

Word Problem 1

Daman was making a polyhedron using cardboard and cellotape. He ended up cutting 10

polygons and decided to make a polyhedron with 10 faces, 20 edges and 15 vertices. Verify if such a polyhedron can be made using Euler’s formula.

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Points to Remember • 3-D shapes have length, breadth and height. • Each solid looks different when looked at from a different angle. • A map is a symbolic representation that emphasises connections between objects or places. • Perspective is not important when drawing a picture but it is a very important factor in a map. • The 3-D shapes with flat polygonal faces, straight edges and vertices are called polyhedrons. • A prism is a three-dimensional polyhedron, in which the end faces or the bases are identical and parallel and the lateral faces are parallelograms. • A pyramid is a polyhedron with a polygonal base and triangular-shaped lateral faces with common vertices which join at a common point called the apex. • For any solid shape, the sum of its faces and vertices is always 2 more than the number of edges. F+V–E=2

Math Lab Finding the Polyhedron Setting: Individual Materials Required: Cards with numbers written on them in a box. Method:

1 Each student will pick up one card for the number of faces of a polyhedron and a second card for the number of vertices of the polyhedron. They will then apply Euler’s formula and find the number of edges, if the polyhedron 2 exists. If the polyhedron exists, write the name of the polyhedron. 3 4

The first one to solve and verify it wins.

Chapter Checkup 1

Draw the top, side and front views of the objects. a

Draw the front, side and top views of the solid figures made with two or three 3-D shapes. a

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SAM 8 cm

HOSPITAL

AMAN ANSH

WELL

PARK

FARM

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3 Draw the front, side and top views of each of the solid figures. Also find the number of squares visible from each view.

PARK

SAM

PARK SAM AMAN ANSH PARKSAM SAM AMAN ANSH PARK SAM AMAN ANSH PARK AMAN ANSH 8 cm

a What is the distance between the dairy and the railway station

WELL FARM HOSPITAL 8 cm8 cm 8 cm8 cm WELL FARM WELL FARM WELL FARM WELL FARM HOSPITAL HOSPITAL HOSPITAL HOSPITAL 3 cm

on the map?

b Who lives nearest to the park?

7.5 cm

c What is the actual distance between the park and the railway station?

7.5 cm 7.5 cm

3 cm 3 cm 7.5 cm 7.5 cm

3 cm3 cm

POST RAILWAY OFFICE POST STATION RAILWAY RAILWAY POST POST DAIRY POST RAILWAY RAILWAY OFFICE OFFICE STATION STATION DAIRY OFFICE OFFICE STATION STATION DAIRY DAIRY DAIRY

d What is the actual distance between the park and Ansh’s house? e Sam cycled from the hospital to the dairy. How far did he cycle?

Assume that the scale of the map given in Q5 is 1 mm = 300 m. Find the distance on the map for an actual

What is the minimum number of faces that a polyhedron can have?

What is the other name for:

1 cm

Read the map and answer the questions.

1 cm

Scale: 1 cm = 4.5 km Scale: 1 cm = 4.5 AMAN ANSH 1 km cm = 4.5 Scale: 1 cm =Scale: 4.5 Scale: 1 km cm = 4.5 kmkm

1 cm 1 cm 4 cm 4 cm 1 cm 4 cm 4 cm

Draw the map of your house using suitable symbols.

4 cm

4 cm 4 cm 4 cm 4 cm

distance of 15 km.

a a triangular pyramid having congruent equilateral triangles as faces?

b a polyhedron whose base is a polygon and whose lateral faces are triangles with a common vertex?

9 Apply the Euler's formula to find the third feature. Name the polyhedron. a F = 5; V = 6

d F = 7; E = 15

b F = 5; V = 5

e V = 8; E = 12

c F = 8; E = 12 f

V = 12; E = 18

10 Can a polyhedron have these faces, edges and vertices? Verify using Euler’s formula. a 8 faces, 14 edges and 6 vertices

b 9 faces, 16 edges and 9 vertices

Word Problems 1 Sudha arranged three cubes next to each other. What shape did she get? Draw the front, side and top views of the new shape.

2 In the map, the distance between the places is shown using the scale 1 cm = 0.5 km. Find the greatest and the smallest actual distance (in km) between the dairy and the post office.

3 Hari has a solid having only triangles as faces. He calculates the

Post Oﬃce

5 cm

Book store 3 cm

4 cm Home

Dairy 6 cm

3 cm

2.5 cm Hospital School

number of vertices as 6. Prove that the solid must contain exactly 8 triangles. Name the shape.

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18 Area of Polygons

Let’s Recall

Triangles are of three types based on the sides: equilateral, isosceles and scalene. Every triangular object has an area and a perimeter. The perimeter depends on the type of triangle. For example, look at the given formulas. Equilateral Triangle

Isosceles Triangle

Scalene Triangle

Perimeter = 3 × side

Perimeter = (2 × side) + third side

Perimeter = sum of lengths of all sides

1 × base × height 2 Let us say that a farmer has a triangular field of sides 9 m, 12 m and 15 m as shown below. But the area of every triangle can be calculated on the basis of a formula:

He wants to protect cows and cattle from entering his field by fencing it all around. He also wants to grow two types of crops in equal areas of the field. How do you think he would do that? Let us see. To fence it all around, we need to calculate the perimeter of the field. To grow crops in the field, we need to calculate its area. Perimeter of the field

Area of the field

= Sum of lengths of all sides of the field

= Area of the triangle 12 15 1 = 9 m + 12 m + 15 m = 36 m = × b (base) × h (height) 2 1 9 = × 9 × 126 = 54 sq. m 2 To grow two types of crops equally in the field, we split the area of the field into two by dividing it by 2. 1 So, the area of the field to grow one type of crop = × 5427 sq. m = 27 sq. m. 2

Letʼs Warm-up Match the following.

1 Perimeter of an equilateral triangle of length 5 cm

40 cm

2 Area of a triangle of length 4 m and height 8 m

24 sq. m

3 Perimeter of a triangle of sides 3 cm, 4 cm, and 5 cm

16 sq. m

4 Area of a right triangle of lengths 6 m, 8 m, and 10 m

15 cm

5 Perimeter of a triangle of sides 20 cm, 10 cm, and 10 cm

12 cm

I scored _________ out of 5.

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Area of Figures Made With Polygons In a construction project allotted to a third-party vendor, an office building is to be built in the shape of a parallelogram with a triangle on the side. A reception area is to be included in the building in the shape of a semicircle, as shown below. The vendor wants to know the area of the land that the building will occupy so that he can calculate the cost of its construction.

16 m

18 m

Real Life Connect

10 m

25 m

Area of Figures Made With Parallelograms, Triangles and Circles Let us find the area of land that the office building will occupy. Let us find the area of the parallelogram and the triangle with the measures that are given to find the area occupied by the office building. Base of the parallelogram = 25 m

Base of the triangle = 10 m

Height of the parallelogram = 16 m

Height of the triangle = 16 m 1 Area of the triangle = × b × h 2 1 = × 10 m × 16 m 2 = 80 sq. m

Area of the parallelogram = base × height

= 25 m × 16 m

= 400 sq. m

Total area of land occupied by the building = Area of the parallelogram + Area of the triangle Example 1

= 400 sq. m + 80 sq. m = 480 sq. m

A rectangle of length 42 cm and breadth 28 cm is covered with non-overlapping circles of the same radius. If the radius of each circle is 7 cm, then find the area of the shaded portion. Length of the rectangle = 42 cm Breadth of the rectangle = 28 cm Area of the rectangle = length × breadth

= 42 × 28 = 1176 sq. cm

Radius of the circle = 7 cm There are 2 full circles, 4 quarter circles, and 6 semicircles. We know that 4 quarters make 1 whole and 2 halves make 1 whole. So, there are 6 circles in total.

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So, the area of 6 circles = πr2 × 6 22 = × 7 × 7 × 6 = 22 × 7 × 6 = 924 sq. cm 7 Therefore, the area of the shaded region = Area of rectangle – Area of 6 circles Example 2

= 1176 sq. cm - 924 sq. cm = 252 sq. cm

The area of a parallelogram is six times the area of a triangle with base 4 cm and height 6 cm. If the height of the parallelogram is 3 cm, then what is the measure of the base of the parallelogram? Base of the triangle = 4 cm; Height of the triangle = 6 cm 1 1 Area of the triangle = × b × h = × 42 × 6 = 12 sq. cm 2 2 Area of the parallelogram = 6 × area of the triangle = 6 × 12 = 72 sq. cm Height of the parallelogram = 3 cm h × b = 72 sq. cm ⇒ 3 × b = 72 sq. cm 72 ⇒b= cm 3 Base of the parallelogram = 24 cm

Example 3

Remember! Area of a triangle =

1 ×b×h 2

Area of a circle = πr

Area of a parallelogram = b × h Circumference of a circle = 2πr

A soccer field is in the shape of a rectangle 200 m long and 150 m wide. The coach asks players to run from one corner to the other diagonally across the field while weaving between cones that are placed 10 m apart. Look at the figure and answer the 2 questions. Rocks 150 m

1 How many cones does the coach need to put out? Diagonal length of the rectangle = 2002 + 1502 = 40,000 + 22,500 = 62,500 = 250 m Since the cones are placed 10 m apart, the number of cones placed are

200 m

Rocks

250 = 25. 10

2 T he diameter of the base of the cone is 0.5 m. A pavement of width equal to thrice the diameter of the cone is being constructed along the diagonal of the field in the shape of a parallelogram, leaving two small triangular regions of equal height and a base of side 1 m at two ends. If the ends are filled with rocks for aesthetic reasons, what is the remaining area of the field? Length of the rectangle = 200 m; Breadth of the rectangle = 150 m Diameter of the base of the cone = 0.5 m. So, width of the pavement = 3 × 0.5 m = 1.5 m Length of the pavement = Diameter of cone × Number of cones + Distance between the cones × Number of gaps between the cones = (0.5 × 25) + (10 × 24) = 12.5 + 240 = 252.5 m Area of a parallelogram = base × height

= length × width = 252.5 m × 1.5 m = 378.75 sq. m

Base and height of the triangular region filled with rocks = 1 m 1 Area of the triangular region = × 1 × 1 × 2 = 1 sq. m 2 Now, area of the rectangle = length × breadth = 200 × 150 = 30,000 sq. m

Chapter 18 • Area of Polygons

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Area of the remaining portion = Area of rectangle – (Area of parallelogram + Area of two triangles) = 30,000 sq. m – (378.75 sq. m + 1 sq. m) = 30,000 – 379.75 sq. m = 29620.25 sq. m Do It Together

The figure is made up of a rectangle, a square, and a triangle. 1 What is the value of x, if the perimeter of the shape is 86 cm? Perimeter = 86 cm

15 cm x

9 cm

9 + 15 + x + 12 + 12 + 12 + 15 = 86 cm ____ + x = 86 x = ____ cm 2 What fraction of the total area is triangular?

12 cm

Length of the rectangle = 15 cm; Breadth of the rectangle = 9 cm Area of the rectangle = length × breadth = ____ × ____ = ____ sq. cm Side of the square = 12 cm Area of the square = side × side = ____ × ____ = ____ sq. cm Base of the triangle = 12 cm; Height of the triangle = 9 cm 1 × base × height 2 1 = × ____ × ____ = ____ sq. cm 2 Total area of the figure = Area of the rectangle + Area of the square + Area of the triangle Area of the triangle =

= ____ + ____ + ____= ____ sq. cm Fraction of the area of the triangle of the total area = ____________

Do It Yourself 18A 1 A rectangle of length 14 cm has an area of 84 sq. cm. The perimeter of a square is the same as the perimeter of the rectangle. What is the side of the square?

2 Find the number of square-shaped tiles of side 100 cm required to tile a

1.5 m wide path around a rectangular park of length 40 m by 32 m.

3 The figure comprises a right-angled triangle and a quarter circle. Find the area of the quarter circle.

4 Every circle in the given figures has a diameter of 32 cm. Find the area of the shaded regions in these squares. a

25 cm

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5 John and David are calculating the area of the parallelogram. John multiplies the length of side CD by the length of line Y. David multiplies the length of side BC by the length of line Z. Who is correct: John or David or both of them? Explain your answer.

6 The ratio of two sides of a parallelogram is 4:3. If its perimeter is 56 cm, and the height of the

Z Y

parallelogram is one-eighth of the longer side which is also the base of the parallelogram, then find its area.

7 A puzzle is composed of identical right-angled triangles with side lengths 6 cm, 8 cm, and 10 cm. These must be arranged into a rectangular wooden frame of length 162 cm and breadth 36 cm. a What is the area of each triangle?

b How many triangles fit perfectly into the wooden frame?

c What is the total perimeter of all the triangles used?

Planet Eden Play Park

8 The Planet Eden Play Park sign is made up of two identical parallelograms and a

7.5 m

triangle. The whole sign has an area of 17 sq. m. What is the area of the triangular part of the sign?

9 The floor of a large room of dimensions 39 m × 15 m is to be paved with square tiles of three different colours: brown tiles on all sides and pink tiles are to be twice the

number of white tiles. If the side of each square tile is 3 m and they are to be arranged as shown in the figure, then find the number of brown, pink, and white tiles.

10 Four identical rectangles are arranged in such a way that they form a square, as shown. Each

rectangle has a perimeter of 48 cm. The area of the square is 5 times the area of each rectangle. What is the area of each rectangle?

Word Problems 1

A park near Aana’s home is in the shape of a rectangle. The park is 8 m long. If Aana walks diagonally across the park, it is a 10 m walk. How much farther would Aana walk if she walked along the length and breadth of the park instead of diagonally across it?

The figure is Raghav’s lawn. He wants to spread fertilizer over the entire

lawn. At the market, the fertilizer is sold only in 10 kg bags costing ₹16 each. If he spreads the fertilizer at a rate of 20 g per sq. m., then find the cost of

the fertilizer which he uses for his lawn.

32 m

Suppose you have a sheet of paper in the shape of a parallelogram. You

fold the paper from the top right-hand corner, as shown in the figure. The shaded area you get is three-eighths of the original area. If you fold the bottom left-hand corner of

the sheet, what is the fraction of the shaded area you obtain in the end compared to the original parallelogram? A

A (B)

Chapter 18 • Area of Polygons

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(D)

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We know that the area of the office building that occupies a part of the land is 480 sq. m. This part of the office building comprises of a parallelogram and a triangle. A parallelogram is a type of quadrilateral.

16 m

Also, notice that both the parallelogram and triangular parts of the building together form a trapezium, which is another type of quadrilateral.

18 m

Area of Quadrilaterals

10 m

25 m

So, area of the building = area of the trapezium = 480 sq. m

Area of a Trapezium and Rhombus If the building were not split into a parallelogram and triangle, it would be a trapezoidal-shaped building with a semicircle mounted on top. To find the part of the area of the building, we need to find the area of the trapezium.

So, let us draw a trapezium PQRS with PQ ǁ SR. Draw perpendiculars PM and QN from P and Q, respectively so that PM = QN = h.

Then, the area of the trapezium PQRS = Area of ∆PMS + Area of rectangle PQNM + Area of ∆QNR 1 1 = × SM × PM + MN × QN + × NR × QN 2 2 =

Think and Tell

1 1 × SM × h + MN × h + × NR × h 2 2

1 1 × SM + MN + × NR 2 2

1 × (SM + NR + 2MN) 2

How can we find the area of a trapezium using the area of the triangle only?

1 1 [(SM + MN + NR) + MN] = h × × (SR + PQ) 2 2 =h×

Hence, Area of a trapezium =

(Since MN = PQ which are opposite sides of rectangle PQNM)

1 × (sum of parallel sides) × height 2

1 1 × (25 + 10 + 25) × 16 = × 60 × 168 = 480 sq. m 2 2 Therefore, the part of the land that the building occupies is 480 sq. m. Area of trapezoidal building =

Remember! The diagonals of a

Now, what if the shape of the building is a rhombus?

rhombus divide it into four

If we know the area of a rhombus, we can find the area of land occupied by the building too.

congruent right triangles.

So, let us draw a rhombus PQRS with PR and QS as its diagonals intersecting each other at O at right angles. Then, area of rhombus = 4 × area of ∆POQ 1 = 4 × × PO × OQ 2 1 2 1 1 1 1 1 1 = 4 × × PR × QS = 4 × × PR × QS 2 2 2 2 2 2

S O

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1 1 × PR × QS = × product of its diagonals 2 2 So, if all sides of the parallelogram in the building are the same and the measures of the diagonals are 18 m and 25 m, then =

the area of the rhombus = Example 4

1 × 189 × 25 = 225 sq. m 2

Think and Tell

How can we find the measure of the sides of a rhombus using the diagonals of a rhombus?

Look at the figure. Is it a trapezium? If yes, find its area. We are given that ∠S = ∠P = 90°. S

Sum of these angles = 180° So, PQ and RS are parallel, and PQRS is a trapezium. 1 Area of trapezium = (a + b) × h 2 1 1 = (22 + 32) × 15 = × 5427 × 15 = 405 sq. cm 2 2 Example 5

22 cm

15 cm P

32 cm

Find the area of a rhombus of which each side is 17 m and one of the diagonals is 16 m. Let ABCD be a rhombus with side 17 cm and AC and BD intersect each other at O, with BD = 16 cm. As the diagonals of a rhombus bisect each other at right angles, ∠AOB is a right angle. So, BO = OD = 8 cm. In ∆AOB, using the Pythagoras theorem,

AO2 + OB2 = AB2 ⇒ AO2 + 82 = 172 ⇒ AO2 + 64 = 289 AO2 = 289 – 64 ⇒ AO2 = 225

AO = 225 = 15 cm AO + OC = 15 + 15 = 30 cm = AC 1 1 Area of rhombus ABCD = × AC × BD = × 3015 × 16 = 240 sq. cm 2 2 Example 6

Rani attends a school in which the playground is in the shape of a trapezium of height 700 cm. Due to a plantation drive at the school, Rani plants seedlings along each of the parallel sides of the playground. If the area of the playground is 91 sq. m, and one of its parallel sides is longer than the other by 8 m, find the length of the parallel sides. Let the length of the shorter parallel side be a m and the longer parallel side be b m. Area of the playground = 91 sq. m; Height of the playground = 700 cm = 7 m 1 Area of the playground = × sum of the parallel sides × h 2 Error alert! 1 91 = × (a + b) × 7 2 Convert the dimensions into the same unit 9113 × 2 a+b= = 26 m before any operation. 7 Given that, b − a = 8 ⇒ b = 8 + a So, a + b = 26 ⇒ a + b = a + 8 + a = 26 ⇒ 2a + 8 = 26 ⇒ 2a = 26 – 8 = 18 ⇒ a = So, b = 8 + a = 8 + 9 = 17 m

700 cm = 700 m

18 =9m 2

700 cm = 700 ÷ 100 = 7 m

Thus, the lengths of the parallel sides of the playground are 9 m and 17 m. Chapter 18 • Area of Polygons

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Do It Together

The area of a rhombus is 336 sq. cm. If one of its diagonals is 14 cm, then find its perimeter. Area of the rhombus = 336 sq. cm 1 × d1 × d2 = 336 sq. cm ⇒ d1 × d2 = 336 × 2 = 672 sq. cm 2 One diagonal = 14 cm So, ____ × d2 = 672 ⇒ d2 = 672 = ____ cm To find the perimeter of the rhombus, we need to find the side of the rhombus. Side of the rhombus =

d1 2

d2 2

Perimeter = 4 × side = 4 × ____ = ____cm

_____ _____ + 2 2

2 2 = _____ + _____

Area of General Quadrilaterals What if during the course of the construction, the design of the building is improved and the shape of the parallelogram changes to the shape of a general quadrilateral whose sides are unknown but the measure of the diagonal is 10 m and the measures of the perpendiculars from its opposite vertices to the diagonal are 5 m and 7 m? To find the area of the new building, we need to find the area of the quadrilateral and add it to the area of the triangle. S

So, let us draw a quadrilateral PQRS with PR as one of its diagonals, and QM and SN as perpendiculars drawn from the vertices Q and S, respectively, on diagonal PR. Area of quadrilateral PQRS = Area of ∆PQR + Area of ∆PRS 1 1 1 = × PR × QM + × PR × SN = × PR × (QM + SN) 2 2 2 = Example 7

M N P

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

Find the area of a quadrilateral where one diagonal measures 20 cm and the lengths of the perpendiculars drawn from the opposite vertices on the diagonal are 16 cm and 15 cm. Diagonal = 20 cm; Lengths of perpendiculars from the opposite vertices = 16 cm and 15 cm 1 Area of the quadrilateral = × diagonal × s um of length of perpendiculars on the diagonal from the 2 opposite vertices 1 1 × 20 × (16 + 15) = × 2010 × 31 = 310 sq. cm 2 2 Shreya wants to place a long rod from one corner of the room to the other across a quadrilateral-shaped floor. If the area of the floor is 2475 sq. m and the perpendicular distances from the opposite corners of the room to the rod are 50 m and 60 m, find the length of the rod. =

Example 8

Lengths of perpendiculars = 50 m and 60 m; Area of quadrilateral-shaped floor = 2475 sq. m Area of quadrilateral-shaped floor =

1 × length of the rod × sum of length of perpendiculars on the rod 2 from the opposite corners

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2475 =

1 1 × length of the rod × (50 + 60) ⇒ 2475 = × length 2 2 of the rod × 11055

2475 = 45 m 55 Find the area of a quadrilateral and the lengths of the perpendiculars drawn from opposite vertices, where the diagonal is 25 cm and the sum of the lengths of the perpendiculars drawn from the opposite vertices to the diagonal is 12 cm and they are in the ratio 1:2. Length of the rod =

Do It Together

Diagonal = 25 cm, Sum of lengths of perpendiculars from the opposite vertices = 12 cm Area of quadrilateral =

1 × diagonal × s um of length of perpendiculars on the diagonal from the opposite 2 vertices

1 × ____ × ____ = ____ sq. cm 2 Lengths of perpendiculars from opposite vertices are ____ cm and ____ cm. =

Do It Yourself 18B 1 Find the area of the figures.

16 cm

6 cm

32 cm

27 cm

25 cm

12 cm

18 cm 10 cm

c 2

12 16

25 cm

2 The diagonal of a quadrilateral is 32 cm and the perpendiculars dropped on it from the remaining opposite vertices are 12 cm and 15 cm. Find the area of the quadrilateral.

3 Two parallel sides of a trapezium are in the ratio 4:6 and the distance between them is 15 cm. If the area of the trapezium is 450 sq. cm, find the lengths of its parallel sides.

4 The area of a rhombus is 672 sq. cm. If the perimeter is 200 cm and the length of one diagonal is 96 cm, find the length of each side of the rhombus as well as the length of the other diagonal.

5 The area of a rhombus is 204 sq. cm. If the length of one of its diagonals is 12 cm, find the length of the other diagonal.

6 The area of an isosceles trapezium is 185 sq. m. If the lengths of the parallel sides are 32 m and 25 m, respectively, then find the lengths of the non-parallel sides.

7 A quadrilateral whose area is 170 sq. cm and the length of its diagonal is 17 cm has a perpendicular on the diagonal of length 15 cm. What is the length of the other perpendicular?

8 A rope of length 80 cm is cut into two pieces. One piece is used to form a rectangle of length 14 cm and width 8 cm. The other piece is bent into a rhombus. What is the length of each side of the rhombus?

9 A floor tile has the shape of a trapezium whose parallel sides are of lengths 24 cm and 20 cm and height is 8 cm. How many such tiles are required to cover a floor of area 2992 sq. m? Also, find the cost of tiling at the rate of ₹7.50 per sq. m.

Chapter 18 • Area of Polygons

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10 A field in the shape of a rhombus has sides of 64 m each and an altitude of 36 m. If the area of a square field is

equal to the area of the rhombus, then find the length of the side of the square field. Also, find the difference of their perimeters.

11 The perimeter of a trapezium is 78 cm. If the ratio of the sides of a trapezium is 6:13:9:11, find the area of the trapezium taking the two largest sides as the parallel sides and height as 10 cm.

Word Problem 1

Rani is sitting on a wall which is 5 m high. The shadow of the wall falls on the road in the form of a rhombus. The shadow covers the width of the road exactly. Help Rani to find: a the width of the road, if it is one and a quarter times the height of the wall. b the area of the road in shade if the area of the wall is 12 sq. m.

m c.

Area of Combined Shapes

For this, we need to calculate the area of all three shapes. Since we already know the total area of the parallelogram and the triangle, it is sufficient to calculate the area of the semicircle only and then add it to the remaining area.

16 m

What if we have to calculate the construction cost? We then have to find the area of the whole building including the reception area.

18 m

We know that the shape of the building comprises a parallelogram, a triangle, and a semicircle. This is called a combined shape. We also know that the area of the office building that occupies the land is 480 sq. m.

25 m

10 m

So, diameter of the semicircle = base of the parallelogram = 25 m Radius of the semicircle = 12.5 m 1 1 2211 Area of the semicircle = × πr2 = × × 12.5 × 12.5 = 245.54 sq. m 2 2 7 So, the total area of the building = 480 + 245.54 = 725.54 sq. m Area of a Polygon In our daily lives, we see that not all buildings are square, rectangular, triangular, circular, or quadrilateral-shaped. They are also found in Pentagon, hexagonal, heptagonal shapes, etc. For example, the famous Pentagon building in the United States, is a regular pentagon with each side equal to 281 m and a height of 23 m. To find the area of a regular polygon, we split it into triangles of equal size and find the area of each triangle. These areas are then added together to get the area of the polygon. For example, let us consider a regular pentagon of side 5 cm. This pentagon is split into equal-sized isosceles triangles. Each triangle has a line segment of length 3 cm that joins the centre of the pentagon to the midpoint of its side. How can we find the area of this polygon? Let us see!

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To find the area of this polygon, we find the area of each triangle and then add them to find the total. 1 1 15 Area of each triangle = ×b×h= ×5×3= = 7.5 sq. cm 2 2 2 There are 5 triangles in the pentagon, so the total area of 5 triangles = 7.5 × 5 = 37.5 sq. cm Therefore, area of the pentagon = area of 5 triangles = 37.5 sq. cm What if the polygons are irregular? To find the area of irregular polygons, we split it into different geometrical shapes such as triangles, squares, rectangles, trapeziums, parallelograms, etc. We then find the area of each shape and add them to get the area of the L polygon. For example, let us consider the given irregular pentagon.

T 6 cm

M 4 cm

Example 9

5 cm I

So, area of the pentagon = area of ∆ LMT + area of trapezium MOHT + area of rectangle IGON + area of triangle LIN 1 1 1 = × LM × MT + × (MT + OH) × MO + GO × ON + × IN × NL 2 2 2 1 1 1 × 6 × 6 + × (6 + 4) × (4 + 4) + 5 × 4 + × 5 × (4 + 6) 2 2 2

4 cm O

N 4 cm

This pentagon comprises two triangles, one trapezium, and one rectangle.

6 cm

Remember!

1 1 1 1 1 1 × 36 + × 10 × 8 + 20 + × 5 × 10 = × 36 + × 80 + 20 + × 50 2 2 2 2 2 2

The number of triangles formed inside a regular

1 1 = × (36 + 80 + 50) + 20 = × 166 + 20 = 83 + 20 = 103 sq. cm 2 2 Find the area of the given figure. (All measures are given in metres.)

polygon is the same as the number of sides.

Area of the figure = area of ∆ ABI + area of trapezium IBCJ + area of trapezium JCDK + area of trapezium FLJG + area of trapezium GJMH + area of triangle HMA 1 1 1 = × AI × IB + × (IB + JC) × IJ + × (DK + JC) × JK 2 2 2 + =

1 1 1 × (GJ + FL) × JL + × (HM + GJ) × MJ + × AM × MH 2 2 2

1 1 1 × 35 × 4020 + × (40 + 10) × 35 + × (35 + 10) × 40 2 2 2 1 + × (50 + 25) × 50 2 +

1 1 × (50 + 35) × 45 + × 2512.5 × 35 2 2

= 700 + +

40 A

25 M

10 35

C 35

35 J

L 25 F

1 1 × 5025 × 35 + × 4020 × 45 2 2

1 1 × 75 × 5025 + × 85 × 4522.5 + 437.5 2 2

Did You Know?

= 700 + 875 + 900 + 1875 + 1912.5 + 437.5

Many natural objects, such as

= 6700 sq. m

minerals, exhibit polygonal shapes

crystals, snowflakes, and certain due to their atomic or molecular arrangements.

Chapter 18 • Area of Polygons

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Example 10

In an octagon-shaped park of side 12 m each, Yamini walks from the centre to the midpoint of a side. If Yamini walks for about 5 m, find the area of the park. Number of sides in an octagon-shaped park = 8; Measure of each side of the octagon = 12 m

Do It Together

Distance travelled by Yamini from the centre to the midpoint of a side = 5 m 1 1 Area of the park = × b × h × number of triangles = × 126 × 5 × 8 = 240 sq. m 2 2 Find the area of the polygon. (All measures in cm.) Draw a line segment VO so that it touches PW at O, forming a parallelogram QVOP. Draw the height of 9 cm from V on OW. Now, area of the polygon = area of rectangle STUR + area of parallelogram RUVQ + area of parallelogram QVOP + area of ∆ VOW 1 Area of the polygon = 22 × 6 + 22 × _____ + 22 × _____ + × 20 × _____ 2 = 22 × (6 + _____ + _____) + _____

R Q

= 22 × _____ + _____ = _____ + _____ = _____

9 9

T 6 U

V O 42

Do It Yourself 18C 1 Find the area of the figures.

15 cm A

E 15 cm

1.3 m

55 cm

10 cm

22 cm

5.7 cm

24 cm

60 cm

16 cm

9 cm

2 Find the area of the polygons.

E 35 m

5 cm

16 cm

A 9 cm B

D 22 m C

60 m

J H

18 m

50 m

G 4 cm 4 cm F I 2 cm

B 5 cm 14 cm

35 cm

3 Find the area of the shaded regions. a

F 5 cm G

1.2 cm

4.2 cm

5.5 cm C

8 cm 12 cm H

3 cm

1 cm

2 cm B

6 cm

2.4 cm

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4 Find the area of a regular hexagon of side 15 cm each. Is there more than one way of finding its area? Draw the figure and find out.

5 Two tiles are fitted together as shown.

2 cm

18 cm

If the length of one tile is 10 cm, then find the area of the tiling pattern obtained by fitting five such tiles together.

6 The measurements of a chimney are given below. Find its area.

2 cm

B 8 cm

7 An agricultural field is in the shape of a polygon. Find its area if XR = 9 m, TR = 15 m and TY = 10 m.

8 cm T

3.4 cm C

Q 4.4 m 2m

3.6 m

8 A floor is in the shape of a polygon. If the floor is tiled with tiles of dimensions 25 cm × 20 cm, then find the

B A

2.5 m

4m 3m

number of tiles that would be required to cover the floor.

P2 m D Q 2.5 m 1m

Word Problem 1

Gaurav has some right-angled triangular tiles whose sides are 13 cm,

12 cm, and 5 cm. He makes the shape using four of these tiles and some square tiles. Find the area of the shape.

Chapter 18 • Area of Polygons

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Points to Remember • • • •

Area of a square = (side)2; Perimeter of a square = 4 × side

Area of a rectangle = length × breadth; Perimeter of a rectangle = 2 (length + breadth) 1 Area of a triangle = × b × h 2 Area of a parallelogram = b × h

Area of a circle = πr2; Circumference = 2πr • Area of a trapezium = 1 × (sum of parallel sides) × height 2 1 • Area of a rhombus = × product of its diagonals or Base × altitude 2 • Area of a general quadrilateral = 1 × diagonal × sum of perpendiculars on the diagonal from 2 the opposite vertices •

• To find the area of a regular polygon, we split it into triangles of equal size and find the area of each triangle. These areas are then added to get the area of the polygon.

Math Lab Polygon Art Gallery Setting: In groups of 3 Materials Required: Cut-outs of polygons of different shapes and sizes from coloured sheets of paper, cardboard, rulers, coloured pencils, markers or crayons, a pair of scissors, glue, large cardboard sheets as a canvas for art gallery, measuring tape Method: All 3 members of each group must follow these steps. ach group must collect a set of coloured sheets of paper and prepare a set of cut-outs 1 E of polygons of different shapes and sizes. (This step should preferably be done prior to the class.)

rrange these polygons in such a way that a beautiful and aesthetically pleasing design 2 A of a scene is created. For example, a mountain landscape in which mountains can be designed using triangles and trapeziums. Lakes can be created using different irregular polygons such as irregular pentagons, etc., and trees can be created using rectangles and triangles. 3 Measure the sides of each polygon used to create the scene and find their area. 4 In the end, find the total area of the sheet of paper used to create the scene.

The group that creates a scene with the least number of polygons wins the game.

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Chapter Checkup 1 Look at the figure. Find the difference of the areas of the regions A and C. (PQRS is a square) m 6c

D P A

8cm

2 cm

9 cm

2 Nishant says, “If the length and breadth of a rectangle are the same as the base and height of a parallelogram, then both the shapes have the same area.” Explain whether Nishant’s statement is always, sometimes, or never true.

3 Find the area of the figures. a

A 6 cm

24 cm 63 cm

B 6 cm

95 cm C

4 Find the area of the shaded rhombus if the length of the rectangle is 9.2 cm and the breadth is 6.4 cm.

5 The parallel sides of a trapezium are 25 cm and 15 cm. Its non-parallel sides are 13 cm each. Find the area of the trapezium.

6 In a rhombus, one diagonal is five times the other. If the area of the rhombus is 1000 sq. cm, find the length of the diagonals.

7 One of the parallel sides of a trapezium is thrice the other. If the area of the trapezium is 350 sq. cm and the height is 14 cm, then find the lengths of the parallel sides. Find the area of the polygons.

58 m 12 m

Chapter 18 • Area of Polygons

15 m D

B 8m C

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95 cm

30 cm

115 cm

22 m

}

90 cm

15 m

95 cm

R100 cm m 6c

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A farmer has a field in the shape of a trapezium where the lengths of its parallel

40 m

sides are 160 m and 240 m. If the distance between the parallel sides is 150 m and the field produces 6500 kg of wheat per hectare, then find the amount of wheat in kg the field produces. (Hint: 1 hectare = 10,000 sq. m)

10 A square garden of side 40 m is being designed in this manner. a 4 congruent triangular regions are carved from all corners of the square.

The triangle has 2 equal sides of length 5 m.

b There is a circular swimming pool of radius 3 m at the centre of the

garden.

c The remaining area of the garden is covered with grass.

If the grass is planted at the rate of ₹15 per sq. m, what is the cost of planting the grass?

Word Problems 1

Mr. Chopra arranges 15 parallelogram-shaped tiles on his

bathroom floor to make a parallelogram-shaped pattern, as given in the image. What area of the floor is not covered by

0.5 m

the tiles?

A floor is in the shape of a trapezium whose parallel sides are of length 12 m and 15 m and the distance between them is 8 m. David wants to paint the floor using a 5 L tin of paint

costing ₹2899 in such a way that 1 L of paint covers an area of 10 sq. m. If he has ₹5400 to spend on paint, has David got enough money to buy all the paint he needs?

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Surface Area and 19 Volume of Solids Let's Recall We have learnt about two-dimensional shapes. Let us see some basic shapes and revise their perimeter and area. Rectangle

Square

Circle

Equilateral triangle

Right triangle

Perimeter = 2(L + B) Area = L × B

Perimeter = 4a Area = a2

Circumference = 2πr Area = πr2

Perimeter = 3a Area = 3 a2 4

Perimeter = Sum of sides 1 Area = b × h 2

Letʼs Warm-up Match the following. 20 cm

12 cm

10 cm

15 cm

Perimeter = 45 cm

Area = 100 cm2

Perimeter = 64 cm

Area = 154 cm2 12 cm 15 cm

Area = 90 cm2

I scored _________ out of 5.

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Surface Area Real Life Connect

An architect has to design a warehouse for storage purposes. Firstly, he prepares a 3-D warehouse model.

Surface Area of Cuboids, Cubes and Cylinders The architect prepares a cuboid-shaped room made of cardboard. The architect plans the dimensions of the model to be 15 units × 12 units × 10 units.

Surface Area of Cuboids He wants to paste coloured paper cut-outs on the faces of the cuboid.

Let us check how the area of the coloured paper required by him is calculated. Let us assume the length to be ‘l’, breadth to be ‘b’ and height to be ‘h’. Total surface area

Length l

Height h

h adt Bre b

Height h

Breadth b

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

Total area of paper required = 2 × ((15 × 12) + (15 × 10) + (12 × 10)) = 2 × (180 + 150 + 120) = 2 × 450 sq. units = 900 sq. units Hence, the architect requires 900 square units of coloured paper. Lateral surface area

What if the architect has to paste coloured paper cut-outs on only the walls of the cuboid? We can find the area of the paper required by finding the lateral surface area. The area of the lateral faces is called the lateral surface area.

Length l lh Length l

Height h h adt Bre b

Height h bh

Breadth b

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Lateral surface area = lh + bh + lh + bh = 2h(l + b)

Total area of the walls = 2 × 10 × (15 + 12) = 20 × 27 units = 540 sq. units Total surface area = Lateral surface area + (2 × Area of base) Example 1

Find the lateral surface area and total surface area of a cuboid with length = 25 cm, breadth = 20 cm and height = 18 cm. Lateral surface area = 2h (l + b) = 2 × 18 cm (25 cm + 20 cm) = 36 cm × 45 cm = 1620 cm2

Error Alert!

Total surface area = 2 × (lb + lh + bh) = 2 × ((25 × 20) + (25 × 18) + (20 × 18)) = 2 × (500 + 450 + 360) = 2 × 1310 = 2620 cm2 Example 2

Lateral surface area is the area of the lateral faces (walls) only. Length

The ratio of the length, breadth and height of a cuboid is 5:4:2. What is the total surface area of the cuboid if the length of the cuboid is 60 cm? Length of the cuboid = 60 cm; Ratio of length, breadth and height = 5:4:2 60 60 Breadth of the cuboid = × 4 = 48 cm; Height of the cuboid = ×2 5 5 = 24 cm

Breadth = 5 m

Height = 3 m

Breadth b

Length

Total surface area = 190 m

Find the length of a cuboid if the breadth, height and total surface area of the cuboid are 5 m, 3 m and 190 m2, respectively. Let length = l

lb lh

Total surface area of the cuboid = 2 × (lb + lh + bh) = 2 × ((60 × 48) + (60 × 24) + (48 × 24)) = 2 × (2880 + 1440 + 1152) = 2 × 5472 cm2 = 10,994 cm2 Do It Together

Height

lh bh

Height h

Breadth b

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

Surface Area of Cubes

A cube is a three-dimensional shape with six square faces. Let us find the area of a cube.

Total surface area

Lateral surface area a a2

a a

a2 a2

a a

a a2

a a

a a2

Total surface area = a2 + a2 + a2 + a2 + a2 + a2 = 6a2

Lateral surface area = a2 + a2 + a2 + a2 = 4a2

Total surface area = Lateral surface area + (2 × Area of the base) Chapter 19 • Surface Area and Volume of Solids

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Example 3

What is the lateral surface area and total surface area of a cube with sides of 80 cm? Side = 80 cm

Lateral surface area = 4 × Side2 = 4 × 80 cm × 80 cm = 25,600 cm2 Total surface area = 6 × Side2 = 6 × 80 cm × 80 cm = 38,400 cm2 Example 4

If the lateral surface area of a cube is 10,000 m2, what is the length of the side of the cube?

Remember! A cube is made up of 6 similar faces.

Lateral surface area = 10,000 m2; Let the side of the cube = ɑ Lateral surface area = 4a2 ⇒ 10,000 = 4a2 ⇒ a2 = 2500 ɑ = 50 m Do It Together

If the length of the side of a cube is increased by 50%, what is the percentage increase in the total surface area of the cube? Let the length of the side of a cube be ɑ. Total surface area of a cube = 6ɑ2

New length of the side of the cube = ɑ + 50% of ɑ = ___________________ ɑ Total surface area of the new cube = ___________________

Increase in the total surface area of the cube = ___________________

Percentage increase in the total surface area of the cube = ___________________

Surface Area of Cylinders A cylinder is a solid shape that has circular faces at the top and bottom and these are joined by a circular cross section. Total surface area r r

Lateral surface area

πr2

r 2πr

2πr h

Area = 2πrh

Area = Area of two circles + Area of rectangle

Area = 2πrh

πr2

Total surface area = πr2 + 2πrh + πr2 = 2πr2 + 2πrh

Lateral surface area = 2πrh

= 2πr (r + h)

Hollow cylinder

Let the outer radius be ‘R’, the inner radius be ‘r’ and the height of the cylinder be ‘h’.

Thickness of the cylinder = R – r

External lateral surface area = 2πRh

Internal lateral surface area = 2πrh

Area of the cross section = π(R2 – r2)

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Lateral surface area

Total surface area

External surface area + Internal surface area

External surface area + Internal surface area + Area of 2

= 2πh (R + r)

= 2πRh + 2πrh + (2 × π (R2 – r2))

= 2πRh + 2πrh

base rings

= 2π (Rh + rh + (R2 – r2)) Example 5

Example 6

(

)

Find the lateral surface area and total surface area of a cylinder if the radius and height of the cylinder 22 are 7 cm and 10 cm, respectively. Take π = 7 22 Lateral surface area = 2πrh = 2 × × 7 × 10 = 440 cm2 7 22 Total surface area = 2πr (r + h) = 2 × × 7(7 + 10) = 44 × 17 = 748 cm2 7 What is the lateral and total surface area of a hollow cylinder whose internal radius, external radius and height are 7 cm, 14 cm and 20 cm, respectively? Internal radius = 7 cm

External radius = 14 cm

Height = 20 cm

22 22 × 20(14 + 7) = 2 × × 20 × 21 = 2640 cm2 7 7 22 Total surface area = 2π (Rh + rh + (R2 – r2)) = 2 × × ((14 × 20) + (7 × 20) + (142 – 72)) 7 22 22 =2× × (280 + 140 + 147) = 2 × × 567 = 3564 cm2 7 7 Lateral surface area = 2πh (R + r) = 2 ×

Do It Together

What is the radius of a cylinder whose lateral surface area and height are 330 m2 and 15 m, respectively? Let the radius of the cylinder be r. Height = 15 m

Lateral surface area = 330 m2

Lateral surface area = 2πrh 22 × r × 15 7 r = __________ m 330 = 2 ×

Do It Yourself 19A 1 Find the lateral surface area and total surface area of a cuboid of the given length, breadth and height. a l = 12 cm, b = 8 cm and h = 15 cm

c l = 55.5 m, b = 11.5 m and h = 9.25 m

b l = 60 cm, b = 7 cm and h = 18 cm

2 Find the lateral surface area and total surface area of the cube for the side given. a ɑ = 21 m

b ɑ = 17 mm

c ɑ = 28 m

d ɑ = 35 dm

e ɑ = 80.5 cm

3 Find the lateral surface area and total surface area of the cylinder for the given height and radius. a r = 12 cm and h = 14 cm

b r=

7 2

m and h = 22 m

c r=

7 4

m and h = 40 cm

4 Find the lateral surface area and total surface area of a hollow cylinder with the given dimensions. (Use π = 3.14) a R = 14 cm, r = 8 cm and h = 1 m

c R = 11 dm, r = 1 m and h = 40 cm

Chapter 19 • Surface Area and Volume of Solids

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b R = 10 dm, r = 80 cm and h = 1.2 m

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5 Identify and find the missing dimensions of the solid shapes. a b

l = 30 m

11 dm 2

r = 2m h= 19 m 2

h=?

R=?

a=?

TSA = 1800 m

LSA = 225 cm

TSA = 605 dm

LSA = 627 m2

6 The perimeter of each face of a cube is 32 cm. What is the total surface area of the cube? 7 A hollow cylindrical tube is open at both ends and is made of iron of 2 cm thickness. The external diameter of the tube is 50 cm and the length of the tube is 140 cm. What is the lateral surface area of the tube?

8 The curved surface area of a cylinder is 1200 sq. cm. If the radius of the base of the cylinder is 10 cm, what is its height?

9 Three similar cubes are placed adjacent to each other. What will be the ratio of the total surface area of the new shape to the sum of the surface areas of the cubes?

10 A cubical box has a side of length 20 cm and another cuboidal box is 25 cm long, 20 cm wide and 16 cm high. Which box has the greater total surface area and by how much?

Word Problems 1

A cuboid-shaped room measures 15 m × 12 m × 6 m. If the cost of whitewashing is

₹19 per m2, then what is the total cost of whitewashing the walls of the room? If the ceiling also has to be whitewashed, what will be the total cost of whitewashing?

The area of the base of a cuboidal tank is 12,800 cm2 and the lateral surface area of the

tank is 1,08,000 cm2. If the length of the base is eight times the breadth, what is the depth of the tank?

45 circular plates each of radius 21 cm and thickness 2 cm are placed one on top of the

A cuboidal tank 12 m long and 4.5 m wide contains water to a depth of 2.25 m. What is the

The inner diameter of a tank is 5.6 m and it is 11 m high. What is the cost of plastering the

There is a cylindrical ion boiler at Raj’s factory. The boiler is open at the top. It is 6.3 m high

other to form a cylinder. What is the total surface area of the cylinder? area of the wet surface?

tank from inside (including the base) at the rate of ₹45 per m2?

and its diameter is 2.8 m. If the boiler gets corroded at a rate of 1.25 m2 per day, then in how many days will the boiler be completely corroded?

The cost of painting is ₹25 per m2. What is the cost of painting a cubical container which is open at the top and has a side length of 3.5 m?

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Volume Remember the architect who made the warehouse model? He now wants to check the storage capacity of the warehouse. How can he do that? Let us see.

Real Life Connect

Volume of Cuboids, Cubes and Cylinders The architect has to calculate the volume to calculate the storage capacity of the warehouse. Volume is the amount of space occupied by a solid shape.

Volume of Cuboids Let us find the volume of the cuboid.

Volume of the cuboid = Area of the base × Height of the cuboid = (Length × Breadth) × Height

Length

= lbh

2 2 2 Length of the diagonal of the cuboid = l + b + h

The dimensions of the warehouse are,

Height h

Breadth

Length = 15 units, Breadth = 12 units and Height = 10 units

Volume of the warehouse = 15 units × 12 units × 10 units = 1800 units3 Thus, the volume of the warehouse is 1800 units3. Units of Volume

Think and Tell

Length is one-dimensional and its standard unit is metre, area is two-dimensional and its standard unit is metre2 while volume is three-dimensional and its standard unit is metre3. The other units for volume are cm3, dm3, mm3, km3 etc. Conversion of volume units 1 m3 = 1 m × 1 m × 1 m

= 100 cm × 100 cm × 100 cm = 10,00,000 cm3 = 10 dm × 10 dm × 10 dm = 1000 dm3 = 1000 litres Example 7

What is the difference between volume and capacity?

1 dm3 = 1 dm × 1 dm × 1 dm

= 0.1 m × 0.1 m × 0.1 m = 0.001 m3

= 10 cm × 10 cm × 10 cm = 1000 cm3 = 1 litre

1 cm3 = 1 cm × 1 cm × 1 cm

= 10 mm × 10 mm × 10 mm = 1000 mm3 = 1 mL

Find the volume of a cuboid of length, breadth and height 16 cm, 12 cm and 15 cm respectively. Also, find the length of the diagonal. Length = 16 cm

Breadth = 12 cm

Volume of cuboid = lbh = 16 cm × 12 cm × 15 cm = 2880 cm

Height = 15 cm

Length of the diagonal = l2 + b2 + h2 = 162 + 122 + 152 = 256 + 144 + 225 = 625 = 25 cm Example 8

What is the capacity of a container (in litres) of length 60 m, with a breadth one-third the length and height twice the breadth? 1 Length = 60 m Breadth = × 60 = 20 m Height = 2 × 20 m = 40 m 3 Volume of a cuboid = lbh = 60 m × 20 m × 40 m = 48,000 m3 We know that, 1 m3 = 1000 L

48,000 m3 = 48,000 × 1000 L = 4,80,00,000 L

Chapter 19 • Surface Area and Volume of Solids

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Do It Together

Find the height of a tank whose capacity is 4,80,000 L if the length and breadth of the tank are 12 m and 800 cm, respectively. Length = 12 m

Breadth = 800 cm = ______________________ m

Capacity of the tank = 4,80,000 L

1000 L = ______________________ m3

4,80,000 L = ______________________ m3

Volume = lbh = l × 12 × ______________________

h = ______________________ m

Volume of Cubes Let us find the volume of a cube. Volume of a cube = Side × Side × Side = a × a × a

= a3

a a

Diagonal of a cube = 3 a Unit cube: A cube with all sides of 1 unit is called a unit cube. Volume of unit cube Volume = Side × Side × Side = 1 unit × 1 unit × 1 unit

1 unit

1 unit 1 unit

= 1 unit Example 9

Find the volume of a cube of side 15 cm. Side of the cube = 15 cm Volume of the cube = Side3 = (15 cm)3 = 3375 cm3

Example 10

The volume of a cube is 8000 cm3. If the side of the cube is increased by 20%, then what is the new volume of the cube? Let the original side of the cube be a. Volume of the cube = 8000 cm3; Volume = Side3 8000 cm3 = Side3 ⇒ Side = 20 cm Now, the side is increased by 20%. New length of the side = 20 cm + 20% of 20 cm = 20 cm + 4 cm = 24 cm New volume of the cube = 24 cm × 24 cm × 24 cm = 13,824 cm3

Do It Together

The side of a cube is 12 cm. What is the capacity of the cube (in mL)? Side of the cube = ________________ cm Volume = Side3 = _________________ cm3 We know that, 1 cm3 = ___________ mL Total capacity of the cube = __________ mL

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Volume of a Cylinder Let us find the volume of a cylinder.

Let the radius of the cylinder be ‘r ’ and the height of the cylinder be ‘h’. Volume of the cylinder = Area of the base × Height of the cylinder = πr2 × h = πr2 h

Hollow Cylinder

Volume = Area of the base × Height of the cylinder

= π(R2 − r2) × h

= π(R2 − r2)h Example 11

What is the volume of a cylinder whose radius and height are 10 cm and 14 cm, respectively? Radius = 10 cm; Height = 14 cm

Volume of the cylinder = πr2 h = Example 12

22 7

× 10 × 10 × 14 = 4400 cm3

The difference of the external radius and internal radius of a hollow cylinder is 3 cm. What is the volume of the hollow cylinder if the internal radius is 12 cm and height is 28 cm? Internal radius (r) = 12 cm; R – r = 3 cm; Height (h) = 28 cm R – r = 3 ⇒ R = 3 + 12 = 15 cm Volume = π(R2 − r2)h =

Do It Together

22 7

× (152 − 122) × 28 = 88 × (225 – 144) = 88 × 81 = 7128 cm3

What is the radius of a cylinder whose volume is 79.2 m3 and height is 70 m? Height = 70 m Volume = πr2h 79.2 =

22 7

Volume = 79.2 m3

Did You Know? The volume of a human brain

× r × 70

varies between approximately 1100 cm3 and 1350 cm3.

r2 = _______ m2 r = ____ m

Do It Yourself 19B 1 Find the volume of the cuboid for the given length, breadth and height. Also find the length of the diagonal. a l = 15 cm, b = 10 cm and h = 8 cm

c l = 45.5 m, b = 8.5 m and h = 6.5 m

b l = 75 cm, b = 5 cm and h = 60 cm

2 Find the volume of the cube for the given side length. Also find the length of the diagonal. a a = 16 cm

b a = 24 mm

c a = 30 m

d a = 50.5 dm

e a = 130.5 cm

3 Find the volume of the cylinder for the given height and radius. a r = 3.5 cm and h = 10 cm

c r = 0.8 dm and h = 35 dm

Chapter 19 • Surface Area and Volume of Solids

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b r = 4.2 m and h = 12 m

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4 Find the volume of a hollow cylinder with the dimensions given. a R = 15 m, r = 8 m and h = 3.5 m c R = 7 mm, r = 2 mm and h =

14 9

b R = 18 cm, r = 7 cm and h = mm

21 5

5 If the volumes of two cubes are in the ratio 1:64 then what is the ratio of their total surface areas? 6 The diagonal of a cube is 8 3 cm. What is the volume of the cube? 7 A cylinder is put inside a cube. The diameter of the cylinder is equal to the side of the cube and its height is the same as the length of the side of the cube. What volume of empty space is left in the cube after placing the cylinder? (Assume the edges of the cube to be a unit.)

8 If each edge of a cube is increased by 50%, what is the percentage increase in the volume of the cube? 9 A rectangular block measures 15 cm × 12 cm × 18 cm. What is the volume of the largest cube that can be cut from this cuboid?

5 = b

10 Identify and find the missing dimension for the given volume of the solid shapes. r = 5.5 mm

l = 17 cm

r = 11m

h=?

d h= ? a=?

R = 21 m

V = 3060 cm3

V = 3993 mm3

V = 352 m3

V = 1331 dm3

11 The length of a cuboid is increased by 25%, its breadth is reduced by 50% and its height is increased by 40%. What is the ratio of the volumes of the original cuboid and the newly formed cuboid?

Word Problems 1

A container can hold 60 L of water. How many ice cubes of side 10 cm can be formed from

The area of the base of a rectangular tank is 1300 cm2. The volume of the tank is 5.2 m3.

How many aluminum rods of length 3.5 m and diameter 4 cm each can be made from

Find the length of the longest perch that can be placed in a hall which is 25 m long, 16 m

A warehouse measures 50 m × 35 m × 20 m. How many cubical wooden crates of length

the water in the container?

What is the depth of the tank? 1.76 m3 of aluminum? wide and 6 m high.

50 cm can fit inside the warehouse?

Water flows out through a circular pipe of internal diameter 4 cm, at the rate of 3 metres per second into a cylindrical tank with a base radius of 90 cm. By how much will the level of water rise in 45 minutes?

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Points to Remember •

Total Surface Area = Area of all the faces of a solid shape.

•

Lateral Surface Area = Area of only lateral or curved faces of a solid shape.

Cuboid – 2(lb + lh + bh); Cube – 6a2; Cylinder – 2πr (r + h); Hollow cylinder – 2π (Rh + rh + (R2– r2)) Cuboid – 2h (l + b); Cube – 4a2; Cylinder – 2πrh; Hollow cylinder – 2π (R + r)h

• Volume = Amount of space occupied by an object.

Cuboid – lbh; Cube – a3; Cylinder – πr2h; Hollow cylinder – πh (R2 – r2)

Math Lab Finding the Volume of Containers!

Aim: To measure the volume of water.

Materials Required: Water bottles filled up to different levels, transparent cuboid box with known base dimensions, scale, paper, pen Setting: Whole class Method: 1 Ask the students to fill their water bottles. 2 They should pour the water into the box and then measure the depth of the water level. 3 They will then calculate the volume of the water and write the volume in millilitres. 4 They could also identify the total capacity of their water bottles. 5 Write the capacity of the water bottles on the paper and see which child brought the bottle with the greatest or lowest capacity.

Chapter Checkup 1 Find the lateral surface area and total surface area of the cuboid for the length, breadth and height. a l = 15 dm, b = 12 dm, h = 90 cm

b l = 12 m, b = 50 cm, h = 11 m

c l = 80 cm, b = 0.5 m, h = 11 dm

d l = 20 m, b = 150 cm, h = 90 dm

2 Find the lateral surface area and total surface area of a cube with the given side length. a a = 15 cm

b a = 3.75 dm

c a=5

m d a = 6.5 m 2 Find the lateral surface area and total surface area of the cylinder for the radius and height given. a r = 2.25 cm, h = 16 cm

b r = 17.5 cm, h = 12 cm

c r = 21 m, h = 160 dm

d r = 161 cm, h = 1.8 m

Chapter 19 • Surface Area and Volume of Solids

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4 Find the lateral surface area and total surface area of the hollow cylinder with the internal radius, external radius and the height. (Take π = 3.14)

a r = 2 cm, R = 5 cm and h = 30 mm

b r = 12 mm, R = 5 cm and h = 0.25 m

c r = 1.5 m, R = 4 m and h = 1250 mm

d r = 11 mm, R = 2.5 cm and h = 0.15 m

5 Find the volume of the cube or cuboid of the dimensions given. Also, find the length of the diagonal. a a = 1.2 m

b l = 12 mm, b = 1 cm, h = 18 mm

c l = 2.5 m, b = 150 cm, h = 600 cm

d l = 20 cm, b = 0.1 m, h = 40 cm

e a = 2.6 mm

l = 15 m, b = 8 m, h = 180 cm

6 Identify how many unit cubes can fit inside the cube or cuboid of the dimensions given. a l = 12 m, b = 10 m, h = 5 m

b a=3m

c l = 10 m, b = 5 m, h = 8 m

7 Find the volume of the cylinder or hollow cylinder of the dimensions given. a r = 1.05 cm, h = 2 cm

b r = 11.2 mm, R = 1.5 cm, h = 12.6 mm

c r = 50 cm, h = 13.3 dm

d r = 12 mm, R = 2 cm, h = 2.1 cm

e r = 14 m, R = 161 dm, h = 45 m

8 Find the capacity (in litres) of the containers of the dimensions given. a Cube (a = 150 cm)

b Cuboid (l = 2 m, b = 15 dm, h = 50 cm)

c Cylinder (r = 8 m, h = 3 m)

d Hollow cylinder (R = 3 m, r = 120 cm, h = 1.4 dm)

9 What is the percentage increase in the total surface area of a cube when each side of the cube is doubled? 10 What is the change in the curved surface area of a cylinder when the radius is doubled and the height is halved? 11 What is the height of a cylinder of radius 5 m and total surface area of 660 m2? 12 The radius of the base of the cylinder is 7 mm and its total surface area is 2640 mm². What is the height of the cylinder?

13 Square holes of dimensions 1 m x 1 m are cut from all the faces of a cube with edges measuring 4 m. What is the total surface area of the faces of the cube now?

14 Three cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new cube. What is the total surface of the new cube?

15 The outer curved surface area of a hollow cylinder is 31,680 cm2 and it can hold 2,49,480 cm3 of air in it. What is the thickness of the cylinder if the height of the cylinder is 180 cm?

16 What is the ratio of the volume of a cylinder of radius X and height Y to the volume of a cylinder of radius Y and height X?

17 Look at the image on the right. What is the volume of the cylinder formed from the given net?

22 cm

18 A wooden pipe is in the shape of a hollow cylinder. The internal diameter of the

pipe is 28 cm and its external diameter is 35 cm. The length of the pipe is 48 cm. Find the weight of the pipe if each cm3 of wood weighs 0.8 g.

4 cm

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19 Find the missing dimensions of the solid shapes. b

b=?

dm l l==66dm

r = 3m

r=? h=

h=8m

h = 45 cm

11 m 2

a=?

R=?

TSA = 9126 m

LSA = 8100 cm

LSA = 88 m

3520 m2 TSA = 7

20 Identify the measure of the missing dimensions for the given volume of the solid shapes. c

m l l==65dm h=?

r = 2.6 mm

r=?

h = 3.5 cm

b b=3m

h=?

R = 14 cm

a=? V = 1728 cm3

V = 63 m3

V = 475.904 mm3

V = 1056 cm3

Word Problems 1 A square sheet of paper of side 30 cm is folded to form a cylinder. What is the lateral surface area of the cylinder formed?

2 The length and breadth of a wooden box are 50 cm and 30 cm respectively. The total cost of packaging the box is ₹26,250 at the rate of ₹0.25 per cm2. What is the height of the box?

3 Jamie wants to decorate a Christmas tree. He wants to wrap a cuboidal wooden box in

coloured paper. The measurements of the wooden box are 60 cm × 40 cm × 30 cm. How many square sheets of coloured paper does he need to buy if the length of each side of a square sheet is 50 cm?

4 A brick measures 15 cm × 5 cm × 5 cm. How many bricks will be required for a wall of dimensions 5 m × 1.5 m × 1.5 m?

5 Water flows into a tank of base measurements of 250 m × 100 m through a square pipe of

measurement 1.5 m × 1.5 m at the rate of 25 km per hour. In how much time (in minutes) will the water level rise to 3 m?

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3 Exponents 20 and Powers Let's Recall We have learnt about numbers that are perfect squares. We have also learnt about perfect cubes. A number that can be arranged in the form of a square using unit squares is called a perfect square, whereas a number that can be arranged in the form of a cube using unit cubes is called a perfect cube. For example, look at the set of unit squares and unit cubes. The unit squares are arranged in such a way that they form a square. So, 64 is a perfect square.

The unit cubes are arranged in such a way that they form a bigger cube. So, 64 is a perfect cube.

82 = 8 × 8 = 64

43 = 4 × 4 × 4 = 64

We can find the last digit of the square and cube of any number by looking at the digit in its ones place. For example, look at the square numbers and cube numbers. Both squares end in 6 142 = 196 and 162 = 256 If a number has 4 or 6 in its ones place, the number's square ends in 6.

This cube ends in 8

This cube ends in 2

123 = 1728 and 183 = 5832 If a number has 2 in its ones place, its cube ends in 8 and vice versa.

Letʼs Warm-up Fill in the blanks.

1 The square of 11 is _________ and its cube is _________. 2 I f the units digit of a number is 3, then the units digit of its square is _________ and that of its cube is _________. 3 The prime factorisation of 48 is _________. a

Is it a perfect square? _________ (Yes/No)

The smallest number that 48 should be divided by to get a perfect square is _________.

4 Reeti says, “125 is a perfect square”, whereas Meeti says, “125 is a perfect cube.”_________ is correct. I scored _________ out of 4.

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Exponents Real Life Connect

Bacteria are very small microorganisms that grow and reproduce very quickly. Under favourable environmental conditions, the bacterial growth shows an increase in the number of bacteria in a population rather than in the size of individual cells. During binary fission, with each generation, the bacteria split into new identical copies every time.

Fission Begins 2 New bacteria Bacteria

Process Repeats

Generation

Population of Bacteria

Exponential value

20 = 1

21 = 2

22 = 2 × 2

Log Number of Cell

The table given below shows how quickly bacteria grows. Based on the table, let us see how can we find the population of bacteria in its 10th generation. Stationary face

23 = 2 × 2 × 2

Log Face 4

Death 5Face

16 Lag Face

24 = 2 × 2 × 2 × 2

32 25 = 2 × 2 × 2 × 2 × 2

Time

Bacteria Growth

Exponents of a Rational Number

We saw that for generation 5 the population of bacteria was 32. This was expressed as given below. It is read as 2 raised to the power 5 or the 5th power of 2. 32 = 2 × 2 × 2 × 2 × 2 = 25 So, we can say that for generation n, the population of bacteria will be 2n. Therefore, for generation 10 the population of bacteria will be 210, that is, 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024. Here, the number 2 is a rational number. What if the numbers are other types of rational numbers such as fractions or decimals? Let us see! Look at the numbers. How would we find their value? 2

6 7

0.5

615 0.718 6

Numbers such as 22, 34, 48 and 615 (given above) are very large numbers and numbers such as 6 , 0.511 7 and 0.718 are very small numbers. To write such numbers, we use exponents. This helps us read such numbers easily and in less time. Power of Exponent

22 34 48 0.511

6 7

615 0.718 823

Base

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A number that can be expressed in exponential form can be either a positive or a negative rational number. For example, a positive rational number 6 when multiplied by itself 6 times is written as: 6 7 7 6 × 6 × 6 × 6 × 6 × 6 = 46,656 . 7 7 7 7 7 7 1,17,649

6 It is read as 6 raised to the power 6 or the 6th power of . 7 7 Similarly, a negative rational number, say −7 when multiplied by itself 7 times is written as: (−7)7 = (−7) × (−7) × (−7) × (−7) × (−7) × (−7) × (−7) = −8,23,543 It is read as −7 raised to the power 7 or the 7th power of −7.

Therefore, if any rational number, say a, is multiplied by itself n times, then a × a × a × a … n times = an It is read as a raised to the power n or the nth power of a.

Remember! A number that is repeatedly multiplied by itself n number of times is expressed in exponential form as an, where a is the base (the number to be multiplied) and n is the exponent or power (number of times the base is multiplied). It is read as a raised to the power n or the nth power of a.

Positive Integral Exponent of a Rational Number

A number that can be expressed in exponential form can also be either a positive or a negative rational 6 number. For example, a positive rational number 5 when multiplied by itself 6 times is written as: 5 = 6 6 5 × 5 × 5 × 5 × 5 × 5 = 56. 6 6 6 6 6 6 66 Similarly, a negative rational number, say −6 when multiplied by itself 7 times is written as: 7 7

−6 −6 −6 −6 −6 −6 −6 (−6)7 −6 = × × × × × × = 7 . 7 7 7 7 7 7 7 7 7

We know that positive and negative integers, fractions and decimal numbers all together constitute rational numbers. So, if any rational number, say a b Example 1

a is multiplied by itself n times, then b

an a = a × a × a … n times = n, where is the base and n is the exponent. b b b b b

Find the value of: 1

−2 3 −2 3

−2048 . (−2)11 −2 −2 −2 −2 −2 −2 −2 −2 −2 −2 −2 = × × × × × × × × × × = 11 1,77,147 3 3 3 3 3 3 3 3 3 3 3 3

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1 3 (−0.02)2 8

1 8

Example 2

13 1 1 1 1 = × × = . 3 8 8 8 512 8

Evaluate: −1 3 8

+ 3 6

(−0.02)2 = (−0.02) × (−0.02) = 0.0004

(−1)8 35 + 5 38 6 −1 −1 −1 −1 −1 −1 −1 −1 3 3 3 3 3 = × × × × × × × + × × × × 3 3 3 3 3 3 3 3 6 6 6 6 6 1 6593 1 32 + 6561 = + = = 6561 32 2,09,952 2,09,952 −1 + 3 3 6

Do It Together

(

Find the value of −3 5 6

) (

)

÷ 2 . 15

6 −3 ÷ 2 = (−3) ÷ 23 6 5 153 5 15

= ________________

Negative Integral Exponent of a Rational Number We have already learnt about how to solve numbers with positive exponents. Now, let us understand how to solve numbers with negative exponents. Look at the given pattern. … … 23 = 2 × 2 × 2 = 8 = 16 × 1 2 2 1 2 =2×2=4=8× 2 21 = 2 = 4 × 1 2 0 2 =1=2×1 2

We know that the set of integers is represented by {…, −3, −2, −1, 0, 1, 2, 3, …}. So, if using integers, continuing this pattern with negative exponents, this is what we get as their values.

Think and Tell

As the exponent decreases by 1, what do you think about the rule of this pattern? Can we extend this pattern further?

Remember! The reciprocal (multiplicative inverse) of a

2−1 = 1 × 1 = 1 2 2 2−2 = 1 × 1 = 1 = 12 2 2 4 2 1 −3 2 = 12 × = 1 × 1 = 1 = 13 and so on. 2 4 2 8 2 2 On observing the pattern, we can say that in each case, exponential numbers are reciprocals of each other.

number is 1 divided by that number. 1 For example, the reciprocal of 3 is . 3

So, if a is any number and n is any positive integer, then its reciprocal is written as a−n. It is read as a raised to the power −n.

represents repeated division. Negative exponents tell

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Remember! When a number has a negative exponent, it us the number of times a number is divided by itself.

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Example 3

Express the numbers with negative exponents. 1 (−8) × (−8) × (−8)

Example 4

(−8) × (−8) × (−8)3 = −1 8 Evaluate the numbers. 1 (−4)−4

(−4)−4 =

1 = 4

−2

−5

5 6

× 2 3

1 4

× 1 2

−3

× 3

= 6 5

1 × 4

1 = 1 4 256

× 3 2

× 1 2

8 8 8 8 8 × × × × 7 7 7 7 7

−2

(1.2)

Example 5

7 8

−5

Think and Tell Is

63 63 3 3 3 3 3 1 1 1 2187 = × × × × × × × × × = 5 5 2 2 2 2 2 2 2 2 1600

3 (1.2)−2

8 8 8 8 8 × × × × = 8 7 7 7 7 7 7

−2

2 3

3 2

? Why?

2 2 1 = = 10 = 102 = 100 = 25 (1.2) 12 144 36 12 −1

6 By what number should 7

be multiplied so that the product is 14 15

−1

Let the required number be a. Then, a× 6 7

−1

= 14 15

−1

⇒a× 7 6

So, the required number is 45. 49 Do It Together

= 15 14

63 45 ⇒ a = 157 × = 7 49 14

Write the expanded form of 23.456 with negative exponents. 23.456 = 20 + 3 + 0.4 + 0.05 + 0.006

= 20 + 3 + 4 + ____ + ____ 10 1 1 1 = 2 × 10 + 3 × 1 + 4 × + ____ × ____ + ____ × ____ 10 = 2 × 101 + 3 × 100 + 4 × 10−1 + ____ × 10 ____ + ____ × 10 ____

Do It Yourself 20A 1 Expand and find the value of the numbers. a 46

b (−7)−3

4 7 5

2 Write the expanded form of the numbers using exponents. a 45.123

b 541.056

c 107.02356

−2 3

−4

e (0.02)5

d 28.54687

e 5003.671

d (−20)−2

f (6.1)−2

3 Find the reciprocal of the numbers. a 5−10

2 3

−4 7

−5

3 5

−6

(1.1)−8

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4 Express the numbers with negative exponents. a d

7 7 7 7 7 × × × × 11 11 11 11 11

8 8 8 8 × × × 13 13 13 13

4 4 4 4 4 4 4 4 × × × × × × × 9 9 9 9 9 9 9 9 3 3 3 × × 7 7 7

(−5) × (−5) × (−5) × (−5)

0.9 × 0.9 × 0.9 × 0.9 × 0.9 × 0.9

5 8

5 Solve and express your answer with negative exponents. 3

2 3

6 11

−3

+ 3 4 ÷

−3

3 121

−2

1 6

−2

× 4 5

−4

e [(4)−5 + (−0.8)−3] ×

6 By what number should 3

−3

7 By what number should −5

−3

8 6

8 What should be added to 3 9 Find the value of x if 5 6

1 2

−3

−2

− 1 5

1 f [(−1.2)−5 ÷ (0.01)3] × 5

be completely divided so that the quotient is 16? be multiplied so that the product is (0.8)4?

−8

−5

× 5 2

to get 2 3

−x= 5 6

10 Shreya solves (0.4)2 × (0.01)2 × 1 ÷ (6)−1 and gets the answer as 12. Is she correct? Explain your answer. 2

Word Problems 1

When a frog larva hatches, it weighs only 5–3 g. But each day, it can eat only 26 times its body weight. How much food can the larva eat on the first day?

A school’s function video was uploaded on the

school’s YouTube channel. Since the time it was

uploaded, the online traffic has been monitored.

The table shows the number of views the video got

Week

Number of Views

125

625

each week.

a By how much are the views increasing each week? b Observe the pattern of several views, and find the rule of the pattern. c How many views will the school get in the 4th week, if it follows the same pattern?

Application of Exponents and Laws of Exponents We know that the table given below shows how quickly bacteria grow. Each generation increases its population to 2 times. Number of Generations Number of Populations Exponential Value

20 = 1

21 = 2

22 = 2 × 2

23 = 2 × 2 × 2

Chapter 20 • Exponents and Powers

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24 = 2 × 2 × 2×2

25 = 2 × 2 × 2×2×2

26 = 2 × 2 × 2 × 2 × 2×2

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In the previous grade, we already studied the laws of exponents for positive powers. We know that, am × an = am + n, where a is a rational number and m, n are two different positive integers as powers. So, using this law we can find the number for the population of bacteria for the 10th generation as; 29 × 2 = 29 + 1 = 210 = 1024.

What if the powers are negative? Let us see!

1 1 = = (1024)−1 210 1024 So, am × an = am + n holds true for any negative integers m and n and rational number a also. If the powers are negative, then 2−9 × 2−1 = 2(−9) + (−1) = 2−9 −1 = 2−10 = Apart from this, we have some other laws of exponents. 1 am ÷ an = am – n 0

4 a =1

(am)n = amn

a b

−m

= b a

am × bm = (ab)m

am = a bm b

Applying Laws of Exponents We can apply the laws of exponents in various forms. For example, let us find the value of 2 3

−2

2 3

−5

Here, like the laws for positive exponents, we have to find the quotient of two rational numbers when the bases are the same but the powers are different. So, in this case, we apply the law: am ÷ an = am – n 2 3

−2

2 ÷ 3

−5

2 = 3

−2 − (−5)

2 = 3

−2 + 5

= 2 3

Similarly, we can apply other laws as well. Example 6

Evaluate: 3 4

−1 6

3 = 4

−1 × 6

3 = 4 –3

−6

4 3

1 40

−3

1 (5−1 + 7−2 + 9−3)−1 = 1 + 12 + 13 5 7 9

−1

2 Example 7

23 8 = 33 27

1 5

−3

× 1 8

−3

1 1 = × 5 8

Find the value of:

6 11

−3

11 4

= −2

11 6

6 4096 = 46 = 729 3

= 403 = 64,000

1 1 1 = + + 5 49 729

35,721 + 3645 + 245 1,78,605 3

−1

39,611 1,78,605

−1

1,78,605 = 39,611

2 2 4 113 42 = 3 × 2 = 11 ×3 4 = 11 ×(2 × 32) 11 6 11 6 (2 × 3)

11 × 22 × 22 11 ×22 11 × 4 44 22 = = = = 23 × 33 2 × 33 2 × 27 54 27

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3 ((−1)−2 + (2.03)−3) × 33 = (−1)2 +

Example 8

−3

100 ×3 = 1+ 203 3

× 33

10,00,000 93,65,427 × 33 = × 27 = 25,28,66,529 83,65,427 83,65,427 83,65,427

= 1+

1 2.03

Simplify: 64 × 5 2 × 3 243 × 2 × 10 64 × 5−3 × 34 43 × 5−3 × 34 43 × 5−3−1 × 34−5 (22)3 × 5−4 × 3−1 26 × 5−4 × 3−1 = 5 = = = 2 2 2+1 3 243 × 2 × 10 3 ×2 ×2×5 2 2 23 = 26−3 × 5−4 × 3−1 = 23 × 5−4 × 3−1 = 23 ×

Example 9

1 1 1 1 8 = 8× × = 4× 5 3 625 3 1875

Find the value of x for which 7x ÷ 7−5 = 78. 7x ÷ 7−5 = 78 ⇒7x−(−5) = 78 ⇒ 7x + 5 = 78 ⇒ x + 5 = 8 ⇒ x = 8 – 5⇒ x = 3

Do It Together

Find the value of 7 9 =

−2

9 7

× 5 7

−2

× 7 5

7 9

−2

5 7

−2

+ (3.5)2 − −5

+ (3.5)2 −

1 4

+ (3.5)2 −

45 ÷ 4−8

1 4

1 −5 1 ÷ 4 4

= = =

Use of Exponents In our daily lives, we come across numbers like 2 cm, 766 km, 3,59,000 km, 18,76,53,00,000 m, 0.000000045 mm, etc. in the fields of science and engineering. Among these, numbers like 3,59,000 km and 18,76,53,00,000 m are very large and a number like 0.000000045 mm is very small. Reading these numbers in the usual form is very difficult and time-consuming. So, to read and write these numbers easily in less time, we express them in exponential form using the standard form or scientific notation. Scientific notation is a way of expressing very large or very small numbers as a product of a decimal number (ranging from 1.0 to 10) and an exponent of 10, where the exponent tells us the number of places the decimal point is moved. For example, 2.9 × 102, 3.6587 × 10−3, etc. Therefore, a number can be expressed in the standard form (or using scientific notation) as k × 10n, where 0 < k < 10 and n is an integer Now, let us see how we can express 3,59,000 km and 0.000000045 mm using scientific notation.

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359000 km

0.000000045 mm

The decimal is moved 5 places to the left.

The decimal is moved 8 places to the right.

359000. = 359 × 1000

0.000000045 =

= 3.59 × 100 × 103 2

= 3.59 × 10 × 10

Example 10

Example 14

1000000000 45

4.5 × 10 109

= 3.59 × 102 + 3

= 4.5 × 101 − 9

= 3.59 × 105 km

= 4.5 × 10−8 mm

Write the numbers in standard form. 1 548900000 = 5489 × 100000 = 5.489 × 103 × 105 = 5.489 × 108 3465 3465 3.465 × 103 2 0.0000000003465 = = = = 3.465 × 103 – 13 = 3.465 × 10−10 10000000000000 1013 1013

Example 11

Write the numbers in the usual form. 1 5.67 × 105 = 567 × 105 = 567 × 105 100 102

= 567 × 105 – 2= 567 × 103 = 5,67,000

8912 1 8912 1 2 8.912 × 10−8 = × = × 8 8 1000 10 103 10

8912

103 + 8

8912 11

10 = 0.00000008912

Example 12

Do It Together

Did You Know? The distance from the Earth to the nearest star, Proxima

Centauri, is about 4 × 1013 km.

8912 100000000000

Which is greater, 7.42 × 104 or 5.6 × 107?

742 × 104 = 742 × 104 – 2 = 742 × 102 = 742 × 100 = 74,200 100 56 5.6 × 107 = × 107 = 56 × 107 – 1 = 56 × 106 = 56 × 10,00,000 = 5,60,00,000 10 Here, 5,60,00,000 > 74,200. 7.42 × 104 =

So, 5.6 × 107 > 7.42 × 104. Example 13

The mass of the Earth is 5.97 × 1024 kg and that of the moon is 7.34 × 1022 kg.

1 What is the combined mass of the two? Express your answer using scientific notation. Total mass = 5.97 × 1024 + 7.34 × 1022 kg = 5.97 × 1022 × 102 + 7.34 × 1022 kg

= (5.97 × 102 + 7.34) × 1022 kg = (597 + 7.34) × 1022 kg = 604.34 × 1022 kg = 6.0434 × 1024 kg 2 H ow many times the mass of the Earth is the mass of the moon? Express your answer using scientific notation. 5.97 × 1024 Number of times the mass of the Earth is the mass of the moon = 7.34 × 1022 24 – 22 5.97 × 10 5.97 597 = = × 102 = 7.34 7.34 7.34 = 81.335 = 8.1335 × 101

Thus, the mass of the Earth is 8.1335 × 101 times the mass of the moon.

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Example 14

The average distance from the sun to the Canis Major Dwarf galaxy is 1.47 × 1017 miles. Light travels at a speed of 5.88 × 1012 miles per year (called a light year). How long does it take (in years) for light to travel from the sun to the Canis Major Dwarf galaxy? Distance from the sun to the Canis Major Dwarf galaxy = 1.47 × 1017 miles Speed of light = 5.88 × 1012 miles per year Duration of the time light takes to travel from the sun to the galaxy = 5

1.47 × 1017 1.47 × 1017 – 12 = 5.88 × 1012 5.88

Error Alert!

1.47 × 10 5.88 1.47 = × 105 = 0.25 × 105 = 2.5 × 10−1 × 105 5.88 =

In scientific notation, ‘× 10’ NEVER means multiplication. It represents the exponent, that is, the number of times we need to move the decimal point towards the left or right to obtain the actual value.

= 2.5 × 10−1 + 5 = 2.5 × 104 years Do It Together

Which is smaller, 5.09 × 10−6 or 2.13 × 10−8 and by how much? 5.09 × 10−6 = 2.13 × 10−8 =

2.5 × 104 = 2.5 ×

2.5 × 104 =

509 1 509 1 × = × = ____________ 100 106 10____ 106

10,000 = 25,000

2.5 × 10 = 25

213 1 213 1 213 × ____ = × 8 = ____ = _____________ ___ 10 ___ 10 10

Do It Yourself 20B 1 Evaluate. 3

–5

–1 –5

–1

–2

4 ÷

–1 –2

2 –3 25 2 ÷ 5 4

–4 2

–4 –5 9

–4 –1 9

3 8

–4

c (5−9 × 5−3) × 155

–2 3

–5

–2 –3 ÷ (2.4)−3 × (0.5)0

3 –4 × 7

e ((−2)−2 + (1.6)−3) × (0.8)3

(–7)–1+

2 Find the value of:

a (3–2 + 8–3 + (–2)–4)0

–2 –4

–5

9 1 –1 3 7

3 Simplify using laws of exponents. a

72 × 3–6 × 42

–2

81 × 16 × 27

20 × 3–3 × 40

36 × 5 × 45

4 Match the following. Explain your reasoning. a (–5)3 × (–5)–5 3 4 b 8 × 2 × 3 3 × 64

611 65 × 66

Chapter 20 • Exponents and Powers

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–5

9 × t–3 –5

18 × 3

–8

×t

16 × 2m + 1 – 4 × 2m

16 × 2m + 1 – 2 × 2m + 2

–2 –3 (–2)5 × 3 4 20 × 30 × 70 625 × (–5) –2 54

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5 Write the numbers in standard form. a 7687000000

b 2300000000000000

e 98.0000324

c 10.446678

f 453.98785

d 65.9084

g 4320000000

h 10980000

c 6.0945 × 10-3

d 4.21 × 104

6 Write the numbers in the usual form. a 1.03 × 108

b 3.467 × 10−5

e 8.7 × 1011

f 1.768 × 104

g 54.1 × 106

b 2.11 × 102, 1.67 × 10−2

c 7.3 × 10−5, 0.7 × 10−8

h 40.857 × 10−2

7 Compare the numbers. a 3.4 × 103, 8.05 × 106

8 Solve as directed. Give your answers in standard form. a 675.7869 × 105 + 980.436 × 103

b (3.87 × 10−7) × (1.02 × 10−3)

c 0.000000000043 − 0.000000000892

d 0.000000071192 ÷ 0.0000000006472

9 Write the missing number in the equation: 10 Find the value of m if

y m+3× ( y 2 ) m–1 y 3m – 9

–1 2 5

–1

–2

___

× 511 = 55

= 1024 and y = −2.

Word Problems 1

An ant carries a small amount of food that is 2.78 × 10–3 inches. Another ant carries an

amount of food that is 4.56 × 10−5 inches. How long are the pieces of food carried by the

ants altogether?

In the population of a country, 7.7 × 103 people spoke French as their first language and

A swarm of mosquitoes contains as many as 80 million mosquitos per square mile on a 1500

3.2 × 104 people spoke German. Which language had more speakers and by how much? square mile area of land. How many mosquitoes should the swarm have to cover the entire area of land?

4 5

What is the mass of 6 million dust particles if a dust particle has a mass of 7.32 × 10−10 mg? Scientists discovered that the size of the Antarctic Ocean is 20,330,000 sq. km. How big 3 is the Arctic Ocean, if it is th of the size of the Antarctic Ocean? Express your answer in 4 standard form.

Points to Remember •

This representation of numbers with a base and exponent is the exponential form of numbers.

• If any integer, say a is multiplied by itself n times, then a × a × a × a … n times = an. It is read as: a raised to the power n or the ath power of n. a n a a a a • If any rational number, say is multiplied by itself n times, then = × × … n times = b b b b b n a × a × a...× n times a a = , where is the base and n is the exponent. b × b × b...× n times bn b • If a is any number and n is any positive integer, then its reciprocal is written as a−n. It is read as: a raised to the power −n 284

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•

Some laws of exponents are: 1

am × an = am + n

(am)n = amn

a0 = 1

am a m m= b b

am ÷ an = am – n

am × bm = (ab)m

b m a –m = a b

• A number can be expressed in the standard form (or scientific notation) as k × 10n, where 0 < k < 10 and n is an integer.

Math Lab Bacterial Growth Culture Setting: In groups of 5 Materials Required: Active dry yeast, warm water, sugar, a plastic bottle with a narrow neck, a balloon, a funnel, a teaspoon, measuring cup, marker, gloves, ruler or tape Method: All 5 members of each group must follow these steps. 1 1 Take a plastic bottle and add 1 packet of active dry yeast about 2 teaspoons to it using 4 the funnel. 2 Use the measuring cup to measure 2-3 tablespoons of warm water (around 110°F or 43°C) and pour it into the bottle with the yeast. 3 Add 1-2 teaspoons of sugar to the bottle. 4 Quickly attach the balloon to the top of the bottle, ensuring it forms a tight seal over the opening. 5 Use a marker to mark the initial height of the balloon on the bottle. Place the bottle in a warm and stable location. 6 Observe the bottle and balloon every 15 minutes. What do you observe? Does your balloon inflate? Why? 7 Measure the diameter of the inflated balloon after every 15 minutes using a ruler or a tape and record it your notebook. 8 Represent the growth of the gas in the balloon with an exponent. The number of times the balloon increases in size can be calculated by dividing the final diameter by the initial diameter. 9 What do you conclude? Is the growth of gas directly related to bacterial growth in the bottle? 10 Find the general rule of the pattern you obtain and find the growth of gas in the balloon when its diameter increases to 20 cm or more.

Chapter Checkup 1

Write the expanded form of the given numbers using exponents. a 65.879

978.9675

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21.4879

4876.9023

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Find the reciprocals of the given expressions. a d

–2 –2 –2 –2 –2 –2 × × × × × 5 5 5 5 5 5 −5

Solve. a

−5

11 2

−

−5

11 6

−5

−2

7 1

5 2 d [(6) × (−1.5) ] ×

3 3 3 3 3 3 3 3 × × × × × × × 8 8 8 8 8 8 8 8 1

−3

−2

10 3

−2

Express 59,049 as a power of 3.

What power of 49 is equal to 710?

Arrange the numbers in ascending and descending order. b

7 9

c f

10 3 2 −2 3 a 2 , 32 , 64 , 128 , 8

−1

2 5

[(−0.2)−4 ÷ (0.03)3] ×

−5

310, 273, 7292, 18−2, 93

7 A teacher asked her students to find the multiplicative inverse of x−2. Tanya said the answer is –x2, whereas Sayani said the answer is x−3. Who do you think is right? Explain the reason for the error.

8 A teacher asked the class to write 55 with a negative exponent so that its value remains the same. Hetal said it was 5−5 whereas Kamya said it was

as we need to take the reciprocal before changing the negative of the 5−5 exponent. Who do you think is right? Explain your answer.

Find the value of: −2 −3 −4 0 a (8 + 5 + (−3) )

−4

–

−4

−2

−3

÷ (1.7)−3 × (0.2)0

−2

((−1)

c −3

+ (1.01) ) × (0.1)

−1

(−8)

–5

−4

7 −1

10 Simplify using laws of exponents. a

128 × 3−6 × 42 243 × 36 × 27−2

40 × 3−3 × 40 81 × 5−5 × 45

11 Find the value of x in: 10 6 −4 3x − 1 a 2 ÷2 =2 ×2

−8x

24 × p−3 × 49 21 × 3−5 × p−8 =

−10

5 8

15 × 5m + 1 + 20 × 5m × 5 3 × 5m + 2 + 20 × 5m + 1

72x + 1 ÷ 343 = 2401

12 Identify if the statements are written in standard form or usual form. If the numbers are in standard form, express them in usual form and vice versa.

−5 a The size of a plant cell is 1.275 × 10 m.

b The thickness of a human hair is in the range 0.005 cm to 0.01 cm. 8 c The distance between the Earth and the moon is 3.84 × 10 m.

9 d The total volume of water on Earth is 1.386 × 10 cu. km.

e The weight of an electron is 0.0000000000000000000000000000911 kg.

13 Which of the following is in usual form? Give reason. 5 a 8.09 × 10

0.000000654879

9.21 × 10−21

0.000122434

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14 Srishti got the value of (1296)2 as 6,04,66,176. Is she correct? Justify your answer. (Hint: Calculate the square root of 1296 first.)

Simplify: x

p+1

×xq+p xl+q (x × x q × x l)2 p

Word Problems 1

A farm that has 6 × 105 m of land produces 2.4 × 106 kg of cotton each year. What is the average

At a storage warehouse, each crate weighs 2.2 × 103 kg. If there are 4 × 102 crates, what is their

amount of cotton planted per metre? combined weight?

The area of the Pacific Ocean is 1.55 × 108 sq. km. The area of the Atlantic Ocean is 1.0646 × 108 sq. km. How many times greater is the area of Pacific Ocean than the Atlantic Ocean? Express your answer in scientific notation.

The size of a plant cell is 0.00002155 m and that of a red blood cell is 0.0000000015 m. Is the

The diameters of the planets Jupiter and Saturn are 13.9822 × 104 km and 1.165 × 105 km,

plant cell double the size of a red blood cell?

respectively. Compare the diameters of these two planets.

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and 321 Direct Inverse Proportions Let's Recall Ratio is defined as the comparison between two quantities of the same unit that indicates how much of one quantity is present in the other quantity. The ratio of quantities a and b can be given as:

a:b = Antecedent

a b Consequent

Ratios are said to be in their simplest form when there are no common factors between the antecedent and consequent except 1. 12 For example, is in its simplest form as the only common factor between 12 and 5 is 1. 5 We can find the simplest form of a rational number by dividing both the antecedent and consequent by their HCF. 81 81 ÷ 3 27 For example, = = 6 6÷3 2 We can also find the equivalent ratio of a ratio by multiplying or dividing the antecedent and consequent by the same natural number. Equivalent ratios represent the same value when reduced to their simplest form 2 2×2 4 2×3 6 2×4 8 For example: Equivalent ratio of = = or = or = 5 5 × 2 10 5×3 15 5 × 4 20 6 8 Hence, 2 = 4 = = 15 20 5 10

Letʼs Warm-up State True or False.

1 There are 45 boys and 39 girls in a hall. The ratio of boys to girls can be given as 2 The simplest form of the ratio 27:6 = 9:2. 11 88 3 is the simplest form of . 2 12 4 A ratio can have infinite equivalent ratios. 5 One of the equivalent ratios of 81:30 is 27:10

39 45

___________ ___________ ___________ ___________ ___________

I scored _________ out of 5.

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Proportion Real Life Connect

Aakash’s birthday is approaching. His parents are busy in the preparation of the party. Father: I talked to the caterer. He is charging ₹250 per guest. Mother: I have prepared the list, there are 20 guests that are coming to the party. Father: Great! I also have a list of 5 people who might join the party.

Direct Proportion We saw that the ratio of number of guests to the amount to be paid is 1:250 Let us make a table and see how much Aakash's parents need to pay to the caterer if the number of guests changes. Number of guests

Amount to be paid

Ratio

₹250

1:250

₹2500

10:2500 = 1:250

₹5000

20:5000 = 1:250

₹6250

25:6250 = 1:250

Increase in the number of guests

Increase in the amount to be paid

Direct Proportion

If the values of two quantities depend on each other, so that a change in one leads to a change in the other, they are said to be in variation. Two quantities a and b are said to be in direct proportion, if the increase or decrease in the value of a leads to a corresponding increase or decrease in the value of b so that their ratio remains constant.

a = c where c is a constant. b Further, if a1 and a2 are the number of guests and b1 and b2 are the corresponding amounts, then Thus, a and b are in direct proportion, if

a1 a2 = =c b1 b2

a1 × b2 = a2 × b1 = c

Let us now find the amount to be paid if there are 30 guests at the party. We know that amount to be paid for 20 guests = ₹5000 Let the amount to be paid for 30 guests = x 20 30 1,50,000 = or 20x = 1,50,000 ⇒ x = = 7500 5000 x 20 Hence, the couple needs to pay ₹7500, if 30 guests join the party. Using the above formula,

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Example 1

If the weight of 25 sheets of paper is 60 grams, how many sheets of the same paper will weigh 3 kilograms? Let the number of sheets be x. Number of Sheets

Weight of Sheets

3000

Remember! The units of ratios in proportion are the same.

25 x = ⇒ 60x = 75,000 ⇒ x = 1250 60 3000 Hence 1250 sheets of the same paper will weigh 3 kg. Example 2

A bus travels 126 km on 9 litres of petrol. How far would it travel on 5 litres? Let the distance travelled be x. Distance Travelled

126

Petrol Needed

126 x = ⇒ 9x = 630 ⇒ x = 70 9 5 Hence the bus can travel 70 km on 5 litres of petrol. Do It Together

The cost of 12 m of cloth is ₹900. Complete the table using this data. Length of Cloth

Cost of Cloth

15 m

₹150

₹600

₹1500

Do It Yourself 21A 1 Which of the given quantities vary directly with each other? Write Yes/No. a The speed of a car and the time taken by the car to cover a particular distance b The number of men required to construct a wall and the time taken by the men c The number of pencils and the total cost of the pencils d The number of vehicles on the road and space on the road

2 Observe the tables to find whether a and b are directly proportional. Also give the constant of variation in the case of direct proportion. a

108

190

282

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117

180

100

3 Complete the table if x and y vary in direct proportion. x

132

176

152

361

4 Find the cost of 52 kg of rice if 9 kg rice costs ₹585. 5 A machine in a chocolate factory makes 450 chocolates in six hours. How many chocolates will it make in four hours?

6 Kamal can type 1020 words in 15 minutes. How many words will he type in 1 hours? 2

7 A worker saved ₹148.20 when he worked for 3 days. How much could he save if he worked for 20 days? 8 Shalini takes 25 minutes to walk a distance of 1.5 km. How much distance will she cover in 40 minutes? 9 A bottling machine manufactures 75 bottles in 40 minutes. How many bottles can it manufacture if it worked continuously for a day?

10 If five cardboard boxes can be put in 1500 cubic centimetres of space, how much space can 150 such boxes occupy?

11 15 workers can dig a 6 m long stretch in one day, how many workers will be able to dig a 42 m long stretch of the same type in a day?

12 A cab driver charges a fare of ₹204 for a distance of 8 km. How much will a customer need to pay if he travelled a distance of 22 km?

13 72 boxes of certain goods require a shelf of length 17.6 m. How many boxes of the same goods would occupy a shelf of length 19.8 m?

14 A worker is paid ₹2000 for 8 days of work. If he works for 12 more days, find the amount he will be paid for those 12 days.

15 A recipe requires 15 tablespoons of sugar to make 22 cookies. If 1 tablespoon weighs 12.5 grams, how much sugar (in grams) is required to make 154 cookies?

16 The recipe shows the ingredients needed to make 20 brownies. 100 g chocolate

3 eggs

75 g plain flour

325 g sugar

1 teaspoon baking powder

2 tablespoons butter

1 4

Rohit wants to make 35 brownies. He has plenty of the other ingredients but 150 g of plain flour and 500 g of sugar. Does he have enough plain flour and sugar?

17 A 4 m 80 cm high vertical pole casts a shadow 1 m 20 cm long. Find the length (in cm) of the shadow cast by another pole 12 m 60 cm high under the same conditions.

18 In 14 days, the earth picks up 2.8 × 1010 kg of dust from the atmosphere. In how many days (approx.) will it pick up 5600 × 109 g of dust?

Chapter 21 • Direct and Inverse Proportions

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Word Problems Example 40 labourers are required to construct a building in 1 year. How many labourers are required to Simran bought bagsdays? of sugar for resale that weigh 90 kg. How many such bags will construct the1same building in5200 weigh 576 kg?

Let the number of labourers required be 2 Karishma takes 112 minutes to walk 7 miles. How many miles can she walk in 4 hours?

Inverse Proportion For Aakash’s birthday party, his parents also ordered 180 ladoos and thought of packing the same number of ladoos in each box as a sweet gesture for each guest.

Problems on Inverse Proportion Number of Guests

Ladoos to be packed in each box

Increase in the number of guests

Decrease in the number of ladoos in each box

Inverse Proportion

Two quantities a and b are said to be in inverse proportion, if the increase or decrease in the value of a leads to a corresponding decrease or increase in the value of b such that their product remains constant. Thus, a and b are in an inverse proportion, if ab = c where c is a constant. Further, if a1 and a2 are the number of guests and b1 and b2 are the corresponding ladoos to be packed, then a1 b2 = =c a2 b1

a1 × b1 = a2 × b2 = c

Let us now find the number of ladoos to be packed if there are 30 guests at the party. We know that for 20 guests, ladoos to be packed = 9 Let the number of ladoos to be packed for 30 guests = x 180 Using the above formula, 20 × 9 = 30 × x ⇒ x = =6 30 Hence, the couple needs to pack 6 ladoos in each box, if 30 guests join the party. Example 3

40 labourers are required to construct a building in 1 year. How many labourers are required to construct the same building in 200 days? Let the number of labourers required be x.

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Number of Labourers

Time

365

200

As the time decreases, the number of labourers required increases, 40 × 365 Using the inverse proportion formula, 40 × 365 = x × 200 ⇒ x = = 73 200 Hence, 73 labourers are required to finish the work in 200 days. Example 4

A train takes 23 hours to complete a journey if it travels at a speed of 100 km/hr. What should the speed of the train be to cover the same journey in 20 hours? Let the speed of the train be x. Number of Hours

Speed

100

As the speed increases, the time taken to cover the distance decreases. 23 × 100 Using the inverse proportion formula, 23 × 100 = 20 × x ⇒ x = = 115 km/hr 20 Hence, the train should travel at a speed of 115 km/hr to cover the distance in 20 hrs. Do It Together

A farmer has enough food to feed 45 cows for 60 days. The farmer buys a few more cows and the food lasts for 54 days. How many cows has he bought? Let the total number of cows after he buys some cows be x. Number of cows

Time

More cows will consume more food and hence it will last for fewer days. Using the inverse proportion formula, 45 ×

=x×

Hence, the farmer bought

more cows.

− 45 =

⇒x=

Time and Work Time is the period in which any activity or work happens while work is a task or series of actions to achieve a certain result. The problems of time and work can be solved using the unitary method or proportions. We can find the work done in a certain period of time or time required to complete the work using certain rules: Rule 1 If a person requires x days to complete a task, then the 1 amount of work done in 1 day = x

Rule 2 1 of work in one day, then the amount x of time required to complete the whole task = x days.

If a person does

For example, A can do a piece of work in 12 days and B can do the same work in 18 days. How many days will they take to complete the work if they work together? Time taken by A to complete the work = 12 days 1 ∴ Work done by A in 1 day = ….(1) 12 Time taken by B to complete the work = 18 days Chapter 21 • Direct and Inverse Proportions

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∴ Work done by B in 1 day =

Example 5

1 18

….(2)

1 1 3+2 5 Work done by A and B in one day = + = = 12 18 36 36 [Using (1) and (2)]

Did You Know?

1 36 1 Time taken by A and B to do the piece of work = 5 = = 7 days. 5 36 5

the Taj Mahal in 20 years.

Suman can do a piece of work in 24 days and Disha in 20 days. If they work on it together for 6 days, what is the fraction of work left? Time taken by Suman to complete the work = 24 days 1 ∴ Work done by Suman in 1 day = ….(1) 24 Time taken by Disha to complete the work = 20 days 1 ∴ Work done by Disha in 1 day = ….(2) 20 1 1 5 + 6 11 Work done by Suman and Disha in one day = + = = 24 20 120 120 11 11 Work done in 6 days = 6 × = 120 20

Example 6

22,000 workers worked to build

[Using (1) and (2)]

11 9 Fraction of work left = 1 – = 20 20 9 So, of the work is left. 20 Karim can lay a railway track between two stations in 16 days and Vishal can do the same job in 12 days. With the help of Mukesh, they did the job in 4 days. In how much time can Mukesh alone do the job? 1 (Karim, Vishal and Mukesh)’s 1 day work = 4 work in 4 days) Karim’s 1 day work

(as they complete the

Error Alert!

1 1 ; Vishal’s 1 day work = 16 12

One divided by a fraction is the reciprocal of that fraction

1 1 1 5 1 7 – = = + – 4 48 48 4 16 12 1 48 3 So, Mukesh can alone do the work in 5 = = 9 days. 5 48 5 ∴ Mukesh’s 1 day work =

Do It Together

1 3 = 3 8 8

8 1 = 3 3 8

Working together, A and B can finish a piece of work in 30 days. They worked together for 10 days and then B left. After another 40 days, A finished the remaining work. In how many days can A finish the work alone? Time taken by A and B to finish the work if they worked together = 30 days (A + B)’s 1 day of work = (A + B)’s 10 days of work = Remaining work = (1 – Now

work is done by A in 40 days.

Therefore, the whole piece of work will be done by A in (40 ×

days

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Pipes and Cisterns Problems The problems on pipes and cisterns are similar to time and work. Portion of a tank filled or emptied

Amount of work done

Time taken to fill or empty a tank

Time taken to do a piece of work Inlet pipes

Work done is positive

Cistern

Work done is negative

Outlet pipes

For example, Tap A fills a tank in 4 hours while tap B empties it in 6 hours. In how much time will the tank be filled if both the taps are opened together? Time taken by tap A to fill the tank = 4 hours 1 ∴ Work done by tap A in 1 hour = 4 Time taken by tap B to empty the tank = 6 hours

–1 6 1 1 3–2 1 Work done by A and B together in 1 hour = – = = 4 6 12 12 1 of the tank gets filled in 1 hour when both the taps are open 12 1 Time taken by A and B to fill the tank together = = 12 hours. 1 12 ∴ Work done by tap B in 1 hour =

Example 7

Think and Tell Is it possible to fill a tank if both the inlet and outlet pipes are opened together and the outlet pipe takes less time compared to the inlet pipe?

Pipes A and B can fill a cistern in 10 and 15 hours, respectively. In how much time will the cistern be filled if both the pipes are opened together? 1 Time taken by pipe A to fill the cistern = 10 hours; ∴ Work done by pipe A in 1 hour = 10 1 Time taken by pipe B to fill the cistern = 15 hours; ∴ Work done by pipe B in 1 hour = 15 1 1 3+2 5 Work done by A and B together in 1 hour = + = = 10 15 30 30

1 30 = = 6 hours. 5 5 30 Pipes A and B can fill a reservoir in 10 hours and 12 hours respectively while a third pipe C empties the full tank in 20 hours. If all three pipes are opened together, how much time will it take to fill the tank completely? 1 Time taken by pipe A to fill the reservoir = 10 hours; ∴ Work done by pipe A in 1 hour = 10 1 Time taken by pipe B to fill the reservoir = 12 hours; ∴ Work done by pipe B in 1 hour = 12 1 Time taken by pipe C to empty the reservoir = 20 hours; ∴ Work done by pipe C in 1 hour = – 20 1 1 1 6+5–3 8 Work done by A and B and C together in 1 hour = + – = = 10 12 20 60 60 Time taken by A and B to fill the cistern together

Example 8

Chapter 21 • Direct and Inverse Proportions

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1 60 15 1 = = = 7 hours. 8 8 2 2 60 1 A pump can fill a tank with water in 2 hours, but due to a leak in the tank, it takes 2 hours to fill the 3 tank. In how much time can the leakage empty the tank? Time taken by A, B and C to fill the reservoir =

Do It Together

Time taken by the pump to fill the tank without leakage = 2 hours Work done by the pump in 1 hour without leakage = Time taken by the pump to fill the tank with leakage = 2 Work done by the pump in 1 hour with leakage = 1 Leakage per hour = – = 2 Time taken by leakage to empty the tank =

1 7 = hours 3 3

hours.

Do It Yourself 21B 1 Which of the quantities vary inversely with each other? Write Yes/No. a Distance travelled by a car and time taken by the car b Number of men required to paint a house and time taken by the men c Number of books and total cost of the books d Distance travelled by a bus and petrol consumed by it

2 If x and y vary inversely, then find the value of a. x

3 42 women can do a piece of work in 15 days. In how many days can 18 women do the same amount of work? 4 If a box of chocolates is divided among 28 children, they get 6 chocolates each. How many more chocolates would each get, if the number of the children was reduced by 7?

5 A person has money to buy 15 pairs of shoes worth ₹450 each. How many fewer pairs of shoes will he be able to buy if each pair of shoes costs ₹300 more?

6 Three people could fit new doors in a house in 2 days. One of the people fell ill before the work started. How long would the job take now?

7 A group of 4 friends, staying together, consume 42 kg of wheat in 30 days. Some more friends join this group, and they find that the same amount of wheat lasts 20 days. How many new members are there in this group now?

8 A car can finish a certain journey in 12 hours at the speed of 60 km/hr. By how much should its speed be increased so that it will take only 9 hours to cover the same distance?

9 Mohit can finish some work in 6 days. How much work can he do in 1 day? 10 Suhani can revise all her subjects in 30 days. What fraction of the course will she revise in 20 days? 11 A can finish a piece of work in 14 days while B can finish the same work in 21 days. In how much time will they take to complete the work if they both work together?

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12 Manish can finish a piece of work in 40 minutes. He works at it for 8 minutes and then Priya finishes it in 16 minutes. How long will they take together to complete the work?

13 X and Y can do a piece of work in 20 hours and 12 hours respectively. X started the work alone and then after 4 hours Y joined him until the work was completed. How long did the work last?

14 Three taps can fill a tank in 2 hours, 3 hours and 5 hours respectively. If all the taps are opened together in an empty tank, how much time will they take to fill the tank?

15 A tap P can empty a full tank in 25 hours and another tank Q can empty the same tank in 30 hours. How much time would both taps take to empty the tank if opened together?

16 There is a leakage at the bottom of a tank. When the tank is repaired, it will take 3.5 hours to fill it completely but now it is taking half an hour more than usual. How long will the leakage take to empty the full tank?

17 Pipe A can fill a tank in 36 hours. Pipe B can fill it in 18 hours. Pipe C can empty the full tank in 133 hours. If all the pipes are opened together, how much time will be needed to fill the tank?

18 A can do a piece of work in 4 days; B and C together can do it in 3 days, while A and C together can do it in 2 days. How long will B alone take to do it?

19 1800 members of the merchant navy on a ship had enough food for 27 days. Some of them were transferred to a boat, so the food lasted 13 more days. How many of them were transferred to the boat?

20 A machine X can recycle 500 plastic bottles in 8 hours. Machine Y can recycle the same number of bottles in

10 hours while machine Z can recycle them in 12 hours. All the machines are started at 11 a.m., machine Z is

closed at 1 p.m. and the remaining two machines complete the process. At approximately what time will the work (to recycle 500 plastic bottles) be finished?

Word Problems 1

Anshi can finish a job in 18 days and Vinita can do the same work in 15 days. Vinita started the work alone and worked for 10 days and then left the job. In how many days can Anshi alone finish the remaining work?

2 Julie and Nimish can make a wedding cake in 15 hours and 10 hours, respectively. They

started doing the work together but after 2 hours Nimish had to leave and Julie completed the cake alone. In how much time was the whole task completed?

3 Suman and Kavya can finish a project in 30 days, while Kavya and Pulkit can finish the

same project in 24 days and Pulkit and Suman in 20 days. They all work together for 10

days when Kavya and Pulkit leave. How many days will Suman take to complete the rest of the project?

Points to Remember • If the values of two quantities depend on each other, so that a change in one leads to a change in the other, they are said to be in variation. • Two quantities a and b are said to be in direct proportion, if the increase or decrease in the value of a leads to a corresponding increase or decrease in the value of b so that their ratio remains constant. Chapter 21 • Direct and Inverse Proportions

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a1 a = 2 = c or a1 × b2 = a2 × b1 = c. b1 b2 • Two quantities a and b are said to be in inverse proportion, if the increase or decrease in the value of a leads to a corresponding decrease or increase in the value of b so that their product remains constant. a1 b • In inverse proportion, = 2 = c or a1 × b1 = a2 × b2. a2 b1 1 • If a person requires x days to complete a job, then the amount of work done in 1 day = . x 1 • If a person does of work in one day, then the amount of time required to complete the x whole job days = x days. • Work done by an inlet pipe is always positive and work done by an outlet pipe is always negative. •

In direct proportion,

Math Lab Experiment Using Inverse Proportion Setting: In groups of 4 Materials Required: Small containers or cups, water, measuring spoons of different sizes, stopwatch or timer, graph paper Method: 1 Each group will have a small container and measuring spoons of different sizes. 2 Instruct each group to pour water from the measuring spoon into the container while timing how long it takes to fill the container to a certain level (e.g., halfway). 3 Have each group perform this experiment with different spoons and record the time taken for each. 4 Ask the students to show their results with the help of a graph. 5 Discuss the results as a class and note that as the spoon size increases (greater flow rate), the time taken decreases (inverse proportion).

Chapter Checkup 1 Identify whether the values are in direct or inverse proportion and find the missing values. a

104

150

240

128

2 Find the cost of 28 markers if 18 markers cost ₹576. 3 A recipe requires 880 g of rice flour for 5 persons. How much rice flour is required if the same recipe is made for 12 persons?

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4 A truck travels 6 km in 30 minutes. What distance will it cover in 5 hours? 5 A labourer is paid ₹4950 for working for 11 days. How much money will he earn if he works for 30 days? 6 Rohan can do a piece of work in a month. What fraction of work can he do in 6 days? 7 Mihir has ₹25,000 which he can use to buy 20 tables. How many tables can he buy for ₹10,000? 8 Priya has enough money to buy 35 boxes of chocolates worth ₹120 each. How many more boxes of chocolates can she buy if she gets a discount of ₹20 on each box?

9 If 12 men can plough a field in 7 days, how many men can plough the same field in two days? 10 A road map shows a distance of 2 m representing 25 km. Seema drove for 600 km. What is the distance covered by her on the map?

11 A water tank of a society is filled in 4 hours by 8 tankers. How many tankers are required to fill the tank in 2 12 20 cows can graze a field for 32 days. How many cows can graze the same field for 128 days?

2 hours? 3

13 Mehar bought 5 kg of apples for ₹1750. How many apples can she buy for ₹2500? 14 A and B can do a job in 8 hours, while B and C can do the same work in 12 hours. A, B and C together can finish the job in 6 hours. In how much time can A and C finish the job together?

15 A lift can carry 90 people in 6 trips. How many people can it carry in 35 trips? 16 A pile of 15 identical tiles weigh 127.5 kg. What is the weight of 50 such tiles? 17 Two pipes P and Q can fill a tank in 25 hours and 35 hours, respectively. How much time will it take to fill the tank if both the pipes are opened simultaneously?

18 The daily consumption of water by 40 students in a hostel is 1.3 × 104 litres. Find the daily consumption of water if the number of students increases by 20.

19 An NGO collected ₹18,900 to distribute equally among 36 children in an orphanage. If 6 more children are admitted to the orphanage, how much money will each child get?

20 A tap can fill a tub in 24 minutes. Due to a leakage at the bottom of the tub, the tap fills the tub in 36 minutes. How much time will the leakage take to empty the tub if the tub is full?

21 3 kg of wheat contains 1.5 × 105 grains. How many grains are there in 5.5 kg of wheat? 22 Pipe M alone can fill a tank in 50 hours while pipe N alone can fill the same tank in 40 hours. Pipe O alone empties the tank in 80 hours. If all three pipes are opened together, how much time will it take to fill the tank completely?

23 Sonu can plough a field in 16 days while Kamal can plough the same field in 12 days. If they work together for 3 days, what is the fraction of work left?

24 Krishvi cycles to her school at a speed of 12 km/hr and reaches the school in 20 minutes. One day her speed was reduced by 2 km/hr due to some traffic on her way. If she left her home at 8:00 a.m., at what time did she reach the school?

25 Two taps P and Q can fill an overhead tank in 8 hours and 12 hours, respectively. Both taps are opened for 3 hours and then Q is turned off. How much time will P take to fill the rest of the tank?

In a photograph of a bacteria which is enlarged 2,00,000 times, the length it attains is 2 cm. If the photograph is reduced by 50,000 times, what would be the bacteria’s enlarged length?

Chapter 21 • Direct and Inverse Proportions

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Word Problems 1 Darsh, Yash and Pooja can complete a project in 24, 6 and 12 days, respectively. In how much time will they complete the work if they work together?

2 Manisha takes tuition classes in 10 batches each with a duration of 45 minutes. How many batches can she take if the duration is increased by 30 minutes?

3 Mohan can finish painting a house in 24 days, Subhash in 9 days and Vipul in 12 days.

Subhash and Vipul start the work but are forced to leave after 3 days. In how much time does Mohan complete the remaining work?

4 X and Y together can do a piece of work in 30 hours. X works for 16 hours; Y finishes the remaining work alone in 44 hours. In how many hours would Y finish the whole piece of work alone?

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Factorisation and Division of Algebraic Expressions

Let's Recall

Terms

Algebraic expressions are mathematical expressions that contain variables, constants, and operations like addition, subtraction, multiplication, and division. These expressions can be categorised into various types based on their characteristics and structure.

5 x Coefficient

Binomials

Monomials

Variable

–

Constant

Trinomials

An algebraic expression with only one term.

An algebraic expression with two unlike terms separated by addition or subtraction.

An algebraic expression with three unlike terms separated by addition or subtraction.

Example: 7, 3x, 2y2

Example: 2x + 5, 2y2 − z

Example: x + y + 2, 3x2 + 2x + 1

Let us see how to add or subtract expressions. Addition of Algebraic Expressions

Subtraction of Algebraic Expressions

While adding algebraic expressions, we collect the like

To subtract algebraic expressions, change the sign of

is the like term whose coefficient is the sum of the

like terms with the first expression, following the rules

terms and add them. The sum of several like terms coefficients of these like terms.

each term in the second expression and then combine of addition and subtraction.

Add (6a + 5cd) and (2a – 2cd).

Subtract (2a – 3cd) from (6a + 5cd)

6a + 5cd + 2a − 2cd

6a + 5cd + 2a − 3cd (−) (+)

= 8a + 3cd

4a + 8cd

Letʼs Warm-up

Solve to find the answer. 1 (6x + 5y) + (12y + 3x) = ______________

2 (8x + 5yx) + (12xy − 2x) = ______________

3 (6z + 5y2) − 12z = ______________

4 (7x − 5y) − (12y + 3x) = ______________

I scored _________ out of 4.

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Factorisation of Algebraic Expressions Real Life Connect

In a cosy study session, Lily was puzzled by factorisation. Her teacher began by explaining how they break down composite numbers, like 24 into 23 × 3. She linked it to solving puzzles with prime factors. As Lily’s understanding grew, her teacher introduced algebraic expressions, like 5x (y + 5), where factorisation meant finding common factors. Lily saw that factorisation in algebra, is much like factorisation of composite numbers.

Factorising Algebraic Expressions Similar to natural numbers, algebraic expressions can also be represented as products of factors, which can include numbers, algebraic variables, or other expressions. This process of finding expressions that when multipied, create a given expression, is called factorisation. It involves identifying the factors that make up the algebraic expression. The expression 5x (y + 5) can be factorised as shown below: Factorisation Algebraic 5x (y + 5) Expression

Did You Know?

Factor 5, x, (y + 5)

Multiplication 5 × x × (y + 5)

The Indian mathematician Brahmagupta developed the rules for operating with negative numbers and zero in his book BrahmaSphuta Siddhanta, which also contained many algebraic problems. He was the first to use the word “Sankalita” to mean equation.

Using Common Factors By identifying a common monomial Factorising by taking out a common monomial involves identifying the largest monomial that divides all the terms in a polynomial expression exactly. This common monomial is then factored out, simplifying the expression. For example, let’s factorise 6x2y + 12xy2 − 18xyz using a factor tree. 6x2y + 12xy2 − 18xyz

Expressions:

6x2y

12xy2

–18xyz

Terms:

2×3×x×x×y

2×2×3×x×y×y

−1 × 2 × 3 × 3 × x × y × z

So, 6x2y + 12xy2 − 18xyz = 6xy(x + 2y – 3z) 302

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Let us factorise −7mn2 − 14m2n − 21m2n2 without using the factor tree. Factors of −7mn2 = −1 × 7 × m × n × n

Think and Tell

Factors of − 14m2n = −1 × 2 × 7 × m × m × n

Does the number of terms in a

Factors of − 21m2n2 = −1 × 3 × 7 × m × m × n × n

polynomial increase if we multiply it by a monomial?

Common factors = −1 × 7 × m × n = − 7mn

So, on factorising, −7mn2 − 14m2n − 21m2n2, we get −7mn (n + 2m + 3mn). By identifying a common binomial: Factorising by taking out a common binomial involves identifying the largest binomial that evenly divides all the terms in a polynomial expression. This common binomial is then factored out, simplifying the expression. For example, to factorise 4x (x + 2) − 3y (x + 2), the common binomial is (x + 2). So, 4x (x + 2) − 3y (x + 2) = (x + 2) (4x − 3y). Example 1

Factorise the expression using a factor tree: 15a2b + 18ab2 − 24ab 15a2b + 18ab2 − 24ab

Expression: 15a2b

Terms:

18ab2 24ab

3×5×a×a×b

3×2×3×a×b×b

–1 × 3 × 2 × 2 × 2 × a × b

So, 15a2b + 18ab2 − 24ab = 3ab(5a + 6b – 8). Example 2

Factorise: 6(2x + 3y) − 17(2x + 3y)2 6 (2x + 3y) − 17 (2x + 3y)2 = (2x + 3y) (6 − 17 (2x + 3y)) = (2x + 3y) (6 − 34x − 51y)

Do It Together

Factorise the expressions. 1 4a3 − 9a2b2

4a3 − 9a2b2 3

2 a (a − 3) − 7xy (3 − a)3 = a × (a − 3) + 7xy (a − 3)3 =

Using Grouping and Regrouping Terms In certain algebraic expressions, not every term can be divided by the same monomial or binomial. However, by organising the terms into groups where each group shares a common factor, factorisation can be simplified and made more straightforward. Chapter 22 • Factorisation and Division of Algebraic Expressions

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Let’s see this through an example by factorising the expression: − m2 − mp + mn − np. Step 1: Group the terms in the

expression so that each pair of

m2 − mp + mn − np

terms shares a common factor.

= m2 + mn − mp − np

Step 3: Factor out the common

= m (m + n) − p (m + n)

factor and write the factored expression. Example 3

Do It Together

Step 2: Factorise each group.

= (m − p) (m + n)

Factorise: a2b − ac2 − ab + c2.

Example 4

Factorise: ap2 + bq2 + bp2 + aq2.

a2b − ac2 − ab + c2 = a2b − ab − ac2 + c2

ap2 + bq2 + bp2 + aq2 = ap2 + bp2 + bq2 + aq2

= ab (a − 1) − c2 (a − 1) = (a − 1) (ab − c2)

= p2 (a + b) + q2 (b + a) = (p2 + q2) (a + b)

Factorise the given expressions. 2 x2 − ax − bx + ab

1 3mn + 2pn + 3mq + 2pq = 3mn + _______ + 2pn + 2pq

= _______ (x − a) − b (______ − _____)

= 3m (_______ +_______) + _______ (n + q)

= (_______ −_______) (x − a)

= (3m + _______) (n + _______)

Do It Yourself 22A 1 Factorise the expressions by identifying the common monomial using a factor tree. a 3x2 + 9x

b 4y3 − 2y2

c 5a2b + 10ab2

d 6x3y − 18xy2

e 8c4 − 12c2

f 9p3q2 − 6pq3

g 12m4n3 − 20m2n2

h 15x3y2 − 25x2y3

20a5 − 30a3b2

2 Solve to factorise the expressions by identifying the common monomial. a 7x2 − 21xy

b 10ab3 − 15a2b2

c 16p4q − 12p3 q2

d 18m3n4 − 6m2n3

e 25x5y2 − 5x4y3

f 30a3b4 − 10a2b3

g 11x3 + 22x2

h 14y4 − 7y3

17a2b3 − 34ab2

19c5 − 38c3

k 23p4q − 46p2q2

27m3n2 − 18mn3

m 29x4y3 − 58x3y4

n 33a3b2 − 22a2b3

o 37x2 − 74xy

3 Factorise the expressions by identifying the common binomial. a (x − 2) (x + 4) − 3 (x + 4)3

b (3x + 2) (2x − 1) − (3x + 2)

c (5y − 1) (3y + 2) − 4 (3y + 2)

d (2a + 5) (a − 3) − (2a + 5)

e (x + 3) (x + 5) − (x + 3)2

f (2y − 4) (3y − 6) − 2 (3y − 6)

g (4a + 7)2 (a + 2) − (4a + 7)

h (4x − 3) (x + 2) − 5 (4x - 3)3

(3y + 2) (y − 1) − 2 (3y + 2)

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4 Factorise the expressions by regrouping the terms. a (2x3 − 3x2 + 4xy − 6y)

b (3ab + 6ac − 4cd − 2bd)

c (x2 − 7x − 2xy + 14y)

d (2ab + 5ac − 4bd − 10cd)

e (x2 − 3x − 2x + 6)

f (3x2− 2xy − 15x + 10y)

g (6xy − 3x2 − 10y + 5x)

h (4x2 − 12xy + 2x − 6y)

a 40ab4 − 60a2b3

b 43p3q2 − 86pq3

c 47m4n3 − 94m2n2

d 50x5y2 − 25x4y3

e 53a5 − 106a3b2

f 57x2 − 114xy

g (2a − 1) (a + 3) − 3 (2a − 1)2

h (3x2 + 6x + 2x + 4)

b 2a3 − 19a2 − 7a

c 15y2z3− 20y3z4 + 35y2z2

(6xy − 9x2 − 4y + 6x)

5 Factorise.

(2a3 − a2 + 4ab − 2b)

6 List the factors in these expressions. a (3a3 + 2a2 − 3ab − 2b)

Word Problem 1

A farmer spent ₹(4x2 + 8xy + x + 2y) on buying plants. If he bought (x + 2y) more plants, what is the cost of each plant? [Factorise to find your answer]

Factorisation Using Identities In the previous section, we were factorising the expressions by spotting the common monomial/binomial term or regrouping the terms. Not all expressions can be factorised this way. In this section we will use algebraic identities to factorise the expressions.

Factorisation When the Expression is a Perfect Square To factorise an expression that is a perfect square, we can use either of the identities Identity 2: a2 − 2ab + b2 = (a − b)2

Identity 1: a2 + 2ab + b2 = (a + b)2

Let us factorise the expression (4m2 + 20mn + 25n2). Step 1: R ecognise that the expression can be written as the

(4m2 + 20mn + 25n2)

Step 2: Determine what squares make up the expression.

= (2m)2 + 2 × 2m × 5n + (5n)2

Step 3: Write the expression by substituting a and b for the

= (2m + 5n)2

square of a binomial.

Here, a = 2m and b = 5n

corresponding terms.

Example 5

Factorise: 1

1 − 6x2 + 9x4

6x2 + 9x4.

Do It Together

= 1 − 2 × 3x2 + (3x2)2 = (1 − 3x2)2

Chapter 22 • Factorisation and Division of Algebraic Expressions

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Factorise the expressions using suitable identities. b2 − 6ab + 9a2

= b2 + 2 × ____ × b + (____)2 = (b + ____)2

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Factorisation When the Expression is a Difference of Two Squares To factorise an expression that is a difference of two squares, we will use the given identity Identity 3: a2 − b2 = (a + b) (a − b)

Step 1: R ecognise the expression that fits the difference of the squares form.

(16x2 − 81y2)

a2 – b2.

= (4x)2 − (9y)2

terms.

= (4x + 9y) (4x − 9y)

Step 2: Identify a and b as squared terms separated by subtraction in Step 3: Write the expression as (a + b) (a – b) with the corresponding

Example 6

Factorise: 64

9x2

Example 7

64 − 9x2 = 82 − (3x)2

20b2

5 − 20 b2 = 5 (1 − 4b2 )

= 5 (1 − (2b)2) = 5 (1 + 2b) (1 − 2b)

= (8 + 3x) (8 − 3x) Do It Together

Factorise: 5

Factorise the expressions using suitable identities. m4 − n4 = (m2)2 − (__)2 = (m2 − n2) (__+__) = (m − n) (__+__) (m2 + n2)

Do It Yourself 22B 1 Factorise the expressions using the difference of squares identity. a a2 − 4

b 25 − x2

c 9b2 − 16

d 49p2 − 16

e a2 − b2

f a2x2 − y2

g 100b2 − 81

h x2y2 − 4

b4 − 100

k 25 − a2x2

1 − d2

144 − a6

2 Factorise the expressions using the square of a binomial identity. a a2 + 8a + 16

b x2 + 10x + 25

c b2 + 4b + 4

d 81 + 18a + a2

3 Find the missing term so that the expression can be easily factorised using the square of a binomial identity. a b2 − 14b + ?

b x2 − ? + 9

c ? − 2cd + d2

d 36 − ? + r2

4 What constant must be added to the expressions to create a perfect square? a a2 + a

b − 8y + y2

c 9d2 − 6d

d x2 + 5x

e − 40b + 25b2

f 24x + x2

g x2 − 14x

h x2 − 16x

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5 Factorise the expressions using a suitable algebraic identity. a 1 − 8ax + 16a2 x2

b 3x3 y − 243xy3

c a4 − b4

d b4 − 81

e 64 − 176x + 121x2

f 49x4 − 168x2y2 + 144y4

g 49y2 − 56y +16

h 64 − 16y + y2

b a2 − b2 − 2b − 1

c a2 + 4b2 − c2− 4ab

6 Factorise. a m4 − 16n4

x2 y2 + 2 + y2 x2

Word Problem 1

Oliver wants to install bookshelves in his room. He has a rectangular wall space available with an area represented by the expression (9y2 − 4x2) sq. cm. Factorise the expression of area and separate the factors as the length and breadth. Determine the length and breadth if x = 2 and y = 3 and the length of the rectangle is more than its breadth.

Factorisation of Quadratic Trinomials A quadratic trinomial is a polynomial of the second degree, which means it is a polynomial of the form: (ax2 + bx + c = 0) where x is the variable and a, b and c are not zero. To factorise a quadratic trinomial, follow the steps given below: Step 1: Begin with the quadratic equation, (ax2 + bx + c = 0)

(3x2 + 7x + 4)

Step 2: D iscover a pair of numbers whose product equals ac and whose sum equals b.

Step 3: Break down the middle term of the expression using these two

(3x2 + 3x + 4x + 4)

numbers, ax2 + (Number 1)x + (Number 2)x + c = 0. Here, Number 1 = 3 and Number 2 = 4

= (3x (x + 1) + 4 (x + 1))

Step 4: Extract common factors. Write the common factors. Example 8

Do It Together

Factorise: 8x2

26x + 15.

Example 9

= (3x + 4) (x + 1)

Factorise: x2 + x

72.

8x2 − 26x + 15 =8x2 − 20x − 6x + 15

x2 + x − 72 = x2 + 9x − 8x − 72

= 4x (2x − 5) − 3 (2x − 5) = (2x − 5) (4x − 3)

= x (x + 9) − 8 (x + 9) = (x − 8) (x + 9)

Factorise the quadratic trinomials given below. 1 2x2 + x − 6

2 x2 − 2x − 15

= 2x2_____________________________

= x2 −_______________________________________

= ________________________________________

= ____________ + ___(x − 5)

= (__x − __)(x + 2)

= (_____________)(x − 5)

Chapter 22 • Factorisation and Division of Algebraic Expressions

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Do It Yourself 22C 1 How should the middle term of the expressions given below be split to factorise the expressions? a x2 − 5x + 4

b 2x2 − 9x + 4

c 6x2 − 11x + 4

d 4x2 − 17x + 4

c x2 + 11x + 10

d x2 − 9x + 14

2 Which of the expressions has (x + 1) as one of its factors? a x2 − 13x + 42

b x2 − 12x + 35

3 If the area of rectangles is given by the trinomials, what could the length and breadth of each of these rectangles be?

a 3x2 − 23x + 14

b 5x2 − 37x + 14

c 4x2 − 18x + 14

d 4x2 − 8x + 3

a 6x2 − 11x + 3

b 2x2 − 7x + 6

c 2x2 − 5x + 2

d 3x2 − 8x + 4

e 9x2 − 9x + 2

f 9x2 − 15x + 6

g 3x2 − 11x + 10

h 15x2 − 11x + 2

4 Factorise the trinomials.

5 Can the trinomials be factorised without taking any monomial term common before splitting the middle term? a b2c3 + 8bc4 + 12c5

b 3c5 − 18c4 − 48c3

Word Problem 1

A small business owner models the profit of her store on the quadratic trinomial

(3x2 − 12x − 15). If each product sold brings equal profit and (x − 5) number of products are sold, what is the algebraic expression for her profit?

Division of Algebraic Expressions Real Life Connect

Teacher: Today’s challenge: Tom has 10x candles to share equally between Mini and Mike. How many candles will each get?

10x ÷ 2

Jim: Is it 20x candles? Teacher: No, we want divide the candles, so it’s 5x each. We divide 10x by 2. Jane: Got it! Tom gives 5x candles to each friend. Teacher: Exactly! Division helps us share things equally, just like equally distributing candles among friends. This is a simple example of algebraic division. Let’s get deeper into algorithms for the division of polynomials.

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Dividing Algebraic Expressions To divide 10x candles between two kids, the teacher divides 10x by 2. Both 2 and 10x are monomials. Division is possible between all kind of polynomials and monomials. Let’s see how we perform these divisions one by one.

Dividing a Monomial by a Monomial To divide one monomial by another monomial, you can follow these steps: Suppose we want to divide the monomial 12a2b3c4 by 3ab2c2. This can be expressed as follows: Step 4: Simplify the expression.

Step 2: Factorise the monomials.

12a2 b3 c4 = 2 × 2 × 3 × a2 × b3 × c4 = 2 × 2 × a(2 − 1) × b(3 − 2) × c(4 − 2) = 4abc2 3ab2 c2 3 × a × b2 × c2 Step 1: Express the monomials to be divided

in numerator by denominator form.

Step 3: Cancel out matching exponents in the numerator and denominator.

So, 12a2b3c4 ÷ 3ab2c2 = 4abc2. Example 10

Divide 15x3y5z4 by 5xy3z3.

Example 11

15 x3 y5 z4 15 × x3 × y5× z4 � 5 × x × y3 ×z3 5xy3z3

14 × a8 × m3× n6 2 × n3 × m2 � 7 × a8 × m3 − 2 × n6 − 3 � 7a8mn3

� 3 × x3−1 × y5−3 × z4−3 � 3x2y2z Do It Together

Divide 14a8m3n6 by 2n3m2

Simplify. 1

21k9e4y7 6e2m2k3y2

�

21 × k9 × e4 × y7 6 × k3 × e2 × y2 × m2

64z9d7c6 12d2z2c3 �

64 × z9 × d7 × c6 12 × z2 × d2 × c3

Dividing a Polynomial by a Monomial To divide a polynomial by a monomial, you divide each term within the polynomial by the monomial. Step 2: Divide each term separately by the monomial. 18x6 + 24x5 − 12x3 18x6 24x5 12x3 � + − � 6x6−2 + 8x5−2 − 4x3−2 � 6x4 + 8x3 − 4x 3x2 3x2 3x2 3x2 Step 1: Express the polynomial and monomial to

be divided in numerator by denominator form. Example 12

Step 3: Simplify

the expression.

Divide 14x2y5 − 35x3y2 − 21x5y3 by 7x2y2. 14x2y5 − 35x3y2 − 21x5y3 14x2y5 35x3y2 21x5y3 � − − � 2y3 − 5x − 3x3y 7x2y2 7x2y2 7x2y2 7x2y2 Chapter 22 • Factorisation and Division of Algebraic Expressions

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Example 13

Divide 12a3b4c5 − 18a5b4c3 + 6a4b3c5 by 6abc. 12a3b4c5− 18a5b4c3 + 6a4b3c5 6abc

Do It Together

12a3b4c5 6abc

−

18a5b4c3 6abc

6a4b3c5 6abc

� 2a2b3c4 − 3a4b3c2 + a3b2c4

Simplify the expressions. 1 (10j3k4l7 − 18j5k9l2 + 25j4k3l5) ÷ 5jk2l 10j3k4l7 − 18j5k9l2 + 25j 4k3l5 =

3 47

10j k l 2

5jk l

5jk2l

−

5 92

18j k l 2

5jk l

4 35

25j k l 5jk2l

2 (12a5b4c7 − 26a2b3c3 + 6a4b4c6) ÷ 4b3a2c3 =

12a5b4c7 − 26a2b3c3 + 6a4b4c6 4b3a2c3

Dividing a Polynomial by a Polynomial To divide one polynomial by another polynomial, the first step is to factorise both polynomials and then eliminate any common factors. Let’s try to divide x3 − 4x2 + 3x by x2− x using factorisation. x3 − 4x2 + 3x � x(x2 − 4x + 3) x2 − x x(x −1) 2 � x(x − 3x − x + 3) (Factorising by splitting the middle term) x(x −1)

Remember!

� x(x(x − 3) − 1(x − 3)) x(x −1) �

The degree of the remainder is always less than the degree of the divisor.

x(x − 1)(x − 3) � x − 3 (By cancelling the common factors) x(x − 1)

One other way of dividing polynomials by polynomials is by using long division. To divide one polynomial by another using the long division method, you should follow the steps. 1

Rearrange the terms in the division house.

egin by determining the first term of the B quotient, by dividing the first term of the dividend by the first term of the divisor.

Multiply all the terms within the divisor by the first term of the quotient, and then subtract this product from the dividend to get the remainder. If the remainder is not zero, treat it as the new dividend, and repeat the same process from step 2 onwards until we get the remainder as zero or a polynomial with a degree lower than that of the divisor.

Let’s divide 3a4 − 3a3 − 4a2 − 5a by a2 − 2a a2 − 2a

3a2 + 3a + 2

3a4 − 3a3 − 4a2 − 5a

−3a4 − 6a3 + 3a3 − 4a2 3 2 − − +3a − + 6a

2a2 − 5a

−

+ 2a2 − 4a + − –a

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Divide 2y2 + 17y + 21 by 2y + 3 using factorisation.

Example 14

2y2 + 17y + 21 2y2 + 14y + 3y + 21 2y (y + 7) + 3(y + 7) (2y + 3) (y + 7) � � � �y+7 2y + 3 2y + 3 2y + 3 2y + 3 Divide 16x3 − 46x2 + 39x − 9 by 8x − 3 using long division.

Example 15

2x2 – 5x + 3

16x3 − 46x2 + 39x − 9

8x − 3

Divide 10x4 + 17x3 − 62x2 + 30x − 3 by 2x2 + 7x − 1 using long division. 2x2 + 7x − 1

2 −40x + + − 15x 24x − 9

−

5x2 − 9x + 3

10x4 + 17x3 − 62x2 + 30x − 3

− 10x4 + 35x3 − 5x2 − + − −18x3 − 57x2 + 30x

− 16x3 − 6x2 + −40x2 + 39x −

Do It Together

Example 16

−

24x − 9 + 0 Divide the polynomial using long division.

2x + 7x

6x2 + 21x − 3 − − + 0

Error Alert!

−

14x4 + 21x3 − 62x2 + 30x by 2x2 + 7x

−18x3 + − 63x2 + + − 9x 2 6x + 21x − 3

14x4 + 21x3 − 62x2 + 30x

Always check the degree of the quotient after division. It should be less than or equal to the dividend.

−10x4 + 49x3

2x2 + 7x − 1

5x5 − 9x + 3

10x4 + 17x3 − 62x2 + 30x − 3 5x2 − 9x + 3

10x4 + 17x3 − 62x2 + 30x − 3

Do It Yourself 22D 1

Divide the algebraic expressions. a 15x3 by 3x

b 15a2b4 by 5ab2

c 24y4z3 by 6yz

d 18x5y 2z by 9xyz2

e 36p4q3r2 by 12p2qr2

f 84j5k7r9 by 7jk3r2

a (2x3 + 4x2 − 6x) by (2x)

b (3y4 − 9y3 + 6y2) by (3y)

c (4a5 − 12a4 + 8a3) by (4a)

d (6m3 − 18m2 + 12m) by (6m)

e (5X6 + 10x5 − 15x4) by (5x2)

f (5a3x − 15a2y2 + 3az3) by (3a)

Divide polynomials by monomials.

Solve the expressions. a 225abc (4a − 8) (5b − 15) ÷ 125 (a − 2) (b − 3)

b 144ab (a + 5) (b − 4) ÷ 36a (b − 4)

c 28 (a + 3) (a2 + 3a + 70) ÷ 7 (a + 3)

d 117abc (a + b) (b + c) (c + a) ÷ 78ab (b + c) (c + a)

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Divide the polynomials by the binomials using long division. a (4x3 − 6x2 + 3x − 9) by (x2 − 4)

b (3y4 − 9y3 + 6y2 − 12y) by (y2 − 4)

c (x5 + 5x4 + 4x3 + 20x2 + 16x) by (x2 + 4)

d (6a4 − 3a3 + 9a2 − 6a) by (a + 2)

e (9b6 + 12b5 − 3b4 − 4b3) by (b2 + 2)

f (2c5 − 8c4 + 6c3 − 12c) by (c + 2)

Divide the polynomials by the trinomials using long division. a (3x3 − 5x2 + 2x − 1) by (x2 − 4x + 4)

b (4y4 + 6y3 − 2y2 + 8y − 12) by (2y2 − 3y − 6)

c (6a4 − 2a3 + 9a2 − 5a + 3) by (2a2 − a + 1)

d (x5 − 3x4 + 5x3 − 2x2 − 4x + 8) by (x3 −2x2 − x + 2)

e (7x3 + 2x2 − 9x − 5) by (x2 − 3x − 2)

f (9y5 + 6y4 + 3y3 − 18y2 + 9y − 6) by (3y3 − 6y2 −3y + 6)

What must be multiplied by x − 7 to obtain x3−12x2 + 38x − 21?

What must be subtracted from (a4 + 2a3 − 2a2 + a − 1) so that the resulting polynomial is exactly divisible by (a2 + 2a − 3)?

What must be added to 3x3 − 8x to make it perfectly divisible by x − 1?

Word Problem 1

Rahul wants to distribute (x3 + 6x2 + 12 + 13x) chocolates among his classmates on his birthday. If there are (x + 3) kids in his class, how many chocolates will each kid get?

Points to Remember • When an expression is the product of two or more expressions, then each of these expressions is called a factor of the expression. • The greatest common factor of two or more monomials is the product of the greatest common factors of the numerical coefficients and the common letters with the smallest powers. • When a binomial is a common factor, we factorise by writing the expression as the product of this binomial and the quotient of the expression divided by the binomial. • •

Difference of Squares Identity: a2 − b2 = (a + b) (a − b)

Perfect Square Identity: a2 + 2ab + b2 = (a + b)2; a2 − 2ab + b2 = (a − b)2

Math Lab Factorisation Bingo Materials Required: • Bingo cards (pre-made or created by the teacher) • Algebraic expressions (to be called out as clues) • Markers or chips for each student 312

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Steps: 1 Prepare Bingo cards with algebraic expressions in a grid format. 2 Explain the rules of Bingo, where players aim to mark a row, column, or diagonal of squares on their card by matching the expressions called-out. 3 Distribute Bingo cards and markers or chips to students. 4 Call out the algebraic expressions one by one, providing context or stories related to each expression. 5 After calling out an expression, ask students to factorise it and check if it matches any expression on their Bingo card. If it matches, they can cover that square. 6 The first student to complete a row, column, or diagonal shouts "Bingo!" and wins the round. 7 Discuss the factorisation process and ask students to explain how they factorised the expressions and share any patterns they noticed. 8 Play multiple rounds with different expressions and factorisations to reinforce the concept.

Chapter Checkup 1 Factorise the expressions by taking out a common monomial. a 2x2 + 6x

b 3ab − 9a

c 4y3 + 8y2

d 5x2 − 15x + 10

e 6m2n + 12mn2

7xy − 14x

2 Factorise the expressions by taking the binomial term common. a (x + y) − (x + y)(2x − 11)

b x (x − 2z) + y (x − 2z) + (2z − x)

c (n − 10)2 + (10 − n)

d (3a − 1)2 − 6a +2

e 5x + 10y − 7 (x + 2y)2

(5x + y) − (5x + y)3

3 Factorise the expressions by grouping/regrouping the terms. a 4a2b − 6ab2 − 2ab + 3b2

b 5x2 − 10xy + 3xy − 6y2

c 3x2 − 2xy + 6x − 4y

d 9x2 − 6x − 12xy + 8y

e 4x3 − 2x2 − 6xy + 3y

6x2 + 7xy − 3x − 21y

4 Factorise the expressions into perfect squares. a (a2 + 6ab + 9b2)

b (x2 − 10xy + 25y2)

c (4x2 + 4xy + y2)

d (25x2 + 40xy + 16y2)

e (x2 − 14xy + 49y2)

(16a2 − 32ab + 16b2)

5 Factorise the expressions by using the difference of two squares identity. a (p2 − 9q2)

b (4x2 − 49y2)

c (a2 − 16b2)

d (9m2 − 25n2)

e (x2 − 36y2)

a 6x2 + 11x − 2

b x2 − 5x + 6

c 3x2 + 12x − 15

d 3x2 − 10x + 7

e x2 + 14x + 48

(25u2 − 4v2)

6 Factorise the quadratic trinomials.

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7 Divide the polynomials by the monomials. a (6x3 + 9x2 − 12x) by 3x

b (4a2b3c2 − 8ab2c2d) by 2abc

c (12a3b2 − 6a2b3 + 18ab4) by 6ab2

d (9x5 − 3x4 + 6x3) by 3x2

e (7a4b3c2 − 14a3b2c3 + 21a2b1c4) by 7a2bc2

(15x6 − 10x5 + 5x4) by 5x3

8 Divide the polynomials by the binomials using long division. a (2x3 + 4x2 − 6x) by (x − 2)

b (3x4 − 9x2 + 6x) by (x + 1)

c (4x3 + 12x2 − 3x) by (2x − 1)

d (5x4 − 13x3 + 9x2) by (5x + 2)

e (6x4 − 12x2 + 6) by (2x − 3)

(7x3 − 7x2 + 14x) by (7x − 4)

9 Divide the polynomials by the trinomials using long division. a (3x3 + 4x2 − 2x + 1) by (x2 − 3x + 2)

b (2x4 − 5x2 + 3) by (x2 − x − 2)

c (6x3 − 8x2 + 2x − 4) by (2x2 − 3x − 2)

d (5x3 − 6x2 + 3x − 2) by (x2 + 2x − 1)

e (x4 + 3x2 − 4) by (x2 + 4x + 4)

(8x3 − 12x2 − 6x + 9) by (4x2 − 6x − 3)

Word Problem 1 Lucas wishes to plant flowers in a rectangular garden. He wants the area of his garden to be (64x3 − 25xy2). What are the possible length and breadth combinations of his garden?

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23 Linear Graphs Let's Recall

10 8 6 4 2

Badminton

Number of Students

Favourite Sport Number of students

Cricket

Hockey

Badminton

Chess

Tennis

Cricket

Hockey

Let us say we conduct a survey in our class, where we ask the students what their favourite sport is and 0 the data set obtained is given below. Favourite Sport Tennis Chess 7

The above data can be represented in the form of different graphs. A histogram is also a pictorial representation of data using bars. It is very similar to the bar graph except that the bars are adjacent to each other. We use a histogram to represent grouped frequency distributions with continuous class intervals. 12

Favourite Sport

4 2 0

Tennis Chess

Chess

Tennis

Badminton

Hockey

Badminton

Cricket

Hockey

Number of Students

Cricket

Number of Students

A bar graph is a pictorial representation of data using bars that can either be horizontal or vertical bars.

Favourite Sport

Letʼs Warm-up Fill in the blanks.

1 The most popular sport among the students is _________________. 2 The least popular sport among the students is _________________. 3 More students like _________________ than hockey. 4 Fewer students like _________________ than cricket. 12

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8 6

I scored _________ out of 4.

Tennis Chess

Cricket

Badminton

4 Hockey

Number of Students

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Line Graphs and Linear Graphs Line Graphs Real Life Connect

A company sells two brands of laptops, A and B. The sales head wants to visualise their daily sales data over a week.

Reading Double Line Graphs He makes graphs to show the sales of both brands.

2800

Laptop A

2800 2400

2400

2200

2200 2000

Number of Laptops

Laptop B

2600

1800

1600 1400

1200 1000 800

2000 1800

1600 1400

1200 1000 800 600

600

400

200

200 0

Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Day 7 Day

We can say that a line graph displays data that changes continuously over a certain period of time. How many Laptops A were sold in total?

Remember!

In total, 1000 + 1400 + 1800 + 1600 + 1600 + 2200 + 2400 = 12,000 Laptops A were sold.

A line graph is a type of graph in which a line connects several data points, showing the relationship between them, hence showing the changes in a single variable or data over a period of time.

How many Laptops B were sold in total?

In total, 1600 + 1000 + 1800 + 1800 + 2000 + 2400 + 2000 = 12,600 Laptops B were sold. What if the data is represented using one graph? A graph with two lines on the same graph is called a double line graph. It is an extension of a regular line graph.

This graph outlines the daily sales data of the company over a particular week.

On which day was the highest sale of Laptop A? From the graph, we see that the greatest number sold of Laptop A was 2400 on Day 7.

On which day was the lowest sale of Laptop B? From the graph, we see that the lowest number sold of Laptop B was 1000 on Day 2.

Laptop A and Laptop B

Scale: 1 division = 200 units

2800 2600 2400 2200 2000

Number of Laptops

A double line graph allows us to compare and analyse two related data series simultaneously on the same graph for better insight and comparison.

1800

Laptop A Laptop B

1600 1400 1200 1000 800 600 400 200 0

Day 1

Day 2

Day 3

Day 4 Day

Day 5

Day 6

Day 7

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100 100

Read the graph carefully and answer the questions:

1 Find the year where the percentage profit is the lowest.

90 90

80 80

70 60

% Profit

ince, in the line graph the lowest point is 40, S the lowest percentage of profit is in the year 2020.

50 2 What is the average profit earned during the years 2017 to 2022?

The average profit earned during the years 2017 to 2022

y12

2 On which day did it snow the most and least in City A?

Snowfall (in inches)

3 On which day did it snow the most and the least in city B? In city B, it snowed the most on Day 9 and Day 10 and the least on Day 2.

City A B City

City B

9 8 7 6 5 4

36 47 The given graph shows the profit gained by two companies in an interval of years. Scale: 1 division = 2000Day units

x 10 69 Company 7 11 8 A 12 9 10 11 Day Company B

1 y1

Read the graph carefully and answer the following questions.

14000

he profit gained in the year 2012 in total was 1 T ________.

he percentage increase in the profit of 4 T company B from 2011 to 2014 is ________.

he company that earned more profit overall is 5 T ________.

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Profit

140008000 120006000 4000

10000

2000

8000 6000

y Company A Company A Scale: 1 division = 2000 units = 2000 units Scale: 1 division Company B Company B

14000 12000 10000

Profit

he percentage decrease in the profit of 3 T company A from 2011 to 2014 is ________.

10000

Profit

he difference between the highest profit 2 T gained by Company A and the lowest profit gained by Company B is ________.

12000

10 11

6 7 Day

Both cities had equal levels of snowfall on day 4 and day 10. Do It Together

y City A Scale: 1 box = Scale: 1 division 1 box = 1 division

4 O n which day did both cities have equal levels of snowfall?

City B

City A

Scale: 1 box = 1 division

In city A, it snowed the most on Day 10 and the least on Day 3.

2017 2018 2019 2020 2021 2022 Year

X 0 2019 2020 2021 2022 2017 2018 2017 2018 2019 2020 2021 2022 Year Year y

he x-axis represents ‘Day’ and the y-axis represents T ‘Snowfall (in inches)’.

Snowfall (in inches)

hat is represented on the x-axis and y-axis 1 W respectively?

Snowfall (in inches)

Example 2

= Total profit percentage Total number of years 60 + 90 + 80 + 40 + 60 + 90 420 = = = 70% 6 6 The graph represents the average snowfall in two different cities, A and B, for 10 consecutive days. Read the graph carefully and answer the following questions.

Y Scale: 1 division = =10 Scale: 1 division 10 units Scale: 1units division = 10 units 100

% Profit

The graph shows the annual percentage of profit earned by the company during the years 2017 to 2022.

% Profit

Example 1

8000 2010 6000

4000

2000

2000 2010

2011

2012 Year

2013

2014

2011201020122011 20132012 20142013 Year Year

x 2014

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Drawing Double Line Graphs

We now know how to read a double line graph. But, how is the graph created? Let us first follow the steps and draw a line graph for the sales of Laptop A and B. y Laptop A and Laptop B

2800

1 Select a suitable scale.

2600

3 Mark the

2400

points.

2200

2000

1600

2600

1400

2400

1200

2200

1000

2000

2 Draw and

800

1800

axis.

400

600

label the

200 0

Day 1

Day 2

Day 3

Number of Laptops

2800

Number of Laptops

1800

Scale: Laptop A and Laptop B 1 division = 200 units Laptop A

Laptop B

4 Draw line

segments to join

1600

the points.

1400 1200

1000 Day 4 800 Day 600

Day 5

Day 6

Day 7

400

Remember!

200

A single line graph represents data from one set of information, while a double line graph represents data Scale:Day 1 division Day 5 Day 6 Day 7 1 Day= 210%Day 3 Day 4 100

from two different sets of information on the same graph.

Day

90 80

The table represents the average percentage scored by the students of a class in each subject. 70 y 60

Subjects

Percentage (%)

Hindi

3050

Mathematics

2090

Science Social Science

English

Scale: 1 division = 10%

100

Hindi

Draw the line graph to represent the above information.

Percentage (%)

Example 3

70 60 50

40 English Mathematics Science Social Science

Subject 30 20 10 0

Hindi

English Mathematics Science Social Science

Subject

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Example 4

A patient’s body temperature was recorded every hour for two days and is given in the table. Time

10 a.m.

11 a.m.

12 noon

1 p.m.

2 p.m.

3 p.m.

Day 2

32°C

35°C

36°C

35°C

37°C

35°C

Day 1

35°C

34°C

37°C

Draw the line graph for the above information. y

34°C

32°C

Day 1

1 division = 1°C

Day 2

Temperature

38 37 36 Temperature

Scale:

Scale: 36 1 division = 1°C

35 34

Day 1 Day 2

35 34 33 32 31

31 30 Do It Together

35°C

10:00 11:00 12:00 1:00 2:00 3:00 a.m. a.m. noon p.m. p.m. p.m.

Time

x in different years is given below. The population (in thousands) of men and women in a village 0

Years

2018

Number of Men

Number of Women

10:00 11:00 12:00 1:00 2:00 3:00 a.m. a.m. noon p.m. p.m. p.m. 2019

Time

y 16

Draw the line graph for the above information. y

Population (in thousands)

2021

2022

16 14

2020

12 10 Number of men

Number of women

8 6

Number of men

Number of women

2 4

2018

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2019

2020 Year

2019

2020 Year

2021

2022

2021

2022

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Do It Yourself 23A 1 State true or false. a A line graph can be a whole unbroken line. b A line graph is used to study how data changes over time. c When interpreting a double line graph, if one line is consistently higher than the other, it means that the data represented by that line is always greater.

d Line graphs can also be called histograms. e In a double line graph, the two lines must always intersect at some point on the graph. the vertical axis represents f The horizontal axis in a line graph represents categories of the given data, while y values or quantities.

Scale: 1 Division = 5 units

50y

Scale: 1 Division = 5 units

50 45

2 The graph shows the number of shirts sold by a shopkeeper in

45 40

a week.

Number of shirts sold

Answer the following questions. a On which day were the most shirts sold? b On which day were the least shirts sold? c What was the percentage increase in the number of shirts

35 30 25 20 15 10

35 30 25 20 15 10 5

sold from Wednesday to Thursday?

d What is the ratio of the number of shirts sold on Tuesday to the average number of shirts sold on all 5 days?

y 18 18

3 The graph shows the population of two cities from 2000 to 2004.

x Monday TuesdayWednesdayThursday Friday

Monday TuesdayWednesdayThursday Friday Week day Week day

Scale: y Scale: 1 division = 2 units 1 division = 2 units

City A

City BCity B

Population (in thousands) Population (in thousands)

16 16

Read the line graph and answer the questions.

14 14 12 12

a When was city A the most and least populated?

10 10

b What percentage of people were in city A in comparison with city B in the year 2001?

c What was the percentage increase in the number of people

in city A from 2000 to 2004?

x 2001 2002 2003 2004 0 2000 2000 2001 2002 2003 2004 Year Year

4 Draw a double line graph for the data on the savings of a sister and her brother for consecutive years. Year

2018

2019

2020

2021

Sister

₹1100

₹1400

₹1750

₹2500

Brother

₹1200

₹1600

₹2050

₹2350

5 The table shows the temperature in Fahrenheit of two cities over the period of June to October. Represent it as a line graph.

Month

June

July

August

September

October

City A

115

105

100

City B

105

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Linear Graph Raman and his father took their dog Bruno for a training session where they saw a chart displaying the cost of training dogs. Cost

₹100

₹200

₹300

₹400

₹500

Time (in hours)

Let us plot a linear graph for the above data. Cost of Dog Training

Cost of Dog Training

y x-axis: 1 division = 1 hour

y-axis: 1 division = ₹100

600

400

(3, 300)

300

100 0

(2, 200)

300

How is this graph different from (2, 200) the line graph?

100 2

Time (Hours)

The graph is a straight line. So, it is called a linear graph.

What is the difference between a line graph and a linear graph? A line graph has line segments joined one after the other, but a linear graph is a continuous line.

(4, 400)

400 Think and Tell (3, 300) 200

(1, 100)

(5, 500)

500 Cost (₹)

Cost (₹)

(4, 400)

y-axis: 1 division = ₹100

600

(5, 500)

500

200

y x-axis: 1 division = 1 hour

(1, 100)

Time (Hours)

Remember! Every point is in the form of coordinates (x, y).

Cartesian Coordinate of a Point Y

A plane that has an x-axis Yand a y-axis is called a Cartesian plane.

The point where the x-axis and the y-axis intersect is called the origin.

P(x, y)

It is represented by O.

Ordinate is the line parallel to the y-axis. Y(Ordinate)

Y(Ordinate)

Abscissa is the line parallel to the x-axis. x

O When we put a point on paper we cannot describe (Abscissa) the position of the point. We have to measure the x-coordinate and y-coordinate to find the position of the point. The point is given as P (x, y).

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The coordinate axes (x-axis and y-axis) divide a plane into y 4 regions, called quadrants. Quadrant

XOY

YOX’

X’OY’

III

(–, –)

Y’OX

IV O

(+, –)

Quadrant 2 (–, +)

x’

Signs of Coordinates

Region

(+, +)

Quadrant 1 (+, +)

(–, +)

x’

x Quadrant 3 (–, –)

Here, x is the positive x-axis and x’ is the negative Quadrant 3 Quadrant 4 x-axis. (–, –)

Quadrant 1 (+, +)

Quadrant 2 (–, +)

Quadrant 4 (+, –)

(+, –)

Similarly, y is the positive y-axis and y’ is the negative y-axis.

y’

The coordinates of O are (0, 0). Every point on the x-axis has its y-coordinate zero and therefore, it is of the form y’ (x, 0) and vice-versa. A point whose x-coordinate is zero and y-coordinate is non-zero will lie on the y-axis. Example 5

In which quadrant do the following points lie?

Did You Know?

1 (4, 3)

Rene Descartes provided a

2 (–4, 3)

system for locating a point with

3 (4, –3)

called the x-axis and y-axis in

the help of two measurements the 17th century.

4 (–4, –3) Which shape is formed when all the points are joined?

Quadrant

1 (4, 3) lies in the first quadrant. 2 (–4, 3) lies in the second quadrant.

Quadrant 2

3 (4, –3) lies in the fourth quadrant.

(–4, 3)

Quadrant 1

(4, 3)

4 (–4, –3) lies in the third quadrant. A quadrilateral (rectangle) is formed when all the points are joined.

(–4, 3)

–5

–3

1 –5

–4

–3

–2

–1 0

–1

5 (–4, –3)

–2 (–4, –3) Quadrant 3

–3 –4 –5

Quadrant

(4, –3) Quadrant 4

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Example 6

Write the abscissa and ordinate of the point: 1 (3, 4)

2 (4, 3)

4 5 3 4

We know that abscissa means the x-coordinate and ordinate is the y-coordinate.

2 3

Therefore,

–5

1 In (3, 4): The abscissa is 3 and the ordinate is 4.

–4 –5

–4

–3

–4

–3

2 In (4, 3): The abscissa is 4 and the ordinate is 3.

Example 7

What is the perpendicular distance of the point (–2, –3) from the: 1 x-axis?

2 y-axis?

4 3 2

(3, 4) (3, 4) (4, 3) (4, 3) (3, 4)

(4, 3)

O 4 2 3 1 5 –1 1 4 2 3 5 –1 O –11 –1 4 2 3 5 –2 –1 O –2 1 –2 –1 –3 –3 –2 –4 –3 –4 –5 –4 –5 –5

–3

–2

1 2

–2

The perpendicular distance of the point (–2, –3) from the: x – axis = 2 units; y – axis = 3 units. y

5 y

3 x’

x’

–5

2 –5

–4

–3 1 –2

x’

–4 –5 –3 –4 –2 –3–1 –2 O –1

–5

Do It Together

3 2 1

–1 O 1 –1

–1 1 O –22 1 –1

–2(–2, –3) –2 –3 (–2, –3) (–2, –3) –4

–3 –4

Remember!

Distance can never be negative.

–3 –4 –5 y’

–5

y’ Write six coordinates y’of the points for which y = x + 2.

The coordinates of the points for which y = x + 2 are: x

___

Drawing a Linear Graph We know what a linear graph is. But, how do we draw a linear graph? Look at the steps below: 1 Based on the data,

choose a suitable scale

Scale: X-axis: 1 division = 2 kg Y-axis: 1 division = 2000 grams Scale: X-axis: 1 division = 2 kg Y-axis: 1 division 2000 grams 1 division = 2=kg Scale: X-axis:

24000 Y Y 22000 24000

and draw the x-axis and y-axis.

Y-axis: 1 division E= (22, 2000 grams 22000)

Weight (in g) Weight (in g)

24000 20000 22000

D (20, 20000) E (22, 22000)

Weight (in g)

22000 18000 20000

3 Draw a line to connect all the points.

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coordinates on the graph.

(16,14000) 16000) AB(14, 12(14,14 16 18 20 22 24 A 14000) Weight (in Kg) 12 14 16 18 20 22 24

12000

mark their respective

C (18, 18000) A (14, 14000) B (16, 16000)

1600014000 12000 1400012000

2 Then, from the data, we

(20,18000) 20000) B (16, 16000) CD(18,

1800016000 14000

Chapter 23 • Linear Graphs

E (22, 22000) C (18, 18000) D (20, 20000)

2000018000 16000

Weight (in Kg) 18 20 22 24

X X

Weight (in Kg)

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Now we know how to make a linear graph!

Let us see some examples based on linear graphs. Example 8

The table shows the relation between simple interest and the period of time. Draw the graph for the same.

₹600

₹800

₹1000

Number of Years

1200 Simple interest (in ₹)

Scale:

Simple interest (in ₹)

X-axis: 1 division 1 year = 1 year X-axis: 1= division We plot the number of years on the x-axis Y-axis: 1 division ₹200 = ₹200 Y-axis: 1= division 1200the1200 and simple interest on the y-axis.

1000 We then1000 plot the points and connect them. 800 600

400

800 600

600 400 200

The scale should always be the same throughout one axis.

2 3 4 5 Number of Years

Error Alert!

Example 9

800

x 2 13 24 35 4 5 Number of Years of Years Number

1000

400

2 3 4 5 200 Number of Years

200

Scale: X-axis: 1 division = 1 year Y-axis: 1 division = ₹200

1 divisionScale: = ₹200

800

200

₹400

Taking the scale,

1000

Simple interest (in ₹)

1200

400

₹200

Scale: X-axis: 1 division = 1 year Y-axis: a1 graph division for = ₹200 Draw the above information.

600

Simple Interest

0 2

0 5

x 0 8

0 4

In the table, we are given the side lengths of a square and its perimeter. Length of Side

1 cm

2 cm

3 cm

4 cm

5 cm

Perimeter

4 cm

8 cm

12 cm

16 cm

20 cm

Draw a graph for it. Also, check if the graph obtained is linear or not. y

16 12

16 12 8 4

y 24

Scale: x-axis: 1 division = 1 cm

y-axis: 1 division = 4 cm

12 8 4

x 1 2 13 24 35 46 5 Length ofLength Side (cm) of Side (cm)

4 1

2 3 4 5 Length of Side (cm)

20 Perimeter (cm)

Perimeter (cm)

Remember!

16 12 8 4

A straight-line graph is called a linear graph. 1

2 3 4 5 Length of Side (cm)

The graph obtained is a linear graph.

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Example 10

In the table, the area of squares with respect to their sides is given. Length of side

1 cm

2 cm

3 cm

4 cm

5 cm

Area

1 cm2

4 cm2

9 cm2

16 cm2

25 cm2

Draw a graph for it. Also, check if the graph obtained is linear or not. yy

Scale:

25 25

25 x-axis: 1 division = 1 cm

Area (cm (cm22)) Area

20 20

20 y-axis: 1 division = 5 cm2 Area (cm2)

15 15 10 10

10 The graph obtained is not a linear graph.

55 11

Do It Together

22 33 44 55 Length Lengthof ofside side(cm) (cm)

2 3 4 5 Length of side (cm)

Sohan was riding his bicycle to school from his home. Distance (in metres)

Time (in seconds)

Complete the graph for the above data. yy

30 30

(__, (__,__) __)

25 25 20 20

(4, 10)

(2, (2,5)5)

(6, __)

(4, (4,10) 10)

10 10

(__, __)

(6, (6,__) __)

15 15

(__, __)

(__, (__,__) __)

(2, 5)

5 44

10 10 12 12

Reading Linear Graphs We learnt about the relationship between the number of training hours for dogs and the cost by plotting the points on the graph. But what exactly can we interpret from this? Cost Costof ofDog DogTraining Training

On the x-axis, we have represented the time taken 600 600 to train the dogs with the scale taken as 1500) division (5, (5,500) 500 500 = 1 hour. (4, (4,400) 400)

(1, (1,100) 100)

The cost 100 100 of training a dog for 1 hour is ₹100.

The cost 00 of training a dog for 2 hours is ₹200. 11

Similarly, the cost of training a dog for 5 hours Time Time(Hours) (Hours) is ₹500.

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(5, 500)

500 Cost (₹)

Cost (₹) (₹) Cost

400 400 On the y-axis, we have represented the cost of (3, (3,300) 300) 300 300 training dogs with the scale taken as 1 division (2, (2,200) 200) = ₹100. 200 200

Cost of Dog Training

600

(4, 400)

400

(3, 300)

300 200 100 0

(2, 200) (1, 100)

Time (Hours)

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Example 11

Look at the graph and answer the questions. y

hat is represented on the x-axis and the 1 W y-axis?

6000

The x-axis represents simple interest and the y-axis represents deposits.

Interest on the deposit of ₹2500 is ₹200 for a year. o get interest of ₹280 per year, how 3 T much money should be deposited?

Deposit (in ₹)

ind the interest on the deposit of ₹2500 2 F for a year.

(400, 5000)

5000

(320, 4000)

4000 3500

(240, 3000)

3000 2500

(160, 2000)

2000

₹3500 should be deposited every year to get interest of ₹280.

(80, 1000)

1000 0

Example 12

160 200 240 280 320

Simple Interest (in ₹)

400

From the following graph, find the perimeter of the square that has the given length of the side:

Perimeter

y 24

Side

Perimeter

1 unit

4 units

2 units

8 units

3 units

12 units

4 units

16 units

5 units

20 units

16 12 8 4 0

Do It Together

2 3 4 5 Length of side

Monica was driving her car. The following graph represents the distance (in km) covered in a particular time (in minutes). y

1 The x-axis represents _________________.

2 The y-axis represents _________________.

he distance covered in 4 minutes is 3 T _________________.

6 Distance

he distance of 5 km was covered in 4 T _________________.

5 4 3 2 1 1

Time

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Do It Yourself 23B 1 State true or false.

a Every point on the y-axis is of the form (x, 0).

______________

c A point whose y-coordinate is zero and x-coordinate is 5 will lie on the y-axis.

______________

b The graph of every linear equation in two variables need not be a line.

______________

d The coordinates of the origin are (0, 0).

e A point whose x-coordinate is zero and y-coordinate is non-zero will lie on the y-axis.

2 Write the abscissa and the ordinate of each of the following points.

______________ ______________

a (9, 4)

b (4, –5)

c (0, 4)

d (–2, –8)

e (0, –7)

a (0, 1)

b (–8, 0)

c (9, 0)

d (0, –5)

e (0, 0)

a (–8, 8)

b (–9, –2)

c (1, –8)

d (8, 5)

e (–5, 0)

3 On which axis do the following points lie? 4 In what quadrants do the following points lie?

5 Plot the given points on a graph sheet and verify if they lie in a straight line. x y

100

144

169

196

6 Draw a line passing through (4, 4) and (–4, –4). Find the coordinates of the points at which the line meets the x-axis and y-axis.

7 Is the graph of y = x2 a linear graph if x = 1, 2, 3, 4, 5? Explain using a graph.

Word Problem 1 The cost of a notebook is ₹50. Draw a graph after making a table showing the cost of 2, 3, 4, … so on up to 10 notebooks. Use it to find, a the cost of 7 notebooks. b the number of notebooks purchased with ₹550.

Points to Remember • A line graph displays data that changes continuously over a period of time. • A double line graph allows us to compare and analyse two related data series simultaneously on the same graph for better insight and comparison. • A single-line graph represents data from one set of information, while a double line graph represents data from two different sets of information on the same graph. • A plane that has an x-axis and y-axis is called a Cartesian plane. • Every point is in the form of coordinates (x, y).

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• The coordinates at the origin are O (0, 0). • A quadrant is a region defined by the two axes (x-axis and y-axis) of the coordinate system. When the two axes, the x-axis and y-axis, intersect each other at 90 degrees, the four regions so formed are called the quadrants. • A straight-line graph is called a linear graph. • A line graph has line segments that are joined consecutively, but a linear graph is a continuous line.

Math Lab Aim: Identify the difference between a line graph and a linear graph. Setting: In groups of 5 members Material: A pen, a piece of paper and a sheet of graph paper Method: 1

Make a list of all the students in the group with their respective heights in cm on the paper.

Mark the points for each student’s respective heights.

Discuss the results and check if the graph is linear or not.

2 On the graph paper, mark the names of each student on the x-axis and heights on the y-axis. 4

Join all the points and form a line graph.

Chapter Checkup 1 Write the abscissa, ordinate and the quadrants of the points: a (5, 4)

b (–8, –2)

c (–9, –18)

d (–5, –10)

e (14, –21)

2 Plot the following on a graph sheet. Verify if they lie on a line. a A (4, 6), B (4, 2.5), C (4, 2), D (4, 0) b A (5, 3), B (2, 3), C (2, 5), D (5, 5) c A (1, 1), B (2, 2), C (3, 3), D (4, 4)

3 Draw the points (5, 4) and (4, 5). Do they represent the same point? 4 Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and the y-axis.

5 Draw a line passing through (1, 3) and (0, 4). Write the coordinates of the point at which this line meets the x-axis and y-axis.

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6 The line joining the points that have the same x-coordinates will be parallel to which axis? 7 To which axis will the line joining the points having a constant y-coordinate be parallel? 8 What figure will you get on joining the points? a X (0, 0), Y (3, 0), Z (0, 3) b A (6, 6), B (4, 4), C (3, 3), D (5, 5)

9 Draw a graph representing five multiples of 2. 10 Look at the figure and complete the following table: y

Point

A B

–5

–4

–3

–2

–1

Coordinates

–6

Ordinate

–7

Abscissa

D 4

–1

–2

–3

–4

11 Look at the graph. y

Answer the following:

a What are the coordinates of A, B, C, D, P, Q, R, S, T, U

and V?

b What types of quadrilaterals are ABCD and STUV?

c What is the length of PR in ∆PQR?

d Find the area of the triangle ∆PQR.

–7

–6

–5

–4

–3

–2

–1 –1

Q 4

–2

–3

–4

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Word Problems 1 The number of electrical appliances manufactured by a factory during five consecutive years is given below.

Years

2000

2001

2002

2003

2004

Number of appliances (in thousands)

Draw a line graph representing the above data.

2 Mr. Gupta owns a company. The graph below

represents the sales data from

2015 to 2022.

Sale (in crores)

What were the sales in:

i 2017? ii 2021? iii 2022? iv 2016? b Compute the difference between the sales

Scale: 1 Division = 2 units

10 8 6 4

of 2017 and 2021.

c In which year was the greatest difference in

the sales compared to the previous year?

3 Raman and Ansh were walking on a road. The

2015 2016 2017 2018 2019 2020 2021 2022 Year

data was collected for the distance they walked in a particular time.

Time (in minutes)

Distance (in m) by Raman

Distance (in m) by Ansh

Draw a double line graph for the data.

The following data represents the interest on deposits Farhan made for a year. Deposit

₹500

₹1000

₹1500

₹2000

₹2500

Simple interest

₹100

₹200

₹300

₹400

₹500

Draw a graph for the above data. Does the graph pass through the origin?

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Playing with Numbers

Let's Recall We have learnt about the Indian Place Value chart. It can be given as:

Ones

Tens

Ones

Hundreds

Thousands

Thousands Ten thousands

Ten Lakhs

Places

Lakhs

Crores

Ten crores

Crores

Lakhs

Periods

Using the above chart, the place value and expanded form of a 4-digit number can be given as:

2 2 ones 3 tens

9 hundreds

6 thousands

6932: Six thousand nine hundred thirty-two Expanded Form = (6 × 1000) + (9 × 100) + (3 × 10) + (2 × 1) 6 thousands

9 hundreds

3 tens

2 ones

Letʼs Warm-up

Write the expanded form of the numbers. 1 235

___________________________________________________________________________________________________

2 362

___________________________________________________________________________________________________

3 894

___________________________________________________________________________________________________

4 1265

___________________________________________________________________________________________________

5 3657

___________________________________________________________________________________________________

I scored _________ out of 5.

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Numbers and Divisibility Rules Real Life Connect

Nisha and Sam are playing a board game called Prime Climb. Each of them has two pawns. They take turns rolling two 10-sided dice and applying the value of the digits on the dice to the numbers on which the pawns are placed using any of the four basic arithmetic operations: addition, subtraction, multiplication, and division. The first to get both the pawns on the 101 circle at the same time wins the game!

Numbers and Number Puzzles The children performed various operations, and crossed one-digit and two-digit numbers to finally reach a 3-digit number that is 101. Let us learn more about numbers and play with them!

Numbers in General Form Let us take the two-digit numbers 28 and 46 and write their expanded form, 28 = 20 + 8 = 2 × 10 + 8 46 = 40 + 6 = 4 × 10 + 6 Any 2-digit number ab can be written in general form as:

ab = a × 10 + b

Similarly, let us take two three-digit numbers 125 and 269 and write their expanded form, 125 = 100 + 20 + 5 = 1 × 100 + 2 × 10 + 5

Here ab or abc does not mean

a × b or a × b × c, respectively.

269 = 200 + 60 + 9 = 2 × 100 + 6 × 10 + 9 Any 3-digit number abc can be written in general form as: Example 1

Remember!

abc = a × 100 + b × 10 + c

What is the generalised form of the number 87? 87 = 8 × 10 + 7

Example 2

What is the value of p in the number pqr in the generalised form? The generalised form of pqr = 100 × p + 10 × q + 1 × r

Therefore, the value of p in generalised form = 100 × p Do It Together

Write the numbers in their usual form. 1 5 × 10 + 9 × 1 = __________________

2 8 × 100 + 7 × 10 + 2 = __________________

Reversing Digits Reversing the digits of a two-digit number

Let’s take a 2-digit number and reverse the digits. Choose any 2-digit number

(a × 10 + b) = 10a + b

Reverse the number

(b × 10 + a) = 10b + a

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Let’s see what happens when we apply mathematical operations to the two numbers. 10a + b + 10b + a

Add the numbers

= 11a + 11b = 11(a + b)

Subtract the numbers

10a + b – (10b + a) = 9a – 9b = 9(a – b)

Based on the above result, we can say that,

1 T he sum of any two-digit number and its reverse is completely divisible by 11 and the quotient is (a + b) or vice versa

2 T he difference of any two-digit number and its reverse is completely divisible by 9 and the quotient is (a – b) or (b – a) or vice versa. Reversing the digits of a three-digit number

Let us now take any three-digit number abc and do some operations on it and see the results! Choose any 3-digit number Form two numbers by changing

the order of the digits.

100a + 10b + c 100c + 10a + b; 100b + 10c + a 100a + 10b + c + 100c + 10a + b + 100b + 10c + a

Add the numbers

= 111a + 111b + 111c = 111(a + b + c) = 37 × 3(a + b + c)

Based on the above result, we can say that, 1 T he sum of any three-digit number and two other numbers formed by interchanging the order of the digits in cyclic order is completely divisible by: • 37 and the quotient is 3(a + b + c).

• Thrice the sum of the digits and the quotient is 37. Example 3

Without performing actual addition and division, find the quotient when the sum of 86 and 68 is divided by 14. 86 is a 2-digit number and 68 is obtained by reversing its digits.

Also, (86 + 68) is completely divisible by 8 + 6 = 14 and the quotient is 11. Do It Together

Without performing actual addition and division, find the quotient when the sum of 123, 312 and 231 is divided by 18. 123 is a 3-digit number and 312 and 231 are obtained by arranging the digits in cyclic order. 18 = 3 (_____ + 2 + _____), hence the quotient = _____.

Letters for Digits In this section, we will learn about puzzles in which letters take the place of digits in an arithmetic sum and the problem is to find out which letter represents which digit. We usually follow the rules below to solve such puzzles. Rule 1:

Each digit is represented by a single letter. Each letter represents just one digit.

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Rule 2:

The first digit of a number can never be zero. For example, fifteen = 15 not 015 or 0015

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For example, let us find the value of: Step 2: 3 + B = A As A = 6 and we have a carry forward from the previous addition;

+ B

1+3+B=6⇒B=2

A 5 1

Step 1: A + 5 = 1 The only possible digit that can satisfy this condition is A = 6 6 + 5 = 11

Error Alert! Do not forget to add the carry forward digit while finding the value of letters. B should be 2 and not 3.

3+B=6⇒B=3 Example 4

1+3+B=6⇒B=2

Think and Tell Why isn't A = – 4 but 6 in the above statement?

Find the value of A in the addition. 3+A=1

A would be a one-digit number; 3 + A = 11

Hence A = 8. Example 5

Find the value of A in the multiplication. 6×A=A

To make the above statement true; A can be 2, 4, or 8

Placing the above values of A in the multiplicand, we get the products as: 42 × 6 = 252; 44 × 6 = 264; 48 × 6 = 288

In all the above results; A = 8 is the only value which satisfies the multiplication sentence. Do It Together

Find the value of A and B in the addition. There are two unknowns _________ and B. We know that: B + A = 8 and 2 + B = A

Substituting the value of A in the first equation, we get B + _________ = 8

2B + _________ = 8

2B = _________ or B = _________ _____________________________ _____________________________

Think and Tell Can the value of A and B = 9?

Thus, A = ________________ and B = ________________.

Number Puzzles and Patterns Magic Square

A magic square contains several distinct whole numbers arranged so that the sum of the numbers is the same in every row, column, and the main diagonal and usually in some or all of the other diagonals. The square could be 3 × 3, 4 × 4 or even bigger. For example: 334

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3 × 3; Sum = 15

4 × 4; Sum = 34

Arithmetic Sequence

5 × 5; Sum = 65

An arithmetic sequence is a sequence of numbers where the difference of every two consecutive terms is the same. For example,

Here, the difference of two consecutive numbers is 2. Geometric Sequence

A geometric sequence is an ordered set of numbers that progresses by multiplying or dividing each term by the same number. For example:

Here, each number is multiplied by 2. Triangular Numbers

Triangular numbers are usually represented as a sequence of numbers created by organising rows of dots into equilateral triangles. For example,

4 × 4 = 16

5 × 5 = 25

Square Numbers

A square number is the result of multiplying a number by itself.

1×1=1

2×2=4

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3×3=9

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Cube Numbers

A cube number is the result of multiplying another number by itself thrice.

1×1×1=1

2×2×2=8

3 × 3 × 3 = 27

Fibonacci Numbers

The Fibonacci sequence is a sequence in which each number is the sum of the two preceding ones. The sequence can be written as: 1, 1, 2, 3, 5, 8, 13, 21, ….. The Fibonacci numbers can also be shown using a triangle. 1

Example 6

1 9

1 8

1 7

1 6

1 5

1 4

1 3

1 2 6

1 3

10 10

1 4

15 20 15

1 5

21 35 35 21

1 6

28 56 70 56 28

1 7

36 84 126 126 84 36

8 13 21 34 55 89144

Did You Know?

The Fibonacci sequence can also be seen in the way tree

1 8

branches form or 1 9

split at many places 1

10 45 120 210 252 210 120 45 10

in nature. 1

11 55 165 330 462 462 330 165 55 11

Fill the magic square so that the sum of each row, column and diagonal is 15.

The sum of each row and diagonal must be 15, hence, 6 + 1 + ? = 15 ⇒ ? = 8

2 + ? + 4 = 15 ⇒ ? = 9

The magic square with the sum of each row, column and diagonal as 15 can be given as:

4 + ? + 8 = 15 ⇒ ? = 3 ? + 5 + 3 = 15 ⇒ ? = 7 ? + 7 + 6 = 15 ⇒ ? = 2

Example 7

What would be the next three terms of the given sequence 10, 4, –2, –8, ________, ________, ________? In the sequence, 4 – 10 = –6; –2 – 4 = –6; –8 – (–2) = –6 The next three terms would be, –8 + (–6) = –14 –14 + (–6) = –20 and –20 + (–6) = –26

Therefore, the sequence would be: 10, 4, –2, –8, –14, –20, –26 Do It Together

What would be the 7th term of the sequence 2, 6, 18, ....?

In the sequence, each successive term is obtained by multiplying the preceding term by ________. Hence, 4th term = 18 × ________ = ________; 5th term = ________ × ________ = ________ 6th term = ________ × ________ = ________; 7th term = ________ × ________ = ________

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Do It Yourself 24A 1

Write the numbers in generalised form. a 75

b 98

c 235

d 954

c 6 × 100 + 0 × 10 + 2

d 9 × 100 + 5 × 10 + 0

Write the numbers in the usual form. a 3 × 10 + 4

b m × 10 + n

Without performing actual addition and division, find the quotient when the sum of 25 and its reverse is divided by 11.

Without performing actual addition and division, find the quotient when the sum of 582, 258 and 825 is divided by 37.

Find the value of the letters in the operations. a +

× 3

3 ×

Fill in the magic square so that each row, column and diagonal adds up to 34.

12 1

Which is not a triangular number? a 21

b 45

c 90

d 105

e 153

d 108

e 236

What will be the units digit of the square of each number? a 32

b 55

c 69

What would be the missing number in the series 21, 34, 55, _____, 144?

10 What will be the next 3 terms in the sequence 9375, 1875, 375, _____, _____, _____? 11 Fill in the magic square so that each row, column and diagonal adds up to 260. 96

32 60

92 16

76 84 12 44 72 Chapter 24 • Playing with Numbers

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Word Problem 1

Suhani filled in the magic square and claimed that each row, column and diagonal of the square adds up to 111. Is the

square filled correctly? If not, make the correction to justify her statement.

34 35

11 27 28

1 30

19 14 16 15 23 24 18 20 22 21 17 13 25 29 10

24 12

Test of Divisibility Remember Nisha and Sam playing Prime Climb? Nisha’s pawns were on the numbers 27 and 87 but she didn’t want to finish the game soon. On rolling the dice, she got the numbers 2 and 3. She wasn’t sure if 87 can be divided by 3 or not. Let us help her out! We can test the divisibility of any number by following certain rules without actual division. These are called divisibility rules, which can be given as: A number is divisible by ….

Example 8

If the last digit is even or zero. For example, 12, 24, 36, 48, 60 etc.

If the sum of the digits is divisible by 3. For example, 87 = 8 + 7 = 15 is divisible by 3.

If the last digit is 0 or 5. For example, 25, 30, 85, 200 etc.

If the sum of the digits is divisible by 9. For example, 324 = 3 + 2 + 4 = 9 is divisible by 9.

If the last digit is 0. For example, 80, 250, 1000 etc.

Which of the numbers is divisible by 3? 1

3510

3+5+1+0=9

Hence, divisible by 3. Example 9

6504

6 + 5 + 0 + 4 = 15 Hence, divisible by 3.

4872

4 + 8 + 7 + 2 = 21

Hence, divisible by 3.

5234

5 + 2 + 3 + 4 = 14

Hence, not divisible by 3.

What would be the smallest value of x to make the number 25x3 divisible by 9? For a number to be divisible by 9, the sum of the digits must be divisible by 9. 2 + 5 + x + 3 = 10 + x

Since x is a digit, it can take the values 0, 1, 2, 3, ….

But the smallest value that would make the number divisible by 9 will be 8 as 10 + 8 = 18 which is divisible by 9.

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Do It Together

Test the divisibility of the number 6,34,530 by 2, 3, 5, 9 and 10.

Divisibility by 2 = The number ends with 0, hence it is ___________ by 2.

Divisibility by 3 = The sum of the digits of the number is ___________, hence it is ___________ by 3. Divisibility by 5 = The number ends with 0, hence it is ___________ by 5.

Divisibility by 9 = The sum of the digits of the number is ___________, hence it is ___________ by 9. Divisibility by 10 = The number ends with 0, hence it is ___________ by 10.

Do It Yourself 24B 1

Which of the numbers are divisible by 2? a 36,126

b 42,365

b 25,341

b 36,152

b 49,365

d 47,322

e 65,246

c 48,630

d 7,02,365

e 9,43,270

c 57,780

d 64,822

e 82,674

Select the option where the first number is divisible by the second number. a 52,365; 10

c 35,924

Which of the numbers is divisible by 9? a 38,855

e 81,658

Test the divisibility of the numbers by 5 and 10. a 25,365

d 72,362

Which of the numbers are not divisible by 3? a 13,653

c 65,254

b 54,395; 3

c 59,490; 9

d 60,589; 3

Match the numbers with their divisors. a 1250

2, 3, 9

b 3645

3, 5, 9

c 5214

2, 3

d 9558

2, 5, 10

Give all the possible values of x in the number 35x27 which makes the number divisible by 3.

What would be the smallest and largest value of z, in the number 4z41, to make it divisible by 9?

What would be the value of x and y to make the number 492x35y divisible by 2, 3, 5, 9 and 10?

Word Problem 1

Rajat and Suman are playing a game of divisibility. Rajat chooses the nearest number to

56,829 which is smaller than it and divisible by 9, and Suman chooses the nearest number

to 56,829 which is greater than it and divisible by 9. What are the numbers chosen by each of them?

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Points to Remember • Any 2-digit number can be written as ab = a × 10 + b and a 3-digit number as abc = a × 100 + b × 10 + c in general form. • The sum of any two-digit number and its reverse is completely divisible by 11 and the quotient is (a + b) and vice versa. • The difference of any two-digit number and its reverse is completely divisible by 9 and the quotient is (a − b) or (b − a) and vice versa. • The sum of any three-digit number and two other numbers formed by interchanging the order of the digits in cyclic order is completely divisible by: • 37 and the quotient is 3(a + b + c). • thrice the sum of the digits and the quotient is 37. • In letter puzzles, each digit is represented by a single letter. Each letter represents just one digit, and the first digit of a number can never be zero.

Math Lab Setting: Individual

Let’s do the divisibility test!

Materials Required: Pen, paper Method:

1 Every student writes his/her date of birth in the format DDMMYYYY. 2 The student now tests the number formed for divisibility by 2, 3, 5, 9 and 10. 3 All the students share their answers with the class.

Chapter Checkup 1 Fill in the blanks to complete the generalised form of the numbers. a 51 = ________ d

________ = 2 × 100 + 3 × 10 + 0

________ = 3 × 10 + 6

e 354 = ________

c pq = ________ f

________ = 7 × 100 + 0 × 10 + 9

2 Without performing actual addition and division, find the quotient when the sum of 84 and its reverse is divided by 12. 3 Without performing actual addition and division, find the quotient when the sum of 631, 163 and 316 is divided by 30.

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4 Find the value of the letters in the operations. a

B 6

3 ×

A B

5 Fill in the magic square so that each row, column and diagonal adds up to 65. 3

21 7

25 12

15 2

1 5

6 Which of the numbers is a triangular number? a 27

b 65

c 91

d 180

e 600

d 119

e 387

7 What will be the units digit of the cubes of the numbers? a 51

b 72

c 85

8 What will be the next 2 terms in the sequence 7, 56, 448, _____, _____? 9 Fill in the table with ‘Yes’ or ‘No’ using the divisibility test rules. Number a

26,350

38,142

48,537

5,69,841

9,61,314

By 2

By 3

By 5

By 9

By 10

10 Give all possible values of z which would make the number 57,3z5 not divisible by 3. 11 What would be the smallest value of x to make the number 47x2x4 divisible by both 3 and 9?

Word Problem 1 Sarah is arranging her students in rows for a school event. If there are a total of 917 students, will she be able to arrange them in rows of 9? Justify your answer.

Chapter 24 • Playing with Numbers

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AAnnswe swers rs Chapter 1

Let's Warm-up 5. 41.75

23 12

2. 113.51

Do It Yourself 1A 1. a.

3. 46

4. −

c. Subtraction property of zero

3 8

0 1 2 3 4 5 5 5 5 5 5

Word Problems 1. 11

1 2 3 4 5 6 7 8 4 4 4 4 4 4 4 4

0 –2

–1

1 3. a. 3

2 c. − 7

1 b. − 8

8 d. 21

4. Answers may vary. Sample answers: 10 15 20 a. ; ; 16 24 32

5. a. <

b. <

c. <

d. =

e. <

c. −2 <

f. =

g. >

h. >

5 7 1 < 1 < 1.25 b. −0.75 < < 9 10 6

8 2 13 < 1.25 < 1 < 9 4 4

d. −

−18 1 16 < −2.5 < 3 < 3 < e. 5 4 3

3 7 2 < < 1.5 < 2 < 4.2 2 5 5

11 39

2. a.

10 11

b. −

67 165

g. −

3. −8.7

c. −

473 80

83 3 or 20 4 4

7. a . Closure property

53 60

g. −27 76

4 −14 or −2 5 5

−17 30

10.

d. <

4 m 21

31 kg 120 −36 5

11 7 , Add 18 18

3. 2.65 kg

h. 1428 5 e. 8

42 25

2. a.

32 −7 b.

−65 108

−3 4

−7 48

4. a.

20 33

−13 6

38 17 or 1 21 21

−56 17 ; 17 −56

−672 1105

108 455

40 1 or 13 3 3

2 2 34,225 740 m or 3802 sq. m; P = m or 246 m 3 9 9 3

55 m 64

8. ₹195

1743 520 ; 520 1743

2. ₹72000

d. −

−31 98

Chapter Checkup 1. −

e. –1.5

147 120

887 105 5. 9

h .

7. 1

7. a, d 8. Answers may vary. Sample answers:

d. 4.5

25 8

f. −368 25

6. A =

−25 3 36 17 < < 2.1 < < f. 2 5 5 2

57 58 59 60 − ,− ,− ,− 36 36 36 36 Word Problem 1. Raj 8 103 10 1B 1. a. b. − c. − 99 15 7

2. 5

10. Subtract

over subtraction. c. S tudents will prove using commutative property of multiplication. d. S tudents will solve using the associative property of subtraction.

−15 −5 −30 ; ; 21 7 42

8 9 1 5 6. a. −3.5 < < < 2 < 7 5 4 2

5 feet 12

403 840

13 2

5 . a. S tudents will solve using the associative property of multiplication. b. S tudents will solve using associative property of division

−3 −18 −27 b. ; ; 36 4 24

12 3 18 ; ; 16 4 24

b. 7 kg

b. −

c. −15

10 9 8 7 6 5 4 3 2 1 − − − − − − − − − − 5 5 5 5 5 5 5 5 5 5

9. −

36 35

1C 1. a.

21 14

b. −

4. 6.325 L

4 3 2 1 − − − − 0 4 4 4 4

d. −

11. a . 38.17 kg

7 2. − 4

5 9

12 7

8. a.

6. ₹277

b. Associative property

f. 23.44

131 630

e. =

2. g. <

Word Problems 1. 55

−27 87

3. a. <

h. >

4. a.

17 8 1 > > 1.75 > −0.85 > −1 3 3 12

5. b

1167 179 m or 1 m 988 988

11. 320 pages

1 4

f. <

10 32

b. =

5 m 64

c. =

5 5 −7 1 > 1.2 > > > −1 2 6 6 2

21 18 3 > > 3.25 > 2 > −5 5 5 6

6. Answer may vary. Sample answer:

−10 20 10 20 ,− , , . 20 20 20 20

7. Students will prove using the given values for m, n, and o. 8. a. zero b. negative c. 0 d. Zero 9. −

11 181 or −2 85 85

12. a.

30 23

56 15

10.

153 28

11. 1489.8 sq. m

6 43

1 9

125 −338

51 43

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144 g. 1003

507 h. 696

1 13. 82 m 4

15. 1 minute 35.2975 seconds Word Problems 1. 14

14. − 31 983 700

16.

23 hours 24

4. c, d, e 5. a. 3 b. 4 6. a. 9

209 29 kg or 3 kg 60 60

7. a. −40

23 pizza 24

5. 75

Chapter 2

6. ₹250

Let's Warm-up

1. x = 2

Do It Yourself 2A 1. a. 3

c. 3

5. x = – 6

2 11

e .

b. 1 f. 6

10. 650

2. x = 6 2. a. 3

3. a. 102

101 f. 24 4. a. 7 b. 4.2 17 13 e. 4.4 f. or 4.33 5. 45, 90 3 7. 15, 45

8. a. 9 cm

b. 5 cm

b. Yes

d. –3 d.

e. –3

5. a .

143

d. 7

23 cm 8

Do It Yourself 3A 1. a. heptagon

3. Polygon

3. 2205

4. a. y = 3

b. y =

1 7

c. 1

1. a .2 e. 9 e. 7

7. 267954

Answers

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b. 4 f. 23 f. 0

c. 3 2. a. 10.5 3. a. 78 b. 18.5

12. 230, 232, and 234

13.

2. Closed curve

b. 9 c. regular polygon d. decagon e. straight; equal length. 2. a. True b. False c. True d. False 3. a. Convex polygon. b. Concave polygon. 4. a. Convex polygon. b. Convex polygon. c. Concave polygon. d. Concave polygon. e. Concave polygon. 5. Convex polygon 6.

x – 13 b. 4y + 11 = 2y + 56 2 c. n + (n + 2) + (n + 4) = 144 d. st = 240; (s + 20)(t – 2) = 240 2. 85, 87, 89, and 91 3. 132, 143, and 154 4. 72 5. 30° 6. 4200, 1800 7. 26 years 8. 30, 15 9. ₹8050 10. Rohit - 92 runs, Virat - 204 runs 11. 20 pencils 12. Daughter - ₹15,00,000 Son - ₹10,00,000 13. 52 14. 12 km/h 15. The number of ₹10 notes = 82, the number of ₹5 notes = 10 16. 25 km/h, 35 km/h.

Chapter Checkup

1. Open curve 4. Non-simple curve

Let's Warm-up

14 Word Problem 1. 24 years; 12 years

2C 1. a . 3x + 5 =

11.

6. 20, 22

b. 5

6. 30°, 60°

Chapter 3

8 27

3. a. −

f. –48

d. 1

b. −

2. a.

d. –19 e. 5

c. 3

c. 11

92 45

c. No

f. 5

d. 18

4. x = – 8

Word Problems 1. 12 years 2. 3 years and 33 years square metres 4. 150, 50, 25 5. 30 years

2B 1. a. No

b. 8

c. 5

b. 15

e . −

c. 14

109

Word Problems 1. 3 years 2. 44 years 3. 1802.25 sq. m 4. The number of ₹50 notes = 60, ₹100 notes = 40, and ₹500 notes = 35. 5. 300 metres by 140 metres 6. 120 goats 5 7. 21 hours 8. 35°, 85°, 95°, and 145° 9. 7 hours 6 10. ₹3000 11. 16 years, 36 years 12. 15 years, 45 years

3. x = 7

b. 12

1 6

d. 8 e.

8. a. 1 b. 3 c. 1 d. 4 e. 3; Phone Number: 8130011423 9. a. + b. −,+ c. +, +; −, + d. +, +; +, −

3. Shweta; 5943.59 sq.m. 4. 6

b. −

c. 7

b. 7 c. 4

c. 6 d. 1 d. −10

7. a. 2, square b. 20, Octagon c. 35, Decagon 8. 5 9. 3 Word Problem 1. Heptagon; 14; Concave polygon 3B 1. a. 360° b. 2n – 4 c. 120° d. 12 2. a. False b. False c. False d. True e. True 3. a. 900° b. 1980° c. 2700° d. 3240° e. 3600° 4. Yes; 12 sides 5. 16; 157.5° Word Problem 1. 135° 360° 3C 1. a. Quadrilateral b. c. 20° d. 135°; 45° n 2. a. True b. False c. True d. False 3. a. 120° b. 72° c. 45° d. 30° e. 15° 4. a. 9 sides b. 8 sides c. 6 sides d. 5 sides e. 4 sides 5. 32°; 16°; 81°; 95°; 27°; 109° Word Problem 1. 51.43°

Chapter Checkup

1. a. irregular polygon d. 5 e. 360° 2. a. 2 b. 0 c. 119

b. 720°

c. Concave

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3. a. Concave polygon. b. Convex polygon. c. Concave polygon. d. Convex polygon. 4. a. 1260° b. 1980° c. 2520° d. 3420° e. 5580° 5. a. 12 sides b. 17 sides c. 20 sides d. 23 sides e. 27 sides 6. a. 40° b. 24° c. 18° d. 10° e. 8° 7. a. 9 b. 8 c. 5 d. 4 e. 3 8. 1:3 9. 92° 10. 108° 11. 30°; 240° 12. 135° 13. No 14. 12 sides 15. 54 sides 16. 10; 36°, 144° 17. 60°, 120° 18. a. 60° b. 120° c. 6 sides Word Problems 1. 9; 20 2. Regular polygon. 3. 54 diagonal; 10 triangles 4. 108° 5. 46.67°, 93.33° and 40° 6. 10 bridges; 35

Chapter 4

8. a. 103° b. 127° c. 110° 9. 53°, 127°, 53°, 127° 10. 25 cm, 45 cm 11. 140°, 40°, 140°, 40° 12. 72°, 108°, 72°, 108° 13. 60°, 120°, 60°, 120°. 14. a. w° = 50°, x° = 125° y° = 10°, z° = 45° b. x = 15°, y = 11° c. x = 101°, y = 19° 15. Use congruence of a triangles 16. ∠ PSR = ∠ PQR = 104° ∠ QPS = ∠ QRS = 76° Word Problems 1. Not a parallelogram 2. 24 m; 120°, 60°, 120° and 60°.

Chapter 5

Let's Warm-up

1. Kite

Do It Yourself 5A 1. a .

7.8 cm

C A

B E

B F

2 . a. OL = 6.3 cm D

G F

7.2

3.8 cm

5.6 cm

3 6.

3. a. DB = 7.1 cm, AC = 8.5 cm 4 cm 95° 75 °

5. a.

5.4 cm

A cm

5°

8 cm

A 4.2 cm

6 cm

90° 2.9 cm

3.6 cm

3.9 cm

7 cm 12 cm

9 cm D

7 cm

B C

6 cm A

60° 9 cm

L 3.1 cm

60°

Word Problem 1. Figures may Vary. Sample answers:

60°

5.3 cm

3.3 cm

3.6 cm

60°

D 3.6 cm

b. T

6 cm

135°

4 cm

6 cm

85°

3. 5

80°

75° 120°

4.9 cm

5.2 cm

110

7 cm

4. a.

4.3 cm

b. ZX = 9.7 cm, WY = 9.6 cm

2. a. Concave Quadrilateral b. Convex Quadrilateral c. Concave Quadrilateral 3. ∠ MJL = 70°; ∠ TJK = 20°. 4. x = 9.8°, y = 27° 5. 40°; 29° 6. a = 4, b = 7 7. 80°, 100°, 80°, 100°. 8. 25 cm, 40 cm 9. 60°, 120°, 60° and 120°. 10. Answer may vary. Sample answers: a. PZ & ZR, ZQ & ZS, PQ & PS, RS & QR b. ∠ QPS & ∠ QRS, ∠ PQR & ∠ PSR, ∠ QSR & ∠ PSQ, ∠ PQS & ∠ RQS c. ∆ QPR ≅ ∆ SPR, ∆ QRS ≅ ∆ QPS, ∆ QZR ≅ ∆ SZP d. Yes e. Yes Word Problem 1. 200 2 m 4B 1. b 2. b 3. a. 114° b. 224° c. 132° d. 200° e. 110° f. 247° 4. a. 107° b. 76° c. 123° d. 123° e. 130° f. 106° 5. a. x° = 97°, y° = 18°, z° = 94° b. y° = 30°, x° = 55° c. x° = 116°, y° =27°, z° = 37° 6. 95°, 85°, 95°, 85° 7. 120°, 60°, 120°, 60° 8. 105°, 75°, 105°, 75° 9. 137°, 64°, 50°, 109° 10. x° = 45°, y° = 28°, z° = 107° Word Problem 1. 36°, 72°, 108° and 144°; convex Chapter Checkup 1. a. True b. True c. False d. False e. False 2. c 3. Interior points: P, Q, R Exterior points: S, T, U 4. d 5. c 6. a. 110°, 90°, 70°, 90° b. 135°, 120°, 80°, 25° c. 100°, 95°, 85°, 80° 7. a. 240° b. 150° c. 123°

b. OW = 5.4 cm

3.4

6.3 cm

4.2 cm

5.6 cm M

4.8 cm 6.7 cm

4.7 cm

7.4 cm

Do It Yourself 4A 1. Figures may vary. Sample figures.

5.2 cm

6.5 cm

3. Rectangle W

3. PS, SR

1. P, Q, R, S 2. PS, SR, QR, PQ 4. PS, QR 5. ∠Q, ∠S

Let's Warm-up

2. Trapezium 4. Rhombus

C 12 cm

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5.6 cm

4. M

cm B

5.6 cm

4.6 cm

m K

N 4.2 cm K

Answers

UM24CB8_Batch 4.indb 345

5 cm

7 cm

45°

70° 90°

6 cm

115°

4.9 cm J

4 cm 4 cm 4 cm

W 4.5

6.1 cm

V 7.3 cm

5.1 cm

M 3.5 cm L

90°

6.5 cm

4. a.

5.8 cm

135°

4.5 cm 100°

7.4

b. KI = 8.4 cm, JL = 7 cm

4°

4 cm

3.9 cm

c. PN = 8.5 cm, OM = 16.2 cm

7.2

60°

Chapter Checkup

6.8 cm

4.2 cm

6.5 cm

120°

5.8 cm

4.5

7.4 cm

77°

Word Problem 1.

1. a.

3.6

2°

67°

m 1c

2.5 cm

8.3

3.9 cm

6 . Largest angle in the trapezium ABCD is 112.

4.7

4.9 cm

6.8

6.7 cm 45° 4.6 cm

6.9 cm

3. a. GE = 4.3 cm, FH = 5.4 cm

6.7 cm

5 cm

c. NE = 6.7 cm

5 . S

5.5 cm

2.7

4.7 cm 7.6 cm P

5 cm

4.5 cm

5 cm

120°

b. NI = 3.7 cm N

.2 cm

5 cm

2. a. EP = 2.8 cm

2 3.

5 cm

3 .

5 cm

2.7

5B 1.

6.2 cm

8.9

Z 150°

80°

S 4.5 cm T

135° 60°

X 5.7 cm

4.4 cm Y

5.6 cm C 4.1 cm A

3.2 cm

100°

5.5 cm

5.2 cm

6.6 cm

4.6 cm

5. a.

Z W

5.7 cm

60°

6.4 cm

3.9 cm

70°

4.6 cm

6 cm

R 3.3 cm

5 cm

80° P

3.9 cm

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5.2 Y

5.8 cm

2.9 cm

7. a.

P cm

6 cm

c 3.7

3.7

6.6 cm

cm R

R D

3.7

Tally

3.3

||||||||

||||

Cricket

Sport

2.9 cm

6 cm

3.7 cm

5 cm

3.7 cm 40° E

40°

4.5 cm

11.

5.5

3.6 cm

2.6 cm

3.6 cm D M

|||| |||| |||| ||||

|||| ||||||

||||

|||| |||||||

b. 14 Ages

c. 60,000 Number of people (Frequency)

10. We can construct only one trapezium with the given measures.

Tennis

2. a. 14 3.

4.6

2.9 cm

60°

Tally Marks

Tennis

3.7 cm

students

(Frequency) 12

sport

Number of

Badminton No favourite

||||||||

Do It Yourself 6A 1.

4 cm

Total

||||||

A 5.6 cm

4.2 cm

Let's Warm-up

B 60°

Chapter 6

3.3

8. a. A

3.7

4.2 cm

6.8 cm

4.2 cm

cm 7.6

B 2.9 cm

120°

Word Problem 1. Diagonals of a rhombus bisect each other. We can use this property to construct the rhombus using ruler and compass.

4.6 c

Tally Marks |

|||

|||| |||

|||

Scores

Number of students

Tally Marks

75 76 77

(Frequency) 1 1 1

| | | |

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|||

16-20

4. a. 10 b. 6 5. a. 20 students e. 71.41% Word Problems 1.

Number of individuals

||||

Digits

Frequency

Tally Marks

holidays

(Frequency)

6B 1. a. 50

||||| |||||

Frequency

30-39

50-59

Math Grade 8 CB Answer Key.indd 347

Class intervals

Frequency

iv. 28

(Number of

Tally marks

140-149

|||

150-159

|||| ||||

160-169

||||||

170-179

||||

Frequency

Tally marks

||||

(Height in cm)

iii. 16

Class intervals

students)

101-200

201-300

||||

301-400

|||

401-500

||||

501-600

601-700

|||

701-800

11-15

ii. 18-20 and 24-26

i. 2

(Electricity bill in ₹)

Answers

c. 10

0-5

6-10

||||

Frequency

Tally marks

(number of people)

students)

|||||||

||||

(ages of people)

Frequency

(Number of

Class intervals

(Weight in kg)

|||

||||||

f. 5

27-29

|||

e. 8

21-25

24-26

d. 79

5 3

||||||

c. 37

||||

11-15 16-20

|||

90-99

||||||

21-23

100-109

|||

80-89

6-10

||||||

||||

70-79

||||

|||||||

||||

60-69

40-49

0-5

Class intervals

Tally marks

students)

15-17

(percentage of crops harvested)

d. 28

(Number of

Class intervals

||||

Frequency

c. 7.5

18-20

b. 100

b. 45; 55

2. a.

||||||

Tally Marks

|||||

Range = 4

c. 26 d. 9 b. 8 students

(Marks obtained)

Number of

|| || || |||| | |

2 2 2 4 1 1

Class intervals

Range = 20

||||

21-25 26-30 30-35 36-40 40-45 46-50

Tally marks ||

|||

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Chapter Checkup Observations

Frequency

Tally Marks

||||||

|||||

|||

Observations

Frequency

Tally Marks

||||||

4 5

2 4

Frequency

Tally Marks

||||

|||||

|||

Class interval

Frequency

Tally Marks

56-60

61-65

||||

66-70

||||

71-75

|||| |||||

76-80

|||||||

Class interval

Frequency

Tally Marks

55-60

60-65

||||

|||||| |||| ||||||

||||

|||||

|||

65-70

|||

70-75

|||| ||||||

75-80

|||||||

||||

|||

||||

Ages

Frequency

Tally Marks

||||

|||| ||||||

||||||||

b. 15.4% c. 63:87 2. a. 13.14% b. 21.37% c. 97:67 3. a. $3500 million b. $600 million d. 53:66 4. Scale: 1 division = 1 crores 20

19.4

19 18 17 16 15

16.7 15.6

Lower limit 0 10 20 30 40

Word Problems 1. a. 5 crores c. Nagaland

Upper limit 10 20 30 40 50

19.8

17.0

14.8

Class limit

d. 71,400 tonnes d. 217:293 c. 12.9%

b. 48

Do It Yourself 7A 1. a. 26.89%

2. 6000 3. C 5. 24,000

4. 2000

1. 4000

Let's Warm-up

||||

Chapter 7

||||

0-10 10-20 20-30 30-40 40-50

||||

interval

||||

Tally Marks

Class

Frequency

5. a. AB

Observations

|| ||

Observations

2 2

Tally Marks

4 cars

||||

8 9

|| ||||

Frequency

registered

20 16

Number of cars

||||

Export (in crores)

1. a.

Year

Class size Class marks 10 10 10 10 10

65 15 25 35 45

b. Maharashtra and Uttarakhand

. 103.3 crores a b. 17.217 crores c. 8.97% d. 2016 and 2019 5. a. 400 students b. 66.67% c. 77.78% d. 14.28% Word Problem 1. a. 48 million b. 5 million c. 20.5 million d. 6 million 7B 1. a. 37.34% b. Fruit c. $11,056 million d. $13,771 million 2. a. 210 tonnes b. 1:1 c. 25% d. 28.57%

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800 700

700 600

600

500

650

50 40 30 20

20052006

a. M essi: 353

4. lavender

11 8

Do It Yourself 8A 1. 112.5°

3 1

a. 4 presidents b. 12 presidents c. 29 presidents Chapter Checkup 1. a. 1,04,000 units b. 83,200 units c. 83,200 units d. 5:4 2. a. 360 students b. 25% c. 40 students d. 360 students 3. a. 760 students b. 21.05% c. 1:1 d. 77.78% 4. Scale: 1 division = 10 students 60

School A School B

50 40

Let's Warm-up

3 20

Tamil Nadu, 620

60° 20°

124°

72° 96°

Others, 100

Karnataka, 440

88°

Chemicals, ₹180

72° 60°

Furniture, ₹150

tre

us ic

Fuels and oil, ₹240

Ar ts

e nc

3 4

Kerala, 300

Import Items Gems, `150

Plastics, ₹180

No. of workers from various states working in a garment factory

68°

2. blue

c. 63:160

2. 162°

et ics

2011- 2012- 20132012 2013 2014

1 5

60°

Activity

20102011

b. Ronaldo

Andhra Pradesh, 340

Age Group

At hl

2008- 20092009 2010

Ronaldo: 355

Chapter 8

42 48 54 60 66 72 above

Number of Students

20072008

5. a. 3400 b. 3600 cars c. 11:14 d. 17.65% 6. a.94.44% b. 82,000 packets of biscuits c. 95.83% d. 86,000 packets of biscuits 7. a. 1,62,000 b. 1,83,000 c. 15:13 d. 27,000 e. 100% 8. Scale: 1 division = 5 units

Sales of Smartphones Brand C, 550

Frequency

20062007

Year

20 12 20 13 20 14 Number of Presidents

a. 3300 mm b. 2016-2017 c. 8.33% decrease d. 2016-17 Word Problem 1. Scale: 1 division = 2 Presidents

Messi Ronaldo

Scale: 1 division = 10 Goals

550

Year

800

9. a. 106 students b. 216 students c. 165:61 d. 40 students 10. a. 515 cars b. 300 cars c. 435 d. 43:16 Word Problems 1. a. 95 players b. 187 players c. 75.22% d. 222 players e. 70.72% f. 157:65

Number of Goals

900

Rainfall (in mm)

3. a. 8.57% b. 1300 people c. 10:19 d. India Gate 4. a. 300 fruit b. 340 fruit c. 7:8 d. Banana 5. a. 380 tickets b. 760 tickets c. 720 tickets d. 88.89% Word Problem 1. a. 8040 runs b. 6940 runs c. 157.14 d. 188:223 7C 1. a. 365 people b. 86.30% c. 75.34% d. 43:18 2. a. 550 workers b. 380 workers c. 82.35% d. 23:45 e. 330 3. a. 81.48% b. 9.25% c. 35:19 d. 34 4. a. 10 employees b. 23:16 c. 84.6% d. 195 5. Scale: 1 division = 100 mm

Brand B, 1050

Brand D, 650

55° 105°

65°

35°

100°

Others, 350 Brand A, 1000

20 15 10 5 5 15 25 35 45 55 65 Class Interval

Answers

UM24CB8_Batch 4.indb 349

349

08-01-2024 14:41:23

% of Buyers Brand P, 30 108°

Brand 0, 20

72°

Brand N, 15

Others, 10

36°

English, 420

Brand M, 25

90°

54°

Hindi, 120 72° Science, 100

6. a. Notebooks

8B 1. Education = 40,000, Arts = 1,00,000, Law = 70,000, b.

1 8

c. Shared cab

. a. 2000 cars 3 b. 360 cars c. 1140 cars 4. a. 600 students b. Apple juice; 250 students 5. a. 700 animals

b. Tigers; 35

6. a. 75 plants 7. a. 12.5%

5 16

c. 21

c. 9:5

b. 33:40

Word Problems 1. 1500 teenagers

2. 15:17

9 10

Items Sold Rice, 35

Spices, 10

36° 45°

157.5°

54°

Bread, 12

67.5°

Pulses, 15

Items Sold Sandwiches, 66 79.2° 90° Shakes, 75

72° 54° 64.8°

Ice creams, 54

3. No. of Families Owning Pets Birds, 6

54° 117°

27°

72° 90°

Others, 16

Cat, 20

Burgers, 60

Frenchfries, 45

1 4

c. 15.08%

7. 900 ice creams

Vanilla, 35

Guava, 50 90°

Chapter 9 3. Equally likely

Others, 8

Fantasy, 280

Favourite Ice Cream Flavour

Let's Warm-up

Chapter Checkup

Dog, 26

Mango, 55

51 c. 100

42°

90°

8. a. 20% b. ₹30,000 c. ₹25% 9. a. 18% b. 1:7 c. 68% 10. a. 21.6° b. 28,800 tonnes c. June, 1800 tonnes d. 25% Word Problems 1. a. ₹2304 b. 86.4° c. Bathroom

Commerce = 1,65,000, Science = 1,25,000

2. a. Shared cab

66°

144°

Adventure, 600

English, 115

Maths, 140

Start.

Biography, 440

EVS, 125

69°

84°

Comedy, 760 48°

75°

60°

Books Read Novels, 320

Ruchika's Score

Rabbit, 12

No. of Students Science, 385 77° 45° Others, 35° 175 84° Maths, 119° 595

Word Problem 1.

1. 82.8° 2. a.

French, 225

Do It Yourself 9A 1. a. Sure

99°

63° 108°

Chocolate, 60

1. Impossible 2. Sure 4. Likely 5. Unlikely

b. Impossible c. Likely d. Unlikely e. Equally likely 2. a. 0 b. 1 c. heads; tails d. 3 3. Spinner A, as the probability of winning a prize is greater for Spinner A than Spinner B. 4 25 4. Basket B, or 20%. 5. 1 6. 7. b 20 26 1 8. 12 contestants 9. a. b. 125 contestants 2 1 1 1 1 10. a. b. c. 4 times 11. a. b. 6 2 6 3 4 1 5 1 c. 50 times 12. a.= b. c. 36 9 36 12 13. a. H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6

b. 12

1 1 1 1 5 5 c. i. ii. iii. iv. v. vi. 12 2 6 3 12 6 14. About 83 people Word Problems 1. a. 30 chips 2 b. 6 contestants 2. a. 18 outfits b. 9 1 2 9B 1. a, b, d, e 2. or 0.16 or 16.67% 3. a. or 0.40 6 5 5 or 40% b. 16 times 4. a. or 0.55 or 55.55% b. 40 times 9

350

UM24CB8_Batch 4.indb 350

08-01-2024 14:41:24

3 10

10 or 90 27 53 35 11.11% b. or 58.88% c. or 38.88% d. or 30% 90 90 90

5. a.

b. 15 Marbles

61 or 48.8% 125 39 d . or 31.2% 125

8. a.

6. 63

7. a.

64 or 51.2% 125

29 or 23.2% 125

25 1 or 20% 9. a. or 0.083 12 125 11 or 8.33%; 50 times b. or 0.22 or 22%; 132 times 50 c. As the number of trials increase the prediction of results tend to be more accurate. Word Problem 1. 125 pairs of jeans

Chapter Checkup 1. a. 1

b. 0

1 7

d. 1

1 6

2. a. Heads, Tails

1 or 0.125 or 12.5%. 8

b . 1, 2, 3, 4, 5, 6 c. a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y and z d. (Heads, Heads), (Heads, Tails), (Tails, Heads), (Tails, Tails) e. (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), 3 1 (6, 5), (6, 6) 3. a. or 0.125 or 12.5% b. or 0.375 or 8 8 37.5%

3 or 0.375 or 37.5% 8

2 1 1 4.= 5. 6. a. Theoretical Probability is greater. 2 12 6 b. Experimental probability is greater. c. Experimental probability is greater. d. Both are equal to 1. 4 1 8 7. a.= or 44.44% b.= or 11.11% c.= or 88.88% 9 9 9 1 5 1 4 ) 8.= d. )= or 44.44% = a. or 50% b.= or 10% 10 10 2 9 4 2 4 2 9 3 = = c.= or 30% d.= or 90% e. or = 40% =f. or 40% 10 5 10 5 10 10 1 1 9. 1 2 pearl bangles 10.= or 50% 11. a. or 25% 4 2 1 3 1 b. or 75% 12. a. or 16.66% b. or 25% 6 4 4 17 2 3 Word Problems 1. a. b. 2. a. or 85% b. 3 or 15% 20 5 5 20

Chapter 10

1. False

Let's Warm-up Do It Yourself 10A 1. a. i. 1 b. i. 1 c. i. 6 d. i. 4 e. i. 9 f. i. 5

4. True

2. True 5. False

3. False

ii. Odd square iii. Perfect square ii. Odd square iii. Perfect square ii. Even square iii. Perfect square ii. Even square iii. Not a perfect square ii. Odd square iii. Not a perfect square ii. Odd square iii. Not a perfect square

49 144 16 c. d. 8,55,625 e. 24 169 25 f. 43,264 3. a. Not a perfect square b. Not a perfect square c. A perfect square, 86 d. Not a perfect square e. A perfect square, 163 4. a. 5, 170 b. 7, 252 c. 3, 99 d. 2, 112 e. 2, 192 f. 15, 810 5. a. 6, 24 b. 7, 35 c. 3, 44 d. 17, 46 e. 106, 30 f. 13, 72 6. 1,76,400 7. 5 2. a. 3025

Answers

UM24CB8_Batch 4.indb 351

Word Problems 1. 2025 plants 2. 3969 students 10B 1. a. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 b. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 c. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 d. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55 + 57 + 59 e. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55 + 57 + 59 + 61 + 63 + 65 + 67 + 69 f. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55 + 57 + 59 + 61 + 63 2. 21, 28, 36, 45; 21 + 28 = 49 = 72, 28 + 36 = 64 = 82, 36 + 45 = 81 = 92 3. a. 25, 24 b. 41, 40 c. 35, 34 d. 51, 50 e. 63, 62 f. 97, 96 4. a. 1624 + 1625 b. 1624 + 1625 c. 2964 + 2965 d. 760 + 761 e. 3612 + 3613 f. 144 + 145 5. a. 5625 b. 101 c. 9025 d. 155 e. 11,025 f. 197 6. a. 5775 b. 4352 c. 2600 d. 8645 e. 1440 7. a. (12, 35, 37) b. (16, 63, 65) c. (28, 195, 197) d. (32, 255, 257) e. (36, 323, 325) f. (48, 575, 577) 8. 11115556, 1111155556, 333334 9. 9999800001, 999998000001 10. a. 64 b. 289 c. 484 Word Problem 1. a. 22, 32, 42, …, n2 + (2n + 1) = (n + 1)2 b. 22, 32, 42, 52, 62, 72 c. 3136 10C 1. a. 3 or 7 b. 4 or 6 c. 2 or 8 d. 1 or 9 e. 2 or 8 f. 4 or 6 2. a. 12 b. 6 c. 21 d. 30 e. 13 f. 10 3. a. 81 b. 94 c. 142 d. 157 e. 173 49 58 69 25 64 b. c. d. e. 50 60 75 34 79 5. a. 3,162 b. 2,102 c. 31,310 d. 39,780 e. 10,200 f. 7,728 6. a. 2, 68 b. 3, 42 c. 5, 54 d. 17, 42 e. 13, 66 f. 5, 286 7. 144 Word Problems 1. 36 students 2. No 10D 1. a. 1 digit b. 2 digits c. 2 digits d. 2 digits e. 3 digits f. 3 digits 2. a. 256 b. 278 c. 324 d. 356 e. 381 3. a. 12, 38 b. 59, 72 c. 76, 94 d. 248, 166 e. 479, 358 f. 84, 560 4. a. 17, 75 b. 4, 68 c. 24, 418 d. 41, 279 e. 24, 532 f. 1353, 738 5. a. 2.65 b. 6.52 c. 8.88 d. 90.01 e. 32.14 6. a. 0.894 b. 9.219 c. 2.645 d. 2.828 e. 3.464 f. 3.098 g. 2.345 7. a. 304 17 39 b. c. 16 d. e. 10.56 8. 9801, 99 16 59 9. 10,000; 100 4. a.

Word Problems 1. 25 m

2. 390 m 9 64 9 81 49 Chapter Checkup 1. a. b. c. d. e. 16 121 49 169 225 2. a. 1769 b. 4355 c. 7225 d. 14,880 3. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 4. (96, 2303, 2305), (24, 143, 145) 5. a. Not a perfect square b. Perfect square, 462 c. Not a perfect square d. Perfect square, 562 2 e. Perfect square, 269 f. Not a perfect square

351

08-01-2024 14:41:25

6. a. 2.81 b. 20.01 c. c. 80.70 d. 21.63

Word Problem 1. 31.62 m

465 302

d. 47.56

14 11

9. 21, 811

7. a. 1.93 b. 9.00 10. 3600

Chapter 11 Let's Warm-up 1. Pineapple

Do It Yourself 11A 1. a . 3375

2. 30

3. 48

4. 12

5. 112

b. 17,576 c. 59,319 d. 1,03,823 e. 4,05,224 f. 6,81,472 g. 1,48,877 h. 7,78,688 i. 17,71,561 2. i. a. 2 b. 3 c. 7 d. 6 e. 1 f. 0 g. 3 h. 0 ii. Odd cubes – b, c, e, g; even cubes – a, d, f, h iii. c, f, g 512 216 64 21,952 3. a. b. − c. d. 729 512 343 3375 35,937 5832 8 125 e. − f. − g. h. − 2197 35,937 125 4913

4. a. 21.952 b. 1.728 c. 0.175616 d. 0.000512 e. 103.823 f. 204.336469 g. 28.094464 h. 753.571 5. a . Not a perfect cube. b. Perfect cube; 24 c. perfect cube; 47 d. Perfect cube; 49 e. Not a perfect cube f. Perfect cube; 95 6. a. 4; 12 b. 4; 8 c. 2; 18 d. 4; 24 e. 9; 45 f. 78; 156 g. 6; 90 7. a. 9; 4 b. 15; 7 c. 21; 12 d. 6; 21 e. 735; 6 f. 5; 43 8. 1331, 3375, 5832, 8000 9. a. 2025 b. 6084 c. 14,400 10. a. 4921 b. 9241 c. 23,497 d. 29,701 Word Problems 1. 13 cm 2. 50 3. 12 11B 1. a. 20 b. 25 c. 28 d. 36 e. 48 2. a.

16 46

b. −

28 51

c. –70

d. 228

3. a. 3 b. 6 c. 8 d. 12 e. 10 f. 11 4. a. 84 cm b. 44 cm c. 62 cm 5. a. 65; 7 b. 18; 13 c. 54; 16 d. 61; 11 e. 34; 19 6. a. 9; 9 b. 44; 14 c. 75; 16 d. 60; 21 e. 21; 18 7. 4, 12, 24 8. 9, 12 Word Problem 1. 42 cm 11C 1. a. 10 b. 19 c. 26 d. 33 e. 43 2. a. 51 b. 52 c. 67 d. 78 3. 92 cm 4. 11 5. 49 Word Problem 1. 57.66 sq. cm 8 2197 Chapter Checkup 1. a. 68,921 b. c. − 729 343

d. 5,92,704 e. 941.192 f. 46.656 2. a. Perfect cube; 94 b. Not a perfect cube c. Not a perfect cube d. Perfect cube; 51 3. 14; 70 4. 17; 68 5. a. 2197 b. 4356 6. a. 63 b. 90 c. 96 d. 47 7. 61 m 14 8. a. 0.8 b. − c. 1600 d. 2.2 9. 45; 14 20 10. 13; 17 11. 5 cu. units 12. 2058 cu. cm Word Problems 1. 3125 cu. cm

2. 38; 57; 95

Chapter 12 Let's Warm-up 1. Red

2. Blue

3. 17%

Do It Yourself 12A 1. a. 56%

4. 31%

5. 68

b. 20% c. 14% d. 50% e. 42% 2. a. 0.135 b. 0.45 c. 1.205 d. 6.75 e. 0.005 1 1 6 5 23 3. a. b. c. d. e. 250 800 5 8 200 4. a . 21:10 b. 11:500 c. 131:200 d. 77:500 e. 11:2000 5. a . 200% b. 30% c. 60% d. 30% Word Problems 1. 1000 students 2. Samaira 3. 300 4. 100 ml 5. 26 students 6. a. 17,500 b. 10,500 c. 4200 7. 200 marks 8. ₹18,000 9. a. 1500 smartphones b. 600 smartphones c. 900 smartphones 12B 1. a. 20% b. 33.33% c. 80% d. 10% e. 200% 2. a. 50% b. 55.55% c. 10.20% d. 33.33% e. 10% 3. a. 84 b. 261 c. 224.4 d. 142.5 e. 152.88 f. 96 4. a. ₹228 b. 105 kg c. ₹2400 d. 875 L e. 753.75 grams 5. 4% increase 6. ₹275 Word Problems 1. 7.5% 2. 1344 3. 1,48,176 4. ₹2,83,500 Chapter Checkup 1. a. 250% b. 33.33% c. 12.5% d. 15.71% e. 64% 2. a. 16% b. 380% c. 266.67% d. 60% e. 8.54% 3. a. 0.3% b. 135% c. 278.9% d. 1660% e. 56.7% 4. a. 23:50 b. 1:6 c. 4:125 d. 7:5

e. 19:2

5. a.

1 12,500

101 47 e. 6. a. 0.018 300 300 d. 0.014 e. 0.04 7. a. 90% d. 20.83% 8. a. ₹6.25 b. 2 L e. 0.9 L 9. a. 2.32% b. 91.11%

123 5000

b. 0.0006

163 25

c. 0.058

b. 20% c. 12.14% c. 30 cm d. 48.05 kg c. 17.77% d. 127.27%

7 c. 20% 15 Word Problems 1. 20% decrease 2. 72 green balls, 108 blue balls, 45 white balls, 36 red balls and 189 yellow balls 3. 2000 students 4. ₹4,41,000 5. 300 marks 10. a. 33.33%

1 7. 37 % 2 8. a. 420 students b. 28% c. 252 votes 9. a. ₹200 b. ₹50 10. ₹20,000 6. 28.29 kg; 14.58 kg

Chapter 13 Let's Warm-up 1. ₹25,000

2. ₹27,000

Do It Yourself 13A 1. a. Loss, ₹5000

3. ₹9000

4. ₹2000

5. ₹1000

b. Profit, ₹2000 c. Loss, ₹2000 d. Profit, ₹5000 2. a. ₹500 b. ₹7490 c. ₹9000 d. ₹1700 e. ₹6,57,500 f. ₹8200 3. Profit, 33.33% 4. 4% 5. Profit percent = 14% 6 . ₹64,000 7. 12.5% Word Problems 1. ₹74,000 2. ₹300 3. Loss percent = 3.45% 4. ₹945 5. ₹1000; ₹1500; Loss = 4% 6. ₹98,000 7. ₹16,500 13B 1. a. ₹1000; ₹3000 b. ₹800; ₹3200 c. ₹500; ₹3500 d. ₹400; ₹3600 2. ₹725, 5% 3. ₹20,000 4. ₹725 5. 6.25% 6. ₹600 7. 34% 8. ₹60,000 9. ₹10,000 10. ₹1152 11. ₹6000 Word Problems 1. ₹13,350 2. 23.33% 3. 16.66% 4. 23.5%

352

UM24CB8_Batch 4.indb 352

08-01-2024 14:41:26

13C 1. ₹49,220

2. ₹16,000 3. ₹13,000 4. ₹4950 6. ₹5 7. ₹9440 Word Problem 1. ₹14,500 Chapter Checkup 1. a. ₹7200, 12% b. ₹7276, 7% c. ₹7500, 10% 2. a. ₹23, 5% b. ₹520, 10% c. ₹6208, 3% 3. a. ₹100 b. ₹1513 4. 50% 5. ₹9600, ₹10,800, ₹1200, 12.5% 6. 24% 7. ₹32,450 8. Loss = ₹1 9. Yes, 20% 10. ₹42,840 Word Problems 1. 12% 2. ₹6,13,149 3. ₹1096 4. ₹1800 5. ₹18,500

Chapter 14 Let's Warm-up 5. ₹360

1. ₹100

2. ₹200

3. ₹140

4. ₹240

Do It Yourself 14A 1. a. ₹460

b. ₹3770 c. ₹122 d. ₹10,000 e. ₹600 f. ₹6055 2. a. ₹112; ₹912 b. ₹1187.5; ₹5937.5 c. ₹8; ₹168 d. ₹722.5; ₹9222.5 e. ₹192; ₹9792 3. a. ₹7500 b. ₹380 c. ₹2582.85 4. a. ₹825; 5.5% b. ₹1800; 4.5% c. ₹16.25; 6.5% 5. a. 2.5 years b. 4 years c. 0.5 years d. 0.6 years 6. a. ₹607.5 b. ₹6500 c. ₹3680 d. ₹903.6 7 . ₹6776 8. ₹74.79 9. P = ₹4000, R = 5% p.a. 10. Scale: 1 cm = ₹1000 on x-axis

Simple Interest (₹)

1 cm = ₹50 on y-axis

Simple Interest on deposits for a year

450 400 350 300 250 200 150 100

50 0

400 320

UM24CB8_Batch 4.indb 353

Let's Warm-up 1. 5p 2. 15 3. 7p 4. 5b Do It Yourself 15A 1. a. –x2 + 2x + 3; –x2, –2x b. –x2 + 5x + 6; –x2, 5x c. –x2 + 4x + 2; –x2, 4x

d. 2x + 7; 2x

b. Terms are 6a and 5. Factors are 6, a; 5. The coefficient of ‘a’ is 6. 1 1 c. Terms are 8a2, –5a and . Factors are 8, a, a; −5, a; . 3 3 The coefficients of ‘a’ are 8a and −5. 2 2 d. Terms are 9ab, − a and –1. Factors are 9, a, b; − , a; −1. 7 7 2 The coefficients of ‘a’ are 9b and − . 7 7 3 4 e. Terms are a , 5abc, − a2 and –10. Factors are 12 9

7 4 , a, a, a; 5, a, b, c ; − , a, a; –10. The coefficients of ‘a’ are 12 9 7 2 4 a , 5bc and − a. 12 9 5 5 3. Monomials: –3y, − ab , stu 3 6

Trinomials:

160

5. −5b

3 3 2. a. Term is − a2bc . Factors are − , a, a, b, c. The coefficient 4 4 3 of ‘a’ is − abc. 4

Binomials: 2 x + 5 ,

240

5 1 p− 2 3

−4x 2 + 1 a2 + 5 , 11 g − 2h + 3 i , 11p2 + 7 q2 − 1 6 4 8 4 2

Quadrinomials:

1000

2000

3000

Deposit (₹)

4000

5000

. Yes a b. ₹200 c. ₹3500 Word Problems 1. ₹720; ₹8720 2. ₹7500 3. 2.5 years 4. 8% per annum 5. ₹16,800 14B 1. SI = ₹500, Difference = ₹25 2. A = ₹2141.16, CI = ₹541.16 3. CI = ₹9270 4. P = ₹4000 5. 20% per annum 6. 3 years Word Problems 1. ₹31,932.23 2. ₹2805.58 3. ₹7981.94 4. ₹21,899.75 5. ₹2,24,643.825 6. ₹60,000 14C 1. 18,30,125 2. 1,03,041 3. ₹2,31,525 4. ₹1,82,250 5. ₹21,600 6. 17,576 units 7. ₹41,070.285 8. ₹2977.54 9. ₹15,614.97 10. ₹6.05 11. ₹14,64,100 12. 4,77,405 13. 20% per year 14. ₹4,80,000 Chapter Checkup 1. a. ₹3900, ₹13,900 b. ₹4.48, ₹564.48 c. ₹97.2, ₹2797.2 2. a. ₹240 b. ₹3200 3. ₹1800, 9% 4. ₹600, 15% 5. 12.5 years 6. ₹3248.7 7. ₹7632 8. ₹1272 9. a. ₹5940.5, ₹940.5 b. ₹85,753.152, ₹13,753.152 10. ₹13,500 11. 2 years 12. 10% per year 13. ₹1164.8 14. ₹3152.5 15. ₹34,481 16. 1.5 years 17. ₹100 Word Problems 1. ₹98.56 2. ₹14,257.2 3. 50,000 4. ₹16,620.49 5. ₹6,50,000

Answers

Chapter 15

9 xy −

5 13 5 pq + 2q + p2, 11pqr − yz + 2 p − 6 8 6

4. Circle a, d and e; cross out b and c 5. a. −

3 4

b. 6 and 5

c. 8, −5 and

1 3

d. 9, −

2 and −1 7

7 4 , 5, − and −10 12 9 5 5 11 5 5 7 3 1 6. 2; −3; −4, ; − ; , −2, ; 11, ; 9, − , 2, 1; ; ; 6 3 6 2 6 8 4 2 e.

11, −

13 ,2 6

7. a.

x+y ; x, y; x, y 8

b. 2 x −

1 y ; x, y; x, y 6

c. (a – b) – ab; a, b, a, b; a, b, ab d. (4a + b); a, b; a, b 8. a , b, d, f, g are polynomials because they have a whole number as the degree. c, e, h are not polynomials because they have negative exponents. Word Problem 1. Figure A: 2x + y Figure B: 2 y + x

15B 1. a. 3 x 2 + 5 x + 6

b. 3 x 2 − 5 x + 3

2. a. 5 x − 4 y − 3 21 5 c. −2 p2 + p+ 5 4

b. 10 xy 2 + 3 xy − 3 12 8 d. pq + p + 7 8 7

3. a. 2 x 2 + x − 8

b. 6mn +

3 3 n −1 2

353

08-01-2024 14:41:27

c. 2 y 2 + 12

d. −1 − 15 pq − 7q

4. a. 33 x − 26

b. −

11 3 7 2 m + m 4 2

c. 6 x − 11 x + 6 x − 4

5. a. −5a 4 + 9 + 8a3

c. −12n2 − 18n − 10 2

6. 15 p + 14p − 84

d. − x + 6 x − 10

b. 16 x 2 + 10 x + 10

d. 2q 4 + 10q2 − 10q

7. 13h − 6ht + 16h

8. −3 y − 7 − 3 y

9. 2 p2 – 1 pq2 – 1 p2 – 2 pq2; 1 p2 + 1 pq2 5 3 5 4 4 3 Word Problems 1. (29m2 − 5m + 14) metres 2. (10s2 + 43s − 1) metres

Chapter Checkup 1. a. Terms: 6pq2 and 8p2q. Factors of 6pq2

are 6, p, q, q. p, q, q are variables. 6 is the numerical coefficient of pq2. Factors of 8p2q are 8, p, p, q. p, p, q are variables. 8 is the numerical coefficient of p2q. b. Terms: –mn, 2n2 and –3m2. Factors of –mn are m and n. m and n are variables. –1 is the numerical coefficient of mn. Factors of 2n2 are 2, n, n. n and n are variables. 2 is the numerical coefficient of n2. Factors of –3m2 are –3, m, m. m and m are variables. –3 is the numerical coefficient of m2. 2. a. Binomial b. Monomial c. Trinomial d. Quadrinomial 3. a 4. Circle a and d; cross out b and c 5. a. 2 x 2 − 5

b. −4x 2 + 6 x + 5

c. 13q2 −

b. 9 z 3 − 2 z 2 − z + 1

6. a. m2 + 7m + 5

6 3 q − 6q − 5 7

15 5 53 z − 6 z3 + 9z2 − z 2 2 5 b. 2 x 4 − 7 x 2 + 3 x c. −11 y 4 + y 2 + 13 4 7. a. 9 z6 −

8. a. 3s + 3s + 8s 2

b. 14t − 16t + 14t

9. 9 x 3 + 3 x 2 + 4x − 2

c. 4k − 4k − 12k − 6

11. −2 p − 14p4 + 5

10. 4k + 9k − 1

12. 6c − 5c 4 + 3 − 8c 2 Word Problems 1. (10 x 2 + 26 x − 30) cm

2. ₹(34x2 − 7x + 28)

3. (6 x 2 + 20 x ) metres

1. 6 × 3 5. 11 × 4

Let's Warm-up Do It Yourself 16A 1. a .

2. 11 × 7

1 1

−×

−×2

−× −× −× −×

−×

−× −× −× −×

−×

−×2

−× −× −× −×

y y y

× × × ×

−1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1

−× −×2

−×2

× × ×

−× −×2

−×2

× × ×

Expression: (3 – 2x) (4x − 5) Product: 22x – 15 − 8x2 2. a. product = −20xy b. product = −10x2 + 35x c. product = 6xy + 8y – 9x – 12 3. a . −180 x4y4z5 b. −6a2b2c + 5abc – 7ab c. 24xy + 15x – 8y – 5 d. 21yz2 + 42yz – 35y + 27z2 + 54z – 45 e. 8a4 + 32a2 + 9a + 77 f. 18z2 – 45z 5 4 3 2 g. 2p + 15p + 57p + 93p + 80p – 60 h. 24p7 + 72p6 + 85p5 + 12p4 + 82p3 + 152p2 + 65p – 55 4. a. −56 b. −3885 c. −18 d. −594 82 702 5. a. b. −24 c. − 6. a. −30x5y3z sq. units 5 7  54  56 99  1723 45  sq. units b.  x2 + x− c.  a3b2c4 + 77 5 5 25 66    99  a2b3c2  sq. units d. (6z3 + 19z2 – 37z + 10) sq. units 5  99  7 7 7. a.  p4q5r5 + p4q4r6  cu. cm b. (8rp2 – 8rq2 + 96rp 2 5  2    9977 77 2 99 + 96rq) cu. cm c.  − p2q3r2 − p q3r  cu. cm 5  2  52

8. a. 27s3 (stu + 1)3 cu. cm b. −27s6t9u15 cu. cm c. (−6ab + 2a2 + 3)3 cu. cm d. 64p9q12 (9r + 2q – 5)3 cu. cm 9. −19 Word Problem 1. (12x2 + 5x – 2) sq. m 16B 1. a. 36a2 + 84ab + 49b2 b. 81a2 − 90ab + 25b2 c. 22.09x2 + 31.96xy + 11.56y2 d.

9 4 7 2 2 49 4 a + a b + b 25 5 36

f. 9p2 −

Chapter 16 4. 12 × 12

× × × ×

1 1 1

Expression: (−3x + 2y) (2x + 4) Product: −6x2 – 12x + 4xy + 8y

3. 8 × 9

12 4 p+ 11 121

h. p16 − 2 + 2. a. 7225

g. i.

e. x8 + 2 +

1 4 1 2 2 1 4 p − p q + q 25 15 36

4 2 4 1 2 x − xy + y 9 15 25

b. 36,864

c. 11,664

e. 20.25x2 + 15.3xy + 2.89y2 g.

1 2 5 25 2 a + ab + b 4 6 36

i. 25p2 −

60 36 p+ 11 121

d. 10,032

f. 35.36

49 4 7 2 16 2 x − x y+ y 64 5 25 961 9 3. a. b. c. 33 100 4 h.

1199 4. a. –20a2 – 74ab + 12b2 900 585 4 16 2 2 625 4 a − a b + b b. c. 0 d. a4 – b4 784 7 576

d. −

81 b. x4 – 2401 256 c. 329 7. a. 13 b. 167

5. a. x 4 −

6. a. 47

b. 2207

c. 27,887

8. a. 5

354

UM24CB8_Batch 4.indb 354

08-01-2024 14:41:28

9. Students will verify using identities.

e. 13,125 5

f. 125

10. 2040 11. a. 57.02 b. 12.38 c. 148 Word Problem 1. ₹(4x2 – 11x) Chapter Checkup 1. a. −198x3y6z7 b. 8yz2 + 20yz − 24y + 14z2 + 5 4 3 35z – 42 c. −2p + 2p + 16p + 97p2 + 75p – 45 2 2 d. −5s t u + 3stu – 9st e. 21a4 + 29a3 + 98a2 + 16a + 96 f. 24p7 + 42p6 + 56p5 – 20p4 + 18p3 + 111p2 + 149p – 12 2. a. −154 b. 7236 c. −56 d. −940 2027 372 3. a. b. 214 c. 4. a. 56x5y3z sq. units 30 7  36 9 3  3  110  263 a2b3c2  sq. units  sq. units c.  b.  x2 + x− a3b2c4 + 26  26  7  14  91 2   4 4 4 99 d. (10z3 + 39z2 − 82z + 33) sq. units 5. a.  p4q5r5 + p q4r6  cu. cm 5 11 11   b. (14p2r – 7pqr + 98pr – 21q2r + 98qr) cu. cm c. (−45p2q3r2 – 225p2q3r) cu. cm 6. a. 125s3 (stu + 2)3 cu. cm b. 64s6t9u15 cu. cm c. (−7ab + 11a2 + 2)3 cu. cm d. 27p9q12 (2r + 5q – 6)3 cu. cm 7. a. 64a2 + 176ab + 121b2 b.

4 4 14 2 2 49 4 a + a b + b 15 25 36

d. 4p2 −

28 49 p+ 11 121

e. x8 −

f. p16 −

j. 9p2 −

30 25 p+ 11 121

9. a.

10. a. −15a2 − 83ab − 27b2

484 81

Let's Warm-up

3. Square pyramid

; Side View:

;

Top View:

5. a. Front View:

; Side View:

Top View:

b. Front View:

; Side View:

Top View:

c. Front View:

Side View:

; Top View:

; ;

;

Side View:

Black Board

c. 2809

Chair Table Window

64 81

ALMIRAHS

113 4 97 4 431 2 2 b. a + b − a b 784 576 84

1. Triangular pyramid 4. Triangular prism

d. Front View:

Door

Window

Bin

 256  11. a.  x 4 − 12. a. 23 b. 527  b. (x4 − 4096)  625   13. a. 17 b. 287 14. a. 3 b. 5 c. 255 d. −1792 e. 65,535 Word Problems 1. ₹100 (10x2 + x – 3) 2. 2x5 + x4 + x3 + 7x2 + 5x

Chapter 17

Top View:

b. Reception c. Library d. Library . a. 199 m 2 b. 36 m c. 200 m d. 265 m 3. a. 68.75 km b. 25.85 km c. 40.15 km 4. 12 mm 5. a. Answers may vary. Sample answer:

9 4 15 2 25 2 i. x − x y+ y 64 121 44

1024 81

;

Word Problem 1. Front View:

1 2 8 64 2 ab + b g. 38,809 h. a + 4 11 121

; Side View:

17B 1. a. Garden

p16

8 16 2 1 2 x − xy + y h. 121a2 − 154ab + 49b2 15 25 9 i. 12.25x2 + 51.8xy + 54.76y2 8. a. 7744 b. 44,944 d. 5580 e. 4.84x2 + 14.52xy + 10.89y2 f. 37.12

c. Front View:

Top View:

9 4 1 2 2 1 4 p − p q + q 5 25 36

Top View:

Desks

d. −2250

Desks

c. 75 5

Desks

House House

AMBULANCE

Market Market

2. Square prism

Do It Yourself 17A 1. a. Cone, Cylinder, Hemisphere. 2. a. Front

Answers

b. Side

b. Cylinder, Cone. d. Sphere, Cone.

3. a. Front View:

; Side View:

Top View:

b. Front View:

; Side View:

Top View:

c. Front View:

; Side View:

Top View:

d. Front View:

; Side View:

Top View:

4. a. Front View:

; Side View:

Top View:

b. Front View:

; Side View:

; ; ;

6. 3.3 cm Word Problem 1. Answers may vary. Sample answer: Super Market Super Market

Sudha's House Sudha's House

; ;

Bus BusStop Stop

Park Park

c. Cylinder, Hemispheres.

Grocery Grocery Shop Shop

;

355

BED ROOM BED ROOM

UM24CB8_Batch 4.indb 355

08-01-2024 14:41:29

Super Market

17C 1. a. Regular

b. Cuboid c. Faces d. Tetrahedron 2. a . No b. Yes c. Yes 3. a. 5; infinite b. 4; infinite 4. A square Grocery Bus Stop 5. F aces: ABCDEF, GHIJKL, ABKL, BCJK, Shop CDIJ, DEHI, EFGH, FALG Edges: AB, BC, CD, DE, EF, FA, GH, HI, IJ, JK, KL, LG, AL, BK, CJ, DI, EH, FG Vertices: A, B, C, D, E, F, G, H, I, J, K, L 6. A nswer may vary. Sample answer: H Park

Sudha's House

d. Front view:

⇒ 5; Side view:

⇒ 4;

Top view: ⇒7 4. Answers may vary. Sample answer: WASHROOM

BED ROOM

D C Faces: ABCDEFG, ABH, BCH, CDH, DEH, EFH, FGH, GAH Edges: AB, BC, CD, DE, EF, FG, GA, AH, BH, CH, DH, EH, FH, GH Vertices: A, B, C, D, E, F, G, H 7. No, because it has curved faces. Word Problem 1. 11 faces, 20 edges and 11 vertices. 17D 1. Prisms have two parallel bases with rectangular faces connecting corresponding vertices, while pyramids have a single base with triangular faces connecting to a common apex. 2. Students will verify using Euler’s formula. 3. a . 6 vertices b. 18 edges c. 9 vertices d. 30 edges 4. 2 + n, 2n, 3n 5. Students will verify using Euler’s formula. Word Problem 1. Such a polyhedron cannot be made.

Chapter Checkup

1. a. Top view:

Side view:

Front view:

b. Top view:

Side view:

Front view:

c. Top view:

Side view:

Front view:

2. a. Front view:

Side view:

Top view:

b. Front view:

Side view:

Top view:

c. Front view:

Side view:

, 6; Side view:

Top view:

,6 b. Front view:

Side view:

c. Front view:

, 4; Top view:

DINING ROOM

KITCHEN

BED ROOM

5. a. 11.5 cm b. Sam c. 87.75 km d. 36 km e. 51.75 km 6. 50 mm 7. 4 faces 8. a. Tetrahedron b. Pyramid 9. a. 9; Triangular prism b. 8; Rectangular pyramid c. 6; Octahedron d. 10; Pentagonal prism e. 6; Cube f. 8; Hexagonal prism 10. a. No b. Yes Word Problems 1. Cuboid; Front view: Side view: Top view:

Chapter 18

1. 15 cm 5. 40 cm

Let's Warm-up 4. 24 sq. m

Do It Yourself 18A 1. 10 cm

, 3; , 5;

⇒ 20; Side view:

⇒ 30; Top view:

WASHROOM

2. 6.25 km and 5.5 km 3. The solid is an octahedron. Students will prove using Euler's formula.

Top view:

3. a. Front view:

LIVING ROOM

BALCONY

2. 16 sq. m

3. 12 cm

2. 225 tiles 3. 35.35 sq. m 4. a. 423.85 sq. cm b. 219.43 sq. cm c. 365.71 sq. cm 5. Both John and David 6. 32 sq. cm 7. a. 24 sq. cm b. 243 triangles c. 5832 cm 8. 2 sq. m 9. 32 brown tiles, 22 pink tiles and 11 white tiles. 10. 115.2 sq. cm Word Problems 1. 4 m 2. ₹71.39 9 3. of the original area 64 18B 1. a. 477 sq. cm b. 485 sq. cm c. 87 sq. units 2. 432 sq. cm 3. 24 cm and 36 cm 4. 50 cm; 14 cm 5. 34 cm 6. 7.38 m 7. 5 cm 8. 9 cm 9. 1,70,000 tiles; ₹22,440 10. 48 m; 64 m 11. 240 sq. cm Word Problem 1. a. 6.25 m b. 15 sq. m

18C 1. a. 501 sq. cm

⇒25

b. 9064.28 sq. cm c. 56.57 sq. cm 2. a. 269 sq. cm b. 1913 sq. m c. 519 sq. cm 3. a. 9 sq. cm b. 10.2 sq. cm 4. 585 sq. cm; Yes 5. 450 sq. cm 6. 26.2 sq. cm 7. 58 sq. m 8. 830 tiles Word Problem 1. 314 sq. cm Chapter Checkup 1. 44 sq. cm 2. Always true. 3. a. 313.71 sq. cm b. 4882.5 sq. cm 4. 29.44 sq. cm

356

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5. 240 sq. cm 6. 20 cm and 100 cm 8. a. 1866 sq. m b. 34,997.5 sq. cm 10. ₹22,825.65 Word Problems 1. 15 sq. m 2. No

Chapter 19 Let's Warm-up 2. Area = 154 cm2 4. Area = 90 cm2

7. 12.5 cm and 37.5 cm. 9. 19,500 kg

1. Perimeter = 64 cm 3. Area = 100 cm2 5. Perimeter = 45 cm

Do It Yourself 19A 1. a. 600 cm2; 792 cm2

b. 2412 cm2; 3252 cm2 c. 1239.5 m2; 2516 m2 2. a. 1764 m2; 2646 m2 b. 1156 mm2; 1734 mm2 c. 3136 m2; 4704 m2; 2 2 d. 4900 dm ; 7350 dm ; e. 25,921 cm2; 38,881.5 cm2 2 2 3. a. 1056 cm ; 1961.14 cm b. 484 m2; 561 m2 c. 4.4 m2; 23.65 m2 4. a. 13,828.57 cm2; 14,658.29 cm2 b. 1,35,771.43 cm2; 1,58,400 cm2 c. 52,800 cm2; 2 66,000 cm 5. a. 10 m b. 7.5 cm c. 12 dm d. 8.5 m 6. 384 cm2 7. 42,240 cm2 8. 19.09 cm 9. 7:9 10. Cuboidal Box; 40 cm2 Word Problems 1. ₹6156, ₹9576 2. 150 cm 3. 14,652 cm2 4. 128.25 m2 5. ₹9820.8 6. 49.28 days 7. ₹1531.25 19B 1. a. 1200 cm3; 19.72 cm b. 22,500 cm3, 96.18 cm c. 2513.875 m3, 46.74 m 2. a. 4096 cm3; 16 3 cm cm

b. 13,824 mm3; 24 3 cm cm

c. 27,000 m3; 30 3 m m d. 1,28,787.625 1,28,787.625dm dm33;;50.5 3 dm dm

d. 5 m 20. a. 12 cm b. 4.2 m c. 22.40 mm d.10 cm Word Problems 1. 900 cm2 2. 637.5 cm 3. 5 approx. 4. 30,000 bricks 5. 80 minutes

Chapter 20 Let's Warm-up a. No

1.728 m 5. a. 1.728 m33; 1.2 3 m m b. 2160 mm3; 23.83 m 3 c. 2,25,00,000 cm , 667.08 cm d. 8000 cm3; 45.82 cm 3 e. 17.576 mm ; 2.6 3 mm f. 216 m3; 17.09 m 6. a. 600 b. 27 c. 400 7. a. 6.93 cm3 b. 3942.576 mm3 c. 10,45,000 cm3 3 3 d. 16,896 mm e. 8939.7 m 8. a. 3375 litre b. 1500 litre c. 6,03,428.6 litre d. 3326.4 litre 9. 300% 10. 0 11. 16 m 12. 53 mm 13. 90 m2 14. 486 cm2 15. 7 cm 16. X:Y 17. 154 cm3 18. 13,305.6 g 19. a. 39 m b. 30 cm c. 2.55 m

Answers

UM24CB8_Batch 4.indb 357

b. −

1 343

e. 0.0000000032

3. 24 × 3

2. 9; 7 16,384 78,125

81 16

100 3721

2. a. 4 × 101 + 5 × 100 + 1 × 10−1 + 2 × 10−2 + 3 × 10−3 b. 5 × 102 + 4 × 101 + 1 × 100 + 0 × 10−1 + 5 × 10−2 + 6 × 10−3 c. 1 × 102 + 0 × 101 + 7 × 100 + 0 × 10−1 + 2 × 10−2 + 3 × 10−3 + 5 × 10−4 + 6 × 10−5 d. 2 × 101 + 8 × 100 + 5 × 10−1 + 4 × 10−2 + 6 × 10−3 + 8 × 10−4 + 7 × 10−5 e. 5 × 103 + 3 × 100 + 6 × 10−1 + 7 × 10−2 + 1 × 10−3 3

3 3. a. 510 b.   2 f. (1.1)8  11  4. a.   7    7 e.   3  

e. 22,22,447.63 cm3; 130.5 3 m cm 3. a. 385 cm3 3 3 b. 665.28 m c. 70.4 dm 4. a. 1771 m3 b. 3630 cm3 c. 220 mm3

 π 3 1 5. 1:16 6. 512cm3 7.  1 −  a 8. 237 % 4 2  9. 1728 cm3 10. a. 36 cm b. 42 mm c. 0.35 m d. 11 cm 11. 8:7 Word Problems 1. 60 2. 4000 cm 3. 400 4. 30.28 m 5. 2,80,000 6. 400 cm Chapter Checkup 1. a. 48,600 cm2; 84,600 cm2 b. 275 m2; 287 m2 c. 28,600 cm2; 36,600 cm2 d. 3 8,70,000 cm2; 44,70,000 cm2 2. a. 900 cm2; 1350 cm2 b. 56.25 dm2; 84.375 dm2 c. 121 m2; 181.5 m2 d. 169 m2; 253.5 m2 3. a. 226.29 cm2; 258.11 cm2 b. 1320 cm2; 3245 cm2 c. 2112 m2; 4884 m2 d. 1,82,160 cm2; 3,45,092 cm2 4. a. 13,200 mm2; 26,400 mm2 b. 97,428.57 mm2; 1,12,237.71 mm2 c. 43.214 m2 ; 129.64 m2 d. 33,942.86 mm2; 4,26,039.43 mm2

b. 3

Do It Yourself 20A 1. a. 4096

1. 121; 1331 4. Meeti

−5

9 b.    4  

−3

 125  c.   319   

 −4  c.    7 

 10  f.   9    −1

−8

−6

−3

Word Problems 1. 0.512 g

20B 1. a. 27

2. a. 1 f.

1,10,592 343

625 × (–5)–2

−4

  5. a.  3  8  

 13  d.   8   

−1

32 27

−4

 10  b.   9   

 128  e.     −1999  7. −

−1

32 135

10. 0.000012; No b. 5n + 1

2. a. 5 times

b. 256 b. −216

 1  c.   −5   

d. 2−3 × 33−1

 6  f. (−1) × 6−2 ×    100  608 −15625 8. 9. 6561 279936

3 e.   5  

d. (−20)2

243 78125

c. c.

729 7

3. a.

8 9

d. −

d. b.

5 4

6561 256

3125 2187

c. 55 46656 e. 15625

e. 0.253 c.

243t 5 2

d. 2

(–2)5 –2 –3 c. 20 × 30 × 70 × 4 3 5 5. a. 7.687 × 109 b. 2.3 × 1015 c. 1.0446678 × 101 1 d. 6.59084 × 10 e. 9.80000324 × 101 f. 4.5398785 × 102 g. 4.32 × 109 h. 1.098 × 107 6. a. 103000000 b. 0.00003467 c. 0.0060945 d. 42100 e. 870000000000 f. 17680 g. 54100000 h. 0.40857 7. a. 8.05 × 106 is greater. b. 2.11 × 102 is greater. c. 7.3 × 10−5 is greater. 8. a. 6.8559126 × 107 b. 3.9474 × 10−10 c. −8.49 × 10−10 2 −2 d. 1.1 × 10 9. x = 5 10. y = 2 4. a.

357

08-01-2024 14:41:32

Word Problems 1. 0.0028256 inches 2. German, 2.43 × 104 3. 120000 million sq. miles 4. 4.392 × 10−3 mg 7 5. 1.5247500 × 10 sq. km Chapter Checkup 1. a. 6 × 101 + 5 × 100 + 8 × 10−1 + 7 × 10−2 + 9 × 10−3 b. 9 × 102 + 7 × 101 + 8 × 100 + 9 × 10−1 + 6 × 10−2 + 7 × 10−3 + 5 × 10−4 c. 3 × 102 + 4 × 101 + 2 × 100 + 4 × 10−1 + 5 × 10−2 + 6 × 10−3 + 7 × 10−4 d. 2 × 101 + 1 × 100 + 4 × 10−1 + 8 × 10−2 + 7 × 10−3 + 9 × 10−4 e. 4 × 103 + 8 × 102 + 7 × 101 + 6 × 100 + 9 × 10−1 + 0 × 10−2 + 2 × 10−3 + 3 × 10−4  5  2. a.   −2    3. a.

−6

8 b.   3

−1105 1764

−8

9 c.   7  

b. 960

−4

25 729

 −11  d.   5    d.

−5

2187 8

e. (10)−6 e.

125 4

f. 4,60,800 4. 310 5. 495 6. a. Ascending order: 128−2, 83, 210, 642, 323; Descending order: 323, 642, 210, 83, 128−2 b. Ascending order: 18−2, 93, 273, 310, 7292 Descending order: 7292, 310, 273, 93, 18−2 7. Nobody is correct; reason: Tanya’s answer is negative x2 and Sayani’s answer has power as negative 2 8. Kamya; reason: the fraction is the reciprocal of the number with the negative exponent 3

 51  b.  −  20   

9. a. 1 f.

−9261 512

7 64

10. a. 29 × 3−7

d. b.

596 1331

25000 19683

e. 2030.301 c. 13608 p5

d. 1

13 c. x = 3 8 12. a. Standard form, 0.00001275 b. U sual form, 5 × 10−3, 1.0 × 10−2 c. S tandard form, 384000000 d. S tandard form, 1386000000 e. U sual form, 9.11 × 10−29 kg 13. b and d; they have not been expressed as k × 10n 11. a. x = 3

(

)

( 1296 ) = 6,04,66,176

Word Problems 1. 0.25 kg 3. 1.46 (approx.) times

Chapter 21 Let's Warm-up 5. True

Do It Yourself 21A 1. a . No

1 of the work 6

11. 8

2 days 5

10.

12. 13

1 of the work 5

7. 8 tablets

b. No

8. 7 boxes

9. 42 men 10. 48 m 11. 12 tankers 12. 5 cows 13. 7.14 kg 14. 8 hours 15. 525 people 16. 425 kg 7 4 17. 14 hours 18. 1.95 × 10 litres 12 19. ₹450 20. 72 minutes 21. 2.75 × 105 grains 10 9 22. 30 hours 23. of work 24. 8:24 a.m. 13 16 25. 3 hours 26. 1.5 cm 3 Word Problems 1. 3 hours 2. 6 batches 3. 10 days 7 4. 60 hours

1. 17y + 9x

2. 17xy + 6x

Do It Yourself

22A 1. a. 3 x ( x + 3) 2

4. True

c. Yes d. No 13 7 2. a. NO b. YES; c. YES; 20 12 3. 88, 228, 304, 209 4. ₹3380 5. 300 chocolates 6. 6120 words 7. ₹ 988 8. 2.4 km 9. 2700 bottles 10. 4 5000 cc 11. 105 workers 12. ₹561 13. 81 boxes 14. ₹3000 15. 1312.5 g 16. Plain Flour: Yes, Sugar: No 17. 3 m 15 cm 18. 2.8 days Word Problems 1. 32 bags 2. 15 miles 21B 1. a. No b. Yes c. No d. No 2. 2 3. 35 days 4. 2 chocolates 5. 6 shoes 6. 3 days 7. 2 members 8. 20 km/h

3. –6z + 5y2

b. 2 y 2 (2 y − 1) 2

d. 6 xy ( x − 3 y )

15. 1

3. False

13. 10 hours

7 30 hour 15. 13 hours 16. 28 hours 11 31 23 17. 13 hours 18. 12 days 19. 585 sailors 121 20. 2:43 p.m. Word Problems 1. 6 days 2. 12 hours 3. 18 days Chapter Checkup 1. a. Direct Proportion; 65 b. Inverse Proportion; 40 c. Inverse Proportion; 100 d. Direct Proportion; 32 2. ₹896 3. 2112 g 4. 60 km 5. ₹13,500

c. 5ab(a + 2b)

f. 3 pq (3 p − 2q)

g. 4m2n2 (3m2n − 5)

h. 5 x 2 y 2 (3 x − 5 y )

i. 10a3 (2a2 − 3b2 )

2. a. 7 x ( x − 3 y ) 3

c. 4p q(4p − 3q)

4. 4x – 17y

e. 4c (2c − 3)

b. 5ab2 (2b − 3a)

e. 5 x 4 y 2 (5 x − y ) 2. True

1 minutes 3

14.

Let's Warm-up

2. 8,80,000 kg 4. No 5. Diameter of Jupiter

1. False

2 of the course 3

Chapter 22

b. x =

5 14. yes; 1296 2 =

g. 11 x 2 ( x + 2) 2

i. 17ab (ab − 2)

d. 6m2n3 (3mn − 1)

h. 7 y 3 (2 y − 1)

k. 23 p2q( p2 − 2q) 3 3

m. 29 x y ( x − 2 y ) o. 37 x ( x − 2 y )

f. 10a2b3 (3ab − 1)

j. 19c 3 (c 2 − 2)

l. 9mn2 (3m2 − 2n)

n. 11a2b2 (3a − 2b)

3. a. ( x + 4){( x − 2) − 3( x + 4)2 } b. 2( x − 1)(3 x + 2)

c. 5(3 y + 2)( y − 1)

d. (2a + 5)(a − 4)

e. 2( x + 3)

f. (3y – 6)(2y – 6)

g. (4a + 7)((4a + 7)(a + 2) − 1)

h. (4x − 3)( x + 2 − 5(4x − 3)2 ) 2

4. a. ( x + 2 y )(2 x − 3)

i. (3 y + 2)( y − 3)

b. (3a − 2d )(b + 2c )

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d. (a − 2d )(2b + 5c )

c. ( x − 2 y )( x − 7)

c. 4(a2 + 3a + 70)

f. ( x − 5)(3 x − 2 y )

e. ( x − 3)( x − 2)

g. (3 x − 5)(2 y − x )

h. (4x + 2)( x − 3 y )

5. a. 20ab (2b − 3a)

i. (3 x − 2)(2 y − 3 x )

b. 43 pq ( p − 2q)

d. 25 x 4 y 2 (2 x − y )

c. 47m2n2 (m2n − 2) e. 53a3 (a2 − 2b2 )

f. 57 x ( x − 2 y )

g. (2a − 1)(−5a + 6)

h. (3 x + 2)( x + 2)

i. (a + 2b)(2a − 1)

6. a. (3a + 2)(a2 − b)

b. a (2a2 − 19a − 7)

c. 5 y 2 z 2 (3 z − 4 yz 2 + 7)

22B 1. a. (a + 2)(a − 2)

b. (5 + x )(5 − x )

c. (3b + 4)(3b − 4)

d. (7 p + 4)(7 p − 4)

e. (a + b)(a − b)

f. (ax + y)(ax – y)

g. (10b + 9)(10b − 9) i. (1 + d )(1 − d )

b. 6x

4. a.

b. 16

1 4 g. 49

c. 1

h. 64

5. a. (4ax − 1)2

c. (b + 2)2

d. 12r 25 d. 4

h. ( y − 8)

b. –8x − x

d. –16x − x

c. (4x − 14); ( x − 1)

d. (2 x − 1); (2 x − 3)

c. (2 x − 1)( x − 2)

d. (3 x − 2)( x − 2)

e. (3 x − 1)(3 x − 2)

f. (9 x − 6)( x − 1)

g. (3 x − 5)( x − 2)

h. (3 x − 1)(5 x − 2)

e. 3p2q2 2. a. x 2 + 2 x − 3 4

c. 4y 3 z 2

f. 12 j 4k 4r 7 b. y 3 − 3 y 2 + 2 y

d. m2 − 3m + 2 5 2 2 3 e. x 4 + 2 x 3 − 3 x 2 f. a x − 5ay + z 3 3. a. 36abc b. 4b(a + 5) c. a − 3a + 2a

Answers

UM24CB8_Batch 4.indb 359

d. 5(x − 3x + 2)

f. 7x(y − 2)

2. a. −2(x + y)(x + 5)

c. (10 − n)(11 − n)

e. 6mn(m + 2n)

b. (x − 2z)(x + y −1)

d. 3(3a − 1)(a −1)

f. (5x + y) (1 – (5x + y)2)

2. c

f. 16(a − b)

b. (2x + 7y)(2x − 7y)

d. (5x + 4y)2

5. a. (p + 3q)(p − 3q)

c. (a + 4b)(a − 4b)

d. (3m + 5n)(3m − 5n) e. (x + 6y)(x − 6y) f. (5u − 2v)(5u + 2v) 6. a. (6x − 1)(x + 2) b. (x − 2)(x − 3) c. 3(x −1)(x + 5) d. (x −1)(3x − 7) e. (x + 6)(x + 8) f. (x + 1)(2x − 5)

c. 2a − ab + 3b f. 3 x 3 − 2 x 2 + x

7. a. 2 x 2 + 3 x − 4 3

d. 3 x − x + 2 x

b. 2ab2c − 4bcd

e. a2b2 − 2abc + 3c 2

8. a. Q = 2x2 + 8x + 10 R = 20 c. Q = 2x2 + 7x + 2

5. a. Yes b. Yes Word Problem 1. 3( x + 1) b. 3ab2

c. 4y (y + 2)

b. (2 x − 3)( x − 2)

b. 3a(b − 3)

e. (2x2 − 3y)(2x −1) f. (x − 3)(6x + 7y) 4. a. (a + 3b)2 b. (x − 5y)2 c. (2x + y)2

b. (a − b − 1)(a + b + 1)

b. (5 x − 2); ( x − 7)

4. a. (3 x − 1)(2 x − 3)

e. (x − 7y)

3. a. (3 x − 2); ( x − 7)

8. 5

Chapter Checkup 1. a. 2 x ( x + 3)

c. –8x − 3x

R = 108y2 + 21y – 126

3. a. b(2a − 1)(2a − 3b) b. (5x + 3y)(x − 2y) c. (x + 2)(3x − 2y) d. (3x − 4y)(3x − 2)

x y d.  +  y x  

Word Problem 1. 13 cm, 5 cm

f. Q = 3y2 + 8y + 20

7. −a + 2

e. (x + 2y)(5 − 7x − 14y)

c. (a − 2b + c )(a − 2b − c )

22D 1. a. 5x 2

f. 144

d. (b2 + 9)(b − 9)(b + 9)

6. a. (m + 4n )(m − 2n)(m + 2n)

22C 1. a. −4x − x

e. 16

f. (12 y 2 − 7 x 2 )2

e. (11 x − 8)2 g. (7 y − 4)

d. (a + 9)2

b. 3 xy ( x + 9 y )( x − 9 y )

c. (a2 + b2 )(a + b)(a − b)

R = 3x2 + 2x

Word Problem 1. ( x 2 + 3 x + 4) chocolates

l. (12 + a3 )(12 − a3 )

c. c2

3. a. 49

6. x 2 − 5 x + 3

j. (b2 + 10)(b2 − 10)

b. ( x + 5)2

2. a. (a + 4)2

d. Q = x2 – x + 4

R = 74x + 41

h. ( xy + 2)( xy − 2)

k. (5 + ax )(5 − ax )

4. a. Q = 4x – 6 b. Q = 3y2 – 9y + 18 R = 19x – 33 R = –48y + 72 c. Q = x3 + 5x2 d. Q = 6a3 – 15a2 + 39a – 84 R = 16x R = 168 e. Q = 9b4 + 12b3 – 21b2 – 28b + 42 R = 56b – 84 f. Q = 2c4 – 12c3 + 30c2 – 60c + 108 R = –216 5. a. Q = 3x + 7 b. Q = 2y2 + 6y + 14 R = 18x – 29 R = 86y + 72 c. Q = 3a2 + a + 13 2 4 R = – 9a – 1 4 4 e. Q = 7x + 23

Word Problem 1. ₹(4x + 1)

3 c (a + b) 2

2x 4 y z

b. Q = 3x3 – 3x2 – 6x + 12 R = – 12 d. Q = x3 – 3x2 + 3x – 6 5 R=2 R = 12 5 2 e. Q = 3x3 + 9x + 3x + 9 4 8 2 R = 75 8 f. Q = x2 – 3x + 86 7 49 R = 344 49 9. a. Q = 3x + 13 b. Q = 2x2 + 2x + 1 R = 31x – 25 R = 5x + 5 c. Q = 3x + 1 d. Q = 5x – 16 2 R = 19x – 3 R = 40x – 18 2

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e. Q = x2 – 4x + 15

f. Q = 2x

R = – 44x – 64 Word Problem 1.

2. a. Abscissa = 9, Ordinate = 4 b. Abscissa = 4, Ordinate = –5 c. Abscissa = 0, Ordinate = 4 d. Abscissa = –2, Ordinate = –8 e. Abscissa = 0, Ordinate = –7 3. a. y-axis b. x-axis c. x-axis d. y-axis e. Origin 4. a. Second b. Third c. Fourth d. First e. Second 5. Scale: x-axis = 1 division = 2 units

R= 9

Length

Breadth

(8x + 5y)(8x − 5y)

(8x + 5y)

x(8x − 5y)

(8x − 5y)

x(8x + 5y)

(8x − 5y)

x(8x − 5y)

(8x + 5y)

(8x + 5y)(8x − 5y)

Y-axis = 1 division = 20 units

200 180 160 140

Chapter 23

120

Let's Warm-up

100

Do It Yourself 23A 1. a. True

1. Badminton 2. Chess 3. Badminton 4. Chess

b. True c. True d. False e. False f. True 2. a. Thursday b. Wednesday c. 125% d. 20:17 3. a. 2004 and 2001 b. 60% c. 40% 4.

40 20 0

2500

Y-axis = 1 division = 1 unit y

Sister

Brother

2000

1500

4 3

1000

2 1

500

x’

2018

2019

2020

2021

–5

–4

–3

–2

–1 O –1

–2

Year

The points do not lie in a straight line 6. Scale: x-axis = 1 division = 1 unit

3000

Savings (in ₹)

–3 –4

–5

140

y’

Temperature (in Fahrenheit)

120

City A City B

100

(0, 0) 7. Scale: x-axis = 1 division = 1 unit

Y-axis = 1 division = 5 units

20 40

15 10

5 0

23B 1. a. False

June

July

b. False

August

Months

c. False

September

d. True

October

e. True

Not a linear graph.

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Word Problems 1. Scale: x-axis = 1 division = 1 unit Y-axis = 1 division = 50 units

c. The points lie on a line. Scale: x-axis: 1 division = 1 unit y-axis: 1 division = 1 unit y

500

450

400 350 Cost

300

250

200

50 1

Number of notebooks

a. ₹350 b. 11 notebooks Chapter Checkup 1. a. Abscissa = 5, Ordinate = 4; first quadrant. b. Abscissa = −8, Ordinate = –2; third quadrant. c. Abscissa = −9, Ordinate = −8; third quadrant. d. Abscissa = –5, Ordinate = –10; third quadrant. e. Abscissa = 14, Ordinate = –21; fourth quadrant. 2. a. The points lie on a line. Scale: x-axis: 1 division = 1 unit y-axis: 1 division = 1 unit y 6

6 B

4 3 2 1 0

b. The points do not lie on a line. Scale: x-axis: 1 division = 1 unit y-axis: 1 division = 1 unit

UM24CB8_Batch 4.indb 361

Q 5

B A

Answers

P(0,5) and Q(5,0) 5. Scale: x-axis: 1 division = 1 unit y-axis: 1 division = 1 unit

1 3

B C

No. 4. Scale: x-axis: 1 division = 1 unit y-axis: 1 division = 1 unit

y-axis: 1 division = 1 unit

100

3. Scale: x-axis: 1 division = 1 unit

150

0 1

C(4, 0) and B(0, 4)

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6. Y-axis 7. X-axis 8. a. Triangle 9. Scale: x-axis: 1 division = 1 unit y-axis: 1 division = 1 unit 10

b. Straight line

4. Scale: x-axis: 1 division = ₹100

Y-axis: 1 division = ₹500

3000

2500 2000

Deposit

7 6 5

1500 1000 500

100 200 300 400 500 600

Simple Interest

Yes.

1 0

10. a. –2; 1.5; (–2, 1.5) b. 3; 1; (3, 1) c. 1; –3; (1, –3) d. –3; –2; (–3, –2) 11. a. (–6, –2); (–2, –2); (2, 2); (–2, 2); (7, 4); (7, 0); (3, 0); (2, –1); (4, –1); (4, –3); (2, –3) b. Parallelogram and rectangle. c. 4 2 units. d. 8 square units. Word Problems 1. Scale: x-axis: 1 division = 1 year Y axis: 1 division = 2 units Number of appliances (in thousands)

Let's Warm-up

1. (2 × 100) + (3 × 10) + (5 × 1) 2. (3 × 100) + (6 × 10) + (2 × 1) 3. (8 × 100) + (9 × 10) + (4 × 1) 4. (1 × 1000) + (2 × 100) + (6 × 10) + (5 × 1) 5. (3 × 1000) + (6 × 100) + (5 × 10) + (7 × 1)

Do It Yourself 24A 1. a. 7 × 10 + 5

b. 9 × 10 + 8 c. 2 × 100 + 3 × 10 + 5 d.9 × 100 + 5 × 10 + 4 2. a. 34 b. 10m + n c. 602 d. 950 3. 7 4. 45 5. a. A = 2 b. A = 2 c. A = 2; B = 5 d. A = 6; B = 2 e. A = 3; B = 5 f. A = 3; B = 5; C = 8 6.

20 18 16 14 12 10 8

6 4 2 0

2000 2001 2002 2003 2004

Years

2. a. i. 5 crores. ii. 10 crores. iii. 15 crores. iv. 6 crores. b. 5 crores. c. In 2017 and 2018 & 2021 and 2022 3. Scale: x-axis: 1 division = 5 minutes Y-axis: 1 division = 5 m y 50 45

7. c. 8. a. 4 10. 75, 15, 3 11.

b. 5

c. 1

d. 4

e. 6

100

Raman Ansh

30 25

10 5 0

9. 89

Word Problem 1. No,

Distance (in m)

Chapter 24

Time (in minutes)

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24B 1. a, c, d and e are divisible by 2

2. Options c and e, are not divisible by 3 while others are divisible by 3 3. a. Only divisible by 5 b. not divisible by 5 and 10 c. divisible by 5 and 10 d. Only divisible by 5 e. divisible by 5 and 10 4. b, c and e are divisible by 9 5. c. 6. a. 2, 5, 10 b. 3, 5, 9 c. 2, 3 d. 2, 3, 9 7. 1, 4, 7 8. 0; 9 9. x = 4 and y = 0 Word Problem 1. 56,826 and 56,835 Chapter Checkup 1. a. 5 × 10 + 1 b. 36 c. p × 10 + q d. 230 e. 3 × 100 + 5 × 10 + 4 f. 709 2. 11 3. 37 4. a. A = 5; B = 2 b. A = 2; B = 5 c. A = 6; B = 8 5.

Answers

UM24CB8_Batch 4.indb 363

6. Only option c is a triangular number, while others are not a triangular number 7. a. 1 b. 8 c. 5 d. 9 e. 3 8. 3584 and 28,672 9. By 2

By 3

By 5

By 9

By 10

Yes

10. 2, 3, 5, 6, 8, 9 11. 5 Word Problem 1. No (9 + 1 + 7 = 17) which is not divisible by 9.

363

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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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