Imagine_Maths_CB_Grade7 by Uolo

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.

Imagine Mathematics

About the Book

Key Features

MATHEMATICS

• Let’s Recall: Helps to revisit students’ prior knowledge to facilitate learning the new chapter • Real Life Connect: Introduces a new concept by relating it to day-to-day life • Examples: Provides the complete solution in a step-by-step manner • Do It Together: Guides learners to solve a problem by giving clues and hints • Think and Tell: Probing questions to stimulate Higher Order Thinking Skills (HOTS) • Error Alert: A simple tip off to help avoid misconceptions and common mistakes • Remember: Key points for easy recollection • Did You Know? Interesting facts related to the application of concept • Math Lab: Fun cross-curricular activities • QR Codes: Digital integration through the app to promote self-learning and practice

About Uolo Uolo partners with K-12 schools to provide technology-based learning programs. We believe pedagogy and technology must come together to deliver scalable learning experiences that generate measurable outcomes. Uolo is trusted by over 10,000 schools across India, South East Asia, and the Middle East.

Singapore

IM_G07_MA_MB_Cover.indd All Pages

Gurugram

Bengaluru

hello@uolo.com �524

NEP 2020 based

NCF compliant

CBSE aligned

01/01/24 3:10 PM

MATHEMATICS Master Mathematical Thinking

Grade 7

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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

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87.89 × 35.5 = ₹3120.095

4 4 3 + 2 6 3 5 3 1 2 Ones

Hence, Naman's father paid ₹3120.095 for the fuel.

� 3 9 6 0

3 9 4 7 0

• decimal poi

WHOLE

whole numbers.

Step 2: Count the total number of decimal places andHundreds place the decimal Tens point in the product accordingly.

PART

5 5 4 5 5 Tenths 0 0 0 9 9 5

Hundredths

Thousandths

While multiplying a decimal number with 10, 100, 1000 etc., we move the decimal point as many places to the right as there are zeroes in the multiplier.

Remember!

For example:

On multiplication 10 number ofOn multiplication with 100 The product should have the same number of decimals place as the sumwith of the decimal

K ey El emen t s o f a C h apt e r— a Q u i c k G lanc e 23.389 � 10 = 2 3.3 8 9 = 2 3 3.8 9

places in both the multiplicand and the multiplier.

23.389 � 100 = 2 3.3 8 9 = 2 3 3 8.9

On multiplication with 1000

Naman saw that one of the auto rickshaw drivers paid ₹1018.75 for filling CNG. He started 23.389 � 1000 = 2 3.3 8 9 = 2 3 3 8 9.0 wondering how much CNG this driver must have put into his autorickshaw tank. Let us find out!

12.5 815 10187.5 Similarly,the while dividing a decimal number by 100, 1000 the decimal point as many Step 1: Make the divisor a natural number by multiplying divisor and the dividend by10, a power of etc., 10. we move – 815 places to the left as there are zeroes in the divisor.

815

81.5 × 10

1018.75 × 10

10187.5

For example:

2037 – 1630 4075 85.3 � 100 = 8 5.3 0.8 5 3 – =4075 0

Step 2: Place the decimal point directly above the decimal point in the dividend. On dividing by 10 Step 3: Divide the whole number part and then the tenth part.

85.3 � 10 = 8 5.3 = 8.5 3

Hence, the auto-rickshaw driver filled the tank with 12.5 litres of CNG.

Example 6

Multiply.

1 457.2 and 132

3 4 6 0 0

� 4 Fill in the blanks. 8 2 3 4 1 351.2542× 100 = _______ 0 5 8 5 + 4 ÷1 101= _______ 7 0 0 3 124.12 4 4 0 5 1 9 5 587.2 ÷ 1000 = _______

Mean, Mean,Median Medianand andMode Mode

introduction

8 0 0 0 0

45.123 × 1000 = _______

39.145 ÷ 100 = _______

Ria’s parents are planning to go on a vacation to Jaipur for the first time.

Real Life Connect

with a real life

Ria: Papa! What is the weather in Jaipur like? Ria’s father checks the online weather forecast for Jaipur for the next 15 days and finds the following:

example

125.36 26 3259.36 – 26 65 – 52 139 – 130 Division of an integer 93 by –1 – 78 dividing an integer by –1, the156 result is the additive – 156 e of the integer. 0

Day 1

Day 2

Day 3

Day 4

Day 5

Day 6

Day 7

Day 8

Day 9

Day 10

Day 11

Day 12

Day 13

Day 14

Day 15

Arithmetic Mean or Mean

I scored _________ out of 5.

One possibility is that Ria and her father want to take a decision based on a single value that represents the entire data. But what will that value be and how is it calculated? Arithmetic mean or mean or average of a data is a value that can represent all of the data. It is defined as the sum of all the observations of data divided by the total number of observations.

2 0.888 by 1.11

1.11 × 100 = 111; 0.888 × 100 = 88.8

Mean =

02_UM24CB0703_R9.indd 17

Sum of all observations Number of observations

But data could be presented in different forms. How do we calculate the mean in all such cases? Let us learn!

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0.8 Mean of Ungrouped Data 111 88.8 In the last chapter, we learnt that when data is listed as individual values or observations it is called ungrouped data. – 00 For example, the temperature forecast data that Ria’s father showed is also ungrouped data. 888 So, in the above case, the average temperature can be calculated by dividing the sum of all the observations (temperature in °C) by the total number of observations (the number of days). Thus, the – 888 mean temperature can be calculated as: 1 If the ratio of red balls to blue balls in a bag is 4:3, find an equivalentMean temperature = 32 + 34 + 32 + 37 + 42 + 40 + 30 + 32 + 36 + 42 + 44 + 44 + 42 + 41 + 42 °C 0 Division of zero by anExample integer

A quick-thinking question

Days Temp (in °C)

Ria: Papa! The temperature is so different every day. What clothes should we pack?

Divide. 1 3259.36 by 26

ny integer a,

2 0 0 0 0

102.925 × 4.28 = 440.51900

457.2 × 132 = 60350.4 Example 7

Concept

Let’s Warm-up 1 0 2 9 2 5

2 4 0 0 4

quick warm-up

8 5.3 = 0.0 8 5 3

2 102.925 and 4.28

page a 4 5 7 with 2 1 1 1 2 5

On dividing by 1000 85.3 � 1000 =

Introductory

� 9 1 3 7 4 5 7 6 0 3

On dividing by 100

Fun fact, related

ratio when there are 24 red balls. to the concept

When dividing zero by an integer, the result is always

Think and Tell

red balls 4 = Do we get a natural number on dividing 2 decimal bluenumbers? balls 3

zero.

= 38°C

Ratio of

For any integer a,

So, the average or mean temperature is 38°C.

a ÷Find (–1) the = (–a) 0 ÷143.75 (a) = 0cm . missing dimension of the given rectangle with area

overall duration of their stay, they should pack their clothes based on the mean temperature, 38°C.

moon and sun is 1:400.

Product ratio of Means = Product Extremes ⇒24:18. 15 × x = 25 × 42 So, the equivalent of 4:3 with 24ofred balls is

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

Remember! The mean of the data may not always be an observation from the data.

The ratio of the sizes of the Therefore, if Ria and her father want to make a decision based on a single temperature that reflects the

4 × 6 24 RatioExample with5 24 redthe balls = of x in=the proportion 25:15 :: x:42 Find value 3 × 6 18

Do It Together

570 °C 15

Did You Know?

11.5 cm 25 × 42 ⇒x= = 70 15 whether the quotient 482 ÷ =3 _________ is an integer. Length =of Area = _________ Breadth ? Example 2 1 Example 2 If A:B If A:B 4:9 bag andof B:C = 2:5, sold by Seema contains = 1:2 and B:C = 6:7, A:B:CIf she sells 117 blue 6 = Each marbles 9 blue marbles and 5 redfind marbles. of the rectangle is _________ cm. 3 = 160.66 whichHence, is not the an length integer. Can you think of special cases wherefind the A:C marbles in a day, how many red marbles has she sold? 04_UM24CB0704V1.indd 46

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Think and Tell

1 but 16 ÷ 256 An= important 16 02_UM24CB0703_R9.indd 24 1 ≠ , integers are not commutative under division. point to keep in 16

To find A:B:C; the value of B must be the same in both ratios. Using4the method of2proportion, 9:5 :: 117:x A:B = 4:9 = ; B:C = 2:5 = Value of B in 1st ratio = 2; Value of B in 2nd ratio = 6 9 5 5 × 117 ⇒ 9 × x = 5 × 117 ⇒ x = ⇒ x = 65 LCM of 2 and 6 = 6 9 A A B 4 2 8 A:C = Hence, = × Seema = ×has=sold 65 red marbles. 1×3 3 6 Pointing out C B C 9 5 45 A:B = 1:2 = = ; B:C = 2×3 6 7 Error Alert! 12/28/2023 7:05:07 PM Hence, A:C =number 8:45 must be added Example 7 What to 8, 21, 13 and 31 so that they are commonly made Hence, A:B:C = 3:6:7

Commutative Property may hold true?

that the integers 256 and 16 are not commutative division.

16 = 16

mind

the blanks.

125) ÷ (–25) = ________

in proportion?

Remember!

10 A survey asked participants to rate a product on a scale of 1 to 10. The ratings received were as follows: 4, 6, 7, 8,

The sign of the operator changes during transposition from LHS to

mistakes and

Let the number to be added be x

9, 9, 10. Which measure of central tendency should be used to represent the typical rating, and why?

Dividing any non-zero integer by RHS and vice versa. 11 What is the median age of a family whose members are 45, 42, 38, 35, 18, and 10 years old? Then, (8 + x):(21 + x) :: (13 + how x):(31 + to x) avoid 3 A piece of string that is 84 inches long is cut into three parts so that the lengths parts zero is undefined inExample mathematics 2x – 3 = 5xof + the 2 2x – 3 =of 5xthe +2 12 The sizes of 10 shirts are as follows: + x ratio 13 + 6:10:5. x 2 0 ÷ 45 = ________ string are in8the Find the length of each part. 80, 85, 90, 80, 80,and 85, 85, does 90, 80, 95not yield a valid result. 7x = –1 3x = –5 them = 21 + x

31 + x

Find the modal size. If the size of one shirt is misread as 80 instead of 85, find the correct modal size.

Total number of parts = 6 + 10 + 5 = 21 4 136 ÷ ________ = 1the mean and median of the following set showing the number of hours of operating Letlifeus subtract 13 Find of 20 flashlight 1 from both the sides to remove x from the numerator,

÷ ________ = –23

6 8+ x = 13×+84 x = 248inches; + x – 21 – x 13 + x – 31 – x Length of 1st part – 1 =21 –1 ⇒ = 21 + x 31 + x 21 + x 31 + x 14 The weights (in kg) of 15 students are as follows: 10 –13 –18 Length of 2nd × 84 40 inches; 27, 30, 42, 43, 36, 34, 35, 37, 28, 29, 31, 44, 41, 32, 33. ⇒ part = = ⇒= –13(31 + x) = –18(21 + x) ⇒ –403 – 13x = –378 – 18x ⇒ 5x = 25 ⇒ x = 5 21 + x 21 + x 31 If 30 kg is replaced by 25 kg and 41 kg by 45 kg, then find the new median. 5 be added to 8, 21, 13 and 31 to make them proportional. Hence, 5 must 18, 18, 19, 21. 15 The ages of a group of students are as follows: 12, 14, 15, 15, 16, 17, Length ofWhich 3rdmeasure partof=central × 84 = 20 inches. tendency should be used to represent the typical age of the students, and why?Do It 21 batteries:

19, 25, 20, 21, 25, 19, 21, 22, 24, 20, 25, 18, 23, 21, 24, 20, 19, 23, 25, 18

Do It Yourself 3C

The first three terms of a proportion are 4.2, 1.2 and 0.7. Find the fourth term.

Together

nd the quotient.

HOTS:

72 ÷ (–4)

b (–56) ÷ 7

344 ÷ 43

f (–984) ÷ 12

Applicative and analytical

olve.

questions

(–88 ÷ 4) ÷ (–1)

b 125 ÷ (125 ÷ 5)

(900 ÷ (18)) ÷ 6

f (–3060 ÷ (–36)) ÷ 5

3 Let the fourth term be of x; 4.2:_______ :: _______:x Sunita purchased two colours ribbons from the market. The red Points ribbon was 2 m long, and the blue to Remember Chapter end 4 5 ___ × ___ • The sumto of supplementary is 180°. x = ______ × ______ ⇒ratio x= _______ of the red ribbon one was 1 4.2 m×long. What is the of the= length that ofangles the blue ribbon? summary 1 The teacher of class 7 found the mean weight of a class consisting of 20 students as 45 kg. 4.2 • The sum of complementary angles is 90°. 8 –585 ÷weighing (–13) 55 kg and 52 kg, respectively, g (–676) ÷ (–26) h and Later, two more students Ankit Suhani join the • Adjacent angles have a common vertex and a common arm. 3 5 11 13 Hence, the fourth term is ________ . class. What is the mean weight of the class now? • A linear pair of angles share a common vertex and a common side, while their non-shared sides together Ratio of the length of the red ribbon to the length of the blue ribboncreate = a2straight :1 = : 2 The ages of a group of people are 22, 28, 30, 29 and 31. If another person, aged 32, joins the 4 line. 8 4 8 • Vertically opposite angles are angles that are opposite to each other when two lines intersect. group, will the median age change? If yes, what will the new median age be? Parallel lines areof two or more lines that are always the same distance apart and never meet, no matter To write theContinued ratios in theirProportion simplest form, multiply both sides with •the LCM the denominator. how far they are extended. Mohan ÷ secured and 76 marks÷in(–18 four tests. What is the lowest number of marks he 4 75, 82, 79 d 324 ÷ 2) c (192 ÷ 3(–16)) • A transversal intersects two or more lines at distinct points. 11 13 can secure in his next test, if he needs to maintain a mean score of 80 marks in five tests. • When two lines are crossed by a transversal, corresponding angles are equal, alternate interior LCM of 4 and 8 three = 8; Hence, × b8 and = 22; × 8 =to13 Any quantities a, c are said be in continued proportion if, and exterior angles are equal, and the sum of interior angles on the same side of the transversal is 4 8 g 0 ÷ (–1508 ÷ 29) h (1000 ÷ (20)) ÷ 100 180 degrees. If these angle conditions are met, the lines are parallel. The required ratio is 22:13. Third Proportional First Proportional a:b :: b:c Word Problems c –176 ÷ (–11)

nd the number which when divided by (–1) gives 145.

Example 4

d 192 ÷ (–12)

Do It Together

ate True or False. Division is commutative for integers.

mbers up to gits Division has the associative property for integers.

The quotient of two negative integers is always negative. A

Math Lab

multidisciplinary Division by zero is defined for integers.

Setting: In groups of 5

Parallel Line Art

1 5 the proportionality 4 2 2 Materials Needed: Drawing paper, ruler, pencils, coloured pencils or markers Using 1:3 = ; 5:7 = ; 4:15 = and 2:5 =condition; b × b = a × c ⇒ b = ac __________________ Instructions: 3 7 15 5 1 Begin by drawing two parallel lines horizontally across the paper, leaving some space between them. is the value of the first proportional if 10 and 4 are in continued proportion? __________________ LCM Example of 3,87,What 15 and 5 = ________ 2

Choose a point near the top of the paper and draw a diagonal line (the transversal) that intersects

Colour the angles using different colours to differentiate them and make your artwork visually

You can also create a legend on the side of the paper explaining the colour code for each type of

As you work on this activity, take the time to explain each angle type to yourself or a friend,

the parallel lines. The as:2 x:10 :: 10:4. 1 5 proportion can 4 be givenQR __________________ Code: Access 3 Label the points where the transversal crosses the parallel lines as A, B, C, D, and so on. = ________; = ________; = ________; = ________; 100 4 3 7 15 5 ⇒ 4x = 10 × 10 ⇒ x = = 25Now, start identifying and measuring different pairs of angles: to interactive __________________ 4 • Measure and draw corresponding angles that are equal across the parallel lines. Hence, the descending order is ________>________>________>________ • Draw alternate interior angles and alternate exterior angles, labeling them as you go. Hence, the value of the first proportional is 25. 63 digital resources __________________ • Measure and label interior angles on the same side of the transversal, ensuring that their

Materials Required: Index cards with different integer expressions written on them, stopwatch or timer. Method:

+23 – 52 = ?

classroom ve two pairs of integers such that a ÷ b = –15.

04_UM24CB0704V1.indd 63

nd the value of (12 × –3) ÷ (–2) + (–4).

3 • Integers

15 × (–5) = ?

sum is 180°.

Chapter 11 • Ratio and Proportion

Place the index cards face down on a table. Start the stopwatch or timer and begin the relay.

The first team member will pick up an index card, read the expression aloud, solve it and return it

The relay continues until all the index cards have been used or until a designated time limit is reached.

The team with the highest number of correct solutions wins the game!

96 � (–4) = ?

to their team to tag the next person.

Solve.

Find the product.

Divide.

96 =? –4

(–15) + (–9) = ?

172

a (–256) + 362

b 214 – (–126)

e (–659) + (–1352)

f (–2698) – (–1236)

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11_UM24CB0712_R6.indd 175

Chapter Checkup

b (–125) × (–26)

e (1245) × (–142)

f (–2365) × (123)

c (–36) × 48

appealing.

175

angle.

discussing how they relate to parallel lines and the transversal.

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130

c 652 + (–129) 11_UM24CB0712_R6.indd 172 d 248 – (–369)

a 25 × (–89)

d (245) × (–57) 08_UM24CB0708V1.indd 130

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iii

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12/28/2023 7:31:51 PM

a 363 by (–11)

b (–2652) by (–51)

e 2852 by 23

f (–9594) by (–41)

c 5525 by (–25)

d (–5472) by 36

Fill in the blanks. a (–156) + 389 = ______

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Math Lab Mean Proportional

Integer Operations Relay

4 • Measures of Central Tendency fun integer Chapter The quotient of zero divided byand any non-zero is zero. 23 – 52 = ?

activity

Arrange the ratios 1:3, 5:7, 4:15 and 2:5 in descending order.

b 524 + ______ = 256

c ______ – (–269) = –698

d 29 × ______ = (–1885)

e ______ � (–17) = 56

f –568 – (–258) = ______

g ______ × (–15) = –5475

h (2356) – ______ = 5886

______ + (–1265) = –4526

k (–1265) × (–352) = ______

(–3562) � (–137)= ______

3654 � ______ = (–87)

Chapter 3 • Integers

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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 Word Problems comprehensive scaffolding—including modelling, guided practice, and ultimately fostering independent 1 Rita went from place X to place Y and then to place Z. X is 14.05 km from Y and Y is 12.7 km application of concepts. from Z. Aman went from place X to place A and then to place Z. A is 29.3 km from X and Z is 21.45 km from A. Who travelled farther and by how much?

GRR, endorsed and promoted by both the NEP 2020 and2NCF, plays a and pivotal role in equipping teachers to Sam ordered a coffee a waffle. Find the change he will receive if he paid $10. 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

Coffee: $1.46

Garlic bread: $3.95 Ice tea: $2.89 Waffle: $0.85

Unit Component

Snapshot

Multiplication and Division of Decimals Naman and his father are patiently waiting for their turn at the petrol pump. Naman notices the displayed prices of various fuels on the board.

Real Life Connect

Naman: Dad, how much fuel are we putting in the car today?

Real Life Connect

Petrol –101.94

Father: We're going to fill our car with 35.5 litres of diesel.

Theoretical explanation

Diesel – 87.89

Naman's curiosity is sparked as he begins to calculate the amount they will need to pay at the pump.

LPG –

77.85

CNG –

81.5

Multiplying and Dividing Decimals Let us find the amount paid by Naman’s father.

Adjacent Angles

X numbers, we ToTwo find the amount paidas byadjacent Naman’s father, need to multiply. To multiply two decimal angles are classified when theywe possess Y follow the given a common arm, asteps. common vertex, and their non-

I do

common sides extend to opposite sides of the common arm. For example: ∠XOY and ∠YOZ are adjacent angles with common vertexon O Decimals and common arm OY. Chapter 2 • Operations Example 4

Common Vertex

arm

Name the adjacent angles of the angles given below. 1 ∠POQ

2 ∠TOS

3 ∠ROS

4 ∠POT

02_UM24CB0703_R9.indd 23

1 Adjacent angles of ∠POQ are ∠POT and ∠QOR.

12/28/2023 7:05:06 PM

2 Adjacent angles of ∠TOS are ∠ROS and ∠POT. 3 Adjacent angles of ∠ROS are ∠TOS and ∠QOR. 4 Adjacent angles of ∠POT are ∠POQ and ∠TOS.

Examples

Example 5

Draw an angle QOS. Draw an adjacent angle to ∠QOS whose shared arm is 1 OQ

2 P

Fill in the blanks with the names of adjacent angles. F

Think and Tell

Is there any adjacent pair of angles on line XY?

1 Adjacent angles of ∠FCE are ∠DCE and ∠FCA. 2 Adjacent angles of ∠ACB are _____ and _____. 3 Adjacent angles of ∠ECD are _____ and _____. 4 Adjacent angles of ∠FCD are _____ and _____.

Linear Pair of Angles A linear pair of angles comprises two adjacent angles whose sum is 180°. These angles share a common vertex and a common side, while their non-shared sides together create a straight line. For example, in the adjoining figure, ∠KOL and ∠MOL are a linear pair of angles.

UM24CB07_FM.indd 4

Do It Together

O T

12/30/2023 5:37:42 PM

Example 9

Komal reads 518 pages of a book in two weeks. In how many days will she read a book 1221 pages? Number of pages read in 2 weeks or 14 days = 518

518 = 37; Number of days for 37 pages = 1 day 14

Number of pages read in 1 day = Number of days for 1 page = Example 10

GRR Steps

Unit Component

1 1 ; Number of days for 1221 pages = × 1221 = 33 days 37 37

30 women can finish a piece of work in 18 days. How many women can finish the same 20 days? In 18 days, the number of women who can do a piece of work = 30 Snapshot ∴ In 1 day, the number of women who can do a piece of work = 30 × 18 (fewer days, more

30 × 18 = 27 (more day 20

∴ In 20 days, the number of women who can do the piece of work = Do It Together

We do

A machine produces 208 toys in a day if it runs for 6.5 hours. How many toys will it prod the machine runs for 20 hours? Number of toys produced in 6.5 hours = 208

Do It Together

Number of toys produced in 1 hour = 208 ÷ _______ Number of toys produced in 20 hours = ______ × ______ = ______

Do It Yourself 11C 2 kg of potatoes cost ₹16. How many potatoes (in kg) can be bought for ₹120?

A car travels Do2It Yourself 8B 150 km in 3 hours. How far does it travel in 9 hours? 1

Do It Yourself Word Problems

If the cost of 15 pens is ₹300, find the cost of 5 such pens.

The human heart beats 375 timesP in 5 minutes. How many times does it beat in an hour?

Identify all the pairs of corresponding angles and pairs of alternate angles in the figure. 1

Word Problem

Chapter 11 • Ratio and Proportion 8 1

n If the rain makes Due to the effect of the wind, rain sometimes falls to the ground at an angle. 12

11 this rain hit the ground? Assume that rain an angle of 60° with the clouds, at what angle would drops travel in a straight line. Illustrate with a rough diagram.

Check whether the lines are parallel in each case, based on the known angles. a

11_UM24CB0712_R6.indd 177 110°

89°

102°

109°

89°

You do

72° Constructing Parallel Lines

99°

109°

Constructing a line parallel to a given line l passing through an external point P.

Chapter Checkup

3 1: Find the value of the indicated angles. PQ Step Mark point Q on Step 2:unknown Draw line segment line l.

radius to intersect line l and line segment PQ at

points B and A, respectively. p° Check whether the given angles are supplementary.

32°

a 14°,x° 166°

m° 50° b 150°, 110°

Chapter Checkup

Step 3: From point Q, draw an arc with a chosen

by connecting points P and Q.

q° c 36°, 135°

n°

105°

y°

Q 56° a 24°,

l 60°

Q b 50°, 16°

e 76°, 109°

h°

dB 44°, 46°

c 36°, 54°

g°

45°

80° Check whether the given angles are complementary.

72°

d 40°, 135°

e 78°, 12°

In a trapezium PQRS, it is known that PQ ꠱ RS. Also, ∠PQR = 150° and ∠QPS = 120°. Find the value of the other Step 6: Connect points PD and extend the line

Step 5: Draw an arc from point

Step 4: Draw another arc from

two the angles. Find supplement of the given angles. 3point P, using the same radius as C with radius equal to AB,

on both sides to obtain the desired line m,

5 A line (DE) parallel to thebside BCthe of previously ΔABC passes through A. The angles the is parallel to ∠DAB the given line l.and ∠EAC = 60°. Find e cutting drawn arc before, line segment a to intersect c which d= 90°

34°

119°

126°

D. PQ atmeasure point C. of all the angles at ofpoint the triangle. P

24°

In the given figure, l ꠱ m. p and qangles. are transversals. the complement ofAlso, the given 4 6 Find

b 20°

Q value of x. Find the

5 7

c 26°

A d

Find the measures of the unknown angles.

Constructing a line parallel to a given line at a b fixed distance a

(x + 9)° 35°

Pearson, P. D., & Gallagher, G. (1983). Contemporary Educational Psychology.

x° 116° Fisher, D., & Frey, N. (2014). Checking for understanding: Formative assessment techniques for your classroom. x°

Gradual Release of Responsibility

(2x + 6)°

3x°

x°

126

4x°

(2x – 10)° A x°

Find the values of ∠ABC, ∠ACB and ∠CAB.

142°

72°

2x° (3x – 10)°

4x° 127

e 86°

Chapter 8 • Angle Pairs and Parallel Lines

08_UM24CB0708V1.indd 127 08_UM24CB0708V1.indd 126

UM24CB07_FM.indd 5

136° of responsibility. Fisher, D., & Frey, N. (2021). Better learning through structured teaching:X A framework A B for Ythe gradual release y°

a°

b Step 2: Construct right angles ∠CAY and ∠DBY at points x° A and B respectively. x° x°

Step 1: Choose any50° two 2x° points A and B on line XY.

166°

b°

62°

40°

C Note that pCis not parallel to q, find the values of a° and b°.

a 47°

2x° x°

In the figure given below, c || d || e. Lines a and b are

28-12-2023 19:43:24 the transversals. Find the value of g°, m°, y° and 28-12-2023 19:43:22h°.

12/30/2023 5:37:44 PM

C o nt e nt s

Operations on Fractions �� 1 • Addition and Subtraction of Fractions • Multiplication and Division of Fractions

2 6

Operations on Decimals �� 17

Integers �� 30

• Addition and Subtraction of Decimals • Multiplication and Division of Decimals

• Addition and Subtraction of Integers • Multiplication and Division of Integers

4 5

31 35

Data Handling �� 64 • Organisation and Representation of Data

Ratio and Proportion �� 170

Percentage, Profit and Loss, Simple Interest �� 183

18 23

Measures of Central Tendency �� 45 • Mean, Median and Mode

• Percentage and Its Applications • Profit and Loss • Simple Interest

Simple Equations �� 102

Angle Pairs and Parallel Lines �� 116

• Linear Equations in One Variable

• Angle Pairs • Parallel Lines

Construction of Triangles �� 225

Perimeter and Area �� 238

Algebraic Expressions �� 263

Exponents and Powers �� 283

Symmetry �� 297

Visualizing Solid Shapes �� 315

103

117 123

Triangles and Their Properties �� 134 • Understanding Triangles and Their Properties 135 • Right-angled Triangles 145

Congruence �� 151 • Congruence in Shapes and Figures • Congruence in Triangles

152 156

184 193 198

Rational Numbers �� 204

171

Introduction to Probability �� 89 • Understanding Probability

• Ratio and Proportion

• Introducing Rational Numbers • Operations on Rational Numbers

• Constructing Triangles

205 214

226

• Squares, Rectangles, Triangles and Parallelograms 239 • Circles 248 • Word Problems on Perimeter and Area 257

• Generating Rules in Formulas and Patterns 264 • Algebraic Expressions and Terms 267 • Addition and Subtraction of Algebraic Expressions 270

• Exponents 284

• Understanding Symmetry

• 3-D Shapes and 2-D Shapes

298

316

Answers �� 331

UM24CB07_FM.indd 6

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Operations on Fractions

Let’s Recall

A fraction represents a part of a whole. It has two parts: a numerator and a denominator. Let’s say Raghav has a chocolate bar, and he divides it into 5 equal parts. So, here, a chocolate bar is considered a whole and each part is considered a part of a whole. He eats 3 parts out of 5 equal parts. To represent this as a fraction, Raghav shades 3 parts out of 5. 3 The shaded part represents . 5

Here, 3 is the numerator and 5 is the denominator.

Types of fractions

Examples

A proper fraction is a fraction where the numerator is smaller than the denominator. 1 3 5 7 , , , A proper fraction is always less than 1. 2 4 9 8 An improper fraction is a fraction where the numerator is equal to or greater than the denominator. An improper fraction is always greater than or equal to 1. A mixed fraction, or a mixed number, is a combination of a whole number and a proper fraction. A mixed fraction is another way of representing an improper fraction and is always greater than 1.

5 7 11 13 5 , , , , 2 4 9 8 5 3 4

3 4

1 9

2 ,5 ,7 ,6

5 11

Let’s Warm-up Fill in the blanks. 1

The numerator is always _________ than the denominator for proper fractions.

A mixed fraction has a __________ part and a fractional part.

3 4 5

5 __________ is the denominator in . 8 17 The numerator in is __________. 19 An improper fraction cannot be __________ than 1.

I scored __________ out of 5.

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Mean, Median and Mode of Fractions Addition and Subtraction Real Life Connect

2 Riya is happy to get her first salary. She spends of her salary on 5 1 house rent, on food and saves the remaining salary. 4

Adding and Subtracting Fractions Riya wanted to check what fraction of her salary she spent. How could she do that? Let us find out! To find the fraction of salary spent, we will add the individual parts of the salaries spent. Addition of Fractions

2 Fraction of salary spent by Riya on house rent = 5 1 Fraction of salary spent by Riya on food = 4 2 1 Fraction of total salary spent by Riya = + 5 4 Take the LCM of 5 and 4. LCM of 5 and 4 is 20. 2 2 4 8 = × = 5 5 4 20

Sum of like fractions =

1 1 5 5 = × = 4 4 5 20

2 1 8 5 8 + 5 13 + = + = = 5 4 20 20 20 20 13 Thus, Riya spends of her salary. 20 Addition of mixed numbers 1 1 Add 4 + 2 5 3 Whole number part → 4 + 2 = 6 Fractional part → 6+

8 8 =6 15 15

Remember! Sum of numerators Common denominator

Did You Know? Musicians use fractions to understand and create

1 1 1 3 1 5 3 5 8 + = × + × = + = 5 3  5 3   3 5  15 15 15

rhythms.

1 1 8 =2 =6 5 3 15 Properties of Addition of Fractions Thus, 4

Commutative Law

Associative Law

a c c a + = + b d d b

a + c  + e = a +  c + e b d f b d f     

1 1 3 2 (3 + 2) 5 + = + = = 2 3 6 6 6 6

 1 + 1  + 1 =  3 + 2  + 1 = 5 + 1 = 10 + 3 = 13  2 3  4  6 6  4 6 4 12 12 12     1 1 1 1  4 3 1 7 6 7 13 + + = + + = + = = + 2  3 4  2  12 12  2 12 12 12 12

1 1 2 3 2+3 5 + = + = = 3 2 6 6 6 6

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Subtraction of Fractions

13 of her salary. What fraction of Riya’s salary is left with her? 20 Fraction of salary left = Total salary – Fraction of salary spent 13 =1− 20 20 13 = − Remember! 20 20 Difference of numerators 7 Difference of like fractions = = Common denominator 20 Subtraction of Mixed Numbers 4 1 Subtract: 9 – 3 5 6 Convert both the fractions into improper fractions. We know that Riya spent

4 49 1 19 9 = 3 = 5 5 6 6 Find the LCM of 5 and 6.

LCM of 5 and 6 = 30 49 294 19 95 = = 5 30 6 30

4 1 294 95 199 19 9 –3 = − = =6 5 6 30 30 30 30

Think and Tell Check if the associative and commutative properties hold true or not for subtraction of fractions.

4 1 19 Thus, 9 – 3 = 6 5 6 30 Example 1

Add. 8 5 1 + 9 6

8 8 × 2 16 = = 9 9 × 2 18 16 15 31 + = 18 18 18

5 5 × 3 15 = = 6 6 × 3 18

31 (1 × 18) + 13 18 13 13 = = + =1 18 18 18 18 18

8 5 13 Thus, + = 1 9 6 18 Example 2

5 1 +1 9 2 Whole number part → 2 + 1 = 3

2 2

Fractional part → =3+

5 1 5 2 1 9 10 9 19 + = × + × = + = 9 2 9 2 2 9 18 18 18

19 1 1 =3+1 =4 18 18 18

5 1 1 Thus, 2 + 1 = 4 9 2 18

1 7 Parth jogged 3 km and Priya jogged km. Who jogged the longer distance? How much more? 3 8 1 7 Parth jogged more since 3 is greater than . 3 8 7 1 To find how much more distance Parth jogged, subtract km from 3 km. 8 3 1 7 10 7 =3 − = − 3 8 3 8 10 8 7 3 80 21 59 11 LCM of 3 and 8 = 24;  × − ×  = − = =2 24  3 8 8 3 24 24 24 Thus, Parth jogged 2

11 km more than Priya. 24

Chapter 1 • Operations on Fractions

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Do It Together

1 1 1 Add: 2 + + 5 2 5 4 (2 + 5) +

7 3 Subtract: 5 − 2 8 7

1 1 1 + + 2 5 4

7 47 5 = 8 8

3 17 2 = 7 3

________ − ________ = ________

= ________ + ________ = ________

Do It Yourself 1A 1

Solve. a

9 + 2 22 11

Simplify. a

2 5 + 3 9 2 + 1 + 3 17 17 17

4 + 12 + 21 9 9 9

7 2 − 13 9

2 − 30

3 +5 40 6

2 + 1 − 3 17 17 17

5 + 1 − 3 12 12 12

15 −

1 2 + 1  + 4 = 1 2 +     3   3 6 5

3 0 + 50 = 8

Fill in the blanks.

9 9 + ________ = 2 11 11

11 21 21 + = + 19 40 40

Compare using appropriate symbols. (>, <, =) a

2 1 2 1 + 3 _____ 7 − 2 3 5 4 3

5 1 7 − 4 _____ 3 7 4 8

11 + 1 1 _____ 23 + 1 9 2 7 7

7 +45 − 3 8 6 12

+4 5

5 2 1 3 − _____ + 8 3 2 4

Answer the given questions. 2 7 a How much is 5 less than 7 ? 3 8 2 1 b What must be added to 2 to get 4 ? 5 4

1 8 4 cm, 1 cm and cm. 4 9 5 7 2 d Find the perimeter of a rectangle which is cm long and cm wide. 11 9 1 2 3 5 e Subtract the difference of 7 and 9 from the sum of 2 and 4 . 4 5 4 6 c What is the perimeter of a triangle with sides of

Word Problems 3 1 and her sister painted of the wall. What portion of the wall did they paint 5 6

Aisha painted

Ram drank

A rectangular field is

together?

3 2 of a bottle of mango juice and Lisa drank of a bottle of orange juice. How 4 9 much juice did they consume in all? 3 2 m long and m wide. Find the perimeter of the rectangular field. 8 5

01_UM24CB0701V1.indd 4

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Multi-step Problems on Adding and Subtracting Fractions

1 Remember, Riya got her first salary? When she gets her next salary, she spends of her salary on house 3 1 1 rent, of her salary on transportation and of her salary on food. 12 4 Let’s see what fraction of salary is left with her. 1 1 1 4 1 3 4+1+3 8 + + = + + = = 3 12 4 12 12 12 12 12 8 12 8 12 − 8 4 1 Fraction of salary left = 1 − = − = = = 12 12 12 12 12 3 1 Thus, next month Riya saves of her salary. 3 Total fraction of salary spent =

Example 3

3 Mary is baking a cake and needs cups of 4 2 flour. She accidentally adds cups extra, but 5 1 then realizes her mistake and removes 8 cups. How much flour does she have now?

3 The initial amount of flour Mary needed = cups 4 2 Extra flour added = cups 5 3 2 15 + 8 23 Total flour = + = = cups 4 5 20 20 1 Flour removed = cups 8 23 1 46 − 5 41 1 Flour left = − = = =1 cups 20 8 40 40 40 1 Thus, Mary has 1 cups of flour. 40

Do It Together

Example 4

3 Sarah baked a pie and shared with her 8 1 neighbours and with her colleagues. Later, 6 1 she ate of the pie. What fraction of the pie 4 was left? The baked pie can be considered as a whole.

3 Pie shared with neighbours = ; pie shared 8 1 with colleagues = 6 1 Pie eaten by Sarah later = 4 3 1 1 9+4+6 Total pie eaten or shared = + + = 8 6 4 24 19 = 24 19 24 − 19 5 Pie left = 1 − = = 24 24 24 5 Thus, of the pie was left. 24

3 1 cups to make fruit punch, 6 cups to make flavoured jelly 4 3 and drank the remaining amount. How much juice did she drink? Pihu brought 17 cups of juice. She used 4 Total juice with Pihu = 17 cups

3 cups 4 1 Juice used to make flavoured jelly = 6 cups 3 Total juice used to make fruit punch and flavoured jelly = _____ + _____ = _____ Juice used to make fruit punch = 4

Juice left for drinking = _____ − _____ = _____

Do It Yourself 1B 1

1 4 gallons of water. If Nishant drinks gallons of water each day from the jug, then what quantity of 4 7 water will be left in the jug after 3 days? A jug has 2

Chapter 1 • Operations on Fractions

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1 1 1 km, Muskan jogged 3 km, Neeta jogged 2 km and Avyaan jogged 3 km. What distance did 4 2 6 they all jog together?

Vishwa jogged 3

Shikha studies for 5

4 5

2 4 hours daily. She devotes 2 hours of this time to science and maths. How much time does 3 5 she devote to other subjects? 6 2 An empty box weighs 20 kg. An iron ball weighing 7 kg is placed inside it. What is the total weight of the box now? 7 3 2 1 Carlton has 30 yards of string. He uses 11 yards of string to tie a parcel and 6 yards of string to tie a box. How 5 4 much length of string is left with Carlton? Kashvi recorded the following amounts of water used for various activities at home. Activity

Water used (in litres)

Activity

Water used (in litres)

Bathing

2 3

Washing car

2 5

Washing utensils

1 5

Other activities

3 4

Laundry

1 4

Answer the given questions. a Which activity uses the most amount of water and which activity uses the least amount of water? What is the difference of the amounts of water used in these two activities?

b How much water in total is used for washing utensils and cars? c What is the total amount of water used? d How much more water is used for bathing than for other activities?

Mean, Median and Multiplication andMode Division of Fractions Real Life Connect

Remember, Riya found out what fraction of her salary she spent and saved in the first month? If she earns ₹60,000 in a month, then let us see how much money she spent and how much she saved.

Multiplying Fractions To find the amount of money spent or saved by Riya, we need to multiply the total amount earned by the fraction of the amount saved or spent.

Multiplication of a Fraction by a Whole Number The fraction of money spent by Riya = Total salary = ₹60,000 3000 13 Amount spent = × 60,000 20 1

13 of her total salary 20

= ₹39,000

Thus, she spent ₹39,000. 6

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The fraction of money saved by Riya =

7 of her total salary 20

Total salary of Riya = ₹60,000 7 7 × 60,000 Amount saved = × 60,000 = = ₹21,000 20 20 Thus, Riya saved ₹21,000. Example 5

Do It Together

7 2 4 × 7 28 = = = 14 2 2 Solve: 4 ×

Example 6

Remember! When the fraction is used with the word “of”, it acts as an operator. “of” means multiplication.

4 5 14 420 = 30 × = = 84 5 5 Solve: 30 × 2

3 Rajeev went shopping and carried ₹5000 with him. He spent of the total amount on shopping. 8 How much money did he spend? Total amount of money with Rajeev = ₹5000 3 Fraction of money spent = 8 3 Total money spent = ₹5000 × = = ₹__________ 8 Thus, Rajeev spent __________.

Multiplication of a Fraction by a Fraction Riya keeps half of her saved salary in her bank account and gives half of her savings to her father for investment. How much money does she give to her father? 1 7 Fraction of savings given to father = of 2 20 1 7 7 = × = 2 20 40 7 7 Amount given to father = of ₹60,000 = × ₹60,000 = ₹10,500 40 40 Thus, Riya gives ₹10,500 to her father for investment. Example 7

Remember! Multiplication of numerators N1 N2 N1 × N2 × = D1 D2 D1 × D2 Multiplication of denominators

1 of it, and Jill consumed the remaining amount. 5 Find out how much glucose water Jack and Jill consumed individually. Jack and Jill had 5 litres of glucose water. Jack consumed Total quantity of glucose water = 5 litres

1 Fraction of glucose water consumed by Jack = 5 1 Glucose water consumed by Jack = × 5 = 1 litre 5 Glucose water consumed by Jill = 5 litres – 1 litre = 4 litres Example 8

Multiply.

3 2 5 5 Multiply the numerator with the numerator and the denominator with the denominator. 3×2 6 = 5 × 5 25

1 ×

1 1 × 5 3 1 Convert 4 into an improper fraction first, then 5 multiply. 1 21 4 = 5 5

2 4

4 Chapter 1 • Operations on Fractions

01_UM24CB0701V1.indd 7

1 1 21 1 7 2 × = × = =1 5 3 5 3 5 5

28-12-2023 18:54:40

Do It Together

Multiply. 1 3

1 × 5 = ______ 2

1 = 2 2

2 2

2 1 × 5 = ______ 3 2

2 1 2 1 = 5 = 2 × 5 = _____ 3 3 2 2 3 2

1 × 5 = _____ 2

Do It Yourself 1C 1

Multiply. a 9×

1 3 4 × × 2 4 5

2 9 × 3 3

d 1

b 1

4 3 × ×6 9 2

1 5 1 × ×3 ×2 4 6 5

2  7 1 × 1 −  5  9 2

1 1 5 + ×1  4 2 6

3 1 1 5 c 2 × ×  + 2 4 2 9 7  

1 of 1000 grams 5

1 of an hour 6

Find. a

1 1 × 9 3

Perform the given calculations. a 2

b 2

Simplify. a

6 5

Answer the given questions. 2 a What is of 18? 9 3 c What is of 2 hours? 4 7 e What is of 5 dozen? 8

9 of 51 17

5 6 ×2 6 7

1 1 5 ×5× × 6 25 2

d 1

2  5 1 × 2 + 3  3  6 3

d 3

1 2 of 7 5

e 5

3 × 21 4

e 2

1 1 1 ×1 ×1 4 3 2

5 of a metre 6

16 9 of ? 27 2 2 d How many days are of 7 weeks? 7 2 f How many days are of a leap year? 3 b How much is

Word Problems 1 2

Ameena has 8 she use?

2 2 litres of oil. She uses only of it in making french fries. How much oil did 5 3

1 1 Ram has 2 litres of milk. He uses only of it and gives the remaining amount to his friend. 5 3 a How much milk did he use? b How much milk did his friend get? 3 pizzas. The total weight of a pizza was 800 g. What weight of 4 pizza did Simran and her friends eat? Simran and her friends ate 1

01_UM24CB0701V1.indd 8

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Pihu made 30 small cookies. She gave two-thirds of these to her sister and one-fifth of

In a library, there are 1500 books. Of these,

these to her neighbours. How many cookies did she distribute? remaining are English books. Find: a Number of English books

1 3 are science books, are novels and the 5 5

b Number of novels

c Number of science books

Dividing Fractions Remember, Riya gave half of her savings to her father and kept the rest in her bank account. Let’s see what fraction of money she kept in her bank account.

Division of a Fraction by a Whole Number

7 7 of her salary and kept half of the savings in her bank account. So, we will divide by 2 to 20 20 find the fraction of the amount kept in the bank account. 7 Let us find ÷2 20 Riya saved

Write the whole number as a fraction.

Write the reciprocal of the fraction formed.

7 7 2 7 1 7×1 7 ÷2= ÷ = × = = 20 20 1 20 2 20 × 2 40 Multiply the fractions to get the answer.

Reverse the '÷' symbol to '×'. Let’s see the Keep, Change and Flip method.

Thus, Riya kept Example 9

Divide 4 5

Keep

Change

Flip

7 20

1 2

Answer

7 40

7 of the savings in her bank account. 20

4 by 2. 5

Example 10

Keep

Change

Flip

4 5

1 2

Divide 2 2

Answer

Chapter 1 • Operations on Fractions

01_UM24CB0701V1.indd 9

7 20

4×1 = 4 =2 5 × 2 10 5

2 12 = 5 5

2 by 3. 5

12 12 3 12 1 12 ÷3= ÷ = × = 5 5 1 5 3 15 Reduce

12 4 = 15 5

28-12-2023 18:54:55

Do It Together

Divide 7

1 by 5. 2 7

1 = 2

÷5=

5 = 1

Error Alert!

Always multiply by the reciprocal while dividing a fraction.

Division of a Whole Number by a Fraction What if we want to divide 2 by half? How many halves do we get? 1 2÷ 2 Write the whole number as a fraction.

2÷

Do It Together

1 2 1 2 2 2×2 4 = ÷ = × = = 2 1 2 1 1 1×1 1 Multiply the fractions to get the answer.

3 Divide 12 by . 4 3 12 3 12 4 12 × 4 48 12 ÷ = ÷ = × = = =16 4 1 4 1 3 1×3 3

2 Divide 9 by 1 . 3 2 1 = 3 9 9 9÷ = ÷ = × 1 1

2 2 1 2 ÷3= × = 3 3 3 9

Write the reciprocal of the fraction formed.

Replace the '÷' symbol with '×'.

Example 11

2 6 ÷3= =2 3 3

Example 12

1 Divide 21 by 2 . 3 1 7 2 = 3 3 7 21 7 21 3 21 × 3 63 21 ÷ = ÷ = × = = =9 3 1 3 1 7 1×7 7

Division of a Fraction by Another Fraction Now, let us learn to divide a fraction by another fraction. 1 1 Divide by . 3 9

Example 13

Divide

1 1 by . 4 3

1 1 1 3 1×3 3 ÷ = × = = 4 3 4 1 4×1 4

1 3

1 9

Keep

Change

Flip

1 3

9 1

Example 14

Answer

Divide 3 3 3

1×9 9 = =3 3×1 3

2 1 by 2 . 5 3

2 17 1 7 = 2 = 5 5 3 3

2 1 17 7 17 3 17 × 3 51 16 ÷2 = ÷ = × = = =1 5 3 5 3 5 7 5 × 7 35 35

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Do It Together

Divide 5 5

2 = 3

2 5 by . 3 7 ÷

5 = 7

Do It Yourself 1D 1

Find the reciprocal of the given numbers. a

7 11

b 2

2 ÷ 6 = _____ 3

3 5

d 2

c 14 ÷

2 ÷ 66 = _____ 11

2 1 ÷ 5 4

67 4

b 2

5 2 ÷1 8 3

Answer the given questions. a What is 96 divided by the sum of

3 40

1 4

e 1

7 8

c 8

1 ÷ 4 = ____ 2

c 1

7 2 by 10 5

Compare using appropriate symbols. (>, <, =) a 3

b 18 ÷

Fill in the blanks. a

c 3

Divide and simplify. a 15 ÷ 35

5 9

d 2÷5

19 4

1 6

d 2

10 ÷ 1 = _____ 13

d 1

1 2 by 2 3 27

19 11

e 6÷9

2 3

5 7

12 8 and ? 8 12

4 2 b The product of two numbers is . If one of the numbers is , find the other. 5 3 3 c The perimeter of an equilateral triangle is m. What is the length of each side? 7

Word Problems 1

The area of a rectangular field is 43

A large can of juice contains 18

3 1 cm2. If its breadth is 12 cm, find its length. 4 2

1 litres of juice. If the entire juice has to be poured into 3 5 small containers, each of capacity 1 litres, then find the number of containers required. 6 5 Seema bought kg of fruits. She packed them equally into 15 packets. 9 a Find the weight of 1 packet of fruit in kg. b She sold 5 packets of fruits. How many kg of fruits did she sell?

Akshay has a piece of cloth that is 12 cm long. He cuts it into small pieces of length each. How many pieces of cloth will he have?

Chapter 1 • Operations on Fractions

01_UM24CB0701V1.indd 11

2 cm 3

28-12-2023 18:55:15

Multi-Step Problems on Operations on Fractions

1 5 of her salary the next month? She gave of the savings to her father for 3 11 investment and kept the remaining savings in her bank account. She earned ₹66,000 that month. Remember, Riya saved

Let us see what amount of money she kept in her bank account. 1 × ₹66,000 = ₹22,000 3 5 5 Savings given to father for investment = of the savings = × ₹22,000 = ₹10,000 11 11 Savings deposited in the bank account = Total savings – Savings given to father Total money saved =

= ₹22,000 − ₹10,000 = ₹12,000 Thus, Riya kept ₹12,000 in her bank account. Example 15

3 5 of the fruits are mangoes. The total number of mangoes is 48, and of the 5 8 remaining fruits are apples. How many apples are there in the market, and what fraction of the total In a fruit market,

fruits are apples?

3 Fraction of fruits that are mangoes = ; Total number of mangoes = 48 5 3 of total fruit = mangoes = 48 5 5 So, Total fruits = 48 × = 80 3 Remaining fruits = 80 – 48 = 32 5 Fraction of remaining fruits that are apples = 8 5 160 Total number of apples = × 32 = = 20 8 8 20 1 Fraction of fruits that are apples = = 80 4 Example 16

4 A man sells orange juice. A medium glass contains 350 mL, whereas the large glass is more than the 5 1 2 medium and the small glass is of the large glass. If three people drink 1 small, 1 medium and of a 3 9 large glass respectively, then find the total quantity of juice drunk by three of them. Quantity of a medium glass of juice = 350 mL 4 4 Quantity of a large glass of juice = 350 mL + of 350 mL = 350 mL + × 350 mL = 350 mL + 280 mL = 630 mL 5 5 1 1 Quantity of a small glass of juice = of large glass of juice = × 630 mL = 210 mL 3 3 2 2 Let us find of a large glass = × 630 mL = 140 mL 9 9 2 Total juice consumed by three people = 1 small glass + 1 medium glass + of large glass 9 = 210 mL + 350 mL + 140 mL = 700 mL

Do It Together

Thus, the total quantity of juice that three people drank is 700 mL. 6 5 Preeti baked a cake and shared with her neighbours and with 16 12 2 her friends. Later, she gave of what was left to her sister. What 5 fraction of the cake was left with Preeti at the end? 6 Cake shared with neighbours = 16 12

01_UM24CB0701V1.indd 12

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5 12 Total cake shared = _____ + _____ = _____ Cake shared with friends =

Fraction of cake remaining after sharing = 1 – _____ = _____ 2 Fraction of cake given to her sister = × _____ = _____ 5 Fraction of cake remaining after sharing with the sister = _____ – _____ = _____

Do It Yourself 1E 3 of the people at a restaurant are males. If the number of females is 28 more than males, then how many 7 females are there at the restaurant?

1 1 of all the shirts at a shop are printed, of the remaining shirts are striped and the rest of the shirts are plain 3 4 shirts. If there are 96 plain shirts, then how many shirts are there in total?

3 of all the players at an event are football players. If there are 72 players in all then how many players at the 8 event are non-football players?

Kevin has 25 m of ribbon. He gives 13 m of ribbon to his sister. He works on a project and uses 1

Rakul spends

A cistern has two holes. One leaks at the rate of

each project. How many projects can he complete using the available ribbon?

1 m of ribbon for 2

1 1 2 of her money on house rent, of the money on food and of the money on other expenses. She 5 4 5 is left with ₹3000. How much money did she have initially? 1 1 gallons per half an hour, whereas the other leaks at the rate of 4 5 gallons every quarter of an hour. In how many hours will the cistern be empty if it contains 500 gallons of water?

Points to Remember a The numbers of the form , where a and b are whole numbers and b = 0 are called fractions. b • To add/subtract unlike fractions, find the LCM of the denominators, convert them into like fractions and then perform addition/subtraction. •

• To multiply one fraction with another, multiply the numerators, multiply the denominators and then write the answer in the lowest form. •

When a fraction is used with the word “Of” it acts as an operator. ”Of“ means multiplication.

•

To divide two or more fractions, multiply the first fraction with the reciprocal of the other fraction.

• A reciprocal of a fraction can be obtained by interchanging the numerator and the denominator. Two fractions are said to be reciprocals of each other if their product is 1.

Chapter 1 • Operations on Fractions

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Math Lab Fraction Collage Addition Aim: Adding unlike fractions using hands-on visual representation. Materials Required: Coloured construction paper, pen, paper, scissors, glue stick Setting: In pairs Method:

1 1 Instruct students to draw and label fractions with various colours. Example: in red, 4 1 in blue etc. 3 2 Students need to carefully cut out the coloured fractional parts. 3 Students need to combine the cut-outs from different colours to create new fractions. For 1 1 7 example, adding (red) and (blue) to get (using coloured pencils for clarity). 4 3 12 4 Instruct students to glue their fraction cutouts representing the addition onto a separate piece of paper, visually representing the addition of unlike fractions.

Chapter Checkup 1

Write true or false. a Any fraction divided by 0 is not defined.

______________________

b The associative law is applicable to the division of fractions.

______________________

c The reciprocal of 1 is 1.

______________________

d The commutative law is applicable to the subtraction of fractions.

______________________

Add the given unlike fractions. a

7 11 + 9 17

5 2 + 13 39

1 5 + 9 6

14 25 + 6 4

12 64 + 7 14

Find the difference of the given mixed numbers and write the answer in its simplest form. a 11

1 1 −2 4 6

b 6

4 1 −2 5 7

c 27

10 7 −2 11 9

d 1

6 1 −1 13 54

Fill in the blanks.

5 and its reciprocal is ______. 6 b A fraction is said to be in its lowest form when the ____________ of the numerator and denominator is 1. a The product of

1 is the same as multiplying by ______. 2 d The reciprocal of proper fractions is always greater than ___________. c Dividing by

Find the reciprocal of the fractions. 1 a 5 b 4

c 101

15 17

18 5

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Simplify. a

1 3 × 2 5

b 3

2 1 ÷ 9 2

Complete the following.

1 of two hours = ________ minutes 9 1 c 5 of a year = ________ days 5

Solve. a 1

2 4 1 +3 +4 3 5 2

b 23

19 5 − 17 23 11

c 2

13 1 +4 15 5

d 2

15 2 − +1 17 34

2 of a day = ________ hours 3 3 d of a metre = ________ cm 4 b

c 2

6 1 5 × × 7 3 7

d 5

3 2 2 ÷4 ÷1 4 3 3

Answer the given questions.

1 1 1 1 and 2 or quotient of 2 and 4 ? 5 4 3 6 4 1 4 b Which is less: The product of 25 and or the sum of 3 and 5 ? 75 4 5 2 2 c What is the difference of the quotient of 4 and and the product of and 15? 7 5 1 1 1 d What is the sum of the product of 9 and 1 and the sum of 9 and 7 ? 2 2 3 a Which is more: The product of 3

10 Solve the given questions. a A train covers a distance of 48

2 of x = 26. 13 2 Find the value of x, if 22 × x = 1800 9 3 2 The product of two numbers is . If one of the numbers is , find the other. 5 3 6 of a number is 180. Find the number. 7 3 47 The product of two numbers is 21 . If one number is 7 , find the other. 5 55

b Find the value of x, if c d e f

3 1 miles in 1 hours. How far does it go in 2 hours? 4 4

Word Problems 3 2 of a cake. She ate of it. 5 3 a Find what fraction of cake she ate. b What fraction of cake is left with her? 3 2 A planet completes 1 turns about its axis each day. 5 a How many turns will it complete in a week? 1 Miya has

b How many turns will it complete in one fortnight? 4 7 full. Find how much water is 3 A tank is full. When 65 litres of water are drawn from it, it is 5 12 left in the tank. 1 1 4 Piyush walks at a speed of 4 km per hour for 5 hours and then at the speed of 3 km per 5 4 hour for the next 3 hours. Find the distance travelled by Piyush in 8 hours.

Chapter 1 • Operations on Fractions

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1 2 of a distance by train, by bus and walks for the remaining distance. If she 3 5 walks for 16 km, then find:

5 Himani travels

a the distance travelled by bus.

b the distance travelled by train.

c the total distance. group of students decided to organise a charity event to raise funds for their local animal 6 A shelter. They planned to sell custom-made bracelets and donate a fraction of the proceeds to the shelter. Here is how their plan unfolds: They created 3 different types of bracelets: A, B and C. • • •

2 of the proceeds to the shelter. 5 3 For every 3 bracelets of type B sold, they would donate of the proceeds to the shelter. 7 5 For every 4 bracelets of type C sold, they would donate of the proceeds to the shelter. 8 During the event: For every 5 bracelets of type A sold, they would donate

a They sold 35 bracelets of type A for ₹70 each. b They sold 21 bracelets of type B for ₹90 each. c They sold 28 bracelets of type C for ₹120 each. Calculate the total amount donated to the animal shelter from the proceeds of each type of bracelet sale.

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Operations

2 on Decimals Let’s Recall Decimals are the numbers which consist of two parts, namely a whole number part and a fractional part separated by a decimal point.

Hundreds

Tens

Ones

PART

• decimal point

WHOLE

Tenths

Hundredths Thousandths

While multiplying a decimal number with 10, 100, 1000 etc., we move the decimal point as many places to the right as there are zeroes in the multiplier. For example:

On multiplication with 10

On multiplication with 100

23.389 � 10 = 2 3.3 8 9 = 2 3 3.8 9

23.389 � 100 = 2 3.3 8 9 = 2 3 3 8.9

On multiplication with 1000 23.389 � 1000 = 2 3.3 8 9 = 2 3 3 8 9.0 Similarly, while dividing a decimal number by 10, 100, 1000 etc., we move the decimal point as many places to the left as there are zeroes in the divisor. For example:

On dividing by 10

On dividing by 100

85.3 � 10 = 8 5.3 = 8.5 3

85.3 � 100 = 8 5.3 = 0.8 5 3

On dividing by 1000 85.3 � 1000 =

8 5.3 = 0.0 8 5 3

Let’s Warm-up Fill in the blanks. 1

3 51.254 × 100 = _______

4 5.123 × 1000 = _______

1 24.12 ÷ 10 = _______

3 9.145 ÷ 100 = _______

5 87.2 ÷ 1000 = _______ I scored _________ out of 5.

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Mean, Median and Mode of Decimals Addition and Subtraction Real Life Connect

Suhani and her mother visited the market to buy fruit and vegetables. The shopkeeper weighed the purchase using a digital scale. They purchased 5.755 kg of vegetables and 7.35 kg of fruits.

Adding and Subtracting Decimals Let us see the total weight of the items purchased by Suhani and her mother. To find the total weight, we need to add. Here, the weights are unlike decimals. Like decimals are decimals with the same number of decimal places after the decimal point. For example, 25.36 and 52.65 are like decimals. Unlike decimals are decimals with a different number of decimal places after the decimal point. For example, 52.45 and 65.124 are unlike decimals.

Error Alert! The misalignment of decimal points leads to an incorrect answer. 23.53 + 1.6 39.53

Let us see how to add decimals that are unlike. Step 1: Convert the decimals into like decimals. Here, 5.755 and 7.35 as 7.350.

Step 3: Add as whole numbers are added. T O .

t h th

5 + 7

7 5 5 3 5 0

1 3

. .

1 0 5

23.53 + 1.6 25.13

Step 2: Write the digits in the column method aligning the decimal points and the digits. T O . 5 . + 7 .

t h th 7 5 5 3 5 0

Step 4: In the sum, put the decimal point directly below the other decimal points. T O .

t h th

5 + 7

7 5 5 3 5 0

1 3

. . .

1 0 5

We can use the same method to subtract decimals. Let us now find the difference of the weights of the fruits and vegetables purchased. Step 1: Convert the decimals into like decimals. Here, 5.755 and 7.35 as 7.350.

Step 2: Write the digits in the column method aligning the decimal points and the digits. T O . 7 . – 5 .

t h th 3 5 0 7 5 5

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Step 4: In the difference, put the decimal point directly below the other decimal points.

Step 3: Subtract as whole numbers are subtracted. O .

t h th

12 14 10

5 9 5

7 – 5

7 – 5 1

Example 1

. .

3 5 0 7 5 5

What is the sum of 125.23, 213.523, 148.58 and 326.125? H T O . 1

1 2 1 + 3

2 1 4 2

5 3 8 6

8 1 3

Example 3

. . . . .

Example 2

3 2 8 2

. . .

h th

12 14 10

3 5 0 7 5 5 5 9 5

Subtract 235.28 from 365. H T O

t h th

2 5 5 1

0 3 0 5

–

3 6 5 2 3 5 1 2 9

. . . .

0 0 2 8 7 2

4 5 8

The length of a rope is 102 m 80 cm. If a piece measuring 59.35 metres is cut from it, what is the length of the remaining rope? Length of rope = 102 m 80 cm As the measure is not in decimal form, we can write it as a decimal by converting to the higher unit as 102.8 m. Length of rope cut = 59.35 m Length of remaining rope = 102.80 – 59.35 = 43.45 m

Do It Together

Add the sum of 367.25, 125.2 and 512 to the difference of 563.24 and 321. Sum = 367.25 + 125.2 + 512 = _____ .... (1) Difference = 563.24 − 321 = ______ ...... (2)

H 3 1 + 5

T 6 2 1

O 7 5 2

. . . .

t 2 2 0

h 5 0 0

H T O . 5 6 3 . – 3 2 1 .

t h 2 4 0 0

Adding (1) and (2) gives ____ + _____ = _____

Do It Yourself 2A 1

Add. a 223.25 and 12.9

b 315.36 and 218.234

c 128.2 and 236.54

d 332, 127.456 and 122.26

e 136.23, 556.14 and 25.125

f 152.214, 235.3 and 365.28

a 322.36 from 438.2

b 132.89 from 325.566

c 118.5 from 732.55

d 438.236 from 752.23

e 856.18 from 998.856

f 1065.235 from 1189.15

Subtract.

Chapter 2 • Operations on Decimals

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Fill in the blanks. a 236.45 + ____ = 568.2

b 514.7 − _____ = 241.23

c ___ + 321 = 512.47

d ____ − 475 = 251.487

e 785.5 + 324.21 = _____

f 987 – 562.26 = _____

What should be added to 389.265 to get 625.56?

By how much should 235.125 be decreased to get 126.24?

Simplify: 126.256 + 325.46 – 195.264 + 326.152 – 256.369

Subtract the difference of 469.251 and 829.23 from the sum of 356.21 and 585.236.

A recipe requires 3.75 cups of flour, 2.5 cups of sugar, and 1.25 cups of butter. How much of these ingredients are

The distance travelled (in km) by Mohit in a week was recorded on a bar graph. 86.00

84.00

84.5

82.75

82.00

81.5

80.5

80.00

79.75

78.25

78.00

76.55

76.00 74.00

ay nd Su

da tur

y da Fri

sd ur Th

da es dn We

da es Tu

nd Mo

72.00 ay

Distance Travelled in km

needed in total for the recipe?

Day

a What is the total distance travelled by Mohit in the week? b What is the difference in the distance travelled on the last 3 days to the initial 4 days of the week?

Word Problems 1

Ravi goes to buy a big box of sweets. He purchases 10.5 kg of sweets, and the weight of

Vineet purchased 2.5 kg of coffee from the market. He used 1 kg 75 g of it. What weight of

A delivery truck had 56.75 gallons of fuel in the morning. After a long trip, it had

Swati bought 7 kg 450 g of sugar, 2.35 kg of rice

the empty box is 1 kg 350 g. What is the total weight of the sweets and of the box? coffee did he have left?

32.5 gallons left. How much fuel did the truck use during the journey?

altogether? 5

The average temperature in Amsterdam for

5 months is given in the line graph. What is the difference of the temperatures of the hottest and coldest months?

Temperature in ℃

and 5 kg 700 g of flour. How much did she buy

17.6

20 15 10 5 0

9.7

20.2

12.8

5.5

Feb

March

April

May

June

Month

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Multi-step Word Problems on Adding and Subtracting Decimals Suhani and her mother spent ₹596.75 on purchasing vegetables and ₹895.45 on purchasing fruit. They also made miscellaneous purchases of ₹1236. If they gave ₹3000 to the shopkeeper, how much change did they receive? Th H T O . 2

5 1 8 + 1 2 2 7

9 9 3 2

6 5 6 8

. . . .

t h 1

7 4 0 2

5 5 0 0

Th H T O . 2

3 0 0 0 – 2 7 2 8 0 2 7 1

. . .

t 10

0 2 8

Amount spent on vegetables = ₹596.75 Amount spent on fruit = ₹895.45

Amount spent on miscellaneous purchases = ₹1236

Total amount spent = 596.75 + 895.45 + 1236 = ₹2728.2

Change received from the shopkeeper = 3000 – 2728.2 = ₹271.80 Example 4

A drum has 59.55 kg of rice in it. A pack of 25.5 kg of rice is emptied into the drum. After this, 312 kg of 5 rice is used. What is the weight of the rice left in the drum? T O . 1

5 9 + 2 5 8 5

. . .

t h 5 5 5 0 0 5

T O . 4

8 5 – 3 1 5 3

. . .

t h 10

0 5 4 0 6 5

Quantity of rice in the drum = 59.55 kg; Quantity of rice in the packet = 25.5 kg Total quantity of rice after emptying the packet = 59.55 + 25.5 = 85.05 kg Quantity of rice used = 312 = 157 = 31.4 kg 5 5 Quantity of rice left in the drum = 85.05 – 31.4 = 53.65 kg Example 5

Mayra took 7 hours to finish her maths homework, 1.2 hours to complete her science homework and 4 7 spent hours less time on her social studies homework compared to the time she spent on the maths 10 homework. What was the total amount of time she spent completing her homework that day? Time to finish maths homework 7 hours = 1.75 hours 4 Time to finish science homework = 1.2 hours

Time to finish social studies homework = 7 = 0.7 hours less than maths homework = 1.75 – 0.7 = 1.05 hours 10 Total time to finish the homework = 1.75 + 1.2 + 1.05 = 4 hours Do It Together

Sushant had 182 cups of sugar. He used 2.25 cups to make milkshake, 63 cups for making a dessert 5 4 and 1.45 cups for baking a cake. How many cups of sugar was left? Total cups of sugar = 182 = 18.4 cups 5 3 Sugar used = 2.25 + 6 + 1.45 = 2.25 + _________ + 1.45 = _________ cups. 4 Sugar left = 18.4 − _________ = _________ cups.

Chapter 2 • Operations on Decimals

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Do It Yourself 2B 1

A palm tree is 42.68 feet tall, whereas a mango tree is 15.23 feet shorter than the palm tree. What is the total

Ronita pays ₹635.80 every month for cable and internet. She came to know about a new plan by another company

height of both trees?

that offered ₹452.75 for internet plus an additional ₹126.55 for cable. If Ronita switches to the plan, how much will she save every month?

Sumit downloaded three applications on his mobile that were 986.48 KB in total. If one of the apps was 256.36 KB

Rhea donated 35.65 kg of sugar, 78 kg 540 g of flour, 17 kg 320 g of rice and some pulses to an NGO. If the total

Sam saved ₹3652.45 in June and ₹4578.58 in July. He then purchased a bicycle worth ₹2468.75 and a helmet worth

There is 2 L 125 mL of juice in jug A, whereas there is 1 L 350 mL juice in jug B. If both jugs together can hold

The amounts deposited and withdrawn by Manish in the initial four months of the year are shown below. What is

and the other was 485.84 KB, how big was the third application?

weight of the items donated was 142.63 kg, what was the weight of the pulses donated?

₹689.48. How much money was he left with at the end of July?

4.75 L of juice, how much more juice can be poured into both the jugs?

the difference in the amounts deposited and withdrawn in the four months?

Amount in ₹

12000 10000 8000 6000

10512.25

5423.58

4000

9684.5

10365

6842.75

3754.36

4365.45

February

March

April

8523

2000 0

January

Month Money deposited Money withdrawn

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Word Problems 1

Rita went from place X to place Y and then to place Z. X is 14.05 km from Y and Y is 12.7 km

from Z. Aman went from place X to place A and then to place Z. A is 29.3 km from X and Z is 21.45 km from A. Who travelled farther and by how much?

Sam ordered a coffee and a waffle. Find the change he will receive if he paid $10.

Coffee: $1.46

Garlic bread: $3.95 Ice tea: $2.89 Waffle: $0.85

Multiplication and Division of Decimals Real Life Connect

Naman and his father are patiently waiting for their turn at the petrol pump. Naman notices the displayed prices of various fuels on the board. Naman: Dad, how much fuel are we putting in the car today? Father: We're going to fill our car with 35.5 litres of diesel. Naman's curiosity is sparked as he begins to calculate the amount they will need to pay at the pump.

Petrol – 101.94 Diesel – 87.89 LPG –

77.85

CNG –

81.5

Multiplying and Dividing Decimals Let us find the amount paid by Naman’s father. To find the amount paid by Naman’s father, we need to multiply. To multiply two decimal numbers, we follow the given steps. Chapter 2 • Operations on Decimals

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Step 1: Multiply the decimal numbers ignoring the decimal points, just like whole numbers.

4 4 3 + 2 6 3 3 1 2

Step 2: Count the total number of decimal places and place the decimal point in the product accordingly. 87.89 × 35.5 = ₹3120.095

Hence, Naman's father paid ₹3120.095 for the fuel.

8 � 3 9 6 0

7 3 9 4 7 0

8 5 4 5 0 9

9 5 5 0 0 5

Remember! The product should have the same number of decimals place as the sum of the number of decimal places in both the multiplicand and the multiplier.

Naman saw that one of the auto rickshaw drivers paid ₹1018.75 for filling CNG. He started wondering how much CNG this driver must have put into his autorickshaw tank. Let us find out! Step 1: Make the divisor a natural number by multiplying the divisor and the dividend by a power of 10. 81.5 × 10

815

1018.75 × 10

10187.5

Step 2: Place the decimal point directly above the decimal point in the dividend. Step 3: Divide the whole number part and then the tenth part.

Hence, the auto-rickshaw driver filled the tank with 12.5 litres of CNG.

Example 6

Multiply. 1 457.2 and 132

4 � 9 1 3 7 4 5 7 6 0 3

5 1 1 1 2 5

7 3 4 6 0 0

2 2 4 0 0 4

457.2 × 132 = 60350.4 Example 7

2 102.925 and 4.28 1 0 2 � 8 2 3 2 0 5 8 + 4 1 1 7 0 4 4 0 5 1

9 4 4 5 0 9

2 2 0 0 0 0

5 8 0 0 0 0

102.925 × 4.28 = 440.51900

Divide. 1 3259.36 by 26 125.36 26 3259.36 – 26 65 – 52 139 – 130 93 – 78 156 – 156 0

Do It Together

12.5 815 10187.5 – 815 2037 – 1630 4075 – 4075 0

2 0.888 by 1.11

1.11 × 100 = 111; 0.888 × 100 = 88.8 0.8 111 88.8 – 00 888 – 888 0

Think and Tell Do we get a natural number on dividing 2 decimal numbers?

Find the missing dimension of the given rectangle with area 143.75 cm2. Area of rectangle = 143.75 cm2; Breadth of rectangle = 11.5 cm Area = _________ = _________ Breadth Hence, the length of the rectangle is _________ cm. Length =

11.5 cm ?

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Do It Yourself 2C 1

Find the product. a 0.12 × 15

b 112.2 × 6.5

c 32.1 × 1.33

d 5.014 × 25

e 52.25 × 5.23

f 122.56 × 5.21

g 322.15 × 2.24

h 21.16 × 12.83

If 365 × 124 = 45,260, then find the following without actual multiplication. a 36.5 × 124

b 3.65 × 12.4

c 36.5 × 1.24

d 3.65 × 1.24

a 0.165 ÷ 1.5

b 42.7 ÷ 14

c 49.08 ÷ 0.02

d 88.8 ÷ 2.22

e 371.68 ÷ 16

f 164.84 ÷ 13

g 23.85 ÷ 1.8

h 487.36 ÷ 3.2

k 2039.96 ÷ 5.2

Find the quotient.

i 4

1117.5 ÷ 12.5

115.542 ÷ 4.9

4756.05 ÷ 58.5

Fill in the blanks. a 68.527 × _____ = 23.63

b _____ × 1.6 = 52

c 2177.35 ÷ _____ = 62.21

d 74.2 × 12.3 = _____

e 984.576 ÷ _____ = 25.6

f 616.54 ÷ 14.5 = _____

The product of two decimals is 3538.29. If one of them is 41.5, find the other.

Find the area and perimeter of the given square. 124.75 cm

3 Find the cost of 45 litres of milk at the rate of ₹54.5 per litre. 4

A cloth of length 2.35 m is needed to make a shirt. How many metres of cloth are needed to make 48 such shirts?

A ribbon 30 m 80 cm long is to be divided into 11 equal pieces. Find the length of each piece.

10 A car moving at a constant speed covers a distance of 452.8 km in 8 hours. Find the distance covered by the car in 4 2 hours. 5

Word Problems 1

Ravi went to the grocery store and bought 12 kg 450 g of rice at a cost of ₹52.75 per

If 2 m 25 cm of cloth is required to stitch a pair of pants, how many pairs of pants can be

A grocery store owner bought some bags of rice each weighing 97.5 kg. If the total weight

Mohan has a rectangular plot of area 875.25 square metres. He wishes to divide it equally

kilogram. What is the total cost of the rice bought by Ravi? stitched with a piece of cloth that is 20.25 m long?

of all the bags is 682.5 kg, find the number of bags of rice bought by the owner. among his 1 son and 3 daughters. How much of the plot will each receive?

Chapter 2 • Operations on Decimals

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Multi-step Word Problems on Operations on Decimals Naman’s father purchased a computer system for ₹60,225 with an additional tax of ₹10,840.5. He made a payment of ₹12,000 and paid the rest of the amount in six equal installments. What is the amount paid by Naman’s father for each installment. Total cost of the computer system = 60,225 + 10,840.5 = ₹71,065.5 Balance amount after paying ₹12,000 = ₹71,065.5 − ₹12,000 = ₹59,065.5 59065.5 Amount paid in each installment = = ₹9844.25 6 Hence Naman’s father paid ₹9844.25 in each installment. Example 8

A fruit vendor is selling oranges at ₹35.50 per kg and apples at ₹86 per kg. If Meera buys 21 kg of oranges and 1.75 kg of 4 apples, what will be the total cost she needs to pay?

Did You Know? The decimal system of currency in India was introduced on 1 April 1957.

21 kg oranges = 2 kg + 0.25 kg = 2.25 kg 4 Cost of 2.25 kg oranges at ₹35.50 per kg = 2.25 × 35.5 = ₹79.875 Cost of 1.75 kg apples at ₹86 per kg = 1.75 × 86 = ₹150.5 Total amount paid = 79.875 + 150.5 = ₹230.375 Example 9

Fifteen individuals visited a restaurant. Each one of them ordered the items shown in the table. While paying the total bill, they found that five members of the group forgot to bring money. To cover the bill, how much additional money did each of the remaining individuals need to contribute? Number of individuals = 15; Cost of meal per person = 25.45 + 34.95 + 70.25 + 15 = ₹145.65 Total bill = 15 × 145.65 = ₹2184.75 As 5 people forgot to bring money, the amount would now be paid by 10 people. Amount paid by each individual = 2184.75 ÷ 10 = ₹218.475

Item

Price

Idli

₹25.45

Vada

₹34.95

Dosa

₹70.25

Tea

₹15

Additional amount to be paid by each of the remaining individuals = 218.475 – 145.65 = ₹72.825 Do It Together

During a visit to a history museum, a group comprising 12 adults and 5 children purchased tickets. With each adult ticket priced at ₹125.5 and each child's ticket priced at ₹75.5, what is the overall difference in the total cost of the tickets for the adults and children combined? Price of 12 adult tickets at ₹125.5 per ticket = 12 × 125.5 =_____ Price of 5 child tickets at ₹75.5 per ticket = 5 × 75.5 = _____ Difference in the cost = _____ − _____ = ₹_______

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Do It Yourself 2D 1

Suhani bought 7 kg 650 g of sugar at ₹38.5 per kg, 2 kg 485 g of rice at ₹62.75 per kg and 5 kg 725 g of flour at

1 Kim bought 25.5 litres of milk. He fills 8 bottles equally with the milk, with 1 litres of milk left. Find the volume of 2 milk in each bottle.

Kunal has a rectangular garden measuring 5.8 m by 7.25 m. What is the cost of fencing the garden if the cost of

Mayra bought 2 pairs of shoes for ₹525.75 a pair, three T-shirts for ₹263.50 each, five pairs of socks for ₹50.25 a

Sunish bought a music system for ₹25,526.75. He made a down payment of ₹10,250 and paid the rest in 15 equal

Sam used some cloth to make 8 banners and 3 tablecloths. He used 2 m 75 cm of cloth for each banner and

₹45 per kg. What is the total amount paid by Suhani?

fencing is ₹58.75 per m?

pair and one bag for ₹755. What was the total cost of the items purchased? installments. What is the amount paid in each installment?

1.55 m of cloth for each tablecloth. Find the total length of cloth used. Also, find the cost of the cloth used if the cost of 1 m of cloth is ₹75.5.

Sharvil purchased 25 notebooks at ₹45.65 each, 3 boxes of pens at ₹520.75 each and 2 boxes of markers at

1 Suman bought 7 kg of tomatoes for ₹250.25. How much will Seema pay for 4 kg of tomatoes? 2

Saumya is making a report on the computer. She wants to place a figure which is 12.75 cm wide so that it appears

₹368.15 each. If the shopkeeper gave ₹560.2 as change, what amount was given to the shopkeeper by Sharvil?

centred on the page. The width of the page is 22.63 cm. How much margin should she leave at both ends so that the figure is centred?

Word Problem 1

Krishvi bought 3 pairs of gloves for ₹234.75 and two pairs of socks for ₹165. How many equal pairs of gloves and socks can be bought for ₹1286?

Points to Remember • Decimals can be added or subtracted by converting them to like decimals and aligning the decimal points in the column method. • To multiply decimal numbers multiply as whole numbers are multiplied count the total number of decimal places place the decimal point at the same number of decimal places in the product. • To divide decimal numbers convert the divisor to a whole number place the decimal point just above the dividend divide as whole numbers are divided.

Chapter 2 • Operations on Decimals

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Math Lab Setting: In groups of 5 Materials Required: Decimal domino cards (create or print cards with a decimal equation on one card and its solution on another card), Timer Method: 1 Distribute an equal number of decimal domino cards (both equation and solution) to each group. 2 Start the timer and ask the groups to match the equations with their corresponding solutions by placing the cards next to each other in a domino-like chain.

3 The team with the highest number of correct solution matches in the least time will win the game!

Chapter Checkup 1

Add. a 122.56 + 36.2

b 321.4 + 12.63

c 16.523 + 132.29

d 225.213 + 88.9

e 536.3 + 311.236

f 524.8 + 333.69

g 525.236 + 226.21

h 756.59 + 413.5 + 623.12

a 132.56 − 45.2

b 225.3 – 152.35

c 267.523 – 159.47

d 347.14 − 178.5

e 336.8 – 285.123

f 586.4 – 387.25

g 658.412 – 247.48

h 745.14 − 398.475

a 212.5 and 35

b 110.52 and 21.5

c 156.52 and 12.29

d 315.13 and 81.2

e 136.3 and 125.2

f 624.82 and 32.55

g 312.1 and 15.24

h 362.12 and 213.5

a 304.8 by 12

b 23.94 by 6.3

c 90.15 by 1.5

d 498.8 by 4

e 100.32 by 80

f 42.7 by 14

g 3.024 by 0.36

h 564.975 by 15.5

Subtract.

Multiply.

Divide.

Aarav scored 452.65 marks out of 600 in the final examination. How many marks did he lose?

There is 0.625 kg of powdered milk in each tin. If a carton contains 12 tins, find the total mass of powdered milk in

Sack A contains 125.35 kg of rice, sack B contains 25.65 kg more rice than sack A and sack C contains 13.49 kg less

The daily consumption of milk in a house is 3.25 litres. How much milk will be consumed in a year?

A rectangular garden measures 13.52 metres in width and 17.36 metres in length. What is the area and perimeter

the carton.

rice than sack B. What is the total weight of the rice in all three sacks?

of the garden?

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10 Considering an average growth rate of 1 cm 8 mm per month, what would be the total length of a strand of hair after three years of growth?

1 11 Kavya earns ₹523.65 per hour for the first 40 hours she works in a week. She earns 1 times that amount per 2 1 hour for each hour beyond 40 hours in a week. How much will she earn if she worked for 52 hours? 2 12 Abhinav went to a restaurant with his family. The family ordered 3 vegetable bowls, 2 apple pies and 4 milkshakes. a How much did the family spend at the restaurant? b How much change will they receive if they paid an amount of ₹1000?

Item

Rate

Vegetable bowl

₹155.26

Apple pie

₹62.37

Milkshake

₹89.49

Word Problems he watermelon bought by Rani is 3 times as heavy as the papaya bought by Manya. If the 1 T watermelon bought by Rani weighs 4.2 kg, what is the weight of the papaya? he weight of a baby elephant was 118.99 kg. After two years, its weight increased 2 T by 109.85 kg. Find the weight of the baby elephant after two years. avi bought 8.6 kg of sugar. He poured the sugar equally into 5 bottles. There was 0.35 kg of 3 R sugar left over. What was the mass of sugar in 1 bottle? 4 The graph shows the amount saved by Monika during 5 weeks.

Scale: 1 division = ₹200

1800 Amount Saved in ₹

1600 1400 1200 1000

1058.25

1254

1365.78

1563.75

1026.65

800 600 400 200 0

Week 1

Week 2

Week 3 Week

Week 4

Week 5

a What is the total amount saved during the five weeks? b What is the difference of the lowest and highest amounts saved? uchi bought 1 dozen cupcakes for ₹519.60. Her mother asked her to buy 5 more cupcakes. 5 R How much did Ruchi pay for the 5 cupcakes? ohit's family went on a road trip. They got to point A on the first day and point B on the second 6 M day. How many more kilometres do they need to drive to reach their destination? If their car travels 11.5 km per litre, how many litres of fuel will they need for the trip? Car

A 150.25 km

B 175.50 km

450.75 km

Chapter 2 • Operations on Decimals

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3 Integers Let’s Recall Integers are a collection of whole numbers and negative numbers. Integers do not include the fractional part, similar to whole numbers. Negative integers

–9

–8

–7

–6

–5

–4

–3

Positive integers

–2

–1

Zero (Origin)

For example, temperatures above 0° are considered positive temperatures and temperature below 0° are considered negative temperatures. We can compare integers using the rules as given below: Rule 1:

Rule 2:

Rule 3:

Positive integer > Negative integer

Positive integer with greater numerical value > Positive integer with smaller numerical value

Negative integer with smaller numerical value > Negative integer with greater numerical value

Example: 45 > 26

Example: –25 > –52

Example: 25 > – 45

We can also arrange integers in ascending or descending order, using the above comparison rules as: Smaller to greater

–9

–8

–7

–6

–5

–4

–3

–2

–1

Greater to smaller

For example, Ascending order: – 5 < –2 < 1 < 5 < 9 Descending order: 8 > 5 > 2 > –4 > –10

Let’s Warm-up Fill in the blanks with the > or < sign. 1

25 _____ 89

–32 _____ –16

15 _____ 8 _____ 0 _____ –12 _____ –36

–14 _____ –9 _____ –1 _____ 13 _____ 20

–12 _____ 5

I scored _________ out of 5.

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Addition and Subtraction of Integers Real Life Connect

Sara writes down the money she receives and spends every week in a notebook. Sara: Mom, I’ve written down the money you and dad gave me this week. I’ve also written down the amount I have spent from it. Money Received

Money Spent

From mom : `90

On notebooks : `60

From dad : `70

On crayons : `50

Mom: Great Sara! How much have you saved this week? Sara quickly calculates her savings and tells her mom.

Addition of Integers We have studied the addition of integers with like and unlike signs in our previous class. Let us recall it. Rule 1: For addition of integers with like signs: Step 1 Add the absolute values of the integers. Step 2 Place the common sign before the sum. In the case of Sara, the total money received and spent can be given as:

Remember! The absolute value (shown by symbol ||) of an integer is the numerical value of the integer regardless of the sign.

|+90| + |+70| = 90 + 70 = ₹160

Think and Tell

|–60|+|–50|= 60 + 50 = ₹110

Why have we not put a

As both the integers have the (–) sign,

(+) sign before 160?

(–60) + (–50) = (–110)

Rule 2: For addition of integers with unlike signs: Step 1 Subtract the smaller absolute value from the greater absolute value.

Remember! Money received is shown by a (+) sign. Money spent is shown by a (–) sign.

Step 2 Place the sign of the integer with the greater absolute value before the difference. The money left could be given as (160) + (–110) or, |+160| – |–110| = 160 – 110 = ₹50 (as the integer with the greater absolute value has a positive sign)

Chapter 3 • Integers

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Properties of Addition of Integers Closure Property If a and b are two integers, then a + b = c, where c will always be an integer. Commutative Property For any two integers a and b, a + b = b + a Associative Property For any three integers a, b and c, (a + b) + c = a + (b + c) Existence of additive identity For any integer a, a + 0 = 0 + a = a Existence of additive inverse For any integer a, another integer –a exists so that: a + (–a) = 0

Example 1

Rohan opened a bank account by depositing `1500 in his account in June 2021. He deposited `750 in July 2021 and withdrew `1100 in August 2021. Find his balance. Amount deposited in June = `1500 Amount deposited in July = `750 Total amount deposited in June and July = `1500 + `750 = `2250 Amount withdrawn = `1100 = –1100 (as an amount withdrawn is denoted by a negative sign) Balance = 2250 + (–1100) = 2250 – 1100 = ₹1150 The balance is ₹1150.

Do It Together

The temperature of water in a bowl is 85℃. It dropped by 33℃ after 20 minutes. The temperature dropped by a further 13℃ in the next 10 minutes. What is the temperature of the water after 30 minutes? Temperature of water in the bowl = 85℃ Temperature drop in 20 minutes = _________ Temperature drop in the next 10 minutes = –13℃ Total temperature drop = (–33) + (–13) = _________ Temperature after 30 minutes = 85℃ + _________ = _________

Did You Know? Indian mathematicians, Aryabhata and Brahmagupta, made significant contributions to the understanding and calculation of integers.

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Subtraction of Integers We have recalled the addition of integers. Let us now recall the subtraction of two integers. If a and b are two integers, then a – b = a + (–b), i.e. to subtract an integer b from another integer a, add the additive inverse of b to a and keep the sign of the integer with the greater absolute value. Let us subtract (–12) from (–15). Here, a = (–15); b = (–12) (–15) – (–12) = (–15) + (additive inverse of (–12)) = (–15) + (12) = (–3)

Remember! Subtracting a negative number is equivalent to adding a positive number.

Properties of Subtraction of Integers Closure Property If a and b are two integers, then a – b = c, where c will always be an integer. Commutative Property For any two integers a and b, a – b ≠ b – a Associative Property For any three integers a, b and c, (a – b) – c ≠ a – (b – c) Subtraction Property of Zero It states that subtracting zero from any integer leaves the integer unchanged. In general, a – 0 = a (where a is an integer).

Example 2

The record high temperature of Canada is 2℃ and the record low is –18℃. What is the difference in the high and low temperatures? Record high = 2℃ Record low = –18℃. Difference in temperature = 2 – (–18) = 2 + (18) = 20℃ The difference in the high and low temperatures is 20℃.

Do It Together

Mt. Everest, the highest peak in Asia, is 29,029 feet above sea level. The Assal lake in Africa is 510 feet below sea level. What is the difference in sea level between these two places? Height of Mt. Everest above sea level: 29,029 feet Depth of Assal lake below sea level = ______ Difference in depth = 29,029 – (–510) = ______

Chapter 3 • Integers

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Do It Yourself 3A 1

Add. a 5 and –9

b 15 and 14

c –28 and 59

d (–23) and 12

e 32 and –122

f –136 and –25

g –152 and 365

h 158 and –125

a –7 + 3

b 12 + (–5)

c –19 + (–14)

d 20 + (–30)

e –69 + 81

f (–123) + (–235)

g 565 + (–125)

h (–485) + (–325)

a 8 from –9

b –12 from 17

c 25 from –32

d 38 from –20

e –36 from –98

f 54 from 223

g 136 from –182

h –214 from 156

Solve the given problems.

Subtract.

Determine the missing numbers in the equations. a 12 + _____ = 18

b _____ – (–17) = 10

c 956 – _____ = 422

d 215 – 136 = _____

e –548 + _____ = –267

f _____ – 192 = 564

g 705 + _____ = 1202

h 815 – (–125) = _____

Fill in the blanks. a The sum of –5 and 8 is ________. b Subtracting 15 from –20 gives ________. c The absolute value of –10 is ________. d If I owe `50 and I receive `30, my new balance is ________. e If I have a gain of `310 and a loss of `125, my net result is ________. f Adding the additive inverse of –8 to –8 gives ________.

Subtract 2369 from the sum of –3652 and 5864.

State True or False.

a The commutative property of addition holds true for integers.

____________________

b The associative property of subtraction holds true for integers.

____________________

c The commutative property of subtraction holds true for integers.

____________________

d The associative property of addition holds true for integers.

____________________

Fill in the blanks using the properties. Also, name the property used. a (21) + (13) = _________ + (21); Property used - ____________________ b _________ + 0 = –3; Property used - ____________________ c (–35 + 13) + _________ = –35 + (_________+ (–16)); Property used - ____________________ d (–23) – _________ = –23; Property used - ____________________

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Word Problems 1

A store had 754 items in stock. They received a shipment of 325 more items. How many items are there now in total?

The altitude of a mountain peak is 2500 m.

A hiker climbs to an altitude of 1200 m and then descends to an altitude of 500 m. What is the total change in altitude experienced by the hiker?

A basketball team scores 80 points in the

first quarter, loses 25 points in the second quarter and then scores 15 points in the

third quarter. In the final quarter, they score 10 more points than they lost in the second quarter. What is the team’s total score?

Multiplication and Division of Integers Real Life Connect

Rahul and Megha are watching a submarine show on T.V. They listen to an announcement that the submarine has descended 14 km in 20 minutes. Rahul: Wow! The speed of the submarine is really fast. I wish I could travel in such a submarine some day. Megha: That’s true, Rahul! Although I’m not really fond of underwater adventures, I’m wondering how deep this submarine will travel in the next 40 minutes. 0

Multiplication of Integers Multiplying integers with unlike signs The submarine descends 14 kilometers in 20 minutes. As the sea level is 0, the distance above the sea is in a positive direction. Similarly, the distance below the surface can be taken as negative. We can say that the submarine travels –14 kilometres in 20 minutes. Therefore it travels –28 kilometres in 40 minutes as shown. Starting from 0, we always move to the negative side (in this case, the downward direction) when multiplying integers of unlike signs.

Chapter 3 • Integers

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(sea level)

20 minutes

–14 kms (distance below sea level) 40 minutes –28 kms (distance below sea level)

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Multiplying integers with like signs Case 1: When both the integers are positive: Assume that there are two people on the same side of a door. The push represents the positive sign, and the pull represents the negative sign. When both people push the door, the door moves in the pushed direction! This signifies that the product of two positive integers is always positive! Case 2: When both the integers are negative: As we know, the negative sign signifies the pull direction. It is the opposite direction to the push direction. Now, when both people try to pull the door, they will still end up moving the door in the same direction and ʺhelpʺ the pull movement. This signifies that the product of two negative integers is always positive! The above discussion leads us to the following two rules of multiplication of integers. Rule 1: The product of two integers with unlike signs is always negative. Rule 2: The product of two integers with like signs is always positive.

Multiplication of more than two negative integers Let us now multiply three integers, say –5, –12 and –7, and see the result. Here, we first multiply the initial two numbers, i.e., –5 and –12. (–5) × (–12) = 60 (The product will be positive as two integers with like signs are multiplied.) Now multiply the result with the third number, i.e., 60 × (–7). 60 × (–7) = –420 (The product will be negative as two integers with unlike signs are multiplied.) The final result has a negative sign when we multiply 3 negative integers. Let us now multiply the above result (– 420) with another negative integer (–2). (–420) × (–2) = 840 Here, the final result is positive when we multiply 4 negative integers. The above discussion leads us to the conclusion that: If the number of negative integers multiplied is even, then the product is positive. If the number of negative integers multiplied is odd, then the product is negative.

Example 3

What will you get on multiplying –5, –6, –2 and –4? We will first multiply the initial two numbers, i.e., –5 and –6. (–5) × (–6) = 30 (as two integers with like signs are multiplied)

Think and Tell What will be the sign of the final product when 99 negative integers are multiplied?

Multiply the result with the third number, i.e., 30 × –2. 36

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30 × (–2) = –60 (as two integers with unlike signs are multiplied) Now, multiply the result with the fourth number i.e. (–60) × (–4). (–60) × (–4) = 240 (as two integers with like signs are multiplied) Do It Together

Error Alert! Don’t forget to place the correct sign (+ or –) before the resultant product.

A store loses ₹125 a day for six days. How much money has been lost in total over these six days? Money lost by the store every day = _____ Money lost in 6 days = (–125) × _____ = _____

Properties of Multiplication of Integers Closure Property

Commutative Property

If a and b are two integers, then

For any two integers a and b,

a×b=c

For any three integers a, b, and c,

a × b = b × a.

where c will always be an integer.

1 13 × (–3) = –39

This means that when multiplying

of the order in which they are

in which the multiplications are

multiplied.

2 (–25) × (–9) = +225 In both the above cases, the result of the multiplication is an integer.

(a × b) × c = a × (b × c)

This means that the product of two integers is the same regardless

For example,

three or more integers, the order

performed does not change the final product.

For example, (15) × (–3) = (–3) × (15) = –45

Distributive property of multiplication of integers over addition For any three integers a, b and c,

For example, (–2 × 3) × (–4) = –2 × (3 × (–4)) = 24 Existence of multiplicative identity Any integer multiplied by 1 will result in the same integer.

a × (b + c) = (a × b) + (a × c) This means that when we multiply an integer by the sum of two other

integers, we get the same result by multiplying each addend separately and then adding the products together. For example,

Associative Property

a×1=a For example, 1 –4 × 1 = –4 2 17 × 1 = 17

–4 × (3 + 2)= –4 × 5 = –20 and (–4 × 3) + (–4 × 2) = (–12) + (–8) = –20

Existence of multiplicative inverse The product of any non-zero integer and its reciprocal is 1.

Multiplicative inverse of (–12) is –

Chapter 3 • Integers

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For any integer a, a×0=0

1 1 = × a = 1 or a a 1 1 (–a) × – = – × (–a) = 1 a a a×

For example,

Multiplication property of zero

1 12

Multiplication of an integer by –1 For any integer a, a × (–1) = –a

For example,

1 –10 × 0 = 0

1 12 × (–1) = –12

2 5×0=0

2 –25 × (–1) = 25

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Example 4

What is the multiplicative inverse of −14 ? Multiplicative inverse of –14 =

Example 5

1 –14

What should be multiplied with 10 to get its additive inverse? –1 should be multiplied with 10 to get its additive inverse.

Think and Tell Why does the concept of the multiplicative inverse not apply to zero?

10 × –1 = (–10) = additive inverse of 10 Do It Together

Fill in the blanks using the correct property of multiplication of integers. 1 (–25) × (–8) = ( _____ ) × ( _____ )

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

2 52 × _____ = 52

4 (17 × 5) × _____ = _____ × (5 × (–9))

Do It Yourself 3B 1

Evaluate the product. a 12 × 15

b –11 × 6

c –21 × 13

d –14 × (–6)

e 25 × 0

f 56 × (–1)

g –15 × (–24)

h 16 × (–8)

a –11 × 5 × (–2)

b –16 × 6 × (–18)

c 51 × (–26) × 6

d 12 × (–1) × (–8)

e 25 × (–12) × 3

f 56 × (–12) × 0

g –45 × (–4) × (–13)

h 11 × 0 × (–28)

Find the product.

Solve. a –23 × 15 × (–2) × 5 d

–6 × (–12) × (–3) × (–10)

b –8 × (–16) × 5 × (–11)

c 13 × (–21) × (–26) × 6

e 34 × (–12) × 20 × 17

f –23 × (–24) × (–43) × 9

What will be the sign of the product if we multiply the integers? a 5 negative integers and 5 positive integers b 12 negative integers and 20 positive integers c 20 negative integers and 12 positive integers d 25 negative integers and 35 positive integers e 33 negative integers and 50 positive integers

Fill in the blanks with the correct property of multiplication. a 36 × 0 = ___________

b (–123) × ___________ = –123

c (–12) × (56) = ___________ × (–12)

d 56 × (–12 + 23) = ___________ × (–12) + 56 × ___________

e ___________ × (–1) = 2198

f (123 × 143) × (–36) = ___________ × (143 × (–36))

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Word Problems 1

A submarine descends 55 feet per minute from sea

level. Where will the submarine be in relation to the sea level 8 minutes after it starts descending?

A shopkeeper earns a profit of ₹8 when selling a large packet of potato chips and a loss of ₹2 when selling a packet of biscuits.

a How much profit will he make on selling 25 packets of potato chips?

b How much loss will he make on selling 115 packets of biscuits?

A test has 25 questions. The test awards 5 points if the answer is correct and takes away 2 points if the answer is incorrect. A student answered 7 questions incorrectly. How many points did the student score?

Division of Integers Remember that the submarine travelled 14 km in 20 minutes? What if we need to find the distance travelled by the submarine in 10 minutes? This can be done with the help of division. Division of two integers with unlike signs We already know that division facts come from multiplication facts. So if the submarine travels –14 kilometres in 20 minutes, we need to find how far it would have travelled in half the time. The above calculation can be represented as below sea level in 10 minutes.

(–14) = –7. This implies that the submarine travelled 7 km 2

Division of two integers with like signs Dividing a positive integer by another positive integer results in a positive quotient. Dividing a negative integer by another negative integer again results in a positive quotient. Rules for division of integers can be given as: Rule 1: The quotient of two integers with unlike signs is always negative. Rule 2: The quotient of two integers with like signs is always positive.

Chapter 3 • Integers

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Example 6

Divide (–625) by 25. –625 = –25 (as Negative ÷ Positive = Negative) 25

Example 7

Ravi participated in a quiz competition where each incorrect answer deducts 3 points. If Ravi’s total negative score is –45, how many questions did he answer incorrectly? Score of each incorrect answer: (–3) Total negative score: (–45) Number of questions answered incorrectly: (–45) ÷ (–3) = 15 Hence, 15 questions were answered incorrectly.

Do It Together

In a basketball match, Team A and Team B played four quarters. If Team A’s total score for the match was –48 points, what was their average score per quarter? Number of quarters played = 4 Team A's total score for the match = __________ Average score per quarter = (–48) ÷ 4 = __________

Properties of Division of Integers Closure Property If a and b are two integers, then

Commutative Property

Associative Property

For any two integers a and b,

a ÷ b = c,

For any three integers a, b, and c,

a÷b≠b÷a

where c may or may not be an

(a ÷ b) ÷ c ≠ a ÷ (b ÷ c)

This means that while dividing two

This means that when dividing three

For example,

integers can change the answer.

the divisions are performed changes

1 –25 ÷ 5 = –5 = integer

For example,

2 –(–18) ÷ (5) = 3.6 = non-integer

(22) ÷ 11 = 2 but,

integer.

integers, a change in the order of the

or more integers, the order in which the final result. For example, (625 ÷ 25) ÷ 5 = 25 ÷ 5 = 5 but,

11 ÷ (22) = 0.5

625 ÷ (25 ÷ 5) = 625 ÷ 5 = 125 Division of an integer by itself

Division of an integer by 1

While dividing a non-zero integer by itself, the result is

When dividing an integer by 1, the result is always the

For any non-zero integer a,

For any integer a,

always 1.

a÷a=1

integer itself.

a÷1=a

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Division of an integer by –1

Division of zero by an integer

When dividing an integer by –1, the result is the additive

When dividing zero by an integer, the result is always

For any integer a,

inverse of the integer.

Example 8

a ÷ (–1) = (–a)

zero.

Check whether the quotient of 482 ÷ 3 is an integer.

Think and Tell

482 ÷ 3 = 160.66 which is not an integer. Example 9

Can you think of special cases where the

Prove that the integers 256 and 16 are not commutative under division.

Commutative Property may hold true?

16 ÷ 256 = 1 16 1 As 16 ≠ , integers are not commutative under division. 16 256 ÷ 16 = 16

Do It Together

0 ÷ (a) = 0

but

Remember!

Fill in the blanks. 1 (–125) ÷ (–25) = ________

2 0 ÷ 45 = ________

3 23 ÷ ________ = –23

4 136 ÷ ________ = 1

Dividing any non-zero integer by zero is undefined in mathematics and does not yield a valid result.

Do It Yourself 3C 1

Find the quotient. a 72 ÷ (–4)

b (–56) ÷ 7

c –176 ÷ (–11)

d 192 ÷ (–12)

e 344 ÷ 43

f (–984) ÷ 12

g (–676) ÷ (–26)

h –585 ÷ (–13)

a (–88 ÷ 4) ÷ (–1)

b 125 ÷ (125 ÷ 5)

c (192 ÷ (–16)) ÷ 4

d 324 ÷ (–18 ÷ 2)

e (900 ÷ (18)) ÷ 6

f (–3060 ÷ (–36)) ÷ 5

g 0 ÷ (–1508 ÷ 29)

h (1000 ÷ (20)) ÷ 100

Solve.

Find the number which when divided by (–1) gives 145.

State True or False. a Division is commutative for integers.

__________________

b Division has the associative property for integers.

__________________

c The quotient of two negative integers is always negative.

__________________

d Division by zero is defined for integers.

__________________

e The quotient of zero divided by any non-zero integer is zero.

__________________

Give two pairs of integers such that a ÷ b = –15.

Find the value of (12 × –3) ÷ (–2) + (–4).

Chapter 3 • Integers

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Word Problems 1

In the first level of a video game, 2 points are deducted for each time a character falls. In the

second level, each fall deducts 4 points. Rahul lost 10 points in the first level and 16 points in the second level. In which level did Rahulʹs character fall more often?

The temperature recorded at 12 noon was 18 ℃ above zero. If it falls at the rate of 3 ℃ per

hour until midnight, at what time will the temperature be 6 ℃ below zero? What will be the temperature at midnight?

Points to Remember

Addition Like signs: Add the absolute values and place the common sign.

Subtraction To subtract an integer b from another integer a, add the additive inverse of b to a.

Unlike signs: Find the difference of the absolute values and use the sign of the integer with the greater absolute value.

Integers {…..–3, –2, –1, 0, 1, 2, 3….}

Multiplication

Division

Like signs: The product of two integers with like signs is positive.

Like signs: The quotient of two integers with like signs is positive.

Unlike signs: The product of two integers with unlike signs is negative.

Unlike signs: The quotient of two integers with unlike signs is negative.

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Math Lab Integer Operations Relay Setting: In groups of 5 Materials Required: Index cards with different integer expressions written on them, stopwatch or timer. Method:

23 – 52 =? = ? +23 – 52

15 × (–5) = ?

96 =? –4

(–15) + (–9) = ?

Place the index cards face down on a table. Start the stopwatch or timer and begin the relay.

The first team member will pick up an index card, read the expression aloud, solve it and return it

The relay continues until all the index cards have been used or until a designated time limit is reached.

The team with the highest number of correct solutions wins the game!

96 � (–4) = ?

to their team to tag the next person.

Chapter Checkup 1

Solve. a (–256) + 362

b 214 – (–126)

e (–659) + (–1352)

f (–2698) – (–1236)

c 652 + (–129)

d 248 – (–369)

c (–36) × 48

d (245) × (–57)

c 5525 by (–25)

d (–5472) by 36

Find the product. a 25 × (–89)

b (–125) × (–26)

e (1245) × (–142)

f (–2365) × (123)

Divide. a 363 by (–11)

b (–2652) by (–51)

e 2852 by 23

f (–9594) by (–41)

Fill in the blanks. a (–156) + 389 = ______

b 524 + ______ = 256

c ______ – (–269) = –698

d 29 × ______ = (–1885)

e ______ � (–17) = 56

f –568 – (–258) = ______

g ______ × (–15) = –5475

h (2356) – ______ = 5886

______ + (–1265) = –4526

k (–1265) × (–352) = ______

(–3562) � (–137)= ______

3654 � ______ = (–87)

Chapter 3 • Integers

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Simplify. a 42 ÷ (–6 + 5)

b –64 ÷ 4 × (2 – 6)

e 7 × (5 + 3) ÷ 4 × (9 – 2)

f (6 + 2) – 15 ÷ (5 × 2)

d 4(–12 + 6) ÷ 3

c –148 + (–157)

d 365 + 124

Solve and write the additive and multiplicative inverse. a 25 + (–63)

c (9 ÷ 3) + 7 × (4 ÷ 2)

b –125 + 47

State True or False. a The associative property of multiplication states that changing the grouping of the numbers being multiplied does not change the product.

b The additive inverse property states that every integer has an additive

inverse such that the sum of the integer and its additive inverse is zero.

c The multiplicative inverse property states that every non-zero integer has a multiplicative inverse such that the product of the integer and its multiplicative inverse is one.

________________ ________________ ________________

d The commutative property holds for division of integers.

________________

e The associative property holds for subtraction of integers.

________________

Add the sum of (135) and (–325) to the difference of 253 and (–528).

What should be multiplied with (–165) to get a product of 9240?

10 Add the product of 25 and (–36) to the quotient of 2380 divided by (–68). 11 What should be divided from 23,072 to get (–412)? 12 An aeroplane is flying at 12,000 feet. The plane climbs 15,000 feet to approach the cruising altitude. After a few minutes at this new altitude, the plane hits turbulence and descends 23,945 feet. Express each increase and decrease in altitude as an integer operation and determine the new altitude of the plane.

Word Problems 1

A mercury thermometer records a temperature of 6 °F at 11 a.m. If the temperature drops by

Kajal gets into the elevator on the fourth floor of a shopping mall. She goes up 10 floors to

3 °F every hour, what will be the temperature by 4 p.m. of the same day?

reach the food court. After an hour, she goes down 5 floors to buy books. At which floor is she now?

A shoe store marked ₹55 off on the price of each pair of shoes in stock. If the total reduction in price is ₹7590, what is the total number of pairs of shoes in the store?

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Measures of Central Tendency

Let’s Recall

Ritika wanted a pet for herself and asked her mother if she could get one. She collected data on the number of pets in her neighborhood and presented it to her mother.

Number of Pets

Number of Houses

How many houses have no pets? 15 houses have no pets.

How many houses have more than 1 pet? Houses with 2 pets + houses with 3 pets + houses with 4 pets =5+2+1=8 So, 8 houses have more than 1 pet.

What is the total number of houses that have pets? No. of houses that were surveyed = 35 Houses with no pets = 15 Therefore, houses with pets = 35 – 15 = 20

Let’s Warm-up Read the data given above and fill in the blanks. 1

________ households have 3 pets.

________ households have less than 2 pets.

The total number of pets in the neighborhood is ____________.

There are _______ households with 1 pet each.

The difference in number between the households which have pets and which do not have pets is _______.

I scored _________ out of 5.

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Mean, Mean,Median Medianand andMode Mode Real Life Connect

Ria’s parents are planning to go on a vacation to Jaipur for the first time. Ria: Papa! What is the weather in Jaipur like? Ria’s father checks the online weather forecast for Jaipur for the next 15 days and finds the following: Days

Day 1

Day 2

Day 3

Day 4

Day 5

Day 6

Day 7

Day 8

Day 9

Day 10

Day 11

Day 12

Day 13

Day 14

Day 15

Temp (in °C)

Ria: P apa! The temperature is so different every day. What clothes should we pack?

Arithmetic Mean or Mean One possibility is that Ria and her father want to take a decision based on a single value that represents the entire data. But what will that value be and how is it calculated? Arithmetic mean or mean or average of a data is a value that can represent all of the data. It is defined as the sum of all the observations of data divided by the total number of observations. Mean =

Sum of all observations Number of observations

But data could be presented in different forms. How do we calculate the mean in all such cases? Let us learn!

Mean of Ungrouped Data In the last chapter, we learnt that when data is listed as individual values or observations it is called ungrouped data. For example, the temperature forecast data that Ria’s father showed is also ungrouped data. So, in the above case, the average temperature can be calculated by dividing the sum of all the observations (temperature in °C) by the total number of observations (the number of days). Thus, the mean temperature can be calculated as: Mean temperature = 32 + 34 + 32 + 37 + 42 + 40 + 30 + 32 + 36 + 42 + 44 + 44 + 42 + 41 + 42 °C 15 =

570 °C 15

= 38°C So, the average or mean temperature is 38°C.

Remember! The mean of the data may not always be an observation from the data.

Therefore, if Ria and her father want to make a decision based on a single temperature that reflects the overall duration of their stay, they should pack their clothes based on the mean temperature, 38°C. 46

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Example 1

Find the mean of the first 5 natural numbers. Step 1

Step 2

Write the first 5 natural numbers.

Find the sum of the observations.

1, 2, 3, 4 and 5.

1 + 2 + 3 + 4 + 5 = 15

Step 3 Count the total number of observations. Here, there are 5.

Think and Tell

Step 4

Could the mean be more than the

Find the mean using the formula. Mean = =

highest value of the data?

Sum of all observations No. of observations 15 =3 5

So, the mean of the first five natural numbers is 3. Example 2

The given table shows the number of goals scored by three lead players in four matches. Players

Vishal

Bharat

Tejas

Goals in Match 1

Goals in Match 2

Goals in Match 3

Did not play

Goals in Match 4

Which player had the highest average score? We know that, mean = Sum of all observations No. of observations Let’s first arrange the data. Players

Vishal

Bharat

Tejas

Goals in Match 1

Goals in Match 2

Goals in Match 3

Did not play

Goals in Match 4

Total Goals Scored

Total No. of Matches - Player Wise

Calculating Averages - Player Wise

16 4

12 4

18 3

Average Goals - Player Wise

Thus, Tejas has the highest average score of 6 goals. Chapter 4 • Measures of Central Tendency

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Do It Together

Find the mean of the first five prime numbers. Step 1 Find the first five prime numbers. 2, 3, _______, _______, _______

Did You Know?

Step 2

Prof. Prasanta Chandra Mahalanobis

Find their sum.

is known as the father of modern

2 + 3 + _______ + _______ + _______ = _______

statistics in India. In 1968, he

Step 3

second highest civilian award

was honoured with India’s - the Padma Vibhushan

Find the mean. Mean =

= _______

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

Mean of Grouped Data We learnt that grouped data is a way of organising data by putting observations into groups. Instead of looking at each number individually, we group them together based on their values. We saw that the temperature data was ungrouped. Let us try to group it and find the mean. Step 1 Create a grouped data table. The observation column is denoted by x, while the frequency by f. Temperature (in °C) (x)

Tally Marks

No. of Days (frequency) ( f )

|||

||||

Total We know that, mean =

15 Sum of all observations No. of observations

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Step 2 Find the sum of all observations. First, multiply each observation by its frequency and list the products in a column as shown, and then find the sum. Temperature (in °C) (x)

Tally Marks

No. of Days (Frequency)

168

570

|||

40 41 42

| |||| ||

f×x

(f)

Total

Sum of all observations = 570 We can in general represent this sum of all

observations by ∑( f × x). The symbol S (called sigma) is a Greek letter that represents summation.

Step 3 Find the total number of observations as the sum of all the frequencies. We see from the table that the total number of observations is 15. In general, the total observations can be denoted as ∑f.

Error Alert!

Step 4

To find the total number of observations, NEVER add all the observations. We ALWAYS add their frequencies.

Calculate the mean. Mean =

Sum of all observations No. of observations

∑( f × x) ∑f

570 15

= 38°C

The mean is 38°C. Example 3

Data on the number of books read by students in a class is represented in the frequency distribution table below: No. of Books

No. of Students

Find the mean of the data. We know that, mean = Sum of all observations = No. of observations

∑( f × x) ∑f

Remember! The summation sign is especially helpful in the case of large data. There is no need to write each observation separately.

Let us find ∑f and ∑( f × x). No. of Books (x)

No. of Students (f)

f×x

Total

Σf =18

Σ( f × x) = 44

Chapter 4 • Measures of Central Tendency

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Thus, the mean =

Σ( f × x) 44 = = 2.44 books Σf 18 49

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Example 4

The table below shows the pocket money of 20 students. Pocket Money

₹100

₹120

₹105

₹125

₹140

No. of Students

Pocket Money (x)

No. of Students ( f )

f×x

₹100

400

₹120

600

₹105

735

₹125

375

₹140

140

Total

∑f = 20

∑( f × x) = 2250

Find the mean pocket money. Let us find ∑f and ∑( f × x).

Mean =

∑( f × x) 2250 = = 112.5 Σf 20

Thus, the mean pocket money is ₹112.50. Example 5

The mean of x, x + 1, x + 3, x + 5, x + 7 is 228.2. What is the value of x? Mean = 228.2 =

Sum of all observations Total number of observations

x + (x + 1) + (x + 3) + (x + 5) + (x + 7) 5

228.2 × 5 = 5x + 16

5x + 16 = 1141 or 5x = 1141 − 16 5x = 1125 x = 225 Do It Together

Find the arithmetic mean or the average of the marks (out of 10) in a class test. Marks (out of 10)

No. of Students

Step 1

Step 2

Let us find ∑f and ∑( f × x).

Find the mean.

Marks (x)

No. of Students ( f )

f×x

_______

4 6 7 8 9

Total

3 5 4

_______ 28

_______

∑f = 30

∑( f × x) = _______

∑( f × x) = = _________ Σf 30 Thus, the arithmetic mean or the average marks of the students is __________. Mean =

_______

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Do It Yourself 4A 1

Find the mean of the first 10 odd numbers.

The temperatures in a city were recorded as 3°C, –7°C, –1°C, –5°C and 2°C. Find the mean temperature for the

The given table shows the number of pages having a certain number of illustrations.

recorded data.

No. of Illustrations (per page)

No. of Pages

Find the mean number of illustrations per page. 4

The mean of 40, 50, 55, a and 58 is 56.2. What is the value of a?

The mean of x, x + 2, x + 3, x + 5 and x + 7 is 92. What is the value of x?

The mean of 5 numbers is 26. One of the observations, which was 6, was removed. Find the new mean. Is the

The mean of 30, 25, x, 50, y, 45 and z is 40 and the sum of y and z is 76. What is the value of x?

new mean greater or less than the original mean?

Word Problems 1

A group of friends went to a restaurant and shared the bill equally among themselves. The total bill

A group of soldiers went on a journey. They had to take a road trip through India passing

came to ₹2080. If there were 8 friends in the group, what was the amount each friend had to pay? through many rivers along the way. The rivers are listed, and the lengths of the rivers (in kilometres) are given. River

Tapi

Yamuna

Cauvery

Ganga

Length (in km)

724

1376

800

2525

Find the mean length of the rivers visited by the troop. 3

The mean attendance of a class from Monday to Saturday was 33. If the mean attendance

from Monday to Thursday was 32 and that from Thursday to Saturday was 34, then find the attendance on Thursday.

Mode We learnt that the average can represent the whole data. But what if Ria’s father wants to take a decision based on the temperature that occurs most frequently? In this case, Ria and her father must choose what is called the mode of the data. A value in the data that occurs most frequently is called the mode. Just like average, mode can also be calculated for both ungrouped and grouped data. Let us understand how. Chapter 4 • Measures of Central Tendency

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Special Case 1: More than one mode It is possible that in a set of data there are several observations with the same highest frequency (more than 1). In such a case, all such observations are the modes. For the given data: 2, 2, 3, 4, 4, 5, 6, 6, 7, 7 The modes are → 2, 4, 6 and 7 Special Case 2: No defined modes If all the values in a set of data occur exactly once, there is no value that can be considered the mode. In this case, the data is said to have no mode. For the given data: 1, 2, 3, 4, 6, 7 The mode does not exist because each observation occurs exactly once.

Mode of Ungrouped Data Let’s once again consider the data gathered by Ria’s father when they were planning to go on a vacation. Days

Day 1

Day 2

Day 3

Day 4

Day 5

Day 6

Day 7

Day 8

Day 9

Day 10

Day 11

Day 12

Day 13

Day 14

Day 15

Temp (in °C)

Let’s find the mode of the above data. Step 1 Arrange the temperatures (in °C) in ascending order.

Remember!

30, 32, 32, 32, 34, 36, 37, 40, 41, 42, 42, 42, 42, 44, 44

Mode will always be an observation from the given data.

Step 2 Find the observation that occurs most frequently.

30°C, 32°C, 32°C, 32°C, 34°C, 36°C, 37°C, 40°C, 41°C, 42°C, 42°C, 42°C, 42°C, 44°C, 44°C Occurs thrice

Occurs 4 times

Occurs twice

We see that the observation 42°C occurs the most, i.e., 4 times. Thus, it is the mode of the data. So, Ria and her father can pack their clothes for the most frequent temperature or the mode of the data, 42°C. Example 6

The marks scored (out of 20) by Vikas in various class tests during a term are given below: 18, 15, 17, 13, 17, 19, 17, 16, 14 Find the mode. Step 1

Step 2

Arrange the marks in ascending order.

Find the observation that occurs most frequently.

13, 14, 15, 16, 17, 17, 17, 18, 19

13, 14, 15, 16, 17, 17, 17, 18, 19 Occurs thrice

The observation 17 occurs the most, i.e., 3 times. Thus, it is the mode of the data.

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Do It Together

The amount spent by Raj in 10 days on commuting to office is given below: ₹120, ₹130, ₹130, ₹135, ₹120, ₹140, ₹130, ₹250, ₹200, ₹120 Find the mode of the data. Step 1 Arrange the observations in ascending order.

₹120, ₹_____, ₹_____, ₹130, ₹_____, ₹_____, ₹135, ₹_____, ₹_____, ₹250 Step 2 Find the observation that occurs most frequently. The observation(s) __________________ occur(s) three times each. Thus, the mode(s) of the data is/are ₹__________________.

Mode of Grouped Data Remember, we converted the above data in the grouped form to find the mean. We obtained the given table. Temperature (in °C)

Tally Marks

No. of Days (frequency)

32 34 36 37 40 41

||| | | | | |

||||

3 1 1 1 1 1 4 2

Error Alert! The observation with the highest frequency is the mode and not the frequency. Temperature

Frequency

42°C

43°C

Mode is 5

Mode is 42°C

The observation with the highest frequency is the mode of the data. From the table, the observation 42°C has a frequency of 4, which is more than any other observation in the data. Thus, the mode of the data is 42°C. Example 7

The following table shows the colours of dresses worn by various guests at a party. Colour of Dresses

No. of Guests

Black

Red

Blue

Brown

Pink

Find the modal colour of the dresses worn by the guests. From the table, it is clear that ‘Red’ is the modal colour as it occurs the most frequently. Chapter 4 • Measures of Central Tendency

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Do It Together

Vivaan wrote a few numbers on a piece of paper. The most frequent number (i.e., mode) was 8. One number got wiped out. Can you find out what that number was? 4, 6, 6, 7, 8, 3, 8, 6, 5, 8 Let us start by arranging the data in ascending order 3, 4, 5, 6, 6, 6, 7, 8, 8, 8 Can the missing number be 3? No, because then the modes will be 8 and 6. Can the missing number be 4? No, because then the modes will be 8 and 6. Can the missing number be 5? _________, because then the mode(s) will be _________. Can the missing number be 6? _________, because then the mode(s) will be _________. Can the missing number be 7? _________, because then the mode(s) will be _________. Can the missing number be 8? _________, because then the mode will be _________.

Do It Yourself 4B 1

State true or false.

he mode is the observation that occurs the least number of times in a data set. _________ a T he mode is always one of the numbers in the data. _________ b T he data 2, 4, 3, 4, 6, 3, 6, 7, 4 has a mode of 3. _________ c T here is only one mode for any given set of data. _________ d T 2

Find the mode of the given years of experience of teachers in a school: 12, 5, 7, 10, 3, 2, 1 ,7, 6, 7, 5, 10, 7, 4, 2, 1, 4, 4

The given data shows the marks obtained (out of 20) by 14 students of a class in a test. Name

Jai

Ajay

Ron

Aru

Ria

Anya

Sia

Shiv

Remo

Dev

Dia

Dino

Sam

Pihu

Marks Obtained

Find the mode of the data. How many students got the highest marks? 4

In a singing competition, points were awarded to 14 participants with a maximum possible score of 60. The allotted points for each participant were as follows: 60, 52, 42, 40, 43, 44, 46, 45, 42, 41, 40, 46, 48, and 46. Determine the mode of the given data.

The number of books read by 30 students of a class is given below: No. of Books

No. of Students

Find the mode of books read by the students.

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Word Problem LIBRARY

A librarian keeps track of the types of books that children

choose from the library. She notices that the last 10 books belonged to the following genres: Fiction, Poetry, Drama,

Drama, Fiction, Fantasy, Poetry, Drama and Fantasy, Drama. Find the modal genre.

Median

We saw how Ria and her father decided to pack clothes based on:

1 A single value for the temperature that represents the whole data set, i.e., the arithmetic mean or

mean, which was 38°C.

2 A single value of temperature that occurred most frequently, i.e., the mode, which was 42°C.

What if Ria and her father wanted to take a ‘balanced’ decision to choose a temperature value that falls exactly in the middle of the temperature extremes? So, they would expect exactly as many hotter days as there were cooler days. The median is the middle value in a set of data when arranged in ascending or descending order. Let’s find the median of the data collected by Ria’s father. We arrange the data in ascending or descending order. 30, 32, 32, 32, 34, 36, 37, 40, 41, 42, 42, 42, 42, 44, 44 Values less than 40

Median

Values more than 40

Here, 40, i.e., the 8th value, has exactly 7 values higher and 7 values lower. Therefore, 40°C is the median of the data. Let’s learn more about median calculation. Case I: When the total number of observations is an odd number, When the total number of observations is odd, the median would be the observation in the middle, n + 1 i.e.,  th observation in the arranged data. Let us consider the following data.  2 

Step 1

Error Alert!

Arrange the data in ascending order. 1

Step 2 Find the total number of observations. Total number of observations is 7.

Chapter 4 • Measures of Central Tendency

04_UM24CB0704V1.indd 55

 n + 1, you find the median  2 

By using 

place and NOT the median itself. The observation in that place is the median.

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Step 3 n + 1th observation in the arranged data is the median. The    2 

Remember!

7 + 1th observation, which is the 4th observation is the median.   2 

Here, the 

When arranged in ascending or descending order, the median has half of the observations lying to its left and the other half to its right.

Step 4 Find the median. 1

Less than median

Median

More than median

Thus, the median value is 10. Exactly three values are more than 10 and three values are less than 10.

Case II: When the total number of observations is even Let us consider the same data as above, but just add another observation to it. So, the new data is: 20

New observation

Step 1

Step 2

Arrange the data in ascending order.

Note the total number of observations.

Step 3

Step 4

Find the two middle observations. 1

The total number of observations is 8.

Two middle observations

n n  These are the  th and the  + 1th observations. 2 2 

The median is the average of the middle two observations. Median = Median is

1  n

n     th+  + 1th observation 2  2 2   (10 + 15) 2

= 12.5

Here, these are the 4th and 5th observations. Example 8

Find the median of the given data: 10, 8, 14, 16, 10, 6, 18, 14, 16 Here, n = 9, which is an odd number. By arranging the data in ascending order, we get:

Think and Tell

6, 8, 10, 10, 14, 14, 16, 16, 18

Is it necessary that the median will

Middle value

always be present in the data?

Thus, the median is 14. We can also use the formula to find the median.

Median =  n + 1 th observation =  9 + 1 th observation =  10 th observation = 5th observation  2   2  2 The 5th observation is 14. Thus, the median is 14.

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Find the median of the first 10 prime numbers.

Example 9

We know that the first 10 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 As they are already arranged in ascending order, we can start finding the median. Here, the number of observations, n = 10, is even.

So, here the median is the average or mean of  n th and  n + 1th observations. 2 2 

That is, n = 10 = 5th observation, and  n + 1 = 10 + 1 = 5 + 1 = 6th observation. 2 2 2  2 The ascending order of observations is: 2, 3, 5, 7, 11, 13 ,17, 19, 23, 29 5th and 6th observations

  Thus, median = 1 nth+ n + 1th observations = 1 {5th + 6th} observations = 1 (11 + 13) = 1 × 24 = 12 2  2  2   2 2 2 Thus, the median of the first 10 prime numbers is 12.

Example 10

The median of the given data is 50. What is the value of a? 15, 20, 25, a, 80, 120 Total observations = 6 (Even number) 3rd term + 4th term 2 25 + a = 50 × 2 Median =

50 =

25 + a 2

a = 100 − 25 = 75 Do It Together

The heights of 7 girls of a class are given in the following table: Name

Radha

Avni

Rita

Jaya

Shikha

Shilpi

Rinki

Height (in cm)

160

164

148

169

171

154

162

Find the median height. To find the median height (in cm): Step 1

Step 2

Arrange the heights (in cm) in ascending order.

Find the total number of observations.

148, _____, _____, _____, _____, _____, 171

The total number of observation = 7, which is an

Step 3

_________ (even/odd) number.

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

Chapter 4 • Measures of Central Tendency

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Do It Yourself 4C 1

Fill in the blanks. a To find the median, the data can be arranged in _______________ or ____________ order.

b If the number of observations is an odd number, then the median is the ____________ observation. c The median of the first 6 odd numbers is _________.

The enrolment in a school during six consecutive years is given below. Find the median of the data. Year

2018

2019

2020

2021

2022

2023

Enrolments

1555

1670

1850

1780

1400

2540

The runs scored by 11 players of a school cricket team are given in the table below. Players Ishan Ravi Murli John Iqbal Arun Vikas Viru Vipin Jai Aman Runs

120

Find the median score of the team. 4

Consider the data arranged in ascending order. 10, 20, x, 30, 40, 50, 70 If the median of the data is 30, then what is the maximum possible value of x?

Consider the data arranged in descending order. 60, 55, 54, 53, 45, 35, x, 25, 20, 17, 15, 10 a If the median of the data is 30, then what is the value of x? b If 60 is replaced by 20 then what is the new median?

The time taken (in minutes) by students in a class to complete a certain task is given below: 7, 21, 12, 15, 18, 9 a Find the median time taken by the students. b If the largest time value is replaced by 35, will it affect the median? Why? c Can new students be added to the above data so that the median does not change?

The numbers 2, 4, 6, 6 , 2x + 1, 9, 9, 10, 11 are written in ascending order. If the median of the given data is 17, find the value of x.

Word Problems 1

A group of friends went to a store to buy snacks. They decide to pool their money to make a

purchase. Each friend contributed a different amount of money. The amounts contributed by each friend are as follows: ₹15, ₹20, ₹10, ₹25, ₹30 Find the median amount contributed by the group of friends.

The front row in a movie theatre has 23 seats. Rohan booked a seat in the front row that occupied the median position. Which seat did Rohan book?

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Choosing the Right Measure of Central Tendency From the previous sections, we learnt about the different ways in which Ria and her father could have made the decision about their packing. The mean, median and mode helped them decide. These are called the measures of central tendency. Measures of central tendency are ways to find the typical or “central” value of data. Mean: The mean, also called the average, is a value that can represent the whole data. Median: The median is the middle observation in the data when arranged in an order. Mode: The mode is the observation that appears most frequently in the data. But which measure of central tendency do we choose and why? We can choose the appropriate measure of central tendency according to the purpose of our exercise. The following table highlights when each measure of central tendency should be used. Mean When to Use

• To find one value that represents the whole data.

Median

Mode

• To find a “middle” value in data that has a sizeable number of observations.

• To recognise the most commonly occurring observation.

• To see which observation divides the entire set of data into two equal halves or parts. Example

Example 11

Sachin Tendulkar’s average score in One-Day International Cricket matches is 44.8 runs. He played 463 matches. This means that if we replace all his scores in individual matches with this average, it would represent his performance throughout his career.

In an exam, if the median score was 75%, this implies that exactly half the number of students who participated scored below 75% and the other half scored above 75%

A shopkeeper sells T-shirts in 5 sizes: Extra Small, Small, Medium, Large and Extra Large. The most frequently sold T-shirt size represents the modal size and is the most important for his business.

A researcher wants to study the duration of the commute of employees in a city. The researcher randomly selects 15 employees and records the duration of their commute in minutes. The dataset consists of the following durations of commute: 25, 30, 33, 40, 42, 50, 53, 60, 62, 70, 74, 78, 85, 90, 93. Which measure of central tendency should the researcher use to describe the average duration of their commute? Using Mean: As the researcher wants to describe the average duration of their commute, the mean can be given as: Mean = (25 + 30 + 33 + …….. + 93) = 885 = 59 minutes 15 15 Using Median:

Remember! One or more measures could possibly be good measures.

Another way of describing the average duration of their commute is by using the median. It can be given as: Median = As the data has an odd number of observations, the median will be the  n + 1 th observation =  2  15 + 1th = 8th observation.    2  Chapter 4 • Measures of Central Tendency

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Hence, the median for (25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95) is 60 minutes. Middle value

Using Mode:

We know that the mode is the most commonly occurring observation but in the above case, none of the data values is repeating. Therefore, the mode cannot be used to describe the average duration of their commute for the above data. Do It Together

Which measure of central tendency should the retailer use to plan his purchase for the upcoming week, considering the sales record of shirts in five sizes during the first week of the month? Shirt Size No. of shirts sold

Extra Small

Small

Medium

Large

Extra Large

104

Using Mean and Median: As the retailer is dealing with _________ (categorical/numerical) data, he _________ (can/can't) use the mean or median to plan his purchase for the upcoming week. Using Mode: As the mode is used to recognise the _________ (most/least) commonly occurring observation, the retailer can use the mode to plan his upcoming purchase. The mode for the given data is _________ . Therefore, only the _________ can be used to plan the purchase for the upcoming week.

Do It Yourself 4D 1

Fill in the blanks. a The measures of central tendency are __________, __________ and __________. b __________ represents one mathematical equivalent for the data. c __________ is used when you need a “middle” value of a data set that has a sizeable number of observations. d __________ cannot be defined for data with no repeated values.

A researcher is studying the heights of trees in a forest. The heights of the trees are as follows (in metres): 5, 7,

The scores of the team members of a golf team are shown below.

10, 12, 15, 18, 22, 30. Which measure of central tendency should not be used by the researcher and why?

72, 73, 72, 76, 72, 74, 73, 75, 76, 74, 72 Which measure of central tendency would be the most appropriate to find the average score of the team members?

A data set has 50 different values. 47 values are between 1 and 5, whereas 3 values are more than 50. Which

A charity organisation collected donations from individuals for a fundraising event. The donations are as follows:

measure of central tendency will better represent the typical value of the data set?

₹10, ₹25, ₹50, ₹100, ₹250, ₹500, ₹1000, ₹2000, ₹5000 and ₹10,000. Which measure of central tendency should be used to represent the typical donation, and why?

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Word Problem 1

The teacher took a surprise test. The marks scored (out of 100) by Ravi and his classmates are shown. 87

Calculate the mean score. Calculate the median score of the class. Explain which of the measures of central tendency better explains the score of the class. Why?

Points to Remember • Mean, median and mode are the three measures of central tendency. •

Mean:

Mean = Sum of all observations No. of Observations  Cannot be more than the highest value of the data. 



•

Mode:    

•

Need not always be an observation from the data. Observation that occurs the greatest number of times in the data. Will always be an observation from the given data.

A set of data can have more than one mode, or even no mode. Is used when the data has many identical values.

Median: The middle value of a set of data when the data is arranged in ascending or descending order.  Odd number of observations, the median is given by the n + 1th observation.  2   Even number of observations, the median is given by the average of the two observations in the middle, i.e.,  n th and  n + 1th observations. 2 2  

Math Lab ShoeTopia Setting: Groups of 6 Materials Required: Chart paper, markers and sticky notes Method: 1 Divide the class into groups of 6 each. Give each group a sticky note. Each child in the group notes down her/his shoe size on the sticky note.

Chapter 4 • Measures of Central Tendency

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2 After all the groups are done, the teacher asks the students to find the following for their respective groups:

a The mean shoe size

b The modal shoe size

c The median shoe size

d The biggest shoe size

e The smallest shoe size 3 The group that finds all the correct answers first, wins.

Chapter Checkup 1

For each set of data, find the mean, median and mode.

Find the mean and median of:

The weights of 5 students are given as 38 kg, 41 kg, 36 kg, 39 kg and 41 kg. Find the mean weight recorded in

Find the median of the data: 3, 2, 4, 2, 5, 6, 2, 7, 3, 4, 2.

The marks obtained (out of 10) by the students in a class test are given below:

a 8, 12, 16, 13, 9, 10, 15, 12

b 12, −15, −32, −11, −15, 14, −30, −18

a the first 10 natural numbers.

c the factors of 36.

b the first 8 prime numbers.

the data.

4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7 Find the mean and median of the data using a frequency distribution table. 6

Find the mode of the given data: 20, 26, 16, 22, 20, 30, 22, 20, 26

The runs scored by 11 players in a cricket match are given below: 6, 15, 120, 50, 100, 80, 20, 15, 8, 10, 15, 90, 90 Prepare a frequency distribution table and find the mode.

The weights of 70 students are given below: Weight (kg)

No. of students

Find the mean, median and mode weight. 9

The recorded rainfall values (in mm) for each of the last 10 days in Meghalaya during the month of June 2023 was: 97

96.5

125.5

52.2

82.1

108.5

35.6

13.6

1.5

2.7

Which measure of central tendency would be most suitable to represent the rainfall and why?

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10 A survey asked participants to rate a product on a scale of 1 to 10. The ratings received were as follows: 4, 6, 7, 8, 9, 9, 10. Which measure of central tendency should be used to represent the typical rating, and why?

11 What is the median age of a family whose members are 45, 42, 38, 35, 18, and 10 years old? 12 The sizes of 10 shirts are as follows: 80, 85, 90, 80, 80, 85, 85, 90, 80, 95 Find the modal size. If the size of one shirt is misread as 80 instead of 85, find the correct modal size. 13 Find the mean and median of the following set showing the number of hours of operating life of 20 flashlight batteries:

19, 25, 20, 21, 25, 19, 21, 22, 24, 20, 25, 18, 23, 21, 24, 20, 19, 23, 25, 18 14 The weights (in kg) of 15 students are as follows: 27, 30, 42, 43, 36, 34, 35, 37, 28, 29, 31, 44, 41, 32, 33. If 30 kg is replaced by 25 kg and 41 kg by 45 kg, then find the new median. 15 The ages of a group of students are as follows: 12, 14, 15, 15, 16, 17, 18, 18, 19, 21. Which measure of central tendency should be used to represent the typical age of the students, and why?

Word Problems he teacher of class 7 found the mean weight of a class consisting of 20 students as 45 kg. 1 T Later, two more students Ankit and Suhani weighing 55 kg and 52 kg, respectively, join the class. What is the mean weight of the class now?

he ages of a group of people are 22, 28, 30, 29 and 31. If another person, aged 32, joins the 2 T group, will the median age change? If yes, what will the new median age be? ohan secured 75, 82, 79 and 76 marks in four tests. What is the lowest number of marks he 3 M can secure in his next test, if he needs to maintain a mean score of 80 marks in five tests.

Chapter 4 • Measures of Central Tendency

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Data Handling

Let’s Recall We can collect data through various methods and organise it in tabular form. Let’s consider an example. Roy collected the data for the most international test runs by cricket players.

He put the data in tabular form. 1

Player Name

Runs

Which player scored the most runs?

Sachin Tendulkar

15,921

Sachin Tendulkar scored the most runs.

Ricky Ponting

13,378

Jacques Kallis

13,289

Rahul Dravid

13,288

Alastair Cook

12,472

Which player scored 13,288 runs? Rahul Dravid scored 13,288 runs.

Let’s Warm-up Study the data given above and fill in the blanks. 1

_______________ scored 1 run more than Rahul Dravid.

The total number of runs made by Ricky Ponting and Alastair Cook is _______________.

The total number of runs scored by the 5 batsmen is _______________.

_______________ scored the least runs of the 5 players.

Sachin Tendulkar scored _______________ more runs than Jacques Kallis.

I scored ____________ out of 5.

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Organisation and Representation of Data Real Life Connect

Annie owned a warehouse. She kept a stock of multiple products in her warehouse. She collected the data of all the products, including the stock available and the sale of the products.

Remember! Numerical information or facts collected are known as data.

Data Organisation Annie collected the data on all the categories of products and different types of products available at her warehouse. Category

Dairy

Number of Products

Fruit and Vegetables 26

Stationery 19

Beauty and Cosmetics 21

Baby Care 15

The process of gathering data and recording information is called data collection. Annie wanted to represent the data using tally marks. Let us see! Category

Tally Marks

Number of Products

Dairy

Fruits and vegetables

Stationery

Beauty and cosmetics

Baby care

15 Important terms

Raw data

When the data is collected in its original form then it is called raw data.

Primary data

Primary data is the data collected by the user or researcher themselves.

Secondary data

Secondary data is the data collected by someone else. The main sources of secondary data are:

Observation

Each item in the raw data is called an observation.

Array

When the raw data is arranged in ascending order or descending order it is called an array.

Frequency

The frequency is the number of times an observation occurs.

Frequency

Frequency distribution is the tabular arrangement of the numerical data which shows

distribution table Range

2 International organisations

3 Government bodies

the frequency of each observation. The table showing the data is called the frequency distribution table.

The difference between the largest and the smallest observation in a set of data is called the range of the data.

Chapter 5 • Data Handling

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Research organisations

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Example 1

The scores obtained in 20 throws of a dice are: 5, 4, 3, 2, 1, 1, 2, 5, 4, 6, 6, 6, 3, 2, 1, 4, 3, 2, 2, 4 Organise the data and represent it in the form of a frequency distribution table. First, arranging the data in ascending order, we get: 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 6 The information can also be represented in tabular form as given below. Scores

Example 2

Tally Marks

Frequency

The heights of 25 students (in cm) of class VII are as follows: 155, 158, 154, 162, 159, 151, 148, 156, 150, 152, 149, 160, 159, 163, 148, 147, 161, 155, 153, 156, 162, 159, 152, 159, 152 Prepare the frequency distribution table for the given data and answer the questions. 1

What is the frequency of 159 cm?

What is the difference in the frequency of 153 cm and 162 cm?

Which observation has the highest frequency?

How many students have a height of more than 154 cm?

First, arrange the above data in ascending order. We get:

Heights

147, 148, 148, 149, 150, 151, 152, 152, 152, 153, 154, 155, 155, 156, 156, 158, 159, 159, 159, 159, 160, 161, 162, 162, 163

147

148

149

The information can also be represented in tabular form as shown.

150

151

152

153

154

155

156

158

159

160

161

162

163

The frequency of 159 cm is 4.

The frequency of 162 cm is 2 and that of 153 cm is 1.

So, the difference in the frequency of 153 cm and 162 cm is 2 – 1 = 1. 3

The height of 159 cm has the highest frequency.

14 students are more than 154 cm in height.

Tally Marks

Number of Students

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Do It Together

A new movie was released last Friday. Ratings (out of 10) by 31 people who saw the movie are shown below: 3, 4, 4, 1, 2, 6, 3, 1, 4, 5, 1, 1, 2, 3, 2, 2, 6, 3, 2, 3, 2, 4, 4, 2, 5, 5, 2, 3, 2, 4, 4 1

Organise the data and represent it in the form of a frequency distribution table.

Find the range of the data.

Arranging the above data in ascending order, we get: 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6 The above information can also be represented in tabular form as given below. Rating

Tally Marks

Frequency

2 3 4 5 6

Range of data = Highest rating – Lowest rating = 6 – 1 = 5

Do It Yourself 5A 1

Which of these steps is the first step in any investigation? a Organisation of data

b Collection of data

c Tabulation of data

d Understanding of the data

A table that lists observations and uses tally marks to record and show the number of times they occur is called a

_______________________________. a frequency

b tally marks table

c frequency distribution table

Write (T) for true and (F) for false. Give reasons for choosing a statement as false. a Interviews, surveys and direct observations are examples of primary data. b Information published in magazines and newspapers after an event is an example of secondary data. c Autobiographies and memoirs are examples of primary data.

The data shows the colours of cushion covers chosen by 25 women from a local shop. Represent the data in a frequency distribution table.

Red, Blue, Red, Green, Yellow, Red, Blue, Blue, Green, Maroon, Black, Brown, Red, Brown, Black, Black, Yellow, Blue, Blue, Red, Brown, Yellow, Red, Green, Green 5

The data gives the number of birthday cakes sold by a bakery for 15 consecutive days. Prepare a frequency distribution table for the data. Find the range of the data given. 5, 4, 3, 2, 1, 1, 2, 5, 4, 6, 6, 6, 5, 2, 1

Chapter 5 • Data Handling

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The marks obtained by 20 students in a mathematics test (out of 100) are given below: 45, 65, 63, 38, 46, 87, 72, 43, 64, 79, 91, 79, 76, 33, 82, 56, 69, 99, 82, 99 a Prepare a frequency distribution table, and find the range of this data. b If 45 is the passing mark, how many students have failed? c How many students have scored less than 60 marks?

Word Problem 1

The students of a class were tested to find their pulse rate. Data was collected on the number of beats per minute.

60, 70, 70, 68, 67, 62, 71, 71, 59, 73, 73, 60, 60, 62, 62, 62, 68, 70, 70, 70, 74, 69. a Create a frequency distribution table for the data, and find the range of the data. b How many students’ pulse rate was 60 or less? c How many students had a pulse rate of more than 65?

Making Bar Graphs Annie recorded the data of the stock available at her warehouse for different categories of products in a table. Annie wondered how to represent the data, using a bar graph. Let us find out. Step 3

Step 1

Draw the rectangular bars.

Draw the x-axis and y-axis.

Category

Stock available

Dairy

1200

Fruit and vegetables

4600

Stationery

2600

Beauty and cosmetics

1800

Baby care

1600

Scale: 1 division = 500

5000 4500 4000

Step 2 Label both the axes.

Stock Available

3500 3000 2500 2000 1500 1000 500 0

Dairy

Fruits and Stationery vegetables Category

Beauty and cosmetics

Baby care

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Double Bar Graphs Ria made a list of pizza toppings, according to the choices of boys and girls. Favorite pizza toppings

Girls

Boys

Cheese

Capsicum

Onions

Remember! The double bar graph is drawn to compare two data sets.

Here, two sets of data are given. So, if we want to represent how many boys and how many girls chose the same type of topping, we need to draw two bars of different colours side by side for boys and girls separately to compare two sets of data simultaneously. This is known as a double bar graph. Let us learn how to draw a double bar graph for the two data sets given above: Step 1

Step 2

Draw two perpendicular axes – horizontal and vertical

Take the scale as 1 division = 5 children.

Step 3

Step 4

Draw the bars of equal width with equal spacing between

Draw two bars for each type of topping, first for the girls

number of children.

on the graph.

and label both axes.

types of pizza toppings and heights corresponding to the

and second for the boys. Place them next to each other

y Scale: 1 division = 5 children

Number of Children

15 Girls 10

Boys

Cheese

Capsicum

Onion

Pizza Toppings

Chapter 5 • Data Handling

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Error Alert! The bars should be of the same width, and there should be equal spaces between them. y 30

The first bar graph

appears in a book

by William Playfair

in 1786.

Example 3

Did You Know?

Pen

Pencil Book

Pen

Pencil

Book

The runs scored by a cricketer in the first six overs are given: Overs

III

Runs

Draw a bar graph for the data. The runs a cricketer scored in the first six overs are shown in the bar graph below. y

Scale: 1 division = 5 runs

Number of Runs

III

Overs Example 4

Create a bar graph representing the number of orders received on a Monday by the food court inside a mall. The sales data is given in the table below. Item No. of orders

Sandwich

Chow mein

Flavoured milk

Idli

300

500

200

550

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From the table, we observe that the values of the sales data are too large to be represented. A different scale would be required for this data set. The bar graph below represents the number of orders received by the food court on a Monday. y

Scale: 1 division = 50 orders

600 550

Number of Items Ordered

500 450 400 350 300 250 200 150 100 50 0

Sandwich

Chow mein

Flavoured milk

Idli

Item Example 5

The data shows the number of motorbikes of the same brand sold by two dealers in the first three months of a year. Draw a double bar graph, choosing an appropriate scale.

Month

January

February

March

Dealer I

Dealer II

The double bar graph shows the number of motorbikes of the same brand sold by two dealers in the first three months of a year. y

Scale: 1 division = 5 motorbikes

Number of Motorbikes Sold

35 30 25 20

Dealer I Dealer II

15 10 5 0

January

February

March

Month

Chapter 5 • Data Handling

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Example 6

Consider this data collected from a survey of a colony. Favourite Sport

Cricket

Basketball

Swimming

Hockey

Athletics

Watching

1000

470

520

540

250

Participating

620

350

270

370

200

Draw a double bar graph by choosing an appropriate scale. This bar graph represents the number of people who are watching and participating in their favourite sports. Let us consider, Scale: 1 division = 100 people y

Scale: 1 division = 100 people

1000 900

Number of People

800 700 600

Watching

500

Participating

400 300 200 100 0

Cricket

Basketball Swimming

Hockey

Athletics

Favourite Sport Do It Together

60 students from a certain locality use different modes of travel to school, as given below: Mode of travel

Car

Bus

Scooter

Bicycle

Rickshaw

Number of students

Draw the bar graph representing the above data. Mode of transport used by students y

Scale: 1 division = 2 Students

Number of Students

18 16 14 12 10 8 6 4 2 0

Car

Bus

Scooter

Bicycle

Rickshaw

Mode of Transport

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Do It Yourself 5B 1

Fill in the blanks. a A bar graph consists of rectangular bars of equal _______________. b The spaces between two consecutive bars are _______________. c In a bar graph, the bars can be represented either _______________ or _______________. d The _______________ of a bar represents the frequency of the corresponding data.

The table shows the number of points scored by Rahul in 10 rounds of a game. Represent the data using a bar graph. Rounds of the Game

Points

The table shows the favourite sport of 300 students of a school. Represent the data using a bar graph. Sport Number of Students

Football

Tennis

Athletics

Swimming

Hockey

Volleyball

The table shows the average intake of nutrients in calories by rural and urban groups in a particular year. Using a suitable scale for the given data, draw a double bar graph to compare the data. Foodstuff

Cricket

Pulses

Leafy vegetables

Other vegetables

Fruit

Milk

Fish and flesh foods

Fats and oils

Sugar

Rural

Urban

The performance of a student in the 1st term and 2nd term is given. Draw a double bar graph by choosing an appropriate scale. Subject

Maths

Science

English

Social Science

Hindi

1st term

2nd term

Word Problems 1

Choose a suitable scale and draw a bar graph for the data below. Day

Monday

Tuesday Wednesday Thursday

Friday

Number of toys manufactured

15,000

25,000

42,000

30,000

The data on the production of paper (in lakh tonnes) by two different companies X and Y over the years is given. Read the data and represent the data in the form of a double bar graph. Years

2018

2019

2020

2021

2022

Company Y

Company X

Chapter 5 • Data Handling

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35,000

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Interpreting Bar Graphs Annie stored the data for the total revenue generated in the last year for the products in the form of a bar graph. y

Scale: 1 division = ₹50,000

500 000 450 000

Revenue Generated (in ₹)

400 000 350 000 300 000 250 000 200 000 150 000 100 000 50 000 0

Dairy

Fruit and Vegetables

Stationery

Beauty and Cosmetics

Baby Care

Category Annie wanted to check which category generated the most revenue. Let’s see. The rectangular bar for fruit and vegetables is the highest. The amount of revenue generated for fruit and vegetables is ₹4,00,000. Horizontal bar graph Unlike vertical bar graphs or column graphs, there is another type of bar graph known as a horizontal bar graph. A horizontal bar graph is drawn with rectangular bars of lengths proportional to the values that they represent, the same as a vertical bar graph. The only difference is that in a horizontal bar graph, the vertical axis shows the data categories that are being compared and the horizontal axis represents the values corresponding to each data category. Let us study an example to understand.

Think and Tell Which type of data can be displayed in a bar graph?

The bar graph represents the data collected on the popularity of the types of music liked by students. After observing the bar graph, answer the given questions:

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1 Is this a vertical or horizontal

bar graph?

2 How many students like rock music?

Scale: 1 division = 2 students

Hip hop

Hip hop? Jazz? Classical?

Type of Music

3 Which type of music is the most

popular?

4 How many students were asked about

their favourite music?

I t is clear from the horizontal bar graph that:

Classical

Rock

Jazz

1 Here, the vertical axis displays the

music category while the horizontal axis represents the corresponding value of each music category. So, this is a horizontal bar graph.

Number of Students

2 7 students like rock music, 12 students like hip hop, 14 students

like Jazz and 8 students like classical music.

Jazz is the most popular type of music among the students.

Total number of students = 12 + 8 + 7 + 14 = 41

Error Alert! Do not forget to label the axes and write the scale.

So, 41 students were asked about their favourite music. Example 7

The bar graph depicts the pass percentage of five different students in a class examination. Observing the bar graph, answer the questions: y Scale: 1 division = 10 % 100 90 80

Marks (in %)

70 60 50 40 30 20 10 0

Suman

Raman

Prem

Ankita

Rosy

Student’s Name Chapter 5 • Data Handling

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Name the student whose pass percentage is more than 90%.

Name the student who has the lowest pass percentage.

Find the number of students whose pass percentage is less than 80%.

Find the number of students whose pass percentage is more than 70%.

It is clear from the bar graph above that: 1

Prem is the student whose pass percentage is 100%, which is more than 90%.

Suman has the lowest pass percentage, which is 50%.

Suman and Raman are the two students whose pass percentage is less than 80%.

4 Raman, Prem, Ankita and Rosy are the students whose pass percentage is more than 70%. Thus,

4 students pass with more than 70%.

Example 8

The double bar graph shows the favourite sports of boys and girls in a school. Study the graph carefully and answer the questions. Favourite Sport in a School

Scale: 1 division = 2 students

Number of Students

14 12 10

Girls

Boys

6 4 2 0

Football

Tennis

Basketball

Cricket

Sports 1

In which sport have the most boys shown interest?

Who has more interest in basketball?

Think and Tell

How many students in total like football?

Does the width of the bar in a bar

Which sports do 12 girls like?

How many girls like cricket?

graph have any significance?

It is clear from the bar graph given above that: 1

The most boys have shown interest in cricket.

More girls have shown interest in basketball.

There are a total of 19 students who like football.

Basketball is liked by 12 girls.

Cricket is liked by 5 girls.

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Example 9

The bar graph shows the favourite fruits of the students in a class. Read the bar graph and answer the questions. y Scale: 1 division = 2 students 1 How many students like papaya? 2 Which fruit is the favourite

Oranges

among students? How many students like it?

Pears

students?

4 How many more students like

bananas than grapes?

Fruits

3 What is the total number of

5 What per cent of the total

students like oranges?

6 What is the ratio of the number

Grapes Papaya Bananas Apples

of students who like papayas to the number of students who like pears?

Number of Students

Papaya is liked by 10 students.

2 The rectangular bar for bananas is the longest. So, the most students like bananas; 12 students like

bananas.

Students who like apples = 8

Students who like bananas = 12

Students who like papaya = 10

Students who like grapes = 7

Students who like pears = 9

Students who like oranges = 6

Total number of students = 8 + 12 + 10 + 7 + 9 + 6 = 52 4

Students who like bananas = 12 Students who like grapes = 7 Difference in the number of students who like bananas and those who like grapes = 12 – 7 = 5

Total number of students = 52 Number of students who like oranges = 6 Percentage =

Total number of students who like oranges × 100% Total number of students 6 × 100% = 11.5% 52

The number of students who like papayas 10 = The number of students who like pears 9 So, the required ratio is 10:9.

Chapter 5 • Data Handling

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Do It Together

The double bar graph represents the data collected on the popularity of Indoor Games and Outdoor Games as participant sports for students in Grades 5 to 8. Popularity Survey: Indoor Games vs Outdoor Games y

Scale: 1 division = 10%

100 90

Percentage of Students

80 70 60 Indoor Games

Outdoor Games

40 30 20 10 0

5th

6th

7th

8th

Grade 1

Is there any grade in which more students prefer indoor games?

What percentage of students prefer indoor games in Grade 6?

What percentage of students prefer outdoor games in Grade 7?

What percentage of students prefer outdoor games in Grade 8?

60% of students of which grade prefer outdoor games?

It is clear from the double bar graph given above that: 1

Yes grade 7 _______________, more _______________ students prefer indoor games.

_______________ of students prefer indoor games in Grade 6.

_______________ of students prefer outdoor games in Grade 7.

_______________ of students prefer outdoor games in Grade 8.

60% of grade _______________ students prefer outdoor games.

Do It Yourself 5C 1

1 cm on a scale of a bar graph represents 5 ft. Find the actual height of the building if this is represented by a bar of 5 cm height in the bar graph.

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The double bar graph shows the performance of a student in the 1st and 2nd terms.

Scale: 1 division = 10 marks

100 90 80

Marks

70 60

1st term

2nd term

40 30 20 10 0

Hindi

English

Maths

Science Drawings

Subject

a In which subject is the performance of the student in the 2nd term poorer than in the 1st term? b In which subject (s) did the student perform the best in term 1? How many marks did he score? c In which subject(s) did the student perform the best in term 2? How many marks did he score? d What was the total of the marks scored in term 1? e What is the difference of the total marks scored in term 1 and term 2? 3

The bar graph shows the heights of six mountain peaks in India. y

Scale: 1 division = 1000 m

12 000 11 000

9 000 8 000 7 000

Aconcagua

2 000

Mount Everest

3 000

Nanda Devi

4 000

Kanchenjunga

5 000

Annapurna

6 000 Nanga Parbat

Height (in metres)

10 000

1 000 x

0 Mountain Peak

Chapter 5 • Data Handling

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a Arrange the mountain peaks according to their heights in descending order. b What is the highest peak, and what is its height? c What is the ratio of the heights of the highest and lowest peaks? d Which mountain peak is the second highest? 4

The table shows the cost price and profits earned by a merchant for 4 different types of merchandise P, Q, R and S. Merchandise

Cost price (in ₹)

Profit (in ₹)

a Choose an appropriate scale and draw a double bar graph for the given data. b Which merchandise gives the most profit on selling? 5

The bar graph shows the monthly expenditure of a family (in ₹) in a year. Read the bar graph and answer the questions. y

Scale: 1 division = ₹200

2 600 2 400

Monthly Expenditure (in ₹)

2 200 2 000 1 800 1 600 1 400 1 200 1 000 800 400 200 x ec D

ov N

ct O

Se p

Au g

l Ju

n Ju

ay M

Ap r

M ar

Fe b

Ja n

0 Month

a What was the family’s expenditure in March? b In which month/s did the family spend the most? How much did they spend? c What was the total expenditure of the family in the full year? d What is the ratio of the expenditure in the first six months and the expenditure in the last 6 months? 6

The data shows the rainfall on different days of the week. Study the data and answer the questions. Days Rainfall (in mm)

Monday

Tuesday

Wednesday

Thursday

Friday

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a Draw a bar graph of the data. b On which day was the rainfall the highest? c What is the ratio of the rainfall on the first two days and the last two days? 7

The bar graph shows the number of people

with their shirt sizes. Read the bar graph and

Scale: 1 division = 100 people

answer the questions.

a How many people wear medium-sized

Extra-Large

b How many people wear extra-large sized shirts?

c What is the ratio of the people who wear

Shirt Size

shirts?

large-sized shirts to those who wear

Large Medium Small

small-sized shirts?

d How many people are there in total?

100

200

300

400

500

Number of People 8

The bar graph shows the number of cars sold by two showrooms in seven months.

Scale: 1 division = 15 cars

120

Number of cars sold

105 90 75

Showroom A

Showroom B

45 30 15 0

Jan

Feb

Mar

April

May

June

July

Month a How many cars in total were sold by showroom A in the first three months? b How many cars in total were sold by showroom B in the last three months? c Which showroom sold more cars in the first 4 months altogether? d What is the ratio of the total number of cars sold by showroom A and showroom B?

Chapter 5 • Data Handling

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Word Problems 1

The data on the number of visitors to a museum for five months is shown. Draw a horizontal bar graph to show the data. Month

January

February

March

April

May

300

250

300

400

350

Number of Visitors

a How many visitors were there in January? b How many more visitors were there in May than in February? c What percentage of the total number of visitors visited in April? 2

The double bar graph shows the percentage of profits earned by two companies A and B over the years. Read the graph and answer the questions.

Profit earned by company A and company B over the years y Scale: 1 division = 10%

Profit earned (in percentage)

60 50 40

Company A

30 Company B

20 10 0

2011

2012

2013

2014

2015

2016

Year

a If the income of company A in 2011 was ₹1,42,500, what was its expenditure in that year? b If the expenditure of Company A in 2010 was ₹70 lakhs and the income of Company

A in 2010 was equal to its expenditure in 2011, what was the total income (in lakhs) of

Company A in 2010 and 2011 together? (Hint: Consider income as the selling price and expenditure as the cost price.)

Points to Remember • The process of gathering required data from relevant sources is called data collection. • Primary data is the information collected directly from a source, and secondary data is the information collected from an external source or agency. • The frequency of a particular data value is the number of times the value occurs in the data set.

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• A frequency table is a table which lists observations and uses tally marks to record and show the number of times they occur.

• The range is the difference between the highest and lowest values of the data. It is also called the spread of the data. • A bar graph is another way of representing numerical data using bars of uniform width drawn vertically or horizontally with equal spacing between them. •

It is important to choose the right scale for the data.

• While choosing the scale, two factors should always be kept in mind, namely, the nature of the data and the purpose for which the measure is required. • A bar graph is a pictorial representation of data using bars (rectangles) of equal width and varying heights with uniform spacing between them.

• A bar graph is used to compare the observations of one set of data, whereas a double bar graph is used to compare two sets of data simultaneously.

Math Lab Objective: To compare the marks obtained in all the subjects by a student in the first and second unit tests in class VII by drawing a double bar graph, using paper cutting and pasting. Setting: Individual Materials Required: Graph paper, origami sheets, pencil, eraser, sketch pens/glitter pens, glue and scissors.

Method: 1

Collect your marks for all the subjects in each term and tabulate the data.

2 Draw two perpendicular axes on the graph paper and label them to show what they represent. 3 Choose an appropriate scale and cut the strips of origami sheets of different lengths (or heights) according to the marks obtained in different subjects in each term, representing one term with one colour say blue and the other term with another colour, say green. 4

Paste these strips of paper adjacent to each other, representing a double bar graph.

Now, refer to the double bar graph and answer the following question:

In which subject did you improve your performance the most?

Chapter 5 • Data Handling

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Chapter Checkup 1

The marks obtained by 21 pupils of a class in a monthly test are given below: 37, 21, 43, 16, 25, 21, 28, 32, 45, 14, 43, 43, 21, 21, 16, 21, 28, 32, 45, 45, 14 How many students obtained more than 28 marks?

Strauss was rolling a die. He rolled it 40 times and noted the number appearing each time, as shown below. 1

a Draw a frequency distribution table for the data. b Which digit appeared the least number of times? c Which digit appeared the most number of times? d Find the digits which appeared the same number of times.

Month

The table shows the number of children who have birthdays in

five different months. Observe the data and answer the questions. a Which month has the highest number of birthdays?

September

c How many more birthdays fall in September than in

November

December?

December

d Which months have an equal number of birthdays?

February July

b What is the total number of birthdays in February and July?

Number of Children

The birth rate per thousand of five countries over some time is shown below. Represent the data with a horizontal bar graph. Countries

India

Australia

Belgium

Canada

Denmark

Birth Rate (per thousand)

The data shows the total population of India based on the census conducted from 1971 to 2021, after every 10 years. Draw a bar graph for the given data. Year

1971

1981

1991

2001

2011

2021

Population (in millions)

420

540

680

1020

1200

1460

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The given bar graph shows the production of food grains in an Indian state for five consecutive years. Read the bar graph and answer the given questions. Production of Food (in million tonnes)

Scale: 1 division = 10 million tonnes

100 90 80 70 60 50 40 30 20 10 0

2014

2015

2016

2017

2018

Year a In which year was the production highest? What was the production in that year? b What was the production in 2016? c What was the difference in production between 2016 and 2017? d What is the ratio of the production in 2015 to the total production? 7

The bar graph represents the number of people who use various electronic appliances. Study the graph and answer the given questions.

Scale: 1 division = 2 lakhs

Number of Users (in lakhs)

12 10 8 6 4 2 0

Computer

Telephone

iPod

Webcam

Electronic Appliance

a How many people use TV compared to telephones? b How many more people use computers than iPods?

c Find the difference between the number of users of iPods and webcams. d Find the total number of people who participated in this survey. e Give the graph a suitable title. Chapter 5 • Data Handling

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The number of electric bulbs sold in a store over a week is depicted in the given bar graph. y

Scale: 1 division = 25 bulbs

250

Number of bulbs sold

225 200 175 150 125 100 75 50 25 0

Monday

Tuesday Wednesday Thursday

Friday

Day a How many bulbs were sold on Monday? b On which day was the highest number of bulbs sold and how many? c How many bulbs were sold in the entire week? d What fraction of the total bulbs were sold on Tuesday? e What is the average number of bulbs sold over the week? 9

In the bar graph, the salaries of 5 employees are shown. Spot the error and explain why this bar graph is misleading. Represent the data in tabular form and redraw the correct graph.

Salaries (in thousands)

Scale: 1 division = ₹1000

28 27 26 25 24 23 22 21 20 19

Raj

Sahil

Nivira

Yuvan

Kabir

Employees

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Look at the bar graph which shows the approximate length of some national highways in India. Read the bar graph and answer the questions. y

Scale: 1 division = 200 km

N.H. 10

National Highways

N.H. 9

N.H. 8

N.H. 3

N.H. 2 x 0

200

400

600

800 1 000 1 200 1 400 1 600 1 800 2 000

Length (in km) a Which national highway is the shortest? b What is the length of National Highway 9? c The length of which National Highway is about three times the length of National Highway 10?

Word Problems 1 The marks obtained by Kunal in his annual examination are shown below. Subject

Maths

Science

Hindi

English

Social Science

Drawings

Marks

Draw a bar graph to represent the given data.

2 The bar graph given below shows the sales of books (in thousands) from six branches of

a publishing company during two consecutive years, 2010 and 2011. Read the double bar graph and answer the questions that follow.

Chapter 5 • Data Handling

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y Scale: 1 division = 20,000

Sales (in thousands)

140 120 100

2010

2011

60 40 20 0

Branch a What information is given in the double bar graph? b What is the ratio of the total sales of branch B2 for both years to the total sales of branch B4 for both years?

c What are the average sales of all the branches (in thousands) for the year 2010? d What are the total sales of branches B1, B3 and B5 together for both years (in thousands)?

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Introduction to Probability

Let’s Recall In a classroom of 14 students, there are 6 boys and 8 girls. We can find the ratio of the number of boys and girls and write it in the simplest form.

Ratio of girls to boys =

Girls in the classroom 8 = Boys in the classroom 6

On reducing to the simplest form, we get

8 8÷2 4 = = 6 6÷2 3

Boys in the classroom Total number of students in the classroom 6 = 14

2 Ratio of boys in the classroom to the total number of students =

On reducing to the simplest form, we get

6 6÷2 3 = = 14 14 ÷ 2 7

3 Ratio of girls in the classroom to the total number of students =

On reducing to the simplest form, we get

8 8÷2 4 = = 14 14 ÷ 2 7

Girls in the classroom Total number of students in the classroom 8 14

Let’s Warm-up Match the ratios to their simplest forms. Ratios

Simplest Form

6:12

1:5

5:25

4:5

3:9

1:2

8:10

1:4

4:16

1:3

I scored ____________ out of 5.

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Understanding Probability Real Life Connect

The twins Ajay and Rocky get a box of marbles for their birthday. They are distributing these marbles among themselves. Ajay: R ocky, how do we distribute these marbles? I want all of the red marbles. Rocky: A jay! Let’s put our hand inside the bag and pick one marble at a time. I’m sure you will get red every time you pick a marble. Ajay: I think it’s impossible to pick a red marble each time, Rocky. I see very few red marbles in the bag. We can infer from the above conversation that Ajay and Rocky are trying to predict whether Ajay will be able to pick all of the red marbles. In various life situations, we can often not accurately predict the outcome, whether the event is sure to happen or impossible. In such cases, we use words like sure, impossible, likely, unlikely or equally likely. These terms convey the uncertainty about the happening of an event. To measure this uncertainty, we use the concept of probability.

Chance Understanding Probability Chance is the likelihood of some event happening. It can be expressed using words such as sure, impossible, unlikely, even chance, likely and certain. Let us see some examples. Impossible Event: Bird flying underwater

Sure Event: The sun rising in the East

When events have more, less or an even a chance of happening, they can be called likely, unlikely or equally likely, respectively. Likely Event: Carrying an umbrella during the monsoon

Unlikely Event: Child going to school on a Sunday

Equally likely Event: Tossing a coin and getting heads

MONSOON SEASON

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The chance that an event will happen is called the probability of that event. Look at the spinner. Let’s look at the chances of the wheel stopping at different colours. Wheel stopping at blue:

Wheel stopping at red:

Equally likely, as 6 out of 12, which is half of the parts, are shaded in blue (light blue and dark blue).

Unlikely, as 3 out of 12 parts are shaded in red.

Wheel stopping at green:

Wheel stopping at blue, red or yellow: Sure, as wherever the wheel stops, it will stop on one of these colours.

Impossible, as the colour is not on the wheel.

Example 1

Example 2

Do It Together

What are the chances of the given events when you roll the dice? 1

Getting a 3, 4 or 5. Equally likely

Getting a 6. Unlikely

Getting a number from 1 to 6. Sure

Getting the number 0. Impossible

Name one event for each. 1

Sure: Birds flying in the air

Impossible: A man flying in the air

Equally likely: Getting an odd or even number on rolling a dice

Describe the chance or likelihood of picking different colour balloons, without looking. 1

Equally likely Green balloon – _______________

White balloon – _______________

Green or orange balloon – _______________

Red balloon – _______________

Do It Yourself 6A 1

Write the chance of the given events happening, using words like sure, impossible, likely, unlikely and equally likely. a Getting a number greater than 5 on rolling a dice b Getting heads on tossing a coin c A lion flying in the sky d A child going to school on Monday e Picking a pencil from a box of pencils

Chapter 6 • Introduction to Probability

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Ramesh spins the spinner as shown. What is the chance of the following events? a Getting an even number. b Getting a number from 1 to 10. c Getting the number 5 or 6. d Getting the number 20.

Ajay spins the spinner. What is the chance of the spinner stopping at: a yellow? b red? c green?

You are playing a game using the spinners shown. Answer the questions.

a You want to move down. On which spinner are you more likely to spin “Down”? Explain.

b You want to move forward. Which spinner would you spin? Explain. 5

Spinner B

Forward

Down

Reverse

Down

Name one event for each. a Likely

Spinner A

b Unlikely

c Impossible

Forward

Up Reverse

d Sure

Word Problem 1

Era draws a card from a set of cards numbered from 1 to 40. Write the chance of drawing these number cards, in words. a Factor of 40

b Even-number card

Theoretical Probability Ajay and Rocky want to know the probability of picking a red marble from the bag of marbles. Let us help Ajay and Rocky understand how to find the probability of getting a red marble. The probability of an event is the number that measures the chance or likelihood that an event will occur. Probabilities are between 0 and 1, including 0 and 1. We can sometimes measure probability using fractions. Sure or Certain Impossible Less likely Equally likely More likely 0

1 4

1 2

3 4

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Let us understand some terms Outcomes and Events Experiment: An action that produces clearly defined results is called an experiment. Flipping a coin and rolling dice are examples of an experiment. Outcome: Each possible result of an experiment is called its outcome. For example, getting a tail when a coin is tossed is an outcome. Event: A collection of one or more outcomes is an event. Favourable Outcome: The outcome that is the same as what we guessed is called a favourable outcome.

Finding the probability of an event When all possible outcomes are equally likely, the probability of an event is the ratio of the number of favourable outcomes to the number of possible outcomes. The probability of an event is written as P(event). P(event) =

number of favourable outcomes number of possible outcomes

This is also known as the theoretical probability. Let us now find the probability of getting a red marble from the bag of 20 marbles. Given below are the marbles that were in the bag.

The bag has a total of 20 marbles. There are only 4 red marbles in the bag. The chance of selecting a red marble is 4 out of 20. We can write the probability of the event as: Number of favourable outcomes Number of possible outcomes Example 3

4 1 = 20 5

Probability of an Event

Ramesh rolls the dice. What is the probability of rolling an odd number? There are 6 numbers on a dice. 3 of them are odd numbers 1, 3 and 5. P(event) = P (odd) =

number of favourable outcomes number of possible outcomes

3 6

1 On simplifying the fraction, we get, P (odd) = 2 Therefore, the probability of rolling an odd 3 1 number is or . 6 2 Chapter 6 • Introduction to Probability

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Did You Know? The modern mathematical theory of probability has its roots in attempts to analyse games of chance by Gerolamo Cardano in the sixteenth century.

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Example 4

Rakesh randomly chooses one of the letters in the word EXPLORE. What is the theoretical probability of choosing a vowel? P(event) =

number of favourable outcomes number of possible outcomes

There are 3 vowels in the word EXPLORE (e, o and e). There are a total of 7 letters. 3 7 There are 10 cards in a box which are marked with distinct numbers from 1 to 10. If a card is drawn without looking, what is the probability of drawing a prime number card?

Therefore, the probability of choosing a vowel is . Do It Together

Cards in the box are 1, 2, _______________ The number of cards that have prime numbers is _______________ Total cards = _______________ Therefore, the probability of drawing a prime number card = _______________.

Do It Yourself 6B 1

Amit randomly chooses a letter from a hat that contains the letters A through K. a What are the possible outcomes? b What are the favourable outcomes of choosing a vowel?

Ajay spins the colour spinner. Answer the questions. a How many possible outcomes are there? b How many times can the arrow land on red? c How many times can the arrow land on all the colours but not purple? Sameer spins the number spinner shown. Answer the questions. a How many possible outcomes are there?

d How many times can the arrow land on an odd number?

c How many times can the arrow land on an even number?

b What are the favourable outcomes of spinning a number greater than 3?

e How many times can the arrow land on a prime number? Use the spinner to find the theoretical probability of the event for the given events.

e Spinning a number less than 7

f Spinning a 9

d Spinning a multiple of 2

c Spinning an odd number

1 3

b Spinning a 1

a Spinning red

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A dodecahedron has 12 faces numbered 1 through 12. Find the probability and describe

the likelihood of each event. a Rolling a 1

b Rolling a number less than 9

c Rolling a multiple of 3

d Rolling a number greater than 6

Each letter of the alphabet is printed on an index card. What is the theoretical probability of randomly choosing

There are 16 females and 20 males in a class. What is the theoretical probability that a randomly chosen student

A box contains 6 black crayons, 4 blue crayons, 5 red crayons, 3 yellow crayons and 2 white crayons. One crayon is

any letter except Z?

will be a female?

chosen at random. Write each probability as a fraction. a P(black)

b P(blue)

c P(not white)

e P(black or blue)

f P(blue, red, or yellow)

d P(pink)

Numbers from 1 to 25 are written on slips of paper and one is selected at random. Write each probability as a fraction.

a P(odd number)

b P(three-digit number)

e P(prime number)

f P(number greater than 19)

c P(not 4)

d P(positive number)

10 What is the probability that a month picked at random ends with the letter y? 11 Amit calculates the probability of getting a number less than 3 when randomly choosing an integer from 1 to 10. Favourable outcomes 3 = Total outcomes 10 Describe Amit’s error and give the correct probability. Write the numbers that are less than 3. 12 What is the probability of guessing the correct answer to a multiple-choice question if there are 5 choices? 13 What is the probability of guessing the correct answer to a true-false question?

Word Problems 1

Anand plays a word board game. He places 98 lettered tiles and 2 blank tiles in a bag.

Players will draw tiles from the bag one at a time without looking. What is the probability that the first tile drawn will be: a A blank tile?

b Labelled with a letter?

There are 12 pieces of fruit in a bowl. Seven of the pieces are apples and two are peaches. What is the probability that a randomly selected piece of fruit will not be an apple or a peach?

Chapter 6 • Introduction to Probability

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Experimental Probability Ajay and Rocky think of checking how many red marbles Ajay gets when he picks marbles at random from the bag. They start picking the marbles one by one from the bag and listing the frequencies in a table. Marbles Picked

Frequency

Red

Blue

Green

Yellow

He is surprised to note that he picked red marbles as many as 7 times in 20 trials. Probability that is based on such observations or repeated trials of an experiment is called experimental probability.

Think and Tell How does the experimental probability compare with the theoretical probability of picking a red marble?

Number of Times the Event Occurs P(event) = Total Number of Trials The table shows the results of spinning a coin 25 times. What is the experimental probability of spinning heads? Heads

As the number of trials increases, the experimental probability gets closer to the theoretical probability.

The experimental probability of a red marble 7 being picked is . 20

Example 5

Remember!

Tails

Example 6

The table shows the results of rolling a dice 50 times. What is the experimental probability of rolling an odd number? Number Rolled Frequency

6 25

Therefore, the experimental probability is Do It Together

The table shows 10 ones, 8 threes, and 11 fives.

So, an odd number was rolled 10 + 8 + 11 = 29 times in a total of 50 rolls.

Heads was spun 6 times in a total of 6 + 19 = 25 spins. number of times the event occurs P(event) = total number of trials P(heads) =

P(event) =

number of times the event occurs total number of trials

29 50 29 Therefore, the experimental probability is . 50 P(odd) =

6 . 25

The spinner shows 3 different colours and 3 numbers, one on each section. The spinner is spun 15 times, and the outcome is recorded in the table below. Find the probability of the spinner stopping at the numbers 6 and 7. Outcome

Number of Times

8 7

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number of times the event occurs total number of trials P(6) = _______________ P(7) = _______________ P(event) =

The probability of the spinner stopping at 6 = _________. Probability of the spinner stopping at 7 = __________.

Do It Yourself 6C 1

Use the bar graph to find the experimental probability of the event. a Spinning a 6

10 Times Spun

b Spinning an even number c Spinning an odd number d Spinning a number less than 3 e Spinning a 7 f Not spinning a 1 2

Spinning a Spinner

12 8 6 4 2 0

3 4 5 Number Spun

A spinner has four sections marked A, B, C and D. The table shows the results of several spins. Find the experimental probability of spinning each letter as a fraction in its simplest form. Letter

Frequency

Mala tossed a coin many times. She got 40 heads and 60 tails. She said the experimental probability of getting 40 heads was . Explain and correct her error. 60

Malati has a bag of marbles. She removes one marble at random, records the colour and then places it back in the bag. She repeats this process several times and records her results in the table. Find the experimental probability of drawing each colour.

Colour

Red

Blue

Green

Yellow

Frequency

One hundred and twenty randomly selected students at a school are asked to name their favourite sport. The results are shown in the table. Sport

Cricket

Baseball

Football

Hockey

Other

Number of Responses

Find the experimental probability that a student selected at random has the response given. a P(cricket)

b P(hockey)

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c P(baseball)

d P(football)

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In a survey, 125 people were asked to choose one card out of five cards labelled 1 to 5. The results are shown in

the table. Compare the theoretical probability and experimental probability of choosing a card with the number 1. Cards Chosen Number

Frequency

The bar graph shows the results of spinning the spinner 200 times.

Spinning a Spinner

2 3 4 Number Spun

40 5

35 30

Times Spun

Compare the theoretical and experimental probabilities of the events. a Spinning a 4

b Spinning a 3

c Spinning a number greater than 4

d Spinning an odd number

There are 700 students at a high school. You survey 75 randomly selected students and find that 60 plan to go to

college after completing high school. How many students are likely to go to college after completing high school?

Word Problem 1

A company produces electronic devices, and it is known that 5 out of 100 of the devices are defective. A quality control inspector randomly selects a device from a batch of 100. What is the probability of selecting a device which is not defective?

Points to Remember •

Probability measures the chance of an event happening.

•

An event with even chances has the same likelihood of happening and not happening.

• • • • • • • • •

An impossible event has no chance of happening.

A good chance event is more likely to happen than not to happen. A certain event has a sure chance of happening.

An experiment is an investigation or a procedure that has varying results. The possible results of an experiment are called outcomes. A collection of one or more outcomes is an event.

The outcomes of a specific event are called favourable outcomes.

The probability of an event is a number that represents the likelihood that the event will occur.

Probability that is based on repeated trials of an experiment is called experimental probability.

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Math Lab Rolling Dice! Setting: In groups of 3. Materials Required: One six-sided dice, paper and pencil. Method: 1

Write all the possible outcomes of rolling a dice.

Roll the dice and note the outcome on paper.

Repeat this experiment 36 times.

Draw a table and list the frequency of outcomes in it.

5 Choose an expression from those in the list below that best describes the probability of each outcome.

Impossible

Unlikely

Even chance

Likely

Certain

We know that all numbers on a fair dice are equally probable. Has your dice turned out to be fair?

Chapter Checkup 1

Write the chance of the given events happening, using words like sure or impossible.

What is the chance of the events happening? Fill in the blanks with words like sure, impossible, likely, unlikely or

a Kitten with wings

b Child wearing woollens in winter

equally likely.

a January coming before February is a/an _______________ event. b The sun setting in the East is a/an _______________ event. c A child going to school on a Wednesday is a/an _______________ event. 3

Use the labels on the right to describe the chance that a dice when rolled will show: a An odd number

b 6

c A number greater than 6

d Zero

e A number less than 10

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Impossible Unlikely

Equally likely

Likely

Sure

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b A number less than 3 is not rolled.

c An even number is rolled.

d An even number is not rolled.

e A 5 is not rolled.

f A 7 is not rolled.

You spin the spinner. Find the favourable outcomes of the event and the number of ways the event can occur. a Spinning a 1

b Spinning a 3

c Spinning an odd number

d Spinning an even number

e Spinning a number greater than 0

f Spinning a number less than 3

1 1

The bar graph shows the results of picking a numbered tile from

a A number less than 3 is rolled.

A dice is rolled once. List the outcomes in each of these events.

Picking a Numbered Tile

a bag 100 times. Use the bar graph to find the experimental

25 Times drawn

probability of the given events. a Picking a 2 b Picking an even number c Not picking a 5

d Picking a number less than 3

15 10 5 0

Number Picked

d Spinning a 4

c Spinning an even number

b Spinning a 1

a Spinning blue

Use the spinner to find the theoretical probability of the events.

Each student in your class voted for his or her favourite day of the week. Their votes are shown in the circle graph. A student from your class is picked at random. What is the probability that this student’s favourite day of the week is Sunday?

Favourite Day of the Week Other, 6

Sunday, 6

Friday, 8

Saturday, 10

Ten coloured discs are placed in a hat. Five are red, three are yellow and two are black. If one disc is drawn from the hat, what is the probability that the disc will be: a Red?

b Black?

e Blue?

f Red, yellow or black?

c Red or black?

d Not black?

10 The 26 letters of the alphabet are written on cards and placed in a box. If one card is picked at random from the box, what is the chance that the letter on the card is: a X?

b A vowel?

c M or N?

d A letter in the word MATHEMATICS?

100

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11 The numbers from 1 to 25 are written on cards. If one card is chosen at random, what is the probability that the number on the card will be: a An odd number?

b A multiple of 5?

c A factor of 24?

d A number that contains the digit 2?

1 12 If the probability of an event is , how many times, on average, would you expect it to occur in 20 trials? 5 13 A bag contains five red balls numbered 1 to 5 and seven blue balls numbered 1 to 7. If a ball is chosen at random, what is the chance of choosing: a A red ball?

b A ball numbered 3?

d The blue ball numbered 1?

e An even-numbered ball?

c A ball numbered 6?

Word Problem 1 A library has 3000 fiction books and 4000 non-fiction books. What is the probability that a book selected at random is fiction?

N O N F I C T I O N

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F I C T I O N

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Simple Equations

Let’s Recall We already know how to frame algebraic expressions. Let us look at these pictures and try to guess the algebraic expressions formed by each of them. We don’t know the amount of money in the bag. So let that amount be ₹x.

We also have a 10-rupee coin with us. So, the total amount we have is ₹(x + 10). Here, x is the variable and 10 is a constant. Now, we have two bags with the same amount of money in each. Let the amount of money in each bag be ₹x.

So, the sum of money in the two bags would be ₹2x. We also have a 10-rupee coin. So, the total amount of money we have is ₹(2x + 10).

Let’s Warm-up Match each statement with its expression. Statement

Expression

1 3 more than twice a number

2n − 5

2 2 less than five times a number

3p + 2

3 2 more than thrice a number

2n + 3

4 5 less than twice a number

3m – 5

5 5 less than thrice a number

5n − 2

I scored __________ out of 5.

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Mean, and in Mode LinearMedian Equations One Variable Real Life Connect

Bhakti is a college student planning a summer vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays ₹75 per hour. She opened a savings account with an initial deposit of ₹500 on January 1. If the summer vacation begins April 1 and the trip will cost approximately ₹11,000, how many hours will she have to work to earn enough to pay for her vacation?

Forming Linear Equations Let the number of hours worked by Bhakti be x. She will be earning ₹75 per hour. So, the money earned by her would be ₹75x. She already has ₹500 in her savings account. So, the total money saved by Bhakti = 75x + 500. We know that the cost of the trip = ₹11,000. So, 75x + 500 = 11,000 This is an equation in x. You can get the minimum number of hours worked by Bhakti to save ₹11,000 by solving this equation. We will solve this equation later in this chapter using different methods. Let us form a few more equations of this type. Consider this statement: ‘Seven added to a number equals 12’. We can write the statement in an equation as: x + 7 = 12. In the expression (x + 7), the exponent of x is 1, so it is a linear expression. Also, (x + 7) has only one variable. Hence, we can call x + 7 = 12 a linear equation in one variable. the variable

equality x + 7 = 12 x + 7 = LHS;

equation 12 = RHS

Remember! The word linear means forming a line. The graph of a one variable equation with a highest power of one is a straight line. That is why such equations are called linear.

In an equation, the left-hand side (LHS) is equal to the right-hand side (RHS). Equations can have one or more variables, and each of those variables can have a power of one or more. However, an equation which involves only one variable with the highest exponent as 1 is called a linear equation in one variable. A linear equation in one variable is an equation that can be written in the form ax + b = 0, where a and b are constants and a ≠ 0. For example: 2x + 3 = 7, x + 2 = 9, and a + 3 = 5. What about x2 + 4 = 13? Is it a linear equation? No, because the variable x has the exponent 2.

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x + 3y = 5 is a linear equation in two variables. 2x + 3y + z = 25 is a linear equation in three variables. Example 1

Write the given mathematical statements as linear equations. 1 The sum of x and 8 gives 12.

2 Three more than 2 times of x gives 17.

The sum of x and 8 is x + 8.

Two times x is 2x. Three more than two times x is 2x + 3.

Equation is x + 8 = 12. Example 2

Equation is 2x + 3 = 17.

Convert the given equations into statements. 1 x + 3 = 8

2 3x + 5 = 26

The sum of x and 3 gives 8. Example 3

Five more than three times x gives 26.

Akash scored 15 out of 41 points for the home team in a basketball game. His teammates scored x points in total. Write an equation to represent the given situation. Points scored by Aakash + points scored by his teammates = 41 15 + x = 41; Thus, 15 + x = 41 represents the given situation.

Do It Together

Sahil scored twice as many runs as Aakash. Together, their runs fell four short of a century. Form an equation to represent the situation. Let the runs scored by Aakash be x. Sahil scored twice as many runs as Aakash = __________ Together, their runs fell four short of 100 = 100 − 4 = 96 This situation can be represented in the form of an equation as: x + _____ = __________. On simplifying both the sides of the equation, we get, ____=____.

Do It Yourself 7A 1

Which of the following are equations? a 5x + 5 = 35

b 3a – 7 + 24

5x 3 18 + = 2 2 5

d 8x −

1 43 = 2 2

Which of these are linear equations? a x + 8 = 13

b x+9

e x2 + 5 = 14

f 9x + 7y + z = 24

c 5x + 6 = 8

d x + 8y = 6

Tick the linear equations in one variable and cross the others. a x + 9 = 14

b x + 12 = 4

e x2 + 7 = 29

f 6x + 8y + 3z = 38

c 4x + 5y = 8

d 7x + 3 = 10

104

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State True or False. a 3p + 2 < 7 is not an equation.

__________________

b 2a + 6 – 11 is an equation.

__________________

c 2x2 + 3 = 19 is a linear equation.

__________________

d 8x + 3y = 9 is a linear equation in one variable.

__________________

8 = 4 denotes a numerical equation. 2

Form equations for each of the given statements. a A number added to half of itself is 9.

b The sum of thrice a number and 5 is 17.

c Half of a number is 15.

d Half of the sum of 32 times of a number and 5 is 27.

e Two more than the number of frogs is 8.

f The number of students increased by 10 is 38.

Write the given equations in statement form. a y – 9 = 81

b 4q – 5 = 15

e 4x + 12 = 28 7

1 y+2=8 f 4

c 30z = 1650

Match each statement with its equation. Statement

Equation

a One-tenth of a number gives 7.

10 – x = 7

b 10 subtracted from a number gives 7.

x + 27 = 77 1 x=7 10 x – 10 = 7

c A number subtracted from 10 gives 7. d The sum of a number and 27 is 77. 8

d 9x + 5 = 50

Write an equation for the given situations. a I have m candies. Lina has 3 less candies than I have. We both have 18 candies altogether.

__________________

b Tina is c years old. Her 12-year-old sister is 3 years older than Tina.

__________________

c Rajat is twice as old as his brother Rakesh. The sum of their ages is 18 years.

__________________

d The length of a rectangle with a perimeter of 18 cm is 3 cm more than its breadth.

__________________

Word Problems 1

The length of a rectangular hall is 4 metres less than 3 times the breadth of the hall. What

A hotel has x rooms on each floor. There are 217 rooms on 7 floors. Write an equation to

is the length if the breadth is b metres? represent this situation.

Solving Equations: Same Operation on Both Sides Earlier in this chapter we learnt how to set up an equation for the given situation. Now, we will learn how to solve the equation formed.

Chapter 7 • Simple Equations

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1 If any number is added to one side of an equation, then the same number must be added to the other

side of the equation to maintain the equality between the LHS and RHS of the equation.

2 If any number is subtracted from one side of an equation, then the same number must be subtracted

from the other side of the equation to maintain the equality of the equation.

3 If one side of an equation is multiplied by a number, then the other side must also be multiplied by the

same number to maintain the equality between the LHS and RHS of the equation.

4 If one side of an equation is divided by a number, then the other side must also be divided by the same

number to maintain the equality between the LHS and RHS of the equation.

In Bhakti’s situation, we wanted to know the number of hours she would work to save for her trip. We had set up an equation as: 75x + 500 = 11,000 where x is the minimum number of hours. Let’s solve this equation to get the value of the variable x. First, subtract 500 from both sides of the equation. 75x + 500 – 500 = 11,000 – 500 On simplifying both sides, we get, 75x = 10,500 Divide both sides by 75. This will give us just x on the LHS.

Remember! While solving a two-step equation like this, we always add or subtract the same number from both sides of the equation, then we perform multiplication or division on both sides of the equation.

75x 10,500 = ; So, x = 140 75 75 Therefore, Bhakti would need to work for a minimum of 140 hours to save enough money for her trip. Example 4

Solve x + 8 = −5. We shall subtract 8 from both sides of the equation. x + 8 – 8 = −5 – 8 On simplifying both sides, we get, x = −13

Example 5

Solve 13a – 7 = 19. We shall add 7 to both sides. 13a – 7 + 7 = 19 + 7 On simplifying both sides, we get, 13a = 26 We will now divide both sides by 13. 13a 26 = ; So, a = 2. 13 13

Do It Together

Sahil has twice as much money as Jayant has. Together they have ₹240. How much money does Jayant have? Let the money with Jayant be ₹x. Money with Sahil = Twice as much as Jayant = _____ So, the required equation is (x + _____) = 240. On simplifying the equation, we get, So, x = _____ Thus, Jayant has ₹_____.

106

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Do It Yourself 7B 1

Write the first step you will use to separate the variable and then solve the equation. a

x–7=0

x+9=0

x−2=6

x+5=3

y − 8 = −4

y−1=1

y+5=5

y + 6 = −6

Solve the equation. a 4a = 60 e

b =9 3 z 6 f = 7 5

c =5 8 p 5 g = 6 18

9y = 12

d 5x = 28 h

30q = −5

30p = 60 7

Write the steps to solve the equation. a −x + 14 = 21

b 2x + 12 = 64

c 4z – 6 = 3

x 1 f =6 x−1=4 2 2 Eleven added to five times a number gives 61. Find the number.

Aakash solved the equation 12 + x = 20 and got 32. Is the answer correct? If not, find the correct answer.

Describe a real-world situation for each of the given equations and solve.

e 5+

a x – 75 = 150 7

b 249 + x = 350

c 9x = 72

What would you do to make equation 1 and equation 2 equivalent?

 1 X Equation 1: x −  = + 5  3 6

Equation 2: 6x – 2 = x + 30

Word Problems 1

Tina saved ₹125 from her pocket money. How much more money does she need to buy a

Sameer has 20 stamps more than three times the number of stamps Rakesh has. If both

Two small jugs and one large jug can hold 8 cups of water. One large jug can hold 2 cups

jewellery set that costs ₹200?

together have 140 stamps, how many stamps does Rakesh have?

more than the small jug. Find how many cups of water each jug can hold.

Solving Equations: Transposition We know the equation, 75x + 500 = 11,000 where x is the minimum number of hours. Instead of subtracting 500 from both sides of the equation, we can move 500 from the LHS to the RHS by reversing the operation from addition to subtraction. 75x = 11,000 – 500 On simplifying the equation, we get, 75x = 10,500 Chapter 7 • Simple Equations

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Instead of dividing both sides by 75, we can simply move 75 from the LHS to the RHS by reversing the operation from multiplication to division. 10,500 x= 75 On further simplification, we get, x = 140 Thus, Bhakti would have to work for a minimum of 140 hours to save enough money for the trip.

Did You Know? The idea of using letters to represent variables was first suggested by French mathematician François Viète (1540–1603) in his work.

We got the same value for the variable despite using different methods. The method used above for solving the equation is called the transposition method. In this method, we move terms from one side of the equation to the other with the purpose of separating the variable on one side and the constants on the other side. However, while moving the terms from one side of the equation to the other, we should change the operator sign as: + to −

− to +

× to ÷

÷ to ×

Remember! Transpose means to change the position or order of two things. Using the transposition method, we shift a term from one side to the other.

Example 6

Solve 5x – 7 = 13.

Example 7

Transposing −7 from LHS to RHS as +7, we get, 5x = 13 + 7 On simplifying the RHS, we get, 5x = 20 On transposing 5 from LHS to RHS, the operation changes from multiplication to division. 20 5 So, x = 4. x=

Do It Together

x+8 = 13. 3 On transposing 3 from LHS to RHS, the operation changes from division to multiplication. Solve

x + 8 = 13 × 3 On simplifying the RHS, we get, x + 8 = 39 On transposing 8 from LHS to RHS, the operation changes from addition to subtraction. x = 39 – 8 So, x = 31.

Solve 0.8(2x − 3) – 0.5x = 6. On opening the brackets and simplifying the LHS, we get, 1.6x – 2.4 – 0.5x = 6 On transposing −2.4 from LHS to RHS, the operation changes from subtraction to addition. 1.6x – 0.5x = 6 + 2.4

Error Alert! The number lying exactly outside the brackets should not be moved to the other side before opening the brackets.

On simplifying both sides of the equation, we get, _________________ 108

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

_________________ On simplifying the RHS, we get, x = ___

Do It Yourself 7C 1

Solve each equation by the transposition method. a x + 4 = 10

b a + 9 = 13

d n + 2.6 = 4

e x–9=3

2 5 = 7 7 f x – 5.4 = 12.8

g −14 + a = 27

h p–

d −7m = −21 g

x – 6.5 = 15

Solve each of the equations. a 4y = 20

5 7 = 9 9

c x+

−9

= −5

b 90 = 15y

c −9a = 108

j =6 8 g h = 20 3.2

x = −11 3 2x = 30 4.5

Solve the given equations. a 6x + 14 = 16

b 7 + 5x – 3x = 21

c 1.2x + 4.8x = 18

d 6(5x − 2) + 2 = 30

e 7(x − 4) + 13 = 6

f 10y – 0.2(7 − y) = 19

What is the number? a I’m thinking of a number. If 60 is subtracted from it, the result is 4.

__________________

b The sum of 13 and an unknown number is 46.

__________________

c A certain number decreased by 21 is 81.

__________________

d The product of 7 and a number is 56.

__________________

e A number when divided by 7 gives the quotient 5.

__________________

f If 10 is subtracted from thrice a number, the answer is 35.

__________________

g A number is multiplied by 3 and then 5 is added to it to get the answer as 20.

__________________

Nine added to thrice a whole number gives 45. Find the number.

Two-thirds of a number is greater than one-third of the number by 3. Find the number.

Divide 38 into two parts so that one part is 18 more than the other.

The length of a rectangle is 5 m greater than its breadth, and its perimeter is 250 m. Find the length and breadth of the rectangle.

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Word Problems 1 of the cost of the bat. Find the 5

A bat and a ball together cost ₹1200. The cost of the ball is

A vegetable seller has 31 kg of onions. He divides the onions in two bags so that one bag

costs of the bat and the ball.

weighs 5 kg more than the other. Find the weight of the bags.

Applications of Simple Equations In general, we come across many real-life problems, which can be solved using the concept of linear equations in one variable. Let us solve a real-world problem using some steps. Kajal takes 10 rounds of a square garden. Find the side of the square if the total distance covered by her is 2560 m. Step 1: Note down what is given and what is required.

Step 2: Denote the unknown by some variable.

Given: The garden is square.

Let the measure of each side of this square garden be x.

Required: Measure of each side of a square garden. Step 3: Translate the statements into algebraic expressions. Taking 1 round of the square garden = Perimeter of the garden = 4x

Step 4: Form an equation as per the condition given in the problem. 10 × 4x = 2560

Step 5: Solve the equation to find the value of the variable. 2560 ; x = 64 m 10 So, each side of the square garden measures 64 m. 4x =

Example 8

The sum of two numbers is 25. If one of the numbers exceeds the other by 7, then find the numbers. Let the first number be x; The second number will be x + 7. As per the given condition, x + x + 7 = 25 On simplifying, we get, 2x + 7 = 25

x = 18; x = 9 2 so, one of the numbers is 9. The other number = x + 7 = 9 + 7 = 16. So, the required numbers are 9 and 16. Example 9

The sum of two consecutive numbers is 13. Find the numbers. Let one number be x; Its consecutive number will be (x + 1). As per the given condition, x + (x + 1) = 13 On simplifying, we get, 2x + 1 = 13

x = 12; x = 6 2 So, one of the numbers is 6. The other number is 6 + 1 = 7.

110

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Example 10

The cost of three pens and four pencils is ₹248. If one pen costs ₹15 more than one pencil, then find the cost of each pen and each pencil. A pen costs ₹15 more than a pencil. So, the cost of the pen is dependent on the cost of the pencil. Hence, let us assume the cost of one pencil to be x. Then the cost of a pen = 15 + x The cost of 4 pencils = 4x and the cost of 3 pens = 3(15 + x) Total cost of 3 pens and 4 pencils = ₹248 3(15 + x) + 4x = 248 On opening the brackets, we get, 45 + 3x + 4x = 248 On simplifying, we get, 45 + 7x = 248

x = 203 ; x = 29 7 Thus, the cost of a pencil is ₹29. And the cost of a pen = 15 + x = 15 + 29 = ₹44. The cost of each pen is ₹44, and the cost of each pencil is ₹29. Example 11

There are 50 paise and 20 paise coins in a purse. The number of 20 paise coins is 4 times that of 50 paise coins. If the total value of the money is ₹26, find the number of coins of each denomination. Let the number of 50 paise coins be x. Then, the number of 20 paise coins will be 4x. Therefore, the value of x coins of 50 paise = 50x paise Also, the value of 4x coins of 20 paise = 20 × 4x paise = 80x paise The total value of the money = ₹26 or 2600 paise As per the given condition, we have, 50x + 80x = 2600 On simplifying, we get,

x = 2600; x = 20 130 Since, x = 20, 4x = 4 × 20 = 80 Hence, there are 20 coins of 50 paise and 80 coins of 20 paise. Do It Together

The length of a rectangular plot exceeds its breadth by 5 metres. If the perimeter of the plot is 142 metres, find the dimensions of the plot. Let the breadth of the plot be x metres. So, the length will be (x + 5) metres. We know that perimeter of a rectangle = 2(Length + Breadth) As per the given condition, 2(x + 5 + x) = 142 metres

Chapter 7 • Simple Equations

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On simplifying the LHS, we get, 2(2x + 5) = 142 OR __________ On transposing 10 from LHS to RHS, the operation changes from addition to subtraction. 4x = 142 − 10 On simplifying, we get,

x = 132; x = _____ 4 Breadth of the rectangular plot is ___ metres. Length of the rectangular plot is ___ metres.

Do It Yourself 7D 1

Twice the number decreased by 22 is 48. Find the number.

A number when added to its half gives 72. Find the number.

The sum of two numbers is 95. If one exceeds the other by 3, find the numbers.

Divide 20 into two parts so that one part is 8 more than the other. What are the two parts?

Find three consecutive numbers so that the sum of the second and the third exceeds the first by 24.

The angles of a triangle are x°, (x + 40)° and (x − 10)°. Find the angles.

Three sides of a triangle measure (2x + 3) units, (2x + 5) units and 3x units. Find the measure of the sides if its perimeter is 148 units.

Word Problems 1

The age of Olivia’s mother is 5 years more than three times Olivia’s age. Find Olivia’s age if

1 Five years from now, Satish will be 1 times his present age. How old is he now? 2

There are 320 books and magazines altogether in a bookshop. There are three times as

The cost of five books and three pens is ₹730. If a books costs ₹50 more than a pen, then

Heena’s mother is 4 times as old as she is. After 20 years, her age will be twice that of

Savita is Tina’s elder sister. The sum of the ages of Savita and Tina is 18 years, and the

her mother is 44 years old.

many books as magazines. How many books are there?

find the cost of the book and the pen.

Heena’s. Find their present ages.

difference of their ages is 4 years. Find the ages of Savita and Tina.

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Points to Remember •

An equation is a statement of equality which has one or more than one variable.

•

An equation in one variable is called a linear equation if the highest power of the variable is one.

•

The value of the variable of an equation is called the solution of the equation.

•

For solving an equation, we can: a Add the same number on both sides. b Subtract the same number from both sides. c Multiply both sides by the same non-zero number. d Divide both sides by the same non-zero number.

• The process of shifting a number from one side of an equation to the other by changing the sign is called transposition.

Math Lab Modelling Simple Equations! Setting: In groups of 3. Materials Required: Rectangular paper strips of blue and square paper strips of yellow and pink. Method:

1 Make rectangular paper strips of blue and square paper strips of yellow and pink from chart paper. 2 A blue rectangular strip represents a variable. A yellow square strip represents a positive constant, and a pink square strip represents a negative constant. 3

A yellow and a pink strip collectively represent zero.

Think of a linear equation, say 3x + 4 = 2x – 5.

5 Represent the linear equation by using rectangular and square strips. 6 Subtract or remove strips to solve the equation. 7 Repeat the same process to find the solution for other linear equations.

Chapter Checkup 1

Find the value of the unknown. 3 3 a x+ =1 8 8 5 3 d x+ = 19 19

Chapter 7 • Simple Equations

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b x−

1 2 = 3 3

e x – 3 = −20

c x+5=8 f x–3=7

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Complete the equations by performing the same operations on both sides. z z a b x + 1 = −6. So, x = _____ − 2 = 26. So, = ______ 2 2 x+1 x+1 c d 2x + 11 = 17. So, 2x = _____ − 3 = 27. So, = ______ 5 5 e 5(x − 3) = 25. So, (x − 3) = _____ f 3(y − 7) – 4 = 14. So, 3(y − 7) = _____

Solve the equations.

a 9x = 18

b 4a = 12

c −3y = −27

d −18 = −6x

e −4x = 8

f −7x = 3

Frame the equations and solve. a One and a half times a number is 150. What is the number? b I am thinking of a number. Five times that number is 65. What is that number? c 3 subtracted from 5 times of p gives 22. What is the value of p? d Double of a number added to 36 gives 48. What is the number? e 31 is 9 more than 4 times the number p. What is the value of p?

Solve. q a = 7 6 9 c x = 54 2 e x + 10 = −10

x =2 −29 −13 d x = 26 2 f x – 17 = −20 b

g y + 7 = 4

h x + 20 = −20

Find the solution of each of the equations. x a − 5 = 6 3 c 12y = 7y – 15

b 3x – 2 = 5x – 12

e 13y = −12y + 100

f 5(x − 3) = 10

g 9(x + 14) = 27

5x + 12 = 2

d −22 = 14 – 9x 3−x =6 5

When you divide a certain number by 13, the quotient is −18 and the remainder is 7. Find the number.

Find three consecutive even numbers whose sum is 96.

Find three consecutive numbers so that the sum of the second and the third number exceeds the first by 14.

10 If one side of a square is represented by 4x – 7 and the adjacent side is represented by 3x + 5, find the value of x. 11 Each of the two equal sides of an isosceles triangle is three times as large as the third side. If the perimeter of the triangle is 28 cm, find each side of the triangle.

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Word Problems 1 1 There are three packets of goods p, q and r. The weight of packet p is 1 kg more than that of 2 1 packet q, and the weight of packet r is 2 kg more than that of packet q. If their total weight is 2 32 kg, find the weight of the packets. he price of 2 notebooks and 3 textbooks is ₹120, but a notebook cost ₹10 more than the 2 T textbook. Find the cost of both the notebook and the textbook. amesh has a certain number of toffees. His sister has double the number of toffees 3 R compared to Ramesh. If together they have 54 toffees, then how many toffees does Ramesh have? man is 30 years older than his son. In 12 years, the man will be three times as old as 4 A his son. Find their present ages.

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Pairs 38 Angle and Parallel Lines Let’s Recall To help understand how much a line segment, line or ray is inclined on another, we have the geometrical idea of angles.

arm vertex Q

Steps to measure an angle using a protractor

P angle R

arm

Place the centre of the protractor over the vertex (the point where the two lines of the angle meet).

ake sure that the baseline of the protractor (the straight line at the bottom) is aligned with one of M the angle's arms.

ook where the other arm of the angle crosses the markings on the protractor. The number where L the arm crosses is the measure of the angle in degrees.

A Measure of the angle in degrees

Baseline

Centre of the protractor

Let’s Warm-up

Measure the angles. Fill in the blanks. 1

I scored _________ out of 4.

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Angle Pairs Real Life Connect

Once upon a time, in a small town, there was a group of curious friends. Their teacher had given them a special list of angles to explore. Teacher: Hey kids! Did you try adding the special list of angles I have given you? Mukesh: I tried adding them but all of them are always adding up to 90°. Saras: The angles I have are always adding up to 180°. Mukesh: Are these some special sets of angles, teacher? Teacher: T hese angles are related to some special geometrical concepts. We will learn more about them today.

Did You Know? Eratosthenes was an ancient Greek scholar

Pairs of Angles

who calculated the size of the Earth by using

Angle pairs are pairs of angles that can be classified on the basis of their properties. They can be complementary, supplementary, adjacent or linear pairs of angles.

different locations, Alexandria and Syrene.

the angles of the sun's rays at noon at two

Complementary and Supplementary Angles Two angles are considered complementary if their measures add up to 90°. If two angles are labelled ∠X and ∠Y, they are complementary if ∠X + ∠Y = 90°. Here, ∠X and ∠Y are also referred to as complements of each other. For example, the pairs of angles given below are complementary pairs of angles.

49°

41° 26°

49° + 41° = 90°

64° 26° + 64° = 90°

Two angles are considered supplementary if their measures add up to 180°. If two angles are labelled ∠X and ∠Y, they are supplementary if ∠X+ ∠Y = 180°. Here, ∠X and ∠Y are also referred to as supplements of each other. For example, the pairs of angles given below are supplementary pairs of angles.

49°

131°

49° + 131° = 180°

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105°

75°

105° + 75° = 180°

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Example 1

Example 2

Check whether the angles are complementary. 1 22°, 66°

2 30°, 46°

3 29°, 61°

22° and 66° are not complementary angles because their sum is 88°.

0° and 46° are not 3 complementary angles because their sum is 76°.

9° and 61° are 2 complementary angles because their sum is 90°.

Check whether the angles are supplementary. 1 79°, 101°

2 168°, 12°

3 126°, 64°

9° and 101° are 7 supplementary angles because their sum is 180°.

68° and 12° are 1 supplementary angles because their sum is 180°.

26° and 64° are not 1 supplementary angles because their sum is 190°.

Remember! Supplementary and complementary angles may or may not have a common arm.

Example 3

Do It Together

Find the complement and supplement of the given angles. 1 17°

2 83°

3 64°

Complement of 17° = 90° – 17° = 73°;

Complement of 83° = 90° – 83° = 7°;

Complement of 64° = 90° – 64° = 26°;

Supplement of 17° = 180° – 17° = 163°

Supplement of 83° = 180° – 83° = 97°

Supplement of 64° = 180° – 64° = 116°

Measure the angles and label them as complementary or supplementary angles. 1

Think and Tell

Are angles AOC and COB supplementary?

∠A = _____ ∠B = _____

B C

_________________________ A

∠C = _____ ∠D = _____

_________________________ 118

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Adjacent Angles

Two angles are classified as adjacent when they possess a common arm, a common vertex, and their noncommon sides extend to opposite sides of the common arm. For example: ∠XOY and ∠YOZ are adjacent angles with common vertex O and common arm OY. Example 4

Common Vertex

Y rm

o mm

Name the adjacent angles of the angles given below. 1 ∠POQ

3 ∠ROS

2 ∠TOS

1 Adjacent angles of ∠POQ are ∠POT and ∠QOR.

4 ∠POT Q

2 Adjacent angles of ∠TOS are ∠ROS and ∠POT. 3 Adjacent angles of ∠ROS are ∠TOS and ∠QOR. T

4 Adjacent angles of ∠POT are ∠POQ and ∠TOS.

Example 5

Draw an angle QOS. Draw an adjacent angle to ∠QOS whose shared arm is 1 OQ

2 P

Do It Together

O S

Fill in the blanks with the names of adjacent angles.

S O

Think and Tell

Is there any adjacent pair of angles on line XY?

1 Adjacent angles of ∠FCE are ∠DCE and ∠FCA. 2 Adjacent angles of ∠ACB are _____ and _____. 3 Adjacent angles of ∠ECD are _____ and _____. 4 Adjacent angles of ∠FCD are _____ and _____.

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Let us now measure the angles in the given figure to find the linear pair of angles. ∠ACG = 60°, ∠ACD = 100°, ∠GCB = 120°, ∠DCB = 80°, ∠ACF = 135°, ∠FCB = 45°.

Here, the linear pairs of angles are:

∠ACG and ∠GCB ∠ACD and ∠DCB ∠ACF and ∠FCB

Example 6

Example 7

Check whether the given adjacent angles are linear pairs of angles. 1 56°, 56°

2 90°, 90°

3 120°, 70°

Sum of angles = 56° + 56° = 112° (Not a linear pair of angles)

Sum of angles 90° + 90° = 180° (Linear pair of angles)

Sum of angles = 120° + 70° = 190° (Not a linear pair of angles)

Solve for x.

(4x + 10)°

4x + 10° + x – 20° = 180° (Linear pair of angles)

(x – 20)°

5x – 10° = 180° 5x = 190° ⇒ x = 38° Do It Together

Find the value of x in each case: 1

x°

2x°

(3x + 7)°

(2x + 18)°

(3x + 7)° + (2x + 18)° = ______

x + 2x = 180° (linear pair) __________ = 180° So, x = __________

5x + ______ = ______ So, x = ______

Vertically Opposite Angles Vertically opposite angles are pairs of angles that are formed by the intersection of two lines, and they share the same vertex while being opposite to each other. These angles are always equal in measure. K

Point of

intersection N

For example, ∠KON and ∠LOM are a pair of vertically opposite angles. Similarly, ∠KOL and ∠NOM are also a pair of vertically opposite angles. 120

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Let us see some important rules. The sum of all angles about a

The sum of all the angles lying on one side of a line at a specific point equals 180°.

point is equal to 360°. T

∠2 ∠3 P

∠4

O ∠1 ∠5

∠2

∠1

∠3 Q

For example: ∠1 + ∠2 + ∠3 + ∠4 + ∠5 = ∠360° Example 8

For example: ∠1 + ∠2 + ∠3 = 180°

Count the number of pairs of vertically opposite angles in the figure shown below. p

1 2

q m n

Two pairs of vertically opposite angles are formed when two lines intersect. 1 3 intersections: 6 pairs Example 9

2 4 intersections: 8 pairs

Look at the figure and answer the questions. 1 Are the following pairs of angles vertically opposite angles?

∠AXT, ∠ YXC: Yes

∠PCQ, ∠ BCQ: No

∠YBC, ∠ YBS: No

∠AYU, ∠ XYB: Yes

2 Find the vertically opposite angle to:

A T

∠PCQ - ∠XCB is vertically opposite to ∠PCQ.

U S

∠YBC - ∠SBR is vertically opposite to ∠YBC. Do It Together

Find the value of x in each case. 1

75°

2 (2x – 10)°

(3x – 15)°

We know that vertically opposite angles are equal. Hence, 3x – 15° = _____________________

60°

________ = ________ 2x = ________ x = ________

3x = 75° + 15° = _____________________ 3x = 90° ⇒ x = _____________________ Chapter 8 • Angle Pairs and Parallel Lines

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Do It Yourself 8A 1

Check whether the given angles are supplementary or not. a 80°, 100°

b 45°, 135°

b 72°

b 40°, 53°

b 19°

e 98°

c 78°, 33°

d 37°, 37°

e 46°, 44°

c 89°

d 41°

e 37°

Find the value of x. a

b (4X + 50)°

d 51°

Identify whether the pairs of angles are complementary, supplementary or neither. a

c 123°

Find the complement of the given angles. a 1°

e 70°, 110°

Check whether the given angles are complementary or not. a 27°, 63°

d 35°, 125°

Find the supplement of the given angles. a 33°

c 60°, 120°

 x °   2

(3x – 50)°

(2x + 15)°

(x – 20)°

 x °   3

(2x + 25)°

Find the value of the unknown variables. a

135° y°

y° y°

2a°

n° m°

30°

3a°

Draw a rectangle ABCD and identify all pairs of supplementary angles.

If the angles (3x + 10)° and (7x – 20)° are complementary angles, find the value of x.

a°

10 If the angles (4y + 13)° and (3y – 8)° are supplementary angles, find the measures of these angles. 11 An angle exceeds its complement by 30°. Find the measure of the angle. 12 The supplement of an angle is thrice its complement. Determine the angle. 13 The supplement of an angle is half its measure. Find the angle and its supplement. 14 Two complementary angles are in the ratio 5:4. Find the measures of the angles.

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15 Two supplementary angles are in the ratio 13:5. Find the measures of the angles. 16 If one angle measures 46° less than its supplement and another angle measures 20° more than its complement, find the measures of both angles.

Word Problem 1

Two intersecting roads form an X shape. If one of the angle formed at the intersection is 70°, what is the measure of the angle directly opposite to it?

Parallel Lines Real Life Connect

While having lunch, some school kids talked about their weekend experiences. Rajni: This weekend, I explored the deep forest near our village. Friend: Wow, that sounds thrilling! What happened next? Rajni: I noticed something interesting about the trees. Some trees had trunks that were equally spaced from one another, whereas others leaned on or crossed over each other.

Parallel Lines and Transversals The trees Rajni spotted in the forest area whose trunks were equidistant from each other are related to the geometry of parallel lines. Let’s understand parallel lines and the associated geometry in detail. Two or more lines are called parallel if they are in the same plane and don't cross even if you keep extending them in both directions. The distance between parallel lines is the same everywhere. We denote two lines p and q to be parallel as p ‖ q. Parallel lines are found all around us in our daily lives. Some of the examples from our daily life include window blinds, zebra crossings, lines on notebook paper and railroad tracks.

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When two or more lines are intersected by another line at distinct points, this line is called their transversal. The lines may or may not be parallel. l

b c

Transversal intersecting non-parallel lines

Transversal intersecting parallel lines

Angles Made by a Transversal k

Exterior angles:

Interior angles:

∠2, ∠3, ∠5, ∠8

∠1, ∠4, ∠6, ∠7 2

Pairs of alternate exterior angles: ∠3 and ∠5, ∠2 and ∠8 Pairs of interior angles on the same side of the transversal: ∠1 and ∠6, ∠4 and ∠7

Pairs of alternate interior angles: ∠4 and ∠6, ∠1 and ∠7 Pairs of corresponding angles: ∠2 and ∠6, ∠3 and ∠7 ∠4 and ∠8, ∠1 and ∠5 X

When two parallel lines are cut by a transversal:

50°

• Corresponding angles are equal. ∠AMX = ∠CLM

• Alternate interior and exterior angles are equal. ∠BML = ∠MLC

Example 10

B L

• The sum of interior angles on the same side of a transversal is 180°. ∠AML + ∠MLC = 180°. Conversely, if two lines intersected by a transversal have equal corresponding angles, equal alternate interior/exterior angles, and same-side interior angles adding up to 180°, then the lines are parallel.

130°

Think and Tell

If line a is parallel to line b and line b is parallel to line c, is line a parallel to line c?

Given q ꠱ r and p is the transversal. Find the value of all the labelled angles if ∠3 = 68°. From the figure,

∠3 + ∠4 = 180° (Linear pair)

68° + ∠4 = 180°

∠4 = 180° – 68° = 112°

∠4 = ∠2 = 112° (vertically opposite angles)

∠1 = ∠3 = 68° (vertically opposite angles) ∠7 = ∠3 = 68° (corresponding angles)

∠4 = ∠8 = 112° (corresponding angles)

3 4

6 5

7 8

∠2 = ∠6 = 112° (corresponding angles) ∠5 = ∠1 = 68° (corresponding angles) 124

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Check whether any of the pairs of lines are parallel or not based on the information given.

Example 11

1 For 2 lines to be parallel, the alternate

exterior angles should be equal. In this figure, the alternate exterior angles are 105° and 115°. So, line p is not parallel to line q.

105°

2 Let the two angles that form the linear pair be x.

Therefore,

x + 65° = 180° (Linear pair) x = 180° – 65° = 115° ince corresponding angles are equal, the S lines m and n are parallel. l

65°

115°

Example 12

Do It Together

115°

In the figure given below, SP ꠱ RQ and QP has been extended to T. Given ∠TPS = 95°, find the value of x°, y° and z°. From the figure, 95° + x° = 180° (Linear pair) T x° = 180° − 95° x° = 85° 95° y° = 95° (corresponding angles) S x° + z° = 180° (interior angles on the same side of the transversal) 85°+ z° = 180° z° = 180° – 85° z° = 95°

In the figure, it is given that PQ ꠱ XR, ∠QPR = 65°and ∠XRY = 60°.

z°

Find 1 ∠PQR

In the figure, ∠PQR = ∠XRY = 60° (_____________________angles)

y° Q

65°

2 ∠PRQ

x°

60° Q

Error Alert! Unless explicitly mentioned or indicated, do not assume lines to be parallel to each other.

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Do It Yourself 8B 1

Identify all the pairs of corresponding angles and pairs of alternate angles in the figure. P

1 5 8

2 6

Check whether the lines are parallel in each case, based on the known angles. a 110°

b c 102° 89°

109°

89° 72°

99°

109°

Find the value of the indicated unknown angles. a x°

32°

c d p°

m° 110°

q°

n°

72°

105°

g°

45°

80°

y°

h°

60°

In a trapezium PQRS, it is known that PQ ꠱ RS. Also, ∠PQR = 150° and ∠QPS = 120°. Find the value of the other

A line (DE) parallel to the side BC of ΔABC passes through A. The angles ∠DAB = 90° and ∠EAC = 60°. Find the

In the given figure, l ꠱ m. Also, p and q are transversals.

two angles.

measure of all the angles of the triangle.

Note that p is not parallel to q, find the values of a° and b°.

p b°

62° l m

a°

72°

Find the measures of the unknown angles. a 50°

35° y° x°

136° x° 116°

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Word Problem 1

Due to the effect of the wind, rain sometimes falls to the ground at an angle. If the rain makes

an angle of 60° with the clouds, at what angle would this rain hit the ground? Assume that rain drops travel in a straight line. Illustrate with a rough diagram.

Constructing Parallel Lines Constructing a line parallel to a given line l passing through an external point P. Step 1: Mark point Q on

Step 2: Draw line segment PQ

line l.

by connecting points P and Q. P

Step 3: From point Q, draw an arc with a chosen radius to intersect line l and line segment PQ at points B and A, respectively. P

A l

Step 4: Draw another arc from

Step 5: Draw an arc from point

Step 6: Connect points PD and extend the line

before, to intersect line segment

cutting the previously drawn arc

which is parallel to the given line l.

point P, using the same radius as

C with radius equal to AB,

PQ at point C.

at point D.

A l

A Q

on both sides to obtain the desired line m,

Constructing a line parallel to a given line at a fixed distance Step 1: Choose any two points A and B on line XY.

Step 2: Construct right angles ∠CAY and ∠DBY at points A and B respectively.

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Step 3: Draw arcs with centres at A and B, each with a 4

Step 4: Connect points EF and extend this line on both

maintaining a distance of 4 cm from it.

cm (say) radius, intersecting CA at point E and DB at point C E

sides to achieve the desired parallel line relative to XY, C

Checking for Parallel Lines Let’s say you have been given two lines p and q. To check whether these two lines are parallel or not, we can follow the steps given below: 1 Draw another transversal line that cuts across both lines p and q. 2 Now, where the transversal crosses lines p and q, you'll see some corresponding and alternate angles. 3 Using a protractor, measure the size of the corresponding angles (or alternate angles). 4 Compare the measurements of the corresponding angles (or alternate angles).

If they have the same measurement, that's awesome! It means the lines p and q are parallel. If the measurements are not the same, the lines are not parallel. Example 13

Draw an ∠EFG = 50°. Construct a line k so that it is parallel to EF and passes through G.

1 Draw an angle ∠EFG = 50° using a protractor. 2 Draw an arc from F with a chosen radius to intersect EF at B and FG at A.

C A

3 Draw an arc with the same radius from G to intersect FG at C.

50°

4 Draw an arc from C with a radius of AB to intersect the previous arc at D. 5 Join GD and extend both sides to get line k, parallel to line segment EF.

Example 14

Draw any ΔABC. Construct a line h, parallel to BA, and positioned 2 cm away from it so that it doesn’t intersect at any part of the triangle. 1 Draw ΔABC with sides of suitable length.

2 Choose any two points P and Q on the line segment AB.

3 Construct right angles ∠XPA and ∠YQA at points P and Q

respectively.

intersecting XP at point E and YQ at point F.

parallel line h at a distance of 2 cm from AB.

F h

4 Draw arcs with centres at P and Q, each with a 2 cm radius, 5 Join EF and extend this line on both sides to achieve the desired

Q C

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Do It Together

This is a right-angled triangle ΔPQR. Construct a line t, parallel to RQ and positioned 2.5 cm away from RQ so that it intersects the triangle.

Do It Yourself 8C 1

Check whether the following sets of lines are parallel or not by drawing a transversal and measuring alternate or corresponding angles. a

b c

Construct a line AB and select a point C that lies outside it. Use a ruler and compass to draw a line through C that

Draw any ΔPQR.

is parallel to AB.

a Draw a line m parallel to QR passing through P. b Draw a line n parallel to PQ passing through R. Check whether m and n are parallel to each other or not. 4

Draw a line parallel to a line p at a distance of 5.9 cm.

Draw a line labelled n and construct a perpendicular line from any point on n. Select a point Q located at a

distance of 5.5 cm from n along the perpendicular line. Finally, draw another line labelled m through point Q so that it is parallel to line n.

Construct ∠XYZ = 45°, ensuring that XY measures 4.2 cm and YZ measures 5.6 cm. Construct a line through

Z that is parallel to YX and another line through X that is parallel to YZ, intersecting at point D. Determine the measurements of XD, YD and ZD.

Word Problem 1

A child has a rectangular birthday greeting card of dimensions 10 cm × 8 cm. She wants to draw a little box in the centre to put in a birthday quote for her friend. All of the sides of

this box are 2 cm inside the respective edges of the greeting card. Suggest the construction steps to neatly draw this box on the greeting card.

[Hint: Construct a line that runs parallel to the edges of the greeting card, maintaining a distance of 2 cm from the edges.]

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Points to Remember • The sum of supplementary angles is 180°. • The sum of complementary angles is 90°.

• Adjacent angles have a common vertex and a common arm.

• A linear pair of angles share a common vertex and a common side, while their non-shared sides together create a straight line. • Vertically opposite angles are angles that are opposite to each other when two lines intersect.

• Parallel lines are two or more lines that are always the same distance apart and never meet, no matter how far they are extended. • A transversal intersects two or more lines at distinct points.

• When two lines are crossed by a transversal, corresponding angles are equal, alternate interior and exterior angles are equal, and the sum of interior angles on the same side of the transversal is 180 degrees. If these angle conditions are met, the lines are parallel.

Math Lab Parallel Line Art Materials Needed: Drawing paper, ruler, pencils, coloured pencils or markers Instructions: 1

Begin by drawing two parallel lines horizontally across the paper, leaving some space between

Choose a point near the top of the paper and draw a diagonal line (the transversal) that intersects

Label the points where the transversal crosses the parallel lines as A, B, C, D, and so on.

Now, start identifying and measuring different pairs of angles:

them.

the parallel lines.

•

Measure and draw corresponding angles that are equal across the parallel lines.

•

Draw alternate interior angles and alternate exterior angles, labeling them as you go.

•

Measure and label interior angles on the same side of the transversal, ensuring that their

sum is 180°.

Colour the angles using different colours to differentiate them and make your artwork visually

You can also create a legend on the side of the paper explaining the colour code for each type of

As you work on this activity, take the time to explain each angle type to yourself or a friend,

appealing. angle.

discussing how they relate to parallel lines and the transversal.

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Chapter Checkup 1

Check whether the given angles are supplementary. a 14°, 166°

b 150°, 50°

b 50°, 16°

b 119°

b 20°

d 44°, 46°

e 78°, 12°

c 126°

d 24°

e 166°

c 26°

e 86°

40°

Find the value of x. a

b 2x°

(x + 9)°

c x°

x°

(3x – 10)°

f 3x°

3x°

x°

4x°

Find the values of ∠ABC, ∠ACB and ∠CAB.

142°

2x°

x°

(2x + 6)°

c 36°, 54°

Find the complement of the given angles. a 47°

e 76°, 109°

Find the supplement of the given angles. a 34°

d 40°, 135°

Check whether the given angles are complementary. a 24°, 56°

c 36°, 135°

(2x – 10)° x°

4x°

In the figure given below, c || d || e. Lines a and b are the transversals. Find the value of g°, m°, y° and h°. b

A 137°

113°

B C

2x° x°

y°

113°

g°

m°

e h°

Identify the figures in which the lines are parallel. a

120°

130°

b p q

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70° 120°

105° s

m n

75°

d 110° 110°

b c

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Find the measure of all the angles created by the transversals if none of the lines in any of the figures are parallel. a

105°

55°

120°

140°

125°

65°

110°

q 130°

10 Find the measure of the unknown angles, as indicated in each figure. a

b 65°

y° c

130°

70° 85°

z°

m°

x° s

11 In the figure given below, l || m || n. Lines j and k are the transversals. Find the value of: a g° + h°

b x° + y°

c g°

d h°

e x°

f y°

g°

153°

y°

l m

x°

h°

53°

12 Find the values of w°, x°, y° and z° in the figure shown below. x° y°

z° 15° 75°

15° w°

75°

13 PQRS is a parallelogram. Given ∠QPS = 115°, find the value of the other angles of the parallelogram. 14 ABCD is a trapezium. The diagonals of the trapezium intersect at Q. Given, ∠QBA = 63° and ∠BAQ = 47°. Find the values of ∠QDC and ∠DCQ.

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Word Problem 1

The yellow lines in a parking space are parallel. What is the measure of ∠m?

115°

Chapter 8 • Angle Pairs and Parallel Lines

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Triangles and Their Properties

Let’s Recall Look at the given triangles. What do you observe about their sides and angles?

Triangles can be classified on the basis of their sides and angles. On the basis of sides Equilateral triangle

Isosceles triangle

Scalene triangle

All sides are equal. All angles are equal.

Two sides are equal. Two angles (base angles) are equal.

No equal sides. No equal angles.

On the basis of angles Acute-angled triangle

Obtuse-angled triangle

Right-angled triangle

All three angles are less than 90°.

One angle is more than 90°, and the rest are less than 90°.

One angle is equal to 90°, and the rest are less than 90°.

Let’s Warm-up Match the following. 1

60°, 70°, 50°

Isosceles triangle

90°, 45°, 45°

Obtuse-angled triangle

AB = BC = CA

Acute-angled triangle

100°, 30°, 50°

Equilateral triangle

AB = BC

Isosceles-right triangle

I scored __________ out of 5.

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Mean, Median and Mode and Their Properties Understanding Triangles Real Life Connect

Raj and Riya are travelling on a train. Suddenly, the train passes by a bridge on the other track. The children watch the bridge as it passes by. Raj: Riya, do you notice something in this triangle-based design? Raj: Just look at how accurate and precise the triangles are! Riya: The end of the straight beam is exactly at the centre of the adjoining ends! Raj: Not only that, the straight beam is also exactly perpendicular to the base!

Medians and Altitudes This straight beam here is the median. Median of a Triangle A median is a line segment that connects a vertex of the triangle to the midpoint of the opposite side. Each triangle has three medians, one from each vertex. The point where the medians intersect is called the centroid of the triangle. The centroid is often denoted by the letter “G”. It is also called the centre of gravity of the triangle.

Midpoint Midpoint

Centroid

Centroids in different types of triangles

Midpoint

G is the centroid of DABC. A

A R B

G Q

R C

Acute-angled Triangle

A G

P Q

Right-angled Triangle

Think and Tell

R B

Obtuse-angled Triangle

How will you show the medians in a triangle on the basis of their sides?

The properties of a median and a centroid • A median always lies inside the triangle.

• All three medians of a triangle are concurrent, i.e., they all intersect at a common point, that is, the centroid. • The centroid of a triangle divides each median in the ratio 2:1. For example, in DABC, AD and CE are the medians to sides BC and AB, respectively. The two medians meet at point G. AG:GD = CG:GE = 2:1

G D

• A centroid always lies inside the triangle irrespective of the type of the triangle. Altitude of a Triangle The straight beam that is exactly perpendicular to the base of a triangle is the altitude. We have learnt about medians of a triangle. Now, let us learn about altitudes.

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An altitude is a line segment perpendicular to a side of the triangle, and it extends from a vertex to the opposite side.

Altitudes

Each triangle has three altitudes, one from each vertex.

orthocentre

The point where the altitudes intersect is called the orthocentre of the triangle. The orthocentre is often denoted by the letter “O”.

Properties of an altitude:

• An altitude may or may not lie inside the triangle.

• All three altitudes of a triangle are concurrent, that is, they intersect at a common point called the orthocentre.

• The orthocentre is not always located inside the triangle. It can be inside, on the triangle, or outside the triangle, depending on the type of triangle. • The orthocentre lies:

Inside in an acute-angled triangle.

ii Outside in an obtuse-angled triangle.

iii On the vertex of a right-angled triangle.

A P

A R

B Q

Acute-angled triangle

Right-angled triangle

Obtuse-angled triangle

Remember! A median ‘bisects’ a side into two equal halves. The altitude is ‘perpendicular’ to the side it is drawn on from a vertex.

Example 1

BE is the median of DABC and CE = 4 cm. Find AC.

BE is the median in DABC, CE = 4 cm.

A median divides the opposite side in two equal parts. Therefore, CE = AE, E is the midpoint of AC

AC = AE + CE Putting the values, AC = AE + CE AC = 4 + 4 = 8 cm. Example 2

In triangle ABC, AD and CE are medians on sides BC and AB, respectively. The medians intersect at point G. AG = 2.4 cm, DC = 3 cm. Find AD and BC.

A E

AD is the median on the opposite side BC.

A median divides the opposite side in two equal parts. Therefore, BD = DC. BC = BD + DC

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BC = BD + DC = 3 + 3 = 6 cm G is the centroid where AD and CE intersect. Therefore, AG:GD = 2:1 (property of a centroid) AG = 2.4 cm GD = 2.4 ÷ 2 = 1.2 cm AD = 2.4 + 1.2 = 3.6, so AD = 3.6 cm. Example 3

Choose the correct option. 1 How many altitudes can a triangle have? a 1

b 2

c 3

d 4

2 The point at which the three altitudes of a triangle intersect is called the __________. a centroid

b centre

c orthocentre

d incentre

Remember! The centroid divides the length of the median in the ratio 2:1. The part of the median that is closer to the vertex is always greater than the one closer to the midpoint of the opposite side.

Do It Together

In DPSQ, PR and QT are medians on the sides SQ and PS, respectively. The medians intersect at point G. If PG = 8 cm and QR = 7 cm, then find SQ and PR. P

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

SQ = SR + ____. So, SQ = _____________ PG:GR = 2: ____ (Property of a centroid) PR = PG + _______ (G lies between P and R)

G R

So, PR = _________

Remember! In equilateral triangles, the median and the altitude are the same. So, the median and the altitude for equilateral triangles are “perpendicular bisectors” from the vertex on the side.

Do It Yourself 9A 1

Fill in the blanks.

a __________________ is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.

b ________ is a line segment perpendicular to a side of the triangle. c The point where the altitudes intersect is called the _________ of the triangle. d An altitude may or _______ lie inside the triangle. Chapter 9 • Triangles and Their Properties

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In a DABC, BD is the median on the opposite side AC. AD = 5 cm. Find the length of AC.

In a DABC, AD and CE are two medians on the opposite sides BC and AB, respectively. The two medians intersect at point G. If AG = 2 cm and BD = 4 cm, find AD and BC.

CF, EB and AD are the medians of a DACE. Use the given figure to answer the following questions. a If BG = 5 cm, find GE and BE.

b If CG = 16 cm, find GF and CF.

c If AD = 30 cm, find AG and GD.

d If GF = x, find GC and CF.

e If AG = 9x and GD = 5x – 1, find x and AD. 5

C D

G F

If the orthocentre and the centroid of the triangle are the same, then what kind of a triangle is it? Choose the correct option. a Scalene triangle

b Right-angled triangle

c Equilateral triangle

d Obtuse-angled triangle

(15x + 42)°

If PS is the median of DPQR, then find the value of ∠PSR.

Point M is inside a DXYZ and is equidistant from points X and Y. On which of the given segments must M be located?

a The line segment XY

b The perpendicular bisector of XY

c The midsegment opposite XY

d The altitude and the median of XY

10x – 7

5x + 3

Word Problem 1

Shyam, a carpenter, gets an order to design a triangular table with one leg. How do you think he would build the table? Choose the correct option. Give reasons for your answer. a Shyam would find the median of the triangular table.

b Shyam would find the altitude of the triangular table. c S hyam would find a point inside the table that is equidistant from the vertices of the base of the triangular table. d Shyam would find the orthocentre of the triangular table. e Shyam would find the centroid of the triangular table.

Properties of Angles of a Triangle Notice that the triangles used to design the triangular structure in the bridge look exactly of the same size and shape. The lengths of the sides and the measure of angles look the same such that when each of these are joined together, they give the bridge a symmetrical design. But how would the engineers be sure of building triangles of a certain measure such that the bridge also looks symmetrical? This is only possible by verifying the measures using some properties of triangles. Let us learn about them.

Angle Sum Property of a Triangle

The sum of all the angles of a triangle is 180°, but how do we know for sure? Let us take a triangular piece of paper and cut out its 3 angles. (∠A, ∠B and ∠C)

C C

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Place them together such that the vertices lie on the same point. We see that the 3 angles form a straight line or a straight angle. So, we can say that the sum of the 3 angles of a triangle is equal to 180°. Here, ∠A + ∠B + ∠C = 180°. Let us prove this using another method. Let ABC be any triangle. Through A, draw a line XAY parallel to BC and mark the angles, as shown in the figure. XY is parallel to BC, and the transversal AB cuts them. So, ∠1 = ∠4 (Alternate interior angles) XY is parallel to BC, and the transversal AC cuts them. So, ∠2 = ∠5 (Alternate interior angles) Adding both sides, we get ∠1 + ∠2 = ∠4 + ∠5.

Adding ∠3 to both sides we get,

A 3

∠1 + ∠2 + ∠3 = ∠3 + ∠4 + ∠5 ∠3 + ∠4 + ∠5 = ∠XAY = 180° (∠XAY is a straight angle)

Therefore, ∠1 + ∠2 + ∠3 = 180° or ∠A + ∠B + ∠C = 180°

Hence, the sum of all the angles of a triangle is 180°. Example 4

What is the measure of the third angle in a triangle where 2 angles measure 45° and 87°?

Do It Together

The angles of a triangle are (x + 5)°, (x + 10)° and (2x + 5)°. Find the measures of all the angles of the triangle. Using the angle sum property of a triangle,

The sum of angles in a triangle is 180°.

∠1 + ∠2 + ∠3 = _____

(x + 5)°

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

Let the missing angle be x.

____ + 20° = 180° So, x = _____

45° + 87° + x = 180° x = 180° – 132° = 48°

Using the value of x,

So, the measure of the third angle is 48°.

(x + 10)° = _____________

(x + 10)°

(2x + 5)°

(x + 5)° = _____________ (2x + 5)° = _____________ The measures of the angles are 45°, _____ and _____.

Exterior Angle Sum Property of a Triangle This property helps to establish the relationship between the interior angles and the exterior angles formed by extending the sides of a triangle. Let’s learn about the exterior angle sum property of a triangle. Consider a triangle as shown below with angles a, b and c. Extend one of the sides to form ∠e. The exterior angle

The two opposite interior angles b

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Notice that the interior opposite angles of the triangle when rearranged and joined together overlap ∠e exactly. So, ∠a + ∠b = ∠e. Statement: If the side of a triangle is extended, then the exterior angle formed is equal to the sum of the two interior opposite angles.

A 1

Proof: Let the side BC be extended to D, forming the exterior ∠ACD, i.e., ∠4. In the DABC, we have ∠CAB + ∠ABC + ∠BCA = 180°

(Sum of the angles of a triangle is 180°.) Step 1: ∠1 + ∠2 + ∠3 = 180°

●

Remember!

Step 2: ∠3 + ∠4 = 180° (Linear pair)

The exterior angle of a triangle is greater than each one of the interior opposite angles. Here, ∠ACD > ∠CAB and ∠ABC.

∠3 + ∠4 = ∠1 + ∠2 + ∠3 (From steps 1 and 2) ∠4 = ∠1 + ∠2 Hence, ∠ACD = ∠CAB + ∠ABC

Example 5

The interior angles of the triangle given on the side AB are in the ratio 2:3. Determine the angles of the triangle, where BC is extended to D and ∠ACD = 100°.

A 2x

Let the given interior angles be 2x and 3x.

Step 1

Step 2

Using the exterior angle property of a triangle,

∠CAB = 2x = 2 × 20° = 40°

∠CAB + ∠ABC = ∠ACD

∠ABC = 3x = 3 × 20° = 60°

Putting the values of the respective angles

∠ACB + ∠ACD = 180° (Linear pair)

2x + 3x = 100°

∠ACB + 100° = 180°

5x = 100° or x = 20°.

∠ACB = 180° – 100° = 80°

100º

●

Hence, the angles of the triangle ABC are 40°, 60° and 80°.

Error Alert! The exterior angle property applies to the exterior angle and the two interior angles. Never apply it to a wrong set of angles.

1 2

1 4

∠3 = ∠2 + ∠4 ∠3 = ∠1 + ∠4

Example 6

∠3 = ∠1 + ∠2

In the adjoining figure, if the line segments PQ and RS intersect at point T so that ∠PRT = 75°, ∠RPT = 60° and ∠TSQ = 75°, find ∠SQT. Step 1 The side RT of a DPRT is extended to S, the exterior angle = sum of the interior angles. ∠PTS = ∠TPR + ∠PRT

60° 75°

S T

75º

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Step 2 Putting the values, ∠PTS = 60° + 75° = 135°

Do It Together

Step 3

Step 4

The side QT of DSQT is extended to P, The exterior angle = the sum of interior angles

Putting the values,

∠PTS = ∠TSQ + ∠SQT

∠SQT = 135° – 75° = 60°

135° = 75° + ∠SQT

In a DPQR, QR is extended to S such that ∠PRS = 140°. Find the measure of ∠PQR and ∠PRQ, if ∠QPR = 80°. Step 1

Step 2

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

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

Putting the values of the respective angles,

Therefore, ∠PQR + ∠QPR + ∠PRQ = _____

∠PQR + 80° = ______

60° + _____ + ∠PRQ = 180°

So, ∠PQR = ______

∠PRQ = ______ The angles of the triangle PQR are ______________________ .

Do It Yourself 9B 1

Select the correct option:

a A triangle can have two ______ angles. acute

ii obtuse

iii right iv reflex

b If two acute angles of a right-angled triangle are equal, then the measure of the acute angle is: 30°

ii 60°

iii 45°

iv 90°

c The exterior angle is ________ than each of the interior opposite angles. greater

iii equal iv twice

Find the values of x and y in the figures. b

40º 35º

0º

30º

0º

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y x

0 30º 12

x+2 3x + 3 2x + 10

50º

y 60º

0º

ii smaller

48º

2x 3x

0º

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One of the exterior angles of a triangle is 130°, and the interior opposite angles are in the ratio 2:3. Find the angles of the triangle.

One angle of a triangle is 60°. Find the remaining two angles if their difference is 20°.

The angles of a triangle are in the ratio 5:6:7. Find the measures of the angles.

In a DABC, ∠A is 50° more than ∠B and 20° less than ∠C. What are the angles of the triangle?

In a DABC, ∠A – ∠B = 15°, ∠B – ∠C = 30°. Find the measure of each angle of the triangle.

The angles of a triangle are arranged in ascending order of their values. If the difference between the two consecutive angles is 10°, find the three angles.

In the adjoining figure, the side BC of a DABC is extended to form a ray BD. CE is drawn parallel to BA. Without using the angle sum property of a triangle:

a Show that ∠ACD = ∠A + ∠B; and

b ∠A + ∠B + ∠C = 180°.

●

Word Problem 1

Shown here is a picture of the Incheon Bridge in South Korea. The bridge was built using cables. There is a vertical pole QT standing on the road SU such that the angles (marked in green) between the cables and the pole are equal. If the sum of the angles (marked in white) made by the same cables with respect to the road is 80°, then find the measure of ∠PQT.

R Q S

Properties of the Sides of a Triangle Any three points that do not fall in a straight line can be joined to form a triangle, as seen in the triangular structure used to build the bridge. But, can we draw a triangle with sides of any given lengths? Let us find out. Triangle Inequality Property A

Triangles are bound by something called “the triangle inequality property”. Let us learn about it. Statement: The sum of the lengths of any two sides of a triangle is always greater than the length of the third side. Consider DABC, as shown in the adjoining figure.

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Let the sides opposite to the vertices, A, B and C be denoted by a, b and c respectively. Since side BC = a, side AC = b, side AB = c

Did You Know?

So, as per the property: a+b>c

a+c>b

The Bermuda Triangle, also known as the “Devil’s Triangle”, is a region in the western part of the North Atlantic Ocean where several aircrafts and ships have disappeared under mysterious circumstances. The three vertices of the Bermuda Triangle on the map are Miami, Bermuda and Puerto Rico.

b+c>a

Let us test this for a triangle that has the following dimensions: a = 2, b = 3, c = 4

Combination to Be Tested

LHS (sum of 2 sides)

RHS (3rd side)

Inequality Property Satisfied?

a+b>c

2+3=5

5 > 4; Yes

a+c>b

2+4=6

6 > 3; Yes

b+c>a

3+4=7

7 > 2; Yes

All possible combinations satisfy the triangle inequality property. Therefore, this triangle satisfies the triangle inequality property. So, such a triangle can be drawn. Now, let us look at this property in another way. We know that a + b > c, a + c > b and b + c > a. So, |a – c| < b. Similarly, |a – b| < c and |b – c| < a. Each of the left-hand sides of the inequalities represents the difference of two sides and the right-hand side, which is the third side. So, the inequality property of triangles can also be rephrased as: The difference of the lengths of any two sides of a triangle is less than the length of the third side. Example 7

Is it possible to draw a triangle with the sides 4 cm, 5 cm and 6 cm? Step 1

Using the triangle inequality property, the sum of the lengths of any two sides of a triangle is greater than the length of the third side. Side 1 + Side 2 > Side 3 Step 2 Let us apply the inequality property. Combination to Be Tested

LHS (sum of 2 sides)

RHS (3rd side)

Inequality Property Satisfied?

a+b>c

4 + 5 = 9 cm

6 cm

9 > 6; Yes

a+c>b

5 + 6 = 11 cm

4 cm

11 > 4; Yes

b+c>a

4 + 6 = 10 cm

5 cm

10 > 5; Yes

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Step 3 Is the inequality property true for all combinations? Yes it is! Therefore, it is possible to draw a triangle with the sides 4 cm, 5 cm and 6 cm. Example 8

The lengths of the two sides of a triangle are 9 cm and 13 cm. What could be the possible values of the third side of the triangle? Step 1 We know that Side 1 + Side 2 > Side 3. (The triangle inequality property) Finding the sum of the two sides, Side 1 + Side 2 = 13 + 9 = 22 cm Step 2 We know that Side 1 – Side 2 < Side 3. (The triangle inequality property) Finding the difference of the two sides, Side 1 – Side 2 = 13 – 9 = 4 cm Therefore, the length of the third side of the triangle has to be more than 4 cm but less than 22 cm.

Do It Together

The lengths of the two sides of a triangle are 4 cm and 12 cm. What could be the possible values of the third side of the triangle? Step 1 Using the triangle inequality property, the sum of the lengths of any two sides of a triangle is _______than the length of the third side. Side 1 + Side 2 > ______. The third side will be ______ than the sum of the two sides. Putting the values of the sides of the triangle, finding the sum of the two sides, Side 1 + Side 2 = _______________________. Step 2 We also know that the difference of the lengths of any two sides of a triangle is _____ than the length of the third side. Finding the difference of the two sides, Side 1 – Side 2 = ___________________ . The length of the third side cannot be _______ than the difference of the two sides. Therefore, the possible length of the third side of the triangle can be between __________.

Error Alert! A triangle cannot be formed if any of the three sides is longer than the sum of the other two sides. For example, in DABC, CA > AB + BC

AB + BC > CA

Do It Yourself 9C 1

State which of the following sets of lengths will form a triangle. a 2 cm, 3 cm and 4 cm

b 4 cm, 9 cm and 12 cm

d 2.3 mm, 3.4 mm and 7.8 mm

e 3.2 cm, 5.6 cm and 8.3 cm

c 1 cm, 5 cm and 10 cm

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If the following are the lengths of the two sides of a triangle, what are the possible values of the third side?

S is any point on the side QR of triangle PQR. Fill in the blanks with the >, < or = signs.

In a DPQR, PS is the median. Is PQ + QR + PR > 2 PS? Give reasons for your answer.

O is any point in a DABC. If AB = 5 cm, BC = 6 cm and AC = 7 cm.

a 3 cm and 6 cm

b 14 cm and 9 cm

a PS ______ PR + SR

c 34 cm and 7 cm

d 23 cm and 27 cm

b PS ______ PQ – QS

Prove that AO + BO + CO > 9 cm. 6

O is any point in a DABC and AO, BO and CO are joined. Show that 2(AO + BO + CO) > AB + BC + AC.

Word Problem 1

Jane travelled from New Delhi to Najafgarh and then from Najafgarh to Dwarka. Describe the relationship between the distance travelled from New Delhi to Dwarka and the other two distances.

New Delhi

Najafgarh

Dwarka

Mean, Median and Mode Right-angled Triangles Real Life Connect

Riya: It is so much fun learning about the properties of triangles. Raj: Yes. But, I still have a question. Remember the railway bridge we saw? One of the angles was 90°. It looked special! What kind of triangles are these? Riya: Such special triangles are called right-angled triangles!

Let us see right-angled triangles in things around us.

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A right-angled triangle is formed when one of the angles in the triangle is a right angle or 90°. It has two shorter sides that are called the legs of the triangle. The side opposite to the right angle is called the hypotenuse, which is the longest side of a right-angled triangle. Here, ∠C is a right angle; AB is the hypotenuse; and sides AC and CB are the legs. Here, AC will be called the perpendicular and BC will be called the base of the DABC.

po c te nu

The Pythagoras Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Base b

In DACB, ∠C is the right angle, AB is the hypotenuse and AC and BC are the legs. Here, AB2 = BC2 + AC2 If side AB = c, side AC = a, Side BC = b Then, c2 = a2 + b2 Let us now look at some properties that the Pythagoras Theorem produces. Property 1: The hypotenuse is the longest side of a right-angled triangle. Proof: We know that, c2 = a2 + b2 (Pythagoras property) Therefore, c2 > a2 and c2 > b2. c > a and c > b. Hence, in a right-angled triangle, the longest side is the hypotenuse. Property 2: Of all the line segments that can be drawn to a given line from a point outside it, the perpendicular line segment is the shortest. Proof: Let l be the given line and P a point outside it. Draw PQ perpendicular to line l.

Let R be a point other than Q on line l. Join PR. Now DPQR is a right-angled triangle, ∠Q = 90°. PR is the hypotenuse, and PQ is one of the legs of the right-angled triangle PQR.

In a right-angled triangle, the hypotenuse is the longest side. Therefore, PQ < PR. This is true for all the points other than Q on line l. Hence, PQ is the shortest of all the line segments that can be drawn from P to line l. Converse of the Pythagoras Property Statement: If the square of one of the sides of a triangle is equal to the sum of the squares of the other two sides, then it is a right-angled triangle with the angle opposite to the longest side being the right angle. Thus, if in DABC, AB2 = AC2 + BC2, then DACB is a right-angled triangle, right-angled at C.

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Pythagorean Triplets Three positive numbers a, b and c are said to form a Pythagorean triplet if they satisfy the Pythagoras property i.e., the square of one of the sides of a triangle is equal to the sum of the squares of the other two sides. c2 = a2 + b2

Think and Tell

Some Pythagorean triplets are (3, 4, 5); (5, 12, 13); (8, 15, 17); (7, 24, 25); (12, 35, 37).

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

Forming a Pythagorean Triplet Consider any two natural numbers, a and b, where a > b. Then, (a2 – b2), (2ab), (a2 + b2) form a Pythagorean triplet. Example 9

a Angle ABC

b Angle BAC

c Angle ACB

d CAB

In the given figure, find the value of H if DABC is a right-angled triangle. Step 1

Step 2

Using the Pythagoras theorem,

Adding the values,

AC2 = AB2 + BC2

H2 = 169 = 13 × 13

Putting the values,

So, H = 13 cm

H2 = 52 + 122

5 cm

12 cm

H = 25 + 144 Example 10

A triangle has sides of the lengths, 3 cm, 6 cm and 10 cm. Is it a right-angled triangle? Step 1

Step 2

The longest side of a right-angled triangle is the hypotenuse.

Comparing the steps 1 and 2, as per the Pythagoras property,

Let us apply the Pythagoras property. The other two sides are 3 cm and 6 cm.

By the converse of the Pythagoras property,

We know that Hypotenuse2 = 102 = 100

So, 32 + 62 = 9 + 36 = 45. Do It Together

102 is not equal to 32 + 62.

the given triangle is not a right-angled triangle.

Two poles of heights 8 m and 16 m respectively are placed at the same level, 6 m apart from each other. What is the distance between their tops? Step 1

Draw the figure and label the vertices to show the right-angled triangle. AB = 8 m, CD = 16 m and BD = 6 m Draw AE || BD meeting CD at E. Then, ABDE is a rectangle. Using the properties of a rectangle, ED = _______ and AE = __________

In a right-angled triangle AEC, we have, AC2 = __________

16 m

Step 2

Putting the values, AC2 = _____________, AC = __________

The distance between the tops of the two poles is ________.

Do It Yourself 9D 1

Which of the following sets of lengths of a triangle belong to right-angled triangles? a 2 cm, 3 cm and 6 cm

b 6 m, 8 m and 10 m

Chapter 9 • Triangles and Their Properties

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c 15 cm, 20 cm and 25 cm

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Find the missing lengths of each of the triangles. a

b 20 m

c 7 cm

30 m

25 cm

24 m

12 m

If (k, 24, 25) is a Pythagorean triplet, find the value of k.

The sides of a rectangle are 12 cm and 16 cm. Find the length of its diagonal.

The hypotenuse of a right-angled triangle is 25 cm. The length of the base is 7 cm. What is the length of the altitude of the triangle? C 13 cm D

Find the values of x and y in the given figure.

Rahul drives 15 m to the north and then, 8 m to the east. Find the distance between the starting point and the ending point of Rahul’s drive.

y A

4 cm

3 cm

Word Problem 1

A tree of height 9 m is broken at a height of 4 m from the ground, and its top touches the ground. Find the distance of the point where the top touches the ground from the base of the tree.

Points to Remember • A median is a line segment that connects the vertex of a triangle to the midpoint of the opposite side. •

The point where the medians intersect is called the centroid of the triangle.

• An altitude is perpendicular to a side of the triangle, and it extends from a vertex to the opposite side. •

The point where the altitudes intersect is called the orthocentre of the triangle.

•

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

•

The exterior angle formed is equal to the sum of the two interior opposite angles.

•

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

• The difference of the lengths of any two sides of a triangle is less than the length of the third side. • In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Math Lab Finding Pythagoras’ Triplets with Squares Objective: To find if the group of given natural numbers is a Pythagorean triplet. Materials Required: Graph paper sheets, pencil, scale, etc. Steps: Find if (4, 4, 4) is a Pythagorean triplet.

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1 Take a graph paper sheet and cut three squares of sides 4, 4 and 4 units. Tip: Cut the squares in such a way that all sides have

4 squares of the graph paper each, which will also be 4 cm as 1 square = 1 cm

2 Arrange the squares to form a triangle. 3 Trace the triangle. 4 Measure all the angles of a triangle. 5 Is this triangle a right-angled triangle? Hint: Do the sides satisfy the Pythagoras theorem?

6 Try this activity for more sets of numbers.

Chapter Checkup 1

Fill in the blanks. a If the three angles of a triangle are equal, then each of them is equal to ______. b In a ______ triangle, two of its sides are also its altitudes. c In an _________ triangle, the median and altitudes are denoted by the same line segment. d The sum of the angles of a triangle is _____.

Write T for True and F for False. a There can be only two acute angles in a triangle. b The Pythagoras theorem is only applicable for a right-angled triangle. c The exterior angles of a triangle are equal to the difference of the opposite interior angles. d The sum of the lengths of any two sides of a triangle is greater than the third side.

Which of the following sets of angles will form a triangle? a 30°, 40° and 110° b 34°, 29° and 120°

Which of the following sets of lengths of a triangle will form a right-angled triangle? a 6 m, 8 m and 9 m b 9 cm, 12 cm and 15 cm c 26 cm, 24 cm and 10 cm

Find the missing angles for the following figures. H P x – 10 a b D 65º

Q 4x – 10

70º

c 50°, 120° and 10°

50º 80º

54º F

d 34°, 28° and 100°

B Q

65º

130º R

The bottom of a ladder must be placed 6 m away from a wall. The ladder is 10 m long. Where above the ground does the ladder touch the wall?

DABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, then will ABC be a right-angled triangle? Give reasons for your answer.

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In a 52-inch plasma television set, the length of the screen’s diagonal is 52 inches. If the screen’s width is 48 inches, find the height of the screen.

The floor of a delivery truck is 48 cm above the ground. The ramp attached to the back of the truck touches the ground at point P, which is 14 cm away from the truck. What is the length of the ramp?

10 The roof of a building is 6 m high. A rope is tied from the roof to a peg on the ground 8 m away from the wall. What could be the shortest length of the rope? 11 In DABC, ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O. Find ∠C and ∠BOC. A

60º

80º

12 In the figure, sides AB and AC of a triangle are produced to P and Q respectively. The bisectors of ∠PBC and ∠QCB 1 intersect at O. Prove that angle BOC = 90° – Angle BAC. 2 A

13 If the sides of a triangle are produced in the way as shown in the figure given below, then prove that the sum of the exterior angles formed is equal to four right angles. E

A 1

2 B

Word Problems soccer field is in the shape of a rectangle that is 9 metres wide and 12 metres long. The 1 A coach asks the players to run from one corner of the field to the corner that is diagonally across. What is the distance that the players run? ladder 13 m long reaches a window which is 5 m above the ground on one side of a street. 2 A Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.

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10 Congruence Let’s Recall

A triangle is a closed figure with three sides. TYPES OF TRIANGLES Based on Sides Scalene triangle

Isosceles Triangle

Equilateral Triangle

All the three sides are unequal.

Two sides are equal.

All three sides are equal.

All the angles are unequal.

Angles opposite to these sides are equal.

All three angles are 60°.

Based on Angles Acute-angled triangle

Right-angled triangle

Obtuse-angled triangle

All angles are less than 90°.

One of the three angles is exactly 90°.

One of the three angles is more than 90°.

We have learnt that the sum of the lengths of the two sides of a triangle is greater than the length of the third side. We also know that the sum of all the angles of a triangle is equal to 180°. The exterior angle of a triangle is always equal to the sum of the interior opposite angles. This property of a triangle is called the exterior angle property. 1 We find the area of a triangle using the formula × Base × Height. 2 We can calculate the perimeter of a triangle by adding the sums of its three sides.

Let’s Warm-up Fill in the blanks. 1

A point where two sides of a triangle meet is known as the _______________ of a triangle.

Each angle of an equilateral triangle is of the measure _______________°.

A _______________ triangle has one angle of 90°.

A triangle has _______________ vertices and _______________ sides.

In an obtuse-angled triangle, the remaining two angles are _______________. I scored ____________ out of 5.

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Congruence in Shapes and Figures Real Life Connect

In a class, the students were reading a book. Teacher: Did you notice how each page of the book covers the next page completely? Rashi: Yes, teacher. This is because the pages are of the same shape and size. Teacher: Yes, Rashi. These two pages are congruent. Sahil: What does congruent mean? When one figure is placed on the top of another and they match exactly in shape and size, then the figures are said to be congruent. This method of comparing shapes is called the method of superposition. Congruence of Line Segments Think of two unsharpened pencils. They have the same lengths. When two line segments have the same or equal lengths, we say that they are congruent. For example, AB and CD are two line segments. We can see that, AB = CD = 9.2 cm

9.2 cm

Therefore, AB and CD are congruent to each other. This can be written as AB ~ = CD. Congruence of Angles Similarly, when the measure of two angles is the same, then we say that the two angles are congruent. Look at the given figure.

B and

Q are two angles.

From the figure, we can see that m

B=m

Therefore,

Q B and

Q are congruent to each other.

This can be written as

B~ = ˜

60°

Congruence of Squares Have you ever played with a Rubik’s cube? If you notice the shape and size of each face of the Rubik’s cube, each face is identical. When the length of the side of one square is exactly equal to the length of the side of another square, we say the squares are congruent to each other.

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So, square ABCD and square PQRS are congruent to each other.

This can be written as ABCD ~ = PQRS.

5.6 cm

Q 5.6 cm

AB = PQ = 5.6 cm

5.6 cm

Sides of square ABCD = Sides of square PQRS

5.6 cm

In the figure given, we can see that ABCD and PQRS are two squares.

5.6 cm

Congruence of Rectangles 6.6 cm

Now, look at a ₹500 note. It is rectangular in shape. Is one ₹500 note congruent to another ₹500 note? 15 cm

Therefore, a ₹500 note is congruent to another ₹500 note.

15 cm

6.6 cm

= ˜

Q 6.6 cm

15 cm

6.6 cm

15 cm

6.6 cm

So, ABCD ~ = PQRS.

15 cm

6.6 cm

Side AD = Side PS = 6.6 cm

6.6 cm

From the figure, we can see that ABCD and PQRS are two rectangles. Side AB = Side PQ = 15 cm

15 cm

6.6 cm

In rectangles, if the length and the breadth of one rectangle are exactly equal to the length and the breadth of another rectangle, we say that the rectangles are congruent to each other.

15 cm

Congruence of Circles A pair of bangles that our mother may wear is always of the same shape and size.

This is because both bangles have the same radii. If the radius of one circle is exactly equal to the radius of another circle, then we say that the circles are congruent. Look at the given figure.

2.8 cm

The radius r1 = r2 = 2.8 cm. Thus, circle C1 ~ = circle C2.

Congruence of Triangles

C1 A

B Chapter 10 • Congruence

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2.8 cm

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Imagine cutting a rectangular sheet diagonally.

We get two triangles. Are these two triangles congruent?

A triangle is said to be congruent to another triangle if and only if there exists a correspondence between all the vertices such that the corresponding sides and the corresponding angles of both triangles are equal. In the figure, ∆ABC and ∆PQR are two triangles, where A (

stands for matching)

If,

P, B

R, (corresponding angles)

and,

Q, C

Think and Tell Compare your hand with that of your family member. Are your hands congruent? Why?

AB = PQ, BC = QR, AC = PR (corresponding sides) then, ∆ ABC ~ = ∆ PQR. Example 1

Show that rectangle ABCD is congruent to rectangle MNOP. In the given figure, we can see that: AD = MN (corresponding sides) and

AB = MP (corresponding sides)

Therefore, Rectangle ABCD ~ = rectangle MNOP under the correspondence ABCD Example 2

Example 3

MPON.

In the given figure, lines a and b are parallel to each other. Which angle is congruent to 6 and 7 if 1 ~ = 3 and 4 ~ = 2? We can see that:

1 and

Similarly,

7 and

Therefore,

7~ =

5 and

4 and

6 and

8 are also opposite angles.

6~ =

If ∆DEF and ∆SUV are congruent to each other under the correspondence: DEF SUV, then write the parts of ∆DEF that are corresponding to: 1

2 are opposite angles.

2 8

7 5

l 4

F U

Given: ∆DEF ~ = ∆SUV The correspondence is DEF D

S and E

U and F

SUV. D

So, 1

DF = SV

F U

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Do It Together

If ∆PQR ~ = ∆XYZ under the correspondence PQR XYZ, write all the corresponding congruent parts of the triangles. We know that ∆ ______________ ~ = ∆ ______________ The correspondence is PQR

______________

Did You Know?

The mirror image of an object is congruent to the

The corresponding angles are: P

____________________________ e

P Q

The corresponding sides are: d

actual object.

____________________________

Do It Yourself 10A 1

A line segment AB = 10 cm. If AB ~ = CD, then CD = _____________?

What is the measurement of x in the following pairs of geometrically congruent figures: O

4. cm

U P

L x

55°

x B

Q 4 cm

8 cm

d S

Rectangle PQRS contains three small congruent rectangles. If the breadth of one of the small

Complete the congruence statement.

rectangles is 5 cm, what is the area of rectangle PQRS in cm2? D

T U 7.2 cmV

Square ABCD ~ = _______________? B

Give a reason. 5 6 7

∆UVW and ∆XYZ are congruent under the correspondence UVW

XYZ. Which parts would correspond to

and WU ?

If ∆ABC ~ = ∆DEF, under the correspondence ABC

DEF and

If, ∆LMN ~ = ∆UVW, then write all the corresponding angles. L

M Chapter 10 • Congruence

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B = 105°, what is the measure of

X, XY

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If the arcs UMV and XNY of the given circle are congruent, then find the ratio of the chords UV and XY. V

U N

X 9

In the given figure, if

PQR =

TQS, then is

= PQS ~

TQR? Justify.

S R Q

10 In the given figure, if a = b = c, name the angle which is congruent to D

c b O

AOC.

B A

Word Problems 1

Raj has two rectangular sheets of paper. The first rectangle has a length of 10 cm and a

breadth of 5 cm. The second rectangle has a length of 5 cm and a breadth of 10 cm. Help Raj figure out, if the two sheets of paper are congruent.

An archer noticed that two circular targets have the same radius. Are the two circular

Shyam has two identical triangles. The sides of one triangle are 5 cm, 12cm and 13 cm.

targets congruent? Why or why not?

If the two triangles are congruent, what are the measurements of the sides of the other triangle?

Congruence in Triangles Real Life Connect

Jeena’s mother made some biscuits at home. Jeena noticed that the biscuits were in the same shape as a triangle. Mother: Jeena, can you find how many triangular biscuits are congruent to each other? Jeena: How can I find that out?

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To help Jeena figure out the triangles that would be congruent to each other, let us first learn about some congruence rules.

Did You Know? The Great Pyramid of

SSS Congruence Criteria

Giza’s sides are nearly perfectly congruent.

SSS stands for Side-Side-Side. When all three sides of one triangle are equal to the corresponding three sides of another triangle, then the two triangles are said to be congruent by SSS congruence criteria.

In ∆ABC and ∆PQR, AB = PQ

(Side)

BC = QR

(Side)

AC = PR

(Side)

Therefore, using SSS congruence criteria, ∆ABC ~ = ∆PQR. Example 4

In the given figure, AD = BC and AC = BD. Show that ∆ABD = ˜ ∆BAC.

In ∆ABD and ∆BAC, AD = BC (given) AC = BD (given) AB = BA (common side) Therefore, by SSS congruence criteria, ∆ABD ~ = ∆BAC. Example 5

In ∆XYZ, XY = 7.5 cm, YZ = 4.3 cm and XZ = 8.7 cm. In ∆WUV, UV = 4.3 cm, VW = 8.7 cm and UW = 7.5 cm. Examine whether the given two triangles are congruent or not. X

In ∆XYZ and ∆WUV,

Therefore, by SSS congruence criteria, ∆XYZ ~ = ∆WUV.

7.5 c m

XZ = WV = 8.7 cm (side)

m 8.7 c

YZ = UV = 4.3 cm (side)

4.3 cm

XY = WU = 7.5 cm (side)

7.5 c m

m 8.7 c

Remember! The corresponding parts must always be equal for two figures to be congruent.

Chapter 10 • Congruence

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Example 6

∆ABC is congruent to ∆PQR by SSS congruence criteria, under the correspondence ABC PQR. If AB = 10 cm, BC = 5 cm and PR = 11 cm. Find AC, PQ and QR.

Given: ∆ABC ~ = ∆PQR; AB = 10 cm; BC = 5 cm; PR = 11 cm To find: AC = ?; PQ = ?; QR = ?

By SSS congruence criteria, AB = PQ = 10 cm; BC = QR = 5 cm; AC = PR = 11 cm Therefore, AC = 11 cm, PQ = 10 cm and QR = 5 cm Do It Together

The triangles in the given figure are congruent by SSS congruence criteria. Write the congruence in symbolic form. A D In ∆ABC and ∆DEC,

AB = DE (side) BC = _________ (__________________) _________ = _________ (__________________)

Therefore, by _________ congruence criteria ∆ _________ ~ = ∆ _________.

Do It Yourself 10B 1

In the given figure, PR = SQ and RS = QP. a Write all the pairs of equal parts in ∆PRS and ∆SQP.

b Is ∆PRS ~ = ∆SQP? Why or why not? 2

In the given figure, LMN is a triangle in which LM = LN. MP and NO are the medians of

the triangle.

Prove that:

a ∆OMN ~ = ∆PNM

b MP = NO

c ∆LPM ~ = ∆LON

In the given figure, ∆PQR and ∆SQR both share the same base QR and are isosceles triangles. Show that ∆PQS and ∆PRS are congruent. P

If ABCDE is a regular pentagon, then show that ∆ABC ~ = ∆AED.

A B

S Q

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In the given figure, AP = BQ, AS = BR and PR = QS. Is ∆APS congruent to ∆BQR? A

In the given figure, AC = 5 cm, BD = 6 cm, AB =

DC = 3 cm and BC = 4 cm. Check if ∆ABC ~ = ∆DCB.

ABCD is a square. The diagonals AC and BD divide the square into four triangles. Show that the two triangles

Show that a diagonal of a parallelogram divides it into two congruent triangles.

∆LMN is an equilateral triangle. O is the midpoint of MN. Prove that ∆LMO ~ = ∆LNO.

opposite to each other are congruent.

Word Problem 1

Jay was hanging his clothes in his cupboard, using a fabric hanger. The inside of the hanger is triangular with two sides of 21 cm

and one side of 30 cm. How can he draw a triangle of these exact

21 cm

measurements on paper? Is the triangle drawn by Jay congruent to

30 cm

the one of the fabric hanger?

SAS Congruence Criteria SAS stands for Side-Angle-Side. If any two sides and the angle included between the two sides in a triangle are equal to the two sides and the angle included between them in another triangle, then the two triangles are said to be congruent by SAS congruence criteria. In ∆ABC and ∆PQR,

If even one pair of corresponding sides and angles are unequal then the triangles are not congruent.

AC = PR (side) CAB =

Remember!

RPQ (angle)

AB = PQ (side) Therefore, ∆ABC ~ = ∆PQR.

Chapter 10 • Congruence

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Given below are the measurements of some parts of two triangles. Examine whether the two triangles are congruent or not.

cm 7

TU = YX = 5 cm (side)

Y = 50º (angle)

50º

5 cm

ST = ZY = 7 cm (side) T=

In ∆STU and ∆ZYX,

5 cm

Example 7

Therefore, by SAS congruence criteria ∆STU ~ = ∆ZYX. Example 8

AB = AC and AD is the bisector of

BAC. Show that ∆ABD ~ = ∆ACD.

In ∆ABD and ∆ACD, AB = AC (given) (side) BAD =

CAD (AD is the bisector of

BAC) (angle)

AD = AD (common side)

Therefore, by SAS congruence criteria, ∆ABD ~ = ∆ACD. Do It Together

Find the measurements of the unknown sides or angles in the following pair of congruent triangles. We know that ∆PQR ~ = ∆XYZ. So, PQ = XY = z

z = _________

PR = _________ = x

x = _________

and

y = _________

R = _________ = y

5 cm

QR = _________

6 cm

50º

z R

m 8c

Do It Yourself 10C 1

In the given figure, show that ∆PRS ~ = ∆PRQ. 8.5 cm P Q 30º

In the given figure, if

30º 8.5 cm PQR =

Show that ∆ADC ~ = ∆ABC.

R B QPS and QR = PS, then show that PR = QS. P Q

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Check if the two triangles are congruent. C A

O D

BX is the bisector of

ABC. If P is any point on BX and the perpendiculars drawn from P to AB and AC are equal,

then show that ∆BMP and ∆BLP are congruent by the SAS rule. M B

P L

ΔABC is isosceles with AB = AC. Line segment AD bisects a Is ΔADB ~ = ΔADC?

A A and meets the base BC at D.

b State the three pairs of matching parts used to answer (a). c Is it true to say that BD = DC? 7

If in triangles ABC and DEF, AB = DE and

Line segments AB and CD bisect each other at O. AC and BD are joined, forming triangles AOC and BOD. State the

congruence criteria?

D, then when will the two triangles be congruent by SAS

three equality relations between the parts of the two triangles that are given or otherwise known. Are the two triangles congruent?

Show that in an isosceles triangle, the angles opposite to the equal sides are equal.

Word Problem 1

Akash had a rectangular sheet of paper which he folded in half diagonally. After folding, two triangles were obtained. Are the triangles congruent to each other?

ASA Congruence Criteria ASA stands for Angle-Side-Angle. If any two angles and the side that is between the two angles in a triangle are equal to the corresponding two angles and the side between the two angles, then the two triangles are said to be congruent by ASA congruence criteria.

Chapter 10 • Congruence

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In ∆ABC and ∆PQR, ABC =

PQR

BC = QR

(Side)

ACB =

PRQ

(Angle) (Angle)

Check if ∆ABC and ∆DEF are congruent to each other.

In ∆ABC and ∆DEF, ABC =

DEF

BC = EF BCA =

Therefore, ∆ABC ~ = ∆PQR. Example 9

EFD

Therefore, by ASA congruence criteria, ∆ABC ~ = ∆DEF. Example 10

Prove that the diagonal of a parallelogram divides it into two congruent triangles. D

In ∆ACD and ∆DBA, CAD =

BDA (alternate interior angles of a parallelogram)

AD = DA (common) CDA =

BAD (alternate interior angles of a parallelogram)

Therefore, by ASA congruence criteria,

∆ACD ~ = ∆DBA.

Error Alert! In ASA, do not include any side but only the side between the angles. A B Do It Together

P C Q

Prove that ∆AOC and ∆DOB are congruent. In ∆AOC and ∆DOB, CAO =

_________ (given)

AO = _________ (given) AOC =

_________ (vertically opposite angles)

Therefore, by _________ congruence criteria,

A R

P C Q

R A

∆ _________ ~ = ∆ _________.

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Do It Yourself 10D 1

In the given figure, AD is the bisector of CPD =

BAC, and

BPD. Prove that ∆CAP ~ = ∆BAP.

In the given figure, AB = BC and BD is the median. Is ∆ABD ~ = ∆CBD?

D B

of triangles ABC and FED are congruent or not. Write the result in symbolic form. 3.5 cm

In the figure, check whether the pair of triangles ADB and BCA are congruent or not.

60º

40º

In the figure, measures of some parts are indicated. By applying the ASA congruence rule, state whether the pair A

3.5 cm

40º

BX is the bisector of

ABC. If P is any point on BX,

then prove that the perpendiculars drawn from P to AB and BC are equal.

45º 30º

60º

45º 30º

M B

a ∆DBC ~ = ∆EAC

B and

DCA =

P L

In the given figure, AC = BC,

A D

ECB. Prove that:

b DC = EC A 7

In the given figure ΔABE ~ = ΔACD.

ABE =

ACD, prove that

In the given figure

ΔABE ~ = ΔACD.

ABE =

ACD. Prove that

A D

E K

Chapter 10 • Congruence

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In the given figure, AD bisects a Is ΔADB ~ = ΔADC?

A and AD

BC.

b State the three pairs of matching parts you have used in (a). c Is it true to say that BD = DC?

= ∆PRQ. 10 In the given figure, if BA || RP, QP || BC and AQ = CR, then prove that ∆ABC ~ P R C

Word Problem 1

Amy is tiling her kitchen floor with triangular tiles. She chooses a tile (say ΔABC) with A = 30°,

D = 30°,

B = 60° and side AB = 6 inches. If her friend Bob uses the tiles (say ∆DEF) with E = 60° and side DE = 6 inches for his kitchen, can they conclude that their

kitchens have congruent tile patterns, based on the ASA property?

RHS Congruence Criteria RHS stands for Right angle-Hypotenuse-Side. If in a given right-angled triangle, the hypotenuse and a side are equal to the hypotenuse and side of another right-angled triangle, then the two triangles are said to be congruent by the RHS congruence criteria. In ∆ABC and ∆PQR, ABC = PQR = 90°

(right angle)

AC = PR

(hypotenuse)

BC = QR

(side)

Therefore, ∆ABC ~ = ∆PQR. Example 11

Show that ∆ABC ~ = ∆BAD. In ∆ABC and ∆BAD, A=

B (right angle)

AB = BA (common) AC = BD = 2 cm (given) Therefore, by RHS congruence criteria, ∆ABC ~ = ∆BAD.

D 2 cm

2 cm A

90°

90° 3.5 cm

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Example 12

Show that the bisector PS divides ∆PQR into two congruent triangles. P

In ∆PQS and ∆PRS, PSQ =

PSR (= 90°) 3 cm

PQ = PR = 3 cm QS = SR (PS is a bisector)

Do It Together

90°

Therefore, by RHS congruence criteria, ∆PQS ~ = ∆PRS.

3 cm R

In the given figure, prove that ∆ABD ~ = ∆ACD. B

ABD = _______

cm 4.9

In ∆ABD and ∆ACD,

_______ = _______ (common side)

4. 9

_______ = _______ = _______ cm

Therefore, by _______ congruence criteria ∆ABD ~ = ∆ACD.

Do It Yourself 10E 1

∆ABC is an isosceles triangle and D is the midpoint of AC. Show that ∆ABD ~ = ∆CBD. B

A 2

In the given figure, AS = BR and PR = QS. Is ∆APS congruent to ∆BQR?

P 3

In the given figure, AB = BC, a ΔABD ~ = ΔCBD

ADB =

A D

c AD = CD B

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CDB = 90°. Prove that:

Chapter 10 • Congruence

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In the given figures, measures of some parts of triangles are given. By applying the RHS congruence rule, state whether the pair of triangles are congruent. Write the result in symbolic form. A D 10 cm 10 cm 4 cm 3 cm 90° 90° C

In the given figure, LM = NM, ML is perpendicular to PQ and MN is perpendicular to PR. Show that LPM =

NPM.

Q 6

In the given pair of triangles, using only the RHS congruence criterion, determine whether the triangles are congruent. In the case of congruence, write the result in symbolic form. A B

E 7

ΔABC is isosceles with AB = AC. Also, AD

State in symbolic form. Which congruence condition do you use? Which side of ADC equals BD? Which angle of ΔADC equals

BC meeting BC at D. Are the two triangles ABD and ACD congruent?

In the given figure, ∆ABC and ∆EFD are right-angled at B and F. Prove that ∆ABC ~ = ΔEFD. A

E 9

Let ΔABC and ΔPQR be right triangles such ΔABC ~ = ΔPQR. If BC = 5 cm, PQ = 12 cm and measures of AB, AC, QR and PR?

B = 90°, what are the

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Word Problem 1

Tree A has a height of 10 m while Tree B has a height of 8 m. They both cast shadows on the ground. A hypotenuse is drawn from the top of each tree to the top of its shadow, making two right triangles. These hypotenuses measure 13 m and 11 m respectively. Determine if these two triangles are congruent or not using the RHS criterion.

Points to Remember •

Two planes are said to be congruent when they are superposed such that they cover each other exactly.

• If the corresponding sides and corresponding angles of two triangles are equal, they are congruent. •

Criteria for Congruence of Triangles: 

SSS congruence criteria (three sides)



SAS congruence criteria (two sides and the included angle)



ASA congruence criteria (two angles and the included side)



RHS congruence criteria (right angle, hypotenuse and side)

Math Lab Finding Congruent Triangles Setting: In pairs Materials required: Sheets of paper and pen 1 Draw different triangles on separate sheets. Each triangle should be drawn two times on two separate sheets of paper.

Distribute one sheet of paper to each of the students.

Instruct students to walk around the classroom and find the identical triangle.

The pair will then have to write the congruence criteria.

The first one to do so wins.

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Chapter Checkup 1

Write True or False. a Any two congruent figures look the same. b Two angles are congruent if the sum of their measures is 180°. c Two lines are congruent if they are parallel. d If two rectangles have equal areas then they are always congruent. e Two circles with the same radii are congruent.

In the given figure,

~ ROP =

QOP ~ =

SVU and

TVU. Which angle is congruent to

U Q

In the given figure, ∆ABC ~ = ∆PQR. Find the values of x and y.

B 4

T R

O 3

ROQ ?

75°

(2y + 5)°

(x − 7)°

LMO?

a Is ∆LMN ~ = ∆OMN?

b Does MN bisect

In the given figure, LM = OM, LN = ON.

70°

M O

If ∆PQR ~ = ∆XYZ and

The line segments DG and EF bisect each other at O. Prove that ∆DOE ~ = ∆GOF.

P = (2x + 15)°;

X = (5x – 60)°, then find the value of x.

F E

O G

If AC = BD, prove that ∆ACB ~ = ∆ADB. D

C 95°

20°

95°

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In the given figure ∆ABC ~ = ∆DCB. Find the value of B

ABC. D

45° 55°

A 9

In the given figure, Z is any point on AD. XZY is perpendicular to AD, and it meets AB at X and AC at Y, such that

AX = AY. Prove that ZX = ZY.

X A Y

10 In the given figure, BD = CD, ED = FD, DE

AB and DF

AC. If

B = 50°, find

A E B

= ∆PQR, then find the value of 11 If ∆ABC ~

50°

F D

BC . QR

12 In the given figure, altitudes PS, QT and RU of ∆PQR are equal. Show that the measure of each angle in ∆PQR is 60°.

P U Q

T S

Word Problems 1 Sam had a rectangular sheet of paper of dimensions 50 cm × 30 cm. He cut four congruent isosceles triangles from the rectangle, as shown in

the figure. If the two sides of each triangle = 10 cm. Find the area of the remaining portion.

2 Vanshika has two right triangles. The first has legs of length 6 cm and

8 cm, and the second has legs of length 10 cm and 6 cm. Determine if these right triangles are congruent.

Chapter 10 • Congruence

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311

Ratio and

Proportion

Let’s Recall Sam has a big farmhouse. He has various kinds of animals on the farm. The farm has horses, sheep, buffaloes, cows, dogs, swans, hens, chickens and many more. There is a total of 100 animals at the farm. There are 3 horses, 10 sheep, 5 buffaloes, 8 cows, 6 dogs, 6 swans, 4 hens and 7 chickens on the farm. We know that ratio is the number or amount of one thing compared to another thing. For example, the ratio of hens to chickens can be given as:

4 hens to 7 chickens or 4:7 or 4 . 7 Similarly, the ratio of horses to sheep Buffaloes to cows

5:8 and

Horses to swans

3:6 = 1:2

3:10

Let’s Warm-up Read the graph and fill in the blanks.

2 3

he ratio of Pooja's income to Arushiʼs T income is ___________. The ratio of the highest income to the lowest income is _____________.

35,000

The ratio of the difference in the income of Rachna and Amit to that of Rohan and Divya is _____________ .

20,000

The ratio of the sum of the incomes of Pooja and Divya to that of Arushi and Rachna is _____________.

30,000 25,000 Income

Income 23,250

22,000

15,000

26,500

29,000

30,750

18,500

10,000 5,000 0

Amit

Pooja

Divya

Name

Arushi

Rohan

Rachna

I scored _________ out of 4.

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Ratio and Proportion Real Life Connect

Aman and Nisha jog every day in a nearby park. Aman jogs 5 km every day, whereas Nisha jogs 7 km every day.

Equivalent Ratios

We know that ratios are the comparison of two quantities of the same unit. The ratio of the distance jogged by Aman to that by Nisha can be given as:

5 7 Aman jogged 35 km in a week, whereas Nisha jogged 49 km in a week. The ratio of distances covered by them in a week can now be given as 35:49. 5:7 =

If the HCF of both the terms of a ratio is 1, then the ratio is said to be in its simplest form. To reduce a ratio to its simplest form, divide the terms by their HCF.

35 ÷ 7 5 = 49 ÷ 7 7

HCF of 35 and 49

We saw that the ratio of distances jogged by Aman and Nisha in a day was 5:7 and the ratio of distances jogged in a week was 35:49 which when simplified gave 5:7. The ratios 5:7 and 35:49 are called equivalent ratios as they can be simplified to the same value. We can find the equivalent ratio of a given ratio by multiplying or dividing the numerator and denominator by the same natural number. 5 5 × 2 10 5 × 3 15 5 × 4 20 For example, = = = = = = . 7 7 × 2 14 7 × 3 21 7 × 4 28 Comparing Ratios: Let us compare 3:5 and 4:12 using the steps.

Think and Tell

How many equivalent ratios does 2:3 have?

Step 1

Step 2

Step 3

Write the ratios in their simplest form.

Convert into like fractions. LCM of 5 and 3 = 15

Compare the numerators of the like fractions.

3 4 1 3:5 = ; 4:12 = = 5 12 3

3 9 1 5 = ; = 5 15 3 15

As 9 > 5;

9 5 > ; hence 3:5 > 4:12 15 15

Using the steps of comparison, we can arrange the rational numbers in ascending or descending order. Let us now learn to divide a certain quantity into two or more ratios. The ratio of apples to oranges to guavas on a farm is 2:3:4. If there are 927 fruits in total, find the quantity of apples. Step 1

Step 2

Step 3

Find the total number of parts by adding the ratios.

Find the number of parts given to the asked quantity.

Multiply the number of parts given to the quantity with the total quantity.

2+3+4=9

Apples =

Hence, there are 206 apples. Chapter 11 • Ratio and Proportion

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2 × 927 = 206 9

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Example 1

If the ratio of red balls to blue balls in a bag is 4:3, find an equivalent ratio when there are 24 red balls. Ratio of

red balls 4 = blue balls 3

Ratio with 24 red balls =

Did You Know? The ratio of the sizes of the

4 × 6 24 = 3 × 6 18

moon and sun is 1:400.

So, the equivalent ratio of 4:3 with 24 red balls is 24:18.

Example 2

A:B = 4:9 and B:C = 2:5, If find A:C

4 2 A:B = 4:9 = ; B:C = 2:5 = 9 5 A:C =

A A B 4 2 8 = × = × = C B C 9 5 45

Hence, A:C = 8:45

Example 3

If A:B = 1:2 and B:C = 6:7, find A:B:C

To find A:B:C; the value of B must be the same in both ratios. Value of B in 1st ratio = 2; Value of B in 2nd ratio = 6 LCM of 2 and 6 = 6 1×3 3 6 A:B = 1:2 = = ; B:C = 2×3 6 7 Hence, A:B:C = 3:6:7

A piece of string that is 84 inches long is cut into three parts so that the lengths of the parts of the string are in the ratio 6:10:5. Find the length of each part. Total number of parts = 6 + 10 + 5 = 21 6 Length of 1st part = × 84 = 24 inches; 21 10 Length of 2nd part = × 84 = 40 inches; 21 5 Length of 3rd part = × 84 = 20 inches. 21

Example 4

Do It Together

3 Sunita purchased two colours of ribbons from the market. The red ribbon was 2 m long, and the blue 4 5 one was 1 m long. What is the ratio of the length of the red ribbon to that of the blue ribbon? 8 3 5 11 13 Ratio of the length of the red ribbon to the length of the blue ribbon = 2 :1 = : 4 8 4 8 To write the ratios in their simplest form, multiply both sides with the LCM of the denominator. 11 13 LCM of 4 and 8 = 8; Hence, × 8 = 22; × 8 = 13 4 8 The required ratio is 22:13. Arrange the ratios 1:3, 5:7, 4:15 and 2:5 in descending order. 1 5 4 2 1:3 = ; 5:7 = ; 4:15 = and 2:5 = 3 7 15 5 LCM of 3, 7, 15 and 5 = ________

1 5 4 2 = ________; = ________; = ________; = ________; 3 7 15 5

Hence, the descending order is ________>________>________>________

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Do It Yourself 11A 1

A bouquet of 45 roses has 32 red roses and the rest are white. What is the ratio of red to white roses?

Show the given ratios in their simplest form. a 120 cm to 3 m

d 240 s to 1 hour

Fill in the blanks to make the given ratios equivalent. a 6:1 = 30: ____

c 5 km to 800 m

b ____ :25 = 8:5

c 5: ____ = 20:16

d 55:60 = ____ :12

Compare the ratios and fill in the boxes with the correct sign (>, <, =). a 1:7

5:21

b 10:5

14:9

c 7:15

6:13

Arrange the ratios 2:5, 7:12, 3:7, 9:2 and 11:14 in ascending order.

If A:B = 5:11 and B:C = 7:13, then find A:C and A:B:C.

Find the ratio of the perimeters of the given figures.

d 8:15

9:20

9 cm 8 cm

12 cm

9 cm

b 6 L to 240 mL

The ratio of two numbers is 7:5. If each number is increased by 9, the ratio so formed is 5:4. Find the

The ratio of pencils to erasers to sharpeners in a classroom is 4:5:3. If there are 20 pencils, find the number of

numbers.

erasers and sharpeners.

10 The table shows the performance of a hockey team over two years. In which year did the team perform better? Year

Number of Matches Won

Number of Matches Lost

2020

2021

11 A recipe requires 1

2 1 cups of sugar and 2 cups of flour. What is the ratio of cups of sugar to cups of flour? 3 6

12 The angles of a triangle are in the ratio 2:3:4. What is the sum of the largest and smallest angles of the triangle? 13 The ratio of the lengths of three pieces of ribbon is 3:4:6. If the shortest piece is 12 inches long, find the total length of the ribbon.

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Word Problems 1

Nikita and Mahi have bats weighing 2.45 pounds and 1.95 pounds respectively. What is the ratio of

A farmer mixes two types of fertilizer in the ratio of 4:3 to create a special blend. If the farmer uses

Three friends, Rehan, Bobby and Heera, share a sum of money in the ratio of 2:3:5. If they receive

2 7 The sides of a rectangle are in the ratio : . If the sum of the sides is 55 cm, find the measure of 5 10 the sides.

The ages of Priya and Surbhi are in the ratio 7:8. After six years the ratio of their ages will be

A recipe for making a fruit salad calls for apples, grapes, and strawberries in the ratio of 2:3:1.

the weights of the bats in its simplest form?

240 g of the first type, what is the weight of the second type of fertilizer used?

a total of ₹3700, how much does each friend receive?

10:11. Find their present ages.

If 450 g of fruit salad is prepared, how many more grams of grapes than apples does the salad include?

Proportional Terms Remember, Aman and Nisha jog in the park in the ratio 5:7. We also saw that the ratio of the distances covered by them in a week was 35:49, which is equivalent to 5:7. The two ratios can also be said to be in proportion as they are equal. Equivalent Ratios

Sign of proportionality

Second term (b) First term (a)

Third term (c) ::

5:7

35:49

Fourth term (d)

Means Extremes

Proportionality Condition Two ratios are said to be in proportion if the increase or decrease in the ratio of one leads to the same increase or decrease in the ratio of other. If a:b :: c:d then, Product of Means = Product of Extremes b×c=a×d

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Example 5

Find the value of x in the proportion 25:15 :: x:42 Product of Means = Product of Extremes ⇒ 15 × x = 25 × 42 ⇒x=

Example 6

25 × 42 = 70 15

Each bag of marbles sold by Seema contains 9 blue marbles and 5 red marbles. If she sells 117 blue marbles in a day, how many red marbles has she sold? Using the method of proportion, 9:5 :: 117:x 5 × 117 ⇒ x = 65 9 Hence, Seema has sold 65 red marbles. ⇒ 9 × x = 5 × 117 ⇒ x =

Example 7

What number must be added to 8, 21, 13 and 31 so that they are in proportion? Let the number to be added be x Then, (8 + x):(21 + x) :: (13 + x):(31 + x) 8 + x 13 + x = 21 + x 31 + x

Error Alert! The sign of the operator changes during transposition from LHS to RHS and vice versa. 2x – 3 = 5x + 2

2x – 3 = 5x + 2

7x = –1

3x = –5

Let us subtract 1 from both the sides to remove x from the numerator, 8+x 13 + x 8 + x – 21 – x 13 + x – 31 – x –1 = –1 ⇒ = 21 + x 31 + x 21 + x 31 + x

–13 –18 = ⇒ –13(31 + x) = –18(21 + x) ⇒ –403 – 13x = –378 – 18x ⇒ 5x = 25 ⇒ x = 5 21 + x 31 + x Hence, 5 must be added to 8, 21, 13 and 31 to make them proportional. ⇒

Do It Together

The first three terms of a proportion are 4.2, 1.2 and 0.7. Find the fourth term. Let the fourth term be x; 4.2:_______ :: _______:x ___ × ___ = _______ 4.2 Hence, the fourth term is ________ . 4.2 × x = ______ × ______ ⇒ x =

Continued Proportion Any three quantities a, b and c are said to be in continued proportion if, First Proportional

a:b :: b:c

Third Proportional Mean Proportional

Using the proportionality condition; b × b = a × c ⇒ b2 = ac Example 8

What is the value of the first proportional if 10 and 4 are in continued proportion? The proportion can be given as: x:10 :: 10:4. ⇒ 4x = 10 × 10 ⇒ x = Hence, the value of the first proportional is 25.

Chapter 11 • Ratio and Proportion

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Do It Together

Find the third proportional to Let the third term = x.

1 2 and if the terms are in continued proportion. 3 9

1 2 1 2 2 1 :_____ :: :x ⇒ × x = × _____ ⇒ x = × _____ ÷ 3 9 3 9 9 3

⇒x=

2 3 × _____ × ⇒ x = ___________ 9 1

Do It Yourself 11B 1

Find whether the numbers are in proportion or not. a 5, 6, 15, 18

1 2 14 1 , , , 5 7 5 4

3 7 3 8 , , , 8 9 21 9

25 4 : :: :x 37 5

31 4 : :: x: 52 9

Find the value of x in each of the proportions. a 40:24 :: 20:x

b 4, 3, 16, 14

b 6:8 :: x:56

Find the third proportional for each set of continued proportions. a 4,6

b 12.5, 2.5

5 2 , 7 5

3 5

, 4.2

If the terms 21, (24 + d), 7 and d are in proportion, then find the value of d.

A boat can travel 126 miles on 18 gallons of gasoline. How much gasoline will it need to go 175 miles?

A worker earns ₹3000 in 5 days. In how many days will he earn ₹5000?

If 4 TV sets cost ₹1,12,000, how much will 23 such TV sets cost?

If 270 kg of grass would feed 42 cows for 21 days, for how many days would 360 kg of grass feed the same

27 m of cloth can be purchased for ₹1518.75. How much cloth can one buy for ₹1068.75?

number of cows?

10 What number should be subtracted from each of the numbers 23, 30, 57 and 78 to make them in proportion?

Word Problems 1

Sam spent $2,307.50 on his science research in 35.5 weeks. How much money did he spend in

The ages of Rohan, Kunal and Vicky are in continued proportion. If the ages of Rohan and Kunal

15 weeks?

are 8 years and 16 years, then what is Vicky’s age?

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Unitary Method Rahul, a friend of Aman and Nisha, also jogs in the same park. He covers a distance of 42 km in 12 days. How much distance does he cover in a week? Let us find out! The distance covered by Rahul can be found by unitary method. In this method, we first find the value of 1 unit and then the number of units asked. Step 1

Step 2

Value of many

Value of one

Distance covered in 1 day = 42 ÷ 12 = 3.5 km

Example 9

Value of one

Value of many

Distance covered in a week = 3.5 × 7 = 24.5 km

Komal reads 518 pages of a book in two weeks. In how many days will she read a book containing 1221 pages? Number of pages read in 2 weeks or 14 days = 518 Number of pages read in 1 day = Number of days for 1 page =

Example 10

518 = 37; Number of days for 37 pages = 1 day 14

1 1 ; Number of days for 1221 pages = × 1221 = 33 days 37 37

30 women can finish a piece of work in 18 days. How many women can finish the same work in 20 days? In 18 days, the number of women who can do a piece of work = 30 ∴ In 1 day, the number of women who can do a piece of work = 30 × 18 (fewer days, more women) ∴ In 20 days, the number of women who can do the piece of work =

Do It Together

30 × 18 = 27 (more days, fewer women) 20

A machine produces 208 toys in a day if it runs for 6.5 hours. How many toys will it produce in a day if the machine runs for 20 hours? Number of toys produced in 6.5 hours = 208 Number of toys produced in 1 hour = 208 ÷ _______ Number of toys produced in 20 hours = ______ × ______ = ______

Do It Yourself 11C 1

2 kg of potatoes cost ₹16. How many potatoes (in kg) can be bought for ₹120?

A car travels 150 km in 3 hours. How far does it travel in 9 hours?

If the cost of 15 pens is ₹300, find the cost of 5 such pens.

The human heart beats 375 times in 5 minutes. How many times does it beat in an hour?

Chapter 11 • Ratio and Proportion

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A 20 cm long candle burns for 50 minutes. How long will another candle of the same thickness and material but

19 bags of wheat of equal weight weigh 332.5 kg. How much will 35 such bags weigh?

33 m of cloth is required to make 15 shirts. How many shirts can be made with 110 m of cloth?

8 similar books put one on top of the other weigh 1840 grams. How many such books will weigh 4.37 kg?

If 12 workers can build a wall in 5 days, how many workers will be required to build the same wall in 3 days?

10 cm longer than the previous one burn?

10 If 12 men can do a piece of work in 18 days, in how many days will 8 men complete it? 11 One box can hold 35 sweets, and 11 teaspoons of flour are needed to make 5 such sweets. How many teaspoons of flour are required to make enough sweets to fill 5 such boxes?

Word Problems 1

Shikha takes 40 minutes to type 12 pages. How many hours will she take to type 90 pages?

Rahul bought a box of 15 pens for ₹172.5 whereas Sunita bought a box of 20 pens for ₹210. a Who bought the pens at a lower price? b Keshav bought a box of 25 pens with the price of each pen 1.5 times that of the pens bought by Rahul. What is the cost of the box of pens bought by Keshav?

Speed, Distance and Time Remember Rahul who jogs in the nearby park. If he takes 35 minutes to cover a distance of 3.5 km, then what is his speed? Let us find out!

S=D÷T

D=S�T

T=D÷S

Speed The distance an object travels per unit of time.

Distance Length of the path between two points.

Time Duration of an event.

Speed of Rahul =

Distance 3.5 km = = 0.1 km per minute. Time 35

Remember! The unit of speed depends on the unit of distance and the time taken.

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Example 11

Richa lives in town A. If Richa pedals her bicycle at a speed of 18 km/hr, how long will it take her to reach Town B? 45 km

Speed = 18 km/hr; Distance = 45 km Time =

Example 12

Distance 45 = = 2.5 hours Speed 18

Town A

A train covers a certain distance at a speed of 110 km/hr in 4 hours 30 minutes. What is the distance covered by the train? We know that distance = speed × time

Think and Tell

Speed = 110 km/hr; time = 4 hours 30 minutes = 4.5 hours

converted to 4.5 hours?

= 110 × 4.5 = 495 km Do It Together

Town B

How is 4 hours 30 minutes

Kunal covers a distance of 4.5 km at the rate of 1.5 km/hr, whereas Soham covers a distance of 8 km at the rate of 2 km/hr. What is the ratio of the time taken by both of them? Distance covered by Kunal = 4.5 km; Kunal’s speed = ________ Time taken by Kunal = Distance ÷ Speed = ________ Distance covered by Soham = _______; Soham’s speed = 2 km/hr; Time taken by Soham = Distance ÷ Speed = ________ Ratio of the time taken by both of them = _______:_______

Do It Yourself 11D 1

Find the speed of a car if it travels 400 m in 25 seconds.

A whale swims at a constant speed of 8.5 m/s for 1 minute. Calculate the distance travelled by the whale.

A train travels at a speed of 20 m/s and travels a distance of 3.2 km. Calculate the time it takes the train to

Kavita walks 500 m in 6 min. Find her speed in m/s to two decimal points.

A vehicle drives a distance of 270 km at a speed of 18 m/s. Calculate the time taken for this journey in minutes.

The distance between Delhi and Uttarakhand is 372 km. If John travels by car at a speed of 48 km/ hr, how many

The earth takes one year to go round the sun. The distance travelled is 940 million km if there are 365 days in

complete this journey in minutes.

minutes will he take to reach Uttarakhand?

a year.

a Calculate the speed the Earth travels at in km per day. b Calculate the speed of the Earth in km per hour.

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Word Problems 1

Kirti travelled 36,000 m at a speed of 8 km/hr. Mihir travelled 45 km at a speed of 10 km/hr. a Whose journey took the shortest time? b What is the time difference in minutes?

Rahul writes down his jogging times for each day. Mon – 17 min; Tue – 12 min; Wed – 15 min; Thu – 8 min; Fri – No jogging. a State on which day he jogs the farthest. b He jogs at a constant speed of 5.5 km/hr. Calculate the distance he jogs each day.

Som and Drishti want to ride their bicycles from their neighbourhood to the school, which is

10.8 km away. It takes Drishti 30 minutes to arrive at the school. Som arrives 15 minutes after Drishti. How much faster (in m/s) is Drishti’s speed for the entire trip?

Points to Remember •

Ratios are the comparison of two quantities of the same unit.

•

If the HCF of both the terms of a ratio is 1, then the ratio is said to be in its simplest form.

•

Equivalent ratios are the ratios that are the same when we compare them.

• Two ratios are said to be in proportion if the increase or decrease in the ratio of one leads to the same increase or decrease in the ratio of the other. •

If a:b :: c:d, then a × d = b × c.

•

In the unitary method, we first find the value of 1 unit and then the number of units asked.

•

We can find the speed, distance or time taken by using the formula Distance = Speed × Time.

Math Lab Artistic Ratios Setting: In groups of 2 Materials Required: Coloured pencils or any markers, simple figures on square grids, large sheets of paper or canvas for each student or group, rulers, pencils and erasers.

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Method: • Distribute the figures on square grids to different groups. • The groups draw a grid on their large sheets of paper or canvas. The size and number of

squares in the grid will depend on the image used as a reference. Emphasise the importance of maintaining the proportions when drawing the grid.

• Explain that each square in their grid represents a corresponding square on the reference image. This process introduces the concept of proportion.

• After the image is scaled and grid lines are drawn, start colouring each square based on the

colours you see in the reference image. Ensure the colours and ratios in the grid are the same as in the original image.

• The groups showcase their artwork and discuss the ratios and proportions they used in the designs.

Chapter Checkup 1

If 51:85 and x:5 are equivalent, then find the value of x.

The ratio of two numbers is 7:8. If each number is reduced by 24; the ratio is 5:6. Find the numbers.

What number should be added to each term of the ratio 3:5, so that the ratio becomes 5:7?

If the terms (49 – y), 56, y and 9 are in proportion, then find the value of y.

A stick 4 feet long is held perpendicular to the ground, casting a shadow 3 feet long. At the same time, an

A pancake recipe requires

A baker uses 3 kg of flour to bake 12 loaves of bread. How much flour will he need to bake 20 loaves of bread at

A bus consumes 22 litres of diesel in covering a distance of 99 km. How much diesel will it consume in covering

A beetle travels at a speed of 0.2 m/s. It travels a distance of 1.09 m before it is caught in a jar. Find the time taken

electrical pole casts a shadow 90 feet long. What is the height of the electrical pole?

1 3 5 tablespoon of baking powder; 1 tablespoons of sugar and 2 tablespoons of flour. 2 4 8 What is the ratio of baking powder to sugar to flour required?

the same rate?

500 km?

by the beetle to fly before getting caught.

10 A machine can produce 32 widgets in 5 minutes. How many widgets can it produce in an hour? 11 A plane travels 2,90,000 metres in 9000 seconds. What is the speed of the plane?

Chapter 11 • Ratio and Proportion

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12 If 25 men can construct a building in 150 days, in how many days will 30 men do the same work? 13 A mouse runs a distance of 5.5 metres in 30 seconds. Calculate the speed of the mouse to two decimal points. 1 14 Suman cycles for 5 minutes at a speed of 4 m/s. Find the distance she travels. 2 15 Vivek estimated that he can run for hours at a steady rate of 8.0 mph. He enters a marathon, a distance of 28 miles. Calculate the time it takes him to complete the race.

16 The present age of a father is 42 years and that of his son is 14 years. Find the ratio of the ages of the father and son after 10 years.

17 What number should be added to each of the numbers 10, 18, 22 and 38 to make them in proportion? 18 The ratio of the ages of Suresh and his son Kartik is 11:5. If the ratio of their ages 5 years ago was 5:2, find their present ages.

Word Problems 1

Sam earns ₹1750 in a week for washing cars. How much money will he earn in the month

The ratio of Sunil’s expenditure:saving is 4:1. If his total income is ₹18,000, how much does

Mr. Sharma drives 62.5 km from work at a speed of 50 km/hr. Mrs. Sharma drives 81.2 km

of October?

he save and spend?

from work at a speed of 58 km/hr. They both leave work at the same time. a Find the time taken for each person to travel home. b State which person arrives home first. c Find the time between the first and the second person getting home.

111 A certain amount was divided among Kirti, Vikas and Avinash in the ratio : : . 345 If the amount received by Avinash is ₹28,800, find the total amount to be distributed.

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Percentage, Profit and Loss, Simple Interest

Let's Recall The employees in Danny’s office come to office either by train, bus, bicycle or car. The pie chart shows the fraction of employees who opt for different types of transportation. Bus Bicycle

10 4

Train

10 Car

3 10 Which transportation is opted for by the highest number of employees? Car What fraction of the employees come to office by bicycle?

Let’s Warm-up Read the pie chart on a few bakery shop items and fill in the blanks. Pastries Donuts

Juice 1

100

Coffee

Tea

1 The most popular bakery item is ______________. 2 The least popular bakery item is ______________. 3 The fraction of people (out of 100) who like coffee is ______________. 4 The fraction of people (out of 100) who like pastries is ______________. 5 The fraction of people (out of 100) who like juice is ______________.

I scored _________ out of 5.

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Percentage and Its Applications Real Life Connect

Rakesh conducts a survey to find the total population of Kota. The total population of the city is about 15,00,000. There are 6,75,000 females, and the rest are males.

Female = 6,75,000

Percentages, Decimals and Fractions

Male = 8,25,000

Rakesh wants to find the percentage of females in the city. How can he do that? Let us see! The term per cent has been derived from a Latin word ‘per centum’ which means ‘out of one hundred’. Per cent is denoted as ‘%’.

37 parts shaded out of 100 As a decimal, 0.37

As a fraction,

37 100

As a percentage, 37%

Fraction to Percentage a a  =  × 100  % b b  To convert a fraction to a percentage, multiply the fraction by 100 and put a percentage symbol (%). For example,

6,75,000 15,00,000 6,75,000 Percentage of females in the city = × 100 15,00,000 = 45% Fraction of females in the city

Percentage to Fraction a 100 To convert a percentage to a fraction, remove the percentage symbol (%) and write 100 in the denominator. a% =

For example, convert 48% into a fraction. 48% =

48 12 = 100 25

Decimal to Percentage

Percentage to Decimal

a = (a × 100)%

a% = a ÷ 100

To convert a decimal to a percentage, multiply the decimal number by 100 and put the percentage symbol (%).

To convert a percentage to a decimal, remove the percentage symbol (%) and divide the number by 100.

For example,

0.25 = (0.25 × 100) % = 25%

81% = (81 ÷ 100) = 0.81

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Convert the given fractions into percentages.

Example 1

4 42 2 10 200 4 4 42 42 = × 100% = 40% = × 100% = 21% 10 10 200 200

Convert the given percentages into fractions.

Example 2

23% 2 125% 23% =

23 100

125% =

125 5 = 100 4

Convert the decimals into percentages.

Example 3

0.58 2 2.5 0.58 = 0.58 × 100% = 58%

2.5 = 2.5 × 100% = 250%

Convert the percentages into decimals.

Example 4

48% 2 302% 48% = 48 ÷ 100 = 0.48

Do It Together

302% = 302 ÷ 100 = 3.02

Convert the percentages into fractions and decimals. 1

56% 56 14 = 100 25 56% = 56 ÷ 100 = _______ 56% =

120%

550%

120% = ______ = ______

550% = ______ = ______

120% = 120 ÷ ______ = ______

550% = ______ ÷ ______ = ______

Do It Yourself 12A 1

Express the percentages as fractions in their simplest form.

e 3000 4 6 j 10

d 0.005%

e 123.5%

c 180%

d 56.4%

f 6.5%

g 0.36%

h 360%

Express the fractions as percentages. 1 5 21 f 20

2 20 1 g 5 4 b

18 25 3 h 12 5 c

d i

Express the percentages as decimals. a 48%

b 1.25%

c 12.64%

f 1.2%

1 g 11 % 5

h 12

Chapter 12 • Percentage, Profit and Loss, Simple Interest

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5 3000 3 4

b 55%

1 58 % 4

1 e 16 % 2 1 j 64 % 3

a 2.25%

4 % 10

3 % 50

1 145 % 8

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Express the given decimals as percentages. a 0.2

b 0.34

c 0.547

d 0.984

f 1.2

g 2.34

h 12.1

e 0.0048

15.23

142.214

Find what fraction of the figure is coloured and also write the percentage of the coloured part. a

Word Problems 1

Vivan goes to eat a pizza. He eats 40% of the pizza. What fraction of the pizza is left?

If 6 out 8 students passed an examination, what percentage of the students did not pass the

Inflation was 4.5% in a year. How do we represent this percentage as a decimal?

In a shipment delivery, 95% of the packages were delivered without any damage. What fraction of

Yash scored 360 marks out of 500, and Jaya scored 450 marks out of 600. Who has the better

The marks scored by Rahul in various subjects (out of 100) are shown using a bar graph. What is

examination?

the shipment was damaged? percentage?

the average percentage of the marks scored by him?

1 division = 10 marks

100 90 80

Marks

70 60 50 40 30 20 10 Mathematics

Science

Hindi Subjects

English

Social Science

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Percentage and Ratios Remember the survey conducted by Rakesh. He found the ratio of the number of males to females is 11:9. He wanted to find the percentage of males in the city. Let us see! Convert ratios into percentages

Did You Know?

To convert a ratio to a percentage, we write the ratio as a fraction and then convert it into a percentage.

About 29% of earth’s

If there are 11x males, then there are 9x females in the city.

and the rest with water.

surface is covered with land

Total number of people in the city = 20x Ratio of males to the total number of people = 11:20  11

Percentage of males in the city = 

 20



× 100  % = 55%

Thus, there are 55% males in the city.

Remember!



The denominator always has the total number of parts.

Convert percentages into ratios There were 65% males in a meeting. Let us find the ratio of males to females in the meeting. Percentage of males in the meeting = 65%

Only 0.006% of earth’s

Percentage of females in the meeting = 100% – 65% = 35% 65 35 Ratio of males to females = 65%:35% = : = 65:35 = 13:7 100 100 Thus, the ratio of males to females in the meeting is 13:7. Example 5

Did You Know? water is available for human use.

In a class, 25% students are enrolled in the science club. What is the ratio of science club members to the total number of students? Percentage of students enrolled in the science club = 25% Ratio of the science club members to the total number of students = 25%:100% = Thus, the ratio of science club members to the total number of students is 1:4.

Example 6

At a bookstore, 30% of the books are fiction. If there are 300 books in total, what is the ratio of fiction to non-fiction books? Percentage of fiction books at the bookstore = 30% Percentage of non-fiction books at the bookstore = 100% – 30% = 70% Ratio of fiction to non-fiction books = 30%:70%

25 100 : = 25:100 = 1:4 100 100

Error Alert! Ratio of fiction books to total books Fiction:Non-fiction

Fiction:(Fiction + Non-fiction)

3:7

3:(3 + 7)

3:10

30 70 : = 30:70 = 3:7 100 100 Thus, the ratio of fiction books to non-fiction books is 3:7. =

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Example 7

In a collection of coins, the ratio of quarters to dimes is 3:17. What percentage of the total coins are dimes? Ratio of quarters to dimes = 3:17 Ratio of dimes to total coins = 17:(3 + 17) = 17:20 17 × 100% = 85% 20 Thus, 85% of the total coins are dimes. Percentage of dimes =

Do It Together

At a zoo, the ratio of monkeys to elephants is 3:5. If there are 64 monkeys and elephants in the zoo, what percentage are monkeys? Ratio of monkeys to elephants = 3:5 Ratio of monkeys to total animals = ________ Percentage of monkeys at the zoo = ________

Think and Tell

What percentage of the animals are elephants if there are 400 animals in the zoo?

Do It Yourself 12B 1

In a class of 30 students, 20% are enrolled in the art club. What is the ratio of art club members to the total

In a jar of candies, 25% are blue and the rest are red. What is the ratio of blue to red candies?

At a party, 30% of the attendees are children. If there are 50 people in total, what is the ratio of children

A recipe calls for 2 cups of flour and 3 cups of sugar. What percentage of the ingredients is sugar?

In a garden, the ratio of roses to tulips is 3:2. If there are 50 flowers in total, what percentage are roses?

In a game, a player successfully completed 8 out of 10 challenges. What percentage of the challenges were

In a group of 80 people, 75% prefer apple pie over donuts. What is the ratio of apple pie lovers to donuts

number of students?

to adults?

completed?

lovers?

Word Problem 1

In a company, the ratio of managers to employees is 1:9. What percentage of total people are managers?

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Finding Percent and Percent Change After a year, 10% of the total number of people migrated to the city. Rakesh wants to know how many people were there after a year. Let us see! Finding percentage of a given quantity Total number of people in the city = 15,00,000. Percentage of people migrated = 10% Let us find 10% of 15,00,000. To find a% of b, we multiply

a by b. 100 a% of b =

a ×b 100

10 × 15,00,000 = 1,50,000 100 Thus, 1,50,000 people migrated to the city that year. 10% of 15,00,000 =

Total number of people after a year = 15,00,000 + 1,50,000 = 16,50,000 Finding a number as a percentage of another The following year 1,65,000 people migrated to another city. What percentage of the people left the city? Let us see! The total number of people = 16,50,000 Number of people migrated = 1,65,000 Let us find what percent of 16,50,000 is 1,65,000. Let a% of 16,50,000 = 1,65,000 Number of parts × 100 Total number of parts  1,65,000  = × 100 % = 10%  16,50,000  Thus, 10% of the people migrated to another city. a% =

Finding a whole number using the percentage Find a number of which 5% is 48. Assume the number to be a. 5% of a = 48 1 1 × a = 48 ⇒ × a = 48 100 20 ⇒ a = 48 × 20 = 960 5×

Thus, the number is 960.

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Percent Change (Increase or Decrease) Find the percent change when 80 is increased to 88.  Amount of increase or decrease  Percent change =  × 100 %   Original amount

 88 – 80  8  Percent change =  × 100 % =  × 100 % = 10% increase  80   80  Example 8

Find: 1

Example 9

What percent of: 1

Example 10

1 12% of 96 2 20 % of 500 2 1 41 12 1152 41 1 41 12% of 96 = × 96 = = 11.52 20 % of 500 = % of 500 = × × 500 = × 5 = 102.5 2 2 100 100 2 100 2

200 is 40?

Let a% of 200 = 40

Let a% of 30 m = 12 m 50 cm

 40  1 a% =  × 100  % = × 100% = 20%  200  5

a% =

1250 cm 5 12 m 50 cm × 100% = × 100% = × 100% 3000 cm 12 30 m 2 = 41 % 3

Find the number of which: 1

5% is 23

Let the number be a.

1 4 % is 34 4 Let the number be a. 17 % of a = 34 4 17 1 4 × × a = 34; a = 34 × 100 × = 800 4 100 17

5% of a = 23 1 × a = 23 100 1 a = 23 × 100 × ; a = 460 5 5×

Example 11

30 m is 12 m 50 cm?

Find the per cent change when: 1

45 is increased to 54  Amount of increase  9 54 − 45 × 100% = 20% × 100 % = × 100% = Per cent increase =  Original amount 45 45  

2 400 is decreased to 360  Amount of decrease  400 − 360 40 × 100% = 10% × 100% = Per cent decrease =  × 100 % = 400 400  Original amount  Do It Together

Find: 1

1 45 % of 2000 2

1 91 45 % of 2000 = % of 2000 2 2 =

The number which is 35% more than 520 35% of 520 = 520 × ______ = ______ Number 35% more than 520 = 520 + ______ = ______

91 1 × × 2000 = ______ 2 100

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Do It Yourself 12C Find.

a 25% of 250

b 30% of 750

c 125% of 45

d 14% of 4800

Express.

a 5 cm as a percentage of 4 m 25 cm

b 450 mL as a percentage of 3 L

Find the number of which:

a 12% is 48.

2 b 66 % is 567. 3

c 125% is 5 hours 20 minutes.

d 20% is 3 m 40 cm.

Find the per cent change.

a 320 is decreased to 272

b 110 is increased to 165

c 150 is increased to 195

Answer the questions.

a What is the value of 3% of 20 minutes in seconds? 1 c What number is 5 % of 24? 2

b What per cent of 350 kg is 3 kg 500 g? d Which number is 12

1 % more than 2500? 2

Word Problems 1

There are 450 women employees in a factory. How many employees are there in total in the

If ₹500 is to be divided among Ravi, Raju and Roy so that Ravi gets two parts, Raju three parts and

factory if the factory has 55% male employees?

Roy five parts, how much money will each get? What will this be in percentages?

Word Problems on Percentage Real Life Connect

After 5 years, the population of Kota city becomes 18,00,000. What is the total percentage increase in the population? We know, initial population of Kota city = 15,00,000 Population of the city after 5 years = 18,00,000 Population increase = 18,00,000 – 15,00,000 = 3,00,000  3,00,000  Percent age increase in the population =  × 100 % = 20%  15,00,000  Thus, the population increased by 20%.

Chapter 12 • Percentage, Profit and Loss, Simple Interest

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Example 12

A school bus stops at a distance of 22 km from the school while travelling to a picnic spot which is 55 km away. What percentage of the distance is remaining for the bus to reach the picnic spot? Fraction of the distance remaining for the bus to reach the picnic spot = Percentage of remaining distance =

55 − 22 33 = 55 55

33 3 × 100% = × 100% = 60% 55 5

Thus, 60% of the distance is remaining for the bus to reach the picnic spot. Example 13

After discarding 15% of his oranges, a farmer has 3400 oranges left. How many oranges did the farmer have initially? Percentage of oranges discarded = 15% Percentage of oranges left in the basket = 100% – 15% = 85% Let the total number of oranges be x. 85% of x = 3400;

85 100 × x = 3400 ⇒ x = 3400 × = 4000 100 85

Thus, the farmer initially had 4000 oranges. Example 14

There are 5000 students in a school. 3600 students participated in the annual function. What percentage of students did not participate in the annual function? Number of students that did not participate in the function = 5000 – 3600 = 1400 Percentage of students who did not participate =

1400 7 × 100% = × 100% = 28% 5000 25

Thus, 28% of the total students did not participate in the annual function. Do It Together

On Sunday, 845 people went to an exhibition. On Monday, only 169 people went. What is the percentage decrease in the number of people visiting the exhibition on Monday? Number of people who went to the exhibition on Sunday = 845 Number of people who went to the exhibition on Monday = 169 Change in the number of people = _____ − ______ = ______ ________________________________ _____________ × 100% = × 100% = _____% ________________________________ _____________ Thus, there was a _____% decrease in the number of people visiting the exhibition on Monday. Decrease in per cent =

Do It Yourself 12D 1

An alloy of copper and zinc contains 26% copper. What quantity of alloy is required to get 1480 g of zinc?

There are 1440 men at a gathering. If 40% of the total people are women, how many people are there at the gathering?

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In an election, Candidate A received 12% of the total votes. The total votes are 1,53,600. If 54% of people did not

In an examination, 35% of the students scored above 80% and half of the students scored between 50% and 80%.

cast any vote, how many votes did Candidate B get?

The rest of the students scored less than 50%. If there were 500 students who appeared for the examination, how many students scored less than 50%?

The population of a city decreased by 8% every year. If the current population of the city is 1,69,280, what was the

In an examination, B scored 20% more than A and A scored 10% less than C. D obtained 20% more marks than C.

In a town, the population was 10,000. After 1 year, the male population increased by 10% and female population

A person’s salary was first increased by 20% and then reduced by 10%. What is the percentage change in the salary of

The cost of a pomegranate is twice that of a pear and the cost of a pear is 25% more than that of an orange.

population of the city 2 years ago?

By what percentage are the marks obtained by D more than those obtained by A?

increased by 12% but the total population increased by 11%. What was the male population initially? the person?

If the cost of each type of fruit is increased by 10% then what is the percentage increase in the cost of 4 pears, 2 pomegranates and 3 oranges?

10 The number 4000 is first increased by 25% and then decreased by 30%. What is the percentage change in the initial value?

11 The cost of a car goes down every year by 20% of its previous value. Find its original cost if the cost of it after 2 years is ₹5,65,000?

Word Problem 1

The price of petrol goes up by 4%. Before the hike, the price of petrol was ₹95. A man

travels 3456 km every month and his car gives a mileage of 16 km per litre. What is the

increase in the monthly expenditure of the man (in nearest ₹) on travelling due to the hike in the petrol prices?

Profit and Loss Real Life Connect

Raju is a fish seller. He has a shop and purchases fish from multiple fishermen and sells it further.

Profit and Loss Percentages Raju purchases the fish for ₹50,000 and sells it all for ₹60,000. How much money does he earn? Let us see! Chapter 12 • Percentage, Profit and Loss, Simple Interest

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Cost Price The price at which goods are purchased.

Price at which Raju purchases the fish = ₹50,000 Price at which Raju sells the fish = ₹60,000

Selling Price The price at which goods are sold.

Money earned by Raju = ₹60,000 – ₹50,000 = ₹10,000 Profit When Selling Price > Cost Price  Profit  Profit% =  × 100  %  CP 

Loss When Cost Price > Selling Price  Loss  Loss% =  × 100  %  CP 

Finding SP when CP and profit/ loss % is given: SP =

Example 15

Finding CP when SP and profit/ loss % is given:

100 + Profit % 100 – Loss% × CP OR × CP 100 100

CP =

CP = ₹1500 and SP = ₹1800 SP > CP Profit% =

1800 − 1500

1500 = 20%

100 100 × SP OR × SP 100 + Profit % 100 – Loss%

CP = ₹2000 and SP = ₹1500 CP > SP

× 100% =

300 × 100% 1500

Loss% =

2000 − 1500

2000 = 25%

× 100% =

500 × 100% 2000

Find. 1

CP when profit% = 20% and SP = ₹1800 CP = =

Example 17

Profit and loss are always calculated on cost price.

Find the profit or loss percentage. 1

Example 16

Remember!

100 100 × SP = × 1800 100 + Profit% 100 + 20 100 5 × 1800 = × 1800 = ₹1500 120 6

SP when profit% = 15% and CP = ₹3000 SP = =

100 + Profit% 100

× CP =

115 × 3000 = ₹3450 100

100 + 15 100

× 3000

A farmer sold 2 bullocks for ₹48,000 each. On one bullock, he gained 25%; and on the other, he lost 20%. Find the percentage of his total profit or loss. SP of each bullock = ₹48,000 CP of the first bullock =

100 100 100 4 × SP = × 48,000 = × 48,000 = × 48,000 = ₹38,400 100 + Profit% 100 + 25 125 5

CP of the second bullock =

100 100 100 5 × SP = × 48,000 = × 48,000 = × 48,000 = ₹60,000 100 − Loss% 100 − 20 80 4

Total cost price = ₹38,400 + ₹60,000 = ₹98,400; Total selling price = ₹48,000 + ₹48,000 = ₹96,000 CP > SP, so a loss was incurred. Loss = CP – SP = ₹98,400 – ₹96,000 = ₹2,400 Loss% =

Loss 2400 × 100% = × 100% = 2.44% CP 98,400

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Do It Together

Calculate: 1

CP when SP = ₹1600 and loss% = 20% 100 × SP = ________ = ________ CP = 100 − Loss%

SP when CP = ₹2500 and profit% = 12% 100 + Profit% SP = × CP = ________ = ________ 100

Marked Price Raju labelled the prices on the fish counters. He labelled a particular type of fish with a price of ₹500. The price on the label of an item is called its marked price. The marked price can be different from the selling price. The marked price is denoted by MP. Discount: When a shopkeeper reduces a certain amount from the marked price of an article and sells it to the customers, the reduction is called a discount. The discount amount is given as a certain percentage of the marked price. If a discount is given then MP > SP. Discount = MP – SP Discount % =

Discount amount × 100% MP

Find the discount if MP = ₹5000 and discount % = 15%. Discount = 15% of 5000 =

15 × 5000 = ₹750 100

Find the discount if discount % = 35% and MP = ₹8000.

Example 18

35 × 8000 = ₹2800 100 Thus, the discount is ₹2800. Discount = 35% of 8000 =

Calculate the discount % if the discount is ₹1700 and the MP = ₹8500.

Example 19

Discount% = Do It Together

Discount amount 1700 × 100% = × 100% = 20% MP 8500

What is the discount % if the MP = ₹1600 and discount = ₹80? Discount% =

Discount amount × 100% = __________ = _________ MP

Do It Yourself 12E 1

Find the gain or loss. a CP = ₹620 and SP = ₹745

b CP = ₹5230 and SP = ₹6000

c CP = ₹50,000 and SP = ₹49,812

Select the correct answer. a CP = ₹405 and SP = ₹510 then gain = ? i

₹100

ii ₹105

iii ₹110

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iv ₹125

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b CP = ₹10,235 and SP = ₹9400 then loss = ? i

₹800

ii ₹835

iii ₹935

iv ₹1000

iii 60%

iv 75%

iii 15%

iv 20%

c CP = ₹2000 and SP = ₹2500 then profit% = ? i

25%

ii 40%

d CP = ₹2000 and SP = ₹1800 then loss% = ? i

10%

ii 12%

Find the missing term.

a SP = ₹500, CP = ₹450 then profit = ?

b CP = ₹2364, loss = ₹364 then SP = ?

c SP = ₹450, MP ₹585 then discount% = ?

d MP = ₹3060, SP = ₹2550 then discount% = ?

Find the profit% or loss%.

a CP = ₹1800 and SP = ₹2400

b CP = ₹500 and SP = ₹480

c CP = ₹2300 and SP = ₹2415

Match the loss % or profit % with their respective CP and SP.

a CP = ₹150 and SP = ₹175 b CP = ₹250 and SP = ₹225

Profit = 25% 1 Loss = 8 % 3

d CP = ₹3000 and SP = ₹2750

2 Profit% = 16 % 3 Loss = 20%

e CP = ₹4500 and SP = ₹3600

Loss = 10%

c CP = ₹200 and SP = ₹250

Word Problems 1

The selling price of a toy car is ₹540. If the profit made by the shopkeeper is 20%, what is the

A man sells two horses for ₹4000 each, neither losing nor gaining on the deal. If he sold one

A shopkeeper purchased 200 bulbs for ₹10 each. However, 5 bulbs were fused and had to be

Alex’s father bought a scooter for ₹49,000. He spent ₹6000 on transportation and ₹5000 on

cost price of this toy?

horse at a gain of 25%, then at what loss was the other horse sold?

thrown away. The remaining were sold at ₹12 each. Find the percentage of gain or loss. repairs. If he sells the scooter for ₹58,000, what is the total profit/loss?

Word Problems on Profit and Loss Remember Raju the fish seller? He buys some more fish for ₹1,00,000. He sells the fish for ₹1,05,000. How much profit does Raju earn? CP of fish = ₹1,00,000; SP of fish = ₹1,05,000 Profit = SP – CP = ₹1,05,000 – ₹1,00,000 = ₹5000 Thus, Raju earns ₹5000 on selling the fish. 196

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Utkarsh sold a plot of land for ₹5,00,000, gaining 25%. For how much did he purchase the plot?

Example 20

SP = ₹5,00,000 and Profit = 25% 100 100 4 × SP = CP = × ₹5,00,000 = × ₹5,00,000 = ₹4,00,000 100 + Profit% 100 + 25 5 Thus, Utkarsh purchased the plot for ₹4,00,000. Samuel buys a laptop at ₹90,000. After three years, he sells it at ₹75,000. Calculate the loss %.

Example 21

CP = ₹90,000 and SP = ₹75,000

Loss 2 × 100% = ₹90,000 – ₹75,000 × 100% = 16 % CP 3 ₹90,000 2 Thus, Samuel sells the laptop at a loss of 16 %. 3 Loss % =

Jessica sold an air conditioner for ₹32,500 at a loss of 20%. Find the cost price of the air conditioner. At what price should she have sold it to gain 20%?

Example 22

SP = ₹32,500 and loss = 20% 100 100 100 × SP = CP = × ₹32,500 = × ₹32,500 = ₹40,625 100 − Loss% 100 − 20 80 Thus, the cost price of the air conditioner was ₹40,625. If the gain was 20%, SP =

100 + Profit% 100

× CP =

100 + 20 100

× ₹40,625 =

120 × ₹40,625 = ₹48,750 100

Thus, Jessica should have sold the air conditioner at ₹48,750 to gain 20%. Do It Together

Jennifer buys an old scooter for ₹47,000 and spends ₹8000 on repairs. If she sells the scooter for ₹58,000, what is her profit %? Total CP of scooter = ₹47,000 + ₹8000 = ₹__________ SP of scooter = ₹__________  Profit  × 100  % = ________ = ________ Profit% =   CP 

Do It Yourself 12F 1

Mia sold a ring which she had bought from an exhibition for ₹9000, to her friend for ₹7600. What is the cost price

Anu bought a sofa for ₹19,700 and sold it for ₹20,000, whereas Manu bought another sofa for ₹21,200 and sold it

The marked price of a water cooler is ₹8350. The shopkeeper offers a discount of 18% on it. Find its selling price.

The percentage profit earned by selling an article for ₹1920 is equal to the percentage loss incurred by selling the

of the ring for Mia’s friend?

for ₹21,400. The transportation cost for Anu was ₹400, while for Manu it was ₹250. Who incurred the greater loss?

same article for ₹1280. At what price should the article be sold to make a 25% profit?

Chapter 12 • Percentage, Profit and Loss, Simple Interest

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On selling 17 balls at ₹720, there is a loss equal to the cost price of 5 balls. What is the cost price of the ball?

A trader mixes 26 kg of rice at ₹20 per kg with 30 kg of rice of another variety at ₹36 per kg and sells the mixture

A dealer incurs a loss of 5% if he sells an article for ₹1805. What price must he sell the article for to gain 5% on

A man sold two watches at the same price, one at a 10% profit and the other at a 10% loss. Find his overall

A shopkeeper marks his products at 35% above the cost price and allows a discount of 20% on the marked price.

at ₹30 per kg. What is the trader’s profit? that article?

percentage of gain or loss.

Find his gain or loss per cent.

10 The cost price of 20 articles is the same as the selling price of a articles. If the profit is 25%, then what is the value of a?

Word Problem 1

In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?

Simple Interest Real Life Connect

Emma had saved $500. She deposited the money in her bank. The manager told her that they would add a certain percentage to her total savings every month. Emma was really happy to hear that.

Using the Formula The manager said that they would give her an extra 2% every month. He said she could get $510 the next month. Emma was confused. Let us see how she gets the extra money. Principal Money borrowed or deposited Simple Interest The extra amount paid by the borrower.

SI =

P×R×T 100

Rate Rate of interest given on the principal amount. Time Period of time for which money is borrowed or deposited.

Money deposited in the bank = $500; Rate of interest = 2% per month; Time = 1 month P×R × T $500 × 2 × 1 SI = = = $10 100 100 Amount = P + SI = $500 + $10 = $510

Thus, Emma gets $510 after a month. 198

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Example 23

If Sanchita pays interest of ₹54 for one year at 9% p.a., what sum had she borrowed? SI = ₹54, R = 9% per annum, T = 1 year P×R × T P ×9×1 5400 SI = ⇒ 54 = ;P= = 600 9 100 100 Thus, Sanchita borrowed ₹600.

Example 24

Do It Together

1 3 Find the total amount a person gets when he deposits ₹40,000 at a rate of 5 % per annum for 4 years 2 4 in a bank. 1 11 3 19 P = ₹40,000, R = 5 % per annum = % per annum and T = 4 years = years 2 4 2 4 11 19 ₹40,000 × × 2 4 = ₹40,000 × 11 × 19 = ₹10,450; A = P + SI = ₹40,000 + ₹10,450 = ₹50,450 SI = 100 × 2 × 4 100 3 Thus, the person gets ₹50,450 from the bank after 4 years. 4 If Harish earns simple interest of ₹1000 by lending money to his friend at an interest rate of 10% per annum for 5 years, find the amount lent by him. SI = ₹1000, R = 10% and T = 5 years SI =

SI × 100 P×R × T ⇒P= = _________ = _________ = _________ R×T 100

Word Problems on Simple Interest Remember, Emma deposits $500 in the bank. How much money does Emma get after a year? Principal = $500, R = 2% per month and T = 1 year = 12 months SI = P × R × T = $500 × 2 × 12 = $120; A = P + SI = $500 + $120 = $620 100 100 Thus, Emma gets $620 after a year. Example 25

Shilpa lends a sum at 5% per annum to a friend who promises to return the sum with the interest after 5 years. What amount will she get if the interest she gets is ₹1250? SI = ₹1250, R = 5% and T = 5 years P = SI × 100 = $1250 × 100 = ₹5000; A = P + SI = ₹5000 + ₹1250 = ₹6250 R×T 5×5 Thus, Shilpa gets ₹6250 back from her friend.

Example 26

Raghav has a certain sum deposited in a bank at 5% per annum. Anshul also has the same sum deposited in the same bank at 9% per annum. The difference in the annual interest received by each of them is ₹440. Find the total sum of the money deposited by them. Let the money deposited by Raghav and Anshul be ₹x. 5x x×5×1 Interest received by Raghav after a year = =₹ ; 100 100 Interest received by Anshul after a year = Difference in the annual interest = ₹440

9x x×9×1 =₹ 100 100

Chapter 12 • Percentage, Profit and Loss, Simple Interest

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So, ₹

9x 5x 4x –₹ = ₹440 ⇒ ₹ = ₹440 ⇒ x = ₹11,000 100 100 100

So, Raghav and Anshul each deposited ₹11,000. Total sum of money deposited by Raghav and Anshul = ₹11,000 + ₹11,000 = ₹22,000 Example 27

Vishesh invested an amount of ₹10,000 divided into two different schemes A and B at the simple interest rate of 10% p.a. and 15% p.a. respectively. If the total amount of simple interest earned in 2 years is ₹2450, what was the difference of the amounts invested in each of the schemes? Let the amount invested in scheme A be x; Amount invested in scheme B = 10,000 – x SI from scheme A + SI from scheme B = Total SI x × 10 × 2 (10,000 − x ) × 15 × 2 20 x 30(10,000 − x ) + = 2450 + = 2450; 100 100 100 100 20 x + 30(10,000 – x ) = 2,45,000 ⇒ 20 x + 3,00,000 – 30 x = 2,45,000

⇒ 30 x – 20 x = 3,00,000 – 2,45,000 ⇒ 10 x = 55,000 ⇒ x = 5500 10,000 – x = 10,000 − 5500 = ₹4500

Difference of the amounts invested in the two schemes = 5500 – 4500 = ₹1000 Do It Together

Mr. Wilson took a loan of ₹1500 with simple interest for as many years as the rate of interest. If he paid ₹375 as interest at the end of the loan period, what was the rate of interest? P = ₹1500, R = T = R, SI = ₹375 P×R×T P×R×R SI = = 100 100 ____ =

_______ × R × R 2 _______ × 100 ;R = = _______ _______ 100

Do It Yourself 12G 1

Find the simple interest. a P = ₹5000, R = 5% per annum and T = 3 years

b P = ₹6000, R = 12% per annum and T = 5 years

c P = ₹2000, R = 10% per annum and T = 3 years

d P = ₹2250, R = 2% per annum and T = 4 years

Find the total amount.

1 years 2 1 1 c P = ₹10,000, R = 4 % per annum and T = 5 years 5 2

a P = ₹2000, R = 4% per annum and T = 2

1 years 3 1 1 d P = ₹20,000, R = 5 % per annum and T = 4 years 2 2

b P = ₹3000, R = 12% per annum and T = 3

Find the time taken.

a P = ₹2000, R = 4% per annum and SI = 1 of P 4 c P = ₹3375, R = 15% per annum and SI = ₹1518.75

Find the principal amount.

a R = 5% per annum, T = 9 1 years and SI = ₹2800 3 1 1 c R = 5 % per annum, T = 3 years and SI = ₹23,100 2 2

b P = ₹7500, R = 3% per annum and SI = ₹2250 1 d P = ₹12,000, R = 4 % per annum and SI = ₹3510 2

b R = 6% per annum, T = 8 years and SI = ₹23,520 d R = 12

3 1 % per annum, T = 1 years and SI = ₹31,500 5 4

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Find the rate of interest.

a P = ₹9500, T = 7 years and SI = ₹1995 c P = ₹20,000, T = 12 years and SI = ₹18,000

1 years and SI = ₹3412.5 5 1 d P = ₹3,60,000, T = 9 years and SI = ₹1111.5 2

b P = ₹12,500, T = 4

Word Problems 1

Akila paid an interest of ₹450 to her friend after a year at a rate of interest of 4.5% per annum.

Preeti took a loan of ₹2,50,000 from a bank at a rate of 8% interest per annum. Calculate the

A principal of ₹40,000 is deposited in a bank which offers a simple interest at the rate of

Nitish earns simple interest of ₹1000 by lending money to his friend at an interest rate of

How much money did she borrow?

simple interest that she has to pay after 4 years.

10% per annum. What is the total amount at the end of 2 years?

10% per annum for 5 years. How much is the amount lent by him? 7 A certain sum of money amounts to of itself in 3 years. What is the rate of interest? 4 A man earns an interest of ₹720 on a sum he lent at 18% per annum for two years. Had he earned

Abella borrowed a certain sum of money at the rate of 5% per annum for the first three years,

12% more interest on the same amount in the same time, what would have been the rate of interest? 8% per annum for the next four years, and 12% per annum for the period beyond 7 years. If she

pays a total amount of interest of ₹17,845 at the end of 10 years, find the amount she borrowed. 8

A sum of ₹1000 was lent to two people, one at the rate of 5% and the other at the rate of 8%. If the simple interest after one year is ₹62, find the amount lent at each rate.

Points to Remember • Percent means out of 100.

 a a =  × 100 % b b  a • To convert percentages to fractions a % = 100 • To convert fractions to percentages

• To convert decimals to percentages, multiply the decimal by 100 and put the percentage symbol (%). • To convert percentages to decimals, remove the percentage symbol (%) and divide by 100.  a a • To convert a ratio to a percentage- a : b = =  × 100 % b b   Amount of increase or decrease  × 100 % • To find the percent change, Percent change =  Original amount    Profit/Loss  × 100 % • Profit/Loss % =  CP   100 + Profit% 100 × CP and CP = × SP • When there is a profit, SP = 100 100 + Profit% 100 − Loss% 100 × CP and × SP • When there is a loss, SP = CP = 100 100 − Loss% P×R × T • Simple interest SI = 100 Chapter 12 • Percentage, Profit and Loss, Simple Interest

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Math Lab Math lab – Discount Shopping Aim: To practise calculating discounts and final prices, using percentages. Setting: In pairs Material required: Paper or pen, list of items with original prices Method: 1

Create a list of items with their original prices. (Example, ₹500, ₹1000, ₹1200 etc.)

Assign a discount percentage to each item. (Example, 10%, 15%, 20%, etc.)

Give the list to each pair and ask them to calculate the sale price after the discount.

Check the answers.

After this, give the students a budget (example ₹5000) and ask them to choose the items from the list within the budget.

Chapter Checkup 1

Express:

4 as a percentage 5 Express: a

b 18.75% as a fraction

a 0.48 as a percentage

Evaluate. a 25% of ₹460

1 4 % as a decimal 2

b 12

e 40% of 6 km 400 m

1 % of 20 L 2

1 % as a fraction 5

d 11

c 5.64 as a percentage

2 as a percentage 5

1 % as a decimal 8

1 50 % of 1000 2

d 5% of 3 kg 200 g

c ₹15 is ₹2 and 10 p?

d 20 km is 1000 m?

What per cent of: a 4.5 kg is 0.9 kg?

b 6 L is 20 mL?

e 3.5 L is 0.25 L?

Find a number of which: 1 a 4 % is 20 8 e 24% is 3 L 250 mL

1 % is 500 2

1 % is ₹933 3

Find the percentage when:

Find the profit or loss for the given cases. Fill in the blank and write the missing term.

a 12 is increased to 18

b 15 is increased by 15

a CP = ₹150, SP = ₹200, Profit = _____________

c CP = ₹5100, SP = ₹5235, Profit = _____________

1 % is 80 m 4 cm 12

c 200 is decreased to 180

b CP = ₹1580, SP = ₹1235, Loss = _____________ d CP = ₹458, SP = ₹369, Loss = _____________

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Find the profit% or loss%.

Find the discount%.

a CP = ₹50, SP = ₹65

a Discount = ₹80, MP = ₹580

b CP = ₹150, SP = ₹125

c CP = ₹1000, SP = ₹896

b Discount = ₹289, MP = ₹1139

c Discount = ₹155, MP = ₹775

10 Find the missing terms in the questions.

a P = ₹500, R = 2% p.a., T = 3 years, A = ?

b P = ₹5500, R = 14

1 1 % p.a., T = 6 years, A = ? 2 2

Word Problems 1 A person buys a pen for ₹100 and sells it for ₹125. What is the gain or loss? What is the gain % or loss %?

2 A cobbler spends ₹80 on leather, ₹10 on strings and adhesives, and ₹10 on polishing to make a pair of shoes. He sells each pair for ₹180. How much profit % or loss % does he make?

3 The price of a company’s stock increases from ₹54 per share to ₹72 per share in a week. By what percent age has the stock’s price increased?

4 In a school of 480 students, the ratio of boys to girls is 7:8. a What percentage of the total students are boys? b

What percentage of the total students are girls?

5 Dharmendra thinks of three numbers. The first two numbers are 20% and 50% more than the third number respectively. What is the ratio of the first two numbers?

6 The ratio of the number of boys and girls in a college is 5:6. If the number of boys and girls increases by 40% and 50% respectively, what will the new ratio be?

7 Kapil’s expenditure is 20% of his total income. His salary increases by 20%, but his savings remain the same. What is the percentage increase in Kapil’s expenditure?

Jon bought 150 shirts at the rate of ₹250 per shirt. The transportation cost was ₹1500. He paid ₹1.2 as tax per shirt and labour charges were ₹500. What should be the SP of each shirt if Jon wants to earn 30%?

A person allowed a discount of 12% and sold an item at ₹880. What is the marked price on the item?

10 Sonu sells a watch at a 5% discount. If he sold the watch at a 6% discount, then he would have earned ₹27 less. What is the marked price of the watch?

11 A man took a loan from the bank at the rate of 14% per annum. After 3 years, he had to pay ₹42,000 to the bank. What was the principal amount that the man borrowed?

1 12 Sansa borrows ₹5000 for 2 years at the rate of 4 p.a. and lends it to Arya at the rate of 6 % p.a. 4 for 2 years. What is the gain earned by Sansa?

13 Seats for Mathematics, Physics and Biology in a school are in the ratio 6:8:12. There is a

proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of the increased seats?

14 The value of a television set depreciates at the rate of 10% every year. If the cost of a television was ₹1,60,000, what will be its value after 2 years?

15 A wholesaler gets a discount of 40% on the marked price of a product. He sells it at the marked price. What was the wholesaler’s profit percent?

Chapter 12 • Percentage, Profit and Loss, Simple Interest

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313 Rational Numbers Let’s Recall We know that a fraction is a part of the whole. It is a way of expressing a number that is the ratio of two natural numbers.

Mixed Number: 2 1 4

If the numerator is less than the denominator, then the fraction is a proper fraction.

Improper Fraction: 9 4

For example:

7 10

If the numerator is greater than or equal to the denominator, then the fraction is an improper fraction. For example:

11 10

When the numerator and the denominator have no common factor other than 1, the fraction is said to be in standard form. For example:

16 4 can be reduced to , where 4 and 5 are prime numbers. 20 5

Equivalent fractions always represent the same part of the whole. For example:

2 6 and are equivalent fractions. 3 9

Let’s Warm-up Fill in the blanks. 1

Fractions with the same denominator are called

A fraction with the denominator greater than the numerator is called a

A fraction with the numerator greater than the denominator is called an

14 is a

7 8

17 is a/an 2

. . .

fraction. fraction.

I scored _________ out of 5.

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Introducing Rational Numbers Real Life Connect

Rahul and Saumya learnt about the conversion of units in the classroom. They are now trying to convert different units in real life. Rahul: Saumya, can you convert 1236 m above sea level into km? What type of number will you get? Saumya: Yes, Rahul! 1236 m above sea level = +1236 m or 1236 + km. These numbers are called fractions. 1000 Rahul: That’s right! What about 1236 m below sea level in km? Saumya got confused as to how to convert the data in km and what these numbers are called. Let’s find out!

Rational Numbers: Representing, Standard Form, Equivalent Representing Rational Numbers Saumya needs to convert 1236 m below sea level into km. We know that:

Are all fractions rational numbers?

1236 m below sea level = –1236 m It can be converted to km as: = –

Denominator

p q

Think and Tell

1236 km 1000

Numerator

p Any number that can be written in the form , where q p and q are both integers and q ≠ 0 is called a rational number. 1236 1236 For example, both + and – are rational 1000 1000 numbers with both numerator and denominator = integer and denominator ≠ 0.

Did You Know? Pythagoras was a Greek mathematician and philosopher who invented rational numbers in the sixth century BCE.

Rational numbers include natural numbers, whole numbers, integers and fractions. There are two types of rational numbers. Let us learn more about them! Positive Rational Numbers p In a rational number , if p and q are both positive or q p negative integers, then is called a positive rational q number. 3 –4 For example, and are both positive rational numbers. 5 –7

Chapter 13 • Rational Numbers

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Negative Rational Numbers p In a rational number , if either p or q is a negative q p integer, then is called a negative rational number. q

–3 –4 4 For example, and and are all negative rational 5 7 –7 numbers.

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Let us represent

3 on a number line. 5

Step 1: Draw a number line by marking 0 as the origin.

0 1 2 3 4 1 Step 2: Identify the numbers between which the given rational 5 5 5 5 3 number lies. Here, = 0.6 which lies between 0 and 1. 5 Step 3: Check the value of the denominator. Divide each unit by the value equal to that of the denominator. Here, we will divide each unit into 5 subunits. 3 Step 4: Mark on the number line. 5

–4 on a number line. 7 Follow the steps to mark the given rational number. Here, we will divide each unit into 7 subunits. Let us now represent

–1

Example 1

–6 7

2 9

Positive rational number

–3 7

–2 7

–1 7

Represent

2 – 9

– 2 –9 Positive rational number

2 on a number line. 3

–3 3

Represent

Negative rational number

–1

Do It Together

–4 7

Identify positive and negative rational numbers. 1

Example 2

–5 7

0 –2 3

–1 3

0 3

1 3

2 3

–2 –1 6 6

1 6

3 3

4 3

5 3

5 –5 and on a number line. 6 6 –1

–4 6

3 6

Rational Numbers in Standard Form

p A rational number , is said to be in standard form if q is a positive integer and the only common factor q between p and q is 1. We can reduce any rational number to its standard form by dividing the numerator and denominator by their HCF. If the denominator is negative, multiply or divide the rational number by –1. Let us reduce

63 63 and to the standard form. The HCF of 63 and 45 is 9. 45 – 45

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63 has a positive denominator; hence, 45 dividing both the numerator and denominator by 9, 63 ÷ 9 7 63 = is the standard form of . 45 ÷ 9 5 45

63 has a negative denominator; hence, make the denominator – 45 positive by multiplying both the numerator and denominator with −1. 63 × (–1) – 63 = – 45 × (–1) 45 Now, divide both the numerator and denominator by 9, – 63 ÷ 9 – 7 63 = is the standard form of 45 ÷ 9 5 – 45

–15 to the standard form. 75 The HCF of 75 and 15 is 15. Reduce

Example 3

Example 4

So, we will divide the numerator and denominator by 15. –15 –15 ÷ 15 –1 = = 75 75 ÷ 15 5

3 3÷3 1 3 = = = standard form of 15 15 ÷ 3 5 15 –3 1 Hence, the standard form of is . –15 5

–15 –1 Hence, the standard form of is . 75 5

Do It Together

Reduce

12 to the standard from. –18

The denominator cannot be negative; so,

–3 to the standard form. –15 –3 The denominator cannot be negative; so, –15 –3 × (–1) 3 = = –15 × (–1) 15 Divide the numerator and denominator by the HCF of 3 and 15, that is, 3.

Reduce

12 __ × __ __ = = –18 __ × __ __

The HCF of ___ and ___ is ___. Hence, the standard form of

12 is ______. –18

Equivalent Rational Numbers

18 27 and – the same? 20 30 If two or more rational numbers have equal simplest forms, then these rational numbers are called equivalent rational numbers. Are the rational numbers –

Let us start by simplifying –18

–18 ÷ 2

20 ÷ 2

Therefore,

–9 10

–18 –27 and to their standard form: 20 30

–27

–27 ÷ 3 –9 = 30 ÷ 3 10

–18 –27 = 20 30

a c = if a × d = b × c b d

Product of the denominator of the first rational number with the numerator of the second rational number

(–18) × (30) = (20) × (–27) = 540 = equivalent rational numbers

Chapter 13 • Rational Numbers

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–18 –27 –9 = = . 20 30 10

–18 –27 = are equivalent. 20 30

Product of the numerator of the first rational number with the denominator of the other rational number

i.e.,

Thus,

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Example 5

Find four rational numbers that are equivalent to

4 . 7

4 4×2 4×3 4×4 4×5 = = = = 7 7×2 7×3 7×4 7×5 4 8 12 16 20 = = = = 7 14 21 28 35

Therefore, the rational numbers that are equivalent to

Example 6

Check whether

4 8 12 16 20 are , , and . 7 14 21 28 35

6 –9 and are equivalent. –18 27

6 –9 and in standard form: –18 27

Expressing

6 6 × (–1) –6 ÷ 3 –1 = = = –18 –18 × (–1) 18 ÷ 3 3

–9

–9 ÷ 3 –1 = 27 ÷ 3 3

6 –9 Since the standard forms of both rational numbers are the same; and are equivalent rational –18 27 numbers. Do It Together

Fill in the blanks. 1

3 ___ 12 = = 2 6 ___

1 5 ___ = = 11 ___ 77

5 35 ___ = = 8 ___ 64

Do It Yourself 13A 1

Express the shaded portion as a rational number. a

Find a rational number whose numerator is (3 × (–2)) and denominator is (7 × (– 4)).

Which of these are negative rational numbers? a

Represent these on a number line. a –

11 5 6

and

5 6

4 9

and

–4 9

–9

–13 3 4

and

–3 4

Reduce the given rational numbers to their standard forms. a

–24 15

–7

12 20

8 1 6

and

27 63

–1 6

–9 5

and

–5

–98

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Fill in the blanks. a

___ 24

–12

___

–4

–7

14 x

–4 9

___

–16

___

–22

7 8

___ 16

–11

___

–45

105

Find the rational numbers to fill in the blanks such that they are equivalent fractions. a

Find the value of x if: a

–3

7 , ______, ______, 28 13 52

Express

–2 7

12 36 , , ______, ______ 20 60

as a rational number with denominator 63.

p q

10 Express the following rational numbers in the form of . a –7

b 1.4

c 0

d –0.5

Word Problems 1

Ryaan ate pudding?

1 3

of a pudding cup. Rayna ate

2 4

of a pudding cup. Did they eat the same amount of

Maya, Riya and Sia shared a pizza. The pizza was cut into equal parts. They each ate one part. No pizza was left. How did they cut the pizza?

Comparing and Properties of Rational Numbers Comparing and Ordering Rational Numbers When we look at two integers, say –2 and 2, we can easily tell that 2 is greater than –2.

Think and Tell

What makes fractions different

But can we do the same for rational numbers? Let us compare the

from rational numbers?

1 4 rational numbers and . 6 9

Step 1: Both the rational numbers must be expressed with a positive denominator. 1 4 and 6 9 Step 3: E xpress both the rational numbers with the same denominator. 1 1×3 3 = = , 6 6 × 3 18

Chapter 13 • Rational Numbers

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4 4×2 8 = = 9 9 × 2 18

Step 2: Calculate the LCM of the two denominators.

LCM of 6 and 9 = 18 Step 4: C ompare the numerators of the rational numbers obtained. Since, 8 > 3,

8 3 4 1 > . Therefore, > . 18 18 9 6

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1 4 –1 –3 Now, let us learn arrange the rational numbers , , and in ascending and descending order. 6 9 5 8 The LCM of 5, 6, 8 and 9 is 360; so, 1 1 × 60 60 = = ; 6 6 × 60 360

Example 7

4 4 × 40 160 = = ; 9 9 × 40 360

–1 –1 × 72 –72 = = ; 5 5 × 72 360

–135 < –72 < 60 < 160

160 > 60 > –72 > –135

–135 –72 60 160 < < < 360 360 306 360

160 60 –72 –135 > > > 360 360 306 360

Ascending order

Descending order

–3 –1 1 4 < < < 8 5 6 9

4 1 –1 –3 > > > 9 6 5 8

Which of the two rational numbers is greater, The two rational numbers are:

–4 –8 or ? 3 7

–4 –8 and 3 7

Remember!

The LCM of 3 and 7 is 21.

Example 8

–3 –3 × 45 –135 = = 8 8 × 45 360

–4 –4 × 7 –28 = = 3 3×7 21

–8 –8 × 3 –24 = = 7 7×3 21

We know that –24 > –28; Therefore,

–24 –28 –8 –4 > ; i.e., > 21 21 7 3

A positive rational number is always greater than a negative rational number.

Arrange the given rational numbers in descending order. –3 –7 3 –11 –13 , , , 5 10 10 15 20 The LCM of 5, 10, 15 and 20 is 60. –3 –3 × (12) –36 = = 5 5 × (12) 60

–7 –7 × 6 –42 = = 10 10 × 6 60

3 3 × 6 18 = = 10 10 × 6 60

–11 –11 × 4 –44 = = 15 15 × 4 60

–13 –13 × 3 –39 = = 20 20 × 3 60

We know that, 18 > –36 > –39 > –42 > –44 So,

18 –36 –39 –42 –44 > > > > 60 60 60 60 60

Therefore, Do It Together

3 –3 –13 –7 –11 > > > > 10 5 20 10 10

Arrange the given rational numbers in ascending order. –13 –3 –2 11 , , , 9 27 9 3

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The LCM of 3, 9 and 27 is _______. –13 –13 × 3 ___ = = 9 9 × ___ ___

–3 –3 × ___ ___ = = 27 27 × ___ 27

–2 –2 × ___ ___ = = 9 9 × ___ ___

11 11 × ___ ___ = = 3 3 × ___ ___

Ascending order:

Rational Numbers Between Two Rational Numbers We know that the number of whole numbers or integers between two given numbers is limited. But the number of rational numbers between any two numbers is infinite. Let us find rational numbers 1 2 between and using the following steps: –2 3 Step 1: B oth the rational numbers must be expressed with a positive denominator. 1 1 × –1 –1 = = –2 –2 × –1 2

Step 2: Calculate the LCM of the two denominators. LCM of 2 and 3 = 6

Step 3: E xpress both the rational numbers with the same denominator. –1 –1 × 3 –3 = = , 2 2×3 6

……

Step 4: Write the rational numbers in increasing order.

2 2×2 4 = = 3 3×2 6

–3 –2 –1 1 2 3 4 < < <0< < < < 6 6 6 6 6 6 6

We can find more rational numbers between the same numbers. We know that:

–3 –30 = 6 60

4 40 = 6 60

and,

So, the rational numbers: –

29 28 27 37 38 39 30 40 1 2 ,– ,– …, , , lie between – and or – and . 60 60 60 60 60 60 60 60 2 3

If we proceed in a similar way, we can find infinite rational numbers between any two given rational numbers!

Example 9

List five rational numbers between –

1 1 and . 2 2

Let us multiply and divide the numerator and denominator of – –

1 × 3 –3 = 2×3 6

1×3 3 = 2×3 6

Therefore, the rational numbers between

Example 10

Write six rational numbers between

–3 3 –2 –1 1 2 and are: , , 0, , . 6 6 6 6 6 6

1 4 and . 3 5

To expand the range, multiply the numbers 1 1×5 5 = = 3 3 × 5 15

4 4 × 3 12 = = 5 5 × 3 15

Chapter 13 • Rational Numbers

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1 1 and by 3 to expand the range. 2 2

1 4 by 5 and by 3. 3 5

Remember! 0 is also a rational number because it has a non-zero denominator.

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Therefore, the rational numbers between Do It Together

1 4 6 7 8 9 10 11 and are , , , , and . 3 5 15 15 15 15 15 15

List some rational numbers between 1 and 3. Let us multiply and divide 1 and 3 by 4 to expand the range. 1×

4 = _____ 4

3×

4 = _____ 4

We know that, _____>_____. So, the rational numbers between 1 and 3 are:

Properties of Rational Numbers If we multiply the numerator and denominator p of a rational number with the same non-zero q integer (say, m ≠ 0), the rational number remains unchanged. p p×m = q q×m

p is a rational number and m ≠ 0 is an integer, q p p÷m = q q÷m

m is a common divisor of p and q, and the rational number remains unchanged.

2 is a rational number and –5 is an 3 2 2 × –5 –10 integer then, = = 3 3 × –5 –15 For example, if

Example 11

–3 with denominator 49. 7 Let us multiply the numerator and denominator by 7. Express

7 is a rational number and 7 is an 77 7 7÷7 1 integer then, = = 77 77 ÷ 7 11

For example, if

Do It Together

–3 –3 × 7 –21 = = 7 7×7 49

35 with numerator 5. 63 Let us divide the numerator and denominator by _____. Express

35 35 ÷ ____ 5 = = 63 63 ÷ ____ ____

Do It Yourself 13B 1

Compare the given rational numbers. a

2 7 and 3 3

–1 –3 and 2 5

Which of the given rational numbers is the greatest? a

4 5 and 7 9

8 –4 –2 , , 1, 3 3 5

1 –3 4 9 , , , 2 5 9 10

Which of the given rational numbers is the smallest? a

1 2 3 4 , , , 8 9 10 11

–8 –4 3 5 , , , 16 12 9 –15

13 9 and –24 –4

2 2 3 1 , , , 3 4 9 24

3 2 4 c 1, , , 4 3 5

4 3 and 5 –5

–8 21 33 5 , , , 11 22 44 11

3 1 d 2, , 0, 1, 2 8

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Fill in the blanks with >, < or =. a

3 ____ 5 5 6

–5 ____ –13 6 4

–10 2 5 7 , , , 9 9 12 18

–14 10 4 –1 , , , 20 –16 –6 3

–1 –3 4 –3 , , , 3 5 –7 6

–3 6 8 8 , , , 5 13 25 –19

3 –2 –4 5 , , , 5 3 5 6

1 2 3 11 , ,– , 9 7 7 15

–1 –3 4 –3 , , , 3 5 –7 6

Arrange the given rational numbers in descending order. –10 2 5 7 , , , 9 9 12 18

–14 10 4 –1 , , , 20 –16 –6 3

List 5 rational numbers between the following. a –1 and 0

12 ____ 2 13 3

Arrange the given rational numbers in ascending order.

a 7

–4 ____ –7 7 15

–4 –2 and 5 3

1 2 and 6 6

d –

2 3 and – 3 5

e 2 and –1

The points P, Q, R, S, T, U, V and W on a number line are such that: VT = TU = UW and PR = RS = SQ. Name the rational numbers represented by R, S, T and U. W U –5

–4

–3

V –2

P –1

Q 2

8 3 4 , – , –1, – 7 9 8

Arrange the temperatures in order from warmest to coldest. a

9 44 3 14 , ,– ,– 3 15 8 7

3 3 7 9 ,– , , 4 4 8 10

c –5.0, –3.9, 4.4, 32.8

10 Which rational number is greater? a The rational number whose numerator is the largest two-digit number and denominator is the smallest three-digit number.

OR b The rational number whose numerator is the smallest three-digit number and denominator is the largest three-digit number.

Word Problems 1

Raman and Farhan were talking about whose bottle can hold more water. They poured the water 11 13 from their bottles into glasses. Raman filled glasses while Farhan filled glasses. Whose 5 7 bottle can hold more water?

Radha and Kashish needed sheets for an art project. Radha bought a packet that has sheets of 1 2 paper that are of an inch thick. Kashish bought a packet that has inches thick sheets. Which 3 7 girl bought the thicker paper?

Chapter 13 • Rational Numbers

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Operations on Rational Numbers Real Life Connect

Missy and her sister run a grocery store. At the store, there is a drum 65 39 full of wheat flour that weighs kg. Missy added kg more flour to 2 2 the drum. How much does the drum of wheat flour weigh now?

Addition and Subtraction of Rational Numbers We have already learnt how to add and subtract integers and fractions. Applying the operations of addition and subtraction on rational numbers is similar to that.

Addition of Rational Numbers When we add two rational numbers, we have to make sure that the denominators are always positive. If not, then we multiply the numerator as well as the denominator with (–1) to make it positive.

Case 1: Addition of rational numbers that have the same denominator. For two rational numbers

a c and , the sum is: b b

a c a+c + = b b b

Let us now help Missy figure out the total weight of the wheat drum after she added more flour to it. We will add

65 39 and , using the given steps: 2 2

Step 1: Add the rational numbers. 65 39 65 + 39 + = 2 2 2 =

∴

Step 2: Reduce the obtained rational number to the standard form. The LCM of 104 and 2 is 2.

104 2

104 104 ÷ 2 = = 52 2 2÷2

65 39 + = 52 kg. The total weight of the wheat drum is 52 kg. 2 2

Case 2: Addition of rational numbers that have different denominators. Let us add

2 1 and . 5 –15

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Step 1: Check the LCM of the denominators. The LCM of 5 and 15 is 15.

Step 2: Add the rational numbers. 6 –1 6 + (–1) 6 – 1 5 + = = = 15 15 15 15 15

2 2×3 6 1 1 × (–1) –1 = = ; = = 5 5 × 3 15 –15 –15 × (–1) 15

Step 3: Reduce the obtained rational number to the standard form. 5 5÷5 1 = = 15 15 ÷ 5 3 2 –1 1 + = 5 15 3

Add the rational numbers:

Example 12

13 –8 and . 15 15

13 –8 13 + (–8) (We can see the denominators are the same.) + = 15 15 15 =

13 – 8 5 = 15 15

5 5÷5 1 = = 15 15 ÷ 5 3

13 –8 1 + = 15 15 3

Example 13

–8 6 and . 7 5 The LCM of the denominators 7 and 5 is 35. So, let us rewrite the rational numbers: Add the rational numbers:

–8 –8 × 5 –40 6 6 × 7 42 = = ; = = 7 7×5 35 5 5 × 7 35 So,

–8 6 –40 42 –40 + 42 2 + = + = = 7 5 35 35 35 35

–8

Do It Together

6 2 = 5 35

2 3 5 + + . 5 4 12 The LCM of the denominators 5, 4 and 12 is _______. Simplify

Rewriting the rational numbers: 2 2 × 12 24 = = 5 5 × 12 ___ So,

3 3 × ___ 45 = = 4 4 × ___ ___

5 5 × ___ ___ = = 12 12 × ___ ___

2 3 5 + + = _____ + _____ + _____ = _____ 5 4 12

Additive Inverse We have already learnt that if adding one integer to another gives the sum as zero, then the integers are said to be the additive inverses of each other. Similarly, if the sum of two rational numbers is zero, then they are said to be the additive inverses of each other. Chapter 13 • Rational Numbers

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p p p p + – = 0. So, and – are additive inverses of each other. q q q q 5 5 The additive inverse of = – 7 7 i.e.,

Example 14

A shopkeeper sold 2 Weight of sugar = 2 Weight of rice

Total weight

1 3 kg of sugar and 3 kg of rice to a customer. How much in total did he sell? 2 5

1 5 kg = kg 2 2

3 18 kg = kg 5 5

5 18 + 2 5

The LCM of 2 and 5 is 10. 5 25 = 2 10

18 36 = 5 10

So, 25 + 36 = 61 10 10 10 Therefore, the total is

Do It Together

61 kg. 10

Write the additive inverse of the following. 1

–6 6 = 7 7

5 = _______ 14

19 = _______ –21

–12 = _______ –17

Subtraction of Rational Numbers At the end of the month, when the drum was emptied out, Missy measured the weight of the empty drum.

35 kg. 3 Sister: Wait, so how much wheat flour did we have when the drum 65 weighed kg? 2 Missy: See! This empty drum itself weighs

Let us say,

a c c a and are two rational numbers. On subtracting from , we will get: b d d b a c – b d

( )

a –c + b d

c This means adding the additive inverse or the negative of . d

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Now, to find the actual weight of the wheat flour, subtract the weight of the empty drum from the weight of the drum when it was filled with wheat flour. 65 Total weight of the drum filled with wheat flour = kg 2 35 Weight of the empty drum = kg 3 65 35 Actual weight of the wheat flour = – kg 2 3 Therefore, the actual weight of the wheat flour is

Example 15

Subtract the rational numbers The LCM of 9 and 5 is 45. –8 –8 × 5 –40 = = 9 9×5 45 We know that So,

125 kg. 6

We can write this as: 65 35 → + – 2 3 65 × 3 –35 × 2 → + 2×3 3×2 195 70 → + – 6 6 195 – 70 125 → = 6 6

( ) ( ) ( )

–8 –3 and . 9 5

–3 –3 × 9 –27 = = 5 5×9 45

–27 –40 > 45 45

–27 (–40) (–27) 40 –27 + 40 13 – = + = = 45 45 45 45 45 45

Therefore, (–3) – (–8) = 13 5 9 45 Example 16

The sum of two rational numbers is Let the number be x. Then,

5 –1 –1 5 +x= ⇒x= – 6 2 2 6

LCM of 2 and 6 = 6. Hence, x =

–3 5 –3 – 5 –8 –4 – ⇒x= = = 6 6 6 6 3

Therefore, the other number is

Do It Together

–4 . 3

What number should be added to Let the required number be x. Then,

–1 5 . If one of the numbers is , find the other. 2 6

–3 6 to get ? 8 7

–3 6 6 (–3) 6 3 +x= ⇒x= – = + 8 7 7 8 7 8

LCM of 8 and 7 = ______. Hence, x = ______ + ______ ⇒ x = ______ ∴ x = _____.

Chapter 13 • Rational Numbers

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Do It Yourself 13C 1

Add the given rational numbers. a

12 14 and 3 5

d –

1 3 1 + + 8 8 –8

8 –3 + 9 9

1 –5 8 + + 13 13 13

b 5+

5 7

11 8

2 3 and 5 5

3 1 and 6 2

Evaluate. a

–1 –2

(–7) (–5) 9 + + 8 12 16

–1 7 and 8 8

–9 2 –5 + + 50 5 6

The sum of two rational numbers is -2. If one of the numbers is

What number should be added to

Evaluate and find the answer. Find its additive inverse. a

15 17 and 7 3

12 15 and 17 17

3 56 59 + + 7 7 7

12 13 –10 + + 17 17 17

–9 11

–5 9

7 2 and 9 5

5 3 and 7 8

2 3 –1 – + 33 22 44

Subtract the first rational number from the second rational number. a

8 –8 and 3 5

Find the additive inverse of the given rational numbers. a

Find the sum. a

2 9 and 5 5

–6 3 + 7 4

b –

2 7 – 9 9

–5 to get –1? 7 c –

9 4 + 7 –7

–14 , find the other number. 5

–3 2 + 5 5

5 8 – 6 4

44 –11

Express each of the given rational numbers as the sum of an integer and a rational number. a

54 17

b –

12 20

–17 7

–80 19

10 Add the given numbers.

11 2 + –4 + 0.5 3

Word Problems 1 2 3

1 1 L of milk. She accidentally forgot to turn off the gas stove on time and L of 2 4 milk evaporated. How much milk was left? 1 10 33 From a rope of 12 metres, two pieces of lengths and metres are cut off. What is the 5 3 9 remaining length of the rope? 680 510 One day, a taxi driver earned ₹800. Out of his earnings, he spent ₹ on tea and snacks, ₹ 9 2 220 on food and ₹ on repairs of the taxi. How much did he save that day? 5 17 A carpenter bought a piece of wood metres long. He then cut off 3.4 metres of the wood. How 4 long is the piece of wood now? Kanak was boiling 2

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Multiplication and Division of Rational Numbers Missy and her sister donate flour to a local charity event. They distribute How much flour is donated in total? How will they find out?

11 kg of flour each to 23 people. 4

Multiplication of Rational Numbers If

a c and are two rational numbers, then: b d

a c a×c × = b d b×d i.e., product of two rational numbers = Total number of people = 23 Total amount of flour donated =

product of the numerators product of the denominators

23 11 23 × 11 253 253 × = = = kg 1 4 1×4 4 4

Therefore, the total amount of flour donated is

Example 17

Multiply

11 3 and . 12 11

11 3 11 × 3 33 1 × = = = 12 11 12 × 11 132 4

1 Therefore, the product is . 4

253 kg. 4 Example 18

Find the area of a square plot of land whose 14 each side measures metres. 3 14 Measure of each side = m 3 We know that the area of a square = Side × Side =

14 14 14 × 14 196 2 × = = m 3 3 3×3 9

Therefore, the area of the square plot is Do It Together

196 2 m . 9

5 Each sheet of metal is inches thick. If Megha stacks 2 sheets one on top of the other, how thick will 7 the stack be? Thickness of each sheet = _______ Thickness of two sheets = _______ × _______ Therefore, the thickness of the stack of two metal sheets will be _______.

Reciprocal or Multiplicative Inverse of a Non-zero Rational Number p is rational number, where p and q are integers and q ≠ 0. q q p If p ≠ 0, then the rational number is called the reciprocal or the multiplicative inverse of such that: p q We now know that

p q q p × =1= × q p p q

Chapter 13 • Rational Numbers

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p q and are called the reciprocals or the multiplicative inverses of each other. q p 11 4 For example, the multiplicative inverse of is . 4 11 The rational numbers

Example 19

Find the product of

3 and its reciprocal. 7

Did You Know? 1 and –1 are the only rational numbers

3 7 3 7 The reciprocal of is . So, × = 1. 7 3 7 3 Example 20

which are their own reciprocal, whereas 0 has no reciprocal.

Find the reciprocal of 15 and justify your answer. The reciprocal of a number is 1 divided by the given number. 1 So, the reciprocal of 15 is . 15

Do It Together

Verification: The product of the given number with its reciprocal must be 1. 1 15 × =1 15 Hence, verified.

p If p = 2m × r and q = n × 2r, then find the reciprocal of . q The given rational number is:

p 2m × r = q _______

So, the reciprocal of

p is: _______ q

Division of Rational Numbers

108 kg of flour to some people in 4 hours. How much flour did she sell in an hour? To find that, 5 108 4 we will have to divide by , i.e., 5 1 Missy sold

Division is the inverse of multiplication.

108 4 ÷ 5 1

When we divide one rational number by another rational number, we can write it as: 108 1 × 5 4 1 4 is the reciprocal or the multiplicative inverse of . 4 1 So, to divide, we can multiply a given rational number with the reciprocal of the other rational number. We can see that

108 1 108 × 1 108 27 × = = = 5 4 5×4 20 5

Therefore, the quantity of flour Missy sold in an hour is Hence, we can say that if

27 kg. 5

a c and are two rational numbers, then: b d a c a d a×d ÷ = × = b d b c b×c

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Example 21

Divide –

3 4 by . 4 3

–3 4 –3 3 –3 × 3 –9 ÷ = × = = 4 3 4 4 4 × 4 16

Example 22

261 2 87 m . If its breadth is 4 16 metres, what is its length? The area of a room is

Do It Together

If ab =

–2 –1 , a = , then find b. 5 8

ab = a × b

We know that area of a rectangle = length × breadth

–2 –1 –2 –1 ___ ___ ___ = ×b→b= ÷ = × = 5 8 5 8 ___ ___ ___

261 87 261 87 261 16 = length × ; length = ÷ = × 4 16 4 16 4 87 1044 = = 12 87 length = 12 metres.

∴b=

___ ___

Do It Yourself 13D 1

Find the product. a

3 4 and 8 7

5 8

b –

6 11 and 11 7

9 11

d –

3 8 and 20 9

c 1

Divide the given rational numbers. a

b –

e –

Write the reciprocal of the given rational numbers. a

3 2 and 5 4

–6 –2 by 5 3

Simplify. a

( )÷( 3 5 – 4 8

–

4 5 × 9 8

)

–4 by –3 5

(÷ ) 4 9

Find (m + n) ÷ (m – n) if:

By what number should we multiply

Find the product of

Divide the reciprocal of

Multiply the sum of

1 3 a m= ;n= 2 2

–5 18 × 12 19

2 13

–8 to get the product as 24? 13

1 7

71 –9

–16 4 by 21 3

( ÷ )÷

3 5 b m= ;n= 5 3

14 26

(

11 19 by 12 12

) (

–3 15 × ÷ –9 ÷ 2 5 2 4 3

)

4 –2 c m= ;n= 7 7

–8 and its reciprocal. 11 2 –4 by . 5 5

6 5 4 and by the product of and 15. 7 9 –9

10 The product of two rational numbers is – 11 Find the multiplicative inverse of

Chapter 13 • Rational Numbers

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–1 3 by 8 4

4 6 and – 7 5

4 8 . If one number is , then find the other. 30 9

85 60 + ; 125 125

5 5 + 4 4

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Word Problems 1

1 3 If the cost of 2 kg of fruit is ₹115 , find the cost of 1 kg of fruit. 2 4

1 1 The cost of 8 m of fabric is ₹75 . Find the cost of cloth per metre. 2 4

A ribbon of length

1 Sambhav is training for a race. He ran 11 km in total in 5 days. How far did he run each day? 4

each piece?

153 m has been cut into 20 pieces of equal length. What is the length of 3

1 3 Rama builds a garden in her backyard that is 4 metres in length and 1 metres in width. 3 4 What is the area of the garden?

Points to Remember p • Any number that can be written in the form , where p and q are both integers and q ≠ 0 is called a q rational number. p • If in a rational number , q is always a positive integer and the only common factor between p and q is 1, q then such a rational number is said to be in standard form. • Two rational numbers are said to be equivalent if their standard forms after reduction are equal. •

There are infinite rational numbers between two rational numbers.

• We can multiply or divide the numerator and denominator of a rational number by a non-zero integer. • The denominator of the given two rational numbers must be the same for adding or subtracting. • If the sum of two rational numbers is zero, then they are said to be the additive inverses of each other.

Math Lab Setting: In groups of 10. Materials Required: Cards with rational numbers from – a box.

9 9 to written on them in 10 10

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Method: 1

Each group will stand in a straight line.

One student will be given a rational number, and he/she will pick up another card and add the two

Depending on the answer, the student will switch the places forward or backward.

The student already at the position would pick up a card and do the same.

Repeat the same for 10 times, and then the group that finishes first, wins.

rational numbers together.

Chapter Checkup 1

Find the rational number whose numerator is (65 – 32) and denominator is 3 × 9.

Represent

Reduce the rational numbers to the standard form. a

1 6

–4

–32

and

5 7

e 1+

36 54

–1 11

–45

4 7

1 2

2 9

15 –

2 4

and

2 5

–

52 13

( )( ) 8 9

13 11

5 4

–

–14 5

1 5

2 –

39 12

–13 11

(

8 9

–

19 13

)

–1

26 –5

× +

Find the additive inverse of the following. a

4 5

6 7

(

10 Find the multiplicative inverse of –5 × Chapter 13 • Rational Numbers

13_UM24CB0714_R5.indd 223

12 15

)(

– –3 ×

2 9

)

42 12 13

4 7

10 8

–10 100

+ 21 ×

e 0 and –

c ×

7 63

50 16

–5

45 12

11 12

d – 1 and 1

2 15

b –7

and –

–55

c –

Simplify the rational expressions. a

b –

Evaluate. a

List five rational numbers between the following. a

b –

Find the value of x. a

Find three rational numbers equivalent to the following. a

3 –3 and on a number line. 2 4

1 2

( ) ( ) –

45 14

13 7

( ) ( ) 1 2

1 4

1 2

×6

13 14

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11 Do –

4 9

and –

16 36

represent the same number? Why or why not?

12 Divide the sum of

65 12

and

8 3

by their difference.

4 1 2 5 13 What is the average of the middle two numbers if , , and are arranged in ascending order? 7 3 5 9 4 1 9 14 What is the difference of the greatest and the least number? , , 8 4 27 14 15 and by their difference. 15 Divide the sum of 15 14 16 Simplify

43 2

÷5–

1 5

(

41 2

–

110 2

)

1 2

17 2

Word Problems 1 2

3 4 5 6

1 3 of a book in 2 hours. How much of the book will he read in 3 hours? 2 7 1 14 9 The length of a thread is 20 cm. From it, cm and cm pieces are cut off. What is the 2 6 2 length of the remaining thread? A student reads

1 2 of the books are in English, are in Hindi and the 6 6 rest are French books, how many French books are there? 11 A truck moves at an average speed of 30 k/hr. How much distance will it cover in hrs? 2 140 50 The length of a park is m and the breadth is m. What is the area of the park? 3 3 A worker at a shop is filling the new fish aquarium. The aquarium has a capacity of There are 18,000 books in a library. If

40 gallons. If he fills to the brim? 7

20 of the acquarium, how many gallons of water will he need to fill it 25

Tanmay is buying a computer on sale. The original price was ₹12,000. If the sale price is of the original price, how much did he pay for the computer?

5 6

1 Sakshi was playing in a park. She noticed an insect that crawled 3 inches in one minute. 6 3 The next minute, the insect crawled 2 inches. How far did the insect crawl in total? 4

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Construction of Triangles

Let’s Recall It is important to know how to construct angles of different measures and their angle bisectors before learning to construct triangles. Let us recall some constructions. Constructing an angle of 90° Let us see the steps of constructing an angle of 90° or a perpendicular to a line segment. 1

Draw a line segment BC.

2 Starting from point B, use any convenient radius to draw an arc that

intersects BC at point P.

3 From point P, using the same radius as in step 2, draw another arc

intersecting the previously drawn arc at point Q.

Q 90°

4 Starting from point Q, with the same radius as before, draw another arc

that intersects the arc drawn in step 2 at point R.

5 Using points R and Q as centres and a radius greater than half of QR, draw arcs above the line,

cutting each other. Mark the point as S.

6 Connect points B and S with a line segment, extending it to form a ray BA. SB⊥BC.

Constructing an Angle Bisector Let us see the steps of constructing an angle bisector to an angle of any given measure. An angle bisector is a line that divides an angle into two equal parts.

1 To bisect, ∠PQR, place the compass point on point Q and draw an arc that

intersects both sides of the angle. Label the points of intersection as A and B.

2 With the same compass width, draw arcs from points A and B cutting each

other. Label the point of intersection as C.

3 Draw a straight line from Q to C. This line is the angle bisector of ∠PQR.

C A

Let’s Warm-up

State whether the statements are true or false. 1

You construct an angle of 90° by bisecting a 120° angle.

___________

An angle bisector bisects the angle into 3 equal parts.

___________

When an angle bisector of a 60° angle is constructed, we get two angles of 30° each.

___________

While bisecting an angle, the compass width can be less than half of the width of the arc.

___________

You can construct an angle of 15° by bisecting an angle of 45°.

___________ I scored _________ out of 5.

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Constructing Triangles Real Life Connect

Once upon a time in a quiet neighborhood, there lived a curious young boy named Max. Max loved to explore and learn about the world around him. His favourite visits were when his engineer uncle, Uncle Charlie, came to stay. Max: Uncle Charlie, what’s inside your geometry box? Uncle Charlie: Well, Max, it’s filled with special tools like compasses and rulers that we can use to draw triangles. Would you like to learn how? Max: Absolutely, Uncle! A triangle has 3 angles and 3 sides. To construct a unique triangle, we need a minimum of three measurements (at least one being a side). You can construct a triangle with the following criteria. 1 Side-Side-Side (SSS) Criterion: Three sides

Did You Know?

2 Side-Angle-Side (SAS) Criterion: Two sides and one

Ancient civilizations such as the

included angle

Babylonians and the Egyptians used

3 Angle-Side-Angle (ASA) Criterion: Two angles and one

concepts associated with triangles

included side

for surveying, and to create perfectly

4 Right Angle-Hypotenuse-Side (RHS) Criterion: Hypotenuse

square corners for their buildings.

and one other side

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

Constructing Triangles: SSS Criterion R

A triangle can be constructed with the lengths of its 3 sides given. Construct a ∆PQR with sides of length 5.3 cm, 4.5 cm and 6 cm.

4.5 cm

First draw a rough outline of ∆PQR with the measures marked. Given below are the steps of construction. 1 Draw a line segment PQ measuring 5.3 cm in

length.

5.3 cm

6 cm

5.3 cm

2 Draw an arc with the centre at point P and a

radius of 4.5 cm.

5.3 cm

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3 Draw a second arc with the centre at point

Q and a radius of 6 cm, intersecting the previous arc at point R.

4 Join PR and QR. ∆PQR is the required triangle. R

R 4.5 cm

5.3 cm

6 cm

5.3 cm

Try constructing a ∆PQR with sides of length 3 cm, 3 cm and 7 cm. We know that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. Clearly, 3 cm + 3 cm = 6 cm < 7 cm. Hence, no triangle with side lengths 3 cm, 3 cm and 7 cm is possible. When we try constructing a triangle with the given measures, we see that the two arcs of radius 3 cm do not meet.

Therefore, a triangle cannot be constructed with the lengths given. Example 1

Construct an equilateral triangle of side 5.8 cm. An equilateral triangle can be constructed when the length of only one of its sides is given. Since the lengths of all 3 sides of an equilateral triangle are equal, we can construct the triangle given the length of 1 side.

Think and Tell

Can a triangle have sides of length 5 cm, 6 cm and 10 cm? C

1 Draw a line segment AB of length 5.8 cm.

second arc with the centre at point B and a radius of 5.8 cm, intersecting the previous arc at point C.

cm 5.8

3 Without changing the extension of the compass, draw a

5.8

2 Draw an arc with the centre at point A and a radius of 5.8 cm.

7 cm

5.8 cm

4 Join AC and BC. ∆ABC is the required equilateral triangle. Example 2

Construct a triangle with sides of length 4.8 cm, 6.3 cm and 6.3 cm. Name the type of triangle. 1 Draw a line segment XY measuring 4.8 cm in length.

4 Join XZ and YZ. ∆XYZ is the required triangle.

Since the lengths of 2 sides are equal, this is an isosceles triangle. Chapter 14 • Construction of Triangles

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6.3

centred at point Y with a radius of 6.3 cm. Ensure that this arc intersects the earlier arc at point Z.

6.3

3 Keeping the compass extension unchanged, draw a second arc

2 Construct an arc centred at point X with a radius of 6.3 cm.

4.8 cm

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Do It Together

Construct triangles of the measures given. Write the type of triangle. 1 ∆ABC with AB = 3.8 cm, BC = 3.2 cm and CA = 4.3 cm

2 ∆PQR with PQ = 3 cm, QR = 4 cm, RP = 3 cm.

∆ABC =

∆PQR =

Do It Yourself 14A 1

Fill in the blanks. a A triangle can be constructed with the length of ____ sides given. b A triangle cannot be constructed if the sum of the lengths of any two sides is ____ than the length of the third side.

c We need the length of ____ side/sides to construct an equilateral triangle. 2

Construct triangles with the length of 3 sides given. a 4 cm, 5 cm, 3.5 cm

b 6 cm, 4.5 cm, 5.1 cm

c 5.2 cm, 4.8 cm, 3.8 cm

The lengths of the three sides of a triangle are given as 6 cm, 8 cm, and 10 cm. Construct the triangle and

Can a triangle be constructed if the sides measure 5 cm, 12 cm, and 13 cm? Explain your answer.

Construct a triangle with all its sides measuring 4.5 cm. Write the type of triangle.

In ∆ABC, AB = 4.3 cm, BC = 7.4 cm, and CA = 6.1 cm. Construct the triangle and find its angles using a protractor.

Using a compass and ruler, construct ∆RST with side lengths of 5 cm, 6 cm, and 7 cm.

The perimeter of a triangle is 18 cm. The lengths of its sides are in the ratio 2:3:4. Construct the triangle using a

determine its type (e.g., scalene, isosceles, or equilateral).

compass.

Word Problem 1

Nimrat is working on a project to create a triangular-shaped small paper shelf. She knows that the three sides of the triangular shelf are 8 cm, 7 cm, and 6 cm in length. Using a ruler and compass, help Nimrat construct the triangular shape for her shelf on paper.

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Constructing Triangles: SAS Criterion A triangle can be constructed if the lengths of any 2 sides and an included angle are given. cm 4.5

Construct a ∆PQR with QR = 7.3 cm, PQ = 4.5 cm, and ∠PQR = 60° using a ruler and a compass. Draw a rough outline of ∆PQR as illustrated in the diagram. 1

raw a line segment QR with a length of D 7.3 cm.

7.3 cm

Q 3

60°

R 4

4.5

Construct a ∆PQR with side QR = 6.8 cm, PQ = 4.3 cm and ∠PQR = 55°. Draw a rough outline of ∆PQR as illustrated in the diagram given below.

7.3 cm

60°

7.3 cm

Error Alert! In the construction of ∆PQR with side QR = 6.8 cm, PQ = 4.3 cm and ∠PQR = 55°, the arc with radius of 4.3 cm should be drawn from Q and not R. Z

cm 3

6.8 cm

Chapter 14 • Construction of Triangles

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cm 55°

55° Q

6.8 cm

4.3

4.3 cm

4. Q

55°

onnect points P and R. ∆PQR is the required C triangle. P

7.3 cm

From point Q, form an ∠ZQR measuring 60°.

60°

sing Q as the centre and a radius of 4.5 cm, U draw an arc that intersects QZ at point P

60°

6.8 cm

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raw a line segment QR with a length of D 6.8 cm.

rom point Q, form ∠ZQR measuring 55° F using a protractor. Z

R 6.8 cm

sing Q as the centre and a radius of 4.3 cm, draw U an arc that intersects QZ at a point named P. Z

onnect points P and R to complete the C construction. P

Example 3

3 4. Q

6.8 cm

55° 6.8 cm

55°

6.8 cm

Construct ∆FGH with side FG = 4.8 cm, FH = 3.3 cm and ∠HFG = 50°. Find the measure of the unknown side. 1 Draw a line segment FG with a length of 4.8 cm.

intersects FZ at a point named H.

4 Join HG. Length of GH = 3.7 cm

50°

4.8 cm

Construct a ∆JKL with side JK = 5.8 cm, JL = 4.5 cm and ∠KJL = 39°. Find the measure of ∠K. 1 Draw a line segment JK with a length of 5.8 cm.

2 From point J, form ∠ ZJK measuring 39° using a protractor.

3 Using J as the centre and a radius of 4.5 cm, draw an arc that intersects

JZ at a point named L.

4 Join LK. Measure of ∠K = 51°. Do It Together

3 Using F as the centre and a radius of 3.3 cm, draw an arc that

Example 4

2 From point F, form ∠ ZFG measuring 50° using a protractor.

39°

51° 5.8 cm

Construct a ∆DEF with side DE = 5.8 cm, EF = 4.2 cm, and ∠DEF = 30° using a ruler and compass. Find the difference of the measures of the two unknown angles.

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Do It Yourself 14B Which of these measures will help you construct triangles using the SAS criterion?

a ∆ABC with ∠A = 60°, AB = 5 cm, BC = 6 cm

b ∆XYZ with YX = 6 cm, ∠X = 45°, XZ = 5 cm

c ∆PQR with PQ = 3 cm, QR = 4 cm, ∠Q = 30°.

d ∆LMN with LM = 7 cm, ∠L = 50°, ∠M = 65°

Construct triangles with the given measures.

a ∆ABC with AB = 6.2 cm, BC = 8.7 cm, ∠CBA = 60°

b ∆XYZ with XY = 5.7 cm, ∠Y = 60°, YZ = 4.8 cm

In ∆XYZ, side XY = 5.4 cm, side XZ = 7.8 cm, and ∠ YXZ = 45°. Construct ∆XYZ and find the measures of ∠XYZ and

∆PQR has sides PQ = 6.3 cm, QR = 7.1 cm, and ∠PQR = 30°. Construct ∆PQR and identify the type of triangle it is

Two sides of a triangle measuring 4.1 cm and 6.5 cm are given, and the included angle is 120°. Construct the

You have information about two sides of a triangle, one with a length of 3.9 cm and the other with a length

∠YZX. Also, classify ∆XYZ based on its angles and sides.

based on its angles and sides. Name the longest side of the triangle.

triangle and find the measures of the other 2 angles.

of 5.9 cm, and the included angle measures 80°. Construct the triangle and label the vertices.

Word Problem 1

Emma is building a cardboard signpost for a hiking trail. She wants to create a triangular sign, and she knows that two sides of the triangle are 12 cm and 8 cm long, with an included angle of 45°. Using a ruler and compass, help Emma construct the triangular-shaped sign.

Constructing Triangles: ASA Criterion A triangle can be constructed with the given measures of two angles and length of the included side. Construct a ∆XYZ with side XY = 5.7 cm, ∠ZXY = 45° and ∠ZYX = 75° using a compass. Draw a rough outline of ∆XYZ as illustrated in the diagram given below. Z

Error Alert! X

75° 45° 5.7 cm

Chapter 14 • Construction of Triangles

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Do not change the compass width when constructing 60° and 120°.

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raw a line D segment XY with a length of 5.7 cm.

t point X, form an ∠PXY A measuring 45°.

t point Y, draw an ∠QYX measuring A 75°. Identify the point of intersection between XP and YQ as Z. Q

P Z

45° X

5.7 cm

45°

5.7 cm

75° Y

5.7 cm

Construct a ∆PQR with side QR = 6.5 cm, ∠PRQ = 43° and ∠PQR = 55° using a protractor.

Draw a rough outline of ∆PQR as illustrated in the diagram. Draw a line segment QR with a length of 6.5 cm.

6.5 cm

55°

6.5 cm

43°

From point Q, form ∠ ZQR measuring 55° using a protractor.

6.5 cm

rom point R, form ∠YRQ measuring 43° F using a protractor.

I dentify point P as the intersection of RY and QZ. ∆PQR is the required triangle. Y

Y P

55°

6.5 cm

55°

6.5 cm

43°

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Example 5

Construct a ∆RST with side ST = 6.8 cm, ∠RST = 47° and ∠RTS = 76°. Find the length of the unknown sides. Identify the type of triangle based on its sides. Y

1 Draw a line segment ST with a length of 6.8 cm. R

2 From point S, draw ∠ZST measuring 47° using a protractor. 7. 8

4 Identify point R as the intersection of SZ and TY.

∆RST is the required triangle. ∆RST is a scalene triangle as all its sides have unequal lengths.

5.9 cm

3 From point T, draw ∠YTS measuring 76° using a protractor.

47°

6.8 cm

76°

Remember! In a triangle, the sides opposite larger angles are always longer than the sides opposite smaller angles. Do It Together

Construct a ∆MNO with side ON = 4.4 cm, ∠MNO = 49° and ∠MON = 79°. Find the difference in the length of the unknown sides.

Do It Yourself 14C 1

In ∆ABC, ∠A = 45°, ∠B = 60°, and side AB = 7.5 cm. Construct ∆ABC with the measures given.

Given that ∠X = 30°, ∠Y = 75°, and side XY = 6.2 cm in ∆XYZ, construct ∆XYZ and find the lengths of sides XZ and YZ.

∆PQR has ∠P = 40°, ∠Q = 70°, and side PQ = 5.4 cm. Perform the construction of ∆PQR and find the measure of ∠R.

In ∆DEF, ∠D = 50°, ∠E = 70°, and side DE = 5.8 cm. Construct ∆DEF and determine the lengths of sides DF and EF.

∆LMN has ∠L = 60°, ∠M = 45°, and side LM = 9.7 cm. Draw ∆LMN and calculate the lengths of sides LN and MN. Determine the kind of triangle based on its sides and angles.

Chapter 14 • Construction of Triangles

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Word Problem 1

Tom is designing a roof for a small playhouse. He knows that the two angles at the base of the triangular roof are 45° and 60°, and the length of the base of the roof is 6 cm. Using a ruler and compass, help Tom construct the triangular pattern for

the roof on paper. What steps should he follow to accurately create the roof for his playhouse?

Constructing Triangles: RHS Criterion A right-angled triangle can be constructed when the lengths of its hypotenuse and one of the other sides are given. One of the angles measures 90°, so this helps us construct a triangle under the RHS criterion. Construct a ∆LMN right-angled at M, whose hypotenuse is 9 cm and one of the shorter sides is 7.5 cm. Draw a rough outline of ∆LMN as illustrated in the diagram. 1

raw a line segment MN, making it 7.5 cm D long.

7.5 cm

rom point M, construct a 90° angle labelled as F ∠XMN. X

90° M 3

7.5 cm

se point N as the centre and a radius of U 9 cm to draw an arc that intersects the line XM at a point named L.

L 4.6 cm

7.5 cm

J oin L and N by drawing a line segment LN. ∆LMN is the required triangle.

90°

7.5 cm

90° 7.5 cm

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Example 6

Construct a right-angled triangle whose sides measure 6.3 cm, x cm, and 4.3 cm. It is also known that ’x’ is less than 6.3 cm.

3 Use point B as the centre and a radius of 6.3 cm to draw an arc that

intersects XA at a point C.

2 From point A, construct a 90° angle labelled ∠ XAB.

3 6.

1 Draw a line segment AB, making it 4.3 cm long.

X 4.6 cm

From the information given, it is clear that the hypotenuse of the triangle measures 6.3 cm, since the hypotenuse is the longest side in a right-angled triangle.

90° A

4.3 cm

4 Join points C and B. ∆ABC is the required triangle. Do It Together

Construct a right-angled triangle whose sides measure 7.3 cm, x cm, and 5.9 cm. It is also known that ’x’ is the smallest side of the triangle.

Do It Yourself 14D 1

Given that one leg of a right-angled triangle is 4.5 cm and the hypotenuse is 6.3 cm, construct the right triangle

In a right triangle, if you know that one leg measures 4.2 cm and the hypotenuse measures 8.2 cm, construct the

∆ABC is a right triangle with hypotenuse AC measuring 9.5 cm, and one leg AB measuring 6.6 cm. Construct ∆ABC

Given that a right triangle has a hypotenuse of 7.1 cm and one leg measuring 3.8 cm, construct the right triangle

In a right triangle ∆XYZ, the hypotenuse XY is 9.4 cm long, and one leg XZ measures 5.7 cm. Construct ∆XYZ and

and determine the length of the unknown side.

right triangle and calculate the length of the remaining side.

and find the length of side BC. Also, measure the value of the two unknown angles.

and determine the measure of the unknown angles.

calculate the measure of the angle opposite to YZ.

Chapter 14 • Construction of Triangles

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Word Problem 1

Sarah has a right-angled triangular flower garden in her backyard. The largest side of this

garden measures 15 m and the smallest side measures 6 m. If 1 cm on paper is equal to 3 m on the ground, construct a right-angled triangle representing Sarah’s garden on paper. What is the approximate length of the unknown side of Sarah’s garden?

Points to Remember •

A triangle can be constructed with the SSS, SAS, ASA and RHS criteria.

• The sum of the lengths of any two sides of the triangle is greater than the length of the remaining side. • Measure and mark lengths accurately using a ruler and a compass. Small errors in measurement can lead to significant differences in the constructed triangle. • If angles are known, use a protractor to construct them accurately. Ensure that the angles match the measurements given as closely as possible.

• After constructing the triangle, verify that it satisfies all the conditions. Check that the angles and sides correspond to the provided information.

Math Lab Triangle Challenge Material Needed: Rulers, compasses, protractors, paper, pencils, worksheet with various triangleconstruction challenges

Instructions for the students: 1

Select one triangle construction challenge from the worksheet provided by your teacher.

Get your ruler, compass, protractor, paper, and pencil ready.

Carefully read the instructions for your selected challenge, noting the conditions given.

Follow the steps in the challenge, using your ruler and compass to construct the required

Review your construction to ensure it meets the challenge requirements.

If time allows and you want to do more, choose another challenge from the worksheet.

Participate in the class discussion when the teacher will go over solutions, common mistakes,

triangle. Pay close attention to measurements and conditions.

and your construction methods.

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Chapter Checkup What triangle construction criterion will be used for constructing triangles with the given measures.

a Length of 3 sides

b Length of 2 sides and included angle

c Length of hypotenuse of a right-angled triangle and one of its sides d Measure of 2 angles and included side 2

Construct a triangle with sides measuring 5 cm, 7 cm, and 8 cm, and identify its type based on its angles.

Perform the construction of a triangle named ∆PQR with side lengths of 4.5 cm, 6.3 cm, and 9.0 cm. Determine

Given ∆ABC where ∠A is 40°, ∠B is 70°, and side AB is 6.0 cm, construct ∆ABC and calculate the lengths of sides AC

Draw ∆XYZ with known angles ∠X = 60°, ∠Y = 45°, and an included side XY measuring 5.2 cm. Find the lengths of

Construct a triangle using two side lengths of 5.0 cm and 6.0 cm, and an included angle of 45°. Determine the

Construct a triangle, ∆DEF with side lengths DE = 7.1 cm and EF = 4.8 cm, and an included angle ∠E of 30°. Identify

Draw a right triangle with a known leg length of 3.5 cm and a hypotenuse of 5.0 cm. Calculate the length of the

Given that the hypotenuse AC measures 7.8 cm and one leg AB is 6.0 cm long, construct ∆ABC and find the length

the classification of ∆PQR based on its sides and angles.

and BC.

sides XZ and YZ through geometric construction.

measures of the remaining two angles in the triangle.

the criterion used to construct the triangle.

other leg using geometric construction.

of the remaining leg.

10 Given ∆UVW, where ∠U is 50°, ∠V is 70°, and side UV is 5.5 cm, construct ∆UVW and determine the sum of the lengths of sides UW and VW.

Word Problem 1

John is planning to build a model bridge for his science project. The actual bridge has a

right-angled triangular shape, with the largest side measuring 45 cm and one leg measuring 30 cm. If John wants to represent the model bridge on paper using a scale of 10 cm on

the model equal to 1 cm on paper, construct a scaled right-angled triangle on paper to

represent John’s model bridge. What is the approximate length of the unknown side of the model bridge?

Chapter 14 • Construction of Triangles

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315 Perimeter and Area Let’s Recall What do you notice about these two shapes?

All angles are equal but not all sides are equal.

One of the shapes has sides of equal length, while the other shape has sides of unequal length.

He also wants to fence it all around. Let us see how we can find the length of the fence required for his garden and the area of the garden on which he can grow fruit and vegetables.

Let us say that Raghu has a small L-shaped kitchen garden. This garden is made by joining a rectangle and a square. He wants to grow vegetables in the rectangular part and fruit in the square part of the garden as shown here.

The shapes that have equal sides and angles are regular shapes whereas shapes with either unequal sides, angles or both are irregular shapes.

10 m

Vegetables

All sides and angles are equal.

Perimeter of the garden

Area of the garden

= sum of lengths of all sides of the garden

= Area of the vegetable garden + Area of the fruit garden

= 10 m + 4 m + 2 m + 2 m + 8 m + 2 m

= Area of rectangle + Area of square

= 28 m

= (10 m × 2 m) + (2 m × 2 m) = 20 sq. m + 4 sq. m = 24 sq. m

So, the length of the fence required for Raghu’s garden is 28 m and the area of the garden on which he can grow fruit and vegetables is 24 sq. m.

Let’s Warm-up Fill in the blanks. 2 cm

12 cm 2 cm

80 mm

2 cm

Perimeter = ______ cm

84.2 cm 208 cm

60 mm Area = _______ sq. cm 14 cm

? 168.4 cm

Perimeter = ______ cm 84.2 cm Area = _______ sq. cm 84.2 cm

250.6 cm

I scored __________ out of 4.

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Mean, Median and Mode Squares, Rectangles, Triangles and Parallelograms Richa and Rahul go to the park daily at 6:00 a.m.

Real Life Connect

Richa: I run around a rectangular park of length 120 m and breadth 60 m. Rahul: I run around a square park of length 40 m. Let us see who covers the greater distance.

Perimeter and Area of Squares and Rectangles Let us find the distance covered by Richa and Rahul around the park. Since Richa runs around the rectangular park, we need to find the perimeter of the rectangle. Perimeter of the rectangular park = Perimeter of rectangle 60 m

= 2 (l + b)

120 m

= 2 (120 + 60) = 2 × 180 = 360 m ince Rahul runs around the square park, we need to find the perimeter S of the square. 40 m

Perimeter of the square park = Perimeter of square = 4s = 4 × 40 = 160 m Here, 360 > 160. So, Richa covers the greater distance. Both parks must be completely covered in grass. How can we determine the amount of grass needed for each park? To find the amount of grass needed for each park, we need to find the area of the parks. For a rectangular park, area = l × b Area of the park = 120 × 60 = 7200 sq. m So, in this case, 7200 sq. m of the rectangular park needs to be covered with grass. For a square park, area = s × s Area of the park = 40 × 40 = 1600 sq. m So, in this case, 1600 sq. m of the square park needs to be covered with grass. Example 1

Remember! Perimeter of a rectangle = 2 (l + b), where l = length, b = breadth Area of a rectangle = l × b Perimeter of a square = 4s, where s = side Area of a square = s × s

The breadth of a rectangular field is double its length. If the area of the field is 338 sq. m, find the cost of fencing it at ₹20 per m.

Chapter 15 • Perimeter and Area

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Given: Area = 338 sq. m; Length = L; Breadth = 2L Area = Length × Breadth

Perimeter of the field = 2 (L + B)

338 = L × 2L

= 2 (13 + 26)

L × 2L = 338 sq. m

= 2 × 39 = 78 m

2 × L × L = 338 sq. m

Cost of fencing 1 m = ₹20

L × L = 169 sq. m = 13 m × 13 m

Cost of fencing 78 m = ₹20 × 78 = ₹1560

L = 13 m So, B = 2 × 13 = 26 m. So, the cost of fencing the field is ₹1560. Example 2

Find the area of a square, the perimeter and area of which are in the ratio 3:4. Let the perimeter of the square be P. Let the area of the square be A. We are given that, P:A = 3:4 This means, 4P = 3A. Now, A = s × s and P = 4s. So, 4P = 3A

4 × 4s = 3 × s × s

16s = 3 × s × s 3s = 16 16 s= = 5.3 (approx.) 3 So, the area of the square = 5.3 × 5.3 = 28.09 sq. units (approx.) Example 3

A hall has four walls, each measuring 16.4 m × 8.2 m. It has two doors of height 2 m and breadth 1 m. It also has three windows of size 1 m × 0.5 m. Find the cost of whitewashing the walls at the rate of ₹75 per square metre. Length of the wall = 16.4 m, breadth of the wall = 8.2 m Area of 1 wall of the hall = 16.4 m × 8.2 m = 134.48 sq. m Area of 4 walls of the room = 4 × 134.48 = 537.92 sq. m Length (height) of the door = 2 m

Length of window = 1 m

Breadth of door = 1 m

Breadth of window = 0.5 m

Area of 1 door = 2 m × 1 m = 2 sq. m

Area of 1 window = 1 × 0.5 = 0.5 sq. m

Area of 2 doors = 2 × 2 = 4 sq. m

Area of 3 windows = 3 × 0.5 = 1.5 sq. m

Area of walls to be whitewashed = Area of 4 walls – (Area of 2 doors + Area of 3 windows) = 537.92 – (4 + 1.5) = 537.92 – 5.5 = 532.42 sq. m Cost of whitewashing 1 sq. m = ₹75 Cost of whitewashing 532.42 sq. m = ₹75 × 532.42 = ₹39931.50 So, the cost of whitewashing 4 walls of the hall is ₹39,931.50. 240

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Do It Together

How many tiles of size 12.5 cm by 10 cm can be cut out of a tile of size 102 cm by 65 cm? Size of 1 tile to be cut = 12.5 cm × 10 cm Area of 1 tile to be cut = _____________________ Size of tile to be used for cutting = 102 cm × 65 cm Area of tile to be used for cutting = _____________________ Number of tiles = _____________________ So, ______ tiles are required to be cut from the tile of size 102 cm × 65 cm.

Do It Yourself 15A Find the perimeter and area of the given figures. 3.7 cm

6.8 cm

4.5 cm

Find the area of the square if its perimeter is given below. a 8 cm

b 16 cm

5.3 cm

8.2 cm

c 32 cm

5.9 cm

d 52 m

e 108 m

Find the area of a rectangle if its perimeter and length are as given below, respectively. a 10 cm, 2 cm

b 18 cm, 5 cm

c 48 cm, 21 cm

d 94 m, 38 m

The perimeter of a rectangle is equal to the perimeter of a square with a side length of 15 m. If the rectangle is 18 m long, what is its breadth?

Find the area of a square, the perimeter and area of which are in the ratio 5 : 8.

A rectangle and a square both have the same area of 49 sq. cm each. If both the figures are of integral dimensions and the length is of the maximum possible measure, then which of the two, the square or rectangle has a greater perimeter than the other?

What are the measures of the rectangle if the length and the breadth are in the ratio of 2:3, and the ratio of its perimeter to area is 5 : 3? Also, find its perimeter and area.

Word Problems 1

The cost of fencing a square field at ₹12 per metre is ₹24,000. Find the area of the field.

A playground is 750 m long and 250 m broad. Find the a cost of levelling at ₹16 per 100 sq. m. b number of rounds along the boundary one should take to run 6 km.

The length and breadth of a rectangular field are in the ratio 3 : 2. If the perimeter of the field is 320 m, find the cost of reaping the field at ₹25 per 50 sq. m.

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The breadth of a rectangular field is thrice its length. If the area of the field is 1083 sq. m, find the cost of fencing it at ₹25 per m.

A piece of wire is in the shape of a square of side 21 cm. If the same wire is rebent into the shape of a rectangle whose length and breadth are in the ratio 3 : 4, what will be the dimensions of the rectangle? Also, find which shape encloses the greater area.

A room has four walls, each measuring 18.4 m x 10.2 m. It has two doors of height 1 m and width 2 m. It also has three windows of size 1 m × 0.6 m. Find the cost of whitewashing the walls at the rate of ₹80 per sq. m.

How many rectangles of 20 cm by 6 cm can be cut out of a sheet that is 105 cm by 45 cm?

A room is 18 m long and 10.5 m wide. If a 10 m square table is placed in the room, find the area which is left unoccupied.

The cost of carpeting a hallway of 15 m width at ₹150 per square metre is ₹45,000. Find the length of the hallway.

Area of Parallelograms We know that Richa runs around the rectangular park of length 120 m and breadth 60 m.

Let us say we have to find the amount of grass in the park. For this, we need to find the area of the park. The park is in the shape of a parallelogram. So, if we know the area of a parallelogram, we can find the area of the park too.

60 m

What if the shape of the park is a parallelogram with the same measures? How would you find its area? 120 m

So, let us draw a parallelogram ABCD of any length (base = BC) and breadth (AB) on a squared sheet of paper, as shown below. Draw a perpendicular AP (height) on BC to obtain a triangle ABP. A

D height

Think and Tell

Can you find the area of the parallelogram by counting squares? C

Cut off the triangle ABP and paste it on the other side of the parallelogram to obtain a rectangle, as shown below. What do you see? A

D (A)

breadth = height B

From the above picture, we can make these observations.

P length = base C (B)

• the length of the rectangle is the same as the base of the parallelogram. • the breadth of the rectangle is the same as the height of the parallelogram.

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So, area of a parallelogram = area of a rectangle =l×b = AD × AP = BC × AP (AD = BC,

these are opposite sides of a rectangle.)

=b×h Now, the area of the park = b × h = 120 × 60 = 7200 sq. m Therefore, 7200 sq. m of grass is grown in the park. Example 4

PQRS is a parallelogram of area 1470 sq. cm. SO and QT are the heights on the sides PQ and SP, respectively. PQ = 30 cm, PS = 25 cm. Find the length of SO and QT. According to the diagram, SO is a height on PQ. Area of the parallelogram = 1470 sq. cm PQ × SO = 1470 sq. cm

25 cm

b × h = 1470 sq. cm

30 × SO = 1470 sq. cm

SO = 1470 ÷ 30 = 49 cm Similarly, QT is a height on PS.

30 cm

So, PS × QT = 1470 sq. cm 25 × QT = 1470 sq. cm QT = 1470 ÷ 25 = 58.8 cm So, SO = 49 cm, QT = 58.8 cm. Example 5

The base of a parallelogram is triple its height. If its area is 363 sq.cm, find the base and height of the parallelogram. Let the base of the parallelogram be b and the height of the parallelogram be h. We are given that, b = 3h Area of the parallelogram = 363 sq. cm b × h = 363 sq. cm 3h × h = 363 sq. cm 3 × h × h = 363 sq. cm h × h = 363 ÷ 3 h × h = 121 h × h = 11 × 11 h = 11 cm So, b = 3h = 3 × 11 = 33 cm

Do It Together

STUV is a parallelogram. VE find VE.

ST. If ST = 40 cm and the area of the parallelogram is 400 sq. cm, then V

We are given that VE is a height on ST.

So, the area of the parallelogram = ST × VE _______________________ VE = _____________________ cm Chapter 15 • Perimeter and Area

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40 cm

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Do It Yourself 15B Find the area of the parallelograms.

6 cm 2

7 cm

d 8 cm

3 cm

b 4 cm

9 cm

7 cm

Find the missing measure. a

b = 11 cm, h = ?, A= 88 sq. cm

b = 17.2 cm, h = ?, A = 349.16 sq. cm

b = ?, h = 16 cm, A = 252.8 sq. cm

b = ?, h = 7.6 cm, A = 70.68 sq. cm

6 cm

The difference in the lengths of two adjacent sides of a parallelogram is 8 cm. If the perimeter of the parallelogram is 64 cm and its height is the same as the difference in the lengths of the adjacent sides, then find its area.

The base of a parallelogram is twice its height. If its area is 648 sq. cm, find the base and height of the parallelogram.

The adjacent sides of a parallelogram HIJK are 15 cm and 8 cm. The height corresponding to side IJ is 10 cm. Find the a

area of the parallelogram

altitude corresponding to the base JK

The adjacent sides of a parallelogram are in the ratio 4 : 3. If the longer side is 8.5 cm, find the perimeter of the parallelogram.

In a parallelogram PQRS, PQ = 25 cm, QR = 15 cm, PL

ABCD is a parallelogram. DE AB such that point E lies inside the parallelogram. If AB = 30 cm, AD = 17 cm and the area of the parallelogram is 450 sq. cm, then find AE and EB.

LMNO is a parallelogram, where NE

LM and MF

SR and PM

QR. If PL = 10 cm, then find PM.

LO.

If LM = 21 cm, LO = 15 cm and NE = 10 cm, then find MF.

If LM = 30 cm, NE = 12 cm and MF = 20 cm, then find LO.

Word Problem 1

A sports company purchased a field in the form of a parallelogram with an area of 576 sq. m. Two pathways A and B are created from the adjacent corners of the field to their opposite sides and are perpendicular to the sides on which they are created. The length of pathway A is 18 m. Find the length of pathway B if its base is three-fourths the length of the other side of the parallelogram.

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Area of Triangles 60 m

We know that Richa runs around the rectangular park of length 120 m and breadth 60 m. What if the residents of a colony decide to construct a shortcut between two parallel roads outside the park?

120 m

For this, they build a path diagonally that divides the park into two triangular halves. Let us now see how we can find the area of the park. To find the area of the park, we need to find the area of the rectangle. The park is divided into two halves. Each half is a triangle. Every rectangle is made up of two triangles. So, if we find the area of each half of the park (triangle), we would get the area of the park. If we know the area of a rectangle with two triangles, we can find the area of the park too. So, let us draw a rectangle ABCD and cut it along the diagonal BD to obtain two triangles ABD and BDC. D

Superimpose one triangle over another so that the vertices A, B, and D of Δ ABD fall on vertices C, D, and B of ΔCDB, respectively. What do you see?

Think and Tell

When two triangles of the same size look the same and on superimposition, they cover each other exactly, then do you know what such triangles are called?

From the above picture, we can see that the two triangles cover each other exactly. So, ΔABD

ΔCDB

Area of ΔABD = Area of ΔCDB

So, area of rectangle ABCD = Area of ΔABD + Area of ΔCDB = Area of ΔABD + Area of ΔABD (As Area of ΔABD = Area of ΔCDB) = 2 × Area of ΔABD

1 × area of rectangle ABCD 2 1 What if the park is in the shape of a square = × (l × b) 2 of side 60 m? How would you find the area 1 of the park? = × (AB × AD) 2 1 = × (base of ΔABD × height of ΔABD) 2 (As length of rectangle = base of ΔABD, breadth of rectangle = height of ΔABD)

Area of ΔABD =

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Think and Tell

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1 × (base of ΔCDB × height of ΔCDB) (As Area of ΔABD = Area of ΔCDB) 2 1 = × b × h, where b = base of a triangle, h = height of a triangle 2 1 So, Area of ΔABD = Area of ΔCDB = × b × h 2 =

Area of a triangle =

1 × b × h where b is the base and h is the height. 2

Now, we find the area of each half of the park (triangle) to get the area of the park. 1 1 The area of one triangular region of the park = × b × h = × 60120 × 60 = 3600 sq. m 2 2 Similarly, the area of the other triangular region of the park = 3600 sq. m So, area of the park (rectangle) = Sum of areas of the two triangular regions = 3600 sq. m + 3600 sq. m = 7200 sq. m. Triangles as Part of Parallelograms

Look at the given parallelogram. Two triangles ABC and PBC are joined b h together to form a parallelogram, where the base and height of the triangle ABC are the same as the base and height of the parallelogram C c ABPC. 1 So, area of ΔABC = × (area of parallelogram) 2 1 = × (base × height) 2 1 = × (c × h), where c = base of the parallelogram, h = height of the parallelogram 2 Example 6

B a A

Find the area of triangle ABC in each of the given figures. C

5 cm

10 cm

Base of the triangle = 10 cm, height = 5 cm Area of the triangle ABC = 2 P

1 1 1 × (base × height) = × (10 × 5) = × 50 = 25 sq. cm 2 2 2

6 cm

12 cm

Base of the parallelogram = 12 cm, height = 6 cm 1 × (area of parallelogram) 2 1 1 1 = × (base × height) = × (12 × 6) = × 72 = 36 sq. cm 2 2 2

Area of ΔABC =

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Example 7

A lawn is split into triangles of different sizes, each of height 10 m, as shown. The base of the smaller triangle is half its height and the base of the bigger triangle is double its height. Find the area of the lawn. Height of each triangle = 10 m = height of the lawn 1 1 Base of the smaller triangle ADC = × height = × 10 = 5 m 2 2 Base of the bigger triangle ABD = 2 × height = 2 × 10 = 20 m

10 m

Base of the lawn = base of the smaller triangle + base of the bigger triangle

= 5 m + 20 m

Do It Together

= 25 m 1 1 1 Area of the lawn = (base × height) = (25 × 10) = × 250 = 125 sq. m 2 2 2

The area of a triangle is half of the area of the square. If the side of the square is 12 m, find the area of the triangle. 1 Area of triangle = × area of square 2 Side of square = 12 m So, area of square = _________ =_____ sq. m Area of triangle =

1 × ____ = ____ sq. m 2

Do It Yourself 15C Find the area of the triangle in the given figures.

10 in

9 cm

7 cm Find the missing measure in the triangle where b is the base, h is the height and A is the area. a b = 11 cm, h = ?, A = 44 sq. cm

b b = 15.8 cm, h = ?, A = 49.77 sq. cm

c b = ?, h = 16 cm, A = 128 sq. cm

d b = ?, h = 7.6 cm, A = 39.90 sq. cm

Find the area of a right-angled triangle of which the base is 16 cm and altitude is 10 cm.

ΔSTU is a right-angled triangle with S = 90°, ST = 6 cm, TU = 10 cm, and SU = 8 cm. OS is perpendicular to TU. Find OS.

In ΔABC (see figure), AB = 15 cm, BC= 8 cm and AD = 7.5 cm. Find: a area of ΔABC

6 7

The base and height of a triangle are in the ratio 3 : 4. If the area of the triangle is 96 sq. cm, find its base and height.

ABCD is a rectangle with dimensions 30 m by 25 m. AOB is a triangle so that O is the point of intersection of the diagonals. Find the area of the shaded region.

Chapter 15 • Perimeter and Area

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8 cm C

b altitude CE

m 5c

7.5 cm

d 5 cm

13 cm

c 8 in

b 16 cm

10 cm

O 30 m

C 25 m B

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Find the area of the shaded region. E

35 cm

A 9

50 cm

15 cm

12 cm B

10 cm 20 cm

20 cm 10 cm

27 cm

EFGH is a rectangle in which EF = 28 cm and EH = 22 cm. P, Q, R, and S are midpoints of the sides of the rectangle EFGH. Find the area of the shaded region.

P E

Word Problem 2m

A rectangular lawn of length 30 m and breadth 15 m with triangular flower beds at each corner is shown in the figure. Find the area of the lawn excluding the flower beds.

Mean, Median and Mode Circles Real Life Connect

Behind a building, there is a circular field of radius 7 m. The field is full of grass but due to the G20 summit, a fountain inside a circular fence of a radius of 2 m is to be built exactly in the centre of the field for aesthetic reasons. Let us find the length of the fence required for the fountain.

Circumference of Circles We know that a circle is a curved path of points equidistant from a fixed point called the centre. The distance of these fixed points from the centre is called the radius, and the curved path forming the boundary of the circle is called the circumference. cumferenc Cir e

Dia

ius Rad

ter

Remember! • All radii of a circle are equal. • The distance between two points on the circle through the centre of the circle is the diameter of the circle. The diameter is always double the radius.

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Now, we know that a fountain inside a circular fence is to be built exactly in the centre of the field. For this, we need to find the length of the fence. Since finding the length of the fence is the same as the finding the circumference of the circular fence, it is sufficient to find the circumference of the circle. If we know the circumference of a circle, we can find the length of the circular fence too. For this, let us first understand the relationship between the circumference and the diameter. So, take some circles of different radii and find their circumference using a thread and a ruler. Also, let us find the ratio of circumference to diameter in each case.

2 cm

3 cm

4 cm

5 cm

Circle

Radius (r)

Diameter (D = 2r)

Circumference (C)

Ratio =

2 cm

4 cm

12.6 cm

3.15

3 cm

6 cm

18.8 cm

3.133

4 cm

8 cm

25.1 cm

3.137

5 cm

10 cm

31.4 cm

3.14

C D

What do you observe in the above table? Are all the ratios approximately the same? In each case, the ratio of circumference to diameter is close to 3 and roughly the same as 3.14. So, we say that the circumference is always more than three times its diameter. Now, using this relationship, we can formulate the rule for the perimeter of the circle.

Circumference = 3.14, this ratio is called a Diameter 22 constant and denoted by π (pi). Its approximate value is or 3.14. 7 C Therefore, = π, where C = circumference, D = diameter D C = Dπ Notice that since

Did You Know? March 14th (3/14) every year is celebrated as Pi Day in honour of the mathematical constant π.

C = π × 2r = 2π r

22 unless stated otherwise.) 7 Now, let us find the length of the fence required for the fountain. Look at the given picture.

Thus, circumference = 2πr

(Note: Take the value of π as

Chapter 15 • Perimeter and Area

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Circumference = 2πr 22 =2× × 2 = 12.57 m 7 So, the length of the circular fence required to build the fountain is 12.57 m.

Radius of the circular fence = 2 m 2m

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Example 8

The difference of the circumference and the diameter of a circle is 40 cm. Find the radius of the circle.

Example 9

Radius of the tyre = 20 cm

Given: C – D = 40 cm

Length of 1 revolution of the tyre = 2πr 22 880 =2× × 20 = cm 7 7 Length of 80 revolutions of the tyre 880 = × 80 = 10,057.14 cm 7 So, the distance covered by the bus tyre is 10,057.14 cm.

2πr – 2r = 40 2r (π – 1) = 40

22 2 × r  – 1 = 40  7

22 – 7  r   = 20  7  15 r = 20 7 4 20 × 7 28 r= = = 9.3 cm (approx.) 153 3

Do It Together

Find the distance covered by a bus tyre of radius 20 cm in 80 revolutions.

A piece of wire is bent in the shape of a square of side 66 cm. If the same wire is bent in the form of a circle, what will the radius of the circle be? Also, find the cost of the wire if 1 cm of wire costs ₹4. (Take π = 3.14) Side of the square (made using wire) = 66 cm So, the perimeter of the square = 4s = 4 × ______ = ______ cm Circumference of the circle = ______ cm ___________________________________ ___________________________________ ___________________________________ r = ______ cm Cost of 1 cm of wire = ₹4 Cost of _____ cm of wire = ₹4 × _____ = ₹_______

Do It Yourself 15D 1

Find the circumference of the circle for the radii. a 5 cm

b 8 cm

c 10 cm

d 12 cm

e 18 cm

d 30 cm

e 40 cm

Find the circumference of the circle for the diameters. a 10 cm

b 15 cm

c 24 cm

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Find the ratio of circumference to diameter for each circle of the given radii. Is it close to 3.14? a r = 16 cm

b r = 20 cm

c r = 24 cm

d r = 36 cm

Find the diameter of the circle for the given circumference. a 75.42 cm

b 603.42 cm

c 502.85 cm

d 352 cm

Find the ratio of the circumferences of two circles whose radii are in the ratio 5:6.

The circumference of a circle exceeds the diameter by 90 cm. Find its radius.

Find the perimeter of a semicircle in which the diameter is 50 cm. (Take π = 3.14)

The circumference of a circular playground is 296 m. Find its diameter.

Find the distance covered by a car tyre of radius 25 cm in 120 revolutions.

10 To fence a circular garden, the total cost is ₹36,500 at the rate of ₹40 per metre. Find the radius of the garden. 11 The diameter of the wheel of a truck is 57.15 cm, How many revolutions will it make to cover 1 km 30 m? 12 The minute hand of a circular clock is 20 cm long. How much distance is covered (in cm) by the tip of the minute hand in one hour?

Word Problems 1

Tara wants to put lace on the edge of a circular cushion of diameter 18 cm. Find the length of the lace required and also find its cost if one metre of lace costs ₹22.

The diameter of a circular field is 96 m. If an adult man walks at the rate of 8 m per minute, how long will it take to walk around the field twice?

A wire is bent into the shape of a square of side 88 cm. If the same wire is looped in the form of a circle, what will the radius of the circle be?

Seema is cycling at a speed of 12 km/hr. How long will she take to complete a round of a circular field of radius 80 m? (Take π = 3.14)

Area of Circles, Semicircles and Quadrants We know that a fountain inside a circular fence is to be built exactly in the centre of the circular field. Now, what if we have to find the area of the field used for building the fountain? To find the area of this region, we need to find the area of the circle. If we know the area of a circle, we can find the area of this region too. For this, let us first understand the relationship between a rectangle, a circle and its area.

Chapter 15 • Perimeter and Area

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So, take a circle of radius r and shade one-half of the circle. Cut the circle into eight sectors and join them together to form a parallelogram. This is not even nearly a rectangle. The greater the number of sectors, the more appropriate the rectangle. So, we again divide the circle into smaller sectors and rearrange them to form an even more perfect-looking rectangle. We keep repeating this step until the lines of the sectors are indistinguishable and we get a perfect rectangle. In each case given here, what do you observe? •

breadth of the rectangle = radius of the circle = r

•

length of the rectangle = length of half the total number of sectors = half of the circumference

So, area of the circle = area of the rectangle =l×b = half of the circumference of the circle × radius of the circle 1 = × 2πr × r = πr2 2 Now, using this formula, let us find the area of the field used Error Alert! for building the fountain. Area of the circle = πr2 22 = ×2×2 7 = 12.57 sq. m

Area of circle = π r2, where r = 5 cm

So, the area of the field used for building the fountain is 12.57 sq. m. Example 10

22 ×5×2 7 = 31.43 cm

Area =

22 ×5×5 7 = 78.57 sq. cm

Area =

Find the area of the shaded region in the figures. 1

7 cm 10 cm

3 cm

Area of the circle = πr2 22 = ×7×7 7 = 154 sq. m

Example 11

1 Area of the semicircle = πr2 2 1 22 = × × 10 × 10 2 7 = 157.14 sq. cm

Area of a quarter of a circle 1 (quadrant) = πr2 4 1 2211 = 2× × 3 × 3 = 7.07 sq. cm 4 7

The ratio of the areas of circle S and circle T is 2 : 3. If the radius of circle T is 12 cm, what is the area of circle S? Radius of circle T = 12 cm; The ratio of areas of circle S to circle T = 2 : 3 Let the area of any circle be x sq. cm.

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So, area of circle S = 2x and area of circle T = 3x Now, the area of circle T = 3x 3x = πr2 22 3x = × 12 × 12 7

22 × 12 × 124 = 150.86 sq. cm (approx.) 7×3 Therefore, the area of circle S = 2x = 2 × 150.86 sq. cm = 301.72 sq. cm x=

Example 12

A horse is tied with a rope affixed to one corner of a square farm of side 24 m. If the length of the rope is 6 m, find the area of the farm on which the horse can graze. Also, find the remaining area of the farm. (Take π = 3.14) Side of the square farm = 24 m, length of the rope = 6 m = radius The area in which the horse can graze is in the shape of a quadrant. 1 So, area of the quadrant = × πr2 4 1 = × 3.14 × 6 × 6 4 1 = × 113.04 = 28.26 sq. m 4 Therefore, the horse can graze in an area of 28.26 sq. m. Now, the remaining area of the farm = Area of the square farm – Area in which the horse grazes = (24 × 24) – 28.26 = 547.74 sq. m

Do It Together

A piece of wire in the shape of a square of side 15 cm is reshaped in the form of a circle. Find the radius and area of the circle. Side of the square = 15 cm Perimeter of the square = 4s So, Perimeter of the square = 4 × 15 cm = ______ Perimeter of the circle = _____ __________________________________

__________________________________

Do It Yourself 15E 1

Find the area of the circle for the radii. a 3 cm

b 6 cm

c 12 cm

d 15 cm

e 25 cm

c 26 cm

d 35 cm

e 48 cm

Find the area of the circle for the diameters. a 10 cm

b 18 cm

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Find the diameter of the circle for the areas. a 50.28 sq. cm

b 154 sq. cm

c 616 sq. cm

d 1257.14 sq. cm

Find the area of the figures. a

18 mm

14.5 cm

42 mm

16 cm

Find the ratio of the areas of two circles whose radii are in the ratio 4:7.

The ratio of the areas of circle P and circle Q is 25:36. If the radius of circle Q is 36 cm, what is the area of circle P?

A wire in the shape of a rectangle of length 26 cm and width 18 cm is reshaped in the form of a circle. Find the radius and the area of the circle.

A small metallic washer of radius 8 mm has a hole of 2.5 mm radius. What is the area of the washer?

A copper wire in the form of a circle encloses an area of 1386 sq. cm. If the same wire is bent to form a square, find its area.

10 A circular flower garden has an area of 628 sq. m. A sprinkler at the centre of the garden can cover an area that has a radius of 15 m. Will the sprinkler water the entire garden? (Take π = 3.14)

Word Problems 1

A cow is tied with a rope affixed to one corner of a rectangular lawn of length 30 m and breadth 20 m. If the length of the rope is 8 m, find the area of the lawn which the cow can graze. Also find the remaining area of the lawn.

Sonal was making a birthday cap using a sheet of golden paper as shown in the figure. Find the area of the sheet of paper used to make the birthday cap.

24 cm

We know that a fountain inside a circular fence is to be built exactly in the centre of the circular field, and the rest of the field is covered with grass.

Combined Shapes 2m

Now, what if we have to find the area of the field covered with grass? To find the area of this region, we need to find the area between two circles. If we know the area of two circles, we can find the area of the field with the grass. So, let us say there is an outer circle of radius R and an inner circle of radius r. Then, the area between the two circles, = Area of outer circle − Area of inner circle

= πR2 − πr2 = π(R2 − r2)

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Now, the area of the field with grass = Area of the total field − Area of the field used for building the fountain Radius (R) of outer circle = 7 m; Radius (r) of inner circle = 2 m Area of the field without grass = Area of the field – Area of the fountain 2

= π (7 − 2 ) = π (49 − 4) 22 = × 45 = 141.43 sq. m 7 Example 13

Think and Tell These circles have the same centre but different radii. What are such circles called? Where else do you find such circles in real life?

A circle fits perfectly in a square of side 25 cm. Find the area of the remaining part of the square outside the circle. We are given that, the side of the square = 25 cm = diameter of the circle. So, the radius of the circle = 12.5 cm Area of the green part of the figure = Area of the square – Area of the circle = s × s − πr2

22 × 12.5 × 12.5 7 = 625 – 491.07 = 133.93 sq. cm (approx.) = 25 × 25 −

Example 14

Shreya has cut out two semi-circular pieces from a rectangular sheet of paper of size 12 cm × 8 cm (see fig.). Find the area and perimeter of the figure obtained. (Take π = 3.14) We are given that, the length of the rectangle = 12 cm, breadth of the rectangle = 8 cm = the diameter of the semicircle. So, the radius of the circle = 4 cm. Now, the perimeter of the green part of the figure = Lengths of the rectangles + Perimeter of 2 semicircles (Perimeter of a circle) = (2 × 12) + 2πr 22 ×4 7 = 24 + 25.14 = 49.14 cm (approx.) = 24 + 2 ×

Area of the green part of the circle = Area of the rectangle – 2 × Area of a semicircle 1 = 12 × 8 – 2 × × πr2 2 22 = 96 – ×4×4 7 = 96 – 50.29 = 45.71 sq. cm (approx.) Do It Together

From a circular sheet of paper of radius 24 cm, two full circles and a quarter circle of radius 1.6 cm each and a square of side 6 cm are cut out. Find the area of the remaining sheet of paper. We are given that, the radius of the circular sheet = R = 24 cm The radii of two full circles = r1 = 1.6 cm each and the radius of a quarter circle = r2 = 1.6 cm Side of the square = 6 cm

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The area of the remaining sheet of paper = Area of the sheet − (Area of 2 full circles + Area of a quarter circle + Area of the square) 1 = πR2 – (2 × πr1 + × πr2 + s × s) 4 = ______________________________________ = ______________________________________ = ______________________________________ = ______________________________________ = ______________________________________

Do It Yourself 15F Find the area of the figures.

30 cm

9.5 cm

22 cm

6 cm

15 cm

b H 4 cm

d 60 m

100 m

30 cm

18 cm

Find the area of the shaded region. a P

25 cm

8.2 cm

22 cm

70 m

PQRS is a diameter of a circle of radius 7.5 cm such that PQ = QR = RS. Semicircles are drawn on PQ and QS as diameters as shown in the given figure. Find the area of the shaded region. (Take π = 3.14)

Three circular sheets of radius 25 cm are cut out of a rectangular metal sheet of length and breadth 1.50 m and 90 cm, respectively. What area of the metal sheet is left? (Take π = 3.14)

A piece of cardboard is in the shape of a square of side 32 cm. A circle with the biggest possible diameter is cut out from the square. Find the area of the remaining portion of the cardboard.

From a circular sheet of metal of radius 8 cm, a quarter circle of radius 5 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

A rectangular sheet of paper is 50 cm long and 38 cm broad. 27 buttons each of radius 2.5 cm have been cut out from it. Find the area of the remaining sheet.

A semicircle of radius 6 cm is cut out of a rectangular sheet of paper of dimensions 8 cm × 12 cm. What is the area of the leftover sheet of paper? (Take π = 3.14)

1 Two circles are drawn inside a big circle of diameter 28 cm. The diameters of the two circles are 4 3 and of the diameter of the big circle as shown in the given figure. Find the ratio of the areas of the 4 unshaded part and the shaded part of the circle.

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Word Problem 1

While making a mask, Ranita removed two circles of radius 6 cm each, and a rectangle of length 5 cm and breadth 2 cm from a circular card of radius 44 cm, as shown in the given figure. Find the area of the remaining mask.

Mean, and Word Median Problems onMode Perimeter and Area Real Life Connect

A society is built on a plot of land of length 200 m and breadth 180 m. It has two kinds of pavement areas given below. 1 A path 2 m wide runs inside along the sides of the plot. 2 Crossroads 3 m wide run at right angles through the centre of the plot and parallel to its sides.

For aesthetic development, four fountains, each enclosed by a circular fence with a radius of 50 cm, are also installed, as shown below. 200 m Pathway 3m

Crossroads

50 cm

Fountain

50 cm

180 m

Fountain

50 cm

Let us find the area of the pathways and crossroads. Area of the Pathway

Think and Tell

Length of the outer rectangle (plot) = 200 m

How would you calculate the area of the

Breadth of the outer rectangle (plot) = 180 m Length of the inner rectangle = 200 – 2 – 2 = 196 m

plot without the pathways, crossroads, and fountains?

Breadth of the inner rectangle = 180 – 2 – 2 = 176 m So, area of the pathway = Area of the outer rectangle – Area of the inner rectangle = (200 × 180) sq. m – (196 × 176) sq. m = 36000 sq. m – 34496 sq. m = 1504 sq. m Area of the Crossroads To find the area of the crossroads, we must first find the area of the two roads and the area of the intersecting square, and then find their difference. Chapter 15 • Perimeter and Area

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Length of the road that is parallel to the length of the plot = 196 m

Length of the road that is parallel to the breadth of the plot = 176 m

Breadth of this road = 3 m

Area of this road = 196 × 3 = 588 sq. m

Area of this road = 176 × 3 = 528 sq. m

The intersection of the crossroads is a square where two squares are overlapping each other. So, to find the area of the crossroads, we need to subtract the area of 1 square from the area of the two roads. Therefore, area of 1 square = 3 × 3 = 9 sq. m Thus, area of the crossroads = 588 + 528 – 9 = 1107 sq. m Example 15

The inner and outer radii of a cylindrical pipe are 5 cm and 6 cm, respectively. Find the area of the cross-section of the pipe. (Take π = 3.14) Let the inner radius of the pipe be r and outer radius of the pipe be R. Inner radius of the pipe (r) = 5 cm, Outer radius of the pipe (R) = 6 cm. Area of the cross-section of the pipe = π (R2 − r2) = 3.14 × (6 × 6 – 5 × 5) = 3.14 × (36 – 25) = 3.14 × 11 = 34.54 sq. cm

Example 16

A path 3.5 m wide runs inside a rectangular piece of land of length 20 m and breadth 15 m. Find the area of the path. Also, find the cost of levelling the path at the rate of ₹100 per 10 sq. m. Length of the rectangle = 20 m, breadth of the rectangle = 15 m. Width of the path = 3.5 m Length of the inner rectangle = 20 – 3.5 – 3.5 = 13 m Breadth of the inner rectangle = 15 – 3.5 – 3.5 = 8 m So, area of the path = Area of the outer rectangle – Area of the inner rectangle = 20 × 15 – (13 × 8) = 300 – 104 = 196 sq. m Cost of levelling 1 sq. m = ₹100 per 10 sq. m 196 Cost of levelling 196 sq. m = × 100 = ₹1960 10 So, the cost of levelling 196 sq. m of the path is ₹1960.

Do It Together

A square plot of land of side 150 m has two roads of width 2 m passing through its centre and parallel to its sides. Find the cost of constructing the roads at the rate of ₹520 per sq. m. Side of the square = 150 m, width of the road = 2 m. Length of road 1 (parallel to the length of the square land) = 150 m Width of road 1 = 2 m

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Area of road 1 = 150 × 2 = _____ sq. m ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ Area of the roads = Area of road 1 + Area of road 2 – Area of the common square __________________________________________________________________ __________________________________________________________________ Cost of constructing 1 sq. m of road = ₹520 Cost of constructing _____ sq. m = ₹520 × _____ = ₹_______

Do It Yourself 15G 1

A rectangular garden of length 8 m and breadth 15 m is built with a 2 m wide path around the edge of the garden. Find the area of the path.

In a circular garden of diameter 160 m, a pond is constructed in the centre in the form of a circle with radius 25 m. Find the area of the remaining garden. (Take π = 3.14)

A rectangular field of dimensions 100 m × 50 m has crossroads running parallel to the two sides and cutting at right angles through the centre of the field. If both roads are 2 m wide, find: a the area covered by the roads. b the cost of grassing the remaining area at the rate of ₹110 per 10 sq. m.

Find the area of a circular ring in which the external and the internal radii are 35 cm and 27 cm, respectively.

The inner circumference of a circular track is 176 m. The track is 8 m wide. Find the area of the track.

Word Problems 1

A swimming pool of length 12 m and breadth 5 m is to be built with a coping 1.5 m wide all around it. Find the area of the coping. Also, find the cost of coping if the rate is ₹50 per sq. m.

A square garden of side 70 m is newly constructed in the district with 12 m two walking pavements. Each pavement is 10 m wide running across the centre and parallel to the sides of the garden. Find the area of the total garden without the pavements.

A sheet of paper of length 25 cm and breadth 10 cm is to be decorated with a ribbon all around it. Find the area of the sheet not covered by the ribbon, if the width of the ribbon is 2 cm. Also, find the area of the sheet covered by the ribbon.

The circumference of a circular park is 264 metres. A 2.5 m road runs on the outside around it. Find the cost of constructing the road at ₹220 per sq. m.

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1.5 m

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In a lawn of breadth 6 m, a path of equal width is built on three sides as shown. If the perimeter of the path is 74 m, find: a width of the path

b area of the lawn

c area of the path

Lawn

10 m

15 m

Points to Remember •

Perimeter of a rectangle = 2 (l + b), where l = length, b = breadth

•

Perimeter of a square = 4s, where s = side

• • •

Area of a rectangle = l × b Area of a square = s × s

Area of a parallelogram = area of a rectangle = b × h

1 × area of a rectangle (or a square or a parallelogram) 2 1 = ×b×h 2

•

Area of a triangle =

•

Circumference of a circle = 2πr

•

Area between two circles = π (R2 − r2)

•

Area of a circle = area of a rectangle = πr2

Math Lab Circle Art Gallery Setting: In groups of 5 Materials Required: Large sheets of paper, rulers, circular objects of various sizes, coloured pencils, markers or paints, a pair of scissors Method: 1 Each group must collect a large sheet of paper from their teacher. Look at and read the circumference written on the board. 2 Each group shall use a circular object to trace a circle onto the paper of the circumference given and find its area. 3 Each group shall then make another circle of area equal to onethird of the previous circle. 4 Repeat step 3 until the area is equal to less than 0. Stop when you get a negative number. 5 Now, use the circles you created so far to overlap and position them creatively to form a beautiful circular design to display in a circular art gallery. One of the beautiful designs is shown above.

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Chapter Checkup b 6 cm

8 cm

4 cm

Find the area of the triangles.

8 cm

9 cm

6 cm 2

The area of a square is 81 sq. cm. Find the area of the square formed by joining the midpoints of the square.

If the length of a rectangle is doubled and its breadth is one-third of the original, then find the ratio of the area of the new rectangle to the area of the original rectangle.

The length of a rectangular hall is 5 times its breadth. If the perimeter of the hall is 42 m, find its length and breadth.

A man wants to acquire a rectangular plot 160 m long and of the same area as that of a square plot of side 108 m. Find the width of the rectangular plot.

A floor 6.68 m long and 8.2 m wide is covered with rectangular tiles of size 64 cm by 32 cm. Find the total cost of the tiles at the rate of 20.50 per tile.

A park is in the shape of a quadrilateral PQSR, where QR is the diagonal, SO is perpendicular to QR, and RX is perpendicular to PQ. If PQ = 25 cm, QR= 24 cm, RX = 6 cm, and SO = 5 cm, find the area of the park.

The area of a parallelogram is equal to the area of a square of side 12 m. Find the height corresponding to the base measuring 20 m in the parallelogram.

ABCD is a parallelogram of area 1520 sq. m. DP and BQ are the heights on the sides AB and DA, respectively. AB = 28 cm, AD = 32 cm. Find the length of DP and BQ.

Find the area of a scalene right-angled triangle whose base is 15 cm and altitude is 8 cm. D

11 Look at the figure. Find the following:

8 cm

a Area of the parallelogram b Area of each triangle c Ratio of the area of the shaded portion to the remaining area of the rectangle

P 12 cm

15 cm

12 The sum of the circumference and the radius of a circle is 153 cm. Find the diameter and the circumference of the circle. Also, find the area A B of the circle.

5 cm

13 Find the area of the shaded region. D

14 The radius of the base of a circular fence is 10 m. Find the circumference of the circular fence. 15 Find the ratio of the areas of two circles of which the diameters are in the ratio 5:7.

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16 A 44 cm long wire is bent to form a square. If the same wire is bent to make a circular ring, which shape would have a greater area and by how much? 17 In a rectangular garden of dimensions 85 m × 60 m, a circular pond is constructed with a radius of 24 m. Find the area of the garden left out. 18 From a circular sheet of radius 27 cm, four circles of radius 1.6 cm, a rectangle of length 5 cm and breadth 3 cm and a square of side 4 cm are removed. Find the area of the remaining sheet. 19 The inner and the outer radii of a cylindrical pipe are 3.5 cm and 4.5 cm, respectively. Find the area of a crosssection of the pipe. [Take π = 3.14] 20 A rectangular park of length 24 m and breadth 15 m has a footpath of width 2 m just inside the boundary. Find the area of the footpath. 21 A field is circular with a circumference of 264 m. Outside the field, a road around it is constructed. If the circumference of the road is 308 m, find the width of the road. 22 The ratio of two adjacent sides of a parallelogram is 7:3. Its perimeter is 120 cm. Find its area if the altitude corresponding to the larger side is 14 cm. 23 The inner and outer circumferences a circular racing track are 308 m and 352 m, respectively. Find the width of the track.

Word Problems hedge row needs to be planted around a rectangular lawn 22 m long and 15 m wide. 1 A If 5 shrubs can be planted to grow in a metre of hedge, how many shrubs will be needed in total? 2 I nside a rectangular garden 25 m long and 15 m wide, a 1.5 m margin is left for planting lilies. If 4 lily stems can be planted in 1 sq. m and each lily stem costs ₹35, find the total expenditure incurred.

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16 Algebraic Expressions Let’s Recall A number pattern is a pattern or sequence of numbers that follows a certain rule or order. For example, in the table of 3, its multiples 3, 6, 9, 12, 15 . . . form a number pattern where the next number is obtained by adding 3 to the preceding one. Let us find the rule and next number for the number pattern 35, 34, 32, 29, 25, 20, 14, ….. On observing the numbers, we can see that the numbers are decreasing by (−1, −2, −3, −4 and so on…)

−2

−1

−3

−4

−6

−5

The next number in the pattern will be 14 − 7 = 7 Let us see another pattern. Here the numbers are multiplied by 4 each time.

×4

256

×4

The next number in the pattern will be 256 × 4 = 1024

Let’s Warm-up Find the pattern and fill in the blanks to find the next number. 1

105, 100, 95, 90, ________

9, 19, 29, 39, 49, ________

15, 28, 41, 54, 67, ________

90, 75, 60, 45, 30, ________

2, 4, 8, 16, 32, 64, ________

I scored __________ out of 5.

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Mean, Median and Mode Generating Rules in Formulas and Patterns Real Life Connect

Priya and Rahul are architects. They need to find the perimeter and area of different shapes to make their drawings. Priya: Rahul, the measures of different shapes are always different. It takes me a lot of time to find the perimeter and area of the shapes. Rahul: You are right Priya! Earlier I also used to struggle with this issue. But now I’ve started using some formulas and rules to find the area and perimeter. Let us see what rules and formulas Rahul is using.

Finding Formulae We know that the perimeter of a shape is defined as the total length of its boundary. Let us look at the perimeter of some regular shapes. a a

a a

Equilateral triangle

Square

a a

Regular pentagon

Regular hexagon

Perimeter = 3a

Perimeter = 4a

Perimeter = 5a

Perimeter = 6a

where a = length of the side

h h

The formula for area of some shapes can be given as: a

a a

a Square

Example 1

Rectangle

Right triangle

Area = a2

Area = a × b

where a is the length of the side.

where a is the length of the rectangle and b is the width of the rectangle.

Find the perimeter and area of a square with side 10.5 cm.

1 ×b×h 2 where b is the length of the base of the triangle and h is its height. Area =

Perimeter of a square = 4a; here a = 10.5 cm

Think and Tell

Perimeter = 4 × 10.5 = 42 cm

Find the measure for which the perimeter

Area of a square = a2 = 10.52 = 110.25 cm2

of a square is equal to its area.

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Example 2

Find the difference in the areas of the given rectangles. 10 cm

12 cm 4 cm

6 cm Length of rectangle, a = 12 cm

Length of rectangle, a = 10 cm

Breadth of rectangle, b = 4 cm

Breadth of rectangle, b = 6 cm

Area of rectangle = a × b = 10 × 6 = 60 sq. cm

Area of rectangle = a × b = 12 × 4 = 48 sq. cm

Difference in area = 60 sq. cm – 48 sq. cm = 12 sq. cm Do It Together

Find the perimeter of the given figures. Which figure has the greatest perimeter?

15.7 cm

Length of the side of the given triangle a = 15.7 cm

Perimeter of triangle = 3a = 3 × ____ = ____ cm

7.5 cm

6.5 cm

Length of the side of the given pentagon a = _____ Perimeter of pentagon = 5a = 5 × ____ = ____ cm

Length of the side of the given hexagon a = 6.5 cm Perimeter of hexagon = 6a = 6 × 6.5 = ____ cm

On comparing, the perimeter of the _____________ is the greatest.

Number Patterns Let us see some rules for number patterns. If n is a natural number, then, Rule

Example

(n + 1) is its successor

If n = 25, then n + 1 = 25 + 1 = 26 is the successor of 25.

2n is an even number

If n = 7, then 2n = 2 × 7 = 14, which is even

(2n + 1) is an odd number

If n = 9, then 2n + 1 = 2 × 9 + 1 = 18 + 1 = 19, which is odd

Sum of n natural numbers =

n(n + 1) 2

n(n + 1) 11(11 + 1) 11 × 12 = = = 66 is the sum of the first 2 2 2 11 natural numbers. If n = 11, then

Let us see some more number pattern rules: Number Pattern

Example 3

Rule

Example

3, 5, 7, 9 ……

Each term of the pattern can be given by (2n + 1), where n is a natural number

When n = 12,

9, 19, 29, 39, 49 ….

Each term of the pattern can be given by (10n − 1), where n is a natural number

When n = 6

12th term = (2 × 12 + 1) = 25 6th term = (10 × 6 − 1) = 59

Find the sum of the first 20 natural numbers. n(n + 1) We know that the sum of n natural numbers = 2 Putting n = 20 in the given formula, we get

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Example 4

Find the rule for the number pattern. Also find its 21st term. 1, 4, 7, 10, 13…. First term = 1 Successor term = Predecessor term + 3 If the rule = 3n, for n = 1 the first term must be 3, but here the first term is 1 Hence the rule for the given number pattern = 3n − 2

Remember! The pattern must be true for all the natural numbers.

21st term of the pattern = 3 × 21 − 2 = 63 − 2 = 61 Do It Together

Find the difference of the sum of the first 15 natural numbers and of the first 10 natural numbers. n(n + 1) We know that the sum of n natural numbers = 2 The sum of the first 15 natural numbers = ____________________________ The sum of the first 10 natural numbers = ____________________________ Difference = __________________

Do It Yourself 16A 1

The perimeter of a square is 62 cm. Find its area.

The height of a triangle is 20 cm. If its area is 175 cm2 find the length of its base.

If an even number is denoted by n; the next even number can be given as _____.

A wire is in the shape of a regular pentagon of perimeter 30 cm. If the same wire is bent in the form of a regular hexagon, what would be the length of each side of the hexagon?

The area of a square and a rectangle are equal. If the side of the square is 35 cm and the length of the rectangle is 45 cm, find the breadth of the rectangle.

If 89 is the successor of a natural number n, then what is the value of n?

Find the sum of the first 12 natural numbers and the first 18 natural numbers.

Find the rule for the given pattern: 1, 3, 5, 7, 9 …. Also find the 100th term of the pattern.

Word Problems 1

Sudha needs to order a shade for a triangular-shaped window that has a base of 8 feet and a height of 5 feet. What is the area of the shade?

Kunal has a square plot of an area double that of Nikhil’s rectangular plot. If the length of Nikhil’s plot if 25 m and its breadth is half of the length, find the area of Kunal’s plot.

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Mean, Median and Mode Algebraic Expressions and Terms Real Life Connect

Kushal went to school after 10 days as he had been ill. He saw some maths problems in the homework and got confused. He saw various letters of the alphabet along with numbers. Let us help him out.

Parts of an Expression An algebraic expression is an expression made up of variables and constants and joined using mathematical operators (like addition, subtraction, multiplication, or division) A variable in an algebraic expression is the letter used to represent a certain quantity.

4x + 2

A constant in an algebraic expression is the number of which the value is fixed.

3x2 − 4y The parts of an algebraic expression separated by + and − signs are called terms. We saw that 3x2 − 4y consists of two terms 3x2 and − 4y. The term 3x2 is a product of 3, x and x; we can say that 3, x and x are the factors of the term 3x2. Hence, a term is a product of its factors. The term 4y is a product of the factors 4 and y. One of the ways to show algebraic expressions is by using algebraic tiles. Let us represent 3x2 + 3x − 4 on the algebraic tiles. 2

3x2

x x x

−1 −1 −1 −1

−4

3x2 + 3x − 4

An expression can be shown with the help of a factor tree as:

Error Alert!

5xy − 6y2 + 12

Expression

Terms

5xy

−6y

Factors

5 x y

−6 y y

The minus sign (−) is included in a term.

For the expression 5y − 3x the terms are 5y and 3x

In a term of an algebraic expression, any of the factors with the sign of the term is called the coefficient of the product of the other factors in that term. A coefficient can be a numerical factor, or an algebraic factor or the product of two or more factors. For example, in the term −5xy, −5 is the coefficient of xy or −5x is the coefficient of y or −5y is the coefficient of x.

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5y and −3x

Think and Tell What is the coefficient of xy in the term −xy?

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Example 5

Example 6

Identify the constant terms in the given expressions. Also give the coefficient of x in each of the terms. Expression

Constant term

Coefficient of x

9x + 5

−5y + 3x + 2

7xy − 2

−2

Express the given expression using a factor tree. 8x − 4x2 + 3y 8x − 4x2 + 3y

Expression

−4x2

Terms

Did You Know?

Algebra was invented in the 9th century.

Factors Do It Together

−4 x x

What are the factors of the term −5a2b in the expression 15ab − 5a2b + 5a? The factor tree for the expression can be given as: 15ab − 5a2b + 5a

Expression Terms

15ab

_____

Factors

__ __ b

−5 __ __ b

_____ 5

The factors of the term −5a2b are −5,____, ____, and _____.

Like and Unlike Terms The terms are called like terms when they have the same algebraic factors. If the algebraic factors of two terms are different, they are called unlike terms. For example, in the expression 15xy + 28x2 − 8xy + y2, the algebraic factors of the terms 15xy and −8xy are the same i.e., xy, hence 15xy and −8xy are like terms. The algebraic factors of 28x2 are x and x and algebraic factors of y2 are y and y, hence 28x2 and y2 are unlike terms. Example 7

List the like terms in the expression 5x2y + 4xy − xy − 9yx2. The like terms are 5x2y and −9yx2 since both have the same algebraic factors i.e., x2y. 4xy and −xy are also like terms since they have the same algebraic factors i.e., xy.

Example 8

Group the like terms from the terms given. −xy2, −y2x2, 5y2x, −10x2y2 and 3xy2 Like terms = −xy2, 5y2x, 3xy2; −y2x2, −10x2y2

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Do It Together

Write the like terms for the given expressions. Expression

Like terms

5a2b − 3ab + 2ab2 − 10a2b

5a2b, −10a2b

2x2y + 5x − 3x − 7y2x 12b − 5ab + 7ba2 + 9a2b

Types of Algebraic Expressions Based upon the number of terms, algebraic expressions are classified as: Monomial: An algebraic expression that contains only one term is called a monomial. For example, 3y, 4a2b, −8 are monomials. Binomial An algebraic expression that contains two terms is called a binomial. For example, 2xy − 2x, a2 + 5ab, 7pq − 8 are binomials. Trinomial: An algebraic expression that contains three terms is called a trinomial. For example, 4ab + 5b − 3c, 4xy2 + 2xy − 3 are trinomials. Quadrinomial: An algebraic expression that contains four terms is called a quadrinomial. For example, 5pq + 7qr − p2 + 8pr, 8x2y − 3xy + 7zx − 8yz are quadrinomials. An expression with more than one term is called a polynomial. Example 9

Which among the following is a binomial expression? 1 m

2 ab + c

3 pq + r − q

4 xy − zx

A binomial has two terms, hence the expressions ab + c and xy − zx are binomials. Example 10

Find the odd one out. 1 x + y + 8

2 2z + y − z

3 x2 − y2

4 2ab + a + 3y

All the given expressions are trinomials except x2 − y2 which is a binomial. Do It Together

Identify the terms in the expressions and classify them as monomial, binomial, trinomial or quadrinomial. Expression

Terms

Type of Expression

2x2, 5

Binomial

9xvy + 7 − 9xy + 3xv 2x2 + 5 −7a + 5 + 6b 3x

Do It Yourself 16B 1

Using constants, variables and mathematical operations, form algebraic expressions for the given statements. a

Addition of x and y.

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One-fourth of a number a added to thrice the number b.

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3 less than z added to the product of x and y.

The product of m and n subtracted from twice the product of p and q.

List the coefficients of a and b in the expressions. a 6a2b + 12bc

b 3ac − 10ba

c 15ba2 + 12ac + 7b2c

d 5ca2 + 3ab − 6ca

e 2.5ab − 3.2a2b

Identify the terms in the following expressions.

1 2 b + 3ab 3

a 5m2 + mn

b 3pq + 5q2p

c 7y3 + 8xy2 − 5xy

d 6ab + 8a2bc − 2bc2 − ca

e 0.5xy + 1.2x2y − 3.3y

Draw a factor tree for the given expressions.

2 5 yz + x2y + 0.6xy 5 6

a 5xy + 8yz

b 3pq2 − 5pr + 2

c 5ab − 8ba2 + 5c

d m + mn2 − 6l + 8nl

e 1.2x2y − 3xyz

a −2xy2, 7xy2, −2x2y, 5xy2

b 8mn, 6mn2, 5nm, nm

c 3p2q2, 5p2q, 8q2p2, q2p2

d ab2, 5a2b, 7ba2, a2b

e 0.2xy, 2yx, 0.5zx, 0.8xy

Find the odd one out.

Group the like and unlike terms.

1 2 ab − 3a2b + 7 2

1 2 2 2 m n, n m, 0.6nm2, 0.4m2n 6 5

a m2n, 3mn, 5m2n, 6nm, 8nm2, 5m, 7n

b 2a2c, 3ac2, 3ca2, 5b2a, c2a, 8b2c

c pqr, pq2, 9qrp, 5pr, 6rpq, 8rp

d 2klm, 5mln, k2lm, 5lmn, 6lk2m

State true or false. a 6mn + 2mn2 + 3 is a binomial.

b 3 × m is a monomial.

c 9pq + 2qr + 5s is a trinomial

d 4a2 + 5ab + 3c + 8 is a trinomial

Word Problem 1

Suman has a rectangle whose area can be expressed using the expression 3a2 − 2a + 5. What is the constant term in the expression?

Mean, Median and Mode of Algebraic Expressions Addition and Subtraction Real Life Connect

Kushal learnt about the basic terms of algebraic expressions. The next day when he went to school, the teacher introduced the class to the addition and subtraction of algebraic expressions.

Addition of Algebraic Expressions We know that we can add or subtract two quantities only if they have the same unit. The same rule applies to algebraic expressions. While adding two algebraic expressions, we add only the like terms. 270

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Let us add the expressions 3x + 2 and 4x − 2y + 1 using algebraic tiles.

x x x

1 1

x +

3x + 2

x x x

−y 1

−y

x =

x x

4x − 2y + 1

−y

7x − 2y + 3

We can also add polynomials using the horizontal or column method. Let us add 6p2 + 9pq + 5q2 and 4p2 + 5pq + q2 using both methods. Horizontal Method: Step 1: Write the expressions to be added with a + sign between them.

Column Method: Write the like terms one below the other and then add. The above expression can be added as:

(6p2 + 9pq + 5q2) + (4p2 + 5pq + q2) Step 2: Write the like terms together.

6p2 + 9pq + 5q2 +

6p2 + 4p2 + 5pq + 9pq + q2 + 5q2 Step 3: Add the numerical coefficients of all the terms and keep the algebraic factor the same.

4p2 + 5pq + q2 (6 + 4)p2 + (9 + 5)pq + (5 + 1)q2

10p2 + 14pq + 6q2

(6 + 4)p2 + (5 + 9)pq + (1 + 5)q2 10p2 + 14pq + 6q2 Hence, (6p2 + 9pq + 5q2) + (4p2 + 5pq + q2) = 10p2 + 14pq + 6q2 Example 11

Add 5xy − 3y + 2z and 3xy − 8z + 2y by both the column and horizontal methods. Horizontal Method

Column Method

5xy − 3y + 2z + 3xy − 8z + 2y = 5xy + 3xy − 3y + 2y + 2z − 8z = (5 + 3)xy + (−3 + 2)y + (2 − 8)z = 8xy − y − 6z Example 12

5xy − 3y + 2z

3xy + 2y − 8z (5 + 3)xy + (−3 + 2)y + (2 − 8)z

8xy − y − 6z

Add the expressions. 4ab + 3b; 5ac − 5b; 12ab − 3ac and ab − 4ac + 12b Writing all the expressions with a + sign between them, we get (4ab + 3b) + (5ac − 5b) + (12ab − 3ac) + (ab − 4ac + 12b) 4ab + 12ab + ab + 3b − 5b + 12b + 5ac − 3ac − 4ac = (4 + 12 + 1)ab + (3 − 5 + 12)b + (5 − 3 − 4)ac = 17ab + 10b −2ac

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Remember! Add the numerical coefficients along with their signs.

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Example 13

Add 4x2 − 5xy2 + 14xy − 3yz to the sum of 2x2 + 7xy2 − 3xy and 5xy − 8yz − 2xy2. We first find the sum of 2x2 + 7xy2 − 3xy and 5xy − 8yz − 2xy2 2x2 + 7xy2 − 3xy + 5xy − 8yz − 2xy2 = 7xy2 − 2xy2 − 3xy + 5xy − 8yz + 2x2 (Writing the like terms together) = (7 − 2)xy2 + (−3 + 5)xy − 8yz + 2x2 = 5xy2 + 2xy − 8yz + 2x2 Adding the above sum to the expression 4x2 − 5xy2 + 14xy − 3yz, we get (5xy2 + 2xy − 8yz + 2x2) + (4x2 − 5xy2 + 14xy − 3yz) 5xy2 − 5xy2 + 2xy + 14xy − 8yz − 3yz + 2x2 + 4x2 2

Error Alert!

= (5 − 5)xy + (2 + 14)xy + (−8 − 3)yz + (2 + 4)x

Do It Together

= 16xy − 11yz + 6x2

Do not leave the unlike terms while adding the like terms.

Add the expressions using the column method.

3xy + 2y + 5xy = 8xy

3xy + 2y + 5xy = 8xy +2y

4pqr + 9pr2 − 3pr and 10pr2 − 12pqr 4pqr +

+ ________ − 3pr

________ + 10pr2 (4 + ____)pqr + (____ + 10)pr2 − 3pr

_________________

Do It Yourself 16C 1

Add the monomials. a 5xy and x

b 3p2 and 5p2

c 4ab and −3ab

d 2ac2 and 5ac2

e 3mn and −5mn

f 10x2y and 7x2y

Add the expressions using the column method. a 2x + 3y and x + y

b 2x + 3y + z and 2x + y − z

c a + 2b − 3 and 5b − 3a + 5

−5a + 4b − 8c, 4a + 7b − 9c, 2a + 4b + 2c

Add the expressions using the horizontal method. a x − 2y, x + y, y − 2x, x + 3y

b 3p2 +q2 − 2r2, 4q2 + 5r2

c 4a2 + 3b2, −2a2 + 13ab2, 4a2 + 3ab2

d 19a + 15b − 11c, 12a − 5b + 6c, 6b + 4c

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Match the expressions with their answers. a 2x + 5xy − 5x + 6xy

7ab2 − 4b + 6ac

b 7x2 − 3y2 + 4x + 3y2 − 4x2

7pq + 4pr

c 5ab2 − 3b + ac + 2ab2 + 5ac − b

11xy − 3x

d 12pq + 9pr − 5rp + 5qp − 10pq

−2pq + 10pr

e 7pq2 + 3pq − 5qp − 7q2p + 10pr

4a2b − 9ab + 3c

f 9a2b − 15ab + 3c − 5ba2 + 6ba

3x2 + 4x

Add 4ab2 − 7a2b + 12bc to the sum of: a 12ab2 − 3a + 5bc and 5a + 3a2b + 3ab2

b ab2 − 2a2b + 6bc and −11a2b + 9bc

Find the perimeter of the given triangle.

a+7

2a + 5

3a − 9 7

Add

1 2 4 6 2 2 1 mn2 − nm + n2 and n − mn2 + mn 2 3 7 11 5 5

Word Problem 1

Sudha has a copper wire of length (5x + 9) cm. Rakesh has a copper wire of length (3x + 4) cm. What is the total length of the two pieces of copper wire?

Subtraction of Algebraic Expressions Like addition, while subtracting two algebraic expressions, we subtract only the like terms. Subtracting a term is the same as adding its inverse. In subtraction, we change the sign of the expression to be subtracted and add it to the expression from which the subtraction is to be done. Let us subtract the expressions −x2 − x + 2 from 2x2 + 3x − 2

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x x x

−1 −1

−

−x2

−x

1 1

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As there is a minus sign in between, we need to flip the sign of the subtrahend, and the above algebraic tiles can be solved as:

x x x

−1 −1

x x x x

−1 −1 −1 −1

We can also subtract polynomials using the horizontal or column method. Let us subtract 5ab2 − 7ab + 6a2 from 14ab2 + 10ab + 10a2 using both methods. Horizontal Method: Step 1: Write the expressions to be subtracted in brackets with a (−) sign between them. (14ab2 + 10ab + 10a2) − (5ab2 − 7ab + 6a2) Step 2: Change the sign of the subtrahend expression. 14ab2 + 10ab + 10a2 − 5ab2 + 7ab − 6a2 Step 3: Write the like terms together. 2

14ab − 5ab + 10ab + 7ab + 10a − 6a

Step 4: Subtract the numerical coefficients of all the terms and keep the algebraic factor the same.

Remember! The signs of algebraic expressions are handled in the same way as the signs of numbers.

(14 − 5)ab2 + (10 + 7)ab + (10 − 6)a2 = 9ab2 + 17ab + 4a2

Column Method: Write the like terms one below the other. Change the sign of the subtrahend and then add. 14ab2 + 10ab + 10a2 5ab2 − 7ab + 6a2 − + − (14 − 5)ab2 + (10 + 7)ab + (10 − 6)a2

Remember! We indicate the change of the sign of each term in the expression to be subtracted below the original sign of each term.

= 9ab2 + 17ab + 4a2 Example 14

Subtract 8xy + 3yz − 16xz from 9yz + 22xz by both the column and horizontal methods. Horizontal Method

Column Method

(9yz + 22xz) − (8xy + 3yz − 16xz)

9yz + 22xz

= 9yz + 22xz − 8xy − 3yz + 16xz 9yz − 3yz + 22xz + 16xz − 8xy = (9 − 3)yz + (22 + 16)xz + (−8)xy = 6yz + 38xz − 8xy

−

3yz − 16xz + 8xy +

−

(9 − 3)yz + (22 + 16)xz + (−8)xy = 6yz + 38xz − 8xy

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Example 15

What should be subtracted from 5mn2 + 6nm − 8n2 to get 12mn2 − 3mn + 5? Let X denote the required algebraic expression, Then, (5mn2 + 6nm − 8n2) − X = (12mn2 − 3mn + 5) X = (5mn2 + 6nm − 8n2) − (12mn2 − 3mn + 5) X = 5mn2 + 6nm − 8n2 − 12mn2 + 3mn − 5 X = 5mn2 − 12mn2 + 6nm + 3mn − 8n2 − 5 X = (5 − 12)mn2 + (6 + 3)mn − 8n2 − 5 X = −7mn2 + 9mn − 8n2 − 5 Hence, −7mn2 + 9mn − 8n2 − 5 must be subtracted from 5mn2 + 6nm − 8n2 to get 12mn2 − 3mn + 5.

Do It Together

Subtract 14x2z + 12xz − 13yz from 24x2z − 13xz + 3yz using both the horizontal and column method. Horizontal Method

Column Method

(24x2z − 13xz + 3yz) − (__________________)

24x2z − 13xz + 3yz

= 24x2z − 13xz + 3yz − __________________ −

= __________________

14x2z + 12xz − 13yz −

______________________________

= __________________

Do It Yourself 16D 1

Subtract using the column method. a (3ab − 5) from (ab − 9)

b 12xy − 3y2 from 9y2 − 5xy

c (9y − 2y2 − 4) from (4y2 + 5y − 7)

d (−b2 + 7ab) from (5b2 − 3ab + 8)

Subtract using the horizontal method. a (10x2 − 7x) from (−10x + 8x2)

b (mn2 + 2mn − 8n2) from (5mn − 6mn2 + 2n2)

c (3p2q + 2q2 + 5) from (p2q + 5q2 − 8)

d (5 − ab − ac) from (5ab + 6ac + 3bc + 8)

How much is 12ab + 5ac − 3 greater than the sum of 5ab + 6ac and 8ac − 7ab + 8?

What must be added to 3a2 + 15a + 8 to get 8a2 − 12a + 6b?

What should be subtracted from 5xz2 + 2xy − 7yz to obtain 7xy − 4x2z + 12yz + 3?

What will you get on subtracting 4x − 2y − 10 from the sum of 4x − y + 11 and − y − 12?

Subtract the sum of 6x2y + 9xy − 3y2 and x2y − 3xy + 5y2 from the sum of 7xy2 + 12xy − 10x2y and 2xy2 + 7y2 + 2x2y.

If P = 6a2 + 12ab + 3ab2; Q = 5ab − 8ab2; R = 15ab − 4a2 + 9, what is the value of P − R + 3Q?

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Word Problem 1

Mohan has a ribbon of length (3m + 3) metres. He used (3m − 5) metres of ribbon from it. How much ribbon does he have left with him?

Finding the Value of an Expression An algebraic expression is formed by using one or more variables. The expression changes with the change in the value of the variables. To find the value of an expression, we need to substitute a specific value of each variable in the expression and then perform the arithmetic operations. Let us substitute the value of x = 5 in the expression x2 − 2x + 3 and find the value. Putting x = 5, we get (5)2 − 2(5) + 3 = 25 − 10 + 3 = 18 Example 16

Find the value of the expression for a = −7. 5a3 + 2a2 − 12a + 3 Putting a = −7, we get 5 × (−7)3 + 2 × (−7)2 − 12 (−7) + 3 = 5 × (−343) + 2 × 49 + 84 + 3 = −1715 + 98 + 87 = −1530

Example 17

Error Alert! Use the correct signs when performing the arithmetic operations. 3 × (−5) = 15

3 × (−5) = −15

Add the expressions 3mn2 + 5m2n + 4 and −4mn2 + 7m2n + 3mn. Find the value of the expression for m = 2 and n = 1. Adding 3mn2 + 5m2n + 4 and −4mn2 + 7m2n + 3mn, we get (3mn2 + 5m2n + 4) + (−4mn2 + 7m2n + 3mn) = (3 − 4)mn2 + (5 + 7)m2n + 4 + 3mn = −mn2 + 12m2n + 4 + 3mn Putting m = 2 and n = 1, we get = −(2) × (1)2 + 12(2)2 × 1 + 4 + (3 × 2 × 1) = −2 + 48 + 4 + 6 = 56

Do It Together

Subtract the expression 5xy + 7x2 from 9x2 − 3xy + 5. Find the value of the expression for x = 1 and y = −1. Subtracting 5xy + 7x2 from 9x2 − 3xy + 5, we get (9x2 − 3xy + 5) − (____________) = 9x2 − 3xy + 5 − ______ − ______ = (9 − ____)x2 + (−3 − ____)xy + 5 = ____x2 − ____xy + 5 Putting x = 1 and y = −1, we get _____________________ = __________________ = ____________

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Do It Yourself 16E 1

If a = 5, find a (a + 6)

b 2m2 + 2m − 3

c m3 + 3m2 − 4

d 4m3 − 4m2 + 2m − 10

c 3x2 − 4xy + 8

d 2x3y + 3x2 − 5x

c 3a3 − 2(a + 2b)

d 3(a3 − ab) + 3a2 − 7b + 8

b 5x2 + 2y2 − 1

Simplify if needed and find the value for a = 2 and b = 3. a 5ab − 2a + 3b

Find the value of the expressions when x = 1, y = −1 a x2 + y2

c (a2 − 3a + 5)

If m = −3; find the value of a 2m + 7

b (2a − 3)

b 2a2 + 3ab − 4a

Add the expressions and find their value for x = −2 and y = 1. a 4x − 3y and 5y + 2x

b x2y + 2xy and 3xy − 4

c y2 − 3y, 2x + 4y and 2y2 − 4x

Find the value of A − 2B + C if A = 3x + 2y; B = x2 − 5x and C = y2 − 6y where x = 2 and y = 2.

If a = 20, find the value of a3 − 2a2 + 5a − 100.

What should be the value of o if the value of m2 + 2nm + o equals −7 for m = −1 and n = 1?

Word Problem 1

The amount of water in a water tank can be given by the expression (250a2 + 100a) litres. If Pooja used (50a) litres of water, find the amount of water left in the tank when a = 3.

Simplifying Algebraic Expressions Simplifying an algebraic expression means to write the expression in simpler form with a fewer number of terms. For this we need to group all the like terms and then solve. We come across different types of brackets while simplifying an algebraic expression. These brackets are removed in a certain order. Let us simplify the expression: 2x2 − [3x − {4x2 − (3y − 2x − 2y)}] Step 1: Solve the expression inside the small brackets ( ) and then remove it. 2x2 − [3x − {4x2 − (3y − 2x − 2y)}] = 2x2 − [3x − {4x2 − (y − 2x)}] = 2x2 − [3x − {4x2 − y + 2x}] Step 2: Solve the expression inside the curly brackets { } and then remove it.

Remember! If there is a negative sign before the brackets, the sign of every term within the brackets will be changed when removing the brackets.

2x2 − [3x − {4x2 − y + 2x}] = 2x2 − [3x − 4x2 + y − 2x]

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Step 3: Solve the expression inside the square brackets [ ] and then remove it. 2x2 − [3x − 4x2 + y − 2x] = 2x2 − [x − 4x2 + y] = 2x2 − x + 4x2 − y Step 4: Solve the rest of the expression. 2x2 − x + 4x2 − y = 6x2 − x − y Example 18

Simplify the expression and find the value for a = 1, b = −1. 2a − [3a − {5ab + (ab − 5a − 3ab)}] Solving and removing the small brackets in the expression, we get 2a − [3a − {5ab + (−5a − 2ab)}] = 2a − [3a − {5ab − 5a − 2ab}] Solving and removing the curly brackets in the expression, we get 2a − [3a − {5ab + (−5a − 2ab)}] = 2a − [3a − {5ab − 5a − 2ab}] 2a − [3a − {3ab − 5a}] = 2a − [3a − 3ab + 5a] Solving and removing the square brackets in the expression, we get 2a − [3a − 3ab + 5a] = 2a − [8a − 3ab] = 2a − 8a + 3ab Solving the rest of the expression, we get 2a − 8a + 3ab = −6a + 3ab Putting a = 1, b = −1 in the simplified expression, we get −6a + 3ab = −6 × 1 + 3 × 1 × −1 = −6 − 3 = −9

Do It Together

Simplify the expression and find the value for x = 2; y = 1. 10x − [5x2 − {4x2 − (4 + 5x + 7x2) − 2x2}] 10x − [5x2 − {4x2 − 4 − ______ − ______ − 2x2}] 10x − [5x2 − {______x2 − 4 − ______}] = 10x − [5x2 + ______ + 4 + ______] 10x − [______x2 + 4 + ______] = 10x − ______x2 + 4 + ______ Solving the expression, we get 10x − ______x2 + 4 + ______ = ___________ Putting x = 2; y = 1,we get ____________ = ____________ = _________

Do It Yourself 16F 1

Simplify the expressions. a 5x − (y − 2x + 8y)

b (10m − 3mn + 4m) + 5mn

c 3a − [5ab − (2a − 3ab)]

d {4x2 − (3x2 − 2x + 5) + 6x}

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Simplify and find the value for m = −2. a (5m − 3m2 − 6m) − 2m2

b 2(m2 − m) + 3(m2 + m)

c {2m − (5m2 + 5m) + 3m2}

d [6m2 − {2m + 5 − (2m2 − 8)} + 3]

Simplify. a 5x + [7x2 + 2 − (2x2 + y2) + (4x − 3y)]

b [2m2 + {5mn − (8n + 2nm) + 6m} − 5m2]

c 2pq + [4pq2 − {3pq + 4p − (2pq2 − 2p)} − 3pq]

d 2a + [7ab − (2a2 + 5ab − a) − {3 + (2a + a2)}]

Simplify and find the value for a = 1, b = −2. a [2a2 − {(3ab + 6a) − (2b2 − 2ab)} + 3a2]

b 5a − [3a2 − {2b2 − (4ab + 5a + 7a2) − 2b2}]

c 7a − [8ab − {5b2 + (−3ab − 8a + 6ab) + 2b2} − 4ab]

d b2 + [(2a2 + 5ab + b2) − (6ab + 2b2) − (5ab + 2a2)]

Find the value of the expression for x = 3 and y = 2. 5(x3 + y3 − 2x2y) − [2(x2 − y2 + 2xy) − {y3 − 4(x + 2)}]

Word Problem 1

Rohan’s age can be given by the expression 2x + [3x − {4x + (4 − x)}]. If x = 5, find Rohan’s age.

Word Problems Let us create and solve algebraic expressions based on some word problems. Kamal has some stamps; Suhani has 8 more than Kamal. Priya has 6 stamps less than the total number of stamps that Kamal and Suhani have. Write an expression showing the total number of stamps that all three of them have. Since the number of stamps Kamal has is not given, we will consider it as x. The number of stamps Suhani has = 8 more than Kamal = x + 8 The number of stamps Priya has = 6 less than the total number of stamps Kamal and Suhani have = x + x + 8 − 6 = 2x + 2 Total number of stamps all of them together have = x + x + 8 + 2x + 2 = (1 + 1 + 2)x + 8 + 2 = 4x + 10 Example 19

Komal has m rose plants and 12 more sunflower plants than rose plants. She gives 3 rose plants to Poonam and 5 sunflower plants to Rohan. How many plants does she have left? Number of rose plants = m; Number of sunflower plants = m + 12 Number of rose plants Komal has left after giving 3 to Poonam = m − 3 Number of sunflower plants Komal has left after giving 5 to Rohan = m + 12 − 5 = m + 7 Total number of plants left with Poonam = m − 3 + m + 7 = 2m + 4

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Example 20

Suhani has ₹(6y2 + 5y + 2) and Dev has ₹(8y2 − 3y − 8). If y = 5; find the total amount of money they have. Money Sohani has = ₹(6y2 + 5y + 2); Money Dev has = ₹(8y2 − 3y − 8) Total amount = 6y2 + 5y + 2 + 8y2 − 3y − 8 = 6y2 + 8y2 + 5y − 3y + 2 − 8 = 14y2 + 2y − 6 Putting y = 5 in 14y2 + 2y − 6 we get, 14 × 52 + 2 × 5 − 6 = 14 × 25 + 10 − 6 = 350 + 4 = 354 Hence, they have ₹354 altogether.

Do It Together

An ice-cream seller sold (x2 + 3x − 5) ice creams on Saturday and (5x + 3x2 + 2) ice creams on Sunday. What is the difference in the number of ice creams sold on Sunday and on Saturday, if x = 10. Number of ice creams sold on Saturday = (x2 + 3x − 5); Number of ice creams sold on Sunday = (5x + 3x2 + 2) Difference in the number of ice creams sold on Sunday and on Saturday = (5x + 3x2 + 2) − (x2 + 3x − 5) = 5x + 3x2 + 2 − ______ − 3x + ______ = (5 − 3)x + (3 − ______)x2 + ______ = 2x + ______ x2 + ______ Putting x = 10 in the above expression, we get 2 × 10 + ____ × (10)2 + ____ = 20 + _____ + _____ = _______ ice creams

Do It Yourself 16G 1

Nikhil, Riya and Rohan shared some apples. Nikhil got 4 more apples than Riya who got twice as many as Rohan. If Rohan got x apples, how many apples did they get altogether?

Shobha’s mother’s present age is twice her age. If the age of her grandmother is 5 years more than the sum of her mother’s age and her age, how old is Shobha’s grandmother?

Sunil travelled (2x2 + 3x + 4) metres by car and (5x2 + 10x + 15) metres by train. How much farther did he travel by train than by car?

Kunal spent ₹(3x2 + 2xy + 9y2) on a shirt and ₹(5x2 − xy + 3y2) on food. How much money did he spend in total?

Manisha has a triangular plot with sides (3x2 + 3y + 15) m, (2x2 − y + 30) m and (7x2 + 2x − 10) m. What is the perimeter of her plot if x = 2 and y = 1.

Word Problem 1

In a week, Rahul spends x minutes on the internet, Vivek spends twice the amount of time spent by Rahul, and Ankit spends 10 more minutes than Vivek. Write the total time spent by them in the form of algebraic expressions. Also find the amount of time spent by each one of them if x = 22.

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Points to Remember • An algebraic expression is formed by applying mathematical operations on variables and constants. •

Expressions are a combination of terms connected by positive and negative signs.

•

A term is a product of the factors.

• The terms are called like terms when they have the same algebraic factors. If the algebraic factors of two terms are different, they are called unlike terms.

• Based upon the number of terms, algebraic expressions are classified as monomial, binomial, trinomial, or quadrinomial. • To add or subtract an algebraic expression, combine the like terms and then perform the operation. •

To find the value of an expression, replace the variable by the given value.

• To simplify algebraic expressions, we first remove the small brackets (), then the curly brackets {} and then the square brackets [].

Math Lab Algebraic Expression Scavenger Hunt Setting: In groups of 3 Materials Required: Markers, index cards, envelopes for index cards, a list of algebraic expressions Method: Instruct each group to choose one expression from their envelope and work together to: •

Identify the key variables and constants in the expression.

•

Simplify the expression if possible.

The group that solves all the expressions first wins.

Chapter Checkup 1

The length of the base of a triangle is 32 cm. If its area is 184 cm2, find its height.

Find the rule for the pattern: 14, 27, 40, 53…. Also find the 15th term of the pattern.

What will be the formula for the perimeter of an isosceles triangle if the length of the two equal sides is a and the length of the unequal side is b?

State whether the pair of terms is like or unlike.

a 5m2n, 6n2m

b 2ab, −3ba

c x2y, −5yx

d 2p2q2, 8q2p2

e 3mn, −5m

f 6ab2, −2ab2

Draw a factor tree for the expressions. a x2y + yx

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b 2cb2 + 6ab + 8c

c pq − 5pq2 + 9

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Classify the expressions as monomial, binomial, trinomial and quadrinomial. ab + 4a − 5ac, 3yx2 + 5xy − 3yx − 8, m × n, 2x2 + y

Add the following. a 5a2b, −3ab2, −7ab2 and 14a2b

b (10x + 12y – 12xy – 12), (18 – 5x – 10y + 5xy) and 9xy

What should be added to x2 + 2xy + y2 to obtain 5x2 + 7xy?

What should be subtracted from 5x + 8y + 12 to get –3x + 9y + 17?

10 Find the perimeter of the rectangle. The lengths are given in metres.

3a + 2 6a – 7 11 What will be the value of y3 + 4y2 – 7y – 2 if y = –3? 12 Find the value of the expression 4x3 – 5y2 + 7 if x = 2, y = 3. 13 Simplify the expression –5(2x – 1) + 7x – 21 and find the value of x = –2. 14 What will be the value of p3 – 2p – 270 if p = –15? 15 What will be the value of a if the value of 4x3 + 2x – 5a equals 5, when x = –2? 16 Simplify the expression –3(x2 + 3xy) + 3 – xy and find its value when x = 5 and y = –4. 17 Find the value of the expression

m3 + 5m – 3 – 5n if m = –5, n = 3. 3

18 Simplify the expression 5(x + 2) + 5x – 10 + 5xy and find the value if x = –3, y = –2. 19 Simplify and find the value for m = −1, n = −1. a [5 − {(4mn + 3m) − (5mn2 − 2m)} + 3mn2] b 5m − [8m2 − {4n2 − (4m + 5mn + 6m2) − 2n2}] c 8n − [5mn − {5m2 + (−3mn + 8n − 7mn) + 2m2} − 4n] d mn2 − [(5nm2 + 4mn − 3mn2) − (6mn − 2nm2) − (5mn − 2mn2)]

Word Problems unal has an equilateral triangular-shaped wooden frame with a perimeter of 96 cm. 1 K He wants a square frame with the length of each side equal to that of the triangular frame. What will be the perimeter of the square frame? unita had ₹(15a2 + 3ab + 3b2). She spent ₹(10a2 − ab + b2) on purchasing books. How much 2 S money does she have left? unil has a rope which is (5x − 2y + 2) metres long and is attached to another rope of length 3 S (3x − xy + 5) metres. Find the total length of the rope if x = 5 and y = 2.

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Exponents and Powers

Let’s Recall

We have learnt about large numbers. Large numbers are the numbers that are significantly larger than the numbers we commonly come across in our daily lives. They are used to represent quantities and values on a much larger scale. Large numbers are crucial in various fields such as mathematics, physics, astronomy, computer science, and finance, where extremely large quantities or measurements are involved. We often encounter lakhs and crores in daily conversations. Let us look at some examples. A ‘lakh’ is equal to 10 ten thousands. We often talk about car prices in lakhs. A ‘crore’ is equal to 100 lakhs. When it comes to houses, we usually mention prices in crores.

Price: ₹10,00,000 (Rupees Ten Lakh Only)

Price: ₹1,00,00,000 (Rupees One Crore Only)

Let’s Warm-up Match the following correctly. 1

100 hundreds

10,00,00,000 (Ten Crore)

1000 hundreds

10,00,000 (Ten Lakh)

100 ten thousands

1,00,00,000 (One Crore)

100 lakhs

10,000 (Ten Thousand)

1000 lakhs

1,00,000 (One Lakh)

I scored ___________ out of 5.

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Mean, Median and Mode Exponents Real Life Connect

A celebrity hired a marketing team that promised to double her social media followers every week. In the first week, she only had 2 followers. In the second week, the team helps her to increase the followers to a total of 4. In the third week, the number of followers reached 8. Each week the number of followers increased in a pattern.

Exponential Form The celebrity wants to understand the pattern. First week: 2 followers

Week 1

Second week: 2 × 2 followers = 4 followers

Week 2

Third week: 2 × 2 × 2 followers = 8 followers

Week 3

This means 2 is multiplied by itself 3 times. This can be written as 23 and read as 2 raised to the power of 3.

Power or Exponent

Base 2 is called the base and 3 is the power or index or exponent. Fourth week: 2 × 2 × 2 × 2 followers = 24 = 16 followers Fifth week: 2 × 2 × 2 × 2 × 2 followers = 25 = 32 followers The number of followers she had in the 5th week is 32. This way of writing a number that is multiplied by itself a number of times is called the exponential form or power notation. Let us generalise for number a. When a is multiplied n times by itself, it is written as

a × ··· × a n times

= a

Exponent Base

This is read as a raised to the power of n. A number raised to the power 1 gives the number itself. 21 = 2 31 = 3 101 = 10 and so on. 284

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Product

Exponential Form

10 × 10

102

10 × 10 × 10

10 × 10 × 10 × 10 × 10

Read as

Base and Exponent

10 squared or 10 raised to the power of 2

Base = 10, Exponent = 2

10 cubed or 10 raised to the power of 3

Base = 10, Exponent = 3

10 raised to the power of 5

Base = 10, Exponent = 5

Think and Tell What will be the exponent value of 7n = 7?

A number raised to the power of 2 is called the square of that number, a2. For example, the square of 5 = 52 = 5 × 5 = 25. A number raised to the power of 3 is called the cube of that number, a3. For example, the cube of 5 = 53 = 5 × 5 × 5 = 125. Power of Positive Integers • A positive integer raised to any number is always positive. 22 = 2 × 2 = 4

23 = 2 × 2 × 2 = 8

Positive

3 = 3 × 3 × 3 × 3 = 81 Power of Negative Integers • A negative integer raised to an odd positive power is negative. (–1)odd positive integer = –1 (–1)1 = –1

Negative

(–1)3 = (–1) × (–1) × (–1) = –1 5

(–2) = (–2) × (–2) × (–2) × (–2) × (–2) = –32 • A negative integer raised to an even positive power is positive. (–1)even positive integer = 1 (–1)2 = (–1) × (–1) = 1

(–1)4 = (–1) × (–1) × (–1) × (–1) = 1

Positive

(–3) = (–3) × (–3) × (–3) × (–3) × (–3) × (–3) = 729

Exponents and Prime Factors

216

We already learnt that a prime factor is a natural number whose factors are 1 and itself. We find the prime factors of any given number with the help of the prime factorisation method.

108

For example, the prime factors of 216 are:

2×2×2×3×3×3 Number 2 is multiplied 3 times, hence 2 × 2 × 2 = 23.

Number 3 is multiplied 3 times, hence 3 × 3 × 3 = 33.

Hence, 216 can be written as the product of prime factors in exponential form as 23 × 33. Chapter 17 • Exponents and Powers

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Example 1

Write the base and power of: 1 58 Base = 5 and Power = 8

2 (–24)15

Base = –24 and Power = 15

Power 8

5 Example 2

−24

Base

2 (–a) × (–a) × (–a) × (–a) × (–a) × (–a) × (–a) ×(–a) × (–a) × (–a) × (–a) × (–a)

= (–a)12

Express 525 as the product of prime factors in exponential form.

Example 4

What is the square and cube of 25? Square of 25 = 252 = 625

The prime factorisation of 525 = 3 × 5 × 5 × 7

Cube of 25 = 253 = 15,625

=3×5 ×7 Do It Together

Base

Write the exponential form for: 1 8×8×8×8×8×8

Example 3

Power

Find the prime factors of 768. Express the prime factors using exponents.

Prime factors of 768 = ________________

768 384

Prime factors in exponential form = __________ 48

Comparing Numbers in Exponential Form The numbers in exponential form can be compared by comparing their expanded forms. 3

Let us compare 15 and 5 . 153 = 15 × 15 × 15 = 3375 and 56 = 15,625

As 15,625 > 3375; hence 56 > 153. Example 5

Which of the given numbers is smaller: 75 or 57? 75 = 7 × 7 × 7 × 7 × 7 = 16,807 and 57 = 5 × 5 × 5 × 5 × 5 × 5 × 5 = 78,125 As 16,807 < 78,125; hence 75 < 57.

Do It Together

Compare and fill in the blanks with the correct sign. 1 83 ______ 74

2 105 ________ 97

3 184 ______ 48

4 213 ______ 321

83 = _________

105 = _________

184 = _______

213 = ______

74 = _______

97 = 4,782,969

48 = _______

321 = ______

Powers of Rational Numbers Rational numbers can also be written in the exponential form. p q

If a rational number , is multiplied n times by itself, it is written as: p p p p n pn × × ... n times =   = n q q q q q

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Product

Example 6

Exponential Form

Read as

2 2 × 3 3

22 3  

2 squared or 3

2 Base = , Exponent = 2 3

−5 −5 −5 × × 3 3 3

 −5 3 3  

−5 cubed or 3

Base =

3 3 3 3 3 × × × × 7 7 7 7 7

35 7  

3 raised to the power of 5 7

3 Base = , Exponent = 5 7

2 raised to the power of 2 3

−5 raised to the power of 3 3

7 7 7 7 × × × 5 5 5 5

Write the following in exponential form. 1 1 2187

Find the prime factors of 2187, we get 2187 = 3 × 3 × 3 × 3 × 3 × 3 × 3 1 So, we can write in exponential form as: 2187 1 1 1 1 1 1 1 1 1 7 = × × × × × × =   2187 3 3 3 3 3 3 3  3 

1 −2 3 Simplify and find the value of ×   . 5 3

 1 −2 −2 −2  =    ×   ×   ×   5 3 3 3          1 −8 −8 =   ×   =  5   27  135 Example 10

Example 9

−32 243

Find the prime factors of the numerator and denominator to write them in exponential form. −32 = −2 × −2 × −2 × −2 × −2 243 = 3 × 3 × 3 × 3 × 3 −32 So, we can write in exponential form as: 243 −32 −2 −2 −2 −2 −2 −25 = × × × × =  243 3 3 3 3 3 3 Find the value of a, when 1 3a = 27

3a = 3 × 3 × 3 = 33 3a = 33 So, a = 3.

2 (−5)a = 625

(−5)a = (−5) × (−5) × (−5) × (−5) = (−5)4 (−5)a = (−5)4 So, a = 4.

Find the value of t, when 2 t 8 1   = 5

125

−3 −81 2   = t

 2  = 2 × 2 × 2 = 2 =  2 3     3 5 5 × 5 × 5 5 5 So, t = 3. t

Do It Together

−3 −3 −3 −3 −3 −3 −3 × × × × × × 5 5 5 5 5 5 5

−3 7 =   5

7 4 =   5

Example 8

−5 , Exponent = 3 3

Write the following in exponential form. 1

Example 7

Base and Exponent

7

2401

4  −3  = −3 × −3 × −3 × −3 = −3 =  −3 4     7×7×7×7 74  7  7 So, t = 4. t

2 3 Simplify and find the value of   × (−5)4. 5   2   2   2  =    ×   ×   × [___________________]   5   5   5  Chapter 17 • Exponents and Powers

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 2   2   2   2  =    × ______ ×    × ______ ×    × ______ ×    × ______ 5 5 5 5              2  ____ =   × ______  5   = (_____)4 = _____

Reciprocal of a Rational Number The reciprocal of a number is the multiplicative inverse of that number. If

p q is a non-zero rational number, then its reciprocal will be . q p

p n qn q n And the reciprocal of   will be n =   . p p q n n p q That is, reciprocal of   =   . q   p Example 11

2 4 Find the reciprocal of   . 5 p n q n We know the reciprocal of   =   . q p

Example 12

The reciprocal of –12 will be

−1 . 12

−1 8 The reciprocal of (–12)8 is   .  12 

2 4 5 4 The reciprocal of   is   . 5 2 Do It Together

Find the reciprocal of (–12)8.

−1 2 3 3 Find the reciprocal of   ×   .  3  2  −1 2 × 33 = −1 × −1 × 3 × 3 × 3 = _____     3 3 2 2 2  3  2

−1 2 3 3 The reciprocal of   ×   is _____.  3  2

Do It Yourself 17A 1

Write the base and power of the following: a 911

c n10

Write the following in the exponential form.

3 5 d   8

−7 8 e    11 

a 5 × 5 × 5 × 5

b (–9) × (–9) × (–9) × (–9) × (–9) × (–9) × (–9)

c 7 × 7 × 7 × 8 × 8 × 8 × 8 × 8

d n×n×n×n×n×p×p×p×p×p

e 3

b (–17)9

5 5 5 5 5 × × × × 3 3 3 3 3

 11 15  18   

 −4  ×  −4  ×  −4  ×  −4  ×  −4  ×  −4  ×  −4   11   11   11   11   11   11   11               

Evaluate: a 151

2 3 e   9

b 173

−11 2 f    12 

c (–5)5

3 3 2 4 g   ×   7 3

d 43 × 34

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Express the numbers as products of powers of prime factors.

Find the square and cube of the numbers.

Which is greater?

Find the reciprocal of each of the following.

a 64

b 105

a 8

a 34 or 43

a 26 8

c 480

d 1038

e 1920

b 11

c 18

d 21

e 30

b (–5)4 or 54

23 34 c   or   3   2

−5 2 72 d   or   7   5

b (–3)4

Simplify.

1 4

−1 5 d   2

a 113 × 122

b 103 × 151

d (–12)2 × 102 – 112

 43   32 23 43 32 e   × 5 ×   f   × (−1)3 +   ×    5 8  5    7   3  

32  2 x Find the value of x if   = 5 3125  

c 73 × (–3)4

Word Problem 1

Vivek sells marble balls of different colours. He sold 78,125 marble balls in the month of July. Express this number as a power of 5.

Laws of Exponents Do you remember the celebrity who has a team that helps her to increase her followers? The celebrity fell ill in the 5th week and was not able to track the increase in her followers for the next 3 weeks. Now, she wants to know the number of followers she has altogether.

Multiplying Powers Multiplying powers with the same base We know in the 5th week there were 32 followers: 25 If the same pattern follows, the number of followers in the next three weeks = 25 × 23 = 256 8

We can also find the number of followers in 8 weeks as 2 = 256

...… (1) …… (2)

From (1) and (2), it can be concluded that, 25 × 23 = 2(5 + 3) = 28 Hence, if a is a non-zero number and m and n are whole numbers then, am × an = a(m + n)

Error Alert!

Remember! The base value a should be the same in this case to add the exponents.

Chapter 17 • Exponents and Powers

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Never use this law for addition of two expressions, that is: 32 + 33 ≠ 32 + 3

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Multiplying powers with the same exponents Let us multiply 24 by 34 and find the result. 24 = 16; 34 = 81 So, 24 × 34 = 1296 Also, 64 = 1296. Hence, if a and b are non-zero numbers and m is a whole number then, am × bm = (a × b)m Example 13

Find the value of –54 × –55. Using the law of exponents, −54 × –55 = –5(4 + 5) = –59 = –19,53,125

Example 14

3 5 3 7 Simplify   ×   4  4  3 =   4

5+7

Do It Together

Simplify and express the result in exponential form. −5 11 −5 5 =   ×   7 7

3 =   4

−5 =   7

Dividing Powers Dividing powers with the same base Let us divide 46 by 42 and find the result, 46 ÷ 42 =

4×4×4×4×4×4 = 4 × 4 × 4 × 4 = 44 4×4 46 ÷ 42 = 46 − 2= 44

Hence, if a is a non-zero number and m and n are whole numbers than, am ÷ an = a(m − n) Dividing powers with the same exponents Let us divide 54 by 64 and find the result. 54 ÷ 64 =

5 × 5 × 5 × 5  5 4 =  6 × 6 × 6 × 6 6

Hence, if a and b are non-zero numbers and m is a whole number then, a m am ÷ bm =   b What if the value of both bases and exponents are equal? Let us find out! Let us divide 46 by 46. Using the above laws; 46 ÷ 46 = 4(6 − 6) = 40 Also,

4 4×4×4×4×4×4 = 1 6 = 4 4×4×4×4×4×4

…… (1) …… (2)

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From 1 and 2, we can conclude that if a is a non-zero number, then a0 = 1

These are called zero exponent numbers. Example 15

3 5 3 3 Simplify and find the value of   ÷   . 4 4 3 5 3 3 3 (5 − 3)  3 2 3 × 3 9 =   ÷   =   =  = = 4 4 4  4  4 × 4 16

Example 16

2 3 Simplify and find the value of (−3)3 ÷   . 5

 −3 × 53  −15 3 −3375 2 3 −33 (−3)3 ÷   = 3 = =   =   = 2 8 5  2   2  5

Do It Together

Simplify and find the value of: −3 6 −3 4 1   ÷   11

−2 5 8 5 2   ÷  

11

3

 −3 6 ÷  −3 4 = ________     11 11

9

 −2 5 ÷  8 5 = ________      3  9

Taking Power of a Power

Let us simplify (53)2 and find the result. 53 = 5 × 5 × 5 = 125 (53)2 = 1252 = 15,625 Also, 56 = 15,625 Hence, if a is a non-zero number and m and n are whole numbers, then (am)n = amn Example 17

Simplify (32)3.

Example 18

Simplify [(k)4]2. = (k)(4 × 2)

Here, 32 is raised to the power of 3.

= (k)8

= 32 × 32 × 32 = 3(2 + 2 + 2) = 36 Example 19

Find the value of n, when 2 5

 2  2 n+5 1    =   5  5    2 2 × 5 =  2 n + 5     5 5 10 2 2 n+5 =   =   5 5 10 = n + 5

−5 2n + 1  −5 n + 6 2   =  7 7 2n + 1 = n + 6 2n − n = 6 − 1 n=5 So, n = 5.

Do It Together

−1 5 2 Simplify and find the value of    .  2   5× −1 =   2 −1 =   2 1 =

n = 10 − 5

So, n = 5.

Chapter 17 • Exponents and Powers

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Simplifying Exponential Expressions We have learnt about the laws of exponents. Let us now learn to simplify expressions using the same laws. Let us simplify

((62)3 × 69) . 65

By using the law of multiplication: Using the law of division: Example 20

Simplify:

38 × t7 92 × t3

((62)3 × 69) (66 × 69) 615 = = 5 65 65 6

615 = 6(15 − 5) = 610 65

38 × t7 38 × t7 38 × t7 = = 92 × t3 (32)2 × t3 34 × t3

(Using the law: (am)n = amn)

38 × t7 = 38 − 4 × t7 − 3 = 34 × t4 34 × t3

34 × t4 = (3t)4 = 81t4 Example 21

Simplify:

(Using the law: am ÷ an = am − n )

(Using the law: am × bm = (a × b)m )

5 × 10 × 12 600 × 8

Do It Together

53 × 104 × 12 53 × (5 × 2)4 × 4 × 3 53 × 54 × 24 × 22 × 3 = = 600 × 8 3 × 2 × 100 × 8 3 × 2 × 52 × 22 × 23 57 × 26 × 3 (Since am × an = am + n) 3 × 52 × 26 am = 57 − 2 × 26 − 6 × 31 − 1 (Since n = am − n) a 5 0 0 =5 ×2 ×3 =

Simplify:

36 × 6 × a × b 216 × a4 × b2

36 = 6 × 6 = 62 and 216 = 6 × 6 × 6 = 63 36 × 64 × a7 × b = ______________ 216 × a4 × b2 Using the law: xa × xb = xa + b ________________ = ______________ Using the law: xa ÷ xb = xa − b

= 5 × 1 × 1 = 5 (Since a = 1)

________________ = ______________

Do It Yourself 17B 1

Simplify using the laws of multiplication of exponents. a 105 × 109

b (–5)2 × (–5)4 × (–5)6

72 75 c   ×   9 9

−2 3 −2 6 −2 9 d   ×   ×   5 5 5

e 24 × 54

f (–9)3 × (–1)3

−410 310 g   ×   3 4

h (−5)3 × 

Solve using the laws of division of exponents and fill in the blanks.

−9 3  2 3  ×   5  3

a 1315 ÷ 139 = __________

b (–11)4 ÷ (–11)1 = __________

−49 −4 3 c   ÷   = __________ 7 7

d 57 ÷ 37 = __________

2 12 3 12 e   ÷   = __________ 5 7

f (–15)8 ÷ (5)8 = __________

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Simplify and express in exponential form. a ((9)2)2

b ((–12)1)3

3 42 c     2  

–1 33 d     8  

b ((7)2)0

3 0 c (−5)0 ×   8

x 0 y 0 d   ÷   y x

Evaluate: a 40

Find the value of (20 + 30) × (60 + 71).

Solve and match with the solution. a 24 × 23

−27

b (−3) ÷ (−3)

128

c (−5)4 × (−4)4

−17

d 50 × (−17)1

1,60,000

Find the value of p for which 6p × 63 = 362.

Simplify: a

212 × m8 (2m)6

62 × 243 44 × 35

255 × 498 × 1213 54 × 75 × 114

(5ab)4 25ab2 × 25a2b

Find the value of n in each of the following.  −7 26 −7 n + 5 a 9 × 3n + 2 = 243 b    =    13   13 



Word Problem 1

Ronita bought 28 kg of rice for ₹42 per kg, while Suman bought 14 kg of rice for ₹63 per kg. Express the product of the amounts spent by both the girls as a product of prime factors only in exponential form.

Expressing Large Numbers in the Standard Form With the help of the social marketing team, the celebrity is able to increase her followers in the same pattern. The number of followers she will have in 10 weeks is 1,024. We can write 1024 as 1.024 × 1000 = 1.024 × 103. We can write 10,48,000 as 1.048 × 10,00,000 = 1.048 × 106. A number in standard form will be K × 10n, where K is a number between 1 and 10 and n is an integer. Example 22

Express 8980000000 in standard form. 8980000000 = 8.98 × 1000000000 = 8.98 × 109

Do It Together

Example 23

Write 7.05 × 106 in usual form. 7.05 × 106 = 7.05 × 1000000 = 7050000

Express 604000000 in standard form. 604000000 = _____ × _______ = _____ × 10 Chapter 17 • Exponents and Powers

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Do It Yourself 17C 1

Express the number in standard form. a 864000

b 5500000

c 4152000000

d 614000000000

c 7.11 × 107

d 9.8 × 109

Write the number in the usual form. a 3 × 104

b 4.55 × 105

What will be the power of 10 when 40,075 is expressed in standard form?

If 8.3279 × 10n = 8,32,79,000, what is the value of n?

Find the number from the following expanded form. Write it in exponential form.

What is the value of 6,43,298 in expanded exponential form?

Express the numbers in the statements in standard form.

7 × 106 + 5 × 105 + 4 × 103 + 6 × 102 + 8 × 101 + 2 × 100

a The distance between India and the United States is 13568000 m.

b The population of India was around 1410000000 in 2021.

c The distance between Venus and Uranus is around 2760000000 km.

d The diameter of Jupiter is around 140000000 m.

Word Problem 1

Sumit read that the earth is around 4.54 × 109 years old. Express the number in the usual form.

Points to Remember • A number multiplied by itself a number of times is called the exponential form or power notation an, a

raised to the power of n.

• If a is a non-zero number and m and n are whole numbers, then am × an = a(m + n). • If a and b are non-zero numbers and m is a whole number, then am × bm = (ab)m.

• If a is a non-zero number, m and n are whole numbers and m > n then am ÷ an = a(m – n). am  a m • If a and b are non-zero numbers and m is a whole number, then am ÷ bm = m =   . b b • If a is a non-zero number, and m and n are whole numbers, then (am)n = amn. • If a is a non-zero number, a raised to the power 0 is 1: a0 = 1.

• A number in standard form will be K × 10n, where K is a number between 1 and 10 and n is an integer.

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Math Lab Exponent Bingo Setting: In groups of 5 Materials Needed: 3

•

Exponent cards (2 , 5 , 10 )

•

Bingo cards with answers to the exponent

•

cards written

512 0 8

100

67 35

Markers or chips

Method: •

Create exponent cards and bingo cards

•

Distribute the bingo cards and markers or chips among the students.

•

Draw the exponent cards one by one and call out the expression.

•

Students find the answer to the exponent and cross out the corresponding answer given

•

on the bingo card.

The first student to complete a row or column shouts “Bingo!” and is the winner.

Chapter Checkup 1

Express 7,776 in exponential form with base 6.

Write the value of the following:

a 123

b 103 × 103

e (–22)0

13 0 f    20 

d (–1)9

c (2p − q)3

d (p − 2q)2

Consider: p = 3 & q = 4 and find: a (p + q)4

c (7 × 2)4

b (p − q)2

Write the following in exponential form. a 8×8×8×8×8×8 b (–15) × (–15) × (–15) × (–15) × (–15) × (–15) × (–15) × (–15) c (–m) × (–m) × (–m) × (–m) × n × n × n × n × n × n × n d 2×2×2×2×a×a×a×a×a×b×b×b×b×b×b×b×b e

3 3 3 3 3 3 3 3 3 × × × × × × × × 7 7 7 7 7 7 7 7 7

−2  −2  −2  −2  −2  −2  4 4 4 4 4 4 4 f   ×   ×   ×   ×   ×   ×   ×   ×   ×   ×   ×   ×   11 11 11 11 11 11 15 15 15 15 15 15 15                          

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Express the numbers as the product of prime factors in exponential form. a 256

b 582

d 2050

e 3418

Compare and place the correct sign. a 7.2 × 102

c 1090

3.8 × 103

b 5.8 × 105

3.6 × 108

c 6.9 × 108

4.5 × 109

Simplify: a

153 × (−2)4 20

n23 × (−1)221

c (170 – 230) × (170 + 230)

100 × n

24 4 23 4 d   ×    23   12 

a  2 2  2  a 3 =   ×  , find the value of   . b 5 5 b

Write the following in standard form. a 4750000

b 70000000

c 12 million

d 289 crore

10 Match the numbers with their other forms. a 9.1 × 106

1.3 × 107

43600000000

b 70600 10

c 4.36 × 10

9100000

d 13000000

7.06 × 104

11 Find the value of n. a (–1)3 × (–1)9 = (–1)n

b 3n + 1 × 81 = (3)8

c 52n + 1 ÷ 25 = (5)3

−3 3 −3 21 −3 3n d   ×   =    11   11   11 

5 43 5 2n e    =    3    3 

−12 62  −12 n + 5 f    =   19    19 

Word Problems 1

According to the census, the male population of a country in 2011 was 623700000. Express this

Find the distance in centimetres for the aircraft moving at the speed of 690 km/hr and covering

in standard form.

the distance in 3 hours 20 minutes.

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18 Symmetry Let’s Recall Look at the given shapes. What do you observe? Do they have anything in common in relation to symmetry? All of these objects exhibit symmetry. Let us understand when a shape exhibits symmetry. When a shape/design cut from a sheet of paper is folded once in such a way that one half of the shape completely overlaps the other, the shape is said to exhibit symmetry and is symmetrical. All parts of the symmetrical figures, objects, and shapes are the same in shape and size. For example, the design on the paper is obtained by folding and cutting triangles. On unfolding the paper, both sides have the same design. So, the design on one side of the paper is symmetrical to the other side of the paper. The line along which you fold a figure to obtain the two equal parts is called the line (axis) of symmetry. The line of symmetry can be vertical, horizontal, and diagonal. One shape can have more than one line of symmetry.

Vertical line of symmetry

Horizontal line of symmetry

Both horizontal and vertical lines of symmetry

Multiple lines of symmetry

Let’s Warm-up Match the following. 1

Human face

Two lines of symmetry

Letter H

One line of symmetry

Plus sign

Infinite lines of symmetry

Sun

Four lines of symmetry

I scored _________ out of 4.

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Mean, Median and Mode Understanding Symmetry Real Life Connect

Sakshi went to a summer carnival. There were many games, swings, rides, and a circus show. She bought a ticket from the ticket counter and entered the carnival. She then went to her favourite rides—the merry-go-round and the Ferris wheel. She concluded her day with the circus show.

Reflection Symmetry Let us look at the ticket counter closely. Is it symmetrical in any way? How can you say? When we insert a line, one half of the ticket counter is exactly the same as the other half. These halves are nothing but flipped images of one another. So, we can say that these flipped images have Reflection Symmetry, and the figure/object is reflectively symmetrical about the given line. The line which divides the figure into two symmetrical parts is called the line of symmetry.

Think and Tell What other structures in the carnival have a line of symmetry? Draw their lines of symmetry.

Remember! Reflection symmetry exists if there is at least one line which divides a figure into halves such that one half is the mirror image of the other half.

We also observe reflection symmetry through mirror reflections. If a plain mirror is placed along/in the centre of the object, the reflection of the object is visible in the mirror. The image looks identical to the object. Therefore, the mirror acts as the line of symmetry between the image and the object. Let us see this through a real-life example. Look at the given picture of Mount Taranaki in New Zealand. What do you see? 298

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The reflection of the mountain in the still water looks exactly the same in size and shape as the mountain itself. The surface of the water is the line of symmetry in the picture.

One Line of Symmetry There are three types of lines of symmetry: horizontal line of symmetry, vertical line of symmetry, and diagonal line of symmetry. Let us look at some examples with different types of lines of symmetry. Vertical line of symmetry

Horizontal line of symmetry

Diagonal line of symmetry

Many figures/objects do not have any lines of symmetry. Such figures/objects are said to be asymmetrical. Look at some of the examples given below.

Example 1

Look at the given figures. Answer the given questions. 1 Which figure is symmetrical?

2 Draw the line of symmetry on the symmetrical figure. 3 What type of line of symmetry does it have? Chapter 18 • Symmetry

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Figure (a) is not symmetrical, but figure (b) is symmetrical about the diagonal line of symmetry. a

No line of symmetry

Example 2

Complete the given patterns using the dotted line as the mirror line. 1

Do It Together

Diagonal line of symmetry

Look at the given figures. Answer the following questions.

1 Are the given figures symmetrical?

2 Draw lines of symmetry if they are symmetrical. 3 What type of line of symmetry do they have?

Figure (a) is symmetrical about the vertical line of symmetry, but figure (b) is ____________ about the ____________ line of symmetry, as shown below. a

More than One Line of Symmetry Some real-life objects, shapes, letters, and numbers have more than one line of symmetry. They may have two, three, four, or more lines of symmetry. Some of the examples are:

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Two Lines of Symmetry

Three Lines of Symmetry

Four Lines of Symmetry

More than Four Lines of Symmetry

Think and Tell How many lines of symmetry are there in the Ferris wheel?

Example 3

Look at the given figures. Answer the given questions.

1 Are the given figures symmetrical? 2 How many lines of symmetry do they have? 3 Draw lines of symmetry on each of the given figures.

Both figures are symmetrical. Figure (a) is symmetrical about each of its diagonals, and figure (b) is symmetrical about each of the meeting points of the circle and the star. The lines of symmetry are shown here for each of the figures.

Example 4

2 lines of symmetry

6 lines of symmetry

Shade as many squares as required such that the given figure is symmetrical about the given lines of symmetry. 1

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Do It Together

Look at the given figures. Answer the given questions. 1

Are the given figures symmetrical? ___________________________________________ .

b How many lines of symmetry do they have? c

Draw the lines of symmetry in each of the given figures.

____________ lines of symmetry

Error Alert! While finding the diagonal line of symmetry, do NOT change the direction of the shape. Keep the shape static, and then mark the diagonal line of symmetry.

On rotation, this shape has a diagonal line of symmetry.

Originally, this shape had a vertical line of symmetry.

Do It Yourself 18A 1

Tick ( ) the figures that show reflection symmetry. a

Is the given dotted line in the figures a line of symmetry? If not, then draw the figure with the correct line of symmetry. a

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Identify and draw the lines of symmetry in the given figures. a

Complete the given figures such that they are symmetrical about the given mirror line of symmetry. a

Draw two ways in which the given number of squares can be shaded along the diagonal line so that the given grid has symmetry about the given line. a two squares

b three squares

How would you shade the squares away from the diagonal line? 6

Which letters of the word ‘MATHEMATICS’ have no line of symmetry?

On the given grid, draw a quadrilateral with an area of 13 square units which has exactly one line of symmetry.

Shade the squares such that the given figures are symmetrical about the given dotted line of symmetry. a

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Word Problem 1

In a park, there is a flowerbed in the shape of a regular hexagon. There is a semi-circular flowerbed on the outside of each edge of the hexagon. a Draw a rough sketch of the flowerbed. b Is the figure symmetrical? If yes, draw the lines of symmetry.

Rotational Symmetry Remember Sakshi enjoyed swinging on the Ferris wheel at the carnival? Let us look at it closely. The Ferris wheel has passenger cabins in four different colours: red, yellow, blue and green. These cabins are placed opposite to each other and have the same colour. For example, the two red cabins shown in the picture here are placed opposite each other. When the wheel rotates about its centre in a particular direction (clockwise or anticlockwise), such that the cabins of the same colour appear to be opposite to each other, and the Ferris wheel looks exactly the same as before. This appears after every half turn, that is, 180°. This is rotational symmetry.

Half turn (180°) A

So, we can say that rotational symmetry is a type of symmetry exhibited by an object when it can be rotated around a central point by a certain angle in a direction and still appear unchanged at specific intervals during a full rotation. The central point around which the object rotates is called the centre of rotation. For example, look at the given ceiling fan. When it is turned on, the blades OA, OB, and OC of the fan start rotating in the clockwise direction, at an angle of 120°. The rotation takes place about the point O. So, O is the centre of rotation.

120°

Order and Angle of Rotational Symmetry Every figure/object that has rotational symmetry looks exactly the same at some point during the 360° rotation. The number of times the figure looks the same as the original during the rotation of 360° is called the order of rotational symmetry, and the minimum measure of the angle through which a figure/an object has to be rotated to get the original figure is called the angle of rotation.

Did You Know? Cyclones and hurricanes rotate either in a clockwise or anticlockwise direction, depending on their geographical location.

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Let us look at some figures. D

1 A 3

C B

We can find their order of rotation and angle of rotation in the given way.

Rotation through 90°

Rotation through 180°

Rotation through 270°

Rotation through 360°

The letter B fits onto itself only once when rotated through 360°. It has a symmetry of order 1 with an angle of rotation of 360° (360° ÷ 1). Such figures do not show rotational symmetry since rotational symmetry of order 1 does not exist.

Rotation through 90°

Rotation through 180°

Rotation through 270°

Rotation through 360°

The letter H fits onto itself only twice when rotated through 360°. So, it has a rotational symmetry of order 2 with an angle of rotation of 180° (360° ÷ 2). 1

Rotation through 120°

Rotation through 240°

Rotation through 360°

Here, the equilateral triangle is a regular polygon. The order of rotational symmetry of a regular polygon is always the same as its number of sides. As the equilateral triangle is rotated through 120° thrice, the order of rotational symmetry of the equilateral triangle is the same as its number of sides, that is, 3, and the angle of rotation is 120° (360° ÷ 3). Here, the windmill is rotated 90° four times. Here, the order of rotational symmetry is 4 and the angle of rotation is 90° (360° ÷ 4). D

A Rotation through 90°

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A A

D Rotation through 180°

D D

C Rotation through 270°

C B

Rotation through 360°

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Therefore, we can say that if we know any one of the following, we can calculate the other on its basis. • Order of rotational symmetry • Angle of rotational symmetry For example, we know the order of rotational symmetry of the equilateral triangle is 3. So, we can calculate 360° the angle of rotational symmetry of the equilateral triangle as . This means, order of rotational symmetry the angle of rotational symmetry of the equilateral triangle is 360° ÷ 3 = 120°.

Remember! Angle of rotational symmetry =

Think and Tell

360° order of rotational symmetry

What is the order and angle of rotational symmetry of

360° Order of rotational symmetry = Angle of rotational symmetry

Example 5

the Ferris wheel?

Write the order of rotational symmetry for each of the given figures. 1 1800

600

he angle of rotational symmetry of the image T is 180°. So, the order of rotational symmetry 360° = Angle of rotational symmetry 360°

180°

So, the order of rotational symmetry 360° = Angle of rotational symmetry =

Thus, the shape will fit onto itself two times in a complete revolution.

Example 6

Angle of rotational symmetry of the image is 60°.

360°

=6 60° Thus, the shape will fit onto itself six times in a complete revolution.

Write the order and angle of rotational symmetry of the given figure.

Rotation through 90°

Rotation through 180°

Rotation through 270°

Rotation through 360°

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This figure fits onto itself four times in a complete revolution. So, the order of rotational symmetry is 4. Now, the angle of rotational symmetry =

Do It Together

360°

order of rotational symmetry

360° 4

= 90°.

Write the order and angle of rotation of the given figure.

Rotation through ____________

This figure fits onto itself _______ times in a complete revolution. So, the order of rotational symmetry is ______. 360° 360° Now, the angle of rotational symmetry = = = ________. order of rotational symmetry ____

Both Reflection and Rotational Symmetry Some figures have both rotational symmetry as well as reflection symmetry. The table given below shows some examples of such figures. Shapes

Reflection Symmetry

Rotational Symmetry 1

Isosceles Trapezium

One line of symmetry

Rotational Symmetry of order 1 does not exist. 2

Rectangle Two lines of symmetry

Rotational symmetry of order 2 3

Equilateral Triangle 2

Three lines of symmetry

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Rotational symmetry of order 3

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Shapes

Reflection Symmetry

Rotational Symmetry 3

Square

Four lines of symmetry

Rotational symmetry of order 4 5 4

Pentagon

Five lines of symmetry

Rotational symmetry of order 5

Infinite lines of symmetry

Rotational symmetry of infinite order

Circle

Here, notice that the square, equilateral triangle, and pentagon have the same order of rotational symmetry and lines of symmetry as the number of sides in the shapes. This is because they are regular polygons.

Remember! In every regular polygon, the order of rotational symmetry and the number of lines of symmetry are the same as the number of sides in the shape.

Example 7

Write the number of lines of symmetry and the order of rotational symmetry of the given figure. Also, draw the lines of symmetry in the given figure. The given figure shows reflective symmetry about the diagonals and the central point. So, the number of lines of symmetry = 4 The given figure is rotationally symmetrical after every quarter turn (90°). So, the order of rotational symmetry = 4

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Example 8

What is the minimum number of squares that can be shaded to get a shape with four lines of symmetry and rotational symmetry of order 4? Draw and show your answer. The minimum number of squares that can be shaded to get a shape with four lines of symmetry and a rotational symmetry of order 4 is 8, as shown below.

Do It Together

Write the number of lines of symmetry and the order of rotational symmetry of the given figure. Also, draw the lines of symmetry in the given figure. This is a regular octagon. _________ Number of lines of symmetry = _________ Order of rotational symmetry = number of lines of symmetry = _______

Do It Yourself 18B 1

Classify the given figures as rotationally symmetrical with order more than 1 and rotationally symmetrical with order equal to 1. a

Given the angle of rotation, find the order of rotational symmetry. a

720

1200

Look at the given figures. Answer the questions. a

900

1800

Mark the centre of rotation in each of the given figures.

ii If each figure is turned up to 360° about the centre point, then how many times will the figure match the original?

iii Find the angle of rotation of the given figures.

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What letters of the English alphabet have: a both lines of symmetry and rotational symmetry?

b rotational symmetry but no reflection symmetry?

c reflection symmetry but no rotational symmetry?

d neither reflection symmetry nor rotational symmetry?

Explain your answer by drawing line(s) of symmetry and rotational symmetry for at least 1 of the letters. 5

Draw a shape with: a Rotational symmetry and reflective symmetry b Rotational symmetry but no reflective symmetry c Reflective symmetry but no rotational symmetry

image

d Rotational symmetry of order 2 but no reflective symmetry 6

Esha rotated the given shape as shown. a Did she rotate it clockwise or anticlockwise? b What degree of rotation did she make?

center of rotation

c Did the figures look identical after rotation? If yes, find the angle of rotation. d What is the order of rotation of the shape? 7

Look at the given grids. Some squares are shaded in each of the grids. Answer the given questions.

Figure 2

Figure 1

a Write the order of rotational symmetry and angle of rotation for the given figures. b Do these figures show reflection symmetry? If yes, draw the lines of symmetry. If no, shade the squares of the grid to make them symmetrical.

Colour the parts of the given figure in such a way that the figure has the given order of rotational symmetry.

Order: 2

Order: 3

Order: 4

Word Problem 1

Anu wants to make rangoli on Diwali. She doesn’t know how to make them, but she knows about rotational symmetry. She uses her knowledge and

the help of her friends to draw the following rangoli designs. Look at these designs and answer the following questions.

a Find the lines of symmetry in each of the given designs. b Write the order of rotational symmetry for each of these designs. Use it to find the angle of rotation as well.

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Points to Remember • When an object is flipped, the size of the image is the same, but the flipped image of the object is exactly the opposite of the original. So, in short, reflection means ‘to flip’. • An object has reflection symmetry if there is at least one line which splits it in half so that the portions on either side of the line are identical to each other. This line is called the line of symmetry or axis of symmetry. • Objects may rotate either in a clockwise or anticlockwise direction around a fixed point. So, in short, rotation means ‘to turn’. • If an object rotates around a fixed point at a certain angle and it looks exactly the same after rotation, then the object is said to have rotational symmetry. This fixed point is called the centre of rotation. • The number of times the figure coincides with itself during the rotation of 360° is called the order of rotation. • The minimum measure of the angle through which a figure has to be rotated to get the original figure is called the angle of rotation.

Math Lab Setting: In groups of 2

Reflections and Rotations!

Materials Required: Origami sheets, a pencil, a ruler, an eraser, a pair of scissors, pushpins or thumbtacks, small beads or buttons, a stick Method: Both the members of each group must follow these steps. 1 Each group must draw and cut out a regular polygon and an irregular polygon from a sheet of paper. Using pushpins or thumbtacks, mark the centre of rotation in the centre of the shapes.

2 Place a small bead or button onto the pin from the back of the shapes, securing it in place. Then,

attach the shapes to a stick by pushing the pin through the centre of the shapes onto the top end of a wooden stick or a pencil eraser. Make sure the shapes are securely attached but still allow them to spin freely.

3 Hold the stick or eraser firmly and gently blow on the shapes, or spin it with your hand to see it rotate.

4 Rotate the shapes to see if they exhibit rotational symmetry. Write the order and angle

of rotational symmetry for each of the shapes. Also, identify if the shapes are reflectively symmetrical and write the number of lines of symmetry.

5 You may repeat the activity with other shapes as well, if time permits.

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Chapter Checkup 1

Is the given dotted line in the given figures a line of symmetry? Draw the correct line of symmetry, wherever required. a

Identify and draw lines of symmetry in the given figures. a

Complete the given figures such that they are symmetrical about the given line(s) of symmetry (mirror line). a

Complete the given table. Use the table to find the order of rotation and angle of rotation for each of the given figures. S. No.

Figure

Rotation by 90°

Rotation by 180°

Rotation by 270°

Rotation by 360°

Number of times a shape looks the same

a b c d e

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Look at each letter of the English alphabet. Some letters have one line of symmetry, and some letters have more than one line of symmetry. A

Find: a at least 5 letters that have a horizontal line of symmetry. b at least 5 letters that have a vertical line of symmetry. c at least 2 letters that have both a vertical and a horizontal line of symmetry. 6

Name and draw. a A triangle that has no rotational symmetry b A quadrilateral that has no reflection symmetry c A triangle that has only 2 lines of symmetry d A polygon that has both reflection symmetry and rotational symmetry of order 9 e A quadrilateral with reflection symmetry but no rotational symmetry

Which digits of the mobile number ‘8432657910’ have both reflective and rotational symmetry?

Satya has drawn an incorrect mirror image of the object. Identify the error and correct the image.

Object Mirror image

If the given shape is turned 270° clockwise and then 45° anticlockwise, then find the new position of point C.

10 Shade the squares such that the given figures are symmetrical about the lines of

B D

symmetry and are also rotationally symmetrical. a

11 Shade the squares such that the given figures have rotational symmetry of the order stated below. a

Order 2

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Order 4

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12 This clock has been reflected in a mirror. What time is it?

13 Which of the given faces of the dice has less than three lines of symmetry?

14 What section of these circular grids should be shaded such that the overall shape in the first grid has one line of symmetry and the second grid has a rotational symmetry of order two? Make sure you mark your line of symmetry.

Reflective symmetry

Rotational symmetry of order 2

15 What is the minimum number of squares that can be added to the given shape such that the final shape has a rotational symmetry of order 4?

Word Problem 1

Seema made a pattern by joining regular hexagons as her art and craft activity. She drew lines passing through the centre of the regular hexagons. A part of the pattern with some shapes shaded is given below.

a Which shape is a vertical reflection of shape P? b How many lines of symmetry does each shape have? c What is the order of rotational symmetry for each shape?

QR T

S U

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Visualizing Solid Shapes

19 Let’s Recall

We see shapes all around us. Shapes can be 2-D or 3-D. A 2-D shape has only two dimensions and a 3-D shape has three dimensions. The basic unit of any shape or figure is a line. A line can either be straight or curved. In this picture, we can see shapes, lines and curves. We also know that we can see lines, curves or 2-D shapes in 3-D shapes. For example, a triangular prism has 2 triangular faces and 3 rectangular faces. The edges of the shapes are straight lines. There are no curves in a triangular prism. Let us now recall some solid shapes that we have learnt about. Cube

Cuboid

Cylinder

Cone

Sphere

Square Pyramid

Hexagonal Prism

Pentagonal Prism

Tetrahedron

Dodecahedron

Let’s Warm-up A packaging company has boxes in different shapes. Name the shapes of these boxes. 1

______________________________ 3

______________________________ 4

______________________________

______________________________ I scored ____________ out of 4.

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3-D Shapes and 2-D Shapes Real Life Connect

Yashvi and her sister Priya were playing with building blocks. Yashvi noticed how the shape of one block was different from another. Yashvi: Priya, have you noticed the shapes of these blocks we are playing with? Priya: Yes! Some blocks are square and some are triangular in shape. Yashvi: Well, these are not exactly squares and triangles. These are called three-dimensional shapes. They have different features. Priya: What are these features? Yashvi: Let us find out!

Features of 3-D Shapes Shapes or figures can be categorised into one-dimensional, two-dimensional and three-dimensional shapes depending on their dimensions and properties. But what are one-dimensional, two-dimensional and three-dimensional shapes or figures?

Remember! Some figures have no dimensions and are called 0-D figures. For example, a point is a 0-D shape. Here A, B, C are points.

One-dimensional figures are shapes that exist only in one dimension, which is typically length.

Examples of 1-D figures include lines, line segments, and curves that have no width or height but only length. Two-dimensional figures are shapes that exist in two dimensions, which are length and breadth. These shapes have a flat surface, like a plane, and do not have depth. For example, squares, rectangles, circles or rectangles. Three-dimensional figures are shapes or objects that exist in three dimensions, which are length, breadth and height. Unlike 2-D figures, 3-D figures have volume and occupy space in three dimensions. In a 3-D shape, every individual flat or curved surface is called a face. The line segment where two faces meet is called an edge. The points where two or more edges meet is called a vertex.

Vertex Face

Edge

Examples of 3-D figures include cubes, cuboids, cones, spheres, cylinders, pyramids and prisms. Let us see the features of some 3-D shapes. 316

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Think and Tell

Remember! A tetrahedron is a type of pyramid. In the word, tetra means four

Is a cone also a pyramid?

and hedra means face. So, it is a pyramid with 4 triangular faces.

We already know the features of a cube, cuboid, cone, sphere and cylinder. Let us differentiate between prisms and pyramids. Now, a prism and a pyramid are both three-dimensional shapes, but they differ in a few key ways. PRISM

PYRAMID

Apex

A prism is a 3-D solid that has two bases that are polygonal in shape and rectangular sides perpendicular to the base. There are two bases in a prism that are joined by the edges. The sides or faces are always rectangular in shape and perpendicular to the bases.

A pyramid is a 3-D solid that has one polygonal base and triangular sides. There is only one base in a pyramid. The sides are triangular and they meet at a point at the top called an apex.

There is no apex in a prism. A glass prism is an example, which is used in science experiments, like splitting a ray of white light.

The pyramids in Egypt, such as the Great Pyramid of Giza, are iconic examples of pyramid-shaped structures.

Think and Tell Remember! Solid shapes with straight edges and flat sides are

Can we say a cylinder is a circular prism?

called polyhedrons. Example 1

Write the number of faces, edges and vertices in a square pyramid. Write the type of faces. A square pyramid has 5 faces, 8 edges and 5 vertices. It has 1 square face and 4 triangular faces.

Example 2

Name the given 3-D shapes. How are these two shapes different? 1

1 is a hexagonal prism and 2 is a hexagonal pyramid. The lateral faces of a hexagonal prism are rectangular and the lateral faces of a hexagonal pyramid are triangular. Chapter 19 • Visualizing Solid Shapes

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Do It Together

Fill in the blank spaces.

Solid Faces

Vertices

Edges

Do It Yourself 19A 1

Fill in the blanks. a The solid that has only one vertex is a _________. b If three cubes are joined side by side, the shape obtained is a _________. c The number of edges in a football is _________. d A square prism is also called a _________. e The polygon region of a solid shape is called a _________.

How many dimensions does a solid have?

Name the prisms or pyramids. a

Write down the number of edges and vertices in the 3-D shapes. a Sphere

b Cone

c Tetrahedron

d Triangular prism

A solid has one curved and two flat faces. Name the 3-D shape.

If a certain number of ₹2 coins are placed one on top of another, what shape will be obtained?

Name two 3-D shapes in which the number of faces, vertices and edges of one shape are equal to the number of

How is an octagonal prism different from an octagonal pyramid? How are the shapes similar?

faces, vertices and edges of the other.

Word Problem 1

Shreya formed a 3-D shape with 4 congruent triangles joining at a point. What 3-D shape did she form?

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Nets of 3-D Shapes Yashvi and Priya got a gift from their mother in the evening. Yashvi: Look Priya, how lovely the gift wrapping is! Priya: Yes Yashvi! We can easily make these wrappings that open like a box! Let us open it and see! The shape of the gift box is a triangular pyramid. When the box is opened, 4 triangles can be seen. This is called the net of the shape. The faces of a solid shape tell us the number of flat or 2-D shapes in their net. The net of a shape is a term used to describe what a 3-D shape would look like if it was opened out and laid flat. Nets look like paper cutouts or frameworks of shapes, and when we fold them, they become real 3-D objects. Let us see the nets of some prisms. A prism has 2 opposite identical and parallel bases that are polygonal. Its net has 2 polygonal bases and the same number of rectangular faces as the edges of the polygonal bases. The exception is the cube where all 6 faces are square. Triangular Prism

Cube

Cuboid

Hexagonal Prism

2 triangular faces

6 square faces

2 square faces

2 hexagonal faces

4 rectangular faces

6 rectangular faces

3 rectangular faces

Remember! The opposite faces of a prism are identical. So, in a net the opposite faces will also be identical.

Given below are the nets of some pyramids. A pyramid has one polygonal base and the same number of triangular faces as the sides of the polygonal base. Rectangular Pyramid

Pentagonal Pyramid

Hexagonal Pyramid

1 rectangular face

1 pentagonal face

1 hexagonal face

4 triangular faces

5 triangular faces

6 triangular faces

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Let us also see the nets of shapes that have curved faces. Cone

A cone has a circular base and one curved surface pointed towards the top, called the apex.

Cylinder

A cylinder has a curved surface that is rectangular in shape and two parallel circular faces.

The net of a cone is a full circle for the base, and a part of The net of a cylinder is formed of two circles, with each another circle for the wall of the cone. attached to one pair of opposite sides of the rectangle.

Example 3

Is the given figure a net of a cube? Why? A cube is a 3-D solid object with six square faces and all the sides of a cube are of the same length.

TOP SIDE

The figure has 6 squares and on folding it will form a cube. Example 4

SIDE BASE SIDE SIDE

Which of these is a net for a square pyramid? 1

Option 1 can’t be folded into a square pyramid as it will result in a pyramid that is open on one side. But option 2 can be folded. Therefore, option b is the correct answer.

Example 5

Draw the nets of an octagonal prism and a triangular pyramid. Octagonal Prism

Do It Together

Triangular Pyramid

List the faces of a heptagonal prism. Draw the net of the shape. Number of faces = _________ Net of a heptagonal prism 2 _____________________ faces ___________ rectangular faces

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Do It Yourself 19B 1

List the number of faces and the type of faces for each of these 3-D shapes. a Octagonal Prism

b Nonagonal Pyramid

c Rectangular Prism

d Cone

Match the solids with their correct nets. a

Look at the nets. Name the 3-D shape for each net. a

Draw the nets of a cone and a triangular pyramid to show how they are different.

Draw the nets of the shapes. a Pentagonal Prism

b Hexagonal Pyramid

Name the different shapes needed to draw the net of a triangular prism.

Do a cube and a square prism have the same nets?

A cylinder has a diameter of 14 cm and length of 15 cm.

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15 cm

44 cm

14 cm

44 cm

15 cm

Which of the following is the correct net for the cylinder?

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Draw the net of a cube that has a side of 4 cm and label all the dimensions.

10 This is the net of a triangular prism. Label the base. 11 Draw the net of a triangular pyramid of which the dimensions of a triangle’s edges are 5 cm × 5 cm × 5 cm.

Word Problem 1

Seema has two bangles and a sheet of paper. If she rolls the paper and attaches the bangles to each end edge-to-edge, what shape will she get?

Oblique and Isometric Sketches Yashvi and Priya are working on their project where they need to make oblique and isometric sketches of different 3-D shapes. Yashvi: Priya, can we draw a cuboid shaped box on a sheet of paper? Priya: I wonder too! A cuboid is a 3-D shape while we can only draw 2-D shapes on paper. Let us help them.

Oblique Sketches An oblique sketch is a simple 2-D drawing of a 3-D shape where the front face is shown as if it’s full size, while the other faces are drawn at a smaller scale and at an angle to give a basic 3-D impression. Let us draw an oblique sketch of a cube of dimensions 4 units × 4 units × 4 units.

Step 1: Draw a square of dimensions 4 units × 4 units.

Step 2: Draw another square of the same dimensions leaving one square unit from each corner as shown.

Step 3: Join the corresponding corners using dotted lines.

This oblique sketch represents a cube of dimensions 4 units × 4 units × 4 units. We can also draw oblique sketches of 2 or more cubes put together. The oblique sketch below shows three cubes each with 2 cm edges placed side by side to form a cuboid.

2 cm 2 cm

2 cm

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Isometric Sketches An isometric sketch is the image of the object which is drawn in such a manner that the original dimensions of the object remain intact irrespective of its position. Now, let us draw an isometric sketch of a cuboid of dimensions 7 units × 2 units × 2 units.

Step 1: Join 7 dots to form the length of the cuboid and 2 dots for the breadth as shown.

Step 2: From the vertices of the rectangle drawn, draw 4 dotted parallel line segments of 2 units each as shown.

This isometric sketch represents a cube of dimensions 7 units × 2 units × 2 units. Example 6

Example 7

Do It Together

Draw an isometric sketch for a cuboid with dimensions 4 units × 4 units × 2 units.

The dimensions of a cuboid are 5 units, 2 units and 3 units. Draw three different isometric sketches of this cuboid.

Step 3: Join all 4 vertices of the line segments to form another rectangle.

Error Alert! While drawing isometric sketches make sure the lines are always parallel and right through the dots.

Example 8

Draw an oblique sketch of a cuboid with dimensions 5 units × 4 units × 1 unit.

Draw an oblique sketch and isometric sketch of a cuboid with dimensions 5 units, 3 units and 2 units.

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Do It Yourself 19C 1

Draw an oblique sketch of a cube that has a side of 5 units.

Make an oblique sketch of the isometric sketch.

Draw an oblique sketch and an isometric sketch of a cube with the dimensions given. a 5 units

b 6 units

Write the dimensions of the given oblique sketches. b

Write the dimensions of the isometric sketches. a

Make an isometric sketch of the oblique sketch.

c 3 units

Draw the oblique and isometric sketches of the given shapes. 2u

4 units

2 units

it un

2 units 6 units

2 units

its

6 units

2 units

nit

its un

2 units 2u

4 units

2 units

nit

2 units

6 units

Word Problem 1

Reema placed three cubes of dimensions 3 cm × 3 cm × 3 cm end to end. What shape did it

form? What are the dimensions of the shape formed? Draw its oblique and isometric sketches.

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Visualising Solid Shapes We now know that any shape that has length, breadth and height is called a solid shape or a 3-D shape. The gift boxes that Yashvi and Priya received were lying on the table. From the side, when Priya looked at them, the shapes appeared to be square. She went over and looked at the box from the top and she realised it looked rectangular in shape. A solid shape, when viewed from different angles, looks different.

Cross Section and Shadows Let us say we slice a sphere in half. Then, the cross section of the sphere will be a circle. A cross section is the shape that we get when a solid shape is intersected by a plane. Let us look at the cross section of a cuboid. Cuboid

Vertical Cross Section We see squares.

Horizontal Cross Section We see rectangles.

Example 9

Draw horizontal and vertical cross sections of a triangular prism. Triangular Prism

Example 10

Horizontal Cross Section

Vertical Cross Section

Draw a horizontal cross section of a square pyramid. So, the horizontal cross section of a square pyramid is a square. Another way to see 2-D shapes in 3-D shapes is in shadows. Let us place a ball on a surface. If we bring a source of light in front of the solid, we can see that a somewhat circular shadow is formed. Similarly, if we place a cone, the shadow obtained will be triangular in shape.

Thus, we can conclude that, if the light of a torch is projected at a 3-D object, the shadow of that object on the plane will be in the form of a 2-D shape. Chapter 19 • Visualizing Solid Shapes

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Example 11

Let us see what the shadow of a cylinder will look like when light falls on it from the side. The shadow of the cylindrical water bottle obtained is rectangular in shape.

Do It Together

ere are the shadows of some objects when seen under the light of a torch. Name the possible H solid that might match each shadow.

A ball 2

____________________

A birthday cap

____________________

ame the shape of the cross section in each of the given solids. What will the shadow of the N cylinder and cone look like when light is shone from the top and side? Draw the shadows.

Rectangle

____________________

Shadow of the cylinder Top

____________________ Shadow of the cone

Side

Top

Side

Views of 3-D Shapes When we see an object from a different angle, it does not look the same. We have different views when we see them from different sides. Each object has a top view, front view and side view. Side View

Front View

Top View

We can also draw the 3 views of a structure made with small cubes.

Front View

Side View

Top View

In the figure we can see all three views. 326

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Example 12

Draw the front, top and side views of the object. Object

Do It Together

Front View

Top View

Side view

Draw the shapes of the three views of the given objects. Object

Front View

Top View

Side View

Do It Yourself 19D 1

What cross sections do you get when you give a vertical and a horizontal cross section to: a A birthday hat

b A dice

c A brick

d A ball

Match the shape with their horizontal cross sections. a

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Draw the shadow of the objects. a

Draw the front, top and side views of the following cube structures.

Draw the front, side and top views of each object.

Can a cube cast a shadow in the shape of a rectangle? Give a reason for your answer.

Name the 3-D shape that could have possibly caused these shadows.

Word Problem 1

Kush has a 3-metre-long cylindrical pipe that has a radius of 7 cm. If he cuts the pipe, what will its cross-sectional area be?

Points to Remember • One-dimensional figures are shapes or objects that exist only in one dimension and have only a length, like lines and curves. •

Two-dimensional figures are shapes or objects that exist in two dimensions, which are length and breadth.

•

In an oblique sketch, we can’t see certain faces of a solid but in isometric sketches we can.

• Three-dimensional figures are shapes or objects that exist in three dimensions, which include length, breadth, and height. •

A cross section is the shape obtained when a solid shape is intersected by a plane.

• If a torchlight is projected at a 3-D object, the shadow of that object on the plane will be in the form of a 2-D shape. •

Each object has a top view, front view and side view.

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Math Lab Setting: In groups of 5 each. Material required: Cards with the name of a solid shape written on them in a box. Method: 1 One student from each group picks a card from the box. 2 The student does not tell anyone the name of the solid shape. He describes the shape with the help of its properties.

3 Other students in the group guess the name of the shape. 4 The group with the most correct answers wins!

Chapter Checkup 1

Name the 3-D shapes. a

Name the solids that have: a 8 triangular faces

b 6 triangular and 2 hexagonal faces

c 5 faces and 5 vertices

d A rectangular face and 2 equal circular faces

Name the solids that can be formed by the nets. a

Name the 3-D shapes of the objects. a

Draw a net of the solids. a Cube

b Pentagonal pyramid

c Square prism

d Triangular pyramid

Draw the net of a cuboid that has the same breadth and height but a length that is double the breadth.

Draw (i) an oblique sketch and (ii) an isometric sketch for a cuboid of dimensions 5 cm, 1 cm and 1 cm.

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What shadow will be formed by the objects when light falls on them from the front? a

What will the horizontal and vertical cross sections of the images in question 8 be?

10 Draw the front, top and side views of the objects. a

11 Name the different plane shapes needed to draw the net of a heptagonal prism and heptagonal pyramid. 12 Make an oblique sketch of the given isometric shapes. a

13 A dice has A, B, C, D written in a clockwise order on the adjacent faces and E and F on the top and bottom faces respectively. When C is on the top, what letter will be on the bottom?

Word Problems 1 Maya was making a cube-shaped box from a sheet of paper. She ended up with the given net. In how many ways can she complete this net to make a cube?

2 Vansh cuts a cuboid cake of dimensions 20 cm × 10 cm × 4 cm exactly in half. What two possible cross sections does he get and what is the cross-sectional area of both sections?

3 Ram placed 5 cubes of 2 cm each edge to edge. He then placed 4 cubes of the same

measurement on top of them and then three, two and one. How many cubes did he place in total? Draw the front, top and side views of the structure formed.

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A n swe rs Chapter 1 Let’s Warm-up 4. 17

5. less

Do It Yourself 1A 1. a. 1 b. 6

7 9

2 9

13 22

c. 0

1. smaller

c. 1 4

37 117 e. 18

2. whole number

3. 8

43 73

6 17

109 120

3. a. 0

d. 1 17 24

2. a. 1 6

11 19

d. 50

2 3 b. 22 turns 3. 175 litres 4. 30 km 5 4 b. 20 km c. 60 km 6. ₹3890

Chapter 2 3 8

5 17 169 b. 1 c. 2 cm 24 20 180 71 13 23 d. 1 cm e. 5 Word Problems 1. of the wall 99 30 30 35 11 2. of a bottle 3. 1 m 36 20 15 11 13 11 1B 1. gallons 2. 11 km 3. 2 hours 4. 28 kg 28 12 15 21 7 5. 12 yards 6. a. Most: Washing car Least: Washing Utensils 20 1 3 4 11 15 litres b. 35 litres c. 90 litres d. 1 litres 5 5 15 12 4 19 5 3 3 1C 1. a. 10 b. c. 2 d. 5 e. 120 2. a. 5 27 21 4 10 4 1 1 1 1 437 b. 13 c. d. e. 4 3. a. 3 b. 1 c. 2 3 12 2 15 6 504 5 d. 10 4. a. 200 grams b. 10 minutes c. 27 18 9 1 8 1 d. 1 e. 83 cm 5. a. 4 b. c. 1 hours d. 14 days 35 3 3 2 1 3 3 e. 52 or 4 dozens f. 244 days Word Problems 1. 5 litres 2 8 5 11 7 2. a. litres b. 1 litres 3. 1400 g 4. 26 cookies 15 15 4. a. > b. < c. > d. <

5. a. 2

5. a. 300 books b. 900 books c. 300 books

11 3 2. a. 19 7 12 1 1 1 b. 30 c. 16 d. 3. a. b. c. 2 31 9 363 8 10 4 d. 2 4. a. < b. < c. < d. < 5. a. 44 13 13 6 1 1 b. c. m Word Problems 1. 3 cm 2. 10 containers 5 7 2 1 5 3. a. kg b. kg 4. 18 pieces 27 27

1D 1. a.

11 7

9 23

1E 1. 112 females 4. 8 projects

1 3

4 9 18 e. 29 d.

e. 1 f.

2. 192 shirts

5. ₹20,000

3. 45 players

6. 384

8 hours 13

Chapter Checkup 1. a. True b. False c. True d. False 17 17 7 2 1 c. d. 8 e. 6 3. a. 9 39 18 12 7 12 23 13 311 b. 4 c. 25 d. 4. a. 1 b. HCF 35 99 702 1 1 17 5 c. 2 d. 1 5. a. b. 4 c. d. e. 5 101 15 18 3 4 1 14 1 6. a. b. 6 c. 7 d. 3 7. a. 13 b. 16 10 9 15 17 3 29 94 100 207 c. 1898 d. 75 8. a. 9 b. 6 c. d. 30 253 147 280 1 1 4 9. a. Product of 3 and 2 b. Product of 25 and 5 4 75 1 9 c. 8 d. 30 10. a. 78 miles b. 169 c. 81 d. 3 10 3 2 1 1 e. 210 f. 2 Word Problems 1. a. b. 2. a. 11 turns 4 5 5 5 2. a. 1

65 153

Answers

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Let’s Warm-up 4. 0.39145

1. 35,125.4 5. 0.5872

2. 45,123

5. a. 24 km

3. 12.412

Do It Yourself 2A 1. a. 236.15 b. 533.594 c. 364.74 d. 581.716 e. 717.495

f. 752.794 2. a. 115.84 b. 192.676 c. 614.05 d. 313.994 e. 142.676 f. 123.915 3. a. 331.75 b. 273.47 c. 191.47 d. 726.487 e. 1109.71 f. 424.74 4. 236.295 5. 108.885 6. 326.235 7. 581.467 8. 7.5 cups 9. a. 563.80 km b. 82.2 km Word Problems 1. 11.85 kg 2. 0.75 kg 3. 24.25 gallons 4. 15.5 kg 5. 14.7 ºC 2B 1. 70.13 feet 2. ₹56.50 3. 244.28 KB 4. 11.12 kg. 5. ₹5072.8 6. 1.275 L 7. ₹18,698.61 Word Problems 1. Aman travelled farther by 24 Km. 2. $7.69 2C 1. a. 1.8 b. 729.3 c. 42.693 d. 125.35 e. 273.2675 f. 638.5376 g. 721.616 h. 271.4828 2. a. 4526 b. 45.26 c. 45.26 d. 4.526 3. a. 0.11 b. 3.05 c. 2454 d. 40 e. 23.23 f. 12.68 g. 13.25 h. 152.3 i. 89.4 j. 23.58 k. 392.3 l. 81.3 4. a. 0.345 b. 32.5 c. 35 d. 912.66 e. 38.46 f. 42.52 5. 85.26 6. 15,562.5625 cm2; 499 cm 7. ₹2493.375 8. 112.8 m 9. 2.8 m 10. 158.48 km Word Problems 1. ₹656.7375 2. 9 pair of pants 3. 7 bags 4. 218.8125 square metres 2D 1. ₹708.08375 2. 3 litres 3. ₹1533.375 4. ₹2848.25 5. ₹1018.45 6. 26.65m, ₹2012.075 7. ₹4000 8. ₹160.875 9. 4.94 cm Word Problem 1. 8 pairs of gloves and 8 pairs of socks Chapter Checkup 1. a. 158.76 b. 334.03 c. 148.813 d. 314.113 e. 847.536 f. 858.49 g. 751.446 h. 1793.21 2. a. 87.36 b. 72.95 c. 108.053 d. 168.64 e. 51.677 f. 199.15 g. 410.932 h. 346.665 3. a. 7437.5 b. 2376.18 c. 1923.6308 d. 25,588.556 e. 17,064.76 f. 20,337.891 g. 4756.404 h. 77,312.62 4. a. 25.4 b. 3.8 c. 60.1 d. 124.7 e. 1.254 f. 3.05 g. 8.4 h. 36.45 5. 147.35 marks 6. 7.5 kg 7. 413.86 kg 8. 1186.25 litres 9. 234.7072 m2; 61.76 m 10. 64.8 cm 11. ₹30,764.4375 12. a. ₹948.48 b. ₹51.52 Word Problems 1. 1.4 kg 2. 228.84 kg 3. 1.65 kg 4. a. ₹6268.43 b. ₹537.1 5. ₹216.50 6. 125 km, 39.2 litres

Chapter 3

Let’s Warm-up 1. < 2. < 3. < 4. >, >, >, > 5. <, <, <, < Do It Yourself 3A 1. a. −4 b. 29 c. 31 d. −11 e. −90 f. −161 g. 213 h. 33 2. a. −4 b. 7 c. −33 d. −10 e. 12 f. −358 g. 440 h. −810 3. a. −17 b. 29 c. −57 d. −58 e. −62 f. 169 g. −318 h. 370 4. a. 6 b. −7 c. 534 d. 79 e. 281 f. 756 g. 497 h. 940 5. a. 3 b. −35 c. 10 d. −₹20 e. ₹185 f. 0 6. −157 7. a. True b. False c. False d. True 8. a. 13, Commutative property b. −3, Existence of additive identity c. −16, 13, Associative property d. 0, Subtraction property of zero Word Problems 1. 1079 items 2. 700 m 3. 105 points

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3B 1. a. 180 b. −66 c. −273 d. 84 e. 0 f. −56 g. 360

h. −128 2. a. 110 b. 1728 c. −7956 d. 96 e. −900 f. 0 g. −2340 h. 0 3. a. 3450 b. −7040 c. 42,588 d. 2160 e. −1,38,720 f. −2,13,624 4. a. negative sign b. positive sign c. positive sign d. negative sign e. negative sign 5. a. 0 b. 1 c. 56 d. 56, 23 e. −2198 f. 123 Word Problems 1. 440 feet below sea level. 2. a. ₹200 b. ₹230 3. 76 points 3C 1. a. −18 b. −8 c. 16 d. −16 e. 8 f. −82 g. 26 h. 45 2. a. 22 b. 5 c. –3 d. −36 e. 8.33 f. 17 g. 0 h. 0.5 3. −145 4. a. False b. False c. False d. False e. True 5. Answers may vary. Sample answers: a = −270, b = 18 a = 30, b = –2 6. 14 Word Problems 1. First level 2. 8 p.m., –18℃ Chapter Checkup 1. a. 106 b. 340 c. 523 d. 617 e. −2011 f. −1462 2. a. −2225 b. 3250 c. −1728 d. −13,965 e. −1,76,790 f. −2,90,895 3. a. −33 b. 52 c. −221 d. −152 e. 124 f. 234 4. a. 233 b. −268 c. −967 d. −65 e. −952 f. −310 g. 365 h. −3530 i. −3261 j. −42 k. 4,45,280 l. 26 5. a. −42 b. 64 c. 17 d. −8 e. 98 f. 6.5 6. a. −38; 38; 1 −38 1 b. −78; 78; 1 c. −305; 305; 1 d. 489; −489; 489 −78 −305 7. a. True b. True c. True d. False e. False 8. 591 9. −56 10. −935 11. −56 12. +12,000 feet; +15,000 feet; −23,945 feet; New altitude = +3055 feet Word Problems 1. –9℉

2. 9th floor

Do It Yourself 5A 1. b 2. c

3. a. T b. T c. F, because autobiographies and memoirs are books of the past. 4. Colour of 5. Range = 5

Let’s Warm-up 1. 2 2. 27 Do It Yourself 4A 1. 10 2. –1.6 °C 3. 4.17

3. 32

4. 12

4. 78

5. 88.6

5. 5

6. 31, greater 7. 54 Word Problems 1. ₹260 2. 1356.25 km 3. 32 4B 1. a. False b. True c. False d. False 2. 7 3. 18; 1 4. 46 5. 4 Word Problem 1. Drama n+1 4C 1. a. ascending; descending b. 2. 1725 2 th c. 6 3. 40 4. 30 5. a. 25 b. 25 6. a. 13.5 minutes b. No; the 3rd and 4th observation do not change so, the median will remain the same. c. No 7. 8 Word Problems 1. ₹20 2. 12th seat 4D 1. a. mean, median, mode b. Mean c. Median d. Mode 2. Mode; there are no repeating values. 3. Mean 4. Mode 5. Median; the data has a sizeable number of observations. Word Problem 1. 75.5; 85.5; Median, as the given data has sizable number of observations.

(

)

Blue

Green

Yellow

Maroon

Black

Brown

Chapter 5 Let’s Warm-up

4. Alastair Cook

Number of birthday cakes sold

Tally Frequency marks

6. a. Range = 66 Marks (out of 100)

Tally marks

Frequency

b. 3 students c. 6 students Word Problems 1. a. Range = 15 Pulse Rate (Number of beats per min.)

Number of students

Tally marks

59 60

b. 4 students c. 14 students

5B 1. a. width b. equal c. horizontally or vertically d. height

Number of points scored in different rounds of the game y

Scale: 1 division = 2 points

20 18 16 14 Points

Chapter Checkup 1. a. 11.875; 12; 12 b. −11.875; −15; −15

2. a. 5.5; 5.5 b. 9.625; 9 c. 10.11; 6 3. 39 kg 4. 3 5. 5; 5 6. 20 7. 15 8. 54.97; 54; 54 9. as the given data has sizable number of observations. 10. Mean, as we need to find one value that represents the whole data. 11. 36.5 years 12. 80; 85 13. 21.6; 21 14. 34 15. Mean and median, since both are equal. Word Problems 1. 45.77 kg 2. Yes; 29.5 years 3. 88 marks

Frequency

Red

3. 138 pairs

Chapter 4

Tally marks

cushion covers

12 10 8 6 4 2 0

Rounds of the game

1. Jacques Kallis 5. 2632

2. 25,850

3. 68,348

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Favourite sport of students of a school y

Scale: 1 division = 10 students

100

Production of paper (in lakh tonnes) by two different companies X and Y over the years

100

60 40 30 20 10 0

Cricket

Football

Tennis

Athletics

Swimming

Hockey

Volleyball

Sport

Average intake of nutrients by rural and urban groups for different foods

Average Intake of Nutrients (in calories)

Scale: 1 division = 10 calories

100

Urban

Rural

70 Company X

Company Y

50 40 30 20 10 x

2018

2020

40 30 20 10 Pulses

Leafy vegetables

Other vegetables

Fruit

Milk

Fish and flesh foods

Fats and Oils

Sugar

Scale: 1 division = 10 marks

100 90 80 70

1st term

2nd term

Scale: 1 division = ₹20

Cost price and profit (in ₹)

Performance of a student in different subjects in two terms

2022

2. a. English b. Maths and Science; 80 marks c. Maths; 96 marks d. 360 marks e. 26 marks 3. a. Mt. Everest > Kanchenjunga > Nanga Parbat > Annapurna > Nanda Devi > Aconcagua b. Mt. Everest; 8800 m c. 88:69 d. Kanchenjunga Cost price and profits earned by a merchant in 4 different types of merchandise 4. a.

2021

Year

5C 1. a. 25 ft.

Cost Price

100

Profit

80 60 40 20 0

x P

Merchandise

b. Merchandise S 5. a. ₹2200 b. January and August, ₹2400 in each c. ₹24,200 d. 60:61 Rainfall on different days of the week 6. a.

40 30 20

10 Maths

Science

English

Social Science

Rainfall (in mm)

Hindi

Subject

Word Problems 1.

Number of toys manufactured on different days y

Scale: 1 division = 5000 toys

15 10 5 0

Monday

Tuesday

Wednesday

Thursday

Friday

b. Friday c. 19:30 7. a. 420 people b. 340 people c. 39:40 d. 1550 people 8. a. 210 cars b. 210 cars c. Showroom A d. 17:16 Word Problems 1.

40 000 35 000 30 000 25 000

Number of visitors to a museum for five months

20 000

15 000

Scale: 1 division = 100 visitors

5 000 Monday

Tuesday

Wednesday Day

Thursday

Friday

Number of Visitors

500

10 000

Scale: 1 division = 5 mm 20

Day

45 000

Number of Toys

2019

Foodstuff

Marks

Scale: 1 division = 10 lakh tonnes of paper

Production of Paper (in lakh tonnes)

Number of students

400 300 200 100 0

January

February

March

April

May

Day

Answers

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Tally marks

rolling a dice

is obtained

5 4

Denmark

Country

Canada

Belgium

Australia

India 10

Birth Rate (per thousands)

Total population of India from 1971 to 2021 y

Scale: 1 division = 200 million

1 600 1 400

Population (in millions)

1 200 1 000 800 600 400 200 0

1971

1981

1991

2001

2011

2021

Year

6. a. 2018, 90 million tonnes b. 69 million tonnes c. 8 million tonnes d. 60:347 7. a. 2 lakhs b. 1 lakh more c. 2 lakhs d. 45 lakhs e. Number of Users of Different Electronic Appliances 8. a. 175 bulbs b. Friday, 225 bulbs 8 c. 875 bulbs d. e. 175 bulbs 9. The graph is misleading; 35 the numbers on the y-axis start at 19. It should always start at 0. The tabular form and correct graph would be: Name of employees Salaries (in thousands)

Raj

Sahil

Nivira

Yuvan

Kabir

Salaries of 5 employees y

Scale: 1 division = 5 thousand

Salaries (in thousands)

30 25 20 15 10 5 0

Raj

Sahil

Nivira Employee

80 60 40 20 Maths

Science

Hindi

English

Social Science

Drawings

2. a. Sales of books (in thousands) from six branches of a publishing company during two consecutive years, 2010 and 2011. b. 217:170 c. 81.67 (Approx.) thousand d. 577 thousand

Chapter 6

Let’s Warm-up 1. 1:2 2. 1:5 3. 1:3 4. 4:5 5. 1:4 Do It Yourself 6A 1. a. Unlikely b. Equally likely c. Impossible d. Likely

Scale: 1 division = 5 thousands

Scale: 1 division = 20 marks

Subject

b. 4 c. 5 d. Digits 1 and 6 3. a. September b. 11 birthdays c. 5 birthdays d. July and November Birth rate per thousand for five countries 4.

100

10. a. NH 10 b. 900 km c. NH 3 Word Problems Marks obtained by Kunal in his annual examination 1. Marks obtained

a. 300 visitors b. 100 visitors c. 25% 2. a. ₹1,05,555.56 b. ₹2,48,055.56 Chapter Checkup 1. 9 students 2. a. Number obtained on Number of times a digit

Yuvan

Kabir

e. Sure 2. a. Equally Likely b. Sure c. Unlikely d. Impossible 3. a. Impossible b. Equally Likely c. Unlikely 4. a. Spinner B, as it has two sections out of six that represent ʻʻDownʼʼ b. Any of the two spinners as each of them has 2 sections out of six that represent ʻʻForwardʼʼ. 5. Answers may vary. Sample answers: a. Wearing mask due to increasing pollution b. Travelling to moon c. Christmas falls in the month of January. d. Sun sets in the west. Word Problem 1. a. Unlikely b. Equally Likely 6B 1. a. A, B, C, D, E, F, G, H, I, J, K b. A, E, I 2. a. 6 b. 3 1 c. 5 3. a. 8 b. 4, 5, 6, 7, 8 c. 4 d. 4 e. 4 4. a. 3 1 1 1 1 2 1 1 b. c. d. e. 1 f. 0 5. a. b. c. d. 6 2 2 12 3 3 2 25 4 3 1 9 1 3 6. 7. 8. a. b. c. d. 0 e. f. 26 9 10 5 10 2 5 13 24 9 6 1 9. a. b. 0 c. d. 1 e. f. 10. 25 25 25 25 3 1 1 1 11. Error: He included number 3 also; ; 1, 2 12. 13. 5 5 2 1 49 1 Word Problems 1. a. b. 2. 50 50 4 7 12 13 7 21 7 7 11 1 6C 1. a. b. c. d. e. 0 f. 2. , , , 50 25 25 25 25 20 40 40 5 3. Mala has found the probability of getting heads over the 2 6 1 3 13 number of times she gets a tail; 4. , , , 5 25 5 10 50 1 1 11 17 5. a. b. c. d. 6. Theoretical probability > 4 6 60 60 experimental probability 7. a. Theoretical probability > Experimental probability b. Theoretical probability > Experimental probability c. Theoretical probability = Experimental probability d. Theoretical probability = Experimental probability 8. 560 students 19 Word Problem 1. 50 Chapter Checkup 1. a. Impossible b. Sure 2. a. Sure b. Impossible c. Likely 3. a. Equally likely b. Unlikely 1 2 1 c. Impossible d. Impossible e. Sure 4. a. b. c. 3 3 2 1 5 d. e. f. 1 5. a. 1,1; 2 b. 3, 3, 3; 3 c. 1, 1, 3, 3, 3; 5 2 6 9 d. 2, 2, 2; 3 e. 1, 1, 2, 2, 2, 3, 3, 3; 8 f. 1, 1, 2, 2, 2; 5 6. a. 50 39 81 19 1 3 5 1 1 b. c. d. 7. a. b. c. d. 8. 100 100 50 4 8 8 8 5

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Chapter 7

1. 2n + 3 5. 3m – 5

Let’s Warm-up 4. 2n – 5

Do It Yourself 7A 1. a, c, d

2. 5n – 2

10. a. d.

8 25

1 26

5 26

12. 4 times

Word Problem 1.

3 7

3. 3p + 2

2. a, c, d, f 3. a.  b.  c.  d.  e.  4. a. True b. False c. False d. False e. True x x 32x + 5 5. a. x + = 9 b. 3x + 5 = 17 c. = 15 d. = 27 2 2 2 e. x + 2 = 8 f. x + 10 = 38 6. Statements may vary. Sample statements. a. 9 subtracted from y gives 81. b. 5 subtracted from 4 times q gives 15. c. 30 times z equals 1650. d. The sum of 9 times x and 5 gives 50. e. 12 added to 4 times x is 28. f. 2 added to one-fourth of y equals 8. 1 7. a. x = 7 b. x – 10 = 7 c. 10 – x = 7 d. x + 27 = 77 10 8. a. m + m – 3 = 18 OR 2m – 3 = 18 b. c + 3 = 12 c. x + 2x = 18 OR 3x = 18 d. 2(2x + 3) = 18 Word Problems 1. (3b – 4) metres 2. 7x = 217 7B 1. a. x = 7 b. x = −9 c. x = 8 d. x = −2 e. y = 4 f. y = 2 g. y = 0 h. y = −12 2. a. a = 15 b. b = 27 c. c = 40 28 3 4 1 42 2 5 d. x = or 5 e. y = or 1 f. z = or 8 g. p = or 5 5 3 3 5 5 3 2 –1 p=1 h. q = 3. a. Step 1: Subtract 14 from both sides 3 6 of the equation. Step 2: Multiply with (−1) on both sides of the equation. b. Step 1: Subtract 12 from both sides of the equation. Step 2: Divide both sides of the equation by 2. c. Step 1: Add 6 to both sides of the equation. Step 2: Divide both sides of the equation by 4. d. Step 1: Multiply both sides of the equation by 7. Step 2: Divide both sides of the equation by 30. e. Step 1: Subtract 5 from both sides of the equation. Step 2: Multiply both sides of the equation by 2. f. Step 1: Add 1 to both sides of the equation. Step 2: Multiply both sides of the equation by 2. 4. 10 5. Incorrect; 8 6. Answers may vary. Sample answers: a. Veena had saved some money in her money bank. She spent ₹75 on buying stationery items. If she is now left with ₹150, how much money did she have to begin with? ₹225 b. Ankur has 249 postcards. He added new postcards to his collection. If he has 350 postcards in all, how many new postcards did he add to his collection? 101 postcards c. The age of Rima’s grandmother is 9 times Rima’s age. If the age of Rima’s grandmother is 72, what is the age of Rima? 8 years 7. We shall either multiply both sides of equation 1 by 6 or divide both sides of equation 2 by 6. Word Problems 1. ₹75 2. 30 stamps 3. Small jug = 2 cups; large jug = 4 cups 3 7C 1. a. x = 6 b. a = 4 c. x = d. n = 1.4 e. x = 12 f. x = 18.2 7 4 g. a = 41 h. p = i. x = 21.5 2. a. y = 5 b. y = 6 c. a = –12 3 d. m = 3 e. j = 48 f. x = –33 g. w = 45 h. g = 64 1 4 1 i. x = 67.5 3. a. x = b. x = 7 c. x = 3 d. x = OR x = 1 3 3 3 e. x = 3 f. y = 2 4. a. 64 b. 33 c. 102 d. 8 e. 35 f. 15 g. 5 5. 12 6. 9 7. 10; 28 8. Length = 65 m; breadth = 60 m. Word Problems 1. Bat = ₹1000; ball = ₹200 2. 13 kg and 18 kg 7D 1. 35 2. 48 3. 46 and 49 4. 6; 14 5. 21, 22, and 23 f. 

Answers

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6. 50°, 90°, and 40° 7. 43 units, 45 units, and 60 units Word Problems 1. 13 years 2. 10 years 3. 240 books 4. Book = ₹110; Pen = ₹60 5. 10 years; 40 years 6. 11 years; 7 years –2 Chapter Checkup 1. a. 1 b. 1 c. 3 d. e. −17 f. 10 19 2. a. 28 b. −7 c. 30 d. 6 e. 5 f. 18 3. a. x = 2 b. a = 3 3 c. y = 9 d. x = 3 e. x = –2 f. x = – 4. a. 100 b. 13 c. 5 7 11 d. 6 e. 5. a. q = 42 b. x = –58 c. x = 12 d. x = –4 2 e. x = –20 f. x = –3 g. y = –3 h. x = –40 6. a. x = 33 b. x = 5 c. y = –3 d. x = 4 e. y = 4 f. x = 5 g. x = –11 h. x = –27 i. x = –2 7. –227 8. 30, 32, and 34 9. 11, 12, and 13 10. 12 11. 4 cm; 12 cm; 12cm 5 1 5 Word Problems 1. 10 kg, 9 kg, 11 kg 2. ₹30; ₹20 6 3 6 3. 18 toffees 4. 3 years; 33 years

Chapter 8

Let’s Warm-up 1. 27° 2. 142° 3. 180° 4. 213° Do It Yourself 8A 1. a. Yes b. Yes c. Yes d. No e. Yes 2. a. 147°

b. 108° c. 57° d. 129° e. 82° 3. a. Yes b. No c. No d. No e. Yes 4. a. 89° b. 71° c. 1° d. 49° e. 53° 5. a. Neither b. Neither c. Complementary d. Supplementary 6. a. 25.7° approx. b. 216° c. 50° 7. a. 45° b. m = 120°, n = 60° c. 30° 8. a. A B ∠A and ∠B, ∠B and ∠C, ∠C and ∠D, ∠D and ∠A, ∠D and 90° 90° ∠B, ∠C and ∠A. 9. 10° 10. 113°, 67° 11. 60° 90° 90° 12. 45° 13. 120°, 60° D C 14. 50°, 40° 15. 130°, 50° 16. x = 67°, y = 55° Word Problem 1. 70° 8B 1. Pairs of corresponding angles: 2 and 6, 2 and 10, 6 and 10, 1 and 5, 1 and 9, 5 and 9, 3 and 7, 7 and 11, 3 and 11, 4 and 8, 8 and 12, 12 and 4. Pairs of alternate angles: 2 and 8, 1 and 7, 3 and 5, 4 and 6, 1 and 11, 2 and 12, 3 and 9, 4 and 10, 6 and 12, 5 and 11, 7 and 9, 8 and 10. 2. a. Not parallel b. Not parallel c. Not parallel d. parallel 3. a. x° = 80°, y° = 32° b. m° = n° = 70° c. p° = 60°, q° = 105° d. h° = 45°, g° = 72° 4. ∠PSR = 60°, ∠QRS = 30° 5. ∠ACB = 60°, ∠ABC = 90°, ∠BAC = 30° 6. a° = 118°, b° = 108° 7. a. y° = 35°, x° = 50° b. 108° Word Problem 1. 60° 60°

60°

8C 1. a. Not parallel b. Not parallel c. Not parallel d. Parallel 2.

R Q A

3. Figures may vary. Sample answer: m is not parallel to n. 4. P n

5.9 cm

1 2

5.9 cm

1 7 4 c. d. e. 0 f. 1 5 10 5 1 4 13 1 8 c. d. 11. a. b. c. 13 13 25 5 25 5 1 1 1 5 13. a. b. c. d. e. 12 6 12 12 12 9. a.

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6. XD = 5.6 cm; YD = 9 cm; ZD = 4.2 cm Z

4.2 cm

5.5 cm

5. 6

Word Problem 1. Steps: i. Draw a rectangle ABCD of length 10 cm × 8 cm. ii. Mark any two points A and B on any edge of the outer rectangle. iii. Construct right angles at points A and B, respectively. iv. Draw arcs with centres at A and B, each with a 2 cm radius, intersecting the perpendiculars at A and B at points E and F. v. Connect points EF and extend this line on both sides to achieve the desired parallel line. vi. Repeat the process for the other three sides. vii. Connect the corresponding points on these lines to form the inner rectangle. Chapter Checkup 1. a. Yes b. No c. No d. No e. No 2. a. No b. No c. Yes d. Yes e. Yes 3. a. 146° b. 61° c. 54° d. 156° e. 14° 4. a. 43° b. 70° c. 64° d. 50° e. 4° 5. a. 57° b. 60° c. 20° d. 29° e. 28° f. 25.7° (approx) 6. ∠ABC = 75°, ∠ACB =67°, ∠CAB = 38° 7. m° = 137°, g° = 137°, y° = 67°, h° = 67° 8. a. not parallel b. not parallel c. parallel d. not parallel 9. a. b. u l 105°

75°

75° 105° 115° 65°

65°

115°

c. 55° 70°

110°

55°

125°

110° 70°

120°

b c

55°

140°

60°

m n

r 40°

140° 40° 130° 50°

50°

125°

120°

60°

a 125°

55°

125°

p q

130°

10. a. 115° b. 130° c. 110° d. 95° 11. a. 180° b. 180° c. 153° d. 27° e. 100° f. 80° 12. x° = 105°, y° = 120°, w° = 105°, z° = 60° 13. ∠PQR = 65°, ∠RSP = 65°, ∠QRS = 115° 14. ∠QDC = 63°, ∠DCQ = 47° Word Problem 1. 115°

Chapter 9

1. Acute-angled triangle 2. Isosceles-right triangle 3. Equilateral triangle 4. Obtuse-angled triangle 5. Isosceles triangle

Let’s Warm-up

Do It Yourself 9A 1. a. Median b. Altitude c. orthocenter d. may not

2. 10 cm 3. AD = 3 cm, BC = 8 cm 4. a. GE = 10 cm, BE = 15 cm b. GF = 8 cm, CF = 24 cm c. AG = 20 cm, GD = 10 cm d. GC = 2x, CF = 3x e. x = 2, AD = 27 5. c 6. 72° 7. b Word Problem 1. Option e; Shyam would find the centroid of the triangular table. 130° 165° 9B 1. a. i b. iii c. i 2. a. 105° b. c. 7 6 d. 60°, 60° e. x = 70°, y = 55° f. 96° g. 80° h. 30°, 30° i. 30° 3. 52°, 78°, 50° 4. 70° and 50° 5. 50°, 60°and 70° 6. 70°, 20° and 90° 7. 80°, 65°, and 35° 8. 50°, 60°, and 70° 9. Prove using the linear pair of angles and the exterior angle property. Word Problem 1. 50° 9C 1. a. a, b, e 2. a. 3 cm < x < 9 cm b. 5 cm < x < 23 cm c. 27 cm < x < 41 cm d. 4 cm < x < 50 cm 3. a. < b. > 4. Yes. As we have shown that the triangle inequality property holds true for this statement, hence, PQ + QR + PR > 2 PS.

5–6. Solve using the Triangle Inequality Theorem. Word Problem 1. The distance travelled from New Delhi to Dwarka is less than the other two distances. 9D 1. b, c 2. a. 16 m b. 24 cm c. 18 m 3. 7 4. 20 cm 5. 24 cm 6. 5 cm; 12 cm 7. 17 m Word Problem 1. 3 m Chapter Checkup 1. a. 60° b. right-angled c. equilateral d. 180° 2. a. F b. T c. F d. T 3. a, c 4. b and c 5. a. ∠P = 16; ∠Q = 94° b. ∠x = 50° c. ∠y = 63° d. ∠x = 65°; ∠y = 50° 6. 8 m 7. Prove using the Pythagoras Theorem. 8. 20 inches 9. 50 cm 10. 10 m 11. 40°; 120° 12. Prove using linear pair and angle sum property of a triangle. 13. Prove using the exterior angle property of a triangle. Word Problems 1. 15 m 2. 17 m

Chapter 10 Let’s Warm-up

3. Right-angled

Do It Yourself 10A 1. 10 cm

1. vertex 4. 3; 3

2. 60 5. acute

2. a. 4.2 cm b. 55° c. 4 cm d. 7.2 cm 3. 150 cm2 4. Square EFCD; as AB = EF; AD = DE; BC = CF and DC = DC 5. ∠U; UV; ZX 6. 105° 7. ∠L and ∠U, ∠M and ∠V, ∠N and ∠W 8. 1:1 9. Yes. Prove using the property of congruency which states that when the measure of two angles is the same, the two angles are congruent. 10. ∠DOB or ∠BOD Word Problems 1. Sheets are congruent. 2. The circular targets are congruent, since if the radius of one circle is equal to the radius of the other circle, then, the circles are congruent. 3. 5 cm, 12 cm and 13 cm. 10B 1. a. PR = SQ RS = QP PS = PS b. Yes; By SSS congruence ΔPRS ≅ ΔSQP. 2–4. Prove using SSS congruence criteria. 5. Yes, ∆APS ≅ ∆BQR 6. ∆ABC ≅ ∆DCB 7–9. Show / Prove using the SSS congruence criteria. Word Problem 1. Students will draw using the given measurements. Yes, the triangle drawn is congruent. 10C 1–3. Prove using SAS congruence criteria. 4. Yes; prove using SAS congruence criteria. 5. Prove using SAS congruence criteria. 6. a. Yes b. AB = AC; ∠BAD = ∠CAD and AD = AD. c. Yes 7. When AC = DF. 8. OC = OD; ∠AOC = ∠BOD and OA = OB; Yes 9. Prove using SAS congruence criteria. Word Problem 1. Yes; Prove using SAS congruence criteria. 10D 1. Prove using ASA congruence criteria. 2. Yes 3. Yes; ΔABC ≅ ΔFED. 4. Yes, ΔADB ≅ ΔBCA 5. Prove using ASA congruence criteria. 6. a. Prove using ASA congruence criteria. b. Prove using corresponding parts of congruent triangles. 7. ∠CAD = ∠BAD, AD = AD, ∠CDA = ∠BDA 8. Prove using ASA congruence criteria. 9. a. Yes, b. ∠DAB = ∠DAC; AD = AD and ∠BDA = ∠CDA c. Yes. 10. Prove using the alternate interior angle property. Word Problem 1. Yes 10E 1. Prove using RHS congruence criteria. 2. Yes 3. a. Prove using RHS congruence criteria. b. Prove using Corresponding parts of Congruent triangles c. Prove using Corresponding parts of Congruent triangles 4. No; ∆ABC ≅ ∆DEF. 5. Prove using RHS congruence criteria. 6. ∆ABC ≅ ∆EBD. 7. Yes; ∆ABD ≅ ∆ACD; RHS congruence rule; CD; ∠C 8. Prove using RHS congruence criteria. 9. AB = 12 cm; AC = 13 cm; QR = 5 cm; PR = 13 cm Word Problem 1. The triangles aren’t congruent.

336

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Chapter Checkup 1. a. True b. False c. False d. False e. True

2. ∠SVT 3. x = 77°; y = 35° 4. a. Yes; 2:5 < 3:7 < 7:12 < 11:14 < 9:2; 20 cm; 35 cm b. Yes 5. 25° 6–7. Prove using SAS congruence criteria. 8. 55° 9. Prove using RHS congruence criteria. 10. 50° 11. 1 12. Show using RHS congruence criteria and angle sum property of triangle. Word Problems 1. 1300 cm2 2. Not congruent; Use SSS congruence criteria.

Chapter 11

Let’s Warm-up 1. 44:53 2. 123:74 3. 5:7 4. 162:229 Do It Yourself 11A 1. 32:13 2. a. 2:5 b. 25:1 c. 25:4 d. 1:15

3. a. 5 b. 40 c. 4 d. 11 4. a. < b. > c. > d. > 5. 2:5 < 3:7 < 7:12 < 11:14 < 9:2 6. A:C = 35:143; A:B:C = 35:77:143 7. 10:9 8. 21; 15 9. 25 erasers, 15 sharpeners 10. Year 2020 11. 10:13 12. 120° 13. 52 inches Word Problems 1. 49:39 2. 180 g 3. ₹740, ₹1110 and ₹1850 4. 20 cm; 35 cm 5. 14 years; 16 years 6. 75 g

6 11B 1. a. Yes b. No c. No d. No 2. a. 12 b. 42 c. 7 24 1 28 d. 3. a. 9 b. c. d. 29.4 4. 12 45 2 125 25 5. 25 gallons 6. days 7. ₹6,44,000 8. 28 days 3 9. 19 metres 10. 6 Word Problems 1. $975 2. 32 years 11C 1. 15 kg 2. 450 km 3. ₹100 4. 4500 times 5. 75 minutes 6. 612.5 kg 7. 50 shirts 8. 19 books 9. 20 workers 10. 27 days 11. 385 teaspoons Word Problems 1. 5 hours 2. a. Sunita b. ₹431.25 11D 1. 16 m/s 2. 510 m 3. 2.67 minutes (approx.) 4. 1.39 m/s 5. 250 minutes 6. 465 minutes 7. a. 25,75,342.47 km/day b. 1,07,305.94 km/hour Word Problems 1. a. Both took same time b. Zero minutes 2. a. Monday b. Monday: 1.558 km Tuesday: 1.1 km, Wednesday: 1.375km Thursday: 0.733 km Friday: 0 km 3. 2 m/s Chapter Checkup 1. 3 2. 84; 96 3. 2 4. 6.785 (approx.) 5. 120 feet 6. 4:14:21 7. 5 kg 8. 111.11 litre 9. 5.45 seconds 10. 384 widgets 11. 32.22 m/s 12. 125 days 13. 0.18 m/s 14. 1320 metres 15. 3.5 hours 16. 13:6 17. 2 18. 55 years, 25 years Word Problems 1. ₹7750 2. ₹3,600; ₹14,400 3. a. 1.25 hours; 1.4 hours b. Mr. Sharma c. 0.15 hour 4. ₹1,12,800

Chapter 12 Let’s Warm-up 5.

5 100

Do It Yourself 12A 1. a.

9 400

1. coffee

11 20

9 5

2. juice

141 250

33 200

38 100

10 100

13 200

9 2500

18 233 193 i. j. 2. a. 20% b. 10% c. 72% 5 400 300 1 d. % e. 75,000% f. 105% g. 525% h. 1260% i. 75% 6 j. 60% 3. a. 0.48 b. 0.0125 c. 0.1264 d. 0.00005 e. 1.235 f. 0.012 g. 0.112 h. 0.124 i. 0.8106 j. 1.45125 4. a. 20% b. 34% c. 54.7% d. 98.4% e. 0.48% f. 120% g. 234% h. 1210% i. 1523%

Answers

UM24CB07_Answer_R3.indd 337

1 5 1 ; 50% b. ; 62.5% c. ; 50% 2 8 2 3 Word Problems 1. of the pizza 2. 25% 3. 0.045 5 1 4. of the shipment 5. Jaya 6. 84% 20 12B 1. 1:5 2. 1:3 3. 3:7 4. 60% 5. 60% 6. 80% 7. 3:1 Word Problems 1. 10% 3 12C 1. a. 62.5 b. 225 c. 56.25 d. 672 2. a. 1 % 17 b. 15% 3. a. 400 b. 850.5 c. 4 hours 16 minutes d. 17 m 4. a. 15% decrease b. 50% increase c. 30% increase 5. a. 36 seconds b. 1% c. 1.32 d. 2812.5 Word Problems 1. 1000 employees 2. ₹100, ₹150, ₹250; 20%, 30%, 50% j. 14221.4%

5. a.

12D 1. 2000 g

2. 2400 people 3. 52,224 votes 4. 75 students 1 5. 2,00,000 6. 33 % 7. 5000 8. 8% increase 9. 10% 3 10. 12.5% decrease 11. ₹8,82,812.5 Word Problem 1. ₹820.8

12E 1. a. Gain of ₹125 b. Gain of ₹770 c. Loss of ₹188 2. a. (ii) b. (ii) c. (i) d. (i) 3. a. ₹50 b. ₹2000 9 2 1 c. 23 % d. 16 % 4. a. Profit = 33 % b. Loss = 4% 117 3 3 2 c. Profit = 5% 5. a. Profit = 16 % b. Loss = 10% 3 1 c. Profit = 25% d. Loss = 8 % e. Loss = 20% 3 2 Word Problems 1. ₹450 2. 16 % 3. Gain = 17% 3 4. Loss = ₹2000

12F 1. a. ₹7600 6. ₹80

2. Anu

7. ₹1995

3. ₹6847 8. 1% loss

4. ₹2000

5. ₹60

9. 8% gain

10. 16

Word Problem 1. 70%

12G 1. a. ₹750 b. ₹3600 c. ₹600 d. ₹180

2. a. ₹2200

b. ₹4200 c. ₹12,340 d. ₹24,950

3. a. 6.25 years

b. 10 years c. 3 years d. 6.5 years

4. a. ₹6000

b. ₹49,000 c. ₹1,20,000 d. ₹2,00,000

5. a. 3%

b. 6.5% c. 7.5% d. 0.0325% Word Problems 1. ₹10,000 2. ₹80,000 3. ₹48,000 4. ₹2000 5. 25% 6. 30% 7. ₹21,500 8. ₹600 at 5% and ₹400 at 8%

3 16 b. 0.045 c. 564% d. 0.08125

11 d. 1140% 2. a. 48% 500 3. a. 115 b. 2.5 L c. 505 1 d. 160 g e. 2 km 560 m 4. a. 20% b. % c. 14% d. 5% 3 1 28 10 32 195 e. 7 % 5. a. 484 b. 9090 c. ₹7564 d. 1605 cm 7 33 11 37 601 2 e. 13,541 mL 6. a. 50% b. 100% c. 10% 7. a. ₹50 3 2 b. ₹345 c. ₹135 d. ₹89 8. a. 30% profit b. 16 % loss 3 23 25 c. 10.4% loss 9. a. 13 % b. 25 % c. 20% 29 67 10. a. ₹530 b. ₹10,683.75

Chapter Checkup 1. a. 80% b.

Word Problems 1. Gain = ₹25, 25%

2. Profit% = 80%

2 1 4. a. 46 % b. 53 % 5. 4:5 6. 7:9 7. 100% 3 3 67 33 8. ₹343 9. ₹1000 10. ₹2700 11. ₹29,577 75 71 2 13. 14:20:35 14. ₹1,29,600 15. 66 % 3

1 3. 33 % 3 12. ₹225

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Chapter 13 Let’s Warm-up

3. Improper fraction

Do It Yourself 13A 1. a. 4. a. b. c.

3 8

2 4

4 8

2 5

–6 –28

3. e.

3 3 –5 c. d. e. 6. a. –9; 32 b. –8; 36 3 5 7 7 7. a. –12 b. –44 c. 21 d. 9.5 e. –15 14 21 8. Answers may vary. Sample answers: a. ; 26 39 60 84 –18 b. ; 9. 10. Answers may vary. 100 140 63 –7 7 0 –1 Sample answers: a. b. c. d. 1 5 5 2 Word Problems 1. No 2. 3 equal parts. 4 5 2 7 –1 –3 13 9 4 –3 13B 1. a. > b. < c. > d. > e. > 7 9 3 3 2 5 –24 –4 5 5 8 9 2 21 1 –8 2 2. a. b. c. d. 3. a. b. c. 3 10 3 22 8 16 3 –10 2 7 5 d. 0 4. a. < b. > c. < d. > 5. a. < < < 9 9 18 12 –14 4 10 –1 –3 4 –3 –1 –3 8 8 6 b. < < < c. < < < d. < < < 20 –6 –16 3 5 –7 6 3 5 –19 25 13 –4 –2 3 5 –3 1 2 11 5 7 2 –10 e. < < < f. < < < 6. a. > > > 5 3 5 6 7 9 7 15 12 18 9 9 –1 10 4 –14 –1 –3 4 –3 b. > > > c. > > > 7. Answers may 3 –16 –6 20 3 6 –7 5 –1 –2 –3 –4 –5 –31 –32 –33 –34 vary. Sample answers: a. , , , , , b. , , , , 6 6 6 6 6 45 45 45 45 –35 7 8 9 10 11 –55 –56 –57 –58 –59 –1 1 , c. , , , , d. , , , , e. , 0, , 45 36 36 36 36 36 90 90 90 90 90 2 2 2 3 4 5 –7 –8 9 44 –3 –14 , 8. R = ; S = ; T = ; U = ; 9. a. > > > 2 2 3 3 3 3 3 15 8 7 9 7 3 –3 8 –3 –4 b. > > > c. 32.8 > 4.4 > –3.9 > –5.0 d. > > > –1 10 8 4 4 7 9 8 10. a Word Problems 1. Raman 2. Radha 11 16 34 74 27 3 50 13C 1. a. b. c. d. e. 2. a. b. 5 15 5 21 17 8 9 4 118 15 –5 –11 –1 9 5 c. d. e. 3. a. b. c. d. e. 13 7 17 7 8 2 11 9 1 5 –13 c. 132

4. a.

–1 1

; 5 5

–1

b. 0 c. 1 d.

–17 45

–19 56 –3 3 8. a. ; 28 28 e.

–46 75 –13 13 b. –1; 1 c. ; 7 7 5. a.

–35 48

4 –2 7. 5 7 –7 7 e. ; 9. Answer may vary. Sample answers: 6 6 6.

–1 –3 –2 –1 4 4 4

153

m 4.

km 5.

3 2

1 4

2 4

3 4

5 4

6 4

7 4

9 4

10 4

1 2 1 1 d. e. – f. 4. Answers may vary. 15 3 5 9 2 3 4 18 27 36 Sample answers: a. ; ; b. – ; – ; – 12 18 24 24 36 48 8 12 16 –2 –3 –4 22 33 44 –20 –30 –40 c. ; ; d. ; ; e. ; ; f. ; ; 14 21 28 22 33 44 24 36 48 200 300 400

3. a.

1 9

301

–3 4

–1 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 1 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11

L 2.

m 3. ₹425.45 4. m 5 20 –3 6 –2 24 8 11 b. c. d. e. 2. a. b. – c. 1 14 7 15 35 5 9 –9 9 4 –1 –4 11 d. 3. a. b. c. d. e. 71 5 15 6 7 19 9 –96 4 –17 1 4. a. – b. c. 2 d. 5. a. –2 b. c. 20 95 3 8 3 –25 –1780 3 25 2 6. –39 7. 1 8. 9. 10. – 11. a. b. 8 189 20 29 5

Chapter Checkup 1.

–1 –5 –4 –3 –2 –1 0 1 2 3 4 5 1 6 6 6 6 6 6 6 6 6 6

–8

Word Problems 1.

Word Problems 1. ₹46.3 2. ₹

–1 –3 –2 –1 0 1 2 3 1 4 4 4 4 4 4

5 c. 14; 40

487 42 3 13D 1. a. 10

–1 –5 –4 –3 –2 –1 0 1 2 3 4 5 1 6 6 6 6 6 6 6 6 6 6

5. a.

   3 –12 3 4  0   e. – 4 +  b. 0 + c. – 2 +  d. – 4 +    17 20 7 19  11 

10.

–1 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 1 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9

d. e.

1 4

1. Like fractions 2. Proper fraction 4. Mixed 5. Improper

a. 3 +

b. –

c. –

5. a. 56 b. 80 c. 4 d. –20 6. Answer may vary. Sample 19 20 21 22 23 13 14 15 16 17 , , , , b. , , , , 30 30 30 30 30 54 54 54 54 54 –4 –5 –6 –7 –8 –2 –1 1 2 –1 –2 –3 –4 –5 c. , , , , d. , , 0, , e. , , , , 6 6 6 6 6 3 3 3 3 12 12 12 12 12 13 –1 3 40 3 7. a. b. c. d. e. 3 f. g. 1 10 10 10 49 4 (–45) 9 25 34, 241 h. 8. a. 1 b. c. d. –13 e. 26 10 8 240 answer: a.

9. a. –

b. –

10. –

lent rational numbers. 15. –

421

16.

Word Problems 1. 5.

7000

11. Yes, as both are equiva97

12.

309

Let’s Warm-up 5. False

7. ₹10,000

1. False

2. False

Do It Yourself 14A 1. a. three b. less c. one 2. a.

A 4 cm B

3.5 cm 5 cm

14.

3. 9000 books

6. 8 gallons

Chapter 14

13.

part

4. 165 km

3. True

inches

4. False

5.1 cm

4.5 cm

C Q

6 cm

338

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30-12-2023 10.57.36 PM

3. Scalene Triangle

R = 70°

J 4.8 cm

3.8 cm

8 cm

5.2 cm

4.5 cm

13 cm

J 89°

56°

12 cm

6.1 cm

7.4 cm

35°

14B 1. b, c

A 6.2 cm

4.5 cm

6.8 cm

44°

8.7 cm

3. XYZ = 91°, YZX = 44° Scalene triangle; Obtuseangled triangle

60°

4.8 cm

46°

ZXY = 54°

5. BAC = 37°, A

7.8 cm

BCA = 23°

30°

7.1 cm

120°

W ord Problem 1.

23°

40° A

2. XZ = 6.2 cm, YZ = 3.2 cm

60°

70°

6 cm

80°

6.2 cm

4.7 cm

60°

45°

5.2 cm

7. SAS criterion.

4.8 cm

30°

45°

5 cm

6 cm

3.2 cm 75°

9 cm

55°

6.3 cm

5. XZ = 3.8 cm; YZ = 4.7 cm

F Z

30°

4.5 cm

4.1 cm

6. 55° and 80°.

12 cm

6.2

UM24CB07_Answer.indd 339

6 cm

8 cm

7.5 cm

8 cm

2 cm

7 cm

5.9 cm

3. Scalene triangle; Obtuseangled triangle (111°, 40°, 28°)

4. AC = 6 cm; BC = 4.1 cm

80°

54°

5 cm

4.6 cm

5.7 cm X

2. Acute-angled triangle

5 cm

6.5 cm

14C 1.

Answers

59°

58°

3.8 cm

Chapter Checkup 1. a. SSS b. SAS c. RHS d. ASA

3.9 cm

45°

5.2 cm

37°

4.1 cm

75°

7.1 cm

Word Problem 1. 13.8 m

44°

45°

32°

3.8

cm 4 5.

6.3

6 cm

9.4 cm

91°

4. 32°, 58°

4. Scalene triangle; Acuteangled triangle; QR

4.2 cm R

6.6 cm

8.2 cm

7 cm

9.5 cm

5.7 cm

60°

8 cm

(Unknown side)

3. BC = 6.8 cm; BAC = 46°, BCA = 44° C

6 cm

2. 7 cm

6.3 cm

7 cm

45° 60°

4.6 cm (Unknown side)

2. a.

7 cm

45°

9.7 cm

14D 1. 4.6 cm

6 cm

5.8 cm

8.7 cm

60°

6 cm

8 cm

5.4 cm

4.5 cm

70°

5. Scalene triangle; Acute angled triangle LN = 7.1 cm; MN = 8.7 cm Word Problem 1. N

Word Problem 1.

6 cm

5 cm

G 4 cm

4.5 cm

5.1 cm

50°

7.1 cm

A 4.3 cm

6.3 cm

70°

40°

5. Equilateral Triangle

2 smaller sides is greater than the length of the largest side. L

70°

10 cm

4. Yes. In a triangle, the sum of the lengths of

5 cm

6 cm

4. DF = 6.3 cm, EF = 5.1 cm

7.1 cm

339

30-12-2023 10.57.51 PM

8. 3.6 cm approx.

3.5 cm

10. Sum of length of sides UW and VW = 6 cm + 4.9 cm = 10.9 cm

6 cm

3.4 cm 4.9 cm

70°

5.5 cm

6.0 cm

Word Problem 1. 34 cm approx.

50°

7.8 cm

5 cm

3.6 cm

4.5 cm

3 cm

Chapter 15

1. 96 cm, 92 sq. cm 2. 1254 sq. cm, 52,625 sq. cm

Let’s Warm-up

Do It Yourself 15A 1. a. P = 21 cm, A = 25.16 sq. cm b. P = 18 cm,

A = 20.25 sq. cm c. P = 27 cm, A = 43.46 sq. cm d. P = 23.6 cm, A = 34.81 sq. cm 2. a. 4 sq. cm b. 16 sq. cm c. 64 sq. cm d. 169 sq. m e. 729 sq. m 3. a. 6 sq. cm b. 20 sq. cm c. 63 sq. cm d. 342 sq. m 4. 12 m 5. 40.96 sq. units 6. Perimeter of the rectangle is greater than the square 7. Length = 2 units; Breadth = 3 units; Perimeter = 10 units; Area = 6 sq. units Word Problems 1. 2,50,000 sq. m 2. a. ₹30,000 b. 3 rounds 3. ₹3072 4. ₹3800 5. Length = 18 cm, breadth = 24 cm, Square 6. ₹59,593.60 7. 39 8. 89 sq. m 9. 20 m 15B 1. a. 24 sq. cm b. 63 sq. cm c. 21 sq. cm d. 48 sq. cm 2. a. 8 cm b. 20.3 cm c. 15.8 cm d. 9.3 cm 3. 160 sq. cm 4. Height = 18 cm, base = 36 cm 5. a. 80 sq. cm b. 5.3 cm 6. 29.75 cm 7. 16.7 cm (approx.) 8. AE = 8 cm, EB = 22 cm 9. a. 14 cm b. 18 cm Word Problem 1. 24 m 15C 1. a. 104 sq. cm b. 35 sq. cm c. 40 sq. in d. 22.5 sq. cm 2. a. 8 cm b. 6.3 cm c. 16 cm d. 10.5 cm 3. 80 sq. cm 4. 4.8 cm 5. a. 30 sq. cm b. 4 cm 6. Base = 12 cm, height = 16 cm 7. 562.5 sq. cm 8. a. 875 sq. cm b. 615 sq. cm 9. 308 sq. cm Word Problem 1. 442 sq. m 15D 1. a. 31.43 cm b. 50.29 cm c. 62.86 cm d. 75.43 cm e. 113.14 cm 2. a. 31.43 cm b. 47.14 cm c. 75.43 cm 100.57 d. 94.29 cm e. 125.71 cm 3. a. Yes, = 3.14285… 32 125.71 150.86 b. Yes, = 3.14285… c. Yes, = 3.14285… d. Yes, 40 48 226.29 = 3.14285… 4. a. 24 cm b. 192 cm c. 160 cm 72 d. 112 cm 5. 5:6 6. 21 cm 7. 128.5 cm 8. 94.18 m 9. 18,857.15 cm or 188.57 m 10. 145.17 m 11. 574 revolutions 12. 125.71 cm Word Problems 1. ₹12.45 2. 75.43 min 3. 56 cm 4. 150.72 s 15E 1. a. 28.29 sq. cm b. 113.14 sq. cm c. 452.57 sq. cm d. 707.14 sq. cm e. 1964.29 sq. cm 2. a. 78.57 sq. cm b. 254.57 sq. cm c. 531.14 sq. cm d. 962.5 sq. cm e. 1810.29 sq. cm 3. a. 8 cm b. 14 cm c. 28 cm d. 40 cm 4. a. 82.60 sq. cm b. 254.57 sq. mm c. 346.5 sq. mm d. 201.14 sq. cm 5. 16:49

6. 2828.57 sq. cm 7. 14 cm, 616 sq. cm 8. 181.5 sq. mm 9. 1089 sq. cm 10. Yes Word Problems 1. 50.29 sq. cm, 549.71 sq. cm 2. 1357.71 sq. cm 15F 1. a. 42.44 sq. cm b. 850.14 sq. cm c. 169.14 sq. cm d. 728.57 sq. cm 2. a. 48.21 sq. cm b. 11.43 sq. cm c. 120.54 sq. cm d. 2021.43 sq. cm 3. 58.875 sq. cm 4. 7612.50 sq. cm 5. 219.43 sq. cm 6. 181.335 sq. cm 7. 1369.64 sq. cm 8. 39.48 sq. cm 9. 3:5 Word Problem 1. 5848.28 sq. cm 15G 1. 108 sq. cm 2. 18,133.50 sq. m 3. a. 296 sq. cm b. ₹51,744 4. 1558.86 sq. cm 5. 1609.15 sq. m Word Problems 1. 42 sq. m, ₹2100 2. 3600 sq. cm 3. 126 sq. cm, 124 sq. cm 4. ₹1,49,523 5. a. 2 m b. 78 sq. m c. 72 sq. m Chapter Checkup 1. a. 18 sq. cm b. 32 sq. cm c. 18 sq. cm 2. 40.5 sq. cm 3. 2:3 4. Length = 17.5 m, breadth = 3.5 m 5. 72.9 m 6. ₹5494 (approx) 7. 135 sq. cm 8. 7.2 m 9. DP = 54.29 m, BQ = 47.5 m 10. 60 sq. cm 11. a. 56 sq. cm b. 20 sq. cm c. 7:5 12. 42 cm, 132 cm, 1386 sq. cm 13. 96.43 sq. cm 14. 62.86 m 15. 25:49 16. Area of circular ring by 33 sq. cm 17. 3289.71 sq. m 18. 2227.96 sq. cm 19. 25.12 sq. cm 20. 140 sq. m 21. 7 m 22. 588 sq. cm 23. 7 m Word Problems 1. 370 shrubs 2. ₹15,540

Chapter 16

Let’s Warm-up 1. 85 2. 59 3. 80 4. 15 5. 128 Do It Yourself 16A 1. 240.25 cm2 2. 17.5 cm 3. n + 2 4. 5 cm 5. 27.22 cm

6. 88 7. 249 8. (2n − 1), 199 Word Problems 1. 20 feet2 2 2. 625 m a 16B 1. a. x + y b. + 3b c. z – 3 + xy d. 2pq − mn 2. a. 6ab; 4 6a2, 12c b. 3c, –10b; –10a c. 15ba, 12c; 15a2, 7bc d. 5ac, 1 3b, –6c; 3a e. 2.5b, –3.2ab; 2.5a, –3.2a2 f. 3b; b, 3a 3 2 2 3 2 3. a. 5m , mn b. 3pq, 5q p c. 7y , 8xy , –5xy d. 6ab, 2 5 8a2bc, –2bc2, –ca e. 0.5xy, 1.2x2y, –3.3y f. yz, x2y, 0.6xy 5 6 4. a. Expression b. Expression 3pq2 – 5pr + 2 5xy + 8yz Terms

5xy

Factors 5 x y

c. Expression

8yz 8

Terms

Factors

5ab

– 8ba2

Factors

5 a b –8 b a a

Terms

1.2x2y

–3xyz

Factors 1.2 x x y

–3 x y z

Expression Terms Factors

Like terms

m + mn2 – 6l + 8nl mn2

– 6l

8nl

mnn

–6l

8nl

1 2 ab – 3a2b + 7 2

1 2 ab 2 1 abb 2

–3a2b

–3 a a b

2 2 n m 5 Unlike terms

c. 5p2q d. ab2 e. 0.5zx f.

a. (m2n, 5m2n, 8nm2), (3mn, 6nm) 2

Factors 3 p q q –5 p r

5ab – 8ba + 5c d. Expression

b. 6mn2

Terms

5. a. –2x2y

3pq2 – 5pr

e. Expression 1.2x2y – 3xyz

Terms

b. (2a c, 3ca ), (3ac , c a)

5m, 7n

5b2a, 8b2c

c. (pqr, 9qrp, 6rpq), (8rp, 5pr)

pq2

d. (5mln, 5lmn), (6lk2m, k2lm)

2klm

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7. a. False b. True

c. True 2

Word Problem 1. 5 2

16C 1. a. 5xy + x b. 8p 2. a. 3x + 4y

d. False

c. ab d. 7ac

b. 4x + 4y

e. –2mn

f. 17x y

c. –2a + 7b + 2 d. a + 15b – 15c

3. a. x + 3y b. 3p2 + 5q2 + 3r2

c. 6a2 + 3b2 + 16ab2

4. a. 11xy – 3x b. 3x2 + 4x

d. 31a + 16b – c 2

c. 7ab + 4b + 6ac d. 7pq + 4pr e. –2pq + 10pr

f. 4a2b – 9ab + 3c 5. a. 19ab2 + 2a + 17bc – 4a2b 1 7 9 b. 5ab2 – 20a2b + 27bc 6. 6a + 3 7. mn2 – mn + 1 n2 10 15 77 Word Problem 1. (8x + 13) cm 2

16D 1. a. –2ab – 4 b. 12y – 17xy c. 6y – 4y – 3 d. 6b2 – 10ab + 8

2. a. –3x – 2x2

3. 14ab – 9ac – 11

4. 5a2 – 27a + 6b – 8

b. 3mn – 7mn2 + 10n2

c. –2p q + 3q – 13 d. 6ab + 7ac + 3bc + 3 2

5. 5xz – 5xy – 19yz + 4x2z – 3

6. 9

7. 9xy2 + 6xy – 15x2y + 5y2 8. 10a2 + 12ab – 21ab2 – 9 Word Problem 1. 8 metres 16E 1. a. 11 b. 7 c. 15 d. 3 2. a. 1 b. 9 c. −4 d. −160 3. a. 2 b. 6 c. 15 d. − 4 4. a. 35 b. 18 c. 8 d. 5 5. a. −10 b. −10 c. 8 6. 14 7. 7200 8. −6 Word Problem 1. 2400 litres

16F 1. a. 7x – 9y b. 14m + 2mn c. 5a – 8ab d. x2 + 8x – 5 2. a. –18 b. 18 c. –2 d. 26 2

b. –3m + 3mn – 8n + 6m

3. a. 5x2 + 9x – y2 – 3y + 2

c. 6pq – 4pq – 6p d. a + 2ab –

3a2 – 3 4. a. 17 b. –2 c. 29 d. 12 Word Problem 1. 6 years

5. –51

16G 1. 5x + 4 2. 3x + 5 3. (3x2 + 7x + 11) metres 4. ₹(8x2 + xy + 12y2) 5. 89 metres 44 minutes; 54 minutes

Word Problem 1. 5x + 10; 22 minutes;

Chapter Checkup 1. 11.5 cm 2. (13n + 1); 196 3. 2a + b

4. a. Unlike b. Like c. Unlike d. Like e. Unlike f. Like 5. a. Expression

b. Expression

x2y + yx 2

Terms

x y

Factors x

x y

c. Expression

yz y

2cb2 + 6ab + 8c

Terms

2cb2

6ab

Factors

2cbb 6ab

pq – 5pq2 + 9

Terms

–5pq2

Factors

p q –5 p q q

Word Problems 1. 128 cm 2. ₹(5a2 + 4ab + 2b2) 3. 33 metres

Chapter 17

1. 10,000 (Ten Thousand) 2. 1,00,000 (One Lakh) 3. 10,00,000 (Ten Lakh) 4. 1,00,00,000 (One Crore) 5. 10,00,00,000 (Ten Crore)

Let’s Warm-up

Do It Yourself 3 –7 11 17A 1. a. 9; 11 b. −17; 9 c. n; 10 d. ; 5 e. ; 8 f. ; 15 8

2. a. 5

Answers

UM24CB07_Answer.indd 341

b. (–9)

 7

 –2 18  e. (10)4 f. 93  5   –4 6  5 7 g. 1 h. (6)3 2. a. 136 b. (−11)3 c.   d.    7  3  14 12  3 8  –1 9 e.   f. (−3)8 3. a. 94 b. (−12)3 c.   d.    15  2  8 

7 17B 1. a. 1014 b. (−5)12 c.   9

d. 

4. a. 1 b. 1 c. 1 d. 1 5. 16 6. a. 128 b. −27 c. 1,60,000 d. −17 7. 1 8. a. 26 × m2 b. 23 6 11 2 c. 5 × 7 × 11 d. ab 9. a. 1 b. 7 Word Problem 1. 24 × 33 × 74 17C 1. a. 8.64 × 105 b. 5.5 × 106 c. 4.152 × 109 d. 6.14 × 1011 2. a. 30000 b. 455000 c. 71100000 d. 9800000000 3. 4 4. 7 5. 7504682; 7.504682 × 106 6. 6 × 105 + 4 3 2 1 0 4 × 10 + 3 × 10 + 2 × 10 + 9 × 10 + 8 × 10 7. a. 1.3568 × 107 b. 1.41 × 109 c. 2.76 × 109 d. 1.4 × 108 Word Problem 1. a. 4540000000 years Chapter Checkup 1. 65 2. a. 1728 b. 10,00,000 c. 38,416 d. −1 e. 1 f. 1 3. a. 2401 b. 1 c. 8 d. 25 4. a. 86 b. (−15)8  3 9  –2 6  4 7 c. (−m)4 × n7 d. 24 × a5 × b8 e.   f.   ×   7  11   15  5. a. 28 b. 2 × 3 × 97 c. 2 × 5 × 109 d. 2 × 52 × 41 e. 2 × 1709 n2 6. a. < b. < c. < 7. a. 2700 b. – 2 c. 0 d. 16 10   512  9. a. 4.75 × 106 b. 7 × 107 c. 1.2 × 107 8.   19,53,125  d. 2.89 × 109 10. a. 9100000 b. 7.06 × 104 c. 43600000000 d. 1.3 × 107 11. a. 12 b. 3 c. 2 d. 8 e. 6 f. 7 Word Problems 1. 6.237 × 108 2. 2.3 × 108 centimeters

Chapter 18

1. One line of symmetry 2. Two lines of symmetry 3. Four lines of symmetry 4. Infinite lines of symmetry

Let’s Warm-up

6. Trinomial, Quadrinomial, Monomial, Binomial 7. a. 19a2b – 10ab2 b. 5x + 2y + 2xy + 6 8. 4x2 + 5xy – y2 9. 8x – y – 5 10. (18a – 10) metres 11. 28 12. –6 13. –10 14. −3615 15. −8.2 16. 128 17. −84.67 18. 0 19. a. −2 b. −18 c. −28 d. 12

8 121 3. a. 1 b. 4913 c. –3125 d. 5184 e. f. 729 144 16 g. 4. a. 26 b. 3 × 5 × 7 c. 25 × 31 × 51 d. 2 × 3 × 1029 173 e. 27 × 31 × 51 5. a. 64; 512 b. 121; 1331 c. 324; 5832 d. 441; 9261 e. 900; 27,000 6. a. 34 b. Both are  3 4  7 2  1 6  –1 4 equal c.   d.   7. a.   b.   c. 4 2 5 2  3  d. (−2)5 8. a. 1,91,664 b. 15,000 c. 27,783 d. 14,279 9 8408 e. f. – 9. 5 Word Problem 1. 57 25 18375

c. 7 × 8

d. n × p

11 18  5 5  –4 7 e.   f.   3  11 

Do It Yourself 18A 1. ,

2. a. Yes b. No

e. No,

c. Yes d. Yes

f. No

g. Yes h. No

3. a.

4. a.

b. c.

5. Answers may vary. Sample answers: a. b.

We would shade the squares along the diagonal line at an equal distance on both sides.

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6. S 7.

8. a.

Word Problems 1. a.

2. Yes, It has 6 lines of symmetry

S. No.

Figure

Rotation by 90°

Rotation by 180°

Rotation by 270°

Rotation by 360°

Number of times a shape looks the same

a. 1; 360° b. 2; 180° c. 4; 90° d. 4; 90° e. 2; 180° 5. a. B, C, D, E, K b. A, T, U, V, W c. X, O 6. a. Scalene triangle b. Parallelogram c. None 1

18B 1. a. more than 1 b. equal to 1 c. more than 1 d. equal to 1

e. more than 1 f. more than 1 g. equal to 1 h. equal to 1 2. a. 5 b. 3 c. 4 d. 2 3. i. a. ii. a. 4 b. 4 c. 4

d. Nonagon,

d. 6 iii. a. 90° b. 90° c. 90° d. 60° 4. a. Letter H, I, O, X Figures may vary. Sample figure. b. Letter N, S, Z c. Letter A, B, C, D, E, K,

7. 0 and 8 8. Error: The marked portion of the mirror image should be flipped on the opposite side. Correct image:

Object Mirror image

he new position of C will T replace H in the original C

Figure 2: Yes

8. Colours may vary. Sample figures. a. b. c.

c. 4,

h. 4,

d. 2,

3. a.

f. No

Lines of symmetry

2. a. 1,

b. 0,

e. 2,

f. 1,

g. 4,

Object Mirror image

10. Answer may vary. Sample answers: a. b.

12. 7:15

13. Each of the given faces have 2 lines of symmetry.

15. Minimum of 6 squares should be added to get a rotational symmetry of order 4.

b. For

Chapter Checkup 1. a. Yes b. Yes

may vary. Sample answer:

11. Answer may vary. Sample answers: a. b.

14. Figure may vary. Sample Figures.

Word Problems 1. a. 4 lines of symmetry in each figure. both the figures, order = 4, angle of rotation = 90° d. Yes e. No,

6. a. Anticlockwise b. 90° c. No d. 1 7. a. Figure 1: Order of rotational symmetry = 4, angle of rotation = 90°; Figure 2: Order of rotational symmetry = 2, angle of rotation = 180° b. Figure 1: No.

c. No,

x x

e. Trapezium

d. Letter F, G, J, L, P, Q, R

5. a.

40°

M, T, U, V, W, Y

140°

Word Problems 1. a. Shape Q b. P - 1, Q - 1, R - 1, S - 1, T - 1, U - 0 c. Rotational symmetry of order 1

Chapter 19

1. Pentagonal prism 2. Cone 3. Triangular prism 4. Cylinder

Let’s Warm-up

Do It Yourself 19A 1. a. cone b. cuboid c. 0 d. cuboid/cube e. face

2. 3 3. a. Pentagonal pyramid b. Square prism or cuboid c. Rectangular pyramid d. Hexagonal prism e. Octagonal pyramid 4. a. E = 0; V = 0 b. E = 1; V = 1 c. E = 6; V = 4 d. E = 9; V = 6 5. Cylinder 6. Cylinder 7. Cube and cuboid 8. Difference: Octagonal prism: 2 octagonal bases and 8 rectangular lateral faces.

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Octagonal pyramid: 1 octagonal base and 8 triangular lateral faces. Both are 3-D shapes with octagonal bases. Word Problem 1. Square pyramid

19B 1. a. 10; 2 octagonal identical faces and 8 rectangular faces. b. 10; 1 nonagonal base and 9 triangular lateral faces c. 6; all rectangular or both square/ rectangular faces. d. 2; 1 circular base and 1 curved lateral face. 2. a. b. c. b.

3. a. Triangular prism b. Rectangular prism/cuboid c. Cone d. Square pyramid e. Cylinder f. Pentagonal pyramid 4. 5. a. b.

6. Triangle and square/rectangle 7. No 4 cm

4 cm 4 cm

4 cm

4 cm 4 cm

4 cm

44 cm

10. The triangles are the bases. 11. 5 cm 5 cm

4 cm

5 cm

Word Problem 1. Cylinder 19C 1. 2.

, Top

5. a. Front

, Side

b. Front

, Side

c. Front

, Side

, Top , Top

Side

, Top

6. Yes, if the light source creating the shadow falls at an angle on the cube. 7. Answers may vary. Sample answer: a. Sphere b. Cone c. Cube/cuboid d. Cuboid Word Problem 1. 4200 cm2

Chapter Checkup 1. a. Cone b. Triangular pyramid c. Triangular 15 cm

d. Front

prism d. Cylinder e. Cube 2. a. Cuboid b. Square pyramid c. Cube d. Cylinder 3. a. Octahedron b. Hexagonal prism c. Square/rectangular pyramid d. Cylinder 4. a. Triangular pyramid b. Cube c. Square pyramid d. Cone e. Cylinder 5. a. b. c. d.

6. Figures may vary. Sample figure.

7. i.

3. ii.

4. a.

8. a. Rectangle b. Rectangle c. Rectangle 9. a. Horizontal:

5. a. 5 units × 4 units × 1 unit b. 3 units × 5 units × 2 units c. 5 units × 3 units × 3 units

6. a. 4 units × 4 units × 4 units 7. a.

b. 3 units × 5 units × 2 units

c. 5 units × 2 units × 2 units

b. Horizontal:

Vertical:

c. Horizontal:

Vertical:

10. a. Front view:

Word Problem 1. Cuboid, 9 cm × 3 cm × 3 cm

square, Vertical: square c. Horizontal: rectangle, Vertical: rectangle d. Horizontal: circle, Vertical: circle 2. a. b. c. d. e.

4. a. Front

, Top

b. Front

, Top

c. Front

Answers

UM24CB07_Answer.indd 343

, Top

, Side

Top view:

Side view:

b. Front view:

Top view:

Side view:

c. Front view:

Top view:

Side view:

d. Front view:

Top view:

e. Front view:

19D 1. a. Vertical: triangle, Horizontal: circle b. Horizontal:

3. Figures may vary. Sample figures: a. b. c. d.

Vertical:

Side view:

Top view:

Side view:

11. 2 heptagons and 7 rectangles; 1 heptagon and 7 triangles 12. a.

13. A nswer may vary. Sample answer:

E D C B A F

Word Problems 1. 11 2. Vertical cross-section and horizontal cross-section; 40 cm2; 200 cm2 3. Figures may vary. Sample figures: 15;

Front

Top

Side

, Side

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Imagine Mathematics

About the Book

Key Features

MATHEMATICS

Singapore

IM_G07_MA_MB_Cover.indd All Pages

Gurugram

Bengaluru

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NEP 2020 based

NCF compliant

CBSE aligned

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